Mechanics Of Materials 5th Beer Johnston

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MECHANICS OF MATERIALS

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Fifth Edition

MECHANICS OF MATERIALS

FERDINAND P. BEER Late of Lehigh University

E. RUSSELL JOHNSTON , JR . University 01 Connecticut

JOHN T. DEWOLF University 01 Connecticut

DAVID F. MAZUREK United States Coast Guard Academy

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MECHANICS OF MATERIALS. FIFfH EDITION Published by McG:aw-Hi IL a business unit of The McGraw-Hill Companies. Inc. , 1221 Avenue of the America~. New York. NY 10020. Copyright 1!;l2009 by The McGraw-Hill Companies. Inc. All rights reserved. Previous editions C 2006. 2002. 1992. and 1981. No pan of this publication may be reproduced or distributed in any foml or by any means. or stored in a database or retrieval system. without the prior written consent of The McGraw-Hill Companies. Inc .. including. but nOl limited to. in any network or other electronic storage or transmission. or broadca~t for distance learning. Some ancillaries. including electronic and print components. may nOl be available to customers outside the United States. This book is printed on acid -free paper. I 23456789 c' QPVIQPV 0 9 8 ISBN 978-0-07- 352938-7 MHID 0-(17- 352938-9 Global Publi sher: R"ghmham"" Sri"im."''' Senior Sponsoring Editor: Bill SIeI!quisi Director of Development: Krisa"e 7ibbeus Developmental Editor: Lorn Neye"s Senior Marketing Manager: Curt Reynolds

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About the Authors As publi ~ hcrs of the books written by Ferd Beer and Russ Johnston. we arc often asked how did they happen to vmlc the books together. with o ne of them at Lehigh and the other at the Uni versity of Connecticut. The answer to this question is simple. Russ Johnston 's /irsl teaching appointment was in the Depanmcl1l of Civi l Eng ineering and Mechanics at Lehigh University. There he met Ferd Beer, who had joined that depaIlment two years earlier and was in charge of the courses in mechanics. Born in France and educated in France and Switzerland (he held all M .S . deg ree from the Sorbonne and an Sc .D. degree in Ihe fi eld of theoret ical mechanics from the University o f Geneva) . Ferd had

come to Ihe United States after serv ing in the French army during the earl y part of World War II and had taught for four years at Williams College in the Williams-M IT joint arts and engineering program. Born in Phil adelphi a. Russ had obtained a B.S. degree ill civil enginecring from the University of Delaware and an SC.D. degree in the lield of structural e ngineering from MIT. Ferd was delighted to discover that the young man who had been hired chielly to teach graduate structural e ngi neeri ng courses was not o nly willing but eager to help him reorganize the mcehani c~ courses. Both believed that these <:ourses should be taught from:1 few basic prin cipleS and th31 the vario us concepts involved wou ld be best understood and remembered by the students if they were presented to them in a graphic way. Together they wrote leclLlre notes in statics and dynamics. to which they later added problems they felt would appeal to future engi neers, and soon they produced the manuscript o f the first edition o f Mechanics fo r Enxilleen. The second edition of Medumics fo r EIIxilleer.\· and the first edition uf Vector Mechanics/or Ellgineers foun d Russ Johnston at Worcester Polytechnic Institute and the next editi ons at the University of Connectic ut. In the meant ime. both Ferd and Russ had assumed administrative responsibilities in their departments . and both were involved in research, consulting. and superv ising gmduale ~ llIdent s-Fe rd III the area of stochastic processes and random vibrations. and Ru.<;s in ule area of clastic stabi lit y and structural analysis and design. However. their interest in improving the teaching of the basic mechanics courses had not subsided. and they both taught sectiom of these co urses as they kept revising their tex ts alld began

v

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vi

Aboutlhe Authors

wriling together the manuscript of the first editi on of Mechonics of Mmer;ol.\'. Ferd. and Russ's contributi ons to e ngi neering education earned them a number l)f honors and awards. They were presented with the Weste m Elec tri c Fund Award for excellence in the instruction of engineering stude nts by their respective regional sections of the American Society for Engi neering Edu cat ion. and they bOlh received the Di stinguished Educator Award from the Mechanics Di vision of the same society. In 1991 Russ received the Outstanding Civil Engineer Award fmm the Connecticut Section of the American Society of Civil Engineers, and in 1995 Ferd was awarded an hunorary Doctor of Engineering degree hy Lehigh University. John T. DeWolf, Professor of Civil Engineering at the University of Connecticut, joined the Beer and Johnston team as an author on the second edition of M eclwll ics o/ MlIIeriah. John holds a B.S . degree in civil engineering from the University of Hawaii and M.E. and Ph.D. degrees in structu ral e ngineering from Cornell Uni versity. His research interests arc in the area of clastic stabi lily, brjdge monitoring, and structuml analysis and design. He is a member of the Connecticllt Board of Examiners fO[ Professional Engineers and was selected as a University of Connecticut Teaching Fellow in 2()()6. David F. Mazurek. Professor of Civil Engi neering at the United States COlL~t Guard Academy, is a new author for this edition. David holds a B.S. degree in ocean engineering and an M.S. degree in civil e ngineering from the Florida LnstitUle of Technology. and a Ph.D. degrec in civit engineering from the University of Connecticut. He has scrveu on the American Railway Engineering & Maintenance of Way Association's Committee IS- Steel Structures for the pa;t seventeen years. Professional interests include bridge engineering. structural forensics, and blast-resistant design.

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PREFACE OBJECTIVES

The main objective of a basic mechanics course should be to develop in the engineering student the ilbility to analyze a given problem in a simple and logical manner and 10 apply to iL~ solution a few fundamental and well-understood principles. This lex! is designed for the tirsl course in mechanics of materials-or strength of malcriais-oflcrcd to engineering students in thc sophomore or junior year. The authors hope that it will help instnlctors achieve this goal in that pan.kular course in the same way thai their other texts may have helped them in statics and dynamics. GENERAL APPROACH

In this tex t the study of the mechanics of materials is based on the unde rstanding of 11 few basic concepts and on the usc of simplified models. This approach makes it possible to develop all the necessary formulas in a rmional and logical manner. and 10 clearl y indicate the conditio ns under which they can be sarely applied 10 thc analysis and design of actual engineering structures and machine components. Free-body Diagrams Are Used Extensively. Throughout thc text free-body diagrams ace used to detennineextemal or internal fo rces. The u~ of "picture equations" will also help the students undersland the superposition of loadings and the rcsuhing stresses and dcfonnalions. Design Concepts Are Discussed Throughout the Text Whenever Appropriate. A discussion of the application of the factor of safety to design can be found in Chap. I. where the concepts of both allowable stcess design and load and resislance factor design arc presented. A Careful Balance Between SI and U.S. Customary Units Is Consistently Maintained. Because il is essemial Ihm students be able 10 handle cffeclively both SI melric units and U.S. customary units. half Ihe exampl es, sample probl ems, and problems ( 0 be assigned have been stated in SI unils and half in U.S. customary units. Since a large number of problems arc available, instructors can assign problems using

xiii

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xiv

Preface

each system of units in whatever proportion they find most desi rable for their class. Optional Sections Offer Advanced or Specialty Topics. Topics such as residual stresses, torsion of noncircular and thin-walled members. bending of curved beams. shearing stresses in non~sy mmetrical members. and fai lure criteria, have been included in optional sections for usc in COllrscs of varying emphases. To preserve the integrity of the subject, these topics are presented in the proper sequence, wherever they logically belong. Thus. even when not covered in the course, Uley are highly visible and can be easi ly referred to by the students ir needed in a later course or in engineering practice. For convenience all optional sections have been indicated by asterisks.

CHAPTER ORGANIZATION

It is expected that students using this text will have completed a COllrse in statics. However, Chap. I is designed to provide them with an opportuni ty to review the concepts learned in that course, whi le shear and bending-moment diagrams are covered in detai l in Secs. 5.2 Hnd 5.3. The properti e.~ of moments and centroids of arcas arc described in Appe ndi x A; thi s material can be used to reinforce the discussion of the determ inat ion of nurmal and shearing stresses in beams (Chaps. 4, 5, and 6). The fir st fOllr chaplers of Ihe lext arc devoted to the analysis of the st resses and of the corresponding deformations in vari ous st ructural members, considering success ively axial load ing, torsion, and pure bending. Eac h anal ysis is based on a few basic concepts, namely, the conditions of equilibrium of the forces exerted on the member, the relations ex isting between stress and strain in the material, and the conditions imposed by the su pports and loading of the member. The study of each type of loading is com plemented by a large number of exam ples, samp le problems, and prob lems to be assigned, all designed to strengt hen the students' understanding of the subject. The concept of stress at a point is introduced in Chap. I, where it is shown that an axial load can produce shearing stresses as well as normal stresses, depending upon the section considered. The fact that stresses depend upon the orientati on of the su rface on which they are com puted is emphasized again in Chaps. 3 and 4 in the cases of torsion and pure bending. However, the di scussion of computational techniques-such as Mohr's circle- used fo r the transformation of stress at a point is delayed until Chap. 7. after student s have had the o pport unity to solve prob lems involving a combination of the basic loadings and have di scovered for themse lves the need for suc h techniques. The discussion in Chap. 2 of the relation between stress and strain in various materials includes fiber-rei nforced composite materials. Also. the study of beams under transverse loads is covered in two

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separate chapters . Chapte r 5 is devoted to the determination of thc normal stresses in a beam and to the design of beams based on the allowable normal stress in the. material used (Sec . 5.4). The chapter begins with a discussion o f the shear and bending-momenr diagm ms (Sees. 5.2 and 5.3) and includes an optional sec tion (In the use of singularity func tions for the determinati on of the shear and bending moment in a beam (Sec. 5.5). The chapter e nds with an optional section on Ilonprismmic beams (Sec. 5.6). Chapter 6 is devoted to the determination of sheari ng stresses in beams and thin -wlIlled me mbers under transverse loadings. The formula for the shear flow, q = VQIl. is deri ved in the traditi onal wlly. More advanced aspects of the design of beams. such as the de termination of the principal stresses at the junction of the fli.lng e and web of a W-beam, arc in Chap. 8, an o pti onal chapter that may be covered after the trans formatio ns of stresses havc been di scu ssed in Chap. 7. T he design of transm ission shafts is in that chapter for the same reason, as well as the de te rmination ()f stresses under combined loadin gs that can now include the determi nation of the princi pnl stresses, principal planes. and maximum sheari ng stre ss at a given point. Statically indctemlinate problems are first discussed in Chap. 2 and considered throughout the text for the various loading conditions encountered. Thus, students are presented at an early stagc with a mcthCKI of solution that combines [he analysis or deformations with [he conventional analysis of forces used in statics. In this way, they will have become thoroughly familiar with this fundamental method by the end of the course. In addition, this approach helps the students realize that stresses themselves are statically indeterminate and can be computed only by considering the corresponding distribution of strains. The concept of plastic deformation is introduced in Chap. 2, where it is applied to the analysis of members under axial loading. Problems involving the plastic deformation of circ ular shafts and of prismatic beams arc also considered in optional sections of Chaps. 3, 4, and 6. While some of this material can be omitted at the choice of the instructor, its inclusion in the body of the text will help slltdents realize the limitations of the assumption of a linear stress-strain relation and serve to caution them against the inappropriate usc of the elastic torsion and nexure formu las. The determination of the deflection of beams is discussed in Chap. 9. The fi rst pan of the chapter is devoted to the integration method and to the method of superposition, with an optional section (Sec . 9.6) based on the use of singularity fu nctions. (Thi s section should be used only if Sec. 5.5 was covered earlier.) The second pan of Chap. 9 is optional. It presents the moment-area method in two lessons. Chapter lOis devoted to columns and contains material on the design of steel, aluminum, and wood columns. Chapter II covers e nergy methods, including Castigliano's theorem.

Preface

xv

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xvi

Preface

PEDAGOG ICAL FEATURES

Each chapter begins with an il1lroductory section setting the purpose and goals of the chapter and describi ng in simple temlS the material to be covered and its application to the solution of engilleeri ng problems. Chapter Lessons. The body of the text has been divided into uni ts. each consisting of one or several theory sections followed by sample problems and a large number oj' problems 10 be assigned. Each unit corresponds to a wel l-defined topic and generally can be covered in one lesson. Examples and Sample Problems. The theory sections include many examples designed to illustmle the material being presented ami L1ci litate its understanding. The sample problems arc intended to show some of the applications of the theory to the solution of engineering problems. Since they have been sct up in much the same foml that students will use in solving the assigned problems, the sample problems serve the double purpose of amplifying the text and demonstr..tting the type of neat and orderly work that students shou ld cultivate in their own solutions. Homework Problem Sets. Most of the problems are of a practical nature and shou ld appeal 10 engineering students. They arc primaril y designed, however, 10 illustrate the material presented in the tex t and help the students unde rstand thc basic principles used in mechanics of materials. Tht: problems have been grouped accordi ng to the portiuns of materi al they illustrate :md have becn arrangcd in order of increasing diffic ulty. Problems requ iring special attention have been indicated by asterisks. Answers to problems are given at the end of the book. except for those with a number set in italics. Chapter Review and Summary. Each chapter ends with 11 review and summary of the mate rial covered in the chapter. Notes in the tnllfgin have been included to help the students organize their review work. and cross references provided to help Ihem fi nd the portions of material requiring their special attention. Review Problems. A set of review problems is included at the end of each chapter. These problems provide students further opportunity to apply the most important concepts introduced in the chapter. Computer Problems. The availabi lity of personal computers makes it possible for engineering students to solve a great number of challenging problems. A group of six or more problems designed to be solved with a computer can be found at the end of each chapter. Developing the algorithm req uired to solve a given problem will beneli t the students in two dilfe rent ways: (I) it will help them gain a better understanding of the mec hanics principles involved ; (2) it will provide them with an opportunity to apply the skills acquired in their computer programmi ng course to the sol ution of a meaningful engineering problem.

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Fundamentals of Engineering Examination . Engineers who seek to be li cefl~ed as Pmfessiona/ Engineers must take two exams. The lirst exam. the Fllndam en ta/.I· of Engineering Examination, includes subject material from Mechanics of Materials. Appendix E li sts the topics in M ech{/ni c~' of Materials that are covered in thi s exam along with problems that can be solved to review this material. SUPPLEMENTAL RESOURCES Instructor's Solutions Manual. The Instructor"s and Solutions Manual that accompanies the lirth edition continues the tradition of exceptional acc ura ~y and keeping solutions contained to a single page for easier reference. The manual also features tables designed to assist instmctors in creating a schedule of assignments for their courses. The various topics covered in the tex t arc listed in Table I. and a suggested number of periods 10 be spent on each topic is indicated. Table II provi des a brief description of all groups of probl~l1l s llnd a classilication of the problems in each group according to the units used. Sample lesson schedules are also found within the manual. McGraw-Hill's ARIS-Assessment, Review, and InstrUction System. ARIS is a complete, on line tutorial. electronic homework

and course management system designed 10 allow instructors to create and grade homework assignments. edit questions and algorithms. impan their own content, create and share course materi als with other instructors. and create announcements and due dates for assignments. ARIS has automatic grading and reporting of easy-to-assign algorithmically generated homework, quizzes. and tests. Students benelit from the unlimited practice via algorithmic problems. Other resources available on ARIS include PowerPoint tiles and images from the tex!. Visit thl:. site at www.mhhe.com!beerjohnston. Hands-On Mechanics. Hands·On Mechanics is a website designed for instructors who arc interested in incorporating threedimensional, hands-on teaching aids into their lectures. Developed through a partnership between McGraw-Hill :md the Department of Civ ii and Mechanical Engineering at the United States Military Academy at West Point, thi s website not only provides detailed instructions for how to build 3- D teaching tools using materials found in allY lab or local hardware 5tor~ but also provides a community where educators can share ideas, trade best practices. and submit their own demonstrations fo r posting on the site. Visit www.handsonmechanics.com to sec how you can put this to use in your classroom. ACKNOWLEDGMENTS

llle allth()rs thank the many companies that provided photogr:lphs for

thi s edition. We also wish to recognize the determined efforts and pat ience of Ollr photo researcher Sabina Dowell.

Preface

xvii

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xvii i

Preface

We are pleased to recogni ze Demlls Ormand of Fine-Line lIlustmtions for the artful illu strations which contributed so much to the effecti veness of the text. Our special thanks go to Professor Dean Updike. of the Department of Mechanica l Engineering and Mechanics, Lehi gh Uni versity [0[ hi s patience and cooperation as he checked the soluti ons and answers of all the problems in this edition. We also gratefully acknow ledge the help. comments and suggestions offered by the many lIsers of previous editions of M t!.c/wll ic.f of M ater ia/s.

E. Russell Johnston. J/: John T De Ht)lj" Dl/Fill F M a:;un'k

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List of Symbols "

A, B. C, ... A. B. C.

c C, d D

E

I F G

"

H H,J. K I. I , . .

1,,_, •

, J

Constant; distance Forces: reactions Points Area Distance: width Conslalll: distance: radius Centroid Constants of in tegration Column stability factor Distance; di ameter: depth Diameter Distance: eccentricity: dilatation Modulus of elasticity f requency: funct ion Force I'actor of safety Modulus of rigidity: shear modulu s Distance; height Force Poi nts MOlll cHI of incnb Product of incn ia Po lar moment of in en ia Spring cOlislal11; shape faclOr; bulk modulus: COIlSlanl

1\

L L,

'"

M M . M ,.

fda AI,

M,

" f1 P P/J

Stress cOllcelllralion fa.:lor. tors io nal spri ng constalll Length; span Length: span Effective length Mass COIJp!c Bending 1ll0lllCfiI Bending moment. dead loud (LRI'O ) Bending Illomcm. live load ( LRFDl Bending moment. ultimate load (L RI'D) Number: ratio of moduli of e lasticity: normal direction Pressure I'orcc; concentrated load Dead load (LRFD)

I"

P, q Q Q

R R

S

,

T T II. "

"

U

V V

w

W. IV r. \','::: -'t. " . :

Z cr , (J ,),

y

y"

,

y,

,n "p ", J, w

Live load (LRFD) Ultimate load (LRFD) Shearing force per unit length; shear flow Force First moment o f area Radius: radius of gyration Force; reaction Radius; modulus of rupture Length Elastic sect io n modulu s Thickness; distance: tangential deviation Torque Tcmpcmtu re Rectangu lar coord inates Strain-energy dens ity Strain energy: work Veloci ty Shearing force Volume; shear Width '. dL~tance: load per unit le ngth Weight. load Rectangu lar coordinates; distancc; displacement s; deflections Coordinates of centroid Plastic section modulus Aagles Coefficient of thermal expansion: influence cccfficient Shearing strain: ,pccitic wcight Load factor. dead load (L RFD ) Load factor. live load (LRFD) Defonnation; displacement Nor mal strain Aolgle; slope DII'Cction cosine Poisson's ratio Radi us o f curvature: d istance: density NDrmal stress Shearing Slress Angle; angle of twi st; resistance faClor A.clgu lar veloc it y

xix

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MECHANICS OF MATERIALS

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C

HAP

T

E

Introduction-Concept of Stress

This chapter Is devoted to the study of Ihe stresses occurring in many of the elements contained in these excavators, such 3S two-force members, axles, bolts, and pins.

R

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2

Introductlon-Concept of Stress

1.1 . INTRODUCTION

The main objective of the study of the mechanics of materials is to provide the future engineer WiUl the means of analyzi ng and designing various machines and load-bearing structures. Both the analysis and the design of a given structure invol\'e the determination of stressl'S llnd dejormatiom. This first Chapter is dcvoted to the concept of streu. Section 1.2 b devoted to a short review of the b:l~ie methods of statics and to their application to Ule determination of the fOKes in the members of a simple .structure consisting of pin-connected members. Section 1.3 will introduce. you to the concept of SIJ"I!JS in ~ member of a structure. and you will be shown how that stress can be determined from the jorce in the member. After a shOlt discussion of engineering analysis and design (Sec. 1.4), you will consider successively the 110fIIUd stresses in II member under axial loading (Sec. 1.5). the shearing ,\·trl'.ues caused by the application of equal and opposite transverse forces (Sec. 1.6), and the bearing stresses created by bolts and pi ns in the members they connect (Sel' . 1.7). TIlese vari· ous concepts will be applied in SL'C. I.X to the detelmimuion of the stresses in the members of the simple stlUcture considered earlier in Sec. 1.2. The lirst part of the chapter cnds with a description of Ihemethod you should use in the solution of an assigned problem (Sec. 1.9) and with a di scu.~sion of the numerical accuracy appropriate in engineering caicul:lIions (Sec. 1.10). In Sec. 1.11 , where a two-force member undcr axial loading is con· s irle rerl 'lg:lin, it will he nh~ervel' IhM Ihl'. ~ tTf~S~e.~ nn :In O/Jlif}lIf' rlJine indude both 1101"11/(/1 and shearing stresses. while in Sec. 1.12 you will note that six cl!lnjJl!Ill'n/s are required to describe the state of stress at a point in a body under the most general loading conditions. Finally, Sec. 1. 13 will he devoted to the detennination from test specimens of the Ilitimate streng'" of a given material and to the use of ajaclor of safety in the computation of the al/()lI'aiJIe load for a structural componcnt made of that material.

1 .2. A SHORT REVIEW OF THE METHODS OF STATICS In this section you will review the basic methods of statics while detennining the forces in the members of a simple structure. Consider the structure shown in Fig. 1.I , which was designed to support a 30-kN load . It consists of a boom AB with a 30 X 50-mm rectangu lar cross section and of a rod Be wi th a 20-mm-diallleter circular cross section. T he boom and the rod arc connected by a pin at B and are sup ported by pins and bmckets at. A and C, respective ly. Our first step should be 10 draw a free-body dillgrmll of the struc ture by detaching it from its supports at A and C, and showing the reactions that these supports exert on the structure (Fig. 1.2). Note that thc sketch of the structure has been simplified by omi tting all unnecessary details. Many of you may have recognized at this poi nt that AB and Be aft: 11\"0force memhers. For those of you who have not, we wi ll pursue our analysis, ignoring that fact and assuming that the directions of the reactions at A and Care unknowll . Each uf these reactions, therefore, will

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4

Introductlon-Concept of Stress

These results cou ld have bet!n anticipated by recognizing that AB Be are two-force members. i.e., members that arc subjected to fo rces at only two points, these points beingA and B for member AB, and Band C fo r member BC. Indeed. for a two-force member the lines of action of the resultants of the forces acting at each of the two points art! e4ual and opposite and pass through both points. Using this pruperty. we could have obtained a simpler solution by considering the freebody diagram of pill B. The forces on pin B are the forces FAR and F BC exerted, respecti vely. by members AB and BC, and the 30- kN load (Fig. 1.4a). We can express that pin B is in equilibrium by drawing the corresponding force triangle (Fig. 1.4/)). Since the force FBC is directed along member BC, its slope is the same as [hat of BC, namely, 3/4. We can, therefore, write the proportion ~m d

'IOLN (II )

Fig. 1.4

from which we obtain FBC = 50kN

The forces F~B and F6C exerted by pin B. respectively. on boom AB and rod BC arc equal and opposite 10 FAll and F8e (Fig. 1.5).

I'J ~

Fig. 1.5

A

Fig. 1.6

Knowing the forces at the ends of each of the members, we can now determine the internal forces in these members. Passing a section at some arbitrary poim 0 of rod BC, we obtain two portions BO and CD (Fig. 1.6). Since 50-kN forces must be applied at D to both portions of the rod to keep [hem in equilibrium, wt! conclude that an internal force of 50 kN is produced in rod BC when a 30-kN load is applied at B. We funher check from the directions of the forces FIIC and F kc in Fig. [.6 that the rod is in tension. A similar procedure would enable us to determine that the internal force in boom AB is 40 kN and that the boom is in compression.

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1.3. Stresses In the Members of a Structure

1.3. STRESSES IN THE MEMBERS OF A STRUCTURE

While the res ults obtained in the preceding section represent a t:irst and necessary step in the analysis of the given structure, they do not tell llS whether lhe given load can be safely supported. Whether rod Be. for example. wilJ break or not under this load ing depends not on ly upon the value round for the internal force FBI.'. but also upon lhe crosssectional area of the rod and the material of which the rod is made. Indeed . the internal rorce FBI.' actuall y represents the resultant of eleme ntary rorces di stributed over the e ntire :trea A of the eross section (Fig. 1. 7) and the average intensity of these di stributed forces is e4uul to the force per unit area, F/Jc/A, in the section. Whether or not lhe rod will break. under the given loading clearly depends upon the abi lity of the material to withstand the corresponding vulue FBC/A of the intensity of the distributed internal forces. It thus depends upon Ule force F Be• the cross-sectional area A . and the material of the rod. The force p<:r unit area. or intensi ty of Ule forces distributed over a given section, is called the stres.l· on that section and is de noted by the Gree k leller u (sigma). The stress in a member or cross-sectional area A subjected to an axial load P (Fi g. 1. 8) is thererore obtained by di viding the magnitude P of the load by the area A : P {T = -

A

t'

( 1.5)

A positive sign will be used to intJicatl! a tensile stress (member in tension) and a negative sign to indicate a compressive stress (member in compression). Since SI metric units are used in this discussion, with P expressed in newtons (N ) and A in s4uare meters (m!). the stress (T will be expressed in N/ m2 . Thi s un it is called a pa.~cal (Pa). However, one li nds that the pascal i~ :tn exceedingly small quantity and thaI. in practice, mult iples of thi s unit must be used, namely, the kilop:tscal (kPa), the megapascal (MPa). and the gigapascal (GPa). We have

I GPa = 10 9 Pa = I O~ N/ 11l ~

Whe n U.S. customary units are llsed, the force P is usually expressed in pounds (Ib ) or kilopounds (ki p). and the cross-secti onal areu A in square inches (in!). The mess (T will then be expressed in pounds per S4Ullfe inch (psi) or kilopounds per s4uare inch (ksi).t tThe principal S t and U.S. cu~tomary units usctJ in mechanics are 1i,led in IlIblc, im.ide rront CO\"Cr of thi! book. From Ihe table on Ihe right·hantl sitlc. we nOle thai I pI] is approximaJcly equal to 7 kPa. and I hi apprnxim:ucly equal to 7 MPa. ~le

Fig. 1.7

A

Fig. 1.8

'"

'"

Ib\

5

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6

Introductlon-Concept of Stress

1.4. ANALYSIS AND DESIGN Considering again the structure of Fig. 1.1, let us assume that rod Be is made of a steel with a maximum allowable stress a .1I '"' 165 MPa. Can rod Be safely support the load 10 which it will be subjected? Thc magnitude of the force F flC in the rod was found earlier 10 be 50 kN. Recalling that the diameter of the rod is 20 mm. we lise Eq. (1.5) to detennine the stress created in the rod by the given loading. We have P == F8C= + 50kN = + 50 X IO-IN

, (20mm)'

A = 7rr = To ~

a

P A

+ 50X103 N 3 14 X 10 (, m~

,

= To(tO X 10- 3 m)- = 3 14 X

+ 159

X Icf' Pa ==

+ 159 MPa

Si nce the value obtained for u is s maller than the va lue U al l of the allowable stress in the sted used, we conclude that rod Be can safely support the load to which it will be subjected. To be complete, our analysis of the given structure shou ld also include the determination of the compressive stress in boomAB, as well as an investigation of the stresses produced in the pins and their bearings. This witl be discussed later in thi s chapter. We should also determine whether the defo rmations produced by the give n loading are acceptable. The study of deformations under axial loads will be the subject of Chap. 2. An additional consideration required for members in compression involves the .ttabifit), of the member, i.e .. its ability to support a given load withom experiencing a sudden change in configuration. This will be discllssed in Chap. 10. The engineer's role is not limited to the analysis of existing structures and machines subjected to given loading conditions. Of even greater importance to the engineer is the de.\·ign of new structures and machines, that is, the selection of appropriate comJXlnents to perfonn a given task. As an example of design, let LIS return to the structure of Fig. 1.1, and assume that aluminum with an allowable stress U . ll "" 100 MPa is to be used. Since the force in rod Be will still be P = F BC == 50 kN under the given loading, we must have, from Eq. (1.5), p

a . 1I =

A

A

p

SOX 103 N

a.1I

100 X 106 Pa

and. since A = 7rr 2 ,

12.62 X 10- 3 m =

12.62 mm

d = 2r == 25.2 mm

We conclude that an aluminum rod 26 mm or more in diameter will be adequate.

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1.5. Axial Loading; Normal Stress

1.5. AXIAL LOADING ; NORMAL STRESS As we have already indicated . rod Be of the example considered in the preceding section is a two- force member and. therefore, the forces F BC and F~c acting on its ends Band C (Fi g. 1.5) are directed along the axis of the rod. We say that the rod is under lIxiallout/illg . An actual example of structural members under axial loading is provided by the members of the bri dge truss shown in Fig. 1.9 ..

Flg.l.9 This bridge truss consists of two-force members that may be in tension or in comp re ssion.

Returning to rod Be o f Fig. 1.5. we recall that the section we passed through the rod 10 determine the internal fo rce in the rO<.! and the corresponding stress was perpend icular to the ax is of the rod: the internal force was therefore normal to the plane of the section (Fig. 1.7) lmd the corresponding stress is described as a normal ~·t ren . Thus, formula ( 1.5) gives us the normal stre.\"s ill II member umler (trial loadillR= p u=A

( 1.5)

We .~hould a.lso note that, in formul a ( 1.5), u is obtained by dividing the magnitude P of the resultant of the inte rnal forces disuibuted over the cross _~cct i on by the area A of the cross section; it represents, therefore, the aW/"lIge vallie of the stress ovcr the cross secti on. rather than thc stress at a s pecific point of the cross scction. To define the stress at a given point Q o f the cross section . we shou ld conside r a small area !lA (Fig. 1. 10). Dividing the m:lgniwde of h.F by !lA , we obtain the average value of the stress over h.A. letti ng 6.A appro'l.(;h zero. we obtain the stress at point Q:

IJ.F u = lim .lA -->O h.A

(1.6)

"

Fig. 1.10

7

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8

Introductlon-Concept o f Stress

In general, the value obtained fo r the stress u at a give n point Q of the section is diffe rent from the va lue of the average stress given by fonnu la ( 1.5). and fT is found to vary across the section. In a slender rod subjected 10 equal and opposite concentrated loads r and P ' (Fig. l. ll a). this variation is small in a section away from the poi nts ofapplication of the concentrated loads (Fig. 1.11 c), but it is quite noticeable ill the neighborhood of these points (Fi g. 1.11" and tI) . It follow s from Eq. (1.6) that the. magnitude of the resulta nt of the distributed internal fo rces is

"

rfnI p'

(, I Fi g. 1.11

Fig. 1.12

l

(b)

BUI the condi tions of equi librium of each of the portions of rod shown in Fig. 1.11 requi re that thi s magnitude be eq ual to the magnitude P of the concentrated loads. We have. therefore. p' (e)

p'

{II)

( 1.7)

which means that the volume unde r each of the stress surfaces in Fig. 1.I t must be equal to the magnitude P of the loads. This, however, is the onl y infonnilti on that we C
t 5ee H:rdinand P. BlXr and E. Ru Sr,c 1l John,t(lil. Jr,. Mulo",,;c,' for E".~;"ur.,. 4!h cd .. McGraw-HilL New YQf~. 19117. Of Ve<'W" Meci",,,"'s jor E".~;"eer,'. 6th ed .. McGraw· HiIl . New Yor~. 1996. 'CC,I ..').1 and ~.3.

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r

1.6. Shearing Stress

p

"

1-"-],

: :, ,,

L

Fi g. 1.1 3

P'

However, if a two-force member is loaded axially, but eccclllricully lL" shown in Fig. 1.14a, we lind from the conditions of equi librium of the ponion of member shown in Fi g. 1.I4b that the internal forces in a given section must be equivalent to a force P applied at the centroid of the sectio n and a couple M of moment M = Pd. llle di stribution of forces- and. thus. the corresponding distribution of stresses- cannor be IIn iform. Nor can the distribution of stresses be symmetric as shown in Fig. !.II. This point will be discussed in detail in Chap. 4.

(a)

(it)

Fig. 1.1 4

1.6, SHEARING STRESS

The internal forces and the corresponding stresses discussed in Sees. 1.2 and 1.3 were normal to the section considered. A very different type of stress is obtained when transverse forces P and P' are applied to a member AB (Fig. 1.15). Passing a section at C between the points of application of the two fo rces (Fig. 1. 16a), weoblain the diagram of portion AC shown in Fig . I .16h. We conclude that ill1emal forces must exist in the plane of the section, and that their resul tant is equal to P. Thesc elementary internal forces Hrc called shearing forces. and the magnitude P of their resultant is the shear in the section. Dividing the shear

T )" )

9

A

P'

Fig. 1.1 5

Fig. 1.1 6

Ib)

9

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10

Introdu<:tlon-Concept of Stress

P by the area A of the cross section, we obtai n the tll'erage shearinx .I·tress in the section. Denoting the shearing stress by the Greek letter T

(tau). we write ( 1.8)

It should be emphasized that the value obtained is an average value of the shearing stress over the entire section. COlltrary to what we said earlier for nonnal stresses. the distriblllion of shearing stresses acros~ the section canllot be assumed uniform As you wi ll see in Chap. 6, the [tetual value T of the shearing stress vnries from zero at the surface of the member to a maximum value T ma.>. that may be much larger than the average value T aYe '

Fig. 1.17 Cutaway view of a connection with a bolt in shear.

Shearing stresses are commonly found in bolts, pins, and rivels used to connect various structural members and machine components (Fig. 1.17). Consider the two plates A md B, which are connected by 11 bolt CD (Fig. 1.18). If the plates are subjected to tension forces ()f magnitude F, stresses will develop.in the section of bolt corresponding to the plane EE'. Drawing the diagrams of the bolt and of the portion located above the plane EE' (Fig. I. [9). we conclude that the shear P in the section is equal to F. The average shearing stress in thc section is obtained. according to formula (1.8), by dividing the shear P = F by the area A of U1C cross section : P

'+

Tm

=

/J

D

Fi g. 1.18

"I

Fig. 1.19

( 1.9)

c

c

E

F

A= A

£.

e'

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11

1.7, Bearing Stress In Connections

't -'-: H

E

r'

/I

K

"

-•

pI.

L'

,_

F[J

-

-

p

I

"

J

(hi

(, (

G

Fig. 1.21

Fig. 1.20

The bolt we have j ust considered is said to be in sillgle shear. Different loading situations may arise, however. For example. if splice plates C and D are used \ 0 connect plates A and B (Fig. 1.20), shear will take place in bo lt H.I in each o f the two planes KK' and LL ' (and simi larly in boll £G). The bolts are said to be in llouhle shear. To determine the average shearing stress in each plane, we draw free-body diagrams o f bolt H.I and of UIC ponion of bolt located between the two planes (F ig. 1.21). Observing that the shear P in each of the sections is P = F/ 2, we conclude that the average sheari ng stress is

T

=

ne

P A

-

=

F/2 A

--

~

F 24

-

(1.1 0)

1.7. BEARING STRESS IN CONNECTIONS Bolts. pillS, :md rivets create stresses in Ihe members they cOllnect. along the bearillg sllIjace, or surface of contact. For example. consider again the two plates A and B cunnected by a bolt CD that we have di scussed in the preceding section (Fig. 1.1 8). The bolt exerts on plate A a force P equal and opposite to the force f exerted by the plate o n the bolt (Fig. 1.22). T he force P represents the resultal1l of elementary forces di stri buted on the inside surface o f a half-cylinder of di ameter d and of length t L"q ual 10 the thickness of the plate. Since the di stributi on of these forces-and of thc corresponding stresses- is quite complicated, one uses in practice an average nominal value lIb of the Siress, called Ihe bearing stress, obtained by dividing lhe load P by the area of the rectanglc repre~nting the projection of the bolt on the plate section (Fig. 1. 23). Si nce this area is equal to td, where t is the plate thickness and d the diameter of the bolt, we have

u b

p P = - = A III

~: ~.

--?;r

~

F'

~

~

Fig. 1.22

(l.ll) Fig. 1.23

D

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12

Introductlon-Concept of Stress

1.8. APPLICATION TO THE ANALYSIS AND DESIGN OF SIMPLE STRUCTURES

We are now in a position to delennine the stresses in the members and connections of various simple two-dimensional structures and, thus. to design such structures. As an example, let us return to the structure of Fig. 1.1 that we have already considered in Sec. 1.2 and let us specify the supports and connections at A, B. and C. As shown ill Fig. 1.24. the 20-mm-diameter rod Be has nat ends of 20 X 40-mm rectangular cross section. while boom AB has a 30 X 50-mm rectangular cross section and is titled with a clevis at end B. Both members are connected at B by a pin from which the 30-kN load is sllspended by means of a U-shaped bracket. Boom AB is supported at A by a pin fitted into a double bracket, while rod BC is connecled at C to a single bracket. AU pins are 25 mm in diameter.

'-1 20 ""n TOP "I EW OF ROI) IJ C

f

600"""

FROl\T VI EW

1

END VIEW 125m",

.1Om'"

r """'J :::::I20 """

~25"""

1---1

Fig. 1.24

TOI' VtEW OF BOO,'.! AB

d _ 25 """

a.

Determination of the Normal Stress In Boom AB and Rod

Be. As we found in Secs. 1.2 and 1.4. the fo rce in rod Be is FBe = 50 kN (tension) and the area of its circular cross section is A = 314 X 10- 6 m"; Ihe corresponding average nonnal stress is aBC = + 159 MPa. However, the nat parts or the rod arc also under

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tension and at the narrowest section, where a hole is located, we have A = (20 mm ){40 mm - 25 mm)

-=;

1.8. AnalysIs and DesIgn of Slmpte StrucllIres

13

300 X 10 - 6 m 2

The corresponding average value of the stress, therefore, is P

(tTHc)
A=

50X 10' N 300 X IO-b

m2

167 MPa

Note thar thi s is an average !'111m': close to thc hole, the stress will actuall y reach a much larger value, as you will see in Sec. 2. I 8. It i .~ clear that, under an increasing load, the rod will fail ncar o ne of the holes rather than in its cylindrical ponion: its design, therefore, could be improved by increasing the width or the thickness of the nat ends of the rod.

Turning now our allemion to boom AB. we recall from Scc. 1.2 that the force in the boom is F il8 = 40 kN (compression), Since the area of the boom's rectangular cross section is A = 30 mm X 50 mm = 1.5 X IO- J m!, the average value o f the normal stress in the main part of the rod, between pins A and B. is 40X IO]N

'0 )

- 26.7 X 106 Pa = - 26.7 MPa

Note that the sections of minimum area at A and B arc not under Stress. since thc boom is in compressio n, and, therefore. pushes on the pins (instead of pufling on the pins as rod Be docs).

(It)

Fia. l.25

b. Determination of the Shearing Stress in Various Connections. To determi ne the shearing stress in a connection such as a boll. pin. or rivet, we first clearly show the forces exened by the various members it connects, T hus, in the case o f pi n C of our example (Fig. 1.25a), we draw Fig. 1.25b, showi ng the 50-kN force ex.erted by member Be on the pin. and the equal and opposite force exerted by the bracket. Drawing now the diagram of the portion of the pin located below the plane DD' where shearing stresses occur (Fig. 1.25("). we conclude that the shear in thm plane is P = 50 kN. Since the crosssectional arca of the pin is

A

-

10 l :o.

mm)' = rr{12.5 X

25A = rrr-, "'" rr ( 2-

'")

we lind that the average value of the shearing stress in the pin at C is

d .. 25mm

P 50 X 10 3 N T,,,e = -A ~ 491 X 10- 6 m-' = 102MPa

H

,,-

Considering now the pin at A (Fig. 1.26). we note that it is in double shear. Drawing the free -body diagrams of the pin and of the portiun of pin located betwee n th~ planes DD' and EE' where shearing stresses occur. we conclude that P = 20 kN and that T " 'e

P 20 kN = - = A 491 X JO 6

_

m~

(I»

40.7 MPa Fig. 1.26

-

!OitN

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14

Int-oduC1lon----Ccncepl cl 51.."..3

Con sideling the pill at B (Fig. 1.27(1), \\Ie 1I01c that Ihe fJili Illay be divided into five portions which are. acted upon by forces exerted by the boom, rod. and bracket. Considering ;uccessive ly the ponions DE (Fig. 1.27b) and DC (Fig. 1.27cl. we conclude that the shear in section E is f'£ = 15 kN, ..... hil e the shear in section G is PG = 25 kN. Smce the kk"lding of the pin IS symmetric, we conclude that the maximum value of lh ~ shear in pin B is PG "" 25 kN .•md that the largest shear· ing stresses occur in sections C and H. where

25 kN

''iO.9 MPll

491XlQ o m!

1.1

~

'

Determination of the Bearing Stresses. To determine the nominal bearing ~trl' SS at A in member AB, we use formula (I II ) of Sec. 1.7. From Fig. 1. 24, we have t - 30 mm and d - 25 mm . Recalling thm P = FAe = 40 kN. we have C.

£

P

D

fJh

=

7d =

40kN (30 rnm)(25 111111 )

53.3 MPa

To obtllin the beari ng stress in the brackf>l at A. we I1se I = 50mm and d = 25 mm: P

fJ h

= ~Id ~

40 kJ'J" ('0 " 5 min ) ~ mm )(L

= 2(25 mm)

= 32.0 MPa

The bearing stresses at B in member AB, at Band C in member BC. and in the bracket at C are fOlmd in a simi lar \.\iay. Ie)

fi g. 1.27

1.9. METHOD OF PROBLEM SOLUTION

YOu should ap proach a problem in mechanics of materials as yo u wo uld approach an actual engineering silLLation. By dra wi ng 011 your own experience and intuition, you will lind it ea.<:ier to understand and fonnu late the problem. Once the problem has been clearly stated, however. there is no place in I ts solution fo r yo ur panicular fancy. Y {lU( solution mlJ.~t he ha~erl on the flindament:!1 princi pl es of , rmi r.,," and on the princi ples you will learn in this cOLirse . Every step yo u lake must be justified on that basis. leaving no roolll for jour " intuition." After an answer bas been obtained, it should be checked . Here agai n. you may call upon your common sense and personal experience. [ f not compl etely satisfied with the result obtained. you should carefully check yo ur formu tatIOn of the problem, the validity o f the methods used in its solution, and thc accuracy of your compulmions. The MaUll/em of the problem ~hou l d be dea r and pn:ciSi:. It should contain the given dam and indicate what information is required. A sim· plified drawing showing all csscntial quantities involved should be included. 111e solution of most of the problems yo u will e ncounter wi ll necessitate thtl.t you first dc!ermine the reactioll.l· lI! ~lIpports and illter

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/lal /()rce.~ and couples. This wi ll rC(luire the drawing of one or several free-hody diagrwlIs. as was done in Sec. 1.2, from which you will write equilibrium eqlUltiQns. These equations can be solved for the unknown forces, from wh.ich the required stresses and de/orma/ions will be computed. Artcr the answer has been obtained, it shou ld be carefillly checked. Mistakes in reQ~'()ning can orten be detected by carrying the units through your computations and checking the units obtained for the answer. For example, in the design of the rod discussed in Sec . 1.4, we found. after carrying the unil~ through our computations. that the re(Iuired diameter of the rod was expressed in millimcters. which is the correct unit for a dimension; if another unit had been found , we would have known that some mistake had been made. Errors in cOIIIPutation will usu:dly be found by substituting the numerical values obtained into an e£luati on which has not yet been used and verifying that the equation is satislied. The importance of correct computations ill engi neeri ng crumm be overemphasized.

1.10. NUMERICAL ACCURACY The accuracy of the solution of a problem depends upon two items: (I) the accuracy of the given data ,md (2) the acc unlcy of the compuullions performed. The solution cannot be more accurate than the less accurate of these two ilcms. For example, if the loading of 11 beam is known to be 75,000 Ib with a possible error of 100 [b either way, the relative error which measures the degree o f accuracy of the data is [00 Ib

75,000 Ib

0.00[3 "" 0.[3%

In computing the reaction at one of the beam supports, it would then be meaningless to record it as 14.322 [b. The accuracy of the solution

cannot be greater than O. [3%, no maller how accurate the computations are, and the possible error in the answer may be as large as (0.1 3/100)( 14,322 Ib) = 20 lb. The answer should be properly recorded as 14,320 ::!: 20 lb. In engineering problems, the data are seldom know n with an accuracy greater than 0.2%. It is therefore seldom justified to write the answers to such problems with an accuracy greater than 0.2 percent. A practical rule is to use 4 figu res to record numbers beginning with a "[" and 3 figu res in all other cases. Unless otherwise indicated. the data given in a problem should be assumed known with a comparable degree of accuracy. A fo rce of 40 Ib, fo r example, should be read 40.0 Ib, and a force of [Sib shou ld be read 15.00 lb. Pocket calculators and computers are widely used by practici ng engineers and engineering students. The speed and accuracy of these devices facilitate the numerical computations in the solution of many problems. However, students should not record more significant ligures than can be justified merely because they are easily obtained. As noted above, an accuracy greater than 0.2% is seldom necessary or meaningful in the sollllion of practical engineering problems.

1.10. Numerical Accuracy

15

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SAMPLE PROBLEM 1.1 In lht hangnshuwn, (tit:: upptr portiun uf linl ABC is i ill. lhkk amJ ltlt:: luwer p.1ftions are each t in, th ick. E)Xlxy resin is used 10 bond the upr;r and lower portions together al B. The pm at A is o rl-in. di amete r while a .-in.-diameter pill is used at C. Detemline (a) the shearing stress in pin A, (b) the shearing stress in pin C, (c) the largest nonnal Slre:;s in link ABC, (d) the average shear in)! Slress on the bonded surfaces at B. (e) the bearin)! stress in the link ;)\ C.

SOLUTION FITI' Hud y: Entin' Hanger. Since the link ABC is a two-force member, the reaction at A is vertical; the reaction a D is repre., entcd by il ~ components D., amJ D .,. WI;! ""rill: 0"

n

+~ IMD =

,~

(5 UU lbXl5 in.) - rAdIO in.) = ()

0: F1C

D,

=

+750 Ib

r.J,C=7501b

Since Ihis ~-jn .-d iameler pin is in single

Il. S hea ring Stress in Pin A. shear. we write l()in .-

I

,

"~IC ~

-

~

A

A

lellliioll

750lb t 1T(O.375 in.)'

t.,

= fi790 psi
Since this !-in.-diamete, pin is in double

IJ. ShearIng SIf't"SS In PIn Co wP. Wrilp.

~hp.~r.

~c =

iF

AC

- -A

375 1b = i",(O.25 in. r

"7('=

7640 l1'> i ....

c. Largcsl Norlllal Siress In Link ABC. The largest stress is found where the area is smallest: this OCCUf3 al the cross seclion at A ",here the i ·in. hole is located We have rAe A .. ,

750lb

750 lb U. 375 in.)

(; In.)(1.25III.

0.32 8 in!


=

2290 p,i
d. An'mge Shellring S trl'SS a l B. We nOle Ihal bonding exisl, on bolh sides of Ihe upper portion of the link and Ihal the shear force on each side is F = (750 Ib)/ 2 = 375 lb. The :werage shearing stress on each surface is thus

I ~ -;n .dian"'I(" F. ,r •

FI

F," 1'2 " ~F~c _ 37,; II

rfj -

A-

375 lb ( 1.25 in.)i:1.75 in. )

F .. 37511,

e. Drnri ngSIi·es.\ in Link a\ C. For each portiOll oflhe link. '"I - 375 lb and the nominal beanng lrea is (0.25 in.)(0.15 in.) = 0.0625 in 2.

FI

fT .

' 1 -ln . diarr,,,,!'r

16

=

A

375 Ib = 0 .0625 inl

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SAMPLE PROBLEM 1.2 The sl<:t:i lit: bar shown is LU hi:: !.It:signed LU carry a l~nsiun furet: uf In
SOLUTION

",

II. Diameter or tJlt' Bolt. ~P =60 kN.

Smce the holl is in lloubJe shear. F ] =

p

no kN = --

F,

T= -

t='

nO ~N 100 MPa = -, - , , 11" J"

huf

A nUl

d = 27.6 111m

We will use AI Ihi! point we check the bearing Ille 28'IllIll-diamett:! boll.

T

•

Slres,~

P

120 iN

Id

(U. I)2U m)(O.U2M m)

=-=

;mn rhar Ih ..

214M Pa<350MPa

\p

We consider one of the end

that the thidmcss or tilt:: stet:! platt: is I = 10 mm ~rre,,-~ mll~1 nnt ex'~f't>.d 17) VIP;" we wrire

60 kN 175 MPa = ,;c=~­ (0.02 m)"

u= ~

'" b = d

+

2(1 = 28mm

c. Dimens i{)n " orlhc Bar. i, r - 20 mm. w: h.we p

0,

OK

b~r. Rt:\:allin~

aV~r.12e r~mile

u =-


betwee n the 2O·mm-thick plate and

b. OiJmllsio n b al Each End of th e Bar. puniuns ur tht:

o/=28 mm

+

a= 17.14mm

2(17.1 4 III Ill)

b

=

62.3

lllHl ~

Recalling that the thidness of the steel plate

120 kN 175 MPa = ~~~C ([I.U20 m)h

il = 34.3 rum We will use

Ii == 35

111111 ~

17

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PROBLEMS

1 .1

Two wlid L),lindrical rods All and Be are we lded together at Band

loaded as shown. Detennine the magnitude of the force P for which the tensile s tress In rod AS is twice the magnitude of the cmnptessive stress in rod BC

-I " 3(.(} ''''''

~

25()

Fig. Pl.1

"u"

I

1 .2

-ii.,

-

L .JlIk.\

[n PlOb. I.

t. kllOwing llial P - 40 kips. dele, mine the ave rage nor-

mal stress III the midsection of (II) rOO AB. (b j rod BC 1 .3 Two w lid cyliniincaJ rods AB and He are we ld ~d together aT B :md loaded a s shown. Knuwillg lhat lh e average normal stress must nOl exceed 175 MP a in rod AS alld 150 r>.I P a in rod Be. delennine the smalles t al lowable

values of J, and d!.

FIg. P1.3 :md P1 .4

1 .4 Two ~Iid cylindrical nlI1.~ A R ~n
"2'"

J::!(~I N

r

1.5 A strain gage located at C on the surface of bone AB indicates that the average normal stress In the bone is 3.80 M Pa wh~n the Ixme is s!l bjecte(i to two 1200·j\" force s as shown. Assuming the cross section of tl1e bone at C t<) be annular and knowin g th~t its outer diameter is 25 mm, det .. m)in ~ the. in. ncr diameter of the bone's cross sec tion at C. 1.6 Two steel plates are 10 bt: held together by means of I G·mm·dia.ueteJ high.strength stee l baits fitting snugly inside cylindrical brass spacer.~. Know· ing that the average normal stress must not exceed 2((1 MPa in the bolts and 131) MPa in the spaccn. detennine the oUler d~1meter 01 the spacers thaI yie lds th~ nlO~t oconomical omd safe design.

B

Fig. P1.6

18

Fig. P1.6

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1 .7 Each of the four ven ical links has an 8 X 36-mm uniform rectangular cross section and each of the four pins has a 16-mm diameter. Detennine the maximum value of the average normal stress in the links connccting (1I) points Sand D. (b) points C and E.

19

Problems

~" H

\l

1\ 1."" \

1.92",

J

c

0_00 '"

Fig. P1.7

Fig. P1 .8

1 .8 Knowing that the central ponion of the link SO has a uniform cross· sectional area of 800 mm'. determine thc magnitude of the load P for which the nomlal stress in that ponion of SO is 50 MPa.

ir

1.9 Knowing that the link DE is ~ ill. thick and I in. widc. detemline the nonnal stress in the central ponion of that link when (1I) 0 = O. (b) (J = 90°.

1in.

1.10 Li nk AC has a unifonn rectangular cross section k in. thick and wide. Detemline the normal stress in the celllral portion of the link.

2,,,

1' 0 L , 'B

12

;"- +1£''''''1' '' o

o C

oS in

-.

~~-~-~-~-~-~~~~ ~ ~

B

3 in.

2AOlh

I

I

.------->

30'

Fi g. P1. 9

6 in

7 in.

.1

\l

A

I

210 Ib

C

Fig. P1.10

1 .11 The rigid bar EFG is supponed by the truss system shown. Knowing that the member CG is a solid circular rod of 0.75-in. diameter. determine the nonnal stress in CG. 1 .12 The rigid bar EFG is supponed by the truss system shown. Determine the cross-sectional area of member AE for which the nomlal stress in the member is 15 ksi.

-

~

F

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20

Introcluctlon-Concept of Stress

1.13

A couple 1\'1 or magnitude 1500 N . m is applied 10 the crank of ~hown. detennine (a) Ihe force P required 10 hold the engine system ill equilibrium. (b) the llverage nomlal s tress ill the COIlnecting rod BC, which has a 450-mm ~ uniform cross section.

an engine. For the position

1.14 An aircmft lOW bar is positioned by mean~ of a single hydraulic cylinder connected by a 25·mm·diameler ,Ieel rod to IWO identical arm-andwheel units DE,... The mass of Ihe entire tow bar is 100 kg. and ils center of gravity is located at G. For the posilion shown. determine the nonllal stress in Ihe rod.

Di",,,,,,,i,,,,.",,,,,,,,

~ ., ~15°-1 \...'" 7 ' • I>

F1g. P1.13

.

850

I I 00

D

r. . ;

c

E

.

"*' . -

'~I

250

.

.

Fig. P1 .14

1.15 The wOCKien members A and Bare 10 bejoined by plywood splice plates thai will be fully glued Oil the surfaces in contacl. As pan 01" the design of the joinl, and knowing that the clearance bel ween the ends of Ihe members is to be 6 mm. determine the smallest allowable lenglh L if Ihe average shearing stress in the glue is 1101 to exceed 700 kPa.

t

l:;h

.<

'1 L

6 nu"

) i5 ,,, m

Fig. PUS

FIg. PUS

1.16 When the force P reached I6CO lb. Ihe wooden specimen s hown faited in shear along the surface indicated by the dashed line. Delermine the average shearing Stress along thai surface al the time or failure.

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Probtems

1. 17 Two wooden planks. each} in. thick and 9 in. wide. are joined by the dry mortise joint shown. Knowing that the wood used s hears off along ils grain when the average shearing stress reaches 1.20 ksi. determine the magnitude P of the axial load that will cause the joint to fail.

LI' i

,

lh

l

""

10"",

-

-

p'

p

12"",, _1 1p

Fig. P1.1B

Fig. P1. 17

1 .18 A load P is applied to a steel rod supported as shown by an aluminum plate into which a 12-mm-diameter hole has been drilled. Knowing that the shearing stress must not exceed 180 MPa in the steel rod and 70 MPa in the aluminum plate. detennine the largest load P that can be applied to the rod. 1 .19 The axial force in the column supponing the timber beam shown is P = 20 kips. Detemline the smallest allowable length L of the bearing plate if the bearing stress in the timber is not to exceed 400 psi. 1 .20 The load P applied to a stl!el rod is distribured to a timber support by an annular washer. The diameter of the rod is 22 mm and the inner diameter of the washer is 25 mm. which is slightly larger than the diameter of the hole. Detennine the smallest allowable outer diameTer d of the washer. know ing thal the axial normal stress in the steel rod is 35 MP·J. and that the average bearing stress between the washer and the timber must not exceed 5 MPa.

FIg. P1.19

i<; d

. ·· ·

C>

~

100 """

Fig. P1 .2G

1.21 ported by maximum which the

A 40-kN axial 10;ld is applied to a short wooden post that is su pa concrete footing resting on undisturbed .,oil. Detennine (a) the bearing stress on the concrete footing. (bl the size of the footing for average bearing s tre~s in the soil is 145 kPa.

Fig. P1 .21

21

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22

Introduction-Concept

1.22 .0\" a.'\ialload P is S Upj>Oll~d by a shOlt wa X 40 COIUlIlll of cross· sectional area A '" 11.7 in .~ and l S distributed to a concrete foundntion b)' a square plc!te as shown Knowing tflat the average normnl s tres~ in the column must nOl exce~d 30 k~j and that the ~aring sire... on Ihe concrete 10undatJOn must nOI c.'\ceed 3 0 ksi. d~termin~ Ihe Hide. " of Ihe plate lhal will provide th e most economical and , nfe design.

01 SIre""

1 .23 .0\ 6-mm-diameter pin is used at connec tion C of Ihe pedal shl)wn Knowing that P = 500 N. determine (Il) the ,n'erage shearing ,tress in the pin. (bi the nominal bearing stress in the pednl at C. (c) the Iltlminal b~aring stre ss in each support brocket at C.

1.24 Kllowing that 3 force I' ofma/;nitude 750 N is applied to the pedal shown. delermine la) Ihe diameter of the pin al C for which lite average s hear· ing stress in the pin is 40 /'.lPn , (b) the corresponding Ix:llring stress in Ihe pedal at C. (d Ihe corresponding bc:J.ring stress in each suppon bracket at C.

Fig. P1. 22 75mm

!l ,n",

1i

,

A

I

I~'"" ''''''-I H P

125 ' 1"01

~f)==::Z~D

1.25 _0\ ~- in, -d iiimetCr steel rod AD is lined to a ro und hole near end C o f the \\'ouden memb<.'r CD. For the Ioad mg shown, determine (It) Ihe maxi. mum average lIonml 51rCS S ill the wood. (b) the di stnnce b for which the nv· erdge shearing stress is 100 psi on the surfaces illdkmed b)' the d~L'i hed lines. k ) the average bearing .'I tn:ss tm the wood.

1 .26 Two identical linkage.and.hydraul ic.,·)'linder sy stem~ control th e position of the forts of a fork-lirl tfuct. The load supponed by the one system shown is 15()() Ib_ Knowin g that the Ih icknes., of member BD is! in_ . deter. mine (11 ) lhe average shearing , Iress in the ~- in ..diameter pin at B. fb) Ihe bear· ing stress at B in member BD , -

Sill'"

Fig. P1.23 and P1.24

t"..Jlb

I I" .

./"'-, 44 / IJ

Fig. P1.2S

Fig. P1.26

1.27 Fnr Ihe ~~semhly all/I in.1< 50-mm uniform lcctllllguiar cross scc1ion. ag~

Fig. P1. 28

1.28 Li nk A8. of width b ~ 50 mill and Ih ic kness I = 6 mm. is used I
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1. 11 . Stress on an Oblique Plane Under /lJl:lal Loading

1.11 . STRESS ON AN OBLIQUE PLANE UNDER AXIAL LOADING In the precedi ng sections. ax ia l fo rces exerted on a two-force member (Fi g. 1.28a) were fo und to cause normal stresses in that member (Fig. 1. 28b), while transverse forces exerted on bolts and pins (Fi g [. 29a) were found to cause shearing stresses in those connections (Fi g. 1.29h). The reason such a relation was observed between axial fo rces and normal stresses on one hand. and transverse forces and shearing stresses un the other, was because stresses were being determined only on planes perpendicular to the axis of the member or connection. As you will see in this section, axial forces cause both normal and shearing stresses on planes which are not perpendicular to the axi s of the member. Si milarly, transverse forces exerted on a bolt or a pin cause both normal and shearing stresses on planes which arc not perpendicular to the axis of the bolt or pi n.

(b)

Fi g. 1.28

(, ) Fi g. 1.29

Consider the two-force member of Fig. 1.28. which is subjected 10 axial rorces P and P '. If we pass a secti on forming an angle () with a nonnal plane (Fig. [ .30a) and draw the free-body di agram of the portion of member located to the left of that section (Fig. 1.30b), we fi nd from the equilibrium conditions of the free body that the distributed forces acting on the section must be equ ivalen t to the force P. Resolving P into componellls F and V, respectively normal and langelllial to the section (Fig. 1.30e). we have F = P cos O

V= P sinO

(, '

( 1.1 2)

The force F represellls the resultant of nomml forces distributed over the section, and the fo rce V the resultalll of shearing forces (Fig. 1.30i/). The average valLles of the corresponding normal and shearing stresses are obtained by divid ing, respectively, F and V by the area A& of the section: (T =

F ~

A,

V

T =-

A.

'"

(1.13)

Substituting for F and V from (1.1 2) into (1.1 3), and observing fro m Fig. [. 3Oc that Au = A9 coS" (). or A ~ = A(/cos O. where An denotes the

(d ) Fig. 1.30

23

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24

Introductlon--Goncept of Stress

arca of a section perpenllicular to the axis of the member. we obtain P sin ()

Pens 0 a

~

7

---

A ~cos

P , u = - cos- {j

~

0

At/cos 0

T

=

A"

P

.

~ Sl n

(j cos 0

{ 1.I4)

A"

We note from the first of Eqs. ( 1.1 4) that the normal stress u is maximum when 0 =: D. i.e. , when the plane of the section is pcrpendi · c ular to the axis of the me mber. and that it approaches zero as 0 ap· proaches 90°. We check that the value of u when 11 =: 0 is p

u",""'A",

( 1.1 5)

as we ro und earlier in Sec. 1.3. The second of Eqs. (1.14) shows Ihat Ihe shearing siress T is zero fo r {j = 0 and 0 = 9(P. and that for = 45 0 il reaches its maximum value

e

r

~U ' P""

L _ _ __

r,,,al'l'l. l,,

=

T In

(r ) Str~'S(·. for fJ

~

P

P

- sin45° cos 45° = - Ao lA n

( 1.1 6)

45"

~T" . PI:!I.,I

~U . PI2A"

The first o f Eqs. (1.14) indicates that. when () = 45°, the norma l stress u ' is also equal to P!lAu: (I. 17)

Ag.1.31

The resu lts obtained in Eqs. ( 1.1 5). (LJ6). and (1.17) arc shown graphically in Fig. 1.31 . We note thai the same loading may produce e ither a normal stress U rn = P! Au and no shearing siress (Fig. l.3 Ib), or a normal and a shearing ~t ress of the same magni tude u' = Tm = PI2Au (Fig. 1.31 c nnd d), depending upon the orientation of the section.

1.12. STRESS UNDER GENERAL LOADING CONDITIONS; COMPONENTS OF STRESS

P,

Ag. 1.32

The examples of the previous sections were limited to members under axial loading and connections under trans verse loading. Most structural members and machine co mpon c nL~ arc: under more invol ved loading conditions. Consider a body subjected to several loads P t , P 2 > etc. (Fig. 1.32). To understand the stress condition created by these loads at some point Q within the body, we shall first pass a section thro ugh Q. using a plane parallel 10 the yz plane. The portion of the body to the left of the sec· tion is slIbjecled to some of the original loads. and to normal and shearing forces distributed over the section. We shall denote by 6.P and 6. V..., respecti vely. the normal and the shearin g forces acti ng on a small

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1.12. SIte"" Under General lOlldlng CorKIltle""

J F'

P,

", I

(/,I

Fig. 1.33

area 6.A ~u rrullmlin g pUint Q (Fig. 1. 33a). Nmc: that the superscripL x is used to indicate that the forces ~F.I" and .6. V' act on a surface perpendicular to Ihe.\ axis. While the. normal force ~l'x h'l~ a wcl l-ddi ned direction. the shearing force ~ V' may have any directi on in the plane of the section. We therefore resolve .6. V" into two component forces. 1. V; and ~ V~, in directions parallel to the)' and :; axes. respecti vely (Fig. 1.33 b). Dividing noVJ the magnitude of each fo rce by th.' area l.A. and letling ili\ np pronch zero. we define the three stress components shown in Fig. 1.34: u .< -

T

.xy

""

t.

( 1.1 8)

V:

lim - -

~->O

/IF' 1l~!I l.A

LlA

T ""

= Rg . 1.3"

We note. th:1I the !irst subscript in (T~. T A" and T J... is used to indicate that the stresses under considf!ration are ex-en ed OIl U slIIface pe'pclldiwlar /0 the .l axis. The second subscript in Txv and T"" identifies ,lie directron Of till! ('ompollellf. The normal s tre s~ U , is positive if the correspo nding arrow points in the posit ive x direction. i.e. , if the body is in tension. and negative otherwise. Similarly. the shearin g stress components 7 " and T.r. are positi ve ir the corresponding arrows point. respectively, in the positi ve y a nd :: directions. The abovc analysis may also bc carried Out by considering lhe portion of body loc:ucd 10 the ri ght of the vertical plane through Q (Fig. 1. 35) . The same magnitu::les. but opposite di rections. arc ob(ain ~d for (he normal and shearing forces ~F '" 6. Vj~' and .J.. Therefore. the Same values arc also obtained fo r the corresponding ~ tress components. bllt SillCC the section ill Fig. 1.35 now t"acc5 the IIfw,,/ive .l 1Oi.l. it 1)I.)$i tive sign for (T ,. will indicate that (he corresponding arrow poinl ~ ill til l' 11('8i1til'e x (lirectiol/. Similnrl y. positive signs fo r r ,y and 7 .... will indicate (hat the corresponding nrrow~ point. respectively. in the negative y and ;. directIOns, 3 5 Shown III !-'ig. 1. 35 .

V:.

Ftg. 1.3S

25

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26 Introducllon----Concepl 01 Stm.3

Pal>sillg a sL"CtiolJ throu gh Q pm'aUdlO the :c.\ pl a l~ e, I\e de flm~ ill Ih.. same maune.r thr s tress componenls. iT" 7", and T " . Finall y. a ~ec · tion through Q parallel to the xy plane yields- the components U e • T.T and T :>,. To faCilita te the Vis uali Zation o r the s tress conditiun at point Q . we shall consider u small eube of side a L-entcred at Q ~tIld Ihe Slrcsses exerted on I!ach o f the six faces nf the cube (Fig. 1.36). TIle stress com· pone nt ~ shown in the Jigure are U n U ,., and U -" wbich re present the no r· mal stress on faces respecti vely perpend icular to the x. y . and z axes. and the six shearing stress componcnl" T,.", T"" etc. We recalJ that, ac· cording to the deflniti on o f the s hearing ~lress components. T ", repre· sents lhc -" component of the shearing stress exerted OD the face per· pendiculnr to the x axi s. while T", re presents the x componelll o f the shearing stress exerted on the face perpendic ular 10 the y axis. Note that o nl y three f,Kes ur ltlt:. :: ube an: a(; tuall y visible- in Fig. 1.36, amJ that eq u,,1 and o ppo~ite stre;;s compo n ~ nt S act on the hidden faces. While the stresses acting: on the faces of the cube differ slightl y from the stresseS:ll Q. the error invol ved is s m.-.JI and vanishes as side u of the cube approochcs zero, Important relations amo ng the shearing stress components wi ll now be. derived_ Let li S consider ttl(' f rce- hody di:1gram o f Ihe s maU cube centered at point Q (Fig. 1.37), The normal and shearing forces acting on the various fnces of the cube are o btained by multiplying the corre· s pondi ng stress components by the area 6.A of euch face. We first wri te the fo llow in,!! three equilibrium equUlions:

SF, - 0

IF,

=

0

( 1.1 9 )

Since ror~es equal and oppOSite to the forces a~ lu a ll y sh(jwn in Fig. 1..l7 arc acting on the hidde n faces of the cube. it is clear that Eqs (1 . 19) are satisfied. Considering now the momenls of the forces abo ut axes Qx', Qr', and Q:;' drawn fro m Q in directi ons respecti vely paml· Id IU tin:

"t.

J. am.! <: Wl.es. we wlilc- the tluel;; mJlIiliuu a l

LMr - 0

c4uali u H ~

,£M y- O

( 1.20)

projection on the .\"y' plane (Fi g, 1. 38), we nOte that Ih" onl y momcnr~ Ilhrll lt Ihe ;: ,Ixis Iliffere ni frnm 7crn arc th e shea r· ing fo rtc~ . 11lese forcc.~ form two couples, o ne of counterclockwise ( positive) moment (or.•" ~A )a. tile other of clockWise (negati ve) moment -( T\.x :lA )a. The last 'o f thc three Eqs, ( 1.20) yieldS, there rore. Using

it

fnrc e!i1 w i lh

I "~ ..:i.,\ _

'.,.. .l)' 7" .l.\

from which \\Ie conclude that Fig. 1.36

( 1.21 )

The relation obtained shows that the y exerted on a lace perpendicular [0 the

co mpo!l~nt

x aXIs

of the shearing stress x -compo·

IS eq ual to the

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nent of the shearing stress exerted on a face perpendicular to the y axis. From the remaining tWQ e4umions ( 1.20). we derive in a similar manner the relations

1.13. Design Considerations

( 1.22) P'

We conclude from Eqs. ( 1.21) and (1.22) that only six strcss components arc required to define the condition of stress at a given point Q, instead of nine as originally assumed. These six components are fT" a ,., CT" 7." .'T"", and 7 ~r \Ve also nQte Ihat, al a given point , shear ellllllOf ((Ike place ill one plane only; an cqual shearing stress must be exerted o n another planc perpendicular to the lirst o ne. For example, considering again the bolt of Fi g. 1.29 and a small cube at the center Q of the bolt (Fig. 1.39a). we find that shearing stresses of eq ual magnitude must be exerted o n the two horizoilial faces of the cube and on the two faces that are perpendicular to the forces P and p i (Fig. 1.39h). Before concludi ng our discussion o f stress components. let us consider again the casc o f a member undcr axial loading. If we consider a small cube with faces respectively parallel to the faces of the member :.md recall the reslllL~ obtained in Sec. 1.11 T we lind that the conditions of stress in the member may be described as shown in Fig. 1.40a: Ihe o nly stresses are normal Slresses CT. exerted on the faces of the cube which are perpendicu lar to the x axis. However, if the small cube is rotated by 45° about the ;: axis so that iL~ new orientation matches the orientation of the sections considered in Fig. 1.31 c and d. we conclude that normal and shearing stresses o f equal magnitude arc exerted on four faces of the cube (Fig. 1.40b). We thus observe that the same loading condition may lead to different imerpretations of the stress situation at a given poim , depending upon the orientation of the e lemem considered. More wi ll be said about this in Chap 7.

1.13. DESIGN CONSIDERATIONS

In the preceding sections you leamcd to determine the stresses in rods, bol ts. and pins under simple lortding conditions. [n later chapters you will learn to delermine stresses in more complex situations. I.n e ngineering applications. howcvcr, the determination of stresses is seldom rtn e nd in itself. Rrtther, the knowledge of strcsses is used by engincers to assist in their most important task. namely, the design of structures and machines thrtt will safely and economically pcrfonn a specified function. a. Determination of the Ultimate Strength of a Material. An import rtnt clement to be considered by a designer is how the materi al that has been selected will behave under a load . For a given matcrial, this is determi ned by performing speci fi c tests on prepared samples of the material. For example, a tcst specimen of steel may be prepared and placed in a laboratory tcsting machine to be subjected to a known ceTltric axial te nsi le force, as described in Sec. 2.3. As the magnitude of the force is increased. various chrtnges in the specimen arc measured. forexrtmple , changes in its length and its diameter. Eventually the largest

,,'

(f, )

Fig. 1.39

--'1- --; ;0

fL-·':~ - - _"'·~-: ,i -' ,,)

"

P'

P

TO' -

Fig. 1.40

i"\

.'

r

27

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28

Introductlon-Concept of Stress

force which may be. appl ied to the spc(;imen is reached , and the specimen either breaks or begins to carry less load. This largest lorce is called the II/timatc load for the test specimen and is denoted by PII' Since the applied load is centJic, we may divide Ihe ultimate load by the ori gi nal cross-sectional area of the rod to obtain the ultimaTe nOn/wI .I·tress of the material used. 111is stress, also know n as the ultimate strcngth in tl'nsion o f the material. is ( 1.23)

Fig. 1.41

Fig. 1.42

Several test procedures are available to detennine the ultimate j'het/ring stress, or ultimale ~'Irellglh ill !ihear; of a material. The one most commonly used invol ves the twisting of a circular tube (Sec. 3.5). A more direct. if less accurme. procedure consists in clampi ng a rectangular or round bar in a shear tool (Fig. 1.4l) and applying an increasing load P until the ultimate load Pu fur single shear is obtained. If the free end of the specime n rests on both of the hardened dies (Fig. 1.42), the ultimate load for double shear is obtained. In either C:L~e. the ultimate shearing stress T U is obtained by div iding the ult imate load by the total area over which shear h:L~ taken place. We recall that. in the case of single shear, thi s area is the cross-sectional area A o f the specimen. while in double shear i1 is equal to twice the cross-sectio nal area. b. Allowable Load and Allowable Stress; Factor of Safety_ The maximum load that a structuml member or a machi nc component will be allowed to carry under normal conditi ons of utilizatio n is considerably smaller than the ultimate load. T his smaller load is referred to as the allowable load and , sometimes, as the working load or de.vigll load. Thus, only a fract ion of the ultimate-lo..1d capacity of the member is uti lized when the allowable load is applied. The remaining ponion of the load-carrying capacity of the me mber is kept in reserve to assure its safe performance. The ratio o f the ultimate load to the allowable load is used to de fine the faclor of .wfety. t We have

Factor o f safet y = F.S. =

ultimate load -';;= ':':'-7=: allowable load

(1.24)

An alternative de fin ition o f the factor o f safety is bascd on the lise of stresses: ultimate stress Factor of safety = F.S. = allowable stress

(\.25)

The two expressions given for the factor of safety in Eqs. (1.24) and (1.25) arc identical when a linear relationship exists between the load and the stress. In most engi neering ap plications, however. this relationship ceases to be linear as the load approaches its ultimate value, and the facto r of safety obtained from Eq . (J .25) does not provide a t In some ficids of engineering. notably aeronautical engineering. the lII"rgi" of safery is used in place of the factor of safety. The margin of safety is defined as the factor of safety minus one: that is. margin of safety = F.S. - 1.00.

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II

ue asscssme nt o f tht

~afety

o f a givc n desig n. Nevert heless. the

1II/owlIble -stresj' metholl of design. based on the use of Eq _ (1.2 5), is

widd y used. c . $election of an Appropriate Factor of Safety. The <;elX Tinn of the fu ctor of s afety to be used for various applications is one of the most importunt e nginee rillg. tas k..~. O n the Olle. hand, if a factor o f safety is c hos
I. \/{lriatiol1.\· tllllt IIIlIy occllr ill Ih l' proper/ie.l· of th e lII ember 1111(/er GOII.I·ideratiOlI. The ClJl1lpusitiu n, s trength. anJ

di ll1t: n~iun ~

o f the membe r arc a ll s ubjC<::t to s mall v.uiatjon ~ duri ng manufac tu re. In addi tio n. mate rial properties may be altered and residual ~tresse s introduccd through heali ng o r defo rmati on Uwt m:ty occur during: manulacture. storage, transpo rtation. or construction. 2. The !lllInber (if Imufin8 .1' Ihm /IIc)" bl' expected d urin g the life of th e Sflllc lWf' or machine. For most materials the ultimate stress deere.ages as the number of load applications is increased. Thi s phenomenon is known aSfariglU! and, Jfi gnore!.l. ma y result in sudde n ftlilure (sec Sec. 2.7). 3 . The typc of [om/illgs that (Iri' planllcd for ill Ihe lfe.l'igl1. or I/:(I{ ml ly ocClI r in IhejullI'"e. Very few loadin gs are known with complete accuracy- most design lo adings arc engi neering estimates. In ~ddJljon. future alt ~rati ons or c hH[lges in usage lIlay introduce changes in the aCl ual 10:lding Larger factu/"!i o f safelY arc al so ro~ quired fu r dynamic. cyclic. or impulsive loadings. 4. Th l'. 'We of flli/lH"I' IlllIt /lilly occur. Brittle ma terials fa il suddenl )', usually with no prior indicatio n that coll apse is immiIl e llt. 011 th e uthel l lilIlJ , uu\;ti k s tc~ l.

llliilCli'lh. ~ llch us s tJU \; tul,tl

normally unde rgo a subs tanT ial defo nnatio n c alled yielding before fail ing. titus providing a waming that overloading exists. However. mOSt buc kling o r stab ility failures ure s udden. whc ther the material IS bn llie or not. Wh <:11 the POSS lbLl ity o t sudde n failure exists. a larger faewr of safety should be used than when faillire i<; pTccc(let1 hy nhviollS wa rning sig:ns S. Uncertainl)' due 10 m elhod.I' of cmal.r.l·i.I·. Al l des ign methods arc based o n cerLain s implifying lIss umptiolls which result in c ~l ­ c ulate d s tresses bei ng ap proximations of :l.ct uJ. I St resse~. 6 . DellJriQralioll rllilt may occllr III Ih e [ II/lire beC{II/.I·t! of poor nwillfenllllce o r bec ause Q/ IlIIprevf'mable III/rural callses. A Inrger factor of !\afety is ncces!w.ry in locations where condilions such as corrosion and decny are dillicult to c ontrol or cven LU d istuvt:r. 1. Til l! importance of {j gil'en member to til" ill1es.rity oflhe wh(}le .\·fntCmre. Bracing. 1111d secondary mc mbers may in man y cases be dcsigned with a fac tor of safety lower thallthat ustd for pri· nKiry me mbers.

1.1 3. Destgn Con3!demtloM

29

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30

Introcluctlon-Concept of Stress

In addition to the above considerations. there is the. additional consideration concerning the risk to life aod property that a fai lure wou ld produce. Where a ["ailure would produce no risk to life and only minimal risk [() property. the usc of a smaller factor ()f safety can be considered. Finally, there is the practical consideration that. unless a careful design with a noncxcessivc factor of safety is used. a structure or machine might not perfonn its design Junction. For example. high fac~ tors of safety may have an unacceptable effect on the weight of an aircmft. For the majority of structural and machine applications. factors of safety arc specilied by design speci fic ations or buildi llg codes written by commillees of experienced engineers working with profess ional socie ties. with ind ustries. or with fe deral. state, or city age ncies. Examples of such design specifications and building codes arc I. Steel: American Institute of Steel Construction, Specilication

for Structural Steel Buildings 2.

Con crete: American Concrete Institute. Building Code Re-

quirement for Structural Concrete 3 . Tim/)er: American Forest and Paper Association. National Design Specification for Wood Construction 4 . Highway /Jridges: American Association of State Highway Officials. Standard Speciiications fo r Highway Bridges 'd.

Load and Resistance Factor Design. As we saw above., the

al lnw:thlc-st rcs<; mcthnfl re(lllirt!s Itllli :1 11 the IIncertainties assncin ted

with the design of a structure or machine clement be grouped into :I single factor of safety. An alternative method of design, which is gaining acceptance chieny among structural engineers. makes it possible through the use of three different factors to di stinguish between the uncertainties associated with the structure itself and those associated with the load it is designed to support. Thi s method, referred to as Load and Re.~i.\'Wnce Factor De.l·ign (LRFD). furt her allows the designer to di stinguish between uncertainties assoc iated with the live load. P L , that is, with the load to be supported by the structure, and the dead load. P D , that is, with the weight of the portion of structure contributing to the total load. When this method of design is used. the ultimate load, P u, of the structure, that is, the load at which the structure ceases to be useful, shou ld tirst be determined. The proposed design is then acceptable if the following inequality is satisfied: (1.26)

The coefficient 4> is referred to as the re.\'i.I·Wnce factor: it accounts for the uncertainties associated with the structure itself and wi ll normally be less than I. The cocfficients 'YD and 'YL are referred to as the load factor.\': they account for the uncertainties associated, respectively, with the dead and live load and will normally be greater than I, with 'YL generally larger than 'YD' While a few examples or assigned problems lIsing LRFD are included in this chapter and in Chaps. 5 and 10, the allowable-stress method of design will be used in this text.

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r:.-,,+fJJI

SAMPLE PROBLEM 1.3

p

B

51! kr.:

0.6",

15 k:-.J

I '-11;;1[-'

0-.8.

c

0

1-03".103

,j D

Two forces are applied to the bracket BCD as shown. (a) Know ing th:l! Ihe control rod AS is to be made of a steel having an ultimate nonna] stress of 600 MPa. detemline the diameter orthe rod for which Ihe factor of safelY with respecl to failure will be 3.3. (IJ) The pin at C is to be made of a steel having an ultimate shearing stress of 350 MPa. Delennine the diameter of the pin C for which Ihe factor of safelY with respect \0 shear will also be 3.3. (d Determine the required thickness of the bracket ,uppons at C knowing liWi the allowable bearing stress of the steel used is 300 MPa.

SOLUTION

r- "

Free Bod,V: Entire Brackel . ponents C, and C,.

p

0.6m

+ 5O kI\

15k/\

I c~~ -"]~~r

The reaction al Cis represented by its com-

~ 'fM c = 0: P(O.6 m) - (50 kN)(G.J m) - ( 15 kN )(O.6 ml = 0

'iF, = 0: 'iF, = 0:

C, = 40 k Cy = 65kN

iI . Collt rol Rod AB , able stress is

VC;

+ C; = 76.3 kN

Since the factor of safety is to be 3.3, the allowif u

if... 1 =

c=

P '" 40 kN

60{) MPa

f'.s. = ~1.3 ~ = 18 1. 8 MPa

For P = 40 kN the cross-secti onal area required is P =-=

A '''I

{f .1I

40kN

181.8MPa

220 X iO- 6 m! tiM! = 16.7-1 mm ....

c h. Shear ill Pin C. 'Taa

For a factor of safety of 3.3, we ha ve 'T u

= --

350 MPa

~ --~

F.S. Sinc~

3.3

= 106.1 MPa

the pin is in dOLible she;Jf. we write C/2

(76.3 kN )/2

A ,,~ = -.;:::;; = 106.1 MPa

ac = 21.4mm

Use: d c = 22 nun ....

The next larger size pin available is of 22-mm diameter and shoLild be used. ~ . Rearing a t C. Using d = 22 mm. the nominal bearing area of each bracket is 221. Since the force carried by each bracket is e/2 and the allowable bearing stress is 300 MPa, wc write

C/2 A "4 = - if,lI

Thus 221 = 127.2

(76. 3 kN )/2 300 MPa

t=5.78mm

Use : 1 = 6 mm ....

31

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SAMPLE PROBLEM 1.4 The rigid beam BCD is auached by bolts [0 a control rod at B. to a hydraulic I:ylinder al C. and to a fixed support al V . The diameters of the bolts used are: dB = d D = in .. de = ~ in. Each bolt aCls in double shear and is made from a steel for which the ultimate shearing stress is 1 U = 40 ksi. The control rod AB has a diameter dA = it, in. :md is made of a steel for which the ultimate tensile stress is ifu = 60 ksi. If the minimum factor of safety is to be 3.0 for the entire unit. determine the largest upward force which may be applied by the hydnmlic cylinder al C.

i

SOLUTION

c

The factor of safety with respect to failure nlll~1 be 3.0 or more in each of the three bolts and in the control rod. These four itldependenl eriterin will be considered separately. Free Body : Beam BCD. We firsl determine the force at C in terms of the force at B and in tenllS of the force at D. +~'E.MD = +~

B(14 in. ) - C(S ill.) = 0 - D(14 in. ) + C(6In.) = Q

0:

IMB = 0:

Cont ro l Rod .

C = 1.750B

(I )

C = 2.33D

(2)

For a factor of safety of 3.0 we have (T U

IT.,.I = - P.S.

~

60 ksi

- - = 20ksi 3.0

The allowable force in the control roo is

B = ifall(A) = (20 ks i) ±1T(fr, ill l

= 3.01 kips

Using Eq. ( 1) we find the largest permined value of C: C = I.7.'iOB = 1.750(3.0 1 kips)

c = 5.27 kips

<.l

Roll a l 8 . Tall = "T uf P.S. = (40 ksi)/3 = 13.33 ksi. Since the bolt is in double shear, the allowable magnitude of the force 8 exerted on the boh is

B

= 2PI = 2(t.nA) =

From Eq. (1):

2( 13.33 ksi)(± 1T)a in.)2 = 2.94 kips

C = 1. 75 0B = 1.750(2.94 kips)

C = 5.15 kips <.l

8 011 at D. Since this boh is the same as bolt B. the allowable force L, D = B = 2.94 kip.<. From Eq. (2):

C = 2.33D = 2.33(2.94 kips ) 8 011 a t C.

C = 6.85 kips <.l

We again have 'T.II = 13 .33 ksi and wri te

C = 2Fl = 2(-T'IIA) = 2(13.33 ksi)(br)(t in. )2

C = 5.23 kips <.l

Summa r y. We have found separately four maximum allowable values of the force C. In order to satisfy all these criteria we must choose the small· est value. namely: C = 5.15 kips ....

32

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PROBLEMS

1.29 The lA-kip load P is supported by two wooden members of uniform cross section that are joined by the simple glued scarf splice shown. Determine the nonnal and shearing stresses in Ihe glued splice.

f-:7

5 .0 in,

3.0~ /

W'

I"

P'

Fig. P1.29 and PI.30

1.30 Two wooden members of unifonn cross section are joined by the simple scarf spIke shown. Knowing that the maximum allowable shearing stress in Ihe glued splice is 60 psi. determine (a) the largeslload P thai can be safely supported. (h) the corresponding tensile slress in the splice.

1 .31 Two wooden members of unifonn reclangular cross section are. joined by the simple glued scarf splice shown. Knowing Ihal P = II kN. determine the nonnal and shearing stresses in the glued splice.

Fig. P1 .31 and P1.32

1,32 Two wooden members of unifonn rcctangularcross section are joined by the s imple glued searf splice shown. Knowing thm the maximum allowable tensile stress in the glued splice is 560 kPa, detennine (u) the largest load P that can be safely applied. (b) the corresponding shearing mess in the s plice.

1 ,33 A ce.ntric load P is applied to the gmnite block shown. Knowing that the resulting maximum value of the shearing stress in the block is 2.5 ksi. detenlline (al the ma.gnitude of P, (b) the orientation of the surface on which the maximum shearing stress occurs, (cl the nonnal stress exerted on the SU[face, (eI) the maximum value of the nonnal stress in the block. 1.34 A 24IJ..kip load P is applied to the granite block s hown. Detennine the resulting mnximum value of (1I) the normal stress, (b) the shearing stress. Specify the oricntmion of the plane on which each of these maximum values occurs.

Fig. P1.33 and P1 .34

39

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34

Introductlon-Goncept of Stress p

I

10"""

---11--2(,l'

1.35 A stee.l pipe of 400-mm oUier diameter is f;)bricated from 10-mmthick plale by welding along a helix that form s an ang le of 20° with a plane perpendicular to the axis of the pipe. Knowing thai the maximum allowable nonnal and shearing slresses in the direclions respectively nonnal and tangentiallO the weld are u ~ 60 MPa and T = 36 M Pa. determine Ihe m<Jgnilude P of the largesl ax.ial force thai can be applied \0 the pipe. 1.36 A steel pipe of 400-mm outeruiameter is fabrica. determine the largest load Q that can be applied if an overall factor of safety of 3 is desired.

FIg. Pl.3S and Plo36

Q

T" ~~

(j.G '"

l l - -+--TOl F

o.fil"---

12 """

0"' 111- 1

Fig. Pl .38 and P1 .39

Q' Flg. Pl.37

I'

1 .38 Member ABC, which is supported by a pin and bracket at C and a cable 8U, was designed to support the 1"6-kN load P as shown. Knowing that the ultimate load for cable BD is 100 kN. determine the factor of s afety with respect to cable failure. 1.39 Knowing Ilmt the ultimate load far cable BD is 100 kN and that a factor of ~afet y of 3. 2 with respect to cable failure i~ r('quired. detemline the magnitude or the largest force P that can b,:, safely applied as shown to member ABC. 1.40 The horiZOlllallink Be is ~ in. thick. has a width 11' = 1.25 in.,
FIg. PHO and Pl.4l

1.41 The horizontal1ink BC is ~ in. thick and is milde of a steel with a 6S-ksi uliinmte strength in tension. What should be the width 11' of the link if the structure shown is to be des igned to suppan a load P = 8 kips with a factor of safety L'qual 1031

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Prob lems

1.42 Link AB is to be made of a steel for which the ultimate Ilonnal is 450 MPa. Detemline the cross-sectional area for AB for which the factor of ~afety will be 3.50. Assume that the link will be adequately reinforced around the pins at A and B. ~tress

r,L ~ 8L=t==;tc~"'D 1 2(lkr..l~

01' I

OA

til

0.4

til

0.4

til

Fig. P1.42 1 .43 The two wooden members shown. which support a 16-kN load. are joined by plywood splices fully glued on the surfaces in contact. The ultimate shearing stress in the glue is 2.5 MPa and the clearance between the members is 6 mm. Detemline the required length L of each splice if a factor of safety of 2.75 is to be achieved. Hi kl\

125 """

> Fig. P1.43 1.44 For the joint and loading of Prab. 1.43. determine the factor of safety, knowing that the length of each splice is L = 180 mm. I !II \...'l

1.45 Three 18-mm-diameter steel bolts are to be used to attach the steel plate shown to a wooden beam. Knowing that the plate will support a 110-kN load and that the ultimate shearing stress for the steel used is 360 MIla. determine the factor of safet y for this design.

Fig. P1.4S and P1.46

1.46 Three steel bolts are to be used to attach the ~teel plate shown to a wOCKien beam. Knowing that the plate will s uppOrt a IIO-kN load. that the ultimate shearing stress for the steel used is 360 M Pa, and that a factor of safet y of 3.35 is desired. determine the required diameter of the bolts. 1.47 A load P is s upported a.'> shown by a steel pin that has been inserted in a short wooden member hanging from the ceiling. The ultimate strength of the wOCKi used is 12 ksi in tension and 1.5 ksi in shear. while the ultimate strength of the steel is 30 ksi in s hear. Knowing that the diameter of the pin is d = i in. and that the magnilllde of the loal.! is P =0 5 kips. determine (a) the factor of safety for the pin, (h) the required values of b :md c if the factor of safety for the wOCK!en member is the same as that found in part (/ for the pin. 1 .48 For th~ su pport of Prob. J.47, knowing that b = 1.6 in .. (" = 2.2 in .• and d = t in. , determine the load P if an overall factor of safety of 3.2 is desired.

FIg. P1 .47

35

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36

Introcluctlon-Concept of Stress

1.49 Each of the two vertica l links CF connec ting Ihe two horizontal members AD and EG has a uniform rectangular cross section i in. thick and I in. wide. and is made of a sleel with an ultimate streng th in tension of 60 ksi. The pins at C and F each ha ve a }.in. diameter and are made of a steel wilh an ultimate streng th in shear of 25 ksi. Dctennine the overall factor of safelY for Ihe links CF and Ihe pins connecting them to the horizonTal members.

10 in

c

.,.

F

H:ips FIg. P1.49

l!" -C~o

V

0" JC t-- 121ll

FIg. P1.51

- - 1 11",

1 .51 Each of the stet'.1 links AB and CD is connected 10 a suppon and to member 8CE by I·in. ·diameter steel pins acting in single shear. Knowing: that the ullimate shearing stress is 30 ksi for the steel used in Lhe pins and that the ultimate nonnal stress is 70 ksi for the steel used in the links. determine the allowable load P if an overall factor of safeLY of 3.0 is desired. (Note that the links are not reinforced around Lhe pin holes.)

1.52 An alternative design is being considered to support member BeE of Proh 1.51 . in which link CD will be replaced by two links. each of ! X 2-i n. cross section. callsing the pins at C and D to l.>e in dOllble shear. AsslIming that all oiller specificati ons remain unchanged, detennine the allowable load P if an overall faclor of safety of 3.0 is desired.

2 in.

H

1. 50 Solve I'rob. 1.49, assuming Ihm the pins at C and F have heen rcpl;lced by pins Wilh a i ·in. diameter.

J

1.53 In the steel structure shown. a 6-mm-diameter pin is used at C and iO"mm·diameter pins are used at Band D. The lIltimate shearing s tress is 150 MPa at all connections. and the ultimate nonnal stress is 4CMl MPa in link BD. Kn ow in g thlll a factor of safety of 3J) is desired. ueiennine the lurgest load P that can be applied at A . Note tha i link BD is not reinforced around the. pin hol es.

Front

"

"('W

- IS"""

A

A

!---I.'''''''-I~I20'''"'1 To~e~ ~

6

'Ill"

Sid" "ew

Fig. P1.S3

1.54 Solve Prob. 1.53, assuming that the struc ture has been redesigned to u»c 12-mm-dkuneter pins at Band D and no olher change has been made.

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1 .55 ILl the SUUClU le shown. all 8-lllm-diamo::tt, pill is used at A, and 12-rnm-dia meler pins are used " I Band D . Knowine tlt.1I lhe ultimate slteMing ~tress is 100 MPa at all connections ami that the ultimate normal s tress is 250 M P"J in each or the two links Joming H and V. deternunc tile all owahle load P if an o\'<'fall fa~'lor of safety of 3.0 i, de.,ired.

f'mblem3

r op vlrw

1- 200 nll '_1_1SO "'"'_I,' nn a )~

,,, , ,
:

A

"•

~

B

el

"

c -

2tJ m ill

, , ~

V

1.55

•

- !l IIlIl' D

! 2m m ~

Frollt ,i(><...

FIg. Pl .5S

""n _

I--'

.s ~k '1L'\\"

In an alter11ltiw design for the structure of Prob. 1.55. a pin of

10-mm-dwlUeter is 10 be used at A. Assuming that all other speci fi cations rernain ull~hangt'd. dctennille the allowable load P if an overall facl«" of ~ 'Ifety of J 0 is desired.

'1 .57 A 40_kg platfoml i ~ an3e hed 10 the <,"l1d H of a 50_kg wooden bearn AB. whi:h is supported as s h.Ol'<TI by a pin at A and by a slender skel rod Be wilh a 12-kN ultimate load. (11) Using Ihe Load ,Uld Re sis tance Factor Design met hod wit h a rc~istall,e factor ri> - 0.90 and load factors "'D - 1.25 an d YL = I .b. determine the largest load tha I ca n be safely placed on tll~ Olaform. Ib) Whm is til e corresJXlndin g conventio nal faclor of safeiy for rod BC7

1.8

,~

I---- 2.,1

III - --

Rg . Pl .S7

"1 . 56 The Lmd and l{esl.\t3flCe F:;ctOi Design method is to be used to ~e­ tho. cwo cables that will rai.., anJ l()\\.er a platform ....... ppl>rting two wiml,,\\' washers. The platform weighs 160 Ib and each of lite window washers is assumed 10 weigh 195 Ib with equipment. Since these workt' rs:rrt' free III move on the platfom!. 75% of their IOlal weig hl and th e weighl of their eiluipmenl will be used as Ihp. fi ...~illn 1;\"t~ I n~d nf I'ilrh r~h l e. (II) A~~lIming a re.~ i~lanc:e f~
=

Rg. P1.58

37

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REVIEW AND SUMMARY FOR CHAPTER 1

Axial loa::ling. Normal stress

r

Thi s chapter was devoted to the concept of stress and to an introduction 10 the methods used fo r the analysis and design of machines and load-bearing stmctures. Section 1.2 presented a short re ... iew of the methods of statics and of their application to the determination of the reactions exerted by its supports on a simple structure consisting of pin-connected members. Emphasis was placed on the usc of a free-bady- diagram to obtain equilibrium equations which were solved for the unknown reactions. Free-body diagrams were also used to find the internal forces in the various members of the structure. The concept of stre.H was fi rst introduced in Sec. 1.3 by considering a two-force member under an axial loading. The normal .rtress in that member was obtained by di viding the magnitude Pol' the load by the cross-sectional area A of the member (Fig. 1.8a). We wrote p u-=-

A

A

(1.5)

Section 1.4 was devoted to a short discussion of the two principal tasks of an engineer. namely. the analysis and the doign of structures and machines. As noted in Sec. 1.5. the value of IT obtained from Eq. ( 1.5) represents the average .flre!;S over the section rather than the stress at a specific point Q of the section. Considering a small area .1A surrounding Q and the magnitude .1F of the force exerted on .1A. we defined the stress at point Q as ( 1.6)

Fig. 1.8a

38

In general, the value obtained for the stress (T at point Q is different (rom the value of the average stress given by formu la ( 1.5) amI is found to vary across the section. However, this variation is small in any section away from the points of application of the loads. In practice, therefore, the distribution of the normal stresses in an axially loaded member is assumed to be uniform, except in the immediate vicinity of the points of apphcation of the loads. However, for the distribution of stresses to be unifonn in a given section. it is necessary that the line of action of the loads P and P ' pass through the centroid C of the section. Such a loading is called a celllric axial loading. In the case of an eccelll ric axial loading, the distribution of stresses is /lot uniform. Stresses in members subjected to an eccentric axial loading will be discussed in Chap 4.

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When equal and opposite tratl.\"verse force.f p and p i of magnitude P are applied 10 a member A B (Fig. 1.16(/), j'hearillg stresses T are created over any section located between the points of application of the two forces [Sec 1.6]. These stresses vary greatly across the section and their distribution canllot be assu med uniform . However dividing the magnitude P-referrcd to as thc shear in the sectionby the cross-sectional area A . we dcfined the al'erage shearing slress over the section:

Review and Summary lor Chapter I

39

Trailsverse Forces. Shearing stress

B

"

p

=

T .,<

A

( 1.8)

p'

Fig. 1.16a

Shearing stresses are found in bo I L~ , pi ns, or rivets connecting two structural members or machine components. For example, in the case of bolt CD (Fig. 1.1 8). which is in sillgle shear. we wrote P

F

A= A

=

Tave

Single and double shear

( 1.9)

while, in the case of bolts EG and HJ (Fig. 1.20), which are both in

D

double j·hear. we had Fig. 1.1 8 T

ave

F

P F/ 2 = - = -A

A

2A

( 1.1 0)

Bolts, pins. and rivets also create stresses in the members they connect, along the bea rillg .\'IIrface. or surface of contact [Sec. 1. 71. The bolt CD of Fi g. 1. 18, for example. creates stresses on the scmicy lindrical surface of plate A with which it is in contact (Fig. 1. 22). Since the distribution of these stresses is quite complicated. onc uses in practice an average nominal value U' b of the stress. called bearing .1·lreH, obtained by diVid ing the load P by the area of the rectangle representing the projection of the bolt on the plate section. De noting by t the thickness of the plate and by d the diameter of the bolt. we wrote P

U' b

= -

A

P

= -

u/

/I

G

Fig. 1.20 Bearing stress

( 1.1 I)

In Sec. 1.8, we applied the concept introd uced in the previous sections 10 the analysis of a simple structure consisting of two pinconnected members supporting a given load. We determined successively the normal stresses in the two members. paying speci al attention 10 their narrowest sections, the shearing stresses in the various pins, and the bearing stress at each connection. The method you should use in solving a problem in mechanics of materials was descri bed in Sec . 1. 9. Your solution should begin with a clear and precise statement of the problem . You will then draw one or several free·body diagrams that you will usc to write equilibriuIII equations. These equations will be solved for IInknoWII forces. from which the requ ired .I·tresses and deformatio/l s can be computed. Once the answer has been obtai ned , it shou ld be ClIrefll lly checked.

E

~'

:}~

p

F'

~

Fig. 1.22 Melhod of solution

D

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40

Introductlon-Concept of Stress

Stresses or an oblique section

Fig. 1.308

The tirst part of the chaptcr cnded with a discussion of lIltlnericalllccuracy in engineering, which ~t ressed the faci Ihal the accuracy of an answer can never be greater than Ihe accuracy of the given data [Sec. 1.1 0]. In Sec. 1.1 I, we considered the stresses created on an oblique se(:tio/l in a two-force member under axial loading. We found that both I/ormal and .I·hearil/g stresses occurred in such a situation. Denoting by 8 the angle fomled by the section with a no rmal plane (Fig. 1.30a ) and by Ao the area of a section perpendicular to the axis of the member. we deri ved the following expressions for the no rmal stress (T and the shearing stress T on the oblique section: P , u= - cos-O Ao

Stress under general loading

Fig. 1.36

Factor of safety

Load and Resistance Factor Design

T

P == - sinOcosO Ao

(1.14)

We observed from these formulas that the normal stress is maximum and equal to U m = PIAo for 8 = O. while the shearing stress is maximum and equal to T n. = PI2Ao for 8 = 45". We also noted that T == 0 when 0 = O. while u = PlUG when 0 = 45°. Next. we discussed the state of stress at a point Q in a body under the most general loading condition rSec. 1.121 . Considering 11 small cube centered at Q (Fig. 1.36), we denoted by U x the normal stTC..'>S exerted on a face of the cube perpendicular to the x axis, and by T . y and T.o::: ' respectively, the y and z components of the shearing stress exerted on the same face of the cube. Repeating this procedure for the other two faces of the cube and observing that T.r;" = T)." T.,.~ = T;:y' and T ::x = T r.:' we concluded that six .I"t,.es.~ COII/prmclIls are required to deline the state of stress at a given point Q, namely, u.r> (F ,., u z, T.n" T,.: , and Tu ' Section I:13 was' devoted to a discussion of the various concepts used in the design of engineering structu res. The ultimate load of a given structur.11 member or machine component is the load at which the member or component is expected to fai l; it is computed from the IIltimate ~·tress or ultimate strength of the material used, as determined by a laboratory test on a specimen of that material. The ultimate load shou ld be considerably large.r than the allowable loall, i.e., the load that the member or component will be allowed to carry under no rmal conditions. The ratio of the ultimate load to the allowable. load is ddined as the factor of safety:

Factor of safety = F.S.

ultimate load allowable load

( 1.26)

The determination of the factor of safety that should be used in the design of a given structure depends upon a number of considerations, some of which were listed in thi s section. Section 1.13 ended with the discussion of an alternative approach to design, known as Loall alld Resistallce Factor Design. which allows the engineer to distinguish between the uncertainties associated with the structure and those associated with the load.

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REVIEW PROBLEMS

'1.59 Link BD consists of a single bar I in. wide and t in. thick. Knowing that each pin has;) ~-i n. diameter. uetennine Ihe maximum value of the avcwge normal stress in link BD if (a) 0 = 0, (b) 0 = 900. -I

kip" \

,

Fig. P1.S9

1 .60 Two horizontalS-kip forces are applied to pin B of the assembly shown. Knowing that a pin of 0.8-in. diameter is used at each connection, determine the maximum val ue of the average normal stress (a) in link AB. (b ) in link Be. 1 .61 For lhe assembly and loading of !'rob. 1.60. determine (a) lhe average shearing siress in the pin at C, (b) Ihe average bearing stress at C in member Be, (t·) the average bearing siress at B in member Be. 1.62 nllo wooden planks. each 22 mm thick and [60 mm wide. are joined by the glued monise joint shown. Knowing that the joint will fail when the average shearing stress in the glue reaches 820 kPa. determine the smallest allowable len),!th d of the cuts if the joint is to withstand an axial load of magnitude P = 7.6 kN.

1 "1 Fig. P1.6{)

L

'"

::r

2J.l " " "

---12()",,,.16.1""n I

"

Fig. P1.62

41

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42

Introduction-Concept 01 SIre""

1 .63 The hydJ aulie cylinde, CF. which partially controls the jJ'OSiliuJl of rod DE. has been locked in Ihe position ~hown. Member BD i$ 15 mm thid and is connecled 10 the vertical rod by a 9-mm-diametel' bolt. Knowing that P = 2 kN and IJ = 75", determine (ll) the average sht!aring stress in the holl. (b ) the bearinH stress at C in member BD.

1.64 The hydraulic cylinder CF. which partiaJJy controls Ih~ position of rud DE. hJ.s b~.:n lU!:lt:d in WI:! PUSiliull ShUWll. Link Ai3 ha~ a uniConn n:cr~hglll~r cro~~ sP£liol1 rof 12 X 2~ mm :lnll ;~ cnnnf'cted ~1 R In m.'miler Rn byan 8-mm-diameterpin. Knowing that the ma.,imum allowable aH':m.ge She
1.65 Two wooden members of 70 x IIO-mm unifolm recl:ulgu]ar cross seclion are joined by Ihe !;imple glued scarf .pliee shown. Knowing th:11 lhe maxin:um alluwable sn.earing stress in the glued splice is 500 kPa. determine the largest mdalload P that can be sarely applied

1-1 45 """

Fig. Pl. 63 lind P1.64

I'

Fi~ .

Pl.6S

1.66 The 2000,lb load c:Jn be moved along lhe bemn fJD 10 any posi. tion between SIOp~ at E and P. Knowing tn.at (' al l = 6 ksi for the steel used in rods AB lind CD. detennine where the SIOp~ sn.ould be plhced if Ihc p~nnille
~)iU '----1 n

C

:}.ill.--. _ ___ '~ diJ'"i'l"r

-',----1 .S

"

~,

I.

..---~ in. dj",,,", t~r

F

0

Fig. Pt.66

Fig, P1 .67

1 .67 A steel plate IU mm thic k is embedded In a honzontal ~oncrete sbb and if used to anchor a high-strength vertical cable ~s ~how n. The dJartleler of the hole in the plate L~ 24 mm. the ultim,ue strength of the steel lJ.'ied is 250 MPa. jnd the ultimate bonding stress between plate and concrete is 2. 1 l\.'IPa. Knowing [hal a faclOT of safelY of 3.6{) is desired when /' = 18 kN. nf"lf'.rminl'. (tI) the re'liliren widlh II of the rl~le . (h) thl'. minim 11m deplh h In which a plate of thaI width should be embedded in the concrete slab. INeglect the lIomml ~ tresse., between the concrete and the lower end of the plde.)

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Computer Probtems

1 .68 The two portions of member AB are glued together along a plane forming an angle (} wit h the horizontal. Knowing that the ultimate stress for the glued joint is 2.5 ksi in tension and 1.3 ksi in shear. determine the range of values of (j for whic h the factor of safety of the members is at least 3.0.

.

t 2 .~ ~Ip

A

Fig. P1.68 and P1.69

1.69 The tll'O ponjons of member AB are g lued togt".lher along a plant" forming an angle (} wi th the horizontal. Knowing that the ultimate stress for the glued joint is 2.5 ksi in tension and 1.3 ksi in s hear, determine (a) the value of (j for whi.:h the factor of safety of the member is maximum. (b) the corre· sponding value of the factor of safety. ( Hinl : Equate the expressions obtained for the factors of ,afety with respect to normal stress and shear stress.) 1.70 A force P is applied as shown to a steel reinforcing bar that has been embedded in a block of concrete. Dctemline the smallest length L for which the full allowable nomlal stress in the bar can be developed. Express the result in terms of the diameter d of the bar. the allowable normal stress " ,II in the steel. and the llverage allowable bond stress 'T.II between the concrete and the cylindrical surface of the bar. (Neglect the normal stresses between the concrete and the end of the bar.)

Fig. PUD

COMPUTER PROBLEMS 1--.-11"- E!c."",nt " Tbe following prob kms an.! desigued to be sol\'td wilh a computtf. 1 .C1 A solid steel rod consisting of Il cylindrical elements welded together is subjected to the loading shown. The diameter of element i is denoted by d i and the load applied to its lower end by Pi. with the magnitude PI of this load being assumed positive if Pi is directed downward as shown and negative otberwise. (a) Write a computer program that can be used with either S I or U.S. customary units to determine the average stress in each e lement of the rod. (b) Usc this program to solve Probs. 1.2 and IA.

P,

Fig. P1.C1

43

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44

Introduction-Concept 01 Stre""

1 .C2 A 20-kN load is al'pli! d as ~ h [jW Ll 10 die hOfi •. oLllal llk:miJeJ ABC ABC has" I(] X 50-mm unifonn recmngular no~s sec tion :md is supported by four veIlical links. each of 8 X 36-mm unifonn rectangular cros~ se~ tioll . bell oJ the lbur pins at A. 8 , C. and D has the same di:unet~r d and is in doub le shear. (a) Write a computer prOb'Tam 10 calculate for values of tl from 10 to 30 mm. using I-mill illcremellt~. ( I) the maximum value of the aver.lge nonnni mess in Ihe links connecting pins Band D, (2) the avemgc normal ~Irt:ss ill tltt: Ii Ilks CUllnt":cling pin~ C al1l.l E, (3) lhe <Jvcr.tge sh~aring Mrt:S~ in pin R. (4) lhe :"?T:liW J(h?arinr. qrl'S~ in pin r . ('i) I hi' ;wernl'." rn-aring ~1Tf"~S m B in member ABC, (6) the average bearing S(fes~ at C in membn ABC (b) ChC'Ck your program by comparing the val~es obtained for d = 16 mm with the I\II3WCrs given for Probs. 1.7 and 1.27. (c) L SC this progmm 10 find th e pcmlissibl~ values of the diameter d of the pins. knowing Lhalthe aUowable vnlues of the nomla). ~hearing, and bearing stresses for the steel Ilsed are. respectiyely. 150 MPa, 90 MPa and 230 MPa. (d) Sol,'e part L". nssuming illal the thi~kne~ ; of member ABC hns Ix",n reduc~ from 10 to II nUll. Memb~r

Fig. P1.C2

0.5 in

1.C3 Two horizot1lal 5-kip forces are applied to pin B of Ihe assembly shown. Each of till: Ilu'eo: pius at A. B. and C has Ihe same uJaLllcteJ J allu i; in double ~hea[. (al Write a computer progmm 10 calculate for values of d from 0.50 to 1.50 in., u.;ing 0.05-in. inae.me nts. (I) the maximum value of the avemge nonna l SLress in member AS. (2) Ihe al'erage oomJaI stress in membel Be, (3) the nvenlge shearing Slre~s in pin (\, (4) the lIvemge shearing stres~ in pin C. (5) Ihe avemge bealing stress at A in member AB. (6) the m'erage bearing s tress at C in member BC l7) the average bearing stress at B in membel 11e. (b) Check your progr.:.m by comp.1ting ille values oblained for d = 0.5 in with lhe an~"",eN givl'n fnr P mh.~. l .nO an:l 1.111 (i") I J~e this rm~ra m tn finrl lhe permissible values of the diameter d of the pins. knowing thai the nllowable values of the normal. shearing. and bearing stresses for the sl~el used are. rc.
_a

A

Fig. P1 .C3

I

•

15 in

-1--- - I S ili

- J 2;1I. -

Fig. P1.C4 1 .C4 A 4-kip force P fonnin,!! :-.n aD,I!.le.a with ille. vertical is applied as shown to member ABC, ""hich is ~u pported by a pin and bmcket at C and by a cable 8D forming an angle f3 with the horizontal. ~a) Knowi ng that the ulLi m~te load of the ('~.ble is 15 kips. write a computer progr.l.m to comtruct a table of the vahles of the f;]ctor of ~afety of the cable for value. of (I and {j from () to 45°. using increments in II and (3 corre~ponding to 0. 1 in cremenl~ in tan 0 aud tall (3. (b) Check thai fOl al1~ given value of u. lhe ma.~illlulil value of the factor of safety IS obtninell for (3 = 38.66~ and explain why. (t) Determine the smalle,t possible val ue of the factor of s~fety for f3 = 3&.66". as well as ille corresponding v~Jue of n. and explain the result obtained

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1.CS A load P is suppoll<:d a~ showli by two wcode Ol members of uni· form rectanguhlr cross section that 3re join~d by a simple glued scarf splict':. VI} Denoting by i f l} and 'rIJ. respecti\'ely. the ultima": strength of the joint in ten sion lnd in shear. write a computer program which. for given v.l lues of lI. b. P,
compute. Pmblem3

tu ruilure in to::n.,iuJI. (4) tht: f
dimensions and loading of the mem bers of Probs. 1.29 and 1. 31. knuwing thaI = 150 psi and TU = 114 psi for the. glue used in Prob. 1.29. and that i fI} = 1.26 MPa nnd 1U ~ 1.50 i\ IPa for Ihc glue used in Prob. 1.3 1. (c ) Verify in each of the!;;) two t'ases that the shearing stre ss is maximum for 0' = 45°.

U IJ

1. C6 Member AliC is supported by a pin and bracket at A. and by two link. th~t are pin.ccnnl'Cled to the mem\x>r:ll B and to a fixed suppon at D. Ill) Writ"- a computer program h) cakubte Ihe allowable lo"d i>,,-, for an y g iven values of ( I) the diameter d, of the pill al A. (2 ) the common diameter d 1 of Ihe jlillS .. t li and D. (3 ) tht:. Ilhimate uOllllal SiI es s Uu iLl e ach of the two liuks. 14 ) the ultimate shearin!) s tress 'T " in end of the three pins. (5 ) the desired over· all f~clor of safety I-: S. Your program should also indicate which of The fol· lowing three stresses IS critical : the normal stress in the linKS. the shearing stress in the pin at A. or the shean ng . Iress in the pin~ at B (lnd D. (b and c) Choc k your program by using the data of Probs. 1.55 and 155, respectively. anll co:n· paring the ans w~rs obTamed for PoJ l wi th those given in the tex!. (il ~ Use your program \0 determine. the aUowJbl e 10aJ i'..,. ..s well a ~ which of Llle stresses i~ c rilic~ l , when ri o = d, = I" mm, " u = 110 MPa fnr aluminum link s, 'TIJ = 100 MPa for steel pins. and F.S = 3.2. Top vww

II-- l!UO

'll"'_1_ 11iU """_I I'_ ",m

~~=_l s .nm < L T ~ ~ A R C I 8

c

p,"O'--_ --j0I-_ _ i' 2P "' '' '

I' d """ -

o

°

F mnl,;'",

Fig. P1 .C6

1\""" D

II

In"'_

p'

Fly. P 1.C5

45

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Stress and Strain-Axial Loading

This chapter Is devoted to the study of deformations occurring In structLa'al components subjected to axial loading. The change in length of the diagonal stays W3S carefully accounted for In Ihe design of Ihis cable-stayed bridge.

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2.1. INTRooucnON In Chap. I we analyzed the stresses created in various members and connections by the loads applied to a structure or machine. We also

learned 10 design simple members and connections so that they would not fail under sp~ cificd loading conditions. Another important aspect of the ana1ysis and design of structures relates to the deformations caused by the loails applied 10 a structure. Clearly, il is important \0 avoid deformations so large that they may prevent the structuft:. from fulfilling the pu rpose for which it W'L'i intended. But the analysis of deformations may also help us in the detcnnination of stresses. Indeed, it is not always possible Lu determine the forces in U1C members of a structure by applying on ly the principles of statics. This is because statics is based un the ass umption of undeformablc, rigid structures. By considering engineering structures as deformable and analyzing the deformations in their various members, it will be possible for us to compute forces that arc _1'faTical(v ;Ildelamillllte. i.e., indeterminate within the framework of statics. Also, as we indicated in Sec. 1.5, the dist ribution of stresses in a given member is statically indeterminate, even when the force in that member is known. To detennine tlle actual distribution of stresses within a member, it is thus necessary to analyze the defonnatiuns that take place in that member. In this chapter, you will consider the defonnations of a structural member such as a rod, bar, or plate under Inial loadillg.

First, the norlllal simill f in a member will be ddined as the de/orIIIntinll of Ilu.. "'I'm/",r fJ"r IIIlit /mlgllL p totting th!! stre ss ,T v~r~lI ~ Ihe .~rr:lin f as the load npplied to the memocr is increased will yield a J/reH -Jlrain diagram for the material used. From such a diagram we can determine some imporlam properties of the material, such as its modlilus o/elas/icily. and whether the material is duclile or brillie (Sees. 2.2 to 2.5). You will also see in Sec. 2.5 that. while the behavior of most materials is independent of the direction in which the load is applied, the response of fiber-reinforced composite materials depends upon the direction of the load. From the stress-strain diagmm, we can also determine whether the strains in the specimen will disappear after the load has been removedin which case the material is said to behave elas/ically---or whether a per!llallelll set or plastic de/orlllalioll will result (Sec. 2.6). Se:tion 2.7 is devoted to the phenomenon of faliglle. which causes structural or machine components to fail after a very large number of repeated loadings. even though the stresses remain in the elastic range. The first pari of the chapter ends with Sec. 2.8, which is devoted to the determination of the deformation of various types of members under various conditions of axial loading. In Sees. 2.9 and 2.10. statically indetermillate problems will be considered, i.e., problems in which the reactions and the internal forces call nOi be determined from statics alone. The equilibrium equations derived from the free-body diagram of the member under consideration must be complemented by relations involving deformations; these relations will be obtained from the geometry of the problem. In Secs. 2.11 to 2.15. additional constants associated with isotropic materinls- i.e., materials with mechanical characteristics independent of direction- will be introduced. They include Poisson's ratio. which relates

2.1.lnlIOduction

47

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48

lateral and axial strain. the bulk !/Iodulu,l. which c haracterizes tbe change in volume of a material under hydrostatic pressure. and the moduflls of rigidify. which relates tbe components of the shearing ~tress and shearing strain. Stress-strain relationships for an isotropic material under a multiaxial loading wi 11 also be derived. In Sec. 2. 16. stress-strain relationships involving several distinct val ues of the modulus of elasticity. Poisson's ratio. and the modulu~ of rigidity, will be de veloped for fiber-reinforced composite materials under a Illultiaxialloading. While these materials Elfe not i~otropic , they u~ually display special properties. known as orllwTrop i£' properties. which fileilitate tbeir study. In the text material described so far. stresses are ,l~sumed uniformly distributed in any given cross .~ection: they arc also assumed to remain witbin the clastic r1mge. The validity of the first assumption is discu.~ sed in Sec. 2.1 7. while .\"Ire.\".\" L"oncentratiollS ncar ('ireular holc.<; and fillets in flat bars arc considered in Scc. 2.18. Sections 2. 19 and 2.20 are devoted to the discussion of stre~ses and deformations in members made of a ductile material when the yield point of the material is exceeded. As you will sec. permanent pla.l"li£' deformations and residual stresses result from such loading conditions.

Stress and S1/aln---Alllal loading

2.2. NORMAL STRAIN UNDER AXIAL LOADING

T

B

Let us \Consider a rod Be. of length L and uniform cross-sectional area A. whic h is s uspended from B (Fig. 2.la). If we apply a load P to end C. thc rod elongates (Fig. 2.lb). Plotting the magnitude P uf the load

I.

j

c

,

c

-1 p

1,1

(Ii )

against the deformation 8 (Greek letter delta), we obtain a certain loaddefonnation diag ram (Fig. 2.2). While this di :lgram contains illfonnation useful to the :lnalysis of the rod under consideralioll. it cannot be used directly to predict the dcfonnation of a rod of the same material but of different dime nsions. Indeed, we observe that. if :l deformation 8 is produced in rod Be by :lIOad P, a load 2P is required to cause the same defo rmation in a rod B'C' of the same length L, but of cross· sectional area 2A (Fig. 2.3). We note that. in both cases. the value of the stress is the same: (T = PIA. On the uther hand. a load P ap plied

Fig. 2.1

I'

"----- - - - - - , Fig. 2.2

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2.2. Nonnal Strain Under Axial Loading

T

B'

B'

B"

L

J, ,

C'

2A

c .!.!l '

Fig. 2.3

to a rod B"C': of the same cross-sectional area A. but of leng th 2L, causes a deformation 25 in thaI rod (Fig. 2.4), i.e., a deformation twice

~~el~:7i~ l~~r \~: :~~::ltii~[I~ ~~~:~;;;~~~~t:~ ;r~l:;~Bi~t t~l: ~~~~e~~~~~ equal to oiL This observation brings us \ 0 imroUucc the. concept of strain: We dcline Ihe l10rmal strain in a rod under axial loading as the dejol111l11ioll per IIl1i! lel/gth of that rod. Denoting the normal strain by f

(G reek letter epsilon). we write

C"

A

p

Rg. 2.4

8

(2, 1)

£ = -

L

Planing the stress if = PIA against the strain E = 8/L, we obtain a curve that is characteri stic of the properties of the material and docs not depend upon the dimensions of the parti cular specimen used. This curve is en !led a .\·fress-~·rrail! l/ill)VlIIfI and will be di scussed in detail in Sec. 2.3. Since the rod Be considcred in the preceding discussion had a uniform cross section of area A, the normal stress IT could be assumed to have a constant value PIA throughollt the rod. Thus, it was appropriate to definc the strai n E as the ratio o f thc total ddonnation 8 ovcr the total length L of thc rod. In the case of a member of variable crosssectional area A, howcvcr. the normal stress a = PIA varies along the member, and it IS ncec.~sary to deline the strain at a givcn point Q by considering a small clemcnt of unddormed length 6..\: (Fig. 2.5) . Denoting by 6. 8 the deformation of the c lcment under the given loading. we define the normal .I·fraill at point Q as

€ =

C" /

/18 d5 lim - = -

4..<--.o /1x

dx

(2,2)

Q

.\.+o~ Ldx+ . 10 Fig. 2.5

49

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50

Stress and S!ralll---Allial loading

Since deformation and length are expressed in the same units. the normal strain E obtained by dividing 8 by L (or do by dx ) is a JiIllCIl.\·iolllc.\·s qualltity. Thus. the same nUll'K!rical value is obtained for the normal strain in a give n member, whether SI me tric units or U.S. c ustomary units are used. Consider. for instance, a bar of length L = 0.600 m and uniform cross secti on, which undergoes a deformation 0 = ISO X W- G m. The eorrcsponding strain is

8

1i50 X 10-(; m

L

0.600 III

E= - =

2S0 X W- ti !HI m = 250 X iO- f;

Note that the dcfonnati on could have been expressed in micrometers: I SO 11m. We would then have written

o=

o = -150-1l1ll - =

to = -

L

0.600 II I

2.50 }-lmJm = 2!)0}-l

and read the answer as "250 micros." If U.S. customary unit s are used, the length lmd deformation of the same. bar are, respectively, L = :23.6 in. and 8 = .'5 .9 1 X l O- J in . The corresponding s train is

o

to

= - = L

S.!)! X 10- 3 in . 23.6 in.

= 250 X 10-6 in / in .

which is the srune value that we found using SI unit ~ . It is customary. however. when lengths and defonnations are expressed in inches or microinches ( /-Lin .), to keep the original units in the express ion obtained for the strain. Thu s, in our example. tile st rain would be recorded as 6 in./in . or. alternati vely. as E = 250 }-lin.iin . E = 2.'5 0 X 10-

2.3 . STRESS-STRAIN DIAGRAM

Fig.2.6 Typlcaltensile-Iest specimen.

We saw in Sec. 2.2 that the diagram represcnting the relation between stress and strain in a given material is an important characteristic of the material. To obtain the stress·straill diagram of a material, one usually conducts a lemile lest on a specimen of the material. One type of specimen commonly used is shown in Fi g. 2.6. TIle cross-sectional area of the cylindrical central ponion of the specimen has been accurately determined and two gage marks have been inscribed on that portion ilt a distance Lo from each ()lher. The distance Lu is known as the-gage length of the specimen.

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2.3. Stress-strain Diagram

Fig.2.7 This machine is used to testtensiie test specimens, such as those shown in this chapter.

P

Fig. 2.8

The test specimen is then placed in a testing machine (Fig. 2.7), which is used to apply a centric load P. As the load P increases, the distance L between the two gage marks also increases (Fig. 2.8). The distance L is measured with a dial gage, and the elongation {) = L - Lo is recorded for each value of P. A second dial gage is often used simultaneously to measu re and record the change in diameter of the specimen. From each pair of readings P and 8, the stress u is computed by dividing P by the original cross-sectional area Au of the specimen, and the strain E by dividing the elongation 8 by the original di stance Lo between the two gage marks. The stress-strain diagram may then be obtained by plotting E as an abscissa and u as an ordinate. Stress-strain diagrams of various materials vary widely. and different tensile tests conducted on the same material may yield different results, dependi ng upon the temperature of the specimen and the speed of loading. It is possible. however, to di stinguish some common characteristics among the stress-strai n diagrams of various groups of materials and to div ide materials into two broad categories on the basis of these characteristics, namely, the ductile materials and the brittle materi als. Ductile materials, which compri se structural steel, as well as many alloys of other metals, are characterized by their ability to yield at normal temperatures. As the specimen is subjccted to an increasing load, its length tirst increases linearly with the load and at a very slow rate. Thus, the initial portion of the stress-strain diagram is a straight line

Test specimen with tensile load.

51

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52

Stress and Slraln----Axlal Loading

-, ~:~~~~~~~"----?-"i:-"

~ ~~: ----------= -------' "1""

~ 0"1

211

i :'-:-+-c---c-c---~

iYi,,'dj

Flg. 2. 10

'"

Tested specimen of a ductile malerial.

Fig. 2.1 1 Stress-slrain diagram for a typica l brittle maleriaL

211

Str~j n - h :ml"ning; ~N'kin ~ 1).2

D.U::'

Fig. 2.9 Stress-slra n diagrams 01 two typical ductile materials.

,

:

11.2

02..'i 0.00.1

0.00 12 (II )

Low-
' 1 ~(1 1

(bl Ai",,,,,,,,,"

~n"r

with a steep slope (Fig. 2.9). However. after 11 critical value u r of the stress has been rcached. thc specimen undergoes 11 large deformatio n with a relati vely small increase in the applied load. Thi s defonllUtion is caused by slippage o r ihe material along oblique surfaces and is due, therefore, primarily 10 shearing sircsscs. As we can note from the stressstrain di agrams of two typical duclile materia ls (Fi g. 2.9), the elo ngation of the spec imen after it has startl'd to yield can be 200 times as large as its deformati on before yield. Arter a certain max imum v;llue of the load has been reached. the diameter o f a porti on of the specimen begins to decrease. because of local in., tabllity (Fig. 2. 10(/). This phenomenon is known as necking. After necking has begun, somewhat lower loads arc su fficient to keep the specimen elo ngating further. Ulltil it linally ruptures (t<"[g. 2. lUb). We note that rupture occurs along a cone-shaped su rface. that forms an angle of ap proximately ·l5 C wit h the original surface of the specime n. This indicates thm shear is primarily re .~po n s ib[ e for the failure o f ducti le m;ltcrials, and confirms the fact that, unde.r an ax ial load, shearing stresses are largest o n surfaces forming an angle- of 4.'5 0 with the load (d. Sec. 1.1 1). The stress 0")" at which yield is initiated is called the yield strength of the material, the s(fe.~s ac corresponding to the maximum load applied to the specimen is known as the IIltimate ~·rrcngth. and the StreS5 0" ~ corresponding to rupture is called the breaking .\·/renglh . Britt le materials, which comprise cast iron, glass, and stone, are char;lcterized by the f;lct that mpture occurs without any noticeable prior change in the rate of e.!ongation (Fig. 2. 11). TItuS, fo r brittle materiah, there is no difference between the ultimate ~ lrength and the breaking strength. Also. the strain at the time of rupture is mm:h smaller for brittle than for ductile materials. From Fi g. 2. [2, we note the absence of any nec king o f the speci men in the case o f a briule material, and obse rve that ruptun: occ urs along a _~ urface perpe ndicular to the load. We conclude from this observation that nonnal stresses arc prim.u-ily fe~ n sib [ e for the failure of brinle matcrials. t

t Thc tensile II."'SIS de.scribed in [hi s ~cti ofl were a~slimed [0 be conducted m normat temperalufC s. However. a material that is dUrli!e at nonna! temperatures may di splay 11m chara'·Irn stics of a brintc matcri al at Vl:ry tow lemperat~rcs. white a flomlalt y brink malerial may behal·c in a duclile fashion at vcry hi gh tcmpcmlufC s. At tempcralUrc.l other Ihan Ilorlllal. [hcrcf<)rc. Olle shoutd refer [0" Imtr~rial ill lJ dllcli!e .llme or In a materilJI ill " brilll~ SlaIC. ralher [han 10 a dli~Iik Of brint c IIlmerial.

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2.3. Stress·straln Diagram

Flg.2.12

Tested specimen 01 a brinle material.

The stress-slrai n diagrams of Fig. 2.9 show that structural steel and aluminum, while both ducti le, have different yie ld characteristics. In the case of structuml steel (Fi g. 2.9a). the stress remains constant over a large range o f values of the strain after the Ollset of yield. Later the stress must be increased to keep elongating the specimen, until the maximum vnlue rJu has been reached . This is due to a property of the material knuwn as strain-hardening . 111e yield strength of structural steel cnn be determineu during the. tensile test by watching the load shown on the d isplay of the testing machine. After increasing steadily, the load is observed 10 suddenly drop to a sli ghtly lower value, which is mni nlained for a certain period while the specimen keeps elongating. In a lIery carefully conducted test, one may be able to distinguish between the upper }'ield poilll. whidl corresponds to the load reached just before yield sta rts. and the lower yieltl pOilll, which corresponds to the load required to maintain yield. Since the upper yield point is transient, the lower yield point shou ld be used to detem linc the yield strength of the material. In the case of aluminum (F ig. 2.9b) nnu of many other ductile materials, the onsel of yield is not characterized by a horizontal portion of the stress-strain curve. Instead, the stress keeps increasing- although not linearly- until the ultimate .~ tre n gth is reached , Necking the n begins. leading eventually to rupturc. For such matcrials, the yield strength a r can be defined by the offsel method. The yield strength al 0,2% orfset , for ex · ample, is obtained by drawing through the point of the horizontal axis of abscissa f = 0.2 % (or f = 0.002), a line parallel to the initial stmight-linc portion of the stress-strain diagram (Fig. 2.13). The stress (Fl' corresponding to the point Y obtained in thi~ bshion is defined as the yield strength at 0.1'* offset.

_I ~ U,2 %ofr<<'1 Flg. 2. 13 Determination of yield offset method.

strength by

53

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54

Stress and Siraln---Allial loading

A standard measure of the ductility of a material is its pacem dongalion. which is defined a."

Percent elongation where Lo and L/i denote, respectively. the initial kngth of the tensile test speci men and its final length at rupture. The specified minimum elongation for a 2-in. gage length for commonly used steels with yield strengths up 10 50 ksi is 2 1%. We note that this means that the average strain at rupture should be at least 0.21 in .lin. Another measure o r ductility which is sometimes used is the I,ercelli redllclioll ill area. defined as Pe rcell t H:dm:tion in are:! =

where Ai) and Alj denotc, respectively. the initial cross-sectio nal area of the specimen and its minimum cross-sectional area at rupture. For structural steel. percent n..-ductions in area or 60 to 70 percent are common. Thus far. we have di sc ussed only tensi le tests. If a specimen made of a ductile material were loaded in compression instead of tension, the stress-strllin curve obtained would be essentially the same through its initial straight-line portion and through the beginning of the portion corresponding to yield and strain-hardening. Particularly noteworthy is the fact thaI for a given steel. the yield strength is the same in both tension and compression. For larger values of the strain , the tension and compression stress-strain curves diverge, and il should be noted that necking cannot occur in compression. For most brillic materials. one find s that the ultimatc strength in compression is much larger than the ulti mate strength in tension. This is due to the presence of fla ws. such as microscopic cracks or eavities. which tend 10 weaken the matcriul in tension. while nOl appreeiably arfecting its resistance to compressive failure.

~-------"- Fig.2.14

-- -- --

Ut' '''"P''''''"'

Stress·strain diagram for concrete.

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2.4. True Stress and True Strain

An example of brittle material with different properties in tension :md compression is provided by concrete. whuse stress-strain diagram is shown in Fig. 2. 14. On the tension side of the di agram, we first observe a linear elastic range in which the strain is proportional to the stress. After the yie ld point has been reached, the strain inc re'l~es fa ster than the stn:ss until ruplllre occurs. The behavior of the material in compression is di ffe rent, First. the linear elHstic range is significantly larger. Second. rupture docs not occu r as the stress reaches its maximum value. Instead, the stress decreases in magnitude while the strain keeps increasing until rupture occurs. Note that the modulus of elastic ity, which is represented by the slope of the stress-strain curve in its linear portion. is the same in tension and compression. This is true o f most bri ttle materials.

*2.4. TRUE STRESS AND mUE STRAIN

We recall that the stress ploued in the diagrams of Figs. 2.9 and 2. 11 was obtained by dividing the load P by the cross-~ec ti o na l area All of the specimen measured before any deformation hfld tflken place. Since the cross-secti onal arca of Ihe specime n deereilses flS P increases. Ihe_ stress plotted in our diagrams docs nOI represent Ihe actual stress in the_ specimen. The difference between the engilleerillg stress fT "" PIAo Ihal we have computed and the true !itre.~·.~ fT, = PIA obtained by di viding P by the cross-sectional area A of the defonned specimen becomes apparent in duct ile materials after yield has started. While the engineering stress u, which is directly proportional to the load P. decreases wi th P during the necking phase. the lrue stress (I,. which is proportional to P but also inver~e l y proport ional to A . is observed 10 kee p incre:lsing lIntil rupture uf the specimen occurs. Many scientists also usc a dcfilliti on o f strain different from Ihat of the engineering "tmi/l I" = {jj~l- Instead of u ~ ing the total elongation" and the origi nal value Lo of the gage length, they use all the successive val ues of L that they have recorded. Dividing each increment !:J.L of the distance between the gage marks. by the corresponding value of L. they obtain the elementary strain !:J.E = !:J.L/L. Adding the successive val ues of !:J.f. they define the true stmin 10,:

With tltt:: summat.ion replaced by an integral, they can also express the true strain as fQllow s:

' ~fLdL = lu i:.. t

4-.. L

~I

Yldd

(2.3)

The diagram obtained by plo tting true streS's versus true strain (Fig. 2. 15) reileets more accurately the behavior of the m:uerial. As we have fl lready noted. there is 110 decrease in true stress during the necking phase . Also, Ihe results obtai ned from tensile and [rom compressive

'-- - - - - - - - - " Flg. 2.15 True stress versus true strain lor a typical ductile material.

55

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56

Stress and Straln---Allial loading

tests wi ll yield essenti all y the same plOt when true stress and true strain are used. Thi s is not the case for large values o f the_slmi n when Ihe engineering stress is plotted versus the eng ineering strain . HowevCI, cngineers, whose rcspo nsibilil Y is to d:::lennine whether a load P will prod uce an accept able stress and all acceptable deformalioll in a given member. will walll to usc a diagram based on the engineering stress a "" PjAu and Ihe engineering strain f = i'Jj4J, since these expressio ns involve data that fire available to them, namely the cross-sectional area Ao and the length 4J of the member in its undcJormed state.

2.5. HOOKE'S LAW: MODULUS OF ELASTICITY

Most engineerillg structures are designed to undergo relatively small deformations. involving only the-straight ·line portion of the corresponding stress-strain diagram. For that ini tial portio n o f the diagram (Fig, 2.9). the stress a is directly proportional to the strai n c. and we Cfin write

fF

Q''''U(~hhL It ''''p',rt:d "II, ~' s:re! (A'iWI

t t;):t,- ~trl'ugtl"to\\"-allor

ot('('\ (,\9')2)

PUrl'ir
Fi g.2. 16 Stress·strain diagrams for iron an d different grades of steet.

= E f.

(2.4)

This relation i .~ known as Hooke '.\· Iaw. after the English mathematic ian Robert Hooke ( 1635- 1703). The coefficient E is called the modullis of elasliciTY of the material im'olved. or [lIsa YOllng 's modllluJ. aft er the English seie nti ~t Th o ma~ Young ( 1773 - 1829) . Since the strain £. is a di mension[C.';s quantity. the modulus E is expressed in the ~am e units a~ the stre!iS if . namely in pascals or one of its multiples if SI units are used. and in psi or ksi if U.S. customary units arc used. The largest value of the stress for which Hooke's law can be used for a given material is known as the proportional lim it of that material. In the case of ductile materials po~se~~ in g a wel l-defined yield point. as in Fig. 2.9(/, the proportional limi t almost coincides with the yield point. For other materials. the proportional limit cannot be defin ed as easily, since it is difiic ull to determi ne with :lccuracy the value of the stress a for which toe relation between a and E ceases to be linear. But fro m this very diflic ulty we can conclude for such matelials that using Hooke's law for values o f the stres~ slightly targer than the actual pro· portional limit will nut result in any signinc:1Jlt error. Some of the physical pro perties of struct ural metals. such as strength, ducti lity. and corrosion resistance. can be greatl y affected by alloyi ng. heat treatment, :lnd the manufacturing process used . For ex· ample, we note fro m the stress· strain diagrams of pure iron and of three diffe re nt grades of steel (Fig. 2. 16) that large v:lri:ltions in the yield strength, ulti mate strength, and linal str:l.in (d uctil ity) ex ist among these four metals. All o f them, however. possess the same modulus of elasticity; ill other words. their "stiffness: ' or :lbili ty to resist a deformati on within the linear range, is the same. Therefore, if a high-stre ngth sted is substituted for a lower-strength steel in a given structure. and if all dimensions are kept tlte same. the struclure will have an increased loadcarrying cap:lc ity. but its stiffness will remai n unchanged .

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2..6. Elastic Versus Plastic

For each of the materials considered so far. the re lation between normal stress and normal strain, II = EE, is independent of the directi on o f loading. This is because the mechanical properties of each material , including its modul us of elasticity E, are independent of the direction considered. Such materjal.~ are said to be isotropic. Materials whose properties depend upon the direction considered are said 10 be lllli.wtmpic. An important Cl;l~S of anisotropic materials consists o fjiber-

Behavior

reinforced composite 1II11ferillis.

These composite materials are obtained by embedding libers of a stro ng, sti ff material into a weaker, softer material. referred to as a 11/(/frix.. Typical materi als used as libers are graphite, glass, and polymen;, while various types of resins arc used as a matrix . Figu re 2.17 shows a layer, or /lIIlIilUl. of a composite material consisting of a large number of parallel fiber.<; embedded in a matrix. An axial load applied to the lamina along the x axis. that is. in a direction parallel to the libers, will create a normal stress II.),. in the lamina and a corresponding nomlal strain E,. which will satisfy Hooke's law as the load is increased and as long as the elastiC limit of the lamina is not ex.::eeded . Simi larly, an axial lOHd applied along the y axis, lhat is, in a direction perpendicular to the lamina, wi ll create a normal -stress 0'" and a normal strain E,. satisfyi ng Hooke's law, and an axial load applied along the z axis wit! creale a normal stress 0'0 and a normal strain Eo: which again satisfy Hooke's law. However, the moduli of elasticity E~, E,,, and E; correspondi ng, respectively, 10 each of the above loadings will be different. Because the libcrs are parallel tt) the x axis. the lamina will offer a much stronger resistance to a loading directed along the x axis than l{) a loading directed along the y or z axis, and E.. will be much larger than eithe r E,. ill~

.

A flat /am"':afe is obtained by superposing a number of layers or laminas. If the laminate is 10 be subjected on ly 10 an axial load causing tension, the fibers in at! layers shou ld have the same orientation as the load in on..lerlo obtainlhe greatest possible strength. But jfthe laminate may be in compression, the matrix material may not be sufficie ntly stro ng 10 prevent the tibers from kinking or buckling. The lateral stability of the laminate may then be inere,L~ed by positioning some of the layers so that their fibers will be perpendicular to the load. Positioning some layers so that their fibers are oriented at 30°, 45 or 60 c to the load may also b~ used 10 increase the resistance of the laminate l() inplane .~ hear. Fiber-reinforced composite materials wi ll be further di sc llssed in Sec. 2.16. where their behavior under muhiaxial loadings will be considered. Q

,

2,6. ELASTIC VERSUS PLASTIC BEHAVIOR OF A MATERIAL If the strains cHlIsed in a test specimen by the application of a given load disappear when the load is removed. the materi al is said to behave ela.tfically. The largest value of the stress for which the material behaves elastically is called the elm'fic limit of the malerial. If the material has a well-defined yield poi nt as in Fig. 2.9(1, the clastic limit, the proportional limit (Sec. 2.5), and the yield poi nt are essentially equal. In other words, the material behaves elasticall y and

'[

~I~

~;,;~~ .~,) "

l~,

Fi1",«

Fig. 2.17 Layer ollibe/·reinIOiced composite material.

57

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58

Stress and Straln---Allial loading

1l1l1'hm'

E\

t)

Fig. 2.18

c

A

I)

Fig. 2.19

R"pt",p

linearly a.!.; long as the stress is kept bdow the yield point. If the yield point is reached. however. yie ld takes place as described in Sec. 2.3 and. when the load is removed. the stress and strain decrease in a linear fashi on. along a line CD parallel to the straight-line portion AB of the loading curve (Fig. 2.18). The fact that E docs not return to zcro after the load has been removed indicates that a remitment sel or plastic rie/ormation o f the material has taken placc. For most materials. the plastic defo rmmion depends not only UPOIl the max.imum value reached by the stress, but also upon the tim!:! elapsed before the load is removed. T he stress-dependent part of the plastic deformation is referred to as slip. and the time-depe ndent part- which is also influenced by the temperature- as (lreep . When a material docs not possess a wen-defined yield point. the clastic limit cannot be determi ned with precision. Howevcr. assuming the elastic limit cqual [0 the yield strength as detincd by the offset method (SC(.~. 2.3) results in only a small error. lndeed. referring 10 Fig. 2.13, we nOie that the ~traight line used to detennine poim Y also represents the unloading curv-e after a maximum s[re~s 0"1' has been reached. While the material docs not behave truly elastically. the resulting plastic strain is as .~ mall as the selected offset. If, after being loaded and unloaded (Fig. 2. 19), the test speeimcn is loaded again. the new loading curve will closcly follow the earlier unloading curve until it almost reaches point C; it will then bend to the right and connect with the Cllrved portion of the original stres~-strain diagram. We note that the straight-line portion of the new loading curve is longer than the cOlTe~pondll1g portion of the tIlltlal one. Thus. the proportional limit and the elastic limit have incre:L~ed a~ a result of the strain-hardening that ocCUlTed during the earlier lo:.!ding of the ~pccilllcn. Howeyer. since the point of rupture R remains unChanged. the ductility oflhe specimen. which should now be measured from poi.nt D, has decreased, We have assumed in our discussion thm the specimen wa" loaded twice in the same direction, i.e .. Ih:tt both loads were tensile loads. Let us now consider the case when the second load is applied in a direction opposite to that of the tirst one. We a.~s ume that the materi:.!l is mild steel. for which the yield strength is the same in tension and in compression. The initial load is tensile and is applied until point C has been reached on the stress-strain diagram (Fig. 2.20). After unloadi ng (point D) , a compressive load is applied. c:.!usi ng the material to reach point H , where the stress is equal to - Uy. We note that portion DH of the stress-strain diagram is curved and does not show any clearly defined yield point . This is referred to as the Ballsch illger eiff!cl. As the compressive load is maintained. the materi:tl yields along line H./ . If the load is removed after point J has been reached. the stress returns to zero alo ng line JK. :tnd we note that the slope of JK is equal to the modulus of elasticity E. TIle resulting permanentsct AK may be positive, negative. or zero, depending upon Ihe lengths of the segments BC and f/J. If a tensile load is :.!pplied again to the test specimen. the portion of the stressstrain diagnun beginning al" K (da~hed line) will enrve up and to the right until the yield stress O" y has been reached.

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, I

I

'"

K I

1

c

" 1"

/

2.7. Repeated Loadings : Fatigue

c·

I

I

I

1Ul'

A

r /I

H

-",

Fig. 2.20

If the inilial loading is large enough 10 cause slrain-hardl!ning of the material (point C'), unloading takes place along line CD'. As the reverse load is applied, Ihe stress becomes compressive, reaching its maximum value at f1' and maintaining i1 as the malerial yields along line /1'1' . We note Ihal while the maximum value of the compressive stress is less Ih:m 0" ),. the IOlal change in stress between C' and H' is still equal to 2CTr. If poinl K or K' coincides with the origin A or the diag ram, the perm,ment sel is eqlllil In zern, mut lhl' specime.n may arreaT In have re-

turned to its original condition. However, internal changes will have taken place and, while the same loading sequence may be repeated, the specimen wi ll rupture without any warning aft er relatively few repetition.~. This indicates that the excessive plastic deformations to which the specimen was subjected have caused a radical change in the characteristics of the material. Reverse loadings into the plastic range, theretore, are seldom allowed, and o nl y under carefully controlled conditions. Such s itll~ti o n s occur in the straightening of damaged material and in the final alignment o f a structure or machine.

2.7. REPEATED LOADINGS; FATIGUE Ln the preceding sections we have considered the behavior o f a test spec.imen subjected to an axial loading. We Tecall that, if the maximum stress in the specimen does not cxceed the elastic limit of the material. the specimen returns to it$. initial condition when the load is removed . You might conclude that il given loading may be repeated many times. provided that the stresses re main in the elastic range. Such a conclusion is correct for IOildi ngs repeated a few dozen or even a few hundred times. However. as you will sec, it is not correct when loadings arc repeated thousands or millions of times. In such cases, rupture will occ ur at a ~trcss much lower than the static breaki ng strength : this phenomenon is known as,fatigue. A fatiguc failu re is of a briule nillUrc. even for materials that arc no rmally ductile.

59

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60

Stress and Straln---Allial loading

St N>! ( 11)20HR)

,

•

,;; 2U -

A!"m;num (202r1 ) 10

1(f1

W!

to>

IO~

I(f

lU~

10"

N mH! wr " f ~u mpl rtd)' ,r\'\'rsr.d c),(·Ir. .•

Fig. 2..21

Fatigue must be considered in the design of all struclural and machine components thai arc subjected to repeated or to Iluctuating loads. The number of loading cycles that may be expected during the useful life of a componenl varies greatly. For example , a beam supporting an industrial crane may be lom.led as many as two million times in 25 years (about 300 loadings per working day), an automobile crankshaft will be loaded about half a billion times if the automobile is driven 200.000 miles. and an individu:11 turbine blade may be. loaded several hundred billion limes during its lifetime. Some loadings arc of a nuctuating nature. For example, the passllge of traffic ove.r a blidge will cause stress levels that will nuctuate about the stress level due 10 the weight of the bridge. A more severe condilion occurs when a complete reversal of the load occurs during the loading cycle. The stresses in the nxlc of a railrond car. for example, are completely reversed after ench half-revolution of the wheel. The number of londing cycles required to cause the failure of a specimen through repeated successive loadings and revcrse loadings may be dctcnnined experimentally for lmy given maximum stress leveL If a series of tests is conducted. using di fferent maximum stress levels, the resulting data may be ploued as a (T -II curve. For each test, the maximum stress (T is plolted as an ordinate and the number of cycles II as an abscissa: because of the large number of cycles required to r rupture, the cycles II are plotted on a logarithmic scale. A typical (I-II curve for steel is shown in Fig. 2.21. We note that. if the applied maximum stress is high. rdatively few cycles aft! required to cause rupture. A s the magnitude of lhe maximum stress is reduced. the number of cycles required to cause rupture increases, unlil a stress. known as the emll/mllCl' limit. is reached. The endurance limit is the siress f()r which failure does not occur, even fur an indefinitely large number of loading cycles. For a low-carbon steel, such as structural steel. the endurance limit is abOlll one·IUllf of the ultimate strength of the steel. For nonferrous metals, such as alu minum and copper, a typical (I-II curve (Fig. 2.21) shows Ihat the stress at failu re contin ues to decrease as the number of loading cycles is increased. For such metals, one defines thcfa/(~lle limit as the stress corresponding 10 failure after a specified number of loading cycles. such as 500 million. Examination of test specimens, of shafts, of springs, and of other components thai havc failed in fati gue shows thtll the failure was iniliatcd at a microscopic crack or at some simi lar imperfection. At each loading. the crac k was vcry slightly enlarged. During successive loading cycles, the crack propagated lhrou!,:h the material until the amount of undamaged material was insu fficient to carry the maximum load, and nn abrupl, brittle failure occurred. Because fatigue failure may be ini tiated at any crack or imperfection. the surface condition of a specimen has an important effect on the value of the endurance limit obtained in testing. The endurance limit for machined and polished specimens is higher than for rolled or forged components. or lor c01l1 ponent~ that arc cOiTOded. In applications in or near seawater. or in other applications where corrosion is expeclCd, a reduction of up to 50% in the endurance limit can be expected.

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2 .8. DEFORMATIONS OF MEMBERS UNDER AX IAL LOADING

2.8. DefonnaUons Under Axial Loading

Consider a homogeneous rod Be o f length L and uni form cross seclion of area A subjected to a centric axial load P (Fig. 2.22). If the resulting axial stress fJ = PIA does not exceed the propofli onui timil of the material, we may apply Hooke's law and write

B

(2.4)

from which it follows thai

P

~

E= - = E AE

Recalling that the strain

E

was defined ill Sec. 2.1 as

8 and, substituting for

E

= E L

(2.5) E =

oiL, we have (2 .6)

from (2.5) into (2.6):

c

,

,

c

I

r 8

=

Fig, 2.22

PL

(2.7)

AE

E(lUHLion (27) may be used only if the rod is homogeneous (constant E), has 11 unifonn cross section of area A. and is loaded al ils ends. If the rod is loaded at other points, or if it consists of several portions o f variOlLs cross sections and poss ibly of difJe rent materials, we must divide it into compo ne nt parts that satisfy individually the required conditions lo r the application of formula (2.7). De noting. respectively. by PI' L,. Ai> and E; Ule. internal force, length. cross-sectional area. and modulus of elasticity corresponding 10 part i, we express the. del"ormation of the entire rod as (2.8)

We recall from Set:. 2.2 that. in the case of a rod of variable cross section (Fig. 2.5), the strain € depends upon the position of the point Q where it is computed and is de fined as IE = d8/dx. Solving for dB and substituting for € from Eq . (2 .5). we express the defo nnation of an cle ment o f length dr as P tLr d8= lEl/X = - -

AE

1lle total dcformmi on 8 of the. rod is obtained by intcgrating thi s cxpress ion ollcr the length L of the rod : (2 .9)

Formula (2.9) should be used in place of (2 .7). no t only when the crosssectional area A is a fU lic tion of x, but also when the internal force. P depends upon x. as is the case for a rod hanging under its own weight.

61

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EXAMPLE 2.01 Determine the deformation of the steel rod shown in Fig. 2.23/1 under the given loads (E '" 29 X Icr psi).

We divide the rod into three component pans s hown in Fig. 2.23b and write L I = L2 = 12 in. A I = A, = 0.9 in!

A ~ 0.3 ill~

A

B

e

I

JJ

L ] = 16 in.

A, = 0.3 in!

To lind the internal forces PI' P2 • and P l ' we must pass sections through each of the component parts. drawing each time the free-body diagram of the portion of rod located 10 the right of the section (Fig. 2.23c). Expressing that each of the free bodies is in equilibrium. we obtain successively

.,

I"

I (b)

'-~

!is I.J p.

P I = 60 kip. = 60 X IOl Ib

e ",5lir~

D

·· ·

P2 = - IS kips = - 15 x 10' lb

10 kil"

IJ 1 = 30 kip. = 30 X IOl Ib Carrying the values obtained inlO Eq. (2.8). we have

·

"

~

'. Ie> Fig. 2.23

e B

Fig. 2.24

'"

" I.';:?;: " e i,

75 kips

E2 • #J k,p.,

·

JJ 3(1

kit"

+

D

lU k,p., I.'}

kip.

2

PiLi = .!.(PIL 1 + P2 L2 + P1Ll ) ; AiE, £ AI Al A] 3 ~ __ 1 _[ (60 X 10 )( 12) 29 X lOt 0.9 (- IS X IO~ ) ( 12 ) (30 x 10])(16) ]

0=

0.9

2.20 X 10 [, 29 X lOt

+

0.3

6

75 .9 X 10 - ] in.

The rod Be of Fig. 2.22, which was used to derive formula (2.7), amI the rod AD of Fig. 2.23. which has just been discussed in Example 2.0 .1. both had one e nd attnched to a fix ed support . In each case, therefore, the deformation 5 of the rod was equal to the displacement of its free end. When both ends of a nJ
where A is the cross-sec tional area of AB and E is its modulus of elasticity.

62

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SAMPLE PROBLEM 2.1

·c

-

o[r '

A

'tilL'

il

The rigid bar HDE i s supponed by two links AB and CD. Link AD i s made of ,,]uminu r.1 (E = 70 CPa) and has a cros~_sectional area of 500 Illml: link CD i.~ nmde of steel (£ = 200 GPa:, :'Illd lw.s a cross-sectional area of 600 mml. For the 3O·kN force shown. determine the deneclion (a) of B. (b) of D , (c) of E.

E

/J

W-"",, - I O.2m

SOLUTION Fret: Body: Har BDE

- po kN)(O.6 m) ....

+,) 2:.\1" = 0;

FcuCO.1 ml

=0

FUJ = + 90 kN Fr;v = 90 kN telJ.I';tJlI - (30 kN)(O.4 m) - Fw(Q.2 111) = 0 F,\/j = - 60 kN F ' \/1

-

611

FMJ = 60 kN

cOlJlprnJio/l

J;.~'

IJ .

"

nl,nl'<'I inn fir R .

Sim:1' th e internal

fnr~e

in link AH is

I:nmprl'.ssi\~ .

we have P = - 60 kN

I L,

II =.Sfl(\",,"!

0 .3111

PL

{j

£ =70Gl'a

t

(- 60 X IO' N)(0.3 m)

=-= AE (500 x lO- b m2)(70 X 109 P
The negative sign indicates a contraction of mel:1ber AB. and. thus. an cp· ward det1ection of end B: t',,~ '" IiIJ~j\

8y =O.5 14mm t ....

c

r-

Ii. Ikncction of D .

Sin.::.:: in nxl CU. f > -

A = (l(J() ",,,,! £ =2r)(iCl'a

OA m

L

')0 kN. we write

(90 X I()' N)(O.4 m)

J'L

0 1) = AE

(6f.o X

lo-~ Ill')(200

X I()'] Pal

00

= O,3(XI nmt ~

....

D I'rn

~

J , = 0..514 "',,. (-;- H' ' 1

liD

II

r. Dcnectioll of E . We denote hy B' and D' the displ
!!I,k:" = 1).30)

line and we wrile

Ill'"

/)

E

BB'

BH

0.514 mm

(200 /lun) - x

x = 73.7mll1

--~~

y "'" 'Ba

(2e1) 1)1I11-~)

:c

200m",

:1 :1 ,

E'

-100"",, - - 1

DD'

HD

EE'

HE

3,

DD'

liD

0.300 mm

0.300

mill

(400 nun )

+ (73.7 mm)

73.7 mm

(it:

J ,n~ mm .). ...

63

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1--- - -

ISin

- - ---I

C~~====nFD F

C

<:tIU-liiiiiiiiiiiiiiiiii~Jj:l 1l 12 in .

SAMPLE PROBLEM 2.2 The rigid castings A and H are connected by two f- in.-diameter steel bolt~ CD and GH and are in conlact with the ends of a I.S-ln.-diameter aluminum rOO EF. Each boll is single. threaded with a pilch of 0. 1 in .. and after being snugly fitled. the nuts at D and H are both tightened one-qlmrter of:l turn. Knowing Ihal E is 29 X IO~ psi for steel and 10.6 X 10" psi for aluminum, determine the nomlai s tress in the rod.

SOLunON Ddormalions

D

C

1'/,

BoilS CD alld GR. Tightening the nuts causes tension in Ihe bolts. Because of symmetry. both are subjected to the same internal force 1\ and undergo the same deformation 0b' We ha\le

E

PJ.b

" ,-_~;;;;;;;;~-- P" II P',

Pb( 18 in.)

+ 1.405 x [0 6 Pb

0=+--= + b Alfih ~ " (0.75 ill.)'(29 X Jot'psi )

F

(I)

Rod EF, The rod is in compression. Denoting by P, the magnitude of the force in the rod and by 0, the defoTmmion of the rod. we wri te

a

,

P,L ,

P,.( 12 in.)

AlO.

lr.( 1.5 in.f(10.6 x IO~ psi )

= -- ~

- 0.6406

x

10 ~ ~

P,

p)

Di.'Ipf(/cemelll 0/ D Re{(/five to 8, Tightening the nllls one-quaner of a turn causes ends D tlnd H of the bolts to undergo a displacement oq(O.1 in.) fe/ulil'<' to Cu.ilillg B. Considering end D. we write

8 0lE = t(O.1 in. ) = 0.025 in.

(3)

Bm aU/II = Ou - 88 , where Ou :md OE represent the displacements of D :md B. If we ass ume that casting A is held in a fixed position while the nUL~ at D and H are being tightened. these displacements are equal to the deformations of the bolts and of the rod. respectil/ely. We hal/e. therefore. BUlB = % -

(4)

fi e

Substituting from ( I). (2). and (3) into (4), we obtain 0.Q25 in. = 1.405 X 10

t

Ph + 0.6406 X 10 6 1',

(5)

Frl'c Hody: Casting B

1:.'5:.P = O:

(6)

Forces i.n Bolls and Rod Sl.lbstitl.lting for P, from (6) into (5). we have 0.025 in. = 1.405 X 1O -6Pb + 0.6406 x 10- 6(21'1.) p~ = 9.307 X loJ lb = 9.307 kips

P, = 2Pb = 2(9.307 kips) = 18.61 kips Str ess in Rod P,

18.61 kips

u"=A,= b r( L5in. )2 64

If.

=

10.53 ksi "III

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PROBLEMS

2.1

Two

g~ge

marks are placed exactly 250 mm apart on a 12-mm·

diameter aluminum rod. Kn ow ing thaL wi th an axial load of 6000 N acting on

the rod. the distance between the gage marks is 250.18 rum. delemline the modulus of elasticity of the aluminum used in the rod.

2.2

A poly.,(yrcne roo of length 12 in. and diameter 0.5 in. is subjected

to an 800-lb tensile load. Knowing that £ = OA5 X l ot' psi , determine (a) the

elongation of the rod. (b) the normal stress "in the rod.

2.3

A 60.m.long slecl wiTe is subjected to 6-kN tensile force. KJlOwing

Ih;!l E = 200 GP<J :md tiW( the length of the rod increases by 48 mm. determine (a) the smallest diameter that may be selected for the wire. tb) the corresponding nomlat stre ss. 2 .4 A 28-ft length of O.25-in.-diameter steel wire is to be u~d in a hanger. It is noted that the wire stretches 0.45 in. when a tensile force P is applied. Knowing tllm E = 2.9 x 10& psi. LleTermine (a) lhe magniTULIe of Llle forL'e 1', (b) the corresponding normal stress in the wire.

2 .5 A c:t.s t·iron tube is used to support a compressive load. Knowing th:u £ = 69 G P
2 .6 A control rod made of yellow brass must not stretch more than 3 mm when the tension in the wire is 4 kN. Knowing that £ = 105 GPa and that Lhe maximum allowable normal stress is 180 MPa, determine (a) the Sm
65

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66

61"'33 and SlrBlr>-A);1aI L
2.1 0 .0\ 1.5-Lll- lung alum inum roJ lilustn()t ~tj"t':lch [!lOIC th un I mm and the normal stress must not exceed 40 M Pa when the rod is s ubjed~d 10 a 3-kf' axiallo.1d. Knowing Ihal £ = 70 GPa. determine [he reqlJired diametet of the rod

2 .11 All aluminum control rod must stretch 0.08 in. wilen 3 500-lb len· sile- load is applied to i1. Knowing Ihm i f.. = 22 hi tlnd E = 10.1 X 1 psi. determine the ;; mallest Jiameler alld shortesllenglh which may be sel~cted for the rod.

r.r

2. 12

A sqllare wUl11innm bar should nO! stn'tch mor;: than 1.4 mm when

;1 is s uhjecled to a !c:nslk . l
tellsiJe ~treng th is 120 -'1 Pa. determine (II) the maximum al lowable length of Ihe pipe. (b) the required dimensions of [he cross section if the ten sile load is 28 kN

2.13 Rod BD i ~ made of steel (E "'- 29 X ICf psi) and is uscJ [0 brace the axiall y compress~J memher ABC The maximum force that ca n be deve loped In mcmher flD is U.02P. lflhe stress muot not exceed Ib ksi and the maximum change in 12ngth of BD nUiSI not e.'(ceed 0.001 times the 12ngth or ABC. determine the smallest-diameter rod thai can tx- used for memix'r 8D, t' ~ !.J4lkip_ '

Fig. P2.13

,\

Hi. I ~'",\ -~I" .

i

P

I

I

\ c

Fig. P2.14

B

FIg. P2.15

TI- -' 2.1> ,1\

:,-,) 111

IU ill.

I -

2.14 The 4-mm-
2.15 A ~-in.-I hick hollow polystyrene cylinder (E = 0.45 X 10" psi) and a rigid ci rcular plale (llnly part of which is shown ) arc used 10 :;upport a IO-in.long ~leeJ rod AS (£ '" 29 X 100 pI;;) of ~_il1_ di~meter_ U an 800.1b load I' is iJPplied at B, d~termine (/I) the elongation of rod AB. (b) the deflection of point B. (c) thc avcnge normal stress in roo AB.

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2.16 The specimen shown is made from a l·in.·diameter cylifl(lricaJ steel roo with two I.S-in.·outer-diameter sleeves bonded to the rod as shown. Knowing that £ = 29 X I()f> psi. determine (a) the load P so that the total defomla· tion is 0.002 in., (b) the corresponding deJorm
Problems

II"'" diamd(', I"

I --- !

~~~,',,~_,~

....

2. 17 Two solid cylindrical rods are joined
J·in. dfamrtcr

P =JOkN

r--=

r

FIg. P2.16 -

30lUl1I

I'

2.'>0 """

3(X)

,m"

IIlkN

r

- so"""

2. 18 For the composite rod of Prob. 2.17, detennine (a) the load P for which the tOlal deformation of the rod is - 0.2 mm. (b) the corresponding de· flection of point B. 2 .19 Both portions of the rod ABC are made of an alum inum for which £ = 70 GPa. Knowing that the m
8 25m

~'~~=~~~~'U 4.0", - -

t

B

Fig. P2.tT

.4.0m - -

20-",m '[",mr ler

() I :n

L~c

Fig. P2.21

A

0.510

rl(J-mm

l~ c

Fig. P2.19 and P2.20

,]""nrl,'r

67

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68

51"'33 and

Slmlrl-A)JaI

Loading

2.22

Foc Ill':

sl~d

min~ the defonnation.~

tlUS" (E

29 X 10' psi) aoo iu;]uillg shuwn, u':lcr,

=

?f the ~e~lbers B~ and DE.

knowing that their cross·

sectional areas are 2 In' and 3 m' , respectively. :"lI I1ir S

I

.~

"

ft

tt-

;j1 1 kip,

B

3u l;r·'

/)

C

Sf,

E

sn

I

Fig. P2.22 2.23 Members AB and CD are I ~,' n.,diameter ~teel rods. and members Be and AD are ~-in. ,diameter steel rods. When the turnbud:le is tightened. the diagonal member AC is put in tensioll. Knowing tha[ £ ~ 29 X 10' psi and" - 4 ft. determine the largest al lowable tell.lioll ill AC Ml tlMt the defO!, m~tions in members All and CD do nO! exceed 0.04 in. 2.24 For tile structure in Prnb. 2.23, detennine ia) the distance h so that lhe deformmions in members AB, BC, CD. and liD are &]UaIIO OJ)4 in., (h) the

corresponding tension in member AC. 2.25 Each of th~ tour ven ical links connecting the two horizontal mem, is marle (If aillminlim (P = 70 (i Pa) ~nn hll ~ II llnifflnn rnctanenlar cms~ section of 10 X 40 mm. For the loading shown, detenn ine the deflectioo of hm.~

(lI)

point f. (h) point F, Ie] point G.

E

J

'!.lkN

Fi g. P2.25

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2.26 Each of IJ,e J jllk~ AB and CD j .~ lIlade. or sleel (E - 29 x I O~ 1"'1) and has a uniform reclangular cro~s seelion of ~ X I 111. Delermine Ihe larg~ sl load which C,ill be su~pe nded from point £ if Ihe ddlection of £ is nOllO <:'xce.e tl (J.UJ m.

, ,,

f'mbl""",

~---\lIlD Sin

-' 'A JC ~ II) ill.

-

-

15 ill .

J

FIg. P2..26 2.27 Each of the link, AD and CD i~ made of al uminum (E - 75 Cra) and Itas a cros s-se.ctional area of 125 rnm'. Knowing thillthey suppot1the ri!!id member Be. determine the dellectkm of poinl E.

I

"

v p-

5 ~N

U'L~~=:! ':;;:===~ I" I ",,_I o..

C

0.20",

FIg. P2.27 2.2B The length of th ~ 2-mm·diameter steel wire. CD has been adjusled so Ihat with no load applied. a gap of 1.5 mm exisls between the end B of Ihe rigill bt::aru ACIJ amJ a cunta~l point E. Knowing that E = 200 GP
0"

Ih .. ~"m

;"

"rdf'r

'0 (,""~f" ,-onbC'

,>

1

1'.2.'i ,,1

between B and E. 2 .29 A homogeneous cable of length L and uniform cross seclion is 3 uSpcnded from one end. (al Denoting by fJ the density (mass per unit \'olumel of the cable ami by E its modulus of elasticity. determine the elongation o f the cable due to its own we.igh[. (b) Show that the .arne elongation would be obt:1ined if the c:1ble were horizo ntal and if a force .:qu:ll to hal f of its weig ht were applied at each end.

O.OS 111 Fig. P2.2B

2 .30 A \'eaicaJ load P i ~ applied .. t tbe: cemer A of tl \t~ uppe' section of a homogeneous frustum o f a circular cone of heighlil. minimum radius (/, and maximum radiu~ b. Denoting by E the modulus of eiasLic,ty of the material and negiectlJlg the clleet 01 Its weIght. determme the detlection of pomt A. 2 .31 Denoting by E the ··engi[[cering >t rain"' in a tensile specimen. show lhat the true strain i ~ E, = In(1 + t ). 2 .32 Thl'; voll1l1ll". of a lemile. ~J1t"cimen i~ ~~~t';ntially cn n.'''ant wh ile. rb~­ tic deformation occ urs. If tlte initial diameter of the specimen is d,. show that when the diameter i., d. the Irue strwn i. c., - 2In(d , /d).

Fig. P2.30

69

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70

Stress and S!raln---Allial loading

2.9. STATICALLY INDETER MINATE PROBLEMS In the problems considered in the preceding section, we could always usc frl!e-body di agrams and equilibrium equations to determine the inlernal forces produced in the various portions of a member under givl!n loading conditions. The values obtained for the internal fo rces were then entered into Eq. (2.8) or (2.9) to obtain the deformation of the member. The re are many problems, however. in which the internal forc es cannot be determined from statics alone. In fac t, in most of these problems the reactions themselves- which are exlernal forces-cannot be detennined by simply drawing a free-body diagram of the member and writ ing the corre~pon din g equilibrium equations. The equi librium equations must be complemented by relations involving deformations obtained by considering the geomct[y of the problem. Because staLics is nOI sufficient 10 detemli ne either the. reuclions or the internal forces. problems of this type arc said to be staficolfy illlielel1l1illate. The following examples wilJ show how 10 handle this type of problem.

EXAMPLE 2.02 A rod of length L, cross-sectional area AI. and modulus of elasticity fl. has been placed inside a tube of the same length L. but of cross-sectional area Al and modulus of elasticity £~ (Fig. 2.25(/). What is the defonnation of the rod and tube when a lorce P is: exerted on a rigid end plate as shown'!

Denoting by PI [uld P2. respectively, the a .. ial forces in the rod and 1n the tube. we draw free·body diagrams of all three elements (Fig. 2.25b , c d). Only the last of the diagrams yields any ~ ignilical1l infonnation, namely: (2.11)

- -I. - - - j

'0'

Clearly. one equation is oot sufficient 10 detennine the twO unknown internal forces PI and P~. The problem is statically indeterminate. However. the geometry of the problem shows thai the defornlations OJ and 02 of the rod and IlIbe must be equal. Recalling Eq. (2.7). we wri te (2.12)

(b)

Equating the deformations .s l and

.s~.

we obtain: (2. 13)

'"

~~-.' P,

(d )

Equations (2.11) and (2.13) can be solved simu ltaneously for PI and P1: P,

Fig. 2.25

Either of Eqs. (2. 12) can then be used to detemline the common deformation of the rod ,md tube.

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EXAMPLE 2.03 A bar AB of lengtb. Land unifoml cross section is attached to Jigid suppons at A and B before being loaded. What are the s t res~es in portion; AC and BC due 10 the app lication of a load P at point C (Fig. 2.2OO)?

n,

r e, t- e, "

I

.-

1-

L B

Drawing the free-body diagram of the bar (Fig. 2.26b). we obtain the eqUilibri um equillion (2.14) Since this equation is nO! suffidem to determine the tWO unknown reactioas R. . and R6 • tb.e problt'm is ~ taticaJJy indeterminate. However. the reactions may be determined if we observe from the geometry that the lotal elongation 0 of the bar must be zero. Denoting by 0 1 and 0". re~pectively, the elong3tions of the ponions AC and BC, we write

"1

"

~

"

e

". I,)

or. expressing 0 1 and 0, in terms of the corresponding internal forces P I and 1'2:

1M

(2.15)

Fig. 2.26

":, £,

e, ------y., I. )

-~:-----?.

BlH we note rrom the free-body diagrams shown respectively in parts band C of Fig. 2.27 that P I = R,I and 1', = - R B. Carrying these values into (2.15), we write (2.16)

II, )

Equations ( 2.14) and (2.16) R.... and Rs: we obtain Rio desired s tresses in AC dividing. respecti\'ely. P I = sectional area of the bar:

"I

H

u = PL~ Fig. 2.27

,

AL

ean be solved s imultaneously for = Plnj L and Nil = PL / L. The and (-'1 in BC are obtained by R,I and P, = - R~ by the cross-

PL,

AL

Superposition Method. We obs~ rve that a structure is statically indeterminate whenever it is held by more supports than are required to maintain its equilibrium. This results in more unknown reactions than available equilibrium equat ions. It is o rte n found convenient to designate one of the reactions as redundant and 10 eliminate the corresponding supp()n. Since the stated condit ions of the problem cannot be arbitrarily changed, the redundant rCliction must be maintained in the solution. But it will be treatcd as an IIl/knOll'1/ load that, together with the other loads, must produce deformations that are compatib le with the origi nal constraints. The actual solution of the problem is carried out by considering separately the deformations caused by the given loads and by the redundant reaction. and by adding- or .mperposil/g- the results obwi ned.'

in

'The generat conditiollS under which the rombincd cff~ct of several toads ('all be obtaincd way JI'(' discmscd in Sec. 2.12.

thi~

71

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EXAMPLE 2.04 Determine the reactions ntA and B for the steel bar and loading s hown in Fig. 2.28, assuming :J. dose fit at bolh supports before the loads arc applied.

Following the same procedure as in Exumple 2.01. we write 1'2 = I:>" = 600 X 103 N 1'4 = 900 X 10] N AI = A ! = 400 x 10 6 m l A] = A4 = 250 X 10 10 m"

PI = 0

L~

L , = L2 = L J =

,----. , ----.

"

'1 =2.')(}rn"'~ '1 (1 ~i\

----.

lSOm",

D

I

150 """ /) _,..j ----.l

'\(~) \..1>.

ISO "''''

C II = otlH I ",,,,L..,.

~ ISOm",

K

,

fdJ],,·:-J

= 0. 150m

3

C

~ ISO",,,,

-

15(/"1'"

~

ISO m'lI

2.

K

---1

I1O(lkN

-T-, ~ 15O

Fig. 2.28

11

---.i "

H Fig. 2.30

Substituting the..'ie values into Eq. (2.8), we obtain We consider the reaclion at B as redundant and release

the bar from that support. The reaction RB is now considered as an lmknowrl load (Fig. 2.29<1) alld will 00 determined from the (oudition Ihnt the deform:lIion 0 of the rod must be equal to zero. The solution is carried out by considering separately the deformation (jL caused by the given loads (Fig. 2.29b) and the deformation 01/ due to the redundant reaction R8 (Fig. 2.29c).

±

_P,_L,

,_ , AlE

X = (0 + c:::6O(;C' c;-c:::"";""N,, 400 X 10 om" 600 X 10~ N X [0 6 m2

+ 150

+

900 X 101 N ) 0.150 m 250 X 10 ~ JlI~ --E-

1.125 x 109

E

(2. i7)

Considering now the deformation (jl/ due to the redundant reaction RB • we di vide the bar into two portions. as shown in Fig. 2.3 1. and write A

" 3(Jj1

k~

lUll! kN

,

, , 3I~J

,

l:'li

PI = p" = - RB A

6 m2 A" =250 X lO " m2 L, = L" = 0.300 m

A l = 4oo X 10

+ "

&)UlN

li=O

". 1"

~

I

R,

0"

C

I

(, )

Fig. 2.29

300 """

~

, The defonnation (jL is obtained from Eq. (2.8) after the bar has been divided into four portions, as shown in Fig. 2.30.

72

300m'"

~

:JJ;;-

". Fig. 2.31

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73

2.9. StaUcaJly tndeteflllJnate Problems

Substituting these values into Eq. l2.8), we obt:.tin

(2.18)

R, Ex:pressing that the total deformation 0 of the b:.tr must be zero. we write

"

(2.19)

:md. s ub~tituting for 8/. and 81'/ from (2.17) and (2.\8) into t2. 19).

.12S X 1O~

( 1.95 X J(P)Re

E

£

c

o B

",

Solving for Nil' we have

R, = 577

Fig. 2.32

X IOl N = 577 kN

The reaction R,.. at the upper support is obTained from the free-botly diagram of the bar (Fig. 2.32). We write

+t "iF, =

R" - 300 kN - 600 kN

0:

+ RII

= 0

R. . = 900 kN - R8 = 900 k.N - 577 kN = 323 kN

Once the real1ions have been determined. the STresses and strains in the bar can easily be obTained. [t should be noted that. while the t01il1 deformation orthe bar is zero. each of its component parts lioes deform under the given lo:.tding and reo straining conditions.

EXAMPLE 2.05 Determine The reactions at A :.tnd B fo r the steel bar and load· ing of Example 2.04. :.tssuming now tlwt a 4.50·mm clear:.tnce exists between the bar :.tnd the ground before the loads are ap· plied (Fig. 1.33). Assume £ = 200 GPa.

We follow the same procedure as in Example 2.1>4. Con· sidering the re:.tction at B as redundant. we compute the deformations OL and OR caused. respectively, by the given loads and by the redundant reaction R8 . However. in th is c:.tse the total deformation is nOl zero, but 8 = 4.5 mm. We wriTe Therefore (2.20)

"

I

,I = 250 'n nl~

3()0

1

A =400m "'~ -

3(1) mill

~,

~

I 4.5 ,nm FIg. 2.33

," m

I

C

B

, ,

"

T

Substituting for OL and ON from (2.17) and (2. 18) into (2.20). and recalling that £ = 200 G r a = 200 X 10~ Pa, we have

.\OOlN

c

l'HlkV

1.125 X [0 9

( 1.95 X IOJ)R8

200 X IO~

200 X 109

4.5 X JO- J III

Solving for R8 • we obtain

Ro = 115.4 X 101 N = IIS.4kN B

The re:.tction at A is obtained from the free-bQdy di;lgram of the bar (Fig. 2.32):

+t ::u', =

0:

R. . - 300 k.N - 600 k.N

+ Ro =

()

RA = 900 kN - R8 = 900 kN - [15.4 k.N = 785 kN

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74

2.10. PROBLEMS INVOLVING TEMPERATURE CHANGES

Stress and Strn.ln---AlIlal loading

All. of the members and structures tha! we have considered so fa r were assumed to remain at the same temperature while they were being loaded . We are now goi ng to consid~r various situations involving changes in temperature. Let us first consider a homogeneolls rod AB of uniform cross section, which rests freely on a smooth horizoillal surface (Fig. 2.34a). If the temperature of the rod is raised by .1 T. we observe that the rod elongates by an amount which is proportional to both the temperature change I1T and the length L of the rod (Fig. 2.34h). We have

'A

or

10) L

(2 .2 I )

8

" 1h) Fig. 2.34

" l-----j

where a is a constant characteristic of the material. called the coefficient of thermal expansirm. Since 0T and L arc buth expressed in units of length. a represents a 4uantity per degree C. or per degree F. depending whether the temperature change is expressed in degrees Celsius or in degrees Fahrenheit. With the defmmation 0T must be associated a strain fir = or/L. Recalling Eq. (2 .2 [), we conclude that (2 .22)

The strai n fEr is referred 10 as a themwl straill. si nce it is caused by- the change in temperature of the rod. In the case we are consi deri ng here. there is 110 stress (I.Hociated with the .l'Imill firLet us nt)W assume that the same rod AB of length L is placed between two lixed supports at a distance L from each other (Fig. 2.35a). Again. there is neither stress nor strain in thi s initial condition. If we rai se the temperature by 11 T. the rod cannot elongate because of the restraint.-; imposed on its ends; the elongation of the rod is thus zero. Since the rod is homogeneolLs and of unifonn cross section, the strain Er at lilly poilll is fi r = 0rlL and, thus, also zero. However, the supports wilJ exert e4ual and opposite forces P and p i on the rod after the temperature has been raised, to keep it from elongating (Fig. 2.35b). It thus follows that a state of stress (with no corrcsponding strain) is creatcd in the rod.

or

II

J-" ,,)

I

.[ B

p'

" 8

A

Ii» Fig. 2.35

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As we prepare to determine the stress (J' created by the temperature change !:J.T. we observe that the problem we have It) solve is statically indeterminate . Therefore, we should first compute the magnitude P o f the reactions at the supports from the condit ion that the elongation of the rod is zero . Using the superposition method descri bed in Sec. 2 .9. we detach the rod fro m its support B (Fig. 2.36a) and le t it elo ngate freely as it undergoes the temperature change !:J.T (Fig. 236bl. According to formul a (2 .2 1), the corresponding elongation is

2. 1". Problems Involving Temperature Changes

I

J~ (II)

., Applying now 10 end B the fo rce P representing the red undan t reaction, and recalling formula (2 .7), we o btain a second dcfonnation (Fig. 2.36c) (/} l

PL

{jp = -

AE

"

L

~

J

I.

-1"1::-

l "L p

Expressing that the total deformation {j must be zero. we have

S = ST + Dp = a (!:J. T)L

PL + -AE =

I-

0

L---J

'"

Fig. 2.36

from which we concilLde thaI

P ~ - AEa (M ') and that the stress in the rod due to the temper:l.\ure change !:J..T is p

,, ~ - ~

A

- Ea ( M )

(2 .23)

It sho uld be kept in mind that the result we have obtai ned here and our earl ier re mark regardi ng the absl!nce o f any strain in the rod apply only ill th e ("a.l·e of {/ hOlllogeneolls rod of IIm[OI1I1 ems.\" section . Any other problem involving a res trained structure undergoing a change in temperature must be analyzed on its own merits. Howevl!r, the same general appf0ach can be used ; i.e., we can consider separately the deformation due to the temperature change and the de format ion due to the redundant reaction and superpose the soluti ons obtained .

EXAMPLE 2.06 Determine the values of the stress ill portions AC and CB of the steel bar show n (Fig. 2.37) when the temperature of the bar is - 50°F. knowing that a dose fit exi sL~ at both of the rigid supports when the temperature is + 75°F. Use the values £ = 29 X 106 psi and a "" (;.5 X I06/ ~ F for .IHeel. We firsl detenn ine the reac tions at the supports. Since lhe problem is statically indetenninate. we detach the bar from its support at B and let it undergo the te mperature change

t\ = 1.2in ~

Fig. 2.37

75

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76

Stress and Straln--tuda! loading

Noting that the forces in the two ponions of the bar are PI = /'! = 18.85 kips, we obtain the following values of the

stress in portions AC and CB

l;

r~ C

"

,, ~ bT

:

,:---bN

r

C

,,'

f. r=

Fig. 2.38

liT = n{il'l)L = (6.5 X lO- bjQF)( - 125°F)(24 in.) = - 19.50 X IO- J in.

18.85 kips \, 2 in"

+ 15.7 Iksi

a!:J.T = (0.5 X 10 ('rF)( - 125°F)

L I ""

'-:!.

Al = 0.6 in l =

R8

E

29 X 10000psi

f. AC =

A z = 1.2 in"

E = 29 X 101' psi

Er

ECS

R, (12in. 12in.) = 29 X lO'psi 0.6 in! + 1.2 in!

Expressing that the total deformation of the bar must be zero as a result of the imposed cons traints. we write

The reaction at A is equal and opposi te.

!T I

Ii ""

- 8 12.S X 10

6

+ 1083.4 X 10

,

= ET

ff ,

+ £- = - 812.5

X

10

b

+ 541.7

X JO . ~

= - 271 X ]() ~ injin.

"" (.1.0345 X 10 ~ in '/lb)R8

Rs = 18.85 X 101ib = 18.85 kips

+

A similar computation yields the strain in ponion CB of the bar:

lin = AlE + AlE

from which we obtain

x 10 hin '/in.

= + 271 X IO- b in.lin.

/'1~

o=lir+on=O = - 19.50 X 10- 1 in. + (1.0345 X 10 ~i n .llb ) RR

+ 1083.4

Adding the two components o f the .;tmin in AC, we ob tain

= 12 in.

into Eq. (2.8). we write 1~I LI

The other component of f.,IC i.> associated with the S tre ~s (T I dut: to the force RII applied to the bar (Fig. 2.38c). From Hooke's la w, we e:<press this component of the strain as (T l = + 31.42 X JO'psi

Applying now the unknown force Rs at end B (Fig. 2.38d, we u~e Eq. (2.8) to express the corresponding deformation ox. Substituting

_

,=

I'

A,

+ 31.42 ksi

= - 812.5 X lO~ b inJ in.

The corresponding deformntion (Fig. 2.3Rb) is

PI = P:!

0.0 in!

We cannOL emphasize too strongly the fact Lhat. while the lolal de/ormation of the bar must be zero. the defonnations of the. portions A C and CB are IWI um. A solution of the problem based on the assumption that thes.e defonnations are zero would therefore be wrong. Neither can the values of the strain in AC or CB be assumed equal 10 zero. To amplify this poim. let us determine the strain E,le in portion AC of the. bar. The strai n EAC can be divided into two component pans; one is th e themlal strain f. r produced in the unrt:stmined bar by the temperature change !:J.T (Fig. 2.38b). From &t. (2.22) we wtile

,

I'>

18.85 kips

= -

if!

:

P

A,

= --.!. =

'"

or th e. bar:

=0

The deformations 0", and OCB of the two portions of the bar are expressed respectively as

O"C = f..ldAC) ~ ( +271 X iO b)(12 in.) = +3.25 X ]() l in.

{jCB = f. cJ.CB) = ( - 271 X 10 b)( 12 in. ) = - 3.25 X 10 j in.

°

We thu s check that. while the sum = o,tc + lics of the two deformations is zero. neither of the defonn:uions is zero.

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SAMPLE PROBLEM 2,3

12ill ' l~

c

/)

12 ill, I

I Sill

"

D,

SOLUTION

,sill

C

D

Statics. Considering the free body of bar ABeD. we note Ihal the reac· tion aJ 11 and the forces exerted by the rods are indetenninaJe. However. using statics. we may write

For

+") 2:MB ~ 0:

D"

H! ];il'~

The i· in .. diameler rod CE and the ~ .in,-diameler rod DF are anached to Ihe rigid bar ABCD as shown. Knowing that Ihe rods are made of aluminum and using E =:0 10.6 X IO~ p~i , dClermine (a) the force in each rod caused by the loading shown. (b) Ihe corresponding detleclioll of point A.

Fcr

(10 kips)(J 8 in. ) - Fn.( 12 in.) - FDF (20 in.) = 0 12Fc/,;' + 20FDF = 180

( I)

G L'OlIIctry. After application of the 10-kip load. the pOSition of the bar is A' Be' D'. From the similar triangles BAA '. BCe', and BDD' we have

~=~ 12 in.

Ddonnations, .5c

,-l

[ 2.-1 in

L

",£ F N · c

-tito.

,

F

0,\

OD

ISin.

20 in.

(3)

Using Eq. (2.7). we have

I

Dr"

Substituting for oe and OIJ into (2). we write F CELU'; An·E

3O;n

:s'"

(2)

20 in.

1

=:0

0.6 FDFLDF ADFE

cE " = 7T(~i nYl Fc£ = 0.6 LIJF A fIJI' 0.6 (30in.)[! , I J. , FDtLeE ADF 24 m. :j 11(4 111.)-

Foree in Each Rod. Substituting for Fct,; into ( I) and recalling that all forces have been expressed in kips, \\e have

12(O.333Fm·) + 20Fm - = 180 = 0.333F/JF = 0.333(7.50];ips)

PDF

DcOcctions.

,

_ _F"_,_ LD_' _ D -

= 7.50 kIps ...

Fef.· = 2.50 ki ps ...

FCE

The deflection of point D is

,-'-(7~.5~0'-cX~I~O~'~lb~)~ (3~O~;'~'.~)c

Am-I:." - ~ 7T(f in.)'( 10.6 X 106 psi)

aD

= 48.0 X 10 l in.

Using (3). we write 0.-1 = 0.90/J = 0.9(48.0 X 10 ~ J in.)

0,\ = 43.2 X 10

1

in . ...

77

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SAMPLE PROBLEM 2.4

TI

(1.3",

The rigid bar eDE is allached to a pin support at £ and resl~ 011 Ihe 30-mmdiameter brass cylinder BD . A 22-mm-diamcter sleel rod AC passes through a hole in the bar and is secured by a nUl which is snugly fined when Ihe temperature of Ihe entire assembly is 20°C. The temperature of Ihe brass cylinder is then raised to 50°C while the steel rod remains
/J

!8

lhe cylinder. Rod AC: Steel E = 200 OPa a =

lU x [0

Cy linder BD: Brass E = 1050Pa cr = 20.9 X [0 6/oC

6rc

SOLUTION 'r;:=~:::;;~?-~

Statics. E,

+1 'fM Ii

= 0:

Considering Ihe free body of the entire assembly. we \-\-T ile RIl (O.75 m) - R/"O.3 m) = 0 R", = D.4RB ( I)

Dcrormatiolls. We use ihe method of superposition. considering Rs as redundant With the support at B removed, the temperature rise of the cylinder causes point B to move down through or' The reaction RR must cause a deflection 0 , equal to OT so that the final deflection of B will be zero (Fig. 3).

Dl'.f1ectiml 8 '1' Because of a temperature rise of 50° - 20° = 300C. the length of the brass cylinder increa.~es by or. or = L (a T)o: = (0.3 m)(30°C)(:W .9 X 1U-~/°C) = 1&'\.1 X lO - ti 111.J. C

°D = lJ".3.7S or = OAIl(.

orr.. ~ ?_

c

E

-'\"----------" Deflectiol1 51 ' We note that 01); OAo c and 0 1 = Ou + OR/D' I(,L

R",(0.9 m)

AE

b r(0.022 m)2(200 G Pa)

oc"='-= of) = OAMe

o 8/J)

= OA( 11.84 X 10 ' RA ) = 4.74 X 1O- 9R",1

= Rei- = Rn(0.3m) = 404 X 10 AE ! 7T(O.03mf( I05GPa) ·

We recall from (I) that R,\

0, = Ou But or = 0 ,:

+ OBID =

~R t s

= OAR s and write

[4.74(OAR8J

+ 4.04R8 j 1W 9

188. 1 X lO- b m =5.94 X IO

Stress ill Cylillder:

RB

if 8

78

Q

11.84 X IO- R,jt

= 5.94 X IO- 9R8

t

9 RH

31.7kN

=-=--A t7T(0.03Y

if B

= 44.8 MJ)a .....

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PROBLEMS

2.33 A 2SD-mm bar of [5 X 30~mm rectangular cross section consists of two aluminum layers. S-mm thick. brazed [0 a center brass layer of the same thickness. If it is subjected to centric forces of m3gnitude P = 30 kN. and knowing that Ea = 70 GPa and E~ = 105 GPa. determine the normal stress (a ) in the aluminum layers. (b) in Ihe brass layer.

p'

Aluminum B rass

p

A)"",h",,,,

Fig . P2.33

2.34 Determine Ihe deformmion orlhe composite bar of Prob. 2.33 if it is subjected to centric forces of magnitude P = 45 kN. 2 .35 Compressive centric forces of 40 kips are appljed at both ends of the assembly shown by mean~ of rigid pi ales. Kn owi ng that E, = 29 X 106 psi and E., = [0.1 X IO~ psi. determine (a ) the normal stresses in \lIt' steel core :U1d the i1luminum sheiL (b) the uefomlation of the assembly.

I in .

•\I.""j""", .,],,,11 2.5 ;"

Fig. P2.35

2.36 The length of Ihe assembly decreases by 0.006 in. when an a:
Fig. P2.36

79

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80

2.37 The 1.5~m concrete post is reinforced with six steel bars. each with a l8-mm diameter. Knowing that = 200 G Pa and E~ = 25 GPa. determine the normal stresses in the steel and in lh ~ concrete when a 1550-kN axial centric force P is applied \0 the post.

Stress and Straln----Axlal loading

r

1.5 ro,

t:

,~50 "''''

2.38 For Ihe post of Prob. 2.37. delemline the maximum centric force which can be applied if the allO\\lable normal stress is 160 MPa in Ihe steel and 18 MPa in the concrete. 2.39 Three sicel rods (£ = 200 GPa) support a 36-kN load P. Each of th e rods AB and CD has a 200-mm~ cross-sectional area and rod EF has a 625mm 1 cross-sectional area. Neglecting the deformation of rod 8ED. determine (u) the change in length o f rod EF. (b) the stress in c;Jeh rod.

2.40 Three wires are used to suspend the plate shown. Aluminum wires are used at A and B with a diameter of in. and a steel wire is used at C with a d iameter of ~ in. Kno wing that the allowable stress for a luminum (£ = 10.4 x [0 6 psi) is [4 ksi and that the allowable stress for steel (£ = 29 X 106 psi) is 18 ks i. determine the mnimum load I' that can be applied .

!

Fig. P2.37

0

"

,

2.41 Two cylindrical rods. one of st e~ [ and the other of brass. are joinet.l at C and restrained by rigid suppons at A and E. For the loading shown and knowing thaT E, = 200 GPa and Eb = [05 GPa. deTenn ine (a) the reactions at A and E, (b) lhe denection of point C.

cr

D;"" "";Ol1<;n """

~IM.-rt 2(1~I-""'-1 .\ I cl D E

.'i(X) ''''''

L E

"-+

-100 ,,,m

, Fig. P2.39

liUk~

o F -I().mm db", .

Fig. P2.40

-

lOlN

30·",,,, dia'"

Fig. P2.41

2.42 Solve Prob. 2.41. assuming thm rod AC is made of brass :md rod C£ is made of steel. 2.43 A steel tube (E = 29 X lOb psi) with a l±.in. outer diameter and a !.in. thickness is placed in a vise that is adjustOO so that its jaws jusltouch the ends of the tube without exening any pressure on them. The two forces shown are then applied to the tube. After these forces are applied. the vise is adjusTed to decrease the distance between its jaws by 0.008 in. Dctennine (til the forces exerted by the vise on the tube atA and D. (b) the change in length of the portion Be of the tube.

Fig. P2.43

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2.44 Solve "lOb. 2.43. a~>umillg lhat after the fOJce~ Ilan: been
Pmblema

2 . 45 Link s BC atld DE are bolh made ofsleel ( E = 2') X IU~ p,i) and are ~ in. wid ~ and 1 in. thi ck. Det~rmine (a ) the force in each link when a 6QO·lb

force P IS applieJ to the rigitl m~mbt":r Af-" ~oWn. lb) lhe correspondi ng den~c. lion of poim A. A ..j

c

B 0

0

I'u. •

_I I

,

;2 i"

f 0

~' 4

ill,--I~-

A

B

C

D

•

L

, in.

Fig. P2.4S

2.46 The rigid bar ABeD is susf)Cnded from four identical wires. Delermine Ihe tension in each wire caused by the load " shown.

Fig. P2.46

2.47 The aluminum shell is fully bonded to the brass core and the 3$sembly 1S unstressed a1 a tempml1ure of I SoC. Considering only a!l:.ial deforIllations. determine the ~tress in the aluminum when the lemperature reaches

9_"> "",, 1I",s., torr E ~ t03CP.

c

I 95 C.

" ~ .!20Q X 10-6/'C

Sohe Prob. 2.47. assuming Ihal the cort' i. made of ~ Ieel (£ = 200 GPa. a = 11.7 X 10 ~t'C) instead of brass.

2 .48

2 .49 A 4-fl concrele post is reinforce d by four slce] bar., each of ;-in. diamelcr. Kno\\ ing thaI E, = 29 X 10" psi. (I, = 6.5 x 10 ',oF and E, = .U'i x IO~ p.1 1 and a t" = :'i.:'i X JO ~," F. delennine Ihe norm~1 slresses induCC
Abun;""", ,]\,,\1 ~ = ... =

70(; Pu Zl.6 X IO..fiI'C

Fig. P2.41

nlln

St",·I"m· 1'. = 2f)() CPa

liT""" .I."I! l U'iCb

F. ~

Rg. P2.49

=

2.50 The brass shell (a b 20.9 x 10 6rC ) is fully bonded 10 the sl~e l oore (" ; - 1 1.7 x 1O- 6I" F) DetenniLtt: the largest allowable itlCte
Fig. P2.50

81

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82

Stress and Straln---Allial loading

2 .51 A rod consisting of two cylindrical portions All and Be is reslrained at bolh ends. Ponion AB is made of steel (E. = 200 GPa, IT" = 11.7 X IW bf'C) and portion Be is made of brass (Eb = 105 GPa. 0 /, = 20.9 X 10 "fC). Knowing that the rod is initially unstressed. determine the comp'ressive force induced in ABC when there is a temperature ri.<;e of 50 C. Q

5-J_n"" dillm"'~r

300",,,,

L Fig. P2.51

2.52 A rod consis-ting of two cy li ndrical portions- AB and BC is restrained at both ends. Portion All is made of brass (Eb = 15 X 10" psi. (I I> = 11.6 X 10 brF) and portion IlC is made of steel (E, = 29 X 10~ psi. a , = 6.5 X IO - ~r F). Knowing that the rod is initially unstressed. determine (a) lhe normal stresses induced in portions All and fJC by a temperature rise of 90°F. (b) the corresponding deflection of point B.

,,'

i~

2.53 Soh'e Pmb. 2.52. as.~uming Ihm portion ,\8 of the composite rod made of steel and ponion BC is made of bmss.

2.54 A steel railroad track (E, = 29 :>( 10~ psi. a , = 6.5 laid out at a temperature of 30°F. Detennine the nomlal stress the temperatore reaches 125°F, assuming that the rails (a) are a cominoous track. (b ) are 39 ft long with (-in . gaps hetween Fig. P2.52

X IO ' ·/°F) was in the rail when welded to fonn them.

2 .5 5 Two steel bars (E, = 200 G Pa and a , = 11.7 X 10 brC) are used to reinforce a brass bar (£b = 105 GPa, a t "" 20.9 X 10 6rC) which is subjected to :l load P = 25 kN. When the stcel bars were Inbricated. the distan(;e between the centers orthe holes which were to fit on the pins wns made 0.5 mm smaller than the 2 m needed. The steel bars were then pJnced in :In ollen to increase their length so th:lt they wou ld just lit on the pins. Following fnbrication. the temperature in the steel bars dropped bnck to fOam temperalllre. Determine (a) the increase in temperalllre that was required to tit the steel b:lrs on the pins. (b) the stress in the hmss bar after the 10:ld is applied to it.

Fig. P2.55

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2 .56 DeTermine The maximum 100Id f> thaI may he applied to the brass bar of !'rob. 2.5S if the allowable stress in the sTeel bars is 30 M Pa and lhe allowable stress in the brass bar is 2S MPa.

2.57 A bra ~s link (Eh ~ 105 G Pa. (i" = 20.9 X 10 bf'C) and a stee l rod (£. = 200 G Pa, 0,. = 11.7 X 10 6f'C) have the dimensio ns shown at a temper the tinal length of the steel rod. Q

50"""

1-:.1 ,37'"""

-j

0. t2

0

~37.5mm

'IlL"--l ~I mm--J

30·1"", dialll('It'r

~',·.,e~'''''Li''i

~

"

Fig. P2.57

2 .58 Knowing that a O.Ol-in. gap exists when the temperature is 7S"F, detennine (a) the temperature at which the normal stres.~ in the aluminum bilT will be equal TO - I I hi, (b) the corresponding exact length of the aillminllm bar.

- -+- - 18;".--'"1

Bmn7.1'

AI"",in"m

,\ = 2.4 in.~ E = ].~ x t o"~~i

A= 2..Si" .~ E= 10.6 x l rt~)i

u

u

= 12 x to- r F

=

t2.\l x ]()- ;OF

Rg. P2.S8 and P2.S9 2 .59 Determine (a) ille compressive force in the bars shown after a temperature rise of 180° F, (b) th e corresponding change in length of the bronze bar.

2.60

At room temperature (10°C) a D.S-mm gap exisTs between the ends rods shown. At a later time when the Temperature has reached 140"C. determine (a) the normal stress in lhe aluminum rod. (b) the change in leng th of lhe aluminum Iud.

or the

AI",,,in,,,,,

St.,i"lr~.\

sit,,·1

A = :!lIU J",,,,>!

,\=~'n", i

E = i5 CI'ot

E

=

19() Gt'~

u

=

17.3 x

0:

=23 x tW"rc

Fig. P2.60

!O ~~fC

Problems

83

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84

Stress and Straln--Allial Loadln9

2.11 . POISSON 'S RATIO We saw in Ihe earlier pan o f this chapter that, when a homogeneous slender bar is axially loaded, the resulting stress and strain satisfy Hooke's law, as lo ng as the elastic limit uf the material is nOI exceeded . Assuming thai the load P is directed alo ng the x axis (Fig. 2.39a), we have u .\. = PIA , where A is the cross-sectional area of the bar. and. from Hooke's law. (2 .24) where E is the modu lus of elasticity of the material. We also note that the nonnal stresses on face s respectively perpendicular 10 the y and z axes arc zero: u ,. = u~ 0: 0 (Fig. 2 .39h). It would be tempting to conclude that the COiTcsponding strains E,. and E: arc al so zero. T his, however. is not the m.w'. In all engineering materials, the dongation produced by an ax.ial tensile force P in the din::ction of the force is accompanied by a contraction in any transverse direction (Fig. 2.40).t In this sect ion and tJle follow ing seclio ns (Sees. 2. 12 through 2. 15), all materials considered will be assumed to be both homogeneOil.l· amI i.\"()/mpic. i.e., their mechanical properties will be assumed independent of both position lIIld direction. II follows that the strain must have the same value for any transverse direction. Therefore. t'or the loading shown in Fi g. 2.39 we must have E,. = € o. This common value is referred to as the lateml stmil/ . An important constant for a given material is its POi.HOII'S ratio, named aft er the Fre nch mathe-

I"i

r

,,_ ~ O

J

",

/'

•

(iI)

Fig. 2.39

mmician Simcon n ellis P oisson ( 17R 1- 1R4() ami de noted hy The Greek

leiter

l'

(nu). It is defined as

" lateral strain axial strain

(2.25)

p

(2 .26)

Fig. 2.40

for the loading condition represented in Fig. 2.39. Note the usc of il minus sign in the above eq umio ns to obtain a positive value fo r v. Ihe axial and lateral strains havi ng opposite signs for all engi neering materiil ls.t Sol ving Eq. (2.26) for ",. and '"' :;' and recalling (2 .24), we write the following relmions. which full y describe the cond ition of strain under il n axiill 10ild applied in il direction parallel to the x axis:

E

(2 .27)

t tl WQuid also !)I' TempTing. bUT equally wrong. TO assume ThaT lhe volum~ Qf The rod re· mains utlChanged a.1 a resulT of lhe combined dft1'T of The a.xial elol1galioll amJ Transwrsc contracTion (SIT. $cc. 2.13). ~ li Qwever. some experimenTal materials. such as polymer foams. expand laTerally when strelehed. Since The axial and laleral STrains have dl~n lhe .Iame sign. the Poisson's raTio of lhese maTerials i.1 nCJ;JTivc . (Sa Rodcrie Lakes. "'Foam STructUf!' ~ WiTh a NegaTi\"~ Poisson 'S RaTio." Sd,,"u. 27 February 1987. Volume 135. pro 1031\- 1040. )

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EXAMPLE 2.G7 A SOO-mm-long. lfi-mm-diameter rod made o f a homogenous, Isotropic material is observed to increase in length by 300 11m. and to decrease in diameter by 2.4 11m when s ubjected to an axial 12-kN load. Determine the modulus of elasticity and Poisson 's ratio of the material.

The cross-sectional area of the rod is A =

7Tr }=

7T(8 x 10 ] m)!=201

P

A=

300 11m

L "" 500mm ~,

f

•

=600 X 10 -~'

- 2AJLm 16mm

J

(T_,

E= ~ =

12"-"'---

E,

fl'"

59.7 MPa

= - - - = - IS0 x 10

= -

From Hooke's law.

,

12 X tO J N

20 1 X to - ~ m2

~,

f , ""

d= Hi",,,,

b m!

10

Choosing the x axis along the axis of the rod (Fig. 2.41). we write fT, =

by = - 2--1

X

b

= Et':, . we obtain

59.7 MPa 600 X 10 ' ~

99.5 G Pa

and. from Eq. (2.26).

Fig.2A1

I'-'='

"

"

- 150 X 10 6 600 X 10- 6

0.25

2.12. MUlTIAXIAllOADING; GENERALIZED HOOKE'S LAW All the examples considered so far in thi s c hapler have dealt with slendcr mcmbers subjected to axial loads. i.e., to force s directed along a si ngle axis. Choosing Ihis axis as lhe x axis. and denoting by P the intc.rnal force at a given location. the corresponding stress components were found 10 be a~. = PIA. rr" = O. and a~ = O. Let us now consider structural clements subj(:cted 10 loads acting in the dircctions of the three coordinate axes and produci ng normal stresses a 1 • a,~ and a ~ which are all different from zero (Fig. 2.42). This condition is referred 10 as a IIIlIlfiaxi{/IIOl/(li,,~. Note that this is nottne general stre..~s condition descri bed in Sec. 1.12. since no shearing ~ Iresses are included among Ihe streSses shown in Fig . 2.42.

u.

u,

Fig. 2.42

85

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86

Consider an element of an isotropic material in the shape of a cube (Fig. 2.43a). We can assume the side of the cube to be equal to unity. si nce it is always possible to select the side of the cube as a unit or length. Under the given mu ltiaxial loading, the eleme nt will defoml into a luwngillar parallelepiped oj sides equal, respectively, to [ + E..., [ + E,.• and [ + Eo. where E,. Ii )•• and E, de note the val ues of the normal strain in the direc ti ons of the three courdinate axes (Fi g. 2 .43h). You should note that, as a result or the defomlations of the other elements of the material, the element under consideration could also undergo a translation. but we are concerned here onl y with the actual dejo/1/wtion of the element. and not with any possible superimposed rigid-hody displacement. In urder to express the strain components E~. (0,., fO, in terms of the stress compone nts 0'" 0'" 0'0' we will consi der separately the effect of each stress component and combine the res ults obtained. The approach we propose here will be used repeatedly in th is tex t, and is based on the principle of .l"UpeI7)()sitioll . This principle states that the effect of a given combined loading on a stfuc(Ure can be obtained by detenllinillg

Stress and Straln---Allial loading

(0 )

.\·epamrely the effects of the various load.l· and comhining th e reslllts ob· tllined. provided that the following conditions are satisfied :

".

(b l Fig. 2.43

I. Each effect is linearly related to the load that produces it. 2. The deformation resulting from any given load is small and dOeS not affect the conditions of application 01' the other loads.

In the case or a multiax.ial loading, the lirst condition will be satisfied if the stresses do not excL'"Cd the proportional limit of the material, and the second condition wi ll also be sati slied if the stress on any given face does not cause derormations of the other race.~ that are large enough to affect (he computation of the stresses on those faces. Consideri ng [irst the effect of the stress component O'~. we recall from Sec. 2.1 I that (T.e causes a strain equal to 0'. /E ill the x direction. and strains equal to - va .J E in each of the y and;;; directions. Similarly, the stress componcnt 0',.. if applied separately. will cause a strain ajE in the .r direction and strains - J)(TjE in the other two directions. Finally, the stress component 0', cause~ -a strain ujE in the:: direction and strai ns - VO' j E in the x and y direc tions. Combining the results obtained. wc conclude that the compone nL~ or strain corresponding to the given multiaxia! loading are

VU e

E,

= - -'

E

a·, va . +-' E E

(2 .28)

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The relation~ (2.28) are referred to as the genemfjzed Hooke·.r lall' for the IIwftillxinllooding of {/ IwmogeneollJ isotropic maferial. As we indicated earlier, the results obtained are valid on ly as long as the stresses do not exceed the pro portional li mit, and as long as the deformatio ns involved remain small. We also recall that a positive value for a stress component signifies tension. and a negative value compression. Similarly. a positive val ue fur a strain component indicates expansion in the corresponding direct ion. and a negative val ue contraction.

87

2.13. Dilatatlon: Bulk Modulus

EXAMPLE 2.08 The s teel block shown (Fig. 2.44) is subjecled 10 a uniform pressure on all il~ races. Knowing thaI the change in length of edge AB is - 1.2 X 10 ' in .. delermine (a) Ihe change in lenglh of the other IWO edges, (b) Ihe pressure p applied 10 Ihe races of the block. A.~s ume £ = 29 X 106 psi and" = 0.29.

(a) Change In Length of Other Edges, Substituling a- , = (T . = if 0 = - p inlo Ihe rei
(2.29) Since E, =

o,/AB

= - ) 00

= (- 1.2 X 10

x 10

6

1

in .)/(4 in.)

in.l in ,

we oblain 1'.,

= t . = t . = - 300 X 10 - 6 in.lin.

from whic h il follows Ihat

,s,. = dBC)

= ( - 300 X 10 6)(2 in.) = - 600 X 10 ~ in . 0, = £,(BD ) = ( - 300 X 10- 6)(3 in.) = - 900 X 10- 6 in.

(b) Pressure. Solving Eq, (2.29) for p. we wri te

Fig. 2.44

E€ , [I

= - I - 2" ""

(29 X 1 0~ ps i)( - 300 X JO -6) I - 0.58 p = 20.7 ksi

'2.13. DILATAllON; BULK MODULUS In this section you will examine the effect of the normal stresses u ,"

U p

and (T c on the vQlume of an cleme ltt of isotropic material. Consider the element shown in Fi g. 2.43. In its unstressed state, it is in the shape of a cube of unit volu me: and under Ihe stresses u x, u ," (T " it deforms into a rectangular parallelepiped of volume

1.1 =( 1 + £,,)(1 + t:»)( 1 + (,J Si nce the straill5 E." E)", Eo are much smaller than unity, their products will be even smaller and may be om itted in the expansion of the product. We have, therefore,

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88

Stress and Straln--Alda! loading

De noting by e the change in volume of our clement. we write e = v -- I = I

+ E-, + € " +

E: -

1

0'

(2 .30)

Sillce the ele ment had originally a unit volume. the quantity e represents Ih e chan ge in vv /ullle per unil vO/lIl1le; it is referred h.) as the di/lIIlIt ion of the material. Substituting for Ex, E,., and Eo from E(ls. (2 .28) into (2 .30), we write e =

E

I - 21' ' ~ -E- (u.t

+ O' r + u J

(2.3 [ )t

A casc of special interest is that of a body ~ lIbjected to a uniform hydrostati c pressure p. Each of the -"tress components is then equal to - p and Eq. (2.3 1) yields 3( [ - 2v ) E

P

(2 .32)

Lntroducing the constal1l E

, ~ -"---~

(2.33)

3(1 - 2p)

we write Eq . (2.32) in the lo on

e

=_ek

(2 .34)

The constant k is know n as the hulk fl/odullU or modulus of compl1!~'­ sio/l of the material. It is expressed in the same unit s as the modulus uf elasticity E, that is, in pascals or in psi. Observati on and common se llse indicate that a stable material subjected to a hydrostatic pressure can only d ecI"l'llse in volume: thus the dilrttalion e in Eq. (2.34) is negative, from which it fo llows that the bulk modulu ~ k is a pos iti ve quantity. Referring to Eq . (2.33), we conclude that I - 21' > 0, or 11 < ~ . On the other hand, we rec,ill from Sec. 2.11 that \1 is pos iti ve for all engineering materials. We thus conclude that. fo r any engineering material ,

o < I' < 4

(2. 35)

We note that an ideal material ha ving a vallie of u eq ual to zero could be stretched in one directio n without any lateral contraction. On the and thus k = {Xl, would other hand, an ideal material for which I' =

t,

t Si nce th~ dilatation ~ reprcscnls a change in volume. it nllL~t be independent of the Ofl entation of the clemen! considered. tt then foltows from Eqs, (2.30) Jnd (2.3 t ) that the quall tit ie~~, + E) - ~, and IT. + iT , + (7_- ar~ atso independent of the orientation of the dement. Thi~ propeny wilt be v~rified in Chap. 7.

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be perfe<:tl y incompress ible (e = 0). Referring to Eq. (2 .3 1) we also note that, since v < ! in the elastic range, .~tret c h i ll g an engineering lllillerial in one directi on. for example. in ihe x direction (O"~ > 0.0",. = (J"~ = 0). will res ult in an increase of its volume (e> O).t

2.14 . Shearing Strain

89

EXAMPLE 2.09 Determine the changc in vo lumc j. I' of the sIcel block shown in ['ig. 2.44. whell it is subjccted to the hydrostatic pressure " = 180 MPa. Usc E = :mo GPa and I' = 0 ..29.

Si nce the m lume V of thc bloc k in ils unstresscd stair is

v=

(80 mm )( 40 mm )(60 mm) = 192 X 10-' mm"

From Eq. (2. 33). wc delcnnine the bu lk modu lus of SH.""el.

E

k =

200 GPa

3-(' - - 2-v) = "3(;;"~O'O58")

:uld since e 158.7 CPa

(Iml. fmm Eq. (2. 341. the dilatatio n,

e =

I'

k

180 MPa 158.7 G P(I

rcprcs~nt s

thc c hange in vo lumc J:K:f ull it vo lumc,

e = .6.V! V. wc haw .6. V = e V ""( - 1.134 X 10 --')( 192 X IO' mm) j

- 1.l34 X 10-]

2.14. SHEARING STRAIN

When we derived ill Sec. 2. 12 the relations (2.28) lx::t ween nonnal stresses and nonnal strains in a homogent.."o us isotropi(; material, we assumed that no shearing stresses were involved. In the more general stress situati on represented in Fig. 2.45. shearing stresses T,xp T ,." and T~, will be present (as well. of course. as the corresponding shearing stresses T,.f' T,..' , and 'I".
FIg. 2..45 t l1owevCf. ill the ptastic range. 'he volu me of the material remains nearty COIl51am.

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90

Stress and Straln---Allial loading

Fig. 2.46

T""

Consider first a c ubic clement o f side one (Fig. 2.46) subjected 10 no other stresses than the shearing strei;SeS T..,. and T " applied 10 faces of the element respectively perpendicu lar 10 the x and y aXes. (We recall from Sec . 1.1 2 that T H = T,ot .) The element is observed to de fomt into a rhomboid of sides equal to one (Fig. 2.47). Two o f the angles formed by the lo ur faces under stress are reduced fro m f \l) ~ - Y.{" while the o ther two are increased from 1 to f + Y H , The small angle 1'n (expressed in radi ans) defi nes the s f/carillg .I·train corresponding 10 the x and " directions. When the deformation involves a reduction or the angle formed by the two faces oriemed respectively toward the positive x and y axcs (as shown in Fig. 2.41), the she:uing strai n I'Xl is said \() be positive: otherwise, it is said to bc negative. We should note that, as a result of the defoml,Hions o f the othcr elements of the material. the element under consideration can also undergo an overall rotation . However, as was the case in ()ur study of normal strai ns, we are concerned here only with the {Iell/al deformation ()f the element , and nOI wi th any pt;)~sib l e superimposed ri gid-body displ aeement.t Ploning successive values of T~,. against the corresponding values of y"., we oblain the shearing stress-strain d iagram for the material under conside ration. This can be accomplished by carrying OUI a torsion test, as you will see in Chap. 3. The diagram obtained is similar \() the normal stress-strain diagram obtained for the same material from the tensile test described earlier in this chapter. However, the values obtained fo r the yield strength. ultimale strength, etc .. of a given material are only about half as large in shear as they are in tension. As was the case for normal stresses and strains, the initial porti on of the shearing stress-strain diagram is a strai ght line. For values of the shearing stress

Fig. 2.47 tin defining thl' stra in -Y.,! some authQrs arbitr:lrily as.\ume thattoc actual ddO!1nation of the ckmclll is accompanied by a rigid·body rotation such that the horizontal facr~ of the cl emClll do not mtJle. The strJin ")'. " is then reprcselU~d by the angle through which the other two fuces have rOlatl'd (Fig. 2.4S). Others a.\SUlllC a rigid ·body rolJlion such thaI the hori zontal faces rotatc through ty., ~ount crclockwise lnd the v~nieal faces throu):h h' ~ .lockwise (Fi g. 1.49). Since both assumptions arc unncct/isary and may lead lQ confusion. we prefh in this tcxtto as,(Il'iutc the shearin): strJin ")'" with the ,·hlm..:" in th~ a"gl~ formed by the tWO faces. rathtr than with the of'' Ki,'"'' )lIce under restricti\"\: condition~.

,lit"''''''

Fi g. 2.48

Fig. 2.49

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2 14. Shearing Strain

that do not exceed thl! proportional limit in shear, wecan therefore write for any homogeneous isotropic material. (2.36)

This relation is known as Hooke's I{lW for ~'he([rillg ,I'tre.u ami strain, and the constant G is called the modllius of rigidity or shear modulll.l· of the material. Since the strain y ", was dclined as an angle in radians, it is dimensionless, and the modulus G is expressed in the same units as 7 ,n' that is, in pascals or in psi. The modulus of rigidity G of any given material is less than one-half. but more than o ne-third of the modulus of elasticity E of that material. t Considering now a small dement of material subjected to shearing stresses 7 ...;: and 7 :" (Fig. 2.50{/). we deline the shearing strain Y ,~ as the change in the angle fomled by the faces under stress. The sheari ng strain Ye.< is defined in a simi lar way by consideri ng an element subjected to shearing stresses T :or and Tx:: (Fig. 2.5011). For values of the stress that do not cxceed the proportional limit , we CHll write the two additional relations

Y:,

,., ) "

(2.37)

where the constant G is the same as in Eq. (2.36). For the general stress condition represented in Fi g. 2.45. and as lo ng as nonc of the stresses involvcd exceeds the corresponding proportional limit, we can apply the principle of superposition and combinc the results obtailled in this section and in Sec. 2. 12. We obtain the followin g group of equations representing the generali zed Hooke's law for a homogeneo us isotropic material under the most general stress condition.

1' 1 Fig, 2.50

f ,

=

a,.

va)"

E

E

Ila ..

u ,'

IIU O

E

E

E

+- - -

vu;;:

- -

E

f'. . = - - + - - \

\2.38)

An exmninmion of Eqs. (2.38) might lead us to believe that three distinct constants, E, i l. and G. must first be determined experimentally. if we are to predict the deformations caused in H gi ven materiHI by an arbitrdry combination of stresses. Actually. onl y two of these const:mts need be detennined experimel11ully for any given material. As you will see in the next s~ction, the third constant can then be obtained through a very simple computation. t Ser Prob. l .9t ,

"

91

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EXAMPLE 2.10 A rectangular block o f ~ mmerial with a modulus of rigidity G = 00 hi is bollded 10 two rigid horizontal plates. The lower plaIt: is li xt:u. whill! lhl! lIppt:r pl~ll! is subjl!(;lt:d lu ~ huri,,:ullt~1 forcl'. P (Fie. 2 . ~1) . Knn·~ine thar till'. 1I)1~r )11:111'. mO\'t\~ dlrough 0.04 in. under the 3clion of the force. detemline (u) the average shearing strain in the rnmerial. (b) the force P exerted on rhe uppe r plale.

(a) Shearing Strain. We sel~ct coordinate axes centered at the midpoint C of edge AB and direcled as shown (Fig. 2 .51). AccordIng to its definition. the shearing s tram )',~ is equal to the angle fomled by the lIenic;!1 and the line CF joining Ihe midpoints of e(]ges AB and DE. NOIing thai thi s is a very s mall angle: and recalling ~\at it shoulJ Ix: ex pl e:s~(] ill r.nlialls. we write 0 .04 in. )'~.

- tan ),,,.

2 in.

)'''1=

, Fig . 2.5 1

0.020 rad

(b) Force Exerted on Upper Plate, We first determine the ~hearing s tre ss T o. in the maleri ~ 1. Using H QOke 's Jaw for shearing , tress aud strain, we have

OJH ill.

T~, = G),,,. = (90 X 10\ p5i)(0.020 md ) = 18oo p5;

The force e xutetl on the upper pbte

i~

thu S"

{' - r .. A - ( 11100 p~ i )( 8 in.)( 2.5 in.) - 36.0 X 10' Ib P - % .0 kip,

Fig . 2. 52

2_H'_ FURTHEFI DISCUS!=i;ION OF DEFOFlMATIONS UNDE.R

AXIAL LOADING ; RELATION ANONG E, II , AND G

p'

101

We saw i n Sec . 2. [ [ that 11 slender bar subjecled 10 an axial tensile load I' directed along thex axis will elongate in the x direction and contract in bOlh of the tra nSllerse y and ~ di rections. If lOx denotes the axial strai n. thc late ral strain is expressed as E, = E_ = - VEX' whe re V is Poisson's ratio. Th us . ~n clement in Ihe shaPe of; cube of s ide equal to onc and one nted as shown in Fig. 2.53a will defonn itlto a rectangular paral· lelepiped of sides I + E,. I - VE" :lnd [ - "E.,. (NOlI' that only one face of the e lement is shown in the ligurc.) On Ihe other hand, if the dt;H1elll is oJielllc d al 4S n 10 Ille 3.{is uf tlte load (Fig. 2.53b), lite face s hown in IhL· fig ure is observed to deform in to a rttornbu ~ . We concl ude Ihat the lL'~ia l load P causes i n this element a shearing strain equal to IhcalllOUnl by which eac h o f the angles shown in Fig. 2.53b increases or decreases.'

"I

(/.) Fi g. 2.53

92

' Note thaI the load P allO produces nonn sl str.:l.lIS in the clctf.cnt shoWll in Fig. !.5Jb i set !>rob. 2.74).

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The fact thac shearing strains, as wet! as nonnal strains, rl!su lt from an axial loading should not come to us as a surpri se, since we already observed at the end of Sec. 1.1 2 that an ax.ial load P causes nomlal and shearing stresses of equal magnitude on four of the faces of an eleme nt orie nted at 4SO 10 the axis ()f the member. This was illustrated in Fi g. 1.40, which. for convenie nce, has been repeated here. It was also shown in Sec. 1.11 thalthc shearing stress is maximum on a plane forming an angle ()f 4SO with the ax.is of the load. It follows from Hooke's law for shearing stress and strain that the shearing strain y' associated with thc clement of Fig. 2:.53h is also maximum : y ' = "Ym' While a more detailed study of the transfomlations of strain will be postponed until Chap. 7, we will derive in thi s section a rdation betwecn lhe maximum shearing strain y ' = Y II , assoc iated with the clement of Fig. 2.53h and the nonnal strain 10.\ in the direction of the load. Let us consider for this purpose the prismatic element obtained by intersecting the cubic clement of Fi g. 2.53a by a diagonal plane ~Fig. 2.54a lmu b). Referring to Fig. 2.53a, we conclude Ihat this new element wi ll deform into the clement shown in Fi g. 2.54c, which has horizontal and vertical sides respectively equal to I + lOx and I - 11(, . But the angle fonned by the obliljue and horizolltal faces of the element of Fig. 2.541J is precisely half ()f o nc of the right angles of the cubic elc-

101 Fig. 2.54

men! considered in Fig. 2.53b. The angle f3 into which this angle deforms must therefore be equal 10 half of tr/2 - "YIII' We write rr 4

Y. 2

~ ~ - - -

Applying the formu la for the tangent of the difiercnce o f two angles. we oblai n

tan ~ - tan "Ym 4 2

tan {3 = - - - ' - - - " + tan'!!.-tan "Ym

4

2

- tan r ill 2 Y,.

+ tan 2

2.15. Further Discussion or Derorma1!ons Under Axial Loading

"I

./

p O'

"~ .1,. '0 )

q'

p' -

p T,""".il

Fig. 1.40 (repeated)

P

93

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94

Stress and Slraln---Alllal loading

or, since Y/llI2 is a very small angle,

>m 2

tanf3 = - - I + "}'/II 2

(2.39)

But , from Fi g. 2.54c, we observe that

tan f3 =

1 - V€ ~~~ , I

+ E ..

(2.40)

Elluating the right-hand members o f (2.39) and (2.40) . and solving for 1m> we write

(1 + v)lt:"

1m =

I - v

+ -2.- , ., Since Ex « I, the denominator in the express ion obtained can be assumed equal to one; we ha ve, therefore, (2.4 1) which is the desired relation between the m:uilllu l11 shearing strai n y ", and the axial strai n Ex ' To obtai n a relation among the constants E, v, and G, we recall that, by Hooke's law. "YnJ = TmIG, and that, for an axial load ing. € J = a- J E. Equatio n (2 .4 1) can therefore be wri uen as T ", (T., ~ ~( I + v )­

G

E

E

- ~( I

G

~.

+ ,,)-

Tm

(2.42)

We now recall fro m Fig . 1.40 that (T , = PI A and T ", = PI 2A where A is the cross-sectional area of the member. It thus follows that (T"I T,. = 2. Substituting this value into (2.42) and dividing both members by 2. we obtain the re lation E

2G

I+v

(2 .4 3)

which can be usC
G

~ ""2(;-;I.":+- v-'-)

(2.43 ' 1

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'2.16. STRESS-STRAIN RELATIONSH IPS FOR FIBER-

2.16. Flber-relnfoJl:ed Composite Mate-rials

REINFORCED COMPOS ITE MATERIALS

Fiber-reinforced composite materi als were brielly discussed in Sec. 2.5. It was shown at that time that these materials arc obtained by embedding tillers of a strong, stiff material into a weaker, softer matcrial, rctcued to as a //latrix. It was also shown that the relationship between thc normal stress and the eorresponding normal strain creatcd in a lamina, or layer. of a composite material depends upon the direction in which the load is applied. Different moduli of elasticity, E~, £ " and Ec. arc therefore required to describe thc relationship between nomlal stress and normal strain. accord ing to whether the load is applied in a direction parallel to the fibers, in a direction perpendicular to the layer, or in a transverse direction. Let us consider agai n the layer of composite materiaJ discussed in Sec. 2.5 and let us subject it to a uniaxial tensile load parallel to its tillers, i.e., in the x direction (Fig. 2.551/). To simplify our analysis, it will be assumed that the properties of the fibers and of the matrix have been combined, or "smeared," into a fic titious equivalent homogeneolls

(/"

Fig. 2.55

material possessing these combined properties, We now consider 11 small element of that layer of smeared material (Fig. 2.55h). We denote by a x the corresponding normal stress and observe that 0- ,. = o-c = O. As indicnted earlier in Sec . 2.5, the corresponding nonnal strain in the x di rection is €., = o-)E" wherc E,. is the modulus of elasticity of the composite material in the x dire,tion. As we saw for isotropic materials. the elongation 01' the. material in the.r direction is accompanied by contractions in the y and:; directions. These contmctions depend upon the placement of the fibers in the matrix and will generally be different. It follows that the lateral strains E, and € o will also be diffe rent. and so will the corresponding Poisson's ratios:

"". = - -"

"

(2.44)

and

"

Note that the fi rst subscript in each of the Poi sson's ratios 1/.') and I/x; in Eqs. (2 .44) refers to the direction of the load. and the second to the direction of the :::ontraction. It fo llows frt)1l1 the nbove that. in the case of the IIIII/ritixitll/otlding of a layer of a composite materiaL equations similar to Eqs. (2.28) of Sec. 2. 12 can bt! used to describe the stress-strain relationship. In the

95

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96

Stress and S!raln---Allial loading

present case, however, three different valut:s of the modulus of elasticity lmd six different values of Poisson's ratio will be involved . We wri te lI.xUy " ,",if~ = -IJ~ - -- -

E ;0:

Ex

£~

E,.

" o.,u, - -- + - - -

"." CT,.

aT

E~

E,.

_ ".,:!I.,. _

" ) ,,<1"1·

Et

E,.

(2 .45)

E~

+

ITo

Eo

E(1uations (2.45) may be considered as defining !.he transformati on of stress illto strain for the given layer. It follow s frOIll a general property of such transformations that the coefficients of the stress components arc symmetric. I.e. , that t2.46)

E,

These equations show Ihat. while different. the Poisson's raljos t'" and 1I,.y; are nut independent; either of them can be obtained from the other if the correspond ing values of Ihe modulus of elasticity are kn()wn. The same is true ()f "'c and ".,;:,., and of II~, and I ',,,, . Consider now the effect of thc presence of shearing stresses on the faces of a small element of smeun:d layer. As pointed out in Sec. 2.14 in Ihe case of isotropic materials, these stresses come in pairs of equal and opposite vectors appl ied to opposite sides of the given element and have no effect on the nonnal strains. Thus, Eqs. (2.45) re main valid . The shearing stresses, however, will create shearing strains which arc defined by equations simi lar to the last three of the equations (2.38) of Sec . 2. 14. except that three different values of the modulus of ri gidity, Gn , GJ'" and G"" must now be used. We have T, y~,.

= -0

,

(2.47)

The fact that the three compone nts of strain Ex, Ep and E, can be expressed in tenns of the nonnal stresses only and do not depend upon any shearing stresses characterizes orfhofropic mute rials and disti nguishes them from other ani sotropic materials. As we saw in Sec. 2.5. a flat I{lminate is obtained by superposing a number of layers or laminas. If the libers in all layers arc given the same orientation to beller withstand an axial tensile load, the laminate itself will be orthotropic. If the lateral stability of the laminate is increased by positioning some of its layers so that their fibers arc at a ri ght angle to the fibe rs of the other layers, the resulting laminate wi ll also be orthotropic. On the other hand, if any of the layers of a laminate arc positioned so that their fibe rs arc neither parallel nor perpendicular to the fibers of other layers. the lamina, generally, will not be orthotropic.t

t For more infomm!ion on fiber-reinforced composite matcrial s. sec Hyer. M. W.• St1't'.'s Analysis of Fiber-Rein/arced Composite Maleri,,/.,. McGraw-HitL New Yor~. t998.

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EXAMPLE 2.11 A 6(j·mm cube is made from layers of graphite epoxy with fibers alig ned in the x directio n. The c ube is s ubjected to a oompressiv~ lood uf 140 kN in t.h~ x dirt:(;liun. "flIt prupt:nks of th~ nU1IIWl~i le maleri~1 are: 1-:., I ~~ (I r.]> ~ > 1-:, r 2 10 r. P~ . E, = 12.10 G Pa. " .. = 0 .24R. " .., = 0.248. and " ,., = 0.458. ~temline the changes in th~ cube dimensions. knowing lhal la) the c ube jg free tocxpund in the )' ami ~ din.--ctions (Fig. 2.S6); Ib) the cube is free to expand in the z direc tion. but is reStrained from e xpanJin g in the)' direc tion by two fixed fri ctionless plates (Fig. 2.37).

=

=

I ~1I1r.N

Fig. 2.56

(a) Free in y and z Directions. We first determine the stre ss

iJ,

111 the dirertlOn o /" loadlllg. We ha ve

- 140X IOlN A = ( U.O()~II11J(O.lXlU m) p

,,..<-;

GO"],,,

- 38.89 MPa

Fi.wd

Since the c ube tS not loaded or restrained in the y and z direc tio ns, we have 0" , - i f , - (J . Thus. the right-hand members of Eqs. (2.45) reduce to th eir firs t tmm. Substituting the given data into these eqlJations. we write

u, E,

", - - *

•

- 38.89 MPa 15S .0 GPa

' I

IJ IkN

/y~,.u-

Fig. 2.57

Poisson' s rat io II~, and. as we saw earlier. lhis rati o is fwl tqllc.l to the ratio I'" . which was amon g til! given dal1l. To tlnd " " wa use th", fi rst of Eq s. (2.4M and writ",

150.9 X 10-'

(0.248)(-38.89 MPa)

fT"Cti(">nl ... ~ pl .• t,'s

+ 6 2.22 X 10- 6

U]55.0O)O.248,1' '" I).0 1936 II,. -; (-E.)"" '" ('-E~

£. =

•

" ....II ,

+62.12

--- ~

E.

x

IO -~

The c hanges in the cube d imen sions are obtained by multipl y in g the corresponding ;;lrnin.1 by th .. length L = 0.060 III o f the side of the c ube:

8. = E,L =( - 250.9 X

IO ~ ')(O. lJ6Om)

= - 15.05

/.till

Making" 0 - 0 in thc tirst und Ih ird of Eq s. (2.45) uno substitutmg in these equations the gillen val ues of E,. E,. ".,.. and II.; . a, well ;JS the va lues obtained for (r •• (r •• a nd II", we have

" ..." ,

T-

- 38.89 ]I.IPa

S, = E,L = (+ 62.2 X ][r~)(o.O&O In) = + 3.7.3 I.un E,L = ( +62.2 x lO -b)( O.06O Ill) = +3.73 j-I.m

IS5.0GPa (0.0 1936)( - 752.9 kf'a) t~.IO GP"

{i =

(b) Free In

z Dlrecl/on, Rest rained In y Direction.

in the x direct io n is the 'mll~ ll~ in part u. n:ltllely. II > '" - 38.1\9 .M Pa. Since the cube is free 10 ex pand in the :: direction a.~ in part u. we ngain have u~ = O. BUI since the cube is now 1I:~uaill.:d in the )' diH::~tioll. "'I!: ~houl.J eJlpect a stre~s u . different from :zero, 011 the o th er Iwnd. s inc~ tlte cube cannot expa nd in Ihe. y dirCl'tion, we must have 8, = 0 and. (I IIllile sec· thu s. £ " = 6.1L = O. ]I.·l ak ing v~ = U ,md £ , o nd of Eqs. (2AS). so h' in~ th:11 eq u:1tion fo" IT y, :lI1d ~u bSli­ tuting tIle gi\'en data. we lUll'" The

s tre~s

=

u. (E,)""u. =

E,

",0"', -",,17, ----

- 149.7:X 10-·

(0248)(-38.89 MPIi) 155.0 GPa (OA51l )[- 751.9 kPa)

11. JOG Pa

+ 9[1.72 X

10-$

The changes in the cube dimensions are o btained by nlultipl yi n!; th;: corrcspondi ng stra ins by Ihc lcngth L - O.flnO In of the side o f th e c uhe: .1, = F) . = ( - ]4Q7 X

x

m-6 nor,o .... )=

- I ~Q~ !I""

;)., = E,L -; (O){O.O6(] m) = 0

5, = E,L = (+90.72 X 1O-·XO.06O In) = + 5.44 flm = ( 12.IO)u.l4X)( _ JlUI<} ]\'11'3) IS5.0 = - 752.l) kPa

No\.\" Ihal Ilw thr~ comf1'>n p. nl.' of ~trt~~~ ha vt' hern determ ined . we call U~ the first and InSI () f Eqs. (2.45) to eompute the s train co mponent s c.~ llJ1d Ct . But the first (]f th~se ~qualiol1s con tains

Comparing the re~ul1s of parts 11 and h. we no te that the difference between lhe values o bta ined fo r the deformmion 8 , in The dire cli o n of Ihe li belS is ne gligible. Howeve r. tile diffelI'nee helwl'.l'n th" vn rll l'~ nhl~inl' rl for th ~ l~tl'r:11 rI .. fnnn~li on 0, is not negligible. Thi S" de fo mmti()1l is clearly larger when the cu be is restrained from d~fom1ing in the ), direction.

97

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SAMPLE PROBLEM 2.5 A circle of diameter d = 9 in. is scribed on an unstressed aluminum plate of thickness I = i in. Forces acting in the plane of the plate later cause normal stressesu , = 12 ksi and(l, = 20ksi. ForE = ]0 x 10° ps i and t' = determine the change in (a) the lenglh of diameter AD. (h) the length of diameter CD. (e) the thickness of the plate. (d) the volume of the plate.

t.

SOLUTION Hooke's La w.

We note that

IJ.

= O. Using Eqs. (2.28) we find the strain in

each of the coordinate directions. 1':,

= +~ E

_ vtT.

=

I

_

E

"(,"

E

.[ ( IHSi ) _ 0 -

IO XlOb psJ II(T,

(T ,

~(2()kSi) ] 3

= + 0.533 x IO-J in.lin.

IIU,

t:" = - £" + E - E =

I "

[0 x ]() psi

[~ 12ksi) +O - !(20ksi) ] = - 1.067 X 3 ]

JO -J in J in.

= _ "fT, _ J'(" + ~

f

,

E =

10

x

E

I " 10

E i

'[ - -3 (12kSi) - O + (20 ksl) ] = + 1.600 x IO-J in.lin. p SI

a. Diameter AB.

The change in length is 0 8 /.'0 =

OS/A = "',d = (+0.S33

", ;{J.

x 10 - 3 in ./in.)(9 in.) 0BIA

= +4.8 X 10 ] in . ....

h. Diameter CD.

OCID = "'od = (+ 1.600 X 10 3 in.!in.)(9 in.) 0 C/D

c. Thickness.

Recalling that

I = ~

=

+ 14.4 X 10 ] in . ....

in .. we have

0, = "' / = ( - 1.067 X IO - J in '/i n.) (~ in. ) 0, = - (1.800 x 10-' in . ...

d. Volume of the I'latc.

Using Eq. (2.30). we write

e = ex + e~ + eo = ( + 0.533 j, V = eV =

98

1.067

+ 1.6(0) 10-.:1

= + 1.067 X 10- 3

+ 1.067 X J() -l[( IS in.X 15 in.:d in.) ) j, V = +0.187 X in] ...

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PROBLEMS

2.6 1 In a standard tensile test. an aluminum rod of 20-mm diameter is subjected to a tcmion force of P = 30 kN. Knowing that v = 0.35 and E = 70 G Pa. determine \a) the elongatiun of the rod in a 150-mlll gage length. (b) the change in diameter uf Ihe rod. 2 .62 A 20-mm-diameter rod made. of an experimental plaslk is subjected to a tensile force of magnitude P = 6 kN. Knowing that an elongation of 14 mm and a decrease in diameter of 0.85 mm are ohser\'cd in a 150-mm length, determine the modulus of elasticity. the modulus of ri gidity and Poisson's ratio for the maletiaJ. 2.63 A 600-1b tensile load is applied 10 a test coupon made from l~- in . Oat steel plate (£ '" 29 X [(/' psi and 1.' = 0.30). Determine Ihe resulting change (a) in Ihe 2-il1. gage length. (b) in the width of portion AS of the test coulxm, (c) in the th icknes; of ponion AB. (d) in the cross-sectional area of portiun AB.

I r--'U;"i HOO n, .----1l___~\:p,·======j.F~!~___J~ f'oOt! 110 "! 'I A B ' r--

r

20·",,,( rlhlnH"I"r

tOO nun

l

I"

Fig. P2.61

,IK' ~'I'"

,

.J..in .

Fig. P2.63

2.64 A 6-ft length of a steel pipe of 12-in. outer diameter and ~- in. wall thickness is used as a shon colu mn to carry a 300-kip centric axiallood. Knowing that E = 29 x 10" psi and v = 0.30. detennine (a) the change in length of the pipe. (b) the change in its outer diameter. (c) the change in its wall thickness. 2.65

The change in diameter of a large steel bolt is carefully measured

as the nut is tightened. Knowing that £ = 200 GPa and v = 0.29. determine

the internal force in the boh. if the diameter is observed to decrease by 13 /-lm. FI g. P2.64 flO nun

I

Fig. P2.65

2.66 An aluminum plate (£ = 74 G Pa and I' = 0.33) is subjected to a centric axial load that causes a normal stress (T . Knowing that. before loading, a line of slope 2:1 is scribed on the plate. determine the slope of the line when (J = 125 MPa.

FIg. P2. 66

99

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100

Stress and Strain-Alolal Loading

2 .67 The aluminum rod AD is fined with a jacket tha i is used to apply a hydroslatic pressure of 6000 psi 10 Ihe 11·in. portion BC of the rod. Know· ing Ihal £ = 10. 1 X 10f' psi and p = 0.36. detemline ({I) the change in the 100al lenglh AD, (b) Ihe change in diamelet al Ihe middle of Ihe rod.

Fig. P2.67

2.68 For the rod of !'roh. 2.67. determine the force s Ihal should be applied to the ends A and D of the rod {a) if the axial slrain in pol1ion BC of the rod IS 10 remain zero as the hydroSlallc press ure is applied, (b ) il the lotal length AD of the rod is to remain unchanged.

Fig. P2.69

2 .69 A fabric used in air·inflated structures is subjected to a biaxial loading thai resu lts in normal s tresSeS if, = 120 MPa and (r~ = 160 MPa. Knowing that the properties of the fabric can be approximated as E = 87 GPa and v = 0.34. delermine Ihe change in length of (a) side AB. (b) side BC, (e) diagonal AC. 2 .70 The block shown is made of a magnesium alloy for which E = 45 GPa and p = 0.35. Knowing Ihat if , = - [80 MPa. detennine ((I) the magni. tude of 1ft for which the change in the heighl of the block will be zero. (h) the corresponding change in the area of the face ABCD. (d the corresponding change in the volume of the block.

Fig. P2.70

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Probtoms

2.71 The homogeneous plate ABeD is subjected to:l bi:lxbllo:lding :IS shown. It is known thm (f~ = flo :lnd Ihm the change in length of the plate in the x direction must be zero. that is. f' . = O. Denoting by E the modulus of elasticity and by ~ Poisson's rmio. determine (a) the required magnitude of fT,. (h) the mtio (fo/ f'. 2.72 For a member under :lxial IO:lding. express the normal strain f" in a direction fanning an lingle of 45° with the axis of the load in ten ns of the axial stmin f', by (a) comparing the hypothenuses of the triangles shown in Fig. 2.54. which represent respectively an element before and after deformation. (b' using the values of the corresponding stresses (T ' and CT, shown in Fig. lAO. and the genemlized Hooke's law.

Ag. P2.71

2 .73 In IT1nny situmions it is known that the normnl s tress in n given direction is zero. for example fT, = 0 in the cnse of the thin plate shown. For this cnse. which is known as plane !ilreJS. s how that if the str:lins to.. nnd e> h:lve been determined experimentally. we cnn express 0",. fT , . and f'o as follows : (T~

u"

f' , + "f',

= E- - , I - v-

Fig. P2.73

2 .74 In mWly situmions physical constraints prevent strain from occurring in :I given direction. for example f', = 0 in the case shown. where longitudinal movement of the long prism is prevented :It every point. Plane sections perpendicul:lr to the longitudinal nxis remain pbne and the s:lme distance apan. Show thm for thi~ situation. which is known as pIa!!/' slrain, we c:ln express (T,. e." :lnd f' v as follows: (T "

= v(fT , + fT,.)

f'. = f',

Fig. P2.74

I

"EJ(I - ,,2)0"..- - ,,(I + "Prj I

= "£:[(1 -

,,2)0-, - ,,(1 +

,'P']

101

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102

Stress and Strain-Alolal Loading

2.75 TIle plastic block shown is bonded to a fixed base and tll a horizontal rigid plme to which a force P is applied. Knowing th:n for the plastic used G = 55 ksi. determine the deflection of the plaLe when P = 9 kips.

I'

"

I I

"

~~ Fig. P2.76

Fig. P2.7S

2.76 A vibration isolation unit consists of two block~ of hard rubber bonded to plate AB and to rigid supports as shown. For the type and grade of rubber used "T~II = 220 psi and G = 1800 pll"i. Knowing that a centric venical force of magnitude P = 3.2 kips must cause a O. I -in. vertical deflection of the plate AB. detennine the smallest allowable dimensions {/ and b of the block.

2.n The pl:lstic block shown is bonded to a rigid support and to a vertical plate to which a 240-kN load P is ~pplied. Knowing that for the plastic used G = 1050 MPa, detemline the deflection of the plate. 2.78 Wh~t load P should be applied to the plate of Prob. 2.77 to produce a l .5-mm deflection? 2.79 Two blocks of rubber with a modulus of rigidity G = 1.75 ksi are bonded to rigid supports and to a plate AB. Knowing that c = 4 in. and P = 10 kips. determine the sm~lIest allowable dimensions {/ ami b of the blocks if the shearing stress in the rubber is not to exceed 200 psi and the deflection of the pl:lte is to be at least rl; in.

Dim",..i""s in mil,

Fig. p2.n

Fig. P2.79 and P2.80

2.80 Two blocks of rubber wi th a modulus of rigidity G = 1.50 ksi are bonded to rigid sUPJX)rts and to a plate AB. Knowing that b = 8 in . and t" = 5 in .. detemline the largest allowable load P and the smallest allowable th ickness a of the blocks if the shearing stress in the rubber is not to exceed 2[0 psi and the deflection of the plate L~ 10 be al [east ±in.

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2.81 An dastollleric bealin£ (G - 0.9 MPa) is ust"d to sU pJlOIt a bridge J::irder as shown to provide nexibility during earthquakes. The beam must not displace more ltian 10 mm wh~n a 22·kN lateral load is applied CIS shown. Knowing that the maximum allowable shearing stre~s is 420 kl'a. detemline Ill) the

~ma llest

Problem.

allowable dime lls ion h. (b) the smallest required thickness n .

2.82 For thl! elastomeric bearing ill Prob. 2.3 1 with b = 22[) 111m and 30 mill. uctcnllillt! thc shc
/I

=

=

·2.83 A 6·in.·diameter solid steel sphere is lowered into the ocean to a point where the pressure is 7. 1 ksi (
t n)

For the a xial loading showlI. determine the change in heif!ht

and the change in volume of the brass cylinder shown. (bi Solve part a. a.sSUIllUl g ,hat the loadill£ is hydlOstalic witll II " - U , v , - - 70 MPa.

Fig. P2.81

/

22-", '11 dj,inll'tN

lO ll\

FIg. P2.84

'2.85

DclCnlline the dilulf.!Iioll e and the chung~ in volume of the 200.. mm

1--200,,,,,,--1 Fig. P2.85

leng!lt of (he rod shown if (a) the rod is made of steel wit h E "" 200 GP;J alltl II = 0.30. (h) the rod is made of aluminum with E ~ 70 GPa ;Jnd II = 0.35.

· 2.86 De lenn;ne thc change in volume of the 2. in. g~ge Icngth segment AB in Pmb. 2.63 (a) by compllting the dilatation of th e material. (b) by sub .. tracting the original vo lume of portion AB [rom its firml \'olume. "2.87 A vibration isolation support consists of a rod A of radius H, 10 mm :l.IId a tlJbe B of inner radiu; R~ = 25 mm bonded [0 an 30.mm-Iong ltollClw rubber cyl inder with a modulus 0]" rigid ity G = 12 Ml-'a. IJetenllllle the largesl allowable force P which can be applied [0 rwA if i(.~ deflection is not to exceed 2.50 mm. - 2 .. 88 A \'ibtation lsolmion $uppcn consi.lls of a rod A of rndius HI and Inh/> R of inner r~din~ R, honrtl"';rt 10 :1 RO.mm .. long hollow nJ h~r cy linl1l"';r with a modulus of rigidity G = 10.93 MPa. Detennine the required vllue of lhe mtlO R!iR , if n 10 .. kN force P i ~ to cau.e n 2~n\n\ deflection of rod A. ~

Fig. P2.871lnd P2.88

103

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104

Stress and Straln-Alolal Load ing

E, '"' .511 GP~

VX' ..

Ev " IS.2 CPa f:, = 15.2Cb

V,y " O.2,5.t V'Y: 0.42S

1l.25-!

· 2.89 A composite cube wi th 40-mm sides and the properties shown is made with glass polymer fibers aligned in the x direction. The cube is can· strained against deformations in the )' and ~ directions and is subjected (0 a Tensile load of 65 kN in the x direction. Delennine (a) the change in the length of the cube in the x direction. (b) the Slres~es fT.,. i f\ _ and (r~. ' 2.90 The composite cube of Prob. 2.89 is constrained agains t deformation in the z dirt:ction ;lnd e longated in the x direction by 0.035 mm tlue to a tensile load in Ihe x direction. Determine (a) the stresses if,. fT, . and IT, . (b) the change in the dimension in the y direction. ' 2.91 The material constants E. G. k. and 1/ are related oy Eqs. (2.33) and (2.43). Show that anyone of these constants may be e.1(presscd in terms of any other two constants. For example. show that (a) k = G£/(9G - 3£) and (b) \' = (3k - 2G)/(6k

+ lG).

' 2.92 Show that for any given material. the ratio G/£ of the modulus of rigidi ty over lhe modulus of ela~ tic i ty is always less than ~ out more than i. [Hint: Refer to Eq. (2.43) and to Sec. 2. 13]

Fig. P2.89

2.17. STRESS AND STRAIN DISTRIBUTION UNDER AXIAL LOADING ; SAINT-VENAN T S PRINCIPLE

r

t.'

Fig. 2.58

We have assumed so far that. in an axially loaded member. the normal stresses are uniformly distributed in any section perpend icular to tlte axis of the member. As we saw in Sec . 1.5 , such an assumption may be quitc in error in the immediate vicinity of the points of application of the loads. H()wever, the (.leterminatioll o f tlte actual stresses in a given sl.:ction of the member rcquires the soluti()n of a staticall y indetemlinate problem. Ln Sec. 2.9, you saw that staticall y indetenninate pwblcms invol ving the determination of forces can be solved by consideri ng the deformatiuns caused by these forces. It is thus reasonable to condudt: that the determinati on of the .I'[re.\·.\'e.~ in a member requires the analysi s ()f the .\·frili m produced by the stresses in the mcmber. This is essemially the approach found in advanct:d textbooks, where the mathematica l thcory o f clasticity is used to determine the distribution o f stresses corresponding 10 vario us mod!!s of application of the loads at the ends o f tlte member. Given the more limitcd mathematical means al our disposal, our analysis of stresses will be restricted tt.) the particular case whcn two rigid plates an! used to lransmit the loads to a member made of a homogeneous isotropic material (Fi g. 2.58). [f the l()llds arc applied at the center of each plate.t the plates will move toward each ()!her without rotating. causing the member to get shorter. while increasing in width and thickness. It is reasonable to assu me that the member wi ll remain straight. that plane sections win ret Morc preci,ely. tlW Wlllmoll line of ;K·tion of the toads shoutd of the Cf{)S~ 5cctioll (d. Sec. t .~ ).

pass through

tlW centroid

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p

2.17. Stress and Strain Under Axial loading

•• (h)

main plane. and that all elements of lhe me mber will deform in the same way. since such an assumption is clearly compatible with the given cnd conditi ons. Thi s is illustrated in Fi g. 2.59. which shows a rubber model before and after loading.t Now. if all elements deform in the same way. the distribution of strains throu g hout the me mber must be unifo rm.

[ 11

other words, the axial strain € J and the lateral strain E'r = - VE., are constant. But. if the stresses do not exceed the proportionailimit, Hooke's law applies and we may write fTr = EE., . from which it follows thai the normal stress u~ is also constant. Thus, the distribution of stresses is uniform throughout Ihc member and, at any poi nt ,

CT ,

=

(a"L,,,

p

=

A

On Ihc other hand, if the loads are concentrated. as illustrated in Fig. 2.60, the clements in the imml!diate vicinity of the points o f application of the loads an: subjected to very large stresses, while othl!r clements ncar the ends of the member are unaffected by the loading. This may be veri lied by observing that strong deformations, and thus large strains and large stresses. occur ncar the points o f application of the loads, while no deformation takes place at the comers. As we consider demcnt~ farther and farther from the ends, however, we no te a progressive equalization of the deformations involved. and thus a more nearly 1I11ifonn distribution o f the strains and stresses ac ross a section of the member. Th is is further ill ustrated in Fi g. 2.6 1, which shows the result o f the calculat ion by advanced mathematical methods of the :j:NOle that for ton~. sknder memllcrs. another configuration is po~siblc. and indeed will !"""iI if 'he '~d i, II," """''''' ",,"/,., ~d "'''''''' , ,,~'" , h,,< Thi' will be diS('usscd in Chap. 10.

"m",,,,,, ''''

I'

'

t P'

Fig. 2.60

1 05

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106

Slress and Straln-Alo lal Load ing

"

"

I"

ll' ~ ~"""" L

__

. , ., 1

CT

,\

"',,,0. "'.,., '" (l,lfj3CT,...

U,,""

= Ut2iCT""

CT,,,., = (l.l9Sa.,,, CT",,,,, a 2.5i5u",~

P' Fig. 2.61

distributi on of stresses across various sections of a thin rectangular plate subjected to concentrated loads. We note that at a distance b from either cnd, whcre b is the width of the platc. the stress distribution is nearly uniform across the section. and the value of the stress !T" at any point of that section can be assumed equal to the average value PIA. Thus, at a distance equal to. or greater than, the width of thc member. the distribution of stresses across a given section is the same, whether the member is loaded as shown in Fig. 2. 58 or Fig. 2.60. In other words, except in the immediate vicinity of the points of application o rthe loads, the stress distribution may be ass umed independent of the actual mode of applicati on of the loads. This stateme nt, which applies not only 10 ax ial loadings, but to practically any type of load. is known as SainlVel/alit's principle , after the French mathematician and engineer Adhcmar Barre de Saim-Venant (1797- 1886). While Saint-Venant 's principle makes it possible to replace a given loading by a simpler one for the purpose of computing the stresses in a structural me mber. you should keep in mind two imponant points when qllil'alent. 2. Stresses cannot be computcd in this manner in the immediate

vicinity of the points of applicatio n of the loads. Advanced theoretical or experimental methods mU SI be used to determine (he di stribution of stresses in these afl:HS. You should also observe that the plates used to o btain a unifoml stress distri buti on in the member of Fig. 2.59 must all ow the member 10 freely cxpand latemlly. Thus, the pl ates cannot be ri gidly iluached to the member; you must ass ume the m to be just in contact with the member. and smooth enough not to impede the lateral expansion of the me mber. While such end conditions can actually be achieved for a me mber in compression. they cannot be ph)sieaUy reali zed in the case of a member in tension. It does not matter, however. whether o r not an acrual fixture can be realized and used to 100ld a member so that the di.~tributi on of stresses in the member is uniform . The important thing is 10 illwginc a mode/ that wil.1 allow such a distribution of stresses. and to keL"p thi s model in mind so that you may later compare it with the actual IO
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2.18. Stress Concentrations

2.18. STRESS CONCENTRATIONS

As you saw in the preceding section, the stresses ncar the poillls of application of concentrated loads can reach values much larger than the average value of the stress in UIC member. When a strucllLral member contai ns a discontinui ty, sllch as a hole or a sudden change in cross section, high localized strcsses can also occur near the discontinu ity. Figures 2.62 and 2.63 show the distribution of stresses in critical sections correspond ing to two such situations. Figure 2.62 refers to a fl at bar with a circllhll· hole and shows the stress distribution in a section passing through the center of the hole. Fi gure 2.63 refers to a nat bar consisting of two portions of different widths connected by fillel.\"; it shows the stress distribution in the narrowest part ofille connection, where the highest stresses occur.

,-----=~-,

~

'"

I

'---~'-~-'fT.... FIg. 2.62 Stress dIstribution near circlliar hole in flat bar under aXial loading.

u~

F

ff

. "

~-~

Ag. 2.63 Stress distribution near fillets in flat bar under axial loading.

1l1ese resu lts were obtained experimelllally through the usc of a photoelast ic method. Fortunatel y for the engineer who has to design a given member and cannot afford to carry out such an analysis, the rc~ suits obtained are independent of the size of the mcmber and of the material used; they depend only lIpon U1e ratios of the geometric par.lmc.ters involved, i.e., upon the ratio rid in the C3..~e of a circular hole, and upon the rati os rill and Dill in the case of fi llets. F1l11hermore. the designer is more interested in the 1II1I.1"i1ll1l1ll mille of the .';tress in a given sec tion, than in the actual di ~tribution of _~tresses in that section, since his main concern is to determine v.-!Jefher the allowable stress wi ll be excecded under a given loadi ng, and not where this value wi ll be excecded. For this reaso n, one defi nes the ralio K ~­ (T.,~

~/' "

(2 .48)

of the maximum stress over the ave'dge stress computed in the critieal (narrowest) section of the discont inuity. This ratio is referred to as the Sf re.,\"S-COllcenl mfioll facIo/" of the given discontinuity. Stress-concentmtion L'lctors can be compu ted once and for all in term ~ of the ratios of the geometrie parameters involved, and the resu lt s obtained can be

107

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108

J

Stress and Strain-Alolal Loading

,

,

~ !:;:~£1' f ~

J .2 J.

expressed in the fonn of tables or of graphs. as shown in Fig. 2.64. To determine the maximum stress occurring ncar a discontinuity in a given member subjected to a given ax ial load P, the designer needs only to compute the average stress fTa.c = PIA in the critical section, and multi· ply the result obtained by the appropriate value of the stress-concentration L1clOr K. You should note, however, that this procedure is valid o nly as long as u""'" does not exceed the propoflionallimit of the material. since the values of K plotted in Fig. 2.64 were obtai ned by assuming a linear relation between stress and strai n.

°1"

28

,

" " ",

2.6

,

2.

l- i--

K 2.2

,

0

,

I 6

2 I 0

0.1

°

0.2

0,3

0.4

0.5

0.'-

06

dd

(a) Aat bars willi holes

"

o QV.

~ D

1\

d

Did = 2 l..S

1\ '\ I'"

1.3

I" 11

1.2

I':

" " ," ,

S I

-

I-

"-

1.1

n I;' dd

(>()II >I)!; (I.OS 0 10 «12 Ill.

0.1,"

>I.~>

u.•"'_ 0.24

OI.'!6

o. 0 ~

H.;JJ

(bl Fla! bars with fillets

Fig. 2.64 Stress concentrati on factors lor !lat bars under axial loading' Note that the average stress must be computed across the narroW1!S\ section: fT ... = Pltd, where lis the th ickness of the bar.

EXAMPLE 2.12 Delennine the largest axial land P that can be safe ly suppottcd by a n,lI steel bar consisling of two pottiolls. both 10 mm thick and, respectively, 40 and 60 mrn w ide, connected by fillets of radius r = 8 mm. A'SUlf.C an allowable norm:!1 stress of 165 MPa. We tirsl compute the mtios D

60mm

d

40mlll

I.S0

r

8 111m

d

40 nun

-~--= O .:W

Carrying Ihis val ue inTO Eq. (2.48) and solving for if, ,,. we have 'T".

=

But if m", cannOl exceed the nllowable slress IT, ll 165 M Pa. Substituting thi s value for ff U"'" we find Ihal The C1vemge stress in the narrower poltion (d ~ 40 mill ) of the bar should lIOt ex· ceed Ihe value 0".",

Using the curve in Fig. 2.64b corresponding TO Old = 1.50. we rmd thaT the value of lile stress·concentration fnctor corre· spondi ng to rid = 0.20 is K = 1.82

"=1.&2

Recalling that

O"~"

=

t65 MPa ----uu-= 90.7 MPa

= PI A. we have

p = A" ,,~ = (40 mrn)(10 mm)(9fJ.7 MPa) = 36.3 x loJ N p = 363 kN

t Wo D. Pilhy. ""1"'$011'$ Srreu c,,,,aMmlioll Facro .... 2"" cd .• John Wiley & S()I)~. New

York. 1997.

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2.19. PlasUe Delormatloos

2.19. PLASTIC DEFORMAT IONS The res ult s obtained in the prcced ing sections wc re based o n the assumptio n of a linear stress-strain re lationship. In o thcr words, we assumed that the proportional timit of the material was never exceeded. Thi s is a reasonable assumption in the case of brillie materials, whic h ntpture wi tho ut yielding. In the case of ductile materials, however. this assumption implies th at the yield strength of the m:tlcrial is not exceede d . The deformations will the n remain withi n the elastic range and the strucillral member unde r cons ideration will regai n its o ri ginal shape after all loads have been removed. If, on the other hand, the s tresses in any pan of the member exceed the yic ld s trength of the material, plastic defo rmatio ns occur and mos t of the results obtained in earlier sections cease 10 be valid. A more invol ved analysis, based o n a nonlinear stress-strain relatio ns hip. must thc n be carried out. While un analysis taking illlo account the actual stress-strain relationship is beyond the scope of Ulis text, we gai n considerable insight illlo plastic behavior by considering a n idealized e/a.\"loplastic material for which the s tr~ss-st rain diagram consists of the two s trai ght- line segments shown in Fig. 2.65. We may notc that the s tress-strain diag ram lo r mild steel in the elas tic and plastic rangcs is similar to this idealizati on. As long as thc stress u is less than the yield strength Uy, the material behavcs elastically and obeys Hooke's law, U = E€.. When (T reaches the value (Tf, the material starts yielding and keeps dcfonning plastically under a constant load. If the load is removed , unl oading takes plflCC a lo ng a stmi ghl - linc segment rn pllTa 11c I In th.~ inilia1 portion A Y of the loadin g c urve. The segment AD of the ho rizontal ax.is re presents the strain corrcsponding to the pennane nt set or plastic deformatio n resu lting from the loading and unloading o f the s pecimen . Whi lc no actual mate ria l behaves exactly as shown in Fi g. 2.65. this stress-s train diag ram will prove usefu l in discussing the plastic t.!cfomlati ons o f ductile ma te ri als s uch as mi lt.! steel .

109

;'_ . _ _ _+C' - _ I-IIII'III .....

/

,,/

,

JJ

}t

Ag.2.65

EXAMPLE 2.13

A rod of length L = 500 mm and cross·sectional area A = 60 rlIrl1 ~ is made of an e lasloplastic material having a modulus of elasticity E = 100 GPa in its ehlst ic range and a = 300 MPa. TIle rod is subjected 10 an axial yield point

0,

300 X tif Pa x tOq Pa

£"; T = 200

1.5 X IO- J

u,.

load lIlIIil il is stretched 7 mm anuthe load is then removed. What is the resulting permanent sen Referring to lhe diagram of Fig. 2.65, we find lhal lhe ma:dm um strain. represented by the abscissa of point C. is tic

7nuTI

£c="""L= 500mLn

l = t4 x 10-'

On the 0 1her hand, the yield strain . represemed by the abscissa of point Y. is

The strain after unloading is represenled by Ihe abscissa poim D. We note irom Fig. 2.65 that £p

= AD == YC =

£c -

£0

of

£r

"" 14 X to-) - 1.5 X to-) = 12.5 x IO" J The permanent set is the deformmion 8/) corresponding 10 Ihe strain £/). We hal'f lJ u == £/,1. == ( t2.5 x

1U ~" )(500

mm) == 6.25 mm

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EXAMPLE 2.14 A 30-in.-Iong cylindrical rod of cross-section;)1 area A , = 0.075 in 2 is pl:lced inside a tube of the same length and of cross-sectional area A, = 0 . 100 inl. 11le ends of the roo and tube are attached to a rigid suppott on one side. and to a rigid plate on the other. as shown in the longitudinal section or Fig. 2.66. The rod and tube are both assumed to be elaslOplastic, with moouli of elasticity E, = 30 X 106 psi and £, = 15 x lOb psi. and yield strengths (ff ,)y = 36 ksi and (IT,h =0 45 ksi. Draw the load-deflection diagram of the rodlUbe assembly when a load P is applied to the plate liS shown. (Il)

l',

4 ..~

-------------~-----,-..,_--

1.8

-.-- - - -

" r TIIIX'

11==~==I1V"b'" ,,,,I "

(/rl

/, (kip' )

Fig. 2.66

We first detemline the internal force and the elongation of the rod as it begins to yield:

=]6 X 10

,.Ill.

Since the material is eiastoplastic-. the force-elongation di· agram £1jthl'. rod aloll!'. consists of an oblil/ue straight line and of a horizontal stT
(P,), = (<7,)tA, = (45 ksi)(O . IOO in~) = 4.5 kips ("/"')1' 45 X 10' psi L = 6 . (30 In.) (8J,. = (t:,)rL = E, 15 X j() pSI = 90 X IO -J in.

110

.Hi -----.-

",

1.-)

Fig. 2.67

The load-deflection diagram of the tube alone is shown in Fig. 2.67b. Ob.,erving that the load and deflection of the rod-tube combination are. respectivel y. P = I',. +P,

8=0,.=8,

we draw. the required load-deflection diagram by adding the ordinates of the diagrants obtained for the rod and for the tube (Fig. 2.67d. Points Y, and Y, correspond to the onset of yield in the rod and in the tube. respectively.

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EXAMPLE 2.15 If the load P applied to the rod·tube assembly of Example 2.14

is increased from zero to 5.7 kips and decreased back to zero. detennine (a) the maximum elongation of the assembly, (b) the permanent set after the load has been remo\'ed.

I' e(Ici [,, )

\a) Maximum Elongation. Referring to Fig. 2.67c, we observe that the IOlJd Pow = 5.7 kips corresponds to a point located on the segment Y,Y/ of the load-deflection diagram of the assembly. Thus. the rod has reached the plastic range. with P, = (Pr),· = 2.7 kips and (T ~ = (u rh· = 36 ksi. while the tube is still in the elastic range. with P,

=p

- P, = 5.7 kips - 2.7 kips = 3.0 kips P,

a, = -

A,

0,

= £!- =

J.O kips

= - - - = 30 ksi 0.1 in!

,,

,-I,

/ 1,

D _

, ,,

0

,

""

;;, ( IO-3 in J

(tJ)

I', (kip., )

1',

-------------

3<1

(-', )0 X !OJ psi - L = _ . (30 in. ) = 60 X 10- 3 in. E, [J X ]06 pSI

The maximum elongation of the assembly. therefore. is

c

r,

2.7

"

c

6\l (h )

l ',

f (lipS)

6 m", = 0, = 60 X lO-l in.

c

(b) Permanent Set. AS" the load P decreases from 5.7 kips 10 zero. the internal forces P, and P, both decrease along a str.tight line, as s hown in Fig. 2.68(/ and b, respec· tivel y. The force Pr decreases along line CD parallel to the initial portion of the loading curve. while the force P, decreases along the original loading curve. since the yield stress was not exceeded in the tube. Their sum 1'. therefore, will decrease along a line C£ parallel to ihe portion 0 Yr of the load-deflection curve of the assembly (Fig. 2.68cI. RefelTing to Fig. 2.67c. lVe find thalthe slope of OY,. and lhus of CEo is

0""" '"

--;C4C·5'c:k2ip~,c-

11/

= 36 X 1O ~ 1 in.

60 X 1O~1 i". (,.)

125 kips/in.

Fig. 2.68

The segment of line FE in Fig. 2.68r represents the defoml3tion o· of the assembl y during the unloading phase. and the segment OE the permanent sct 051, after The load P has been re o moved. From triangle CEF we have The permanent set is 8''''' - P...... ; III

5.7 kips

115 kips/in.

- 45.6 X 10- 1 in.

61> ; fj'"A'

+

tllU~

6' = 60 X 10 - ·' - 45.(, X 10- 1 = l-tA x 10-·' in.

111

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112

Stress and Straln-Alolal Loading

~'=!-:_-----'~ '::;s ~ i

~IT)--l

, ,

:"'--1 1«..., = IT)

, ,

Q , ,

,

(, )

(/'1 \'

lei

g==,-----'r--'."c 1==1 l1u" =

iF)-

We recall that the di scussion of stress concentrations of Sec. 2.18 was carricd OLlt under the assumpti on of a linear stress-strai n relati onship. The stress di stributions shown in Figs. 2.62 and 2.63, and the va llies of the stress·concentration factors ploued in Fig. 2.64 can not be used. thcrefore, when plastic defomlations take place, i.e .. when the value o f U "'"" obtained from these figures exceeds the yield strength 0")', Let liS consider again the nat bar with a circular hole of Fi g. 2.62. ,md let us assume that the material is dastoplastic, i.e ., that its strcssstmin diagram is as shown in Fig. 2.65. As lo ng as no plastic deformation takes place. the distribution of stresses is as indicated in Sec. 2.18 (Fig. 2.69a). We ()bserve that the nrea under the stress-distribution curve represents the intcgral f if riA , which is equal 10 the load P. Thus this area. and the value of Un..,' must increase as the load P increases. As long as O" n"', S (F ,' aLi the slIccessive stress distributions obtained as P increases wi ll have the shape shown in Fi g. 2.62 and repeated in Fig. 2.69a. However, as P is increased beyond the value P y corresponding to O"'Rll = (J" y (Fig. 2.69h), the stress-distribution curve must flatten in the vicinity of the hole (Fig. 2.69c). si nce the stress in the material considered cannot exceed the value rT y . This indicates that the material is yielding in the vicinity of the hole. As the load P is further increased, the plastic zone where yield takes place keeps expanding, unti l it reaches the edges of the plate (Fig. 2.69d). At that point, the distribution of stresses across the plate is uniform, (T "" ITy, and the corresponding value P "" Pu of tile load is the largest which may be applied to the bar without causing rupture. It is interesting to compare the maximum value P l' ofthe load which can be applied if no permanent deformation is to be produced in the bar, with the value Pu which will cause rupture. Recalling the delinition of the average stress, (J"m"" Pi A , where A is the net crosssec tional area, and the defi nition of the stress concentrat ion factor. K = O"ma~/(Ta,< ' we write

(1/ )

Fig. 2.69 Distribution of stresses in elastoplastic material under increasing load.

(2.49)

for any val ue of u n"", that does not exceed

up

Whe n

O"m",

=

0"1'

(Fig.

2 .69h). we have P = PI" and Eq . (2.49) yie lds

u,A

Py = -K

On the other hand, when P "" Pu (Fi g. 2.69d), we hayc

(2.50)

u~".

""

IT)'

and

(2 .51 )

Comparing Eqs. (2.50) and (2.5 I), we conclude that

(2.52)

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2.20. Res!dual Stresses

'2.20 RESIDUAL STRESSES

In Example 2. 13 of the preceding section, we considered a roo Ulat was stretched beyond thc yield point. As the load was removed , the rod did not regai n its original length : it had been pemlanently deformed. However, after the load was removed, all stresses disappeared. You shou ld not assume that this will always be the case. Indeed. when on ly some of the pans of an indeterminate structurc undergo plastic deformations, as in Example 2. I 5, or when different pans of the structure undergo different plastic deformations, Ule stresses in the vari ous parts of the structure will not, in geneml, return to zero aft er the load has been removed. Stresses, called residual .l·tresses. will remain in the various parts of the structure. While the computation of the residual stresses in an actual structure can be quite involved, the fo ll owi ng example wi ll provide YO lL wi th a general understanding of the method to be used fo r their determination.

EXAMPLE 2.16 Determine the residual stresses in the rod and tube of Examples 2.14 and 2.15 after the load P h~lg been increased from zero 10 5.7 kips and decreased back 10 zero. We observe from Ihe diagrams of Fig. 2.70 that after the to.'ld P has returned 10 zero. Ihe internal hm:es P ,. :lnu P , Me 110/ equal 10 zero. Their values have been indicated by point E in parts {/ and b. lespeclively. of Fig. 2.10. It followS" that Ihe corresponding stresses are not equal 10 zero either after the assembly has been unloaded. To determine these residual stresses, we shall determine the reverse stresses rI ; and caused by the unloading and add them 10 the maximum stres~eS" rI r = 36 ksi and rI, = 30 kgi found in part 1I of Example 2.15. The s train caused by Ihe unloading is the same in the rod and in the tube. It is equal to 0'/ L. where 0 ' is the defonnation oflhe assembly during unloading. which was found in Example 2. 15. We have

c

2.7

,, " j

, ,,

I) ,

, ,

,,,

(0'

,

y

(I ;

£'

0' = l.

- 45 .6 X lQ - ) in. =

] 0 in.

- 1.52 x 1O-) inJi n.

The corresponding reverse stresses in Ihe rod and tube are

=

<1"; = ~ ' E, ( - 1.52 x 10- 3)(30 x 106 pSi) = - 45 .6 ksi ,,; = £' E, = ( - 1.52 x 10- ')( 15 x 10" psi) = - 22.8 ksi

, ,

I' (kips \ 5 .7

-------------

(<1",)"" = <1", +

IT; =

]0 ksi - 22.8 ksi = +1.2 ksi

C

, ,, "

Y, .l .S

,,

I

,,

-,

"

(,'I

p" ....

, ,,

The residual stres~e s are found by superposing the stresses due to loading and the reverse stresses due to unloading. We have (<1",)..,. = <1",+ u; = 36ksi - 45.6ksi = - 9.6ksi

0, ( 11'1.,1 in .\

r~

0

'£

Uk-I Fig. 2.70

'"

F

ljlt ~in .\

"

113

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114

Stress and Straln-Alolal Loading

Plastic deformations caused by temperature changes com also result 1I1 residual stresses. For example , consider a small plug that is l() be welded to a large plate. For discussion purposes the plug will be considered as a small rod A B that is to be welded acro$S a small hole in the plate (Fig. 2.7 J). During the welding process the temperature of the rod wi ll be raised to over IOOOOC, at which temperature its modu lus of elasticity. and hence its stiffness and stress. will be almost zero. Since the plate is large, its temperature wi ll not be increased significantly abow room tempemlllrc (20cC). T hus. when the we lding is completed. we have rod AB at T = IOOOcC. with no stre;,s. auached to the plate whieh is at 200C.

Flg.2.7t

As the rod cools. it~ modulus o f elasticity increases and . at about 500°C, will approach its l1o nl1al vallie of about 200 GPit. As Ihe. temperature o f the rod dec reases funher, we have a siltlation similar to that considered in Sec. 2. JO and illustrated in Fi g. 2.35. So lving Eq. (2.23) fo r ~ T and making u equal to the yield strength, Uy = 300 MPa, of average steel, and a. = 12 X 10- 6;oC, we find the temperature change that will cause the rod to yield:

~T = - ..!!..... =

Ea

300 MPa (200 GPa)(12 X 10 6;oq

Thi s means that the rod wi ll stan yielding at about 37SOC and will keep yielding at a fairly constant stress level, as it cools down to room temperature . As a result of the welding operation, a residual stress approximately equal to the yield strength o f the steel llsed is thus created in the plug and in the weld. Residual stresses also occur as a result of the cooling o f metals which have been cast or hot rolled. In these cases, the outer laye rs cool more rapidly than the inner core. This causes the outer layers to reacquire their stiffness (£ returns to its normal value) faste r than the inner core. When the entire specimen has returned to room temperature, the inner core will have contracted more than the outer layers. The result is residual longitudinal tensile stresses in the inner core and residual compressive stresses in the outer laye rs. Residual stresses due to welding, casting, and hot rolling can be quite large (of the order of magnitude of the yield strength). These stresses can be removed. when necessary, by reheating the entire specimen to about 6000C, and then allowing it to cool slowly over a period of 12 10 24 hours.

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SAMPLE PROBLEM 2.6

_",~a.,'

AD =

£

400 In",~

CE = 500

,-

0

"",2

5",

?/J

2",

H

"

1-

1 J

C

L,,, +,..,j

The rigid beam ABC is suspended from two steel rods as shown and is initially horizontal. The midpoinl B of the beam is dellected 10 mm downward by the slow application of the force Q, after which the force is slowly removed. Knowing that the steel used for the rods is elastoplastic with E = 200 GPa and U r = 300 MPa. determine (II) th~ required maximum value of Q and the corresponding position of the beam. (/)) the final position of the beam.

SOLUTION Statics.

Since Q is applied at the midpoint of the beam . we have

,md

Q=

2P Ilu

Elastic Action. The maximum value of Q and the maximum elastic deIlection of point J\ occur when" =:= " \. in rod AD.

(/',\IJl.,,, = IT )"'! = (300 Ml'a)(400 mm 2 ) = 120 kN Qm", = 2(PIlD l m.., = 2(120 kN ) Q"... = 240 kN ....

I'

1'r:£ (kN)l [0'

/I

120 ";- - :;

120 --

""

if .

lL = -L =

E

(300 f.,'IPa) - - - - (2111) = 200 CPa

3n1111

Y'

,

I ,' :

oA, =

O!-3C---':I~I-C14-;-'-'-'"

: 0 6"H"

lIod ,.\D

Hoi! CE

Lo;td·.kneeti"" '\iagm,ns

Since

P eE = PAD =

120 kN. the stress in rod CE is if .

(I,

= Pv ; = 120kN = 240Ml'a A 500 mm 1

The corresponding deflection of point G IS 0e, = lL =

(240 MPa) 200 GPa (5 m) = 6mm

{Tel"

T L=

The corresponding deflection of point B is

3",111

0 8 , = h OA,

C, Q _!!,tIIJ.: N

+ oc,) =

t(3 mm + 6 mm ) =

4.5 mm

Since we must have 0 8 = 10 mm. we: conclude that plastic deformation will occur. Plastic Defor mation. For Q = 240 kN, plastic: deformation occurs in rod AD. where if,w = tT y = 300 Ml'a. Since the stress in rod CE is within the clastic range. () c remains equal to fi mm. The deflection (),l for which ()IJ = 10 mm IS obtained by writing ()e, =

c,

1

6m

A I ~_ ... ,~--'c~ J 3,,,,,, ,.' 1>0 ,\~

Q-

If

".

10 mm = ~( OA ,

+ 6 mm )

() A, =

14 mm

Unloading. A s force Q is slowly removed, the force P AD decreases along line Hi parallel to the initial ponion of the load-deflection diagram of rod AD. The final deflection of poinl 1\ is 0A. = 14mm - 3mm = II mm

Since the stress in rod CE remained within the elaslic range, we note that the final deflection of point G is zero.

11 5

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PROBLEMS

2.93 Knowing that P = 10 kips. determine the maximum stress when 0.50 in .. (b ) r = 0.625 in.

(a) r =

2.94 Knowing thaI. for the plate shown. the allowable stress is 16 ksi. detemline the maximum allowable value of P when (a) r = ~ in .. (b) r = t in. 2.95 For P = 8.5 kips. determine the minimum plate th ickness 1 required if the allowable stress is 18 ksi.

r,1 =

Fig. P2.93 and P2.94

'II =

~

+

in.

ill .

B 1.6 ill

p

Fig. P2.95

,.

2 .96 Knowing that the hole has a diameter of ~ in .. determine (a ) the radius of the fillet s for which the same max imum stre~s occur~ at the hole A and at the fillets. (b) the corresponding maxi mum allowable load P if the allowable stress is 15 ksi.

r,

:S ""

4 ill

Fig. P2.96

·\~tiU.

2 .97 A hole is to be drilled in the plate at A. The diameters of Ihe bits available to drill the hole range from 12 to 24 mm in 3· mm increments. ( a) Determine the diameter d of the largest bit that can be used if the allowable load at the hole is to exceed that at the fillets . (b) If the allowable stress in the plate is 145 M Pa. what is the corresponding allowable load P?

112.5",,,,

Fig. P2.97 and P2.98 2.98 (tl) For P = 58 kN and d = 12 mm. determine the maximum stress in the platt' shown. (b) Solve part a. assuming that the hole at A is not drilled.

116

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2.99 The aluminum test specimen shown is subjected to two equal and upposite centric a~ial forces of magnitude P. (a ) Knowing that £ = 70 GPa and { r .>ll = 200 MPa. determine the maximum allowable value uf P and the corresponding total elongation of the specimen. (b) Solve part a. assuming that the specimen has been replaced by an aluminum bar uf the same length and 3 unifonll 60 x IS-mm rectangular cross section.

Dh" r l>~i()n ~

Problems

,I> Illlll

Fig. P2.99

A

2 .100 For the test ~cimen of Prob. 2.99. determine the maximum value of the noml;11 s tress corresponding to a total elongation o f 0.75 rum.

c

2.101 Rod AB is made o f n mild steel tlmt is assumed to be ela.~topbs­ tic with £ = 29 x 1(1) psi nnd iff = 36 ksi. After the rod has been attached to the rigid lever CD. It is found that eml C is ~ in. too high. A I'ertical force Q is then applied at C unti l this poim has moved to position C. Determine the required magnitude of Q and the detlection /) , if the lever is to sl/ap back to a horizontal position after Q is removed. 2.102 is 50 ksl".

Solve Prob. 2. 101. assuming that the yield point of the mild steci

2.103 TIle 30-mm square bar AB has a length L = 2.2 m: it is made of a mild steelth;1t is assumed to be elastopl:tstic with E = 200 GPa and ifl" = 345 MPa. A force P is applied to the bar until end A has moved down by an amount /),.. Detennine the maximum value of Ihe force P and the permanent sel orlhe bM after the force has been removed. knowing that (l/) ,s., = 4.5 mm. (b)

if

___

'~J... i·~--

0'1tt ill

I

" '"

Fig. P2.101

r

B

L

8," = 8 mill.

2. 104 The 30-mm square bar AB hns n length L = 2.5 m: it is made of a mild steel thm is a.~sumed 10 be ela~toplastic with £ ~ 200 GPa and ify = 345 MPa. A forc~ P is applied to the bar and thell removed TO give it a permanelll set/)p' Determine the maximum value of the force P and the maximum amount /)", by which the bar should be sTretched if the desired value of /)1' is (a)

B

"'.+

1 ."

3.5 mm. (b) 6.5 mill.

P Fig. P2. 103 and P2. 104

J

D 0

117

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118

6!mM and

Sm.n-!\xlnl Londing

2.105 Rod AIlC cOIlsi5ls of IWO cyli ntJfi ..
I

.LO·",,,,

!.2m

t-

0.8

In

~

dialn<'t~r

3.lJ.
" P

Fi g. P2.1 05 and P2.106

2.106 Rod ABC cOlIsi.q~ of 1"''0 cylindrkal porliOIl~ AB and BC: il i, made a mild steel thill is assumed to be elastoplastk with £ = 200 G r a and lTy = 250 .W'a. A foft.""!! P is applied ill the rod until ilJ; ent! A h:L~ moved down by :Ill a.mount 6~, - 5 111m. Dttc'llline the ma.~imum va lu e of tilt fOlee P (Hid lhe pcnllaJlellt set of the rod after the force has moen removed.

or

r

"

Ill0 ,nl"

t-

1!lO"""

1

p

Fig. P2.107

D

E

1_

2 .107 Rod AIJ (Qn~lsts 01 two cyl..indrical portions A(." and Ill', each with a cro~~·ge,'lional ~rca of 1750 mm"!. ['onion .~C is nude of a mild steel with E = 200 Gl'a and tTl = 250 MPa. and poction en is made of a l,igh.stn:ngth steel with E = 200 OPa and fry = 345 M Pa. A load P is applied ilt C as shown A.-.~ullling buth s t.:eb lu btl el:lstupla ..... il:. tlelcrm ine (Il) the 1ll.",inlUlIi tlenel:' linn "f ,. if P i~ !'.,,,d,,,,l1 y i""r"~'f'(1 frnm ',,...,, I" 'I,e, .... N and Ih .. " r... du,·;'" back to Zl'fC·. (/» the maximum stre~!l" in each portion of the rod. (t·) the per· manelll denection of C.

2 .108 For the mmposite rod of Pmb. 2.107. if P is gradua ll y increased from zero until tlte deflecLion of point C reaches a maximum value of 0", = 0.03 mm :md then decreased back to zero. determine. fa ) the maximum v:llue of P, (h) the ma.>::imuJll ~tres~ in each portion of the rod. (d the perlll~tleni denection of C after the load is removed.

2 .109 Each cabk: Ius a Cl"O~s-scctiolial ~lca of Ino mm~ and is Itlade of an ela>lOptaslic m~terial fur which (J"I' = 345 MPa and E = 2[)() GPa. A force Q is applied at C to dle rigid bar ABC and i~ gradually increased from 0 \(J 5U kN :md then reduL""!!u 10 '1ero. K.nowlng th]t the cables were lIIilTally laut. de:effi1ine (a) Ihe ma.~imtlm Slress thaI ()ceur~ ill cabb IJD. (b) Ihe IroximulT. denection of paim C. (d the final displacement of point C. (Hir:l: In part c. cable CE is nm taut. )

Fig. P2. 109

2.110 .l\fll\'e Proh 2. 109, a<,lIIn in!! Ih~t Ih~ ca hl ..'1 .1fC rerl~cf'rl hy rf'l{h of th e smm: ~mss·section al area and material. Funher assume that the rods are brac<,d ,;0 thaI the~ can carry compressive forces.

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2.111 Two tem pered-steel bars. each k -in. thick. are bonded [0 a Hn. mild-s[eel bar. This co mposite bar is subjected as shown [0 a centric axial load of magnitude P. Both s[eels arc elastoplastic with E = 29 x 1O~ psi :U1d with yield streng ths equal to 100 ksi and 50 ksi. respedively. for th e tempered and mild sTeel. The lmd P is grad ually increased from zero unTillhe deformation of the bar reac hes a maxim um value 8," = 0 .04 in. and then decreased back to zero. Determ ine (a) the maximum va lue of P. (b) the maxi mu m stress in Ihe tempered-steel bars. (c) the permanenT se t after th e load is removed. 2 .112 For the composile bar of Prob. 2.1 I I, if P is gradual ly increased from Zero to 98 k!ps and then decreased back to zero. determine {a) Ihe maximum deform:uion of the bar. (b) Ihe maximum stress in Ihe tempered-sleel bars. {e) the permanent sel after Ihe load is re mo ved.

2 .113 The rigid bar ABC is supported by twO lin ks. AD and BE. of un iform 37.5 X 6-mm rectangular cross section and made of a mild s[eellhal is assumed [0 be elastoplaslic with E = 200 G l'a and er r = 250 M Pa. nle mag· niTude of Ihe force Q applied al B is graduall y increased frum zero 10 260 kN. Knowing Ihal a 0.640 m. dClenn ine (0) Ihe value o f Ihe nonnal stress in each link. (b) the maximum defleclion of point B.

=

Probtems I"

f

;.

ii;"'.

14 in .

l

21);..

I'

Fig. P2. 111

i -I- O

_" I

t. Tm

t",

c

°

"l

o

o

1-,--1 --- 2("~~ B

Fig. P2.11 3

2.114

Solve !'rob. 2.113. knowing Ihat (/ = 1.76 m and th aT the magnitude of the force Q applied at B is gradually increased frum zero 10 135 kN. "2.115 Solve Prob. 2. 11 3, assuming that the magnitude of Ihe force Q appJfed at B is gradually increased from zero [0 260 k'N and then decreased back 10 zero. Kn owing that a = 0.640 Ill. determine (a) Ihe residual stress in eac h link. (b) Ihe final deflection of point B. Assu me that The links are braced so that they can c:trry compressive forces wi lhoul buckling. 2 .11 6 A unifonn steel rod of cross-sectio nal area A is anac hed 10 rigid suppo rts and is unstressed at a temperalure o f 45° F. The STeel is ass umed 10 be elas tllplaslic WiTh i f )" = 36 ksi and E = 29 X 10" psi. Knowing thai {l' = 6.5 x 10 6;oF. del ermi ne The stress in the bar (0) when the Temperature is mised 10 320°F. (b) after Ihe temperature has relurned 10 45°F. A

1---- L - ---I Fig. P2.11 6

f,;in_

119

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120

Stress and Straln-Alolal Loading

2 .117 The sted rod ABC is nuached to rigid supports and is unstressed at a temperature of 15°C. The-steel is assUlned elasloplastic. with £ = 200 Gl'a and (r l · = 250 M I'u. The temperature of both portions of the rod is then mised to 150°C. Knowing that u = 1l.7 X 10 bFC. determine (a) the stress in both portions of the rod. (b) the dcnection uf point C.

A

=

500 mll\~

A'\

A= :moli\m~

C

'\

Fig. P2.117

(I

-

12Q""n

"2:.1 18 Solve Prob. 2.117. ass uming that the temperature of the- rod i.s raised to 150°C and then returned to 15°C.

I'

I

1----- .\.mllllll - - ! Fig. P2.119

' 2.119 Bar All has a cross·sectional area of 1200 mm 1 and is made of a steel that is ass umed to be elastoplastic with £ = 200 GPa and (f y = 250 MPa. Knowing that the force t' incre-ases from 0 to 520 kN and then decreases to zero. detennine la) the permanent denection of point C, (b) the res idual stress in the bar. '2.120

Solve Prob. 2.119.

a.~suming

thm a = 180 mm.

·2. 121 For the. composite. bar of Prob. l .llJ. determine the residual stresses in the tempered,steel bars if P is gradually increased from zero 10 98 kips and then decreased back to zero. '2. 122 For the composite bar in Prob. 2. 11l. determine the res idual stresses in the tempered,steel bars if P is gradually increased from zero until the drJonnation of the bar reaches a maximum va lue 0", = 0.04 in. and is then decreased back to zero. "2.12 3 Narrow bars of aluminum are bonded to the two sides of a thick steel plate as shown. Initially, at TI = 70°F. all stresses are zero. Know ing that the temperature will be slowly raised 10 7; and then reduced to r l • determine (a) the highest kmperature 1; that dOt:s lIai result ill rt:sidual stresses. (b) the temperature 7; that will result in a residual stress in the aluminum equal (0 58 ksi. Assume Cl'a = 12.8 X IO" ~/"F for the aluminum and u. = 6.5 x \O -~rF for the steel. Further assume that the alurninum is elastoplastic, wilh £ = 10.9 X 10~ psi and (1")" = 58 ksi. (Hillt: Neglect the small stresses in the plate.)

Fig. P2.123

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REVIEW AND SUMMARY FOR CHAPTER 2

This chapter was devoted to the introduction of the concept of sTrain, to the discussion of the relationship between stress and strain in various types of materials. and to the determination of the deformations of structural components unde r axial loadi ng. Considering a rod of length L and uniform cross section and denoting by 8 its deformation under an axial load P (Fig. 2.1), we defined the nonnal Jlrain € in the rod as the deformatio/l per IInit length [Sec. 2.21: €

=

8

Normal strain

T

B

(2.1)

~

L

L

I.n the case o f a rod of variable cross section. the normal strain was defined at any given point Q by considering a small element of rod at Q. Denoting by ~x the length of the element and by 6.8 its de-

fonnation under the given load, we wrote

,=

(2.2)

.... K"l'tun'

--------- -- ------_-_--~

I ".

i

:ndd: , : Str.iin-h:m l,'u;ng \) .02

:~",,~ I 0.2

1),0012 L"w.-carho" stf'd

(h )

I" ) Fig. 2.1

Stress-strain diagram

J~~ --------- -c-~---- --, K "l'tull'

1".

20

0.2.5

Fig. 2. 11

0.004 ((I )

c p

11 20

,

I

"

Plotting the stress u versus the strain E as the load increased. we obtained a xtre.u-.rtraill diagram for the material used {Sec. 2.31. From such a diagram, we were able to distinguish between iJrit/le and ductile mate rials: A specimen made of a brittle material ruptures without any noticeable priur change in the rate of elongation (Fig. 2.11). while a specimen made of a ductile material yield.r after a critical stress O'y, called the yield strength. has been reached, i.e., the specimen undergoes a large deformation berore rupturing, w ith a relatively small increase in the applied load (Fig. 2.9). An example or brittle material with difrerent propenies in tension and in compression was provided by concrete.

,w,

J

c

.1.8 dB lim = .,h--+O Llx dx

B

(h ) AI",,,,,,,,,,,

alloy

Fig. 2.9

121

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122

Stress and Straln-Alolal Loading

Hooke's law Modulus of elasticity

fT

= EE

(2.4)

This relation is known as Hooke's law and the coefticient E as the modulll.l· of elasticity of the materi al. The largest stress for which Eq. (2.4) applies is the proportional limit of the material. Materi als considered up to this point were imtropic, i.e., their properties were independent o f direction. In Sec. 2.5 we also considered a class o f anisotropic materials, i.e., materials whose properties depend upon direction. They were fiber-reinforced composite materials, made of fibers of a s trong. stiff material embedded in layers of a weaker, softer material (Fi g. 2. [7). We saw that different moduli o f elasticity had to be used. depending upon the direction of loadi ng.

Fig. 2.17 Elastic limit. Plastic deformation

R"ptn n'

A

We noted in Sec. 2.5 that the initial portion of the stress-strain diagram is a strai ght line. This mean.~ that for small deformations, the stress is directly proportional to the strain:

D

Fig.2.t8 Fatl~e .

Endurance limit

Elastic deformation under axial loading

T

B

B

If the strains caused in a test specimen by the application of a given load disappear when thc load is removed. the material is said to behave elastically, and the largest stress for which this occurs is called the ela.\·tic lilllit o f the material [Sec . 2.6]. If the clastic limit is exceeded . the stress and strain decrease in a linear fashion when the load is removed and the strain does no t return to zero (Fi g. 2.18), indicating that a permanent set or pla.l·tic de/ormation of the material has taken place. In Sec. 2.7. we discussed the phenomenon of fatigue, which causes the fai lure of structural or machine components after a very large number of repeated loadings, even though the stresses remain in the clastic range. A standard fatigue test consists in determining the number n of successive loading-and-unloading cycles required to cause the failure of a specimcn for any given maximum stress level fT, and plouing the resulting fT -n curve. The val ue o f fT for which failure does not occur, even for an indefinitely large number o f cycles, is known as the endurance limit of the material used in the test. Section 2.8 was devoted to the determination o f the elastic deformations of various types o f machine and structural components under vario us conditions o f axial loading. We saw that if a rod of length L and uniform cross section of area A is subjected at its end to a centric axial load P (Fig. 2.22). the corresponding deformation is

PL

8 ~ ­

AE

L

(2.7)

If the rod is loaded at several points or consists of several parts of various cross sections and possibly of different materials, the deformation {j of the rod must be expressed as the sum of the deformations of its component pans [Example 2.01 ]:

c

c

,\

8 p

Fig. 2.22

~

'" -P,L; ",, ; A,E;

(2.R)

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Review and Summary for Chapter 2

p

Statically indeterminate problems EnlJ plal<'

I---

L- --I

Fig.2.2Sa

H,

Section 2.9 was devoted to the solution of ,I"taticall), illt/efenni· nate problems, i.e.. problems in which the reactions and the internal forces can l/ot be determined from statics alone. The equilibrium equations deri ved from the free-body diagram of the member under consideration were complemented by relations involving deformations and obtained from the geometry of the problem. The forces in the rod and in the tube of Fig. 2.25a, for instance. were determined by observing, on one hand. that their sum is equal to P, and on the other, that the) cause equal deformations in the rod and in the tube [Example 2.02]. Similarly, the reactions at the supports of the bar of Fig. 2.26 could not be obtai ned from the free-body diagram of the bar alone rExample 2.031: but they cou ld be determined by expressing that the total elongation of the bar must be equal to zero. In Sec. 2.\0, we considered problems involving tell/pen/turf! changes. We first observed that if the temperature o f an Ill1restrained n)(1 AB of length L is increased by 6. T. its elongation is

ST = (1'(6. T) L

(h )

10)

Fig. 2.26

Problems with temperature changes

(2.21 )

where a is the coejJiciem of thermal e~pallSi(J1I of the material . We noted that the corresponding strain. called thermal straill, is (2.22) and that 110 stress is associated with this strain. However, if the rod AB is restrained by fixed supportS (Fig. 2.35a). stresses develop in

, A

'I

B

B

Fig. 2.35a

the rod as the temperature increases. because of the reactions at the supports. Todetermine the magnitude P of the reactions, we detaehed the rod from its support at B (Fig. 2.36) and considered separately the defonnation ST of the rod as it expands freely because of the temperature change, and the deformation 8p caused by the force P required to bring it back to its original length, so that it may be reattached to the support at B. Writing that the total deformation S = ST + Sp is equal to zero, we obtained an equation that could be solved for P. While the tinal strain in rod AB is clearly zero, this will generally not be the case for rods and bars consisting of elements o f different cross secti ons or matcrials, since the defonnations of the various elements will usually not be zero rExample 2.06].

I")

,

J"r

l "L

(h )

"

,-----J (,)

Fig. 2.36

r

123

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124

Stress and Straln-Alolal Loading

Fi g . 2.39a

Lateral strain. Poisson's ratio

When an axial load P is applied to a homogeneous, slender bar (Fig. 1.39a), it causes a strain, not only along the axis of the bar but in any transverse direction as well [Sec. 2.11]. Thi s strain is referred to as the lateral straill. and the ratio of the lateral strain over the axial strain is called Poil'son's ratio and is denoted by v (Greek letter nu). We wrote (2.25)

axial strain

Recalling that the axial strain in the bar is 1:':., = (Tex/ E. we expressed as follows the condition of strain under an axial loading in the x direction: I:': ,.= €~=

Multiaxialloading

E

(2 .27)

This result was extended in Sec. 2.12 to the case of a multiaxial loat/ing causing the state of stress shown in Fig. 2.42. The resulting strain condition was described by the following relations, referred to as the generalized Hooke's law for a multiaxial loading.

(2.28) "U_<

Fig. 2."2

1:': . =

".

--

E

I)U, ·

- -

E

U ~

+-

E

an element of material is subjected to the stresses U -" u .... u ,' it will deform and a certain change of volume will result [Sec. 2.131The change ill I'Olullle per ullit I'oillme is referred to as the dilatation of the material and is denoted b) e. We showed that [f

Dilatation

[ - 2v

' ~-E- (u,,-

When a material is subjected

to

e

+ u ) + uJ

a hyurostatic pressure p. we have

=_ I!. !

Bulk modulus

(2.31 )

(2.34)

where k is known as the bulk moduills of the material: ! ~

E

==-:C-c 3(1 2v)

(2.33)

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,

'.

Re\llew and Summary for Chapter 2

Fig. 2.45

Fig. 2.47

As we saw in Chap. I. the state of stress in a material under the most general loading condition involves shearing stresses. as well as normal stresses (Fig. 2.45). The shearing stresses tend to deform a c ubic element of material into an oblique parallelepiped ISec. 2.141. Consideri ng. for instance. the stresses Txr und 'Tv", shown in Fi g. 2.47 (which, we recall, arc equal in magnitude), we noted thaI they cause the angles ronne
'Y~"

we wro te

the relations T .;;<

= Oy",

(2.36, 37)

which are valid for any homogeneous isotropic material within its proportional limit in shear. The constant G is called the modulus of rigidity of the material and the relations obtained express Hooke's law for .vhearillg .wren and strain . Together with Eqs. (2 .28), they form a group of equations representing the generalized Hooke's law for a homogeneous isotropic material under the most general stress condition. We observed in Sec. 2.15 that while an axial load exerted on a slender bar produces only normal strains- both axial and transverseon an element of material oriented along the axis of the bar, it will produce both normal and shearing strains on an element rotated through 45° (Fig. 2.53). We also noted that the three constants E.

P'

I"

1+

Fig. 2.53

'"I

£ ,

(h)

Shearing strain. Modulus of rigidity

125

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126

Stress and Straln-Alo lal Load ing

v, and G are not independent; they s
I + 1'

~=

2G

Flber-remforced composite materials

Saint-Venant's principle

Stress concentrations

(2.43)

which may be used to determine any of the three constants in terms of the other two. Stress-strain relationships for liber-reinfo rced composite materials were discussed in an optional section (Sec. 2.16). Equ
(2 .4R)

(Taw

Plastic deformations

Values of K fo r circular holes and fillets in nat bars were given in Fig. 2.64 on p. lOS. In Sec. 2 .19, we discussed the plastic defol1l/(/fioll.l· which occur in structural members made of
}

+0'-_

c· _ _ _

,/

/ ,,

Rupture

, ,,

, A

D

Fig, 2.65

rExamples 2. 13, 2 .14, and 2.15]. Finall y, in Sec, 2 .20, we observed that when an indetenninate structure undergoes plastic deformations, the stresses do not, in general, return to zero after the load has been re moved. The stresses remaining in the various parts of the structure are called residual stresses and may be detern1ined by adding the maximum stresses reached during the loading phase and the reverse stresses corresponding to the unlo
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REVIEW PROBLEMS

2.124 The aluminum rod ABC (£ = 10. 1 X [00 psi). which consists of cylindrical ponions AB and BC, is \0 be repklced with a cylindrical steel rod DE t£ = 29 X [0" psi) of the same overull length. Determine the minimum required dimllcler d of the steel rod if its vertical deformation is not \0 eXI.."ecd the deformation of the aluminum roo under the same load and if the allowable siress in the steel rod is not 10 exceed 24 ksi.

2t; kil'~

2fi kips

(WO

2.125

The brass strip AB has been attached to a fixed support at A and re SL~ on a rough support at B. Knowing thai the coefficienT of fricti on is 0.60 between the strip and the support at B. detennine the decrease in temperature for which slipping will impend.

I

r

I2 in. -

D

" -

I

L5 in.

B

- d

- 2.25 i".

IS iu. Bmss :

E 0.

= m~ (; p" = 20 x lO -~fC

JOI}

kg

i

c

E

Fig. P2.1 24

Fig. P2.12S

2. 126 Two solid cylindrical rods are joined al B and loaded as shown. Rod AB is made of steel ( £ = 29 X lOb psi), and rod Be of brass (£ = 15 X IO~ psi). Determine (a) th e lOtal deformation of the composite rod ABC, (b) the deflection of poil1l B.

2 .127 Link BD is mad e of brass (£ = IS x lOb psi) and has a crosssectional area of 0.40 in!. Link C£ is made of aluminum (£ = lOA x 106 psi) and has a <..-'foss·sectional Mea of 0.50 inl. Determine the ma:Jmum furce P that can be applied venically at point A if the deflection of A is IItlt to exceed 0.014 in. D

1

9.0,".

~~c"Ll OB

p

U -

.'i.Oill.

-fl .O

0

= -to kip. Fig. P2. 126

oU in. -----I

Fig. P2.127

127

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128

Stress and Straln-Alolal Loading

;'10 "un

2 . 128 A l50.mm. long aluminum tube (£ "" 70 G Pa) of 36-mm outer uiameter and 28-mm inner diameter may be dosed at both ends by means of single-threaded screw-on cOllers of I .S-mm pitch. With one cover screwed on tight. a solid brass rod (£ = 105 GPa) of 25-rnm diameter is placed inside the lube and the second cover is screwed on. Since the rod is slightly longer than the tube. it is obserlleu that the cOlier must be forcL'(\ against the rod by rotating it one-quaner of a turn before it can be tightl y closed. Determine «(I) the average nomlal stress in the tube
21> "un

FIg. P2.12B

2 .129 nle unifonn wire ABC, of unstretched length 2/. is all
8=/ i"P"

VAi

2 .130 nlc rigid bar AD is supported by two s teel wires of t6· in. diameter (£ = 29 X ](ji psi) and a pin and bracket at A. Knowing that the wires were initially taut. determine (a) the additional tension in each wire when a 220-1b load P is applied a1 D. (b) the corresponding uellection of point D.

r F1g. P2.t29

1 l.,s",

,

I> in.

.~.

I

J

~'E

10 in .

.L.c-A -- " o

c

.•

,

F1g. P2.130

Fig. P2.13 t

2.131 The concrete post (Ee = 25 GPa and IT, = 9.9 x 10 6rC) is reinforced with six steel bars. each of 22-mm diameter (£, = 200 G Pa and cr, = 11.7 X 1O ~6rC). Determine the normal stresses induced in the steel and in the concrete by a temperature rise of 35°C. 2..132 A lIibration isolation unit consists of two blocks of hard rubber w ith a modulus of rigidity G = 2.15 ksi bonded to a plateAB and to rigid sup· pons ;JS shown. Denoting by P the magnitude of the force app lied to the plate and by 0 the corresponding deflection, determine the effective spring, constant. k = PIS. of the system. 2.133 Knowing that 11'.11 "" 120 MPa. determine the maximum allowab le lIalue of the centric axial load P. t5,,,,,,

Fig. P2.132 Fig. P2.133

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2.134 Ru:l AS cOilsistsoftwo cylllldrical poniollsACanJBC e .. ~h with a cross-iec tional ~rea of 2950 mm'. Ponion AC is made of a mild steel with £ "" 200 GPa and lIy = 250 MPa. and ponion eB is made of a high-strength steel with E = 2lMJ UPa and tT r = 345 MPa. A load P is apphed at C as shown. Assum ing both ~ 1{! .. ls to be elastoplaslit. deterr.1inll (:.irnum denection of C if P ill" gradually increased from 2ero to 1625 kN and Illen reduced bad 10 zero. (b illie pennanent deflection of C

2.135 Th~ IHlifnrm r()it flr h~~ ~ cr'm~_<;/!~ ri nn " l ~re:l A and 1<;' made nf a mild steel which can be a.<sumed to be c[;;slOplastic ",it11 a modulus of elasticity £ and a yield strength if,. Using the block-and-spring sys tem shown. it i~ dC3ircd to simulate the. deflection Df end C of the rod as the IIxinl force P is waduall y applied ~n~ re:noved. that is, the dellection of points C and C' ,hould be the &ame for all values of P. Deaotiag by p. the coefficient of friction between the block and the horizontal sulface, derive :m expression for (II) th e ~4uired mOli S m of rhe \'tlod:, (b; Ihe requir<,d ~ons!:.mi k of rhe ~pritlg.

rB

L

Ie

CompUU>r Problem.

32'1

"Ull

320)

lOll"

t-

1

'

Fig, P2.134

p

, , , ,

IJ'

Ie

~

Ie

p

Fig. P2.135

COMPUTER PROBLEMS

Tbe followin g problems are designed to be s oln!d with:] comput('r. \ Vrite

each program so lhal it ('an Ix! us('d ",illt

~ilhtr

SI or US cuslomary unils

and in :o.uch a \fay thai solid cylindrical clcmt'nts may bc defincd by eilht'r thei r diameter ur tlldr uUSIj -!;I.'CliulUll an:a.

2.Cl A rod consisting oi /I elemen ts. e~ch of which is homogeneous and 01 unitorm cro;;s section. is subjected 1.0 the JOadillg !hown. The length of elemenl ; is denoted by L,. its cross-sectional area by A,. modulus of e laslicity b) E,. and the load applied to its right end by P;, the magnitude P J of thi s load bemg assumed to be posi tive if P; is directed to the right .. nd negalive Olher'Wise. (a) Write a compUler program Ihal can be used \0 delermine rh e aven~e nnrrnal SIre,s in each elemenT. The def()rmarion of e~ch element . and the lOtn l deformation of the rod. tb) Use this program to solve Probs. 2.20 nnd 2. t 26.

EI~mcut

Po<

Fig, P2.C1

II

,

Elell,eo:t t

129

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130

Stress and Straln-Alolal Loading

EIc"wnt 1

\

B

El!'mrnt I

_1<\'

Fig. P2.C2

Eli'In\"t

n

~ A ~B[ Fig. P2.C3

Fig. P2.C4

2.C2 Rod AB is horizontal with both ends fixed: it consists of /I el<"ments. each of which is homogeneous and ()f uniform cross section, and is sub· jecteil to the loading shown . The length of element i is denoted by L,. it.> cross-sectional area by A" its modulus of elasticit y by E,. and the load applied to its right end by P i' the magnitude P, of {his load being assumed 10 be positive if PI is directed to the right and negmive otherwise. (Note that P I = n.) (a) Write a computer program that can be used to determine the reactions at A and B, the average nomml stress in eitch element, ,md the deformation of ead! element. (b) Use th is program to solve l:>robs. 2.41 and 2.42. 2 .C3 Rod AB consists of IJ elements. eac h of which is homogeneous and of uniform cro~~ section. End A is fixed , while initially there i ~ a gap 00 between end JJ and the fixed vertical ~urface on the right. The length of element i is denoted by L,. its cross-sectional area by A ~ its modulus of elasticity by E;. and its coefficient of thermal exp,msion by (r;. After the temperalllre of the rod ha~ been increased by dT. the gap at B js closed and the vertical s urfaces exert equal and opposite forces on the rod. (a ) Write a computer progrtlm that can be used to detenlline the magnitude of the reactions at A and B. the nor· mal stress in each element. and the deformation of each element. (b) Use this program to solve Probs. 2.51. 2. 59. itnd 2.60. 2. C4 B:lr AB has a length L and is made of two different materials of given cross-seclional area, modulus of elasticiTY. and yield strength. The bar is su bjected as shown 10 a load P that is graduall y increased from zero until the deformation of die bar has reached II maximum valu e 0", and then decreased back to zero. (a) Write a co mputer progr..l.nl that. for each of 25 values of 0", eq uall y spaced over a range extending from 0 to a value equal to [20% of the deformation causing both nwteria[s to yield. can be used to determine the mallim um value p ,. of the load. the maximum normal stress in each material. the pemlanellt deformation Sp of the bar. and the residual stress in each material. (b) Usc this program to solve Probs_ 2. 1 [I lmd 2.1 [2.

r" Fig. P2.C5

2 .C5 TIle plate has a hole centered across the w idth. TIle stress concentr::ttion factor for a nat bar under a:;:ial loading with a centric hole is: K ;- 3.00 - 3.[3

t' + (2')' D - (2')' D ("") 3.66

l.53

where r is the radius of the hole and D is the width of the bar. Write a computer program to determine the allowable load P for the given values of r. D. the thickness I of the bar. arId the allowitble streSS if,,11 of the material. Knowing that I = J in., D = 3.0 in. and ifaU = 16 ksi .. detcnnine lhe allowable load P for Iflllues of r from 0. 125 in. to 0.75 in .. usin g 0. 125 in. incremellts.

Fig. P2.C6

2 .C6 A solid tnmcated cone is subjected to an ax ial force P as shown. The exact elongation is (PLY( 2rrc l £) . By replacing the cone by /I circular cylinders of equal thickness. write a computer program that can be used to calculate the elongation o f the truncated cone. What is the percentage error in the answer obtained from the program using (a) /I = 6, (b) /I = [2. (c)'l = 60?

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C

HAP

T

E

R

Torsion

This chapter is devoted to the study of torsion and of the stresses and deformations it causes. In the jet engine shown here, the central shaft links Ihe components of the engine to develop Ihe thrust that propels the plane.

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132

3.1. INTRODUCTION

Torsion

[n the two preceding chapters you studied how to calculate the stresses and strains in stmctural members subjected 10 axial loads. that is, to forces directed along the axis o f the member. In this chapter structural members and machine parts that are in torsio/l will be considered . More specificall y. you will analyze the stresses and strains in members of circular cross section subjected to twisting couples. o[ torques. T and T' (Fig. 3. 1). These couples have a common magnitude T, and opposite senses. They are: vector quantities and can be represented either by curved arrows as in Fig. 3. la , or by couple vectors as in Fig. 3. 1b.

(II)

-AC"' Q'"

Fi g. 3.1

Members in torsion are encountered in many engineeri ng applications. TIle most common application is provided by tramlllis.I·lon .1·ha/t.I·. which are used to transmit power from one point to another. For example, the shaft shown in Fi g. 3.2 is used to transmit power from the engine to the rear wheels of an automobile. These shafts can be either solid, as shown in Fig. 3. [, or hollow.

Fig. 3.2 In the automotive power tra in shown, the shafltransmits power from the engine to the rear wheels.

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3.1. Introduction

('",nerolor

Iii/ Fig. 3.3

Consider the system ~ hown in Fi g. 3.311. whic h consists of a steam turbine A and an e lectric generator B connected by a tralL~mi ss i o n shaft AB. By breaking the system int o its three component pa rt ~ (Fig. 3.31». you clln see Ihat Ihe turb me exerts n tw isting couple or torque T 011 the sh3ft and that thc shaft C(eflS an equal torque on the generator. The genermor reacts by exening the equal and o pposite to rque T ' on the shaft, and the shaft by ex.erling the torquc T' on the turbine. YOIi will fi rst analyze the stresses and deformations thaI take place in circu lar shafts. In Sec. 3.3. an imponant property of circular shafts is demonstrated: Whell {/ circular ~'I/{!ft i.~ subjected to torsioll, every cron sectioll remaill.l· plO'I /! and IIIlllistorted. Ln other words. while the I'ari olls CI"O~S sect io n s along the shaft ml
133

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134

Torsion

Considering defomlations in the elas/ic range and using Hooke's law for sheari ng stress and strain. you will determine the distriiJutifm of shearing .vfreSl·es in a ci rcular shaft arld derive the elastic torsioll formula.l· (Sec . 3.4). In Sec. 3.5. you will learn how to lind the angle of nvist of a circular shaft subjected to a given torque. assuming again elastic dcformaLions. The solution of problems im'olving statically illdnermin(l/e .1'lwfts is considered in Sec. 3.6. In Sec. 3.7, you will study the design o/rmmmissio/l .l'hqfts. I.n 0(der to accomplish the design. you wi lJ learn to determine thl:: required physical churacteristics of a shuft in terms of its speed of rotation and the power to be transmitted. The torsion formulas cannot bI! used to determine stresses ncar sections where the loading coup les are applied or ncar a section where an abrupt change in the diameter of the shaft occurs. Moreover. these formulas apply ollly within the clastic range of the material. In Sec. 3.8, you will learn how to account for stress concentrations where lUi abrupt change in diameter of the shaft occurs. In Secs. 3.9 to 3. 11. you wi ll consider stresses and deformations in circular shaft s made of a ductile material when the yield point of the material is exceeded. You wi ll then learn how to deICrmine the permanent plastic de/orll/lItirms and I'f!~'itllllli strnses that remain in a shaft after it has been loaded beyond the yield point of the material. In the last sections of this chapter. you wi.ll slUdy the torsion of noncircu lar members (Sec. 3. I 2) and analyze the distribution ()f stresses in thin-walled hollow noncirclliar shafl~ (Sec. 3 .(3).

3.2. PRELIMINARY DISCUSSION OF THE STRESSES IN A SHAFT

Considering a shaft AB subjected at A and B to eqllal and opposite torqlles T and T' . we pass a section perpendicular to the axis of the shaft through some arbitrary point C (Fig. 3.4). The free-body diagram of the portion BC of the shaft mllst include the elementary shearing forces dF, perpendicll iar to the radius of the shaft, that portion AC ex-

Fig. 3.4

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erts on Be as the shaft is twisted (Fig. 3.5{j). But the conditions of equ ilibrium for Be require that the syste m of these ele mental)' forces be equivalent to an illtemaltof(1ue T, equal and opposite to T' (Fig. 3.5h ). Denoting by p the perpendic ular distance from the force dF to the axi s of the shaft. and express ing [hat the sum of the moments of the shearing forces dF about the axis of the shaft is equal in magnitude to the torque T. we write

3.2. Discussio n 01 Stresses In a Shalt

dF

fpdF = T

or, since tfF = area dA,

T

tfA, where

T

is the shearing stress on the clement of (3 .1 )

Whi le the relation obtained expresses an important condition that must be satisfied by the shearing stresses in any given cross section of the shaft, it docs not tell us how these stresses arc distribmed in the cross section. We thus observe, as we already did in Sec. 1.5, that the actual distribut ion of stresses under a given load is statically illtfeter/IIin lltl'. i.e., thi s di stributi on co. l!llOf be de terminClI hy the method.l· oj .\"to.tics. However, having ass umed in Sec. 1.5 that the normal stresses produced by an axial centric load were uniformly distributed, we found later (Sec . 2.17) that this assumption was j ustilied. except in the neighborhood of concentmted loads. A similar assumption with respect to the

" ', r , "

\

II

)-l...J_ A~L' or . IHlft

....... - .

Fig. 3.6

di stribution of shearing stresses in an dastic shaft WO IlItI he wrong. We must withhold any judgment regarding the distribut ion of stresses in a shaft until we have analyzed the dejorll/lltirm s that are produced in thc shaft. This will be done in the next section. One more observation should be made at this point. As was indicated in See. 1. 12, shear cunnOl take place in one plane only. Consider the very small element of shaft shown in Fig. 3.6 . We know that the torque applied to the shaft produces shearing stresses T on the faces perpendicular to the axis of the shaft . But the c{md iti ons of equilibrium di scussed in Sec. 1.1 2 requ ire the existence of equal stresses on thc faces formed by the two planes containing the axis of lhe shaft. That such shearing stresses actually occur in torsion can be demonstrated.

Fig. 3.5

135

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136

To."lon

Fig , 3.7

by considering u "shaft'· made of separate slats pinned at bOlh ends to disks as shown in Fi g. 3.70. If marki ngs have been painted o n two adjoining sluts. it is obscn cd that the slalS slide with rc~pt:ct to each other when eq ual and opJXlsitc to rq ues are applied to the ends of the ··shaft .. (Fig. 3.7b). While sliding will nOi actually take place in a shaft made of a homogeneolls and cohesive material, the tendency for diding will exist. showing that stresses occur on lo ngitudinal planes as well as on planes perpendicular to the ax is of the shaft.t

3.3. DEFORMATIONS IN A CIRCULAR SHAFT

Fi g . 3.B

'" ,,,

~ ..

,.

Consider a circular ~haft that is atmched to a fixed ~up po rt at one end (rig. 3.&1). U a torqu e 'J" is ap plied to the other end, the shall WIll twist. with its frcc cnd rotating th ro ugh an angle ¢ callcd tht: a llgle: of twi.\·1 (Fig. 1.Rn) . Ohservation shows Ihm, wilhi n :l ccrr,lin ra nee of val ll c.~ of T, the angle of [wist cp is pro portional to T. It also s hows that ¢ is proportiooal to Ihe length L of the shaft. In other words. the angle of twist for a shai'!. ot" the same. material and same cross section, but twice as long. will be twice l S lMge under the same torque T. One p llrpOS~ of o ur analysis will be to lind the specific re lation ex j ~ (in z amon2 dJ. L. and T: another purpose will be to de-ten ni ne the distri bLltion of shearing stresses in the shaft, which we were unable to obtain in the pre(;t:liing sl.."t: ti un on the basis of stali{;s alune . AI this point, lUl important pro pert y of circu lar shaft s should be noted : When a circular shaft is subjected to torsion. <"I'CI), ero.I·.\" sectioll remains plane and IIlldistorted. In other words. whi le the va.rious cross sections along the shaft rotate thro ugh dilTcrcnt amounts. each cross section rotr.ICS as a solid rigtd slab. This is illustrated 111 Pig. J .9£1, which ~ h ows the defonnalions in a ru bber model ~u bjccloo to tors ion. The propcny we arc discussi ng is characteri stic of c ircular Shaft s, whether solid or hollow; it is nOI e.njoyed by membe r~ of nonc ircular cross section. Fo r exampl e, when a bar o f square cross section is subjected to lorsion. ils various cross sections warp and do nOI remain plane (Fig. 3.9b).

"

If,)

Fi g . 3 .9

tThe twisting of a canboard lute thiJI hao; Ix:cn ~tntion

stil1cngth\\i~c.

ITO\"[des


a.'oth~r

dcnlOn·

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TIle CIOS~ sec ti o ll ~ ofa dn;ulul' ~h aft IC1l 1iti11 [.llallC aud UlIuistOl tt:d

because 1I circ ula r shaft is oxil'J·mmelric. i.c .. its appe arance remains the same when it is viewed from a fixed position and JOtflted about its axis through an arbitrary angle. (Square bars. on the other hand , retain the. same appearance only j f they are rotated thro ugh 90" or 1800.) As .....c will sec presentl y. the:: axisymmc try of circular $halh may be used to plOVC theoreticall y that their cross sections remai n plane and undistortcd. Consider the point~ Cand D located o n the circumference of a given cross section of the shaft. and le\ C' ami D ' be the positiOns they will occupy after the shaft has bee n twisted (Fig. 3.10a). T he axisymmetry of the ~ h aft and of the loadi ng require~ that the rotation v.- hich wou ld have brought D into C should now bring D' into C'. T hus C' and D' must lie on the ci rcumfe.rence of a circle, and the urc C' D' mus t be equal to tbe an: CD (Fi g. 3.lOb). We will !lOW exami!le whc Lhcr the cirdc on which C' and D' lie is diffe rent from the o riginal circ le. Lt't us ;tssume that C' and 0 ' do lie on a di frerent circle and that the new circle is located to the left of the original circle, as shown ill Fi g. 3. lOb. The same SIIUitllOn wtll prevail lor any other cross sectIOn, Sillce all the cross sections of th~ shaft are subjected to the same internal torque T. imd
i~ tw i~ led.

BUl jf limt well:

~U, iI

givl: lI L1iallielCI uf lilt:

3 .3. Oelorm&llon. In a ClrelJlar Shlll1

B

,K

D' •

Jj:

"D

C' . {

' C

1"1 B

,K c:r:u,~ (/,)

Flg. 3.tO

Fig. 3.11

l: I Ulo~

section wO Llld be distorted into a curve whic h might look as shown in Fig. 3. [2:(1. An ob~eT\'er looking at this curve from A would conclude that the oUler [ayers or the shaft gel more twisted than the inner ones, while an observer looki ng from fJ would reach theopposlle conclusion (Fig. 3. l 2b). Thi s inco nSistency leads us to concl ude that a ny diamcte r ,)1' II given cross ~ect ion remllin~ ~tr:li g ht (Fig. 1. 121") line!, rh e refore, that allY given cross section I)f a circular ~ haft remams plane aDd undis toned.

T~

~,' T'~""'" ,

___ r,;ot ...

, "JI"'''

I" Fig. 3.1 2

137

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138

Torsion

Fig. 3.13

Our discussion so far has ignored the mode of application of the twisting couples T and 'I'. If all sections of the shaft, from one end t t) the other. are to remain plane ,md undi.mm ed, we must make sure that the couples are applied in such a way that the ends of the shaft themselves remain plane and undi storted. This may be accomplished by applying the couples T and T' to rigid plates. which an! solidly atlached to the ends of the shaft (Fig. 3.1 3a). We ean then be sure that all sections will re main plane and undistorted when the Itlading is applied, and that the resulting de formation s will occur in a unifornl fashioll throughout the e ntire length of the shaft. All of the L"tlually spaced circles shown in Fig. 3. 1311 will rotate by the same amount relati ve to their neighbors, and eac h tlf the straight lines will be transfomled into a cu rve (helix) intersecting the vari ous circles at the same angle (Fi g. 3.13b). The derivations given ill thi s and the fo ll owing sections will be based on the assumption of rigid end plates. Loading condititlns e ncountered in practice may diffe r appreciably from those correspondi ng to the model of Fi g. 3.13. The chief merit of this model is that it helps us de line a torsion problem for which we can o btain an exact solution. just as the rigid-end-plates model of Sec. 2.17 made it possible for us to define an axial-load problem which could be easily and accurately solved. By virtue of Saint-Venallt's principle, the results obtained fo r our ideali zed model may be extended 10 most engineering applications. However. we should kee p these results associated ill our mind with the specifil' model shown in Fig. 3.13. We will now determine the distribution of .I·hearing .I·trains in a circular shaft of length L and radius c which has been twisted through an angle ¢ (Fig. 3. t4a). Detaching from the shaft a cylinder of radius p, we cOllsider the small square element formed by two adjacent c ircles and two adjacent strai ght lines traced on the surface of the cyli nder befo re any load is applied (Fig. 3. 14b). As the shaft is subjected to a tOfsionalload. the element deforms into a rhombus (Fig. 3.14d. We now recall from Sec. 2.14 that the .~ hearing strain y in a give n element is measured by the change in the angles formed by the sides of that element. Since the circles de lining Iwo of the sides of the element considered here remain unchanged, the shearing strain y must be equal 10 the angle between lines AB and A ' B. (We recall that y should be expressed in radians.) We observe from Fig. 3. 14c that , for small values of y. we can express the arc length AA I as AA I = L y. But, on the other hand , we have AA ' = p¢. It follows that Ly = PrP, or p¢

y = -

F1g.3.14

L

(3 .2)

where y and ¢ are both expressed in radi ans. The equation obtained shows. as we could have anticipated. that the shearing strain y at a given point of a shaft in torsion is proportional to the angle of twist cPo It also shows that y is proponio nallO the distance p from the axis of the shaft 10 the point under considermion. Thus. the shearing .\·train in a cilrular .rllaf t )'(Iries linearly \IIitll the dis/{lIIce from the axi.\· of Th e shaft.

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the

3.4. Stresses In the Elastic Range

It follows from Eg. (3.2) that the sheari ng strain is maximum on o f the. shaft, where p "" c. We have

s url~1ce

cd,

(3.3)

Eliminating !/J from Eqs. (3.2) and (3.3) , we can express the shearing strain), at a distance p rrom the ax.is of the shaft as (3.4)

3.4. STRESSES IN THE ELASTlC RANGE

No particular stress-strain relationship has been assumed so far in our di scllssion or circular shafts in torsion . Let us now consider the case when the torque T is such that all shearing stresses in the shaft remain below the yield strength T y. We know from Chap. 2: that. for all pn.tctical purposes, this means that the stresses in the shaft wi ll remain below the proportional limit and below the clastic limit as well. Thus, Hooke's law will apply and there will be no permanent deformation. Recalling Hooke's law for shearing stress and strain from Sec. 2. 14. we write 7"

G-y

=

(3.5)

where G is the. modulus of rigidity or shear modulus of thc material. Mul tiplying both members of Eg. (3.4) by G, we write.

p

,", or, making usc of Eq. (3,5), r, .,•• T

=

P

~1",,,,,

.,.""., (3.6)

The equation obtained shows that, as long as the yield strength (or proportionallimit) is not ex.ceeded in any part of a circular shaft , fhe .I·hearin!; .flre!i.\" in fhe shaft varie.\" linearly with the distance p from the axis ofrhe shaff. Fi gllre 3.15a shows the stress distribution in a solid circular shaft of radius c, and Fig. 3. 15b in a hollow circular shaft of inner mdius c i and outer radius c 2 . From Eq. (3.6), we find that, in the latter case. C, TmUl ="G T m:t.\

(3.7)

(h)

Flg. 3. 15

139

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140

Torsion

.,.-""

~ ".,

(,, )

Ag. 3.15 (repeated)

We now recall from Sec. 3.2 that the sum o f the moments o f the elementary forces exerted on ,my cross section of the shaft must be equal to Ihe magnitude T of the torque exerted on the shaft: Jp( TdA)~ T

Substituting for

7

(3 . 1)

from (3 .6) into (3 .0. we write

But the integral in Ihe last member represents the polar moment o f inert ia J of Ihe cross section with respect to its center O. We have there fore (3.8) or, solving for 'T moX' Tc

'Tm .. ""

J

(3 .9)

Substituting for 7 ma.t from (3 .9) into (3 .6), we express the shearing stress 3t any distance p from the axis o f the sh3ft as

Tp

7 =-

J

(3 .10)

Equations 0 .9) and (3 .1 0) are know n as the elastic to rsionjimllllllls. We recall fro m statics that the polar moment of inert ia of a circle of radius c is J = 1 1TC~. I.n the case of a holl ow circular shuft of inner radius (."1 and outer rad ius c~, the polar moment ()f inert ia is

(3 ." ) We note that, if 51 metric units are lIsed in Eq. (3.9) or (3. 10). T will be expressed in N . 111, C or p in meters. and J in m.j; we check that the resulting shearing stress will be expressed in N/ ml, that is, pascals (Pa). If U.S. customary units arc used, T should be expressed in Ill · i ll. , c or p in inches, and J in in4, with the resulting shearing stress expressed in psi.

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EXAMPLE 3 .01 A hollow cylindrical steel shaft is 1.5 m long and has inner Rnd outer diameters respectively equal to 40 and 60 mm (Fig . 3. 16). (a ) Wh,\1 is the largestlOrquethat can be applied to the shaft if the shearing s tress is not to exceed 120 MP a? (b ) What is the corresponding minimum value of the shearing stress in the shaft?

.~ .j GO"""

' '''.........I..:~

(3.12) Recalling Ihat the polar moment of ineMia 1 of the cross sec· tion is given by Eq. (3.11). where (;1 =0 H40 mm) = 0.02 m and c~ = nun ) = 0.03 m, we write

WlO

Subslilllting for j and = 0.03 m. we havt!

into (3.12). and lelling

C

= c1

r = 1 _"_.,,_, ~ ~(I~.O~2~I_X---,I~O~·'~m ;::;.")~(l~2~O_X---,I~ O '_r~",,) ('

·10",,,,

",.. ~

T ,,,,,,,

0.03 111

= 4.08kN·1II

Fig. 3.16

(b) Minimum Shearing Stress. The minimum value

(a) Largest Permissible Torque. The largest torque T that can be applied to Ihe shaft is the torque for which = 120 MPa. Since this value is less [han the yield strength tbr steel. we can use Eq. (3.Y). Solving this equation tor T. we

of the shearing str~ss occurs on the. inner surface of the shaft. It is obtained from Eq. (3.7). wh ich expresses !hat l' mul and Tn...

are respectively plOportionalto

1'""",

have

c,

T""o = -Tn ..... =

The IOr~ion fOmlUI[L~ (3.9) and (3.10) were derived for 3 sh3ft of uniform circular cross section subjected to torques at its e nds. However, they can also be used for a shaft of variable cross scction or fo r a sh3ft subjected to torques at loc3tions other than its ends (Fig. 3.17a). The distribution of she3ring stresses in a given cross section S of the shaft is obtained from Eq. (3.9). where J denotes the polar moment of inertia of th31 section. and where T represents the intemal torque in that sect ion. The value of T is obtained by drawing the free-body diagram of thc portion of shaft located on one side of the section (Fig. 3. 17h) and writing that the sum of the torques applied to that portion. including the internal torque T, is lero (sec S3mpk Prob. 3.1). Up 10 this pOint. our analysis of stresses in a shaft has been limited 10 shearing ,tresses. Thi s is due to the factth3t the element we had selected W3S oriented in such a way that its f3ees were either parallel or perpendicular to the axis of the shaft (Fig. 3.6). We know from earlier discussions (Sccs. 1.11 and 1.12) that normal stresses. shearing stresses. or a combination of both m3Y be found under the same loading condition. depending upon the orientation of the element which has

C1

(" I

and

("1:

0.02 III -003 ( llOMPa) = 80MPa ..

m

101

,bI Fig. 3.17

141

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142

Tors!on

T I?~--------T-~ - '''~ i--J?f:; T

10 1 ~

Fig. 3.18

been chosen. Consider the two clements {/ and h located 011 the surface of a circular shaft subjected 10 tursion (Fig. 3.18). Since the fac es of element (j are respectively panlUci and perpendicular to the axis of the shaft, the only slrcsst:s on the clement will be the sheari ng stresses delined by formul a (3.9), namely T ma~ == Tel). On the other hand, the L1ces o f element b. which form arbitrary angles with the axis of the shaft. wi ll be subjected [0 a combination of normal and shearing stresses. Let LIS consider the particu lar case of an clement c (not shown) at 45° to the axis o f the shaft. In order 10 dete rmine the stresses on the fac es of this clement. we consider the two triangu lar eleme!1ls shown in Fig. 3.19 and draw their free-body diagrams. In the case o f the e leme!1l of Fig. 3.19a, we know that the stresses cxened 0 11 the faces BC and ED arc the shearing stresses 7 n .." ::= Tc/ .!. The magnitude of the corresponding shearing forces is thus 7 ", •.,A ~. where Au denotes the area of the faec. Observing that the compone!1ls along DC of the IWO shearing force s arc equ:d and opposite. we conclude that the force F exen cd on DC must be perpendicular 10 that face. II is a te nsile force, and its magnitude is

Fig. 3.19

(3. [ 3)

The corresponding stress is obtained by di viding the force F by the area A o f face DC. Observing that A = Ao we write

v2,

F

TnuxAO\/2

A

AoV2

a==~ =

_r:

7 ",,,-,

(3 .14)

A similar anal ysis of the cle ment o f Fi g. 3. 19/J shows that the stress on the face BE is a = - T m..,. We conclude that the stresses exened on the faces of an e le ment c at 4Y to the axis of the shaJi (Fi g. 3.20) am nOTmal stresses equal to ::!:T rna.. Thus, while the element (/ in Fi g. 3.20 is in pure shear. the ele ment c in the same fi gure is subjected to a te nsile stress on two of its faces, and to a compressive stress on the other two . We also note that all the stresses involved have the same magnitude. Te/.!. t As yo u learned in Sec. 2.3, ductile materials generally fail in shear. Therefore, when subjected 10 IOrsion, a specimen .I made of a ductile material breaks along a plane perpendicular 10 its longitudinal axis (Fig. 3.2Ia). On the other hand , britlle materials arc weaker in tensio n than in shear. Thus, when subjected 10 torsiun, a specimen made of a brittle material tends to break along surfaces which are perpendicular 10 the direction in which tension is maximum. i.c., along surfaces fonning a 45 0 angle with the lo ngitudinal axis of the specimen (Fig. 3.2 1/).

Ag.3.20

T II' )

Fig. 3.21 t Stressc£ on ckm~n1S of arbitrary orie ntation. such as clement b of fig. J . J8. will be discussed in Chap. 7.

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SAMPLE PROBLEM 3.1 IL9 III

Shaft BC is hollow with inner and outer diameters of 90 mm and 120 mm. respectively. Shafts AB and CD are sulid and of diameter d. For the loading shown, determine (a) the maximum and minimum shearing stress in shaft BC, (b) the required diameter d of shafts AS and CD if the allowable shearing Slress in these shafts is 65 MPa.

U, j "'

Tj! - I ll" · ,,,

c

SOLUTION Equations or Statics. Denoting by T AN the torque in shaft AB. we pass a section through shaft AB and. for the free body shown. we write ~M,

7~N=6kN·m

= 0:

We now pass a section through shaft BC and. for the free body shown. we have ~M, =

0:

(6kN· m)

u. Shaft Be.

+ (l4kN

- 1'oc=

0

TBC =20kN·m

For this hollow shaft we have

MlU:;11111111 Shl!lIring Siress. <., =4511""

·m )

TSCC2

Tn"" = T2 = - J-

On the outer surface, we have

(20 kN· m)(O.06O m) 13.92 X 10 ~ m~

T ....

= 86.2M l'a
,,~ = 6()"""

Minimlllll Shearillg Stress. We write that the stresses are proportional to the distance from the axis of the shaft.

"T.,...

45 mill

86.2 MPa

60 mm

----

Ii k:-..i

T"".

= 64.7 MPa
b. Shaft s AS a nd CD. We note that in both of these shafts the magnitude of the torque is T = 6 kN . m and T.II = 65 MPa. Denoting by c the radius of the shafts. we write

' In

To

T= -

65 MPa = (6 kN . m)e

J

rr

,

-c

2

B

(J

= 58.8 X 10- 6 m J

e = 38.9 X 10- 1 III

d = 2e = 2(38.9 mm)

d = 77.8 mm
143

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SAMPLE PROBLEM 3.2 The preliminary design of a large shaft connecting a motor (0 a generator calls for the use of a hollow shaft with inner and outer diameters of 4 in. and 6 in., respectively. Knowing that the allowable shearing stress is 12 ksi. determine the maximum torque that can be transmitted (a) by the shaft as designed. (b) by a solid shaft of the same weight. (c) by a hollow shaft of the !'!ame weight and of 8-io. otHer diameter.

SOLUTION u. Hollow Shaft as Designed. J = ~(("} -

For the hollow shaft we have

~[(3 in.)~ - (2 iO.)4J

cD =

= 102.1

in~

Using Eq. (3.9). we write 1'(3 in.) 12ksi= - - -

T

102. 1 in~

=

408 kip ' in. ...

T

b . Solid Shaft of [qual Weight. For the shaft as designed and this solid shaft to have the same weight and length. their cross-sectional areas must be equaL =

A (b )

rr [(3 in.)' - (2 in.)!] =

rrd

A (a)

C1

= 2.24 in.

Since,...... = 12ksi. we write 1"("3

Tn""

T

=

12 ksi = 1"(2.24 in .)

J

·r=

2 11 k.ip · in.
171"(2.24 .In.).

c . Hollow Shan of 8-in. Diamcier. For equal weight. the cross-sectional areas again must be equal. We determine the inside diameter of the shaft by writing

..

A (d)

, ,,,...

rr[(3 in.)! - (2 in n

~

For

e )

With

= 3.317 in. and

Toll

C4

=

dJ

(·5

= 3.317in.

= 4 in .•

= 12ksi ande. = 4 in ..

Te.

'r Lnal

144

A (e)

= rr[ (4 in·f -

=

J

7t4 in.) 12 ksi = 212 in'

T = 636 kip' in.
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PROBLEMS

3.1 For the cylindrical shaft shown. detemline the maximum shearing stress caused by 3 torque of magnitude T = 1.5 kN . m.

Fig. P3.1 and P3.2

3.2 Determine the torque T that causes 80 MPa in the sled cylind rica l shaft .shown. 3.3

11

d

maximum shearing stress of

1.6 ;n.

Knowing that the internal diameter of the hollow s haft shown is

Fig. P3.3 and P3.4

d = 0.9 in .. delennine the maximum shearing stress caused by a torque of mag· nitude T "'- 9 kip . in.

3.4 Knowing [hat d = J.2 in .. determine the torque T that imum shearing slIess of 7.5 ksi in the hollow shan shown. 3.5

(u) Delemline the torque that can

ca llSCS

a max-

be applied to a solid shaft of 20·mm

diameter without exceeding an allowable shearing Slress of 80 MPn. (b) Solve part rI, assuming that the ~olid shaft has been replaced by a hollow shaft of the same cross-$ectional area and with an inner diame ter equal to half of its own outer diameter. 3.6 A torque T = 3 kN . m is applied 10 the solid bronze eylimler shown. Determine (a) the maximum shearing stress. (b) the shearing stress at point D which lies on a IS-mm-mdius circle dmwn on the end of the cylinder. (c) the percent of the tonlue carried by the portion of the cylinder within the IS-nllll radius. 3.7 The solid spindle AB is made of a steel with ,m allowable shearing stress of 12 ks i. and sleeve CD is made of a brnss with an allowable shearing stress of 7 ksi. Determine (a) the largest torque T tlwt can be applied at A if the allowable shearing stress is not 10 be exceeded in sleeve CD, (b) the corresponding required value of the diameter ii, of spi ndle AB . 3.8 The solid spindle AB has a diameter d, = 1.5 in. and is mCtde of a steel with an allowable shearing stress of 12 ksi. while sleeve CD is made of a brnss with an allowable shearing stress of 7 ksi. Determine the largest torque T that can be applied at A.

Fig. P3.6

T

t

'-~

'1--" -lin

t

S in

L

-,J,

D

"

. 0.2.5 in.

C ~~

FIg. P3.7 and P3.8

145

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146

Torsion

3.9 The torques shown are exene
T 1=~ ~ )N

'"

II ~ )N

fll

T H=

"

"

Fig. P3.9

3.10 In order to redllce the tot:tl mass of the assembly of Prob. 3.9. a new design is being considered in which the diameter of shaft BC will be smaller. Detenlline the smallest diameter oi shaft BC for which the m:tximum value of the shc:tring stress in the assembly will not increase. 3.11 Under norm:tl operating cOllllilions. the electric motor exerts a torque of 2.8 kN . m on shaft AB. Knowing that each shaft is solid. determine the maximum shearing stress in (a) shaft AB. (b) shaft Be. (c) shaft CD.

Fig. P3.11

3. 12 In order to reduce the total mass of the assembly of Prob. 3.1 1. a new design is being considered in which the diameter of shaft BC will be smaller. Detemline the smallest diameter oi shaft BC for which the maximllm value of the shearing stress in the assembl)' will not be increased.

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3.13 The torques shown are exerted on pulleys A. B, and e. Knowing that both shafts are solid. detemline the maximum shearing stress in (al shan AS. (b) shaft Be.

Probtems

Fig. P3.13 and P3.14

3.14 The ~afts of the pulley assembly shown are to be redesigned. Knowing that the allowable shearing stress in each 5haft is 8.5 ksi. delennine the smallest allowable diameter of (lI) shaft AB, (b) shaft Be. 3.15 The allowable shearing stress is 15 ksi in the steel rod AB and 8 ksi in the brass rod Be. Knowing that a torque of magnitude T = 10 kip' in. is applied at A, detennine the required diameter of (a) rod AB. (0) rod Be

Fig. P3.1S and P3.1 6

3 .16 The allowable shearing stresS is 15 ksi in the 1.5-in.-diameter steel rod AB and 8 ksi in the 1.8-in.-diameter rod Be Neglecting the effect of stress concentrations. detenlline the largest torque that can be applied at A . 3. 17 The solid shaft shown is formed of ,\ brass for which the allowable shearing stress is 55 MPa. Neglecting the elTect of stress concentrations. determine the smallest diamete rs d AB and dBC for which the allowable shearing stress is no\ exceeded. 3 . 18

Solve Prob. 3. 17. assuming that the direction of T c is re\'ersed. Fig. P3. 17

3.19 The allowable stress is 50 MPa ill the brass rod AB and 25 MPa in the aluminum rod Be. Knowing that a torque of magnitude T = 125 N · III is applied at A. determine the_required diameter of (a) rod AB. (b) rod Be. 3 .20 The solid rod Be has a diameter of 30 mm and is made of an aluminum for which the allowable shearing s tres~ is 25 MPa. Rod AB is iloilo\!' and has an outer diameter of 25 mill; it is made of a brass for which the allowable s hearing ~ tress is 50 MPa. Detennille (lI) the largest inner diameter of rod AB for which the factor of safety is the same for eac,h rod, (b) the largest torque that can be applied at A.

T A

Fig. P3.19 and P3.20

147

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148

Tornlon

3.21 .-\ tOlque of Illhg nitud e T - 8 kip· in. is applied at D a! shuwn Know ing that the allowable sheari ng £tre ss is 7.5 ksi ill each shan. determin e the required diameter nf (II) shaft AB. (b) shaft CD.

..

iu

Fig. Pl.21 a nd P3.22

=

3.22 A torque of magnitude T 8 kip' in. is applied;}! D a~ shown Knowing that the diameter of shaft AB is 2.25 in. and Ihat the diametero!" shaft CD i~ 1.75 in .. determine the maximum .~ hearing ~ tre~s in (il ) s haft AB. (b ) shaft CD. 3.23 Two so lid sted sll3fts are ;oLlllected by th>":: gears ~ hQwlI. A tmqu e of magnitude T = 9(10 N . m is applied to shaft AB. Knowing that II-e :tllow. able sheanng. stfCOS is 50 MPa and considering on ly ;;tresses due 10 twi sting. detemtine the required diruneter 01 (a) shaft AB. (b ) shaH CD.

Fi g. P3.2:l a nd P3.24

3.24 Shafl CD is made from a 66-mm-diameler rod and is conneck>d til the 4~·ntm-dttreRS is 60 MPa for each sh aft. de!ennine the larges t torque T that can be applied.

3.25 The tWO solid shnfts are connec ted by gear. as shown and are l11Jde fm w hich th~ illlow~hle ~ hl'..;jriIl E !Mr_~~ i~ 7rM fl~i Knnwin:1 rhp d i. ameters of the tWIl shafts are. respectl\le ly. dJC = J.6 in. and d a = J .25 in uetennine the larg,!S1 torque T c that can be applied til C.

flf~ ~t~el

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Problems

Fig. P3.25 and P3.26 3 .26 The two solid shafts are connected by gears as shown and are made of a steel for which the allowable shearing stress is 8500 psi. Knowing that a torque of magnitude Tc = 5 kip' in. is applied at C and that the assembly is in equilibrium. determine the required diameter of (a) shaft BC, (b) shaft EF. 3.27 A torque of magnitude T = 120 N . m is applied to shaft AB of the gear train shown. Knowing that the allowable shearing stress is 75 MPa in each of the three wlid slwfts, determine the required diameter of (a) sh:lft AB, (b) shaft CD. (c) f.ilaft EF. 3 .28 A torque of m:lgnitude T = 100 N . m i~ applied to shaft AD of the geM tr:lin shown. Knowing th:lt the diameters of the three solid sh:lfts are, respectively, d,lB = 21 rum, del) = 30 mm, and dE} = 40 mill, detemline the m<1X.imum shearing stress in (a) shaft AB, (/1) sh:lft CD, (c) shaft EF.

2,5 """

Fig. P3.27 and P3.2B

3.29 While the exact distribution of the shearing stresses in a hollowcylindrical sh:lft is as shown in Fig. P3.29a. an approximate v:llue can be obtained for 1"mJU by assuming that the .'Iresses are unifonnly distributed over the area A of the cross section. :lS shown in Fig. P3.29b. and then further assuming that a]] of the elementary shearing forces act at a distance from 0 equal to the mean radius !,:c 1 + c~ ) of the cross section. This approximate value TO = ·l jAr",. where T is the applied torque. Determine the ratio T"",.ITo of the true value of the maximum sh~aring stress and its approximate value Tn for values of CI /C l respectively equal to 1.00.0.95,0.75,0.50, and O.

'"I Fig. P3.29

3.30 (a) For a given allowable shearing stress, JetelTI1ine the ratio 7j w of the. maximum allowable torque'" and the weight per unit length IV for the hollow shan shO\\n. (b) Denoting by (1'/w)o the value of this ratio for a solid shaft of the same mdius c,. express the mtio T/w for the hollow ~haft in tenns of P,/II')(I :md CI!<"1'

Fig. P3.30

149

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150

3.5. ANGLE OF TWIST IN THE ELASTIC RANGE

Tors!on

Lo this section, a relation will be derived between the angle of twist cP of a ci rcu lar shaft and the torque T cxened on the shaft. The entire shaft will be assumed to remai n clastic. Considering first the case of a shaft of length L and of uniform cross section of radius c subjecled to 11 torque T al its free end (Fig. 3.22), we reca ll from Sec. 3.3 thai the angle of twist cP and the maximum sheari ng strain "Ym.x are related as fo ll ows:

cq,

"Ym.. =

L

(3.3)

But , in Ihe clastic range, the yield stress is not exceeded anywhere in the .c;haft , Hooke's law applies, and we have "Ym." = T "",jG or, recalli"g Eq. (3.9),

Fig. 3.22

(3 .15) Equating the righi-hand members of Eqs. (3 .3) and (3. 15). and solving for tP. we write (3 .1 6)

where tP is expressed in radians. The relati on obtained shows that, within the e lastic range, Ihe angle of tW;j·/ tP i.v proporriO/uli to rhe torque T applied to the shaft. lllis is in accordance with the experimental evidence cited at Ihe beginni ng of Sec. 3.3. Eq uation (3 .16) provides us with a convenient method fo r determining the modulus of rigid ity of a given material. A specimen of the material, in the form of a cy lindrical rod of known diameter and length, is placed in a torsion te.l·ring machine (Fig. 3.23). Torq ues of increasing magnitude T are applied to the specimen, and the corresponding values of the angle of twist cP in a length L of the spccimen arc recorded. As long as the yield stress of the material is not exceeded, the points obtained by plotting cP against T will fall o n a straight line. The slope of this line represents the quant ity JG/ L, from which the modulus of ri gidity G may be computed.

Fig.3.23

Torsion testing machine.

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EXAMPLE 3.02 What torque should be applied to the end of the shaft of Examp leJ.OI to produce a twist 0 1"2°? Usc the value G = 77 GPa for the modu lus of rigidity of steel. Solving Eq.

(]. 1 6 ~

and recalling from Example 3.0 1 thaI. for lhe gillen section. J = 1.021 X 10 "m~

we howe

for"f". we write

lG 7= [" rf

lG

T"" - L dJ.

=

( 1.021

X

Substituting the w"en values 9

G = 77 X 10 Pa <$J =

2rr C..d) 2°(- - = 360"

cros~

L = 1.5m 34.9 X 10

J

10 - 0 m~)(77 X 1.5 m

IO~

Pal (34.9 X

T = 1. 829 X IO J N . m = 1.829 kN . III

rad

EXAMPLE 3.03 What angle of twi,t will create a sheari ng stress of70 MPa on the inner stuface oJ the hollow steel ~haft of EXollllples 3.0 1 and 3.02"1 The method of attack for solvi ng this problem that first comes to mind is to use Eq. (3.10) to find the torque T corre~ponding to the give n value of 1 , and Eq. (3. 16) to detemline the angle of tWist .p corresJX!ndin g 10 the value of T just found . A more direct solution, however. may be used. From Hooke's law, we first com pute the shearing stra in on the inner surface of the shaft:

70x 10" Pa

"1 _

1'.,1·=(;= 77X I09 Pa

909

x 10

0

Recalling Eq. (3 .2), which was obtained by expressing the length of arc AA ' in Fig. 3. 14(" in terms of bot h l' and cp. we 1m lie

.p = LiB"" = ISOOITlITl (909 X 1O~) = 68.2 x [0 " rad {"I

20mm

To obtai n the angle of twist in degrees. we wri te

cp

Fomlula (3 .16) for the angle of twist can be used only if the shan is homogeneous (cons[ant G), has a uniform cross section. and is loaded only at its ends. If the shaft is subjected to torques at locatio ns other than its ends, or if it consists of several portions with various cross sections and possibly of differe nt materials. we must di vide it into compone nt parts that satisfy individually the required condit ions for the application of formu la (3 .16). Ln the case of the shaft AB shown in Fi g. 3.2:4, for example, four different parts should be considered: AC. CD , DE, and EB. The total angle of twist of the shaft , i.e .. the angle through which end A rot3.tes with respect to end B, is obtained by adding (118('Ilraicully the angles of twist of each component part. De noting, respective ly, by Ti • L,. i i' and G, the internallorque, length. cross-sectional polar moment ()f inertia, and modu lus of ri gidity corresponding to part i, the IOtal angle of twi st of the shaft is expressed as

= (68.2

x

360" 10 3 rad{ -) =3 .91 °

211" rad

Fig. 3.24

(3. 17)

lltc internal torque T, in any given part of the shaft is obtai ned by passing a section through that part and drawing the free-body diagram of

151

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152

Tors!on

the portion of shaft located on one side of the sec tio n. This procedure. which has already been explained in Sec. 3.4 a nd ill uslrttled in Fig. 3.1 7, is applied in Sample Pro b. 3.3. In the case of a shaft with a variable circ ular cross section, as shown in Fig. 3.25. fonnula (3 .16) may be applied to a disk of thickness dr. The angle by which o ne face of the di sk rotates with respect to the other is thus Tdx

d~ ~-

1G

where J is a function of x which may be determined. Integrating in x from 0 to L. we obtain the total angle of twist of the shaft : Fig. 3.25

0 =

L Tdx

r

10 JG

(3 .18)

The shaft shown in Fig. 3.22, which was usc of the shaft. was equal to the angle of rotation of its frec end. When both ends of a shaft rotate, however, the angle of twist of the shaft is equal to the angle through which one end of the shaft rotates with re.ypect to the olher. Consider, for inst ance. the assembly shown in Fig. 3.26u, consisti ng of two clastic !ihafts AD and BE. each of length L. radius c, and modulus of rigidity G. which arc auached to gears meshed at C. If a torque T is applied at E (Fi g. 3.26b), both shafts will be twisted. Since the e nd 0 of shaft AD is fixed, the angle of twist of AD is measured by the angle of rotation cPA of end A. On the other hand, since both ends of shaft BE rotate, the angle of twist of BE is equal to the difference between the angles of rotation rPB and rPE' i.e., the angle of twist is equal to the angle through which end E rotates with reS"pect to end B. DcnOling this relative angle of rotation by I/>E/ H' we wri te

Fig. 3.26

(~ )

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EXAMPLE 3.04 For the ussembly of Fig. 3.26, knowing thm f ,I = 21"8' determine the angle of rot3tion of end £ of shOlft BE when the torque T is applied at E. We first determine the torque TAl) exerted on shaft AD. Observing that equal and opposite forces F and F' 3re applied on the two geOlfs at C (Fig. 3.27). and rec3lling thm r ,I = 2re, we conclude that the torque exerted on shaft AD is twice as large 3S the torque exerted on shaft BE; thus. T,w = 2T.

Observing that the arcs CC' und CC· in Fig. 3.26b must be equal. we write I"ArPA = rlJfPe and obt3in

We have, therefore.

cPe

= 21>,1 =

4TL

"""JG

Considering now .shuft BE. we recall that the angle of twist uf the shaft is equal to the angle CPE{B through whic h end E rotales with respecl to end B. We have

Fig. 3.27

The angle of rotmion of end E is obt3ined by writing Since t.he end D of shaft AD is fL"(ed. the 3ngle of rotation CP,I of gear A is eq ulll \() the Wigle of twist of the shaft lind is obtained by writing 4TL JG

TL .lG

5rL .lG

~-- + -=--

3.6 . STATICALLY INDETERM INATE SHAFTS

You saw in Sec. 3.4 thaI. in order to determine the stresses in a shaft. it was n eces~ary to first calculate the internal torques in the various parts of the shaft. These torques were obtained from statics by drawing the free.-body diagrnm of the portioll of shnft located on one side of a given sec lioll and Writing that the su m of the torques exerted on that portion was zero. lllere arc situations. however. where the internal torques cannot be determined from statics alone. In fact. in such cases the external torques themselves, i.e., the torques exerted on the shaft by the sUPlX'rts ttnd connections, cannot be determined from the free-body diagram of the entire. shaft. llle equilibrium equations must be complemented by rchuions involving the deformations of the shaft 'a nd obtained by considering the geometry of the problem. Becttuse statics is not sufficient to detcrtnine the external and internal torques. the shafts are said to be .rflltically indeterminate. The foll ow ing example, as well as Sample Prob. 3.5 . will show how to analyze sratically indeterminate shafts.

153

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EXAMPLE 3.05 A circular shaft AB consl;;tS of a 1O.ln.-long. i· in,.diameter Sleel cylinder. in which a 5.in.-long, ~- in.-diameter cavity has been drilled from end S . The s hafl is allached to fixed sup· ports at bOlh ends. and a 90 lb . fltorque is applied aI its midsection (Fig. 3.28). Detenuine the torque exerted on the slwft by each of the supports.

Drawing the free-body diagram of the shaft and denoting by T... and T B the torqnes e.~ened by the supports (Fig. 3.29(1), We obtain the equilibrium equation

Since this equ
B

From the free-body diagram of a small portion of shaft including end A (Fig. J.29b), we note that the internal torque 1'1 in AC is equal to 1~: from the free-body diagram of a small portion of shaft including end B (Fig. 3.29(,), we note that the internal torque 1~ in CB is equal to Til' Recall ing Eq. 0 .16) and observing thai portions AC and CB of the shaft are twisted ill opposite senses. we w li te

Fig. 3.28

'I ,

~

G

A

T, B

Sol ving for To. we have

T,

~

T,

T1

(bJ

B

Substituting the numerical dala

(c)

Fig. 3.29

LI =

~

=

5 in.

I t = br(lt; in. )"' = 57.6 X 10 3 in" J, = ~7Tl(~ in. )4 - (f6 in.)"] = 42.6 X 10

J

in 4

we obtain To = 0.740

1~

Substituting this expression into the original equilibrium eqmtlion. we write L7401~ =

1',..= 51.71b · fl

154

901b · ft

"B

= 38.3 lb . ft

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SAMPLE PROBLEM 3.3 60 ",,"

The horizontal shaft AD is attached to
SOLUTION Since the shaft consists of three portions AB. Be. and CD, each of uniform cross section and each with
(250N ' m) - ·1i.8 = 0

0:

Passi ng now a section between Band

7;18 = 250N · m

e. we have

= 0: (250 N . m) + (2000 N . m) - Tec = 0

~M,

Since no torque is applied at

Tec = 2250N'm

e.

TCD = T8c= 2250N · m A

30 """ 15 """

AB

Be

G) CD 22 """

Polar MUII/ellts oj II/ ertill 4 J,1.8 = 2:. = 00795 X IO- ~ m· 2 c = 2:(0015m)' 2' .

lee =

f c = f (0.030 m)4 = 1.272 4

X

[0

6

m 4

leo = f(ci - (·i) = f[(0.030m)4 - (0'{)22m)4] = 0.904 X 1O-~ I11·1 Anglr ufTwist . entire shaft, we have

rP

=

T;1-1 = ~(TM)-A8 , 1;G G lAe

2,

" ~ _

4>"

USing Eq . (3. 17) and recalling that G = 77 Gra for the

+ Tt"CLec + TcJ.-co) lee

[ (250N . m)(OA Ill) 77 Gra 0.0795 X 10- 6 m4 ,_

+

leo

(2250)(0.2 ) 1.272 X IO- b

+

(2250)(0.6) 0.904 X 10- 6

1

= 0.01634 + 0.00459 + 0.01939 = 0.0403 rad 360 = (0.0403 rad)-Q

4>.1

27< rad

155

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SAMPLE PROBLEM 3.4 Two solid steel shafts are connected by the gems shown. Knowing thai for each

shafl G = [1.2 X lOb psi and that the allowable shearing stress is 8 ksi. determine (a) the largest torque To that may be applied to end A of shaft A8. (b) the corresponding angle through which end A of shaft AB rotates.

SOLUT ION Stalics.

Denoting by F the magnitude of the tangential force between

gear teeth. we have

~

(8. ~r

= 2.451)1 .

r~ . T'

arllr B . ":i.M II = 0:

F(0.875 in.) - To = 0

Gf!lIr C. 'f M c = 0:

F(2.45 in.) - TeD = 0

Killcma lic.~.

TCD

= 2.81~

(I )

Not ing Ih:lI the peripheral motions of the gears are equal.

we write (2)

'II

= ().1i75 in.

tI .

Torque Tf

Wilh 7~B = 1~ and c = 0.375 in .. together wilh a maximum permissible shearing stress of 8000 psi. we wrile

Shaft AB.

,

T~BC

~--

ll(X)()

J S haft CD . 'T;l1

pSI =

Td O.375 in.)

"I i) = 663 Jb . in.

1

<J

211"(0.375 in. t

From (I) we have TeD = 2.8Tu. With c = 0.5 in. and

= 8(x)() psi. we write

. 2.81"0(0.5 in.) 8(x)() pSI = 1 ~ ;'17(0.5 in.) i\1'lxilllllln Perm issible Torque. for To

""0

We choose the

= 561 Ib . in.

~ maller


value obtained

T'I = 56 1 lb · in . ....

b. Angle of Rotation al End A . each shaft. S hrift A B .

4>1I/B

For

"I~8

=

1'0

We first compute the angle of twist for

= 561 lb· in .. we have

TA/;L

(56 1 Jb· in.)(24 in.)

= Jii =

t 7T(0.375 in /( 11.2 X 106 psi)

S hllft CD .

TeD = 2.8"1;) = 2.8(561 lb· in. )

7L·vL 4>CfD =

0.0387 rad = 2.2r

lG =

2.8(561 lb · in. )(36 in.)

~7r(0.5 in.)~(I1.2

X

10~ psi)

Since end D of shaft CD is fixed. we have we find the angle of rotation of gear B to be

W8

4>(."

0.514 rod = 2.95 " =

w C(1)

= 2.95 Q • Using (2).

= 2.8wc = 2.8(2.95°) = 8.26 Q

For end A of shaft AB. we have

W. . = W8 + 4>11/8

156

= 8.26°

+ 2.22°

WI =

I0.48 ~

....

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SAMPLE PROBLEM 3.5 A steel shaft and an aluminum rube are connected 10 a tixed SlIpJXlrl and \0 a

rigid disk as shown in the cross section. Knowing that Ihe initial stresses are zero, determine Ihe maxi mum torque To that can be applied to the disk if Ihe allowable stresses are 110 M Pa in the steel shaft and 70 MPa in the aluminum tube. Use G = 77 G Pa for steel and G = 27 G Pa for aluminum.

SOLUTION Stalics. Fret Budy 1)/ Disk. Denoting by T , the torque exerted by the tube on [he disk and by T! the torque exerted by the shaft. we find

To= 1', + 12

DeForma t ions.

( I)

Since bolh Ihe tube and Ihe shaft are connected \0 Ihe

rigid di sk. we have

TIL l "" T2L2 l lG [ l IG!

rPl "" rPl: (2 .003 x

7 1(0.5 m) 1O -~ m ')(27 O Pal

T 2 (0.5 m) (0.6 14 x 10

6

m ' )(77 G Pa) (2)

S ha:urin g S t rcSst'S. We a.\·.l"lIlIIe lhat the requirement critical. For the aluminum lUbe. we hJve

(70 MPa)(2.\')I)3 X JO ~ 1)\4)

T _ T-..1"",11 1 -

A ~ 'minllm

r: j =

27(; 1'" } = H (3Ii 1I1111 )~ - (30 "'m )~l = 2.003 x 1O-~ m~

0.038

C,

III

1"ohJm.£

70 M Pa is

369ON'm

Using Eq (2). we compute the corresponding value T2 and then tind the imum shearing stress in the s teel shaft

mml~

T2 = 0.814'1'1 = 0.8/4(3690) = 3225 N· m li ("~ (322.'5 N . 111)(0.02.5 111) . T"'ed = - - = !; -I = 13 1.3 MPa J2 0 .6 14 x 10 III We note that the alluwable s teel s tres.~ of 120 MPa is exceeded: our assumption was 1I"/lJllg. T hus the maximum torque T" will be obtained by making 1">1,.." = 120 MPa. We tirst det emline the torque T !. 1"uffi J

2

T, = - - = -

( 120 MPa)(0.614 X 10 ~ m-l) 0.025 m

( '2

1950N· m

From Eq. (2 ). we have

25"",,

G , "" iiGl'u JI '" H(2~ "un )~l "' 0.614 X lll-"' m~

2950 N . m = 0.8747",

'1', "" 3315 N · m

Using Eq. ll ). we obtain the maxi mum permissible torque

To = T,

+ Tl

= 3315 N . m

+ 2950 N . m T" = 6.325 kN •

I II

<4

157

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PROBLEMS

40",,,,

r~~"

50,,,,,,

3.31 (a) For the aluminum pipe shown (G = 27 GPa). determine the torque T o causing an ang le of twist of 2°. (b) Detemline the angle of twisT if the same torque To is applied to a solid cylindrkal shaft of the same length and cross-secTional area. 3.32 (a) For The solid steel shan shown (G = 77 GPa). deTermine the angle of twisT at A. (b) Solve pm1 il. assuming That the steel shaft is hollow with a 30-mm-outer diameter and a 20-mm-inner diameter.

2_5 ",

3.33 The ship at A Iws just started to drill for oil on the ocean floor at a depth of 5000 fl. Know ing that The top of the 8-in.-diameter steel drill pipe (G = 11.2 X l Ob psi) rotates through TWO complete revolutions before the drill bit at B starts TO operate. delermine the maximum shearing stress caused in the pipe by torsion. FIg. P3.31

3.34 (G -

Determine th e largest allowable diameter of a 10-ft-Iong steel rod

11.2 X lOb psi) if Ihe rod is 10 be twistcd th rough 300 without e.~ cccd­

ing a shearing stress of 12 ksi.

~I.S ",

Fig. P3.32

1 ----.J •

Y 9' " y2m till

5000 ft

Fig. P3.35

,--.4 "

•

Fig. P3.33

3.35 The electric motor exerts a 500 ~ . m torque on the aluminum shaft A BCD when il is roTating at a constant speed. Knowing that G = 27 GPa and th:lt the torq ues exerted on pulleys Band C are as shown. determine th e angle of twist between (a) B and C. (b) Band D.

158

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3.36 The torques shown are exerted on pulleys A. B. and C. Knowing that both shalh are solid and made of bra~s (G = 39 GPa). determine the an· gle of twist between (a ) A and B. (b) A and C. <;0(. /\

Problems

tn

B

300 mm

Fig. P3.36

3 .37

The aluminum rod

Be (G

"" 26 GPa) is bonded to the brass rod

AB (G = 39 Gra). Knowing that each rod i~ solid and has a diameter of 12 mm. detennine the angle o f twist (a) at B. (b) at

c.

LJ1

Fig. P3.37

3.38 The aluminum rod A B (G - 27 GPa) is bonded to the brnss rod SD (G = 39 GPa). Know ing that portion CD of the brass rod is hollow and has· an inner diameter of 40 mm. detemline the angle of twist at A.

3fl - - -

Fig. P 3.38

Fig. P3.39

3.39 Three solid s hafts. each of t- in. diameter. are connected by the gears shown. Know ing that G = 11 .2 X 106 psi. determine (a) the angle through which end A of sllaft AB rotates. (b ) the angle through which end E of shaft EF rotates.

Ah"";",,,,,

\110 N ".

159

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160

Torsion

3.40 Two shafts. each of lin. diameter. are connected by the gears ~hown . Knowing that G = 11.2 X 10" psi and that the shaft at F is fixed. determine the angle through which end A rotates when a 1.2 kip ' in. torque is applied at A. 3.41 Two solid shafts are connected by gears as shown. Knowing that G = 77.2 GPa for each shaft. determine the angle through which end A rotates when TA = 1200 N . m.

24() Ill'"

\

\

<\

B

so",,,,

4~L2m

/Y'~6m Fig. P3.40

Fig. P3.41

3.42

Solve Prob. 3.41. assuming that the dianleter of each shaft is 54 mm.

3.43 A coder F. used to record in digital form the rotation of shaft A, is connected to the shaft by means of the gear train shown. whkh consists of four gears and three solid steel shafts each of diameter d. Two of the gears have :l radius f and the otherlwo a mdius IIf. If the rotation of the eooer Fis prevented. detemline in terms of T. I. G. 1. and Il the angle through which end A rOiates.

Fig. P3.43

3.44 For the gear train described in Prob. 3.43. determine the angle through which end A rotates when T = 5 Ib . in .. 1 = 2.4 in., d = ft; in .. G = 11.2 X IO~psi.[]ndli = 2.

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3 .45 The d~sign spedfications of a 1.2.m-long solid transmission shaft require that the nnglc of twist of the shaft not exceed 4° when a torque of 750 N . m is applied. Determine the required diameter of the shnli, knowing that the shafl is made of a steel wi th an allowable shearing stress of 90 MPa and a modulus of rigidity of 77.2 OPa. 3 .46 A hole is punched at A in a plastic sheet by applying a 600-N force P to end D of lever CD. which is rigidly
Fig. P3.48

3 .47 The design specitications for the gear.and-shaft sysTem shown require thaI the same diameTer be used for boTh shafts and That the angle through which pulley A will rotate when subjected to a 2-kip . in. TOrque TA. while pul. ley D is held fixal will not exceed 7.5°. Determine the required diameler of Ihe shafts if both shafts are made of a steel with G = 11.2 X 10" psi and 1".>11 = 12 ks i.

Fig. P3.47

3.48 Solve Prob. 3.47. assuming that both shafts are made of a brass with G = 5.6 X 10" psi and T.1l = 8 ksi.

Probloms

161

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, 62

3 .49 The design of The gear-amJ.shafT sysTem shown requ ires That steel shafTs of the same diameter bc used for both AB and CD. [t is further required that 'Tn"", ::S: 60 M Pa and that the angle.po through which end D of shaft CD rotates nOi exceed 1.5°. Knowing that G = 77 GPa. determine the reqllired diameter of the shafK

Tors!on

r A

v

IOOn\rn~ 6(11) \l\1l\

Fig. P3.49

3.50 The electric motor exerts a torqlle of 800 N . m on the steel slmft ABeD when it is rotating at conSi[lnt speed. Design specifications require that the diameter of the shaft be uniform from A to D and that the angle of twist between A to D not exceed 1.5 Knowin g that Tn". ::s: fiO MPa and G "" 71 GPa. determine the minimum diameter shaft thai can be u sed. Q

•

AI"",!",nn ---....

T = 12..5 kil' , m

12 in.

~

Fig. P 3.SO

IS ;".

_l,Uin

Fig. P3. 51

3.51 The solid cyl inders AB and BC are bonded togeth er at B and [Ire [llIached to fixed supports at A and C. Knowing that the modulus of rigidity is 3.7 X 106 psi for aluminum and 5.6 x lOb psi for brass. delennine the maximum shearing stress (lI) in cylinder AB. (b) in cylinder Be. 3.52 Solve Prob. 3.51. assuming that cylinderAB is made of steel. for which G = 11.2 X I06 psi.

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Problems

3.53 The composite shaft shown consists of a 5-mnHhick brass jacket (Clio'" = 39 GPa) bonded to a 40-mm-diameter steel core (C"«I = 77.2 GPa). Knowing that the shaft is subjected to a 600 N . III torque. determine (a) the ma.~imum shearing stress in the brass jacket, (b) the maximum shearing stress in the steel core. (el the angle o f twist of B relative to A. 3.54 For th~ composite shaft of Prob. 3.53, tht' allowable shearing $lfes); in the brass jacket i.'i 20 MPa and 45 MPa in the steel core. Determine (a) the largest torque which can be applied to the shaft. (b) the corresponding angle of twist of B relative to A. 3 .55 At a time when rotation is prevented at the lower end of each shaft. a 50-N . m torque is applied to end A of shaft AB. Knowing that C = 77.2 GPa for both shafts. determine. (a) the ma.~imum shearing stress in shaft CD, tb) the angle of rotation at A.

"""

:; Fig. P3.S3 and P3.S4

3.56 Solve Prob. 3.55. assuming that the 80-N . m torque is applied to end C of shaft CD. 3 .57 and 3.58 Two solid steel shafts are lined with flallges that are then connected b) bolts as showli. The bolts are sli ghlly undersized and permit a 1.5 rotation of one flange with respect to the olher before the flnnges begin to rotate as a single unit. Knowing that C = 11.2 X lOb psi. determine the. maximum shearing stress ill each shaft when a torque of T of magnitude 420 kip' ft is applied to the flange indicated. 3 .57 The torque. T Is applied 10 flange B. 3.58 The torque T is applie(i to flange C. Q

Fig. P3.55

Fig. P3.S7 and P3.SS

3 .59 The steel jacket CD has been attached to the 40-mm-diameter steel shaft AE by means of rigid flanges welded to lhe jacket and to the rod. The ollier diameter of the ja:ket is 80 mm and its wall thickness is 4 mm. If 500 N . m torques are appliC\l as shown. detennine the maximum shearing stress in the jacket.

T

Fig. P3.S9

163

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164

Torsion

3 .60 ll1e mass moment of inertia of a gear is to be determined experimentally by using a torsional pendulum consisting of a 6-ft steel wire. Knowing that G = 11 .2 X 106 psi. determine the diameter of the wire for which the torsional s pring constant will be 4.27 Ib . ft/rad . 3.61 An annular plate of thickness t and modulus G is used to connect shaft AB of radius r 1 tll tube CD of radius '2. Know ing that a torque T is applied to end A of shaft AB and that end D of tube CD is fixed, (a) determine the magnitude and location of the maximum shearing stress in the annular plate. (b) show that the angle through which end B of the shaft rotates with respect to end C of the tube is

(I '2I)

epee = -T- -:; - -:;

Ag. P3.60

4TrGt'i

Fig. P3.61 and P3.62

3 .62 An annular brass plate (G = 39 GPa). of thickness t = 6 mm, i.'i used to connect the bruss shaft AB. of length L I = 50 nun and rauills r l = 30 mm. to the brass lUbe CD. of length L..;. = 125 mm. inner radius r"l = 75 mm. :1110 thickness t = 3 mm. Knowing Lhat a 2.8 kN . m torque T is applied to end A of shaft AB and that end D of tube CD is fixed. determine (a) the maximum shearing stress in the shaft-plate-tube system. (b) the angle through which end A rotates. ( Hilll: Use the formula derived in !'rob. 3.6 1 to solve part b.) 3 .63 A solid shaft and II hollow s haft are made of the same material and are of the same weight and length. Denoting by II the ratio (·I!t: , . show that the ratio 7:/ T~ of the torque ., : in the solid shaft to the torque Th in the II~Y( I + II~ ) if the maximum shearing stress is the hollow shaft is (a) same in each shaft, (b) ( I - /1 )1(1 + ,,1) if the angle of twist is the same for each shaft.

V(l -

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3.7. Design 01 Transmission Shans

3.7. DESIGN OF TRANSM ISSION SHAFTS

The principal speciJications to be met in the desi gn of a transmission shaft are the power to be transmitted and the speed of mlatioll of the shaft. The role of the designer is to select the material and the dimensions of the cross section of the shaft, so that Ule maximum shearing stress allowable in the material. wi ll not be exceeded when the shaft is transmilling the required power at the specified speed. To determi ne the torque exerted on the shaft, we recall from elementary dynamics that the power P associated wi th the rotation o/" a rigid body subjected to a torque T is P = Tw

(3 .1 9)

where w is the angular velocity of the body expressed in radians per second. But w = 27Tt where f is the freq uency of the rotation , i.e., the number of revolutions per second. TIle unit of freq ue ncy is thus I S -I and is called a herr?. (Hz). Substituting for w into Eq . (3 .19). we write P = 2 rr fT

(3.20)

Lf SI Llnits are used we verify thm , withf ex pressed in Hz and Ti n N . m. the power will be expressed in N . mls. that is. in waft.l· (W). Solving Eq. (3.20) fo r T. we obtain the torque exerted on il shaft transmitting the power P at a frequency of rotatiOll f P

T ~ -

27T f

(3 .2 1)

where p ,/, and T are expressed in the units indicatL'-d above. After having. detennined the torque T that wilJ be applied tll the shaft and Illlving selected Ule material tn be used, the designer will carry the values of T and of the maximum allowable stress into the elastic torsion formula (3.9). Solving for lie, we have J

,

T

(3 .22)

and obtain in this way the minimum value illlowable for the parameter l ie. We check that. if SI units arc used. T will be expre..<1.sed in N . rn. T rna, in Pa (or N/ m2), and l ie will be obtained in mJ . In the case of a solid circular sh:1ft. 1 "" !7TC4. and l ie = ~ 7T~: substituting this value for lll: into Eq. (3.22) and solving for e yields the minimum allowable value for the radius of the shafi. In the case of a hollow cir· cular shaft. the critical parameter is lle;-.. where ("2 is the outer radius of the shaft; the value of this parameter may be computed from Eq. (3. 11 ) of Se~. 3.4 to dctennine whether a given cross secti on will be ilcceptable. When U.S. ~ u sto mary lInits are used, the frequency is usually ex· pressed ill rplll and the power in horsepower (hp). It is then necessary. before applying fomuli:1 (3.21 ), to converlthe frequency into revolutions

165

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166

per second (i.e., hcr1.Zcs) and the power illtll ft . Ibis or in . IbIs through the use of the fo llowing relations:

TorSIon

I 1 I rpm = -s-J = - Hz

60

6()

I hp = 550 ft • Ibis = 6600 in . IbIs

If we express {he power ill in . IbIs, fll rmula (3 .2 1) will yield the value of the torque Tin Ib . in. Carryi ng this value of T into Eq. (3 .22), and expressing T rna in psi, we obtain the value of thc paramcter J/e in in 1.

EXAMPLE 3 .06 Wh:l\ size of shaft should be used for the rotor of a 5-hp motor operat ing at 3600 rp m if the shearing Stress is not to exceed 8500 psi in the s haft t

Substituting for T and

We first express the power of the motor in in . Ibis and its frequency in cycles per ~col1d (or hertzes). BUI

)(6600Iin'hp IbiS)

/J = (5 hP

r

l /c

= t rr c 1 for a sol id shaft. We ha ve. therefore.

= 33 .000 in ' Ibis

I Hz

}7T
d = 2c = 0.374 in.

S- I

A ~- in . shaft should be u.
The torque exened o n the shaft is given by Eq. (3.21): P 33 .OCIO in . Ibis T= - - = 27r J 21T (60 s I)

= 10.30 X 10- 3 in l

c = 0.1 872 in.

= (3600 rpm) - - = 60 Hz"" 60

f>Orpm

into Eq. (3.22). we write

87.54 Ib · in.

T

J

Tn ..,

87.54Ib ·in.

EXAMPLE 3.07 A shaft consisting of a steel tube o f 50-mm OUTer diameter is 10 transmiT 100 kW of power while rotating at a frequency of

20 Hz. Determine the lU~ thickness which should be the shearing stress is not to e.\ceed 60 MPa.

u~d

if

The torque e:<erted on the shaft is given by Eq. (3.2 1): T

~

/~

-- = 27r J

100 X 10" W

21r (20 Hz)

J

T

795.8 N . III 60 X !Ofi N/ml

(3.24)

Equuting the right-hand members of Eqs. (3. 23) and (3.24):

(O.025)~ - c~ .: O.()50 (13.26 X 10 795.8 N · m

<:;t == 390.6 x IW 9 -

From &t. (3.22) we conclude th;)t the parameter lll'! must be at least equal to

-~-~

But. from Eq. (3.10) we have

c[ = 20.6 X

10 ~ = 179.6

1»

x

10.6mm

The corresponding tu be thickness is ( '1

0.23)

(0 ) 111 =

" 2 11.0 X

-

('[

= 25 mill - 20.6 nun = 4.4 mill

A tube thic kness of 5 mm shou ld be used.

10 ~ m"

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l.8. SITfISS ConcentratJons in Circular Shans

3.B. STRESS CONCENTRATIONS IN CIRCULAR SHAFTS

The IOrsion fonuu la Tma.>. = Tel ) was derived in Sec. 3.4 for a circular shaft of uniform cross section. Moreover. we had assumed earlier in Sec. 3.3 that the shaft was loaded at its ends through rigid e nd plates solidly auached to it. In practice. however, the torques are usually applied 10 the shaft through !lange coupl ings (Fig. 3.30a) or through gears connected to the shaft by keys lined into keyways (Fig. 3.30h) . In both cases one should expect the distribution of stresses, in and ncar the section where the torques arc applied, to be different from that given by the torsion fonnu la. High concentrations of stresses, for example. wi ll occur in the neighborhood of the keyway shown in Fig. 3.30b. The detemlination of these localized stresses may be carried out by experimental stress analysis methods or. in some cases, through the usc of the mathematical theory of elasticity. As we indicated in Sec. 3.4, the torsion formula can also be used for a shan of variable circular cross section. In Ule case of a shaft WiUl an abrupt chanb,'C in the diameter of its cross section. however, stress concentrations will occur Ilcar the discolltinuity, with the highest stresses occurring at A (Fig. 3.31 ). These stresses may be reduced

Fig.

I")

1M Fig. 3.30

3.31

through the use of a fill et, and the maxjmum value orth!.' Shearing stress at the fillet can be ex pres~ed as

To

T m• •

= K-

.I

(3 .25)

where the stress Tell is the. stress computed for the smaller-diameter ~ haft, and where K is a stress-concentration factor. Since the factor K depe nds only upon the ratio of the two diameters and the ratio of the radius of the fill et to the di ameter of the smaller shaft, it may be computed once and for all and recorded in ihe fonn of a table or a graph, !IS shown in Fi g. 3.32. We should note, however, that thi s procedu re for determining localized shearing stresse ~ is valid only as long as the value of j ma< given by Eq. (3.25) does not exceed the proportional li mit of the material, since the values of K ploncd in Fig. 3.32 were obtai ned under the a s ~umpt ion of a linear relat ion between shearing stress and shearing strain. If plastic deformation s occur, they will result in valuc..~ of the m:lXimum stress lower than those indicatcd by Eq. (3.25). t Wo D . Pitke y. i'elerSOJl·f SI ,""SS O", cell1mlioJl F ar ror$. 2nd ed • John Wile y &. Sam. New York. 1997.

Lj

Htt~+---l

" , '-"'--'-

1.6

Hc\t-I-;;'-c:!

D

~ ~ l.lll

J. :!.5

15 1-~---l1-;;c'c=-,--,

12 1-+---11--P~

" 1-+---1I-+ --+- +---1 10 '-~~I-~~c-c-'c-c-'c o O.O.'} 0. 10 0.15 0.20 0.25 0.30 rid

Fig. 3.32 Stress-concentratlon factors for fi ltets in circular shafts.t

167

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SAMPLE PROBLEM 3.6

3.7S ill.

r =

•

Iii"',

The stepped shafl shown is 10 rotate at 900 rpm as it Ir,U"lsmiIS power from a turbine to a generator. The grade of steel speclfied in the design has an allowable shearing slress of 8 ksi. (1I) For the preliminary design shown. determine the maximum power that can be transmitted. (b) If in the final design the radius of the Iillet is increased so that,. = ti in .. whal will be the percent change. relative to the preliminary design. in the power Ihal can be transmitted?

SOLUTION u. Pn>liminary Design. D = 7.50in .. d = 3.75 in .. r =

D

Using the notation of Fig. 3.32. we have: = 0.5625 in.

ft; in.

7.50 in. 3.75 in.

0.5625 in. 3.75 in.

-~--~2

d

d

K '" 1.33 is found from Fig. 3.32.

A stress-concentration factor

Torque.

0.15

Recalling Eq. (3.25). we wrile ( 1)

To., =

:r,;?', =

6.02 bi

where l ie refers to the smaller-diameter shaft:

lie = t rrd = ~ rr ( 1.875 in.)' = 10.35 in' T "''':IX 8 ksi - - ~ - - = 6.02 ksi

and where

K

1.33

Substituting into Eq. ( I). we find T = ( 10.35 in 3)(6.02 ksi ) = 62.3 kip ' In. Po wer.

1 H, Since /= (900rpm j - - - = 15 Hz = 60 rpm

15 s ~ l.

p" = 2tr [1' = 2tr( 15 s 1)(62.3 kip' in.) = 5.87

1'" = (5.87

b . Final Design.

10~i n' Ib/s)( 1 hp/ 6600in . Ibis)

x

For

r = ~

o

- =2 d To. , =

"T,~"," = 6.67 hi

"T1UlI.'

K

T =

1'_ -- = , K

0.250

K = 1.20

~

8 ksi

- - = 6.67 ksi

1.20

(10.35 in J )(6.67 ksi ) = 69.0 kip' in.

Po = 211" I T = 211'" ( 15 flU.O lip ;"

in' Ibis

Following the procedure used above. we write --

'r" =

10~

Pol = 890 hp ...

in. = 0.9375 in ..

0.9375 in. 3.75 in.

d

x

we write

s ~ I)(69. 0

kip ' in.) = 6.50 X lot' in . Ibis

Ph = (6.50 X 10~ in' IbJs)( 1 hp/6600 in' Ibis) = 985 hp Percelll e lla/fge ill Power

Percent change =

168

+ 11 % ....

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PROBLEMS

3 .64 Determine the maximum shenring Stress in a solid shafl of LS-in. diameter as it transmits 75 hp at a s)l<'cd of (u) 750 rpm, (IJ) 1500 rpm. 3 .65 Determine the maximum s hearing SIl\':SS in a solid shaft of 12-mm diameter as it Uilosmils 2.5 kW 'It a frequency of ((/) 25 Hz. (b) 50 Hz.

shaft

3 .6 6 Using an allowable shearing stress of SO Mi'a, design a solid sicel \0 transmit IS kW at a frequency of (a) 30 Hz, (b.) 60 Hz.

shaft

3.67 Using an allowable shearing stress of 4.5 ksi. design a solid steel \0 transmit 12 hp at a speed of (a) 1200 rpm. (b) 2400 rpm.

3 .68 As the hollow s leel s haft shown rolmes at J 80 rpm. :l stroboscopic measurement indi:ates that U1C angle of twist of the shaft is 3°. Kn owing Ih:lt G eo. 77.2 OPa. determine ta) the power being Iransmined. (IJ) the maximum shearing stress in the s hafl.

25 nnll

Fig. P3.68 and P3.69

3 .69 The hDlIow steel shaft shown (G = 77.2 Gl'a."7,lI = 50 MPa ) rotates at 240 rpm. Determine {u) the max.imum power that can be transmined. (b) the corresponding angle of twist of the shaft.

3.70 One of two hollow drive shafts of a cruise ship is 125 ft long. and its outer and inner diameters are [6 in. and 8 in., respectively. The shaft is made of a steel for which,.all = 8500 psi and G = I [.2 X [0 6 psi. Knowing that the maximum speed of mtation of the shan is 165 rpm. determine (u) the maximum power that can be transmined by the onr shaft tll its propeller. (b ) the corresponding angle of tw ist of th e shafl. 3 .71 A hollow steel drive shaft (G = [1.2 X [0 6 psi) is 8 ft long and its outer and inner diameters are respectively eq ual 10 2.50 in. and 1.25 in. Knowing that the , haft transmits 200 hp while rotating at [500 rpm. determine (1I) the maximum shearing stress. (b) the angle of twist of the shali.

169

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170

3 .72 A s1ed pifJt": of 72·mm OUlC! diameler is 10 !Je used 10 llallsmit ~ torque of 2500 N . m without e~ceedillg an allowable shearing stress of SS M Pa A series of 72·mm-oUler·tiiamcter pipes is avai lable for use. Knowing Ihalthe wall thickness o r ille avai lable pipes varies trom 4 111m 10 10 mm III 2·mm increments. dlOOse Ihe tightes1 pipe tha1 can be lI,ed.

TOI'3I«1

fI( i2"1111

~~ Fig. P3.72

3.73 The design of a machine element calls for a 40.. mm·outer· diameter ",haft tu tr;,II;lII il 45 kW. (u) If tilt: spt:~t.l wf m iatiull i~ 720 rpm. (ktt:rl!lill~ th~ 1lI~~imum ~hl'.aring I:rre.~ .~ in dmft fl. (h) If 1he. ~[\<"e.d of roralion ~ an tw incre.Jw·c\ 50% to IO~O rpm. de lemline the largesl inner diameter of shaft b for which the TIlaximum s hearing stress will be Ihe same in each shafl. 3 .74 A ISm-long solid steel , haft of 22·mm diameler is to tram.mil 12 kW. Determine Ihe minimum frequency al which the shaft can mlate, know· ing Ihat G = 71.2 O r a. tital Ihe allowable shearing stress is 30 MPa. and Ihal 1he ang le of1wis1 mllS1 not exc<-'Cd 3.5°.

(.,

1/,1

Fig. P3.73

3.75 A 2.5 .. m. long solid steel shaft is 10 transmit 10 kW al a frequenC) of25 Hz. Determinc the u:quiled diallle1erof the shaft, knowiug tha1 G - 77.2 G Pa. that the. allowab le shearing stres. is 30 MPa. and thm the angle of twist must not exceed 4'. 3 .76 The two solid shafts llnd geaJ"!; shown ar~ used III Iransmi1 t6 hp from the motor al A operaling at a speed cf 1260 rpm. 10 a machine 1001 at D Knowing that the maximum allowable shearing slress is g ksi. determine the requ ired di~Ulleler (a ) of shaft AB. (b) of shafl CD.

r =

1 ~ 1".

Fig. PUS and P3.77 A

3 .77 Tile tlllO salid shafts and gears shown are us~d 10 transmit 16 hp Irom the motor at A operming al a speed of 1260 rpm, 10 a machine 1001 at D KHowing 1h~1 each shaft h~s a diameler of I in. determine the lllallimUIIl shear .. jng stress (a) in shaft 04.8, (b) in shaft CD. 3 .78 The shafl-diSJ.:-lX'h arrangemem shown is used 10 trnnSC"lit ) hp rrnm min r A 10 ["\Oinr I(J) 1/,,;n3 nn allowahle she.arinJl ~rTl"~~ nf ()'OO psi detennine the required speed of shafl AB. (h) Solve pnn II. nss uming Ihat th e d;~meters of shaft ~ liB and CD ate. rt:' spccr..iv~ l y. 0.75 in. and 0.625 in.

n.

D

FIg. P3.78

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Probtoms

3 .79 A 5-ft-long solid sleel shaft of 0.875-in. diameter is to transmit 18 hp. Determine the minimum speed at which the shaft can rotate, know ing that G = 11.2 X 10" psi. that the allowable shearing stress is 4.5 ksi.
•

3.81 A steel sh
Fig. P3.82 and P3.83 3 .83 A 1.5.m-long tubular steel shaft of 38-mm outer diameter til is to be madeofa steel for Which 1.11 = 65 MPa and G = 77.2 GPa. Knowing that the angle of twist must not exceed 40 when the shaft is subjected to a torque of 600 N . m. detemline the largest inner diameter il2 that can be specified in the design. 3.84 The stepped s haft shown rotates at 450 rpm. Knowing thlll r = 0.2 in., detennine the largest torque T that cml be transmitted without exceeding an allowable shearing stress of 75{)O psi. 3.85 The stepped shaft shown rotMeS:1t 450 rpm. Knowing thm r = 0.5 in .. determine the maximum power thm can be tmns-mitted without exceeding un allowable shearing stress of 75{)O psi.

.-_--r--,- , 6 in

5 in

L ........~

="-I.----.l

Fig. P3.84 and P3.85

3 .86 The stepped shaft shown mllst rotate at a frequency of 50 Hz. Knowing that the radim of the fillet is r = 8 mm and the allowable shearing stress is 45 MPa. determine the ma.ximum power thal can be transmitted. 3 .87 Knowing lhat the stepped shaft s hown must transmit 45 kW at a speed of 2100 rpm, determine the minimum radius r of the fillet if an allowable shearing stress of 50 MPa is not to be exceeded 3.88 The stepped shaft shown must transmit -1-5 kW. Knowing that the allowable shearing stress in the shaft is 40 MPa and that the radius of the lillet is r = 6 mm. detennine the smallest permissible speed of the shaft.

Fig. P3.86, P3.87, and P3.88

171

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172

3 .89 [n Ihe stepped shaft shown . which has a full quaner-circll[ar fillet. the allowable shearing stress is 80 MPa. Knowing Ihat D = 30 mm. determine the largest allowable torqul' thai can be applied 10 the shaft if (a; d = 26 mm.

Torsion

(b) d = 24 mm.

= i w - rl )

3.90

[n the stepped shafl shown. which has a full quaner.drcular fillet ,

D = 1.25 in. and d = J in. Knowing Ulal the speed of the shafT is 2400 rpm and Ihat the allowab[e shearing STress is 7500 psi. detemline the maximum power that can be transmitted by the shaft.

3.91 A torque of magnitude T = 200 Jb . in. is applied to the stepped shaft shown. which has a full quaner.circular fillet. Know ing that D = [ in., determine the maximum shearing stress in Ihe shafT when (tl) d = 0.8 in .. (b) d = 0.9 in. Full

'l',artl'r-ton-'lLl"r mid tn ."d~.' orla.)(."" .,hali

('xh'lIIb

Fig. P3,89, P3.90, and P3.9t

*3.9. PLASTIC DEFORMATIONS IN CIRCULAR SHAFTS

When we derived Eqs. (3.10) and (3 .16), which defin e, respecti vely, the stress distribution and the angle of twist for a circular s haft s ubjec ted to a torque 1~ we assumed that Hooke's law ap plied throughout the shaft. If the yield strength is exceeded in some portion of the shaft, o r if the mate rial inyolvcd is a brittle material with a nonlinear shearingstress-strain diagram. these relations cease to be valid . The purpose of th is section is to develop a more general method - which may be used whcn Hookc's law does not apply- for dctcJ1l1ining the distribution of s trcs~cs in a solid circular shaft , and for computing the torque requirl!d 10 produce a given angle of twist. We fi rst recall that no specific stress-strain re lations hip was assumed in Sec. 3.3, whe n we proved thal the s hearing strain y varies lin-

Fig. 3.33

carly with the distance p from the axis of the shaft (Fig. 3.33). Thus. we may sti ll use this prope rty ill uur present ;malysis a nd write "Y =

where c is the radius of the shaft.

P

~

Y,.,a:<

(3.4)

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Assuming that the maximum value T m.~ of the shcaring stress T has bcen specified. the plot of T versus p may be obtai ned as follow s. We ti rst de teffili nc from the sheari ng-stress-strain diagram the val uc of Ym." corresponding \Q T rna, (Fig. 3.34), and carry this value into Eq. (3.4).

3.9. Plastic DefofTl1aUons In Clrc Lllar Shafts

Fig. 3.34

The n. for each value of p, we deteffiline the corresponding value of Y from Eq. (3.4) or Fig. 3.33 and obtain from the stress-strai n diagram of Fig. 3.34 the shearing stress T corresponding to this value of y . Plolting T against p yiclds the dc~i re d di stributi on of stresses (Fig. 3.35). We now recall that, when we. deri ved Eq . (3.1) in Sec. 3.2, we assumed no particular rciMion between shearing stress and strain. Wc may therefore usc Eq. (3. 1) 10 determine the tor(lue T corresponding to the shearing-s lress distributiun obtained in Fig. 3.35. Considering an annular clement of radius p and thickness dp. we express the clement of area in Eq. (3. 1) as (fA = 27rPllp and write T =

r

pT(27Tp dp )

o

0'

T == 27T

r

p2T

dp

(3 .26)

o

where T is the function of p ploned in Fi g. 3.35. If T is a known analytical function of )" Eq. (3 .4) may be used to express T as a funct ion of p, and the integral in (3.26) may be determined analytically. Otherwise, the torque T may be obtained through a numerical integration. Thi s computation becomes more meaningful if we note that the integral in Eq. (3.26) represents the second moment, or moment of inertia, with respect to the vertical axis of the area in Fig. 3.35 located above the horizontal axis and bounded by the stressdistri bution curve. An important value of the torque is the ultimate torque T u which causes fai lure of the shaft. This value may be determined from the ultimate sheari ng stress TU of the materi al by choosing Tmax == T U and carrying out the computations indicated earlier. However, it is found more convenient in practice to determine T u experimentally by twisting a specimen of a given materi al until it breaks. Assumi ng a fictitious linear distri bution of stresses, Eq. (3 .9) is then used to determine the corresponding maximum shearing stress RT : (3 .27)

r" ,.

o lllilll4

Fig. 3.35

---,p

173

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174

Tors!on

The fictitiou s stress Rr is called the modulus of I"IIptllr.. in tor.firm of the given material. It may be used tQ dt!tenninc thc ultimate IQrque Tu of a shaft made of thc smne material , but of different dimensiolls, by solving Eq . (3.27) lo r Tu. Since the act ual and the fictitiou s linear stress distributions shown in Fig. 3.36 must yield the same value Tu fo r the

" li T ,

f"

p

Fig. 3.36

ultimate torque, the areas they define must have the same moment of inertia with respect to thc vertical axis. It is thus clear that the modulus of rupturc RT will always be larger than the actual ultimate shcaring stress T u' Ln some cases, we may wish to detcrminc the stress distribution and the tOlTjllC T cnrres pnniling In :1 g ivcn :mg le o f tw ist $ . Th is mlly hc done by recalling the expression ubtai llcd in Sec. 3.3 for the shearing strain "y in tcnns of 4>, p, and the length L uf the shaft: p¢

y=-

L

(3.2)

With .p a nd L given. we may de termine from Eq. (3.2) the value of y corresponding to any given value of p. Using the stress-strain diagram of the material. we may then ubtai n the correspondi ng value of the shearing stress T and plot T against p. Once the shearing-stress distribution has been obtained, the torque T may be determined analytically or numericall y as explai ned earlier. "3.10. CIRCULAR SHAFTS MADE OF AN ELA$TOPLASTIC MATERIAL

, Fig. 3.37

Further insight into the plastic behavior of a shaft in torsion is obtai ned by considering the idealized case of a .w l id circular shaft made of (III elaslOplaslic material. The shearing-stress-stmin diagram of such a material is shown in Fig. 3.37. Usi ng thi s diagram. we can proceed as indicated earlier and lind the stress di stribution across a se,ction of the shaft for any value of the torque T. As long. as the shearing stress T does not exceed the yield strength Tr, Hooke's law applies, and the stress distribution across the sec tion is linear (Fig . 3.3Sa). with T ma, given by Eq. (3.9): Tc

Tm...

=j

(3.9)

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As the torque increases. T"',,, eventually reaches the va lue Ty (Fig.

3.10 . Circular Shahs Made o f an E1astoplasUc Mater1al

3.38b). Substituting this value into Eq . (3.9). and solving for the cor-

responding value of T, we obtain the value Ty of the torque at the onset of yield : (3.28)

The value obtained is referred to as the maximum elastic torque. since il is the largest torque for which the deformation remains full y elastic. Recalling that for a solid circular shan lIe = ~ 1T C], we have ( 3.29)

As the tOHlue is further increased, a plastic region develops in the shaft, around an elastic core of radius Pr (Fig. 3.38c). l,n the plastic region the stress is uniformly equal to Ty, while in the clastic core the stress varies linearly with p and may be expressed as 7, T= - P p,.

(3.30)

As T is increase:! . the plastic region expands until, al the limit , the defonnmi on is fully plastic (Fig. J.38d). Equation (3 .26) will be used to determine the val ue of the torque T correspond ing to a given radius Pr of the clastic core. Recalling that T is given by Eq. (3.30) fo r 0 :S P S Pr, and is equal to Ty for p y :S P $ c , we wrile T :=:

27T

I

r () r 0

' Tr p- p)' p dp

~

2 ~

"" 2-7Tp yTy + i-7T

Ty -

+

277'

"I

"

,

p-T ydp

"c 2 ] 3'lT p yT y

T == !.7TC]T.( I _ 1 p t ) 3 1 4 (..3

Q

0

(3.31) (,I

or. in view of &1. (3.29),

T =~Tr(1 - ,-p,) 4 ,J

c, (3.32)

where Ty is the maximum clastic torque. We note thai, as p y approaches zero, the torque approaches the limiting value 4

T" =

'3 Ty

0

p

(3 .33)

This value of the torclue, which corresponds to a fully plastic deformation (Fig. 3.38d), is called the ,,1(lslic torque of the shaft considered. We note that Eq. (3.33) is valid only lo r a solid eircufar shaft made of a ll

p

p,

e/astoplaslic material.

Since the distribution of ~·trail! across the section remains linear after the Ollsct of yield. Eg . (3.2) remai ns valid and can be used 10 express the radius p y of the clastic core in tenns of the angle of twist fIJ. If fIJ is large enough to cause a plastic deformation, the radius Pl' of the

(il)

Fig. 3.38

175

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176

elastic core is obtained by making y equal to the yield strain yy in Eq. (3.2) and solving fO[ the corresponding value py of the distance p. We have

Torsion

L yy

py

=----;;;

(3 .34)

Let us denote by rPr the angle of twist at the onset of yield. i.e .. when Pr = Co Making rP = rPr and PI' = (' in Eq. (3.34), we have Ly, c

~-

.p,

(3 .35)

Di viding (3 .34) by (3.35), member by member. we obtain the following rci:Hion :t (3 .36)

If we carry into Eq. (3.32) the expression obtained for py/c, we express the torque T as a function of the angle of twist dJ. 4 Tr ( 1 "';) T ="3 14(p1

r r

Fig. 3.39

•

13·37)

where Ty and rf y represent. respectively, the torque and the angle of twist at the onset of yield. Note that Eq . (3.37) may be used only fo r values of rP larger than cp) .. For cP < cPr, the relation between T and cP is linear and given by Eq. (3 . [6). Combining both equations, we obtain the plOt of T against 1> represented in FIg. 3.39. We check. that, as rP increases indefinitely. T approaches the limiting value Tn = jTy corre.. sponding to the case of a fully developed piasl.ic zone (Fig. 3.38d). While the value Tp cannot actually bc reached. we note from Eq. 0 .37) that il is rapidly apprmlched as rP increases. For rP = 2cPy. T is within about 3% of T il' and for .p = 30/). within about 1%. Since the plot of T against rP that we have obtained for an idealized elastoplastic material (Fig. 3.39) differs gremly from the she,uing-stressstrain diagmm of that material (Fig. 3.37), it is dear that Ihe shearingstress-strain diagram of an actual materi al cannot be obtained directly from a torsion test carried Ollt on a solid circular rod made of that material. However, a fairly accurate diagram may be obtained from a torsion test if the specimen used incorporates a portion consisting 01' a thin circular tube.t lndeed, we may assume that the sheari llg stress will have a constant value T in that portion . Equation (3.1) thus reduces to T= pk

where p denotes the average radius of the lube and A its cross-sectional area. The shearing stress is thus proportional to the torque, and successive values of T can be easily computed from the corresponding values of T. On the other hand, the val ues of the shearing strain y may be obtained from Eq. (3 .2) and from the values of.p and L measured on the tubular portion of the speci men. t EqU3tioll (3.36) JlljlJi~~ to any du~)1Ic material with a well·ddined yield point. sinte Its (\('rivJtiQn is inilqJendcm of the shape of !he .ltrc's-SU'ain diagram !>ryolld the yield p
..., lh"l th" I... n~ot], nf Iii" I"hular l"lrlion i. no lon~"r thun it~ d"unNn

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EXAMPLE 3.08 A solid circular sllaft. 1.2 m long and SO mm in diameter. is subjected to a 4.60 kN . m torque at each end (Fig. 3.40). Assuming the shaft to be made of an elastoplastic m:lterial wiLh a yield strength in sheM of ISO M Pa and a modulus of rigidity of 77 GPa, detennine (u) the mdius of the elastic core. (P) the angle of twist of the s haft.

Solving Eq. (3.32) for (pJe )' and substituting the values of T and T r . we have

= 4 _ 31" = (~)l c Ty

!!.r. c

= 0.630

4 _ 3(4.60 kN . m) 3.68 kN . m

0.250

PI' = 0.630(25 mm } = [S.S

mm

(b) Angle of Twist. We first detem\ine the angle of twi st <$J,. at the onset of yield from Eq. (3. 16) :

(3. 6R X 10' N·m )(1.2m)

TyL

<$Jy = JG = (61 4 x 10 9 m4 )(77 X 109 Pal = 93.4 X IO - J rad Fig. 3 .40

(a) Radius of Elastic Core. We first de termine the torq ue Ty at the onset of yield. USing Eq. (3.28) with Ty = ISO MPa. c =:0 25 mm.:lnd J = ~7TC4 = ht(2S X 10 ' m)4 = 614 X 10

'J

m'

Solving Eq. (3.36) for f/J and substitu ting the values obtained for ch y and pJc. we write

.p

~

<$Jy

-p y/c

=

93.4 X IO- J rad 0.6 30

148.3 X 10 1 rad

we write .

iT ,.

1 ,. =~

(6 14 X JO ~ m~)( 150 X [(f' Pa) 2SX 10 ·3 m

3.68 kN· m

I/J

= ( 148.3 X ]()

'3.11 . RESIDUAL STRESS.ES IN CIRCULAR SHAFTS

In the two prcceding sect ions, we saw that a plastic region wi ll develop in a shaft subjected to a large enough torque, and that the shearing stress 7 at any given point in the plastic region may be obtained from the sheari ng-stress· strain diagram 01" Fig. 3.34. If the torque is removed. the resulting reduction of stress and strain at the point considered will take place along a straight line (Fig. 3.4 1). As you wi ll see further in this section. the final val ue of the stress will not. in ge neral. be zero. There will be a residual stress al most points. and that stress may be either positive or negative. We note that. as was the case for the normal stress, the shearing stress will keep decreasing until it has reached a value equal to its maximum val ue at C minus twice the yield streng th 01" the material.

c

of-+ --+---: Fig. 3.41

J

360· rad) ( -) = 8.50· 211" rad

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178

Tors!on

T

T

Consider again the idealized case of lhe c1astoplastic materinl chllracterized by the shearing-stress-strain diagram of Fi g. 3.37. Assuming that the relalion between T and "y at any point of the shaft remai ns linear as long as the stress does not decrease by more thnn lTy. we can use Eq. (3. 16) to obtain lhe angle through which lhe shaft untwists as the torque decreases back to zero. As n result, the unloading of the shaft wi ll be represented by a straight line on the T-r:p diagr.ull (Fig. 3.42). We note that lhe angle of twist does not return to zero after the torque has becn removed. Indeed, lhe loading and unloading of the shaft resu lt in a permanent deformation characteri zed by the angle (3.3')

where r:p corresponds to the loading phase and may be obtai ned from T by solving Ell . (3.38). and where 1>' corresponds to lhe unloading phase and may be obtained from Eq. (3.[6). The. residual stresses in an elastoplastic material are obtained by applying the principle of superposition in a manner similar to thai described in Sec. 2.20 for an axial loading. We consider, on one hand, the stresses due to lhe application of the given torque T :.md. on the other, the stresses due tt) the equ:d and opposile 10r(lue which is applied to unload Ihe shaft. The tirs! group of stresses renecls the elastoplastic behavior of Ihe material during the loading phase (Fig. 3.43a). and Ihe

Fig. 3.42

,

p,

"

p

"

,.. (0 )

(Ii)

(d

Fig. 3.43

second group the linear behavior of the same material during Ihe unloading phase (Fig. 3.43b). Adding the two groups of stresscs, we obtain the di stribution of the residual stresses in the shaft (Fig. 3.43c). We note from Fig. 3.43c that some rc..~ idual stresses have the same sense as the original stresses. while others have the opposite se nse. TIlis was to be expected since. according to Eq. (3 .1), the relation fp (TdA) ~ O

must be verified after Ihe torque has been removed.

(3 .39)

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EXAMPLE 3.09 For the shaft of Example 3.08 determine (a) the permanent

The per manent twi.;( is therefo re

twi st, (b l the distribution of residual stresses. after the 4 .60 kN . m torque has been removed.

(a) Permanent Twist. We recall from Example 3.08 tlwt the ang le of twist corresponding to the given torq ue i~ tjJ = 8.50°. The angle tjJ' through which the shaft untwists a.~ the torque is removed is obtained from Eq. (3. 16). Substitut ing the given dat;L

L = 1.2m

tP l ' = tP - 4>' = 8.s0~ -

1' :""" = T' J

,U1d the value J ", 614 X IO- " m.j obtaineJ in Ihe solution of

JG

(614 X 10

9

= 116.8 X 10 -) rad

~ ~(4~.6O =-X~'~O~'~N:,.~'~n~)(~25,-,;X:..:.:'O"--',,":!.,) 6 14 X 10

Q

m'

= 187.3 MPa

Ex.ample 3.08, we ha ve

1lJ')(77 X 109 Pa)

"'" 1.81 0

(b) Residual Stresses. We recall from Exampl cJ.OR th:ll the yie ld strength is Ty = 150 MPa and that the radius of the elastic core co rresponding to the gi ve n torque is PI' = 15.8 mm. The distribution of the stresses in the loaded shaft is thus as s hown in Fig. 3.4411. The distribution of stresses due to !lIe opposite 4.60 kN . 111 torque requireJ to unload the shaft is linear and as shown in Fig. 3.44"- The ma.1(imum stress in the dis tribution of the reve rse stresses is obtained front Eq. (3.9):

G=77X IO· Pa

,p' = TL = c:'("4."60" ,, X,,'CO C_'N ~.:,:""')"(,,,.2:;";:')'-c

6. 69~

Superposing the two di stributi ons of SU'esses, we obtain the residual stresses shown in Fig. 3.44c We check tlmt. even though the reverse stresse~ exceed the yield ~trength T y, the assumption of a linear distriblllion of these stresses i ~ valid. si nce they do 110t exceed 21')".

360' tjJ' = ( 116.8 x 10 - 1 rad) - - - = 6.69" 21t rad

,. (.\lPa)

;(~ lPa)

[50

/

p

,,,

P -3i.3

,,

IS .limm

-118.4 ----

- -:

!5.1i nlln

2~ nllll

-18i.3 ------(" )

(c)

Fig. 3.44

179

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SAMPLE PROBLEM 3.7 Shaft AB is made of a mild steel which is assumed to be eklstoplastic with G = 11.2 x 10° psi and T)" = 21 ksi. A torque T is applied and gradually in~ l.1"eased in magnitude. Detemline the magnitude of T and the corresponding angle of twist (a) when yield first occurs. (b) when the deformation has become fully pklstic.

SOLUTION Geometric Propt' rtil'S

The geometric properties of the cross section are C,

, II.

T r * 3'i'i ki l" 'Ji

=

1( 1.5 in.) "" 0.75 in.

("2

O nset ofYicld. For

Tn""

1")" = ]7.7 kip· in. "IiI Cl

and y = yy in Eq. (3.2) and solving for

4J. we obtain the value

(2 1 X 103 psi )(60 in.)

(', = 0.75 ill.

( 1.125 in.)(I1.2

... /

T ,• - Il.l lop i l'

" = 21 los;

x

106 psi)

0.100 rad

h. Fully Plastic Ddo rm ation. When the plastic zone reaches the inner surface. the stresses are uniforml y distributed as shown. Using Eq. (].26). we write

.......

b

= 1 1T(2 1 ksi)[( 1.125 in.)3 - (0.75 in.)J]

I\ 1-\

"'I'" 11.59' T

·l~

T,

= 44. 1 ki p · in . ....

When yield first occurs on the inner surface. the deformation is fully plastic: we have from Eq. (3.2): y,L

TI'L

(2 1 X 10' psiX60 in.)

C,

C,G

(0.75in.)( 11 .2X loopsi)

rPr =-~-=

T,.

0.150rad

4J, =

>, 180

= 1.125 in.

= Ty = 21 ksi. we find

. Making p = of 4J,':

;;::

= !(2.25 in.)

1 = i1T(C~ - c1) = ~ 1T [ (1. 125 in.)4 - (0.75 in. )4] = 2.02 in 4

>1

8.59~

"IiI

For larger angles of twist. the torque remains constanl: the 1"-th diagram of the shaft is as shown.

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SAMPLE PROBLEM 3.8 For the shaft of Sample Prob. 3.7, determine the. residual stresses and the permanent angle of twist after the IOrque T" = 44.1 kip ' in. h:L~ been removed.

SOLUTION Referring [0 Sample Prob. 3.7, we rec~llihal when the plastic zone first reachl"{/ the inner surface. the applied torque was 7;. = 44. [ kip' in. and the corresponding angle of twist was CPr = 8.59°. These values are shown in Fig. J. Elastir Unloading. We unload the shaft hy applying a 44.1 kip ' in. torque in the sense shown in Fig. 2. During thi s nnloading. the behavior of Ihe material is linear. Recalling from Sample Prob. 3.7 Ihe values found for CI. ("l, and J, we obtain the following stresses and angle of twist:

7C-,

Tn..... =

J

T MID

T

(44.1 kip ' [11.)(1.125 in.) 2.02 in'

= Cl

""

m'" -

11 = -

JG

=

0.75 in. (24.56 ksi )- - -.- = 16.37 ksi 1.125 tn . (44. 1 X lOJ psi)(60 in.) O.1170r:ld = 6.70 0 4

~

t.:1

cp'

24.56 ksi

(2.02 in )( [ 1.2 X lOb psi)

Residual Stresses and Permanl' nt Th·isl. The results of the loading (Fig. \) and the unloading (Fig. 2) are superposed (Fig. 3) to obtain the residual stresses and the pennanent angle of twist CPr

+ (2)

(3)

11, .37 k
~ - H~~;

,I

W ','

3.'~";

~~

T,. .. ~ t l ~ip·i"

/ = 8.59"

q,' = 6.70"

2456 h i


181

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PROBLEMS

~ c = 3:2 """

T'

3.92 The solid circular shaft shown is made of a steel that is assumed to he elastoplastic with 1"). = 145 MPa. Delt!rmine the magnitude T of the applied torque when the plastic zone is (a) 16 mm deep. (h) 24 mm deep. 3 .93 A 1.25-in.-diameter solid rod is made of an elastoplastic material with 1")" = 5 ksi. Knowing th:ltlhe elastic core of the rod is of diameter I in .. determine the magnitude of Ihe torque :lpp!ied to Ihe roo.

Fig. P3.92

3 .94 A 2-in.-diameter solid shaft L'i made of a mild steel that is assumed to be elastoplastic with 1"1' "" 20 ksi. Determine the ma."(imum shearing stress and the radius of the elastic core caused by the application of a torque of magnitude (a ) 30 kip ' in .. (b) 40 kip' in.

3.95 The solid shaft shown is made of a mild steel that is assumed to be elastoplastic with G = 77.2 GPa and 1")' = 145 MPa. Determine the milxi mum shearing stress and the radius of the elastk core caused by the application of II torque of m
Fig. P3.95 and P3.96

3.96

The solid shaft shown is made of a mild steel that is assumed to T)" = 145 MPa. Determine the radius of the elastic core caused by the application of a torque equal to J. I Tr . where Ty is the magnitude of Ihe torque al the onset of yield.

be eiastoplastic with

3 .9 7 It is observed Ihat a straightened paper clip can be twisted through several revolutions by the application of a torque of approximately 60 mN . Ill. Knowing that Ihe diameter of the wire in the paper clip is 0.9 mm. detemline the approximate value of the yield stress of the steel.

Fig. P3.98

182

3 .98 The solid circular shaft shown is made of a steel Ihat is assumed to be elastoplastic wilh 1" )" = 21 ksi and G = J 1.2 X 10~ psi. Delermine the angle of twist caused by the application of a torque of magnitude (a) T = 80 kip' in .. (b) T = 1]0 kip' in.

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3 .99 FOf the solid cifcular shaft of Prob. 3.95, determine the angle of twis t caused by the applicaTion of a torque of magnitude. (a) T = 600 N . m. (b) T = 1000 N . m.

Probtoms

3 .100 The shaft AB is m,lde of a material that is elastoplastic with Ty = 90 MPa and G = 30 GPa. For the loading shown . determine (al the radius of the elastic core of the shaft. (b) the angle of twist at end B .

.,

Fig. P3.tOO

3.101 A 1.1S-in.-diameter solid circular shaft is made of a mntenal that is assumed to be eiastoplastic with Ty = 18 ksi and G = 11.2 X I O~ psi. For an 8-ft length of th.e shaft. determine the maximum shearing stress and the angle of twist caused by a 7.5 kip ' in. torque.

3.102 A 1l.75-in.-diameter solid circular shall is made uf a material til;}t is assumed to be elastoplastic with 1't' ="20 ksi and G = 11.2 X 10" psi. For a 4-ft length of the ~haft. de termine the maximum shearing stress and the angle of twist caused by a 1800 lb· in. torque. 3.103 A s olid circular rOO is made of a matenalthat is assumed to be clastoplastic. Denoting by T y and tPy, respectively. the torque and the angle of twist tit the. onset of yield. detenllille the angle of twist if the torque is increased to (a) T = 1.1 Tn (hI T -"" L25 Tt', (tl T = 1.3 Ty.

3.104 A 3-ft-Iong solid shaft has a diameter of 2.5 in. and is made of a mild s teel th:!t is assumed to be clastoplastk with Ty = 21 ksi and G = 1l.2 x 1O~ psi. Determine the torque required to [wist the shaft through an angle of (a) 2S. (b) 5°. 3.105 For the solid shaft of Proh. 3.95. determine (a) the magnitude of the torque T required to twist the shaft through an angle of 15G • (b) the radius of the corre;;ponding clastic core. 3.106 A hollow shaft is 0.9 m long and has the cross section shown. The steel is assumed to be eiastoplaslic with Ty = 180 MP'd and G = 77.2 G Po. Determine the lIpplied torque and the corresJXjnding angle of twist (iI) ;1t the onSet of yield. (b) when the plastic zone i ~ 10 mm deep. 3.107 The stul is Determine plastic. (hi

A hollow shaft is 0.9 m long and has the cross section shown. assum!d to be elastoplastic with Tr = 180 fliPa and G = 77.2 GPl1. the la) angle of twist at which the section first becomes fully the corresponding magnitude of the applied torque.

@\;]"I" _

-.l

F1g. P3.106 and P3.107

183

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184

TorsIon

3 .108 A steel roo is machined to the sh
3.110 Using Ihe slress-slrnin diagram shown. determine (a) the torque that causes a maximum shearing stress of J5 ksi in a 0.8"in.-diameter .'iOlid rod, (b) the corresponding angle of twist in a 20-in. length of Ihe roo. I

----------------

-

:v(1 -

().0020.<1040.ll.tl11l

()

y

Fig. P3. 110and P3.111

3 . 111 A hollow shaft of outer and inner diameters respectively equal to 0.6 in. and 0.2 in. is fabricated from an aluminum alloy for which the stressstrain dia&rnm is given in the diagrnm shown. Detemline the torqne required to twist a 9-in. length of the shaft through 10°. 3.112 A so lid aluminum roo of 40-mm diameter is su bjected to a torque that produces in the rod a maximum shearing strain of 0.008. lJsing the 1"-"'1 diagram shown for the aluminum alloy used. determine. (aj the magnitude of the torque applied to the rod. (b) the angle of twist in a 75Q..mm length of the rocL

f-

,.

/'

)

75

/

/

/ O.()02

O.()(14

C·.OO6

0 .008

o.Ow y

Fig. P3.112

3.1 13 relation

The curve shown in Fig. P3.1 [2 can he
Using this relation and

= 27.8 X 109")' Eq~.

1.390 X IOl 2y2

(3.2) and (3.26). sollie Proh. 3.1 [2.

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Probloms

3 .114 The solid circular drill rod AB is made of a steel that is assumed to be elastoplastic with 'Tr = 22 ksi and G = 11.2 X lif psi. Knowing that a torque T = 75 kip' in. is applied to the rod and then removed. determine the maximum residual shearing stress in the rod. 3.11 5

In Prob. 3.1 [4. determine the permanent angle of twist of the rod.

3. 116 The hollow shaft shown is made of a steel thaI is assulIlt:d to be elastoplastic wIth 'Ty = 145 M Pa and G = 77.2 G Pa. The magnitude T of the torques is slowly increased until the plastic zone first reaches the inner surface of the shaft: the torques are then removed. Determine the magnitude and location of the maximum residual shearing stress in the rod.

Fig. Pl.114

00 "lin

Fig. P3.116 3. 117 In Prob. 3. [ 16. detennine the permanent angle o f twist of the rod. 3.118 The solid shaft shown is made of a steel thaI is assumed to be el:lstoplastic with '1). = 145 MPa and G = 77.2 GPa. The torque is increased in magnitude lmtll the shaft has been twisted through 6°: the torque is then reo moved. Detemline (a) the magnitude and location of the maximum residua l shearing stress. (b) the permanent angle of twist.

lr ,\ L 0.6 '" ______

(1(; nnn

F[g. Pl.118 3.119 A torque T applied to a solid rod made of an elastoplastic material is increased lmtil the rod is fully plilstic and then removed. ({I) Show that the distribution of residual shearing stresses is as represented in the figure. (b) Detemline the magnitude of the torque due to the stresses acting on the portion of the rod located within a circle of mdius cu. 3.120 After the solid shaft of Prob. 3.1 18 has been loaded and unloaded as described in that problem. a torque T I of sense opposite to the original torque T is applied to the shaft. Assuming no change in the value of Ifr. determine the angle of twist Ifl for which yield is' initiated in this second loading and compare it lVith the angle Ifr for which the shaft started to yield in the original loading.

tv

Fig. Pl.119

y

J

tt,

185

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186

'3.12. TORSION OF NONCIRCULAR MEMBERS

TorsIon

T

Fig. 3.45

The formulas obtained in Secs. 3 .3 and 3.4 fo r the distributi ons o f strain and stress under a torsional loading appl y only to members with a circular cross section. Indeed , their deri vation was based on the assumption that the cross section of the member remained plane and Ulldi storled, and we saw in Sec. 3.3 that Ihe validi ty of this assum plion depends upon the lIXi.\·Ylllllletry of the member. i.e., upon the faci that its appearance remains the same when it is viewed from a fi xed position and rotated about its axis through an arbitrary angle. A square bar, on the other hand. retains the same appearance on ly when il is rotated through 90° or 180°. Followi ng a line of reasoning similar to lllat used in Sec. 3.3, one cou ld show Ihal the diagonals of the square cross sel:tion of the bar and the lines joining the midpoints of the sides of thaI. section remai n straighl (Fig. 3.45). However. because of the lack ofaxisymmetry of the bar, any other line drawn in ils cross section will de.fonn when the bar is twisted , and the cross sec tion itself will be warped out of its original plane. It follows that Eqs. (3.4) and (3.6). which define. respectively, the distributions of strai n and stress in an elastic circular shaft. cannot be used for no nci rcular members. For example, it would be wrong to assume that the shearing stress in the cross sectio n of a square bar varies linearly with the dislance from the axis of the bar and is, therefore. largest at the corners of the cross section . As you wiJl sce presently. the shearing stress is actually zero al these points. C:onsirier a smal l c uhic e le rnenl locaTed:l T a c() rn,~ r nf The Crf)<; S section of a square bar in torsion and select coordinate axes parallel to Ihe edges of the element (Fig. 3.4oa). Since the face of Ihe element perpendicu lar to the y axis is part of Ihe free surface of the bar, all stresses on this face must be zero. Referring 10 Fi g. 3.46b. we write (3.40)

For the same reason. aIL stresses on the face of the element perpendi cular 10 the :: axi s must be zero, and we write 7"

= 0

(3.4 ] )

It follows from the first of Eqs. (3.40) and the first of Eqs. (3.41) that Fig. 3.46

Fig. 3.47

7\~

= 0

(3.42)

Thus. both components of the shearing stress 011 the fac e o f the clement perpendicular to the axis of the bar are zero. We conclude that there is no shearing stress at the comers of the cross section of the bar. By twisting a rubber model of a sqllare bar. one easily verities that no defonnations- and, Thus, no STresse,- occur along the edges of the bar, while the IargcsT deformations- :md. Thus, the largest ~tresses - occur along the center [inc of each of the faces of the bar (Fig. 3.47). The detennination of the STre..<;ses in noncircu[ar members subjected 10 :I torsional loading is beyond the scope of this text. However, results obtained from the mathemntieal Theory of d:lsticity for straight bars wiTh a IIlIiform rectangular cross section will be indicated here for convenience.t t Scc s . P. Timoshcnko and 1. N. Goodier. Tllemy o!Eilw,-dt)", Jd cd .. McGraw-Hili. New York. t969. sec. 109.

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Denoting by L the length of the bar. by (/ and b. respectively. the wider and narrower side of its cross section. and by T the magnitude of the torques applied to the bar (Fig. 3.48). we lind that the maxim um s hearing stress

Fig. 3.48

occurs along the center line of the wider f ace of the bar and is equal to T

7!tl,U

(3.43)

= --,

cllIlr

The angle of tWi St. on the other hand. may be expressed as TL

¢~--

(3.44)

cl l1b3G

The cocflicients CI and C2 depend oilly upon the ratio alb and arc given in Table 3.1 fo r a number of values of that mtio. Note that Eqs. (3 .43) and (3.44) arc valid onl y within the clastic range. TABLE 3 .1. Coefficients for Reclan ular Bars in Torsion

a/ b

c,

c,

1.0 1.2 1.5 1.0 2.5 3.0 4.0 5.0 10.0

0.208 0.219 0.231 0.246 0.258 0.267 0.281 0.291 0.312 0.333

0.1406 0. 1661 0. 1.958 0.229 0.249 0.263 0.281 0.291 0.312 0.333

00

We note from Table 3. 1 that fo r (jIb ~ 5. the. coefficients CI and Cl are equal. It l11ay be shown that for sllch values of alh, we have. (for alb 2" 5 Dilly)

(3.45)

The distribution of shearing stresses in a l10ncircular member may be visualized more easi ly by using the memhran£' analogy. A homogeneous elastic membrane attached to a fixed frame and subjected 10 a uniform pressure 011 Olle of its sides happens to constitute an lInalog of the bar in torsion, i.e., the d(!terminatiOIl of the defonnation of the membrane

3. t2. Torsion 01 Nonclrculaf MembcB

187

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188

d epcnus upon tbe suluLio ll of thc .aUk: panial ull"fercnt ial cquatiou a~ the determination of the s hearin g strl.'sses in thl' bar. t More ~pec ifi­

TDrnlcn

"

.

-\"

Fig. 3.49

cally. if Q is :1 point of the cross .\;ection o f the bar and Q' the corn> sponding point of the membrane (Fig. 3.49). the shcaring stress 7 at Q wil l have the s ame dircction as the horizontal tange nt to thc membrane al Q', lmd ils magnitude w ill be proportiona l to the maximum slope of the membrane at Q'.t Furthennore. the applied torque wi ll be proportional to th ~ volume between the membrane and the plane of the fixe d frame. In the case of the membrane uf Fig. 3.49, which is attached to a reCtangu lar frame, Lhc Stcepcs t slope oceurs at the midpoint N' of tile larger side of th!! frame. Thus. we verify that the maxim um sheari ng stress in a bar of rectangular cross section will occ ur at the m.idpoitlt N of the larger side o f that section. The membrane analogy ma y be used j ust as effectiv ely to vis ualize the shearing SlrC~SCS ill ,IllY;;lraight bar u r unifunn, nondn.;ular c:rus ~ section. In p:u1.icul:"l r, let us con s id ~r J;eve ral thin-wal led members with the cross sections shown in Fig. 3.50. which are s ubjected tu (he SJme torque. Using (he membrane analogy to help llS visuali ze the shearing stresses, we no(e tha!, since the same to"lue IS ap plied tu each member. the s..1 mc: volu me will be located under each membrane,
Fig. 3.50

c,

t Scc ibid sc"c. 107. ~Thi.\ is die slope ITlCaJurcd in ~ dim::tion pcrpcndirut:r to Ihe horizontal umg::nt ~t Q' ~rl {"""M al'" hI' ,hnwft Ih,,. Ih,' "neli' nfTw;'Trn~)' hr o1Nrrn,;n ro1 fmm F'l (1 M ) w;lh = 0.333 .

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3.1 3. Thin-walled Hollow Shafts

'3.13. THIN-WALLED HOLLOW SHAFTS In the preceding section we saw that the determination of stresses ill noncircular members generally requires the use of advanced mathematical methods. In the case of thin-walled hollow noncircular shafts, however. a good approxi mation of the distribution of stresses in the shaft can be obtained by a simple computation. Consider a hollow cylindrical member of lIollcirclllar section subjected to a tOfsionalloading (Fig. 3.5 1).t While the thickness 1 of the wall may vary wilhin a transverse section. it wi ll be assumed that it rcmains small compared 10 Ihe o lher di mensions of the member. We now detach from the member the colored portion of wall AB bounded by two transverse planes at a distance fI..r from each olher, and by two longilUdinal planes perpendicular to the wall. Si nce the portion AB is in equilibrium. the sum of Ihe forc es exerted on it in the lo ngitudinal x direction must be zero (Fi g. 3.52). But the only fo rces involved arc the shearing force s FA and FB exerted on the ends of portion AB. We have therefore FA - Ffl = 0 T,I

(3.46)

We now express FA as the product of the longitudinal shearing stress on the small face at A and of the arca 1..1 fl.x of that fac e:

Fig. 3.51

F

~"

F,,'l _ ~~.\
Fig. 3.52

FA = "TA(t,l .1.x )

We nole that. whi le the shearing stress is independent o f the x coordimHe nf Ihe

rOinT

c(jn ~ i(tcretl ,

iT ma y vary l1cross lhe wa ll : Ihlts , T,I rer-

resents the average value of Ihe stress computed across Ihe waiL Expressing F B in a simi lar way and substituting for FA and Ffl illlo (3.46), we write

(3.47)

Since A and B were c hosen arbitmrily, Eq. (3.47) expresses that the product Tt of the longitudinal shearing stress T and of the wall thickness 1 is constant throughout the member. Denoting thi s prO
(3.48)

Fig. 3.53

We now dctllch a small clement from the wall portion AB (Fig. 3.53). Since the upper and lower faces of this element are part of the [ree surface of the hollow member, the stresses on these faces are equal to z~ro. Recalling relations ( 1.21) and ( 1.22) of Sec. 1.12, it foll ows that the stress components indicated on the other faces by dashed arrows are alst) zero, while those represented by solid arrows are equal. Th us. the shearing stress at any point of a transverse section of the hollow member is parallel to the wa ll surface (Fig. 3.54) and its averagc value computed across the wall satisfies Eq . (3.48). Fig. 3.54 i"Th~

wall of the Incmbcr must cnclose a single cavily and mU.II n01 be slit ope n. wordl. Ihe member shou ld be lopologically equivalent 10 a hollow ei~ular shaft.

tn olher

189

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190

TDrnlcn

At thi s po illt w t call IIOle: an ruml ogy U\;'[ WI!CII tht: dlsuiiJutiou of the s hearing nr~.,ses T in the l.J"an ~v('rsl.' sec tion or a thin-walled hull ow shaft :Ind the di stributio n o f Ihe veloc ities )I in waler n owing through a clost!tl channel of unit de plh and vari;lble willth. While the velocity \' of the wate r van es from point 10 pOint o n account Df the va riati on in the wid th I of the channel, Ihe rale of n uw. q = 1'1. re mai ns conslanl thro ugho ut the cha nnel. jU51 as 7/ in Eq . (3.48) . 8 eeaust! of thi s anal· ogy. the proouct q "" 71 is referred 10 a~ l.he shear j l(Jw in the wall 0 1" the hollow shafl. We will no w deri ve a re lmio n betwee n the torque T applied 10 a hollow membe r and the s he.1f n ow q in ils wall. We consider a s mall element the wall seclion. of length ds (Fig. 3.55) . The a rea of the cl· e me nt is dA = / tis, and the magnilude of the shearing force dF e xerted o n the cleme nt is

or

FIg. 3.55

,IF

= T tlA

=

T(t ,.'s ) = (Tt) ,1.1·

= '7

rls

(1.49)

The moment liMo of thi s force abo ut :m arbitrary point 0 wilhin l.he cav ity o f Ihe mern ber lllay be obtained by multiplying dF by Ihe perpendicular dl~t a n ce p hu m U to Ihc Ime 01 ac.tlon o t dt'. We have

dM o = p tlF

= p( qds)

=

q(pds)

(3.50)

BUI Ihe produci (J d.r is equal to twice the arC;:l e/(f of the colored tria ngle in Fig. 3.56. We thus have

tlMo

=

q(2e!11)

(3.5 1)

Fig. 3.56

Since tJle integral a round the wall sectio n o f the le tt-hand me mber of Eq. 0 .5 1) rcpre~e nl s Ihe stun of Ihc momellls of all Ihe eleme ntary sli(!aring forces exerted on Ihe wall sectio n. and since this s um is equal to the rorque T llpplied 10 the hollow me mber, we ha ve

T

=== <j:i

rlMo

= <j:i

q(2 dd)

The shear n ow q being a co nSlant we write

T

=.

2ct.t

(3. 52 )

whe re If ig the area bounded by the center line of the wall c ross seclion (Fig. 3.57) .

Fig. 3.57

The shearin g stress T ,11 any give n po int o f Ihe wall nlay be expressed in te rms of lhe torque T if we s ubstiltlte fo r q from (3 .48) into (3.52) uml sulvt fur T the clluutiu n u btaine d. We ha vc r = -

T

2111

("l.11)

where t is Ihe wall thickness at the point cons idered and (f Ihe area bounde d by the center line. We recallth nt T represellts lhe a ve ra5e vaJuc

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oftlie shearing stress across the wall. However, for clastic defonnations the distribution of stresses across the wall may be assumed unifonn, and Eg . (3.53) wi ll yiekl the actua l value of the shearing stress at a given poim of the wall. The angle of twist of a thin-walled hollow shaft may be obtained by using the mclhod of energy (Chap. II). Assuming an elastic deformation, it may be shownt that the angle of twist of a thin-walled shaft of Ic ngth L and modulus of rigidity G is

3.13. Thin-wailed Hollow Shafls

191

(3.54)

whcre the integral is computed along the center line of the wall section.

EXAMPLE 3.10 Structural aluminum tub ing of 2.5 x 4-in. rectangular cro~s section wns fabricated by extrusion. Determine the shearing slre~s in each of the four walls of a ponion of slich lUbing when it is subjected to a torque of 24 kip' in., assuming (II) a unifonn O.I60-in. wall thickness (Fig. 3.5811). (b) that, a.~ a resull of defective fabrication. wa ll s AB and AC are 0.120-in. thick. and walls ED and CD are O.lOO-in. Thick (Fig. 3.58b) .

(a) Tubing of Uniform Wall Thickness. bounded by the center line (Fig ..).59) is (t

1----:-- 4 ill

= {3.84 in.)(2.34 in.) = 8.986 in!

Since the thickness of each of the four walls i ~ I "" 0.160 in .. we find from Eq. (3.53) that The shearing stress in each wall

;, T

A

The area

T = -=

2m

24 kip . in. 2(0. J 60 in.) (8.986 in!)

8.35ksi

----II

r: 2L L~====O=.=IG=.~;'='=-=.~

B

t

O. too in

C

J)

C

(, )

, ~ ."".

I-

-

---11

8

0.t20il1

(b) Tubing with Variable Wall Thickness. Observing that the area (f bounded hy the center line is the same as in part a. and substituting successively 1 = 0.120 in. and I = 0.200 in. into Eq. (3.53). we have

2.5 ill.

LL===o=.200::;:t'=",-:::Ji-c

D

I

Fig. 3.59

1

D

TAB

=

1"A C

=

24 kip' in.

,

2(0. 120 in.)(8.986 in-)

and 24 kip' in.

Fig. 3.58

11.13ksi

r 8D

=

T eD

= 2(0.200 in.)(8.986 in'}

6.68 ksi

We note that lhe stress in a given wall dept:nds on ly upon its thickness.

t Sce Prob. 11.70.

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SAMPLE PROBLEM 3.9 Using 1'.1 1 = 40 MPa, uetennille the largest lorque lhal may be applied to each of the brass bars and 10 the brass lube shown. NOle thai lhe lWO solid bars have lhe same cross-sectional area. and that the square bar and square lube have lhe same oUlside dimensions.

(3)

SOLUTION L Har with Square Cross Scclion. For a solid bar of rectangular cross scclion thc maximum shearing stress is given b} Eq. (3.43) T Tn .. ,

where the coefficienl

Cl

= c lab!

is oblained from Table ].1 in Sec. 3. 12. We have u b

a=b=0.()40m

For

Tn""

1.00

T. =

40 Ml'a =

( ' lab l

::-;=c::T".~c-" 0.208(0,040 m)}

2. Har with Rl'ctangu lar Cross Section. (/ = O.064m

1 = 6 ",,,,

I

.IL :['. ~~:] ]'.~ ", I

~ 3""",-I1

~ . iO""l ~

40MPa

c lab-

Th~.

u - = 2.56 b

= 0.259

T,

~-'-,

3. Squal1' Eq. (].53)

T, 0.259(0.(}64 m)(O.025 m)l

For a lube of lhickness t. the shearing stress is given by

,

~

T 2/(1

~-

where (1 is the area bounded by lhe center line of lhe cross section. We have {/ = (0.034 m)(0.034 m) = 1.156 X 10- ) m!

We substitute torque: 1' = -

T

2111

192

Cl

T1 = 5]2N ' m oOil l

We now have

b = 0.025 m

IlIlerpolating in Table 3.1: =

= 0.208

= Toll = 40 MPa, we have

1'm.u

T """

Cl

T

=

1', 11

= 40 MPa and / = 0.006 m and solve for the allowable

40 MPa =

:;;;:c::;:c-,:;-T~,,::-c:-:::-,,, 2(0.t>06m)( 1.156 X 10

J

m !)

7j=555N·m .

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PROBLEMS 45"' 111 (1/ )

3 .121 Using roll = 70 MPa and G = 27 GP:!. determine roreach of the aluminum bars shown the largest torque T t1wt can be applied anu the corre· sponding angle of twist at end B. 3 .122 Knowing that the magnitude of the torque T is 200 N . m :lI1d G = 27 GPa. detennine for each of the aluminum bars s hown the maximum shearing stress and the angle of twist at end B.

tlW\

25 """

(b)

251111 11

I

900 3.123 Using Tal l = 7.5 ksi :lfld knowing that G = 5.6 X [O~ psi, determine for each of (he cold-rolled yellow brass bars shown the [argesllOrque T Ihal can be applied and the corresponding angle of (wist at end B. 3 .124 Knowing that T = 7 kip ' in. and that G = 5.6 X IO~ psi. determine for each of the cold-rolled yellow brass bars shown the maximum shearing stress and the angle of twist of end B. 3 .1 25

"""------I

"E T

Fig. P3.121 and P3.122

<..L..,

"Ill(: lorque T causes a lO\al;OJI of 2 0 al eud 11 of the sta;uJess Slee]

bar shown. Knowmg that b = 10 mill and G = 75 GP,., detemline the maximum shearing stress in the bar.

(b)

Fig. P3.123 and P3.124

Fig. P3.12S and P3.126

3 .126 'nlt torque T causes a rotation of 0.60 at end B of the aluminum bar shown. KnowlIlg that h = 15 mm and G = 26 GPa. determine the maxi· mum shearing stress in the bar. 3.127 Detemline the largest allowable square cross section of a steel shaft of length 20 fT if the maximum shearing stress is not to exceed 10 ksi when the shaft is twisted through one complete revolution. Use G = 11.2 X 10~ psi. 3. 128 Determine the largest allowable length of a stainless steel shaft of ~ X ~ -in . cross section if the shearing stress is not to exceed 15 ksi when the shaft is twisted through 15°. Use G = 11.2 X 106 psi.

193

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194

Torsion

3.129 Each of the three steel bars shown is subjected to a torque of magnitude T = 275 N . m. Knowing that the allowable shearing stress is 50 MPa. determine the rcquired dimension b for each bar. 3 .130 Each of the three aluminum bars shown is to be twisted through an angle of 2°. Knowing that b = 30 mm, T ,II = 50 M Pa. and G = 27 G Pa. determine the shortest allowable length of each bar. 3.131 Each of the three steel bars is subjected to a torque as shown. Knowing that the allowable shearing stres, is 8 ksi and that b = 1.4 in .. determine the maximum torque T that can be applied to each bar. 3.132 Each of the three aluminum bars shown is to be twisted through an angle of 1.25°. Knowing that b = 1.5 in .. T = 7.5 ksi.and G = ].7 X 10~ psi. determine the shortest allowable length for each bar. 3.133 ShafL~ A and B are made of the same material and have the same cross-sectional area. but A has a circular cross section and 8 has a square cross se.ction. Determine the ratio of the maxinRim torques TA and 1's that can be safely applied to A and 8, respectively.

Fig. P3.t29, P3.t30, P3.t 3t , and P3.t 32

Fig. P3.133 and P3.134

3_134 Shans A and 8 are made of the same material and have the same length and cross-sectional area, but A has a circular cross section and B has a square cross section. Determine the ratio of the maximum values of the angles cP,\ and
Fig. P3.135

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3 .1 36

A

J~lll·loJlgMeel

angle

ha~

an L203 X 152 X 12./ CIOl\S

~ectiun.

Problem.

From Appelldix C \ve find that the thickness of the section is 12.7 nun and lh
T

Fig. Pl.136

" 3.137 An 8-ft-long steel member with a W8 X JI cross section is subjected to as-kip' in, torque. The properties the rolled-steel section are given in Appendix C. Knowing that G = 112 X 10' pSI. determine (a) ihe maxi-

or

mum shearing stress along

l im~

Q

{I. (b) the max imum shearing

stre.,,~

alon.g line

bob, (d the angle of twi,\. (Hinl: consider the web and flan!,!;cs sepamtely allll obtain a re lation between the torques ex~rted 011 the weh and a nange. rive ly. by expressing Ihm the .resulting :mgles of (wlsi are equal.)

resp~c­

Fig. Pl.138

3 .1 38 A 3-m-long steel member has a W250 x 58 cross f,ection. Knowing that G = 77.2 GPa and that the allowable shearing stress is 35 M P
\\"1

x: 31

Fig. P3.137

195

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196

3.139 A tOlque T - 5 kN . III is applied to a hollow , haft having the cross ~ection shown. Neglecting the effect of stress concetUralions. determin e th~ shearing stresS at poinL~ u and b,

TDrnlon

10m",

I f--I\nnn

125 "'"

3 , 140 A torque T = 750 kN . m is applied \0 tlltl hollow shaft ~hown that has ~ uniform 8-mm wallthkkness. Neglecting the effect of stress con· cen tral iom. determine the shelling Slress at points u and b.

!

-

'\ W

-lim".

Y.

I -

I, I 7Smm - - j

Fig. P3.139

I,

Fig. P3.14D

3 .141 A 75[)·N . III torque is applied \() a hollow shaft having the cross section shown and a unifonn 6-mm wall thic kness. Neglecting the effect uf stres_~ concenlrailons. de termine the shearing s tre~~ at poillts a and h.

341 nnn

r

."

00 UlU'

l

3 .142 section

A 9()' N .

.~ ll\)wlL.

III

lorqlle i. appJieJ to a hollow shaft having the cross

NeglectiLlI,! the effect of Mress com;elllraliom. dete, JIliLle the

shearing stress at poilus (/ and b. 3O,mn

Fig. P3.141

Fig. P3.142

A hulluw br~ss shaft has the en~s s~ctiun shuwn. Knuw ing that mllst no1 ..,Xce.en 12 k~i :lIld ncglet tin g the effect of ~I re~~ concentrations. determine the largest torque tha t can be app lied to the shaft 3 ,143

th~ s he~rin g s tre~s

J,5;" !i;I>.

'

~t,,," --11--" ", "".------1 J 5111

~

O~;'"

Fill. PS.143

O.!!1ll

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3 .144 A hollow member having the cross section shown is formed from sheet metal of 2-mm thickness. Knowing that the shearing stress must not ex· ceed 3 M Pa. de1emline the largest torque that can be applied to the member. 3 .145 and 3.146 A hollow member having the cross section shown is 10 be formed from sheet metal of 0.06 in. thickness. Kn owing that a 1250 Ib . in . torque will be applied to the member. deTermine the smallesT dimension d That can be used if The shearing stress is not to exceed 750 psi.

r;=====;'1 2 ;1\

=J-r ~

=,11

:,=====~~[ ~3 ill.--l Fig. P3.145

,rt~ - Ill .

"

Probloms

1-50 """1

-r-r=l'" 1--1

10 m", Fig. P3.144

J

'=

-LL~ 1

1--",, --1 Fig. P3.146

(I) Fig. P3.147

3.147 A hulinw cylindric;11 ~h:lfT w,,~ rle~ien ..rI wilh Ihl' cm~~ ~P.I: lion shown in Fig. U) to withstand a maximum torqlle TQ • Defective fabrication. however. resulted in a slight eccentricity I: between the inner and outer cylindrical s urfaces of the shaft as shown in Fig. (2). (n) Express the maximllm torqlle Tthat can be safely applied to the defective shaft in temlS of To> 1:, and I. (b) Calculate the percent decrease in the allowable TOrque for vailles of the ratio 4 1equal TO n.l. 0.5. and 0.9. 3.148 A cooling tube having the cross section shown is formed from a sheet of stainless steel of 3·mm thickness. 11le radii c 1 = 150 mm and c! = 100 mm are measured to the center line of the sheel melal. Knowing thai a lorqne of magnitude T = 3 kN . m is applied TO the Illbe. detennine (a) the maximum shearing slress in Ihe tube. (b) Ihe magnilllde of the lorque carried by the oUler circu lar s hell. Negieci Ihe dimension of Ihe small opening where Ihe oUler and inner shells are connected. 3, 149 &]ual TOrques are applied to thin-walled tubes of the same length L. same thickness I. illld same radius c. One of the lUbes has been slit lengthwise as shown. Determine (u) the ratio Th/r" of the maximum shearing stresses in the rubes. (b) the ratio CPb/r/J. of the angles of tWist of the shafts. 3.150 A hollow cylindrical shaft of length L. meilll radius f.,. and uniform th ickness I is subjected to a torque of magnitude T. Consider, on the one hand. the values of the average s hearing stress T ",.~ and the angle of twist rP obtained from the elastic TOrsion formulas developed in Secs. 3.4 and 3.5 and, on the other hand. th~ corresponding \'alues obtained from the formulas developed in Sec. 3.13 for thin·walled shafts. (a) Show tl!
197

Fig. P3.148

(o)

Fig P3.149

Fig. P3.150

(II)

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REVIEW AND SUMMARY FOR CHAPTER 3

Deformations in circular shafts

This chapter was devoted to the analysis and design of shafts subjected to twisting couples, or torque.l. Except for the last two sections of the chapter. our discussion was limited to circlllar shafts. In a preliminary discussion [Sec. 3.21. it was pointed out that the distribution of stresses in the cross section of a ci rcular shaft is statically indeterminate. The determination of these stresses, therefore. requi res a prior analysis of the dlformatiolls occurring in the shaft [Sec. 3.31. Having demonstrated that m a circular shaft subjected to torsion, evel}' cross sectioll remaill.l· plane and IIndistorted. we derived the fo llowing expression for the shearing strain in a small element with sides parallel and perpendicular to the axis of the shaft and at a distance p from that ax is: p~ y = -

(3.2)

L

where dJ is the angle of twist for a length L of the shaft (Fig. 3.14). Equation (3.2) shows that the .I·hcaring .I·train ill a circular .fhaft varic.I·linearly with the distance from the axis ofthc shaft. It follows that the strain is maximum at the surface of the shaft. where p is equal to the radius c of the shaft. We wrote y""", =

c~

p y = ~ Yma,

L

(3.3. 4)

Considering shearing stre.ucs in a circular shaft within the elastic range [Sec. 3.4] and recalling Hooke's law for shearing stress and strain, 7 = Gy. we derived the relation 7 =

Fig. 3.14

Shearing stresses In elastic range

P

(3.6)

- 7 ...." C

which shows that within the elastic range, the shearing .1·treJ·,v 7 in {/ circular shaft also varies linearly with the di.l·tancefrom the axi.l· of the .I·haft. Equating the sum of the moments of the elementary forces exerted on any section of the shaft to the magnitude T of the torque applied to the shaft, we derived the clastic torsioll formulas Tc

Tmax = j

Tp T =

-

j

(3.9, 10)

where c is the radius of the cross section and J its centroidal polar moment of inertia. We noted that J = ! 7TC 4 for a solid shaft and J = ~ 7T(C~ - c1) for a hollow shaft of inner radius c) and outeT radius c2 .

198

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Review and Summary for Chapter 3

Tc

''''''' - Y Fig. 3.20

We noted that while the clement {/ in Fig. 3.20 is in pure shear. the element c 10 the same figure is subjected to normal stresses of the same magnitude, TeIJ. two of the no rmal stresses being tensile and two compressive. This explains why in a torsion lest ductile materials, which generally fail in shear, will break along a plane perpendicular to the ax is of the specimen. while brittle materials. which arc weaker in tension than in shear, witl break along surfaces fo rming a 45° angle with that axis. In Sec. 3.5, we found that within the elastic range, Ihc angle of (wist 1> of a c ircular shaft is proportional to the torque T applied to it (Fig. 3.22). Expressing 1> in rtll/iUIIS, we wrote

An!lle of twist

(3.16) where

L = length of shaft J === polar moment of inertia of cross section G = modulus of rigidity of material

If the shaft is subjected to torques at locations other than its ends or consists of several parts of various cross sections and possibly of different materials. the angle of twist of the shaft must be expressed as the alsebraic sum of the angles of twist o f its component parts [Sample Prob. 3.3\:

FIg. 3.22

(3.17) We obsen'ed that when both ends of a shaft BE rotate (Fig. 3.26b), the angle of twist of the shaft is equal to the difference between the angles of rotation cPH and ¢c of its ends. We also noted that when two shafts AD and BE are connected by gears A and B. the torques applied. respectively, by gear A on shaft AD and by gear B on shaft BE arc directly proportional to the radii r,J and re of the two gears- since the forces applied on each other by the gear teeth at C arc equal and opposite. On the other hand. the angles q,,! and cPethrough which the two gears rotate are inverse/y proportional to r,! and rn--since the arcs CC' and CC~ described by the gear teeth are equal [Example 3J>4 and Sample Prob. 3A]. If the reactions m the supports of a shaft or the internal torques cannot be detennined from statics alone, the shaft is said to be static(llly illdelermintlle [Sec. 3.6]. The equilibrium equations obtained from freebody diagrams must then be complemented by relations involving the deformations of the shaft and obtained from the geometry of the problem [Example 3.05, Sample Prob. 3.5].

(Ii )

Fig.3.26b

Statically indeterminate shafts

199

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200

Torsion

In Sec. 3.7, we discussed the design ojtrallsmi.uioll shafts. We first observed that the power P transmitted by a shaft is

P = 2njT

Transmission shafts

(3.20)

where T is the torque exerted at each end o f the shaft and f the frequency or speed of rotation o f the shaft. The unit of frequency is the revolution per second (S-I) or hertz (Hz). If SI unil~ arc used, Tis expressed in newton-meters (N . m) and P in watt.l· (W). If U.S. CLlStomary units are used, T is expressed in Ib . ft or Ih . in. , and P in IbIs or in . Ib/s~ the power may then be converted into horsepower (hp) through the LIse of the relation

rt .

[ hp = 550 ft • IbIs = 6600 in . Ibis

Stress concentrations

To design a shaft to transmit a given power P at a frequency f, you should first solve Eq. (3.20) for T. Carrying this value and the maximum allowable value o f r for the material used into the elastic formula (3.9), you will obtain the corresponding value of the parameter l ie, from which the required diameter of the shaft may be calculated [Examples 3.06 and 3.07]. In Sec. 3.8. we di scussed .rlre.I·.\· concentration.l· in ci rcular shafts. We saw that the stress concentrations r.::sulting from an abru pt chang.:: in the diameter of a shaft can be red uced through the use of ajillef (Fig. 3.31). The maximum value of the shearing stress at the fillet is

r,

(3.25)

rm"x = Kj

FIg. 3.31

P!asUc defonTIat!ons

where the strc.~s Tell is computed for [he smaller-diameter shaft, and where K is a stress-concentration factor. Values of K were ploued in Fig. 3.32 on p. 167 against the ratio r/d. where r is the radius o f the fillet , for variolLs values of Did. Sections 3.9 through 3.11 were devoted to the discussion of pia.1'fie deformatiolls and residuul SlrCiSt!i in circu lar shafts. We first recalled that even when Hooke's law does not apply. the di stribution of .rfmins in a circu lar shaft is always linear [Sec. 3.9]. If the shearing-stress-strai n diagram for the material is known, it is then possible to plot the shearing stress T against the distance p from the axis of the shaft for any given value of r rna.< (Fig. 3.35). Summing

o llllllll+c---c" Fig. 3.35

the contributions to the torque of annular clements of radius p and thickness dp, we expressed the torque T as T = f pT(2np dp) == 2n !'p2r lip

,

,

where r is the function of p ploned in Fig. 3.35.

(3.26)

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Review and Summary for Chapter 3

An important value of the torque is the ultimate torque T u which causes failure of the shaft. This value can be determined. either experimentally. or by carryi ng out the computations indicated above with 7" nIaX chosen equal to the ultimate shearing stress 7" u of the materi al. From T fJ, and assuming a linear stress distribution (Fig 3.36), we determined the corresponding fictitious stress Rr = Tuell, known as the //lOllI/Ius oj rupture ill tor.violl of the given material. Considering the idealized case of a solid cirr:ular shaft made of an ela.\·toplastic material [Sec. 3.101, we first noted that, as long as 7" rna, does not exceed the yield strength 7"y of the material, the stress distribution across a section of the shaft is linear (Fig. 3380). The torque Ty corresponding to 7"nIa>. = Ty (Fig. 3.38b) is known as the maximulII elaslic torque; for a solid circular shaft of rad ius c. we have

MOOulus of rupture

0' - --+---::,

Fig. 3.36

(3.29)

As the torque increases, a plastic region develops in the shaft around an elastic core of radius Pl" 1lle torque T corresponding to a given value of Pl' was found to be

Solid shaft of elastoplastic material

(3.32)

" o ""'l.llllj-_---"

frrTTl-m "

,. (h )

I, )

(rI )

Fig. 3.38

We noted that as Pr approaches zero. the torque approaches a limi ting value ~,. called the plastic torqlle of the shaft considered: 4 T, = -T 3 y

T

(3.33)

Plotting the torque T against the angle of twist 1> of a solid circu lar shaft (Fig. 3.39), we obtained the segment of straight line OY dcl'ined by Eq. (3.16), followed by a curve approaching the straight line T = Tp and defined by the equation

(3.37) Fig. 3.39

201

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202

Torsion

Permanent deformatior . Residual stresses

Loading a circular shaft beyond thc onset of yield and unloading it [Sec. 3.1 1] results in a permallent de/ormatio/l characterized by the angle of twist ¢p = ¢ - ¢', where ¢ corresponds to the loading phase described in the previous paragraph, and ¢' to the unloading phase represented by a straight line in Fig. 3.42. There will T

T

°1--°"_1--"_1 1- - . - Fi g. 3.42

Torsion of noncircular members

(i

circular member.\". We first recalled that the derivatiun 0[" the fonnuT

r Fi g. 3.45

Bars of rectangular cross section

TV"/ ' ~ I-AT "!J ~

Fi g. 3.48

alw be residual .,·Ire.ue.,· in the shaft, which can be determined by adding the max im um stresses reached during the loading phase and the reverse stresses correspond ing to thc unloading phase ["Example 3.091· The last two sections of the chapter dealt with the torsion of 110/1-

'" _ _ _ _ _ I

"i Thin-wailed hollow shafts

las for the distribution of strain and stress in circu lar shaft s was based on the fact that due to the axisymmetry of these members, cross sections remai n plane and undistorted. Since this property does not hold for noncircular members, such as the square bar of Fig. 3.45, none of the formulas derived earlier can be used in thei r analys is rSec.3.12J. It was indicated in Sec. 3. 12 that in the case of straight bars with a uniform rectansutor cm.1".f section (Fig. 3.48), the maximum shearing stress occurs along the center line of the wider face of the bar. Fonnulas for the maximum shearing stress and the angle of twist were given without proof. The memhrane a/lalos)' for visuali zing the distribution of stresses in a noncircular member was also discussed. We next analyzed the distribution of stresses in /loncircular thinwalled ho//ow shaft.l· rSec. 3.13J. We saw that the shearing stress is parallel to the wall surface and varies both across the wall and along the wall cross section. Denoti ng by T [he average value of the shearing stress computed across the wall at a given point of the cross section, and by f the thickness of the wall at that point (Fig. 3.57), we showed that the product q = Tt, called the shear flow, is constant along the cross section. Furthermore, denoting by Tlhe torque applied to the hollow shaft and by (f the area bounded by the cellter line of the wall cross section. we expressed as follow s the average shearing stress T at any given point of the cross section: T T

Fi g. 3.57

~

-

2K,

(3.53)

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REVIEW PROBLEMS

3 .151 Knowing that a OAO- in.-diameter hole has been drilled through each of the shaftsAB. Be. and CD. determine (ul the shaft in which the maximum shearing stre ss occurs. (b) the magnitude of that stress.

;;« ()

Ih. ill

Fig. P3.151 3. 152 A steel pipe of 12-in. outer diameter is fabricated from i -in.-thick plate by weld ing along a helix which forms an angle of 45 0 with a pl:lne perpendicul:lr to the axis of the pipe. Kn owing that the maximum allowable tensile stress in the we ld is 12 hi. detemline the largest torque that can be applied to the pipe.

1i ll.

T

Fig. P3.152

3 .153 For the :lluminum shaft shown (G = 27 GPa). determine (u) the torque T that causes an angle of twist of 4°. (b) the angle of twist caused by the same torque T in a solid cylindrical shaft of the same length and cross-sectional area. 3 .154 The .olid cylindrical rod Be is :lttached to the rigid leller AB and 10 the fixed suppon at C. The vertica l force P applied at A causes a small displacement at point A. Show that the corresponding m:lx imum shearing stress in the rod is T

= Cd

~

2L" where d is the diameter of the rod and G its modulus of rigidity.

FIg. P3.153

Fig. P3.1 54

203

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204

Torsion

3. 155 Two solid steel shafts (G "" 77.2 OPa) are connected to a coupling disk IJ and to fix.ed suppol1S at A and C. For the 10'lding shown, determine (ll) the reaction at each suppol1, (b) the max.imum shearing stress in shaft AB. (c) the maximum shearing stress in shaft Be.

/2.50 ' "''''

~m4 v.\" " ..

3.156 The composite s haft shown is twisted hy applying a torque T at endA. Knowing that the maximum shearing stress in the s teel shell is 150 MPa, determine the corresponding maximum shearing stress in the aluminum core.. Use G "" 77.2 OPa for s teel and G =oc. 27 GPa for aluminum.

\3S nu"

Fig. P3.155

,

31J m",

~ -X')

,

AI"",;"",n

v'S"" / lII.

Fig. P3.1S6

3. 157 In the gear-and-shaft system shown. the shaft diameters arc dA/J = 2 in. and tlCD = 1.5 in. Kn owing that G "" 1l.2 X lOb psi. determine (he an)),Ie throuf!h which end D of shaft CD rotat~s.

c

1.6;"

T ,. 5 kip ' IJl /J

Fig. P3.157

Fig. P3.1 S8

3.158 A torque T is applied as shown to a solid tapered shaft AB. Show hy integrJ.tion that the angle of twist at A i3 712 cjJ = 121TGc'

3.159 A 1.5-in.-diameter steel shaft of length 4 ft will he used to transmit 60 hp between a motor and a pump. Knowing that G = 1l.2 X l ot' psi. determine the lowest s peed of rotation at which the stress does not ex.ceed 8500 psi and the angle of (wist docs not exceed 2°.

Fig. P3.160

3.160 Two sol id hrass rods AB and CD are brazed to a brass sleeve EP. Determine the ratio d./d l for which the same max.imum shearing stress occurs in the rods and in the sleeve.

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3. 161 One of the two hollow steel drive shafts of an oce:m liner is 75 m long and has the cross section shown. Knowing Ihat G = 77.2 G Pa and that the. shaft transmits 44 MW to iTS propeller when rotating at 144 rpm. determine (a) the ma."(imum shc.aring stress in the shaft. (b) the angle of tw ist of the s haft.

Computer

FIg. P3.161

3.162 Two shafts are made of the s.ame material. The cross section of shaft A is a square of side band lhat of shaft B is a.circle of diameter b. Knowing that the shafts are subjected to The same torque, determine the ratio fAiTH of the maximum shearing stresses occurring in the shafts.

,\

8

Fig. P3.162

COMPUTER PROBLEMS

The followin g problems arc d esigned to he soh·oo wilh a mmputer. Wrile t'3 ch program so that it rail be used with either SI or U.S. customary units. Et ~ m{'nt

3 .C l Shaft AB consists of II homogeneous cylindrical elements. which can be sol id or hollow. Its end A is fixed. while its end B is free. and it is subjected 10 The. loading shown. The lengTh of element i is denoted by L i • iTS outer diameter by OD i • ils inner diameter by IV!. its modu lus of rigidity by C /. and the torque applied to its right end by T i • the magnilllde Ti of th is torque being assumed TO be posiTive if T , is observed as counterclockwise from end Band negaTive otherwise. (Note that JD; = 0 if the element is solid.) (a) Write a computer program thm can be used 10 determine the maximum shearing stress in each e lement. the angle of twist of each element. and the angle. of twist of the entire shafl. (b) Use this program to solve Probs. 3.35 and 3.38.

FIg. P3.CI

"

Probtems

205

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206

3.C2 The ;membly shown consisls or II cylmdncal shafts, Which can be snlirl or hollow, coon~,:red hy !:f'.ar~ ~nrl "lIpporled hy hr:1Ckt':ls (nor ~h{Jwn) EmJ A, of the first shaft is frre and is subjected to a torque To. while end B, of the last shaft is fixed. TIle length of shaft A,B, is denoted by Li• its oUler di ameter by OD,. il5 inner diameler by ID i• Ilnd it~ modulus of rigidily by G, (Note lhat W i = (] if the element is solid.) The radius of gear A, is denoted b} a i • and Lile radius of gear B; by bi' (a) Write a cornpuler program Ihal can be used 10 determine the maximum shearing stress in eaeh shaft. the angle of Iwist of each shaft. and the ang le throu gh which end A , rola tes. (b) Use Ihis program to solve Prohs ).40 and 3.44.

TDrnl«l

3.C3 Shafl An consislS of n homogen=u .~ cylindrical eJemenu. which can be soUd or hollow. BOlh of it~ ends are fixed. and it is subjeclaJ 10 the loading shown. TIre length of element i is denoted by 4. its u ut e.r diamete.r by QUi. its in tle r diameter by ID;. ilS modulus o f rigidity by Gi • and the torque :lpplied 10 its right elld hy T,. the magnilude T, of Ihi ~ lorque being assumed ]() be positive if T,- is observed as counterdockwise from end JJ and negative otherwise. Note th~t lDi = 0 if the element is solid and also that 1'1 "" O. Write a Lurnputer pmgram thal (;all bt: uscll tu uctnmine Ille It:actium at A al lu B. tht maximum shearing stress in eJch element. and Ihe. angle uf Iwist of each ele· ment. Use this progrum (
Fig. P3.C3

Fig. P3.C4

3.C4 The hamo.c:eneous. solid cylilldrical shaft AB has a length L. a diameter 11, a modulus ur rigidity G. and a yield slreng th TJ. It JS subjected 10 a torque T mal is gradually increased from zero until Ihe :mgk of IWI~l of Ihe ~hafl has reached a max imum \'alue 4>,,, ::lJ1d th~n deC"reased had.: 10 zero (a ) Write a l'ODlputer program that. fUl each of 16 values uf rjJ,. eq uall y SPJced ov~r a range extending from 0 10 a value 3 times as brge as the angle of [wist at the onset of yield. can be used 10 dctcmline the m
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3 .C5 Tht! e.>.:act expression is given in Prob. 3.61 for the angle of twist of the solid tapered shaft AB when a torque T is applied as shown. Derille an approximate expression for the angle of Iwist by replacing the t
,.

Fig. P3.C5

3 .eB

A torque T is applied as shown to the long. hollow. tapered shaft

AB of unifonn thicknes.~ I. The eXilct expression for the angle of twist of the

shaft can be obtained from the expression given in Prob. 3.156. Derive an approximate expression for the angle of twist by replacing the tapered shan by Il cylindrical rings of equal lenglh and of radius r, = (II + i - !)(dn). where i = J. 2, ... ,II. Using for T. L. C. (. and I values of your choice. determine the percentage error in the approximate expression when (a) Il = 4. (b) Il = 8, (C) fl "" 20. (d ) n = 100.

~ ,T L

Fig. P3.C6

Computer Problems

207

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Pure Bending

The athlete shown holds the barbell with his hands placed at equal distances from the weights. This results in pure bending in the center portion of the bar. The normal stresses and the curvature resulting from pure bending will be determined in this chapter.

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4.1. INTRODUCTION

4.1. Introduction

In the preceding chapters you studied how tu determine the stresses ill

prismatic members subjected to axial loads or to twisting couples. [11 this chapter and in the following two you wi ll analyze the stresses and strains in prism~tic members subjected to hending . Bending is a major concept used inlhe design of many machine and structural components. such as beams ,md girders. This chapter wiU be devoted to the analysis of prismatic members subjected to equal and opposite couples M and M' aCling in the same lungitudinal plane. Such members arc said to be in pure belllling. In most of the chapter, the members will be assumed to possess a plane of symmetry and the couples M and M' to be acting in that plane (Fig. 4.1).

".

Fig. 4.1

An example of pure bending is provided by the bar of a typical barbell as it is held overhead by a weight lifter lL<; shown on the opposite pagc. The bar carries equal weights at cqual distances from the hands of the weighllifter. Because of the symmetry of the frcc -body diagram of the bar (Fig. 4.2a), Ihe reactions at the hands must be equal and opposi te to the weights. Therefore. as far as the middle portion CD of the bar is concerned. the weights lmd the reactions can be replaced by two equal and opposite 960-lb . in. couples (Fig. 4.2h), showing that the middle portion of the bar is in pure bending. A similar analysis of the axle of a small trailer (Fi g. 4.3) would show that. between the two points where it is attached 10 the trailer. the axle is in pure bending. As interesti ng as the direct applications of pure bending may be, devoting an e ntire chapter tu its st udy would not be j ustilied if it were not for the fact Ihat the results obtained wi ll be used in the analy .~is of other types of loadings as well, s uch as ecallfric axial IO(l(lings and transverse loadil/gs.

Fig. 4.3 For the sport buggy shown, lt1e center portion 01 the rear axte is in pure bending.

".

'iO III

12 in

" Rc

Ie

t

~o

26;1> .

E! ;Il.

"I Rn - su ll ,

III (Il )

Q M

w

9

00) Iii

11( ' ,. 00 ) )1 ,

(iI)

Fig. 4.2

B

209

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21 0

Pure Bending

Fig ure 4.4 shows a 12-in. steel bar clamp used to exert 150-lb forces on two pieces of lumber as they are being glued together. Figure 4.5(/ shows the equal and opposite fo rces exerted by the lumber on the clamp. These forces result in an eccelllric loading of the straight portion of the clamp. In Fig. 4.5b a section CC' has been passed through the clamp and a free-

I

I

c

~

-.

c·

tP'=

t

150lh ;"II =

IP = ISO lh

,r-"''--..--'lJ

p ~

P '=L50lh

'50lh .;"

L.'iOlh

L

(0'

(b)

Fi g . 4.5

Fi g. 4.4

PI - -"+

"

c,

(II I

Ihl Fig. 4.6

I

body diagram has been drawn of the upper half of the damp. from which we conclude that the internal forces in the sect ion arc equi vale nt to a 150Ib axial tensile force P and a 750·lb . in . couple M. We can thus combine our knowledge of the stresses under a celli ric load and the results of our forthcoming analysis of ~ tre sses in pure bending to o btain the distribution of stressc~ under an ecct!llIric load. This will be further discussed in Sec. 4.1 2. The study of pure bending will al~o play an essential role in the siudy of beams. i.e" the stud y of prismatic members subjected to various- types of Iral!Sl'eru loadl' . Consider. for instance. a canti lever beam AB supporting a concentmted load P at its free end (Fig.. 4.6a ). If we P
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4.2. Symmetric Member In Pure Bendtng

lIlain plane in a lIlember subjected to pure bending. while in Sec. 4.4 for· mulas will be deyclopcd that can be used to determine the lIormal .I·tress ...\·. a.~

well a.~ the radil/.I· of ClllWllllre for that member within the clastic range. In Sec. 4.6. yo u will study the stresses and deformations in composite mCllliJerJ made of more than one materiaL such as rei nforced-concrete beams. which utilize the best features of steel and concrete and are exten· sively used in the construction of buildings ~Uld bridges. You will learn to draw a lral!!;forllled Jcctiul! representing the section of a member made of a homogeneous material that undergoes the same deformations as the composite member Wlder the same loading. The transformed scnion will be used to find the stresses :Uld deformations in the original composite member. Section 4.7 is devoted to the detennination of the Jlres.~ conCt'lIIratiom' occurring at locations where the cross section of a member undergoes a sudde n change. In the next part of the chapter yo u will slUdy plastic dejurlllllfioflJ in bending. i.e" the deformations of members which arc made of a material which docs not follow Hooke's law and arc subjected to bending. After a general discussion of the deformations of such mcmbers (Scc. 4.8). you will investigate the stresses and deformations in members made of an ela.I'to plastic lIIaterial (Sec. 4.9). Starti ng with the II/axinl/wl elwitic mOllll!nt M I" which corresponds to the onset of yidd, you will consider the eHects of im:reasingly larg~r moments until the plastic //IOIIIl!lIt M " is reached. at which time the member has yielded fully. You will also learn to determIne thepcrll1(Jllel11 deforlllaliom' and residual .\·trl!.ue.l' that result from such loadings (Sec. 4.1 I). It should be noll'd that dllring the past half-century the elastoplastic propcny of stccl has been widely u ~ed to produce design~ re· suIting in both improved safety and economy. In See. 4.12, you will learn to analyze an f!l'celllric (lxiII/loading in a plane of symmetry, suc h as the one shown in Fig. 4.4. by superposing the stresses due to pure bending and the stresse.~ due to a centric :Lxial loading. Your study of the bending of prismatic members will conclude with the ~Ulal ysis of IInsyl/Jlnl!tric bEllding (Sec. 4. 13), and the study of the general case of eccentric axia/loading (Sec. 4.14). The final section of the chapter will be devoted to the detennination of the stresses in L'uH'ed mem hEr.l· (Sec. 4. 15).

c

4.2. SYMMETRIC MEM BER IN PURE BENDING Consider a prismatic member AB possessing a plane o f symmetry and subjected to equal and opposite couples M and M ' acti ng in that plane (Fig. 4.7a). We observe that if a section is passed through the member AB at some arbi tra ry point C. the condit ions of equi librium of the portion AC of the member require that the internal forces in the section be equivalent to the couple M (Fig. 4.7b). Thus. the internal forces in any cross section of a symmetric member in pure bendi ng are equi valent to a couple. The moment M of that couplc is referred to as the bendinK moml!nt in the section . Fo llowing the usual conve ntion. a positive sign will be assig ned to M when the member is bent as shown in Fig. 4.7a . i.e., when the concavity of the beam 1:1CeS upward, and a negative sign otherwise.

'M

Flg. 4.7

21 1

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212

Pure Bending

Deno ting by U ., the normal stress at a given point of the cross section and by T.n and T xc the component. of the shearing stress, we express that the system o f the elementary internal forces exerted on the section is equivalent to the couple M (Fi g. 4.8).

Fig. 4.8

We recall from statics that a couple M actually consists of two equal and opposite forces. The sum of the components of these fo rces in any direction is therefore equal to zero. Moreover, the moment of the couple is the same about (ln y axis perpendicular to its plane, and is zero about any axis contained in that plane . Selecti ng arbitrarily the z axi s as shown in Fig. 4.8 , we express the equivalence of the elementary internal fo rces and of the couple M by writing that the sums of the components and of the moments of the elementary fo rces are equal to the corresponding components and moments of the couple M : x components:

(4.1 )

moments about y axis:

(4.2)

moments about

z axis:

f( - ya, dA )

~

M

(4.3)

T hree additional equations could be obtained by setting equal to zero the sums of the y components, z components. and moments about the x axis, but these equations would involve only the components of the shearing stress and, as you will see in the next section, the components of the shearing stress are both equal to zero. Two remarks should be made at this point: (I) The minus sign in Eq. (4.3) is due to the fact that a tensile stress (u x > 0) leads to a negative moment (clockwise) of the normal fo rce U x tiA about the z axis. (2) Equation (4.2) could have been anticipated, since the application of couples in the plane of symmetry of member AB wi ll result in a distribution of normal stresses that is symmetric about the y axis. Once more, we note that the actual distribution of stresses in a given cross section cannot be determined from statics alone. It is .1·tatic(llly indeterminate and may be obtained only by analyzing the de/ormation.l· produced in the member.

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4.3. DEFORMATIONS IN A SYMMETR IC MEMBER IN PURE BENDING

4.3 . Deformations In a Symmetric Member In Pum BendIng

Let us now analyze the deformations of a prismatic member possessing a plane of symmetry and subjected at its ends to equ al and opposite couples 1\1 and M ' acting ill the plane of symmetry. The member will bend under the action of the couples, but wi ll remain symmetric with respect to thm plane (Fig. 4.9). Moreover, since the bending mo-

Fig. 4.9

ment M is the smne in any cross section, thc member will bend uniformly. Thus, the line AB along which the upper face of the member intersects the plane of the coupJes will have a constant curvature. Ln other words. the li ne AB. which was originall y a straight line, will be transformed into a circle of center C. and so wi ll the line A' B' (not shown in the tlgure) along which the tower face of the member intersects the plane of symmetry. We also nolC that (he line AB will decrease in length when Ihe member is bent 'L'> shown in the fi gure. i.e., when M > 0, whileA' B' will bccome longer. Next we wi ll prove that any cross section perpendicular to the axis of the member remai ns plane. and that the plane of the section passc..'> through C. If thi ~ were not the case. we cou ld lind a point E of the original section through D (Fig. 4.10a) which, after the member has been bent. would lIot lie in the plane perpendicular to the plane of symmetry that contains line CD (Fig. 4. 10h). But, because of the symmetry of the member, there would be another point E' that would be transformed .exactly in UIC same way. Let us assume that. after the beam has been bent. both point ~ wou ld be located to the left of lhe plane defined by CD, as shown in Fig. 4. lOh. Since the bendi ng moment M is the same throughout the member, a similar si tuation wou ld prevail in any other cross section, and the points corresponding to E and E' wou ld also move to the left. Thus, an observer at A would conclude that the loading causes the points E and E' in the various cross sections to move forward (toward the observer). But an observer at B. to whom the loading looks the same. and who observes the points E and E' in the same positions (except that they are now invcrted) would re,tch the oppositc conclusion. This inconsistency leads us h.) conclude that E and E' will lie in the plane defi ned by CD and. therefore , that the section remains plane and passes through C. We should note. however, that thi s discussion does not rule out the possi bility of dero nnations lVithill the plane of the section (see Scc. 4.5).

_~-. B

(bl

Fig. 4. 10

213

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214

Pure Bending

(II ) 1.J>Il)"'i!lIdill.J, "f' rtiml 5t."Clioll

(1'1..11" of ~~l!mll'l ry)

".

(bl t.J,,,,,iI, ,,lmal.l lI)rizlInlal ""dinn Fig. 4.11

Suppose that the member is divided into a large number of small cubic clements with faces respecti vely parallel to the three coordinate planes. The property we have establi shed re(luires that these elements be transrormed as shown in Fig. 4.11 when the member is subjected to the couples 1\1 and 1\1 '. Since all the faces represented in the two projections of Fi g. 4.1 [ arc at 90° to each other. we conelude that 1'.') = 'Y", = 0 and. thus. that T n = T .,;: = O. Regarding the three stress components that we have not yet discussed , namely. u " u ;. and T,.o' we note that they must be zero on the surface of the member. Since, on the other hand, the deformations involved do not require any interaction between the elements of a given transverse cross section, we can assume thal these three stress components aTe equal to zero throughoUlthe member. This assumption is veri lied, both (rom experi mental evidence and from the theory of elasticity, lar slender members undergoing small deformatioll s.t We conclude that the only nonzero stress companelll exe rted on any of the small cubic elements considered here is the normal component u ". Thus, at any poilll of a slender member in pure bending. we have a state of uniaxial ~·tn'.u . Recalling that, ror M > 0, lines AB and A' B' are observed, respectively. to decrease and increase in length, we note that the strain fE._ and the stress u .. arc negative in the upper portion of the member (compressioll) and positive in the lower portion (ten.I'ion). It follow s from the above that there must exist a surface parallel to the upper and lower races of the member. where fE., and U x arc zero. This surface is called the nelllral ,Iwface. The neutral surface intersects the plane of symmetry along an arc of circle DE (Fig. 4. 12a), and it intersects a transverse section along a straight line called the neutral axi.I' of the section (Fig. 4. [2b). The origin of coordinates will now be sc-

(II i L,mt,.~t"di" .IL wrlical _ ' ''diOl'

(pl"u"

.,f <\" ""elrv)

Fig. 4.12

leeted on the ne utral surface, rather than on the lower fac e of the member as done earlier, so that the di stance from any point to the J1etllral surface will be measured by its coordinate y. t Also see Prob. ·U8.

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Denoting by p the radius of arc DE (Fig. 4.l2a), by " the central angle corresponding to DE, and observing th;lI. the length of DE is equal to the length L of the undeformed member, we write

L

~

pO

(4.4 )

Considering now the arc JK located at a di stance y above the neutral surl:'u;e, we note that its kngth L' is

(p - )')0

L' ~

(4.5 )

Since the origiml length of arc JK was equal 10 L. the defomlation of JK is

o = [' - L

(4.6)

or. if we substitute from (4.4) lmd (4.5) inlO (4.6),

8

~

(p - ,.)0 - pO

~

- ,.

(4 .7)

The longitudinal strain lOx in lhe elements of JK is obtained by di viding {j by the original length L of J K. We write

8 L

- yO pO

f'...= - = - -

u,

,,

(4.8)

~

p

The minus sign is due to the fat't that we have assumed the bending momcnt to be positive and, thus. the bcam to bc concave upward.

Because of the requirement that transverse sections remai n plane, identical deformations will occur in all planes parallel to the plane of symmetry. Thus the value of the strai n given by Eg. (4.8) is valid anywhere, and we conclude that the longitll(lillal norma! .I·train IO~, varies linearly with th t: distance y from the nellfml .mr/ace.

The strain €f reaches its maximum absolute value when )" itself is largest. Denoting by C lhe largest distance from the neutral surface (which corresponds to either the upper or the lower surface of the member). and by €' /Ir [he lIuu·imum lIbs(Jlule vallie of the strain. we have C

€.

",

=-

(4.9)

p

obta.i n~tI

Solving (4.9) for p llntl s ubstituting the value can also write

into (4.8), we (4.10)

We conclude. our analyS"is of the deformations of a member in pure bending by observing that we ar~ _~ till unable 10 compute the strain or stress nt a given point of the member, _~i nce we have not yet located lhe nell tral surface in lhe member. In order 10 locate thi s surfacc, we musl first speCify the stress-strain relation of the material uscd. t t Lct U.< nov.-,

howfl'~f.

that if the mcmocr possesses both

J

vrnical and a horizontal

plan~

of symmetry (e.g .. a nlcmbt-r with a rectangular cross ~cction •. and if thc :;tress·Slrain curv~

is th~ same in temioa and compression. the IIeutrJl symmetry ~d. $cc . .t.R;.

~ urla~c

will coincide with the planc of

4.3. Deformations In a Symmetric Member In Pure Bending

215

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216

4.4. STRESSES AND DEFORMATIONS IN THE

Pure Bending

ELASTIC RANGE

We now consider the case when the bending moment M is such that the normal stresses in the member remain below the yield streng th (Ty. Thi s means that, for all practical purposes, the stresses in the member will remain below the proportional limit and the clastic limit as well. There will be no permanent deformation. and Hooke's law for uniaxial stress applies. Assuming the material to be homogeneous, and denoting by E its modulus of elasticity. we have in the longitudinal x direction (4.11 )

Recalling Eq. (4.10), and multiplying both members of that equation by E, we write

or, using (4. t 1), (T

•

'l---'---
=

y -~ u

C '"

(4. [2)

where (I ", denotes the maximulII absolute value o r the stress. Thi s resuit shows that, ill the elastic range, the 1I0r/1U!l stress varies linearly with the distance from the lIelltral .I·ulface (Fig. 4.13). It should be noted that. al this poinl, we do nOI know the location of the neutral surface, nor the maximum value (I m of the stress. Both can be found if we recall the relations (4.1) and (4.3) which were obtai ned earlier from statics. Substituting fi rst for (T. , from (4.12) into (4.1), we write

from which it follows that (4. [3)

This equation shows that the fi rst moment of the cross section about its neutral axis must be zero.t In other words, for a member subjected to pure bending, and as 10llg as the .I"Iresse.l· remaill in the ela.l·tic mllge. the lIeutml axis passe.v through the centroid of the sectioll. We now recall Eq. (4.3), which was derived in Sec. 4.2 with respectlO an arbitmry horizontal z axis, (4.3)

Specifying that the z axis should coincide with the neutral axis of the cross section. we substitute for (I x from (4.12) into (4.3) and write t Sec Appendix A for a discussion of the moments of arca~ .

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4.4. Stresses an d Deformations In the EJIIS1.Ic Range

M

I'

-~,.

,

(4. 14)

y-dA = M

Recalling thai in the case o f pure bend ing the neutral axis passes rbrough the centroid of the noss section. we nOle that 1 is the moment of inertia, or second moment. of the cross section with respect \0 a centroida l axis perpendicular to the plane o f the couple M . Solving (4.14) for (Tm_ we write thereforet Me if = (4.[5) .. I Substituting for (T m from (4.15) into (4. 12), we obtain the nonna[ stress (J'~ at any distance y from the neutral axis:

My

(1

,

(4. 16)

= - -----'---

I

Equations (4.15) and (4. 16) are called the elastic flexure forlllllla.~. and the normal stres~ IT, _caused by the bending or " flexing" of the member is often referred!O as thcjlexllral .\'/ress. We verify Ihat the siress is comprc.<;sive (a . < 0) ahove the neutral axis lr > 0) when the bending mo At is positive, and tensilc (tT, > 0) whcn M is negative

mCl1l

Rc turning to Eq. (4. 15). we note that the ratio lie depe nds; o nly upon the geometry of the cross secti on. This rat io is called the elcl.\"tic secrioll modulm' and is denoted by S. We have Elastic section modulus = S =

!.-,

(4.17)

Substituting S for lie into Eq. (4.15). we wri te thi s equation ill the alternati ve form

M

u", =S

(4.18)

Since the maximum stres;s (J" m is inversely proportional to the elastic section modulus S, it is clear that beams should be designed with as large a value of S as practicable. For example, in the case o f a wooden beam with a rectangular cros;s section of width b and depth h, we have I

S= - =

r,IJh'

--~

ch/2

1 ' = 6 hh- 6 = 1All

where A is; the cross-sectional area o f the beam. 111is shows that, of two beams with the same cross-sectional arei'! A (Fig. 4. 14), the beam with the larger depth II will have the larger section modulus and, thus, will be the more effective in reSisting bending.t t We recalt tha( (he bending moment was as.sumed (0 be pos i(iv~. tf (he !lending momml I. ncga(i'·c. M should!)c replaced in Eq. (4. t5) by i(s abS()lu(~ vatue IMI . t Hom'""r. larg•• \"llliws of tli" rJ!i... hjb t"t ,,,ld ",~"It in bter.11 ill~!;l.hilit,. of thl~ beam.

1 J

II = 8 in

(4.19)

~-I

I, Fig. 4.14

=

3 in .

217

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218

Pure 8"n"lng

1.11 the ease of struc tural steel, AnlClicall stmldanl bealll ~ (S-bemns) and wide-Ilange be.ams (W-beam s). Fig _ 4 _15. arc p referred to oth(-r

Fig. 4.15 Wi~e-I!ange steel beam310nn tM Irameo! many buildino;l;.

,./\

Fig. 4.1 6

shapes because a large portIOn 01 their cross section IS located tar trom the Ileutral axis (Pig. 4.16), T hus, fo r a given cross-sectional area and a given clcrth, their design provicles large Ylt h lCS of' amI, c on _~cf]lIently. of 5 . Value, of the clas tIC section modulus o f coml1l o nl ~ manufnclurcd beams com be obtained from lables listing the various geomclric prope rties or such beams. To Clele nninc the maxi mum stress u tt, in a gIven ~cc lion of u standard beam. the engi n::er needs only to re ad the value of the e l1l~tic set1ioll modulus S in a table. and divide Lhe bending moment M in the section by S. The deformation of the member cauSL"'
'.

(4.20)

p

Rut , in the e last ic Tlll1gl!,

W~

h;lve.

f' ,,, = rT",I F.

SlLhsTiltlti ng for

f' ",

in10

(4.10 ). :J.nd recalling (4.15). we wri le _ = u ",=":" A1r p

Ec

Er: I

0' M p

EI

(4.1 1)

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EXAMPLE 4.01 A steel bar of 0.8 X 2.5-in. rectnngular cross section is subjected to two equal and opposite couples acting in the vertical plane of symmetry of the b:u- (Fig. 4.17). Determine the va lue of the bending moment M dwt causes the bar to yield. Assume O"y = 36 ksi.

(I S

\1 ', , -_ _- - ,"

I = f-Ibh"' =

1 r-

J ) ~~,

(I

Since the neutral axis must pass through the centroid C of the cross ;;ectioo. we hnve (. = 1.25 in. (Fig. 4.18). On the other hand. the centroidal moment of inerti
in .

Fig. 4.17

M O.8 in.)(2.5 in Y = 1.042 in ~

Solving Eq. (4.15) for M. and substituting the above data, we have

At

=

1.042 in 4 1.25m. (' M = 30 kip . in.

I -0-,. =

-, -(36 ksi )

Fig. 4.18

EXAMPLE 4.02 An aluminum rod with a semicircular cross section of radius 12 mm (Fig. 4.19) is bent into the shape of a circular nrc uf mean radius p = 2.5 m. Knowing that the Ilat face of the rod is tumed IOw"rd the center of curVaTure of the arc. determine the maximum tensile and compressive stress in the rod. Use E 70 Gl'a.

r =

=

0

Fig. 4.19

1""""

The ordinate section is

_ l"

.

vof the centroid C of the semicircular cross 4r

= - = 3r.

4 ( 12 mm)

3r.

5 .093 mm

The neutml :lXis passes through C (Fig. 4.20) itLId the distance c; to the point of the cross section farthes t awny from the neu tral axis is c

= r - Y.eo.

12 mm - 5.093 mm = 6 .907 mm

Using Eq. (4.9). we write

c

f ,. =

P=

6.907 X 10- 3 m 2.5 m

2.763 X 10- '

and. app lying Hooke' s Inw. We cou ld use Eq. (4.21) to determine the bending moment M corresponding to the given radius of curvature p , and then Eq. ~4.15) to determine (T "' . However, it is simpler to use Eq. (4.9) to determine IE.. , and Hooke's law to obtain 0".,.

;~\ N'A Fig. 4.20

0-., =

Ef,. = (70 X \09 Pa)(2.763 X 10 1) = 193.4 M Pa

Since this side of the rod faces away from the center of curvature. the stress obtained is a tens ile stress. The m:ur.imum compressive stress occurs on the flat side of the rod. Using the fact that the stress is proportionaito the di stance from the neu . tral axis. we write

5.093 mm ( 193.4 MPa) 6.907 mm = - 142 .6 M Pa

219

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220

Pure Bending

4.5. DEFORMATIONS IN A TRANSVERSE CROSS SECTION

When we proved in Sec . 4.3 that the transverse cross section of a member in pure bending remains plane, we did not rule out the possibility of deformations within the plane of the section. That such deformations wi ll exist is evident, if we recall from Sec. 2.11 that c1c ment,~ in a state of uniaxial stress, U"J 0, 17,. = U"~ = O. are defonncd in the transverse y and.;; directions, as well as in the axial x direction. The normal strains E,. and Eo depend UPOIl Poisson's ratio I' for the. mrttcrial used and arc expressed as

*"

or, recalling Eel. (4.8),

I"·

E = -

,

P

ry E- =

-

(4.22 )

-

p

The relations we have obtained show that the elements located rtho ve the neutral surface (v > 0) wi ll expand in both the y and;:: directions, while the clements locrtted below the neutral surface (y < 0) will contract. In the case of a member ot" rectangular cross section. the expansion and contraction of the vmious elements in the vertical direction wi ll compensate. and no change in the verticrtl dimension of the c ross section w ill be o b served. A s far as the. deformatio ns in the hori-

zontal transverse:: direction arc concerned. however. the expansion of the clements located above the neutral surface and the corrcgponding contraction of the eleme/lls located below that surface will result in the various horizontrtl lines in the section being bent into rtrcs of circle (Fig. 4.21). The situation observed here is similar to thm observed earlier in a longilUdinal cross section. Comparing the sccond of Eqs. (4.22) with Eq. (4.8). wc conclude that the neutral axis of the trrtnsversc section will be bent into a circle of radius p' = p/ ,'. The center C' of this circle is located below the neutral surface (assuming M > 0), i.e., on the side opposite to the center of curvature C of the member. The reciprocal of the radius of curvature p ' represe nL~ the curvmure of the tf:lnsverse cross section and is callcd the antir;/asli c c un'{JlUre. We have ..

I

v

Anuclastlc curvat ure = - "" p' p

(4. 23)

III ()ur di scussion of the dcfonnations of a sy mmetric member in pure bending. in thi s section and in the precedi ng ones. we have ignored the manner in which the couples M and 1\'1' were actually applied to the. member. If all transverSI! sections of lhe member. from one end to the other, are 10 remain planc and free of shearing stresses. we

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4.5. Deformations In a Transyerse Cross Section

p

,-,:".. 11,,1

I

SIIrf;lCr

N~"tr~ 1 ax;~ tr""~"'~r'"

of se{'f;n"

fl'

p

,, ,, , ,

, ,,, ,I ,,,

= pi"

c Rg.4.21

must make sure Ihat the couples are ~ppli cd in such a way that the ends of the member themselves re main plane and free of shearing stresses . This can be an:omplished by applying the coupks M and M ' to the member through the use of rigid and smooth plates (Fig. 4.22). The elementary forces exerted by the plates on the member will be nonnal 10 the e nd sections. and these sections, while remaining plane, will be free to deform as described earlier in this section. We should note thai these loading cond itions cannol be aCllLaJly realized , si nce they re4u ire each plate to exert tensile forc es on the corresponding end Sl."Ction below its neutral axis, while allowing the section 10 frecly defonn in its own plane. The fact that the ri gid-end-pl:ttes model of Fig. 4.22 can not be physically realized, however, does not detract from ils importance, which is to allow us 10 visualize the loading conditions correspond ing to the relations derived in the preceding sections. Act ua l 100lding conditions may diller appreciably from this idealized model. By virtue of Saint-Venllnt's princi ple, however. the relations obtained can be used ill compute stresses in enginecring situations, as long as the section considered is not too d ose to the points where the couples are. applied.

Fig. 4.22

221

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SAMPLE PROBLEM 4.1 The rectangular lube shown is extruded from an aluminum alloy for which 6 ify = 40 ksi . ( T U = 60 ksi, and £ = 10.6 X 10 psi. Neglecting the effect of tillets, detennine (a) the bending moment /11 for which the factor of safety will be 3.00. (b) the corresponding radius of curvature of the tube.

Sin,

SOLUTION

[t D -'- 1-1~ ' •

_,

4.5 ill

~

am.

X

Moment of Iner t ia. Considering the cross-sectional area of the tube as the dilfcrcncc between the two rectangles shown and recalling the formula for the cell [roidal moment of inertia of a rCClanglc. we writ~

J = fi(3.2 5)(5)1 - 6:(2.75)(4.5)l

1--1- ' .1 .2.') in

2.i5 ill

Allowable Stress.

1 = 12.97 in 4

For a factor of safety of 3.00 and an ultimmc stress of

60 ksi. we have i f !;

60 ksi

F.S.

3.00

=--~--=20 ksi

II

. 11

Since 0-.11 < tT l'. the lUbe remains in the elastic range and we can apply the results of Sec. 4.4. II . , ,

/0 p

, , , , , ,

Bend ing Momen t.

Me

Uol 1

=/

M "='

With

I -u" = c l

L"

= h5 in.) = 2.5 in .. we write

12.97 in· . (20 1.:51) 2.5 Ifl .

- -.-

M = 103.S kip' in . ....

h. Radius of C u n ·3turc. Recalling that E = 10.6 X lOb psi. we ~ubsti ­ tute this value and the values obtained for I and M illlo Eq. (4.21) and find 103.S X 10J lb · in. ( 10.6 X 106 psi)( 12.97 in4) p = 1325 in.

M p

EJ

,

0.755 X 10 J in {/ =

I

lJOAft ....

AfterlUtliI'e SOillli{)II_ Since we know that the maximum stress is U>lt = 20 ksi. we can determine the maximum strain t'", and then usc: Eq. (4.9). IT.U

t' ..

=

E

cc'2~Ock~'~i;-~ = 10.6 X lCf' psi c p

= -;:

1.887 X 10- 3 in./in.

2.5 in. I.S87 X 10 - 3 in./in.

p = 1325 in.

222

p = 1100411 ....

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SAMPLE PROBLEM 4.2 A cMt-iron machine pan is acted upon by the 3 kN . m couple shown. Knowing 111:It E = 165 GPa and neglecting the effect of fillets. determine (a) the maxi· mum tensile and compressivc strcsses in the casting, (b) the radius of curvature of the casting.

t- 00 """ -----j

n

i20mm 4() """

-----.l

1-1 3Om,"

SOLUTION

'--::rr='=r' t-~.n"" -----j

T

~ .- _

_

!II =5O,"m40 !'11'"

c _ e

Cent roid. We dil'ide the T·shapcd cross seclion into the two rectangles shown and write

.....l.,

I "'m,,, yo

,

2

y~=20mm ~I

, 2

Area, mm 2

Y, mm

(20)(90) ~ 1800 (40)(30) = 1200

SO

LA

30,"",

YA, mm'

20

3000

SyA

90 x '0' 24 x '0' 114 x IO~

Y~A

= ~vA

[(3000) =- 114 x 106 y = 38mm

Cenlroidall\'1olllcnl of Inertia, TIle parallel-axis theorem is used to detcmline the moment of inettkl of each rectangle with respecllo the a:ds x' thai passes through the centroid of the composite section. Adding the moments of inertia of the rectangles, we write

" ~~' I"""" C· ! ,. ' """"+-,,= IHmmL _ I . _ e

Y=3SulIl\

2 ~

A

.

I., = ~(i

+ Ad 2 )

= h(90)(20)~

= ~(i2 bh J

= 868 X 10J mm 4

1 = 868 x 10 - ~ m~

u. Ma."imum Tensile Stress. Since the applied couple bends the cast· ing downward. the center of curvature is located below the cross section. The maximum lensi le stress occurs at point A. which is farthest from the center of curvature.

le,\ = 0.022 III , .

Me"

(3 kN . m)(O.022 m) 868 X 10- 9 m~

e ~cl!=O,031i1ll p

B

+ Ad!)

+ (90 X 20)(12)2 + li(30)(40)1 + (30 X 40)(18)'

fT,,=+76.UM Pa ....

Maxim uUl COlUpr£S."ive StITss. This occurs al point B: we have ITB

=

MCIJ =

(3 kN . m)(0.038 m) 868 x 10 ~m~

b. Radius of Cun·alure.

fI B

= - 13 L3 MPa ....

From Eq. (4.2 1). we have

I M 3 kN ' m - = - = p EI ( 165 GPa)(868 x 10 ~9 m~) = 20.95 X 10 ] m ~ J

P "" 47.7 m ....

223

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PROBLEMS

--( i"l 2 il , IIi"ti

4 .1 and 4.2 Knowing that rhe couple ~hown termine the >tress at (a) point A. (b) point B. ~ 1 =lW kiJl . )"

B

fQll

(

aCl~

in a ve.rtical plane. de.

,..

----- --- I.:

..

~1.2i". -"""

J.lin.

I

B

Fig. P4.1

M =2.5\';:'1' )))

(~

~ 4,.iin.~

Fig. P4.2

4.3 A beam of the cros.~ sectiOtl soown IS extruded from an alumitlum alloy for which ", = 250 MPa and (1 10 = 1\50 MP::t. Using a factor of s:lfel) of 3.00, determ ine the largest couple that can be applied to the be
Fig. P4.3

4.4

Soh·e ]"rob. 4.3. as., umrng that the beam

i~

ben t "bout th e

y axis

T ile steel beam shown is made of a grade of s:eel for which u)" = rIu = 400 M Pa. USing a factor of safety ofl.50. determi ne the larges t couple Ihm ca n be IlppUed to we beam when it is bent aboul thex axi s 4 .5

2S[) MPa and

4 .6 Fig. P4.5

224

a.~j S

Solve Prob. 4.5. as,; uming that the steel beam is bent about th.e y

by a couple of moment /II ".

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4 .7 through 4.9 Twu venk:aJ run:es an: applied IU a beilJT1 ur Ihe (;ros~ section ,hOIlIO. Detennine the maximum tensik and compressive stresses in pllrtion Be of the beam

Problem.

;r"", ill

l in,-

---..l

If..

~"

-

1--1

,

Jill,

'\ in,

C

"

l)

~-w;,,-~ 20in ,

20 in,

Fig. P4. B

Fig, P4.7

to '","

to '"'"

lSOm",

L2·)(I"'IIl-+COC-~

Flg. P4.9

4.10 Two equal and opposite couples of magnitude M = 25 tN . III are applied to the channel-shaped ream AB. Obsen'ing that the couples cause the beam 10 bend in a horizolllaJ plane, delcnnine Ihe Slress al ( 11) point C. (b) point D. (c\ point E.

'"

120,","

clio 1 ~ ---s..)mlll I - _ J6n:T 11>()

n,,"

L

E

,

~

:>u 111111

'I

A

1

~++-_J C

Fig, P4.10

4 .11 KnO\\lillg th]t a bean) of the cross secllon shown is bent aholll a horizontal axis and that the bending moment is 8 kip' in., detennine the total force aCli ng 011 lhe shaded ponioll of the beam.

4.12 Solve Prob. 4.1 1. assuming thaI the beam is bent about a axis and that the bending mDlllfnt i. 8 kip · in.

~ertical

---.i

1°'"' Ulill

J

~===~ O.l in , _ I 1--1 1.-::1 0.3 ,". / Fig. P4.1 1

1.2, .. ,

\. 0..1 ill

225

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226

Pure Bending

4.13 Knowi ng thai a beam of the cross section shown is belli about:l horizonlal axis and Lhal Ihe bending moment is 6 kN . m. determine the 10lal force acting on Ihe top nange.

-t-.... ~m

s.t nlln

---r

IO~ nlltl

1--1

;2

~

4. 14 Knowing Lhal a beam of the cross seclion shown is benl aboul a horizonlal axis and Ihat the bending momenl is 6 kN . m. determine Ihe 10lal forc e acting on Ihe shaded ponion of the web. 4 .15 Knowing that for the casting shown the allowable stress is 6 ksi in lension and 15 ksi in compression. determine the largesl couple l\1 Ihat can be appl ied.

nlln

Fig. P4.13 and P4. 14

r

.ju,n,n -----j

I v ~I

[s nlln

"~r"""

..L

~ Fig. P4.16

Fig. P4.1S

4 .16 The beam shown is made of a nylon for whkh Ihe allowable stress is 24 MPa in lension and 30 MPa in compression. Determine the largest couple 1\1 Ihal can be applied 10 the beam. 4. 17 Solve [>rob. 4. 16. assuming Ihal d = 40 mm.

4 .18 a nd 4.19 Kn owing Ihal for th~ eXlruded beam shown Ihe allow able stress is 120 M Pa in tension and 150 MPa in compression . delermine Ihe largest couple M that can be applied.

~fi'" I 50 ,n",

I---c--I -1O",,,, Fig. P4.18

,~ ~

k-I [5U,,,,,, Fig. P4.19

,Ie

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4.20 Knowing that for the beam shown the allowable stress is 12 k.. i in tension and 16 ksi in compression, determine the. largest couple 1\1 that can be applied.

r---2A

i,\,

Probloms

------j

[&J i

1.2in

_

_ .7~

~

Fig. P4.20 4.21 A steel band blade. that was originall y straight. pa~ses over 8-in.diame ter pulleys when mounted on a band saw. Detcnlline the ma ...imum stress in the blade. knowing that it is 0.0 18 in. thick and 0.625 in . wide. Use £ = 29 X lOb psi.

oms in. Fig. P4.21

4.22 Knowing that (fal' = 24 hi for the steel strip AB. determine (a) the largest couple M that can be applied. ( b) the corresponding radius of curvature. Use E = 29 X 106 p~i.

Fig. P4.22 4.23 Straight rods of 6-mm di,lIneter and 30-m length are siored by coiling the rods inside a drum of l.25-m inside diameter. Assuming that the yield strengt h is not e... :eeded, determine (a) the ma ... im um stress in a coiled rod. (b) the corresponding bending moment in the rod. Use E = 200 G Pa.

Fig. P4.23

4.24 A 200-kip . in. couple is applied to the \V8 X 31 rolled-steel beam shown. (a) Assuming that the couple is applied about the t axis as shown. determine the maximum stress and the radius of curvature of the beam. (b) Solve pan a. assuming that the couple i ~ applied about the y axis. Use E = 29 X 1Cf' psi.

: ______

Fig. P4.24

I~­

227

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228

Purn

Bending

4.25 A 6O-N - m couple is applied to the sleei bar shown. (u) Assuming that the couple is applied about the t. axis as shown. detenninc the maximum stress and the radius of curvature of Ihe bar. (b) Solve pari u. assuming thai Ihe couple is applied about the J' axis. Use E = 200 GP,!.

Fig. P4.2S

4.26 A couple of magnitude M is a~lied 10 a square bar of side a. For each of the olientalions shown. delermine Ihe maximum stress and the curvalure of the bar.

,.,

" (b )

Fig. P4.26

4 .27 A portion of a square bar is removed by milling. so thai its cross section is as shown. The bar is then bent nboUi its horizontal axis by a couple l\.I. Considering the case where h = 0.9hu. express the maximum stress in the bnr in Ihe foml u .. = kfl u where f ' lI is the maximum s tress Ihat would have occurred if the origi nal square bar had been benl by Ihe same couple M, and detemline the value of k.

I

/"

"

I

I I

IIU

Fig. P4.27

I

--t" /,

I

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4.28 In Prob. 4.27. determine (a) the value of h for which the maximum stress IT,. is as small as JXlssible, (b) the wrresporuling value o f k. 4.29 A W200 X 31.3 rolled-steel beam is subjected to a couple M of moment 45 kN . m. Knowing that £ = 200 GPa and v = 0.29. determine (tI) the radius of curvature p, (b) the r..tdius of curvature p' of a transverse cross section.

Fig. P4.29

4.30 For the bar and loading of Example 4.01. determine (a) the radius of curvature p. (b) the radius of curvature p ' of n trans\'erse cross section, (el (he angle between (he ~ides of the bar (hat were originally verticnl. U~e E = 29 X 106 psi and v = 0.29. 4 .31

For tl£ aluminum bar and loading of Sample Prob. 4.1 , detennine

(a) the radiLls of curvature p' of n transverse cross section. (b) the angle between the sides of the bar that were originally vertical. Use £ = 10.6 x IO~ psi

and v = 0.33. 4 .3 2 It was assumed in Sec. 4.3 that ihe nonnal stresses (r,. in a member in pure bending are negligible. For :m initially straight elasti c member of rectangular cross ~ectioll. (a) deri ve an approximate expression for (J" . as a function of y. [b) show Ihat (If ,)m.u = - lcj 2p)(lf, l...., and. thus. that Ir, can be neglected in all practical situations. (Hilll : Consider the free-body diagmm of the pOl1ion of beam located below the surface of ordinate y iUld assume that the di stribution of the Slres.~ IT., i.~ st.iIJ linear.)

'J Fig. P4.32

= - c

,.

Probloms

229

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230

4.6. BENDING OF MEMBERS MADE OF SEVERAL MATERIALS

Pure Bending

.--e Fig. 4.23

2

The deri vations given in Sec . 4.4 were based on the assumption of 11 homogeneous material with a given modulus of elasticity E. If the member subjected to pure bend ing is made of two or more materials with different moduli of c I;L~ t icity, our approach to the determi nation of the stresses in the member must be modi lied. Consider, for instance, a bar consisting of two portions of different materials bonded together as shown in cross sectio n in Fi g. 4.23. Thi s composite bar will deform as described in Sec. 4.3, since its cross scction remains the same throu ghout its entire length, and since no assumption was made in Sec. 4.3 regarding the stress-strain rel ationship of the material or materi als involved. Thus, the nonnal strain € ... still varies linearly wilh the distance y from Ihe neutral axis of the section (Fig. 4.24a and b), and formula (4.8) holds: )'

(4.')

p

'"

'o j

Fig. 4.24

Strain and stress distribution in bar made of two malerials.

However, we cannot assume that the ne utral axis passes through the centroid of the composite secti on. and one uf the goals of the present analysis will be to detennine the location of this axis. Since the moduli of elasticity E I and E2 of the two materials are different , the expressions obtained for the nonnal stress in each material will also be different. We: write al =

E lY

EI€,= - p

(4.24) a! = E2 €. ,

E i)'

=-p

and obtain a stress-distribut ion curve consisting of two segments of str:.light line (Fig. 4.24c). It fo llows from Eqs. (4.24) thaI the force dF I exerted on lin element of area dA of the lIpper portion of the. cross seclion is (4.25)

while lhe force

dF~

exerted on :tn clement of the same area

(fA

of the

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lower portion is (4.26) But. denoting by n the ratio Ez/E! of the two modu li o f elasticity. we can express dF1 as (n E !)v ElY dF, = - ~~ - dA "'" - - (n dA) (4.27 ) . p p Comparing Eqs. (4.25) and (4.27). we note that the same force dF 2 would be exerted on M clement of area n dA of the fi rst material. In other words, the resist:1nce to bending of the bar wou ld remain the same if both portions were made o f the first material, provided that the width of each element of the lower ponion were multiplied by the. factor II. Note that thi s ....·idening (iell > I). or n:1rrowing (if II < I). must be effected in (/ dil"f'ction parallel 10 lhe neufral axil' of the sec tio/J. since it is essenti al that the distance)' o f each clement from the neutral axis remai n the same. TIle new cross section obtained in this way is called the mll/.I!arlllell section o f the member (Fig. 4.25). Si nce the transfonned section represents the cross section of a member made o f a IlOlIIogcncolI.l· material with a modulus of elasticity E !. the method described in Sec. 4.4 can be used to dete rmine the neutral ax is of the section and the nOTmal stress at various points o f the section. TIle neutral ax is will be drawn ,hrollS" lhe centroi(/ of the tramfarmcd .I·ectio/l (Fig. 4.26), and the stress U _f at any point of the corresponding

Fig. 4.26 Distribution of stresses fn transformed section .

fictiti o us homogeneoLls member will be obtained from Eq. (4.16)

u'=-,

My

(4.1 6)

where), is the distance from the neutral surface, and I the mOlllclI1 of incrtia of the Inll!.lformcd section with respect to its ccntroidal axis. To obtain the stress U! at a point located in the upper portion of the cross se<:tion of the original composite bar. we simply compute the stress (1'., at the corresponding poi nt of the transformed section. However, to obtain the stress Ul at a point in the lower ponion o f the cross section, we must IIIlIltiply by II the stress U x computed at the corresponding point of the transfonned sectio n. Indeed. as we saw earlier,

- - - --j,,-

,~\ .. ····· ,..."..:-·_\c£:;;''''_ _ ~ ~+-l I---,,/, ----l Ag. 4.25

Transformed

sec~on

lor composite bar.

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232

Pure Bending

the same elementary force dF2 is applied 10 an clement of area /I tilt uf the tmnsformed section and to an element of area tilt of Ihe origi nal section. ThllS, Ihe siress U" 2 at a poinl ()f the ori gi nal section must be II times larger than the stress at the corresponding point oflhe transformed section . The deformations of a composile member can also be determined by lIsing the transfonned section. We recall thai the transfomled section represents the cross section of a me:mber. made. of a homogeneous material of modulus £1. which deforms in the same man ner as the composite member. Therefore, using Eq . (4.2 1), we. write that the curvature of the composite me.mber is

where 1 is the moment of inen ia of the transform ed seclion with respect to ils neutral axis. EXAMPLE 4.03 A

bar obtained by bending tOgether pieces of steel 29 x 1O~ psi) and brass (Eh = IS x IO~ psi) has the cross section shown (Fig. 4 21). Detennine the maximum stress "i n Ihe steel and in the bra.s when the bar is in pure bendin g with a bending momen t M = 40 kip' in.

(Et

"""

formed section nbout its ccntroid<11 axis is

I = fi b}? = -&(2.25 in. )(3 in.)' = 5.063 in"' and the maximum distnnce from the neUlrnl axis is c = 1.5 in. Using Eq. (4. 15), we find the mnximum stress in the transformed section:

The tr:ln~formed section correl'ponding to an equivalent bar made entirely of br<1~s is shown in Fig. 4.28. Since

Me

IT ",

E,

29 X 106 psi

Eb

IS X 10~ psi

= -=

=

I

(40 ki p' in.)( 1.S in. ) 5.063 in.

11.85 ksi

(0.15 in.)( 1.933) = lAS in.

The va lue obtained also represents the m
Note lhntthis change in d!mension occurs in a direction p
(If'''''',) ..... = 11.85ksi (cr,'....I)..... = (1.933 )( 11.85 ksi) = 22.9 ksi

/I

1.933

the width of Ihe central portion of brass. which rep laces the original steel ponion. is obtained by ffillitiplying the original wid th by 1.933. we have

U.; 5 'u.

I r-------J 1- 0.4 'n.

r

004 in

3,".

L~ St,..,1 \

Hms<

Fig. 4.27

Brass

Fig. 4.28

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An important example of stmctural members made of two different materials is furni shed by reinforce d COl/crete belmls (Fig. 4.29). These beams, when subjected to positive bending moments. are reinforccd by steel rods placed a short di stance above their lowcr face (Fig. 4.30a). Since concre te is very weak in tension, it wi ll crack below the neutral surface and the steel rods will carry the entire tensile load, while the upper part of the concrete beam will C:UTY the compressive load. To obtain the transformed section of a reinforced concrete beam, we replace the total cross-sectional area A. of tile steel bars by an equivalent area I/A .. where 1/ is the ratio £ .1£ ,. of the moduli of elasticity of steel and cOllcrcte (Fig. 4.30b). On the other hand. since the concrete in the beam acts effectively only in compression, only the porti on of the cross section located above the neutral axis should be used in the transformed section. The position of the neutral axis is obtained by determining the di stance x from the upper face of the beam to the centroid C of the transfomled sectioll . Denoting by b the width of the beam. and by d the di stance from the upper face to the center line of the steel rods, we write that the first moment of the transform ed secti on with respect to the neu-

4.6. Bending 01 Members Made 01 Several Materials

Fig. 4.29

c

"

I ,..,. d- .~

=:===-.1 "A, ,h)

(,,)

Fig. 4.30

F

'd

tral axis must be zero. Since. the first moment of each of the two portions of the tramformed section is obtained by multiplying its area by the distance of its own centroid from the neutral axis, we have x

(bx)'2 - IIA , (d - x) = 0 0'

I

,

2"hr + IlA ,x - IIA,d = 0

(4.28)

Solving this quadratic equation for x. we obtain both the position of the neutral axis in the beam. and the. portion of the cross section of the concre te beam which is elTectivcly used. The determination of the stresses in the transformed section is carried ou t as explained earl ier in this secti on (see Sample Prob. 4.4). The di stri bution of the compressive stresses in the concrete and the resultant f . of the tensi le fo rces in the steel rods are shown in Fi g. 4.30c.

233

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234

4.7. STRESS CONCENTRATIONS

Pure Bending

The formula if", = Me/I was deri ved in Sec. 4.4 for a member with a plane of symmetry and a uniform cross section. and we saw in Sec. 4.5 that it was accurate throughout the entire length of the member only if the couples M and M ' were appl ied through the use of rigid and smooth plates. Under other condi tions of application of the loads, stress COIlcentrations will exist ncar the points where the loads are applied . Higher stresses will also occlir if the cross section of the member undergoes a sudden change. Two particular cases of interest have been studied,t the case of a !lat bar with a sudden change in width. and the case of a nat bar with grooves. Since the distribution of stresses in the nitical cross sections depe nds only upon the geometry of the members. stress-concentratioll factors can be detennined for various ratios of the parameters involved and recorded as shown in Figs. 4.3 [ and 4.32. The

,

0

2.8 2.6

, 2

K 2.0

I\\, , 1\\ ,,\ 'J 2 ll ~

~ \5 ~~~2

I"

I.8 I.6

,

I.

2

0

"(~)"

II \11

-

"-

-

0.05

,

2.6

:1..2

,

!

I.U2 LOt

O.tO

I ' ,_ ,\ II ( [I3§J) M \\ ,\\ b=!_ W \,[ 1 ~~ h \'

OR

2.4

;:--.. r::;;;;~"

"-

'.0

-

Oh

,\ ,-<::

\:

1.8

~ I 0.20

1.6

1

1.2

,)

II

"'- 1"-' K::--- ~5

~~

1.2

0.2.5

0.3

rid

Fig. 4.31 Stress-concentration factors for naf bars with fillets under pure bending.t

1.0

o

-

-

0.20 0.25 0.30 0,\" ..td Fig.4.32 Stress·concenlralion factors for nat bars with grooves under pure bending.t 0 .10

t Wo D. Pi lkcy. /'e lerso,,-, Slress COIICl'mraliOl' FaClOrs. 2d ed .. Jotm Wiley & Sons. New York. t997 .

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4.7. Stress Concentratloos

yalue of th(: ma.,
•

M ,' = KI

235

(4.29)

....,here K is the stress-concentration factor. and where c and I refer to the cri tical seclion. i.e .. to the section o f width d in both of the cases considered here. An exami nation of Figs . 4.3 [ and 4.32 clearly shows the impo rt ance of usin g tillets and grooves o f radius r as large as practical. Finally. we , hould point OUt that. as was the case for axial loading and torsion. the values of the. factors K have been computed lInder the assumption of a linear rehltio n between stress and Strain. In many appli cations, plastic defo nnalions will occur and result in values of the maximum stress [ower than those indicated by Eq. (4.29).

EXAMPLE 4.04 Grooves 10 mm deep nre to be CUI ill a steel b
The va lue of the .>tress Mcl l is thus Me;

( IRON 'm)(20X 10

48)( 10

l

m)

' m~

75 MPa

SubsTituting thi s "alue for Mel l into Eq. (4.29) allll makin g = 150 MPa. \\e wri te

(f in

150 Ml'a = K(75 MPa)

K= 2 We have. on the other hand.

D d

60mm

- ~ --- =

/i=9mm

Fig. 4.33

'"

,

- = 0.13

d = 60 mm - 2( 10 mm) = 40 mm ~ II = 20 mm

1.5

Using Ihe curve of Fig. 4.32 correspo nd ing to 0111 = 1.5. we lind that the va lue K = 2 corresponds 10 a value of rid equal to 0. 13. We have. th erefore.

(bl

We nOTe from Fi g. 4.33u that

c=

40mm

d

b = 9 mITt

The moment of inenia of The critical cross seclion about its neutral axis is

I = fibd~ = -b(9 x 10 3 m)(40 X 10 =48X 10 ~ m4

J

m) l

r = 0 13d = 0.13(40 mm ) = 5.2 mm The smallest allowable wid th of the grooves is Ihus 2r = 2(5.2 mill) = 10.4 mm

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r

SAMPLE PROBLEM 4.3 2()Omm -

-.l

1"'' ' '

1: 300 """

~J

Two steel plates have been welded together to form a beam in the shape of a T that has been strengthened by securely bolting to it the two oak timbers shown. The modulus of elasticity is 12.5 GPa for the wood and 200 GPa for the steel. Knowing that a bending moment M = 50 kN . m is applied to the composite beam. detemline (a ) dIe maximum stress in the wood . (b) the stress in the steel along the top edge.

75"""TI~75m' " ,Ill" 20

SOLUTION Transfon ncd Section. 11 =

We first compute lhe mlio

E, = 200 GPa = 16 E.. 12.5 GPa

Multiplying the horizontal dimensions of the steel portion of the section by 16. we obtain a tmnsformed section made entirely of wood.

11 =

Ncut ral Axis. The neutral axis passes through the centroid of the transfOnlled section. Since the .~ction consists of two rectangles. we have

y= _ ~3_ 'A = ~"(O".I"60 :;;:;m,,)(::,J:::.2:::m:;-:;;Xc;O::;.:;:O 2::,o,,:m::!),,+,,,,O= 3.2 m X 0.020 m + 0.470 m X 0.300 m

2:A

Ccntroidal MOIlll'" llt of IIll'rlia.

0.050 m

Using the parallel-a:;:is theorem:

J = 6 (0.470)(0.300)3 + (0.470 X Q.300)(0.050j2

+fd3.2)(0.020)) 1 = 2.19 X

+ (3.2

X

0.(20)(0.160 - 0.05of

IO w 3 m~

a. Maximum StITss ill " 'ood. The wood farthest from the neutral axis is located along the bonom edge. where c1 = 0.200 m. Mel lr

(50 X 10] N· m)(0.200 m)

=--~

2.19XIO -3 m~

"I

b. St rt'ss in S t('cl. Along the top edge c, = 0.120 m. From the transformed section we obtain an equivalent stress in wood. which must be multiplied by II to obtain the stress in steel. M e!

O"s

=

1 -,-

. (50 X 103 N . m)(O. [20 m) = (16)

2.19 x IO -J m"' fT, =

236

43 .8 MPa ....

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SAMPLE PROBLEM 4.4 A concrete floor slab is reinforced by i-ill.-diameter steel rods placed 1.5 in. above the lower face of the slab and spaced 6 in. on centers. The modulus of elasticity is 3.6 X 106 psi for the concrete used and 29 X 10" psi for the steel. Knowing that a bending moment of 40 kip' in. is applied to each l-fl width of the sl:lb. determine (u) the maximum stress in the concrete. (b) the stress in the steel.

SOLUTION

r--

-I

4'\..-, --.-;

1+

12

ill---j

c

Trunsfonncd Mctioll. We comidcr a ponion of the slab 12 in. wide, in which there arc two i -ln. -diameter rod~ having a to tal cmss-sectional area

I"

\ lILI, - 4.95 in '-

Since concrete aCls only in compression. all tensile forces are carried by the sleel rods. and the transformed section consists of the two areas shown. One

is the ponion of concrete in compression (located above the neutral axis), and the other is the transfomled steel area /lA,. We have

t.:

29 X 10" psi 8.06 Eo 3.fi x 106 psi 1 1L'l., = 8.06(0.614 in ) = 4.95 in! 11 = - =

I "hi

il

4

Ill. -

1,

I~wo'" . !

Neutral Axis. The neutral axis of the slab passes through the centroid of the transformed section. Summing moments of the transfonned area about the neutral axis, we write

tl=4 - .~ '=2.5.'iin

12\{~) -

,- - L

4.95(4 -

xl

=

0

x = 1.450 in.

1\10111t'nt of lncrtia. The centroidal moment of inenia of the transfonned area

IS

1 = }(l2)(1.450)l

+ 4.95(4 - J.45W

Maxim um Stn.'Ss in Concrete. c, = 1.450 in. and ll.

u c

= Mc 1 =(40 kip · in.)( 1.450in.) I 44.4 in 4

b. Stn.'SS ill Steel.

if.

=

= 44.4 in-l

At the top of the slab. we have

(f ,

= 1.]06 k~i ....

For the steel. we have Cl = 2.55 in., II

Mc~ c(4~O~k~;~P~·~;'~!.~)(S2~ .5~5~;~ ".") 11-,= S.06 44.4 in'

if,

= 8.06 and

= IS.52 ksi ....

237

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PROBLEMS

10,,,rn 1lI "OJ"

H

H-.l

Al,u"rnuu' = r g 10",,,,

I

Ilr:O$s

4.33 and 4.34 A bar having Ihe cross section shown has been formed by securely bonding brass and aluminum stock. Using the data g iven below. determine the largest permissible bending moment when the composite bar is bent about a horizontal axis.

40 un"

~ 10",,,, 1-

40

m",---l!

Modulus of ela.~t ic ity Allowable stress

Aluminum

Brass

7(] GPn 100 M Pa

105 G Pa 160 M Pa

Fig. P4.33

SUlm lin.m H--32mm - -tl- -I1

1

3 2.nH n

Y------~J "- lira"

\...AI""""'''<1

Fig. P4.34

4 .35 and 4.36 For the composite bar indicated, detemlille the largest permissible bending moment whe n the bar is benl abollt a vertical axis. 4.35 Bar of Prob. 4.]3. 4.36 Bar of Prob. 4.34.

1

J ~EF

u...ut:wJ.o....

!

In.

FIg. P4.37

238

4.37 Three wooden beams and two sleel plates are securely bolted II.) · gether to form the composite member shl.)wn. Using the data given below. de· termine the largest permissible bending moment when the member is benl aboul a horizontal axis. Wood

Modulus of elastidty Allowable s (re~s

2

x

1O~

2000 psi

Steel psi

30

x 106 psi

22.000 psi

4.38 For the composite member of Prob. 4.37, dcterminc the largest permissible bending moment when the member is bent about a vertical axis.

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4 .39 and 4.40 A steel bal (E. - 210GPa) and an alumillum bal ( E" 10 GPa) are bonded IOgether 10 forr:! the composite blr shown. Detenmne the maximum stress in (a ) the a1uminUl:l. (b) the steel. when the bar is bcm about a horizOIltal axis. with /11 = 2lKl N . m.

Fig. F4.39

Problem.

Fig. PUO

4.41 and 4.42 The 6 X 12-in. timber beam has been strengthened by bolting to it the steel re inforcem~nt shown. TIle modulm of elasticity for wood is 1.8 X 10' psi and for steel 29 X lot' psi. Knowing that the beam is bent about a horizonlal a\is by a couple of moment M = 450 kip . in .. determine Ihe maximum stress in (a) the wood. (b) [he steel.

Fig. P4. 41

Fig. P4.42

4 .43 and 4.44 For the composite bar indkated. determine the radius of curvature caused by the couple of mom~n1 200 N . m. 4 .43 Bar of Prob. 4 .39. 4.44 Bar of Prob. 4.40. 4 .45 and 4.46

For tlte cOlnJXlsite helm indi cated, determine [he mdius

of curvalun:: caust::u by Iht:: l'oup lt:: of ITlUITlt::[\l 4 50

~ip

. in.

4.45 Beam of Prob. 4.4 1. 4.46 Beam of PlOb. 4.4! . 4 .47 .'\ co ncrete beam is reinforced by three Hee l rod, placed as shown. The modulw; of ela.>licily is 3 X 10" psi for [he co ncrete and 3(1 X I O~ psi for Ihe steel Using an allowable stress of 1350 psi for the concrete and 20 ksi for the Heel, determine lht: largest allowable positive bending momem in lhe beam.

1!lln

LL-/_~

0---,1

I-S i!O.-j

Rg . P4 .47

t ·in. ~h"n 'tN

=':"

239

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240

Pure Bending

I

54()mm

_ I ~ Fig. P4.48

4.48 The reinforceu concreto:: beam shown is subjected to a posili\"e bending moment of 175 kN . In. Knowing thai the modulus of el;lsticity i.'i 25 G Pa for Ihe concrete and 200 GPa for the sleei. delennine (11) the stress in the steel, (b) Ihe maximum siress in the concrele.

2.5-"""

d;am~\t'T

4 .49 350 mm.

Solve Prob. 4.48. assuming Ihat Ihe 300-mm widlh is increased 10

4.50 A connete slab is reinforced by ~- in .. diameter rods placed on .5.5·in. celllers as shown. The modulus of elasticity is 3 x 10" psi for the concrete and 29 X lOb psi for the steel. Using an allowable stress of 1400 psi for the concrete and 20 ksi for the steel. del~rmine the largest bending moment per foot of widlh Ihal can be safely applied to Ihe slab.

Fig. P4.50

4 .51 Knowing Ilwt the bending mOIl~nl in the reinforceu concrete beam is + 150 kip· f1 ;lnd Ilwt Ihe modulus of elasticilY is 3.75 X IO~ psi for the concrete ,md 30 X 106 psi for Ihe sleel. determine ([I) the slress in the steel. (b) tile maximum stress in tile concreTe.

3Ui".

24 in .

J-m diHU,dp,

...'\

'"~i"

1-I'2il\<- 1 Fig. P4.S1

I

~

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4.52 The design of a reinforced concrete beam is said to be balanced if the maximum stresses in the steel and concrete are equal. respectively, to the allowable stresses (T, and IT, . Show thm to achieve a balanced design the dis· tance x from the tup of the beam to the neutml axi~ must be

ProblOms

where E. and E, are the moduli of elastidty of concrete and steel. respectively. and d is the distarlce from the top of the beam to ule reinforcing steel.

r

"

••• 1--,,---1

_I

Fig. P4.52

4 .53 For the concrere beam shown. the modulus of e lasticity is 25 GPa for the concrete and 200 GPa for the steel. Knowing thm b "" 200 mm and d = 450 mm. and using an allowable s tres~ of 12.5 M.Pa for the concrete and 140 M P~I for the steel. detennine (rr) the required nre<\. A, of the steel reinforcement if the benm is to be balanced. (b) the largest allow:lble bending moment. (See Prob. 4.52 for definition of a balanced beam.)

4.54 For the concrete beam shown. the modulus of elasticity is

3.5 X 106 psi fO! the concrete and 29 X ](1 psi for the steel. Knowin g lhat b "" 8 in. and d "" 22 in .. and using an allowable stress of 1800 psi for the concrete and 20 ksi for the steel. determine (1I) the required area A, of the steel reinforcement if the beam is to be balanced, (b) the largest nllowable bending moment. (See Prob. 4.52 for definition of:l balanced beam.)

1

"

• ••

Fig. P4.S3 and P4.54

4.55 and 4 .56 Fille metal strips. each of 15 X 45-mm cross section. are bonded togelh~r to form the composite beam shown. The modulus of elasticity is 210 GPn ior the s leel. 105 GPa for the brass. and 10 GPa for the aluminum. Knowing that the bcnm is bent about a horizontal aXIs by :I couple of moment 1400 N . m. determine (1I) the maximum stress in c:lch of the three metals. (hI the radius of curvature of the compositc beam.

t--45 "",,--I Fig. P4.S5

J

Fi g . P4.56

241

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242

Pure

Bcn()tng

4 .57

A steel pipe ,md an alumlnuLIl p ipe are. secuH:ly bonded to£elhel

to forlll Ihe compesite beam showJI. The modulus of elaS[lcity is 30 X 10" psi for the sleel and ][) X Hf psi for the aluminum. KnO'.'.'ing thaI Ihe composite beJm l~ bent by a couple ()f moment 5 kip · in.. determine !he m:vamum Slfess (a ) in dte a.luminum. (b) in the s teel.

Aj," "UH,m

S I""]

tin OA in

Fig. P4.57

~ :C" 1-1

J

~ +"

F" " ~E,

"

,.

50mm

Fig. P4.59

4 .58 Solve Prob. 4.57. ll.~suming that lite ~.ill.-lhjd;: inner pipe is made of aJulnmum and thatlhe :-in.-Ihi ck outer pipe is made 01 sleei. 4 .59 The rectangular b~am shown is mad" of a plastic for wllich the value of the modulus of elaSllcity in tension is otle-half of its v:;.J uc in compression. For a bending moment AI == 600 N . m, determine the maximu m (.1 ) tensile s tress. (b) compressive s tres~.

'4.60 A rectangular bc::.m is made of material for ",hich the tlHxlulus of i, E, in tensiun and E, in compression. Show that the curvature of Ihe bellll in pure bending is

el~slicil)'

I

M

P

E,./

where

:\1 /

Q FI g. P4. 61 and P4. 62

4 .61 Knowing ill:lt the aJlow:lble SlTeS$ for the beam ~hown i> 12 I.::si. de!ennine the allowable bending mOlr.ent AI when Ihe rndius 'of Ihe fillels is (a } in. , (0) ~ in.

!

4 .62 Kllowing th:ll M == 3 kip . in.. determine the maximum stress in the beam shown when the rndius r oflhe iillets is (u ) in .. (b) ~ in.

1

I lOS """

!

4.63 Semicircular grooves Df radius r must be milled as shown in Ihe st~eI member. Knowing thm M - 450 N . m, dete:nnine: the: maxi mum slress in the member when the radius r of the semicircular grooves i ~ (a) r = 9 mm. (b) r = 18 mm. sides of a

4 .64 Semicircu lar grooves of radius r must bc milled as shown in the fir a ~ rl'el meml"Wr. lJ~in~ ~n allowahl.. ~Ire<..~ of WI MP,l , dr.lf'mlin r lhr largest bending momen t (hat can be applied 10 the memberwlien (c:) r ~ 9 mm. tb ) ,. - 18 mm. ~icie~

Fig. P4.63 :md P4.64

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4.65 A couple of moment M "" 2 kN . m is to be applied to the end of a sleel bar. Determine the maximum Stress in the bar (a) if the bar is designed with grooves having semicircular portions of radius r = 10 mm. as shown in Fig. P4.65(1. (b) if the bar is redesigned by removing the material above the grooves as shown in Fig. P4.65b.

."

y; ,

I

o

~" 15U }lI111

, oJ

---

tt;

1111 "

(I»

Fig. P4.65 and P4.66

4 .66 The allowable stress used in the design of a s teel bar is 80 MPa. Determine the largest couple M that can be applied to the bar (a) if the bar is designed with grooves ha ving semicircu lar portions of radius r = 15 mill. as shown in Fig. P4.6511. (b, if the bar is redesigned by removing the material above the grooves as shown in Fig. P4.65b.

*4.8. PLASTIC DEFORMATIONS

Whe n we derived the fundamental relat ion u , = - My/I in Sec. 4.4. we assumed that Hooke's law applied throughout the member. If the yield strength is exceeded in some portion of the member, or if the material involved is a briltle material with a nonlinear stress-strain diagram. this relation ceases to be valid. The purpose of this section is 10 develop a more general method for the determination of the distribution of stresses in a me mber in pure be nding, which can be used when Hooke's law docs not apply. We fir~ t recnll that no specific stress-strain relationship wus assu med in Sec. 4.3. when we proved that the normal strain E~ varies linearly with the distance y from the neUiral surface. Thus, we can still usc this property in Ollr present anulysis and write

where y represents the distance of the point considered ff()m the neutral Sllrface, and c the maxi mum va lue of y.

4. 8. Ptasttc Deformattons

243

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244

Pure Bending

- t ,.•

"

)

( Fig. 4.34

However. we can not assume anymore thaL in a given section. the neutral
( "

( ,

Fig. 4.35 'I

Fig. 4.36

The distribution of stresses in the cross sec tion of the member. i.e., the plot of U ., versus y, is obtained as lollow.~. Assuming that am"" has been specified, we fi rst detennine the corresponding value of E", from thc stress-s train diagram and carry th is value into Eq. (4.10). Then. for each value of )', we delcnni nc the cOlTespondi ng value of 10 , from Eq. (4. 10) or Fi g. 4.34. and obtain from the stress-strain diagram of Fi g. 4.35 the stress u~ correspondi ng to this value of ~\. Plotting u , against y yields the desired distribution of stresses (Fig. 4.36). We now recall that , when we derived Eq. (4.3) in Sec. 4.2. we assumed no particular rc1atiull between strcss and strain. We can therefo rc lise Eq. (4.3) to determinc the bendi ng moment M corresponding to thc stress distribution obtained in Fig. 4.36. Considering the particular C;L~e of 11 member with a re(;tangular cross section of width b, we express the element of area in Elj . (4.3) li S dA = h d)' and write M = - b

where u ,

i.~

r -,

yu .• dy

(4.30)

the fu nct ion ()f y plotted ill Fi g. 4.36. Since a .• is an odd

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function of y. we can write Eq . (4.30) in [he. alternative foml

M = - 2/) frfT' ely

4.8. PlasUc Delormatloos

(4.3 1)

o If fT.,' is a known analyt ical function o f E.... Eq . (4. 10) can be uscd to express fT" as a function of y. and the integral in (4.31) can be detennined analytically. Otherwisc. thc bending moment M can be obtained through a numerical integration. Thi s computation bccomes more meaningful if we note thatthc integral in Eq. (4.3 1) represents the ti rst moment with respect to thc horizontal axis of the area in Fig. 4.36 that is located above the horizontal axis and is bounded by the stressdistribution cune and the vertical axis. A n important v:Iluc of the bending moment is the ultimatc bending moment M y that causes rai lurc o f the member. Thi s value can be delennined from the ultimate stfCngth fT u of the material by choosing fT m., = fT u and carrying out the computations indicated earlier. However, it is found more convenient in practice to determine M u experimelllully for a specimen of a given material. Assuming a fictitiou s linear distriblilion of stresses, Eq. (4.1 5) is then used 10 determine the corresponding maximum stress Rn:

M uc RB = - , -

(4.32)

The lictitious stress RB is called the /IIi1dU/II.\" of mpture in bendillg of the given material. It can be used h.) determine the ultimate bcnding moment M y o f a membcr made of tht! same material and having a cross

--

--1--.,---1 I

1--,.- Ag.4.37

section of the Sllmc shape, bUl o f differe nt dimensions, by solving Eq. (4 .32) for M u. Since. in the case of a member with a rcct:mgular cross scction, the actual and the licti tious linear stress distributions shown in Fig. 4.37 must yield the same value M u for the ultimate bending moment, the areas they del'ine must have the same first moment with respect to the horizontal axis. It is thus clear that the mod ulus of ruptu re R8 will always be larger lhan the actual ultimate strength a u.

245

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246

Pure Bending

' 4.9. MEMBERS MADE OF AN ELASTOPLASTIC MATERIAL In order to gai n a better insight into the plastic behavior of a member in bending, let us consider the case of a member made of an elasto{Jlm·tic material and first assume the member 10 have a rectallgll larcros.f .I·cetioll of width b and depth 2c (Fig. 4.38). We recall from Sec. 2.1 7 that the stress-strain diagram for an ideali zed eiastoplastic material. is as shown in Fig. 4.39.

Fig. 4.38 Ag. 4.39

As long as the normal stress (T, do~s not exceed thc yield stre ngth Hooke's law applies, and the stress distribution across the section is lincar (Fig. 4.40a ). The maximum value of the stress is

(Ty,

Me (T,~

(4.15)

= - ,-

As the bendi ng moment increases, (T", eventually reaches the value (T r (Fig. 4.40h). Substituting this value inlo Eq. (4.15 ). and solvi ng for thc corresponding value of M , we obtain the value M y of the bending mome nt at the onset of yield: (4.3 3)

The moment M y is referred to as the maximllm elastic mOll/elll, since it is the largest moment fo r which the deformatio n remains fully elastic. Recalling that, for the rectangular cross section considered here, I

b(2c)3

2

c

12c

3

,

-~~~=- bc-

(4.34)

we write My=

, "32 bc-(Ty

(4.35)

As the bending moment further increases, plastic zones develop in the member, with the stress uniformly equal to - (Tr in the upper zone, and to + (Ty in the lower zone (Fig. 4.40c). Between the plastic zones, an elastic core subsists, in which the stress (Tx varies linearly with y, ~, .

(T.,. = - - ) '

(4.36)

Y,

where )'r represents half the thickness of the clastic core. As M increases, the plastic zones expand until. at the limit, the deformation is full y plastic (Fig. 4.40d). Equation (4.31) will be used 10 determine the value of the bending moment M corresponding to a given thickness 2)'r of the elastic core.

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Recalling 1hat (T , is given by Eq. (4.36) for 0 ::s y S Yr, and is equal to - IT },

for )'y

:s y ::S:

4.9. M embers Made o f an Elastoplastlc Material

we write

C,

f"'y( - ~»

M = - 2b

o = M = hc1 (T

\

dy - 2b f-r( - rTy) dY

-

~ ~ "32 "Yr(Tr + berT r -

'. by;.(Ty

( J - ~4)

(4.37)

3 c-

or. in view o f Eq. (4.35).

3(

M =-M [ - -I )"~) 2 Y 3 c2

(4.38) {n) .II <: M t

where M y is the maxi mum clastic moment. Note that as)'r approaches zero, the bending moment approac hes the limi ting va lue t 4.39)

Th is value o f the bending moment, which corresponds to a full y plastic defannation (Fig. 4.40d). is called the pla.I"tic mo ment of the member considered. Note thai Eq. (4.39) is valid onl y fo r n rectangula r mcmbel' lIIade of an e!astopla.\'ric material. You should kee p in mind that Ihe distribution of straill across the section remains linear after the onset of yield. Therefore, Eq . (4. 8) of Sec . 4 .3 re mai n ~ valid and can be used to determine the half-thic kness }'y o f the elastic core. We have (4.40) when: Ey is the yield strai n and p the radius of cur vat ure com:sponding to a bending moment M 2: M y. Whe n the bending moment is eq ual to M r , we have),y = c and Eq. (4.40) yields

-c

1' 1 ~,,!>'T 1 C

ELASTIC

l'L . \ST IC

(4.4 1) where py is the mdius of curvature corresponding to the maximum elastic moment M y- Di viding (4 .40) by (4.4 1) member by member, we obtai n the relationt J'r

P

c

py

(4.42)

Substituting for Yr/c from (4.42) into Eq. (4 .38). we express the bending moment M :'ts a functio n o f the r:'td ius of c urvature r of thc nc utral surface:

r')

1M = -3 M ( 1 - 2 y 3

rt

Fig. 4.40

(4.43)

Note that Eq. (4.43) is valid onl y afte r the onset of yield. i.e ., for values of M larger than Mr. For M < M y, Eq. (4.2 1) of Sec. 4.4 sho uld be used . t Equ3tion (4..t2) ~lpties to any membtr made of any ductite material with J wetl ·ddincd yield point. sinc e its derivation is independent of the shape of the cross 5eL1iotl and of the sJupc of the stress· strain di3~raJII beyond the yirtd point

1T~."""' u,,, - tJ l

(b) M=,ut ~ (1)

!I

-"'1""'---,,",

247

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248

Pure Bending

We observe from Eq. (4.43) that the bending moment reaches the value M" == 1M y o nly whe n p = O. Since we dearly cannot have a zero radius of' curvature at every point of the ne utral surface, we conclude that a full y plastic deformatio n canno t develop ill pure bending. As you will sce in Chap. 5. however, such a situat ion may occur at one point in the case of a beam under a transverse loading. The stress distributions in a rectangular me mber corresponding respecti vely to the maximum elastic moment M ), and to the limiting case of the plastic moment Mp have been represented in thrL"e dimensions in Fig. 4.4 1. Since, in both cases. the resultants of the elementary te nsile and co mpressive forces must pass through the centroids of the volumes representing the stress distributions and be equal ill magnilllde to these volumes. we eheck that


and that the mo ments of the corresponding couples are. respectively. (4.44)

nnd (4.45)

(/))

Fig. 4.41

We thus veri fy that, for a rectangular member. MJ, = ! Mr as required by Eq. (4.39). For beams of JIOIJ J"fiC f{lIl g u /ar cm.I·.I· section, thc computation of the maximum elastic moment Mr and of the plastic moment Mp will USllall y be simpl ified if a gmphical methoo of nnal ysis is used. as shown in Sample Prob. 4.5. It wit! be fo und in thi s more geneml C:"lSC th:"lt the mtio k = M,/M y is generall y not equal to For m ucluml shapes such as wide-Bange beams. fo r example. thi s ratio vari es approximatel y from 1.08 to 1.14. Because it depends o nl y upon the Sh3pe of thc cross ~ec ­ tion. the ratio k = M"JM y is referred to as the .I"'wp e fac to r of the c ros~ section. We note that, if the shape factor k and the max imum cl:"lstic moment M y of a beam :"Irc know n. the plastic moment Ml' of the benm c:"ln be obtained by multiplying M )' by k:

1.

(4.46)

The ratio M"Jur obtained by dividing the plastic mo ment Ml' of a member by the yield strength U r of its mlllerial is called the p la stic .fec(ion modJl{II.fofthe memlx:r and is denot~d by Z. When the plastic section moduluS' Z and the yield strength U y of a beam are known. the plastic moment M" of the beam C3n be obt 3ined by multipl ying Ur by Z: (4.4))

Recalling from Eq . (4 .1 8) that M )' = SU )', and comparing this relatio n with Eq. (4.47). we note that the shape factor k "" M"JM r of a given

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cross section can bt: expressed as the mtio of the plastic and clastic seclion moduli: M, W I' Z k ~ - I ~ ~ (4.48) M)' SU I' S Considering the particu lar case of a rectangular beam of width h and depth II, we nOle from Eqs. (4.45) and (4.47) that the plaslie sectilm modulus o f a rectangu lar beam is Mp

belu )'

~

1

4.9. Members Made of an Elastoplastlc Matenal

249

'

Z= = - - = be- = " bllfTr ur

On the other hand. we recall from Eq. (4.19) of Sec. 4.4 that the elm'lie section lIIodl'/US of the. same beam is S = ~hh l

Substituting inlO Eq. (4.48) the values obtained fo r Z and S, we verify that the shape factor of a rectangular beam is Z

~ bh 2

3

S

~ bh 2

2

k =- =- ~-

EXAMPLE 4.05 A member of uniform rectangular cross section 50 by 120 mm (Fig. 4.42) is subjected 10 a bending momenT M = 36.8 kN . 111. Ass uming thaT the member is made of an e1astoplastic material with a yield strength of 240 MPa and a mCKiulus of elasticity of 200 GPa. determine (a) the thickness of the elastic core. (b) the radius of cUf\lature of the neutral surface.

and carrying Ihis value . as well Eq. (4.33).

I

I

Substituting the

"alue~

of M and My into Eq. (4.38). we have

m =~(28.8kN.m
36.8kN

~ ,. = 0.666

,· = f,O m",

I

and. since c = 60 mm. Y)' = 0.666(60 mm ) = 40 mm

,. =1,, , ,

L

The Thickness 2.1'y of the elastic core is thus 80 mm.

(b) Radius of Curvature. stnlin is

L- -- _- - ----' --

Fig. 4.42

2

- = ~b(;2 = - (50 X IO ~ J m)(60 X 10

J

= 120 x 10

We nOie Ihat the yield

240 X lcf Pa E = 200 X I09 Pa

tT e

(a) ThIckness of Elastic Core. We tirst delermine the maximum elastic moment M y. Substituting Ihe given data into Eq. (4.34). we have

I' , 3

= 240 MPa. into

M r = ;tTy = (120 X 10 ~ ml)(240 MPa) = 28.8 kN . m

h = .'lQ ,,,,,,

I

a.~ rT y

J

m

lmr

E y=

Solving Eq. (4.40,1 for (J and SUbstituting the values obtained for }'y and E r. we write y

p =....! = IEI'

40 X 10 - 1 m _) - 31.3m 1.2 X [0

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250

*4.10. PLASTIC DEFORMATIONS OF MEMBERS WITH A SINGLE PLANE OF SYMMETRY

Pum Bending

In our discussion of plastic deformations. we have assumed so far that the member in bending had two planes of symmctry, one containing thc couples 1\1 and M ' . and the other perpendicular 10 that planc. Let us now consider the more gcnera l case when the member posscsses on ly one planc of symmctry containing the couplcs 1\1 and M ' . However, o ur analysis wi ll be limited to the si tuation where the deformation is fully plastic. with the normal stress unifOimly cqual to - U y abovc thc nCUlral surface. and to + U y below that surface (Fig. 4.43a). As indicatcd in Sec. 4.8. the Ilcutrill axis cannot be ass umed to coincide with the centroidal axis o f thc cross section whcn the cross scction is not symmctric with respect to that axis. To locate thc nc utral axis, we consider thc resultant R I o f the elemcntary compressive fo rces exen ed on the ponion Al of the cross s~tion located above the neutral axis, and the resultant R2 of the tensile force s exerted o n the portion A1 located below the nelltral axis (Fig. 4.43b). Since the forces R I and R1 fonn a couple equ ivalent to the couple appl icd [0 thc me mber, thcy must have the same magnitude. We have therdore RI = R 1 • or Aju y = A zuy, from which we conclude that AI "" A 1. In mher words, the neutral ax i.l· d ivides the cross sectioll into portiolls of equal areas. Note that the axi s obtained in this fashion wilJ /lOl. in general, be a centroidal axis of thc section. We also observe that the lines of action of the resultants R I and R2

I")

pn ss Ihrollgh Ihe centroids t:'! a nn £:'1 o f rhe IWO portions we h;lvej llsr

delincd. Denoting by d the distance between C I and C!, ilnd by A thc IOtal area of thc cross section, we express the plastic momc.nt of thc member as

FIg. 4.43

An example of the actual computati on of the plastic moment of 11 membcr with onl y one plane of symmetry iii givcn in Sample Prob. 4.6.

u,

*4.11 . RESIDUAL STRESSES

"

-u, Fig. 4.44

We saw in thc preceding sections that plastic zones will deve lop in a member made of an elastoplaslic mat~ria l if the bcnding momelll is large e no ugh. When the bending mo ment is decreased back to zcro, thc corresponding reduction in stress and strain at any givcn point can bc repr~sc ntcd by a straight line on the stress-strain diagram, as sl1()wn in Fig. 4.44. As you will see presentl y, the linal value o f the stress at a point will not, in general, be zero. There wi ll be a residual stress al most points, and that stress mayor may not have thc same sign as the maximum stress reached at the end of the loading phase. Si nce Ihe linear relation between " .• and E, applies at all poi nts o f the member during the unloading phase, Eq . (4.16) can be used 10 obtain the change in slress at any given point. In other words, the unloadi ng phase can be handled by assuming the member 10 be fu ll y elastic.

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4.11. Res!dual Stresses

The residual stresses are obtained by applying the principle of superposition in a man ner similar to that described in Sec. 2.20 for an axial centric loading and used again in Sec. 3.11 fo r torsion. We consider, on one hand , the stresses due to the application of the given bending moment M and. on the other. the reverse stresses due to the equal and opposite bending moment - M which is applied to unl oad the member. The lirs! group of stresses renect the e/asfoplaslic behavior of the material during the loadi ng phase, and the second !,'fO Up the linea/" behavi or of the same material during the unloading phase. Adding the two groups of stresses. we obtain the distribution of residual stresses in the member.

251

EXAMPLE 4.06 For lhe member of Example 4.05. delermine (ll) the distribution of the residual stresses. [b) the rad ius of curvature, after the bending moment has been decreased from its maximum value of 36.8 kN m back to zero.

(a) Distribution of Residual Stresses. We recall from Example 4.05 that the yield strength i.~ if!" = 240 MPa and that the thickness of the elastic core is 2y y = 80 mm. The distribution of th~ stresses in the loaded member is t1lU S as shown in Fig, 4.45(/. The distribution of the reverse stres ses due to the opposite 36.8 kN . m bending moment required to unload the member is linear and as shown in Fig. 4.45b. The maxim um stress u~ in that distribution is obtained from Eq. (4.15 ). Recalling from Ex.ample 4.05 that I/ e -'=' 120 X 10- 6 m3. we write u ' = M t"= 36.8kN ·m '" I 120 X 10 "m)

306.7 " ·I Pa

Superposing the two distributions of stresses. we obtain the residual stresses shown in Fig. 4.45c. We check that. even y( mm )

though the reverse stresses exceed the yield strength 0" , .• lhe assumplion of a linear distribution of the reverse stresses is valid. since they do not exceed 2u y.

(b) Radius of Curvature After Unloading. We call apply Hooke 's law at any point OftilC core lvl < 40 mm, since no plastic deformation has occurred in that poJ1ion of the member. Thus. the residu
Solving Eq. (4,8) for p and

of )' and

f "

p =

the appropriate val ues

3

_1:..

= 40 x 10- m

fA

177.5 X 10- 6

225

III

The vallie obtained for p after the load has been removed represents a permanent deformation of the member.

y(mm)

y(mm)

2o.t.5 306,;

~ -60 '")

~ubsti tuti ll g

we wri te

+

Fig. 4.45

- 177.5 X lO - b

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SAMPLE PROBLEM 4,5 Beam AB has been fabricated from a high-strength low-alloy sleel that is assumed to be el:lSIoplastlc wilh £ = 29 X 100 psi and ify = 50 ksi. Neglecting the effect of fillets. detennine the bending moment M and the corresponding radius of curvature (u) when yield first occurs. (b) when the tlanges have jusl become fully pl
SOLUTION n. Onset or Yield.

The centroidal moment of inenia of the section is

I = M il ill.)(16 illY - n( 12 in. - 0.75 in.)(14 in)' = 1524 in" Bemling M Olnl'lll.

For

fT II...

=

if y

= 50 ksi and c = R in .. we have

(50 ksi )( 1524 in~)

(J)/

My = ~ =

.<\1 1'

8 in.

c

= 9525 kip ' in . ...

Radi/H' !JICllfl'lIllIr/!. No ting that. a! C = 8 in., the stmin is " I' = ITr/£ = (5 0 ksi )/(29 X [0(, psi ) = OJJOI724. we have from Eq. (4.41)

8 in. = O.OOI724/Jr

' -",WI"'"

c

u,

~--- ~----LI Sill

\

---- ---8 in.

-- ----

Slr.!i" di.\trihutio"

PI' = 4640 in. .....

b. Flanges Fully Plastic.

When [he flanges twve just become fully plastic. the strains and stresses in the section are as shown in the figure below. We replace the elemenlary compressive for~e s exerted on the top flange and on the top half of the web by their resu ltants R I and ~ . and s imilarly replace the tensile forces by R] and R 4. R I = R4 = (50 ksi )( [2 in.)( l in.) = 600 kips

S!r~.s.'

distrih"tiol1

R, = R] = ~ (50 ksi)(7 in.)(0.75 in.) = 131.3 kips 1 ill

f' \

=== -1 T ~ i" .7 ill. c

= 1I.U!tJ724

I

it..

===+

J ill.

St"·,, dislrih"tim,

di
Hemlillg Momelli.

R~suh"nt

rurL."

Summing the moments of R I , R" R], and

R. about

the :: axis, we write M = 2[Rp.5 in. ) + R,(4.67 in. )] ~

2[(600)(7. 5)

Radius oj Curl'oillre,

+ ( 131.3)(4.67)1

Since

Jy

M = 10,230 ki p ' in . .....

= 7 in. for this loading, we have from

Eq. (4.40)

7 in. = (0.OO[724 )p

252

p = 4060 in. = 338ft ...

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1

100

1E

20n'''' -

SAMPLE PROBLEM 4.6

"'111 - 1 , 20moo

1

_

Determine the plastic moment Mpof II beam with the cross sec tion shown when th e beam is benl about a horizontal axis. Assume Ihat the material is elaslOpiaslic with a yield strength of 240 MPa.

&:l nlln

~20"""

SOLUTION Neutral Axis. When the defonnation is fully plastic. the neutral axis divides the cross section into two portions of equal areas. Since the total area is A ~ ( 100)(20)

+ (80)(20) + (60)(20)

4800 mm! the area located 3bove the neutral axi s must be 2400 mm 2. We write

(20)( 100) + 20y = 2400

1-- 100 """ - 1

Note that the neutral axis does

,

II(}/

=0

y=20mm

pass through the centroid of the cross section.

Plastic Moment. The resultant R, of the elementary forces exerted on the partial ;1re;1 AI is equal to

Rf = A it7 y

2011"" _

and passes through the L"entroid of that are;1. We have = [(0.100 mHO. 020 m) J240 MPa = 480 kN

RI =

A10" y

= =

A!ify

= [(U.010 m)(O.U20 m))240 11.11'.1 = Y6 kN

A ] if y

= [(0.010 m)(O.060 m) ]240 MPa = 288 kN

R! R)

R4 = A4if)' = [(0.060 m)(O.020 m) J240 MPa = 288 kN

r

:::::::::f r..o"'111

201ll1l~

lOO """

---j

:\ ~~~~~~~~E ~~~ -

20 Ill'"~

'\1 ' ---------" 4'

"I

tTl'" 2-1U \ 11' .•

________

==

R,

30 "'''' ----I'--{?j---.'~.,."-,l_-- x It) ~30 Ill'" -"1- - iO",,,,

_ _______

I

,_'-_~K,

I

~ The plastic moment Mr is obtained by summing the moments of the forces about the z axis.

Mp"" (0.030 m)R)

+ (0.010 m)R! + (0.030 m)R]

= (0.030 m)(480 kN )

+ (0.Q70 m)R4

+ (O.GiO m)(96 kN )

+(0.030 m)(288 kN) + (0.Q70 m)(288 kN)

= 44.16kN·

m

M,. =

44 .2 kN · rn ....

Note: Since the cross section is 1/01 symmetric ilbout the :: axis. the slim of the mo ments of R] and R2 is I/ot equal to the sum of the momems of R3 and R 4 •

253

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SAMPLE PROBLEM 4.7 For the beam of Sample Prob. 4.5. determine the residual stresses and the permanent radiu s of curv:llure after the IO.230-kip . in. couple 1\1 has been removed.

SOLUTION Loading. In Sample Prob. 4.5 a couple of mome nt M = IO.:nOkip· in. was applied and the stresses shown in rig. I were obtained.

Elast ic Unloading. The beam is unloaded by the application of a couple of moment M = - 10.230 kip' in. (which is equal and opposite \0 the couple originally applied), Outing this unloading. the action of Ihe beam L'i fully elastic; recalling from Sample Prob. 4.5 that I = 1524 in4, we compute the maximum stress

Mr

fT'

or

= -

I

=

( lO.230 kip' in.l(8 in. ) 1524 in';

= 53.70 ksi

The stresses caused by the unloading are shown in Fig. 2. Residual Stresses. We superpose the stresses due (0 (he loading (Fig. I) and to (he unlooding (Fig. 2) and obt:!in (he residual stresses in (he beam (Fig. 3).

+ M

= !O.2.'W kip

IV.23iJ kip .

- 3.Ul h i

+3.7Hk.i

I

E~~~ -------4':§~~3 :::.:.:.:::.:.:.:.:.:::.:::.:.:.:.:.-

/

(2)

( II

+J.tH I.:.i

- 3.70 hI (3 )

Permanent Radius of Curva t ure, At )" = 7 in. the residual stress is = - 3.01 ksi. Since no plastic deformation occurred althis point. Hooke's law can be used and we h:!ve f A = uiE. Recalling Eq. (4.8). we write

p

if

p =

L" 254

=

- J .7IJ l..,; i""""p""_"o".1

y

_ )'1:." =

fA

if

(7in. )(29 x 10" psi ) 3.01 ksi

+ 67.400 in.

p = 5620 fl ....

We note that Ihe residual stress is lensile on the upper face of Ihe beam and compressive on the lower face. even though the beam is concave upward.

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PROBLEMS

4.67 The prismatic bar shown is made of a sleel thai is assumed 10 be elastoplaslic with rTy = 300 MPa and is subjected to a couple M parallel 10 the x axis. Determine the moment M of Ihe couple for which (a) yield first occurs. (b) the ebslic core of the bar is 4 mm thick.

"

Fig. P4.67

4 .68

Solve Prob. 4.67.

a.~suming

that the couple 1\1 is parallel 10 the z axis.

4 .69 A bar having the cross section ~hown is made of a sleellhm is assumed to be elasloplastic with £ = 30 X 106 psi and u r = 48 ksl. Determine the thickness of the plastic zones m the lOp and bottom of the bar when (u) M =

250 Ib . in., (b l M = 300 Ib . in. 0.3 in ,

Fig. P4.69

4.70 For the steel bar of Prob. 4.69. delennine the bending moment M at which (u) yield firsl occurs, (b) the plastic zones at the lOp and bottom of the bar are OJ)/) in. thick.

4.71 The prismmic bar s hown. made of II steel that is assumed 10 be elastoplastlc with E = 29 X 106 psi and {T y = 36 ksi, is subjected to a couple of 1350 Ib . in. parallel [0 the ;; axis. Delcnnine (a) the thickness of [he clastic core. (b) [he mdius of curvature of the bar.

\0

4.72 Solve Prob. 4.71. assuming that the l350·lb' in. couple is parallel the y axis.

Fig. P4.71

255

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256

4.73 and 4.74 A beam of the cross section shown 'is made of a steel I1wt is assumed to be elastoplastic WiLh E = 200 GPa and tT,. = 240 MPa. For rending about the z axis. detennine the bending moment at which (a) yield first occurs. (b) the plastic zones at the top and bonom of the bar are 30 nun thick.

Pure Bending

1~ 60I",,, -1 Fig. P4.73

Fig. P4.74

4.75 and 4.76 A beam of the cross section shown i~ made of a sted Ihat is assumed to be elasloplastic with £ z 29 X 10h psi and fh = 42 ksi. For bending about the~. axis. determine the bending moment at which (a) yield first occurs. (b) the plastic zones at the tcp and bonom of the bar are I in. thick

o~ ~ ' 1;" C

I

=r

lln .

O.5ill.--l

Fig. P4.7S

~ ~ U5i".

t ill.--I

!.:III .

I---l 1 ill.

~ 1 ill

Fig. P4.76

4.n through 4.80 For the beam indicated. detennine (a) the fully tic moment MI" (b) Ihe shape factor of Lhe cross section. 4.n Beam of Prob. 4.73. 4.78 Beam of Prob. 4.74. 4.79 Beam of Prob. 4.75. 4 .80 Beam of Prob. 4.76.

pla~-

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4.81 and 4.82 Detennine The plasTic moment Mp of a sted beam of the cross section shown, assuming The steel to be elastoplastic with a yield strength uf 240 MPn .

Fig. P4.81

4.83 A thick-walled pipe or the ero.>s section shown is made of a steel th:lt is :lssumed 10 be elaslOplastic with a yield sTrength (T y . Derive an expres· sion for the plastic moment M" of the pipe in Terms of c i • ("2 ' and rTf.

Fig. P4.83 and P4.84

4 .84 Determine the plastic moment MI' of a thick-walled pipe of the cross section ~hown. knowing that C1 = 60 mill. C:, = 40 mm. and (T y = 240 MPa. 4 .85 Determine the plasTic moment Mp of the cross sh:tion shown. assuming the ~teelto be elustoplastic with a yield strengTh of 48 ksi.

Fig. P4.85

4 .86 Determine the plasTic moment ki p of the cross seCTion shown. assuming the sTeelt[) be elastoplastic with a yield strength of 36 ksi.

Fig. P4.86

Problems

Fi g. P4.82

257

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258

Pure

Bending

4 .87 and 4.88 For the heam indicated . a couple of momenT equal (0 the full pla.~tic momenT Mp is applied and Ihen removed. Using a yield strenglh of 240 MPa. delennine Ihe residual stress al y = 45 mm. 4.87 Beam of Prob. 4.73. 4.88 Beam of Prob. 4.74.

4.89 and 4.90 For the heam indicaled . a couple of moment equal 10 the full plastic momelll MI' is applied and Then removed. Using a yield strengllr of 42 ksi. determine the residual stress
-'- ~ ~{l - il'[l - ~(l')'l } 2 3 PR

p

Pr

Pr

4.94 A sol id bar of rectangular cross section is made of a material that is
Fig. P4.95

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4.96 The prismaric bar AB is m
Probloms

l~nsile

(T

( ~l~ )

~II "'Ill

200

8

r++-l7f-t++lH-1 A

"" Hi't++I+H-tl <)

O.{JlO

0.005

Fig. P4.96

4 .97 The prismatic bar A B is made of a bronze alloy for which the tensile slress-strain diagram is as shown. Assuming that the if-f diagr.lm is the same in compression as in tension. detennine (a) the maximum stress in the bar when the radius of curvature of lhe bar is 100 in .. (b) the corresponding value of the bending moment. (See him given in Prob. 4.96.)

)

" 1.2 in.

-.....

A

f-

)

/'

)

,

/

/ 0.00.1

(JOOIi

Fig. P4.97

4 .98 A prismatic bar ofrecrangular cross section is made of an alloy for which the stress-strain diagram can be represented by the relation f = krT" for if > 0 and f = -11:cr"1for if < O. If a couple 1\'1 is applied to the bar. show that the maximum stress is 1 + 2/l Me

u

~----

'"

3/l

J

Fig. P4.98

259

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260

Pure

Bcnclng

4.12. ECCENTRIC AXIAL LOADING IN A PLANE OF SYMMETRY We saw in Sec. 1.5 that thc distribution of stresses in the eross section of
line of action of the loads r and P' passes through the cemroid of th e cross section . Such a loading is said 10 be cc'llric. Lei us now analyze the distribution of stresses when the line of action of the loads docs /lot pass through thc centroid of the cross section. i.e., when the landing is eccelllric. Two c:\amples of an eccentric loading are shown in Figs. 4.46 and 4.47. III the case o f the hig hway light, thc weigh I of the lamp causes an cceenuie loading on the post. Likewise. Ihe ve rtical forces exerted on the press cause an eccentric loading on the b!lck column of the presS

D d

"

~

A

E

Fig. 4.46

- - - -!-- ---B

"I

III

li l b

Fig. 4.47

~cuiull. lIU I ;1I H1ly~i~

-

P'

A

Fig. 4.48

F=P (Oi l

-~ 'I !;c,"\ " -

~~ ", p (h)

Fig. 4.49

will be lilllilCll lU

Il It:lIllx:J~

wh idl

pu~­

sess
M = Pd

(4.49)

We now observe thai the internal fo rces 111 the section would have bt.'cn re pn:senlcu by tht same force anu cuupl!: i f the straig ht ponion DE o f member .4.B had been detached from A B :md ~ ubjectcd ~ i muha­ Ilcously 10 the centric loads P and P ' and 10 the bendlllg couples M and 1\'1 ' ( Fig. 4.49). Thus. the SIres .. distribution due to the original eccelltric loading t;an be obtained by s uperposing the uniform Slres~ distri-

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bution corresponding to the centric loads P and P' and the linear di stributi on corresponding to the bending co uples 1\1 and 1\1 ' (Fig. 4 .50). We write

4, 12. Eccen tr1c Axial Loading In a Plane of Symmetry

"I Fig. 4.50

or. recalling Eqs. (1 .5) and (4.1 6): P

My

(T'= "A - ,

(4. 50)

where A is the area of the cross section and J its centroidal moment of inertia, and where y is measured from the centroidal axi s of the cross section. The relation obtairK'd shows that the distribution of stresses across the section is linea r blllnOl unifo rm. Depending upon the geometry of thc cross section and the eccentrici ty of the load, the combined stresses may all have the same sign. as shown in Fi g. 4.50, or some may be pos itive and others negative, as shown in Fi g. 4.5 1. In the latter case, there will be a line in the secti on, along which 0",\ = O. This line represents the neutral axi.l· of the secti on. We note that the neutral axis does lI o t coi ncide with the centroidal axis of the section, since fJ.r'* O for ), "" O.

'I c

'I

~~ >

Fig. 4.51

Thc resul ts obtained arc valid only to the extent that the conditions of applicability of the superposition principle (Sec . 2. 12) and of SaintVenunt's principle (Sec. 2.17) are met. Thi s means that the stresses involvcd must not exceed the proportio nal limit of the materiaL that the deformat ions due to bending must not appreciably affect the di stance d in Fig. 4 .48a, .mil that the cross section where the stresses are computed must not be too close to points D or E in the same fi gure. The tirst of these requirements clearly shows that the superposition method cannot be applied to plas tic deformati ons.

261

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EXAMPLE 4.07 An open-link chain is obtained by bending low-carbon steel rods of 0.5-in. diameter into the shape shown (Fig. 4.52). Knowing that the chain carries a load of 160 lb. determine (a) the largest tcnsile and compressive stresses in the straight portion of a link. (b) the distance betwcen the centroidal and the neutral axis of a cross section.

The corresponding s tress distributions are shown in parts a and b of Fig. 4.54. The distribution due to the centric force P is unifonn and equal to f' ll = PIA. We have A = 1n;! = 7f(0.25 in.f = 0.1963 in!

p 1601b tTo=A"= 0.1963in2

815psi

"J 51'''

lit 'It! / c

+ (, I

lei

Ih )

Fig. 4.54

Fig. 4. 52

The distribution due to the bending couple M is linear with a maximum stress if," = Me/I. We write I = !rn::~ = t 7f(0.25 in.)~ = 3.068 X IW ) in4

(a) Largest Tensile and Compressive Stresses. The internal forces in the cross section are equivalent to;l centric force P and a bending couple M (Fig. 4.5]) of magnitudes /' = 160lb M

=:0

Pd = ( 160 lb)(O.65 in.) = 1041b· in.

Me

U ",

=

f

=

( 1041b' ;n.)(0.25 in.) 3.068 X 10 1 in~

8475 psi

Superposing the two distribution~ , we obtain the stress distribution corresponding to the given eccentric loading (Fig. 4.54c). The largest tensile and compressive stresses in the section are found to be. respecti vely. = Uo + (T", = 815 + 8475 = 9290 psi tTc = ifo - if", =:0 815 - 8475 = - 7660 psi

fl ,

(b) Distance Between Centroidal and Neutral Axes. The distance )"0 from the centroidallO the neutral axis of the section is obtained by selling (I, = 0 in Eq. (4.50) and solving for Yu:

J f~ l lI,

Fig. 4.53

262

)"0 =

(A"P)(Ai

O = ~ - Myu A I

J)

= (815 psi)

3.068 X 10- ' ;,' 104lb. in.

)"u = Q.0240 in.

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SAMPLE PROBLEM 4.8 Knuwillg lhat fUf lln: l:;tSl iron link shuwn the.
· , M

1~~m' l

)'

L

I Ifl''''''

SOLUTION

=120"""

A = 3000 mm" = 3 X 10

,{u mn,

I)

_.

From J

Furce and Couple a t C. lhe. centroid C

nun

We repla~e P by an equivalenl force-couple

c,, - (H'~m

I J)

"n

=

I OWS I

,r" =

/'

A=

P 3 X 10 " = 111P

= Me" '" (O.028P)(O.022)

868 X 10-' Me"

(O.028P)(O.038)

if 2=I- ~

c

.,

D B

1, -

-

,

(rflmrri'S~ifln)

7

868 x 10- 0

7 10P

(Tension)

l226P

(Compression)

Superposi ti on. The total stress distribution (Fig. 3) is found by superposing the stress distributiom caused by the centric force P and by the couple l\ L Sincc lensioll is posilivl'. and compression n~galiVl' . WI' have

;II if

P

0.028 P

,11

if ,

.,

= 1'(0.028 m) =

The force P acting at the centroid causes a uniform mes.' diwibution (Hg. I J. The bending couple 1\1 causes a linear Slress dist.ribution (Fig. 2).

r--~'i-'-~ --. I

Y = 38 mm = 0.038 m

s~'stem at

M = P(d)

:

Prob. 4.2. we have

d= (0.038m) - (O.O Il1m) = 0.028m

We now wrile:

St·l'tilJn 11-11

- t'

m"

Samp!~

J = 868 X IO ' Ym'

j

I b"--I 30

Properties of Cross Section.

"

0B

=

,

=

- I:..A + MeA 1

- A"P -

MC B

- 1-

""

- 333P

= - 333P -

+ 7101' =

+3771'

1226P = - 15.59P

(Tension) (Compre.~siOl1)

LargesT Allowllblc Force. The mngnitude of P for which the tensile point A is equal 10 the allO\\~ble lensile Slres~ of 3(] MPa is found hy writing

Slre~s ~l

if"

Wt!

~ISrl

df".lt!rm;nl". Ihl".

P = 79.6tN
= 377P = 30 MPa

m~gnilllrle

of P for whic:hlht!

~lr""'s

al R is t':(jlJallo lhl".

allowable compressive stress of 120 MPa. if E

i"

(2'

(.3)

= - 1559P = - 1lOMPa

P = 77.0l:N <J

The magnitude of the largest force P that can be :Jpplied without exceeding either of the allowable stresses is the smaller of llie tlVO \a!tles we have found.

P = nO l:N ....

263

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PROBLEMS

4 .99 De termine the s tre ss :It poinlsA a nd B. (tI) for the loadin!> shown. (b ) if the IS.kip loads are applied at poinls I and 2 only.

Fig. P4.99 and P4.100

4 .100 Determine the ~Hess at poims A md B. ((I) for Ihe loading shown. (h ) if the l'i_kip lo~rI~ ~r(1lied ~I (lOims 2 and 1 ~ re removed 4 .101

Two forces P can be applied separately or at th e same lime to a

plate th.at is welded to a solid cir.:ular bar of radius r. Determine the la rgest compressive stress in the circu lar bar. (a) when boill forces are applied. (b) when

only otle (lft he forces is applied.

, '

fin,!"

~ /' /""

Y'

Rg. P4.102

264

Fig. P4.101

)2 ,,,,,,

12,,,111

4.102 termine tha

Kn owing thm the 1ll2gnilud e of The '1enical (a) po int 11, (0) poi nt B.

~tre5S at

fOft'!!

P is 2 kN. de·

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4.103 The venical ponion of the press shown consists of a reclangular lUbe of walllhickness I = 10 mm. Knowing Iha! the press has been tightened on wooden planks being glued together until P = 20 kN. determine the stress a! (a) point A. (b) point B.

£J~.m ", ,B J-SO mm-j Sed;o" II -i/

200 mm Ml nit" Fig. P4.103

4. 104

Solve Prob. 4.103, assuming thaI I = 8 mm.

4 .105 Portions of a } X !-in. square bar have becn bent to form the two ma<.:hine componcnts shown. Knowing that the allowable STress is 15 ksi. deTermine the maximum load lhal can be applied to each component.

Fig. P4.105

4.106 The four forces shown are applied to a rigid plate supported by a solid steel poST of radius a. Kn owing lhal" = 100 kN and a = 40 mm. deTermine the maximum STress in the post when (a) the force a! D is removed. (b) the forces at C and D are removed.

Fig. P4.106

Probloms

265

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266

Pure

Bcn"tng

4.1 07 A milling operation was u~d to remo ve a pol'lioll of a ~o li rl bal of sq uare cross seetiot\. Knowin g that (/ == 30 mm. d = 20 mm. amI O".It = 60 MP.1. determine the magnitude P of Ih.e largesl forces Ih31. Cln be safel) applied at Ill
Fill. P4.107 and P4.108 4 .108 A milling ope-ration was ll~d to remov.... :l ponion of a wlid bar of ~q llare ~ross ~ec tio". For~es of magni lude P = t& kN al1.' applied at the centers of tlte ends of the b:lr. Knowing thm 11 = .10 rum and (Tloll "" 135 MPa. uelennine the smallest allowable depth d of the milled portion of Ille bar. An offset II musl be !Illroduced inlo a solid Circ ular rod of dJa mm~l.ximum stress after the offset is Introduced must not ex,eed Slimes the sIre,s in the rod when it is straight. detemline Ihe largest offset thaI can be used. 4. T09

eI~r

d. Knowing Ih:1t the

d

I

P'

•

P'

I'

d

Fig. P4.109and P4.1 10 4 .110 An offset h must be intrOOuced into a metal rube ofO.75-in. OUler diameter and 0.08-in. walilhickness. Knowing that tlie maximum Strt rorce P thOlt can be "pplieJ to the bracket shown

Ji". "

() ,': ;n.

Fig. Pot. 111

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4. 112 A shon column is made by nailing four I X 4-in. planks to il 4 X 4-in. timber. Determine the largest compressive stress created in the column by a 16-kip load applied as shown in the center of the top section of the timber if (a) the column is as described. (b) plank I is removed. (c) planks I and 2 are removed. (ell planks I. 2, and 3 are removed. (f') all planks are removed.

Probloms

4.113 Knowin!; that the clamp shown h;JS been tightened on wooden planks being glued together until P "" 400 N. determine in section a-(/ (a ) the stress at point A. (b) the stress at point D. (e) the location of the neutral axis.

rJ~ 14"""

FI g. P4.112

~

"B

-•

•

20"""

D_ I

Fig. P4.113

-I

r--" ' ' '

4.114 Three steel plates, each of 15 x 150-mm cross section. are welded togethcr to form a short H-shapcd column. Latcr. for architectural reasons. a 25-mm strip is removed from each s ide of one of the nanges. Knowing that lhc load remains centric with respect to the original cross section. and that the allowable stress is 100 MPa. dctemline the largest force P (a) that could be applied to the original column. (b) lhat can be applied to the modified column.

50,:"" 4i"

At:

~ ' 1--.;. .-1

--,~ .129 i,\,

c

-i2:"

----..l.

Fig. P4.11S

4. 115 In order to provide access to the interior of a hollow square tube of 0.25-in. wall thickness. the portion CD of one side of the tube has been removcd. Knowing lhat the loading of the tube is equivalcnt to two cqual and opposite IS-kip forces ,Icting at the geomctric centers A and E of the ends of the tube. determine (a) the maximum slress in section a-a. (b) the stress at point F. Gil"en: the centroid of the cross section is at C and I, "" 4.81 in4.

Fig. P4.114

267

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268

Pure

BanC1lng

4.11 6 Knowjn~ Ihm the allowable stress jn secTion u·u of the hydraullc press sho\\"n is 6 hi in temion and 12 ksi in compression. Iletermine th~ largest force P lhat CEn be exerted by the press.

p

12 in.

I-"'-=-j

I"

"r

1"

Fig. P4.11G

4.11 7 The four ba~ ~hown have Ihe same ~ros~_se~tlonal area. For the given lo:tdings. ~how that (a) the maximum ccmpressi\'e stresses are in the ra· tlO 4:5:7:9, (b) the maximum t~n~ile stresses are in the ratio 2:3:5:3. (N()/e: the cross ~eCliou of the lJianguhu bar is 'lil tquilateraltriangle.)

Fig. P4.117

4.11B Knowing lhal the allowable s tress is 150 MPa in section (Hi of the hanger shown, detenmne (u) the largest ~ enjcal force P that call be applied al poml A. (b) Ihe corre~po nding localion of the neutral axis of section (/-il.

r

]

2U """

"

I--c---I----., "",,-1--1 '\Ollila 4(J'11I1

60

1.-60 ,"",- HE

I

'''"I

I

20111111

5,'<1;011("'"

Fig. P4.118

4.11 9 at point B.

Solve Prob. 4. 118. a.suming that the vertical force P is applied

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4.120 The C-shaped steel bar is used as a dynamometer to determine the m;lgnitude }' of the forces shown. Kn owing thaI the cruss section of the bar is a square of side 1.6 in. and the strain on the inner edge was measured and found to be 450 p., determine the magnitude}' of the forces. Use £ = 29 X 10" psi.

Probloms

p'

\. 6 in .

Fig. P4.120

4.121 A shon length of a rolled-steel column supports a rigid plate on which two loads " and Q are applied as shown. The strains at twO points A and B on the centerline of the outer faces of the flanges have been measured and found to be £ 11

= - 400

X

10 - 6 in.lin.

i 8

= - 300 X 10 - 6 inJ in.

Knowing that E = 29 X l()'" psi. determine the magnitude of each load.

P

(jill .

6 in. Q

,\ = 10.0 in 2 '- = Zi3 in4

Fig. P4.121

4 .122 An eccentric force P is applied as shown to a steel bar of 25 X 90-mm cro,s section. TIIC strains at A and B have been measured and found to be f8

= - 70JL

Knowing that E = 200 GPa. detemline (a) the distance d, (b) the magnitude of the force P.

4.123 Solve Prob. 4.122. assuming that the measured strains are

Fig. P4.122

269

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270

4. 124 The eccentric axial force P acts at point D. which must be located 25 mm below the lOp surface of the steel bar shown. For P = 60 kN. determine (a ) the depth d of Ihe bar for which Ihe tensile stress at point A is maximum. (b) the corresponding stress at point A.

Pure Bending

b

=

40" u"

Fig. P4.124

4.125 For Ihe bar and loading of Prob. 4.124. determine (a) the depth d of the bar for which the compressive stress ~l point 13 is maximum, (b) the cor· responding stress al point B.

4.13. UNSYMMETRIC BENDING :-L\

C

" ru)


Fig. 4.55

Our analysis of pure bending has been limited so far to members possessing at leasl one plane of symmetry and subjected to couples aCling in that plane. Because of the symmetry of such members and of their loadings, we concluded that the members would remain symmetric with respect to the plane of the couples and thus bend in that plane (Sec. 4.3). This is illustrated in Fig. 4.55: part (I shows the cross section of a member possessing two planes of symmetry, one vertical and one horizontal. and part h thc cross section of a member with a single, vertical plane of symmetry. In both cases the couple exerted on the section acts in the vertical plane of symmetry of the member and is represented by the horizontal couple vector M, and in both cases the neutral axis of the cross section is found to coincide with the axis of the coup le. Let us now consider situations where the bending coup les do 1I0r act in a plane of symmetry of the member. ei ther because they act in a different plane, or because the member docs not possess any plane of symmetry. In such situations, we cannot assume that the member will bend in the plane of the couples. Thi s is illustrated in Fig. 4.56. In each part of the figure, the couple exertcd on the section has again been ,l~y

y

"

~ ("' Fig. 4.56

c

(h)

~

c

" «,

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sumed to act in a vertical plane and has been represented by ;'\ hanzomal couple ye: tar M. However, since the vertical plane is nO! a plane of sy mmetry, we cal/I/ot expect the member to hend ill that plane. or Ihe /lel/lml axis of the sectio/l to coincitle with The axis oj the couple.

We propose to detemline the precise conditions under which the neutral axis of a cross section of arbitrary shape coincides with the axis of the couple M representing the force s acting on that section. Such a section is shown in Fig. 4.57. and both the couple yector 1\'1 and the neutral axis have been assu med to be directed along the z axis. We re-

Fig. 4.57

call fro m Sec. 4.2 that, if wc then express thaI the elementary internal forces a ., llA form 11 system equivalem to the couple M, we obtain '\' components: moments about y axis: moments about z axis:

f(T~ dA

- 0

J zu xdA == 0

f( - yu,dA)

~

M

(4. 1) (4.2) (4.3)

As we saw earlier, when all the stresses are within the proportional limit , the fi rst of these equations leads to the requiremem that the neutral axis be a centroidal axis, and the last to the fundamental relation U x = - My/I. Since we had assumed in Sec. 4.2 that the cross section was symmetric with res pect to the y axis, Eq. (4.2) was dismissed as trivial at that time. Now that we are considering a cross section of arbitrary shape, Eq. (4.2) becomes highly significant. Assuming the stresse;; to remain within the proportional limit of the material, we can substitLlte u , = - u"sic into Eq. (4.2) and write JYZl/A = 0

(4.5 1)

The inlegral JyzdA represents the product of inertia IF of the cross section with respect to the y and z axes, and will be zero if these axes are the principal cellfmidal axe.v of the emu seetioll .t We thus conclude that the neutral axis of the cross section will coincide with the axis of the couple M representing the forces acting on that section if. and ollly if. the ample Fector M i.v directed alollg olle of the principal eemmidal {ue.l· of the ems.l· section.

t See Fr.rdinand P. Beer and E , Russell Johnston. Jr.. Mechanics for Engilleers. 4th ed .. Yor~. 1987. or Vector Mec/",,,ic,, for Engillars. 7th cd .. McGraw-Hi ll. New Vorl;:. 200+. sees. 9.8- 9.10.

McGraw·Hill. New

4.13. Unsymmetr1c Bending

271

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272

Pure Bcn"lng

We IlQte that (he CIQSS sccliollS showu ill Fi g. 4.55 ru"\:: SY "llm:tt1c with respe.ct (0 a( lea.~t one of the courdinati' axes . It follows thm, in each case. tilt J and z. a."(es :Ire the principal centroidal axes of the section. Si nce the couple vec tor M is directed
, ~

l\'A 0

'I

"<.A

C

"

~ G " l_;~ I

{Ii ,

,,)

Fig. 4.58

tJ l1:: a.\i s of the couple, e vc llthough

ill

cast: U Ih e CUlIplt:

doe..~ I/v/a l:!

in

n plilnc. of iymmc.try of the mcmbec

In Fig. 456. on the other hand, neither of the coordinate axes is an axis of symmetry for the .~ectioll S shown, and the coordinate axc.<; ure nut prinCIpal axes. T llU" the couple vector M is nOi direc{ed along a pnncipaJ ccntroidal a xis, and the neutral axis docs not coincide with the axis of the couple. However. any given section posses~ s principal centroid:!l axes, even if it is u n ~ymmetr ic, as the section shown in Fig. 4.56(", and these axes may be determ ined analytically a rb y using M ohr's ci rcl e.t If the couple veCLOr M is di rected along one of the pri ncipal centroidal axes of the ~ect i{J n. the neutral axis wi ll coinc ide wi th the axis of the COliple (Fi g. 4.59) and the eqllutiOilS derived in Sees. 4.3 and 4.4 for symmetric members ca n be tl~ed to determine the stresses in thi s case as well

,

(r, )

(hi

Fig, 4 .59

As yo u wilt see presently. the principle of superposi tion can be used to detennine stresses in the most general case of unsymmetrie bending. t Scc renlinand P. Beer an d E. Russett Johnston. Jr.• Mrchallic. Jor £11!;;.'ee'·.•• ~lh ~d .. McGraw-Hill, Nrw Yor~. 19l17. Of Vecldr "'echo"ics fi>r £nllir.,ws. 7th ed .. McGraw· Hill N~w Y{)f~ .~. ""cs. 9.8-9.tO.

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Consider first a member with a vertical plane of symmetry, which is subjected 10 bending couples M and M ' acting in a plane forming an angle 8 with the vertical plane (Fig. 4.60). The couple vector M representing tnc forces acting on a given cross section witl form the same

4.13. Unsymmetrlc Bending

y '\I__ _ ___ , , , , ,

l

.

l"

~

, '

-.. - ~ - {-'L1'!'b-C ~L

Fig. 4.60

Fig. 4.61

angle 8 with the horizontal :: axis (Fi g. 4.61 ). Resolving the vector M into component vectors M : and M , along the:: and )' axes, respectively, we write M~ =

Mcosfl

My = M sin 8

(4.52)

Since the)' and ~ axes are Ihe principal centroidal axes of the cross section, we can usc Eq. (4.16) to determine the stresses resulting from the applicati on of cithcr of the couples reprcscnled by M z and 1\1,. TIlC couple Mo acts in a vertical plane and be nds the member in · that plane (Fig. 4.62). The resulting stresses arc

Fig. 4.62

(4.53)

where 1, is the moment of inertia of the section about the principal centroidal ;: axis. The negative sign is due to the fact that we have eompression above the X l; plane (y > 0) and tension below (y < 0). On the other hand , the couple M,. acts in a horizontal plane and bends Ihe member in that plane (Fig. 4.63). The resulting stresses are found tu be (4.54)

where f,. is the moment of inertia of the section abowthe principal centTOidal .\' axis, and where the positive sign is due to the fact that we have tension to the lefl of the vertical .t). plane (z > 0) and compre..,sion 10 its right (~ < 0). The di stribution of the stresses caused by the original couple l\1 is obtai ned by superposing the stress distributions defined by Eqs. (4.53) and (4.54). respectively. We have M .y

M }z

Ie..

I).

--' + --

(4. 55)

-'I '~

C

FIg. 4.63

273

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274

We note that the expression obtain ~d can also be used to computc the stresses in an unsyrnmdric scction, such as the one shown in Fig. 4.64, once the principal centroi dal .I' and z axes havc been dctermi ned.

Pure Bending

'\ft/I ~

Fig. 4.64

On the other hand, Eq. (4.55) is valid only if the conditions of applicabil ity of the principle of superposition are mel. In other words. it shou ld nOl be used if the combi ned stresses exceed the pro portional limit of the material. or if the de formations caused by one of the compone nt couples apprec iably affcct the distribution of the stresses due to the other. Equation (4.55) shows that the di stri bm io n of st resses caused by unsymmetric bendi ng is linear. Howev~ r. as we have ind icated earlier in this scction, the neutral axis of the cross sectio n wi ll no t, in ge neral. coi ncide with the axis of the bendi ng couple. Since the normal stress is ...ero al any point of the ncutra l ax is. the equation defi ning that axis can be obtaincd by setting fT" = 0 in Eq . (4.55). We wri te

M.\'

M,,z

I~

1\

- ---=:.... + - . = 0 or. solving for ), and substituting for Me and M, rrom Eqs. (4.52), (4.56)

I '\.,

The equati on obtained is that of a straightli nc of slope III = (/;11,,) tan O. Thus, the angle 1> that the neutral axis forms with the;; axis (Fig. 4.65) is dclined by the relation tan

1> =

,.

"":' Ian 9

(4.57)

" F1g. 4.65

where 0 is the angle that the couple vector M forms with the same axis. Si nce 10 and I,. are both positive, 1> and 0 have the sam~ sign. Furthermore. we note that I/> > 0 when I , > l j' anu q., < (J when I , < I v' Thus. the neutral axi s is always locnted between (he couple vector 1\1 and the principal axis corresponding to the minimum momenl of inertia.

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EXAMPLE 4.08 A 1600-lb . in. cou ple is app lied 10 a wooden beam, of rec-

tangular cross section 1.5 by 3.5 in .. in a plane forming an an· gle of 30° with the venical (Fig. 4.66). Determine (u) the maximum stress in the beam. (b) the angle that the neutral surface forms with the horizontal plane.

The largest tensile stress due to M, occurs along AB and is ( 1386Ib' in.)( 1.75 in.) 5.359 in'

111:>'

(' 1 =

T

=

452.6 psi

The largest tensile stress due to 1\1, occurs along AD and is

M,z

{f~ =

7; ""

(800 lb· ;n.)(0.75 in.) 0.9844 in'

609.5 psi

The largest tensile stress due to the combined loading. there· fore. occurs al A and is ifnw.

=

if l

+ if"

= 452.6

+

609.5 = 1062 psi

The largest eompres~ i ve s tress has the same magnitude and occurs at E.

(b) Angle of Neutral Surface with Horizontal Plane. The angle I/J that the neutral sUlface fonns with the horizontal plane (Fig. 4.68) is obtained from Eq. (4.57):

,,

Fig. 4.66

J) 7

(a) Maximum Stress. The components Me and Mj • of the couple vector are first de termined (Fig. 4.67): M , - ( 1600

lb · ill .) cos

30~

I~

- 13861b· in.

,

M , = ( 1600 lb · in. ) sin 30° = 800 lb· in.

E

c

~~H

,

Fig. 4.68

I, 5.359 in' tan.p = - tan 8 = tan 30° = 3.143 I~ 0.9844 in4 .p = 72.4°

M, II = 3if

\. i5 in .

"I-.~--'.-l

The distriblllion of the stresses across the section is shown in Fig. 4.69.

0.7S in

Fig. 4.67

,l

-,-'

:1 -',

We also compute the moments of inerti:! of the cross section with res'peet to the z and J axes:

......-" -:1 ..... ;~ J

( = -fi( 1.5 in. )(3.5 in.)) = 5.359 in 4

...

~/ ~

--~~~~}:~~~

11 = -fi(3.5 in. )(J.5 in.)3 = 0.9844 in 4 Fig. 4.69

[(t6:!11S1

8

275

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276

4.14. GENERAL CASE OF ECCENTRIC AXIAL LOADING

Pure Bending

,\

Ln Sec. 4.12 you analyzed the stresses produced in a member by an eccentric axial load appl ied in a plane 01 symmetry of the member. You will now st udy the more general case when the axial load is not applied in a plane of symmetry. Consider a straight member AB subjected Lo equal and opposite eccentric axial forces P and p i (Fig. 4.70a), and let a and IJ denote the distances from the line of action of the forces to the principal centroidal axes of the cross scction of the me mber. The eccentric force P is statically equivalent to the system consisting of a centric force P and of the two couples My and M: of moments MJ = Pa and Me = PI) represented in Fig. 4 .70h. Similarl y, the eCfentric force p i is equi valent to the centric lo rce P ' and the couples and M~. By vio uc of Saint-Venunt's principle (Sec. 2. (7). we can replace the original loading of Fig. 4 .70a by the stat ically equivalent lauding of Fig. 4.70b in order 10 determine the distributi on of stresses in a scction S of the member, as long as that section is not too close to either end of the member. Furthermore, the stresses due La the loading of Fi g. 4. 70b can be obtained by superposing the stresses correspond ing to the centric axial load P and to the bending couples M,. and M" as long as the conditi ons of applicability of the principle of superposition are satislied (Sec. 2. 12). The stresses due to the centric load P are given by Eq. (1 .5), and the stresses due to the bending couples by Eq. (4.55), since the corresponding couple vectors are directed along the principal cenlroi da! a);"es of Ihe seclinn. We wrile, Ih crer() r.~ ,

M:

,,'

f

l-, ,\

"I

(J~

(M Fig. 4.70

P M~y = - - -A I~

My<-

+ -I ,.

(4.58)

where )' and z arc· measured from the principal centroidal axes of the section. The relation obtai ned shows that the dis tri bution of stresses across the sect ion is IiI/ear. In computing the combined stress if" from 5l (4.58), care should be taken to correctly determi ne the sign of each of the three terms in the right·hand member. since each of these tenns can be positive or negati ve. dependi ng upon the sense of the loads P and P ' and the location of their line of action with respect to the princi pal eentroidal axes of the cross section. Depending upon the geometry of the cross section and the locati on of the line of action of P and P: the combined stresses if., obtained from Eq. (4.58) at varioU'i points of the section may all have. the same sign, or some may be positi ve and ot hers negative. [n the latter casc, there will be a line in the section, along which the stresses are "Zero . Setting IJL = 0 in Eq . (4.58). we obtain the equat ion of a straight line. which re prese n L~ the nellfral axi.l· of the section:

M_ M. P --"v-- - = " -

I,. "

A

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EXAMPLE 4.09 A vertical 4.S0·kN load is applied as shown on a wooden post of rectangular cross section, 80 by 120 mm (Fig. 4.71). (a) Determine the stress at points A. B. C. and D. (b) Locate the neutral axis of the cross section. , I

The stresses at the comers of the section are U ).

=

0"0

==

0" , ::!: O"~

where the signs must be detemlined from Fig. 4.72. Noting that the stresses due to ]\'1, are positive at C
(f,," ""

O"B (f c

(T D

FIg. 4.71

= - 0.5 - 1.5

Fig. 4.72

(a) Stresses. The given eccentric load is replaced by an equivalent system consisting of a centric load P and two

couples M .l and 1\.1. represented by vectors directed along the principal centroidal axes of the section (Fig. 4.72). We have M. = (4.80kN)(40mm)= 192N'm M; - (4.S0kN)(601111ll - 35 mm) -

L20N'1ll

We also compute the area and the centroidal moments of inertia of the cross section: A = (0.080m)(0.1 20m) = 9.60 X 10 I., =

h(o.no

1m

l

m)(0.080 ml = 5.12 X IO ~ ~ m4

'"

FIg. 4.73

'"

(b) Neutral Axis. We note that the stress will be zero at a point G between Band C. and at a point H between D and A (Fig. 4.73). Since the stress distribution is linear. we write

BG SO mm

1.375 1.625 + 1.375

BG = 36.7 mm

HA 80 mm

2.625 2.625 + 0.375

HA

= 70mm

3

J~ = /!(0.030 m)(O. 120 m)l "" 11.52 X 10- ~ m ~

The neutral axis can be drawn through points G and H (Fig. 4.74).

The stress if 0 due to the centric load P is negative and uniform across the section We ha ve

P

"u=

- 4.80kN

A=

9.60 X la- 3 m!

- 0.5 MP
11le stresses due to the bending couples M , and M . are lin· early distributed across the section . with maximum values equal. respectively. to

The di~tribLHion of the stresses across the section is shown in Fig. 4.75. I " 17:; ~!1' .

M ,Zrna:.. 0" ,

= - -/,-

( 192N · m)(40mm ) = 5. 12 X 10 - 6 m~

1.5MPa

0.625 MPa

FIg. 4.75

, I

"!~ .\ II '~

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SAMPLE PROBLEM 4.9 A horizontal load P is applied as shown to a short !;Cction of an 510 X 25.4 rolled-steel member. Knowing that the compre;;sive stress in the member is not to exceed 12 ksi. determine the largest permissible load P_

l.5 in,

SOLUTION

"I

'fI~' H

Propertit'S of Cross Sectiun. The following data are taken from Appendix C. Area: A = 7.46 in'

SectioIJ moduli: S , = 24.7 in)

S" = 2.91 in J

Force and Couple al C. We replace P by an equivalent force-couple system at the centroid C of the cross section. M., = (4.75 in.)I>

4.66 in ,

M" = ( 1.5 in.)P

Note that the couple vectors 1\1, and My are directed along the principal axes of the cross section. Normal SIrt!ssl's. The absolute values of the stresses at points A. B. D. and E due, respectively. to the centric load P and to the couples 1\1, and M" are

P

/'

(I I

=A" = 7.46in2 =

(12

= --'- = -, - - .- l = 0.1923(>

(I

= -

M,

J

S~

0.13401'

4.751' _4.7 m

M" J.5P = - - - l = 0.51551' .'I" 2.91 in

Superposition. The total stress at each point is found by superposing the stresses due to P, M~, and 1\'1,.. We determine the sign of eJch stress by carefully examining the sketch of the force-couple system. D

(I I

+

(1 1

(I I!

=

- (I I

+

(1 2 -

(TD

=

-

fT E

=

- (I I -

uA =

-

{I I -

fJ",

+

(l l

= - 0.13401'

(l l

= - 0.13401'

+ fT]

fT2 -

fTl

+ 0.1923P + 0.51551' = +0.5741'

+ O.1923P - 0.51551' = = - 0.1340P - 0.19231' + 0.5155P =

- 0.4571' + 0.1891' = - 0.1340/' - 0.1923/' - 0.5155P = - 0.8421'

Largest P('rmissihll'. Load, The maximum compressive stress occurs at point E. Recalling that fT all = - 12 ksi, we write (I, ll

278

=

fTf.'

- 12 ksi = - 0.8421'

I' = 14.3 kips
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' SAMPLE PROBLEM 4.10

, I I iIO nlln I ,\

I

j

, I

J1 """ (

"',,= L~kN

III

I

1211"n

C

A couple of magnitude Mo = I.S kN . m acting in :l vertical plane is applied (0 a beam having the Z-shaped cross section shown. Detemline (a) the stress at point A. (b) the angle that the neutral axis forms with the horizontal plane. The moments and product of inertia of the section with respect 10 the y and ~. axes h
100m'"

12 "'''' I

1. = 4.18 X lO- b m4

~

I ,., = 2.87 X 10

b

m~

SOLUTION Y(3.2.~.

2.lm

P ri ncipal A.u's. We draw Mohr's circle and determine the orientation of the principal axes and the corresponding principal mOIllCIllS of incliia. I~, I, ( J(I-4< ",, )

PZ 2.87 tan 28", = EF = 0.465

+ (F"'L;<

2Hm = 80.8

0

1:1,. = 40.4

0

+ (2.87)'

R = 2.91 X IO-~ m4 b '. = = au = J..~ - R = 3.72 - 2.91 = 0.810 X IO- m· 4 ' . = lou,. = Ot! = I..~ + R = 3.72 + 2.91 = 6.63 x 10-" m

R2 = (EFf

'mi.

= (0.465)"

Loadi n g. TIle applied couple l\-1v IS resolved into comJXlnellts parallcl to the principal axes. M . = Mo sin 8". = 1500sin40Ao = 972N· m M , = Mocos8", = 1500 cos 40.4° = 1142N· m

a. Si n'ss al A . JXlint A arc ~,\

y,\

=50ml~ I

IIA = YA cos 0", + .tA sin 0" = 50 cos 40A o + 74 sin 40A o = 86.0 mm o V A = - YAsi nO., + ::ACOSfJ",= - 50 sin 40Ao t 74cos40A = 219mm

= 74 """

)=/Uc "A

The perpendicular distances from each principal axis to

I/., l'''' II",

Considering separately the bending aoout each principal axis. we note that 1\"1. produces a tensile stress at JXlint A while Mo produces a compressive stress at the same JXlint.

\ (f A

=

+_ M_""_, _ I"

M _ "'_A = +c(9~7~2~N=.~m2)(~o~.O~22 3~ 9 2m2) I" 0.810 X 10- 6 m4

+-(28.68 MPa) - (14.81 M Pa) h. I\'culralAxis. axis forms with the v

tan rh = The angle

f3

(1142 N . 111)(0.0860

6.63 x 10 (fA

m)

m4 = + 13.87 MPa ....

Using Eq. (4.57), we find the angle

~

6

that the neutral

a.~is.

I:' " tan 0,. =

o 6.63 0.810 tan 40A

formed by the neutral axis and the horizontal is

279

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PROBLEMS

4 .126 through 4.128 The couple 1\1 is applied to a beam of the cross section shown in ;) plane fanning an angle {3 wilh the vertical. Determine the stress at (a) point A. (b) point B. (el point D.

, f3

f3 = 30'

=30'

A---1-----., ~ 50 '"'"

--t

5O",m

__I

M ~ 251)NI"

M =OOkil'.lu

A

, 'I -'-1---"-'ID

M ~ WJlh

I I u.7S in , _1-

c 3 n. D

Ii",

= 2U'

Fig. P4. 128

4 .129 through 4.131 The couple M is applied to a beam of the cross section shown in a plane forming an angle {3 with the vertical. Determine the stress at (al point A. (b) point B. (el point D. {J = 30'

A~p--'B c M - 251!~11'

Fig. P4.129

Fig. P4.130

M = If; 1,;/\ - lll

Fig. P4.132

280

=Fl;" 10 ill.

- -03;"D[ I

I---

W310 x 3S"

"

Ii".

U,S in

Fig. P4.127

f3

_ . - "'' r'-",-i-F--, B 3 -" .

H

40 """ 40tHlI'

Fig. P4.126

In

(l.iS in .

S ill . _ _

I

M =L5lN ' !II

'I------= .--= t;:=,p~Y 11_--= ' = i..'=""""c=:::::J/J 120 ...... - il

70",,,50mm

0.5 ill

Fig. P4.131

t

'40 Inm----J

4 .132 The couple 1\1 aels in a vertical plane and is applied to;) beam oriented as shown. Determine (a) the angle that the neutral axis fomls with the horizontal. (b) the maximum tensile streSS in the beam.

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4.133 and 4.134 TIle couple 1\'1 acts in a ve rtical plane and is applied to a beam ori ented as shown. Determine tal the an.gle thal the neutral axi s form s with the horizontal. (b) Ihe maximum tensi le stress in the beam.

C200x 17.1

281

Problem.

'"

Fig. P4.f33

Fig. P4.134

4.135 and 4. 136 The couple M acts in a w rtical plane and is applied 10 a beam o nentoo as sho wn. Dele rmine (a ) the an g le thm the neUiralaxi s fo rm s with the hori zo maL (b) Ihe maximum tensile stress in the beam.

y

" ~I =

!S Lip , ill

:\t ~ 1J211'l! . 1I

'" =ti74",'

I.,. = 14 .Ti X 11.)3 "'m~

1,, =2 1 .4in~

t.,=53.6X

Fig. P4.135

Fig. P4.136

10-1,,, ,,,4

.

\

~

M - I ~; kil' ill.

\3 1~

~'

Wmm

y

Fig. P4.137

,--

~t ~ I ~"~"H l l

lo""n~

C

I 40,mn

..l.

,

I!"t n'",

~"II

jO,n,n--lI-Ju,un

1~ - l.I>9~ x .tf",,, , ~ I, = 0.614 x li P m "'~ I" = -.-Q,SUO)( JO~,,",,'

Fig. P4. 138

t y ~ S. ~ i l1~

t_=

24.:} ;11 1

['I' = +/-i .3 in4

Fig. P4.139

-+ -+ -+ --=-:\'' -

I2.4 . in.·1·2.4 " ;.1

10 nnn

'4.137 through ' 4. 139 The couple M acts in a venical plane and is applied (0 a beam orieJlled as shown. Determine the mess al point A.

" zM

2.4 ill.

,--+We

---'\ Wmm

/"

"I --"

2.4 ill . 2 .4 ill.

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282

Pure Bending

4.140 For the loading shown. determine {a) Ihe stress at points A ami B. (b) the point where the neutral axis iniersecis line ABO.

-I

i ll .

A

150n,

5UU U,

Fig. P4.140

4.141 Solve Prob. 4.140. assuming Ihallhe magnitude of Ihe force applied al G is increased from 250 Ib 10 400 lb. 4.142 The tube shown has a uniform walilhickness of 12 mm. For the IO;lding given . determine (a) the stress al poin!.> A and B. (b) the poinl where the neutral axis iniersecis line ABD.

,-n,

D fI

II k:N

1

2b kN 12.5 Illl ll

"

F

'lIi

~N

~"' '

Fig. P4.142

4. 143 Solve Prob. 4.142. assuming Ihallhe 28-kN force al point £ is removed.

4.144 An axial load P of magnitude SO kN is applied as shown \0 a short seclion of a W I SO X 24 rolled-steel member. Detennine Ihe largesl dislance a for which Ihe maximum compressive stress does nOI exceed 90 M.Pa.

Fig. P4.144

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4.145 A horiwLltal load r of magn itude 100 kN' is aj:>plied to the beam shown. Determine lhe largest di:it ::lr1~e (/ for which the maximum te.nsile stress in tile Ix-am does not exceed 75 MFa.

Problem.

Fig. P4. 145

4.146

A beam having lite cross section shown is s ubjected to a couple in a vertic::!1 pl:lne. Delennine the Ia.rgest permIssible va lll e of the moment Mu of the couple if the maximum Hres.> in the beam is not to exceed 11 ksi. GiI'l'll: I , I~ = I I J in'. A =0 4,75 in 1 • I:mjn = 0.983 in. (Hint: By rea~ :iOn uf sjm metry. lh~ prim..:ipal axes fUTIIJ . = I , + I:»

Mu that aCI~

=

,

U.S in .

(),'i;,,-

1.43;" j~. I L;:" n- Fig. P4.146

4.147 Solve Prob. 4.146. assuming that the couJlle .\'1" act$ in a hori wntnl plane.

~ -

4.148

TIlt! Z se<: lion shown is ~ubjecleJ

to a couple ~ acting in a ven ical

p\;:me. Delemline the I::tf'J est pemlissible value oi the moment M.) of the couple if the maximum ,tre~s is 1I0tlO exceed 80 MPa. Girer:: I"", = 2.28 X 10 5 IIlm~. 1,.," = 0.23 X W - 6 mm' , principal axe~ 25.7"""'<: and 64.Y.::i::'.

4 .149 zontal plane.

Solve Prob. 4.148 assuming that lhe couple M{I acts in a hori -

'

~ M.

!t)nnn-

Fig. P4.148

r

- 70",,,,..1

40!mn

J,"",,,,

! 40 """ J

~!on"r

283

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284

Pure Bending

4.150 A beam having the cross section shown is subjected to a couple acting in a venical plane. Detennine the largest permissible value of the moment Mn of lhe couple if the maximum stress is nol to exceed 100 MPa. GiI'ell: I,. = I. = b4 / 36 and /" = b 4/ 72. l\~)

!I 20

mill

Yr

I N~mm I

('

L;j2UIIl'" IJ = r-.Onllil

Fig. P4.150

FI!I. P4.151

4.151 A couple J\olu acting in a venical plane is applied to a WI] X 16 rolled-steel beam. whose web forms an angle (J with the verticaL Denoting by u () the maximum Stress in the beam when (J = 0, detenlline the angle of inclination fi of the beam for which the maximum stress is ]ifu. 4 .152 A beam of unsymmetric cross section is subjected to a couple M 'l acting in the venical plane xy. Show that the stress at poim A, of coordin,\le~ )' and t . is

where I" I"~ and I\~ denote the moments and product of inertia of the cruss section with respect \t) the coordin:lIe a.'lt'!S, and M~ lhe. moment of the couple.

Fig. P4.152 and P4.153

4.153 A beam of unsymmeLric cross section is subjected 10 a couplt! M il acting in Ihe huriwntal plane x.:. Show thai the stress at point A. of coordinates y and:;, is

::.1, - y l"

, M. u ... = - - 1, ( - I :',

where f ,.I;.. and 1.: denote the moments and product of inertia of the cross st!ction with respect 10 the coordinate axes, and M \. the moment of the couple.

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4.154 (u ) Show thm the s u es~ at corner A oflhe pl·j ~m
4.1S. Bending 01 Curved Membl!....

x

-+---~

b/6

1I!6

(b) Further ~how that. if no tensile stress is to occur in the membel. the force P must lit applied al a point localed w ithin tile Clea Uoumkd by Iht: lint: found

in pan a and three similar lines corresponding to the conditio n of zero stress at B. C. and D. respecti\ely. TIti > area. shown in Fig. P4.154b. is known as the tern of the cross section.

(II )

Fig. P4.154

4 .1S5 (a ) Show Ihal.if a vert ical force P is :lpplied at point A o f tile section shown. the equatiun of the lIeUlral axis BD is

(~)x+(~)z ~

- I

Is and k. denote the radius of gyration of the cross section with respo!ct Ihe: 8..xis and the x axiS. respe;:tively. (b) FUl1her show th:ll. if J vert ical force Q j g Ilpplied alllny point locllted online BD. the stres, III point A. W ill bo:o zero. where

10

'4.15. BENDING OF CURVED MEMBERS analy~is of stresses dut: to bcm.ling has been restricted su far to stmigh t members. In this sec tion we will conside r the st resses caused

Our

by the 3pplicmion o f equal and opposite couples to members that arc initially curved. Our discussion will be limited to curved memlx:rs of IInifonn cross section pos se ~sing a plane o f Symmelry in which the bending couples are applied, and il will be assumed that all s tresses remain ~I ow the proportio nal limit. If the. initial curvature. o f the mcmber is small, i.e.. if its radius of curvature is large compared to the depth of its £TOSS section, a good approxim:nion can be obtained for the distribution of stresses by :l.~sum­ ing the member to be slroight rtnd using Ihe formulas de rived in Secs . 4.3 and 4.4.t However. when thc radIUS of curvature and thc dimel1sioos of the cross section of the member are o f the same order of mngnilUde. we must use a di llerent method of analysis, whi(;h was lirst introduced by !he German engineer E. Wink.ler (1835-1 888). ;- See !'rob. Ul\5 .

Fig . P<4.15 5

285

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286

Consider the curved member of unifoml cross section shown in Fig. 4.76. Its transvcrse section is symmetric with respect to the y axis (Fig. 4.76b) and. in its unstressed state. its upper and lower surfaces intersect the vertical x)' planc along arcs of circle AB ,md FG centcred at C (Fig. 4.76a).

Pum Bending

,

rl c

R

,

t, G x

F

Fig. 4.76

'"'

F'

X .\

(II)

'"

We now lIpply two equlll lind opposite couples M lind M ' in the pl1lne of symmetl)' of the member (Fig. 4.76c). A reasoning similllr to that of Sec. 4.3 would show thllt lIny tmnsverse plllile section containing C will rcmllin plane, llild thllt the vllrious lIrcs of circle indicllted in Fig. 4.700 will be tmnsformed into circular and concentric lIres with a centcr C' different from C. Morc specifically, if the couples M llild M' arc directed as shown, the curvllture of the various arcs of circle will increase; that is A' C' < AC. We 1Ilso note that the couples M lind M ' will clluse the length of the upper surfllce of the member to decrease (A' B' < AB) and the length of the lower surface to increase (F'G' > FG). We conclude that a neutral surface must exist in the member, the length of which remains constllnt. The intersection of the neutral surfllce with the x)' plllne has been represented in Fig. 4.76a by the 1I/"C DE of mdius R, and in Fig. 4.76c by the arc D' E' of mdius R'. Denoting by 0 lind 0' the centml angles corresponding respectively to DE llild D' E', we express the fllct that the length of the neutral surface remains constllnt by writing

R8 = R'O'

(4,59)

Consideri ng now the arc o f circle lK located at a distance y above the neutral surface , and denoting respectively by rand r' the radius of this arc before and after the bending couples have been applied, we express the deformation of lK as

8

=

r'O' - rfJ

(4,60)

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4.15. Be-nding 01 Curved Membl!fS

Observing from Fig. 4.76 thaI r' = R' - y

r = R- Y

(4.6 1)

and substituting these ex pressions illLO Eg . (4.60), we wri te

8

~

(R' - .,,)11' - (R - ,')0

or. recalling Eg . (4.59) and setting () ' -

MJ.

(J =

(4.62)

8 "" - y flO

The normal stram € ... in the elements of JK is obtained by dividing the defonnation fj by the origi nallt:ngth rO of arc JK. We write E

..

fj = -=

rO

y flO rO

or, recalling the first of the relations (4.61 ), f'.

•

6.0 Y =- - - OR - y

(4 .63)

The relation obtained shows that, whil e each transverse section remains plane, the 110rm:1l strain 1:'.,. (Ioc!i 1101 va ry linearly with the distance )" from the neutral surface. The norma.l Stress a x can 110W be obtained from Hooke 's I:IW. fTr "" £E ,. by substituting for 1:'.< from Eq. (4.63). We have fT••

=

_ £!J.8 _ .1_· _ 8 R - y

(4.64)

or. alternatively, recalling the fi rst of Eqs. (4.61). a ,. =

E!l.O R -

r

(4.65)

II

Equatio n (4.64) shows that, like f'.x. the nonnal stress fT ., doc!i n ot WII)" linearly wiLh the distance y from the neutral surt:1ce. Plotting fT.,' versus y, we obtain an arc of hyperbola (Fig. 4.77).

Fig. 4.77

287

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288

In order to determine the location of the ne utral surface in the member and the value of the coefficient E nOl O used in Eqs. (4.64) and (4.65). we no w recall thai the eleme ntary forces aCling on any transverse section musl be statically equi valent to Ihe bending couple M. Expressing, as we did in Sec. 4.2 fo r a straight member, lhat the sum of the d ementary fo rces acting on the section must be zero, and Ihat the slim of lheir momenlS about the tr:msverse z axis must be e(lual 10 the bending momenl M , we write the equations

Pure Bending

(4.1 )

Iu, dA '" 0 Hnd

f( - vu.f dA) Substituting for

u~

-==:

M

(4.3)

from (4. 65) into Eq. (4.0, we write -

EMJR - r

Io

- - - -dA = O

~I R

,.

R - , - -dA = 0

I ,. jI7dA

dA = O

from wilkh il fo llows Ihal the dis lAnc e R from lilc ('cnler o f (' ur vflturc

,

I<

C 10

the neutral surface is defined by the relalio n A

R = --

f {~\

l\ .\

'/

.

1

,I

(4.66)

We note thai the value o btained for R is not equal to the distance ,. from C 10 Ihe celllroid of the cross section, since r is de fin ed by a diOe rc nt relation, namely.

I

A

-r = 1

rdA

(4.67)

Fig. 4.78 {/xi~· of {/ tnm.l"I'erse sectioll does no t pas.\" th ro ugh th e centroil/ oftlull section (Fig. 4.78).t Expressions for the radius R of the ne utral surface will be deri ved for

\Ve thus conclude that, ill a curved member, tile I!elllm l

some specifi c cross-scclional shapes in Example 4.10 and in Probs. 4.207 through 4.209. For convenience . tnese expressions are show n in Fig. 4 .79 .

t l·lowcvcr. an intcrcstin g propcn y of the neutralmrfoCi: enn be noted if we writ~ Eq. H .6Ii ) in the allematil"c form

(4.66' 1 EquJti on l4.66·I .'\holl" s lhat. i f the memocr is divided i!!lo a b~ c numl.ll'r of fioc l> of cross5e(tional area dll. the CUlvaturc ! IR of the neutral iurfa~e will be equal \0 the averJ.g,· \·aluc of the curvature I lr of the \·arious fioc f1l .

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c

'j'

c

r.

,

,

'l ~I

fv r

I

TO r

--- ,---

~ b

" ".,! an~l('

Tml'i'l~)jd

C lrd e

Il = - "-

Il =~ ( r +V .. 2 -r.~)

" "

R~

In ~

Fig.4.79

i J.!(b J + h } - - " ' - ' ' ' - ; , ' 2' ' - - - (1)1'-2 - (, " "I) III ~ - I.(h, - /;0• ) Fj ~

Radius of neutral surtace for various cross-sectional shapes.

Substituting now for

from (4.65) into Eq. (4.3), we write

if ,

Io

E !l O R - r

- - - - - y dA = M

or. since ),

= R -

,.

r, E !!.OI (R - , )' - - - - - -dA = M

o

,

Expanding the !>quare:. in the integrand, we obtain aft er reductions

E!!.O [Rl

- 0-

IdA -;- -

2RA+

Ir dA 1= M

Recall ing Eqs. (4.66) and (4.67). we notc that the fin;[ term ill th e brackets i ~ equal 10 RA. while Ihe last tern) i ~ equal \o,A We have. there fore, E

ao o

- - (RA - 2RA

+ , A) ~ M

and, solving for E 6.0/ fl, E 6. 0

M

o

A(, - R)

(4.68)

Referring to Fig 4.76, we notc lhu\ fl O > 0 for M > O. II follows 1hal R > 0, or R < regard less of the shape o f the section. Thus. the neutral axis of a transverse section is always located between the centroid of the section and the center o f c urvature of the me mber (Fig. 4 .78). Selling R = e. we write EtJ. (4.68) in the form

r-

r,

r-

E!l.O

M

o

M

(4.69)

289

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290

Pure Bending

Substituting II OW for E flfJ /8 from (4.69) into Eqs. (4.64) and (4.65), we obtain the foll owing alternative expressio ns for the normal stress u " in a curved beam : (4.70)

(4.71 )

We should note that the paramC\er e in the above eq uati o n ~ is a small quantity obtai ned by subtracti ng two lengths of comparable size. Rand r. In order to detcnnine u •. with a reasonable degree of acc uracy. it is therefore necessary to com pute R and r very accurately, particularly when both of these quamities arc large , i.c .. when the curvature of the member is small. However. as we indi cated earlier. it is possible in such 11 C;1se to obtai n a good apprOXimation fo r u " by Llsing the fo rmula u " "" ~My// dc.veloped fo r straight members. Let us now determine the change in curvature of the neutral suri:1ce c;1used by the bend ing moment M. Solving Eq. (4.59) for the curvature I/ R' of the neutral surface in tbe deformed me mber, we write I "

R' or,

SC lt lllg

fJ' - fJ

R 0

+ fl l) alld fccaUjllg Eq. (4.69),

...!... R'

~ ~(I + R

flU) = {j

~(I + ~) R

EAt!

from which it follows that the change in curvature of the neutral surface is I

I

M

--- = -R' R EAeR

EXAMPLE 4.10 A curved rectangular bar has a mean radius r = 6 in. and a cross seclion of width b = 2.5 in. and deplh II = 1.5 in. (Fig. 4.80). Determ ine the distance e between the centroid and th e. neulral axis of the cross section.

We first derive the. e.~pression for the radius R of the neutral surface. Denoting by I"J and 1"2. respecli vely. the inner and

Fig. 4.80

(4.72)

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outer radius of the bar (Fig. 4.81), we use Eq. (4.66) and write

".,

R ~ --

(4.73)

In ~

" Fig. 4.81

For the given data, we have

r - ~ h "" 6 - 0.75 = =r + ill = 6 + 0.75 =

1", = Y2

5.25 in. 6.75 in.

c

Substituting for II, r" and 1"2 into Eq. (4.73). we have

R

~

II

-r, In "';' r,

~

1.5 in. - - - = 5.9686 in. 6.75 In - 5.25

=

r

-

I(.

d= 0.03t../ ill.J - L

= 0 - 5.9680 = OJJ31 4in.

J

1'= (; ' n.

,,,,,,"I~;, __

The distance between the centroid and the neutral axis of the cross section (Fig 4.82) is thus f!

I

1I= .~ . 9686 '"

-,,<--, C("ntroid

We note that it was necessary to calcul!!te R with five significant figures in order to obtain t: with the usual degree of accuracy.

Fig. 4.82

EXAMPLE 4. 11 For the bar of Example 4.10. Jetennine the largest tensile lmJ compressive stresses. knowing that the bending momen t in the bar is M = 8 kip' in.

Making now

I"

=

(T n,in

=

R = 5.969

M(., - R) =

= - 9.30ksi

Remark. Let u~ compure the values obtained for 0"","-, and rJ"nWl with the result we would get for a ~trai£h l bar. Us· ing Eq. (4. 15) of Sec. 4.4. we wri te Me

Aerl

(8 kip' in.)(6.75 in. - 5.969 in. ) (3.75 i n1 )(O.031 4 in. )( 6.75 in.) IT" , ....

(3.75 in2)(0.0314 in.)(5.25 i n.) 0"", ,"

e = 0.0314 in.

Making first r = r2 = 6.75 in. in Eli. (4.71). we write U ruu

(8 kip' in. )(5:25 in. - 5.969 in. )

A = bh = (2.5 in.)( 1.5 in.) = 3.75 in l

mllJ the values obtained in Example 4. 10 for Rand c.

= 7.86 ksi

= 5.25 in. in Eq . (4.71). we have

M(Y1 - R)

We use Eq. (4.71) with the given data. M = 8 kip ' in.

r,

ff m",. n""

=

J

(8 kip' in.)(0.75 in.) - ,':1 (2.5 in.)(1.5 in.)l

:!: 8.53 ksi

291

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20 '''IIl -~1 f-

n ~mm

c:::::l::::J .~o

, -

6(1,,,,,,

L

SAMPLE PROBLEM 4.11

.

A

111111

\-'iO _._.'1'",-1- ,'_.:;0 _.mm

- -

lo~ded

co mponent has a T.,haped cross secti on and is as shown. Knowing Ihal lhe rulowabJe compressive

~tre~s

l :;rge ~1

machi~e

plied

S,'£:linn a-a

is 50 MPa. dCI<'rmi ne the the component.

force P th:Jt can be ap-

\0

p

p'

SOLUTION We locale the centroid D of the cross section

Centroid of the Cross Stction.

A 1, mm 1 2

2

r ~ mm

f lA I'

rum'

64 X 103 56 X 103 ';' , / Ai "" 120 x 1 0~

40 70

(20)(00 ) " 1600 (40)(20) .. 800 L AI " 2400

(12400) = 120 X 10' ," z SOmm z O.050m

Force and Couple lit D. The internal forces in section (I-U lent to a force P acting at D and a couple 1\'1 of moment

0±=r~mm

W~60"'lU

"...

r

"j:

aT!

equiva-

M = p(50mm + 60mm) = (Q.llOm)P

SupuposiUolI. The centric force P causes a unifonn compressive stress on section (I-n. The I>.!nding couple M causes a varying stress distrlbmion [Eq. (4.7111. We note that the coup le 1\1 tend~ to incre:l!>e the curvature of the mem_ ber and is therefcre positive (d. Fig. 4.76). The tOlal stress at a point of section (HI located at distance /" from th~ center of curvature Cis

.1/ 1.-- m


p M(r - H) = - - 04-

A

lT ~~

B

Radius or Neutral Surfa ce..

(I)

At r

We now determine the radius R of the

neutral surtace by usmg t.q. (4.bt!).

R~

,I:,

-f- ~ A d:-

2400mm! -f"'C'(C8UCC"C,:C"C,'lC'"""""+-"":fCO C(",UO"c,;c"c'lCdC'c.

'.

'.

2400 90 50 80111 - + 20 111 -

,

c·

30

2400 40.8(,(/ + 11 .75"

45 .6 1 mm

50

= 0.0456 1 m

We

al~o

compute" =

r-

A lIo"'abl c Load.

R = 0.0 5000 m - 0.1)4561 m = 0 .00439 m

We observe thai the laJgesl comp, cssil'c SlJess w ill If"" = Sf) MPa and us ing

occ ur at JIOinl A where r = 0.030 m . Recalling that Eq. (I). we \\rile P

- 50 x 10" Pa =

10- 3 m! - 50 X lOb = - 4 171' - 54J2P 2.4

292

f(

T

(0. 110 r)(().030 m - 0.0456 1 m)

=-',","""C'..'l::'~""_.'OC,:"::,""5'-

(2.4 X IO- J ml)(O.OO439 m)(O.030 m) " ~ 8.55 kN -
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PROBLEMS

4.156 For {he curved bar and loading shown. determine the stress at point A when (a) r[ = 1.2 in .. (b) r, = 2 in.

Fig. P4.1S6 and P4.t57

4 .157

For the curved bar and loading shown. determine the stress at

points A and B when

4 .158

'I =

1.6 in.

For Ihe curved bar and loading shown. determine the stress at

pointS A and B when II = 55 mm.

24 mIl'

Fig. P4.1 S8 and P4. 1S9

4. 159

For the curved bar and loading shown. determine the stress at

poinl A when (a ) I, = 50 rum. (b) It = 60 mm. 4 .1 60 The curved pottion oflhe bar shown has an inner radius of 20 mm. Knowing Ihat the allownble stress in the bar is ISO MPa. determine the largest

permissible distance (/ from the line of action of the 3-kN force to the vertical plane containing the center of curvature of the bar.

4. 161 The curved portion ofthc bar shown ha~ an innermdius of20 mm. Knowing ih<1tthe line of action of the 3-kN force is located at a distance (/ = 60 mm from the vertical planc containing the center of curvalllre of the bar. determine the largest compressi\'e stress in the bar.

Fig. P4.160 and P4.161

293

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294

Pum Bending

4.162

For the split ring shown. delermine the stress al (a) point A.

(b) point B.

Fig. P4.162

4.163 Sleellinks having the cross secl ion shown are available with d if~ ferenl central angles {3. Knowing Ilwt the allowable s tress is 12 ksi . determine the largesl force P thai can be applied 10 a link for which {3 = 90°.

'" Fig. P4.163

4. 164

Solve Prob. 4.163. assuming Ihm {3 = 60°.

4. 165 Three plates are welded logcther 10 form the curved beam shown. For the given loading. delermine the distance t' between the neutral axis and the centroid of the cross seclion. 2. in.

C Fig. P4.165 and P4.166

4 .1 66 111ree plmes are welded together to fonn the curved beam shown. For M = 8 kip' in .. determine the stress at (a) point A, (b) point B. (c) the centroid of the cross section.

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4.167 a nd 4.168

Knowing that M = 20 kN . m. determine the stress

Problems

at (a) point A. (b) point 8 .

~/~,c,-,-;:""-ii: --'-' '-,- IT r ~ l I l~"'m

1SO """

~ H

-'-"5"""

I.!i7I T 36mm

Fig. P4.167

Fig. P4.168

4.169 The split ring shown has an inner mdius", = 20 mm and a ci,.mlar cross section of diameter d = 32 mm. For the loading sho wn. determine the stress at (a) point A. (b) point 8.

Fig. P4.169 and P4.170

4 .170 The spilt ring shown has an inner r::ldius r, = 16 mm and a ci,.ctlhll" cross section of diameter d = 32 mill. For the loading shown. determine the stress at (a) point A. (b) point B. 4 .171 The split ring shown ha~ an inner radius r , = 0.8 in. and a cirl"tIlar cross section of diame ter d = 0.6 in. Knowing that each of the 120-lb force s is applied at the centroid of Ihe cross seclion, determine the Slress at (a) point A. (b ) point 8 .

B

Fig. P4.171

4 .172

Solve Prob. 4.171. assuming that (he ring has an inner mdius

r, "" 0.6 in. and a cross-sectional diameter d = 0.8 in.

295

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296

Pum Bending

4. 173 For the crane hook shown. delermine the largesl tensile slrcss in seclion (/.(1.

35

25,mn

n,,,,

1 Sect;,,,, a -a

J51<."

Fig. P4.173

4.174 For the curved beam and loadin g shown. delermine the stress at (al point A. (b) poinl B.

'-' '"

250/\

Sectinn a-;/

Fig. P4.174

4. 175 Knowing thai the m
-, A

" _ -L

6;11 4 in.

FIg. P4.17S and P4.176

4.176 Knowing that the machine component shown has a trapezoidal cross section with {/ = 2.5 in. and b = 3.5 in. detennine the stress
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at

4 .177 and 4. 178 Knowing that M point A. (b) point B.

-'='

Probtems

5 kip ' in .. detemline the stress

(a )

2.5 ill

2.5j"

Fig. p4.1n Fig. P4.178

4.179 Show that if the (ross section of a curved beam consists of two or more rectangle •. the radius R o f the neutral surface (an he expressed as

where A is the total area of the cross section.

4.180 through 4.182 Using Eq. (4.66). derive the expression for R given in Fig. 4.79 for '4.180 A circu lar cross section. 4.181 A trapezoidal cross section. 4.182 A triangular cross section. "4.183 For a curved bar of rectangular cross section subjected 10 a bending couple M, show that the radial stress at the neutral surface is

M( I - "R

u = '

Ae

R)

~- ln I"j

and compute the val ue of If, for the curved bar of Examples 4. 10 and 4.11. (Hint: consider the free-body diagram of the portion of the beam located above

the neutral surfa(e.)

Fig. P4.183

FIg. P4.179

297

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REVIEW AND SUMMARY FOR CHAPTER 4 M'

c "

,

Fig. 4. 1

Normal strain in bending

Thi s chapter was devoted to the analysis of members in pure bending. That is, we considered the stresses and defo rmation in members subjected to equal and opposite couples M and M ' acting in the same longitudinal plane (Fig. 4.1). We lirst studied members possessing a plane of symmetry and subjected to couples acting in that plane. Considcring possible defo rmations of the member, we proved that tran.l·verse .I·ectiol/.v remaill pIal/I! as a member is deformed [Sec. 4.3[ . We then noted that a member in pure bending has a neutral surface along which normal strains and stresses are zero and that the iongitudinall/ormal strail/ E~ varies linearly with the distance y from the neutral surface: E~

J' = - -

p

(4.8)

where p is the rUllius of cllrvlllllre of the neutral surface (Fig. 4.12a). The intersection of the neutral suri:1cc with a transverse section is known as the neutral axis of the section. For members made of a material that fo llows Hooke's law [Sec. 4.41, we found that the nomwlstreH (J~ varieJ'linearly with the distance from the neUlral axis (Fig. 4.13). Denoting by U III the maximum stress we wrote Fig. 4.12a

)'

u <= --; cr m

(4.12)

Normal stress in elastic range

N.. utr~1 surfa.,-

\ - --'-- - "

where c is the largest distance from the neutral axis to a point in the section. By setting the sum of the elementary forces, u~ dA. equal to zero, we proved that the neulral axis pa.vse.v ll1mugh the centmid of the cross section of a member in pure bending. Then by setting the sum of the moments of the elementary forces equal to the bending moment, we de rived the elm·tic flexure formilla for the maximum normal stress u

Fig. 4.1 3

M,

m

= -

I

(4.15)

where I is the moment of inertia of the cross section with respect to the neu tral axis. We also obtained the normal stress at any d istance )' from the neUlral axis: Elastic flexure formula

My

I

298

(4.16)

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Noting th
S ~ ­

(4. 17)

c

Review and Summary for Chapter 4

Elastic section modulus

and then used the section modulus to write an alternative expression for the maximum normal stress: a

M

m

= -

(4.18)

S

Recalling that the curvature of a member is the reciproc
Curvature of member

(4.21 )

£1

In Sec. 4.5, we completed our study of the bending of homogeneous members possessing a plane of symmetry by noting that deformations occur in the plane of a transverse cross section and result in allficlil.l"tic curvature of the members. Next we oonsidered the bending of members made of several materials with differellf moduli oj elasticity [Sec. 4.6J. While transverse sections remain plane. we found that. in general, the neutral axis does not pa.l".\· through the cellfroid of the composite cross section (Fig. 4.24). Using the ratio of the moduli of elasticity of the ma-

- +--\-- N A. - -t---..,

Anticlastic curvature

Mellbers made of several materials

~-----C"'------~ ~. A

Fig. 4.26 (0)

(h)

(c)

Fig. 4.24

terials, we obtained a frall.Vormed section correspondi ng to an equivalent member made entirely of one material. We then used the methods previously developed 10 determine the stresses in this equ ivalent homogeneous member (Fig. 4.26) and then again used the ratio of the moduli of elasticity to determine the stresses in the composite beam rSam ple Probs. 4.3 and 4.4J. In Sec. 4.7, .\"tres~· cOllcentratioll.V that occur in members in pure bend ing were discussed and charts giving stress-concentration factors fo r Ilat bars with tillets and grooves were presented in Fi gs. 4.31 and 4.32.

Stress concentrations

299

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300

Pure Bending

We next investigated members made of materials that do no t follow Hooke's law [Sec. 4.8[. A rectangular beam made of an elastoplastic material (Fig. 4.39) was analyzed as the magnitude of the bending moment was increased. The lIIaximum elastic moment My occurred when yielding was initiated in the beam (Fig. 4.40). As the bending moment was further increased. plastic zones developed and the size of the elastic core of the member decreased [Sec. 4.91. Fi nally the beam became fu lly plastic and we obtained the maximum or plastic moment Mp. In Sec. 4.1 1, we fou nd that permanent deformations and resitlila/ .I"tresses remain in a member alier the loads that caused yielding have been removed.

u

Fig. 4.39

Plastic deformations ELASTIC

--\-- -u';-,

ra) M < M)

P LA ST IC

, ~ "}

u,

Fig. 4.40

Eccentric axial loading

"F

~ .,

P_ '

~

~~~~~~~~~TI

In Sec. 4. 12, we studied the stresses in members loaded eccentrically ill {/ plane of .1)"IIl11letry. Our analysis made use of methods developed earlier. We replaced the eccelltric load by a force-couple system located at the centroid of the cross section (Fig. 4.48b) and then superposed stresses due to the centric load and the bending couple (Fig. 4.5 1):

P

Fig. 4.4Sb

(Ix

=

'I

My

A - -,-

(4.50)

'I

KA

c Fig. 4.51

u,

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llle bending of members o f IInsymmetric cross .I·cetion was considered next [Sec. 4.1 31. We found that the flexure formula may be used, provided that the couple vector M is directed along one of the principal centroidal axes of the cross section. When necessary we re-

Review and Summary for Chapter 4

301

Unsymmetrlc bending

'I

'~ t ----- 1M,! , , , ,

~

,,

-- M:..- {- ~ --.~'~C~-Z

Fig. 4.60 Fi g. 4.61

solved 1\1 into components along the principal axes and superposed the stresses due to the component couples (Figs . 4.60 and 4.6 1). M .y

M y:'

I:

1.1'

- - -- + _ .-

(4.55)

For the couple 1\1 shown in Fig. 4.65, we determined the orientation of the neutral axis by writing tancp =

I.

~ ta nO

(4.57)

I,.

Fia.4.65

The general case of eccentric axial loadin g was considered in Sec. 4.1 4. where we agai n repl aced the load by a fo rce-couple system located at the centroid. We then superposed the stresses due to the centric load and two component couples directed along the principal axes: (Ix

P M ,y = - - -A Ie

M ).:.

+ -1)

General eccentric axial loading

(4.58)

The chapter concluded with the analysis o f stresses in cllrwd mcm her.f (Fi g. 4.760). While transverse sections remain pl ane when the member is subjected to bending, we fo und that the stresses do not wIry linea rly and the neutral surface does not pass through the

Curved members

centroid o f the section. The distance R from the center o f c urvature o f the member to the ne utral surface was found to be A

R ~ --

I ,.

dA

(4. 66)

where A is the area of the cross section. The normal stress at a distance y fro m the neutral surface was expressed as My

Ae(R - y)

(4.70)

where M is the bending moment and e the distance from the centroid o f the section to the neutral surface.

F

Fig. 4.76a

G '

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REVIEW PROBLEMS

4 .184 and 4.185 Two W4 x I3 rolled seclions arc welded together as shown. Knowing Ihat for the steel alloy used fT r = 36 ksi and fT U = 58 ksi and using II factor of safety of 3.0. determine lh~ largest couple that can be applied when the assembly is bent about the z axis.

- ·c lH16 i" ,

--1 H

,;J.::-=- ~" r~" ,

Fig. P4.186

.

Fig. P4.184

Fig. P4.185

4 .186 II is observed that a Ihin stee l strip of OJ)6- in. width can be bent into a circle of ~- in. diameter without :my resulting pennanenl deformation. Knowin g that t: = 29 X l Ob psi. delemline (a) Ihe maximum stre ss in [he benl strip. (b) the magn itude of the couples required 10 bend the s trip.

4 .187

A bar having the cross sectiun shown has been formed by

se~

cure ly bonding brass :lI1d alum inum stocK. Using the data given below. determine th e largest permissible bending moment when the .:ompositc bar is bent about a horizontal ax is.

Modulus of elasticity Allowable stress

t) m",

Aluminum

Brass

70GPa

105 G Pa 160 MPa

100 MI).J

X ,nm

1 1--+1-

32 """ ---+ 1-II

~,- ,

,

r

32 "un

l

Fig. P4.187

302

~ ]~nll"

'- AI"",,,,,,,,,

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4.188 For the composite bal of !'fob, 4. [87. determine the largest pt:Lmissible bending moment "hen the bar is bent about a IIcrtical axis.

Rm/lew Problem.

4.189 As many a, thrce axial loads each or magnitude P = 10 kips can be applied to the end of:L \\is x 1 [ rolled-steel shape. DetemLiue the stress at point A, (
I'

3. ""(:~/

3..i iIJ .......

1- - 12l1IUIIL - j - L

l lOmm

Fig. P4.18Q

J

!:Wmlll

4 .190 Three 120 X 10-mm >Ieel plaes have been welded togeLher to lonn the beam shown. Assuming thm the steel is eJastopJastic with E - 200 GPa and fT ,. = )00 MPa. detemline (a) the bendin)! momen! for which the plastic zones al the top and bDltom of the bt:am are 40 mm thic k. (b) the corresponding radiu s of curvature of the beam. 4. 191 A vertical force P of magni tude 20 kips is app lied al a point C locmeJ on the axis of symnlNry of the cross section of a shon L"Olumn. Knowillg lIU1L ) · - 5 iu. , d c tcnni"c (u ) Lhe MfC~~ n~Ulral axis.


point A. ( {.) d, ~ ~t,e~~ at point

Ie) the location of the

,

,

- lh-I B

J'"

U

1,),

A

, -.-I..+,c-I _lll .

..

( )

Fig.

P4.191

\(

I in (1)1

_Ill.

I

,

n.

-lOmm

_o;;;;;;t''::=::o -----r 10 Fig. P4.190

III 'U

303

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304

Pure Bending

4 .192 The shape shown was fomlCd by bending:J Ihin steel plate. Assuming thai the thickness I is small compared to the length a of a side of the shil pe. detemline the stress (u) at A. (b) at B. (c) at C.

p

Fig. P4.192

, =20""" Fig. P4.193

'"

4 .193 TIle couple 1\"1 is applied to a beam of the cros~ section shown in a plane fomling an angle (3 with the vertical. Detemline the stress at (a) poin! A. (b) point B. (c) point D.

4. 194 A ngid circular plate 01 125-mm radius is allached to a solid 150 x 200-mm rectangular pos!. with the center of the plate directly above the center of the pos\. If a 4-kN force P is applied at E with 8 = 30°. determine (u) the stress at point A. (h) the stress at point B. (e) the point where the neutral axis intersects line ABD.

Fig. P4.194

4 .195 The curved bar shown has a cross section of 40 X 60 mm and an inner mdius = 15 mm. For the loading shown, determine the largest tensile and compressive stresses in the bar.

'I

FIg. P4.195

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COMPUTER PROBLEMS

TIll" following problems are rlfSigncd h) 00 s(lh'cll II-jUt 0 c UIIlpulcr. 4 .Cl TWo aluminum >lripS and a steel smp are 10 be bonded logf:'lher \0 fflrm ~ Cf)mrlll~ire memooT '1f wirhh h = tlO mm ~nd depth" = 40 mm The modulus of elasticity is 200 GPa for the! steel Jnd 75 GPa for the aluminum.

KnowinglhaM"" lSooN·

m.wrileacompllierprogramiocalculatelh~max­

imum strc~s in the nluminum and in lhe steel for valm:s of" from 0 to 20 mm using 2·mm increment.;. Using appropriate smaller increments. dClennine Ill) the largest stress Ihal can occur in the Sleel. (b) the corresponding value ur u

,\ \",nin1l1ll

~

s"el~-T I l::~:d-t' ~ {, - (i(IIIH,,-1

Fig.

zr' ' "

P~.C1

4 .C2 A beam of the cross section shown. made of a steel thai is assumed to be elasloplaslic with a yield strength lIf :U1d a modulus of elasticity E. is bent ahoU( lhe -' axis. (u) Denoting by Yr the lmlr thicklless of (he cbs(i: COle, write a. computer program to calculate the bending moment M and the radius Ill' curvature p for VJlues of .1'1' from ~ d to ~ d using decrements equal to I,. Neg!ect the effect ot tillets. (b) Use thj ~ program to sollie l'rob, 4.II)U.

i

Fig. P4.C2

4,C3

An 8- kip . In. couple i\1 is applied to a beam of the cross section in u plane fonning an angle {j wi(h (he venical. Koung (hm (he centroid ()f the cross section is located :It C and thai the )' and .:. axes are principal axes. write a computer program to c::.lc.ulate the siress al A. B. C, and D for values of j3 from 0 \0 180 using 100 incremems. (GiI'en: Iv 6.23 in ' and I = 1.481 in' .) ~ho\\'n

=

Q

0.4

--j A

1-

~'

n.4

-ir-

,.

. .

J .2

--..iU

I~ ~I Iu./~

D, 4

1.6

u,~

"

0,11

Fig. P4.C3

305

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306

Pure Bendtng

4.C4 Couples of moment M = 2 kN . m are applied as shown 10 a curved bar having a rectangular cross seclion with Ir = 100 mm and h = 25 mm. Write a compuler program and use it to calcuktle Ihe stresses al points A and B for values of Ihe ralio r/h from 10 to I using decremenls of I. and from I 10 0. 1 using decremenls of D.1. Using appropriate smaller in· crements. determine Ihe ratio r/h for which Ihe maximum siress in the curved bar is 5D% larger than the maximum slress in a straight bar of Ihe same cross seclion.

Fig. P4.C4

Ag. P4.CS

4.C5 The couple M is applied 10 a beam of Ihe cross section shown. Write a computer program Ihal. for loads expressed in either SI or U.S . customary units. can be used 10 calculate Ihe maximum tensile and compressive stresses in Iht" beam. (b) Use Ihis program 10 solve Probs. 4.7. 4.R. and 4.9.

(a)

Fig. P4.C6

4 .C6 A solid rod of radius c = 1.2 In. is made of a steel lhat is assumed to be elasloplastic with £ = 19.000 ksi and i f y = 42 ksi. The rod is subjected 10 a couple of momenT ,'I4lhat increases from zero 10 Ihe maximum elastic mo· menT Mr and then to the plastic moment Mr Denoting by yy Ihe half thickness of the elastic core, write a computer program and use it to c~lculate Ihe bending moment A1 and the radius of curvalure p for values of Yr from 1.2 in. to 0 using D.l-in . decrements. ,Hilll: Divide Iht" cross section inlo 80 horizontal elemenTs of 0.03· in. height.) 4.C7 11lc machine elemem uf Prob. 4.178 is to be redesigned by removing part of Ihe tri~ngular cross section. II is believed Ih~t Ihe remov~1 of a small tri~ngular ~rea of width a willluwer the maximum stress in Ihe element. In order to verify this design concept. write a computer program 10 calculate the ma.~imum stress in the element for values of II from 0101 in. using O.I·in. increments. Using ~ppropri~te smaller inCrel11enL~, delenninc Ihe dist ance a for which the maximum slre..~s is as small as possible and Ihe corresponding value of the maximum stress.

Fig. P4.C7

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C

HAP

T

E

R

Analysis and Design of Beams for Bending

The beams supporting the multiple overhead cranes system shown in this picture are subjected to transverse loads causing the beams to bend. The normal stresses resulting from such loadings will be determined in this chaptcr.

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308

Analysis and Design of Beams for Bending

5.1. INTRODUCTION This chapter and most of the next one will be devoted to the analysis and the design of beams, i.e .. structural members supporting loads applied at variolls poi nts along the member. Beams arc usually long, straight pri smatic members, as shown in the photo on the previous page. Steel and aluminum beams play important part in both stmctural and mechanical engineering. Timber beams are wide ly used in home construction (Fig. 5.1). In most cases, the loads are perpendicular to the axis of the beam. Such a lral!Sl"en·c loading causes o nly bending and shear in the beam. When the loads are not at a right angle to the beam. they also produce axial forces in the beam.

,m

t

Ie c""::",,_1:''::::;oo!

A'i='_ _

Fig. 5.1

D

,\, ••!!::=.:••r::.'::lll, c Il

(t,) D;
Fig. 5.2

The transverse loading of a beam may consist of cOllcelllmled loads P I' P b ... , expressed in newtons. pounds, or their multiples, kilonewtons and kips (Fig. 5,2a). of a dislribuleJ /oaJlt1. expressed in N/m, kN/m, [b/ft, or kips/ft (Fig. 5.2b), or of a combi nation of both. When the load 'W per unit length h3S a constant value over part of the bcmll (as between A and B in Fig. 5.2b), the load is said to be IIl1ifonllly diSTrihuted over thai part of the bellm. Beams are classified according to the way in which they are supported. Several types of beams frequently used are s hown in Fig. 5.3, The distance L shown in the various PllnS of the figure is called the span. No te that the reactions at the supports of the beams in parts a, h. and c of the tigure in-

volve a total of only three unknowns and. therefore. can be determined by

St.'Ii('al ~· D~t" nnfnat" B, ~ am ,-

TI--L-

--I

f.:-----L-

1--- -1.- -

(I» ) O\"t'rl'''"gin )!; 1)(~lIn

1 ,"lt· t,, " ~~ ;n '~t ...

Bcams

(ii I Continll"".' bea" ,

Fi g. S.3

1-- - 1. ----1

I-- - L- ---I

(10) Ill'am rued at One 1',,,[ .",,1 sj,,,p~' ""ppurt",1 al thr ",h,·,. ,.nd

( f ) Fi.\,.,II~'"",

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5.t . Introduction

lIlt: 1Ilt:IIHxh uf )I iltil·~. Sudl be;IIU~ ilIt: ~;Ji Ll tu ut: ~lUlit"Ully dt'lem/i,wle am.l wil l be di~("ussl.'d in thi ~ chapte r and the nexI On the otner hand, Ihc re~

ac tio ns at th e supports of the bcall1~ in part~ d. e, and f of Fig. 5.3 involvc more than three unknow ns and ca nnOI be determined by the melhods of stilllCS alone. The prOpcl1lc..~ 01 the ~all1s wnh rcg ard to theIr [CSlstilnce to dcfcmllltiollll must be mkcn into con~ l dcrnt.ion. Such bcam.~ arc said to be srlllically i/lderen/linGle antllh~ir anai llsis will be postponed until Chap. 9. where deformation .1 of be:mls will be discuss~d. Sometimes two Of more beams are. connected by hi nges to form :I si ngle co nlinu ous structurc. Two exwnples of beams hi nged :II :I poi nt Hare shown in Fig. 5.4. It will be noted that the Ie acti otl.~ al the supports involve four unknowns and c annot be determined from Ihe free ·body diagram of the two-beam system. They can be detemlined. however, by co nside ring the frce-body diagram of each bcam ~parn t cly : six unknow ns arc involved (incl uding two force wmponcnt:i al the hitlge). ilnd six ~{luations arc 3vaJlabt'"'. It was shown in Sec. 4. 1 that if we pass 11 I'tCcti[lIl throllJ:h 11 point C of a cantilever beam supporti ng a ~o n ee ntrated load P at it~ I!nd (Fig. 4.6). the inlernal forces in the section are found to consist of a ~ h ear force p ' eq ual and opposite to the load P and il bending c(J uple .M of moment equal to the moment of P about C. A simi lar silllation prevails for othe r types of su pports and loadings. Consider, for example. a simply su pported beam AB carrying two co ncentrated loads and a llnifonnly di stributed load (Fig. ~.5c). '10 delermtne the tnl ernal forces 111 a ~cctlon through pomt C wc fIrst draw th;: free -hody di agram of the entire beam 10 obtuin the rcm::tions QI the IiUpporiS (Fig. 5.5b). Passing a section through C. we thcn d raw the free -body diagram of AC (Fig, 5.51.'), from which we determi ne the shear (orce V an d the bending co upl e !\I. The bending coupl e M creates lIor/1lI11 Slre.I'ses in the cross soctlQIl, while the shear force V creates shearing STr"s.~.u in that section. In most cases the dominant criterion in the design of a beam for st rength is Ihe maximum value of the normal stress in the beam. The detennination of the normal stresscs in :1 beam will be the subje,cl of this chapte r. while sheariug .,II"t:~M::~ will be L1iM:us~u ill elm].!. 6. Since the distribution of the n0l1n31 Slr<,sses in a given section depends nnl ) upon the va lue o flhe be nding momcnt AI in that section antllne geometry of thc section,t the d astic tlexlirc formula;; derived in Sec. 4.4 can be tL<;cd to de termine the maximum stress. as well as Ihe ~t res.~ at any given point, in the 5cctioll. We writc:j:

IMlc

IT", = - 1 -

a, = -

My 1

t Il is

a~SUlm.·d l.1al th~ distribution of the nonna! SUl:sses ir, a gil'CIl cross !iITtiOil is not by the dcbrmatiom cau:;ed by the .shearing stresse.~, This assumption will tlc I'cri-

fiNl~, S~".

6.5

d~pencing upon whrthcr the concovily of the beam ,,' the point roL"id,r"d roc", UpI!.'aJ"ti or downward. Thu<, in the CJ<" <.onlirl~n~1 h~fI' nf" Ir~"IVNSI' Irnrlitl ~ IhI- I'!:" nf M ~M ,"My nll'lIr Ih~ lIram nn Ih~ "'!wr h~nd, <7 .. is a posi tL ve qJantity. ttoc absolUic I'alm' of AI is u~rd in Eq, {~.l ~

j Wc recall from Sec. 4.2 thJt .If can be positil'c or ncgalil'c.

Fig. 5.4

(5. 1,5.2)

where / is Ihc mo mct\l of inertia Qf the cross sectio n with res pect to a ce ntroiLla l axis perpendi cu lar 10 the plane o f tile. co up le. )' is the d islance from th e neulral s urface. an d c i ~ the maximum vul ue of tha t d iStance (Fig. 4.13). We a lso recall from Sec. 4.4 that. introducing the.

lrr~cled

-

Fiy.5.5

309

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310

Analysl!l and Design 01 Beams lor Bending

clastic section modulus S = l/c of the beam, the maximum value a ", of tht': normal stress in the section can be ex pressed as

(5.3)

The fact that a ," is inversely proportional to S underlines the importanCe of selecting beams with tI large sct:lioll modulus. Section moduli of various rolled-steel shapes are· given in Appendix C, whi le the section modulus of a rectangular shape can be expressed. as shown in Sec'. 4.4. a..<; (5.4)

where b and II are, respectively, the width and Ihe depth of the cross section. Equtllion (5 .3) also shows that. for a beam of uniform cross section, a .. is proportional to IMI: Thus, the mru:illlulll value of the normal stress in the beam occurs in the section where IMI is largest. It follows Il1at one of the mOSl impoltant parts of the design of a beam for a given loading condition is the determination of the location and magnitude of the largest bending momem. This task is made easier if a bending-mOil/em diagram is drawn. i.c., if the value of the bending moment M is determined at various points of the beam and plOlted against lh{"" distance x m e asured from one. end of the beam. It is further facilitated if a shear diagram is drawn althe same time by plotting the shear V against x. The sign convention to be used to record the values of the shear and bending moment will be discussed in Sec. 5.2. The values of Vand M will then be obtained at various points of the beam by drawing free-body diagrams of successive portions of the beam. In Sec. 5.3 relations among load, shear, and bending moment will be derived and used to obtain the shear and bending-moment diagrams. This approach facilitates the detennination of the largest absolute value of the bending moment and, thus. the determination of the maximum nonnal stress in the beam. In Sec. 5,4 you will learn to design a beam for bending, i.e., so that the maximum normal stress in the beam will not exceed its allowable value. As indicated earlier, this is the dominant criterion in the design of a beam. Another method for the determination of the maximum values of the shear and bending moment, based on expressing V and M in tenns of singularity jJlnctions. will be discussed in Sec. 5.5. This approach lends itself well to the use of computers and will be expanded in Chap. 9 to facilitate the determination of the slope and deflection of beams. Finally, the design of IwnprislI1aric beatl/s. i.e., beams with a variable cross section, will be discussed in Sec. 5.6. By selecting the shape and size of the variable cross section so that its elastic section modulus S = Ilc varies along the length of the beam in the same way as IMI , it is possible to design beams for which the maximum nonnal stre.;;s in each section is equal to the allowable stress of the material. Such beams are said to be of COIlStalll strength.

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5.2. Shear and Bendlng-Momoot Diagrams

5.2. SHEAR AND BE NDING-MOMENT DIAGRAMS

As indicated in Sec. 5. 1. the determination of the maximum absolute values of the s h~ar and of the bending moment in a beam are greatly facilitated if V and M are plotted against the distance x measured from one end of the beam. Besides, as you wilJ see in Chap. 9. the knowledge of M as a function of.r is essential to the determination of the de!lection of a beam. In the examples and sample problems of this section, the shear and bending-moment diagrams wi ll be obtained by determining the values of V and M at selected points o f the beam. These values wi lJ be found in the usual way. i.c .. by passing a section through the point where they are to be determined (Fi g. 5.6a) and considering the equilibrium of the portion o f beam located on either side o f the section (Fig. 5.6b). Since the shear force s V and V' have opposite senses, recording the shear at point C with an up or down arrow would be meaningless, unless wc indicated at the same time which of the free bodies AC and eB we are considering. For thi s reason. the shear V will be recorded with a sign: a plll.I' ~'ig l! if the shearing forces are directed as shown in Fig. 5.6b, and a millll.l· sign otherwise. A si milar convention wi ll apply for the bending moment M. It wi ll be considered as positive if the bending couples are directed as shown in that figure, and negative otherwise.t Summarizing the sign conventions we have presented, we state: The ~'hear V lIIul the belldillg moment M at a gil'ell poilll of a heam (Ire ~'lI id to he positive whell the internal foren alld couples acting Oil 1'(1('11 portion oj till' 1I1'{1I1I aI''' dirl'rll'fl IH .1·IIml·" in Fig . 5. 7f1 These convelllit)fls can be more easily remembered if we note that l.

2.

",

I' .,

B

A

1,,1

" " V

II! )

B, l'!

V'

~t '

(L

B

R, Fig, 5.6

The shear tit till)' given point of II beam is 1}()Jirive whell the external force.\' (loat!.l· and reactiolls) acring Oil/he heall//end to ~'hetlr off rh{' beam at (IUlt point as indicated in Fig. 5.7b. Th e he/ulin!: mO/llent al allY given point of a beam i.l· p(uitive II'hell Ihe exlernal fOIT'es acting 011 the bell/II telld to bend the beam m that point as indicated ill Fig. 5.7c.

It is also of help to note that the situation described in Fig. 5.7, in which the va lu e~ of the shear and of the bending mome nt are positive, is preci sely the si tuation that occurs in the left half of a simply supported beam carrying a single concentrated load at its midpoin!. llli s particu lar case is fu lly disc ussed in the next example.

~lj'Jl~ "

(II ) 1"I{' rnal fnr<'Cs (rosin\'!• • hf'lIr aud I'olitkf' hf'n,lin); """'](' n!)

(h ) E1T"el nfexl" mal foret's

Fig. 5.7

t Notr that thi s ('oDYell1ion is the same Ihut we usl'd earlil'T in SCl'. ~.2

Ip(Jsili\'i'sh"a d

If')

Elr{'~~

ot' " xlc m ~l fOI(.>"

' pmitivo,I", ,,,!;ng"''''''''nl )

311

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EXAMPLE 5.01 Draw ihe shear and bending-moment diagrams for a simply supponed beam AB of span L subjected to a single concentrated load P m it midpoint C (Fig. 5.8).

"

Fig. 5.8

We first determine Lhe. reactions at the suppons from the free-body diagram of the entire beam (Fig. 5.9a): we find that the magnitude of each reaction is equal to 1'/2.

Next we cut the beam at a point D between A and C and draw the free-body diagrams of AD and DB (Fig. 5.9b). AJ.mmillg Ihal shearund bending mamenl are pIJSilil'e. we direct dIe internal forces V and V' and the internal couples M and 1\1 ' as indicated in Fi g. 5.7(/. Considering the free body AD and writing that the sum of the venical components and the sum of the moments about D of the forces Ileting on the free body are zero. we find V = + 1'/ 2 lUld M = + Px/2. BOlh the

shear and the bending moment are therefore positive: dlis may be checked by observing dlat the reaction at A tends to shear ofT and to bend the beam at D as indicated in Figs. 5.7b and ('. We now plO! V and M between A and C (Figs. 5.9d and e): Ihe shear has a constant value V = 1'/2 , while the bending moment increases linearly from M = 0 al x = 0 10 M = PL/ 4 at x = L/2.

CUlling. now. the beam at a point £ between C and Band considering the free body EB (Fig. 5.9('), we write that the sum of the vertical components and the sum of the moments about E of the forces acting on the free body are zero. We obtain V = - p/2 and M = P(L - x )/2. The shear is therefore neg· ative and the bending moment positive: this can be checked by observing that the reaction at B bends the beam at E as in· dicated in Fig. 5.7(" but lends to shear it off in a manner opposite 10 that shown in Fig. S.7b. We can complele. now. Ihe shear and bending-momelll diagrams of Figs. 5.9d and e; the shear has a conslant value V = - 1'/ 2 between C and B, while the bending moment decreases linearly fmm M = PL/4 at x = L/ 21OM = Oatx = L.

v E .r '?I_=.C-=~

I)"

"(JI~ _ ,~r (d

tP~" ---+L I --!I LL I ___ (iI)

Fig. 5.9

312

Rn'" ~I'

'"'

-1 1'

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5.2. Shear IIn
We Ilote flOllltile fOlegoiug l:Xalilplc that. whe u a beam is subj ectcll Llnly to concentrated IGads. th e shear is ~o n it ant between loads and the bending moment varies linearl y between loads. In Sl1ch situations , therefore. the shear and bending-moment di agrams can easily be d rawn, once the va lue!; of V a nd At ha ve been obtained at sections selected just to the left a nd juSt to the right o f the point~ whe re the load s and reactions are ap plied (sec. Sample Prob. 5.1).

EXAMPLE 5.0 2 Draw the shear and bending.momelll diagrams for a cumilever beam AB of span L ~upponing a uniformly dislfiomcd load 10

tFiB "_10)

J1111 [1111 (illllW III[ I·

L

B

•

Fig. 5.10 Wt: l'UI Ult: bealll al a poilll C ht:lweell .4 alltJ JJ amJ tJrJw

Ihe free.horly niaeram nf Ar (Fie_ 'i 1111). rlin~Clin e \' :Illrl M as indicated in Fig. 5.70. Denoting by x the distance from A to C and re placing the dis lrI buted load ovcr AC by its resultun! wx applied allho::. midlxlilll of AC. we wrile

II r I

.L

•I .._T I

III)

+ t:iF, =O:

- mr - V""'O

V=

- W .l"

" + ~!i:Mc -{) :

wx(f)

+ AI - ()

AI -

_ ~'l'.,.2

IL,

"

2

II,

We IIOl e Ih ..1 die shear diagIalll is leplc.5cllle d by all obl ique stmight line (Fig. S. ll hj and the bending-moment diagram by a parabola (Fig. S. lle). The maximum vaJu e~ of V and .M both OCCI!f at 11 , where we h~ve

Fig. 5.11

313

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SAMPLE PROBLEM 5.1 For the timber tJeam and loading sho'Nn. draw the shear and bending-moment diagram~ and determine the maximum normal stress due \0 bending.

SOLUTION Reactions. Considering the entire beam as a free body. we find

RD = 14 kN

t

Sltl'ar a nd Bending.i\Ioment Diagram.. . We first determine the inter. nal forces jusl10 the right of the 2O-kN load at A. Considering the stub of beam

10 the left of section I as a free body and a;;sUI:ling V and At IG be positive (accOJdillg to [he staJldanl convention), we wlite

+tIf".=U : + 1::£M, = 0

- 20 kN -

VI = U

(20 kN)(O ru)

+ M,

0

=

th~

We next ccnsidcr as a Irre booy the portion 01 beam \0

leli 01 section 2

and write

+t:u--. == + ~ :£.\12

0:

- 0

v!

- 20 kN - V1 = 0

(20 kN )(2.5 m) + M~ - 0

= - 10 k)l - 50kN· m

/If] -

TIlt: shear and bt:mling murnt:nt a1 sectiuns J. 4, 5, and 6 an: lklerlllim:li in a similar way from the free-body diagf
v3 v4 =

+26kN

Mj

- so kN·

J\l

+ 26kN

1'.1 .. = +28 kN .

Ell

V, = - 14 kl\ V6"='

- ]4kl\

-

M,=+28kN·m Mo = 0

For several of the ku tcr secti ons. the results ma}' be more easily obtained by considering a:; a free body the pcrtion of the beam to the right of the section. For example. for the portion of the beam to the right of section 4, we have

+ t:!: F) = 0

v. -

+ ~~ M~"'"

- M..

0:

40 kN + 14 kN = + ( 14 kN)(2 m) =

0 0

V4 = + 26 kN +28kN'm

,114=

We can now plot the six points ~holVn 011 the shear and bending-moment diagrams. As ilxlicmecl earlier in this .eelion, the ~he:!r is of constant value beI\\!p.r,n cOl1cp.ntr.' kd lo~d s, ~nil rh e ht:nding momCOI v:"lrir.s lin p. Hriy; \liP' ohl:1in therefore the ~he~r and bending-moment diagrams shown. l\IaXinlUnl Normal Stress. It occurs at B, where IMI is largest. W~ use Eq. (5 .4 ) to determine the sC1:tion mooulus of the beam:

S = ~ bh2 = ~(O.O~O m)(0.2S0mf = 833.33 X 10 Substitutinj.! this value amI

IM"I

m.!

X 10J N 'm into Eq. (5.]):

(i OX IOl N · m )

60.00 X IO"Pa 833.33 X 10 6 Max im um no nml .>tress in the beam = 60.0 Ml'a ...

G," = - , - =

314

IMI = IMRI = so

6

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SAMPLE PROBLEM 5.2

W"p.

' '" 'I.sr' I 3Cl"rrT'"'l'r~ nT~

f

n

3[,

'I

3 kl,w1t

1 I

I

~lk~l . HI

2:

C

.I,

3 J ~ klp II

'f! : ,)

\l ll lj l ll ,\

t 11~~ii :

:3 B t

-, D

III L,p' : :U ki .....

:

~jll l l ll lIi :: ,'

,

[)ia g nnL~

' -'II,.

3x V - 0 3.t(lx) + M = 0

I tYr. - 0: 0:

- P:MJ= M

~,---Ix jl l ll l lI l C'e r':" L, I":''P'~ _ ) I" JI"" ,Gr, - .j -, - -,

~i-I_...JI '

,If

B is determined by cons ider-

I

4

f

:

S ]w a r a nd Bend ing_MoJlll" nl

lit

IIITHe

'

T -

The IO-kip loud is rcplnccd by nn Cqlllv-

Pro", A to C. We delenniJlc th~ in lernal forces al adistance_l from POllit A by considering the poltion of beam to the left of section I. That part of the distributed load acting on the free body is replaced by its resultant, and we

i

' -I v

2./ !:in, I

1 1 kip"

Elluh"a lcllt Lo adin g or Bl"am .

alent force-coup le s~s[em at D. The reaction ing the beam as 1 free body. /l.

c,,;,.''''.)''

.

SOLUnON

,00

1

1

,-- t-

.,UlJCIUre sh()wll consist. uf a W10 X 11 2 rulk'tl-Mcd bt:~1T1 AD anu uf 1'.... 0 short membt>rs wellled togelher and 10 the beam. (a) Draw Ihe shear :md bending-moment diagrams for the beam and the given loading. (b) Determine the maximum namlal stress in sections just 10 Ih.e left and just 10 the right of poinlD.

c

A

Th~

-MT

'----=---;---;----'--T-' ft 141> kil"

Since the frcc·lxxly diagmm , hown can be used for all vclues of x smaller than 8 ft . Ihe expressions obtained for Valid M are valid in thc re}! ion 0 < x < 8 fL Pru", C I" D. Considering th" portion of beam 10 the left of sec ti on 2 and again replacing Ihe distnbuted load by its resultant. we obtain

+ t LF, = 0: 0:

- 24 - V = 0 24(.\' - 4) +M = O

- ~ IM! =

V = - 24 kips /It = 96 - 24x

Thcse expressions are valid in the region 8 ft

<x <

- 3 15 kip . ft /

kip·ft

I I ft.

FW III D IIJ /J.. Us ing the position of be'U11to the left of ,;ection 3. we· ob· tain for the region I I ft < _\ < I (j ft

V = - 34 kips

M = 226 - 34x

kip· ft

~ h~ar and "",ndin~_mom~nt diagram~ fm thee.ntire heanl c an nnw he ]'llmled. We nOle thai Ihe couple of moment 20 kip' ft applied at poi III D intro· duces a discontinuit) into the bending-moment diagram.

The

b. MaximuDI Normal St ress 10 Ihe Lt' rt :lnd Rigbt or Poinl D. From AppendIX C we find Ihal for the W JO x 11 2 rolled·sted ~ hape, S = 116 in '

:lbom the X-X :lxis.

'ra II,,, IfJI fJf D : Wt: Ildvt: IMI = 168 kip' ft tuting for IMI and S i11l0 Eq. (5.3). we write

- 168 kir

3->, kip!'

\I -

M = - 15 .,.2kip· ft

IMI u ,,

= -S - ~

2016 kip' in 126 in'

16.01l ksi

=

20 16 kip' in. Sub~ li ­

u « = 16.0()lsi ....

Ta l lurigh t flJD: W"ha\'C IMI 148kip·ft = 1776kip · in. Sub_ sti tuting for All and S into Eq, (5.3). we write

IMI

IT..

=5 =

1776 kip' in. 126 in ~

"" 14.1\1 kst

u~

= 14. 10 t SJ ..

315

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PROBLEMS

5 .1 through 5 .6 For the beam and loading show~. (m drnw the shear and bending. moment diagram~. (b) determine the equations of the s hear an
"

~ IIIIIIIIIIIII ~ 1 Flg. PS.1

L

I

c

"

t:=:=L"-

I

Fig. PS.2

'::::5BIITIIR I-

Fig. PS.3

Fig. ps.s

L-

B

Fig. PS. 4

Fig. PS.6 5 .7 and 5.8 Dmw [ftc shellr lind bending-moment dillgram s for the bellIn and loadin,~ , hown. and detennine the max imum absolute value (iI) of the shear. (b) of lhe ht'nding moment 14k'l2.1k,\ 24.:-'

!!Ikr..

t~"L,l iJ' 0.;:;",

Fig. PS.7

316

Fig. PS.S

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5 .9 and 5.10 Draw the shea] a ll d loending-momem diagram s tor the beam and loading shown. and detennine the maximum absolu te v~lue (a) of the shear. (b) of the bending moment.

Problem.

•

1111I111 Ie "."" - - - -6ft--- -3ft -

Ora\\, the

1

Fig. P5.10

Fig. PS.9

5.11 and 5.1 ::Z

•

~he:.r

and bending-moment

di~gralll~

for the

beam
ttlll l!]

1001Ib

31.,\

31.:-<

'- ~8 ; .. _______

~IU

A

D, ' I ' "'''''h ~I~.-I

--=--!-;; . :!oJO 300 t!lin .

n

•

CI

20\}

.>CAl

Oi",,'n.i<><,. inlnm

l2.i ll .

Fill. PS.11

Fill. PS.t2

5 .1 3 and 5. 14 Assuming tlwt the reaction of the ground [0 be un ifonllly dis tributed. tlr;)\\' the shear and bending-moment diagr~ms for the beam A D and determine the maximum absolute value (a" of the Shelf. (b) of the bending moment

I~kl'l

C

t

!i~1\

D

A

8

1- 1-0., . , 1 1 0.3111

Fig. PS.1 3

0.3m

"

(iTI1c

i!lllr~

3[,

Fig. PS.14

3fi

Jffi

D

L J 1

2. kh,/r,

3f,

3n

B

1

317

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318

""'01),,13 ond De31gn o f B ....ms fo r Bonding

5 .15 and 5. '6 Fur the b.:am anu k.KIUlIlg ,hown. delt:rmine tht: maxi· mum normal sires.; due 10 bending on a trans verse seclion :It C. 3k~

Fig. PS.tS

Jk~

Fig. P5.16

5 .17 Foc the beam and loading shown. determine the maximum nonnal stress due to bendmg on a transverse section at C.

.1",p~·11

A~C~D?'1E:1'::::l"llirl" I

IVl6 X 7i

B

t) ~ft-. ~ 1:-~5 ft

,sr,

Fig. PS.17

5 .18 Foc the beam and loading shown. determine the maximum nonnal stress due te bendlllg on section (I-U.

" 1!ii!~=~~!!=~!!!~ '~'~ ;n? 52 ~ ~ 1,1

'm---l 5Ci'I}.Snl=4111 - --

I

Fig. P5.18

5 .19 and 5 .20 For rhe. he.~m ~nrl In.1rl1l1E ~h()wn. dt';re.rmine. rhe mui. mum normal stres; due to bending on a transverse section at C. 2.1 2.5 H.

~.'\'

c A

[ Fig. P5.19

ill

h)

~l\

~:-.

Iv l,IFIe

6 @0.3jS

FIg. PS.20

10 I...'\'

m~ 2.2.5 m.:..r

B

I

S2()(1 X

27. ~

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5 .21 Draw lht: :;Ilt:ar am] bt:mJing-rnurm:1ll L1iagrdllls fur the bc,uTl anLl loading ~holVn and determine tlle maximum nonnal stress due to bending.

1.;

,

n

I

r,

211

319

~!"

"

J L",---J-

Problem.

I

St2 X 35

HI

Fig. PS.21

5.22 and 5.23 Dra\\ the shear and bending-moment diagrams for the beam and loading shown and uetennine the maximum normal stress due [0 bending

W3J ()

x 3/i. 7

;k

Fi g . PS.l2

Ii--o,c,c,,- + ---',c,c,,-

.... ~ [----J

\\'200 X :!:! . ~

:! "'

Fig. PS.23

5 .24 and 5.25 Draw Ihe sheill" and bending-moment diagnmls for Ihe beal!"! and loading shown and delennine the maximum normal stress due to bending

; ~il'"'

I

\\,1 4 X 21 Fig. PS.24

Fig. PS.25

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320

Analysl!l and Design of Beams for Benalng

5.26 Knowing thai W = 12 kN, draw Ihe shear and bending-momenl diagrams for beam AH and determine Ihe maximum nonnal stres> due to bending.

5 .27 Detemline (tI) Ihe magnilude of the counterweighl Wfor which the maximum absolute value of the bending moment in the beam is as small as possible. (b) Ihe corres)XJnding maximum nonnal stress due to bending. (Hillt: Draw Ihe bending-moment diagram andequJle the absolule values oflhe largest posilive and negative bending moments oblained.)

C

E

D

\V31O X 23.1'1 B

t\

Im

U-Ur 1m

1m

J:

1m

Fig. PS.26 and PS.27

5.28 Knowing Ihal P = Q = 480 N. determine (a) Ihe distance (/ fur whkh thc absolule value uf the bending moment ill the beam is as small as possible. (b) the corres)XJl1ding maximnm normal stress dne 10 bending. (See hint of Prob. 5.27.)

Fig. PS.28

5.29

Solve Prob. 5.28. assnming that P = 480 Nand Q = 320 N.

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5.30 Determine (a) the distance (/ for which the absolute value of the bending moment in the beam is as small as possible. (b) the corresponding maximum normal stress due to bending. (See hint of Prob. 5.27.)

WI4

x 22

Fig. PS.30

5.31 Determine (a) the distance (/ for which the absolute value of the bending moment in the beam is as small as possible. (b) the corresponding maximum normal stress due to bending. (See hint of Prob. 5.27.)

" 1 - - - - -1$

f t - - ---I

Fig. P5.31

5.32 A solid steel rod of diameter J is supported as s hown. Knowing tiwi for steel y = 490 lb/ ft]. determine the smallest diameter J that can be used if the noml::d stress due to bending is not \0 exceed 4 ksi. d A

B

H

----t~-1U-n----1+ ~

;;(1

Fig. P5.32

5 .33 A solid sleel bar has a square cross section of side b and is supported as shown. Knowing Ihm for sled p = 7860 kg/m] , determine the dimension b for whkh the maximum normal stress due to bending is (u) 10 MP:l. (h) 50 MPa.

C

Fig. P5.33

D

B

Probloms

321

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322

Analysis and Design of Beams for Bend ing

I -""' , f---i i11'

5.3. RELATIONS AMONG LOAD, SHEAR, AND BENDING MOMENT

\Vhen a beam carries more than two or three concentrated loads, or when it carries distributed loads, the method outlined in Sec. 5.2 for pl aning ~ he:lr and bending moment can prove quite c umbersome. The construction of the shear diagram and. especially, of the bendingmoment diagram wi ll be greatl y facilitated if certain relations existing among load. shear, and bending moment are taken into consideration. Let us consider a simply supported beam AB carrying a distributed load IV per unit length (Fi g. 5. 12a), and let C and C' be two point s of the beam at a di stance a.x from each othe r. The shear and bend ing moment at C will be denoted by V and M, respectively, and will be assumed positive: the shear lind bending moment at C' will be denoted by V + a. V and M + a.M. We now detach the portion of beam ec' and draw its free-body diagram (Fi g. 5. l 2b). The forces exerted on the free body include a load of magnitude w a.x and internal force s and couplcs at C and C' . Since shear and bending mome nt have been assumed positive. the forces and couples will be directed as shown in the fi gure. Relations between Load a nd Shear. Writing that the sum of the vertical components of the forcc .~ acti ng on the free body CC' is zero, we have

a.x = 0 - w a.x members of the e(luation by a.x and then letti ng a.x :JpV - (V + LI. V) - w 6.V =

Dividing both prom: h zero, we obtain Fig. 5.12

(1\1 ~

,b

=

-·w

(5.5)

Equation (5 .5) indicates that. for a beam loaded as show n in Fig. 5. 12£1, the slope dVldx of the shear curve is negative; the numerical value of the slope at any point is equal to the load per unit length at that point. lntegraling (5 .5) between points C and D. we write VD

VD

-

-

Vc "" -

'" r"',' wdx

Vc = - (area under load curve between Cand D )

(5,6')

Note that thi s result cou ld also have been obtained by considering the C(luilibrium of the portion of beam CD, since the area under the load curve represents the tOlal loud applied between C and D. It sh()uld be observed Ihat Eq. (5.5) is not valid at 11 p()int where a concentrated load is applied; the shear curve is discontinllOlls at such a point, as seen in Sec . 5.2. Similarl y, Eqs. (5.6) and (5 .6') cease 10 be valid when concentrated loads are applied between C and D, si nce they do not take into account the sudden change in shear caused by a concentrated load. Equations (5.6) and (5.6'), therefore, should be applied only be tween successive concentrated loads.

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RelaliolLs between S hea r alld Heut.ling MOllIelll. Rt:tumillg to the frcc-body diagram o r Fi g_ 5 _12b, and writing now that tht' s um Qf thf' moments about C' is zero. we hlve

(.\1 + AM ) - M - V
= V 6.x -

AI"

2

5.3. Relnllon3 among Load, Shen', and Bending McmlHll

= 0

ltO (A.lV

2

'

Dividing both members o r thf' equation by .1.x and then lelling .ix approac h zero. we obtain

tiM = V dx

(5,7)

Equution (5.7) indicates thnt the slope dM/dx of the bcnding-mClment curve i~ equal to the value o f the shear. This is true al allY iJOillt wl n:re the shear has a wcll -de litled valuc. i.e ., m any point where nu coneentnned load is applied. Equal1011 (5.7) also ,how~ thm V "" 0 at points whl're M is maximum. Thi s prupt!Tly fm:ilitates the lletcnninatiun uf the points where the beam is likely to rail under b ~ lldin g . Integnning (5.7) between points C and D. we write

(5 ,8) Mn - AIr = area under shear eurve between C and D

(5 ,8 ')

~o te thm the area under (he shear curve s hould be considered positive whe re the shelr is positive and negative where the shear is negative. &:juations 1.5 .8) anll (5 .8 ') are v;t lid even when concentr;lted loads are ap plied between C and D. as long as the shear clin e hls bcen correcil y dra wn. The equations cease to be valid. however. if a couple is a pplied at a point between C and D, since they do not take into account the stldden c hange in bending moment callscd by a couple (see Sample Prob. 5.6).

EXAMPLE 5.03 Draw the shenr and bcnding·m()menl dilgrams for the ~imply .upponed beam shown in Fi~. 5.13 and delerniine the maxi ilium value of the belld ing momen1. A

I 111111111111 H

- - -L- -- r

From the free-body dia)1;ram of the entire beam, we delemu nc the magnitu<Je of the reactior$ at the ~upporLs.

"I -'Sex t. we draw the. shear diagmm. Close to the cn<J A of the beam, the shear i ~ equal to R" . that i•. to ~l/)L. as we can eheck by cOllsideri ng as u free body a \ery", mall portioll of the beam.

I 1111111 11111I"

L \,,"~ u L

Fig. 5.13

I'l n<= {IIJ

323

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U~ill!!

Eq. (5.6), We thell dete rmine ule s heru Va l ruly diSlilli ce

x from A: we wri te

V""

v- V" '" - fWd.' = - 1Il\" v" - IV.\" = i wL" - 1f'X"" w(}L

- x)

The shear cw,'e is thus '.n obliq ue straight li ne which crosses thexaxls at .l· = LIZ (Fig. :'i . 1411). Considering. now. the bending

mom~nt.

we fi rs t observe th:ll M ~ = O. The \-.llue M of tile

bending mo ment at any distnnce x from A may then be ob-

ta.i ned from Eq. (5.8): we Iwve

I'Vd.t " 111 = f\IJ(lL - .f) ,l... = } w(Lx _

- tIC'/.

M - M ,,=

,J)

" curve i<; a parnbola The maxim um value 'nle bending-moment of the bending 1!1On"lenl occurs When .r = LIZ. si n~e V (and Ihu~ dM /dr);~ "2ero for th~'1 value of x. Sub.tituting x = LI 2 in the Imil cLJuation . we obtai n Mow.. = wL'/8 ( Fi g. 5. 14b).

i''''~ ~L

I. (b )

Fig. 5.14

In mOSI engineerin g. applications. one needs to know the value. of Ihe bending moment onl y ;.it a rew specilic poi n l~ . Once the s hear diagram has been drawn. and after M has been de termined at one of the ends or the beam. the value of the bendtng moment can then be obtained at any g i ~c n point by computing the arca unde r thc .~ hc:l r cur ve find lIii llg Eq. (5 .8'). FOJ insuUlcc, sillce MA - 0 for the beau) of Exam ple 5 .03, the m a}..imulll va lue of Ihe be nding nlomc nt fo r th at beam c:.m be obt:lined , imply by mells uring lhc area of the shaded triangle in the shear diagram of Fig . 5.1 4a. We have 'VI ·ml.:<

I L wL 222

w L! S

= --- ~ -

We. nOlf: thil\. in lhi ~ ex ~ mrl e, The 10:114 curve is il hori].!)nl:11 ~ tr:l i g h l line. the shear c urve an oblique str.light line, and the bendi ng-momem curvc
324

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SAMPLE PROBLEM 5.3 Draw the shear and bending-momem diagrams for the beam and loading shown.

SOLUTION Reacti ons.

Considering the entire beam a.l a free body.

v,'e,

write

- "iHI A = 0: 0(24 f\ ) - (20 kips) (6 0) - ( II kips)( 14 ft) - ( llklps )( 28 0) = 0 [J

-t 'iF, =

0:

"":"'''i.F, = O

D = 2 6kip~t

= +2(> kips

A, - 20 kips - 12 kips + 26 kips - 12 kips = 0 AI = + 18 kips A " = 18 kipst A,= O A, = O

We .lisa note that at hoth A and £ the bencfing moment is zero: thus. two points lmtiicat,!d by dots, are obtained on the bending-moment diagram. Shear Diagram. Since dV/dx = - w . we find that between concentratoo and rea.."tiolls the slope of the shear diagram is zero (i.e , the shear i, COlIstanll. The shear at any point is determined by dividing the beam inlO tWO pal S and considering either p:lIt as a free body. For example. using the punlon of beam tll (he leli of section I , we obtatn the , hear between Band C' load~

-i

+ 18

~ips -

2Okips -

V= 0

V = - 2kips

We also lind that the shear is + 12 kips jlL~t to the right of D 111d zero at end E. Since the slope tlV!dx = - tv is constant between D and E. the sJtear diagram belween these two points is a stlaig~1 Ii lie.

V ~ l-; r·!l

,,'

Dcnd ing-i\l()IIlt'lIt Diagn llll. (-1- 108)

~ 1~

1"' ~I!!

/

- 16)

-,

-I \l l l~

'iF,. = 0:

'"

j

,

].10)

the same two points. For convenience. the area of each portion of the shear diagram is computed and l S indicated in parentheses on the diagram. Since tbe bendm!; moment M A m the bft end is known to be zero, we wrile

MB - M.\ = + 108

,liB = + IOBkip'

M , - M tJ = - 16

M, =

Mv - Me = - 140

Itl.'l

M E

- 4S

W..:; ,ecali Illat the area unue, the s hea,

r:lIT\'e helween lwn tylints is eqml tfl th~ chaneI' in hemline mnment hctwf"en

- .M D = + 48

Mv

~

n

+ 92 kip ' fl -

48 kip . ft

,WE = 0

Since Me is known 10 be zero. a check of the l:ompUlalions is obtaint!d. Between the concemraleu loads and reactions Ihe shear is constant Ihus, the slope dM Idx is constant Jnd the bending-moment diagram IS drawn by con· necting Ihe known points with .maigh l Ijne~. Belwecn D and E where Ihe shelr diagram IS all oblique straight line, the bending-moment diagram is a parabola. From the V an d M diagrams we note that VIII.,. = 18 kips anu M,,,"-, = 108 l ip' fl.

325

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2U).;N/m

SAMPLE PROBLEM 5.4

Al'~i!~~~.i~=B~"""~i'C

6m ---l-3m

The W360 X 79 rolled-steel beamAC is simply s upponed and canies the uniformly distributed load shown. Draw the shear and bending.moment diagrams for the beam and determine the location and magnitude of the maximum normal stress due to bending.

SOLUTION Reactions.

I'"

R., ""- 80kNt

)~I;;;!;I;;:;,1"", I i;;;! 1I~1~1"""""=IiC t M.lk-"

B

I(d
Re=40kNt

Shtar Diagram. The shear just to the right of A is v.~ = + 80 kN . Since the change in shear between two points is eqllal to minus the area under the load curve between the same two points. we obtain VB by writing VB - V" = - (20 kN/ m)( 6 m) = - 120 kN VB = - 120 + VA = - 120 + 80 = - 40 kN

I

V Ml kN

a

Considering the entire beam as a free body. we find

A 1+1 60 )

D

---l

F

B I,

C (- 120)

(-,W)

I 6m H

The slope dV/dx = - 10 being constant between A and B. the shear diagram between these two points is represented by a ,traight line. Between Band C. the area under the load curve is zero: thcrefore. - -1O kN

"

' '~ ''''I~ I I II

r 12U kN . III

and the shear is constant between Band C. Bending-Moment Diagram. We note that the bending moment at each end of the beam is zero. In order to dctcnnine the maximum bending moment. we locate the section D of the beam where V = O. We wri te Vo - V,\ = - U}x 0 - 80 kN = - (20 kN/ m) x

',' ''------'----'-----''''--,.

= 4m 4

and. solving for x:

The maximum bending momen t occurs at point D. where we have d.H / dx = V = O. The areas of the various portions of thc shear diagram are computcd and arc givcn (in parentheses) on the diagr
Mn - MA = + .160kN· m MB - Mo = - 40 kN . m M c - M il = - .120kN· m

M v =+ 160kN'm

MB ~ +120 kN' m Me

=0

The bending. moment diagram consists of an atC of parabola followed by a segment of straight line: the slope of the parabola at Ais equal to the value of V at that point. Maximu m Nonn al Str('ss. It occurs at D , where IMI is largest. From Appendix C we find that for a W360 X 79 rolled-steel slmpe. S = 1280 mm 3 aoollt a horizont:!1 axis. Subslitllting thi ~ value and IMI = IMol = 160 x IO~ N . minto Eq. (5.3). we ""rite

[Mol 160 X 10J N . T1l a-.,=-S-= 1280 X ](j 6 m ~ Ml,,(i lll lI ITJ nomlal

326

~tres"

125.0 X

IO~

Pa

in Ihe beam = 125 .0 MPll ...

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SAMPLE PROBLEM 5.5 Skt:ICh llie sht:ar muJ bt:lllling-lTlulTlt:1ll diagram.'> fur Iht: call1ilt:ver bt:arrl .'>llUwn.

SOLUTION Shear Diagraln Al Ihe free end of Ihe beam. we find VA = O. Between A and B. Ihe area under Ihe load clirve i., t W<}il : we find V" by writing

\,

/

Between 8 and C the beam i~ not loaded: thus Vc = VB' AtA . wehav~ II! = !vo and, acccrding to Eq. (5.5). thO' dope of thO' ~hear curve i~
Bending.~ laIlUlll

of

th~

My -

= - b('oa2

filA

MB

=

M, - M il = - iwua(L - il)

Me = - broa(JL - a) The s letch of the b..-mdmg_moment di::.gram i~ completed by recalling that dM /dx = V. We find that retween A. aad B the diagram is represented by a wbit curve with zero slope al A. and betwe~n Band C by a straight line.

SAMPLE PROBLEM 5.6 The ,imple beam AC is loaded by a couple of moment T [jppli~d m fJ(lint B. Dmw the shear and bending-moment diagnuns of the beam.

SOLUTION The entire beam is taken as

J.

R,

free hotly, and we obtain T

=- t L

The ';hear at lUI} section is constant and equal 10 '/ / L. Since a couple i ~ ~p­ plied aI B. th~ bending-moment diagram is discontinuous atH: it is reprerenled by tWO oblique straight Jines and decreases suddenly at !J b)' an amount equal to 1.

327

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PROBLEMS

5.34

Using the method of See. 5.3. m lve Prob. 5.1a.

5.35

Using the method of Sec. 5.3. solve Prob. 5.2a.

5.36

Using the method of See. 5.3. solve Prob. 5.3(/.

5.37

Using the method of Sec. 5.3. solve Prob. 5.4(/.

5.38

Using the method of See. 5.3. solve Prob. S.5a.

5.39

Using the method of Sec. 5.3. solve Prob. 5.6<1.

5.40

Using the method of See. 5.3. solve Prob. 5.7.

5 .41

Using the method of Sec. 5.3. solve Prob. 5.8.

5.42

Using the method of Sec. 5.3. solve Prob. 5.9.

5 .43

Using the method of Sec. 5.3. solve Prob. 5.10.

5.44 and 5.45 Draw the shear and bending-momel1l diagrams for the beam and loading shown. and determine the maximum abso lute value (a) of the shear. (b) of the bending moment.

Fig. PS.44

328

Fig. PS.4S

5.46

Using the method of Sec. 5.3. solve Prob. 5.15.

5.47

Using the method of Sec. 5.3. solve Prob. 5. 16.

5.48

Using the method of Sec. 5.3. solve Prob. 5.17.

5.49

Using the method of Sec. 5.3. solve Prob. S. IS.

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5.50 and 5.51 Delermine (a) the equations of Ihe shear and bending. momem curves for the beam and loading shown, (b) the maximum absolute value of the bending momem in Ihe beam.

Problems

~r I

L

Fig. PS.51

Fig. PS.SO

5.52 For the beam and loading shown, delermine the equations of Ihe shear and bending.momem curves and the maximum absolule vallie of the bending moment in the beam, knowing Ihat (a) k ~ L (b) k = 0.5.

'"I

w

--------:::.-1'

Orl

-k ll" ~I_ _ _ L _ _ _ _

L

Fig. PS.S2

5.53 Determine (a) Ihe equalions of Ihe shear and bending.momem curves for the beam and loading shown, (b) the maximum absolule value of the bending moment in the beam.

A

I----L--Fig. PS.S3

329

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330

Analysis and Design of Beams for Bending

5.54 and 5.55 Draw the shear and bending-moment diagrams for the beam and loading shown and determine the maximum normal stress due 10 bending.

161.:-.11",

3 kip~ft

I

SI50 X IS6

1IIIIIIIIIe

I -

Fig. PS.54

'Ok'1' "

,\~~~") g}o;" - 8 £1- --

Fig. P S.SS

L

4 1"1 -

-:1 il~

5.56 and 5.57 Draw the shear and bending-moment diagrams for the beam and loading shown and determine the maximum normal stress due [0 bending.

W250 X ./\l. 1

Fig. P S.56

Fig. P5. 57

5.58 and 5.59 Draw the shear and bending-moment diagrams for the beam and loading shown and detemline the maximum normal slress due 10 bending.

2LN J4011"lI

h kil'~f!

A

,

~C

.

0

~

2 ft - - - fi 11------=-1 2 fl

Fig. P S.58

:r \\'12><26

,\

ell II 1 1ITfllil I

B

O LI

~ - , ",----'-I I~ I",

Fig. PS.S9

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5 .60 and 5.61 Knowing that beam AB is in equilibrium untler the loatl~ ing shown. draw me shear and bending. moment diagr,uns and determine the maximum normal stress due to bentling.

tOO

" " ~ 5u 1i >lU

1tr..1" ,

cI_ I I_ I III

A

'"

A8

(b

I

I I I I I I I I I I L"11200 '

A

22-'

1--0.311,-1--0.4 ",--!-0.3 "'~~

Fig. PS.62

"5.63 The beam AiJ su pports two concent rated loads P and Q. The normal stress due 10 bending on the bonom edge of the beanl is +55 MPa at D and +37.5 MPa at P. ({I) Draw the shear and bending-mome11l diagrams for the beam. (b) Determ ine the maximum nomlal stres~ due 10 bending that occurs in the beam.

0.2 m

i0-5'' l ITI""'' '-IQ c

V

~ 11.4

III

Fig. PS.63

L

Fig. PS.61

"5.62 Beam AB supportS a uniformly distributed load of 2 kN/m and two concentrated loads P and Q. Il has been experimentall y determined that the nomlal stress due to bending in the bottom edge of the beam is - 56.9 MPa at A and - 29.9 MPa at C. Draw the shear and bending-moment diagrams for the beam and determine the magnitudes of the loads P and Q.

F.

2.J nllll

I~

F

lj 0.3 "'

f)

!60ll""

I

till.

;""

1.2

Fig. PS.60

,\

Problems

il<8 rt

~ ".

C - l - - 1.2 rt

--l

1::"1

331

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332

Analysis and Design of Beams for Bending

'5.64 The beam AB suppons a uniformly dis tributed load of 480 lb/n and IWO concentrated loads P and Q. The Jl0rmal stress due to bending on the bollom edge of the lower flange is +14.&5 ksi at D and +10.65 ksi al E. (a) Draw the shear and bending-momem diagrams for Ihe beam. (b) Determine the maximum nomlal stress due to bending Ihat occurs in the beam. Q

I'

I

WS

x

31

In

J Ii 1.5 ft

L5 1l

f - - - sn - -"" Fig. PS.64

5.4. DESIGN OF PRISMATIC BEAMS FOR BENDING

As indicated in Sec. 5.1, the des ign ofa beam is usually controlled by .. of the bending moment that will the maximum absolute value IMlm occur in the be3m. T he largest normal stress am in the beam is found at the surface of the beam in Ihe critical section where IMlmax occurs and can be obtained by substituting IMlmax fo r IMI in Eq. (5 .1 ) or Eq. (5 .3).t We wri te

IMlm"c

~

(5. 1',5.3')

~---

m

I

A safe design requires that am S U . tb where U . tt is lhe allowable stress for the materialuscd. Substituting Uatt for U m in (5.3') and solving for S yields the minimum allowable value of the section modulus for the beam being designed:

s. mm

~

IMI_, Uatt

(5.9)

The design of common types of beams, such as timber beams of rectangu lar cross section and rolled-steel beams of various crosssectional shapes. will be considered in this section. A proper procedure shou ld lead to the most economical design. This means that, among beams of the same type and the same material, and other thi ngs being equal, the beam with the smallest weight per unit length- and, thus, the smallest cross-sectional area- should be selected, si nce this beam will be the least expensive. t ,""{)r beams that are not symmetricat with rcspc<:t to their neutral surface. the largest of the distances from the neutral surface to the surfaces of the beam should be used for e in Eq. (5. 1) and in the computation of the section modulus S = lie.

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The. design procedure. wlll include the following stepst: 1.

2.

3. oJ.

S,

First determine the value of a all for the material selec ted from a table of properties of materials or frum design speci lications. You can also compute this value by dividing the ultimate strength a u of the. material by an appropriate factor of safety (Sec. 1.13). Assuming for the time being that the value of a all is the same in tension and in compression, proceed as follows. Draw the shear and bending·moment diagrams corresponding to the speci lied loading conditions, lmd determine. the maximum absolute vallie IMlmax or the bending moment in the beam. Determine from Eq. (5 .9) the minimllm allowabl e value Sn,in of the section modulus of the beam. For a timber beam. the depth" of the beam, its width b. or the ratio hlb characterizing the shape of its cross sec tion will probably have been specified. The lInknown dimensions may then be selected by recalling from Eq. (4.19) of Sec. 4.4 that band It must sati sfy the relation ~ bh1 = S 2: Sn,in. For a rolled-steel beam. consu lt the appropriate. table in Appendi x C. Of the available beam sections, consider onl y those with a section modulus S 2: SlIUn and select frum this group the section with the smallest weight per lInit length. This is the most economical of the sections for which S 2: Smin. Note that this is not nL'Cessarily the section with the smallest vallie of S (see Example 5.04 ). In some cases. the selec ti on of a sec tion may be limited by other considerati ons. slich as the allowable depth of the cross section, or the allowable deflection of the beam (d. Chap. 9).

The foregoing discussion was limited 10 materials for which a . 1I is the same in tension and in compression. If a .1I is differe nt in tension and in compression, you should make sure 10 select the beam section in such a way that a m ::5 a .1I for both tensile and compressive stresses. If the cross section is not symmetric about its ne utral axis. the largest tensile and the largest compressive stresses will not necessarily occur in the section where IMI is maximum. One may occur where M is maximum and the other where M is minimum. Thus, step 2 should include the determination of both Mmax and Mmin , and step 3 should be modified to take inlO account both tensile and compressive stresses. Finally, keep in mind that the design procedure described in thi s section takes into account only the normal stresses occurring on the surface of the beam. Short beams, especially those made of timber, may fail in shear under a transverse loading. The determination of shearing st resse~ in beams will be discussed in Chap. 6. Also, in the case of rolled- ~ teel beams, normal stresses larger than those considered here may occur at the junction of the web with the nanges. Thi s will be di sc ussed in Chap. 8. t We assume that all beams considered in this chapler arc adequately braced to prevent lat· eral buckting, and thai bearing plmes are provided under concentraled loads applied to rolled· steel beams 10 prevent tocal buckting (crippling) of Ihe web.

5.4. Design 0' Prismatic Beams 'or Bending

333

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EXAMPLE 5.04 Select a wide-lhngc beam to suppon the 15· kip load as shown in Fig. 5.15 . The allowable normal stress for the steel used is 24 ksi.

-'.

Referring to the table of Properties of Rolled·Sleei Shupe.l· in Appendi x C. we nOle thaI the shapes are arranged in group;; of the same depth and that in each group they are listed in order of decreasing weight. We choose in each group the lightest beam having a section modulus S "" 1/t: at least as large as SOIl" and record the results in the follow ing tab le.

I- - - - sft - - - - I

Sha pe Fi g. 5.1 5

l.

The all owable normal stress is given:

"2.

The shear is constant and equal to 15 kips. The bend· ing momeJ1l is maximum at B. We have

ifall

= 24 ksi.

W21 W18 W16 Wt4 W12 W10

X 44 X 50 X 40 x 43 X 50 X 54

S, in 3 81.6 88.9 64.7 62.7 64.7 60.0

IM I",,, = ( 15 kips)(8 ft) = 120 kip · ft ~ 144fJ kip· in.

.1.

The min imum allow
The most economical is the W 16 X 40 .~hape SHlce It we Ighs only 40 Ib/ ft, even though it ha ~ a I:lrger section modulus than two of the ol her shapes. \\"e also note that the lutal weigh! of th e beam will be (8 rt) X (40 Ib) ~ 320 lb. Thi$ weight i.'i small compared to the 15.000- 1b load and can be neglected in our analysi S.

" Load and Resistance Factor Design. This alternative method of design was brielly described in Sec. 1.13 and applied to members under axial loading. II can readi ly be applied to the design of beams in bending. Replacing in Eq. ( 1.26) the loads PD' PL! and Pu. respectivcl y. by the bending momems M D. M L. and M u< we write (S.IO) The coefficient s "YD and YL arc referred to as the load flu:ton· and the coefficient q, as the re.l";.I·tallcefactor. The moments M D and M I. arc the bending moments due. respecti vely, to the dead and the live loads, while Mu is equlIl to the product of the ultimate slrength Uu of the material and lhe section modulus S of the beam: Mu === SU u.

334

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SAMPLE PROBLEM 5.7 A 12-ft-long overhanging timber beam AC wilh an s-n span AB is 10 he de~ienen 10 ~1J[1[lOrl the di,rrihnre41 ml(l conce.nrr:llell In~[l ~

shown. Knowing that timber of 4·in. (Iominal width (]Sin. actual width) with a 1.75-ksi allowable stres~ is 10 be used. determine lh.e minimum reqUIred depth II of the bellm.

SOLUTION Considering the entire beam a s a rree body. we write

Readiun.5.

11 111 111,11,

-t- 1:YM A = 0: 8 (8 fl) - (3.2

~ips )(4

=.. 'f.F, - O.

B = &.35 kips A, - ()

t:1:l', ~ 0 : A, r 8.35 kip~

"

lip'

-"'t-__...J."C-- ,

A!-_ _ _ _ _ -o. (~

k;p,

(1 ) - C4.5 kips )(12 f(1 = 0 B = IUS kip~

i

3.2 kip~ 4.5 kip~ - Q A . "., - 0 ,65 I:ips A '" (l ,65 kipd-.

Shea r Diag ram. The shear just to tile righl of A is VA = A,. "" - 0.65 kips . Since Ihe change in shear between A and H i s equal 10/1lllWS the area under the load curve between these tlVO points. we obtain V~ by writing VE

(400 lblft )(8 ft ) -

V,, -

32001b - - J.20kips

VB = V~ - 3.20 kips = - (l.65 kips - 3.20 kip;; = - 3.85 kips.

The rcJc!ion at B produccs

u

sudden increase. of 8.35 kips in V. res ulting in u

value of the shear equailo 4.50 kips 10 the right of B. Since no load is ;)pplied between Band C. th<'- she;)]" remains constant between these two points.

I>clrr m lll >, Uo n of 1 "11.... , We lirst obscn·c that the b.::nding moment is equal 10 lew at both ends of the beam: ,1.1" = Me = O. Between A and B the bending moment decrea;cs 0)' an amount equal 10 thc area under the sh~~r curve. and between lJ and C it increases by a corresponding amOtlnt. Thus. the maximum

absolut~

va lu e of the bending moment is IM I.,,, = 18.00 kip· ft.

Minim um Allowublc Secti on Modul us. Substituting illlo Eq. (59) the J,;iven v;)luc of u .u and Ihe value of MI... , that we have found. we wrill!

_ IMlw. . _ (18 kip · ft )( 12 in.lft) Su,,,. -

U . II

-

I •75 k St.

123.43 in l

.\li nim um Req u ind Depth of B "~lIn . Re~:llling Ihe fomlUla de veloped in P;)fl 4 of LIte design procedure described in Sec. 5.4 and substi tuting the val· ues of b ~nd Snu • . W~ have

h The minimum required deplh of the beam is

~

14.546 in. II = 14 .55 itl . ...

335

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SAMPLE PROBLEM 5.8

1111111111,

tJl~

A 5-rn-lung. simp!y suppuntxl sled bt:am AD is IU cany

t1imibutcd ,UU] CUIl-

centr~led

IOllds shown. KnOWUlg lhal the allowable nonnal stress for the grnde of sleello be used is 160 MPa. select the wide.l1ange 5]mpe thai should be used.

1m

SOLUTION R('udion!l. -t~:£MA

'" 0: D(5

Cons idering the entire beam as a free body, we wrile

1ll:1-

(60 kN)( I..'i m) - (50 tN ){4 m) = 0

D - 58.0 kN ~LF,= O:

D - 58 .0 kN

t

A ,= O

+t:EF, - O:A ) I- 58.0kN - 60kN - 50kN - 0 ~

,

- 1.5",- - 1.5 "'

ii, "" 'i2.0 kN I ,"

1m

"

I

,\

= 'i 2 0 10:\1 t

Shl'ar Diagram. The s he ar jusl to lite right of A is VI - A v - t 52.0 kN . Since the clmngc in shear between A and B is equal (0 lIlillll.l Ihe area under the load curve between these two points. we have

VB -'" 52.0kN - 60kN = - SkN 52kN

\

~,

I"" I

J

3

2Jilll

The shear rem;)im constant between Band C. where il drops [0 - 58 kN, and keeps thi s value between C and D. We locate the section £ of the beam where V = 0 by writing

"

1 ':SkN

r

n

v."

Vc = - lIH o - 52.0 kN = -(20 kN / m ).\'

J

Solving lor x we lind x = 2.60

'":7 -,~.~ kN

F.l.

or

Determination I M I _~. The bending moment is maximum at E, where V "" O. Since /II is zero m the SUPPOll A , its maxi nlllm value at E is e
to

IMI",,,,,

Ifle are" under tfle nd E. \Ve h:lv.... 1herefore,

= ME = 67.6 kN· m.

Mi nimum A1lo\\"ab l ~ Sct'li(J1I l\'1odulus. Subs!ilutmg mlo Eq. (5.9) 11.e given va lu e of (',.1 and the va lu e of 1 MI",,, thaI we have found. lI'e wri le

St""

_ IMln'" _ 67.6 kN· a .n

m _ 422 .5 X

160 MPa

1O-~ m) _ 422.5

x 10J mmJ

Sdedion of Wide· Flange Shape. From Appendix C we compile a list o f shapes Ih al h;) ve a seclion modulus larger Ihan S",. alld are also the lighle~t

shape: ; ,1 a gi Vl: 11 de pth g"oup.

Shape

S , mm 3

W410 x 38.8 W360 x 32.9 W310 x 38.7 W250 X 44.8 W200 x 46.1

637 474 540 535 448

We select Ihe lightest s hape uv ailable, namel y

336

WJ60 X 319 ..

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PROBLEMS

5.65 and 5.66 For the beam and loading shown. des ign the cross sectionofthebeam, knowing tlwt the grade of timber used has an allowable nomla] stress of [2 MPa. 3 . 6"~

16kN

f

II

1 1 C

1

U,llm

1

(I,Sm

15kNI" ,

AI III IIIIIII~ ':f"~ AD}'

.tOm",

- I l-

o.S",

I

¥

I.

~I _ B 0.9

III

2m

C I~ 0.9m

Fig. PS.66

Fig. PS.65

5.67 and 5 .68 Fur the beS st:(;lion of the beam. knowing thallhe grade of timber used has an allowable nonnal stress of 1750 psi

1.5 Lips/I!

A

5.l) ,.. ,

,11111111 --;1-

·~~~C

}-

1-3.S r.-L35r.-J Fig. P5.67

Fig. PS.68

5.69 and 5.70 For the be.un and loading shown, design the cross seclion of the beam. knowing that the grade of limber lISed has an allowable namml stress of 12 MPa.

2.5

2..5 LN

k.,'1

"'",

A

~ 1-L'-3m- d B

(1.6",

Fig. PS.69

tOO ",,,,

1-1

C

0.6

/J

}-

III

Fig. P5.70

337

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338

""'01),,13 ond 0&319n o f B ....

=

5.71 and 5.72 Knowing tI,at die allowab le S1J1.:SS fOJ ,he ,teel used is 24 ksi. select the most economical wide-flange beam to support the loatling shown

for Bending

2f1 ~,ip. B

A

11111 111 111 11

•

~I~J- " ----'f 2ft

Fig. PS.71

~

A

1i",L 15,,----"-l Ag. PS.72

5 .73 an d 5 .74 Knowing th~t th~ ~lInw~hl ~ ~rTl'~~ for I h~ . r~el 1I~
l'li/",

1------ (} m - - - -- L Fig. PS.73

Fig. PS.74

5 .75 and 5.76 Knuwing that die alluwabh: Stress rur lh!! ,Ie!!] USl:U i~ 24 ksi. seleci lite most economical S·sltape beam 10 support the loading shown

S kipol/t

A~'~~q

i'lLr; f'~ 2f:

Fig. PS. 7S

2 fl

2f1 2 f1

Fig. PS.76

5.77 and 5.78 Knowing that die allowable stress for the ." eel used is 160 l\1Pa. select the most economical S-shape beam to suppon the loading shown

70 kN

Fig. p S.n

'oil kilO}".

Fig. PS.78

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Problem.

5.79 !\ sleel pipe of 4·in. dimneter is [() suppon Ihe I03ding shown. Knnwin.f lh~l rh E"; mod: of Jl;pe~ ~vadah ~ h ~s rhict.:: nes~es v~rying fmm ~ in ro I in. in , .in. inc re ments, and that the allow3hle normal stresg for the steel used is 24 ksi, determine the minimum wall ihicl.:ness f th~t can be used.

339

5 .80 Three s teel pl3t
Fig. P5.79

n

t:

A

1~"I\-L"J-11

~ .'i

!l .~

It Fig. PS.80

If

5 .81 Two metri c rolled-s tcel channels are to be welded al ong the ir edges and w;ed to s upport the IOJding s hown. Knowing th3t the 3110wable nonn31 n re ss for rhe steel used is 1SO MP a. delemline the most economical c hannels ih31 can be used. 152

"'In

1- -1 lIf'T 10:",,,,

1- - 4 @o.6i5 "=

2.; 1H ~

Fig. PS.81

Fig. P5.82

5.8 2 Two Ll 02 x 76 ro lled-,teel
5.83 Assuming th~ upward reaction of the ground to be umformly tli strib-ute
T"t.J I ,,~n - 2 ~I ~

,lllllll e

.,

r

() .~5 Hl

1

, n

0 .7.5 '"

5.84 A. ssum in e rh p. ttp wiml rellclinn of Ihp. gmu nd rn hP. ttnifnrmly dis_ tributed and knowing that the allowabl e nonllai s tre ss forthe steel used is 24 ksl. seleci the most eco no mic31 wide n:mge belm 10 s uppon the ILludin g, shown.

c D

"

lm -

Fig. PS.83

2n. r ~Ir'

Fill. PS.84

,J-"J

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340

Analysis and Design of Beams for Bending

5 .85 Delcrmine the largest permissible distributed load I\' for the beam shown, knowing Ihal the allowable normal siress is + 80 MPa in te,nsion and - 130 MPa in compression. 60",,,,

I-I ~

"

1 11~fll llll~rl l ~~~20"''' .

A

8

J.:i'

1~llfll'

Fig. P5.87

III

Ii"

1-10 ;" " 10 '''.-1 -11"~~~~~E "

I:::-1-H.S "'0.2

r.

V

I - --

ri)",m ~

I - I 1-""",,, -

0.2,"

Fig. ps.as

5 .86 Solve Prob. 5.85. assuming thai the cross ~ec tion of the beam i.'i reversed. with the flange of the beam resting 011 Ihe suppons aT Band C.

5.87 Determine the allowable value of P for the loading shown. knowing that the allowable normal stress is + 8 ksi in tension and - 18 ksi in compression. 5 .BB

Solve Prob. 5.87, ass uming thai Ihe T.shaped beam is invened.

5 .89 Beams AB. Be, and CD have the cross section shown and are pinconnected at B and Knowing Ihal the allDwable normal siress is + 110 MPa in tension and - 150 MPa in compression. delermine (a) the largesl permissible. value of IV if beam BC is nOlIO be overstressed. (b) Ihe corresponding ma'{imum dislan<.:c a for which thc crullilc.vcr beams AB nnd CD arc nOI O\·crstrcssoo.

e.

r-- 2

1.1U

" I'

cL

/J 1-,,--- j.2m--- ,,-1

12.5"""

Ill'"

1

L 'l" -I~

12.5"'111

Fig. PS.a9

5 .90 Beams AB. Be. and CD have the cross section shown and are pinL'Olmecled at B :md Knowing Ihal the allowable normal stress is + 110 MPa in tension and - 150 MPa in compression. detennine (a) Ihe largest permissible. value of P if beam BC is nOlto be overstressed. (b) the corresponding maximum dislance 1I for which the cantilever beams AB and CD are nOI overstressed.

e.

12.5

Ag. PS.90

""ll

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5.91 A 240-kN load is to be supported at the center of the 5-m span shown. Knowing lhalthe allowable normal stress for the llteel used is 165 MPa. determine (a) the smallest allowable length I of beam CD if the W310 X 74 beam AB is not 10 be overstressed. (b) the most economical W shape that can be used for beam CD. Neglect the weight of both beams. l..lllkN

1/12-1-1121

CF

f

f f. =S

III

it

\\'3 10

x 74

----.j!

Fig. PS. 91

5.92. Beam ABC is bohed to beams DBE and rCG. Knowing that the allowable normal mess is 24 ksi. select the most economical wide-flange shape that can be llsed (a) for beam ABC, (b) for beam DBE, (d for beam FCC.

Fig. P5.92

5.93 A uniformly distributed load of 66 kN/m is to be supponed over the 6-m span shown. Knowing that the allowable nonnal stress for the steet used is 140 M Pa. determine (a) the smallest allowable length 1 of beam CD if the W460 X 74 beam AB is not 10 be overstressed, (b) the most economical W shape that can be used for beam CD. Neglect the weight of bolh beams. hfi J.~lm

Fig. PS.93

Probloms

341

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342

Analysl!l and Design of Beams for Benalng

'5.94 A roof structure consists of plywood and roofing malerial supported by several timber beams of length L = 16 m. The dead load carried by each beam. induding Ihe estimaled weight of Ihe beam. can be represemed by a uniformly distributed loadwv = 350 N/m. The live. load.:onsists of a snow load. represented by a uniformly distribUied load w L = 600 N/m. and a 6-kN concentrated load P applied allhe midpoinl C of cadi beam. Knowing that the ultimate strength for the limber used is IT u = 50 MPa and Ulat the width of the beam is b = 75 mm, determine the minimum allowable deplh " of the beams, using LRFD with the load factors Yo = 1.2. YL = 1.6 and the resistance factor .p = 0.9.

A

11111111 1111111 11R p

Fig. PS.94

"5.95 Sol lie Prob. 5.94, assuming thatlhe 6-kN concentrated load P applied to e~h beam is replaced by 3-kN concentrated loads PI and P~ applied aI a dislance of 4 m from each end o f the beams. "5.96 A bridge of length L = 48 fl is to be built on a secondary road whose acce~~ to trucks is limited to two-axle vehicles of medium weight. [t will consist of a concrete !;Jab and of simply supported steel beams with an ultimate strength ITu = 60 ksi. The combined weight of the slab and beams can be approxim
1I--L- 3"' Fig. PS.96

"5.9 7 Assuming that the froll! and lear axle loads remain in the same ratio as for the truck of Prob. 5.96, detennine how much heavier a truck could safely cross the bridge designed in that problem.

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5.5. U3ng Singulllrlty FUn<:tlon,

-5.5. USING SINGULARITY FUNCTIONS TO DETERMINE SHEAR AND BENDING MOMENT IN A BEAM Reviewi ug tlie wm k t1 UIlt: ill till; pl ccetlillg SL'C tiOllS. we !lote tlmt the and be ndi ng momcnt cou ld onl y rarel y be described by ~ in8 k analytical fu nctions. In the case of the cantilever beam or Example 5.02 (Fig. 5.1 0). which supported a uniforml y di stribu ted load w. the shear and bending momenl could be represented by single anal yllcal [u nctions. namely, V ~ lUX ilnd AI Jw.,2: thi s was due to the fac t that i lr? dismn/inlliTy ~~ i s u:d in th r. IOMline o f th r- hemn . On the mhr-r h ~ n d, In the case of the ~ i mp ly s uppon ed beam of Example 5.01 . which \\iUS loaded only ill ils midpoint C, the load P applied at C rcpresellled a .1"111 gularity in the beam loading. This singularity rcs LLlred m discontin uities 111 the shear and bending moment and reqUi red the use of diffe rent :m.'llytical functions to represent V and A1 in the portions of beam located. re.~ pec ti vely, to the left and to the right of point C. [n Sample. Prob. 5,2, lhe beam had 10 be di vided into th ree portio ns, in each of which different fun ctio ns were used lU represent the shear and (he bending mome nt. TIlis situation led u .~ to rely on (he graphical rcprcsc Iltatl0 n of the functions V and .\1 provided by the shear and bend ingmoment diagrams and , late r in Sec. 5.3. on a gra phica l method of integra tion to determi ne V and M from the distributed load 1V. The purpose of th is section is to show how the use of ,I'illgula rity lill ie/ions makes it possible [0 represent the shear Vand thi' bend illg moment M by si ngle mathematical expressions. Consider the simply ~ uppo rtcd beam AB , of length 211. which carries a uniformly t!istnbuted load 'We extending from its midpoint C 10 Its right-hlUld support B (F ig. 5. 16). We tirst dmw the frce-body diugram o i the cnti re beam (Fig. 5 .1 7(1): replaCing the di.m ibutcd load by an equ ivalent concentrated IQad and , s umming moments about B, we write .~ hear

" It

cllllllllllil

A~~~~ll

L.- I- .---J

Fig. 5.16

:'--Iexl we e ul the beam a l a po inl D belwee n A Imd C. F ro m the free-

body diagra m of A D (Fig. 5. 17b) we conclude that. over the interval II. the shear and be llding moment arc exprcs., ed, respecti vel y. by the funct ions

o< x <

, nd Cuttillg, now, the beam at Il point E between C and B, we draw the freebody diagram of ponion AE (Fig. 5. [7c) . Rep[acing the distributed load by an equivalent concentrated load, we write 11Voa - w~lr - a) -

+tLF , = 0: +~:EME=O:

-1w lIx + Q

Vl = 0

WlI(X - l. )[~(X - u)J + M ~ = 0

and conclude that, over the interval ll < x < 211. the shear and bendmg moment arc expressed. respectively. by the fun ctions

Vll)

0:=

iw(){/ - woCr - a)

allCl

Mlr)

= :~Vu{U - hvoCr -

lI)~

Fig. 5.17

343

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344

Analysl!l and Design 01 Beams lor Bend ing

As we poilJted out earlier in this section, the facI lhat the shear and Ix:nding mument are represented by different funct ions of x, depending upon whether x is smaller or larger than (', is due to the discuntinuity in the loading of the beam. However, the functi ons V,{x) and V!(x) can be represented by the single expression

V(x )

=

1w oll - w~x -

a)

(5.11 )

if we specify Ihat the st:cund tenn should be induded 'in our cumputations when x 2" a and ignored when x < o. In other words. lite brackets < >~'h()1I1d be rep/aced iJy ordill(// )' parentllf:~'f'J ( ) when x 2:. a alld by zero when x < a . With the same convent.ion, the bending moment can be represented at any point of the beam by the single ex press ion (5 .12)

From the convention we have adopted, it follows that brackets < ) can be diffe rentiated or integrated as ordinary parentheses. Instead of calculating the bending moment from free-body diagrams. we could have used the method indicated in Sec. 5.3 and inlegr.:lled the expressiun obtained for V{x):

M(x) - M(O) = f V(x) dx "" \l

r

l woa dx -

(I

r

"Wo(x - (J} dx

I,)

After integration, and observing that M(O) = 0, we obtain as before

Furthennore, using the same convention again, we note that the distributed load at any point of the beam can be expressed as (5.13)

Indeed, the brackets should be replaced by zero for x < (/ and by parentheses for x 2:. a; we thus check that w(x) = 0 for x < a and, defining the zero power of any number as unity, that (x - a)O = (x - a)O = I and w(x) = Wo for x 2:. a. From Sec. 5.3 we recall that the shear cou ld have been obtained by integrating the functio n - 1£:(X). Observing that V = i woa for x = 0, we write

V{x) - V(O) "" -

r

w {x) dx

"

V{x) - ~ wo a

=

- wo{r - a)1

Sol ving fo r V(x) and dropping the exponent I, we obtain again

V(x)

=

1woa - wd.,x - a)

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The expressions (x - a)(), (r - a), (x - a)2 are called singularity fill/ctions . By definition, we have( for n 2: 0, wbenx when.x

2:

a

(5 .14)


We also note that whenever the quantity between brackets is positive or zero, the. brackets should be replaced by ordinary parentheses, and whenever that quantity is negative, the bracket itself is equal to zero.

o

o (,,) ,,=1)

(h ) "

/,.\ " = 2:

=J

Fig. 5.1 8

The three singularity fun ctions corresponding respectively to II

= O. n == I. and n == 2 have been ploned in Fig. 5. [8. We note that

the function (x - a? is di scontinuous at x = {/ and is in the shape of a "step." For that reason it is referred to as the step junction. According to (5. [4), and with the zero power of any number defined as unity, we ha\'et

(x - a)"

~ {~

whenx whenx

2:

<

a a

(5 .15)

It follows from the definiti on of singularity functions that

f (x -

ar t/x =

_ '_ (x 11 + 1

a,(,+1

fo r n

fo r n

2: [

2:

0

(5 .[6)

(5 .17)

Most of the beam loadings encountered in engineering practice can be broken down into the basic loadings shown in Fi g. 5.19. Whenever applicable, the corresponding functions w(x), Vex), and M(x) have been

expressed in terms of singularity functions and plotted against a color background. A heav ier color background was used to indicate for each loading the expression that is most easily derived or remembered and from which the other function s can be obtained by integration. tSioce (x - "t is discontinuous at x - a. it can be argued that th is function should be len undefined for x = "or that it should be assigned hath of the values 0 and I for x = a. However. defining (x - a)Oas !:Guat to I when x = ". as stated in (5.15). has the advamage of being unambiguous and. thus. readily applicable to computer programming (d. page 348).

5.5. Using Singularity FUlK:tlons

345

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1 "71

o ~~ ." ... !------ .~ M" (0 )

M k

Rending Moment

Shear

Lo"ding

"~ "

o

o

..

- _It" :

"h,-

o

,, , , "-- - : V(,) =_p< .. _

°f--

,

"

M L-

'

~ '

0 1

.If(~ )=

(1)0

_ 1' <,_,, rel="nofollow"> 1

M L-

--'
~'

0 1

(c)

,,;\x) ="n<'- - ">~ M

"oL ~x (.I)

U' (x)= k<;< _,,> i

1t'~x)=k<x -a>"

(I

>~

O~'

1'(.tl = -i<x-ll>l

V(x) = -;h<.r - (I:>'"

-t '"',)< x -

M~ 1/ (x) =

"01L.~ -' (,-)

~x ) =

-A < .r -

a:>J

M~

O~ '

I

U(;t )=-i~<X-">""~

Flg.5.19 Basic load.ngs and cor responding shears and bending moments expressed in terms of singularity functions .

After a given beam loading has been broken down into the basic loadings of Fig. 5. 19. the fun ctions V(x) and M(x) representing thc shear and bending moment at any point of the beam can be obtained by adding the correspondi ng functions associated with each of the basic loadings and reactions. Si nce all the distributcd loadings shown in Fig. 5. 19 arc

346

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5.5. Using Singularity FUlK:tlons

open-ended to the righ t, a distributed loading that docs not extend to the ri ght e nd o r the beam or that is discontinuous should be re placed as shown in Fig. 5.20 by an equivalent combination o f open-ended loadings. (See also Exam ple 5.05 and Sample Prob. 5.9.) As you will see in Sec. 9.6. the usc of singularity fUllct ions also greatly simplifies the determination o f beam de Oections. It was in connection with that problem that the approach used in thi s section was first suggestcd in 1862 by the Ge nnan mathematician A. Clebsch (1833 1872). However. the British mathematician and engineer W. H. Macaulay ( 1853 - 1936) is usually given credit for introducing the singularity functions in thc foml used hcre. and thc brackets ( ) arc generally referred 10 as Macau/a)' '.~ /Jmckets. t

t Wo ~I . Macaulay,

'NOI~

on the DeflcC!ion of

B cam.~:· Meuell~er

of Mmhemmic.<, vol. 48.

pp. t29- 130. t9t9

Fig. 5.20

EXAMPLE 5.05

"=1.2 kN

For the beam and loading shown (Fig. 5.2Ia) and using singularity lunctions. express the shear and bending momenl as functions of the distance x from the suppa" at A.

C

We first determine Ihe reaction at A by drawing the frecbody diagram of Ihe beam (Fig. 5.2 Jb) and writing

tt,...J- LJ

(" I ,,±,H~ =

0:

0.6'n

A, = 0 "

- A,.(3 .6 m) + (1.2 kN)(3 m) + ( 1.8 kN )(2.4 m) + 1.44 kN . m = 0 Ay = 2.60 kN

Next. we replace the given dLmibwed loading by 1WO equivalenl open-ended loadings (Fig. 5.2 k) and express the dislJibuted load w(x) as Ihe sum of the correspond ing step functions:

/J

A'lIi1I!~::~~~:illiB

r

1.2 k:>/

0.8

mm

!.lllll

I UH.;/\

fol"

I

A C

III

= I

II

D

k~

. "I

8

~·ic::::::::::=C~~:I~t (b )

l- 2Aml---3"'----1 B

A.

1---- - -3.6 III

The function Vex) is obtained by integrating w(x). reversing Ihe + and - signs. and adding 10 the result the constants At and - p(x - 0.6)° representing the respective contributions to Ihe shear of Ihe reaction at A and of the concentrated load. (No other conSlant of integration is required.) Since Ihe concentrated couple does not direclly affect the s hear. it shoukl be ignored in Ihis computation. We write 1\ ., = 2.h kl>. Fig. 5.21

-'I:,,~

- 1.5L,,\/m

347

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348

Analysis and Design of Beams for Bend ing

In 3 si mibr way. the funclJOn M( () is obtained by integrating

V(x ) :tnd adding 10 the result the conSI3n t - Mo(x - 2.6}O represenling the conlribulion of th e concenlr31ed couple 10 the bending momenl. We lw ve

M(_() = - ~ w~x - 0.6? + f wJ"x - 1.8Y

+ A,x - p(x - 0 .6)1 - Ml r - 2.6)" Subslituting the numerical val ues of Ihe reaction and loads into the expressions obtained for V(x) and M(x) 3nd being careful !wl to compute ;my product or expand any square invo lving a bracke t. we obt3in the following expressions for the s he3r :U1d bending moment 31 any poinl of the beam: .\~ =

2.fi

Vex) = - 1.5(x - 0.6)1 + 1.5(x - 1.8)1 +2.6 - 1.2(',r - 0 .6)°

k~

Fig. 5.21 c (repeated)

M(x) = - 0.75(x - 0.6)" + 0 .75(.1 - 1.8}" +2. 6x - 1.2(x - 0 .6)1 - 1.44{x - 2.6)"

EXAMPLE 5.06

V( 1.8) = - 1.5(1.2)1 + 1.5(0)1 + 2.6 - 1.2(1.2)1)

For the beam and 10
+ 1.5(0) + 2.6 - 1.2( 1) = - 1.8 + 0 +2.6 - 1.2

= - 1.5(J.2)

Making x = 1.8 m In the expressions found for \l(x) and Af(\") in Example 5.05. we obTain

V( 1.8)

~

- 0.4 kN

and

V( 1.8) = - 1.5(1.2)1

+

1.5 (0)1

+ 2.6

- 1.2 ( 1.2)°

M(l.') ~ - 0.75 ( 1.2)' + 0.75 (0)'

+ 2.6(1.8) -

M( 1. 8)

= - 0 .75( 1.2 )'

+ O.75(0)~

+ 2.6( 1.8) -

1.2( 1.2 )1 - 1.44(0)

= - 1.08 + 0 + 4.68 - 1.44 - 0

1.2 ( 1.2)1 - 1.44(- 0.8)0

Rel:alling that whenever a quantity beTween br3ckets is posit ive or zero. the brackets should be replaced by ordinary parentheses. and wheneveTlhe quantit y is neg3ti ve, the bracket iTself is equal to zero. we wri le

M( 1.8 ) = +2. 16kN · 1I1

Application 10 Com pUler Programmin g. Singularity fu nctions arc particu larly well suited to the use of computers. First wc note that the step function {x - a }I). which will be represented by the symbol STP. can be defin ed by an lFrrHENIELSE stftlemcnt :IS being equal to I for X ~ A and to 0 otherwise. Any other singularity function {x - a)", with n ~ I. can then be expressed as the product of the ordinary algebraic function (x - at and the step fun ction (x - (I }I). When k different singularity functions 3re involved. such as where i = I. 2. ... . k . the n the correspond ing step functions STP(I), where J = 1. 2, .. .. K. C3n be defined by a loop colllaining a single lFrrHEN/ELSE statement.

(x- a;r.

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iJ1b Uol

L'"

1./4

SAMPLE PROBLEM 5.9 For the beam and loading shown, determine (a) Ihe equations defining the shear :'lnri ""'n!line rnom<>nl m ~ny minI, (h) the ~h~ilr and Iw,n.lin g moment ill [\O;nr~

C. D, and E.

1.14

SOLUnON Reac tions. The tOlal load is kwnL: because of symmetry, eadl reaction is equal 10 half that value. namely. £WIJ L.

Distribu ted Load. equiyulem

open-~ndoo

The given distributed loading is replaced by

loadings as

\1'10

L' sing a !ingularity function to

s h.OWtl.

e.l '

press the second loading. we write (1) U.

E4uutiulls fur Shear alltl

8tlldin~

We ubwin Ii(x) by irJ-

MUIIICIII.

legr:lring l[). ch ~ nf',;ng the ~ ;f',n~ , and aodinE a mn~I:1I11 ef]II~1 III R"

We obtain M(x) by integraling (2): since there is no concentrated couple. no constant of integration is needed:

.'1C-')

v ",."" L

=

+ -2w,,( 3L I

'n',\

3 L "r

--

-

iLl' + ~l

~

(3) "II

it; L I 0

b. Sht'ar and Rcnding .\1omcnl at C. D, and E AI Poi", C : Making x = tL in Eqs. (2) and (3) and r~calling that wherlever a quantity between brackets is positive or zero, the brackets may be repklced by parentheses. we have 1Vu ,

Vc -

,

T \lL)-

_ 1vl' (~L1J + 3L-

j

21VI, (), T O-

•

j

1'(:"-

l !lloL

2u'o(O'~ + 111). L(' L) 3L

,

IJ

(I

I • ·\1 e = 11 weL-

2

.. ..

iL

Ai Poill l D : Making x = in Eqs_ (2) and (3) and recalling that a brackel containing a negative quantity is equal 10 zero. we write j

Vv= 16 l£'uL "

M" = u

Makine ,- = f l. in F.'l~ (2) and (i).

A ll'flilll F.: A

v

,.

VE =

1V.)

I.

10

u'L' "II I

havl'.

r .

- + -2wo AL + 4'toU L I. 0

')' ~

- - C~L

\\'1'.

II

192

M t . = ---.!I.. 3L ('4L)'

+ _2wo l w LC' L) 3L (LL\' 41 + ~U4

349

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SAMPLE PROBLEM 5.10 The rigi\J b
i~

wt:ltkd
sl~eI

bt:
lh~

luading

sb.own. d ~lermlne (al the equations defining Ihe shear and bending moment at any point of the beam. (b) the ioc ntioll and magnitude of the largest bending moment.

SOLUTION Reactions. p ,. D

~~

lIIu - j.'tjllb It

F

l~llh

I

\Ve cO[lskl~r

the

~am

and bar as a free booy and obsene

Ihat lhe [olalload is 960 lb. Because of s ymmelry. each reaction is equal to 480 lb.

at'"

Modified Load in!,! Diaemm. We repillee the 16G-lb load applied by an equi\lalent force-couple system at D. We thus obtain a loading diagram consisting of a concentrated couple, three concentrated loads (mduding Ihe two reaclion». :md a unifonnly distributed load

F

w(x) - 50 lb/ft

w,n,

(I>

II . Equations for Shear and Benllillg Moment. We obtain V(x) by integrating (I ). changing the sign. and adding con.<;tallt~ representing the res pec_ tive contribUlions of RA and P to the shear. Sinc~ P affects V(x) only for val-

ues of .t

larg~r

than II ft. we use a step function to express its contribution.

V(x) = - SOx

t

~ I V=4'>(' lb. H

1-- - - I H I

- --1-

(2)

111-'

H~

p = IGI)lb

I'l l"" Il>Olh

f- 4S0 - 16Il{.r -

5

ft --l

~

We obtain M(x) by integrating (2) and using a !tep function to represent tre t'Ontribution of tlte concentrated couple M/): (3)

~

b. Largrst Bt.'llrling !\lornI'D!. Since /II is maximum or minimum when ~' = U. we set V = (j in (2) and solve that equation lor x to tlnd the locatioo of the lurgest bimding moment. Considering first values of x less than II ft and noting that for such values the bracket is eqnal to zero. we write T

= I).M ft

Conllidering now values of .t larger than II ft. for which the br..!cket is to I. we have

~qual

- 50 .\' + 480 - 160 == 0 Since thi~ value is " 0 / brgerthan II ft, it must be rejected Thus, !lIe value of x corresponding 10 the largest bending moment is ,\ .. = 9.60 ft .... I 2:JO.I

Sllhs! itlltm~ !hi~

Ih · ft

_ _- ...., +2255lh .ti

,, , ,,

-+ t77~l h·ft

" ~ I -------7I 7,J)---','-'

' ... - ' .00 n

val ue for,. intn FI I (l) Wf" nh!ain

M •." = - 25(9.60]2

+ 4S0(9.cO) - 160( - 1.'10)1 - 4S0(- 1.40r

amI , recillliuJ; lilal brack::ts ,ontaining a neplti"e quanuty are equal to zero.

M""" = - 25(9.60f + 480(9.60)

M,,,,,, =

230-l1 b . ft ...

The bending-moment diagram has bren plotted. Note the di scontinuity al poinl D du;:. to the COt1CelHr
and tepl~cing 1he step func1ion (r - I I ~' by 0 and I. re~pectively.

350

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PROBLEMS

5 .98 through 5.100 (a) Using singularity functions . write the equations defining the shear and bending moment for the beam and loading shown. (b) Use the equation obtained for M to detennine the bending moment at poilU C and check your answer by dmwing the free-body diagram of the entire beam.

JIIITII IIB

c

1--,,----1-,,Fi g. P5.99

Fig. PS.98

Fig. P5.100

5.101 t hrough 5. 103 (a) Using singularity functions. write the equations defining the shear and bending moment for the beam and loading shown. (b) Use the equation obtained for AI to detennille the bendinj.! momelll at poilU E

and check your answer by drawin g the free-body diagram of the ponion of the beam \0 the right of £. I'

p

JIIIIIIIIIB

Fig. P5.1 01

E

c

Fig. P5. 103

Fig. P5.102

5.104 (a) Using singulariTY functions. write the equations for th e shear and bending moment for beam ABC under the loading shown. (b) Use the equation obtained for M to detenlline the bending mome nt just to the righ t of point D. p

r

p

,\ ~=~I!C~IE=!B

1-U3J-U3J- UJ-I

Fig. P5.104

A

I

1--"

8

c

1

Fig. P5.105

5.105 (u) Using ~ ingularity functions. write the equations for the shear and bending moment for beam ABC under the looding shown. (b) Use the equation obtained for M to detemline the bending moment just to the righ t of point B.

351

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352

Analysts and Destgn of Beams for Bendtng

5.106 through 5.109 (a) Using singularit y function s. write the equations for the shear and bending moment for the beam and loading shown. (b) Determine the maximum value of the bending moment in the beam. fi kip.

"j~1I1I~":::lIlI:::~~'I)::i E

L", lL5"'-~ 0_6 m 0 .9 "'

Fig. P5.106

j,i(lIt N/nl

IIIIIIIIIIII I c J Jlj "

'[ t U.S

2.<",

0.8

til

III

Fig. P5.109

Fig. P5.108

5 .110 and 5. 111 (a ) Using singu larity functions. wri te the equations for the shear and bending moment for the beam and loading shown. (b) Determine the maxi mum normal stress due to bending. 2-1 ~N

2-1 k~

B

UlN

2 1 ~N

C

D

E

l

F

A

X

I

W25U X Z8 .4

SI.5\)

x

HW

C-'@ 075'" ~3",-'J-J

U.7!im

Fig. PS.111

Fig. PS.110

5.112 and 5. 113 (a) Using singularity functions. find the magnilllde and location of the maximum bending moment for the beam and loading shown. (b) Determine the maximum normal stress due to bending.

<;.#1 .IIIIIIIIIIIIV: if{

ISkN",

7

.""

I

S310 X 52 11'530 X 66

L.,,,,-L ,,,,,--J Fig. PS. 112

Fig. PS.113

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5 .114 and 5.115 A beam i.~ being designed to be supported and loaded shown. (a) Using singularity functions. find the magnitude and location of the maximum bending moment in the beam. (b) Knowing Ihm the allowable normal stress for the steel to be used is 14 ksi, find the most economical widenange shape that can be used.

Problems

~s

12 kips

t2 kips

B

c

D

A

E

L,;,t,l,, -hJ Fig. PS.114

Fig. PS.11S

5 .116 and S.1 17 A timber beam is being designed to be supported and loaded as shown. (a ) Using singularity functions. find the magnitude and 10· cation of the maximum bending moment in the beam. (b ) Knowing that the available stock consists of beams with an allowable stress of 12 MPa and a rectangular cross section of 30-mm width and depth II varying from 80 mm to 160 mm in lO·mm incremcnts. determine the most economical cross section that can be used.

AIIII IIII ~ ~";

;t; ~l l-'-L5 ",J-25 "'~ B

5()(J ~""

C

Fig. PS.116

Fig. PS.117

5 .118 through 5.1 21 Using a computer and step functions. calculate the shcar and bending moment for the beam and loading shown. Use the specified increment tiL starting at point A and ending at the right.hand support. J2 kN

~L

= 0 .4 111

t6 kN/",

,1111 11 111111111

A

I

1.2 111

C

I

Fig. PS.118

.'Hl~ ..\

Fig. PS.119

~ ~~~lllrnffiTIn

,,,l6r.--

Fig. PS.120

t.L

11L tSl-ipVt1 =U.5 ft

C

Fig. PS.t21

= U.5 rt

353

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354

Analysis and Design of Beams for Bending

5.122 and 5.123 For the beam and loading shown. and using a computer and step fun ctions. (a) tabulate the ~hear, bending moment. and maxi · mum normal stress in sections of the beam from _\ = 0 to x = L, using the increments !J.L indicated, (b) us ing smaller increments if necessary. determine with a 2% accurac y the maximum normal stress in the beam. Place the origin of the x axis at end A of the beam HN

W"JX) X 22 ,5

I. = .5 III !J.L = 0.25,.,

I. = 6,., = 0.5

!lL

III

J ill

Fig. PS.1 22

Fig. PS.123

5.124 and 5. 125 For the beam and loading shown. and using a computer and step fun ctions. (a) tabulate the ~hear. bending moment. and maximum normal stress in sections of the beam from x = 0 to x = L using the increments !J.L indicated. (b) using smaller increments if necessary. determine with a 2% accurac y the maximum normal mess in the beam. Place the origin of the x axis at end A of the beam. 2 ki

ft

•

D

1,

"1

I-I..~ ft- l - 2 ft -

C

-----.:...r 1.5n

" ~~~

L =5fl dL

1

= ().25 It

' rft CI

WI!!. x 3jf

I.

~--- llJll 2,'; ft 2, ,~ n

=

t 5 ft 1.25

~L =

rt

Fig. P5.125

Fig. PS.124

*5.6 . NON PRISMATIC BEAMS

Our analysis has been limited so far to prismatic OO:.lms, i.e.. to beams of uniform cross section. As we saw in Sec . 5.4. prismatic beams are designed so that the normal strcsses in their critical sec ti ons are at most c(lua! to (he allowable value of the nomlal strcss for the material being uscd. It follows that. in all other scctiuns, the normal stresses will be smaller, possibly much smalle.r, than th ~i r allowable value. A pri smatic beam, therefore. is almost always overdesigned, and considerablc savings of material can be realized by using nonprismatic beams. i.e. , beams of variable cross section . The cantilever beams shown in the bridge during construction in Fi g. 5.22 arc examples of nonprismatic beams. Since the maximum nonnal stresses if", usually control the design of a beam. the design of a nonprismalic beam will be optimum jf the section modulus S = lie of every cross sectioll satisfi es Eq. (5.3) of Scc. 5. L Solving that equation for S, we write

IMI

S~-

(J ot!

(5 .' 8)

A beam designed in thi s man lier is referred to as a heam oJ constallt .\·frel/g th .

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5.6. Nonpnsmatlc Beams

355

Fig. 5.22

For a forged or cast structural or machine component, it is possible to vary the cross section of the component along its length and to eliminate most of the unnecessary material (sec Example 5.07). For a timber beam or a rolled-steel beam, however, it is not possible to vary the cross section of the beam. But considerable savings of material can be achieved by gl ui ng wooden planks of appropriate lengths to a timber beam (sec Sample Prob. 5.11) and using cover plates in ponions of a rolled-steel beam where the bending moment is large (see Sample Prob.5. 12).

EXAMPLE 5.07 A cast-alum inum plate of uniform thic kness b is 10 .~ uppon a uniformly distributed load tv as shown in Fig. 5.23. (a ) Determi ne the shape of the plate that will yield the most economical design. (b) Kn owing that the allowable nonnal stress for the alumin um used is 72 MPa and that b = 40 mm. L = 800 mm. and w = 135 kN/ m, determine the maximum dep th 110 of the plate. FIg. 5.23

Bending Moment. Measuring the diSTance x from A and observing thm VA = /itA = 0, we use Eqs. (5.6) and (5.8) of Sec. 5.3 and write

V(x) = -

M(x) =

r

i,

\ vdx

V(x)d.\· = -

"

:lI1d. after s ubstituting

= - 1VX

,

11- = - b(f all

rWXdX = -~ w.r (I

(a) Shape of Plate. We reca ll from Sec. 5.4 that the modulus S of a recTangular cross secTion of width b and depth Ii is S = i bil'. Carrying this value into Eq. (5. IR) and solving for h~. we have 1

61MI

h ""-b if:ll i

IMI "" } 1v.r .

]lVX"

(5.19)

Jw 11 = ( -) ';', bif>l i

(5.20)

Since the re lation between h and x is linear. the lowe r edge of The plate is a straight line. Thus. th e plate providing The mo~t economical design is of Irial!8u1ar slwpl'.

(b) Maximum Depth ho- Making x = Lin Eq. (5.20) and substitu ting the g ive n dat a, we obTain hu =

3( 135 kN/ m) [ (0.040 m)(72 MPa)

]'fi

(800 mill ) = 300 mm

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SAMPLE PROBLEM 5.11 A 12.fl-]ong beam m;lde of a timber with an allowable normal stress of 2.40 ksi and an allowable shearing stress of 0.40 ksi is [0 carry (WO 4.8-kip loads located at its third points. As shown in Chap. 6, a beam of uniform rectangular cross section. 4 in. wide and 4.5 in. deep, wauld satisfy the allowable shearing stress requirement. Since such a beam would not satisfy the allowable nOTmal stress requirement, it will be reinforced by gluing planks orlhe same timber. 4 in. wide and 1.25 in. thick. to the top and boHom of the beam in a symmetric manner. Determine (a) the required number of pairs of planks, ( b) the length of the planks in each pair lhal will yield the most economical design.

SOLunON 8l'lIdin g l\'iomc nl. We dmw the free-body diagr:lln of the beam and find the following expressions for the bending moment:

-/.6 \,i[rl

A

\'

,-~

F

,)'

From A

\0

B (O :S x :s 48 in.):

From B

\0

C (48 in. :s x :s: 96 in.) : At = (4.80 kips ) x - (4.80 kips )(x - 48 in.) = 230.4 kip' in.

A1 = (4.80 kips)...-

iI. Number of Pairs of Pllmks. We first determine the required 101al depth of the reinforced beam between Band C. We recall from Sec. SA lhal S = bl/ for a beam with a rectangular cross section of width b and depth h. Substituting Ihis value into Eq. (5. [7) and solving for It", we have

!

I .'IJ..I)):(

r

61MI

o

11- = - -

-L~ ~-jp~

(I)

bUill 48 i n , -~

t- ,--1 "

A

,P I "'P"

B )

Substituting the value obtained for M from B and f! all> we write h! =

:6"(2::.'"0.,, , ;;-,,:;;·p,,·,,;, ,.,1 ~ (4 io.)(2.40 ksi )

\0

144 in."

C and the given values of b

h = 12.00 in.

Since Ihe original beam has a depth of 4.50 in .. the planks must provide an addit ional depth of 7.50 in Recalling that each p.1ir of planks is 2.50 in. thick: Rcyuired n\unber of pain. of pl::tnks = :1
b. L(,lIgt h of P l:tnk....

The bending moment was found to be At = (4.80 kips )x in the portion AB of the beam. Substituting this e.xpression and the given values of b and (Ta ll. into Eq. (I) and solving for x. we have

x =

(4 in.)(2.40 hi )

112

~

11-

x = -

( 2)

-

] in.

6 (4.80 kips )

Equation (2) defines the maximum distancl' x from end A at which a given depth II of the cross section is acceptable. Making" 4.50 in .. we find the distance X I from A :It which the original prismatic beam is safe: .(1 = 6.75 in. From that point on. the original beam should be reinforced by the tirst pair of planks. Making II = 4 .50 in. + 2.50 in. = 7.00 in. yields thl' distance x 2 = 16.33 in. from which the second pair of plank.~ should be used. and making II = 9.50 in. yields the distance x] '" 30.08 in. from which the thild pair of pl:lnks should be used. The length /; of the pl:lnk.~ of the pair i. where i = 1.2.3. is obtained by subtracting h i from the 144-in. length of the beam. We lind

=

II = 130.5 in .. I! = 11 [ .3 in.. The corners of thl' various

356

plank.~

I~

= 83.8 in . ....

lie on the parabola defined by Eq. (2).

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16 j)

C

SAMPLE PROBLEM 5.12

L

Twu slt:c1 plates. c
"~

W6!J() x 125 of Ih~ pla tes.

J I- ~ HI --~4In

I' 1-1/-1-~/-

SOLUTION Bendin g " 'IQmcnl. We first firld the r~ac uoLls. Fro m the free body of D JXlrtion of beam of length x $ 4m, we obtain M between A and C:

l et

•••!"::jil

"r:::J r • ••'::~=:::;

,I)

M = (250 kN J_(

u. Rc quir«l Lcllg l b or PI:,tu . We filst determine the ma.( iIllUIil allowab le letlgth x .. of the portion AD of the unremforced beam. From Appendix C we lind Ihm Ihe section modulus of a Wo90 x J25 beam is 5 = 3510 X IO~mmJ. or S '" 3.5 1 X [0 3 m3 , Substituting for S amI lT m into Eq. (5.17) ,md solving for M. we write

M = Str.,1 = (3.51 X IO- J m l)060 X 10J kN/m1) = 561.6 kN . m Su bs!i!u!mg for M ill Eq, (I ). we have 561.6 kN . m = (250 kN)x.,

x'" = 2.246m

The reqllired le nelh I of rh e pl:nes i ~ ohw ine rl h y <;nhl mc. rine 2 \... fmlll lh e. length of the beam: 1= ] .5 1 m
11. Rcqui rtd W id th of Plilles. The max imu m bending rmmenl occ ~rs in Ihe mid sec tion Cof the beam. Mnki!\g x - 4 m in Eq. (1). we obl!un Ill e bendinJ< moment in 111m seclion: fit = (250 kN )(4 m )

I,

IT

I r'i'ld

1', - -- 1

1 , N \. ,

= 1000 kN . m

[n order 10 use Eq. (5. 1) of Sec. 5 . 1. we now delermine lhe moment of in· ertia of Ihe cross section of the reinforced beam with r~s]X'Ct 10 a cenlroidal axis ami the dislance c from that axi s to !lIe outer surfaces of the p[mes. From Appendix C we find that the mo ment of inerlia of u W690 X 125 beam is I., = 1190 X I fY' mm 4 and its depth is d = 678 mm. On the other hand. d ~· noting by f the thickness of one plate. by bits widlh, :l11d b) y the di stance of itS ce11!roid from Ihe neulral axis. we express lhe moment of inertia 1/1 Df Ihe two plate~ with res pecl 10 Ihe neutral ax,s:

I,! = 1f,tih~ + Ayl) = (f ll)b + 2 b~1d + 11)2 SUbSlilutmg 1 - 16 mm and d - 678 mm. we obtain I~ - (3 .854 X 1(]I> mmJ) h. The mom ~l1t of inenia 1 of the beam and plates i,

!

=" +

Ip = 1190 X 10" mm 4

+ (3 .854

X 101 mmJ)b

m

:uuJ the di slance from lht' neutral axi s to Ihe surface is c = ~ d + I = 355 mm. Solving Eq. (5.1) for 1 and ~ubstitl1ling the v;l lues of M . (rali • and (., wewri le

IMk

( lOoo k..~ . m)(J55 mm )

if.!!

16()MPJ

1~ -- =

2.219 X 10

j

m4 = 2219 X 10" mm 4

Replacing I by th is value in Eq. (2) and ;mlving for b. we have 2219 X 10(· mm 4 11 90 X 106 mm 4 + (3.854 X IO~ mm 1)b

=

b= 267 nll11 ....

357

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PROBLEMS

5.126 and 5.127 The beam AiJ. consisting of a cast-iron plate of uniform thickness b and length L. is to support the load shown. (0) Knowing thm the beam is to be of constant strenglh. express h in lerms of x. L, and 110 , (b) Determine the maximum allowable load if L = 36 in .. 110 = 12 in., b = 1.15 in .. ami (T..!I = 24 ksi.

Fig. PS.127

Ag. PS.126

5.128 and 5.129 The beam AB. consisting of an aluminum plate of unironll thickness b and length L , is to support Ihe load shown. (a) Knowin g that the beam is to be of constam strength. express Ii in terms of x. L. and " I) for portion AC of the beam. (b) Detennine the maximum allowable load if L = 800 mm, 110 = 200 mill, b = 25 mm. and (loll = 72 MPa. p

Fig. PS.12B

Fig. P5.129 5.130 and 5.131 The beam AB. consisting of a cast-iron plme of unifornl thickness b and length L. is to suppon Ihe distributed load 1V(X) shown. (a) Knowing Ihm the beam is 10 be of constalll strength. express" in lenns of x, L. and hlr (11) Detemline the smallest value of if L = 750 mm. b = 30 mm. Wu = 300 kN/m. and (1",., = 100 MPa.

"0

..

~. I ' l~Ul ~

"

~" ,----I

Ag. P5.130

358

"

_ I I,"

' '1 I

l~t" ,----I

FIg. PS.131

_ I /"

, =-T I

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5.132 and 5. 133 A prelimillary de.o; lgn unth.: u,e ofa calililever pJismatic timber beam indicaled Ihal a beam wilh a rectan gu lar cross section '2 in. wide anti 10 in. deep would be required 10 safely gUppott the load shown in patt a of the figure. It was Ihen decIded 10 replace thaI beam with a built-up beam obtained by glu ing togeill
Problem.

J - - - - f d . 5 f t - - - - I_

,

/J

c

- - - - 6.2.5

r'----L

D A

A

1M

(b)

Fig. P5. 132

Fig. PS.133

5.134 and 5.135 A preliminary llesign on the use of a simply supported prismatic limber beam indicatell illat a beam with a rectangular cross sec tion 50 mm wide and 200 mm deep would be required 10 safely support Ihe load ~hown in pan 1I of the figure. It wa~ then decided 10 replace that beam with a built-up beam ootained by gluing toge ther. as showll ill part b of the ligure. four pieces of the same timber as the onginal beam and of 50 x 50-mm cross ~ecliun.

Ihe

DcltTmille lile Itngtll I uf

~aml'.

faclilr nf

s~feTy a~

Ih~

IWO OUleT

p iec ~~

of limlJcr Ihw will yklU

Ihe mi£in.11 rlesign

("'

(0 1

,

.,

,\

- (- I

1--,-1

(II)

,"I

Fig. P5.134

Fig. PS.135

,

359

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360

Analysis and Design of Beams for Bending

5. 1:36 a nd 5.137 A machine elemen! of casl aluminum and in Ihe slwpc of a solid of revolUlion of vruiable diameter d is being designed 10 supporl the lo;!d shown. Know ing th;!1 the machine element is to be of conslanl sirength. express d in terms of x. L. and duo

11111 11111111

Fig. PS.136

Fig. P5.137

5.138

A canl ilever beam AD consisting of a sleet plate of uniform deplh

II and variable width b is to support Ihe concentrated load P at poinl A. (u) Knowing that the bc;!m is 10 be of constant strength. express b in lerms of 1". L and bu. (b) Determine the smallest allowable value of II if L = 300 mm. bu = 375 mm. P = 14.4 kN. and 17," 1 = 160 MPa.

'-"

Fig. P5.138

5.139 A transverse force P is applied as shown al end A of the conical laper AB. Denoting by do the diameler of Ihe taper at A, show thai the maximum normal stres~ occurs at poinl H, which is conlained in a trans\'erse section of diameler d = 1.5 do.

"t!pI

A

Fig. PS.139

5.140 Assuming thm Ihe length and width of the cover plmes used wilh the beam of Sample Prob. 5.12 are. respectively, I = 4 m ;!nd b "" 285 mm. and recalling Ihal Ihe thickness of each plate is 16 mm, delermine the maxiilium nonnal stress on a transverse seclion [u) through the l'enter of Ihe beam. (b) jusl to the left of D.

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5. 141 Two cover plates. each ~ in. thick. are welded to a W27 X 84 beam as shown. Knowing that I = 10 ft and h = 10.5 in .• determine the maximum nomlal stress on a transverse section (u) through the center of Ihe beam. (h ) just [0 the left of D. J!~I

klpo;

Fig. PS. 141 and PS.142

5 .142 Two cover plates, each ti n. Ihick. are welded to a W27 X 84 beam as shown. Knowing Ihat rr 01, = 24 ksi for bolh the beam and the plales. determine the required value of (u) the length of the plates, (b) Ihe width of Ihe plales. 5.143 Knowing thai !T.n = 150 MPa. determine the largest concentrated load P Ihal can be applied a[ end E of the beam shown.

Fig. PS.143

5 .144 Two cover plates. each 7.5 mm thick. are welded [0 a W460 X 74 beam as shown. Knowing that I = 5 m and b = 200 mm. detemline the maximum nonnal stress on a transverse section (u) through the center of the beam. (b) just [0 Ihe lef[ of D. 10 kNIll ,

f-- - - -8,,, - - --1 Fig. PS.144 and PS.14S

5.145 Two cover plates. each 7.5 mm thick. are welded loa W460 X 74 beam as shown. Knowing that (r ali = I SO MPa for both Ihe beam and [he plates. detennine Ihe required value of (a) Ihe length of the pla[es. (b) Ihe width of lhe plates.

Probloms

361

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362

Analysis and Destgn of Beams for Bendtng

5 .146 Two cover plates. each ~ in. th ick. are welded to a W30 X 99 beam as shown. Knowing that I = 9 fI and b = 12 in .. detennine the maxi~ mum nonnal stress on a transverse section (a ) through the center of the beam. (b) just to the left of D.

Fig. PS.146 and PS.147

5.147 Two cover plates. each ~ in. th ick. are welded to a W30 X 99 beam as shown. Knowing th;]t (fall = 22 ksi for bOlh the beam and the plates. detennine the required value of (a) the length of the plates. (b) th e width of the pl;]tes. 5 .148 For the tapered beam shown. detennine (a) the transverse section in which the maximum normal stress occurs. (b) the largest distributed load IV that can be applied. knowing th;]t (T. U = 24 ks i.

Fig. PS.148

Fig. PS.149

5 .149 For the tapered beam shown. detennine (a) the transverse section in which the maximum nonnal stress OCCUIS. (b) the brgest concentrated load P that can be applied. knowing that 0"011 "" 24 ksi.

5.150 For the tapered beam shown. detennine \a) the transverse section in which the maximum normal stress occurs. (b) the largest distributed load 'IV that can be applied. knowing that O".oJ! = 140 M Pa.

Fig. PS. 1S0 and PS.151

5 .151 For the Tapered beam shown. knowing that w = 160 kN/m , deTermine (a) the transverse section in wh ich the maximum norm;]l stress occurs. (b) the correspondin g value of the normal ~tress.

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REVIEW AND SUMMARY FOR CHAPTER 5

This chapler was devoted to the analysis and design of beams under transverse loadings. Such loadings can consist of concentrated loads or distributed loads and the beams themselves arc classified according to the way they are supported (Fig. 5.3). Only statically determinate beams were considered in this chapter, the analysis of statically indeterminate beams being postponed until Chap. 9.

Considerations for the design of prismatic beams

SMkaliy Dckrminatf'

Be"ms

11 -- . --I

I---

L

(Ii )

Simply snpportrd 1><;:11"

St"tkally l,uteternlin l.!" Bf'am,

I-- - L- ---I (d ) Conti,,,,,,,,", heam

- L- ---I

(c)

(to) I'\e"", fm"lat on," end ~nd ., jmply "'pportf'd

Cantil(" '~r

hf'am

~ L ---I (f) Fixed h<'am

at the oll,('r end

Fig. 5.3

While transverse loadings cause both bending and shear in a beam, the normal stresses caused by bending are the dominant criterion in the design of a beam for strength [Sec. 5.1 [. Therefore, this chapter dealt only wi th the determination of the normal stresses in a beam. the efrect of shearing stresses being examined in the next one. We recalled from Sec. 4.4 the nexure formula for the determination of the maximum value (T", of the nonnal stress in a given section of the beam,

Normal stresses due to bending

(5.1) where I is the moment of inertia of the cross section with respect to a centroidal ax.ls perpendicular to the plane uf the bending couple M and c is the maximum distance from the neutral surface (Fig. 4.13).

'1---'-- ". Fi g. 4.13

363

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364

Analysis and Design of Beams for Bending

We also recalled from Sec. 4.4 that, Introduci ng the clastic section modulus S =: Ilc of the beam, the maximum value (T~, of the normal stress in the section can be expressed as (5.3)

Shear and bending-moment diagrams

(a )

Inkmal foro's

(posi!i\'<' sh,'ar and pm;!i,"" l)('nding 1Il0m"nt)

Fig.5.7a

Relations among load, shear, and bending moment

It follows from Eq. (5.1) that the maximum normal stress occurs in the section where IMI is largest, at the point farthe st from the neural ax is. The determination of the maximum value of IMI and of the critical section of the beam in which it occurs is greatly simplified if we draw a shear diagram and a belldillg-momellf diagram. These diagrams represent, respectively, the variation of the shear and of the bending moment along the beam and were obtained by determining the values of Vand M at selected points of the beam [Sec. 5.21. These values were found by pass ing a section through the point where they were to be determined and drawing the freebody diagram of either of the portions of beam obtained in this fa shion. To avoid any confusion regarding the sense of the shearing force V and of the bending couple M (which act in opposite sense on the two portions of the beam). we followed the sign convention adopted earlier in the text and illustrated in Fig. 5.7a [Examples S.OI and 5.02. Sample Probs. 5.1 and 5.21.

The construction of the shear and bending-moment diagrams is facilitated if the following relations are taken into account [Sec. 5.3]. Denoting by 'W the distributed load per unit length (assumed positive if directed downward), we wrote dV

= - w dx

dM

dx

v

(5.5,5.7)

or, in integrated form.

VD

-

MD ~

Vc "" - (area under load curve between Cand D ) M c = area under shear curve between Cand D

(5.6') (5.8')

Equation (5.6') makes it possible to draw the shear diagram of a beam from the curve representing the distributed load on that beam and the value of Vat one end oFthe beam. Similarly, Eq. (5.8') makes it possible to dra.w the bending-moment diagram from the shear diagram and the value of M at one end of the beam. However, concentrated loads introduce discontinuities in the shear diagram and concentrated couples in the bending-moment diagram, none of which is accounted for in these equations [Sample Probs. 5.3 and 5.6]. Finally. we noted from Eq. (5 .7) that the points of the beam where the bending moment is maximum or minimum are also the points where the shear is zero [Sample Prob. 5.4l

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A proper procedure for the design of a prismatic beam was described in Sec. 5.4 and is summarized here: Having detennined (Toll for the material used and assuming that the design of the beam is controlled by the maximum normal stress in the beam, compute the minimum allowable value of the section modulus:

1MI " . .

S"' i"= ~~­

Review and Summary for Chapter 5

Deslgn of prismatJc beams

(5.9)

(Tall

t

For a timber beam of rectangular cross section,S = bh 2 , where b is the width o f the beam and" its depth. The dimensions of the section. therefore. must be selected so that ~ bh 2 2: 5,,,,,,. For a rolled-steel beam, consult the appropriate table in Appendix C. Of the available beam sections, consider only those with a section modulus S 2: S"." and select from this group the section with the smallest weight per unit length. This is the most economical of the sections for which 5 2: 5,,,,". In Sec. 5.5. we discussed an alternative method for the determination of the maximum values of the shear and bending moment based on the use of the sillgulariry/unctiolls \x - a'r. By definition. and for 1/ 2:: 0, we had

(x - a'r

~

{b' - a)"

when x

2:

a

when x

<

a

(5. [4)

We noted that whenever the quantity between brackets is positive or zero. the brackets should be replaced by ordinary parentheses. and whenever that quantity is negative, the bracket itself is equal to zero. We also noted that singularity functi ons can be integrated and differentiated as ordinary binomials. Finally. we observed that the singu larity function corresponding to 1/ = 0 is discontinuous at x = a (Fig. 5.ISa). This function is called the .I·tep jlll/ction. We wrote

(x - a)'~ {~

when x

2:

a

when x

<

a

"

o (a )"

Fig. 5.18a

=0

Singularity functions

(5.15)

Step functlon

365

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366

Analysis and Design of Beams for Bending

Using singularity lunctions to express shear and bending moment

p

r- tL--J-c-- tL - -A

The use of singu larity functions makes it possible to represent the shear or the bending moment in a beam by a single expression, valid at any point of the beam. For example, the contribution to the shear of the concentrated load P applied at the midpoint C of a simply supported beam (Fig. 5.8) can be represented by - p(x - ~ L)O, since this expression is equal to zero to the left of C, and to - P to the right of C. Adding the contribution of the reaction RA = tp at A, we express the shear at any point of the beam as V(x ) "" ~P - P{x -

}L)O

The bending moment is obtained by integrating this expression: ,

M(x) = iPx Fig. 5.8

Equivalent open-ended loadings

P(x -

~L)l

The singularity functions representing, respectively, the load, shear, and bending moment corresponding to various basic loadings were given in Fig. 5.19 on page 346. We noted that a distributed loading that does not extend to the right end of the beam, or which is discontinuous. should be replaced by an equivalent combination of open-ended loadings. For instance, a unifonnly distributed load extending from x = a to x = b (Fig. 5.20) should be expressed as

w(x) = wo(x - a)O - wJ..x - b)O

Fig. 5.20

Nonprismatic beams

Beams of constant strength

The contribution of this load to the shear and 10 the bending moment can be obtai ned through two successive integrations. Care should be taken, however. to also include in the expression for V(x) the contribution of concentrated loads and reactions, and 10 include in the expression fo r M(x) the contribution of concentrated couples [Examples 5.05 and 5.06, Sample Probs. 5.9 and 5.101. We also observed that si ngularity functions are particu larly well suited to the use of computers. We were concerned so fa r only with prismatic beams, i.e .. beams of uniform cross section. Considering in Sec. 5.6 the design of nonprismatic beams, i.e., beams of variable cross section, we saw that by selccting the shape and size of the cross section so that its elastic section modulus S = lIe varied along the beam in the same way as the bending moment M, we were able to design beams for which Um at each section was equal to U all' Such beams, called bewlIS of COllsta/lf strellgth. clearly provide a more effective use of the material than prismatic beams. Their section modulus at any section along the beam was defined by the relation M

S~ ­

U .!I

(5. 18)

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REVIEW PROBLEMS 251) nUll

5 .152 Dr:lw the shear and bending-moment diagrams for the beam and loading shown. and determine the maximum absolute value (a) of the shear. (b) of the !>ending moment

7:;:-.J

250 ""11

2S()

"un

,.'j r-.

Fig. PS.152

5.153 Determine (a) the magnitude of the upward force P for which the maximum absolute value of the bending moment in the beam is as small as possible. (b) the corresponding maximum normal stress due \0 bending. (Hillt: See hint of Prab. 5.27.) 5 .154 Draw the shear and bending-moment diagrams for the beam and loading shown. :lIId determine the maximum abso lute value (a) of the shear, (b) oflhe bend ing moment.

\\'310 X 23.t!

X I", I "' Fig. PS.1S3

I",

j III

W8 X 3 1

Fig. P5.154

Fig. PS.1SS

5 .155 Draw the shear :md bending-moment diagrams for the beam and loading s hown and de termine the maximum nonnal stress due to bending. 5 .156 Beam A B. of length L and square cross section of side a. is supponed by a pillot at C and loaded as ~hown. (a) Check that the beam is in equilibrium. (b) Show that the maximum stress due to bending occurs at C and is equal to 1vl)i.~/(1.5(/)1.

II/kNIll'

121) mm

IIIIIIIIIIIIIII -j

+

Fig. P5.157

5",

¥

5. 157 and 5.158 For the beam and loading shown. design the cross section of the beam . knowing that the gmde of timber used has an allowable nomlal stress of 12 MPa.

Fig. PS.1S8

367

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368

Analysis and Design 01Beams for Benalng

'-iB~~=.. Cq

A~

5. 159 Knowing that the allowable stress for the steelllsed is 24 ksi. Si:lect the most economical wide-flange beam to suppon the loading ~hown.

5.160 Determine the largest pemlis.ible value of P for the beam and loading shown. knowing thai the allowable normal stress is +80 MPa in tellsion and - 140 MPa in compression.

D

U-12" 5f<

Fig. P5.159

(I.2.'1m

BCD E

1\

LU.- 6[, --.:..J

1ft 2ft Fig. P 5.161

2fr

0. 15 ".

Fig. P 5.160

5.161 «(I) Using singularity functions. write Ihe equations for the shear and bending moment for the beam and loading shown. (h) Determine the maximum value of the bending moment in the be:1I11. 5. 162 The beam AB. consisting of an aluminum pbte of uniform thickness IJ and length L is 10 s uppon the load shown. (al Know ing that the beam is to be of constant strength. express fr in temlS of x. L. and fro for ponion AC of the beam. (h) Determine the maximum allowable IOlldif L = 32 in., Irll = 8 in .. b = I in .• and lT~u = 10 ksi.

Fig. P5.162

5.163 A cantilever beam AB consisting of II steel plate of uniform depth " and variable width IJ is \0 suppon the distributed load )II along its centerline AB. (a) Knowing that the beam is to be of constant strength. express b in temlS of x. L. and boo (h) Detemline the maximum allowable value of w if L = 15 in .. bQ = 8 in .. It = 0.75 in .. and U",11 = 24 ksi.

Fig. P5.163

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COMPUTER PROBLEMS

T h .. following problems n re dfsign cd t o be. s()in' t1

~ilh

II co mput er.

5 .Cl Sev~ral concentmled loads 1"; ( I = 1. 2 ..... /I ) can be applied 10 a beam !l.~ s l!ow['\. Write a computer program thaI can ~ used to calculate the shear. bt:nding moment. and normal stress al any point cf the beam for a g iven IO;Jding of tn.e beam and a given value of ils section modulus. Use this program to sol\c Pcobs. 5.18, 5.21. ami 5.25. (H;'d. Maximuill valu es wil l oeCU]

ffi'11 _

-l2

J', -

.l:

~

PI

p~

~

I

P,

1'"

at a support or under a load.) 5 .C2 A timber beam is to be designed [0 support a dIstributed load and up to two concentrn ted load~ ru; shown. One of lite di me nsions of its uniform reclan),!ular tross section has ben specified and the olher is 10 be detemlined Kl that the maximum normal Slres~ in the beam 'Will nOI exceed a given 31lowable value (foil. Wrile a complller program 111m can be used 10 calcul:lle a\ given imerval~ ilL the ~hear, the bending moment. and the mlalleSI acceplable value of Ihe unknown dimension. Appl y Ihis program 10 solve Ihe following problem~. using the intervnls !J.L indicated: (a) Prob. 5.6.5 (aL = 0.1 r.l). Ib) Pmb. 5.69 (!J.L - 0.) mI. (e) Pmb. 5.70 (!J.L - 0 .2 mI.

Fig. P5.C1

Fig. P5.C2

" S.C3

1\vo cover plates. eJch of thic.kness I. are to be welded to a widetlange be:lll! of length L thai i~ to gupport a uniformly di stributed lead w. IN_ noting by if. u th~ allowable normal stre;s in th e beam and in the plmes. by ,/ the depth o[ lhe beam. and by I/,a nt:! Sh. respec thel y. the moment of menia and 1111: ,el:liull rmxlulus uf the l"fU~S St:,",liull uf the uIJrcinfu[l;eu ueam alluUl a horizontal cenlroidal axis. wril~ a ClIITlputel program titat can bl'- used to l:alcu l.1le the reqUired value of (a) the lenglh " of the plates. (b) the width b of the pl.1tcs. Use thi s program 10 501'11.' Prob. 5.115.

Fig. P5.C3

369

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370

Analysis and Design 01 Beams for Bending

Fig. P5.C4

1

"

5.C4 Two 25-kip loads are maintained 6 [t apart as Ihey are moyed slowly across the 18-[1 beam AB. Wrile a complller program and use ilia calculate Ihe bending momenl under each load and at the midpainl C o[ Ihe beam [or yalues o[ x [rom 0 to 24 ft al inlervals ix = J.5 fl.

1

1111 11111 11

Fig. P5.C5

t

5 .C5 Write a computer program Ihal can be lIsed 10 pial the ~hear and bending-moment diagrams [or Ihe beam and loading shown. Apply this program wilh a planing interval tJ.L = 0.2 ft 10 Ihe beam and loading of (a) Proh. 5.72. (b) Prob. 5.115.

j, ~ F"111111111

"

~1111111.!::::':-=-~

AI .

I- -L- -II

Fig. P5.C6

S.C6 Write a compulCr program Ihal can he used 10 plot Ihe shear ami bending-momem diagmms For Ihe beam and loading shown. Apply Ihis program wilh a plollingimerval tJ.L "" 0.025 m 10 Ihe beam and JO
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C

HAP

T

E

Shearing Stresses in Beams and Thin-Walled Members

A reinlorced concrete deck will be aHached to each of the steel sections shown to form a composite box girder bridge. In this chapter the shearing stresses will be determined In various types of beams and girders.

R

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372

Shearing Stresses In Beams and Tllin-Walled Members

6.1. INTRODUCTION You saw in Sec. 5. [ that a transverse loading applied to a beam will result in normal and shearing stresses in any given transverse section of the beam. The nomlal stresses arc created by the bending couple M in that section and the shearing stresses by the shear V. Since lhe dominanl criterion in the design of a beam for strength is the maximum value of the normal stress in the beam, our analysis was limited in Chap. 5 to the determination of the nonnal stresses. Shearing stresses, however. can be important, particularly in the design of short, stubby beams, and their analysis will be the subject of the first part of this chapter.

Fig. 6. 1

Figure 6.1 expresses graphically that the elementary nonnal and loro..;es cxcrted 011 a givclI trans vclSc scction of a pris matic beam with a vertical pilUle of symmetry arc equivalent to the bending couple 1\1 and the shearing force V. Six equations can be wriuen 10 express thut fact. Three of these equations involve o nl y the normal forces u _. dA and have already been discussed in Sec 4.2; they art! Eqs. (4.1), (4.2), and (4.3), which express that the sum of the normal forces is zero and thut the sums uf their moments about the y and:; axes are elJual to -zero and M. respectively. Three more equations invo lving the shearing forces 7 xv dA and 7 .<.: dA can now be written. One of them expresses that the su m of the moments of the shearing force s about the x axis is zero and can be dismissed as trivia l in view of the symmetry of the beam with respect to the X}' plane. The other two involve the y and z components of the elementary forces and are ~hearj llg

)' components: :: components:

u,

Fig. 6,2

fr.' )dA =- V =0

fr~,dA

(6. 1) (6.2)

The first of these equations shows that vertical shearing stresses must ex ist in a transverse section of a beam under transverse loading. T he second equation indicates that the average horizontal shearing stress in any sect ion is zero. However, this does not mean that the shearing stress 7 .r. is zero everywhere. Let us now consider a small cubic element l o~~ated in the vertical plane of symmetry of the beam (where we know that T,-, must be zero) and exami ne the stresses exerted on its faces (Fig. 6.2). As we have just seen, a normal stress IT_< and a shearing stress T .t). are exened on each of the two faces perpendicular to the .r axis. BUI we know from Chap. I that, when shearing stresses 7.~ are exerted on the vertical faces of an

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clement. equ:11 stresses must be exerted on the horizontal faces of the same element. We thus conclude that longitudinal shearing stresses must exist in any member subjected to a transverse loadi ng. This can be verilied by considering a cantilever beam made of separate planks clamped together at one cnd (Fig. 6.3a). When a transverse load P is applied to the free end of this composite ocam, the planks are observed to slide with respect to each other (Fig. 6.3b). [n contrast. if a couple M is applied to the free end of the same composite bemn (Fig. 6.3c), the various planks will bend into concentric arcs of circle and will not sl.ide with respect to each other. thus verifyi ng the fact that shear docs nol occur in a beam subjected to pure bending (d. Sec. 4.3). While sliding docs not actually take place when a transverse load P is applied to a beam made of a homogeneous and cohesive material such as steel. the tendency to slide does exist, showing that stresses occur on horizontal longitudinal planes as well as on vertical transverse planes. In the case oftimbcr beams, whose resistance to shear is weaker between fibers. failure due to shear will occur along a longillldinal plane rather than a transverse plane (Fig. 6.4). [n Sec. 6.2, a beam element of length 6.x bounded by two transverse planes and a horizontal one will be considered and the shearing force 6. H exerted on its horizontal face will be determined. as well as the shear per uni t length. q, also known as shear flow. A formula for the shearing stress in a beam with a vertical plane of symmetry will be derived in See. 6.3 and used in Sec. 6.4!() determine the shearing stresses in common types of beams. The distribution ()f s tres~es in a narrow rectangular bt:am will be funher discussed in Sec. 6.5. The deri vation given in Sec. 6.2 will be extended in Sec. 6.6 to cover the case of a beam clement bounded by two transverse planes and a curved surface. Thi s will allow us in Sec. 6.7 to determine the shearing stresses at any point of a symmetric thin-walled member, such as the flanges of wide-flange beams and box beams. The effect of plastic deformations on the magnitude and distri bution of shearing stresses will be di scussed in Sec. 6.8. In the last section of the chapter (Sec . 6.9), the unsymmetric loading of thin-walled members will be considered and the concept of shear center will be introduced. You will then learn to determine the distribution of shearing stresses in such members.

Fig. 6.4

6.1. Introduction

,", I'

Fig. 6.3

373

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374

Shearing Stresses In Beams and Tllin-Walled Membe~

6 .2. SHEAR ON THE HOR IZONTAL FACE OF A BEAM ELEMENT

Fig. 6.5

Consider a prismatic beam AB with a vertical plane of symmetry that supports various concentrated and distnbuted loads (Fig. 6.5). At a distance x from end A we detach from the beam an element CDD'C' of length ~x extending across the width of the beam from the upper surface o r the beam to a horizontal plane located at a distance YI from the ncutral axis (Fi g. 6.6). The forces exerted 011 this element consist o f

Fig. 6.6

';tmr" "

C

urrlA

D

if,,".'

- ~n

Fig. 6.7

vertical shearing fo rces V 'c and V'o, a horizontal shearing force .1. H excr1ed on the lower face of the element , elementary horizontal nonnal fo rces ere (fA and (TD (fA. and possibly a toad w ~x (Fig. 6.7). We write the equilibrium equ;:uion

~ 2F., = O:

i1H

+

J((TD -

lTddA = 0

" where the integral extends over the shaded area (f of the section located above the line y = h Solving this equation for t1H amI using Eq. (5.2) of Sec. 5.1. IT = My/I, to express the normal stresses ill terms o f the bending moments at C and D . we have

(6.3)

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The integral in (6.3) represents thejirxt m Oll/elll with respcetto the neutral axis of the portion U of the cross section of the beam that is located above the line)' = )'1 and w ill be denoted by Q. On thc other hand. recalling Eq . (5.7) o f Sec. 5.5, we can express the incremc.nt M D - M e of the bending momelll as MD - M e = fl.M = (dM / dr) fl.x = V fl.x

Substituting into (6.3), we obtai n the following expression for the horizolllal shear exerted on the beam element fl.H =

VQ -, - d,

(6.4)

The same result wo uld have been obtained if we had used as a free body the lower clemelll C' D' D"C" , rather than the upper cle melll

Fig. 6.8

e DD'C' (Fig. 6,8). since the shearing fo rces fl." and fl.H ' exerted by the two e l emenl~ on each other are equal and opposite. This leads us to observe that the first moment Q of the portion H' of the cross section located be low the line J' = .\'1 (Fi g. 6.8) is equal in magnitude and opposite in sign l() the first moment of the portion (f located above that line (Fig. 6.6). Indeed. the sum of these two momel11S is equal 10 the momel11 of the area of the entire cross secti on with respect to its centroidal axis and, thus, must be zero. Th is pmpert y can sometimes be used to simplify the computatio n of Q. We also note that Q is maximum .\'1 = 0, since the elements of the cross section located above the neutral axis cont ribute positively to the integral (5 .5) that de fine... Q, while the c lements located below that axis contribute negatively. The IWl"jzontal .l"/l etll" pt'r IIIlitlength, which will be denoted by the letter (/. is obtained by di viding both me mbers of Eq. (6.4) by fl. .\":

ror

~H

q=

n..r

=-,VQ

(6.5)

We recall that Q is the lirst moment with respect to the neutral axis of the portion of the cross sec tion located either above or below the point at which q is being computed, and that I is the centroidal moment of inerti a of the {'mire cross-secti onal area. For a reason that will become apparent later (Sec. 6.7). the horizontal shear per unit length q is also referred to as the ~·he(J/"J1o\l".

6.2. Shear on the Horizontal Face of a Beam Element

375

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EXAMPLE 6.01

A beam is made of three planks. 20 by 100 mm in cross sec· tion. nailed together (Fi'g. 6.9), Knowing that the spacing between nails is 25 mm and that the .... ertical shear in the bc:lZll is V = 500 N, determine the shearing force in e;Jch nail.

r-J l)jjllllll l ---L

I "'"'''

Q ""

Fig. 6.9

.LJ- ~,:~) "'1 .-

]-;;.

J

=

16.20 X 10 (, m4

.

T0.000 ",

1 I

I

0. 100 ,,1

\

1 -l

~

~ 0,020lH

("'

Substituting into Eq . (6.5), we write q =

,,- =

0 .U20 III

1 = # 0.020 m)(fJ.IOQm f

(500 N)( 120 X 10 . 6 Ill l )

, 0'"""

o.

= 120 X 1O- ~ m l

VQ

~

~~t:::::I4111'

Ay """ (0.020 m X 0.100 m)(O.060 Ill)

+ 2[# 0.100 111)(0.020 Ill )] + (0.020 m X 0.100 m)(0.060 m)l ] = 1.667 X 10 6 + 2(0.0667 + 7. 2)10 "" 16.20 X 10 ~ nl'

1011 Ill'"

l!.tJ Ill"' -

We first determine the horiwntal force per unit length, q, exerted on the lower face of the upper plank. We use Eq. (6.5). where Q represents the fim moment with respect to the neu· tral axis of the shaded area A shown in Fig. 6.IOu, and where I is the moment of inenia about the same axis of the entire cross·sectional area (Fig. 6. IOb), Recalling thaI lhe first moment of an area with respect to a gi ven axis is equal to the proouct o f the arell and of the distance from its centroid 10 the axi s.t we have

Fig. 6.10 ] 704 N/ m

Since the spac ing between the nails is 25 mm. the shearing

force in each nail is

F = (0.025 m)q = (0.025 m)(3704 N/ m) = 92.6 N

6 .3 . DETERMINATION OF THE SHEARING STRESSES IN A BEAM Consider again a beam with a vertical plane of symmetry. subjected to various concentrated or distributed loads applied in that plane. We saw in the preceding section that if, through two vcnical cuts and one horizontal cut. we detach from the beam an element of length dx (Fig. 6. 1 [), thc magnitude dH of the sheari ng force exerted on the horizontal face of the element can be obtained from Eq. (6.4). The (ll'erage .I·hearing .I·tress T a l·c on that face of the element is obtained by dividing dH by the area dA of the face . Observing that dA "'" t dx , where 1 is the width of the clement at the cut, we write !1H

r.,," "'" !1A

VQ!1.r = - ,-

f

d.r

(6.6)

376

t See Appendi x A .

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We note that, since the shearing stresses T ,r\' and T ;:< exerted respecti vely o n a transverse and a horizontal plane through D' are equal , the expression obtai ned also represents the average value o r T " ,' a long lhe line D 'I D'! (Fig. 6. 12).

c;'

6.4. Shear1ng Stresses Txy In Common "TYpes 01 Beams

OJ'

Fig. 6.12 FIg. 6. 13

We observe that Tn- = 0 on the upper and lower faces o r the beam. since no forces are exerted on these faces. It fo llows that T;q = 0 along the upper and lower edges o f the transverse section (Fig. 6.1 3). We also note that , while Q is maximum lo r ), = 0 (see Sec. 6.2), we cannot conclude that T m will be maximum along the neutral ax i .~, since T aY< depends upon the width t o f the section as well as upon Q. As long as the width of the beam cross section remains small compared to its depth, the shearing stress varies o nly sli ghtly along the line D'I D'2 (F ig. 6.12) and Eq. (6.6) can be used to compute T,,,' at any poi nt along D'I D'2' Actually. T.•! is larger at points D'I and D '2 than at D', but the theory o f elasticity showst that, for a beam of rectangular section of width b and depth h. and as long as b ::s: 11/4, the value of the shearing stress at points CI and ~ (Fi g. 6.14) does 1I0t exceed by more [han 0.8% Ihe average va lu ~ of the stress computed alo ng the neutral axis..:j: 6.4. SHEARING STRESSES TYPES OF BEAMS

Txy

IN COMMON

We saw in the preceding section that, for lInlirrolV rectangular beam. i.e. , for a beam of rectangular section of width IJ and depth h with IJ S ! II. the vari atio n of the shearing stress T" across the width of the beam is less than 0.8 % of T . ,c. We can, there fore. use Eq . (6.6) in practical app li cation ~ to determine the shearing stress at any poi nt of the cross section of a narrow rect:mgul ar beam and write VQ

T ry

=

It

(6.7)

where I is equal to the widlh b of the beam, and where Q is the first t Scc S. P. Till1l)shcnko and J. N. Good icr, T/'M'), Q[ Ela.
"-./T••~

,...... /T.. ,

0.25

,.,

I.()()S 0.996

1.033 [).983

J.J26 0.9 .. 0

1.]96 0.856

1.988 0.805

1.51:12 (),8oo

"

3.770 0.800

6.740 0.800

"

15.65 0.800

I 'I.

FIg. 6.14

377

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378

Shearing Stresses In Beams and

moment with respect to the neutral axis of the shaded arca A (Fig. 6.15 ). Observing rnal the distance from tht:: ne utral axi s III the centroid C ' of A is Y = He + .v), and rcc:Illing that Q = AY. we write

Thin-Wailed Membel15

Q ""

Ay = b(e - y)!{c + y) =

t b(c~

- /)

Recalling. on the other hand. that J = bl,Jj12 "" ! hc T o:v

= VQ =

Ib

~r

-

4

he l

3

,

(6.S)

we have

J'~ V

or, noting thar the cross-sectional area of the beam is A = 2/1£', T <.l'

=

~~ ([

2 A

-

i.)

(6.9)

('2

E(1u:.ttion (6.9) shows that the distribution of shearing stresses in 11 transverse section of a rectangular beam is parabolic (Fig. 6.16). As we have already observed ill the preceding section. the shearing stresses are zero at the top and bottom of the cross section (y "" ± c). Making y = 0 in Eq. (6.9), we obtain the value of the maximum shcaring stress in a gi ven section of a narrow rectan gular beam:

Fig. 6.15

J l u~---l.,_,-·

3V

= --

T

(6.10)

2A

rna.>.

The relation obtained shows that the maximum value of the shearing stress in a beam of rectangular eross section is 50% larger than the valuc V!A titat would be obtained by wrongly a~ s umjn g a uniform stress diMribution across the e ntire cross seelion. In the case o f an American .I·f(/mlunl beam (S-beam) or a witlefla nge beam (W-beam), Eq. (6.6) can be used to determine the average value of the shearing stress T ..... over 11 section (la ' or bb' of the transverse cross section of the beam (Figs. 6. 17a and b). We wri te

I Fig. 6.16

VQ

T a"e=

h

(6.6)

where V is the vertical shear, t the width or the section at the e levation considered, Q the first moment o f the shaded area with respect to the nelltral axi s ee', and Ithe moment of inertia of the enti re cross-sectional area about ec '. Pl ouing T~", against the vertical distance)', we obtain the curve shown in Fi g. 6.17c . We note the di scontinuities existing in this curve. which re n eet the dilTerencc between the values of t corresponding respecti vely to the Ilanges ABGD and A'B'G'D' and 10 the web EFF' E' .

Fig. 6.17

(hi

(e)

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In the case ()f the web. the shearing stress T ,.... varies only very across the section hlJ', and can be assumed equal to its average value T ow . This is not true. however, for the llanges. For example, considering the horizontal line DEFG, we note that "T.rr is zero between D and E and bdween F and G. since these. two segments arc pan of the rree surface of the beam. On the other hand the val ue ()r T~"I' between E and F can be obtai ned by making 1 == EF in Eq. (6.61. In pmctice. one usu:llly assumes that the entire shear load is carried by the web, and that a good approximation ()r the maximum value or the shearing stress in the cros~ section can be obtained by di viding V by the Cf()ss-sectional area of lhe web.

6.4. Shearing Slresses 'T'1 In Common Types 01 Beams

~ Ii ghtly

v

379

(6. 11 )

We shou ld note. however, that while the verticlIl component "T"'. of the sheming stress in the llanges can be neglected, its horizontal component T~~ hw; a signi ticant value that wil! be determined in Sec. 6.7.

EXAMPLE 6.02 Knowing lhm Ihc al lowable shearing siress for Ihe timber beam Tall = 0.250 ksi. check Ihat Ihe design obtained in that sample problem is accep tablc from Ihe point of view of the shearing stresses. of Sample Prab. 5.7 is

Ir = 14 ..'i5 in. Using Eq. l6.I O) for Ihe maximum shearing

stress 111 a n
.3 V t"m",

We recall from the shear diagram of Sample Prob. 5.7 thai V"m = 4.50 kips. The actual width of the beam wa~ given
=

3 \'

2A ="1 hI! =

3(4.50 kips) 2(3.5 in.)( 14.55 in.)

0.1325 ksi

Since 1', .... < 1"all.the design Oblained in Sample llrob. 5.7 is acceptable.

EXAMPLE 6 .03 Knowing thm the allowablc shearing stress for the steel beam of Sample Pro b. 5.8 i~ 1",11 = 90 M Pa. check thai the W360 x 32.9 stwpe obtained in that sample problem is ac· ceptable from the poinl o f view of the shearing stresses.

We recall from the shear diagram of Sample Prob. 5.8 that the maximum absolu tc value of the shear in the beam is WI...... "" 58 kN. As we saw in Sec. 6.4. il may be a~sumed in practice thaI the entire shear load is carried by lhe web and that the maximum value of lhc shearing stress in the beam can be obtained from Eq . (6. 1I). From Appendix C we lind that for a W360 X 32.9 .~hapc the deplh of the beam and thc

thickness of its web are. respecli"cly. d = 349 111m and 1" = 5.8 mill. We thus have A"o~ "" d I" = (349 mm)(5.8

Substituting thc values of obtain 1',,,,,",

=

Win....

mm) = 2024 10m!

and A""j, into Eq. l6.11). we

58kN 2024 mm!

18.7 MPa

Since 1"",", < 1".11. the design obtained in Sample Prob. 5.8 is acceptable.

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380

Shearing Stresses In Beams and Thin-Wailed Members

*6.5. FURTHER DISCUSSION OF THE DISTRIBUTION OF STRESSES IN A NARROW RECTANGULAR BEAM

Fig. 6.18

Consider a narrow cantilever beam o f rectangular cross sect ion o f width b and depth It subjected to a load P at its free end (Fig. 6.18). Since the shear V in the beam is constant and equal in magnitude to Ihe load P, Eq. (6.9) yields

(6.12)

D

D'

'111!/!l Fig. 6.19

We note from Eq. (6. 12) that the shearing stresses depend onl y upon the distance )' from the neut ral surface. They afe independent , the refore. of the distance from the point of applicmion of the load; il follows that all e lements located at the same dista nce from the neut ral surface undergo the same shear deformation (Fig. 6.(9). While plane sections do nol remain plane, the distance between two corresponding points D and D' located in diffcrcnI sect ions remai ns the same. Th is indicates that the nonnal strai ns fO ,. and thus th ~ nonnal s tresses 0"". arc unaffected by the shearing stresses. and tha t the assumpt ion made in Sec. 5. I is justified for the loading condition o f Fig. 6. 18. We conclude thai our a nalysis of the s tresses in a cant ileve r beam of rectangu lar cross section. subjected to a concentrated load P at its free e nd. is valid. The corrcct values of the shearing stresses in the beam are given by Eq. (6. I 2), and the normal stresses at a distance x from the free e nd are obtai ned by making M = - Pxin Eq. (5 .2) of Sec. 5. 1. We ha ve

Pry

+I

(6, [3)

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The validityofthe above statement. however, depends upon the end conditions. If Eq . (6.12) is to apply everywhere. then the load P must be distributed parabolically over the free-end section. Moreover. the fixed-end support must be of such a nature thnt it wi ll allow the type of shear deformation indicated in Fig. 6.19. The resulting model (Fig. 6.20) is highly lI nlikely to bc encountered in practice. However. it 10 1lows from Saint-Venant's principlc that, for other modes of applicat ion of the load and for other types of lixed-cnd suppor1s. Eqs. (6. I 2) and (6.13) still provide us with the correct di st ri bution of stresses. except close to either end of the be:l.m.

"

6.5. Distribution of Stresses In a Narrow Rectangular Beam

I! i I i ..I ' ' I ,I I

Fig. 6.20

" When a beam of rectangular cross section is subjected to several concentrated loads (Fig. 6.2 I). the principle of superposition can be used to determine the normal and shearing stresses in sections located between the points of application of the loads. However, since the loads P2' P~. etc .. are applied on the surface of the beam and cannot be assumed to be distributed parabolically throughQlu the cross section, the results obtained cease to be valid in the immediate vici nity of the points of appl ication of the loads. When the beam is subjected [ 0 a di stributed load (Fig. 6.22), the shear varies with [he di stance from the end of [he beam, and so docs the shearing stress at a given elevation y. The resulting shear deformations are such that the di stance between two corresponding points of different cross sections, such as 0 1 and 0;. or O2 and 0;. wi ll depend upon their elevation. Thi .~ indicates that the assumption that plane sections remain pl:me. under which Eqs. (6. 12) and (6. 13) were derived. must be rejected for the loading condition of Fig. 6.22. The error involved. however, is small for the values of the span-depth ratio encountered in practice. We should also note that, in portions of the beam located under a concentrated or distributed load, normal stresses (T) wi ll be exerted on the horizontal Iilccs of a cubic element of material. in addition to the stresses 7 ,). show n in Fig. 6.2.

Fig. 6.21

Fig. 6.22

381

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!5k'1 '" e

SAMPLE PROBLEM 6.1

,

Beam AB is made of three planks glued together and is subjected. in its plane of symmetry. 10 Ihe loading shown. Knowing that the width of each g lued joint is 20 mm. determine the average shearing stress in each joint at section /1-11 of the beam. The location of Ihe centroid of the section is given in Ihe sketch and the centroida] moment of inertia is known to be I = 8.63 X 10 " m~.

,." Ll O.4"' ~

OAm

0.2m

~I !IJ()1ll"'1 20

",m

r-

Joint 11

SOm",

I

C

20m", 2()

Joint b

C,s.3 ",m

mlll. ~=== 1:::l1~::::::::"':::':11

SOLUTION

W

Verlkal Shear a1 Section 11-11. Since lhe beam and loading are both symmetric with respeci to the center of the beam. we have A = B = 1.5 kN

60",,,,

t.

1\=)51\1'

Considering the portion of the beam to the letl of section we write

+t ~F., =

1.SkN - V=O

0:

Il-Il

as a free body.

V = I.S kN

Shearing Stress in J oint a. We pa~s the section a-a through the glued joint :md separate Ihe cross-sectional area into two parTs. We (;hoose to determine Q by compltling the first moment with respect 10 the neutral axis of Ihe area above seclion a-a.

Q=

AY I =

!(0. 100 m)(O.020 m) J(0.0417 m) =

83.4 X 10 w ~ mj

Rec;}lling that The width of the g lued joint is I = 0.020 m. we use Eq. (6.7) to de lemline the average she;}ring stress in the joint. VQ

,._---, .

7>1 <

, ,

=

h

( 1500 N)(83.4 x low6 m') = (8.63

X

10

b

m4)(O.020 m)

-r ... = 725 kP;} •

,-- ~ -,--- ,

~'"·e"""'e"·JC""""''__ _i'~c l--_.--IJ

b

... _:

:

1---1 OU60",

x'

'll=O.05/lJ",

--.l

om.QrllL

382

'

I

Shearing Stl't'SS in Joint b. We now using the area below The section .

Q = A y! VQ 7",.

=

It"

pa~s

section bob and compute Q by

= [(0 .060 m)(O.020 m)](0.OS83 m) = 70.0 X JO ~

( ISOON)(70.0 X lo- b m' ) = (8.63 X 10 b m~)(O.020 m)

't • ••

m'

= 60S kPa •

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SAMPLE PROBLEM 6,2 A timber beam All of span 10 fI and nominal width 4 in. (actual width = 3.5 in.) i ~ to s upport the three concentrated load ~ ~hown KllOwing that for the grade of limber used (! . II = 1800 psi and T . II = 120 psi. delennine the minimum required depth d of the beam.

SOLUTION Maxi mum Shear and Bending MOlllent. After dnlwing the shear and bending-Illomem diagrams. we nOlI' that

"

=-

7.5 klp · Ii = YO kip

VOl ..

=

3

" In .

kip.~

"llap<

{I .S)

1m

u.s kip - 0,]

Desj~n

~Ip

Based

.

~ f' ,-

Allowa ble No rmal St ress. We firs! express the

s_

t bd l

16 kip

0 11

"

_____ I.:; lio · _ ft 6\..1p ft

I _ I

s=

IT all

bd~ _ I (3.5)J' _ O.S83:k/ !

6

C

For MI!\ll = 90 kip ' in. and

~---------------'--- ,

6

= 1800 psi we write

9oO,,-,X~1c O '~I,bc"-';"' " O.58.33d , - =-

M,,,,,,

1800 psi

ir:.ll

J! = 85.7

d = 9.26 in.

We have satisfied the requireme nt that (J ., Ch,,~I..'

T

/. - 3,5;" ,

r"1

ela.~­

tic seCTion modulus S in terms of the depth ,I. We have

(- 1.51

"

M~

Jllw"""G

..

=

Slrt1"'~.

"3 V..... --~

2

"3

:s: 1800 psi.

For V,,,",, - 3

kjp,~

0

A

and J - 9.16 ;'n ..

3000lb

2 (3. 5 in.)(9.26 in. )

1 ..

w,",

find

= 138.8 psi

Since 7". 11 "" 120 psi. the deplh d = 9.26 in. IS /lot acceptable and we mu s t rede si!!n the beam Oil Ihe basi. of the R'quirement that T ,. S 110 psi.

I L "

DC~j ~ D Based un All uwa uh: S hc a,.ill~ S l.r~s . Since we nuw kno',y 1hal the a1Jo\.\able shearing stre ss control.; the de:iign. we write

3

1 .. ="1 111

3.5

\I,.,.,

=2A

3 3000 Ib

120 psi = - - - - -

1

- I

in.),1

,, - 10.7 1 ill . ...

1 J

11.:25 ill

.\ j " ,

(3. ~

j!l

X 12;" ,

Norll; "U .i:»'

The normal s tress I ~. oj course. less than 10. 7 1 in. is fully acceptnble.

(T

>II "" I XOO pSI. and the deplh of

Commell /. Since limber i, norm211y available in depth increments of 2 in., a 4 )( 12 . in _ n(l lninal l i:le timber .hould be med The actual ('fOSS se~_ lion would then be 3.5 X I 1.25 in.

383

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PROBLEMS

<

~ <

I

<

20"""~

Sll l m t

•

•

20 """

1-1

!2~~

Fig. PS.1 and PS.2

6 .1 A ~quare box be,\m is made of two 20 X 80-mm planks and tWO 20 X 120-mm planks nailed together as shown. Knowin g that the spacing between the nails is s = 50 mm and that [he allowable shearing force in each nail is 300 N. determine (II) the largest allow:lble vertical shear in the be;).m. (b) the corresponding maximum shearing stress in the beam. 6.2 A square box beam is made of IWO 20 X 80-mm planks and IWO 20 X 120-mlll plank ~ nailed together as ~hown. Knowin g that the sp
6 .3 Three boards. each 2 in. thick. arc nailed together 10 foml a beam that is subjected to a vertical shear. Knowing that the allowable shearing force in each nail is 150 lb. determine the allowable shear if the spacing s between the nails is 3 in.

Fig. PS.3 and P6.4

6.4 Three boards. each 2 in. thick. tre nailed together to form a beam that is subjected to a venical shear of 300 lb. Knowing that the allowable shearing force in each nail is 100 lb. determine the largest longitudinal spacing s of the nails that can be used. 6 .5 Thecomposite beam shown is fabricated by connecting two W6 X 20 rolled-steel members. using bolts of ~-in. diameter spaced longitudinallY every 6 in. Kn owi ng that the average allowable shearing stress in the bolts is 10.5 ksi. determine the largest allowable vertical shear in the beam.

384

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Probtems

6.6 The beam shown is fabricated by connecting two channe l shapes and two plaTes, using holts of ~-in. diameter spaced longitudinally every 7.5 in. Determine the average shearing stress in the bolts caused by a shearing force of 2S kips parallel 10 the J' axis.

6.7 The American Standard ro lled·steel beam shown has been reinforced by attach ing to it two 16 X 200·mm plates. using IS·mm·diameter bolts spaced longitudinally every 120 nun. Knowing that the average allowable. shearing stress in the bolts is 90 MPa. determine the largest permissi ble vettkal shearing force. Solve Prob. 6.7. assuming that the reinforcing plates are only 12 rnm

6.8

thick.

Fig. P6.6

6.9 through 6.12 For the beam and loading shown . consider section and determine (a) the largest shearing stress in that seetion. (b) the shearmg stress at point a.

IHI

16

x 200 mm

S3 10 X 52

Fig. P6.7

Dimf'nsions in ",m

Fig. P6.9 0.3 U1

t----I

,"I Ii;=l==' ~...,=j :...: L , '"

I

I()k~

41l",,,,T ~

,

IJ",,,

~-. "

I Ii"' ' I

1- 200"'m ~

f------ l .5 m - - - --oj

Fig. P6.10

I

r-IO'''--Id-,

1 ±'~1.1 {-l

J in ,

0,0"';1),,,.

I '"'

0.3751". -

(1.6 in

Fig. P6.11

I

150 m",

Fig. P6.12

385

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386

Shearing Stresses In Beams and Thin-Wailed MembeN;

6 .13 Two Sleel plates of 12 X 220·mm rectangular cross section are welded to the W250 X 58 beam as shown. Determine the largest allowable venical s hear if the shearing stress in Ihe beam is not to exceed 90 MPa.

1

220

'''''' 1

1-.. .

~Siii"c~ -.1 " ~

11'2.50 X 51'1

2-'52 nll n

Fig. P6.1 3

"

( \\'27

x 116

A'1"!ii!i!=;:~!ii!i!!!'=~!!!!!!:':l'C

~_B

_ _ IHt

- - --I

6. 14 Solve Prob. 6.1], assuming that Ihe IWO steel plates are (a) replaced by 8 X 220-mm plates. (b) removed.

6.15 For Ihe wide.t1ange beam wilh the loading shown. determine Ihe largest load P that can be applied, knowing I1wt the maximum normal stress i.'i 24 ksi and Ihe largest shearing stress using Ihe approximation T", = V/ A ....
HI

Fig. P6.1S

14,5 ksi. p

\\'36'.1 )( 122 E

lJ.{i '"

- L-L". . \

OJ;

6.16 For the wide.nange beam with the loading shown. dctermine the largest load I' that can be applied . knowing that the maximum normal stress is 160 MPa and the largest shearing stress using the approximation 'Tm = V/A"..b is [00 MPa.

6.17 For the beam and loading shown. determine the minimum requiral wid th b, knowing tl1m for the grade of timber used. (T.!I = 11 MPa and 7;lj1 = &15 kPa.

III

().6," Fig. P6.16

2.H :\'

!I. .~

1-- - -16 ft - - ---II

E

50 """

0.>

iii

Fig. P6.17

6.18 For the beam and loading showtl. determitle the mitlimum required depth h. ktlowing thal for the grade of timber used. tT,1I .:0 1750 psi and 'fall

Fig. P6.19

! I ~ ', ~1

BCD

~, . ,J,,"J, . 1J

in

~ 1111TI''i' 1111 1 I}, Fig. PS.18

7-'!k~

Hk:'>l

= 130 psi.

6.19 A timber beam AB of length L and rectangular cross section carries a single concentrated load P al its mld]Xlint C. (a) Show thal the ratio 'f .. /u.. of the maximum values of the shearing and normal stresses in the beam is equal to 2/i/L where" and L are, respectively, the depth and the length of the beam. (b) Determine the depth" and the width b of the beam. knowing tlwt L = 2 m. P = 40 kN. 7 ,. = 960 kPa, and (f,. = [2 MPa.

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G.20 A timbe, beam AB of length Land rcciangllla, ClOSS section CaIries a uniformly d islributed load U" and is supponed as shown. (a) Show that Ihe ratio T.. /eT," of the maximum values of the shearing and nOffilai stre.;ses in lhe beam is equ.lI 10 Ih/L. where II and L are, res ~ctively. the deplh and the hmglh of the beam. (b) D."llermintl Ihe depth II and the wi dth b o f Lhe beam. knowing thai L = 5 rn.1V = 8 kN/lll. T.~ = l.OS MPa. and IT," = II MPa.

6 .21 and 6.22 and ril'.trm);nt'

Ih~

~1 1UW[].

For lilt: beam ami luauing

she.1rin2

' In., ~.,

at (r/) poiTt I

{J ,

IMI !.;" "

H

r

387

'~f I';IJJUHl l'

~(;tiLJn ,;./1

1../4

(h) poinl h

I

, •

(;L'llsid<:r

Problem.

l- u2 - [ 1../4

Fig. P6.20

lij)rnml __1 10'JI'"

'''''''''' I

.

li

11

TI3"",,,,

I..H - -11-1

3O",,,, ~3()"H" 20

IllH·

Fig. PG.21 and PG.23

i r.-=:,.. j.2jjH .

~ ill

~ I 1", --1 1-- .

,/,

3

],5i" I,Sln.

-~ 1 -+'

L sm.- -I

FIg. P6.22 and P6.24

6.23 and 6.24 For the beam and lo&ding shown, dClcnninc the largesT shearin!! stress in sa:lio(! IHI . 6.25 through 6.211

A beam having the cross section .,holl'n is subjected

10 a vertical shear V. Detennine (£I) the horizontal line along which the shear_

ing s tre~s is maximum. (b) Ihe constant k in the following expression for the rnaximum shearing stress

'.I!here A is the cross-sectional area of the beam

1-' - 1

\7J

Fi g. PS.25

'"

Fig. P6.26

G

Fig . 1'6.27

0 Fig. P6.28

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388

Shearing Stresses In Beams and Thin-Wailed Membe~

(II )

Fig . 6.23

6 .6 . LONGITUDINAL SHEAR ON A BEAM ELEMENT OF ARBITRARY SHAPE Consider a box beam obtained by naiHng together four planks. as shown in Fig. 6.23u. You learned in Sec . 6.2 how to determine the shear per unit length. q. on the horizontal surfaces along which the planks arc joined. BUI cou ld you determine q if the planks had been joined along vertical surfaces, as shown in Fig. 6.23b? We examined in Sec. 6.4 the distribution of the vertical components T" of the stresses on a transverse secti on of a W-beam or an S-bcam and found that these stresses had a fa irly constant value in the web of Ihc beam and were negli gible in ils n angcs. But what abo ut the IlOr;:onfal components Txz of the stresses in the /langes? To answer these questio ns we must ex tend the procedu re developed in Sec. 6.2 for the dctenni nati on o f the shear per unit length, q, so that it will ap ply to the cases j ust descri bed .

Fig. 6.5 (repea ted )

Consider the prismatic beam AS of Fi g. 6.5, which has a vert ical pl ane of symmetry and supports the loads shown. AI a di stance x from end A we detach again an e leme nt eDD' e' of length Llx. Thi s c lement , however. wi ll now ex tend from two sides of the beam 10 an arbitrary curved surface (Fig. 6 .24). The forces exerted on the e lement include

Fig. 6.24

vertical sheari ng forces V'c and V 'D' ele mentary horizontal lIonnal fo rces lTc dA and lTv dA. possibly a load w Llx. and a longi tudinal shearing force Ll H representing Ihe resultant o f the elementary 10ngillLdinai shearing forces exerted on the curved surface (Fig. 6.25). We write the equilibrium equat ion 6.H Fig. 6.25

+

I "

((TD - ITe) dA

= ()

where the integral is 10 be computed mer Ihe shaded area (f o f the section. We observe thaI the equation obtained is the same as the one we

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6.6. Longitudinal Shear on a Beam Element of Arb!lrary Shape

obtained in Sec. 6.2, but that the shaded area (f over which the integral is to bt! computed now extends to the curved _~u rfac e. The remainder of the derivation is the same as in Sec. 6.2. We find that the longitudinal shear exerted o n the beam element is

,

VQ !:lH = - i l x

(6.4)

where 1 is the eentroid,,1 moment of inertia of the entire section, Q the first moment of the shaded area l1' with respect 10 the ne utral axis. and V the vertical shear in the section. Di vidi ng both members of Eq. (6.4) by ,1x , we obtain the hori zontal shear per unit length. or shear now : uH

VQ

q = 6.x = - '-

(6.5)

EXAMPLE 6.04 O_ili h~- I

A square box beam is made of IWO 0.75 X 3-in. planks ami 1\\10 0.75 X 4.5· in. planks. nailed togcther as shown (Fig. 6.26). Knowing thallhe spacing bel\\leen nails is t .75 ill. and Ihal the beam is subjccled to a IIcnical shear of magnilude V = 600 lb. determine the shearing force in each nail. We isolalc the upper plank and consider Ihe 100ai force per unit length. q. ~ .,cncd on its two edges. W~ lIStl Eq. (1,,5 ). where Q represen:s the firs t moment \\lith respect to the neu· tral axis of thtl shaded area A ' shown in Fig. 6.27a. and where I is the moment of inenia about thc same axis of the entire cross·sectional area of the box beam (Fig. 6.27b). We have

Q = A'y

1 1iiF ~'f,iIO 75 ,,, ,15

in.

L~ Fig. 6.26

= (0.75 in.)(3 in.)( 1.875 in.) = 4 .22 in J

Recalling Ihat the moment ofittenia of a square of side a abou t a centroidal axis is I = T!U4, we write

1-3 in'-1tits;n.

N,

=

U i7Sin

I

1 = M 4.5 in.)~ - Tz(3 in.)~ = 27.42 in ~

,,

,,,

1'''' :, ,

0.i5 ill.

--L

!

r

'1.5 in.

L

1~ ·1i n .

I

3 in.

I

Substituting into Eq. (6.5). we obtain In)

(600 Ib)(4.22 inl)

VQ

(I =

J

=

27.42 in ~

Ag. 6.27

92.31b1in.

Bl!cause both the beam and the upper plank are symmetric with respect to the vellkal plane of loading, equal forces are e.~· crted on both edges of the pl.mk. The force per unit length on each of these edges is thus!q = 1(92.3 ) = 46.15Ib/in. Since the spacing between nails is 1.75 in .. the shearing force in each nail is F = ( 1.75 in. )(46. 151b1in.) = 80.81b

("1

-I

389

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390

Shearing Stresses In Beams and Ttlln-Walled Members

6.7 . SHEARING STRESSES IN THIN·WALLED MEMBERS We saw in the preceding section Ihat Eq. (6.4) may be llsed to determine the longitudinal shear 6." exerted on [he walls of a beam clement of arbitrary shape and Eq. (6.5) to determine the corresponding shear now q. These equations will be llsed in thi s section 10 calculate both the shear Oow and the average shearing stress in thin-walled members such as the tlanges of wide-tlallge beams (Fig. 6.28) and box beams, or the walls of structural tubes (Fig. 6.29).

FIg. 6.28

Fig. 6.29

Cons ider, for instance, !) segment of length .1.x of fl widc-Ilrmgc beam (Fig. 6.30a) and let V be Ihc vertical shear in the transverse seclion shown. LeI us detach an clement ABB'A' of the upper Ilange (Fig. 6.30b). The longitudinal shear ~ H exerted on that clement can be obtained from Eq. (6.4):

VQ

~H = - ~x

I

Fig. 6.30

Dividing ~H by the area aA = I ~x of the cu t, we obtain for the average shearing stress exerted on the clement the smne expression that we had obtained in Sec. 6.3 in the case of a horizontal cut: VQ

Tave

==

Ir

(6.6)

Note that T n< now represents the average value of the shearing stress over a vertical cut, but si nce the thickness t of the n ange is small . there is very little vruiation of T ,", across the cuI. Recalling that T .,; == T o. (Fig. 6.3 t), we conclude that the horizontal component T .", of the shearing stress at any point of a transverse section of the !l ange C'.m be obtained from Eq . (6.6), where Q is the first moment of the shaded area about the neutral axis (Fig. 6.32a). We recall that a similar result was obtained in Sec. 6.4 for the vertical component T ') of the shearing stress in the web (Fig. 6.32b). Equation (6.6) can be used to determine shear-

T ;:r

Fi g. 6.31

(6.4)

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6.7.

S~earing

Stresses In Thin-Wailed Members

,

391

, I

,,-

"

j

"N A N .A

N.A.

N.A

(,Il

Fig. 6 .32

ing stresses in box beams (Fig. 6.33), half pipes (Fig. 6.34), and other thin-walled members, as long as the loads ar!! applied in a plan!! o[ symmetry of the member. In each case. Ihe cui must be perpendicular to the surface of the member, and Eq . (6.6) wi ll yield the component of the shearing stress in the di rection of the tangent 10 that surface. (The other component may be assumed equal to zero, in view of the proximity of the two free surfaces.) Comparing Eqs. (6.5) and (6.6), we note that the product of the shearing stress T at a given point of the section and of the thickness I of the sect ion at that point is equal to q. Since V and I are constant in any given section. q depends only upon the first moment Q and. thus. can easi ly be sketched on the section. In the case of a box beam, for example (Fig. 6.35), we nOle that q grows smoothly from zem at A tll a maximum value at C and C' on the neutral axis, and then decreases back 10 zero as E is reached. We als() note that there is no sudden variati()11 in the magnitude of q as we pass a corner at 8 , D, 8 ' . or D ' , and that the sense of q ill the horizontal portions of the section may be easily obtained from its sense in the venical portions (which is the same as the sense of the shear V ). In the case of a wide-llange section (Fig. 6.36), the values of q in porti ons All and A' B of the upper llange are distributed symmetrically. As we turn at B into the web, the values of lJ corresponding to the two halves of the llange must be combincd to obtain the value of q at the top of the web. After reaching a maximum lIalue at C on the neutral axis, q decreases, and at D splits into two eljual parts corresponding to the two hal ves of the lower nange. The name of shear J1011' commonly used to refer to the shear per un it length, q. renects the similarity between the properties o[ q that we halle just described and some of the characteri stics ot' a lluid now through an open channel or pipe. t So far we halle assumed that all the loads were applied in a plane of symmetry of the member. In the case of members possessi ng two planes of syml11dry, such as the wide-tlange beam of Fi g. 6.32 or the box beam of Fig. 6.33, any load applied through the centroid of a given

'01 Fig. 6.33

Fi g. 6.34

B

,\

B'

t- ... .. - -.

t

'I

t '/

C

C"

N.A.

__ .. _ _ t E

D

D

Variation of q in box-beam section

Fig. 6.35

r

'"\

" .-

'I~

B

A'

'/ ~'I j '" 'I!!

c N .A. 'I

I

"C:::g~t~f)::::::::J E'

E lL: t We recall thaI lhr conrept of shear 110\\' was used to analyze tlK' distribution of shearing stresses in thin-walled hollOW shafts (Sec. 3. t.1). Howc\"cr. while the shear flow in a hollow shaft is con.'tant. thc shear now in J mcmocr undc.r a tr.In swI'\C loading is not.

'/ 1

'I~

Fig. 6.36 Variation of q in wide-flange beam section.

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Shcnrtng &1"'''"''''

392

Thin-Wailed

I~

B.... m ~ and

Ci'OSS sc:ctiu n ca.n be Icsolvcd 11111) CQlILf-'O lIe Jl IS alo ng the tw o aACS o f sym me try o f th e st',cti o n. Elic h ('om po nen t will calise Ihe me.mber ItJ bel1d m a pltillc of symmetry, and the corresponding s hearing 51re~!;eS elm be o btained from Eq. (6.6). 1lle pri nc iple of superposition can then be llsed 10 delcrmine Ihe resulting S lrc!;)(c~. However, if the me mber cons idered possesses no plane o f synunctry. or if It possesse.. a single pla ne symiJ'letry a nd is s ubjeclcd to a load tha t is not contained in that pl ane. the member is o bserved 10 hen d muf twist m lhe same time. except when the load is appl ied m a ~ pe­ cilic point. called Ihe .I·/tear center. Note that the shear ce nter generall y docs IIO! coincide with the cent roid of the cross section. The de termi-

Me mbe~

or

nation uf the shear center of vn rious thin-walled shapes is discussed in Sec . 6.9.

~6_8_

Clmside r a cantilever beam AD 0 1' length L and rectangul:1I cross section. subjec led at i l ~ free end it 10 a conce ntmled load P (Fig. 6 _3 7) The largest value of the bending moment occurs al tbe li xed end B and is equal to At = PL. As long as thi s value docs not exceed the. maximu m clastic moment M y, that is, as long as PL ::s: M b the normal stress (T~ will not exceed the yield stre ngth (Tr anyw here; in the beam. However. as P is inc reased beyo nd the val ue M :JL. yield is in itiated at ]lOints B and B ' and s preads IOward the free e nd of the beam. Assuming the material to be elas toplastic, and considering a cross section CC' located aJ a disrance.r from the frcc end A o f the beam (Fig. 6.38), we obtain the hal f thIckness Yr of the elastic eorc .in that section by making .M = Px in Eq . (4 .38) of Sec. 4 .9. We have

I'

.,

R

W Fig. 6.37

I

L

(PL :s

M~)

r.

"I

'lI,r-'-

"

e-

Fig. 6.38

. r. ,, L

n

(6. t 4")

(PL "> My'

IL ,J ~

I'

PLASTIC DEFO RMATIONS

Fia.6.39 (PL= M~

= ~M.)

"!lr= O 8'

where C 1£ th ~ hll l-depth 01 the be:1.m. PIOtllflg)'r :1. ~:1.lIlS t x . we obtam the boundary between the clastic and plastic zones. As Illng n .~ PI . < ; M v, th ... pamhn l;1 defi nel! hy F.lJ- (11_ 14) intersecL~ the line BB ' . as s h O\~ n in F ig . 6.38. However. when PL reaches the value ~ M y, that is. when PL = M {I" where M p is the plastic mo ment d Clincd in Sec. 4 .9 . Eq. (6. 14) y ield..: Yr = 0 for x = L. which show s that the vertex of th8 parabola is now located in section BB '. and th m this section has become full y plaStic (Fig. 6.39), Recalling Eq. (4.40) o f Sec. 4 .9, we also note that the radIus of curv.lturc p of the. ne utral surface at th:!t point is eq ual to zero, indit'ating the presence of a sharp bend 111 tht: bea m al ils li xcd e nd. We say Ihill a p last ic h ill ge has tleveloped aI that point. The load P = Mp!L is die largest load that can be supported by the beam.

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The above discussion was based on ly on the analysis of the normal stresses in the bcam. Let us now exami ne thc dislriblllion of the shearing stresses in a section that has bccome partly plastic. Consider the portion o f beam eC"D"D located betwC\!n the transvcrse sections ec' and DD'. and abovc the horizontal plane D"C" (Fig. 6.401/). If this portion is located cl11ircly in the plastic zonc, the normal stresses exerted on the flccs CC" :md DD" will be uni form ly distributed and equal to thc yic ld strength (Tl' (Fig. 6.40b). Thc equilibrium of the Free

:-

i'LASTtCe "I

-D.:.:,J-=k'- '" :,c)DIliC~~'" , >I, D C

~

D" _

,, , ,,

, , , ,

IJ

C

-----

C"

ELo\STt C

(ill ~

~~

PLASTIC

Fig. 6.41

I")

body eC"D"D thus requires that the horizontal sheari ng force ll H exerted on its lower t:'1CC be eq ual 10 zero. It follows that the averagc value of the horizontal shearing stress T,. . across the beam at C" is zero, as well as the average value o f the vertical sheari ng stress T ,O" We thus co nclllde Ihm th t! vertical s hear V = P in secTi o n cr' mllST he (ti .~ Irih­ llled elllircly over the portion EE' of that section that is located within the elastic zone (Fig. 6.4 J). It can be shownt that the distribution o f the shearing stresses over EE' is the same as in an clastic rectangular beam of the same width b as beam AB, and o f depth equal 10 the thickness 2yy of the elastic zone. Denoting by A I the area 2by y of the clastic portion o f the cross section, we have

i) y}

(6.1 5)

The maximum value o f the shearing stress occurs for y = 0 and is 3 P

T ma,

=

2:47

(6.16)

As the area A' of the elastic portion of the section decreases, T max increases and evel11ualty reaches the yield strength in shear T y . Thus, shear contribllles to the ultimate fa ilure orthe beam. A more exact analysis of this mode of failure shou ld take into account the combined effect o f the normal and shearing stresses.

PlastIc Oelonnations

C E

~

/

f '

, ,

Fig. 6.40

t See Prob. 6.60.

6.B.

C'

~'"

""'

393

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SAMPLE PROBLEM 6.3

-\.31 ill.

I~~J[7=O="'==~~i n~'['2=_'o"','~ 'HI.; , , lOA in.

Knowing that the vertical shear is 50 kips in a WIO X 68 rolled-steel beam. determine the horizontal shearing stress in the top flange at a point a located 4.31 in. from the edge of the beam. The dimellsions and other geometric data of the rolled-steel section are given in Appendix C.

SOLUTION We isolate the shaded ponion of the flange by cUlling along the dashed line. that passes through point a.

c

Q = (4.31 in.)(0.770 in.)(4.815 in.) = 15.98 VQ

(50 kips)( 15.98 inJ)

II

(394 in~)(0.770 in.)

7 =~=

0. , 5 ill. X 12 ill

U

~. 3 1

in~

T= 2.63 ksi "

SAMPLE PROBLEM 6.4 Solve Sample Prob. 6.3. assuming tlwt 0.75 X [2-in. plates have been allached to the flanges of the WI 0 x 68 beam by continuous fillet welds as shown.

ill.

SOLUTION For the composite beam the centroidal moment of inertia is

1 = 394 in.j 1 = 954in ~

+ 2[-6( [2

in. )(0.75 in. f

+ ( [2 in.)(0.75 in.)(5.575 in.)!J

Since the top plate and the flange are connected only at the welds. we lind the shearing stress at a by passing a section through the flange at a. bl'fwcen the pl:lIe and the flange. and again through the flange at the symmetric point a '.

0.7511~.r-- 1 211l --1 I I • _ __I

T

WAin .

, - - "'''. - - - 1

5i~'f lKI 'T

[ 11.3,.5,i ll '

I.~57r '"

·/1:' 4, 3 1'n.

r--- :11]"

: :

4 31m

I815

,,,[

0.7,0 ill Ie' -'-______ :. : _ _ _ _L-"-

C

1~ii2.J \),751"

0.75 ill.

,

'

'-'

For the shaded area that we ha ve isolmed. we have I =

21r = 2(0.770 in.)

= J .540 in.

Q = 2[(4.31 in.)(0.770 in.)(4.815 in. )] + ( 12 in.)(0.75 in.)(5.575 in. ) Q = 82.1 in) T

394

VQ

(50 kips )(82. [ in.!)

II

(954 in 4 )(1.540 in.)

= ~ =

T

= 2.79 k:;-i ..

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SAMPLE PROBLEM 6.5 The thin-walled extruded beam shown is made of aluminum and has a uniform 3-mm wall thickness. Knowing th;}! the shear in the beam is 5 kN, determine (a ) the shearing stress at point A. (b) the maximum shearing stress in the beam. NOIe: The dimensions given are to lines midway between the outer and inner surfaces of the beam. D 1

li B 25 tmn 25 "n"

SOLUTION Centroid.

_

We note that AB = AD = 65 mm.

~=

2:

yA

2[(fi5 mm)(3 rnm)(30 mm))

-'-A- ~ °2["("65;-m " 'm -;-; )(::3 -m-"m")J;-+ .,-'(~50;-m-m-;')("'3-m-m"")

Y = 2 1.67 mm

/) 1--1_ _ 1B 15 min 2."i m ",

CClllroida l ~'lomclIl of Ine.rtia. E:lch side of the thin- walled beam can be considered as a parallelogram and we recall that for the case shown /"" = bh 3 j l 2 where b is measured parallel to Ihe axis 1111. 3.25 mm

f-

- oj

J=

b = (3 mm)/cos {J = (3 mm)/( 12/ 13 ) = 3.25 mm 2lfi(3 .25 mm )(60 mm )' + (3. 25 mm)(60 mm )( 8.33 mmf] + [h(50 mm )(3 mm )! + (50 mm )(3 mm)(21.67 mm f ] / = 0.2146 X to ~ m4 / =2 14.6 X 10.l mm'

'5:. (i + A(f ) =

u. Shearing Stress at A . If a shearing stress T.! occ urs :It A. the shear fl o\\l will be qA = 'tA l and muSt be directed in one of the two ways shown. But the cross secti on and the loadin g are symmetric about a vertical line through A. and thu s the shear flow must also be sy mmetric. Since neither of the possi't .. = 0 .... ble shear flows is symmetric. we conclude that b. ~ l al.illlum Shta rin g Stress. Since the wall thic kness is constant, Lhe maximum shearing stress occ urs at the neutral axis. where Q is maxi mum. Since we know that the shearing s tre ~s at A is zero. we cut the section along the dashed line shown and isolme the shaded portion of the beam. In order to obtain the largest shearing s tre ss. the cut at the ne utral axis is made perpendic ular to die sides. and is of length t = 3 mm.

Q = [(3 .25 mm )(38.33 mm) ] e8.3~ mm )

= 2387 mm '

Q = 2.387 X 10- 6 mJ VQ (5 kN )(2.387 X [0 bml)

7 f:

=

h

= (0 .2 146 X 10- 6 m.j)(O.003 Ill )

T " ...

= fA = 18.54 MP("J ....

395

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PROBLEMS

6 .29 The buill-up limbt:r Dt:J.m is subjt:cttu lU a \'crllc
largest pemlissible spacing .I' of the nails. 6 .30 TYlo 20 x lOO·mm and 1WO 20 X lSO-mm boards arc glued to}!"elher as ~hown 10 form a 120 x 2oo·mm 00)( ~am. Knowin}'! thm me beJnl is ;ubjected to II vertical shear o f 3.5 kN. de(ermin~ the average sh~arin g stress in Ihe g lued johll (a) al A. (b) al B.

r IOO""n-i

20""" j

r

"

2 in

Fig. P6.29

8

ISO n""

L

20",,,,

'-C

D

FIg. P6.30

6 .31

2 ill

6 111.

The builH1P limber beam is subjected to

it

Knowing th:tllhe longitudinal spacing of lhe nails is .1

r-

nail

1 in ,

L 2 In


I

_ 1_ .

1;'0.1.) .1;,;1 2b. Fig. PG.31

i~

~he~,;nll

3.5 in. long. determine the

1500-lb vertical she;]! =

2.5 in.

that each

6 .32 The built-up wooden beam shown is subjected to a vertical shear of 8 k:"l. KllOwill!! thm the nails ille spa~-~d longitudinally tvery GO 111m at A anj ev~ry 25 mm at 8. dell!rmine the ~heari n g force ;11 the nails (1I) at A. (h) at B. (Given: I, = 1.504 X 1O~ 1!lm~.)

'l'l

r

.'00

l

""', '1"'1'

~

A ;j(1

"" --L

,

-I

-

,

50

I :'.l

,

:Of!

D;'''('11si"".1 ,n ''''''

Fig . P6.32

396

~Uld

forc .. in "ach nail.

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Probtoms

6.33 The built.up beam was made by gluing together several wooden phmks. Knowing that the beam IS ~ubjected to a 5-kN shear. determine the average shearing stress in the glued joint (II ) at A. (b) at B.

3f1 16

H

397

16 3()

r-""- -j r-i J

"":

"""1

B

r-...l'"

re

16

I----l 11 2 >lU ll If)

Di llll!n.,iolls ill mHlill"'t<-r~

Fig. P6.33

Fig. P6.34

6 .34 The composite beam shown Is made by welding C200 X 17. 1 rolled-steel channels to the l1ange ~ of a W250 X 80 wide-flange rolled-stet'.1 shape. Knowing that the beam is subjected to a vertical shear of 200 kN. determine ta) the horizo ntal shearing force per meter at each weld. (b) the shearing stress at point ( 4 of the flange of the wide-flange shape. 6 .35 Knowing that a given vertical s.hear V causes a maximum shearing stress of 10 k~i in the hat-~h aped extru~ion shown. determine the corresponding shearing s tre~s at (al point a. (b) point h.

r-I

, ''' j-.1 tOO'"

0.3,"._

_0 3 m

3 ill

0.; ill.

....10 .2 ill. I--f----l---jl t 1 il L

1.-1 ill.

1 ill.

Fig. P6.35

rL

-ILS ill

-0.5 ill.

I,

O·l

i

".

1.2 in

~I::-",.,._*,,_:: :;-I~I-1

Fig. P6.36

6.36 An e.xtmded aiumitlllm beam has the cross section shown. Knowing that the vertical shear in the beam is 10 kips. determine the shearing stress at (II) point a, {h) point b. 6.37 An extruded beam has the cross section shown and a uniform wall thickness of 0.20 in. Knowing that a given vertical shear V causes a maximum shearing stre~~ T = 9 ksi. determine the shearing Stress at the four points indicated. 6.38 Solve Prob. 6.37 assuming that the beam is subjected 10 a hon2ont.'l1 shear V.

I -

2.4 ill

Fig. P6.37

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398

Shearing Stresses In Beams and Thin-Wailed Members

6.39 Knowing that a given vertical shear V causes a maximum shearing stress of 75 MPa in an extruded beam having The cross section shown. determine the shearing: STress at the three poims indicated.

6.40

Solve Prob. 6.39 assuming that the beam is subjected 10 a hori-

zontal shear V.

6.41 The vertical shear is 25 kN in a beam h
$ ",",

~~+r 120m,,,

Din,..".;",,-,II\ In'"

Fig. PS.39

l--d-~i2"""+d-l "m,,, ~ _ Fig. PS.41 and PS.42

6.42 shown.

The vertical shear is 25 kN in a beam h:lving the cross seclion

D~lermillt:

(£I) lilt: tlistanl:t: d fUf which T~ = To_ (b) the curre~pulH..ling

shearing stress at points a and h. 6.43 Three pl
[00,,,,,,

F~H 1

2.50,,,,,,

I Fig. P6.43

r

2 ill

I

2 ill.

4

6 iB ,

l!

:1: i' L

c -_-

~

I

Fig. PS.44

6 .44 A beam consiSls of three planks connected as shown by i-i n.diameter bolts spaced every 12 in. along !he longitudinal axi~ of the benm. Knowing that the beam is subjected to a 2500-lb vertical shellr. determine the average shearing stress in the bolts.

-I\lI'lmul

l~

Fig. PS.45

6.45 Four L102 x 102 X 9.5 steel angle shapes and u 12 x 400-Olm steel plate are bolted together to form a bellm with the cross section shown. The bolts llre of 22-mm dinmeter nnd are spaced longitudinnlJy every 120 mm. Knowing th
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5 .45

Thlcl!:. I X 18-ill. steel

platc~ aH~

bolted 10 foul L6 X 6 X I allgle~

Problem.

10 form a beam wilh Ihe cro~s section shown. The boh~ have a ~-in. diame~er and are spaced longitudinally c\'ery 5 in. Knowing that the aliOW;lble average shearing stress in the bolts h; 12 ksi, determille the largest pcmllssible ~enica l ~.hear in the beam. ( Gi...,n: J, = 6123 ;n~.)

~

'",,, L:\ -

!

HE

1-2ill'~

2 ;,, _

I.!! i, •. I.:! in .

Fig. P6.4 7

Fig. P6.46

6.47 A plate of Hn. Ihid.:ness is cOmJgaled as shown and Ih~n used as a beam. For a vertical shear of 1.2 tips, delenmne \u) Ihe maximum shearing stress in Ihe section, lb) the shearing s tress flow in the cross section.

6.48

An extruded

be~m

(l(

point B. Also sketch the shear

has the cross secti on shown and a unifonn wall

thid:ness of 3 mm _ For a vertical ~hear of 10 kN.
shear 110w in th e cross

~ection.

r-----

tiO

rnm --------j

I

A

:!OO n.m

., , "'[ -- I------IJ to.I .... "

Fig. PS.4Q

Rg. PS.4R

6 .49

Three

flbte~,

ea r.h 11 mm Ihi d ,

:m~

welded tn/: ...II",.r In f(lnn Ihe

section ~hown. For a venical shear of 100 kK determine Ihe . hear 110w through Ihe wt!lded surfaces and sketch the ,; hear flow in thi! cross sec.tion.

6.50 A plate of thickness I is bent as shown an d then used as a beam. For a vertic~l sh~ar of 600 Ib, determine (a) the thickness I for which the m~x­ Imum slletress Is j()(J psi. (h) the corresponding shearing. stress at point E. AlS() s ketch the

~he:l.r

flow in the cmss ,eclion.

Fi~.

P8.50

10" ..... ,

399

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400

St>eQ~ng

Stma..,3

I~

6.51 and 6.52 Au e.'lfudcd il<:;:UIl Ins a uniform ....all u,;cKness I Denoting by V the vertical shear and by A the eros i-sec tional area of the beam, e.~pres~ the maximum shearing s tress as T""" "" k{ V/A) and delt':ntline the con·

Beama nnd

Tl1ln-Walled Memban;

s tant k lor e1ch 01 Ihe 1\110 orientations show n.

0 ~ 1"- 1

1'0 '0 '"

'"

(1/ \

(I»

Fi g. P6.52

Fig. P6.51

6 .53 w ) Detennine the shearin)!. s uess a\ point P of a Ihin·waUed pipe of the cross ~ection shuwn cau,cd by a venical shear V. (b) Sho\\l that the maximum . hearing s tress occurs for IJ = 9if and is equal 10 2V!A, where A is Ihe cro~s_~ecljonal area of lhe pip~ _ 6,54 The design of a beam require;; welding four horizontal plates 10 .a vertic ..} 0.5 X S-in. plate as shown. ror a vertical sh~ar V, dcter.ni ne the di mension II for which the shear flow through the welded surface iI maximllm

FIg. P6,53

O..'}m .

O..'i In .

Fig. P6 .S4

6 .55 FIT a beam m:ille of two or more materials with different moollli of elasticity, show that Eq (6.6)

VQ

-r"· =h

~"";"",, nJ" ""I

D

.=1"''

1 --1 ,1Dmm

Fig. PS.56

remains v~lid pro'fided dmt both Q and' are computed by lIsing the tran sformed section oflhe beat.'llste S~C. 4.6J and provided further Ih.:.t I i.~ tht lIClu~l Wldth o r the beam where 1",,~ i ~ computed.

6.56 A steel bar and ;)n aluminum bar are bonded together as .'.ho.... n [() rUnTl a (;ulTI(lillit<:: beam. Knuwing that ule vertkal shtm iu Ole bt::UII is 2{J IJ\

and th~t tile modulus of el:Jstkity is 210 GPa for lhe sled and 70 GPa for the aluminum. detennine (a) the :wera;]:e stress at dIe bonded StIrface. (b) the m:l..1(iilium she:lJi~g ;tre~~ in ille beanl. (Hi,,!: Use the method indic:lted i~ !'rob. 6.55.)

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6 .57 A steel har and an aluminum bar are bontled together as shown [0 fOfm a co mposite beam. Knowing th:llthe vertical shear in till.' beam is 4 kips and that lhe modulus of elasticity 'is 29 X 10" psi for Ihe sleel and 10.6 X 10~ psi for the aluminum. delermine (a) the average stress :lI the bonded surface. (b) the maximum sh!!aring stress in the beam. (Hilll: Use the method indicated in Proh. 6.55.)

Ll ()''''''1 t5

"""'r

I i"

P

o

SI<><'I

Problems

-

2511 """

---t 1 in -..l

AI",,,i,,,,,n

1--------1 1.5,11

12,,,,,

~"==iiP

Fig. P6.S8

Fi g. P6.57

6.58 A composite beam is made by attaching the timber and steel portions shown with boils of 12-mm diameter spaced longi tudinally every 200 mm. The modulus of elasticity is 10 GPa for the wood and 200 G i'a for the steel. For a vertical shear of 4 kN. determine (a) the average sh!!aring stress in the bolts. (b) the shearing stress at the center of the cross section. (Hint: Use the mdhod indicat!!d in Prob. 6.55';

6.59 A composite beam is made by attaching the timber and stc!!l portions shown wilh bolts of ~- in. diameter spnced longitudinally every & in. The. modulus of elasticity is 1.9 x 10" psi for the wood and 29 X 10" psi for the steel. For a vertical shear of 4000 Ih. delennine (ll) the average shearing stress in the bolts. (b) the shearing stress al the center of the <-TOSS section. (H illt: Use the me thod ind icated in Prob. 6.55.) 6 .60 Consider the cantilever beam AB discussed in Sec. 6.8 and the por· tion ACKJ of the oeam Ihat is located to the left of the transverse section CC' and above the horizontal planeJK. where K is a point at a distance )'
l")

I ( 2c -), y - H = -birr

2

J' r

(b) Ohserving Ihat the shearing stress at K is

tJ.H

'T

"

and recalling that

)'y

i tJ.H

I iJH

= lim-- = lim - - - = - .L\ -o tJ.A .l.,-o b tJ.x box

is a function of x ddined hy Eq. (6.14). derive Eq. (6.15).

Fi g. P6.60

Fi g. P6.S9

401

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402

· 6 .9 . UNSYMMETRIC LOADING OF THIN-WALLED MEMBERS; SHEAR CENTER

Shearing Stresses In Beams and Thln·Walled Members

Our analysis of the effects of transverse loadings in Chap. 5 and in the preceding sections of this chaptcr was limited to members possessing a vertical plane of symmetry and to loads applied in that plane. TIIC members were observed 10 bend in the plane of loading (Fig. 6.42) and, in any gillen cross section, the bending couple 1\1 and the shear V (Fi g. 6.43) were foun d to res ult in normal and sheari ng strc.~ses defined, respectively, by the fonnulas

My

(4. 16)

I

and Fig. 6.42 T

In this section, the

"~~:::~:~~_

l- ,,\

\I

C' (1' =P, M= Px\

Fig. 6.43

M

(\'= /', '\/

Fig. 6.44

=

I'r)

effcct~

memhers Ihat do /lot pm',l'ess

fl

.,e_

(6.6)

II

of tr.U1sversc loadings on thin-walled be ex-

I'ertical plane of S}'IIII11NI)' will

amincd. Let us assume, fur example, that the channel member of Fig. 6.42 has ~en rotated through 90 0 and that the linc of action of P still passes through the centroid of the end section. TIle couple veclor M representing the bending moment in 11 given cross section is still directed along a principal ax is of the section (Fig. 6.44), and the nelltral axis will coincide wilh thai axis (d. Sec. 4. [3). Equati on (4.16), therefore. is applicable and can be used 10 compute the normal stresses in the section. However, Eq. (6.6) cannot be llsed to dctcmline the shearing stresses in the section, since Ihis equation was derived lo r a member possessing a vertical plane o f symmetry (cf. Sec. 6.7). Actually, the member will be observed to bend ami twij·t under the applied load (Fig. 6.45), and the resulting di stribution of shearing stresses will be quite difJerent from that dcfined by Eq. (6.6). The fo llowing {1uestion now arises: Is it possible to apply the vertical load P in sllch a way that the channel member of Fig. 6.45 will bend withollt twistillg and, if so, where should the load P be applied? If the member bends without twisting, then the shearing stress at any point of a given cross section can be obtained from Eq. (6.6), where Q is the first moment of the shaded area with respect to the neutral axis (Fig. 6.460), and the di stributio n of stresses wi ll look as shown in Fig. 6.46h, with T = 0
,

"

----- --liD D

E (1/)

Fig. 6.45

VQ

= -

Fig. 6.46

! " 'E

~~

D

~~~~~ ~;;A

__ (b )

£:

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crted on a small element of cross-sectional area dA = I (Is is dF = nlA = 71 d.I·, or dF = q lh (Fig. 6.47a). where q is the shear flow q = TI = VQ/I at the point considered. The resultant of the shearing forces exerted 0 11 the c1el11elll~ of the upper flange AB of the channel is found to be a horizontal force F (Fig. 6.47b) of magnitude

(6.17)

Because of the ~ymmetry of the channel section about its neutral axis. the resultant of the shearing forces exerteU on the lower nange DE is a force F' of the same magn itude as F but of opposite sense. We conclude that the resultant of the shearing lorces exerted on the web liD must be e4ual ({) the vertical shear V in the section:

403

6.9. Unsymmetfk: Loading of Thln·Walied Members; Shear Center

[ FA .~: B

oL E

V E

1)

F' (h )

(11 )

Fig. 6.47

(6.18)

We now observe that the forces f and

r ' form

a couple of moment

Fl!. where h is the distance between the cente r lilles of the flanges AB and DE (Fig. 6.48a). 111i s couple can be eliminated if the vertical shear

V is moved to the left through a distance e such that the moment of V about B is equa l to Fh (Fig. 6.48h). We write Ve = Fh or

"~F t=,r.i[ , , "I

:'",.!!!'::II

Fh

V

1 \' E

b"

J)'-

e=-

A

/,

I

,,)

(6.19)

E

J)

(M

Fig. 6.48

and conclude that, when the fo rce P i~ applied at a distance e to the left of the center line of the web BD. the member bend ~ in a vertical plane without twisting (Fig. 6.49). The point 0 where the line of action of P intersects the axis of symmetry of the end section is called the shear rellfer of that section. We note that, in the case of an oblique load P (Fi g. 6.50a). the member will also be free of any twi.~t if the load P is applied at the shear center of the sect ion. Indeed, the load P can then be resolved into two components P. and P v (Fig. 6.50h) corresponding respectively to the loading conditi~ns of Figs. 6.42 and 6.49. neither of which causes the member to twi s\.

\ 1-'

I

1

-

--·--E------:- - ·-E------o

~

I, ) Fig. 6.50

0

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EXAMPLE 6.05 Determine the shear center 0 of a channel sec tion of uniform thickness (Fig. 6.5 1). knowing that b = 4 in., h = 6 in .. and t = 0. 15 in.

H

VQ

q= -

I

Vstl!

= -

- - o-tt---

(6.20)

21

Fig. 6.51

where V is the vertical shear and I the moment of inertia of the sl.'Ction with respect to the lIeutral axi s. Recalling Eq. (6. 17). we detemline the magnitude of the shearing force F exerted on flange AS by integrnting the shear flow q fmm A to S :

,.TI __________ L Id~

I

VtJrb ]

r '--11

H

,,, Vstl! Vlhf ' " .1· d.l· Jo qds = u - ,_1- d.l· =,_1 0 F =-41

rb ---j :

r' "1" DL!=:=)

Assuming that the member does 1I0t twist, we first determine the shear now q in flange AB at a distance .1' from A (Fig. 6.52). Recalling Eq. (6.5) and observing that the first moment Q of the shaded area with respect to the neutral axis is Q = (st)(h/2 ). we write

Dt.'::::::=:::JE Fig. 6.52

(6.21)

The di stance t' from the center Ii lie of the web BD 10 the shear center 0 can now be obtained from Eq. (6. 19):

f11 t'=

Vlhb~"

- = --- ~

V

41

V

,h"b! --

41

Substituting this e.~pres~ i on into (6.22" we write (6. 22)

1

~

-3b -6b

+ II

~

" --

It

Jh

The mome nt of inertia 1 of the channel section can be expressed as follows:

J

= 1... 01>

12

We not e that the dL~tance I' does not depend upon, and can vary from 0 to b/2. depending upon the value of the mtio Jr/Jb. For the given channel section. we have

+ 2 /11 ""~

= - I Ih l

(b)']

+ 2 [ - I htl + bl 12

It 3h

2

/:, in.

- ~ -. - = OS

Neglecting the teml cOInaining I J, which is very small. we have

l6.23 )

404

(6.24)

+-

3(411"1.)

.

and I'

.t in.

= - - - = 1.6 in.

2

+

0.5

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EXAMPLE 6.06 For the channel section of Example 6.05 detemline the dislri~ bUlion of the shearing stresses caused hy a 2.S-kip vertical shear V applied al the shear center 0 (Fig. 6.53)

Shearing stresses in flanges. Since V is applied at the shear center. there is no torsion. and Ihe stresses in nange AB are obtained from Eq. ,6.20) of Example 6.05. We have q

VQ

VII

f

II

21

(6.25)

T=-=-=-S

which shows that the stress distribution in flange AB is linear. Lelling .l" "" b and substituting for I from Eq. (6.23), we obtain the value of the shearing stress a\ B:

,

Vhb

~

2(rr tfl ')(6b

II

+ Ii)

6Vb rh(6b + Ii)

r

= 1.6 j" .

Fig. 6.53

(6.26)

Letting V = 2.5 kips. and using the given dimensions, we have 6(2.5 kips)(4 in.) oriJ = (0.15 in.)(6in.)(6 X 4 in.

+ [] in.)

= 2.22 ksi

Shearing stresses in web. The distribution of the shearing stresses in the web BD is pamoolic, as in the case of a \V-beam. and the maximum stress occurs at the neutral axis. Computing the first moment of the upper half o f the cross section with respect to the neutral axis (Fig. 6.54). we wri te

Q = hl(~Ii) + ! hI Gh) = ! hl(4b + 11)

Fig. 6.54

(6.27)

Substituting for I and Q from (6.23) and (6.27). respectively. into the expression for the shearing stress. we have VQ 1".,",

=

h

+ iI)

3V(4b

+ iI)

+ 1i)1

2111(6b

+ II)

V(ihl)( 4b = rrlh!(6b

or, willt the given data. 3(2.5 kips)(4 x 4 in. + 6 in.) 1."" = 2(0.15 in. )(6 in.)(6 X 4 in. + 6 in. ) = 3 .06 ksi

Distribution of stresses over the section. The distribution of the shearing stresses over the entire channel section has been ploued in Fig. 6.55.

Fig. 6.55

405

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EXAMPLE 6.07 For the channel section of Example 6.05. and neglecting stress concentrations, determine the maximum shearing stress caused by a l.5-kip venical shear V applieJ at the centroid C of the section. which is located 1.143 in. to the right of the center line of the web BD (Fig. 6.56). Equivalent forc e-couple system a t shear center. The shear center 0 of the cross section was determined in Example 6.05 and found 10 be at a distance e "" 1.6 in. to the left of the center line or the web BD. We replace the shear V (Fig. 6.57a) by an equivalent force-couple system at the shear center 0 Wig. 6.57b). This system consisL~ of a l.5-kip force V and of a torque T of magnitude

Fig. 6.56

T :o V(OC ) = (2.5 kips)( J.6 in. + 1.1 43 in.) = 6.86 kip' in.

Ute edge of the section. Using Eqs. (3.45) and (3.43) wlIh

Stresses due to bending. The l.5-kip force V c;)uses the member 10 bend, and the corresponding distribution of shearing stresses in the section (Fig. 6.5h-) was detennined in Example 6.06. We recall that the maximum value of the stress due to this force was found 10 be

Stresses due to twisting. The torque T causes the member to twis\, and the corres]Xlnding distribution of stresses is shown in Fig. 6.57d. Wereea]] from Sec. 3.12 that the membrane analogy shows that. in a thin-walled member of uniform thickness. the stress caused by a torque T is maximum along

v

B

01

1=" E

406

Combined stresses. The maximum stress due to the combined bending and twi~ling occurs al the neutral axis. on the inside surface of the web, and is

'Tn"", = 3.06 ksi

~

+ 65.8 ksi

= 68.9 ksi

2.5 kip,

+

n E

T",is!in)(

\ \.1.f3 ;11.

("' Fig. 6.57

L':==""

=

J -t.t-" - ,.

.---

c

D

HI - 0.630b/(I ) = t(1 -

0.630 X 0.0107) = 0.33 1 T 6.86 kip . in. (T o"" ),,,,,,,,"& = club' = (0.33 1)( 14 in.)(U.15 in.)' 65.8 ksi CI

\'

l~'> e = L6in .

(/ = 4 in. + 6 in. + 4 in. ~ 14 in. b=I=0.15in b/
(b )

1<,

(d )

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Turning our attention to thin-walled membt:rs possessing llO plane of sy mmetry. we now consider the case of an angle shape subjected to a vertical load P. If the member is oriented in such a way that the load P is perpendicular to one of tht: principal centroidal axes Cz of the cross section. the couple vector M representing the bending moment in a given section will be directed along Cz (Fig. 6.58). and the neUlml axi s will coincide wilh that ax.is (cf. Sec. 4.1 3). Equation (4.[6). therefore. is applicable and can be used to compUie the normal stresses in the section. We now propose to determine where the load P should be apptit:d if Eq. (6.6) is to define the shearing stresses in the section. i.e. , if the mt:mber is to iJend withou//wistillg . LeI us (/S!iUme thaI the shearing stresses in the section are de fined by Eq . (6.6). As in the case of the channel member considered earlier, the elementary shearing forces exerted on the section can be expressed as dF = q tis, with q = VQI I, where Q represents a first moment with respect to the neutral ax.is (Fig. 6.59a ). We note that Ihe resultant ()f the

6.9. Unsymmetfk: Loading of Thin-Wailed Members; Shear Center

"

Ag. 6.58

o

" 8 lor )

Fig. 6.59

shearing forces exerted o n port ion OA of the cross section is a force F J directed along OA. and thaI the resultant of the shearing forces exertetl o n port ion OB i ~ a force F2 along OB (Fig. 6.59h). Since both Fl and F2 p:1 ~S through point a at the corner of the angle. it foll ows that their o ....'n resultant, which is the shear V in the section. must al ~o pass through (Fig, 6.S9c). We conclude that the member will not be twisted if the line of aClion of the [o:ld " passes through the corner 0 of the section in which it is applied. The sftme rcasOIling cftn be appli ed when the load P i ~ perpendicular to the other principal centroidal axi s Cy of the angle ~ec tiotl. And. sinei.'" :lny load " :lpplied at the corner 0 of a cross section can be resolved into components perpendicular to the principal axes. it follows thm the membel will nOl be twisted if each load is applied at the cor· ncr 0 of a cross section. We Ihus conclude that a is the shear cente r of Ihe section. Angle shftpcs v.tith one ve.nical and o ne horizontal leg arc e nCOlLntered in many structures. It follows from the preceding discussion thm such members willllOt be twisted if vertical loads a.re applied :lIang the

o

407

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408

Shorutng Slre3 • .,3

I~

SeBm3 and

Thin-Wailed MembH!;

o

A

Fig. 6.60

,

!!

"

cente r line of thei r venica] leg. We nOle from Fig. 6,60 Iha l the result-

6

,,

al1l of the elementary shearing forces exerted 011 lhe vertical portion OA

,,

of a given section will be equal to Ihe shear V, while the resultant of the shearing fo rces on the ho rizontal portion on wi ll be zero:

0

,

", £

D

/I '

Fig. 6.61

Thi s docs I/ot mean, however, that there wi ll be no s hearin g Sires." in the horizontal leg of the membe r. By resolving Ihe :;hear V inlo com·

, '\~ "

\7

pllnents perpcndicuhlr to the principal centroidal axes of the rection and

tJ

D

,

(\

/I

~

,

10 £

Fig_ 6 .62

/I

A tf.~

tlF

..,

,

Fig. S.63

6

computing the s bearing stress at every point, we would ve rify Ihat T i ~ Lcro m unl y une pOint between 0 and B (see Sample Prob. 6.6l. Another type of thin-walled me mber rrC
a

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SAMPLE PROBLEM 6.6 Delemline the distribution of shearing stresse~ irL th~ loading shown.

Ih~

th in-walled angle shape

DE of uniform thickness I for

SOLUTION Sht'lIr Cl'.111er. secl;lJn ur ;! i, il[lfllip.
We recall from Sec. 6.9 Ihm the shear center of Ihe cm.""S angle shapt! is lucalcu ~l il~ cumer. SiJl(:e tilt: luau P

lhill·wall~u

I),

henrlin ~

il C:lIISP.S

nl1l1vi~rin~

hili

of Ihp. sha[w..

Prindpal A.,,~. We IOClllC the centroid C of a giv:n cross seclion AOlJ Since the y' axis is an axis of symmetry, Ihe ),' and :;' axes are the principal Oil I~ 'Ii ~ r-,- '"-- ~~

L O~~

cenlroidaJ axes of the section. We recall that for the parallelogram shown I"" = h bh 3 and 'n"" = ~ b113 . Considering each leg of the section as a parallelogram. we now determine the c~ntroidal momenlS of inertia I .. ami 10:

'"

__ _

"

d

' 8Y~ ' / +~ .,. /

~-A,, =

P

-

C

0 -;;. /

C

0

\',.;/''''... <1".

:,/

SIIJH'rpusilion. The. ~hear V in the ~e('tion i.. equal to the load P. We re_ solve it into components pamllelto the principal axes.

0

V,,=I'~n' 45°

e

We det~rmine dte shearinl! slre., s at point

Shearing Siresses Dur 10 \'... )':

of~oordinate

= Ha + y) cos 45° - ta cos 45° = ~Y cos 45° Q = t(ll - .r)Y' = ~ t(a - y ), cos 45° V,. Q (I' cos 45")[ ~ ~a - y).\. cos 45'] 31'(a - Yh' ,.- , Lt (-I! ta l)1 tal

y'

=

--

~

The , hearing stress at point f is represented by a similar function of z. Shearin!! Stresses Due to V". ~ ' = ~("

-I--

y) CO" 45

Q =

(a - y)tT

,

V"Q

- -- -

2 -

I,)

=

We :lJ.!.ain consider JXlint e

0

!':a2

-

)'1)/

cos 45 0

(Pco~45 °)[H
31'(l/ -

dta )).t

-

i)

4ta l

The , hearing stress at point f is represented by a similar fUllction of:::. Co mbined S In'sst'"S.

,H ollg th e \'crrica l Leg.

The shearing stress at

point e is A

31'(a - y) =-~,-"- l(a

/

41a

+» + 4y]

3P(11 - ;'-)(11 + 5.1') T,. =

A lO Ng 1/,,: Hurho llial Leg.

The shean ng

3!'(a - :::)z

.,

'"'

stfe~s

4 /,,3

....

at point/is

31'(a - t ) ~~4~W,,-"Ir(a +

z-I - 4z1

3P(t' - '::)(11 - k) ... 41a'

409

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PROBLEMS

6 .61 ttlrougn 6 .64 Determine the locmiort of the shear center 0 of lhin-walled beam of uniform lhicknes.i having the cross seclion sllown.

:q

D

1

1-"-1;;-"1

'I J E

C

C-;C;L-_

"

F

£-+ -+

Fig. P6.63

Rg. P6.61

"

!

D.

1--,

~

"

~

-.l

- - 1;, Fi~.

~

-----Ie

P6.64

6.65 and 6.66 An extruded beam has the cro~s section shown. Deter· mine (a) the location of Ih;! shear center 0 , (b) (he distribution of the shearing S lr~sses (;au~t:d by the 2.7.'i -kip verlil:al sht:aring fun.:!: applicu al O.

" 2 i lL

"l

°f---- , r

6in.

,J

£ " '" 2.75 kip.

2; " .

G

I-- ~ ill.--j , =

i

ill

t =titl

Fig. P6.66

410

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6.67 and 6.68 FOJ an e.\lruJeu bt:alll havi Jl<,; th e ClOSS ,ection shown. determine (a) tlte local ion of the shear center O. (b) the dislriblllion of the shearing stresse ~ caused by Ihe vertical shearing fore!." V show n ilpjllied al O.

Problems

Um",

Ii"""

JJf"""~,I

'ri+-.

Ii"""

"1

12"""

c \' - I \0 ~N

o

J 1\;2 " ""

I2mm

I:::::;:~

c

lil2 m,,,

v~ ' ' '~'~I_=' l=m'~IJ

D~' 71m",

;2",",

Fig, P6.67

Fig. PS.68

6 .69 through 6.74 Detennine the locati on of the mear center 0 of a thin-walled beam of uniform Ihickness having the cross section show n.

J ~rl/O.J ~ l

JJ

,/

B

I

ISin

H

i

3S nllll

! ';--f--. 35mm

q

I ,.L~' W

E

"I hi) """

0

•

~.

I

0

fl()llllll

E SOmm

F

--1 .KJmlll

Fig_ P6.72

_I

I-

0

1.5 ill

I ~' (i) . ,

-.-1 Fig. P6.73

F

Fig. PS.71

Fig. P6.70

JJ

IU

1- .,,£1

~ Fig. P6.69

211,

0

-

(' "\,

" B .--1

Rg. P6. 74

411

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412

Sh"Q~ng &1...,3."3 I~ Boam3 and Thln·Walied MembH!;

r L Till.--j

6 .7 5 and G.76

CJO~S

nos~

S..:ctiOIl shown. De..

section.

I-

:::,'-1

1

0

Sill

Fig. Pti. 7S

A th in·walled beam has the

tennine. Iht! location of lh t: shear center 0 of the

Oill ,

t-",,~

J Fig. P6 .76

s.n and 6.78

A thin-walled beam of uniform thickness has Ihe cross

section shown. Determine Ihe dimension f; for wluch the shear center 0 of Ihe cross sec tion I, located
i[

fv

ji(I ,"",

-I ' -

I

"

V

1

E

!t;[) ,,,,,,

no 111,,1

t-:-

W L !-,.(.

H

A ~O "'iii

L

0

J E

:!O".."

J l-,, -"-I---V "

OCI

Fig. PS.77

.200 ' '''''

0

n,,"

FIg. PI>.7Q

6.79 For

Jq d;.;

th~

allgle shape and loading of Sample I'rob. 0.6. check that

- 0 along the h orizontal leg of the angle and

f" dy

- P along ils vertical

Ie!:. 6 .80 Foc the angle shape and loading of Sample Prob. 6,6, (i/) determine the po;nl~ where Ihe sheanng stres~ i ~ ma..:im um and the oorre~ponding \'~Ill{,~ of the stress, (b) verify thatth~ paints obtained ar!!. located on the. neutral axi ~ corresponding to the given lo;':ding. 1.25,".

.jO() III

Fi g . P6 .81

'6.8 1 The cantilever beam A.B, consisting of half of a thirHvalled pipe of 1.25·in. mean radiu s and ~-in . wa.ll thickness. is ~ ubjectcd to a 500·lh venical load. K.Jl0Wll1g that the hne of action of the load passes through the centroid C of the ::ros~ . cction of the beam, det<>mline (tJ) the eqllivalent force..coup lc sys. tem at the shenr cemerof the cross section. (b) the maximum shearing stress in the be::.m. (Him: Th.e shear cenler 0 of this cross section was shewn in Prob. 6.74 tu bt: localell twi<.:c a'i rar frum ils v<::rti(;,JI diall1cler as ils ~cntruid C.) ~ .82

dll ~ed

10

Solve !'rob. 0.81, as.>U ming Ihm th e \hid.:ncs.s of the beam i .. reo

i in.

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Probtoms

'6.83 The cantilever beam shown consists of a Z slwpe of 1-in. thickness. For (he given loading. determine (he distribution of the shearing stresses along line A '8' in the upper horizon tal leg of the Z shape. The x' and y' axes are the principal cenu'oidal axes of the cross section and the corresponding moments of inertia are = 166.3 in~ and I"~ = 13.61 in~.

413

'A'

'I

0.5966

(,)

y'

(/; )

Fig. P6.83

"6.84 For the cantilever beam and loading of Prob. 6.83, delenlline the distribution of the shearing stress along line B'D' in the vertical web of the Z shape.

".=iA2&,,:1 "

·S,85

Determine the distribution of the shearing stresses along line D 'B ' in the horizontal leg of the angle shnpe for the loading shown. The x ' and y' axes are the principal centroida! axes or the cross ~ectlon.

"S.86 For the angle shape and loading of Prob. 6.85, determine the dis· tribution of the shearing stresses along line D'A' in the venicalleg. ~S . 87 A steel plate. 160 mill wide and 8 mm thick. is bent to fonn the channel shown. Knowing that the lIenicalload P acts at a point in the midplane of the web of the channel. determine (a) the torque T thm would cause the channel to twist in the same way that it does under the load P, tb) the maximum shearing stress in the channel caused by the load P.

~

Hill",,,,

( /)

P "'-I5lil\

~E

30 ,,,,,, Fig. P6.87

"S.88 Solve Prob. 6.87. assuming that a 6·mm·thick plme is bent to fom} the channel shown.

Fig. P6.85

I. = 0. 155; 1".1

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REVIEW AND SUMMARY FOR CHAPTER 6

This chapler was devoted to the analysis of beams amI thin-walled lII t: mbc l s UUl.h::1 t rallSV~ l se

Stre3ses on a beam element

1,---

-,,

loadiJlgs.

In Sec. 6. t we considered a small element located in the vertical plane of symmetry of a beam under a transverse loading (Rg. 6.2) and found thai normal stres ses a, and shearin g stresses T~ ,. were exerted on the transverse faces of that cleme nt. whi le shearing stresses T,c<' equal in magni tude \0 T , }" were exerted on its horizontal faces. In St:t:. 6.2 we cunsiden:d a prismalk beam AB with a ve:rtkal plan" o f symmetry supporting varioll s conce ntrated and distriblllcd loads (Fig. 6.5). AI a dis tance x from end A we detached from the

Fig. 6.2

Fig. 6.5

beam an element CDD'C' of length 6.x extending across the width of the beam from the upper surface of the beam to a horizontal plane loca1ed at a di stance )'1 from the neUlr:l1 axis (Fig. 6.6). We found Ilorizontal "'hear in a

,

be~m

-I cl

1-.' D

I !h ~'-'~--+---\- N.A

Fig. 6.6

that the magni1ude of the shearing force tlH exerted on the lower face of Ihe beam cleme nt was

VQ I'1H = - ,- " where V Q

~

(" .4)

ve rtical shear in the given transverse section fi rst mo ment with respec1 to the neutral axis of the shad l:d portion (f of the sl:ction I = centroidal momenl of inertia of the e ntire c rosssectional area

414

=:

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Reylew and Summary lor Chapter 6

The horizontal shear per unit length. or .I·hear flow. which was denoted by the letter q. was obtained by di viding both members of Eq. (6.4) by"''' "'H VQ q= = ~x I

(6.5)

Di viding both members of Eq. (6.4) by the area 6.A o f the horizontal face of the element and observing that ~A = 1 6. x. where t is the width of the clement at the cut , we obtai ned in Sec. 6.3 the fo llowing expression for the average .I·hearing stress on the hori zontal face o f the element VQ

7 a ,< =

Ir

Shear !fow

Shearing stresses in a beam

(6.6)

We further noted that, since the shearing stresses 7 Xl and 7 )"< exerted, respectively, on a transverse and a horizontal plane through 0 ' are equal, the expression in (6.6) also re presents the average value of 7 " alo ng the line 0; 0; (Fig. 6. 12).

Di' Fig. 6.12

In Secs. 6.4 and 6.5 we analyzed the shearing stresses in a beam of rectangular cross section. We found that the distribution of stresses is parabolic and that the maximum stress, which occurs at the center of the section. is

3V

7 m . , ""

2A

(6.10)

where A is the area o f the rectangular section. For wide-Ilange beams, we found that a gooo approximati on of the maximum shearing stress can be obtained by dividing the shear V by the cross-sectional area o f the web. In Sec . 6.6 we showed that Eqs. (6.4) and (6.5) could sti ll be used to detcrmine, respectively. the longitudinal shearing force ~H and the shear flow q exerted on a beam clement if the e lement was bounded by an arbitrary curved surl:1ce instead of a horizontal plane (Fig. 6.24). This made it possible for us in Scc. 6.7 to extend the use

Fig. 6.24

Shearing stresses In a beam of rectangular cross section

Longitudinal shear on curved surface

415

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416

Shearing Stresses In Beams and Thin-Wailed Members

of Eq. (6.6) to the detcrminalion of the average shearing stress in thin-walled members such as wide-fl ange beams and box beams. in the flan ges o f such members. and in the ir webs (Fig. 6.32).

y

Shearing stresses in thin-walled members

N.A.

N. A

(")

(h )

Fig. 6.32

Plastic deformatlons

Unsymmetric loading shear center

In Sec. 6.8 we considered the effect o f pl astic defo rmations on the magnitude and d istribution of shearing stresses. From Chap. 4 we recalled that o nce plastic deformatio n has been initiated, additional loading causes plastic zones to penetrate into the clastic core of a beam. After de monstrating that shearing stresses can occur only in the clastic core o f a beam, we noted that both an increase in loading and the resulting decrease in the size of the clastic core contribute to an increase in shearing stresses. In Sec. 6.9 we considered prismatic members that are not loaded in thei r plane of symmetry and observed that, in general, both bending and twisting will occur. Yo u learned to locate the point 0 of the cross section, known as the .I·hea r cellfer. where the loads should be applied if the member is to bend wilhout twisting (Fig. 6.49) and found that if the loads are applied at that poi nt. the following equations remain valid:

(4.16, 6.6) Using the principle o f superposition, you also learned to determine the stresses in unsy mmetric thin-walled members such as channels, angles, and extruded beams [Example 6.07 and Sample Prob. 6.61

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REVIEW PROBLEMS

6 .89 Three boards. each of 1.5 x 3.5-in. rect:lJlgu]ar cross section, are nailed together to form a beam that is subjected to a vertical shear of 250 lb. Knowing that the spacing between e;)ch pair of nails is 2.5 in .. determine the shearing force in each naiL

k:' 3.5in~ Fig. P6.90

Fig. P6.89

6.90 A column is fabricated by connecting the rolled-steel members shown by bolts of i·in. diameter spaced longitudinally C\'cry 5 in. Determine the average shearing s tress in the bolts caused by a shearing force of ]0 kips parallel to the y axis.

6.91 For the beam and loading llhown. consider section 11-11 and delermine (a) the largest shearing stress in that seclion. (b) the shearing stress al point

lI.

,''''I t I

D;,,,e,,,;,,,,, in

., I

111m

Fig. P6.91

417

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418

Shearing Stresses In Beams and Thin-Wailed Members

6.92 For the beam and loading shown. consider section mine the shearin g stress at (a) point {/. (b) point b.

11-11

lin'--j

and deter-

1-

I -,. =r:! ::: :

4 'n .

b .

-! "".IIII-,

2 in .

-'-!'-~

1---4in.---J

Fig. P6.92

6.93 FOfthe beam and loading show n in Prob. 6.92. determine the larges! shearing stress in section 11-11. 6.94 Several planks are glued together to form the box beam shown. Knowing thai the beam is su bjected 10 a ve nical shear of 3 kN. determine the average shearing stress in the gl ued joint (a) at A, (b) at B. 6.95 Knowin g Lhat a W360 X 122 ro lled-steel beam is su bjected to a lSO-leN vertical shear. determine the shearing stress (aJ at point a. (b) at the centroid C of the section.

:

I lt~ "'"'

H

Dim e n. lon s ito """

Fig. P6.94

I

c

Fig. P6.95

6 .96 A beam consists of fi\'(' planks of 1.5 X 6-in. cross section connected by steel bolts with a longitudinal sp:lcing of9 in. Knowing thaI the shear in the beam is vertical and equal to 2000 lb and that the allowable average shearing stress in each bolt is 7500 psi. detennine the s mallest permissible bolt diameter that may be used.

I

(lin .

I 1,1 -, I i ll

Fig. PS.96

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5 .97 A plal e of ..j.-nlln Ll, icJ..u e~~ is lJeLli as ~ h(/" n and ~H: ll U500 il.I a beam. For a I'erueal .;hear of 12 kN, determine (a) the shearing stress a1 point A. ~b) the maximu r.1 sh~ ari ng slres~ in the beam. Also ~etch the she ar 110"' in the eros> section.

Rm/ tew Problem.

~-I

1 -'--- JI~ ~

~) I~

;()

..

20

L)lmpmioll, i" nltn

Fig. P6.97

5 .9B and 6,9 9 For an extruded ooam h,winS the c ros~ section ~hown . determine (11) the location of the shear ceOler O. (b) the distribution of lhe . he arillg SIl t,5ses cau sed by tht: velli c al she ariu); fOJ\:e- V showLi applie.J at O.

16 ,,, ,,,

Ar="""" ~I

--14IUII,--+~tI'"'"

fj,m,, _

D

o

B ,r _ , I

t

E

.. 1

\ ' =15kN

I

JOm,,.

p,nltl-+ Co

'30mm

III Inltl.-.-l

~

J

30 " "11

l. = 1.14~ X

lit '"11,4

Fi g. P6.98

6 .100 of unifOnll

Determ ine the location of the shear ct!nter 0 of a th in-walled beam hav ing the cross section shown.

thic knes~

'''''1AI [

3",

, o •

I

I /)ll,.- - - -TII 5

l

"-

I. = il,~33 X lO' IIlIn'

Fig. P6.99

I

Fig. P6.100

ill ,

3 in.

E

_ I

419

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COMPUTER PROBLEMS

The following problcnL'I

:In'

designed 10 be soh'cd witll

:I

compulcr,

6.C1 A timber beam is to be designed to support a distributed load and up to two concentrated loads as shown. One of the dimensions of its uniform rect.'lilgular cross section has been specified and the other is to be detennincd so thaI the maximum normal stress and the maximum shearing stress in the beam will not exceed given allowable values 0',"1 and T;lJ. Measuring x from end A and using either 5 1 or U.S. customary units. write a computer program to calculate for s uccessi ve cross sections. from x = 0 to.r '" L and using given increments fl.x. the shear. the bending moment. and the s mallest value of the unknown dimension that satisfies in that section ( I) the allowable nonnal stress requirement. t2} the allowable shearing stress requirement. Use dlis program to design Ihe beams of uniform cross i;ection of the following problems. assuming (Tall = 12 MPa and Tal l = 825 kPa. and using the increments indicated: (al Prob. 5.65 ( fl.x = 0.1 m), (IJ) Prob. 5.157 ( fl.x = 0.2 m).

Fig. P6.C1

p

'\ I=~~~ii;j'H I- - - L- -..I Fig. P6.C2

6 .C2 A cantilever limber beam AB of length L and of unifornl rectangular section shown supports a concentrated load P at its free end and a uniformly distributed load 1V along its entire length. Write a computer program to deternline the length L and the width b of the beam for which both the maximum nonnal stress and the maximum .shearing stress in the beam reach their largest al lowable values. Assuming if . n = 1.8 ksi and T oJ l = 120 psi. use this program to detennine the dimensions Land b when (tI) P = 1000 lb and tv = O. (b) P = 0 and 'W = 12.5 lblin .. (c) P = 500 Ib and 'W = 12.5 lb/in S.C3 A beam having the cross seclion shown is subjected to ;J vertical shear V. Write a computer program that. for loads and dimensions expressed in either SI or U.S. customary units. can be used to calculate the shearing stress along the line between any two adjacent rectangular areas fonning the cross section. Use UllS program to solve (al Prob. 6.10. (I,) Prob. 6. 12, tel Prob. 6.21.

Fig. P6.C3

420

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G.C4

A plate of unifollll

thicknc~s

I is bc: nt as

a vertical plane of s~mmetry anri is then usctl as

~how l\

into ;) shape with

Compute, Problem.

~

beam. Wrile a cOlnput~r pro· expressed in eilher SI or U.S. customtry

gram Ihat , for IMds and dimension~ units. call be used \0 uelemune th e distnbutlOn of slleann g stre,ses ca u,ed by a ... ertical ~hear V. Use this program (ll) 10 so l... e Prob. 6..17. (b ) 10 lind Ihe shearing Stl"i:SS at a poil\! E for the s hape and 10.ld of Prob 6.50. assuming a thickness I = ~ in.

"I-"1

I

II ,

-,,--1 - -

' ,- - I

1

Fig. PS.C4

6 .C5 The cross stction of an extruded beam is symmetric wi lh respect the .t· axis and co ns isl<; of sevual straight segments as shown. Write a computer program th:lI. for loads cs caused by a vertical fort:e appl ied al 0. Use Ihis program [0 solve Probs. 6.66 and 6.70 . \0

1/

Fig. P6.C5

6 .C6 A [hiu-walleu beam has thl! cru~s se~ [iun shown. Write a cumputer program Ih;ll. fIX loads ami dimensions expressed in either 51 ur U.S. custom· ary lInits. can be used to de lennine the localio n of the $hear center 0 of the eros;; se cti on. Use Ihe program lO w ive Prob. 675.

Fig. P6.C6

421

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Transformations of Stress and Strain

The plane shown is being tested to determine how the torces due to lift would be distributed over the wing. This chapter deals with stresses and strains In structures and machine components.

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7.1. Introduction

7.1. INTRODUCTION We saw in Sec. 1.12 lhat the most general state o f stress at a given point Q may be represented by six components. Three. of these components. (T., . u ' and (T" del'ine the nOfmal stresses exerted on the faces of a small Y cubic element centered at Q and of the same orientatio n as the coordi nate axes (Fig. 1. la ), and the other three, 7-,,", 7 and 7~"t the components of the shearing stresses on the same element. As we remarked at the time, the same state of stress will be represented by a different set of components if the coordinate axes are rotated (Fi g. 7. lh). We propose in the first part of this chapter 10 determine how the components of stress are trallsfomled under a rotation of the coordinate axes. The second part of the chapter will be devoted to a similar analysis of the transformation ()f the components o f strai n. y ,.

"

o (b)

{II.'

FIg. 7. 1

Our discussion o f the transformation o f stress will deal mainly with plalle slre.V!>. i.e., with a silUation ill which two of the faces of the cubic element arc free of any stress. If the z axis is chosen perpendicu lar to these faces, we have a_ = T:;;x = T,.. = 0, and the only remaining stress components are a x' a ,., a~d 7...,. (Fig~ 7.2). Such a situation occurs in a thin plate subjected 10 forces acting in the midplane of the plate (Fig. 7.3). It also occurs on the free surface of a structural e lement or machine component. i.e.. at any point of the surface of that element or component that is not subjected 10 an external force (Fig. 7.4) .

Flg. 7.2

.., Fig. 7.3 ! We recall (ha(

T,.,

=

T" . T"

Fig. 7.4

=

T".

and

T "O

=

T.....

423

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424

Tnmslormatlons 01 Stress and StraIn

Considering in Sec. 7.2 a state of planc stress at a given point Q characterized by the stress components fT, . fT" and T.". associated with the clement shown in Fig. 7.5(1, yo u will learn to detcm li ne the components fT.," fT ,i . and T ., .-' associated with that element after it has been rotated through an angle () about the z axis (Fig. 7.Sb). In Sec . 7.3, you will determine the value Or of f} for which the stresses fT.r' and fTv are, respectively. maximum and minimum: these values of the nonnal stress are the principal .l"fres.l"e.~ at point Q, and the races of the corresponding elemelll define the principal plane.l· of Mre.l"s at thaI point. You will alsQ determine the value fl, of the angle of rotation for which the shearing stress is maximum. as well as the value of that stress.

(b)

FIg. 7.5

In Sec. 7.4, an alternative method for the solution of problems involving the transformation of plane stress, based on the use of Mohr's circle. will be presented. In Sec. 7.5, the three-dimensional state oj .vtre.u at a given point will be considered and a fo rmula fo r the determination of the normal stress on a plane of arbitrary orientation at that poim will be de veloped. In Sec. 7.6, you will consider the rotations of a cubic clement about each of the principal axes of stress and note that the corresponding transfo rmations of stress can be described by three different Mohr's circles. You will also observe that. in the case of a state of plane stre.vs at a given point, the maximum value of the shearing stress obtained earlier by considering rotations in the plane of stress docs not necessarily represent the maximum shearing stress at that point. This will bring you to d isti nguish between in-plane and ollt-oj-plane maximum shearing stresses. Yield criteria fo r ductile materials under plane stress will be developed in Sec. 7.7. To predict whether a material will yield at some critical point unde r given loading conditions. you will determine the principal stresses fTa and fT b at that point and check whether fTa' fT b' and the yield strength fT y of the material satisfy some criterion. Two criteria in common use are: the maximllm-shearillg-.\·trength criterioll and the mcuimllm-di.vtortioll-energy criterioll. In Sec. 7.S,jracture criteria fo r brittle materials unde r plane stress will be developed in a simi lar fa shio n; they will involve the principal stresses fTa and fTb at some critical point and the ultimate strength fT u of the material. Two criteria will be di scussed: the maximlllll-llormal-stre.u criterion and Moh r's criterion .

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Thin-walled pressure vesse!.\· provide an important ap plication of thc anal ysis of plane stress. In Sec. 7.9, we will discuss stresses in both cylindrical and spherical pressure vessels (Figs. 7.6 and 7.7).

Fig. 7.7

Sections 7.10 and 7. [I will be de voted 10 a discussion o f tile tl"lIllSformatioll of plalle straill and to Mohr's cire/e fo r plane slrain. In Sec . 7. 12, we wi ll consider the three-dime nsional analysis uf strai n and see how Mohr's circles can be used to determine the maximum shearing strain al a given point. Two particu lar cases are o f special interest ,md shou ld not be confused : the case o f plane .I·!lnin ,U1d the case o f plalle stress. Finally. in Sec. 7. 13, we di scu s~ the use of strain gage.f 10 measure the nonnal strain on the surface o f a structural clement or machine component. You wi ll see how the components CI , f \., and y" characterizi ng the slale of strain at a gi ven point can be computed from [he measurements made with three strain gages forming a slmill roselle.

7.2. T RANSFORMATION OF PLANE STRES S Let us ass ume that a state of plane stress exists at point Q (with fT~ = T~, = T "" = 0), and thaI it is de lined by the stress components fT.<" fT,.. and T.". associated with the element shown in Fig. 7.5a. We propose to determine the stress components fT." fT", and T .N associated with the elemem after il has been rotated through an angle 0 about the .:; axi s (Fig. 7.5b) , and to express these components in terms of 1.7.<" fT,., T .." and {J .

7.2. Trans/onnatlon 01 Plane Strass

425

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426

Transformations 01 Stress and Strain

'0 )

Fig. 7.5 (re pealed)

In order [0 dctcnnine the lloTmal stress U x' and the shearing stress T,-',,' exerted on Ihc face perpendic ular to Ihc x ' axis, we consider a prismatic clement with faces respectively perpendic ular to thc x . y, and x' axes (Fig. 7.8a). We observe thilt, if the aTca of the oblique face is denoted by ~A , the areas of the vertical and horizontal 1~1CCS are respectively eq ual 10 dA cosO and dA sinO. II follows that the forces exerted on the three faces are a~ shown in Fig. 7.8b . (No forces are cx-

" M~iI\O

(II )

Fig. 7.8

crted on the uiangular faces of the clement, since the corresponding nannal and shearing stresses have all been assumed equal to zero.) Using components along the x' and y' axes, we wri te the. followin g eq uilibrium equations: z.F,<' = 0:

cr,,' 6.A

(T

T.<'y' 6.A

6.A ~os 0) ~os 0_- T,'J'( 6.A ~os 8) s~n (}_ ,( 6.A Sin 8) Sin () T.",~6.A Sin 8) cos e - 0

-=- cri l

+ cr..{ 6.A cos 0) sin 0 - 7,'l( 6.A cos 0) cos 0 , - cr,,(6.A si n 0) cos 0 + T,wl 6.A sin 0) sin (} ~ 0

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Solving the. tirst equalion for if.,' =

ifx COS

2

f7,r'

and lhe second t'or T,,' ,,'. we have 2

0 + if , sin 0

+ 2T.,.. sin 8CDS H

(7. 1)

2

(7.2)

2

T.,y = - (if, - if,.) sin (j cos 0 + T.".(cos 0 - si n 8)

Recalling the trigonometric relalions sin 20 = 2 sin (fens H

(7.3)

, =-.'_+,-,,~0~;~2:-0 cos-(J 2

---='O".c',2~" sin-, 0 = -.,'-2

(7.4)

we write Eq . (7. 1) lL'; follows:

+ cos 20

if;' = if"

2

- cos 20 2 + T.,. sin 20

+ if ,

0' a~

+ fT. 2

(T, -

tTy

+ - -,- - cos 20 +

T (,

sin 20

(7.5)

Using the relations (7.3). we write Eq. (7.2) as a~

T I,.' =

-

-

{T ,

- -2--· sin 20

+ T ~ cos 28

(7.6)

The expression fo r the nonnal stress fT v' is obtained by replacing 0 in Eq . (7.5) by the angle 0 + 90° that the y' axis forms with the x axis. Since cos (20 + 180°) = - cos 28 and sin (2fJ + 180a ) = - sin 20 , we ha \lc

a x + u ,' 2

a ... - a , 2

u " '"' - - _ . - - - - cos 20 -

T n ' sin

'.

20

(7.7)

Adding Eqs. (7.5) and (7.7) member to member. we obtain (7.8)

Since a : = fT;> = O. we thus verify in the eMe of plane stress that the sum of the normal stresses exerted on a cubic element of material is independent of the orientation of that elemenLt ICf. fOOlllO!I."" on pagl."" II!\.

7.2. Transformation 01 Plane Stress

427

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428

Transformations 01 Stress and Strain

7 .3 . PRINCIPAL STRESSES; MAXIMUM SHEARING STRESS

The equations (7.5) and (7.6) obtained in the precedi ng section arc the parametric equ:lIions of a circle. This means th::lI. if we choose a set of rectangular axes and plol a point M of abscissa and ordinate T, ,>,, for any given value of the parameter 0, all the points thus obtained wi U lie on a circle. To establish thi s propen y we c1iminate 0 from Eqs. (7.5) and (7.6): thi s is done by first transposing (u.. + u ,,)/2 in Eq. (7.5) and squaring both members of the eq uati on, then squ~rin g both members of Eq. (7.6), and linall y adding member 10 member the two equati ons obtai ned in thi ~ fashion. W e have

u,.

( U+U)' + T;',' "" (u.-U)' + T;,. ;r

,

,

•

--2--

fT.", -

,

--2--

,

(7.9)

Scuing

u r + u )' u ..c == - -2- -

and

we write the ide nlity (7.9) in the form (U"i -

u .. ~y

+ T.~V

= R2

(7 .11 )

which is thc equation of a circle of radius R ccntered at thc point C of abscissa (Jm and ordinate 0 (Fig. 7.9). It can be observed that. due 10 the symmctry of the circle about the hori zontal axi s, the same result would have been obtained if, instead of plouing M, we had ploncd a point N of abscissa u .r' and ordinate - T x'y' (Fig. 7. 10). TIlis pro perty will be lIsed in Sec. 7.4.

f,'"

"'N" I

(' I

D

{I

I I

.'''_l:j : E

f---- .,~

Fig. 7.9

T,"I

,

a .1

~( I- - - ur . - J Ag.7.10

Thc two points A and B whcre the circle of Fig. 7.9 intcrsects the horizontal axis are of special interest: Point A corresponds to the maximum value of the normal stress (If' , while point B corresponds 10 its

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minimum value. Besides, both points correspond 10 a zero value o f the shearing stress T.N ' Thu s. the values 6" of the parameter fJ which correspond to points A and B can be obtained by setting 1"i',.' = 0 in Eq . (7 .6). We writet 21"", lan20 = - -'"

(T.. -

(7, 12)

U ,

°

This equation defines two values 201' Ihal are I SOQ apart. and thus IWO values 1, that are 90" apan. Either of these values can be used to deten nine the orientati on of the corresponding element (Fi g. 7.11 ). The planes contai ning the faces of the element obtained in this way arc called

"\

(T.. .",

Fig. 7.11

the: principal plane~' of stress at poi nt Q, and the corresponding values and a nun of the normal stress exerted o n these planes arc cal led the prillcipal.rfreue.\· at Q. Since the two values 01' defined by Eq. (7 .1 2) were obtained by sening 7 ...,. , = 0 in Eq. (7.6), it is clear that no shearing stress is exened on the principal planes. We observe fro m Fig.. 7.9 that U m3J<

and Substitllling for

(T.,.c

=

II .... - R

0 , 13)

and R fro m Eq. (7. 10). we write

.• + =U ~ (T \

(T m b . min

a mi n

::!:: \

I(U--2-~ UY)' + 1";"

(7 ,14)

Unless it is pos,ible to tell by inspection which of the two principal pl anes is subjected to U ma < and which is subjected 10 lTm,n' it is necesSfirY 10 substitule one of the values 01' into Eq. (7.5) in orde r 10 determine which of the two co rre ~po nds to the maximum value o f the normal s tre~s . Referring agai n 10 the c ircle of Fig. 7.9, \ve note thm the point s D and E located on the vertical diameter o f the circl e correspond 10 the largest numerical \'al uc of the shearing stress 7 N. Since the abscissa of poi nts D and E is U ' Vf: = (17 r + IT,.)/2 . the values 0, of the parameter (J corresponding to these poin l~ arc obt ained by setting II ,. = (u r + 17).)/2 JThis relatio n can also beobtninrd by riv:ltivl' I'llua) to zero: dor, /d8 = O.

dllfc~ntiating

t1 _

in Eq. (7.5) and setting the dr·

7.3. Principal Stresses; Maximum Shearing Stress

429

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430

Transformations 01 Stress and Strain

in Eq. (7 .5). It follows that the slim of the last two terms in that equation must be zero. Thus, for 0 = 0., we writet U , -

U .

- ' -2--cOS 28,

+ T"

sin 20, = 0

_ U, -

u )"

(7.15)

tan 28., =

/' .'

Th is equation defin es two values 28, that are 180" apart , and thus two values 8, that are 900 apan. Either of these values can be lIsed Lo determine the orientation of the clement corresponding to the maximum shearing stress (Fig. 7. 12). Observing from Fig. 7.9 that the maximum value of the shearing stress is equal to the radius R of the circle, and recalling the second of Eqs. (7.10). we write

7rna.' =

J(U, if,), + T.;,, - -2- -

(7 . [6)

Ag .7.12

As observed earlier, the normal stress corresponding to the condition of maximum shearing stress is

(7 . 17)

Comparing Eqs. (7. 12) and (7.15), we note that tan 28~ is the negative rcciprocal of tan 281" Thi s means that the angles 28, and 281' are 90" apart and, therefore, that the angles 8, and {II' are 45° apan. We thus conclude that the plane.\" of maximum .I·hearillg .I·tress are at 45 0 to the principal planes. This confirms the results obtained earlier in Sec. 1.12 in the case of a centric axial loading (Fig. 1.40) and in Scc. 3.4 in the case of a torsional loading (Fig. 3.20.) We should be aware that ollr analysis of the transformation of plane stress has been limited to rotations in the plane of.vtress. If the cubic clement of Fig. 7.7 is rotated about an axis other than the z axis, its faces may be subjected to shearing stresses larger than the stress defined by Eq. (7.16). As you wi ll sec in Sec. 7.5, this occurs when the principal stresses defined by Eq. (7.14) have the same sign, i.e., when they are either both tensile or both compressive. In such cases, the value given by Eq. (7 .16) is referred to as the maximum ill -plane shearing stress. t Thi s relation may al.lo be obtained by differentiating rivative equal to zero: dT..., /dO = O.

Txy

in Eq (7.6) and setting the de·

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EXAMPLE 7.01 For the sTate of Idane

stre.~s

shown in Fig. 7.13 , determine

(II) the principal plane~. (b) the principal stresses. (c) the max-

(c) Maximum Shearing Stress. yields

Formula (7.16)

imum shearing STress and the corresponding nonnal stress.

Ag. 7.13

(a) Principal Planes. Following the usual sign convention. we write the stress components as 0" , =

+50MPa

0", =

Since 0" .,,, and IT mon have opposite signs. the value obt:lined for '1,"" :lctu::llly represents the maximum value of the ~heari ng StresS at the point con ~ idered . The orientation of the planes of maximum shearing s tress and the sense of The shearing STresses are best deTennined by passing a secTion along the diagonal planeAC of the element of Fig. 7. 14. Since the face.<; AB and BC of the element are contained in the principal planes. the di::lgonal planeAC mUST be one of the planes of maxim um shearing stress (Fig. 7. 15). Furthermore, the equi librium conditions for the prismatic element A BC require that the shear-

- IOMl'a

SUbSTiTuting into Eq. (7. 12). we have

H~= 2G.G.

"" \

2'1.". 2( +40) 80 tan 20 == - - - ~ cc''-'~'cc I' 0"> - IT , 50 - ( - 10) 60 and 180° + 53 . 1° = 233. 1° 21J,. ~ 53. 1" Or = 26.6" and 116N

,,r,_~ .,. C",~

II, = II" _ 45"

or'

(b) Principal Stresses. IT, II """. min

+

II,

= - -2- - ,

Formula 0. 14) yields

J(~)' 2 + '1.",

+

V(30)' + (40)' 50 = 70M P;}

= 20

IT "u .

= 20 - 50 = - 30 MPa

TIle principal planes and principal stresses are sketched in Fig. 7. 14. Making IJ = 26.6° in Eq. (7.5), we check thaT the nomtal stress exerted on face BC of the e lemelll is The maximum stress: u~,

50 -

10

= --2--

20

u ' ==

=

IT ~I<

50 + 10 + --,-- cos 53.1 0 + 40 sin 53.1 0

+ 30 cos 53. 1" + 40 si n 53.1 °

A

0" = 211.11'

c Fig. 7.1 4

= 70M Pa =

_ 18.4"

ing stress exerted on AC be directed as shown. The cubic element corresponding to the maximum shearing Stress is s hown in Fig. 7. 16. The nonnal stress on e::lch of The four faces of the element is give n by Eq. (7. 17):

~ 20 , IT n"",

=

Fig. 7.1 5

0" ,

+ IT, 2

= 50 -

2

10 = 20 MP::l

IT ' = 20 ,\ ll ' .. IT ....

, Fig. 7. 16

\ 431

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, 1---" ' '''iiii:::: to

SAMPLE PROBLEM 7.1 A smgle hOrlzolltal force P of magnitude 150 It! is applied 10 end 1J of lever ABD. Knowing that portion AB of the lever ha~ :l diameter of 1.1 in .. rleler_ mine ((I) the nornlai and shearing stresses on an element located at po~nt H

ill

J)

,

,Ii .. ,

anu having sides parallel to the.x and y axes. (b) the principal planes and Ihe PJincijJal 5tre.'Sc':s at point H.

. ~

SOLUTION ro rce-Couplc System. We replace the force P by an equivalent forcecouple sy;;lem allhe cenler C oflhe transven;e section containing poil1l H:

P=1501b

T=(150Ib)(18in.)=2.7kip'in. M, = (150 Ib)([(J in.) = 1.5 kip ' in.

fl . StresS{'!; fr ,. l 'f ,,, T " al Point H. Using the sign convention shown in Fig. 7.2. we det~rmine Ihe sense and Ihe sign of each SIres., component by care-

fully

~xamiJliJlg

tht:

(f ,, =

I to". = I,S4 h i

+Me

ur tht:

"'-

fur~t:-l:llUplt: ~)stt:m


+ (1.5 kip' ;n. )(0.6 in. :1 ~7T (0.6

I

I"'

- 101~",

~kt:tl:h

Tc

+-J = +

in.)'

(2.7 kip' ;n.)(0.6 in.) 1" "

i1T (0.6 i~.)4

= +7.96 hi <41

We note that the shearing force P does not cause any shearing stress at point H. b. PrincipII I Planes and Princi pal St l"CSl'ics. SubSiituting the values of the stress components into &t. (7.12). we determine the orientation of the mindpal planes: tan 28" -

-1- " "::'"

2"'~"

~~~ ff , - fl f

20" = - 61.0 0

2(7.96) - - - - - - - 1.80 0 - 8.g4 0 and 18[10 - 61.0 =

+ 119 0 and

+ 59.5" ...

Substituting into bJ. (7.14), we determine the magnitudes 01 the priTlcipal stresses:

= 0

\~.84

±

JC

8 84 2 . )' + (7.96f

= H.42

± 9.10

" "",. - + 13.52 hi ... " ""' = - 1.58 ksi .... Consideri Ll);; face "b of tlr eltment showlJ. we m"ke (Jp - - )0.5' ill Eq. (7.5 ) ~nd find " " = - 4.ti~ ks i. WP. conclllflp. l1L11 rhp. rtincir~l Slrp.«<;p.s arp. .1~ ~hown

432

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PROBLEMS

7 .1 through 7 .4 For the given slate of siress. determine the nomlal and shearing stresses exerted on the oblique face of the shaded triangular element shown. Use a method of analysis based on the equilibrium of that element. as was done in the derivations of Sec. 7.2. JO hi

t2l,i

fill .\ IPu

;:;:t:~ 4,~ .\[P" 'if

2. MP"

120

'==4=="

til

Fig. P7.t

~t P., Fig. P7.2

7.5 through 7.8 For the given state of slress. determine pal planes. (b) the principal stresses.

Fig. P7.S and P7.9

.\1f'~

Fig. P7.6 and P7.tO

Fig. P7.3 (a )

Fig. P7.4

the princi-

Fig. P7.7 and P7.11

Fig. P7.8 and P7.12

7 .9 through 7.12 For the given state of stress. determine (a) the ori· entation of the planes of maximum in-plane shearing stress. (b) the corre· sponding normal stress. 7. 13 through 7 .16 For the given state of siress. detennine the nonnal and shearing stres,es after the element shown has been rotated through (0) 25° clockwise. (b) 10° counterclockwise.

6'P ;

Sh i

,...--'-_

5 k.
1 50~I P3 Fig. P7.13

Fig. P7.14

Fig. P7.1S

Fig. P7.16

433

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434

Transformations of Stress and Strain

7.17 and 7.18 The grain of a wooden member forms an angle of 15° with the vertical. For the ~tate of stress s ho""n. determine (a) the in-plane shearing stress paraliellO the grain. (b ) the normal Stress perpendicular to the grain.

11.5 Ml'a

Fig. P7.17

Fig. P7.18

7. 19 Two members of uniform cross section 50 X 80 mm are glued together along plane (HI that forms an angle of 25~ with the horizontal. Knowing that the allowable stresses for the glued joint are if = 800 kl'a and r = 600 kPa. determine the largest centric load P that can be applied.

5()""u_

Fig. P7.19

Fig. P7.20

7.20 A steel pipe of 12-il1. outer dial1leter is fabricated frOI1l t-i n.-thick plate by welding along a heli.~ which fonns an angle of 22.5" with a plane perpendicular 10 the axis of the pipe. Knowing that a 40-kip axi'al force P and an SO-kip' in. torque T, each directed as shown. are applied to the pipe. determine u and r in directions. respe.::ti vely. normal and tangential to the weld.

Fig. P7.21 and P7.22

7.21 Two wooden l1lemlx.-rs of 80 X 120-I1lm uniform rectangular cross section are joined by the si mple glued scarf splice shown. Know ing that (3 = 25" and that centrk loads of l1lagnitude P = I{) kN are applied 10 the members as shown, determine (a) the in-plane shearing stres..~ parallel to ule splice. (b) the nonnal stress perpendicular to the splice. 7 .22 Two wooden members of gO X 120-111111 uniform rectangular cross section are joined by the simple glued scarf splice shown. Know ing that (3 = 22" and that the ma,imum allowable s tresses in the joint are. respectively. 400 kPa in tension (perpendicular to the splice) and 600 kPa in shear (parallel to the splice). detemline the largest centric loal.l P that can be applied. 7 .23 A 19.5-kN force is applied at point D of the cast-iron post shown. Knowing that the post has a dirtmeter of 60 nlnl. determine the principal stresses and the maximum shearing stress nt point H.

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Probloms

150 "nn

Fig. P7.23 and P7.24

7 .24 A 19.5-kN force is applied at poim D of the cast-iron post shown. Knowing that the post has a diameter of 60 mm. determine the principal stresses and the maximum shearing stress at point K. 7 .25 The axle of an automobile is acted upon by the forces and couple shown. Knowing that the diameter of the solid axle i.~ 1.25 in., determine (u) the principal planes and principal stresses at point H located on top of the axle. (b) the maximum shearing stress at the same point 7 .26 Several forces are applied to the pipe assembly shown. Knowing that the inner and Olller diameters of the pipe are equal to 1.50 in. and 1.75 in. , respectively, detemline (u) the principal planes and the principal stresses at poim H located at the top of the outside surface of the pipe. (b) the maximum shearing stress at the same poin!.

',- in. (l

12 in lOi n .

/)

I

3IJ 11>

II in

C

., " in.

C

Fig. P7.26

I

3fI III

liOO III FIg. P7.2S

435

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436

Transformations of Stress and Strain

7 .27 For the siale of plane stress shown. delemline the largesl value of for which the maximum in-plane shearing stress is equal 10 or less Ihan 75 MPa.

IT,

;.=t=~ :20 .\11'11 (;0 \11'"

Fig. P7.27

Fig. P7.2B

7.28 For Ihe siale o f plane siress shown. delennine (a) Ihe largest value of 1".,~ for which the maximum in-plane shearing stress is equal to or less Ihan 12 ksi . (b) the corresponding principal slre.;ses. 7.29 For the state of plane stress shown. delermine VI) Ihe value of T" for which the in-plane shearing stress para llel to Ihe weld is zero. (b) the corresponding principal stresses. 15

!3 .\IPa

Fig. P7.29

k~i

Fig. P7.30

7.30 Delermine the range of values of plane shearing stress is equal

10

CT. for which the maximum illor less than 10 ksi.

7.4. MOHR 'S CIRCLE FOR PLANE STRESS The circle used in the preceding section to derive some of the basic formulas relating to the transformation of plane stress was first introduced by the Gennan engineer Ouo Mohr (1835- 1918) and is known as Mohr's circle for plane stress. As you will see presently, this circle can be used to obtain an alternative method for the so lution of the v:uious problems considered in Sees. 7.2 and 7.3. This method is based on simple geometric considerations and docs not require the usc of specialized fonnulas. While originally designed lor graphical solutions. it lends itself well to the use of a calculator.

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Co nsider a , quare cleme nt of a matelial subjected to pl ane stress (Fig. 7.17a), and let fT" a " and T,
7.4. Moh r's CIrc le lor Plane Stress

., o

cI

12111,

:/

"'' 'J

~ X(
.r,,"',

t(", - " 1) (b )

Fig. 7.17

We also note that, since tan (XCA ) = 2T../( a r - ay). the angle XCA is equal in magnilUde to one of the angles 2fJp that smi sfy Eq. (7. 12). Thus, the angle OJ! that defines in Fig. 7. 17a the orielllation of the principal pl ane corresponding to point A in Fig. 7. 17b can be ob-

tained by dividi ng in half the angle XCA measllred on Mohr's circle. We funher observe that if a x > IT ,. and T n- > 0, as in the case co nsid· ered here, the rotat ion that brings into CA is countercloc kwi se. But, in that case, the angle 0" obtained from Eq. (7. 11) and dcJ'ining the direction of the no nnal Oll to the principal plnne is positi ve ; thus, the rotati on bringing Ox into Ou is also coul1lc rclockwisc. We concl ude that the senses of rotation in bo th parts of Fig. 7. I7 are the same; if a coun!ercl ockwise rotation through 28j! is rcqu ired to bring CX il1lo CA on Mohr's circle, a c:o ul1Icrcloc kwise rotation through H" witt bring Ox il1lo Oll in Fig. 7. 17a. t

ex

IThis is du e 10 Ihe far lthat we are

Fig. 7,9 as M ohr's circ le.

u~i ng

toc circl e of Fi g 7. \0 rather thall !he circle of

437

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438

Si nce Mohr's circle is uniquely defined, the same circle can be obtained by considering the stress compone nts 0'Y;' , 0'y" and r <,y', corresponding to the x' and )" axes shown in Fig. 7. 180. The point X' of coordi nates 0'.<, and - T x '/ , and the pointY' of coordinates 0'1 and + TY;'y" arc therefore located on Mohr's circle. and the angle X'CA in Fi g. 7. 18b must be equal to twice the angle x'Ga in Fig. 7.ISa. Since. as noted

Transformations of Stress and Strain

o

Iii)

Fig. 7.18

before, the angle XCA is twice the angle .x.Ga, it fo llows that the angle XCX' in Fig. 7. 18b is twice Ihe angle _lOX' in Fig. 7. 180. Thus the diameter X' Y' defining the nonnal and shearing stresses O'~" 0'/ . and T <,.' can be obtained by rotaling the diameter XY through an angle equal to twice the angle () formed by the x'aml x axes in Fig. 7. 18a. We note that the rotation that brings the diamet ~r XY into the diameter X' Y' in Fig. 7. ISb has the same sense iLS the rotation that bri ngs the.\)' axes into the x'y ' axes in Fi g. 7. 18(1. The property we have just indicated can be used to verify the fact that the planes of maximum sheming stress are at 4SO 10 the princi pal planes. Indeed, we recall that points D and E on Mohr's circle correspond to the planes of maximum shearing stress, while A and B correspond to the principal planes (Fig. 7. 190). Since the di ameters AB and DE of Mohr·s circle arc at 90 ~ to each other, it follows that the faces of the corresponding clements arc at 45° to each other (Fig. 7.19(1). ~-

". '=

"·"· - ~I D

----+"-=---.,

,

f", ..

00"

o

8

c

A

£ (0 1

Fig. 7.19

(iJ)

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7.4. Mohr':> Ctr::le lor Plane Slrn3.

TIle COllStJUctioll of Motu"s cLIC\e fOI plwlI::' SUl:SS is greatl y sim plilil."U if we eo nsidt' r scpamlc ly each fate of the e lement usees are co ncerned.

we

(~)

Clod", ..."" - _ .'\ \K''''

Rg . 7.20

the usual co nvention hold~. i.e .. a tensile stress is co nsidered :IS positive and is plotted \0 the right. whi le a compressive stress is conside red as negative 1100 is plmtcd to thc left.

! The foU"wingjin gk i . h,tpruJ in r<:mcmhc, in g !hi. CD"'·COU,,", ·'In 'he kitch-OIL. the d.xl<

is nb<J\'c. and Ihe

CiJ",J1o'

Is b.;>low:·

EXAMPLE 7.0 2 For Ihe stille of plane

~tn,!»

already

7.01. «(/1 construct Mohr's circlO'.

clln sid~reJ

Ex~ mple

in

(b) uelennine the principal

stress es. (e) determine the maXJIlIUln shearing "tress and the corresponding nonn:!.] siress.

(a) Construction of Mohr's Circle. Wc

nOl~ from

Fig. 7.2 1a that the normal stress exened on the fac~ onented IOWMd Ihe x axis is len:
(T, +a"

50 + ( - 10)

= - -,- - ~

1

I

Sincc the sides of the shndd lriangle Me CF = 50 - 20 = JOMPa

the radius o j lhe cin: le

R-

ex -

and

,.-x =

- ",

)x

"

10 MPa

50- 40 MPa

i.~

V(JO)l I (40f - 50 MPa

Fig. 7 .21

43 9

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440

Transformations 01 Stress and Strain

(b) Principal Planes and Principal Stresses. The principal stresses are IT"",-,

rT",in

= OA = OC

y

+

50 = 10M Pa

= OB = OC - BC = 20 - 50 = - 30 MPa

Ii

FX 40 tan 28" = CF ~ 30

40

8

= 20

Recallin g that the angle ACX represenls 2ti,. (Fig. 7.2Ib). we wnte

10-1

L

+ CA

c

c 0 20

F

A

I " -"---1 x

.!If

5() -

(/,1

FIg. 7.21b (repeated)

-

I

,, (~ Ib )

28" = 53.1 0 Since the rotation which brings ex into CA in Fig. 7.22bis counterclockwise. the rotation that brings 0.\' into the ax.is corresponding to it"," , in Fig. 7.22a is also counterclockwise.

au

(c) Maximum Shearing Stress. Since u rurther rotation or 90 0 counterclodwise brings CA into CD in Fig. 1.22h, a fnrther rotation of 45 ° counterclockwise will bring the axi s Oa into the axis Odcorresponding 10 the maximum shearing stress in Fig. 7.22a. We note from Fig. 7.22b that 7",>., = R ~ 50 M Pa and that the corresponding normal stress is u' = u •• , = 20 MPa. Since point D is located above the if axis in Fig. 7.22h. the shearing stresses exerted on the faces perpendicular 10 ad in Fig. 7.22a must be directed so that they will lend to rotate the element clockwise.

T( MPa))

2 11" = 53.1

U"".

= -

(1) )

Fig. 7.22

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Mohr's circle provides a convenient way of checking the resulL~ obtained earlier for stresses under a centric axial loading (Sec. 1.12) and under a torsionlliloading (Sec. 3.4). In the lirst case (Fig. 7.23a), we have ax = PIA,fT,. "" O. and T XI. = O. The corresponding poinl~ X and Y deline a circle of radius R = Pl2A that passes through the origin of

7.4. Mohr's Cln:le lor Plane Stress

o

(,I)

Fig.7.23

(iI )

Mohr's circle lor centric axial loading.

coordinates (Fig. 7.23b). Points D and E yield the orientation of the planes of maximum sheari ng stress (Fig. 7.23e), as well as the values of Tnt", and of the corresponding nonnal stresses a': T

T"

rna,

~

a'

~

R

- .2A !. . . .

(7 .1 B)

In the case of torsion (Fig. 7.24a), we have ax "" a J = 0 and Tell. Points X and Y. therefore, are located on the Taxis,

= T roo. =

(,, )

Fig.7.24

{d

(I»

Mohr's circle for lorsiooalloading.

and Mohr's ci rcle is a circle of radius R = Tell centered at the origi n (Fig. 7.24/). Points A and B dc/ine the principal planes (Fig. 7.24c) and the principal stresses: T, am.,. min = .::!: R = +- -J

(7.19)

441

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, I

t

SAMPLE PROBLEM 7.2

tijI Mi'J

For tile

-lu lC'!:'''''-, - , -

of plane stress sllOwn. determine Vl,Ilhe prin.ciJXll planes and the

~tf<,,~ses,

(b) the s tress t'Omponenls exerted on lhe. element o btained

by mlllling the gil-en element counterclockwise Ihrough 30°.

IS 111'.•

d ~IPa l)

SOLUTION

IJ." ="'O ~ll'.,

Construction IIf Mohr's C ircle. We note Ihm on a face perpendicular (() thex axis. the normal ,tress is tensile and the , hearing stress tends to rotme

~O+~'t---~'f--'-;;----'-t.~1 t" -

u"." 1. ~

st~te

principal

= \ Jpc

52

~ lPa

the element clockwi se; thus. we plol X at a point tOO unil.'l to Ihe right of the vertical axis and 48 un its above Ihe horizomul axis. In a si milar fa~hion. we cxammc lhe stress componcllIs on the upper fae;! and plot point Y(60 , - 48) . Joining poinu X and t' by a slraight line. we define tJle cCllIer C of Mohr's circle. nIl' abscissa of C. which rcpresems If "'~, and Ihe radiu s R. of [he c ircl~ can he nw~sllre/t rlill'ctly or cakLil.1rffi as follows :

U",," -

1'100. - iS I

DC -

!(o-, + <7 ,) - !( IOO + 60) - 80 MP.L

R - V(CF)l + (FXf

52 ~IP ,L

V:20)1 + (48)1

II. Principal Planes a nd Principal St l't'sses. We rotate thediarnekr XY clockwise through 2{Jr until it coincides with the diameter AB. We h~ ve

tall 20 = XI-' '" 48 = 2.4 p CF 20

2IJ" = 67.4 0

IJ,. = 33.7 v J <4

J

The r:rincipal Slresses are represented by the abscissas of points A and B:

+

+

(/" .., = OA '" OC CA = 80 52 = OB = OC - Be = 80 - 52

(f, .. "

"H..., =

+ 132 1\I P~L

fT ,.,", =

+ 28 MP,l <4

"""iIII

Since the rotmion that brings XY into AB is clockwise. the rotation that brings Ox into (he axis 01.1 l"Orre,ponding to iT "... is also clockwl~e: we ub(ain th ~ oti,. (,\ tPa) )

u,. · I, ';~·

." =

r .~ fI·

'" =

52.6'

_ hn° _ (;7

_ ""X,'

--.--+--el~ '-',*+,.~ "" ", :L ;-; .~

l

j(

C

~\

r ______ Y·

0- •.•

h.

. /u ,

= 1 ~ 4 \ 11'"

/~ ~ -" . =.IU;\II'J

-IEJP '' ~ :::...30· 442

phne~ .

lRO" - fi(]0 -

=

OL = OC

_ __

'1~ = 'i 2 h C

fi7 .4°

= OK = OC - KC = 80 - 52 <..~.s 52.6 0

+ CL

= 80 + 52l"()S 52.6°

Th · = KX' = 52 sill 52.6°

[T~.=\ 1( 6 ~ Jt'a /.'·

T

4> = (T ...

1-----"11

t

.. nl'lIi"n "wwn f"r Ih" pnncip:J.1

1; . Stress Com pon ents 011 [Iement Rotated 30" ~. Points X' and Y' on Mohr· s circle Ihm corres pond 10 the stress components on the rotnted clement i f (\IPa) are ob tained by rotating XY counterclockwise through 211 = 60°. We find

)....

OB

j '

iT, IT .

...

+

-18.4 "l l'a <4 =+ 1IJ.6 r-. IPa <4

-I 1J MPa <4

Since X' is located above (he horizontal axi~, the shearillg SIres., on the faee perpendicular to Ox' tends to rot~te the element clockwise.

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SAMPLE PROBLEM 7.3 A slate or plane stress consists of a tenslle stress (To = 8 ksi exened on veni-

"~I(BI ~ " Ioi-- '..

cal sllrfaces and of unk.nown shearing

stres~es_

Delemline (a ) the magnitude

of Ih.! shearing stress To for whi:h the largest nonmi stress is 10 hi. (b) the corresponding maximum shearing stress

SOLUTION C omlructi oll of Mohr 's

We

Cird~ .

as.~ume

that the s hearing s tres.les

acl in the senses shown. Thu.~. the shearing stress Tu on a face perpendicu lar to Ihe .r axis tends [() mille the element clockwise and we plot [he point X Olf coordinates 8 ksi and TQ above (he b.orizollial a:{is. Considering a horizontal face o f Ihe e lement. we observe th,,'

(T \' -

0 and that TUtends

\0

rolat e the d-

emenl counterclock ..... ise; thus. we plot point Y at a distance TO below 0 We note that Ihe abscissa of the center C of Mohr's circle is lI,"" = ~(u"

;""" 8

n

'" L-" l'

c

JOksi

,

+ u .) =

~( 8 + 0) = 4 ksi

The mditls If of the circle IS uetennined by obser ving that ttle maximum normal ~ress . lI", ... = J() ksi. is represented by ttle abscissa oi point A and wntmg

u. Shearing St ress r". cos 28~ =

CF

ex

= 4ksi +

R = 6ksi

Considering the right triangle CFX.

CF

4ksi

R

6 ksi

't~ = fX = R sin 20p =

20 p = 48.2'

Wi!-

find

l

(6 ksi ) sin 4S.ZO

b. Maximum Shearing Stress. ci~l e

If

The coordinates of )Xlim V of tl.lohr's

""pr.. ~"nl th .. maximum "-hearing "-lreW and Ihe

corr"~ pond;t\S

nom",1

Slress. T m....

28. = 90" - 26,! = 90 0

-

= R=6ksi 48.2" = 4l.g o "\

T ......

(J,t =

= 6I.:si ...

20.9°"\

The maximum shearing stress is exened on an element that is oriented::ls shown in Fig. (I. (The elem~nt upon which the principJI stresse.~ are exerted is also shown.)

,vIJIt . If our original a~sumpljon regarding the oon ~e of 'TQ was reversed. we would obtain the same circle and the same ans ....'Crs. but the oriel11ation of the elelllclIIs would be as shown in Fig. b .

443

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PROBLEMS

7.31 7 .32 7 .33

Solve Probs. 7.5 and 7.9, using Mohr's circle. Solve Probs. 7.6 and 7.10. llsing Mohr's circle. Solve Prob. 7.11. using Mohr's circle.

7.34 Solve Prob. 7.12. using Mohr's circle. 7.35 7 .36 7.37 7 .38 7.39

Solve Prob. 7.13, using Mohr's circle. Solve Prob. 7.14. using Mohr's circle. Solve Prob. 7.15, using Mohr's circle. Solve Prob. 7.16. using Mohr's circle. Solve Prob. 7.17. using Mohr's circle.

7 .40 Solve l>cob. 7.18, using Mohr's circle. 7 .41

Solve Prob. 7.19, using Mohr's circle.

7.42 Solve Prob. 7.20, using Mohr's circle. 7 .43

Solve Prob. 7.21. using Mohr's circle.

7.44 Solve Prob. 7.22. using Mohr's circle. 7.45

Solve Prob. 7.23, using Mohr's circle.

7.46 Solve Prob. 7.24. using Mohr's circle. 7 .47 Solve Prob. 7.25 , using Mohr's circle. 7.48 7.49 7.50 7.51 7.52

Solve Prob. 7.26. using Mohr's circle. Solve Prob. 7.27 . using Mohr's circle. Solve Prob. 7.28. using Mohr's circle. Solve Prob. 7.29. using Mohr's circle. Solve Prob. 7.30. using Mohr's circle.

7.53 Knowing that the bracket AB has a uniform thickness of i in .. determine (a) the principal pklnes and principal stresses at point H, (b) the maximum shearing stress at point H.

U.;5 in.

,.

,

II ·

..

I--

5

i".~ 1.25 in.

Fig. P7. S3

7.54

444

Solve P rob. 7.53. considering point K.

B

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7 .55 through 7 .sa DelellllUle Ihe plillCipal pl~lle s ami the prUlcipal stresses for the Hate of plane stress resu l1ing from the superposition of th.e two states of stress sho\\rt.

511

Problem.

~ l l'a

Fig. PT.SS

Fig. P7.5S

u"

u"

u"

+ Fig.

Fig. PT.se

PT. ~T

7.59 For the clement shown. determine the range of values OfT.". for which the maximum tensile stress is equal to or lesi than 60 MPa.

120MI'..

7.60 For the elel:1ent shown. determine the range of values of.,. Q' for which the maximum in-plane shearing stres. is equal to or less than 150 M l'a. 7 .61 Fm the ,!
fj

Fig. pr.S9 and pr.60

_ __

12h.l

Fig. P7.S1 and PT.S2

lo r

7 .62 For the Slate of siress shown. determine the range of \a1ues of which the nonn:l.l SlreSS
e

445

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446

TrD"'·....mallo"" 01 Slmss ""~ SImi"

7 .63

FOI the Slale of Slles~ ~ liowLi il is kllown Ihal the llorJllal ami

shearing stresses arc directed as shown and thm II, = 14 ksi. (T, = 9 ksi. and if..... = 5 bi. Determine ((II thc orientation Of the prim:ipal planes. (bl the principal slfess 0"1""'" (e) the maximum in·plane sheanng stress.

Fig. P7.63

u!--+---="'~=ol-''-----'I-'' T,,~

I f---",--I

1 - -", - Fig. P7.G4

"

7 .64 The Muhr'JS \:in:k shuwll (;um:spmuh IU LIlt: stalt: uf Slress given in Fig 7 ~f1 ~nrl b Nmin3 Ih~1 " A' = or + (rX') eM (lAp - 2A) ~nd Ihal r , 1 = (eX') Sill (21J" - 20i. dt!l"ivc the expressions for rf,' and r ,'1 given in Eqs (75) and (76). respectively. [Hill t: Use sin (A + 8) = sin A eo~ 8 + eos A si n Band oos (A + 8 ) - wsA co~ B - si n!. sin B.]

7 .65 (tI } Prove that the expres~ion if., II ,'! - ,;;,,, . where {T,', fYi ' and 1',(y are eOlllponems of the stress along the rectan gular axes .\' and i . is imlepen· derll of the orienl
7.5, GENERAL STATE OF STRESS In the preceding sections. we have assumed 11 .stale of plane stress with = T ; .r = T;:.y = D. and have considered onl y transformations of stress il~~l)\;iU lell Wllh ,I JU lat i\! 1J auullt Llle .<; il:\.b, We will HUW (;ulI~il.lel tin: ge neral ~tale of ~tre.,s rep resented in Fig. 7. la :Uld the transformation of stress associated with the rotat ion of axes , hol'Jn m Fig. 7.lb. How· ever. our ,uwlysis will be. limited to the. tle tenmnation of the IIQ/11W/ stress {Tn on !.I plane 01 arbitrary onentallon . Consider the te trahedron shown in Fig. 7.25. Threc of its faces Ilrc rar:lll~1 In th e coordinate planes. whi le il s fOllrrh face, ARC: i~ rcrre n· diclIlar to th e line QN. DenOling by 6.A the area of face ABC. illld by A.I • A" Ac the direc tion cosines of line QN, we find thut the area~ of the faces perpe ndicular to til e x, y, and z axes are, respective ly, ( :1A ))'~" ( ~ )A " and (M )Ac. If lhe state of stress at point Q is ddin ed by the stress components fT , . fT, .. {To. T '.!, . "T1'" and -;- v,o ' then the!orn'.\· ex· cned on the faces parallel to the coordinate. pl al1~s can be obtained b~ multipl y ing Ihe. llppropriate stress cO l11poncTll ~ by the area of each fac e ( Fig. 7.26). On the other hanll. the IUKes extn e d un face ABC w nsist of a normal force o f mag n itude Ir" Ll..A. directed along QN, and of :J. s hearing force of magnit ude T LiA perpendicular to QN but of otherw ise un known direction. Note that. since QBe. QCA. and QAB. reo s pet·t ivc.ly, face the neg:llive x, y, and ::; axes, the force s c.~erted on them must be .'l h ow n with negative ...cn sc.~ . fr ,

)o ;c-------- ,

Fig. 7.25

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7.5. Gene ral State of Stmss

)oc-- - - - - - - - ,

Fig. 7.26

We now express that the sum of the component'> along QN of all the forces acting on the tetmhedron is zero. Observing that the component along QN of a force parallel to the x axis is obtained by multi plying the magnitude of that force by the direction cosine Ap and that the components of forces parallel to the}' and z axes are obtained in 11 similar way. we write

'2.Fn = 0:

Un dA - (o-•. LiA A,)A..- - (T..... 6.A A,)A), - (T,; 6.A AAA: -(Tu M A,.)A, - (a , '" A,.)A,. - (T" M A,.)A, -(T" M A,)A, - (T". M A,)A, - (a, M A,)A, ~ 0

Dividing through by LiA and sol ving for (Tn

=

U n_

we have

fT ,..\.; + a" A; + (T, A~ + 2 'T.~,\A r + 27).,A)...\: + 2T:..
0 .20)

We note that the expression obillined fo r the nonnal stress u " is a qmulratic f o rm in A" A", and A,. It fo llows that we can select the coordinate axes in such a way that the righi-hand member of Eq. (7.20) reduces to the three lerms containing the squares of the direction cosincs.t Denoting Ihese axes by u, IJ, and c, the corresponding normal stresses by a a bo and 0",., and the direction cosines of QN with respect to Ihese axes by A ~, Ab • and A,.• we write Q'

(7 .21 )

The coordinate axes 1I. iJ. c are referred to as Ihe principal axe.f of Str ess. Since their orie ntatio n depends upon the slate of stress at Q. and thus upon the pos ition of Q, they have been rt! prcsented in Fi g. 7.'27 as attached to Q. The corresponding coordinate planes are known as the principal pltm e.\"(~r.l"fres.l", and the correspondingn ormal stresses an. O"b. and a e as the principal .ttres.W~.I" at Q. * tIn Sec. 9.11i of r p. B~cr and E. R. Johnston. VeCfor M ec/UIII;('A for En/:;necr•• 7th ed .• McGraw- Hi ll Boo).: Company. lOOt a similar quadratic fonn is round to represent tile moment of in ~nia 01 a ngid bod} With respLxt to an EU"bitrary axi •. tl i, .;;hown in Sec. 9 .17 thal mi., fbrm is associated with a quadric SlIlfilU. and that reducinJ; the quadratic fonn to terms containing only the squares of the Jirection co.\ ines i, equivnlent to octemlining the principal axe. of that ~urfacc. f Fo,. a ,li...·n ....'j"" ~f II,,· dl'\('nnin"liu" of Ihe I'rin"il'"l l'bne~ of .1",", lUlll ,,[ th .. 1";"";_ 1'.,I .tn·.<;<,·,. """ S. P limosh,'"ko '(lui J. N. ('~ ~ ~li"r. T"mry of EIIl,,·ricil,. 3tl ''1l ~l<-c.rJ,,"_ lIill Book C""ll"" " '. 1970, .' '''.".77.

Fig. 7.27

447

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448

Transformations of Stress and Strain

7.6. APPLICATION OF MOHR'S CIRCLE TO THE THREEDIMENSIONAL ANALYSIS OF STRESS

If the element show n in Fig. 7.27 is rotated about one of the principal axes at Q, say the c axjs (Fig. 7.28), the corresponding transformation of stress can be analyzed by means of Mohr's circle as if it were a transformation of plane stress. indeed, the shearing stresses exerted on the faces perpendicular to the c axis remain equal 10 zero, and the normal stress 17,.. is perpendicu lar 10 the plane ab in which the transformation lakes place and. thus, does not affect this transformation. We therefore usc the circle of diameter AB 10 determine the normal and shearing stresses exerted on the face s of the element as it is rotated about the c axis (Fig. 7.29). Similarly, ci rclc.~ of diamctcr Be and CA can be used to determine the stresses on the element as it is rotated about the a and IJ axes, respectively. While our analysis will be limited to rotations about

Fig. 7.28

,)

c

8

o

Fig. 7.29

the principal axes, it could be shown that any other transformation of axes would I(!ad to stresses represented in Fig. 7.29 by a point locatcd within the shaded area. Thus, the radill~ of the largest of the three circles yields the maximum va lue of the shearing stress at point Q. Noting that the diameter of that circle is equal to the dilference between 17 "'ax and 17 min' we write (7.22)

D

,,

/ ~----

z=o

E

Fig. 7.30

......

,"

I

, A

'~I'

where 17rna, and (J" min represent the algebraic values of the maximum and minimum stresses at point Q. Let us now return tQ the particular case of pilUle stress. which was discussed in Secs. 7.2 through 7.4. We recall that. if the.r and)' axes arc se lected in the plane of stress. we have (J~ = T;x = T :'" = O. Thi s means that the z axis. i.e., the axis perpendiculu to the plane of stress, is one of the three principal axes of stress. In a Mohr-circle diagram. this axis corresponds to the origi n 0, whc:re 17 = T "" O. We alS() recall that the other two principal axes correspond to points.4. and B where Mohr's circle for the Joy plane intersects the 17 axis. If A and B arc located on opposite sides of the otigin 0 (Fig. 7.30). the corresponding principal stresses represent

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the ma ximum at lU milli llllllll lIo lmal stresses at fJOint Q, atul the maxi-

mum she;mng ~tre~s is e£JuHllo the maximum "in-pilUle" sheanng ~Iress_ A~ noted in Sec. 7.}, the planes of maximum shearing ;tre,s corre~po rld to points D and E of Mohr's circle and are at 45 ~ to the principal planes com spondmg to points A and B. They are, therefore, the shaded diagonal plancs shown in Figs. 7.31a and b,

"\

~

--

",

\\~

.~. '.- .-

7.6. hppllc"tlo.~ 01 Mohr'," Circle to the Thll!e-Dmemlonal Analysis of Stress

~,

'0

.

ff"

I, )

,)

Fig. 7.31

D'

" [f, on the other hand, A and B are on Ihe same side of 0, that is, if

u " and

have the same sign, then the circle defining if""" , U m;" , and T mv; is /lot the circle corresponding to a transfonnmion o f stress within the xy plane. If U~ rel="nofollow"> (J"b > U, as assumed in Fig. 7.32, we have U m3.l - (T il ' (T Illin - 0, and 1" onx is equal to the radius of the circle detinen hy rninl~ () and A , th:lt is, T,_ = ~ I"m .... - We also n nt,~ that the nomlals Qd' and Qe ' to the planes of maximum shearing stre;s are obtained by rotating the axis Qa through 45" within the :;(1 plane. Thus, the planes of maximum shearing stress are the shaded diagonal planes shown

Uh

111 Fi g~. 7.33~,

,

"\

\ ",

\

",

",

I' ) Fig. 7.33

/

(IJ)

T" .., ""

z=o t ... --~

k_

" j' "

'_

E'

<>;,,,, +u - · _--

- - - fTo,,, - a;, Fia.7.32

and h.

--.---:-. n'

_""

~

1...

449

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EXAMPLE 7.03 For the stale of plane stress s hown in Fig. 7.34. detennine luJ the three principal planes and principal stresses. (b) the maximum s hearing stress.

'.5 l>'

I"

I

Since the faces of the element thaL ilre perpendicular to the

z 1I:;:i5 arc free of Stress. these faces define one of the principal planes. and the corresponding principal stress is u , = O. The other two principal planes are defined by points A and B on Mohr's circle. The angle Bp through which the element should be rotated about the:; axis to bring its faces to coincide with the..~e planes (Fig. 7.36) is h.alf the angle ACX. We have

FX

~:;-'

3

tan20p= CF """ 1.25 2f1r = 67.4 0

J

ti" = 33Y

J

Fig. 7.34

(a) Principal Planes and Principal Stresses. We construe( Mohr's circle for the tmnsfOmlalion of stress in the xy plane (Fig. 7.35). Point X is ploued 6 units to the right of the'T axis and 3 units above the (T axi s (si nce the correspolIding shearing stress tends to rotate the element clockwise). (b) Maximum Shearing Stress. We now draw the

6ks'- -1

",@),+. 3..5 kJ,;

drdes ur diameter DB um.l OA. which currespond res pectively

to rotations of the element about the a and b axes (Fig. 7.37). We note that the maximum shearing stress is equal to the ra dius of the circle of diameler ~A. We thus have 'T " .. ,

=

!u" =

~(8.00 ksi ) = 4.00 ksi

,)

" , ,,

, l Fig. 7.35

,,

Point Y is plotted 3.5 units to the right of the 'T axis and 3 units below the if axis. Drawing the line XY. we obtain the center C of Mohr's circle for the X)' plane: its abscissa is if..

if .... ""

+

(1",

6 + 15

--2-- = ~ =4.75ksi

Since the sidesoftherighttrhmgle CFXare CF = 6 - 4.75 = 1.25 ksi and FX = 3 hi. th.e radius of the circle is

R = CX = V(1.25f

+ (3)!

3. 25 ksi

The principal s tresses in the plane of stress are (1"~ = ITI>

450

OA = OC

+

CA = 4.75

+ 3.25

= 8.00 ks!

= 08 = OC - BC = 4.75 - 3.25 = 1.50ksi

Fig. 7.37

Since points D ' and £ ' . which de fine the planes of maxim um shearing stress, are located at the ends of the ve rtical diameter of the circle corresponding to a rotation ahoul the h axis. the faces of the element of Fig. 7.36 can be brought to COIUcide Wilh the planes of ma.\imum shearing stress through a fUtalion of 45 0 about the b axi.•.

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7 .7. YIELD CRITERIA FOR DUCTILE MATERIALS UNDER PLANE STRESS

7.7. 'flald Criteria for OLtCtiic MlIlorials under Pllne Stress

Structuml eh:me lll~ aud lHilchim' ClllL q.JUIlCl!tS IiI:Uk: of a uuclile lLIlIICrial are usually de.signcd so tbm the m:Ut'rial wi ll not yield under thc expected loading conditions. When the element orcomponenl is under uniax.ial stress (Fig. 7.38), the value of the normal stress IT.. Lbat wi ll cause the material to yieltl can be obtained readily frolll a tensile test conducted

Fig. 7.38

on a specimen of the $ame material , since the test specimen lmd the structural clement or machine component are in the same stute of ~ trcss . Thus. regardlc ~s of the actual mechanism that causes the material to yield. we can state that the element or component will be safe as long as a" < a y. where a y b; the yield strength o f the test specimen. On the other hand, when a structural elemeJll or machine compone nt is in;} st:.lIe ()f plane stress (Fig. 7.39a). il lS I"ounu convenie nt to use one of the mcthods developed earlier to determine the principal messes all and a~ at any given poim (Fig. 7.39b) . The mrtterial Clin then be regarded as being in a sute of biaxial stress at that point. Since dli s state i5 di.lTeft:lIt frum Lht staLe of uuiaxial stress found ill a spt::dmen subjected to a t L'n ~i l e test. it is c!e::trl y not possible to predict directly from such a test whether or not the structural cl ement or machine component under investi gation will fail. Some criterion regarding the actual mechani sm o f fa il ure of the material t11U;t first he establi shed, which will make it possible to compare the cUee.!s of both states o f stress 0 11 the t11aterial. The purpose of thi s section is to presenttbe two yield criteria most frequently used for ductile nlJlerials.

Maximum.Shearing·Stress C riterion. This criteri on is based on the obse rvation that yield in ductile materials is caused by Sli ppage o f the material along oblique surfaces and is due primarily to shearing stresses (cf. Sec. 2.3 ). According (() this criterion. a given sUlletural component is ~aJe as long as the maximum value Tn"" OJ" the shc;}ring stress in that component remains smaller than the correspondi ng value of the shearing stress ill a tensile-test specimen of the same material as the specimen starts to yield. Rc>;ul.lillg rl om Sec. I .11 that the Itlaxilt tUlt] value of the slteat iug S Lre~s under;} CClltrie axi:!1 load is equal to hall' lhe V:U lIC of the eorre~ pondi ng non nal. axial stress, we conclude that the maximum shearing ~trcss in a ten ~ ilc -leS t specime n is as the specimen stuns to yield. On the o ther hand, we saw in Sec. 7.6 Ihat. fo r plane s tre.~s. the maximum value T mu of thc shcaring.sLTCss i, eGual to : IfTtl> •• 1if the princir:"t l strMScs :"Ire. ei ther horh posiri ve ()f Mth n,~g:"ttive.. :"t nrl to ~ lfT <;La, - a"lInl if the maximum stress is positi ve and the minimum stress

1fT,.

."

--D-

.~

I. (,1

, 'I

. II" FIg. 7.39

-i/i-,. ., (/,1

451

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452

Transformations 01 St""ess and Strain

negative. TIlU S, if the principal stresses (T" and thi: maximum-shearing-stress criterion give.'>

u" have the same sign. (7 .23)

[f the principal stresses (Td and (Tb have opposite signs, the maximumshearing-stress criterion yields

..,;'1---'1 -_ == "-,fC--O~---,L.C"C,,--- ".

'-------t::- O"r Fig. 7.40

(7 .24) The relations obtained have been represented graphically in Fig. 7.40. Any given state of stress will be repre~ented in that figure by a point of coordinates (T" and (T h. where U and r.r" are the two principal stresses. Jf this poi nt fall s within the area shown in the figure , the struc tural component is safe. If it fall s outside thi s area, the component wi ll fail as a result or yield in the material. The hexagon associated with the initiation of yie ld in the material is known as Trcsc{/ ',~ lU!.wgoll after the French engineer Henri Edouard Tresca (1814- 1885). Q

MaximUJtI " Disl o rlioll - ~ II~ rgy Cr ite rion. This cri terion is based on the determination of the distortion energy in a given material , i.e .. of the energy associated with changes in shape in that material (as opposed 10 the energy associated with changes in vol ume in the same maICrial). According 10 this criterion, also known as the V{)/I Mises criterioll . after the German-America.n applied mathematician Richard von Mi sc..~ (1 883- 1953). a given structural component is safe as long as the maximum va.lue or the distortion e nergy per unit vol ume in that mate-

ri :l l

re rnain .~ ~m nller

than rhe (li.~r{) rt inn ene rgy

p,~r

unit vnlilme rCfJllirL"fl

tt) cause yield in a tensile-test specimen o r the same material. As you wilJ see in Sec. 11.6, the distortion energy per unit volume in an isotropic material under plane stress is (7 .25) where (T~ and U b are the princi pal stresses ami G the modulus of ri gidity. In the particular case of a te llsi le-test specimen that is starting 10 yield, we have 0'" = O'y, O' ~ = 0, and (uJ)Y = ui/6G . Thus, the maximum-distortion-energy <:riterion indicates that the structural component is sa fe as long as lid < (ll'lh. or (7 .26)

i.e., as long as the poi nt of coordinates O"a and U~ falls within the area shown in Fi g. 7.41. This area is bounded by the e llipse of equati on + 'TY V-~_ A

c

----"C'f_---,l:/,~'")--f_,___-"" +'Tr D

Ag. 7.41

(7 .27) which illiersects the coordinate axes at U a = ± O"y and O"b = :t Uy. We can verify that the major axis of the ellipse bisects the fi rst and third quadrants and extends rrom A (0-" = O"b = Uy) to B (u a = (Tb = - Uy ), while its minor axis extends from C (u a = - U b == - O.577u y) 10 D «(T" = - Ub = O.577O"y). The maximum-shearing-slfess criterion and the maximum-distortionenergy criterion are compared in Fig. 7.42. We note that the ellipse passes through the vertices of the hexagon. Thus, for the states of stress re presented by these six points, the two criteria give the same resu lts.

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For any other state of stress, the maximum-shearing-stress criterion is more conservative than the maximum-diswrtion-energy criterion, since the hexagon is located within the ell ipse. A state of stress o f particular interest is that associated with yield in a lorsion les\. We recall from Fig. 7.24 of Sec. 7.4 thai, for torsion, rTmin = - U m "'; thus. Ihe corresponding points in Fi g. 7.42 are located on Ihe bisector of the second and fourth quadrants. It follows that yield occurs in a turslon test when u " "" - U" = ± 0.5uy according to the maximum-sheanng-stress crilerion, and when fT~ "" - U" = ±: 0.577u y according [0 Ihe maximum-distortion-energy criterion. But recalling again Fi g. 7.24, we note Ihat (T~ and (Tb must be equal in magnitude to T m " " that is, tt) the value obtai ned from a torsion lest for Ihe yield strength T y of the malerial. Si nce Ihe values of the yield strength O"y in tension and of the yield strenglh Ty in shear are given for various ductile materials in Appendix B, we can compute Ihe ralio Ty/ Uy fo r Ihese malerials and veri fy that the values oblained range from 0.55 10 0.60. Thus, the maximum-distortion-energy criterion appears somewhat more accurate than the maximum-shearing-stress criterion as ['Ir as predicting yie ld in torsion is concerned.

7.8. Fracture Crlterla lor Brlttle Maler1als under Plane Stress

",

Fig. 7.42

' 7.8 . FRACTURE CRITERIA FOR BRITTLE MATERIALS UNDER PLANE STRESS

As we saw in Chap. 2. brillie materials are characterized by the fact that, when subjected to a tensile test they fail suddenly through ruplure ~ Uf rral'lurc ~ wilhuul ,Ill Y priUI yit:luiu b. Whcll a ~Irul:lural cklllcnl ur machine compollent made of a brittle material is under uniaxial tensile stress. the value of the normal strc.~s that causes it to hlil is equal 10 the ultimate strength a u of the material as detennined from a tensile test , ;;ince both the tensile-te;;t specimen and the element or component under inves· tigmion are in the ;;ame sUIte of stress. However. when a stnlclUraJ element or machine component is in a state of plane stre,~s. it is found convenient to tirst determine the principal stresses V a :U1d u b at any given point. and to use one of the criteria indicllled in this section 10 predici whether or no\" the structural element or machine component will faii. Maximum·~ormal.Stress Crit er ion. According to this criterion, a given structural component fails when Ihe maximum normal stress in that component reaches the ultimate stre ngth (T u obtained from the tensi le lest of a specimen of the same materiaL Thus, Ihe structural component wi ll be sufc as long as lite absolute values of the princi pal stresses (T" and (T b are bolh less than (T u:

lu,,1<

o" u

(7.28)

llie maximum-normal-stress criterion can be expressed gmphically as shown in Fig. 7.43. If Ihe point obtained by plotting Ihe values (T" and (T" of the principal stresses fall s within the square area shown in the ligure, the structural component is safe. If it falls outside that area, the component will fail. The maximum-normal-stress criterion, al;;o known as Cou/olllb's ai/erilm. after the French physici.~t Charles Augustin de Cou lomb ( 1736- 1806). sulTers from an important shortcoming, since it is bllsed

~=n"c,.+--+---I""c"--~,

Fig. 7.43

453

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454

Transformations of St""ess and Strain

on the assumption that the ultimate strength of the material is lhe same in tension and in compression. As we [loled in Sec . 2.3 , this is seldom the case, because of the presence o f fla ws in the material, such as microscopic eracks or cavities, which tend to weaken the material in tension. whi le 1I0t appreciably affecting its resistance to compressive fa ilure. Besides, this criterion makes no allowance for effects other than those o f the normal stresses on the fail ure mechanis m of the materinl. t

". "( '(;

."

•

(II )

"I, "
I



L-------i·'e (/11

Fig. 7.45

",

M ohr 's C r iterion, Thi s criterion. suggested by the German engineer Otto Mohr. can be used to predict the effect of a given state of plane stress on a brinlc material, when results of vari oliS types o f tes ts are available for that material. Let us li rst a.~sume that a te nsi le test and a compressive test have been conducted on a given material. and thm the values a VT and (Tue of the. uitimnte sLrength in tension and in compression have been determi ned for thm material . The state of stress correspo nding to lhe fUpture of the tensile-test specimen can be represented on a Mohr-ci rcle di agram by the circle illlersecting the horizontal axis at 0 and U UT (Fig . 7.45(/). Simi larly, the state of stress corresponding to the failure of the compressive-test s pecime n can be represented by the circle inlersecting the horizontal axis at 0 and (T ue. Clearly, a state ur stress represenled by a circle entire ly contained in ei ther of these circles wi ll be safe. Thus. if bmh princi pal stresses are positive, the state of stress is safe as long as (T" < a llT and a~ < u ur; if both principal stresses are negative, the state o f stress is safe as long as luQ I < [u uel and IUbl < lauel. Plottin g the poillt nf conrd in:Hc~ (r~ :lnd (r b (F ig. 7.45/J), we ve ri fy thnt th e state of stfess is safe as long as thaI point falls within one o f the square areas shown in Lhat li gu fe. In order to analyze the cases when (T" and U b have opposi te signs, we n()w assume that a torsion test has been conducted on the material and that its ultimate strength in shear, 'fu . has been detemlined. Drawing the c.ircle centered at 0 representing the state of stress corresponding to the failure of the torsion-test specimen (Fig. 7.46(/), we observe that any state of stress represented by a circle entirely contained in Ihm circle is also safe. Mohr's criteri on is a logical extens ion of this ubser-

t Another failure criterion known as the minimum' Morma/,sImi" criter/"". or SaintVcnam ·s criterion. WllS widely used during the nineteenth century. Aceordin£ (0 thi s cri terion. a gil·en ,tructural component is sale JS lon~ as (he ma.\ imum I'alue of the nonnal strain in that 'omponent remains ~ma llcr (han tIm v.... tue ~ I of (he strain at which a tensilc·te~t sped· men of the same material will fail. Bm. as willb<' shown in See. 7. !2. the str.J.in is maximum aLong one of (he principal axes of str~ss . if the dcionnation is clastic and Ihe matcrial homogeneous and i.lotropic. 1l1U~. denoting by ~. and ~. the values of the nomml ~(rain along (he princip;il axe!; in the plane of stre.<s. we write (1. 211 )

-
FIg. 7.44

Making usc of the genCfalized H ookt-~ law tSec. 2..t2). we could express these. relmion!; in tcmlS of the principal ,tre.sc,1 Q", and fT, and the ~ltill1ate st rength (~, of the; material. We would find that. according to (he maximum· normal-strain criterion. the wuctural comvoncm is safr as long a~ the point obtained by plott ing IT. J,nd IT , falls WIthin the area shown in fig. 7.44 where I' is Poi SSQn·s rntio for the giI"Cn matl-nal.

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7.8. Fracture Criteria lor BrIttle Materials under Plane Stress

u'r f-------,D + --tUc,O u- ,u,

L ------t;; ff!'C ('I

(hi

Fig. 7.46 vation: According to Mohr' s criterion. a state of stress is safe if it is represented by a circle located entirely within the area bounded by the envelope of the circles corresponding to the available data. The remaining portions of the princi pal-stress diagram can now be obtained by drawing various circles tangent to this envelope, determining the corresponding values of U I! and (Tb, and plouing the poinL~ of coordinates (Ta and U b (Fig. 7.46h) . More accurate diagrams can be drawn when additional test results, corresponding to various states of stress, are available. If. on the o ther hand, the only available data consists of the ultimate strengths U IJT and (T uc. the envelope in Fig. 7.46(1 is replaced by the tangel1ls AB and A ' B' to the circles corresponding respectivcly to failure in tension and failure in compression (Fig. 7 .47a). From the similar triangles drawn in that fi gure, we note that the abscissa of the ce nter C of a circle tangent to AB and A' B' is linearly related to its radius R. Since u" = OC + R and (T b = OC - R. it fo llows that u " ami (T b are also linearly related. Thus. the .~ haded arell corresponding to this simplified Mohr 's criterion is bounded by straight lines ill the second and fo urth quadrants (Fi g. 7.47b). NOle that in order 10 determine whether a structural component will be safe under a given loadi ng, the state of stress shou ld be calcu lated at all critical points of the component, i.e., at all point s where stress concentrations are likely to occur. T his clln be done in a number of cases by using the stress-concelllration factors given in Figs. 2.64. 3.32,4.3 1. and 4.32. There are many instances. however. when the theory o f elasticity must be used to detennine the state of stress at a critical point. Special care should be taken whcn macroscopic cracks have been detccted in a structural componcnt. While it can be assumed that the test specimen used to determine the ultimate tensile strength of the material contained lhe same type of flaw s (i.e .. micmscopic cracks or cavities) as the structurdl component under investigation, the specimen was certainly free of any detectable macroscopic cracks. When a crack is detccted in a structural compone nt, it is necessary to determine whe ther that crack will tend to propagate under the expected loadi ng condition and cause the component to fail , or whether it will remain .stable. This requires an analysis involving the energy associated with the growth of the crack. Such an analysis is beyond the scope of thi s text and should be cmTied out by the methods of fracture mechanics.

~------("rc

(Ii )

Fig. 7.47

455

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'I

SAMPLE PROBLEM 7.4

~\II"

The Slate of plane stress shO\\l n occurs al a critical point of a steel machine component. As a res ult of seve ral tensile test~. it has been found that the tl'll_ sile Yield stre llg th is if y = 250 M Pa for the grade of >leel used. Determine the factor of ;;afel), with respect to yield, usi ng la) Ihe maximum-shearing-stress criterion. and tb) the ma;\imum-dis tort ion-e nergy criterion.

SOLUTION lUohr"s Circle. We construel Mohr's circle for the given state of stress 40

:md lind

~1I'.,


\'

--,I

1

I, ,

,

25 MPu

!r:

o

T ,"

'. F

A

~

I

J25~lPa

I 'R

OC - ~( '" + if ,) - 1(80 - 40) - 20M Pa

= R = V{ CFY-

,r

U

- -0.-1

+ (25)" -

'I/,: 6Oj'

u

= OC

+ CA

= 20

+ 65 = + 85 MPa

iI .

l\Iaxinmm·Shl'aring.Slnss Criterion. Since for [h.e grade of sleel

used the tensile !.i:.engLh is u r - 250 MPa. the cOHespondi llg s hearing ~ U't:ss

at yield is 1y

= ~ u y = HZ5() MPa) = 115 ~I Pa or

1-'.5.= -

FlIr1'" = 65 MPa:

125 Mila 65 MPa

= ---

1 ",

h. MliximuDI · Di1Ilortion· E nc rgy Cril t rion. safet}' into Eq. (7.26), we wrile

Fur

iT "

= +85 M Pa. if " = - 45 M Pa. and

(85)' - (85)(- 45 ) IT, =

65 MPa

(f b = OC: - BC= 20 - b5 = - 45M Pa

X

"""" ./ I

+ (FX f! -

Princip:lI St rcsSl'S

250.\fPa

lI y

FS, '" 1.92
= 250 Mi'a. we have

+ (45)' ~

(250Y I-.S.,

114.3 = 250 F.S.

r ..'). -

2. 19 ....

Comment. For a duc tile matet;al with O" y = 250 MPJ. we hJve dr
OT

(a) F.S. = OH = 1.92

456

OM

(b)F.S.= OH = 2. 19

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PROBLEMS

7 .66 For the state of plane stress shown. detennine the maximum shearing siress when (a) i f , = 0 and (1, = 10 ksi. (b) u, = 18 ksi and If,. = 8 ksi. (Hin l: Consider bolh in-plane and out.of-plane shearing stresses.)

7 k, i

Fig. P7.G6 and P7.G7

7.67 For the state of plane stress shown. determine the maximum shearing siress when (a) IT, = 5 ksi and fT ,. = [5 ksi. (b ) fT , = 12 ksi and IT, = 2 ksi. (Hillt: Consider both in.plane and out-of-plane shearin g stresses. ) 7.68 For the state of stress shown. determine the maximum shearing stress when (a) IT, = 40 MPa. (b) IT,. = 120 MPa. (Hill t: Consider bolh inplane and out-of-plane shearing stresses.) 7.69 For the state of stress shown. determine the maximum shearing stress when (a) ITf = 20 MPa. (b) IT,. = 140 MPa. (Hill t: Consider bolh inplane and oUI-of-plane shearing stresses.)

t-!l I MP"

Fig. P7.68 and P7.69

7.70 and 7.71 For Ihe state of stress shown. determine Ihe maximum shearing stress when (tI) (T, = +24 MPa. (b) fT, = - 24 M Pa. (L") i f, = O.

'I

12 Ml'u

Fig. P7.70

Fig. P7.71

457

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458

Tmn,·ormo.tion. 01 Sire",

an~

SWln

7 .72 a nd 7.73 FOJ tlle state of stless shown. Ilctermillc the ma.\imum shearing stress when (a) 1"..,. = 17.5 ksi. (b) T" = 8 ksi. (I") T" = O.

12 ksi

3 lSI

Fig. P7.72

Fig. pr.73 7 .74 For the state of plane stress shown. determine the value of which the llla.\imum shcaling stress h (u) GO M Pa. (L) 7ll MPa.

T"

for

T~I

for

!

, I

IU ~lI'LI

l~

I UI..u l P~

,~

Fig. PT.75

Fig. P7.74

7 .75 For thi'; ~t~l~ of phn... s tre~~ shown. flNi';Tl'Iline the v"llIe of which the maximum shearing s tre ~s i~ (a) 10 ksi. (b) 8.25 ksi.

7 .76 r or the ~tme o f Mrc..'~ ,hown. cctcrminc two va lue; of
IU k.,

7U ~I Pu

.-...............1"

Fig. PH6

FiiJ. p r.77 7. 77 For the state of stress shown. determine two va lues of II, for \\ilich the Lll uximuLIl shem;ille: ",'r\":>5 L~ 75 t'.{Pa.

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7 .78 For the state of stress shown. determine the mnge of values of for which the maximum shearing stress is equal 10 or less than 90 MPa.

Problems

T"

"I

, I

,~

Fig. P7.7B

Fig. P7.79

7.79 For thl! state of stress shown. determine two values of O"y for which the maximum shearing stress is 64 MPa.

"7.80 For the stale of stress of Prob. 7.69. determine (a) the value of u , for which the maximum shearing stress is as small as possible. (b) the corresponding value of the shearing stress. 7.81 The s mte of plane s tress shown occurs in II machine componenl made of a steel with (Ty = 325 MPa. Using the maximum-distortion-energy criterion. determine whe ther yield will occur when (11) (To = 200 MP:l. (b) (To = 240 MPa. (c ) Uo = 280 MPa. If yield does not occur. detemline the corresponding f:lctor of s:lfety. 7 .82

Fig. P7.8t

Solve Prob. 7.81. using the maximum-shearing-stress criterion.

7.83 The s t
Fig. P7.83

7 .84

Solve Prob. 7.83. using the maximum-shearing-stress criterion.

7 .85 The J .75-in.-diameter shaft AB is made of a grade of steel for which the yield strength is U y = 36 ksi. Using the maximum-shearing-stress criterion. determine the magnitude of the force P for which yield occurs when T = 15 kip· in. 7 .86

Solve Prob. 7.85, using the maximum-distortion-energy criterion.

Fig. P7.85

MP...

459

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460

Transformations 01 Stress and Strain

7.8 7 The 36-mm-diameter shaft is made of a grade of steel with a 250MPa tensile yield stress. Using the maximum-shearing-stress criterion. determine the magnitude of the torque T for which yield occurs when P = 200 kN.

Fig. P7.87

7.88

Solve Prob. 7.87. using the mnximum-distortion-energy criterion.

7 .89 and 7 .90

1

15

~;

- 10'\:', ;

The state of plane stress shown is expected to occur in

all aluminum casting. Knowi ng that for the aluminum alloy used U UT = JO ksi and u (!C = 30 ksi and using Mohr's criterion. determine whether rupture of the

comIXlllenl will occur.

1 Fig. P7.89

Fig. P7.90

7 .91 a nd 7.92 The state of plane stress shown is expected in an aluminum cast ing. Knowing that for the alumitlUm alloy used /fur = 80 MPa and u uc = 200 M Pa and using Mohr's criterion, detenlline whether rupture of the cnsting will occur.

rW1MI'a ___

_

i5" t P~

fJO.\ tb

- ID l72AA

-1Dl~

Fig. P7.91

Fig. P7.92

T

Fig. P7. 93

7.93 The state of plane stress shown will occur at a critical point in an aluminum casting that is made of an alloy for which {fUT = 10 ksi and (fvc = 25 ksi. Using Mohr's criterion. detenlli ne the shearing stress TO for which failure should be expected.

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7 .94 The state of plane stress shown will occur at a critical point in a pipe made of an aluminum alloy for which {fur = 75 Ml'a and {rue = ISO MPa. Using Moh(s criterion. determine the shearing stress To for which failure should be expected.

-1 [ ]j-=I\' '"

Fig. P7.94

7 .95 The cast-aluminum rod shown is m,lde of an alloy for which 70 MPa and U ue = 175 MPa. Knowing that the magnitude Tofthe applied torques is slowly increased and using Mohr's criterion. determine the shearing stress 1"0 that should be expected at rupture.

U rn =

Fig. P7.95

7 .96 The cast-aluminum rod shown is made of an alloy for whIch = 60 Mr'a and u [''C = 120 MPa. USing Mohr's criterion. determine the magnitude of the torque T for which failure shou ld be expected.

U IfT

32

Ill '" ~

I

I Fig. P7.96

7.97 A machine component is made of a grade of cast iron for whkh U Iff

= 8 ksi and (rue = 20 ksi. For each of the states of stress shown, and using

Mohr's crilerion, determine the normal stres s component should be expected.

(TO

-'"

Fig. P7.97

ibl

:" ¢

at which rupture of the

--u"

(d

Probtoms

461

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462

Transformations of Stress and Strain

7.9. STRESSES IN THIN-WALLED PRESSURE VESSELS

Fig. 7.48

Thin-walled pressure vessels provide an important application of the analysis of plane stress . Since their walls offer little resistance to bending, it can be assumed that the internal forces exerted on a given portion of wall arc tangent to the surface of the vessel (Fig. 7.48). The resu lting stresses on an element of wall will thus be contained in a plane tangent to the surface of the vessel. Our analysis of stresses in thin-walled pressure vessels will be limited to the two types of vessels most frequently encountered: cy lindri cal pressure vessels and spherical pressure vessels (Figs. 7.49 and 7.50).

Fig. 7.49

Fig. 7.51

Fig. 7.52

Fig. 7.50

Consider a cylindrical vessel of inner radius r and wall thickness 1 containing a fluid under pressure (Fig. 7.5 1). We propose to de tennine the stresses exerted on a small clement of wall with sides respectively parallel and perpendicular to the axis of the cylinder. Because of the axisymmetry of the vessel and its contents. it is dear that no shearing st ress is exerted on the element. TIle normal stresses a) and a 1 shown in Fig. 7.5 1 arc therefore principal stresses . The stress a) is known as the hoop sIren. because it is the type of stress found ill hoops used to hold together the various slats of a wooden barrel. and the stress is called the longitlldillal .ftreS!i. LI1 order to determine the hoop stress aj. we detach a portion of the vessel and its contents bounded by the .\)' plane and by two planes parallel to the y:: plane at a distance n.x from each other (Fig. 7.52). TIle forces parallel to the.: axis acting on the free body defined in this fashion consist of the eleme ntary internal forces IT) dA on the wall sect i on.~ . and of the elementary pressure forces p dA exerted on the portion of fluid included in the free body. Note that p denotes the gage preuure of the fluid, i.e., the excess of the inside pressure over the outside atmospheric pressure. TIle resultant of the internal forces a) dA is equal to the product of a j and of the cross-sectional area 21 l1 x of the wall, while the resu ltant of the pressure forces p dA is equal to the product of p and of the area 2r n.x. Writing the equilibrium equation "'iF:;: = O. we have

a,

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7.9. Stresses In ThIn-Wailed Pressure Vessels

and, solving for the hoop stress

f7j,

(7.30)

To determine the longitudinal stress (7 1. we now pass a secti on perpendicu lar to the x axis and consider the fre e body consisting o f the portion of the vesse l and its contents located to the left of the section

I,,/A

Fig. 7.53

(Fig. 7.53). The fo rces acting on thi s free body are the elementary internal forces (T2 dA on the wall section anu the elementary pressure forces p dA exened on the portion of tlu id included in the free body. Nn rin g rhnr rhe are a nf rh(~ nu irl sect in n is 7Trl and thnr rhe area of th e wall section can be oblained by mullipl ying the circ umference 2 7fT of the cy linder by its wall thickness t , we write the equilibrium equation:t

"E F,

= 0:

amI. solving for the longitudinal stress IT..,

-

(7! ,

=P' 21

We note from Eqs. (7.30) and (7.31) Ihat the hoop stress large a5i the lo ngitudi nal stress (/ 2:

(7 .3 1) (7 r

is twice

llS

(7.321

I Using (he mean f:wius of (he walt .....'Cr ion. r~ = r + 11. in computing (he re,ultam of (he torres on tha( !;Ce(iO]. we wool(] ob(ain a more aceura(e value of (he longitudinal me;;s. narrn'ly.

pr

!

fT , = - - - 1

-

21

(7.3 I ')

1+ 1,

However. for a (hin-nalled pres~urc vc;;se l. lhr (crill 1/2r i.1 sufficiemly SlllJlt 10 allow (he u,e of Eq. i 7.31) for enJ;in~ring design and analy~i~" If a pres.lure \"c;;,.. l is llot (hin-waliL'(\ (i.e .. if 111 ,. is not smatJ ). lhr 'Ire.sse~ fTL and
463

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464

Transformations 01 Stress and Strain

Drawing Mohr's ci rcle through the points A and B that correspond respectively to the principal stresses tT l and 171 (Fig. 7.54), and recalling that the maximum in-plane shearing stress is equal to the radius of this circle, we have (7.33)

O ,~,

\- _ _

'- ~

_ '"

This stress corresponds to points D and E and is exerted Oil an element obtained by mlating the original element of Fig. 7.5 1 through 45 0 within the phllle tangent to the surface of the vessel. The maximum shearing stress in the wall of the vessel, however. is larger. Il is equal to Ihe radius of the circle of diameter OA and corresponds to a rotation of 45° about a longitudinal axis and out of the plane of stress.t We have

~, : ~/' I

,

:

. . __ ~

E /'

~ lc:'

__ /'

_ I__ u,

'r~ _

0" 1 = 2"2

P'

1" rna., =: (J" , = -

•

Fig. 7.54

21

(7 .34)

We now consider a spherical vessel or inner radius r and wall thickness t, containing a fluid under a gage pressure p. For reasons of symmetry, the stresses exened on the four faces o f a small element of wall musl be equal (Fig. 7.55). We have (7.35) Rg . 7.S5

~

I,dl\

Fig. 7.56

,,

(7.36)

Since the. principal stresses fTI and (J"l are equal, Mohr'.~ circle for transfonTIations of stress wilhin the plrme tangent to the s urface of the vessel reduces to a point (Fig. 7.57); we conclude that Ihe in-plane normal st ress is constant and Ihal the in-plane maximum shearing stress is zero. The maximum shearing stress in the wail of the vessel. however, is nOi zero; it is equal to the radius o f the circle of diameter OA and corresponds to a rotation of 45° OUI of the plane of stress. We have

P'

o,

,

To determine the value of the stress, we pass a section through the center C of the vessel and consider the fre e body consisting of the portion of the vessel and its contents located to the left of the section (Fig. 7.56). The equation of equilibrium for this free body is the same as for the free body of Fig. 7.53. We thus conclude that. for a spherical vessel.

41

(7.37)

,

'.

" 1 =<12.--

Fig. 7.57

t il should be. obSl.""I"\·ed thaI, while the third principal 5tre.~s is zero on the outer surface of lhe vessel. it i.~ equal to - I' on the inner surface , :nd is represented by a point c( - p, 0) on a Mohr-c irele diagram. 11lU>. close to the iMide surface of the ,·csse l. the maximum shearing stress is equal to the radiu s of a circle of diarnrtcr CA . and we have

p,( ')

I . .... =-(0',+ ,,)=1+ 2 1, ,.

For a thi~ - waned vessel. however. the term fir is small. and we can neglect Ihe variation Qf ,. _ , across the wall sec lion . This remark also applies to spherical pressure vcssel.~ .

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SAMPLE PROBLEM 7.5 A compressed-air tank is ~upported by two cradles as shown: one of the cra· dies is designed so that 1\ does not exen any longitudinal force on the tank. The cylindrical body of the lank has a 30-in. outer diameter and is fabricated from a ~- in. steel plate by bUll welding along a helix that forms an angle of 25° wilh a transverse plane. The end caps are spherical and have a uniform wall thickness of Ii in. For iln internal gage pressure of 180 psi, determine (a) the [lomlal stress and the maximum shearing stress in the spherical caps. (b) the stresses in directions perpendicular and parallel to the helical weld.

SOLUT ION

"I

S pherical C a p.

c/ .

~

f' = ISO psi. I =

Using Eq. (7.36), we write

fi; in. =

0.3125 in .. r = 15 - 0.3[25 = 14.68S in. ( [SOpsi)( [4.688 in.) = 2(0.3125 in.) U = 4230 psi .. f'r

0" 1 =

I/~'~O---

(T 2

27

We note that for stres ses in a plane t~ngent 10 the c~p. Mohr's circle reduces to a point (A. B) on the horizontal axis and that all in-plane shearing stresses are zero. On the surf~ce of the cap the third principal STress is zero and carre· sponds 10 poil1l O. On a Mohr's circle of diameter AO. point D' represents the maximum shearing stress: it occurs on planes at 45° to the plane langentTo the cap. 7 m :u

,,

C

,,

I

fl, /j

= f(4230 psi)

7 0u•

= 2 [ [5 psi ..

h . Cylindrical Rud y of thi' Tllnk. We first determine The hoop stress (T I and the longitudinal stress 0" 2' Using Eqs. (7.30) and (7.32). we write f' = 180 psi. I = ~ in. = 0.375 in ..

([SOpsi)(14.625 in.) 0" 1= ---;- = 0.375 in.

T

=

[5 -

0.37.5 = 14.625 in.

f'r

"I t

IT I

'

---- b

"2 ~ 1510 [Hi

"

St resses al the WC'ld. Noting that both the hoop stress and the [ongitudinar s tress are principal stresses, we draw Mohr's circle as shown. An element hailing a face parallel to the weld is obtained by rotating the face perpendicular to the a:ds Ob counterclockwise Through 25°. Therefore. on Mohr's circle we locate the point X' corresponding to the stress components on the weld by rotating radius CB counterclockwise through 20 = 50°. = tJ,,, , - Rcos50° = 5265 - 1755 cos 50° 7 U ' = R sin .50° = 1755 sin 50°

(T il'

"' ~7020P"N

" ..., = .'i~('5IlS1 -_1


lJ

0"2 =1fTI = 35JOpsi

~ ,0201";

~ o.

7020 psi

Since X' is below the horizontal axis. counterclockwise.

T~.

u •.

=

7"

=

+ 4140 psi ... lJ44 p SI ...

tends to rotate the element

I

C

,\

2f1=5W

465

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PROBLEMS

7 .98 A basketball has a 300·0101 outa diameter and a 3·mm wall thick· nr,ss. Determine the normal stress in the wall when the basketball is infbted to a 120-kPa gage pressure. 7. 99 A spherical pressure vessel of 1.2-m outer diameter is to be fabricated from II steel having an ultimate stress U u = 450 MPa. Knowing that a factor of safety of 4.0 is desired and that thl' gagr pressure can reach 3 MPa. determine the smallest wall thickness that should be used. 7 . 100 A spherical gas container made of steel has a 2()...ft outer diameter and a wall thickness of in. Knowing that the internal pressure is 75 psi. determine th e maximum nonnal stress and lhe max imum shearing stress in Ihe co nlainer.

k

7 .101 A spherical pressure vessel Ills an outer diameter of 3 III and;} wall Ihickness of 11 film . Knowing that for Ihe steel used (T"'I = 80 M Pa, E = 200 GPa. and v = 0.29. determine «(I) the allowable gage pressure. (b) the corresponding increase in Ihe diameler of Ihe vessel

I

7 . 102 A spherical gas container having an outer diametrr of 15 ft and a wall thickness of 0.90 in. is made of a steel fur which E = 29 X [O~ pSI and I' = 0.29. Knowing that the gage press ure in the container is increased from zero to 250 psi. determine (a) the maximum normal stress in the co ntainer. (b) the increase in the diameter of the conminer. /,

I Fig. P7. 104

7. 103 The maximum gage pressure is known to be 10 MPa in a spherical steel pressure vessel having a 200-mln outer diameter and a 6-mm wall thickness. Knowin g that die ultimate stress in the steel used is (Tv = 400 Mpa, determine the factor of safety with respect to tensile failure. 7 . 104 The unpress urized cylindrical ,torage tank shown has a 1;;. in_ wall thicknt:ss and is made of sted having a 6O-ksi ultimate strength in tension. Detennine the maximum hei ght Ii to which 1t can be filled with water if a factor of safety of 4.0 is desired. (Specific weight of water = 62.4 lblf~.) 7 .105 For the storage tank of l'rob. 'i. 104. detemline the maximulII nOTlIIal stress and the maximum sharing stress in the cylindrical wall when thr take is filled to CilPilcity (/1 = 48 ft) .

7. 106 A standard-weight steel pipe of 12-in. nominal diameter carries water under a pressure of 40() psi. (a) Knowing that the outside diameter is 12.75 in_ and the Willi thicknrss is 0.375 in ., determine the maximum tensile stress in the pipe. (b) Sol\'e part a. assuming an extra-strong pipe is used. of 12.75-i n. outside diameter and 0 .500-in. wall thickness. 7. 10 7 A storage tank contains liquified propane under a pressure of 2 10 psi ilt il temperature of lOO°F. Kllowing that the tank hilS iln outer diameter of 12.6 in. ,md a wall thickness of 0.1 [ in .. detennine the maximum normal stress and the maxinillm shearing stress in the tank.

466

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7 .108 The bulk storage tank shown in Fig. 7.49 has an outer diameter of 3.5 m and a wall thickness of 20 mm. At a time when the internal pressure of the tank is 1.2 MPa. determine the maximum normal stress and the maxi· mum shearing stress in the tank.

Probloms

7 .109 Determine lhe largest inTernal pressure thaI can be applied to a cylindrical tank of 1.75-m outer diameter and 16-mm wall thickness if the ultimate normal stress of the steel used is 450 MPa and a factor of safety of 5.0 is desired. 7.110 A st:!el penstock has a 750·mm outer diameter. a 12-mm wall thickness, and conneCL~ a reserlloir at A with a generating station at B. Knowing that the density of water is 1000 kglm 1. determine the maximum nonnal stress and the maximum shearing stress in the penstock under static conditions.

7 .111 A steel penstock has a 7S0-mm outer diameter and connects a reserlloir at A with a generating station at B. Know ing that the density of water is 1000 kg/m 3 and that the allowable normal stress in the steel is 85 MPa. detemline the smallest thickness that can be lIsed for the penstock 7.112 The steel pressure tank shown has a 3D- in. inner diameter and a O.37S-in. wall thickness. Kno wing that the. butt welded seams fonn an angle. f3 = 50° with the longitudinal axis of ihe tank and thaI the gage pressure in the tank is 200 psi, determine (a) the nonnal stress perpendicular to the weld. (b) the shearing stress parallel to the weld. 7.113 The pressurized tank shown was fabricated by welding strips of plme along a helix fonning an angle {J with a tmnsverse plane. Detemline the largest value of f3 thm can be used if the 1I0mlal stress [l<"!rpcndicular to the weld is not be. larger than 85 percent of the ma.\imllm stress in the tank. 7. 114 The cylindric:!l por1ion of the compressed air tank shown is fabricated of 0.25-in.-thick plate welded along a helix forming an angle {3 = 30° wi th the horizontal. Kn owing 1hat the allowable stress 1I0rm:!1 to the weld is IO.S ksi. detennine the largest gage pressure that can be used in the tank.

Fig. P7.1 14

7. 115 For the compressed.air tank of Prob. 7. 114. determine the gage pressure that will cause. a shearing stress parallcl to the weld of 4 ksi.

3()() u.

L 751)",,,, Fig. P7.11 0 and P1.111

fl Fig. P7.112 and P1.113

467

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468

Transformations of Stress and Strain

7 .11 6 Square plates. each of 16-mm thickness. call be bent and welded together in either of the two ways shown to form the cylindrical ponion of a compressed air tank. Knowi ng that the allow:lble normal stress perpendicular to the weld is 65 MPa. deTermine the brgcst allowable g:lge pressure in c:lch case.

(0 )

(II)

Fig. P7.116 7 _11 7 The pressure tank shown has:ln S-mm wall thic kness :lnd bUll welded seams forming an angle f3 "" 20° with a transverse planc. For a gage pressure of 600 kP:l. deTermine (a) thc normal stress perpendicular 10 Ihe weld. (b) Ihe shearing stress parulielTo Ihe weld.

Fig. P7.117

7 .11 8 For Ihe tank of Prob. 7.1 17. determine Ihe largest allowable gage pressure. knowing tll:l1 Ihe allowable normal stress perpendicular 10 Ihe weld IS 120 MPa and thc allowable shearin g stress parallel 10 Ihe weld is SO MPa. 7.119 For the lank of Prob. 7. I 17. delemline Ihe range of va lues of f3 111:11 can be used if the shearing siress parallel 10 the weld is nOI 10 exceed 12 MPa when the gage pressure is 600 kPa. 7. 120 A IOrque of magnitude T = 12 kN . m is applied TO the end of a lank co ruainin g compressed air under a pressure. of S MPa. Knowing Ihal the lank has a ISO-mm inner diameTer and a 12-mm wallihickness. determine The m:l.\imum normal siress and Ihe ma.ximum she:lring siress in ule tank.

Fig. P7. 120 and P7. 121 7_121 The Tank s hown has a JSO-mm inner diameler and a 12-llIm wall Ihickness. Knowing Ihal Ihe lank contains com pressed air under a press ure of 8 MPa. determine lhe magnitude T of Ihe applied lorque for which the. maximum normal slress is 75 MPa.

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7.122 A pressure vessel of I O-in. innt!r diameter and IJ.25-in. wallihickness i~ fabricated from a 4-ft section of spirally welded pipe All and is equipped with two rigid end plmes. The gage pressure inside the vessel is 300 psi and lO-kip centric axial forces I) and P' are applied to lhe end plates. Detemline \ll) the normal str~s s perpendicular to Ihe weld. (b) the shearing stress paralleI to the weld.

Probloms

p

Fig. P7.122

7.123 Solve !>rob. 7.122. assuming that the magnitude I' of lhe two forces is increased to 30 kips. 7 .124 nlc .;:ompressed-ai r tank AB has a 250-mm outside diameter and an 8·mm wall thi"kness. It is fitted with a collar by which a 40-kN force P is applied at 11 in the horizontal direction. Knowing thm the gage pressure inside the tank is 5 MPa. determine tlte maximum normal stress and the maximum shearing stress at point K. 7 .125 In I'rob. 7. 124. determine the maximum normal stress and the maximum shearing stress at point L. 7.126 A brass ring of 5-in. ouler diameter and 0.25-in. thickness filS exactly inside a steel ring of 5-in. inner diameter and O.12S·ill. thickness when the tempemture of both rings is SO"F. Knowing that the temperature of both rings is then raised to 125°F. determine (a) the tensile stress in the steel ring, \b) the corresponding pressure exerled by the brass ring on the steel ring.

,.

STE£L

, = ;;

111.

= 2!lX 1QlillSl

, = 6.5 X to-'l/' F BRA SS

h,·

= 15 x Itt 1.5] a,, =tI 6X I ~I"F

Fig. P7.126

7 .127 Solve Prob. 7.126, assuming lhat the bmss ring is 0. 125 in. thick and the steel ring is 0.25 in. thick.

Fig. P7.1 24

469

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470

Transformations 01 Stress and Strain

*7.10. TRANSFORMATION OF PLANE STRAIN

1/

I

/

,

tt

Fi ..."i""l'l'ort Fig. 7.58

Transfonnatio ns of strain under a rota tion of the coordinate axes will now be considered. OUf analysis will tirst be limited to stales of plane .I'tmin. i.e .. to situations where the defomlalions of the material take place within parallel planes. and arc the same in each o f these planes. Lf the::: ax is is chosen perpendic ular to the planes in which the deformations take place, we have l':~ = 'Y.:x == 'YO'. = 0, and the only remaining strain compone n L~ arc €x, €,., and ,),,,.. Such a situation occurs in a plate subjected along its edges to uni formly distributed loads and restrai ned from expanding or contracting laterall y by smooth, rigid. and fixed supports (Fig. 7.58). It wou ld also be found in a bar of infinite length subjected on its sides to uniformly distributed loads since, by reason of symmetry, the clements located in a given transverse plane cannot move out of that plane. This idealized model shows that. in the actual case of a long bar subjected to uni fonnly distributed transverse loads (Fig. 7.59), a state of plane strain exists in any given transverse seclion that is not located too close to ~i ther end of the bar.t

Ag.7.59

'~ 0<

Fig. 7.60

'f+Y,y

,. ¥-"'~)f

" o Fig. 7.61

Let us assume that a state of plane strain exists at point Q (with 'Yo., "" 'Yo, = 0). and that it is delined by the strain components f 2 f ". and 'Y.n" associated with the x and y axes. As we know from Sec.s . 2.12 [Uld 2. 14. this means that a sqllare clement of center Q, with sides of lellg th !ls respectively parallel to the x and y axes. is deformed into a parallelogram with side..<; of length respectively equal to 6..1' (1 + f_,) and !l.r ( 1 + f ,.). fonning angles of f - 'Y,.,' and ~ + 'Y.", with each other (Fig. 7.60). We recall that. as a result of the deformatlolls of the other c1emelll~ located in the x)' plane, the element considered may also undergo a rigid-body motion. but such a motion is irrelcva1ll to the determination of the strai ns at point Q and will be ignored in this analysi.~. Our purpose is to determine ill terms of fx' € " 'Y-'J' and 0 the stress CO Inpo n ent~ €,_', f,.., and 'Y-,'.' associated with the frame of refcrence x'y' obw,incd by rotati llg the x and y axe~ through the angle O. As shown in Fig. 7.6 1. these new strain components define the parallelogram into which a square with sides respecti vely parallel to the x' and y' axes is defo rmed. f~

.l..,( t -+ "..,)

"

""

Itt Should be ob,crved that n ~tatc of I"Wir .nmi" and a ~talC of /,/m,,: sm' .... (d. Sec, 7. t1 do nOi occur !,imuttancou\]y. except lor id~Jl materiats with 3 Poi~son mtio equal to zero. The constraim~ ptaced on the ctemcnt~ of the ptate of Fig. 7.58 and of the bar of Fig. 7.~9 resutt in a stress (T . different from zero. On the other hane!. in the ca.~ of the ptale of Fi g. 7.J.llIC abse nce of any laterat n:!,traim resut ts in (T, = 0 []J)d "0"" 0,

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We first derive an expression for the normal strain E(R) along a line AB fonning an arbitrary angle R with the x axis. To do so. we con.';ider the ri ght triangle ABC. which has AB for hypothcnuse (Fig. 7.62a). and the obli{lue triangle A' B 'C' int o which triangle ABC is deformed (Fig. 7.62h). Dcnoting by 6. s the length of A8. we express the length of A' B' as 6..\' [I + E(R) !. Similarly. denoting by 6.x and 6.)" the lengths of sides AC and CB , we express the lengths o f A ' e' and e' B' as 6.x (I + E,) and 6.)' ( I + f) . respel:tillcly. Recalling from Fig. 7.60 that the right ang'le at C in Fi g. 7.6211 defonl1 ~ into an angle equal to t + Y.,. in Fig. 7.62b, and applying thc law of cosines 10 triang le A' B'C', we wri te

(A ' B')' ( ~ .,)' [l

~ (A 'C,), + (C' B')' -

+ , (O) i'

2(A'C')( C' B') cos (~

~ (tu )'( 1

+ ,J' + (" y)'( 1 + " )'

- 2(6.x)( 1

+

EX 6.y)( 1

+

E,.)L'Os(i +y.... )

+

7.10. TranslomtaUon 01 Plane Strain

o

'"

Yn)

(7.38)

"

But from Fi g. 7.620 we hallc

Fig. 7.62

dx = (6..\") cos 0

ay = (as ) sin 0

(7.39)

and we note that, since y ,", is very small ,

cos

(% +

yw) = - sin Y')" = - "), ,,,

(7.40)

Substituting from Eqs. (7.39) and (7.40) into Eq. (7.38), recalling Lhat cos1 (} + sin":! 0 = I, and neglecti ng second-order terms in E(O), E.,., E, .• and Y'J' we wri te

E(O)

=

E" cos1 0

+ E,. sin1 0 + Y,,.,. sin 0 cos 0

(7.41 )

Equation (7.4 1) enables us to de termine the normal strain E(O) in any directio n AB in terms of the strain compone nts E..., f l" Y'J" and the angle 0 that AB forms with the x axis. We eheck that, for () = O. Eq. (7 .41) yields E(O) = E.• and that. for 0 = 90°. it yield s E(900) = ElO" On the other hand. making fi = 4Y in Eq . (7.4 1), we obtain the nomlal strain in the direction of the bi sector OB o f the angle formed by the x and y axes (Fig. 7.63). Denoting this stTllin by E OB, we wri te

o (7.42) Solving Eq. (7 .42) for

r H"' we

have (7.43)

This relation makes it possible 10 express the ~'hel1rillg .I·traill associated with a given pair of rectangu lar axes in tcnns o f the lIormal straill.l· measured along these axes and their bisector. It wi l.l playa fundamental role in our prescnt derivation and will also be used .in Sec. 7.13 in connection with the experimental determination of shearing strains.

Fig. 7.63

471

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472

Translol1T\iltlons 01 Stress and Strain

Recalling that the main purpose of thi s sectio n is to express the strain components associated with the frmne of reference x' y' of Fig. 7.61 in terms of the angle 0 and the strain components E" En and 1'" associated with the x and y axes. we note that the normal strain Ex along the x' axis is given by Eq. (7.4 1). Using the trigonometric relati ons (7.3) and (7 .4). we write thi s equation in the aitemalivc fonn f',.

+

Ell

E . "" - - -

-'

2

E~ -

E).

+ - -2

eos 20

y".

+ ---'sill 20 2

(7 .44)

Replacing 0 by () + 90°, we obtain the normal strain along the y' axis. Since cos (20 + 180°) = - cos 20 and sill (20 + 180°) = - sill 20, we have t\

+

E,.

~ --- -

,

-"

E... -

Ey

- --

2

{."Os

2

1.'1 20 - -sin20 2

(7 .45)

Adding Eqs. (7 .44) find (7.45) member to member. we obtain E ~.

+

E,J

=

E.•

+ E,.

(7.46)

Since Ee = E", = O. we thus verify in the case of plane strain that the sum of the normal strains associated with a cubic clement of material is independent of the orientation of thnt elemenLt Replacing now 0 by 0 + 45 D in Eq. (7.44), we obtain an expressioll for the 110nnal strain along the bisector OB' of the angle formed by the x' and y ' axes. Since cos (20 + 90°) = - sin 20 and sin (20 + 90°) ~ l:OS 20. we have E,

EOIl ' = -

+

-

Ell

-

E.. -

E).

- - - - sin

20

Y...."

+-

222

cos 20

(7.471

Writing E4 . (7.43) with respect to the x ' and )" axes, we express the shearing strain 1'.... ,' in terms of the nomlal strains measured along the .r ' ami y' axes and the bisector OB': 1'N = 2EoB' - (Eo<,

+ €..I)

(7.4")

Substilllting from Eqs . (7.46) and (7.47) into (7.48). we obtain y ..... = - (E" - €\") sin '!.(J

+ Yn

cos 20

(7.49)

Equations (7.44), (7.45), and (7.49) are the desired equations defining the transformation of plane strain under a rotation of axes in the plant! of strain. Dividing all terms in Eq (7.49) by 2, we write this equation in the alternative form

0.49')

2

and observe that Eqs. (7.44), (7.45), and (7.49') for the transfonnation of plane strain closely resemble the equations deri ved in Sec. 7.2 for the transfonnation o f plane stress. While the former may be obtained from the laller by replacing the normal stresses by the corresponding normal strains, it should be noted, however. that the sheari ng stresses T .•~. and T .N sholiid be replaced by half of the corresponding shearing strains, i.e., by and respectively.

h..,.

ICf. footnote 0 11 PJgc 88.

h,y.

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'7.11. MOHR 'S CIRCLE FOR PLANE STRAIN

Since the equations for the transfomlation of phmc. strain arc of the SUliK': fOlH I as the cquutio lls [01 the tHUlSfOl"lllULiOlJ of p lulII:' slJess. the ulle of Mohr's ei rc1e can be extended to the an:.l1ysis of plane strain_ Given the 5train components f:.• , f: . , and )'1}' defining the defonnation rep resented in Fig. 7.60. we plot n' poinL X(~ " - h,Y} of abscissa equal to the normal strain ~-' and of ordinate equal 10 minus half the shearing stram Tn'" and II point Y(II'" .I'" I h '.y) (Fig. 7.64). Drawing the: di ameter XY, we define the center r Ill" Mohr's circle for rlant: ~Irni n . T he ahscis~ a of C and the rooius R of the circle are respectively equal to

and

N

~ J("

2 ' , )'

473

7, 11. ,",01\(3 Cll<:t" lor Plone Stmtn

+ C~,)'

(7.SO)

o

•

c

h) Fig. 7.64

We note that if 1'<1' is positive, as assumed in Fig. 7.60. points X alld Y 3fe plotled. respectively, below and above thc horizontal axis in rig. 7.64. Ilut. in the r.bs.:ncc of any overall ri gid·bodyrotation, the side. of the element in Fig. 7.60 that is associated with E" is ubserved to rotate counterclockwise, whil e the side associ~ted with f', is observcd to rotate clockwi se . Thus, if the shear defomlation callses a given side to rotate c/oc/ol'ise, the corre~ponding fXIint on Mohr's circle for plane stram i5 plottcd alxmt:. the horizontal axi.' , and if thc deformation causcs the sidc to rotate cOllllferclocJ..wi.\·e. the correspond in~ point is planed be/ow Ihe horizol1lal axis. We note Ilk'll Ihis convcntion matches the convention used to draw Mohr's circle for plane S lres~ . Poims A <'HId B where Mohr's circle imersccts the horizontal axis correspond to the (}rillcipal .!lrain.f Em., and f:" m (Fig. 7.650). We find and where

(7 .51 )

ami R ure defined by Eq!;. (7.50). The corresJXlnding value :rngtc i:I is ubtained by ubservln g lhat tile shearin g strain i~ zero for A and B. Set1ing )" ,' = 0 in Eq. (7.49). we have flp

--;,c,t----;; t---~~ccr::_t ;-~-.

t"" .. <

u f the

/'

,

(7.52)

"

°

The corresponding axes li llnd b in Fig. 7.65h are the principal {lXi'S of :ifraill . The angle 1" Which tlefines the direction of the principal axis Ou m Fig. 7.65b corresponding to point A in Fi g. 7.65u. is equal 10 half of the ungle XCA measured on Mohr's circle. and the rotation that bring ~ Ox into Oa ha.' the same sense as the rotation that brings the diameter KYoI' Mohr's circle into the diameter AB. We recall from Sec. 2.1 ... tlmt, in the case of the e la.>tic defonnation of a ho mogeneolls, isotropic material, Hooke's law 1'01 slltm illg Hress ;md slrain applies and yields T , }, = Gl'n for :my pair of rect:mgular x and y axes. Thus, y ,.. = 0 when T ", = O. which indi:ates that the principal axes of stmin mincide with the principal axes o f stress.

Ftg.7.65

",

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474

111t: ulaximullL ill-plauc shcali ng stlUili i~ ddi ueJ by poi uts Dam:] L~ cljual tu the diamcter of Mohr's circle Rrcalling

rm"3'ormallo,," 01 :;1"'33 ""~ SImi"

E in Fig_ 7 6SlL It lh~

second of Eqs. (7.50). we write (7.';1)

Finall y, we !Iotc that the points X' and Y' that dcfin~ the components of strain corre~llond i ng to a rotation ofthc coordinate axes througll an angle e(Fig. 7.6 1) are obtai ned by rotating the diameter XY ofMohr\ circle in the same sense through an angle 20 (Fig. 7.66)

"

"

" c

Fig. 7.66

Fig, 7,61 (repeated)

EXAMPLE 7.04 In a malerial in astate of plane strain, it is known that the horizontal side of a 10 )( I O.mm square elongates by 4 fo'm, while its vertical s ide remains unchanged, and that the angle at the lower left corner inc teases by 0 .4 X 10 ] rad (Fig. 7.(7). Deli:, mine (u) the p,incipal axes aJld prillcipal shain ..... (/~) themaximum shearing strain and the corresponding nOlmal strain.

'I

",, ,, ,

8f-~OC-~:~C-"~--+t~

'I

,

I (U, - :.'!)(J !

~u~""' fl _ In"u"

r

~

/ 1() """ + 4 !-
Fig. 7.68

,

¥+ OA X ]()~' r.l
tates i"OWI/I'ILlodwi.1"e, point Y is ploned directly bdo)\o the origin (Fig. 7.68). Drawing the diameter XY, we determine the centel C of Mohr's circle and its radius 11.. W~ have

Fig. 7,67 ra) Principal Axes and Principal Strains. We first detennino: the coordinates of points X and Yon 'vtohr's circle

~,

+4 X 10 ' m 10 X IOJ m

+4001--'-

E"

= 0

1"2'''1

= 200~

Since the side of the square associated with € , rotates dod". wi" 'e, point X of coord;n3te~ £, and b"./21 ;~ plotted "bOl'/' tl~ horizont~1 axis. Since € ,. = 0 and the corresponding side ro-

+

OC = - 2-

fOJ strain. We have

"

X(4C(1, 200)

" j

'"

200~

or =

200~

R = Y(OC)2 + (OYf _ V(200fo')2 + (2001-'-)1 - 2831-'The princLpal stmins are detined by the 3bsCLo;S3S of po ints A

and B. We write

nA

=

nr+ R =

200/,.

+ 2.1:1/,. =

4R1" .

Fu

=

~~

'" DB = OC - R = 2oo I-'- - 283 IJ- = - 831J-

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The principal axe. 011 and Ob are shown in Fig. 7.69. Since OC = OY. the angle at C in triangle OCY is 45°. Thus. the an· gle 28 that brings X Y into AB is 45~ J and the angle IJ" bring. ing Ox" into Oll is 22.5° J. (b) Ma)timum Shearing Strain. Points D and £ detinc the maximum in.plane shearing strain which, since the prin. cipal strains have opposite signs . is also the actual maximum shearing strain (see Sec. 7.12). We have

Yn"" = R = 2831.l 2

"y,... , =

7.12. Three-Dtmenslonal Analysis

01 Strain

/

5661.l Fig. 7.69

The corresponding nonnal stroins are both equal to f'

= OC =2001.l

The axes of maximum shearing stroin are shown in Fig. 7.70.

a Fig. 7.70

-7.12. THREE-DIMENSIONAL ANALYSIS OF STRAIN We saw in Sec. 7.5 that. in the most general case of stress. we can determine three coordinate axes {/, b, and c, called the principal axes of stress. A small cubic element with faces respectively perpendicular 10 these axes is free of shearing stresses (Fig. 7.27): i.e .. we have T ab = T Ix- = T ,." = O. As recalled in the preceding section, Hooke's law for sheari ng stress and strain applies when the deformation is e hl~tie and the material homogeneous and isotropic. It fo llows that, in slLch a case, "Yah = "Y1x- = ""Yea = O. i.e., the axes a, b, and c are also principal (nes of strain. A small cube of side eel ual to unity, cen tered at Q and with faces respectively perpendicular \0 the principal axes, is deformed into a rectangu lar parallelepiped of sides I + fEa' I + fEb, and I + fE, (Fig. 7.7 1).

Fig. 7.71

Fig. 7.27 (repealed)

475

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476

Transformations 01 Stress and Strain

/.

Fi g . 1.72

Fig.7.7 1 (repe ated) [I' the

element o f Fig. 7.7 1 is rotated about one of the pri nci pal axes at Q, say the c axis (Fig. 7.72 ), the method of analys is developed earlier for the transformation of plane strain can be used to determine the stmi n component.. E... E,. and 1',1,. associated with the faces perpendicular 10 the (' axis. since the deri vation o f this method did not involve any of the other strain eomponents.t We can. therefore. draw Mohr's circle through the points A and B corresponding to the principal axes a and b (Fig. 7.73). Similarl y, ci rcles o f di ameters Be and CA can be used to analyze-the transformation of strain lL<; the element is ro tated about the 11 and b axes, respectively.

o

r

- •......~ Fi g . 7.73

The three-di mensional analys is of strain by means o f Mohr's circle is limitcd here to rotations about principal axes (as was thc case for the analysis o f stress) and is used to determine the maximum shearing strai n 1'ma< at point Q. Si nce 'Y m;U is cqtllli to thc di ameter o f thc largest of the three circles shown in Fig. 7.73. we have (7 .54) where E max and Emin reprcsent the- algebraic values of the maximu m and min imum strains at poi nt Q. Returning to the p:1rticular case of plalle strain, and selecting the x and y axes in the plane of strain. we have E, = 'Y:;r = 'Y.q. = O. Thus, the:: axis is one o f the three princip:1l axes at Q, and the corresponding t We note that the olhcr four face.~ of the element remain rectanJ;ular and that the edges paralkllo the c axis remain unchanged.

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point in the Mohr-circle diagram is the origin 0, where € =- 'Y =- O. If the points A and B that define the principal axes within the plane of strain fall on opposite sides of 0 (Fig. 7.74(1), the corresponding principal strains represent the maximum and minimum normal strains al point Q. and the maximum shearing strain is cclual to the maximum inplane shearing strain corresponding to poi nts D and E. [1', on the other hand. A and B are on the same side of 0 (Fig. 7.74h). that is. if €~ and Eb have the same sign, then the maximum shearing strain is defined by points D' and E' on the circle of diameter OA , Hnd we have 1m., = E""". We now consider the particu lar case of plane stress encountered in a thin plate or on the free surface of a structural ele ment or machine component (Sec. 7. [). Selecting the x and y axes in the plane of stress, we have a : = T o, "" T :y = 0 and vcrify that the z ax is is a principal axis of stress. As we saw earlier, if the deformation is clastic and if the material is homogeneous and isotropic, it follows from Hooke's law thaI 'Y,,-, = 'Y'J' = 0: tlIUS, the.::: axis is also a principal axis of strain. and Mohr's circle can be used to ,malyze the transfonnation of strain in the x)' plane. However, as we s hall sec prcsenlly, it docs not fo ll ow from Hooke's law that € : "" 0: indeed , a state of plane stress docs nol, in general, result in a statc of plane slrain. t Denoting by a lmd b the principal axcs within the plane of stress, and by c the principal axis perpendicular to that plane, we let a. = a~, (T,. "" a b, and a , = () in Eqs . (2.2 8) for the generalized Hooke's law (Sec . 2. 12) and write E

"

D

,,/ ,Z=O

B

~ ~ ....

£ ",,"

, /~-, ,

,

\--- .. j'

Z ~ O ,'

'-

£

'~~~

---

--~

E'

~ ,.'"'.

=

(7.56)

"

= - "E(a"

+ a b)

(7.57)

Adding Eqs. (7.55) and (7 .56 ) member to member, we have €a

Solving Eq. (7 .58) for write

+ a"

Eh

I - I' = -E-(a"

+ a lt

+ a n)

(7.58)

and substituting into Eq. (7.57), we

e: , = - -[ _"-

{€ + Eb )

,, "

_-_

(7.59)

The rel ation obtained de fines the third principal strain in tenns of the "in-plane" princi pal strains. We note thal, if B is located between A and C on the Mohr-circle diagram (Fig . 7.75). the maximum shearing strain is equal to the diameter CA of the circle corresponding to a rotation about the b axis. OUI of the plane of stress. tSee f()()lnote nn page 470.

,/ ,,' , I

E ~ ­

£' Fig. 7.75

/'

(b) ~. ,

Fig. 7.74

E,.

A

,

,

(7.55)

E

A

\'---/1

' 0..,,, ~ (!

aa VUb = - - -E

7.12. Three-Dimensional Analysis 01 Strain

477

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EXAMPLE 7.05 As a result of measurements made on the s UJi"ace of a machine comJXment with strain gaJ;es oriented In \'urious ways. it has been established that the principal strains on the free surface are f . = +400 X 10 w ~ inl in. and lO b = - 50 X 10 b in./in. Knowing thaI Poisson's mtiofor the given material is v = 0.30. dctennine (a) the maximwn in· plane shearing main. (Ii) the true value of the maximum shearing strain near the surface of the component.

(a) Maximum In-Plane Shearing Strain. We draw Mohr's circle through the points A and B corresponding to the given principal strains (Fig. 7.76). 'nle maximum in-plane shearing strain is defined by points D and E and is equal to the diameter of Mohr's cirde: "YL""'("' Po.n.) =

400 x 1O - ~

+ 50 x 10

b =

450 x 10

6

rad

(b) Maximum Shearing StraIn. We first detemline the third principal strain !':,. Since we have a state of pJ:me .\"tres~· on the sUlface of the machine component. we use Eq. (7.59) and write f', =

- " + "b)

- -, - " -( ""

= - 0.30 (400 X 10 - " - 50 X lO- fi) = - 150

0.70

x to- bin.j in.

Drawing Mohr's circles through A and C and through Band C (Fig. 7.77). we lind th~t the maximum shearing strain is equal to the diameter of the circle of diameter CA: "y"""

= 400 X 10

6

+ 150

X 10

h

= 550 X 10 - 6 rad

We note that. even though ". and lO b have opposite signs. the rna:
:h ( J O- ~rad l)

D

~Y,,,., "" I'''~

~-,H'j-+C~_f-__-+,,,_--,I___~ { t U-~ I"Ji". ) _ 50

0

- ,:,j FIg. 7.76

Fig. 7.77

~ . '3 .

8

~.:# Fig. 7.78

478

MEASUREMENTS OF STRAIN; STRAIN ROSEnE

The nomlal strain can be de tcnnine d in an y given direc tion o n the surL1ce of 11 structural element o r machine compone nt by scribing two gage marks A and B across a line draw n in the desired direc tio n a nd measuring the length o f the segment AB before and after the load has been ap plie d . II" L is the un dcformed length of AB and 8 its de formation, the no rmal strain alo ng AB is f An = 8/ L. A mo re convenient a nd more accurate method fo r the measurement of no rmal strains is provided by e lectrical strain gages. A ty pical electrical strain gage consists o f a length of thi n wire arranged as shown in Fig. 7.78 and cemented 10 tWO pieces of paper. In orde r to measu re the strain f' Ml of a given material in the d irection AB, the gage is cemel1led to the s urface of the material, wi th the wi re folds run ni ng paralle l 10 AB. As the material elongates. the wire increases in lc ngth and decreases in diameter, causing the e lectrical resistance of the g;Jge to inc re;Jse. By measuri ng the clirren t passi ng thro ugh a properly calibrated gage, the

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strain € All can be determined
(7 .43)

8

" III " Fig. 7.79

It shou ld be noted that the strai n compo nent s IE. .. , 10,., and Y,,, at a given point could be obtained from no rmal strain measure ments made alo ng allY three lilies drawn through that point (Fig. 7.80). Denoling res pectively by 0 ]. 0 1, and OJ the angle each o f Ihe three lines fo rms with thex axis. by IE.] , 1E.1, and E} the corresponding strain measurements. and s ubstituting into Eq. (7.41). we write the three equati ons 2

+

2

+

IE.]

= € j COS 0)

IE.l

cos O2 1 = €.! cos O}

IE. J

E.!

+

E,.

s in ~ 0 1

+

Y... v sin 0 1 cos 0 1

si n! O2 + y"" sill O~ cos 82 E)" s in ~ OJ + Y.,·' sin OJ cos 8 J lOy

(7.60)

which can be solved simultaneously for E.," E ) . and "Y <).. t The arrangement o f strain gages used to measure the three nonnal strai ns IE. I, 1E.1' and E] is know n as a straill rosette. The rosette used to measure no rmal strains along the x and y axes and their bisector is referred 10 as a 45° roselle. Another rosette frequentl y used is the 60° rose tte (see Sam ple Prob. 7.7).

FI9·7.80 III shoukl be nOIl>d thallhc [TN) ,urf:lCC on which lhe 'IrJin llIeaSUTi'n~m s arc made is in of pllili f .m"i'SS. while Eq~. (7.41) and (7.4.1) were derived for a Slale of ,!Iilile SImi". However. as observed ~arlier. the nonnal 10 Ihe flT~ surface is a principal axis of SIfJ.i n ;U1d IIIe derivation s J;iI"Cn in Sec. 7.10 remain \·alid.

a stlle

7.1 3. Measurements of Strain: Strain Rosette

479

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l -------r~

___

24 ill.

SAMPLE PROBLEM 7.6 A cy hnrlncaJ stonge

tan~

used to transpon gas under pressure.ha;. an inner di-

anlel .. r of 24 in. and a wall l1li d::n<"ss of ~ in. Slr...;n gage, al1ach ..d \0 the ~ur·

t-.

face of the lank in transverse and longimdina! directions indicate strains of 255 X [0 band 60 X lO- b in./il1. respectively. Knowing that a torsion test has shown thaI the modu lus of Tij;id ity of the nmterial used in (h~ lank is G = [1.2 X ](1' psi. delennine (a) the .cage pressure inside the tank. (b) the

or the tank.

principnl ,tresses and Ihe maximum .hearing siress in the wall

SOLunON II. "age Prcssurt' Inside Tank. We Ilote that Ihe given strains are the princIpal "trains;)[ the surlaee artlle tank. PI oiling Ihe correspondin g points A and B. we dr::.w Mohr's c ircle for

~ i lO .... rad) )

o

" C"r,-,,-----r ,

B

~train.

The maximum in-plane s hearing strain

is eq ual 10 lhe diameler of Ihe circle. )" .... ('nr ......1 = € , -

From Hooke'; law foc

€J

~ 255 X I() - ~ - 60 X 10 ~ = 195 X 10 (, rad

s h~aring

SIreS! :l!Id

~Irajn.

we ha ve

C' T""" ('DpI:ill<) - GY"""- (inP''''''')

= ( [1.:2

x 10"psiX 195

X

10

~r:l,j )

=2 184psi=:2. 184ksi Sllhstilllrin ~

Ihi,

v~[lle

and th e given

d~t;1

in Eq. (7 11). we write

2[84 p,j = Solving

fOT

p{l2 in.) 4(0.75 in.)

the gage pressure p . we have

p-

5~6psi

....

IJ. P rincipal St!"('~s :md Ma!\;imum Shearing Stress. l{eca[hng Ihal. fer n thin wo lled cyli ndric al pressure \'es~ [ . tT l - 2o! . \\e dlllw Mohr's circle

for stress and obuln IT,

=

2T m . , ( .. pI=<)

IT,

=

2tf2

= 2(2. 184 ksi) = 4.368 ksi

= 2(4.368 ksl)

1Tl

= ..U7 Itsi ....

~I

= 8.74 l si ....

The maximum shearing stres;; is equ~l to Ihe mdius of the circle of diameter OA and corre.>ponds to a rotation of 45° about a longitudinal axi;;. r .... '" 1.37 h i ....

480

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SAMPLE PROBLEM 7.7 Using a 60 0 roselle. [he following strains have been determined at poinl Q on the surface of a steel machine base:

Using the .:oordinute axes shown. determine m point Q. (a) Ihe strain l:omponents f A' £,_ and y". (b) the principal strains. (c) Ihe maximum shearing strain. (Use v = 0.29.)

SOLUTION II .

St rain COmpOIll'lIts

£~ ,

£,_r~_..

For [he coordinate :txes shown

Substituting these values into Eqs. (7.60), we have S60/L~

o

W"

l·F Ie

1' 1

= £~(I)

£~ = £j o.500f

""" I\' \I

€, = £~( - O.5oof

+ £,(0) + )",,(0)(1) + f ,.(O.366f + )" .,(0.866)(0.500) + £,.(0366)1 + 1.,,(0.866)(- 0.500)

Solving these equations for f, . f ,., and y... we obtain B

£2 -

1... = Substituting the given values for f ,

= -101.1.

f ,.

f l . f,.

and 6. we have f ,=+ 860JL ....

= H 2(980) + 2(330) - 40]

r ,. =

11'3

0.866

(980 - 330)/0.866

'Y .. = 750 JL ....

These strains are indicated on the element shown.

,....

h . Pri ncipa l St ra ins. We note that the side of the e lement associated with f A rotates counterclockwise; tim, . we plot point X below the horizontal axis. i.e .. X(40. - 375). We then plot Y(860, + 375) and draw Mohr's circle,

~ {, ..( I

' •

./... , L~ ' ~21.2" I ::{ ~" .! y ) ,

..

,,

D'

,

+ (4 10JL f -

5561L

375 JL

tan28 = - I' 410 JL

-. , . ,I , 1",." Since

\,1 , -I' " - --lUi)!)..,

3fi...'i 1'

= !(860 IL + 40 JL) = 450 IL

Points A and B correspond to the principal strains, We have

c' A ;

-

E,.e

N = Y(375 JLf

H fc

.

,

=

iT,

N = 450IL - 5561L

f"

=

fo

= t " 'e + N = 450IL + 5561L

t ... -

E. = - I06JL .... = + IOO6JL ....

E)

= 0 on the surface. we use Eq. (7.59) to find the principal strain E,,:

,J - --(Eu

I - "

+

f b)

0.29 = - - - - ( - I061L

I - 0.29

+

lOO6j1)

If' ,

=

- 368p. ....

c. Maxim u lll Shearing St rai n. Plolling point C and drawing Mohr's circle through points Band C. we obtain point D' and write

y, ..,

= 1374 p.

....

481

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PROBLEMS

7.128 through 7.131 For the give n state of plane strain. use the method of Sec. 7. [0 \0 detennine the state of plane strain associated with axes x' and y' rotated through the given angle tJ.

"I

0 _,

Fig. P7.128 through P7.135

,

<.

<,

Y.,

- 240~

+32011 + 16011

- 33011

65a~

+15011

60 0 ) 15°) 30°1

7.128 and 7.132 7 .129 and 7.133

+ 240 11

7 . 130 and 7. 134

+ 35O fL

7.131 and 7.135

0

0

+ 32011

+ 120fL

- 100,:,t

7.132 through 7.135 For the give n state of plane strain, use Mohr's circle \0 determine the state of plane strain associated with axes and .1" roo lated through the given angle II.

x

7.136 through 7.139 TIle foll owing Siale of strain has been measufet.l on th e surface of a thin plate. KJlowin g tlwt the surfac e of the plate is unstressed, delemline (u) the direction and magnitude of the principal strains.
h

7 .136 7.137

7 .138 7.139

<.

<.

Y.,

+ 30jL + 160jL - 600jL - 260jL

+ 570jL - 480jL - 400jL - 60jL

+ 720jL - 600jL + 350jL + 480jL

7.140 through 7.143 For the given slate of plane strain. use Mohr's ci rcle 10 determine V,) 1he orientation and magnilllde of the principal strains. (b) the maxi mum in-plane strain. (e) the maxi mum sbearing s train.

7 .140 7.141

7.142 7.143

482

<.

<.

'Y_..

+ 400jL - l BOjL + 60jL + 300jL

+ 200j-l - 260jL + 240jL + 60j.L

+ 375jL + 3 15jL - 50jL + tOOj.L

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7. 144 Determine the strain E , knowing that the follow ing strains have been determined by use of the roselle shown:

Probloms

~

I~:\~r

" '5%£J15

0

I

Fig. P7.144

7.145 The ;;trains detemlined by the use of the rosette shown during Ihe test of a machine element are

Determine ing strain.

(u)

the in-plane principal strains. (b) the in.plane maximum shear-

3t~~ l ~"

Fig. P7.145

7.146 The rosette shown has been used to determine the following strains al a point on the surface of a crane hook: 1' ( ""

+420 x 10 E4

6

=

injin.

+ 165

IE l

= - 45 x 10

6

in jin.

X 10 (, in.lin.

What should be Ihl' rl'ading of gagl' 3? (b) Delerminl' Ihe principal and the maximum in.plane shearing strain.

(1I)

~Irain s

Fig. P7.146

7.147 The mains determined by the use of the rosette attached as shown durin g the lest of a machine element are 1' (

= - 93. 1 x 10 " ~ in.lin. IE )

f~ =

+385 X 10 ~ in'/in.

= + 210 X 1O ~ 6 in.lin.

Determine (a) the orientation and magnitude of the principal s trains in the plane of Ihe rosette. (Il) the maximum in· plane s hearing stress.

Fig. P7.147

483

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484

'frDn3'ormallon. 01 Slre33

on~

7 .148

Slmln

Show

Ih~1

the

~uUl

of Ihe Ilu t e Mrail l meaSUJemtLlts made with

il

6(f rorette i , independent of the orielllation of the rorette
where .. ".~

i~

Ihe

tlb~ei5Stl

of Ihe ecnlCr of the

~onc sponditlg

Mohr'.'! circle .

Fig. P7.148

7 .149 U ~ing a A.5~ roselle.. Ihe ~train,<; £1' £1. and £3 have been delermine.:i at a given point. Usin g Mohr' .. circle. show that the principal strains are: f"",,-,,,," = "2I (11' 1 + 11',) ::!:.

I 1(£, - £, j-' + v'2

(f, - E ~ )-'J''

(Hillt: The shaded triangle s are congruent.)

, a l---"+"---'~~

__-r~A'---..

~ ... ;: -~

Rg. P7.149

7 .150 A centric axial force P aud a horizontal force. Q., are Dolll applied al IlOim C uf Ille rect,mguiar bar showlI. A 45· SIJ-aill rosette 011 Ihe s Ulfa.::e of the b:lI at point A indicates Ihe following , trains:

"J '"

Fig. F7.1S0

- 60 x 10

6

in.lill.

'" = +2.l0

x

](J - b

ill.lin.

' 3 = + 200 X ][) 6 in.!in

Knowing that E = 29 x 10" ps i and " = 0.30. delermine the magniludes of P and Q".

7 ,151 Ihe

Solve Prob. 7.150. a,sumillg lhal Ihe ro:;Clle at paim A "" - -30 x 10 b in.lin.

(, '" + 100 X

2;11.

tl -

+250 X lQ _ b ill./in.

II) 6

in.lin.

7.152 A single gage is ccmemed \() a solid 4-in.-diameter steel shaft at nn!,le {l = 2 .~ o with it line pilrallel to the axis of Ih", ShilfL KnO\~ ing IhilT G = 11.5 X 106 psi. detemli!te the torque T indicated by a gage reading of 300 X lO- b in.lin. ~n

Fig. P7. 152

inui<:ate~

fnl1owl11~ ~tr:lin s'

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7.153 SolII;: Prob. 7. 152. assuming that Ihe gage fonns an ang.le {3 "" 35° wilh a line parallel to the axis of Ihe shaft.

7 .154 A single strain gage forming an angle {3 "" 18° wi th a horizontal plane is lIsed 10 determine the gage pressure in the cylindrical steel lank shown. The cylindrical wall of the tank is 6 mm thick. has
Fig. P7.154

7 .155 Sollie Prob. 7. 154. assuming that the gage forms an angle f3 = 35" with a horizontal piane. 7 .156 The given state of plane strain is known to exist on lhe surface of a machine. component. Kn owing thaT £ = 200 GPa and G = 77.2 GPa, determine the direction and magnitude of Ihe three principal strains (II) by determining the cDrresponding state of strain luse Eq. (2.43) and Eq. (2.38)J :U1d then using Mohr's circle for sTrain. (b) by using Moltr's circle for stress 10 determine the principal planes and principal stresses and then determining the l"Orresponding strains.

XMJ'"

lD I I

;S~J f'"

Fig. P7.156

7 .157 of a

ca~ t -iron

The foll owing st.:.te of stmin has been de termined on the surl:1ce machine part: t:~

= - 710p.

"), ,, = + 660p.

Knowing that E "" 69 GPa and G = 28 GPa, determine the principal planes and principal SLIes<;cs (II) by detemlining the corresponding st.:.te of plane stress [use Eq. (2.36). Bl (2.43), and the first two eqllations of Prob. 2.73J and then using Mohr's cirde for stress, (b) by llsing Mohr's circle for s train 10 determine the orientntion and mngnitude of the principal stmins and then detennine the corres]Xlnding SLIesses.

Probtoms

4aS

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REVIEW AND SUMMARY FOR CHAPTER 7

The fi rst part of this chapter was devoted to a study of the tralls/ormatio/l of streH under a rotation of axes and to its application to the solution of engineering problems, and the second part to a similar study of the trails/ormation of .I·traill.

'I

if.t

.~t.-< Transformation of plane stress

(h )

'"I

Fig. 7.5

Considering fi rst a state of plalle stress at a given point Q [Sec. 7.2[ and denoting by 0""" (T p and"TXl' the stress components associated with the clement shown in Fig. 7.5a, we derived the following formulas del'ining the components 0"," (r.v" and 7"..,>", associated with that element after it had been rotated through an angle 8 about the z axis (Fig. 7.5h): 0".,"

+

0" ),

(r,' = - - -

2

0" .• -

(T,

+ - - - cos 28 + "T.no sin 28 2

0"]; +0"),

O" x - O",

2

2

(T . , = - - - - ---cos 28 -

'1 \

-

0".. -

Tn

si n28

(7.5) (7.7)

0",.

7"..,.. = - - - - sin28 + 7" ~ cos28 2 ."

(7.6)

In Sec. 7.3, we determined the values Op of the angle or rotation which correspond to the maximum and minimum values of the normal stress at point Q. We wrote (7 .12)

Fig. 7.11

486

The two values obtained for 8p arc 90° apart (Fig. 7.11) and define the principal plalles of stre.u at point Q. The corresponding values

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of the normal stress are called the principal obtained

a. . + a,.

a m< • . min = --2--

~

stre~·se.f

J(U'- -" , -a)-.)' + ,

T;y

Review and Summary for Chapter 7

at Q; we

(7.14)

Principal planes. Principal stresses

We also noted that the corresponding value of the shearing stress is zero. Next, we determined the values 8. of the angle 0 for which the largest value of the shearing stress occurs. We wrote tan 28

/' u'

a.

(T... =: - - - -

(7.15)

27~y

•

487

The two values obtained for 8.. are 90 0 apan (Fig. 7. 12). We also noted that the planes of maximum shearing stress are at 4SO to the principal planes. The maximum value of the shearing stress for a rotation in the plane of .l"tre.fS i.f

if,), + ,

c;-c----cc-;c;--

7 max

=

J( - -2- ~'

T~"

(7.16)

and the corresponding value of the nonnal stresses is a.,.

+

a.

Fig. 7.12

Maximum in-plane shearing stress

(7.17)

2

We saw in Sec. 7.4 that Mohr's circle provides an alternative method, based on simple geometric considerations. fo r the analysis

Mom' s circle for stress

o

(1) )

Fig. 7.17

of the transformation of plane stress. Given the state of stress shown in black in Fig. 7.17a. we plot point X of coordinates a ...,- T,,. and point Y o f coordinates (T,., +7 n (Fig. 7.l7b). Drawing the circle of diameter Xv. we obtai n Mohr's·ci rcle. The ab.~cissas of the points o f intersection A and B of the ci rcle with the horizontal axis represent the principal stresses, and the angle of rotation bri nging the di ameter XY into AB is twice the angle Op defining the principal planes in Fig. 7.170, with both angles having the same sense. We also noted that diameter DE delines the maximum shearing stress and the ori entation ofthc corresponding plane (Fig. 7. 19b) [Example 7.02, Sample Probs. 7.2 and 7.3J.

u' = u

",I_ -f'D,--::-__,

I ,I

~ ol-~"t--'C+--'--jAcc-'--~

Fig. 7.19b

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488

Transformations of Stress and Strain

General state of stress

o

Considering a general state of sIren" characterized by six stress components [Sec. 7.51, we showed that the normal stress on a plane of arbitrary orientation can be expres,ed as a quadratic form of the direction cosines of thc nonnal to that plane. This proves the ex istence of three principal axes of slrc.u and three prillcipal .\"tre.\".\"CS at any given point. Rotating a small cubic element about each of the three principal axes [Sec. 7.6], we drew the corresponding Mohr's circles that yield the values o f if max'Umin' and T""", (Fig. 7.29). In the panicular case of plane slre.v.I", and if Ihc x and )' axes are selected in the plane of stress. poi nt C coi ncides with the origin O. If A and B are located on opposite sides of O. the maximum shearing stress is equal 10 the maximum "in-plane" shearing stress as determined in Sees. 7.3 or 7.4. If A and B are located on the same side of 0, this will not be the casco If (J" a > (J" b > 0, for instance the maximum shearing stress is equal to ! a a and corresponds to a rotation out of the plane of stress (Fi g. 7.32).

Fig. 7.29

Fig. 7.32

Yield criteria for ductile materials

Yield criteria for ductile materials under plane stress were developed in Sec . 7.7. To predict whether a structural or machine component will fail at some critical point due to yield in the material, we first detcnnine the principal stresses a a and (J"b at that point for the given loading condition. We then plot the poi nt of coordinates a a and a bo If thiS" point fa lls within a certain area. the component is safe; if it falls outs ide, the component wilt fail. The area used with the maximum-shearing-strength criterion is shown in Fi g. 7.40 and the area used with the maximum-distortion-energy criterion in Fi g. 7.41. We note that both areas depend upon the value of the yield strength (J"f of the material.

+U')'f-_ _•

c

-u,

H

Fi g. 7.40

Fig. 7.41

D

/

+ (7)"

D _ fT )"

u.

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Fracture criteria for brittle materials under plane stress were developed in Sec. 7.8 in a similar fashion. The most commonly used is Mohr's criterioll, which utilizes the results of various types of test available for a given material. The shaded area shown in Fig. 7.47h is used when the ulti mate strengths alrr and a ile have been deler-

Review and Summary for Chapter 7

Fracture criteria fo r brittle materials

a"

L _____{uc< Ih)

Flg.7.47b

mined , respectively. from a tension and a compression test. Again, the principal stresses a a and a b are determined at a givcn point of the structural or machine component being investigated. If the corresponding point falls within the shaded area, the component is safe; if it falls outs ide. the component will rupture. In Sec. 7.9, we discussed the stresses in thin·walletl pressure venels and derived formulas relating the stresses in the walls of the vessels and the gage pressure p in the nu id they contain. In the case of a cylindrical veHei o f inside radius r and thickness I (Fig. 7.5 1), we obtained the following expressions for the hoop slreH a l and the IOllgitudinal Slress a!:

P' 21

(7.30,7.3 1)

We also found that the maximum .I·hearillg stress occurs out of the plane o f stress and is

Fig. 7.51

(7.34)

T m ax

[n the case of a .Ipherical vessel of inside radius r and thickness (Fig. 7.55), we found that the two principal stresses are equal: a,

Cylindrical pressure vessels

a, = I" .

21

I

(7.36)

Spherical pressure vessels

a,

Again. the mmillllllll .vhearillg stress occurs out of the plane of stress: it is

P' 41

(7.37)

Fig. 7.55

489

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490

rmn,·ormo.tlon. 01 SIre..

an~

Slmln

Transformation of plane strain

The last pan of the chaplcr was dcvoted to the tmmfo rmatioll of .~Imi!l. In Secs. "7.10 and 7.11, we discussed tile transfonnation o f plane .\ ·tmi/l and introduced Afollr's circle for 1,Ialle straill. The discusSion \\las similar to the corresponding di ~ u ss ion of the tnmsformatinn of stres~, cxccpt that, where the shearing stress T was used , we HOW used 1}', that is, hili/the ~hf'II ";lIg ~Imill. TIle fOllllulas obtained forthc tfansformation of strain under a rotation ofaxes through an angle (J were Ex- t') t:;r- I!', )'.1). ' r - - - - + - - - cos 28 + - sill 28 (7 .44)

222

)'.1). .

Tsm 28

(7.45)

(7.49) Mohl's circle for strain

n )'

tan 2()

/

,

0

Using Mohr's circle for strain (Fig. 7.65 ). we also obtained the following relations denning the angle of rotation Op corresponding to the principal (He.l· of straill and the values of the IJrincip<11 straUB· I!' m.u and I!'rn;,:

Yo t'. - E,.

---

~

P

1!'"",,= t'a,< + R

c

I!'mi, = Ea,.<- R

(7 .51 )

R ~ J(¥)' +C;'),

(7.50)

where

,,", J

x

' ,,, - --

and

(7 .52)

and

E

f---- , ~. ---l

,.)

The II/luimlIIn shearilJ8 was fo und to be

.~tm ill

for a rotati on in the plane of strain (7.53)

Sec[ion 7. 12 was devoted to the three-dimensional analysis of strain, with ap pliclllion 10 lhe determination of the maximltm shearing strain in the particular cases of plane strain and plane stress. [n

y

the

ca.~c

c f {' Ia " ", ..tr"'.H, we
a direction perpendicular to the plane of stress could be expressed as follows in terms of the "in-plane'· principal strains t'a and f: b: (7 .59) Fi g. 7.65

Strdin gages. SIIi3.in rosette

\L,

Fi nally. we disclissed in Sec.7. [3 the use of strain gagt.l· to measure the nonnal strain o n the surface of a structural element or machine component. Considering a STrain roselle consisting of three gages ali gned along lines fo nning respectively, angles 8 9z, and OJ " with the x ax is (Fig . 7.80), we wrote the following relations among the measurements I!' ,. t'2' f:3 o f the gages and the components f x, f J' ~x:o characterizing the state of strai n at thai point:

\11 "

I!' ,

~ t'xCOS~8 ,

~2

:

~ 3 - t: xcosJ83

Fig. 7 .80

£j

+

t'xcosl tt2 +

+

f ,. sin ~ I!' ,. sin

2

tt, 82

~y s ill '83

+ 1,) sin 8 , cos f/, + "YX) sin ttl cos f/ 1 +

"YX)

(7.60)

sill 8 3 cos (/3

These equations can be solved for Ex. E,.• and have been d e le rmin~d.

1 <;1'

once I!',. 1!'2' and

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REVIEW PROBLEMS

7.158 The grain of a wooden member forms an angle of 15° with the vertical. For the state of siress shown. determine (a) the in-plane shearing siress parallel (0 the grain. (b) the normal stress perpendicular to the grain.

Fig. P7.1S8

7.159 The :entric force P is applied [0 a short post as shown. Knowing Ill::!! the stresses on plane a·a are if = - 15 ksi and T = 5 ksi, delennine (<<) the :lngle (3 that plane a-a forms with the horizontal. (b) the maximum compressive stress in the post.

r !I

Fig. P7. 159

7 .160 l11c ,teel pipe AB hils a 102-mm outer diameter and;) 6-mm wall thickness. Knowing that arm CD is rigid ly :luached to the pipe. detemline the principal slre.~ses and the maximum shearing slres ~ at point H. 7 .161 l11c ~tecl pipeAB hils a I Ol-mm outer diameter and;) 6-mm wall thickness. Knowing that arm CD is rigldly attached to the pipe. detemline the principal slre.~ses and the maximum shearing slres~ at point K.

Fig. P7.160 and P7.161

491

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492

Transformations of Stress and Strain

7.162 Delermine the principal planes and the principal stresses for the slate of plane stress resulting from the superposition of the IWO slates of stress shown.

+ Fig. P7.162

7. 163 For the state of stress shown. delermine the range of values of f} for which Ihe magnitude of the shearing stress TN is equal to or less than 40 MPa.

'I

t ~o '\ ]1'••

~

,~

u,

120,\II'a

Fig. Pl.163

7 .164 FOrlhe state of stress shown. delemline Ihe value of,." for which the maximum shearing stress is 80 MPa.

Fig. P7.164

11-1 1.:.,

-lor", , j

Flg. Pl . 165

7. 165 l1lt": Slate of plant stress showlI occurs in a lIlachine component made of a steel with (fl ' = 30 ksi. Using tht: maximum.distortion.energy crite· rion. determine whether yield will occur wren (oj 1"i~ = 6 hi. (b) 1',. = 12 k..~ i. (") 1"" . = 14 ksi. 1£ yield does not occur. determine the corresponding factor of safety.

7 .166 When filled to capacity. the unpressurized stomge tank shown oont
Fig. P7.166

Fig. P7.167

7. 167 The compressed.air tank AB has an inner diameter of 450 mm and a uniform wall thickness of 6 mm. Knowing thnt the gage pressure in the tank is 1.2 MPa. determine the maximum normal stress and the maximum in· plane shearing stress m point a on the top of the tank.

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Computer Probtems

7.168 The brass pipe AD is filled with a jacket used to apply a hydro. static pressure of SOO psi to portion Be of the pipe. Knowing that the pressure inside the pipe is 100 psi . determine the maximum normal stress in the pipe. 7.169 Determine the largest in-plane nonnal strain. knowing that the following strains h.a ve been obtained by the use of the rosette shown: "1 =

- 50

X

10- 6 in.lin. £J

IE, =

+ 360

X 10 - 6 in.lin .

= +3 15 X IW 6 in.lin.

I

Fig. P7.168

COMPUTER PROBLEMS Thl' followin g problems are 10 be solvoo wilh II computer. 7 .C1 A state of plane stress is defined by the stress components fT, . fT,. and T." a.~sociated with the element shown in Fig. P7.Cla. (fj) Write a computer program tha! can be used to calculate the stress components {f"

if,.. and T ,' l' associated with the element after it has rotated throllgh an angle about the z axis (Fig. P.7Clb). (b) Use this program to solve Probs. 7.13 through 7.16.

f}

'I

::l

.~ Fig. P7.C1

I"'

b. ~ /J I

~~IU~

Fig. P7.169

(h)

493

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494

Transl ormatlons of St""ess and Strain

7.C2 A slate of plane. stress is defined by the ~Iress components IT ,. V, .• and T " . the principal stresses. the maximum in-plane Jl hearing stress, and the maximum shearing stress. ~b ) Use this progm/l1 to solve Probs. 7.5. 7.9. 7.68. and 7.69. 7.C3 (a) Write a computer program that. for a given sta~ of plane stress and a gi\'en yield strength of a ductile material. can be used to detemLint: whether the matt:rial will yield. The program should use both the maximum shearing.stre.ngth criterion and the maximum-distortion-energy criterion. II should a11;o prinT the values 01" the principal stresses and. if the material does not yield, ca1cul:lte Ihe faclOr of safelY. (bi Use this program 10 solve Probs. 7.81. 7.82. and 7. 165. 7.C4 (a) Write a computer program based 011 Mohr's fraclUre criterion for brillie materials that, for:l given stale of pl:lne stress and give n values of the ultimate strength of the material in tension and compression. can be usctl 10 dctemline whether ruplUre will occur. The program should also print the values of the principal stresses. (b) Use thi, progr,un to solve Probs. 7.89 and 7.90 and 10 check the answers to Probs. 7.93 and 7.94. 7 .C5 A 5t::Jte of plane slrain is defin~d by Ihe strain components f,. f, . and "'I,~ as~ociated wi th die _t and y axes. (a) Write ::J computer pro!,'TIlm Illat can be used to c::Jlculme the strain comP,Onent.<; f ," f of. and y"" ::Jssociated with the frame of reference x'/ obtained by TOtating the .( and .l' axes through an angle 11. (b) Use this progmm 10 solve Probs. 7. 129 and 7.IJL

D_, "'\0

'I

/

~

.

.\'

Fig. P7.C5

7 .C6 A state of strain is defined by the strain components f " f,. ::Jnd y" associated with the x and y axes. (a) Writ~ ::J computer program tlwt c;}n be used to delermine the orientation ::Jnd magnitude of the princip::JI strains. the maximum in-plane shearing strain. and the maximum shearing strain. (b) Use this progrrunto solve Probs. 7.136 through 7.1J9.

7.e7 A state of pl::Jne strain is defined by the strain components f " f ,., and "'11) measured ::Jt a point. (a) Write a compuler program ih;}t can be used to detemtinethe orientation and mllgnitude of the principal str:tins, the ma.~imllln in-pbne shearing strain . and the magnitude of the shearing SIl-:-tin. (b) U!k! Ihis program to solve Probs. 7.140 through 7.143. 7.C8 A roselle consisting of three g::Jges fonning, respect ively. angles of 8[. II!. and Il, with the x axis is ::Jltache:l 10 the free surfnce of n machine component mnde of n materi::JI with a given Poisson's ratio v. (n) Write a computer progmm th::J!. for given readings f l . E2. and f } of the gages. can be used to calculme the stmin component~
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C

HAP

T

E

R

Principal Stresses under a Given Loading

Due to gravity and wind load, the post supporting the sign shown Is subjected simu ltaneously to compression, bending, and torsion. in this chapter you will learn to determine the stresses created by such combined loadings in structures and machine components.

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496

PrincIpal Stresses under a GIven Load !ng

"S.1. INTRODUCTION In the first part of this chapter, you will apply to the design of beams and shafts the knowledge Ulat you acquired in Chap. 7 on the transformation of stresses. In the second part of the chapter, you will1cam how to determine the principal stresses in structural members and machine clements under given loading conditions.

j l/ )

(hi

Fig. 8.1

("' Fig. 8.2

(M

In Chap. 5 you learncd to calculate the maximum nonnal ~ tress (J" m occurring in a beam under a transverse loadillg (Fig. S.la) and check whether this val ue exceeded the allowable stress f1 311 for the given material. If it did, the design of the beam was nOi acceptable. While the danger for a brillie material is actually 10 fail in tension. thc danger for a ductile matcrial is to fail in shear (Fig. !:Ub). The fact that (J"", > (J" ..l indicates that IMlm',", is too large for the cross section selected, but does nOi provide any information on the actual mcchanism of failurc. Similarly. the tilet that Tm > 7.11 simply indicates that I vi""" is too large for thccross scction selected. While UlC dangcr for a ductile matcrial is actually to fail in shear (Fig. S.2a). the dangcr for a brittle material is to fail in tension under the principal stresses (Fig. R.2b). Thc distribution of the principal stresses in a beam will be discussed in Sec. JU. Depending upon the shape of the cross section of the beam and the value of the ~hear V in the critical section where IM I = IMln"'" it may happen that the hU'gest value of the normal stress will not occur at the lOp or bottom of the section. but at some other point within the section. As you will sce in Sec. 8.2, a combi nation of large val ues of (J"" and T." near thc junction of the weh Imrlthe tllm~e~ nf!l \V-hellm or an S-hcam Clm res ult in a value of the principal stress u "''''' (Fig. 1(3) that is larger than the val ue of am on the surface of the beam.

uI I

Fig. 8.3

Section 8.3 wi.ll be devotcd 10 the d~sign of transmission shafts su[)jected to transverse loads as well :t<; to torques. The effect of both the normal stresses due to bending mId the shearing stresses due to torsion will be taken into account In Scc. 8.4 you will learn 10 determine the stresses at a given point K of a body of arbitrary shape. subjectcd to a combined toading. First. yOll will reduce the given loading to forces alld couples in the section containing K. Ncxt, you will calculate the nomml and s hearing strc.l;ses at K. Finally, using one of the methods for the transfonnation of stresses that you leamed in Chap. 7. you will determine the principal pl:mes, principal stresses. and maximum shearing ~tress at K.

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8.2. Principal Stresses In a Beam

' 8.2. PRINCIPAL STRESSES IN A BEAM Consider a pri smatic beam AB subjected to some arbitrary transverse loading (Fig. 8.4). We de note by V and M , respectively, the shear and bending moment in a section through a given point C. We recall from Chaps. 5 and 6 that. wi thin [he clastic limit. the stresses exen ed on a small element with faces perpendic ular. respectively, 10 the x and y axes reduce to the normal stresses i fm = M e/I if the element is at the free surL1ce of the beam. and to the shearing stresses 7 ", = VQ! IJ if the e leme nt is at the ne utral surface (Fi g. 8.5). Fig. 8.4

, ,

U'.' _~ U'"

<1,=-__ ,r,,,.. , ~" ' 0; •.• ..

o

, Fig. 8.6

Fig. 8.5

At any other point of the cross section, an element of material is subjected simultaneously to the normal stresses My

if

, =- - /

(8. 1)

where), is the distance from the neutral surface and I the centroidal moment of inertia of the section, and to the shearing stresses 7 9·

VQ ==- If

(8.2)

where Q is the fi rst moment about the neutral axis o f the ponion o f the cross-sectional area located above the point where the stresses are computed, and f the width of the cross section at that point. Using either o f the methods of analysis presented in Chap. 7, we can obtain the principal stresses at any point o f the cross section (Fig. 8.6). The following question now arises: Can the maximum normal stress if max at some poi nt within the cross section be larger than the value o f if m = Me/I computed at the surface o f the beam? If it can, then the determi nation o f the largest normal stress in the beam will involve a great deal more than the computation of I MIrna> and the use o f Eq. (8. 1). We can obtai n an answer to this question by investi gati ng the distribution

'T", __

. _I'J".

[,

,

497

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498

Principal Stresses under a Given Loading

of the principal stresses in a IllUTOW rectangular cantilevcr beam subjcctedto a concentrated load P al ils free end (Fig. 8.7) . We recall from Sec. 6.5 Ihal the normal and shearing Mresses al a distance x from Ihe load P and a distance )' above the neutral surfacc are given, respectively. by Eq. (6.13) and Eq. (6.12). Since the moment of inertia of Ihe cross section is / ~ _b/_l"l = ,,(bcc hl,,(2<::c·,)l

12

1

Ac

12

3

where A is the cross-sectional area and (. the half-depth of the beam. we wri te

l-r-,-I ';;-t--

(8.3)

Fig. 8.7

T

..,.

=

~2 !..( 1 A

-i )

(8.4)

cl

Using the method of Sec. 7.3 or Sec. 7.4, the value of (T"",.~ can be determined at any point of the beam. Figure 8.8 shows Ihe results of the computation of the ralios a "",","ju,,, and u 'niju m in two scctions of the beam, correspond ing respectively to x = 2(" and x =- 8c . In each

,

J" =1r

yle 1.0

r

:~ ::fn-i----=-=i---{ .l=2c

.l =oc

{roll"fu m

"

0.8

- 0.010

0.6

- 0.&10

um./u",

, -:::=0--

--D--

~ \\

0'

- 0.090

0.2

- 0. 160

0

- 0 .2.')0

- 0.2

- 0.3(,0

- 0.4

- 0.490

- 0.6

- 0 ,&40

- 0.8

- 0.8 10

- 1.0

- 11~)()

~

X ):(

1.000

"

0.1>10

- 0.001

0.640

- 0.003 - OOOi

0.3(,0

- O.O l i

0.2..~

- 0.003

0. 160

- 0.2 1i

O.()g()

- OA Oi

-~ o.o-to

- 0.603

jJ,

-¢:::::::---0---

(r" ,,)u m

--D--

-=9-:::b--

1.000

0.$0 1 0.603

I

0.400

):( )1,

&

II"""/II,,,

0.0 10

- 0 .80 1

0

- LOOU -

---6-I

~ ~

-p-

-9_¢:::-.

OAOi 0.2 1i 0.063 0.0 17 O.OOi 0.003

_¢===-O JK)[ 0 -

0

Fig. 8.8 Distribution of principal stresses in two transverse sections of a rectangular cantile'ler beam supporting a single concentrated load.

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section, these rat ios have been determined at [I different points, and the Olientati on llf the principal axes has been indicated at each poinct It is d ear that (Till.' does not exceed (T,,, in eithl!r of the two sections considered in Fi g. 8.8 anu that. if it docs exceed (TffI elsewhere, it wi ll be in secti ons close \0 the load P, where (T", is small cumpared to 'Ton.:} Bu t. fur sectiuns close to the luad P, Saint-Venant 's principle docs not appl y. Eqs. (8 .3) and (8 .4) cease to be valid. except in the very unlikely case of a load distributed parabo lically over lhe end section (cf. Sec. 6.5), and more advanced methods uf analysis taking into accollnt the effect of stress concentrations should be ust:d. We Lhus concl ude thaI. to r beams ofrec langular cross section, and within the scupe of the theory presented in this text, the maxi mulll nonnal stress can be ubwined from Eq. (8 .1). In Fig. 8.8 the directions of the principal axes were de temtined at [I points in each of the two sections considered. I f this analysis were ex tended to a larger nu mber of sections and a larger number of poillls in each section, it would be possibl e to draw two orthogonal systems of c urves on the side of the beam (Fig. 8.9). One system would consist of curves tangent to the principal axes corresponding to (T m:n and the other of curves tangent 10 the principal axes curresponding to U min ' The curves obtained in this manner arc known as lhe stre.U trajectories. A trajectory of the first group (solid lines) defines at each of its points the direction of the largest tensile stress. while a trajectory of the second group (dnshed lines) delines the direction of the largest compressive stress .§ Theconclusion we have reached for bt!ams of rectangular cross section. that the maximum normal stress in the beam can be obtaincd from Eq. (8. 1). remains valid for many beams of nonrectangular cross section. However, when the width of the cross section varies in such a way thaI large sltcMing stressc..~ Tn will occ ur at points close to the surface of the beam. whf!re (T., is also"lnrge, a value of tlte principal stress a max larger than a", may result at such points. One should be particularly aware ()f this po~sib i li t y whe n selecting W-beams or S-beams, and calculate the principal stress a m"", at the j unctions band d or the web with the fl anges of the beam (Fig. 8. 10). Thi ~ is done by determining a x and 1" 'Y at that point from Eqs. (8.1 ) and (8 .2). respective ly. and using either of the methods of analysis of Chap. 7 to obtain a m.... (see Sample Prob. 8. 1). An alternati ve procedure, used in design to se lect an acceptable section, consists of using for 'T.n the maximum value of the shearing stress in the scction, T m ., = VIA,,~~. given by Eq . (6. 1 [) of Sec. 6.4. This leads to a sli ghtly larger, and thus conservative, value of the principal stress am .., at the juncti on of the web with the Ilanges of the beam (sec SJ.mp1c Prob. 8.2). t Sce Prob. 8.Cl. which refer. to the. progmm u!iCd to obtrun the ~As

will be:

~crificd

in Prob. g.Cl,


Il'sult~

shown in Fig.

if ." ", O.j.J4c.

~A brittlc matcrial. such a. COlJ.Cretc. will fail in \em.ion along planes that are p"rpcndku· lar to the ten~il~-~tres~ traj&toric..~ . Thus. to ~ effe.ctil"e., steel rcinfolting bars should be placed so that they imefliect these pl1U1CS. On the other hand . stitfeners attached to the wcb of a plme girda will tic cffectil'e in prcvcnling buckting only iF they Inlersect planes perpen· dicular to the comprcss.il·~·.I.tress trajectories.

~ . ~.

8.2. Principal Stresses In a Beam

C""'prc",ivr

Fig. 8.9

Fig. 8.10

Stress trajectories.

499

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500

Principal Stresses under a Given Load ing

'S.3. DESIGN OF TRANSMISSION SHAFTS

When we discussed the design of transmission shafts in Sec. 3.7, we considered on ly the stresses due to thc torques exerted on the shafts. However, if the power is transferred to and from the shaft by means of gears or sprcx;:ket wheels (Fig. 8.1 Ill), the forces exerted on the gear teeth or sprcx;:kets arc equivalent to force-coupJe systems applied at the centers of the corresponding cross sections (Fig. 8.llb). This means that the shaft is subjected to a transverse loadi ng, as well as to a torsional loading.

A

Fig. 8.1 1

The shearing stresses produced in the shaft by the transverse loads arc usually much smaller thrm those produced by the torques and will be neglected in thi s analysis.t The normal stresses due to the transverse loads. however, may be quite large and, as you will see presently, their contribution to the maximum shearing stress 7 01,,, should be taken into account.

t For an application wh~re the shearing .Itresscs produced by the transverse considered. see Probs. 8.2 t and 8.22.

toads must he

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Consider the cross section of the shaft at some point C. We reprethe tor4ue 'r and the bending couples My and M e acting, respecti vely, in a hori zontal and a vertical plane by the couple vectors shown (Fig. 8. l2a). Since any di ameter of the section is a principal axis of inertia fo r the section, we can replace M , and M . by their resultant M (Fig. S.12h) in order to compute the normal stre~ses cr., exerted on the section. We thus li nd that fTx is maximum at the end of the diameter perpendicu lar to the vector representing M (Fig. 8.13). Recalling that the values o f the normal stresses at that point are, respectively, fT", = Mcl! and zero, while the shearing stress is T m = TeIJ , we plot the corresponding points X and Y on a Mohr-circle diagram (Fig. 8.14) and determine the value of the maximum shearing stress:

8.3. Design 01 Transmission Shalts

~ent

( 2CT",)'

T"",, = R =

+ (Tm) L

10 )

Fig. 8.12

- -yI(M 2iC)' + (T')' J

Recalling that. for a circular or annulftf cross section. 2! == J, we write Fig. 8.13 (8.5)

D

It fol!ow~ that the minimum allowable value o f the ratio Jle for the cross section o f the shaft is

B

j

,

(8.6)

where the numerator in the right-hand member o f the expression obtained represents the maximum value of V M ' + r 1 in the shaft , and Tall the allowable shearing stress. Expressing the bending moment M in terms of its components in the two coordinate planes, we can also write

j

c

(8 .7)

Equations (S .6) and (S.7) can be used to design both solid and hollow circular shafts and shou ld be compared with Eq. (3.22) of Sec. 3.7. which was obtained under the assumption of a torsional loadin only. The determination orthe maximum value of M ; + M; + T1 will be faci litated if the bending-moment diagrams correspondi ng to My and Mo arc drawn, a, weI! as a third diagram representing the values of T along the shaft (see Sample Prob. 8.3).

Fig. 8.14

501

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SAMPLE PROBLEM 8.1

",

-L = 375 nnn- I

A J60-kN force is applied as shown at the end of a W200 x .'\2 roIled-sleel beam Ne-glecting the effect of fillets and o f ~lreg S cOllcC'ntration., . dl'lermi(lf'

,,

whether the normal stresses in the beam satisfy 11 design specificalion that they be equal 10 or less than 150 MPa at ~ection A -A'.

" SOLUT ION

!hU ki\

-0.3i5m-1

Sh e ar !lnd Bend in g M om e nt.

t)M,

M il = (160kN)(O.375

",

V~

Normal

!

•

f

on Transn'.rsf PI:me.

Sl~ss es

M",

%=100. ! '1 '1'111~

h


""" - 5 -

203 Inn

.L- - -

m) = 60 kN · m

- 160kN

Referrmg to the table of Prop-

atie.,- of Rolled-S/tl,'l Shapes in Appendix C. we obtain Ihe data s hown :md then determine the stresses (T" and (J" 0 ' ""& At point t1

f-- 2lJ.j IEIll-j 12.6 """1 I " I

1,· I= ~m'" .

AI sectIO n A.-A'. w e;; h:wc

;.9 " ""

60kN-m Sl2 X 10 ~ m )

[17.2 MPa

AI point b:

1 = 52.7 X lO-('m' ~ = .';11': )( 111-,\,,1

90A

I'b

1:J1ll

ab= IT~~= ( I 17.2 Ml'a ) I03mm = 102.IJ M I'a

Sh earin g S l ~.'ise.'l o n Trans ....~I'".'IC'- Plant' AI point a:

Q= O

T,

= 0

Al poim b

Q = :204 X 12.6)(96.7) = 248.6 x [0' mm! = 248.6 Ii. IO- b Ill' ( l6()kN)(248.6 X to- bmJ )

t'AQ

"-"'---1 rr,,,,,

"

c

"

I H~

-:J

u

T

~

-

It

-

(52.! X IO - ~ m4)(O.OO79 m)

_ -

95.:> MPa

Principal Stress at POill1 h . The Slale of stresS :II point b consists of the normal stress if. = [02.4 MPa and the shearing stress T o = 95.5 MPa. Wedmw Mohr's dn:lt:
= 10;.9

+

J(1O;.9Y+ (95.5f

,r"".., = 159.9 MP" TIl l' spedliL:alion. (T rn ••

~

150 Ml'a.

j.

11m "ali sli t u ....

Commellt. For this beam and loading. the principal stress at point b is 36% larger than the Ilomml s tress at point {I. For L ~ 874 mm , the mllximlJm nomlal stress ",ould occur at point II.

502

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2() k~r' 1--- 9 f l -

I

SAMPLE PROBLEM 8.2

3.2 klp~ft

,\ ._=?_?pr~ B ~ C 1)% 1

1L_- wr, -

_+"--I 5 ft

,- -

The overhanging beam AB supports a unifonnly distributed load of 3.2 kips/ft and a concentrated load of 20 kips at C. Knowing that for the grade of steel to be used (T.n = 24 ksi and T oll = 14.5 ksi . select the wide-flange shape that should be used.

SOLUTION !!(I

kl p.

Reactions at A and D. We dtaw the free-body diagram of the beam. From the equilibrium equations ';;.Mv = () and IM.. = 0 we find the val ues of R,\ and No shown in the diagram.

"",pr

A

t·n

CI

"'I"

f.. - 9t1 - -- Jl ft

B

Sf! I

I' .11 kip.'

: : I

16 kip_' ( 23\1..1 ) ~

Shear and Rending-Moment Diagrams. Using the methods of Sees. 5.2 and 5.3, we draw the diagrams and observe that

1

12.2 kips

Ivl" .., =

IM I.." = 239.4 kip' ft = 2873 kip ' in.

43 kips

Sl'Clion Modulus. For IMlo= = 2873 kip' in. and (T>l1 = 24 ksi. lhe minimum acceptable section modulus of lhe rolled·sleel shape is

7,S kips

S,m"

=

1M!,..,

2873 kip' ill. 24 ksi

if.U

-43 kips

119.7 in l

Selection or Wide-Flange ShaJH". From the table of Proper/in. of Rol/ed·Steel Shape.l· in Appendix C. we compile a list of the lightes t sh;)pes of a given depth that h
Shape

I d=2J in. L

r

," = (J ,401) in. \\,21 X fi2

S= 1::!7;,,] A·"".h

~=O .6J5il1.

105

'"I ,JT

W24 x 68 W21 X 62 WI8 x 76 Wl6 x 77 WI4 X 82 Wl2 X 96

~.~",. ~

= t" d = 8..10 ill Z

(I

=

= 22.6 hi

~;,., 21 3k" -~

154

127 146 134 123 131

We now select the lightest shape available. namel y

W2 1 1< 62 ....

Shearin g Siress. Since we are designing the beam. we will conservatively ass ume that the maximum sheilf is uniformly distributed over the web area of a W21 X 62. We write T ..

=

Vn...

43 kips

~- ~ ~~ . -, A"eIl 8.40 III"

= 5.12 ksi

<

14.5 ksi

(O K)

Principal Str ess at Point h. We check that the maximum principal stress at point b in the critical section 'Where M is maximum does not exceed 0" .11 = 24 ksi. We write M n ""

0""

= ~ S- =

2873 kip' in. 127 inl

22.6 ksi

9.88 in. II h = (T. = (22.6 ksi)- - -.- = 21.3 ksi c 10.50 1Il. V 12.2 kips 1.45 ksi Conservatively. 1'b = - - ~ - - - , A".b 8.40 in Yb

We draw Mohr's circle and find if o,",-,

=

I

'i ifb

21.3ksi

+ R = --,- +

., V' (2I.3kSi)' --,- + ( IA5ksl)0-........ =

2 1.4ksi :::::; 14 ks i

(OK) "

503

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SAMPLE PROBLEM 8.3 TIle solid shaft AB rolatcs at 480 rpm and Ir.msmits 30 kW from the motor M to machine tools connected \0 gears G and H: 10 kW is tal:en off nt gear G aud 10 kW a1 gear H. Knu wiJ lg Ihm "'.JI = 50 MPJ., uetelilliue Ihe smalksi pelmissible diamelel for shafl AB.

SOLUTION Torques Enrted on G ea rs. Observing that f"" 480 rpm = 8 Hz. we determine the toTtjue exened on gear E:

P 30kW - - - - - 597N ·m 211[ 2711:8 Hz)

T - ~

The corresponding tangential force acting on Ihe gear is

597 N ·

T" Ft, - -

III

3.73 kN

O,16m

rE

A similar atwl)si.<; of gears C and D yields

20kW

Tr;:= - - - = 398:"J' m

Fe = 6.63 kN

I OkW Tv = 2':1(8 Hz) = 199 -"I .

Fo = 2.49 kN

2':7(8 Hz)

III

We now replace Ihe forces all the gears by equivalent force-couple systems.

Bendi ng-Moment and Torque Diagrams

, +

).ii;Aiiiiiiiiii..Eiliii;;;j' :-,

t!- o.6m-J-.-jt J.il:E kN

I

+

2.1>0 kN

TI"m

(U,n

'" A

C

DEB -

A

C

DEB

V~ 5SON l Z4 1 N III

'I

A

.II

597N '"

DE

C

B

III

liGON ·m

Cr it ical Tran.'iYusc SCl,:lio n. Dy computing VMJ t AI~ I 1' 2"I "II p0tentiall y critical sections. we find that it;;; maximum value occurs just to the right of D:

YC" M";-+-""" 1;-'+- '''')""" Oifl nll"l l'r nf S hlift.

v(1 1601 + (373f + (597)~ For -r.. = ,0 MP:l. Eq (712)

- 1357

N. m

y ie lll~

1357 N . m

J

,.

=

t

.u

50MPa

For a solid circular shaft of radius c. we have c - 0.02585 m - 25.85 mm Diameter = ll' = 5 1.7 mm ...

504

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PROBLEMS

8 .1 A WlO X 39 rolled-steel beam supports 11 load P as shown. Know ing that P = 45 kips. a = [0 in .. and 0 -.11 = 18 ksi. detennine Ca) the maximum value of the normal stress i fm in the beam. (b) the maximum value of Ihe principal stress fT, .., at the jUlldion of the flange and web. (c) whether the specified shape is acceptable as far as these two stresses are concerned. 8.2

Fig. PS.1

Solve Prob. 8.1. assuming that P = 22.5 kips. a = 20 in.

8 .3 An overhanging W920 X 446 rolled-steel heam supports a load P shown. Knowing that P = 700 kN. a = 2.5 m. and (Toll = ]00 MPa, determine ~a) the maxi:num value of the nomla] stress (T" , in the beam. (b) the maximum value of the principal stress fTo,", at the junction of the flange and web, (e) whether the specified shape is acceptable as far as these two stresses are concerned. a.~

8 .4

Solve Prob. 8.3. assuming that P = 850 kN and a = 2.0 m.

8.5 and 8.6 (a) Knowing that IT..1 = 24 ksi and 1'..1 = 14.5 ksi. select the most economical wide-flange shape that should be used to support the load· ing shown. (n) Determine the values to be expected for (T,. . 1'.,. and the princi . pal stress if""" at the junctio n of a flange and the web of the selected beam.

FIg. P B.3

atnTII.' IIIa 9 kill>

A

111111 111 111 1111c

~ ft - -'- I - fl-I t2

6

FIg. PB.S

LJ!ift --L~rt

FIg. PB.6

B.7 and B.8 (a) Knowing that if. 1I = 160 MPa and 1'.. 1 = 100 MPa. select the most economical metric wide-flange shape that should be used to support the loading shown. (b) Determine the values to be expected for if,•• 1'",. and the principal stress IT n"", at the junction of a flange and the web of the selected beam. 2S0 kl\ 250 k>l

J

"

U.9

III

FIg. P B.7

.2!io

kN

c

O.g m

F.

o.g

III

0.9

III

Fig. PB.B

505

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506

Principal Stresses under a Given Load ing

'}IX) i\

Il'iko

nun

,\

8.9 through 8 .14 Each of the following problems refers to a rolledsteel shape selected in a problem of Chap 5 to support a given loading at a minimal cost while satisfying the requirement fT",:S;: u all . For the selected design. determine (a) the actual value of fT., in the beam. (b) Ihe mnimum value of the principal stress fTm•• at the junction of a !lange and the web. 8 .9 Loading of Prob. 5.74 and selected W53() x 66 shape. 8 .10 Loading of Prob. 5.73 and sele;::ted W250 X 28.4 shape. 8 . 11 Loading of Prob. 5.78 and selected S310 x 47.3 slwpe. 8 .12 Loading of Prob. 5.75 and sele:ted 520 x 66 shape. 8 . 13 Loading of Prob. 5.77 and selected 5510 X 98.3 shape. 8 .14 Loading of Prob. 5.76 and sele;::ted 520 X 66 shape. 8 .15 Neglecting the effect of fillets and of stress concentrations. determine the smallest pemlissible diameters {)f the solid rods BC and CD . Use Toll = 60 MPa.

Fig. PS.1 S an d PS.1 6

8 .16 Knowing that rods Be and CD are of diameter 24 mm and 36 mm. respectively. detennine the m:lxirnum shearing stress in each rod. Neglectlhe effect of tillels and of stress concel1lrations. 8 . 17 Detemline the smallest allowable diameter of the solid shaftABCD. knowing that Tal, = 60 MPa and that the radius of disk B is r = 80 mill.

c

__

ISH

Im,',

8 . 18 The 4-kN force is parallel 10 the.r axi s, and the force Q is parallel to the;: axis. The shaft AD is hollow. Kn owing Ihal the inner diameler is half the outer diameter and Ihat T>I' = 60 MPa, deu.:rmine th .. sm:lllesl permissible oUIer diameter of the shaft .

~T~ FIg. PB. 17

6O,,,m

flO "''''

140

7 in.

"n"

7 in.

71" . 4 in

7m Fig. PB.1B

8 .19

The two 500-lb forces are vertical :lnd the force P is parallel 10 the 101, = 8 ksi. determine the smallest permissible diameler of the solid shaft At.".

z axis. Knowing lhat

Fig. PS.19

8 .20 For the gcar-and-shaft systcm and loading of Prob. 8.19. determine the smallest pemlissible diameler of shaft A£. knowing that the shaft is hollow :lnd h:ls :In inner diameter tlwt is f the oUler diameter.

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8.21 Us ing the llotatloll of Sec. 8.3 and llegle~til1g the effect of shea!"ing stresses C"
Problem.

+ M:I' + (/11~ + M: + T'll """ = S(Ml / " . ,. -""'"

8.22 It was stated in Sec. 8.3 thut the shearing stresses produeed in >l slmft by the tmnsve[se IO tre ss in a given seclion occu rred at poll11 H (Fig. P8.22a) and was eq ual [0 the e.xpres,ion oblaint".d in Eg. (8.5). namely, T

H

= ~ VMl

J

+ T~

Show dlat die maximum sheuin~ ~tres~ ~' I poinl K (Fig. P8.22 b). of th .. .~ hear II is grt'ate~a. can Ix: expressed a..~

wher~

die effect

(I.)

where f3 is the angle between the vectors \' and M. It is clear that Ihe eJfecl of the shear V cannot be ignored when "TJ( ~ TIJ. (H illt· Only the component of M along V contributes to the shearing stress at K. )

Ag. PS.22

8.23 The solid shafts ABC and DEF and the gears shown are used to transmit 20 hp from the motor M to a machme tool cOl1tlccted 10 shaft VEF. Knowing that til" m010J rol1tes at 240 rpm ancl tll:lt T"'I = 7 .5 k ~ i. clelermine the smallest pemlissi ble diameter of (a) shafl ABC, (0) shaft DEF. 8.24

Sohe PIOU. 8.23, Cls>uming that the JIIOior

L\)tatl!'~


8.25 The solid shaft A.B rot
"'zs"","'-" 5\),",,, lOOmm

~J ') ,~

""11

Fig. PB.2S 8.26 gear G.

Sohe ?rob. lUS. ass uming that the enlire 20 I:.W is t::ken off at

Fig. PB.23

507

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508

Pr11'1C1p
8 .27 The solid shaft ABC and the gears sllO\\l1 an: USN to u nll~ ll1it l(] kW from the motor /It to a machine tool connected to g~ar D. Know ing that th e motor rotates at 240 rpm and thal 1.11 "'" 60 MPa. Jet~mline the s mallest permi~sible

diameter 01 shaft A.He.

8.28 Assuming that shaft ABC of £>rob. 8.27 is hollow and hilS iln outer diameter of 50 mm delemline the IilCgest permissible inner diameter of the shaft 8 .29

TI ll': SC11i rt ~h:, fI AFrnTare:s m 000 rpm and r r.1n~mi,,; fiO hp fTOm rlw

motor.lll to machine tools connected to gears G and H. Knowing that T.. = 8 ksi ,U1d that 40 hp is taken off at gear G and 20 hp is lllken off at gear H, determine lhe smlllle,r p.:rmissiblc dmmcrcr of shaft AE.

E

Fig. PS.2.7

Fig.

8.30

P8~

Solve Prob. 8.29. assum ing lhat :30 hp is taken off m gear G and

30 hp is taken off at A,..-:ar 11.

·8.4. STRESSES UNDER COMBINED LOADINGS

Fig. 8.15

In Chaps. I and 2 you learned to dete rmine the stresses caused by II cellt] ic axial load. III Chap. 3, you analyzed the dislLilJUtioll of .'i lJ"e,sc.o; in a cylindricaimt'mbcr subjected to a twisting couple . In Chap. 4, you determi ned the stresses caused by bending couples and , in Chap;. 5 and 6, the stresse~ produced by transverse loads. As you will see presen tl y. you ca.n combine the knowledge you have acquired \ 0 dete rmine the stresscs in s lender structural members o r mac hine component:;; undcr fair ly general loading conditions. Consider, for example. the bent member ABDE of circular cross section, th
for<"",oJptc sy~lcm

,hol>orlion 0/'/1<

m<'~lh<"F

mC

c~ n

/N::aJed to

u.s"

,h~

be defined

rig''''

!IS


,;,~

fOl"e .• acti,x " "

"f th," SO'·';0I1 <<<'C E""mple 8.0t).

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509

8.4. Stresses under Comb ined loadings

+ 'r

v (,, )

(IJ)

Fig. 8.17

Fig. 8.16

in the section and, in generaL consists of three force components and three couple vectors that will be assumed directed as shown (Fig. 8. 16). The force P is a centric axial force that produces normal stresses in the section. T he couple vectors My and 1\1 .. cause the me mber to bend and also produce normal stresses in the section. They have there fore been grouped with the fo rce P in part (j o f Fig. 8.17 and the sums u , of the normal stresses they produce at points Hand K have been shown in part a o f Fig. 8.18. These stresses can be determined as shown in Sec. 4. 14 . On the other hand , the twisting couple T and the shearing forces V, and V, produce shearing stresses ill the section. The sums 1." and 1 ,-, of the componenls of the shearing stresses they produce at poillts H and K have been shown in part b of Fig. 8.18 and can be detennined as indicated in Secs. 3.4 and 6.3. t The normal and shearing stresses shown in parts a ami /; of Fig. 8. 18 can now be combined and displayed at points H and K on the surface of the member (Fig. 8.19). The principal stresses and the o rientation of the principal planes at points Hand K ~a n be determined from the va lues of fT, . T .... and 'r." at each o f these points by one of the methods presellled in Chap. 7 (Fig. 8.20). The values of the maximum shearing stress at each of these points and the correspond ing planes can be found in 11 similar way. The results obtained in this section arc valid only 10 the extent Lhat the conditions of applicability of the superposition principle (Sec. 2. 12) and of Saint-Venant's principle (Sec. 2.17) are met. This means that the stresses involved must not exceed the proportional limit of the materi aL that the deformations due to one of the loadi ngs must not affect the determination of the stresses due to the others. and that the section used in you r analysis must not be 100 close to the points of application of the given forces. It is clear from the lirst o r thesc re(luiremcnts that the method presented here cannot be applied to plastic defonnatioliS.

H

(1)

tJ,

Fig. 8.19

Fig. 8.20

t N"()lC Ihal your pr?r.cm knowledge allows YOIl to drtrnnine Iht' effect of the twisting cou· ple T only in the cas~s of circul:u shaft,. of membe~ with a rectanguLar ~TOS1; sl'{;tiotl (Sec. 3.11). or ofthio -waLl~d hollow members (Sec. 3.13).

(I»

Fig. 8.18

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EXAMPLE 8.01 Two forces P , and P,. of magnitude ,>, = 15 kN and P! 18 kN. are applied as shown 10 the end A of bar AB. whiclt is weidel! to a cylindrical member 80 of mdius c = 10 m m (Fi g , 8.21). Knowing that the di~tatlCe from A to tlte a.xis of member fJD is {/ = 50 mm and ass wning that all stre sses remain below the proponional limi t of the material. Uctelllline ( u ) the HOIllml and shearin g ,tresses at poilll K of tlte transverse section of member BD locmed at a distance b "" 60 mm irom end B. (b) tlte principli axes and principal stresses ~ t K, (c) the maxtmum !hearing stress :\I K.

=

Fig. 8.21

Internal Forces in Given Section. We first replace the farces P and P~ by an equi valem system of force~ and roup les applied at the cemer C of the section comalning point K (Flf: ~ 12) Thi ~ ~ys tem . which re rre~e nl ~ the int e rnal fnrcf'J;

in the sec tion. consi.,ts of the following fOices and couples:

I. A centric axial force F cqual to thc lOfce. P ,. of magnitude Fig. 8.22

F = P t = 15kN 1. <\ s hearin g fnrc? V eqnal to th r. for':e p, . !If

'I

maenitl,, ~

3. !\ tWtstlllg couple T of torque J cqual to the mom ent of P! about the axi~ of m
=

Pp

=

( IR kN )(.50 nun) =900N · m

4. A bending couple M.." of moment M" equal to the moment of P , abou t a venical x< is throu gh C: M " = P ,a = ([j kN)(50 mm ) "" 750 N .

III

5 . /\ bending cou ple M o. of moment M , cqual to tlte moment ,~f p " about " t,.... n~"en:Ol, horizontal a>::;. throu!1-h C,

M,

The

re ~u l t s

=

P!J =(18kN)(6(lmm)

=

Geometr ic Properties of the Section.

We. have

A = 7Tf.l = 7T(0 020 m )! = 1. 2.'i7 X 111 - 3 m 1

=

=

1~ '" J,1Tt.-l '" h·(O.o20m)4 [25.7 X [0 ~ m-l Jr = ~ 1Tt4 = ~ 1r(o.mo m)" = 25 1.3 X 10 j 1114

ann I -

2c - 2(0.020 m) - 0.040 m

Normal Stresses. We observe lhat nonnal stre sses are proouced at K by the centric force F and tile bending w upl~ M),. but that the (·ouple M, doe s no t produce any s tre~ s at K. ~i nce K is located on lhe neutral axi s corre!;ponding to that couple. D~temlin ing each sign from Fig. 8.23. we wtite F

M \. £"

A

1\

- - + -~Iso delen11ine the first monlent Q and tilt" width I of the area of the cross section located above the z axis. Recalling that ~ = 4c/37T for a semicircle of radius L", we bave

\V.,

510

!~ ~ N

obtained are shown in Fig. 8.23.

8. Normal and Shearing Stresses at Point K. Each of the forces and couple s shown in Fig. 8.23 can prouuce a nOlma l 01 s healing stress ill point K. OUI puqtOse is to rompute sepaflltel y each of these stre sses. and then to
I,

' ·_

Fig. 8.23

1080N · m

=

- 11.9 M I ~..t

- 11.9MPa + 119.3M Pa IT, '"

+ 107.4 MPa

+

(750 N . m)(O.020 Ill) 125.7 x IO -~ m"

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Shearing Stresses . These consi ST of tJl<~ shearing mess (1,.,)v due to the vertical shear V and of Ihe shearing Slress ( T.,y)'~i" caused by the torque T. Recalling Ihe values oblained lor Q. r.l,. and l c- we wrile

Stre3""3 und .... Com:>IMd Loadrn9"

( 18 X IO"'N)(S.33 X 10 ' m~)

VQ

+1,1 "" +

(T, ),. =

SA.

(1 25.7 X 10

~ m' )(0 040 ml

- + 19. 1 MPa

Tr (1, ),""LO. = - J{. =

(900N . m)(O.020 m) 25 1.3 X ]()-~ m~

Adding Ihcse two expre,sions. we obtain

T•• = (T,y)., + (T,,.),,,,,,, = T .. = - 52.5 MPa

+ 19.1

T_,~

lij k'<

- 7L6MPa at point K.

t..-IPa - 71.6 MPa

Fig, 8.24

In Fig. 8.24. the normal stress if, and the shearing Stresses and

have been show n aCling 011 a square e lement located ai K on the su rface of the cylindrical membi!r. Note that ,hearing .tresse s acting on the lon gitudina l sidc5 of the element haye been included. T.,

b . Principal Planes and Principal Stresses at Point K. We can lise eith ~r of the two m~lhoJ~ of Chap_7 to determine the principal planes and principal StreSSes at K. Selecting Mohr's circle. we plot point X of coordinates rr, ~ 1- 107.4 MPa and T " - ! 52.5 MPaand poilll Yof coordi nates if. = 0 and + T•• = - 52,5 MPa and draw the circle of d iameter XY (Fig. 8.25). Observing that DC = CD

=0-

we deremline till1

~( 107.4 ) = 53.~ MPa

tl~

nrienmion nf 1he

B 0

,,

,/ r

Fig. 8.25

DX = 52.5 MPa

princip~l [llane ~~

20 -= DX = 52.5 = 0.97765 P CD 53.7 Qp -

We nuw

u ~l ~ll mne

22.2°

A

J 151:~

lhe r
R = \1(53.7)' - (52.5f

75 .1 MPa

and the principal stresses,

OC + R = 53.7 + 75. I = 128.8 MPa if,,,," = OC - R = 53.7 - 75,1 = - 2 1.4 MPl


IT,,,, = - 21

l,\ tP"

FIg. 8.26

0"",., ""

The results obtained are show n in Fig, 8.26.

c . Maximum Shearing Stress at Point K. This st re>s corresponds to points E and F in Fig. 8.25. We have T",,,, = CE = R = 75. 1 MPa 6 28~ = 90~ - 44.4" = 45.6 , W~ cnnc1une 1h~1 1he rlanc~ of mHimum ~hearine ,rr..,'~ ffll"m an angle tip = 12.8' ~ with the horizontal. The corres ponding el-

Obst":fvi n~

thaI 28, = 90 0

ement is shown in Fig. 8.27. Note that the normal stre.ses act· ing on this clemen1 arc TLl'rcscntcd by OC in Fi g. 8.25 and arc Ihus equal to + 53.7 MPa.

Fig. 8.27

I:2H, .l t P'1

511

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-I.Si".

SAMPLE PROBLEM 8.4 A horizontal 500-lb force acts at point D of crankshaft AB which is held in static equilibrium by a twisting couple T and by reactions at A and B. Knowing that the bearings are self-aligning and exert no couples on the shaft. determine the normal and shearing stresses at points H. 1. K. and L located at the ends of the venical and horizontal diameters of a transverse section located 2.5 in. to the left of bearing B.

SOLUTION Free Body. Enlire Crankshal\.

+T=

- (500 Ib)( 1.8 in.)

T

250 Ib

A = B =

0

T = 9OOlb · in.

IntrnJai Forces in Transl'rrsc St'Ction. We replace the reaction B and the twisting couple T by an equivalent force-couple system at the center C of the transverse section containing H. 1. K. and L.

V = B = 250lb

T = 900lb· in.

M . = (250 Ib)(2.5 in.) = 625 lb · in.

£

The geometric propenies of the 0.9-in.-diameter section are 1 = ! 7T(OA 5 in . )~ = 32.2 x IW) in 4

A = 7T(0.45 inJ = 0.636 in'

J = ~1T(OA5 in. )4 = 64A X ]0 l in~

G

T

62\)01'''

=

StresSl'S Produced by Twisting Cou[lic T. Using Eq. (3.8), we determine the shearing stresses at points H. 1. K. and L and show them in Fig. (a). Tc

r

= 62\)0

T= p
J

=

T ~

5241's;

Q=

G

2 1TC ) ( : : )

VQ ".=-= II '" X7J(I P"

It

~t! = ~(0.45 inY

(250 Ib)(60.7 X 10 (32.2

IM"k

IT

I. T

512

=

x

10

1

J

in 3)

in')(0.9 in. )

1

= 60.7 X 10- in

l

524 psi

Stresses Produced by Ihe Bending Couple 1\1,. Since the bending couple M) acts in a horizontal plane. it produces no stresses at Hand K. Using Eq. (4. 15). we determine the normal stresses at points 1 and L and show them in Fig. (d.

(,)

= fi2YO psi

6290 psi

Strc..'ises Produerd by Shearing Force V. The shearing force V pro· duces no shearing stresses at points 1 and L. At points H and K we first compute Q for a semicircle about a venical diameter and then determine the ~hear­ ing stress produced by the shear force V = 250 lb. These stresses are shown in Fig. (b).

L

r

(900 Ib . in.)(OA5 in.) 64.4 X 10 J in~

=

=

"',30 psi

&iWl',i

u ~ ~/~ =

(625 lb· in.)(0.45 in. ) 32.2 X 10- 3 in~

8730 psi

Sununary. We add the stresses shown and obtain the total nonnal and shearing stresses at points H . 1. K. and L.

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SAMPLE PROBLEM 8.5 130

",,,, - 1-..--'1

Three forces are applied as shown at points A. B. and D of a shon steel post. Knowing that the horizontal cross section of the post is a 40 x 140-mm rectangle. determine the principal stresses. principal planes and maximum shearing stress at point H.

B

A. ,"

SOLUTION [nterna l Forces ill Section EFG. We replace the three applied forces by an equivalent force-couple system at the center C of the rcctangular section EFG. We have V~ = - 30kN P = 50kN V, = - 75kN M , = (50 kN )(O.130 m) - (75 kN)(0.200 m) = - 8.5 kN . m

M. = 0

M, = (30kN)(0.IOOm) = 3kN· m

We note that there is no twisting couple about the y a:
I, = h (0.040m)(0.140m)J = 9.15 X 10 - b m4 ( = M O.140 m)(0.040 m» = 0.747 X 10 b m~

" = 0.020 "r'+,:~","G'------T

~ yb =

---

1

OJ1ZSm 0.140",

j

M~~3kN

' I-W

F

Nor mal St ress a t H . We note that normal stresses U'\ are produced by the centric force P and by the bending couples M , and M,. We detennine the sign of each stress by carefully examining the sketch of the force-couple system at C.

(W.JO", tT , =

I=OJI40", t---j

"m5ml ~' .-11 T~, ~ (_, . ~ I =(LO.J __ ,.~ ",

O. L,

"'r

c.-

", I.,--7',,,

d Ml'a)) ~~ = 6/l() ~IPa

10 1

33.0

-. =

l'

o

T~,

c

B

_

_

(T, ..

I, 50 kN

I~

+ (3 kN . m)(0.020 m)

(8.5 kN . m)(0.025 m)

5.6 X IO - l m2 0.747 X 10 b m 4 = 8.93 MPa + 80.3 MPa - 23.2 MPa

,

li. 52 :\\P"

Q "" A1 YI = [(0.040 m)(0.045 m)J(0.0475 m) V,Q (75 kN)(85.5 X 10 b ml ) 1' ,! = = (9.15 X 10- 6 m4)(O.040 m)

Ii

b m~

9.15X 10 (F~

= 66.0 MPa <J

/)

,\

tT ( ~IP,, )

-'JI ~o/~ ~

tan 20r =

R

= V(33.of

X 10- 6 ml

1'"

= 17.52 M Pa

<:J

17.52 no

+ ( 17.52f = 37.4 MPa

_: : -

I ",."

= 85.5

P rinci pal S tn:sscs, Pr incipal I' la oc.'i, a od MaximulII Shearing Sl n.'ss at H_ We draw Mohr's circle for the stresses at point H

I ?/J

z

A

Shea rin g Stress at H . Considering first the shearing force V.,. we note that Q = () with respect to the z axis, since H is on the edge of the cross section. Thus V, produces no shearing stress at H. The shearing force Y, does produce a shearing ~tress :lI H and we write

V

33.0

tT ,.

/' +IM,I" 1M,I" +-- - -

+R

+ 37.4

U'," ..

= OA = OC

(F." .

= OB = DC - R = 33.0 - 37.4

= no

1',,>4).

= n.4 MPa ...

tT, 1W, = tT"".

70.4 Mf'u ...

= -7.4 MP;1 ...

513

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PROBLEMS 100

""L

"

100"",[ /J

I 0.75 III

"I----j 150 mill

n\

~~ "t

8 .31

The cmllilevl.,..

b~um

AB has a

r"'~langulur

.'ross

~cliQn

of

150 X 200 mm. KnowinR Ihal the tension in Ill(' cable BD is 10.4 kN and neglecting the weight oflhe beam. d~(cnnine Ihe normal an d shearing stresses at the three pOIntS intiicmed.

J" ,~:

8 .32 Two ].2_kip forces are applied to an L_maped machine element AB as s ho\\"n. Determine the normal and shearing stresses at (lI) point (I . (b) point h, (.:;) pomt c.

L I'""-'' '--f''-" '" 0.3 m 0.6 In Fig. P8.31

D~"·

I-I 1.0 iu

FIg. P8.32 and P8.33 8 .33

Two 1.2·kip fun.:o::s are
All ~< .hm... " 1)'>'Prminl' rh .. n,-,nn,,1 "nti .h,,;,",,!,; (b ) point 1'. (el point !

~ rr' .. ,
fll
demen t

,,' tn) pnin, tI

8 .34 t h rough 8 .36 Member AD has a llnifoml rectangular cros~ section of 10 X 24 mill. For the loading shown, determ ine the nonml and shearing str?ss at (a) point H, (b) JXlint K.

Fig. PS.34

514

Fig. PB.35

Fig. PS. 3S

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51 5

Problems

8.37 The ade of a small truck is acted upon by the forces and couple shown. Knowing that the diameter of the axle is 1.42 in .. detennine the normal and shearing stresses at point H located on the top of the axle.

10 in.

!! in .

Fig. PB.37

8.38 A thin strap is wrapped [lround a solid rod of radius c = 20 mm as shown. Knowing that 1= 100 mm [lnd F = 5 kN. determine the normal and shearing stresses at (a) point H. (b ) point K. 8.39

Scvcral

rvn:c~

Fig. PB.as

an:: applied tv the pipe assembly shuwn. Knvwing

lhatlhe pipe has inner and oUier dimneters equal \0 1.6! and 1.90 in .. respectively. delermine the nomwl and shearing stresses at (a) point H. (b) point K.

200 III 45mm J5{J

lh

121)1) ~

Fig. PB.39

8.40 Two forces are [lpplied \0 the pipe AS as shown. Knowing that the pipe has inlier and oUier diameters equal to 35 and 42 mm. respectively. determine the nonnal and shearing stresses at (a) point a. (b) point h.

751111ll

211 "''''

Fig. PB.40

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516

Principal Stresses under a GIYen Loading

8 .41 A IO-kN force and a I A-kN . m couple are applied at the top of the 65-mm diameter cast-iron post shown. Determine the principal stresses and m;l ximu m shearing stress at (a) poim H. (t) painl K. 8 .42 Three forces are applied to a 4-in.-diameter plate that is attachaJ to the solid 1.8-in. diameter sh;lft AB. At poim H. determine (a) the principal stresses and principal planes. (b) the maximum shearing stress.

240,,,,,,

~ Fig. PS.41

1 in

Fig. PS.42

Fig. PB.43

8.43 Forces are applied :.t points A and B of the solid cast-iron bracket shown. Knowing that the bracket has a diaoeler of 0.8 in .. detemline the principal siresses and the maximum shearing stress (a) poim H. (b) point K . 8 .44 The steel pipe AD has a 72-mm outer diameter and a 5-mm wall thickness. Knowing that the ann CDE is rigidly attached to the pipe. determine the principal stresses. principal planes. and the maximum shearing stress at point H .

:3 l.'l C ,

<

120 nllll

Fig. PS.44

t

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8.45 Three forces are applied to the bar shown. Determine the nomlal and shearing stresses at (a) point (I. (b) point b. (e) point C. 8.46

517

Problems

Solve Prob. 8.45 . assuming that 11 = 12 in. 2.4 ill.

8.47 Three forces are applied 10 the bar shown. Determine the normal and shearing stresses at (a ) point a. (b) point b, (d point c.

.2 in .

16",,,, J.S in

Fig. PB.45

Fig. PB.47

8 .48 Solve Prob. 8.47. assuming that the 7S0-N force is directed venically upward. 8 .49 Three forces are applied to the machine componentABD as shown. Knowing that the ~ross section containing point H is a 20 X 40-mm rectangle, detemline the principal stresses and the maximum shearing stress at point H.

!I

2.3

50 "''''

t'.,

Fig. P8.49

8.50 Solve Prob. 8.49. assuming thm Ihe magnitude of the 2.S-kN force is increased to 10 kN.

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518

Principal Stresses under a GIYen Loading

8 .51 Three forces are appl ied to the camilever beam shown. Determine the principal stresses and the maximum shearing stress at point H.

Fig. PB.51

8 .52 For the beam and loading of " rob. 8.5 1, determine the principal stresses and maximum shearing stress CIt poim K. 8.53 Three steel plates. each 13 mm thick. are welded together to form a cantilever beam. For the loading shown. dt":lermine the normal and shearing stresses at points (/ and b.

"I ISO """ 1=13",,"

IlkN Fig. PB.S3 and PB.54

8.54 Three steel plates. each 13 mm thick. arc welded together to form a cantilever beam. For the loading shown. determine the normal and shearing stresses at points d and I!.

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B.S5

Two forces P J

~nd

P 2 ale applied

a5

shown in dila:-liullS perpen-

Problem.

dicular to the longitudinal axis of a W 12 X 40 beam. Knowing that P, = 5 kips and P1 = 3 kips. determine the principal stresses and the ma:dmum shearing stress al point II

"t:.

~ / ,,-:t5.?

j

P,

~

\\'12 X

to

Fig. Pi .55 a nd PS.56

8 .56 Two forces P, nnd P1 are applied as stlOwn in direclions J=erpendiclltar 10 the longittx:linal axis of a WI2 x 40 beam_ Knowing tbal PI = 5 L:.ip~ and 1' 1 = 3 kips. determine the principal siresses and the mnximum shearing stress m point b

8 .57 Four forces nre app lied to a W2()() x 41.7 rolled beam as shown. Determitte the prineipnl stresses and maximl.lm shearing stress at point a.

'I

L , ~

l75,,,,,, -I

Fi g. P8.57 ilInd PB.58

8 .58 Dct~rmil1e

8 .59

FOl.lr forces nre applied 10 a W200 x 41.7 rolled beam ns ;;hown. the principal stresses and m"xinmlll shearing str.>ss at point b. A force P is applied to a call1ilever beam by rnean~ of a cable at·

to a bolt located al ttte center of the free end of the beam. Knowing lhal P acl~ in a direction p€rpendicuhr 10 the longi tud inal axi, of Ihe beam. detennine (01 tTh: nonml stress at poinl (/ in terms of P. b, II, I, tmd (3. (bl the vnlll:S of f3 ror which the normul stress at n is zero. tach~d

Fig. PS.S9

519

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520

Principal Stresses under a Given Loading

8 .60 A vertical force P is applied at the center of the free end of cantilever beam AB. (a) If the beam is installed with the web vertical (13 = 0) and with ils longitudinal axis AB horizontal. determine the magnitude of the force P for which the normal stress at point a is + 120 MPa. (b) Solve part a. assum· ing tha t the beam is installed with 13 = 3°.

Fig. PB.60

Fig. PB.61

' 8.61

A S-kN force P is applied to a wire that is wrapped around bar

AB as shown. Knowing that the cross seclion of the bar is a square of side d = 40 mm. detem1ine the principal stresses and the maximum shearing stress at point (/.

!!in.

l=:

..:::II . . . . . 0. 15 in.

10 in ..........

Y kip>

Fig. PB.62

' 8 .62 Knnwine lhallhe ~1T1IClllnl llllhl' ~ hown h.1~ ~ Ilnifonll W.1 111hid:. ness of 0.3 in .. determine the principal stresses. principal planes. and maximum shearing stress al (a) point H, (h) point K.

*8.63 TIle s tructural tube s hown has II unifonn wall th ickness of 0.] in. Knowing that the I S-kip load is applied 0.1 S in. above the base of the tube. detennine the shearing stress at (al point u. (h) point b.

Fig. PB.63

'8.64 For the tu be and loading of Prob. 8.63. determine the principal stresses and the maximum shearing stress at point b.

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REVIEW AND SUMMARY FOR CHAPTER 8

This chapter was devoted to the determination of the principal stresses in beams. transmission shafts. and bOOics of arbitrary shape subjected to combi ned loading~. We first recalled in Sec. 8.2 the two fundamental relations derived in Chaps. 5 and 6 for the normal stress a , and the shearing stress 7". at allY given point o f a ClOSS section o f a f'rismutic \.Jeam, My VQ u , =- ~ (8.1.8.2) 1 h

where V = shear in the section M == bending moment in the section )' = distance of the point from the neutral surface J .= centroidal moment of ineniu of the cross sect ion Q = first mumcllI about Lhl: neutral
Using one of the methods presented in Chap. 7 for the transformation of str~sses. we were able to obtain the principal planes and pnnclpal stresses at the gIven poml (I--Ig. ~.6). We investigated the distribution o f the principal stresses in a narrow rectangular cantilever beam subjected to a conc~ntrated load P at its free end and found thai in any given transverse scctiol1-

Principal pl.<JnP.!\ ann prinr:iJ1al stresses In a beam

0;"

.. lJ"" ..... did not I'xceed lhl' m,uimum normal ~Ire~~ lJ""", occurring at The surface of The beam. While this conclusion remains valid for many beams of nonrectangular cross section. it may not hold for W-beams or S-bcams, where a ....., at the junctions band d of The \veb with the nanges of Ihe beam (Fig. 8.10) may exceed the value o f am occurring at points a and e. Therefore. the des i!!n of a rolled-steel beam should include the computation of the maxi~lUm principal stress:u these poillls. (See Sample Probs. 8.1 and 8.2.) principal

~t r ;:!O

-~-

"',

C~-" "'."-' t--.

excepl duse 10 the point of applil:lllion o r the load- the mltximullI

Of-+~X-:.-.-+-,-1". , Fi g. 8.6

,"

Fi g. 8. 10

521

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522

Principal Stresses under a Given Loading

Design of transmission shafts under transverse loads

In Sec. 8.3, we considered the design of trall.I"1/1issioll sluift.f subjected to trall.fl'crse /OUd.f as well as 10 torques. Taking into account the effect of both the normal stresses due to the bending moment M and the shearing stresses due to the torque T in any given transverse section of a cylindrical shaft (either solid or hollow). we found that the minimum allowable value of the ralio Jlc for the cross section was j

c

Stresses under general loading conditions

C/M'

+ T')~

(8.6)

In preceding chaplers, you learned to determine the stresses in prismatic members caused by axial loadings (Chaps. I and 2), torsion (Chap. 3). bending (Chap. 4). and transverse loadings (Chaps. 5 and 6). In the second pan of this chapter (Sec. 8.4). we combined this knowledge to determine stresses under more general loading condi tions.

E

Fig. S.IS

Fig.S.16

For instance. to determine the stresses at point H or K of the bem member shown in Fi g. 8.15. we passed a section Ihrough these points and replaced the applied loads by an equivalent force-couple system at the centroid C of the section (Fig. 8.16). The normal and shearing stresses produced al H or K by eaeh of the forces and couples applied at C were determined and then combined to obtain the resulting normal stress (T" and Ihe resulting shearing stresses T...,. and T,o at H or K. Finally, the principal ~t resses. the orientation o"f the principal planes. and the maximum shearing stress at point H or K were determined by one of the methods presented in Chap. 7 from the values obtained for (T." T ;ry, and T x '

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REVIEW PROBLEMS

8.65 (a) Knowing that !Toll = 24 ksi and T all = [4.5 ksi. select the most economical wide· flan ge shape that should be used to support the loading shown. (b) Determine the values to be expected for if",. 1'",. and the principal st ress {ton» at the junction of a flange and the web of the selected beam.

B

C

A

D

~6nlnLl2n--J Fig. PB.65

B.66 The vertical force P I and the horizontal force P! are applied as shown (0 disks we lded 10 the solid shafl AD. Kn owing Ih:1I the diameter of the shaft is 40 mm and thal Tol l = 55 MPa. determine the largesl pemlissible magnitude of the force P~.

Fig. PB.66

8.67 The solid shaft AB ro(al<~S at 360 rpm and transmiTs 20 kW from the motor M \0 machine tools co nn ected \0 gears £ and F. Kn owi ng that T:oJI = 45 MPa and assuming lhal [0 kW is laken off at c3c h gear. determine th e smalles t perITllssible diameter of shaft AB.

Fig. P8.67

523

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524

8.G8 FOI the: bracke t ami lomllng SI ,()Wll. detel'lnine. the normal anu shem · ing stresses al (a) poiltl lI. (b ) poinl b.

Prtnclpol 51rn3",,9 undo. II G:van LDIIDlng

", 3 ft IUI~ l lb

HI Fig. PB.68

8 .69

Fig. PB.69

~ 175

The billboard shown weighs 8()()(llb and

i~

slippoited by a

~ Iruc·

[ura l tube that has :l 15. in. outer diameter and a O.S.in. W:lU thicl:.nesg. At a time when the resultant of the wind pressure is 3 k.ip~ located 3tthe center C of the billboard. deterlllin~ the normal and shearing stresses at point H.

I

8.70 Several forces are applkd to the pipe assembly sito',',n. Knowing that each secti on of pipe ha~ inner and oUler diameters equal to 36 and 42 mm. respectively. d:!tennine Ihe nomlal and shearing stresses at poilll H located at the top of lhe OU ler surface of Ihe pipe.

-.... ..

8.71 A dose· coiled spring is made of a circular wire of radius r tlt;.lt i.1 fUJ1Ilt:<.i intu a hd i\ uf r.uliu s R. Dele J1niJl ~ the maxim um s hearing stres~ pm· nll~en hy the two 1''l1 1 ~ 1 ann npJ1
15(] ~

I'

Di",,.nsi,Xl.' in "lin

Fig. PS.70

,

25"'' 'l

I'

A

E

Fig. PB.71

8 .72 .'\ verti cal force P of magnitude 250 N is applied [0 the crank at point A_ Knowing thm th l' ~h:1ft RnF h a_~ ~ n;~mf'J f'r of 1R mm . nm~r",ine the

principal stresses and (he maximum shearing stress at pcint H located at the Fi g. PS.72

top of the shaft. 50 mm to the right of suppol1 D.

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8.73 A 2 8-kip forc e IS llpplied as showil to the 2A·ill.-dialllt:tel casliron post A BD. At point H. determine (a ) Ih.e principal s tresses and prmcipal planes. (b) th.e maximum shearing stress.

RevIew Problem.

,

Fig. PB.73

8.74 For the post and loading showli. determine the principal stresses. principal planes. and maximum shearing stress at point H.

Fig. PB. 74

8.75 Knowing lhat the structural tube shown has a uniform wall thickness of 0.25 in .. detemline Ihe no rmal llnd shellJing stresses at the three points indicated.

~~ (lOU III

..,.,.,

f,(.olb

-...........~ !500 h

_. am

1

!,"'~ l lh

2.;'5 in

, FIg. PB.7S

8.76 The cantile\er beam AB will be installed so Ihal the 6O-mm side fonns an angle (3 belween 0 ,md 90" with [he verticnl. Knowing thm lhe 6OO·N 'Ienical force is applied at lhe cemer of the free end of the beam. delermine Ih~ nnrm~l ~tres~ ~t point (/ whf'l1 (11) (3 = fl , (h ) (3 = ()(}O (r) AI~(l . rl~t~rmin~ th e '/ahl~ of /3 for which the. nonn~1 slIess at point /I is a max imum and the corre ~PQnding v:ll ue o f Ihm strt';&

Fig. P8.76

525

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COMPUTER PROBLEMS

The followin g problems are dl'signcd to tK' solved with

Il

computer.

S.C 1 Let us assume that the shear \' and the bending moment M have been determined in a gi\'en section of a rolled-steel beam. Write a computer program \0 calculate in that seclion. from the data available in Appendix C. (£I) the maximum normal stress if.. . (b) the principal stress 0"",., at the junction of a flange and the web. Use this program 10 solve parts a and b of the following problems: (1) Prob. 8.1 (Use V = 45 kips and M = 450 kip ' in.) (2) Prob. 8.2 (Use V = 22.5 kips and M = 450 kip' in.) (3) Prob. 8.3 (Use V = 700 kN and M = 1750 kN . m) (4) Prob. 8.4 (U se V = 850 kN and M = 1700 kN . m) 8.C2 A cantilever beam AB wilh :l roctangular cross section of width IJ :U1d depth 2(" supports a single concentrated load P at its end A . Write a computer program to calculate, for any val ues of xlc and .vIc, (1I) the ratios u m.,lu", ,md " ml,/"",' where u ',,,",, and u, ... are the plincipal stresses at point K (.• , y ) and (T,. th e maximum nonnal stress in the same Lransverse section, (b) the angle 8 , 1 that the principal planes at K form with a wUlsverse and a horizontal pl,me through K. Use thi s program to check the values show n ;n Fig. 8.8 and to verify that 0-".,. exceeds (T.. if _t' $ 0.544(", a si ntlic~ted in !lIe second fooUlOte on p
Fig, PB.C2

8.C3 Disks D 1• D2 , ••• • D. are attached as shown in Fig. S.C3 to the solid shaft All of length L. uniform dillineter d. and ~llowabl e. shearing stress 1">ll ' Forces Pl . P2' . ..• p. of known magnitude lexcept for one. of them)
526

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Computer Problem,

, Fig. P8.C3

8 .C4 Thesoljd shaft AB of length L. uniform diarnt'ler d. and allowable shearing Slress 1. 11 rotates at a given speed expressed ill rpm (Fig. 8.C4). Gears G I • G,. , . . . G. are allache<.i to the shah amI each of tht'se gears meshes wIth another gear lnol ~howl\). eahe r at the lOp or bonom of its venical diameter. or at Ihe left or right end of ils horizotllal diameler. One of these gears is connected 10 a mo\()r and the rest or" them 10 various machine look Denoting by r, the radius of disk Gi • by c, liS dislance from Ille support .. t A. and by 1-'/ the power trammillcd 10 lhat gear ( + sign) or taken of that gf'ar (- sign). write a compu ter program 10 calculate the smallesl permissible value of the dbmeter d of shaft AB. L"sc thi~ program to solve Probs. 8.25 and 827.

Fi g . PH.C4

527

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528

Principal Stresses under a Given Loading

S.C5 Write a computer program that can be used to cakulate the normal and shearing stresses at points with given coordinates y and z loc~ted on the surface of a machine part having a reclangular cross section. The il1lemal forces are known to be equivalent to Ihe force-couple system shown. Write the progr~m so that the loads Clnd dimensions can be expressed in e ither Sl or US. customary units. Use this program to solve (a) !'rob. 8ASh. (b) Prob. 8.47a .

"I

/

I

'' A ~I". I 'il

t,~\[ yJ P "" / "~ Fig. P8.C5

D.C6 Member AB has a rectangular cross section of 10 X 24 mill. For the loading shown. write a compuler progr~m that can be used to detennine the nonnal and shearing stresses at points Hand K for values of d from 0 to 120 mm. using IS-mm increments. Use lhis program 10 solve Prob. 8.3S.

,

Fig. P8.C6

,

Fig. P8.C7

"S.C7 The structural tube shown has a unifonn wall thickness of 0.3 in. A 9-kip force is applied at a bar {not shown) that is welded to the end of the tube. Write a computer program that can be used to determine, for any given value of c, the principal stresses. principal planes. and maximum shearing stress at point H for values of d from - 3 in. to] in .. using one-inch increments. Use this program to solve Prob. 8.62(1.

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Deflection of Beams

'.

The photo shows a multiple-girder bridge during construction. The design of the steel girders is based on both strength considerations and deflection evaluations.

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530

Deflection 01 Beams

9.1. INTRODUCTION Ln the precedi ng chapler we learned to design beams for strength. In this chapter we will be concerned with anot her a;:;pect in the design of beams, namely, the detenninalion of the defleclion. Of particular intcrest is the detenninmion of the l/1aximwlI (iej7ecliol! or a beam under a given loading, si nce the de;:;ign spccificat ions of a beam will generally includc a maximum allowable va lue for il ~ deflection. Also of interest is that a knowledge of the de fl ections is required 10 analyze illdefenni!late beall/s. These ;'Ire beams in whic h the number of reactions at the supports exceeds the number of equilibrium Ctl uations available 10 detenlline these unknowns. We saw in Sec. 4.4 that a prismatic beam subjected 10 pure bending is bent into an arc of ci rcle. and that. within the. clastic range, the curvature of the neutral surface can be cltpressed as M p

EI

(4.2 1)

where M is Ihe bending momcnt, E the modulus of e lastici ty, Ulld i!.he moment of inertia of the cross ...ection about its neutral axi.... Whcn a beam is ;:;ubjccted to a transverse loading. Eq. (4.21) remains valid fo r any given transverse sect ion, providcd that SainiVenant's principle applics. However. both !.he bendi ng moment and !.he curvature of thc neutral surface will vary from section to section. Denoting by x the diiHance of the section from thc IcJt cnd of the beam. we Write

_ ~ M(x) p EI

(9. 1)

"I

A~~BX

(b)

Fig. 9.1

Simp!)' "HI'P'JI'ipd b.,,,,,,

The knowledge of the curvature at various points of the beam will e nable us 10 draw some ge neral conclusions regarding the deformation of the beam under loading (Sec . 9.2). To determine. the slope. and deflection of the beam at any given point. we fi rst derive the following second-order linear differential equation. which governs the elm·ric cun'e characterizi ng the shape of lhe defonned beam (Sec. 9.3): (f y

M (x)

dx l

EJ

If the bend ing moment can be represented for all values of x by a single function M (x). as in the case of the beams and loadings shown in Fig. 9.1 , the slope (J = dy/dr and the deflection ), al any point of !.he beam may be obtained thro ugh two successive integrations. The two constants of integration introduced in lhe process will be de temlined from the boundary c0nditions indicated in the figure. However. if different analytical functions are required 10 represenl the bending moment in various portio ns of the beam, different differ-

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ential equations will also be required. leading to different functions defining the cla.tic curve in the various portions of the beam. In the case of the beam and loading of Fi g. 9.2, for example, two differential equ:uions are required, one for the portion of beam AD and the other for the portion DB. The lirst equation yields the fUJlctions 0 1 and )'1. and the second the functions O~ and )"2. Altogether. four constants of integration must be determined: two will be obtained by writing thaI the dclleclion is zero at A and B, and the other two by expressing thaI the portions of beam AD and DB have the same slope and the same dellecLion at D. You will observe in Sec. 9.4 that in the case of a beam supporting a distributed load 'W(x), the clastic curve can be obtained directly from ·w (x) through four successive integrations. The const;Ults introduced in this process wil l be determined from the boundary values of V, M, O. and y . In Sec. 9.5, we will discuss statically illdetel1l1inllfl! bea/lls where the reactions at the suppurts involve four or more unknowns. The three equilibrium equJ.tions must be supplemented with c£luations obtained rrom the boundary conditions imposed by the supports. The method described earlier for the determination of the elastic curve when several func tions arc re(luired to represent the bending moment M can be quite laborious. since it requires matching slopes and deflections at every transition point. YOll will see in Sec. 9.6 thaI the usc of .,·iugular;t)' fUllction .l· (previously discussed in Sec. 5.5) considerab ly simplifies the dctermination of 0 and y at any point of the hemn. The next part of the chapter (Secs. 9.7 and 9.8) is devoted to the method of .mperpositioll, which consists of dctermining separatcly, and then adding. the slope and deflection caused by the various loads applied to a beam. This procedure can be facilitated by the use of the table in Appendix D, which gives the slopes and deflections of beams fo r various loadings and types of support. In Sec. 9.9, certain geometric properties of the elastic curve will be used to dctermine the dellcction and slope of a beam at a givcn point. Instead of expressing the bcnding momcnt as a function M(x) and intcgrating this fun ction analytically, thc diagram representing the vari ation of M/E/ over the length of the beam will be drawn and two momentarea theorems will be dcrived. The fir.l·t moment·area theorem will enable us to calcu late the angle between the tangents to the beam at two points: the .1·eClmd momem·area theorem will be used to calculate the vcnical distance from a point on the beam to a tangent at a second point. The moment-area theorems will be used in Sec. 9.10 to dctermine the slope and dellection at selected points of cantilever beams and beams with symmetric loadings. In Sec. 9.1 I you will find that in many cases the areas and moments of areas ddined by the M/E/ diagram may be more casily determined if you draw the bending-mo/llem diagram by part.v. As you study the moment-area method. you will observe that this method is particularly elTective in the case of beam.v of variable emu .1·ectiOlI.

9.1. Introduction

0'

ti l ~ IIJ

II ~ ~L Fig. 9.2

'I! =

'!1)

531

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532

Beams with unsymmetric loadings and overhanging beams will be considered in Sec . 9.1 2. Since for an l1nsymmetric loading the maximum dellection docs not occur at the center of a beam, you will learn in Sec. 9.1 3 how to locate the point where the tangent is horizontal in order to determine the IIwxillllllll deflection. Section 9.14 wi ll be devoted to the solution ()f problems inv()lvi ng ,\·tatiClilly indetermillate

Doflectlon 01 Beams

iJelims. 9.2. DEFORMATION OF A BEAM UNDER TRANSVERSE LOADING

At the beginning of this chapter. we recalled Eq . (4.2 1) of Sec. 4.4, which rdates the clITvature of the ncutral surface and the bcnding moment in a beam in pure bending. We pointed out that this equati on re~ mains valid for any given transverse section of a beam subjected to a transverse loading, provided that Saint-Venant"s principle applies. However. both the bending moment and the curvature of the neutral surface will vary from section to section. Denoti ng by x the distance of the section from the left end of the beam. we write

.' B

"

(9. 1)

.. (II)

Consider. for example, a cantilever beam AB of length L subjected to a concentrated load I' at its free end A (Fig. 9.3a). We have M(x) = - Px and, subs tituting intu (9 . 1) .

•

Px

H

p

A

P. ,'" -,

", Fig. 9.3

EI

which shows that the curvature of the nelltral surface varies linearly with x, from zero at A. where PA itselfis infinite. to - PL/ EJ at B , where IPnl = EJ/ PL (Fig. 9.3b). Consider now the overhanging beam AD of Fig. 9.40 that supports two concentrated loads as shown. From the free- body diagram of the beam (Fig . 9.4b). we lind that the reactions at the supports are RA = I kN and Rc = 5 kN, respectively, and draw the corresponding bending-moment diagram (Fi g. 9.5a ). We note from the diagram that M , and thus the c urvature of the beam. are boLh zero at each end of the beam , and al so at a point E located at x = 4 ITI . Between A and E the bending mome nt is positive and the beam is concave upward ; between

HN

.jJ,;X

HN 3 111

D

C

Fig. 9.4

'0'

D

"

'"

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.If

9.3. Equation of the Elastic Curve

1kN .2k"l

I \.:~

" 4m~

£

~6 J,.N .

Iffl

D

C

~ '"

I"

C

~

(Ii)

a::;t

E

Fig. 9.5

E and D the bending moment is negati ve and the beam is concave downward (Fig. 9.5h). We also note that the largest value of the curvature (i.e., the smallest value of the radius or curvature) occurs at the suppon C, where IMI is maximum. From the information obtained 011 its curvature, we get a fairly good idea of the shape of the defonned beam. However, the analysis and design of a beam usuall y require more precise information on the dej1ectio/l and the .1·lope of the beam at various points. or particular imporlance is the knowledge or the maximum llej1ecfio/l of the beam. In the next section Eq . (9.1) will be used to obtain a relation between the

!J

dellection y measured at a given point Q on the axi s of the beam and the distance x of that point from some fixed origin (Fig. 9.6). T he relation obtai ned is the equation of the eia.l·ric CIIrve, i.e., the equation of the curve into which the axis of the beam is transformed under the given loading (F ig. 9.6b).t

""!~~~ijilE"!~~!I!"':J ...

9.3. EQUATION OF THE ELASTIC CURVE

Fig. 9.6

We first recall from elementary calculus that the curvature o f a plane c urve al a point Q(x. y) o f the cu rve can be expressed as d 2)"

dx 1 p

[I + (;;)'r

(9.2)

1

where dyldx and ,Pyld.r arc the first and second derivatives of the fUllcti on y(x) represented by that curve. But. in the case of Ihe elastic curve of a beam. the slope dyldx is very small. and its square is negligible compared 10 unit y. We write , therefore. -

=

(Py

p

(9.3)

Substituting for l i p from (9.3) into (9.1). we have

d'y dx 1 =

M(x)

E1

(9.4)

The equation obtained is a semnd-order linear differential equation : it is the governing differential equation for the clastic curve. t it should be noted !hat. in Ihi.1 chapter.)' reprCStnL' a vcnical di.~placcmcm. while tt was used in prcviou.1 chapters 10 rcprcoent the distance of a glvcn point in J trJns\"crxc section from the neutral axis of that $Celion,

533

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534

The product EI is known as the flexuml rigidity and, if it varies along the beam. as in the case o f a beam of varyi ng depth. we must express it as a function of x before proceedillg to integrate E.l . (9.4). However. in the case of a pri smatic beam, which is the case considered here, the tlexural rigidity is constalli. We may thus multi ply both members of Eq. (8.4) by El and integrate in x. We write

Doflectlon 01 Booms

dy El-

i'

=

d.

'1

O "I~;:---"-'''-'I'oj-_-_-'_~_:O",,~ x

,

M(x) (Ix + C 1

(9.5)

where C[ is a constant o f integration. Delloting by 8(x) the angle, measured in radians, that ihe tangent to the elaStic eurve at Q fonns with the horizontal (Fig. 9.7), and recalling thm this angle is very small, we have

Fig. 9.7

dy dx = tall H = H(x)

Thus. we write Eq. (9.5) in the alternative foml

El0(x)

I'

=

M(x) dx

+ C1

(9.5')

"

Integrat ing both members or Eq. (9.5) in x, we have EJ y ==

El y

,I

=

r[{ " ,

rr

~

J ~n=() 1

{II) S;"'plys "I'Pllrl.i\.j

1""""

'I A~~B~

p -

(b) o.·l · rh n "~"~ be"'"

,>

Ii

(.,1 Can lil" "n lwalll

Fig. 9.8 Boundary conditions for staticalty determinate beams.

.\.

+ C1 J dX + C1

M(x) (/.r + C1x + C1

d.r

"

~LlLUWk

M(x)dx

(9.6)

0

where C1 is a second constant, and where the tirst [eml in the righthand member represenl~ the function of x obtained by integrating twice in x the bend ing moment M(x ). If it were not for the fact that the constants C[ and C1 are as yet undetermined. Eq. (9.6) would dcfine the dellection of the beam at any given point Q, and Eq. (9.5) or (9.5' ) would similarly defi ne the slope of the beam at Q. The constants C[ and C1 are determi ned from the boulldary emu/i· tiolls or. more precisely, from the conditions imposed on the beam by its su pport~ . Li miti ng OU T analysis in this section to slatiwlly determi· nate beams. i.e.. to beams supponed in such a way thaI the reactions at the supports can be obtained by the methods of statics, we note that only three types of beams need to be ccnsidered here (Fig. 9.8): (a) the .I·imply supported bew". (h) the OI'erltullgillg /)eam. and (e) the ClIlltilever bealll. In Ihe tirst two cases. the supports consist of a pin and bracket at A and of a roller at B, and require thaI Ihe delleclion be zero al each of these points. Letting lirsl x = XAd' = YA = 0 in Eq. (9.6), and then .r = x/j.)' = )'8 = 0 in the same equation. we obtain [wo equations that C'Ul be solved for C[ and C!. In the case of the C".mtilever beam (Fig. 9Jlc). we note that both the de llection and the slope al A must be zero. Lettillg x = XA')' = )'A = () in Eq. (9.6), and x =: X A' 0 = OA = 0 in Eq. (9.5'). we obtain again IWO equations which can be solved fo r C[ and C!.

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EXAMPLE 9 .01 The cantilever beam AB is of uniform cross section and carries a load P at ilS free end A (F ig. 9.9). Detennine the equation of die elnstic curve and the deflection and slope at A.

(9.9)

Intcgrating bolh members of Eq. (9.9), we write

+ iPL~x + C~ = o. Substilllting

£1 .\'= - ~Pr

p

BUI. at B we have x = L. y

into (9.10). we

have

8

0= - ~PL] + ; I'LJ + C1 C2 = - }pe

1--- - 1.- -- 1 Fig. 9.9

(9.10)

Fig. 9.10

Using the free-body diagram of the portion AC of the beam (Fig. 9.10). where C is locmed at a distan.::e x from end A. we find

Carrying the value of C2 back into &j . (9.10). we obtain the equatfon of the e lastic curve: £/ .1' = - ~P1J

+ ;PL!x - ; PL'

oc

AI = - Px

(9.7)

(9.11 )

Substituting for Minto Eq. (9.4) and mUltiplying both members by the constant EI. we write

Thedeflection and ~Iope alA are obtained by letting x = 0 in &js. (9. 11) and (9.9). We lind

d~r

EI dx'~ "" - Px

Integrating in

_t'.

8 "

and

we obtain (9.8 )

~

(h ,\

C, =

tPL"

io

L , 'I "" II]

,.f~ 1 _~~'jW!~ A

which we carry back i1ll0 (9.8):

PL~ 2£1

I..-=[.I) = OJ L,

We now observe that at the fixed end B we have x = Land (J = dY/1Lr = 0 (Fig. 9. 11). Substitllling Ihese values i1ll0 (9.8) and solving for C . we have

(rlY) =

1--__

8

1. . _ _-"

Fig. 9.11

EXAMPLE 9 .02 The simply supported prismatic beam AB carries a un iformly distr ibuted load 1V per unit length (Fig. 9.12). Determine the equation of lhc e lastic curve and the maximum deflection of the beam.

" k"[I- " 'l"' ~ .tWig!) - ,--I Fig. 9.12

1.-

V

-1

III

~

tu:L

M = ~wLt - ~ 1VX ~

(9.12)

Substituting for Minto Eq. (9.4) and multiplying both members of this et]uatiun by the constal1l EI. we write

'n i~

-

DrJwing the free-bOOy diagram of the ponion AD of the beam (Fig. 9. 13) and taking moments about D. we find thm

(9. 13 )

I.ntegrating twice in x. we have d.' 1 £ / -:" "" - - lI't' dx 6 ·-

+

I , -4 WLC

+ C!

(9. 14)

Fig. 9.13

535

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O bserving that y "" 0 at both ends of tJl(~ beam (Fig. 9. 14), we ti na lel.r = [) and y = (J in Eq . (9. 15) and oblain C, = (J. We then mal.:e x Land y = 0 in the same equation and wri te

=

0 = _ j,wL4 + f:; w L4 + C,L C , = - ;~ ~j)LJ

I~--t~--I

Carrying Ihe va l~es of C, :lnd C: b:lck into Eq. lain the <'quhlion of Ihe ~la.~ lic C"llrw:

(9.1~).

we ob-

£1)" - - h lO.' ~ + 'l W L.,..! - !~ wLJ.•

Fig. 9.14

'I .1

\

.

~

W

- - (_x + 2u.· 4

J -

24£1

Uri

(9. 16)

SU b;ai tuting into &j. (9.[ 4) the. value oblai ned for Cj, we check Ih~ 1 the s lope of Ihe beam is zero for x - U2 and thll Ihe elastic curve has a minimum at tI le midpo int C of the beam (Fig. 9.15). Letting x = LI2 in Eij. (9. 16). we have W

( L'+

J'r - 24£1 - [6

L' -

2L g

L)

L ;l2

-

,

1 1.12- -1 c

Fig. 9.15

The maximum deflection or. more precisely. the tlI:lx imum absolute v:llue of the deflection. is thus

SwL"

1.1'1 """ -

38 4EJ

SlOL· 384El

In each of the tlVO examples considered so f:lr. onl y one frcc-body di agrJm was requ irl!d todctenninc the bending moment in the beam. As a result. a ~ il1 g lc fu nction of x was used to reprcselll M throughout tite Uc..11H. l1,is, ho wl::vcr, is Hot ge"er.Jly tlte casc:o CQIlCe ntfated load s. ICal' ti olL~ at supports. or disconlinuities in a distri bl,ted load will make it necessary (0 di vide the beam into several portiom. and \0 re pre.~e n t the t>cnding mument by 11 different function M(x) in each oJ" these portlons o f beam (Fig. 9.16). Each o f the functlons Mt.x) will then lead \0 a d if· ferent CxplT!I:!io n forthc .'I lopc 0(.,) and for Ih e J efleclion J(x}. Since eneh of theex rre.~;;in n~ nh taincil for th e de n cclin n mlJ.~t cont ain tWO cnn ~tant .~

Ftg.9.16 A different functton MIl) is required in each portion of the cant~eve~ arms.

of integration, a large number of constant, will have to be detennined. As you wi ll sec in the next example. Ute rcql.lired additional boundary condition~ can be obtained by observi ng that, while the shear and bend · ing moment c:.n be discontinuous at several points in a beam. the de flec tioll and the !}lope of the beam cannot be (iiscontillllOIlS at any point.

EXAMPLE 9.03 For the prism... tk beam and th e loading shown (Fig. 9. 17). determine th e , lope and de.f1ection al point D.

Wp m"~1 Ili l'illc Ih c ~am into two (loninl1 ~ . AI) :uul /)H. and dete,rmjne th! function ),(x ) which defi ne .. the e lasti c cur\'e for e~ch of these portions.

536

p

u,

--3U'---1

"jI·~D;:::lII!!'::lIII!!~B Fig. 9.17

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1. From A to D (x < U4) . We uraw the free-body ui:lgr:lTn of :I portion of beam AE of length x < L/4 (Fig. 9.18). Taking moments :lbout E. we IWlle

Determination of the Constants of Integration. The condilions th~t musl be satisfied by the conSI:mL<; of inte· gration have been summarized in Fig. 9.20, At the support A,

(9 .17)

p

I r" 4

or, recalling Eq. (9.4),

d'

3

dx~

4

E/ ~ =- Px

'I Z= tI J

B -,

(9.18 )

I~

where )' I(X) is !lIe function which defines the elastic curve for porlioll AD of Ihe hew/!. Integrating in x. we write

<=

\1.III = II~ I

I.I ={ L. Y I ~ !ll l Fig. 9.20

(9.19) (9.20)

where the deflectiull is defined by Eq. (9.20), we must have x = 0 and )'1 = O. At the support B. where the deflection is defined by Eq. (9,24). we must have x = Land )'" = O. Also, the fact that there can be no sudden change in deflection or in slope at point D requires that YI = Y" and 8 1 = IJ l when x = L/4. We IWlle thercfore:

"

I) k 't- .-'I-,

[x = O' YI

,-h-I

f)

[x =

L. )'1 =

[x = L/4, 0 1 = H! j. Eqs. (9.19 ) and (9.23 ): 2- pL1 + C = ~ "Ll 128 I 128

Ix =

2. From D to 8 (x > Ll4). We now dr:lw the freebody di:lgram of a ponion ofbcam AEof length x > L/4 (Fig. 9. 19) and write (9.21)

11I'L3

"" 1536

3

(9.22)

£18, = £1 dh = -

dx

£I ,y,- = - i..pxl 24

-.!8 pr 4 +1. PLr + -

C -l

,9.23)

+ .!"L\-~ 8 - + CJ- r + C4

19.24)

(9.28)

C4

,

7PL 2

i Pr - 128 1

£1,v1 = Sf'x

where y:{ rj is the funclion which defines the elastic curvefiH porliOf! D8 oflhe bealll. Integraling in x, we wrile

L

+ Cf 4 +

Substituting for C] and C2 into Eqs. (9. 19) and (9.20). we write. thaI for _t :$ L/ 4. £fIJ I =

4

(9.27)

J

Soll'ing these equations simultaneously. we find

or. recalling Eq. (9.4) and rearranging terms.

4-

+C

L/4' )"1 = .I'l j. Eqs. (9.20) and (9.24):

pe L ill + C l4

d\" 1 1 £ 1·-- = - - Pr + - PL

+ C4 (9.26)

M,

Fi g. 9.19

dr-

o = I~ PLJ + C3L

0], &j. (9.24):

,

11' Fig. 9. 18

(9.25)

= OJ . Eq. (9.20):

3

7PL

(9.29)

2

- lli-t

(9.30)

Lctting .r ""- L/4 in each of these equations. we find that lhe slope and deflection at point D are. re~pectively,

PL' 32£1

= - -

(J D

*'

and

We note that. since IJf) 0, the deflection al D is imum deflection of the beam.

I/Of

the max·

537

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538

'9.4. DIRECT DETERMINATION OFTHE ELASTIC CURVE FROM THE LOAD DISTRIBUTION

Doflectlon 01 Beams

We saw in Sec. 9.3 that the equation of the clastic curve can be obtained by integrating twice the diffcrential equation (9.4) where M(x ) is the bendi ng moment in the beam. We now recall from Sec. 5.3 thai, when a beam supports a dislributed load w (x). we have. dMl dx = Vand (1Vl dr == - w at any point of Ihe beam. Differentiating both members of Eq. (9.4) wilh respect 10 x and assuming EJ 10 be constant, we have therefore lily

[£1M

V(x )

dlJ

EI £Ix

EI

(9.31 )

and, differcnliating again ,

We conclude that , when a prismatic beam s upporL~ a distributed load its clastic curve is governed by the fourth-order linear diffe rential equation

'W(x).

(9.32)

Multiplying both members of Eq. (9.32) by the constant EI and integrating four times, we write

! !I,\,"'II] [1/\ ~t\ l (ll ) C'mlil~~ ,'r

Iwall '

Ed'y / -, = M(x) = (!J...-

I !I,I eo 0] [.11\=11

I

[ !/J<=

(I ]

I M~=O

I

(h ) Simpl)' ,uppOl1<'d bt'anl

Fig. 9.21 Boundary conditions for beams carrying a distributed load.

J dx

I

w(x) dx

+c

lx

+ C1

(9.33)

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9.4. Direct Determination of the Elastic Curve

Thc four constants of integration can be de temli ned from the boundary conditio ns. These conditions include to) the conditions imposed on the dellcctioll or s lo pe of the beam by its supports (cf. Sec. 9.3), and (b) the condit ion that V and M be zero at the free e nd of a cantilever beam, or that M be zcro m both ends of a simply supported beam (d". Scc. 5.3). llllS has been illustrated in Fig. 9.21. The method presented here can be used elTectivcly with cantilever o r simply supported beams carrying a distributed load. In the case of overhanging beams, however, the reactions at the SUPP011S will cause discontinuities in the sbear, i.e., in the third derivati ve of y. and different functions would be re
539

EXAMPLE 9.04 The simply supported prismatic beam AB carries a unifonnly distributed load "1(' per unit length (Fig. 9.22). Delermine the equation of the elastic curve and the maximum deflection of the beam. (This is the same beam :lI1U loading as in Example 9.02.) Since

"IV =

constalll, the first three of Eqs. (9.33 ) yield

d' y E! d fl = V(.() = d ~)'

I

Fig. 9.22

1 - - -,'"- --' - "lUX

,

EI dX2 = M(x) = - 2 ~vx-

+ CI B

+

Cjx

+

C~

(9.34) [r~ I

Noting that the boundary conditions require that M = 0 at both ends of the beam (Fig. 9.23). we first let x = 0 and M = 0 in Eq. (9.34) and obtain C~ = O. We then make x = Land M = 0 in the !>arne equallon and obtain C I = t wL. Canying the values of C1 and C2 back into Eq. (9.34 ). and integrating lI"ice. we write

..I/~()

1

[ ~"'-L.\I =j)1

rI ~I-.'/~ III

f!=t.!I~ o !

Fig. 9.23

we write

0= -

~ WL4 + BwL4 C.l = - fi ·wL J

+ C)L

Carrying the valles of C] and C. back into Eq. (9.35) and dividing bolh members by 1:.'1. we obtain the equation of the elastic eurve: (9.36)

(9.35) But the boundary conditions also require that y = 0 at bolh ends of the beam. Letting.t = 0 and )' = 0 in Eq. (9.35). we obtain C. = 0: letting x = Land )' = 0 in The same equation.

TIle value of the maximum denection is obtained by making.t = L/ 2 in Eq. (9.36). We have 5wL~

lY l....... =

384l:."1

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540

9,5. STATICALLY INDETERMINATE BEAMS

O"""Cllon of Brulm .

III the prccedmg sectionS, ollr
(9.37)

"iF, =0

( II '

Since o nly A, can be de termined from theseequatiuns. we conclude that the beam is staticaliy ill{/elerminate. I-Iowevcr. we rc c:all from Chaps. 2 and 3 Ihm. in a stallcall y indetcmlinatc problem, Ihe reactions can be obtained by considering the lie· jorma/ioll.f of the str uct ure involved. \\ic should. the refore. proceed with the cornpllrarion of the ~ lope ;Jrul dcJonnmion ;JIang the he:Jm. Following the method used ill Scc. 9.3. we first express the bC'I1ding mo me nt M (.l) at ,my g ivc lI poiut of AS iJi te rms of Un:: dislallcc.\ flOW A. Il rc g ive n load, and the. lInknown reac lio n ~_ ln tegrating in x, we obtain ex· pr
(/, )

Fig. 9. 24

f.1" =O. ~ = I) 1 (r = o, ¥= lll

[f =L",=H /

Fi g. 9. 25

EXAMPLE 9.05 Detemline the of Fig. 9.24cl.

reaction~ ~( th ~ suppon~

for the prigmatic beam

Equilibrium EqulItions. From the free-body di
t . :£F, +t :£ F, =

iJ:

A, - 0

().

A,

+R

- w i. = 0

(9 . 1 ~ )

Fi g. 9.26

M,, + BL - ~ !I,)L1-0

(9.39)

Equation of Elastic Curve. Drawing the free-body diagram of a ponion of beam AC (Fig. 9.26). we wl ite

Solving Eq. (9.39) for M

~nd

:arrying in to Eq. (9.4). we write

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Eq. (9.4 1) as follows: (9.42) Integrating in x. we ha\le EI II = £I

dy = dx

Bul the third boundary cond ition req ui res Ihat y = () for x = L. Carrying these \lalues into (9.41), we wrile

- ~wr' +- ~A _l~ 6 2 '

- M x .i

+ CI

(9.40)

0=

- tlOe +

tA ,LJ - t M,\L2

0' (9.4 1) Referri ng to the boundtlry conditions indicated in Fig. 9.25. we make x = 0,9 = 0 in Eq. (9.40). x = o. y = 0 in Eq. (9.4 1), and conclude that C l = C~ = O. Thus. we rewri te

(9.43) Sol \ling Ihis eq uation s i"multallcous]y wi th Ihe three eq uilibri um equ mions {9.38J, we obt:Jin Ihe reaclions al lhe su pports:

A,

=

! wL

In lhe example we htl\le j usl co nsidered, there waS one redundant reaction, i.e., there was one. more reaction than could be determined from lhe equilibrium equaLions alone. The corresponding beam is said to be statically i'ldetermillate to the fi rst l/cgree . Another example of a beam indeterminate to the lirst degree is provided in Sample Prob. 9.3. If the beam suppon s are such that two reactio ns arc redundant (Fig. 9.27a ), the beam is said to be indeterminaTe {(J the second l/egree. VVhile there arc now fi ve unknown reactions (Fig. 9.27b ), we find that f OUT equations may be. obtained from the boundary conditions (Fig. 9.27c). Thus, altogether seven equations are ;:wuilable. to de tennine Ihe li ve rt::_u.:tio ns a nd the two constants o f integra tio n.

Frictionl,'" ~ u rfa,-,..

Fixed elld

\ h--r--r-T--T-TB

A

,- -L- --L.. (")

~ IUIIlll}),," (b)

'1- - L- ------1 B

"

~~- '

[x = 0, 11

[, =L, o- ol

=01

[x =o,y= ol

[ l = L , Y~ () 1

k) Fig. 9.27

541

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r

l"--""="'-"'i""-~ c

--I-J

SAMPLE PROBLEM 9.1 The overhanging steel beam ABC carries a cOI\centrated load P at end C. For ponion AB of Ihe beam. (til derive the equation of Ihe elastic curve.
J = 723 in 4

£ = 29X 106 psi

P = 50 kips

L = ISft = 180in.

{1 = 4ft = 48in.

SOLUTION Fn..~- Bod y Diagrams. Reactions: RA = Pu/Li Rs = P(l + lI/L) t Using the frce-body diagram of the portion of beam AD of length x. we find

(0

< x < L)

Uiffcrcnlial Equation or lhl' Elasti c C un'c.

We use Eq. (9.4) and write

d'y (/ EI dx' = - PT.x Noting Ihal the flexural rigidity £1 is constant, we imcgnllc twice and find

d lt 1 U E/ -"- = - -P- Y + C dx 2 L 1 £1 Y =

I " - "(/i:iJ +(

lktcrminalion IIf COllst:U1l~.

[x = 0.), = OJ : [x = L. y= OJ:

'/ 11 ' ~ "" / ~ 1 1

l " q ~ "1

~

L;z~

11---- 1.-

fl.

(I )

+ C1

jX

(2)

For the boundary conditions shown. we have

C, = 0

From Eq. (2 ), we find

Again usin g &j. (2). we wriTe

Eq ua ti on of the Elastic C urve.

Sub£lilUling for C] and C 2 inlo &js.

( 1) and (2). we h:t\le

1 , --I"

£/y = dx

.-!-P~.t" + !PaL 2 L

6

I"

I

El y = - "6PLr

+ "6PaL1

h. I\laxim ulII o "n cction ill Portion AH. TIle m::tximum deflection ),,,,,,,, occurs at point £ where the slope of the elastic curve is zero. Setting dy/dx = 0 in Eq. (3 ). we determine the abscissa .f", of poinl E:

P"L[ 6£1

O ~ -

1

- 3

(x_ )'] L -

L

x'" =

Vi =

0.577L

We subslitUTe x,/L = 0.577 into Eq. (4) and h::tve

PaL'[

Yn"" = 6£1 (0.577) - (0.577) C.

Evaluation of y""".

'I

For the dma given, Ihe value of Ymn is (50 kips )(48 in. )(180 in y

)'m ...

542

= 0.0642

(29 X lot' psi)(723 in-l)

)"nw.

= 0 .238 in.

....

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SAMPLE PROBLEM 9.2 For the beam and loading shown, detennine (a) the equation of the elastic curve, (b) the slope at end A. (c) the maximum deflection.

SOLUTION Diffcrenti:11 Equ:Jtion of the Elastic Curn "

From Eq. (9.32).

d 4y

11"X

£1 dx~ = - w(x) = - w o s i nT

( I)

Integrate Eq. (1) twice: • d 3y L 11".t El----:;: = V = + 1Vo-COS- + Cl dr 1T L d!y

L2

( 2)

1TX

El -;-:; = M = +wo---; sin - + C l·t + C2 dr 1TL

(3)

BOl/lldary CIJIlllitiollS :

[x = O.M !..-~Q.·\I =lI l ! ..- ~ u.y~O

{ .t * L . .II= OI { .•' ,. 1~!J ~ 0 I

<;tB I- -L--II

~

= 0 ]: [x = L.M = O]:

From Eq. (3). we find Again using Eq. (3), we write

o=

L' U\,.2 sin 11"

+

CIL

Thus: (4)

In tegrate Eq. (4) (wice: dy

L]

1TX

+ C,

(5)

+ C;X + C4

(6)

EI ;.. =EltJ= - wo )" cos 1T

(u

L4

El y = - 1vo -

~'

11"X

sinL

L

BOlllldary COllditioll s:

[x = O.y = OJ : [x= L. )" = 01 :

Using Eq. (6). we find C~ = 0 Again using Eq. (6). we find C} = 0 £/)' =

a. E(luation of Elastic Cun'.. b. SIOIJe a t End A.

- IV

L4 1T..f - sin .... II 11"4 L

For.r = 0, we have

L' EIB" = - 1Vo- cosO ~'

C.

Maximum Deflection.

For x = ~ L

L' ~ ELYm", = - Wo11"~ sin "2

543

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~: ~

L

SAMPLE PROBLEM 9.3 For the uniform beam An. (lI) determine the reaction at A. (b) derive the equa-

tion of the elastic curve. (d determine the slcpe at A. (Note Ihat the beam is statically indeterminate to the first degree.)

I

SOLUTION Using Ihe free body shown. we write

Bending Moment.

Rx _ .1

~(1Vo-II)~ 2

L

3

_M 0 =

DifTr rcntial Equation orthe [Iastic e un'c.

We use Eq. (9.4) and write

, d"y VJo-l.J E I - = Rx - - d).~ 6L Noting thai the nexuml rigidity EI is conSTant. we integrate twice and find

I

dy

2WOX"

E/J; = £18 = "iR"x - 24L + C , I

1VoX~

j

EI }' = (jRAX - 120L - G,x +

(I)

c;

(2)

Boundary Conditions. The three boundary conditions thaI must be satisfied are shown on the sketch

l.x

=

I.., 1/=

111

[\'~L.!I=UJ [x~n·'/~() 1

~ ~

0/

-={B

[x = 0. ),= OJ: C2 "'- 0 I 2 [x=L.O=O J: -RL 2 A

[x

(3)

tooL1 - --+c 24 I

I

"6 R,\LJ

= Lo y = OJ:

-

WJ.4 120

+

=0

GIL

(4)

+ C!

(5)

= 0

u. Reaction at A. Multiplying Eq. (4) by L, subtracting Eq. (5) member by member from the equation obtained. and noting that G1 = 0, we have

R,,= tuwuLt .. We note that the reaction is independent of E and I. Substituting R" = into Eq. (4), we have

h. Equation or the E lastic Cu r \'{'. Eq. (2). we have

Substituting for R,I' C I, and C! into

1(1 ).il - -"'."- - (I

El y = -

-tI)

6 10

L

)

-!OJ.} x

l20L

!)

120

It =

c. Slope a t A .

I.e~ ( -.1-'+'L" 120EIL _ A

-

L' .r) •

We ditTerentiate the abO\'e equation with respect 10 x:

dy

(J = -

dx

w!) = ---( - 5x~

120EIL

= ---

{J A

,

+ 6L2[ _ L4)

w';}

Making x = 0, we hm'e

544

k wuL

120EI

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PROBLEMS

[11 thl' following problems assume tha i the fl exural rigidity EI of each beam is constant. 9.1 through 9.4 For Ihe loading shown. determine (a) the equation of the elastir curve for the cantilever beam An. (b) the deflection at the free end. (£") the slope at the free end.

'I "

.\In

"'"

"

B

f.-------I--------j!' Fig. P9.1

,Ji-TI-rlIrTlnl'jrTl"II'-I"1r

Bt

e'- - -I-- - Fig. P9.3

9.5 and 9.6

l-

I

Fig. P9.2

'1 I

r

'I

'I

c---nl1""

" ~-=-n- '

""I--U,l

U2- J

Fig. P9.4

For the cantilever beam and loading shown. determine

(u) the equation of Ihe eJa~tic curve for portion AB of the beam. (b) Ihe deflection at B. (c) the slope at B.

Fi g. P9. S

Fig. P9.6

9.7 For the beam and loading shown. determine (a) Ihe equation of Ihe elastic curve for JXlrtion BC of Ihe beam. (b) the deflection al midspan. (cl Ihe slope at B.

Fig. P9.7

545

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546

9.8 For the beam auu loading shown. Uelermine (,,) the equatioll or" the curve for portion AB of the beam. (b) the slope at A. (d thO! slope at B

O""CCllon o f Beam .

cl~stic

F1g. P9.B

Fig. P9.9 9 .9

Knowing that beam All is an SS X 1S.4 rolled shnpe lnd that

wI.' = -1 kips/fl. L "" 9 fl. and E = 19 X IO~ psi, determine (lI) Ihe slope al A , (b ) the deflect ion at C.

9 .1 0

Klluwillg lhal beam AB

i~ U

Wl30 )( 23.8 rulku

sh~pt:

am.llhul

p = 'i0 kl\. I. = 1.2'i m . ~nrl F = 200 r.Pil .ll f'.I,~mlinf'. (fl) Ihe ~ l orf':11 A. ( h) the

deflection at C.

'I

. IP

~

.

WJi

L,~l-u,

J:w Fig. P9. t t

Fig. P9.10

9 .1 1 For lh ~ beam and loadin.l: shown, (a) express the magn itude and locatio n of the maximum defl ection in terms of It'u. L. E. and I. (b) Calcul:tte th .. value ofmi' maximum dellection. ass wning Ihal beam AB is" a W460 X 74 rolled shape and that Wil = 60 kN/m. L == 6 Ill, and E = 200 G Pa. 9 .12 (tIl Detennine Ih( locati on and ougnitude of Ihe maximum abdeflection in AH betwe~n A and thO! cente r of the beam. (II) A.sumin £ that beam All is a WIS X 76. Mo = 150 kip' ft and E = 29 X lOb psi. determine the maximulll allowable length L so tltat tile maximum dellec tio n doe s not ex~eed 0.05 in. ~olute

r

Fig. P9.12

9 .13

For th e beam :md loading shown. deternine the tlellection at point C [0' psi.

U~E=29X

'I

(> =1,) kip,

*. .

:x:

C

A 'F 11 .. .1 =="1"- ,:X

E'1. =

\V 100

x .1(1,3

W14x30

15 fr - --

Fig . P9.13

Rg. pg.14

9.14 For the beam and loading shown, detennine the tlellectiotl at I>oi ni C Use E - 200 GPa.

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9.15 Knuwing thlt b~am AE is a W3(j() x 101 rullt:d ~hapt: and that Al,, -"=' 310kN · m. L "" 1.4 tn. a = 0.5 m. and £ = 200 GPa. detemline La) the equmion of the elastic curve for portion BD. (bi the denection at point C.

Problem.

,

Fig. P9.15

r r

Fig. P9.1G

9.16

Knowing tint beam AE is an S200 X 21.4 rolled .hape and that := 2.5 m," = 0.8 m and £ = 200 GPa. det~nll;ne I<J) the equation of the elastic curve for portion HD . (b ) the deflection at the center C (.of tilt: beam. P = 17.5 kN. L

9.17 For the beam and loading shown. detemtine (a) the equmion of the elastic curve, (h) the detlection at the l"ree end.

"I \" ~ " . [, -~:l "O!I I I !~ " 1---- - 1. ----I J~

t '" Fig. P9.17

9.18

Fig. P9.18

For lhe beam and loading shown, detennine La) the equmion of the

elastic curve. (b) the
9.19 through 9.22

«.) the deflection at the m ,Jpoint of tile ~pan.

For the benm and loading shown. determine the

leaction at the wile] suftport.

j) I III 'j I ! IllJi I

1

I

1---- - 1. - - FIg. P9.2ll

Fig. P9.19

".

~pz-r~ , I Fig. P9.21

-1 ~--1i

Fig. P9.22

547

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548

Dcnectlon 01 Beam.

9 .23 F... Ih>;: bc
1

B

L=4"' ---~

r~
11 Iht: roller suppurt

dDr[

--L

1--1

~

IOn

I

Fig. P9.24

FIg. P9.23

9.24 when

tOo

For the beam shown. detennine the reactlOn at the roller support = 1.4 kipslft.

925 through 9.28 DNennine the reaction at the roller ~uppon and draw the bending moment diagram for the beam and IO:lding 5hoIVn. :\

G

P=":::::#====J'

L , ,-I-U2- 1

Fig. P9.2S

Fig. 1'9.26

~c

t- ' L-----I

B[ I

L---

Fig. P9.27

Fig. P9.2B

929 and 9,30 flection at point C.

Fig. P9.29

Detenlline Ihe reaclion at the roller su pporl and the de·

Fig. P9.30

9.31 and 9 ,32 p

Detenlline the reaclion at the roller support and the de·

nc.::tiOI1 m poin t D if a is cqu<o U3.

Ap!I~ D ~L::;:]Ictt

~L-.I

Fig. P9.31

Fig. P9.32

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!l.S. Singularity Function. lor Slope

9.33 and 9.34 iklt:[[uint: lht: rCJl:lion al A JIKl L1raw thc bclllJing IllUment diagram for the beam and loading shown.

}' J

I-- U, - L,R -

and eena-cllal

jJ!I !!!!!!!!1 "

'[ I

I

Fig. P9.33

Fig. P9.34

'9.6 . USING SINGULARITY FUNCTIONS TO DETERMINE THE SLOPE AND DEFLECTION OF A BEAM Reviewing the work L10ne !
gration method provides a convenient and elTective way of detcnnining the slope and dcllcclion at any point of a pris matic beam, as /ollg liS the bellding m(lment cun bl! represellled by 11 sil/gll'. ulIlIlytrcaljwlcliull M(A). Howe ver. when lh~ loading of tilt: !Jcam is such that two different functions are n ~eded to represent the bendin g lIloment over the entire leng th of [he beam. as in Example 9.03 (Fig. 9.17), four conStants of integration are required. and an equal n umber of equations, expressing continuity conditions at point D. as well as boundary conditions at the ~upports A and B , mu ~t bc: used to uc tcnninc these COIlstants. If three o r more functions were nee ded to represe nt the bending mome nt. additional constants a nd a corresponding !lumber of addi tional equntiofls would be required. resu lting in rather lengthy computations. Such would be the case for the beam shown in Fig. 9.28. III this seclion th ese cump utiltion s will be s implified through the

u ~e

ur the s in

gularity functions discussed in Sec. 5.5

Fig. 9.28 [n this loof structure, eEch of the Jolsts applies a concentrated load to the beam

that 5IJPFCrlll it.

,

I

549

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550

Let us consider again the beam and loading of Example 9.03 (Fi g. 9. 17) and draw the free-body di agmm of that beam (Fig. 9.29). Using

Deflection 01 Beams

y

"

p

u' 1--3U'--1

.,~.E==:iiilll"=3I"=l"'/J Fig. 9.17 (repeated)

tP Fig. 9.29

the appropriate singu lmity function, as explained in Sec. 5.5, to n:present the contri hllti on to the shear of the concentrated load P1 we write

Integrati ng in x and recall ing fro m Sec. 5.5 that in the absence of any concentrated COLIple, the expression obtai ned for the bending moment will not contain any constant term, we have 3P

M(x) ~ - x 4

p(, - l L)

(9.44)

Substitut ing for M(x) from (9.44) into Eq. (9.4), we write

£y 3P £1 - , = - x dr 4

P(r - SL)

(9.45)

and. integrating in x, (9.46)

(9.47)t The constants C I and C z can be dctw ni ned from the bou ndary COll ditions shown in Fig. 9.30. Letting x = 0,-," = 0 in Eq. (9.47). we have

I. "'- 1..1} "'- 01 .' ~ Fig. 9.30

,

which reduces to C2 = 0, since any bracket contai ning a negative quantity is equal to zero. Le tting now x = L, .\' = 0, and C2 =: 0 in Eq. (9.47). we write

G), +CL

1 1P ~ L O = -PLJ--

8

6

'

tThc continuity "onditions for the slope and deflection at D arc " built-in" in Eqs. (9.4b) and (9.47). Indced. the difference between the c.~pression~ for the slope H, in ,If) and the slopc H, in DB is represe nted by the tcnn -! P(~' - ! Lr in Eq. (9.46). and tbis term is equal to zero at D . Similarly. th~ differenc e between thc expr~ss ion~ for tbe defle.."\ion )' 1 in AD and the de· flcctiol1 J" in DB is rrpresented by the term - ~p(.\- -iL)' in Eq. (9.·H). and thi5 term is al so equal to urn at D.

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~.6.

Since:. the qua lllity bctWCl' ll bt'ad:ets is positi ve, thl: b , ackel~ call be Je-placed by ordi nary pnrellt hest"s_ Solving for C h we hlive

Slngulnrity FunC'1.lon3 lor Slope anc Oel:ectlon

551

7PL2

C,

=- ill

We check that the exp ressions obtai ned for the constants eland ~ are Ihe slime Ihllt were found curlier in Sec. I) 3. BUI Ihe need for 00ditio llaJ constan ts C J and C 4 has now been e limin ated, and we do not have 10 wrile Ctjuations express ing thai {he .. lo pe and the dellection arc conlinllolls at poillt D.

EXAMPLE 9.06 For the beam and loadilll! shown (Fig. 9.31a~ and using sin· :,iularilY functions. (a) e"press the slope and deilection a.~ fufl('lions oflhe distan ce.x from Ihe support at A , tb) determine the dcn"clion 3t the midpoint D. LJs~ E = 1()() GP3 and 1 =

6.87 X 10 ; m",

Integr.lling the i:l;;t exprt.'Ssioll mice . we obtain

-0.25(.1 - 0.6)1 - O.25{.. - 1.8)1 - O.6(x - 0.6}.'. - 1.44 (x - 2.6)'

EJ(J -

+ I.3x:

+ C,

(9.48)

Elf = - O.062S({ - 0.6)4 + O.062S{x - 1.8)4 + 0.4333x' - 0.2{x - 0.6:rl - O.72(r - 2.6r

+ C t.\" +

19.49)

C~

The constants C] and (."~ can tle detennined [rom th~ boundary conditions ,hOWl! in Fig. 9.]2. Lellmg x = O. Y = fl in Eq. (9.49) and notmg that allihe brackets contain negati ve

c

'I

jx- II, !I-I, I

jr->'6, '1- J]

Fig. 9.32

quantities and, therefore. are equal to zero, we conclude that C l - O. Lcuingno\\!x - 3.6. )' - O. J.ndC~ - (lin Eq.(9.4?), \~e write

() =

- ()flh? ~(Hl)·

+ ()flfl? o;.(l R)4

+ 0.4333(3.6)3

(b )

Fig. 9.31 (u) We LInt", thm the iJ<:alll is looJeJ and supported ill the

. ame manner as the beam of Example 5.0S, Referring 10 that example. we recall that the given di stributed IU:lding was reo placed by the two equivale!lt open·ended loadings ~own in Fig. 9.3lb and that the following expressions were obtained for the shear and bending moment:

V(x) = - 15(~ - (J.6)! + 1.5( , - I.S)I

+ 2.6 ll(x ) = - O.7S(I - 0.6f

+ 0.75(t

- 0.2(3.0)' - O.72{I.O? + C t(3.6).,. 0

Since all the tJuanti lies between brackets are positive. the brackets can be replaced by ordinary parentheses. Solving for C 1. we find C] 2.692. (b) Subslitutmg ror C J IIlId C, inlo Ell. (9.49, and makino,! .x = IV "" 1.&m, we find thai the detlection at poilll D is defined by the relation

1.2(.t - 0.6)°

- 1.8)"

+ 2.6x - l.2(x - 0.6)' - 1.44<-, - 2.6}U

Elyv = - O.0625{1.2t 1 0.4]J)( 1.8)

+ 0.0625(0)4 (J.2( 1.2)1

(J.72{- O.8)1

2692( 1.8)

The last br..cI.:et contains a neg .. tive quantity and. therefore, is equ ..lto zero. All the. other brackets contain positive quruui· ties and can be replaced by ordinary parenth~ ses. We have

EIYI> =

- OJ1625 ( 1.1)~

+ 0.4333( 1.8)1

+ 0.052S(0)4

- 0.2(1.2)1 - 0 - 2.692( 1.8) = - 2.79-\

Recalling the gi ven numeric,ll values of E and I. we write (200 GPa)(6.87 X 10 6 m4)yv = - 2.794 kN . mJ Y,,= - 13.61 x 10 ' Ill = - 2.03mm

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SAMPLE PROBLEM 9.4 For the prismatic beam and loading shown. delemline (0) the equation of the elastic curve, (b) the slope at A. (c) the maximum deflection.

SOLUTION Bendin g 1\'l ol1l('nt. The equation defining the bending moment of the beam was obtained in Sample Prob. 5.9. Using the modified loading diagram shown. we had [Eq. (3)1: tvo -I..l +~~ 2wO(x M(x)= --

3L

3L

' L)l + -, wLx 2

4

U

Usi ng Eq. (9.4), we write

II. Equa ti on of Ihe Elastic Cu r ve,

too 3 2wO( r - ~ L~ £ d~y /- = _ _ x + ~~ dx23L 3L" -

+ I n'

~o

Lx

( I)

and. integrating [wiee in .(,

"I ! ,,- "- U.I/ =() ]

~~

1 '= I~!I=OI

c

~

L---II

r

EI B =

lOU 4 lOU( - illA + 6L x

EI J' =

tvu _ -.>:5

60L

'

)4

- iL

100 ( + -30L" t

lVoL .1 +~ 8~ -\ + C 1

w,J- )..3 I L) S + ~_

-

24

1

(2)

+ C1-t + C.,-

BOIIl/dary Co nd itions. [x = O. Y = 0 ]: Using Eq. (3) and noting that each bracket ( a ncg,ltive quantity and. thus. is equal to zero. we find C, = O.

(3)

) contains

[x = L, y = OJ: Again using Eq. (3), we write 4 O = _ w()L + Wo('c)S+ WoL4+ CL 6030L2

24

I

SubstiTuting C , and C1 into Eqs. (2) and (3), we ha\le

EI/J = _ _W u

\"4

12L"

El y =

Wo

- ~~ x j

60L

h. Slope al A.

'I

A

~

Y,. j

1'1··_"

~_ U2 _I C

+ -W()( X 6L

-

.) + ~-r woL, - - 5 wuL 3

~L 4 l

I ~ + -WO(x - ,L/

30L

.

8

waL

+ ~~ ~ -

24

- 5 wLJ_\" 192

(5) 4

0

Substituting x = 0 into Eq. (4), we lind

r. Maxim um Detleelioll, Because of The symmetry of The supports and loading, the maximum deflection occurs at point C. where x = ~L. SubstiTut· ing into Eq. (5). we obTain . E1 ),,,,,,,

.

,[

= woL -

I 60(32)

+0+

I 5 24(8) - 192(2)

1= - 120 w,J-' \'

......

552

(4)

192

=

L4 It'u !

120EI

..

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50 nr.ln

1 in.

SAMPLE PROBLEM 9.5

11111 111111 ,111 ,1111 1 1 A

L

Fe

El"

3f<

Sf' -----+"'_+_

The rigi~ bar D£F i, wddctl at puilll D tu the uniform MIX! bt:~m AB. For the loading :; hown. determine ((J) the eqllation of the elaslie OJfve orthe beam. (b) the deflection at the rnidpoint C of the beam. Use E -'" 29 X I 0" psi.

J (~ l lh

SOLUTION Bending !\Iomt!)!.

'''I

,e"u -5U lh/fl

The equation defining the bending moment of the

beam wa, obtained in Sample Pmb.•"i.10. Using the modified loading diagram shown and expressing.( in feel, we had [Eg. (3)1:

rnTTT-rrTon"TrITTTTn

+ 48Qr

M(x) = - 2jx" ll.

- 160\.\ - Ill' - 48o(x - II)"

Equation of the Elastic Cun'c.

E/(d1y!ilr) = - 25x"

+ 480x -

,md. integrating

in x.

lwic~

lb· n

Using Eq. (8.4). we wnte

lJ )O

]6(Xx - 11)1 - 48CI(( -

Ih· ft

(1 )

£10= - 8.333x' + 240r' - 80(1: - [I ? - 480(.( - 11)1 + C, lb· ft2 m EI y = - 2.0H(' + 80x' - 26.67( ( - 11 )3 - 24o(x - 11 )2 + Clx + C, lb· fl l (3) Bowl/lary ( olldif;IJIIS. [x = 0. )' = 0]: Uoing E4. (J) and mlling Ihal ~ad, brackel < ) l:ol1lains a 1H~8a T iv~ !]lInnTily anrt Ihll <;, ;~ ~(Illal In ?~m, \\~ li nd (:, = () [.t = 16 ft. J = [)]: Again usi ng Eq. (J) and noting that eadl bracket contains a positive quanlity and. thus. ca n be replaced by a parenthesis. we write

- 2 .08J(16~· + 80(16)3 - 26.67(5: 3 - 240(5)' C] = - 11.36X 10]

o=

Substitutmg the

valu~

+ CJ( 16)

found for CJ and C2 into Eq. (3 ). we have

£ 1 \ = - 2.0 Bx4 + ~O.r~ - 26.67(.1' - II }l - 240{.1' - I If - 11.36 X 10 1.1

Jb'

re

(3')
To determine EI. we recall that £ = 29 X JO~ p. i a.nd compute

I = h,hh J = hrJ in.)(J in. f = 2.25 in' EI = (29 X lOb psiX2.25 in') 65.25 X J06 1b . in!

=

However. since all previous computations have been carried out with feet as the unit of length. we write

EI = (65.25 x I06 Ib· in")(1 ftIl2i u. )' = 453.[ X b. Dcn tct ion al M idpo int C.

. .: -C

"I

El.k - - 2.083(8)4

[oJ lb.

ft 2

Making.( = 8 I"t in Eq. (3'). we write

+ 80(8)] - 26.67(- Jy - 24o(- 3)! - 11.36 x 101(8)

r Noting lhal each brackel is equal zero and s ubstilulin g for EJ its numerical !'{(••ii~~~"'lii ~ II:1lue. we. ha ve e I (453. 1 x IOJlb·ft!)yc - -~8.4S x 1011b'ft] .Ht - --j., I"-- Sf, - --

'\ I~t: ~-~iiil.~:;r~

\0

B

and. solving for rc: Not~

Yc = -{).1190 fl

Ihm the deflect Io n obtained i.\·

/101

\'("~

- 1.54Ilm . ...

Ihe maximum deflection.

553

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r

.\

~

r-"-L

SAMPLE PROBLEM 9 .6 For th e uniforr.1 beam ABC. (al express Ihe reaction at A in terms of P. L. 11 • E. and I. (b) determine'. the reaction al A :l.IId lite denection under the load when {/ = L/2.

8

1

SOLUTION Reactions.

For (he given venkal bad P the reactions are as shown. We

note lhat Ihey are statically

indel~rmmale.

Shear an d Bending l\ loment. U$ing a step function 10 represent the contnbution of P \0 Ihe ~hear. we write

a)O

VI» = RA - I'{x -

Integrating in x. we obtain the bending moment: It"

I- - - L - - - I

p(x - a)1

M(x) = RA-{ -

Rc

Using Eq. (9.4). we wrile

Equation of Ihe EI:.Istic C Url'c.

d 2y £1 dx 2 - N" x

a)1

I'Y--

I.ntegrating twice in x. dy

EI-;;; - EI

£/ y =

'I

l .t - u '1- il I f r= I 1/=11 1 [ \ =

0" =0 1

~ " I - -~~> - _ L_ _cl

[I -

I

,

2R~x -

I I

i>\;~

-

~R,x.J - ~P(.\ -

a)J

- ill\."! + C

+ C,x +

BUII/u/ary COlldll;ORS. NOI ing Ihal the br~cket for x = O. alld to (L - u) for x = L. we write

[.1:=0. ), =0]:

Cz = 0

[.r

~ RA L'

= L. f! = 0]:

[ r= I .. -,,= Il} tI.

C, a) is equlJ 10 zero

(I)

- :,P(L - u)l + C, = 0

~ /{,1.3 _ tP{1 - a)l

Reaction lit A .

(I" -

1

+

r ,l .

(2 )

+ r",

= n

(1)

Muhiplyillg Eq. (2) by L. s ubtracting Eq. (3) rnember ~nd nOling that C, = O. we ha\'e

by member from the equation ohtained.

\ 1 ]R.\L3 - '6P(L - af(3 L - (L - a)] - 0

R," P(I- ")'(1+ "-)t . L

We [1U1 t: lhal Iht: ft:;JdiUJI is im.lt:pt:lHknl

b. Reaction al A a nd Dcneclioll lit B wh(,11 a in the expression obuincd for R.I' we have 1
2L ,

ur t' ;JIlLl I .

+ :\)

=

jL. Making (/ = i L

= 5P/l6

Substituting 11 = Ll2 and RA = 51'/ 16 into Eq. (2) and solving for C[, we find C 1 = - Fe /32 . Making x = L/2, C 1 = - PL'!32. ;Jnu C~ = 0 ill lill.< t:..'prt:s~;fln flhr:1in~rl

for y , W~ hav~

7fL) 768£1

1PL1 1'. = 768 l:.'/-! ...

r-,'ote that the deflection obtained is not the maximum deflection,

554

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PROBLEMS

US(' singularity fun ctions to soh'!' the foUowing pro blems and ass ume tha t the nl'.'iura l ri gidity HI of each bcam is constant . 9.35 and 9.36 For the be
J

'1

"

~

;.t;

~i h- j

Fig. P9.3S

Fig. P9.36

9.37 and 9.38 For the beam and loading shown. delennine the deflection al (a) point B. (b) point C. (c) point D.

'I ]r\

r r r - Jj

C

J)

Fig. P9.37

Fig. PS.38

9 .39 and 9.40 For the beam and loading show n, detennine (a) the slope al end A, (b) the deflection at point B, (c) the deflection a1 end V .

Fig. P9.39

Fig. P9.40

555

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556

9.41 For the beam and loading shown. determine (a) the equation of the elastic curve. (b) the denection at point B. (d the dellection at point C.

Deflection 01 Beams

'Ir-·

J'I--

B

'I

;.g

~ C"" --,

- + - 1)2

~

~'--l

....... ..- U2

III IIIIIJ

-J

Fig. P9.41

Fig. P9.42

9.42 For the beam and loading shown. detemline (1I) the equation of the elastic curve, (b) the slope at point A. (d the denection at point C. 9.43 For the beam and loading shown. determine (u) the equation ofth~ elastic curve. (b ) the denection at point B. (c) the denection at point D.

9.44 For the beam and loading shown. determine (a) the equation of the elastic curve. (b) the denection at the midpoint C.

Fig. P9.43

'I

",p rm ~C~rm ~B ~J-,, -LJ-,,-=-1

A

"'i i :il~"l: !ilcl

' I!!l.: '( !l!! 1"I! i'/) iiil!!!

r

1l i5ft Il.i 5 ftI

Fig. P9.44

3..', in

I-J 3..'' ift J

,

.5.5 in .

--1

Fig. P9.4S

at

9.45 For the timber beam and lo..1ding shown. de termine (a) the slope end A. (b) the deneetion at the midpoint C. Use £ = 1.6 X \O~ psi. 9.46

For the beam and IlXlding shown. determine (u) the slope at end A. 10" psi.

(b ) the denection at point B. Use £ = 29 X 2UU II,

3 k.'\'/m

X

W310 X 60 1.8 III

Fig. P9.46

-1- -1--1-

Fig. P9.47

9.47 For the beam and loading shown. determine (u) the slope at end A. (b ) the deflection at the midpoint C. Use E = 200 GPa.

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Problem.

9.49 FOJ the lleam iOld IOJding s hown. deteJllline(u ) the >lope at end A. Ih) the deflection althe midpoint C. Use £ = 200 GPa

557

."n-~r,;;,,C ~ ="'\" ~ ' o J-,,,-'J

Fig, P9.48

9.49 and 9.50 For the beam
(ll)

the

cf

11---1.1,--1.1,-1~ I

Fig. P9.49

Fig. P9.SO

9.51 and 9.52 For the beam ,lIId Ioadmg sho"'n. detennine (a) th e reaclion at the roller suppan, (b) the deflection at poinl B.

Fi g.

9.5 3

P9.51

For the beam and loading

poim C. (b) the deflection

al

Fig. P9.52

~hown.

determine

(II)

the reaction at

peNm 11. Use £ - 200 GPa.

50kK

J J

;£i 9.54 For the beam and loading shown. determine point A. (b ) the defl ection at point B. Use E - 200 C Pa.

(ll)

the reaction at

SO kl\

D[ \\,200 X 32

"" I

Fig. P9.S4

'2m

I

1.2

III

I

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558

9.55 and 9.56 For the beam and loading shown. determine (u) the reliction at point A. (b) Ihe deflection at point C. Usc E = 29 X IO~ psi.

Deflection 01 Beams

" = 4 5 klp..l!!

BIIIIIII D ' [ ,t.;' 2.5ft I 2..')[1 CI 2.5 ft ~.5fl A

W IO

x

LJ

22

Fig. PS.S5

WI4

x 22

Fig. P9.56

;1) Illl.t

1A

I -U2-1-W--l Fig. P9.S7 p A

B

c

[)

9.57 and 9.58 For the beam and loading shown. determine (a) the reaction at point A, (b) the deflection at midpoint C. 9.59 through 9.6 2 For the beam and loading indicated. determine the magnitude and location of the largest downward deflection. 9 .59 Beam and loading of Prob. 9.45. 9.60 Beam and loading of Proll. 9.46. 9.61 Beam and loading of Prob. 9.47. 9.62 Beam and loading of Proll. 9.48. 9.63 The rigid bar BDEis welded at point B to the rolled-steel beam AC For the loading shown. determine (u) the slope at point A. (b) the deflection at point H. Usc E = 200 GPa.

Fig. P9.58 0.5 '" (1.3

II,

0,3", 0.5

II.

I I 1-----1 W UJu X 19.3

G

0,15", Fig. PS.64

9.64 The rigid bars HF and D H are welded to the rolled·steel beam AE shown. Determine for the loading shown (a) the deflection at point B, (b) the deflection at midpoint C of the beam. Use E = 200 GPa. a.~

9 .7 . METHOD OF SUPERPOSITION

When a beam is subjected 10 several concentrated or di stributed loads. il is often found convenient 10 compute separately the slope and de!lection caused by each of the given loads. The slope and deflection due to the combined loads are then obtained by applying the princi ple of superposition (Sec. 2. [2) and adding the values of the slope or deflection corresponding 10 the various loads.

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EXAMPLE 9 .07 Determine the slope and deflection at D for the beam and loading shown (Fig. 9.33). knowing that the flexural rigidity of the beam is £'1 = I()() MN . m!. The slope and deflection at any point of the beam can be obtained by superposing the slopes and deflections caused re· spectively by the concentrated load and by the distributed load (Fi g. 9.34). Fig. 9.33 ~11n"/n,

+

8

r=2':~

I

kJ

(b)

(1/)

I

1"=f>"' ~

Fig. 9.34

Since the concentrated load in Fig. 9.34b is applied !It quarter span. we can llse the results obtained for the beam and loading of Example 9.03 and wri te PL'

(OD),. = - 32£1 =

3J>Li ('yD)1' = - 256EI =

( 150

x IO')(8f

- 3 X IO- J rad

32( I00 X 10") 3( 150 X 10 1)(8)1

dy ll! = - - (- 4x 3 (Lt 24£1

0= -

( l:Io) ~.

=

'='

256( 100 X 10~)

~( -.I'" + 2L x:l - L:lx) 24E1

and. differentiating with respect

10

x.

(9.51)

20xlO) _ 24(100 X 106)( - 3~2) ~ - 2.93 X 10-J rad 20x 10J

(1'1)) ,, = 24(1()() X IOb)(- 912) = - 7.00 X 10- l m

On the other hand. recalling the equation of the e lastic curve oblained for .1 uniromlly distributed load in Example 9.02. we express the deflection in Fi g. 9.34c as l'

LJ)

Makingw = 20kN/m.x = 2m. and L = 8 m in Eqs. (9.5 1) and (9.50). we obtain

= - 9mm

.

+ 6Lx-, -

(9.50)

= - 7.60mm

Combining the slopes and deflections proouced by the con· centrated and the distributed loads. we have ti~

= (Oo)p + (OJ)", = - 3 X 10

J -

2.93 X 10 l

= - 5.93 X 10 - 3 rad

Yo = (Yo)p + (YD),,' = - 9 mm - 7.60 mm = - 16.60 111m

To facilitate lhe task of praclicing engineers, most structural and mechanical c ngineering handlx>oks include tables giving the deileCtions and slopes of beams fo r vario us loadings and types of suppon. Such a table will be found ill Appendix D. We note lhat the slope and defl ection of the beam of Fi g 9.33 could have been determined from that table. In· deed. using the mfonnati on given under cases 5 and 6. we could have expressed the dcilection of the beam for any value x :s L/4. Taking the derivativc of the expressio n obtained in this way would havc yie lded the slo pe of the beam over the same interval. We also note that the slope at both e nds of the beam can be obtained by simpl y addi ng the

559

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560

con c~fJOlidillg values giwil ill the table. !iowcvel, the lllaXililUlI1 deflection of the beam of Fig. 9.33 CIII1IIOI be obtained by addillg the maximum dellections of cases 5 and 6. since these deflections occur at dirferent points of the beam.t

O""""llon of Brulm.

9.8. APPLICATION OF SUPERPOSITION TO STATICAllY INDETERMINATE BEAM S

often lind it convenient 10 use the method of superposition to detennine the renction ~ althe suppons of a slatjeally indeterminate beam. Considering tirst the ca ~e of n beam indeterminate to the first degree (d. Sec. 9.5), such as the beam shown in Fig. 9.35. we fo llow the approach described in See. 2.9. We designate one of the reactions as redu ndant lind eliminate or modify accordingly the corresponding support. The redundant reaction is then treated as an unknown load that, logelher \.\Iilh Ihe O1her loacl"" mll.~t prCldlice defr,rmations that arc compatible with the anginal supports. The slope or deflection at the We

paill1

Fig. 9.35 The Gontinuou~ beams supporting this highway overpass have Ih'ee supports and are thus indeterminate.

WIICH;

tile su ppoil

lta ~ UcCII

lllolilt1cd

UI

diuLiliated is oiJtaim;d iJy

computing separat e.ly the defonnations caused by the given loads a nd by the redunJ ant reaction, and by superposi ng the results obtained. Once the reactions at the supports havt! been found, th~ slope :lIId deflection e1m be determmed in the usual way al any other point 01 the beam. t An Il pprMirml~ ",III~Clf Ihe ","xin",m ,lrn~("linn oflh" Mnn, ~n n Iv oh,,;,rrl hy plnuint the valles of J" corresponding 10 "arious "alU('s oJ .t. TII<."' d:lwllinaLon rj the cxaCI k"alion

and ma~niludc OJ Ihe millumulU denCClion would reqUlre Sf iling equilio zero obtained for tile ItOpe of the bram md !olviD~ lhi~ cquauen 1'01 ...

Ih~

expression

EXAMPLE 9.08 Determine the reactions atth ~ suppons for the prismatic beam and loadm g . hown in Fig. 9.36. (This is Ihc same beam aoo Ifl~din ~ a~ in F. x~ mrl p. I) 0<; nf S~e. q <; ) We comider the reaclion at B as redundant and release. the beam from Ihe s upport. The reaction R 8 L< now considered as an unknown load IFig. 9.37(1) and will be determined from the condition that the delleclion of thi! beam at B must be zero.

I'I~ =

1\

I

JIL

"

Fig. 9.36

,

k,

(c.l

(I,)

(c,

Fig. 9.37

The sulu liun is l'arrkd oul by I:unsiuerillg sep<.nalt:ly Ihe d<:flection (YB) ... caused at B by the unifonnly distribllted load "IV (Fig. 9.37b) and the deflection (V t )1I produced at the same point by Ihe r
Frumlhe labk uf App.::ndlA D (ca>t:~ 2 anu I). we [lud Ulal

RL' + - '3£1

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Writing that the deflection at B is the sum of these two quantities and that it must be zero. we have

9 .B. Superpositlon for Stallcally

and. solving for RD'

III QB

Drawing the free-body diagram of the be:ml (Fig. 9.38) and writing the cocresponding equilibrium equations. we have

+ t 'iF,.

I.,

R,\ + R8 - 'WL = 0 (9.52) R,\ = 1vL - R8 = wL - i1VL = i wL

= 0:

I---L----I

R.I = iwL t +~ LM,\ = 0:

561

Indeterminate Beams

Fig. 9.38

M", + R8L - (1vL)(~L) = 0 (9.53) MA = ~lvL2 - R8L = !wL2 - ~WL2 = f1VU M",=~1VL1~

Alternative Solution. We may consider the couple exerted at the fixed end A as redundant and replace the fixed end by a pin-and-bracket sup)XJn. The couple M", is now considered as an unknown load (Fig. 9.39a) and will be deter-

(h )

Fig. 9.39

mined from the condition Ihat Ihe slope of Ihe beant at A must be zero. The solution is "arried out by considering separalely Ihe slope (8,1)'" caused at A by Ihe uniformity distributed load "II) (Fig. 9.39b) and the slope (6"')M produced at the same point by the unknown couple M,\ (Fig 9.39c). Using the table of Appendix D (cases 6 and 7). and noting Ihal in case 7. A and B must be interchanged. we find Lhal

and . solving for M,I'

Writing that the slope at A is the sum of these rwo quantities and th,\I it mUSt ~ zero, we have

The values of R... and R8 may Ihell be foulld front the equilibrium equations (952) and (9.53).

'wLJ

M AL

25£1

3£1

- --+--~O

ll lC beam considered in the preceding example was indetenninate 10 the tirst degree. In the case of a beam indelcnninatc to the second degree (d. Sec. 9.5). two reactions must be dcsigna!cd as redundant. and the corrcspvnding supports mllS! be climin;)tcd or modified accordingly. The redundant rc;)ctions arc then treated :l5i unk.Jl0wn loads which. sim ultmtcousIy and together with the other loads. must produce deformations which arc compatible with the original supports. (See Sample Prob. 9.9.)

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SAMPLE PROBLEM 9.7 For the beam and loading shown. determine the slDpe and deflection ill POUlI B.

SO LUTION Prindple of Superposition. The given loading can be obtained by superposing the Joadillgs shown in the following "piclure equatioll." The beam AB is, of wurse. Lhe ~ am~ in each pan of the figure.

J

A

r:+ ,

UB

I "I i

1--U2-l-U,--J

'I

"

"

1

I

'I

1

""'~ 1" ,-

B ~08

1

I

'I

~ 1(,,"1,+ JA Il =::r(~H~'

For each of Ihe loadillgs I and II, we now determine Ihe slope and defleclion al B by using the table of Bellm Dejhc/i/lflJ and Slopt'.\' in Appendix D.

Loading I

J ,\

" ItI III,Ii I

I

I.

Un I

'I

,

]"

l'

LlHUlillg II

In portion elJ. the bending moment for loadmg !l is zero and thu s the elastic curve i~ a straight lin~.

Lom/ine II

c

IIIII t "

"

1

I-- U2- 1- 1.I2-1 'I

1

Siopt, at Point B

Dcneclion al B wL~

8EI

562

71Ve 384EI

."B -

4 luL'l J,. ] 84£1


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SAMPLE PROBLEM 9.8 For the unifonn beam and loading shown. determine (a) the reaction at each support. (b) the slope at end A.

SOLUTION Principle of SUIK' rposilion. The reaction R/I is designated as redundant and considered as an unknown load. The deflections due 10 the distributed load and to the reaction Rn are considered separately as shown below.

'1

;£i~C3i''''''j.\~H-=i 1:' l '/8

= ll)

For each loading the deflection at point B is found by using the table of Beam Defiections and Slopes in Appendix D. DI slribured Load/llg.

AI point B. x =

iL:

\' = -

We use case 6, Appendix 0

- ~(x' - 2Lxl 24£1

+ LJx )

().) ~ ---"'-[('L)' - 2L('L)' +L "\3 L)]~ - 001132 wL' 24EI3 J £/ Boo

Redllllda/ll Reaction LIll/ding. and b = ~ L. we have

Po'b'

(YB)R = - 3EIL =

From case 5, Appendix D. wilh

R, (23L )'(L)' "3

+ 3EIL

II

=

lL

R,L'

= O.OI646£i

u. Reactions at Supporls. Recalling that Ye "" O. we write Ye = (Ye)", + (ve)" wL 4 Rfi-.l Re = 0 .688wL t
E1

Since the reaction Re is now known. we may use the methods of st:Jtics to detemline the other reactions: RA = 0.271 wL t R(" = O.0413wL t "III b. Slope III End A . Referring again to Appendix D. we have Di~·tribllted Loadil1g.

Redll l/dulII Rellctioll Louclillg .

+

we

lOLl

(B,')M. = - 24£1 = - O.04167£i

For P = - He = - 0.688lOLandb = ~ L

O. fi88WL(~)[L' - (~)' l 6E1L

3

lvLl

(8,t)R = 0.03398£i

3

Finally. B" = (OA)U. + (BA)R lOLl

wLJ

1vLJ

OA = - 0.04167£1 + O.03398 T! = - O.OO769 T!

0,\

=

lI·e ~

0.(10769 B

~

563

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SAMPLE PROBLEM 9.9 "or the: beam ilnd loading shown. determine the reaction al the

fix~d

sl.Ippon C.

SOLUTION Principle of S upcrposilloll. A.sUI:I;ng the axial force in the bt:am to be zero, ute beam ABC is indeterminate 10 [he second degre<: and W~ choose two real.'tion components as redundant, namely. the venical f()rcc R c and the cou pie Mr. The deformations caused by (he gh'ell load P, 1h.c force R r . amI lhe coupl~ j\·lc will re considered separatelY:ls s1,0\\n.

~ T)({Jr'I" I",..IM

For each load. the slope and deflectiotl at point C will be found by lIsing the table of Bl'illl\ Deflections ilnd Slopes in Appendix D.

wad P.

We note that. lor this loading, portion

I'll = - 3£1 -

Be of the beam is straight PI1'

2El =

Pa' - 6E,c'ln

+ 3b )

Force Rc COl/ple Me BOllndary Conditions. At end C Ihe slope
Et

r'~p"

M'~q R

_ __ a

.,

I

I

L

b- R c I

[-I

=

L. Ye =0 ]: (2)

Reactio n C()mponents al C. SO]lImg simultaneomly Eqs. (I) :md (2), we find after reductions Pi!' plJ ' RC = + [ l (a + 3b) ~ ~ L}( U + Jb)t ~

Using the rnelhoos of stalics. we c;ln now de!ennine the reaction at A.

564

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PROBLEMS

Usc till.' nltthod of

~upc rp osi lion

to

9O I\'C

the f{lllflwiog proble ms lind

assullItllhat tht flexural rigidity £1 of carll heam is constant. 9 .65 through 9 .68 For the beam ana loading deflection at C, (b) the slope .11 end A.

~hown.

dCl<'rmine (a) the

p

, A

t.rJ1UJ1UJ~

D

Fig. P9.65

Fig. P9.66

Fig. PQ .S7

Fig. P9.68

9.69 and 9.70 For the cantilever beam and loading shown. determine the slope and deflection at the free end. p p

.,

p

8 C

~

U2

Fig. P9.69

9.71 and 9.72

- -1-- ---J

"' I'"'?"

~

c

~"

I

1.11

, L

Fig. P9.70

For the cantilever beam and loading shown. determine

the slope and deflection at point B.

jl

ITI! I I ! P ~"L

" 8 JC

Fig. P9.71

- u2 - - I -- LO

Fig. P9.72

565

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566

Deflectton 01 Beams

9.73 For the cantilever beam and loading shown. determine the slope and deflection at end C. Use E = 200 G Pa.

r Jt-""",--J_o,,,J J"

X

StOOx 11..5

Fig. P9.73 and P9.74

1

125 11>

j 11111111' 15 1Mn

~30

in. - J Jilin.

Fig. P9.7S and P9.76

I --t

e ~.

r-

9.74 For the cantilever beam and loading shown, determine the slope and deflection at point B. Use E = 200 GPa. 9.75 For the cantilever beam and loading s hown. determine the slope and deflection at end C. Use E = 29 X I rJ psi. 9.76 For the cantilever beam :md loading shown, detennine the slope and deflection at point B. Use E = 29 X 106 psi. 9.77 and 9.78 For the beam and loading shown. determine (a) the slope at end A, (b) the deflection at point C. Use E = 200 GPa.

HOk-'"

l;()lN (

,

J

c

B

~2..5", -

-L

Fig. P9.n

foI) n., . n1

~". -=.r

2..

I

L ." e

W·tlox4fi.l

\\' t5O X 24

u,,, --l n.1i,"'

Fig. P9.7B

9.79 and 9.80 For the uniform beam shown. determine (al the reaction at A. (b) the re
Fig. P9.79

Fig, P9.80

9.81 and 9.82 For the uniform beam shown. detennine the reaction at each of the three supports.

Fig. P9.81

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9 .83 and 9.84

For the lxam shown.

det~rtllin~ ~u:

leaction at fl.

~,~,

A

567

Problem.

,

I--- Uz. ~ U2--1

-- U'!. -+- Ul ~

Rg. P9.83

Rg. P9.B4

9 .85 Bo::am DE [O::M~ Oil lho:: l'amilevo::r bO::,UIl AC a~ ~hown, KnL>wing lhat wll1He rod of ~ine 10 mm i~ men for ~~~h Iwmn. neJermine rh.~ d.. fl<'Ct;on ~t end C if the 25-N . m couple is applie d (a) 10 end E of beam DE, (b ) to end C of beam AC. Use E = 200 GPa. ~

IOUln,

0.75 In.

'~ A ~D~E ) H-L ~J[l1'1I1I ___ Ble.--___CJ 12\1111111

H-.l U.7S b.

d,,"- '

ISH 1111"

Fig. P9.85

9 .86 Benm 8D rests on thc cnnrik\'cr bcnm AE ns shown. Knowing Ihat a square rod of side 0 .75 in. is used for each beam, determine for the loading shown (a ) the deflection at point C, (b) tire ddlec tion ~ t point E. L:se 1:: = 29 X lOb psi. 9.87 The tWO beams shown have tire same cross section and are joilX.-d by a hinge at C. For the loading shown, determine 1(1 ) the slope at poin t A. fb) the deflection at JXli nt n. Usc E - 29 >< [0' psI.

, ~ t2 tJl.-. Fig. P9.87

/illO [h

c

/)

1'C"''''"J ;J>. L ~1:! fi ;"

} 11 UIIIU U (lJlJ;t D~ A

iB

E

C

1"'' ' ' 1 1 1' '"'''1 0.4111

U.411\

[Hill

--L

m~llm ,

1--1 ""n ~4

OAII1

Fig. P9.88

9.8B A central beam 8D is joined:ll hinges to two cantilever beams AB and DE. All beams h:l ve the cross section show n. For the loading shown, determine the largest 10 so Lhnt the deflect ion at C does not exceed 3 mm. LJ~e E = 200 G Pa. 9 .89 Before the uniformly distribUied load w is applied, a gap. (10 = 1.2 mm . ex;~I~ hclwl':en lhe..ends ofthl' c~ nti[ evi';rhars A R ~nd rf) Kno..v_ in g that E = 105 Gl'a and w = 30 kN/m determine (a) th e. reaction at A. (b) the reaction at D.

51)

m,"

H-.J.

~5UlRm

I

Rg. P9.99

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568

Deflection 01 Beams

9.90 Before Ihe 2-kipJft load is applied. a gap. /)u = 0.8 in .. exisis be.tween the W 16 X 40 beam and the support m C. Knowing thai E = 29 X I if psi. delermine the reaclion at each s uppott afler the uniformly disttibUied load is applied.

I

W!6 X 411

Fig. P9.90

9.91 For the loading shown . and knowing thai beams AB and DE have the same flexural rigidilY. delemline the reaclion (a) aI B. (b) at E.

P .....-A

= 2.5.~ 1I1I1I~

"

3 III

20 kN/I"

l lrTTT"i-TTl-nl r B ,~~~

D

1-.-6 ", ___c_

~r}=5ft Fig. P9.91

w ·tlO

x 46.1

Fig P9.92

9.92 The cantilever beam BC is allached to Ihe steel cable AS as shown. Kn owing thai the cable is initially taut. detelmine the lension in the cable caused by the dislributed load shown. Use £ = 200 GPa. 9.93 A ~ -in.-diameter rod BC is allached to the lever AB and to the rIXed support at C. Lever AS has a uniform cross section ~ in. thick and I in. deep. Forthe loading shown, determine the deflection of poimA. Use E = 29 X Ief' psi and G = 11.2 X lOb psi.

so 1111"'-

20iJl !II ill

"

Fig. P9.93

200 N

Fig. P9.94

9 .94 A 16-mm-diameter rod has been bent into the shape shown. Determine the deflection of end C afler the 200-N force is applied. Use E = 200 O Pa and G = 800 P:l.

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D.!). MDment ·Are" Th.,.,,,,m.

'9 .9. MOMENT·AREA THEOREMS In Sec. 1).2 through Sec. 9.0 we used a mathe matical method based on the integration of a Jifferential eq ulltion to determi ne the dellcction and ~ l nf1C of a ncam lit any e ive n [loin!. The hemli ng mome ot was i!x pre.<;s~1 1 as il function M(x) o f the di stance ." meas ured do ng the beam. and t wo sliccessive integrations Jet! to the fun ctions £i().') and y(x) represellting. res pec ti Vely, the s lope and de n ection at any poi nt of the beam. In this secti o n you will see how geometric properties of the elastic curve e£ul be used 10 de tennine the d eflectio n and s lope of a beam at a specilic paim (Fi g. 9.40). Consider a beam AB subjected 10 some arbitrary loading (Fig. 9.4la). We draw [he diag ram representing Ihe variation a long the beam of the q uantity M/EI obtained by d ividing the bending moment M by Ihe Ilex ural ri gi dit y EI (Fig. 9.4 1b). We note that. except for a diffe re nce in the ~c~ I ('_~ of [mlinares . thi .~ cli aeram wi ll he Ihe sam e. ;'I .~ Ihe be nding-mo me nt d iagram if the Ilexural ri gidit ), of the beam is CUII Slalll.

Recall ing Eq (9.4) o f Sec . 9.3, and the write M till (py dx=dx 2 =EJ

i:lC"t

thm tly/ d.l = t:(

M

w('

(9.54)!

dO = - d x

EI

Fig. 9.40 The deflections of the beams sup~rt· Ing the IkKlIS ul a tuilding stlOuld ~ I
,.,

Considering two arb itrary poinL~ C and D on the bcum an
Jr"

£jdX

~c

A

,~

Un - Or

=[

''' M

".

= area undcr(MiEl) di agram

hclwee n

r: ;'Inc! n

(9.56)

tThi~ lelation can alw be deri\'cd ...·ittlOut fl'f~rring te the rc;uks ohtained in Sec. '1.3. by DQting Ihm Ihe angle Ufi fOrulCd by lloe taJIgcm, to Ihe rla"i, ell"'c a: l' and l" i~ dbo Ihe mgle formed hy the corrcsponding normal> to that curve ~Fig. 9 .421. We tllU; ha\C dtj = d,/p I\'hcre ds is tht length of the arc PP' and (> the radills of cur\ature at P. Substituting for lip from Eq. (-t.ll). aJld nOling Ihal. .ince the ,lore all' is very lnlJlI . <1.,. i~
the horizontal distallc'c

dx t>ctwrcn P and p ' .

.'\

D

B

we write (9 54)

IJ / .

~'>K"

(,II

Ag_9.41

c

Thi s is thefil'st m oment-area themeI/!.

proximaticm

C

(9.55)

EI dx

where Oc and ltv d enote the slope at C and D, respective ly (Fig. 9.4 1c) . But the right- hand member of Eq. (9.55) represenls the area under the (MI Ef) d iagram between C flnd D, Wld the left·hand membefthe an gle between the tangents to the e lastic curve at C and D (Fi g. 9.41d). Dc noting this angle by ODIC, we have ODIC

f, ~

['" M

(10 -

569

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570

We. nUle thatlhe an glc emf: aml lhc area untler the (MIE1) tliagriu n haw the s:lme ~ign. In o ther w"rd~. a positi vC" rtre".l. (i.e. , an arC"a loealed a bove the.l.· ax is) corresponds to a countcrd ockwise ralation of the tangent to the clastic curve as we move from C to D. lUld a negative area correspond, to a clockw ise rotation. Let us now considc: r two points r and I" locat c:d betwc:en C and D , and ;"II il rl islance (/.I" from each O1hcr (Fig. 9.41). The Tan ee n1.~ In Ihe clastic curve drawn al P and P' intercept it segment of length (lI on the ve rtical lhrough point C. Since the slope 0 at P and the an gle dB formed by the tangents at P and P' are both small quantities, we can assume that (/: is equl\l to th~ llrc o f circle 01' radi us x s ubtc nding Ihe angle d/) . We have. thereforc.

O""""llon of 8/lnm.

Fig. 9.43

iii -

lciL .-\

C

/,,..

J)

.\1

or. s ubstituting for i/O from Eq. (9.54). (9.571 lJ

We nuw intcgra rc Eq. (9.57) from C to D. We notc thaI. a.~ point P the el2Stic curve from C to D , the tangenl llt P sweeps Ihe vertical through C fro m C to E. Thc intcgrul o f the le rt-hand member is thus e4ual to the ve rtical distance from C 10 the tangent ilt D. Thi s distance i ~ denoted by 100 and is clllled the IlIngelliial l/el'ialiOIl of C with respect to D. We haw, therefore,

Fig. 9.44

d c~c ribes

'. Ion = [

.'

c·

riO

M x - dx

(9. 58 )

I F.!

"

We now observe Ihal (MIEI) tLT: represe nLS an element of area under the (M( E/ ) diagram, IIf1d XI (MIEI) tlx lhe first moment of that element wi th respect to a vcrtic.tI axis throu gh C (Fig. 9.44). The right hand member in Eq. (9.58). th us. rep resentS the first moment wilh respect to Ihm axis ofthc area locllted under Ihc ( Ali Ef) diagram betwecn C and D. We UIn, thc n:furc. stale the Sf!COIUJ mom ent- area tileorem as fu llows: Tlu IlIllgell lilll (/l! via/ion 1(:1lJ of C with reSIJert 10 D is l!qua/lo Ihe fir~·t mome nt wilh respect to a ['erfical (uis throll}:.h C of the tJrea lIIuler the (MI EI ) diagrlllll belweell C lIlll! D. . Recalling that thc first moment of an area with respect to an llxi s is equal to th ~ product o f tile area and o f Ihe di stilnec from its centro id to that axis. we may a l~o expf{·.s~ the second moment-area theorem as follows: felD "" (arca between C and D ) i

l

(9.59)

whe re the area reft:n; to the an::a under the (fIII Ef) diag ram, and where Xl is the distance from the centroid of the a rea to the velikal a)ds thro ugh Fig. 9.45

C (Fi )!,. 9.4SlI).

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Care should be taken to distinguish between the tangential deviation of C with respect to D. de noted by {em, and the tangemial deviati on of D with respect 10 C, which is denoted by IOIC' The tangential deviation 'DiC represe nts the vert ical distance from D to the tange nt to the elastic curve at C, and is obtained by multipl ying the area under the (MIEI) diagram by the dist ance X2 from its centroid to the vertical axis through D (Fig. 9.45b): tox "" (area between C and

D) X2

9.10. Cantilever Beams and Beams with Symmetric Loadings

(9.60)

We note that, if an area under the (MIEJ) diagram is located above the x axis, its first moment with respect to a vertical axis wi n be positive: if it is located below the x axis, its lirst mume lll will be negative. We check from Fig. 9.45, that a poim with a pmitive tangential deviation is located ablJl'e the corresponding tangem, while a poi nt with a negative tangemial deviation would be located below that tangent . ' 9.10. APPLICATION TO CANTILEVER BEAMS AND BEAMS WITH SYMMETRIC LOADINGS

We recall that the first momellt·area theorem deri ved in the preceding section defines the angle ODIC beflt'eell the tan ge n l~ at two poi nts C and D of the clastic curve. Thus, the angle aD that the tangent at D fomls with the horizontal, i.e., the slope at D, can be obtained only if the slope at C is known. Similarl y. the second moment-area theorem deli nes the vertical distance of one poim of the clastic curve from the tangent at another poin!. The tangential deviation ' DlC , therefore, will help us locate point D only if the tangent at C is known. We conclude that the two moment -area theorems can be applied effectively to the detennination of slopes and deileetions only if a certain reference tangent to the elastic c urve has first been determi ned. tIl the case of a cantilever beam (Fig. 9.46), the tangent to the elastic curve at the fi xed end A is known and can be used as the reference tangent. Since 0A = 0, the slope of the beam at any poi nt D is On "" 0NA and can be obwined by the fi rst moment-area theorem. On the other hand, lhe deilection YD of point D is equal to the tangential deviation IDfA meas ured from the horizontal reference tangent at A and can be obtained by the second moment-area theorem . Ln the ctlse of a simply supported beam A B with a symmetric load· illg (Fig. 9.47(1) or in the case of an overhanging symmetric beam with a sym met ric loading (sec Sample Prob. 9.11 ), the tangent atthc ccnter C of the beam must be horizontal by reason of symmetry and can be used as the reference tangent (Fig. 9.47h). Since Oc "" 0, the slope at the support B is all = flBlc and can be obtained by the first mome ntarea theorem. We also notc that b'lmlU. is equal to the tangenti al deviation IB/C and can , therefore. be obtained by the second moment-area theorem. The slvpe at any other point D of the beam (Fig. 9.47c) is found in a similar fashion. and the deilect ion at D can be expressed as Yo =

I DiC -

fBlc·

(rl

Fig. 9.47

571

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EXAMPLE 9.09 Determine the slope and deflect ion ilt end B of the pris matic L"ltn til e\'ef beam AB when it is load~d as shown (Fig. 9AS). knowing that the llexural rigidity of the beam IS

'l~I,N -",

1:.'1 = 10 MN · rn 2 •

We first draw the free-body diagram of the beam (Fig.

A

"" , _ _ _

9.49,,). Summ ing vt:ni<:~ 1 compullt:I1lS and mUlllt:n ts
WI': linrl1hm Ihl': rl':;Jc1inn

~11hl':

liter!

~nd

A con .~is1~ nf

~ ~O

Fig. 9 .48

kN

upward vertical force R.t and a 60 tN . m ;::Cluntcrdockwise couple M A . Next. wedmw the bending.morrent diagram (Fig. 'J.49b) and dctcmlinc from s imilur triang1c~ the di.tancc ,,'D from Ihe end A 10 Ih ~ point D of the beam where M = 0: XI)

3-

bO

3

x ;>

\)()

15U

XI> ""

\ 1, ,,,, lill l '

(.

1.2 m

nl

J '''11.1\ '"

Dividing by Ihe fle.turaJ rigidil y EI Ih~ values obtained for M. we draw the (MI£J) d i~gram (Fig. 9.50) and compute Ihe areas corresponding respective ly 10 the segment;; AD and DB. assigniug a positi ve ~ign 10 Ihe ,uea loc
"' ~l

+ A2 m I) +W .8 m)(9 >< 10 !o m "" - 3.6 X I() 1 + 8.1 X JO 3

k.'< . ,..

BJ'II = B~ - 1111 -'" area from A loB = A I

- - !( 1.2m)(6 X

1O~ 1

Fi g. Q.4Q

= + 4.5 X 10 l rad

l'.t. '"

and. since (J~ - Q,

I" -I+!J X

10-; ", -1

Os = +4.5

:x 10 . '

rad

Using now the second momenl·area theorem. we write tllUt the tung~ nrial deviation tn!,\ is equal to the tics1 moment :Jbout a vertical axis UlfO Jgh B of the tOlal
I 2

"' -;;1

"

.I

"1--,,·,~!i'L "------'R'- ' I

(

0 8 ",

L~m -~

I

- 6 X lO -l ", - ! ISlA

=

A,(2.6 m)

+ A1(0.6 m)

Fig. 9.50

'" ( - 3.6 X JO ')(2.6m) + (8. 1 X 10 ' )(O.6m) =

- 9.36 mm

+ 4.86

nUll =

- 4.50 mIll

Since th e reference tangent ~t A is horilOntal. the defieelion at B is equal 10 IWA and we have

The deflecled beam has been sketched in Fig. 9.51 .

572

lis "

tiJ;;,l

== -

1.5 X

\·rT r.d

r:~"'' ' Jjt Fi g . 9.51

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'9.11 . BENDlNG·MOMENT DIAGRAMS BY PARTS In many applicmions the determi nati on of the angle ODiC and of the tangential deviati on t Dle is simplified if the effed of each load is evaluated independently. A separate (M IEI) diagram is draw n for each l(lad. and the angle (IDle is (lbtllined by adding algebraically the areas under the vari ous diagrams. Similarly. the tangential de viali(ln fmc is obtained by adding the fi rst moments of these areas about a vertical axis through D. A bend ing-mUll'k!nl or (MIEI) diagram plolted in this fashion is said to be lfmwn by parts. When a bend ing-moment or (MIEI) di agram is drawn by parts. the various areas deli ned by the diagram consist of simple geometric shapes, such as rectangles. triangles , and parabolic spandrels. For convenience , the area~ and centroids of these variolLs shapes have been indicated in Fig. 9.52.

Shap..

An' ~

,

~ h ---1

Ik e,,,,,,,I.,

[1]]

b{.

,"

1-,-1

Ch~

Tri"rlle;I.,

~] ,

, M,

~b_1 Par"bo li." JPamlr/'[

281 ~h~ Y J%l1 , ,-M

,"

,"

-;-I

C ubk

spa"' lrd

'\

~

°C

It!

,,

-I,· f--

(',,'ne ra[

spand rel

rh~

'4] e

Fig. 9.52 shapes.

Me '"

I

"~

"+

I-

Areas and centroids of common

9.11. Bendlng-Momoot Diagrams by pans

573

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EXAMPLE 9.10 Determine the slope and detlection at end B of the prismatic beam of Example 9.09. drawing the bending-moment diagram by parts.

90 k N.

'" 1-- ------, " ~------------J8c'

+ ,' ~----------~Bc' - 1.50 kN.",

, 3m 9 X 10- 1 ",- '

fj

f----------1 1\,

A

1.5

11"1-1 B

- IS X

A

I;

---I

J

x+ ---

I

J"

In

.

ilr' ",- '

:2111

Fig. 9.53

We rep lace Ihe given l o~ding by the IWO equivalent loadings shown in Fig. 9.53, ~nd draw Ihe corres ponding bendingmomen t and (MIE/) diagrams from right to left. ~Iarting at the free end B. Applying Ihe first moment-area theorem. and recalling that /J" = 0, we write all = 081A = A ,

+

Al

= (9

X 10 - 1 m ')(3 Ill) - 1( 15 x 10 l m ')(3111 ) = 27 x 10 J - 22.5 x 10 J = 4.5 x 10 .I rad Applying the second moment-area theorem. we compute the firs t momen t of each area about a vertical axis through Band wrile t BiA = A ,( 1.5 m) + A l (2 m) = (27 x 10 3)( 1.5 m) - (22.5 X 10 3)(2 m)

)'8 =

A

= 40.5 mm - 45 mm = - 4.5 mm - 15 X 10-.1 11, - '

II is convenient. in practice. to group into a single draw ing the two portions of the (MIE/) d iagram (Fig. 9.54).

574

Fig. 9.54

"

H

c • 2,n--

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EXAMPLE 9.11 For me prismatic beam AB and the loading shown (Fig. 9.55). determine the slope at a SUppofl and the maximum denectioll.

A

_"1-"1-"-1-'-1 1 DIIIIIIII: B

I-'--

C

L

~ ,,,-1

Fig. 9.55

We first sketch the deflected beam (Fig. 9.56). Since Ihe tangent al the center C of Ihe beam is horizontal. it will he used as the reference tangem. ;,nd \\Ie have ll'l,n:u = 1,1.'r:. On Ihl'. nthl'.r hann . ,in!:e 8 c = O. WI'. writl'. /JL /A -

fh - 0,\ - - 0,\

"'

dj~11i1l1

Flom the fl<x-loo), lind tltnt

of the beam ( Fig. 9.57). we

R,\ = RB = W(I Nex t. we draw Ihe shear and bending-moment diagrams for lhe portion AC of Ih~ beam. We draw these diagram.> by pnns. comideJing sepa.rately the dfecl~ of the leaction R,t and of the distribuTed lond. H owe~er. for cOllwnience. Ihe two p~flS of each diagrnm h~ve been ploned together (Fig. 9.58). We recall (rom Se~. 5.3 that. the distrilJuted lOld being uniform. the COfres pondmg parts of the shear and bending-moment rliag:rams will be, respectivel}, linear
h~ ,

FIg. 9.57

~ - II -r- IJ-l

A

I (2 .; ;) _;;1 l'

A J = ;(20)

=

.,

H \ '"

u" -

A~ = - }"(
,,,, -

"

Applying the fir~t moment·aTea theorem, we write 2 11w l

Be.A = A 1 +A 2 =

T! -

I llll(l ~

11Jr/ l

.

AI!

= A 4(1

+ A ,7a

13

-4

c

~ {-tlfll')1

1L

and

l 1-wC 6£1

384£1

Applying now the second momet1l·area tht':orem. !

II ""

" -----j

-,, -~I -""

61:.."1 = ~

Recalling from Figs. 9.55 and 9.56 Ihat ilA = - ilctA ' WI: havt

OA =

lUI

\'

.1

(1(1(12) = - 6£1 If'll"

I

~

C

D

Tf':'pt'.<:riwly .

"

= (2.wa

J

)4a + ( _ .Wl1

£1 ) 3

l

W~

I,

write

)7l1 = 19wa 6£1 , 481:."1

,\

c

4

and Fi g . fl .5S

575

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J~ "~J -

0

, 0 '1

[

- - .- -1

SAMPLE PROBLEM 9.10 TIle prisnmlic rods AD and DB are welded IOgclher 10 form Ihe cami leller t>eam ADR Kn c.wiot] that Ihp. flex ur:ll ri!',irlil y I!: F.I in portion AD of rhe /learn and 2£1 in portion DIJ. determine. for the loading shown. the slope and delleclioo 31 eml A.

SOLlITlON (M/ E/) Diagram. We first draw th e bending-momcm dj ~gr.tm for the beam ;lnd then obtain Ihe (MIEl) diagram by di\'iding the value of M at each poi nt o f the beam by the co rresponding va lue or the flexu ral rigidity.

Rcl"crcnre Tan gl' nl. We ch.oose the horizonTal tangent at lhe fixed end B as the refNence tangent. Since Oil '" 0 and y" = 0 , we note tlnl y~ =

'IIIB

1l , 1;'n 'lC' I,",!;,"t - IJ~,_,

\

Slope at A.

'\ .6

.--- \

Dividing the (MIff) diagmm into the three trbngular por-

tions shown. \\'e write p{J.~

1 Pa

A, =

- "2 1::1 a ~

- ZEI

H I' I Pa A~ '" - 22El u = - 4EI 11PfI

A)"' - 22El a

,PfI"

=- W

Using the lint moment _:lCea theorem. we have

Pu!

p,,"

3Pu 2

-------- ~

2£1

4£1

41,.,

3 Pa~

+ -2EI

Of'necliol! al A .

576

1 ;~i nE

tht' second

mome;nt_:lr~:l

theorem. we h:lve

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SAMPLE PROBLEM 9.11 For the prismatic

be~m

and loading shown. determine Ihe slope and

den~ction

at end E

SOLUTION !AIIF.I ) nia~r:am . Froln a fref'.\1.ody 'hae.ram of rh~ N-A1m, \\If' (lett'Tminf' the rea: lions and then draw Ihe shear and bending-moment d iagrams. Since Ihe flexuml rigidit), of the beam is canslant. we divide each value of M by 1::1 aluj obtain Ihe (MIEI) diagram show".

Rcftrence Tangi'lIl.

Since the beam ;md its loading are S) mmetric wi th \0 (he midpo int C. the tangent at C is horizontal and is used l S Ihe ret~ erence langent. Referring to Ihe sketch. we observe lhal. since Or = O. re~pecl

(JE =

{h +

II)

BEle'" (J /!/c

(2)

J£ = IEle - IDle SlolK' II I E .

Referring

[0

the (AllfJ) diagram and u., iog lhe fin;l mome nt·

area theorem. we write

"

"

Usin!): Eq. (I). .....e have

_ A ~ _ ( _ 1f(l>L) ~ __ 11)(/'lL1 14 4£1 415£/

I

[)/C -

'm: = A,

4 ( +"4L) + A, (3,,) 1I

_( ""'L)(, I L) I ( ""')(3") 4EI

flEI

4

4

wa' 4EI

Using Eq. (2).

\\.'e

16EI

8EI

have 1/ ,'{jJ / .

11,,,4

4EI

8EI

-----)"t' -

;;;

(2 L

u)

we? "£ -

8EI (lL

+

aH· ...

sn

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PROBLEMS

Use the

m Olllent · a~a

method to solve the following

problcnL~ .

9.95 through 9.98 For the uniform cantilever beam and loading shown. determine (a) the slope at the free end, (b) the deflection at the free end

" 1,1

(-w::::::::..~B -

1 - 1

L- - i

Fig. P9.95

r

B

A

L

1

Fig. P9.96

"n

U.l lllll llll t IA

BI

I.

Fig. P9.97

Fig. P9.98 P '"' JKfI

'] IiI!ll li

j

9.99 and 9.100 For the uniform cantilever beam and loading shown. determine (a) the slope and deflection at (al point B. (b) point C.

1..-",-1- " Fig. P9.99

Fig. P9.100

9.101 For the cantilever beam and loading shown. detennine (al the slope at point B. (b) the deflection at point B. Use £ = 29 X 10" psi.

IO()llolILl

lIA nIITlI"1 n11"-"I~'i'JI"~' ~

A ,~ ~::';;j B C

"'' d

11'360

x

{j·1

1 -2. 1",

- - -30"' - - -Fig. P9.101

1- - - - 3 m

Fig. P9.102

9.102 For the cantilever beam and loading shown. detemline ((j) the slope at point A. (b) the deflection at point A. Use £ = 200 O Pa.

578

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Problem.

9 .103 For the cautilever b<:am and loading show il. detennim: the slope and deflection at (lIl point A. (b) point B. Use E = 29 >< 10" p,i.

f:i)LlII,

f>l.1O III

I I

j !!!!!!!!!!! [

li t

Lt B

(

1 I--~"~;"'I +-30;,,----1

)~

S4 X T.T

r

O,5n,

C

kl>

I---- 2.2 ",--------1

9 .104 For the cMtilever beam and loading shown. determine (a) the slop<' at poiDt A. (b) the deflection at point A. Use E = 200 OPa. 9. 105 Pm ule cantile\'cr /x,am and loading shown. dctennine the deflection and slope at end A caused by the momelll Mo.

1 --

Fig. P9.105

9.106

For the cantilever beam and loading shown. detemlllle (a) the

9.107 '£Wocover plates art' welded to the roIled,steel beam as shown. Using E - 200 Gra. determine (u) the slope at end A. (b ) the dene.:lioll al end .4.

ii X

~ L:N'·~"_ _"iiiJ~~i;;i;i~~~~

'I l·' U.S m~ I,2m -.! C

t20,nm

t

1V250 X

~2, 3

Fig. P9.107

9.108 T\~o cover plates are welded to the rolled·steel beam as shown. Us ing E = 29 X ](f psi. delerr.line Ihe slope and derleclion ::II end C.

Fig. P9,1OS

LIZ

Fig. P9.1Ofi

at point C. (b) the deflection at point C.

(

L5El

rI' uo- I:I £1

J: \\'.'250 X .'2>;. 4

I

Fig. P9.t04

Fig. P9.103

slo~

579

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580

9.109 through 9.114 For the prismatic beam and loading shown. determine (a) the slope at end A. (b) the deflection at the center C of the beam.

Deflection 01 Beams

p

" p

Ap~=+:=_oq'B C

L U2-11---U2-f Fig. P9.109

Fig. P9.110

"L:cl,1"1"~E 4

of

4

4

Fig. P9. 111 II -

fl.

Fig. P9.11 3

I(

rTm , C DrTm

E

Fig. P9.114

9.115 and 9. 116 For the beam and IO
B

C

D

~qE

Fig. P9.11 S

Fig. P9.116

9.11 7

Knowing that the magnitude of the load P is 7 kips, determine

(a) the slope at end A. (h ) the deflection mend A. (t·) the deflection
t5

ki["

"io"'!",=,C ii'''"'''=IIjD ,,,.!E SG X 12.5

Fig. P9.117

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Problom3

9 .118 and 9 .119 For Ih::. beam and loauing sJlown. delel'llllm: (u) tlu: slope at endA. (h) th~ deflect ion at ille midpoint of tile beam. Use £ '" 200 G Pa.

"''" r.;;;"81I 1111111110'"'' ~"' I

~

t;J

r:-;;

ff)

S2.'5I.J :<378

~3.€1tl~

Fig. PS.118

kN .11

9. 120 For the beam and loading of Pmb. 9.1 17. dctcnnlne (
9.121 For Ihe beam and loadmg shown and kno ..... ing Ihat 1(1 ' " II kN/rli. detennine (II) the slope ;)I end A. (hi th e deflection al midpoint C. L:se E = 200 GPa.

( * "" ;111

til

I~) X

1--5 , , --J- 5 ''' --~

W3 1O X 60

Fig. PS.121 and P9.122

9 .122

For Ih" beam and l."uling s hown. delermino the vo.luc o f • .., for

which the ddleclion is zero:"l.l the midpoint C 01 the beam. Use E = 200 G Pa. "'9 .123 Determine the

\0 be supponed at IWO poims 8 and D. for which the slope {II ends A and E i ~ zero.

A uniform rocl AE is di~l ::lnCe"

1- ;J2- lc

l,,

t 511 \.:'\

~5m l-2'" Fig. P9.119

In 1<.." . to,,'TTTT,"": ' " , ,-i-W I.N

15t'L~" 1

D

"-I_j~ I ~J3=,,_1E t- -_ I Fig. PS.12 3 and PS.124

"9 .12 4 A nniform rod A~~ ;.<; TO he ~ nr]lflnl".rl :11 Iwn pninl!; R anrl f) Determine the distance II from Iht' ends of the roo to the points of su ppoJ1, if Ihe downward deneclions of pom ts A. C, and E are 10 be equnl

581

OOk:-- . m

;:";

\\'460 X 74

I

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582

· 9.12. APPLICATION OF MOMENT-AREATHEOAEMS TO BEAMS WITH UNSVMMETRIC LOADINGS

Dellectlon 01 Be .......

I,,)

I --

L

1/,)

Fig. 9 .69

We SilW ill Sec . 9 . [0 111111, wht1l il simply suppunct.! 01 QVCl haugillg ocilln carries a symmetric load, lhe. langenl .tl lhe center C of lhe. beam i~ ho r· izontal and can be used as the reference tangent. When a simpl y supported or overhanging beam carries an unsymmetric load, it is generillly not possible to determi ne by inspection the point of \he beam where the langent is horizontaL Other means must then be fou nd for locating a reference Inngenl, i.c., a I~n ge nl of kno\\ln slope. 111 he IIsNI in aprlying either of the two moment-area theorems. It is usua.lIy mo:;t convenient \0 select the reference tangent at one of the beam supports. Consideting, for example. the tangent at the supper! A o rthe simply supported beam A B (Fig. 9.59a), we determine its slope by computing the tangential deviation tS 110 o f the support B with rc~ pcc.1 to A, lint.! di viding fBi... by the di stance L belween the supports Recalling that the tangential deviatio n of a point located above the tangt:flt i~ positive, we. wriie

(9.61 )

---- - --------/J A f)

.\

Oncc thc slope of the reference tangent has been found , the slope

OD of the be
(9.62)

Fig. 9.60

H

"

D

I 'n,

The tangential deviation IO}'" of D with respect to the support A can ueuulaiutci frollllilc seculld Illotllcltt-area theurclJl. We nOlC lhal/IlI'" is c{lualto the segment ED (Fig 9.61) and represents the vertkal distance o f D fro m the referellce /aligelil. On the other hand. the den ection YD o f point D rep resenB the vertical di stance o f D fro m the ho riwntallillf' AB (Fig. 9.62). Since YD is equal in magnitude to the segme nt FD, il

R" I" '"f"JK~' 'I"~

h,,,~, ·" t

I "~B F

Fig. 9.61

D

Fig. 9.62

"

I- <~ -- , D

EF

HB

x

L


E Fig. 9.63

can be expressed as the di ffere nce between EF and ED (Fig. 9.63). Observ ing from the similar triallgles AFE lind ABH that

"

X

.I'D = ED - EF = lOlA - i.fBIA

(9.63)

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EXAMPLE 9.12

r

For the prismatic beam and loading shown (Fig. 9.64). deter~ mine the slope ami deflection at poil1l D.

I'

1/

A

• J)

f - - - L---J[ Reference Tangent at Support A. We compute the reactions at the suppons and draw the (MIEI) diagram (Fig. 9.65). We detenmne the tangential deviation IBM of the sup· port B with respect to the support A by applying the second momel1l·area theorem and computing the moments about a vcr· tic;)] axis through B of the areas A I and A~. We Iwve 1 L 3PL AI

=:2 '4

16£/

3PL'-

A~

= 128£1

Fig. 9.64

L--1

1 3L 3PL 9PL~ 16EI = 128EI

= :2"4

I\. ~ '" I

=A(~2 + 34L) +A2 (~)

IBIA

I

3/JL2 IOL 12

9PL' L

7pe

128£1 2

128EI

~ --- + --- ~ --

128EI

The slope of the leference tangent ;)t A (Fig. 9.(6) is

181..1 = _ 71'L" 128El

L

-,I

Slope at D. Applying the first momelll-area theorem from A to D. we write

3PL'"

J)

I - ~J-~.

•

FIg. 9.65

128EI

Thus. the slope at D is 7PL~

3fJL 2

---+-- ~

128EI

128E1

Deffeclion at D. We first detemline the tangential deviation DE = IIJIA by computing the moment of the area A j tlbout a vertical axis through 0 :

L)

DE

= l/)tll = A I ( 12

1 1 3PL L PL = 128EI12 = 5121:"/

The deflection at D is equal to the difference between the segmenls DE and EF (Fig. 9.66). We haw

1\<'I'·r"'I("1.110>«·,,1

Fig. 9.66

J'v =: DE - EF = IIJIIl - ~ IIII.~ fJL 1 1 7fJL 1

)'

v

-- -

----

512£1

4 128El

3PL J

=: - - - =

256El

- 0.01 I 72/J L' IEI

583

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584

*9.13. MAXIMUM DEFLECTION

Deflection 01 Beams

When a simply supported or overhanging beam carries an unsymmetric load. the maximum dellection generally docs not occur at the center of the beam. Thi s will be the case for the beams used in the bridge shown in Fig. 9.67, which is being crossed by the truck.

Fi g. 9.67

The defi ections of the beams used for the bridge must be reviewed for different possible positions of the load.

,F- L---l

-'-

~

_

_

~

_____

~~~_

'" < 0 IJ~ =

n

B

I

I~I.I

I\crcn'''~c' /

To detennine the maximum dcllection of such a beam, we should locate the poi nt K orthe beam where the tangent is horizontal, and compute the dellection at that poi nt. Our analysis must begin with the determination of a reference tangent at one of the supports. If support A is selected. the slope OA of the tangent at A is obtained by the method indicated in the preceding section, i.e., by computing the tangential deviation fBi... of support B with rcspect to A and dividing that quantity by the distance L between the two supports. Since the slope 8 K at point K is zero (Fig. 9.68(/), we must have

~...,w·t

(h)

Fig. 9.68

B

Recall ing the lirsl moment-area theorem, we concl ude that point K may be determined by measuring under the (MIE!) diagram an area equal to OK.fA = - 8... (Fig. 9.68h). Observing that the maximum uellection Iyln"" is equal to the tangential deviation tAlK of support A with respect to K (Fig. 9.68a), we can obtain b'ln.. , by compllting the first moment with respect to the vertical ax.is through A of the area between A and K (Fi g. 9.68b).

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EXAMPLE 9.13 Detemline the maximum dellectioll of the beam ofEx.unple 9. [2. Determination of Point K Where Slope Is Zero. We recall from Example 9.12 that the slope at point D. where the load is applied. is negative. It follows that poillt K. where the slope is zero. is located between D and the support B (Fig. 9.69). Our computations. therefore. will be simplified if we relate the slope at K to the slope at B. rather than to the slope at A. Since the slope at A has already been determined in Ex· ample 9.12. the slope at B is obtained by writing ON = /J"

+ ON!J.

= 0",

7PL2

ON = - 128E/

+

+ A , + A,

3PL1

9PL'

5PL 2

[28£1

+ 12REI

= 12REI

p

Observing that the bending moment at a distance II from end B is M = ! PII (Fig. 9.70a). we express the arcolA' located be· tween K and B under the (MIEJ) diagram (Fig. 9.70b) as I PlI /J1/ A' = - - - II = - 24EI REI

A~~Q~;~U~

By the first moment·area theorem. we have

Ih u

(lIlIK = (IN - OK = A'

and. since 8 K = O.

fiB

E

88 = A'

1/1".• ,=18'1:

Fig. 9.69

Substituting the values obtained for 8 8 and A'. we write PII 2

5PL2 --

and. solving for

~

--

12R£1

REI

7v'S L =

O.559L

II.

II

=

Thus, the distance from the support A to point K is AK = L - O.559L = O.44 IL

Maximum

Deflection. The maximum deflection t&X and, tlllls, to the first moment of the area A' about a vertical axis through B (Fig. 9.70b). We write

Iylrna. is equal to the tangential deviation

~' I ",.. =

IVK

= A'

(---:32,)

p,,'

= REI

Substituting the value obtained for

II.

(2") "3

p,,'

= 12EI D

we have

I" I-

,, -

I"

-

Fig. 9.70

585

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586

Doflectlon 01 Beams

'9.14. USE OF MOMENT-AREA THEOREMS WITH STATICALLY INDETERM INATE BEAMS

The reactitlflS at the supports o f a stiltically inde terminate beam can be dctennined by the moment-area method in much the same way that was described in Sec. 9.8. In Ule case uf a beam indeterminate to the lirst degree. fo r example. we designate one of the reactions as redundant and d imin;1tc or modify accordingly the correspondi ng SUppOT!. The redundant reaction is then treated as an unknown load. which, together with the uther loads, must produce deformations that are compatible with the ori gi nal supports. The compati bility condit ion is usually expressed by writing that the tangential d ~via l ion of onc support with respect to another e ither is zero or has a predetermined value. Two separate free- body di agram~ of the beam are drawn. One shows the giren lollC/,,' {llId the corresponding reactiolls al. the supports that have not been eliminated : the o ther shows the redlllldalll reaction and the cOlTc.I"pmlllinl!, reactioll.\· at the same supports (sec Example 9.1 4 ). An M/ EI diagram is then draw n lo r each of the two loadi ngs, and the desired tangential dev iations arc ohtained by the second moment-area theore m. Superposing the resul ts o btained. we express the req uired compatibility condition and de termi ne the red undant reaction. The other reactions :Ire obtained from the free-body diagram of beam. Once the reactions at the su pports have been determined, the slope and de n ection may be obtained by the moment-area method at an y other point uf the beam. EXAMPLE 9. 14

Delermine the reaction :llihe supports for the prismalic beam :lnd loading shown (Fig. 9.71). We consider the couple exerted at the fixed end A as redundant and replace the ftxed end by a pin-and-brackel suppot1. The couple J\.1,t is now considered as an unknown load (Fig. 9.72a) and will be delennined from the condition thai the langent 10 Ihe beam at A must be horizontal. IL fo llows that this tangenl must pass Ihmugh the suppot1 B and. thus, that the tangential devialion IBJA of B with respect to A must be zero. The solution is carried OUI by computing sep:lralely the langential deviation (IB/".)u.caused by the unifonnly distribllled load 10 (Fig. 9.72b) and Ihe tangential deviation (1 81,1).11 produced by Ihe un known couple 1\.1'1 (Fig. 9.72c).

(II)

FIg. 9.72

Ihl

Fig. 9.71

",

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Consklering tirst the free-body di:!gram of the beam un
"

1111111111 II

{ut

l .."

(9.64)

We can now dmw the corresponding shear and (MI EJ) diagrams (Figs. 9.73b and c). Observing that MI ff is represented fo r by an arc of parabola. and recalling the formula. A =0 the area under a parabola. we compute the first moment of this area ... bout a vertical [I.~i s through B and write

twL"

(9.65)

,

I

;bIr.

I

/,..(* !l·L ~1 B

~-I

(i!)

Consi
,

(9.66)

"r

-1- IL'L

"

'~I

Drawing the corresponding (MIEI) diagram (Fig. 9.74b). we apply again the second moment-area theorem and write

l ."

I.

------

,

r-t~

~

B

\1

ie, I.

Fig. 9.73

Combining the results obtained in (9.65) and (9.67). WId expressing that the resulting tangential deviation 1111" must be zero (Fig. 9.72 ), we ha ve I~I~ = (I BIA)"

+ (Ie",)",

1VL~

MAC·

24EI

3El

=0

)1 , (II )

0

---- ~ O

I

"

,.

R~ = (R')I + ( R~)~ = ~~oL + kl0 L = ~ "WL Re = (Rs)1 + (Re)~ = }w L - ~w L = ~ wL

I

I.

~

and, solving for MA •

Substituting for M~ into \9.66). and recalling (9.64). we obtain the values of R~ and Re:

l .,

(1."

(IJ)

-~

"

B

I,

I-~

,

Fig. 9.74

In the example we have JUSI considered, there was a single redundant reaction. i.e.. the beam was ~·tl!tically indeterminmc /0 Ihe first degree . The mOll/elll-area Iheorems can also be used when there arc additional redundant reactions. As discussed in Sec. 9.5, it is then necessm)' to write additional equations. Thus for a beam thaI is stalimfly indetenllillllie to Ihe .I·eeo/Ill degree. it would be necessary to se!ecttwQ rcdundants and write two equations considering the dejonllllfioll.l" of the stmcture involved.

587

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SAMPLE PROBLEM 9.12 For the beam and IO:Jding shown, (a) detennine the deflection at end A. (b) evaluate )'A. for the following data: IV]OX)3: / = 170in4 a = 3 fl = 36 in. 'IV = D.S kipslft = [l25lh/in.

E = 29x lOb psi L = 5.5 ft = 66 in.

SOLUTION

(MIEI ) Diagram. We firsl draw the bending~moment diagram. Since the fle.~uraJ rigidity EI is constant. we obtain the (M I Ef) diagram shown. which consists of a parabolic spandrel of area AI an d a triangle of area Al . A

R,

"

"'ii A

;"1

A, =

1(- ~;/2)

A2 =

('/Oa 2EI

I

1

2" -

~;:

a = -

)

lIH/'L

L= -

4EI

Reference Tangent at B.

R

c

se~'ond

s hown. Using the

The reference tangent is drawn at point B as moment -area theorem. we delennine the tangential

deviation of C with respect

\0

B:

I~~~)

L lOB = A 2 23 = ( _

2£/

W:;~l

L

23 = _

From the similar triangles A"A'B and CCB. we find

,. ,.

A A =

1m

(")

L

= -

wa' L' (")

L

6£1

wa ' L 6£1

= -

Again using the second moment-area theorem, we write ' AiB = 0..

J"

"4 =

A!

(

wa -

J

6£1

)

_wa

30.

"4 =

-

4

8EI

Dcncclion at End A

~L'(/~

(

.I'·= 8EI

b. Emluati on of Y..t'

8(29 X

4

L)

+ J-;;J
Substituting the d2ta given. we write

( 1125 lblin. )(36 in. t )'A =

I

IO~ Iblin 2)( 170 in~)

466in_)

( I

+ "3

36 in.

Y,I = 0. 1650 in,,J- ...

588

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,~

"" 25 kN/1Il

1111111111

A

SAMPLE PROBLEM 9.13

:I

R

\x

\\"230

"'

't'

"" I111111

j R,~ f I ", I -"~I' -I -

L

For the beam :lnd louling showl\. delermine the magnimJe and location of Ih e lare"",r ddlt~c l;OIl . IJo;e F. == J()or; p~ ~3

SOlUTlON USing the free-body

ReactiOll'i.

R,\

~

16,81 kN

dl~grnm

t

R~

of the entire beam. we rmd

= 38.2 kN

t

(AilE/) Diagram, We draw the (Mi ff) diagr.:rlll hy parts. considering s~paraldy

lht: t:ni::(;I~ of lIlt: re
-

"" "t-= '---- -=:::: --:-:-t'"- .' ~ ,\c

~

I

"liZ

' T:l-

Rl"frn>ll('(' Tan gent. The l~nf:p. nI f() rhe he.,1m:11 ~lIp[lOn A ;~ c hmen ~~ the reference tangem. U. ing the second moment-area theorem, we detennine the tnngential devi atio n I NI " of support 8 Wilh respect to SUppOrl A:

2£1

L b IB/A= A' 3+ A'"4

,

"' ~l

(HAL") L

=

lEI

(

3+ -

1 11Jb )

6£1

b

"4=

4

Rj,U

wb

6£1 - 24£1

Slope at A (I )

, '11..,.

1--- -1..- ---1

J

Largcsi Dcncction. The largest d~nectioll occurs a1 point K. where the slope_of the beam is zero. We wrile ther~fore " )

8m

(3 )

We substitute for 0" and

(1';1),

from Eqs. ( I) and (3) mto Eq. (2):

2;:J+[:": -61~1

_ (:~~l _

(x"

_a)Jj ~O

Substituting the nllll"klncal dntn. we hnve

- 29.53

10'

Ei +

,

](Y

10'

8.40S_t;;, EI - 4.167(x., - 1.4)1- = 0 EI

Sol vi ng by trial and erroc for xm • we find

.\'''' "" 1.8% rn
Computing the moments of A1 and A4 alxllil a ver1icaJ axi s through A, we ha l'e

lY l..

= ""'; =

l<. + Al) R".r.,

A 4

W!I

[ + "43 U

(x., - (/ 1

)1

1V

4

= 3EI - fiEI (x", - a) - St.1 (x .. - a)

Using the gi ven data. R" = J6.8 1 kN. and I = 28.9 X 10 • nl' . we find y,, = 6 .J9 Ill m~
589

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SAMPLE PROBLEM 9.14 For Ihe uniform beam and loading ShOWII. determine the reaction at B.

SOLUTION The beam is indetermin<:tte to the lint degree. We choose the reaction RB as redundam and consider separatel) Ihe distributed loading and lhe redundalll re-

£

tk

,

"

I I I I I If .1 1_ ~ __ ~-1


, A

'" "

the s imilar triangles ASH' and ACe'. we lind Ihal

11111,lldll IR , I,=

:x

"',

c (1t"'1

1-' i ~

For eJell loading we draw tile (Mi ff ) diagram
[)btri hu l~d

wilIJ in g. CUllsilkrinl! lht (MIEI; diagram frum emJ A lU an arhirrary poinl X, Wf'. wr;lf

~--;'I...X '~'1

1-1- --:::--

/ ,," A]

(I)

ge nti al d ev iatiolls of lJ and C widl respect

I ' _. ..

• ,\

A

c

,<,,2 CF./

uuing succe;sively x = L and x =

;L. we have

Red undant Reaction Loading

L L (lR~L)L (IRsL)L 4R L '2 3£1 '3 '9 + - '2 3£1 L "3 = - 81 £I 2L [ 1 lRsL (2L)]2L 4 RsLl (18/")R - A~ 9 - - 2" 9EI 3 9 - - 243 D B

(Ic/A)R = A ] '9 + A4 '3 =

Combi ned Load ing. l ei.'

WL4 = 24£1 -

Reac t ion al B.

Adding the resuhs obtained. we

4 ReL l

81 B

I~", = 243

wril~

(wL' - R~})

EI

Substituting for I q ,\ and 'BiA into Eq. (I). we ha ve

No = U.6!HSwL

590

4

3

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PROBLEMS

U~C

th e m o m e nt -arcH me thod to soh -t th e fo llo v.i ng problems.

9 .125 through 9. 12B For the prisma1ic beam and Joadillg shown. dctennine (ul the deflection III point D, (b) the dope a1 end A.

it J

Fig. P9.125

Fig. P9.t2fi

" D

B

A

1-.

w~

J

Fig. P9.127

9 .129

Fig. PU28

For Ihe be.11ll and loading shown, delel1nine (a) the slope at point

A. (b) the deflection at po int D. Use E - 200 GP3.

(' ,111': 1 1:.' 20

WI2)(?fi

B

A ~~!!!!!!!!!!!!!!"'l'

~1. .,.!.os-, -,~-3.0"'~

\\12.50 " 14.8

Fig. P9.t30

Fig. pg.129

9 .130

For the beam and IO:lding shown. detemline (ll ) IIII' slope m point A , (h) Ihe deflection;)( poim £. U se £ eo 29 x 10' poi.

For th e timber beam aJ1d loading ~hown . determine (a) Ihe slupe m point A. (1)) the d~lkdion m po ill! C. Use £ - 1.7 x 10" psi. 9. 131

~ :l

;). in .

:uu llt-f!

cl !! !!

D

"

"

I :! fl I

oIn ~

-It-

IJ

Fig. P9. 131

591

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592

Deflection 01 Beams

9.132 For the beam and loading shown. delermine (u) the slope al end A. (b) the deflection at point D. Use £ = 200 GPa.

l:

B

\\'ISO X 24

Fig. P9.132

Fig. P9.133

9.133 For the beam and loading shown. delermine (a) Ihe slope at point A. (b) the deflection at point A. 9.134 For the beam and loading shown. delennine (a) Ihe slope at point C, (b) Ihe deflection at point C

15(111.

A~IIIIr:=:"II==~H~I[:ic

,,!oiDioo;___~E~.B

10" i

I - - - - L - -+_ Fig. P9.134

U!kN

A

Fig. P9.135

9.135 Knowing that the beam AB is made of a solid steel rod of diam(,-Ier d = 0.75 in., delermine for the loading shown (a) the slope at poinl D. (h) the deflection at point A. Use E = 29 X 100 psi.

1 k:>'/II '

b••"~;:':::l• .r~C:::"1Il

-L

[lJ~"m

_I f.-' 3()

Fig. P9.136

nn"

9.136 Knowing thai the beam AD is made of a solid steel bar. determine (u) the slope al point B. (b) the deflection at poinl A. Use E = 200 GP3. 9 .137 For the be.1m and loading ~own. delennine (a) the slope at point C, (b) the deflection at point D. Use E = 29 X 10" psi.

B

C

A

D

~6f.-l-,,,~

\\,12

x 30

Fig. P9.137

Fig. P9.138

9. 138 For the beam and loading shown, delermine (a) the slope at point B. (b) Ihe deflection at point V. Use E = 200 GPa.

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9.139 For the beam and loading shown. determine (a) the slope at end A. (b) the slope at end B. (c) the deflection at the midpoint C.

Problems

p

I'

D

Ai:~~E~/~~c!i1ic=='~E~/i!t:i

L U2

E

A

B

1_:~_2:J- :3~

U2

Fig. P9.139

Fig. P9.140

9 .140 For the beam and loading shown. determine the detlection (a) at point D. (b) at po~nt t:. 9.141 through 9.144 For the beam and loading shown. determine the magnitude and location of the largest downward dellection. 9 .141 Beam and loading of Prob. 9.126 9. 142 Beamandloadingof Prob.9.128 9. 143 Beam and loading of Prob. 9.129 9.144 Beam and loading of Prob. 9.130 9.145 For the beam and loading of Prob. 9.135. determine the upward detlection in span DE.

large~t

9.146 For the beam and loading of Prob. 9.138. detemline the largest upward detlection in span AB. 9.147 through 9.150

Fur lhe beam ;mu Juauing shown. ut:lt:nnine lht:

reaction at the roller suppon.

Fig. P9.148

Fig. P9.147

]"I ~1 L

Fig. P9.150

Fig. P9.149

9.151 a nd 9.152 reaction at each suppon

For the beam and loading shown. determine the p

B

- -1..- --+ FIg. P9.151

Fig. P9.152

593

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594

9 .153 A l,ydraulic jack Cd n be used IQ raise poi ,u B of the call1iJel'e, be:nn ABC. ·Ilt ~ be:nn was originally straigh t. horizonTal. and unloaded. A 20.k!\' load was thtn applied at poilll C, causi ng this poinllo mOlle down. Determine (al how much point B should be raised to return poiDt C 10 its onginal posi·

o"nOCllon of Btlon,.

lion, (0) the fina l

v alu~

"

of Ihe reaclion at B.

U ~e

E = 2()() G Pa.

"'"' 'I D

\\' 1311 x: 23.S

1V14

Fig. PQ.153

x

31l

Fig. pg.154 9 .154 Determine th~ reaction althe roller suppon and dra w lhe bendin g. moment diagram for Ihe beam and loading s hown. 9.1 55

For Ih e beam and looding

~hown .

demmu ne the

~ prillg

conSlant k

for wluch the force in the s pring is eljual to one·third of tlte 100al bad on Ihe beam.

"

;£! ! ! ! IJ ! I ! ! ! Ie i&i "

BH

!-- l"

L- -

Fig. P9.155 and P9.156

9.156 For Ihe beam and loaling shown delCrmine the spring constant k for which the bending moment al n i ~ Ale wL!/IO.

REVIEW AND SUMMARY FOR CHAPTER 9

Thi s c hapter was devoted 10 tlte delenninalion of slo pes and detlcctions of beams under tmnsverse loadings. Two approaches were used. First we used a mathematical method b,L<;ed o n the methcxJ of integl aliuJi of a diffel cl ltial eq uation 10 gel tlte slopt:s and ddlcctiOlIS at :my poi nt alnng the he~m. We rh r. n 1J~C{llhe IIU1I11I'II/'(lf"(>(1 IIIl'flm(/ In lind the slopes and deneclion~ at a given poi nt along thc beam. Particular emphasIs was placed on the computation 01 the maXlIllum denec ti o n of (t beam under a given loading. We also applied these melhcxJs fo r dc tennini ng de n ections 10 the analysis of indeterminate healllS, [hose in whic h lhe number of reactio ns al the supports exceeds the number o f equilibrium equatIons avai lable to de termine thcJ;c unknowns.

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595

Review and Summary for Chapter 9

We noted in Sec. 9.2 that Eq. (4 .2 1) of Sec. 4.4, whieh relates the curvalUre II p of the ne utral surface and the bending moment M in a prismatic beam in pure bending. can be applied to a beam under a transverse loading, but that both M and lip will vary from section 10 secti on. Dcnoting by x the distance from the left end of the beam. we wrote _ ~ M(x)

P

Deformation 01 a beam under transverse loading

(9. 1)

EI

This equation enabled us to determine the radius of curvature of the neUlral surface for any value of x and to draw some general conclusions regard ing the shape of the deformed beam. In Sec . 9.3. we discussed how to obtain a relat ion between the deflection y of a beam, measured at a given poi nt Q, and the di stance x of that poilll from some fixed origin (Fig. 9.6b). Such a relation defines the elastic cun'c of a beam . Expressing the curvature li p in terms of the derivati ves of thc function y(x ) and substituting into (9. 1), we obtained the following second-order linear differential equation: 2

dy

M(x )

dx.1

EI

P,

Fig.9.6b

(9.4)

illlegrating this eq uation twice. we obtained the following expressions de fiH illg the slope O(A) - dy/ch aud the defl ection Y(A) , respectively: EJ-dy = dx

El y =

i'

M(x) dx + C 1

(9.5)

0

r

dxfM(X ) dX + C1x + c;

o

(9.6)

0

The product El is known as theflexuml rigidity of the beam; C 1 and C2 are two constants of integration that can be determined from the houndal)' ("ouditiom imposed on the beam by its supports (Fig. 9.8) [Example 9.01]. The maxi mum deflection can then be obtained by determining the value of x for which the slope is zero and the corres ponding value of )' [Example 9.02, Sample Prob. 9.11.

'I

B

A

I !/8=O I (a ) Simpl)' su pported I)('~m

I ~_\=O 1 (h ) OIuhallging

Fig. 9.8 Boundary conditions for statically determinate beams

I ~;I ~ lI I h"~Hl

Boundary conditions

P

'I

P

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596

Deflection 01 Beams

Elastic curve defined by different functions I'

It =t.!i~=() J B -,

When the loading is such thai differe nt analytical functions are required to represent the bending moment in various portions of the beam, then different differential equations are also required, leading to different functions representing the slope O(x) and the denection y(x) in the various portions of the beam. In Ihe case of the beam and loading considered in Example 9.03 (Fig. 9.20). two differential equations were required, one for Ihe portion of beam AD and the other for the portion DB. The first equation yielded Ihe fun ctions OJ and YI' and the second the functions 82 and )'1' Altogcther, four constants of integration had to be determined ; two were obtained by writing that the denections at A and B were zero, and the other two by expressing that Ihe portions of beam AD and DB had the same slope and the same dellection at D.

1_' ~ :L~,=tJ21

We observed in Sec. 9.4 Ihal in the case of a beam supporting a distributed load w(x) , Ihc elastic curve can be determi ned directly from w(x) through four successive integrations yielding V, M, 0, and y in that order. For the cantilever beam of Fig. 9.210 and the simply supported beam of Fig. 9.2Ib, the resulting four conslants of integralion can be determined from the four boundary conditions indicated in each part of the figure [Example 9.04, Sample Prob. 9.2].

II =t t '!l1 =~hl Fig. 9.20

'I

A~B 1'11=0] If/.I =0]

A

'

/ 1',,- "" 0] !.I18=0]

[ '1.I ~t!1

[!i8=

[Ml=O]

1M ~ =

(n ) r:a!lti~'\'<'r l ~'''lll

Fig. 9.21

11 1 uJ

(b) Simply Sl1Pt)(lftt'J [)('alll

Boondary conditions for beams carrying a distributed load.

Statically indeterminate beams

In Sec. 9.5, we discussed statically indeterminate beams, i.e .. beams supported in such a way that the reactions al the supports involved four or more unknowns. Since only three equilibrium equations are avai lable 10 determi ne these unknowns, the equilibrium equations had to be supplemented by equations obtained from Ihe boundary condit ions imposed by the supports. In the case of the beam of Fig 9.24, we noted Ihat the reactions at the supports involved four unknowns. namely, M A • Ax> A}, and B. Such a beam is said 10 be indeterminate to the first (legree. (If five unknowns were involved, the

I

,,,d.

U2 ~

;:-e .\I ~

" L

'"'

Fig. 9.24

A, (bl

B L M

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beam wou ld be indeterminate to the second degree.) Expressing the bending moment M(x) in terms of the four unknowns and integrating twice [Example 9.051. we determined the slope 8(x) and the deflection y(x) in terms of the same unknowns and the constants of integration C] and C2 • The six unknowns involved in thi s computation were obtained by solving simultaneously the three equilibrium equations for the free body of Fig. 9. 24b and the three equations expressing that 8 = 0.), = 0 for x = O. and that y = 0 for x = L (Fig. 9.25) [sec also Sample Prob. 9.3].

Review and Summary for Chapter 9

[ .., = 0, II = U

I

[:r =o.y = 01 The integralion method provides an clfective way for determining the slope and deflection at any point of a prismatic beam. as long as the bending moment M can be represented by a single analytical function. However, when several functions arc required to represent M over the enllre length of the beam, this method can become quite laborious. since it requires matching slopes and deflections at every transition point. We saw in Sec. 9.6 that the use of singularity fimctions (previously introduced in Sec. 5.5) considerably simplifies the determination of 0 and)" at any point of the beam. Considering agai n

I, = L.y - 01

Fig. 9.25

Use of singularity functions

p

p

I.

A

D

~P

"P

Fig. 9.17

Fig. 9.29

the beam of Example 9.03 (Fig. 9.17) and drawi ng its free-body diagram (Fi g. 9.29), we expressed the shear at any point of the beam as

V(x) =

3: -

p(x -

~ L)O

where the step function <x - ~L)U is equal to zero when the quantity inside the brackets ( ) is negal ive, and equal to one otherwise. Integrating three times. we obtai ned successivel y M(x) = -3P x - P (x - "4'L) 4 dy EI(J = £1 - = lS pr _l2 p(x - 4! L)' dX

(9.44)

+ C]

(9.46) (9.47)

where the brackets ( ) should be replaced by zero when the quantity inside is negative, and by ordinary parentheses otherwise. The constants C 1 and C1 were determined from the boundary conditions shown in Fig. 9.30 rExample 9.06: Sample Probs. 9.4. 9.5. and 9.6].

.... ~ II··1

Fig. 9.30

=('1

597

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598

Deflection 01 Beams

Method of superposition

Statically iooeterminate beams by superposition

The next ~ec tion was devoted 10 the method of superposition, which consists of determining separately, and then adding, the slope and deflecti on callsed by the variolls loads applied to a beam [Sec. 9.71. This procedure was facilitated by the usc of the table of Appendix D, which gives the slopes and deflections of beams for various loadings and types of support [Example 9.07. Sample Prob. 9. 7] . The method of superposition can be used effectively with statically indeterminate heams rSee. 9.8]. In the case of the beam of Example 9.08 (Fig. 9.36), which involves four unknown reactions and is thus indeterminate 10 the first (legree, the reaction at B was con-

"

Fig. 9.36

sidered as retlllndanl amI the beam was released from that support. Treating the reaction RBas an unknown load and considering separately the deflections caused at B by the given di stributed load and by R B , we wrote that the su m of these defl ections was zero (Fig. 9.37). The equation obtaincd was solved for Rn [sec also Sample Prob. 9.81. In the case of a beam indctcnninate to the .I·econd degree, i.e .. with reactions at the supports involving fi ve unknowns. two reactions must be designated as redundant, and the corresponding supports must be eliminated or modified accord ingly [Sample Prob. 9.91 .

A

Ie)

(bl

Fig. 9.37

First moment-area theorem

We next studied the detennination of deflections and slopes of beams using the moment-urnl m et/u)(l. In order 10 derive the lIIomelll-lIrea theorem.\" rSec. 9.91. we first drew a di agram representing the variation along the beam of the quantity MIEI obtained by dividing the bending moment M by the flexural rigidi ty EI (Fig. 9.41). We then derived the.fir.~t !/Ioment-area theorem, which may be stated as fo llows: The area under the (MI Ei) diagra m between two poilll.~ i.\" eqllal to the angle between the /lIngent.\" to the ela.~tic CIIn'e drawn at these point.I". Considering tangents at C and D. we wrote OD/C

= area under (MIEI) diagram between C and D

(9.56)

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Rovlew lind Summllry for Chnpter

~

(o j A

.-1.

(h )

C

D

,, I

Di,

,, :

Ii

B

~ ,.j ' c=~

tc~ ,\

1": ,;

C'

B

M

A

";eQ

B

~ u - ~=ffiHI1

,\

----y---

Fi g. 9.41

c

First moment-area theorem

= (area between C and D) XI

1.9.59)

and the tangential deviation of D wilh respect to C (Fig. 9.45h):

t Die

= (area bi=lween C and D).\2

B

(OJ

0' Fig. 9.45 Second moment·area thEorem

Again us ing the (MIEI ) diagra m and a sketch of the dcnecled beam (Fig. 9.45), we drew a tangent at point D and consid~red the vern eal distance felD. whIch IS called the tangential devIation 01 C with respect \0 D. \Ve then d eri ved the second moment-a rea thea rem, which may be stated as follows: The tangential deviation frm of C with respect to D IS equal to the first moment with respect to 11 vertical axis through C of the area under the (MIEI) diagmm between C and D_ We well' careflll to disting ui sh bet ween the tangen· tial devimioTl of C with respect to D (Fi g. 9.45(/). t a l)

,,, ,

D'

~ ~1~

'~_ _ _ _ _ _~ ___ ~~~'C (ill

,, ,,

C'

19.(0)

Second moment·
599

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600

Dcnectlon 01 B
In Sec 9.10 we learned to determine the slope and de llection al Cantilever Beams Beams with symmetric loadings

,

poinl'> of emili/ever beams and beams with symmetric loadillJ;s. For

cantilever beams. the tangent at the fixed support is horizontal (Fig.

9.46): and for symmetrically loaded beams. the tangent is horizontal at the midpoint C of the beam (Fig. 9.47). Using the hori zontal langelll ru. a refl;:/"/!lIce

IUI/SCIII,

we we re able to detCllJIillC slopc~ ami

de nections by using. respectively. the first and second moment-area theorems [Exampk 9.09, Sample Probs. 9.10 and 9.11]. We noted

that to lind :l. del1ectlon thai IS not a tangential deviallon (Fig. lJ.47c). it is necessilry to firs t determine which lungentinl deviations can be combined to obtain the desired deflection. Fi g . 9 .46

~ C

I")

~Inrimnl"l

'"

~mSr~:"c I

Jld("It',,,,... (''''ifni (h )

Ii D-

(j

1.1.«"

f

o.r

(c)

Fi g. 9.47

In many cases the application of the: moment-area theorems is simBending-moment diagram by parts

Unsymmetric loadings

plified if we consider the effect of each load separately [Sec. 9.11]. To do this we drew the (MIEf) diagralll by lJarb by drawing a separate (MIE1) diagram for each load. The areas and the moments of areas under the several diagrams could then be added to delennine slopes and tan ge ntial deviations for lhe original b~am and loading lExamplcs 9. [0 and 9.111. In St:<:. 9.12 we ex panded the usc of the moment-area melhud 10 cover ~ams w ith III/symmelric loatliI/8.!. Observing Ihat locating a

horizontal tangent is usually nol pmsible. we selected a re ference tangent at one of the beam supports, si nce the slope of that tangent can be readil y determined. f'o r exa mple , for the beam l1 nd loading shown in Fig. 9.59, the slope of the tangent at A can be obtained by

computing the tangential de\iiation

and dividing it by the di s-

r~

" mn

X I·

IB/A

L

(" )

·1

(b)

Fig. 9.59

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Rovtew lind Summllry

tance L between the supportS A and B. Then, usi ng both mo mentarea theorems and simple geometry. we could detennine the slope and deflection at any point of the beam [Example 9.12. Sample Prob. 9.121The ml/ximulII deflection o r an unsymmetrically loaded beam generall y does not occur at midspan. The approach indicated in the pre· ceding paragra ph was used to detcnnine point K where the maxi· mum dcllection occurs and the magnitude of thrll dellection [Sec. 9.131. ObscI"Ving that till: slope at K is ze to (Fig. 9.68), we cOllciudell ___________

mn "

p

"

IlJl < O IK

~L

1

Il!f•.

MaxImum defiection

Ii

I I

11K '*' 11

Ii!;;.,

B

for ehnple,

If. '

rt'''''''/

I ~ry:(rl

Fig. 9.68

that 0 ,'(1,0. .= - 0,0.' Recalling the first moment·area theorem, we de· termined the location of K by mea~uring unde r the (MIEI) diagram an area equal to 0KI.... The maximum deflection was then obtained by computing the tangentlUl devialion tAlK [Sample Probs. 9.12 and 9.131. In the last section of the chapter [Sec. 9 .141 we applied the moment -area method to the analysis of statimlly indeterminate beams. Since the reactions for the beam and loadi ng shown in Fi g. 9.71 cannot be detemlined by statics alone, we de~ignatcd o ne of the reactions of the beam as redundant (M A in Fig 9.72a) and consid· ered the redundant reaction as an unknown load. The tangential de· viation of D wi th respect toA was co n ~ide red separately fo r the dis· Irih tllccl load (Fi e. Q.72h) mwi f()f the redlln(i:J rll re:lc l inn (Fie. 9.72,-) Expre5si ng that under the combined action of the dislribuled load and 01 the couple M A the tangential deViation o j I:J with respect to A must be zero, we wrote tWA

(tWA),,' - (tYA)M

Statically indeterminate heams

Fig. 9.71

=0

From this expression we detenni ned the magnitude of the redundant reaction MA IExample 9. 14, Sample Prob. 9. 141.

("' R g . 9.72

(hi

k)

~

601

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REVIEW PROBLEMS

9 .157 For tftc loading >flOwn. determine (,,) the equation of the e lastic curve for th e cantilever beam AB. (b ) the defleclion at the free end. (c) th e slope at the free end.

'I "i Af

'('1'

:1T!fITDT\[ L

1

HI

Fig. P9.158

FIg. P9.1ST

'IV

9.158 For the beam and loading shown. know ing thm {j = 2 m. = 50 kN/m . and £ = 100 GPa. determine (iI) the slope III support A. (M tlte

deflection at point C.

For the beam and loading shown. determine (a ) the equation of th e CUr>'t:. (/1) ilt: slupt: al t:nd A. (c) Iht: dt:nt:CLiull
9 .159 cla~til;

"

B

Fig. P9.IS9 Fig. pg.160

9 .160 Determine the reaction at A and draw the bending moment dia· gram for the beam and loading shown. 9 .161

For the timber beam and loadin}! shown. determine (a) tfte slope C. Use £ = 12 GPa.

at end A. (b ) the deflection m the midpoint

J JI lnrI

P = ·lkK

f

(15m

oosm l

Fig. P9.1S1

602

~ ll"~

!,o """

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9.162 The rigid bar DEF is welded at point D to the rolled-steel beam AB. For Ihe loading shown, determine (a) the slope at point A. (b) the deflection at midpoint C of the beam Use E = 200 G Pa

Review Problems

3[) kl\/111

603

'II) kip"

111111

D

c

F. IV 10 X 33

Fig. P9.162

Fig. P9.163

9 .163 Beam CE rests on beamAB. as shown . Knowing that a WIO X 33 rolled-steel shape is used for each beam. determine for Ihe loading shown the deflection at point D. Use E = 29 X JO~ psi.

9.164 For the loading shown, knowing Ihal beams AC and BD have Ihe same flexural rigidity. determine Ihe reaction at B.

('/I

Fig. P9.164

9 .165 Two C6 X 8.2 channels are welded back to back and loaded as shown. Knowing thaI £ = 29 X 106 psi. determine (a) the s lope at point D, (b) the deflection at point D .

at

9.166 For the pri smatic beam and loading shown. determine (a) the slope end A. (b) the deflection at the center C of the beam.

Ii'.!

Fig. P9.165 p

B

E

A

12LLL_bl b 4

.j

9.167 For the beam and loa.ding shown. determine the magnitude.1nd location of the largest downward denection.

x

I'

.j

.j

Fig. P9.166

;5 kN

III k:-ll,,,

A'~~~

/JI L':j"" 1-0.3,,,

~

8

U,9 III

9.168 Determine the reaction at the roller suppon and drnw the bendingmoment diagram for the beam and loading shown.

Fig. P9.168

3.6 III

\\'3 10

x 44 .,5

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COMPUTER PROBLEMS

TIn.' following pnlblenlS IU "t'. designed to be sohed with D romputer.

9 .C1 Sever.J concentrated loads can be applied to the cantile\'er beamAB Write a comp-uter program to caicuia[(> the slope and denecl.Jon of beam AB fro m x "" 0 to x = L. using givetl increments .1x. Apply this program with increments .1->:: ~ 50 mm (0 the beam and loading of l'roh 9.73 and Prob. 9.74. 9 .C2 The 22-ft beam AB consiSis of a W21 x 62 rolled·steel shape and SUppollS a ].5.kipift dlstributeLiload a. shown. Write a computer program and use It to c::.[culate for values of a from 0 to 22 n, uSllIg t · lI increments. lll ~ the s lope and deflection ~t D. (b) the location and magnitude of the maximu m deOection. Use E = 29 X 106 psi.

Fig. P9.C1

:t

I

W25() 'X 32.7

1 - - - ":'. ft - -- j Fig. P9.C3

Fig. P9.C2

9 .C3

The cantilever beam AS carries the distributed londs shown. Write

a co mput(:r program to calculate the slope and denection of beam AB frorr, I; = 0 to x = L using /;i\,~LI in~n:111ent s .1_LApply lhi~ pmgralll with jw,:n::mt:nl.1 1\ c = t()() mm . ~~."m;ng lh~1 I = ?.:l m . '" ""' 11i i:N/ n, . ~nr\ (n)" = or.. m (b) (/ = 1.2 nt . (t") II = 1.8 m. Use E = 200 OPa.

9 .C4 The s imple beam AS is of consta nt nexural rigidity £1 and carnCll several concentrated loads as sho .... n. Using th e Melh(J(/ of /llIrgrulhm. write a computer progra m that can be used to calculate the slope and denection at points along the beam from x = OIO.x = L using given increments .1x. Apply thi s program to th ... be3m and loading o f (a) Prob. 9. [3 with ;lx = t ft, (}J) Prob 9. 16 \\ith j.x = 0.05 Ill, te) Prob. 9.129 j.x = 0.25 nl.

~-- L ---IT Fi g . P9.C4

604

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Computer Probtems

Fig. P9.C5

g.CS The supports of beam AB consist uf a fixed suppon at end A and a roller supponlocated at point D. Write a computer program that can be used to calculate the slope and denection at the free end of the beam for values of 11 from 0 to L using given increments !lao Apply Ihis program 10 calcu late the slope and denection at point B for each uf Ihe following cases:

,'J

11 ft

(b)

3m

O.S fT 0.1 m

1.6

w

E

kin

19X1o" psi

WJ6

ZOO GPa

W460 x IlJ

IS kNIm

Shape

x

57

g .e6 For the beam and loading shown. use the Moment -Area M ethod [0 write a computer program [0 calculate the slope and denectiun at points along the beam from x "" 0 10 .\. = L using given incremems Ilx. Apply this program [0 calculate the slope and denection at each concentrated load for the beam of (11) Prob. 9.77 with .0..\' = 0.5 m. (b ) Prob. 9.1 19 with 11." = 0.5 m.

9.C7 Two 52-kN loads are maintained 2.5 m apart as they are moved slowly across beam AB. Wrile a computer program 10 calcu l,lIe the denec[ion at the midpoint C of the beam for values of.( from 0 to 9 m. using 0.5-m increments. Use t: = 200 GPa.

5:!kN

,

J

L" 5m~

I \\'460

! -- - L- -Fig. P9.C6

x tJ 3

I----H m ~ Fig. P9.C1

9.es A uniformly distributed load ·w and several distributed loads PI may be applied to beam AB. Write a computer prugmm 10 determine the reaction at the roller support and apply this program 10 the beam and loading uf (a) l'rob. 9.53(/. (b) Prob. 9. 154. Fig. P9.C8

605

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Columns

A steel wide-flange column is being tested in the five-mil lion-pound universal testing machine at Lehigh University, Bethlehem, Pennsylvania. The analysis and design of members supporting axial compressive loads will be discussed In this chapter.

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'0.1. INTRODUCTION In the prcccdingchapters. we had two primary concerns: (I) the strcngth

of the stmctufe. i.e. , its ability to support a specilied load without experiencing excess ive stress; (2) the abi lity of the Structure 10 support a specified load without undergoing unacceptable deformations. In this chapter. our concern will be with the stability of Ihe structu re. i.e. , with its abili ty to support a given load without experiencing a sudden change in il~ configuration. Our discussion will relate chit:ny to columns, i.e., to the analysis and design of vertical prismatic members supporting axial loads. In Sec. 10.2. the stability of a simpliijed model of a column. consisting of two ri gid rods connected by a pin and a spring and supporting a load P, will tirst be considered. You will observe that if its equilibrium is disturbed , this system will relurn to its original equilibrium posilion as long as P does not exceed a certain value Pcr> called the critical load. However, if P > Pw the system wi ll move away from its original position and seule in a new position of equilibrium. In the (irst case. the system is said to be stable. and in the sccond case. it is said to be un.l·table. In Sec. 10.3. you will begin the study of the stability of eioslic columns by considering a pin-ended column subjected 10 a centric axial load. Ellfer"s[ormllla for the critical load of the column will be derived and from that formula the correspond ing critical nonna l stress in the column wi ll be detennined. By applying a factor of safety to the cril ie;}l lO;l(I , you wil l he :lhle 10 delermine the llllnwllhie In:u l lhat can be applied to ,1 pin-ended column. In Sec. 10.4, the analysis of the stability of columns with different end conditions will be considered. You will simplify these analyses by learning how to detennine the effective lel/gth of a column, i.e., the length of a pin-ended column having the same critical load. In Sec. 10.5, you will consider columns supporting eccentric axial loads ; these columns have transverse dellections for all magnitudes of the load. An expression for the maximum dellection under a given load will be derived and used to determine the maximum normal stress in the column. Finally, the .I·ecwlt formula which relates the average and maximum stresses in a column will be developed. In the first sections of the chapter, each column is initially assumed to be a strai ght homogeneous prism. In the last pan of the chapter, you will consider real columns which are designed and analyzed using empirical fonnulas set fonh by professional organizations. In Sec. 10.6, formulas will be presented for the allowable stress in columns made of steel, aluminum, or wood and subjected to a centric axial load . In the last section of the chapter (Sec. 10.7), the design of columns under an eccentric axial load will be considered .

10.1. Introduction

607

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608

Columns

10.2. STABIUTY OF STRUCTURES

Suppose we arc to design a col umn AB of length L to suppon a give n load P (Fig. 10.1). The column will be pin·connected at both ends and we assume that P is a centric axial load. If the cross-sectional area A of the column i ~ selccted so that the value u = Pi A of the stress on a transverse section is l es~ than the allowable stress (Tall for the material used, and if the deformation {j = PL/AE falls within the given specifi cations. we might conclude that the column has been properl y designed. However, it may happen that, as the load i.~ applied. the column will buckle; instead of remai ning straight, it wi ll suddenly become sharply c urved (Fig. 10.2). Figure 10.3 shows a column similar to that in the opening photo of this chapter after it has been loaded so that it is no longer straight ; the col umn has buckled. Clearly, a column that buc k· les under the load it is to suppon is not properl y designed. F1g. 10.1

Fig. 10.2

Fig. 10.3

Buckled column

Before getting into the actual discussion of the stability of elastic columns. some insight wi ll be gained on the problem by considering a simplified model consisting of two rigid rods AC and BC connected at C by a pin and a torsional spring of constant K (Fig. 10.4).

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10.2. Sta bility 01 Structures

r

I

"

r

",

"

U'2.

t-L OB~'"" "'"

U2

C

o

K

,

B

P' {u J

Fig. 10.4

, , 2llO ,, , , C 10 , , ~8 '

I"

(/J )

Fig. 10.5

If the two rods and the two force s P and P ' arc perfectly al igned. the system wi ll remain in the position of equilibrium shown in Fig. 10.5a as long as it is nOI distu rbed. But sup pose thai we move C slighlly to the right. so thai each rod now forms a small angle 6.0 wi th the vertical (Fig. [0.5/)). Will the system return to ils ori gi nal equilibrium position. or will il move furthe r away from that position? In the fi rst case. the system is said to be .I,table, and in the second case, it is said to be wI.Illlh'e.

To uelCrmine whether the two-rod system is stable or unstable. we consider the rorces acting on rod AC (F ig. 10.6). These forces consist of two couples, namely the couple fo rmed by P and P' , of m()ment P(L/2) sin flO , which tends to move the rod away from the vertical , ami the couple M exerted by the spring, which tenlls to bring the rod back into its original vertical position. Since the angle of de ileclion of Ihe spring is 2 tl(J , the moment of lhe couple M is M = K(2 MJ ). If the moment of the second coup le is larger than the moment of the first couple, the syste m tends to return to its original equi librium positio n; the system is stable. If the moment of the first couple is larger thlln the moment of the second couple, the system tends to move away from its original equilibrium positi on; the system is unstable.. The value of the load fo r which the IWO couples balllnce each olher is called the cril ic"(l/ load lind is denoted by P'T' We have

Pc,(L!2) sin tlO = K(2 tl/J)

(10.1 )

or, since sin tl(J = fl O, (10.2)

Clearly, the system is siable for P < P,,, Ihat is, for values ot' the load smaller than the critical value, and unslable fo r P > pC!. Let us ass ume Ihat a load P > PCT has been ap pl ied to the Iwo rods o f Fig . 10.4 and Ihat Ihe system has been disturbed . Since P > p~" the system win move further away from the vertical and, after some

I'

I"

Fig. 10.6

609

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610

oscillations, will settle into a new equilibrium position (Fi g. [0.7(1). Co nsidering the equilibrium of the free body AC (Fig. [0.7b) , we obtain an '-'"
Columns

P{L/ 2) sin 0 ~ K(20)

I"

(II )

Fig. 10.7

PL

f)

4K

sin O

(10.3)

The value of 8 corresponding to the equilibrium positio n represented in Fig. [0.7 is obtained by solving Eq. (1 0 .3 ) by trial and error. But we observe that. for any positive value of f}, we have sin (} < 0. Thus, Eq. (10. 3) yields a value of 8 different from zero only when the left-hand member of the equation is larger than one. Recalling Eq. ( [0.2). we note that this is illdeed the case here. since we have ass umed P > Per. But, if we had lL<;sumed P < Pc" the second equilibrium position shown in Fig. 10.7 wo uld not ex ist and the only possible equi librium position would be the position corresponding to 0 = O. We thus check that, for P < Per> the position 8 = 0 must be stable. This observation applies to structures and mechanical systems ill general. and will be used in the next section, where the stability of clastic columns will be discussed. 10.3 . EU LER 'S FORMULA FOR PIN-ENDED COLUMNS Re turning to thc column A B cons idered in the prcceding sectio n

(Fig. 10.1). we propose to determine the critical value of the load P, i.e., the value PC! of the load for which the position shown in Fig. 10. 1 ceases to be stable. If P > Per. the sli ghtest misalignmellt or disturbance will cause the. column to buckle. i.e.. to assume a curved shape as shown in Fig. 10.2.

Fig. 10.1

Fig. 10.2

(repeated)

(repeated)

Our approach will bc to determinc the conditions under which the configuration of Fig. 10. 2 is possible . Since a column can be co nsidered as a beam placed in a vertical po:; iti on and subjected 10 an axial load. we proceed as in Chap. 9 and de no te by .r the distance from end

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A of the co lum n to a given point Q of its clastic curve, and by )' the de!lection of that point (Fig. 10.8a). It fo llows that the x axis will be vertical and directed downward , and the J axis horizontal and directed to the ri ght. Considering the equilibrium of the free body AQ (Fig. 10.8h), we lind that the bending moment at Q is M = - P),. Substituting thi s value for Min Eq . (9.4) of Sec. 9.3, we write (p)'

At

dx 2

EI

10.3. EtJler"s Formula lor Pln·ended Columns

I' b", u" /~Hl

Er

( 10.4)

Q

or, transposing the last term, (/ ~' P - 1+ -),= 0 dr EI

L

( 10.5)

This equation is a linear, ho mogeneous differential equation of the second order with ~onstant coefficients. Selling

,

P

p = EI

(10.6)

we write &j . ( 10.5) in the fo rm

dx

( 10.7)

which is the sam~ as that of the differential equation fo r simple harmonic motion except, of course. that the independent variable is now the distance x instead of lhe lime t. The geneml sollllion of Eq. (10.7) is

)" = A sinpx

[J ~ I·· 'I ~ul

(,I Fig. 10.8

1

-d '"2 + /1\ = 0

+ BCOS I}X

(10. 8)

1

as we easily check by computing lfy/dx and substitllling fo r y and d 2y/dx l into Eq. (10.7).

Recalling the boundary conditions that must be satis fi ed at ends II and B o f the column (Fig. 1O.8([), we first make .r = 0 , J = 0 in E(l. (10.8) and fi nd that B = O. Substituting next x = L, y = 0, we obtain A sinpL = O

'1-

A

rr' L_\t I

p - ~"

(10.9)

This equation is satisfi ed either if .A = O. or if si n pL = O. If the lirst o f these conditions is satisfi ed, Eq. (10.8) redllces to y = 0 and the column is straight (Fig. 10. 1). For the second condi tion 10 be satislied, we must have pL = 1l7T or. substituting fo r p from ( 10.6) and solving for P. ( 10.10)

The smallest of the values of P dc!ined by Eq . ( 10.10) is that corresponding 10 /I = I. We thus have ( 10. 11 )

The expression obtained is known as Eule,.·~· jomwla. after the Swiss mathematician Leonhard Euler (J 707 - 1783). SubstitUling this

611

lJ

A

'i -1- '

. . ."

~t

Ix "

8

'" I,

(/,i

Q

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612

Columns

expression for Pinto Eq . (lO.6) and the value o btained for p into Eq. (10.8), and recalling that B = 0, we write y = A sill

1IX

L

( 10. 12)

which is the equation of the elasti c curve after the column has buckled (Fig. 10.2). We no te that the value of the maximum deflection, Ym = A. is indetenninate. This is due to the fact that the differential equation 00.5) is a linearized approximation o f the actual goveming differential equation fo r the elastic curve.!" If P < Pw the condition sin pL "" 0 canno t be satisfi ed. and the solution given by Eq. (10.1 2) does 1I0t exi sl. We must then have A = O. and the only po~ s ib le configuration for the column i .~ a straight one. Thus, fo r P < Per the straight configuration of Fi g. 10.1 is stable. In the case of a column with a circular or syuare cross secti on. the moment of inertia' of the cross secti on is the same about any centroidal axis. and the column is as likely to buckle in one plane as another. except for the restraints that can be imposed by the end connections. For other shapes of cross section, the critical load should be computed by making 1 = I min in Eq. ( 10. [ I ); if buckling occurs. it will take place in a plane perpendinLlar to the corresponding principal axis o f inertia. The value of the stress corres ponding to the critical load is called the critical stress and is de no ted by U,~. Recalling Eq. ( 10.11 ) and selting I = Arl. where A is the cross-sectional area and r its radius of gyraLion. we have U

2 P~r r, EA r "l = - = --"A A{ l

~

~

"

.".-1£ (LI')'

--

(10.1 3)

The ([uaJltity Llr is called the .I'l cn dc m f'.H ratio of the col umn . It is clear, in view of the remark of ihe preceding paragraph, that the mini mum value of the radius of gyration r should be used in computing the slenderness ratio and the critical stress in 11 column. E{[uati oll ( 10. I 3) _~ hows that the critical stress is proportio nal 10 the modul us of elastic ity of the. material, and inversely proporti onal to the square of the slende rness ratio of the column . The plot of U or versus LI ,. is shown in Fig. 10.9 fo r s truc tural steel, a..~s urnin g E = 200 OPa and U r = 250 MPa. We should keep irl mind that no factor of safety has been used in ploning ITer. We also note that , if the value obtained fo r U c< from Eq. ( 10.1 3) or from the curve of Fig. lO.9 is larger than the yield strength U \". thi s value is of no interest to LI S, since the collIInn will yield in compression and cease 10 be elastic before it has a chance t9 buckJe. t Wc rce~ t1lhal lhe' equalion lI'y/<1,' = MIE/was oblainrd in Sec. \1.3 by assuming lhal Ihr .tope dy/ d.< of lhr beam could I'l"' I,rgleeled and lhal the CX
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u {.\ IPa)

10.3. Euler's Fonnula lor Pin-ended Columns

,

y u)"

613

= 1Slf.\IPll

0;0 1----"' / ::....,"

£ ...

2()U CP.l

,., "

Ur

200

k!lJJOO

Fig. 10.9

Our analysis of the behavior of a column has been based so far on the assumption of a perfectly ali gned n'ntric load. In practice. this is seldom the case, and in Scc. 10.5 the effect of the eccentricity of the loading is taken into account. Thi s approach will lead to a smoother transi tion from the buckling failure of long, slender columns to the compression failure of short, stubby columns. It wi ll al so provide us with a more realistic view of the relation between the slenderness ratio of a column and the load that causes it to rail.

EXAMPLE 10.01 A 2-m-long pin-ended column of square cross section is to be made of wood. Assuming E = 13 GPa. u . u = 12 MPu, and using u fuctor of SUfNY of 2.5 in computing Euler's critical load for buckl ing. determine the size of the cross section if the column is to safely support (£I) a l00-kN load. (b) a 200-kN load.

(a) For the l00-kN Load. iiafety. we make

j'n Euler's formu la (10.11) and I

"
sol~e

E = 13GPa

for I. We have

mf

(250 X ]OJ N)(2 7T 2(13X 10') l'a)

1 ~ 15.588 X 10 ~ nf'

7.794 X 10

~:

= 15.588 X 10

~

P

m"'

{]" = - =

A

')

A=

200 kN (0.1 1695m)'

14.62 MPa

= 98.3 mm '" IOOmm

A

~

a = 116.95mm

Since this value is larger than the allowable stress, the dimension obtained is not acceptable, :md we must select the cross section on the basis of its resis tance to compression. We write

We check the value of the normal stress in the column:

if

~

The vulue of the Ilomlal s tress is

Recalling that, fOl a square of side a, we. have 1 = £1' /12 , we write /J

(b) For the 200-kN Load. Solving itgain Eq. (10.11 ) for I . but making tlOW P" = 2.5(200) = 500 kN. we hitve

Using the given factor of

L = 2m

P" = 2.5(100 kN) = 250 kN

Since if is smaJlerthan the allowable stress. a 100 X IOO-mm cross section is acceptable.

100 kN (0.100 m)l

/l Z

10MPa

= ~'" 200kN = 16.67

ud 12 Ml'a = 16.67 X 10- 3 Ill'

X

10

] m~

a = 119. 1 mm

A 1]0 X I ]O-mm cro% section is acceptable.

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614

10.4. EXTENSION OF EULER 'S FORMULA TO COLUMNS WITH OTHER END CONDITIONS

Columns

Euler's fo rmu la (10.11) was derived in the preceding section for a column that was pin-connected at both ends. Now the critical load Pc, will be determined for columns with different end conditions. In the case of a column with one free end A supporting a load P and one fixed end B (Fig. 10.IOa). we observe that the column will behave as the upper half of a pin-connected column (Fig. 10. lOb). TIle critical load for the column of Fig. 10. IDa is tituS the same as for the pin-ended column of Fig. 10. lOb and can be obtained from Euler's

p

A

A

1 L

J

H

8

,, ,

I ••

= ~L

'

"" "

(I» \\

I")

,'

"

"'f-,, Fig. 10.10

IOmlula (10.11) by using a column length equal to twice the actual length L of the given column. We say that the effecth'e length L, of the column of Fi g. 10.10 is equal to 2L and substitute L, = 2L in Euler's fo rmula: I'

7r

~~

2

EI

- L;

~-­

( )0.11 ')

The critical stress is found in a similar way from the fo rmul a

lTc, ""

n-1.E (L~/rr

( 10.13' )

The quantity L
Fig. 10.11

Consider next a column with two tixed ends A and B supporting a load P (Fig. 10. 11). The symmetry of the supports and of the loading about a horizontal axis through the midpoint C requires that the shear at C and the horizolltal components of (he reactions at A and B be zero (Fig. 10.12). It rollows thai the restraints imposed upon Ihe upper half AC of the column by the support at A and by the lower half CB arc

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iuellticlll (Fi g . 10.13). Portiol! AC must thus be .ymmeui c aboui its midpoint D . a nd this fXlint must be a point of innectiol1. where the be.l1ding moment il> zero. A similar reasoning shows that the bending moment al the midpoil1t E of the lower haH of the col umn must also be zero (Mg. 1O.14l1). Si nce the be nding mo ment al the ends of a pin-ended column is zero, It foUows that the portion DE o f the: column of Fig. 1O.14a must behave as a pin-ended column (Fig. J O. J 4b). We thus conclude that the effective length or a column wilh two lixed ends i ~ L. : L/2.

10.4. Column 3

w~h

Olher End Condition,

p

Do '"l

T

u·,

C

----'-

OC

P'

,

,, '-./

l

8

, /M'

(.)

P'

Fig. 10.12

Fig . 10.13

Fig. 10.14

In the case of a column \\-ilh llne fixed end B and one pin-connected end A supporting a load P (Fig. 10.15), we must write and solve the ili ffcre mial equation of the ci;Jstil- curve to {h::te"1I1ill~ the clTective leu gtll of the column. From the frec-body diagram of the enlire column

(Fig. 10.16), 'We lirst note mill a transverse fo rce V is exerted at end A, in addition 10 the axial load P, and that V is statically indetermi nate. Consldcnng now the Iree-body dIagram 01 a poruon A{l 01 the column \Fig. 10.1 7), we find that the bending moment at Q is

- r y - V.I

M =

,

, ! [= II

"

A

,

'I = OJ

"

T

A

L

\"

,,I

D

'"

[, =1"',=(11 [1 = L.':ylr/:x =

t!J

I, Fi g. 10.15

Fig.l 0.1Ei

I

r ~"

I

.'I1

II! IQ

Fig. 10.17

, V'

615

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616

Substilllting this value into Eq. (9.4) of Sec. 9.3. we write

Columns

d~ r

lb.:~

M

P

V

= EI "" - E/'>' -

El

x

Transposing the. term cOlllaining v lind setting p

o

( 10.6)

p - = £1

as we did in Sec. 10.3. we write

v

- - x EI

( 10.14)

This equatio n is a linear. nonhomogeneous differential equation of the second order with constant coeffi cients. Observing that the ]eft-hand members o f Ec]s. (to.7) and ( 10.14) are identical, we condude that the general solutio n o f Eq. (10.1 4) can be obtained by adding a particular solution of Eq. (10.14) to the so luti on (10. 8) obtained for Eq . (10.7). Such a particular solution is easily seen to be

r =

-

V -

- , -

/rEI

X

or, recalling ( 10.6),

vx

(10.15 )

p

Ix~

\'

r

I)

'I - ul

Adding the solutions (1 0.8) and (1 0.15), we write the general solutio n of Eq. ( 10.14) as

y

A

y = A sinp.r

+ B eospx -

v pX

( 10.1 6)

The constants A and B, and the magnitude V o r the unknown transverse force V arc obtained from the boundary conditions indicated in Fig. ( 10.16). Making first x = 0, y = 0 in Eq. (1 0 .16), we find that B = O. Making next x = L .)" == 0 , we obtain

L

B

\"

MN"-.:./ P'

I x~ I",. d'jMx= 111

I, Fig.10.1 6

. ASlI1pL

I, ~ L.'I=II I

(repeated)

V =pL

(10.17)

Finally, co mputing dy dx = Apcos px -

V

p

and making x = L. dy/ dx = 0, we have A" cospL =

V

P

(10.18)

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10.4. Columns with Other End Conditions

Dividing ( 10.17) by (10. 18) member by member, we conclude that a so luti on of the form ( 10.16) can exi st only if ta n pL = pL

( 10.19)

Solving this equalion by trial and error, we find thai Ihe smalleSI va luc o f pL which satisfies ( 10.1 9) is pL = 4.4934

(10.20)

Canying the value of p dclined by Eq. ( 10.20) into Eq . ( 10.6) and solving for P. we obtai n the (;rit ical load for the column of Fig . 10. 15 20. 19£1

( 10.21 )

P",.= --L-'-

Theeffe.ctive length oflh e column is obtai ned by equating the righthand members of Eqs. ( 10.11 ') and (10.2 1): -rr2El L;

Fig. 10.15

20. 19£1 L~

Solv ing fo r L,. we find that the effective lengt h o f j col umn with o ne fixed e nd and o ne pin-connectcd end is L, = 0.699L = D.7L. The elTcctive lcngths con csponding to the vari ous end conditio ns considercd in this section are shown ill Fig. 10.1 8.

(II) On,' fiu'd end. On p fr"" l"1~d

Ib) S" lh em).,; pin l}"d

(c) On(' fixp,! ,·nd, o ne pin n,·d end

p

Fig.10.18

Effecti\e length of column for various end conditions.

(d) Both pml,

ftwd

(repeated)

617

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"

SAMPLE PROBLEM 10.1

"

An alummum column of length Land rectanguJarcrms section has a fixed end B and ~uppon~ :l centric ioa.J al A . Two smooth and rowldeJ fi.\ed p l,lte~ re_ st min end A [lUll mo'~il1g in one of the vertical planes of symmelfY of the column. blll allow it to move in the other plane. (u) Determine the ratio alb of

the two Slues of the cross so:~tioll cOHes~lHliLlg to the illOst effn;iellt lksign against bucklmg. (b ) De~ ign Ihe most efficie nt cross section for the column. knowmg that L .:0 20 in .. £ = 10, I X let psi. P = 5 kips. aJld that a f:Jclor of safet), of 2.5 is required.

SOLUTION Rue U in g in .l)' Plant'.

Referring to Fig. 10.18. we note that the effec-

tive length of the co lumn wilh respecl 10 buckling in this plane is L. - O.7L

The radius of gyration ,"_of the cross section is obtained by writing J" = {, b"J

aud. since I: = Ar;.

,

r :=

-

A = lIb

• hlx,3 "

r, = a!v12

- = -- ~ -

A

ab

12

r~spt!ct

The effective sle nderness ratio of ule column with .ry plf:.ne i ~ L, 07L

"

to buckli ng in the (' )

a/V12

Duckling in x:: Pla ne. The effecti\'c length of the column widl respeCT to buckling in dli s plane IS L, = 2L. a.nd the corresponding radius of gYr:ltiOII is r, - b/v'T2. TItus. L, lL (2) ~ - 1J/V12 fl. l\'los l Efficie nt ~i gll . The most efficient design Is thaI for which lhe critic,,] Slrdses corresponding to Ihe IWO po~sible mode~ of buckling are equal Referring to Etj. (1O.13'). we nOTe thm This will be the case if the IWO values obtained abov~ for the effective slendemess ratio are equal. We write 1l.IL

a/m

lL

=

b/v'TI

and. solving for Ihe ratio a/b. ll. D tslgD ro r Gfn- n DUIII.

('

0. 7

b

2

"b =

IUS ..

Since FS. = 25 is required.

Po, = (F.S.)P = (2.5)(5 kips) = 12.5 kips Using {/ = OJ5b. we have A = ab = 0.351J 2 <.Ind fT "

-

Pc, 12.500lb A - 0.35b"

Making L = '20 in. in Eq. (2). we have L,If, = 138.6/b. SubSTiTUting for E. L,/ f. and ( f
w'E CT., = (L./f)2

12,500 lb 0.351J' =

'"110.1

b = 1.611l tn .

618

X 1000psi)

( 138.6/b): Ii =

0.35b = 1l,567 Ill. ..

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PROBLEMS

1 0 .1 Knowing th3( th~ torsional spring a1 B is cf constant K and thai the bar AB is rigid, determine the critical load Per.

r

'1 L

,. K

r

r,

N

.

A

-~,

, I B

• B

Ag. Pl0.2

,

t l. C

'/II

L

J l

Fig. Pl0.1

.,

-t l"

~

Fig. Pl0.J

10 .2 Knowing Ihal the spring at A. is of constant k aad thallhe bar AB is ri gid. de termin e. th e criLical luad POf " 1 0.3 Two rigid bau AC and He are connected ,u ~ hown 10 a spring of constant k. Knowing Ihal Ih ~ spring can act in eilher tension or cornpr~ss ion . delemline Ihe critical load Pc, for Ihe s)s tem. 1 0 .4 Two r;Eirl h~r~ Knowing that the torsional load Per for lhe .ystem.

Ar ann Hr. ~prillg al

art'.. connec rl'd Ity a (lin at r:\, dmwn . B is of cons tant K. determine lhe criti cal

,

p

"

B k

'V

,

w,

---,

- .,"I

c

I

"-4 /,

A • Fig. Pl0.4

Fig. PIO.5

10.5 The ri gid rC1f1 All i .~ mr~chl'll In a hinge ~I A and In two ,!,rin g~ , eac h of constant k '" 2 kips/in. Ihal can aCI in either tension or compression. Knowing t.h~t II - 2 ft. determine the critic).1 load.

619

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620

Columns

I'

10.6 The rigid bar AD is allached \0 (WO springs of constant k and is in equilibrium in Ihe position shown. Knowing that the equal:Uld opposite load.> J' and P ' relllaill /lOriWlllal, determine the magnitude P,., of the critical load for th e system.

10.7 The rigid rod AB is attached \0 a hinge at A and to two springs. each of constant k. If Il = 450 rum. d = 300 mm, and II! 200 kg. determine Ihe range of values of k for which the equilibrium of rod AB is stable in th e position shown. E
=

Fig. P10.6

Fig. P10.7

10.8 A frame consisls of fOllf L-shaped members connected by four IOrsianal springs. each of constant K. Knowing lhal equal loads P are app lied al points A and D as shown . delemline the critical value P", of the loads applied to the frame.

Fig. P10.8

10.9 Determine the critical load of a wooden meter stick which has a 7 X 24-mm rectangular cross section. Use E "" 12 GPa.

10, 10 Detennine the critical load of a round wooden dowel that is 0.9 m long
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1 0.11 Determine the dimension d so that the aluminum and steel struts will have the same weight. and compute the critical load for each strut.

Problems

r

:v

c

D

AIoIlIl ;""".

E = 10. 1 = JiJil"; r = 171111o/fr1

Fig. P10.11 and P10.12 15",Il,

10.12 Dt:lermine (1I) lht: critil:alluad fur lhe slt:d MrUI. (b ) lhedilllell~il)n tJ for which the aluminum strut will have the same critical load. (c) Express the

weight of the aluminum strut as a percent of the weight of the steel Strut. 10.13 A compression member of 1.5-m effective length consists of;] so lid 30-mm-diameter bmss rod. In order to reduce the weight of the member by 25%, the sol id rod is replaced by a hollow roo of the cross section shown. Determine (1I) the percent reduc tion in the critical load. (b) the value of the critical load for the hQllow rod. Use E = 105 GPa.

I

Q~ 30 Inm

3) nun

Fig. P10.13

10.14 A column of effective length L C!lJl be made by gluing together identical planks in either of the arrangements shown. Detennine the mtio of the critical load using the ammgement 1I to the critical load using the arrangement h.

fl.5

r-

l[:l] -I I(,(

tlllll

-'-

12O",,,,

d/3

(Il l

Fig. P10.14

L

-19.5

10. 15 A column of 6-m effective length is to be made from three plates as shown. Using £ = 200 G Pa. determine the factor of safety with respect to buckling for a centri c load of 16 kN .

- 101.5",,,,

.1lUl

-tUnill.

Fig. P10.15

621

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622

10.16 and 10.17 A compression member of 12-ft effective lengLh is ll1:lde by welding together twO 4 x 3 x loin. steel angles as shown. Using E = 29 X lot' psi. determ ine the allowable centric load for the membe r if a factor of safe ty 01'2.5 is required.

Columns

I

""I""

I

"f T

Fig. Pl0.16

Fig. PtO.17

10.18 A single compres~ion member of S.l-m etTective length is obtained by connecting twoC200 X 17.1 steel channels wi th lacing bars a~ shown. Knowing th,\I the factor of safety is 1.85. detennine the allowable centric load for the member. Use E = 200 GPa ilnd d = 100 mm.

Fi g . P10.18

10. 19 Members AB and CD are 30-mm-diameter steel rods. and members Be and AD are 22-mm-diameter steel rods. When the turn buckle is tigh t...ned . the diagonal member AC is put in tension. Knowing that a factor or safet y with respect to buckl ing of 2.75 is requ ired. determine the I:lTgest illlowable tension in AC. Use E = 200 G r a and consider only buckling in the plane of the structure.

r

B

3.5 .n

20·""" dla"'['I ~ r B

15-n'''' ,l;""'cl,'.

I

/

I

0_5",

.\

L-I-I ' ' ~ 0.5 ".

Fig. PtO.20

U

-

-

2.25 ",

FIg. P1 0.19

10.20 Kn owing that a factor of s"feTY of2.6 is required. deterntine the largest load P that can be applied to the structllre s hown. Use E = 100 GPa and consider only buckling in the plm]e o f the structure.

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10.2 1 Th~ unifunn aluminum ba AB has a 20 x )6'1I111l It:l:t;ulgular cros. section and is supported by pins and bradets as shown. Each end of the har nmy rotate free l)· about a horizo ntal axi s thrcugh the pin. bu t rotation about a vert ical axis is prevented by the brack\!ts. Using E = 70 GPa. determine the allowable centric load P if u f,\clor of safet y of2.5 is required.

Problem,

Fig. P1C.21

10.22 Column AB carries a centric lood P of magnitude IS kips. Cables Be and BD are tJ.ut anu pleVe tll Ulotiollof f'OiJll B inlhe .. ,. plane. Us ing Eu h":l··s

formula and a factor of iafe ty of 2 .2. anJ neglecting the tension in the cables. detemline the ma >:.imu m allowable le ngt h L Use E = 29 X 10' psi.

D

r

~ Li;IJ

Rg. P1 0.22

~ c

10.23 A W8 X 21 rolleJ·steel sll,lpe is used with the support and cable. arrangement shown in Prob. 1022. Knowing that L = 24 ft. detemline the allow· able centri c load P ;1 a l ~ctor 0 ]" safety of 2.2 is required. Use E = 2Y X ](J'I pSI.

10.24 A I- ill.-sq uare aluminum s trut is ma intained in the posi tio n shown by OJ pin suppun al A 300 by sets of rollers at Band C that preve nt rotation of the Strut in tile. plane Qfthe iig ure. Know ing that L /l3 '" 3 fl . delemline (a) the brges t \·ahl es nf I .ne ~nd I ....u Ih~t may ke 1I~f'd If th~ .~llr"lwahlf' lnad P io; tn he as large as possible. (h) ule nlJgn itudt' of Ihe comspcnJin g allowable load. COIl.iider only bLlc kling in the plane of the flgur~ and use E - 10.4 x 10" ps i.

" Rg. P10.24

623

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624

Columns

10.25 Column ABC has a unifonnrectangularcross section withb = 12 mm and d = 22 mm. The column is braced in the x:; plane at its midpoint C and carries a celllric load P of magnitude 3.8 kN. Knowing thaI a factor of safelY of 3.2 is required. detenlline the largest allowable length L Use E = 200 G Pa.

, I

Fig. P10.25 and P10.26

10.26 Column ABC has a unifornl rect;.mgularcross section and is braced in the x:;. plane at its midpoint C. (u) Detenlline the ratio bid for which the factor of snfety is the snme with respect to buckling in the xz and yz planes. (b) Using the mtio found in part u. design the cross section of the column so that the factor of s afelY will be 3.0 when J' = 4.4 kN, L ~ I m, and £ =0 200 G Pa. 10 .27 Each of Ihe live struts consi5l.s of an aluminum tube that has n 32-mm outer diameter and a 4-mm wall thickness. Using. £ = 70 GPa and a factor of snfety of 2.3. determine Ihe allowable load P I1 for each support COLI· dition shown.

r

r

211m

L

::..lL-r",

II I Fig. P10.27

(2)

(3\

141

(5)

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10.28 Two columns arc used to suppon a hlock weighing 3.25 ki ps in each of the four ways shown. (a) Knowing that the column of Fi g. ( I ) is made of stecl wi th a 1.25-in .-diameter. determine the factor of safety with respect to buckling for the loading shown. (b) Detcnnine thc diameter of each of the olher columns for which the factor of safety is the same as the factor of safety obtained in pan ll. Use £ == 29 x IO~ psi.

10.5. Eccentric Loading; Secant Formula

"I.

~r

k

Sft

L

;:;!

II)

(21

141

(3)

Fig. Pl0.28

*10.5 . ECCENTRIC LOADING ; THE SECANT FORMULA

In this sectio n the problem of column buckli ng wi ll be approached in a di fferent way, by observing that the load P applied 10 a column is never perfectly centric. Denoti ng by e the eccentricity of the load, i.e ., the distance between the line of action P and the axis of the col umn (Fig. IO.19a), we replace the given eccentric load by a centric fo rce P and a couple M ). o f moment MA = Pc (Fig. IO.19b). It is clear that, no mailer how small !.he load P and the eccentricity e, the couple MA will cause some bending of the colum n (Fig. 10.20). As the eccentric load

,

p

r

.,

~

,

~ t \ ~ I'I'

.,

~l"" ~ ., M"

L

L'

8

8

'- ./

MJj ~

"

I,) Fig. 10.19

f"

P' (Ii)

Fig. l0.2G

P.,

625

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626

is iIlC! t:;1~eu. ooth tlJe couple M A and tllc axial fQ lcc P i uCfe
Colu""",

lem of buckling is not a question of dctcnnining how long the column elm remain straight lind stable under an increasing load. but rather how nlllch the column can be permilled to bcnd under the increasing load, if the allowable stress i .• nm to be exceeded and if the dc ncction Ym,,, is not (0 become excessive. We lirst write and solve the dinercntial equation of Lhc clastic curve, proceeding in the same manner as we did earlier in Sees. 10.3 and 10.4. Drawing tile rrre-body diagram of a portion AQ of the column and choosing t hecoordi ll D.te axes as shown (Fig, [O.:!I ). we fi nd thai th e bending momellt at Q is M = - P.l' - M Ao = - Py -

{I 0.22)

Pe

Substituting the value u f Minto E4. (9.4) u f SC"{;. 9.3. we wri te Fig. 10.21

d2.,'

M

P

Pe

dx-

EI

El -

EI

------":;- "" ~= - -\' - -

Trn nsposing the term containing)" and setting

,r ,

P EJ

= -

as dom: cru li t l. we wE ile

( 10.6)

110.23)

I

j

=~. 'I '" ~ Ij

A

--.-,-' c

1I

Since the lefl-hand member of this equation is thc same as thal of Eq , pO.7), which was solvcd in Sec. to.3 , we write the general solutiun of Eq. (1O.23) as )' = A

Sin px

- ('

(10.24)

where Ule l:lSI term is a particu lar solulion of Eq. (10.23). T hc constants A and B are obtained from Ihe boundary cOlldit lons ~hown

in Fig. 10.22. Milking fi mt

J:-

~

0, ), - 0 inBlj. (10.24), we havil

B =€

Ul

~I

+ H emp.l

Making nC."i.t .l

= L. y = 0, we write A sin pL = e(l

- cospL)

Recalling tha Fig. 10.22

sinpL

=

pl.

fll.

25in ""2 cos ""2

ond ~

cospL =

, pL 28in-""2

and substituting into Eq. (10.25), we obtain, after reductions,

pL A =e tan "2

(10.25)

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Substituting for A and B into Eq . (10.24), we write the equation of the clastic curve:

,{t:In P2L s inp.r + COSpx -

)' =

I)

( 10.26)

The value of the maximum dellection is obtained by setting.r = L/2 ill Eq. (10.26). We have )

"""

~ ,(tan pL si n pL + cos f1L 2

= e

)m..

=

(

2

.~ PL

sm-"2

+

2

_

I)

., pL cos-

2

pL

cos 2

e(secl~

- I)

( 10.27)

Recalling Eq. (10.6), we wri te

( [0.28)

We note from the

cxp rcs~ ion

obtained that Ym,,-, becomes infini te when

J£~ = ~

( 10.29)

Whi le the deflection does not aClllully become infinite, it nevertheless becomes unacceptably large, and P should not be allowed to reach the critical value which satislies Eq. ( 10.29). Solving ( 10.29) for P, we have

( [0.30)

which is the yalue that we obtained in Sec. 10.3 for a column under u centric load. Sol ving (10.30) fo r EI and substituting into (10. 28), we can express the maximum deflectio n in the alternatiyc foml

)'rna, ~ ,(sec"-2 \j!P p:. _ [)

( [0.3 ))

The maximum stress a rn ..... occurs in the section of the (;oiUffin where the bending moment is maximum. i.e .. in the transverse section th ro ugh the midpoint C, Elnd can be obtained by adding the normal stresses due. respectively. to the axial force and the bending couple exerted on that section (c f. Sec. 4. 12). We have ( [0.32)

10.S. ECl:entrlc Loading: Secant formula

627

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628

~+~

Columns M \ ~

f',

:

\I

1.)2

i~ .;... / M".., ,

~~

P'

, \.!J""", Fig. 10.23

From the free-body diagram of the portiOli AC of the column (Fig. [0.23). we fi nd thm Mm.. =

p )'ma,

+ MA = P(y"",. + c)

Substituting th i ~ value into (10.32) and recalling that I (T"",x =

= Arl.

P [ ()'m" + ,)c] A I + r2

we wri te (10.33)

Substituti ng for Yrr= the value obtained in (10.28), we write

0"

P[

= ~

m""A

[

c,

+~sec

rl

(J£ L) ] -~

E12

( 10.34)

An alternative fonn for (T""" is obtained by substituting for y""", from (1 0.3 1) into ( 10.33) . We have (10.35)

The eq uation obtained can be used wi th any end conditions. as long as the appropriate vlllue is used for the cri tical load (cf. Sec. 10.4). We note that, si nce a m., docs not vary linearly with the load P, the principle of sLlperposition docs not apply to the determination of the stress due to the simultaneous application of several loads: the resultant load must first be computed. and then Eq. ( 10.34) or Eq . (10. 35) call be used to detennine the corresponding stress. For the same reason. any given fac tor of safety should be applied to the load, and not to the stress. Maki ng I = Ar2 in Eq . (10. 34) and solving fur the flui o PIA in from of the bracket. we write P A

+

'~ "", ('-2\f"EA IP ~) r-

(10.36)

r

where the efti!clive length is used to make the formu la applicable to varioLls end condi tions. This formula is refe rred to ( 1.1" the sectm' formula: it deli nes the force per unit area. PIA , that causes a specified

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10.5. Eccentric Loading; Secant Formula

maximum stress rTmax in a column of given effecti ve slenderness ratio, Li r. for a givcn value of the rat io eclrl, where e is the eccentricity of the applied load. We note that. since PI A appears in both members, it is necessary to solve a transcendental equati on by trial and error to obtain the value of PIA corresponding to a given column and load ing condi tion. Equation (10.36) was used to draw the curves shown in Fig. IO.24a and IJ for a steel column, assuming the values o f E and O"y shown in the figure . These curves make it possible to determine the load per unit area PIA. which causes the column to yield for given values of the ratios L~ /r and eclrl.

629

300

i% " 0

36

tIl"

0. 1

"

0.2

30

250

= 36 ksi = 29 X J(J~ psi

£t ,- :

0

0.1 0.2 200

,

0.< 0.6

]!

,

= 250 ,\II'" E "' 200Gl'a

tTl"

"

"

""

(U\

20

"

.11:. :

.'

n.4 U.6

150

~

I

100

n.s

orr : I

10

50

II

Fig. 10.24

150

'"

200

'0'

Load per unit area, PIA. causing yield in column.

We nOle that. for small values of L~ /r, the secam is to 1 in Eq . (10.36), and PIA can be assumed equal to A

"

al lll o~t

equal

(10.37)

1 +-

"

a value that cou ld be obtained by neglecting Ihe effect of the lateral defl ection of the column and lIsing the method of Sec. 4. 12. O n the other hand. we note from Fig. 10.24 that , for large values of L, lr, Ihe c urves corresponding to the various values of the ratio eclrl get very close to Euler's curve defi ned by Eg . ( 10.13 '), and thus thaI the effect of the eccentricity of the loading on the value of PIA becomes negligible. T he SL'C Hnt fonnu la is chielly lIseful for intermediate valucs o f LJr. However. 10 usc it e nc ctivcly, we shou ld know the value of the eccentricity e of the loading, and this quantity, unfonunately. is seldom knuwn with any degree of at:curacy.

200

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SAMPLE PROBLEM 10.2 The unifonll column AB consists of an 8-ft section of struclUraltubing having the cross section shown. (a) Using Euler's formula and a factor of safety of IWO. delermine Ihe allowable centric load for the column and the corresponding nonnal siress. (b) Assuming that the allowable load. found in pan a. L'i applied as shown at a poilll 0.75 in. from the geometric axis of the column. determine the horizontal deflection of the lOp of the column and the maximum normal stress in the column. Use E = 29 X 106 psi.

'I I~~ cL~,

~- ",,=.j

(h )

J\ ~

3.S.1 in" 1= 1>.00 ill~ \ ,50i l!.

,.= <:'"

2.00 ill.

SOLUTION Effccl h'c Length. its cffective lenglh is Critical Load.

.,

t',JJ""3 J.J

kiP'l-q-c : = (I.'S;n. I - -- 1/1

L ..

= 2(8 ft) =

16 fl

= 192 in.

Using Euler's fonllula. we write

~ _"_"_ '1 = ~tr~'(~2~9_X-iclO~·~P~'~i)(~8~.IlO~;~O'~)

p

II .

Since the column has one end fixed and one end free_

L~

( 192 inY

Allowable Load a nd St ress.

P" = 62.1 kips

For a factor of safety of 2. we find

1'«

62.1 kips

F.5.

2

Pol) = 3 L] kips ..

P:ill =--~---

p. u

tI = -

A

31.1 kips

= --2

IT =

3.54 in

8.79 ksi ..

h. Eccentric Load. We observe that column AB and its loading are identical to the upper half of the column of Fig. 10.19 which was used in the derivation of the secant formulas: we conclude that the formulas of Sec. 10.5 apply directly to the case considered here. Recalling Ihal Pol IPe, = ~ and using Eq. ( 10.31). we compute the horizontal deflection of point A:

PI rtL ,.,,~093""

" = 0.75 ill.-.

I'.. = 0 .939 in.
= (0.75 in.)(2.252 - I )

The maximum normal siress is obtained from Eq. (10.35): " sec (" tI,. = -p [ I +2' J_ A r

8

31.! kips [ I ~--3.54 in 2 = (8.79 ksi)[l

630

+

ff.-;-' ). 1« _

(0.75 in.)(2 in.) sec (")] ( 1.50 in·r

+ 0.667(2.252) J

2V2

ITII< =

22.0 ksi ..

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tl .6 in

PROBLEMS

10 .29 The line of action of the 75-kip axial load is parallel to the geometric axis of the column AB and intersects the x a.~is at x = 0.6 in. Using E = 29 X [0" psi. determine (tl) the horizontal denection of the midpoint C of the column. (b) the maximum stress in the column. 10.30 An axial load P of magnitude 560 kN is applied al a point on the axis at a distance (' = 6 mm from the geometric axis of the W200 X 46.1 rolled-steel column Be. Using £ = 200 Gr a. determine (a) the horizontal deflection of end C, (b) the maximum stress in the column.

20 ft

\VSx 35

.t

75 kip.

Fig. P10.29

4 """

Fig. P1 0.3 1

Fig. P10.30

10.31 An axi:!] load P = 15 kN is npplied at point D thaI is 4 mm from the geometric axis of the square aluminum bar Be. Using E = 70 Gra.. detemline (a) the horizontal deflection of end C. (h) the maximum stress in the column.

10.32 An axiallo,ld P is applied to the 1.375-in.-diameter steel rod All as shown. When P = 21 kips. it is observed that the horizontal deflection of the mid(Xlint C is 0.03 in. Using E = 29 X 106 psi, detennine (a) the eccentricity e of the load. (b) the maximum stress in the rod.

C-f l~

r

1.375-1n. di anwtt'F

~[

~· -! tI

1"

Fig. P10.32

631

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632

Columns

10.33 An axiallolld P is applied to the 32-mm-square aluminum bar BC shown. When P = 24 kN. the horizontal deflection at end Cis 4 mn!. Using E = 70 GPa, delermine (lI) the eccentricity e of the 10
Fig. P10.33

10.34 TIle axial load P is applied at
1 J !Wn

p'

FIg. P1 0. 34

Fig. P10.35

10.35 A bmss pipe having the cross section s hown has an axial load P applied 0.15 in. from its geometric axis. Using E = 17 X Icf' psi. determine (u) the load P for which the horizontal deflection althe midpoint C is 0.20 in., (b) the corresponding maximum s tress in the column. 10.36 Solve Prob. 10.35. assuming that the axial load P is applied 0.3 in. from the geometric axis of the column

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10.37 An axial load P is applied at a point located on the x axis at a distance e = 0.5 in. from the geometric axis of the WIO X 39 rolled·steel column Be. Using £ = 29 X ]()' psi. determine (a ) the load I) for wh ich the horizontal deflection of the top of the column is 0.0 in .. (b) the corresponding maximum stress in the column.

Problems

\\,10 X 39

lOft

c 22 n

\\ '8 X 4U

Fig. P1O.37 10.38 The line of action of an axial load P is parallel to the geometric axis of the column AB and intersects the x axis at x = 0.8 in. Using E = 29 X 10" psi. determine (a) the load P for which the horizontal deflection of the midpoint C of the column is 0.5 in., (b) the corresponding maximum stress in the column.

10.39 An axial load P is applied at a point located on the l axis at a distance e = 12 mm from the geometric axis of the W310 X 60 rolled-sleel column BC. Assuming Ihat L = 3.5 m and using E = 200 G Pa. determine (a) the load P for which the horizontal deflection aL end C is 15 mm, (b) the correspond ing maximum stress in the col umn.

Fig. P10. 39

10.40

Solve Prob. 10.39. assuming: that L is 4.5 m.

p'

Fig. P1O.38

633

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634

Cotumns

10.41 TIle steel bar AB has a 10 X 10-mm square cross section and is held by pins that are a fixed distance apan and are located at a distance I! =:0 0.9 mm from the geometric axis of the bar. Knowing tlwt at temperature '/~ the pins are in contact with the bar and that the force in the bar is zero, determine the increase in temperature for which the bar will just make contact with point C if d = 0.3 nun. Use E = 200 G Pa and the coefficient of themlal expansion a = 11.7 X 10 bloC.

rI ~

('

0. 1111

10.42 For the bar of Prob. 10.41. detemline the required distance d for which the bar will just make contact with point C when the temperature increases by 600 e.

0.1 III

10.43 A pipe having the cross sedien shown is used as a 10-ft column. For the grade of steel nsed fT •. = 36 ksi and E = 29 X lif r.si. Know ing that a factor of safety of 2.8 with respett to permanent defonnation is required. determine the allowable load I) when the eccentricity e is (a) 0.6 in. , (b) 0.3 in. (Hilll: Since the fattor of safelY must be applied to the load P. not to the stress. uSt! Fig. 10.24 to deTermine I'r).

jl---'-'l~

10"" n ~

~

" --jl---

p

=

0_9 Illln

Fig. P10.41

10.44 Solve Prob. 10.4]. assuming thm the length of the column is increased to 14 fl.

,--I

a 5.563 in

"

p

1 J 10 ft

I = 0.2.'>8 in

B

,

'l

10.45 An axial load P is app lied to The W250 x 44.8 rolled-steel column BCthat is free at iTS top C and fixed at its base B. Knowing thai the eccentrici ty of the load is e = 12 mm and that for the g-rade of sTeel used I T." = 250 MPa and E = 200 GPa. determine (a) The magniTude of P of the allowable. load when a factor of safety of 2.4 wi th respett topermanent defonn:lIion is required. (b) the Tmio orlhe load found In pan a to the rnagnilUdeofthe allowable centric load for The column. (See hint of Prob. 10.43).

p'

Fig. P10.43 G

Fig. P10.45

10.46 Sol\"e Prob. 10.45. assuming thai the lenglh of the column is reduced to 1.6 m.

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10.47 A lOO-kN axial load P is applied to the Wl50 X 18 rolled-steel column BC that is free at its top C and fixed at its base B. Knowing that the eccentricit y of the load is I' = 6 mm. determine the largest pemliss ible length L if th e allowable stress in t.he column is 80 M Pa. Use E = 200 GPa.

Probloms

10.48

A 55-kip axial load P is applied to a W8 X 24 rolled-steel column a! its base B. Kn owing that the eccentricity of the load is e = 0.25 in .. determine the largest permissible lengt.h L if the allowable stress in the colu mn is 14 ksi. Use £ 29 X 10~ psi.

BC that is free at its top C and fixed

=

10.49 Adal loads of magnilllde P = 135 kips are app lied parallel to th e geo metric axis of the WI0 X 54 rolled-steel column AB and intersect the x axis at a distance e from the geometric axis. Knowing tiwt 0-..1 = 12 ksi and E = 29 x 10" psi. determine tht' largest pennissible length L when (a) I' = 0.25 in., (b) I' = 0.5 in.

Fig. P10.47 and P1 0, 48

c

'"

Fig. P10.49 and P10.S0

10.50 Axial loads of magnitude P = 580 kI\I" are applied parallel to the gcomctric axis of the W250 x 80 rolled-steel column AB and intersect the x axis a! a distance e from tile geometric axis. Know ing that (Jail = 75 MPa and E = 200 GPa. determine the largest permissible length L when (a) I' = 5 mill (b ) f' = IOmm.

Fig. P1 0.51

10.51 An axial load of magnitULle P = 220 kN is appl ieJ at a point located on the. x axis at a distanc e. e = 6 mm from the geometric ax is of the wide-flange column Be. Knowing that £ = 200 0 1'a. choose th e ligh tes t W200 sha pe that can be used if (1:011 = 120 Mi'a. 10.52 Solve Prob. 10.5 1. assuming that the magnitude. of tile axial load is P = 345 kN. 10 .53 A 12-kip axial load is applied wi th an eccentricity e = 0.375 in. to the circular steel roo Be that is free at its top C and fixed a! its base B. Knowing thaI the stoc k of rods available for use have diameters in increments of f in. from 1.5 in. to 3.0 in .. determine the li ghtest rod that may be used if 6 0".11 = 15 ksi. Use E = 29 X 10 psi. 10.54 Solve Prob. 10.53, assuming iha! ule 12- kip axial load will be applieJ to the rod with an eccentricity e = d

I

.t,HII

Fig. P1 0.53

635

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636

Columns

10.55 Axial loads of magniTude P = 175 kN are applied parallel to the geometric axis of a W250 X 44.8 rolled-stu l column A B and inTersecl the axi.> at a distance e = 12 mm from iTS geometric axis. Knowing that {T} , = 250 MPa and E = 200 GPa. determine the facTor of safety with respect to yield. (Hin t: Since the factor of safety mUST be applied 10 the load P. not TO The stresses. use Fig. 10.24 10 delemline P .".)

'I

., c 3.8 m

I"

Fig. P10.55

10.56

Sol\'e !'rob. 10.55, assunling thaT e = 0.16 mm and P = 155 kN.

10.6. DESIGN OF COLUMNS UNDER A CENTRIC LOAD In the preceding scctions, we have determined the critical load of a column by using Euler's formula. and we have investi gated the deformations and Stresses in eccentrically loaded columns by using the secant formula. In eac h ca.~e we assumed that all stresses remai ned below the proponional limit and that the column was initially a straight homogeneous prism. Real columns fall shon of such an idealizati on, and in practice the design of col umns is based 01 1 empirical formulas that retled the results of numerous laboratory tests . Over the last century, many steel columns have been tested by applyi ng to them a centric axial load and increasing the load until fai lure occurred. The results of such tests arc represented in Fig. 10.25 where. fo r each of many tests. a point has been plotted with its ordinate equal to the normal stress {Icr at 1:1ilure, and its abscissa equal to the corrc-sponding value of the effective slenderness ratio, Lj ,.. Although there is considerable scatter in the test results, regions corresponding to three types ()f failure can be observed. For long columns, where Lj ,. is large, failure is closely predicted by Euler's fu rmu la. and the value of (T cr is

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10.6. Design 01 Columns under a Cenlfk: Load

,

u,'

r.-.;-:-.o.~.-:k·.: . ,' . ,

Shmt

""llIn",. Fig. 10.2.5

Loll~

m l","ns

I,.h

obser ved to depend on the modulus of ehL<;ticil Y E of the SLed used, but

not on its yield strength fTy. For very short col umns and compression blocks. failure occurs essentially as a result of yield. and we have fTcr = U y. Columns of intemlcdiate length comprise those cases where failure is dependent on both fT y and E. In this range, column failure is an extremely complex phenomenon, and lest data have been lIsed extensively to guide the development of specifi cations and design formu las. Empirical formulas that ex press lin allowable stress or c ritical slre~s

in terms of the effective slenderness rati o were tirs\ introduced over a century ago, and since then have undergone a continuous process of refineme nt and improvement. Typical empirical rormulas previously used to approx imHte test data are shuwn in Fig. 10.26. It is not always feasible to use a single formula for ull values of LJr. Most design spcci-

l.,1r Fig. 10.26

tications usc dil1e rent fo rmulas, each with a delinite mnge or applicabi lity. In each t:ase we must check that {he fo rmu la we propose to use is applicable ror the value of L,lr fur the column involved. Furthermure. we must determine whether the formula provides the val ue of the critical stress for the column. in which case we must apply the appropriatc fuctor ot' safety, or whether it prtlvides directly an all()wable stress.

637

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638

Spo.;cific fonnu las [01 the desig u of steel. a lu millulJI alll! wood columns under centric loading will [J[)W be cO I1 ~ idere d_ Fig ure 10_ 27 shows examples of columns that would be designed using these fomlula s . Th1! d esign [or the three din'erent materials using AllolI'ilhle Stress De.I'lglI is first presented. This is followed with the formula s needed for the design of s teel columns based on L m ld (lnd R e:.I;.\"f(ll/ct: Factor D e.ligll. t

ColumM

Fig. 10.27 The water tank in (il) is supported by steel columns and the building in conslruction in (b) is framed with wood cDlumns.

",'

Structural Slefl-Allowable- Siress Design. The formulas mosl widely used for the a lJowablc stress desig n of steel columns LInder a eelllrie load arc fou nd in the Spcci ii cution fClf Structural Steel Build in2s of the American Ins titute of Stee l Construction.:!: As we s hall sec. :111 expone ntial ex pressio n IS u;ed to predic t U all for C01UIlUl S of shon and intemlediate lengths. and an Euler-based re lmiun is used for long o.;ulu[JlIlS. TIle dc,.'·;ign relations d f e devdupcLl in lwu s le ps:

A

Q.39u,- _________ _

, r:

"

I. First a curve rl.' presemi ng the variatio n of Ucr with L/r is ob· taillc d (Fig. 10.28). It is im polttuJ! to lIote lhat this cune dves 1101 i JIcorporate all)' f:1ttor of safet y.§ T h:! portion AB of this Cllrve is detined by the equatio n

Ur

(10.38)

Fig. 10.28

t tn

Sf"'<'ili~

drsilln fpnnutal.

th~

lea.r 1_ will

~Iw-"y>

.... fer

In

the drOdl\·. length of {he

column.

tMIlnor,,/ 'J/Slre/ C"mrrucIII'''. I Jth ~ .. Amaican IA_ltjtute orSt~el Construction. Chkago,

'21111'i ~In

l1e Sperif~"tio" for Strucmral Sleel jor BNildl~~.'. Ihe

symbol F is usrd for strcSSe!i

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where 7r!£ fT ,

The portion

Be is defined

( 10.391

= (L/r) 2

by the equation ( IOAO)

ITIT = O.877IT.

We note that when Llr "" O. U or "" u y in E(l. (10.38). At point B. Eq. ( 10.38) joins Eq. (10.40). T he value o f slenderness Ll r at the junction between the two equations is

~=4.71 [E r V-;;;'

([0.41 )

If L/r is smaller than the value in Eq. (10.4 1). U er is detennined from Eq. (10.38). and if Ur is greater. IT"r is determined from Eq. (10.40). At the value o f the slenderness L/r specified in Eq. ( 10.41 ). the stress II, = 0.44 U r. Using Eq. (l0.40), lIor = 0.877 (0.44 Uy) = 0.39 (II" 2. A f[t cto r o f safety must be introduced to obtain the fini"l.l AISC design formulas. The factor of safety specified by the specification is 1. 67. Thus ( IOA2)

The formu las obtained can be used with SI or U.S. customary units . We observe lhat, by using Eqs. {IO.38), (lOAD), (IOAI). and (IOA2). we can determine the allowable axial stress for a given grade o f stee l and any given value of L/ r. The procedure is to first compute the va lue of L/ r at the intersection between the two equations from Eq. (10.41). For given values of L/ ,. smaller than that in Eq . (10.41), we use &15. ( 10.38) and ( 10.42) to calcu late (T' iI ' and for values greater than that in Eq. (10.41). we usc Eqs. (10.40) and (10.42) to calculate (Till)' Figure 10.29 provides a general illustration of how u
,, \, ,, ,, ,

o

50

100

Lk

Fig. 10.29

(54)

200

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EXAMPLE 10.02 We must compute the crilk-al slress fT,.,. Assum ing L/ r is larger than th e slenderness specified by Eq. ( 10.4 1). we use Eq. ( IOAO) wi th (10.39) and wtite

Determine the longest unsupported length L for which the S 100 X J 1.5 ro lled-steel compre~si{ln member A B can safely carry the centric load shown (Fig. 10.30). Assume ( f r = 250 MPa and £ = 200 G Pa.

1-"'"

(/"" =

131 I

=

7T'E 0.877 q~ = 0.877 (L/ rf

0.877

1.73 1 X 10 '~ POl

11\ 200 X 10" Pal

, (LI<)-

(L/,),

Using this expression in Eq. (10.42) fur

"r

0""11 .

(LI,)'

L

l/

Equll!lng Ihis expression 10 the required value of rr .11. we wri le

1.037 X 10 '2 Pa (LI,)'

/~

4l. 1 X IO~ Pa

The slendcrness ralio from

&t.

z 41.6 mm

f,

p 60 x lOi N (ToJl=A"= 1460 X 10 - 6 m I

Qur assumption thll! L/ r i ~ greater than this slenderness ratio was correct. Choosing the smal ler of the t IVO rad ii of gyration. we have

= 14.8 mm

If the 60-kN IO,ld is to be safely supported. we

m ll ~t

have

L

41.1 X

C,
,u ,.r

/ Ur

Fig. 10.31

= 158.8

133.2

From Appendix C we fintlthat. for an S 100 X 1 1.5 shape, r~

L/r

( 10.41 ) is

Fi g. 10.30

A = 1460mm'

we wrile

1.037 X 10 '2Pa

1 0~

Pa

T.

L 14.8 X lO- J m

158.8

L = 2.35m

Aluminum. Many aluminum alloys are available for use in structural and machinc construction. For most columns the spec ifications of the Alumi num Assoc iationt provide twO formulas for the allowable stress in columns under centri c loading. The ,ariation o f (Tall with L/r definl.'{[ by these fon n ulas is shown in Fi g. 10.3 1. We note that for short columns a li near relalion between q "1I with Ljr is used and for long col umns an Euler-type fo rmula is used. Spec ilie formulas for use in the des ign o f buildi ngs and similar stmcturcs arc given below in both 51 and U.S. customary units for two commonly used alloys. Alloy 606 1-T6:

L/r

<

66:

(Tall

= [20. 2 - 0.126(Ljr) ] ksi ~

Ur 2 66:

(Tdll

=

[ 139 - 0.868(L;,.) J MP,

5 1,000 ks i

35 1 X 10·\ MPa

(L/,,),

(LM'

(1 0.43)

( 10.43') (1 0.44)

t SIJt!cijiclll;OJJS a",1 Gu;delines fo, .... ,um;"''''' Stmcwru. A luminum ASSOCiation. tnc •. D.C.. 2ao'!.

Wlt~hlngton

640

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Alloy 20l4-T6:

Llr

<

55:

10.6. Design 01 Columns under a

Centfk: Load (TiilJ

= [30.7 - 0.23(Llr) ] ksi ~

Ur

[21 2 - 1.585(L;') 1 MPa 54.000 ks i 372 X IOJ MP:l

2:

55:

O"all

~

(LIe)'

(Lie)'

([0.45)

( IOAY ) ( 10.46)

Wood. For the design of wood columns the specifications of the American Forest & Paper Associationt provides it single equation that can be used to obtain the allowable stress for short, intennediate, and long columns under centric loading. For a column with a reC/(IlIgu{ar cross section of sides band d, where d < b, the variation of 0"111 with Li ll is shown in Fig. 10.32.

.~ o

"

Lid

Fig. 10.32

For .w lid columns made from a single piece of wood or made by gluing laminations together, the allowable stress 0" . 11 is ([0.47)

where O"e is the adjusted allowable stress for compression parallel to the grain.* Adjustments used to obtain Uc arc included in the spccilications to account for different variations, such as in the load duration. The column stability factor Cp accounts for the column length and is defined by the following equation:

J"[~[---:-+-(~~-~-."/~-d"l~'~--_u-_c-,=:=~::c

([0.48)

The parameter c accounts fo r the type of column, and it is equal to 0.8 for sawn lumber columns and 0.90 fo r glued laminated wood columns. The value of u CE is defined as 0.822£ O"eE

=

(Ud)!

( [0.49)

Where E is an adjusted modulus of elasticity for column buckling. Columns in which Lid exceeds 50 arc not permitted by the National De.l·igl) Specificatioll jor Wood Constructioll. t Natimwl Design Specijicatim' f or 1\'00<1 Construction. American Forest & Papcr Association. American Wood Council. Washin gton. D.C.. 2005. U n the National Desi!;n Specification f or IVOO(I Construction. the symbol F is used for stresses.

641

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EXAMPLE 10.03 Knowing thilt column AB (Fig. 10.33) has an effective leng/h of 14 ft. ruJd !hilt it must safely carry a 32-kip load. design !he column using a square glued laminated cross section. The adjusted modulus of elasticit y for the wood is E = 800 X 10.1 psi. ;:md the adju sted allow3ble stress for compression pamllel to the grain is If C 1060 psi.

=

We note that i: = 0.90 for glued laminated wood columns. We must compute the value of U C!". Using Eq. ( 10.49) we write

0.822(800 X IO J psi)

0.822£

(fCE = (L/d)l

=

(168 in./lI)l

23.299d! psi

We then use Eq. ( 10.48) III express the column stability factor in telms of 11, with (u n/(f cl = (23 .299d ~/ 1.060 X 10J) = 2 1.98 X 10 · J d 2 •

I t 21.98 X 10 3d 1

J[l +

2(0 .90)

21.98 X 10 2(0.00)

3

d

!]'

Fig. 10.33

21.98 X 10

3

d!

0.90

Since the column must carry 32 kip~ . which is equilllO fTc {(. we use. Eq. (10.47) to write

Solving this equiltion for C p and substituting the value obt:]ined into the previou ~ equation. we write 30 . 19 = I + 2 1.98 X 10 3 d! _ /[ 1 d1 2(0.90) \

+

1 2 21.98 X 1O- d ]'

2(0.90)

Solving for d by trial and error y ields d = 6.45 in.

642

2 1.98 X 10 3 d ' 0.90

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· Struclural Steel- L.oad and Resistunce Factor Design. As we ~aw

in Sec. 1.13. an alternative method of design is based on the de termination o f the load at which the strucillre ceases to be usefu l. Design is based on the inequality given by Eq. (1.26): (1.26)

The approach useJ for the design o f steel columns undcr a celllric load using Load and Resistancc FaclOr Design with the AISC Specification is similar to thaI for Al.Iowable Stress Design. Using the critical stress (J..,. the ultimate load P u is deli ned as (10.50)

The determination of the cri tical stress U ,r follow s the same approach used for Allowable Stress Design. This re(luires using Eq . ( 10.41 ) to determine. the slenderness at the junction between Eqs. ( 10.38) and Eel. (10.40). If the specified slenderness L/,. is smaller than the value from Eq . (10.4 1). Eq . ( 10.38) governs. and if it is larger. Eq . ( 10.40) governs. The e(luations can be used with SI or U.S. customary units . We observe that, by using Eq. (10.50) with Eq. (1 .26), we can determine if the design is acceptable. The procedure is tt) first determine the slenderness ratio from Eq . ( 10.41). For values of L/,. smaller than this slenderness. the IIltill/lIIe load P fl for usc with Eq. ( 1.26) is obtained from E4 . (10.501. using a ct· determined from Eq. (10.38). For values of L/,. larger than this sle nderness. the ultimate laad Pv is obtained by using Eq. (10.50) with Eq. (10.40). The Load and Resistancc Factor Design Spcci lication of lhe American Institute of Steel Construction specifics that the resistance facto,. cp is 0.90. Nnte: The design formula s presented in this section are intcnded 10 provide examples of different design approaches. These formulas do not provide all the require ments that are nceded for many designs. and the student should refer to the appro priate design specifications bcfon: attempting actual designs.

10.6. DesIgn 01 Columns under a Cenlfk: Load

643

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\\' 10 X 39 A=lJ ,5i,, 2 r, = 4. 2i in

I'

I- '

'"1= L9.'lin

I'

r{

SAMPLE PROBLEM 10.3 Column AB consiwi of a WIO X 39 rolled-steel shape marleof a grade of steel for which (f r = 36 ksi and E = 29 X 106 psi. Determine the allowable centric IO
SOLUTION

24ft

We first compute the value of the slenderness mlio from Eq. IOAl corresponding \0 the given yield strength if,. = 36 ksi.

12ft

l

/!I ~ ~,

L -;:= 4.71

c

~.

~29

X let 36X IOJ= 133.7

U . Effccti\"c Length = 24 ft , Since r., < r., buckling will lake place in the .\'~ plane. For L = 24 fl and r = /"1 = 1.98 in .. the slenderness ratio is

12ft

(24 X 12) in. 1.98 in.

L

r,

288 in. = [45.5 1.98 in.

Since L/ r > 133.7, we use Eq. (10.39) in Eq. (IOAO) to determine

tr 2£ (f e,

p~,

= 0.877 fT , = 0.877

=

(fa

1 [.86ksi

The allowable stress. detemlined using Eq. ( IOA2). and P-.JI are

u .... = ~ = 11.86 k.~ i = 7. 10 ksi

24ft

, ,,

( L/r) ~

tr!(29 X [(}l ksi) 0.877 (145.5)1

[.67

,

p.1!

=

(f.IIA

1.67

= (7. 10 ksi)( 11.5

in 1)

=

81.7 kips

...

b. Bracing at l\I idpoint C. Since bracing prevents movement of point C in the xz plane but not in the yz plane, we must compute the slenderness ratio correspoinding 10 buckling in each plane and determine which is larger.

B

x: Pial/I':

Effective length

=

[2 fl

=

144 in .. r = r,

=

1.98 in.

L/r = ( 144 in.)l(1.98 in.) = 72.7

yz Pial/eo'

Effective length

= 24 fl = 2~8 in.. r = r, = 4.27 in.

L/ r = (288 in.)l(4.27 in.) = 67A Since the larger slenderness ratio corresponds 10 a smaller a[lowab[e load, we choose L/ r = 72.7. Since this is sma[lerlhan L/ r = 145.5, we use Eqs. ( 10.39) and ( 10.38) 10 detennine If,., B Il'j(·k!in~

in.r::: plan,·

" B,lt'klin~

jn !F pianp

if •.,

tr"£ 1(2(29 X 103 ksi) - -- -

(72.7)'

• - (L/, )' -

54.1 ksi

= [ 0.658("J",)] Fy = [0.658(Jbl.\I~. jl>.i)1 36 ksi = 27.3 ksi

We now calculate Ihe allowable stress using Eq. ( IO A2) and the allowable load. U:>II

=

u,.,

l.67 =

27.3 ksi

~ =

16.32 ksi

P ;l]1 = fJ"aliA = ( 16.32 ksi)( 11 .5 in!)

644

f'.u = Iln~ 1 k.'1 ...

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J p =bHkN

n

A

SAMPLE PROBLEM 10.4 Using the aluminum alloy 20 14·T6. detennine the smallest diameter rod which can he used to suppon the centric load P = 60 kN if (a) L = 750 mm. (b) L = 300 mm.

L

L

,

SOLUTION For the cross section of a solid circular rod. we have

u. Ll' ngth of 750 111m. Since the diameter of the rod is not known. a value of Llr must be assumed: we aSJ W II(! that Llr > 55 and lise Eq. (10.46). For the centric load P, we have (r = PIA and write p

A=

372 X !OJ MPa (r, 1I

(L/r)!

=

372 X J09 Pa

m)'

0.750 el 2 ('4 = J 15.5 X 1O- 9 m' (

c = 18.44 mm

For (. = 18.44 mm. the slendemess ralio is L

Our

as ~umption

L

750mm

el2

(18.44 mm)/2

81.3 > 55

is correct, and for L = 750 mm, the required diameter is

d = 2e

= 2( 18.44 mm )

II = 30.9 111111 ...

b. Length of JOO 0101, We again (Il'.\·lI/lI(! that L/r> 55. Using Eq. (10.46). and following Ihe pro: edure used in pan a. we find Ihat L' = I J .66 mm and Ll r = 51.5. Since Llr is less than 55. our assumption is wrong; we now as ~me that Llr < 55 and use Eq. (10.45') for the design of this rod. ; fi()

=

fT.J1

= [ 212 - 1.585

::~] N

[212 _ 1.585

(~)1MPa (O~~~;l) J IO~ Pa

(. = 11.00 mm

For c = 12.00 mm. the slenderness ratio is L

- = -

L

L'/2

Our second assumption that Ll r diameter is

=

300mm (12.00 mm)/ 2

<

55 is correct. For L = 300 mm. the required

d = 2c = 2(12.00 mm )

50

Ii = 24.0mm ...

645

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PROBLEMS ~lo;"",,~

'~l;" " -O Fi g . P1 D.57

10.57 A ~'1eel pipe having the cross section shown is used as a column. Using the AISC allowable stress design formubs. determine the allowable cenlnc load if the efTectivc length of the column i~ (a) 6 m. (b) 4 m. Use U y -"" 250 MPa and E = 200 GPa. 10 .58 A column with the cross section shown has a 13.5·fl effecti ve length. Using allowable stress design. determine the largest centric load thaI can be applied \0 the column. Use O"y = 36 ksi and E = 29 X lot' psi.

Fig. P10.58

10 .59 Using allowable stress design. delemlinc the allowable cenlric load for a column of 6-m effective length !lla! is made from the following rolled-sleel sh:lpe: (a) \V200 X 35.9. (b) W200 X 86. u~ (Ty = 250 MPa :md E = 200 GPa.

10.60 A W8 X 31 rolled-sleel shape is used for a column of 21-ft effective length. Using allowable SLIe~s deSign. determine the allowable centric load if the yield strength of the grade of steel used is (oj (1y = 36 ksi. (b) (T y = 50 k.~ i. Use £ = 29 X lif p~i. 10.61 A column hailing a 15-m effective length is made of sawn lulUber with a 114 X 140-mm cross section. KnOWing that for the grade of wood used the adjusted allowable Stress for compression parallel to the grain is (1c = 7.6 MP.1 and the adjusted modulus E = 2.8 G Pa. determine the maximum allowable centric load for the collmlll.

10.62 A sawn lumbercolullll1 with a 7.5 X 5.5-iLl. cross section has an 18-ft effective length. Knowing that for the grade of wood used the adj u ~ted allowable stress for compression paralle l to the grain is fTc = 1200 psi and that the adjusted modulus £ = 470 X 10] psi. detennine the maximum allowable centric load for the column.

646

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10.63 Bar ,1B is free at its end A and fixed at it'i base 11. Determine the allowable .:emric load P if the aluminum alloy is (a) 6061 ·T6, (b) 2014·T6.

1>5 ,,,'"

Fig. P10.63

10.64 A compression member has the cross section shown and an effective length of 5 ft. Knowing lhat the aluminum alloy used is 606 1-T6. detennine the allllWable centric load.

I "" I~

.

IJE' "·' ' L

jill.

-

-0.-1111

~ . IO.lim.

FIg. P10.64

10.65 and 10.66 A compression member of IJ-m effective length is obtained by welding two IO-mm-thick steel plates 10 a W250 X 80 rolled-steel shape as shown.. Knowing lhal IT)" = 345 MPa and £ = 200 GP
DDI Fig. P10.65

Fig. P10. 66

10.67 A ccmpression member of 2.3-m elTecli\'e length is oblained by bolting toge ther two 127 x 76 X l 2.7-mm steel angles as shown. Using allowable stress deSign. determine the a llowable centnc load for the column. Use if,. = 250 MPa and £ = 200 GPa.

Fig. P10.67

Probloms

647

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648

Columns

10.68 A column of 21-ft etrective lenglh is obtained by connecting two CIO x 20 steel channels with lacing bars as shown. Using allowable stress design. detennine the allowable centric load for the column. Use IT y = 36 ksi and £ = 29 x 10" psi. 10 .69 A rectangular column with a 4.4-m effective length is trude of glued laminated wood. Knowing that for the grade of wood used the adjusted allowable stress for compression parallel to the grain is (Fe = 8.3 MPa and thi:. adjusted modulus £ = 4.6 GPa. determine the maximum allowable centric load for the column.

1

Fig. P10.68

216m

"' l

I

I-I(J nuu

~~~~ Fig. P10.69

10.70 An aluminum structural tube is reinforced by bolting two plates to it as shown for use as a column of 5.6·fl effective length. Knowing that all material is aluminum alloy 2014-T6. determine the maximum allowable centric load. 10.71 An IS-kip centric loud is applied to II rccillnguluf sawn lumocr column of 22-ft effective length. Using sawn lumber for which the adjusted al100vabie stress for compression parallel 10 the grain is if c = 1050 psi and th e adjusted modulus is E = 440 X IOl psi. determine the smallest cross section that may be used. Use b = 2d.

r b

,

' B

Fi g. P10.71

Fig. P1 0.72

10. 72 A column of 2.I-m effectiw length is to be made by gluing together laminated wood pieces of 25 x 150·mm cross section. Knowing that for the grade of wood used the adjusted allowable stress for compression parallelto the grain is ife = 7.7 MPa and the adjusted modulus is E = 5.4 GPa. detemtine the number of wood pieces that must be used to snpport the concentric load shown when (a) P = 52 kN. (b) P = 108 kN.

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10 .73 The glued laminated column shown is free at its top A and fixed at its base B. Using wood that has an adjusted allowable stress for compres· sion parallel to the grain (J" c = 9.2 MPa and an adjusted modulus of elasticity E = 5.7 G Pa. determine the smallest cross section !lUlt c:m suppon 11 centric

Problems

load of 62 kN.

r

2,"

,/

d

FIg. P10.73

10.74 A 16-kip centric load must be supported by an aluminum column as shown. Using the aluminum alloy 6061·T6. determine the minimum dimension b that can be used.

n

r

18 in.

l OJ' 8

Fig. P10.74

Fig. P10.75

2 .2,.';",

!IO·n,n, m,t , i']c dbm ct(>r

10.75 A 2110-kN centric load is applied to the column shown, which is free at its top A and fixed at its base B. Using aluminum alloy 20 l 4-T6. se lect the smallest square section that can be used. 10. 76 An aluminum tube of 90-mm Quter diameter is to carry a centric load of 120 kN. Knowing that the stock of tubes available for use are made of alloy 20 14-T6 and with wall thicknesses in increments of 3 mm from 6 rum to 15 mill, determine the lightest tube that can be used.

B

FIg. P10.76

649

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650

Columns

10.77 A centric load P must be supponed by Ihe steel bar AB. Using allowable stress design. determine Ihe smallest dimension d of the cross section Ihal can be used when (11) " ~ 24 kips. (b) P = 36 kips. Use tTl' = 36 ksi and E~ 29 X 10" psi.

3d

.5_2 fl

, ", Fig. P10.77

U

10.78 A column of 4.5-m effective lenglh must carry a centric load of kN. Knowing Ihat tTy = 345 MPa and £ = 200 GPa. use allowable. stress design to select the wide-flange shape of 250-mm nominal depth thai should be used.

I)()()

10.79 A column of 22.5-ft effeclive lenglh musl carry a centric load of 288 kips. Using allowable stress design. select the wide-flange shape of 14·in. nominal depth that should be used. Use fT y = 50 ksi and £ = 29 X ]06 psi. 1 0.80 A column of 4.6-m effeclive length must carry a centric load of 525 kN. Knowing that tTy = 345 MPa and E = 200 G Pa. use allowable stress design to select the wide-flange shape. of 2(K)..mm nominal depth thal should be used.

Fig. P10.81

10.81 Two 89 X 64·mm angles are bolted together as shown lor use as a column of 2.4·m e.ffective length to carr~ a cenlric load of 180 kN. Know · ing thai the angles available twve lhicknesses of 6.4 mm. 9.5 mill. and 12.7 mm. use allowable stress design to detemline the lightest angles Ihat can be used. Use tTy = 250 MPa and £ = 200 GPa. 10.82 Two 89 X 64-mm angles are bolted together as shown for use a., a column of 2.4·m effective length to carr~ a centric load of 325 kN. Knowing that the angles available have thicknesses of6.4 mm. 9.5 mm. ,md 12.7 mm. use allowable stress design to detemline the lightest angles that can be used. Use fTl' = 250 MPa and E = 200 GPa.

I

64"'' ' l

fi4 ''HIl

I

1[2".I FIg. P10.82

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Probtoms

10.83 A square steel tube having the cross section shown is used as a wlumn of 26-ft effective length to carry a centric load of 65 kips. Knowing that the tubes available for use are made with wall thkknesses ranging from 1in. to 1 in. in increments of I~ in .. use allowable stress design to determine the lightest tube that can be used. Use ITI"" = ]6 ksi and £ = 29 x 10~ psi .

OJ 6tll. ---1

Ag. P10.83

Fig. PtO.84

10.84 Two]! x 2l.-in. angles are bolted toge ther as shown for use as a column of 6-ft effective length to cany a centric load of 54 kipll. Knowing that the angles ;}I'iiilable have thic knesses of ! . {. ;lnd 1in .. use ;lllowable stress design to detel1l1ine the lightest angles that C;ln be used. Use (TI =]6 ksi and t.' = 29 X 106 psi. ' 10.85 A rect::l llgubr steel tube having the cross section ~h owll is used as a column of 14.5-ft effecth'c length. Knowing that IT y = 36 ksi nnd E = 29 X 106 psi. use load and resistance f;lctor design to determine the largest ceutric Jive load that can be applied if the centric dead load is 54 kips. Use :J. dead load factor "YD = 1.2, a live load factor "Y~ = 1.6 and the resistance factor.p = 0.90. '10 .86 A Lulumn with a 5.8-m effective length $ UPPOI1 S a centric load. with ratio of dead to live load equal to 1.35. The dead load factor is "YD = 1.2, the live load factor "h = 1.6. and the resistance fnctor q, = 0.90. Use load and resistance factor design to determine the allowable centric £lend and live loads if the column is mnde or the following rolled-steel shape' lu) W250 X 67. lb) W360 X 101. Use if)' = ]45 MPn and E = 200 GP;}. "10 .87 The steel lUbe having the cross section s hown is used as
to f..-,;,,-i

Fig. PtO.87

' 10 .88 A steel column of 5.5-01 effective length must carry a centric dead load of 310 kN and;} centric live lond of 375 kN. Knowing thutlT)" = 250 MP;} nnd E = 200 OPa. use load and resistance factor design to $elect the widefl;}ngc ~hnpe of 31O-mm nominal depth thm should be used. The dead load factor )'L> = 1.2, the li ve load factor h = 1.6, Mid the resistance factor dJ = 0.90.

1"-' ;".----j 1=~ln. -

1 7;n.

~J

Fig. PtO.8S

651

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652

Columns

Fi g. 10.34

10.7. DESIGN OF COLUMNS UNDER AN ECCENTRIC LOAD

In this section. the design of columns subjected to an eccentric load will be considered. You will see how the empirical fonnula s developed in the preceding secti on for columns under a centric load can be modified and used when the load P applied to the column has an eccentricity e which is known. We ti rst recall from Sec. 4.12 that an eccentric ax ial load P applied in a plane of symmetry of the column can be replaced by an equi valent syste m consisting of a celllric load P and a couple M of moment M = Pe, where e i.~ the distance from the line of action of the load to the longitudin al axis of the column (Fig. [0.34). The normal stresses exerted on a tran ~ve rse section of the column can then be obtaincd by superposi ng the stresses due, respecti vely. to the cenlfic load P and to the couple M (Fig. 10.35), provid ed that the section cons idcred is not too close 10 either e nd of the col umn , and as long as the stresses involved do not e(ceed the propon.ional limit of the material. The normal stresses due to the eccentric load P can thu s be ex pressed as (10.51 )

Recalling the results obtained in Sec. 4. I 2, we find that the maximum compressive stress ill the column is

a

P

m.a~

Me

= -A + - I

(10.52)

Fig . 10.35

In a properly designed co lumn, the maxim um stress defi ned by Eq. ([0.52) should not exceed the allowable Slress for the column . Two altcrnmive approaches can be used 10 satisfy thi s requireme nt, namely, the allowahle·stress m ethod and the. imemctio/l metliud.

n. A lloWllblc-Stress Method . This method is based on the assumption that the allowable stress for an eccentrically loaded column is the same as if the column were ct:ntrically loaded. We must have, therefore, 0" ma.. :S: (Tall> where a- all is the allowable stress under a centric load. or substituting fo r (Tm.~ from Eq. ( [0.52) P

Me

A + - ,-

S O" "II

(10. 53)

The allowable stress is obtained from the formulas of Sec. 10.6 which. lo r a given material , express rT~1I as 11 function of the slenderness ratio of the column. The major engineering codes require that the largesl value of the s lenderness ratio of the column be used to determine the allowable stress, whether or not this value corresponds to the actual plane of bending. Thi s re(luiremcnt sometimes rl!s ults in an overly conser vative design.

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EXAMPLE 10.04 A column with a 2.in .. square cross section and 28-in. effective length is mad~ of the aluminum alloy 2014- To. Using the allowable-stress meu]()(L determine the maximum lo'ld P thal can be safely sllpp()[ted with an eccentricity of 0.8 in.

We next wmpute L/r = (2S in.)/(0.5773 in.) = 48.50. Since L/r < 55, we useEq. (10.48) to deTennine the al· lowable strt!Ss for the aluminum column subjected to a centric load. We have

We first compUle the rndius of gyration r using: the given

We now use Eqo\10.53) with M = I'e and {' = ~ (2 in.) = I in. to determine the allowahle load:

0->11

data

p 4 in~

I = I~ (2 in.)4 = 1.333 in 4

A = (2 in.)' ~ 4 in 2

= [30.7 - 0.23(48.50)) = 19.55 hi

+

P(O.S In.)( I in.) 1.333 in~

<:

. 19.55 kS I

p :s; 22.3 kips

1.333 in 4

~ =

0.5773 in.

The maximum load (haL can he safely applied is P = 22.3 kips.

h. Inte-raetion

~te-th od . We recalJ thm the allowable stress for a colum n subjected to fI centric load (Fig. 10.30a) is generall y smaller than the allowflble stress fo r a colum n in pu re bending (Fig. IO.36h). since. the former takes into account the possibility o f buckling. Therefore. when we use the allowable-stress method to des ign fin eccentrically loaded column and write that the sum o f the stresses due to the centric load P and the bending couple M (Fig. 10.3(0) must not exceed the allowable stress for a centrically loaded column, the result ing design is generall y overl y conserVative. An improved method of design Can be developed by rewriting Eq. 10.53 in the fonn

PIA

-

(Tall

Mcl!

+ - - :s;

I

( 10.54)

(T.II

and substituting for (T . II in the fi rst and second terms the values of the allowable ~t res~ which correspond, res pectively, to the Centric 10'lding of Fig. 10.36a:md to the pure bendi ng of Fig. 1O.36h. We have

I" .".-xl '1""

L 101 Fig. 10.36

~I ' (Il)

'"

(10.55) The type of fonnu la obtained is know n as an interaction formula. We note thut. when M = 0, the use of this formula resu lts in the design of a centrically loaded column by the method of Sec. 10.6. O n the other hand. when P = O. the use of the formu la results ill the design o f a beam ill pure bending by the method of Chap. 4. When P and At arc both different from zero. the interaction formu la resu lts in a design that takes into account the capacity o f the member to resist bending as well 'L~ axial loading. In all cases. (0" .1I)«n";~ will be detemlined by using the largest sle nde rness ratio of the column, regardless of the plane in which bending takes place.t t This procedure i.s required by aIL major codr~ for Ihe design of .lll'cl. aluminum. and timber compression members. tn additiOIl. many s~cifieatioll.l call for Ihe use of all :ldditiollal faCIQr in lhe sc~ond Icrm of Eq. (LO.60): Ihis factor taJ.:n iniO aCCQUni Ihe additional Sifes!oC.\ rnulling from Ibl' d~noction of the column due to bending.

653

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654

Columns

When the eccentric load P is no t applied in a pla ne o f symmetry of the colu mn, it causes bend ing abo ut both of the principal axes o f the cross sectio n. We recall from S ec. 4. 14 tha t the load P can then be replaced by a centric load P and two couples re presented by the couple

p

"

Fig. 10.37

vectors Mx and M e shown in Fig. 10. 37. The interaction fo rm ula to be used in thi s case is

(10.56)

EXAMPLE 10.05 Use the interaction me thod to delermine the ma."(imum load P tho:!t can be safely supponed by the column of Example 10.04 with an eccentricity of O.S in. The allowable s tress in bending is 24 ksl. The value of (u ..ll t< •• ", lws alrco:!dy been dctermined in Example 10.04. We have

Using the numerical data from Exmllple lOJ)4. we write

P/4 P(0.8)( 1.0)/1.333 --- + :s 1.0 19.55 ksi

24 ksi

P :::;: 26.5 kips

The maximum load Iltm can be safely applied is thus P = 26.5 kips. Substilllting these val uesjnto Eq. (10.55). we write P/A Me/I - - - + --::5::1.0 19.55 k~ 24 ksi

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SAMPLE PROBLEM 10.5 Using the allowable-stress method. determine the largest load P that can be safely carried by a W310 X 74 steel column of 4.S-m effective length. Use E = 200 GPa and i f )' = 250 MPa. A

= f).\$O """~

r, = 132 r~ = 49"

111m

II"n

S~ = 10fl0 X l(}l 'm" ~

SOLUTION

,-

200

c

The largest slenderness ratio of the column is L/J", = (4.5 m)!(O.0497 rn ) = 90.5. Using Eq. ( 10AI) wi th £ = 200 GPa and i f y = 250 MPa. we lind that the slenderness ratio at Ihe junction between Ihe two equations for (T" is LI T = 133.2. Thus. we use Eqs. \I 0.38) anj (10.39) and find that ff ,~ = 161.9 MPa. Using Eq. (10.42). the allowable stress is

p

111m

(if,1jI).""" ic = ]61.9/1.67 = 96.9 MPa

c

For the given column and loading. we have

P

/'

A

9.48 X 10 - 1 m 2

Me

-

AI

= -

S

=

P(O.200 m) 1.060 X 10- ) m1

Subslillliing into Eq. (10.58), we write P

Me

A+ , S ff.U " 9.48 X iO ) m !

P (0.200 m)

+ 'IC.06O ;:-;;:-X-;CIOC-~,Cm'J -< 96.9 MPa

p

~

330kN

P = J30kN.1. ..

The largest allowable load P is thus

SAMPLE PROBLEM 10.6 Using the interaction method. solve Sample Prob. 10.5. Ass ume (fT"I)ben~' "~ = 150 MPa.

SOLUTION Using Eq. (10.60). we write

~ + Mel l <: J (cr,uL,nu", ( frall h..'>di.~ Substituting the given allowable bending stress and the allowable centric stress found in Sanlple Prob. 10.5. as well as the other given d:lta. we have PI(9.48 X iO- 3 m]) P (0.200 m)/(I.060 X JO- 1 Ill')

~~9"6C.9"'XCC:IO'O;-;pC,~ +

150 X

10~ Pa

-<

I

P :$ 426 kN The IMgest allowable load P is thus

p ~4 26kN.1. ..

655

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SAMPLE PROBLEM 10.7 A steel column having an effective length of 16 fl i~ loaded eccentrically as sbuwn. Using the intt:ranion methud . sd<:<:\ L.h~ widt!-flaJl),:t: Shilpt!uf 8-in.numinal depth that should be used Assume £ = 29 X 106 psi and II y = 3fi ksi. and

use an allowable ~Iress In bending of 22 ksi.

SOLUTION So lhal we can selecl a trial $ection. we lise the allowable-stress method wilh = 22 ksi and write

(T&II

5lr .

,

, I~ /~ ~"k'P'

=

/cr-~

,

/y

c

From Appendix C we observe for shapes of 8-in. nominal depth lhm ~ 3.5 in, Subs tituting in1 0 Eq. (I). we have

C ...

4 in.

and r.

. 85 lips!. 4~2~ 5~k~ ip~.~i~ " .~)(~4~i"C1') 22bl= - - - + -

....,

A ". 10.2 in~

A(J .5 iny

A

M = 10;.5 J,;ip,,( ~ in I We select for a first trial shape: W8 ><. 35. =.8.5 lop . h Trial I : \\ '8 x 35. The allowable ~Ire>ses arc AlIlJl'o/lible BI"IJin/t Slrl's~: (see data) (IT..tI)""odll... = 22 ksi

I"

!:[- ,

\\ ')j

F,

Al/o\ml,l~ Cfmc""lrk SIren:

35

>(

A =

[(l.3

i"~

= 3.51 in .

,.-~ =

2'<13 in.

S.., =3J21U' L = 10ft = 192,,,,

large-~l ~Iendeme>s

The

ralio of the column is

Ur, = ( 192 ill.)/(2.03 in.) = 94.6. Using Eq. (10.41) with E = 29 X lOb psi and {} y = 36 ksi, we iind tllat the slenderness ratio at the junction between the two eqlliltions for fT ,., is L/ ,. - ])).7. Thus. we U.'>e Eqs ( to.J 8-1 ilnd ( 10.39) aud find that if" = 22.5 ksi. Using Eq. (10.42). lhe allowabl;! suess is (U..J«nL""= 22.5/1.67 = 13.46ksi For the W8 X 35 tri:ll shape. we lla ve P

A= With

I'

~- ,

WIj )( -tti A - I~ I luI

F, = 3.El in .

"'1

=

2.Min.

S~~ "33in ' I. = 16ft = 192in

I'

~-,

11'" X 4(J ,\ = 11.7'n' r, '" 3.53 in r~ =2.().Iin

S~ - 35.5 in l L = 16ft ~ 1921n.

mi .

M..

85 kip' 10.3 in1 = 8.25 kSI d"t~

we find th.u the

teft _h~nd

PIA

Md l

(u ••).....<

(tT ""I)'->J'",

--- + S iner. 1.2U

>

1.1100. rhl'

425 kip' in.

S,

3 l.2in l

13.62 ksi

..nember of Eq. ( 10.60 )

8.25 ksi 13.52 ks; -- + -- = 13.46 ksi

22 ksi

r'~'lttirem~rll ~xrrl'_'~cd

hy (he

i~

1.232

inr~r:lC: rinn

fornlttln is

not satisfied: we mu sl select .1 larger trial ~h~pe. Trial 2: W8 X ~8. Following the pro..'"edure used in tri31 l. we write L 192 in. -~--=92.3 r, 2.08 in. P 85 kips Me A = 14. 1 in2 = 6.()3 ksi

(u .tlL. ...., = 13.76 ksi AI

425 kip' in.

S,

43.3 in)

9.82 ksi

Substi tuting into Eq. ( 10.60) gilles ~+

('Tal.l.~ I "'.' T he W8 X .J.8

Mell

(fT.tI)I>dIilJ"t

sh~pe

is

6m ksi + 9.82 k ~i = 0.884 22ksi 13.76ksi

s ati sf~ctory

Trial 3: \\'8 X -ID. Following interaction formu la is not s.atisfied.

Sdedio ll or Shape.

656

M

<

1.000

but m:.y be unne.:e ssaril) lurge. ~gain

the same procedure, we find thatthc

Tire shape to be UliOO is

\V 8 x .J.8 •

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PROBLEMS !,slnlll

10 .89

A co lumn of 5.S-1ll effec tive length is made of dIe aluminu m alloy

10l4·T6. for which the allowable mess in bending is 22[) MPa. Using the interaction method. detenninc the allowable load P, knowing that Ihe eccen(ridly i ~ (a) f' = O. (b) e = 40 mill.

5.5

1-- 152nll,,--I

10.90 Solve Proh 10_89. a~s llming that the dTecli ve length of the co l. H

umn is).O m.

10 .91

A , awn-lumber column or5.0 X 7,5-in. cross section lms an cf-

1M

J

Fig. P10.89

feclive len gth of8.5 fL Th e grade of woOO used has an adjust~d a1Joy,.able stress for compres.>ion parallel 10 the grain fTc = 1180 psi an d
=

Fig. Pl0.91

10.92 Solve Prob. 10.91 using the stress in bending of 1300 p~i. 10.93

A

~ te,,"l

imerac[jo~

method and an :Jllowable

compressi()11 member of 2.75 m effective

I ~ngth

supports

an eccentric load as shown. Using th e allowahle-stress medlOd and assuming t = 40 mill. de termine the maximum allowahle load P. U.;e fIr = 250 MPa and E "" 200 GI' a.

Fig. P10.93

10.94

Solve Prob. 10.93. using " - 60 rnm.

657

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658

Columns

10.95 A rectangular column is made of a gr
.:

<

p

=

II;

kip'

" c ~ ;"

JOi~:::; """"'"

71"

Fig. P10.95

10.96

501\'e Prob. 10.95. assuming that P = 24 kips.

10.97 An eccentric load P = 48 kN is ;)pplieJ at a point 20 mill from the geometric axis of a 50-mm-diameter rod mnde of the aluminum alloy fi061-T6. Using the. interaction method and an allowable stress in bending of 145 MPa. determine the largest allowable effective length L th,\I can be used.

21} """ - .

5tJ..,,,m dialnt!t('r

1 J L

B

Fig. P10.97

10.98 Sol ve Prob. 10.97. assuming that the aluminum alloy llsed is 2014-T6 and that the allowable stress in bending is 180 MPa. 10.99

50""" (T y

75 u,lll -

The compression member AB is made of a steel for which

= 250 MPa ;)nd £ = 200 OPa. IL is free at its top A and fi xed at its base B.

Using the allowable-stress methoo_ determille the largest allowable eccentricitye .... knowing that (a) 1',. "" 0, (b) 1' . = 8 mm. The compression member AD is made of ;) steel for which or;). It is free;)t its top A and fixed at its base B. Using the interaction method with an allowable bending stress equal to 120 MPa ;)nd knowing that the eccentricities e.• and T \. are equal. determine their largest allowable common value. 10.100

U y = 250 MPa ;)nd £ = 200

Fig. P10.99 and P10.100

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10. 101 A column of 14-ft effective length cOl\!;ists of a section of steel tub· ing Iwving the cross section shown. Using the allowable·stress method, detemline. the ma.~il11um allowable psi.

Probloms

Fig. P10.101

10.102 Solve Prob. 10.101. assuming Ihat ilte effective length of the. column is increased to 18 ft and that (a) P = 28 kips.lb) P = 18 kips. 10.103 A sawn lumber column of rectangu lar cross section has a 7.2·ft effective len,gth and s upports a 9.2·kip load as shown. The s izes available for use have b equal to 3.5. 5.5. 75. and 9.5 in. The grade of wood has an adjusted allowilble stress for compression parallel to the grain rYe ~ 1180 psi and an adjusted modulus E = 440 X 101 psi. Using the allowable·stress method. detemline the lightest section that can be used.

/,

FIg. P10.103

"

'-I

10.104 Solve Prob. 10.103. assuming that e = 3.2 in. 10.105 The eccentric load P has a magnitude of 85 kN and is applied at a point located at a distance e = 30 mm fmm the geometric axis of a rod made of the aluminum alloy 6Oi6-T6. Using the intemction melhCKl with a 140-MPa allowable stress in bending. determine the smallest diameter d Ihat can be used. 10.106 Solve Prob. 10.105. using the allowable·stress meiltCKl and assuming that Ihe aluminum alloy uset:l is 2014-T6.

"

t

B

FIg. P10.105

5 '"

J

659

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660

10.107 A compression member of rectangular cross section has an effective lengt h of 36 in . :H1d is made of the aluminum alloy 20l4 -T6 for which the al lowab le stress in bending is 24 ksi. Usi ng the interaction method. deTemline the smallest dimension d of the cross section that can be used w hen /' "" 0.4 in.

Columns

2.2.5 in.

Fig. P10.107

10.10B

Fig. P10.109

Sulve "rob. 10.107, assuming that /' = 0.2 in.

10.109 A compression member made of steel has a 720-nun effecti ve length and must suppon the 198-kN load P as shown. For the malerial used (T y = 250 MPa and £ = 200 Gl'a. Using the interaction melhod with an allowable bending stress equal to 150 MPa, determine the smallest dimension d of The cross section Ihat can be used. 10.1 10 Solve Pmb. 10.109. assuming that the effective length is 1.62 m and that the magni tude P of the ec..:enTric IDad is 128 kN. 10.1 11 A stee1tube of 80-mm OUTer diameter is 10 carry a 93-kN load P with an eccentricity o f 20 mm. The tubes available for use are made with wall thicknesse..~ in increments of 3 mm from 6 mm to 15 mm. Using the allowable-stress method. determine the lighlesTlUbe thai can be used. A_'isume £ = 200 GPa and (r r = 250 MPa.

= 20 ,,,,,,--I I

Fig. P10.111

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t! ~

Probloms

10.112 Solve Prob.10.III.usingtheinteractionmethodwithP =:o 165kN. 15 mm, and an allowable slress in bending of 150 MPa,

10.113 A steel column having a 24-ft effective length is loaded eccentrically as shown. Using the allowable-stress method. select thc wideflange shape of 14-in. nomin"l depth that should be used. Use IT y = 36 ksi and E=29XI06p~i.

Fig. P10.11 3

10.114 Solve Prob. 10.11 3 using the intenlction method. assuming that Uy

= 50 ksi and the "llowablc siress in bending is 30 ksi.

10.115 A S1eel compression member of 5.8-m effective length is to suppon a 296-kN eccentric load P. Using thc inleraction method, select the wideflange shape of200-mm nominal depth that should be used_ Use £ = 200 GPu. rTy = 250 MPa. and rToIl = 150 MPa in bending.

"-

12.5

I'Z P

p

e,

Fig. P10.115

c D

10. 116 A steel column of 7.2-m effective length is to suppon an 83-kN eccentric load P at a point D , located on the x axis as s hown. Using the allowablestress method, select thc wide-flange shape of 250·mm nominal depth that should be used. Use E .:= 200 G Pa and if y ':= 250 MPa.

Fig. P10.116

= 70

nun

661

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REVIEW AND SUMMARY FOR CHAPTER 10

Critical load

This chapter was devoted to the design and analysis of columns. i.e., pri smatic members supporting axial loads. In order to gai n insight into the behav ior o f co lumns, we first considered in Sec . 10.2 the equilibrium o f a si mple model and found that for values o f the load P exceeding a certain value PIT' called the critical load. two equi librium positions of the model were possible: the original position with zero transverse defl ections and a second position involving detlections that could be quite large. Thi s led us to conelude that the first equilibrium position was unstable fo r P > Pc'" and stable fo r P < Pcp si nce in the latter case it was the only possible equilibrium position. In Sec. 10.3, we considered a pin-ended column o f length L and o f constant fl exural rigidity EI subjected 10 an axial centric load P. Assuming that the column had buckled (Fi g. 10.8), we noted that the bend ing moment at point Q was equal to - Pv and wro te

p - El Y

( 10.4)

p'

("'

I,

(b)

Fig. 10.8

Euler's form lila u (M PaI

'"

,,

(1 0.11 )

~ Ul

= 250,\ 1 p" £ =200(:1'"

\

2.')0

Solving this d ifferential equation. subject to the boundary conditions corresponding to a pin-ended col umn, we detennined the smallest load P fo r which buckling can take place. This load. known as the criticalloat! and denoted by Pw is given by Euler's formula:

2()()

a"~

/

where L is the length o f the column. For this load or any larger load, the equil ibrium of the column is unstable and transverse de llections will occur.

-~ ( Ur l~

Denoting the cross-sectional area of the column by A and its radius of gyration by r, we determined the critical stress O'er corresponding \0 the critical load Per:

100

S9

J

00

200

10,

( 10. 13)

Fig. 10.9

Slenderness ratio

The quantity Ll r is called the slendeme.u ratio and we plotted

0'", as a function of Llr (Fig. 10.9). Since o ur analysis was based on stresses remaining below the yield strength o f the material, we noted that the column would fai l by yield ing when O' O' y.

662

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Review and Summary for Chapter 10

In Sec. 10.4. we disc ussed the critical load of columns with various end condi tions and wrote 2 7r EJ P
(ID.II ')

where Le is the effectil'e length o f the column. i.e., the length of an equi valent pin-ended column. The effective lengths o f several columns with vari ous end conditions were calcu lated and shown in Fig. 10.1 8 on page 61 7. In Sec. 10. 5, we considered co lumns suppon ing an eccentric axialload. For a pin-ended column subjected to a load P applied with an eccentricity e, we replaced the load by a centric axial load and a couple of moment MA = Pe (Fi gs. 10.19a and 10.20 ) and derived the fo llowing expression for Ihe maxi mum transverse defl ectio n: Yma, =

e[sec ( If~) - I ]

(1 0.28)

We then determined the maximum stress in the column, and fTO m the expression obtained for that stress, we deri ved the secantfor mula: P A

U rn'"

+

e~ sec (~ (P Le) tV £;'

r-

P

Me

::=;:

(u all)",n,,;"

+

I'

p

,..1"

r

A

<

M,

~

I',

L

't./",·

B

( 10.55)

B

p,.

P'

Fig. 10.19a

P'

Fig. 10.20

Design of real columns Centrically loaded columns

Eccentrically loaded columns Allowable-stress method

(10.53)

U all

M c/1 (u all) ""ooing

Eccentric axial load. Secafll formula .

r

The second method was the interaction method. a method used in most modern specifications. In this method the allowable stress for a centricall y loaded column is used for the portion of the total stress due to the ax ial load and the allowable stresS in bendi ng for the stress d ue to bending. Thus, the inequality to be sat isfi ed is -,-Pc: cc- i A

Effective length

( 10.36)

Thi s equation can be solved for the force per unit area, PIA. that causes a specified max imum stress (T ,nax in a pin-ended column or any other column of effective slenderness ratio L.lr. In the fi rst pan of the chapter we considered each column as a straight homogeneous pri sm. Since imperfections ex ist in all real columns. the de.l·ign of real cO/limns is done by using empirical formulas based on laboratory tests and set forth in specifications codes issued by professional organizations. LTl Sec. 10.6. we discussed the design of centrically loaded COIWlIll.f made of steel, aluminum. or wood. For each material, the design o f the column was based on fo rmulas expressing Ihe allowable stress as a function o f the slenderness ratio Llr of the column . For structural steel. we also discussed the alternative method of Load and Resi.rtance Factor Design. In the last sectio n of the chapter [Sec. 10.7 1, we studied two methods used for the design of colu mns under an eccentric load. The first method was the allo\\,able-.I"freD· me/holl. a conservati ve method in which it is assumed thaI the allowable stress is the same as if the col umn were centrically loaded. The allowble-stress method requ ires that the fo llowing inequality be satisfied:

A + - I-

663

Interaction method

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REVIEW PROBLEMS

10.117 The steel rod Be is attached \0 the rigid bar AB and 10 the fixed Sllpport at C. Knowing Ihat G = Il.l x [0" psi. d~tennjne Ihe critkalloml Pc of the sys tem when d = ~ in.

I'

A

ISilJ

l

r

----,,: ~

B

.,- -

L~ "

Fig. Pl0.117

10. 118

TIle rig.!d bar AD is anached to IWO spnng.. of constant k and is

in equilibrium in the position shown. Kno",ing that the equal and opposite londs P Jnd P' remllil! l'enital, detennine the magllilude f " of the criTical IO<.ld for the sy>tcm. Each "pIing IOdll act in cjllt"" ("usiaH VI

fig . P1 0 .118

cOlllptc~sJ on.

10.119 A column of 3-m effective length i; \0 be made by welding together two C130 x LI rolled·steel cl\;lnnels. Using £ = 200 OPJ. determine for e;jeh arran/;cm~n( .; hown the allo ..... able centric load if a fa~tor of safety of 2.4 is requir~d.

I 0", '" ) Fig. Pl0.119

Ag. P10.120

10.120 Member AB consis[~ of a single Cl30 x lOA steel channd Of ll'.rlglh 2, III KnflWing Ih~r The rins A finn U rfl~~ Ihmll~h rht' ~enrTfl,,1 of rhl'. cm~~ secLion of {he channel. determine the fxtOfof ~ety for the load shown with reo Spocl 10 buck.ling in Ire pl>l.ne of Ihe iigure when I) - 300. U ~ t.' - 200 GPn

664

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10.121 A l-ill.-squalt aluminum su utl~ mailllained iuthe. posidon shown by a pin suppon at A a nd by sels of rollers al Band C that pre\ent rotaliun of Ihe Slrul in Ihe plane of Ihe figure . Knowing that L .IB = 3 Ii. L BC ~ 4 ft, and LC/) = I ft. delermine th~ allowable load P uSlIlg a factor 01 safety \.\·jlh respect to buckling of 3.1. Comider only buckling in the plane of the figure and use E = 10.4 X 10" psi

R..... lnw Problem.

8

" Fig. P10.121

Fig. P IO.122

=

10.122 Knowing that P 5.2 kN. detennine the factor of safelY for the structure s hown. Use E = 100 GPu and comider only buckling in the plane of the nrutture 10.123 An aX.i:!lloud P is applied to the 125·in.-Jiameter 'iteel rod AB as shown. For J' - 8.6 kips and t - I~ in., delemline (a) the derl ection at Ihe m ,dpoint C of the rod. (b) the ntll:;:imunt ~tn'!!;,.~ lt1 the rod. Use E = 29 X 10" psi.

'~r-

r L

1.2.5. " . '~''''''~''r

C

B

"

' +

,..

Fig. P10.123

I 0.124 A column is made fTom half or II W36() x 216 roll~d-sleel shape, wilh Ihe' geome'lric propel'ie~ as shown. Using allowable Sh'ess deSi g n. deler_ tnine the allowable centric lood if the effoclive length of the l'O lumn h (u) 4.0 111. Ib) 6.5 m. Usc O" r .. 345 MPII :;nd £ - 20() G i'l!.

.\. ,\ =

13.S X 10 1 Ul"'~

I'

1~ =26\J" W~"",,!

::

1'1 =

"

'------:..-:.:_':..--- ----..! FIg. P 10. 124

1~1.1) X

lU"""h'

665

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666

Columns

10.125 A compression member has the cross section shown and an effective le,ngth of 5 fT. Knowing that the aluminum alloy uired is 2014-T6, tJetennine the alluwable centric load.

Fig. P1O.12S

10.126 A column of 17-11 effective length must carry a centric load or 235 kips. Using allowable stress design. select the wide-flange shape of 10-in. nominal depth that should be used. Use {J"j = 36 ksi and E = 29 X IO~ psi.

10.127 A 32-kN vertical load P is applied itt the midpoint of one edge of the square cross section of the aluminLlm compression member AB that is free at its top A and fixed at its base B.. Knowing that the alloy used is 6061-T6 and using the allowable-stress method. determine the smallest allowable dimension d.

1.2

III

l Fig. 10.127

Fig. P10.128

10.128 A 43-kip axial load P is applied [0 the rolled-steel column Be at a point on the x axis at a distance e "" 2.5 in. from the geometric axis of [he column. Using the allowable-stress method. select the wide-flange shape of 8-in. nominal tJepth Ihat should be used. Use £ = 29 X 10" psi and fT y = 36 ksi.

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COMPUTER PROBLEMS

Th e foIlO\\·ing proble ms are rl u ig ncd to IK" s(lh 'c tl II-jUt

(I

cOlllpul c r.

A solid steel rod having an elfecllve length of 500 mm t~ 10 be ,rnll 10 '~ ~IT)' ~ cenrric lo~rl P. For I h~ gril
11~l'It n ~ ~ ctlm(1rf'_~~ion

is required and using

Euler'~

formula, write;j computer program and me it to

calculate the allowable centric load 1'. 11 for \'a!ucs of th e mdius of lh~ rod from Ii mill to 24 mm. using 2-mm incremems. 10.C2 An aluminum bar l S tixed at end A amJ supported at end B so that i! is free to ro!ate :'00111 .. horizontal axis through the pin. ROI~ljon about a veni_ cal axis at end B is prc\'ented by the brackets. Knowing Ihat E = 10. 1 X 106 psi.

"

1,5 ;n_

use Euler's formul a wilh a faclOr of safelY of 2.5 to determine lhe allow"ble ceJ1llic load I' for

\'alue~

of b flom 0.75 ill.

10

1.5 iu .. using D.IZ5-in.

increments.

I'

Fig. P10.C2

10.C3 The pin-ended members All :.nd BC : onslsl 01 seclions 01 aluminum pipe of 120 mm OUler diameter :md 10 mm wall thickness Knowing Ihat a foctor of , afety of 3.5 is required. determine the nt
Jlbn~

of the SlmCII1rt'_

n

1:l111 -1-':lm-;1

,

1

P

'-

/.

A

Ll Fig. P10.C3

10.C4 An axial load P is applied at a point located on the x axis at a distance e = 0.5 in. fro1llth~ geometric axi~ of the \V8 X 40 rolled-s teel column AB. Using E = 29 x 10" psi , write a computer program and use it \0 caleil iale fm

v a h lt~s

of P frum 2'i In 7'i kips .

u~ine

'i-kip inerenll'rlls. (/I) the P'

horizontal denection al the midpoint C, (b) the maximum stress in lhe column.

Fig. P10.C4

667

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668

Columns

10.CS A column of effective length L is made from a rolled-steel slw pe and carri es a centric a.\ ialload P. The yield strength for Ihe grade of sleel used is denoled by tT ,.. the modulus of elasticity by E. the cross-sectional area of the selected shape by A. and its smallest radius of gyration by r. Usi ng the AISC design formulas for allowable stress design. write a computer progr,ml that ca n be. used wi th either SI or U.S . customary ull iL~ to detennine the allowable load P. Use th is program to solve (a) Prob. 10.59. (b) Prob. IO.6{I. (el Prob. 10.124. 10.C& A column o f effective Icngtb L is made from a rolled-steel shape and is loaded eccentrica ll y as shown. The yield strength of the g f
Fig. P10.C6

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C

HAP

T

E

Energy Methods

;

As the diver comes down on the diving board the potential energy due to his elevation above the board will be converted into strain energy due to the bending of the board. The normal and shearing stresses resulting from energy loadings will be determined in this chapler.

R

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670

11 .1. INTRODUCTION

Energy M!!Ihods

Ln the previous chapter we were concerned with the relations existing between forces and deformations under variou:; loading cond itions . Our analysis was based on two fundamental concepts, the eonct:pt of stress (Chap. [) and the concept of Slrain (Chap. 2). A third important concept. the concept o /" slrain i'/!el"MY. will now be introduced. In Sec. 11.2. the slmin energy of a member will be. defined a:; the increase in energy associated with the deformation of the member. You will see that the strain energy is equal to the work done by a slowly increasing load applied to the member. The slrain.ellergy demit)' of a material will be defined as the strain energy per unit volume: it will be seen that it is equal to the area under the stres~ -s train diagflllll of the material (Sec. J 1.3). From the stress-strain diagram of a material two additional properties will be defined, namely. the modulus of IOlIghnc.I's and the I/Jodldu.~ of resiliencf' of the material. In Sec. IIA the elastic strain energy associated with normal stresses will be discussed. first in members under axial loading lind then in members in bending. Later you will consider the elastic strain energy associated with shearing stresses such a" occur in torsional loadings of shafts and in transverse loadings of beams (Sec. 11.5). Strain energy for a gr!llerl11 stare of stress wi.ll be considered in Sec. 11.6, where the maximllm.distonio/l ,cllcrgy criterioll for yielding will be derived. The. effect of impacl loadillg on members will be considered in Sec. 11 .7. You willieam to calculate both the maximum stress and the mllXi· II/lilll d,pI'f'lirm cllIIsefi hy II moving mass impacting Oil a memhe r. Properties that increase the ability of a structure to withstand impact load~ elTeetivcly will be disc ussed in Sec. IIX 1n Sec. 11.9 the clastic strai n of it member subjected to a single COII celltmled load will be calculated , and in Sec. 11 . 10 the deflection at the point of application of a si ngle load will be detennined. The last portion of the chapter will be devoted to the detennination of the strain energy of structures subjected to several loads (Sec. 11.11). Cas· tiglial/o's theorem will be derived in Sec. II. J 2 and used in Sec. J 1.1 3 to detcnnine the dellcction at a give ll point of a structure subjected to several loads. In the last section Castigliano's theorem will be applied to the analysis of indetemlinate structures (Sec. J 1.14).

L~ 'I

'1 .2. STRAIN ENERGY

p

c Fig. 11 .1

Consider a rod Be orlcngth L and uniform cross-sectional area A, which is attached at B 10 a fi xed s upport . and s ubjected at C to a slowly increasing axial load r (Fig. 11.1 ). As \\Ie noted in Sec. 2.2, by plo tting the magnitude P of the load against the defom mtion x of the rod. we obtain a certai n load-deformation diagram (F ig. 11.2) thaI is characteristic of the rod Be.

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I'

11.2. StraIn Energy

FIg. 11.2

Lei us now consider the work dU done by the load P as the rod elongates by a small amount dx. This elemcllflJry work is equal to the product of the magnitude P o f the load and of the small elongation dx. We write JU "" Pdx

p

1

(ll.l)

and note that tht expression obtained is equal to the clement of area of wid th dx located under the load-deformation diagram (Fig. [1 .3). The fotaf work U done by the load as Ute rod undergoes a deformation

is thus U ""

l' = AN'''

XI

I""

P dr

Fig. 11.3

and is c(llial to the area under the load-de formation diugram between x - 0 and x -

Xl"

The work done by the load P as it is slow ly applied to the rod must result in the incrc:L~c of some energy associuled with the deformation of the rou. This energy is referred to as the slrain t:'nergy o f Ihe rod. We have, by definition, Strain ent! rgy = U =

l"

f' dx

( 11.2)

"

We recall that work and energy shou ld be expressed in units obtained by multiplying units o f length by units of force. Thus. if S I metri c units are used, work and energy arc expressed in N . In : thi s unit is called ajol/le (1). If U.S. customary units arc used. work and energy tire expressed in ft . Ib or in in . lb . In the case of a linear and e lastic deformation. the portio n of the load-deformation di agram involved can be represented by a straight line of equation P = kx (Fig. I 1.4). Substituting fo r P in Eq. (I 1.2). we have p

l',,<, h

U= fhdX=1kxf

,

1' [

( 11.3)

wher\! PI is tht! value of the load correspond ing

to

the deformation

Xl.

------ -

o ~----;"C-,- --; Fig. 11.4

671

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672

The concept of strain energy is particularl y uscful in the determi nation of the effects of impact l oad ing~ on structu res or machine components. Consider, fur example, a body ()f mass III movi ng with a velocity Vo which stri kes the end B of a rod AB (Fig. 11.5(/). Neglecting the inertia of the elements ()f the rod, and assuming no dissipat ion of energy duri ng the impact. we tind that the maximum ,~tmin energy Urn acq uired by the rod (Fig. 11.5b) is equal to the ori ginal kinet ic energy T = ~ IIIV& of the moving body. We thell determine the va lue P", of the static load which would have produced the same strain energy in the rod, and obtain the val ue U Itt of the largest stress occllrring in the rod by dividing Pm by the cross-sec tional area ()f the rod.

Energy Methods

\=0

Fig. 11 .5

T=O

11 .3. STRAIN-ENERGY DENSITY As we noted in Sec. 2.2, the load-defonnation diagram for a rod Be depends upon the length L and the cross-sectional area A of the rod. The strain energy U defi ned by Eq. (1 1.2). therefore, will al so depend upon the dimensions of the rod. Ln orde r to eliminate the effect of size from our di scussion and direct our attention to the properties of the material, the strain energy per uni t volume will be considered. Di viding the strain energy U by the v()Jume V = AL o r the rod (Fig. 11 . 1). and using Eq . ( 11.2), we have

u V

f

<' !!...dx

o A L

Recalling that PIA represents the normal stress xlL the normal strai n € ,. we wri te

u I.",

- = V

U"

"

(f "

in the rod. and

dE

where E I de notes the value of the strain c()rresponding to the e longati() n XI . The strain energy per un it volume, Ul V. is re ferred to as the strain -energy density and wi ll be denoted by the letter II. We have. theref()re.

Srraill-encrg), denSity

~ II

=

" f"

u .• d£\

( 11 .4)

The strain-energy density 11 is expressed in units obwined by di vidi ng units of e nergy by units ()f volume. T hus, if SI metric units arc used. the strain-energy density is expressed in JIm] or its multiples kJ /m j and MJ/ml; if U.S. c ustomary units arc used , it is expressed in in . Jb/in 3.t t Wc not~ lhal t JIm' and ) Pa arc bolh equa) l~ t Nlm ' . while I in . tblin ' am] ! psi arc bolh cquatlO ] tblin ' . Thu s. .\!rain-energy density and :Slress arc J imeosionally equat :Iml cou ld be expre ssed in Ihe !
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11.3. Strain· Energy Oenslty

Rcfe ni ng to Fig. 11 .6. we note that the strain-energy density II is equal to the area under the stress-strain curve. mt:as ured from E~ = 0 to E~ = E, . Lf the material is unloaded, the stress returns to zero, but there is a pemlanent defo mlation represented by the strain Ep ' and onl y the portion o f the strai n energy per unit volume corresponding to the triangular area i, recovered. The re mainder of the energy spent in deformin g the material is dissipated in the form o f heat.

673

l. t0d"h,!; ,\f t o " )(!III(,SS

Rllp tll rl'

0 '-----7,,", ----O,L , -----------;

" ~-------:: " C------'

Fig. 11 .6

Fig. 11.7

The value of the strain-energy density obtained by .~e tL i ng 1", = fE ll in Eq. (11.4). where E N is the strain at rupture, is known a.~ the mot/r/Ius of lOulj llll c,u of the material. It is equal 10 the area under the entire stress-strain di ag ram (Fig. 11 .7) and represents the e nergy per unit volume required to cause the material to rupture . LL is clear that the toughness of a mate ri al is related 10 its ductilit y as well as to its ultimate stre ngth (Sec. 2 .3). and that the capacity of a structure to withstand an impact load depends upon the toughness of the mate ri;)l used (Fig . 11. 8). if the stress a A remains within the proporti onal limit of the mate· rial, Hooke's law applies and we write ( 11 .5)

Substituting for a J from ( 11.5) into 01.4). wc ha ve

(1 1.6)

or, using Eq. ( 11.5) to express 1" 1 in terms o f the corresponding stress

,

~,

1/ = -

2£

( 11 .7)

The value Uy o f the strain-energy density obtained by setting == a y in Eq. (1 1.7), w here a y is the yield stre ngth. is called the lIIodulus of rej'iliem:1' of the material. We have

a,

( II .8)

Fig. 11.8 The railroad coupler is made of a due· lile steel which has a large modulus 0/ toug hness.

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674

Energy Methods

The modulus of resilience is equal to the area under the straight-line portion OY of the stress-strain diagram (Fig. 11.9) and represents the energy per unit volume that the material can absorb without yielding. The capacity of a structure to withstand an impact load without being permanently defonned clearly depends upon the resilience of the material used . Since the modulus of toughness and the modulus of res ilience represent chamcteristic values of the strain-energy density of the material considered, they are both expressed in J/m 3 or its multiples if SI units are used, and in in . Ib/in 3 if U.S. clistomary units an: used. t 11 .4. ELASTIC STRAIN ENERGY FOR NORMAL STRESSES

Si nce the rod considered in the preceding section was subjected !O uniformly di stribmed stresses (J"~, the strai n-energy density was constant throughom the rod and could be defined as the ratio UIV of the strain e nergy U and the volume V of the rod In a structural clement or machine part with a nonuniform stress distribution, the strain-energy density If can be defi ned by considering the strain energy of a small element of material of volume !:J. V and writing ~u

1/ =

lim-t.y-->o!:J.V

dU u =dV

( 11.9)

The expression obtained fo r If in Sec. 11.3 in terms of mains valid, i.e .. we still have 11 =

f"u,d€~

U

J

and

E".\

re-

(11.1 0)

"

bm the stress u " the strain €." ami the strain-energy density u will generally vary from point to point. Fur values of u .• within the proponionallimit, we may set cr., = E€.t in Eq. (11.10) and write 11 = "!" Et:2 = "!"cr t: 2. ' 2. x J

I U->7 2. E

(11.11 )

The yftlue of the straill energy U of a body subjected to uniaxial normal stresses can be obtai ned by subslitLJting for II from Eq . (11. 11 ) intO Eq. ( 11.9) and integrating both members. We have (11.12)

The expression obtai ned is valid only for elastic deformations and is referred to as the elastic straill energy of the body. tl-Iowcvcr. referring 10 thc footnOte ou pagl' 672. Wit flotlt thm thc modu lus of toughucss ;md the modulus of res ilience C{)uld be expressed in the s ame units a.s stress.

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tt.4. Elastic Strain Energy lor Nonnal Stresses

S train Energy under Axial Loading. We recall Jrom Sec. 2.1 7 that. when a rod is subjected to a centlic axial loading, the norillal stresses u , can be assumed unifurmly distributed in any given transverse sectio n. Denoting by A the area of the section located at a distance x from the end B of the rod (Fig. 11.10), and by P the il11emal force in that section. we write U x == PIA . Substituting fur U x into Eq. (11.12), we have

675

or. seILing dV = A dx,

U~

' f

P'

- - dx o 2AE

(11.13)

In the case of a rod o f unifonn cross section subjected at it~ c nds to equal and opposite forces of magnitude P (Fig. II.I I). Eq. ( 11.1 3) yields

Fig. 11 .10

I"

( 11.14)

" Fig. 11.1 1

EXAMPLE 11.01 A rod consists of two ponions Be and CD of Ihe same material and same length. but of different cross se(,lions (Fig. 11.12 ). Determine the strain energy of the rod when itjs sub-jected 10 a centric axial load P, expressing the result in terms of P. L. E. the cross-section;)l area A of pDl1i'o n CD, and the mTio II of the two diameters.

1) +II~

(]1.I5)

We check lhat. for

p

Fig. 11.12

"

We use Eq. (11.14) 10 compule the strain energy o f each of The two port iol1s. and add the expressions obtained:

II

I . we Iwve

which is lhe expression give n in &1. ( 11.14) for a rod of length L and uniform cross section of area A. We also note that. for II > I. we have V. < VI: for example. when II == 2 . we have VI = (i)U I . Since the maximum slress occurs in portion CD of the rod and is equal 10 U,n.o' = PIA. it follows that. for a given allowable STress. increasing Ihe diamete.r of ponion BC of the rod results in a decrease of the overall energy-absorbing capaciTY of the rod. Unnecessary changes in cross-secTional area sho uld therefore be avoided in the desig n of members dwt may be subjected 10 loadings. suc h as impact loadings. where the energy-absorbing c
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EXAMPLE 11 .02 A lo:!d P is supponed at B by two rods of the same m
HllI we note from Fig. I l.lJ that Be = 0.61

BD "" 0.81

and from the free.OCx]y diagram of pin B and the

r

torrespon~

c F~f'

.~.

H

I

,

1

5

,

I'

Fig. 11.14

/)

Fig. 11.13

ding force tri:!ngle (Fig. 11.14) that

Denoting by FDC and FBD' respectively. the forces in mem ~ bers Be;md BD, and recalling Eq. ( 11.14). we express Ihe strain energy of the system as F1d BC)

Fi.r,(BD)

2<\£

ZAE

u~ --- + ---

FDe = + 0.61'

FIJI) = - 0.81'

Substituting into Eq. (11.16). we have

( 11.16)

U =

p2/[(0.6 )J + (O.WJ lAE

1"1 AE

0.364 -

Strain Energy in Bend ing. Consider a beam AB subjected to
mm : lmm Fig. 11.15

U=

f

~, --"5...dV = 2£

f M' y'

--dV 2£/ 2

Settmg dV = dA dx, whe re dA represents an eleme nl of the c rossst."Ctional area, and recalling thai M 1 /2£( is a fUI1l;tion of x alone, we have

Recalling thai the integral within the parentheses represents the moment of inertia J of the cross section about its neutral axis. we wri te

U=

676

' M'

I

-

o 2EJ

dx

(11.17)

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EXAMPLE 11 .03 Determine the strain energy of the prismatic cantilever beam AB (Fig. 11.16). taking into account only the effect of the nor-

mal stresses.

1 - - - /' - --1.

The bending moment at a distance x from end A is - Px. Substituting this expression into &J. ( 11.1 1), wewtite

M

~

Fig. 11.16

11 .5. ELASTIC STRAIN ENERGY FOR SHEARING STRESSES When a material is subjected 10 plane shearing stresses energy densi ty al a given POI III can be expressed as

7 .,,",

the strain-

1.1 ( 11.18)

where 'YX\" is the shearing strain correspondi ng to 7n' (Fig. 11.17u). We note that 'lhe strain-energy densi ty

/I

is equal 10 the area unde r the shear-

ing-stress-strain diagram (Fig. 11.17h). For values of 7X\" within tlte proportional limit. we. havc 7n' = G'Y~.• where G is the modulus of ri gidity of tlte material. Substituting for 7 n into Eq. (11.18) and performing thc inlegrmion . we write

o '---_ _ _L.-----::(b)

'"

Fig. 11 .17

2G

(ll.lg)

llle value of the strain energy U of 11 body subjected to plane shearing stresses can be obtained by recalling from Sec. 11.4 that

dV

( 11.9)

1/= -

dV

Substituting for II rrom Eq . (1 1. 19) into Eq. (11.9) and integrating both members. we. have

U=

I

,.;.

-

2G

dV

(11.20)

Thi s ex pre ssion defines Ihe clastic strain associated with the shenr de rormations of tlte body. Like the simil ar expression obtained in Scc . 11.4 ror uniaxial normal stresses, it is valid onl y ror elastic de rormations.

677

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678

S train Enerl-:Y in TOnlioll . Consider a shaft Be of length L subjected to one or several twisting couples. Denoting by J the polar moment of inertia of the cross section located at a J istance x from B (Fig. 11. 1S), and by T the internal lOrque in that section, we recall that the sheari ng stresses in the section are T~ \ Tp/J. Substituting for 7 " into Eq. ( 11.20), we have .

Energy Methods

U =

T'.

--.!2..t/V =

f 2G

J. ~-dV T' p' 2GJ

1

Setting t/V = dA {h , where (iA rcprc ~e n L~ an element of the crosssectional area, and ob~ervin g that r2/ 2GJ 1 is a function 01 x alone. we write Fig. 11.18

Recalling that the integral with in the parentheses represents the polar moment of inert ia J of the cross sect ion, we have (11.21 )

T

In the case of a shaft of uniform cross section subjected al its ends equal and opposite couples of magnitude T (Fi g. 11.19), Eq. (11.2 1) yields (0

(11.22) Fig. 11 .19

EXAMPLE 11 .0 4 A circular shaft consists of two portions Be and CD of the same material and same length. but of ditrerent cross sections (Fig. 11.20). Determine the strain energy of the shaft when il is subjected 10 a twisting couple T at end D. expressing lhe resul1 in terms of T, L. C. the polar momcrn of inenia J of lhe smaller cross section, and the rmio /I of lhe IWO diameters.

We check that. for II

=:

I. we have

u,

Fig. 11 .20

We llse Eq. ( 1 1.22) to compllte the strain energy of each of the two portions of shaft. and add the expressions obtained. Noting that the polar moment ofincrtia of portion BC is equal to 1/4.1. we write

which is the expression given in Eq. ( 11.22) for a shan of length L and uniform cross section. We also note. that. for /I > I. we have U. < VI: for example. when Ii = 2. we have U! =: (R) VI' Since the maximum shearing stress occurs in the portion CD of the , h3ft and is proportional to the torque T. we note as we did earlier in the case of the axial loading of a rod thaI. for a given allowable stress. increasing the diameter of portion Be of the shaft results in a decrell.l"/! of the overall energy-absorbing capacity of the shaft.

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Strain Energy under Transverse Loading. In Sec. 11.4 we obtained an expression for the strain energy of a beam subjet:ted to a transverse loading. However, in deriving that expression we took into account only the effect of the normal stresses due I t I bending and neglected the effe<:t of the shearing stresses. In Example 11.05 both Iypes of stresses will be laken into accou nt.

11.5. Elastic Strain Energy lor Shearing Slresses

679

EXAMPLE 11 .05 Determine the strain energy of the rectangular cant ilever beam AB (Fig. 1l.21). taking into account the effect of both normal

and shearing

s tres.~es.

We first recall from Example 11.03 that the strain energy due to the nonnal s tresses ('~ is [ilL'

U "

~­

6£1

To determine the main energy U, due to the shearing stresses T n. we recall Eq. (6.9) of Sec. 6.4 and find that. for a beam with a rectangulal cross section of width b and depth h. ~L

1. = ~~(I -~)=l~(I -~) lA

Substituting for

T :j

,-

" ~

c::-

2bll

into Eq. ( 11.20), We write

1(l )'f( /)' 1-,

V , = - - J> 2G2bh

c-

I.

"

FIg. 11.21

dV

or, sel1 ing dV = b dy dt. and afler reducTioM.

I' (

>' ,.')

U - 1 - 2 , + -4 dy OP' , - 8Gbh~ _ ~ cc

I'

IIx

Il

Perfomling the il1tegmtions. and recnl1ing lhal have

l:

= !l/2. we

The total STrain energy of the beam is thus

or, noting that I/ A = 1?/12 and fnctoring The expressIon for V".

p'L' ( 1+ lEh' lEI,' -) =U ( 1 + -)

U~ --

6EI

lOGe

"

\o(jL~

(11.24)

Recalling frolll Sec. 2.14 that G ~ £/3. we conclude that the parenthe~is in the expression obttlined is less than I + 0.9(h/L)~ and. thus, that the relative error is less than 0.9(III L)' when tIle effect of shear is neglected. For a beam with a ratio h/L less than the percentage error is less than 0.9%. II is therefore customnry in engineering practice to neglcct the effect of shear in computing the stmin encrgy of slender beams.

to.

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680

Energy Methods

11 .6. STRAIN ENERGY FOR A GENERAL STATE OF STRESS In the preceding sections, we detennined the strain energy of a body in a state of uniaxial stress (Sec. 11.4) and in a Slate of plane shearing stress (Sec. 11.5). In the case uf a body in a general state of stress chara(;terizcd by the six stress components (T. , fT ,. a " T.lI • T '0' and T :.t . the strain-energy density can be obtained by adding ule expressions given in Eqs. tIl. 10) and ( II.18), as well as the four other expressions obtained through a pcnnutation of the subscripts. In the case ()f the elastic defonnatiCln of an isotropic body. each of the six stress-strain relations involved is linear. and the strain-energy density c:m be expressed as (11.25)

Recalling the relations (2.38) obtained in Sec. 2.14. and substi tuting for the strain componellls inlO (1 1.25). we have. for the mosl general state of stress at a given poi 111 of an clastic isotropic body, I

II

..,

= 2E[fT;

,

..,

+ 0"; + q; -

2"(0",0" ,

+ upo + I

I

0":0".. )] ~

+ 2(}Tt). + r;:, +

,

~,)

(11.20)

If the principa l axes at the given point are used as coordinate axes, the ~ht:ari ug s(res~es

bet:oule zero aud Eq . ( t t .2ei) reduces to

(11.27) where (T a> U b, and O"c arc the principal stresses at the given point. We now recall from Sec. 7.7 that one of the criteria used 10 predict whether a given state of stress will cause a ductile material to yield, namely, the maximum-distort ion-energy criterion, is based on the determination of the energy per unit volume associated with the dislOrtion, or change in shape. of that materi al. Let us. therefore. attempt 10 separate the strain-energy density II at a given point into two parts, a part lI u associated with a change in volume of the material at that point, and a part lid associated with a di stortion, or change in shape, o f the material at the same point. We write (11.28) In order to determine U u and Ud, we introduce the average value o f the principal stresses at the point considered,

3

(j

(11.29)

and sct ( 11.30)

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11.6. Strain Energy lor a General

State 01 Stress

(,I

Fig. 11 .22

Thus. the given state of stress (Fig. I 1.22a) can be obtained by superposing the states of stress shown in Fi g. I 1.22h and c. We nOle that the state or stress described in Fig. 11.22h tends to change the volume of the element of material. but not its shape, since all the faces of the dement are subjected to the same stress ri. 011 the other halld. it follows from Eqs. (11.29) and (.l 1.30) that ( lUI )

which indicates that some of the stresses shown in Fig. I 1.22c are tensile and others compressive. Thus, this state of stress tends 10 change the shape o f the e lemen1. Howeve r, it does nOl tend to change ilS volume. Indeed, recalling Eq. (2.31) of Sec. 2. I 3, we note that the dilatation e (i.e. , the challge in volume per unit volume) caused by this state of stress is

,

I - 2v

~ -E-( O"~

+

O"~

+ 0";)

or e = 0, in view of Eq. (1 1.31). We conclude from these observations that the portion /I v of the strain-energy density must be associated with the state of stress shown in Fig. 11.22b, while the portion lid must be associated with the state of stress shown in Fig. I 1.22c. It follows that the portion /Iv of the strain-energy density corresponding 10 a change in vol ume of the element can be obtained by substituting (i for each of the principal stresses in Eq. (11.27). We have

or, recalling Eq. (11.29),

(11.32)

681

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682

Energy M!!Ihods

The portion of the strain-energy de nsity correspond ing to the distortion of the clement is ob\'ained by solving Eq . (11.28) fo r lid and substituting tor II and II " from Eqs. ( 11.27) and ( 11.32), respectively. We write IIJ

= II -

11"

=

I [3( O"~' + 6£

,

O"i,

+

(r~')

-

6V(U aO"h

+

(T bO""

+ u "O" a)

- ( I - 211)(0"~ + (r ~ +

uen

Expanding the square and rearranging terms, we have

Noting that each of the parentheses inside the bracket is a pc.rfect square. and recall ing from Eq. (2.43) of Sec. 2.1 5 that the coefficient in fro nt of the brac ket is equal to l/ 12G. we obtain the following expression for the portion IIJ o f the strain-energy dem~ity. i.e., for the distortion energy per unit volume, ( 11.33) In the elise of p/alle s lrel·S. and assumi ng that the c axis is perpendic u-

lar to the plane of stress, we have

O"c

=

0 and Eq. (1 1. 33) reduces to (11 .34)

Consideri ng the particular case of a tensile-test specimen, we note that, at yield. we have U = O"y, O"b = 0, and thus (lldh = 0";./ 6G. The maximum-distort ion-energy cri terio n for plane stress indicates that a given state o f stress is safe as long as lid < ( lId) Y or, substit uting for lid from Eq. (11.34), as long as (J

(7 .26)

which is the condit ion stated in Sec . 7.7 and represented graphicall y by the elli pse of Fig . 7.41. In the case of a general state of stress. the ex pression (11.33) obtained for lid shou ld be used. The maximumdistortion-energy cri terion is then expressed by the cond ition. (11.35)

which indicates that a given state of stress is safe if the point of coordinates O"a' O"b, O"c is located within the surface defined by the equation (11 .36) Thi s surface is a circular cylinder of radius V2l3 O"y with an axis of symmetry forming equal angles with the three principal axes o f stress.

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i -i ll. diaIllL'!cr

SAMPLE PROBLEM 11 .1 During a routine manufacturing operalion, rod AB must acquire an e lastic SLmin

energy of 120 in ' lb. Using £ = 29 X 100 psi, determine the required yield strength of the steel if the factor of s afety with respect to pemlanenl defannation is to be five.

SOLUTION Factor of SaC"'y. Since a fnclor of safety of five is required. the rod should be designed for a strain energy of U = 5(1201n - lb ) = 600 in · lb

TIle volume of the rod is

Strain-Energy D('nsily.

"

V = AL = 4(0.75 m.)'(60 in.) = 26.5 in]

Since the rod is o f uniform cross section. the required sirain-energy density is U

II

=

V=

6OCtin · Jb 2 6.5

inl

22.6 in . [b/in]

Yield Strength. We recalilhal the modulus of resilience is equal to the strain-energy density when the maximum stress is equal to IT r . Using Eq. ( 11 .8), we write

u,

: : .-;-

,

~l ()d"l", of

r~jllirl~L1:

a}

1/ = -

JE fTy

= 36.2 k~j ...

C V/IIllltl/t. It is imponant to note thaI. since energy loads are not linearly related 10 the stresses they proouce. factors of safet y associated with energy loads should be applied to the energy loads and nOi to the stresses.

683

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SAMPLE PROBLEM 11 .2

~Lb_f

(a) Taking into :!ccount only the effect of normal stresses due to bending. determine the simin energy of the prismatic beam AiJ for the loading shown. (b) Evaluate the strain energy. knowing Ihal the beam is a WID X 45.1' = 40 kips. L = 12 ft. a = 3 ft. b = 9 fl. and E = 29 X 106 psi.

SOLUTION

.,

,

B

Bendi ng l\1olUl'n t. determine the reactions

I--"-!-- b- I H

.\

I

I

= If.

I.

I

It ~!'!. U I l.

"'~'

~i

Using the free-body diagram of the entire beam. we Ph

R = fa

R, ~-t L

,

t

For portion AD of the beam. the bending moment is Ph

M , = -L x

For portion DB. the bending moment

:II

M2 =

FnJlJlAtoD-

L

a distance 'v from end B is

p"

LV

n . S tra in E ne r gy. Since strain energy ;~ a scalar quantit y. WI' add Ihe strain energy of ponion AD (0 Ihat of portion DB 10 obtain the total strain energy of the beam. Usi ng Eq. ( [ 1.17). we write

U=UAD + UD8

FmlllBtoD

or. since (a

+ b) =

pJa'"b'"

U ~--

L.

6£IL

•

b. EVlIlunlio n of Ihe Strnill Energy. The moment of inertia of a WIO X 4S rolled-steel shape is obtained from Appendix C and the given data is restated using units of kips and inches.

P = 40 kips a =3 fl =36 in. £ = 29 X [O~ psi = 29 X 10J ksi

L = 12 f1 = 144 in. b=9ft= [08 in. I = 248 in 4

Substituting into the expression for U. we have

-c~ ( 4~O~k~;I~ "~ )'~ (3~6~;'~L~ )'~ ( I~O~ 8 ~;'~ '.)~'-c

U= --;.

6(29 x [oj ksi)(248 in-l)( 144 in .)

684

U = 3,89 in .

kip~


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PROBLEMS

11 .1

Determine the modulus of resilience for c;)ch of the following

aluminum alloys: (a) 11OO- HI4: tb) 2014-T6 (I.") 6061 -T6

£ = 70GPa £ = 72 GPa:

V I' =

E = 69 GPa.:

( r),

55 MPa 220 MPa = 150 MPn

If y=

11.2 Determine the modulus of resilienL'"e grades of s\rl.lclUrai sleel: Ir r ~ (a ) ASTM A709 Gradt' 50: (b) ASTM A91 3 Grade 65: if) = (c) ASTM A709 Grade 100: 0-1"= 11.3

for each of the fo llowing 50 ksi

65 ksi looksi

Determine the mooulus of resiiiem.'"e for each of the follow ing

alloys: (a) Titanium: tb) Magnesium: (c )

11 .4 metals :

CupronickeJ (annealed)

£ = 165 X IO~ psi (T r= ]20ksi E = 6.5 X lOb psi IT r = 29 ksi £ = 20 X IOb p~i {rr = lfi ksi

Delf'rm ine lhe moon!lI <;- n f re!':ilience for each of 1he fnllnw in g

Stainless steel AISI 302 (annealed): (b ) Stainless steel2014-T6 A ISI 302 (cold-rolled): (c ) Malleable cast iron: (a)

E = 190 GPa IT r = 260 MPa

E = I90GPa E = 165 GPa

if r

= 520MPa

if r

= 230 MPa

11 .5 The stress-strain diagram shown h
."

1l~1

·tsn

300

n 'l-~Oc!O",Cl-----l")~'-~ll"'"'-0.002

150

Fig. P1 1.S

11.6 The stress-strain diagram shown has been drawn from data obtaincd during a tensile test of an aluminum alloy. Using E = 72 G Pa, dctennine (a) the modulus of resilience of the alloy. (b) the mooulus of toughness of lhe alloy.

O,l ·t 0.1$ 0.006 Fig. P11 .6

685

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686

Energy Methods

11 .7 TIte load-deformation dillgram shown has been drawn from data obtained during a tensile te.<;t of structural steel. Knowing that the crosssectional area of the specimen is 250 mm 2 and Illat the deformation was measured using a 500-mm gage length. determine (al the modulus of resilience uf the steel. (b) the modulus of toughne ~s of the steel.

r ikN ) 100

.1

5OUm ,,,

:1'

50

u,---;cS."(;--------"7~'--C"" :!:c,-

fj (m ,n )

0.6

Fig. P11 .7

11.8 The load-deformation di3)!.r.::tm shown has been dr.::twn from data obtained during a tensile test of a 0.875-in.-diameter rod of an aluminum alloy. Knowing that the deformation was measured using a IS-in. gllge length. detemline (lI) the modulus of resilience of the all oy. (b) the modulus o f toug hness of the alloy.

P (kit" )

I'

,w

I 15 in

30

-.1, I

2.0

I"

IU

- ~ in . 1.&'5

fj ( ;IL.)

(1[(~1

B

Fig. P11.B

A I'

Ag. P11.9

11 .9 Using E = 29 >< IO~ psi. determine (a) the strain energy of the steel rod ABC when P = 8 kips. (b) the corresponding strain energy densiLY in portions AB and Be of the rod.

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11 .10 Using £ = 200 GPa. determine ((I) the strain energy of the steel rod ABC when P = 25 kN. (b) the corresponding strain.energy density in poniolls AB and Be of the rod.

Problems

11 . 11 A 30-in. length of illuminum pipe of eross-sectionill area .85 in~ is welded to ;l fixed ~uppon A and to a rigid cap B . The steel rod f P. of 0.75-in. diameter, is welded to cap B. Knowing that the modulus of elasticity is 29 X IO~ psi for the steel and 10.6 x lOb psi for the 3luminum, determine ((I) the total strain energy of the system when P = 10 kips, (b) the corre~ponding strain-energy density of the pipe CD and in the rod EP.

/

Fig. P11 .11

'11. 12 Rod AB is made of a stet'.1 for which the yield strength is /'- y = 450 MPa and E = 200 GPa: roo BC is made of an aluminum alloy for which U y = 28() MPa ,llld E = 73 GPa. Determine the maximum strain energ y that can be acquired by the composite rod ABC withou t causing any permanent deformations.

p

)

14-,,,,,, diamd .. r

10-""" ,jiarnf'lt'r Fi g. P11.12

11 .13 Rods A lJ and BC are made of a steel for which the yield strength is fl y = 300 MPa and the modulus of elasticity is E = 200 GPa. Determine the maximum strain energy that can be acquired by !lIe assembl y without causing pennanent deformation when the length a of rod AB is (a) 2 m. (b) 4 m.

i II

Fig. P11.13

l2-mm dhll" .. !('r B

I

S-""" dim",,!('r

20-m", ,Iiu,"rtcr H i

1fi-m",di,u"rtrr C

Fig. P11.10

687

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688

Energy Methods

11. 14 Rod Be is made of a steel for which the yield strength is ifr = 300 MPa and the modulus of elasticity is £ = 100 GPa. Knowing that a strain energy of [0 J must be acquired b} the rod when the axial load P is applied . determine the diameter of the rod for which the factor of safelY with respect to permanent deformation is six.

I .S

p

Il ------l

Fig. P11.14

, 1.1 5 Using £ = 10.6 X 10~ psi, determine by approximate means the maximum strain energy that can be acql.llred hy the aluminum rod shown if the allowable normal s tress is (f. 1l = 12 ksi.

p 3 in .

A

l-ll-'r-cJ~ I B

L "@ 1.5i".=6in.-! Fig. P11 .1S

11 .16

Show by integration that the s train energy of the tapered rod

AB is I P~L

U = ---

4 EA"in where Am,.

i~

the cross-sectional area at end B.

Fig. P11.16

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11 .17 through 11 .20 in the truss shown, all members are made of the same material lind halle lhe uniform cross·sectional area indicated. Determine the strain energy of the truss when lhe load l' is applied.

Problems

689

1 '-1 ,> I

0)

~B--'='---{('

r

B

B

cA------,

p

f'-]

Fig. P11.17

Fig. P11.1 8

Fig. P11.19

Fig. P11 .20

11 .21 In the truss shown, all members lire made of aluminum and halle lhe uniform cross·sectional area shown. Using £ = 72 GPlI. determine the strain energy of the trus, for the loading shown. 11.22

120lN

Sollie Prob. 11.21. assuming that the 120·kN load is removed.

11.23 through 11 .26 Taking into accounl only the effecl of nomlal stresses. delemline the stmin energy of the prismatic beam AB for the loading shown.

Fig. P11.23

Fig. P11 .24

i"{'

J"1 r

B

"

Fig. P11.25

Fig. P11 .21 and P11 .22

'"

Fig. P11.26

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690

Energy

Methods

11.27 Assuming th;:ll the prismatic beam AB has a rectangular cross section. show that for the given loading the maximum vulue of the strain.energy density in the beam is

45 U

1" ""' ' ' 8\1 where U is the strain energy of the beam and V is its volume.

III I II I II

j lllll l lllil L (

[

I

Fig. P11.28

Fig. P11. 27

11.28 Assuming that the prismatic beam AB has a rectangular cross section. show that for the given loading the maximum vulue of the strain.energy density in the beam is

where U

i~

the strain erH'rgy of the beam and V is iUi volume

11 .29 and 11 .31 Using £ = 200 OPa. determine the strain energy due to bending for the steel beam and loading shown.

D'

"'' I

~" 1- 6fi - ).:3".J Fig. P11.29

S/iX I.sA

Fig. P1 1.30

11 .30 and 1 f .32 Using £ = 29 X I(f> psi. determine the strain eoergy due to bending for the steel beam and loading shown .

2 liJl~

.2 kip<

H

c

1\

1)

tt-m ", --!!!J

15 in.

FIg. P11.31

1.5 in

Fig. P11.32

15 in.

H --, 3 in.

--.l

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Probtoms

11.33 The ship ilt A has just started to drill for oil on the ocean noor at n depth of 5000 fl. The steel drill pipe h;!s an outer diameter of 8 in. ilnd a unifortll wall thickness of 0.5 in. Knowing thill th~ top of the drill pipe rotates through two complete revolutions before the drill bit at B starts 10 operate and using G ~ 11.2 :x 10" psi. determine the maximum slrain energy acquired by the drill pipe.

1

5000 {j

io---'"ir--; . ,J B

Fig. P11.33

11 .34 Rod AC is made of aluminum and is s ubjected to a torque T applied at C. Knowing that G = 73 GPa and that portion Be of the rod is hollow and has an inner diameter of 16 mm, detennine the strain energy of the rod for a maximum s hearing stress of 120 MPa.

Fig. P1t.34

11 .35

Fig. P11 .35

Show by integration that the strain energy in the tapered rod

AB is

7 T'L U=---

48 GJ,,"" where I n,," is the polar moment of inertia of the rod at end B. 11.36 The state of stress shown occurs in a machine component made of a brass for wh ich i f y = 160 M Pa. Using the maximum.distortion-energy criterion. determine whether yield occurs when (a) fT, = + 45 Mila. (b) (r . = - 45 MPa.

11 .31 l11e state of stress shown occurs in a machine component made of a brass for which IT!, = 160 MPa. Using the maximum.distortion.energy criterion, detertlline the range of values of fT, for which yield does not occur.

'I

t 20 ~ I l'.,

~ ,---Fig. P11.36 and P11.37

,5

\I l'~

691

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692

Energy Methods

11.38 The state of stress shown occurs in a machine component made of a grade of steel for which {T r = 65 ksi. USl1lg the maximum-di stortion.energy criterion, determine the faclor of safety associated with the yield strength when (£1) (',= + Ifiksi. (b) u ) = - 16ksi.

'I "" Shi

------ ,

14 hI Fig, P11.38 and P11.39

11.39 The stale of stress shown occurs in a machine component made of a grade of steel for which (Tr = 65 ksi. Using the maximum-distortion-energy criterion, determine the range of values of if r for which the factor of safety a~­ sociated with the yield s trength is equal to or larger th
.\I "

G~gB -

1-1

1,, -

H-.

-'-"

-'-1

Fig_ P11.40

"11,41 A vibration i50l:.lion s upport is made by bonding a rod A. of radius R I • and a tube B. of inner radius R;. to a hollow rubber cylinder. DcnOling by G the mooulus of rigidity of the rubber, detennine the s train energy of the hollow rubber cylinder for the loading shown.

B

Q

(" )

Fig. P11.41

(I»

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11.7. Impact LoadIng

11 .7. IMPACT LOADING

Consider a rod BD of uniform cross section which is hit at its end B by a body of mass III moving with a veloci ty Vo (Fig. 11. 23{1). As the roJ deforms under the impact (Fig. 11. 23b). stresses develop within the roJ and reach a max.imum value (T,n. After vi brating for a while, the rod will come to rest, and all stresses will disappear. Such a sequence of events is referred to as an impact loadill!; (Fi g. 11.24). In order 10 detennine the maximum value (T m of the stress occurring at a given poinl o f a struclUre subjected to an impact loading, we arc going 10 make several simplifyi ng assumptions. First, we assume that the kinetic energy T = ~l1rv~ of the stri king body is transferred entirely to the struclUre and. thus, that the strain energy Um corresponding to the maximum deformation xm is

V", = t lllV~ This assumption leads I.

to

, =\1

693

B

Fi g. 11.23

( 11.37)

the following two specific requirements:

No energy shou ld be di ssipated during the impact.

2. The striki ng body shou ld no t bounce o ff the structure and retain part of its energy. Thi s. in tum. necessitates that the inertia of the structure be negligible. compared to the inertia of the striki ng body. In practice. neither of these require ments is satisfied, and only part of the kinetic energy o f the striking body is actually transfe.rred to the structure. Thus, assuming that all of the kinetic e ne rgy of lhe striking body is transferred to the structure leads to a conservative design of that structure. We further assume that the stress-strain diagram obtained from a static test o f the material is also valid under impact loading. Thus, for an elastic deformation of the structure, we can express the maximum value of the strain energy as U = m

I' ~m

- dV 2£

(\1.38)

In the case of the uniform rod o f Fi g. 11.23, the maximum stress (T m has the same value throughout the rod. and we write Vrn == a~ V/2£. Solving for a m and substituting for Um from Eq. (1 1. 37), we write u

~

m

~2UmE

--~

V

~mvi,E -V

( \1.39)

We note from the express ion obtained that selecting a rod with a large volume V and a low modulus of elasticity £ will result in a smaller value of the maximum stress (T m fo r a given impact loading. In most problems, the distribution o f stresses in the structure is not uniform, and formula (11. 39) does not apply. It is then convenient to determine the static load Pm, which would produce the same strain energy as the impact loading, and compute from Pm the corresponding value Urn o f the largest stress occurring in the structure.

Fi g. l l.24 Steam alternately lifts a weight inside the pile driver and then propels it downward. This delivers a large impact load to the pile which is being driven into the ground

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EXAMPLE 11.06

v"

A body of mass II! moving with a velocity hits the end B of the nonuniform rod BCD (Fig. 11.25). Knowing that the diameter of ponion Be i~ twice the diameter of portion CD. determine the maxi mum va lue if .. of the stre~s in the roo.

produces in the rod the same stra in energy as the given impact 10.'lding is J

j.!!i U",AE L

_

' ,,, - \ 5

where U ... is given by Eq. (I 1.37). T he largest stress occurs in ponion CD of the rod. Di viding 1'", by the are a A of that portion . we h.'lve = p ..

(T

'"

A

~ J~

U• .E

5 AL

( 11.41 )

or, substitut ing for U,. from Eq. (11.37). Fig. P11.2S

U.,

Making II ~ 2 in the expression ( 11.1 5) obtained in Example Il.Ot. we find thai when rod BCD is subjected to a static load 1'",. its strain energy is

u. =

5P~.L 16A£

( I 1.40)

or

where A is u1e cro~s-sectio nal area portion CD of the rod. Solving Eq. (11.40) for Pol' we lind thaI the static load that

=

Comparing this value with the val ue obtained for u '" in the C.'lSt' of the un iform rod of Fig. 1 I .24 and making V = AL in Eq . ( 11. 39). we note that the max im um stress in the rod of va riable cross section is 26.5% larger than in the lighter uniform rod. TItus . .'lS we observed earlier in our discussion of Example 11 .01 . increasing the di ameter of ponion BC of Ihe md resulTs in a dl'L'rease of the energy-absorbing capacity of the roo.

EXAMPLE 11 .07 A block of weight II' is dropped from a height h onto Ihe free end of the cantilever beam AB (Fig. 11.26). Detennine the ma ximum value of the stress in Ihe beam.

r

Recalling the expression obtained for th e stmin energy uf the ca ntile ver beam AB in Example 11 .03 and neglecting rhe effect of shear. we write 1" LJ U ~ -'-'..

/,

L

" 1-1-

""'"

L-

6£1

Solving this equat ion for P", . we find that th e static force that produces in the beam Ihe same strain energy is

~l

/'

=

•

'

Fig. 11 .26

J6U",EI L'

(1 1.43)

The maximum stress fT ", occurs althe fixed end B and is

As it falls through the distance h. the potelltial energy \VII

of the block is transfomled into kinetic energy. As a result of the impac t. the kinetic energy in tum is transformed into strain energy. We ha ve. Iherefore.t

U", = Wh

Substitutin g for P'" from ( I I .43), we wrile

( 11.42, ( 11.44)

'~hich thc blOCk: drops is actually It + _~"" the maximum dcncctiol1 of the end "rlhe beam. Thus. a more lICCUrJIC c!lprcssion for u,. (see Sample Prob. 11.3) is

tThe total distance through

whcre _,.~

i~

U. = IV(h+y.. ) ( 11.42' ) Howc,'Cr, \\hcn II » Y... we nlly neglect y", and usc Eq. ( t J ,42).

694

or, recalling (11.42),

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11.8. Design for Impact Loads

1 1.8. DESIGN FOR IM PACT LOADS

Let us now compare the val ues obtained in the preceding section for the maxi mum stress (Tm (a) in the rod of uniform cross section of Fig. 11.23, (/)) in the rod of variable. cross secti on of Example 11.06, and (c) in tlte canti k'\'er beam of Example 11 .07. assumi ng that the last has a circu lar cross .;;ecti on of radius c. (a) We first recall from Eq. ( 11 .39) that, if Um denotes the amount of energy tran s r~rred to the rod as a res ult of the impact loading, the maximum stress in the rod of uni form cross section is

_ f2UV E m

am - \

( 11.45a)

where V is the \oolume of the rod. (b) Considering next the rod of Example 11.06 and observing that tlte volume of the rod is

v ~ 4A(Li2) + A(Li2) we substitute AL

= 2V/5

~

5ALi2

il1lo Eq. ( 11.4 1) and write (T ffl

~c) Finally, recalling that' tioll. we note that

=

pU;E

= ~ 1TC"

for

:1

(11.45h)

beam of circular cross sec-

L(l/c 2) =- Ln1TC~/r) = 1(7H·2 L) = 1v where V de notes the volume or the beam. Substituting into Eq. ( I [.44), we express the maximum stress in the cantilever beam of Example [1.07 as (l1.45c) We note that, in each case, the maximum stress U m is proportional to the square root of the modulus of elasticity of the material and inversely proportional to the square root of the volume of the member. Assuming all three members to have the same volume and to be of the same material, we also note that, for a given value of the absorbed energy, the uniform rod will experience the lowest maximum stress, and the cantilever beam the highest one. This observation can be explained by the fact that, the di stribut ion of stresses being uniform in case a, the strain energy will be uniformly distributed throughout the rod. In case h. on the other hand, the stresses in portion BC of the rod are only 25% as large as the stresses in portion CD. This uneven di stribution of the stresses and of the strain energy results in a maximum stress U twice as large as the corresponding stress in the uniform rod. Finall y, in case c, where the cantilever beam is subjected to a transverse impact loading, the stresses vary linearly along the beam as well as across a transverse section. The very uneven resulting di stribution of strain energy causes the maximum stress U m to be 3.46 times larger than if the same member had been loaded axially as in case a. III

695

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696

The propenies noted in thc three specific cases di scussed in this section are 4uite general and can be observed in all types of structures and impact loadings. We thus conclude that a structure designed to withstand effectively an impact load shou ld

Energy Methods

I. 2.

3.

Have a large volume Be made of a material with a low modulus o f elasticity and a high yie ld strength Be shaped so that the stresses are distributed as evenly as possible throughout the structure

11 .9. WORK AND ENERGY UNDER A SINGLE LOAD

When we !irst introduced the concept of strain energy at the beginning of this chaptcr. we considered the work do ne by an ax.ialload P applied to the end of a rod uf unifonn cross section (Fig. 11.1). We del1ned the strain energy of the rod for an elongation x I as the work of the load P as it is slowly increased from 0 10 the value P I corresponding to XI' We wrote Strain energy = U ""

f~lp dx

( 11.2)

"

Ln the case o f an clastic defontlation, the work of the load P, and thus the strain energy of the rod. were expressed as ( 11.3)

"1- " [

':C ~ ~:;::=;;\8 ,\

Lat er. in Sees. llA and I 1.5, we computed the main energy of structuml membe~ under various loading conditions by detennining the strain·energy density If at every point of the member and integrating 1/ over the entire member. However, when a Structure or member is subje<:ted 10 a ~'illgle COli' cell/rated loall, it is possible to usc Eq. (11.3) to evaluarc its clastic strain energy, provided, of course, that the relation between the load and the resulting deformati on is known. For instance, in the case o f the cantilever beam of Example 11.03 (Fig. I 1.27), we write

Fig. 11 .27

U = tPIYI

and, substituting fo r )'1 the yalue obtained from the table of Belllli Deflectio/B" lIlid SlopeJ of Appendix. D,

1

U=

(

~:; -,,-

A

1\1 ,

o[

(PILl) PiLl

"2 P1

3£1

= 6£1

(J 1.46)

A similar approach can be used to detenni ne the strain energy of a structu re or member subjected to " j'ill!;le couple. Recalling that the elementary work of a couple of moment !vi is !vi dO, where dO is a small angle, we find. since M and R are linearly reJ.lIed, that the clastic strain energy of a cantilc.ver beam AB subjected 10 a single couple M I at its e nd A (Fig. 11.28) can be expressed as

Fig. 11 .28

( 11.47)

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wheTe 0 1 is the slope of the beam at A. Substituting for 0 1 the value obt!lined from Appendix D. we write

U= ~M (M ,L) = MiL 2 1 £/2EI

(1 1.48)

In a similar way, the elastic strain energy of a uniform circular shaft AB of length L subjected at its end B to a single torque T J (F ig. 11.29) can be expressed as

( 11 .49)

Fig. 11.29

Substituting for the angle of twist

4>1 from Eq. (3 .16), we veri fy that

(T'L) TiL

I U = 2TI lG

= 2JG

as previously obtained in Sec. 11.5. The method presented in this section may simplify the solution o f many impact-loading problems. In Example 11.08. the crash of an automobile into a barrier (Fig. 11.30) is considered by usi ng a simplified model consisting of a block and a simple beam.

Fig. 11.30 As the automobile crashed into the barrier considerable energy was dissipated as heat during the permanent deformation of the automObile and the barrier.

11.9. Work and Energy under a Single Load

697

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EXAMPLE 11 .08 bloc k of mass m moving with a velocity Vl' hiu squarely the prismatic member AB at its mi dpoint C (Fig. 11. 31). Determine (tl) the equivalenT static load r •. (bl the maximum me ~s " ., in the member. and « ) the m:IXimum denection .f.. at point C.

"

A

(a) Equivd /ent Static; L.oad. The llIaximulll SllaLt' CLIergy of the membe.r is equal 10 the kinetic. energy of the block before impacl. We h~ve

,,mv"'

( 11.51) where x., is the d~nection of C corre sponding to Ihe static load i'.... From th~ [able of Bellll! Drjlet·/ilJII.\· lllld Slopu· of Append ix D. we find that ( 11.52)

-t -

~c

"•

'"

-J:l.

}I"

A

A

FIg. 11 .3 1

-L K,- !-I'..

Fig. 11.32

(b) Max/mum Stress. Dmwing the free·bcKJydi:lgram o f the member (Fig. 11.32). we find th at the maximum value of the bemJin/: momenl occurs at C and is M"IJ.' "" P. ,1../ 4. The maxir."lUm s tre~s. therefore, occurs in a trt nsverse section !luougl, C and Is equal to M .... c ~

~

--

..

~

J

I pl L) "'

-I

C

P,.,L c -II

--

Substituting for p... from tl 1.53 ). we write

Subsli tutlng for .r.. from ( 11.52) into ( 11.5 I ). we write U

II R ~~ I' ..

~ /"

(11.50)

On the other hand. expres~lng /.-',. as the wDrk Dr the equlvalenl horizol1l'11 static load ;IS it is slowly applied at the midPOUII C of lhe member. we wri te

'-

~

3mvJ EI

= -'"2 48£1

L(J lc ) ~


Solving for P", and recalling Eq. \ 11.50). we find that the static load equivalent \0 the gi ven impJct Ioadlllg is (11.53)

(c) Maximum Deflection. Substituting ilHO Eq ( 11 .52) the expression obtained ror P", 111 (1 1.53) . we have LJ

48I1l v~E1

.',,, = 48£1 \ -L -'~ =

II/wiLl

V48£1

11.10. DEFLECTION UNDER A SINGLE LOAD BY THE WORK-ENERGY METHOD We saw ill Ih e preceding section th at. if th e tlcncclion XI o f a structure o r member under a si n gle co nce ntnlled load P I is known, the corresponding stra.in ene rgy U is obtained by writing

A simi lar express io n for the strain e nergy of a st ru ctu ral mcmbe r un· der a single co upl e M I is:

U = ~M ,O I

698

(IL47)

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Converse ly, if lhe stmin energy U of a structure or member subjected to a single concentrated load PI or couple MI is known, Eq . (11.3) or ( 11.47) can be used to determine the correspondi ng deflection XI or angle 0 1, In order h.) de termine the detlect ion under a single load applied 10 a structure consisting of severa l component parts. it is easier, rather than use one of the methods ofehap. 9. to firs t compute the strain energy of the struClure by integrati ng the. strain-energy density over ils various parts, as was done in Sees. 11.4 and 11.5, and then lise eilher Eq. (1 1. 3) or Eq. ( 11.47 ) 10 obtai n the desired detlecti on. Similarly, the angle of twist q;l of a composi te shaft can be obtai ned by integrating the strain-e nergy density over the vari ous parts of the shaft and solving Eel. ( I 1.49) for rPl' It should be kept in mind that the method presented in th is sect ion can bt: lIsed ollly if the givell structllre is .w/JjiYfea fo a sing l e Cl)llCi!IIfmled load or cOllple. The strain energy of a structure subjected to several loads ctlllll o f bt: detemli ncd by computing the work of each load as if it were applied independent ly to the structure (sec Sec. II. I I). We can also observe that, even if it were possible to compute the strain energy (I f the S(fm;ture in thi s manner, only one e(luati un would be availab le to detemlinc the de tleclions correspondi ng to lhe various loads. In Sees. 11 .12 and 1 1. 13, another method based on the concept of strain energy is prese nted , one that can be used 10 determine the deflection or slope al 11 given point of a structure, even when that structure is subjected simultanc:=ously to several concentrated loads. distributed loads. or couples.

11.10. Deflection urlder a SIngle Load by the Work-Energy Method

EXAMPLE 11 .09 A load P is suppurted al B by 1\\"0 uniform rods of the. same cross-sectional area A (Fig. I 1.33). Determine the venical deflection of point B. The s train energy of the system llnder the given load was tletenninedin Example 11.02. Equaling the expression ubtained for U to the work of the load, ..... e write

c 0,

1'"/ I U = 0.364 A£ = i P>,s B

and. solving for the I'ertical deflection of B.

Remark. We should note that. once the forces in the two rods h,we been obtained (see E:wmple 11.02). the deform;l(ions OBle and OBiD of the rod~ could be obtained by the meulod of Chap. 2 Determining the vel1ical deflection of point B from these. deformations. however, wOllld require a careful geometric analysis of the various displacements involved. The ~train-energy method used here makes such an IInalysis unnecessary.

J

l

J'

D

Fig. 11 .33

699

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EXAMPLE 11 .10 Determine the deflection {If end A of the cantilever beam AB (Fig. 11.34). taking into account the elreet of (a) the normal stresses only. (b) both the nonnal and shearing stresses.

the normal stresses was considered. we write

and. solving for J'A ,

(b) Effect of Normal and Shearing Stresses. We now substitute for U the expression (1 1.24) obtained in Example 11.05. where the effects of both the normal and shearing stresses were taken into account. We have

3£,,2) -I1'\'

+ --10CU

Fig. 11 .34

=

1 ~A

and. solving for )'", (a) Effect of Normal Stresses. force (. as it is slowly applied to A is

The work of the

U = ~ PYA Substituting for U the expression obtained for the strain energy of the beam in Example 11.03. where only the effect of

PL' ,-( I )" - 3f1

+3£h --

1 )

10GL"

We lIote that the relative error when the elreet of shear is ne· glected is the same that was obtained in Example 11.05. i.e" less than 0.9(h IL)!. As we indicated then. this is less than 0.9% for a be:lm with a ratio h l L less than to.

EXAMPLE 11 .11 A torque T is applied at the end D of shan BCD (Fig. 11.35 ). !<J\owing that both portions of the shaft are of the same rna· terial and same length. but that the diameter of Be is twice the diameter of CD. determine the angle of twist for the entire

111e strain energy of a ,imilar shaft was determined in Ex ample 11.04 by breaking the shaft into its component pans Be and CD. Making 1/ = 2 in Eq. ( 11.23). we hal'e

~I:1ft.

where G is the modulus of rigidity of the material and i the polar moment of inenia of portion CD of the shaft. Selling U equal to the work of the torque as it is slowly applied to end D. and recalling Eq. (I 1,49). we write

~"",£)" . dium.

Fig. 11 .35

700

= II

17 TIL I 32 lGi = l?ifJvlB and. solving for the angle of twist

tPD//j'

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SAMPLE PROBLEM 11.3 40 """

The block D of mass //I is released from rest and falls a distance II before iT strikes the midpoinT C of the aluminum beam AB. Using E = 73 GPa, determine (a) the ma:dmum denection of point C, (b) The maximum stress thaT occurs in the beam.

/,=4U".",

1.=1", -

-[

SOLUTION

·""--T

""_~~"""'iJi ~/,

,..

D

A

B

A " ~,b..""""''''_~ '.I..

I'mitiOll !

I'". itio" 2

Prindple of Work and Energy. Since the block is released from rest. we note that in position 1 both the kinetic energy and the strain energy are zero. In position 2. where the max.imum deflection Y., occurs. the kinetic energy is again zero. Referring to the Table of Bellm Dej7e£"liUlIs alld Siopel' of Appendix D. we find The expression for Y.. shown. "nle strain energy of The beam in position 2 is 1148£1

Uz =

"21'",)"., =

27 y", 2

Fmm .0\1'1" ' "'''' D p

J'L'

,\

l

.mE l

",'" L:1'I",

We observe that the work done by the weight W of the block is IV(h + y,.). Equating the ~T rain energy of the beam 10 The work done by W, we have

H

'.I"'= t~EI~ C

(I)

o. Maximum Den ection or Point C. From lhe given tlata we have EI = (73 >< lOYPa)n(O.04 m)4 = 15.573 X 10J N " m" II = 0.040 III IV = I'lg = (80 kg)(9.8 1 O1/s1) = 784.8 N

L = 1m

Substituting into Eq. (I). we obtain and solve lhe quadratic equation

(373.8 X 10l)y~ - 784.8)".. - 31.39 = 0 b . Maximum Stress.

48EI

p ..

=----;yy", =

...

48(15.573 x IO l ~ . m) ( Im )l (O.O I027m )

= (t P,.L)c = 1

10.27 mm ..

The value of p., is

Recalling Ihat u '" = M,,..,,c11 and Mm:., = ~ TT

\' .. =

1(7677 NFl

P.. =7677N

/J".L. we write

m)(O.020 m) i2(O.040 m)~

/1"..

= 179.9 MPa ...

An apprmimation for the work done by the weight of the blo.:k can be obtained by omitting Yo. from the expres,ion for the work and from the right-hand member of Eq. (I), as was done in Example 11.07 . If thi.> approximmion is used here. we lind .\'.. = 9.16 mm: Ihe error is 10.8%. However. if an 8.kg block is dropped from a height of 400 mm. producing Ihe same value of IVh. omitting .1'.. from the right-hand member of Eq. ( I) results in an error of only 1.2%. A further di sc ussion ofl.his approximation is given in Prob. 11 .70.

701

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SAMPLE PROBLEM 11.4 Members of the truss shown consist of sections of aluminum pipe with the cross-sectional areas indicated. Using £ = 73 G Pa. determine Ihe lIenicai deflection of point E caused by Ihe 10:11.1 P.

SOLUTION Axial Forces in Truss Members. The reactions are found by using Ihe free.body diagram of the entire truss. We then consider in sequence the equilibrium of joints. E. C. D. and B. At each joint we determine the forces indicated by dashed lines. AI joint B. the equation 'iF, = 0 pro \l ides a check of our compllIations.

Noting lhm E is the same for all members. we express

St rain Energy.

the slmin energy of the

[ru~s

U=

as follows

2: f1 L j =..2... 2A;E

2£

~ FiLI

(I )

Aj

where F, is the force in a ghell member as indicmed in Ihe following table and where the summation is extended over all members o f Ihe LntSS. Member

F,

L" m

A" m'

0.8 0.6

500 X 10

AB

0

AC

+ 151'18

AD

+5P/4 - 211'18 0 + 151'/ 8 - J7P/8

BD

CD CE DE

1.0 0.6

0.8 1.5 1.7

Ft L, A,

• • •

FoL 2: _,_J =

0

500 X 10 6 500 X 10 1000 x 10 1000 X 10- 6 500 X 10 6 lOOO X 10 b

4219P" 3 l25f'"' 4134?,

Returning 10 Eq. ( I) . we have

0 10 5471'" 7677/'"

Principl{> of Work.Encrgy.

29 700P"

A,

U = ( 1I2E)(29.7 X IO' ?').

We recall Itml Ihe work done by Ihe to;jd P

a~ il is graduatly applied is ~ PYE. Equating Ihe work done by P 10 Ihe strain energy U ;jnd recalling Ihal E = 73 GPa and f> = 40 kN. we have

I

)'1:

Yf. = 16.27 X 10- ] m

702

(29.7 X JOl)(40 X W )

= £ (29.7 X 10' 1') = '-~~7=3~X~I~0""~~ \'£=

J 6.27 mm ~ '"

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PROBLEMS

11.42 A G-k£ coll;:u has
A

B A

/J I - - - - - - - U ! '"

-------1

Fig. P11.42 and P11.43

11.43 A 5-kg collar D moves along the utlifonn rOO All :md hllS ~ s peed "0 = 6 mls when il s trikes a s mall plate anached to end A of the rod. Using E - 200 OPa amJ knowlllg dml the allowable s tress in the 1'00 is 250 MPa. d.:(ermine the 'imalles t diameter thai mil be used for the rod.

11 .44 nl~ [00 Ih collar G is reicllSed fmm rest in the position shown and is sloprd hy plate BDP that is attached \0 the ~.in .-diamelfr s teel rod CD and to Ihe ~-in"Lhameter SIC ... ] rods AB::od EF. Knowing lhat for the grade of steel used u.u = 24 ksi and E = 29 x 1O~ psi. dC lcmline ule Imgcsl allowable llistance It

r

c

E

'"

L~ c ~, B

D

F

Fig. P1 1.44

, 1 .45 Sol ve J'rob. [1.44. a.<.s urning that tflc: } in.-diame ter st~el rod CD is replaced by it ~-in.-dian"t:ter rod made of an aluminum alloy for which 6 IT.II '" 20 k~ and £ = 10.6 X 10 psi. 11 .46 Collar D is rdeasetl from rest in Ihe position shown and is SIOP)x''il by a small plate attached at end C of the venical rod ABC. Delemline the mass of Ihe collar for which Ihe maxImum normal slress in ponion BC is [25 Mf'll.

r

A

.Im

]2 .",,,, ,~:u"ph'r

~

B

Alll nill""l E = "TllGP"

2.5 m D

L

Brm= £ - l().jGP"

~

Cl

0.6 1n

Fig. P I1.46

11.4 7 So lve ?rob. 11.46. assumin g th
nl;j{J~

703

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704

En~rgy

MOlho
--~ c

11.48 "l~ post AS ("oll~ists of a ~teel pipe of 3.5-ilL dlnJJ1el~1 amIO.J-in wdl ihid:ne,s. A 15-lb block C moving horizontally with a velocity II" hit ~ the post squarely at A. Using E = 29 X lIl" p~i . detemline the largest sl*ed ") fo r which the maximum normal stress in lhe pipe dOi!s not C.l.cee(] 14 k.~L

11.49 Solve Prob. l l.4 B. assuming thaI the post A8l'ons;sts of a solin stee l rod of ].5-in. oUier diameter. 11 .50 Tht>. '1ee1 ~am AR is ~(nd: <:fJ11:tr~1y m iJ.~ midfM"\int r hy a 4~.kf block mO\'ing horizontall y with II speed "0 = 2 mho Using E = 200 GPa. detenni ne the equivalent static load, (b) the maximum normal stress in the beJm, (c) th~ maximum deflection of the nlidpoin t C of Ihc beam,

Fig. Pl1.48

Fig, Pl1.50

11.51 Solve Prob. 11.50. assuming that tie W150 X 13.5 rolled·>teel helm (s rotr,ted by 9(f about its longitudinal ax is so thaI its web is verti cal 1 1.52 The -l-5-lb block D is dropped frJm ~ height" = 0 .6 ft ontD the steel beam AB. Knowing that E = 29 X 10" psi. detemline (0) tite maximu m deflection at point E, (b) the maximum nonnal stress in the beam.

Fig. P11.52

11,53 and 11.54 The 2·k~ blOi.'k D is dropped from the position shown onto the end of a 16-mm-di:lmeterrod. Knowing that E = 200 GPa, detennine la) the mll.ximum deflection of end A. (0) the maximum bending moment in the rod. (c) the maximum nonna l ~tre>s in the rod

/J III 2kp;

4u",,"l

,\ -

!l

r

~').6",--l

Fig. P11.53

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11. 55 A 160·lb diver jumps from a heighl of 20 in. onto end C of a diving hoard having Ihe uniform cros~ section shown. Assuming lhatlhe diver's legs remain rigid and using E = 1.8 X 10" pSI. determine (a) the maximum deflection at poim C. (b) the maximum nom1al stress in the board, (c ) the ellulv. alent stmi c load. 1 1.56 A block of weighl IV is dropped trom a heighl II 01110 the hori· zontal beamAB ;[[Id hits iT al point D. (a) Show that the maximum deflect"ion ),,, at point D can be expressed as J' = \' ~,

."

(I + JI + 2h) y"

I

~

8

;;;., r I -

]20;". _

-i-1'flI-----,Ie

, 2.5 ft

where )'" represents the deflection at D caused by a stalic load IV applied at that point :tnd where Ihe quantity in parenthesis is referred 10 as the impu("/jtICtor. (b) Compute the impact factor for the beam and the imp
11' r2.l

Problems

1~

1J,5ft~

Hi in .

Fig. PH .55

,\p;~~D~~I~==~~r ' __'G. __ /

B

-...... ------D'

Fig. P11.S6 and P11 .57

11 .S7 A block uf weigh t IV is dropped from a height h onto the horizon· tal beam AB and hiL~ point D . (1I) Denoting by ),,,, the exaCl value of the maxi· mum deflection aT U and by y;, the value obtained by neglecting the effect of this tJeflection on lhe change in potential energy of the block. show lhm lhe absoluTe value of lhe rebll\'e error is (y~ - y.. )ly.... never exceeding }:"!2h. (b) Check the result obtained in pan {j by solving pan a of Prob. 11.53 without taking y.. into account when deTermining the change in potential energy of the load. and com· paring lhe answer obtained in thi s way with the exacl answer to lhm problem. 11 .58 and 11 .59 Using the method of work and energy. determine lhe deflection at point D caused by the load P. p

Fig. P11.S9

11.60 and 11 .61 Using the method of work and energy. detemline the slope at point D caused by the couple 1\10.

~~1-==-I!::i'i~B-~'" f - - - 1. Fig. P11.60

Fig. P11.61

2.65 ;n.

-1

1-1'

/.

Fig. P11.S8

705

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706

Energy Methods

FIg. P11.62

11.62 and 11 .63 Using the method of work and energy. de temline the deflection at point C caused by the load P.

Fig. P1 1.63

11 .64 Using the method of work and energy, determine the slope at point B c[lused by the couple M(J.

11.65 Using the method of work and energy, detennine the slope at point A caused by the couple l\'I t)-

Fig, P11 .65

11.66 The 20-mlll-diameter steel rod BC is attached to the lever A B and to the fixed support C. The unifonll steel leve r is 10 mm thic k and 30 mill deep. Using the method of work and energy,detemline the deflection of poin t A when L = 600 mm, Use £ = 200 G Pa and G = 77.2 GP[I.

Fig. P11.66 and P11.67

11 .67 The 20-mm-diameter steel rod BC is attached to the lever A B :md to the fixed support C. The un iform stee l le\er is 10 mill th ick and 30 mm deep, Using the method of work and energy. determine Ihe length L of the rod Be for which thc dcflection al point A is 40 mm. Use E = 200 GPa and G = 77.2 GPa.

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Probtems

11.68 Two steel shafts. each of O.75-in diameter. are connected by the gears shown. Knowing that G = 1l.2 X [0" psi and that shaflLJF is fixed at F. determine the
FIg. P11 .68

11 .69 TIle 20-mm-diameter steel rod CD is welded to the 20·mmdiameter steel sh
where d~' is an element of the center line of the wall cross section
FIg. P11 .70

Fig. P11.69

707

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708

11 .71 and 11.72 Each member of the truss shown has a uniform crosssectional area A . Using the method of work and energy. determine the horizontal deflection of the poinl of applicatio n of the luad P.

Energy Mothods

B),'

A

~ C

T~ " B

"I

L

f)

C

f)

-r.:-I

!-'--I-

Fig. P11.71

Fig. P11.72

11 .73 Each member of the truss sho .... n is made of steel ilnd has il uniform cross-sectional area of 3 in". Using E = 29 X 10" psi. detemline the vertical deflection of joint A caused by the application of the 24-kip load.

Fig. P11.73

Fig. P11.74

11.74 Each member of the truss shO\~n is madeof stccl and has a uniform cross-sectional area of 5 inl. Using E = 29 X 106 psi. delemline the vertical deflection of joint C caused by the application of the IS-kip load. 11 .75 Members of the truss shown are made of steel and have the c ros~­ sectional areas shown. Using E = 200 Gila. determine the vertical deflection of joint C caused by the application of the 210-kN load.

( 1 2()O"'m~

t- r:==-;,C

1360 nnn

h~==~~c

:>(..0",,,,

I Fig. P11.7S

Fig. P11.76

11.76 The ~teel rod Be Ilns a 24-mm diameter and the steel cable ABDC4. as a 12-mm diameter. Using £ = ::!OO GPa. detennine the dc/lection of joint D caused by the 12-kN load.

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*11 .11 . WORK ANO ENERGY UNOER SEVERAL LOAOS

11.11. Work .. nd Energ)' under

Several Loads

In thiS section. the ~t raLLl energy 01 a structure subjeCted to sevemlloads will be considered and will be e~prcsscd in tenns of the loads and the

,\

resl illing (let1ec. iorK

Consider an clastic beam AB subjcC:lcd 10 two concentrated loads PI :.md P2' The strain energy of the beam is Nlual 10 Iht! work of PI and 1'2 as they are slowly app lied to the beam at C! and C 2 , respeCtively (Fig. 11.36). Howevcr. in o rdcr to cyalufI!e this work, we must jir.~1 el(press the denec:tions XI and x~ in leons of the lo;.Ids P I and P 2' Let us ass ume thai unly 1'1 is applied to the beam (Fi g. 11 .37). We note th:11 both CI ,md C1 arc ocnected and that their dcflecli o tl~ are proportional to the load P I' Denoting these det1ections by XII and x 2l • respectively, we wrile

' .

I'}

Fig. 11.36

XII

-"2 1

-~:rV-----114

(1 1.S4)

where UJ I and (:1'21 are constants called influence coefficients. These constants represent Ihe dcllections 0 1 C! and C! , respectively, when a Ul111 load is appli ed at C 1 and are c h uracteri stic ~ of the beam AD. Let U~ now assume thai onl y l'2 I~ applic'd to the bC'am ( F ig. 11. 38)_ Denoting by XII and X~ 2' respectively. the resulting dcncclion~ of C I and C;. we write

-~l£---n';F

I

'!

"

Fig.ll .37

_"~___'r~

(11.55)

Ie; , .,

Cj'

where all :md O!!~ are the influence coefficient.> re presellti ng lhe de tlections o f C 1 and C1 , respectively, when a unit load is applied at Cl . Applyi ng me principle of superpos ition, we express the dellections XI and X2 o f C j and C1 whc lI both loads arc applied (Fig. [1.36) as -lol - _' II Xl -

'~ll

+ All

- (fIIP I

I "':1 -


+

UI!Pl

(11.5{1)

I anf'l

(1 1. 57)

To compute thc work done by P I and P" and thus the strain energy of Ihe beam, it is convenient to a~Sllme that PJ is !'i rst applied hlowly at C I (Fig. I U9a). Recalling the li rs t of Eqs. (1 1.54). we c).prcss the wo rk of P I as

Fig. 11.38

.tll

"

:J

~l Cj

(Il)

(11 .58) and note thai P2 does no wo rk while C 2 moves Ihroug h .r~I' since it has yel been applied to the beam. N()w we slowly apply P 1 at C1 (Fig. I j .39M: recalling the second of Eqs. (I L55~. we eJl.p re~~ Ille wo rk of 1'2 as nOl

(11.59) Bu\. as P 2 is s lowly appli~d m C!> the point of ap plication of P I moves th rough X I! from CLIO C" and the load P I d oe~ work. Since P I h/ll!!Y

ij,: FIg. 11 .39

PI

~!I

\

'

~ C2

709

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71 0

Energy Methods p

P, I'~

c;

°L,,, CL . J

~~

C ,

(Il) L"ad-{H'pl"'-~·"."nt

(bl Ln"d.disl'i:K'''H''' HI di agram for C~

diugnun fnrC I

Fig. 11.40

applied during this displacement (Fig. 11 .40), its work is eq ual to or. recalling the first of Eqs. ( 11 .55),

PI Xll

(11.60)

l ' ll

Adding the expressions obtained in (1 1.58), ( 11.59), and ( 11.60), we express the strain energy o f the beam under the loads P I and P 2 as

X22

'l..-onJ-m -- ~r:o.J CL

C2

p~

ill )

C A

.::: ,-CL ... {

' ••• 0 - B

~- ----~- --

(11.6 1)

Lfthe load P2 had first been appl ied to the beam (Fig. 11.4 Ia ), and then the load P I (Fig. I 1.4 1h), the work done by each load wou ld have becn as shown in Fig. 11.42. Calculations similar to those we have just carried out would lead to the follo wing alternative expression for the strain e nergy of the beam: (11.62)

-I'll

Equating the ri ght-hand members of Eqs. (1 1.6 1) and ( 11.62), we find that (I' l l = (\'1 ]' and thus conclude that the denection prod uced at C I by a unit load applied at C2 is equal to the deilection produced at C1 by a unit load appl ied at C I . This is known a; Maxwell'l" reciprocalrheol1!lII. after the British physicist James Clerk Maxwell (183 1- 1879).

(b)

Fig. 11.41

p

P

I', f'~

- --

°1_, __ I:"', . -.J ~ l"~~_ ~_X~] _

'a) L""d.di .pl "l~· ""'"1 diagmm forC]

Fig. 11.42

_

(I,) LI)~d.di_'l'i:K""lO'tll di~~rdm for C~

C

'

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While we are now able to express Ihe strai n energy U of a s tru t~ lure subjected to several loads as a function of these loads, we cannm lise the method of Sec. I 1.10 to detennine the delleclion of sueh a struclure. Indeed. computing the strain energy U by integrating the strain· energy density Ii over the structure and substituting the expression ob· tained inll) ( I 1.61 ) would yie ld only one equation. which d early could nOl be solved for the various coe fli cient s a. '11 .12. CASTIGLlANO 'S THEOREM

We recall the expression obtained in the preceding section for the strain energy of an cla~ti c structurc subjected Lo two loads PI and p!: «

1.6( )

where 0' 11,0'11, and a n arc the inlluem:e coc llicients assoc iated with the points of application C I and C1 of tht:. two !()llds. Differentiating both members of Eq . (11.61) with respect to PI and reca lli ng Eq. (11.56), we wri te

au

oP = al lP I

+ al2 Pl

=

XI

« 1.63)

I

Differentiating both members 01' ~ .. (J 1.61 ) with respect to p~ . recall· ing E(l' ( 11. 57), and keeping in mind that a ll = 0.2 1. we have ( 11.64)

More generally. if an elastic struc ture is subjected to /I loads PN • the dellectio n "\j of the point of application of Pl' measured along the line of :lction of Pj . can be expressed as the partial deri vative of the strain energy of the structure with respect to the load PJ" We write

PI' P"1,""

x

}

,u

(l1.65)

=-

UP;

lllis is Castiglitmo's theorem. named after the Ltalian engineer Albertu Casligliano (1847- [884) who first staled it .t t in lhe case of an ~13~tk ~ Iruclure subj~eled to II io.1ds I', . P, ..... p •. the dclkclion of the poinl of applicatiDn of PJ' mcalunxl along Ihe Ifnr of aellon of Po' can be uprcsscd a.o; .r j

and the strain energy of the .<;lruClure.

i~

=

found

L, a"P

( 11 .66)

j

!O be

U= ~ L L ff(\P,l',

( 11 .67 )

Ditfcrcntiatiug U wun r~spcct 10 1',. an
-

i1l',

~ ,- L al,I" .,. I

2: a " I',

2 ,

or. since a,! = ali' ,)U I '" I " ;-p =;;- ..t.; al.P,+; l a l/" ! • ( •

~

Recallin g Eq. (1 1.661, we vcri fy Ihal

,U

r = -

-,

01',

( 11.65)

11.1 2. Casligllano's Theorem

711

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712

Energy Mothods

Reca lling that the work of a couple M is ~MO. where {J is the angie of rotation at the point where the couple is slowly applied. we note that Casti gliano's theorem may be used to detennine the slope o r a beam at the point of application of a couple _Vl j . We have

au

o· ~ ­

j

aM,

( 11. 68)

Si milarly, the angle of twist rPJ in a section of a shaft where a torque T I is slowly applied is obtained by differentiating the strain energy of the shaft with respect to T1:

(11.69)

· 11 .13. DEFLECTIONS BY CASTIGLIANO'S THEOREM

We saw in the preceding sect ion that the deflectio n Xj of :1 structure at the poi nt of application of a load Pi can be detcmlined by computing the partial deri vative au/ap; o r the strai n energy U of the structure. As we recall from Sees. 11.4 and 11.5, the strain e nergy U is obt~lined by integrating or summ ing over the structure the strain energy of each element of the structure. The calculation by Castigliano's theore m o r the deflectio n Xi is simplified if the differentiation with respect to the load P, is carried out before the integration or su mmation. [n the case of a beam. for example. we recall from Sec. I [.4 that ( 11 . 17)

and determine the deflection .1) of the point of rtpplic:1tion of the load p) by writing (II. 70)

In the case of a truss consisting of II uniform members of length L" cross-sectional area Aj • and internal force F j • we recall Eq. (I [. (4) and express the strain energy U of the truss as

(1l.7 1)

u ~

The deflection XI of the point of application of the load Pj is obtained by differentiatin g with respect 10 Pj eJch term of the sum. We write

au

r = -

'J

uP,

=

(11.71)

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EXAMPLE 11.1 2 The cantilever be~m AD suppol1S a uniformly distributed load IV nnd a concentrated load P as shown (Fig. 11.43). Knowing that L = 2m. HI "" 4kN/m, P = 6 kN. and EI= 5 MN ' m~. detemline the deflection al A.

and its derivative with respect to P is aM

-

aP

= - ..r

Substituting for M and ilM/ilP into

&to

(11.73). we write

(1 1.75 ) Fig. 11.43

Substituting the given daln. we have The dellection )".1 of the point A where the load P is <Jpplied is obtained from Eq. (1 1.70). Since P is vel1ical and directed downwnrd. YA represems~, vel1ical deflection and is positive downward. We have

I"

L

M aM

~- d.f

Ef ill'

The bending moment M at a distance M = - (p,r

+

,r

~w..?)

,)'.1=

( 11.73)

from A is

5X I06N 'm1

r(6 X J0

1

YA = 4.8

x

N)(2 m)' + (4 x 10 ~/m)(2 m)"' ] 3 3

10

'm

}A

= 4.8 mm

t

We note thal the computation of the partial derivative aM /il l' 1101 11(11'(' been curried o ut if the numerical value of I' had been substituted for I' in the expression {I 1.74) for the bending moment.

could (1 1.74)

We can observe that the deflect ion Xj of 11 structurt! at 11 given point Cj can be obtained by the direct application 01' Castigli ano's theorem only if 11 load PJ happens to be applied at Cj in the direction in which .\j is to be determined. When no load is applied at Cj' or when a load is applied in a direction other than the desired one, we can still obtain the de ncction Xj by Castigliano's theorem if we. use the following procedure: We apply a fi ctitious or "dummy" load Qj at Cj in the direction in whi ch the denection '\j is to be determined and use C1Istigliano·s theore m to obtai n the de nection (11.76) due tt) Qj and the actuul loads. Making Q; c:= 0 in Eq. (11.76) yields the dellection at Cj in the desired direction under the given loading. 111c slope 0: of a beam al a point Cj can be detcnnined in a simiJar manner by applying a fictitious couple M; HI Cj , computing the partial derivative iJU/iJM)' and making Mi = 0 in the expression obtained.

713

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EXAMPLE 11 .13 The cantilever beam AS suppons a uniformly distributed load (Fig. 11.44). Detemline the deflection and ~Iope at A .

"IV

I I- - -L- -

Deflection at A. We apply a dummy downward load Q \ aLA (Fig. 11.45) and write

fL

au y" = aQ" =

At aM £1 aQA dx

0

( 11.77)

Fig. 11.44

The bending moment M
- QAX -

M =

~wr

( 11.18)

and its derivative WiLh respect 10 QA is

aM ~-

=

- x

( 11.79)

ilQ,\

Substituting for M and aMlaQ" from ( 11.78) and (11.79) into (11.77). and making QA = O. we obtain the deflection at A for Lhe given loading: )'A =

I £/

fL 0

!

'

( - zwr)( - x)dx=

A [' 1LL'~ 11111,[ Q" Fig. 11 .45

wL'!

+ 8£1

Since the dummy load was direcled downward. the positive sign indicates Lhat 'wL4

j J'A -- 8£1 Slope at A. We apply a dummy counterclockwise couple MA
l{ecaJling

Fig. 11.46

Substituling for M and aM/aM" from (11.81) :Uld ( 11.82) inlo (11.80). and making M'I = O. we obtain the slope at A for the given loading:

Elf. (1117), \\'ehaw

f

L

M aM

~~-d,

(11.80)

o £1 (lM A

The bending moment M al a distance x from A is M = - /11" -

and

iL~

( 11.81 )

derivative with respect 10 M" is

aM

-

ilM~

714

!w?

= - 1

(11.82)

Since the dummy couple was counterclockwise. the posi ti ve sign indicates thai the angle {J A is also L'Ourllerclockwise:

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EXAMPLE 11 .14 A load P is supported at B by two rods of the same material and of the ~ame cross·sectional area A (Fig. I 1.47). Delemline the horizontal and vertical deflection of point B.

Differentiating these expressions with respect write ---.!!!!. = 0.6

OQ

'Q

f

Of'

c

~ = O.6 aP

o

f

F"it

/J

Fig. 11 .48

Fig. 11 .49

We apply a dummy horizontal load Q at B (Fig. 11.48). From Castigliano's (heorem we have

Substituting from ( 11.85) and (11.86) into both (11.83) and ( 11.84). making Q = O. and noting Ihal BC = 0.6/ and lJU = 0.8/. we obtain the horizontal and vertical deflections of pOInt lJ under the given load P:

(0.6/')(0.6/) Recalling from Sec. 11.4 (he expression (II [4) for Ihe strain energy of a rod. we write

FidBC)

U=

----::v:E" +

ap

Q

1

Fig. 11.47

( 11.86)

aFeu = - 0.8

8

I

D

Q and p. we

'F

aree = 0.8

c

(0

),"e =

=

Fb (BD) 2A E

where FlIC and Fyu represent Ihe forces in BC and BD , respectively. We have, therefore.

>'e =

AE

(0.8 )

+

. (0.6)

+

(- O.8/')(O.RI) At:

(0.6)

- 0.096 .fl.. AE (0.61')(0.61) A£

(- 0.RP)(0.81) AE

(- 0.8 )

(11.83 ) Referring to (he directions of the loads Q and p. we couclude that

and (11.84)

PI

,(n

= 0.0')6 A£ (-

PI

" , = 0.728 ~ 1 . A£

From the free-hody diagmm of pin B (Fig. 11.49). we obtain

r'BC =

0.61'

+ 0.8Q

Feu = - 0.81'

+ 0.6Q

(11.85)

We check that the expression obtained for the vertical deflec· lion of lJ is lhe saille that was fo und in Ex::.mple. 11.09.

715

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7 16

Energy

Mothods

' 11.14. STATICALLY INDETERM INATE STRUCTURES

The reactions at the supports of a statically indetermi nate clastic struclUre can be delemlined by Castigliano's theorem. In the case of a strw.:lUre indetenninate to the lirst degree, fu r example. we designate one of the reactions as redundant and eliminate or modify accordingly the correspondi ng support. The redundant reaction is then trented as an UI1known load that. tOgether with the other loads, must produce deformations thai are compatible with the ori ginal supports. We lirsl caku late the strain energy U of the structure due. to the combined action ()f the given loads and the redundant reactioll. Observing that the partial derivative of U with respect to the redundant reaction represents the dellectiol1 (or slope) at the support that has been eliminated or modified. we thell set this deri vative equal to zero and so lve the equation obtained lor the redundant reactiol1.t The remaining reactions can be obtained from the equations of statics. t Thi.1 i~ in tlK> ca~ of n rigid $uppon allowing no dctlcction. For othcr types of ~uppon. the panial den\'atil"e of U ,hould be sct ~u81 to the allowed dcflection.

EXAMPLE 11.15 Detennine the reactions at ule supports for the prismalk beam and loading shown (Fig. 11.50). TIle beam is sllIIically indelemlinale 10 the nrst degree. We consider the reaction al A as redundant and release Ihe beam from Ihal support. The reaction RA is now considered as an unknown load (Fig. 11.51) and will be determined from the cond ition dlat th e deflection YA at A mu~t be zero. B y Castigliano's theorem YA = fjUla NA, whe re U is Ihe strain energy of die bt-am under Ihe distributed load and the redundanl reaction. Recall ing Eq. (l t.70), we write \. "" aU = . ,\ aR"

IL !:!... aM dx () £1 ARA

Fig. 1150

(11.87)

We now e.~press th e bending momenl M for the loading of Fig. 11.51. The bending moment al a distance x from A is (11.88) lind its derivative with respect 10 RA i~

ft ,

Fig. 11 .51

YA= ~{ [(R*r - ~WX1)dX= ~J(R.;Ll _'1l!8L~) Setting)'A = 0 and solving for NA , we have

(11.89) Substituting for M and r1M/<1RA from (11.88) and ( 11.89) into (11.87). we write

From the conditions of equilibrium for the beam. we lind that the reaclion at B consisls of the following force and couple:

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EXAMPLE 11.16 A load P is supponed at B by three rods of the same material and the same cros.~-sectional area A (Fig. J 1.52). Detemline the force in each rod. The structure is statically indeterminate to the llrst degree. We consider the reaction at H as redundant and release rod BH from its support at H . The reaction Ru is now considered a~ an unknown load (Fig. 11.53) and will be detemlined from the condition that the deflection )"u of point H must be zero. By Cllstig liano's theore m ."H =:0 fJU/iIR/I ' where U is th .. strain en ('rgy of the three-rod system under the load P and the redllndanl reaction R/I' RecllJJing Eq. (J 1.72). we write

(11.90)

r

o /I

c

0••'>1 lUll

r:

Fig. 11 .52

We note IImt the force in rod BH is equal to Rtf and write ( 11.91 )

Then, from the obtain

f~e-body

diagram of pin B (Fig. J 1.54). we FBI) - 0.8R/I - O.8P

FBI: - 0.61' - O.6R II

11

~

(11.92)

Differentiating with respect to Nil the force in each rod. we wnte aF BD

- - = 0.8

r:

( 11.93)

(IN/I

Subslilllting from (11.91 ). ( 11.92), and ( 11.93) into ( 11.90), and noting that the lengths Be. BD, and BH arc. re~pectivel y.

equal to 0.6/. 0.8!. and 0.5/. we write

I

YII = AE [ (0.6P - 0.6R/I)(O.!l!)( - 0.6)

+ (0 .8R/I St!uing

)"11 =

Fig. 11.53

- 0.81')(0.81) (0.8 )

+ R/I(O.5f)( I))

O. we obtain

1.228R/I - n.72Sp = 0 and. solving for Rlf. Rlf = O.S93P

Carrying thi ~ val ue into Eqs. (J 1.91) and (11.92). we obtain the force s in the three rods: F Bc = + 0.2441'

FBI)

= - 0.3261'

F B/I =

Fig. 11.54

+0.593 /~

717

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W\:Nl

P=

~-L4-~

____~E

~

__ ~IB~lID0100J , , , ' 0_6",

SAMPLE PROBLEM 11 .5

- - 1.5",

1

For the truss and loading of Sample Prob. 11.-1. determine the \lenient deflection of joint C.

J

SOLUTION Cast igliallo's Tlu.'orcm. Sillce no vertical load is applied al joint C. we intnxluce the dummy load Q as shown. Using Castigliano's theorem. and denot ing by Fi the force in a given member i caused by the combined loading of P and Q , we have, since £ = constant. (I)

Force in Me mbers. Considering in sequence the equilibrium of joints E. C. B. and D. we determine the force in each member caused by load Q. Jui,,! J)

'Q1f7 ' t

I' 0 .8111

-'-- .!Q ,

Q3 ~

B

,~

H n.6

Member

A8 AC

AD BD CD CE DE

4

E

Joint E:

F a; = PDF. =

0

Joint C:

FAC = O:Fcv

=- Q

Joinl B:

FilS = 0: FeD =

0 + 15P/ 8 + 5P14 + 5Q I4 - 21P/8 - 3Q I4

-Q T 15 1'/8 - 17P18

AV\ ,

.

F CI>~Q

1

"

I'CI) ~

,

-iQ

-

Ql\\F U,=}9 ,

[)

"

F ~D ~ ~('

Fnu=!"<>

III

F,

:J '

F '

V

Fme{' tri""gie

The force in each member caused by the load P was previously found in Sample Prob. 11.4. The lotal force in each member under the combined action of Q and P is shown in the following lable. Fomling ()F/ aQ for each member. we then compute (FjL/ A,)(r1F/.JQ) M indicated in the table.

(F,AIL ,)"', i/O

L/, m

AJ, m2

, • - 1

0.8 0 .6 1.0 0.6 0.8

500 X 10 500 X 1O - ~ 500 X 10 " 1000 X IO "

0 0

1.5 1.7

1000 X 10 "

aF,!aQ

0 0

_l

"

0 0 +3 J25P +3 125Q + I 18.lP + 338Q + 800Q 0 0

lO00 X I O ~ 500xlO~

F,I.) :...... 'F= 2 (--' A aQ

4306P

+ 4263Q

J

Dent'clion

or C.

)'

("

Substituting into Eq. (1). we have I ,, (PILj ) aFi I -- = - (4306P A, aQ E

= - £..J E

+ 4263Q)

Since Ihe load Q is not pan of the original loading. we set Q = n. Substituting the given .hla. P "" 40 kN and £ = 73 G l'a. we lind )'r

718

=

4306 (40 X 101 N) 73 X JO~ ra

\'~' ~ 2.36

tllll l !

....

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SAMPLE PROBLEM 11.6

I' ~ Ui k[l~lfl

IIIIIIII ~

X ~f "~~'5n D

For th e beam and

£-29 X I06 p,i.

loadi~g

shown. detenlline the

ddl~tion

at point D. r ,e

8

a = .t." I

L - t:!r;

Q

SOLUTION ClL'itigliall u's Th ..·tm .:m . Sin(!t: tht: giv~n luadi'l/,; Uu.::s nUl indude a v~r~ mint IJ Wf'. intmd,lCf'. til f'. nllmmy IO!lli Q .1 ~ s hown _ I I~i n g ('a~_ ligliano's theorem and noting th3t EI is constant. we wri te t; c ~1 IC1~n ~t

"

Yv

=

f ~(:)dX

=

i f J/G~)dX

,I)

The integration will be performoo separJtel) for portions AD and DB.

1---- \ ----I

Using !he free,body

KeaClloR.f .

Portioll AD 0/ Bellm,

diagr~m

of the entire beam. we lind

wb(u + J:b) R o= L

wb 2 b R ~ -+ Q - l .1 2L L

a ,. 1

+ Qi.

Uging the free body ghQwn , we lind

Substituting into Eq. (1) and integrating from A to D

r

gi~es

~ J ,w, ~~' dx ~ ~ RxX(~) dr = ~~/~

""t'" ",

<

K,

U j,)",

- (x ,:,,;, ,, )

'

We substi tute for Rio and then set the dummy load Q equal to zero. -i

£1

f

j

j

.111, 1IJu b M - J.x = -2 1

oQ

(2)

6£lL

PonilJ/I DB uJ DetllR. Using the free bCKiy shown, we find that the bending mmnellt at a distance 'V from end B is aM 2 =

+ (H'

'Q

L

Substituting into Eq . ( I) and integrming from point B where 'v = O. to POlllt D where v '" b. we write iJM2 d ll = ~ ('" (RV _ E{ aQ EI Jo R Sllbsti tuting for RB and .setting Q = D.

-'- J' M,

-'- f HI

' 1_aM l d." = [1Oh((/ + ~b) " "au " L

DejleflilJlI )'J)

(It

Point D.

WUZ)("v) dv = 2

L

1 13t:JL ab _ wah"' 'IlEIL

Re~alling

= 5(/b

4

RII/b ) _ wab 3EIL &£IL

4

+ ab

24EfL'

s If

<3)

Eqs. (I). (2), and (3j. we have

, 'wab l , wub j wahl = 24EIL2 (4 a- + 5ab + iJ' ) = 24£IL~ (4a + b)(a + b) = 24£lL ('Ia + b)

From Appendix C we find that I = 68.9in 4 for a W[ O x 15. Substituting for 1, 1O. a. h. and L their nllmerical values, we obtain l'C -

'1.262 in. ~ ..

719

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A

11 11111 11 1111C

SAMPLE PROBLEM 11 .7 For the uniform beam and loading shown. determine the reactions al the supports.

SOLUTION Casligliano's Theorem. The beam is indetemlina!e (0 the first degree and we choose the reaction RA as redundant. Using Casligliano's theorem. we determine the deneclion at A due to the combined action of RlI and the djs~ tributed load Since 1:."1 is constant. we write }'11

~ f ~e~)dX = ~I JM;~ dx

( 1)

The integwtion will be performed separately for portions AD and BC of Ihe beam. Finally, RA is obtained by setting YA equal 10 zero.

III IIII I I II

A

R,

Free Bod}': Elltire Belli/I . of Ril and the dislribmed load C

Re = ~wL - 3R"

f.,

' IR"

We express Ihe reactions at Band C in lemlS

Port i(}/! AD

He = 2R" - hoL

(2)

II! 8e(lll/ . Using Ihe free.bo:!y diagram shown. we find

I---L- I -- ~---J

Substituting into Eqo (1) and integrating from A to B. we ha\'e

From A 108 '~J

) I

I

~

t-"i t Illl}',

£/

<: t

M

fj

~

I "~

.!!.

2

EI

f MzaR,., aM"

L (

0----1

< /0 )

1:."1

=

du =

Jo

(

EI

2

C - I

I •

'I)

= 0 , to

3)

(U2

6

(4R v 2 ,.,

-

~wLv~ -

2

B. where

WV1) do

16

64

1:'1

Rellclloll a t A . Adding the expressions obtamed termine \,., and set II equal to zero

, =-.!...(R'ILl _ 1I)L4)+~( R ILl EI 3 8 EI 6

I,.,

0_ "2

Solving for R.1 • Rellctions at Band C.

13 R,., = 32 wL

6

In

64

4 ( )

(3) and (4). we de-

_ 5WL4)~O 64

n

R", = 32 wL 1

...

Substituting for RA into Eqs. (2). we obtain

wL

RC= 16 t

720

(3)

wo' -11)L v - 4 2

A

EIJu

1:."1

R " ~Il _l id

8

We have

2R -

~

3

~ ~(R,L' - wL' - WL') ~ ~(R,L' - 5WL')

I

V.

~ (L(R~ _ W~) JX = ~(R,.,L] _ 1I)L')

Substituting into Eq. ( I) and integrating from C. where u=t Lwehave

)

'0' ! \l~ (t te

F mm C 1<>

dx =

ORA

I

Portio" Be of Berllll.

R ,L ,_Y (x

f M aM

...

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PROBLEMS

11.77 through 11 . 79 Using the infonnali o n 1I1 Appendix D. campute the work of the loads as they are applied 10 th.e beam (a) if the load P is applied firsl, (b) if the couple M is applied first. p

~

c

A

4-

U2

Fig. PII .n

-I --u, - -

'[ 1 Fig. P11 .78

1 1-

1..12-

Fig. Pl1 .79

,I' 0 J 1.12 - 1

f 1.80 through 11 .82 For the beam !lnd loading ~ho...m. (,,) compute the work of the loads as Ihey are applied successively 10 tile beam. usin): the information provided in Appendix D. (b) compUl~ the simin ~nergy of the beam by the method of Sec. It.4 and show 1111t it Is equal 10 the work obtained in pan {/

J

y 1-- 1.0 Fig. P11 ,80

Fig. Plt.a!

11.53 and 11 .84 d"flecti()n of point D .

-J-

1; U'- ---j

Fig . P l t .82

For the prismatIc beam sho\\n. delcmline Ihe

r

"

111 111 111 111

AL =~ IJIJ I ~B

A

Fig. Pl1.83 and P11.85

Fig. P 11.84 ami Pl l .86

U2 ~ U2 -j

L ", j -Inj

11 .85 and 11 ,86 Fer the prismatic beam shown. determine the slope at point D. 11 .87

For Ihe prismatic beam shown. detennine thc slope at point B. M"

~=--",= C ~ql'

LV2l-In]

Fig. P11.87

721

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722

11.88 and 11 .89 tion at point D.

Energy Methods

*.

For the prismatic beam shown. detemline the denec-

11i'1 11

~ J.:- Ll2 -I-Ul ~" E

D

A~l~ /J ~ E Ii£jB

J [

[ /R

Fig. P11.88 and P" .90

Ll2

U1J

Fig. P11.89 and P11 .91 11.90 and 11 .91

For the prismatic beam shown. determine the slope

at point D.

11 .92 For the beam and loading shown . determine the dencction Of point A. Use E = 29 X 106 psi.

r

, 1.5ki PS

5 kip'

A L

8

X

C[ --.~ Il--l

ws x

1

5fl -

-

13

Fig. P11.92 and P11.93

11.93 For the beam and 10;Jding shown. determine the denection of point B. Use E = 29 X 106 psi. 11 .94 and 11 .95 For the beam and loading shown. determine the denection at point B. Use E = 200 G Pa. H~()'n'"

"I

so"'''' ~

W2S()

x

22.3

Ag. P11.95

\\"310 X 74

A~_='C'' :l-='\"

1."-

2.4 '"

-L '" '"

X

Use

11 .96 For the beam and loading shown. detenlline the slope at end A. 200 G Pa.

t: =

11.97 For the beam and loading shown. determine the denection point C. Use E = 29 X 106 psi.

a1

1----4.8,,,

Fig. P11 .96

i41i J'B

6/. -1..:"

ss x

IS4

Fig. PH.97and P11.98

11 .98 For the beam and loading shown. determine the slope at end A. Use E = 200 G Pa.

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Problem.

11.99 and 11.100 Each memlxL of the lru5~ shown is mad:ing £ = 200 GPa. detemline the de nection indicated. 1 1.99 VeniC'dl d~nection ofjoinl C. 11. 100 Horizontal denection of joint C.

7.S kip"

c

1.5

~ -' ' ~~II ~

ill

L

[ 4ft

2m Fig. Pl1.99 and PlUM

11 .101 and 11 .102 Each member of Ih~ truss ..hown is made of steel has Ihe cross-sectiolt3.l Mea shown. Using E -'"' 29 X IO~ psi, delennine the defl ec ti on indicated. 11 .101 Venlcal dcnection of joint C. 11.102 Horizontal dencclion of joint C. 11 .103 and 1 I . 104 Each member of the tru~~ shown i ~ made of steel :md h:l.~ a cross-sectional area of 500 mm 2 • Using E '" 200 GPa. delemlille Ihe den~.cti(\1t indicated . 11.103 Vt:rtica l lJcn ~cLiun uf juint B. 11.104 Horizontal deflection of joint B.

L""'J~ ] ,2

,

,

-I IlL

B

1,2 m

c l-------':~~

akN

tJ

~ 2_511l ---

Fig. P11.103 and P11 .f04

B

50

Fig. Pll.101 and Pl1.102

~nd

[

'~

-

III

Fig. P11 .1OS

11 .105 For the beam and loading shuwn and using Casligl iano's theorem. detennine (u) the horizontal de n~clion of point B. (b) the venical delleclion of point B. 11 .106 For Ihe unifoffil rod :lnd loading shown and using Casligliano's Iheorem. delennine the Jellecliun uf poinl B.

P

Rg. P11.106

3 751\

723

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724

11 .107 'fu-o roos AS altd BC of the ~all;e flexulal rigidity EI are welded together at B. For Ihe loading shown. detemtine (II) the denection of poi hi C, (b j the slope of m~mber Be al point C.

Energy MClhoca

Fig. Pl1.1 07

11.108 A U
L

IB

r /J "

c

I

1 Fig. PH.1OB and Fll.109

~.~==-I H L ,_ ~r. "Iy. PII . I' O

11 . '09 A uniform rod of flexural rigidity £1 is ~nt and loaded w; shown Detennine (tl) the vertical deflection of poim D. tb) the slope of BC at poinl C

11. 110 A uniforn! rod of flexural rigidity £1 is ~m and load~d a.> shown. Delennine (II) Ihe venica l deflection of poinl A . (b) !he horizontal deneclion uC puilll A

11 .111 through 11 .114

Delermine the reaction
and draw the bending.momenl diagram tor the beam and loading shown. p

B

C-Ifl- -LU2-

Fig. Pl1.lll

Fig. Pl1.112

;t;,

~ 111,l

- - U2

FIg. P11 .113

Fig. Pll.114

1.J2~

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Probtems

11. 115 Determine the reaction at the roller suppon and draw the bending-moment diagram for the beam and load shown.

;S -~ II III ~ ~C J:U2

L------I

Fig. P1 1.1 t 6

Fig. P" .1'S

11 .116 For the unifonn beam and loading shown, detennine the reaction at each support. 11 .117 through 11.120 Three members of the same material and same cross-sectional area are used to suppon the load P. Detemline the force in member Be.

I

,

C

D[

/)

1

3Cf

~I

L

E B p

--

I

e

"

1--,

1--1

P

Fig. Pt 1.1 17

Fig. Pt1 .1tB

C

~ "

E

,

/)

45" E

P

I'

Fig. Pt1.tt 9

Fig. P" .' 20

11.121 and 11.122 Knowing that the eight members of the indeterminate truss shown have the same uniform cross-sect ional area. determine the force in member AB.

,

p

,

l!

A

It C

I

G ~,------':I

Fig. P11.121

c

f'

I!'--,

E I'

Fi g. P11.1 22

725

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REVIEW AND SUMMARY FOR CHAPTER 11

This chapter was devoted to the study of strain energy and to thc ways in which it can be used 10 dete rmine Ihc stresses and deformations in structures subjected to both static and impact loadi ngs. In Sec. 11.2 we considered a unifo rm rod subjected to a slow ly increasing axial load P (Fig [ 1. 1). We noted Ihat the area under thc

c Fig. 11.3

Fig. 11 .1

Strain energy

load-deformation diagram (Fig 11.3) represents thc work done by P. Thi s work is equal to the .I"lmin energy o f thc rod assoc iated with thc deformation caused by thc load P: Strain energy = U = {" P dx

( 11 .2)

o

Strain-energy density

Since the stress is unifann throughout the rod, we were able to di vide the strain energy by thc volume of the rod and obtain the strain energy per unit volume. which we defined as the strain-ellergv density of the material [Sec. 11.31. We found that Strain-energy density

~ II

=

Lr'

a~ d€..\

( 11 .4 )

o

O<-____"."L -__~.L , __________-; Fig. 11.6 Modulus of toughness

726

and noted that the strdin-energy densi ty is equal to the area under the stress-strain diagram of the material (Fig . 11.6). As we saw in Sec. 11.4. Eq. (11.4) re mains valid when the stresses are not unifonnl y distributed. but the strain-energy density will then vary from point to point. If the material is unloaded. the re is a permanent strain €.p and onl y the strain-energy density correspondi ng to the tri angu lar area is recovered. the remainder of the energy having been dissipated in the lonn o f heat during the deformation of the material. The area under the entire stress-strain diagram was defined as the modulus of toughness and is a measure of the total energy that can be acquired by the material.

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If the normal stress u remains within the pro portio nal limit of the materi al, the strain-energy density II is expressed as

Review and Summary for Chapter t 1

a' 1/ :

-

2£

The area under the stress-strain curve from zero strain to the strain Er at yield (Fi g. 11.9) is referred to as the " lOl lll/llS of resilience of the materi al and represents the energy per unit vol ume that the material can absorb without yielding. We wrote

Mooulus of resilience

,

a"

Ily

= 2E

( II ,8)

In Sec. 11.4 we considered the strain energy associated with I!ormal stresse.I·. We saw that if a rod of length L and varitlhle emssj·ectiol!al area A is subjected at its end to a centric axial load P, the strain energy of the rod is

u ~ fio

p1 dx

( 11 13)

o 2A E

If the rod is of

Ill/ifO ml

eros.l· uctiol! of area A , the strai n energy is

Strain energy under axial load

( II ,14) We saw that for a beam subjected to transverse loads (Fig. 11.1 5) the strain energy assoc iated with the normal stresses is U ~

' M'

J -2E1d x

( 11,1 7)

IJ

where M is the bending moment and Elth e fl exural rigidity o f the beam. The strain energy associated with shearing stre.ue.l· was considered in Sec. 11.5. We found that the strain-energy density for a material in pure shear is

T;' 1/ =

-

2G

Strain energy due to bending

mm : lmm Fig. 11.15

Strain energy due to shearing stresses

(1 1.1 9)

where T no is the shearing stress and G the mod ul us of rigidity of the material. Strain energy due to torsion

For 11 shaft of length L and uniform cross section subjected at its ends to coupl es of magnitude T (Fi g. 11.1 9) the strain energy was found to be ( tl .22)

whcre J is the polar moment of inertia of the cross-sectional area of the shaft.

T

Fig. 11 .19

727

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728

Energy Methods

General state of stress

Ln Sec. 11.6 we considered the strain energy of an elastic isotropic material under a general state of stress and expressed the strai ne nergy density at a given point in terms of the principal stresses U a • Ub, and (T e at that point:

The strain-energy density at a given point was divided into two parts: /( .', associated with a change in volume of the material attha! point, and Ud, associated with a distortion of the material at the same poin!. We wrote /I = II ,. + lid , where (11.32)

and

Using the expression obtained for lid, we derived the maximumdistortion-energy criterion, which was used in Sec. 7.7 10 predict whether a ductile material would yield under a given slate of plane stress. Impact loading

Equivalent static load

Ln Sec. 11.7 we considered the impact loading of an elastic structure being hit by a mass moving with a given velocity. We assumed that the kinetic energy of the mass is transferred entirely to the structure and defin ed the equivalent static load as the load that wou ld cause the same deformations and stresses as are caused by the impact loading.

After discussi ng several examples. we noted that a structure designed to withstand effectively an impact load should be shaped in such a way that stresses are evenly distributed throughout the structure, and that the material used shou ld have a low modulus of elasticity and a hi gh yield strength rSec. 11.81. Members subjected to a single load

P'1

L~I[

",[ ~ A

Fig. 11 .27

~

The strain energy of stmctural members subjected to a sillgle load was considered in Sec. 11.9. Ln the case of the beam and loading of Fig. 11.27 we fou nd that the strain energy of the beam is ( 11.46)

Observing that the work done by the load P is equal to tPtJ'l> we equated the work of the load and the strain energy of the beam and determined the dellection )'1 at the point of application of the load [Sec. 11.10 and Example 11.101.

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Tile metllodjust described is of li mited value. since it is restricted to structures subjected to a single concentrated load and 10 the determination of the deflection at the point of application of that load. In the remai ning sections of the chapter. we presented a more general method. \\-hich can be used to determine deflections at various points of structures subjected to several loads. In Sec. 11.11 we di sc ussed the strain energy of a structure subjected to several loads. and in Sec . 11.1 2 introduced Castigliano's theorem, which states that the deflection Xj' of the point of application of a load Pj measured along the ll ne of act ion of Pj is equal to the partial derivative of the strain energy of the structure with respect to the load Pi' We wrote

Review and Summary for Chapter t t

Castiglfano's theorem

(11.65)

We also found that we could use Castigliano's theorem to determine the slope of a beam at the point of application of a couple MI by writing

o =~ I

( 11.68)

aM,

and the angle OfTWi.\·T in a section of a shaft where a torque TJ is appl ied by writing (1 1.69 )

Ln Sec. 11.13, Castigliano's theorem was applied to the determination of deflections and slopes at various points of a given structure. The use of "d ummy" loads enabled us to include points where no actual load was applied. We also observed that the calc ulation of a deflection Xi was simplified if the dilTerentiation with respect to the load Pj was carried out before the inlegration. I.n the case of a beam, recalling Eg. (1 1.17), we wrote

au Xj

= iJP = j

IL M

aM

0 £ 1 (JP dx

(11.70)

j

Simi larl y, for a truss consisting of n members. the defl ection x; at the point of application of the load Pj was found by writing (11.72)

The chapter concluded [Sec. Il.14[ with the application of Castigliano's theorem 10 the analysis of statically indeterminate .rln/("tltres [Sample Prob. 11.7. Examples 11.15 and 11.1 61.

Indeterminate structures

729

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REVIEW PROBLEMS

11 .123 Roo AB is made of a steel for which the yield strength is = 65 ksi and the modulus of elasticity is E = 29 x IO~ psi. Know ing that a strain energy of 60 in . Ib must be acquired by the rod as the axial load P is

U y

.1

applied, determine the factor of safclY of the rod with respect to perm,ment deformation.

~ -;'L dian"'! (,f

t -itLdbn"'tcr

11 111 111 111 11

<

B

A

4li in~

p

I - -- - L

- --
Fig. P11.124

Fig. P1 1.123

11. 124 Taking inlo account only the effect of normal stresses. determine T.\

= 3I1U N

the strain energ y of the prismatic beam AB for the loading shown.

III

11 .1 25 T~

= hill K ,,1

11. 126 A single 6-mm-diameter steel pin B is used to connect the stee l strip DE to two aluminum strips. each of 20-mm width ilnd 5-mm thickness. The modulus of elnsticity is 200 GI\, for the steel and 70 GPa for the aluminum. Knowing that for the pin ::It B the ::Illowable shearing stress is fTall = 85 MPa. determine. for the loading shown. the maxi mum strain energy that can be acquired by the assembled strips.

Fig. P11.12S

E

- , _.....~

1 . 25rn~ Fig. P1' .126

In the assembly shown torques T", and T8 are exened on disks

A and B, res pecti vel y. Knowing Ihm both Sh3fls are solid and made of aluminum lG = 73 GPa ). determine the total strain energy acquired by the assembly.

I'

I

.- ~ Llr

1.-3.,"-1 Fig. P11 .127

11 .12 7 The cy lindrical block E has a sJ?Ced Vu = J 6 ftls when it strikes squarely the yoke BD that is attached to the s-in.-diameter rods AIJ and CD . Knowing that the rods ::Ire made of a steel for which i f )' = SO ksi and E = 29 X 10" psi. detemline the weight of block E for which the factor of safet y is fi ve with respect to pemlanent deformation of the rods.

730

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11.128 A block of weight IV is placed in contact with a beam at some given point D and released. Show that the resulTing maximum dellection at point D is twice a~ large as the dellection due 10 a statk load IV applied at D. 11.129 The 12,mm·diameter steel roo ABC has been bent into the shape shown. Kn owing that E = 200 GPa and G = 77.2 G Pa. determine the dellection of end C caused by the 150·N force.

Revtew Problems

I

lJ 0.2-;1}. ,liam!,!pr

25;",

l~-===::::J: 1_1_ 8

_

30;"

\0 ill

Fig. P11.129

-J

Fig. P11 .130

11. 130 The steel bar AIlC has a square cross section of side 0.75 in. and is subjected to a 50-1b load ('. Using E = 29 X 10" psi. detemline the detlection of point C.

11 .131 E
L

FIg. P11.131

11.132 A tJisk of radius a has been welded to end B of the solid steel shaft AB. A cable is then wrapped around the disk and a ~'ertical force P is applied to end C of the cable. Knowing that the radius of the shaft is r and neglecting the deformations of the disk and of the cable. show that the deflection of point C caused by the application of P is

8

c

PL'(

~ -

3£1

J

Fig. P11.132

E?)

+- 1.5 GL1

11. 133 Forthe prismatic beam shown, detennine the detlection of point D.

p

B

C

A

D

,1 FIg. P11.133

11 .134 'l1uee rods. each of the s ame flexural rigidity £1. are welded to form the frame ABCD. For the loading shown. detemline the angle fonned by the frame at point D.

l

L L- J Fig. P11.134

731

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COMPUTER PROBLEMS

The followin g problems an' designi'd to be soh'cd with a compull'r. 11 .C1 A rod consisting of 11 elements, each of which is homogeneous and of uniform cross section. is subjected to a load P applied at its free end. The length of element i is denoted by Li and its diameter by d,. (al Denoting by £ the modulus of elasticity of the material used in the rod. write a computer program that can be used to determine the strain energy acquired by the rod and the deformation measured at its free end. (b) Use this program to deter· mine the strain energy and deformnlion for the rods of Probs. I L9 and II.JO.

~

Ek,,"'nt I

(

6 in

Elf'",,'n! I

(

W8X IS

" FIg. P11.C1

x

rll) i",

- --1-- - r~J in,

Fig. Ptt .C2

11 ,C2 Two 0.75 X (j..in. cover plme, are welded (() a W8 X 18 rolledsteel beam as shown. The ISOO-lb block is te be dropped from;) height II = 2 in. onlD the bearn. (a) Write II computer progmm to caJcubte the maximum normal stress on transverse sections just to the left of D and at the center of the beam for values of II from 0 to 60 in. using 5-in. increments. (b) From the values considered in pan a, select ule distance (J for which the maximum normal stress is as small as jXlssible, Use E = 29 X JOb p~i. 11 .C3 11le In-kg block D is dropped from a height It onto the free end of the ~teel bar AH. For the steel used (r... = 120 MPa and £ = 200 Gp.-a. (£I) Write a computer program to calculate lhe mllximum allowable height II for values of the length L from 100 mm to 1.2 Ill. using lOO-mm increments. (b) From the value..<; considered in pan a. seiectthe length corresponding to the largest allowable height.

~24"'111 Ii

Fig. Ptt .C3

732

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'1 .C4 The block D of IllU SS //I ~ 8 kg is dropped from a height It = 750 mm onto the rolled-steel be:tm AB. Knowing that E = 200 G Pa, write a computer program 10 culculatc the maximum denectioll of poim E and the maximum normal stress in the beam for values of (/ from JOO to 900 m. using loo·mm increments.

\\,150

x

1.3.5

Fig. P11 .C4

IT)"

11 .CS The sleel rods AB and BC are made of a steel for which = 300 MPa and E = 21Xl Gl'a. (u) Write a computer program 10 calculate for

\"alue.~ of 1I from 0106 m. using I·m increments. the. maximum slrain energy Ihat can be acquired by Ihe assembly withoul c
~ 10·""" dbn,,"l ~c

"t:~"·~ ~~~'..~/~".'~ "~""~d~hln"'tl'r '''' ~ Fig. P11.CS

11 .C6 A 160·1b diver jumps from a height of 20 in. ontu end C of a div· ing board having the uniform cross section shown. Write a computer program to calculate for values of 11 rrom 10 to 50 in., using IO-in. increments. (a) the maximum deflection of point C, (b) Ihe ma."(imum bending moment in the board, (c) the equivalent 5Iatic load. Assume that the di ve r's legs remain rigid and use £ = 1.8 X lOb psi.

1.-" I----l2fl

Fig. P11.C6

16;1\

Computer

Problems

733

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Appendices

APPENDIX A

Moments of Areas

APPENDIX B

Typical Properties of Sclcctt.'d Materials

736

Used i ll ElI glllCcrill);;

746

APPENDIX C

Propemes 01 Kolled-Steel Shapes,

750

APPENDIX 0

Beam Dellections and Slopes

762

APPENDIX E

Fundamentals of Engineering Examination

763

735

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A

P

PEN

0

I

X

A Moments of Areas

A.1. FIRST MOMENT OF AN AREA; CENTROID OF AN AREA

"---1 A.

dA

Consider an area A located in thc x )' plane (Fig. A.I). Denoting by x and)' the coordinates of an element of area dA. we dcline thcftrxi moment (~f 'he area A lVilh re.lpeci to the x axis as the integral

"I!I

(A. ] )

()

Fig. A. !

Simi larly, the first mOll/ent oj the area A with reJpec/ to The J' defined as the integral

Q,. ~

r ,

UlA

(Lri~'

is

(A.2)

We nOll' that each of these integrals may be positive, negative. or zero, depending on the position of the coordinate axes. [f SI uniL~ arc used , the tirst moments Q. and Qy arc expressed in m) or mm' : if U.S. customary units arc used. they arc expressed in fl 3 or in 3. Thc centlVitl of the area A is delined as the point C of coordinate... and y (Fig. A.2). which satisfy thc rdations

x

f, Fig. A.2

x dA

=Ax

,

(A.3)

Comparing EelS. (A.l) and (A.2) with Eqs. (A.3). we note that the first moments of the area A can be expressed as the products of the area and of the coordinates of its centroid :

Q. = Ay

736

J y dA = Ay

(A.4)

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When an area possesses all lLri.f of symmeli}'. the tirst moment of the area with respect to that axis is "Zero. Indeed, considering the area A of Fig. A.3, which is symmetric wi th respect to the y axis. we observe that to every element of area dA of abscissa x corresponds an element of area dA I of abscissa - x . It follows that the integral in Eq . (A .2) js "Zero and , thus. that Q). = O. Lt also follow s from the fi rst of the relat ions (A3) that = O. Thus. if an area A possesses an axis of symmetry. its centroid C is located on that axis.

A..I First Moment 01 an Area

,

x

,\

o

.,

Fig. A.3 I

,,' Fig. A.4

Since a rectangle possesses two axes o f sy mmetry (Fig. A.4a), the centroid C of a rectangular area coi ncides with its geometric center. Similarly, the centroid of a circular area coincides with the center of the circle (Fig. A.4hJ. When an area possesses a cell/a of symmeli)" 0 , the tirst moment of the an.~a abo ut any axis through 0 is zero. Indeed. considering the area A of Fi g. A.5. we observe that to every clement o f area dA of coordinates x and )' corresponds an e lement of area ciA I o[ coordinates - x and - y . It follows that the integrals in Eqs. (A. I) and (A.2) are both zero, and that Q, = Q,. = 0. It also follows from Eqs. (A.3) that = y == 0, that is, the centroid of the area coincides with its center of symmetry. When the centroid C of an area can be located by symmetry, the tirst moment of that area with respect to any given axis can be readily obtained from Eqs. (A.4). For example, in the case of the rectangular area of Fig. A.6. we have

x

Fig. A.S

Q,. = Ay "" (hII)(~h) = ~hh2 and

In most cases. Ix)wever. it is necessary to perfonn the illlegrat ions indicated in El.js. (A. I) through (A.3) IQ determine the tirst moments and the cent roid of a given area. While each of the illlegrals involved is actually a double illlegraL it is possible in many applications to select elements of area {/A in the shape of thin horizontal or lIenical strips. and thus to reduce the computations to integrations in a single variable. This is illustrated in Example A.OI. Celllroids of common geometric shapes are indicated in a table inside the back cover of this book.

F1g. A.6

737

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EXAMPLE A.01

,

For the trian gular are:! of Fig. A.7. determine (a) the fIrst moment Q, of the area with respect to the x axis. (b) the ordinate .Ii of the centroi d of the area.

/\1

~ L,

(a) First Moment Q... \lie select as an element or" area a hori zontal ~ trip of length /I and thickness d.\'. and note that all the poiJlt~ within the elemelll an: at the same dis tance J' from the x axis (Fi g. A.8 ). From similar triangles. we ha\'e

1-----" ---I Fig . A.7

lI=b " - -J

"

dA = lid) =

h- ,

b --~

/,

d)·

The rirst moment of the area with respect to the x a.{ls is Q~. =

=

f.

y dA=

f"" /, -)

b

ylJ-,,-dy=-;;

il"

~["i -3 I! 2

f" ,

(h)' - .r-)dy

"

~

(b) Ordinate of Centroid. Recalling th e fi rst of Eqs. (A.4J and observing that A = ~bll, we have

Q.

=

Ay

Fig. A.S

tbh" = (~bh)y

y = jll

A.'2. DETERMINATION OF THE FIRST MOMENT AND

CENTROID OF A COMPOSITE AREA Conside r an area A , such as the tr:!p,-,zOldal area show n in Fig. A.9,

r

0'-------'----'

which may be divided into simple geometric .~hapes, As we saw in the preceding section. the lirst moment QJ of the area with respect to (hex aXIs is represented by the mtegra l f y dA , whIch extends over the e n tin: area A Di viding A into its component part. AI' Az, A j , we write Qx =

I)' dA = fYdA + IYdA + ,I

,I

.1 ,

jYdA .1 ,

or. recalling the second of Eqs. (A.3).

0'------'-------' - , FIg. A.9

738

where "I. rl' and v~ represenl the onlinates of the cent roids of the-component area... , Exte nd ing thi s res ult 10 :lll arbitrary number of "Compo-

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"'.2. Centroid

IICIll areas, and IIOlillg Ihal a ~ i Hl..il al ex pl'c ssio ll lJla y be o iJmined fOi Q" we wrik

rA II Composlt" "''''''

739

(A.S)

To obtain the coordinate.-< X and Y of the centroid C or the CO l11posite arcaA . we substitu te Q~ = A}' and Q,. = AX into Eqs. (A.5). We have

AY ~ Sol ~ in g for

L

A,Y,

AX ~

L Ax,

X and Yand recall ing that the area A is the sum o f the CO Ill -

fJO IiCHl ,li e ns A ,. .... e wlile

(A.b)

EXAMPLE A.02 Locate the cenlroid C of the area A show n in Fig. A.1O

I

•

~u

I

I

·c

"\ 1

~-

f~

00

L

I

SU

,I

. 1.

-

C------'

211

1- -4l!- '"

n

11'0

l"~ "

:W

D h l{'miomil L n' n L

i)i"""""""I""",,

Fig. A.10

Fig. A.11

Area, mm' (20)(80) Selecting the coo rdinale axes shown in Fig . A.lt . \\'<' not e Ihal the cen troid C must be located on Ihe )' a(i ~ . ~ ince ul is ;uis is an llXi ~ of sy mm etry: th us . X - O. . Di viding A into il~ component parIS A) and A ~. we use the of th e second Dr Eqs. lA 6) to Ile tenn ine the unlinate c<)l1t m id Th" :Ictu a ] ('QmputatioJl i ~ bes l c:lrried OUI in ta.bular fonn.

r

~

y,. mm .600

70

(40)(GO) - 2400

30

A,f ,. mm' II I X 10.1 72 X 10'

L A, = 4000

LA

i :::,

Y =-'-- ~

~ A.

,

184 X 10 l nlln' 4 >< [0) mm"

46mm

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EXAMPLE A.03 Referrin g 10 the area A of Example A02. weco lBiu cr th e horx' axiS throu gh its cent roid C. (Such [tn Jxis is called
I'

1- - ·'

1

I

.\ =

2~

"

I

J1"

.' .

IL

.',

-

-1

c

I

'1'1 =7

L

'-----+--' _

"

L

,

Fig. A.12 DI"'<'1\,I",,, In 111"1

Fig. A.13

Solution. We divide the area A' into ils componen ts ! \ J (Fig. A.. 13). Rccnllin g from Example A.02 thm C;~ loc at~d 46 mm abov~ Ihe lower edge of A. we detennine the ordi nates y; .. nd v~ of Al and A] and express the first moment

il l and

Q;. of A ' with respecllO.I·' as follow s: Q;' == AI 5'; + A.0'3 - (20 x aO)(24) + (I .j. x 40)(7)

- 42.3 x 10'

1Il1ll'

Altl!rnativl! Solution. We fi .. , nole lh,,' "; n,,~ lh~ ,,"" _ tIO id C of A is located on the x' axis, th e first mo ment Qx' of the emir!' urw A with respec t to that axis is zero:

••

Q~. = Av' = A(O ) = 0

Denoti ng by A" the ponion of A located be low the .r' axis and by Q:, lIS t"irst momenT wi th respec t 10 Ihnt aXIs. we have there· f()n!

0. which sh()w~ Ih"l Iht: 1ir~1 IllUlTlcm. of A '
Qi

--'I~j---:c+---1-T,-_.-.~-23---,-

1 A "~~­ L , , ---J

= Aj J = (40 x 46)( - 23) = - 42.3 X IO-lmlll"' Fig, A.14

740

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A.3. SECOND MOMENT, OR MOMENT OF INERTIA,

11.3 Second Moment or Moment ollnenla. of an Area

OF AN AREA; RADIUS OF GYRATION

Consider again an area A located in the xy plane (Fig. A I) and the element of area dA of coordinates x and y. The .\·ecolld mom elli. or m oment of il/erlia. of the area A with respect to the x axis. and the second moment. or moment of inertia, of A with respect to the y axi s arc defin ed, respectively, as

l·=Ii,

o dA

(A .7) FIg. A.1

These integrals are referred to as rectal/ gular moments of illertia. since they arc computed from the rectangular coordinates of the element dA. While each integral is actually a double integral. it is possible in many applicati ons to select clement s of area dA in the shape of thin horizontal or vertical strips. and thus reduce the computations 10 integrations in a single variable. Thi s is illustrated in Example A.04. We now define the pola r //Iomem of ine rtia of the area A with respect to point 0 (Fig. A IS) as the integral

(repeated)

!I

(A.S)

when:' p is the distance from 0 to the clement dA Whi.le this integral is again a double integral, it is possible in the case of a circular area to select elements of area dA in the shape of thin circular rings, and thus reduce the complltmion 01'.10 to a singh:: integration (see Example A.05). We note from Eqs. (A7) and (A.8) that the moments of inertia of an area arc positi ve quantities. If SI units are used, moments of inertia arc expressed in m4 or mm": if U.S. customary units arc used. they arc expressed in ft 4 or in". An important rel ati on may be established between the polar mome nt of inertia Jo of a given area and the rectangular moments of inertia I,. and I,. of the same area. Not ing that p I = x 2 + / , \v e write

'" Jo :=

~,_ +

I,

(A .9J

The mdill.~ of 8vratioll of an area A with respect to the x axis is fi ned as the quantity r .. that satisfies the relation

/ = r!A "

d e-~

(A .IO)

Fig. A.1S

741

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742

where 1~ is the mome nt of inerti a of A with respect 10 the x axis. Solving Eq. (A. to) for ':" we have

Moml!f1ts of Areas

r,·=fi

tA.l I )

In a similar way. we defin e the radi i of gyration with respec t to the J' axis and the origin 0. We write (A. 12) (A. 13)

Substituting for .In. 1.,. and 1, in terms of the corresponding radii of gyration in Eq. (A. 9), we observe thar

(A. 14)

EXAMPLE A_04 For the rectang ul ar area of Fig. A .16, determine (a) the mome nl of inertia I. of the

~re~

wi lh res["If';cl 10 th ... centmi.b l ..

axis. tb) the corresponding radius of gyration r•. (a) Moment of Inertial.. We select as an element of arcn a horizontal strip of length b and thickness dy (Fig. A. I? ). Since all the points within the stri p are at the same distance )' from the x axis. the moment of inenia of the strip with respect to that axis is dl , == l dA == /( bd.l')

r

/, - f-o;o t---t- -

L~ l- b ~

Integrating from y = - 11/ 2 to )' = +h/2, we wri te I, =

f

"'"

/ dA = J

J'"'( b dy ) =

1b[l J'ZI1

Fig. A.1 6

,

II/!

.-1

+ Id2

"-

0'

-

b 0

(b) Radius of Gyration

and. solving for

'It '

From Eq. (A. IO). we have - h!2/

F"

r., = h/W

Fig. A.1 7

-

--.l d~1

I,

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EXAMPLE A.OS For the circular area of Fig. A.18. deTermine (a) the polar moment of inertia l n, (b) the rectangular rnomenl~ of inertia I, and I ,..

(a) Polar Moment of Inertia. We select as an element of area a ring of radius p and thickness dp (Fig. A.19). Since all the points within the ring are at the same distance p from the origin O. the polar moment of inertia of the ring is (Uo = p2 dA = p\27Tp dp )

Integrating in p fWIll 0 to Jo =

f

p2 dA =

C,

i"

we write

P 2(2'J7'(Jd{J) = 21T I"p,ldP

"

'"

o

Fig.A..18

(b) Rectangular Moments ot Inertia. Because of the symmeTry of the circular area. we have I , = I ,,, Recal1ing Eq. lA.9). we write

Fig. A.19

and . thus.

The r.._'mh~ ohrain~([ in rhe precerlin g rwo cX:lmple~ . :lnrt rhe moment" of II1<'.r_

tia of other common geometric shapes, are listed in a t3blc inside the back cover of this book.

A.4. PARAL LEL-AXIS THEOREM

Consider the moment of inertia ' ..., of an area A with respec t to an arbitrary x axis (Fi g. A.20). Denoting by )' the di stance from an ele ment of area dA to that axis. we recall from Sec. A.3 that

.--.-----.1'-- - 11 riA

"I ,/

Let us now draw the cellfroidul x' (lxis. i.e., the ax is parallel to the x axis which passes through the centroid C of the area. Denoting by y' the di ~ tance from the clement dA to that axis. we write y = y' + d, where d is the diStance between the two axes. Substituti ng for J' in the integral representing I" we write

'.< = f l dA = ,

I., ==

fY'~dA ,

f(.v' +

I

r .\

Ag. A.20

dj2llA

A

+ 2d

J.v

,

l

dA

+ d 2 fdA

,

(A.15)

The fi rst integml in Eq. (A.tS) represents the moment of inerti" I... of the area with respect to the ccntroiJ al x' axis. The second integral represents

743

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744

Momel1lS of Areas

the iirst moment Q.... of thl! area with respect to the x' axis and is equal to zero, since the centroid C of the area is located on that axis. Indeed. we recall [rom Sec. A I lhat

Q,

~

Aji = AIO) ~ 0

Finally, we observe that the last integral in Eq. (A.15) is cqualto the total area A. We have. therefo re. (A. 16)

This formula expresses that the moment o f inertia Ix of an area with respect to an arbitrary x axi s is equal to the moment o f inertia Ix' o f the area with respect to the centroidal x ' axi s pamllcllO the x axi s, plus the product Ad ~ o f the area A and of the square of the distance d between the two axes . T his result is known li S the parallel'(lxis th eorem. It makes it possible to determine the moment of inertia of an area with respect to a given axi s. when its moment" of inertia with respect to a centroidal axis of the same direction is known. Conversely. it mllkes it possible to determine the moment o r inertia Z ., o f an area A with respect to a centro idal axis x; when the moment of inertia Ie of A with respect to a parallel axis is known, by subtractin g from Ix the product Ad ". We should note that the parallel-axis theorem may be llsed only if one of th e t1l"0 ax e.l· im'o lved is a cenlroidal axis. A similar formula may be derived. which relates the polar momelll o f inertia J o of an area with respect to :m arbitrary point 0 and the polar mome nt of inertia l c o f the same area with respect to its centroid C. Denoting by d the di stance between 0 and C. we write ( A.

I 7)

A.S. DETERMINATION OF THE MOMENT OF INERTIA OF A COMPOSITE AREA

Consider a composite area A made o f several co mponent parts A I' A l . and so forth. Since the integral represellting the moment o f inertia of A may be subdi vided into inlegrals extending over A I .A 2 • and so fortb. the moment of inertia of A with respect to a given ax is will be obtained by adding tbe moments of inertia of the areas AI , A 2• and so forth, with respect to the same axis. llle moment of inertia of an area made of several of the common shapes shown in the table inside the back cover of this book may thus be obtained from the formulas given in that 13ble. Before adding the moments of inerlia of the co mponent areas, however. the parallel -ax is theorem should be used to transfer each moment o f inCrTi;) to the desired axi s. This is shown in Example A.06.

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EXAMPLE A.OS

,

Determine the moment of inenia I, of the area shown with respect to the centroidal x axis (Fig. A.21).

,-

L I

t .)

Location of Centroid. The centroid C of the area must first be located. However. this has already been done in Example A.02 for the given area. We recall from that example that C is located 46 mm above the lower edge of the area A.

~

r:

-[

Computation of Moment a/Inertia. We divide the area A into the two rectangular areas A I and Al lFig. A.221. lIIld compute the moment of inettia of each area with re.<>pect to the x axis.

,

J,

om

DilllOom iolls ;" mm

Fi g. A.21

Rectangular Area AI' To obtain the moment of inertia (1..)1 of A I with respect to the x axis. we first compute the moment of inertia of A I with respecllo il~' 011"11 ulliroidalllxi.i· x'. Recalling the formula derived in pan 1I of Example A.04 for the centroidal moment of inenia of a rectangular area, we have ( ~ ')I = Tibh j

'"

1- 110 I'

h(80 mm)(20 mm)J "" 53.3 X 10l mm"'

Using the parallel-axis theorem. we transfer the moment orin· ertia of A I from its centroidal a.~is x' to the parallel axis x:

d,

~ 24101 T

C,

"I

""

'\1

Id~

c

1

(J~)I = (1.. )1 + A id ; = 53.3 X lOJ + (80 X 20)(24)1 = 975 X 10' mm 4

Rectangular Area A 1 • Computing the moment of inertia of A l with respeclto its centroidal axis x". and using the parallel-axis theorem to transfer it to the x axis. wc have

1

[01

1

I I =

16

I

r~

'"

3C)

-.l

Di", ... "si (}m ill nll tt

Fig. A.22

(/" )l = 1ib hJ = t(40)(MW = no X IO} mm~ (I,), ~ (1,.), + A,di ~ 720 x 10' + (40 x 60)(16)' = 1334 X lWmrn4

Entire Area A. Adding the values computed for the moments of inettia of A I and ~~ with respect to the x axis. we obtain the moment of inertia I , of the entire area: ~ = (/. )1 + (/ J2 = 975 X 10J + 1334 X IOl 2.31 X 1()6 1l1m~

I" =

745

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746

Appendix B.

Typical Properties of Selected Materials Used in Engineering, ·5 (U S Customary Units) Ultimate Strength C.om~res.

Specific

Weight, Tension. Slon, Material

Sleel Structural (ASTM·A36) High.strength.] ow.alloy ASTM· A 709 Grade 50 ASTM · A913 Grade 65 ASTM· A992 Gwde 50 Quenched & tempered ASTM-A709 Grade 100 Stainless. A I51 302

Iblin'

ksi

ksi

Yield Strength'

Modulus Modulus Coefficient Ductilltv, 01 01 of Thermal Percenl Shear, Tension, Shear, El a sticity. R,idity, Expansion, Elongation ks; In 2 in. ksi ksl 10"' psi 1 psi 1041{"F

0.2 84

58

36

0 .284 0 .284 0.284

65 80 65

50 65 50

29 29

0 .284

110

100

0.286 0.286

125

75

Reinforcing Steel Medium strength

0.283

High strength

p.28J

;::

0 .260

25

95

J5

0. 2.6 4

50

90

48

33

0.098

16 66 68 46

10 40 41

14 58 47

27

]3

19

lOA

it

24

20

10.1

48

35 73

lOA

0 .322 0 .322

32 57

22

10

29

rl-I

17 17

0.306 0 .306

74

43

6Q

46

.12

15

85

46 31

Cold·rol led Annealed

Cast Iron Gray Cast Iron 4.5% C, ASTM A-48 Malle
0.10 1 0.101 0.095 OJ)98 0 . 101

0.316 ..0.31 0 .3 18

Tin bronze (88 Cu. 8Sn. -IZn) Mangan ese bronze 0 .302 (63 Cu. 25 Zn. 6 AI. 3 Mn. 3 Fe) Aluminum bronze 0 .301 (81 Cu. 4 Ni. 4 Fe. II AI)

95

"

21

22

40 6Q

29

[ 1.2

6.5

21

[ 1.2

29

6.5 6.5 6.5

21

[ 1.2 [ 1.2

11

29

11.2

6.5

18

28 28

10.8

9.6 9.6

12 50

~

II II

10. 8

17

6.5 ..f>~

10

4.1

6.7

24

9.3

6.7

10.1

:'.7 3 .• 3.7

13.1 12.8 12.9 13.3 13.1

4

13. 1

13 19 16 17 11

6.4 6.4

9.4 9.4

45 4

15 15

5.6 5.6

11.6 11.6

65

63

17

6.4

17

6~

45

,0 21

14

lOA 10 10

3 4' 30

95

48

15

12

20

40

16

9

6

"

90

130

8 33

10.9 10.6

36 9

6. 1

(TlIi>lt! t;omltwed

0.5 10

•

8

(II '

I'{'I;~

741l}

No part of any book may be reproduced or transmitted by any means without the publisher’s prior permission. Use (other than qualified fair use) in violation of the law or Terms of Service is prohibited. Violators will be prosecuted to the full extent of the law.

747

Appendix B. Typical Properties of Selected Materials Used In Englneerlng 1.S (81 Units) UltImate Strength C.om~res-

Material

Denllity Tension, ilion. MP, kg/m1 "P,

YIeld Slrength'

Modulus Modulus Coefficient Duclility, of of of Thermal Percent Te nsion, Elasticity, Shllar, Shear, Rigidi~, EKpanllion, Elongation MP, MP, GP, GP, 10--"'C M.. in 50 mm

StCf]

Structural (ASTM -A36) H igh-stn:ngth-Iow-alloy ASTM-A 109 Grade 345 ASTM-A913 Grade 450 A!:ITM-AlJlJ2 Grade 345 Qucnche<.l & tempered ASTM-A 109 Grade 690 Stainless. AtSI 302 Cukl-rulkiJ Anneall'n Reinforcing Steel MediU!n s1renglh High streng th

1860

400

250

7860 7860

345

7li60

450 150 45U

1860 7920 7()20

145

200

11.2

11.1

21

200 200

11.2. 11.2

345

2(')

77.2

11.1 11.1 11.7

21

450

21

'60

69f)

200

11.2

11.1

18

8(j(J

520

190 100

75

\1.3

7'

17.1

12 '0

200 200

7J 77

11.7 11.7

(f)

28

12. 1

165

65

12. 1

10

23.6 23.0 23.2 23.9

9 13 19 16

1\0

17

7860

'"480

"" 275

1860

620

415

7200

170

655

24U

7300

345

620

330

AlIo)-54S6- HI 16 All oy 606 1-16 All o} 7075-T6

21 10 2800 2800 2630 27 10

110 455 470 315 26<)

185 165

70

26

13.6

I7

2&()()

570

330

500

72

II

Copper Oxygen-free copper (99 .9% Cu) Annealed Han:/_<:lrown

"

23.6

8910 8910

no

150

44 44

16.9 16:9

45

200

70 265

120

39()

8470 8470

, 10 320

300

410

22U

100

Il~

39 39

20.9 20.9

65

8740 8140 8800

585

320

435

120

44

270 310

'10

70 145

120 95

44

18.7 18.1 18.0

3 48 ]0

8360 3 Fe) 8330

655

330

lOS

21.6

20

215

110

16.2

6

Casl Iro n ilray r~
230

70

95

55

70

26

275 280

400

230

27

325 230

75 73

130

72

2"0

1"0

120

O.S

4

Yellow-Brass 1:65% Cu. 35% Zn) Cold-rolled Annealed Red B rn~s l!l5% C u. 15% Zn) Cold ro lle<.l Annealed Ti n bronze (88 Cu. 8Sn. 4Znj Manganest" brome (63 Cu. 25 Zn. 6 AI, 3 Mn. Aluminum bronze (8 1 CU,4 Ni, 4- Pc. II AI)

620

900

250 60

105

42

8

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7 48

AppendiX B.

•

Typ ical Properties of Selected Materials Used In Engineering" (U.S. Customary Units) Continued from page 746 Yield Strength'

Ultimate Strength

Modulus Modulus COlltficillnt Ductility, Specific Compres· at of1hermal Percent Weight. Te nsion. sion,' Sheaf, Tenshm. Shellf, Eluticlty. R,'dit y, ~pilln5 ion. I!!:langation

Malerlal

Iblln'

Mu g nl'_" il lm Allnys Alloy AZ80 (Forging) Alloy AZ31 (Exlrmion)

0.065

ksi

.,

0'

h'

ksl

hi

lot

psi

1

psi

2.4 2.4

10- '!"F

14 14

in 2 in.

13

36

i9

29

6.5 6.5

130

120

16.5

5.3

10

0.319 0.319

98 80

"32

26 26

7.7

1.7

22 46

U.323 0.323

53 S5

16 79

0.017

IS

0.064

50 37

0. 16 1

6

12

t itanium All(')Yw(5% AI 4% V)

Monel Alloy 400(N i·Cu) Cold-worked Annea l ~d

Cupronil'kcJ (90% C u, 10% Ni) Annealed Co ld worked

50

is

7.5 7.5

2U

20

Y.5 0.5

35 J

T imb.::,.., ail dry

Douglas fir Spruce, Sitka ShortJeal I?me We~l e rn whitt pine Ponderosa pine Whi te oak

U.UH! 0.0 14 0.1115 0.025

.Reu oak

0.024

We,rp.rn hf'm lod.: Shagbark hick()ry RedwoOO

0.01(; 0.026 0.015

Icon"".

Medi um SHeng!lt Hi,\lh SlrcnSlh

O.O[S

8.6

8.

11

9.4

7.2 5.6 7.3 5.0

5.3 7.4 6.8 7.2 9.2 6.1

II II 1.4 10 II 2O is

1.9 1.5

Varies l.7 to 2.5

-

1.7 I~

13 1.8 1.8 1< 2.2 I.J

"

24 U9

~O

O.OM 0.084

.1 .07

,-

I

'.j

3.6 4.5

6.0

5.5

Plnstic-s

Nylon, type 6/6. (nlolding compound)

0.0412

Polycarbon~le

0.0433 O.04!W

Polyesll'r. PBT (thermoplastIc) Pulyester el!lst O fl~r Polystyrene Vill'f1..n2:id PVC

Rubber Gmnile (Avg. valllcs) Marble (Av g. va l ue~) Sandstone (Avg. values) Glass. 98% silica

0.0433 0.0374 0.0520 0.033

O.JOll 0. 100 0.083 0.079

"

14

6.5

0.4

80

SO

95

12.5

9

II

8

0.35 0.35

68 75

110

8

IJ

6

0

2 .l

,

35

S

I

" 12

4 2

7

0045

711

,

0.45

7~

40

,

90

600

2 4. 1

5 44

'"'01)3

"""lS-

65 8

8 6.5

ISO

,

SOO

,

10

6 9.<

4 6

'Propenln of me ,al, vary WJddy as a fClUIl of \':rJatwns m rompo,nlOlI, h~at \rCalmrr.l. an:! mechamcal workmg. ' For dunilc nl~ta!. tho compression strc"glh i. ~e""rnlJy lIS.umrd 10 00 ,,
"r

No part of any book may be reproduced or transmitted by any means without the publisher’s prior permission. Use (other than qualified fair use) in violation of the law or Terms of Service is prohibited. Violators will be prosecuted to the full extent of the law.

Append ix B.

•

Yie ld Sirenglh3

Ultimate Strength

Modulus Ccmpres· of Density Ten5ion. sicn,' Shear, Tension, Shear. Elasticity, .P, MP, MP, MP, MP, GP, kgfm'

Modulus of Rigidity, G..

1800

345

Hi(]

1770

255

130

16 16

Titanium AlIu)' (6% A I. 4 % V)

4730

Monel Alloy 400(Ni· C u) Cold worked Annealed C upronickcl (90% Cu. 10% Ni) Annealed Cold-worked

Material

Ma gnl'sium All oys Alloy AZ80 I,Forging) Alloy AZ31 (Extrusion)

7 49

l'yplcal Properties 01 Selected Materials Used in Englneenng" S (SI Units) Continued from page 747

250 200

45 45

900

830

8830 8830

675 550

585 220

8940 8940

365 585

545

470 415 500 390 415 690 660 440

100

IlO

345 125

Coefficiont of Thermal EKpansion. 10- l r C

Ductility, Percent EIOflgation In 50 mm

25.2 25.2

6 12

11 5

9.5

10

"I

13.9 13.9

22 46

17.1 l7.1

J5

100

14(1 14(1

52 52

3

Timber, a ir dry 1)n 1J8 1 ~ s

fir

Spruce. Sitka Shorl leaf pine Western white pine Ponderosa pine W hi t ~ oak Red oak Wi.'~li.'m hemloc k Sh:lgoark hickory Redwood

6(1

55

51 9{J

no 415

C un c retc Medium StrellgUl

2j20

Hig h str~ngth

2320

'"

39 50 34 36

65

47 50 63 42

76 7.6 9.7 7.0 7.6 [3.8 11.4

11

10

Pol y~tyft'lle

"Inyi. rigid PVC Rubber umnite (A\g. \~t1ues) Marble (Avg. YJ.luef» Sands tone (Avg. values ] Gla, s. 98% silica

V~tit'~1

3.0 to 4.5

12 10

9 12 12

I(J.O

II

[6.5 6.2

IS

-

9

"

9.9

'"'"

40

P la st iCll

Nylon. type 6/6. (mold ing compound) I'olycflroonate Po lyes tet, I'B T (tnermoplustjq Polyestet' e lastomer

0 .7 0.5

9.9

1140

75

95

45

2.8

144

50

1200 1340

65 55

"

55

"

2.4 2.4

122 135

I HI ISO

1200 1030 1440

45

910 27711 2770 2300 2190

75

40

55 40 15

90 )0

20

'l4U

35

15

1:!:5

7

85

"

50

"

45

14

500

0,2 J. I 3, 1

7U 55 40 65

of VarJ2liOnl til ConljXl!;ll1on. heal treatn~m, and ~tccha11lnl worklllJ;. ",rcnS,h i. gcncrnlly a •• umed ,~ !It equat '" 'ho ,en. ion s!rcn),'ili .

4 3 2 4. 1

'" 115

40

162

600

2

7 .2 10.8 9.0

80

'Prop:mcs of nK"ta1~ I'Cry wi(]cly as a rc,ult

'For ductile 11I0tOI., tile comp"''''' i~'' 'Offsel 010.2 percent ' I ,mI'<", P"'l"'noc.,.,.,. I". lo.,lIn £ po r.II,,1 In thll I:rnln 'Scc Iiso ,11mb' M~dli1";tal E1Igi1lu"JlJ1 !-ilmd/JoQt. 100Jt cd .. McGrJ'~'-1 lilL New Yor\.:. 199(); AII","'/ Hoo~ <)fASTM. American Society for TCI1in g Mmeri;]15. P~il:tdclphla. Pl.; M~/a'" HIII.' I1OO;.k, American Society of MClal'. Melais Puk . OhiO; amJ A"m';"III.'1 De,(~" MJllllw/. file Aluminum Associlllion. Wash· ington. DC ,

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750 Appendix C.

Properties of Rolled-Steel Shapes (U.S. Customary Units)

W Shapes (Wide-Flange Shapes)

......

-

w_

Thlck-

4In'

..... d, In.

b,.ln.

.... ~

88.3 39.7

36.74 35.55

16.655 11 .950

1.680 0.790

E3~~'

59.1

33.68

.:;l!~.

W30 X 173

50.8 29.1

30.44 29.65

14.985 10.450

1.065 0.670

42.9

27.38

~~~ ,

0.975'

12.750 8.965

0.750

12 .~~ ·

DHlgnaUont W36 X 300

I3S

r r

99 7 .... 146 ·

W24 X 104

30.6

68

20.1

24.06 23.73

21X I01

-,"18.3 .1'. 20.99

4,J

l3Jl • .-20. M

62

WI8 X 106

76 50 35

I"'· '

~;

31.1 22.3 14.7 10.3

18.73 18.21 17.99 17.70

n~ .

1~.s2

4O

16.8 11 .8

;~

.;:~; :;!~

16.01

.

1.150

Axl. X-X.

t hick·

AxI'

I~, In·

$.,Inl

r.. ln.

I,. In·

s~,lnI

r,o In.

0.945

20300

1110

15.2

1300

1S6

0.600

7800 11500 '

439

14.0

225

37.7

3.83 1.38

6"-

14.rr

74. • "7.

95.2

356 1

0.655 0.520

8200 }990

539

12.7

598

269

11.7

128

79.8 24.5

3.43 1.10

0.605

5630 '

411 _21

11 .4

~,

259

0.715

. 0.7411. . 0.5;0 ,

, 0 .4611 ,

0.500 OAI5

3100

258 1S4

HU

1830

0,800 0.(; 15

· 0,500

'l4~~"

227

- 9,02

1330

• ,s.OO• .Jl,<.\O

01<"

127 '1.6204

8. UO

0.585

OAOO

'"

9.55 g.54

8,M

0.590 0.425 0.355 0.300

191Q 1330

6000 - 10.295

0.940 0.680 0.570 0.425

800 SlO

88.9 57.6

704

0,760

7.12() 0.995

0.715 0.505

. ~.45' OA3tl 0.:105

" 758 ~~

,."92.2

5.72

5 1M

64.7

~7

47.2

(,.63 5.41

"4

_ 6.2.

11 .200 11.035 7.495

;;~ -~~ , ~;;~

146

7.84 7.73 7.38 7~ ·

1.655 0.680 0.510

607

1710

232

82

2.660 1.090 0.855

5440

42.7 24.1

16.475 15.500 10.1 30

882

I2J

7.07 6.33 6.05

~~.

;~~.

I:~

,O.oJ5

00415

- ~Ol

II.>

72.1 541 4;!S

,03

43

0.720 0.660 0.5:\0 -"SI'

."30 26 22

12.6 8.85 7.69 6.49

D .92 13.66 '4.10 13.84 13.91 13. 74

8.!160 7.995

..6.770 6.730 5.025

5.000

0 .385 0.420 0.335

r·Y

t".ln.

17.92 14.78 14.31

Wl4 X 370 145

109

16.43

in.

--

0.3711

O.]()S

""" 0.270 0.255 0.230

77.1S 62.7

.'" 291

'4,· 42.0

245 199

35.3 29.0

t A wlde·flange shape IS desIgnated by Ihe leiter W followed by lhe nommal deplh m mehes a.nd!he

.n •

63.'- - 3,21 1

70A 248

;~~ 220

1S2 40.1 15. 3

138

I;!o 677 148 12 1

39.4 27.6 10.7 5.12 12.1

,.7

5.73 5.65 5.54

19.6 8.91

7,00

- ;~ 1.57

4A~

. :::~.

· '42

<,07

2.66 2.61 1.65 1.22

8.25

87.3 29.3

57.7 45.2

1.87

..8'

241

1990

2.91

'0.; 13.9 1.'77 ·,)6 • 1.26.

· '6,9

4.1I 28.9

5.89 5.82

~elghl In

40.7 15.7

14.3

ILJ '" 5.82 3.54 2.80

4.27 3.98 2.48

- 2.46 1.92

1.'9 1,55 1.49 1.08

104

pounds per loot. ( Tt,hl~

com",,,ed

Oil

pOlle 751/

No part of any book may be reproduced or transmitted by any means without the publisher’s prior permission. Use (other than qualified fair use) in violation of the law or Terms of Service is prohibited. Violators will be prosecuted to the full extent of the law.

751 Appendix C.

Properties of Rolled·Steel Shapes (5 1 Units)

W Shapes (Wide-Flange Shapes)

I

Flange

W.b Area A,mm'

Depth d, mm

Width b" mm

57000 15600

423

38 [()(} 22400

933 903 855 835

32600 18700

173 753

381

W690X217

27700

695

355

125

16000

678

W610 X 155

[9700 13()])()

Designationt

W920 X 446

20 1 W840 X 299

176 W760 X 257

147

101 W530

x 150 66

19200 11800 8370

x 158

20[00

113 74 52

14400

\.\1410 X 114

14600

85 60

10BOO 7580 5890

92 W460

46.1 38.8 W360 X 55 1

216 122

101 79 64 ~

57.8

44 39 32.9 tA

Wld~ ·f1Bnge

9450 6630

4990 70[ 00 17600 15500 12900 10[00

8140 7220 5730 4980 4170

shape' IS

Thick· ness '" mm

Thick· ness t. , mm

42,70 20. 10

Axis y.y

AxisX·X

1

I, 108 mm 4

5, 103 mm'

"mm

24.0 15.2

8470 3250

18200

540

29.20 [8.80

18.2 14.0

4790

[ 12(X)

2460

5890

385 356 355 331

27.10 [7.00

16.6 13.2

3420

8850

324

250

1660

4410

15.4 11.7

2340

6730

253

14.80 16.30

298 291

1190

3510

273

611 603 543

314 228

19.00 [4.90

12.7 10.5

1290

4220 2530

256 242

108

312

12. 7 iO.2

229

103

209 165

20.30 [5.60 [1.40

1010

533 525

476 463 457 450 420 417 407 403 399 455

284

23.90 [7.30 14.50 10.80

15.0 10.8

552 351 796 556

9.0 7.6

1 12

[9.30 [8.20 12.80 [1.20 8.80

11.6 10,9

67.60 27.70 2 1.70 18.30 [6.80 13.50 [3.10 9.80 [0.70 &.50

42.0 17.3 13.0 10,5

375

363 357 354 347 358 352

353 349

de~lgn31~d

304 400 292 265

280 190 152 261

181 178 140 14() 418 394 257 255

205 203 172

In 128 127

89

7.7 7.0 6.4

9.4 7.7 7.9

764

333

462 315 216

7200

3720 2070 1340

3340 2400 1460 942 2200 1510

712

1060 774 637 9930 3800

365

2010

302 227 178 i61

1690 1280 1030 899 693 578 474

156 127 2260

6.9 6.5

5.'

by the letter W followed by thc ttomlltal deptl]

[22 [02.0 82,7 Itt

216 205 199 196 188 179 178 171 169 163 160 180 161 153 153 150 148 149 146 143 141

nuliunclers aDd the rna,s

In

108 mm 4

94.4 312

78.2 52.9

185 44.[ 29.5 23.8 8.57

S. 10' mm:l

2550 62[ 1560

536

59. 1

1310

87.6 53.2

399 1040 349 667 259 660 228 104 644

57.2 18.0 12.[ 5.[4

438 199 136

61.5 50,6 24.2 IS.9 ILl S.18 3.75 2.9 1

97.3

60.7 90.5

91.4 63.3 16.6 6.34

4.04 825 283

'.mm

452 175 83.4

73.4 57.7 3950 1440

479 397 236 186 129 95.7 58.6 45.8

81.7 52.5 74.0 47.6 73.2

44.' 32.0 67.4 66.3 4 1.9 30.9 62.6

40.8 40.0 29.5 28.5

10' 101 63.0 62.6 48.9 48.2 39.2 37.8 27.4 26.4

kIlograms per meter. (Ti,bl~

com;n"eJ on P"Xe 753)

No part of any book may be reproduced or transmitted by any means without the publisher’s prior permission. Use (other than qualified fair use) in violation of the law or Terms of Service is prohibited. Violators will be prosecuted to the full extent of the law.

752 Appendix C.

Properties of Rolled-Steel Shapes (U.S. Customary Units) Continued from page 750

W Shapes (Wide-Flange Shapes)

16

.~~

131

12.25 12.1 9

11.94

8.005

0.515

" In.

IZ;. :;";? 7.65 6.48 4.7 1

12.22

12.3 1 11.99

,2 '

I !.J6· 10.40 10.00..

;<;.~ 13.3

45 39

3J 30

11.5 9.7 1 B.84

21

6.4.

I'15

~:~~

58

40 35 31

10.3 . ~:~~

18

"21

'08 6.16 5.26

.

Thick·

2'

'"

~.;~

;~.;

33.4 25.4

5.17

17.3

0.260

156

0.220

103 716· 394

97.4

17.1

126

4.9 1 4.67

45.3

:~

26.4

~:;;~ .m,

248

49. 1 42 . 1 35.0 32.4

4.32 4.27 4.19 4.38

10 ~ :

~~;~ .~:;~ ~;;;:

!!~

~J _ 18.1!

4. 14

1)..8

'-',

52.0 43.3 35 .5 31.2

3.65 3.6 1 3.53 3.5 1

OA35 0.510

209

170 170

9.99

4.000

Q.'lQ..

0.2.'0

~n

B.75 8.50 8.25 8. 12

8.220 8. 110 8.070 8.020

0.810 0.685 0.560

0.510

228

OAOO

184 146

~.OO

8.06 -J..91

OA95

~;~~ -~:;;~

0.360 0.3 10

0.2. 5

127

· , 10

0.285

98.0

-6.49,,-

M Oll..

0.245

5.270 5.250 4.0 15

OA()()

0.250 0.230 0.245 0.230

'"

- 0.455

b.20

6.U20

16

4.74

6.28

·'-030

0.365 0.405

I~

;~;

~?3

19

i6

5.54 4.68

5. 15 5.01

13

3.83

4.1 6

~~:~ . --'fl.'

,.i,

,.21

..;W

18.2 15.2 11.8 9.9 1

3.49 3.43 3.29 3.2 1

n~:o-

53.4 4 1A

0.260

J2 .1 12 .1

·16.7 13.4 to.:!

1.70

0.26H

-~:~ . -g~;~~ J:.:' _ l1>.' .

5}.4 45.0 36.6 16.7

".,4.29 2.1l'I

75. 1 60.9 49.1 42.6

-;:; .;;;

75.3 61.9 48.0 39.6

.

0.847 0.773

'"1'6

0.350 0.3 15 0.290 0.300

0.620 0.530

l.S I

2.3 1 IAI

"-44

8.020 7.985 7.960 5.8 10

lQ,O~Jij'i

4.66 2 .82

'" 9.77 7.97 3.4 1 2.73

-;;!

''''-

17.1 13.J 4.43 2.9g 21' .

2.66

2.oQ 2A~

1.96

5.34

4.00·

10. 10 9.92 9.73 10.47

:~~

~ h apc

LOr 1.54

;;~

204

5.87

t A wide-flange

.1.0 7.47

4".1

0.230

4.000

x::20

't, ln. 3.09 3.04

5. 13

6.490 4.030 3.990

· ~O.O

4A4 3.84

SI' Ins 44.4 32.4 13.9

195 56.3

;~;

6.38

IS

' ... In'

5."

~;~ ~.~ ~.~

10.41 ' · - 1:250 · 0.755 10.130 0.770

rOlln.

5.3 1 5.18

0.295

0.380 0.425 0.265

Axis V·V

270

64.7 ,1. • ..

0.330 0.315 0.255

13

x-x

597 394 , 10

8.28 B. 14 8. 11 7.99

18

AxitI

Th...•

;:,.~

lO.24

17. 1 14. 1 11.7

48

x

833

. 1..

WIO )( 112

W4

S~. 1n3

0.550 0.430 0.370

40

22

x

I.. ln'

0.900 0.670 0.640

50

26

W5

t... ln.

[2. 160 12.040 8.080

..""

. ~~

W.

t" In .

-

12.7 1

A, In'

W I2 X 96 72

x

.... ....

28.2 21.1 14.7

Ceo_I

W8

WIdIh b,,11L

....

F1.....

'". 13.3 11.3 9.20 5 .75

;~~ JA;.. lB.3 15.0 12.2 10.6

. ~~ . 5 "'

2.68 2.59 .oJ.5" 2.01 1.98 1.94 1.37

'- . 3

j::~4 . II 2. 10 2.08 2.04 2.03 -:Z.OZ ·

1.621.61

3.71 3.04 1.70 1. 37

1.26 1.23 0 .876 0. B43

5,61 441

I.S ~

2 .20 1.50

0.966

1.50 O.gll~

_ 0 . 9!l~

5.030

OA30

5.000 4.060

0.360

0.270 0.240

26.2 2LJ

10.2 8.5 1

2. 17 2. 13

9.13 7.5 1

300

1.28 1.27

0.345

0.280

11.3

5A6

1.72

3. 86

1.90

1.00

is designated by the letter W folloll"ed by the nominal depth in indIC, and the weight in pounds per foot.

3.63

No part of any book may be reproduced or transmitted by any means without the publisher’s prior permission. Use (other than qualified fair use) in violation of the law or Terms of Service is prohibited. Violators will be prosecuted to the full extent of the law.

753 Appendix C .

Properties of Rolled ~Steel Shapes (SI Unils) Continued fro m page 751

W Shapes (Wide· Flange Shapes)

Flange W.b Thick· ness t. , mm

Am. A.mm'

Depth d. mm

Width b~ mm

Thi ckness r" mm

[~200

13600 9480

323 311 310

309 306 205

22.9 17.') 16.3

14.0 10.9 94

60 52 44.5

7590 (,670 5690

303 31B 313

203 167 166

13.[ 13.2 11.2

7.5 7.6 6.6

38.7 32.7

4940 4180 304U

310 313 305

165 102 lUI

9.7

5.B

10.3 6.7

6.6 5.6

W25i) X 167 101

21300 12900

2~Y

265

80

[()2oo

264 2sti

257 255

3 I.!! 19.6 15.0

IY.2 11.9 94

67

257 252 247 266

204 203 202 148

15.7 13.5 11.0 13.0

8.9 8.0

49.1 44.8

8580 7420 6250 5720

32.7 28.4 12.3

4180 3630 2lO:SU

258

260

146 102

Designationt

\V3 10 x 143 t07 74

23.1;

58

W2UO X

"

I](JOU 9100 i560

"

6660

I, 10· mm 4

S, 10' mm'

"mm

'"

2[50 1590 lOW

138 135 132

113 8t.2

731 53 [

23A

22'

78.8 77.3 49.7

129 119 99.2

8S[ 74B 634

130 134 132

18.3 10.3 8.55

180 123 103

85.1 65.0 42.7

549 415 2!:'O

131 125 IIY

7.27 1.92 LIb

2U!!O 1240 984

119 113 III

553

746 132

43.1

338

65.6 65.0

7.6

104 87.3 70.6 7 1.1

809 693 572 535

110 108 106 III

22.2 18.8 15. 1 7.03

218 185 150 950

51.0 50.3 49.2 35.1

6.1 6.4 5.8

48.9 40.0 28.Y

3i9 308 228

108 105 lUI

4.73

648 349 241

33.7 22.1 2U.K

Y4.7 76.6 6 1.1 52.7 45.5 40.9 34.4 3 1.4 25.8 20.0 16.6

'"

7A

2().0 17.4 14.2 12.6 11.0 11 .'0: 10.2 10.2 8.4 8.0 6.5

13.0 10.2 9.1 7.9 7.2 72 6.2

154 iS3 102

11.6 9.3 10.3 7.1

8. 1 6.6 6.6 5.8

5.5 10.9

W150 X 37.1 29.& 24.0 18.0 [3,5

4730 3790 3060 2290 1730

162 157 160 153 150

102 1110

W [30 X 28.l 23.1\

3580 JO IO

13 I 127

128 127

W 100 X

2480

106

103

19.3

'.

W2

4580 4000 33')0 1860 2480

~lln

mm

2119 206 205 204 203 I
.:1860

.....

S. 10' mm'

254

46.1 41.7 35.9 31.3 26.6 22.5 19.3

r--

I. 10" mm4

10.0 6.9

222 216 210 206 203 205 201 210 207 206 203

71 59

9.1

A xis Y· Y

A xis X·X

248 l65

300 164 126

YK~

1.78 123

448

Y2.4 91.7 89.9 89.0 87.9

1\)9 342 299 249 194 164

86.7 88.6 87.2 83.6 81.8

274 21' 168 120 9 1.6

68.5 67.4 662 63.3 63.0

7.07 5.56 1.83

4.J

2:!.2 17-2 13.4 9.17 6.87

9.1

6.9 6. 1

10.9 8.80

166 1:39

8.3

7. [

4.77

6.4 5.8 6.2 5.8

709 582 5 12

900

TA 1I·ldc· Jlange shape IJ dcsi:;nmcd h)' (he ICl(!r W follclI'l'd by (he- nominJI dcp(h 1.1 millilllClcrs

Ind!h~

~7.R

88 I 376 23.U

300 247 199 175 151 109

31A 25.4

20A 17.8 15.3 9.01 7.64 4.1 33 1.42 1.15

926 61.2 496 278 225

019.1 39.3 38.8

38A 21.4 [1) .5

61U

53.2 52.8 51.9 51.7 51.1 41 2 40.8 32.0 31.2

11.3 21.5

0,918

918 727 359 247 184

38.7 38.J 24.5 23.5 23.0

55.2 54.1

l81 3. 11

595 49.0

32.6 32. 1

43.9

1.6[

31 3

25.5

mass In

U6

~ilogram s

p.:T

me(ef.

No part of any book may be reproduced or transmitted by any means without the publisher’s prior permission. Use (other than qualified fair use) in violation of the law or Terms of Service is prohibited. Violators will be prosecuted to the full extent of the law.

754 Appendi x C.

rt"~tx

Properties of Rolled·Steel Shapes (U.S. Customary Units)

d

S Shapes

I

(American Standard Shapes)

......

.... ....

~'" L:;,J

.... .....

IlopIh

WIdth

d, lh.

b,.In.

ft, in.

t.... in.

'•• in'

S., in'

r.,in.

',. In'

5,. In'

x 121

35.6

24.50

1.090

24.50

258 240

29.3

26.5

7.245 7.125

9.21

44.9

lO.? 19.6 13.2 12.6

23.5

0.870 0.810 0.870

47.7

90 80

24.00 24.00 24.00

9.43 9.71 9.02

83.3

31.2

0.800 0.620

3160

106 100

8.050 7.870

9.47

42.2

12.1

1.34

r LO x 8.96

~8.2

20.30

165

7.71

50.2

13.9

1.33

25.3

ZO.3()

155

7.89

46.8

13.3

1.36

-:1

22.0 19.4

20.00

128

7.62

70 54.7

20.6 16.[

50

14.7

42.9 50

n..,....'lont S24

SIS

SIS S12

x X

x

40.8

35 31.8

l SIO x 58

x

56 x S5 x

r

S4X

53

x

Web

Thlck-

2!l.OO 18.00

7.000 7:00 7.060 6.385 6.255

18.00

6.25 1 6.00 1

12.6

.5.00 J.5.00

14. 7 12.0 10.3 9.35

12.00 12.00 [2.00 12.00

10.3

35 25.4

7.46

- 10.00 ..JO.OO

23 18.4

6.77 5.4[

8.00 8.00

17.25 2.

5.07

10

6.00 , 6 l - §.OO

1.()90

0.920 0.920

Allis

Th...•

0.745

2940 2390

0.625 0.500

2250 2100

0.800

1670 1580 1280

0.660

0.795

0.635 . OJ9S. . n .5OS 0.69\ O.711

...t!2f\-

"3

103

1.30

9.32

1.16

. s.s~

J. U~

7.72 6.94

1.08 1.14

557 .5.23

1.07

24.1

20.8

64.8 59.6

575 5.95

14.4

218

50.8 45.4 38.2 ]6.4

4,55 4.77 4.72 4.83

15.7 13.6 9.87 9.36

5.74 5.16 3.89 3.74

147 124

29.4 24.7

3.78

1--s:J6

3.32

4.07

6.79

2.9J

0.954

16.2 14.4

3. 10 ].26

4.31 3.73

2.07 1.86

0.798 0.8] I

2.28

2,3 [ $2

•.U\9

5.640 .5.50J

0.622 0.622.

0.550 OAIL

,147

5.477 5.252 5.078

5000

0.659 0.659 0.544 0.544

0.687 0.462 0.428 0.350

4.944

0.491

0.594

4.66 1

0.491

0.31L

4.171

0.425 0.425

0.441 0.271

0.359 0.465 .35.9_ c- o.Z12~

.Il'

29.8 . 27.,

1.53 1.57 1.27

7.07

926 804

3.565

77.1

' " In.

6.7 [

0.461

,....l..33.2

199 187 175

obis y.y

89.4

0.691

4.001

x-x

486 305 272 229

64.9 57.6

26.3 8.77 2.2..1.-1,37

2&...

15,7

1.30

1. 03

1

1.03

1.06 0.980

1.00

O.90! ~

0.675

Q.lll

2.94

5.00

3.004

0.326

0.214

12.3

4.92

2.05

1.22

0.809

9.5 7.7

2.79 2.26

4.00 4.00

2.796

0.326

. 6.79

3.39

1.~6 ·

0.90]

0.646

2.663

0.293 0.223

0,)93

6"08

3.04-.1M.-

0.764

0.574

0.51U

7.5

2.21 1.67

3.00 3.00

2.509 2.330

0.260 0.260

0.349 0 . [70

2.93 2.52

1.95 1.68

0.586 0.455

0.468 0.390

0.5 16 0.522

5.7

1.1 5 1.23

t All Arnencan Standard Bcam IS dc'[gnate<.l b~ the lcllcr S lolloll'cd by the nominal depth III fnches and the weight

In

pou nds ~r fOOl.

0.643

0.56~~

No part of any book may be reproduced or transmitted by any means without the publisher’s prior permission. Use (other than qualified fair use) in violation of the law or Terms of Service is prohibited. Violators will be prosecuted to the full extent of the law.

Appendix C.

755

~ r

"~tr Ll'" \. I

Properties of Rolled· Sleel Shapes

(SI Units)

S Shapes (American Standard Shapes)

• /'J •

r

Flange

Web

Designationt

S610 X

180 158

149 134

110 S, IO X 141 128 112 98.3 S460

x II)j 81.4

S380 X

74 64

S310 X

74 60.7

"

47.3 S250 X

"

d , mm

Width b" mm

Thickness f" mm

Thick· ness t•., mm

622

20<1

27.7

622

200

27.7

610 610

184

/lW

'"

Alea A. mm·

Depth

22900 20 !(X1 19000 1,100 1~200

5, 101 mm '

22.1 22.1 22.1

20.3 15.7 18.9 15.9 12.7

1320 1230 995

41-10 3951) 3260 3080 2lUlil

214

201

23.4 20.2 20.2

16.8 16.1 12.8

700 658 495

271() 2550 2090 1<)50

IS.I 11 .7

m

1685

152

17.6 17.6

33]

143 140

15.6 15.8

14.0 10.4

201

1460 1060

139

16.7 16.7 J"U 13 .8

17.4 11.7

lSI

1)1.200 16400 14200 12500

516

179

508 50,

162

13300 10400

451 457

159

95(Xl 8150

381 381 305 305 10\

133 120

305

127

9480 ,730 f,f"O 0040

6670

'"

37.8

4820

254 254

S200 X 34 27.4

4370 3500

203 20J

SI50 X 25.7 tB.6

3270 2370

152 152

'"

159

126

118 106 102 91 85 7.

SI30 X

15 14.1 11.5

1890 1800 1460

127

SIOO X

102 102

68

11.2 8.5

1430 1070

76

64

7G

59

575 X

T,\n 'Imcrlcan

~taDdard

Allis X·X

I, 10" mm'

71

licam IS dcsignatrd b) tilc Jetter

938

'"

530

185 126 11 3

971

Axis Y· Y

"

I, 10" mml

5, 10' mm'

"mm

240 247

34'

39.0 39.9 32.3 33.0

240

311.9 32.5 20.1 19.0 17.9

\%

211

ZOO 193 199

19 .7 12.6 I 1.8

170 179

lOA 8.83

145

6.65 6. 15

90.8 85.7

26.1 27.1

93.2 83.6 1l1/l 61.1

26.1 26.8 24.8 25.3

mm

229

234

lS I

152

145 127

113

34.4 29.5 30.4 27.5 28.8

9, l

62'i

]20

8.9

90.5

122

12.5 12.5

15.1

95.8 103

3.59 2.86

55.7

7.9

61.2 51.1

593 482

475

22.9 24.[

10.8

1 1.2 6.9

26.8 23.9

78.3 82.6

1.83 1.60

33.8 30.6

20.2 21.1

11.8

10 .8 9. 11

17.2 18.0

7.4 7.4 6.6 6.(, ~

5.8 5.4 8.3 4.9 8.9

4.3

121

'" 216

~4. 0

110

\D9

9.1 9.1 8.3

115

741

206 19'

6.69 5.73 4.1 9 3.97

IO.S

826

321 215

402

264 235 142

57.5

1.00

120

.20

0.782

21.3 18.0

5.07

79.8

5 1.8

0.513

13.2

16.3

2.82 2.53

55.3

0 .383 0.328

10.5

49.6

39,6 41.6

9.41

14.4 14.8

1.20 1.03

31.6 27.1

29.0 31.0

0.254 0 . 190

7.72 6.44

13.1 13.3

to llowcd by the nominal depth In nllll lm ctw; and the ma'~ In

~lln6rall"6

per nleler.

No part of any book may be reproduced or transmitted by any means without the publisher’s prior permission. Use (other than qualified fair use) in violation of the law or Terms of Service is prohibited. Violators will be prosecuted to the full extent of the law.

756

Appendix C.

Properties of Rolled-Steel Shapes (U.S. Customary Units)

C Shapes (American Standard Channels)

L

Flange Thick·

''' ' ' ",UO"t

Oop" d, ln.

Width b" in.

00. .

t" ln.

"n. t .., ln.

3.716 3.520

9.96

15.00 15.00 15.00

].400

0.650 0.650 0.650

8.82

12.00

lTIO

Are. A,

in ~

CIS X 50

14.7

40

11.8

33.9

ell x)O 2S

2QL CIO

x

30 25

21' IS.)

C9 X 20 IS 13.4

C8 X 18. 75 13.75 11.5

ox

12.25

9.8 C6 x 13

3.047 2.942

8.82 7.35 5.88 4.49

10.00 10.00 10.00

3.033

S •• in

f l'

0.716 0.520 0.400

4()4

53.8

349

46.5 42.0

5.24 5.44

0301 0.510 \l,SOI 0.387 0.501 0.282

16, 144

210

l;!9

2U

103

20.7

3.42

18.2 15.8

3.52 3.66

13.5

3.87

2.81 2.28

0.763

1.88

0.780

o0.674 67;J

1.71 •

0.799

.0.698

1.65 1.48 1.32 1.16

0.669 0.676

0.649

0.692

0.713

0.606 0.634

1.11

0.642 0.661

0.586

• J.88 3.94 3.36

~.88

9.1'0

4.41

9.00 . 5WO 8.00 8.00 800

;~~

0.413 0.4 13 0.413__

1'448 0.285 0.233

51.0 47.9

tJ5 11.3

. '.22 '.40

2.42 1.93

J .6

3~ 48

1.7.2

1.01 0J.l1i2

0.390 0.390 0.390

0.487

44.1'

11.0

1.98

0.303 0.220

36.1

9.03

32.6

8.14

2.82 2.99 3 .11

OJI4 0.210

24".2 21.3

· 6.93

17.4 15.2 13.1

5.51 4.04

3.38

2.527

2.343 2.260

~r,

;:~

2. 194

2.090

· 0.166 0.366

3.83 3.09 2.40

600

0.343 0.343 0.]43

0.437 0.314 0.200

5.00

0.7

..1 .97

S.UO

...blli-

0.320-

7.25

2.13 1.59

4.00 4.00

1.72! 1.584

1.76 1.47 1.') 1

3.00

1.596 1,..1.98 \.410

t An American Standard

Chann~1

6.00

3.(10

3.00

0.777 0.187

2.06

0.436

~33

0.798

0.886 0.904

51~

1.600

3.94

0.867

4.47

10.00

60.9

Jr,l n.

].78 3.37 3.11

0.379 0.240

91.2 78.9 67.4

r" ln.

9.23 8.13

[l.O

0.436

0,325

4.1

4.19 4.43 4.61

in~

2.739

0,320

5

5.62

' 1,

0.673 0.526

1.885

Clx 6

2-l-.1

in.

S,. in'

0.436 0.436

6oo

5.4

31S

3

2.886

2.64

8.2

C5x 9

x

7.35 12.(>0 6.09 .1;/.00

Axis y.y

Axis X-X I .. in~

2.157 2.034 1.920

10.5

C4

W.b Thick·

0.617

OS8~1

Q,6.9

.601

LOt

0.599

0.565

1.53

0.854 0.781

0.615 0.625

0.553

1.32

;/6(1

1.17

6.08

2.72

0.968

5.80 5.06

2.13 2.22 2.34

1.05

0.703 0.625 0.642

0.866 0.693

0.564 0.492

0.537

0.632

0.450

0.489

1).479

11.378

0.493

0.484

4.38

0.571

• 0.57' - 0.525 0.581 0.540 0.525

0.529

0.514 0.499 0.511

O.4~tl

0.190

8.90 JA9

3.00

1.83 1.95

0.296 0.296

0.321 0.184

4.59 3.85

2.29 1.93

1.47 1.56

0.433 0.319

0.343 0.283

0.450 0.449

0.459 0.457

0.273 0.273 0.')73

0.356 0.258 0.170

2.07 1.85 1.66

1.38

1.08 1.1 2

1.17

0.268 0.233 0.202

0.455

1.10

0.305 0.247 0.197

0.416

1.24

0.410 0.4()4

0.438 0.436

3.56

IS deSIgnated by (he kncr C followed by (he nommaJ dcpth III IIlches and toc wCl gh(

In

pound s per fool.

No part of any book may be reproduced or transmitted by any means without the publisher’s prior permission. Use (other than qualified fair use) in violation of the law or Terms of Service is prohibited. Violators will be prosecuted to the full extent of the law.

757

A ppendix C.

Properties of Rolled·Steel Shapes (5 1 Unil s)

C Shapes (American Standard Channels)

I

Flange

Am.

mm

18.2 13.2 10.2

167 144 134

877 756 688

133 138 143

4.54 3.79 3.34

61.5 54.7 50.5

22.4

86

[6.5 16.5 [6.5

80 77 74

12.7 12.7 12.7

13.0 9.8 7.2

67.2 59.7 53.4

441 39 1 350

109 112 117

2.09

33.2 30.5 27.7

\9.2 19.7 20.0

76

11. 1

17.1

73

11.1 11.1

336 298 257 218

16.7 17.0 17.4 [7.8

16.3 15.6

69 65

42.7 37.9 32.6 27.7

86.8 89.3

254

13.4 9.6 6.1

229 229 229

67 63 61

10.5 10.5

11.4 7.2 5.9

25.4 21.2 19.8

222 185

64 59

12.4 7.7 5.6

18.2 14.9 13.4

C380 X 74 60 50.4

9480 7570 6430

381 381 381

94 89

J7

5690 4720

30,8

3920

305 305 305

3780

254 254 254

2880

3800

30 22.8

C230 X 30

4750

20.2 19.7 19.9

10& mm~ 103 mm'

Width b" mm

5670

21.9

"mm

I,

10' mm'

Depth d,mm

e250 x 45 J7

x mm

10' mm·

A, mm'

x 45

"

ness t,., mm

Ooslgnationt

C310

Thickness !" mm

I

Axis y.y

Axis X-X

W.b Thick·

I 1.1

10.S

I,

S,

un 1.57 1.58 1.38

S,

22.8

\7.0

17.0 17.4

92.9

1.14

98.1

0.912

26.5 24.0 21.2 18.5

I7J

81.8 86.4 88.5

0.997 0.796 0.708

19.1 16.5 15.4

16.2 16.7 16.7

14.7 14.9 15.0

179 147 132

71.5 75.7 78.6

0.81 7 0.620 0.538

16.4 13.7 12.6

15.1 15.4 15.7

14.3 13.9 14.4

15.3 15.8

22 19.9

2840 2530

27.9 20.5 17.1

3560 2170

203 203 203

57

9.9 9.9 9.9

CI80 X 18.2 14.6

2310 1850

178 178

55 53

9.3 9.3

8.0 5.3

10.0 8.83

112 99.2

65.8 69.1

0.470 0.400

11.2 10.2

14.3 14.7

13.1

CI50

x

19.3 15.6 12.2

2450 1980 1540

152 152 152

54 51 48

8.7 8.7 8.7

11.1 8.0 5.1

7.11 6.21 5.35

93.6 81.7 70.4

53.9 56.0 58.9

OA20 0.347 0.276

10.2 9.01 7.82

13.1 13.2 13.4

12.9 12.5 12.7

C130

x

13 lOA

1710 1310

127 127

48 47

8. 1 8. 1

8.3 4.8

3.70 3.25

58.3 51.2

46.5 -19.8

0.264 0.229

7.37 6.74

12.4 13.2

12.2 13.0

ClOO

x

10.8 8.0

1370 1020

102 102

43 40

7.5 7.5

8.2 4.7

1.90 1.61

37.3 31.6

37.2 39.7

0.172 0. 130

5A4 4.56

11.2 11.3

11.4 11.5

8.9 7.4 6. 1

1130 936 765

40 J7

6.9 6.9 6.9

9.0 6.6 4.3

0.850 0.75 1 0.671

22.3 19.7 17.6

27.4 28.3 29.6

0.122 0.0948 0.0765

4.25 3.62 3. 16

10.4 10.1 10.0

11.3 10.8 10.8

C200

C75

x

x

2660

76.2 76.2 76.2

35

t An Amencan StandJrd Channel IS deSignated by the letter C followed by the nonllnaJ depth

In

IJ.7

millimeter.; and the lIIass III kilograms per meter.

No part of any book may be reproduced or transmitted by any means without the publisher’s prior permission. Use (other than qualified fair use) in violation of the law or Terms of Service is prohibited. Violators will be prosecuted to the full extent of the law.

758 )'

I' Appendix C.

z

Properties of Rolled-Steel Shapes (U.S. Customary Units)

,

Angles

y

Equal Legs l'

I

x

Z

Axis X·X and Axis y.y

Size and Thickness, in.

L8 X 8X J

Y.

Weight per Foot, ibltt

51.0 38.9

Y,

26.4

r L6X6 X I X

37.4 28.7

.

24.2 19.6

'I.

X

Axis

z,z r, in.

Area, in

I. In-

S, in'

r, In.

)for y. In.

15.0

89.0 69.7 48.6

15.8 12.2

2.44 2.47 2.50

2.37

8.57 6.66 5.66

1.80 1. 83

1.86

1.84

1.73

4.61 ' ,53

1.86

1.68

1.18 1.18

1.8 11

,64

~

1.51 1.52 1.54 1.56

1.52 1.48 1.43

0.975 0.978

1.19 1.20

l.21

0.778

1.23

0.719 0.782

2

11.4 7.75

11.0 8,44 7.11

35.5

5.75

19.9

28.2 24.2

8.36

2.28 2.19

1.78

1.56 1.58 1.59

1.17 1.17

. J4,9

,36

1 5~4

L5 X 5X X ';I.

23.6

6.94

15.7

y, Y.

20.0

5.86

13.6

16.2

4.75

11.3

12.3

3.61

8.74

;.

185 15.7

5.44

7.67

2.81

4.61

2.4()

~

12.8

6.66 5.56

X

y.

9.8 6,.

1.52 1.05

1.22 1.23

1.I4

1.94

4,36 3 ,04

1.25

1,09

l :.

L3 ! X3 ~XYl

ILl

3.25 2.48 1.69

3.64 2.87 2.01

1.49 1.15 0 .794

1.06 1.07 1.09

1.06 1.01 0.968

0.683 0.687 0.694

2.75 2.11 1.44

2 .22 1.76

1.07 0.833

0.898 0.':113

0.9J2

0.888

0.584 0.587

L2A

0.577

.0.930

0.842

.....llli1.

2.25 1.73 1.19 0.902

1.23 0.984 0.703 0.547

0 .724 0.566 0. 394 0.303

0.739 0. 753 0. 769 0.778

0.806 0.762 0.717 0.694

0.487 0.487 0.491 0.495

1."36 0.938 0.484

0.479

- 0.351

- 0.636

- 0.389

0.348 0.190

0.247

0,609

0.131

0.626

0.592 0.546

0.391 0.398

L4X4X "!-

y, y.

L3X3X ~

X

8.5 5.8 9.4 7.2

Yo

4.9

L24 x 2~XYl

7.7

Yo

Y. 'f,. ,- L'1X'1 X~

Y. X

5.9 4.1 3.07 4.7

3.19 1.65

3.75 2.86

4.53 3.86 3 .16

2.42

1.97

059.-

1.39

1.18

0.983 0.990

O. 95

No part of any book may be reproduced or transmitted by any means without the publisher’s prior permission. Use (other than qualified fair use) in violation of the law or Terms of Service is prohibited. Violators will be prosecuted to the full extent of the law.

759

Appendix C.

z

Properties of Rolled-Steel Shapes (SI Units)

x ~x

Angles Equal Legs

~

!I

z Axis X·X and Axis y.y

Size and Thickness. rom

L203 X 203 X 25.4 19.0

12.7 L1 S2 X 152 X 25.4 19,0 15.9 12.7 9.5 LI27 X 127 X 19.0 15.9

12.7 9.5 L lO2 X 102 X 19.0 15,9

Mass per Meier, Kg/ m

I 8

5 10' mm'

Area, mm'

10 rom'

75.9 57,9 39.3

9670

258

4990

36.9 28.9 20.2

55.7

7080

14.6

139

42.7

5420 4580

11.6

108 92.5

7350

199

[37

Axis

,

xor y

,

mm

mm

mm

Z·Z

61.8 62.7 63.6

60.0 57.8 55.5

39.7 40.0 40.4

47.2 44.9 43.9 42.7 41.5

29.5 29.7 29.9 30.0 30.2

29.2 22.2

]700

8.22

2800

6.3..J.

75,2 57.4

45.4 46.3 46.7 47.1 47.6

35.1 29.8 24.1 18.3

4470 3060 2320

6.54 5.66 4.68 3.63

14.0 63.2 51.7 39,6

38.3 38.6 39.1 39.6

38.6 37.5 36.5 35.3

24.7 24.S 25.0 25.1

27.5

3520 2990 2430 1850 1260

3.23 2.81 2.34 1.83 1.29

46,3 39.1 32.6 25.1 17.4

30.3 30.7 31.0 31.5 32.0

32.3 31.3 30.2 29.0 28.0

19.9 19.9 19.9 20.0 20.3

24.5 18.8 13.1

26.9 27.3 27.7

26.9 25.8 24.6

17.4 17.4 17.6

36,0

3790

10.0

9.5 6.4

23.4 19.0 14.6 9.8

L89 X 89X 12.7 9.5 6.4

16.5 12 .6 8.6

2l()()

ll()()

1.52 1.19 0.845

L76X76X 12.7 9.S 6.4

14.0 10.7 7.3

1770 1350 932

0.915 0.725 0.517

17.5 13.6 9.50

22.7 23.2 23.6

23.6 22.5 21.4

14.8 14.9 15.0

L64 X 64 X 12.7 9.5 6.4 4.8

11.4 8.7 6.1 4.6

1460 Jl30 778 591

0.524 0.419 0.302 0.235

12.1 9.40 6.62 5.09

18.9 19.3 19.7 19.9

20.6 19.4 18.4 17.8

12.5 12.5 12.6 12.7

L5I X SIX9.5 6.4 3.2

7.0 4.7 2.4

879 612 3 16

0.202 0.147 0.0806

5.80 4.09 2. 17

15.2 15.5 16.0

16.2 15.1 1].9

9.95 9.94 10.1

12.7

1600

No part of any book may be reproduced or transmitted by any means without the publisher’s prior permission. Use (other than qualified fair use) in violation of the law or Terms of Service is prohibited. Violators will be prosecuted to the full extent of the law.

760 r

--I' Appendix C.

z

Properties of Rolled-Steel Shapes (U.S. Customary Units)

Angles Unequal Legs

x

r

Axis Size and Thickness, in.

x-x

z

Axis y. y

Weight per

Axis Z·Z

Foot, lb/tt.

Area, in'

I.,in·

S. , in'

r.. in.

]/. in.

',. in·

S"in'

'y. in.

}C,

~

44.2 33 .8

13.0 9 .94 6.75

80.8 63.4 44.3

IS.l 1l.7 8.02

2.49 2.53 2.56

2.65 2.56 2.47

38.8 30.7 21.7

8.92 6.92 4.79

1.73 1.76 1.79

24.5 11.4 13.5

6.25 4.33 3.32

1.88

2.08 1.99 1.94

'.9U,. ).60

1.59 1.62

1.75 1.70 1.66

2.58 2.04 1.44

1.25 1.26

1.33 1.28

2.42 1.92

. 1.2&.

1.24

I J b _ O~9'l

1.09 1.10 1.12

1.20 1.1 6 1.11

1.36 0.760 1.09 0 .592 0.777 OAI2

0.924 0.940

1.08 1.04

0.672 OA74 0.543 0.371

L8x6 X I

!I,

23.0

L6X 4X Y. y,:

23.6 16.2

Y.

12.3

4.75 J.61

L5 x3x!l, Y. Y.

12.8 9.8 6.6

3.75 2.86 1.94

9.45 7.37 5. [I

l.9! 2.24 1.53

I L4 X 3 X~

11.1

3.25

1.89

8.5

2A8 .69

5.05 3.96

.;;. uix 2J X Y, l< Y. L3 X 2:XY,

Y. .li. L2~X 2 xi;

Y.

"

694"

1.46

_2J1-ljXI

1.91 1.93

l.6t

9 .4 7.2 4.9

2.75 2. 11 1.44

3.24 2.56 1.80

7.7 5.9 4.

2.25 1.73

•lO

1.92 1.53 )'09

0.542. . 0.951 .• 0993

5 .3 3.62

1.55 l.Ob

0 .912 0.654

0.547 0.381

IAI

1.09 0.755

I'/XI 0.781

0.768 0.784

0.831 0.787

8.68 6.27

J!,J~

in.

I

'z, in.

tan 0

1.65 1.56 1.47

1.28 1.29 1.30

0.543 0.551 0.55 8

Ll2

1.08

0.860

0.428

1.15 I. J 7

0.987 0.94 1

0.870 0.877

0.440 0.446

l.IS 0.888 0 .614

0.829 0.845 0.861

0.750 0.704 0 .657

0.648 0 .654 0.663

0.357 0.364 0 .371

1.12 0.866

0.864 0.827 0.639 0.879 0.782 0.644 ~~ 1Jl,§51 · 0.704 0.705 0.5 34 0.71 9 0.660 0 .537 0.735 0.614 0.544

0.543 0.551 0 .558

OA28 OA30 0.435 .

0.414 0.428 0 .440

OA20 0.424

0.614 0.626

2.97 2.08

0.546 0.583 0.559 0.539 260 JQ 74 . 0.49J

0.514 0.363 0.372 0.254

0.577 0.592

0.581 0.537

0.486 0.496 0.506

No part of any book may be reproduced or transmitted by any means without the publisher’s prior permission. Use (other than qualified fair use) in violation of the law or Terms of Service is prohibited. Violators will be prosecuted to the full extent of the law.

761 l'

Appendix C.

z

Properties of Rolled-Steel Shapes (SI Units)

Ang les

x -II+---.--x

Unequal Legs

..-j~1

z

l'

Axis y.y

Axis X·X Size and Thickness,

Mass per Meter

Area

mm

kg l m

mm'

L203 X 152 X 25.4 19.0 12.7

65.5 50. 1 34. [

8370 6380 4350

33.5 26.2

L1S2>< 102 X 19.0 12.7 9,5

35.0 24.0 18.2

4470 3060 2320

10. 1 7.20

x 12.7

2420

9,5 6.4

19.0 14.5 9.8

L1 02 X 76 X 12.7 9.5 6.4

x 64 x 12.7

I, 10ft mm"

S,

"

67.4 16.0 65.1 12.7 62.7 8.96

145 113 78.1

43.7 41.9 44.6 39.6 45.4 37.3

47,5 52.5 48.5 50.3 49.0 49.1

3.65 2.64 2.06

49.0 34.4 26.4

28.6 29.4 29.8

27.5 25.3 24.1

22.4

0.435 0.446 0 .452

47.6 36.6 25.2

40.3 40.8 41.2

44.4 43.3 42.[

1.06 0.841 0.598

18.6 14.5 10.1

20.9 2 1.4 21.8

19.0 17.8 16.6

16.3 J6.6 16.8

0.355 0.362 0.369

2.12 1.66 1.17

31.1 24.0 [6.6

31.S 33.9 32.2 32.8 32.6 31.6

I.llIl 0.792 0.564

IS.1 14.1 9.83

21.8 20.9 22.2 19.8 22.6 J8.6

16.2 16.] 16.5

0.536 0.5-1-5 0.552

1780 1360 938

1.36 1.07 0.759

23.3 IS.O 12.5

27.6 30.6 28.0 29.5 28.4 28.3

0.581 0.463 0.]33

12.7 9.83 6.91

IS.I IS.5 IS.8

18.1 16.9 15.8

13.7 13.8 13.9

0.491 0.503 0.512

11.5 8.8 6.1

1450 1120 772

0.795 0.632 0.453

16.4 12.7 8.90

23.4 27.4 23.S 26.2 24.2 25.1

0.283 0.228 0.166

7.84 6.11 4.32

14.0 14.3 J4 .7

14~

13.7 12.6

10.9 10.9 11.1

0.420 0.434 0.446

7.9 5.4

1/lOIl

0.388 0.280

9. 10 6.39

19.5 21.3 20. 1 20.2

0.217 0.158

5.99 4.24

14.7 15.1

14.8 13.7

10.8 10.8

0.610 0 .621

Y

190 131

63.3 64.1 65 .0

5.56

102 70.8 54.0

1840 1260

3.93 3.06 2.14

16.4 12.6 8.6

2100 16O/l 1100

9.5 6.4

13.9 10.7 7.3

L76 X 5 1 X 12.7 9.5 6.4 L64 X 51 X 9.5 6.4

L89

,

S,

"mm

Ll2? X 76

I,

Axis Z·Z

695

18.4

10"mm3

247

mm

10' mm·

1iJ3 mm' mm

mm

"mm

to" "

32.4 Q.54! 32.7 D.5S! 33.0 0.556 21.9 22.2

No part of any book may be reproduced or transmitted by any means without the publisher’s prior permission. Use (other than qualified fair use) in violation of the law or Terms of Service is prohibited. Violators will be prosecuted to the full extent of the law.

AppendiJil D.

762

Beam Deflections and Slopes

Maximum Bea m ilnd Loadi ng

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E Fundamentals of Engineering Examination

Engineers are required 10 be licensed when their work directly affects the public health. safety. and welfare. TIle intent is to ensure that e ngineers have met minimum qualifications, involving competence, ability. experience, and character. T.he licensing process involves an initial exam . called the Flllu/alllentals of Eng ineering examinatIOn. pro fessional experience, and a second exam, called the Principles and Pmcfiet' of EII/iillcering. Those who success fully complete these requirements are lice.nsed as a Professiollal £/l gineer. The exams arc deve loped under the auspi~e..~ o f the National Cmmcil of Examiners for EII~i­ IIcerillg mul SlIIveyillg. The firsl exam, the Flindamental.I' of Engillcering Emlllillatioll, can be taken JUS! before or after gmdutttion from a four-year accredited engineeri ng program. The exam stresses subject material in a typical undergraduate engineering program, including Mcch{/lIi c.~ of Mal erials. The topics included in the exam cover much of the material in this book. The following i~ a list of the main topic areas. with rderences to the appropriate sectiolls in this book. Al so included are probkms that can be solved h.) review this material. Stresses (1 .3- 1.8; 1.11- 1.11) Problems: 1.3, 1.7, 1.32, 1.40 Strains (2.2-2.3; 2.5-2.6; 2.8-1.11; 2.14-2.15) Problems: 2.6, 2.19, 2.4\. 2.50, 2.62, 2.69 Torsion (3.2-3.6; 3.13) Problems: 3.5. 3.27, 3.36. 3.52, 3. 134. 3.138 Bending (4.2-4.6; 4.12) Problems: 4.9, 4.22. 4.37. 4.48, 4.1O\. 4. 108

763

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764

Appendix E

Shear and Bending-Moment Diagrams (5.2-5.3) Problems: 5.6, 5.8, 5.43 , 5.49 Normal Stresses in Beams (5.1 - 5.3) Problems: 5. 17, 5.22, 5.54, 5.58 Shear (6.2-6.4; 6.6-6.7) Problems: 6.3, 6. 12, 6.30. 6.36 Transformation of Stresses and Strains (7.2-7.4; 7.7- 7"J) Problems: 7.8, 7. 18. 7.34. 7.39, 7.8 1,7.85.7. 10 1,7. 104 Deflection 01" Beams <9.2-9.4; 9.7, Problems: 9.7. 9.10, 9.66. 9.75 Columns (10.2-10.4) Problems: 10.1 L 10.26, 10.28 Strain Energy (11.2- 11.4) Problems: 11.10. lJ.D. Il.lS

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Photo Credits

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765

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Index

Accuracy, Ilumerical, 15- 22,40 Actual deformation, 36. 90 Allowable load .Uld allowable stress. 2 faclOr of safety, 28- 29, 40 shearing stresses, 147- 149 Allowable-stress method. 222 design of columns under an eccentric loo.d. 652- 653, 663 Aluminum design of columns under a centric load. 640-64 J properties of. 746-747 American Forest & Paper Association, 641 American InslitUic of Steel Construction. 638, 643 American standard channel steel (C shapes). 2 1X. 378 properties of. 756-757 American standard shape steel (5 shapes), 2 18. 378 properties of. 754--755 Analysis and design of beams for bending. 307- 370 computer problems, 369- 370 design of prismatic beams for bending. 332- 342,

363. 365 introduction. 308- 310 nonprismatic beams. 354- 362, 366 relations among load. shear; and bending moment. 322- 332.364 review problems. 367- 368 shear and bending-moment diagrams, 31 1- 321. 363- 364 summary, 363- 366 using s ingularity function s to detcmlinc shear and bending moment in a beam, 343- 354, 365- 366 Amllysis and design of simple structures. 12- 14 determination of the bearing slresses. 14 determination of the normal stress, 12- 13 determination of the shearing SLress, 13- 14 Angle of twist. 134, 136, 177 adding algebraically, 15 1 in elastic- range, 150-153, 199 Angle steel equal legs, 758- 759 properties of, 758- 761 unequal leg$, 760-761 Anisotropic materials, 57 Anticlastie curva ture, 220, 299 Areas, St!e Moments of areas

Average value, of Stresses. 7, 38 Axes centroidal. 740, 743- 744 of symmetry, j:'7 Axial loading bearing stress in connections, II, 39 centric. 38 deformations under, 6 1--69. 92- 94 eccentric, 38 , 276-285, 30 1 normal s tres.~. 7-9, 3M shearing stress. 9-11. 39 slowly increasing, 670 stress and strain distribution under. 46--130 stress and strain in. 129--130 Axisymmctry, of circular shafts. 137, 186 Beam deflections and sJope~. 564, 696. 762 Beam clements of arbitrary shape. longitudinal shear on, 388- 389. 415-416 shear on the horizontal face of, 374--376, 414-415 Beams. St!e also Analysi.~ and design of beams for bending of constant Strength, 366 nonprismatic, 310, 354- 362. 366 shearing stresse.~ in, 420-421 statically indetemlinate, 540- 549. 596-597 Bearing stresscs. 2. II. 14. 16, 39 average, 23 ill connections . 11. 39 detennination of. 14 Bearing surfaces, II. 39 Bend and twist, a02. 407 Bending. See also Pure bending anal ysis and design of bcam.~ for, 307- 370 of curved membcrs, 285- 197, 301 of mcmber.; nlJde of seveml materials. 230-233 . 299 stresses due to. 406. 5 12, 656 Bending moment, 21 1,221. 252 relation to shear. 323- 324 Bending-moment diagrams. 3 10-32 1, 325- 327, 363- 364 by parts, 531. 573- 58 1. 600 Bou ndary conditions. 534, 543- 544. 552- 554, 595 Breaki ng strength. 52

767

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768

Index

Brittle materials, 47, 51. 121 under plane stress, ffllcture criteria for, 453--461, 489 sudden failure of. 29 Bulk modulus. 48,87- 89. 124 C shapes, See Standard shape steel channels Cantilever beams, 534.571. 600 and beams with symm ~ tric loadings, 571-572, 600 Cast iron. properties of. 746- 747 Castigliano. Alberto. 711 Castigliano's theore m. 6iO. 711-712, 729 deflections by, 712- 719 Center of symmetry. 408. 737 Centric loading. 8. 260 axial. 38 design of columns under. 636--651. 663 Centric stress, 656 Centroid, 223 of an area. 136--738 of a composite area, 738- 740 Centroidal axis. 740. 743--744 Centroidal moment of inertia. 223, 389, 395.497 Circular sbafts as axisymmetric, 137, 186 deformations in. 135- 139, 172-1 74, 198,200 made of an elastoplastic material. 174-1 77, 201 Ciebsch. A .. 347 Coefficients influence. 709 of thennal expansion. 74, 123 Columns. 606--668 computer problems. 667--668 critical load. 662 design of under a centric load. 636--651, 663 design of under an eccentric load. 652--661. 663 eccentric loading. 625--636, 663 effective length, 607, 663 Euler's formula for pin-ended columns, 610--613, 662 extension of Euler's fonnula to columns with other end conditions. 614--625 imroduction, 607 review problems, 664--666 the secant formula, 607. 625--636, 663 slenderness ratio. 662 stability of structures. 608--610 summary, 662--fl63 Combined loadings, stresses under, 508- 520, 522. 590 Combined stresses, 406--408 Components of stress, 2, 24-27

Composite materials. 2 11 fiber-reinforced. stress-strain relationships for. 95- 104.1 26 Compression, 214 modulus of, 88 Computations. 15 errors in. 15 Computer problems analysis and design of beams for bending. 369- 370 applying singularity functions to dctenuine she:lr and bending moment in a beam, 348 ilxialloading. 129- 130 columns, 667--668 concept of stress, 43--45 de,fl!:Ction of beams, 604-605 energy methods, 732- i 33 principal stresses under a given loading. 526--528 pure bending, 305- 306 shearing stresses in beams and thin-walled members. 420-421 torsion, 205- 207 tr.msfomJations of stress and strain, 493--494 Concentmted loads. 308 single, 691'1 Concept of stress, 1--45 computer problems. 43--45 Concrete maximum stress in, 237 properties of, 748- 749 reinforced beams of. 233 Constant strength, 310, 354, 366 Constants of integration. determination of, 537 Copper. properties of. 746--747 Coulomb. Charles Augustin de, 453 Coulomb's criterion, 453 Creep, 58 Critical load, on columns, 662 Critical stress, 612 Cupronickel, properties of, 748- 749 Curvature, 218 anticlastic. 220. 299 radius of, 2 11. 252 Curved members, bending of, 285- 297, 30 1 Cylindrical thin-walled pressure vessels, stresses in, 489 Deflection of beams, 63--64, 77- 78, 529--605 application of moment-area theorems to beams with unsymmetric loadings. 582- 583, 600--601 application of superposition to statically indetenninate beams, 560--568. 598

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Index

Dellectlon of beams- Cun/. application to cantilever beams alld beams with sYllllllctric loadings, 57 1-572. 600 bending-moment diagrams by parts, 573- 58 1. 6CM) boundary cond ilion ~. 595 by Castigliano's theorem, 7 12-7 19. 729 computer problems, 604-605 direct determination of the clastic curve from the load di ~tribution . 538- 539 equation of the elastic curve, 533- 537, 596 introduction. 530-532 maximum, 584-585, 601. 698. 701. 762 method of ~up~ rposi tio n . 558-560, 598 moment-area theore m ~. 569- 571 . 598- 599 review problems. 602-603 unde.r a single load, 698- 7OS statically indctemlinate beams. 540-549. 596--597 summary. 594--601 under transvene loading. 532- 533. 595 usc of mOlllcnt-area theorems with statically indetcrminate beams, 586- 594. 601 using singularit y functions to determine.. 549- 558. 597 by the work-cnergy method, 698- 708 Defomlations, 47, 77- 78, 104, 157.2 12.540, 587. See alJu Elastic deformations: Plastic deformations actual, 86, 90 under axial loading. 6 1--69, 92-94 of a beam under transverse loading, 532- 533. 595 in a circular shaft, 135- 139, 198 computing, 15 max imum, 693 penmment, 21 1 in a symmetric member in pure bending, 2 13- 2 15 total. 76 in a transverse cross section. 220--229, 299 Design considerations, 27- 37 . See also Analysis and design allowable load and allowable stress, 28- 29. 40 determination of the ultimate strength of a material. 27- 28 factor of safety. 28- 29, 40 for impact loads, 695--696 load llild resistance factors, 30, 40. 334-336 for loads. 28 of prismatic beams for bending, 332- 342, 363 , 365 selection of an appropriate factor of safety. 29- 30 specifications of, 30 oftr..msmi ssion shafts, 134. 165- 166, 200,500--508,522 Design of columns allowable-stress method, 638--640. 652--653. 663 aluminum, 640--641

Design of columns-Cum. unde r a centric load. 636- 651, 663 under an eccentric load, 652- 66 1. 663 fo r greatest efficiency, 618 interaction method. 653--654. 663 with load and resistance factor design, 643 ~ truc tural ~tecJ. 638--640. 643 wood. 641 - 642 Detcrioration.29 Detcnnination of the bearing st resses. 14 of const,Ult ~ of integration, 537 of clastic curve, 538- 539 of first moment. 738- 740 of force~. 104. 426 of the moment of illertia of a composite area, 744-745 of the [Jormal .I tress. 12-1 3 of the shearing stress. 13-14 of the sheari ng stresses III a hewn, 376-377 , 415 of the ultimate strength of a material. 27- 28.40 Deviation. tangential. 570 Diagonal stays, 46 Diagrams free-body, 2, 15- 16. 3 1-32, 38, 63--64 loading. 350 of shear, 311 - 321 , 325- 327. 335- 336. 363- 364 of shear and bending-moment, 311 - 321 , 363- 364, 573- 581. 600 of stress-strain relationships. 47, 49- 55, 12 1, 176, 693 Dilatation. 88, 124 bulk modulus, 87- 89, 124 Dimensionless qua ntities, 50 Discontinuit y, 343 Displacement, relati ve, 62 Distributed loading. 308, 563, 590 Distribution of stresses in a narrow rectangular beam. 380--387. 405,415 over the section, 405-406 statically indeterminate, 8 Double shear, II Ductile materials, 47, 5 1,121 under plane stress. yield criteria for, 451-453, 488 Eccentric axial loading. 38, 211 general case of, 276-285. 301 in a plane of symmetry. 260--270, 300

769

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no

Index

Eccentric 10:.ldiug, 210. 260 columm under. 625- 636. 663 design of columns under. 652--66!. 663 Effective length. of columns. 607. 663 Efficient design , for columns. 618 Elastic action. lIS Elastic core. mdius of. 177 Elastic curve direct determination from the load distribution. 538- 539 equation of. 533- 537,542- 543.552-554.596.762 Elastic defonnations. 2 16-219. 298- 299 under a"ialloading. 122 Elastic flexure formula. 217 . 298 Elastic limit. 57.122 Elasti c rangc. 216 illlglc of twi~t in. 150-153, 199 Elastic section modulus. 217. 249. 299 Elastic simin energy under axial loading. 675--676, 727 in bending. 676- 677. 727 for uonnal stresses, 674--677 for shearing stres~s. 677-679, 727 in lorsion. 678. 727 under tmnsvcrse loading, 679 Elastic torque formulas for, 140 maximum, 175 Elastic unloading, lSI, 254 Elastic versus plastic behavior of a material, 57- 59. 122 Elasticity, modllius of, 47, 56- 57, 122 Elastomeric m3terials. 103 Elastoplastic materials, 109, 126.211,246-247,300 circular shafls made of. 174--177,201 members mooe of, 246--249 Elementary work. 671 Elongation maximum, III percent, 54 Endurance limit, 60, 122 Energy method... 669- 733 Castigliano's theorem. 7 11-7 12, 729 computer problems. 732- 733 deflection under a single load by the work-energy method, 698- 708 deflections by Castigliano's theorem. 712- 715, 729 design for impact loads, 695--696 clastic strain energy for normal stresses, 674--677 clastic strai n energy for shearing stresses, 677--679, 727

Energy methods-Colli. equiv:t1ent stalic load. 728 impaci loading, 693--694. 728 introduction. 670 modulus of resilience, 727 modulus of toughness. 72(}-727 review problcms. 730- 73 1 statically indctenninate structures. 716- 725. 729 strain encrgy. 670--672. 726 strain-encrgy density. 672--{174. 726 strain energy for a gcn~ral slate of .~tress. 680-692. 728 summary. 726--729 work ,Uld energy under ,) single load. 696-698.

nS-729 work and ene rgy undel several loads. 709- 71 1 Engineering stress. 55 Equal-kg angle steel. 758-759 Equations of the clastic curve. 533-537. 542-543. 552- 554. 596. 762 equilibrium. 39 of statics. 143 Equilibrium equations. J'J Equivalent force-couple system, at shear center, 406 Equivalent open-ended loadings, 366 Equivalent static load, 698. 728 Euler, Leonhard. 611 Euler's formula. 607, 611, 630 extension to columns with other end conditions, 6 14--625 for pin-ended columns, 610--613, 662 Factor of safety. 28- 29, 40, 683 selection of appropriate, 29- 30 Fatigue, from repeated loadings, 47. 59--60. 122 Fatigue limit, 60 Fiber-reinforced composite materials, ~tress-strain relationships for, 95-1 04. 122, 126 First moment, 375, 736-740 detennination of, 738- 740 First moment-area theorem, 53!. 569, 574--577, 584. 599 Flexural rigidity. 534, 572, 595 Flexural stress. 2 17 Force-couple system, at shear center, equivalent, 406 Forces detennination of, 104,426 unknown, 39

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tndex Formulas clastic Ilexure. 217. 298 clastic torsion. 140 Euler's. 607. 610-625. 630 Gordon-Rankine. 637 interaction. (5)--654 ~ccant. 607. 625- 636. 663 Fmcture criteria for brittle materials under plane stress. 424. 453-461.489 maximum -nomlal-strcss cri terion. 453-454 Mohr's criterion. 454--455 Free-body diagrams. 2.15-16. 3 1-32.38.63--64 Flindalllt!l!Ta/J oj Ellgilwering E.mlllillmimi. 763-764 Gages length. 50 pre..~ sufe. 462. 4!W strain. 425 Gordon-Rankine fonnula. 637 Gymtion. nldius of. 741-743 HeItz (Hz), 165 Homogeneous materials. 84 Hooke. Robert, 50 Hooke's law, 98, 109, 125, 139. 174 generalized,84--87, 124 modulus of elasticity. 56--57, 122 Horizontal shear, 375 Hydrostatic pressure, 88 Hz. See Hertz IFffHENIELSE statements. 348 Impact loading, 670. 693--694, 728 Inertia. See Moments of inertia Inlluence coefficients, 709 Integration constants of, 537 methods of, 604 Interaction formula, 653--654 Interaction methoo. design of columns under an eccentric load, 653--654, 663 Internal torq ues, 141, 153 Isotropic materials. 57, 84 Joule (J), 67 1 Ki netic energy, 693 Lamina. 57, 96 Lateral strain, 84, 124

771

Load and Rcsisl1mee Factor Design (LRFU). 30. 37. 40. 334--336. See al.l"o Allowable load and allowable stress Load distribution. direct determination of the clastic curve from. 538- 539 Loading diagram, modified, 350 Loadings. Set! also Unloading axiaL 7- 11. 38-39. 46--130. 276--285, 30 1 centric, 8, 38, 260. 636--651. 663 combined, 508-520. 522, 590 concentrated, 308 distributed, 30&. 563, 590 eccentric. 210, 260-270, 276-285, 300--301 , 625- 636. 652--66 1,663 general conditions of. 24--27. 40 impact. 670, 693-694, 728 multiaxial. 85-87,124 open-ended. 366 redundant reaction, 563, 590 relation to shear. 322 repeated, 59--fi). 122 symmetric. 571 - 572, 600 torsionaL 50 I transverse. 2 10. J 08. 532- 5.33. 595 ultimate, 28, 30, 37, 643 unknown, 71 - 72 unsymmetrie, 402-413, 4 16. 582- 583, 600--601 visualizing, 22 1 Longitudinal shear, on a beam element of arbitral)' shape, 388- 389,415-416 Longitudinal stress, 462-463 Lower yield point, 53 LRFD. See Load and resistance factor design Macaulay, W. H., 347 Macaulay's brackets, 347 Magnesi um alloys, properties of, 748- 749 Margin of safety, 28 Materials. See alJo Anisotropic materials: Brittle materials: Composite materials: Ductile materials: Elastomeric materials: Elastoplastic mmenals: Homogeneous materials: Isotropic materials: Orthotropie materials bending of members made of several, 230--233, 299 determining ultimate strength of, 27- 28 clastic versus plastic behavior 01',57- 59, 122 Materials used in engineering, 746--749 aluminum, 746--747 cast iron, 746--747 concrete, 748- 749 copper, 746--747

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TI2

Index

Materials used in engineering- Colli. eupronickel.748- 749 magnesium alloys. 74li- 749 Monel alloy 400. 748-749 plastics, 748- 749 steel. 746-747 timber. 748-749 titanium. 748- 749 Matrix. 57. 95 Maximum absolute strain. 2 15 Maximum absolute s tre s~ . 2 16 Maximum deflection. 533. 584- 585. 601. 10 1. 762 Maximum defonnation. 693 Maximum-distonion-energy criterion, 424. 452-453. 456. 670 Maximum el:L~tie moment. 211 Maximum ehL~tic torquc. 175 Maximum elongation. I I I Maximum - normal - strc~s critcrion. 424. 453-454 Maximum shearing strain. 475. 478 Maximum·shearing ·stress criterion. 424. 430- 431. 440. 450-452.456.489 Maximum stress. 695. 698. 701 Maxwell. James Clerk. 7 10 Maxwel['s reciprocal theorem, 710 Measurements of strain, strain rosette, 478-485, 490 Members curved, 285- 297. 301 made of an elastoplastic material. 246-249 noncircu[ar. 186- 188, 202 with a single plane of symmetry, 250 stability of. 6 symmetric. 211 - 212, 298 thin-walled, 402-4[3. 416 two-foree, 2-4 Membrane analogy, 187- 188 Methods of integration, 604 of problem solution. 14-15,39 of statics, review of. 2-4 of superposition, 558- 560, 598 Minimum shearing stresses, 141, 143 Mistakes, error, in, 15 Modu[ us bulk, 48.87- 89, 124 of compression, 88 elastic section, 2 17. 249, 299 of elasticity, 47, 56-57. 122 plastic section. 248- 249 of resilience, 670, 673-674, 727

Modu[us- Cont. of rigidity. 48. 91. 96. 125 of rupture. 174. 201. 245 of toughness. 670. 673, 726-727 Mohr, OLIo. 436, 454 Mohr's circle application to the three-dimensional analysis of stress. 448-450 creating. 439. 442-443. 456, 464. 477 for plane strain. 425. 490-491 for plane stress. 424. 436-446. 473-475. 4S7. 490 Mohr's criterion. 424. 454-455. 489 Moment ·area theorems. 569-571. 587.594.598- 599 application to beams \~ith unsymmetric loadings. 582- 583. 600- 60J using with statically indeterminate beams. 586-594. 60] Moments of areas. 736- 745 centroid of a composite area. 738-740 centroid of an area. 736-738 determination of the first moment. 738--740 determination of the moment of inertia of a composite area. 744-745 first moment of an area. 736-738 parallel ·axis theorem. 743-744 radi us of gyration, 741 - 743 second moment or moment of inertia of an area, 741 - 743 Moments of inertia, 222. See also Bending moment centroidal, 223. 389. 395, 497 of a composite area, determining, 744--745 polar, 155,741 Monel alloy 400. properties of, 748- 749 Multiaxial loading, 95 generalized Hooke's law. 85- 87, 124

National COllncil of Examiners for Engineering and Surveying. 763 National Design Specification for Wood Construction. 641 Necking, 52- 53 Neutral surface, 214, 216, 286. 298 Noncircular sections, 189 Nonprismatic beams, 310. 354-362, 366 beams of constant strength, 366 Norma[ strain, 471 under axial loading, 48- 50,121 Norma[ stresses, 2, 7- 9, 16, 18.38,211,309.446,510, 513,670.700,727. See also Maximum- normalstress criterion detennination of, 12- 13 elastic strain energy for, 674-677 Numerical accuracy, 15- 22.40

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Index

Oblique parallelepipeds, 89 Oblique plane, stresses on . 2. 40 Open -ended loadings. equi valent. 366 Orthotrupic matcrials. 48. 96 Pa. See Pascals Pamllel-axis theorem. 743- 744 Paml\elcpipeds oblique. 89 rectangular, 8e Pascals (Pa), 5 Permanent dcl"ormations. 211 Permanent set. 58. Ill. 121 Pen11
773

Pure bending, 20~-J06 computer problems. 305-306 of curved members. 285- 297. 30 I deformations in a symmetric member. 213--215 deformations in a transvef$e cross section, 220-229. 299 eccentric axial loading in a plane of ~ymmetry. 260-270. 300 geneml case of eccentric axial loading. 27(}-285. 301 introouction.209- 211 members made of an elastoplaslic materia\. 246-249 of members m:lde of scveral materials, 230-233.299 plastic deformations. 243--245. 250. 3(1) residual stresses. 250-259 review problems. 302- 304 stress cOllcentratiolls. 234--143. 299 stresses and deformations in the clastic range. 216-219.298-299 summary. 298- 301 symmetric member in. 211 - 2 12. 298 unsymll1{'tric. 270-275. 301 KadiLL~

of curvature, 2 11. 252 pennanent, 254 Radius of gyration, 741 - 743 Rectangular beams, narrow, distribution of stresses in, 380--387,415 Rectangular cross section bars, torsion of, 187, 202 Rectangular parallelepipeds, 86 Redundant reaction loading, 563. 590 Redundant reactions. 71 Reference tangent, 571, 577. 582- 583, 588, 600 Relative displacement, 62 Repeated loadings, fatigue from, 59-60, 122 Residual stresses, 48, 11 3-1 20, 126.211,250-259 in circular shafts, 134. 177- 185,200,202 Resilience, modulus of, 670, 673--674. 727 Resistance factor, 643. See al.w Load and resistance factor design Review problems analysis llild design of beams for bending, 367- 368 axial loading, 127- 129 columns, 664--f-.66 concept of stress, 41-43 deflection of beams, 602--603 energy methods, 730-731 principal stresses under a given loading, 523- 525

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TI4

tndex

Review problems- Corll. pure bending. 302- 304 shearing stresses- in beams and thin-walled members. 417-419 torsion. 203-205 transformations of str~ s- and strain. 49 1-493 Rigidity flexural. 534. 572. 595 modulus of. 48. 91. 96.125 Rolled-steel shapes. 750-76 1 American standard channels. 756-757 American standard sh;lpes. 754--755 angles. 758- 761 wide.-flange shapes. 750-753 Rotation. speed of. 165 Rupture. modulus of. 174.201,245 Safety factor. See Factor of safety ; Margin of safety Suint-Yenant. Adhemar BalTe de. 106 Sui nt-Yenant"s prinei pic, 104--106. 126. 138. 221. 276. 38 1.499.509.532 Secant fonnula. 607. 625-fJ36. 628. 663 Second momenl. of area~. 741-743 Second moment-area theorem. 53 I. 570. 574- 577,599 Shafts, statically indeterminate, 153- 164, 199 Shape factor, 248 Shear double, II horizontal, 375 relation to bending moment, 323- 324 relation to load, 322 single, II. 39 Shear center, 373, 392, 402-413, 416 equivalent force-couple system at. 406 Shear diagrams, 311 - 321. 325- 327, 335- 336, 363- 364 Shear flow, 190,373,375,391 Shearing strain." 89- 92, 125.471 distribution of, 133-1 34, 138 Shearing stresses, 2, 9- 11, 16,26--27.39,309,382- 383, 405, 700. See also Maximum-shearing-stress criterion allowable, 147- 149. 157 average, 16,39,376,415 in beams, 376--379.415 in a circular shaft. 139 components of, 26 computer problems. 420-421 detenninatioil of, 13- 14.376--377.415

clastic strain energy for. 677--679. 727 in Ilwlges, 405 on the horizontal face of 1l beam clement. 374--376. 414-415 imroduction,37 1- 373 longitudinal. on a beam element of arbitrary shape, 388- 389.4 15-416 minimum. 141. 143 in a narrow rectangular beam. 380--387. 415 plastic deformations, 392--40 I. 4 16 review problems. 417--4 I 9 summary. 4 14--416 in thin-walled members. 390-392. 416 unsymmetric loading of thin-walled members. 402--413.416 in wcbs. 405 Simple structures. analysis and design of. 12- 14 Single shear, II. 39 Singularity function~. 310. 531 application to computer programmi ng. 348 equivalent open-ended loadings. 366 step function. 365 using to determine shelr and bending moment in a beam. 343- 354 . .165-.166 using to determine the slope and deflection of a beam, 549- 558. 597 Slenderness ratio. 612, 645, 662 Speed of rotation. 165 Spherical thin-walled pressure vessels. stresses in. 489 Stability of members. 6 Stability of structures, in columns, 607--610 Standard shape steel beams (S shapes). 218, 378 properties of, 754- 755 Standard shape steel channels (C shapes). 218, 378 properties of, 756--757 Statically detenninate problems, 309. 363, 534 Statically indeterminate problems. 47, 70--73, 123.212, 309. 531 beams, 540--549. 596- 597 distribution of stresses. 8 to the first degree, 541, 587. 596, 598 to the second degree, 541, 587, 597- 598 shafts, 134--135, 153- 164. 198- 199 superposition method, 71 - 73 use of moment-area theorems with, 586--594,601 Statically indeterminate structures, energy methods for, 716--725.729 Statics. 77- 78 equations of, 143 review of methods. 2--4

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Index

Steel. See 1I1JV Rolled-steel shapes: Structural 5teel properties oC 146--147 stresses in. 236 Step fun ction (STP). 345. 365 Strain energy. 684, 694 , 696. 702 under axial loading. 675--676. 727 in bending. 67&--{)77. 727 and energy methods, 670-672. 726 for a general Slate of stress. 680--692. 728 in torsion. 678. 727 under transverse loading. 679 Strain-energy density. 670. 683 energ y methods. 672- 674. 726 Strain gage~. 42~ Strain roselle. 425. 478-485. 490 Stmins. See also Stre~ s and strain distribution under axial loading: Stress-strain relationships: True stress and true strain analysis of. 104distribution of, 175 laleral. 84. 124 normaL under axial loading. 48- 50. 121 plane. 101 thermal. 123 three-dimensional analysis oC 475-478 Strength. See also Ultimate strength of a material breaking. 52 constam, 3 10. 354, 366 yield, 683 Stress and strain distribution under axial loading, 46--130 deformations under, 6 1--69.92- 94, 122 dilatation, 124 clastic versus plastic behavior of a matcrial, 57- 59. 122 Hooke's law, 56--57, 122 introouction.47-48 modulus of rigidit y, 125 multiaxialloading, 85- 87, 124 normal strai n under, 48- 50, 12 1 plastic deformations, 109- 112, 126 Poisson's ratio, 84- 85, 124 problems involving tcmperature changes, 74- 83. 123 repeated loadings. fatigue, 59--60, 122 residual stresses, 11 3- 120. 126 review problems. 127- 129 Saint-Venant's principlc. 104-106, 126 under Saint-Venant"s principle. 104-106, 126 shearing strain, 89- 92, 125 statically indeterminate problems. 70--73, 123 stress concentrations, 107- 108, 126

775

Stress and strain distributi on under a."tial loading- Cullf. 5ummary. 121-126 true stress and true stmin. 55- 56 Stress concentrations. 48. 107- 108. 126. 167. 2 11 . 234--243.299 in circular shaft~. 167-1 72. 200 Stress-strain relationships. 172. See alw True stress and true strain diagrams of. 47. 49- 55. 121. 176.693 for fiber-reinforced composite material ~ . 95- 104. 126 Stress traj ec to rie~. 499 Stresses. See alJo Allowable load and allowable stress: Distribution of stresses: Principal stresses : Shearing stresses analysis and design. 6 application to the. anal ysiS and design of simple structures. 12-1 4 average value of. 7.38 bearing, 2, II. 14. 39 under combined loadings. 508- 520. 522 computing. 15 concept of. 1--45 critical. oI l design considerations, 27-37 determination of. 104 due to bending, 406 due to twisting, 406 in the elastic range, 139-1 49, 198- 199 engineering, 55 flexural , 21 7 under general loading conditions, 40 general state of, 446-447, 488 introduction, 2 longitudinal ,462-463 maximum, 695, 698. 701 in the members of a structure. 5 method of problem solution, 14- 15. 39 normal , 2, 7- 9,16, 18,38,309.446,510,5 13.670, 700.727 numerical accuracy. 15- 22,40 on an oblique plane under axial loading. 23- 24, 40 resid ual, 113-1 20, 126, 177- 185.202,250--259 review of methods of statics, 2-4 review problems, 41-43 in a shaft. 134- 136 in steel, 236 summary, 38-40 in thin-walled pressure vessels, 462-469 uniaxial,214

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n6

Index

Stresses and defonnations in the elastic range. 2 16--2 19.

298- 299 clastic flexure formula. 298 Structural steel allowable stress des ign. for columns under a centric load, 638- 640 load and re s i.~t
560-568, 598 method of. 7 1-73, 263, 278. 292. 409, 598 principle of. 86 Symmetric loadings, eamilever beams and beams with.

57 1-572.600 Symmetric members. in pure bending of. 211 - 215. 298 Symmetry axis of. 737 center of. 408. 737

Transformations of stress and strain. 422--494 application of Mohr's circle to the three-dimensional anal ys i.~ of stress. 448--450 computer problems. 493--494 fracture criteria for brilt1c materials under plane stress.

453--461 . 489 general state of stress. 446--447. 488 introduction. 423--425 maximum shearing stress. 428--436. 487 measurements of strain. 478--485. 490 Mohr's circle for plane .~ tres s. 436-446. 473--475.

487. 490 of plane stress. 425--427. 470--472. 486. 490 principal stresses. 487 review proble ms. 491--493 stresses in thin-walled pressure vessels. 462--469 summary. 486--490 Ihree-dimcnsional analys is of ~train. 475--478 yield c ri teria for ductile Illme rials under plane ~tre ss .

451--453.488 Tangential deviation, 570 Temperature changes. problems involving. 74--83. 123 Tensile tesl. 50 Tension. 214 Thennal expansion. coefficient of. 74. 123 Thennal strain, 123 Thin-walled hollow shafts, 189-197.202 Thin-walled members, shearing stresses in, 420--421 Thin-walled pressure vessels, 425, 489 Three-dimensional analysis of strain. 475--478 Three-dimensional state of stress, 424 Timber, propenies of. 748- 749 Titanium. properties 01'.748--749 Torques. 132. See also Elastic torque: Plastic torque internal. 141. 153 largest pennissible. 141. 156 Torsion. 131 - 207 of bars of rectangular cross section. 202 computer problems, 205-207 introduction. 132-1 34 modulus of rupture in. 174 of noncircular members. 186--188. 202 plastic deformations in circular shafts. 172-1 74.200 review problems, 203- 205 summary, 198- 202 Torsion testing machine, 150 Torsional loading, 501 Total deformation. 76 Total work. 67 1 Toughness, modulus of. 670, 673. 726- 727

Transmission sh:Jfts. 132 des ign of. 134 Tmnsver~ c ross section. defomlaliollS in ,

220--229. 299 Transverse loading. 210. 308 deformations of a beam under. 39, 532- 533. 595 True stress and true strain. 55- 56 Twisting. See also Angle of twist; Pcrm:Jnent twist stresses due to. 406, 512 Two-force mcmbers, 2--4 Ultimate loads, 28. 30. 37, 643 Ultimate strength of a m:Jterial, 2. 52 detcnnination of. 27- 28.40 Unequal-leg angle steel. 760--761 Uniaxial stress. 214 Unknown forces. 39 Unknown loads. 71-72 Unloading, 115 elastic, 18 I Unsymmetric bending. 211. 270--275,30 1 Unsymmetric loadings combined stresses, 406--408 distribution of stresses over the section.

405--406 equivalent force-couple system at shear center, 406 shear center, 402--413. 416 shearing stresses in flanges. 405 shearing stresses in webs, 405

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Index

Unsy mmetric loadings- Colli. ~tresses due to bending. 406 ~tresses due to twisting. 406 of thin-walled members. 402-41) Upper yield point. 53

Work and energy principle of. 701 - 702 undcr sevcral loads. 709- 7 11 under a single load. 696-698. 728- 729 Working load, 28

Wide-flange sbaped stcel (W ~h apes). properties of. 750-753 Winkler, E. , 285 Wood. See alJo Timber design of col umns under a centric load . 641-642 Work elementary, 67 I tOtal. 67 1

Yield criteria for ductile materiab under plane stress. 424. 451-45), 488 maximum-distonion-energy criterion, 452-453 maximum·sbearillg- ~tress criterion. 45 1-452 Yield points. upper and lower. 53 Yield strength. 52, 121 , 683 You ng. Thomas. 56 Young's modulus. 56

777

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Answers to Problems

Answers to problclllS with a number sc t in stmight Iype arc given on this and the following pages. Answers 10 problems with a number SCI in italjc arc not listed.

CHAPTER 1

1.1

28.2 kips.

1.2

(a) 12.73 ksi. (b) - 2.83 ksi.

1.3

d l = 22.6 mill: d 2 = 15.96 1lI1li. (a) 35.7 MPa. (b ) 42.4 MPa. (a) 101.6 MPa. (b) - 21.7 MPa. 33. 1 kN. 13.58 ksi. 0.400 in", (a) 17.86 kN. (b) - 41.4 MPu.

1.4

1.7 1.8 1.11 1.12 1.13 1.14

1.15 1.16 1.18 1.19

1.20 1.21

1.23 1.25 1.26 1.27 1.29

1.31 1.32 1.33

1.35 1.36 1.37 1.38 1.40 1.41 1.43 1.44 1.47

1.48 1.49 1.51 1.53 1.54 1.55 1.56 1.57

- 4.97 MPa. 292 1lI1ll. 889psi. 67.9 kN. 8.33 in. 63.3 IlItll. (a) J.:B MPa. (b) 525 mm. (a) 23.0 MPa. (b) 24.1 MPa. (e) 21.7 MPa. (a) 444 psi. (b) 7.50 in. (e) 24()() psi. (a) 9.94 ksi. (h) 6.25 ksi. (a) 80.8 /I.'IPa. (b) 127.0 MPa. (e) 203 MPa. u = 70.0 psi: T'" 40.4 psi. 0" '; 489 kPa: T = 489 kPa. (a) 12.60 kN. (b) 560 kPa. (a) 180.0 kips. (b) 45°. (e) - 2.5 ksL (d) - 5.0 ks i. 833 kN. - 21.6 MPa: 7.87 MPa. 15.08 k.N. HS. 2.35. 1.279 in. 146.8 mm. HO. (a) 3.68. (b) 1.392 in. (e) .3.07 ill. 3.68 kips. 3.02. 3.1 4 kips. 1.683 kN. 2.06 kN. 3.72 kN. 3.97 kN. (a) 362 k.g. (b) 1.718.

1.58 1.59 1.60 1.61 1.63

1.65 1.67

1.68 1.70

1.C2 1.C3 1.C4

629 lb. (b) 1.689. 11.09 hi. (b) - 12.00 ksi. 14.64 lsi. (b) - 9.96 ks i. 8.92 kli. (b) 22.4 ksL. (e ) 11.21 ksi . 94.1 MPu. (b) 44.3 r.."IPa. 11.98kN. (a) 49.9 mm. (1)) 257 mm. 21.3° 5 1/ s 32.3°. L",," = U... J /4 T;oJI' (e) 16n1Ill ::s d s12 1l1n1.(d) 18mm ::s;ds221ll1ll. (e) 0.70 in. :5: d :5: l.IO in. (d) 0.85 in. :5: J:5: 1.25 in. (b) For 13 = 38.66°. tan 13 = 0.8: BD is perpendicular (a) (a) (a) (a) (a)

10

BC

(e) F.S. = 3.58 for a = 26.6°: line AC

P is perpendicular 10

1.C5 (bJ Member of Fig. P 1.29,foru =

1.C6

6(/':

(I) 70.0 psi: (2) 40.4 psi: (3) 2.14: (4) 5.30: (5) 2.14. Member of Fig. P I.JJ.jora = 45": ( I) 489 kPa: (2) 489 kPa: (3) 2.58: (4) 3.07: (5) 2.58. (d) Poll = 5.79 kN: stress in links is critica!.

CHAPTER 2

2.1 2.3 2.5 2.6 2.7 2 .8

73.7 GPa. (/I) 6.91 mill. (b) 160.0 MPa. (/I) 17.25 MPu. (b) 2.82 nun. (/I) 5.32 Illnl. (b) USO Ill. (
2.10 9.77lllm. 2.11 d""0 = O.1701 in .. Lrruo= 36.7 in. 2.13 0.429 in. 2.14 1.988 kN. 2.15 (
2.24 2.25 2.27

1-.

(e) 16.30 ksi .

(a) 0.0 1819 Illill t. (b) OJ)919 mill! , = - 2.1 I mill. DAD = 2.03 mill. 30.0 kips. (/I) 46.3 in. (b) 31.6 kips. (/I) 80A /-till t. (b) 209 11m -.I.. (e) 390 /-lill 0.1095 mm~ . DA B

!.

779

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2.28 2.29 2 .30 2 .35 2.36 2.37 2 .38

2.39 2 .41 2.42

2.43 2.44 2.46 2.47 2.48 2.50 2 .51 2.52 2 .54 2.55 2 .56

2.58 2.59 2 .62 2 .63

2.64 2.65 2.67 2.68

2.69 2.70 2.75 2.76

2.n 2.78 2.81 2.82

2.83 2.84 2.86 2 .88 2.89

2.90 2.95 2.96

2.97 2.98

780

x < 92.6 nun. (a) 5 = p!{L2I2E. (b:, F = ~ IV. 8, = Phh TEab t. (a) a , = - litol hi, U d = - 6.27 lsi. (b) - 6.21 X IO- J in. (a) 23.9 kips. (b) 14.50 kips. " , = 67.1 MPa.", = 8.38 MPa. 3330 kN. (a) 0.0762 mm. (b) U "'8 = 30.5 MPa. Ut'~' = 3R.I MPa. (a) 31 A: 62.& kN ~: 31 E: 37.2 kN ~. (b) 46.3 p.m -to (a) al A: 45.5 kN ~; m E: 54.5 kN <:-. (b) 48.& ftm-. . (a) a\ A: 14.72 ki ps __ ; al D: 12.72 kips ~ . (b) - 1.574 X 10-) m. (a) al A: 9.03 kips ..... : AI D: 7.03 kips ~. (b) - 24 1 x 10 - ~ in

if/" TH = ~ P. Te = ~iJ P. Tn = ~ P. - 8.15 MP:!. - 56.2 MPa. 75.4°C. 142.6 kN. (a) 'TAB = - 10.72 k, i. 'T R(" = - 24.1 lsi. (b) 5R 9.88 X lO- J in.!. (a) - 17 .91 ksi. (bl - 2.42 ksi. (a) 21.4°C. (b) 3.6R MPa. 5.70 kN. (a) 201.6°F. (b) 18.0 ](17 in. (a) 52.3 kips. (b) 9.9 1 X lO - ' in. E - 205 MP~ : u - 0.455: G - 70.3 MPa. (a) 1.324 X IO- J in. (b) - 99.3 X 1O- ~ in. (e) - 12.41 in. (d) - 12.41 X 10- 6 inl . (a) - 41.2 X IO- J in. (b) 2.06 X IO- J in. (c) 85 .9 X 10- 6 in. 422 kN. (a) 5.13 X IO- J in. (b) - 0.570 X j()-) in. (a) 7630 lb. compTe"sion. (b) 45RO lb. compression. (a) 0.0754 mm. (b) 0.1028 mm. (e) 0.1220 mm. (a) - 63 .0 MP:!. (b) - 13 .50 mm 1. (e) - 540 mm). 0.0 187 in. a = 0.R18 in; b = 2.42 in. l.OO1 mm! . 302 kN . (a) 262 mm. (b) 214 mill. G = 1.080 MPa: T = 431 kPa. (a) 588 X lO- o in. (b) 33.2 X 10- ) in). (e) 0.0294%. (a) :il! = - 0.074(, nun; :iV = - 143.9 mm). (b) :ill = - 0.0306 tlun::iV = - 521 mml . (a) 16.55 X 10- 6 in' . (b) 16.54 X 10- 6 in'. 3.00. (a) O.oJ03 mm. (b) u . = 40.6 MPa: u ,. = (J", = 5.48 MPa. (a) '1. = 44.6 MPa; '1, = 0; (J"< = 3.45 MPa. (b) - 0.0129 mm. 0.874 in. (a) 0.425 in. (b) 7.23 ki p~. (a) 12 mm. (b) 62.1 kN. (a) U4.7 MPa. (b) U5 .3 MPa. T" =

=

2.99

(a) 92.3 kN , 0.79 1 mm. (bl 180.0 kN. 1.714 mm.

2 .100 2 .101 2.102 2 .105 2 .106 2 .107

189.6 MPa. 2.65 kips,0. 111 7in. 3.68 kips. 0. 1552 in. 176.7 kN , 3.84 mm. 176.7 kN , 3. 16 mm. (a) 0.292 mm. (bJ (J",\e = 250 MPa, O'eR = - 307 MPa (e) 0.027 mm. (a) 99() kN. (b) {TA(' = 250 MPa. {Te8 = - 3 16 MPa. (e) O.OJ I 111m. (a) 112.1 kips. (b) 82.9 ksi. (c) 0.00906 in. (II) 0.0309 in. (h ) 64.0 ksi. «") 0.(XB87 in, (a) AD: 250 MPa. 8E: 124.3 MPa. (b) 0.622 nun t. (a) AD: 2:13 MPa. 8E: 250 MPa. (b) 1.322 mm !. (a) AD: - 4 .70 MPa. BE: 19.34 MPa. (b) 0.0967 mill L (a) - 36.0 ksi. (b) 15.84 ksi. (a) AC: - ISO.O MPa, CB: - 250 MPa, (b ) 0.1069 IlIm ....... (a) AC = 56.5 MPa, CB:94. 1 MPa. (1))0.0424I11m -+. (a ) 0. 1042 mm. (h) AC: - 65.2 MPa. CB: - 65.2 MPa. (a) 0.00788 mm. (b) AC: - 6.06 MPa, CB: - 6.06 MPa, (a) 915'1'. (b) 175gcF. 1.2 19 in. 4.67"C. 3.5 1 kips. fT, = - 9.47 MPa. u, = 0..191 MPa. 105.6 X l(P lb/i n. 41.7 kN.

2 .108 2 .111 2 .112 2 .113 2 .114 2 .115 2.116 2 .117

2.118 2.119

2.120

2 .123 2 .124 2.125 2 .127 2 .131 2 .132 2 .133 2 .135 (a) Au r / /L8. (b) EAIL. 2 .C1 Proh. 2.126: (a) 11.90 X ]()-3 in . .}. (b ) 5.66 X 10-" in. t. 2. C3 Prob. 2.(j(): (a) - 116.2 MPa. (b) 0.363 mm. 2 .C5 r = 0.25 in.: 3.89 kips r = 0.75 in.: 2.78 kips 2 .C6 (a) - 0,40083. (b) - 0.101 00. (c) - 0.00405

CHAPTER 3

3.1 3.2 3.3 3.S 3.6 3.8 3.' 3.11 3.13 3.14 3.15 3.16 3.19 3.20 3.21 3.23 3 .24

87.7 MPa. 133.8 kN· m. 12,44 ksi. ("l 125.7 N' m. (b) 181.4 N' m. (a) 7().7 MPa. (h) 35.4 MPa. (el 6.25%. 7.95 kip ' in. (a) 56.6 MPa. (bl 36.6 MPa. (II) 81.2 MPa. (b) b4.5 MPa. (e) 23.0 MPa. ("l 8.35 ksi. (b) 5.94 ksi. ("l 1.292 in. (bl 1.597 in. ("l 1.503 in. (b) I.S53 in. 9. 16 kip' in. (a) 5(U mm. (M 63.4 nlln. ("l 15.18 nlln. (b) 132.5 N· m. (a) 2.39 in. (b) 1.758 in. (II) 45.1 mm. (b) 65.0 mm. 1.129 kN . m.

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3.26 3.27 3.28

(a) l.442 in.

(b)

l.2:n

in.

(a) 20.1 mnl. (h) 26.9 mm. (e) 36.6 111m. (a) 55.0 MPa. (h) 453 MPa. (e ) 47.7 MPa

3.29

1.000. 1.025. 1.120. 1.200. l.OOO.

3.30

(a ) (TIW) '"

(d +

t!j)Tol.j 2pge l .

(b) (TIW) = (T/w"lo(I

+ dId).

3.31

(a) 2.19kN · m.(b) 9.11 °.

3.33 3.34

9.38 ksi. 0.491 in. (a) 8.87°. (b) 0.450". (a) 14.43". (h) 46 .9°, 6.020. 3.77". 12.22°. 13.23". 53.8°. 36. 1 Illlll. 22.5 111m. 1.285 in. 1.483 in. 62.9 Illill. 42.1 111m. (a) 17.45 MPa. (b) 27.6 MPa. (e) 2.05°. (a ) 688 N . Ill. (b) 2.35°. TAB = 9.95 ksi. TeD = 1.849 ksl. TAB = 1.086 ksi, TCIJ = 6.98 ksi. 12.24 MPa. 0.241 in. (a) T """ = Tllm? at p = rl' (a) 82.5 MPa. (b) 0.273°. (a) 9.51 ksi. (b) 4.76 ksi. (Il) 46.9 MPa. (b) :B.5 MPa, (Il) 20.1 mm. (b ) 15.94 Illill. (a) 18.80 kW (b) 24.3 MPa. (a) 51.7 kW. (b) 6.lr. (Il) 2.92 ksi. (b) 1.148°. 1 = 81l11l1. 30.4 Hz. (a) n.799 in. (b) 0.947 in. (Il) 4.08 ksi. (b) 6.79 ksi. (a) 16.02 Ih. (b) 27.2 Hz. 1917 rpill. 50.0 kW 2].4 1lI11l. 63.5 kW 5.1 1lI111. 42.6 Hz. (Il) 203 N· Ill. (b) 165.8 N· m. 42.8 hp. (a) 2.61 hi. (b) 2.01 hi. (a) 9.64 kN · Ill . (b) 9.91 kN · m. 22]0 Ib . in. (a) 19.10 hi, I.()(X) in. (b) 20.n ksi. 0.565 in. 13.32 1lI1l1 , (a) 2.47". (b) 4.34°. (Il) 6.72". (b) 18.71 °.

3.36 3.37 3.38 3 .40 3.41 3.42 3.44 3.45 3.46 3.47 3.48 3.49 3.50

3.53 3.54 3.57

3.58 3.59 3.60

3.61 3.62 3.64 3.65 3 .66

3.68 3.69 3.71 3.72 3.74

3.76 3.77 3 .78 3.79 3.80 3.83 3.86 3.87 3.88 3.89 3 .90

3.91 3.92 3.93

3.94 3.96

3.98 3.99

3 .100 (aJ 8.17 nIDI. (b) 42.1 ". 3 .101 (al 18 ksi. (b) 15 .63°. 3 .103 (al 1.126
4.1 4.3 4.4

(a) - 6.52 ksi. (b) 9.78 ksi. 5.28 kN . m. 4.51 kN· m.

781

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4 .5 4 .6 4.7 4 .9 4 .11 4 .12 4.13 4.15 4 .16 4 .18 4.20 4 .21

4.22 4.23 4.25 4.27 4 .28

4.29 4.30 4.31 4.32 4.33 4.35 4.37 4 .38

4.39 4.40 4.41

4.43 4.44 4.45 4.47

4.48 4.49 4.51 4.53 4.55

4.56 4.57 4.58 4.59 4.61 4.62 4.64 4.66 4.67 4.68 4.69 4.71 4.73 4.74

4.75 4.76 4.n

782

129.6 kN . m. 34.2 kN . Ill . 8.82 ksi, - 14.71 ksi. 73.2 MPa. - 102.4 MPa. 2.17 kips. 2.89 kips. 58.8kN. 41.9kip·in. ]()6.IN·m. 3.79 kN· m. 4.63 kip' in. 65.1 ksi. (a) 250 lb · in. (b) 151.0 in. (a) 965 MPa. (b) 20.5 N . m. (a) 75.0 MPa. 26.7 m. (b) 125.0 MPa. 9.60 m. 0.950. 0.949. (a) 139.6 m. (b) 481 m. (a) 1007 in. (b) 3470 in. (e) O.0132Q". (a) 334 fl. ~b) 0.()4(,4". (a) (0) ...,(1 - e!Y2pc. (b) -«T,)"""cl2p. 2.22 kN . m. 1.933 kN . m. 330 kip' in. 195.6 kip' in. (a) 44.5 MPa. (b) - 80.1 MPa. (a) 51.2 MPa. (b) - 119.5 MPa. lu) - 1.921 k.si . (hI 12.73 k.!;i. 23.6 m. 18.45 m. 692 ft. 32.4 kip' ft. (a) 212 MPa. (bl - 15.59 MPa. (a) 210 MPa. (bl - 14.08 MPa. (a) 29.0 ksi. (bl - 1.163 hi. (a) 1674 Illlll c. (b) 90.8 kN · Ill. (II) 29.6 MPa (al uminum). 26.7 MPa (brass). 17.78 MPa (s teel). (bJ 88.6 m. (a) 38.7 MPa (steel), 7.74 MPa (aluminum). 3.87 MPa (brass). (h) 203 Ill. (a) 7.70 h i. (b) 19.37 ksi. (a) 5.91 ksi. (b) 21.2 ks i. (a) 6.15 MPa. (b) - 8.69 MPa. (a) 4.17 kip ' in. (hl 4.63 kip' In. (a) 10.77 ksi. (b) 8.64 ksi. (a) 704 N . Ill. (b) 580 N . m. (a) 1.25 kN· m. (hl 1.53 kN· m. (a) 38.4 N . m. (b) 52.8 N . m. (a) 57.6 N . m. (b) 83.2 N . m. (a) 0.0258 in. (b) 00793 in. (a) 0.521 in. (bl 1750 ft. (0) 19.44 kN· m. (h) 28.1 kN· m. (II) 10.08 kN . m. (h) 16.12 kN . Ill . (a) 322 kip' ill. (b) 434 kip' in. (a) 308 kip' in. (b) 406 kip . in. (a) 29.2 kN· Ill . (b) 1.500.

4.79 4.80

(a)

462 kip ' in. (b) 1.435.

(a) 420 kip' ill. (b) 1.364.

4.82 4 .84 4 .85 4 .86 4 .87 4.88

91IN·Ill. 48.6 kN . Ill . 111.0 kip' In . 211.5 kip' in. 120 MPa. 145.7 MPa. 4.89 (a) - 11 .87 ks i. (bl 18.26 ksi. 4.90 (a) - 13.36 ks i. (b) 15.27 ksi. 4.91 (a) 106.7 MPa. (b) - 31.15 Illlll , O. 31.15 (e) 24.1 Ill . 4.95 (II) 292 MPa. (b) 7.01 mm. 4.97 (a) 43 hi. (b) 10.75 kip' in . 4.99 (a) (TA =IT B = - 1.875 hi . (b) (TA ; - 3.59 ksL (T il = 1.094 ksi. 4 .100 (II) (TA ; (T il = - 1.875ksi. (b) ITA = - 2.97 hL (Til 1.719 ksi.

Illlll.

=

4.101 (a) - 2Plrrr. (b) - 5Plrrr. 4.103 (II) 112.8 MPa. (b) - 96.0 MPa. 4.105 (II) 288 lb. (b) 201J lb. 4.106 (a) - 79.6 MPa. (b) - 139.3 MPa. (e) - 152.3 MPa. 4.107 14,40 kN. 4 .108 16.04 mill. 4.110 (l,455 in. 4.113 (a) 52.7 MPa. (b) - 67.2 MPa. (e) 11.20 Illlll above D. 4.114 (a) 1125 kN. (b) 817 kN. 4.116 23.0 k.ips. 4.118 (a) 40.3 kN. (b) 56.3 Illlll from left face. 4.119 (a) 69.6 kN. (b) 41.9 nlln frolll left face. 4 .120 2.09 kips. 4.121 P = 44.2 kips, Q"'" 57.3 kips. 4 .122 (II) 30 11I1l1. (b) 94.5 kN. 4.125 (a) 150 Illlll . (b) - 10 MPa. 4.126 (a) - 2.80 MPa. (b) 0.452 MPa. (e) 2.80 MPa. 4.127 (a) 7.57 ksi. (b) - 1.03 ksi. (e) - 7.57 hi. 4.129 (a) 0.321 ksi. (b) - (l.107 ksi. (c) 0.427 hi. 4.131 (a) 65.8 MPa. (b) - 164.5 MP'..I. (C) - 65 .8 MPa. 4.132 (a) 57Y (b) 75.1 MPa. 4.134 (a) 1&'28'. (b) 13.59 hi. 4 .135 (II) 27.5". (b) (T o = 5.07 ksi. 4.136 (a) 32.9°. (b) (T E = 61.4 MPa. 4 .137 - 2.32 ksi. 4.138 113.0 MPa. 4.140 (a) (TA =41.7 psi. u B =292 psi. (b) Imcrsecls AB at 0.500 in. from A and imersccts BD 0.750 in. from D. 4 .141 (II) (T" = - 62.5 psi. (T 8 = - 271 psi. (b) Docs not imcrsecl AB. Imcrsects BD al 0.780 in . from B. 4.142 (a) (TA = 31.5 MPa. (T 8 = - 10.39 MPn (b) 94.0 nun above poim A. 4 .144 36.8 mill. 4 .146 29.1 kip' in. 4 .148 733 N · Ill . 4 .149 1.]23 kN· m.

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4. 150 4. 151 4.156 4.157 4.158 4. 160 4.162 4.163 4.166

4.Hi7 4.168 4.169

4.170 4.171 4.172 4.174 4.176

4.177 4.184 4.1flS 4.187 4.189 4.190

900 N' Ill. 470 0 (0 ) - 5.62 ksl. (b) j.27 hi. 0 ... = - 5.40 hi. ITII = 4 .12 ksi. 0 .1 65.1 MPa. 0 8 - 39.7 MPa 60.9 mm. (a) - 82.4 MPa. (b) 36.6 MPa. 655 ]0. (tI ) 3.06 hi. ih) - 2.8] hi. «() 0.529 hi . (a) - M.I MPa. (b) 65.2 MPI. (a) - 106.1 !\IPa. (b) 38.9 MPa. (tI) - 45.2 MPa. (b) 17.40 MPa. (a) - 43.3 MPa. (h) 14.43 MPa. (a) 1t..Oj h i. (b) - 9.84 hi. (a) 7.07 hi.lb) - 3.37 hi. (a) 6~.9 MPn. (b) - 52.6 MPa. (a) 6.71 hi. (hi 3.24 b i. (a) - 3.65 ksi. (b l 3.72 hi. 259 kip' in. (n) IQ11 hi (h) n (}.t!!l It-. . in 887 N · ITI. (a) 4.87ksi.(b) 5.17k5i. (tI) 5&.7 kN· Ll1 . ib) lO.O m.

4.192 (0) - Pl2nt. (b) 2 Plat. (e) - PI2a1. 4.193 (a) 57.& MPa (b) - 56.8 MPa. (c) 25.') MPa. 4.195 5.22 MP~, - 12.49 MPa. 4.Cl (/ = J 111m: a', = 50.6 MPa. 0 , = 107.9 MPa; (/ = 14 mm: IT. '" 89.7 MPa. rJ, '" 7 1.8 MPa. (u) II 1.6 MPa. (b) 6.6]

f3

111111.

= J(f': u ... - 7.&3 ksi, a n = - 5.27 ksi. 0 c = 7.19 h i. 0 0 = 5.9 1 ks i: f3 = 12if: u A = 1.557 ks i. u ll = {..t)1 ksi , '7C - 2.67 hi, lTD - - 4.8') ks i. 4.C4 f ilII = 0.529 for 50% increase in ITm.. ' 4.C5 Prob.4.8. 15.40 ksi: - 10,38 hi. 4 .GII )'y = D./} in.: 76.9 I.:lp . In . '51 in: )",. =0. 1 ;n.: 95.5 kip' in . 13&.lm 4.C7 (/ = 0.2 itr.: - 7.27 ksi , (/ = ON ill.: - 6.6 1 hi. r-or 1/ = 0.625 in .. " = - 6.51 ksi

4.C3

5.6

5.7 5.8 5 .• 5.10

5.2

(II) WI ...... = 11'//2 , IMI _ = l1>,-'"/ R (0) V = W(U2 - x). M = W(Lr - .t)Y2. (,I) and (0) (0 < x < a). V ;- PhiL.

M - Pbx/L; (a < ., < L). V - PaiL. M = Pa{L - x )lL. (a) and (D) (0 < x < (I), V ;- - p , M = - Px; «(I <: .r <: 2a), V = - 2P. M = - 2P.< + Pa. (a) 1\1..., = waU2.IMI ....... = 1V,j..2i6. (D) V = - WC-rl/ 2L, At = - w o-r·'/ 6L. A lu B: V - P.M - P.'; B tuC: V - O, M - Pu ; CloD: V =- - P.M = P{L - x ). 0;

5.3

5.4 5.5

no

5.12

lo) 3 45 kN . (b) 1125 N·IlI. (a ) 900 N. (h) 11 2.5 N .

5.15 5.17 5.18 5.20

] O]J hi.

5.21 5.22

5.24 5.25 5.27

5.28 5.29 5.31

5.32 5.33 5.50

5.54 5.55

5.56

5.59 5.60 5.62 5.64

75.& MPa IVlmII' - 27.5 kips , IM I..., - 45.0 kip ' 1'1 , Urn ... ]4. ]4 hi. WI",", = f.6.8 IN. jAl l_ = 30.7 kN . In, '7 m ", = 56.0 II1P3. iV l.,~, = 142 N, IMI-.... = ~ I (,\J. m. ...,,,,... ;- 17.19MPa. WI""" = 5.77 kips , IMI..., = 25.0 kip ' ft. '7,""" - ]0.3 4 hi. (a) 10.67 kN . (b) 9.52 MPa. (a ) 866 1U1ll. (b) 99.2 MPa. (a) X]'" mill. (b) S9.5 MP'.I. (tI) 309 ft. (h) 12.95 hi. 102 1 in. (a) 33.3 !1l1ll. (b) 66.6 Illm. (0) V - w o(L' - 3.,lY6L. M - !(iri,L, - .,:;IL )l6.

,

(a) V = wo(x - xlL ). !of = JV~fn - x l /3L). w(lL~/6. (b) V = 1VJxi2 - 3.r'i4L). /If - w o(.i/4 - ...)/4L). 11',)..'/27 . (a) IVI ....., = I(,.SO kN . JAI\.,.., = 8.82 kN ' Ill. (b) 73.5 MPa to) WI=>. = 15.00 I.:ips, IMI ..., = 37.' I.: ip , !l . (b) 900 h i (a) IVI..., = 128.0 kN , jAl ~, ;- 81).6 kN . Ill. (h) 1'(' .6 MPa. (u) WI".., - 30.6 kip~. !hl l ..., - (,(J.O k...p' fl. (h) 21.6 hi. Wlb .., = f..50 IN, jAl l"", = 5.~ kN . In. 30.3 MPa. IVI""" = 4X.U kN .IMlm" = ]:V,JU I.:N ' 111.61.') MPa. P = 500 N. Q = 250 N, W I."" = 1150 N. !AIl".., = 221 N ' m. (a) WI""" = 24.5 kips, IMI - = 36.3 kip . ft.

5.73

(b) 15.82 k,i. II = 173.2 mm . II = 15.06 in. = 203 mill. b - 48.0 mill. W16X40. W27 X 84. WZ50 x 1&.4

5.74

W5 30 x 66

5.65 5.68 5.69 5.70 5.71 5.72

Ill.

10.49 hi 139.0 MPa.

(h) 00642 WJ} , 5.52

s.sa

S. l

(a) 42.0 kN . (b) 27.0 kN ' Ill . (a ) kN . (h) 96.0 tN · Ill. (,,) 30.0 kip •. (h) 90.0 kip' fl.

5.13

0;

CHAPTER 5

ta) IVI..... = '/O( L - 2a )/2, !h11..." '" lL~ elh - a'/~) (i>l A In R: V = 1/1(1. - 1.n)/2. M = m(f , - 211) <12' 810 C: V = w(li2 - .f). M = W[(L - 2a).r - (x - (I)' J12. C 10 D: V !V(L 2ay Z. !of - w(L 2a X L .~y2 . (a) 700 lb. (b ) 57.0 Ib . ill.

'I

783

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5.76

S20 X 66.

s.n

S~10X[)R

5.79

318 in , b = 11.74 in. CI80 x 1~.6.

5.BO 5.81 5.82 5.83 5.84 5.85

5.B6 5.89

Ll02 X 76 X 12.7. W610 X IOJ. W 14 x 68. 1,6.8 kN/m. IOS.s kN/m. (a) lABS kN/Ill.

(0' 1.935

S.'" 5.92

5.94 5.95 5.96 5.97 S.98

5.110 (al V=30 - 24( t - 0.75't - 2 4(r- l..:5 r - 24(.T - 2.25)0 .... 66{~ - 3)l} m. M = ]Ox - 2~x - 0.75)1 - 24{x - 1.5)' - 14(r - 2.25) + 66(x - 3)) kN . III (b) 87.7 !'.IPa. 5.114 (al 80.0 kip . it at point C (b) W 14 X 30. 5,115 (a) 121.5 kip' ft at 6.00 ft fmlll A. (b) WI6 X 40. 5. 116 (a) OXll kN . m at 2.1)<) III from A. (hi Ii "" lJll mill. 5.118 1"",..."'35.6kN· IMI-, =25.0kN.m 5.119 iV'""" = 89.0 kN, IM!- = 178,0 kN . m. 5.121 IVI"", '" I 0.75 kips. ~Im.. = 52.5 kip· fL 5. 122 (a) IVI.... 1J.80kN , IM ,- = 1616 kN . m.

l

I .....

= 12.7111111.

Ill.

;0

(a) 4 .01 kN. tb) 3.27 m. (1I) Wl6 X 40. (b) W21 X 44. (e) Wl4 x 30.

383 336

(I,) Rll

~IP.l

5.123 (al IVim., = 40.0 kN.1A1 ~ .... = 30.0 kN . nl (b) <W.O MPa. 5.124 (a) IVI""" - 3.84 kips, ~1 ,," .. - :l.80 kip · ft. (bl 0.951 ks i. 5.126 (al II = hn[x(L - x)/L! ] lI2 .

nlllL

nUll .

W21 x 84. 1U'l, (a) V = - wox + 1V o(X - fl)'. M = - Wot'12 + 1Vo(x - a'l12. (b) -31/J~12 5.100 (a) V = - we)' + wfj-r-/2a - wix - d~/2a. M = - wox 1/ 2 + 1I.'1~\.l/6a - wJ"x - a'/I60. (b) - 510#,16

(bl 4 .44 kip/in. 5.127 (a) h..(.l IL)If1. . (b) 10.0 kiF' 5.128 (a) II = ",,(luLl!" forO < x < L'2. (b) 60.0 kN. 5.131 (u) II - /,o(.)L) " ". (v) 167.7111111. 5.132 " = 6.00 ft. I, = 4.00 f1. 5.134 I"" 1.&00 m.

5.101 (a) V= 1.15P - p{x - a)IJ- p\). - 1.a)o. M = 1.25 P.I - P\l - a)' - ~x - la)'.

5.137 d = d,,[ 4x, L - x )/L2]1'.'

,,=

5.135 1= 1.900 m. 5.138 (a) b""b~I - X/L) . (b) 20.S 111111. 5.140 (a) 155.2 MPa. (b) 142.4 \iPa

(b) 0.750 Pa.

5.102 (u) V - - wox + 11'0<1" - 211)' + 3wo. - 1V3)' - P4r - 1.Ll3)'h (b) - 4PUJ. 5. 105 la) V= - p(x - d,". M = - p(x - a)' - Fa(r -

wlr -

5.142 (a) 71~f,. 5.143 )93.8kN.

a/".

(b) - Fa

5.106

(a)

V= 40 - 4S(x - 1.5)0 - 6O(x - 3.0)n + 60(x - 3.6)" kN.

M - 40x - 48(x - 1.5)' - 60(\ - 3.0)' + 60{.\ - 3.6)' kN . (bl 60.0 kN . m.

5.10" (a) V = - 3

- 6(x -

+ <J.75(x - j)O -

Ill.

6(x - 7'f

11)° bps.

M = - 3x + 9.75(x -

3)'

- 6{x - 7)' - 6(.f - 1I)' kip (b) 21.0 J..ip . fl . S.108 (a) V = - 1.5x + M = - 0.75x'

' 1'1.

J(x - O,S)11 + 3\:>. - J.i')l kN. + 3{.l - (UW + 3{1 - 3.2)' kN ' Ill.

(b) 600N·1ll

6 . 1M MV-13 - h + * _ »I _ ~ _ ~

- 3(x - II)lki?s. , , M = !J\'- 15r+ I. ~,x - J)· - 8(x - 7)1 - 1.5 ~l - I It kip . fl. (bl 41. 5 kip . fI

784

(hl

Il'i~ ;n

5.14-4 (al 152.6 MPa. (bl 133.6 -"IPa. 5.145 (al 4,49 m. (hi 1.11 nlill. 5.146 (a) 24.0 ksi. (b) 211.3 l si. 5.148 (a) x .. = 15 .00 in. (b) .120Ib/in. ::1.149 lal x,. = ~u.u ilL (/.» ILl:! k,ps. 5.151 (a) x., = 0.140 m . onlh,'j" S.C1 Prob. 5. 18: Au = 2 m: V = O. M = 1()4.0 kN . n:. fT = 139.0 MFa. 5.C4 F(J r x - U .S ft: M, - 13 1.25 kip· f,; M , = 15('.25 kip· fl: Mr = 150.0 kip · fl. S.CS Prob. 5.72: VA = 48 kips. Ms = 320.6 kip' ft. 5.C6 Prob. 5.112: VA = 29.5 kN, M _ "" 28.3 kN· m, :11 1.938 m from A

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CHAPTER 6 6.1 6.2

(a) 1.387 kN. (b) 3&0 kPa. (a) 155.& N. (b) 329 kPa.

6.3 6.5 6.7 6.8

326 lb. 11.54 kips. 193.5 kN. 217 kN. (a) 8.97 MPa. (b) 8. 15 MPa. (a) 13.32 hi. (b) 12.07 hi. (a) 3.17 ksi.(b) 2.40 hi. 177.9 kN. 300 kips. 178.7 kN . 14.05 in. (a) 320 mm. (b) 97.7 n1l1l . (a) 31.0 MPa. (b) 23.2 MPa. (a ) 1.745 hi. (b) 2.82 hi . 32.7 MPa. 3.21 hi. 1.500. 1.333. 1.672 in. 189.6 lb. (a) 239 N. (b) 549 N. «(I) 146.1 kN/m.
6 .9 6 .11 6.12

6.13 6.15 6.16 6.18 6.19 6.21 6.22

6.23 6.24 6.25 6.27 6.29 6.31 6.32 6.34 6.35 6.37 6.39 6.40 6.41 6.42 6.43 6.44 6.45 6.46 6.48 6.49 6.51 6.52 6.53 6.54 6.57 6.59 6.61 6.62 6.63 6.64 6.67 6.68 6.69 6.70

6.71 6.72 6.75 6.76

6.n

(a) Vsin{ifrrr.. ,

0.774 in. (a) 0.&88 hi. (b) 1.453 ksi. (a) 2.59 h i. (b) 967 psi. 3(b 2 - a' )(6((1 + b) + Is J. 1.250 a. 0.3 45 a. 0.71 4 a. (
6.78 6.81 6.82 6.85 6.86 6.87

6.88 6.89 6 .91 6.92 6 .94

6.96 6.97 6 .98

6.100 6.C1 6.C2

6.C3 6.C4 6.C5

75.0 nUll. (a ) V = 500 lb. M o = 398 lb· in. (b) 2980 p~i. (a) V = 500 lb. M o = 398 lb · in. (b) 6090 psi.

=

Tm Plat . Tm = 1.333 Pla/. (a) 144.6 N . m. (b) 65.9 MPa. (a) 144.6 N • m. (b ) 106.6 MPa 92.6 lb. (a) 17.63 MPa. (b ) 13.0 1 ~·I Pa. (a) 1.313 l si. (b) 2.25 hi. (a) 379 kFa. (b) O. 0.371 in. (a) 23.2 MPa. (bl 35.2 MPa. (a) 10.22 mm. (h ) 81.1 MPa (max imum) 1.265 in. (a) II = 173.2 mm. (b) Ir = 379 mm. «(I) L = 37.5 in.; b = 1.250 in. (b) L = 70.3 in.: b = 1.172 in. (e) L = 59.8 in.: b = 1.396 in. Proh. 6. 10: (a) 920 kFa. lb) 7()s kPa. «(I) Tm.o, = 2.03 hi: TB = 1.800 hi. (bl 194 psi. Prob. 6.66: (a) 2.67 in. (b) T8 = 0.917 hi; TlJ = ].36 ksi: T.."" = 4.28 ksi.

CHAPTER 7 7 .1

7.2 7.3 7.4 7.5 7.8 7.9

7.10 7.11

7.12 7.14 7.16 7.17 7 .18

7.20 7.22 7.23 7.24 7.25

7.26 7.27 7 .29

7.55 7.56 7.57

49.2 MPa. T - 2.41 MPa. = 5.49 k~i , T = I UB hi .

0" (1'

14. 19 MPa.T = 15. 19 MPa. - 0.078 ksi. T = 8.46 hi. (a) - 37.0'.53.0°. (b) - 13.60 MPa. - 86.4 MPa. (
(1' ""

785

No part of any book may be reproduced or transmitted by any means without the publisher’s prior permission. Use (other than qualified fair use) in violation of the law or Terms of Service is prohibited. Violators will be prosecuted to the full extent of the law.

7 .58 7 .S1 7 .62

0, 90.0°; 1.732 u[>. - 1.732 U o. _ 5.2° ~ 0 oS 132.0°.

7 .63 7 .65

(a) 33.7", 123.7°.

7 .66

la) S.60 hi. (h) IO.RO ksi .

7 .68 7 .69 7 .70 7 .71 7 .72 7.74 7 .75

(a) 94.J MPa. (b) 105.J ks i. (a) 100.0 MPa. (b) 110.0 MPa. (a) 39.0 MPa. (b) 54.0 MPa. (e) 42.0 MPa. (a) 39.0 MPa. (b) 45.0 MPa. (e) 39.0 MPa. (a) IS.50 hi. (h) 11.00 hi. (el I J.(Xl hi . (a) 40.0 MPa. (b) 72.0 MPa. (a) 8.00 hi. (b ) 4 ..50 hi .

7 .76 7 .78

1.000 ksi; 7.80 ksi. - 60.0 MPa 10 60.n MPa. (a ) 45.7 MPa. (b) 92.9 MPa. (a) 1.228. (h) 1.098. (c) yiclding. (a) 1.083. (b) yielding. (e) yiclding. (a) 1.279. (h) 1.091. (e) yiclding. (a ) 1.149. (b) yielding. (c) yielding. 52.9 kips. 63.0 kips. ruplllrc. no ruplllrc. ruplllrc. ruplUre. TO = :!:49. 1 MPa. TO - 1. 50.0 MPIi. 196.9N · m. 2.94 MPa. 5. 12 ksi. (a ) 1.290 /..·l Pa. (b) 0.853 nun. (a) 12.J8 hi . (b) 0.0545 in. 43.J fl. 16.62 hi. 8.31 hi. 103.5 MPa. 5 1.8 MPa. 1.676 MPa. 89.0 MPa. 44.5 MP~ 12.55111111. 5('sOpsi.1970psi . 56.8°. (a ) 419 kPa. (b) 558 kPa. (a ) 33.2 MPa. (b) 9.55 MPa. 2. 17 MPa. - 27.0" 10 27.0" and 63 .0° \ 0 I 17.0°. (a ) 3. 15 hi. (b) 1.993 hi . (a) 1.486 ksi. (h) 3.16 hi. 77.4 MPa. 38.7 MPa. 73.1 MPa. 51.9 MPa. (a) 5.64 ksi. (b) 282 p,i. (a ) 2.28 ksi. (b) 22& psi. 93.61-<. - 13.58 fL; 641 1-<. 115.01J-.285/J.; - 5.721J36.7 fl. 283 /J.: 227 J.L . 93.61J-. - 13 .6fL;64IIJ.. 115.0/J.. 285/1 : - 5.7 1J..

7 .80 7 .81 7 .82 7 .83 7.84 7 .85 7 .86 7 .89 7 .90 7 .91 7 .92 7 .94 7 .95 7 .96 7.98 7.100

7.101 7 .102 7 .104 7 .105 7.108 7.109 7.110 7 .111 7.112 7 .113

7 .116 7 .117 7.118 7 .119

7.122 7.123 7.124 7 .125

7 .126 7 .127 7.128 7 .129 7.1 31 7.132 7.133

786

7 .135 36.7 IJ.. 283 IJ.; 227 /1-.

7 .136 (a) 64.4°. - 26.6°; ~50 IJ. . - 150.0 fl. - 300 IJ..

16.5" 5. 0 ~ 110. l o.

(b) IT,~ =

V(T, U,

(b) 900 fl . (e) 1050 IJ..

(h) IR.OO hi. (e ) 6.50 ksi .

7 .138 (a ) - 30.1°.59.9°; - 298 /J.. - 702 fl. 500 /J.. (h) 500 fl . (e) 12021J..

UO'U,'T"" • .

7 .139

(a)

- 33.7".56.3°; 100 IJ. . - 420/1-.160 /1.

(b) 520 fl. (e) 580/J..

7.140 (a) JUf. 121.(f: 513 /J.. 87.5 /J.. o. (bl 425 /J.. (C) 513/1. 7.142 (a) 97.8' , 7.8°: 24311-. 56.61J.. O. (h) 186.8/1. (el 243 fl . 7 .143 (a ) 11 .3°. 101.3°; 3lO /1-.50IJ. . 0.(b) 2601J..(e) 3101J.. 7 .146 (a) - 300 X 1O -~ iJl./ in . (b) 435 X IO- ~ in jin .. - 315

X

10- n in.lin.; 750

X

IO- o in./in.

7 .147 (a) 30.0°. 120.0°: 560 X 1O- ~ in.lin .. - 140.0

X

]()-Ii" in./in.

(b) 700 X 1O - ~ in./in.

7 .150 7 .151 7 .154 7 .155

7 .156 7 .157 7 .158 7 .160 7 .161 7 .162 7 .164

7 .166 7 .168 7.169 7 .Cl

7 .C3 7 .C4 7 .C7

7 .C8

69.6 kips. 30.3 kips 34.8 kips. 38.4 kips 1.421 MPa 1.76 1 MPa (a ) - 22.5°.67.5"; 426 /J.. - 9521J.. - 2241J.. (b) Sallic as pan a. (a) 57.9°. - 32.1 °: - 29.8 MPa. - 70.9 MPa. (b) Sallic as part (J. (a ) - 0.300 MPa. (b) - 2.92 MPa. 35.4 MPa. - 35.4 MPa; 35.4 MPa. 12. 18 MPa, - 48.7 MPa: 30.5 MPa. 'To(l + cos 0), 'TOO - cos 0). :!: 60.0 ks'. (a) 4.97 ksi. (b ) 2.49 ksi. 392 psi (tension), 3130 psi (comprcssion). 415 X Io- ~ ;n./in. Pmh. 7.13: (a) 24.0 MPa., - H)'t.O MPa .. - 1.50 MPa. (b ) - 19.5l MPa. - 60.5 MPa. - 60.7 MPa. Pmb. 7.15: (a ) - 56.2 MPa .. 86.2 MPa.. - 38.2 MPa. (b) - 45.2 MPa, 75.2 MPa. 53 .S MPa. Proh. 7.165: (a) 1.286. (b) 1.018. le) yielding. Pmb. 7.93: Ru pmrc occurs at TO = 3.67 h i. Pmb. 7.141: Ufi = 37.9°: E. = - 57.51J.: Eh = - 3831-1: Ec = O. )'..., = 325 micro radians (in plane); )'m., = 383 micro radians Pmb. 7.144: E, = 25311--; E , = 3()7:"y., = - 893. Eo = 727fL; E. = - 167.2; "Ym..... = - 894. Pmh. 7.145: E.• = 72511--: E , = - 75.0;),,, = 173.2. E. = 734/J.; Eo -= - 84.3; y ..... = 819.

CHAPTER 8

8.1 8.2 8.3 8.4

8.7 8.8

(a) (a)

(a) (a) (a) (b) (a) (h)

10.69 ksi. (h) 19. 17 hi . (e) not acceptable. 10.69 ksi. (b) 13.08 hi. (el aeceplablc. 96.2 MPa. (b) 95.4 ksi. (e) acceptable. 93.4 MPa. (b) 96.6 MPa. (e) neccptablc. \\'690 X 125 . 128.2 MPa; 47.3 MPa; 124.0 MPa. W360 X 32.9. 146. 1 MPa: 27.6 MPa; 118.4 MPa.

No part of any book may be reproduced or transmitted by any means without the publisher’s prior permission. Use (other than qualified fair use) in violation of the law or Terms of Service is prohibited. Violators will be prosecuted to the full extent of the law.

8.9 8.10 8.12 8.14 8.15 8.16 8.19 8.20 8.25 8 .26 8.27 8 .28 8.29 8.30 8.31

(a) 134.3 MPa. (b) 129.5 MPa. (a) 155.8 MPa. (h) 143.8 MPa.

(a) 18.97 ksi. (b) 19.92 ksi. (lI) 22.3 ksi. (b) 20.6 ksi at midspan. 16.38 ksi at C and D. 21.7 n1lll for Be. 33.4 Illlll for CD. 44.4 MPa in BC, 48.0 MPa in CD. 1.578 in. 1.698 in.

46.5 Illlll. 45.9 Illlll. 37.0 Illlll. 43.9 Illlll. 1.822 in. 1.792 in. 2.08 MPa a:K! 0 at a; - 2.72 MPa and 0 at b: - 3.20 MPa and 0.500 MPa at c. 8.32 (a) 11.06 ksi. O. (b) - 0.537 ksi. 1.610 hi. (e ) - 12.13 ksi. O. 8.33 (a) - 12.34 ksi. O. (b) - 1.073 ksL 0.805 ksi. (e) 10.20 ksi. O. 8.35 (a) - 37.9 \lIPa. 14.06 MPa. (b) - 131.6 MPa. O. 8.37 - 21.3 ksi. 5.23 ksi. 8.38 (a) 79.6 MPa. 7.96 MPa. (h) O. 13.26 MPa. 8.40 (lI) 20.4 MPa. 14.34 !l.IPa. (h) - 21.5 MPa. 19.98 MPa. 8.41 (a) 30.0 MPa. - 30.0 MPa. 30.0 MPa. (h) 7.02 MPa. - 96.0 MPa. 51.5 MPa. 8.42 (a) 3.79 ksi. - 8.50 ksi. (h) 6.15 ksi. 8.44 55.0 MPa. - 55.0 MPa. 55.0 MPa. 8.46 (a) 3.47 ksi, 1.042 ksi. (b) 7.81 hi, 0.7S1 ksi. k) 12.15 ksi. O. 8.47 (a) IS.39 MPa. 0.391 MPa. (b) 21.3 MPa. 0.293 MPa. (e) 24.1 MPa. O. 8.48 (a) - 7.98 \lIPa. 0.391 MPa. (b) - 5.11 .\lPa. 0.293 MPa. (c) - 2.25 MPa. O. 8.49 25.2 MPa. - 0.87 MPa. 13.06 MPa. 8.50 34.6 MPa. - 10.18 MPa: 22.4 MPa. 8.52 1.798 ksi, - 0.006 ksi: 0.902 ksi. 8.53 86.5 MPa and 0 at a: 57.0 MPa and 9.47 MPa at h. 8.55 3.68 ksi. - D.0 15 ksi: 1.845 ksi. 8.56 0.252 ksi, - 6.59 ks i: 3.42 ksi. 8.57 29.S MPa. - 0.09 MPa: 14.92 MPa. 8.60 (a) 51.4 kN. (b) 39.7 kN. 8.62 (a) 12.90 ksi, - 0.32 ksi: 6.61 ksi. (h) 6.43 ksi, - 6.43 ksi: 6.43 ksi. 8.64 0.48 ksi. - 44.7 ksi: 22.6 ksi . 8.65 (a) WIS X 35. (h) 21.9 ksi: 3.06 ks i: 21.0 ksi aL C, 19.81 hi at B. 8.67 41.3 mm. 8.69 - 3.96 ksi, 0.938 ksi. 8.71 P(2R + 4r13)1m.l. 8.72 65.5 MPa. - 21.8 MPa; 43.8 MPa. 8.7 4 30.1 MPa, - 0.62 MPa. 15.37 MPa. 8. 75 - 16.41 hi and 0 at a; - 15.63 ksi and 0.047 ksi at b; - 7.10 ksi and 1.256 ksi at L". 8.76 (u) 7.50 MPa. (h) 11.25 MPa. (e) 56.3 6 , 13.52 MPa. 8 .C5 Prob. 8.45: a = 6.00 ksi: 7" ~ 0.781 ksi.

CHAPTER 9

i. (e)

' .1

(a) )' = M~\112 El. lh) MrJ..?/2 EI

' .2

( a) )' = - P.r2 (JL - x)l6 EI. (h) PL '/3 EI

' .3 ..4

. .S ' .7

•••

9.10 9 .11 9.12 9.13 9.16 9.17 9.18 9 .19 9.20 9.23 9.24 9 .25 9.26

MJ.1EI.d..

-l-.

(e) I'L2/2 EI""". (a) )' = - WU4 - 4LlX + 3L")124 EI. (h) 11)L4/8 El l (e) 'WL% El.d.. (lI) )' = - wJ'h~ - 5Lr4 + IOL4 x - 7L~)/120 ElL. (b) 7wr/.!/ 120 EI t. (e) Wi.'/ 12 E/~. (a) - 'W(t~ - 4LtJ + 4L'l:(1)124 £1, (b) 1OL4/24 t:J -.1,. (e) O. (lI) Y = W(12L~ - 5x' - 6L'x' )j120 El.

e.

(h) 13wL~.'I920 Ell (e) wL1/ 120 EI """<:;. (a) 6.55 X lQ-l raJ ""{,;. (h) 0.226 in. t. (a) 2.77 X 10-' md~. (h ) 1.l56 Illllll (a) 0.(lO652 wL'IEI J.. at x = 0.519 L. (1)) 7.6 1 Illlll t. (a) O.Q]6{l..1 MrJ.}IEI at.r = 0.211 L. (b) 21.5 fL 0.398 in. -.I, (a) )' = P(3nr - 3aLr + al J/6 EI. (h) 1.976 mm t. (a) y = - 1!'rf,.5L2x~ - 4L~ + x" - 5L4x')I2 4 ElL". (b) WrJ..·/4D EI-.I,. (a» )' = wJ.1!' - 15L2.\4 + 25L'.r' - IIL~xV3(,O EIL2. (b) I hll,j}!3()(] El ""{,;. (c) O.0091611JrJ..~/El l 3wU8 i. 3Mr.f2L i. 14.44 kN 1. 3.03 kips 1. Rs = 5P!l6i: M ,\ = - 3PUI6. M ,. = 5PUJ2. Ms = 0, RA = 411OU128i: M,I. = O. M", = 0.05 131(1L1, Me = 0.0351 lOLl. Ms = - 7wL'/ 128.

9.27

~~;4(~t~~~(~.~g;;:4'1~(1~~;1~~" ( max. pos.).

9.28

Ro = 9MrJ8L; MJ8 at A. - 71110/ 16 just to the left of C. 9MJI6 just to the right of C. 0 at B. 7wU I28 t, I3lFL~/6144 Ell 5M,/6L .1-. 7MrJ.."/486 EI RA = PI2 i. 111,.. = PU8 ~: - PUS at A and C. PUS at B, R" = wlfl t, M,.. = WLl/ 1 2~; M = 'w(6.tL - 6.1" - elll2. (a) \' = P[bxl - L(r: - ar' - b(L 2 - b~:UV6 ElL. (h) 'Pb(L 1 - b1)/6 ElL "'ll.. (c) Pa"b"/3 EIL.J... (/I) Y = Me[.!J - 3L{.l - af + (3h 2 - L")X1/6 ElL. (b) M J3h" - L1)/(, EIL """<:i:. (e) Mol/h(b - a)/3 ElL i. (a) 5Pa l /2 EI -.I,. (b) 49 Pa 3/6 El l (c) 15 Pif/EI l. (a) Pa 2/ 12 El.d.. (b) Hr'/12 EI i. (d J a/14 EI L (/I) )' = 11'd - 5et.2/48 + L 1.2124 U2) s/OO1/EIL. (h) 1OrJ..4/4SEI.J... (e) 12 IWrJ..4/ 1920EI.J... (II) \' = 'W[Lr:1/27 - {t. - UJt12 4 - 7L 1.t12 43 [lEI. (h) 71O L]12 43 EI""<',;. (e) 2u;L'/243 EI-l-,

9.29 9 .32 9.33 9 .34 9.35 9.36 9.37 9 .39 9.41 9.42 9.44 9.45 9 .47 9.48 9 .49 9.50 9.51 9.53 9.54

t

'r -

+ (.t - a)~124 - 'r: - 3(/)4/2 4 - 5a x/61 IEI . (h)

(1I) W[{/~/6 - .1'4/24

2JWa~/24 £/ -.1, . ' (a) 14.1Xl X IO -J rad""C. (b) 0.3 40 in.l. (a) 0.873 X 1O- 3 rad~(b) 1.64IIllIll-.l,. (1I) 5.46 X lO -l rad -«:;. (b) 3.09 mill l.

t.

(b) 7PL J/ 168 EI J... (b) MrJ.}1I28 Ell (a) 2PI3 t (b) 5pe/486 El l. (1I) 11.54 kN i. (b) 4.18 1l1I1l-.l,. (a) 33.3 kN i. (h) 3.19 Illlll-.l,. (a) 51'116

(II) 9Mr/SL

i.

787

No part of any book may be reproduced or transmitted by any means without the publisher’s prior permission. Use (other than qualified fair use) in violation of the law or Terms of Service is prohibited. Violators will be prosecuted to the full extent of the law.

t. (b) 0.0526 in. l. t. 5'wL 2/l92 ~ (b) wL4n61i EJt.

9.56

(a) 7.38 kips

9.57

(al JwU32

9. 58 9.59

(a) 201'/27

9.60 9.61 9.62 9.55 9.66

0.1520 in.l u126.4 in. from left end. I.M8 Illlll ~ at 2.86 m from left end. 3.10 mOl" al 0.94201 from left end. ((I) 5Pv'/ 162 EI 1.-. (b) PL'f<) £1 "'l.;:. in) Pa1(L - n.V6 ElL t.

9.67 9.68

9.n 9.72

9.73 9.75 9.76

t.

4PL/27 ~. (b) 5PL l /129li Ell. (a) 0.341 in.l at 3.34 fl fmm left ~nd .

(b) Pa~(3L - iI)/6 ElL.d.. (a) M./.}/8 Ell (b) M0U2 EI..d.'.. (a) wL ~r.:lg4 Ef../.. (b) O. 13Wa'/6 El ""l",;; 29wa 4(24 EI ~ 'w e/48 E/~; wL1/384 E/~. (dU X IU- 1 rad ""<;: 5.j:l Illill t 12.55 X ]() ~J ",d"7,:; 0.3M in. L

(cl SwL4 /512 EI!-

9.140 (a) 17PLJ ,'972 EI !-. (b) 19PL' I972 EI !-. 9. 141 0.01792PL 'l EI../. a! 0 ..544 L from left end . 9.144 0.21 2 in . --1 'u 5.15 f! from lefl ~nd .

9.79

12.08 X IQ -J rod "'<>:;:; O.24{) in. 1.-. (a) 0.501 X IO - J tad ~. (b) 3.67 mm-1.. (a) R" = 3MJ.L" - a 1Y2L' t: M ,\ ;- O.

9.80

(b) R~ = 3M,JJ..' - al}12Lj ..J.- : !I1~;" Md..L2 - ::I,hllL!~. (a) R~ = 7wUI28 MA. '" O.

9.77

9.81 9.83 9.85 9.86 9.87

9.88 9.89 9.9 2 9.93 9.94 9.95 9.96 9.97

t.

(b) R, = 57wU128 i:MB =9wL'.' 128.J.. RA - 3l'/8 1. He - 7P,'S t. Ro - P.'4t R, = 3M(~'2L l: Mu = MoI4 \ (a) 5.94 mlll~ . (b) 6.75 Illlll l. (a) 0.221 in. l. (b) 0.368 III. l. (tI) 5.06 Y 10- 1 ,......t ""'t". (1)) 0 .0477 in. t. 12 1.5 N/m. (a) R~ = 10.86 kN t ; .H" = 1.942 kN· III 'I' ib) Rp - 1.144 kNt M /) - O.28GkN· mi. 4".9 kN. O.27S in. t

',1..5 I

!11m

T.

(tI) MoLiEI "1;. (h) M"Ll l2 El

t.

(a) PL"12 EI d. . (h) PL /3 EJ 1. (a) wL /6 EI d. (h) wL~i% EJ 1. ' 9. 100 \a)3Pa'f2EI'"'>;, 11/'{i'/ oEIJ.. (h) Pa 2I FI""t . Pa'l2 FI !9.101 (ll) 16.56 X IO - J md ~. (b) 0.379 in. !-. 9.102 (a) 2.55 X 10- " raJ d. (bl 6.25 mill l 9. 103 (a) 4 .98 X IO-J rnd d: O.157!) in . ./.-. j

(b) 4.59 X IO - l roo d: 0.0842 in. !-.

t.

9.104 (a) 5.22 X 10- ) roo d. (bl 10.88 mm 9.1 06 ( tI) J WL'124 EI "... (bJ J j PL'/36 EI !.

g.107 (ll) 6. 10 X IO - l roo 2. (hi 6.03 mm ! 9.109 (a) Pa(L - <1)/2 Ef"""<;, (b) P(I(3L 1 - 4a 1 )124 Ef-.1--. 9.11 0 (a) PL:/16EI-:x;..(b) PL'/48EJl. 9.1 11 (a) 5rl}f"J2 n....-.;. (b) 19nj IJS4 EJ l 9.112 (a) MtlL - 21;)I2EI<:::.(b) M ,jL' - Llh/ 8EI --!.. g.1 14 (a) w.11(3L - 2a)/12 EJ"""Z (b) w.r(JLl - 2(1z)i48 £/!-. 9.11 5 (a) 51'a'/ i! H ""t. (b) JJ'a l /4 t.1 ../.. ; .11 8 «(I) ,J.72 X 10- 1 roo -<:t;. {h) 5.&5 Illill I. 9.119 (a) 4.50 X IO - l roo ~. (h) 8.26 mm 1. 9.121 (a) 5.17 X IO - J rad '""'<;. (bl 21.0 mill l. g.1 22 J.~4 kN l m. 9.1 23 n.l ll I ,

788

9.12-4 0.223 L. 9.125 (a) 3Mo IL'/64 EI .... (b) SMu U32 EI""<;. 9.126 (a) 4PL J1243 Ell. (b) 4PL:/8 1 EI"'{. . 9,127 (a) 5wLY 'oll t.1 J.. {bl j'lce/ llr, 1:.:1 """'C. 9.129 (tI) 8.70)< lO-l ,.,.d""<;.. (hi 15,03 nun 1. 9.130 taj 5.3 I X IO - J raj """<;. (h) 0.204 in. !9.132 (ll) 7.48)( 10-] raj "l.;:. (b) 5.35 mill .J.. 9.134 (u) P,,(2L + ),<)16 EI-q;. (I~) P",(L + uYJ EI-J.. 9.135 (a) 5.33 X 10-' raj ..d.'... (bi 0.01421 in l 9.136 (u) 3.61)( 1O- 1 rad-q;.(b) O.960mmt. 9.137 (a) 2.34;X 1O- 1 md "l.;:. (b) 0.1763 in. t 9.139 (ll) <)lIIL)1256 EJ '""C. (b) 7wL·'/256 EJ ..d.'..

9.145 0.1049 in. 9.146 1.841 in. 9.147 9McI8L i. 9. 148 5P/ 16i. 9.150 7wUI28 t 9.152 JI'/j2 t atA. 1:1 /' 1:12 f ot H. I WilD f at C;, 9.153 (tI) 6.95 mm 1. (b) 46.3 k'>I 9.154 R£ = 10. 1& kips 1; - 87.9 kip ' fl at A. 46.'1 kip· ft at D. 45.8 kip . ft at E. 0 at B.

t.

9.155 4~ anL' . 9.156 144 Ellt'. 9.157 (a) y = - w()(~..--I - 5ex

+ 4(5)1 120 ElL. wJ.'124 EI L. 9. 169 (tI) y - IV l (.," - JL,.'f + 5L V - 3L5x ).llJO EIL1. (b) w"L'/30 Ef~. (0:;) 61 ,ftJ"I}15760 Ei (b) w~~/3() Ef

t (d

l

9.160 R~ = Ph~(3a + MIL' t. M ,I = Pub 1/L! ~; - Pali:L1 ~t A. ZPa'b'fLJ at D. - Pa'bfL' al B. 9. 161 (tI) 9.51 X IO -J rad"""l;;. (b) 5.&0 mill ! 9,163 0.210 in. ~ . 9.165 (a) 5.84 X 10- 1 rad ~. (b) 0.300 in . ../.. 9. 1 G1 0.00677 rL 'fEI J.- ~I 0.4')3 L flUJU kfl un!. 9 .168 R" = 65.2 kN 1: OatA. 5S.7 k.~. maID. 55.8 kN . m at E. - 82.8 kN . m at B. 9.C 1 I'mb. '1./4: 5.~b X W -] rad '"": 2.5U mill ~ . 9.C2 II = 6/t: {a) :>.11 X W -j!"ad~. 0.292 in . ~ : (h) 0.397 in. !- a\ 11.27 fl [0 the right cf A. 9.C3 X = 1.6 /JI' ((I) 7.90 X IO - J rad oq,:, 8.16 mlll!-: (b)

6.05 X 10-" rsd"""l:;, 5.79 mm y;

(el 1.021 X IO-l rad"C. (L114mm J.-. 9.e5 (a) {I = 3ft: 1.586 X 10- 1 rad oq;;: 0.1369 in. (b ) a = I.a m: 0.293 X IO - J rad~. 0.479 mill../., 9.C7 x - 2.5 'II" 5.31 Illlll l: x - 5.0 III: 12.28 mm 1

J:

CHAPTE R 10 10. 1 10.2 10.3 10.4

KIL. kL. 2/..U9. KIL.

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10.7 10.8 10.9 10.11 10.12 10.13 10.15 10.16 10.17 10.20 10.21 10 .24 10.25 10.26 10.27 10.28 10.29 10.30 10.32 10.33 10.35 10.36 10.39 10.40 10.41 10.42 10.43 10.45 10.46 10.47 10.48 10.50 10.51 10.52 10.53 10.54 10.57 10.58 10.59 10.61 10.63 10.64 10.65 10.68 10.69 10.70 10.71 10.73 10.74 10.75 10.78 10.79

/.: > 4.91 tN/m. RKIL. 81.2 N. d .; 0.&49 in.: sled PIT = 647 lb. aluminum PIT = 1872 lb. (a) 641 lb. (b) 0.651 in . (e) 58.R';{ (al 6.25%. (bl 17.17 kN. 2.81. 15.02 kips. 25.1 kips. 4.(X) kN . 5.37 kN . (a) LIJC = 4.1 fl. LC~ = 1.05 ft. (b) 4.21 kips. 657 mm. (a) bid = 0.500. (b) d=28 .3 m. b;- 14.151ll1ll. (I ) 2.64 kN. (2) 0.661 kN. (3) 10.51 kN. (4) 5.39 kN, (5) 2.64 kN. (a) 2.19. (b) (2) 1.168 in.: (3) 1.250 in.: (4) 1.046 in. (a ) 0.4lO in. (b) 14.43 ksi. (a) 4.84 Illlll. (b) 135.1 MPa. (a) 0.0399 in. (bl 19.89 ksi. (a) 1.552 mm. (bl 47.8 MPa. (a) 18.2 kips. (b) 27.5 ks i. (a) 52.9 kips. (b) 20.4 ksi. (a) 36R kN. (b) lO3.8 MPa. «I) 223 kN. (h) 62.8 MPa. 31.2°C. 5.8 1 111m. (a) 32.1 kips. (b) 39.4 kips. (a) 195 .4 kN. (b) 0.596 . (a) 247 kN. (b) 0.437. 1.J37 Ill. 9.57 f1. (a) 8.31 Ill . (b) 2.54 m. W200 x 26.6. W200 X 35.9. 2.125 in. 2.625 in. (a) 114.7 kN. (b) 208 kN. 95 .8 kips. (a) 220 kN. (b) IB9 kN. 35.9 kN. (a) 26.4 kN. (b) 32.4 kN. 16.6 kips. 1600 kN. 113 .8 kips. 101.7 kN. 38.9 kips. 6.53 in. 123. 1 11I1ll. 0.884 in. 44.9 111m. W250 x 67. WI4 X 82.

10.80 10 .81 10.82 10.84 10.85 10.86 10 .87 10.88 10.89 10.91 10.92 10.93 10.95 10.97 10 .98 10.99 10.100 10.102 10.103 10.105 10.106 10.107 10 .109 10.110 10.111 10.113 10.114 10.115 10.117 10.119 10.120 10.121 10.123 10.124 10.126 10.128 10.Cl 10.C2 10.C3 10.C4 10.C5 10 .C6

W200 X ~6.1 . L89 x 64 X 12.7 mm. L89 x 64 x 12.7 mm. L3 ~ X 2 1 X ~ in. 56.1 kips (a) 433 kN. 321 kN.
CHAPTER 11 11 .1

(a) 21.6 kllml. (b) 336 kJ/m).

11.2

(a) 43.1 in ·Ib.lin l . (b) 72.8 in ·lb/in 3 .

11 .4 11 .5 11 .6 11 .7 11 .9

(a) 111.9 kllml. (b) 112 kJ/m l . (e) 160.3 kl/m J .

(c) 163.0 kJ/m). (e) 172.4 in ·Iblin l .

(a) 58 in · lb.lin l . (b) 20 kip' in/inl. (a) 1296kJ/m J .(b) 9OM1Iml.

(a) 150.() kl/ml. (b) 63 M1/ml. (a) 116.2 in· lb. (b) 11 .72 in ·lbiin·' ,5.65 in·lblin J •

789

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1 1.10 11.11 11 .13 1 1.15 11.17 11.18 11 .20 11 .21 11.23 11 .24 11.26 11 .29 11 .30 11.31 11.33 11 .34 11.37 11 .38 11 .40 11.41 11.42 1 1.44 11.45 11 .46 11 .48 11.50 11 .51 1 1.52 11 .53 11 .54 11 .56 11 .57 11 .58 11 .59 11.61 11 .62 11 .63 11 .64 11 .67 11.69 11.71 11.73 11.74 1 1.75 11.77 11.79 11 .81 11 .82 11 .83

790

(al 12.18 1. (b) 15.83 kJ/mJ • 38.6 kJ/mJ . (a) 264 in . Ih. (b) 1.378 in . Iblin l . 8.83 in . Iblin l . (a) 44.0 1. (b) 31.4 1. 102.7 in . lb. 1.398 1" I/EA . 3.8 1 I"IIEA. 2.37 I" l/EA. 1015 1. 'w ' L 3 /40 EI.

M~(al

+b

~al (a

+

J

)/6 ElL'. L)/6 EI.

1048 J. 894 in . lb. 662 J. 2.45 X 10" in . lb. 14.70 1. - 2.65 MPa to 122.6 MPa. (a) 2.33. (b) 2'()2. 2 MAL (I + 3 EtfllO GL')/ Ebd' . (Q ' /41T GL ) In (R,iRI)' (al 79.8 kN. (b) 254 MPa. (e) 1.523 mm. 11.50 in. 7.87 in. 4.76 kg. 8.50 ftls. (a) 2 1.0 kN. lb) 171.7 MPa. lel 8.S8Imn. (a) 7.67 kN. (b) 312 MPa. (e) 23.5 mm. (al 0.1064 ill. (b) 20. ] hi . (a) IS.63 mm. (b) 83.8 N · m. (c) 208 MPa. (a) 23.6 nun. (b) 64.4 N . m. (el IS7.6 MPa. (b) 7.12 (b) re lati vc crror O. IS2 . .v~/2h = 0. 166. l'a' b l (3 El L !. Pa'(a + L)/3 EI !. Mo (L + 3a).!3 EI "{;:. 31'L '/16 E/~. 31'aJ/4 EI !. 5MoL/16 E/ ~.

386 mm. S.28°. 2.37S I'IIEA ~ .

0.0447 in. ~. 0.366 in. t. 3.19 mm.1.. (a) and (b) P'L J/96 EI - PM ,.j...'1l6 EI + M fj../6 EI . (a) and (b) I'"'L" 48 EI + PMrP/R EI + M~12 EI. (a) and (b) P'L J;48 EI . (a) and (b) 7I'"L'124 EI . SPL ~/48 EI .1..

11 .84

0.0443wL41E1 ~.

11 .85 11 .87 11 .88 11 .89

3PL ' ;& El .d.. M oLl6 EI -q;:. PaL 2/ 16EIt. IUL~/ 1 28 Eli. wLJ/ 192 El .d.. 0.987 in. ~.

11 .91 11 .92 11.94 11 .95 11 .96 11 .98 11 .99 11 .101 11 .102 11 .103 11 .105 11 .106 11 .107 11 .109 11 .111 11 .11 2 11 .114

11 .116 11 .117 11 .118 11 .119 11 .121

7.25 mm .1.. 5.08 mm .1.. 0.698 X ]()- ' rad~. 2.07 X ]() -J rad L. 3.19 mill J.. 0.233 in. ~. 0.1504 in. ~. 0.1 459 mm!. (a) PRJI2 E/~. (b) TrPR]/4EI!.

TrPR]12 EI .1.. (a) 2PI1/3 EI_.(b) I'P/6 EI Lf.. (a) P/'/EI i. (b) 31'P12 El .d.. SPII6 - 3PUlli at A. 0 at B. SPU32 at C. 3M,j2L M ol2 JlA . - M (j at B. 7wLl l 28 t ; 7wu' 128 - WIX - U2)lfl. wLl6 .1. at A. 3wL'4 t at B, 5wU12 i at C. 71'/8. 0.652 P . 1'/( 1 +2cosJ.p). 2 1'/3. 7 P1l2. 2.09. 12.70 J. 9. 1:2 lb.

t

t

11 .122 11 .123 11.125 1 1.127 11.129 11.57mm .1.. 11 .131 1.030mm --+.

11 .134 PLl/6 El \ 11 .C2 ta) a = 15 ill.: (TD = 17. 19 ks i, (Tc = 21.() ksi; a = 45 in.: IT n "" 36.2 ksi. (Tc "" 14.74 ksi. (b) a = 18.34 in .. (1" = 20.67 hi . 1 1.C3 (a) L =< 200 mm:h = 2.27 mm; L = 800 mm: 11 = 1.076 rnm. (b) L = 44U mill: II = 3.23 n U ll . 11.C4 a"" 300111111: 1.795 mm. 179.46 MPa; a = 600 1Il1ll: 2.87 111m . 179.59 MPa. 11 .C5 a = 2m: (a) 30.0 J; (b) 7.57 mm. 60.8 J. a = 4 Ill: (a) 2 1.9 J; (b) 8.87 mm. 83.4 J. 11.C6 a = 20 in: (a ) 13 .26 in.; (b ) 99.5 kip ' in.; (c) 803 lb. a = 50 in: (a) 9.46 in.; (b ) 93.7 kip ' in.; ~e) 996 lb.

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CHAPTER 6 6.1 6.2

(a) 1.387 kN. (b) 3&0 kPa. (a) 155.& N. (b) 329 kPa.

6.3 6.5 6.7 6.8

326 lb. 11.54 kips. 193.5 kN. 217 kN. (a) 8.97 MPa. (b) 8. 15 MPa. (a) 13.32 hi. (b) 12.07 hi. (a) 3.17 ksi.(b) 2.40 hi. 177.9 kN. 300 kips. 178.7 kN . 14.05 in. (a) 320 mm. (b) 97.7 n1l1l . (a) 31.0 MPa. (b) 23.2 MPa. (a ) 1.745 hi. (b) 2.82 hi . 32.7 MPa. 3.21 hi. 1.500. 1.333. 1.672 in. 189.6 lb. (a) 239 N. (b) 549 N. «(I) 146.1 kN/m.
6 .9 6 .11 6.12

6.13 6.15 6.16 6.18 6.19 6.21 6.22

6.23 6.24 6.25 6.27 6.29 6.31 6.32 6.34 6.35 6.37 6.39 6.40 6.41 6.42 6.43 6.44 6.45 6.46 6.48 6.49 6.51 6.52 6.53 6.54 6.57 6.59 6.61 6.62 6.63 6.64 6.67 6.68 6.69 6.70

6.71 6.72 6.75 6.76

6.n

(a) Vsin{ifrrr.. ,

0.774 in. (a) 0.&88 hi. (b) 1.453 ksi. (a) 2.59 h i. (b) 967 psi. 3(b 2 - a' )(6((1 + b) + Is J. 1.250 a. 0.3 45 a. 0.71 4 a. (
6.78 6.81 6.82 6.85 6.86 6.87

6.88 6.89 6 .91 6.92 6 .94

6.96 6.97 6 .98

6.100 6.C1 6.C2

6.C3 6.C4 6.C5

75.0 nUll. (a ) V = 500 lb. M o = 398 lb· in. (b) 2980 p~i. (a) V = 500 lb. M o = 398 lb · in. (b) 6090 psi.

=

Tm Plat . Tm = 1.333 Pla/. (a) 144.6 N . m. (b) 65.9 MPa. (a) 144.6 N • m. (b ) 106.6 MPa 92.6 lb. (a) 17.63 MPa. (b ) 13.0 1 ~·I Pa. (a) 1.313 l si. (b) 2.25 hi. (a) 379 kFa. (b) O. 0.371 in. (a) 23.2 MPa. (bl 35.2 MPa. (a) 10.22 mm. (h ) 81.1 MPa (max imum) 1.265 in. (a) II = 173.2 mm. (b) Ir = 379 mm. «(I) L = 37.5 in.; b = 1.250 in. (b) L = 70.3 in.: b = 1.172 in. (e) L = 59.8 in.: b = 1.396 in. Proh. 6. 10: (a) 920 kFa. lb) 7()s kPa. «(I) Tm.o, = 2.03 hi: TB = 1.800 hi. (bl 194 psi. Prob. 6.66: (a) 2.67 in. (b) T8 = 0.917 hi; TlJ = ].36 ksi: T.."" = 4.28 ksi.

CHAPTER 7 7 .1

7.2 7.3 7.4 7.5 7.8 7.9

7.10 7.11

7.12 7.14 7.16 7.17 7 .18

7.20 7.22 7.23 7.24 7.25

7.26 7.27 7 .29

7.55 7.56 7.57

49.2 MPa. T - 2.41 MPa. = 5.49 k~i , T = I UB hi .

0" (1'

14. 19 MPa.T = 15. 19 MPa. - 0.078 ksi. T = 8.46 hi. (a) - 37.0'.53.0°. (b) - 13.60 MPa. - 86.4 MPa. (
(1' ""

785

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7 .58 7 .S1 7 .62

0, 90.0°; 1.732 u[>. - 1.732 U o. _ 5.2° ~ 0 oS 132.0°.

7 .63 7 .65

(a) 33.7", 123.7°.

7 .66

la) S.60 hi. (h) IO.RO ksi .

7 .68 7 .69 7 .70 7 .71 7 .72 7.74 7 .75

(a) 94.J MPa. (b) 105.J ks i. (a) 100.0 MPa. (b) 110.0 MPa. (a) 39.0 MPa. (b) 54.0 MPa. (e) 42.0 MPa. (a) 39.0 MPa. (b) 45.0 MPa. (e) 39.0 MPa. (a) IS.50 hi. (h) 11.00 hi. (el I J.(Xl hi . (a) 40.0 MPa. (b) 72.0 MPa. (a) 8.00 hi. (b ) 4 ..50 hi .

7 .76 7 .78

1.000 ksi; 7.80 ksi. - 60.0 MPa 10 60.n MPa. (a ) 45.7 MPa. (b) 92.9 MPa. (a) 1.228. (h) 1.098. (c) yiclding. (a) 1.083. (b) yielding. (e) yiclding. (a) 1.279. (h) 1.091. (e) yiclding. (a ) 1.149. (b) yielding. (c) yielding. 52.9 kips. 63.0 kips. ruplllrc. no ruplllrc. ruplllrc. ruplUre. TO = :!:49. 1 MPa. TO - 1. 50.0 MPIi. 196.9N · m. 2.94 MPa. 5. 12 ksi. (a ) 1.290 /..·l Pa. (b) 0.853 nun. (a) 12.J8 hi . (b) 0.0545 in. 43.J fl. 16.62 hi. 8.31 hi. 103.5 MPa. 5 1.8 MPa. 1.676 MPa. 89.0 MPa. 44.5 MP~ 12.55111111. 5('sOpsi.1970psi . 56.8°. (a ) 419 kPa. (b) 558 kPa. (a ) 33.2 MPa. (b) 9.55 MPa. 2. 17 MPa. - 27.0" 10 27.0" and 63 .0° \ 0 I 17.0°. (a ) 3. 15 hi. (b) 1.993 hi . (a) 1.486 ksi. (h) 3.16 hi. 77.4 MPa. 38.7 MPa. 73.1 MPa. 51.9 MPa. (a) 5.64 ksi. (b) 282 p,i. (a ) 2.28 ksi. (b) 22& psi. 93.61-<. - 13.58 fL; 641 1-<. 115.01J-.285/J.; - 5.721J36.7 fl. 283 /J.: 227 J.L . 93.61J-. - 13 .6fL;64IIJ.. 115.0/J.. 285/1 : - 5.7 1J..

7 .80 7 .81 7 .82 7 .83 7.84 7 .85 7 .86 7 .89 7 .90 7 .91 7 .92 7 .94 7 .95 7 .96 7.98 7.100

7.101 7 .102 7 .104 7 .105 7.108 7.109 7.110 7 .111 7.112 7 .113

7 .116 7 .117 7.118 7 .119

7.122 7.123 7.124 7 .125

7 .126 7 .127 7.128 7 .129 7.1 31 7.132 7.133

786

7 .135 36.7 IJ.. 283 IJ.; 227 /1-.

7 .136 (a) 64.4°. - 26.6°; ~50 IJ. . - 150.0 fl. - 300 IJ..

16.5" 5. 0 ~ 110. l o.

(b) IT,~ =

V(T, U,

(b) 900 fl . (e) 1050 IJ..

(h) IR.OO hi. (e ) 6.50 ksi .

7 .138 (a ) - 30.1°.59.9°; - 298 /J.. - 702 fl. 500 /J.. (h) 500 fl . (e) 12021J..

UO'U,'T"" • .

7 .139

(a)

- 33.7".56.3°; 100 IJ. . - 420/1-.160 /1.

(b) 520 fl. (e) 580/J..

7.140 (a) JUf. 121.(f: 513 /J.. 87.5 /J.. o. (bl 425 /J.. (C) 513/1. 7.142 (a) 97.8' , 7.8°: 24311-. 56.61J.. O. (h) 186.8/1. (el 243 fl . 7 .143 (a ) 11 .3°. 101.3°; 3lO /1-.50IJ. . 0.(b) 2601J..(e) 3101J.. 7 .146 (a) - 300 X 1O -~ iJl./ in . (b) 435 X IO- ~ in jin .. - 315

X

10- n in.lin.; 750

X

IO- o in./in.

7 .147 (a) 30.0°. 120.0°: 560 X 1O- ~ in.lin .. - 140.0

X

]()-Ii" in./in.

(b) 700 X 1O - ~ in./in.

7 .150 7 .151 7 .154 7 .155

7 .156 7 .157 7 .158 7 .160 7 .161 7 .162 7 .164

7 .166 7 .168 7.169 7 .Cl

7 .C3 7 .C4 7 .C7

7 .C8

69.6 kips. 30.3 kips 34.8 kips. 38.4 kips 1.421 MPa 1.76 1 MPa (a ) - 22.5°.67.5"; 426 /J.. - 9521J.. - 2241J.. (b) Sallic as pan a. (a) 57.9°. - 32.1 °: - 29.8 MPa. - 70.9 MPa. (b) Sallic as part (J. (a ) - 0.300 MPa. (b) - 2.92 MPa. 35.4 MPa. - 35.4 MPa; 35.4 MPa. 12. 18 MPa, - 48.7 MPa: 30.5 MPa. 'To(l + cos 0), 'TOO - cos 0). :!: 60.0 ks'. (a) 4.97 ksi. (b ) 2.49 ksi. 392 psi (tension), 3130 psi (comprcssion). 415 X Io- ~ ;n./in. Pmh. 7.13: (a) 24.0 MPa., - H)'t.O MPa .. - 1.50 MPa. (b ) - 19.5l MPa. - 60.5 MPa. - 60.7 MPa. Pmb. 7.15: (a ) - 56.2 MPa .. 86.2 MPa.. - 38.2 MPa. (b) - 45.2 MPa, 75.2 MPa. 53 .S MPa. Proh. 7.165: (a) 1.286. (b) 1.018. le) yielding. Pmb. 7.93: Ru pmrc occurs at TO = 3.67 h i. Pmb. 7.141: Ufi = 37.9°: E. = - 57.51J.: Eh = - 3831-1: Ec = O. )'..., = 325 micro radians (in plane); )'m., = 383 micro radians Pmb. 7.144: E, = 25311--; E , = 3()7:"y., = - 893. Eo = 727fL; E. = - 167.2; "Ym..... = - 894. Pmh. 7.145: E.• = 72511--: E , = - 75.0;),,, = 173.2. E. = 734/J.; Eo -= - 84.3; y ..... = 819.

CHAPTER 8

8.1 8.2 8.3 8.4

8.7 8.8

(a) (a)

(a) (a) (a) (b) (a) (h)

10.69 ksi. (h) 19. 17 hi . (e) not acceptable. 10.69 ksi. (b) 13.08 hi. (el aeceplablc. 96.2 MPa. (b) 95.4 ksi. (e) acceptable. 93.4 MPa. (b) 96.6 MPa. (e) neccptablc. \\'690 X 125 . 128.2 MPa; 47.3 MPa; 124.0 MPa. W360 X 32.9. 146. 1 MPa: 27.6 MPa; 118.4 MPa.

No part of any book may be reproduced or transmitted by any means without the publisher’s prior permission. Use (other than qualified fair use) in violation of the law or Terms of Service is prohibited. Violators will be prosecuted to the full extent of the law.

8.9 8.10 8.12 8.14 8.15 8.16 8.19 8.20 8.25 8 .26 8.27 8 .28 8.29 8.30 8.31

(a) 134.3 MPa. (b) 129.5 MPa. (a) 155.8 MPa. (h) 143.8 MPa.

(a) 18.97 ksi. (b) 19.92 ksi. (lI) 22.3 ksi. (b) 20.6 ksi at midspan. 16.38 ksi at C and D. 21.7 n1lll for Be. 33.4 Illlll for CD. 44.4 MPa in BC, 48.0 MPa in CD. 1.578 in. 1.698 in.

46.5 Illlll. 45.9 Illlll. 37.0 Illlll. 43.9 Illlll. 1.822 in. 1.792 in. 2.08 MPa a:K! 0 at a; - 2.72 MPa and 0 at b: - 3.20 MPa and 0.500 MPa at c. 8.32 (a) 11.06 ksi. O. (b) - 0.537 ksi. 1.610 hi. (e ) - 12.13 ksi. O. 8.33 (a) - 12.34 ksi. O. (b) - 1.073 ksL 0.805 ksi. (e) 10.20 ksi. O. 8.35 (a) - 37.9 \lIPa. 14.06 MPa. (b) - 131.6 MPa. O. 8.37 - 21.3 ksi. 5.23 ksi. 8.38 (a) 79.6 MPa. 7.96 MPa. (h) O. 13.26 MPa. 8.40 (lI) 20.4 MPa. 14.34 !l.IPa. (h) - 21.5 MPa. 19.98 MPa. 8.41 (a) 30.0 MPa. - 30.0 MPa. 30.0 MPa. (h) 7.02 MPa. - 96.0 MPa. 51.5 MPa. 8.42 (a) 3.79 ksi. - 8.50 ksi. (h) 6.15 ksi. 8.44 55.0 MPa. - 55.0 MPa. 55.0 MPa. 8.46 (a) 3.47 ksi, 1.042 ksi. (b) 7.81 hi, 0.7S1 ksi. k) 12.15 ksi. O. 8.47 (a) IS.39 MPa. 0.391 MPa. (b) 21.3 MPa. 0.293 MPa. (e) 24.1 MPa. O. 8.48 (a) - 7.98 \lIPa. 0.391 MPa. (b) - 5.11 .\lPa. 0.293 MPa. (c) - 2.25 MPa. O. 8.49 25.2 MPa. - 0.87 MPa. 13.06 MPa. 8.50 34.6 MPa. - 10.18 MPa: 22.4 MPa. 8.52 1.798 ksi, - 0.006 ksi: 0.902 ksi. 8.53 86.5 MPa and 0 at a: 57.0 MPa and 9.47 MPa at h. 8.55 3.68 ksi. - D.0 15 ksi: 1.845 ksi. 8.56 0.252 ksi, - 6.59 ks i: 3.42 ksi. 8.57 29.S MPa. - 0.09 MPa: 14.92 MPa. 8.60 (a) 51.4 kN. (b) 39.7 kN. 8.62 (a) 12.90 ksi, - 0.32 ksi: 6.61 ksi. (h) 6.43 ksi, - 6.43 ksi: 6.43 ksi. 8.64 0.48 ksi. - 44.7 ksi: 22.6 ksi . 8.65 (a) WIS X 35. (h) 21.9 ksi: 3.06 ks i: 21.0 ksi aL C, 19.81 hi at B. 8.67 41.3 mm. 8.69 - 3.96 ksi, 0.938 ksi. 8.71 P(2R + 4r13)1m.l. 8.72 65.5 MPa. - 21.8 MPa; 43.8 MPa. 8.7 4 30.1 MPa, - 0.62 MPa. 15.37 MPa. 8. 75 - 16.41 hi and 0 at a; - 15.63 ksi and 0.047 ksi at b; - 7.10 ksi and 1.256 ksi at L". 8.76 (u) 7.50 MPa. (h) 11.25 MPa. (e) 56.3 6 , 13.52 MPa. 8 .C5 Prob. 8.45: a = 6.00 ksi: 7" ~ 0.781 ksi.

CHAPTER 9

i. (e)

' .1

(a) )' = M~\112 El. lh) MrJ..?/2 EI

' .2

( a) )' = - P.r2 (JL - x)l6 EI. (h) PL '/3 EI

' .3 ..4

. .S ' .7

•••

9.10 9 .11 9.12 9.13 9.16 9.17 9.18 9 .19 9.20 9.23 9.24 9 .25 9.26

MJ.1EI.d..

-l-.

(e) I'L2/2 EI""". (a) )' = - WU4 - 4LlX + 3L")124 EI. (h) 11)L4/8 El l (e) 'WL% El.d.. (lI) )' = - wJ'h~ - 5Lr4 + IOL4 x - 7L~)/120 ElL. (b) 7wr/.!/ 120 EI t. (e) Wi.'/ 12 E/~. (a) - 'W(t~ - 4LtJ + 4L'l:(1)124 £1, (b) 1OL4/24 t:J -.1,. (e) O. (lI) Y = W(12L~ - 5x' - 6L'x' )j120 El.

e.

(h) 13wL~.'I920 Ell (e) wL1/ 120 EI """<:;. (a) 6.55 X lQ-l raJ ""{,;. (h) 0.226 in. t. (a) 2.77 X 10-' md~. (h ) 1.l56 Illllll (a) 0.(lO652 wL'IEI J.. at x = 0.519 L. (1)) 7.6 1 Illlll t. (a) O.Q]6{l..1 MrJ.}IEI at.r = 0.211 L. (b) 21.5 fL 0.398 in. -.I, (a) )' = P(3nr - 3aLr + al J/6 EI. (h) 1.976 mm t. (a) y = - 1!'rf,.5L2x~ - 4L~ + x" - 5L4x')I2 4 ElL". (b) WrJ..·/4D EI-.I,. (a» )' = wJ.1!' - 15L2.\4 + 25L'.r' - IIL~xV3(,O EIL2. (b) I hll,j}!3()(] El ""{,;. (c) O.0091611JrJ..~/El l 3wU8 i. 3Mr.f2L i. 14.44 kN 1. 3.03 kips 1. Rs = 5P!l6i: M ,\ = - 3PUI6. M ,. = 5PUJ2. Ms = 0, RA = 411OU128i: M,I. = O. M", = 0.05 131(1L1, Me = 0.0351 lOLl. Ms = - 7wL'/ 128.

9.27

~~;4(~t~~~(~.~g;;:4'1~(1~~;1~~" ( max. pos.).

9.28

Ro = 9MrJ8L; MJ8 at A. - 71110/ 16 just to the left of C. 9MJI6 just to the right of C. 0 at B. 7wU I28 t, I3lFL~/6144 Ell 5M,/6L .1-. 7MrJ.."/486 EI RA = PI2 i. 111,.. = PU8 ~: - PUS at A and C. PUS at B, R" = wlfl t, M,.. = WLl/ 1 2~; M = 'w(6.tL - 6.1" - elll2. (a) \' = P[bxl - L(r: - ar' - b(L 2 - b~:UV6 ElL. (h) 'Pb(L 1 - b1)/6 ElL "'ll.. (c) Pa"b"/3 EIL.J... (/I) Y = Me[.!J - 3L{.l - af + (3h 2 - L")X1/6 ElL. (b) M J3h" - L1)/(, EIL """<:i:. (e) Mol/h(b - a)/3 ElL i. (a) 5Pa l /2 EI -.I,. (b) 49 Pa 3/6 El l (c) 15 Pif/EI l. (a) Pa 2/ 12 El.d.. (b) Hr'/12 EI i. (d J a/14 EI L (/I) )' = 11'd - 5et.2/48 + L 1.2124 U2) s/OO1/EIL. (h) 1OrJ..4/4SEI.J... (e) 12 IWrJ..4/ 1920EI.J... (II) \' = 'W[Lr:1/27 - {t. - UJt12 4 - 7L 1.t12 43 [lEI. (h) 71O L]12 43 EI""<',;. (e) 2u;L'/243 EI-l-,

9.29 9 .32 9.33 9 .34 9.35 9.36 9.37 9 .39 9.41 9.42 9.44 9.45 9 .47 9.48 9 .49 9.50 9.51 9.53 9.54

t

'r -

+ (.t - a)~124 - 'r: - 3(/)4/2 4 - 5a x/61 IEI . (h)

(1I) W[{/~/6 - .1'4/24

2JWa~/24 £/ -.1, . ' (a) 14.1Xl X IO -J rad""C. (b) 0.3 40 in.l. (a) 0.873 X 1O- 3 rad~(b) 1.64IIllIll-.l,. (1I) 5.46 X lO -l rad -«:;. (b) 3.09 mill l.

t.

(b) 7PL J/ 168 EI J... (b) MrJ.}1I28 Ell (a) 2PI3 t (b) 5pe/486 El l. (1I) 11.54 kN i. (b) 4.18 1l1I1l-.l,. (a) 33.3 kN i. (h) 3.19 Illlll-.l,. (a) 51'116

(II) 9Mr/SL

i.

787

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t. (b) 0.0526 in. l. t. 5'wL 2/l92 ~ (b) wL4n61i EJt.

9.56

(a) 7.38 kips

9.57

(al JwU32

9. 58 9.59

(a) 201'/27

9.60 9.61 9.62 9.55 9.66

0.1520 in.l u126.4 in. from left end. I.M8 Illlll ~ at 2.86 m from left end. 3.10 mOl" al 0.94201 from left end. ((I) 5Pv'/ 162 EI 1.-. (b) PL'f<) £1 "'l.;:. in) Pa1(L - n.V6 ElL t.

9.67 9.68

9.n 9.72

9.73 9.75 9.76

t.

4PL/27 ~. (b) 5PL l /129li Ell. (a) 0.341 in.l at 3.34 fl fmm left ~nd .

(b) Pa~(3L - iI)/6 ElL.d.. (a) M./.}/8 Ell (b) M0U2 EI..d.'.. (a) wL ~r.:lg4 Ef../.. (b) O. 13Wa'/6 El ""l",;; 29wa 4(24 EI ~ 'w e/48 E/~; wL1/384 E/~. (dU X IU- 1 rad ""<;: 5.j:l Illill t 12.55 X ]() ~J ",d"7,:; 0.3M in. L

(cl SwL4 /512 EI!-

9.140 (a) 17PLJ ,'972 EI !-. (b) 19PL' I972 EI !-. 9. 141 0.01792PL 'l EI../. a! 0 ..544 L from left end . 9.144 0.21 2 in . --1 'u 5.15 f! from lefl ~nd .

9.79

12.08 X IQ -J rod "'<>:;:; O.24{) in. 1.-. (a) 0.501 X IO - J tad ~. (b) 3.67 mm-1.. (a) R" = 3MJ.L" - a 1Y2L' t: M ,\ ;- O.

9.80

(b) R~ = 3M,JJ..' - al}12Lj ..J.- : !I1~;" Md..L2 - ::I,hllL!~. (a) R~ = 7wUI28 MA. '" O.

9.77

9.81 9.83 9.85 9.86 9.87

9.88 9.89 9.9 2 9.93 9.94 9.95 9.96 9.97

t.

(b) R, = 57wU128 i:MB =9wL'.' 128.J.. RA - 3l'/8 1. He - 7P,'S t. Ro - P.'4t R, = 3M(~'2L l: Mu = MoI4 \ (a) 5.94 mlll~ . (b) 6.75 Illlll l. (a) 0.221 in. l. (b) 0.368 III. l. (tI) 5.06 Y 10- 1 ,......t ""'t". (1)) 0 .0477 in. t. 12 1.5 N/m. (a) R~ = 10.86 kN t ; .H" = 1.942 kN· III 'I' ib) Rp - 1.144 kNt M /) - O.28GkN· mi. 4".9 kN. O.27S in. t

',1..5 I

!11m

T.

(tI) MoLiEI "1;. (h) M"Ll l2 El

t.

(a) PL"12 EI d. . (h) PL /3 EJ 1. (a) wL /6 EI d. (h) wL~i% EJ 1. ' 9. 100 \a)3Pa'f2EI'"'>;, 11/'{i'/ oEIJ.. (h) Pa 2I FI""t . Pa'l2 FI !9.101 (ll) 16.56 X IO - J md ~. (b) 0.379 in. !-. 9.102 (a) 2.55 X 10- " raJ d. (bl 6.25 mill l 9. 103 (a) 4 .98 X IO-J rnd d: O.157!) in . ./.-. j

(b) 4.59 X IO - l roo d: 0.0842 in. !-.

t.

9.104 (a) 5.22 X 10- ) roo d. (bl 10.88 mm 9.1 06 ( tI) J WL'124 EI "... (bJ J j PL'/36 EI !.

g.107 (ll) 6. 10 X IO - l roo 2. (hi 6.03 mm ! 9.109 (a) Pa(L - <1)/2 Ef"""<;, (b) P(I(3L 1 - 4a 1 )124 Ef-.1--. 9.11 0 (a) PL:/16EI-:x;..(b) PL'/48EJl. 9.1 11 (a) 5rl}f"J2 n....-.;. (b) 19nj IJS4 EJ l 9.112 (a) MtlL - 21;)I2EI<:::.(b) M ,jL' - Llh/ 8EI --!.. g.1 14 (a) w.11(3L - 2a)/12 EJ"""Z (b) w.r(JLl - 2(1z)i48 £/!-. 9.11 5 (a) 51'a'/ i! H ""t. (b) JJ'a l /4 t.1 ../.. ; .11 8 «(I) ,J.72 X 10- 1 roo -<:t;. {h) 5.&5 Illill I. 9.119 (a) 4.50 X IO - l roo ~. (h) 8.26 mm 1. 9.121 (a) 5.17 X IO - J rad '""'<;. (bl 21.0 mill l. g.1 22 J.~4 kN l m. 9.1 23 n.l ll I ,

788

9.12-4 0.223 L. 9.125 (a) 3Mo IL'/64 EI .... (b) SMu U32 EI""<;. 9.126 (a) 4PL J1243 Ell. (b) 4PL:/8 1 EI"'{. . 9,127 (a) 5wLY 'oll t.1 J.. {bl j'lce/ llr, 1:.:1 """'C. 9.129 (tI) 8.70)< lO-l ,.,.d""<;.. (hi 15,03 nun 1. 9.130 taj 5.3 I X IO - J raj """<;. (h) 0.204 in. !9.132 (ll) 7.48)( 10-] raj "l.;:. (b) 5.35 mill .J.. 9.134 (u) P,,(2L + ),<)16 EI-q;. (I~) P",(L + uYJ EI-J.. 9.135 (a) 5.33 X 10-' raj ..d.'... (bi 0.01421 in l 9.136 (u) 3.61)( 1O- 1 rad-q;.(b) O.960mmt. 9.137 (a) 2.34;X 1O- 1 md "l.;:. (b) 0.1763 in. t 9.139 (ll) <)lIIL)1256 EJ '""C. (b) 7wL·'/256 EJ ..d.'..

9.145 0.1049 in. 9.146 1.841 in. 9.147 9McI8L i. 9. 148 5P/ 16i. 9.150 7wUI28 t 9.152 JI'/j2 t atA. 1:1 /' 1:12 f ot H. I WilD f at C;, 9.153 (tI) 6.95 mm 1. (b) 46.3 k'>I 9.154 R£ = 10. 1& kips 1; - 87.9 kip ' fl at A. 46.'1 kip· ft at D. 45.8 kip . ft at E. 0 at B.

t.

9.155 4~ anL' . 9.156 144 Ellt'. 9.157 (a) y = - w()(~..--I - 5ex

+ 4(5)1 120 ElL. wJ.'124 EI L. 9. 169 (tI) y - IV l (.," - JL,.'f + 5L V - 3L5x ).llJO EIL1. (b) w"L'/30 Ef~. (0:;) 61 ,ftJ"I}15760 Ei (b) w~~/3() Ef

t (d

l

9.160 R~ = Ph~(3a + MIL' t. M ,I = Pub 1/L! ~; - Pali:L1 ~t A. ZPa'b'fLJ at D. - Pa'bfL' al B. 9. 161 (tI) 9.51 X IO -J rad"""l;;. (b) 5.&0 mill ! 9,163 0.210 in. ~ . 9.165 (a) 5.84 X 10- 1 rad ~. (b) 0.300 in . ../.. 9. 1 G1 0.00677 rL 'fEI J.- ~I 0.4')3 L flUJU kfl un!. 9 .168 R" = 65.2 kN 1: OatA. 5S.7 k.~. maID. 55.8 kN . m at E. - 82.8 kN . m at B. 9.C 1 I'mb. '1./4: 5.~b X W -] rad '"": 2.5U mill ~ . 9.C2 II = 6/t: {a) :>.11 X W -j!"ad~. 0.292 in . ~ : (h) 0.397 in. !- a\ 11.27 fl [0 the right cf A. 9.C3 X = 1.6 /JI' ((I) 7.90 X IO - J rad oq,:, 8.16 mlll!-: (b)

6.05 X 10-" rsd"""l:;, 5.79 mm y;

(el 1.021 X IO-l rad"C. (L114mm J.-. 9.e5 (a) {I = 3ft: 1.586 X 10- 1 rad oq;;: 0.1369 in. (b ) a = I.a m: 0.293 X IO - J rad~. 0.479 mill../., 9.C7 x - 2.5 'II" 5.31 Illlll l: x - 5.0 III: 12.28 mm 1

J:

CHAPTE R 10 10. 1 10.2 10.3 10.4

KIL. kL. 2/..U9. KIL.

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10.7 10.8 10.9 10.11 10.12 10.13 10.15 10.16 10.17 10.20 10.21 10 .24 10.25 10.26 10.27 10.28 10.29 10.30 10.32 10.33 10.35 10.36 10.39 10.40 10.41 10.42 10.43 10.45 10.46 10.47 10.48 10.50 10.51 10.52 10.53 10.54 10.57 10.58 10.59 10.61 10.63 10.64 10.65 10.68 10.69 10.70 10.71 10.73 10.74 10.75 10.78 10.79

/.: > 4.91 tN/m. RKIL. 81.2 N. d .; 0.&49 in.: sled PIT = 647 lb. aluminum PIT = 1872 lb. (a) 641 lb. (b) 0.651 in . (e) 58.R';{ (al 6.25%. (bl 17.17 kN. 2.81. 15.02 kips. 25.1 kips. 4.(X) kN . 5.37 kN . (a) LIJC = 4.1 fl. LC~ = 1.05 ft. (b) 4.21 kips. 657 mm. (a) bid = 0.500. (b) d=28 .3 m. b;- 14.151ll1ll. (I ) 2.64 kN. (2) 0.661 kN. (3) 10.51 kN. (4) 5.39 kN, (5) 2.64 kN. (a) 2.19. (b) (2) 1.168 in.: (3) 1.250 in.: (4) 1.046 in. (a ) 0.4lO in. (b) 14.43 ksi. (a) 4.84 Illlll. (b) 135.1 MPa. (a) 0.0399 in. (bl 19.89 ksi. (a) 1.552 mm. (bl 47.8 MPa. (a) 18.2 kips. (b) 27.5 ks i. (a) 52.9 kips. (b) 20.4 ksi. (a) 36R kN. (b) lO3.8 MPa. «I) 223 kN. (h) 62.8 MPa. 31.2°C. 5.8 1 111m. (a) 32.1 kips. (b) 39.4 kips. (a) 195 .4 kN. (b) 0.596 . (a) 247 kN. (b) 0.437. 1.J37 Ill. 9.57 f1. (a) 8.31 Ill . (b) 2.54 m. W200 x 26.6. W200 X 35.9. 2.125 in. 2.625 in. (a) 114.7 kN. (b) 208 kN. 95 .8 kips. (a) 220 kN. (b) IB9 kN. 35.9 kN. (a) 26.4 kN. (b) 32.4 kN. 16.6 kips. 1600 kN. 113 .8 kips. 101.7 kN. 38.9 kips. 6.53 in. 123. 1 11I1ll. 0.884 in. 44.9 111m. W250 x 67. WI4 X 82.

10.80 10 .81 10.82 10.84 10.85 10.86 10 .87 10.88 10.89 10.91 10.92 10.93 10.95 10.97 10 .98 10.99 10.100 10.102 10.103 10.105 10.106 10.107 10 .109 10.110 10.111 10.113 10.114 10.115 10.117 10.119 10.120 10.121 10.123 10.124 10.126 10.128 10.Cl 10.C2 10.C3 10.C4 10.C5 10 .C6

W200 X ~6.1 . L89 x 64 X 12.7 mm. L89 x 64 x 12.7 mm. L3 ~ X 2 1 X ~ in. 56.1 kips (a) 433 kN. 321 kN.
CHAPTER 11 11 .1

(a) 21.6 kllml. (b) 336 kJ/m).

11.2

(a) 43.1 in ·Ib.lin l . (b) 72.8 in ·lb/in 3 .

11 .4 11 .5 11 .6 11 .7 11 .9

(a) 111.9 kllml. (b) 112 kJ/m l . (e) 160.3 kl/m J .

(c) 163.0 kJ/m). (e) 172.4 in ·Iblin l .

(a) 58 in · lb.lin l . (b) 20 kip' in/inl. (a) 1296kJ/m J .(b) 9OM1Iml.

(a) 150.() kl/ml. (b) 63 M1/ml. (a) 116.2 in· lb. (b) 11 .72 in ·lbiin·' ,5.65 in·lblin J •

789

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1 1.10 11.11 11 .13 1 1.15 11.17 11.18 11 .20 11 .21 11.23 11 .24 11.26 11 .29 11 .30 11.31 11.33 11 .34 11.37 11 .38 11 .40 11.41 11.42 1 1.44 11.45 11 .46 11 .48 11.50 11 .51 1 1.52 11 .53 11 .54 11 .56 11 .57 11 .58 11 .59 11.61 11 .62 11 .63 11 .64 11 .67 11.69 11.71 11.73 11.74 1 1.75 11.77 11.79 11 .81 11 .82 11 .83

790

(al 12.18 1. (b) 15.83 kJ/mJ • 38.6 kJ/mJ . (a) 264 in . Ih. (b) 1.378 in . Iblin l . 8.83 in . Iblin l . (a) 44.0 1. (b) 31.4 1. 102.7 in . lb. 1.398 1" I/EA . 3.8 1 I"IIEA. 2.37 I" l/EA. 1015 1. 'w ' L 3 /40 EI.

M~(al

+b

~al (a

+

J

)/6 ElL'. L)/6 EI.

1048 J. 894 in . lb. 662 J. 2.45 X 10" in . lb. 14.70 1. - 2.65 MPa to 122.6 MPa. (a) 2.33. (b) 2'()2. 2 MAL (I + 3 EtfllO GL')/ Ebd' . (Q ' /41T GL ) In (R,iRI)' (al 79.8 kN. (b) 254 MPa. (e) 1.523 mm. 11.50 in. 7.87 in. 4.76 kg. 8.50 ftls. (a) 2 1.0 kN. lb) 171.7 MPa. lel 8.S8Imn. (a) 7.67 kN. (b) 312 MPa. (e) 23.5 mm. (al 0.1064 ill. (b) 20. ] hi . (a) IS.63 mm. (b) 83.8 N · m. (c) 208 MPa. (a) 23.6 nun. (b) 64.4 N . m. (el IS7.6 MPa. (b) 7.12 (b) re lati vc crror O. IS2 . .v~/2h = 0. 166. l'a' b l (3 El L !. Pa'(a + L)/3 EI !. Mo (L + 3a).!3 EI "{;:. 31'L '/16 E/~. 31'aJ/4 EI !. 5MoL/16 E/ ~.

386 mm. S.28°. 2.37S I'IIEA ~ .

0.0447 in. ~. 0.366 in. t. 3.19 mm.1.. (a) and (b) P'L J/96 EI - PM ,.j...'1l6 EI + M fj../6 EI . (a) and (b) I'"'L" 48 EI + PMrP/R EI + M~12 EI. (a) and (b) P'L J;48 EI . (a) and (b) 7I'"L'124 EI . SPL ~/48 EI .1..

11 .84

0.0443wL41E1 ~.

11 .85 11 .87 11 .88 11 .89

3PL ' ;& El .d.. M oLl6 EI -q;:. PaL 2/ 16EIt. IUL~/ 1 28 Eli. wLJ/ 192 El .d.. 0.987 in. ~.

11 .91 11 .92 11.94 11 .95 11 .96 11 .98 11 .99 11 .101 11 .102 11 .103 11 .105 11 .106 11 .107 11 .109 11 .111 11 .11 2 11 .114

11 .116 11 .117 11 .118 11 .119 11 .121

7.25 mm .1.. 5.08 mm .1.. 0.698 X ]()- ' rad~. 2.07 X ]() -J rad L. 3.19 mill J.. 0.233 in. ~. 0.1504 in. ~. 0.1 459 mm!. (a) PRJI2 E/~. (b) TrPR]/4EI!.

TrPR]12 EI .1.. (a) 2PI1/3 EI_.(b) I'P/6 EI Lf.. (a) P/'/EI i. (b) 31'P12 El .d.. SPII6 - 3PUlli at A. 0 at B. SPU32 at C. 3M,j2L M ol2 JlA . - M (j at B. 7wLl l 28 t ; 7wu' 128 - WIX - U2)lfl. wLl6 .1. at A. 3wL'4 t at B, 5wU12 i at C. 71'/8. 0.652 P . 1'/( 1 +2cosJ.p). 2 1'/3. 7 P1l2. 2.09. 12.70 J. 9. 1:2 lb.

t

t

11 .122 11 .123 11.125 1 1.127 11.129 11.57mm .1.. 11 .131 1.030mm --+.

11 .134 PLl/6 El \ 11 .C2 ta) a = 15 ill.: (TD = 17. 19 ks i, (Tc = 21.() ksi; a = 45 in.: IT n "" 36.2 ksi. (Tc "" 14.74 ksi. (b) a = 18.34 in .. (1" = 20.67 hi . 1 1.C3 (a) L =< 200 mm:h = 2.27 mm; L = 800 mm: 11 = 1.076 rnm. (b) L = 44U mill: II = 3.23 n U ll . 11.C4 a"" 300111111: 1.795 mm. 179.46 MPa; a = 600 1Il1ll: 2.87 111m . 179.59 MPa. 11 .C5 a = 2m: (a) 30.0 J; (b) 7.57 mm. 60.8 J. a = 4 Ill: (a) 2 1.9 J; (b) 8.87 mm. 83.4 J. 11.C6 a = 20 in: (a ) 13 .26 in.; (b ) 99.5 kip ' in.; (c) 803 lb. a = 50 in: (a) 9.46 in.; (b ) 93.7 kip ' in.; ~e) 996 lb.

No part of any book may be reproduced or transmitted by any means without the publisher’s prior permission. Use (other than qualified fair use) in violation of the law or Terms of Service is prohibited. Violators will be prosecuted to the full extent of the law.

No part of any book may be reproduced or transmitted by any means without the publisher’s prior permission. Use (other than qualified fair use) in violation of the law or Terms of Service is prohibited. Violators will be prosecuted to the full extent of the law.

No part of any book may be reproduced or transmitted by any means without the publisher’s prior permission. Use (other than qualified fair use) in violation of the law or Terms of Service is prohibited. Violators will be prosecuted to the full extent of the law.

No part of any book may be reproduced or transmitted by any means without the publisher’s prior permission. Use (other than qualified fair use) in violation of the law or Terms of Service is prohibited. Violators will be prosecuted to the full extent of the law.