Solutions Manual For: Mechanical Design Of Machine Components
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SOLUTIONS MANUAL FOR MECHANICAL DESIGN OF MACHINE COMPONENTS
SECOND EDITION: SI VERSION by
ANSEL C. UGURAL
SOLUTIONS MANUAL FOR MECHANICAL DESIGN OF MACHINE COMPONENTS SECOND EDITION: SI VERSION
by
ANSEL C. UGURAL
Boca Raton London New York
CRC Press is an imprint of the Taylor & Francis Group, an informa business
MATLAB® is a trademark of The MathWorks, Inc. and is used with permission. The MathWorks does not warrant the accuracy of the text or exercises in this book. This book’s use or discussion of MATLAB® software or related products does not constitute endorsement or sponsorship by The MathWorks of a particular pedagogical approach or particular use of the MATLAB® software.
Taylor & Francis Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2016 by Taylor & Francis Group, LLC Taylor & Francis is an Informa business No claim to original U.S. Government works Printed on acid-free paper Version Date: 20160303 International Standard Book Number-13: 978-1-4987-3541-4 (Ancillary) This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com
CONTENTS
Part I
BASICS
Chapter 1
INTRODUCTION
1
Chapter 2
MATERIALS
16
Chapter 3
STRESS AND STRAIN
24
Chapter 4
DEFLECTION AND IMPACT
48
Chapter 5
ENERGY METHODS AND STABILITY
68
Part II
FAILURE PREVENTION
Chapter 6
STATIC FAILURE CRITERIA AND RELIABILITY
100
Chapter 7
FATIGUE FAILURE CRITERIA
117
Chapter 8
SURFACE FAILURE
135
Part III
APPLICATIONS
Chapter 9
SHAFTS AND ASSOCIATED PARTS
145
Chapter 10
BEARINGS AND LUBRICATION
164
Chapter 11
SPUR GEARS
176
Chapter 12
HELICAL, BEVEL, AND WORM GEARS
194
Chapter 13
BELTS, CHAINS, CLUTCHES, AND BRAKES
208
Chapter 14
MECHANICAL SPRINGS
225
Chapter 15
POWER SCREWS, FASTENERS, AND CONNECTIONS
240
Chapter 16
MISCELLANEOUS MACHINE COMPONENTS
261
Chapter 17
FINITE ELEMENT ANALYSIS IN DESIGN
278
Chapter 18
CASE STUDIES IN MACHINE DESIGN
308
vi
NOTES TO THE INSTRUCTOR
The Solutions Manual to accompany the text MECHANICAL DESIGN of Machine Components supplements the study of machine design developed in the book. The main objective of the manual is to provide efficient solutions for problems in design and analysis of variously loaded mechanical components. In addition, this manual can serve to guide the instructor in the assignment of problems, in grading these problems, and in preparing lecture materials as well as examination questions. Every effort has been made to have a solutions manual that cuts through the clutter and is self –explanatory as possible thus reducing the work on the instructor. It is written and class tested by the author. As indicated in its preface, the text is designed for the junior-senior courses in machine or mechanical design. However, because of the number of optional sections which have been included, MECHANICAL DESIGN of Machine Components may also be used to teach an upper level course. In order to accommodate courses of varying emphases, considerably more material has been presented in the book than can be covered effectively in a single three-credit-hour course. Machine/mechanical design is one of the student’s first courses in professional engineering, as distinct from basic science and mathematics. There is never enough time to discuss all of the required material in details. To assist the instructor in making up a schedule that will best fit his classes, major topics that will probably be covered in every machine design course and secondary topics which may be selected to complement this core to form courses of various emphases are indicated in the following Sample Assignment Schedule. The major topics should be covered in some depth. The secondary topics, because of time limitations and/or treatment on other courses, are suggested for brief coverage. We note that the topics which may be used with more advanced students are marked with asterisks in the textbook. The problems in the sample schedule have been listed according to the portions of material they illustrate. Instructor will easily find additional problems in the text to amplify a particular subject in discussing a problem assigned for homework. Answers to selected problems are given at the end of the text. Space limitations preclude our including solutions to open-ended web problems. Since the integrated approach used in this text differs from that used in other texts, the instructor is advised to read its preface, where the author has outlined his general philosophy. A brief description of the topics covered in each chapter throughout the text is given in the following. It is hoped that this material will help the instructor in organizing his course to best fit the needs of, his students. Ansel C. Ugural Holmdel, N.J.
vii
DESCRIPTION OF THE MATERIAL CONTAINED IN “MECHANICAL DESIGN of Machine Components”
Chapter 1 attempts to present the basic concepts and an overview of the subject. Sections 1.1 through 1.8 discuss the scope of treatment, machine and mechanical design, problem formulation, factor of safety, and units. The load analysis is normally the critical step in designing any machine or structural member (Secs. 1.8 through 1.9). The determination of loads is encountered repeatedly in subsequent chapters. Case studies provide a number of machine or component projects throughout the book. These show that the members must function in combination to produce a useful device. Section 1.10 review the work, energy, and power. The foregoing basic considerations need to be understood in order to appreciate the loading applied to a member. The last two sections emphasize the fact that stress and strain are concepts of great importance to a comprehension of design analysis. Chapter 2 reviews the general properties of materials and some processes to improve the strength of metals. Sections 2.3 through 2.14 introduce stress-strain relationships, material behavior under various loads, modulus of resilience and toughness, and hardness, selecting materials. Since students have previously taken materials courses, little time can be justified in covering this chapter. Much of the material included in Chapters 3 through 5 is also a review for students. Of particular significance are the Mohr’s circle representation of state of stress, a clear understanding of the three-dimensional aspects of stress, influence of impact force on stress and deformation within a component, applications of Castigliano’s theorem, energy of distortion, and Euler’s formula. Stress concentration is introduced in here, but little applications made of it until studying fatigue (Chap.7). The first section of Chapter 6 attempts to provide an overview of the broad subject of “failure”, against which all machine/mechanical elements must be designed. The discipline of fracture mechanics is introduced in Secs. 6.2 through 6.4. Yield and fracture criteria for static failure are discussed in Secs. 6.4 through 6.12. The last 3 sections deal with the method of reliability prediction in design. Chapter 7 is devoted to the fatigue and behavior of materials under repeated loadings. The emphasis is on the Goodman failure criterion. Surface failure is discussed in Chapter 8. Sections 8.1 through 8.3 briefly review the corrosion and friction. Following these the surface wear is discussed. Sections 8.6 through 8.10 deal with the surfaces contact stresses and the surface fatigue failure and its prevention. The background provided here is directly applied to representative common machine elements in later chapters. Sections 9.1 through 9.4 of Chapter 9 treat the stresses and design of shafts under static loads. Emphasis is on design of shafts for fluctuating loading (Secs. 9.6 and 9.7). The last 5 sections introduce common parts associated with shafting. Chapter 10 introduces the lubrication as well as both journal and roller bearings. As pointed out in Sec. 8.9, rolling element bearings provide interesting applications of contact stress and fatigue. Much of the material covered in Secs. 11.1 through 11.7 of Chapter 11 introduce nomenclature, tooth systems, and fundamentals of general gearing. Gear trains and spur gear force analysis are taken up in Secs. 11.6 and 11.7. The remaining sections concern with gear design, material, and manufacture. Non-spur gearing is considered in Chapter 12. Spur gears are merely a special case of helical gears (Secs. 12.2 through 12.5) having zero helix angle. Sections 12.6 through 12.8 deal with bevel gears. Worm gears are fundamentally different from other gears, but have much in common with power screws to be taken up in Chap. 15.
viii
Chapter 13 is devoted to the design of belts, chains, clutches, and brakes. Only a few different analyses are needed, with surface forms effecting the equations more than the functions of these devices. Belts, clutches, and brakes are machine elements depending upon friction for their function. Design of various springs is considered in Chapter 14. The emphasis is on helical coil springs (Secs. 14.3 through 14.9) that provide good illustrations of the static load analysis and torsional fatigue loading. Leaf springs (Sec. 14.11) illustrate primarily bending fatigue loading. Chapter 15 attempts to present screws and connections. Of particular importance is the load analysis of power screws and a clear understanding of the fatigue stresses in threaded fasteners. There are alternatives to threaded fasteners and riveted or welded joints. Modern adhesives (Secs. 15.17 and 15.18) can change traditional preferred choices. It is important to assign at least portions of the analysis and design of miscellaneous mechanical members treated in Chapter 16. Sections 16.3 through 16.7 concern with thick-walled cylinders, press or shrink fits, and disk flywheels. The remaining sections concerns with the bending of curved frames, plate and shells-like machine and structural components, and pressure vessels. Buckling of thin-walled cylinders and spheres is also briefly discussed. Chapter 17 represents an addition to the material traditionally covered in “Machine/Mechanical Design” textbooks. It attempts to provide an introduction to the finite element analysis in design. Some practical case studies illustrate solutions of problems involving structural assemblies, deflection of beams, and stress concentration factors in plates. Finally, case studies in preliminary design of the entire crane with winch and a high-speed cutting machine are introduced in Chapter 18.
ix
SAMPLE ASSIGNMENT SCHEDULE
MACHINE/MECHANICAL DESIGN (3 credits.)
Prerequisites:
A. C. Ugural, MECHANICAL DESIGN of Machine Components, 2nd SI Version, CRC Press (T & F Group). Courses on Mechanics of Materials and Engineering Materials.
WEEK
TOPICS
TEXT:
SECTIONS
PROBLEMS
1
Introduction Materials*
1.1 to 1.12 2.1 to 2.5,2.8 to 2.11
1.6,1.17,1.26,1.34 2.7,2.11,2.15,2.20
2
A Review of Stress Analysis*
3.1 to 3.14, 4.1 to 4.9 5.1 to 5.12
3.12,3.34,3.45,4.24,4.37 5.20,5.26,5.30,5.54, 5.73
3
Static Failure Criteria & Reliability
6.1 to 6.15
6.7,6.14,6.25,6.29,6.40
4
Fatigue Criteria and Surface Failure
7.1 to 7.15, 8.1 to 8.10
7.6,7.17,7.28,7.32,7.34 8.3,8.8, 8.14, 8.21
5
EXAM # 1 Shafts and Associated Parts
- - - - 9.1 to 9.12
- - - - 9.7,9.14,9.19,9.24,9.28
6
Lubrication and Bearings
10.1 to 10.16
10.4,10.10,10.16,10.31,10.36
7
Spur Gears
11.1 to 11.12
11.15,11.18,11.22,11.33,11.38
8
Helical, Bevel, and Worm Gears*
12.1 to 12.10
12.10,12.12,12.18,12.21,12.32
9
EXAM # 2 Belt and Chain Drives
- - - - 13.1 to 13.6
- - - - 13.1,13.8,13.9, 13.10,13.13
10
Clutches and Brakes
13.8 to 13.15
13.21,13.27,13.34,13.39,13.46
11
Mechanical Springs
14.1 to 14.11
14.8,14.12,14.22,14.25,14.35
12
EXAM # 3 Power Screws and Fasteners
- - - - 15.1 to 15.12
- - - - 15.2,15.6,15.14,15.19,15.30
13
Connections*
15.13 to 15.16
15.34,15.40,15.48,15.54
14
Miscellaneous Machine Components*
16.1 to 16.5, 16.8 to 16.11
16.6,16.17,16.25,16.38
15
FEA in Design and Case Studies in Machine Design* FINAL EXAM
17.1 to 17.3,17.5,17.7 18.1 and 18.2 - - - - -
17.2,17.4,17.8,17.12 18.2,18.4,18.6,18.10 - - - - -
* Secondary topics. The remaining, major topics constitute the “main stream” of the machine design course.
x
Section I
BASICS
CHAPTER 1
INTRODUCTION
SOLUTION (1.1) Free Body: Angle Bracket (Fig. S1.1) (a)
(b)
M
B
0;
F ( 0 .5 ) 6 8 ( 0 .2 5 ) 2 8 .8 ( 0 .5 ) 0 ,
Fx 0 :
R B x 2 8 .8 F 0 ,
Fy 0 :
R B y 6 8 2 1 .6 0 ,
F 6 2 .8 k N
R Bx 3 4 k N R B y 8 9 .6 k N
Thus RB
(3 4 ) (8 9 .6 ) 2
2
9 5 .8 k N
and ta n
1
34 8 9 .6
2 0 .8
o
F
A 0.5
34
21.6 kN
68 kN
B 89.6
R Bx
Figure S1.1
28,8 kN
B R By
C
RB
0.25
0.25
SOLUTION (1.2) Free Body: Beam ADE (Fig. S1.2)
M
A
0:
W (3 .7 5 a ) F B D ( 2 a ) 0 ,
E D
W
FBD
A 2a
1.75a
R Ay
a
Fx 0 :
R Ax 0
Fy 0 :
R Ay FBD W 0 ,
Free-Body: Entire structure (Fig. S1.2)
V
M
A
0:
R C (3 a ) W (3 .7 5 a ) 0 , R C 1 .2 5 W
F O
0.875 kN
R A y 0 .8 7 5W
R Ax
A
F B D 1 .8 7 5 W
M”
Free Body: Part AO (Fig. S1.2)
Figure S1.2
V 0 .8 7 5W F 0
M 0 .8 7 5 a W
1
SOLUTION (1.3) 29 kN/m
C
D
32 kN m A
0.6 m
B
E
RA
1,2 m
M
A
0:
Fy 0 :
R B 1 1 .0 2 k N R A 4 5 .8 2 k N
RB
2.4 m
2.4 m
Segment CD 29 kN/m
M
C
M
0.6 m D V D
D
( 2 9 )( 0 .6 ) 5 .2 2 k N m 2
1 2
V D 1 7 .4 k N
D
Segment CE 3 4 .8 k N
C
VE
M
0.6 m
M
1.2 m A
E
1.2 m E
3 4 .8 (1 .8 ) 4 5 .8 2 (1 .2 ) 7 .6 5 6 k N m
E
V E 1 1 .0 2 k N
45 . 82
SOLUTION (1.4)
(a) 8 kN
1m
m
M
B
0:
2m
0 .8 R C ( 6 ) 0 .6 R C ( 2 ) 2 4 ( 4 ) 0
R C 26 . 667
10 kN
kN
R Cx 16 kN , R Cy 21 . 334 kN
C 3m D
RC
B 2m R Bx
4 3
R By
Then
2m
A
F x 0 : R Bx 16 kN F y 0 : R By 12 . 66 kN
( b ) Segment CD 8
M
4 M
16 C
3m D V D 21.334
D
21 . 334 ( 3 ) 12 (1 . 5 ) 6 ( 2 ) 34 kN m
F D 16 kN
D
FD
V D 21 . 334 18 3 . 334
2
kN
SOLUTION (1.5) 3m
(a) A 2
M
A
0:
RCx
3 2
RCy
C RCy
RCx
4 3
R Cy
R By RCx
FD
40 kN
C
0:
4 0 (5 ) 4 R C y 0
M
RCy 5 0 k N
R Cx 75 kN
Then
R By 1 0 k N ,
R Bx 7 5 k N
D
0.75
VD
B
1 m
4m
(b)
M
2m
3
R Bx B
D
F D 75 ( 53 ) 50 ( 54 ) 85 kN
1.0
V D 75 ( 54 ) 50 ( 53 ) 30 kN
75
C 5 50 4 3
M
D
7 5 (1) 5 0 ( 0 .7 5 ) 3 7 .5 k N m
SOLUTION (1.6) (a)
Free body entire connection
B
P A
T
M
C
0 :
0,5 m
T 0 .7 R A
0.2 m
Segment AB A
AB
F AB
B
0.15 m
M
( 0 .5 ) ( 0 .1 5 ) 2
B
0 :
18
0 .5 2 2 m
18 ( 0 . 15 ) R A ( 0 . 5 ) 0 T 3 .7 8 k N m
and Fx 0 :
2
R A 5 .4 k N
0.5 m RA
(b)
R A ( 0 .7 ) T 0
C
RA
18 kN
0.15 m
0 .5 0 .5 2 2
FAB 0,
F A B 1 8 .7 2 9 k N
SOLUTION (1.7) 120 mm
120 mm
y
A R A 2 kN
B
RB 2 kN
50 mm
4 kN
70 mm
D z
Free Body: Entire Crankshaft (Fig. S1.7a) ( a ) From symmetry: R A R B
T
Fz 0 : R A R B 2 k N M
x
0 : 4 ( 0 .0 5 ) T 0 ,
T 0 .2 k N m 2 0 0 N m
x
C (a)
(CONT.)
3
1.7 (CONT.) M
y
( b ) Cross Section at D (Fig. S1.7b)
D
T
Vz
Vz 2 kN
(b)
T 200 N m
M
Figure S 1.7
y
2 ( 0 .0 7 ) 0 .1 4 k N m 140 N m
SOLUTION (1.8) 30 kN Free-Body Diagram, Beam AB
B 1.8 m
4 7
C
FC D
1.2 m
Fx 0 :
Fy 0 : R A
M
A
0:
7 65
FC D 6 0 0 , 4 65
FC D 3 0 0 ,
6 0 (1 .8 )
M
60 kN
A
F C D 6 9 .1 1 k N
7 65
R A 6 4 .3 k N
FC D ( 3 ) M
A
0,
72 kN m
1.8 m A M
A
RA
SOLUTION (1.9) B 1.2 m C 129.6 kN 0.9 m 0.9 m 1.2 m 2
A
Free body entire frame
2.4 m
1 1 2
D
M
A
0 :
129 . 6 ( 0 . 9 )
R D y 4 8 .6 k N ,
1 2
R Dy (1 . 2 ) R Dy ( 3 ) 0
R D x 2 4 .3 k N
R Dy
R Dy R D
B
R By
R B x 1.2 m
D
C
Free body BCD
2.4 m
24.3
Fx 0 :
R B x 2 4 .3 k N
Fy 0 :
R B y 4 8 .6 k N
RB
2 4 .3 4 6 .6 2
48.6
4
2
5 2 .6 k N
SOLUTION (1.10)
y 4 kN 3 kN
R Ay A
0.3 m
R Az
T
R Ey
5 kN
z C E
1m
D
R Ez
1m
2 kN x
B
0.5 m 0.5 m 0.3 m
(a)
M
x
3 ( 0 .1 5 ) 4 ( 0 .1 5 ) T 5 ( 0 .1 5 ) 2 ( 0 .1 5 ) 0
T 0 . 6 kN m
or (b)
0:
M
z
0 : ( 4 3 )( 1 ) R Ey ( 2 . 5 ) 0 , R Ey 2 . 8 kN
M
y
0 : R Ez ( 2 . 5 ) ( 5 2 )( 3 ) 0 , R Ez 8 . 4 kN
F y 0 : R Ay 4 3 R Ey 0 , R Ay 4 . 2 kN
F z 0 : R Az R Ez 5 2 0 , R Az 1 . 4 kN
Thus
RA
4 .2
RE
2 .8
2
1 .4
2
4 . 427
2
8 .4
2
8 . 854 kN
kN
SOLUTION (1.11) (a) Free-body Diagrams, Arm BC and shaft AB V
z
T
C
y B
x
M T
V
A T
Figure S1.11
M
V
M
V B
T
(b) At C: V 2 kN
T 50 N m
At end B of arm BC: V 2 kN
T 50 N m
M 200 N m
At end B of shaft AB: V 2 kN
At A:
V 2 kN
T 200 N m T 200 N m
M 50 N m M 300 N m
5
SOLUTION (1.12) Free Body: Entire Pipe 200 N
0.15 m
D
y
36 N m
C Ry
T Rx
A
0.2 m
Rz Mz
My
z
0.3 m
B
x
Reactional forces at point A:
Fx 0 :
Rx 0
Fy 0 :
R y 200 0,
Fz 0 :
Rz 0
R y 200 N
Moments about point A:
M
x
0:
T 2 0 0 ( 0 .1 5 ) 0 ,
M
y
0:
M
y
M
z
0:
M
z
T 30 N m
0 2 0 0 ( 0 .3 ) 3 6 0 ,
M
z
96 N m
The reactions act in the directions shown on the free-body diagram.
SOLUTION (1.13) Free Body : Entire Pipe 0.15 m
y Rx
Mz z
36 N m
C WCD
T Rz
D
0.075 m
Ry
200 N
WAB
0.2 m
WBC
A
My 0.15 m
0.3 m
B
x
(CONT.)
6
1.13 (CONT.) We have 1 lb/ft=14.5939 N/m (Table A.2). Thus, for 3 in. or 75-mm pipe (Table A.4): 14.5939(7.58)=110.62 N/m Total weights of each part acting at midlength are: W A B 1 1 0 .6 2 ( 0 .3) 3 3 .2 N W B C 1 1 0 .6 2 ( 0 .2 ) 2 2 .1 N W C D 1 1 0 .6 2 ( 0 .1 5 ) 1 6 .6 N
Reactional forces at point A:
Fx 0 : Fy 0 :
Rx 0 R y W AB W BC W CD 2 0 0 0,
Fz 0 :
R y 2 7 1 .9 N
Rz 0
Moments about point A:
M
x
0:
M
y
0:
M
z
0:
T W C D ( 0 .0 7 5 ) 1 0 ( 0 .1 5 ) 0 ,
T 2 .7 4 5 N m M
M
z
y
0
( 2 0 0 W C D W B C )( 0 .3 ) W A B ( 0 .1 5 ) 3 6 0
M
z
1 1 2 .6 N m .
SOLUTION (1.14) 1.6 kN
RCy
(a) R Ay
D
R Ax A
C
Dy
0.5 m
RCx
1m
Dx
Bx
E
By
D
1.6 kN 0.15 m
Dx
1.6 kN By
Dy
Free body pulley B
B
Bx
0.25 m B
150 mm
F x 0 : B x 1 . 6 kN F y 0 : B y 1 . 6 kN
Free body CED
M
D
0 : R Cx ( 0 . 4 ) 1 . 6 ( 0 . 15 ) 0 ,
F x 0 : D x 0 .6 1 .6 0 , F y 0:
R Cy D
R Cx 0 . 6 kN
D x 1 kN
y
Free body ADB
M
A
0 : D y ( 0 . 5 ) B y (1 . 5 ) 0 , D y 4 . 8 kN , R Cy 4 . 8 kN
F x 0 : R Ay D y B y 0 ,
R Ay 3 . 2 kN
(CONT.)
7
1.14 (CONT.)
(b)
M
F x 0 : R Ax D x B x 0 ,
V G 0.6 m B
G
FG
G
M
1.6
G
R Ax 0 . 6 kN
1 . 6 ( 0 . 6 ) 960
N m , V G 1 . 6 kN
F G 1 . 6 kN
1.6
SOLUTION (1.15) Free body entire rod
M
0 : R Dy ( 0 . 25 ) 300 ( 0 . 1 ),
x
M
z
0:
R Dy 120
N
2 0 0 ( 0 .3 5 ) R B y ( 0 .2 5 ) ( 3 0 0 1 2 0 ) ( 0 .2 ) 0
C
R By 136
N
Free body ABE y B
A 200 N
Vy
100
175
M
z
E
0: M
z
x
E
136 N
M
z 200 ( 0 . 275 ) 136 ( 0 . 175 ) 0 , M
z
z
31 . 2 N m
F y 0 : V y 200 136 64 N
SOLUTION (1.16) Free body entire rod:
M 0: M
R D y ( 0 .2 5 ) 4 0 0 ( 0 .1) 0 ,
x
z
R B y ( 0 .2 5 ) ( 4 0 0 1 6 0 ) ( 0 .2 ) 0 ,
0:
C
Segment ABE A
y
B
Vy
0.175 m
192 N
z
x
E M
z
At point E: M
z
192 ( 0 . 175 ) 33 . 6 N m
V y 192
RDy 160 N
N
8
R By 1 9 2 N
SOLUTION (1.17) Side view
Top view
F2 50 mm
F2 B C
D
Td
50 mm
100 mm
Figure (c)
Figure (a)
F1
150 N m A 50 mm
Fig. (b): Fig. (a): Fig. (c):
F1
Figure (b)
M
A
0 : F 1 ( 0 . 05 ) 150 0 , F 1 3 kN
M
B
0 : F 1 ( 0 . 1 ) F 2 ( 0 . 05 ) 0 , F 2 6 kN
M
D
0:
F 2 ( 0 .0 5 ) T d 0 ,
T d 0 .3
kN m
SOLUTION (1.18) 18 kN
13.5 kN
1m
C
3m B A
Ry
6m
Rx 3 4
4m
R
Free body-entire frame
M
A
0:
R y (1 0 ) 1 3 .5 ( 6 ) 1 8 ( 4 ) 0 ,
Free body-member BC
M
C
0:
R x (3) R y ( 4 ) 0
and Rx
4 3
(1 5 .3) 2 0 .4 k N
Thus FBC R
( 2 0 .4 ) (1 5 .3 ) 2
2
2 5 .5 k N
9
R y 1 5 .3 k N
SOLUTION (1.19) P
Free body-member AB B
R Ax
A
E 40
o
a
0: o
o
R E 1.1 5 6 P
RE
R Ay
A
R E ( 4 a ) P c o s 4 0 ( a ) P s in 4 0 ( 6 a ) 0
2a
4a
M
F x 0:
R Ax P co s 4 0
0 .7 6 6 P
F y 0:
R A y 1.1 5 6 P P s in 4 0
o
o
0,
R A y 0 .5 1 3 P
Free body-member CD RE
RCx
C 2a E
RD
RCy
M
C
D
4a
0:
30
M
D
0:
R C y 0 .7 7 1 P
o
R D s in 3 0 ( 6 a ) R E ( 2 a ) 0 , o
F x 0:
RCx R D co s 3 0
R E ( 4 a ) RCy ( 6 a ) 0
o
R D 0 .7 7 1 P
0 .6 6 8 P
SOLUTION(1.20) ( a ) Power= P = ( p A ) ( L ) ( n /6 0 ) (1 .2 )( 2 1 0 0 )( 0 .0 6 )( 1 56 00 0 ) 3 .7 8 k W
Power required
P e
3 .7 8 0 .9
4 .2 k W
( b ) Use Eq.(1.15), T
9549 kW n
9 5 4 9 ( 4 .2 ) 1500
2 6 .7 4 N m
SOLUTION (1.21) a=1.5 m, b=0.55 m, c=0.625 m, L=2.7 m, V=29 m/s, W=14.4 kN, kW=14 ( a ) From Eq. (1.15), the drag force equals, Fd
1000 kW V
1 0 0 0 (1 4 ) 29
4 8 2 .8 N
See: Fig. P1.21:
Fx 0 :
F d 4 8 2 .8 N
It follows that
M
A
0:
W a Fd c R f L 0
or 1 4 4 0 0 (1 .5 ) ( 4 8 2 .8 )( 0 .6 2 5 ) R f ( 2 .7 ) 0
(CONT.)
10
1.21 (CONT.) Solving, R f 7 .8 8 8 k N
and
Fy 0 :
R r 1 4 .4 7 .8 8 8 0
or R r 6 .5 1 2 k N
(b) We have V 0 ,
Fd 0 ,
F 0.
See Fig. P1.21:
M
A
Rf
a L
0:
Wa Rf L 0
Thus
So,
W
1 .5 2 .7
(1 4 .4 ) 8 k N
F y 0 gives
R r W R f 1 4 .4 8 6 .4 k N
SOLUTION (1.22) Refer to Solution of Prob. 1.21. a=1.5 m, b=0.55 m, c=0.625 m, L=2.7 m, V=29 m/s, W=14.4 kN, kW=14 Now we have W t 1 4 .4 5 .4 1 9 .8 k N
( a ) From Eq. (1.15), the drag force equals, Fd
1000 kW
V
1 0 0 0 (1 4 ) 29
4 8 2 .8 N
See: Fig. 1.21 (with W W t ):
M
A
0:
W t a Fd c R f L 0
1 9 8 0 0 (1 .5 ) ( 4 8 2 .8 )( 0 .6 2 5 ) R f ( 2 .7 ) 0
from which R f 1 0 .8 8 8 k N
and
Fy 0 :
R r 1 9 .8 1 0 .8 8 8 0
R r 8 .9 1 2 k N
( b ) V 0,
Fd 0 ,
M
A
Then
Rf
a L
So,
F 0 , as before,
0: W
Wta R f L 0 1 .5 2 .7
(1 9 .8 ) 1 1 k N
F y 0 gives
R r W t R f 1 9 .8 1 1 8 .8 k N
11
SOLUTION (1.23) ( a ) Free-Body Diagram: Gears (Fig. S1.23). Applying Eq. (1.15): T AC
9550 P n
9550 (35 )
6 6 8 .5 N m
500
Therefore, TA
F
rA
5 .3 4 8 k N
6 6 8 .5 0 .1 2 5
( b ) T D E F rD 5, 3 4 8 ( 0 .0 7 5 ) 4 0 1 .1 N m TDE
D rD
T AC
A
F rA
F
Figure S1.23 SOLUTION (1.24) n1 1 8 0 0 rp m ,
n 2 4 2 5 rp m ,
kW 20
From Eq. (1.15), we obtain T = 9549 kW/n. Thus For input shaft T
9549 ( 23) 1800
122 N m
For output shaft T
9549 ( 20 ) 425
4 4 9 .4 N m
Equation (1.14) gives e
20 23
100 87 %
SOLUTION (1.25) N=150, F=2.25 kN, s=62.5 mm, e=88% ( a ) Referring to Eq. (1.12): power output F s ( 6N0 ) 2 2 5 0 0 .0 6 2 5 ( 16500 ) 3 5 1 .6 W
( b ) Using Eq. (1.14), power transmitted by the shaft: power input 3 5 1 .6 ( 0 .8 8 ) 3 9 9 .5 W SOLUTION (1.26) Equation (1.10) becomes Ek
1 2
I (
2 max
2 min
(1)
)
Here, mass moment inertia with 5 percent added: I ( 1 . 05 )
32
( d o d i ) l 4
4
(Table A.5) (CONT.)
12
1.26 (CONT.)
1 . 05
( 0 .4
32
4
1 . 299 kg m
0 . 3 )( 0 . 1 )( 7 , 200 ) 4
2
m a x 1 2 0 0 ( 610 ) 2 0 r p s 1 2 5 .7 ra d s
m in 1 1 0 0 ( 610 ) 1 8 .3 r p s 1 1 5 ra d s
Equation (1) is therefore Ek
1 2
( 1 . 299 )( 125 . 7
115
2
2
)
1, 6 7 3 J
SOLUTION (1.27) Final length of the wire: L AC '
( 2 ) (1 .2 6 ) 2
2
2 .3 6 3 8 m
Initial length of the wire is L AC
( 2 ) (1 2 .5 ) 2
2
2 .3 5 8 5 m
Hence, Eq. (1.20): AC
L AC ' L AC L AC
2 .3 6 3 8 2 .3 5 8 5 2 .3 5 8 5
0 .0 0 2 2 5 2 2 5 0 μ
SOLUTION (1.28) (a) c
2 ( r r ) 2 r 2 r
( c ) i
0 .3 150
( c ) o
(b) r
r r
2000
μ
800
0 .2 250
r o ri
r o ri
μ
0 .3 0 .2 250 150
1000
μ
SOLUTION (1.29) LO B d ,
( a ) OB
L AB L BC d
0 .0 0 1 2 d d
1200 μ
( b ) L AB ' L CB ' d AB BC
( c ) C A B ta n
1
2 1 .4 1 4 2 1 d
2
(1 . 0012 d )
1 .4 1 5 0 6 1 .4 1 4 2 1 1 .4 1 4 2 1
1 .0 0d1 2 d
2
1 2
1 . 41506 d
601 μ 4 5 .0 3 4 4
o
Increase in angle C A B is 4 5 .0 3 4 4 4 5 0 .0 3 4 4 . Thus o
0 .0 3 4 4 18 0 6 0 0 μ
13
SOLUTION (1.30) (a) x
0 .8 0 .5 250
1200
μ
y
0 .4 0 200
2000 μ
( b ) L ' AD L AD x L AD L AD (1 x ) 250 (1 . 0012 ) 250 . 3 mm
SOLUTION (1.31) L AB 800 (10
6
)150 120 (10
6
L AD 1000 (10
3
) 200 200 (10
) mm 3
) mm
We have L BD L AB L AD 2
2
2
2 L BD L BD 2 L AB L AB 2 L AD L AD
or L AB
L BD
LBD
150 250
L AB
L AD LBD
L AD
(120 )
200 250
( 200 ) 10
(1) 3
0 . 232
mm
SOLUTION (1.32) AC BD
300
2
300
2
424 . 26 mm
B ' D ' 4 2 4 .2 6 0 .5 4 2 3 .7 6 m m ,
A ' C ' 4 2 4 .2 6 0 .2 4 2 4 .4 6 m m
Geometry: A ' B ' A ' D ' x
C'
B'
D'
A'
xy
y
A ' D ' AD AD
4 2 3 .7 6 2 4 2 4 .4 6 2 2 2
1 2
300
363 μ
300
2
2
2 tan
1 423 . 76 2 424 . 46 2
1651
μ
SOLUTION (1.33) 0.5 m A' 0.00064 m A
F
1m B
C
B
800 10
6
( 0 .8 )
0 .0 0 0 6 4 m
B'
x
We have AD AD L AD
0.005 m
0.33 m C' (CONT.)
14
1.33 (CONT.)
From triangles A ' A F and C ' C F : 0 .0 0 0 6 4 x
0 .0 0 5 1 .5 x
,
x 0 .1 7 m
From triangles B ' B F and C ' C F : 0 .3 3 01.0.3035 , B 0 .0 0 1 2 4 m B E (contraction) B
Therefore BE
BE LBE
0 .0 0 1 2 4 1
1 2 4 0 (1 0
6
)
1, 2 4 0
SOLUTION (1.34) (a) x
xy
0 . 006 50
120
μ
y
1000 200 800
0 . 004 25
160
μ
μ
( b ) L ' A B L A B (1 y ) 2 5 (1 0 .0 0 0 1 6 ) 2 4 .9 9 6 m m L ' A D L A D (1 x ) 5 0 (1 0 .0 0 0 1 2 ) 5 0 .0 0 6 m m
End of Chapter 1
15
CHAPTER 2
MATERIALS
SOLUTION (2.1) A0
4
(1 2 .5 ) 1 2 2 .7 m m , 2
We have a
2
1500 μ ,
0 .3 200
Af
t
0 .0 0 6 1 2 .5
(1 2 .5 0 .0 0 6 ) 1 2 2 .6 m m 2
4
480 μ
Thus S
p
E
S
p
a
3
1 8 (1 0 )
P A0
1 4 6 .8 M P a
1 2 2 .7 6
1 4 6 .7 (1 0 ) 1 5 0 0 (1 0
6
9 7 .8 G P a ,
)
t a
0 .3 2
Also % e lo n g a tio n
0 .3 200
(1 0 0 ) 0 .1 5
in area
% reduction
122 . 7 122 . 6 122 . 7
(100 ) 0 . 082
SOLUTION (2.2) Normal stress is
P A
2 8 6 .8 M P a
2200
( 3 .1 2 5 )
2
4
This is below the yield strength of 350 MPa (Table B.1). We have L 576.50 0 0 .0 0 1 3 3 9 1 3 3 9 μ Hence E
6
2 8 6 .8 (1 0 ) 1 3 3 9 (1 0
6
2 1 4 .2 G P a
)
SOLUTION (2.3) The cross-sectional area: A w o t o 1 2 .7 ( 6 .1) 7 7 .4 7 m m ( a ) Axial strain and axial stress are a 0 6.038.54 1 0 .0 0 1 3 2 4 1 3 2 4 μ
a
Because
a
S y (See Table B.1), Hooke's Law is valid.
P A
2 1 ,5 0 0 7 7 .4 7 (1 0
6
)
2 7 7 .5 M P a
( b ) Modulus of elasticity, E
a
a
6
2 7 7 .5 (1 0 ) 1 3 2 4 (1 0
6
)
2 0 9 .6 G P a
( c ) Decrease in the width and thickness w w o 0 .3(1 2 .7 ) 3 .8 1 m m t t o 0 .3( 6 .1) 1 .8 3 m m
16
2
2
SOLUTION (2.4) Assume Hooke's Law applies. We have t 15.5 3 0 0 μ a
t
300 0 .3 4
822 μ
Thus, E a (1 0 5 1 0 )(8 2 2 1 0 9
9
) 9 2 .6 1 M P a
S y , our assumption is valid.
Since So
P A (9 2 .6 1)( 4 )(5 ) 1 .8 1 8 k N 2
SOLUTION (2.5) We obtain L AC L BD
L AC
x
L BD
y
(a) E
L AC
L BD
x
x
y
(b)
x
(c) G
15
15
2
21 . 17 21 . 21 21 . 21
1886
21 . 22 21 . 21 21 . 21
471 μ
100 ( 10
6
1886 ( 10
471 1886
53 2 ( 1 0 . 25 )
) 6
2
21 . 21 mm
μ
53 GPa
)
0 .2 5
21 . 2 GPa
SOLUTION (2.6) Use generalized Hooke’s law: x y z
1 2 E
( x
y
For a constant triaxial state of stress: x y z , x
y
Then, Eq. (1) becomes 1 2 0
or
1 2 E
z)
(1)
z
. Since and
must have identical signs:
1 2
SOLUTION (2.7) We have
x
3
4 5 0 (1 0 ) 50 (75 )
120
M Pa
(CONT.)
17
2.7 (CONT.) (a) x
(b) E
x
x
(c) z
6
1 2 0 (1 0 )
2 0 0 0 (1 0
6
500 μ
0 .0 2 5 50
60 G Pa
)
6
1 2 0 (1 0 )
0 .2 5
x
E
6
a 5 0 0 (1 0
(d) G
y
0 .2 5
500 2000
2000 μ,
0 .5 250
9
6 0 (1 0 )
500 μ
9
6 0 (1 0 )
) 7 5 3 7 .5 (1 0
3
a ' 7 5 0 .0 3 7 5 7 4 .9 6 2 5 m m
) mm;
24 G Pa
2 (1 0 .2 5 )
SOLUTION (2.8) We have y z 0
x
3
25 ( 10 ) 20 10 ( 10
6
)
125
MPa
Thus y 0
1 E
[
( x z )]
(1)
z 0
1 E
[ z ( x y )]
(2)
x
y
[ x (
1 E
z )]
y
(3)
Equations (1) and (2) become y z x z
Adding : (
(1’)
x
y
(2’)
z ) 2 x (1 ) . Then, Eq. (3): 2
y
x
2
1 2 1
x
E
Substituting the data:
x
6
1 0 . 3 0 . 18 125 ( 10 ) 9 0 .7 70 ( 10 )
1327
μ
SOLUTION (2.9) Hooke's Law. We have x
10
9
x
E
10
6
7 2 1 0
z
E
6
7 2 1 0
y
x
E
9
x
E
10
6
7 2 1 0
9
y
0 and
y
z
E
[(8 0 ) 0 0 .3 (1 4 0 )] 0 .0 0 0 5 2 8 5 2 8 μ
y
E
z
E
[ 0 .3 (8 0 ) 0 0 .3 (1 4 0 )] 9 1 7 μ
E
y
z
E
[ 0 .3 (8 0 ) 0 1 4 0 ] 1 6 1 1 μ
(CONT.)
18
2.9 (CONT.) (a)
L AB x a ,
Change in length
6
L A B (5 2 8 1 0
(b)
)(3 2 0 ) 0 .1 6 9 m m
Change in thickness t yt (917 10
(c)
6
)(1 5 ) 0 .0 1 4 m m
Change in volume, e x
z 5 2 8 9 1 7 1 6 1 1 1 .2 2 2
y
V e V o 1 .2 2 2 (3 2 0 3 2 0 1 5 ) 1 .8 7 7 m m
3
SOLUTION (2.10) By assumptions, rubber in triaxial stress:
x
z
p,
y
F
d
2
4
4F
d
2
Stains are x z 0 . Hooke's law gives x 0
1 E
[
(
x
y
y
z )]
or 0 p
4F
Solving, p
x
2
d (1 )
4 F
z
Q.E.D.
2
d (1 )
Substitute the data: p
3
4 ( 0 .5 )(1 0 1 0 ) 2
( 6 2 .5 ) (1 0 .5 )
3 .2 6 M P a (C)
SOLUTION (2.11) Hooke’s law gives 90 MPa y 10 mm
150 MPa
x
1 E
(
x
y
1 E
(
y
z
x 100 mm
E
(
y
)
x ) x
y
10
6
100 ( 10 10
9
6 9
1 0 0 (1 0 )
)
)
(150
(90
( 1 3 ) 10 100 ( 10
6
9
)
L 1800 (100 ) 180 μm a 1400 ( 50 ) 70 μm b 200 (10 ) 2 μm
and mm ,
a ' 49 . 993
19
mm ,
b ' 9 . 9998
150 3
) 1800
μ
) 1400 μ
(150 90 ) 200 μ
Thus
L ' 100 . 018
90 3
mm
SOLUTION (2.12) We have x
y
z p
Gen. Hooke’s law:
x
y
z
(1 2 )
p E
120 ( 10 100 ( 10
6 9
) 1 ) 3
400 μ
Thus L 400 (100 ) 40 μm a 400 ( 50 ) 20 μm b 400 (10 ) 4 μm
and L ' 99 . 96 mm ,
a ' 49 . 98 mm ,
b ' 9 . 996
SOLUTION (2.13) We have
x
Vo
(a) x
r 3
p . The volume is
z
4 3
(1 2 5 ) 8 .1 8 1(1 0 ) m m 3
[ ( )]
1 E
6
1 6 8 (1 0 )
4 3
y
9
7 0 (1 0 )
E
6
3
(1 2 )
(1 0 .5 ) 1 2 0 0 μ
Change in diameter, d x d 1 2 0 0 (1 0
6
) 2 5 0 0 .3 m m
Decrease in circumference: ( d ) 0 .3 0 .9 4 2 5 m m
( b ) V e V o (1 2 ) xV o ( 0 .5 )( 1 2 0 0 1 0
6
)(8 .1 8 1 1 0 ) 4 9 0 9 m m 6
SOLUTION (2.14) From Fig.2.3b and Eq.2.20: U
t
S y Su 2
f
250 440 2
( 0 . 27 ) 93 MPa
We have L f 50 50 ( 0 . 27 ) 63 . 5 mm Using Eq.(2.1): % e lo n g a tio n
6 3 .5 5 0 50
(1 0 0 ) 2 7 %
20
3
mm
SOLUTION (2.15) Table B.1:
S
260
y
MPa , E 70 GPa
We have V AL
For U
U
r
U
app
S
2 y
2E
( 0 . 005 ) ( 3 ) 58 . 9 10 2
4
( 260 10
6
2
) 9
2 ( 70 10
)
482 . 9 kJ
m
6
m
3
3
U r V 482 . 9 10 ( 58 . 9 10 3
6
) 28 . 44 J
9 J :
app
n
2 8 .4 4 9
3.1 6
SOLUTION (2.16) ( a ) ASTM-A242. E 2 0 0 G P a and U
o
Sy
2
2E
6
( 3 4 5 1 0 )
2
9
2 ( 2 0 0 1 0 )
y
298
345 M Pa kJ m
3
( b ) Stainless (302). E 1 9 0 G P a and S y 5 2 0 M P a U
o
Sy
2
2E
6
( 5 2 0 1 0 )
2
9
2 ( 2 0 0 1 0 )
712
kJ m
3
SOLUTION (2.17) ( a ) Aluminum 2014-T6. E 7 2 G P a and Uo
Sy
2
2E
6
( 4 1 0 1 0 )
y
410 M Pa
2
9
2 ( 7 2 1 0 )
1167
kJ m
3
( b ) Annealed yellow brass. E 1 0 5 G P a and S y 1 0 5 M P a Uo
Sy
2
2E
6
(1 0 5 1 0 )
9
2
2 (1 0 5 1 0 )
5 2 .5
kJ m
3
SOLUTION (2.18) Referring to Fig. P2.18: E (a) Uo
Sy
2
2E
6
(1 9 2 .5 1 0 ) 9
2 ( 4 2 1 0 )
2
6
1 0 5 (1 0 ) 0 .0 0 2 5
42 G Pa,
4 4 1 .5 k J m
S y 1 9 2 .5 M P a
3
( b ) Total area under diagram: U
2 6 2 .5 (1 0 )( 0 .1 7 6 ) 4 6 .2 M J m 6
t
21
3
SOLUTION (2.19) ( a ) V 50 50 1, 500 3 . 75 (10 ) mm 6
Thus
nU
or
S
S
V
2E
[
y
2 y
1
2 EnU V
]2 9
[ 2 200 10
r
S
2 y
6
( 253 10
2E
)
2
9
2 ( 200 10
)
1
1 . 5 400
3 . 75 ( 10
(b) U
3
3
] 2 253
)
160
MPa
kPa
SOLUTION (2.20) S y 250
Table B.1:
E 200
M Pa,
GPa
We have U nU
U
r
S
app
2 y
6
( 2 5 0 1 0 )
2E
5 (1 7 ) 8 5 N m 2
1 5 6 .2 5
9
2 ( 2 0 0 1 0 )
kN m m
3
Therefore V
U Ur
85 3
1 5 6 .2 5 (1 0 )
V AL :
Also
3
0 .5 4 4 (1 0
0 . 544 (10
3
)
) m
3
2
d ( 2 .4 )
4
or d 0 .0 1 7 m = 1 7 m m
SOLUTION (2.21) Refer to Fig. P2.21. We have E
(a) Uo
Sy
2
2E
6
1 9 0 (1 0 ) 0 .0 0 1 3
( 2 4 5 1 0 )
2
9
2 (1 9 0 1 0 )
190 G Pa,
158
S y 245 M Pa
kJ m
3
( b ) Total area under diagram: 3 5 0 1 0 ( 0 .2 8 ) 9 8 3
U
t
MJ m
3
SOLUTION (2.22) V ( 0 .0 5 )( 0 .0 5 )(1 .2 ) 0 .0 0 3 m
( a ) n U
S
3
and S y S p
2 p
2E
V,
S
2 p
nU ( 2 E ) V
Substituting the data given, (CONT.)
22
2.22 (CONT.) Sp 2
9
1 .8 (1 5 0 )( 2 2 1 0 1 0 ) 0 .0 0 3
3 7 .8 1 0
15
or S p 1 9 4 .4 M P a
(b) Uo
Sp
2
2E
3 7 .8 1 0
15 9
2 ( 2 1 0 1 0 )
90
kJ m
3
SOLUTION (2.23) Applying Eq. (2.22), we find S u 3 .4 5 H
B
M P a 3 .4 5 (1 4 9 ) 5 1 4 M P a
Equation (2.24): S y 3 .6 2 (1 4 9 ) 2 0 7 3 3 2 .4 M P a
SOLUTION (2.24) Using Eq. (2.22), S u 3 .4 5 H
B
M P a 3 .4 5 (1 7 9 ) 6 1 8 M P a
Formula (2.24): S y 3 .6 2 (1 7 9 ) 2 0 7 4 4 1 M P a
SOLUTION (2.25) Formula (2.22), S u 3 .4 5 H
B
M P a 3 .4 5 (1 5 6 ) 5 3 8 M P a
Equation (2.24): S y 3 .6 2 (1 5 6 ) 2 0 7 3 7 8 M P a
SOLUTION (2.26) Equation (2.22) gives S u 3 .4 5 H
B
M P a 3 .4 5 ( 2 9 3) 1 0 1 0 .9 M P a
Formula (2.24): S y 3 .6 2 ( 2 9 3 ) 2 0 7 8 5 3 .7 M P a
End of Chapter 2
23
CHAPTER 3
STRESS AND STRAIN
SOLUTION (3.1) (a)
s
3
4 5 (1 0 )
( 2 5 1 0
3
)
9 1 .6 7
4
3
(b) h
( 2 5 .7 8 1 0
(c) t
( 2 1 .1 2 1 0
(d) b
2
4 5 (1 0 ) 3
3
)(1 0 .9 4 1 0
3
3
)(1 2 .5 1 0
3
5 0 .7 9 M P a
)
4 5 (1 0 )
M Pa
5 4 .2 6
)
3
4 5 (1 0 )
[( 5 0 1 0
3
2
) ( 2 5 .7 8 1 0
M Pa
3 1 .2 2 M P a
3
2
) ]
4
SOLUTION (3.2) A [( a
a
9 16
Thus
a 4
) a ] 2
2
2
Pn Sy
a
9 16
,
a
2
2
16 Pn 9 S y
all
a
,
S
y
n
A
,
P
all
Pn Sy
Pn
4 3
Sy
Substituting the data given: a
1 . 2 ( 10
4 3
6
)( 2 . 2 ) 6
( 280 10 )
73 mm a min
SOLUTION (3.3) Free body-Member ACD
R Ay
A
R Ax
0 : 45 sin 15 ( 0 . 8 ) o
A
FBC
0.4 m 3
0.1 m 0.4 m
Fx 0 :
3 5
F BC 45 sin 15
R A x 5 .8 2 5
D o
RA
Resultant is
F y 0 : 45 cos 15
(5 .8 2 5 ) ( 6 6 .7 6 3 ) 2
We therefore have
n
RA 2
d
2
4
,
2
d (
o
6 6 .9 9 2 R An
1
)2
Hence 3
dA [
2 ( 6 6 .7 6 3 1 0 )( 2 )
dB [
2 ( 2 9 .1 2 1 0 )( 2 )
6
(1 9 6 1 0 ) 3
6
(1 9 6 1 0 )
1
] 2 0 .0 2 0 8 m = 2 0 .8 m m 1
] 2 0 .0 1 3 7 5 m = 1 3 .7 5 m m
24
F BC ( 0 . 3 )
kN o
R Ax 0
kN R Ay
R A y 6 6 .7 6 3
45 kN
3 5
F BC ( 0 . 1) 0
4 5
F B C R B 2 9 .1 2
4
C
15
M
kN
kN
4 5
F BC 0
SOLUTION (3.4) R Ay
5 kN B 1m 1m
C 0.75 m
R Ax A 1 m
1.5 m
E 3
A
0:
5( 3 )
RE
RA
3 5
R E ( 0 .7 5 )
F x 0 : R Ax 7 . 82 kN
F y 0 : R Ay 5 . 43
7 .8 2 5 .4 3 2
2
kN
9 .5 2 k N
Free body-beam ABC 7.82 A 1m
(b)
A
1 . 018
8 ,145 3
8 ( 10
)
3
9 .5 2 (1 0 ) 2
( 0 .0 2 5 )
2
M
C
0 : F BD 8 . 145
RCy
FBD
BD
2m
5.43
(a)
RCx
C
B
MPa
9 .6 9 7 M P a
4
SOLUTION (3.5) RA
A
L cos
L ta n
RB
B
0:
A
F y 0:
R B FBC R A F AC
P s in
A AC
sin
L AC
L cos
A BC
P
P
tan
L BC L
Total weight, W ( A A C L A C A B C L B C ) : W
PL
( sin 1cos
cos sin
)
PL
2
1 cos sin cos
Therefore dW d
P L
2
[
(s in c o s )
2
The foregoing reduces to 2
or
5 4 .7
2
2
s in c o s ( 2 c o s )( s in ) (1 c o s )( s in c o s )
3 cos 1
P ta n
We have
C P
L
M
or
cos
1 3
o
25
4 5
R E 13 . 04 kN
or
4 D
M
] 0
kN ( C )
RE (2) 0
SOLUTION (3.6) Refer to Table 3.1 (Case C) a b 0 .0 9 0 .0 4 5 4 .0 5 1 0
t
3
m
2
45 mm t
T
;
30 10 6
2abt
90 mm
1 .2 1 0
2 ( 4 .0 5 1 0
Solving, t 4 .9 4 m m
We have 1 .5 0 .0 2 6 ra d and a ll L 0 .0 2 6 0 .8 0 .0 3 3 ra d m . o
Hence
( a b ) tT 2
2
2
2t a b G
( a b )T 2
2
2 ta b G
0 .1 3 5 (1 .2 1 0 ) 3
0 .0 3
2 ( 0 .0 4 5 ) ( 0 .0 9 ) ( 2 8 1 0 ) t 2
2
9
Solving: t 5 .3 4 m m Use t a ll 5 .5 m m
SOLUTION (3.7) Refer to Solution of Prob. 3.6. We now have a=b.
T 2
;
30 10 6
2a t 70 mm
t
Solving:
70 mm
Similarly,
T 3
;
0 .0 3 3
a Gt
Solving: Use
1 .2 1 0
3
( 0 .0 7 ) ( 2 8 1 0 ) t 3
9
t 3 .8 m m
t a ll 4 .1 m m
26
1 .2 1 0
3 2
2 ( 0 .0 7 ) t t 4 .1 m m
3 3
)t
SOLUTION (3.8) Refer to Case C, Table 3.1. We now have a=b=31.25 mm, a ll 1 2 2 .4 5 4 .9 0 .0 8 6 ra d m . o
T
a ll
2
105 10
;
6
2a t
T 2 ( 3 1 .2 5 1 0
3
) ( 4 .7 1 0 2
3
,
T 9 6 3 .8 7 N m
,
T 3 5 4 .0 2 N m
)
Likewise, a ll
T 3
0 .0 8 6
;
a tG
T ( 3 1 .2 5 1 0
3
) ( 4 .7 1 0 3
3
) ( 2 8 .7 1 0 ) 9
T a ll 3 5 4 .0 2 N m
Thus, SOLUTION (3.9)
We have a b ( ro ri ) 2 1 1 .7 m m Case E, Table 3.1). From Eq. (3.11), m ax
Tr J
T (1 2 .4 )(1 0 4
3
4
)
( 2 )(1 2 .4 1 1 )(1 0
8 7 7 (1 0 ) T 3
12
)
From Table 3.1: m ax
T 2 abt
8 3 0 .5 (1 0 ) T 3
T 2
2 (1 1 .7 ) (1 .4 )(1 0
9
)
Therefore, error in the thin-wall estimate: 8 7 7 8 3 0 .5 877
0 .0 5 3 5 .3 %
SOLUTION (3.10) ( a ) Maximum moment occurs at midlenth B. Thus
60 kN
a ll
d
V(,kN) A
3
30 kN 30 mm
14 mm
B
14 mm
30 E
D
32 M
m ax
d
3
3
32 M
m ax a ll
32 ( 435 )
2 6 .1 m m
6
( 2 5 0 1 0 )
( b ) Maximum shear is at D and E. Hence, from Table 3.2: x
-30
M, (N m) 435 210
Mc I
or
d 30 kN
a ll
4 V m ax 3 A
d
1 6 V m ax
4 V m ax 3 d2 4
1 6 V m ax 3 d
2
or 2
x
Use d a ll 2 6 .1 m m
27
3 a ll 3
2
1 6 ( 3 0 1 0 ) 6
3 (1 5 0 1 0 )
1 8 .4 3 m m
SOLUTION (3.11) 12 kN
24 kN
21 kN 2m
2m
B 3m 4m
RB
D
2m
4m
A
C
RD
RA
RC
Reactions
M
D
M F
0 : R B 21 kN
F y 0 : R D 15 kN
0 : R A 14 kN
C
0 : R C 7 kN
y
21 kN A
C 2m
14 V (kN)
4m
7
14
+
x
_ 7
M ( kN m )
28 +
x
SOLUTION (3.12) 4 2.7 1m 1m 1m
A
B
3.13
V (kN)
x 3.57
_
0:
A
Fy 0 :
a ll
M
( kN m )
R B 3 .5 7 k N m R A 3 .1 3 k N m
m ax c
:
I
1 2 .5 (1 0 ) 6
3
3 .5 6 (1 0 )( h 2 ) ( 0 .0 5 )( h
3
12 )
3
3 .5 6 1 0 6 2
0 .0 5 ( h )
+
max
3 V 2 A
3 3 . 57 2 0 . 05 ( 0 . 185 )
578 . 9 700 OK
x SOLUTION (3.13) b h 2
2
d , 2
h
2
d
2
b
S
2
We obtain dS db
Thus
b
2
2
d 6
d
and h d
3
b 2
0:
,
h 0 .1 8 5 m
3.56
3.13
M
0.43
+
M
b
2
2
d 3
2 3
28
bh 6
2
b 6
(d
2
b ) 2
bd 6
2
3
b 6
SOLUTION (3.14) c I
200 ( 25 )( 162 . 5 ) 150 ( 15 )( 75 )
135 . 35 mm
200 ( 25 ) 150 ( 15 ) 1 12
200 25 ( 27 . 15 )
3
( 200 )( 25 )
135 . 35 15 ( 1352. 35 )
2
13 . 24 10
2
We have at N.A.: Q m a x 1 3 5 .3 5 (1 5 )( 1 3 52.3 5 ) 1 3 7 .4 1 0
VQ
max
6
3
22 ( 10 )( 137 . 4 10
Ib
13 . 24 ( 10
6
)
3
3
mm
15 . 22 MPa
)( 0 . 015 )
SOLUTION (3.15) I
1 12
[ 200 ( 300 )
100 ( 200 ) ] 383 . 3 (10
3
3
3
Q A y 200 50 (100 25 ) 1 . 25 (10 *
q
VQ I
F (2) q s,
,
2F s
6
) m
) m
4
3
VQ I
Thus V a ll
3
2 (1 5 1 0 )( 3 8 3 .3 1 0
2 FI sQ
0 .1 (1 .2 5 1 0
3
6
)
)
92 kN
SOLUTION (3.16) V max 2 . 4 kN
M
max
Design is based on
all
12 MPa :
all
M
max
c
1 2
)
6
: 12 (10
I
wL
2
1 . 44 kN m
3
1 . 44 ( 10 ) b 2b
4
3
2 ,160 b
3
, b 56 . 5 mm
Check all 0 . 18 MPa :
max
3
3 V max 2 A
:
2 . 4 ( 10 ) 3 2 2 ( 0 . 0565 ) 2
564 kPa 810
kPa
OK .
SOLUTION (3.17) V max
1 2
( 50 6 ) 150
Design based on S min
M
all max
kN ,
170
all
max
1 8
( 50 )( 6 )
2
225
kN m
MPa : 3
225 ( 10 ) 170 ( 10
M
6
Table A.7: S 4 6 0 8 1 .
)
1 . 324 (10
6
) mm
3
(smallest possible, others will work)
Shear stress S 460 81 : A web d ( t w ) 457 (11 . 7 ) 5 . 347 (10 ) mm 3
max
V max A web
3
150 ( 10 ) 5 . 347 ( 10
3
)
28 . 1 MPa
29
100
MPa
OK .
2
6
mm
4
SOLUTION (3.18) M
(w0 x
x
M
S
x
bh 6
;
all
2 L )( x 3 ) w 0 x
2
2
w0x
6L
6L
(1)
all
bh 1
h h1
At x=L:
3
3
w0x
6
3
6L
(2)
all
Divide Eq.(1) by (2): 2
h
2
h1
3
x L
h h 1 ( Lx )
or
3
3 2
SOLUTION (3.19) RA RB
1 2
wL,
M
x
wLx
1 2
bh 6
Equation (3.27), at a distance x : At x
L 2
2 bh 1
:
6
w 2
2
(
h
Divide Eq.(1) by Eq.(2): or h 2 h 1
x L
( Lx )
SOLUTION (3.20)
4
L 2
L 4
)
2
2
h1
2
1 2
w 2
( Lx x ) 2
Lx x L
2
2
1
(1)
all
2
4
y Es
n
(20)200=40(103)
3
( 4 1 0 )( 3 2 5 )
2 .8 9 1 0
126 10 6
20 M
s
( 0 .1 6 2 5 )
2 .8 9 1 0
3
3
12
9
3
( 3 .8 1 0 )( 3 0 0 )
mm 6
;
1 4 1 .6
w
M
3
12
7 .3 5 1 0
w
M ;
20
Ew
It
12.5 mm
I
bh 6
(2)
300 mm
nM y
2
z
x
a ll
all
12.5 mm
s
1
200 mm
M
S
2
wx ,
4
My I
2 .8 9 1 0
M
w
( 0 .1 5 )
2 .8 9 1 0
3
kN m
112 kN m
s
Stress in steel governs: M 112 kN m
SOLUTION (3.21) Es
n
I
Ew
1 12
M
20
[( 7 5 4 0 t )( 2 2 5 )] 3
1 8
wL 2
[ 7 5 (1 0 0 )] ( 6 4 .9 4 1 3 7 .9 6 9 t ) 1 0 3
1 12
[3 5 1 0 ( 2 .5 )(1 2 )] 2 7 .3 4 3
1 8
kN m
2
Therefore, we write
s
w
nM y I
;
135 10
6
3
2 0 ( 2 7 .3 4 1 0 )( 0 .1 1 2 5 ) ( 6 4 .9 4 1 3 7 .9 6 9 t ) 1 0
6
,
t 1 0 .2 9 m m
Similarly
My I
;
8 10
6
3
( 2 7 .3 4 1 0 )( 0 .1 1 2 5 ) ( 6 4 .9 4 1 3 7 .9 6 9 t ) 1 0
Stress in steel governs: t 1 0 .2 9 m m
30
6
,
t 8 .4 1 5 m m
6
m
4
3
m
4
SOLUTION (3.22) Transformed Section of Brass y Es
n
It [
89.1 mm C1
z
3
A d ]1 [ 2
bh 36
3
bh 12
[ 3 6 ( 2 0 0 )(1 2 0 ) 3
1
9.1 mm 43.4 mm
y 55.9 mm
2
Eb
A d ]2 2
200 2
1 2 0 ( 9 .1 ) ]1 2
[ 1 2 (1 0 0 )( 2 5 ) 1 0 0 ( 2 5 )( 4 3.4 ) ] 2 3
1
C2
15 . 43 10
1100mm
6
mm
2
4
200 mm Thus 120 (10
6
3
)
M ( 55 . 9 10
)
2 M ( 89 . 1 10
)
6
15 . 43 10
M 33 . 12 kN m
,
Similarly, 140 (10
6
15 . 43 10
3
)
6
M 12 . 12 kN m
,
(governs)
SOLUTION (3.23) n
Es Eb
2
I t I b nI
s
[
30 ( 50 )
3
12
( 7 .5 )
4
4
( 7 .5 )
] 2[
4
4
] 314 , 985
mm
4
Thus, we write ( b ) max
M ( 25 10
Mc It
314 , 985 ( 10
3
)
12
120 (10
)
6
M 1 . 512
),
kN m
Similarly, ( s ) max
3
2 M ( 7 . 5 10
nMc It
314 , 985 ( 10
12
)
140 (10
)
6
M 2 . 94 kN m
),
SOLUTION (3.24) n
Eb Ea
1.5 Ib
Transform to brass:
64
[d
2
( d2 ) ] 4
15 d
4
1024
Aluminum core: Each element of area of area has its width reduced in ratio n . Ia
1 n 64
( d2 )
4
d
4
1024 n
and It Ib Ia
d
4
1024
( 15
1 n
)
We have
b
Mc It
M
,
bIt d 2
d 512
3
b ( 15
1 n
)
Substituting numerical values: M
( 0 .0 5 ) 512
3
(3 5 0 1 0 )(1 5 6
1 1 .5
) 4 2 0 5 .7
31
N m = 4 .2 1 k N m
(governs)
SOLUTION (3.25) 1 0 5 . Apply Eq.(3.31): o
( a ) x'
1 2
(25 15)
1 2
( 2 5 1 5 ) c o s 2 1 0 1 0 sin 2 1 0 o
o
x’
5 17 . 32 5 17 . 32 MPa
x'y'
1 2
( 2 5 1 5 ) s in 2 1 0
10 cos 210
o
(20) (10) 2
x'y' 15
o
o
x
y’
10 8 . 66 1 . 34 MPa
( b ) 1, 2 5
x'
5 2 2 .3 6
2
or 1 17 . 36 MPa
p ' ' 13 . 28
27 . 36 MPa
2
o
1
2
p''
SOLUTION (3.26) 70 (a)
y’ m ax
x
30 o
x’ ( b ) max
1 2
30 70 2
cos 2( 55 ) o
20 17 . 1 37 . 1 kPa
x 55
30 70 2
y
xy
20 17 . 1 2 . 9 kPa
30 70 2
sin 2 ( 55 ) 47 kPa o
[ 30 ( 70 )] 50 kPa
SOLUTION (3.27) y 50(0.866) (0.866)
35 90 125
50(1)
x’ x
60
x
,
y
0,
xy
50 MPa
0 .8 6 6 x 0 .5 ( 5 0 ) 0 .5 ( 5 0 )
F x 0:
o
o
x
57 . 74 MPa
(C )
50(0.5) ( a ) Equations (3.31) with 1 2 5 : o
x'
x'y'
2 8 .8 7 2 8 .8 7 c o s 2 5 0 5 0 s in 2 5 0 o
28 . 87 sin 250
o
50 cos 250
o
o
6 5 .9 8
M Pa
44 . 23 MPa
(CONT.)
32
3.27 (CONT.) 1 2
( b ) 1 , 2 2 8 .8 7 [( 2 8 .8 7 ) 5 0 ] 2 8 .8 7 5 7 .7 4 2
1 2 8 .8 7 M P a ,
2
2
p ''
8 6 .6 1 M P a ,
1 2
ta n
1
x’
50 2 8 .8 7
30
o
x
44.23
p''
35
x’
o
x y’
1
2
65.99
SOLUTION (3.28) From Solution of Prob. 3.27:
x
xy
6
57 . 74 ( 10 200 ( 10 2 (1 0 .3 )
200 ( 10
9
)
9
)
)
289 μ ,
( 50 10
6
5 7 .7 4 M P a ,
x
y
x
y
0 , and
50 M P a.
xy
87 μ
) 650 μ
( x ) AC
C
B
40
34.6
L A C 52.915
60
A
40 4 0 .8 6
289 87 2 650 2
28987 2
o
c o s 2 ( 4 0 .8 6 ) o
s in 2 ( 4 0 .8 6 )
1 0 1 2 7 .0 7 3 2 1.6 1 194 μ
o
D
20 o
( a ) ( x ' ) A B 1 0 1 1 8 8 c o s 2 (1 2 0 ) 3 2 5 s in 2 (1 2 0 ) 2 8 8 μ o
o
L A B 4 0 ( 2 8 8 ) 0 .0 1 1 5 m m
( b ) L A C 5 2 .9 1 5 (1 9 4 1 0
6
) 0 .0 1 0 3 m m
SOLUTION (3.29)
y (a)
y
y
7 0 ( 0 .5 ) 3 5 M P a
60.6(0.866)=52.5
x y 7 0 ( 0 .8 6 6 ) 6 0 .6 M P a
( 0 .5 )
30
o
140(1)
x
x
0 . 5
Fx 0 :
(b) 1
175 35 2
x
[( 175 2 35 )
140 52 . 5 0
x
175
M Pa
p'
1
2
( 60 . 6 ) ] 2 70 121 . 23 191 . 23 MPa 2
2 5 1 .2 3 M P a p'
1 2
ta n
1
2 ( 6 0 .6 ) 175 35
15
x x’
o
1
33
2
SOLUTION (3.30) From Solution of Prob. 3.29:
9
2 0 0 1 0 2 (1 0 .2 5 )
G
x
y
x'
175 M Pa,
x
(
1 E
10
80 10 y )
x
6 9
2 0 0 (1 0 )
3 5 M P a , and
Pa, 10
6 9
2 0 0 (1 0 )
xy
xy
G
xy
6 0 .6 1 0 8 0 1 0
6
9
6 0 .6 M P a .
758
(1 7 5 3 .7 5 ) 9 1 9 ,
( 3 5 4 3 .7 5 ) 3 9 4
919 394 2
9
y
919 394 2
cos 2 (141 . 5 ) o
o
758 2
sin 2 (141 . 5 ) 2 6 2 .5 1 4 7 .6 8 3 6 9 .2 9 7 7 9 .4 7
x’
L B D 0.0416 m
B
C
0.05 m A
=141.5o x
0.075 m D
Thus L B D 0 .0 4 1 6 ( 7 7 9 .4 7 1 0
6
) 0 .0 3 1 3 m m
SOLUTION (3.31) 40
x
xy
o
y
100 cos 45
100 sin 45
o
o
70 . 71 MPa
70 . 71 MPa
Equations (3.31):
x'
y'
70 . 71 0 69 . 64 1 . 07 MPa
x'y '
70 . 71 70 . 71 2
0 70 . 71 sin 80
0 70 . 71 cos 80
y’
o
x'
40
140 . 3 MPa
12 . 28 MPa
x’
x'y'
o
o
x y'
34
SOLUTION (3.32) 40 90 130
y
o
40 MPa ,
50 MPa ,
x
25 MPa
xy
( a ) Equations (3.31): 50 40 2
x'
50 40 2
c o s 2 (1 3 0 ) 2 5 sin 2 (1 3 0 ) o
o
5 7 . 814 24 . 62 27 . 43 MPa
(b)
x'y'
y'
[(
max
p'
1 2
45 sin 260
o
5 7 . 814 24 . 62 37 . 434
50 40 2
44 . 316 4 . 341 39 . 975
MPa
MPa
1
)
25 ] 2 51 . 48 MPa
2
2
1 25 45
tan
25 cos 260
o
14 . 53
o
x’
x'
y'
x’
x'y'
m ax
x
p'
x x
y’ SOLUTION (3.33) x”
B
AC BD 90 . 139
x’
C
75 mm
G E 2 .6
x
50 mm 3 3 .6 9
1 4 6 .3
o
xy
1 E
( xy G
o
1 E
( x y )
10 E
6
[ 5 0 0 .3 ( 4 0 ) ]
6 2 1 0 E
6
x
A y
mm
D y
x ) 6
2 5 1 0 E 2 .6
x'
10 E
6
6 5 1 0 E
[ 4 0 ( 0 .3 ) ( 5 0 ) ]
5 5 1 0 E
6
6
We have
6
10 2E
[( 62 55 ) ( 62 55 ) cos 2 ( 33 . 69
( 65 ) sin 2 ( 33 . 69
x"
6
10 2E
o
)]
89 . 85 E
(10
6
o
)
)
[117 ( 7 ) cos 2 (146 . 3 ) ( 65 ) sin 2 (146 . 3 )] o
o
Thus L A C x ' (9 0 .1 3 9 ) L B D x " (9 0 .1 3 9 )
8 9 .8 5 1 0 2 1 0 1 0
6
9
2 9 .8 4 1 0 2 1 0 1 0
9
6
(9 0 .1 3 9 ) 0 .0 3 8 6
mm
(9 0 .1 3 9 ) 0 .0 1 2 8
mm
35
29 . 84 E
6
(10 )
SOLUTION (3.34)
Vessel is thin walled.
1.2 kN/m B
0.9 m
M
V 1.5 m
VQ
xy
A
0 at A and C
Ib
V 2 .7 1 .2 (1 .5 ) 0 .9 k N
M 2 .7 (1 .5 )
C
2.7 kN
2 .7
1 2
(1 .2 )(1 .5 )
2
kN m
Point A
x'
Mc I
(a)
y
(b)
m ax
1 2
2 .7 ( 0 .4 5 ) 3
( 0 .4 5 ) ( 3 1 0
2
a
3
)
6 .3 M P a 1 ,
( 1 2 )
1 2
1 .4 1 5 M P a ,
2
a
pr 2t
4 2 ( 0 .4 5 ) 2 ( 3 1 0
3
)
3 .1 5 M P a
3 .1 5 1 .4 1 5 1 .7 3 5
( 6 .3 1 .7 3 5 ) 2 .2 8 2 5
M Pa
M Pa
Point C (a)
(b)
m ax
x'
2 1 .4 1 5 3 .1 5 4 .5 6 5 M P a ,
1 .4 1 5 M P a , 1 2
( 1 2 )
1 2
( 6 .3 4 .5 6 5 ) 0 .8 6 8
1 6 .3 M P a
M Pa
SOLUTION (3.35) Refer to Solution of Prob. 3.34.
(a)
x
xy
1, 2
Thus
x'
VQ Ib 0 .9
rt
a
0 3 .1 5 3 .1 5 M P a ,
0 .9 ( r t )(
2r
0 .9
( 0 .4 5 ) ( 3 1 0
3
)
0 .2 1 2 2 M P a
1
[( 6 .3 23 .1 5 ) ( 2 .1 2 2 ) ] 2 4 .7 2 5 1 .5 8 9 2
1 6 .3 1 4 M P a 1 2
6 .3 M P a
(see Table A-3)
2
or
( b ) m ax
y
)
3
r t(2t)
6 .3 3 .1 5 2
2 3 .1 3 6 M P a
( 1 2 ) 1 .5 8 9 M P a
36
SOLUTION (3.36)
x’
P A
Tr J
3
1 1 0 (1 0 ) 2
2
( 0 .0 2 5 0 .0 1 2 5 )
7 4 .7 M P a
74.7
x
3
2 .3 (1 0 )( 0 .0 2 5 )
4
9 9 .9 6 M P a 4
( 0 .0 2 5 0 .0 1 2 5 )
2
99.96
50 90 140
o
x’
Equations (3.31): x'
140
o
x'y'
x'
7 4 .7 2
7 4 .7 2
c o s 2 8 0 9 9 .9 6 s in 2 8 0 o
o
1 4 2 .3 M P a
x
x'y'
7 4 .7 2
s in 2 8 0 9 9 .9 6 c o s 2 8 0 o
o
1 9 .4 2 M P a
y’ SOLUTION (3.37) 410 mm
y A
120 18
z
24 kN
M
40
50
120
24
3 4
18 kN
y
A 40 (120 ) 4 . 8 10
x
A
I
x
xy
1 12
( 40 )( 120 )
3
3
mm
5 . 76 10
2 6
mm
4
M 18 ( 0 . 12 ) 24 ( 0 . 41 ) 12 kN m
We have at A: x
P A
Mc I
3
1 8 (1 0 )
4 . 8 (1 0
3
)
( 3 . 75 104 . 17 ) 10
xy
VQ Ib
3
2 4 1 0 ( 2 2 1 0 5 .7 6 1 0
6
100 . 4 2
2
6
)
( 0 .0 4 )
3
1 2 (1 0 )( 0 . 0 5 )
5 . 7 6 (1 0
6
6
)
Q 4 0 (1 0 )(5 5 ) 2 2 (1 0 ) m m
100 . 4 MPa
2 .2 9 M P a
Thus
or
1,2
100 . 4 2
[(
)
1 1 0 0 .5 M P a
2
2 0 .0 5 M P a
m a x 5 0 .2 8 M P a
and tan 2 p '
or
p ' 1 . 31
2
xy
x
o
1
( 2 . 29 ) ] 2 50 . 2 50 . 25
2 ( 2 . 29 ) 100 . 4
1 . 31
37
3
3
SOLUTION (3.38)
y y
B
B
16
M P
yo P=40 kN
C
62 z
A Z
10 mm
y
A
8
45
From Z axis: 6 2 1 6 (3 9 ) 4 5 8 ( 4 )
y
2 9 .7 m m
62 16 45 8
y 0 1 0 y 3 9 .7 m m A 1 6 6 2 4 5 8 1 .3 5 (1 0 ) m m 3
2
About the z axis: I
1 6 (6 2 )
3
1 6 6 2 ( 9 .3 ) 2
4 5 (8 )
12
3
4 5 8 ( 2 5 .7 )
2
12
1 0 ( 0 .3 1 8 0 .0 8 6 0 .0 0 2 0 .2 3 8 ) 0 .6 4 4 1 0 6
6
mm
4
M 4 0 0 .0 3 9 7 1 .5 8 8 k N m
Thus,
a
P
A
40 10
Mc
( b ) B
3
1 .3 5 1 0
I
( b ) A 9 9 .4
3
2 9 .6 M P a
1 5 8 8 ( 0 .0 4 0 3 ) 0 .6 4 4 1 0
2 9 .7
6
9 9 .4 M P a
7 3 .3 M P a
4 0 .3
A
2 6 .3 7 3 .3 9 9 .6 M P a
B
2 6 .3 9 9 .4 7 3 .1 M P a
SOLUTION (3.39) A b h 2 5 (1 0 0 ) 2 .5 (1 0 ) m m 3
I
2
2 5 (1 0 0 ) 12
P 50 kN
M 5 0 1 0 ( 0 .0 5 ) 2 .5 k N m 3
( a ) At the top fibers: t
P A
Mc I
5 10 2 .5 (1 0
3
3
2 .5 1 0 ( 0 .0 5 ) 3
)
2 .0 8 (1 0
6
20 60 40 M Pa
At the bottom fibers: b 20 60 80 M Pa
38
)
40.3
3
2 .0 8 (1 0 ) m m 6
4
SOLUTION (3.40) A b h 2 5 (1 5 0 ) 3 .7 5 (1 0 ) m m 3
I bh
1 2 2 5 (1 5 0 )
3
2
1 2 7 .0 3(1 0 ) m m
3
6
4
At the bottom fibers:
P
b
Mc
A
P
I
3 .7 5 (1 0
3
0 .0 7 5 P ( 0 .0 7 5 )
)
7 .0 3 (1 0
6
)
Solving, 120 10
2 6 6 .7 P 8 0 0 .1 4 P
6
Pa ll 1 1 2 .5 k N
SOLUTION (3.41) J
[( 0 .0 6 ) ( 0 .0 5 0 ) ] 1 0 .5 4 (1 0 4
2
c 0 .0 6
4
m,
45
6
) m
8 0 M P a m ax
o
4
Tc J
Thus T ( 0 .0 6 )
8 0 (1 0 ) 6
1 0 .5 4 (1 0
6
)
T 1 4 .0 5 k N m
,
SOLUTION (3.42) ( a ) 1, 2
500 800 2
1
[( 5 0 0 2 8 0 0 ) ( 3 52 0 ) ] 2 6 5 0 2 3 0 μ 2
2
or 1 880 μ ,
( b ) max G
max
2 420 μ , 3
70 ( 10 )
2 (1 0 .3 )
max
460 μ
( 460 ) 12 . 38 MPa
SOLUTION (3.43) (a)
max
2 [(
200 600 2
A
1
)
2
( 400 ) ] 2 566 μ 2 2
149
( b ) x'
200 600 2
200 600 2
400 2
o
c o s 2 (1 4 9 )
x
C
s in 2 (1 4 9 ) 1 3 0 μ o
o
L A . C 2 7 .3 3 3 m m
Thus L A C x ' L A C 1 3 0 (1 0
6
)( 2 7 .3 3 3 ) 3 .5 5 1 0
3
mm
SOLUTION (3.44) 1, 2
50 250 2
[( 5 0 22 5 0 ) ( 2
150 2
1
) ]2 150 μ 125 μ 2
or 1 275 2 25
(CONT.)
39
3.44 (CONT.) Apply Hooke’s Law (with z 0 ): 6
2 7 5 (1 0
and
2 5 (1 0
6
)
)
( 1 2 )
1 9
2 1 0 (1 0 )
(1)
( 2 1 )
1 9
2 1 0 (1 0 )
(2)
Solving 1 65 . 2 MPa ,
24 . 8 MPa
2
SOLUTION (3.45)
a
pr 2t pr t
p (500 )
P A
2 (1 0 )
0 .0 2 2 ( 0 .5 )( 0 .0 1 )
25 p 2 M Pa
x
50 p
y
50 20
s in 4 0
A=1
40
o
cos 40
5 0 ( 2 5 p 2 ) s in
2
40
o
50 p cos 40 2
p all 1 . 281 MPa
or
25p-2
o
F y 0:
F x 0:
2 0 ( 2 5 p 2 ) s in 4 0 c o s 4 0 o
5 0 p c o s 4 0 s in 4 0
o
o
50 p
p all 1 . 546
or
o
MPa
SOLUTION (3.46) ( a ) x y ( x ' y ' ) s in ( 2 ) x ' y ' c o s ( 2 ) o
o
0 ( 2 4 0 4 1 0 ) s in ( 6 8 ) o
Then
1
x
2 1
( x '
y'
)
1 2
( x ' 1
(240 410)
2
cos(68 ) , o
x'y'
1
) c o s ( 2 ) o
y'
x'y'
2
( 2 4 0 4 1 0 ) co s( 6 8 ) o
2
421 μ
s in ( 2 ) o
x'y'
1
( 4 2 1) s in ( 6 8 ) o
2
9 8 .2 μ
y
1
( x '
2 1
y'
)
1 2
(240 410)
2
( x ' 1
1
) c o s ( 2 ) o
y'
2
( 2 4 0 4 1 0 ) co s( 6 8 ) o
2
( b ) Hooke’s Law, with t y , a x , and t 2 a
So
y
x
x
(1 2 )
E
552 9 8 .2
5 .6 2
y
2 1 2
x
(2 )
E
0 .3 5
,
40
s in ( 2 ) o
x'y'
1 2
552 μ
x
o
( 4 2 1) s in ( 6 8 )
o
o
SOLUTION (3.47) a 880
( a 0 ),
b 320
o
( b 6 0 ),
a x c o s a y s in a 2
Thus
8 8 0 (1 0
6
xy
) x c o s 0 y s in 0 o
( c 1 2 0 ) o
s in a c o s a
2
2
c 60
o
2
o
o
x 880 μ
o
s in 0 c o s 0 ,
xy
Likewise, b x c o s b y s in b x y s in b c o s b 2
3 2 0 (1 0 3 2 0 (1 0
6
6
2
) x c o s ( 6 0 ) y s in ( 6 0 ) 2
o
2
) 0 .2 5 x 0 .7 5
y
0 .4 4 3
c x c o s c y s in c 2
and,
6 0 (1 0
6
6 0 (1 0
6
o
2
xy
s in ( 6 0 ) c o s ( 6 0 ) o
xy
(1)
xy
s in c c o s c
) x c o s ( 1 2 0 ) y s in ( 1 2 0 ) 2
o
2
) 0 .2 5 x 0 .7 5
y
o
0 .4 4 3
o
s in ( 1 2 0 ) c o s ( 1 2 0 ) o
xy
o
(2)
xy
Subtract Eq. (2) from Eq. (1): So
3 8 0 μ 0 .8 8 6
xy
and then from Eq. (1):
xy
4 2 8 .9 μ
y
1 0 2 .1 μ
2
880 A
388
2
O
avg
x p
288.1
'
()
C R
y
1
1 2
( x y )
(8 8 0 1 0 2 .1) 3 8 9 μ
2 R [ (8 8 0 3 8 9 ) ( 2
4 2 8 .9 2
5 3 5 .8 μ
1,2 a vg R
1 3 8 9 5 3 5 .8 9 2 4 .8 μ 2 1 4 6 .8 μ ta n 2
p
'
2 8 8 .1 880 389
p ' 1 6 .8
,
x’
o
x 16.8o
924.8 y
A
y’ 146.8
41
2
1
) ]2
SOLUTION (3.48) y xy
A
x
Tc J
y
2
( x y ) 0
a v g R s in 2 xy
c
3
2
0
xy
xy G
x y 0
Mohr’s circle for strain: avg
25
o
x
1
J
Tc
s i n 2
2G
R
1
2
1
2
xy
2
x 2
xy
A’ x’ R
s in 2
O
s i n 2
y
2G J
This gives 2G J
T
c s in 2 Substitute the given data:
(8 0 .5 1 0 )( 0 .0 4 4 ) ( 6 0 0 1 0 9
T
3
( 0 .0 4 4 ) s in 5 0
6
o
)
2 2 5 .0 3 8 k N m
SOLUTION (3.49) Using Eqs. (3.39), we have x 300 μ,
y
150 μ,
xy
2(375) (300 150) 600 μ
Equations (3.38) are thus, 300 150
1, 2
p
(
450
2
1 300 μ ,
1 2
ta n
1
) ( 2
2
600
)
2
175 375
2
2 450 μ [
600 300 150
] 2 6 .6
o
x ' 1 7 5 2 2 5 c o s 5 3 .1 3 0 0 s in 5 3 .1 4 5 0 2 o
o
Thus, p " 2 6 .6
y’ 300
o
450 x’ 26.6o x 42
SOLUTION (3.50) Given: D 5 0 m m , t 1 5 m m , P 2 5 k N . Refer to Figs. P3.50 and C.1: d
r
r d
D d
K
32 33 34 35
9 8.5 8 7.5
0.28 0.26 0.24 0.21
1.56 1.52 1.47 1.43
1.64 1.66 1.62 1.7
t
We have 6
a ll K t
P A
;
150 (10 ) 1.9
Kt
3
25 (10 ) 0 .0 1 5 h
In this equation, minimum h is reached when K t is minimum. Thus, use h 34 mm
SOLUTION (3.51) D d
45 30
1.5 ,
r d
6 30
0 .2
Figure C.1, K t 1.7 2 .
Hence
and
Pall A
nom
max
1 . 72
nom
210 1 . 5
81 . 4 MPa
1 . 72
( 30 12 )( 81 . 4 ) 29 . 3 kN
SOLUTION (3.52) At the notch and hole:
nom
P ( D d h 2 r )t
12 10
3
( 0 .0 9 0 .0 1 5 2 0 .0 0 7 5 ) ( 0 .0 1)
20 M Pa
For the notch : (see Fig. C.1): D
d
90
1 .2
75
r
7 .5
d
0 .1,
75
K t 1 .7 8
m a x K t n o m 1 .7 8 ( 2 0 ) 3 5 .6 M P a
For the hole (see Fig. C.5): dh
D
15
0 .1 6 7 ,
90
K t 2 .5
m a x K t n o m 2 .5 ( 2 0 ) 5 0 M P a
Hence
SOLUTION (3.53) Kt
m ax
nom
180
1 .6 3 6 and D d 1 .5
110
From Fig. C.1: r d 0 .2 4 . Then D 2r d
(CONT.)
43
3.53 (CONT.) 4 0 .6 2 5 2 ( 0 .2 4 d ) d 1 .4 8 d
gives or
d 2 7 .4 5 m m r 6 .5 9 m m
SOLUTION (3.54) For
r
0 .2 0 :
d D 2r d ;
4 0 .6 2 5 2 ( 0 .2 d ) d 1 .4 d
or d 2 9 .0 2 m m We thus have D
1 .4
d
Figure C.1 gives K t 1 .7 . Hence,
3 5 0 (1 0 ) 1 .7
Pa ll
6
m ax
(1 2 .5 1 0
3
) ( 4 0 .6 2 5 1 0
3
)
Pa ll 1 0 4 .5 k N
or
SOLUTION (3.55) (ksi)
avg
1
(
2
1
x
y)
(1 6 8 8 4 ) 1 2 6 M P a
2
3
O
R
C
(
x
y
1 (MPa)
(
168 84 2
5 9 .4 M P a
avg =59.4 ( a ) 1,2
R
avg
1 1 2 6 5 9 .4 1 8 5 .4 M P a 2 6 6 .6 M P a
3 21 M Pa
( b ) ( m a x ) a
1 2
1
( 1 3 )
[1 8 5 .4 ( 2 1) ] 1 0 3 .2 M P a
2
44
) 2
2
2 xy
) (42) 2
2
SOLUTION (3.56) (MPa)
avg
avg
1
(
2
1
x
y)
(5 0 0 ) 2 5 M P a
2
R 3
O
1
C
R
(
x
y
) 2
2
(MPa)
50 0
(
2 xy
) (25) 2
2
2 3 5 .3 6 P a
( a ) 1,2
R
avg
1 2 5 3 5 .3 6 6 0 .3 6 M P a
2 1 0 .3 6 M P a 3 60 M Pa
( b ) ( m a x ) a
1
( 1 3 )
2
1
[ 6 0 .3 6 ( 6 0 ) ] 6 0 .1 8 M P a
2
SOLUTION (3.57) (ksi) avg =42
avg
1
(
2
35
1
x
y)
(7 0 1 4 ) 4 2 M P a
2
3
C
1
(MPa)
R
(
x
y
) 2
2
(
70 14
2 xy
) (56) 2
2
2 6 2 .6 M P a
( a ) 1,3
avg
R
1 4 2 6 2 .6 1 0 4 .6 M P a
3 4 2 6 2 .6 2 0 .6 M P a 2 35 M Pa
(CONT.)
45
3.57 (CONT.) ( b ) ( m a x ) a
1 2 1
( 1 3 ) [1 0 4 .6 ( 2 0 .6 ) ] 6 2 .6 M P a
2
SOLUTION (3.58) ( a ) In the yz plane: ' =35
70 MPa
(MPa)
y
21 MPa
z
21
C
3
R
(MPa)
O 1
70 We have R
35 21 2
2
4 0 .8 2 M P a
Thus 1 R ' 4 0 .8 2 3 5 5 .8 2 M P a
3 R ' 4 0 .8 2 3 5 7 5 .8 2 M P a 2 28 M Pa
( b ) Using Eqs. (3.58) : oct
1 3
1
[ ( 5 .8 2 2 8 ) ( 2 8 7 5 .8 2 ) ( 7 5 .8 2 5 .8 2 ) ] 2 2
2
2
3 3 .4 9 M P a
oct
1 3
(5 .8 2 2 8 7 5 .8 2 ) 3 2 .6 7 M P a
From Eq. (3.50), m a x 12 (5 .8 2 7 5 .8 2 ) 4 0 .8 2 M P a SOLUTION (3.59) ( a ) Using Eq. (3.47): (70 i )
0
56
0
(1 4 i )
0
56
0
(1 4 i )
0
(CONT.)
46
3.59 (CONT.) Expanding, (1 4 i )[( i 7 0 )( i 1 4 ) 3 1 3 6 ] 0
or
1 42 M Pa,
2 14 M Pa,
3 98 M Pa
( b ) From Eq. (3.50), m a x 12 ( 4 2 9 8 ) 7 0 M P a acts on planes shown in Fig. 3.41. ( c ) Using Eqs. (3.52), we have o c t 13 ( 4 2 1 4 9 8 ) 1 4 M P a oct
1 3
1
[( 4 2 1 4 ) (1 4 9 8 ) ( 9 8 4 2 ) ] 2 2
2
2
6 0 .4 9 M P a
They act on planes depicted in Fig. 3.43. SOLUTION (3.60) The principal stresses are 1 8 4 M P a , 2 6 3 M P a , and 3 1 2 6 M P a . ( a ) By Eq. (3.50), m ax
1 2
(8 4 1 2 6 ) 1 0 5 M P a
acts on planes shown in Fig. 3.41. ( b ) Applying Eqs. (3.52): o c t 13 (8 4 6 3 1 2 6 ) 7 M P a oct
1 3
1
[ (8 4 6 3 ) ( 6 3 1 2 6 ) ( 1 2 6 8 4 ) ] 2 9 4 .4 4 M P a 2
2
2
They act on planes shown in Fig. 3.43. SOLUTION (3.61) The principal stresses are 1 2 9 7 .5 M P a , 2 3 6 .8 2 M P a , and 3 5 4 .7 4 M P a . The direction cosines: l cos 40
o
m cos 60
n cos 66 . 2
o
o
Thus, by Eqs. (3.51), we obtain 2 9 7 .5 (c o s 4 0 ) 3 6 .8 2 (c o s 6 0 ) 5 4 .7 4 (c o s 6 6 .2 ) o
2
o
2
o
2
= 174.9 MPA [( 2 9 7 .5 3 6 .8 2 ) (c o s 4 0 ) (c o s 6 0 ) 2
o
2
o
2
(3 6 .8 2 5 4 .7 4 ) (c o s 6 0 ) (c o s 6 6 .2 ) 2
o
2
o
2 1
( 5 4 .7 4 2 9 7 .5 ) (c o s 6 6 .2 ) (c o s 4 0 ) ] 2 2
o
2
o
2
1 4 8 .9 M P a
End of Chapter 3
47
CHAPTER 4
DEFLECTION AND IMPACT
SOLUTION (4.1) ( a ) We have A req . A req .
and
48 mm
3
3
10 ( 10 )( 6 10 )
PL
)
250 1 . 2
all
E
3
10 ( 10
P
5 ( 200 10
3
2
60 mm
)
2
Since 60 48 mm , d
(b) k
AE L
( 6 0 1 0
6
4 ( 60 )
4A
8 . 74 mm .
9
)( 2 0 0 1 0 )
2 (1 0 ) k N m 3
6
SOLUTION (4.2) Refer to Example 4.4. Use numerical values for bar AB. Cross-sectional area: A A B 1 2 8 9 6 m m . 2
Stress:
AB
F A
4 0 1 0 9 6 (1 0
3
6
4 1 6 .7 M P a 4 3 5 M P a
)
OK.
Deflection: AB
FL AE
k AB
2 .3 8 m m
3
3
4 0 (1 0 )
F
3
4 0 (1 0 )( 0 .6 ) ( 9 6 )(1 0 5 )(1 0 )
2 .3 8 (1 0
3
1 6 .8 1(1 0 ) k N m 3
)
SOLUTION (4.3) The cross-sectional area:
A
4
(D
2
d ) 2
D 4
2
[1 ( Dd ) ] 0 .5 0 2 7 D 2
2
Also A
7 5 1 0
P
3
1 4 0 1 0
6
0 .5 3 6 1 0
3
m
2
Equating these, D
2
1 .0 6 6 2 1 0
3
D 0 .0 3 2 7 m
2
m ,
It follows that k
PL AE P
3
7 5 1 0 ( 0 .3 7 5 ) ( 0 .5 3 6 1 0 7 5 1 0
3
0 .7 2 9 1 0
3
3
9
)( 7 2 1 0 )
0 .7 2 9 1 0
3
m
1 0 2 .8 8 M N m
SOLUTION (4.4) Statics: RA
36 kN
R A RB 36
RB
kN
(1)
Deformations and Compatibility: Assume gap closes. (CONT.)
48
4.4 (CONT.) 0 .3 5 1 0
3
1
(1 2 .5 1 0
3
2
9
) (1 2 0 1 0 )
( 0 .2 R A 0 .2 5 R B )
4
R A 1 .2 5 R B 2 5, 7 7 0 .8 7 7
or
(2)
Solving Eqs.(1) and (2): R B 4 .5 4 6
kN
R A 3 1 .4 5 4
kN
Since the answer for R B is positive, the gap closes, as assumed. SOLUTION (4.5) Increase in length due to T (unrestrained): t
(T )L
(12 10
6
)( 120
o
)( 250 ) ( 23 10
6
)( 120
o
)( 300 ) 1 . 188 mm
( a ) Compressive axial force P P t 1 1 . 188 1 0 . 188 mm
(1)
But
P
PL AE
P ( 0 .2 5 )
5 0 0 (1 0
6
9
)( 2 1 0 1 0 )
P ( 0 .3 )
1 0 0 0 (1 0
6
(2)
9
)( 7 0 1 0 )
Equating Eqs.(1) and (2): 0 .1 8 8 (1 0
3
) 2 .3 8 1(1 0
9
) P 4 .2 8 6 ( 1 0
9
)P
P 28 . 2 kN
or
( b ) Change in length of aluminum bar a ( t ) a ( P ) a a ( T ) L a 4 .2 8 6 ( 1 0 ( 23 10
6
)( 120
o
)( 0 . 3 ) 4 . 28 (10
9
9
)P
)( 28 . 2 10 ) 0 . 707 3
mm
SOLUTION (4.6) Refer to Solution of Prob.4.5: ( a ) P t 1 . 188 mm and
1 . 188 (10
( b ) a 0 .8 2 8 (1 0
3
3
) ( 2 . 381 4 . 286 )10
) 4 .2 8 (1 0
9
9
P , P 178 . 2 kN
)(1 7 8 .2 1 0 ) 0 .0 6 5 3 m m 3
SOLUTION (4.7) Let the compressive axial force in the pipe be R. Moment equilibrium about point A: R
Pa b
1 2 (1 .3 )
4 4 .5 7 k N
0 .3 5
a P
b
C
A R
C
B
B
(CONT.)
49
4.7 (CONT.) The beam can rotate only about A and the deflections B and C can be described by the rotational angle as B a and C b a
B
Hence,
b
C
(1)
The contraction of pipe C is as follows: C
RL
RL
( 4 )( D
AE
2
(2)
d )E 2
Inserting Eq.(2) into Eq.(1) gives B
b(D d )E 2
2
0 .3 2 9 1 0
4 ( 4 4 .5 7 1 0 )( 0 .6 2 5 )(1 .3 ) 3
4RLa 3
( 0 .3 5 )( 0 .1 0 5 0 .0 9 5 )( 2 0 0 1 0 ) 2
2
9
m = 0 .3 2 9 m m
SOLUTION (4.8) (a) b
P Lb
P La
a
Eb A La
a b;
Ea
Ea A
Lb
(1)
Eb
La Lb L
(2)
Solving, Eb 1 Lb L ( ) L E Ea Eb 1 a Eb
,
La L Lb
Introducing the given data: 1 L b 0 .6 0 .3 6 7 m , 70 1 110
(b)
PL
P
A
AE 3
1 0 0 (1 0 ) ( 4 ) ( 0 .0 4 )
2
(
Li Ei
L a 0 .2 3 3 m
a b
0 .3 6 7 110
0 .2 3 3 70
)
1 10
9
0 .2 6 5 0 .2 6 5 0 .5 3 m m
50
SOLUTION (4.9)
C FD
Q=12 kN A
2b
A 2 D
b
PCD = -20 kN D
FD =24 kN
D
B
D 0.15 m
PDE =4 kN
M
B
0 : 2b Q b FD
E P=4 kN
FD 2 Q 2 4 k N
(a) D
A
0.3 m
2 0 (1 0 )(3 0 0 ) 3
PL
AE
130 10
6
(7 0 1 0 ) 9
0 .6 6 m m
2 D 1 .2 2 m m
(b) E
PL
AE 1
130 70
10 130 10
6
3
(7 0 1 0 ) 9
( 0 .6 6 ) 0 .5 9 1 0
[ 4 0 .1 5 2 0 0 .3 ] 3
m 0 .5 9 m m
SOLUTION (4.10) PD E 4 k N
(a) D
A
PD C 2 8 k N 2 8 (1 0 )(3 0 0 ) 3
PL
AE
130 10
6
(7 0 1 0 ) 9
0 .9 2 m m
2 D 1 .8 4 m m
(b) E
PL
AE 1
130 70
10 130 10
6
3
(7 0 1 0 ) 9
[ 4 0 .1 5 2 8 0 .3 ]
( 0 .6 8 .4 ) 9 8 9 1 0
3
m 0 .9 9 m m
SOLUTION (4.11) F C
F B
120 mm
180 mm (CONT.)
51
4.11 (CONT.) T CD F ( 0 . 12 ) 500
N m
or F 4 . 167
( a ) AB
7 5 0 (1 .2 )
TL GJ
T AB 4 . 167 ( 0 . 18 ) 750
kN
9
7 9 (1 0 )
N m
0 .0 2 8 ra d
( 0 .0 2 2 5 )
4
2
CD
5 0 0 (1 .8 ) 9
( 7 9 1 0 )
0 .0 7 7 ra d
( 0 .0 1 7 5 )
4
2
D CD 1 . 5 AB 0 . 119
Thus (b)
AB
2TAB
c
2 (750 )
3
( 0 .0 2 2 5 )
rad 6 . 82
o
4 1 .9 2 M P a
3
SOLUTION (4.12) all TC D FC
125
150 1 .2
3 2
a ll ( d )
T CD
6
3
1 2 5 1 0 ( 0 .0 6 5 )
16
( r gear ) C
MPa 6 .7 4 k N m
16
56 . 167
6 . 74 0 . 24 2
kN m
Hence, we have T AB 56 . 167 ( 0 .236 ) 10 . 11 kN m
all
Thus
16 T AB
3 d1
d1 3
,
3
16 ( 10 . 11 10 ) 6
( 125 10 )
or d 1 74 . 4 mm
SOLUTION (4.13) ( a ) Polar moment of inertia for a hollow cylinder is J
(c b ) 4
2
4
c
4
2
[1 ( bc ) ] 4
(1)
From Eq. (4.10), we have
J c
T
m ax
4 .5 1 0
3
1 4 0 1 0
6
3 2 .1 4 3 1 0
6
m
2
3 2 .1 4 3 m m
Note that, b 0 .5 c . Using Eqs. (1) and (2) then c 2
3
[1 ( 0 .5 ) ] 3 2 .1 4 3,
c 2 .7 9 5 m m
4
So, D 2 c 5 .5 9 m m ,
(b)
TL GJ
(c) k
T
d 0 .5 D 2 .7 9 5 m m
3
( 4 .5 1 0 )( 0 .2 5 ) 9
( 7 9 1 0 )( 3 2 .1 4 3 1 0 3
4 .5 (1 0 ) 0 .1 5 8 5
6
)( 2 .7 9 5 1 0
3
)
0 .1 5 8 5 ra d 9 .0 8 1
2 8 .3 9 k N m ra d
52
o
2
(2)
SOLUTION (4.14) TBC 5 6 0 N m
TC D 8 4 0 N m
( a ) Shaft BC J BC
d
4
32
(3 4 ) 1 3 1, 1 9 4 m m 4
LBC 6 2 5 m m
4
32
TBC 5 6 0 N m
So
BC
TBC L BC
G J BC
5 6 0 (6 2 5 1 0
3
)
( 7 9 1 0 ) (1 3 1, 1 9 4 1 0 9
0 .0 3 4 r a d 1 .9 5
12
o
)
( b ) Shaft CD J CD
d2 4
32
(2 5) 3 8, 3 5 0 m m 4
4
32
LCD 7 5 0 m m
TC D 8 4 0 N m
CD
TC D LC D
G J CD
8 4 0 (7 5 0 1 0
3
)
(7 9 1 0 )(3 8, 3 5 0 1 0 9
12
0 .2 0 8 r a d 1 1 .9 2
o
)
Hence B D B C C D 0 .0 3 4 0 .2 0 8 0 .1 7 4 ra d 9 .9 7
o
SOLUTION (4.15) The polar moment of inertia for the shaft is J
32
d )
4
(D
4
(5 0 3 5 ) 4 6 6 , 2 6 9 m m 4
32
4
4
Equilibrium Condition. From the free-body diagram of appropriate portions of the shaft: T A C 1 .1 k N m
T C D 1 .9 k N m
T D E 2 .6 k N m
The results are shown on the torque diagram in Fig. S4.15 ( a ) Angle of twist. The shear modulus of elasticity G is a constant for the entire shaft. Through the use of Eq. (4.9), we obtain A
TL GJ
1 GJ
10
( T A C L1 T C D L 2 T D E L 3 )
3
9
( 7 9 1 0 ) ( 4 6 6 , 2 6 9 1 0
12
)
[ ( 1 .1) ( 0 .4 5 ) (1 .9 ) ( 0 .3 7 5 ) ( 2 .6 ) ( 0 .6 2 5 ) ]
0 .0 5 0 0 ra d = 2 .8 6
o
Comments: A positive result means that the gear will rotate in the direction of the applied torque at free end. ( b ) Maximum shear stress. The largest stress takes place in region DE, where magnitude of the highest torque occurs (Fig. S4.15) and J is a constant for the shaft. Applying the torsion formula: m ax
TD E ( J
D 2
)
3
( 2 .6 1 0 )( 2 5 1 0 4 6 6 , 2 6 9 1 0
3
12
)
1 3 9 .4 M P a
(CONT.)
53
4.15 (CONT.)
Hence, n
y
m ax
210 1 3 9 .4
1 .5
Comments: For the situation under consideration, a safety factor of 1.5 to 2 is to be selected (see Sec. 1.6) 2.6
T (kN m)
1.9
A
C
D
-1.1
x
E
Figure S4.15
SOLUTION (4.16)
or
TL GJ
1.5 180
;
0 . 02618
T ( 0 .5 h ) GJs
1000 79 ( 10
9
)
[
Th GJh
0 .5 h ( 0 . 02 )
4
h 4
( 0 . 02 ) ( 0 . 011 )
4
]
2
Solving h 1 9 7 m m SOLUTION (4.17) T Assume T A as redundant.
TA
x
TB
Tx TA
x
T 0
1
d x T A T1 x
Deformation:
A
Tx dx GJ
TA GJ
A 0;
Geometry:
T A T B T1 L ,
Statics:
L
dx
T1
GJ
0
TA
1 2
L
xdx
0
TAL GJ
T1 L
2
2GJ
T1 L
TB
1 2
T1 L
SOLUTION (4.18) Conditions of equilibrium gives: R A P and M For portion AC E Iv1 '' P x E I v1 ' E I v1
1 2 1 6
B
P L 2 as shown in Fig. S4.18.
P A
Px
RA=P
x
P x C1
B
C Px
PL/2
x
2
L/2
P x C1x C 2 3
L/2 Figure S4.18 (CONT.)
54
4.18 (CONT.) For portion CB 1
E I v 2 '' E I v1 '
1
PLx C3
2 1
E Iv 2
PL
2
PLx C3x C4 2
4
Boundary conditions lead to v1 ( 0 ) 0 :
C2 0
v2 (0 ) 0 :
C3 0
Then v1 (
L
) v2 (
2
v 1 '(
L
L 2
) v 2 '(
2
P
):
(
6 L
2 P
):
2
Solving C 1
L
(
2 3
PL
) C1( 3
L 2
L 2
2
11
C4
8
L
PL(
4
) C1
2
P
)
4
P
L(
2
L
) C4 2
)
2
PL
3
48
At B(x=0): vB v2 (0 )
11
11
PL 3
48
PL 3
48
SOLUTION (4.19) 4
EI
d v dx
4
EIv ' ' ' ' w 0
x L
w0 x
EIv ' ' '
,
2L
2
c1 ,
EIv ' '
Boundary Conditions: v ' ' (0) 0 : c2 0,
v ' ' ( L ) 0 : c1
w0L 6
Then EIv ' E Iv
w0 x
4
24 L w0x
5
120 L
w 0 Lx
2
12 w0Lx
3
36
c3 c3 x c4
Boundary Conditions: v ( 0 ) 0:
c4 0,
v ( L ) 0:
c3
Thus, we obtain
and Solving,
w0x
v
3 6 0 E IL
v'
360 EIL
w0
(7 L 10 L x 4
2
2
3x ) 4
( 7 L 30 L x 1 15 x 1 ) 0
x1 L 1
4
8 15
2
2
4
0 .5 1 9 3 L
Hence v m a x v ( x 1 ) 0 .0 0 6 5 2 2
w0L EI
4
55
7 w0L 360
3
w0 x 6L
3
c1 x c 2
SOLUTION (4.20) M
Segment AC: EIv 1 ' '
0
2
EIv 1 '
,
M
Segment CB: EIv 2 ' ' EIv
x
L
0
L
M
'
0
x
2
c1
2L
y
(L x)
A
2
Mo/L
( Lx
0
L
M
x 2
) c2
a
Mo C
B
x
Mo/L
L
Boundary Conditions: v1 ' ( a ) v 2 ' ( a ) : M 0x
E Iv1 Then Also, we have EIv
2
3
c 2 c1 M 0 a
c1 x c 3 ,
6L
M
' M 0x
0
2
x
2L
v 1 ( 0 ) 0:
c3 0
c1 M 0 a ,
EIv
2
M
0
x
2
2
M
0
x
6L
3
c 1 x M 0 ax c 4
Boundary Conditions: v 2 ( L ) 0;
c4 M 0 L(a
v 1 ( a ) v 2 ( a ); M ox
v1
Thus
M
c1
0
6L
2
(3a
L 3
) c1 L
6 aL 2 L ) 2
(6 aL 3a 2 L x ) 2
6 E IL
2
2
SOLUTION (4.21) We have m a x 1 6 I
w
or
(wL
Mc I
2
8 )( h 2 ) I
m ax
2
L h
Therefore 4
v m ax
5
5L 384 EI
(
1 6 I
5
)
m ax
2
L h
m ax
L
2
24 hE
1
2 4 h E v m ax
Solving, L (
4
5wL 384 EI
)2
m ax
Introducing the given data: L [
9
2 4 ( 0 .3 1 5 )( 2 0 0 1 0 )( 3 .1 1 0
3
)
6
5 ( 7 0 1 0 )
1
] 2 3 .6 6 m
SOLUTION (4.22)
x
A
x1
E
P
C B R B P (1
( a ) EIv ' ' P
a L
x,
EIv '
1 2
a L
RC P )
P
a L
a L
y x
2
c1 ,
EIv
1 6
P
a L
x
3
c1 x c 2
1 6
PaL
Boundary Conditions: v (0) 0 :
Thus
v
PaLx 6 EI
[(
c 2 0, x L
)
v(L) 0 :
c1
1]
2
( b ) The v m a x occurs at E where v ' 0 . Hence 0
PaL 6 EI
[3(
x1 L
)
2
1 ],
x1
1 3
L 0 . 577 L
(CONT.)
56
4.22 (CONT.) v m ax
and
( c ) v m a x 0 .0 6 4 1
2
PaL 6 EI
[( 0 .5 7 7 ) 0 .5 7 7 ] 0 .0 6 4 1 3
3
2 5 (1 0 )( 0 .5 )( 2 ) 9
2 0 0 (1 0 )( 5 .1 2 1 0
2 6
)
PaL EI
2
3 .1 3 m m
SOLUTION (4.23) F
y
B
A
L
x
C
L/2 3F/2
F/2 Figure S4.23
Free-body diagram of shaft is in Fig. S4.23. Since reactions at supports A and B, two differential equations must be written for portion AB and BC. ( a ) We have M1 M
(0 x L )
P 2
Px
2
(L x
3 PL 2
3 PL 2
)
Integrating twice, the preceding leads to the following expression: For segment AB E Iv1 '' E Iv 1 ' E Iv 1
For segment AB E Iv 2 '' P x
Px 2 Px 4
2
3
Px 12
C1
E Iv 2 '
C1 x C 2
E Iv 2
Px 2 Px 6
2
3
3 PL 2 3 PLx 2
3 PLx 4
2
C3 C3x C4
Using the boundary and the continuity conditions, we find v1 ( 0 ) 0 :
C2 0
v1 '( L ) v 2 '( L ) :
C3
5 PL 6
2
2
v1 ( L ) 0 :
C1
v2 ( L ) 0 :
C4
PL 12
PL 4
3
The result elastic curves of the beam are v1
Px 12 EI
(L x )
v2
P 12 EI
(3L 10 L x 9 L x 2 x )
2
(0 x L )
2
3
2
2
3
(L x
(P4.23a) 3L 2
)
( b ) The deflection at the free end of the beam is readily found by introducing x=3L/2 into the second of Eqs. (P4.23a): vc
3
PL 8 EI
3
PL 8 EI
(P4.23b)
Comment: Observe that the deflection of the shaft is downward between B and C and upward between A and B.
57
SOLUTION (4.24)
Refer to Table A.8:
P
B
E 1 I1
A
Beam AB: L/2
L/2
R
c'
D
3
5 PL 48 E 1 I 1
Beam CD:
3
RL 24 E 1 I 1
c''
3
RL 24 E 2 I 2
C E 2 I2
Condition of compatibility, c ' c ' ' : 3
5 PL 48 E 1 I 1
3
RL 24 E 1 I 1
3
RL 24 E 2 I 2
; R
5P (E2I2 ) 2 ( E1 I1 E 2 I 2 )
SOLUTION (4.25) From a free body diagram of beam BC, we observe that it has vertical reactions 2P/3 and P/3 at ends B and C, respectively. Thus, beam AB is in the condition of a cantilever beam under a uniform load of intensity w and a concentrated load B equal to 2P/3. The deflection of the hinge: vB
4
wa 8 EI
2 Pa 9 EI
3
as obtained by Cases 3 and 1 of Table A.8, respectively.
SOLUTION (4.26)
W
A M
C C
A
RA
F
A C
v 'C
B
(a)
W
(b)
B
C
A (c)
F
B v ''C
Figure S4.26: (a) Free-body diagram of beam; (b) deflection due to P; (c) deflection due to reactive force F.
Consider F as redundant and to release the rod from the beam at C (Fig. S4.26a) and then reapplied (Figs. S4.26 b and c). (CONT.)
58
4.26 (CONT.) Refer to Table A.8: 2
vc '
(2 L 3a )
WL 6 EI
3
v c ''
FL 3 EI
Equation of compatibility at point C: c v c ' v c ''
F k
or
2
(2 L 3a )
WL 6 EI
3
FL 3 EI
Solving, 2
WL k (2 L3a )
F
Q.E.D.
3
2 ( kL 3 E I )
SOLUTION (4.27) y
w x
M A
B
F y 0:
M
A
RA RB wL
0:
M
A
RB L
(1) 1 2
wL
2
RB
RA
E Iv " R A x
1 2
E Iv '
RAx
E Iv
1 6
RAx
2
2
wx
1 2
3
v ' ( 0 ) 0:
1 6
M
A
w x M A x c1
1 24
3
wx
4
c1 0 ,
1 2
c1 x c 2
2
M Ax
v ( 0 ) 0:
c2 0
and E Iv
1 6
RAx 3
1 24
The condition that v ( L ) 0
4
wx
1 2
M Ax
2
(2)
RAL
gives
1 4
wL
2
3M
A
0
(3)
Solving Eqs.(1) and (3): RA
5 8
wL
M
A
1 8
wL
RB
2
3 8
wL
SOLUTION (4.28) w0L
Due to symmetry: R A R B
,
4
M
A
M
B
We have, for 0 x L 2 : M RAx M
A
w0x
3
w0Lx
3L
4
M
A
w0x
3
3L
Therefore E Iv " M E Iv ' M
A
x A
w0Lx 4 w0Lx
2
8
Boundary conditions: v ' (0) 0 : v ' ( L2 ) 0 :
w0x
3
3L
w0x
4
c1
12 L
c1 0 M
A
5 w0L
2
96
(CONT.)
59
4.28 (CONT.) Hence 2
5 w0L x
EIv
2
3
w 0 Lx
192
w0 x
24
5
60 L
c2
Boundary condition: v ( 0 ) 0:
c2 0
Thus, we obtain v
w0 x
2
( 25 L 40 L x 16 x ) 3
960 LEI
2
( for 0 x L 2 )
3
and v max v ( L2 )
7 w0L
4
3840 EI
SOLUTION (4.29) y M
P
A
By virtue of symmetry:
B
M
A
C
L/2
x
B
M
A
M
RA RB
L/2
B
P 2
RB
RA
Segment AC EIv ' ' ' ' 0 ,
E Iv '
1 2
c1 x
EIv ' ' ' c 1 , 2
c2 x c3 ,
EIv ' ' c 1 x c 2
E Iv
1 6
c1 x 3
1 2
c2 x
2
c3 x c4
Boundary Conditions: EIv ' ' ' ( 0 ) c 1 V v ' (0) 0 :
P 2
c3 0,
v ( 0 ) 0:
EIv ' ' ( 0 ) M
,
v' (L 2) 0 :
A
c2 M
,
M
A
M
B
A
PL 8
c 4 0.
Equation (1) is thus v
2
(3L 4 x )
Px 48 EI
SOLUTION (4.30) Let the reaction R B is selected as redundant and the corresponding constraint is removed. Then v ' B w L 8 E I , from Case 3 of Table A.8. 4
The deflection caused by the redundant is v"B R B L
3
3EI
The total deflection must be zero; vB
4
wL 8 EI
RB L
3
3EI
0
or RB
3wL 8
Now, applying equations of statics, we obtain RA
5wL 8
M
A
wL 8
2
60
(1)
SOLUTION (4.31) w A
a
C
a
D
M EI
wa
2
3w a
2
a
B
x2 x1
2EI
A2
2EI
A3
x
Spandrel parabola
A1 x3
Tangent at A A
vB
D
B
B A1
bh 3EI
A2
3
3a 4
2
a
3
wa 2 EI
A3
wa 2 EI
x1 a
1 3
wa 6 EI
a(wa )
1 2
x2
7a 4
(App. A.3) 2
1 wa 2 EI
5a 2
x3
3
8a 3
( a ) B A B A A1 A 2 A 3 B
3
wa 12 EI
(2 6 6)
7 wa 6 EI
3
( b ) t B A v B A1 x 1 A 2 x 2 A 3 x 3 vB
4
wa 24 EI
(7 30 32)
69 w a 24 EI
4
23 w a 8 EI
4
SOLUTION (4.32) P
4EI (a)
EI
A L/2
B
L/2
C
P 2
P 2
M/EI PL/4EI
Px/2E I
PL/16EI A
A2
A1
C
x B
D x B
A v m ax
B t BD
Tangent at B
1 PL 2 16 EI
(2)
A2
1 PL 2 4 EI
(2)
L
L
2
1 PL 64 EI 2
1 PL 16 EI
t A B A1 ( 3 ) A 2 ( L
D
t AB
A1
Tangent at D
61
B
1 L
t AB
2L 3
)
3
3 PL 64 EI
2
3 PL 64 EI
(CONT.)
4.32 (CONT.)
( b ) BD B D B 0 .
B
Also
1 Px 2 2 EI
2
(x)
Px 4 EI
Hence 2
3 PL 64 EI
2
Px 4 EI
x
,
3 4
L
2x 3
)
Thus, v m ax t B D
2
1 Px 2 2 EI
3 3L P 6 EI 64
3
(
3
Px 6 EI
3 P L3 128 EI
SOLUTION (4.33)
M
Tangent at A
A
2a
A
P Let R B be redundant.
a
B
RA
A 0
C
RB
A1
M M
1 2
M A (2a ) M Aa
A2
1 2
Pa(2a ) Pa
A3
1 2
Pa
2
2
4a/3 v B t BA 0
A
B
A1
A
A2
or
C x
A3
Pa
M
A
Statics: R B
2a/3
1 EI 1 2
7 4
4a 3
[ A1 (
) A2 (
2a 3
)]
Pa
P
RA
3 4
P
SOLUTION (4.34) Assume R c as redundant. w A
tCB
Tangent at B
RA
M wL
2
B L/2
L
t AB
C RC
RB
L/2 L/3
8
A1
A1 A2
2L/3
x
A3
R C L/2
2 wL 3 8
2
(L)
A2
1 2
Rc L
A3
1 2
Rc L
2
2
wL 12
3
(App. A.3)
(L)
Rc L
(2)
Rc L
L
2
4 2
8
(CONT.)
62
4.34 (CONT.) E I t A B A1 ( 2 ) A 2 ( L
2L 3
Rc L
E It CB A3 ( 3 ) L
4
)
Rc L
wL 24
3
6
3
24
Since
2 t CB t AB :
or
Rc
1 6
wL
Statics: R B
3 4
wL
RcL
3
12
wL 24
RA
4
RcL
3
6
wL
5 12
SOLUTION (4.35) S
m ax
y
n
2
2
W g
v E AL
gS y AL
W a ll
,
2
2
n v E
Substitute given data: 6
2
9 .8 1 ( 2 5 0 1 0 ) [
W a ll
2
( 0 .0 2 ) ( 2 )]
1 6 .6 N
4 2
2
9
( 3 ) ( 3 .5 ) ( 2 1 0 1 0 )
SOLUTION (4.36) P=mg 1.25=a
0 .7 5 2
x
B
P
st
Table A.8 ( Case 6 ).
b=0.75
A
1. 2 5 2
L=2 (L b 2
Pbx 6 LEI
20 ( 9 . 81 )( 0 . 75 )( 1 . 25 )
x )
2
P
2
6 ( 2 )( 210 10
9
)( 0 . 06 0 . 08
3
12 )
(2
2
0 . 75
2
1 . 25 ) 0 . 0535 2
mm
Max. moment is under load. Thus
M
st
m ax
c
We have K 1
( 0 .6 2 5 0 .7 5 2 0 9 .8 1 ) ( 0 .0 4 )
I
0 .0 6 0 .0 8
1
3
12
1 [1
2h
st
1 .4 3 7 M P a 1
2 ( 0 .5 ) 0 . 0535 10
3
] 2 137 . 7
( a ) m a x s t K 7 .3 7 m m ( b ) m ax st K 1 9 8 M P a SOLUTION (4.37)
A
4
max
( 2 5 ) 1 5 6 .2 5 m m 2
W A
[1
1
from which
2h
W
Solving W 2(
hE L
[
st
2 2 m ax A 2
2 m ax
]; 2
Since s t W L A E , thus:
2
m ax A
W
m ax
W
A
1]
2
1 1
1 2 hAE WL
A
6
)
2
( 2 5 0 1 0 ) (1 5 6 .2 5 1 0
6
)
9
2[
st
(1) (P4.37)
m ax
Substituting given data: W
2h
(1 .1 )( 2 0 0 1 0 )
3 1 9 .0 7 N
6
2 5 0 1 0 ]
4 .6
63
SOLUTION (4.38) Refer to Solution of Prob. 4.37. From Eq.(1), we obtain A
2W
[
2 m ax
hE L
6
(P4.38)
]
m ax
9
2 (90 )
(1 2 5 1 0 )
2
[
1 .2 ( 2 0 0 1 0 )
1 2 5 1 0 ] 1 .8 4 5 1 0 6
1 .5
3
m
2
Thus d
2
1 .8 4 5 1 0
4
3
d 0 .0 4 8 5 m = 4 8 .5 m m
,
SOLUTION (4.39) A (10 )
314 . 2 mm
2
From Eq. (P4.37): L
Solving, h
2W (
m ax ) m ax A 2
hE L
( A m ax 2W )
m ax
2W E
3 ( 350 10
2
6
)
2 ( 500 )( 170 10
9
)
(P4.39)
[ 314 . 2 350 2 500 ] 673
mm
SOLUTION (4.40) We have W 2 4 9 .8 1 2 3 5 .4 N and M I
1 12
6
( 0 .0 4 )( 0 .0 6 ) 0 .7 2 1 0 3
m
m ax
1 4
WL
4
W h
C
A L/2
B
k
L/2
W=196.2 N
R st
R
R k
Using Table A.9: st
or
(W R ) L
3
3
R
48EI
650 10
6
R(
1
L
R
k
) 10
6
R[
48EI
7 8 .2 180 10
k
3
6 5 0 R ( 5 3 .3 0 7 ) ,
st
RL
48EI
k
or Hence
3
0 .6 5 1 0
3
1 0 .2
R 7 8 .2 N
(2)
3
]
4 8 ( 0 .0 7 )( 0 .7 2 )
and
W R 1 5 7 .2 N
0 .4 3 4 m m
Maximum stress occurs at midspan:
st
Mc I
1 5 7 .2 ( 2 ) ( 0 .0 3 ) 4 ( 0 .7 2 1 0
6
3 .2 8 M P a
)
(CONT.)
64
4.40 (CONT.)
K 1
2 ( 0 .0 5 )
1
0 .4 3 4 1 0
1 6 .2
3
( a ) v m a x 1 6 .2 ( 0 .4 3 4 ) 7 .0 m m ( b ) m a x 1 6 .2 (3 .2 8 ) 5 3 .1 M P a SOLUTION (4.41) Refer to Example 4.14. EkG
m ax 2
or
AL
4 1 6 .6 7 9 1 0
2 1 0 1 0 2[ 6
;
44100 10
12
4 .4 1 8 1 0
3
9
L m in
]
12
4 ( 7 4 4 9L.3 9 1 0 ) m im
or L m in 0 .6 7 6 m
Thus, we obtain 2 Ek L
m ax
2 ( 4 1 6 .6 )( 0 .6 7 6 )
GJ
0 .0 4 8 ra d = 2 .7 5
3
7 9 1 0 ( 3 .1 1 )
o
SOLUTION (4.42) The E k in wheel B must be absorbed by the shaft. We have J
32
(1 2 5 ) 2 .4 (1 0 4
5
) m
4
Substitute Eqs.(4.42a) and (4.42b) into (4.41): Ek
1 4
b t 4
2 Ek L
(a)
m ax
(b)
m ax 2
GJ
EkG AL
2
[
( 0 . 075 ) ( 0 . 025 )( 1800 )( 150060 2 ) 4
4
1
2 ( 2 7 .5 9 )( 0 .3 ) 9
(1 9 1 0 )( 2 .4 1 0
2[
5
)
9
( 2 7 .5 9 )(1 9 1 0 )
] 2 0 .0 0 6 0 2 5
27 . 59 N m
2
ra d 0 .3 5
o
1
] 2 1 1 9 .3 3 M P a
2
( 0 .0 2 5 ) ( 0 .3 )
4
SOLUTION (4.43) The E k in wheel A must be absorbed by the shaft. Refer to Solution of Prob.4.42. We have Ek
1 4
b t
(a)
m ax
(b)
m ax 2
4
2 Ek L GJ
EkG AL
2
[
( 0 . 0625 ) ( 0 . 025 )( 1800 )( 120060 2 ) 4
4
1
2 ( 8 .5 2 )( 0 .3 ) 9
(1 9 1 0 )( 2 .4 1 0
5
)
9
( 8 .5 2 )(1 9 1 0 )
2[
] 2 0 .0 0 3 3 5 ra d 0 .1 9
1
] 2 6 6 .3 1 M P a
2
( 0 .0 2 5 ) ( 0 .3 )
4
65
o
2
8 . 52 N m
SOLUTION (4.44) Let r r y , then r x r xy , Hence, Eq. (4.46) for z
max
Et 2 ( 1
2
t 2
1 ) r
M M
,
y
M
M
x
xy
0.
:
;
126 (10
6
200 10
)
9
( 3 . 2 10
3
) 1 r
2
2 (1 0 .3 )
or r 2 .7 9 1 m and D 5 .5 8 2 m
Equations (4.50) lead to
6M
max
t
m ax
;
2
6
126 (10
6M
)
m ax
( 3 . 2 10
3
)
2
or M
2 1 5 .0 4
m ax
N m m
SOLUTION (4.45) (a) y
Dw ' ' ' ' p 0 sin
Dw ' ' ' ( b ) p 0 cos
b y
Dw ' ' ( ) p 0 sin 2
b
c1 y c 2
b
y b
c1
Dw ' ( b ) p 0 cos 3
y
b
1 2
c1 y
2
c2 y c3
and y
D w ( ) p 0 s in 4
b
b
1 6
3
c1 y
1 2
c2 y
2
c3 y c4
(1)
Boundary conditions: w ' ( 0 ) 0:
c 3 ( ) P0 ,
w ( b ) 0:
c2
1 3
w ( 0 ) 0:
3
b
c1 b
2
p0b ,
3
c4 0
w ' ( b ) 0:
2
c1 0 ,
c2
2
3
p0b
2
Equation (1) becomes p0b
w
D
4 4
( s in
y
b
b
2
y
2
b
(2)
y)
( b ) At y=0: m ax
(c) D
Et
3
1 2 (1
2
)
6M t
m ax 2
9
7 0 1 0 (1 0 1 0
2
6 t
3
)
d w
[D
2
3
dy
2
]y0
6 t
2
[0 0
2 p0b
3
2
]
12 p0
3
b
(t)
2
6 .4 1 k N m
2
1 2 (1 0 .3 )
Thus w m a x w ( b2 )
p0b 4
4
D
(1
) 4
3
3 5 1 0 ( 0 .5 )
4
3
( 6 .4 1 1 0 )
and
m ax
3
1 2 ( 3 5 1 0 )
3
( 00.0.51 ) 3 3 .8 6 2
4
M Pa
66
(1
4
) 0 .7 5 2 1 0
3
m = 0 .7 5 2 m m
SOLUTION (4.46) ( a ) Dw ' ' ' ' p 0 , Dw'
Dw ' ' ' p 0 y c 1 ,
p0 y 3
1 6
1 2
2
c2 y c3
4
c1 y
Dw ' '
1 2
p0 y
2
c1 y c 2
and Dw
1 24
p0 y
c1 y 3
1 6
1 2
c2 y
2
c3 y c4
(1)
Boundary Conditions: w (0) 0 : c4 0,
w ' ' (0) 0 : c2 0
w ( b ) 0 a n d w ' ( b ) 0:
c1
3 8
p0 b,
c3
1 48
p0b
3
Equation (1) is therefore p0b
w 2
(b)
d w dy
2
p0 2D
(y
4
48 D
2
y
y
y
[ ( b ) 3( b ) 2 ( b ) ]
3 yb 4
3
4
(2)
)
At y=b: M
m ax
2
D
p0
d w dy
2
2
b 8
m ax
,
6M t
m ax 2
0 .7 5 p 0 ( t ) b
2
( c ) From Solution of Prob. 4.45, D 6 4 .1 k N m Thus w m ax w ( 2 ) b
p0b
4
192 D
p0b
4
[ ( 2 ) 3( 2 ) 2 ( 2 ) ] 1
48 D
1
3
( 3 5 1 0 )( 0 .5 ) 3
4
1 9 2 ( 6 .4 1 1 0 )
3
1
4
0 .0 0 1 8 m = 1 .8 m m
and
0 .7 5 p 0 ( bt ) 0 .7 5 (3 5 1 0 )( 51000 ) 2
m ax
3
2
6 5 .6 3 M P a
End of Chapter 4
67
CHAPTER 5
ENERGY METHODS AND STABILITY
SOLUTION (5.1) Axial strain energy
Use Eq. (5.11), circular Part:
2
U
P L
C
2
2
P L
2 AE
d
2(
2P L
2
d E 2
)E
4
Square Part: 2
U
P L
S
2
2
P (L 2)
2
2 AE
P L
2a E
2
4a E
Requirement:: 2
U
US;
C
d
2
2P L
P L
d E 2
2
4a E
8
a
SOLUTION (5.2) Refer to solution 5.1: U
2
P L 2
4a E
2
2P L 2
d E
2
P L E
(
1 4a
2
2
d
(1)
)
2
Using Eq. (5.27), U
1 2
P
(2)
Equating Eqs. (1) and (2), we obtain PEL ( 2 4 ) a
2
d
2
SOLUTION (5.3) T A B 2 .5 k N m
T B C 1 .5 k N m
Segment AB J AB
( 4 5 ) 4 0 2 .5 8 1 0 4
AB
m
4
( 2 .5 1 0 ) ( 0 .5 4 )
2
U
9
32
T L
3
2G J
2
2 ( 4 0 1 0 )( 4 0 2 .5 8 1 0 9
9
1 0 4 .8 J
)
Segment BC J BC
(3 0 ) 7 9 .5 2 1 0 4
BC
m
4
32 2
U
9
T L 2G J
(1 .5 1 0 ) ( 0 .3 6 ) 3
2
2 ( 4 0 1 0 )( 7 9 .5 2 1 0 9
9
7 5 .8 3 J
)
(CONT.)
68
5.3 (CONT.) Total strain energy U 1 0 4 .8 1 2 7 .3 2 3 2 .1 J
SOLUTION (5.4) T AB 3 k N m
TBC 5 k N m
30 mm
20 mm
A
B TB=2 kN m
45 mm
C
TC=5 kN m
0.36 m
0.54 m
Segment AB J AB
( 4 5 2 0 ) 3 8 6 .9 1 0 4
4
AB
m
4
(3 1 0 ) ( 0 .5 4 )
2
U
9
32 3
T L
2
2 ( 4 0 1 0 )(3 8 6 .9 1 0 9
2G J
9
1 5 7 .0 2 J )
Segment BC J BC
(3 0 ) 7 9 .5 2 1 0 4
BC
m
4
32
(5 1 0 ) ( 0 .3 6 ) 3
U
9
2
2 ( 4 0 1 0 )( 7 9 .5 2 1 0 9
9
1, 4 1 5 J )
Total strain energy U 1 5 7 .0 2 1 4 1 5 1 5 7 2 J
SOLUTION (5.5) See solution of Prob. 5.3: J A B 4 0 2 .5 8 1 0
J B C 7 9 .5 2 1 0
9
9
m
m
4
4
Therefore 2
2
U
C
T L
2G J TL
TC
[
0 .5 4
2 ( 2 8 ) 4 0 2 .5 8
GJ
TC
0 .5 4
[
2 8 4 0 2 .5 8
0 .3 6
] 1 0 4 .8 1 0
6
7 9 .5 2
0 .3 6
] 2 0 9 .6 1 0
7 9 .5 2
6
2
TC
TC
or C (1 0 ) 6
TC
2 0 9 .6
Then U
1 0 4 .8 1 0
6
( C 1 0
( 2 0 9 .6 )
2
2
12
)
2 3 8 5 C 2 3 8 5 ( 0 .0 6 ) 2
69
2
8 .5 8 6 J
SOLUTION (5.6) See solution of Prob. 5.3: J A B 4 0 2 .5 8 1 0 TC
U
Thus,
2
2G
L AB
(
J
3
(1 .4 1 0 )
LBC
2
2 (80 )
m
J B C 7 9 .5 2 1 0
4
9
m
4
)
J BC
AB
9
( 4 00 .52 .54 8
0 .3 6 7 9 .5 2
)
7 1 .8 9 J
Equation (5.29): U
1 2
TC C ;
1 4 0 0 C
7 1 .8 9
2
from which C 0 .1 0 2 7 r a d 5 .8 8
o
SOLUTION (5.7) P
B
A
C
a
V AB P ,
6 5
V BC
Pa L
L Pa/L
V
x P
U
s
L
2
Vx
dx
2 AG
0
3 5 AG
[
P dx 2
0
L
2
P a L
0
2
2
dx ]
2
3P a 5 AGL
(L a)
SOLUTION (5.8) Vx U
s
wx
wL 2 L 0
2
Vx
2 AG 2
2
3w 5 AG
L 4
2
dx
x
3w 5 AG 2
x L 2
6 5
L
( 0
L
3
x 3
x ) dx 2
L 2
2
3
3w L 5 AG 12
0
2
1 w L 20 AG
3
SOLUTION (5.9) I bh
P A
B
C
P
U
a
L
(a) U U
b
t
1 2 EI
1 2E
[
a
M 0
2 AB
dx
2
dx
dA
L
M 0
1 2E
2 CB
(
P A
M 2
dx' ]
My I
12
2
P (La ) 2 AE
2
P (La ) 2 Ebh
We have
Pa/L
x’
x
a
3
2
P a 6 EI
2
) dA
AB
(a L )
Px 2
2P a Ebh
2
3
M
CB
Pa L
x'
(a L )
(1) (CONT.)
70
5.9 (CONT.) But
Thus
U
2
P (La )
t
bh
U
3
12
0 ,m ax
U b.
a
2
a
6 Pa bh
U
t
[1 4 ( h ) ]
2 Ebh
Pa ( h 2 )
( b ) b ,m ax Hence
y d A 0 , and Eq. (1) becomes U
2
a ,m ax
,
2 m ax
2E
2
P
2
2 Eb h
2
(1
P bh
6a h
)
m ax
,
P bh
(1
6a h
)
2
SOLUTION (5.10) M
( Lx x ) 2
w 2
A x
wL 2
U
0 ,m ax
U
b
w B
1 2 EI
2 m ax
2E
2
9w L
2
m ax
I bh
1 8
3
m ax
wL
2
c h 2
12 , M
m ax c
I
3wL 4 bh
2 2
4
2
32 Eb h
M dx
h
wL 2
L
M
(1)
4
1 2 EI
L w 2
( 0
) ( Lx x ) dx 2
2
2
2
w L
5
20 Ebh
(2)
3
It is required to obtain C : U 0 , m a x C U V , or C U
V 0 ,m ax U
(3)
Substituting Eqs.(1) and (2) into (3): C 4 5 8 . Thus
U
0 ,m ax
45 U 8 V
SOLUTION (5.11) From Solutions of Probs. 5.9 and 5.7: U
b
2
2
(a L )
P a 6 EI
U
Thus or
2
p a
2
6 EI
U
2
s
(a L )
3P a 5 AGL 2
(a L )
(a L )
3P a 5 AGL
v A 2 P a ( a L )[ 6 E I a
3 5 AGL
1 2
Pv A
]
SOLUTION (5.12) x
w
x A 2
U
M dx 2EI
1 2EI
L
L 0
(
1 2
B 2
w x ) dx 2
2
1 w L 40
71
EI
5
M
x
1 2
wx
2
SOLUTION (5.13) P
P A
a
2a
C
B
D a
MAC=Px MCD=Pa
P
P Pa
M
x Segment AC U
AB
a
2
M dx
0
P
2EI
2
2EI
2
a
x dx 2
P a
0
3
6EI
Segment CD U
CD
2
P a
3a 0
2
2
dx
2EI
By symmetry: U
3
P a EI
U
AC
BD
.
Total strain energy : Pa
U 2
3
2
P a
6EI
3
2
4 P a
EI
3
3
EI
SOLUTION (5.14) x
w
b h
A
L
6
We have
Vx wx
5 U
s
L 0
V
2
6
dx
2 AG
3 w
2
x
5 AG
B
1
5 2 AG 3
L
2
3
1 w L
L
( w x) dx 2
0
3
5 AG
0
SOLUTION (5.15) We have V AC V BD P
6 5
VCD 0
Thus U
s
V
2
2 AG
a
dx 2 0
2
P dx
2(
2 AG
2
6 P a 5 AG
72
6 5
2
)
P a 2 AG
SOLUTION (5.16) See solution of Probs. 5.13 and 5.15: 2
U
b
3
3
4 P a 3
s
5 AG
2
6 P a
EI
vC 2 P a (
or
U
6 P a
(due to symmetry) 2
U
2
EI
vC v D
Thus
3
4 P a
1
5 AG 2a
2
3
3EI
1
P (vCb vCs )
2
P vC
)
5G A
SOLUTION (5.17) ( a ) Axial strain energy in bolt. 2
U
b
2
P L
T L
2 AE
2 AE 2
T ( 0 .0 3 )
2[
2 .6 5 2 6 (1 0
9
)T
2
(6 ) ( 2 0 0 1 0 )] 2
3
4
2 .6 5 2 6 (1 0 Ub
Thus
1 2
T ;
9
)( 6 3 0 ) 1 .0 5 2 8 J 2
1 .0 5 2 8 N m
1 2
( 6 3 0 )
3 .3 4 2 m m
( b ) Bending strain energy in link. I bh U
b
L
3
12 216 m m
2
0
2EI
2EI
{[
0 .0 2 5
(420 x) dx 2
0
0 .0 5
2
( 2 1 0 x ') d x '} 0
A
EI
B
Substituting the data, U
b
50 mm
25 mm
1 .3 8
4
M dx
1
1 2 (1 2 )( 6 )
3
3
6
630
420
( 2 0 0 1 0 )( 2 1 6 )
3 1 9 (1 0
x’
x
1 .3 8
210
) J
SOLUTION (5.18) a x Q
P B
A
C
M
AC
Qx
M
CB
Qx P(x a )
L (CONT.)
73
5.18 (CONT.) Thus
M
vA
1 EI
1 EI
[ ( Q x )( x )d x
M
i
dx
Q
i
a
L
[Q x
0
P ( x a ) ]( x ) d x
a
Set Q 0 , and integrate: vA
1 EI
L
P ( x a )xdx
( 2 L 3a L a ) 3
P 6 EI
a
2
3
SOLUTION (5.19) M
BC
Px
M
B
Thus
1 EI
[
P L P R s in
CA
L
M 0
M
BC
P
BC
dx
2
M
M
Rd ]
CA
P
CA
0
( 4 L 6 R L 2 4 R L 3 R ) 3
P 12 EI
2
2
3
SOLUTION (5.20)
A a
P a
a
Pa/2
C
P/2
M
B
D P/2
Pa/2
M
+
AD
P 2
vD
1 EI
vD
P 2 EI
( x a ), M
[
M i
2a
i
P (xa ) 2
0
M
BD
P 2
dx 2
dx
a 0
2
x 2
dx
x
Pa/2 Integrating, 3
vD
Pa 4 EI
SOLUTION (5.21) B
C
M
A RA
L/2
AB
Consider R A as redundant.
L RB
x
M
0
RAx
M
CB
RC
x’ (
M
0
L
RA
2
M
0
L
RA 2
) x ' M 0 ,
vA 0
1 2
Thus L
vA
2
0
( R A x )xdx
L
( 0
M
0
L
RA 2
) x ' M 0 ]
x' 2
After integrating RA
Then
RC
2 3
M
4 3
M
0
0
L
L
For the entire beam,
F y 0:
RB 2
74
M L
0
dx 0
M
M
i
i R A
dx
x
SOLUTION (5.22) U
a
D
2
P (2L) 2 AE
U P
U
a
2 PL AEa
PL 3EsI
2 PL AEa
or
Solving P
3
3
wL 8
1 EsI
2
M 2Es
0
L
(Px
0 4
0
a
)
wL 8 EsI
3 AE
(
Ls
s
2
AE a L 6 E s I
M Px
dx wx 2
2
wx 2
2
U U
a
U
s
)xdx 0
SOLUTION (5.23) w
M
AB
1 2
M
BC
3x 5
1 EI
M
Q x
x A
B
A
4
3
C
1 EI
L AB 2 a
2
wx
( 2 wa ) 2 wa
{
M Q
i
i
4x 5
Q
2
L BC 5 a
dx
2a
2
1
( 2 w x )( 0 )d x
0
2
5a
0
( 65x wa 2 wa
4x 5
Q )(
4x 5
) dx }
Set Q 0 and integrate:
A
60
wa EI
4
SOLUTION (5.24)
C
M
P
x
A
AB
Px
M
BC
Pa
a
B
x L
A
1 EI
M
M i
dx
i
P
a
1 EI
{ ( P x ) xd x 0
L
( P a )( a )d x} 0
Integrating, A
2
(a 3L )
Pa 3EI
SOLUTION (5.25) Introduce a fictitious horizontal force Q at end B. Vertical reactions at supports are P/2. We have M Thus
BC
R (1 c o s ) 2
B
2 EI
2 EI
M 0
2
M BC
BC
Q
P 2
( R sin ) Q
P 2
M
BC
( R s in ) Q ] [ R s in ] R d
Setting Q=0 and integrating: B
CA
Rd
[ R (1 c o s )
0
M
3
PR 2 EI
75
SOLUTION (5.26) (a)
Let fictitious couple C=0:
x C
B a
2a
a
x
A
P
M
AB
Px
M
BC
Pa
M
DC
Px Pa
C x
D
P Pa+C A
Thus
M
M
i
dx
P
i
a
P x dx Pa 2
2
0
4 3
2 Pa EI
3
P
0
2a
(x
2ax a )dx
2
2
0
Pa ( 3 4 2)
3
Pa
a
dx
3
8
( b ) We now have M
AB
A
Px C
[ M
1 EI
M
M
i
M
DC
Px Pa C
dx ]
C
i
Pa C
BC
Hence, after setting C=0: A
a
P EI
a
[ xdx a dx 0
0
2a
(x
a )dx ]
0
3 Pa 2 EI
2
SOLUTION (5.27) M
(a)
A
AB
Pz,
M
a
1 EI
[ M
P EI
(
0
3
a 3
M
P
AB 3
L 3
AB
)
Px,
BC
dz
L
0
1 GJ
(
T0 L 2
T B C T 0 (1
M
M
BC
P
BC
dx ]
1 GJ
x L
L
0
) Pa
[ T 0 (1
x L
) Pa ]( a ) dx
Pa L) 2
( b ) Introduce a fictitious couple C about x axis at point B. T B C C T 0 (1
x L
) Pa
Hence B
1 EI
[
a
M 0
0 0
M
1 GJ
AB
C
AB L 0
dz
L
M 0
[ C T 0 (1
x L
M BC
1 GJ
(
T0 L 2
dx ]
) P a ] (1 ) d x
Setting C=0 and integrating: B
BC
C
PaL )
76
1 GJ
L 0
TBC
TBC C
dx
SOLUTION (5.28) Due to Symmetry: F A B F C D , Statics: R A x 6 0 k N
F AC FBD
,
R A y 1 2 .5
Method of joints: Joint A
kN
,
R D y 1 2 .5
Joint B
F AC 30 kN ( T )
F BC 12 . 5 kN ( C )
F AB 32 . 5 kN ( T )
Total strain energy: U
2
Fi L i
6
2 AE
[ 3 2 .5 (1.3 ) 2 3 0 (1.2 ) 2 1 2 .5 ( 0 .5 ) ] 2
10 2 AE
2
2
This gives, substituting the given data, U 4 7 .4 7 N m
Hence U W;
4 7 .4 7
mm
1 2
( 6 0 1 0 ) D 3
or D
1 . 582
SOLUTION (5.29)
2P
B
Introduce a fictitious force Q at C.
P
Statics:
0.9 m A
R Ax P Q
C
R Ax
R A y 0 .2 4 P
Q R Ay
1.2 m
0.675 m
R C 1.7 6 P
RC
Apply method of joints at C and B: F B C 2 .2 P
(C )
F A B 0 .4 P
(C )
F A C 1.3 2 P Q
Thus C
1 AE
Fi
Fi Q
Li
Differentiating and setting Q=0: C
1 . 32 P ( 1 ) AE
(1 . 875 )
2 . 475 P AE
77
(T )
kN
SOLUTION (5.30) A
P
B
Introduce Q at C. By method of joints:
L
F AB 0,
L
F AC ( P Q )
P
C
D
FBC P
FC D Q
2,
Q
We write ( C ) v
1 AE
Fi
Fi Q
Li
[ 0 ( P )( 0 ) ( P Q )
1 AE
2(
2)
2 ( Q )( 1 )] L
This yields, for Q 0: ( C ) v
2
( C ) h
1 AE
( C ) h
1 AE
[0 ( P ) ( 1 ) ( P )
2PL AE
2 .8 2 8
PL AE
Similarly
or
1 2 2 AE
Fi
Fi P
Li
P L 3.8 2 8
PL AE
2(
2)
2 ]L
SOLUTION (5.31) Introduce Q at point C. F.b.d. - Entire truss ( from
M 0,
F 0 ):
R A y 1.2 5 P 0 .9 3 7 5 Q
R Ax Q
R B y 2 .2 5 P 0 .9 3 7 5 Q
Joint A:
F AC
13 Q
12
F A C 3 .2 5 P 2 .4 3 7 5 Q
5
F A B 3 P 1. 2 5 Q
F AB
A
(T )
(C )
R Ay F B C 3 .7 5 P 1.5 6 2 5 Q
Joint B:
(C )
Thus, we have ( C ) h
1 AE
Fi L i
( C ) v
1 AE
Fi Li
Fi Q
P AE
[1 2 6 1.7 9 1 2 9 .2 9 6 9 ] 1 0 3.0 8 8
P AE
[ 3 ( 3 . 2 )( 3 ) 3 . 25 ( 7 . 8 )( 3 . 25 )
P AE
and Fi P
( 3 . 75 )( 3 . 75 )( 5 )] 181 . 5
78
P AE
SOLUTION (5.32) F BD 0 .
C
32
Q D
B 32
Introduce Q at point D. Reactions, as found by statics, are shown in the figure. We shall apply the method of joints, as needed. 32 kN C Joint C
15
FBC
42-Q/2
Joint E
17
E
A
8
42+Q/2
E
4
FDE 7 0
5 6
F AE 5 6
Q
Thus
D
(C )
F AD
Q
A
3
56+2Q/3
42-Q/2
(C ) F AD 3 0
(T )
1 AE
10 AE
2 8 6 .1 3 3 1 0 AE
3
5 4
32
42+Q/2
2 3
(32 ) 68 kN
F BC 60 kN ( T )
60
5 3
17 8
Joint A
FDE
F AE
FC D
FC D
F jL j
Q
(C )
F j Q
[0 0 0 0 2 ( 7 0 ) ( 3
5 6
) 3 .2 ( 5 6 ) ( 3 ) 2 ( 3 0 ) (
5 6
2
5 6
)]
SOLUTION (5.33) Joint B FBD
FBC 3
FBA
4
FBD
5 4
FBA
(1)
B
FBC P
3 4
FBA
P A 0
1 AE
1 AE
[( P
FjL j 3 4
F j FBA
F B A )(
3L 4
)( 4 ) ( 3
0 .5 6 2 5 P L 3 .3 7 5 F B A L 0 ,
5 4
F B A )(
5 24
P
FBC
7 8
)(
5 4
) F B A ( L )(1 )]
F B A 0 .1 6 6 P P 6
Then Eqs.(1) give FBD
5L 4
P
79
SOLUTION (5.34) M F R s in P R (1 c o s ) ,
v 0
1 EI
M
M F
dx
Therefore v
1 EI
[ FR sin PR ( 1 cos )]( R sin ) Rd
0
3
FR
2 EI
2 PR EI
3
0,
F
4P
4P
SOLUTION (5.35)
P
C
M
Px 2
c o s Q x s in
L x
A
Q B
P/2
P/2 Line of symmetry
( a ) With Q 0: B
2 EI 3
L
c o s ) ( x s in ) d x
Px 2
( 0
s in c o s
PL 3EI
3
PL 6 EI
s in 2
( b ) With Q R B h : B 0
2 EI
L
c o s R B h x s in ) ( x s in ) d x
Px 2
( 0
or R Bh
P cot
1 2
SOLUTION (5.36) Statics: F B C
2P
FC D
,
3
P
F AB
,
3
P 3
,
M
AB
Px
Thus U
B A
2
FAB 2 AE
1 2 AE
dx
2L 0
( 6 AE 11L
2
P 3
M
)P
2 AB
2 EI
A
dx
3
4L 3EI
B
1 2 RI
dx
2L
C D
2
FBC 2 AE
P x dx 2
2
0
2
We have C
U P
PL 3E
( 11A
8L I
2
dx
)
80
1 2 AE
D
2
FC D
dx
2 AE
C
2L 0
4 3
P dx 2
1 2 AE
L 0
2
P 3
dx
SOLUTION (5.37) Let the load at B be designated by Q. Locate origin of coordinates at A: V AB P
M
Px
AB
V AB
1,
P
M
AB
P
x
Locate origin of coordinates at B: V BC P Q
M
P(x
BC
L 2
) Qx
V BC
1,
P
M
BC
P
x
L 2
( a ) Equation(5.38), with Q=P, gives vA
M
1 EI
M
i
P
i
dx
L
L 2
0 3
7 PL 16 EI
( x )dx
Px EI
V
1 AG
P(2xL 2)
2
EI
0
Vi
dx
P
L
(x
L 2
)dx
L 2
0
1.2 P AG
(1 ) d x
2
0
1.2 ( 2 P ) AG
1 .8 P L AG
(1 ) d x
(P5.37)
( b ) From Table B.1: E=200 GPa and G=79 GPa. Given L/h=5. We have A=bh and I b h 1 2 . Eq.(P5.25) is rewritten as 3
Ebv A P
3
7 (1 2 ) L
7 (1 2 )( 5 )
16 h
3
3
1 .8 E L Gh
16
1.8 ( 2 0 0 ) ( 5 ) 79
656 . 25 22 . 78 679 . 03
Error:
22 . 78 679 . 03
(100 ) 3 . 35 %
SOLUTION (5.38) Inasmuch as horizontal displacement at C, h is zero, Eq.(5.41) gives h
U H
0
1 EI
[ HR sin FR ( 1 cos )] R (sin ) Rd
0 2R
1 EI
[( H
P ) x 2 F R ]x d x
0
H ( 2
8 3
) 2F
8P 3
or 4 . 2375 H 2 F 2 . 6667 P 0
(1)
Similarly, vertical displacement at C is zero: v
U F
0
1 EI
[ HR sin FR ( 1 cos )][ R ( 1 cos ) Rd
0 2R
1 EI
[( H
P ) x 2 F R ]2 R d x
0
2H F(
3 2
8) 4P
or 2 H 1 2 .7 1 2 4 F 4 P 0
(2)
Solving Eqs.(1) and (2): H 0 . 5193 P
F 0 . 2329 P
81
SOLUTION (5.39) The structure is statically indeterminate to the first degree. Select R, the reaction at B, as redundant. From the equilibrium of forces at joint D, R with F A D F D C F : F F 4 3
F
D
5 8
(W R )
(1)
W Substituting Eq.(1), into Eq.(5.45), together with Eq. (5.38) we have B 0
L
F R
[2 F
1 AE
0
1 AE
h
(W R ) L
25 32 AE
R R
R 0
dx ]
(W R )( 85 ) L 0 . 8 RL ]
5 8
[( 2 )
dx
0 .8 R L AE
Solving, R 0 .4 9 4 W Thus F A D F C D 0 .3 1 6 W
F B D 0 .4 9 4 W
and
SOLUTION (5.40) We write M M Thus
U R A
vA
U M A
A
6.
M R A
M
dx
1 EI
L
( M
0
RAx
A
1 6
kx )xdx 0
(1)
k x )( 1) d x 0
(2)
3
0
1 EI
3
0
1 EI
R A x kx
A
M M A
M
dx
1 EI
L
( M
0
A
RAx
1 6
3
Integrating and simplifying Eqs.(1) and (2) we obtain, respectively:
1 2
M
A
1 3
RAL
M
A
1 2
RAL
1 30
kL 1 24
3
kL
3
Solving RA
3 20
kL
2
M
A
1 30
kL
3
SOLUTION (5.41) U W
Thus
EI 2
L
(d
2
dx
0
v 2
) dx 2
L 0
EI 2
L
0
L
w v d x w 0 a s in 0
U W :
E I 2L
3
4
a
( L ) a sin 4
2 x L
w0L 2
dx
w0 EI
L
4
( ) s in
w0L 2
a
,
We have v
2 x L
2
x L
82
dx
a
w0 EI
L
( )
4
4
EI 4L
3
a
2
SOLUTION (5.42) L A D L C D 3 .4 6 m U
AL i E
1 2
2
1 2
AL i E ( v cos / L i )
2
Vertical load of the joint, by Eq.(5.46): 3
U v
W
E jAj Lj
v cos 2
1
or
W E A v [
2
cos 30 L AD
o
2
o
E A v [ c o3s .4360
2
cos 30 LC D 2
o
cos 30 3 .4 6
o
1 LBD
]
13 ] 0 .7 6 6 9 E A v
Substitute the given data W 0 .7 6 6 9 ( 0 .0 0 6 )( 2 0 0 1 0 )( 6 2 5 1 0 9
6
) 5 7 5 .2 k N m
SOLUTION (5.43) v
(a)
2
ax
2L
U
3
EI 2
(3L x )
L
v" 3
(v" ) dx 2
0
9 EI 2L
6
a
2
a L
3
(L x)
L
( L x ) dx 2
0
3EI 2L
3
a
2
(1)
We have W P v A
(2)
Virtual work principle, U W , is thus P a
and
3EI 2L
3
(2a a )
a PL 3EI 3
v
2
Px 6 EI
(3L x )
(3)
( b ) At x=L, Eq.(3) gives v m ax
3
PL 3EI
Using Eq.(3): and at x=L: m ax
(4) dv dx
Px 2 EI
(2 L x )
2
PL 2 EI
(5)
SOLUTION (5.44) We have v ax ( L x ) axL ax ,
v' aL 2ax,
2
So, Also
U
EI 2
L
( v " ) d x 2 a E IL 2
2
0
W P v A P (acL ac ) 2
From u W , it follows that E IL ( 4 a a ) ( P c L P c ) a , 2
a
Pc ( L c ) 4 E IL
Hence, at x=c: vA
2
Pc ( Lc )
2
4 E IL
83
v" 2a
SOLUTION (5.45) I
d
4
64
2
d
A
,
r
,
4
I A d 4 , Le L
We have
cr
2
E
P A
4P
;
2
( Le r )
d
2
2
2
Ed
16 L
2
d
,
4
64 PL
2
3
E
Substituting given data: d [
Check:
L r
3
64 ( 50 10 )( 1 . 2 ) 3
( 210 10
4 ( 1. 2 )
4L d
9
0 .0 2 9
1
2
] 4 29 mm
)
1 6 5 .5 . Also
P A
3
4 ( 50 10 )
( 0 . 029 )
2
75 . 7 MPa
600
MPa.
OK.
SOLUTION (5.46) Substituting the given data: ( L e r ) c
E S y 45 . 7
Try Johnson’s formula Pc r
S
A
S
y
2 y 2
4 E
(
Le r
)
2
or 3
2 2 0 (1 0 )
d
Check
Le
2
d 4
5 2 0 (1 0 )
6
( 5 2 0 1 0 )
6
4
4
2
2
2
( 0 .2 5 ) 1 6
9
(1 1 0 1 0 )
d 2 5 .7 m m
;
2
d
38 . 9 OK.
250 ( 4 ) 25 . 7
Try Euler formula: d [
and
4 Le
d
2
6 4 P Le 3
E
1
]4 [
3
6 4 ( 2 2 0 1 0 )( 0 .2 5 ) 3
2
1
] 4 2 2 .5
9
(1 1 0 1 0 )
44 . 4 45 . 7
4 ( 250 ) 22 . 5
mm
does not apply.
SOLUTION (5.47) ( a ) By Eq.(5.61) with A d d [
64
2 Pc r L e 2
E
]
4 and r d 4 :
2
1 4
( b ) Equation (5.61) with A b h , b
I bh
3
12 ,
r
2
h
2
1 2:
2
1 2 Pc r L e 2
Eh
3
SOLUTION (5.48) ( a ) Same area:
(D
4
2
d ) ao ai 2
ai ao 2
2
4
2
(D
2
2
d ) 50 2
2
(5 0 3 5 ) 2
2
4
(CONT.)
84
5.48 (CONT.) a i 3 8 .7 m m ,
or
1
t
2
( a o a i ) 1 1 .3 m m
( b ) Circular bar I
4
(D
d ) 4
64
( 5 0 3 5 ) 2 3 3 .1 1 0 4
4
9
m
4
64
EI
( 7 2 ) ( 2 3 3 .1)
2
Pc r
2
2
Le
3 4 .2 k N
2
( 2 .2 )
Square bar I
1
4
12
1
(ao ai ) 4
( 5 0 3 8 .7 ) 3 3 3 .9 1 0 4
9
m
4
12
EI
( 7 2 ) ( 3 3 3 .9 )
2
Pc r
4
2
2
Le
2
( 2 .2 )
49 kN
SOLUTION (5.49) I
d
(8 )
4
64 A
d
4
2 0 1 .0 6 m m
4
64 2
(8 )
4
2
5 0 .2 7 m m
2
4
EI 2
P A Pc r
L
2
( 2 0 0 1 0 )( 2 0 1 .0 6 1 0 2
9
( 0 .4 )
2
12
)
2 .4 8 k N
The corresponding stress is
cr
Pc r
A
2480 5 0 .2 7 1 0
6
4 9 .3 M P a 2 5 0 M P a OK .
We have
M
0:
C
( a b ) Q b Pc r
1 8 0 Q 3 0 ( 2 4 8 0 ), or Therefore
Q a ll
Q
n
4 1 3 .3
Q 413 N
295 N
1 .4
SOLUTION (5.50) Pc r n P 2 .6 ( 2 2 ) 5 7 .2 k N , I
d
A
4
d
64
2
r
4
L e 0 .7 L 0 .7 (1) 0 .7 m d 4
Equation (5.61) gives 2
d
4
6 4 Pc r L e
E 3
6 4 (5 7 .2 1 0 )( 0 .7 ) 3
(200 10 ) 3
9
2
,
d 0 .0 2 3 m = 2 3 m m
(CONT.)
85
5.50 (CONT.) Hence Le
0 .7
r
1 2 1 .7
0 .0 2 3 4
Equation (5.62): Le
(
r
E
)c
Sy
200 10
9
250 10
6
8 8 .8 6 1 2 1 .7
Euler formula is valid, Therefore d 23 m m
SOLUTION (5.51) Refer to Solution of Prob. 5.50. Now we have L e 0 .7 ( 6 2 5 ) 4 3 7 .5 m m and Pc r 2 .6 (1 2 5 ) 3 2 5 k N . Equation (5.61): 6 4 (3 2 5 1 0 )( 0 .4 3 7 5 )
2
d
4
6 4 Pc r L e
E 3
3
2
3
d 0 .0 2 8 m
,
(200 10 ) 9
and Le
r
4 3 7 .5
6 2 .5 1 2 1 .7
28 4
Euler formula does not apply. Apply Johnson formula, Eq. (5.66): d 2(
or
2
Pc
Sy
S y Le
E 2
1 2
)
325 10
2[
3
250 10
2 5 0 1 0 ( 0 .4 3 7 5 ) 6
6
(200 10 ) 2
9
2
1
]
2
d 0 .0 4 1 9 m = 4 1 .9 m m
SOLUTION (5.52)
L B C 0 .6 5 m
F AB
5 12
FBC
P,
13 12
P
P B
F AB
12 13 5 FBC
Bar AB ( F A B ) cr
2
EI 2
Le
2
9
( 2 1 0 1 0 )[
( 5 1 0
3
( 0 .4 )
4
) ]
4 2
6 .3 5 9 k N
5 12
Pc r ,
Pc r 1 5 .2 6 k N
Bar BC ( F B C ) cr
2
9
( 2 1 0 1 0 )[
( 7 .5 1 0
4 ( 0 .6 5 )
2
3
4
) ]
1 2 .1 9 k N
Choose the small value, Pc r 1 1.2 5 with n 2 .5 . Thus Pall
Pcr n
11 . 25 2 .5
4 . 5 kN
86
13 12
Pc r ,
Pc r 1 1 .2 5 k N
SOLUTION (5.53) ( a ) Applying the method of joints at A: F AB 40 kN ( C ) and F B C 2 2 0 k N ( C ) L A B 2 .5 m .
cr
FAB
We have r
cr
3
4 0 (1 0 )
2 5 0 (1 0 ) 6
;
A
d
2
d 1 4 .3 m m
,
4
I A d 4 and Euler’s formula: 2
E
(L r)
2
; 250 (10
)
6
2
( 210 10
9
( 2 . 5 0 . 25 d )
)
, d 109 . 8 mm
2
Use, a commercial size of : d 110 mm diameter ( b ) L B C 1 .8 7 5 m
FBC
cr
3
2 2 0 (1 0 )
2 5 0 (1 0 ) 6
;
A
d
2
d 3 3 .5 m m
,
4
Euler formula:
cr
2
E
(L r)
2
6
; 250 (10
)
2
( 210 10
9
)
( 1 . 875 0 . 25 d )
2
, d 82 . 4 mm
Use 8 3 m m diameter
SOLUTION (5.54) I r
64
( 0 . 04 ) I A
d 4
9
125 . 66 10
4
10 mm
L r
m 1600 10
4
160
Euler’s formula: 2
P cr
EI
F a ll
1 .0 0 .5
nL
2
2
( 200 125 . 66 ) 1 .5 (1 .6 )
2
64 . 6 kN
Thus Pc r 2 ( 6 4 .6 ) 1 2 9 .2 k N
SOLUTION (5.55) Two angles: I x 2 I x 2 ( 4 2 6 1 0 ) 8 5 2 (1 0 ) m m 3
3
4
I y 2 ( I y A x ) 2[1 5 3 (1 0 ) 7 4 4 (1 0 .5 ) ] 4 7 0 , 0 5 2 m m 2
I m in 4 7 0 , 0 5 2
Use smaller I : 2
Pc r
Pa ll
3
EI y 2
Le
Pc r n
2
1 2 2 .7 2 .5
mm
9
4
( 2 0 0 1 0 )( 4 7 0 , 0 5 2 1 0 ( 2 .7 5 )
2
We have L e 2 .7 5 m Thus
12
)
2
4 9 .1 k N
87
1 2 2 .7 k N
4
SOLUTION (5.56) A d I
2
4 ( 60 )
(60 )
4
d 64
P cr
2
L
4
64
EI 2
2
2
4 2 ,827
6 3 6 ,1 7 3 m m ( 210 10
9
1
)( 636 10
12
)
mm
4
r
1319
2
2
I A
d 4
60 4
15 mm
kN
Using Eq.(5.72): Py
350 ( 10 ) 6
3
2 . 827 10
0 . 002 ( 2 . 827 10
[1
636 ,173 ( 10
12
3
)
1
) 0 . 03 1
]
Py 3
1319 ( 10 )
or P y 2 6 6 0 1 0 4 .4 P y 1.3 0 5 1 0 2
Solving, this quadratic gives: P y 649
12
0
kN . Thus, Pa ll 6 4 9 3 2 1 6 .3 k N
SOLUTION (5.57) I m in
1
(1 6 0 8 0 1 3 0 5 0 ) 5 .4 7 2 5 1 0 3
3
A 1 6 0 8 0 1 3 0 5 0 6 .3 1 0 rm in
6
mm
4
12
A 2 9 .5 m m
I m in
3
mm
3
L e 0 .5 L 2 .7 5 m
L e r 2 7 5 0 2 9 .5 9 3 .2
Hence, E
(7 2 1 0 )
2
cr
( Le r )
2
2
9
( 9 3 .2 )
2
8 1 .8 M P a
SOLUTION (5.58) L e 0 .7 L 3 .8 5 m . From solution of Prob.5.57: rm in 2 9 .5 m m .
We now have L e r 3 8 5 0 2 9 .5 1 3 0 .5
Hence, E 2
cr
(7 2 1 0 ) 2
( Le r )
9
(1 3 0 .5 )
2
4 1 .7 M P a
SOLUTION (5.59) A (5 0 4 4 ) 1, 7 7 2 2
2
mm
2
Equation (5.68): r
1 4
D
2
d
2
1 4
100
2
88
2
33 . 3 mm ,
L r
2000 33 . 3
60 . 1
(CONT.)
88
5.59 (CONT.)
r
12(50 )
ec
(a)
2
( 3 3 .3 )
Py
Refer to Fig.5.23b:
r
2
( 3 3 .3 )
Refer to Fig.5.23b:
160 M Pa,
A
9( 50 )
ec
(b)
0 .5 4 1
2
P y 1 6 0 (1, 7 7 2 ) 2 8 3 .5
kN
0 .4 0 6
2
Py
175 M Pa,
A
P y 1 7 5 (1, 7 7 2 ) 3 1 0 .1 k N
SOLUTION (5.60) We have
I
(D
d ) and A
4
4
64
(D
2
d ); 2
r
1
I A
4
D
2
d
2
4
A
P n ( 4 4 0 ) 1 .5 ( 4 4 0 ) 6 6 0 k N ,
(200 175 ) 7363 m m 2
2
2
4 I
( 2 0 0 1 7 5 ) 3 2 .5 (1 0 ) m m 4
4
6
4
64 r
1
200 175 2
2
6 6 .4 4 m m
4
Hence,
P
A ec r
2
660 10 7363 10
3 6
r
0 .1 e ( 6 6 .4 4 1 0
L
8 9 .6 M P a
3
)
2
3 .6 6 6 6 .4 4 1 0
5 5 .1
3
2 2 .6 5 e
Substitute these into Eq. (5.74a): 2 1 0 8 9 .6 1 2 2 .6 5 e s e c 2 7 .5 5
8 9 .6 1 0 190 10
6
9
from which e 0 .0 4 9 0 2 m = 4 9 .0 2 m m
SOLUTION (5.61) From solution of Prob. 5.60:
P
A 7363 m m
2
I 3 2 .5 (1 0 ) m m 6
4
Hence EI 2
Pc r
v m ax
L
2
( 2 0 0 1 0 )(3 2 .5 1 0 2
9
( 4 .6 )
e
6
)
2
3 .0 2 3 M N
Figure S5.61
(CONT.)
P
89
5.61 (CONT.) ( a ) Using Eq. (5.73): 1 .2 5 e s e c 2
45 10
3
3 .0 3 2 1 0
6
o 1 e s e c (1 0 .9 6 ) 1 ,
e 6 7 .2 8 m m
( b ) Referring to Fig S5.61: M P ( v m a x e ) 4 5 (1 .2 5 6 7 .2 8 ) 3 0 8 0 N m
Hence,
P
m ax
Mc
A
I
45 10
3
7363 10
6
3 0 8 0 ( 0 .1) 3 2 .5 1 0
6
1 5 .5 9 M P a
SOLUTION (5.62) L e 2 ( 4 .6 ) 9 .2 m . Refer to solution of Prob. 5.61
EI 2
Pc r
( 2 0 0 1 0 )(3 2 .5 1 0 2
2
9
Le
(9 .2 )
2
6
)
7 5 7 .9 k N
( a ) Equation (5.73): 1 .2 5 e s e c 2
o 1 e [s e c ( 2 1 .9 3 ) 1] , 7 5 7 .9 45
e 1 6 .0 2 m m
( b ) M P ( v m a x e ) 4 5 (1 .2 5 1 6 .0 2 ) 7 7 7 .2 N m Therefore,
m ax
P A
Mc I
45 10
3
7363 10
6
7 7 7 .2 ( 0 .1) 3 2 .5 1 0
6
6 .1 1 2 2 .3 9 1 8 .5 M P a
SOLUTION (5.63) A
L e 2 L 3 .6 m
75 mm C 15 mm
A 150 75 120 45
5 .8 5 1 0
P
150 mm D
B
1
I
3
mm
2
(1 5 0 7 5 1 2 0 4 5 ) 3
3
12
P
4 .3 6 2 1 0 r
L
6
mm
4
I A 2 7 .3 m m
e c 3 7 .5 m m
Thus, ec r
2
3 7 .5 3 7 .5 ( 2 7 .3 )
2
1 .8 8 7
Le
6 5 .9 3
2r
(CONT.)
90
5.63 (CONT.) Use Eq.(5.74b) with L L e :
m ax
160 10 5 .8 5 1 0
1 1 .8 8 7 s e c 6 5 .9 3
3 3
9 9 .3 M P a 9 3 2 0 0 1 0 (5 .8 5 1 0 ) 160 10
3
SOLUTION (5.64) 150 mm
C
L e 2 L 3 .6 m
D
P
A 150 75 120 45
75 mm B
A P
5 .8 5 1 0
3
mm
2
15 mm 1
I
(7 5 1 5 0 4 5 1 2 0 ) 3
3
12
1 4 .6 1 1 0
6
mm
4
e c 75 m m
r
I A 4 9 .9 7 m m
Therefore, ec r
2
75 75
( 4 9 .9 7 )
2 .2 5
2
Le
3 6 .0 2
2r
Apply Eq.(5.74b) with L L e :
m ax
160 10 5 .8 5 1 0
3 3
1 2 .2 5 s e c 3 6 .0 2
9 4 .8 M P a 9 3 2 0 0 1 0 (5 .8 5 1 0 ) 160 10
3
SOLUTION (5.65) We have A ao ai 120 100 2
1
I
2
(ao ai ) 4
12
ao
3
2
(1 2 0 1 0 0 ) 8 .9 5 (1 0 ) m m 4
4
6
4
12
8 .9 5 (1 0 )
A c
1
4
4 .4 (1 0 ) m m
2
3
I
r
2
4 5 .1 m m
4 .4
60 m m
2
Le 2 L 2 (2 ) 4 m
Hence ec r
2
(5 5 )(6 0 ) ( 4 5 .1)
EI
2
2
Pc r
2
Le
3
A
( 7 0 1 0 )8 .9 5 (1 0 2
P
1 .6 2 2 , 9
(4)
2
2 0 0 (1 0 ) 4 .4 (1 0 6
)
6
4 5 .5 M P a
)
3 8 6 .5 k N
(CONT.)
91
5.65 (CONT.) P
200
Pc r
0 .5 1 7
3 8 6 .5
( a) Apply Eq. (5.73): v m ax e[ s e c (
P
2
Pc r
) 1]
(5 5 )[se c (
0 .5 1 7 ) 1] 7 3 .7 6 m m
2
( b ) Use Eq.(5.74a)
m ax
P
[1
A
ec r
sec(
2
P
2
Pc r
)]
( 4 5 .5 ) [1 (1 .6 2 2 ) s e c (
0 .5 1 7 ) ] 2 1 8 .3 M P a
2
SOLUTION (5.66) From solution of Prob. 5.65: A a o b i 4 .4 (1 0 ) m m 2
1
I
2
3
( a o a i ) 8 .9 5 (1 0 ) m m 4
12
2
4
6
L e 2 ( L ) 2 (1 .9 ) 3 .8 m
EI 2
Pc r
P
2
Pc r
9
Le
( 3 .8 )
300
c 60 m m
( 2 0 0 1 0 ) (8 .9 5 1 0 2
4
6
)
2
1, 2 2 3 k N
0 .2 4 5
1223
( a ) Equation (5.73): v m ax e[ s e c (
P
2
Pc r
) 1]
or 1 5 e[ s e c (
0 .2 4 5 ) 1] ,
e 3 7 .2 m m
2
(b) M
m ax
m ax
P ( e v m a x ) 3 0 0 (3 7 .2 1 5 ) 1 5 .6 6 k N m
P A
M
m ax
I
c
300 10 4 .4 (1 0
3
3
)
1 5 .6 6 1 0 ( 0 .0 6 ) 3
4 .9 5 (1 0
1 7 3 .2 M P a
92
6
)
SOLUTION (5.67) Figure 5.17a: L e 2 L 2 ( 2 ) 4 m . ( a ) Cylindrical tube: A 500 m m
2
4
2
(D
d ) 2
or d
D
2
4A
40 2
4 (500 )
31 m m
Thus I
64
(D
d )
4
4
[ 4 0 3 1 ] 8 .0 3 3 (1 0 ) m m 4
64
4
4
4
and r
4
( 8 .0 3 3 )(1 0 )
I A
1 2 .6 8 m m
500
It follows that L e r 4 0 0 0 1 2 .6 8 3 1 5 .5
Since L e r 2 0 0 , the Euler formula applies. Hence Pc r
2
EI 2
Le
2
9
(1 0 5 1 0 )( 8 .0 3 3 1 0 (4)
8
)
5 .2 0 3 k N
2
( b ) Square tube. The cross-sectional area: A a o a i . Inner diameter is 2
ai
ao A 2
2
4 0 5 0 0 3 3 .1 7 m m 2
Then I
1 12
( a o bi ) 4
1 12
r
I A
10
1 1 .2 5 500
15 m m ,
4
( 4 0 3 3 .1 7 ) 1 1 .2 5 1 0 4
4
4
mm
4
and 2
Le r 4 0 0 1 5 2 6 7
Since L e r 2 0 0 , the Euler formula is valid. Therefore Pc r
2
EI 2
Le
2
9
(1 0 5 1 0 )( 0 .1 1 2 5 1 0 (4)
2
6
)
7287 N
Comment: Hollow square has a critical load that is 1.4 times more than for a hollow circular section.
SOLUTION (5.68) From Table B.1: E 2 0 0 G P a
S y 250 M Pa
The properties of area are A b h (3 5 )(1 0 ) 3 5 0 m m , 2
I
1 12
bh 3
1
(3 5 )(1 0 ) 2 9 1 7 m m 3
4
12
(CONT.)
93
5.68 (CONT.) I
r
2917
A
2 .8 8 7 m m ,
(
Le r
350
E
)c
200 10
Sy
3
8 8 .9
250
( a ) From Fig. 5.17c, L e 0 .7 L 0 .7 (1 8 0 ) 1 2 6 m m . Hence Le
126
r
4 3 .6
2 .8 8 7
Since 4 3 .6 8 8 .9 , Johnson Formula should be used. Thus: S y ( Le r )
Pc r A S y [1
2
]
4 E 2
2
2 5 0 ( 4 3 .6 )
( 3 5 0 ) ( 2 5 0 ) [1
4
200 10
2
3
] 8 2 .2 3 k N
( b ) Now we have Le
0 .7 ( 5 0 0 )
r
1 2 1 .2 8 8 .9
2 .8 8 7
Euler formula applies. So EI
( 2 0 0 1 0 )( 2 9 1 7 1 0
2
Pc r
2
2
9
Le
( 0 .3 5 )
12
2
)
47 kN
SOLUTION (5.69) ( a ) Cross-sectional area: A
4
2
(D
d ) 2
( 6 2 .5 6 0 ) 2 4 0 .5 m m 2
4
2
2
Moment of inertia:
I
64
4
(D
d ) 4
64
( 6 2 .5 6 0 ) 1 1 2 , 8 4 1 .5 m m 4
4
4
and r Le
I A
r
750 2 1 .7
2 1 .7 m m
1 1 2 ,8 4 1 .5 2 4 0 .5
3 4 .5 6
Also (
Le r
2
E
)c
Sy
2
9
( 7 0 1 0 )
2 7 0 1 0
6
5 0 .5 8
Since L e r 5 0 .9 6 , the Johnson formula applies. Thus Pc r A S y [1
Sy 2
4 E
(
Le r
) ] 2 4 0 .5 1 0 2
6
( 2 7 0 1 0 )[1 6
6
2 7 0 1 0 ( 3 4 .5 6 ) 4
2
9
( 7 0 1 0 )
2
]
5 7 .3 6 k N
( b ) We have C D 2 and: ce r
2
3125 (3) ( 2 1 .7 )
2
0 .2
A 2 4 0 .5 m m
2
P 16 kN
(CONT.)
94
5.69 (CONT.)
Equation (5.74a) gives then
m ax
[1
P A
2
3
2 4 0 .5 1 0
P Pc r
)]
[1 0 .2 s e c ( 2
1 6 (1 0 )
s e c ( 2
ec r
6
16 5 7 .3 6
)]
7 9 .8 M P a
SOLUTION (5.70) From Prob.5.55 for both angles: I m in 4 7 0 , 0 5 2 m m r
1,4 8 8 2
2 E Sy
Cc
Le
1 7 .7 7 m m ,
4 7 0 ,0 5 2
2
2
( 200 10
240 10
9
)
6
r
4
and A 7 4 4 2 1 4 8 8 m m
2 .7 5 1 0 1 7 .7 7
3
1 5 4 .8
128
Since C c 1 5 4 .8, use Eqs. (5.77b):
a ll
2
9
( 2 0 0 1 0 )
1 .9 2 (1 5 4 .8 )
4 2 .9 M P a
2
Hence Pa ll
a ll
A 4 2 .9 (1 4 8 8 ) 6 3 .8 4 k N
SOLUTION (5.71) Cc
2
2 E Sy
[
2
2
1
3
( 200 10 )
Le
] 2 106 ,
350
r
0 . 65 ( 3 ) d 4
7 .8 d
(Eq.c of Sec. 6.2)
Equation (5.77b):
Le
Check:
3
5 0 (1 0 )
a ll
d
r
2
4
6 0 . 0441
2
9
( 2 0 0 1 0 )
1 .9 2 ( 7 .8 d )
2
d 4 4 .1 m m
,
136 C c
OK.
SOLUTION (5.72) L e 0 . 5 L 2 m . Assume 1 1 L e d 2 6 and use Eq.(5.80b). Thus
a ll
P A
3
1 0 0 (1 0 ) d
2
8 .2 7 [1
1 3
(
Le d 26
2
) ]1 0
6
This gives 1 0 0 (1 0 ) 8 .2 7 d 3
2
0 .0 2 4 5,
d 109 m m
Then
P A
3
1 0 0 (1 0 ) ( 0 .1 0 9 )
Check: 8 .4 2 1 0
2
8 .4 2 M P a
O K .,
L e d 2 0 .1 0 9 1 8 .3
95
OK.
2
Thus
SOLUTION (5.73) A
(3 5 0 3 0 0 ) 2 5 , 5 2 5 m m , 2
4
2
L e 0 .7 L 0 .7 ( 6 .1) 4 .2 7 m .
2
Equation (5.68): r
1 4
d
2
D
2
350 300 2
1 4
Le
1 1 5 .2 4 m m ,
2
r
4 .2 7 1 0 1 1 5 .2 4
3
3 7 .0 5
Using Eq.(5.79b): a ll [1 4 0 0 .8 7 (3 7 .0 5 )] 1 0 7 .7 7 M P a Hence Pa ll 1 0 7 .7 7 1 0 ( 2 5 , 5 2 5 1 0 6
6
) 2751 kPa
SOLUTION (5.74) 0 .1 2 0 .0 8 12
I m in
3
6
5 .1 2 (1 0
A 0 .0 0 9 6 m ,
rm in
2
Cc
4
) m , I m in
2
2 E Sy
A 2 3 .0 9 m m ,
1 2 1 .7 Le r
3,5 0 0 2 3.0 9
1 5 1.6
By Eq.(5.77b):
a ll
2
E
1 .9 2 ( L e r )
2
2
9
( 2 1 0 1 0 )
1 .9 2 (1 5 1 .6 )
2
4 6 .9 7 M P a
We have 600 9 .6
62 . 5 MPa
280
MPa
SOLUTION (5.75) Table A.6: A 27 . 5 10
3
mm
r z 101 . 1 mm
2
Buckling in xy plane: C c [ 2
2
( 200 10
1
3)
280 ] 2 119 , L e 0 . 7 L 4 . 2 m
and L e r z 4 .2 0 .1 0 1 1 4 1.5 C c . Apply Eq.(5.77a):
n
5 3
all
3 8
( 119 )
1 8
( 1 1 9 ) 1.7 9
1 2
.5 ( 41 ) ] 146 . 9 MPa 119
4 1.5
280 10 1 . 79
6
[1
3
4 1 .5
2
and Pall 146 . 9 ( 27 . 5 ) 4 , 040
kN
Buckling in xz plane: L e 0 . 5 (12 ) 6 m n
5 3
a ll
3 8
( 3171.59 )
2 8 0 1 0 1 .7 7
6
[1
r y 161
mm ;
1 8
( 31 71 .39 ) 1 .7 7
1 2
( 31 71 .39 ) ] 1 5 0 .4 M P a ,
L r y 37 . 3 C c
2
2
96
Pa ll 1 5 0 .4 ( 2 7 .5 ) 4 ,1 3 6 k N
SOLUTION (5.76) A a
r
2
a
I A 2
a
I a L
a
4
r
12 0 .5
3
a (2
1 .7 3 2 a
3)
Assume: 9 .5 L e r 6 6 . Using Eq. (5.79b):
a ll
250 10
a
3
[1 4 0 0 .8 7 (
2
1 .7 3 2
) ]1 0
6
a
or 3
a 1 0 .7 6 1 0 2
a 1 .7 8 6 1 0
3
0,
a 48 m m .
So, r a 2
3 48 2
3 1 3 .9 m m
L r 5 0 0 1 3 .9 3 6 6 6
Our assumption was correct. Use a 4 8 m m
SOLUTION (5.77) A 200 100 160 60 10, 400 m m I m in rm in
1
2
( 2 0 0 1 0 0 1 6 0 6 0 ) 1 3, 7 8 6 , 6 6 6 .6 7 m m 3
3
4
12 A 3 6 .4 1 m m ,
I m in
L e rm in 4 .3 1 0
3
3 6 .4 1 1 1 8 .1
Use Eq. (5.79c):
a ll
350 10 (1 1 8 .1)
3
2 5 .0 9 M P a
2
Pa ll 2 5 .0 9 (1 0 , 4 0 0 ) 2 6 0 .9 4 k N
SOLUTION (5.78) For the situation described L e 2 L (see Fig. 5.17a) and d=62 mm So, Le
2 (1 .2 2 1 0 ) 3
d
3 9 .4
62
Since 39.4 > 26, use Eq. (5.80c). 0 .3 (1 2 1 0 ) 9
a ll
(3 9 .4 )
2
2 .3 2 M P a
and Pa ll
a ll
A ( 2 .3 2 ) (8 8 6 2 ) 1 2 .6 6 k N
97
SOLUTION (5.79) We now have L e 2 L 2 ( 7 5 0 ) 1 5 0 0 m m and d=62 mm Le
1500
d
2 4 .2
62
Since 24 .2< 26, apply Eq. (5.80b).
1 2 4 .2 2 ( ) ] 5 .8 8 M P a 3 26
8 .2 7[1
a ll
Thus Pa ll
a ll
A ( 5 .8 8 ) (8 8 6 2 ) 3 2 .0 8 k N
SOLUTION (5.80) Boundary conditions: v ( 0 ) 0 , v ( L ) 0 , With w=0, the solution of Eq.(5.82b):
M (0) M 0 ,
M (L) M
0
v A sin k x B c o s k x C x D
(1)
v " A k s in k x B k
(2)
2
2
cos kx
Substituting boundary conditions into these equations, we have v (0 ) B D 0,
v ( L ) A s in k L B c o s k L C L D 0
and since E Iv " M , M ( 0 ) B E Ik
Solving, and setting k M
A
M 0,
M ( L ) A E Ik sin k L B E Ik c o s k L M 2
2
P E I as needed,
2
1 c o s k L s in k L
0
P
2
B D
,
M
0
P
C 0
,
Equation (1) is thus M
v
0
1 c o s k L s in k L
(
P
s in k x c o s k x 1 )
SOLUTION (5.81) Apply Eq.(4.14) 2
EI
dx
M Pv
M
d v 2
2
dx
2
(x
2
Lx )
P
or d v
w 2
k x 2
w 2
(x
2
wL 2
Lx )
A
B
x
wL 2
Figure S5.81 We have, the deflection: v vh vP
(1)
v h A s in k x B c o s k x
(2)
Here It can be shown that, for the case of uniform loading w (Fig. S5.81): vP
w 2 E Ik
2
(x
2
Lx
2 k
2
(3)
)
Boundary conditions v h ( 0 ) 0 and v h ( L ) 0 give A and B. In doing so, Eq.(1) results in: v
w E Ik
4
[(1 c o s k L )
s in k x s in k L
cos kx
98
2
k 2
(x
2
L x ) 1]
0
SOLUTION (5.82) Refer to Example 5.23.
v
m x L
a m sin
(1)
m 1
Hence U W
EI 2
1 2
L
(d
0
2
dx
v 2
L
4
EI
) dx 2
P ( v ') d x 2
0
4L
L
3
4
w vdx
0
2
m am
2
P 4L
m am 2
2
2wL
am
1 m
Applying U W , we obtain am
4 wL 5
4
EI
1 3
2
m (m b)
where b
PL
2
2
EI
Substitution of this into Eq.(1) gives the solution.
End of Chapter 5
99
Section II FAILURE PREVENTATION CHAPTER 6
STATIC FAILURE CRITERIA AND RELIBILITY
SOLUTION (6.1) Table 6.2: K c 5 9 1 0 0 0 M P a
mm
S y 1503 M Pa
and
1 . 01
Case A of Table 6.1: with a w 0 . 1 , From Eq.(6.3), with n 1 :
Kc
a
59
2 0 8 .4
1000
(1 .0 1 )
( 25 )
M Pa
Thus, we have P ( 2 w t ) 2 0 8 .4 ( 0 .5 0 .0 2 5 )1 0 2 , 6 0 5 6
kN
The nominal stress at fracture
6
2 .6 0 5 (1 0 )
2 3 1 .6
0 .0 2 5 ( 0 .5 0 .0 5 )
M Pa
This is well below the yield strength of 1503 MPa.
SOLUTION (6.2) Py
all
A net
( 350 15 ) 2 . 844
650 1 .2
Case B of Table 6.1: with a w 0 .0 7 ,
By Eq.(6.3),
Kc n
a
MN
1.1 2
100 1000 1 .2 (1 .1 2 )
( 25 )
2 6 5 .5 M P a
Since 265.5 < 650, fracture controls. Thus P f 265 . 5 ( 350 15 ) 1 , 394
kN
SOLUTION (6.3) From Table 6.2: K c 23 1000
MPa
Case B of Table 6.1: a w 0 .1 6 ,
mm
and S
1.1 2
By Eq.(6.3), with n 1 :
Kc
a
8 1 .9 3 M P a
23 1000 1 .1 2
( 20 )
It follows that P ( wt ) 81 . 93 (125 25 ) 256
Then
P ( w a )t
3
256 ( 10 ) ( 0 . 125 0 . 02 ) 0 . 025
97 . 5 MPa
S
97 . 5 MPa
y
100
kN
y
444
MPa
SOLUTION (6.4) Table 6.1, Case A: 1 .0 3 . Table 6.2, K c 5 9 M P a
S y 1503 M Pa .
m,
We have
Sy n
6 0 1 .2 M P a
1503 2 .5
Equation (6.1) gives K
a (1 .0 3 )( 6 0 1 .2 ) ( 2 1 0
3
) 4 9 .0 8 M P a
Using Eq. (6.2), we find n
Kc
K
59 4 9 .0 8
1 .2
SOLUTION (6.5) Table 6.1, Case B: 1 .3 7 . Table 6.2, K c 3 1 M P a
m,
S y 392 M Pa .
We have
Sy n
140 M Pa
392 2 .8
Equation (6.1): K
a 1 .3 7 (1 4 0 ) ( 0 .0 0 5 ) 2 4 .0 4 M P a
Applying Eq. (6.2), n
Kc
K
31 2 4 .0 5
1 .2 9
SOLUTION (6.6) Table 6.1: a w 5 0 5 0 0 0 .1, Table 6.2: K c 1 1 1 M P a
m,
1 .0 1 . S y 798 M Pa .
( a ) Equation (6.1), K
a 1 .0 1(1 5 0 ) ( 0 .0 5 ) 6 0 M P a
Equation (6.2) gives the safety factor for fracture as, n
Kc
K
111 60
1 .8 5
Safety factor for yielding is n
Sy
798 150
5 .3 2
( b ) Using Eq. (6.3) with n=1, fracture stress is
f
Kc
a
111 (1 .0 1 )
( 0 .0 5 )
2 7 7 .3 M P a
101
m
m
m
SOLUTION (6.7) Case A of Table 6.1 and Table 6.2: K
c
59
1000
1 . 01
MPa
S y 1503 M Pa
mm
(assumed)
By Eq.(6.3):
Kc n
(a)
pfr
(b)
pfr
2t
t
a
59
3 1 0 .7
1000
(5)
(1 .5 )(1 .0 1 )
M Pa S y
,
p f 2 ( 4 9 .7 1) 9 9 .4 2 M P a
,
pf
t r
3 1 0 .7 ( 4 )
25
4 9 .7 1 M P a
SOLUTION (6.8) Table 6.2: K c 31 MPa Case D of Table 6.1:
m
y
392
MPa
1.3 2
0 .4
a w
S
Using Eq.(6.3) with n 1 and 6 M tw : 2
Kc
a ;
6M tw
2
3 1(1 0 ) 1.3 2 6
6M 0 . 0 2 5 ( 0 .1 )
2
( 0 .0 4 )
Solving M 2 . 76 kN m
SOLUTION (6.9) ( a ) K c 59 M Pa
m
S y 1 5 0 3 M P a (Table 6.2).
and
Case B of Table 6.1: with a=12.5 mm and w=125 mm:
a w
0 . 1 1 . 12
Eq.(6.3) with n 1 : 5 9 (1 .1 2 )
(1 2 .5 1 0
3
),
2 6 5 .8 M P a
Therefore Pa ll ( w t ) (1 2 5 2 5 )( 2 6 5 .8 ) 8 3 0 .6 k N
Note that the nominal stress at fracture
8 3 0 .6 2 5 (1 2 5 1 2 .5 )
2 9 5 .3 M P a S y
( b ) Table 6.2: K c 6 6 M P a
m
and S y 1149 M P a
Thus 66 10
6
(1 . 12 )
a (1 . 12 )(
Solving a 0 .0 1 5 m = 1 5 .6 m m
Comment: This value of a satisfies Table 6.2
102
830 . 6 10 125 25 10
3 6
) a
SOLUTION (6.10) a w
Refer to Example 6.3. We now have a 2 .1 1
0 .4 :
15 37 . 5
b 1.3 2
and
Equation (6.5) is therefore ( 2 . 11 )
Table 6.2: K c 7 7 M P a
m
6 ( 0 . 175 P )
(1 . 32 )
P 0 . 0375 ( 0 . 0125 )
0 . 0125 ( 0 . 0375 )
2
83 , 349 P
and S y 690 M P a
Note that both a and t satisfy Table 6.2. Using Eq.(6.3),
Kc
a
n
7 7 1 0
8 3, 3 4 9 P
:
2
3
( 0 .0 1 5 )
Solving P 2 .1 2 8 k N
The nominal stress at fracture, ( 2 .1 2 8 1 0 ) [ 0 .0 1 2 5 ( 0 .0 3 7 5 0 .0 1 5 )] 7 .5 7 M P a S y . 3
SOLUTION (6.11) By Table 6.2, K c 5 9 M P a
m and S y 1503
M P a . Note that values of a and t satisfy
Table 6.2. a w
From Table 6.1:
a ll
Kc
n
a
1.1 2 . Therefore,
0 .1
5 9 1 0
(1 .4 )(1 .1 2 )
6
( 0 .0 0 5 )
3 0 0 .2 2 M P a
and Pa ll
a ll
( w t ) 3 0 0 .2 2 ( 0 .0 5 0 .0 2 5 ) 0 .3 7 5 M N = 3 7 5 k N
Comment: The nominal stress at the fracture 375 kN t(2w2a)
3
3 7 5 1 0 ( 0 .0 2 5 )( 0 .1 0 .0 1 )
1 6 6 .7 M P a S y
SOLUTION (6.12)
(a) (b)
x
xy
S
y
n
S
y
n
32 M 3
D
16 T
D
3
(
x
(
x
2
2
4P
D
2
3
32 ( 5 10 )
( 0 .1 ) 3
16 ( 8 10 )
( 0 .1 )
3
2
3
4 ( 50 10 )
( 0 .1 )
2
57 . 3 MPa
1
260 n
[ 5 7 .3 3 ( 4 0 .7 4 ) ] 2 ,
260 n
[ 5 7 .3 4 ( 4 0 .7 4 ) ] 2 ,
1
4 xy ) 2 ;
40 . 74 MPa
1
3 x y ) 2 ; 2
3
2
2
103
2
2
1
n 2 .8 6
n 2 .6 1
SOLUTION (6.13) Table B.1: S u 2 4 0 M P a and S u ' 6 5 0 M P a From the solution of Prob. 6.12, we have
5 7 .3 M P a and
x
xy
4 0 .7 5 M P a .
Thus
x
2
1 ,2
(
x
2
) 2
2 xy
2 8 .6 5
( 2 8 .6 5 ) ( 4 0 .7 5 ) 2
2
or 1 7 8 .4 6 M P a
2 2 1 .1 6 M P a
Thus 1
u n
1;
2
u n
7 8 .4 6 240
2 1 .1 6 650
1 n
or n 2 .7 8
SOLUTION (6.14) y A
F
B
0.8 m
A
P T
We have
T
D
M
P=20F,
T=0.4F,
P
x
M=0.8F
Stresses at fixed end:
b
32 M
a
D
3
32 ( 0 . 8 F )
( 0 . 04 )
20 F
( 0 .0 4 )
2
4
3
127 , 324 F
1 5 , 9 1 5 .5 F
x
16 T
D
3
16 ( 0 . 4 F )
( 0 . 04 )
3
31 , 831 F
a b 1 4 3 , 2 3 9 .5 F
Apply Eq.(6.16): S
y
n
1
[ x 3 ] 2 ; 2
2
6
2 5 0 (1 0 ) 1.4
2
or F 1 . 163
kN
SOLUTION (6.15) Refer to solution of Prob. 6.14. We have m ax
(
[(
x
2
) 2
1 4 3 , 2 3 9 .5 F 2
2
) (3 1, 8 3 1 F ) ] 2
2
Hence, m ax
Sy 2n
;
7 8 , 3 7 4 .7 F
1
F [(1 4 3 , 2 3 9 .5 ) 3 ( 3 1, 8 3 1 ) ] 2
6
2 5 0 (1 0 ) 2 (1 .4 )
or F 1 .1 3 9 k N
104
1 2
7 8 , 3 7 4 .7 F
2
x
SOLUTION (6.16) T 0 .4 F
At the fixed end A:
M
0 .8 F
z
Vy F
Thus,
x
32M
D
A 1 6T
D
( 0 .0 6 )
( 0 .0 6 )
3 7 .7 2 6 F
3
1 6 ( 0 .4 F )
3
3 2 ( 0 .8 F )
3
9 .4 3 1 F
3
Then 1, 2
3 7 .7 2 6 F
3 7 .7 2 6 F
) ( 9 .4 3 1 F ) 2
2
2
1 3 9 .9 5 1 0 F 140 10
(
2
3
(a)
m a x 2 1 .0 9 1 0 F
2 .2 3 1 0 F 3
2
3
6
2 1 .0 9 1 0 F ,
F 4 .1 5 k N
3
1 .6
( b ) 1 1 2 2 ( S y n ) 2
2
2
[3 9 .9 5 3 9 .9 5 ( 2 .2 3 ) ( 2 .2 3 ) ] 2
2
1 2
(1 0 ) F ( 2 6 0 1 .6 )1 0 3
6
or F 3 .9 5 k N
SOLUTION (6.17) Q B 102 (10 . 3 )( 74 . 85 ) 78 . 6 10
3
3
mm
Q C Q B 6 . 6 ( 69 . 7 )( 34 . 85 ) 94 . 6 10
3
mm
3
We have
A
Mc I
VQB
3
24 ( 10 )( 0 . 08 ) 13 . 4 ( 10
6
)
143 . 3 MPa ,
B
(143 . 3 ) 124 . 9 MPa
69 . 7 80
and B
Ib
3
6 0 (1 0 )( 7 8 .6 1 0 1 3 .4 (1 0
6
6
)
)( 0 .0 0 6 6 )
5 3 .3 M P a ,
c
VQc Ib
6 4 .2 M P a
Thus ( 1, 2 ) B
1 2 4 .9 2
[(
1 2 4 .9 2
1 B 1 4 4 .6 M P a
or
(
max
1
) 5 3.3 ] 2 6 2 .5 8 2 .1 2
2
2 B 1 9 .6 M P a
) B 82 . 1 MPa
Hence ( m ax ) B
S
y
2(2 )
320 4
8 2 .1 8 0
;
F a ils
Alternatively, using Eq.(6.11), we have [1 2 4 .9
2
1
4 ( 5 3.3 ) ] 2 2
320 2
;
1 6 4 .2 1 6 0
105
F a ils
SOLUTION (6.18) From Solution of Prob.6.17, at point B: 1 1 4 4 .6 M P a 2 1 9 .6 M P a Thus 1
S
[ 1 1 2 2 ] 2 2
2
y
n
320 2
160 1
or
2
[( 144 . 6 )
(144 . 6 )( 19 . 6 ) (19 . 6 ) ] 2 155 . 3 160 2
No failure
Alternatively, by Eq.(6.16): 1
3 ( 53 . 3 ) ] 2
2
[124 . 9
2
; 155 . 3 160
320 2
No failure
SOLUTION (6.19) 1
(a)
1
(b)
1 1
3
pr
t
Sy n
2
2
3 .5 ( 0 .2 5 )
0 .8 7 5 t
;
t
250 1 .5
S
(
2 2
0 .8 7 5 t
,
2
0 .4 3 7 5 t
t 5 .2 5 1 0
,
3 0
, 3
m = 5 .2 5 m m 1
2
y
) ; [(
n
0 . 875 t
)
2
( 0 . 875 20 . 4375 ) ( 0 . 4375 ) ]2 t 2
t
or 1
1 t
[ 0 .7 5 8 ] 2
250 1 .5
t 4 .5 4 8 1 0
,
3
m = 4 .5 4 8 m m
SOLUTION (6.20) From Solution of Prob.6.19: 1 0 . 875 , 2 t 1
(a)
Su
;
n
0 .8 7 5 t
0 . 4375 t
350 1 .5
,
3
0
t 3 .7 5 1 0
,
3
3 .7 5 m m
( b ) Using Eq.(6.26): 1
Su
Su
0 .8 7 5 t
;
n
350 1 .5
t 3 .7 5 1 0
,
3
m = 3 .7 5 m m
and 2
n
0 .4 7 5 t
;
350 1 .5
,
t 1 .8 7 5 1 0
3
SOLUTION (6.21) From Solution of Prob.6.17, at point B: 1 1 4 4 .6 M P a 2 1 9 .6 5 M P a (a)
1
( b ) By Eq.(6.25),
Su n
,
n
n
Su
1
280 144 . 6
280 144 . 6 19 . 65 ( 280 620 )
1 . 94
1 . 82
106
m = 1 .8 7 5 m m
250 1 .5
SOLUTION (6.22)
M
A
0:
Fy 0 :
1 7 5 ( 2 .4 6 )( 2 .4 6 2 ) R B (1 .7 ) 0 ,
R B 3 1 1 .5 k N
R A 119 kN
y
175 kN/m z
A 1.7 m
119 kN
b
311.5 kN 133
V, kN
2b
0.76 m C
B
3 ft
119
x -178.5
M, kN m x -50.54 At B at the upper outermost fiber:
m ax
M
m ax
c
5 0 .5 4 1 0 b 3
1
I
b (2b )
7 5 .8 1 1 0 b
3
3
3
12
Thus,
m ax
7 5 .8 1 1 0
; a ll
b
3
140 10 , 6
3
b 0 .0 8 2 m = 8 2 m m
At B at the neutral axis axis: m ax
3V
2A
3 1 7 8 .5 1 0 2
2b
3
2
1 3 3 .8 7 5 1 0 b
3
2
And m ax 1 2
Thus, m ax
; a ll
1 3 3 .8 7 5 1 0 b
2
3
140 10 , 6
Use a 8 2 m m by 1 6 4 m m rectangular beam.
107
b 0 .0 3 1 m = 3 1 m m
SOLUTION (6.23) We now have
N.A.
A 0 .3 0 .1 2 0 .0 3 6 m
B
120 mm
A
I
1 12
( 0 .1 2 )( 0 .3 ) 0 .2 7 1 0 3
M 0 .4 0 .1 5 0 .5 5 P
300 mm
c A c B 0 .1 5 m
3 ft Referring to Example 6.9:
P A
McA
A
P A
M cB
B
2 7 .7 7 8 P 3 0 5 .5 5 6 P
I
2 7 .7 7 8 P 3 0 5 .5 5 6 P
I
Thus 1 3 3 3 .3 3 4 P
2 0
2 2 7 7 .7 7 8 P
1 0
It follows that 3 3 3 .3 3 4 P
6
1 7 0 (1 0 ) 2 .5
P 204 kN
, 6
6 5 0 (1 0 )
2 7 7 .7 7 8 P
P 936 kN
,
2 .5
SOLUTION (6.24) S u 170
S uc 650
MPa
(Table B.1).
MPa
Thus 1, 2
and (a)
100 50 2
[(
1 127 . 6 MPa 127 . 6 , n 1 . 33
170 n
7 7 .6 ,
1
) (70 ) ]2 2
2
170 n
n
(b)
100 50 2
2
77 . 6 MPa
n 2 .1 9
170 1 2 7 .6 7 7 .6 (1 7 0 6 5 0 )
1.1 5
SOLUTION (6.25) Table B.1: S u 170 1, 2
or (a)
(b)
120 60 2
1 140
S uc 650
MPa [(
MPa ,
120 60 2
) 40 ] 2
2
140 ,
n 1 . 21
170 n
80 ,
n 2 . 13
170 1 4 0 8 0 (1 7 0 6 5 0 )
MPa 1 2
40 MPa ,
170 n
n
2
1.0 6
108
2
3
80 MPa
3
m
3
SOLUTION (6.26) The circumferential, axial, and radial stresses are given by 1
pr
24 p
t
2
pr
2t
3 0
12 p
Insertion of these expression into Eqs. (6.6) and (6.14) provide the critical pressures. ( a ) For the maximum shearing stress theory: 24 p 0
(250 10 ) 6
1 1 .4
p 7 .4 4 M P a
( b ) For the maximum energy of distortion theory: p (24 24 12 12 ) 2
2
1 2
1 1 .2
(250 10 ) 6
p 1 0 .0 2 M P a
Comment: The permissible value of the internal pressure is conservatively limited to 7.44 MPa.
SOLUTION (6.27) Maximum shear stress criterion, substituting the given expressions for n
Sy
1
2
S yt ( 3 . 56 1 . 70 ) a
0 . 19
2
1
and
2
into Eq.(6.6):
S yt
a
2
Maximum distortion energy criterion, using Eq.(6.14) together with the given expressions for 1 and 2 : Sy
n 2
[ 1 1
2
2 2
S yt
1
0 . 215
1 2
2
[ 3 . 56 ( 3 . 56 )( 1 . 70 ) ( 1 . 70 ) ] 2 a
]2
2
S yt
a
2
The factor of safety based on energy of distortion theory is therefore 11.6 % larger than that based on the maximum shear stress theory. This indicates that the maximum energy of distortion criterion is less conservative, as expected.
SOLUTION (6.28) Principle stress criterion, carrying the given expressions for n
Su
1
Su
3 . 56 a
2
t
0 . 281
1
and
2
into Eq.(6.22):
Sut
a
2
Coulomb-Mohr criterion, applying Eq.(6.25) together with the given expressions for
1
and 2 : n
Su
1
2
S u S uc
Sut [ 3 . 56 ( 1 . 7 )( 1 5 )] a
2
0 . 256
Sut
a
2
The n according to the principle stress criterion is thus 8.9 % larger than that on the basis of The Coulomb-Mohr criterion. This indicates that the Coulomb-Mohr theory is more conservative, particularly when S u c S u .
109
SOLUTION (6.29) Table B.1; S y 2 5 0 M P a
x
4P
,
2
D
xy
16T 3
D
Equation (6.11) results in xy
1 2
Sy
[(
1
) x ]2 2
n
2
6
{ ( 2 5 01 .51 0 ) [
1 2
2
3
4 ( 4 5 1 0 )
( 0 .0 5 )
2
1
] } 2 8 2 .5 M P a 2
Thus T
D
3
( 0 .0 5 )
xy
16
3
16
(8 2 .5 1 0 ) 2 .0 2 5 k N m 6
SOLUTION (6.30) Refer to solution of Prob.6.29, Equation (6.16) gives xy
1
[(
3
Sy n
) x] 2
2
1 3
6
{ ( 2 5 01 .51 0 ) [ 2
3
4 ( 4 5 1 0 )
( 0 .0 5 )
2
1
] } 2 9 5 .3 1 M P a 2
and T
D
3
xy
16
(2)
3
16
(9 5 .3 1 1 0 ) 2 .3 3 9 k N m 6
SOLUTION (6.31) We have 1 ,
2 .
Table B.2: S u 1 5 0 M P a ,
S uc 5 7 5 M P a
( a ) Equation (6.22):
Su
n
107 . 1 MPa
150 1 .4
( b ) Equation (6.24) is thus 150
575
1 1 .4
or 84 . 98 MPa
SOLUTION (6.32) We have n=2, Table B.1: S u 340 MPa ,
1, 2
1 2
( 102 0 )
( 10 0 )
2
S uc 620
Equation (6.24): 1 Su
2
S uc
1 n
Thus 5
25 340
2
5
25 620
2
1 2
Solving 111 . 1 MPa
110
2
5
MPa 25
2
SOLUTION (6.33) We have n=2. Table B.1: S y 3 4 5 M P a ( a ) Equation (6.11): S
(
y
n 345 2
or
1
4 xy ) 2
2
2
x
1
8 6 .1 M P a
1
9 9 .4 3 M P a
[( 1 0 ) 4 ] 2 , 2
2
( b ) Equation (6.16): S
(
y
n 345 2
or
1
3 x y ) 2
2
2
x
[ ( 1 0 ) 3 ] 2 , 2
2
SOLUTION (6.34)
We have
y
xy 4 5 M P a
30 M Pa
x
0.
Thus
1, 2
y
y
(
2
) 2
2
1 3 2 .4 M P a
2 xy
30
2
(
30
) (45) 2
2
2
2 6 2 .4 M P a
( a ) Maximum principal stress theory: 1 1;
3 2 .4 5 5
1;
6 2 .4 5 5
Su
But
2
(failure occurs)
Su
( b ) Coulomb-Mohr Theory: 1 2
Su
32
1;
S uc
55
6 2 .4
1
160
gives 0 .5 8 0 .3 9 0 .9 7 1
(no fracture)
SOLUTION (6.35)
x
32M
D
3
xy
1 6T
D
3
M 7 5 0 ( 0 .3 ) 2 2 5 N m
(CONT.)
111
6.35 (CONT.) 350
( S u ) a ll ( S u c ) a ll
Also
140 M Pa
2 .5 630
252 M Pa
2 .5
1, 2
x
(
2
x
) 2
2
2 xy
16
D
3
(M
M
T
2
2
)
Substituting the given data 16
1,2
or
D
(225
3
2
3
D
2
D
4 7 9 3 .7 9
1
1
225 680 )
2
3
(1 1 4 5 .9 2 3 6 4 7 .8 7 )
2 5 0 1 .9 5 D
3
( a ) Maximum Principal Stress Theory 4 7 9 3 .7 9
1 4 0 (1 0 ) 6
3
D
( b ) Coulomb-Mohr Theory: 1 2
( S u ) a ll
or
or
D 0 .0 3 2 5 m = 3 2 .5 m m
4 7 9 3 .7 9
1;
( S u c ) a ll
140 D
3
2 5 0 1 .9 5 252 D
3
10
6
D 0 .0 3 5 3 m = 3 4 .3 m m
SOLUTION (6.36) M (W F ) L ( 9 2 ) 0 .2 5 2 .7 5 k N m
T W a 9 0 .3 2 .7 k N m
An element at point A:
x
32M
t
D 1 6T
D
3
( 0 .0 5 ) 1 6 ( 2 .7 )
3
3 2 ( 2 .7 5 )
( 0 .0 5 )
3
224 M Pa
3
y A T
110 M Pa
z
So m ax
(
x
2 (
224
) t 2
2
Sy
) (1 1 0 ) 2
C
B
W+F
n 2
157 M Pa
y
2
210
x
n
A
Solving, n 1 .3 4
B x
t
t
An element at point B:
z
x
Figure (a) (CONT.)
112
6.36 (CONT.) 4 (W F )
m ax d t
1 6T
D
3A
4 (1 1 1 0 )
3
3
3 ( 0 .0 2 5 )
2
1 1 0 (1 0 ) 1 1 7 .4 7 M P a 6
210 n
or n 1 .7 9
SOLUTION (6.37) Table B.1: S y 250
and 7 . 86 Mg
MPa
m , n 2 .1 3
State of stress, at a point C at bottom surface on midspan:
x
C
32 M
D
3
16T
D
3
We have w 7 . 86 ( 9 . 81 )( D M
Thus
wL
m ax
2
32( w L
x
D
2
D
8)
2 .7 8 1 0 D
6
7 .7 2 8 1 0
6
2
8D
2
3
3
6
2 .7 8 (1 0 ) D
3
D
3
2
) 3(
2
) ( 6 )1 0
2 .0 4 1 0 D
2
3
3
(S ) ( 2
2
y
n) :
2 5 0 1 0 2 .1
6
or D
2
1 4 1.7 1 0
1 2 .4 8 4 8 D
6
8
Solving, by trial and error: D 34 . 34 mm
Use a 35-mm diameter shaft.
SOLUTION (6.38) Applying Eq.(6.34), we have z
From Fig.6.15:
s l
2 s
2 l
kN
2 . 0 4 (1 0 )
x 3
Equation (6.16), (
3 2 ( 6 0 .6 D
3
2
8
3
16( 400 )
4 ) 60 . 6 D
2
400 250 2
30 35
2
3 . 25
R 9 9 .9 4 % .
113
)
2
m
SOLUTION (6.39) The state of stress is
d
d
4P
. Thus
2
d
l
4 ( 200 )
2
l
4 (30 )
2
4
4
d
d
2
2
2 5 4 .6 d 3 8 .2 d
kPa
2
2
kPa
Figure 6.16, for R=99.7 %: z 2 . 75 Equation (6.34), in rearranged form: z (
2 s
2
1
)2 s 1
2 .7 5[ ( 3 5 1 0 ) ( 3 8 2.2 ) ] 2 3 5 0 1 0 3
2
2
3
2 5 4 .6
d
d
2
from which (3 5 1 0 ) ( 3 8 2.2 ) 3
2
2
d
3
(1 2 7 .2 7 3 1 0 3
9 2 .5 8 2 d
2
)
2
6
or d 1 .5 7 4 (1 0 ) d 0 .4 7 5 (1 0 ) 0 Solving, d 0 .0 3 4 2 m and d 3 4 .2 m m 4
2
SOLUTION (6.40) Apply Eqs.(6.30a) and (6.30b): 12
x
1 12
12
xi
912 12
y
76,
i 1
1 12
yi
936 12
78
i 1
12
x [
1 11
1
1
( x i x ) ] 2 [ 111 ( 2 2 0 4 )] 2 1 4 .1 5 5 2
i 1
12
[ 11 1
y
1
1
( y i y ) ] 2 [ 1 1 (1, 3 5 1.3 4 )] 2 1 1.0 8 4 2
1
i 1
SOLUTION (6.41) (a) x
n( xi )
2
n
nx
520
7
3640
7( -44.78 )
2
460
2
920
2(-104.78 )
2
430
5
2150
5(-134.78 )
2
545
5
2725
5( -19.78 )
2
570
10
5700
10(
5.22 )
2
575
18
10350
18( 10.22 )
2
595
8
4760
8( 30.22 )
2
600
3
1800
3( 35.22 )
2
620
6
3720
6( 55.22 )
2
660 -----
4 ----68
2640 4( 95.22 ) ------------ ---------------38405 196521.69
2
(CONT.)
114
6.41 (CONT.) Thus, Eq.(6.30a):
5 6 4 .7 8 M P a
38 ,405 68
Eq.(6.30b): n
[ 617
1
( x i 5 6 4 .7 8 ) ] 2 5 4 .1 5 9 M P a 2
i 1
( b ) Eq.(6.34), z
Figure 6.16:
s l
5 2 5 5 6 4 .7 8 5 4 .1 5 9
m
0 .7 3 5
R 76 %
SOLUTION (6.42) Maximum load: l 2 5 k N ,
l 3 kN
Strength of part: s 3 0 k N ,
s 2 kN
Equation (6.33a): m s l 3 0 2 5 5 k N Equation (6.33b):
m
2 s
2 l
2 3 2
2
3 .6 k N
Thus, failure impends at m
z
Figure 6.16:
m
5 3 .6
1 .3 8 9
R 92 %
SOLUTION (6.43) ( a ) We have
nom
Pn o m
A
4 (35 )
( 0 .0 1 2 5 )
2
35 1 .2 2 7 2 1 0
4
2 8 5 .2 M P a
So n
350 2 8 5 .2
1 .2 3
( b ) Mean stress value equals l n o m 2 8 5 .2 M P a . Estimated standard deviation equals 2 .5 l 2 0 .3 7 2 M P a 1 .2 2 7 2 1 0
4
The margin of safety, Eq. (6.34), is thus z
s l
2 s
2
l
3 5 0 2 8 5 .2 2
2 8 2 0 .3 7 2
2
1 .8 7
(CONT.)
115
6.43 (CONT.) From Fig. 6.16, reliability corresponding to z=1.87 is R 97 %
Hence, failure percentage equals 100-97=3%. Comment: In foregoing calculations, statistical variability of dimensions is omitted.
SOLUTION (6.44) Equation (6.34): Figure 6.16:
z
25 20 2
2 .5 3
2
1 . 28
R 90 %
Thus failure percentage is 10 %. End of Chapter 6
116
CHAPTER 7
FATIGUE FAILURE CRITERIA
SOLUTION (7.1) Use Figure 7.5 for steel (1020). (a)
At infinite life, S e ' 2 3 0 M P a The maximum stress in the beam is
m ax
M
m ax c
I
32 M
D
m ax 3
from which D
3
32 M
(1)
m ax
m ax
Letting m a x S e ' , D
(b)
3
3
3 2 ( 4 1 0 ) 6
( 2 3 0 1 0 )
5 6 .2 m m
5
At 1 0 cycles, S 3 1 0 M P a . Equation (1) gives Then D
3
3
3 2 ( 4 1 0 ) 6
( 3 1 0 1 0 )
5 0 .8 m m
SOLUTION (7.2) Use Figure 7.5 for aluminum alloy (2024). (a)
Endurance strength, S n ' 1 3 5 M P a The maximum stress in the beam:
m ax
32 M
D
or D
m ax 3
3
32 M
(1)
m ax
m ax
With m a x S n ' 1 3 5 M P a , D
(b)
3
3
3 2 (1 .5 1 0 ) 6
(1 3 5 1 0 )
4 8 .4 m m
7
At 1 0 cycles, S 1 6 5 M P a . Equation (1) results in D
3
3
3 2 (1 .5 1 0 ) 6
(1 6 5 1 0 )
4 5 .2 m m
SOLUTION (7.3) To determine the K t , we use Fig.C.1. Structural steel: S u 400
MPa
(Table B.1).
At section C: D d
38 30
1.2 6 7
(a)
r d
4 30
0 .1 3 3
( b ) Figure 7.9a:
max
1 .7
3
15 ( 10 ) 0 . 03 ( 0 . 01 )
85 MPa
(CONT.)
117
7.3 (CONT.) K
t
1. 7
r 4 mm : q 0 . 78
K
1 0 .7 8 (1.7 1 ) 1.5 5
f
(Eq.7.13b)
Similarly, at D: D d
r d
38 34 2 34
K
t
1.1 1 8
(a)
0 .0 5 9
( b ) Figure 7.9a:
1.8
3
15 ( 10 )
1 .8
max
79 . 41 MPa
0 . 034 ( 0 . 01 )
r 2 mm : q 0 . 72
K
1 0 .7 2 (1.8 1 ) 1.5 8
f
SOLUTION (7.4) S u 400
Table B.1:
MPa ,
S e C f C rC sC t
where
K
f
S Se
1.5 8 at D (from Solution of Prob.7.3)
S e 0 .4 5 S u 1 8 0
M Pa
A S u 272( 400 ) b
f
MPa
'
1 K f
'
C
250
y
0 .9 9 5
0 .7
C r 0 .8 7 (Table 7.3) C s 1 (axial loading)
C t 1 0 .0 0 5 8 ( 4 7 5 4 5 0 ) 0 .8 5 5
Thus S e ( 0 .7 )( 0 .8 7 )(1)( 0 .8 5 5 )
1 1 .5 8
(1 8 0 ) 5 9 .3 M P a
SOLUTION (7.5) S e C f C r C s C t (1 K f ) S e '
We have
r d
t
2 .5 (Fig.C.3)
K
2 26
(1)
0 .0 7 7 ,
1.1 5 4
D d
Table B.4: S u 655 MPa , S e 0 .4 5 S u 2 9 4 .8 '
H
B
197
M Pa
Table 7.3: C r 0 .8 9 Fig.7.9a: q 0 . 8 ,
K
f
1 0 .8 ( 2 .5 1) 2 .2
Table 7.2: C f A S u 4 .5 1( 6 5 5 ) b
Use C s 1 (axial loading)
0 .2 6 5
0 .8 0 9
Ct 1
Equation (1) is therefore S e ( 0 .8 0 9 )( 0 .8 9 )(1)(1)( 21.2 )( 2 9 4 .8 ) 9 6 .4 8
118
M Pa
SOLUTION (7.6) S u 626
Table B.4:
MPa
H
B
179
From Eq.(7.1): S e 0 . 5 S u 313 '
MPa
(Note: by Eq.(2.22): S u 3500 (179 ) 626 . 5 MPa. ) C s 0 .8 5
From Eq.(7.9): By Eq.(7.7):
Ct 1
C r 0 .8 9
Using Table 7.3:
C
A S u 4 .5 1( 6 2 6 b
f
0 .2 6 5
) 0 .8 1 9
For Fillet: r d
4 25
0 .1 6
D d
35 25
1.4
K t 1.4 5 Hence, from Fig.C.9: From Fig.7.9a, q=0.82 Equation (7.13b): K f 1 0 .8 2 (1.4 5 1 ) 1.3 7
Thus S e C f C r C s C t ( K1 ) S e ( 0 . 819 )( 0 . 89 )( 0 . 85 )( 1 )( 1 .137 ) 313 141 . 6 MPa '
f
SOLUTION (7.7) Table B.3: S u 4 7 0 M P a . We apply S e C f C r C sC t (
1 K f
'
)Se
where K
f
2 .5 ,
S e 0 .5 S u 2 3 5 '
C s 0 .7 ,
C r 0 .8 4 (Table 7.3),
C
f
AS
b u
M Pa
57 . 7 ( 470 )
0 . 718
Ct 1
0 . 696
Thus S e ( 0 .6 9 6 )( 0 .8 4 )( 0 .7 )(1)( 21.5 )( 2 3 5 ) 3 8 .5 M P a
SOLUTION (7.8) S e C f C r C s C t (1 K f ) S e '
(1)
We have D d
30 25
1 .2
1 d
2 25
0 .0 8
Hence, from Fig. C.12, K t 1 .9 5 Table B.4: S u 6 5 8 M P a ,
H
B
192
Equation (7.1): S e ' 0 .5 S u 3 2 9 M P a Figure 7.9a: r 2 m m ; Therefore, C
A S u 4 .5 1( 6 5 8 b
f
q 0 .8 2 0 .2 6 5
) 0 .8 0 8
(CONT.)
119
7.8 (CONT.) Table 7.3: C r 0 .8 7 K
1 q ( K t 1) 1 0 .8 2 (1 .9 5 1) 1 .7 8
f
Equation (7.9): C s 0 .8 5 Ct 1
(T 4 5 0 C ) o
Equation (1) results in then S e ( 0 .8 0 8 )( 0 .8 7 )( 0 .8 5 )(1)( 1 .71 8 )(3 2 9 ) 1 1 0 .4 M P a
SOLUTION (7.9) Refer to solution of Prob. 7.8, Now, Fig. C.11 give 1 .2 ,
D d
0 .0 8,
r d
Table B.3: S u 7 7 0 M P a ; r 2 mm;
Figure 7.9b:
K t 1 .5 H
B
229
q 0 .9 8
K
f
1 q ( K t 1) 1 0 .9 8 (1 .5 1) 1 .4 9
C
f
A S u 5 7 .7 ( 7 7 0
Also b
C r 0 .8 7 ,
0 .7 1 8
C t 0 .5 6 5,
) 0 .4 8 8
C s 0 .8 5
S e ' 0 .5 S u 3 8 5 M P a
Thus S e C f C r C s C t (1 K f ) S e '
( 0 .4 8 8 )( 0 .8 7 )( 0 .8 5 )( 0 .5 6 5 )( 1 .41 9 )(3 8 5 ) 5 2 .7 M P a
SOLUTION (7.10) Refer to definitions given by Eqs. (7.14). ( a ) m 12 ( m a x m in ) 12 (8 4 8 4 ) 0 a
1 2
(
R
Thus,
m ax
m in
m ax
m in
)
84 84
1 2
(8 4 8 4 ) 8 4 M P a
1
and
A
a m
12 0
(b) m
1 2
(8 4 1 4 ) 3 5 M P a
a
1 2
(8 4 1 4 ) 4 9 M P a
R
14 84
(c) m a R
0 84
0
A
42 42
1
1 6
1 2
,
A
7 5
(8 4 0 ) 4 2 M P a
120
SOLUTION (7.11) M
max
M
M ,
min
m
0,
a
651 , 898 M
32 M
( 0 . 025 )
3
Equation (7.7): C
f
AS
b u
4 . 51 ( 700 )
0 . 265
0 . 794
Also Ct 1
C r 0 .8 7
Table 7.3:
Equation (7.9):
C s 0 .8 5
Equation (7.1):
S e 0 .5 ( 7 0 0 ) 3 5 0 '
M Pa
From Fig. C.9, with D d 1 . 5 , r d 0 . 05 : K t 2 .1 q 0 .7 7
By Fig.7.9a: and K Hence
1 0 . 77 ( 2 . 1 1 ) 1 . 85
f
S e C f C r C sC t (
1 K f
'
)Se
( 0 .7 9 4 )( 0 .8 7 )( 0 .8 5 )(1)( 1 .81 5 )(3 5 0 ) 1 1 1 .1 M P a
Thus, Eq.(7.24): n
Se
6
1 1 1 .1 1 0 6 5 1 ,8 9 8 .6 M
1 .5
;
a
or M 1 1 3 .6 N m
SOLUTION (7.12) Table B.3: S u 4 7 0 M P a
H
B
131
S e 0 .4 5 S u 2 1 1 .5 '
M Pa
Tensile area through the hole: 2 ( R r ) t 2 (10 4 )( 2 . 5 ) 30 mm
m a
and
F 2A
F 2( 30 )
2
F 60
(1)
We have C r 0 .7 0 (Table 7.3)
C
A S u 4 .5 1( 4 7 0 ) b
f
Ct 1 0 .2 6 5
0 .8 8
C s 1 (axial loading)
From Fig.C.5: d D
0 .4 , q 0 .8
By Fig.7.9a: Hence
K
f
K t 2 .8
1 0 . 8 ( 2 . 8 1 ) 2 . 44
Therefore, S e C f C r C s C t (1 K f ) S e '
( 0 .8 8 )( 0 .7 )(1)(1)(1 2 .4 4 )( 2 1 1 .5 ) 5 3 .4
M Pa
(CONT.)
121
7.12 (CONT.) By Eq.(7.20):
m
4 7 0 1 .4
470
(1 )(
3 4 .2 5
(2)
M Pa
) 1
5 3 .4
From Eqs.(1) & (2): 34 . 25
F 60
or F 2 .0 6
kN
SOLUTION (7.13) Refer to solution of Prob. 7.12. We have, m a F 6 0 and C r 0 .8 7
(Table 7.3)
C t 1 0 .0 0 5 8 ( T 4 5 0 )
(Eq. 7.11)
1 0 .0 0 5 8 ( 5 4 5 4 5 0 ) 0 .4 5
Endurance limit becomes S e 5 3 .4 ( 00.8.77 )( 0 .4 5 ) 2 9 .8 7 M P a
The SAE criterion from Eq. (7.20) with S u S
m
S
a m
n
f S
f
Se
1
4 1 5 1 .2 (1)
415 2 9 .8 7
1
f
is
2 3 .2 2 M P a
Thus F 6 0 2 3 .2 2 ,
F 1 .3 9 k N
SOLUTION (7.14) Table B.3: S y 3 9 0 M P a
Refer to solution of Prob. 7.12. We have, m a F 6 0 and C r 0 .8 9
(Table 7.3)
C t 1 0 .0 0 5 8 ( T 4 5 0 )
(Eq. 7.11)
1 0 .0 0 5 8 ( 5 4 0 4 5 0 ) 0 .4 8
Endurance limit is then S e 5 3 .4 ( 00.8.79 )( 0 .4 8 ) 3 2 .5 9 M P a
The Soderberg criterion from Eq. (7.20) with S u S f :
m
Sy n
a
Sy
m
Se
1
3 9 0 2 .2 (1)
390 3 2 .5 9
1
1 3 .6 7 M P a
Thus F 60
1 3 .6 7 ,
F 8 2 0 .2 N
122
SOLUTION (7.15) A 10 ( 25 5 ) 200
Pm
1 2
mm
( 5 25 ) 15 kN ,
Pa 10 kN
( a ) Stress concentration factor is neglected for ductile materials under static loading. Thus
max
Pmax
3
25 ( 10 )
A
S
n
6
200 ( 10
y
m ax
)
580 125
125
MPa
4 .6 4
( b ) We now have 0 .2 ,
d D
S u 690
K t 2 .4 5 (Fig.C.5)
MPa ,
S
y
580
MPa ,
H
197
B
(Table B.3)
q 0 .8 3 (Fig 7.9a)
K
1 0 .8 3 ( 2 .4 5 1 ) 2 .2
f
Ct 1
C r 0 .8 4 (Table 7.3)
C s 1 (axial loading)
A S u 4 .5 1( 6 9 0 ) b
C
f
0 .2 6 5
0 .7 9 8
S e 0 .4 5 S u 3 1 0 .5 M P a '
Hence S e ( 0 .7 9 8 )( 0 .8 4 )(1)(1)( 21.2 )(3 1 0 .5 ) 9 4 .6 1 M P a
We have
m
Pm A
3
15 ( 10 )
200 ( 10
6
)
75 MPa ,
a
50 MPa
Equation (7.22) gives n 75
1 . 57
690 690
( 50 )
94 . 61
SOLUTION (7.16) Refer to Solution of Prob.7.15 (a)
n
S
y
m ax
580 125
4 .6 4
( b ) We now have Pm
1 2
[ 25 ( 5 )] 10 kN ,
Pa 15 kN
Hence
a
75 MPa ,
m
50 MPa
Thus n 50
690 690
1 . 16 ( 75 )
94 . 61
123
SOLUTION (7.17) D
A
4
2
( 0 .0 5 3 1 2 5 )
2
2 .2 1 7 1 0
4
( a ) Su 670 M Pa
H
3
m
1 9 7 (Table B.4)
B
P S u A 6 7 0 ( 2 .2 1 7 1 0
and
2
3
) 1 .4 8 5 M N
( b ) S e C f C r C s C t (1 K f ) S e '
where K
f
1
C r 0 .8 7 (Table 7.3)
C s 1 (axial load)
C
f
AS
4 . 51 ( 670 )
b u
0 . 265
0 . 804
C t 1 0 . 0058 ( 480 450 ) 0 . 826 S e ' 0 .4 5 S u 0 .4 5 ( 6 7 0 ) 3 0 1 .5
M Pa
S e ( 0 .8 0 4 )( 0 .8 7 )(1)( 0 .8 2 6 )(1 / 1)(3 0 1 .5 ) 1 7 4 .2
and
M Pa
Thus P A S e ( 2 .2 1 7 1 0
3
)(1 7 4 .2 ) 3 8 6 .2 k N
SOLUTION (7.18) Refer to Solution of Prob.7.17. We now have A d
2
4 ( 0 .0 5 )
2
4 1 .9 6 3 1 0
3
m
2
( a ) For a static fracture of a ductile material, the groove has little effect. Hence, P S u A ( 6 7 0 1 0 ) (1 .9 6 3 1 0 6
(b)
r d
0 .0 2 5,
D d
1 .0 6 2 5
3
) 1 .3 1 5 M N
K t 2 .6 (Fig.C.10)
From Fig.7.9a, with S u 6 7 0 M P a and r 1 .2 5 m m and
K
f
q 0 .7 5
1 q ( K t 1 ) 1 0 . 75 ( 2 . 6 1 ) 2 . 2
We now have S e 1 7 4 .2 2 .2 7 9 .1 8
M Pa
Thus P A S e (1 .9 6 3 1 0
3
)( 7 9 .1 8 1 0 ) 1 5 5 .4 3 k N 6
SOLUTION (7.19) From Table B.4: S u 519
MPa ,
S
y
353
MPa ,
H
B
149
By Eq.(6.20), S ys 0 . 577 S y 203 . 7 MPa ( a ) Thus, S y s 1 6 T d : 3
T
3
( 0 . 025 ) ( 203 . 7 10 16
6
)
624 . 9 N m
(CONT.)
124
7.19 (CONT.) ( b ) S es C f C r C s C t (1 K f ) S es '
where Ct 1
C r 0 .8 4 (Table 7.3)
C s 0 .8 5 (Eq.7.9)
C
f
AS
1 . 58 ( 519 )
b u
S e 0 .2 9 S u 1 5 0 .5 '
From Fig.C.8, with
From Fig.7.9b: q 0 .9 ,
K
f
0 . 929
D d
(Eq.7.7)
(Eq.7.4)
M Pa
0 . 05 and
r d
0 . 085
2
K t 1 . 72
1 q ( K t 1 ) 1.6 5
Hence S e s ( 0 .9 2 9 )( 0 .8 4 )( 0 .8 5 )(1)( 1 .61 5 )(1 5 0 .5 ) 6 0 .5
M Pa
Refer to Eq.(7.24): S es 16 T d . 3
Thus T
3
6
( 0 .0 2 5 ) ( 6 0 .5 1 0 ) 16
1 8 5 .6
N m
SOLUTION (7.20) Refer to Solution of Prob.7.19. ( a ) A d
2
4 ( 25 )
4 490 . 874
2
mm
2
, and S
y
353
MPa . Thus
P S y A 490 . 874 ( 353 ) 173 . 3 kN
( b ) We now have with r d
0 .0 5 ,
D d
2
From Fig. C.7: K t 2 .5 2 Figure 7.9a: q 0 .7 and K
f
1 q ( K t 1 ) 1 0 .7 ( 2 .5 2 1 ) 2 .0 6
By Eq.(7.3): S e 0 .4 5 S u 2 3 3 .6 M P a Hence S e ( 0 .9 2 9 )( 0 .8 4 )( 0 .8 5 )(1)( 2 .01 6 )( 2 3 3 .6 ) 7 5 .2 2
M Pa
Thus P S e A ( 7 5 .2 2 )( 4 9 0 .8 7 4 ) 3 6 .9 2
kN
SOLUTION (7.21) From Table B.3: S u 830
MPa ,
S
y
460
MPa ,
H
B
248
By Eq.(6.20), S
ys
0 . 577 S
y
265 . 4 MPa
Refer to Solution of Prob.7.19. (CONT.)
125
7.21 (CONT.) 3
d S ys
(a) T
3
6
( 0 .0 2 5 ) ( 2 6 5 .4 1 0 )
16
16
( b ) C f A S u 4 .5 1( 8 3 0 ) b
8 1 4 .2 N m
0 .2 6 5
0 .7 6
S e s 0 .2 9 S u 2 4 0 .7 M P a '
S es C f C r C s C t (1 K
and
'
f
) S es
( 0 .7 6 )( 0 .8 4 )( 0 .8 5 )(1)( 1 .61 5 )( 2 4 0 .7 ) 7 9 .1 6 M P a
Therefore T
3
d S es
3
6
( 0 .0 2 5 ) ( 7 9 .1 6 1 0 )
16
16
2 4 2 .9 k N m
SOLUTION (7.22) Su 658 M Pa
H
1 9 2 (Table B.4)
B
S e C f C r C s C t (1 K f ) S e '
where C r 0 .8 1 (Table 7.3)
C
AS
f
b u
C s 0 .8 5 (Eq.7.9)
1 . 58 ( 658 )
0 . 085
0 . 91
(Table 7.2)
S e 0 .5 S u 3 2 9 M P a '
0 . 125 ,
d D
and
K t 2 . 18
From Fig.7.9a: q 0 . 79 ,
K
f
Fig.C.13) 1 0 . 79 ( 2 . 18 1 ) 1 . 93
Thus S e ( 0 .9 1)( 0 .8 1)( 0 .8 5 )(1)( 1 .91 3 )(3 2 9 ) 1 0 6 .8 M P a
We have M
m
m
1 2
(1 5 8 .2 5 6 .5 ) 1 0 7 .3 5 N m , M
( D
3
m
32 ) ( dD
2
6)
M
1 0 7 .3 5
( 0 .0 2 5 )
3
3 2 [ ( 0 .0 0 3 1 2 5 ) ( 0 .0 2 5 )
2
6]
a
5 0 .8 5 N m
8 8 .8 3 M P a
(Fig.C.13)
a 8 8 .8 3( 15007.8.355 ) 4 2 .0 8 M P a
Equation (7.22): n 8 8 .8 3
658 658
1 .8 9 ( 4 2 .0 8 )
1 0 6 .8
SOLUTION (7.23) Refer to Solution of Prob.7.22. We now have Su 680 M Pa,
H
B
2 0 1 (Table B.3)
q 0 . 8 (Fig.7.9a)
Also
d D
0 .1 2 5
K
f
K t 2 .6 5 (Fig.C.13)
1 0 . 8 ( 2 . 65 1 ) 2 . 32
S e 0 .4 5 S u 0 .4 5 ( 6 8 0 ) 3 0 6 M P a '
(CONT.)
126
7.23 (CONT.) S e ( 0 .9 1)( 0 .8 1)( 0 .8 5 )(1)( 2 .31 2 )(3 0 6 ) 8 2 .6 4 M P a
and
We write Pm
1 2
(60 20) 40 kN ,
Pa 2 0 k N
A D
2
4 Dd
(Fig.C.13)
and
m
9 6 .9 1 M P a ,
40
( 0 .0 2 5 )
2
4 [( 0 .0 2 5 )( 0 .0 0 3 1 2 5 )]
Equation (7.22) is therefore n
1 .3 7
680 680
9 6 .9 1
( 4 8 .4 6 )
8 2 .6 4
SOLUTION (7.24) From Table B.2: S u 5 2 0 M P a ,
H
B
149
C r 0 .7
Table 7.3:
Equation (7.7) and Table 7.2: C
f
4 .5 1(5 2 0
0 .2 6 5
) 0 .8 6
S e ' 0 .5 S u 2 6 0 M P a
Thus
S e C f C r C s C t (1 K f ) S e '
( 0 .8 6 )( 0 .7 )(1)(1)(1 1)( 2 6 0 ) 1 5 6 .5 M P a
We have
32 M
a
Pm
m
d
( 0 .0 2 5 ) 3
4 (1 5 1 0 )
A
3 2 (1 5 0 )
3
( 0 .0 2 5 )
2
3
9 7 .7 8 M P a 3 0 .5 6 M P a
Goodman criterion
a
Se
m
Su
1 n
9 7 .7 8 1 5 6 .5
;
3 0 .5 6 520
1 n
Solving, n 1 .4 6
SOLUTION (7.25) Refer to solution of Prob. 7.22. We have S y 380 M Pa (Table B.4) C r 0 .8 9
(Table 7.3)
C t 1 0 .0 0 5 8 ( T 4 5 0 )
(Eq. 7.11)
1 0 .0 0 5 8 ( 4 5 5 4 5 0 ) 0 .9 7
Endurance limit is therefore S e 1 0 6 .8 ( 00 .8.8 91 )( 0 .9 7 ) 1 1 3 .8 3 M P a
The Soderberg criterion (by Table 7.4)
a
Se
m
Sy
1 n
Inserting the data, we have 4 2 .0 8 1 1 3 .8 3
8 8 .8 3 380
1 n
,
n 1 .6 6
127
a
9 6 .9 1
20 40
4 8 .4 6 M P a
SOLUTION (7.26)
1
pr t
, a
pa
m
pr
2
pm
2 3
,
2t
0.
3
pm 1260 kPa,
pa 840 kPa
.
Replace S u with S y in Eq.(7.30): S
[1
y
n
1e
or
1
1 2
] 2 1 e 0 .8 6 6 1 e
1 4
Sy
(1)
0 .8 6 6 n
Similarly, Eq.(7.25):
1e
S
1m
y
Se
(1260
1 t
1a
280 210
r t
S
( pm
840 )
y
pa )
Se
2380 t
(2)
By Eq.(1) and (2): 280 ( 0 .8 6 6 ) 2 .5
2380 t
t 0 .0 1 8 4 m = 1 8 .4 m m
,
SOLUTION (7.27) 1
pr t
,
2
pr 2t
,
3
0.
p m 1 . 7 MPa ,
p a 1 . 1 MPa
Refer to Solution of Prob.7.26: 1e
Su
(1)
0 .8 6 6 n
Use Eq.(7.25): Su
1e 1m
Se
1a
r t
( pm
Su Se
p a ) 3 0[1 .7
From Eqs.(1) and (2): 128,
350 0 .8 6 6 n
n 3.1 6
SOLUTION (7.28) b 0 .0 1 m ,
M
max
M
a
1 4
M
M PL 4.
( 20 )( 0 . 1) 0 . 5 N m ,
0 . 25 N m ,
m
6M
m
We have,
L 0 .1 m ,
m 2
bh
6 ( 0 .2 5 ) 0 .0 1 h
2
m
150 h
2
Equation (7.20) gives
m
980 4 (1 )
98
7 1 .0 1 M P a
1
40
Thus h
150 71 . 01 10
6
1 . 45 mm
128
M a
min
0
35 15
(1 .1)] 1 2 8 M P a
(2)
SOLUTION (7.29) From Solution of Prob.7.28, we have a m
150 n
.
2
(1)
Equation (7.20) by replacing S u with S y :
m
S
a
m
n
y
S
y
1
Se
Substituting the given data gives
m
620 4
62
(1 )(
6 0 .7 8 M P a
(2)
) 1
40
By Eqs.(1) and (2), 6 0 .7 8 ( 1 0 ) 6
150 h
2
or h 1 . 57 mm
SOLUTION (7.30) At fixed end M
M M
m
P L 2 .2 5 ( 0 .0 3 7 5 ) 0 .0 8 4 4 N m Hence
m ax
M
m
a
bh
2
Equation (7.20):
m ax
M
m in
2 6 ( 0 .0 4 2 2 )
6M
M
a
( 0 .0 0 3 1 2 5 ) h
m
2
1 0 5 0 1 .2 (1 )
1050
0 .0 4 2 2 N m
81 h
2
2 8 3 .7 8 M P a
1
504
Thus 2 8 3 .7 8 1 0
81 h
2
6
Pa,
h 0 .5 3 4 1 0
3
m = 0 .5 3 4 m m
SOLUTION (7.31) Refer to solution of Prob. 7.30. Equation (7.11): C t 1 0 .0 0 5 8 ( T 4 5 0 ) 1 0 .0 0 5 8 ( 4 7 0 4 5 0 ) 0 .8 8
and S e 5 0 4 ( 0 .8 8 ) 4 4 3 .5 2 M P a
Equation (7.20), replacing S u by S f , becomes
m
S
a
m
n
f S
f
Se
1
Inserting numerical values:
m
6 9 0 1 .5 (1)( 4 4639.50 2 ) 1
1 7 9 .9 9 M P a
Therefore, 81 h
2
1 7 9 .9 9 1 0 , 6
h 0 .6 7 1 1 0
129
3
m = 0 .6 7 1 m m
SOLUTION (7.32) P0
Pm Pa
Eq.(7.20):
,
2
m
M 850 4
85
(1 )
0 . 2 Pm 0 . 1 P0 ,
m
Pa Pm 1
a
3 6 .2 8 M P a
1
1 7 .5 6M
m
Also
bh
6 ( 0 .1 ) P0
m 2
0 . 0 0 5 1 0 0 (1 0
1.2 (1 0 ) P0 6
6
)
Thus 1 . 2 P0 36 . 28 ,
P0 30 . 23 N
SOLUTION (7.33) M
M
0 .2 P0 ,
m ax
m
P0
M
m in
0 .2 ( 0 .5 P0 ) 0 .1 P0
( 0 .2 0 .1 ) 0 .0 5 P0 ,
2
Equation (7.20): m
850 2 1 .5 ( 3 )
85
M
3 5 .6 2 9
a
P0 2
( 0 .2 0 .1 ) 0 .1 5 P0
M Pa
1
35 6M
m
Also
bh
m 2
6 ( 0 . 0 5 P0 )
0 . 0 0 5 1 0 0 (1 0
0 .6 (1 0 ) P0 6
6
)
Thus 0 . 6 P0 35 . 629 ,
P0 59 . 38 N
SOLUTION (7.34) I
bh 12
3
24 ( 4 )
3
128
12
mm
4
. Table B.3: S u 690
MPa
For a wide cantilever beam (see Secs 4.4 and 4.10, and Case 1 of Table A.8): 3
(1 ) 2
This gives Pmin
3 EI 0 . 91 L
0 .9 1
PL 3EI
3
min
3
PL 3EI
9
3 ( 200 10 )( 128 10 0 . 91 ( 0 . 3 )
3
12
)
( 0 . 01 )
and hence Pmax 62 . 52 N
31 . 26 N
Thus Pm 46 . 89 N ,
Pa 15 . 63 N and
46 . 89 ( 0 . 3 )( 2 10
3
m
a
. 63 219 . 8 ( 15 ) 73 . 27 MPa 46 . 89
0 . 128 ( 10
9
)
)
219 . 8 MPa
Eq. (7.22): Su
n
m
Su Se
a
219 . 8
690 690
1 . 63 ( 73 . 27 )
250
130
m
1
SOLUTION (7.35) Table B.4: S
500
y
MPa
Refer to Solution of Prob.7.34. Replacing S u by S y in Eq.(7.22): S
n
m
y
S
y
Se
219 . 8
a
500 500
1 . 36 ( 73 . 27 )
250
SOLUTION (7.36) Refer to solution of Prob. 7.34. We now have (Table 7.3) C r 0 .7 5 C t 1 0 .0 0 5 8 ( T 4 5 0 )
(Eq. 7.11)
1 0 .0 0 5 8 ( 4 9 0 4 5 0 ) 0 .7 7
Endurance limit becomes S e 2 5 0 ( 0 .7 5 )( 0 .7 7 ) 1 4 4 .4 M P a
The Gerber criterion (Table 7.4):
(
a
Se
) 1 2
m
Su
Substitution of the numerical values gives ( 2 1699.80 n ) 1 2
7 3 .2 7 n 1 4 4 .4
or 0 .5 0 7 4 n 0 .3 1 8 6 n 1 0 2
Solving this quadratic equation: 0 .5 0 7 4
n
2
( 0 .5 0 7 4 ) 4 ( 0 .3 1 8 6 )( 1 ) 2 ( 0 .3 1 8 6 )
or n 1 .1 5
SOLUTION (7.37) I
4
( 0 . 03 )
4
0 . 636 10
m
a
23 . 6 MPa
M
m
c
I
6
m
2000 ( 0 . 5 )( 0 . 03 )
0 . 636 10
6
4
Pm 4 kN ,
,
Pa 2 kN
47 . 2 MPa
Eq. (7.16), with replacing S u by S y : S
y
n
m
S
y
Se
a;
280 2 .5
4 7 .2
280 150 K
f
( 2 3 .6 )
(CONT.)
131
7.37 (CONT.) Solving K
f
1.4 7 .
From Fig.7.9a: q=0.8 and
K
f
1 q ( K t 1 );
1.4 7 1 0 .8 ( K t 1 ),
K t 1.5 9
Then from Fig. C.9: K
t
r d
1.5 9 9 60
0 .1 5
D
3
d
and D 3 d 3 ( 60 ) 180
mm
SOLUTION (7.38) Table B.4: S u 634 I
bh 12
3
MPa ,
0 . 02 ( 0 . 06 )
3
1 . 8 kN m
D d
120 60
r d
4 60
192
B
0 . 36 10
12
m
M
H
M
6
m
4
1 . 2 kN m
a
Figure C.2: 2
0 .0 6 7
K t 2 .1
Figure 7.9a: q 0 . 82 and K
f
1 0 .8 2 ( 2 .1 1) 1 .9 0 2 .
Se
400 1 .9 0 2
2 1 0 .3 M P a
We have
m
a
150 ( 11 .. 28 ) 100
M
m
c
I
1800 ( 0 . 03 ) 0 . 36 10
6
150
MPa
MPa
Equation (7.22): Su
n
m
Su Se
a
150
634 634
1 .4 ( 100 )
210 . 3
SOLUTION (7.39) Refer to solution of Prob. 7.38. We now have M
m
1 2
(2200 200) 1200 N m
M
a
1 2
(2200 200) 1000 N m
Thus, m 1 5 0 ( 11 28 00 00 ) 1 0 0 M P a ,
a 1 0 0 ( 11 02 00 00 ) 8 3 .3 M P a
(CONT.)
132
7.39 (CONT.) The Gerber criterion (by Table 7.4).
(
a
Se
m
Su
) 2
1 n
Inserting the numerical values, results in 8 3 .3 n 2 1 0 .3
( 160304n ) 1,
0 .3 9 6 1 n 0 .0 2 4 9 n 1 0
2
2
Solving this quadratic equation: 0 .0 2 4 9
n
2
( 0 .0 2 4 9 ) 4 ( 0 .3 9 6 1 )( 1 ) 2 ( 0 .3 9 6 1 )
or n 1 .5 6
SOLUTION (7.40) Refer to solution of Prob. 7.38. We now have S
434 M Pa
f
(from Table B.4)
C t 1 0 .0 0 5 8 ( 4 7 5 4 5 0 ) 0 .8 6
(by Eq. 7.11))
1 0 .0 0 5 8 ( 4 9 0 4 5 0 ) 0 .7 7
C r 0 .8 7
(Table 7.3)
and S e 2 1 0 .3( 0 .8 7 )( 0 .8 6 ) 1 5 7 .3 M P a
Also M
m
1 .5 k N m
M
a
0 .5 k N m
and m 1 5 0 ( 11 58 00 00 ) 1 2 5 M P a ,
a 1 0 0 ( 1520000 ) 4 1 .6 7 M P a
The SAE criterion from Table 7.4:
a
Se
m
S
f
1 n
Introducing the data, we obtain 4 1 .6 7 1 5 7 .3
125 434
1 n
Solving, n 1 .8 1
SOLUTION (7.41) See Tables 7.5, 6.2, and 6.1. We have A 5 .6 1 0
12
n 3 .2 5
K c 77 M Pa
m
S y 690 M Pa
1 .1 4 for a w 0 .4 (Case B, Table 6.1)
(CONT.)
133
7.41 (CONT.) Note that values of a and t satisfy Table 6.2. The stresses are:
m ax
Pm a x
2wt
3
9 5 0 (1 0 ) 2 ( 0 .0 8 )( 0 .0 3 4 )
1 7 4 .6
m in
7 9 .4 M P a
The range stress is then 1 7 4 .6 7 9 .4 9 5 .2 M P a The final crack length at fracture, by Eq. (7.39): af
1
Kc
(
) 2
m ax
( 1 .1 4 7 17 7 4 .6 ) 0 .0 4 7 6 m 4 7 .6 m m 2
1
Equation (7.41) results in then N
0 .0 4 7 6 5 .6 (1 0
12
0 .6 2 5
0 .0 3 2
0 .6 2 5
) ( 0 .6 2 5 ) [ (1 .7 7 ) (1 .1 4 ) ( 9 5 .2 ) ]
3 .2 5
2 0 , 4 5 4 cycles
For a period of 20 second, approximate fatigue life L is L
20 ,454 ( 20 ) 60 (60 )
1 1 3 .6 h
End of Chapter 7
134
CHAPTER 8
SURFACE FAILURE
SOLUTION (8.1) From Table 8.2, we have K 10
2
(steel on steel) 5
K 2 10 K 10
(lead on steel)
3
(brass on steel)
7
K 10 (polyurethylene on steel) The Brinell hardness numbers are in MPa: Steel
Su 58 10 H
s
3
58 10
(Table B.1)
psi 3
500 116 Bhn
(Eq. 2.22)
1 1 6 ( 9 .8 1) 1 1 3 8 M P a
Lead H l 3(9 .8 1) 2 9 .4 M P a
Brass H b 8 (9 .8 1) 7 8 .5 M P a
Polyurethylene H
p
7 (9 .8 1) 6 8 .7 M P a
Thus referring to Eq. (a) of Sec. 8.5: Lead
H K 2 9 .4 (1 0 ) 2 (1 0
Brass
H K 7 8 .5 (1 0 ) 1 0
6
6
3
5
) 1470 G Pa
7 8 .5 G P a
Polyurenthylene H K 6 8 .7 (1 0 ) 1 0 6
7
687, 000 G Pa
2
Steel H K 1 1 3 8 (1 0 ) 1 0 1 1 3 .8 G P a Comment: Polyurethylene gives much longer life than any other material. 6
SOLUTION (8.2) Through the use of Eq. (8.2), wear volume is V K
WL H
K 10
4
where W 40 N
(by Table 8.3),
H b 9 .8 1( 6 0 ) 5 8 9 M P a , L 80 (2
mm c y c le
) (1 5 0 0
c y c le s s ix m o n th s
H
s
) 240, 000 m m
Therefore Vb
(1 1 0
Vs
(1 1 0
4
)( 4 0 )( 2 4 0 ) 6
5 8 9 (1 0 ) 4
)( 4 0 )( 2 4 0 ) 6
1 0 3 0 (1 0 )
9 .8 1(1 0 5 ) 1 0 3 0 M P a
1 .6 3 m m
3
0 .9 3 m m
3
135
SOLUTION (8.3) From Eq. (8.2), we have V K
WL H
Here K 3 10
H
s
H
f
5
W 45 N
(by Table 8.3),
9 .8 1(1 6 0 ) 1 5 6 9 .6 M P a 9 .8 1( 4 5 0 ) 4 4 1 4 .5 M P a
L (3 7 .5 2
mm c yc le
)( 6 0 0 0
c yc le s m o n th
)(1 2 m o n th s ) 5 , 4 0 0 , 0 0 0 m m
Hence, V s (3 1 0
V f ( 4 .6 )
5
)
( 4 5 )( 5 , 4 0 0 ) 6
1 5 6 9 .6 (1 0 )
1 5 6 9 .6 4 4 1 4 .5
4 .6 m m
1 .6 4 m m
3
3
SOLUTION (8.4) Follow procedure of Example 8.1. We have Copper disk: 110 Vickers hardness, V c 0 .9 8 m m Aluminum pin: 95 Brinell hardness, V a 4 .1 m m
3
3
Contact force: W 2 5 N at a radius R 2 4 m m Test duration: t 1 8 0 m in at a speed n 1 2 0 r p m Total length of sliding is then L 2 R n t
2 ( 2 4 )(1 2 0 )(1 8 0 ) 3 .2 6 1 0
6
mm
The hardness of pin and disk are H a 9 .8 1(9 5 ) 9 3 2 M P a H c 9 .8 1(1 1 0 ) 1 0 7 9 M P a
From Eq. (8.2); K V H W L . Thus 4 .1 ( 9 3 2 )
Ka
2 5 ( 3 .2 6 1 0 )
Kc
2 5 ( 3 .2 6 1 0 )
6
0 .9 8 (1 0 7 9 ) 6
4 .9 6 1 0 1 .3 1 0
5
5
SOLUTION (8.5) Follow procedure of Example 8.1. Given data: Steel disk: 248 Brinell hardness, V s 0 .9 8 m m Copper pin: 85 Vickers hardness, V c 4 .1 m m
3
3
Contact force: P 3 5 N at a radius R 2 4 m m Test duration: t 1 8 0 m in at a speed n 1 2 0 r p m Total length of sliding equals L 2 R n t
2 ( 2 4 )(1 2 0 )(1 8 0 ) 3 .2 6 1 0
6
mm
(CONT.)
136
8.5 (CONT.) Hardness of pin and disks are H c 9 .8 1(8 5 ) 8 3 4 M P a ,
H s 9 .8 1( 2 4 8 ) 2 4 3 3 M P a
Using Eq. (8.2); K V H W L . Therefore 4 .1 ( 8 3 4 )
Kc
3 5 ( 3 .2 6 1 0 )
Ks
3 5 ( 3 .2 6 1 0 )
3 10
6
0 .9 8 ( 2 4 3 3 )
5
2 .1 1 0
6
5
SOLUTION (8.6) Case B ( 1st column ), Table 8.4 with r1 r2 , (a)
1 E
1 E
a 0 . 88
3
( b ) p 0 1 .5
0 . 88
r E
F
1 .5
P
a
m
2 E
2
1 r
210 ( 10
500 2
( 0 . 624 ) ( 10
2 r
0 . 624
0 . 15
500
3
1 r
6
)
E1 E 2 .
9
)
mm
613 . 1 MPa
( c ) Equations (8.4) at z=0:
x
z
p 0 613 . 1 MPa
yz
p 0 [( 1 )
y
xz
1 2
(
1 2
]
z)
x
p0 2
1 2 2
p 0 0 . 8 ( 613 . 1 ) 490 . 5 MPa
( 0 . 8 1 ) 61 . 31 MPa
SOLUTION (8.7) Refer to Table 8.4 (Case C, Column 2). n
( a ) a 0 .8 8 3 F where
2 E 1(1 0 n
1 r1
1 r2
11
1 0 .0 0 6
F 2 .2 k N
)
1 0 .0 0 6 0 5
)
1
1 .3 7 7 4
Thus
and
a 0 .8 8[
2 2 0 0 (1 0
p o 1 .5
F
11
1 .3 7 7 4
a
2
1 .5
]
3
2 .2 1 6 2 m m 3
2 .2 (1 0 ) 2
( 2 .2 1 6 2 ) (1 0
Since 2 1 4 M P a 4 3 6 M P a
6
214 M Pa
)
OK
( b ) 0 .7 7 5 F n 3
2
2
0 .7 7 5[( 2 2 0 0 ) (1 0 2
11
2
1
) (1 .3 7 7 4 )]
3
0 .0 0 6 7 7 m m
( c ) From Fig. 8.9a, m a x is at about z 0 .4 a and m a x p o 0 .3 2 . Therefore z 0 .4 ( 2 .2 1 6 2 ) 0 .8 8 6 m m
m a x 0 .3 2 ( 2 1 4 ) 6 8 .4 8 M P a
137
SOLUTION (8.8) See Table 8.4 (Case B, Column 3). We have 11
2 E 1(1 0
( a ) a 1 .0 7 6
m
),
1 0 .0 2 5
1 0 .0 7 5
5 3 .3 3 3 3
Lm
F
2 2 0 (1 0
11
)
1 .0 7 6[ ( 0 .0 2 5 )( 5 3 .3 3 3 3 ) ]
1
4 .3 7 0 7 (1 0
2
5
) m 0 .0 4 3 7 m m
Then po
2
F aL
220
2
5
( 4 .3 7 0 7 1 0
)( 0 .0 2 5 )
1 2 8 .2 M P a 3 2 0 M P a
( b ) From Fig. 8.9b,
OK
is at z 0 .7 5 a and
yz ,m ax
p o 0 .3 .
yz ,m ax
Hence, z 0 .7 5 ( 0 .0 4 3 7 ) 0 .0 3 2 8 m m
yz ,m ax
0 .3 (1 2 8 .2 ) 3 8 .4 6 M P a
SOLUTION (8.9) Refer 2nd column of C, Table 8.4. E 1 E 2 E 210
GPa
1
r1 7 mm
r 2 45 mm
F 1 200
kN
m
Hence
( a ) a 1 . 076 [ ( b ) p0
2
F aL
2 E
2 9
2 1 0 (1 0 )
105 ( 10
)( 120 . 635 )
3
0 . 135 ( 10
n
,
] 2 0 . 135
200 ( 10 )
2
9
1 0 5 (1 0 )
1 0 .0 0 7
1 0 .0 4 5
1 2 0 .6 3 5
1
3
200 ( 10 ) 9
3
mm
943 . 1 MPa
)
SOLUTION (8.10) We have r1 ' . Substitution of the numerical values into Eqs.(8.6) through (8.10)gives m A
4 2 0 .2 5
2 0 .0 3 4 9
n
0 .0 3 4 9
1 0 .0 0 9 3 7 5
5 7 .3 0 6 6 ,
B
1 2
9
4 ( 2 0 0 1 0 ) 3 ( 0 .9 1 )
2 9 3 .0 4 (1 0 ) 9
1
[( 0 .0 019 3 7 5 0 ) ( 0 ) ( 0 ) ] 2 5 3 .3 3 3 3 2
2
2
From Eq.(8.9) cos
5 3 .3 3 3 3 5 7 .3 0 6 6
2 1 .4 6
0 .9 3 0 7 ,
o
Then, interpolating in Table 8.5, we have c a 3 .6 2 5 1 and c b 0 .4 2 0 4 . Through the use of Eq.(8.7): 3
a 3 .6 2 5 1[
2 .2 5 1 0 ( 0 .0 3 4 9 )
b 0 .4 2 0 4[
2 .2 5 1 0 ( 0 .0 3 4 9 )
9
2 9 3 .0 4 (1 0 ) 3
9
2 9 3 .0 4 (1 0 )
1
] 3 2 .3 3 7 1 m m 1
] 3 0 .2 7 1 0 m m
(CONT.)
138
8.10 (CONT.) The maximum contact pressure, by Eq.(8.6), is thus 2 .2 5 1 0
p 0 1 .5
3
1 6 9 6 .2 M P a
( 2 .3 3 7 1 )( 0 .2 7 1 0 )
Thus stress may be satisfactory for the steel used. SOLUTION (8.11) In this case: r2 r2 ' as well as r1 ' . Substituting w=L=6.25 mm. and given data into the equations on the second column of Case A of Table 8.4: a 1 .0 7 6 p0
Thus
2
F aw
3
2 .2 5 1 0 0 .0 0 6 2 5
r1 1 .0 7 6
F w
3
2 ( 2 .2 5 1 0 )
(1 .9 7 6 7 1 0
4
)( 0 .0 0 6 2 5 )
( 0 .0 0 9 3 7 5 )(
2 2 0 0 1 0
9
) 1 .9 7 6 7 1 0
4
m
1 1 5 9 .4 2 3 M P a
Comments: With the flat-faced follower inaccuracies in machining, misalignment, and shaft deflection may cause edge contact and hence higher actual stress than 1159.423. This cannot occur with a spherical face follower, however.
SOLUTION (8.12) Refer to Table 8.4 (Case A, Column 2). r2 and 2 E a 0 .8 8 3 F r1
Thus
0 .8 8[3 6 0 ( 0 .0 5 6 2 5 )(
(a)
p o 1 .5 1 .5
1
2 2 0 0 1 0
)]
9
3
5 .1 6 7 6 1 0
4
m = 0 .5 1 6 7 6 m m
F
a
2
360
( 5 .1 6 7 6 1 0
4
)
2
6 4 3 .7 M P a
2 2 r1
( b ) 0 .7 7 5 3 F
0 .7 7 5[(3 6 0 ) ( 2
4 .7 5 1 1 1 0
6
1
2
2 2 0 0 1 0
9
) ( 0 .0 516 2 5 )]
m = 4 .7 5 1 1 0
3
3
mm
SOLUTION (8.13) See Table 8.4 (Case C, Column 3). 2 E and n 1 r1 1 r2 1 0 .0 1 5 1 0 .0 1 6 2 5 5 .1 2 8 2 ( a ) a 1 .0 7 6
F Ln
9 .0 1 5 1 0 po
2F aL
1 .0 7 6[ 4
3
1 3 .5 1 0 ( 2 )
1
9
( 2 0 0 1 0 )( 0 .0 3 7 5 )( 5 .1 2 8 2 )
]
2
m = 0 .9 0 1 5 m m 3
2 (1 3 .5 1 0 )
( 0 .9 0 1 5 )( 0 .0 3 7 5 )
2 5 4 .2 M P a
2 5 4 .2 M P a 3 1 7 M P a
OK. (CONT.)
139
8.13 (CONT.) ( b ) Equations (8.5) at z=0: x 2 p o 2 ( 0 .3)( 2 5 4 .2 ) 1 5 2 .5 M P a
y
p o 2 5 4 .2 M P a
z
p o 2 5 4 .2 M P a
xy
1 2
( 1 5 2 .5 2 5 4 .2 ) 5 0 .9 M P a
xz
1 2
(
x
z ) 5 0 .9 M P a
1 2
(
y
z) 0
yz
( c ) From Fig. 8.9b,
yz ,m ax
is at z 0 .7 5 a and
p o 0 .3 .
yz ,m ax
So, z 0 .7 5 (9 .0 1 5 5 1 0
4
) 6 .7 6 2 1 0
4
m = 0 .6 7 2 m m
0 .3 ( 2 5 4 .2 ) M P a 7 6 .2 6 M P a
yz ,m ax
SOLUTION (8.14) Refer to Table 8.4 (Case C, Column 2). 2 E 2 (200 10 ) 1 10 9
n
( a ) a 0 .8 8 3 F
1 r1 n
F
a
2
1 0 .0 5
1 0 .0 5 5
0 .8 8[5 6 0
1 .2 8 0 3 6 1 0
( b ) p o 1 .5
1 r2
3
1 .5
11
1 .8 1 8 2 11
1 1 0 1 .8 1 8 2
1
]
3
m = 1 .2 8 0 3 6 m m 560
(1 .2 8 0 3 6 1 0
3
)
2
1 6 3 .1 M P a
( c ) 0 .7 7 5 F n 0 .7 7 5[(5 6 0 ) (1 1 0 3
2 .9 8 3 1 0
2
2
2
6
m = 2 .9 8 3 1 0
3
11
1
2
mm 3
( d ) At z 0 .3 7 5 a 0 .3 7 5 (1 .2 8 0 3 6 1 0 ) 0 .4 8 0 1 1 0
yz ,m ax
3
) (1 .8 1 8 2 )]
3
m = 0 .4 8 0 1 m m (Fig. 8.9a):
0 .3 2 p o 0 .3 2 (1 6 3 .1) 5 2 .1 9 2 M P a
SOLUTION (8.15) See Table 8.4 (Column 2). We have 2 E 2 ( 2 0 0 1 0 ) 1 1 0 9
11
( a ) Case A: r2 . Thus a 0 .8 8 3 F r1 0 .8 8[(5 6 0 )( 0 .0 5 )(1 1 0
5 .7 5 7 1 1 0 p o 1 .5
F
a
2
4
1 .5
11
1
)]
3
m 560
( 5 .7 5 7 1 1 0
4
)
2
8 0 6 .7 1 8 M P a
(CONT.)
140
8.15 (CONT.) ( b ) Case B: r1 r2 1 0 0 m m , m
a 0 .8 8 3 F p o 1 .5
0 .8 8[(5 6 0 )
1 .5
F
a
2
m 2 r 2 0 .1 2 0
560
( 5 .7 5 7 1 0
4
)
11
1 1 0 20
2
1
]
3
5 .7 5 7 1 0
4
m = 0 .5 7 5 7 m m
8 0 6 .7 M P a
SOLUTION (8.16) Use Table 8.4 (Case A, Column 3). 2 E 2 (200 10 ) 1 10 9
a 1 .0 7 6
11
3
r 1 .0 7 6[ 1 .80.11 0 ( 0 .0 1 2 5 )(1 1 0
F L
5 .1 0 3 9 1 0
5
11
1
)]
2
m = 0 .0 5 1 0 4 m m 4
2 a 1 .0 2 0 7 8 1 0
m = 0 .1 0 2 1 m m
Therefore
2 r1
( 13 ln
0 .5 7 9 F EL
3
0 .5 7 9 (1 .8 1 0 ) 9
( 2 0 0 1 0 )( 0 .1 )
a
) 2 ( 0 .0 1 2 5 )
[ 13 ln
3 .4 0 1 4 1 0
7
5
( 5 .1 0 3 9 1 0
)
]
m = 3 .4 0 1 4 1 0
-4
mm
SOLUTION (8.17) We have r1 r2 r . Thus, Eqs. (8.8) and (8.9) become m
2 r 2 ( 0 .2 2 ) 0 .4 4
4 (1 r ) (1 r )
(1 r ) (1 r )
cos
(1 r ) (1 r )
90
0,
o
From Table 8.5 it can be concluded that surface of contact has a circular boundary: c a c b 1 . Then n
9
4 ( 2 0 6 1 0 ) 2
3 (1 0 .2 5 )
2 .9 2 9 7 8 (1 0 ) 11
Eqs. (8.7) gives, a b 1[
3
2 (1 0 ) ( 0 .4 4 ) 2 .9 2 9 7 8 1 0
11
1 3
]
1 .4 4 3 m m
Hence, Eq. (8.6): p o 1 .5
6
2 (1 0 )
(1 .4 4 3 )(1 0
3
)
6 6 1 .8 M P a
SOLUTION (8.18) We have r1 0 .5 m , r2 0 .3 5 m , r1 ' , r2 ' , and 90 . Thus, using Eqs.(8.6) through (8.10): o
m A
4 1 0 .5 2 m
,
1 0 . 35
n
0 . 824
B
1 2
( r1 1
1 r2
9
4 ( 206 10 ) 3 ( 1 0 . 09 )
3 . 0183 (10
11
)
)
(CONT.)
141
8.18 (CONT.) cos
B A
1 0 .5 1 0 .3 5 1 0 .5 1 0 .3 5
7 9 .8 6
0 .1 7 6 ,
Interpolating from Table 8.5: c a 1.1 2 8 , 3
5 ( 10 )( 0 . 824 )
a 1 . 128 [
3 . 0183 ( 10
p 0 1 .5
)
3
5 ( 10 )( 0 . 824 )
b 0 . 893 [
Thus
11
3 . 0183 ( 10
11
c b 0 .8 9 3 . Hence
1
] 3 2 . 6958
mm
1
] 3 2 . 1342
mm
3
5 ( 10 )
1 .5
F
ab
)
o
( 2 . 6958 2 . 1342 10
6
414 . 9 MPa
)
SOLUTION (8.19) We now have F
300
600 2
S
r2 5 .2 m m ,
and r2 ' 3 0 m m .
r1 r1 ' 5 m m ,
and
1500
N for each row,
y
MPa
We proceed as in Example 8.3. (a) m
4
2 0 . 005
1 0 . 0052
1 0 . 030
0 . 0229 ,
n 293 . 0403 10
9
and A
87 . 3362
2 0 . 0229
B
1 2
[( 0 )
(
2
1 0 . 0052
1
1 0 . 03
)
2
2 ( 0 ) ] 2 79 . 4872 2
Using Eq.(8.9), cos
Table 8.5: c a 3.3 3 3 0 Hence
b 0 . 4441 [
o
c b 0 .4 4 4 1 1
300 0 . 0229
a 3 . 3330 [
24 . 48
0 . 9101
79 . 4872 87 . 3362
293 . 0403 10
9
mm
1
300 0 . 0229 293 . 0403 10
] 3 0 . 9539
9
] 3 0 . 1271
mm
1 ,181
MPa
Then p 0 1 .5
(b) n
300
( 0 . 9539 0 . 1271 10
6
)
Fy
(1)
F
It is required that S y 1.5 F y a b S
y
1 .5
or
Fy
3
cacb (m n ) Substituting the data given
2 3
1 .5
1500 ( 10 ) 6
3
( 3 . 3330 0 . 4441 )(
Fy
Solving F y 614
N
Equation (1): n
614 300
2
0 . 0229 293 . 0403 10
2 .0 5
142
9
)
3
SOLUTION (8.20) Refer to Table 8.4 (Case A, Column 3). a 1 .0 7 6
r1
F L
where F 5 kN ,
2 E
L 25 m m ,
2 0 6 1 0
r1 5 0 0 m m
0 .9 7 1(1 0
2 9
11
)
Therefore 3
( 5 1 0 )
a 1 .0 7 6[
( 0 .5 ) ( 0 .9 7 1 1 0
0 .0 2 5
11
1
]
1 .0 6 0 3 m m
2
Hence, po
2
3
2 ( 5 1 0 )
F aL
(1 .0 6 0 3 1 0
3
120 M Pa
)( 0 .0 2 5 )
SOLUTION (8.21) See Table 8.4 (Case C, Column 3).
2
F aL
F Ln
( a ) a 1 .0 7 6 ( b ) po
2 E
1
1 .0 7 6[
1 0 .0 1
3 0 0 1 0
3
9
1 0 0 (1 0 )( 8 4 )
1 0 .0 6 2 5
1
0 .2 0 3 3 m m
]
2
84,
0 .2 0 3 3
( c ) From Fig. 8.9b, z 0 .7 5 a 0 .1 5 2 5 m m
F L 300 kN
9 3 9 .4 M P a
300
2
n
,
9
1 0 0 (1 0 )
yz ,m ax
and
0 .3 p o 0 .3 (9 3 9 .4 ) 2 8 2 M P a
SOLUTION (8.22) Given: r1 r1 ' 0 .0 2 m , Equation (8.8), m
4 1 0 .0 2
1 0 .0 2
1 0 .0 2 5
1 0 .1
r2 0 .0 2 5 m , 0 .0 8 ,
n
0 .3,
r2 ' 0 .1 m , 9
4 ( 2 0 0 1 0 ) 2
3 (1 0 .3 )
E 200 G Pa
2 .9 3 (1 0 ) 11
Also A
2 0 .0 8
cos
1 15 25
2 m
2 5,
B
5 3 .1 3 0 1
1 2
[( 0 ) ( 2
1 0 .0 2 5
1 0 .1
2
o
From Table 8.5, we find c a 1 .6 6 4 5
c b 0 .6 6 4 2
The semiaxes are then a 1 .6 6 4 5[
(1 2 0 0 ) ( 0 .0 8 )
b 0 .6 6 4 2[
(1 2 0 0 ) ( 0 .0 8 )
2 .9 3 (1 0
11
2 .9 3 (1 0
11
)
)
1
]
3
0 .0 0 1 1 4 7 m 1 .1 5 m m
3
0 .0 0 0 4 6 m 0 .4 6 m m
1
]
Thus p o 1 .5
F
ab
1 .5
1200
(1 .1 5 0 .4 6 )(1 0
6
)
1083 M Pa
143
1
) 2 ( 0 )] 2 1 5
SOLUTION (8.23) We have r1 r1 ' 0 .0 1 8 m ,
r2 0 .0 2 m ,
r2 ' 0 .1 m ,
0 .3,
E 200 G Pa
Equations (8.8), m
4 1 0 .0 1 8
1 0 .0 1 8
1 0 .0 2
1 0 .1
0 .0 7 8 3,
n
9
4 ( 2 0 0 1 0 ) 2
3 (1 0 .3 )
2 .9 3 (1 0 ) 11
Equations(8.10) A
2 m
B
2 5 .5 4 2 8 1 2
cos
[( 1
1 0 .0 2
20 2 5 .5 4 2 8
1 0 .1
2
3 8 .4 6
From Table 8.5, c a 2 .2 1 6 4 Equations (8.7): a 2 .2 1 6 4[
b 0 .5 5 5 6[
1
) ]2 20
c b 0 .5 5 5 6
( 9 0 0 ) ( 0 .0 7 8 4 ) 2 .9 3 (1 0
11
)
( 9 0 0 ) ( 0 .0 7 8 3 ) 2 .9 3 (1 0
11
o
)
1
]
3
0 .0 0 1 3 7 9 m 1 .3 8 m m
3
0 .0 0 0 3 5 m 0 .3 5 m m
1
]
It follows that p o 1 .5
F
ab
1 .5
900
(1 .3 8 0 .3 5 )(1 0
6
)
890 M Pa
End of Chapter 8
144
Section III
APPLICATIONS
CHAPTER 9
SHAFTS AND ASSOCIATED PARTS
SOLUTION (9.1) T
9549 kW n
9 5 4 9 (1 0 )
3 8 .1 9 6 N m
2500
and
Also
c 3
2
a ll L
T
a ll
T GJ
3 8 .1 9 6 6
1 5 0 1 0
3
c 7 .8 6 4
,
2 ( 1 8 0 )
;
3 8 .1 9 6 ( 2 ) 9
8 0 (1 0 ) c
4
mm c 9 .6 6 m m
,
A 20-mm dia. shaft would probably be selected.
SOLUTION (9.2) Refer to Example 9.1. We now have S u 3 6 5 M P a (T a b le B .2 ), n 2 .5 , and S y is replaced by S u . Equation (9.9) gives D [ 1 6Sn ( M u
c
M
2 c
Tc ] 2
1 3
Substituting the given data, D [
1 6 ( 2 .5 ) 6
( 3 6 5 1 0 )
( 2 0 2 .5
( 2 0 2 .5 ) (1 6 2 ) ] 2
2
1 3
0 .0 2 5 2 5 6 m = 2 5 .2 6 m m
SOLUTION (9.3) T AC
(a)
9 5 4 9 ( 7 .5 )
9549 kW n
1200
5 9 .6 8 N m
and 2
c 3
T
a ll
5 9 .6 8 6
2 1 0 (1 0 ) 2
c 7 .1 2 6
,
D A C 1 4 .2 5
mm
Similarly TC B 2
c
9 5 4 9 ( 2 2 .5 ) 1200
3
179
179
c 1 0 .2 8
,
6
2 1 0 (1 0 ) 2
N m
( b ) We have T L G J , with J c Thus AC
5 9 .6 8 ( 2 )
4
mm
D C B 2 0 .5 6
2.
0 .3 5 9 6 ra d 2 0 .6
4
9
( 0 .0 0 7 1 2 5 ) ( 8 2 1 0 )
2
BC
179 ( 4 )
4
0 .4 9 8 ra d 2 8 .5 3 9
( 0 .0 1 0 2 8 ) ( 8 2 1 0 )
2
Hence A B B C A C 7 .9 3
o
145
o
o
mm
mm
SOLUTION (9.4) G a 28 GPa ,
Table B.1:
S u 520
Table B.3:
MPa ,
We have a s but L a L s , Js Ja
Ga
0 . 354
Gs
G s 79 GPa S
y
440
MPa
Ta Ts
where J
D
32
4
Hence Ds Da
Wa
Also
Ws
0 .7 7 1
or
Va a
( D a 4 ) a
V s
s
D a 1. 2 9 6 D s
4
2
(Ds
4 )
s
2
Da a 2
Ds
s
a 2 .8 M g m , 3
Table B.1:
s 7 .8 6 M g m
3
Thus Wa Ws
2
(1. 2 9 6 D s ) ( 2 . 8 )
2 Ds
0 .5 9 8
( 7 .8 6 )
SOLUTION (9.5) ( a ) Use Eq.(9.5): 6
2 6 0 (1 0 ) n
3
4 (1 0 )
( 0 .1 )
3
1
[( 8 5 5 0 0 .1 ) ( 8 8 ) ] 2 2
2
n 2 .6 1
or
( b ) Apply Eq.(9.6): 6
2 6 0 (1 0 ) n
3
4 (1 0 )
( 0 .1 )
3
1
[ ( 8 5 5 0 0 .1 ) 4 8 ( 8 ) ] 2 2
2
or n 2 .8 6
SOLUTION (9.6)
F y 0 : 0 and
M
A
0 yield
960
Mz (N m)
480 x A
My (N m) A
R By 8 0 0 N
R Az 5 0 0 N
R Bz 1 0 0 0 N
M
960 300
2
C
1006 N m
M
600 480
2
D
7 6 8 .4 k N m
B
C 300
R Ay 1 6 0 0 N
600
2
T 1 .2 5 k N m
x D
2
B
Critical section is at C.
(CONT.)
146
9.6 (CONT.) We have 1 6T
(a)
n
32M
m ax
D
m ax
Sy
( b ) m ax n
D
(
1 6 (1, 2 5 0 )
( 0 .0 4 5 )
( 0 .0 4 5 )
3
3
6 9 .9 M P a
1 1 2 .5 M P a
3 .0 7
345 1 1 2 .5
3 2 (1, 0 0 6 )
3
3
) 2
2
(5 6 .2 5 ) ( 6 9 .9 ) 2
8 9 .7 M P a
2
2
m ax
S ys
2 .3 4
210 8 9 .7
SOLUTION (9.7) y
0 . 6 kN m
0 . 6 kN m
x
A
B
C z
6 kN
3.75 kN
Critical point is just to the right of C.
2.25 kN
1 . 125
kN m
M x 0 . 6 kN m
T
x S
(a)
y
n
32
D
3
2
M
T
2
6
2 5 0 (1 0 )
;
1 .5
3
3 2 (1 0 )
D
3
[1.1 2 5
2
1
0 .6 ] 2 2
or D 42 . 71 mm 250 ( 10
(b)
1 .5
6
)
3
32 ( 10 )
D
3
[1 . 125
2
3 4
2
1
( 0 .6 ) ] 2
Solving D 4 2 .3 m m SOLUTION (9.8) We have S y 3 4 0 M P a and n 1 .5 . The reactions and the torques, as found by the equations of statics, are marked in Fig. S9.8a. Moment and torque diagrams are shown in Fig. S9.89b. Critical sections is at C. (CONT.)
147
9.8 (CONT.) y A
2m
90 N
C
7.86N m
2m z
360 N
D
7.86N m.
450 N
360 N
1m
720 N
B
x
360 N
(a)
720
Mz (N m) A
C
My (N m)
x
D
Figure S9.8
360
180
B
x
T (N m) -7.86
C
x
B
(b) Eq. (9.7) is thus D
32n 3
Sy
M
M
2 z
2 y
T
2
Inserting the numerical values: [
or
3 2 (1 .5 )
(3 4 0 1 0 ) 6
(
7 2 0 1 8 0 ( 7 .8 6 ) ) ] 2
2
2
1 3
D 0 .0 3 2 2 m = 3 2 .2 m m
SOLUTION (9.9) Refer to solution of Prob. 9.8. Now apply Eq. (9.8): D
[
32n 3
Sy
M
3 2 (1 .2 )
(3 4 0 1 0 ) 6
M
2 z
(
2 y
3
T
2
4
720 180 2
2
3 4
or D 0 .0 2 9 9 m = 2 9 .9 m m
148
( 7 .8 6 ) ) ] 2
1 3
SOLUTION (9.10) Table B.1: S
520
y
xy
We have M 0 ,
MPa
16T
D
3
By Eq.(1.15), T
9549 kW n
9549 (70 ) 110
6 .0 7 7 k N m
( a ) Equation (6.11) Sy n
2
xy
,
D
3
32Tn Sy
32 (6077 ) 4
6
( 5 2 0 1 0 )
D 7 8 .1 m m
,
( b ) Equation (6.16): Sy n
3
xy
,
D
3
16
3T n
Sy
16
3 (6077 ) 4 6
( 5 2 0 1 0 )
D 7 4 .4 m m
or
SOLUTION (9.11) We have S u 5 9 0 M P a and n 1 .4 Ay B y 8 kN
From statics:
8 kN
y
8 kN
0.5 m
A z
1m
0.5 m
C 0.48 kN m
Ay
D 0.48 kN m
B
x
(a)
By
8 0.5=4 kN m
Mz
x T
0.48 kN m
x
(b)
(c)
Figure S9.11 The critical section is between C and D. Thus M
2 z
T
2
4 0 .4 8 2
2
4 .0 2 9 k N m
Apply Eq. (9.9): 1 6 1 0 (1 .4 ) 3
D
3
(5 9 0 1 0 ) 6
( 4 4 .0 2 9 ) 0 .0 4 6 m
Use a 4 6 m m diameter shaft. SOLUTION (9.12) Now Fig. S9.11a (see: Solution of Prob. 9.11) becomes as shown below. From statics: (CONT.)
149
9.12 (CONT.) Ay 2 kN
B y 6 kN
Az 6 k N
Bz 2 kN
y A
1m
0.5 m
D 0.48 kN m
C 0.48 kN m
Az Ay
z
8 kN
8 kN 0.5 m
B
x
(a)
x
(b)
x
(c)
x
(d)
By
Bz
Mz 2 0.5=1 kN m
2 1.5=3 kN in.
3 kN m My
1 kN m
T
0.48 kN m
The critical section is just to the right of C (or just to the left of D). Thus M
2 z
M
2 y
T
2
3 1 0 .4 8 2
2
2
3 .1 9 9 k N m
Applying Eq. (9.9): 1 6 1 0 (1 .5 ) 3
D
3
(5 9 0 1 0 ) 6
( 3 3 .1 9 9 ) 0 .0 4 3 1 m
Use a 4 4 m m diameter shaft. SOLUTION (9.13) y A Az
0 . 6 kN m
0.3 m
C 5.2 kN
3 kN
Ay
z
x
0 . 6 kN m
0.5 m
Bz
30o
975
M
B y 1 . 95 kN ,
B
B z 1 . 125
A y 3 . 25 kN
kN , A z 1 . 875
kN
By
N m
z
x M
563
Critical point is just to the right of C.
N m
M
y
x T
600
1
[ 9 7 5 5 6 3 ] 2 1 .1 2 6 k N m 2
C
2
N m
x
150
(CONT.)
9.13 (CONT.) Thus
M
0
m
M
1 . 126
a
kN m
T m 600
N m
From Sec. 8.6: C t 1 0 .0 0 5 8 ( 5 0 0 4 5 0 ) 0 .7 1 S e 0 . 5 S u 260 '
MPa
C
f
AS
b u
4 . 51 ( 520
0 . 265
) 0 . 86
C r 0 .8 9
Assume D 51 mm and C s 0 .7 0 . Thus
S e C f C r C s C t ( K1 ) S e ( 0 . 86 )( 0 . 89 )( 0 . 70 )( 0 . 71 )( 11. 2 )( 260 ) 82 . 42 MPa '
f
Substitute the data and K s b K s t 1.5 (Table 9.1) into Eq.(9.13): 6
520 (10 ) 1 .5
[ 1.5 ( 8 2 . 4 2 1 ,1 2 6 )
32
D
520
3
2
1
1.5 ( 6 0 0 ) ] 2 2
Solving, D 63 . 5 mm SOLUTION (9.14) Statics: R A 9 .4 5 k N ,
R B 6 .7 5
kN ,
M
D
2 .5 3 k N m M
m ax
Refer to Secs. 7.6 and 7.7: C s 0 .7 K
C r 0 .8 1 (Table 7.3)
C
AS
f
b u
0 . 085
1 . 58 (1260
1, S e 0 . 5 S u 630 '
f
) 0 . 861
S e ( 0 .8 6 1)( 0 .8 1)( 0 .7 )(1)( 6 3 0 ) 3 0 7 .5 6 M P a
and
We have M
m
0
M
a
2 .5 3 k N m
T m 2 .2 6 k N m
T a 0 .2 2 6 k N m
Using Eq.(9.12): 6
1 2 6 0 (1 0 ) n
3
3 2 (1 0 )
( 6 2 .5 1 0
3
[( 310276.506 2 .5 3 ) 2
)
3
3 4
( 2 .2 6
1260 3 0 7 .5 6
1
0 .2 2 6 ) ] 2 2
n 2 .8 2
or
SOLUTION (9.15) Statics: R A 5 .2 2 k N , M
m
0,
R B 8 .2 8 k N ,
T m 3 .4
kN m ,
From Secs. 7.6 and 7.7: C r 0 .8 9
M
2 .6 1 k N m M
C
a
. We have
T a 0 .6 8 k N m
C s 0 .7
K
f
1
C t 1 0 . 0058 ( 510 450 ) 0 . 652 S e 0 .5 S u 7 0 0 '
M Pa
C
A S u 5 7 .7 (1 4 0 0 b
f
0 .7 1 8
S e ( 0 .3 1 8 )( 0 .8 9 )( 0 .7 )( 0 .6 5 2 )(1)( 7 0 0 ) 9 0 .4 M P a
and
Apply Eq.(9.13), with replacing S u by S y and K s t 2 (Table 9.1): 910 ( 10 n
Solving,
6
)
3
32 ( 10 )
( 87 . 5 10
3
)
3
[(
910 90 . 4
2 . 61 )
2
n 1 .9 9
151
2 (3 .4
1
910 90 . 4
0 . 68 ) ] 2 2
) 0 .3 1 8
SOLUTION (9.16) From Fig. C.13, for d D 0 .2 : K t 1 .5 (torsion)
K t 2 .2 (bending)
Figures (7.9) : q 0 .9 5 (torsion) q 0 .8 (bending) The endurance limit is From Eq. 7.13b we have K
fb
1 q ( K t 1) 1 0 .8 ( 2 .2 1) 1 .9 6
K
ft
1 0 .9 5 (1 .5 1) 1 .4 8
Endurance limit: S e C f C r C s C t (1 K t ) S e '
where C
4 .5 1(5 2 0
f
0 .2 6 5
) 0 .8 6
C r 0 .8 1
(Table 7.3)
C s 0 .8 5
(Eq. 7.9)
C t 1 0 .0 0 5 8 ( 4 9 0 4 5 0 ) 0 .7 7 S e ' 0 .5 S u 2 6 0 M P a
(Eq. 7.11)
(Eq. 7.1)
Thus, S e ( 0 .8 6 )( 0 .8 1)( 0 .8 5 )( 0 .7 7 )(1 1 .9 6 )( 2 6 0 ) 6 0 .5 M P a
Also
M
a
120 N m ,
M
0,
m
Tm 6 0 0 N m ,
Ta 0
Equation (9.16a): D
3
32 n
Su
S
[( S u M a ) 2
3 4
e
2
1 2
Tm ]
Substituting the given data, we have 3
( 4 0 ) (1 0
9
)
[ ( 6502.50 1 2 0 ) 2
32 n 6
( 5 2 0 1 0 )
2
1
2
1
3 4
(6 0 0 ) ]
3 4
(6 0 0 ) ]
2
Solving, n 2 .8 3
SOLUTION (9.17) Refer to solution of Prob.9.16. Apply Eq. (9.16b): D
3
32 n
Sy
[(
Sy Se
M a) 2
3 4
2
Tm ]
1 2
Inserting the given numerical values, 3
( 4 0 ) (1 0
9
)
[ ( 6404.50 1 2 0 ) 2
32 n 6
( 4 4 0 1 0 )
From which n 2 .7 2
152
2
SOLUTION (9.18) M S
200
a
N m,
290
y
T m 500
N m,
S u 455
MPa ,
M
0,
m
Ta 0,
K
st
2
MPa
Refer to Secs. 7.6 and 7.7: C
4 . 51 ( 455
f
0 . 265
Assume C s 0 . 7 ,
) 0 . 89 ,
C r 0 . 87 ,
S e ( 0 . 89 )( 0 . 87 )( 0 . 7 )( 1)(
S e 0 . 5 S u 227 . 5 MPa '
1 2 .2
)( 227 . 5 ) 56 . 05 MPa
Through the use of Eq.(9.14), we have 6
4 5 5 (1 0 ) 1 .5
32
D
[1( 5 6 . 0 5 2 0 0 ) 2
455
3
D 38 . 8 mm 51 mm
or
3 4
1
2
( 2 )(5 0 0 ) ] 2
Assumption is incorrect.
Assume C s 0 .8 5 : Se
0 . 85 0 .7
( 56 . 05 ) 68 . 06 MPa
Equation (9.14): 6
4 5 5 (1 0 ) 1.5
32
D
[1( 6 8 . 0 6 2 0 0 ) 2
455
3
D 36 . 7 mm
or
3 4
1
2
( 2 )(5 0 0 ) ] 2
Assumption is correct.
SOLUTION (9.19) Refer to Secs. 7.6 and 7.7: q 0 .9 2 K
1 q ( K t 1 ) 1 0 .9 2 (1.8 1 ) 1.7 4
f
S e 0 . 5 S u 500 '
C
Thus
AS
f
b u
MPa ,
57 . 7 (1000
0 . 718
C s 0 . 85 ,
C r 0 . 84
) 0 . 405
S e ( 0 .4 0 5 )( 0 .8 4 )( 0 .8 5 )(1)( 1 .71 4 )(5 0 0 ) 8 3 .0 9 M P a
We have: M
0,
m
M
a
500
N m,
T m 600
N m,
T a 90 N m
Apply Eq.(9.11), with replacing S u by S y : 6
600 ( 10 ) n
32
( 0 . 05 )
3
[(
600 83 . 09
500 ) ( 600 2
600 83 . 09
1
90 ) ] 2 2
or n 1 . 93
SOLUTION (9.20) The endurance limit is expressed as S e C f C r C s C t (1 K f ) S e '
Here C
f
4 .5 1( 6 9 0
0 .2 6 5
) 0 .7 9 8
(Eq. 7.7 and Table 7.2)
C r 0 .8 7
(Table 7.3),
C t 1 (Eq. 7.11)
C s 0 .7 0
(Assuming D > 51 mm, Eq. 7.9) (CONT.)
153
9.20 (CONT.) S e ' 0 .5 S u 0 .5 ( 6 9 0 ) 3 4 5 M P a
(Eq. 7.1)
Therefore S e ( 0 .7 9 8 )( 0 .8 7 )( 0 .7 )(1)(1 1 .2 )(3 4 5 ) 1 3 9 .7 M P a
Equation (9.13) with T a 0 becomes
3
D
32 n
Sy
[K s (M
m
Su Se
M a ) K sT m ] 2
2
1 2
Introducing the given numerical values, we have 3 2 ( 3 .5 )
3
D
6
( 6 9 0 1 0 )
6
( 6 9 0 1 0 )
{1 .5[ 2 0 0
6
(1 3 9 .7 1 0 )
( 6 0 0 )] 1 .5 (3 6 0 ) } 2
2
1 2
Solving, D 0 .0 5 8 6 m 5 8 .6 m m
Since D 5 1 m m , our assumption is correct. SOLUTION (9.21) We now have K s 2 (from Table 9.1) and S u replaced by S y . See solution of Prob. 9.20. Equation (9.14) becomes
3
D
32 n
Sy
[K s (M
m
Sy Se
M a ) K s ( 34 T m )] 2
2
1 2
or 3 2 ( 3 .5 )
3
D
6
( 5 8 0 1 0 )
{ 2[ 2 0 0
580 1 3 9 .7
(6 0 0 )] 2
3 2
1
2
(3 6 0 ) }
2
Solution is D 0 .0 6 1 7 4 m 6 1 .7 m m
Since D 5 1 m m , our assumption is correct. SOLUTION (9.22) 2 kN m
M
0 . 8 kN m
y
Critical section is just to the left of point D
x C
A M
D
B
z
600
T
2 .5 7 k N m
M
D
[ 2 . 57
x
M
a
2 . 692
T m 600
N m
1
2
0 . 8 ] 2 2 . 692 2
kN m ,
N m,
M
0
m
T a 30 N m
x Refer to Secs. 7.6 and 7.7: S e 0 . 5 S u 330 '
S e C f C rC sC t (
MPa , 1 K f
C
f
AS
b u
4 . 51 ( 660
0 . 265
) 0 . 807
) S e ( 0 . 807 )( 1 )( 0 . 7 )( 1 )( 1 )( 330 ) 186 . 4 MPa '
Equation (9.14): 6
6 6 0 (1 0 ) n
32
( 0 .0 7 5 )
[1.5 ( 1 8 6 . 4 2 , 6 9 2 ) 2
660
3
Solving n 2 .3 4
154
3 4
(1.5 ) ( 6 0 0
660 1 8 6 .4
kN m
1
30 ) ]2 2
SOLUTION (9.23) From Solution of Prob. 9.22 M
0,
m
M
2 . 692
a
kN m ,
T a 30 N m ,
T m 600
N m
S e 1 8 6 .4 M P a
Now we have C r 0 .8 9 . Thus, S e 1 8 6 .4 ( 0 .8 9 ) 1 6 5 .9 M P a Through the use of Eq.(9.13), with replacing S u by S y : 6
550 ( 10 ) n
550 [1 . 5 ( 165 2 , 692 ) 1 . 5 ( 600 .9 2
32
( 0 . 075 )
3
550 165 . 9
1
30 ) ] 2 2
n 2 . 08
or
SOLUTION (9.24) R B y 1 .0 3 8 k N ,
0.45 kN
y
R A y 6 .1 6 2 k N R B z 3 .7 3 8 k N ,
1.35 kN
z x
A C
R Ay M
D
R A z 0 .1 3 8 k N
B
M
( kN m)
B 6.3 kN R B z
2.7 kN
9 kN x
D
C
R Az
R By
z
x
A
y
(kN m)
0.208
x 0.021
0.748
0.924
We have T 0 .9 k N m and 1
M
D
[ 0 .2 0 8 0 .7 4 8 ] 2 0 .7 7 6 k N m
M
C
[ 0 .9 2 4 0 .0 2 1 ] 2 0 .9 2 4 k N m
M
m
0,
2
2
2
2
1
Hence M
a
0 .9 2 4 k N m ,
Ta 0 ,
T m 0 .9 k N m
From Table B.3: S u 5 9 0 M P a . Refer to Secs. 7.6 and 7.7: C r 0 .8 7 ,
C
and
f
AS
C s 0 .7 (assumed D>50 mm)
4 . 51 ( 590 )
b u
0 . 265
0 . 83
S e ( 0 .8 3)( 0 .8 7 )( 0 .7 )(1)( 11.8 )( 0 .5 5 9 0 ) 8 2 .8 4 M P a
Applying Eq.(9.12): 5 9 0 1 0 1 .6
6
[ ( 8529.804 9 2 4 ) 2
32
D
3
3 4
2
1
(9 0 0 ) ] 2
D 0 .0 5 6 8 m = 5 6 .8 m m
155
SOLUTION (9.25) R B y 2 .0 8 k N ,
R A y 6 .9 2 k N
9 kN
y
x C
6.92 kN
M
x
( kN m )
We have T 0 .9 k N m and M
C
M
x 0.021
D
1 2
[ (1 .0 3 8 ) ( 0 .0 2 1) ] 1 .0 3 8 k N m M 2
3.74 kN
y
(kN m )
0.416
B
D 0.748
2.08 kN
1.038
C
C
0.14 kN
z
M
x
A
B
D
R A z 0 .1 4 k N
6.3 kN
2.7 kN
z
A
M
R B z 3 .7 4 k N ,
2
Refer to Solution of Prob. 9.24: C f 0 .8 3 ,
C s 0 .7
a
From Table B-3: S y 4 9 0 M P a
From Sec. 7.6: S e C f C r C s C t ( K1 ) S e ( 0 .8 3 )( 0 .7 5 )( 0 .7 )(1)( 11.2 )( 0 .5 5 9 0 ) 1 0 7 .1 2 M P a '
f
Using Eq.(9.11), with S u replaced by S y : D
or
3
32 n
Sy
[(
Sy Se
1
32 ( 2 )
M a ) Tm ] 2 2
2
D 0 .0 5 8 6 m = 5 8 .5 m m
SOLUTION (9.26) Refer to Case 6 of Table A.8:
We have I
Pbx 6 LEI
4
(L b x ) 2
(12 . 5 )
2
4
2
19 . 175 10
3
mm
4
W D W E 15 9 . 81 147 . 2 N
Deflection at D: D '
D ''
2
2
2
1 4 7 .2 (1 )( 0 .4 )(1 .4 1 0 .4 ) 6 (1 4 0 0 )( 2 1 0 1 9 .1 7 5 ) 2
1 .3 9 3 m m
2
2
1 4 7 .2 ( 0 .4 )( 0 .4 )(1 .4 0 .4 0 .4 ) 6 (1 4 0 0 )( 2 1 0 1 9 .1 7 5 )
1 .1 4 2 m m
and D E 1 .3 9 3 1 .1 4 2 2 .5 3 5 m m
Equation (9.18) results in n cr
1 2
594
2 gW 2W
2
1 2
g
1 2
9 . 81 2 . 535 10
rpm
156
1
[( 1 0479.10 2 1, 0 3 8 ) 9 0 0 ] 2 2
6
( 4 9 0 1 0 )
3
9 . 901 cps
2
SOLUTION (9.27) Refer to Case 6 of Table A.8:
Pbx 6 LEI
(L b x ) 2
2
2
We have W D W E 15 9 . 81 147 . 2 N At midspan C: C
0 . 5 ( 10
3
)
2
147 . 2 ( 0 . 4 )( 0 . 7 )
(1 . 4
6 ( 1 . 4 ) EI
2
0 .4
2
0 .7 ) 2
or EI 25 . 711 (10 ) N m 3
2
Deflection at D: D ' D ''
2
2
2
1 4 7 .2 (1 )( 0 .4 )(1 .4 1 0 .4 )
0 .2 1 8
3
6 (1 .4 )( 2 5 .7 1 1 1 0 ) 2
2
2
1 4 7 .2 ( 0 .4 )( 0 .4 )(1 .4 0 .4 0 .4 ) 3
6 (1 .4 )( 2 5 .7 1 1 1 0 )
mm
0 .1 7 9
mm
and
D
E
0 . 218 0 . 179 0 . 397
mm
Equation (9.18) gives n cr
g
1 2
9 .8 1
1 2
0 .3 9 7 1 0
3
2 5 .0 2
cps
SOLUTION (9.28) We have I
4
(15 )
39 . 76 10
4
3
mm
4
Refer to Case10 of Table A.8: a
P C
b
A B L 2
C
Pb L 3 EI
C
150 ( 0 . 4 )( 1000 )
Thus 2
3 ( 210 39 . 76 )
0 . 958
mm
Equation (9.18) results in n cr
1 2
1 2
0 .9 5 8 1 0
g
9 .8 1 3
1 6 .1 1 c p s 9 6 7
157
rp m
6 . 428 EI
SOLUTION (9.29) See Table A.8 (Case 6) and Example 9.5. We have I
4
d
( 0 .0 5 6 2 5 )
64
4
0 .4 9 1 4 (1 0
64
6
) m
4
( a ) Deflection at C owing to 450 N: C '
4 5 0 ( 2 .2 5 )( 0 .6 2 5 )
( 2 .8 7 5 0 .6 2 5 2 .2 5 ) 1 .0 5 m m 2
9
6 ( 2 .8 7 5 )( 2 0 0 1 0 )( 0 .4 9 1 4 1 0
6
)
2
2
Deflection at C due to 270 N: C "
2 7 0 (1 )( 0 .6 2 5 )
( 2 .8 7 5 0 .6 2 5 1 ) 0 .6 8 4 m m 2
9
6 ( 2 .8 7 5 )( 2 0 0 1 0 )( 0 .4 9 1 4 1 0
6
)
2
2
Total deflection at C: C 1 .0 5 0 .6 8 4 1 .7 3 4 m m Deflection at D owing to 450 N: 4 5 0 ( 0 .6 2 5 )( 2 .8 7 5 1 .8 7 5 )
D '
9
6 ( 2 .8 7 5 )( 2 0 0 1 0 )( 0 .4 9 1 4 1 0
[ 2 ( 2 .8 7 5 )(1 .8 7 5 ) 0 .6 2 5 1 .8 7 5 ] 1 .1 4 1 m m 2
6
)
2
Deflection at D due to 270 N: D "
2 7 0 (1 )(1 .8 7 5 )
( 2 .8 7 5 1 .8 7 5 1 ) 1 .1 2 m m 2
9
6 ( 2 .8 7 5 )( 2 0 0 1 0 )( 0 .4 9 1 4 1 0
6
)
2
2
Total deflection at D: D 1 .1 4 1 1 .1 2 2 .2 6 1 m m Applying Eq. (9.18), we obtain n cr
1 2
[
9 .8 1[( 4 5 0 )(1 .7 3 4 1 0 4 5 0 (1 .7 3 4 1 0
3
3
) ( 2 7 0 )( 2 .2 6 1 1 0
2
) 2 7 0 ( 2 .2 6 1 1 0
3
)
3
)]
2
1
]
2
1 1 .2 4 c p s 6 7 4 .4 r p m
( b ) Apply Eq. (9.21): n c r ,C n c r , D
n cr
where
Hence
2
2
n c r ,C n c r , D
g
n cr ,C
1 2
C '
n cr , D
1 2
D "
n cr
g
( 9 2 3 )( 8 9 4 ) 2
1 2
(923) (894 )
2
1 2
9 .8 1 1 .0 5 1 0
3
9 .8 1 1 .1 2 1 0
3
1 5 .3 8 c p s 9 2 3 rp m 1 4 .9 c p s 8 9 4 rp m
6 4 2 rp m
SOLUTION (9.30) Refer to Table A.8 (Case 6) and Solution of Prob. 9.29. We now have I D
4
6 4 ( 0 .0 8 7 5 )
4
6 4 2 .8 7 7 4 1 0
6
m
4
( a ) Deflection at C owing to 450 N: C '
4 5 0 ( 2 .2 5 )( 0 .6 2 5 ) 9
( 2 .8 7 5 0 .6 2 5 2 .2 5 ) = 0 .1 7 9 3 m m 2
6 ( 2 .8 7 5 )( 2 0 0 1 0 )( 2 .8 7 7 4 4 1 0
6
)
2
2
Deflection at C due to 270 N: C "
2 7 0 (1 )( 0 .6 2 5 ) 9
6 ( 2 .8 7 5 )( 2 0 0 1 0 )( 2 .8 7 7 4 1 0
( 2 .8 7 5 0 .6 2 5 1 ) 0 .1 1 6 9 m m 2
6
)
2
2
Total deflection at C: C 0 .1 7 9 3 0 .1 1 6 9 0 .2 9 6 2 m m (CONT.)
158
9.30 (CONT.) Deflection at D owing to 450 N: D '
4 5 0 ( 0 .6 2 5 )( 2 .8 7 5 1 .8 7 5 )
[ 2 ( 2 .8 7 5 )(1 .8 7 5 ) 0 .6 2 5 1 .8 7 5 ] 0 .1 9 4 8 m m 2
9
6 ( 2 .8 7 5 )( 2 0 0 1 0 )( 2 .8 7 7 4 4 1 0
6
)
2
Deflection at D due to 270 N: D "
2 7 0 (1 )(1 .8 7 5 )
( 2 .8 7 5 1 .8 7 5 1 ) 0 .1 9 1 2 m m 2
9
6 ( 2 .8 7 5 )( 2 0 0 1 0 )( 2 .8 7 7 4 4 1 0
6
)
2
2
Total deflection at D: D 0 .1 9 4 8 0 .1 9 1 2 0 .3 8 6 m m Applying Eq. (9.18), we find n cr
1 2
9 .8 1[( 4 5 0 )( 0 .2 9 6 2 1 0
[
4 5 0 ( 0 .2 9 6 2 1 0
3
3
) ( 2 7 0 )( 0 .3 8 6 1 0 3
2
) 2 7 0 ( 0 .3 8 6 1 0
)
3
)]
2
1
]
2
2 7 .2 1 c p s 1 6 3 3 r p m
( b ) Apply Eq. (9.21): n c r ,C n c r , D
n cr
where
Hence
2
2
n c r ,C n c r , D
g
n c r ,C
1 2
C '
n cr , D
1 2
D "
g
2
( 2234 ) ( 2163)
0 .1 7 9 3 1 0
2
3
9 .8 1
1 2
( 2 2 3 4 )( 2 1 6 3 )
n cr
3 7 .2 3 c p s 2 2 3 4 rp m
9 .8 1
1 2
0 .1 9 1 2 1 0
3
3 6 .0 5 c p s 2 1 6 3 rp m
1 5 5 4 rp m
SOLUTION (9.31) Refer to Table A.8 (Case 6) and Example 9.5. We have I
4
d
64
( 0 .0 4 6 9 )
4
64
0 .2 3 7 5 1 0
6
m
4
( a ) Deflection at C owing to 340 N: C '
3 4 0 (1 .1 2 5 )( 2 5 0 .6 2 5 )
(1 .7 5 1 .1 2 5 0 .6 2 5 ) 1 .2 8 3 9 m m 2
9
6 (1 .7 5 )(1 0 5 1 0 )( 0 .2 3 7 5 1 0
6
)
2
2
Deflection at C due to 500 N: C "
5 0 0 ( 0 .3 7 5 )( 0 .6 2 5 )
(1 .7 5 0 .3 7 5 0 .6 2 5 ) 1 .1 3 2 9 m m 2
9
6 (1 .7 5 )(1 0 5 1 0 )( 0 .2 3 7 5 1 0
6
)
2
2
Total deflection at C: C 1 .2 8 3 9 1 .1 3 2 9 2 .4 1 6 8 m m Deflection at D owing to 340 N: D '
3 4 0 ( 0 .6 2 5 )(1 .7 5 1 .3 7 5 ) 9
6 (1 .7 5 )(1 0 5 1 0 )( 0 .2 3 7 5 1 0
[ 2 (1 .7 5 )(1 .3 7 5 ) 0 .6 2 5 1 .3 7 5 ] 0 .7 7 0 3 m m 2
6
)
2
Deflection at D due to 500 N: D "
5 0 0 ( 0 .3 7 5 )(1 .3 7 5 ) 9
(1 .7 5 1 .3 7 5 0 .3 7 5 ) 1 .0 1 5 4 m m 2
6 (1 .7 5 )(1 0 5 1 0 )( 0 .2 3 7 5 1 0
6
)
2
2
Total deflection at D: D 0 .7 7 0 3 1 .0 1 5 4 1 .7 8 5 7 m m Applying Eq. (9.18), we obtain n cr
1 2
[
9 .8 1 ( 3 4 0 2 .4 1 6 8 1 0 3 4 0 ( 2 .4 1 6 8 1 0
3
3
5 0 0 1 .0 1 5 4 1 0
2
) 1 1 0 (1 .0 1 5 4 1 0
3
3
)
) 2
1
]
2
1 1 .4 9 c p s 6 9 0 r p m
(CONT.)
159
9.31 (CONT.) ( b ) From Eq. (9.21), we have n c r ,C n c r , D
n cr
where
2
2
n c r ,C n c r , D
g
n c r ,C
1 2
C '
n cr , D
1 2
D "
g
9 .8 1
1 2
1 .2 8 3 9 1 0
3
9 .8 1
1 2
1 .0 1 5 4 1 0
3
1 5 .9 1 c p s 8 3 5 rp m 1 5 .6 5 c p s 9 3 9 rp m
It follows that ( 8 3 5 )( 9 3 9 )
n cr
2
6 2 4 rp m
2
(835 ) (939 )
SOLUTION (9.32) Shaft
Sy
a ll
Key
Sy
a ll
a ll
n
n S ys n
580 2 .3
2 5 2 .2 M P a
3 0 .8 2 .3
154 2 .3
1 3 3 .9 1 M P a 6 6 .9 5 M P a
Torque: 9549 kW
T
9549 (90 )
n
9 5 4 .9 N m
900
Force at the shaft surface is then F
(a) L (b) L (c) L
F (w 2)
a ll
F (w 2)
a ll
F
a ll ( w )
T r
9 5 4 .9 1 8 .7 5
5 0 .9 3 k N
5 0 .9 3 1 0
3
2 5 2 .2 ( 9 .3 5 1 0
3
5 0 .9 3 1 0
2) 3
6
1 3 3 .9 1 1 0 ( 9 .3 5 1 0 5 0 .9 3 1 0
3
6
6 6 .9 5 1 0 ( 9 .3 5 1 0
0 .0 4 3 2 m = 4 3 .2 m m
3
)
3
2)
0 .0 8 1 4 m = 8 1 .4 m m
0 .0 8 1 4 m = 8 1 .4 m m
SOLUTION (9.33) From Table B.3: For shaft: S y S y c 4 4 0 M P a For key: S y S y c 3 9 0 M P a n 2 . Shaft:
a ll T
and
F
0 .5 8 S y n
a ll J r
1 0 .5 7 0 .0 7 5 2
0 .5 8 ( 4 4 0 ) 2
a ll ( D
4
D 2
32 )
1 2 7 .6 M P a
(1 2 7 .6 ) ( 0 .0 7 5 )
3
16
1 0 .5 7
kN m
2 8 1 .9 k N
Key length: Based on bearing on shaft: L
2 8 1 .9 1 0
F ( S y n )( h 2 )
(
4 4 0 1 0
6
)(
3
9 .3 7 5 1 0
2
Based on bearing on key: L
F ( S y n )( h 2 )
2 8 1 .9 1 0
(
3 9 0 1 0
6
Based on shear in key: L
)(
(
)w
n
(
0 .5 8 3 9 0 1 0 2
160
0 .3 0 8 4 m = 3 0 8 .4 m m
3
)
2 2 8 1 .9 1 0
)
3
9 .3 7 5 1 0
2 F 0 .5 8 S y
0 .2 7 3 4 m = 2 7 3 .4 m m
3
2
3
6
)(1 8 .7 5 1 0
0 .1 3 3 m = 1 3 3 m m 3
)
SOLUTION (9.34) We have F
T r
0 .4 1 0 3 7 .5 1 0
a ll w L
By Eq.(9.23): n
3
3
6
( 7 0 1 0 ) ( 9 .3 7 5 1 0
2F
2 1 .3 3 k N
2
3
) ( 7 5 1 0
3
)
3
2 ( 2 1 .3 3 1 0 )
1 .1 5
SOLUTION (9.35)
Shaft
Sy
a ll
460 4
460 2
340 4
n S ys
a ll
n
115 M Pa 5 7 .5 M P a
4
Key
Sy
a ll
n Sy 2
a ll
85 M Pa
4 2 .5 M P a
n
Torque in shaft. We have J D
3 2 (6 0 )
4
4
3 2 1 .2 7 2 1 0
6
mm
4
Hence, a ll J
T
r
5 7 .5 (1 .2 7 2 )
2 .4 4 k N m
0 .0 3
Force at the shaft surface: F
(a) L (b) L (c) L
F (w 2)
a ll
F (w 2)
a ll
F
a ll ( w 2 )
T r
2 .4 4 0 .0 3
2 .4 4 k N m 7 0 .7 m m
81300 6
1 1 5 (1 0 )( 0 .0 1 )
9 5 .6 m m
81300 6
8 5 (1 0 )( 0 .0 1 ) 81300 6
4 2 .5 (1 0 )( 0 .0 2 )
9 5 .6 m m
SOLUTION (9.36) Shear force in each bolt: F
T RN
Average shear stress in each bolt, a ll
Sy n
F A
T RN
db 4
4T 2
db RN
Solving db
4Tn
(P9.36)
RNS y
Introducing the given data: db [
3
4 ( 5 1 0 )(1 .2 )
1 6
( 0 .0 8 )( 6 )( 2 6 0 )(1 0 )
]
2
7 .8 2 m m
161
SOLUTION (9.37) From Eq. (1.15), 9549 kW
T
n
9549 (30 )
2 .3 8 7 k N m
120
Force at shaft surface F
T r
F L(w 2)
7 9 .6 k N
2 .3 8 7 0 .0 3
( a ) Key
7 5 ( 5 )(1 0
6
3
7 9 .6 (1 0 )
F L(w)
3
7 9 .6 (1 0 )
7 5 (1 0 )(1 0
6
)
212 M Pa;
)
n
106 M Pa;
n
S ys
Sy
1 .9 8
420 212
210 106
1 .9 8
210 3 1 .3
6 .7 1
420 2 4 .6
1 7 .1
( b ) Bolts Force at bolt circle F
T Db 2
2 .3 8 7 0 .0 7 2
3 3 .1 5 k N
Sear Stress in bolts
F 2
6 ( d b
4)
3 1 .3 M P a ;
n
S ys
2 4 .6 M P a ;
n
Sy
3 3 .1 5 ( 4 ) 6 ( 0 .0 1 5 )
2
( c ) Bearing on bolts in flange
F 6 d b Ft
3 3 .1 5 6 (1 5 )(1 5 )1 0
6
SOLUTION (9.38)
(a)
T
9549 kW n
F
T r
9549 ( 45 )
200
2148 0 .0 2 5
2 .1 4 8 k N m
8 5 .9 2 k N 3
3
Area in shear in key (1 4 .0 6 2 5 1 0 )(8 7 .5 1 0 ) 1 .2 3 1 0 3
3
3
m
2
Area in bearing for key [(1 4 .0 6 2 5 2 ) 1 0 ](8 7 .5 1 0 ) 6 .1 5 1 0
4
m
2
Shear stress in key:
8 5 .9 2 1 0 1 .2 3 1 0
3
3
6 9 .8 5 M P a
Bearing stress in key: (b)
8 5 .9 2 1 0 6 .1 5 1 0
3
4
1 3 9 .7 M P a
Area in shear for bolts 6 ( 4 )(9 .3 7 5 ) 4 1 4 .1 7 m m 2
2
Force at bolt circle: F
2 .1 4 8 0 .0 7 5
2 8 .6 k N
Shear stress in bolts:
28600 4 1 4 .1 7
6 9 .0 5 M P a
(CONT.)
162
9.38 (CONT.) ( c ) Area in bearing for bolts 6 (9 .3 7 5 )( 2 1 .8 7 5 ) 1 2 3 0 .5 m m
2
Bearing stress in bolts:
28600 1 2 3 0 .5 (1 0
6
)
2 3 .2 M P a
( d ) Area in shear at edge of hub D h t f (1 0 0 )( 2 1 .8 7 5 ) 6 8 7 2 .2 m m Force at edge of hub: F Shear stress in web:
T Dh 2
4 2 .9 6 0 1 0 6 8 7 2 .2 (1 0
2148 0 .0 5
3
6
)
2
4 2 .9 6 k N
6 .2 5 M P a
End of Chapter 9
163
CHAPTER 10
BEARINGS AND LUBRICATION
SOLUTION (10.1) We have H 9 .8 1( 6 5 ) 6 3 8 M P a
Equation (8.3') will be used.
KWl HAp
(1)
where A p D L 2 4 (1 2 ) 2 8 8 m m
2
l n D t 1 8 ( 2 4 )( 6 0 1 2 0 0 ) 9 7 .7 1 0
( a ) Good Lub: K 2 1 0
( 2 1 0
6
6
6
mm
(Table 8.3), Equation (1) gives: 6
)(1 5 0 )( 9 7 .7 1 0 ) 6
6
( 6 3 8 1 0 )( 2 8 8 1 0
)
0 .1 6 m m
( b ) Excellent Lub: K 1 1 0
(1 1 0
7
7
(Table 8.3), Equation (1) is then 6
)(1 5 0 )( 9 7 .7 1 0 ) 6
( 6 3 8 1 0 )( 2 8 8 1 0
6
)
0 .0 0 8 m m
SOLUTION (10.2) Refer to solution of Prob. 10.1. H 9 .8 1(1 0 5 ) 1 0 3 0 M P a
(Table B.6 and Sec. 8.5)
Use Eq. (8.3'),
KWl HAp
(1)
Here A p D L ( 2 5 )( 2 5 ) 6 2 5 m m
2
l n D t 1 8 ( 2 5 )( 6 0 1 5 0 0 ) 1 2 7 .2 1 0
( a ) Good Lub: K 2 1 0
( 2 1 0
6
6
6
(Table 8.3), Then, Eq. (1) result in 6
)(1 1 5 )(1 2 7 .2 1 0 ) 6
(1 0 3 0 1 0 )( 6 2 5 1 0
6
)
0 .0 4 5 m m
( b ) Excellent Lub: K 1 1 0
(1 1 0
7
7
(Table 8.3), Equation (1) gives 6
)(1 1 5 )(1 2 7 .2 1 0 ) 6
(1 0 3 0 1 0 )( 6 2 5 1 0
6
)
0 .0 0 2 m m
164
mm
SOLUTION (10.3) We have H 9 .8 1( 6 0 ) 5 8 8 .6 M P a
Equation (8.3) will be used, KH WA l
(1)
a
where 0 .1 5 m m , 7
K 1 10
W 450 N
(Table 8.3)
l n D t n ( 2 5 )( 6 0 2 4 3 6 5 1 .2 ) 4 9 .5 4 1 0 n m m 6
A p ( 2 5 )( 2 5 ) 6 2 5 m m
2
Equation (1) is thus (1 1 0
0 .1 5
7
6
)( 4 5 0 )( 4 9 .5 4 1 0 n ) 6
( 5 8 8 .6 1 0 )( 6 2 5 1 0
6
)
Solving, n 2 4 .8 r p m
SOLUTION (10.4) Equation (10.10): Tf
159 kW n
1 5 9 (1 .5 ) 1200 60
1 1 .9 N m
Equation (10.7): Tfc
4
2
3
Lr n
11 . 9 ( 0 . 09 ) 10 4
t 80
From Fig.10.8:
o
2
3 3
( 0 . 22 )( 0 . 08 ) ( 20 )
12 . 04 mPa s
C
SOLUTION (10.5) (a) Tf
2
f
c
Tf n
( b ) kW (c)
3
4 Lr n
159
Tf
Wr
4
2
( 4 .1 4 1 0
3
3
)( 0 .1 )( 0 .0 3 7 5 ) ( 2 4 0 0 0 6 0 ) 0 .0 7 5 (1 0
4 .6 ( 4 0 0 ) 159
4 .6 2 2 5 0 ( 0 .0 3 7 5 )
3
)
4 .6 4 N m
1 1 .6
0 .0 5 5
SOLUTION (10.6) T
f
(a)
fWr 0 . 01 ( 8000 )( 0 . 06 ) 4 . 8 N m Tfc 4
2
3
Lr n
4 . 8 ( 0 . 05 ) 10 4
2
3 3
( 0 . 12 )( 0 . 06 ) ( 10 )
23 . 45 mPa s
( b ) Figure 10.8: SAE 40 oil
165
SOLUTION (10.7) P
W DL
12 ( 10
3
)
0 . 768
( 0 . 125 )( 0 . 125 )
f
MPa,
F W
80 12 10
3
0 . 667 (10
2
)
Equation (10.9):
fP 2
2
0 . 667 ( 10
( cr )
n
2
)( 0 . 768 )( 10 2
2
(4)
6
)
( 0 . 0004 ) 25 . 95 mPa s
SOLUTION (10.8) From Fig. 10.8: 1 1 m P a s (a) P
W DL
400
2250 ( 0 .1 5 )( 0 .0 3 7 5 )
n
kPa
1500 60
25
rp s
Equation (10.9): 2 n r P c
f 2
2
2 11 ( 10
3
)( 25 )
400 10
3
1 0 . 001
0 . 014
( b ) T f fW D 2 ( 0 .0 1 4 )( 2 .2 5 )( 7 5 ) 2 .3 6 Tf n
kW
2 .3 6 ( 2 5 )
159
159
N m
0 .3 7
SOLUTION (10.9) ( a ) From Eq. (10.10): Tf
159 kW n
1 5 9 (1 .4 ) 2100 60
6 .3 6 N m
Equation (10.7) gives then
Tfc 2
3
4 Lr n
( 6 .3 6 )( 0 .1 7 5 ) 4
2
3
( 0 .2 )( 0 .0 7 5 ) ( 3 5 )
9 .5 5 m P a s
Using Fig. 10.8: t 90 C o
( b ) Equation (10.8) results in f
Tf
Wr
6 .3 6 2 (75 )
0 .0 4 2
SOLUTION (10.10) Figure 10.8: 8 .4 m P a s , Figure 10.14:
h0 c
0 . 025 0 . 05
0 .5
10
6 8 .4 (1 0
c 0 .0 0 1 r 0 .0 0 1(5 0 ) 0 .0 0 5 m m ,
S 0 . 52
Equation (10.17): S ( cr )
2 n P
3
P
)( 3 0 )
0 .5 2 ,
and W P D L 4 8 4 .6 ( 0 .1) ( 0 .0 5 ) 2 .4 2 k N
166
P 4 8 4 .6 k P a
L D 1 2
SOLUTION (10.11) P
W DL
3
1 2 (1 0 )
0 .7 6 8
( 0 .1 2 5 )( 0 .1 2 5 )
f
M Pa,
F W
80 3
1 2 (1 0 )
0 .6 6 7 (1 0
Thus f
r c
( 0 .6 6 7 1 0
1 0 .0 0 0 4
2
) 1 6 .7
From Fig. 10.15: S 0 .8 . The viscosity is therefore
S P n ( r c )2
0 .8 ( 0 .7 6 8 )1 0 4 (1 0 .0 0 0 4 )
6
2
0 .0 2 4 5 8
P a s 2 4 .5
mPa s
SOLUTION (10.12) ( a ) From Solution of Prob.10.8: 1 1 m P a s , P 4 0 0 k P a , n 2 5 rp s , L D 1 4 We thus have S ( cr )
2 n P
1 ( 0 . 001 )
2 11 ( 10
3
)( 25 )
0 . 69
3
400 ( 10 )
From Fig. 10.15: f 20,
r c
( b ) Thus
fW D
Tf
2 Tfn
kW
159
f 2 0 ( r ) 2 0 ( 0 .0 0 1 ) 0 .0 2 c
0 .0 2 ( 2 .2 5 )( 0 .0 7 5 ) 3 .3 7 5 N m
3 . 375 ( 25 ) 159
0 . 531
SOLUTION (10.13) ( a ) We have P
W DL
1 .4 5 ( 0 .0 3 )( 0 .0 3 )
1 .6 1 M P a
L D 30 30 1
Somerfield number: S ( cr ) ( 2
n P
) ( 01.05 3 ) [ 2
2 0 .7 (1 0
3
)( 4 0 ) 6
1 .6 1 (1 0 )
] 0 .1 2 9
Then, Fig. 10.14 gives, h o c 0 .4 2 and 1 h o c 0 .5 8 . It follows that e c 0 .5 8 ( 0 .0 3 ) 0 .0 1 7 4 m m
( b ) Figure 10.15. With S 0 .1 2 9 , L D 1;
( r c ) f 3 .5 .
So, f 3 .5 ( c r ) 3 .5 ( 0 .0 3 ) (1 5 ) 0 .0 0 7 N m Friction torque, T fW ( D 2 ) 0 .0 0 7 (1 .4 5 )(1 5 ) 0 .1 5 2 N m
Thus,
kW
Tn 159
0 .1 5 2 ( 4 0 ) 159
0 .0 3 8 2
( c ) Figure 10.16 ( S 0 .1 2 9 with L D 1 ) gives P P
0 .4 2
m ax
or Pm a x P 0 .4 2 1 .6 1(1 0 ) 0 .4 2 3 .8 3 M P a 6
167
2
),
L D 1
SOLUTION (10.14) P
W DL
3
1 5 (1 0 )
2 .0 8 3 M P a
6
(1 2 0 )( 6 0 )1 0
Figure 10.8: 1 6 m P a s Apply Eq. (10.17): n
S ( cr ) ( 2
P
o
(for SAE 40 oil at 8 0 C )
)
or 0 .1 5 (
120 2 c
)
2 (1 6 1 0
3
)(1 5 0 0 6 0 ) 6
2 .0 8 3 (1 0 )
Solving, c 0 .0 6 7 9 m m
It follows that c r 0 .0 6 7 9 6 0 0 .0 0 1 1
OK (See Sec. 10.8)
Figure 10.14: (for S 0 .1 5 and L D 1 2 )
h o c 0 .2 8
from which h o 0 .2 8 ( 0 .0 6 7 9 ) 0 .0 1 9 m m
SOLUTION (10.15) h0 c
S ( cr )
Figure 10.14: S 0 . 13 ,
0 .4 ,
0 . 0025 0 . 00625 2 n P
( 0 .05006 2 5 )
( a ) Fig.10.8:
t 80 C
( b ) Fig.10.15:
r c
2 (900 60 ) 3
7 0 0 (1 0 )
,
L D 1
9 .4 5 m P a s
o
f 3 , f 3 ( 0 . 00625 ) 0 . 00375 50
( c ) T f fW r 0 .0 0 3 7 5 ( 7 0 0 0 .1 0 .1)( 0 .5 ) 1 3 .1 N m kW
Tfn 1050
13 . 1 ( 15 ) 159
1 . 24
SOLUTION (10.16) P
W DL
S ( cr )
( a ) Figure 10.14: ( b ) Figure 10.15: T
f
1500 0 .0 2 5 0 .0 2 5
2 n P
h0 c
r c
2 .4 M P a ,
( 0 .010 0 8 )
2 5 0 (1 0
3
6
Tf n 159
f 1 1,
0 .1 6 5 (1 6 . 6 7 ) 159
0 .5 4 3,
0 . 75 ; h 0 0 . 75 ( 0 . 01 ) 0 . 008
2 ) 0 . 165
0 .0 1 7
168
L D 1
c 0 .0 1 m m
mm
f 1 1( 0 .0 0 0 8 ) 0 .0 0 8 8
fWr ( 0 . 0088 )( 1500 )( 0 . 025
kW
)(1 6 .6 7 )
2 .4 (1 0 )
n 1 6 .6 7 rp s ,
N m
SOLUTION (10.17) P
W DL
1 .2 5 M P a ,
4000 0 .0 8 0 .0 4
S (c) r
( a ) Figure 10.14: ( b ) Figure 10.16:
h0 c
c 0 .0 8
rp s ,
mm,
L D 1 2
30 mPa s
Fig.10.8: Thus
n 10
2 n P
( 0 .0 0 2 ) 1
0 .1 6 ,
3
)(1 0 ) 6
1 . 2 5 (1 0 )
0 .0 6
h 0 0 .1 6 ( 0 .0 8 ) 0 .0 1 3 m m
0 .2 6 ,
P p m ax
2 3 0 (1 0
p m ax
1 .2 5 0 .2 6
4 .8 0 8 M P a
SOLUTION (10.18) 8 mPa s
Figure 10.8: T
119 hp
n
Try S=0.03: Fig.10.15:
1 1 9 ( 0 .1 6 ) 2 7 .2
0 .7 N m ,
f 1 . 45 ,
r c
n 1 6 3 0 6 0 2 7 .2 F
T r
0 .7 0 .0 2 5
rp s ,
L D 1
28 N
f 1 . 45 ( 0 . 001 ) 0 . 00145
From Eq.(10.17): P ( cr )
Thus
2 n S
2 8 1 0
( 0 .01 0 1 )
3
( 2 7 .2 )
0 .0 3
7 .2 5 M P a
W P D L 7 .2 5 (1 0 )( 0 .0 5 )( 0 .0 5 ) 1 8 .1 3 k N 6
F fW 0 .0 0 1 4 5 (1 8 .1 3 1 0 ) 2 6 .3 N 3
Since
28 > 26.3, assumption S=0.03 is incorrect.
Try S=0.025: Fig.10.15: P ( 0 .01 0 1 )
r c
f 1.3 ,
2 ( 8 1 0
3
)( 2 7 .2 )
0 .0 2 5
f 1.3 ( 0 .0 0 1 ) 0 .0 0 1 3 8 .7 M P a
W P D L 8 .7 (1 0 )( 0 .0 5 )( 0 .0 5 ) 2 1 .8 6
kN
F fW 0 .0 0 1 3( 2 1 .8 1 0 ) 2 8 .3 N , 3
Assumption is correct.
( a ) W 2 1 .8 k N ( b ) Fig.10.14:
h0 c
0 . 13 , h 0 0 . 13 ( 0 . 001 25 ) 0 . 0033
( c ) Equation (10.20): 1
h0 c
mm
1 0 . 13 0 . 87
SOLUTION (10.19) h0
c 0 .0 0 1 5 (5 0 ) 0 .0 7 5 m m ,
Figure 10.14: P
c
0 .0 2 5 0 .0 7 5
1 3
,
L D 1 2
S=0.022 W DL
8000 ( 0 . 1 )( 0 . 05 )
1 . 6 MPa
n 900
60 15 rps
(CONT.)
169
10.19 (CONT.) ( a ) Equation (10.17):
SP n
0 . 22 ( 1 . 6 10
6
)
15
f 6 .5 ,
r c
( b ) Figure 10.15:
2
( cr )
( 0 . 0015 )
kW
( 80 0 . 050 ) 15
159
52 . 8 mPa s
f 6 .5 ( 0 .0 0 1 5 ) 0 .0 1
F fW 0 . 01 ( 8000 ) 80 Tfn
2
N
0 . 377
159
SOLUTION (10.20) P
W DL
1 .2 M P a ,
6000 0 .1 0 .0 5
r 50 m m ,
From Sec. 10.10: A 1 2 .5 D L 1 2 .5 (1 0 0 5 0 ) 1 0 Table 10.3: C 7 . 4 watts Equation (10.17): S ( cr )
2 n P
1 0 .0 0 1
)
L D 1 2
0 .0 6 2 5
2
m ,
C o
2 ( 2 0 1 0
3
)( 5 )
6
1.2 (1 0 )
f 3.2 ,
r c
Figure 10.15:
(
m
2
6
0 .0 8 3
f 3 .2 ( 0 .0 0 1 ) 0 . 0 0 3 2
Through the use of Eq.(10.24), we have H fW ( 2 r n ) 0 .0 0 3 2 ( 6 , 0 0 0 ) ( 2 0 .0 5 5 ) 3 0 .1 6 w a tts
Thus, by Eq.(10.23): t0 ta
20
H CA
30 . 16 7 . 4 ( 0 . 0625 )
20 65 . 2 85 . 2
o
C
SOLUTION (10.21) Refer to Solution of Prob.10.20. We have C 8 . 5 watts
m
Using Eq.(10.23): t 0 t a
2
C (Table 10.3) o
30
H CA
30 . 16 8 . 5 ( 0 . 0625 )
30 56 . 8 86 . 8
o
SOLUTION (10.22) C 14 kN
Table 10.5:
Fa
Table 10.7:
V Fr
3 1.2 ( 2 )
C s 6 . 95 kN Fa
1.2 5 e
X 0 .5 6
Cs
n 1500 3 6 .9 5
0 .4 3 2
Y 1.0 3 7 (by interpolation)
Thus, we obtain P XVF
r
YF a 0 . 56 (1 . 2 )( 2 ) 1 . 037 ( 3 ) 4 . 455
P VF
1 . 2 ( 2 ) 2 . 4 kN
or r
Hence L1 0
6
10 60 n
( CP )
a
6
10 6 0 (1 5 0 0 )
( 4 .41 45 5 ) 3 4 4 .8 h 3
170
kN
C
n 5 rp s
SOLUTION (10.23) Table 10.5:
C 1 4 .8 k N
Table 10.8:
V Fr
Fa
C s 7 .6 5
1.2 5 e 0 .9 5 ;
X 0 .3 7 ,
Y 0 .6 6
We have then P XVF
r
YF a 0 . 37 (1 . 2 )( 2 ) 0 . 66 ( 3 ) 2 . 868
P VF
1 . 2 ( 2 ) 2 . 4 kN
kN
or r
Thus L1 0
6
10 60 n
( CP )
a
6
10 6 0 (1 5 0 0 )
( 21.84 .86 8 ) 1 5 2 6 .9 3
h
SOLUTION (10.24) Refer to Solution of Prob. 10.22: C 14 kN ,
C s 6 .9 5
Fa
kN ,
Cs
0 .4 3 2
Table 10.7: Fa VF
r
3 1( 2 )
1 .5 e ;
X 0 . 56
Y 1 . 037
(by interpolation)
It follows that P K s [ X V F r Y F a ] 1 .5[( 0 .5 6 )1( 2 ) 1 .0 3 7 (3)] 6 .3 4 7 k N
or P K s VF r (1 . 5 )1 ( 2 ) 3 kN
Thus L1 0
6
10 60 n
( CP )
a
6
10 6 0 (1 5 0 0 )
( 6 .31 44 7 ) 1 1 9 .2 3
h
SOLUTION (10.25) Table 10.5: C 5 5 .9 k N
C s 3 5 .5 k N
Table 10.8: Fa V Fr
6 .7 5 (1 )( 2 2 .5 )
iF a
0 .3 e ,
Cs
1 ( 6 .7 5 ) 3 5 .5
0 .1 9
and X 1,
Y 0 .9 2
Thus P X V F r Y F a 1(1)( 2 2 .5 ) ( 0 .9 2 )( 6 .7 5 ) 2 9 k N
or P V F r (1)( 2 2 .5 ) 2 2 .5 k N
Hence L1 0
6
10 60 n
( CP ) 3
6
10 60 (700 )
( 5259.9 ) 1 7 0 .5 h 3
171
SOLUTION (10.26) Table 10.5: C 1 4 k N
C s 6 .9 5 k N
Table 10.7: Fa Cs
Fa
0 .2 9 e ,
2 6 .9 5
V Fr
2 (1 )( 4 )
0 .5 e
and X 0 .5 6 ,
Y 1 .1 4 2
(by interpolation)
Hence P X V F r Y F a 0 .5 6 (1)( 4 ) (1 .1 4 2 )( 2 ) 4 .5 2 k N
or P V F r (1)( 4 ) 4 k N
Thus L1 0
6
10 60 n
( CP )
a
6
10 60 (3500 )
( 41.542 ) 1 4 1 .5 h 3
From Fig. 10.26: K r 0 .6 2 . So L 5 K r L1 0 0 .6 2 (1 4 1 .5 ) 8 7 .7 h
SOLUTION (10.27) Variation factor: V 1 (Eq. 10.25) C 1 9 .5 k N
Table 10.5:
C s 10 kN
Table 10.7: Fa Cs
1 .7 10
0 .1 7 ,
Fa
e 0 .3 4 ;
V Fr
1 .7 (1 )( 4 .5 )
0 .3 8 e
and X 0 .5 6 ,
Y 1 .3 1
Thus P X V F r Y F a 0 .5 6 (1)( 4 .5 ) (1 .3 1)(1 .7 ) 4 .7 4 7 k N
P V F r (1)( 4 .5 ) 4 .5 k N
or
It follows that L1 0
6
10 60 n
( CP )
a
6
10 60 (600 )
( 41.79 .54 7 ) 1 9 2 6 h 3
SOLUTION (10.28) K s 2 .5
Table 10.9:
V 1 .2
and Fa V Fr
1 .7 (1 .2 )( 4 .5 )
0 .3 1 5 e ;
X 0 .5 6 ,
Y 1 .3 1
Therefore, Eq. (10.27): P K s ( X V F r Y F a ) 2 .5[( 0 .5 6 )(1 .2 )( 4 .5 ) (1 .3 1)(1 .7 )] 1 3 .1 3 k N
(CONT.)
172
10.28 (CONT.) or P K sV F r 2 .5 (1 .2 )( 4 .5 ) 1 3 .5 k N
Hence L1 0
6
10 60 n
a
( CP )
6
10 60 (600 )
( 11 39 .5.5 ) 8 3 .7 h 3
From Fig. 10.26: K r 0 .6 2 . So, L 5 K r L1 0 0 .7 (8 3 .7 ) 5 8 .6 h
SOLUTION (10.29) P2 3
3
L '10 P1 2 L " 10
0 . 5 P1 ,
P 2 0 . 794 P1
3
20 . 6 %
SOLUTION (10.30) 10
10 3
P2
L '10 P1 3 2 L " 10
10
0 . 5 P1 3 ,
P 2 0 . 812 P1
18 . 8 %
SOLUTION (10.31) Table 10.5:
C 5 5 .9 k N ,
Table 10.8:
VF
Fa
1 .5 1 .2 ( 5 )
r
C s 3 5 .5 k N iF a
0 . 25 e ,
Cs
2 ( 1 .5 ) 35 . 5
0 . 085
and X 1,
Y 1.3 8 6 (by interpolation)
Then we obtain P XVF
r
YF a 1(1 . 2 )( 5 ) 1 . 386 (1 . 5 ) 8 . 079
or P V F r 1 .2 (5 ) 6 k N
Thus L1 0
6
10 60 n
( CP )
a
6
10 6 0 (1 0 0 0 )
( 85.05 7.99 ) 5 5 2 1 h 3
and 5 L1 0 5 (5 5 2 1) 2 7 , 6 0 5 h
SOLUTION (10.32) P K s [V X F r 0 ] 1 .7[1 .2 (1)5 ] 1 0 .2
35 mm-02-series (Table 10.6): L1 0
6
10 60 n
( CP ) a
6
10 60 ( 2400 )
10
[ 13 01 .2.9 ] 3 3 1 0 .6 5 h
35 mm-03-series (Table 10.6): L1 0
6
10 60 ( 2400 )
10
[ 14 04 .2.6 ] 3 9 4 9 .3 h
173
kN
kN
SOLUTION (10.33) P K sV F r 1(1)( 2 5 ) 2 5
(Eq.10.26)
kN
75 mm-02-series (Table 10.6): L1 0
6
10 60 n
( CP )
a
6
10
10 60 ( 2000 )
[ 9215.3 ] 3 6 2 5 .1 h
and 5 L1 0 3,1 2 5 h r 75 mm-03-series (Table 10.6): L1 0
10
6
10 60 ( 2000 )
[ 12853 ] 3 6 3 4 6 h
and 5 L1 0 3 1, 7 3 0 h
SOLUTION (10.34) P K sV F r 2 (1 1 1 2 .5 ) 2 5
(Eq.10.26)
kN
We have L1 0
6
10 60 n
6
24
a
( CP ) :
02-series (Table 10.6): 03-series (Table 10.6):
10 60 ( 2400 )
3
[ 2C4 ] ,
C 3 6 .2 8 6 k N
40-mm-bore bearing 30-mm-bore bearing
SOLUTION (10.35) Fa
Table 10.7:
Cs
0 , use 0.014: X 1 ,
Y 0
We have, by Eq.(10.27), P K s XVF
r
2 . 5 (1)( 1 . 2 ) 8 24 kN
Thus L 10
6
10 60 n
( CP )
From Table 10.5:
a
40 5
:
6
10 60 ( 900 )
C [ 24 ] , C 18 . 1 kN 3
30-mm-bore bearing
SOLUTION (10.36) Refer to Solution of Prob.10.22: C 14 kN , Figure 10.26: K r 0 .6 2 .
P 4 . 455
We now have n 1200
rpm .
Hence L5 K r
6
10 60 n
( CP ) 0 .6 2 3
6
10 6 0 (1 2 0 0 )
( 4 .41 45 5 ) 2 6 7 h 3
174
kN
SOLUTION (10.37) Figure 10.26:
K
r
0 .3 2 ,
Refer to Solution of Prob.10.32:
35 mm-02-series: L 5 K r L1 0 0 .3 2 (3 1 0 .6 5 ) 9 9 .4 1 h
35 mm-03-series: L 5 K r L1 0 0 .3 2 (9 4 9 .3) 3 0 3 .8 h
End of Chapter 10
175
CHAPTER 11
SPUR GEARS
SOLUTION (11.1) Equation (11.4): m
d N
d 6 (3 2 ) 1 9 2 m m ,
,
r
d 2
96 m m
Table 11.1: a m 6 mm ,
h 2 .2 5 ( 6 ) 1 3 .5 m m ,
hk 2 ( 6 ) 1 2 m m
Equation (11.9) of Sec.11.4: rb r c o s 9 6 c o s 2 0
9 0 .2 1 1 m m ,
o
r0 r a 9 6 .2 1 m m
SOLUTION (11.2) N1
Equation (11.8): r s
N
2
or N 1
1 3
N
2
3
Equations (11.6) and (11.5b): N1 N 2
c r1 r2
2P
m 2
(N1 N 2)
3 2
(
N
2
3
N2)
3 2
(
4 3
N 2 ) 360
or N
1 8 0 teeth.
2
Thus N 1
180 3
6 0 teeth
SOLUTION (11.3) ( a ) Pitch circle is P m 1 0 3 1 .4 2 m m
Pitch diameters are d
N
p
p
1 8 (1 0 ) 1 8 0 m m ,
d g 4 2 (1 0 ) 4 2 0 m m
The center distance: d
c
p
dg 2
1 2
(1 8 0 4 2 0 ) 3 0 0 m m
Base circle are Pb p r p c o s 9 0 c o s 2 0 Pb g 2 1 0 c o s 2 0
(b) c
d
p
dg
o
o
8 4 .5 7 m m
1 9 7 .3 4 m m
300 8 308 m m
2
(1)
The velocity ratio does not change. Hence d
p
dg
18 42
(2)
Solving Eqs. (1) and (2), d
p
1 8 4 .8 2 m m
d g 4 3 1 .1 9 m m
Since rb r c o s , the new pressure angle is new c o s
1
rb p d
p
2
cos
1
8 4 .5 7 1 8 4 .8 2 2
2 3 .7 7
176
o
SOLUTION (11.4) Addendum and circular pitch: a 1 m 1 (4) 4 m m
(Table 11.1)
p m ( 4 ) 1 2 .5 5 6 m m ; (Eq. 11.5a)
Tooth thickness, t 1 .1 5 7 1 m 1 .5 7 1( 4 ) 6 .2 8 4 m m
(Table 11.1)
SOLUTION (11.5) Refer to Table 11.1;
m 1 P :
with
Dedendum,
b d 1 .2 5 m 1 .2 5 (1 2 ) 1 5 m m
Clearance,
f 0 .2 5 m 0 .2 5 (1 2 ) 3 m m
h k 2 m 2 (1 2 ) 2 4 m m
Working depth,
t 1 .5 7 1 m 1 .5 7 1(1 2 ) 1 8 .8 5 2 0 m m
Thickness,
Circular pitch, by Eq. (11.3): p m (1 2 ) 3 7 .6 9 9 1 m m
SOLUTION (11.6) d g 4d
p
c
p
(Eq. 11.8)
We have 1 2
(d
d g ) 2 0 0;
5d
p
400
or
and m d
p
N
4N
p
p
,
N
p
d
p
m 80 5 16
Thus, N
g
64
SOLUTION (11.7) Equation (11.8), rs
N
p
N
g
1200 2400
1 2
,
N
g
2N
p
2 (1 8 ) 3 6
Equation (11.4), d
p
N p m 1 8 (3 ) 5 4 m m
and c
1 2
(d
p
dg)
1 2
(5 4 3 6 ) 4 5 m m
Equation (11.5a) gives p m ( 3 ) 9 .4 2 4 8 m m
177
d
p
80 m m
SOLUTION (11.8) rs
Equation (11.8): m d
p
p
0 . 5234
mN
p
N
p
N
g
1 4
N
p
84
,
N
p
2 1 teeth
4
13 . 1942
4 ( 2 1) 8 4 m m ,
dg mN
p
4 (8 4 ) 3 3 6 m m
Therefore c
1 2
(d g d p )
1 2
(3 3 6 8 4 ) 2 1 0 m m
SOLUTION (11.9) ( a ) From Eq. (11.4), pitch diameters are d g N g m 6 0 (8 ) 4 8 0 m m d
p
2 4 (8 ) 1 9 2 m m
The center distance c
1 2
(d
p
dg)
1 2
(1 9 2 4 8 0 ) 3 3 6 m m
The base circles are rb p
rb g
d
p
2
480 2
cos
cos 20
192 2
o
cos 20
o
9 0 .2 1 m m
2 2 5 .5 2 m m
( b ) When center distance is increased by 6 mm, we have c=342. This corresponds to a 1.79% increased in center distance. However, d
p
d g 2 (3 4 2 ) 6 8 4 m m
(1)
and N
p
d
p
N
g
dg
;
24 dp
60 dg
(2)
Solving Eqs. (1) and (2): d g 4 8 8 .6 m m
d
p
1 9 5 .4 m m
We have m
d
p
N
p
1 9 5 .4 24
8 .1 4 2 m m
Circular pithch p m 8 .1 4 2 2 5 .5 8 m m
Changing center distance does not affect the base circle. Thus, new c o s
1
(
rb p rp
) cos
1
( 1 99 50 .4.2 12 ) 2 2 .6
178
o
SOLUTION (11.10) Equation (11.8): rs
N
p
N
g
1 4
N
4N
g
p
4 ( 22 ) 88 teeth
Equation (11.4): d g N g m 88(4) 352
d
p
mm
N p m 22 ( 4 ) 88 mm
Hence, c
(3 5 2 8 8 ) 2 2 0
1 2
mm
SOLUTION (11.11)
From Eq. (11.5b), the modulus is 1 m
N
p
d
p
or d p N p m 2 5 (3 ) 7 5 m m
The speed at the contact point, V p V g V : V w p rb p w p
d
p
2
cos 20
o
( 3 46 00 0 ) 2 ( 725 ) c o s 2 0
o
1 2 , 5 4 7 m m s 1 2 .5 5 m s
SOLUTION (11.12)
From Eq. (11.5a), the circular pitch, p m ( 2 ) 6 .2 8 m m
By Eqs. (11.5b) and (11.6), the center distance: c
1 2
(N
p
N g )m
1 2
(3 0 1 0 0 )( 2 ) 1 3 0 m m
The pitch circular radii are rp
1 2
N pm
1 2
(3 0 )( 2 ) 3 0 m m
rg
1 2
N gm
1 2
(1 0 0 )( 2 ) 1 0 0 m m
Using Eq. (11.9), The base radii: rb p r p c o s 3 0 c o s 2 0
o
2 8 .1 9 m m
rb g rg c o s 1 0 0 c o s 2 0 9 3 .9 7 m m o
The addendum is equal to a p
1 p
m 2 mm
(CONT.)
179
11.12 (CONT.) The order radii are therefore ro p r p a 3 0 2 3 2 m m
ro g 1 0 0 2 1 0 2 m m
and
ro p ro g 1 3 4
SOLUTION (11.13)
Refer to Solution of Prob. 11.9. We have p m (8 ) 2 5 .1 3 3 m m , rg 4 8 0 2 2 4 0 m m , rb p 9 0 .2 1 m m ,
20
o
rp 1 9 2 2 9 6 m m
rb g 2 2 5 , 5 2 m m ,
c 336 m m
From Table 11.1, the addendum for gears are a 1 p 8 mm
The outside radii are then ro p r p a 9 6 8 1 0 4 m m ro g 2 4 0 8 2 4 8 m m
Substituting the data into Eq. (11.14), the contact ratio: Cr
[ 1 0 4 9 0 .2 1 2
1 2 5 .1 3 3 c o s 2 0
o
2
1 .6 9 4
SOLUTION (11.14)
We have e (
N1 N3
N
)( N 3 ) 4
24 96
0 .2 5
Substituting into Eq. (11.18) 0 .2 5
0 n2 120 n2
,
n 2 2 4 rp m
Direction is the same as that of the sun gear.
180
2 4 8 2 2 5 .5 2 ] 2
2
3 3 6 ta n 2 0 2 5 .1 3 3
o
SOLUTION (11.15) kW
Tn 9549
T1
,
22 (9549 )
d 1 2 7 (8 ) 2 1 6 m m ,
( a ) Ft1
T1 r1
5 2 .5 2 0 .1 0 8
d 2 4 8 (8 ) 3 8 4 m m ,
d 3 3 6 (8 ) 2 8 8 m m
(Eq.11.2)
486 N
Equation (11.19): F r F t ta n ( b ) RC
5 2 .5 2 N m
4000
486 227 2
2
o
4 8 6 ta n 2 5
o
2 2 7 lb
T C 4 8 6 ( 0 .1 4 4 ) 7 0 N m
536 N ,
486 227
A
486
B
486
2 2 .9
o
C RC
486 TC
SOLUTION (11.16) T1
9549 kW n
22 (9549 ) 4000
5 2 .5 2 N m
d 1 2 7 (5 ) 1 3 5 m m ,
( a ) Ft 1
T1 r1
5 2 .5 2 0 .0 6 7 5
d 2 4 8 (5 ) 2 4 0 m m ,
d 3 3 6 (5 ) 1 8 0 m m
778 N
Equation (11.19): F r 1 F t 1 ta n 2 0 2 8 3 N o
( b ) RC
778 283 2
2
T C 7 7 8 ( 0 .0 9 ) 7 0 N m
828 N ,
778 B
A
45
283
o
778
778
20
778
o
C RC TC
181
(Eq.11.4)
SOLUTION (11.17) Equation (11.4): d 1 2 0 ( 6 ) 1 2 0 m m ,
d 2 40(6) 240 m m ,
d 3 20(6) 120 m m ,
( a ) T1
9549 kW n
Ft 1
9549 (37 )
2 9 4 .4 N m
1200
F r 1 4 .9 1 ta n 2 0
4 .9 1 k N ,
2 9 4 .4 0 .0 6
d 4 60(6) 360 m m
We have 4 .9 1( 0 .1 2 ) 0 .0 6 F t 3 ,
o
1 .7 9 k N
F t 3 9 .8 2 k N ,
(Eq.11.19)
F r 3 9 .8 2 ta n 2 0 3 .5 7 k N o
2 ( b ) F x 0 : F x 4 .9 1 3 .5 7 1 .3 4 k N
B
3
3.57
9.82 o
9 .5
RB
4.91 RB
F y 0 : F y 9 .8 2 1 .7 9 8 .0 3 k N 8 .0 3 1 .3 4 2
8 .1 4 k N
2
TB 0
1.79
SOLUTION (11.18) ( a ) T1
9549 kW n T1
F t1
r1
9549 ( 80 )
1600
477 . 5 0 . 054
8 . 843
T1
2
1
F r 1 8 . 843 tan 20
kN ,
3
B
A
mN
r1
2
8843
1
477 . 5 N m ,
TC
8843
20
2 o
54 mm
3 . 219
( b ) r3
C
3219
6 ( 18 )
(Eq.11.4) (Eq.11.19)
kN
mN 2
3
6 ( 50 )
150
2
mm
T C 8 . 843 ( 0 . 15 )
o
1 .3 2 6 k N m
RC
RC
8 . 843
9 . 411
2
3 . 219
2
kN
SOLUTION (11.19) 9549 kW
( a ) T1
n
9549 ( 80 )
1600
Equation (11.4): r1 F r 1 6632 tan 25
( b ) r3
mN3
RC
2
8 (50 ) 2
o
477 . 5 N m mN
1
2
8 ( 18 ) 2
3 . 093
200
6632 3093 2
72 mm .
F t1
r1
477 . 5 0 . 072
6 . 632
kN
kN
mm 2
7 .3 1 8
T C 6 6 3 2 ( 0 .2 0 0 ) 1 .3 2 6 k N m
kN ,
6632 B
A
T1
C
T1
TC
3093
25
o
6632 RC
182
SOLUTION (11.20) Equation (11.4): d 1 N 1 m 1 5 (5 .2 ) 7 8 m m , d 3 2 0 (5 .2 ) 1 0 4 m m ,
( a ) T1
9549 kW n
Ft 1
9 5 4 9 ( 7 .5 ) 1500
d 4 4 0 (5 .2 ) 2 0 8 m m
4 7 .5 7 N m F r 1 1 .2 2 ta n 2 5
1 .2 2 k N F t 2 ,
4 7 .5 7 0 .0 3 9
d 2 3 5 (5 .2 ) 1 8 2 m m ,
o
0 .5 6 k N F r 2
T 2 1 .2 2 ( 0 .0 9 1) 1 1 1 N m Ft 3
2 .1 3 k N ,
111 0 .0 5 2
F r 3 2 .1 3 ta n 2 5
o
0 .9 9 k N
(b) 4
2 3
2.13
RC TC
B
65
0.99
0.56
C
0.99
1.22
o
T C 2 .1 3( 0 .1 0 4 ) 2 2 1 .5 N m
RC
SOLUTION (11.21)
d 3 18(4) 72 m m ,
( a ) T1
9549 kW n
Ft 1 Ft 2
9 5 4 9 (1 5 ) 1800
7 9 .5 8 0 .0 4 8
7 9 .5 8
d 2 36(4) 144 m m , d 4 42(4) 168 m m
N m
1 .6 6 k N
F r 1 1 .6 6 ta n 2 5
o
0 .7 7
kN
T 2 T 3 1 .6 6 ( 0 .0 7 2 ) 1 1 9 .5 N m Ft 3
3 .3 2
1 1 9 .5 0 .0 3 6
kN ,
F r 3 3 .3 2 ta n 2 5
( b ) R x 1 . 66 1 . 55 0 . 11 kN RB
R x R y 2 .5 5 2
2
0.77
RB
1.66
2 2 .5
o
B
o
1 .5 5 k N
R y 3 . 32 0 . 77 2 . 55 kN
kN
1.55
3 3.32
183
2
2 .3 5 k N
2.13
Equation (11.4): d 1 2 4 ( 4 ) 9 6 m m ,
0 .9 9 2 .1 3
2
SOLUTION (11.22) Equations (11.2) and (11.8): d 1 1 5 (5 .2 ) 7 8 m m , n 2 n1
N1
1 5 0 0 ( 13 55 ) 6 4 2 .9
N2
0b Ym K
1
f
rp m
Table 11.4: 0 1 7 2 M P a and 2 5 0 B h n
( a ) Table 11.3: Y=0.443, Fb
d 2 3 5 (5 .2 ) 1 8 2 m m ,
6
1 7 2 1 0 ( 0 .0 1 2 ) 0 .4 4 3 ( 0 .0 0 5 2 ) 1 .6
1
( b ) Table 11.10: K 1 . 117 ,
Q
2 .9 7 2
2 ( 35 )
1 .4
15 35
(Eq.11.33)
kN
(Eq.11.40)
F w d p b Q K 7 8 (1 2 )(1 .4 )(1 .1 1 7 ) 1 .4 6 4 k N
( c ) V2
dn
12
( 0 .1 8 2 )( 6 4 2 .9 ) 60
6 .1 2 m s ,
Fd
Thus
2 .9 7 2 3 .0 0 7 F t ,
Ft 9 8 8
and
1 .4 6 4 3 .0 0 7 F t ,
Ft 4 8 7 N
(Eq.11.38) 3 .0 5 6 .1 2 3 .0 5
F t 3 .0 0 7 F t
(Eq.11.24a)
N
SOLUTION (11.23) Equations (11.4) and (11.8): d 3 mN
V3
3
dn
5 ( 20 ) 100
( 0 .1)( 1 16 20 5 ) 5 .8 9
60
( a ) Table 11.3: Y 0 .3 2 0 , Fb
mm ,
0 bYm K
f
1 1 .5
( c ) Fd
3 . 0 5 5 .8 9 3.0 5
1500 ( 34 ) 1125
N3
rpm
m s
Table 11.4:
0
172
and 250 Bhn
MPa
(172 )( 15 )( 0 . 320 )( 5 ) 2 . 75 kN
( b ) Table 11.10: K 0 .9 0 3 M P a , F w d 3 bQK
N1
n 3 n1
Q
2N4 N3 N4
(Eq.11.33)
100 (15 )( 43 )( 0 . 903 ) 1 . 81 kN
2 ( 40 ) 20 40
4 3
(Eq.11.40)
(Eq.11.38)
F t 2 .9 3 1 F t
Thus
2 . 75 10
and
1 . 81 10
3
3
2 . 931 F t ,
F t 938
2 . 931 F t ,
F t 617 . 5 N
N
SOLUTION (11.24) Refer to Solution of Prob. 11.23. Equation (11.4) and (11.8): d 1 m N 1 5 (1 5 ) 7 5 m m ,
V1
dn 60
n 2 n1
N1 N2
1 5 0 0 ( 13 55 ) 6 4 2 .9 rp m
( 0 .0 7 5 )( 1 56 00 0 ) 5 .8 9 m s
(CONT.)
184
11.24 (CONT.) ( a ) Table 11.3: Y 0 .2 8 9 , Fb
obYm K
1 1 .5
f
Table 11.4: 1 7 2 M P a at 2 5 0 B h n (1 7 2 ) (1 5 ) ( 0 .2 8 9 ) ( 5 ) 2 .4 9 k N
( b ) Table 11.10: K 0 .9 0 3 M P a ,
2N2
(Eq. 11.40)
F w d 1 b Q K 7 5 (1 5 )(1 .4 )( 0 .9 0 3) 1 .4 2 k N
(Eq. 11.38)
N1 N 2
2 (35 )
1 .4
3 .0 5 5 .8 9 3 .0 5
( c ) Fd
Q
(Eq. 11.33)
15 35
F t 2 .9 3 1 F t
Thus 2 .4 9 1 0 2 .9 3 1 F t ,
F t 8 4 9 .5 N
1 .4 2 1 0 2 .9 3 1 F t ,
F t 4 8 4 .5 N
3
and 3
SOLUTION (11.25)
We have d N m 2 2 ( 2 .5 ) 5 5 m m ( a ) o 2 2 1 M P a (Table 11.4), Y 0 .3 3 (Table 11.3) Equation (11.33): Fb
(b) V
dn 60
obYm K
2 2 1 ( 3 0 ) ( 0 .3 3 ) ( 2 .5 )
1 .5
f
( 0 .0 5 5 )(1 5 0 0 )
4 .3 2 m p s
60
Fd
Equation (11.24a):
3 .6 5 k N
3 .0 5 4 .3 2 3 .0 5
F t 2 .4 2 F t
Hence 3 .6 5 2 .4 2 F t ,
F t 1 .5 0 8 k N
Equation (11.22) is then kW
Ft V 1000
1 5 0 8 ( 4 .3 2 ) 1000
6 .5 1
SOLUTION (11.26) Equations (11.4), (11.8), and (11.20'): d 2 36(4) 144 m m , n 2 1 8 0 0 ( 23 ) 1 2 0 0
rpm ,
d 3 18(4) 72 V
dn2 60
mm
( 0 .1 4 4 )(1 2 0 0 ) 60
9 .0 5
m ps
(CONT.)
185
11.26 (CONT.) ( a ) Table 11.3: Y 0 .4 4 6 (by interpolation), Table 11.4: 0 1 7 2 M P a (for steel of 200 Bhn) 0b Ym
Fb
K
1
f
1 7 2 (1 0 ) ( 0 .4 4 6 ) ( 4 ) 1 .4
2 .1 9 2 k N (Eq.11.33)
( b ) Table 11.10: K 0 .6 7 6 M P a ,
2 ( 42 )
Q
18 42
F w d 3 b Q K 7 2 (1 0 )( 2 1 1 5 )( 0 .6 7 6 ) 6 8 1 .4
( c ) Fd
3 . 05 9 . 05 3 . 05
F t 3 . 97 F t
(Eq.11.40) (Eq.11.38)
N
( Eq.11.24a)
2 .1 9 2 3 .9 7 F t ,
Thus
21 15
Ft 5 5 2
N
SOLUTION (11.27) Equation (11.4): d 1 24 (10 ) 240 ( a ) Table 11.3: Y 0 .3 9 3 (interpolated). 0 bYm
Fb
K
f
( b ) Table 11.10: F w d 1 bQK
1 1 .5
mm
Table 11.4:
0
172
[172 (15 )( 0 . 393 )( 10 )] 6 . 76 kN
K 0 . 545
Q
MPa,
2 ( 36 ) 24 36
(for steel of 200 Bhn)
MPa
(Eq.11.33)
6 5
(Eq.11.40)
240 (15 )( 65 )( 0 . 545 ) 2 . 35 kN
(Eq.11.38)
SOLUTION (11.28)
Refer to Solution of Prob. 11.27. Equation (11.4): d 1 2 4 (1 0 ) 2 4 0 m m , ( a ) Table 11.3: Y 0 .3 3 7 , Table 11.4: Fb
obYm K
f
1 1 .5
d 3 1 8 (1 0 ) 1 8 0 m m
o 1 7 2 M P a (for steel of 200 Bhn)
[1 7 2 (1 5 ) ( 0 .3 3 7 ) (1 0 ) ] 5 .8 k N (Eq. 11.33)
( b ) Table 11.10: K 0 .5 4 5 M P a ,
Q
2 N3 N3 N4
2 (1 8 ) 18 42
0 .6
(Eq. 11.40)
F w d 3 b Q K 1 8 0 (1 5 )( 0 .5 4 5 ) 8 8 3 N
(Eq. 11.38)
SOLUTION (11.29) d
p
N p m 24(2) 48
V
d pnp
Kv
60
6 . 1 4 . 02 6 .1
( 0 .0 4 8 )(1 6 0 0 ) 60
1 . 659
mm,
d g 60(2) 120
4 .0 2 m p s
mm
(Eq.11.4)
(Eq.11.20')
(Fig.11.15) (CONT.)
186
11.29 (CONT.) Ft
1000 kW V
1 0 0 0 ( 0 .9 )
2 2 3 .9
4 .0 2
N
J 0 .2 6
Pinion: (Fig.11.16):
Table 11.7: S t 1 9 7 .5 M P a
(mid-point).
Equation (11.35'): Ft K 0 K v
KsKm
1 bm
2 2 3 .9 (1) (1 .6 5 9 )
J
1 (1 .6 ) 1 0 .0 1 5 ( 0 .0 0 2 ) 0 .2 6
7 6 .2 M P a
Equation (11.36):
a ll
St K L
KT K R
1 9 7 .5 (1 )
1 (1 .2 5 )
158
J 0 .3 0 ,
Gear: (Fig.11.16):
2 2 3 .9 (1)(1 .6 5 9 )
M Pa
1 0 .0 1 5 ( 0 .0 0 2 )
0 .3
,
Yes.
S t 58 . 6 ksi
Table 11.7: 1 (1 .6 )
a ll
66 M Pa
and
a ll
5 8 .6 (1 ) 1 (1 .2 5 )
4 6 .9
M Pa
a ll
,
No.
SOLUTION (11.30) From Solution of Prob.11.29: d p 4 8 m m , F t 2 2 3 .9 N ,
C H 1 . 001
mG d g d
p
d g 120 m m ,
2 .5 ,
ng np
N
p
N
g
V 4 .0 2 m p s , 24 1600 ( 60 ) 640
(from Eq.11.46)
Pinion: c C p ( Ft K 0 K v
where
C
p
I
166
Ks
KmC
bd
I
f
1
)
(Eq.11.42)
2
M P a (Table 11.11)
sin cos
mG
2
m G 1
0 . 161
60 84
0 . 115
Thus c 1 6 6 1 0 [ 2 2 3 .9 (1)(1 .6 5 9 ) 3
S c 620 . 5 MPa
c , a ll
ScC LC H
KT K R
1 .6 (1 ) 1 ( 0 .0 1 5 )( 0 .0 4 8 ) 0 .1 1 5
(Table 11.12)
6 2 0 .5 (1 )(1 .0 0 1 ) 1 (1 .2 5 )
1
] 2 4 4 4 .7 M P a
(mid-point)
4 9 6 .9 M P a
(Eq.11.44)
Hence c , a ll c
Gear: Table 11.12:
Yes.
S c 4 8 5 .5
(mid-point)
M Pa
c 1 6 6 1 0 [ 2 2 3 .9 (1)(1 .6 5 9 ) 3
c , a ll
4 8 2 .5 (1 )(1 ) 1 (1 .2 5 )
1 .6 (1 ) 1 ( 0 .0 1 5 )( 0 .1 2 ) 0 .1 1 5
386 M Pa
Hence c , a ll c
Yes.
187
1
] 2 2 8 1 .3 M P a
rpm
SOLUTION (11.31) d
p
N p m 2 0 (3 ) 6 0 m m ,
V
d pnp
( 0 .0 6 )(1 2 0 0 )
60
Ft
1000 kW V
(Eq. 11.4)
3 .7 7 m p s
60
K 1 .6 1 8
d g 5 0 (3 ) 1 5 0 m m
(Fig. 11.15)
1 0 0 0 (1 .0 )
2 6 5 .3 N
3 .7 7
J 0 .2 4 5 (mid-point)
Pinion, (Fig. 11.16):
Table 11.7: S t 2 8 6 M P a
Equation (11.35'): Ft K o K v
KsKm
1 bm
2 6 5 .3 (1)(1 .6 1 8 )
J
1 (1 .7 ) 1 ( 0 .0 1 5 )( 0 .0 0 3 ) 0 .2 4 5
6 6 .2 M P a
Equation (11.36):
a ll
St K L
KT K R
2 8 6 (1 ) 1 (1 .3 )
Table 11.7: S t 8 9 .6 M P a
2 6 5 .3 (1)(1 .6 1 8 )
8 9 .6 (1 )
a ll
1 (1 .3 )
, Yes. a ll
Gear, (Fig. 11.16): J 0 .3 ,
and
220 M Pa
1 (1 .7 ) 1 ( 0 .0 1 5 )( 0 .0 0 3 ) 0 .3
6 8 .9 M P a
a ll
5 4 .1 M P a
, Yes.
SOLUTION (11.32) Refer to Solution of Prob. 11.31 d
p
60 m m ,
ng n p
N
p
N
g
d g 150 m m ,
V 3 .7 7 m p s ,
1 2 0 0 ( 52 00 ) 4 8 0 rp m ,
C H 1 .0 0 8
F t 2 9 6 .7 N ,
mG d g d
p
2 .5
K 4 .2
(from Eq. 11.46).
Pinion:
where
C
p
I
Then
c
Ks
C p [ Ft K o K v
166 10
3
s in c o s
mG
2
m G 1
1 .6 1
50 70
KT K R
7 8 9 (1 )(1 .0 0 8 ) 1 (1 .3 )
1
]
2
(Eq. 11.42)
1 .1 5
S c 8 7 9 M P a (Table 11.12) ScC LC H
f
(Table 11.11)
1 6 6 1 0 [ 2 9 6 .7 (1) ( 4 .2 )
c , a ll
I
M Pa
3
c
K mC
bd
1 .7 (1 ) 1 ( 0 .0 1 5 )( 0 .0 6 ) 1 .1 5
1
]
2
2 3 7 .5 M P a
(mid-point)
6 8 1 .6 M P a
(Eq.11.44) (CONT.)
188
11.32 (CONT.) Hence, c , a ll c
Yes.
Gear: Table 11.12: S c 5 5 1 .5 M P a
1 6 6 1 0 [ 2 9 6 .7 (1) ( 4 .2 ) 3
c
c , a ll
So
(mid-point)
5 5 1 .5 (1 )(1 )
1
]
2
1 5 0 .2 M P a
4 2 4 .2 M P a
1 (1 .3 )
c , a ll
1 .7 (1 ) 1 ( 0 .0 1 5 )( 0 .1 5 ) 1 .1 5
Yes.
c
SOLUTION (11.33) N
p
20, 2( 40 )
Q
20 40
Thus
N
4 3
40,
g
d
2 0 (5 ) 1 0 0
p
Table 11.10: S e 6 2 1 k s i,
,
b 50 m m
mm,
K 1 .8 2 1 M P a
F w d p b Q K 1 0 0 (5 0 )( 43 )(1 .8 2 1) 1 2 .1 4
(Eq.11.38)
kN
Equations (11.20') and (11.24a): dn
V
60
( 0 .1 )( 9 0 )
60
0 .4 7 m p s ,
Then, we obtain 1 2 .1 4 1 .1 5 4 F t ,
Fd
3 .0 5 0 .4 7 3 .0 5
F t 1 .1 5 4 F t
F t 1 0 .5 2 k N
Hence kW
Ft V 1000
1 0 ,5 2 0 ( 0 .4 7 )
1000
4 .9 5
SOLUTION (11.34) From Solution of Prob.11.33: N
p
20, N
g
40, d ScCLC H
c , a ll
KT K
1 0 0 m m , b 5 0 m m , V 0 .4 7 m p s , m G N
p
where S c 5 5 1 .5 M P a (Table 11.12)
,
R
C
L
Then
5 5 1 .5 (1 .1 )(1 )
6
F t ( 6 0 6 C.7 1 0 )
2
1 K0Kv
p
where
C
p
149
Kv
M Pa
5 . 56 0 . 47 5 . 56
K 0 1,
I
6 0 6 .7
(1 )1
K
1,
sin cos
mG
2
m G 1
I K mC
p
2
(mid-point)
K T 1,
K
R
1
M Pa
(Eq.11.42) f
(Table 11.11)
1 . 08 s
bd Ks
N
1.1 (Fig.11.19)
C H 1,
c , a ll
g
(Curve B, Fig.11.15) K m 1 .3,
C
f
1
0 . 107
(CONT.)
189
11.34 (CONT.) 6
F t ( 6 0 6 .7 1 03 )
Thus
0 .0 5 ( 0 .1 ) 0 .1 0 7 1 1 (1 .0 8 ) 1 1 .3 (1 )
2
1 4 9 1 0
6 .3 1 8 k N
and Ft V
kW
1000
6 3 1 8 ( 0 .4 7 )
2 .9 7
1000
SOLUTION (11.35)
N
2 5,
p
2 (50 )
Q
25 50
50,
g
4 3
, Table 11.10: S e 6 2 1 M P a ,
d
N m 2 5 (5 ) 1 2 5 m m ,
N
p
b 40 m m
K 1 .8 2 1 M P a
Thus, F w d p b Q K 0 .1 2 5 ( 0 .0 4 )( 43 )(1 .8 2 1 1 0 ) 1 2 .1 4 k N 6
(Eq. 11.38)
Equations (11.20') and (11.24a) with 6 0 0 fp m 6 0 0 1 9 6 .8 3 .0 5 m s : dn
V
60
( 0 .1 2 5 )(1 2 0 )
0 .7 8 5 m s ,
60
Fd
3 .0 5 0 .7 8 5 3 .0 5
F t 1 .2 5 7 F t
Then, we have 1 2 .1 4 1 .2 5 7 F t ,
F t 9 .6 6 k N
Hence, kW
Ft V 1000
9 6 6 0 ( 0 .7 8 5 ) 1000
7 .5 8
SOLUTION (11.36) N
N
p
ng g
60
np
Table 11.4:
240 600
24,
8 7 .2
0g
Y p 0 .3 3 7 ,
M Pa,
0 p
Y g 0 .4 2 1
172
M Pa
Hence Y p
0 p
0 .3 3 7 (1 7 2 ) 5 8
Yg
0g
0 .4 2 1(8 2 .7 ) 3 4 .8 2
Gear is weaker
Thus Fb
0bYm K
8 2 .7 ( 9 0 ) ( 0 .4 2 1 ) ( 4 ) 1 .4
f
8 .9 5 3
kN
We have d g 60(4) 240 m m ,
V
dn 60
( 0 .2 4 )( 2 4 0 ) 60
3 .0 2 m p s
Hence Fd
3 .0 5 3 .0 2 3 .0 5
F t 1 .9 9 F t
8 .9 5 3 1 .9 9 F t ,
and kW
Ft V 1000
4 5 0 0 ( 3 .0 2 ) 1000
1 3 .5 9
190
F t 4 .5 k N
SOLUTION (11.37) N
N
p
ng
60
np
g
Table 11.4:
24,
240 600
Y p 0 .3 3 7 ,
82 . 7 MPa ,
0g
Y p 0 p Yg 0 g
Y g 0 .4 2 1
172
0 p
MPa
the gear is weaker
Thus 0 bYm
Fb
K
82 . 7 ( 80 )( 0 . 421 )( 4 )
7 . 958
1 .4
f
kN
We have the quantities: d
g
N g m 60 ( 4 ) 240
mm
V d n 6 0 ( 0 .2 4 )( 2 4 0 6 0 ) 3 .0 2 m p s 3 .0 5 3 .0 2 3 .0 5
Fd
F t 1 .9 9 F t
(from Eq.11.24a)
and 7 , 9 5 8 1 .9 9 F t ,
Ft 4 k N
Equation (1.15) results in Ft V
kW
1000
4 0 0 0 ( 3 .0 2 ) 1000
1 2 .1
SOLUTION (11.38) Gear is weaker and can transmit lower hp. d g 60(4) 240 m m , K
KT K
L
s
1,
Table 11.7: S t 3 9 .3 M P a , V
d g ng 60
( 0 .2 4 ) 2 4 0 60
3 .0 2 m p s
We have
St K L
a ll
KtKR
3 9 .3 (1 ) 0 .8 5 (1 )
4 6 .2
M Pa
Also K
v
3 . 56
3 . 02
3 . 56
1 . 49 (Curve C, Fig.11.15)
m 4 m m , b 9 0 m m . (given)
J 0 .3 0 (from Fig.11.16, N K
0
1.2 5 (Table 11.5)
K
s
1 (standard gear),
K
m
g
60 ) 1.7 (Table 11.6)
Thus, Eq.(11.35'): Ft
4 6 .2 ( 9 0 )( 4 )( 0 .3 0 ) 1 .2 5 (1 .0 )(1 .7 )(1 .4 9 )
1 .5 7 6 k N
Hence, kW
Ft V 1000
1 5 7 6 ( 3 .0 2 ) 1000
4 .7 6
191
Table 11.9: K
R
0 .8 5
SOLUTION (11.39) N
N
g
Q
V
p
rs
2N N
p
dn 60
g
N
3 5,
28 4 5
70 28 35
g
d
70 63
( 0 .1 6 8 )( 6 0 0 )
N p m 2 8(6 ) 1 6 8 m m ,
p
K 1 .8 6 2
,
b 50 m m
M P a (Table 11.10)
5 .2 8 m p s
60
We calculate that F w d p b Q K 1 6 8 (5 0 )( 76 03 )(1 .8 6 2 ) 1 7 .3 8
kN
and 3 .0 5 5 .2 8 3 .0 5
Fd
F t 2 .7 3 1 F t ,
1 7 .3 8 2 .7 3 1 F t
F t 6 .3 6 4 k N
Hence Ft V
kW
1000
6 3 6 4 ( 5 .2 8 )
1000
3 3 .6
SOLUTION (11.40) N
g
35 , 2 ( 35 )
Q
63
N
70 63
28 ,
p
p
K 1 . 862
,
F w d p bQK
d
28 ( 6 ) 168
168 ( 60 )( 1 . 862 )(
b 60 mm
(Table 11.10)
MPa 70 63
mm ,
) 20 . 85 kN
We obtain V d n 6 0 ( 0 .1 6 8 )( 6 0 0 6 0 ) 5 .2 8 m p s Using Eq.(11.24a), 3 .0 5 5 .2 8 3 .0 5
Fd
F t 2 .7 3 F t
2 0 .8 5 1 0 2 .7 3 F t
F t 7 .6 3 7 k N
3
Equation (1.15) is thus Ft V
kW
1000
7 6 3 7 ( 5 .2 8 )
1000
4 0 .3
SOLUTION (11.41) From Solution of Prob. 11.39: N
g
3 5,
V 5 .2 8
N
p
2 8,
d
p
168 m m ,
b 50 m m ,
mG N
g
N
p
1 .2 5
m ps
Apply Eq. (11.44): c , a ll
ScCLC H KT K
R
Here S c 1 0 2 0 M P a (mid-point, Table 11.12)
K
v
5 . 56
5 . 28
5 . 56
1 . 19
K 0 K s K T 1,
CL C
f
1,
C H 1 (Eq.11.46)
(Curve A, Fig.11.15) K
R
1. 2 5
(Table 11.9) (CONT.)
192
11.41 (CONT.) Thus c , a ll
1 0 2 0 (1 )(1 )
816 M Pa
(1 )(1 .2 5 )
(Eq.11.44)
We find that C
p
191 M Pa
C
f
1, K m 1 . 3
I
sin cos
mG
2
m G 1
(from Table 11.11) (from Table 11.6)
0 . 161
35 28 35
0 . 089
Hence, from Eq.(11.45): Ft (
c , a ll
C
p
)
2
6
( 8 1 6 1 03 ) 1 9 1 1 0
1 KsKv
2
bd Ks
I KmC
f
( 0 .0 5 )( 0 .1 6 8 ) 0 .0 8 9 1 1 (1 .1 9 ) 1 1 .3 (1 )
8 .8 2 k N
and kW
Ft V 1000
8 8 2 0 ( 5 .2 8 ) 1000
4 6 .6
End of Chapter 11
193
CHAPTER 12
HELICAL, BEVEL, AND WORM GEARS
SOLUTION (12.1) mn
( a ) Use Eqs.(12.1) and (12.2'). m
c o s
p m ( 4 .6 1 9 ) 1 4 .5 1 1 m m ,
4 cos 30
o
p a p c o t 1 4 .5 1 1 c o t 3 0
ta n
( b ) P 1 m 1 4 .6 1 9 0 .2 1 6 m m 5 .4 8 6 in , -1
(c) dp
Nmn
c o s
20 ( 4 ) cos 30
9 2 .4 m m ,
o
p n m n 1 2 .5 6 6 m m
4 .6 1 9 m m ,
40 ( 4 )
dg
cos 30
o
1 ta n n c o s
o
ta n
2 5 .1 3 4 m m 1 ta n 2 5 o cos 30
o
2 8 .3
o
1 8 4 .7 5 m m
Gear (d) 30
Thrust
o
Pinion, R.H. SOLUTION (12.2) ( a ) Apply Eqs.(12.1) through (12.2'). m
mn c o s
3 .1 7 5 cos 30
o
3 .6 6 6 m m , p n m n 9 .9 7 5 m m ,
p m (3 .6 6 6 ) 1 1 .5 1 7 m m , p a p c o t 1 1 .5 1 7 c o t 2 0
( b ) P 1 m 1 3 .6 6 6 0 .2 7 3 m m 6 .9 3 4 in . , ta n -1
(c) dp
Nmn c o s
1 8 ( 3 .1 7 5 ) cos 20
o
6 0 .8 1 8 m m ,
dg
5 5 ( 3 .1 7 5 ) cos 20
o
1 ta n n c o s
o
ta n
3 1 .6 4 m m 1 ta n 1 4 .5 o cos 20
o
1 5 .3 9
o
1 8 5 .8 3 2 m m
Gear (d)
Pinion, L.H. SOLUTION (12.3) ( a ) d N m 3 0 (3 .1 7 5 ) 9 5 .2 5 m m , m n m c o s 3 0 2 .7 5 m m , o
p m 9 .9 7 5 m m
p n p c o s 9 .9 7 5 c o s 4 0
o
7 .6 4 m m
p a p c o t 9 .9 7 5 c o t 4 0
o
1 1 .8 9 m m
(CONT.)
194
12.3 (CONT.) ( b ) Pn 1 m n 1 2 .7 5 0 .3 6 4 m m 9 .2 5 in . ta n
1
ta n n
(
) ta n
c o s
1
o
( ta n 1 4 .5o ) 1 8 .6 5
-1
o
cos 40
SOLUTION (12.4) rs
c
1 4
Pn
N
2n
N
p
N
g
N
p
g
c o s
18 Ng
;
N
g
1 5 .6 2 5 1 8 7 2 2 c o s
250
;
72
Solving, c o s 0 .8 9 5,
2 6 .5
Comment: The helix angle of 2 6 .5
o
o o
o
and
P N d . Thus
is in usual range of 1 5 to 3 0 .
SOLUTION (12.5) cos P Pn where m n 1 Pn
Equation (12.2): Nmn
cos
d
32( 6 ) 260
4 2 .4
0 .7 3 8 5 ,
o
F t F n c o s n c o s 1 0 (c o s 2 0 )(c o s 4 2 .4 ) 6 .9 4 N o
kW
Thus
( dn ) F t 60
o
( 0 . 26 )( 800 )( 6 . 94 ) 60
(Eq.12.10)
75 . 58
SOLUTION (12.6) 1
( a ) n ta n
(ta n c o s ) ta n
1
o
o
(ta n 2 0 c o s 3 0 );
n 1 7 .5
o
N ' N c o s 3 5 c o s 3 0 5 3 .8 9 3
3
o
( b ) A super gear of equal strength would have 53.89 teeth and 1 7 .5
o
SOLUTION (12.7) 1 3
N
p
N
g
40 Ng
,
N
Equation (12.5) c
Pn 2
g
3(4 0 ) 1 2 0
P Pn c o s :
with
N
p
N
c o s
g
14 2
40 120 cos15
o
3 6 9 .1 m m
SOLUTION (12.8) ( a ) n ta n
1
(ta n c o s ) ta n
1
o
o
(ta n 2 0 c o s 2 2 );
n 1 8 .6
N ' N c o s 3 5 c o s 2 2 4 3 .9 1 teeth 3
3
o
( b ) A super gear of equal strength would have 43.91 teeth and 1 8 .6
195
o
o
SOLUTION (12.9) a p a g a 1 P m 1 .5 m m
( a ) Addendum:
p m 1 .5 4 .7 1 2 m m
Circular Pitch: c
1 2
rp
(d 1 2
dg)
p
N Pm
1 2
N
1 2
p
N
g
P
1 2
(N
p
N g )m
( 2 0 )(1 .5 ) 1 5 m m ,
rg
1 2
1 2
( 2 0 1 2 0 )(1 .5 ) 1 0 5 m m
(1 2 0 )(1 .5 ) 9 0 m m
The radii of base circles: rb p r p c o s 1 5 c o s 2 0
1 4 .0 9 5 m m
o
rb g rg c o s 9 0 c o s 2 0 8 4 .5 7 2 m m o
Outside radii: ro p r p a r p m 1 5 1 .5 1 6 .5 m m ro g r g a 9 0 1 .5 9 1 .5 m m
From Eq. (11.14) the contact ratio is therefore Cr
1 p cos
[
( rp a ) ( rp c o s ) 2
1 ( 4 .7 1 2 ) c o s 2 0
[
o
2
(1 6 .5 ) (1 4 .0 9 5 ) 2
2
( rg a ) ( rg c o s ) ] 2
2
(9 1 .5 ) (8 4 .5 7 2 ) ] 2
2
c ta n p
1 0 5 ta n 2 0 4 .7 1 2
o
1 .7 1 5
The total contact ratio is C r t C r rr a 1 .7 1 5
4 0 ta n 3 0 4 .7 1 2
o
6 .6 1 6
( b ) To obtain a total contact ratio of 4.0, the helix angle has to be C ra
b ta n p
ta n
4 .0 1 .7 1 5 2 .2 8 5
2 .2 8 5 ( 4 .7 1 2 ) 40
1 5 .0 7
;
o
SOLUTION (12.10) 299
(a)
153.9
tan
299
1 tan n cos
tan
1 tan 20 o cos 45
27 . 24
o
o
(Eq. 12.3)
Ft F n c o s n c o s 4 5 0 c o s 2 0 c o s 4 5 o
F a F t ta n 2 9 9 ta n 4 5
o
o
299 N
F r F t ta n 2 9 9 ta n 2 7 .2 4
o
(Eqs.12.10)
1 5 3 .9 N
( b ) m m n c o s 2 .5 c o s 4 5 3 .5 3 6 m m o
d g N g m 6 0 (3 .5 3 6 ) 2 1 2 .1 6 m m , d
Thus
299 N
T p 2 9 9 ( 0 .1 123 1 5 ) 1 6 .9 2 N m
T g 2 9 9 ( 0 .2 122 1 6 ) 3 1 .7 2 N m
196
p
(Eq.12.2') 3 2 (3 .5 3 6 ) 1 1 3 .1 5 m m
(Eq.12.3)
SOLUTION (12.11) T2
2
(a)
1178 T3 B
549
C 1
549
1178
429 3
2
T1
429 T1
9549 kW n1
9 5 4 9 (1 5 )
9 5 .4 9 N m
1500
m m n c o s 6 .3 5 c o s 2 0
o
6 .7 5 8 m m ,
d 1 N 1 m 2 4 ( 6 .7 5 8 ) 1 6 2 .1 9 m m
(Eq.12.2)
We have ( rs ) 1 2
24 36
d1
( rs ) 1 3 0 .5 Ft1
T1
d1 2
d1
2 3
d3
d2
d2
,
d3
,
1 6 2 .1 9 2 3
1 6 2 .1 9 0 .5
2 4 3 .3 m m
3 2 4 .4 m m
1 .1 7 8 k N F t 2 F t 3
9 5 .4 9 0 .1 6 2 .1 9 2
F r 1 F t 1 ta n 1 .1 7 8 ta n 2 5
F a 1 F t 1 ta n 1 .1 7 8 ta n 2 0
( b ) T1 9 5 .4 9 N m ,
5 4 9 N Fr 2 Fr 3
o
o
429 N
T2 0 ,
(Eqs.12.10)
T 3 9 5 .4 9 ( 12 ) 1 9 1 N m
SOLUTION (12.12) Equations (12.7b) and (12.2'): N ' d
p
N
p 3
cos
2 9 .6 , m m n c o s 3 .1 2 c o s 2 5
22 3
cos 25
o
2 2 ( 3 .4 4 )
N pm
3
cos 25
o
o
3 .4 4 m m
1 0 1 .6 6 m m
Table 11.4: o 1 7 2 M P a
Table 11.3: Y 0 .3 5 7 , We have 0b Ym n
Fb
K
1
f
d pnp
V
60
1 7 2 ( 5 0 ) 0 .3 5 7 ( 3 .1 2 ) 1 .5
1
( 0 .1 0 1 6 6 )(1 8 0 0 ) 60
6 .3 8 6
kN
(Eq.11.33, modified)
9 .5 8 m p s
Also Fd
or and
5 .5 6
9 .5 8
5 .5 6
Fb F d : Ft
1000 kW V
F t 1 .5 5 7 F t
(1)
6 .3 8 6 1 .5 5 7 F t ,
1000 ( 22 ) 9 .5 8
2 .2 9 6 k N
Hence, by Eqs.(2) and (3): n
4 .1 0 1 2 .2 9 6
1 .7 9
197
F t 4 .1 0 1 k N
(2) (3)
SOLUTION (12.13) ( a ) Speed ratio: 37 05
5
50 25
Center distance is c m(N
p
mn
Ng)
c o s
(N
p
Ng)
(1)
Thus, for equal center distance: (3 0 7 5 )
4 c o s 2 9 .8
(b)
4 c o s 2 9 .8
o
(3 0 7 5 )
5 .5 c o s
3 1 .5
( 2 5 5 0 );
5 .5 c o s
5 3 .8
( 2 0 3 2 );
o
o
SOLUTION (12.14) ( a ) ta n
1 ta n n
ta n
c o s
1 ta n 2 5 o cos 30
m m n co s 4 co s 3 0 d
p
2 8 .3
o
o
o
4 .6 2 m m
N p m 2 2 ( 4 .6 2 ) 1 0 1 .6 m m
( b ) V d n ( 0 .1 0 1 6 )( 2 4 0 0 6 0 ) 1 2 .8 m p s Ft
1000 kW V
1 0 0 0 (1 .5 )
117 N
1 2 .8
( c ) F r F t ta n 1 1 7 ta n 2 8 .3 6 3 N o
F a F t ta n 1 1 7 ta n 3 0
Fn
Ft c o s n c o s
117 o
cos 25 cos 30
o
o
6 7 .5 N
1 4 9 .1 N
SOLUTION (12.15) Refer to Solution of Prob.12.12. We now have Equation (11.40): Q and Fw
d pbQ K 2
cos
2( 40 ) 22 40
40 31
, Table 11.10: K 0 .9 0 3 M P a
(1 0 1 .6 6 )(5 0 )( 43 01 )( 0 .9 0 3 )(
Equation (1) for F w F d :
) 7 .2 1 k N
1 2
cos 25
o
7 .2 1 1 .5 5 7 F t ;
(Eq.11.38, modified)
F t 4 .6 3 1 k N
(4)
Thus, by Eqs.(3) and (4): n
4 .6 3 1 2 .2 9 6
2 .0 2
SOLUTION (12.16) Equation (12.2'), (12.4) and (12.7b): m m n c o s 4 .2 c o s 3 5
o
5 .1 2 7 m m ,
d 2 6 5 ( 5 .1 2 7 ) 3 3 3 .3 m m ,
N1
N1
'
3
cos
d 1 N 1 m 3 0 (5 .1 2 7 ) 1 5 3 .8 m m
30 3
cos 35
o
5 4 .5 8
(CONT.)
198
12.16 (CONT.) and Table 11.4: 0 1 2 4 M P a
Table 11.3: Y 0 .4 8 3 , Thus Ym n
Fb 0 b
0 .4 8 3 ( 4 .2 )
1 2 4 (3 8 )
1
9 .5 6 k N
1
(Eq.11.33, modified)
From Table 11.10: K 0 . 3 5 2 M P a We have the quantities: 2N
Q
N
5 . 56
Fd
2
cos 35
5 . 56
4 .1 9 k N
o
(Eq.11.38, modified)
1 9 .3 3 m p s
60 19 . 33
F t 1 . 791 F t
4 .1 9 1 .7 9 1 F t ,
Hence
(Eq.11.40)
1 5 3 .8 ( 3 8 ) (1 .3 6 8 ) ( 0 .3 5 2 )
( 0 .1 5 3 8 )( 2 4 0 0 )
60
1.3 6 8
30 65
2
cos
d 1 n1
V
g
d 1b Q K
Fw
and
2( 65 )
g
N
p
(Eq.11.24c, modified)
F t 2 .3 4 k N
Therefore Ft V
kW
2 ,3 4 0 (1 9 .3 3 )
1000
1000
4 5 .2
SOLUTION (12.17) ( a ) Equations (12.2'), (12.5), and (12.7b): rs
1 3
N1
N
d1
c
,
d2
2
(N1 N 2 )
m 2
or
N 1 44
and
d 1 4 4 (1 .7 ) 7 4 .8 m m ,
N1 '
N1 cos
3
and N
3
30
1 .7 2
(N1 3N1)
132
2
d 2 2 2 4 .4 m m
74 . 8 mm , m n m cos (1 . 7 ) cos 30
44 cos
150
o
Using 69.28 teeth, Y 0 .4 2 8 (interpolated, Table 11.3). Fb
0b Ym n K
1
f
1 7 2 ( 6 4 ) 0 .4 2 8 (1 .4 7 ) 1 .4
1
4 .9 5 k N
o
1 . 47 mm
By Table 11.4: 0 1 7 2 M P a (Eq.11.33, modified)
Similarly, Table 11.10: K 0 .5 4 5 M P a and Q
2N N
g
N
g
d 1b Q K
Fw
V
p
2
cos
d 1 n1 60
5 .5 6
2 ( 132 ) 44 132
1 .5
( 7 4 .8 ) ( 6 4 ) (1 .5 ) ( 0 .5 4 5 ) 2
cos 30
( 0 .0 7 4 8 )( 9 0 0 ) 60
Also
Fd
and
4 .9 5 1 .3 3 7 F t ,
3 .5 2
5 .5 6
o
(Eq.11.40) 5 .2 2 k N
(Eq.11.38, modified)
3 .5 2 m p s
F t 1 .3 3 7 F t
(Eq.11.24c, modified)
F t 3 .7 k N
Thus kW
Ft V 1000
3 7 0 0 ( 3 .5 2 ) 1000
1 3 .0 2
(CONT.)
199
12.17 (CONT.) ( b ) Equation (11.36): St K
a ll
L
KT KR
We have S t 2 1 0 M P a (by interpolation of mid-point values, Table 11.7) K
L
1
K
R
0 . 85
KT 1
(indefinite life) (Table 11.9)
and
a ll
2 1 0 (1 )
247 M Pa
(1 )( 0 .8 5 )
Equation (11.35): bJm K0K sK mKv
Ft
(where a ll )
where K 0 1.2 5
Ks 1
3 .5 6
Kv
3 .5 2
3 .5 6
K m 1.4
1 .2 4 (Fig.11.15)
(Table 11.6)
J 0 .4 8 (Fig.12.6a) J multiplier 1 . 01
(Fig.12.6b)
and J 0 .4 8 ( 1.0 1 ) 0 .4 8 5
Hence
Ft
2 4 7 ( 6 4 )( 0 .4 8 5 )(1 .7 ) (1 .2 5 )(1 )(1 .4 )(1 .2 4 )
6 kN
Then kW
Ft V 1000
6 0 0 0 ( 3 .5 2 ) 1000
2 1 .1
SOLUTION (12.18) ( a ) d p N p m 2 0 (3 ) 6 0 m m , ( b ) p ta n
1 20 42
2 5 .4 6 , o
g ta n
d g 4 2 (3 ) 1 2 6 m m 1 42 20
6 4 .5 4
o
(Eq.12.12a)
(Eq.12.12b)
(c) 2
60
1
L [ 6 0 1 2 6 ] 2 2 1 3 9 .6 m m
L
2
Equation ( a ) of Sec.12.6:
p
g
L 3
46 . 5 mm ;
Thus
b 30 m m
126 ( d ) Equation ( b ) of Sec.12.6: c 0 .1 8 8 m 0 .0 0 5 0 .1 8 8 ( 3 ) 0 .0 0 5 0 .5 6 9 m m
200
10 m 30
SOLUTION (12.19) Equation (12.13): d p 2 0 0 m m , ta n
g
N
g
N
p
2;
1 rs
d
dg
g 6 3 .4 3 ,
o
p
rs
p
200 0 .5
400 m m
9 0 6 3 .4 3 2 6 .5 7
o
Therefore rp ,avg rp
b 2
s in
r g , a v g rg
b 2
s in g 2 0 0 3 2 .5 s in 6 3 .4 3 1 7 0 .9 m m
Vp
d
p ,a v g
np
o
8 5 .4 6 m m
o
60
1 0 0 3 2 .5 s in 2 6 .5 7
p
( 0 .1 7 0 9 )( 5 0 0 ) 60
4 .4 7 m p s
Hence Ft
1 0 0 0 (1 1 )
1000 kW Vp
4 .4 7
2 .4 6 k N
Equations (12.17): F a p F t ta n s in
p
2 .4 6 (ta n 2 0 )(s in 2 6 .5 7 ) 4 0 0 N F r g
F r p F t ta n c o s
p
2 .4 6 (ta n 2 0 )(c o s 2 6 .5 7 ) 8 0 0 .8 N F a g
o
o
2460
800.8
o
o
Gear
400 800.8
Pinion SOLUTION (12.20)
From Solution of Prob.12.19: V p 2 .4 7 m p s ,
F t 2 .4 6 k N ,
b 65 m m
Thus Fd N
p
3 .0 5 4 .4 7 3 .0 5
d
( 2 .4 6 ) 6 .0 7 k N
m 2 0 0 3 .5 5 7 ,
p
Hence, from Table 11.3:
(Eq.11.24a) N
p
'
N
p
cos
p
57 c o s 2 6 .5 7
(using N p ' 6 3 .7 , interpolated)
Y 0 .4 2 4
0 124 M Pa
Also, from Table 11.4: Hence, Fb
Since
0b Ym K
f
Fb Fd ,
1
1 2 4 ( 6 5 ) 0 .4 2 4 ( 3 .5 ) 1 .4
1
o
8 .5 4 k N
the gears are safe.
201
(Eq.11.33)
6 3 .7
SOLUTION (12.21) From Solution 12.20:
F d 6 .0 7
kN ,
114,
N
N
p
57,
N
p
' 6 3 .7
Hence N
g
N
p
rs 2N
Q '
N
p
g
57 0 .5
'
' N
2 ( 2 5 4 .9 )
'
g
1 .6
6 3 .7 2 5 4 .9
g
'
N
g
cos
114 c o s 6 3 .4 3
g
o
2 5 4 .9
(Eq.11.40, modified)
From Table 11.10: K 0 .6 9 M P a Thus 0 .7 5 d p b K Q '
Fw
Since
cos
0 .7 5 ( 2 0 0 )( 6 5 )( 0 .6 9 )(1 .6 ) c o s 2 6 .5 7
p
Fw Fd ,
o
1 2 .0 4 k N
(Eq.11.38, modified)
gears are safe.
SOLUTION (12.22) Table 11.10: K 0 .5 4 5 M P a , Table 11.4: 0 1 7 2 M P a ,
N
p
' N
c o s 1 3 0 c o s 2 0 3 1 .9 3 o
p
Table 11.3: Y 0 . 364 (using N p ' 31 . 93 , interpolated)
We have r1
N 1m
2
1 tan
30 (8 )
1 120 240
r1 , a v g 1 2 0 V1
d 1 , a v g n1
N2
26 . 6 ,
2
60 (8 ) 2
60
60 cos 63 . 4
o
240 m m
90 26 . 6 63 . 4
s in 1 1 0 4 .3 m m ,
( 0 .2 0 8 7 )( 7 2 0 )
cos 2
r2
o
70 2
60
N2'
120 m m ,
2
o
(Eq.12.12b)
r2 , a v g 2 4 0
70 2
s in 2 2 0 8 .7 m m
7 .8 7 m p s
134 ,
N 1'
30 cos 26 . 6
o
33 . 6
(Eq.12.15)
Hence Fb
Q'
0b Ym K
1 7 2 ( 7 0 ) 0 .3 6 4 ( 8 )
1
f
2N2' N 1 ' N 2 '
1 .5
1 .6
1
2 3 .4 k N
(Eq.11.33)
(Eq.11.40, modified)
It follows that Fw
0 .7 5 d 1 b K Q ' cos 1
0 .7 5 ( 2 4 0 )( 7 0 )( 0 .5 4 5 )(1 .6 ) c o s 2 6 .6
o
1 2 .2 9 k N
Since F w F b , power capacity depends on F w . We find that Fd
and
3 . 05 7 . 87 3 . 05
F t 3 . 58 F t
1 2 .2 9 3 .5 8 F t ,
(Eq.11.24a)
F t 3 .4 3 k N
Therefore kW
Ft V 1000
3 4 3 0 ( 7 .8 7 ) 1000
27
202
(Eq.11.38, modified)
SOLUTION (12.23) ( a ) d p N p m 30(4) 120 m m , d
g
ta n
1
1
dg
( d ) ta n p
p ,a v
d
p
V av d
Ft
b s in
o
1 2 0 4 5 s in 1 8 .4
p
7 4 5 .7 ( 4 0 )
Vav
( 13 26 00 ) 7 1 .6 ,
1500 500
360 m m
9 0 7 1 .6 o
p
o
o
1 8 .4
o
1 0 5 .8 m m
n ( 0 .1 0 5 8 )(1 5 0 0 6 0 ) 8 .3 1 m p s
p ,av
7 4 5 .7 h p
d g (1 2 0 )
3 .5 9 k N
8 .3 1
( b ) F a F t ta n s in
p
3 .5 9 ta n 2 0 s in 1 8 .4
F r F t ta n c o s
p
3 .5 9 ta n 2 0 c o s 1 8 .4
o
o
0 .4 1 2 k N
o
o
1 .2 4 k N
Equation (1.16): Tp
9549 kW
Tg
9549 kW ng
np
9549 (30 )
9549 (30 )
1500
500
191 N m
573 N m
SOLUTION (12.24) r1
1 2
r2
1 2
mN mN
1 ta n
1
2
( 8 . 5 )( 30 ) 127 . 5 mm
1 2
1 1 2 7 .5 255
o
Table 11.4:
70 2
70 2
sin 26 . 6
sin 63 . 4
172
0
o
(Eq.11.4)
mm
2 6 3.4
2 6 .6 ,
r1 , av 127 . 5
r 2 , av 255
( 8 . 5 )( 60 ) 255
1 2
o
(Eq.12.12b)
111 . 8 mm
o
223 . 7 mm
Table 11.10: K 0 . 545
MPa ,
MPa
Note that number of teeth is as same as in Prob.12.14. Hence, from Solution of Prob.12.22: Q ' 1.6 , Y 0 . 364 Therefore, we have: Fb
0b K
Ym
0 . 75 d 1 bKQ '
Fw
6
1 7 2 (1 0 )( 0 .0 7 ) 1 .5
f
cos 1
( 0 .3 6 4 )( 0 .0 0 8 5 ) 2 4 .8 3 k N
0 . 75 ( 0 . 255 )( 0 . 07 )( 0 . 545 10 cos 26 . 6
o
6
)( 1 . 6 )
13 . 06 kN
(Eq.11.33) (Eq.11.38, modified)
Also V 1 d 1 n1 6 0 ( 0 .2 2 3 6 )( 7 2 0 6 0 ) 8 .4 3 m s Ft
3 .0 5 8 .4 3 3 .0 5
F t 3 .7 6 4 F t ;
1 3 .0 6 3 .7 6 4 F t ,
Hence kW
F V1 1000
3 , 4 7 0 ( 8 .4 3 ) 1000
2 9 .2 5
203
F t 3 .4 7
kN
(Eq.11.24a)
SOLUTION (12.25) Worm: Vg
ta n
d
tan w
g
c
1 2
( d g 2 )wg
Vw
(d
2 )w
p
dw
tan 35
rs
dg dw
o
(Eq.12.21)
rs dw
28 d w
0 . 025
( d w d g ) 150 ,
29 d w 300,
and
p
dw d
300
g
d w 1 0 .3 4 5 m m
d g 2 8 (1 0 .3 4 5 ) 2 8 9 .7 m m
Then L d w ta n w (1 0 .3 4 5 )(ta n 3 5 ) 2 2 .7 6 m m o
Gear: pg
p ng
c o s
1 1 .4 8 m m (Eq.12.1)
9 .4 cos 35
o
But p g p w . Hence
Nw
L pw
2 2 .7 6 1 1 .4 8
1 .9 8 or 2 teeth
(Eq.12.20)
SOLUTION (12.26) (T w ) i
9549 kW n
9549 (30 )
1800
1 5 9 .2 N m
( T w ) o 1 5 9 .2 0 .9 0 1 4 3 .3 N m ng nw
N
w
N
g
1800 ( 802 ) 45 rpm
(T g ) o (T w ) o
nw
(Eq.12.18)
1 4 3 .3 ( 1 84 05 0 ) 5 .7 3 2
ng
kN m
Thus (Tg )0
Ft
d 2
5 .7 3 2 0 .2 5 2
4 5 .8 6 k N
SOLUTION (12.27) n 3 ( 22 00 )( 550 )( 2 5 0 ) 2 5 rp m
Gear 3 rotates
counterclockwise (negative) at 25 rpm.
SOLUTION (12.28) ( a ) N g N w 40 2 20,
d g 4 0 (3 .2 ) 1 2 8 m m ,
L 2 (1 0 .0 5 ) 2 0 .1 m m ,
(b)
ta n
c
1 2
L
dw
d w 3 .5 (1 0 .0 5 ) 3 5 .1 8 m m
2 0 .1
( 3 5 .1 8 )
(d w d g )
1 2
p (3 .2 ) 1 0 .0 5 m m
,
1 0 .3
o
(3 5 .1 8 1 2 8 ) 8 1 .5 9 m m
204
SOLUTION (12.29) ( a ) dw c
2 210
0 .8 7 5
ta n
L
dw
0 .8 7 5
m ( 5 3 .8 )
2 5 3 .8 m m ,
6 5 3 .8
6 .3 6
;
d g 2 c d w 4 2 0 5 3 .8 3 6 6 .2 m m
o
( b ) V w d w n w ( 0 .0 5 3 8 )(1 2 0 0 6 0 ) 3 .3 8 m s where d g 3 6 6 .2 m m ,
b 25 m m
K w 1 5 0 p s i 1 5 0 ( 6 8 9 5 ) 1 .0 3 4 M P a
(Table 12.2)
Thus, F w (3 6 6 .2 )( 2 5 )(1 .0 3 4 ) 9 .4 7 k N
SOLUTION (12.30) p g m ( 4 ) 1 2 .5 7 m m p w
( a ) L N w p w 4 (1 2 .5 7 ) 5 0 .2 8 m m ( b ) ta n
1
L
dw
ta n
1 5 0 .2 8 (60 )
1 4 .9 4
(Eq.12.20)
o
(Eq.12.21)
( c ) d g N g m 90(4) 360 m m c
(d w d g )
1 2
1 2
(6 0 3 6 0 ) 2 1 0 m m
SOLUTION (12.31) (a) rs c
1 20 1 2
pg
N
w
N
g
,
N
g
3( 2 0 ) 6 0
(d g d w ) 175 d g N
( 275 )
(d g 75)
d g 275 m m
14 . 4 mm p w
60
g
1 2
Then, Eqs.(12.20) and (12.21): L p w N w 1 4 .4 (3) 4 3 .2 m m
tan
1
L
d w
tan
1
43 . 2
( 75 )
10 . 39
o
(b) Vw
d w nw 60
Fw t F ga
( 0 .0 7 5 )(1 0 0 0 ) 60 1000 kW Vw
3 .9 3 m p s
1 0 0 0 ( 7 .5 ) 3 .9 3
1 .9 1 k N
(CONT.)
205
12.31 (CONT.) (c) Vs
Vw
cos
4 m ps
3 .9 3 c o s 1 0 .3 9
o
f 0 . 024
Hence
(Eq.12.28)
(from Table 12.3)
Thus, Eq.(12.27): c o s n f ta n
e
o
o
o
o
c o s 2 0 0 . 0 2 4 ( ta n 1 0 . 3 9 )
cos n f cot
c o s 2 0 0 . 0 2 4 ( c o t 1 0 .3 9 )
0 .8 7 4
or
8 7 .4 %
Power delivered to machine ( k W ) m 0 .8 7 4 ( 7 .5 ) 6 .5 6 k W
SOLUTION (12.32) p w 1 4 .4 m m ,
From Solution of Prob.12.31:
pn
We have Table 11.4:
Fb
p w c o s
1 4 .4 c o s 1 0 .3 9
0 172 M Pa 0b Ym n K
1
f
o
0 .2 2 2
o
mn
d g 275 m m
4 .5 1 m m
1 pn
Table 12.1: Y 0 .3 9 2
1 7 2 ( 3 8 ) 0 .3 9 2 ( 4 .5 1 ) 1 .4
1 0 .3 9 ,
1
8 .2 5 k N
(Eq.11.33, modified)
Also Vg
T Ft
Hence
Fd
d g ng 60
9540 kW ng
T dg 2
6 .1 V g 6 .1
( 0 .2 7 5 )( 5 0 )
9 5 4 0 ( 7 .5 )
60
50
1 .4 3 0 .2 7 5 2
0 .7 2 m p s
1 .4 3 k N m
1 0 .4 k N
6 .1 0 .7 2 6 .1
F t 1 .1 2 (1 0 .4 ) 1 1 .6 5 k N
(Eq.11.24b)
Since Fb Fd
gears are not safe.
SOLUTION (12.33) From Solution of Prob.12.32: Table 12.2:
K
w
F d 1 1 .6 5 k N
1 . 295 MPa
Therefore F w d g b K w 2 7 5 (3 8 )(1 .2 9 5 ) 1 3 .5 3 k N
Since
Fw Fd
gears are safe.
206
(Eq.12.22)
SOLUTION (12.34) We have c 1 7 5 m m 7 in . and p o w e r 7 .5 k W 1 0 h p Equation (12.24): A 0 .3( 7 )
1 .7
8 .2 ft
2
e 8 7 .4 % , (from Solution of Prob.12.31)
By Fig.12.17:
o C 5 6 ft lb m in ft F 2
Hence, Eqs. (12.25) and (12.26):
Since
(hp )d
C At 3 3 ,0 0 0
( hp ) i
( hp ) d 1 e
5 6 ( 8 .2 )(1 0 0 ) 3 3 ,0 0 0
1 . 392 1 0 . 874
1 .3 9 2 h p = 1 .0 4 4 k W
11 . 05 8 . 288 kW
( hp ) i 10 hp or ( k W ) i 7 .5 k W , overheating will not be a problem.
End of Chapter 12
207
CHAPTER 13
BELTS, CHAINS, CLUTCHES, AND BRAKES
SOLUTION (13.1) ( a ) w (5 )(1 0 0 )(1 0 , 8 0 0 ) 5 .4 N m V
dn
( 0 .1 2 5 )(1 5 0 0 )
60
60
F1 F 2
Fc
w g
2
V
5 .4 9 .8 1
[
7 6 3 .7 N
9 .8 2
( 0 .1 2 5 )(1 5 0 0 ) 60
2 1 8 0 2 s in o
e Thus
f
e
0 .3 ( 2 .9 7 8 )
1500
4 7 .7 5 N m
9 .8 2 m p s
1 0 0 0 ( 7 .5 )
1000 kW V
9 5 4 9 ( 7 .5 )
T1
] 5 3 .0 5 N
1
2
T1 r1
(Eq.13.13)
( 1 8 71.55 2 56 2 .5 ) 1 7 0 .6
2 .4 4 3
F1 F c ( 1 )
(Eq.13.3)
o
2 .9 7 8 ra d
(Eq.13.21)
5 3 .0 5 ( 12 .4.4 44 33 )
4 7 .7 5 0 .0 6 2 5
1 .3 4 7 k N
(Eq.13.20)
F 2 1, 3 4 7 7 6 3 .7 5 8 3 .3 N
( b ) L 2 c ( r1 r2 )
( r2 r1 )
2
(Eq.13.9)
c
2 (1, 5 2 4 ) ( 6 2 .5 1 8 7 .5 )
(1 2 5 )
2
1525
3 8 4 5 .6 m m .
SOLUTION (13.2) ( a ) T1
9549 kW
9549 ( 10 )
n1
2800
( b ) T1 ( F1 F 2 ) r1 , r1 r2
n1 n2
34 . 1 N m
F1 F 2
T1 r1
F 2 2 2 7 .3 F 2
3 4 .1 0 .1 5
n
r2 r1 ( n 1 ) 0 .1 5 ( 12 68 00 00 ) 0 .2 6 2 5
,
(Eq.13.19)
m
2
r2 r1
s in
c
0 . 2 6 2 5 0 .1 5 0 .7
2 2 . 819
9 .2 5
0 .1 6 1 ,
o
(Eq.13.6)
rad
( c ) V d n 6 0 ( 0 .3)( 2 68 00 0 ) 4 3 .9 8 m p s w 25 , 000 ( 0 . 06 0 . 0005 ) 0 . 75 N m
Fc
w g
V
2
0 . 75 9 . 81
T1
F1 F c ( 1 )
r1
( 43 . 98 )
2
147 . 9 N ,
1 4 7 .9 ( 10 .7.7 55 77 )
3 4 .1 0 .1 5
e
0 . 2 ( 2 . 819 )
6 7 5 .5 N
1 . 757
(Eq.13.20)
F 2 6 7 5 .5 2 2 7 .3 4 4 8 .2 N
We have K s 1.5 (Table 13.5). Therefore F max 1 . 5 ( 675 . 5 ) 1 . 013
max
1 , 013 60 0 . 5
kN
33 . 77 MPa
208
(Eq.13.22)
SOLUTION (13.3) We have r2 2 0 0 m m and f 0 .2 5 . The remaining data are the same. Equation (13.6) gives 1
s in [
r2 r1 c
] s in
1
[ 2 01 09 506 2 ] 4 .0 5 8
o
Then 2 1 8 0 2 ( 4 .0 5 8 ) 1 7 1 .9 o
e
f
e
o
( 0 .2 5 ) (1 7 1 .9 ) ( 1 8 0 )
o
2 .1 1 7
and F1 2 1 4 F2 2 1 4
2 .1 1 7
or F1 2 .1 1 7 F 2 2 3 9
(1)
Also F1 F 2 6 2 5 N
Solving, F1 1, 3 9 8 .5 N ,
F 2 7 7 3 .5 N
Length of the belt, by Eq. (13.9): L 2 c ( r2 r1 )
1 C
( r2 r1 )
2
2 (1 .9 5 ) ( 0 .2 0 0 0 .0 6 2 )
1 1 .9 5
( 0 .2 0 0 0 .0 6 2 )
2
4 .7 3 3 m
SOLUTION (13.4) Pulley A Using Eq. (13.16), with F1 2 .5 k N , F1 F2
e
f
,
2 .5 F2
e
0 .1 5 (1 2 0 )( 1 8 0 )
f 0 .1 5 ,
1 .3 6 9 ;
Fc 0 ,
120 :
F 2 1 .8 2 6 k N
Pulley B Then, equilibrium condition
T
0 applied to free-body diagram of
pulley B gives
T
B
0;
T B 2 .5 ( 0 .1 5 ) 1 .8 2 6 ( 0 .1 5 ) 0
or TB 1 0 1 N m
B TB
F1 2 .5 k N
0.15 m F 2 1 .8 2 6 k N
Similarly, we have
T
A
0; T A 2 .5 ( 0 .0 2 ) 1 .8 2 6 ( 0 .0 2 ) 0 ,
209
TA 13 N m
o
SOLUTION (13.5) ( a ) V1
2 r1 n1
60
2 ( 0 .0 3 7 5 )( 2 5 0 0 )
9 .8 2 m p s
60
From Eq. (1.15); with F F1 F 2 : kW
FV 1000
1 .5
;
( F1 F 2 ) 9 .8 2 1000
F1 F 2 1 5 2 .7 5 N
(1)
( b ) Evaluate V 2 as, V1
or
2 r2 n 2 60
9 .8 2
;
r2 9 3 .8 in .,
2 r2 (1 0 0 0 ) 60
d 2 1 8 7 .6 m m
and 1
2 2 s in [ 2 s in
1
r2 r1 c
]
[ 9 3 .86 2 53 7 .5 ] 2 .9 6 ra d
( c ) Centrifugal force, using Eq. (13.13), Fc
w g
V
2
1 .7 5 9 .8 1
(9 .8 2 ) 1 7 .2 N 2
Then, Eq. (13.16): F1 1 7 .2
e
F 2 1 7 .2
0 .3 5 ( 2 .9 6 )
2 .8 1 8
(2)
Solving Eqs. (1) and (2), we have F1 2 5 4 N
F 2 1 0 1 .2 N
SOLUTION (13.6) We have w g 8 .8 9 .8 1 0 .8 9 7
and P V ( F1
w g
V ) V ( 2 5 0 0 0 .8 9 7 V ) 2
2
Thus P V
0 5 0 0 0 .8 9 7 (3 )V ; 2
So V 2 r n ; Solving
V 3 0 .4 8 m p s
3 0 .4 8 2 ( r )( 4 2 0 0 6 0 )
r 6 9 .3 m m
SOLUTION (13.7) By Eq. (13.13): Fc
w g
V
2
1 .4 9 .8 1
[ 2 ( 0 .0 8 )(3 0 0 0 6 0 )] 9 0 .1 N 2
From Eq. (13.16), 1 1 0 0 9 0 .1 F 2 9 0 .1
e
o
( 0 .2 5 ) ( s in 1 8 ) (1 6 0 1 8 0 )
1 .2 4 1
Solving F 2 9 0 3 .9 N
(CONT.)
210
13.7 (CONT.) Then T A ( F1 F 2 ) r1 (1 1 0 0 9 0 3 .9 )( 0 .0 8 ) 1 5 .7 N m
and P
3 0 0 0 (1 5 .7 )
VT 9549
9549
4 .9 3 k W
SOLUTION (13.8) Fc
w g
2
V
F1 F c F2 Fc
8 9 .8 1
[ ( 0 .2 )( 1 66 00 0 )] 2 2 8 .9 N 2
f s in
e
3 , 0 0 0 2 2 8 .9
;
F2 2 2 8 .9
e
1 7 0 2 .9 6 7 ra d o
0 .1 5 ( 2 . 9 6 7 ) s in 1 9
o
3.9 2 4
Solving F 2 935 . 1 N Thus T ( F1 F 2 ) r (3 0 .9 3 1) 1 0 ( 0 .1) 2 0 6 .9
N m
3
kW
Tn 9549
2 0 6 .9 (1 6 0 0 )
3 4 .7
9549
SOLUTION (13.9) ( a ) Fc
w g
F1 F c
2
V
3 9 . 81
f s in
e
8 0 0 Fc
[ ( 0 . 3 )(
e
3000 60
)]
2
0 . 2 5 ( 2 . 7 9 3 ) s in 1 9
o
679 . 1 N ,
160
o
2 . 793
rad
8 .5 4
or F1 679 . 1
8 . 54 , F 1 1, 711 . 6 N
800 679 . 1
Also T ( F 1 F 2 ) r1 (1711 . 6 800 )( 0 . 15 ) 136 . 7 N m kW
Tn 9549
136 . 7 ( 3000 ) 9549
42 . 95
(b)
F1
max
A
1 , 711 . 6 6
145 ( 10
)
11 . 8 MPa
SOLUTION (13.10) r2 r1
n1 n2
2700 , r 2 0 . 1 ( 1800 ) 0 . 15 m
sin
T1
9549 kW
e Fc
0 . 15 0 . 1 0 .5
2700
w g
V
o
9 5 4 9 (1 5 ) 2700
0 . 2 ( 2 . 9 4 1 ) s in 1 7
2
5 . 74 ;
,
2 .5 9 . 81
o
180
5 3 .0 5 N m
o
2 ( 5 . 74 ) 2 . 941
rad
V d n ( 0 .2 )( 2 67 00 0 ) 2 8 .2 7 m p s
7 .4 7 7
( 28 . 27 )
2
203 . 7 N
(CONT.)
211
13.10 (CONT.)
Thus F 1 F c ( 1 )
T1 r1
203 . 7 ( 76 .. 477 ) 477
53 . 05 0 .1
816 . 1 N
Table 13.5: K s 1.4 and F max K s F 1 1 . 4 ( 816 . 1) 1 . 143
kN
SOLUTION (13.11) N 1 n1
(a) N2
2 2 (1 4 0 0 )
n2
4 4 teeth
700
( b ) H d H r K 1K 2 where H r 2 6 .6 h p = 1 9 .8 k N K 1 1 .5
N 1 P n1
1000 H
60 1 0 0 0 ( 5 0 .6 )
d
V1
So,
2 2 ( 0 .0 1 9 0 5 )(1 4 0 0 )
60
F1
K 2 1 .7 by Table 13.10)
(from Table 13.9,
H d 2 6 .6 (1 .5 )(1 .7 ) 6 7 .8 h p = 5 0 .6 k W
Hence ( c ) V1
(by Table 13.8)
9 .7 8
n
F a ll
n
6 2 .6 5 .1 7
9 .7 8 m p s
5 .1 7 k N
F a ll 3 1 .3 ( 2 ) 6 2 .6 k N
F1
(by Table 13.7)
and 1 2 .1
SOLUTION (13.12) N 1 n1
(a) N2
n2
1 8 (1 6 0 0 ) 640
4 5 teeth
( b ) H d H r K 1K 2 where H r 1 6 .1 h p = 1 2 k W K 1 1 .2
(from Table 13.8)
(by Table 13.9),
K 2 3 .3 (from Table 13.10)
and H d 1 6 .1(1 .2 )(3 .3) 6 3 .8 h p = 4 7 .6 k W
( c ) V1 N 1 P n1 1 8 ( 0 .0 1 9 0 5 )(1 6 0 0 6 0 ) 9 .1 4 m p s 7 4 5 .7 H
F1
(d) n
d
V1
F a ll F1
,
7 4 5 .7 ( 6 3 .8 ) 9 .1 4
5 .2 1 k N
where F a ll 2 (3 1 .3) 6 2 .6 k N
Thus, r2
6 2 .6 5 .2 1
12
212
(by Table 13.7)
SOLUTION (13.13) n1 n2
4 1
3
Use c 2 ( r2 r1 )
(Sec.13.6)
We have r1
N1P
2
2 2 (1 6 ) 2
5 6 .0 2 m m ,
r2
n1 n2
r1
4 1
(5 6 .0 2 ) 2 2 4 .0 8 m m
Thus c 2 ( 2 2 4 .0 8 5 6 .0 2 ) 3 3 6 .1 2 m m
r1 r2 2 8 0 .1 m m
Since
sprocket will clear.
SOLUTION (13.14) n1 n2
r1
2 . 19 3 Use c 2 ( r1 r2 )
4600 2100 N1P
2
1 4 (1 4 ) 2
3 1 .1 9 m m ,
n1
r2
n2
(Sec. 13.6)
r1 2 .1 9 (3 1 .1 9 ) 6 3 .3 1 m m
c ( r1 r2 ) 3 1 .1 9 6 3 .3 1 9 4 .5 m m c 2 ( 9 4 .5 ) 1 8 9 m m
Use c 1 9 0 m m
SOLUTION (13.15) ( a ) Speed ratio is ; 3:1. Thus N 2 3( 2 3) 6 9 teeth
( b ) H d H r K 1K 2 where H r 1 9 .5 h p
(from Table 13.8)
K 1 1 .3
(from Table 13.9),
K 2 1 .7 (by Table 13.10)
H d 1 9 .5 (1 .3)(1 .7 ) 4 3 .1 h p
and
( c ) V 1 N 1 P n1 2 3( 0 .0 1 9 )(1 8 0 0 6 0 ) 1 .3 1 m p s 7 4 5 .7 H
F1
(d) n
d
V1
F a ll F1
7 4 5 .7 ( 4 3 .1 ) 1 3 .1
2 .4 5 k N
where F a ll 4 (3 1 .3) 1 2 5 .2 k N
,
(Table 13.7)
So, n
1 2 5 .2 2 .4 5
5 1 .1
SOLUTION (13.16) ( a ) N 2 3(3 5 ) 1 0 5 teeth ( b ) H d H r K 1K 2 where
H r 3 6 .6 h p = 2 7 .1 k W
(by Table 13.8) (CONT.)
213
13.16 (CONT.) K 1 1 .3 ,
K 2 1 .7
(by Table 13.9 & 13.10)
and H d 3 5 (1 .3)(1 .7 ) 7 7 .3 5 h p = 5 7 .6 8 k W
( c ) V 1 N 1 P n1 3 0 ( 0 .0 1 9 0 5 )(9 0 0 6 0 ) 8 .5 7 m p s F1
(d) n
7 4 5 .7 H
F a ll F1
7 4 5 .7 ( 7 7 .3 )
d
V1
6 .7 3 k N
8 .5 2
where F a ll 2 (3 1 .3) 6 2 .6 k N
,
Hence n
9 .3
6 2 .6 6 .7 3
SOLUTION (13.17) (a)
p m ax
T
2 Fa
2(6)
d (D d )
( 0 .1 5 )( 0 .2 5 0 .1 5 )
Fa f ( D d )
1 4
1 4
2 5 4 .6 k P a
(Eq.13.28)
( 6 , 000 )( 0 . 3 )( 0 . 25 0 . 15 ) 180
N m
(Eq.13.30)
(b) p max
T
1 3
4 Fa
(D
2
d
2
3
Fa f
D d D
2
)
d
3
4(6) 2
( 0 . 25 0 . 15
2
2
)
191
3
1 3
( 6 , 000 )( 0 . 3 )
(Eq.13.32)
kPa
0 . 25 0 . 15 0 . 25
2
0 . 15
3 2
183 . 8 N m
SOLUTION (13.18) (a) T
9549 kW n
9549 (30 )
500
5 7 2 .9 N m
N 2
and T
1 12
fp
max
3
(D
577 , 267 . 7 d
d ) 3
3
( 0 . 25 )( 140 10 )[ 64 d 3
12
3
d ] 3
(Eq.13.33)
572 . 9 2
Solving d 7 9 .2 m m and D 4 d 3 1 6 .8 m m ( b ) Fa
1 4
p m ax ( D d ) 2
2
4
(1 4 0 1 0 )[1 6 d 3
2
d ] 1 .6 4 9 (1 0 ) d 2
6
2
1 0 .3 4
kN
SOLUTION (13.19) ( a ) T 5 7 2 .9 N m We now have T
1 8
N 2 (from solution of Prob. 13.18)
fp m a x d ( D d )
206 ,167 d
2
3
2
8
( 0 .2 5 )(1 4 0 1 0 ) d (1 6 d 3
2
d ) 2
(Eq.13.29)
572 . 9 2
Solving, d 1 1 1 .6 m m and D 4 d 4 4 6 .4 m m (CONT.)
214
13.19 (CONT.) ( b ) Fa
1 2
p max d ( D d )
( 0 .1 4 )(1 1 1 .6 )[ 4 4 6 .4 1 1 1 .6 ] 8 .2 2 k N
2
(Eq.13.28)
SOLUTION (13.20) ( a ) From Eq. (13.29), with a factor of safety n : d (D
d )
2
2
8 nT
;
fPm a x
2
0 .0 5 ( D
0 .0 0 2 5 )
8 (1 .6 )(1 3 5 .6 ) 6
( 0 .3 )(1 .6 1 0 )
Solving D 1 5 9 .8 m m ( b ) From Eq. (13.30), we have Fa
4 nT f (Dd )
4 (1 .6 )(1 3 5 .6 )
( 0 .3 )(1 5 9 .8 5 0 )
1 3 .8 k N
SOLUTION (13.21) (a) T
9549 kW n
9 5 4 9 ( 3 7 .5 )
8 9 5 .2 N m
400
and T
fp
1 8
0 .8 (1 0
Fa
2
( b ) p avg
d (D
max 3
Fa 2
(D d
2
)
d ) 2
) p m ax
p m ax d ( D d )
2
2
8
( 0 . 2 ) p max 0 . 15 [ 0 . 3
2
0 . 15
2
2
( 0 .3 0 .1 5 )
1 8 6 .6 k P a
4
SOLUTION (13.22) ( a ) Use Eq. (13.29) by multiplying 3 6 0 : T
360
[ 81 fPm a x d ( D
2
d )] 2
from which p m ax
8 (360 )T 2
fd ( D d
2
)
8 ( 3 6 0 8 0 ) (1 8 0 0 )
2
2
( 0 .3 ) ( 0 .2 ) ( 0 .2 8 0 .2 )
8 9 5 2 .5 k P a
( b ) From Eq. (13.28) by multiplying 3 6 0 : Fa
360
[ 12 p m a x d ( D d )]
80 360
[ 2 (8 9 5 2 .5 )( 0 .2 )( 0 .4 8 )]
300 kN
( c ) Each cylinder supplies a force F a 2 . Thus p hyd
Fa 2
d
2
4
3 0 0 ,0 0 0 2
( 0 .2 )
2
4
]
(Eq.13.29)
p m a x 2 7 9 .8 k P a
,
( 0 .2 7 9 8 )(1 5 0 )(1 5 0 ) 9 .8 9 k N
9890 ( 4 ) 2
8 9 5 .2 4
4 .7 7 M P a
215
(Eq.13.28)
SOLUTION (13.23) ( a ) Use Eq. (13.33) by multiplying 3 6 0 :
T
[ 12 fPm a x ( D d )] 3
360
3
from which 12 (360 )T
p m ax
3
3
f (D d )
1 2 ( 3 6 0 8 0 ) (1 8 0 0 ) 3
3
( 0 .3 ) ( 0 .2 ) ( 0 .2 8 0 .2 )
7 .3 9 M P a
( b ) From Eq. (13.32), by multiplying by 3 6 0 :
Fa
(c)
p hyd
Fa 2
d
2
360
4
[ 4 p m a x ( D
2 ( 49500 )
( 0 .0 4 )
d )]
2
2
80 360
[ 4 ( 7 , 3 9 0 )( 0 .2 8 0 .2 )] 4 9 .5 k N 2
2
1 9 .7 M P a
2
SOLUTION (13.24) T
9549 (35 ) 800
and T
4 1 7 .8 N m
fp m a x 1 2 s in
d )
3
(D
3
R is e 3
( 0 .3 ) 4 2 0 (1 0 ) 1 2 s in 8
o
1 2
( D d ) w s in
[ 0 .2 5 d ] 4 1 7 .8 3
3
(Eq.13.41)
Solving d 2 4 0 .2 m m Therefore w
1 D d 2 s in
2 5 0 2 4 0 .2 2 s in 8
3 5 .2 1 m m
o
SOLUTION (13.25) T 4 1 7 .8 N m (from Solution of Prob. 13.24)
Now we have T
fp m a x d 8 s in
(D
Solving 0 .0 6 2 5 d d
2
d ) 2
0 .0 0 1 2
3
3
( 0 .3 ) 4 2 0 (1 0 ) 8 s in 8
o
[ 0 .0 6 2 5 d d ] 4 1 7 .8 3
or d 2 4 0 m m (by trial and error)
Thus, we have w
1 D d 2 s in
250 240 2 s in 8
o
3 5 .9 m m
SOLUTION (13.26) ( a ) Rise w sin 80 sin 10
o
13 . 89 mm
D 5 0 0 1 3 .8 9 5 1 3 .8 9 m m d 5 0 0 1 3 .8 9 4 8 6 .1 1 m m
Equation (13.38): T
( 0 . 2 )( 500 )( 0 . 48611 ) 8 sin 10
o
[ 0 . 51389
2
0 . 48611
2
] 3 . 05 kN m
Equation (13.37): F a 12 (5 0 0 )( 0 .4 8 6 1 1)[ 0 .5 1 3 8 9 0 .4 8 6 1 1] 1 0 .6 1 k N ( b ) kW
Tn 9549
3 ,0 5 0 ( 5 0 0 ) 9549
1 5 9 .7
216
(Eq.13.38)
SOLUTION (13.27) Fa
4
2
p max ( D
d ), 2
D
2
d
4 Fa
2
p max
0 . 016
4(5)
( 400 )
We have 1 2
( D d ) 0 .2 5
D d 0 .5
or
(1)
Thus d
2
D
2
( D d )( D d ) 0 .5 ( D d ) 0 .0 1 6
or D d 0 .0 3 2
(2) D 266
From Equation (1) & (2):
d 234
mm ,
mm
Equation (13.41): T
5000 ( 0 . 2 ) 0 . 266
3
0 . 234
3
o
2
0 . 234
2
3 sin 12
0 . 266
602
N m
SOLUTION (13.28) d Fn
Fn
d/2
D/2
r
D 2
d 2
p max
F h 0:
2 rdr sin
F a F n s in
D 2
d 2
p max (
2 rdr sin
) sin
Fa
This, after integration, yields Eq.(13.40).
T fF n r
Similarly,
D 2
d 2
fp
max
(
2 rdr sin
)r
Integrating gives Eq.(13.41).
SOLUTION (13.29) F 1 wrp
max
( 0 . 1)( 0 . 2 )( 700 ) 14 kN
f 0 .3 ( 2 4 0
2 360
(Eq.13.45)
) 1.2 5 7
We have F2
F1 e
f
14 e
1 .2 5 7
3 .9 8 3 k N
Thus T ( F 1 F 2 ) r (14 3 . 983 )( 0 . 2 ) 2 . 003
and
kW
Tn 9549
2 , 0 0 3(1 5 0 ) 9549
3 1.4 6
217
kN m
SOLUTION (13.30) 2 1 0 3 .6 6 5 ra d
T I 2 .3( 2 0 0 ) 4 6 0 N m
o
We have F1 F 2 F1 F2
e
f
T r
e
460 0 .1 2 5
3 .6 8 k N
0 .3 ( 3 . 6 6 5 )
(Eq.13.42)
3 .0 0 3
Solving F 1 3 . 003 F 2 3 . 003 ( F 1 3 . 68 )
or F1 5 .5 2 k N
F 2 1 .8 4 k N
Thus Fa F2
1, 8 4 0 ( 13 20 50 ) 7 6 7 N
r a
(Eq.13.43)
SOLUTION (13.31) T
9549 kW n
9549 ( 40 )
636 . 6 N m
600
(1)
Also F1 F 2 e
f
F2 e
5 .7 2 7 F 2
0 . 4 ( 4 .3 6 3 )
and T r ( F1 F 2 ) 0 .2 5 ( 5 .7 2 7 F 2 F 2 ) 1.1 8 2 F 2
Equation (1) and (2) give F 2 538 . 6 N
F 1 3 , 085
N
SOLUTION (13.32) 240
o
4 . 189
rad
(a) F 1 p max wr 600 ( 0 . 075 )( 0 . 15 ) 6 . 75 kN
Also F1 F2
e
f
e
0 . 4 ( 4 .1 8 9 )
5 .3 4 2
Thus F1 5 .3 4 2 F 2
Fa F2
r a
and
F 2 1 . 264
(1 . 264 10 ) 3
150 400
474
( b ) T r ( F 1 F 2 ) 0 . 15 ( 6 . 75 1 . 264 ) 10 kW
Tn 9549
823( 200 ) 9549
kN
1 7 .2 4
218
3
N
823
N m
(2)
SOLUTION (13.33) (a) F 1 p max wr 800 ( 0 . 06 )( 0 . 15 ) 7 . 2 kN T
9549 ( 10 )
9549 kW n
434
220
T ( F1 F 2 ) r ,
N m
F 2 F1
7 .2
T r
0 . 434 0 . 15
4 . 307
kN
We have e
f
F1 F2
f ln 1.6 7 2 0 .5 1 4
1.6 7 2 ,
7 .2 4 .3 0 7
Thus
0 .5 1 4 0 .1 4
(b)
3 .6 7 1 ra d 2 1 0 .3
o
129.4 From triangle ABC: 3 0 .4
s 2 0 6 .6 c o s 3 0 .4 178 . 2 mm
o
o
O 150
75.9
200 124.1 3 0 .4
o
C
A s B 206.6
72.8
SOLUTION (13.34) e
f
e
F1 e
0 . 12 ( 3 . 665 )
f
1 . 552
210
o
3 . 665
rad
F 2 1.5 5 2 F 2
F2
A
M
A
0:
F1 s F 2 c F a a
F1
or 1 . 552 F 2 ( 80 ) 50 F 2 1 . 5 ( 300 )
from which F 2 6 .0 7 k N , and F1 9 .4 2 k N
Thus kW
( F1 F 2 ) r n 9549
( 9 .4 2 6 .0 7 )( 4 )(1 0 0 ) 9549
1 0 .5 2
219
S
Fa
c a
SOLUTION (13.35) T
F1 F 2
F2
900
T r
F1
Also
9 5 4 9 (1 5 )
9549 kW n
f
o
ra d
7 9 5 .7 5 N
1 5 9 .1 5 0 .2
e
2 1 0 3 .6 6 5
1 5 9 .1 5 N m
e
0 .4 ( 3.6 6 5 )
(1)
4 .3 3 2
or F1 4 .3 3 2 F 2
(2)
From Eqs.(1) and (2): F 2 2 3 8 .8 N
F1 1 0 3 4 .6
N
We therefore obtain Fa
( cF 2 sF 1 )
1 a
1 250
111 . 4 N
[100 ( 238 . 8 ) 50 (1034 . 6 )]
Yes. Self-locking
SOLUTION (13.36) From Solution of Prob.13.35: T 159 . 15 ,
F 2 238 . 8 N ,
F 1 1034 . 5 N
Now we have
A F2
F1
c
Fa
M
A
0:
a
( a ) From Eq. (13.45): F1 p m a x w r 5 0 0 (1 0 )( 0 .0 2 )( 0 .1) 1 k N 3
Equation (13.44), 1(1 0 ) e 3
0 .2 5 ( 2 6 5 1 8 0 )
315 N
Equation (13.42): T r ( F1 F 2 ) ( 0 .1)(1 0 0 0 3 1 5 ) 6 8 .5 N m
( b ) Fa
c F 2 s F1 a
4 0 ( 3 1 5 ) 1 0 (1 0 0 0 ) 200
13 N
If F a 0 , the brake will self-lock: s
c F2 F1
[ c F1 s F 2 ]
No. Not self-locking
SOLUTION (13.37)
f
1 a
366 . 04
S
F 2 F1 e
Fa
40 (315 ) 1000
1 2 .6 m m
220
N
SOLUTION (13.38) ( a ) From Eq. (13.45): F1 w r p m a x 7 5 ( 2 5 0 )( 0 .4 9 ) 9 .1 9 k N
Equation (13.44), F 2 F1 e
0 .2 5 ( 4 .5 3 8 )
9 .1 9 ( 0 .3 2 2 ) 2 .9 6 k N
Using Eq. (13.42): T ( 9 .1 9 2 .9 6 ) ( 2 5 0 ) 1 .5 5 8 N m
( b ) Using Eq. (13.46): 2 .9 6 (1 5 0 ) 9 .1 9 ( 3 5 )
Fa
1 9 5 .8 N
0 .6 2 5
From Eq.(13.46): F a 0 for s 2 .9 6 (1 5 0 ) 9 .1 9 4 8 .3 m m The brake is self-locking (for f 0 .2 5 ), if: s 4 8 .3 m m
SOLUTION (13.39) (a) T
9549 kN n
Fn
T fr
9549 ( 25 )
800
298 . 4 0 . 25 ( 0 . 3 )
Hence F a
Fn
298 . 4 N m
3979
N
( b fc )
a
3979 1
( 0 . 4 0 . 25 0 . 05 ) 1 . 542
kN
No. Not self-locking ( b ) R Ax fF n 0 . 25 ( 3979 ) 994 . 8 N R Ay F n F a 3979 1542 R A [ 994 . 8
2437
2
2437
N
1
2
] 2 2 . 632
kN
SOLUTION (13.40) ( a ) From Solution of Prob.13.39: F n 3979
N
We now have Fa
Fn a
( b fc )
3979 1
( 0 . 4 0 . 25 0 . 05 ) 1, 641
N
No. Not self-locking ( b ) From Solution of Prob.13.39: R A 2 . 632
kN
SOLUTION (13.41)
M
A
0:
Fn
4 ( 0 .4 5 ) 0 .2
9 kN
We now use p avg
Fn 2 ( r s in
)w
9 ,0 0 0 o
2 ( 0 .1 5 )(s in 4 5 )( 0 .0 7 5 )
0 .5 6 6 M P a
2
(CONT.)
221
13.41 (CONT.) T fF n r ( 0 .3 5 )(9 1 0 )( 0 .1 5 ) 4 7 2 .5 N m 3
Thus and
kW
Tn 9549
12 . 37
472 . 5 ( 250 ) 9549
Comment: Short-Shoe analysis overestimates pressure, torque, and power capacity of the brake.
SOLUTION (13.42) ( a ) From Eq.(13.47): F n Pm a x [ 2 ( r s in
2
)] w (8 0 0 )(1 0 )[ 2 ( 0 .1 5 s in 3
and
fF n 0 .2 5 ( 4 4 9 5 ) 1 1 2 4 N
So
M
0:
A
44 2
o
)]( 0 .0 5 ) 4 4 9 5 N
0 .5 F a 0 .0 2 5 (1 1 2 4 ) 4 4 9 5 ( 0 .2 ) 0 ,
F a 1 .7 4 2 k N
and T fF n r 1 1 2 4 ( 0 .1 5 ) 1 6 8 .6 N
(b) R A [( 4 4 9 5 1 7 4 2 ) 1 1 2 4 ] 2
2
1
2 .9 7 4 k N
2
Fa
300
200 A
25
fF n
Fn
SOLUTION (13.43) ( a ) From Eq. (13.48): T fF n r ,
(b)
or
M
Fn
T fr
250 ( 0 .4 )( 0 .3 5 )
1 .7 9 k N
( 0 .3 5 )[ 0 .4 1 .7 9 (1 0 )] ( 0 .9 F a ) 0 .3 2[1 .7 9 (1 0 )] 0 3
o
3
Fa 9 1 5 N
SOLUTION (13.44) Equation (13.47): F n p m a x [ 2 ( r s in
2
)] w 7 0 0[ 2 ( 0 .1 s in
30 2
o
1 .4 5 k N
Then, Eg.(13.49): Fa
Fn a
( b fc )
3
1 .4 5 (1 0 ) 0 .2
[ 0 .1 2 0 .2 ( 0 .0 3 ) ]
827 N
222
)]( 0 .0 4 )
SOLUTION (13.45) ( a ) From Eq. (13.57), T ( s in ) m
p m ax
(1)
2
fw r ( c o s 1 c o s 2 )
Substituting Eq. (13.52) and the data: p m ax
1 3 6 s in 9 0
o
2
o
2 .7 6 M P a
o
( 0 .3 5 ) ( 0 .0 2 5 ) ( 0 .0 7 5 ) ( c o s 0 c o s 9 0 )
( b ) Introducing Eq. (13.52) and the data into Eq. (1); we have p m ax
1 3 6 s in 6 5
o
2
o
o
( 0 .3 5 ) ( 0 .0 2 5 ) ( 0 .0 7 5 ) ( c o s 2 0 c o s 6 5 )
4 .8 4 M P a
SOLUTION (13.46) c (cos 2 2 cos 2 1 ) 4 r (cos 2 cos 1 )
Equation (13.54): (1)
where 1 0 , 2 9 0 Equation (1) thus o
AO c
o
c ( 1 1) 4 r ( 1)
c 2r
or
and b c cos 45
o
2 r cos 45
o
1 . 414 r
SOLUTION (13.47) Refer to Figs. P13.47, 13.22, and 13.23. 1
c [ 2 5 0 1 7 5 ] 2 3 0 5 .2 m m 2
tan
2
1 250 175
1 10 . 01 , 2 100 . 01
55 . 01 ; o
We have 2 9 0
o
o
o
hence ( s in ) m 1.
( a ) Equation (13.53): M
n
6
( 0 .0 5 )( 0 .2 )( 0 .3 0 5 2 )( 0 .9 1 0 )
[ 2 ( 2 ) s in 2 0 0 .0 2 s in 2 0 .0 2 ] 2 6 2 7 .5 N m o
4 (1 )
o
Equation (13.54): M
f
6
( 0 . 3 )( 0 . 05 )( 0 . 2 )( 0 . 9 10 ) 4 (1 )
[ 0 . 3052 (cos 200 . 02
o
cos 20 . 02 ) o
4 ( 0 .2 )(c o s 1 0 0 .0 1 c o s 1 0 .0 1 )] 2 3 8 .5 N m o
o
Equation (13.55): Fa
1 a
(M
n
M
f
)
1 0 .5 5
[ 2 3 8 9 ] 4 .3 4 4 k N
( b ) Equation (13.57): T
2
6
( 0 .3 )( 0 .0 5 )( 0 .2 )( 0 .9 1 0 ) 1
[ c o s 1 0 .0 1 c o s 1 0 0 .0 1 ] 6 2 5 .6 N m o
Thus kW
Tn 9549
625 . 6 ( 600 ) 9549
39 . 31
223
o
SOLUTION (13.48) Refer to Figs.P13.48 and 13.22. 1 1 5 ,
2 120 90
o
c
150 s in 3 0
o
o
300
o
and (s in ) m 1,
105
a 300 cos 30 250 510 o
m,
o
mm
( a ) Equations (13.53), (13.54), and (13.55): 6
M
( 0 .0 6 )( 0 .2 )( 0 .3 )( 0 .8 1 0 )
n
M
( 0 .3 )( 0 .0 6 )( 0 .2 )( 0 .8 1 0 )
f
4 (1 ) 6
4 (1 )
[ 2 (1 .8 3 3 ) s in 2 4 0 s in 3 0 ] 3, 6 2 3 N m o
o
[ 0 .3 ( 0 .5 0 .8 6 6 ) 4 0 .2 ( 0 .5 0 .9 6 6 )] 5 4 9 N m
Thus Fa
(b) T
1 a
(M
n
M
2
6
( 0 .3 )( 0 .0 6 )( 0 .2 )( 0 .8 1 0 ) 1
f
)
1 0 .5 1
[3, 6 2 3 5 4 9 ] 6 .0 2 7 k N
[ 0 .9 6 6 0 .5 ] 8 4 4 N m
(Eq.13.57)
Hence kW
Tn 9549
( 844 )( 500 ) 9549
44 . 19
SOLUTION (13.49) ( a ) We have 2 4 5 , Apply Eq. (13.58) to obtain o
a
4 r s in
2
2 2 s in 2
4 (1 2 5 ) s in 4 5
o
2 ( 4 5 )( 1 8 0 ) s in 4 5
o
1 5 5 .2 m m
( b ) Applying Eqs. (13.59) and (13.60): R Ax
2wr a
2
Pm a x s in 2
2 ( 0 .0 5 )( 0 .1 2 5 ) 0 .1 5 5 2
2
(1 .5 5 1 0 ) s in 4 5 6
o
1 1 .0 3 k N
R A y fR A x 0 .3 (1 1 .0 3 ) 3 .3 1 k N
( c ) From Eq. (13.61b): T R A y a (3 .3 1)( 0 .1 5 5 2 ) 0 .5 1 4 k N m
End of Chapter 13
224
CHAPTER 14
MECHANICAL SPRINGS
SOLUTION (14.1)
(a) J
32
TL GJ
(8 )
4
; 80
402 . 124
mm
1 . 396
rad
o
2
T ( 1 . 25 ) 79 ( 10
9
)( 402 . 124 10
12
)
or T 35 . 48 N m
(b)
16T
d
3
1 6 ( 3 5 .4 8 )
( 8 1 0
3
)
353 M Pa
3
SOLUTION (14.2) (a) d
3
3
1 6 ( 2 1 0 )( 0 .1 5 )
16 PR
a ll
( 3 5 0 1 0
6
d 1 8 .7 1 m m
,
1 .5 )
( b ) Equation (14.2): L
4
d G
4
(1 8 .7 1 ) ( 7 9 )( 4 0 )
2
32 PR
3
3 2 ( 2 1 0 )(1 5 0 )
0 .8 4 5 m 8 4 5 m m
2
SOLUTION (14.3) Angle of twist, T L J G ;
T J G L , where
2 0 ( 1 8 0 ) 0 .3 4 9 ra d o
J d
3 2 (1 2 )
4
3 2 2 .0 3 6 (1 0 ) m m
4
3
4
2 .0 3 6 (1 0
Thus, 0 .3 4 9 ( 2 .0 3 6 1 0
T
9
9
)( 7 9 1 0 )
4 6 .8 N m
1 .2
From Eq.(14.1): 1 6T d
3
1 6 ( 4 6 .8 )
( 0 .0 1 2 )
3
1 3 7 .9 M P a
SOLUTION (14.4) D Cd 5 ( 7 ) 35 mm
a ll K s
8 PC
d
2
450
;
8 P2 ( 5 )
(7 )
2
(1
0 .6 1 5 5
),
Thus k
P2 P1
2 1
1 5 4 2 1 0 0 0 20
2 7 .1 N m m
and N
a
dG 3
8C k
3
7 ( 29 10 ) 3
8 ( 5 ) ( 27 . 1 )
7 . 49 coils
225
P2 1 .5 4 2 k N
9
) m
4
SOLUTION (14.5) Apply Eq. (14.10): P
4
d G 3
8D Na
where N a p d 1 2 .5 9 .5 3 m m
So,
P
4
9
( 0 .0 0 9 5 ) ( 7 9 1 0 )( 0 .1 2 5 ) 8 ( 0 .0 5 )
1 .9 3 k N
3
By Eq. (14.6), 8 PD K s d
For
3
C D d 5 0 9 .5 5 .2 6 3
Ks 1
1 .1 1 7
0 .6 1 5 5 .2 6 3
(Eq. 14.7)
It follows that
8 (1 .9 3 )( 0 .0 5 )
( 0 .0 0 9 5 )
(1 .1 1 7 ) 3 2 0 .1 M P a
3
From Table (14.3), S y s 0 .4 S u By Eq. (14.12), Su Ad
1 6 1 0 (9 .5 )
b
0 .1 9 3
1 0 4 2 .6 M P a
Hence S y s 0 .4 5 (1 0 4 2 .6 ) 4 6 9 .2 M P a
Since 3 2 0 .1 4 6 9 .2
Yes.
SOLUTION (14.6) From Eq. (14.11), k
4
P
d G 3
8D Na
As d =constant and k proportional to d
4
3
D , The largest active coil will have the smallest
value of k . That is, the bottom coil will deflect to zero pitch first. Using Eq. (14.10), with N a p 6 m m : P
4
d G
3
8D Na
4
9
( 0 .0 4 ) ( 7 9 1 0 )( 0 .0 0 6 ) 3
8 ( 0 .0 6 ) (1 )
7 0 .2 N
The total deflection is P N a 6 (5 ) 3 0 m m SOLUTION (14.7) ( a ) From Eq. (14.11), k
4
d G 3
8D Na
Outer spring: ko
4
(7 ) (79000 ) 3
8 ( 40 ) ( 4 )
9 2 .6 2 N m m
Inner spring: ki
4
( 4 .5 ) ( 7 9 0 0 0 ) 3
8 ( 22 ) (8 )
4 7 .5 4 N m m
(CONT.)
226
14.7 (CONT.) By Eq. (14.11), k W or W k . Hence
1 4 .2 7 m m
2000 ( 4 7 .5 4 9 2 .6 2 )
( b ) Force on each spring: W o k o (9 2 .6 2 )(1 4 .2 7 ) 1 3 2 2 N W i k i ( 4 7 .5 4 )(1 4 .2 7 ) 6 7 8 N
For outer spring: C 4 0 7 5 .7 1, K s 1
0 .6 1 5 5 .7 1
For inner spring: C 2 2 4 .5 4 .8 9 , K s 1
1 .1 1
0 .6 1 5 4 .8 9
1 .1 3
(8 w D d ) K s 3
Apply Eq. (14.6): i
8 (1 3 2 2 )( 4 0 )
i
8 ( 6 7 8 )( 2 2 )
(7 )
3
( 4 .5 )
(1 .1 1) 4 3 5 M P a (1 .1 3 ) 4 7 1 M P a
3
SOLUTION (14.8) C 15 3 5
S u Ad
b
1510 ( 3
Table 14.3: ( a)
k
3
Nt N
S
0 . 42 S u 509
MPa
3
A 1510
23 . 7 N mm
3
2 12 , h s ( N t 1) d 13 ( 3 ) 39 mm 8 Pmax C
( b ) S ys K s
d
2
Pm a x
Thus
MPa
8 ( 5 ) ( 10 )
a
a
ys
0 . 201
) 1 , 211
3 ( 79 10 )
dG 8C N
b 0 .2 0 1 ,
Table 14.2:
Pmax
,
( 5 0 9 )( 3 ) 8 ( 5 )(1
2
0 .6 1 5
2
S ys d 8 CK
s
3 2 0 .4 N )
5
SOLUTION (14.9) (a) d
2
(1
8 PCn S ys
0 . 615 C
3
8 ( 2 10 )( 5 )( 1 . 3 )
)
6
( 500 10 )
(1
0 . 615 5
)
or d 8 . 62 mm
( b ) D 8 .6 2 ( 5 ) 4 3 .1 m m , N
a
dG
hs (N
(c)
c hf
3
8C K
a
2 4 .4 4 106
3
( 8 .6 2 )( 7 9 1 0 ) 3
8(5 ) (90 )
c 1 .1 0
hf D
90
2 4 .4 4 m m
7 .5 7
2 ) d 82 . 49 mm,
0 .2 3
3
2 (1 0 )
1 0 6 .9 4 3 .1
h f 24 . 44 82 . 49 106 . 9 mm
2 .4 8
Spring is safe (Curve A, Fig.14.10)
227
SOLUTION (14.10) (a) Nt
hs
d
12
2 1.6 1.8
C
N a 12 2 10 ( S ys n ) d
P k
2
8 K sC
3
( 9 0 0 2 )(1 .8 )
( 1 . 8 )( 79 10 )
1 .1 5
P k
1.0 7 4
0 .6 1 5 8 .3 3
64 N
3 . 075
3
8 ( 8 . 33 ) ( 10 )
a
2
8 (1 .0 7 4 )( 8 .3 3 )
s 1 .1 5
Thus
Ks 1
3
dG 8C N
8 .3 3
15 1 .8
N mm
64 3 .0 7 5
2 3 .9 3 m m
h f h s s 2 1 .6 2 3 .9 3 4 5 .5 3
(b)
s h
h
0 . 53
3 . 04
f
D
f
mm
Spring is safe (Curve B, Fig.14.10)
SOLUTION (14.11) d
2
8CP
S ys n
(1
0 .6 1 5 C
)
8 ( 8 )( 2 0 0 ) 6
( 4 2 0 1 0 ) 2 .5
(1
0 .6 1 5 8
),
d 5 .1 1 m m
D 5 .1 1(8 ) 4 0 .8 8 m m
N
a
5 . 11 ( 79 10
dG 3
8C k
3
8 ( 8 ) ( 9 10
6 3
) )
10 . 95
N t 1 0 .9 5 2 1 2 .9 5,
s 1 .2 0
h s (1 2 .9 5 )(5 .1 1) 6 6 .2 m m
2 6 .7 m m
200 9000
h f 6 6 .2 2 6 .7 9 2 .9 m m
Therefore s h
h
0 . 287 ,
f
D
f
2 . 23 Spring is safe (Curve A, Fig.14.10)
SOLUTION (14.12) ( a ) a ll
8 PD
d
3
,
d
3
8 PD
a ll
8 ( 2 )75 6
( 5 2 5 1 0 )
,
d 8 .9 9 m m
( b ) C D d 7 5 8 .9 9 8 .3 4 Na
dG 3
8C k
s 1 .1 0
(c)
s hf
0 .5 3,
P k
3
8 .9 9 ( 7 9 1 0 )
7 .2 9 ,
3
8 ( 8 .3 4 ) ( 2 1 )
1 .1 0 hf D
2 21
h s ( N a 2 1)8 .9 9 9 2 .5 1 m m
1 0 4 .8 m m ,
2 .6 3
h f h s s 1 9 7 .3 m m
Spring is safe (Curve B, Fig.14.10)
228
SOLUTION (14.13) We have C D d 2 0 2 .5 8 .0 ( a ) From Table 14.3: S y s 0 .4 0 S u where Su Ad
b
2 0 6 0 ( 2 .5 )
0 .1 6 3
1774 M Pa
and m a x 0 .4 0 (1 7 7 4 ) 7 0 9 .6 M P a
By Eq. (14.8): 2
d m ax
Pm a x
8CK
where K w 1
w
2
( 2 .5 ) ( 7 0 9 .6 )
0 .6 1 5 8
1 .0 7 7
(Eq. 14.7)
2 0 2 .1
8 ( 8 )(1 .0 7 7 )
Use Eq. (14.11), k
4
d G 3
8D Na
4
( 2 .5 ) ( 7 9 0 0 0 ) 3
8 ( 2 0 ) (1 1 )
4 .3 8 N m
( b ) Pm a x k 2 0 2 .1 4 .3 8 4 6 .1 4 m m By Fig. 14.8c: h s N t d h s ( N a 2 ) d (1 1 2 ) 3 2 .5 m m
Thus, h f 3 2 .5 4 6 .1 4 7 8 .6 4 m m
( c ) h f D 7 8 .6 4 2 0 3 .9 3 2 h f 4 6 .1 4 7 8 .6 4 0 .5 8 7
From Fig. 14.10, for Case B:
Yes, Buckling occurs.
SOLUTION (14.14) ( a ) Use Eq. (14.12) and Table 14.2. Su Ad
b
1 6 1 0 ( 0 .9 )
0 .1 9 3
1643 M Pa
Table (14.3): S y s 0 .4 5 S u 0 .4 5 (1 6 4 3 ) 7 3 9 .4 M P a
( b ) Spring index, with C D d 1 0 0 .9 1 1 .1 : Ks 1
0 .6 1 5 C
1
0 .6 1 5 1 1 .1
1 .0 5 5
(Eq. 14.7)
Rearrange Eq. (14.6), let S y s , and solve for P . 3
P
d S ys 8KsD
3
( 0 .9 ) ( 7 3 9 .9 ) 8 (1 .0 5 5 )(1 0 )
2 0 .1 N
From Fig. 14.7, N a 1 4 .5 2 1 2 .5 . Equation (14.11) is then k
Gd 3
4
8D Na
( 7 9 , 0 0 0 )( 0 .9 ) 3
8 (1 0 ) (1 2 .5 )
4
0 .5 1 8 N m m
(CONT.)
229
14.14 (CONT.) (c) s
P k
3 8 .8 m m
2 0 .1 0 .5 1 8
From Fig. 14.7c: h s ( N a 3) d (1 2 .5 3)( 0 .9 ) 1 3 .9 5 m m h f s h s 3 8 .8 1 3 .9 5 5 2 .7 5 m m
Figure 14.7: p
(d)
s hf
3 8 .8 1 3 .9 5
hf 2d
5 2 .7 5 2 ( 0 .9 )
Na
1 2 .5
hf
2 .7 8 ,
D
5 2 .7 5 10
4 .0 7 6 m m
5 .2 7 5
Figure 14.10, Curve A: No buckling failure.
SOLUTION (14.15) ( a ) Equation (14.4), C D d 1 4 1 .5 9 .3 3 3 Equation (14.7): Ks 1
0 .6 1 5 9 .3 3 3
1 .0 6 6
From Fig. 14.7: N t N a 2 1 6 2 1 8,
h s d N t 1 .5 (1 8 ) 2 7 m m
The pitch equals P ( h f 2 d ) N a [3 5 2 (1 .5 )] 1 6 2 m m
Solid deflection s h f hs 3 5 2 7 8 m m
( b ) Equation (14.12) and Table 14.2: S u 1 6 1 0 (1 .5 )
0 .1 9 3
1498 M Pa
From Table (14.3): S y s 0 .4 5 S u 0 .4 5 (1 4 8 9 ) 6 7 0 M P a
From Eq. (14.11): P
Gds 3
8C N a
( 2 9 , 0 0 0 ) (1 .5 ) ( 8 )
3 .3 4 4 N
3
8 ( 9 .3 3 3 ) (1 6 )
and m ax
8DKsP
8 (1 4 ) (1 .0 6 6 ) ( 3 .3 4 4 )
d
3
670 3 7 .6 5
(1 .5 )
3
3 7 .6 5 M P a
Then n
(c)
s hf
8 3 .5
a ll m ax
0 .2 2 9 ,
hf D
1 7 .8
35 14
2 .5
From Fig. 14.10 No buckling.
230
SOLUTION (14.16) ( a ) Pm
42 2
4 C 1 4C4
Kw a
m
3 kN
Kw
Pa
Ks
Pm
42 2
Pa
0 .6 1 5 C
1.3 1 1
1 . 311 1 . 123
1 3
1 kN
C
Ks 1
15 3
5
1.1 2 3
0 .6 1 5 C
0 . 389
Equation (14.24), with replacing S u s by S y s : m
5 0 0 1 .3 ( 0 .3 8 9 )( 2 5 0 0 2 8 0 )
1 9 2 .3 M P a 1
280
and d
2
8 Pm C
Ks
3
8 ( 3 1 0 )( 5 )
1 .1 2 3
m
6
(1 9 2 .3 1 0 )
s 1 . 10
( b ) D 14 . 94 ( 5 ) 74 . 7 mm N
a
dG
hs (N
(c)
s h
f
3
8C k
a
3
1 4 .9 4 ( 7 9 1 0 )
h
f
D
mm
48 . 89 mm
4000 90
1 3.1 1
3
8( 5 ) ( 90 )
2 ) d 225 . 7 mm
0 .1 8 ,
d 1 4 .9 4
,
3 .6 8
h f 274 . 6 mm
Spring is safe (Curve A, Fig.14.10)
Also fn
356000 d 2
D N
356 , 620 ( 14 . 94 )
2
( 74 . 7 ) ( 13 . 11 )
a
72 . 8 cps 4 , 370
cpm
SOLUTION (14.17) From Eq. (14.20): S e s ' 4 6 5 M P a ,
P (400 0) 2 200 N
k P 200 10 20 N m m
By Eq. (14.8), m a x
8 Pm a x D
d
Kw;
3
465
8 ( 4 0 0 )( 4 0 )
d
3
(1 .3 ) .
Solving d 4 .8 5 m m
Equation (14.11): k d G 8 D N a or 4
Na
4
d G 3
8D k
3
4
( 4 .8 5 ) ( 7 9 , 0 0 0 ) 3
8 ( 40 ) ( 20 )
4 .2 7
Crash allowance =15 % s 1 .1 5 42000 2 3 m m h s ( N a 3) d ( 4 .2 7 3)( 4 .8 5 ) 3 5 .2 6 m m h f 3 5 .2 6 2 3 5 8 .2 6 m m
231
SOLUTION (14.18) From Eq. (14.20): S e s ' 3 1 0 M P a k P 200 10 20 N m m
From Eq. (14.8): 8 Pm a x D
m ax
310
Kw;
3
d
8 ( 4 0 0 )( 4 0 )
d
3
(1 .3 )
Solving d 5 .5 5 m m
Equation (14.11): k d G 8 D N a or 4
4
Na
4
( 5 .5 5 ) ( 7 9 , 0 0 0 )
d G 3
8D k
3
3
8 ( 40 ) ( 20 )
7 .3 2
Crash allowance =8 % s 1 .0 8 42000 2 1 .6 m m h s ( N a 3) d ( 7 .3 2 3)(5 .5 5 ) 5 7 .2 8 m m h f 5 7 .2 8 2 1 .6 7 8 .8 8 m m
SOLUTION (14.19) Refer to Solution of Prob. 14.15. ( a ) We now have 4 C 1 4C 4
Kw
0 .6 1 5 C
Pm a x Pm in
Pa
14 4 2
2
4 ( 9 .3 3 3 ) 1
4 ( 9 .3 3 3 ) 4
5 N,
0 .6 1 5 9 .3 3 3
Pm
1 .1 5 6 14 4 2
9 N
From Eq. (14.8): 8CK
a Pm
m a(
(b) n
Pa
S ys
Pa
2
8 ( 9 .3 3 3 ) (1 .1 5 6 ) ( 5 )
(1 .5 )
2
6 1 .0 5 M P a
) 6 1 .0 5 ( 95 ) 1 0 9 .9 M P a
a m
w
d
3 .9 2
670 6 1 .0 5 1 0 9 .9
( c ) From Table 14.3 the modified endurance limit is S e s ' 0 .2 2 S u 0 .2 2 (1 4 8 9 ) 3 2 7 .6 M P a
Thus S es '
n
a
3 2 7 .6 6 1 .0 5
5 .3 7
Pm
180 30 2
SOLUTION (14.20) ( a ) C 15 3 5 Kw a m
4 C 1 4C 4
Kw
Pa
Ks
Pm
0 . 615 C
1 . 311
105
Pa
N
Ks 1
0 . 615 C
180 30 2
75 N
1 . 123
0 . 834
1 . 311 75 1 . 123 105
Thus
m
K
8 Pm C s
d
2
1 . 123
8 ( 105 )( 5 )
( 3 10
3
)
2
166 . 8 MPa
a 0 . 834 (166 . 8 ) 139 . 1 MPa
(CONT.)
232
14.20 (CONT.) Table 14.2 and Eq.(14.12): S u Ad
b
2060 ( 3
0 . 163
Table 14.3: S es 0 . 23 S u 396 '
) 1 , 722
MPa
Equation (7.5a): S us 0 . 67 S u 1,154
MPa
Equation (14.25) results in 1 ,1 5 4 ( 3 9 6 )
n
( b ) hs ( N a 2 ) d 7 2 m m , s 1 .1 0
(c)
fn
(d)
h
s
1.3 8
1 3 9 .1 ( 2 1 ,1 5 4 3 9 6 ) 1 6 6 . 8 3 9 6
k
1 8 .3 8 m m ,
180 1 0 .7 7
1 356 , 620 d 2 2 D Na
0 .2 ,
356 , 620 ( 3 )
h
6 . 03
f
D
f
108
2
2 ( 15 ) ( 22 )
dG 3
8C N a
3
3 ( 7 9 1 0 ) 3
8 (5 ) ( 22 )
1 0 .7 7 N m m
h f 7 2 1 8 .3 8 9 0 .4
mm
cps 6 , 480
(for fixed-free ends)
cpm
Buckling will occur (Curve B, Fig.14.10)
SOLUTION (14.21) D C d 8 (5 ) 4 0 m m Kw
4 C 1 4C4
m Ks
0 .6 1 5 C
8 Pm C
d
0 .6 1 5 8
1 .0 7 7
1.1 8 4 8 ( 5 0 0 )( 8 )
1 .0 7 7
2
Ks 1
(5)
2
4 3 8 .8 M P a
Table 14.3 and Eq.(14.12): Su Ad
b
2 0 6 0 (5
0 .1 6 3
) 1585 M Pa
Equation (7.5a): S u s 0 .6 7 (1 5 8 5 ) 1 0 6 2 M P a Table 14.3: S e s 0 .2 3 (1 5 8 5 ) 3 6 4 .6 M P a '
Equation (14.23) gives a
1 2
( 3 6 4 .6 )(1 0 6 2 1 .2 4 3 8 .8 ) 1062 (
1 2
)( 3 6 4 .6 )
9 2 .4 7 M P a
Thus Pa
Ks
a
Kw m
Pm
1 .0 7 7 9 2 .4 7 1 .1 8 4 4 3 8 .8
(5 0 0 ) 9 5 .8 N
and Pm in 5 0 0 9 5 .8 4 0 4 .2 N Pm a x 5 0 0 9 5 .8 5 9 5 .8 N
233
MPa
SOLUTION (14.22) ( a ) C 24 5 4 . 8 4 C 1 4C4
Kw
Pm
320
640 2
1.3 2 5
0 .6 1 5 C
Pa
N
Ks 1
0 .6 1 5 C
160 2
80 N
1.1 2 8
Table 14.2 and Eq.(14.12): S u Ad
0 . 163
2060 ( 5
b
S es 0 . 23 S u 365 '
Table 14.3:
) 1 , 585
MPa
MPa
Equation (7.5a): S us 0 . 67 S u 1, 062
MPa
Therefore we obtain
m
K
a
8 Pm C s
K
w
Pa
K
s
Pm
8 ( 320 )( 4 . 8 )
1 . 128
2
d
m
1 . 325 1 . 128
( 0 . 005 ) 80 320
176 . 5 MPa
2
(176 . 5 ) 51 . 83 MPa
Equation (14.25): n
1 , 062 ( 365 ) 51 . 83 ( 2 1 , 062 365 ) 176 . 5 ( 365 )
2 . 49
( b ) Load of 160 N causes deflection of 72-65=7 mm. N
G d
a
8 PC
3
7 ( 7 9 , 0 0 0 )( 5 ) 8 (1 6 0 )( 5 )
1 7 .3
3
SOLUTION (14.23) 90 20 2
( a ) Pa
4 C 1 4C4
Kw a m
35 N
K
w
Pa
K
s
Pm
0 .6 1 5 C
1 . 253 1 . 103
90 20 2
Pm 1.2 5 3
55 N
Ks 1
0 .6 1 5 C
1.1 0 3
0 . 723
35 55
Use Eq.(14.24) with replacing S u s by S y s . m
Thus
5 6 0 1 .6 0 .7 2 3 ( 2 5 6 0 3 1 5 )
1 2 2 .9 M P a 1
315
d
2
K
8 Pm C
m
s
1 .1 0 3
8 ( 5 5 )( 6 ) 6
(1 2 2 .9 1 0 )
,
d 2 .7 5 m m
( b ) D d C 2 .7 5 ( 6 ) 1 6 .5 m m k
P
Na
90 20 0 .0 7 5 0 .0 6 5
dG 3
8C k
7 kN m 3
2 .7 5 ( 2 9 1 0 ) 3
8(6) (7 )
6 .6 ,
h s 8 .6 ( 2 .7 5 ) 2 3 .7 m m ,
N t 6 .6 2 8 .6 h f 0 .0 7 5 1 .1 0 ( 7 02 00 0 ) 7 8 .1 4 m m
s h f h s 7 8 .1 4 2 3 .7 5 4 .4 m m
(c)
(d)
fn s hf
3 5 6 ,6 2 0 d 2
D Na
0 .7 ,
3 5 6 , 6 2 0 ( 2 .7 5 ) 2
(1 6 .5 ) ( 6 .6 )
hf D
4 .7
5 4 5 .8
cps 32, 748
cpm
Spring is safe (Curve A, Fig.14.10)
234
SOLUTION (14.24) 4 C 1 4C4
(a) Kw
1.2 5 3
0 .6 1 5 C
Pa
5 0 1 0 2
20 N
a
K
m
w Pa
K s Pm
1 . 253 1 . 103
m
Thus
Ks 1 5 0 1 0 2
Pm
0 .6 1 5 C
1.1 0 3
30 N
0 . 757
20 30
5 6 0 1 .6
1 1 9 .3 M P a
0 .7 5 7 ( 2 5 6 0 3 1 5 )
1
315
Hence 2
d
8 Pm C
K
1 .1 0 3
m
s
8 ( 3 0 )( 6 ) 6
(1 1 9 .3 1 0 )
d 2 .0 6 m m
or
( b ) D 2 .0 6 ( 6 ) 1 2 .3 6 m m 5 0 1 0 0 .1 2 5 0 .1 0 5
k
P
N
dG
a
3
2 . 06 ( 29 10 )
3
8C k
2 kN m 17 . 29
3
8(6) (2)
N t 17 . 29 2 19 . 29 h s 1 9 .2 9 ( 2 .0 6 ) 3 9 .7 4 m m h f 1 2 5 1 .1 0 ( 2 10 00 0 ) 1 2 5 .0 0 6 m m
s h f h s 1 2 5 .0 0 6 3 9 .7 4 8 5 .2 7 m m
(c)
(d)
fn s
3 5 6 , 6 2 0 ( 2 .0 6 )
hf
0 .6 8 ,
hf
2 7 8 .1 c p s 1 6 , 6 8 6 c p m
2
(1 2 .3 6 ) (1 7 .2 9 )
1 0 .1
D
Spring will buckle (Curve A, Fig.14.10)
SOLUTION (14.25) ( a ) Pm 400 Kw
Pa 100
N
4 C 1 4C4
a
K
1
s
0 . 615 C
1.2 5 3
0 .6 1 5 C
m K s
N
8 Pm C
d
K
w
Pa
K
s
Pm
2
1.1 0 3
m
1.2 5 3 1 .1 0 3
8 ( 4 0 0 )( 6 )
1
( 4 )(
6,741 d
2
6,741
2
(d )
d
)
2
1,9 1 4 d
2
Equation (14.23) leads 1,9 1 4 d
2
(720
330 2
)
330 2
720 1. 6
(
6,741 d
2
)
or 6 , 438 d
2
450
6 , 741 d
2
,
d 5 . 41 mm
( b ) D 5 . 41 ( 6 ) 32 . 46 mm N
a
dG 8 PC
3
5 .4 1( 7 9 , 0 0 0 )( 8 ) 8 ( 5 0 0 3 0 0 )( 6 )
3
9 .8 9
235
1 . 103
SOLUTION (14.26) ( a ) Pm 470
Pa 130
N,
C 30 6 5
N,
Table 14.2 and Eq.(14.12): S u Ad
b
0 . 201
1510 ( 6
) 1053
MPa
Table 14.3: S
0 . 42 (1053 ) 442
ys
Ks 1 P2 P1
We have k N
a
1G d 8 P1 C
K
m
a
3
2 1
d
Kw
Pa
Ks
Pm
600 340 13
1 7 ( 2 9 0 0 0 )( 6 )
8 ( 3 4 0 )( 5 )
8 Pm C s
1.1 2 3 ,
0 .6 1 5 C
3
m
1 .3 1 1 1 .1 2 3
'
Kw
4 C 1 4C4
0 .6 1 5 C
1
20 N mm ,
340 20
MPa
1.3 1 1 17 mm
8 .7 8 ( 470 )( 5 )
1 . 123
2
S es 0 . 21 (1053 ) 221
MPa ,
(6)
2
186 . 7 MPa
( 14 37 )(1 8 6 .7 ) 6 0 .2 9
M Pa
It follows that n
(b)
hs ( N k
s
3
a
1 . 20
fn
(d)
hf
6 ( 79000 )
(Eq.14.25)
54 . 48 N mm
3
8 ( 5 ) ( 8 .7 )
13 . 22 mm ,
600 54 . 48
h
356 , 620 ( 6 )
1 356 , 620 d 2 2 D Na
0 .1 7 ,
1.2
2 ) d (10 . 7 ) 6 64 . 2 mm
dG 8C N
(c)
s
a
442( 221) 6 0 .2 9 ( 2 4 4 2 2 2 1 ) 1 8 6 .7 ( 2 2 1 )
2
2 ( 30 ) ( 8 . 7 )
2 .6
f
D
h f 64 . 2 13 . 22 77 . 42 mm
136 . 6 cps 8 ,196
cpm
(fixed-free ends)
Spring is safe (Curve B, Fig.14.10)
SOLUTION (14.27) Refer to Example 14.6. The critical torsional shear stress in the hook is from Eq. (14.34b): B (
rm ri
)
8 PD
d
3
S ys
where S y s 0 .4 0 S u 0 .4 0 (1 2 5 6 ) 5 0 2 .4 M P a
rm 3 .7 5 m m D 7 .5 m m
ri 3 .7 5 ( 2 .5 2 ) 2 .5 m m d 2 .5 m m
Thus, 5 0 2 .4 ( 32.7.55 )
8 P (1 2 .5 )
( 2 .5 )
3
,
P 1 6 4 .4 N
The larger load is PA 2 3 2 .1 N , as found in Part (b) of Example 14.6. This shows that failure will occur first by shear stress in the hook.
236
SOLUTION (14.28) k 4 d1 3 D1
Gd
4
Since k 1 k 2 :
3
8D Na
d
4 2
3
D2
d 2 d1 4
,
3 D2
4
3
D1
3 ( 81 ) , 4
d 2 14 . 27 mm
3
SOLUTION (14.29) C
4 C 1 4C 4
Kw a m
10
5 0 .5
K
w
Pa
K
s
Pm
K
1
s
0 .6 1 5 C
1 . 0615
0 . 615 C
1 .1 4 5,
Pm 3 N ,
m Ks
0 .7 1 9
8 Pm C
d
2
Pa 2 N
1 .0 6 1 5
8 ( 3 ) (1 0 )
( 0 .5 )
2
3 0 5 .6 M P a
a 0 .7 1 9 (3 0 5 .6 ) 2 1 9 .7 M P a
Table 14.2 and Eq.(14.12): Su Ad
2 0 6 0 ( 0 .5
b
0 .1 6 3
) 2306
M Pa
Table 14.3: S e s 0 .2 3 S u 5 3 0 .4 M P a
Equation (7.5a): S u s 0 .6 7 S u 1 5 4 5 M P a
'
Equation (14.25) is therefore n
1545 ( 530 . 4 )
1 . 13
219 . 7 ( 2 1545 530 . 4 ) 305 . 6 ( 530 . 4 )
SOLUTION (14.30) ( a ) We have C D d 2 .4 0 .6 4 . So, 4 C 1 4C 4
Kw
0 .6 1 5 C
4 ( 4 ) 1
4(4)4
0 .6 1 5 4
1 .2 5 0 .1 5 3 7 5 1 .4 0 3 7 5
From Eq. (14.8): m ax K w
8 PD
d
3
8 ( 6 )( 2 .4 )(1 0
1 .4 0 3 7 5
( 0 .6 )(1 0
3
9
)
)
2 3 8 .3 M P a
( b ) Use Eq.(14.34). Here ( ri ) A ( rm ) A d 2 1 .2 0 .3 0 .9 m m . Hence (
A
)c
3
1 .2 1 6 ( 6 )( 2 .4 1 0 ) 9 0 .9 ( 0 .6 )(1 0 )
4 (5) 2
( 0 .6 ) (1 0
6
)
4 5 2 .7 1 7 .6 8 4 7 0 .3 8 M P a
From Eq. (14.12) and Table 14.2 for music wire: Su Ad
2 0 6 0 ( 0 .6 )
b
0 .1 6 3
2239 M Pa
We also have S y S y s 0 .5 7 7 ( 0 .4 0 S u 0 .5 7 7 0 .6 9 3 S u 0 .6 9 3 ( 2 2 3 9 ) 1, 5 5 2 M P a
Thus n
Sy (
A
)c
1552 4 7 0 .3 8
3 .3
SOLUTION (14.31) ( a ) I bh
ML EI
Hence
3
12 (1 . 2 ) L
,
N
a
L
D
EI M
4
12 0 . 1728 ( 16 2 )( 207 10
14 ,983
(1 8 )
240
mm 3
)( 0 . 1728 )
4
,
C D d 18 1 . 2 15
14 , 983
mm
265
(CONT.)
237
14.31 (CONT.) ( b ) P a 1 5 (1 6 ) 2 4 0 N m m Ki
2
3 C C 0 .8 3 C ( C 1)
i Ki
1 .0 4 6 ,
6 Pa bh
2
1 .0 4 6
6 ( 240 ) (1 .2 )
3
8 7 1 .7 M P a
SOLUTION (14.32) The spring index is C D d 1 5 1 .5 1 0 . From Eq. (14.36): 2
4 C C 1 4 C ( C 1 )
Ki
2
4 (1 0 ) 1 0 1
1 .0 8 1
4 (1 0 )(1 0 1 )
Equation (14.12) and Table 14.2 for hard-drawn wire: Su Ad
1 5 1 0 (1 .5 )
b
0 .2 0 1
1392 M Pa
S y S y S y s 0 .5 7 7 ( 0 .4 2 0 .5 7 7 ) S u 1 0 1 3 M P a 1 0 1 3 N m m
From Eq. (14.39): 3
M
d Sy
3
(1 .5 ) (1 0 1 3 )
32 K i
3 2 (1 .0 8 1 )
3 1 0 .5 N m m
We have I d
6 4 (1 .5 )
4
4
6 4 0 .2 4 9 m m
4
Equation (14.41), with r a d 1 .2 , is thus M lw
1 .2
EI
3 1 0 .5 [ (1 5 ) N a ] 3
( 2 1 0 1 0 ) ( 0 .2 4 9 )
N a 4 .2 8 8
,
SOLUTION (14.33) From Eq.(14.12) and Table 14.2 for oil-tempered wire: Su Ad
b
1610(2)
0 .1 9 3
1 4 0 8 .4 M P a
( a ) Equation (7.5b) and Table 14.3: S y S y s 0 .5 7 7 0 .4 5 S u 0 .5 7 7 0 .7 8 S u 0 .7 8 (1 4 0 8 .4 ) 1 0 9 8 .6 M P a
Spring index: C D d 1 2 .5 2 6 .2 5 . Equation (14.36) : 4 ( 6 .2 5 ) 6 .2 5 1
Ki
4 ( 6 .2 5 )( 6 .2 6 1 )
Equation (14.39) with i
32 Pa
d
3
a ll
0 .1 3 5 S y n 1 0 9 8 .6 1 .8 6 1 0 .3 M P a :
6 1 0 .3 (1 0 ) 6
K i;
3 2 P ( 0 .0 2 8 )
( 0 .0 0 2 )
3
( 0 .1 3 5 )
Solving, P 1 2 6 .8 N ( b ) L w D N a (1 2 .5 )(3 .5 ) 1 3 7 .4 m m I d
4
64 (2)
4
6 4 0 .7 8 5 4 m m
4
Equation (14.41), with M P a 1 2 6 .8 ( 0 .0 2 8 ) 3 .5 5 N m Hence rad
M lw EI
( 3 .5 5 ) ( 0 .1 3 7 4 ) 9
2 0 0 1 0 ( 0 .7 8 5 4 1 0
12
)
3 .1 1 r a d
238
2
SOLUTION (14.34) ( a ) We have a ll m a x n 8 0 0 2 .5 3 2 0 M P a From Eq. (14.42):
a ll
6 PL nbh
3 2 0 (1 0 ) 6
;
2
3
6 ( 4 0 1 0 )( 0 .7 ) 8 ( 6 )( 0 .0 2 2 )
b 136 m m
,
2
( b ) Eq. (14.43): (1 ) 2
( hl ) (1 0 .3 ) 3
6P Enb
3
6 ( 4 0 1 0 )
2
9
2 0 0 (1 0 )( 8 )( 0 .1 3 6 )
( 0 0.0.72 2 )
3
0 .0 3 2 3 m 3 2 .3 m m
SOLUTION (14.35) ( a ) C 0 .6 0 .0 8 7 .5 Table 14.2 and Eq.(14.12): Su Ad
S ys
Table 14.3: S y
0 .1 9 3
1610(2
b
0 .4 5 S u
0 .5 7 7
1 0 9 8 .4 M P a
0 .5 7 7 2
4 C C 1 4 C ( C 1 )
Equation (14.36): K i
) 1 4 0 8 .4 M P a
1 . 11
Equation (14.39) is therefore 3
( 2 ) (1 .0 9 8 4 ) 1 .4
M Pa
3 2 (1 .1 1 )
( b ) L (1 5 ) 6 2 8 2 .7 m m ,
ML EI
0 .5 5 5 ( 0 .2 8 2 7 )
9
( 2 0 0 1 0 )( 0 .7 8 5 4 1 0
12
)
0 .5 5 5 N m
I
4
d 64
0 .7 8 5 4 m m
0 .9 9 9 ra d 5 7 .3
4
o
SOLUTION (14.36) ( a ) P m 700 m
6 Pm L bh
6 ( 7 0 0 )( 0 . 4 )
2
Also
P a 400
N
m
2
40 h ( h )
h
1
'
m
( Se K
f
a m
Pa Pm
4 7
3
1 4 0 0 1 .2
Su
a
42
Su n
N
(
4 7
)
1400
)(
3 6 0 .1 M P a ) 1
5 0 0 1 .4
Thus we obtain 360 . 1 (10
)
42 h
3
,
h 4 . 89 mm
b 195 . 6 mm
Hence (b) k
6
Ebh 3
3
3 L ( 1
2
)
3
( 207 10 )( 195 . 6 )( 4 . 89 ) 3
3 ( 400 ) ( 0 . 91 )
3
27 . 1 N mm
End of Chapter 14
239
CHAPTER 15
POWER SCREWS, FASTENERS, AND CONNECTIONS
SOLUTION (15.1) ( a ) By Fig.15.4b: Thus
thread depth=7.5 mm,
thread width=7.5 mm
d m 75 7 . 5 67 . 5 mm ,
( b ) By Fig.15.4a: Hence
d r 75 15 60 mm ,
thread depth=7.5 mm,
d m 67 . 5 mm ,
L p 15 mm
thread width at pitch line=7.5 mm
d r 60 mm ,
L p 15 mm
SOLUTION (15.2) 4 , p 6 . 35
25 . 4 p
Table 15.3:
( a ) L n p 2 ( 6 .3 5 ) 1 2 .7 m m p
dm d ta n
2
Fig.15.4a: 1 4 .5
3 8 .1 3 .1 7 5 3 4 .9 2 5 m m
L
dm
1 2 .7
( 3 4 .9 2 5 )
6 .6
;
o
ta n n c o s ta n c o s 6 .6 (ta n 1 4 .5 );
n 1 4 .4 1
o
( b ) For starting, we have f Tu Td
(c) e (d) F
Wd
f cos n tan
m
cos n f tan
2
Wf
4 3
( 0 .1 ) 0 .1 3 ,
o
o
o
(c o s 1 4 .4 1 ) ( 0 .1 3 ) ta n 6 .6
7 .5 ( 3 4 .9 2 5 )
0 .1 3 (c o s 1 4 .4 1 ) ta n 6 .6
a
5 4 .2 0 .1 9
7 .5 ( 0 .1 1 )( 5 0 .8 )
7 .5 ( 0 .1 1 ) ( 5 0 .8 )
o
o
(c o s 1 4 .4 1 ) ( 0 .1 3 ) ta n 6 .6
cos n f cot
Tu
o
o
c o s n f ta n
o
o
o
o
o
c o s 1 4 .4 1 ( 0 .1 3 ) ta n 6 .6
4 3
( 0 .0 8 ) 0 .1 1
2
2
2
fc
o
cd c
0 .1 3 (c o s 1 4 .4 1 ) ta n 6 .6
7 .5 ( 3 4 .9 2 5 )
o
c o s 1 4 .4 1 ( 0 .1 3 ) c o t 6 .6
2
2
5 4 .2 N m 2 3 .3 N m
0 .4 6 4 6 %
2 8 5 .3 N
SOLUTION (15.3) f cos n tan
T o 0 d m [ cos
Solving, ta n 1 2 .5 or Then, Eq.(15.12):
n
f tan
] d c fc
f fc cos n ( d c d m ) cos n f fc ( d c d m )
dm [
o
0 .1 0 .0 8 (c o s 1 4 .4 1 )(1 .4 5 4 5 ) o
c o s 1 4 .4 1 ( 0 .1 )( 0 .0 8 )(1 .4 5 4 5 )
o
d m ta n
e
(Eq.15.11)
f c o s n ta n c o s n f ta n
o
3 4 .9 2 5 (ta n 1 2 .5 )
] d c fc
o
3 4 .9 2 5 [
o
o
c o s 1 4 .4 1 0 .1 (ta n 1 2 .5 )
240
0 .4 9 4 4 9 .4 %
o
0 .1 (c o s 1 4 .4 1 )(ta n 1 2 .5 )
] 5 0 .8 ( 0 .0 8 )
SOLUTION (15.4) d m 32 2 30 mm ,
ta n
1
1
ta n
L
d m
Wd
Tu
4 .8 5
8
( 30 )
f tan
m
1 f tan
2
n 0
L 2 ( 4 ) 8 mm ,
Wf
cdc
2
o
6 ( 30 )
2
[
0 . 1 tan 4 . 85 1 0 . 1 tan 4 . 85
o o
]
6 ( 0 . 15 ) 50 2
39 . 28 N m
We have n
V L
40 8
5 rps 300
rpm
Thus kW
Tn 9549
1 . 23
39 . 28 ( 300 ) 9549
SOLUTION (15.5) n 0,
c o s n 1,
L 25 m m ,
V 0 .1 5 6 0 9 m m in T
9549 kW n
9549 ( 4 ) 360
n
V L
9 0 .0 2 5
ta n
360
L
dm
25
( 45 )
rp m
1 0 6 .1 N m
Hence, by Eq.(15.9): Tu
Wd
f tan
m
1 f tan
2
; 106 . 1
10 ( 45 )
f 0 . 1768
2
1 f ( 0 . 1768 )
Solving f 0 . 272
SOLUTION (15.6) N 2,
( a ) Table 15.3:
p
2 5 .4 2
1 2 .7
m m th re a d ,
V 0 .0 1 m p s 0 .6 m m in ,
( b ) We have
fc 0,
ta n
dm d
1
p
ta n
dm
1
p 2
n
70
1 2 .7 2
[ 1( 623.7.6 ) ] 3 .6 4
0 .6 0 .0 1 2 7
n 0
4 7 .2 rp m
6 3 .6 m m o
Then, we obtain Tu
d mW
kW
2
f c o s n ta n
[ cos
Tu n 9549
n
f ta n
]
1 7 1 5 ( 4 7 .2 ) 9549
6 3 .6 ( 2 5 0 ) 2
[
0 .1 5 (1 ) ta n 3 .6 4
o
1 ( 0 .1 5 ) ta n 3 .6 4
o
8 .4 8
Hence ( k W ) req
8 .4 8 0 .8 5
9 .9 8
241
] 1 .7 1 5 k N m
0 .1 7 6 8
SOLUTION (15.7) ( a ) From Fig. 15.4b, d dm
P 4
24
2 5 .5 m m
6 4
Equation (15.6), with n 0 , Wdm
T
f dm L
(
2
d m fL
100 ( 24 )
2
)
c o s n 1, ta n L d m ,
l p , becomes:
W fc d c 2
( 0 .0 1 ) ( 2 4 ) 6
[ ( 2 4 ) ( 0 .1 1 )( 6 ) ]
1 0 0 ( 0 .1 1 )( 3 6 ) 2
2 2 9 .5 1 9 8 4 2 7 .5 N m
( b ) The screw is self locking, if: f
L
dm
6
( 24 )
0 .0 8
SOLUTION (15.8) ( a ) Because of the triple-threaded screw, L 3 p 3 (8 ) 2 4 m m
From Fig. 15.4a: thread depth 0 .5 p 0 .5 (8 ) 4 m m d m d 0 .5 p 5 0 4 4 6 m m
From Eq. (15.2), ta n
1
L
dm
ta n
1
( 4264 ) 9 .4 3
o
( b ) For starting, we have fc
( 0 .1 2 ) 0 .1 6
4 3
f
4 3
( 0 .1 3) 0 .1 7
From Eq. (15.8): n ta n
1
(ta n c o s ) ta n
1
(ta n 1 4 .5 c o s 9 .4 3 ) 1 4 .3 1 o
o
o
By Eq. (15.7): Td
Wdm
f c o s n ta n
2
c o s n f ta n
15 ( 46 ) 2
W fc d c
2
o
0 .1 7 c o s 1 4 .3 1 t a n 9 .4 3
o
o
o
c o s 1 4 .3 1 ( 0 .1 7 ) ta n 9 .4 3
1 5 ( 0 .1 6 ) ( 6 8 ) 2
3 4 5 ( 0 .0 0 9 1) 8 1 .6 8 4 .7 N m
SOLUTION (15.9) ( a ) We have n
0 .0 2 ( 6 0 )( m m in )
5 0 rp m
0 .0 2 4 ( m r e v )
From Eq. (15.6), with f c 0 .1 2 and f 0 .1 3 : Tu
Wdm
f c o s n ta n
2
c o s n f ta n
W fc d c
2 o
o
1 5 ( 4 6 ) ( 0 .1 2 ) c o s 1 4 .3 1 ta n 9 .4 3 2
o
c o s 1 4 .3 1 ( 0 .1 3 ) ta n 9 .4 3
o
1 5 ( 0 .1 2 )( 6 8 ) 2
3 4 5 ( 0 .2 9 6 5 ) 6 1 .2
1 6 3 .5 N m
(CONT.)
242
15.9 (CONT.) Power: ( h p ) out ( h p ) in
FV 7 4 5 .7 Tn 7121
1 5 0 0 0 ( 0 .0 2 )
0 .4 0 2 h p
7 4 5 .7
1 6 9 .4 ( 5 0 ) 7121
1 .1 8 9 h p
Efficiency is thus e
0 .3 4 3 4 %
0 .4 0 2 1 .1 8 9
( b ) From Eq. (15.10): f c o s n ta n c o s 1 4 .3 1 ta n 9 .4 3 o
o
0 .1 6 1
Since f 0 .1 3 0 .1 6 1 the screw is not self-locking and overhauling
SOLUTION (15.10) n 0, e
or
ta n
c o s n 1.
1 f ta n
1 0 .1 2 ta n 1 0 .1 2 c o t
0 .7 0
;
1 f c o t
[
1 0 .1 2 ta n ta n 0 .1 2
] ta n
2 .5 ta n 0 .7 0 0
2
Solving tan 0 . 3213 or 2 . 1783 ; 17 . 81
o
or 65 . 34
o
Thus Wd
T0
m
2
f tan
[ 1 f
tan
]
50 ( 30 )
o
[ 0 . 12 tan 17 . 81o ] 145 . 3 N m 1 0 . 12 tan 17 . 8
2
SOLUTION (15.11) n 0, ta n
p
dm
cos n 1
p d m ta n ( 4 6 .8 ) ta n
,
V 0 .2 ( 6 0 ) 1 2 m m in n p
(n in rp m )
(1) (2)
From Eqs.(1) and (2): n
V p
12
( 0 . 0468 ) tan
We have the torque expressed as
and
Tu
9549 kW n
Tu
Wd 2
m
9549 ( 3 . 75 ) ( 0 . 0468 ) tan
f tan 1 f tan
12
438 . 7 tan
438 . 7 tan
Thus 438 . 7 tan 12 ( 23 . 4 )[ 10. 015. 15 tan ] tan
or
438 . 7 tan 65 . 8 tan
2
42 . 12 280 . 8 tan
or
157 . 9 tan 65 . 8 tan
2
42 . 12
or
tan
2
2 . 4 tan 0 . 64 0
(CONT.)
243
15.11 (CONT.) Solving ta n 1 , 2
2
b
b 4 ac
2a
2 .4 1 .7 9 2
Use ta n 0 .3 0 5 . Then, Eq.(1) gives p 4 6 .8 ( 0 .3 0 5 ) 4 4 .8 4 m m
SOLUTION (15.12) Table 15.1 and Fig.15.3: A t 3 5 3 m m , 2
h d d r 3 .6 8 m m , b
(a) (b) b
P A
2 h ta n 3 0
3 8
3
6 0 (1 0 )
3 5 3 (1 0
P
d m hne
6
170
)
ne
;
p 3 mm,
dm
d dr 2
2 4 2 0 .3 2 2
0 .3 7 5 2 (3 .6 8 ) ta n 3 0
o
d r d - 1 .2 2 7 p = 2 0 .3 2 m m
o
2 2 .1 6 m m
4 .6 2 m m
M Pa
P
d m h
b
3
6 0 (1 0 )
( 2 2 .1 6 ) ( 3 .6 8 ) 7 0
3 .3 4 6
th r e a d engaged
Minimum nut length is thus L n p n e (3 )(3 .3 4 6 ) 1 0 .0 4 m m e
( c ) Screw:
Nut:
3P 2 d r neb
3P 2 d neb
p
8 2
3
3 ( 6 0 1 0 )
2 ( 2 2 .1 6 )( 4 .6 2 )(1 0
6
)( 3 .3 4 6 )
3
3 ( 6 0 1 0 ) 2 ( 2 4 )( 4 .6 2 )(1 0
6
8 3 .6 M P a
7 7 .2 M P a
)( 3 .3 4 6 )
SOLUTION (15.13) d m 46 m m ,
(a) (b) b
W A
h
2
W 2
dr
W
d m h ne
;
4
4 mm, 3
4 (1 5 1 0 )
( 0 .0 4 2 )
ne
2
d r d p 50 8 42 m m ,
1 0 .8 M P a
W
b d m h
3
1 5 (1 0 ) 6
1 0 (1 0 ) ( 0 .0 4 6 )( 0 .0 0 4 )
2 .6 th re a d engaged
L n p n e 8 ( 2 .6 ) 2 0 .8
Thus, minimum length of nut:
e
( c ) Screw: Nut:
3W 2 d r neb
3W 2 d neb
b
3
3 (1 5 1 0 ) 2 ( 0 .0 4 2 ) ( 2 .6 ) ( 0 .0 0 4 ) 3
3 (1 5 1 0 ) 2 ( 0 .0 5 ) ( 2 .6 ) ( 0 .0 0 4 )
1 6 .4 M P a
1 3 .8 M P a
244
mm
p 2
8 2
4 mm
SOLUTION (15.14) kb
kp
2
d E s
0 . 58 50 0 . 5 15 0 . 58 50 2 . 5 15
kb
kb k
3.5 3 4 (1 0
4 ( 0 .0 5 )
0 . 58 ( E s 3 )( 0 . 015 ) 2 ln[ 5
C
2
( 0 .0 1 5 ) E s
4L
4 . 512 (10
3
)E s
(Eq.15.31a)
)E s
(Eq.15.34)
]
0 . 439
3 . 534 3 . 534 4 . 512
p
3
We have the preload: Fi
T 0 .2 d
72 0 . 2 ( 0 . 015 )
24 kN
Thus F b C P F i 0;
0 .4 3 9 P 2 4 ,
P 5 4 .6 7
kN
SOLUTION (15.15) At 3 5 3 m m , 2
Table 15.1:
Table 15.4: S u 5 2 0 M P a We have Pm 6 0 k N ,
C r 0 .8 7
Table 7.3: Table 15.6: K
f
Pa 2 0 k N ,
3 n 1 .4
Equation (15.39) gives S e ( 0 .8 7 )( 13 )( 0 .4 5 5 2 0 ) 6 7 .8 6
( a ) kb
2
d Es
2
( 0 .0 2 4 ) E s
0 .0 0 9 E s
4 ( 0 .0 5 )
0 .5 8 ( E s 3 )( 0 .0 2 4 )
kp
2 ln [ 5
C
4L
0 .5 8 ( 0 .0 5 ) 0 .5 ( 0 .0 2 4 ) 0 .5 8 ( 0 .0 5 ) 2 .5 ( 0 .0 2 4 )
kb
kb k p
0 .0 0 9 0 .0 0 9 0 .0 0 8 7
M Pa
0 .0 0 8 7 E s ]
0 .5 0 8
F b a C Pa 0 .5 0 8 ( 2 0 ) 1 0 .1 6
ba
kN ,
1 0 ,1 7 0 353
Therefore we obtain Su n
or
520 1 .4
bm
bm
Su Se
520 67 . 86
ba
( 28 . 8 ),
bm
150 . 74 MPa
F b m b m A t 1 5 0 .7 4 (3 5 3) 5 3 .2 1 k N
Also
F bm CP m F i ;
53 . 21 0 . 509 ( 60 ) F i
or F i 2 2 .7 k N
( b ) T K d F i 0 .2 ( 2 0 )( 2 2 .7 ) 9 0 .8 N m
245
2 8 .8 M P a
SOLUTION (15.16) Table 15.1: A t 245 mm
2
,
K
3 . 8 (Table 15.6),
f
C r 0 .8 9 (Table 7.3),
Ct 1
Using Eq.(15.39): S e ( 0 . 89 )( 1)(
1 3 .8
)( 0 . 45 750 ) 79 . 05 MPa
( a ) F bm 160 ( 245 ) 39 . 2 kN kb k
p
2
d Es 4L
2
( 0 . 02 ) E s
0 . 58 ( 50 ) 0 . 5 ( 20 )
2 ln[ 5
0 . 58 ( 50 ) 2 . 5 ( 20 )
kb
kb k p
6 .2 8 3 6 .2 8 3 6 .7 2 2
3
6 . 283 (10
4 ( 0 . 05 )
0 . 58 ( E s 3 )( 0 . 02 )
C
6 . 722 ( 10
)E s
3
)E s
]
0 .4 8 3
Hence, we obtain F b m C Pm F i 0 . 4 8 3 P m 2 5
or
39 . 2 0 . 483 Pm 25 , Pm 29 . 4 kN
and S
y
n
bm
or
Also
F ba CP a ;
ba
S
y
Se
ba ;
620 2 .2
15 . 53 MPa
and
160
620 7 9 .0 5
ba
F ba 15 . 53 ( 245 ) 3 . 8 kN
3 . 8 0 . 483 Pa ,
Pa 7 . 867
kN
Then Pmax Pm Pa 29 . 4 7 . 867 37 . 27 kN Pmin Pm Pa 29 . 4 7 . 867 21 . 53 kN
( b ) T 0 . 15 ( 20 )( 25 ) 75 N m SOLUTION (15.17) ( a ) From Eq. (15.23): Fb C P Fi ( k
kb b
4 kb
)5 4 .2 5 .2 k N
By Eq. (15.24): 4k
F P (1 C ) P F i ( 5 k b )5 4 .2 0 .2 k N b
( b ) There is a compression (of- 0 .2 k N ) in parts. Thus they do not separate under 5 k N load. SOLUTION (15.18) ( a ) Each bolt supports P 1 1 .6 2 5 .8 k N . From Eqs. (15.23) and (15.24) Fb C P Fi ( 3 k
kb b
kp
) ( 5 .8 ) F i 1 4 .5 F i 2 kb
F p (1 C ) P F i ( 3 k
b
kp
) ( 5 .8 ) F i 2 .9 F i
(1) (2) (CONT.)
246
15.18 (CONT.) Joint separates when F p 0 . Equation (2) is then F p 2 .9 F i 0 ,
F i 2 .9 k N
( b ) Using Eq. (1): F b 1 .4 5 2 .9 4 .3 5 k N Fb
From Table 15.4: S p
3 1 0 (1 0 ) 6
;
At
4 ,3 5 0 At
,
A t 1 4 .0 3 m m
2
So, by Table 15.1, select: I S O 5 0 .8 steel bolt (with tensile stress area closest to A t 1 4 .0 3 m m .) 2
SOLUTION (15.19) ( a ) kb
2
d Es 4L
3
4 ( 0 .0 3 2 )
0 .5 8 E c d
kp
2 ln [ 5
C
2
(1 8 ) ( 2 0 0 1 0 )
0 .5 8 L 0 .5 d 0 .5 8 L 2 .5 d
kb kb k
p
N m
0 .5 8 (1 6 5 1 0 )(1 8 ) 2 ln [ 5
1 . 59 1 . 59 3 . 49
9
6
]
1 .5 9 1 0
0 .5 8 ( 0 .0 3 2 ) 0 .5 ( 0 .0 1 8 ) 0 .5 8 ( 0 .0 3 2 ) 2 .5 ( 0 .0 1 8 )
3 .4 9 1 0
9
N m
]
0 . 313
At 2 1 6 m m , 2
Tables 15.1 and 15.4:
S p 600 M Pa
F i 0 .9 S p A t 0 .9 ( 6 0 0 )( 2 1 6 ) 1 1 6 .6 4 k N
Equation (15.20): Then
F b C P F i 0 .3 1 3( 860 ) 1 1 6 .6 4 1 2 0 .8 k N
b
(b)
Fb At
120800 216
5 5 9 .3 M P a
T 0 .2 F i d 0 .2 (1 1 6 .6 4 )(1 8 ) 4 1 9 .9 N m
SOLUTION (15.20) Pm Pa 1 0 k N ,
Su 830 M Pa ,
Table 15.6: K
f
3,
Equation (7.11): C t 1 0 .0 0 3 2 ( 9 0 0 8 4 0 ) 0 .8 1 Refer to Solution of Prob.15.19:
C 0 .3 1 3
At 2 1 6 m m
2
Equation (15.39): S e ( 0 .8 4 )( 0 .8 1)( 13 )( 0 .4 5 8 3 0 ) 8 4 .7 M P a Thus, Eq.(15.40) results in n
( 8 3 0 )( 2 1 6 ) 1 1 6 , 6 4 0 ( 0 .3 1 3 )(1 0 , 0 0 0 )[(
830
1 .8 5
) 1]
8 4 .7
247
C r 0 .8 4 (Table 7.3)
SOLUTION (15.21) ( a ) Compression of the parts is lost when F p 0 . Thus, from Eq. (15.24): F i (1 C ) P F P ( k
(k
4 kb b
kp kp
b
) P FP
)(3 5 ) 0 2 8 k N
4 kb
( b ) Minimum force in parts occurs when fluctuating load is maximum. Using Eq. (15.24): Fp ( k
kp b
) P Fi
kp
4 5
(3 5 ) 3 8 1 0 k N
SOLUTION (15.22) ( a ) Compression of the parts is lost when F p 0 . Therefore, by Eq. (15.24): Fi ( k
(k
kp b
)P Fp
kp
2 kb b
)(3 5 ) 0 1 7 .5 k N
3 kb
( b ) Minimum force in parts takes place when fluctuating load is maximum. From Eq. (15.24): Fp ( k
kp b
kp
)P Fp
1 2
(3 5 ) 3 8 2 0 .5 k N
SOLUTION (15.23) ( a ) From Eq. (15.24): F p (1 C ) P F i ( k
2 .5 ( 3 3 1 ) P 1 3,
kp kp
b
) P Fi
P 2 0 .6 7 k N
Fb Fi 1 3 k N
( b ) Load off: Load on: F b
1 4
( 2 0 .6 7 ) 1 3 1 8 .1 7 k N
Hence, Pm
1 3 1 8 .1 7 2
1 5 .5 9 k N
Pa
1 8 .1 7 1 3 2
2 .5 8 5 k N
SOLUTION (15.24) ( a ) From Eq. (15.24): F p (1 C ) P F i ( k 600 ( 2k
kp p
kp
kp p
) P 8000,
kb
) P Fi P 2 5, 8 0 0 N
(CONT.)
248
15.24 (CONT.) Fb Fi 8 0 0 0 N
( b ) Load off: Load on: F b
1 3
( 2 5, 8 0 0 ) 8 0 0 0 1 6 , 6 0 0 N
So, 8 0 0 0 1 6 ,6 0 0
Pm
1 2 .3 k N
2 1 6 ,6 0 0 8 0 0 0
Pa
4 .3 k N
2
SOLUTION (15.25) ( a ) By Eq. (15.24): F p (1 C ) P F i ( k
1800
4 5
P 6000,
kp p
kb
) P Fi
P 9700 N
Fb Fi 6 0 0 0 N
( b ) Load off:
Load on: F b 6 0 0 0
(9 7 5 0 ) 7 9 5 0 N
1 5
Thus Fm
6000 7950 2
Fa
7950 6000 2
6975 N 975 N
SOLUTION (15.26) Table 15.4:
d 20 mm ,
A t 245
2
mm
( a ) We have F b 0 . 9 S y A t 0 . 9 ( 630 )( 245 ) 138 . 915 and
kb
kp
AE
s
L
2
d Es
4L
2
( 20 ) E s 4 ( 60 )
0 . 58 ( E s 2 )( 20 ) 2 ln[ 5
0 . 58 ( 60 ) 0 . 5 ( 20 ) 0 . 58 ( 60 ) 2 . 5 ( 20 )
]
F i 0 . 75 F b 104 . 2 kN
kN
5 . 236 E s
9 . 379 E s
C
5 . 236 5 . 236 9 . 379
0 . 358
Thus Pb C P F i 0 .3 5 8 ( 4 0 ) 1 0 4 .2 1 1 8 .5
(b)
T KdF
i
kN
0 . 15 ( 20 )( 104 . 2 ) 312 . 6 N m
SOLUTION (15.27) Tables 15.1, 15.4, and 15.6:
A t 8 4 .3 m m ,
K
S p 380 M Pa,
S y 420
2
f
2 .2 M Pa ,
Su 520 M Pa
Table 7.3: C r 0 .8 9 Equation (15.39): S e ( 0 .8 9 )(1)( 21.2 )( 0 .4 5 5 2 0 ) 9 4 .7 M P a (CONT.)
249
15.27 (CONT.) 20 4 2
( a ) We have Pa
8 kN ,
Pm
20 4 2
12 kN
Thus a ( MPa )
m
9 4 .9 M P a ,
8 8 4 .3
12 8 4 .3
1 4 2 .3 M P a
Not safe. (Fig. S15.27)
a
(a) No preload
140
Soderberg Line
a
Se
105
Modified Goodman Line
70
ba
Figure S15.27
35
0
(b) With preload m
210 280
140
70
45o
0
S
420 490 520
350 bm
Su
m
( MPa )
y
( b ) F i 0 .7 5 (3 8 0 8 4 .3) 2 4 k N kb
2
d Es
2
( 0 .0 1 2 ) E s
4L
0 .0 0 2 3 E s
4 ( 0 .0 5 )
0 .5 8 ( E s 2 )( 0 .0 1 2 )
kp
2 ln [ 5
0 .5 8 ( 0 .0 5 ) 0 .5 ( 0 .0 1 2 ) 0 .5 8 ( 0 .0 5 ) 2 .5 ( 0 .0 1 2 )
0 .0 0 5 E s ,
C
]
0 .0 0 2 3 0 .0 0 2 3 0 .0 0 5
We have F b m C Pm F i 0 .3 1 5 (1 2 ) 2 4 2 7 .8 k N , ba
F b a 0 .3 1 5 (8 ) 2 .5 2 k N ,
2520 8 4 .3
0 .3 1 5
bm
2 7 ,8 0 0 8 4 .3
3 2 9 .8 M P a
2 9 .9 M P a
It is seen from Fig. S15.27 that the joint fails according to the Soderberg theory, while it is safe on the basis of Goodman criteria. ( c ) Use Eq.(15.37): 5 2 0 ( 8 4 .3 ) 2 4 , 0 0 0
n
520
( 0 .3 1 5 )[ 8 0 0 0 (
1 .1 3
) 1 2 ,0 0 0 ]
9 4 .7
( d ) Applying Eq.(15.28), we have ns
1 . 75
24 20 ( 1 0 . 315 )
SOLUTION (15.28) We have S
p
600
A t 115
i
mm
0 . 75 S
Pa Pm
a
m
(Table 15.5)
MPa
(Table 15.2)
0 . 75 ( 600 ) 450
p
MPa
5 kN
10 2
2
3
5 ( 10 ) 115 ( 10
6
)
43 . 48 MPa
Equation (15.39): Se CrCt (
1 K f
) ( 0 .4 5 S u )
(CONT.)
250
15.28 (CONT.) where
C r 0 .8 7
(Table 7.3)
C t 1 0 .0 0 5 8 ( 4 9 0 4 5 0 ) 0 .7 7
K
3 .8
f
(Table 15.6)
S u 830
Hence
(Table 15.5)
MPa
S e ( 0 . 87 )( 0 . 77 )(
We have
a
S u
n C
a
[(
(Eq.7.11)
1 3 .8
)( 0 . 45 830 ) 65 . 84 MPa
. With preload, Eq.(15.38) gives
m
i
Su
) 1]
Sy 830 450
( 43 . 48 )( 0 . 31 )(
2 . 07
830
1)
65 . 84
Without preload, C=1 and i 0 : n
1 . 40
830 830
1)
( 43 . 48 )(
65 . 84
Comment: The presence of preload is beneficial.
SOLUTION (15.29) Refer to Solution of Prob.15.28. We have A t 1 1 5 m m , 2
Hence,
C 0 .3 1,
S p 600 M Pa.
F i 0 . 75 ( A t S p ) 0 . 75 (115 600 ) 51 . 75 kN
Apply Eq.(15.27), S p At Fi
P
Cn
600 ( 115 ) 51 . 75 10 ( 0 . 31 )( 2 )
3
27 . 82 kN
Using Eq.(15.28b): P
Fi n s (1 C )
37 . 5 kN
51 . 75 2 ( 1 0 . 31 )
Comment: Failure owing to separation will not take place before bolt failure. SOLUTION (15.30) ( a ) We have 380
MPa
At 1 5 7
mm
Pa Pm
12 2
S
p
S u 520 2
MPa
Se 100 M Pa
(Table 15.1),
6
(Table 15.4)
k N b o lt
F i 0 .9 S p A t 0 .9 (3 8 0 )(1 5 7 ) 5 3 .6 9 k N
a
i
0 .9 S
m
6 1 0
1 5 7 (1 0
p
3 6
)
3 8 .2 2
M Pa
0 . 9 ( 380 ) 342
MPa
(CONT.)
251
15.30 (CONT.) kb
C
We have
S u C
a
m
. Eq.(15.38), with preload:
i
Su
[(
0 . 167
kb 5kb
a
n
kb
kb kb
) 1]
Sy 520 342
( 0 . 167 )( 38 . 22 )(
520
4 .5 1)
100
Without preload, C=1 and i 0 : n
2 . 19
520 520
1)
( 38 . 22 )(
100
( b ) Equation (15.28b): Fi
ns
P (1 C )
5 . 37
53 . 69 12 ( 1 0 . 167 )
Comments: The presence of preload is very beneficial. Joint will separate before bolts fail.
SOLUTION (15.31) Repeating section L P
P
We have L=w=50 mm and total number of rivets in the joint n=2. Therefore
P n (d
2
4)
b
P ndt
t
P ( w d e )t
3
4 ( 32 10 )
2 ( )( 0 . 019 ) 3
32 ( 10 ) 2 ( 0 . 019 )( 0 . 01 )
56 . 43 MPa
2
84 . 21 MPa 3
32 ( 10 )
[ 50 ( 19 3 )] 10 ( 10
6
)
114 . 3 MPa
SOLUTION (15.32) Sketch is the same as that given in Solution of Prob.15.31 and n=2. The allowable loads are: F s a ll n d Fb
b , a ll
Ft
t , a ll
2
4 1 0 5 ( 2 )( )(1 8 )
4 5 3 .4 4 k N
n d t 3 3 0 ( 2 )(1 8 )(1 0 ) 1 1 8 .8
kN
( w d e ) t 1 5 0[ 6 0 (1 8 1 .5 )]1 0 6 0 .7 5
Thus e
2
6 0 .7 5 (1 5 0 )( 6 0 )(1 0 )
6 7 .5 %
252
kN
SOLUTION (15.33) For members, by Table B.3: S y 4 6 0 M P a ,
S y s 0 .5 7 7 ( 4 6 0 ) 2 6 5 .4
M Pa
For bolts, from Table 15.4: S y 9 4 0 M P a ,
S y s 0 .5 7 7 (9 4 0 ) 5 4 2 .4
M Pa
from Table 15.1: d r 1 8 1 .2 2 7 ( 2 .5 ) 1 4 .9 3 m m (1 4 .9 3 )
As 2
Shear on bolts:
As S ys
Fs
3 5 0 .1 4 ( 5 4 2 .4 )
n
2
Ab S y
360 (940 )
n
3 5 0 .1 4 m m
2
95 kN
A b 2 (1 8 )(1 0 ) 3 6 0 m m
Bearing on bolts: Fb
2
4
2
1 1 2 .8 k N
3
Bearing on members: 360 ( 460 )
Fb
6 6 .2
2 .5
A t ( 6 0 1 8 )1 0 4 2 0 m m
Tension in members: 420 ( 460 )
Ft
kN 2
5 5 .2 k N Pa ll
3 .5
SOLUTION (15.34) For members, using Table B.3:
S
210
y
MPa
For bolts, from Table 15.1: d r 1 4 1 .2 2 7 ( 2 ) 1 1 .5 5 m m
2 (1 1 .5 5 )
As
Shear of bolt:
P As
2 0 ,0 0 0 2 0 9 .5
2 0 9 .5 m m
119 M Pa,
2 0 ,0 0 0 168
2
9 5 .4 7 M P a ,
n
A b 2 ( 6 )( 14 ) 168
Bearing on bolt: b
2
4
mm
n
640 119
3 .8 8
370 9 5 .4 7
2
5 .3 8
Bearing on members: n
1 . 76
210 119
A t ( 60 2 14 ) 6 192
Tension on members:
t
20 , 000 192
104 . 2 MPa ,
n
210 104 . 2
mm
2
2 . 02
SOLUTION (15.35) Table 15.1: A t 8 4 .3 m m ,
d r 1 2 1 .2 2 7 (1 .7 5 ) 9 .8 5 3 m m
2
Hence, A s
4
(9 .8 5 3 ) 7 6 .2 5 m m 2
2
P 4 As
P 4 ( 7 6 .2 5 )
P 305
Pivots about point A. Row 1 is the highest loaded. Apply Eq.(15.43) with j=2 and e=250 mm: F1
Thus
t ,m ax
M r1 2
2
r1 r2
2
t
250 P ( 225 ) 2
2
2 ( 225 ) 2 (75 )
(
2
t
) 2
t
0 .5 P , 2
P 3 3 7 .2
0 .5 P 8 4 .3
( 3 3P7 .2 ) ( 3 P0 5 ) 2
( 0 .0 0 3 0 .0 0 4 4 ) P 0 .0 0 7 4 P
Hence, 0 .0 0 7 4 P 1 4 0; or
m a x 0 .0 0 4 4 P 8 4 :
Pa ll 1 8 .9 2 k N P 1 9 .0 9
253
P 1 6 8 .6
kN
2
SOLUTION (15.36) The maximum design load is Pm a x n P ( 2 .3)( 2 5 ) 5 7 .5 k N . From geometry: F 2 ( 4 1 5 ) F1 . Here F1 and F 2 are tensile forces in bolts 1 and 2, respectively. We have
M
0:
A
1 0 0 0 (5 7 .5 ) 1 0 0 ( 145 F1 ) 3 7 5 F1 ;
F1 1 4 3 .2 k N
The required tensile stress area is then At
143,200 600
239 m m
2
From Table 15.1: The required thread size is about M 2 0 2 .5 C . SOLUTION (15.37) ( a ) Using Table 15.4, S p 3 1 0 M P a ,
S y 340 M Pa
S y s 0 .5 7 7 S y 1 9 6 .2 M P a
The tensile stress area: ( F o r c e )( n )
At
S
2 ,5 0 0 ( 3 ) 310
p
2 4 .2 m m
2
From Table 15.1, select M 7 1 C thread with A t 2 8 .9 m m
2
( b ) Table 15.1 and Fig. 15.3: p 1 mm, h
1 2
d 7 mm,
d r d 1 .2 2 7 p 5 .7 7 m m ,
( d d r ) 0 .6 1 5 m m
d m d 0 .6 5 p 6 .3 5 m m ,
b
p 8
2 h ta n 3 0
o
0 .8 3 5 m m
Apply Eq. (15.19a): 3K tPp 2 dbLn
S ys
3 ( 4 )( 2 5 0 0 )(1 )
n
2 ( 7 )( 0 .8 3 5 ) L n
1 9 6 .2 3
Solving, L n 1 2 .5 m m
SOLUTION (15.38)
(a)
Refer to Solution of Prob. 15.37. Using Table 15.4: S p 310 M Pa,
S y 340 M Pa ,
F o rce ( n )
6 4 .5 2 m m
At
Sp
4000 (5 ) 310
S y s 0 .5 7 7 S y 1 9 6 .2 M P a
2
From Table 15.1: Select M 1 2 1 .7 5 C thread with A t 8 4 .3 m m (b)
2
Table 15.1 and Fig. 15.3: d 12 m m ,
p 1 .7 5 m m
d m d 0 .6 5 p 1 0 .8 6 m m d r d 1 .2 2 7 p 9 .8 5 m m h
1 2
( d d r ) 1 .0 7 5 m m ,
b 1 .7 5 8 2 h ta n 3 0
o
1 .4 6 m m
(CONT.)
254
15.38 (CONT.) Using Eq. (15.19a): 3 ( 4 ) ( 4 0 0 0 ) (1 .7 5 )
2 (1 2 ) (1 .4 6 ) L n
1 9 6 .2 5
from which L n 1 9 .4 m m
SOLUTION (15.39) Table 15.1:
At 2 4 5 m m ,
Shear area:
As
d r 2 0 1 .2 2 7 ( 2 .5 ) 1 6 .9 3 m m
2
P 3
As
(1 6 .9 3 ) 2 2 5 2
4
2
1 , 481 P
P 3 ( 225 ) 10
mm
6
Pivots about point A. Bolt 1 is the highest loaded. M r1
F1
r1 r2 r3
t
2 4 5 (1 0
2
2
)
From Mohr’s circle:
2
2
140 90 40
2
0 .5 9 7 P
2 ,4 3 7 P
0 .5 9 7 P 6
1 2 5 P (1 4 0 )
2
t , max
t
2
(
t
2
) 2
2
2 , 437 P 2
(
2 , 437 P 2
) ( 1 , 481 P ) 2
2
( 1 , 219 1 , 918 ) P 3 ,137 P
Hence 3,1 3 7 P 1 4 5 1 0 ,
P 4 6 .2 2
6
or
1, 9 1 8 P 8 0 1 0 ,
kN
Pa ll 4 1 .7 1 k N
6
SOLUTION (15.40) Vertical and horizontal components of the force P=10 kN are 8 kN and 6 kN at B, respectively. Rivet B is most heavily loaded. The centroid of the group of rivets is at C. We have FB
M rB
8 3 0 (1 5 0 )
2
rj
2
2
2
2[ 3 0 9 0 1 5 0 ]
0 .5 7 1 k N
6/6=1 kN B
1
V B [1 1 . 904 2
2
] 2 2 . 15 kN
(Fig.a) 8/6+0.571 =1.904 kN
Therefore B
B
VB
d
2
VB
4
dt
2 ,1 5 0
( 20 )
2 ,1 5 0 2 0 1 5
2
4
6 .8 4 4
7 .1 6 7
M Pa
VB
M Pa
Figure S15.40
SOLUTION (15.41) Rivet A is the most heavily loaded. As
Vd
P 5
4
(15 )
50 5
2
176 . 715
10
kN
mm
2
V a ll 1 0 0 (1 7 6 .7 1 5 ) 1 7 .6 7 2
kN
Centroid C of the line AB, with respect to A, is determined from: 5 0 x 7 0 (1 0 ) 2 5 0 (1 0 ) 3 2 0 (1 0 ) 3 9 0 (1 0 ) ,
x
A
x 206
mm
C d
e
B (CONT.)
P
255
15.41 (CONT.) Thus Mr
FA
A 2
rj
50 , 000 e ( 206 ) 206
2
2
136
2
44 114
2
2
184
93 . 875 e
V A 1 0 , 0 0 0 9 3 .8 7 5 e 1 7 , 6 7 2 ,
e 8 1 .7 m m
and d 54 . 3 mm
SOLUTION (15.42) The A is the highest loaded point. A s (15 )
4 176 . 715
2
Thus
FA
A 2
rj
46 P ( 206 ) 206
2
136
2
2
x 206
Refer to Solution of Prob. 15.41: Per
mm
2
44 114
2
2
184
F
,
mm,
P 5
0 .2 P
e 206 70 90 46 mm
0 . 086 P
Hence V A ( 0 .2 0 .0 8 6 ) P 0 .2 8 6 P and
A
100 10 ;
P 6 1 .7 9 k N
6
0 .2 8 6 P 1 7 6 .7 1 5 (1 0
6
)
SOLUTION (15.43) A 0 . 707 hL 0 . 707 ( 7 )( 60 ) 297 P
S
ys
A
n
200 ( 297 ) 2 .5
mm
2
23 . 76 kN
Cross-sectional area of one plate A p 40 (10 ) 400
mm
2
. Then
P ( S y n ) A p ( 2 5 0 2 .5 ) 4 0 0 4 0 k N
Hence the capacity of the plate significantly exceeds that of the weld.
SOLUTION (15.44) Table B.4:
for plates,
Table 15.8:
for weld,
S y 427 M Pa . S y 345 M Pa .
Since 3 4 5 4 2 7 , the weld should yield first. The maximum load that can be applied equals P
SyA n
3 4 5 (1 5 8 7 .5 ) 4
1 1 3 .2 k N
SOLUTION (15.45) Refer to Solution of Prob. 15.44. S y 0 .5 S y 0 .5 ( 4 2 7 ) 2 1 3 .5 M P a From Eq. (15.44):
(for plate)
S y s 0 .5 S y 0 .5 ( 3 4 5 ) 1 7 2 .5 M P a
(for weld) (CONT.)
256
15.45 (CONT.) Since 1 7 2 .5 2 1 3 .5
The weld should yield first.
Hence, S ys A
P
1 7 2 .5 (1 5 8 7 .5 )
n
3
7 5 .4 7 k N
SOLUTION (15.46) For plates: S y 4 6 0 M P a
(Table B.3), S y s 0 .5 S y 2 3 0 M P a
For weld: S y 3 7 9 M P a
(Table 15.8), S y s 0 .5 S y 1 8 9 .5 M P a
(Eq. 15.44) (Eq. 15.44)
Since 1 8 9 .5 2 3 0
weld would yield first.
Thus, Fig. 15.27a: S ys
P
n
1 8 9 .5 ( 0 .7 0 7 )( 8 )( 2 7 0 )
4 2 .8 7 k N
3 .5
SOLUTION (15.47) For plates: S y 5 3 0 M P a
(Table B.3),
S y s 0 .5 S y 2 7 5 M P a
For weld: S y 4 1 4 M P a
(Table 15.8),
S y s 0 .5 S y 2 1 2 M P a
(Eq. 15.44) (Eq. 15.44)
Since 212 275
weld would yield first.
Thus, Fig. 15.26a: S ys
P
n
2 1 2 ( 0 .7 0 7 )( 5 )( 2 6 0 ) 4
2 2 .5 k N
SOLUTION (15.48) A p 75 10 750
mm
2
P A
,
all
750 (140 ) 105
kN
R 1 21 kN
(along AB)
M
D
0:
R 1 ( 75 ) 105 (15 ) 0 ;
M
A
0:
R 2 ( 75 ) 105 ( 75 15 ) 0 ;
Check: Hence
R 1 R 2 105 L1
R1 Sw
21 1 .2
R 2 84 kN
(along DE)
kN
17 . 5 mm ,
L2
R2 Sw
84 1 .2
70 mm
SOLUTION (15.49) I x 2 (1 0 0 t )( 6 2 .5 ) 2
3
2 (1 2 5 ) t 12
1 .1 0 7 (1 0 ) t
A 2 (1 0 0 t 1 2 5 t ) 4 5 0 t m m
We have
6
1 2
[( 1455000t0 ) ( 3 .71 .15 0672t .5 ) ]
Thus
a ll ;
and
h
0 .1 3 4 0 .7 0 7
2
2 1 4 .3 t
5 5,
a ll 5 5 M P a
M 1 5 ( 2 5 0 ) 3 .7 5 M N m m
2
2
4
mm ,
2 1 4 .3 t
t 3 .9 m m
5 .5 2 m m
257
SOLUTION (15.50) I x 1 .1 0 7 (1 0 ) t m m , 6
From Solution of Prob.15.49: We have Pm 1 5 k N ,
Pa 5 k N ,
K
Refer to Example 15.12. We obtain C
f
A 450t m m
4
1.5
(Table 15.9)
A Su 272(420) b
f
2
0 .9 9 5
0 .6 6 7 .
Equation (7.5a): S u s 0 .6 7 ( S u ) 0 .6 7 ( 4 2 0 ) 2 8 1 .4 M P a S e C s C f (1 K f ) S e ( 0 .7 )( 0 .6 6 7 )( 11.5 )( 0 .5 4 2 0 ) 6 5 .3 7 M P a '
Then, from Eq.(7.21): S us n
m (
a
)
m
Su
2 8 1 .4 2 .5
1
3 5 .8 3 M P a
1
420 ) 1 3 6 5 .3 7
(
Se
Also, from Solution of Prob.15.49: m Thus
2 1 4 .3 t
3 5 .8 3
and
h
5 .9 8 0 .7 0 7
1 . 072 t
t 5 .9 8 m m
or
8 .4 6 m m
SOLUTION (15.51) Table 15.8: S y 4 1 4 M P a
S y s 0 .5 S y 2 0 7 M P a
We have t 0 . 707 (12 ) 8 . 484
mm
A one weld 8 . 484 L
T 100 ( 60 ) 6 MN mm
Total
3
J 2
2
tL 12
8 .4 8 4 L 12
3
1.4 1 4 L
3
At point A: A S ys
and
P A
n
207 2 .5
Tr J
100 ,000
1 6 .9 6 8 L
82 . 8
6
6 (1 0 )( L 2 ) 1. 4 1 4 L
3
5 ,893 L
2 ,1 2 2 , 0 0 0
L
(1)
2
(2)
A
From Eqs.(1) and (2): 82 . 8 L 5 ,893 L 2 ,122 , 000 0 ; 2
L 71 . 17 L 25 , 628 0 2
Solving L
1 2
[ 7 1 .1 7
7 1 .1 7 4 ( 2 5 , 6 2 8 ) ] 1 9 9 .6 2
mm
SOLUTION (15.52) Pm 100
Pa 20 kN
kN ,
Table 15.9: K
f
1.5
Refer to Solution of Prob.15.51: m
Also
5 ,893
L
S u 496
C
f
S
ys
AS
2 ,1 2 2 , 0 0 0 L
MPa , b u
(1)
2
S
y
272 ( 496 )
0 . 5 ( 414 ) 207
414 0 . 995
MPa
MPa
(Table 15.8),
0 . 566
Refer to Example 15.12:
S e (0 .7 )(0 .5 6 6 )( 11.5 )(0 .5 4 9 6 ) 6 5 .5 1 M P a
(CONT.)
258
15.52 (CONT.) We have, from Eq.(7.21) with S y s and S y replacing S u s and S u : S ys n
m (
a m
)
Sy
2 0 7 2 .5
1
(
Se
3 6 .5 7 M P a
1
414 ) 1 5 6 5 .5 1
(2)
Equations (1) and (2) are therefore L 161 . 1 L 58 , 025 . 7 0 2
L
or
1 2
[1 6 1 .1
1 6 1 .1 4 (5 8 , 0 2 5 .7 ) ] 3 3 4 .5 2
mm
SOLUTION (15.53) S u s 0 .6 7 S u 2 8 6 .1 M P a ;
Table 15.8: S u 4 2 7 M P a; We have C
f
1 .5
A b o th 2[ 0 .7 0 7 ( 6 .5 ) 2 5 0 ] 2 2 9 7 .8
mm
A Su 272(427) b
f
0 .9 9 5
K
(Table 15.9)
0 .6 5 7
Refer to Example 15.12: S e ( 0 .7 )( 0 .6 5 7 )( 11.5 )( 0 .5 4 2 7 ) 6 5 .5 M P a
We have Pm Pa 3
J 2
tL 12
Pm a x , AL 12
2
2 , 2 9 7 .8 ( 2 5 0 )
2
12
1 2 (1 0 ) 6
We have Pm 0 .5 Pm a x ,
At point A: m
and
1 2
Pm
A
Tm r J
Pm a x
4
T m 7 5 Pm 3 7 .5 Pm a x
3 7 .5 Pm a x (1 2 5 )
2 ( 2 2 9 7 .8 )
mm
2
6
1 2 (1 0 )
0 .0 0 0 6 Pm a x
Also, from Eq.(7.21): S us n
m (
a m
)
Su
2 8 6 .1 2
1
(1 )
427
1 9 .0 2 M P a
1
6 5 .5
Se
Thus 0 .0 0 0 6 Pm a x 1 9 .0 2 ,
Pm a x 3 1 .7 k N
SOLUTION (15.54) 150 mm
y A
B 8
200 mm
C
6
x
T
30 ( 10 A
D
E
40 ( 10 A
3
)
3
)
T ( 200 )
Figure S15.54
J
T ( 75 ) J
Inspection of Fig. S15.54 shows that point E has the highest stress. We write T 4 0 (1 7 5 ) 3 0 (1 0 0 ) 4 k N m
A b o th 2 (1 5 0 t ) 3 0 0 t m m
2
(CONT.)
259
15.54 (CONT.) t (1 5 0 )
I x 2[
3
1 5 0 t (1 0 0 ) ] 2 ( 0 .2 8 1 t 1 .5 t )(1 0 ) 3 .5 6 2 (1 0 ) t m m 2
12
6
I y 2[ 0 1 5 0 t ( 7 5 ) ] 1 .6 8 7 (1 0 ) t m m , 2
6
6
J 5 .2 4 9 (1 0 ) t m m
4
6
4
4
We have 3
v
and
4 0 (1 0 ) 300 t
4 (75 )
6
5 .2 4 9 (1 0 ) t 1 2
E [ v h ] 2
2
1 3 3 .3 t
1 6 6 .6 t
h
,
3
3 0 (1 0 ) 300 t
4 (1 0 0 ) 6
5 .2 4 9 (1 0 ) t
100 t
M Pa
Therefore, by Eq.(15.44), n E 0 .5 S y ;
3(
1 6 6 ,6 t
) 0 .5 (3 5 0 ),
t 2 .8 6 m m
and h
4 .0 5 m m
2 .8 6 0 .7 0 7
SOLUTION (15.55) J 5 .2 4 9 (1 0 ) t m m , 6
From Solution of Prob.15.54: From Table 15.9: C
K
2 .7 ;
f
A Su 272(427) b
f
A b o th 3 0 0 t m m
4
2
S u s 0 .6 7 S u 0 .6 7 ( 4 2 7 ) 2 8 6 M P a
0 .9 9 5
0 .6 5 7
Refer Example 15.12: S e ( 0 .7 )( 0 .6 5 7 )( 21.7 )( 0 .5 4 2 7 ) 3 6 .4 M P a We have Pm 3 0 k N ,
Pa 2 0
a m 2 3
kN ,
Thus, by Eq.(7.21): S us n
m (
a m
)
Su
2 8 6 1 .5
1
(
Se
2
427 ) 1 3 3 6 .4
2 1 .6 M P a
(1)
At point E: Use the method of Solution of Prob.15.54, with P Pm 3 0 k N Then Pm v 2 4 k N
Pm h 1 8 k N
T m 2 4 (1 7 5 ) 1 8 (1 0 0 ) 2 .4
kN m
Therefore
and
3
vm
2 4 (1 0 )
hm
1 8 (1 0 )
300 t 3
300 t
2 .4 ( 7 5 )
6
5 .2 4 9 (1 0 ) t 2 .4 (1 0 0 )
6
5 .2 4 9 (1 0 ) t 1 2
m ( v m h m ) 2
2
80 t
60 t
100 t
(2)
Equations (1) and (2) give 100 t
2 1 .6 ,
t 4 .6 m m
and h
4 .6 0 .7 0 7
6 .5 1 m m
End of Chapter 15
260
CHAPTER 16
MISSELLANEOUS MACHINE COMPONENTS
SOLUTION (16.1) Equation (16.17) and Eq.(16.16a) at r=a: 1 , max p i
(a) 1
Sy;
2
2
2
2
2
a b b a
5 4
pi
2
p i ( p i ) 260 ,
5 4
pi
r , max
p i 115 . 6 MPa
( b ) 1 1 2 2 S y 2
2
2
1
p i [( 54 )
( 54 )( 1 ) ( 1 ) ] 2 260 ,
2
p i 133 . 2 MPa
2
SOLUTION (16.2) T
b
3
2
( 100 ) b
3
157 . 08 b
2
3
Also
T 2 b fp l 2 b fp ( 3b ) 6 fb p
and
6 fb p 157 . 08 b ,
2
2
3
Hence
p
3
h
We have E h E s E ,
3
2 bpc
2
2
2
2 bp ( 4 b )
2
E (c b )
2
0 . 706 (10
8 ( 55 . 56 ) b 3
3 ( 210 10 )
(Eq.16.27b)
55 . 56 MPa
, and a=0. Equation (16.25) becomes
s
8 pb
2
E (4b b )
157 . 08 6 ( 0 . 15 )
3E 3
)b
SOLUTION (16.3) ( a ) From Table B.1: S y 2 5 0 M P a ,
0 .3 .
E 200 G Pa,
Applying Eq. (16.17), p i ,m ax
2
2
2
2
b a b a
250
2
2
2
2
0 .1 8 0 .1 2 0 .1 8 0 .1 2
9 6 .1 5 M P a
Equation (16.16c) at r a : u m ax
2
Pi a
2
( b2 a2 ) b a
E
6
9 6 .1 5 (1 0 ) 9
2 0 0 (1 0 )
2
2
( 0 .1 2 )( 0 .1 8 2 0 .1 2 2 0 .3 ) 0 .1 8 0 .1 2
0 .1 6 7 m m
( b ) Equation (16.19): 2
Po , m a x
b a 2b
2
2
250
2
0 .1 8 0 .1 2 2
2 ( 0 .1 8 )
2
6 9 .4 4 M P a
SOLUTION (16.4) ,m ax p i
(a) Su and
5 3
pi ,
2
2
2
2
b a b a
pi
Su pi ,
pi 1,
5 3
3 5
r ,m ax p i 2
( 350 ) 210 p i 350
MPa
(governs)
MPa
(CONT.)
261
16.4 (CONT.) (b)
1 Su
2
S uc
5 pi
1;
pi
3( 350 )
1
650
p i 158 . 7 MPa
or
SOLUTION (16.5) ( a ) Use Eq.(16.17) with p i p , ,m ax p
2
2
2
2
c b c b
a b,
b c:
6 2 .5 p 1 0 0 1 0
9
[
6 0 1 0 p
2
2
c b
By Eq.(16.25), with a 0 , 0 .0 5
2
c b
60 M Pa,
6
2
(Fig. 16.6b) 6
6 0 (1 0 )
0 . 05 mm :
b 62 . 5 ,
0 .3 ]
6 2 .5 p 2 0 0 1 0
(1)
p
[1 0 .3 ]
9
or 50 10
6
37 . 5 10
6
0 . 188 p 0 . 219 p ,
p 30 . 71 MPa
( b ) Equation (1) becomes 2
2
2
2
c 0 .0 6 2 5
60 3 0 .7 1
c 0 .0 6 2 5
c 110 m m
,
2c 220 m m
SOLUTION (16.6) ( a ) a=0, b=12.5 mm, c=50 mm F
T b
12 kN
150 0 . 0125
Thus
1 2 ( 1 0 ) 2 b fp l 2 ( 0 .0 1 2 5 )( 0 .1 5 )( 0 .0 5 ) p
or
p 20 . 37 MPa
3
(Eq.16.27a)
Equation (16.25) gives then
2
3
1 2 .5 ( 2 0 .3 7 )
5 0 1 2 .5
1 0 0 (1 0 )
( b ) ,m ax p
2
[ 5 0 2 1 2 .5 2 0 .3 ]
1 2 .5 ( 2 0 .3 7 )
2
2
2
2
c b c b
2 0 0 1 0
2
3
(1 0 .3 ) ( 3 .6 5 0 .8 9 1)1 0
3
0 .0 0 5
mm
2
2 0 .3 7[ 5 0 2 1 2 .5 2 ] 2 3 .0 9
M Pa
5 0 1 2 .5
SOLUTION (16.7) We have a=15 mm, b=25 mm, c=50 mm. Equation (16.25): 0 . 025
2
25 p 210 10
9
2
[ 50 2 25 2 0 . 3 ] 50 25
2
25 p 105 10
9
2
[ 25 2 15 2 0 . 3 ], 25 15
Steel, Eq.(16.17): ,m ax p
2
2
2
2
c b c b
3 7 .3 9
2
2
2
2
50 25 50 25
6 2 .3 2
M Pa
Bronze, Eq.(16.19): ,m ax 2 p
b 2
2
b a
2
2 (3 7 .3 9 )
25 2
2
2 5 1 5
262
2
1 1 6 .8
M Pa
p 37 . 39 MPa
SOLUTION (16.8) Equation (16.25) with a=0, b=50 mm, c=150 mm.
bp Ec
2
2
b c
bp
[ c2 b2 c ] 2
150 50
50 p
0 .0 3
12010
9
Es
[1 s ]
2
50 p
[ 1 5 0 2 5 0 2 0 .2 5 ]
21010
9
( 0 .7 )
Solving p 3 7 .8 9 M P a Shaft: r p 3 7 .8 9 M P a Cylinder: ,m ax p
2
2
2
2
c b c b
2
150 50
p 3 7 .8 9
r ,m ax
2
3 7 .8 9[ 1 5 0 2 5 0 2 ] 4 7 .3 6
M Pa
M Pa
SOLUTION (16.9) Equation (16.23):
(u d ) rb
2
bp E
2
b c
2
bp
[ c2 b2 ]
E
( 3 0 .3 ) 1.9 6 7 5
bp E
or p
E 3 .9 3 4 b
Then, Eq.(16.24) with a=0, us
bp E
(1 )
0 .7 3 .9 3 4
0 .1 7 8
Therefore, d s 0 .3 5 6
SOLUTION (16.10) Equation (16.31a) with ( r ) p 0 and b c :
r
3 8
(a b 2
3 0 .3 4 8
2
2
a b r
2
2
r ) 2
[( 0 .0 2 ) ( 0 .0 4 ) 2
2
2
p 2
( 0 .0 2 ) ( 0 .0 4 ) ( 0 .0 3 )
2
2
]8 5 0 0
2
9 0 (1 0 ) 6
Solving, 8 , 0 7 5 .5 tp s 4 8 , 4 5 3 r p m
SOLUTION (16.11) m ax
3600 60
2 1 2 0 ra d s ,
m in 1 1 4
ra d s ,
p 0
( a ) Equation (16.32), with p 0 ; ,m ax
4
2
[ (1 ) a
2
( 3 )b ] 2
(CONT.)
263
16.11 (CONT.) Let a=b and b=c=4b: 75 10
7 , 800 ( 120 )
6
2
3 .3( 4 b ) ]
2
[ 0 .7 b
4
2
Solving, b 7 1 .1 2 m m ,
c 2 8 4 .4 m m
( b ) Equation (16.37) of Sec.16.5: I
l
7 .8 ( 5 0 )
(c b ) 4
2
4
( 0 .2 8 4 4 0 .0 7 1 1 2 ) 3 .9 9 2 1 N m s 4
2
4
2
Thus Ek
I ( m a x m in ) 2
1 2
2
(3 .9 9 2 1)(1 2 0 1 1 4 ) 2
1 2
2
2
2 7 .6 5 9
kN m
SOLUTION (16.12) 2 4 0 0 ( 2 6 0 ) 2 5 1 .3 ra d s
( a ) Equation (16.35) with a=b and b=c:
bp E
2
b
2
[ b2 c 2 1] c b
20 ( 10
20 E
3
2
4E
2
[ 0 .7 b
2
[ 100 2 20 2 1 ]
)p
E
2
20 ( 7 . 8 )
20
100
3 .3 c ]
2
[ 2 . 083 p 64 . 896 ]( 10 2
For p 3 . 2 MPa 1 .0 2 5 1 0
[ 0 . 7 ( 0 . 02 ) 3
2
3 .3( 0 .1) ] 2
(1)
)
2 5 1 .3 ra d s , Eq.(1) gives.
and 3
2
4E
mm
We have at 0: 3
1 .0 2 5 1 0
20 2 1 0 1 0
p 5 .1 6 7 M P a
( 2 .0 8 3 p ) ,
9
( b ) Thus p
2
2
2
2
b c c b
5 .1 6 7 (1 .0 8 3 ) 5 .5 9 6 M P a
SOLUTION (16.13) ( a ) Equation (16.26) with a=0: p
2
E b
c b
2
2
2c
200 ( 10
9
2
)( 0 . 02 ) 300
50
50
2 ( 300 )
2
2
38 . 89 MPa
Then
, max
p
2
2
2
2
c b c b
38 . 89
300 300
2
50
2
2
50
2
41 . 11 MPa
( b ) Equation (16.35) with b=0.05 m, c=0.3 m, p=0, 7 .8 k N m , and E 2 0 0 G P a : 0 . 02 ( 10
3
)
b 4E
2
[ 0 .7 b
2
3 . 3 c ] 145 . 64 10 2
Solving, 3 7 0 .6
ra d s . Thus
n 3 7 0 .6 ( 6 0 2 ) 3 5 3 9 rp m
264
2
12
SOLUTION (16.14) ( a ) From Eq. (16.17), we have p ,m ax
2
2
2
2
c b b c
30
2
2
2
2
( 0 .1 2 ) ( 0 .0 6 ) ( 0 .0 6 ) ( 0 .1 2 )
18 M Pa
( b ) Using Eq. (16.27a): F 2 b p fl 2 ( 0 .0 6 )(1 8 1 0 )( 0 .1 8 )( 0 .2 ) 6
2 4 4 .3 k N
( c ) By Eq. (16.27b): T F b 2 4 4 .3 ( 0 .0 6 ) 1 4 .6 6 k N m
SOLUTION (16.15) 4 I
2
m a x 2 4 0 0 ( 2 6 0 ) 2 5 1 .3 ra d s
(b a )l 4
4
2
m in 1 2 5 .7 ra d s
[ 0 .2 0 .0 5 ]( 6 0 )( 7 .8 ) 1 .1 7 2 N m s 4
4
2
Therefore T
I ( m a x m in ) 2
1 2
2
or T
2
2
1 .1 7 2 ( 2 5 1 .3 1 2 5 .7 ) 2 (4 )
2 .2 0 8 k N m
SOLUTION (16.16) (a) I
2
(b a )l
m ax
4
4
3000 60
[ 0 .2 5 0 .0 2 5 ]( 7 .8 )( 6 0 ) 2 .8 7 1 N m s 4
2
4
( 2 ) 1 0 0 ,
2
m in 0 .9 (1 0 0 ) 9 0 ra d s
Equation (16.32) with p=0: , max
( b ) E k
1 2
4
2
[( 1 ) a
7 ,8 0 0 (1 0 0 )
2
2
(3 )b ] 2
[ 0 .7 ( 0 .0 2 5 ) 3 .3 ( 0 .2 5 ) ] 3 9 .7 8 M P a 2
4
I ( max min ) 2
2
1 2
2
( 2 . 871 )( 100
2
90
2
)
26 . 919
2
kN m
SOLUTION (16.17)
Dimensions are in millimeters.
A2
25 50
B
A
25
A1
r A ri 2 1 5 m m , A 12 , 500 r
150
50
A1 r1 2 A 2 r2 A1 2 A 2
mm
rB 4 1 5 m m
2
( 5 0 1 0 0 )( 2 1 5 2 5 ) 2 ( 2 5 1 5 0 )( 4 1 5 7 5 ) 5 0 1 0 0 2 ( 2 5 1 5 0 )
300 m m
r
(CONT.)
265
16.17 (CONT.) Equation (16.50): R
A
12, 500
dA r
265 dr r
100 215
2 8 8 .4 3 9 6 m m
415
25 265
dr r
e r R 3 0 0 2 8 8 .4 3 9 6 1 1 .6 0 4 m m
Equations (16.55a): ( ) A
100
P A
[1
P 1 2 ,5 0 0
r ( R rA ) e rA
[1
]
3 0 0 ( 2 8 8 .4 3 9 6 2 1 5 ) (1 1 .5 6 0 4 )( 2 1 5 )
]
or P 1 2 6 .7 k N
( ) B
100
P A
[1
P 1 2 ,5 0 0
r ( R rB )
[1
e rB
]
3 0 0 ( 2 8 8 .4 3 9 6 4 1 5 ) (1 1 .5 6 0 4 )( 4 1 5 )
]
or P 1 8 0 .8 k N
SOLUTION (16.18) ro ri h 5 0 4 0 9 0 m m A b h ( 2 0 )( 4 0 ) 8 0 0 m m
2
We have h
R
ln
ro
40
ln
1
6 8 .0 5 1 9
50
ri
r
90
( ri r ) 7 0 m m
2
e r R 1 .9 4 8 1 m m
( a ) Using Eq.(16.52): i
(b)
7 0 0 ( 6 8 .5 1 9 5 0 ) 8 0 0 (1 0
6
7 0 0 ( 6 8 .0 5 1 9 9 0 )
o
o
i
8 0 0 (1 0
6
Mc I
1 6 2 .2 M P a
) (1 .9 4 8 1) ( 0 .0 5 )
1 0 9 .5 M P a
) (1 .9 4 8 1) ( 0 .0 9 )
7 0 0 ( 0 .0 2 ) 1
( 0 .0 2 ) ( 0 .0 4 )
1 3 1 .3 M P a 3
12
266
SOLUTION (16.19) h
ri r
2 0 0 3 2 .5 1 6 7 .5 m m
2 h
ro r
2 0 0 3 2 .5 2 3 2 .5 m m
2
A b h ( 4 5 )( 6 5 ) 2 9 2 5 m m
2
Therefore h
R
ro
ln
65
ln
2 3 2 .5
1 9 8 .2 2 7 m m
1 6 7 .5
ri
e r R 2 0 0 1 9 8 .2 2 7 1 .7 7 3 m m
( a ) Applying Eq.(16.52): i
M ( R ri )
1 .5 1 0 (1 9 8 .2 2 7 1 6 7 .5 ) 3
A e ri
5 3 .0 6 M P a
( 2 9 2 5 ) (1 .7 7 3 ) (1 6 7 .5 )
( b ) Use Eq. (16.52):
o
M ( R ro )
1 .5 1 0 (1 9 8 .2 2 7 2 3 2 .5 ) 3
A e ro
4 2 .6 4 M P a
( 2 9 2 5 ) (1 .7 7 3 ) ( 2 3 2 .5 )
SOLUTION (16.20)
45 mm
rA 1 2 5 m m ,
5 mm
5 mm
A 325
A 25 mm
B
r
A1 A 2
2
r
Equation (16.50):
A1 r1 A 2 r2
mm
A1
A2
R
rB 1 7 0 m m
A
325
dA r
130
25
125
dr r
2 5 ( 5 )(1 2 7 .5 ) ( 4 0 5 )(1 5 0 ) 125 200
1 3 9 .9 7 5 3 m m
170
5
130
dr r
1 4 1 .3 4 6 2 m m
e r R 1 .3 7 9 8 m m
Equations (16.53), with M P ( r 2 5 ) 1 6 6 .3 4 6 2 P and P P : ( ) B
P A
120
P 325
[1
[1
( r 2 5 )( R rB ) e rB
]
1 6 6 .3 4 6 2 (1 3 9 .9 7 5 3 1 7 0 ) (1 .3 7 9 8 )(1 7 0 )
]
or P 1 .9 2 2 k N
267
SOLUTION (16.21) h
R
ln
60
ro
ln
1 2
2 4 8 .7 9 5 4 m m
220
ri
r
280
1
( ri ro )
(220 280) 250 m m
2
A ( 6 0 )( 6 0 ) 3 6 0 0 m m
2
e r R 2 5 0 2 4 8 .7 9 5 4 1 .2 0 4 6 m m
Using Eq.(16.52): i
M ( R ri )
M ( 2 4 8 .7 9 5 4 2 2 0 )
1 5 0 (1 0 ) 6
;
A e ri
3 6 0 0 (1 0
6
, M 4 .9 7 k N m
)(1 .2 0 4 6 )( 0 .2 2 )
Similarly,
o
M ( R ro )
M ( 2 4 8 .7 9 5 4 2 8 0 )
; 1 5 0 (1 0 ) 6
A e ro
3 6 0 0 (1 0
6
,
M 5 .8 4 k N m
)(1 .2 0 4 6 )( 0 .2 8 )
Therefore M
a ll
4 .9 7 k N m
SOLUTION (16.22) A 5 0 b 2 (1 5 0 2 5 ) 5 0 b 7 5 0 0
We have rA 1 5 0 m m and rB 3 5 0 m m . Applying Eq. (16.52):
A
B
( M )( R r A )
( M )( R rB )
A e rA
A e rB
from which rB ( R r A ) r A ( R rB ) ,
or
R 210 m m
Then
R
A
or
350(R 150) 150(R 350)
A 210
dA r
200 b d r 1 5 0 r
350 200
2(25)dr 210 r
6 0 .4 1 3 2 b 5 8 7 5 .9 6 5 8
Hence or
5 0 b 7 5 0 0 6 0 .4 1 3 2 b 5 8 7 5 .9 6 5 8 b 156 m m
268
4 7 b ln 4 0 ln 3 4
SOLUTION (16.23) r 150 m m
A 5000 m m
2
M ( R d )P
Table 16.1, Case A: R
h ln
ro r2
100 ln
2 1
1 4 4 .2 6 9 5 m m
e r R 1 5 0 1 4 4 .2 6 9 5 5 .7 3 0 5 m m
Equation (16.53): ( ) A
P A
( r d ) P ( R rA ) A e rA
]
P A
( r d )( R r A )
[1
e rA
]
or 3
2 5 (1 0 )
80
5000
[1
(1 5 0 d )(1 4 4 .2 6 9 5 1 0 0 ) ( 5 .7 3 0 5 )(1 0 0 )
]
1 5 ( 5 7 3 .0 5 ) (1 5 0 d ) ( 4 4 .2 6 9 5 ) 6 , 6 4 0 .4 2 5 4 4 .2 6 9 5 d
Solving, d 4 4 .1 7 m m
SOLUTION (16.24) Locate centroid : 120 mm 50 mm
A2
r
C A1
O
75 mm
80 mm
A1 r1 A 2 r2 A1 A 2 4 5 0 0 (1 2 0 ) ( 3 0 0 0 ) (1 6 0 ) 4500 3000
136 m m
r
1 2
R
h ( b1 b 2 ) 2
( b1 ro b 2 ri ) ln
ro ri
h ( b1 b 2 )
( 0 .5 ) (1 2 0 ) ( 7 5 5 0 ) 2
[ ( 7 5 ) ( 2 0 0 ) ( 5 0 ) (8 0 ) ] ln
200
1 2 7 .1 3 3 0 m m
P
(1 2 0 ) ( 7 5 5 0 )
80 e r R 8 .8 6 7 m m
( a ) Use Eq.(16.55a),
A
P
[1
r ( R rA )
A
A
B ]
O
M=P r
e rA
P
r
(1 3 6 ) (1 2 7 .1 3 3 0 8 0 ) 1 1 0 0 .4 M P a 6 7 5 0 0 (1 0 ) (8 .8 6 7 ) (8 0 ) 75000
(CONT.)
269
16.24 (CONT.) ( b ) From Eq.(16.55b):
P
B
r ( R rB )
[1
A 75000
]
e rB
7 5 0 0 (1 0
6
(1 3 6 ) (1 2 7 .1 3 3 0 2 0 0 ) 1 4 5 .9 M P a ) (8 .8 6 7 ) ( 2 0 0 )
SOLUTION (16.25) A d
r 100 m m
4 7854 m m
2
2
From Table 16.1, Case B: R
A 2 (r
r
2
c
2
9 3 .3 0 1 5
7854 2 (1 0 0
)
2
100 50
2
)
e r R 6 .6 9 8 5 m m
(a)
Equations (16.55a): ( ) A
P A
[1
r ( R rA ) e rA
]
or 150
P 7854
( ) B
(b)
P A
1 0 0 ( 9 3 .3 0 1 5 5 0 )
[1
( 6 .6 9 8 5 )( 5 0 )
[1
r ( R rB ) e rB
]
P 8 4 .5 8 k N
], 3
8 4 .5 8 (1 0 ) 7 8 5 4 (1 0
6
)
[1
1 0 0 ( 9 3 .3 0 1 5 1 5 0 ) ( 6 .6 9 8 5 )(1 5 0 )
]
50 M Pa
SOLUTION (16.26) rA 1 2 5 m m , r
1
rB 2 2 9 m m , 1
( ro ri )
2
a 100 m m ,
b 50 m m
(229 125) 177 m m
2
From Table 16.1 A a b (1 0 0 )(5 0 ) 5 0 0 0 m m
dA
r
2 b
[r
r
2
2
P
a ] 2
a
2 (5 0 )
[1 7 7
1 7 7 1 0 0 ] 9 7 .2 4 9 5 m m 2
2
r
A
100
Hence A
R
dA A
5 0 0 0
1 6 1 .5 2 1 5 m m
9 7 .2 5
P M=-P r B
r
(CONT.)
270
16.26 (CONT.) e r R 1 5 .4 7 8 5 m m
Equation (16.55a) with M P r :
A
P
[1
r ( R rA )
A
3
1 7 7 (1 6 1 .5 2 1 5 1 2 5 ) 1 ) (1 5 .4 7 8 5 ) (1 2 5 )
1 2 5 (1 0 )
]
e rA
5 0 0 0 (1 0
6
3 4 .5 4 M P a
B
P
[1
r ( R rB )
A
3
1 7 7 (1 6 1 .5 2 1 5 2 2 9 ) 1 ) (1 5 .4 7 8 5 ) ( 2 2 9 )
1 2 5 (1 0 )
]
e rB
5 0 0 0 (1 0
6
1 8 .8 6 M P a
SOLUTION (16.27) rA 1 3 0 m m r
1
rB 2 0 0 m m
( ro ri ) 1 6 5 m m
2
A a b ( 7 0 )(3 5 ) 2 4 5 0 m m
dA
2 b
r
[r
r
2
2
a ] 2
P
a
2 (3 5 )
[1 6 5
1 6 5 7 0 ] 4 8 .9 6 0 1 m m 2
2
70
A
R
dA A
r
2 4 5 0
A
1 5 7 .2 0 7 6 m m
4 8 .9 6 0 1
P
r
e r R 7 .7 9 2 4 m m
M=-P r B
Using Eq.(16.55) with M P r :
B
P
[1
A
r ( R rB )
]
e rB
9 0 (1 0 ) 6
Pa ll 2 4 5 0 (1 0
6
(1 6 5 ) (1 5 7 .2 0 7 6 2 0 0 ) 1 ) ( 7 .7 9 2 4 ) ( 2 0 0 )
Solving Pa ll 1 9 6 .2 k N
271
SOLUTION (16.28) 1
A
bh
2
The section width w varies linearly with r. Thus (1) w c 0 c1 r Since w b
( a t r ri )
w 0
dr (2)
( a t r ro )
r
Substituting Eq.(1) into Eq.(2); c1
b
c0
h
w
O
b
b ro h
ri
h r
Then
dA A
r
ro ri
w
c 0 c1 r
dr
r
dr
r
Inserting c1 and c 0 into this, after integrating and rearranging, we have dA
b(
ro
r
ln
h
ro
1)
ri
Therefore 1
A
R
dA r
2
ro
ln
h
h ro
1
ri
SOLUTION (16.29) r A c
2
r
Through use of the polar coordinates we write: w 2 c s in
r r c cos
d r c s in d
C
d A w d r 2 c s in d 2
2
w
O
c
dA r
2 c s in
0
r c cos
2
2
2
d
c (1 c o s )
0
r c cos
2
ccos
2
2
c cos (r c ) 2
2
2
2
r c cos
0
2 (r c cos )d 2(r 0
2
r
dr
2
c )
2
0
d r c cos
(CONT.)
272
16.29 (CONT.)
r c 2
2r
0
2 c s in
2(r
0
2
2
c ) 2
ta n
r c 2
2
ta n
2
1
r c
2
0
This gives
dA
2 ( r
r c ) 2
2
r
SOLUTION (16.30) A
1 2
( b1 b 2 ) h
ri
h w
The section width w varies linearly with r as
b2
w c 0 c1 r
(1)
( a t r ri )
w b2
( a t r ro )
O
r
dr
We have w b1
b1
r
(2)
Introduce Eq.(1) into Eq.(2), then solve for c 0 and c1 c0
ro b1 ri b 2
c1
h
b1 b 2
(3)
h
Then, we write
dA r
ro
w
ri
dr
r
ro
c 0 c1 r
ri
r
d r c 0 ln
ro ri
c 1 ( ro ri )
This gives, substituting Eqs.(3):
dA
ro b1 ri b 2
r
h
ln
ro ri
( b1 b 2 )
Hence 1 A
R
dA r
2
h ( b1 b 2 ) 2
( ro b1 ri b 2 ) ln
ro ri
( b1 b 2 )
SOLUTION (16.31) As in Example 16.7, we take c 1 c 2 0 in Eq.(16.60). Then, substituting Eq.(16.60) into (16.56a) we obtain an expression for the radial moment M r . (CONT.)
273
16.31 (CONT.) Boundary conditions: w 0
M
0
r
(r a )
now give p0a
c4
2
c3a
4
0
64 D
p0a
c3
2
32 D
3 1
from which c 4 p 0 a ( 5 ) 6 4 (1 ) D 4
The plate deflection is thus p0a
w
4
r
p0a
w m ax
At r=0:
4
3 1
( a4 2
64 D
4
r
2 2
a
5 1
(1)
)
5 1
64 D
Substituting Eq.(1) into Eq.(16.56b): p0
M
[( 3 ) a
16
2
( 1 3 ) r ] 2
At r=a: 6 M
,m ax
t
3 (1 ) p 0
2
4
( at )
2
SOLUTION (16.32) Refer to Example 16.7. At r=a:
r
6M t
r
2
S
( a ) 1 2 or
y
n
1 2
p0a
3 4
2
2,
2
t
p0 ( t ) (
a
S
2
2
2
p 0 ( t ) [ 16 4
y
1
S
y
)
n
1 4
3 4
6M t
S
) S
n 2
,
n
( b ) 1 1 2 2 ( or a
2
a
;
p0 ( t )
2
y
4t
2
2
a
2
p0a
1
y
n
t
(a)
P0
2
2
2
3 16
t
2
9 16
] (
Sy
2
7 4
) ;
n
p0 ( t )
Sy n
Solving, n 1.5 1 2
S
y
p0
(a)
SOLUTION (16.33) Table 16.2: w m ax 1 .2 5
Also
or
m ax
3 16
p0a
(1 ) ( 5 )
Et
2
3
6
3 16
( 0 .7 )(5 .3 )
Sy
3 8
3 .5 (1 0 )( 4 0 ) 9
2 0 0 (1 0 ) t
(3 ) p 0
a t
2
3
2
t 0 .2 5 m m
,
5 6 0 1 .2 3 8
;
t 0 .5 6 m m
274
3 .5 ( 4 0 ) t
2
SOLUTION (16.34) From Example 16.7 1
p0
a
(t)
4
2
2
3 4
a
p0 ( t ) S
We have 1 1 2 2 ( 2
2
p 0 ( at ) [ 161 2
or
4
4 ( 10
or
10
4
)a
( 10 ) ( 10
4
12
3 16
y
)
n
2 6
] ( 150 2 10 ) ;
9 16
12 , 857 . 14 (10
)
2
12
p 0 ( at ) ( 167 ) 5 , 625 ( 10
2
2
4
12
)
), a 238 . 1 mm
SOLUTION (16.35) r 1 7 5 1 1 7 4
r 7 5 1 7 4
F
mm,
mm,
F ri p z ,
p pz,
2
t 2 mm
ri 7 3 m m
90 , o
p 100 kPa
N
Equation (16.64b): N
and
2
( 0 .0 7 3 ) ( 1 0 0 )
F 2 r s in
2 ( 0 .0 7 4 ) (1 )
3 .6
kN
1 . 8 MPa
3 , 600 0 . 002
Then, Eq.(16.64a): r
or
r
6
pz
1 .8 ( 1 0 )
;
t
0 .1 7 4
0 .0 7 4
3
1 0 0 (1 0 ) 0 .0 0 2
2 . 93 MPa
SOLUTION (16.36) N is maximum at A ( 9 0 ): o
,m ax
pa ( b a 2 ) ( b a )t
S
y
n
240 1. 2
;
2 . 2 ( 5 0 )( 2 5 0 2 5 )
( 25050 )t
or t 0 . 619
Similarly
n
mm pr
240 1 .2
;
2t
2 . 2 ( 50 )
, t 0 . 275
2t
mm
SOLUTION (16.37) 1
pa
2 x
t
Pa 2t
( a ) Since 1 and 2 are of the same sign ( 3 0 ) : 1 0
S
y
n
;
pa t
S
y
n
,
t
pan S
y
(CONT.)
275
16.37 (CONT.) S
( b ) 1 1 2 2 ( 2
(
(c)
2
pa
) [1 2
t
1
Su
2
S uc
1 2
1 n
1 4
S
] (
y
n
pa
;
y
)
n
) , t 2
pa
tS u
2
3 2
4 tS u
1 n
pan S
y
3 pan
t
,
4 Su
SOLUTION (16.38) (a)
p 200 x 200 15 (16 ) 440
all
pa t
pr
; t
3
440 ( 10 )( 5 )
150 ( 10
all
6
)
14 . 67 mm
( b ) p 200 15 [ 3 (16 4 )] 380 t
pr
3
380 ( 10 )( 5 )
all
150 ( 10
6
kPa
12 . 67 mm
)
( c ) Top-end plate, with p 200 kPa m ax
3 4
t 158
or
a
(Fig.16.18):
1 5 0 (1 0 )
2
6
p( t ) ;
3 4
( 2 0 0 1 0 )( t )
2
3 4
( 4 4 0 1 0 )( t )
2
3
5
mm
Bottom-end plate, with p 440 m ax
kPa
3 4
a t
kPa :
1 5 0 (1 0 )
2
6
p( ) ;
3
5
or t 235
mm
SOLUTION(16.39) We have r a . The loading is p p z a ( 1 c o s ) . The resultant force F for the part intercepted by : F 2 a
2
a ( 1 c o s ) s in c o s d
0
2a [ 6 3
1
1 2
c o s (1 2
2 3
c o s )]
Substitution of this into Eqs.(16.65) and rearranging yield the equations quoted in this problem.
SOLUTION (16.40) ( a ) L w 1 .7 2 a t 1 .7 2 7 5 0 1 2 1 6 3 .2 m m (b)
cr
0 . 605
Et a
0 . 605
210 10
9
750
( 12 )
2 , 033
MPa
No failure in buckling, since c r S y and material would yield.
276
SOLUTION (16.41)
cr
0 .6 0 5
Et a
0 .6 0 5
9
2 0 0 1 0 (1 0 ) 600
2, 017 M P a
Thus Pc r 2 a t c r
2 ( 6 0 0 ) (1 0 ) ( 2 0 1 7 ) 7 6 , 0 3 9 k N
End of Chapter 16
277
CHAPTER 17
FINITE ELEMENT ANALYSIS IN DESIGN
SOLUTION (17.1) We have (
AE L
)1,2
A(2 E )
6 AE
L 3
(
L
AE L
)3
2 A(E )
L 3
6 AE L
There are four displacement components ( u 1 , u 2 , u 3 , u 4 ) and so the order of the system matrix is 4x4. Using Eq. (17.1): [ k ]1 [ k ] 2 [ k ] 3
1 1
6 AE 1 L 1
( a ) System matrix, [ K ] [ k ]1 [ k ] 2 [ k ] 3 , by superposition is thus u1
u2
1 6 AE 1 [K ] L 0 0
u3
1
0
(1 1)
1
1
(1 1)
0
1
u4
u1
u2
u3
u4
0 0 6 AE 1 L 1
1 1 0 0
1
0
2
1
1
2
0
1
0 0 1 1
( b ) Refer to Eq. (17.7b): F1 x F2 x F 3x F 4x
AE L
1 1 0 0
1
0
2
1
1
2
0
1
0 0 1 1
u1 u2 u 3 u 4
(1)
Boundary conditions are u 1 u 4 0 and F x 3 P . Equation (1) is then 0 AE 2 L 1 P
Solving u 2
PL 9 AE
1 u 2 2 u3
u3
PL 18 AE
( c ) Equations (1) result in F1 x F2 x F 3x F 4x
6 AE L
1 1 0 0
1
0
2
1
1
2
0
1
0 0 1 1
The reactions are R1
2 3
P
R4
1
P
3
278
PL PL
2 P 3 9 AE L P 18 AE AE 0 0 P 3 0
SOLUTION (17.2) We have (
AE L
)1
AE
(
AE
L
L
[ k ]1 [ k ] 2
AE 1 L 1
AE
)2
(
AE
L
L
)3
4 AE
0 .8
5L
AE L
Equation (17.1): 1 1
[ k ]3
A E 0 .8 L 0 .8
0 .8 0 .8
( a ) System matrix, [ K ] [ k ]1 [ k ] 2 [ k ] 3 , by superposition is then 1 AE 1 [K ] L 0 0
1
0
(1 1)
1
1
(1 0 .8 )
0
0 .8
0 0 .8 0 .8 0
( b ) Refer to Eq. (17.7b): F1 x F2 x F 3x F 4x
AE L
u1
u2
u3
1 1 0 0
1
0
2
1
1
1 .8
0
0 .8
u4 0 0 .8 0 .8 0
u1 u2 u 3 u 4
Boundary conditions are u 1 u 4 0 , We have F x 3 P . Thus 0 AE 2 L 1 P
1 u2 1 .8 u 3
Solving PL
u2
2 .6 A E
u3
PL 1 .3 A E
( c ) Equations (1) yield F1 x F2 x F 3x F 4x
AE L
1 1 0 0
1
0
2
1
1
1 .8
0
0 .8
0 0 .8 0 .8 0
The reactions are R1
1 2 .6
P
R4
0 .8
P
1 .3
279
P P
P 2 .6 2 .6 L 0 1 .3 A E P 0 1 .6 P 2 .6
0
(1)
* SOLUTION (17.3) We have A E L the same for all elements. Thus, Eq. (17.1): 1 1
AE 1 L 1
[ k ]1 [ k ] 2 [ k ] 3 [ k ] 4
( a ) By superposition, we have 1 1 AE 0 [K ] L 0 0
1
0
0
2
1
0
1
2
1
0
1
2
0
0
1
0 0 0 1 1
(b) F1 x F 2x F3 x F 4x F 5 x
AE L
1 1 0 0 0
1
0
0
2
1
0
1
2
1
0
1
2
0
0
1
0 0 0 1 1
u1 u 2 u3
(1)
u 4 u 5
Boundary conditions: u 1 0 and u 5 . We have F 2 x F 3 x F 4 x 0 . Hence, 0 AE 20 L 0
1 0 0
2
1
0
1
2
1
0
1
2
0 0 1
0 u 2 u3 u 4
This equation may be rewritten, after transposing the product of the appropriate stiffness coefficients by the known displacement ( ) to the left side. In so doing, 0 0 AE L
AE L
2 1 0
1 2 1
0 1 2
u2 u3 u4
Solving u2
4
u3
2
u4
3 4
(CONT.)
280
17.3 (CONT.) (c)
Equations (1) give then F1 x F 2x F3 x
AE L
F 4x F 5 x
1 1 0 0 0
1
0
0
2
1
0
1
2
1
0
1
2
0
0
1
0 4 4 0 AE 2 0 L 3 4 0 4
0 0 0 1 1
The reaction is R1
1
AE
4L
SOLUTION (17.4) We have AE L
1
(1 2 0 0 1 0
6
)(7 2 1 0 ) 5 4 1 0 9
6
N m
1 .6
c cos 30
o
s s in 3 0
3 2
o
1 2
3 4
( a ) Through the use of Eq. (17.14), we have 3 4 3 4 6 [ k ]1 5 4 (1 0 ) 3 4 3 4
3 4
1 4
3
3 4 3 4
4
3
1 4
4
14 3 4 1 4 3
4
or 0 .7 5 0 .4 3 3 6 [ k ]1 5 4 (1 0 ) 0 .7 5 0 .4 3 3
0 .4 3 3
0 .7 5
0 .2 5
0 .4 3 3
0 .4 3 3
0 .7 5
0 .2 5
0 .4 3 3
0 .4 3 3 0 .2 5 N m 0 .4 3 3 0 .2 5
( b ) Equation (17.11b), { } e [ T ]{ } e : u1 0 .8 6 6 0 .5 v1 0 u 3 v 0 3
0 .5
0
0 .8 6 6
0
0
0 .8 6 6
0
0 .5
0 0 .5 0 .8 6 6 0
1 .5 1 .8 9 9 1 .2 0 .2 8 9 mm 2 .2 1 .9 0 5 0 1 .1 0 0
(CONT.)
281
17.4 (CONT.) ( c ) Inserting the given data into (17.15) with i=1 and j=3, we obtain the axial force 6 F1 F1 3 5 4 (1 0 )
3
1 2
2
2 .2 1 .5 3 (1 0 ) 2 0 5 .4 k N 0 1 .2
So, axial stress in the element equals 1 F1 A 1 7 1 .2 M P a Comment: The negative sign denotes compression.
SOLUTION (17.5) We have E 2 0 0 G P a (by Table B.1). Refer to Solution of Prob. 17.4. AE L
1
(1 2 0 0 1 0
6
)( 2 0 0 1 0 ) 1 5 0 1 0 9
6
N m
1 .6
c cos 60
o
1 2
s s in 6 0
o
3 2
( a ) From Eq. (17.14): 1 4 3 4 6 [ k ]1 1 5 0 (1 0 ) 1 4 3 4
3 4 3 4 3 4 3 4
0 .2 5 0 .4 3 3 6 1 5 0 (1 0 ) 0 .2 5 0 .4 3 3
1 4
43 3 4 3 4
3 4 1 4 3 4
3
4
0 .4 3 3
0 .2 5
0 .7 5
0 .4 3 3
0 .4 3 3
0 .2 5
0 .7 5
0 .4 3 3
0 .4 3 3 0 .7 5 0 .4 3 3 0 .7 5
( b ) Equation (17.11b), { } e [ T ]{ } e : u1 0 .5 0 .8 6 6 v1 u 0 3 v 0 3
( c ) F1 F1 3 1 5 (1 0 ) 12 6
3 2
0 .8 6 6
0
0 .5
0
0
0 .5
0
0 .8 6 6
0 0 .8 6 6 0 .5 0
1 .5 1 .7 8 9 1 .2 0 .6 9 9 mm 2 .2 1 .1 0 1 .9 0 5
2 .2 1 .5 3 (1 0 ) 4 3 3 .4 k N 0 1 .2
1 F1 A 3 6 1 .2 M P a
282
SOLUTION (17.6)
Table 17.6 Data for truss of Fig. P17.6. E le m e n t
o
1
45
2
315 o
3
0
4
90
5
90
o
o
o
c
s
2 2
2 2
2 2
c
2 2
2
s
2
cs
1 2
1 2
1 2
1 2
1 2
1 2
1
0
1
0
0
0
1
0
1
0
0
1
0
1
0
Equation (17.14) u1
[ k ]1
1 1 2 AE 2L 1 1
v1
u2
2
1 2
1 2
2
1 2
1 2
2
1 2
1 2
2
1 2
1 2
v2 1 2 1 2 1 2 1 2
Similarly, for elements 2, 3, 4, and 5, we obtain
u1
[ k ]2
1 2 AE 1 2L 1 1
u1 1 0 AE [ k ]3 L 1 0
v1
v1
u3
2
1 2
1 2
2
1 2
1 2
2
1 2
1 2
2
1 2
1 2
u4
v4
0
1
0
0
0
1
0
0
0 0 0 0
u4
v4
u2
v2
0 AE 0 [ k ]4 L 0 0
0
0
1
0
0
0
1
0
0 1 0 1
v3 1 2 1 2 1 2 1 2
u3 0 AE 0 [ k ]5 L 0 0
283
v3
u4
0
0
1
0
0
0
1
0
v4 0 1 0 1
SOLUTION (17.7) Table P17.7 Data for the truss of Fig P17.7 E le m e n t
L e n g th ( m )
1
7 .5
3 6 .9
2
6
0
3
4 .5
90
4 .5
0
4 .5
2
s
0 .8
0 .6
1
0
o
4 5
o
c
o
135
2
2
cs
s
0 .6 3 9
0 .4 8
0 .3 6
1
0
0
0
1
0
0
1
1
0
1
0
0
0 .7 0 7
0 .7 0 7
0 .5
0 .5
0 .5
o
o
c
Apply Eq.(17.14): AE [ k ]1 7 .5
AE [ k ]2 6 .0
AE [ k ]4 4 .5
u1
v1
u2
v2
0 . 639
0 . 48
0 . 639
0 . 48
0 . 361
0 . 48
0 . 639
0 . 48
0 . 639
0 . 48
0 . 361
0 . 48
u1
v1
u3
1
0
1
0
0
0
1
0
1
0
0
0
u3
v3
1
0
1
0
0
0
1
0
1
0
0
0
u4
0 . 48 u 1 0 . 361 v 1 0 . 48 u 2 0 . 361 v 2
v3
u2
0 0 0 0
0 AE 0 [ k ]3 4 .5 0 0
u1 v1 u3 v3
v4
v2
u3
v3
0
0
1
0
0
0
1
0
0 1 0 1
u2
0 0 0 0
AE [ k ]5 4 .5 2
u3 v3 u4 v4
v2
u2 v2 u3 v3
u4
0 .5
0 .5
0 .5
0 .5
0 .5
0 .5
0 .5
0 .5
0 .5
0 .5
0 .5
0 .5
v4 0 .5 u 2 0 .5 v 2 0 .5 u 4 0 .5 v 4
SOLUTION (17.8) Table P17.8 Data for the truss of Fig.P17.8 E le m e n t
1
0
2
o
60
o
3
120
4
0
5
o
o
60
o
c
s
c
1
0
0 .5
2
2
cs
s
1
0
0
0 .8 6 6
0 .2 5
0 .4 3 3
0 .7 5
0 .5
0 .8 6 6
0 .2 5
0 .4 3 3
0 .7 5
1
0
1
0
0
0 .5
0 .8 6 6
0 .2 5
0 .4 3 3
0 .7 5
(CONT.)
284
17.8 (CONT.) ( a ) Use Eq.(17.14):
AE [ k ]1 L
u1
v1
u4
1
0
1
0
0
0
1
0
1
0
0
0
u1
AE [ k ]2 L
AE [ k ]3 L
AE [ k ]4 L
v4
0 0 0 0
u1 v1 u4 v4
v1
u2
v2
0 .2 5
0 .4 3 3
0 .2 5
0 .4 3 3
0 .7 5
0 .4 3 3
0 .2 5
0 .4 3 3
0 .2 5
0 .4 3 3
0 .7 5
0 .4 3 3
0 .4 3 3 0 .7 5 0 .4 3 3 0 .7 5
u4
v4
u2
v2
0 .2 5
0 .4 3 3
0 .2 5
0 .4 3 3
0 .7 5
0 .4 3 3
0 .2 5
0 .4 3 3
0 .2 5
0 .7 5
0 .4 3 3
u2
v2
u3
1
0
1
0
0
0
1
0
1
0
0
0
v1 u2 v2
0 .4 3 3 0 .7 5 0 .4 3 3 0 .7 5
0 .4 3 3
v3
0 0 0 0
u1
u2 v2 u4 v4
u3
u2 v2 u3 v3
AE [ k ]5 L
v3
u4
v4
0 .2 5
0 .4 3 3
0 .2 5
0 .4 3 3
0 .7 5
0 .4 3 3
0 .2 5
0 .4 3 3
0 .2 5
0 .4 3 3
0 .7 5
0 .4 3 3
0 .4 3 3 0 .7 5 0 .4 3 3 0 .7 5
u3 v3 u4 v4
( b ) Global Stiffness Matrix [ K ] u1 AE L
v1
u2
v2
u3
v3
u4
1. 2 5
0 .4 3 3
0 .2 5
0 .4 3 3
0
0
1
0 .4 3 3
0 .7 5
0 .4 3 3
0 .7 5
0
0
0
0 .2 5
0 .4 3 3
1.5
0
1
0
0 .4 3 3
0 .7 5
0
1.5
0
0
0 .2 5 0 .4 3 3
0
0
1
0
1. 2 5
0 .4 3 3
0 .2 5
0
0
0
0
0 .4 3 3
0 .7 5
0 .4 3 3
1
0
0 .2 5
0 .4 3 3
1.5
0
0
0 .4 3 3
0 .7 5
0
0 .2 5 0 .4 3 3
0 .4 3 3 0 .7 5
v4 0 0 .4 3 3 0 .7 5 0 .4 3 3 0 .7 5 0 1.5 0
u1 v1 u2 v2 u3 v3 u4 v4
(CONT.)
285
17.8 (CONT.) R1 x R 1y 0 P Q 0 R 4x R 4y
{ F } [ K ]{ };
0 0 u2 v2 K u 3 v 3 0 0
SOLUTION (17.9) Table P17.9 Data for the truss of Fig.P17.9 E le m e n t
L e n g th ( in . )
1
7 .2 1 1
5 6 .3 1
2
5
3
o
o
3
90
5
3 6 .8 7
o
7 .2 1 1
5 6 .3 1
o
4 5
3 6 .8 7
o
c
s
0 .5 5 5
0 .8 3 2
0 .8
c
2
2
cs
s
0 .3 0 8
0 .4 6 2
0 .6 9 2
0 .6
0 .6 4
0 .4 8
0 .3 6
0
1
0
0
1
0 .8
0 .6
0 .6 4
0 .4 8
0 .3 6
0 .5 5 5
0 .8 3 2
0 .3 0 8
0 .4 6 2
0 .6 9 2
( a ) Use Eq. (17.14): u1
v1
0 .3 0 8
0 .4 6 2
0 .3 0 8
0 .4 6 2
0 .6 9 2
0 .4 6 2
0 .3 0 8
0 .4 6 2
0 .3 0 8
0 .4 6 2
0 .6 9 2
0 .4 6 2
u1
v1
0 .6 4
0 .4 8
0 .6 4
0 .4 8
0 .3 6
0 .4 8
0 .6 4
0 .4 8
0 .6 4
0 .4 8
0 .3 6
0 .4 8
AE [ k ]1 7 .2 1 1
AE [ k ]2 5 u2
0 AE 0 [ k ]3 3 0 0
u2
u3
v2
u3
v3
0
0
1
0
0
0
1
0
0 1 0 1
v2
0 .4 6 2 0 .6 9 2 0 .4 6 2 0 .6 9 2
u1 v1 u2 v2
v3
0 .4 8 0 .3 6 0 .4 8 0 .3 6
u1 v1 u3 v3
u2 v2 u3 v3
(CONT.)
286
17.9 (CONT.) u3
[k ]4
AE 5
v3
u4
0 . 64
0 . 48
0 . 64
0 . 48
0 . 36
0 . 48
0 . 64
0 . 48
0 . 64
0 . 48
0 . 36
0 . 48
u2
AE [ k ]5 7 .2 1 1
v4
0 . 48 u 3 0 . 36 v 3 0 . 48 u 4 0 . 36 v 4
v2
u4
0 .3 0 8
0 .4 6 2
0 .3 0 8
0 .4 6 2
0 .6 9 2
0 .4 6 2
0 .3 0 8
0 .4 6 2
0 .3 0 8
0 .4 6 2
0 .6 9 2
0 .4 6 2
v4
0 .4 6 2 0 .6 9 2 0 .4 6 2 0 .6 9 2
u2 v2 u4 v4
( b ) Global Stiffness Matrix [ K ]
AE
u1
v1
u2
v2
0 .1 7 1
0 .1 6
0 .0 4 3
0 .0 6 4
0 .1 2 8
0 .0 9 6
0
0 .1 6
0 .1 6 8
0 .0 6 4
0 .0 9 6
0 .0 9 6
0 .0 7 2
0
0 .0 4 3
0 .0 6 4
0 .0 8 5
0
0
0 .0 6 4
0 .0 9 6
0
0 .5 2 5
0
0 .1 2 8
0 .0 9 6
0
0
0 .2 5 6
0
0 .0 9 6
0 .0 7 2
0
0
0 .4 7 7
0 .0 9 6 0 .1 7 1
0 .3 3 3
u3
v3
0 0 .3 3 3
0
0
0 .0 4 3
0 .0 6 4
0 .1 2 8
0 .0 9 6
0
0
0 .0 6 4
0 .0 9 6
0 .0 9 6
0 .0 7 2
{ F } [ K ]{ };
R1 x R 1y P 0 0 Q R 4x R 4y
0 0 u2 v2 K u 3 v 3 0 0
287
u4
0 .0 4 3 0 .0 6 4 0 .1 2 8
0 .1 6
v4
0 0 .0 6 4 0 .0 9 6 0 .0 9 6 0 .0 7 2 0 .1 6 0 .1 6 8 0
u1 v1 u2 v2 u3 v3 u4 v4
SOLUTION (17.10)
Table P17.10 Data for the truss of Fig.P17.10 E le m e n t
L e n g th
1
1. 2
0
2 3
c
s
c
1
0
0 .9 2 3 0
o
1.3
2 2 .6 2
0 .5
90
o
o
o
4
1. 2
0
5
1.3
2 2 .6 2
o
2
2
cs
s
1
0
0
0 .3 8 5
0 .8 5 2
0 .3 5 5
0 .1 4 8
1
0
0
1
1
0
1
0
0
0 .9 2 3
0 .3 8 5
0 .8 5 2
0 .3 5 5
0 .1 4 8
( a ) Use Eq.(17.14); AE [ k ]1 1.2
uA
vA
uC
1
0
1
0
0
0
1
0
1
0
0
0
uA
AE [ k ]2 1.3
0 0 0 0
uA vA uC vC
vA
uB
vB
0 .3 5 5 0 .1 4 8 0 .3 5 5 0 .1 4 8
0 .8 5 2
0 .3 5 5
0 .8 5 2
0 .3 5 5
0 .1 4 8
0 .3 5 5
0 .8 5 2
0 .3 5 5
0 .8 5 2
0 .3 5 5
0 .1 4 8
0 .3 5 5
uB
0 AE 0 [ k ]3 0 .5 0 0 uB
AE [ k ]4 1.2
vC
vB
uC
vC
0
0
1
0
0
0
1
0
0 1 0 1
vB
uD
1
0
1
0
0
0
1
0
1
0
0
0
uA vA uB vB
uB vB uC vC
vD
0 0 0 0
uB vB uD vD
(CONT.)
288
17.10 (CONT.)
AE [ k ]5 1.3
uC
vC
uD
vD
0 .8 5 2
0 .3 5 5
0 .8 5 2
0 .3 5 5
0 .1 4 8
0 .3 5 5
0 .8 5 2
0 .3 5 5
0 .8 5 2
0 .3 5 5
0 .1 4 8
0 .3 5 5
0 .3 5 5 0 .1 4 8 0 .3 5 5 0 .1 4 8
uC vC uD vD
( b ) Global Stiffness Matrix [ K ] AE
uA
vA
uB
vB
uC
vC
uD
1. 4 8 9
0 .2 7 3
0 .6 5 5
0 .2 7 3
0
0
0 .2 7 3
0 .1 1 4
0 .2 7 3
0 .1 1 4
0
0
0
0 .6 5 5
0 .2 7 3
1. 4 8 9
0 .2 7 3
0
0
0 .2 7 3
0 .1 1 4
0 .2 7 3
2 .1 1 4
0
2
0 .8 3 3
0 .8 3 3 0
0 .8 3 3
0
0
0
1. 4 8 9
0 .2 7 3
0 .6 5 5
0
0
0
2
0 .2 7 3
2 .1 1 4
0 .2 7 3
0
0
0
0 .6 5 5
0 .2 7 3
1. 4 8 9
0
0
0
0 .2 7 3
0 .1 1 4
0 .2 7 3
0 .8 3 3 0
R Ax R Ay 0 0 0 0 P R Dy
{ F } [ K ]{ };
0 0 uB vB K u C v C uD 0
vD
0 0 0 0 .2 7 3 0 .1 1 4 0 .2 7 3 0 .1 1 4 0
SOLUTION (17.11) We have AE=30 MN. Table P17.11 Data for the truss of Fig.P17.11 E le m e n t
L e n g th
1
5
5 3 .1 3
2
4
0
o
o
c
s
0 .6
0 .8
1
0
c
2
2
cs
s
0 .3 6
0 .4 8
0 .6 4
1
0
0
( a ) Use Eq.(17.14): (CONT.)
289
uA vA uB vB uC vC uD vD
17.11 (CONT.) AE [ k ]1 5
AE [ k ]2 4
u1
v1
0 .3 6
0 .4 8
0 .3 6
0 .4 8
0 .6 4
0 .4 8
0 .3 6
0 .4 8
0 .3 6
0 .4 8
0 .6 4
0 .4 8
u2
v2
u3
1
0
1
0
0
0
1
0
1
0
0
0
u2
v2
0 .4 8 0 .6 4 0 .4 8 0 .6 4
u1 v1 u2 v2
v3
0 0 0 0
u2 v2 u3 v3
( b ) Global Stiffness Matrix [K ] AE
u1
v1
u2
v2
u3
0 .0 7 2
0 .0 9 6
0 .0 7 2
0 .0 9 6
0
0 .0 9 6
0 .1 2 8
0 .0 9 6
0 .1 2 8
0
0 .0 7 2
0 .0 9 6
0 .3 2 2
0 .0 9 6
0 .0 9 6
0 .1 2 8
0 .0 9 6
0 .1 2 8
0
0
0 .2 5
0
0
0
0
0
0
0 .2 5 0 0
v3
0 .2 5
( c ) Boundary Conditions: u 1 v 1 u 3 v 3 0 . F2 x AE F 2y
0 .3 2 2 0 .0 9 6
u2 1 4 AE 3 v2
F1 x F1 y (d) F 3x F3 y
AE
0 .0 9 6 u 2 0 .1 2 8 v 2 3
0 0 .0 0 1 0 m 1 0 .0 6 3 1 0 , 0 0 0 0 .0 0 3 4
0 .0 7 2 0 .0 9 6 0 .2 5 0 0
0 .0 9 6 7 .6 3 2 0 .1 2 8 0 .0 0 1 0 1 0 .1 7 6 kN 0 .0 0 3 4 0 7 .5 0 0 0 0
( e ) Use Eq.(17.16) F1 2
F23
AE 5 AE 4
0 .6 1
0 .0 0 1 0 .8 1 2 .7 2 k N 0 .0 0 3 4 0 .0 0 1 0 7 .5 k N 0 .0 0 3 4
(C )
290
(C )
0 0 0 0 0 0
u1 v1 u2 v2 u3 v3
SOLUTION (17.12) We have E 2 0 0 G P a
A 2 (1 0 ) m m 3
2
Table P17.12 Data for the truss of Fig.P17.12 E le m e n t
L e n g th ( m m )
1
3000
90
2
3000
3
2
45
3000
0
c
s
c
0
1
0 .7 0 7 1
o
o
o
2
2
cs
s
0
0
1
0 .7 0 7
0 .5
0 .5
0 .5
0
1
0
0
( a ) Use Eq.(17.14) u1
[ k ]1
v1
0 0 0 0
AE 3000
u2
0
0
1
0
0
0
1
0
u1
2
AE
[k ]2
3000
u1
[k ]3
AE 3000
v2
0 u1 1 v1 0 u2 1 v2
v1
0 .5
0 .5
0 .5
0 .5
0 .5
0 .5
0 .5
0 .5
0 .5
0 .5
0 .5
u4
0
1
0
0
0
1
0
1
0
0
0
(10
11
0 u1 0 v1 0 u4 0 v4
) . Adding zero’s in proper locations and adding :
u1
[K ]
4 3
(10
11
1 . 354 0 . 354 0 0 ) 0 . 354 0 . 354 1 0
0 .5 u 1 0 .5 v 1 0 .5 u 3 0 .5 v 3
v4
1
4 3
v3
0 .5
v1
( b ) The common factor
u3
v1
u2
v2
u3
v3
u4
0 . 354
0
0
0 . 354
0 . 354
1
1 . 354
0
1
0 . 354
0 . 354
0
0
0
0
0
0
0
1
0
1
0
0
0
0 . 354
0
0
0 . 354
0 . 354
0
0 . 354
0
0
0 . 354
0 . 354
0
0
0
0
0
0
1
0
0
0
0
0
0
v4
0 u1 0 v1 0 u2 0 v2 0 u3 0 v3 0 u4 0 v 4
(CONT.)
291
17.12 (CONT.)
(c)
u1 25000 v 1 F2x 0 F2 y 0 K F3x 0 0 F3 y F4x 0 0 F4 y 0
Eliminate rows and columns in [K] corresponding to zero displacements: 0 4 (10 3 25000
11
u1 1 11 ( 4 3 ) (1 0 ) v1
0 . 354 u 1 1 . 354 v 1
1 . 354 ) 0 . 354
0 .2 0 7 0 0 .0 3 8 9 6 (1 0 ) m m 0 .7 9 3 2 5 0 0 0 0 .1 4 8 6
0 .7 9 3 0 .2 0 7
(d) R2 y R3x R 3y R 4x
4 11 (1 0 ) 3
1
0 0 .3 5 4 0 .3 5 4 1
1 9 .8 1 3 0 .3 5 4 u 1 5 .1 7 8 kN 0 .3 5 4 v 1 5 .1 7 8 0 5 .1 8 7
( e ) Use Eq.(17.16): AE
F1 2
L1
F1 3
u1 2 (1 0 ) 2 0 0 (1 0 ) 1 0 3000 v1 3
0
AE L2
2 (1 0 ) 2 0 0 (1 0 )
F1 4
AE L3
0 .0 3 8 9 6 0 .7 0 7 (1 0 ) 0 .1 4 8 6
9
3000
0 .7 0 7
2
u1 2 (1 0 ) 2 0 0 (1 0 ) 0 1 3000 v1 3
1
0 .0 3 8 9 6 1 (1 0 ) 1 9 .8 1 3 k N ( T ) 0 .1 4 8 6
u1 0 .7 0 7 v1
0 .7 0 7 3
9
7 .3 1 2 2 k N
(T )
0 .0 3 8 9 6 0 (1 0 ) 5 .1 8 6 7 k N ( C ) 0 .1 4 8 6
9
SOLUTION (17.13) Table P17.13 Data for the truss of Fig.P17.13 E le m e n t
L e n g th
1
5
5 3 .1 3
2
4
180
o
o
c
s
0 .6
0 .8
1
0
c
2
2
cs
s
0 .3 6
0 .4 8
0 .6 4
1
0
0
(CONT.)
292
17.13 (CONT.) ( a ) Use Eq.(17.14): AE [ k ]1 5
AE [ k ]2 4
u1
v1
0 .3 6
0 .4 8
0 .3 6
0 .4 8
0 .6 4
0 .4 8
0 .3 6
0 .4 8
0 .3 6
0 .4 8
0 .6 4
0 .4 8
u2
v2
1
0
1
0
0
0
1
0
1
0
0
0
u2
u3
v2
0 .4 8 0 .6 4 0 .4 8 0 .6 4
u1 v1 u2 v2
v3
0 0 0 0
u2 v2 u3 v3
( b ) Global Stiffness Matrix [K ] AE
u1
v1
u2
v2
u3
0 .0 7 2
0 .0 9 6
0 .0 7 2
0 .0 9 6
0
0 .0 9 6
0 .1 2 8
0 .0 9 6
0 .1 2 8
0
0 .0 7 2
0 .0 9 6
0 .3 2 2
0 .0 9 6
0 .0 9 6
0 .1 2 8
0 .0 9 6
0 .1 2 8
0
0
0 .2 5
0
0
0
0
0
0
0 .2 5 0
0 .2 5
v3
0 0 0 0 0 0
u1 v1 u2 v2 u3 v3
(c) 0 . 096 u 2 0 . 128 v 2
0 0 . 322 AE 60000 0 . 096 u2 1 6 v 2 1 0 (1 0 )
4 3
3
0 18 3 (1 0 ) 1 0 .0 6 3 6 0 0 0 0 6 0 .4
m
(d) R1 x R1 y R 3x
AE
0 .0 7 2 0 .0 9 6 0 .2 5
0 .0 9 6 4 5 .0 2 4 u2 0 .1 2 8 6 0 .0 3 2 v 2 0 45.
kN
( e ) Use Eq.(17.16) F1 2 A E 0 .6
u2 6 0 .8 1 0 (1 0 ) 0 .6 v 2
0 .0 1 8 0 .8 3 7 5 .2 0 .0 6 0 4
kN
(C )
(CONT.)
293
17.13 (CONT.) u2 6) 0 1 0 (1 0 1 v 2
F23 A E 1
0 .0 1 8 0 180 0 .0 6 0 4
kN
(T )
SOLUTION (17.14) We have AE 125
MN .
Table P17.14 Data for the truss of Fig.P17.14 E le m e n t
L e n g th
1
5
1 4 3 .1 3
2
5
3
o
3 6 .8 7
5
1 4 3 .1 3
o
o
c
s
c
0 .8
0 .6
0 .8 0 .8
2
2
cs
s
0 .6 4
0 .4 8
0 .3 6
0 .6
0 .6 4
0 .4 8
0 .3 6
0 .6
0 .6 4
0 .4 8
0 .3 6
( a ) Apply Eq.(17.14): AE [ k ]1 25L
AE [ k ]2 25L
u1
v1
u2
16
12
16
12
9
12
16
12
16
12
9
12
u2
v2
u3
16
12
16
12
9
12
16
12
16
12
9
12
v2
12 9 12 9
u1 v1 u2 v2
v3 12 9 12 9
AE [ k ]3 25 L
u2 v2 u3 v3
u2
v2
u4
16
12
16
12
9
12
16
12
16
12
9
12
v4
12 9 12 9
u2 v2 u4 v4
( b ) Global Stiffness Matrix AE [K ] 25L
u1
v1
u2
v2
u3
v3
u4
v4
16
12
16
12
0
0
0
9
12
9
0
0
0
16
12
48
12
16
12
16
12
9
12
27
12
9
12
0
0
16
12
16
12
0
0
0
12
9
12
9
0
0
0
16
12
0
0
16
0
0
12
9
0
0
12
0 0 12 9 0 0 12 9
12
25000 AE 48 25L 12 40000
(c)
u1 v1 u2 v2 u3 v3 u4 v4
12 u2 27 v2
(CONT.)
294
17.14 (CONT.) 12 25000 1 . 0026 (10 48 40000 1 . 927
u 2 27 25 L AE (1152 ) 12 v2
3
) m
(d) F1 x F 1y F 3 x F 3y F4x F 4 y
1 1152
16 12 16 12 16 12
( e ) Use Eq.(17.16): We have u 1 v 1 0 ;
F 12
F 32
F 42
AE
L
AE
u 2 0 . 6 v2
0 . 8
L
AE
u3 v 3 0;
u 2 0 . 6 v2
0 .8
L
1 1152
1 1152
kN
u 4 v 4 0.
27 0 . 6 12
0 .8
27 0 . 6 12
0 .8
12 25000 8 . 854 48 40000
12 25000 48 . 958 48 40000
27 0 . 6 12
0 . 8
1 1152
u 2 0 . 6 v2
0 .8
7 . 083 5 . 313 39 . 167 12 25000 48 40000 29 . 375 7 . 083 5 . 313
12 9 12 27 9 12 12 9
kN
12 25000 8 . 854 48 40000
kN
(C )
(C )
kN
(C )
SOLUTION (17.15) We have E 210 GPa
4
A 5 10
m
2
Table P17.15 Data for the truss of Fig.P17.15 E le m e n t
L e n g th
1
5
5 3 .1 3
2
4
90
o
c
s
0 .6
0 .8
0
1
o
c
2
2
cs
s
0 .3 6
0 .4 8
0 .6 4
0
0
1
( a ) Apply Eq.(17.14): AE [ k ]1 5
u1
v1
u2
0 .3 6
0 .4 8
0 .3 6
0 .4 8
0 .6 4
0 .4 8
0 .3 6
0 .4 8
0 .3 6
0 .4 8
0 .6 4
0 .4 8
v2
0 .4 8 0 .6 4 0 .4 8 0 .6 4
u1 v1 u2 v2
(CONT.)
295
17.15 (CONT.) u1
v1
u3
v3
0 AE 0 [ k ]2 4 0 0
0
0
1
0
0
0
1
0
0 1 0 1
u1 v1 u3 v3
( b ) Global Stiffness Matrix u1
v1
0 .7 5 6 1.0 0 8 0 .7 5 6 7 [K ] 10 1.0 0 8 0 0
u2
v2
u3
1.0 0 8
0 .7 5 6
1.0 0 8
0
3 .9 6 9
1. 0 0 8
1.3 4 4
0
1.0 0 8
0 .7 5 6
1.0 0 8
0
1.3 4 4
1. 0 0 8
1.3 4 4
0
0
0
0
0
0
0
0 2 .6 2 5
v3
2 .6 2 5 0 0 0 2 .6 2 5 0
u1 v1 u2 v2 u3 v3
(c) F1 x 100000
10
100 10
3
1 . 008 0 . 025 v1 3 . 969
0 . 756 1 . 008
7
(1 . 008 10 )( 0 . 025 ) 3 . 696 10 v 1 7
F 1 x ( 0 . 756 10
7
7
)( 0 . 025 ) 1 . 008 10
7
v1
or
or
v 1 0 . 00887
m
F 1 x 99 . 6 kN
( d ) Support reactions: F2x F2 y F 3x F3 y
10
7
0 . 756 1 . 008 0 0
1 . 008 99 . 59 1 . 344 0 . 025 132 . 79 0 . 00887 0 0 232 . 84 2 . 625
kN
( e ) Use Eq.(17.16): We have u 2 v 2 0 ,
F 12
F 13
AE
0 . 6
L1
AE L2
0
u3 v 3 0.
u1 105 10 0 . 8 5 v1 u1 105 10 1 4 v1
6
0
6
0 . 6
0 . 025 0 . 8 166 0 . 00887
0 . 025 1 232 . 8 kN 0 . 00887
296
kN
(C )
(T )
SOLUTION (17.16) We have AE 20 MN . Table P17.16 Data for the truss of Fig.P17.16 E le m e n t
L e n g th
1
5
3 6 .8 7
2 3
8
0
5
c
s
0 .8
0 .6
1 0 .8
o
o
1 4 3 .1 3
o
c
2
2
cs
s
0 .6 4
0 .4 8
0 .3 6
0
1
0
0
0 .6
0 .6 4
0 .4 8
0 .3 6
( a ) Apply Eq.(17.14): AE [ k ]1 5
AE [ k ]2 8
u1
v1
0 .6 4
0 .4 8
0 .6 4
0 .4 8
0 .3 6
0 .4 8
0 .6 4
0 .4 8
0 .6 4
0 .4 8
0 .3 6
0 .4 8
u1
v1
1
0
1
0
0
0
1
0
1
0
0
0
u2
AE [ k ]3 5
u3
u2
v2
0 .4 8 0 .3 6 0 .4 8 0 .3 6
u1 v1 u2 v2
v3
v2
0 0 0 0
u1 v1 u3 v3 u3
v3
0 .6 4
0 .4 8
0 .6 4
0 .4 8
0 .3 6
0 .4 8
0 .6 4
0 .4 8
0 .6 4
0 .4 8
0 .3 6
0 .4 8
0 .4 8 0 .3 6 0 .4 8 0 .3 6
u2
v2
u3
0 .1 2 5
u2 v2 u3 v3
( b ) Global Stiffness Matrix [K ] AE
u1
v1
0 .2 5 3
0 .0 9 6
0 .1 2 8
0 .0 9 6
0 .0 9 6
0 .0 7 2
0 .0 9 6
0 .0 7 2
0 .1 2 8
0 .0 9 6
0 .2 5 6
0
0 .0 9 6
0 .0 7 2
0
0 .1 1 4
0 .0 9 6
0 0 .1 2 8
0 .1 2 5
0
0 .1 2 8
0 .0 9 6
0 .2 5 3
0
0
0 .0 9 6
0 .0 7 2
0 .0 9 6
v3
0 0 .0 9 6 0 .0 7 2 0 .0 9 6 0 .0 7 2 0
u1 v1 u2 v2 u3 v3
(CONT.)
297
17.16 (CONT.)
(c)
F2 x F2 y F3 x
AE
0 0 .1 1 4 0 .0 9 6
0 .1 2 8 u 2 0 .0 9 6 v 2 0 .2 5 3 u 3
1
K F
1 AE
u2 v2 u 3
0 .2 5 6 0 0 .1 2 8
1 2 0 1 0
6
F1 x ( d ) F1 y A E F 3x
3 .8 9 2 9
6 .2 1 7 7 3 .8 9 2 7 4 .6 2 2 9 0 .1 2 8 0 .0 9 6 0 .0 9 6
1 5 .3 2 8 5 7 .7 8 5 9 0 .0 9 6 0 .0 7 2 0 .0 7 2
4 .6 2 2 9 7 .7 8 5 9 9 .2 4 5 2
0 .1 2 5 0. 0 .0 9 6
40000 80000 0
0 .0 2 8 0 0 .0 6 9 1 m 0 .0 4 0 4
0 .0 2 8 0 4 0 .0 0 8 0 .0 6 9 1 4 5 .7 4 4 0 .0 4 0 4 7 5 .6 9 6
kN
( e ) Use Eq.(17.16): We have u 1 v 1 v 3 0 . F1 2
F1 3
F3 2
AE
0 .0 2 8 0 0 .6 7 6 .2 4 0 .0 6 9 1
0 .8
5 AE
1
8 AE 5
0 .0 4 0 4 0 1 0 1 .0 0
0 .8
kN
kN
(C )
(T )
0 .0 2 8 0 0 .0 4 0 4 0 .6 1 2 6 .1 6 0 0 .0 6 9 1
kN
(C )
SOLUTION (17.17) Due to symmetry, only one-half of the beam need be considered. 12 6L EI [ k ]1 3 L 12 6L
P/2
L
2
6L 4L
2
12 6L
6 L 2L
12
2
6L
0
0
0
0
0
0
0
0
6L 2 2L 6 L 2 4L
1 1 2
k/2
0 3 E I kL E I [ k ]2 3 0 L 0
0 0 0 0
(CONT.)
298
17.17 (CONT.) Therefore 12 6L EI [K ] 3 L 12 6 L
12
6L 4L
6L 2 2L 6L 2 4 L
6 L
2
6L
kL
12
3
EI 2L
6 L
2
( a ) Boundary conditions are v1 0 and 2 0 . Equation (17.19a) with F 2 y P 2 and M 1 0 : 4L 0 EI 3 L 6L P 2
6L 1 3 kL 12 v2 E I
2
Introduce the data and solve: 1 5 .1(1 0
3
v 2 1 3 .5 (1 0
) ra d
3
) m
(b) 12 F1 y 6L 0 EI 3 F L 12 2 y M 2 6 L
12
6L 4L
6 L
2
6L
12
kL
3
EI 2L
6 L
2
6L 2 2L 6L 2 4 L
0 5 .1 3 (1 0 ) 1 3 .5 0
or F1 y 7 .4 2 5 k N F 2 y 9 .8 5 5 k N M 3 0 .1 5 0 k N m 2 F s p r in g 1 8 0 (1 3 .5 ) 2 .4 3 k N
(C )
From symmetry: F1 y F 3 y 7 .4 2 5 k N
SOLUTION (17.18) L 6 .7 m ,
P = 9 kN , v1
12 6L EI [ k ]1 3 L 12 6L
1 2
6 L 2L
5
u2
6L 4L
E I = 6 5 (1 0 ) N m ,
2
12 6L 12 6L
4
2
k 210 kN m v2
6L 2 2L 6 L 2 4L
0 3 E I kL E I [ k ]2 3 0 L 0
2
u3 3
0
0
0
0
0
0
0
0
0 0 0 0
(CONT.)
299
17.18 (CONT.) Thus 12 6L EI [K ] 3 L 12 6 L
12
6L 4L
6L
6L 2 2L 6L 2 4 L
6 L
2
kL
12
3
EI 2L
6 L
2
(1)
( a ) Boundary conditions are v1 0 and 1 0 . Equation (17.19a), with F 2 y P and M
2
0: kL P E I 1 2 EI 3 L 0 6 L
6 L v2 2 2 4 L
3
Substituting the given numerical values and solving, we have v 2 0 .0 3 2 7 m
2 0 .0 0 7 3 3 ra d
(b) 12 F1 y 6L EI M1 3 F L 12 2 y M 2 6 L
12
6L 4L
6L
2
6L
12
kL
3
EI 2L
6L
2
6L 2 2L 6 L 2 4 L
0 0 0 .0 3 2 7 0 .0 0 7 3 3
Introducing the data and multiplying: 2 .1 3 8 k N F1 y M 1 1 4 .2 4 5 k N m F2 y 9 .0 0 5 k N M 0 .0 8 1 k N m 2
The spring force is Ps p r in g 2 1 0 ( 0 .0 3 2 7 ) 6 .8 6 7 k N
(C )
SOLUTION (17.19) We have E I 7 0 1 0 N m , ( a ) Use Eqs.(17.19): v1 1 v2 4
12 18 EI [ k ]1 3 L 12 18
2
18
12
36
18
18
12
18
18
L 3 m,
P 50 kN
2 18 18 18 36
v1
1 v2
2
(CONT.)
300
17.19 (CONT.) v2
2
v3
12 18 EI [ k ]2 3 L 12 18
18
12
36
18
18
12
18
18
( b ) Global Stiffness Matrix v1 1 12 18 EI 12 [K ] 3 L 18 0 0
3 18 18 18 36
v2
2 v3
3
v2
2
v3
3
18
12
18
0
36
18
18
0
18
24
0
12
18
0
72
18
0
12
18
12
0
18
18
18
0 0 18 18 18 36
v1
1 v2
2 v3
3
Use Eq.(17.20a): F2 y 24 EI M 2 3 0 L M 18 3
3
L EI
18 v 2 18 2 36 3
0 72 18
1
K F
v2 3 L 2 EI 3
0 .0 7 2 9 1 7 0 .0 1 0 4 1 7 0 .0 4 1 6 6 7
0 .0 1 0 4 1 7 0 .0 1 7 3 6 1 0 .0 1 3 8 8 9
0 .0 4 1 6 6 7 0 .0 1 3 8 8 9 0 .0 5 5 5 5 6
50 10 0 0
3
0 .1 4 0 6 m 0 .0 2 0 1 r a d 0 .0 8 0 4 r a d
(c) F1 y EI M1 3 L F 3 y
12 18 1 2
18 18 18
0 0 1 8
v 2 3 4 .3 6 2 k N 2 5 6 .2 3 3 k N m 3 1 5 .6 0 2 k N
From a free-body diagram of element 2: ( M 2 ) 2 4 6 .8 0 6 k N m and ( F 2 ) 2 1 5 .6 0 2 k N
(CONT.)
301
17.19 (CONT.)
(d) 50 kN 2
1 1
1.5 m
3
1.5 m
2
34.362
V
+
(kN )
x 15.602
M
46.806
(kN m )
+
x
56.28
SOLUTION (17.20) ( a ) The element stiffness matrices are, from Eq. (17.19a): v1
1
v2
2
v3
12 6L EI 12 [ k ]1 3 L 6L 0 0
6L
12
6L
0
6 L
2L
12
6 L
6 L
4L
0
0
0
0
0
0
0
0
0 0 0 EI [ k ]2 3 0 L 0 0
4L
2
6 L 2L
2
2
0
2
0
0
0
0
0
0
0
0
0
12
6L
12
0
6L
4L
12
12
kL
3
EI 0
6L
2L
6 L
2
0 6L 2 2L 6 L 2 4 L 0
6 L
2
6 L
0 0 0 0 0 0
0
0
0
3
(CONT.)
302
17.20 (CONT.) (b) 12 6L 12 [K ] 6L 0 0
12
6L 4L
2
6L
2L
24
0
0
8L
6L 2L
2
12
0
6L
0
2
0 12 6L
2
6 L
12
kL
3
EI 0
6L
2L
2
0 6L 2 2L 6 L 2 4 L 0
6 L
( c ) Boundary conditions are: v1 0 , 1 0 , v 2 0 System governing equations, by Eqs. (17.19a), after rearrangement: 8 L2 0 EI 2 0 3 2L L P 6L
2L 4L
2
2
6L
6L 6L 3 kL 12 EI
2 3 v 3
Solving, v3
2
where k 1 k L
7PL
3
(
EI 3PL
3
EI
1 1 2 7 k1
2
(
1 1 2 7 k1
)
),
3 3 2
EI
SOLUTION (17.21) Boundary conditions are v1 1 v 3 3
v 2 (given)
Equation (17.19): 24 EI 12 L [ k ]1 3 L 24 1 2 L
12 L 8L
2
12 L 4L
2
24 12 L 24 12 L
12 L 2 4L 12 L 2 8L
12 6L EI [ k ]2 3 L 12 6L
6L 4L
2
6L 2L
2
12 6L 12 6L
6L 2 2L 6L 2 4L
After assembling [ K ] and considering the boundary conditions, the pertinent equations are found as (CONT.)
303
17.21 (CONT.) P EI 3 L 0
24 12 1 2 L 6 L
12 L 6 L 2 2 8L 4L 2
Multiplying, EI
P
( 3 6 6 L 2 )
(1)
(6 L 12 L 2 )
(2)
L 0
EI L
3
3
2
( a ) Equation (1) is then P 33
EI L
3
( b ) Equation (2) gives 2
2L
SOLUTION (17.22) through (17.26) It is important to take into account any conditions of symmetry which may exist. Use a 2-D finite element program such as ANSYS. End of Chapter 17
304
CHAPTER 18
CASE STUDIES IN MACHINE DESIGN
SOLUTION (18.1) a2=0.16 m a1=2.6 m
B
P=15 kN
L=2.5 m
C
H
D 40o
a3=1.0 m
80o FBG
A
FC F
60
o
o
50
Figure S18.1 Free Body diagram of loader arm FAE
( a ) Member forces Dismember the arm ABD. It is assumed that the links and hydraulic cylinder are all in tension. The conditions of moment equilibrium are applied to Fig. S18.1:
M
0:
B
F C F c o s 1 0 ( 0 .1 6 ) F A E s in 5 0 (1 .0 ) 1 5 ( 2 .7 6 ) 0 o
o
F A E 0 .2 0 6 F C F 5 4 .0
Fx 0 :
(1)
F A E c o s 6 0 F B G c o s 4 0 F C F s in 1 0 0 o
o
o
F B G 3 5 .2 4 6 0 .3 6 1 F C F
Fy 0 :
(2)
F A E s in 6 0 F B G c o s 4 0 F C F c o s 1 0 1 5 0 o
o
o
FC F 4 2 k N ( C )
Substitution of this into Eqs. (1) and (2) result in F A E 4 5 .3 5 k N ( T )
F B G 2 0 .0 8 k N ( C )
Comments: Since the result obtained for F A E is positive. A negative sign means that the sense of the force is opposite to that taken originally. ( b ) Diameters of pins at A, B, and C in double shear. We have S ys n
F 2
d
2
4
d [ 2 SF n ]
,
1 2
ys
Thus, dA [
2 ( 4 5 .3 5 ) ( 2 .4 )
dB [
2 ( 2 0 .0 8 ) ( 2 .4 )
dC [
3
(1 5 0 1 0 )
3
(1 5 0 1 0 ) 2 ( 4 2 ) ( 2 .4 ) 3
(1 5 0 1 0 )
1
]
2 1 .5 m m
2
1 4 .3 m m
1
] 1
]
2
2
2 0 .7 m m
305
SOLUTION (18.2) The location of the critical point is at K, where the maximum moment and the shear force are M P L 1 5 ( 2 .5 ) 3 7 .5 N m
V P 15 kN
The cross-sectional area properties: A ( c 2 c 1 ) ( 7 5 5 0 ) 9 .8 1 7 5 1 0 2
I
4
2
2
( c 2 c1 ) 4
4
2
3
( 7 5 5 0 ) 1 9 .9 4 1 8 1 0 4
4
4
mm 6
2
mm
4
The maximum tensile stress due to the bending occurs at point K. Therefore,
m ax
M c2
I
3
3 7 .5 (1 0 )( 0 .0 7 5 ) 1 9 .9 4 1 8 (1 0
6
)
141 M Pa
The shearing stress is zero, 0 , at point K. The maximum shearing stress is at the neutral axis z and parallel to y axis. From third case, Table 3.2: m ax 2
V A
3
1 5 (1 0 )
2
9 .8 1 7 5 (1 0
3
)
3 .0 6 M P a
This is a very low stress for the specified material. The bending stress vanishes at the neutral axis, H 0 . The factor of safety is thus n
3 .4
480 141
SOLUTION (18.3) A sketch of Mohr's circle is shown in Fig. S18.3 constructed by obtaining the position of point C at ( x y ) 2 4 0 0 μ on the horizontal axis and of point at ( x ,
xy
2 ) (1 0 0 0 μ , 3 5 0 μ ) from the origin O.
The principal strains are represented by points A1 and B1 . Thus, referring to the figure:
( 1 0 0 02 2 0 0 ) 3 5 0
400
2
1 ,2
2
'=400 D
y B(-200, 350) B1 O
”s C
’p
A1
A(1000, -350) x
E Figure S18.3 or 1 1 0 9 4 .6 μ ,
2 2 9 4 .6 μ
(CONT.)
306
18.3 (CONT.) As a check, note that x y 1 2 8 0 0 . The planes of principal strains are 2
1 350 600
' ta n
p
3 0 .3
o
2 p " 3 0 .3 1 8 0 2 1 0 .3
and
o
and p ' 1 5 .1 5
o
p " 1 0 5 .1 5
and
o
(1)
From Mohr's circle, p ' locates the 1 direction. The maximum shearing strains are given by points D and E: m ax 2
( 1 0 0 02 2 0 0 ) 3 5 0 2
1389
2
Alternatively, 1 2 1 0 9 4 .6 2 9 4 .6 1 3 8 9 μ . SOLUTION (18.4) From Prob. 18.3, we have 1 1 0 9 4 .6 μ , 2 2 9 4 .6 μ , m a x 1 3 8 9 μ The first two of Eqs. (2.7) together with (1) give 1
2
2 1 0 1 0
3
1 ( 0 .2 8 )
2
2 1 0 1 0
3
1 ( 0 .2 8 )
(1)
[1 0 9 4 .6 0 .2 8 ( 2 9 4 .6 ) ] 2 3 1 M P a
2
[ 2 9 4 .6 0 .2 8 (1 0 9 4 .6 ) ] 2 .7 1 M P a
From the last of Eqs. (2.7): m ax
E 2 (1 )
m ax
3
2 1 0 1 0 2 (1 0 .2 8 )
p ' 1 5 .1 5
From Prob. 18.3:
o
(1 3 8 9 ) 1 1 4 M P a
s 6 0 .1 5
and
o
Using Eq. (3.35), ' 12 ( 2 3 1 2 .7 1) 1 1 6 .9 M P a SOLUTION (18.5) Use Eq. (3.40): a x c o s a y s in a 2
1 1 0 4 (1 0
6
2
xy
s in a c o s a
) x c o s 0 y s in 0 2
o
2
o
o
xy
x 1104 μ
o
s in 0 c o s 0 ,
Similarly, b x c o s b y s in b 2
4 3 2 (1 0
4 3 2 (1 0
6
6
2
xy
s in b c o s b
) x c o s ( 6 0 ) y s in ( 6 0 ) 2
o
) 0 .2 5 x 0 .7 5
2
y
0 .4 3 3
o
xy
s in ( 6 0 ) c o s ( 6 0 ) o
xy
o
(1)
and c x c o s c y s in c 2
9 6 (1 0
6
2
) 0 .2 5 x 0 .7 5
y
xy
s in c c o s c
0 .4 3 3
Subtract Eq. (2) from Eq. (1): 5 2 8 0 .8 6 6
Thus
xy
610 μ ,
xy
y
144 μ
307
xy
(2)
SOLUTION (18.6) Equation (5.78a): 2 E Sy
Cc
2
2
3
( 2 1 0 1 0 )
1 2 8 .8
250
For the 1.6 m link column, L B G r 1 6 0 0 1 0 .6 9 1 4 9 .7 C c
and Eq. (5.77b) is used. Hence,
a ll
2
E 1 .9 2 ( L B G r )
2
9
( 2 1 0 1 0 )
1 .9 2 (1 4 9 .7 )
2
4 8 .2 M P a
Comment: This stress is very low compared to 250 MPa; The link will not yield.
SOLUTION (18.7) From Prob. 18.6: A 3 1 8 m m ,
E 200 G Pa ,
2
r 1 0 .6 9 m m .
The required value of a ll :
a ll
FB G
A
3
1 1 .3 4 (1 0 ) 3 1 8 (1 0
6
3 5 .7 M P a
)
(1)
Assume L m r C c . Equation (5.77b):
a ll
2
E
1 .9 2 ( L m r )
2
2
9
( 2 0 0 1 0 )
1 .9 2 ( L m r )
Lm
Equating Eqs. (1) and (2), we obtain
(2)
2
r
1 6 9 .7
Since L m r C c , our assumption is OK. Thus Lm r
Lm 1 0 .6 9
1 6 9 .7
or L m 1 .8 1 4 m
Comment: If this link is more than 1.814 m in length, it will buckle.
SOLUTION (18.8) ( a ) Central cross brace. Stain energy due to bending. Equation (5.18) with M W x 2 : U 2[
c 2 M 2 EI
0
dx]
1 EI
c 2 0
W 4
2
x dx 2
2
3
W c 96 EI
Due to shear, Eq. (5.23) with V W 2 : U
c 0
2
3V 5GA
dx
2
3W c 20 G A
Side supports. Strain energy owing to bending, with M W x 2 : U 4[
b 2 0
2
M 2 EI
dx]
2 EI
b 2 0
2
W 16
x dx 2
2
3
W b 384 EI
(CONT.)
308
18.8 (CONT.) Due to shear with V W 4 : U 2
b
2
2
dx
3V 5GA
0
3W b 40 G A
Total strain energy is then Ut
2
3
W c 96 EI
2
3
W b 384 EI
2
2
3W c 20 G A
2W b 40 G A
or 3
3
U t W [ E1I ( 9c 6 2
b 384
)
(c
3 20 G A
b 2
Q.E.D.
)]
(P18.8a)
( b ) Introducing Eq. (P18.8a) into Eq. (5.31), we obtain st
U t
3
W [ 4 81E I ( c 3
W
b 4
)
(c
3 10 G A
b 2
(P18.8b)
)]
SOLUTION (18.9) ( a ) Substituting the given data into Eq. (P18.8b) result in s t W [ 4 81E I ( c 3
W{
)
3
4 8 ( 2 0 0 1 0 0 .2 7 8 )
6
6
(c
3 10 G A
[( 0 .8 ) 3
1
W (1 0 10
3
b 4
b 2
)]
(1 .2 )
3
4
]
3 3
1 0 ( 7 9 1 0 8 1 9 )
[( 0 .8 )
1 .2 2
]}
){0 .3 7 4 7[ 0 .5 1 2 0 0 .4 3 2 ] [ 0 .0 0 4 6 4[1 .4 ]}
W {0 .3 5 3 7 0 .0 0 6 5} 1 0
6
W ( 0 .3 6 0 2 )
or s t 0 .3 6 0 2 (1 0 5 .4 0 3(1 0
3
6
) W ( 0 .3 6 0 2 1 0
6
)(1 5 1 0 ) 3
) m
( b ) The impact factor, by Eq. (4.32): K 1
1
2h
st
1
2 ( 250 )
1
1 0 .6 7
5 .4 0 3
Equation (4.35) is therefore m a x K s t 1 0 .6 7 (5 .4 0 3) 5 7 .7 m m
SOLUTION (18.10) ( a ) Thin-walled cylinder is in biaxial stress ( 3 0 ) with 1 and a 2 . From Eqs. (3.6) at r a : 1
pa t
20 (350 )
t
7000 t
,
Sy
2
2
1 2
3500 t
(1)
Equation (6.14): 1 1 2
2
2 2
(
n
)
(2)
Substituting Eqs. (1) into (2), we have 6
[( 7 ) ( 7 )(3 .5 ) (3 .5 ) ] 1 02 ( 5 55 2 ) , 2
2
2
t
t 55 m m
Hence, b a t 405 m m
( b ) We have a t 6 .3 6 : the thin-walled analysis does not apply. So, the solution is not valid.
309
SOLUTION (18.11) Refer to Prob. 18.10. For a closed-ended thick-walled cylinder under internal pressure, the critical section where the maximum stresses occur is at r a (see Fig. 16.3). The stresses, from Eqs. (16.16a), 16.16b), and (16.15) with p p i and p o 0 : r p p
p
a
2
2
2
2
b a b a a
(1)
2
2
b a
2
We observe that 1 , a 2 , and r 3 , where algebraically 1 2 3 . With a safety factor included, Eq. (6.13) appears as [ ( 1 2 ) ( 2
3 ) ( 2
2
1) ] 2( 2
3
Sy n
(2)
)
Substituting Eq. (1) into (2), we have [(b a a ) ( a b a ) ( b a b a ) ]( 2
or
4
3b p
2
2
2
2
2
(b a ) ( 2
2
2
Sy n
2
)
2
2
2
2
2
2
p
2
2
b a
) 2( 2
2
2
Sy n
)
2
(3)
By introducing the given data ( a 0 .3 5 m , S y 5 5 2 M P a , p 2 0 M P a , n 5 ) and simplifying Eq. (3) becomes: 9 .1 5 7 b 2 .4 8 8 b 0 .1 5 2 0 4
2
Let x b and solve the resulting quadratic equation to find x 0 .1 7 7 8 . The outer radius is then 2
b 0 .4 2 1 7 m = 4 2 1 .7 m m
Hence, t b a 4 2 1 .7 3 5 0 7 1 .7 m m Comment: The outer diameter equals 2 b 8 4 4 m m . A standard cylinder with about 8 4 5 m m outer diameter and 7 0 0 m m inner diameter should be selected.
SOLUTION (18.12) ( a ) Element Stiffness Matrix. Referring to Fig. P18.12 we sketch Fig. S18.12. F1 y , v 1
L
1 F1 x , u 1
45o
F3 x , u 3
2
F2 x , u 2
F2 y , v 2
3 F3 y , v 3
Figure S18.12 Finite element model (CONT.)
310
18.12 (CONT.) Using Eq. (17.4), Fig. S18.12, Table P18.1, and the given data: u1
v1
c cs 7 [ k ]1 4 (1 0 ) c2 c s 2
u1 0 0 7 [ k ] 2 4 (1 0 ) 0 0
u2 c
cs s
cs
c
s
cs
2
u3
0
0
1
0
0
0
1
0
u1
cs 2 s cs 2 s
2
cs
2
v1
v2
2
1 v1 0 7 4 (1 0 ) 1 u2 v2 0 u1
v3
v1
u2
v2
0
1
0
0
0
1
0
0
0 0 0 0
u2
0 1 0 1
0 .5 0 .5 7 2 (1 0 ) 0 .5 0 .5
u1 v1
[ k ]3 2
u3 v3
u1 v1 u2 v2
v2
u3
v3
0 .5
0 .5
0 .5
0 .5
0 .5
0 .5
0 .5
0 .5
0 .5 0 .5 0 .5 0 .5
u2 v2 u3 v3
In the foregoing, the column and row of each stiffness matrix are labeled according to the nodal displacements associated with them. Observe that displacements u 3 and v 3 are not involved in element 1; u 2 and v 2 are not involved in element 2; u 1 and v1 are not involved in element 3. Thus, before adding [ k ]1 , [ k ] 2 , and [ k ] 3 to form the system matrix, two row and columns of zero must be added to each of the element matrices to account for the absence of these displacements. 7
In so doing, and using a common factor 1 0 , the element stiffness matrices become: u1
v1
u2
v2
u3
4 0 4 7 [ k ]1 (1 0 ) 0 0 0
0
4
0
0
0
0
0
0
0
4
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0 0 0 0 0 0
u3
v3
u1
v1
u2
v2
0 0 0 7 [ k ] 2 (1 0 ) 0 0 0
0
0
0
0
4
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
4
0
0
0
v3
0 4 0 0 0 4
u1 v1 u2 v2 u3 v3
u1 v1 u2 v2 u3 v3
(CONT.)
311
18.12 (CONT.) u1 0 0 0 7 [ k ] 3 (1 0 ) 0 0 0
v1
u2
v2
u3
0
0
0
0
0
0
0
0
0
0
0
2
2
v3
2
2
0
2
2
2
2
0
2
2
2
2
0
2
2
2
2
u1 v1 u2 v2 u3 v3
System Stiffness Matrix. There are a total of six components of displacement for the truss before boundary constraints are imposed. Hence, the order of the truss stiffness matrix must be 6 6 . Before addition of the terms from each element stiffness matrix into their corresponding locations in [ K ] , we obtain the global matrix for the truss: u1 4 0 4 7 [ K ] (1 0 ) 0 0 0
v1
u2
v2
u3
0
4
0
0
4
0
0
0
4
0
2
2
2
0
2
2
2
4
2
2
2
4 2 2 2 4 2 0
0
2
v3
2
2
u1 v1 u2
(1)
v2 u3 v3
System Force-Displacement Relationship. From Fig. P18.12, the boundary conditions are u 1 0 , v1 0 , and u 3 0 . In addition F 3 y 0 . Equation (17.17 is therefore R1 x R 1y P W R 3x 0
[K
0 0 u 2 ] v 2 0 v 3
(2)
where [ K ] is given by Eq. (1). ( b ) Displacements. To find u 2 , v 2 , and v 3 only part of Eq. (2) relating to these displacements is considered: (CONT.)
312
18.12 (CONT.) (4 2 ) 24, 000 7 3 6 , 0 0 0 (1 0 ) 2 0 2
P W 0
2 2
2 (4 2 ) 2
2
u2 v2 v 3
(3)
Inverting, u2 8 v 2 2 .5 (1 0 ) v 3
1
1 1 0
0 1 1
4 .8 2 8 1
24 1 .5 3 3 6 (1 0 ) 4 .9 4 m m 0 0 .9
Reactions. Inserting of the preceding values of u 2 , v 2 , and v 3 into Eq. (2) gives the reactional forces: R1 x R1 y R 3x
4 0 2
7 10
0 0 2
0 1 .5 60 3 4 4 .9 4 (1 0 ) 3 6 k N 36 2 0 .9
The results may be verified by applying the equations of equilibrium to the free-body diagram of the entire truss, Fig. SP18.12. Axial Forces in Bars. Using Eqs. (17.16), (3) and Table P18.12, we obtain F1 F1 2
AE L
[1
F 2 F1 3
AE L
[0
F3 F 2 3
AE 2L
u2 7 0 ] 4 (1 0 ) [1 v2
1 .5 3 0] (1 0 ) 6 0 k N 4 .9 4
0 7 1] 4 (1 0 ) [ 0 v 3 u2 1] 2 v3 v2
[1
0 3 1] (1 0 ) 3 6 k N 0 .9 2 (1 0 ) [ 1 7
1 .5 3 1] (1 0 ) 0 .9 4 .0 4
7 1 .8 k N
Stresses in Bars. Driving the element forces by the cross-sectional area results in 1
3
6 0 (1 0 ) 4 8 0 (1 0
6
)
125 M Pa
2
1 2 5 ( 36 60 ) 7 5 M P a
3 1 2 5 ( 7610.8 ) 1 4 9 .6 M P a
The negative sign means a compressive stress. Factor of safety against yielding. Dividing the yield strength of S y 2 5 0 M P a (from Table B.1) by each stress, we obtain n1
250 125
2
n2
250 75
3 .3 3
n3
250 1 4 9 .6
1 .6 7
Comments: The bar axial stresses found are relatively low for the well known material considered (see Sec. 1.6). End of Chapter 18
313
SECOND EDITION: SI VERSION by
ANSEL C. UGURAL
SOLUTIONS MANUAL FOR MECHANICAL DESIGN OF MACHINE COMPONENTS SECOND EDITION: SI VERSION
by
ANSEL C. UGURAL
Boca Raton London New York
CRC Press is an imprint of the Taylor & Francis Group, an informa business
MATLAB® is a trademark of The MathWorks, Inc. and is used with permission. The MathWorks does not warrant the accuracy of the text or exercises in this book. This book’s use or discussion of MATLAB® software or related products does not constitute endorsement or sponsorship by The MathWorks of a particular pedagogical approach or particular use of the MATLAB® software.
Taylor & Francis Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2016 by Taylor & Francis Group, LLC Taylor & Francis is an Informa business No claim to original U.S. Government works Printed on acid-free paper Version Date: 20160303 International Standard Book Number-13: 978-1-4987-3541-4 (Ancillary) This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com
CONTENTS
Part I
BASICS
Chapter 1
INTRODUCTION
1
Chapter 2
MATERIALS
16
Chapter 3
STRESS AND STRAIN
24
Chapter 4
DEFLECTION AND IMPACT
48
Chapter 5
ENERGY METHODS AND STABILITY
68
Part II
FAILURE PREVENTION
Chapter 6
STATIC FAILURE CRITERIA AND RELIABILITY
100
Chapter 7
FATIGUE FAILURE CRITERIA
117
Chapter 8
SURFACE FAILURE
135
Part III
APPLICATIONS
Chapter 9
SHAFTS AND ASSOCIATED PARTS
145
Chapter 10
BEARINGS AND LUBRICATION
164
Chapter 11
SPUR GEARS
176
Chapter 12
HELICAL, BEVEL, AND WORM GEARS
194
Chapter 13
BELTS, CHAINS, CLUTCHES, AND BRAKES
208
Chapter 14
MECHANICAL SPRINGS
225
Chapter 15
POWER SCREWS, FASTENERS, AND CONNECTIONS
240
Chapter 16
MISCELLANEOUS MACHINE COMPONENTS
261
Chapter 17
FINITE ELEMENT ANALYSIS IN DESIGN
278
Chapter 18
CASE STUDIES IN MACHINE DESIGN
308
vi
NOTES TO THE INSTRUCTOR
The Solutions Manual to accompany the text MECHANICAL DESIGN of Machine Components supplements the study of machine design developed in the book. The main objective of the manual is to provide efficient solutions for problems in design and analysis of variously loaded mechanical components. In addition, this manual can serve to guide the instructor in the assignment of problems, in grading these problems, and in preparing lecture materials as well as examination questions. Every effort has been made to have a solutions manual that cuts through the clutter and is self –explanatory as possible thus reducing the work on the instructor. It is written and class tested by the author. As indicated in its preface, the text is designed for the junior-senior courses in machine or mechanical design. However, because of the number of optional sections which have been included, MECHANICAL DESIGN of Machine Components may also be used to teach an upper level course. In order to accommodate courses of varying emphases, considerably more material has been presented in the book than can be covered effectively in a single three-credit-hour course. Machine/mechanical design is one of the student’s first courses in professional engineering, as distinct from basic science and mathematics. There is never enough time to discuss all of the required material in details. To assist the instructor in making up a schedule that will best fit his classes, major topics that will probably be covered in every machine design course and secondary topics which may be selected to complement this core to form courses of various emphases are indicated in the following Sample Assignment Schedule. The major topics should be covered in some depth. The secondary topics, because of time limitations and/or treatment on other courses, are suggested for brief coverage. We note that the topics which may be used with more advanced students are marked with asterisks in the textbook. The problems in the sample schedule have been listed according to the portions of material they illustrate. Instructor will easily find additional problems in the text to amplify a particular subject in discussing a problem assigned for homework. Answers to selected problems are given at the end of the text. Space limitations preclude our including solutions to open-ended web problems. Since the integrated approach used in this text differs from that used in other texts, the instructor is advised to read its preface, where the author has outlined his general philosophy. A brief description of the topics covered in each chapter throughout the text is given in the following. It is hoped that this material will help the instructor in organizing his course to best fit the needs of, his students. Ansel C. Ugural Holmdel, N.J.
vii
DESCRIPTION OF THE MATERIAL CONTAINED IN “MECHANICAL DESIGN of Machine Components”
Chapter 1 attempts to present the basic concepts and an overview of the subject. Sections 1.1 through 1.8 discuss the scope of treatment, machine and mechanical design, problem formulation, factor of safety, and units. The load analysis is normally the critical step in designing any machine or structural member (Secs. 1.8 through 1.9). The determination of loads is encountered repeatedly in subsequent chapters. Case studies provide a number of machine or component projects throughout the book. These show that the members must function in combination to produce a useful device. Section 1.10 review the work, energy, and power. The foregoing basic considerations need to be understood in order to appreciate the loading applied to a member. The last two sections emphasize the fact that stress and strain are concepts of great importance to a comprehension of design analysis. Chapter 2 reviews the general properties of materials and some processes to improve the strength of metals. Sections 2.3 through 2.14 introduce stress-strain relationships, material behavior under various loads, modulus of resilience and toughness, and hardness, selecting materials. Since students have previously taken materials courses, little time can be justified in covering this chapter. Much of the material included in Chapters 3 through 5 is also a review for students. Of particular significance are the Mohr’s circle representation of state of stress, a clear understanding of the three-dimensional aspects of stress, influence of impact force on stress and deformation within a component, applications of Castigliano’s theorem, energy of distortion, and Euler’s formula. Stress concentration is introduced in here, but little applications made of it until studying fatigue (Chap.7). The first section of Chapter 6 attempts to provide an overview of the broad subject of “failure”, against which all machine/mechanical elements must be designed. The discipline of fracture mechanics is introduced in Secs. 6.2 through 6.4. Yield and fracture criteria for static failure are discussed in Secs. 6.4 through 6.12. The last 3 sections deal with the method of reliability prediction in design. Chapter 7 is devoted to the fatigue and behavior of materials under repeated loadings. The emphasis is on the Goodman failure criterion. Surface failure is discussed in Chapter 8. Sections 8.1 through 8.3 briefly review the corrosion and friction. Following these the surface wear is discussed. Sections 8.6 through 8.10 deal with the surfaces contact stresses and the surface fatigue failure and its prevention. The background provided here is directly applied to representative common machine elements in later chapters. Sections 9.1 through 9.4 of Chapter 9 treat the stresses and design of shafts under static loads. Emphasis is on design of shafts for fluctuating loading (Secs. 9.6 and 9.7). The last 5 sections introduce common parts associated with shafting. Chapter 10 introduces the lubrication as well as both journal and roller bearings. As pointed out in Sec. 8.9, rolling element bearings provide interesting applications of contact stress and fatigue. Much of the material covered in Secs. 11.1 through 11.7 of Chapter 11 introduce nomenclature, tooth systems, and fundamentals of general gearing. Gear trains and spur gear force analysis are taken up in Secs. 11.6 and 11.7. The remaining sections concern with gear design, material, and manufacture. Non-spur gearing is considered in Chapter 12. Spur gears are merely a special case of helical gears (Secs. 12.2 through 12.5) having zero helix angle. Sections 12.6 through 12.8 deal with bevel gears. Worm gears are fundamentally different from other gears, but have much in common with power screws to be taken up in Chap. 15.
viii
Chapter 13 is devoted to the design of belts, chains, clutches, and brakes. Only a few different analyses are needed, with surface forms effecting the equations more than the functions of these devices. Belts, clutches, and brakes are machine elements depending upon friction for their function. Design of various springs is considered in Chapter 14. The emphasis is on helical coil springs (Secs. 14.3 through 14.9) that provide good illustrations of the static load analysis and torsional fatigue loading. Leaf springs (Sec. 14.11) illustrate primarily bending fatigue loading. Chapter 15 attempts to present screws and connections. Of particular importance is the load analysis of power screws and a clear understanding of the fatigue stresses in threaded fasteners. There are alternatives to threaded fasteners and riveted or welded joints. Modern adhesives (Secs. 15.17 and 15.18) can change traditional preferred choices. It is important to assign at least portions of the analysis and design of miscellaneous mechanical members treated in Chapter 16. Sections 16.3 through 16.7 concern with thick-walled cylinders, press or shrink fits, and disk flywheels. The remaining sections concerns with the bending of curved frames, plate and shells-like machine and structural components, and pressure vessels. Buckling of thin-walled cylinders and spheres is also briefly discussed. Chapter 17 represents an addition to the material traditionally covered in “Machine/Mechanical Design” textbooks. It attempts to provide an introduction to the finite element analysis in design. Some practical case studies illustrate solutions of problems involving structural assemblies, deflection of beams, and stress concentration factors in plates. Finally, case studies in preliminary design of the entire crane with winch and a high-speed cutting machine are introduced in Chapter 18.
ix
SAMPLE ASSIGNMENT SCHEDULE
MACHINE/MECHANICAL DESIGN (3 credits.)
Prerequisites:
A. C. Ugural, MECHANICAL DESIGN of Machine Components, 2nd SI Version, CRC Press (T & F Group). Courses on Mechanics of Materials and Engineering Materials.
WEEK
TOPICS
TEXT:
SECTIONS
PROBLEMS
1
Introduction Materials*
1.1 to 1.12 2.1 to 2.5,2.8 to 2.11
1.6,1.17,1.26,1.34 2.7,2.11,2.15,2.20
2
A Review of Stress Analysis*
3.1 to 3.14, 4.1 to 4.9 5.1 to 5.12
3.12,3.34,3.45,4.24,4.37 5.20,5.26,5.30,5.54, 5.73
3
Static Failure Criteria & Reliability
6.1 to 6.15
6.7,6.14,6.25,6.29,6.40
4
Fatigue Criteria and Surface Failure
7.1 to 7.15, 8.1 to 8.10
7.6,7.17,7.28,7.32,7.34 8.3,8.8, 8.14, 8.21
5
EXAM # 1 Shafts and Associated Parts
- - - - 9.1 to 9.12
- - - - 9.7,9.14,9.19,9.24,9.28
6
Lubrication and Bearings
10.1 to 10.16
10.4,10.10,10.16,10.31,10.36
7
Spur Gears
11.1 to 11.12
11.15,11.18,11.22,11.33,11.38
8
Helical, Bevel, and Worm Gears*
12.1 to 12.10
12.10,12.12,12.18,12.21,12.32
9
EXAM # 2 Belt and Chain Drives
- - - - 13.1 to 13.6
- - - - 13.1,13.8,13.9, 13.10,13.13
10
Clutches and Brakes
13.8 to 13.15
13.21,13.27,13.34,13.39,13.46
11
Mechanical Springs
14.1 to 14.11
14.8,14.12,14.22,14.25,14.35
12
EXAM # 3 Power Screws and Fasteners
- - - - 15.1 to 15.12
- - - - 15.2,15.6,15.14,15.19,15.30
13
Connections*
15.13 to 15.16
15.34,15.40,15.48,15.54
14
Miscellaneous Machine Components*
16.1 to 16.5, 16.8 to 16.11
16.6,16.17,16.25,16.38
15
FEA in Design and Case Studies in Machine Design* FINAL EXAM
17.1 to 17.3,17.5,17.7 18.1 and 18.2 - - - - -
17.2,17.4,17.8,17.12 18.2,18.4,18.6,18.10 - - - - -
* Secondary topics. The remaining, major topics constitute the “main stream” of the machine design course.
x
Section I
BASICS
CHAPTER 1
INTRODUCTION
SOLUTION (1.1) Free Body: Angle Bracket (Fig. S1.1) (a)
(b)
M
B
0;
F ( 0 .5 ) 6 8 ( 0 .2 5 ) 2 8 .8 ( 0 .5 ) 0 ,
Fx 0 :
R B x 2 8 .8 F 0 ,
Fy 0 :
R B y 6 8 2 1 .6 0 ,
F 6 2 .8 k N
R Bx 3 4 k N R B y 8 9 .6 k N
Thus RB
(3 4 ) (8 9 .6 ) 2
2
9 5 .8 k N
and ta n
1
34 8 9 .6
2 0 .8
o
F
A 0.5
34
21.6 kN
68 kN
B 89.6
R Bx
Figure S1.1
28,8 kN
B R By
C
RB
0.25
0.25
SOLUTION (1.2) Free Body: Beam ADE (Fig. S1.2)
M
A
0:
W (3 .7 5 a ) F B D ( 2 a ) 0 ,
E D
W
FBD
A 2a
1.75a
R Ay
a
Fx 0 :
R Ax 0
Fy 0 :
R Ay FBD W 0 ,
Free-Body: Entire structure (Fig. S1.2)
V
M
A
0:
R C (3 a ) W (3 .7 5 a ) 0 , R C 1 .2 5 W
F O
0.875 kN
R A y 0 .8 7 5W
R Ax
A
F B D 1 .8 7 5 W
M”
Free Body: Part AO (Fig. S1.2)
Figure S1.2
V 0 .8 7 5W F 0
M 0 .8 7 5 a W
1
SOLUTION (1.3) 29 kN/m
C
D
32 kN m A
0.6 m
B
E
RA
1,2 m
M
A
0:
Fy 0 :
R B 1 1 .0 2 k N R A 4 5 .8 2 k N
RB
2.4 m
2.4 m
Segment CD 29 kN/m
M
C
M
0.6 m D V D
D
( 2 9 )( 0 .6 ) 5 .2 2 k N m 2
1 2
V D 1 7 .4 k N
D
Segment CE 3 4 .8 k N
C
VE
M
0.6 m
M
1.2 m A
E
1.2 m E
3 4 .8 (1 .8 ) 4 5 .8 2 (1 .2 ) 7 .6 5 6 k N m
E
V E 1 1 .0 2 k N
45 . 82
SOLUTION (1.4)
(a) 8 kN
1m
m
M
B
0:
2m
0 .8 R C ( 6 ) 0 .6 R C ( 2 ) 2 4 ( 4 ) 0
R C 26 . 667
10 kN
kN
R Cx 16 kN , R Cy 21 . 334 kN
C 3m D
RC
B 2m R Bx
4 3
R By
Then
2m
A
F x 0 : R Bx 16 kN F y 0 : R By 12 . 66 kN
( b ) Segment CD 8
M
4 M
16 C
3m D V D 21.334
D
21 . 334 ( 3 ) 12 (1 . 5 ) 6 ( 2 ) 34 kN m
F D 16 kN
D
FD
V D 21 . 334 18 3 . 334
2
kN
SOLUTION (1.5) 3m
(a) A 2
M
A
0:
RCx
3 2
RCy
C RCy
RCx
4 3
R Cy
R By RCx
FD
40 kN
C
0:
4 0 (5 ) 4 R C y 0
M
RCy 5 0 k N
R Cx 75 kN
Then
R By 1 0 k N ,
R Bx 7 5 k N
D
0.75
VD
B
1 m
4m
(b)
M
2m
3
R Bx B
D
F D 75 ( 53 ) 50 ( 54 ) 85 kN
1.0
V D 75 ( 54 ) 50 ( 53 ) 30 kN
75
C 5 50 4 3
M
D
7 5 (1) 5 0 ( 0 .7 5 ) 3 7 .5 k N m
SOLUTION (1.6) (a)
Free body entire connection
B
P A
T
M
C
0 :
0,5 m
T 0 .7 R A
0.2 m
Segment AB A
AB
F AB
B
0.15 m
M
( 0 .5 ) ( 0 .1 5 ) 2
B
0 :
18
0 .5 2 2 m
18 ( 0 . 15 ) R A ( 0 . 5 ) 0 T 3 .7 8 k N m
and Fx 0 :
2
R A 5 .4 k N
0.5 m RA
(b)
R A ( 0 .7 ) T 0
C
RA
18 kN
0.15 m
0 .5 0 .5 2 2
FAB 0,
F A B 1 8 .7 2 9 k N
SOLUTION (1.7) 120 mm
120 mm
y
A R A 2 kN
B
RB 2 kN
50 mm
4 kN
70 mm
D z
Free Body: Entire Crankshaft (Fig. S1.7a) ( a ) From symmetry: R A R B
T
Fz 0 : R A R B 2 k N M
x
0 : 4 ( 0 .0 5 ) T 0 ,
T 0 .2 k N m 2 0 0 N m
x
C (a)
(CONT.)
3
1.7 (CONT.) M
y
( b ) Cross Section at D (Fig. S1.7b)
D
T
Vz
Vz 2 kN
(b)
T 200 N m
M
Figure S 1.7
y
2 ( 0 .0 7 ) 0 .1 4 k N m 140 N m
SOLUTION (1.8) 30 kN Free-Body Diagram, Beam AB
B 1.8 m
4 7
C
FC D
1.2 m
Fx 0 :
Fy 0 : R A
M
A
0:
7 65
FC D 6 0 0 , 4 65
FC D 3 0 0 ,
6 0 (1 .8 )
M
60 kN
A
F C D 6 9 .1 1 k N
7 65
R A 6 4 .3 k N
FC D ( 3 ) M
A
0,
72 kN m
1.8 m A M
A
RA
SOLUTION (1.9) B 1.2 m C 129.6 kN 0.9 m 0.9 m 1.2 m 2
A
Free body entire frame
2.4 m
1 1 2
D
M
A
0 :
129 . 6 ( 0 . 9 )
R D y 4 8 .6 k N ,
1 2
R Dy (1 . 2 ) R Dy ( 3 ) 0
R D x 2 4 .3 k N
R Dy
R Dy R D
B
R By
R B x 1.2 m
D
C
Free body BCD
2.4 m
24.3
Fx 0 :
R B x 2 4 .3 k N
Fy 0 :
R B y 4 8 .6 k N
RB
2 4 .3 4 6 .6 2
48.6
4
2
5 2 .6 k N
SOLUTION (1.10)
y 4 kN 3 kN
R Ay A
0.3 m
R Az
T
R Ey
5 kN
z C E
1m
D
R Ez
1m
2 kN x
B
0.5 m 0.5 m 0.3 m
(a)
M
x
3 ( 0 .1 5 ) 4 ( 0 .1 5 ) T 5 ( 0 .1 5 ) 2 ( 0 .1 5 ) 0
T 0 . 6 kN m
or (b)
0:
M
z
0 : ( 4 3 )( 1 ) R Ey ( 2 . 5 ) 0 , R Ey 2 . 8 kN
M
y
0 : R Ez ( 2 . 5 ) ( 5 2 )( 3 ) 0 , R Ez 8 . 4 kN
F y 0 : R Ay 4 3 R Ey 0 , R Ay 4 . 2 kN
F z 0 : R Az R Ez 5 2 0 , R Az 1 . 4 kN
Thus
RA
4 .2
RE
2 .8
2
1 .4
2
4 . 427
2
8 .4
2
8 . 854 kN
kN
SOLUTION (1.11) (a) Free-body Diagrams, Arm BC and shaft AB V
z
T
C
y B
x
M T
V
A T
Figure S1.11
M
V
M
V B
T
(b) At C: V 2 kN
T 50 N m
At end B of arm BC: V 2 kN
T 50 N m
M 200 N m
At end B of shaft AB: V 2 kN
At A:
V 2 kN
T 200 N m T 200 N m
M 50 N m M 300 N m
5
SOLUTION (1.12) Free Body: Entire Pipe 200 N
0.15 m
D
y
36 N m
C Ry
T Rx
A
0.2 m
Rz Mz
My
z
0.3 m
B
x
Reactional forces at point A:
Fx 0 :
Rx 0
Fy 0 :
R y 200 0,
Fz 0 :
Rz 0
R y 200 N
Moments about point A:
M
x
0:
T 2 0 0 ( 0 .1 5 ) 0 ,
M
y
0:
M
y
M
z
0:
M
z
T 30 N m
0 2 0 0 ( 0 .3 ) 3 6 0 ,
M
z
96 N m
The reactions act in the directions shown on the free-body diagram.
SOLUTION (1.13) Free Body : Entire Pipe 0.15 m
y Rx
Mz z
36 N m
C WCD
T Rz
D
0.075 m
Ry
200 N
WAB
0.2 m
WBC
A
My 0.15 m
0.3 m
B
x
(CONT.)
6
1.13 (CONT.) We have 1 lb/ft=14.5939 N/m (Table A.2). Thus, for 3 in. or 75-mm pipe (Table A.4): 14.5939(7.58)=110.62 N/m Total weights of each part acting at midlength are: W A B 1 1 0 .6 2 ( 0 .3) 3 3 .2 N W B C 1 1 0 .6 2 ( 0 .2 ) 2 2 .1 N W C D 1 1 0 .6 2 ( 0 .1 5 ) 1 6 .6 N
Reactional forces at point A:
Fx 0 : Fy 0 :
Rx 0 R y W AB W BC W CD 2 0 0 0,
Fz 0 :
R y 2 7 1 .9 N
Rz 0
Moments about point A:
M
x
0:
M
y
0:
M
z
0:
T W C D ( 0 .0 7 5 ) 1 0 ( 0 .1 5 ) 0 ,
T 2 .7 4 5 N m M
M
z
y
0
( 2 0 0 W C D W B C )( 0 .3 ) W A B ( 0 .1 5 ) 3 6 0
M
z
1 1 2 .6 N m .
SOLUTION (1.14) 1.6 kN
RCy
(a) R Ay
D
R Ax A
C
Dy
0.5 m
RCx
1m
Dx
Bx
E
By
D
1.6 kN 0.15 m
Dx
1.6 kN By
Dy
Free body pulley B
B
Bx
0.25 m B
150 mm
F x 0 : B x 1 . 6 kN F y 0 : B y 1 . 6 kN
Free body CED
M
D
0 : R Cx ( 0 . 4 ) 1 . 6 ( 0 . 15 ) 0 ,
F x 0 : D x 0 .6 1 .6 0 , F y 0:
R Cy D
R Cx 0 . 6 kN
D x 1 kN
y
Free body ADB
M
A
0 : D y ( 0 . 5 ) B y (1 . 5 ) 0 , D y 4 . 8 kN , R Cy 4 . 8 kN
F x 0 : R Ay D y B y 0 ,
R Ay 3 . 2 kN
(CONT.)
7
1.14 (CONT.)
(b)
M
F x 0 : R Ax D x B x 0 ,
V G 0.6 m B
G
FG
G
M
1.6
G
R Ax 0 . 6 kN
1 . 6 ( 0 . 6 ) 960
N m , V G 1 . 6 kN
F G 1 . 6 kN
1.6
SOLUTION (1.15) Free body entire rod
M
0 : R Dy ( 0 . 25 ) 300 ( 0 . 1 ),
x
M
z
0:
R Dy 120
N
2 0 0 ( 0 .3 5 ) R B y ( 0 .2 5 ) ( 3 0 0 1 2 0 ) ( 0 .2 ) 0
C
R By 136
N
Free body ABE y B
A 200 N
Vy
100
175
M
z
E
0: M
z
x
E
136 N
M
z 200 ( 0 . 275 ) 136 ( 0 . 175 ) 0 , M
z
z
31 . 2 N m
F y 0 : V y 200 136 64 N
SOLUTION (1.16) Free body entire rod:
M 0: M
R D y ( 0 .2 5 ) 4 0 0 ( 0 .1) 0 ,
x
z
R B y ( 0 .2 5 ) ( 4 0 0 1 6 0 ) ( 0 .2 ) 0 ,
0:
C
Segment ABE A
y
B
Vy
0.175 m
192 N
z
x
E M
z
At point E: M
z
192 ( 0 . 175 ) 33 . 6 N m
V y 192
RDy 160 N
N
8
R By 1 9 2 N
SOLUTION (1.17) Side view
Top view
F2 50 mm
F2 B C
D
Td
50 mm
100 mm
Figure (c)
Figure (a)
F1
150 N m A 50 mm
Fig. (b): Fig. (a): Fig. (c):
F1
Figure (b)
M
A
0 : F 1 ( 0 . 05 ) 150 0 , F 1 3 kN
M
B
0 : F 1 ( 0 . 1 ) F 2 ( 0 . 05 ) 0 , F 2 6 kN
M
D
0:
F 2 ( 0 .0 5 ) T d 0 ,
T d 0 .3
kN m
SOLUTION (1.18) 18 kN
13.5 kN
1m
C
3m B A
Ry
6m
Rx 3 4
4m
R
Free body-entire frame
M
A
0:
R y (1 0 ) 1 3 .5 ( 6 ) 1 8 ( 4 ) 0 ,
Free body-member BC
M
C
0:
R x (3) R y ( 4 ) 0
and Rx
4 3
(1 5 .3) 2 0 .4 k N
Thus FBC R
( 2 0 .4 ) (1 5 .3 ) 2
2
2 5 .5 k N
9
R y 1 5 .3 k N
SOLUTION (1.19) P
Free body-member AB B
R Ax
A
E 40
o
a
0: o
o
R E 1.1 5 6 P
RE
R Ay
A
R E ( 4 a ) P c o s 4 0 ( a ) P s in 4 0 ( 6 a ) 0
2a
4a
M
F x 0:
R Ax P co s 4 0
0 .7 6 6 P
F y 0:
R A y 1.1 5 6 P P s in 4 0
o
o
0,
R A y 0 .5 1 3 P
Free body-member CD RE
RCx
C 2a E
RD
RCy
M
C
D
4a
0:
30
M
D
0:
R C y 0 .7 7 1 P
o
R D s in 3 0 ( 6 a ) R E ( 2 a ) 0 , o
F x 0:
RCx R D co s 3 0
R E ( 4 a ) RCy ( 6 a ) 0
o
R D 0 .7 7 1 P
0 .6 6 8 P
SOLUTION(1.20) ( a ) Power= P = ( p A ) ( L ) ( n /6 0 ) (1 .2 )( 2 1 0 0 )( 0 .0 6 )( 1 56 00 0 ) 3 .7 8 k W
Power required
P e
3 .7 8 0 .9
4 .2 k W
( b ) Use Eq.(1.15), T
9549 kW n
9 5 4 9 ( 4 .2 ) 1500
2 6 .7 4 N m
SOLUTION (1.21) a=1.5 m, b=0.55 m, c=0.625 m, L=2.7 m, V=29 m/s, W=14.4 kN, kW=14 ( a ) From Eq. (1.15), the drag force equals, Fd
1000 kW V
1 0 0 0 (1 4 ) 29
4 8 2 .8 N
See: Fig. P1.21:
Fx 0 :
F d 4 8 2 .8 N
It follows that
M
A
0:
W a Fd c R f L 0
or 1 4 4 0 0 (1 .5 ) ( 4 8 2 .8 )( 0 .6 2 5 ) R f ( 2 .7 ) 0
(CONT.)
10
1.21 (CONT.) Solving, R f 7 .8 8 8 k N
and
Fy 0 :
R r 1 4 .4 7 .8 8 8 0
or R r 6 .5 1 2 k N
(b) We have V 0 ,
Fd 0 ,
F 0.
See Fig. P1.21:
M
A
Rf
a L
0:
Wa Rf L 0
Thus
So,
W
1 .5 2 .7
(1 4 .4 ) 8 k N
F y 0 gives
R r W R f 1 4 .4 8 6 .4 k N
SOLUTION (1.22) Refer to Solution of Prob. 1.21. a=1.5 m, b=0.55 m, c=0.625 m, L=2.7 m, V=29 m/s, W=14.4 kN, kW=14 Now we have W t 1 4 .4 5 .4 1 9 .8 k N
( a ) From Eq. (1.15), the drag force equals, Fd
1000 kW
V
1 0 0 0 (1 4 ) 29
4 8 2 .8 N
See: Fig. 1.21 (with W W t ):
M
A
0:
W t a Fd c R f L 0
1 9 8 0 0 (1 .5 ) ( 4 8 2 .8 )( 0 .6 2 5 ) R f ( 2 .7 ) 0
from which R f 1 0 .8 8 8 k N
and
Fy 0 :
R r 1 9 .8 1 0 .8 8 8 0
R r 8 .9 1 2 k N
( b ) V 0,
Fd 0 ,
M
A
Then
Rf
a L
So,
F 0 , as before,
0: W
Wta R f L 0 1 .5 2 .7
(1 9 .8 ) 1 1 k N
F y 0 gives
R r W t R f 1 9 .8 1 1 8 .8 k N
11
SOLUTION (1.23) ( a ) Free-Body Diagram: Gears (Fig. S1.23). Applying Eq. (1.15): T AC
9550 P n
9550 (35 )
6 6 8 .5 N m
500
Therefore, TA
F
rA
5 .3 4 8 k N
6 6 8 .5 0 .1 2 5
( b ) T D E F rD 5, 3 4 8 ( 0 .0 7 5 ) 4 0 1 .1 N m TDE
D rD
T AC
A
F rA
F
Figure S1.23 SOLUTION (1.24) n1 1 8 0 0 rp m ,
n 2 4 2 5 rp m ,
kW 20
From Eq. (1.15), we obtain T = 9549 kW/n. Thus For input shaft T
9549 ( 23) 1800
122 N m
For output shaft T
9549 ( 20 ) 425
4 4 9 .4 N m
Equation (1.14) gives e
20 23
100 87 %
SOLUTION (1.25) N=150, F=2.25 kN, s=62.5 mm, e=88% ( a ) Referring to Eq. (1.12): power output F s ( 6N0 ) 2 2 5 0 0 .0 6 2 5 ( 16500 ) 3 5 1 .6 W
( b ) Using Eq. (1.14), power transmitted by the shaft: power input 3 5 1 .6 ( 0 .8 8 ) 3 9 9 .5 W SOLUTION (1.26) Equation (1.10) becomes Ek
1 2
I (
2 max
2 min
(1)
)
Here, mass moment inertia with 5 percent added: I ( 1 . 05 )
32
( d o d i ) l 4
4
(Table A.5) (CONT.)
12
1.26 (CONT.)
1 . 05
( 0 .4
32
4
1 . 299 kg m
0 . 3 )( 0 . 1 )( 7 , 200 ) 4
2
m a x 1 2 0 0 ( 610 ) 2 0 r p s 1 2 5 .7 ra d s
m in 1 1 0 0 ( 610 ) 1 8 .3 r p s 1 1 5 ra d s
Equation (1) is therefore Ek
1 2
( 1 . 299 )( 125 . 7
115
2
2
)
1, 6 7 3 J
SOLUTION (1.27) Final length of the wire: L AC '
( 2 ) (1 .2 6 ) 2
2
2 .3 6 3 8 m
Initial length of the wire is L AC
( 2 ) (1 2 .5 ) 2
2
2 .3 5 8 5 m
Hence, Eq. (1.20): AC
L AC ' L AC L AC
2 .3 6 3 8 2 .3 5 8 5 2 .3 5 8 5
0 .0 0 2 2 5 2 2 5 0 μ
SOLUTION (1.28) (a) c
2 ( r r ) 2 r 2 r
( c ) i
0 .3 150
( c ) o
(b) r
r r
2000
μ
800
0 .2 250
r o ri
r o ri
μ
0 .3 0 .2 250 150
1000
μ
SOLUTION (1.29) LO B d ,
( a ) OB
L AB L BC d
0 .0 0 1 2 d d
1200 μ
( b ) L AB ' L CB ' d AB BC
( c ) C A B ta n
1
2 1 .4 1 4 2 1 d
2
(1 . 0012 d )
1 .4 1 5 0 6 1 .4 1 4 2 1 1 .4 1 4 2 1
1 .0 0d1 2 d
2
1 2
1 . 41506 d
601 μ 4 5 .0 3 4 4
o
Increase in angle C A B is 4 5 .0 3 4 4 4 5 0 .0 3 4 4 . Thus o
0 .0 3 4 4 18 0 6 0 0 μ
13
SOLUTION (1.30) (a) x
0 .8 0 .5 250
1200
μ
y
0 .4 0 200
2000 μ
( b ) L ' AD L AD x L AD L AD (1 x ) 250 (1 . 0012 ) 250 . 3 mm
SOLUTION (1.31) L AB 800 (10
6
)150 120 (10
6
L AD 1000 (10
3
) 200 200 (10
) mm 3
) mm
We have L BD L AB L AD 2
2
2
2 L BD L BD 2 L AB L AB 2 L AD L AD
or L AB
L BD
LBD
150 250
L AB
L AD LBD
L AD
(120 )
200 250
( 200 ) 10
(1) 3
0 . 232
mm
SOLUTION (1.32) AC BD
300
2
300
2
424 . 26 mm
B ' D ' 4 2 4 .2 6 0 .5 4 2 3 .7 6 m m ,
A ' C ' 4 2 4 .2 6 0 .2 4 2 4 .4 6 m m
Geometry: A ' B ' A ' D ' x
C'
B'
D'
A'
xy
y
A ' D ' AD AD
4 2 3 .7 6 2 4 2 4 .4 6 2 2 2
1 2
300
363 μ
300
2
2
2 tan
1 423 . 76 2 424 . 46 2
1651
μ
SOLUTION (1.33) 0.5 m A' 0.00064 m A
F
1m B
C
B
800 10
6
( 0 .8 )
0 .0 0 0 6 4 m
B'
x
We have AD AD L AD
0.005 m
0.33 m C' (CONT.)
14
1.33 (CONT.)
From triangles A ' A F and C ' C F : 0 .0 0 0 6 4 x
0 .0 0 5 1 .5 x
,
x 0 .1 7 m
From triangles B ' B F and C ' C F : 0 .3 3 01.0.3035 , B 0 .0 0 1 2 4 m B E (contraction) B
Therefore BE
BE LBE
0 .0 0 1 2 4 1
1 2 4 0 (1 0
6
)
1, 2 4 0
SOLUTION (1.34) (a) x
xy
0 . 006 50
120
μ
y
1000 200 800
0 . 004 25
160
μ
μ
( b ) L ' A B L A B (1 y ) 2 5 (1 0 .0 0 0 1 6 ) 2 4 .9 9 6 m m L ' A D L A D (1 x ) 5 0 (1 0 .0 0 0 1 2 ) 5 0 .0 0 6 m m
End of Chapter 1
15
CHAPTER 2
MATERIALS
SOLUTION (2.1) A0
4
(1 2 .5 ) 1 2 2 .7 m m , 2
We have a
2
1500 μ ,
0 .3 200
Af
t
0 .0 0 6 1 2 .5
(1 2 .5 0 .0 0 6 ) 1 2 2 .6 m m 2
4
480 μ
Thus S
p
E
S
p
a
3
1 8 (1 0 )
P A0
1 4 6 .8 M P a
1 2 2 .7 6
1 4 6 .7 (1 0 ) 1 5 0 0 (1 0
6
9 7 .8 G P a ,
)
t a
0 .3 2
Also % e lo n g a tio n
0 .3 200
(1 0 0 ) 0 .1 5
in area
% reduction
122 . 7 122 . 6 122 . 7
(100 ) 0 . 082
SOLUTION (2.2) Normal stress is
P A
2 8 6 .8 M P a
2200
( 3 .1 2 5 )
2
4
This is below the yield strength of 350 MPa (Table B.1). We have L 576.50 0 0 .0 0 1 3 3 9 1 3 3 9 μ Hence E
6
2 8 6 .8 (1 0 ) 1 3 3 9 (1 0
6
2 1 4 .2 G P a
)
SOLUTION (2.3) The cross-sectional area: A w o t o 1 2 .7 ( 6 .1) 7 7 .4 7 m m ( a ) Axial strain and axial stress are a 0 6.038.54 1 0 .0 0 1 3 2 4 1 3 2 4 μ
a
Because
a
S y (See Table B.1), Hooke's Law is valid.
P A
2 1 ,5 0 0 7 7 .4 7 (1 0
6
)
2 7 7 .5 M P a
( b ) Modulus of elasticity, E
a
a
6
2 7 7 .5 (1 0 ) 1 3 2 4 (1 0
6
)
2 0 9 .6 G P a
( c ) Decrease in the width and thickness w w o 0 .3(1 2 .7 ) 3 .8 1 m m t t o 0 .3( 6 .1) 1 .8 3 m m
16
2
2
SOLUTION (2.4) Assume Hooke's Law applies. We have t 15.5 3 0 0 μ a
t
300 0 .3 4
822 μ
Thus, E a (1 0 5 1 0 )(8 2 2 1 0 9
9
) 9 2 .6 1 M P a
S y , our assumption is valid.
Since So
P A (9 2 .6 1)( 4 )(5 ) 1 .8 1 8 k N 2
SOLUTION (2.5) We obtain L AC L BD
L AC
x
L BD
y
(a) E
L AC
L BD
x
x
y
(b)
x
(c) G
15
15
2
21 . 17 21 . 21 21 . 21
1886
21 . 22 21 . 21 21 . 21
471 μ
100 ( 10
6
1886 ( 10
471 1886
53 2 ( 1 0 . 25 )
) 6
2
21 . 21 mm
μ
53 GPa
)
0 .2 5
21 . 2 GPa
SOLUTION (2.6) Use generalized Hooke’s law: x y z
1 2 E
( x
y
For a constant triaxial state of stress: x y z , x
y
Then, Eq. (1) becomes 1 2 0
or
1 2 E
z)
(1)
z
. Since and
must have identical signs:
1 2
SOLUTION (2.7) We have
x
3
4 5 0 (1 0 ) 50 (75 )
120
M Pa
(CONT.)
17
2.7 (CONT.) (a) x
(b) E
x
x
(c) z
6
1 2 0 (1 0 )
2 0 0 0 (1 0
6
500 μ
0 .0 2 5 50
60 G Pa
)
6
1 2 0 (1 0 )
0 .2 5
x
E
6
a 5 0 0 (1 0
(d) G
y
0 .2 5
500 2000
2000 μ,
0 .5 250
9
6 0 (1 0 )
500 μ
9
6 0 (1 0 )
) 7 5 3 7 .5 (1 0
3
a ' 7 5 0 .0 3 7 5 7 4 .9 6 2 5 m m
) mm;
24 G Pa
2 (1 0 .2 5 )
SOLUTION (2.8) We have y z 0
x
3
25 ( 10 ) 20 10 ( 10
6
)
125
MPa
Thus y 0
1 E
[
( x z )]
(1)
z 0
1 E
[ z ( x y )]
(2)
x
y
[ x (
1 E
z )]
y
(3)
Equations (1) and (2) become y z x z
Adding : (
(1’)
x
y
(2’)
z ) 2 x (1 ) . Then, Eq. (3): 2
y
x
2
1 2 1
x
E
Substituting the data:
x
6
1 0 . 3 0 . 18 125 ( 10 ) 9 0 .7 70 ( 10 )
1327
μ
SOLUTION (2.9) Hooke's Law. We have x
10
9
x
E
10
6
7 2 1 0
z
E
6
7 2 1 0
y
x
E
9
x
E
10
6
7 2 1 0
9
y
0 and
y
z
E
[(8 0 ) 0 0 .3 (1 4 0 )] 0 .0 0 0 5 2 8 5 2 8 μ
y
E
z
E
[ 0 .3 (8 0 ) 0 0 .3 (1 4 0 )] 9 1 7 μ
E
y
z
E
[ 0 .3 (8 0 ) 0 1 4 0 ] 1 6 1 1 μ
(CONT.)
18
2.9 (CONT.) (a)
L AB x a ,
Change in length
6
L A B (5 2 8 1 0
(b)
)(3 2 0 ) 0 .1 6 9 m m
Change in thickness t yt (917 10
(c)
6
)(1 5 ) 0 .0 1 4 m m
Change in volume, e x
z 5 2 8 9 1 7 1 6 1 1 1 .2 2 2
y
V e V o 1 .2 2 2 (3 2 0 3 2 0 1 5 ) 1 .8 7 7 m m
3
SOLUTION (2.10) By assumptions, rubber in triaxial stress:
x
z
p,
y
F
d
2
4
4F
d
2
Stains are x z 0 . Hooke's law gives x 0
1 E
[
(
x
y
y
z )]
or 0 p
4F
Solving, p
x
2
d (1 )
4 F
z
Q.E.D.
2
d (1 )
Substitute the data: p
3
4 ( 0 .5 )(1 0 1 0 ) 2
( 6 2 .5 ) (1 0 .5 )
3 .2 6 M P a (C)
SOLUTION (2.11) Hooke’s law gives 90 MPa y 10 mm
150 MPa
x
1 E
(
x
y
1 E
(
y
z
x 100 mm
E
(
y
)
x ) x
y
10
6
100 ( 10 10
9
6 9
1 0 0 (1 0 )
)
)
(150
(90
( 1 3 ) 10 100 ( 10
6
9
)
L 1800 (100 ) 180 μm a 1400 ( 50 ) 70 μm b 200 (10 ) 2 μm
and mm ,
a ' 49 . 993
19
mm ,
b ' 9 . 9998
150 3
) 1800
μ
) 1400 μ
(150 90 ) 200 μ
Thus
L ' 100 . 018
90 3
mm
SOLUTION (2.12) We have x
y
z p
Gen. Hooke’s law:
x
y
z
(1 2 )
p E
120 ( 10 100 ( 10
6 9
) 1 ) 3
400 μ
Thus L 400 (100 ) 40 μm a 400 ( 50 ) 20 μm b 400 (10 ) 4 μm
and L ' 99 . 96 mm ,
a ' 49 . 98 mm ,
b ' 9 . 996
SOLUTION (2.13) We have
x
Vo
(a) x
r 3
p . The volume is
z
4 3
(1 2 5 ) 8 .1 8 1(1 0 ) m m 3
[ ( )]
1 E
6
1 6 8 (1 0 )
4 3
y
9
7 0 (1 0 )
E
6
3
(1 2 )
(1 0 .5 ) 1 2 0 0 μ
Change in diameter, d x d 1 2 0 0 (1 0
6
) 2 5 0 0 .3 m m
Decrease in circumference: ( d ) 0 .3 0 .9 4 2 5 m m
( b ) V e V o (1 2 ) xV o ( 0 .5 )( 1 2 0 0 1 0
6
)(8 .1 8 1 1 0 ) 4 9 0 9 m m 6
SOLUTION (2.14) From Fig.2.3b and Eq.2.20: U
t
S y Su 2
f
250 440 2
( 0 . 27 ) 93 MPa
We have L f 50 50 ( 0 . 27 ) 63 . 5 mm Using Eq.(2.1): % e lo n g a tio n
6 3 .5 5 0 50
(1 0 0 ) 2 7 %
20
3
mm
SOLUTION (2.15) Table B.1:
S
260
y
MPa , E 70 GPa
We have V AL
For U
U
r
U
app
S
2 y
2E
( 0 . 005 ) ( 3 ) 58 . 9 10 2
4
( 260 10
6
2
) 9
2 ( 70 10
)
482 . 9 kJ
m
6
m
3
3
U r V 482 . 9 10 ( 58 . 9 10 3
6
) 28 . 44 J
9 J :
app
n
2 8 .4 4 9
3.1 6
SOLUTION (2.16) ( a ) ASTM-A242. E 2 0 0 G P a and U
o
Sy
2
2E
6
( 3 4 5 1 0 )
2
9
2 ( 2 0 0 1 0 )
y
298
345 M Pa kJ m
3
( b ) Stainless (302). E 1 9 0 G P a and S y 5 2 0 M P a U
o
Sy
2
2E
6
( 5 2 0 1 0 )
2
9
2 ( 2 0 0 1 0 )
712
kJ m
3
SOLUTION (2.17) ( a ) Aluminum 2014-T6. E 7 2 G P a and Uo
Sy
2
2E
6
( 4 1 0 1 0 )
y
410 M Pa
2
9
2 ( 7 2 1 0 )
1167
kJ m
3
( b ) Annealed yellow brass. E 1 0 5 G P a and S y 1 0 5 M P a Uo
Sy
2
2E
6
(1 0 5 1 0 )
9
2
2 (1 0 5 1 0 )
5 2 .5
kJ m
3
SOLUTION (2.18) Referring to Fig. P2.18: E (a) Uo
Sy
2
2E
6
(1 9 2 .5 1 0 ) 9
2 ( 4 2 1 0 )
2
6
1 0 5 (1 0 ) 0 .0 0 2 5
42 G Pa,
4 4 1 .5 k J m
S y 1 9 2 .5 M P a
3
( b ) Total area under diagram: U
2 6 2 .5 (1 0 )( 0 .1 7 6 ) 4 6 .2 M J m 6
t
21
3
SOLUTION (2.19) ( a ) V 50 50 1, 500 3 . 75 (10 ) mm 6
Thus
nU
or
S
S
V
2E
[
y
2 y
1
2 EnU V
]2 9
[ 2 200 10
r
S
2 y
6
( 253 10
2E
)
2
9
2 ( 200 10
)
1
1 . 5 400
3 . 75 ( 10
(b) U
3
3
] 2 253
)
160
MPa
kPa
SOLUTION (2.20) S y 250
Table B.1:
E 200
M Pa,
GPa
We have U nU
U
r
S
app
2 y
6
( 2 5 0 1 0 )
2E
5 (1 7 ) 8 5 N m 2
1 5 6 .2 5
9
2 ( 2 0 0 1 0 )
kN m m
3
Therefore V
U Ur
85 3
1 5 6 .2 5 (1 0 )
V AL :
Also
3
0 .5 4 4 (1 0
0 . 544 (10
3
)
) m
3
2
d ( 2 .4 )
4
or d 0 .0 1 7 m = 1 7 m m
SOLUTION (2.21) Refer to Fig. P2.21. We have E
(a) Uo
Sy
2
2E
6
1 9 0 (1 0 ) 0 .0 0 1 3
( 2 4 5 1 0 )
2
9
2 (1 9 0 1 0 )
190 G Pa,
158
S y 245 M Pa
kJ m
3
( b ) Total area under diagram: 3 5 0 1 0 ( 0 .2 8 ) 9 8 3
U
t
MJ m
3
SOLUTION (2.22) V ( 0 .0 5 )( 0 .0 5 )(1 .2 ) 0 .0 0 3 m
( a ) n U
S
3
and S y S p
2 p
2E
V,
S
2 p
nU ( 2 E ) V
Substituting the data given, (CONT.)
22
2.22 (CONT.) Sp 2
9
1 .8 (1 5 0 )( 2 2 1 0 1 0 ) 0 .0 0 3
3 7 .8 1 0
15
or S p 1 9 4 .4 M P a
(b) Uo
Sp
2
2E
3 7 .8 1 0
15 9
2 ( 2 1 0 1 0 )
90
kJ m
3
SOLUTION (2.23) Applying Eq. (2.22), we find S u 3 .4 5 H
B
M P a 3 .4 5 (1 4 9 ) 5 1 4 M P a
Equation (2.24): S y 3 .6 2 (1 4 9 ) 2 0 7 3 3 2 .4 M P a
SOLUTION (2.24) Using Eq. (2.22), S u 3 .4 5 H
B
M P a 3 .4 5 (1 7 9 ) 6 1 8 M P a
Formula (2.24): S y 3 .6 2 (1 7 9 ) 2 0 7 4 4 1 M P a
SOLUTION (2.25) Formula (2.22), S u 3 .4 5 H
B
M P a 3 .4 5 (1 5 6 ) 5 3 8 M P a
Equation (2.24): S y 3 .6 2 (1 5 6 ) 2 0 7 3 7 8 M P a
SOLUTION (2.26) Equation (2.22) gives S u 3 .4 5 H
B
M P a 3 .4 5 ( 2 9 3) 1 0 1 0 .9 M P a
Formula (2.24): S y 3 .6 2 ( 2 9 3 ) 2 0 7 8 5 3 .7 M P a
End of Chapter 2
23
CHAPTER 3
STRESS AND STRAIN
SOLUTION (3.1) (a)
s
3
4 5 (1 0 )
( 2 5 1 0
3
)
9 1 .6 7
4
3
(b) h
( 2 5 .7 8 1 0
(c) t
( 2 1 .1 2 1 0
(d) b
2
4 5 (1 0 ) 3
3
)(1 0 .9 4 1 0
3
3
)(1 2 .5 1 0
3
5 0 .7 9 M P a
)
4 5 (1 0 )
M Pa
5 4 .2 6
)
3
4 5 (1 0 )
[( 5 0 1 0
3
2
) ( 2 5 .7 8 1 0
M Pa
3 1 .2 2 M P a
3
2
) ]
4
SOLUTION (3.2) A [( a
a
9 16
Thus
a 4
) a ] 2
2
2
Pn Sy
a
9 16
,
a
2
2
16 Pn 9 S y
all
a
,
S
y
n
A
,
P
all
Pn Sy
Pn
4 3
Sy
Substituting the data given: a
1 . 2 ( 10
4 3
6
)( 2 . 2 ) 6
( 280 10 )
73 mm a min
SOLUTION (3.3) Free body-Member ACD
R Ay
A
R Ax
0 : 45 sin 15 ( 0 . 8 ) o
A
FBC
0.4 m 3
0.1 m 0.4 m
Fx 0 :
3 5
F BC 45 sin 15
R A x 5 .8 2 5
D o
RA
Resultant is
F y 0 : 45 cos 15
(5 .8 2 5 ) ( 6 6 .7 6 3 ) 2
We therefore have
n
RA 2
d
2
4
,
2
d (
o
6 6 .9 9 2 R An
1
)2
Hence 3
dA [
2 ( 6 6 .7 6 3 1 0 )( 2 )
dB [
2 ( 2 9 .1 2 1 0 )( 2 )
6
(1 9 6 1 0 ) 3
6
(1 9 6 1 0 )
1
] 2 0 .0 2 0 8 m = 2 0 .8 m m 1
] 2 0 .0 1 3 7 5 m = 1 3 .7 5 m m
24
F BC ( 0 . 3 )
kN o
R Ax 0
kN R Ay
R A y 6 6 .7 6 3
45 kN
3 5
F BC ( 0 . 1) 0
4 5
F B C R B 2 9 .1 2
4
C
15
M
kN
kN
4 5
F BC 0
SOLUTION (3.4) R Ay
5 kN B 1m 1m
C 0.75 m
R Ax A 1 m
1.5 m
E 3
A
0:
5( 3 )
RE
RA
3 5
R E ( 0 .7 5 )
F x 0 : R Ax 7 . 82 kN
F y 0 : R Ay 5 . 43
7 .8 2 5 .4 3 2
2
kN
9 .5 2 k N
Free body-beam ABC 7.82 A 1m
(b)
A
1 . 018
8 ,145 3
8 ( 10
)
3
9 .5 2 (1 0 ) 2
( 0 .0 2 5 )
2
M
C
0 : F BD 8 . 145
RCy
FBD
BD
2m
5.43
(a)
RCx
C
B
MPa
9 .6 9 7 M P a
4
SOLUTION (3.5) RA
A
L cos
L ta n
RB
B
0:
A
F y 0:
R B FBC R A F AC
P s in
A AC
sin
L AC
L cos
A BC
P
P
tan
L BC L
Total weight, W ( A A C L A C A B C L B C ) : W
PL
( sin 1cos
cos sin
)
PL
2
1 cos sin cos
Therefore dW d
P L
2
[
(s in c o s )
2
The foregoing reduces to 2
or
5 4 .7
2
2
s in c o s ( 2 c o s )( s in ) (1 c o s )( s in c o s )
3 cos 1
P ta n
We have
C P
L
M
or
cos
1 3
o
25
4 5
R E 13 . 04 kN
or
4 D
M
] 0
kN ( C )
RE (2) 0
SOLUTION (3.6) Refer to Table 3.1 (Case C) a b 0 .0 9 0 .0 4 5 4 .0 5 1 0
t
3
m
2
45 mm t
T
;
30 10 6
2abt
90 mm
1 .2 1 0
2 ( 4 .0 5 1 0
Solving, t 4 .9 4 m m
We have 1 .5 0 .0 2 6 ra d and a ll L 0 .0 2 6 0 .8 0 .0 3 3 ra d m . o
Hence
( a b ) tT 2
2
2
2t a b G
( a b )T 2
2
2 ta b G
0 .1 3 5 (1 .2 1 0 ) 3
0 .0 3
2 ( 0 .0 4 5 ) ( 0 .0 9 ) ( 2 8 1 0 ) t 2
2
9
Solving: t 5 .3 4 m m Use t a ll 5 .5 m m
SOLUTION (3.7) Refer to Solution of Prob. 3.6. We now have a=b.
T 2
;
30 10 6
2a t 70 mm
t
Solving:
70 mm
Similarly,
T 3
;
0 .0 3 3
a Gt
Solving: Use
1 .2 1 0
3
( 0 .0 7 ) ( 2 8 1 0 ) t 3
9
t 3 .8 m m
t a ll 4 .1 m m
26
1 .2 1 0
3 2
2 ( 0 .0 7 ) t t 4 .1 m m
3 3
)t
SOLUTION (3.8) Refer to Case C, Table 3.1. We now have a=b=31.25 mm, a ll 1 2 2 .4 5 4 .9 0 .0 8 6 ra d m . o
T
a ll
2
105 10
;
6
2a t
T 2 ( 3 1 .2 5 1 0
3
) ( 4 .7 1 0 2
3
,
T 9 6 3 .8 7 N m
,
T 3 5 4 .0 2 N m
)
Likewise, a ll
T 3
0 .0 8 6
;
a tG
T ( 3 1 .2 5 1 0
3
) ( 4 .7 1 0 3
3
) ( 2 8 .7 1 0 ) 9
T a ll 3 5 4 .0 2 N m
Thus, SOLUTION (3.9)
We have a b ( ro ri ) 2 1 1 .7 m m Case E, Table 3.1). From Eq. (3.11), m ax
Tr J
T (1 2 .4 )(1 0 4
3
4
)
( 2 )(1 2 .4 1 1 )(1 0
8 7 7 (1 0 ) T 3
12
)
From Table 3.1: m ax
T 2 abt
8 3 0 .5 (1 0 ) T 3
T 2
2 (1 1 .7 ) (1 .4 )(1 0
9
)
Therefore, error in the thin-wall estimate: 8 7 7 8 3 0 .5 877
0 .0 5 3 5 .3 %
SOLUTION (3.10) ( a ) Maximum moment occurs at midlenth B. Thus
60 kN
a ll
d
V(,kN) A
3
30 kN 30 mm
14 mm
B
14 mm
30 E
D
32 M
m ax
d
3
3
32 M
m ax a ll
32 ( 435 )
2 6 .1 m m
6
( 2 5 0 1 0 )
( b ) Maximum shear is at D and E. Hence, from Table 3.2: x
-30
M, (N m) 435 210
Mc I
or
d 30 kN
a ll
4 V m ax 3 A
d
1 6 V m ax
4 V m ax 3 d2 4
1 6 V m ax 3 d
2
or 2
x
Use d a ll 2 6 .1 m m
27
3 a ll 3
2
1 6 ( 3 0 1 0 ) 6
3 (1 5 0 1 0 )
1 8 .4 3 m m
SOLUTION (3.11) 12 kN
24 kN
21 kN 2m
2m
B 3m 4m
RB
D
2m
4m
A
C
RD
RA
RC
Reactions
M
D
M F
0 : R B 21 kN
F y 0 : R D 15 kN
0 : R A 14 kN
C
0 : R C 7 kN
y
21 kN A
C 2m
14 V (kN)
4m
7
14
+
x
_ 7
M ( kN m )
28 +
x
SOLUTION (3.12) 4 2.7 1m 1m 1m
A
B
3.13
V (kN)
x 3.57
_
0:
A
Fy 0 :
a ll
M
( kN m )
R B 3 .5 7 k N m R A 3 .1 3 k N m
m ax c
:
I
1 2 .5 (1 0 ) 6
3
3 .5 6 (1 0 )( h 2 ) ( 0 .0 5 )( h
3
12 )
3
3 .5 6 1 0 6 2
0 .0 5 ( h )
+
max
3 V 2 A
3 3 . 57 2 0 . 05 ( 0 . 185 )
578 . 9 700 OK
x SOLUTION (3.13) b h 2
2
d , 2
h
2
d
2
b
S
2
We obtain dS db
Thus
b
2
2
d 6
d
and h d
3
b 2
0:
,
h 0 .1 8 5 m
3.56
3.13
M
0.43
+
M
b
2
2
d 3
2 3
28
bh 6
2
b 6
(d
2
b ) 2
bd 6
2
3
b 6
SOLUTION (3.14) c I
200 ( 25 )( 162 . 5 ) 150 ( 15 )( 75 )
135 . 35 mm
200 ( 25 ) 150 ( 15 ) 1 12
200 25 ( 27 . 15 )
3
( 200 )( 25 )
135 . 35 15 ( 1352. 35 )
2
13 . 24 10
2
We have at N.A.: Q m a x 1 3 5 .3 5 (1 5 )( 1 3 52.3 5 ) 1 3 7 .4 1 0
VQ
max
6
3
22 ( 10 )( 137 . 4 10
Ib
13 . 24 ( 10
6
)
3
3
mm
15 . 22 MPa
)( 0 . 015 )
SOLUTION (3.15) I
1 12
[ 200 ( 300 )
100 ( 200 ) ] 383 . 3 (10
3
3
3
Q A y 200 50 (100 25 ) 1 . 25 (10 *
q
VQ I
F (2) q s,
,
2F s
6
) m
) m
4
3
VQ I
Thus V a ll
3
2 (1 5 1 0 )( 3 8 3 .3 1 0
2 FI sQ
0 .1 (1 .2 5 1 0
3
6
)
)
92 kN
SOLUTION (3.16) V max 2 . 4 kN
M
max
Design is based on
all
12 MPa :
all
M
max
c
1 2
)
6
: 12 (10
I
wL
2
1 . 44 kN m
3
1 . 44 ( 10 ) b 2b
4
3
2 ,160 b
3
, b 56 . 5 mm
Check all 0 . 18 MPa :
max
3
3 V max 2 A
:
2 . 4 ( 10 ) 3 2 2 ( 0 . 0565 ) 2
564 kPa 810
kPa
OK .
SOLUTION (3.17) V max
1 2
( 50 6 ) 150
Design based on S min
M
all max
kN ,
170
all
max
1 8
( 50 )( 6 )
2
225
kN m
MPa : 3
225 ( 10 ) 170 ( 10
M
6
Table A.7: S 4 6 0 8 1 .
)
1 . 324 (10
6
) mm
3
(smallest possible, others will work)
Shear stress S 460 81 : A web d ( t w ) 457 (11 . 7 ) 5 . 347 (10 ) mm 3
max
V max A web
3
150 ( 10 ) 5 . 347 ( 10
3
)
28 . 1 MPa
29
100
MPa
OK .
2
6
mm
4
SOLUTION (3.18) M
(w0 x
x
M
S
x
bh 6
;
all
2 L )( x 3 ) w 0 x
2
2
w0x
6L
6L
(1)
all
bh 1
h h1
At x=L:
3
3
w0x
6
3
6L
(2)
all
Divide Eq.(1) by (2): 2
h
2
h1
3
x L
h h 1 ( Lx )
or
3
3 2
SOLUTION (3.19) RA RB
1 2
wL,
M
x
wLx
1 2
bh 6
Equation (3.27), at a distance x : At x
L 2
2 bh 1
:
6
w 2
2
(
h
Divide Eq.(1) by Eq.(2): or h 2 h 1
x L
( Lx )
SOLUTION (3.20)
4
L 2
L 4
)
2
2
h1
2
1 2
w 2
( Lx x ) 2
Lx x L
2
2
1
(1)
all
2
4
y Es
n
(20)200=40(103)
3
( 4 1 0 )( 3 2 5 )
2 .8 9 1 0
126 10 6
20 M
s
( 0 .1 6 2 5 )
2 .8 9 1 0
3
3
12
9
3
( 3 .8 1 0 )( 3 0 0 )
mm 6
;
1 4 1 .6
w
M
3
12
7 .3 5 1 0
w
M ;
20
Ew
It
12.5 mm
I
bh 6
(2)
300 mm
nM y
2
z
x
a ll
all
12.5 mm
s
1
200 mm
M
S
2
wx ,
4
My I
2 .8 9 1 0
M
w
( 0 .1 5 )
2 .8 9 1 0
3
kN m
112 kN m
s
Stress in steel governs: M 112 kN m
SOLUTION (3.21) Es
n
I
Ew
1 12
M
20
[( 7 5 4 0 t )( 2 2 5 )] 3
1 8
wL 2
[ 7 5 (1 0 0 )] ( 6 4 .9 4 1 3 7 .9 6 9 t ) 1 0 3
1 12
[3 5 1 0 ( 2 .5 )(1 2 )] 2 7 .3 4 3
1 8
kN m
2
Therefore, we write
s
w
nM y I
;
135 10
6
3
2 0 ( 2 7 .3 4 1 0 )( 0 .1 1 2 5 ) ( 6 4 .9 4 1 3 7 .9 6 9 t ) 1 0
6
,
t 1 0 .2 9 m m
Similarly
My I
;
8 10
6
3
( 2 7 .3 4 1 0 )( 0 .1 1 2 5 ) ( 6 4 .9 4 1 3 7 .9 6 9 t ) 1 0
Stress in steel governs: t 1 0 .2 9 m m
30
6
,
t 8 .4 1 5 m m
6
m
4
3
m
4
SOLUTION (3.22) Transformed Section of Brass y Es
n
It [
89.1 mm C1
z
3
A d ]1 [ 2
bh 36
3
bh 12
[ 3 6 ( 2 0 0 )(1 2 0 ) 3
1
9.1 mm 43.4 mm
y 55.9 mm
2
Eb
A d ]2 2
200 2
1 2 0 ( 9 .1 ) ]1 2
[ 1 2 (1 0 0 )( 2 5 ) 1 0 0 ( 2 5 )( 4 3.4 ) ] 2 3
1
C2
15 . 43 10
1100mm
6
mm
2
4
200 mm Thus 120 (10
6
3
)
M ( 55 . 9 10
)
2 M ( 89 . 1 10
)
6
15 . 43 10
M 33 . 12 kN m
,
Similarly, 140 (10
6
15 . 43 10
3
)
6
M 12 . 12 kN m
,
(governs)
SOLUTION (3.23) n
Es Eb
2
I t I b nI
s
[
30 ( 50 )
3
12
( 7 .5 )
4
4
( 7 .5 )
] 2[
4
4
] 314 , 985
mm
4
Thus, we write ( b ) max
M ( 25 10
Mc It
314 , 985 ( 10
3
)
12
120 (10
)
6
M 1 . 512
),
kN m
Similarly, ( s ) max
3
2 M ( 7 . 5 10
nMc It
314 , 985 ( 10
12
)
140 (10
)
6
M 2 . 94 kN m
),
SOLUTION (3.24) n
Eb Ea
1.5 Ib
Transform to brass:
64
[d
2
( d2 ) ] 4
15 d
4
1024
Aluminum core: Each element of area of area has its width reduced in ratio n . Ia
1 n 64
( d2 )
4
d
4
1024 n
and It Ib Ia
d
4
1024
( 15
1 n
)
We have
b
Mc It
M
,
bIt d 2
d 512
3
b ( 15
1 n
)
Substituting numerical values: M
( 0 .0 5 ) 512
3
(3 5 0 1 0 )(1 5 6
1 1 .5
) 4 2 0 5 .7
31
N m = 4 .2 1 k N m
(governs)
SOLUTION (3.25) 1 0 5 . Apply Eq.(3.31): o
( a ) x'
1 2
(25 15)
1 2
( 2 5 1 5 ) c o s 2 1 0 1 0 sin 2 1 0 o
o
x’
5 17 . 32 5 17 . 32 MPa
x'y'
1 2
( 2 5 1 5 ) s in 2 1 0
10 cos 210
o
(20) (10) 2
x'y' 15
o
o
x
y’
10 8 . 66 1 . 34 MPa
( b ) 1, 2 5
x'
5 2 2 .3 6
2
or 1 17 . 36 MPa
p ' ' 13 . 28
27 . 36 MPa
2
o
1
2
p''
SOLUTION (3.26) 70 (a)
y’ m ax
x
30 o
x’ ( b ) max
1 2
30 70 2
cos 2( 55 ) o
20 17 . 1 37 . 1 kPa
x 55
30 70 2
y
xy
20 17 . 1 2 . 9 kPa
30 70 2
sin 2 ( 55 ) 47 kPa o
[ 30 ( 70 )] 50 kPa
SOLUTION (3.27) y 50(0.866) (0.866)
35 90 125
50(1)
x’ x
60
x
,
y
0,
xy
50 MPa
0 .8 6 6 x 0 .5 ( 5 0 ) 0 .5 ( 5 0 )
F x 0:
o
o
x
57 . 74 MPa
(C )
50(0.5) ( a ) Equations (3.31) with 1 2 5 : o
x'
x'y'
2 8 .8 7 2 8 .8 7 c o s 2 5 0 5 0 s in 2 5 0 o
28 . 87 sin 250
o
50 cos 250
o
o
6 5 .9 8
M Pa
44 . 23 MPa
(CONT.)
32
3.27 (CONT.) 1 2
( b ) 1 , 2 2 8 .8 7 [( 2 8 .8 7 ) 5 0 ] 2 8 .8 7 5 7 .7 4 2
1 2 8 .8 7 M P a ,
2
2
p ''
8 6 .6 1 M P a ,
1 2
ta n
1
x’
50 2 8 .8 7
30
o
x
44.23
p''
35
x’
o
x y’
1
2
65.99
SOLUTION (3.28) From Solution of Prob. 3.27:
x
xy
6
57 . 74 ( 10 200 ( 10 2 (1 0 .3 )
200 ( 10
9
)
9
)
)
289 μ ,
( 50 10
6
5 7 .7 4 M P a ,
x
y
x
y
0 , and
50 M P a.
xy
87 μ
) 650 μ
( x ) AC
C
B
40
34.6
L A C 52.915
60
A
40 4 0 .8 6
289 87 2 650 2
28987 2
o
c o s 2 ( 4 0 .8 6 ) o
s in 2 ( 4 0 .8 6 )
1 0 1 2 7 .0 7 3 2 1.6 1 194 μ
o
D
20 o
( a ) ( x ' ) A B 1 0 1 1 8 8 c o s 2 (1 2 0 ) 3 2 5 s in 2 (1 2 0 ) 2 8 8 μ o
o
L A B 4 0 ( 2 8 8 ) 0 .0 1 1 5 m m
( b ) L A C 5 2 .9 1 5 (1 9 4 1 0
6
) 0 .0 1 0 3 m m
SOLUTION (3.29)
y (a)
y
y
7 0 ( 0 .5 ) 3 5 M P a
60.6(0.866)=52.5
x y 7 0 ( 0 .8 6 6 ) 6 0 .6 M P a
( 0 .5 )
30
o
140(1)
x
x
0 . 5
Fx 0 :
(b) 1
175 35 2
x
[( 175 2 35 )
140 52 . 5 0
x
175
M Pa
p'
1
2
( 60 . 6 ) ] 2 70 121 . 23 191 . 23 MPa 2
2 5 1 .2 3 M P a p'
1 2
ta n
1
2 ( 6 0 .6 ) 175 35
15
x x’
o
1
33
2
SOLUTION (3.30) From Solution of Prob. 3.29:
9
2 0 0 1 0 2 (1 0 .2 5 )
G
x
y
x'
175 M Pa,
x
(
1 E
10
80 10 y )
x
6 9
2 0 0 (1 0 )
3 5 M P a , and
Pa, 10
6 9
2 0 0 (1 0 )
xy
xy
G
xy
6 0 .6 1 0 8 0 1 0
6
9
6 0 .6 M P a .
758
(1 7 5 3 .7 5 ) 9 1 9 ,
( 3 5 4 3 .7 5 ) 3 9 4
919 394 2
9
y
919 394 2
cos 2 (141 . 5 ) o
o
758 2
sin 2 (141 . 5 ) 2 6 2 .5 1 4 7 .6 8 3 6 9 .2 9 7 7 9 .4 7
x’
L B D 0.0416 m
B
C
0.05 m A
=141.5o x
0.075 m D
Thus L B D 0 .0 4 1 6 ( 7 7 9 .4 7 1 0
6
) 0 .0 3 1 3 m m
SOLUTION (3.31) 40
x
xy
o
y
100 cos 45
100 sin 45
o
o
70 . 71 MPa
70 . 71 MPa
Equations (3.31):
x'
y'
70 . 71 0 69 . 64 1 . 07 MPa
x'y '
70 . 71 70 . 71 2
0 70 . 71 sin 80
0 70 . 71 cos 80
y’
o
x'
40
140 . 3 MPa
12 . 28 MPa
x’
x'y'
o
o
x y'
34
SOLUTION (3.32) 40 90 130
y
o
40 MPa ,
50 MPa ,
x
25 MPa
xy
( a ) Equations (3.31): 50 40 2
x'
50 40 2
c o s 2 (1 3 0 ) 2 5 sin 2 (1 3 0 ) o
o
5 7 . 814 24 . 62 27 . 43 MPa
(b)
x'y'
y'
[(
max
p'
1 2
45 sin 260
o
5 7 . 814 24 . 62 37 . 434
50 40 2
44 . 316 4 . 341 39 . 975
MPa
MPa
1
)
25 ] 2 51 . 48 MPa
2
2
1 25 45
tan
25 cos 260
o
14 . 53
o
x’
x'
y'
x’
x'y'
m ax
x
p'
x x
y’ SOLUTION (3.33) x”
B
AC BD 90 . 139
x’
C
75 mm
G E 2 .6
x
50 mm 3 3 .6 9
1 4 6 .3
o
xy
1 E
( xy G
o
1 E
( x y )
10 E
6
[ 5 0 0 .3 ( 4 0 ) ]
6 2 1 0 E
6
x
A y
mm
D y
x ) 6
2 5 1 0 E 2 .6
x'
10 E
6
6 5 1 0 E
[ 4 0 ( 0 .3 ) ( 5 0 ) ]
5 5 1 0 E
6
6
We have
6
10 2E
[( 62 55 ) ( 62 55 ) cos 2 ( 33 . 69
( 65 ) sin 2 ( 33 . 69
x"
6
10 2E
o
)]
89 . 85 E
(10
6
o
)
)
[117 ( 7 ) cos 2 (146 . 3 ) ( 65 ) sin 2 (146 . 3 )] o
o
Thus L A C x ' (9 0 .1 3 9 ) L B D x " (9 0 .1 3 9 )
8 9 .8 5 1 0 2 1 0 1 0
6
9
2 9 .8 4 1 0 2 1 0 1 0
9
6
(9 0 .1 3 9 ) 0 .0 3 8 6
mm
(9 0 .1 3 9 ) 0 .0 1 2 8
mm
35
29 . 84 E
6
(10 )
SOLUTION (3.34)
Vessel is thin walled.
1.2 kN/m B
0.9 m
M
V 1.5 m
VQ
xy
A
0 at A and C
Ib
V 2 .7 1 .2 (1 .5 ) 0 .9 k N
M 2 .7 (1 .5 )
C
2.7 kN
2 .7
1 2
(1 .2 )(1 .5 )
2
kN m
Point A
x'
Mc I
(a)
y
(b)
m ax
1 2
2 .7 ( 0 .4 5 ) 3
( 0 .4 5 ) ( 3 1 0
2
a
3
)
6 .3 M P a 1 ,
( 1 2 )
1 2
1 .4 1 5 M P a ,
2
a
pr 2t
4 2 ( 0 .4 5 ) 2 ( 3 1 0
3
)
3 .1 5 M P a
3 .1 5 1 .4 1 5 1 .7 3 5
( 6 .3 1 .7 3 5 ) 2 .2 8 2 5
M Pa
M Pa
Point C (a)
(b)
m ax
x'
2 1 .4 1 5 3 .1 5 4 .5 6 5 M P a ,
1 .4 1 5 M P a , 1 2
( 1 2 )
1 2
( 6 .3 4 .5 6 5 ) 0 .8 6 8
1 6 .3 M P a
M Pa
SOLUTION (3.35) Refer to Solution of Prob. 3.34.
(a)
x
xy
1, 2
Thus
x'
VQ Ib 0 .9
rt
a
0 3 .1 5 3 .1 5 M P a ,
0 .9 ( r t )(
2r
0 .9
( 0 .4 5 ) ( 3 1 0
3
)
0 .2 1 2 2 M P a
1
[( 6 .3 23 .1 5 ) ( 2 .1 2 2 ) ] 2 4 .7 2 5 1 .5 8 9 2
1 6 .3 1 4 M P a 1 2
6 .3 M P a
(see Table A-3)
2
or
( b ) m ax
y
)
3
r t(2t)
6 .3 3 .1 5 2
2 3 .1 3 6 M P a
( 1 2 ) 1 .5 8 9 M P a
36
SOLUTION (3.36)
x’
P A
Tr J
3
1 1 0 (1 0 ) 2
2
( 0 .0 2 5 0 .0 1 2 5 )
7 4 .7 M P a
74.7
x
3
2 .3 (1 0 )( 0 .0 2 5 )
4
9 9 .9 6 M P a 4
( 0 .0 2 5 0 .0 1 2 5 )
2
99.96
50 90 140
o
x’
Equations (3.31): x'
140
o
x'y'
x'
7 4 .7 2
7 4 .7 2
c o s 2 8 0 9 9 .9 6 s in 2 8 0 o
o
1 4 2 .3 M P a
x
x'y'
7 4 .7 2
s in 2 8 0 9 9 .9 6 c o s 2 8 0 o
o
1 9 .4 2 M P a
y’ SOLUTION (3.37) 410 mm
y A
120 18
z
24 kN
M
40
50
120
24
3 4
18 kN
y
A 40 (120 ) 4 . 8 10
x
A
I
x
xy
1 12
( 40 )( 120 )
3
3
mm
5 . 76 10
2 6
mm
4
M 18 ( 0 . 12 ) 24 ( 0 . 41 ) 12 kN m
We have at A: x
P A
Mc I
3
1 8 (1 0 )
4 . 8 (1 0
3
)
( 3 . 75 104 . 17 ) 10
xy
VQ Ib
3
2 4 1 0 ( 2 2 1 0 5 .7 6 1 0
6
100 . 4 2
2
6
)
( 0 .0 4 )
3
1 2 (1 0 )( 0 . 0 5 )
5 . 7 6 (1 0
6
6
)
Q 4 0 (1 0 )(5 5 ) 2 2 (1 0 ) m m
100 . 4 MPa
2 .2 9 M P a
Thus
or
1,2
100 . 4 2
[(
)
1 1 0 0 .5 M P a
2
2 0 .0 5 M P a
m a x 5 0 .2 8 M P a
and tan 2 p '
or
p ' 1 . 31
2
xy
x
o
1
( 2 . 29 ) ] 2 50 . 2 50 . 25
2 ( 2 . 29 ) 100 . 4
1 . 31
37
3
3
SOLUTION (3.38)
y y
B
B
16
M P
yo P=40 kN
C
62 z
A Z
10 mm
y
A
8
45
From Z axis: 6 2 1 6 (3 9 ) 4 5 8 ( 4 )
y
2 9 .7 m m
62 16 45 8
y 0 1 0 y 3 9 .7 m m A 1 6 6 2 4 5 8 1 .3 5 (1 0 ) m m 3
2
About the z axis: I
1 6 (6 2 )
3
1 6 6 2 ( 9 .3 ) 2
4 5 (8 )
12
3
4 5 8 ( 2 5 .7 )
2
12
1 0 ( 0 .3 1 8 0 .0 8 6 0 .0 0 2 0 .2 3 8 ) 0 .6 4 4 1 0 6
6
mm
4
M 4 0 0 .0 3 9 7 1 .5 8 8 k N m
Thus,
a
P
A
40 10
Mc
( b ) B
3
1 .3 5 1 0
I
( b ) A 9 9 .4
3
2 9 .6 M P a
1 5 8 8 ( 0 .0 4 0 3 ) 0 .6 4 4 1 0
2 9 .7
6
9 9 .4 M P a
7 3 .3 M P a
4 0 .3
A
2 6 .3 7 3 .3 9 9 .6 M P a
B
2 6 .3 9 9 .4 7 3 .1 M P a
SOLUTION (3.39) A b h 2 5 (1 0 0 ) 2 .5 (1 0 ) m m 3
I
2
2 5 (1 0 0 ) 12
P 50 kN
M 5 0 1 0 ( 0 .0 5 ) 2 .5 k N m 3
( a ) At the top fibers: t
P A
Mc I
5 10 2 .5 (1 0
3
3
2 .5 1 0 ( 0 .0 5 ) 3
)
2 .0 8 (1 0
6
20 60 40 M Pa
At the bottom fibers: b 20 60 80 M Pa
38
)
40.3
3
2 .0 8 (1 0 ) m m 6
4
SOLUTION (3.40) A b h 2 5 (1 5 0 ) 3 .7 5 (1 0 ) m m 3
I bh
1 2 2 5 (1 5 0 )
3
2
1 2 7 .0 3(1 0 ) m m
3
6
4
At the bottom fibers:
P
b
Mc
A
P
I
3 .7 5 (1 0
3
0 .0 7 5 P ( 0 .0 7 5 )
)
7 .0 3 (1 0
6
)
Solving, 120 10
2 6 6 .7 P 8 0 0 .1 4 P
6
Pa ll 1 1 2 .5 k N
SOLUTION (3.41) J
[( 0 .0 6 ) ( 0 .0 5 0 ) ] 1 0 .5 4 (1 0 4
2
c 0 .0 6
4
m,
45
6
) m
8 0 M P a m ax
o
4
Tc J
Thus T ( 0 .0 6 )
8 0 (1 0 ) 6
1 0 .5 4 (1 0
6
)
T 1 4 .0 5 k N m
,
SOLUTION (3.42) ( a ) 1, 2
500 800 2
1
[( 5 0 0 2 8 0 0 ) ( 3 52 0 ) ] 2 6 5 0 2 3 0 μ 2
2
or 1 880 μ ,
( b ) max G
max
2 420 μ , 3
70 ( 10 )
2 (1 0 .3 )
max
460 μ
( 460 ) 12 . 38 MPa
SOLUTION (3.43) (a)
max
2 [(
200 600 2
A
1
)
2
( 400 ) ] 2 566 μ 2 2
149
( b ) x'
200 600 2
200 600 2
400 2
o
c o s 2 (1 4 9 )
x
C
s in 2 (1 4 9 ) 1 3 0 μ o
o
L A . C 2 7 .3 3 3 m m
Thus L A C x ' L A C 1 3 0 (1 0
6
)( 2 7 .3 3 3 ) 3 .5 5 1 0
3
mm
SOLUTION (3.44) 1, 2
50 250 2
[( 5 0 22 5 0 ) ( 2
150 2
1
) ]2 150 μ 125 μ 2
or 1 275 2 25
(CONT.)
39
3.44 (CONT.) Apply Hooke’s Law (with z 0 ): 6
2 7 5 (1 0
and
2 5 (1 0
6
)
)
( 1 2 )
1 9
2 1 0 (1 0 )
(1)
( 2 1 )
1 9
2 1 0 (1 0 )
(2)
Solving 1 65 . 2 MPa ,
24 . 8 MPa
2
SOLUTION (3.45)
a
pr 2t pr t
p (500 )
P A
2 (1 0 )
0 .0 2 2 ( 0 .5 )( 0 .0 1 )
25 p 2 M Pa
x
50 p
y
50 20
s in 4 0
A=1
40
o
cos 40
5 0 ( 2 5 p 2 ) s in
2
40
o
50 p cos 40 2
p all 1 . 281 MPa
or
25p-2
o
F y 0:
F x 0:
2 0 ( 2 5 p 2 ) s in 4 0 c o s 4 0 o
5 0 p c o s 4 0 s in 4 0
o
o
50 p
p all 1 . 546
or
o
MPa
SOLUTION (3.46) ( a ) x y ( x ' y ' ) s in ( 2 ) x ' y ' c o s ( 2 ) o
o
0 ( 2 4 0 4 1 0 ) s in ( 6 8 ) o
Then
1
x
2 1
( x '
y'
)
1 2
( x ' 1
(240 410)
2
cos(68 ) , o
x'y'
1
) c o s ( 2 ) o
y'
x'y'
2
( 2 4 0 4 1 0 ) co s( 6 8 ) o
2
421 μ
s in ( 2 ) o
x'y'
1
( 4 2 1) s in ( 6 8 ) o
2
9 8 .2 μ
y
1
( x '
2 1
y'
)
1 2
(240 410)
2
( x ' 1
1
) c o s ( 2 ) o
y'
2
( 2 4 0 4 1 0 ) co s( 6 8 ) o
2
( b ) Hooke’s Law, with t y , a x , and t 2 a
So
y
x
x
(1 2 )
E
552 9 8 .2
5 .6 2
y
2 1 2
x
(2 )
E
0 .3 5
,
40
s in ( 2 ) o
x'y'
1 2
552 μ
x
o
( 4 2 1) s in ( 6 8 )
o
o
SOLUTION (3.47) a 880
( a 0 ),
b 320
o
( b 6 0 ),
a x c o s a y s in a 2
Thus
8 8 0 (1 0
6
xy
) x c o s 0 y s in 0 o
( c 1 2 0 ) o
s in a c o s a
2
2
c 60
o
2
o
o
x 880 μ
o
s in 0 c o s 0 ,
xy
Likewise, b x c o s b y s in b x y s in b c o s b 2
3 2 0 (1 0 3 2 0 (1 0
6
6
2
) x c o s ( 6 0 ) y s in ( 6 0 ) 2
o
2
) 0 .2 5 x 0 .7 5
y
0 .4 4 3
c x c o s c y s in c 2
and,
6 0 (1 0
6
6 0 (1 0
6
o
2
xy
s in ( 6 0 ) c o s ( 6 0 ) o
xy
(1)
xy
s in c c o s c
) x c o s ( 1 2 0 ) y s in ( 1 2 0 ) 2
o
2
) 0 .2 5 x 0 .7 5
y
o
0 .4 4 3
o
s in ( 1 2 0 ) c o s ( 1 2 0 ) o
xy
o
(2)
xy
Subtract Eq. (2) from Eq. (1): So
3 8 0 μ 0 .8 8 6
xy
and then from Eq. (1):
xy
4 2 8 .9 μ
y
1 0 2 .1 μ
2
880 A
388
2
O
avg
x p
288.1
'
()
C R
y
1
1 2
( x y )
(8 8 0 1 0 2 .1) 3 8 9 μ
2 R [ (8 8 0 3 8 9 ) ( 2
4 2 8 .9 2
5 3 5 .8 μ
1,2 a vg R
1 3 8 9 5 3 5 .8 9 2 4 .8 μ 2 1 4 6 .8 μ ta n 2
p
'
2 8 8 .1 880 389
p ' 1 6 .8
,
x’
o
x 16.8o
924.8 y
A
y’ 146.8
41
2
1
) ]2
SOLUTION (3.48) y xy
A
x
Tc J
y
2
( x y ) 0
a v g R s in 2 xy
c
3
2
0
xy
xy G
x y 0
Mohr’s circle for strain: avg
25
o
x
1
J
Tc
s i n 2
2G
R
1
2
1
2
xy
2
x 2
xy
A’ x’ R
s in 2
O
s i n 2
y
2G J
This gives 2G J
T
c s in 2 Substitute the given data:
(8 0 .5 1 0 )( 0 .0 4 4 ) ( 6 0 0 1 0 9
T
3
( 0 .0 4 4 ) s in 5 0
6
o
)
2 2 5 .0 3 8 k N m
SOLUTION (3.49) Using Eqs. (3.39), we have x 300 μ,
y
150 μ,
xy
2(375) (300 150) 600 μ
Equations (3.38) are thus, 300 150
1, 2
p
(
450
2
1 300 μ ,
1 2
ta n
1
) ( 2
2
600
)
2
175 375
2
2 450 μ [
600 300 150
] 2 6 .6
o
x ' 1 7 5 2 2 5 c o s 5 3 .1 3 0 0 s in 5 3 .1 4 5 0 2 o
o
Thus, p " 2 6 .6
y’ 300
o
450 x’ 26.6o x 42
SOLUTION (3.50) Given: D 5 0 m m , t 1 5 m m , P 2 5 k N . Refer to Figs. P3.50 and C.1: d
r
r d
D d
K
32 33 34 35
9 8.5 8 7.5
0.28 0.26 0.24 0.21
1.56 1.52 1.47 1.43
1.64 1.66 1.62 1.7
t
We have 6
a ll K t
P A
;
150 (10 ) 1.9
Kt
3
25 (10 ) 0 .0 1 5 h
In this equation, minimum h is reached when K t is minimum. Thus, use h 34 mm
SOLUTION (3.51) D d
45 30
1.5 ,
r d
6 30
0 .2
Figure C.1, K t 1.7 2 .
Hence
and
Pall A
nom
max
1 . 72
nom
210 1 . 5
81 . 4 MPa
1 . 72
( 30 12 )( 81 . 4 ) 29 . 3 kN
SOLUTION (3.52) At the notch and hole:
nom
P ( D d h 2 r )t
12 10
3
( 0 .0 9 0 .0 1 5 2 0 .0 0 7 5 ) ( 0 .0 1)
20 M Pa
For the notch : (see Fig. C.1): D
d
90
1 .2
75
r
7 .5
d
0 .1,
75
K t 1 .7 8
m a x K t n o m 1 .7 8 ( 2 0 ) 3 5 .6 M P a
For the hole (see Fig. C.5): dh
D
15
0 .1 6 7 ,
90
K t 2 .5
m a x K t n o m 2 .5 ( 2 0 ) 5 0 M P a
Hence
SOLUTION (3.53) Kt
m ax
nom
180
1 .6 3 6 and D d 1 .5
110
From Fig. C.1: r d 0 .2 4 . Then D 2r d
(CONT.)
43
3.53 (CONT.) 4 0 .6 2 5 2 ( 0 .2 4 d ) d 1 .4 8 d
gives or
d 2 7 .4 5 m m r 6 .5 9 m m
SOLUTION (3.54) For
r
0 .2 0 :
d D 2r d ;
4 0 .6 2 5 2 ( 0 .2 d ) d 1 .4 d
or d 2 9 .0 2 m m We thus have D
1 .4
d
Figure C.1 gives K t 1 .7 . Hence,
3 5 0 (1 0 ) 1 .7
Pa ll
6
m ax
(1 2 .5 1 0
3
) ( 4 0 .6 2 5 1 0
3
)
Pa ll 1 0 4 .5 k N
or
SOLUTION (3.55) (ksi)
avg
1
(
2
1
x
y)
(1 6 8 8 4 ) 1 2 6 M P a
2
3
O
R
C
(
x
y
1 (MPa)
(
168 84 2
5 9 .4 M P a
avg =59.4 ( a ) 1,2
R
avg
1 1 2 6 5 9 .4 1 8 5 .4 M P a 2 6 6 .6 M P a
3 21 M Pa
( b ) ( m a x ) a
1 2
1
( 1 3 )
[1 8 5 .4 ( 2 1) ] 1 0 3 .2 M P a
2
44
) 2
2
2 xy
) (42) 2
2
SOLUTION (3.56) (MPa)
avg
avg
1
(
2
1
x
y)
(5 0 0 ) 2 5 M P a
2
R 3
O
1
C
R
(
x
y
) 2
2
(MPa)
50 0
(
2 xy
) (25) 2
2
2 3 5 .3 6 P a
( a ) 1,2
R
avg
1 2 5 3 5 .3 6 6 0 .3 6 M P a
2 1 0 .3 6 M P a 3 60 M Pa
( b ) ( m a x ) a
1
( 1 3 )
2
1
[ 6 0 .3 6 ( 6 0 ) ] 6 0 .1 8 M P a
2
SOLUTION (3.57) (ksi) avg =42
avg
1
(
2
35
1
x
y)
(7 0 1 4 ) 4 2 M P a
2
3
C
1
(MPa)
R
(
x
y
) 2
2
(
70 14
2 xy
) (56) 2
2
2 6 2 .6 M P a
( a ) 1,3
avg
R
1 4 2 6 2 .6 1 0 4 .6 M P a
3 4 2 6 2 .6 2 0 .6 M P a 2 35 M Pa
(CONT.)
45
3.57 (CONT.) ( b ) ( m a x ) a
1 2 1
( 1 3 ) [1 0 4 .6 ( 2 0 .6 ) ] 6 2 .6 M P a
2
SOLUTION (3.58) ( a ) In the yz plane: ' =35
70 MPa
(MPa)
y
21 MPa
z
21
C
3
R
(MPa)
O 1
70 We have R
35 21 2
2
4 0 .8 2 M P a
Thus 1 R ' 4 0 .8 2 3 5 5 .8 2 M P a
3 R ' 4 0 .8 2 3 5 7 5 .8 2 M P a 2 28 M Pa
( b ) Using Eqs. (3.58) : oct
1 3
1
[ ( 5 .8 2 2 8 ) ( 2 8 7 5 .8 2 ) ( 7 5 .8 2 5 .8 2 ) ] 2 2
2
2
3 3 .4 9 M P a
oct
1 3
(5 .8 2 2 8 7 5 .8 2 ) 3 2 .6 7 M P a
From Eq. (3.50), m a x 12 (5 .8 2 7 5 .8 2 ) 4 0 .8 2 M P a SOLUTION (3.59) ( a ) Using Eq. (3.47): (70 i )
0
56
0
(1 4 i )
0
56
0
(1 4 i )
0
(CONT.)
46
3.59 (CONT.) Expanding, (1 4 i )[( i 7 0 )( i 1 4 ) 3 1 3 6 ] 0
or
1 42 M Pa,
2 14 M Pa,
3 98 M Pa
( b ) From Eq. (3.50), m a x 12 ( 4 2 9 8 ) 7 0 M P a acts on planes shown in Fig. 3.41. ( c ) Using Eqs. (3.52), we have o c t 13 ( 4 2 1 4 9 8 ) 1 4 M P a oct
1 3
1
[( 4 2 1 4 ) (1 4 9 8 ) ( 9 8 4 2 ) ] 2 2
2
2
6 0 .4 9 M P a
They act on planes depicted in Fig. 3.43. SOLUTION (3.60) The principal stresses are 1 8 4 M P a , 2 6 3 M P a , and 3 1 2 6 M P a . ( a ) By Eq. (3.50), m ax
1 2
(8 4 1 2 6 ) 1 0 5 M P a
acts on planes shown in Fig. 3.41. ( b ) Applying Eqs. (3.52): o c t 13 (8 4 6 3 1 2 6 ) 7 M P a oct
1 3
1
[ (8 4 6 3 ) ( 6 3 1 2 6 ) ( 1 2 6 8 4 ) ] 2 9 4 .4 4 M P a 2
2
2
They act on planes shown in Fig. 3.43. SOLUTION (3.61) The principal stresses are 1 2 9 7 .5 M P a , 2 3 6 .8 2 M P a , and 3 5 4 .7 4 M P a . The direction cosines: l cos 40
o
m cos 60
n cos 66 . 2
o
o
Thus, by Eqs. (3.51), we obtain 2 9 7 .5 (c o s 4 0 ) 3 6 .8 2 (c o s 6 0 ) 5 4 .7 4 (c o s 6 6 .2 ) o
2
o
2
o
2
= 174.9 MPA [( 2 9 7 .5 3 6 .8 2 ) (c o s 4 0 ) (c o s 6 0 ) 2
o
2
o
2
(3 6 .8 2 5 4 .7 4 ) (c o s 6 0 ) (c o s 6 6 .2 ) 2
o
2
o
2 1
( 5 4 .7 4 2 9 7 .5 ) (c o s 6 6 .2 ) (c o s 4 0 ) ] 2 2
o
2
o
2
1 4 8 .9 M P a
End of Chapter 3
47
CHAPTER 4
DEFLECTION AND IMPACT
SOLUTION (4.1) ( a ) We have A req . A req .
and
48 mm
3
3
10 ( 10 )( 6 10 )
PL
)
250 1 . 2
all
E
3
10 ( 10
P
5 ( 200 10
3
2
60 mm
)
2
Since 60 48 mm , d
(b) k
AE L
( 6 0 1 0
6
4 ( 60 )
4A
8 . 74 mm .
9
)( 2 0 0 1 0 )
2 (1 0 ) k N m 3
6
SOLUTION (4.2) Refer to Example 4.4. Use numerical values for bar AB. Cross-sectional area: A A B 1 2 8 9 6 m m . 2
Stress:
AB
F A
4 0 1 0 9 6 (1 0
3
6
4 1 6 .7 M P a 4 3 5 M P a
)
OK.
Deflection: AB
FL AE
k AB
2 .3 8 m m
3
3
4 0 (1 0 )
F
3
4 0 (1 0 )( 0 .6 ) ( 9 6 )(1 0 5 )(1 0 )
2 .3 8 (1 0
3
1 6 .8 1(1 0 ) k N m 3
)
SOLUTION (4.3) The cross-sectional area:
A
4
(D
2
d ) 2
D 4
2
[1 ( Dd ) ] 0 .5 0 2 7 D 2
2
Also A
7 5 1 0
P
3
1 4 0 1 0
6
0 .5 3 6 1 0
3
m
2
Equating these, D
2
1 .0 6 6 2 1 0
3
D 0 .0 3 2 7 m
2
m ,
It follows that k
PL AE P
3
7 5 1 0 ( 0 .3 7 5 ) ( 0 .5 3 6 1 0 7 5 1 0
3
0 .7 2 9 1 0
3
3
9
)( 7 2 1 0 )
0 .7 2 9 1 0
3
m
1 0 2 .8 8 M N m
SOLUTION (4.4) Statics: RA
36 kN
R A RB 36
RB
kN
(1)
Deformations and Compatibility: Assume gap closes. (CONT.)
48
4.4 (CONT.) 0 .3 5 1 0
3
1
(1 2 .5 1 0
3
2
9
) (1 2 0 1 0 )
( 0 .2 R A 0 .2 5 R B )
4
R A 1 .2 5 R B 2 5, 7 7 0 .8 7 7
or
(2)
Solving Eqs.(1) and (2): R B 4 .5 4 6
kN
R A 3 1 .4 5 4
kN
Since the answer for R B is positive, the gap closes, as assumed. SOLUTION (4.5) Increase in length due to T (unrestrained): t
(T )L
(12 10
6
)( 120
o
)( 250 ) ( 23 10
6
)( 120
o
)( 300 ) 1 . 188 mm
( a ) Compressive axial force P P t 1 1 . 188 1 0 . 188 mm
(1)
But
P
PL AE
P ( 0 .2 5 )
5 0 0 (1 0
6
9
)( 2 1 0 1 0 )
P ( 0 .3 )
1 0 0 0 (1 0
6
(2)
9
)( 7 0 1 0 )
Equating Eqs.(1) and (2): 0 .1 8 8 (1 0
3
) 2 .3 8 1(1 0
9
) P 4 .2 8 6 ( 1 0
9
)P
P 28 . 2 kN
or
( b ) Change in length of aluminum bar a ( t ) a ( P ) a a ( T ) L a 4 .2 8 6 ( 1 0 ( 23 10
6
)( 120
o
)( 0 . 3 ) 4 . 28 (10
9
9
)P
)( 28 . 2 10 ) 0 . 707 3
mm
SOLUTION (4.6) Refer to Solution of Prob.4.5: ( a ) P t 1 . 188 mm and
1 . 188 (10
( b ) a 0 .8 2 8 (1 0
3
3
) ( 2 . 381 4 . 286 )10
) 4 .2 8 (1 0
9
9
P , P 178 . 2 kN
)(1 7 8 .2 1 0 ) 0 .0 6 5 3 m m 3
SOLUTION (4.7) Let the compressive axial force in the pipe be R. Moment equilibrium about point A: R
Pa b
1 2 (1 .3 )
4 4 .5 7 k N
0 .3 5
a P
b
C
A R
C
B
B
(CONT.)
49
4.7 (CONT.) The beam can rotate only about A and the deflections B and C can be described by the rotational angle as B a and C b a
B
Hence,
b
C
(1)
The contraction of pipe C is as follows: C
RL
RL
( 4 )( D
AE
2
(2)
d )E 2
Inserting Eq.(2) into Eq.(1) gives B
b(D d )E 2
2
0 .3 2 9 1 0
4 ( 4 4 .5 7 1 0 )( 0 .6 2 5 )(1 .3 ) 3
4RLa 3
( 0 .3 5 )( 0 .1 0 5 0 .0 9 5 )( 2 0 0 1 0 ) 2
2
9
m = 0 .3 2 9 m m
SOLUTION (4.8) (a) b
P Lb
P La
a
Eb A La
a b;
Ea
Ea A
Lb
(1)
Eb
La Lb L
(2)
Solving, Eb 1 Lb L ( ) L E Ea Eb 1 a Eb
,
La L Lb
Introducing the given data: 1 L b 0 .6 0 .3 6 7 m , 70 1 110
(b)
PL
P
A
AE 3
1 0 0 (1 0 ) ( 4 ) ( 0 .0 4 )
2
(
Li Ei
L a 0 .2 3 3 m
a b
0 .3 6 7 110
0 .2 3 3 70
)
1 10
9
0 .2 6 5 0 .2 6 5 0 .5 3 m m
50
SOLUTION (4.9)
C FD
Q=12 kN A
2b
A 2 D
b
PCD = -20 kN D
FD =24 kN
D
B
D 0.15 m
PDE =4 kN
M
B
0 : 2b Q b FD
E P=4 kN
FD 2 Q 2 4 k N
(a) D
A
0.3 m
2 0 (1 0 )(3 0 0 ) 3
PL
AE
130 10
6
(7 0 1 0 ) 9
0 .6 6 m m
2 D 1 .2 2 m m
(b) E
PL
AE 1
130 70
10 130 10
6
3
(7 0 1 0 ) 9
( 0 .6 6 ) 0 .5 9 1 0
[ 4 0 .1 5 2 0 0 .3 ] 3
m 0 .5 9 m m
SOLUTION (4.10) PD E 4 k N
(a) D
A
PD C 2 8 k N 2 8 (1 0 )(3 0 0 ) 3
PL
AE
130 10
6
(7 0 1 0 ) 9
0 .9 2 m m
2 D 1 .8 4 m m
(b) E
PL
AE 1
130 70
10 130 10
6
3
(7 0 1 0 ) 9
[ 4 0 .1 5 2 8 0 .3 ]
( 0 .6 8 .4 ) 9 8 9 1 0
3
m 0 .9 9 m m
SOLUTION (4.11) F C
F B
120 mm
180 mm (CONT.)
51
4.11 (CONT.) T CD F ( 0 . 12 ) 500
N m
or F 4 . 167
( a ) AB
7 5 0 (1 .2 )
TL GJ
T AB 4 . 167 ( 0 . 18 ) 750
kN
9
7 9 (1 0 )
N m
0 .0 2 8 ra d
( 0 .0 2 2 5 )
4
2
CD
5 0 0 (1 .8 ) 9
( 7 9 1 0 )
0 .0 7 7 ra d
( 0 .0 1 7 5 )
4
2
D CD 1 . 5 AB 0 . 119
Thus (b)
AB
2TAB
c
2 (750 )
3
( 0 .0 2 2 5 )
rad 6 . 82
o
4 1 .9 2 M P a
3
SOLUTION (4.12) all TC D FC
125
150 1 .2
3 2
a ll ( d )
T CD
6
3
1 2 5 1 0 ( 0 .0 6 5 )
16
( r gear ) C
MPa 6 .7 4 k N m
16
56 . 167
6 . 74 0 . 24 2
kN m
Hence, we have T AB 56 . 167 ( 0 .236 ) 10 . 11 kN m
all
Thus
16 T AB
3 d1
d1 3
,
3
16 ( 10 . 11 10 ) 6
( 125 10 )
or d 1 74 . 4 mm
SOLUTION (4.13) ( a ) Polar moment of inertia for a hollow cylinder is J
(c b ) 4
2
4
c
4
2
[1 ( bc ) ] 4
(1)
From Eq. (4.10), we have
J c
T
m ax
4 .5 1 0
3
1 4 0 1 0
6
3 2 .1 4 3 1 0
6
m
2
3 2 .1 4 3 m m
Note that, b 0 .5 c . Using Eqs. (1) and (2) then c 2
3
[1 ( 0 .5 ) ] 3 2 .1 4 3,
c 2 .7 9 5 m m
4
So, D 2 c 5 .5 9 m m ,
(b)
TL GJ
(c) k
T
d 0 .5 D 2 .7 9 5 m m
3
( 4 .5 1 0 )( 0 .2 5 ) 9
( 7 9 1 0 )( 3 2 .1 4 3 1 0 3
4 .5 (1 0 ) 0 .1 5 8 5
6
)( 2 .7 9 5 1 0
3
)
0 .1 5 8 5 ra d 9 .0 8 1
2 8 .3 9 k N m ra d
52
o
2
(2)
SOLUTION (4.14) TBC 5 6 0 N m
TC D 8 4 0 N m
( a ) Shaft BC J BC
d
4
32
(3 4 ) 1 3 1, 1 9 4 m m 4
LBC 6 2 5 m m
4
32
TBC 5 6 0 N m
So
BC
TBC L BC
G J BC
5 6 0 (6 2 5 1 0
3
)
( 7 9 1 0 ) (1 3 1, 1 9 4 1 0 9
0 .0 3 4 r a d 1 .9 5
12
o
)
( b ) Shaft CD J CD
d2 4
32
(2 5) 3 8, 3 5 0 m m 4
4
32
LCD 7 5 0 m m
TC D 8 4 0 N m
CD
TC D LC D
G J CD
8 4 0 (7 5 0 1 0
3
)
(7 9 1 0 )(3 8, 3 5 0 1 0 9
12
0 .2 0 8 r a d 1 1 .9 2
o
)
Hence B D B C C D 0 .0 3 4 0 .2 0 8 0 .1 7 4 ra d 9 .9 7
o
SOLUTION (4.15) The polar moment of inertia for the shaft is J
32
d )
4
(D
4
(5 0 3 5 ) 4 6 6 , 2 6 9 m m 4
32
4
4
Equilibrium Condition. From the free-body diagram of appropriate portions of the shaft: T A C 1 .1 k N m
T C D 1 .9 k N m
T D E 2 .6 k N m
The results are shown on the torque diagram in Fig. S4.15 ( a ) Angle of twist. The shear modulus of elasticity G is a constant for the entire shaft. Through the use of Eq. (4.9), we obtain A
TL GJ
1 GJ
10
( T A C L1 T C D L 2 T D E L 3 )
3
9
( 7 9 1 0 ) ( 4 6 6 , 2 6 9 1 0
12
)
[ ( 1 .1) ( 0 .4 5 ) (1 .9 ) ( 0 .3 7 5 ) ( 2 .6 ) ( 0 .6 2 5 ) ]
0 .0 5 0 0 ra d = 2 .8 6
o
Comments: A positive result means that the gear will rotate in the direction of the applied torque at free end. ( b ) Maximum shear stress. The largest stress takes place in region DE, where magnitude of the highest torque occurs (Fig. S4.15) and J is a constant for the shaft. Applying the torsion formula: m ax
TD E ( J
D 2
)
3
( 2 .6 1 0 )( 2 5 1 0 4 6 6 , 2 6 9 1 0
3
12
)
1 3 9 .4 M P a
(CONT.)
53
4.15 (CONT.)
Hence, n
y
m ax
210 1 3 9 .4
1 .5
Comments: For the situation under consideration, a safety factor of 1.5 to 2 is to be selected (see Sec. 1.6) 2.6
T (kN m)
1.9
A
C
D
-1.1
x
E
Figure S4.15
SOLUTION (4.16)
or
TL GJ
1.5 180
;
0 . 02618
T ( 0 .5 h ) GJs
1000 79 ( 10
9
)
[
Th GJh
0 .5 h ( 0 . 02 )
4
h 4
( 0 . 02 ) ( 0 . 011 )
4
]
2
Solving h 1 9 7 m m SOLUTION (4.17) T Assume T A as redundant.
TA
x
TB
Tx TA
x
T 0
1
d x T A T1 x
Deformation:
A
Tx dx GJ
TA GJ
A 0;
Geometry:
T A T B T1 L ,
Statics:
L
dx
T1
GJ
0
TA
1 2
L
xdx
0
TAL GJ
T1 L
2
2GJ
T1 L
TB
1 2
T1 L
SOLUTION (4.18) Conditions of equilibrium gives: R A P and M For portion AC E Iv1 '' P x E I v1 ' E I v1
1 2 1 6
B
P L 2 as shown in Fig. S4.18.
P A
Px
RA=P
x
P x C1
B
C Px
PL/2
x
2
L/2
P x C1x C 2 3
L/2 Figure S4.18 (CONT.)
54
4.18 (CONT.) For portion CB 1
E I v 2 '' E I v1 '
1
PLx C3
2 1
E Iv 2
PL
2
PLx C3x C4 2
4
Boundary conditions lead to v1 ( 0 ) 0 :
C2 0
v2 (0 ) 0 :
C3 0
Then v1 (
L
) v2 (
2
v 1 '(
L
L 2
) v 2 '(
2
P
):
(
6 L
2 P
):
2
Solving C 1
L
(
2 3
PL
) C1( 3
L 2
L 2
2
11
C4
8
L
PL(
4
) C1
2
P
)
4
P
L(
2
L
) C4 2
)
2
PL
3
48
At B(x=0): vB v2 (0 )
11
11
PL 3
48
PL 3
48
SOLUTION (4.19) 4
EI
d v dx
4
EIv ' ' ' ' w 0
x L
w0 x
EIv ' ' '
,
2L
2
c1 ,
EIv ' '
Boundary Conditions: v ' ' (0) 0 : c2 0,
v ' ' ( L ) 0 : c1
w0L 6
Then EIv ' E Iv
w0 x
4
24 L w0x
5
120 L
w 0 Lx
2
12 w0Lx
3
36
c3 c3 x c4
Boundary Conditions: v ( 0 ) 0:
c4 0,
v ( L ) 0:
c3
Thus, we obtain
and Solving,
w0x
v
3 6 0 E IL
v'
360 EIL
w0
(7 L 10 L x 4
2
2
3x ) 4
( 7 L 30 L x 1 15 x 1 ) 0
x1 L 1
4
8 15
2
2
4
0 .5 1 9 3 L
Hence v m a x v ( x 1 ) 0 .0 0 6 5 2 2
w0L EI
4
55
7 w0L 360
3
w0 x 6L
3
c1 x c 2
SOLUTION (4.20) M
Segment AC: EIv 1 ' '
0
2
EIv 1 '
,
M
Segment CB: EIv 2 ' ' EIv
x
L
0
L
M
'
0
x
2
c1
2L
y
(L x)
A
2
Mo/L
( Lx
0
L
M
x 2
) c2
a
Mo C
B
x
Mo/L
L
Boundary Conditions: v1 ' ( a ) v 2 ' ( a ) : M 0x
E Iv1 Then Also, we have EIv
2
3
c 2 c1 M 0 a
c1 x c 3 ,
6L
M
' M 0x
0
2
x
2L
v 1 ( 0 ) 0:
c3 0
c1 M 0 a ,
EIv
2
M
0
x
2
2
M
0
x
6L
3
c 1 x M 0 ax c 4
Boundary Conditions: v 2 ( L ) 0;
c4 M 0 L(a
v 1 ( a ) v 2 ( a ); M ox
v1
Thus
M
c1
0
6L
2
(3a
L 3
) c1 L
6 aL 2 L ) 2
(6 aL 3a 2 L x ) 2
6 E IL
2
2
SOLUTION (4.21) We have m a x 1 6 I
w
or
(wL
Mc I
2
8 )( h 2 ) I
m ax
2
L h
Therefore 4
v m ax
5
5L 384 EI
(
1 6 I
5
)
m ax
2
L h
m ax
L
2
24 hE
1
2 4 h E v m ax
Solving, L (
4
5wL 384 EI
)2
m ax
Introducing the given data: L [
9
2 4 ( 0 .3 1 5 )( 2 0 0 1 0 )( 3 .1 1 0
3
)
6
5 ( 7 0 1 0 )
1
] 2 3 .6 6 m
SOLUTION (4.22)
x
A
x1
E
P
C B R B P (1
( a ) EIv ' ' P
a L
x,
EIv '
1 2
a L
RC P )
P
a L
a L
y x
2
c1 ,
EIv
1 6
P
a L
x
3
c1 x c 2
1 6
PaL
Boundary Conditions: v (0) 0 :
Thus
v
PaLx 6 EI
[(
c 2 0, x L
)
v(L) 0 :
c1
1]
2
( b ) The v m a x occurs at E where v ' 0 . Hence 0
PaL 6 EI
[3(
x1 L
)
2
1 ],
x1
1 3
L 0 . 577 L
(CONT.)
56
4.22 (CONT.) v m ax
and
( c ) v m a x 0 .0 6 4 1
2
PaL 6 EI
[( 0 .5 7 7 ) 0 .5 7 7 ] 0 .0 6 4 1 3
3
2 5 (1 0 )( 0 .5 )( 2 ) 9
2 0 0 (1 0 )( 5 .1 2 1 0
2 6
)
PaL EI
2
3 .1 3 m m
SOLUTION (4.23) F
y
B
A
L
x
C
L/2 3F/2
F/2 Figure S4.23
Free-body diagram of shaft is in Fig. S4.23. Since reactions at supports A and B, two differential equations must be written for portion AB and BC. ( a ) We have M1 M
(0 x L )
P 2
Px
2
(L x
3 PL 2
3 PL 2
)
Integrating twice, the preceding leads to the following expression: For segment AB E Iv1 '' E Iv 1 ' E Iv 1
For segment AB E Iv 2 '' P x
Px 2 Px 4
2
3
Px 12
C1
E Iv 2 '
C1 x C 2
E Iv 2
Px 2 Px 6
2
3
3 PL 2 3 PLx 2
3 PLx 4
2
C3 C3x C4
Using the boundary and the continuity conditions, we find v1 ( 0 ) 0 :
C2 0
v1 '( L ) v 2 '( L ) :
C3
5 PL 6
2
2
v1 ( L ) 0 :
C1
v2 ( L ) 0 :
C4
PL 12
PL 4
3
The result elastic curves of the beam are v1
Px 12 EI
(L x )
v2
P 12 EI
(3L 10 L x 9 L x 2 x )
2
(0 x L )
2
3
2
2
3
(L x
(P4.23a) 3L 2
)
( b ) The deflection at the free end of the beam is readily found by introducing x=3L/2 into the second of Eqs. (P4.23a): vc
3
PL 8 EI
3
PL 8 EI
(P4.23b)
Comment: Observe that the deflection of the shaft is downward between B and C and upward between A and B.
57
SOLUTION (4.24)
Refer to Table A.8:
P
B
E 1 I1
A
Beam AB: L/2
L/2
R
c'
D
3
5 PL 48 E 1 I 1
Beam CD:
3
RL 24 E 1 I 1
c''
3
RL 24 E 2 I 2
C E 2 I2
Condition of compatibility, c ' c ' ' : 3
5 PL 48 E 1 I 1
3
RL 24 E 1 I 1
3
RL 24 E 2 I 2
; R
5P (E2I2 ) 2 ( E1 I1 E 2 I 2 )
SOLUTION (4.25) From a free body diagram of beam BC, we observe that it has vertical reactions 2P/3 and P/3 at ends B and C, respectively. Thus, beam AB is in the condition of a cantilever beam under a uniform load of intensity w and a concentrated load B equal to 2P/3. The deflection of the hinge: vB
4
wa 8 EI
2 Pa 9 EI
3
as obtained by Cases 3 and 1 of Table A.8, respectively.
SOLUTION (4.26)
W
A M
C C
A
RA
F
A C
v 'C
B
(a)
W
(b)
B
C
A (c)
F
B v ''C
Figure S4.26: (a) Free-body diagram of beam; (b) deflection due to P; (c) deflection due to reactive force F.
Consider F as redundant and to release the rod from the beam at C (Fig. S4.26a) and then reapplied (Figs. S4.26 b and c). (CONT.)
58
4.26 (CONT.) Refer to Table A.8: 2
vc '
(2 L 3a )
WL 6 EI
3
v c ''
FL 3 EI
Equation of compatibility at point C: c v c ' v c ''
F k
or
2
(2 L 3a )
WL 6 EI
3
FL 3 EI
Solving, 2
WL k (2 L3a )
F
Q.E.D.
3
2 ( kL 3 E I )
SOLUTION (4.27) y
w x
M A
B
F y 0:
M
A
RA RB wL
0:
M
A
RB L
(1) 1 2
wL
2
RB
RA
E Iv " R A x
1 2
E Iv '
RAx
E Iv
1 6
RAx
2
2
wx
1 2
3
v ' ( 0 ) 0:
1 6
M
A
w x M A x c1
1 24
3
wx
4
c1 0 ,
1 2
c1 x c 2
2
M Ax
v ( 0 ) 0:
c2 0
and E Iv
1 6
RAx 3
1 24
The condition that v ( L ) 0
4
wx
1 2
M Ax
2
(2)
RAL
gives
1 4
wL
2
3M
A
0
(3)
Solving Eqs.(1) and (3): RA
5 8
wL
M
A
1 8
wL
RB
2
3 8
wL
SOLUTION (4.28) w0L
Due to symmetry: R A R B
,
4
M
A
M
B
We have, for 0 x L 2 : M RAx M
A
w0x
3
w0Lx
3L
4
M
A
w0x
3
3L
Therefore E Iv " M E Iv ' M
A
x A
w0Lx 4 w0Lx
2
8
Boundary conditions: v ' (0) 0 : v ' ( L2 ) 0 :
w0x
3
3L
w0x
4
c1
12 L
c1 0 M
A
5 w0L
2
96
(CONT.)
59
4.28 (CONT.) Hence 2
5 w0L x
EIv
2
3
w 0 Lx
192
w0 x
24
5
60 L
c2
Boundary condition: v ( 0 ) 0:
c2 0
Thus, we obtain v
w0 x
2
( 25 L 40 L x 16 x ) 3
960 LEI
2
( for 0 x L 2 )
3
and v max v ( L2 )
7 w0L
4
3840 EI
SOLUTION (4.29) y M
P
A
By virtue of symmetry:
B
M
A
C
L/2
x
B
M
A
M
RA RB
L/2
B
P 2
RB
RA
Segment AC EIv ' ' ' ' 0 ,
E Iv '
1 2
c1 x
EIv ' ' ' c 1 , 2
c2 x c3 ,
EIv ' ' c 1 x c 2
E Iv
1 6
c1 x 3
1 2
c2 x
2
c3 x c4
Boundary Conditions: EIv ' ' ' ( 0 ) c 1 V v ' (0) 0 :
P 2
c3 0,
v ( 0 ) 0:
EIv ' ' ( 0 ) M
,
v' (L 2) 0 :
A
c2 M
,
M
A
M
B
A
PL 8
c 4 0.
Equation (1) is thus v
2
(3L 4 x )
Px 48 EI
SOLUTION (4.30) Let the reaction R B is selected as redundant and the corresponding constraint is removed. Then v ' B w L 8 E I , from Case 3 of Table A.8. 4
The deflection caused by the redundant is v"B R B L
3
3EI
The total deflection must be zero; vB
4
wL 8 EI
RB L
3
3EI
0
or RB
3wL 8
Now, applying equations of statics, we obtain RA
5wL 8
M
A
wL 8
2
60
(1)
SOLUTION (4.31) w A
a
C
a
D
M EI
wa
2
3w a
2
a
B
x2 x1
2EI
A2
2EI
A3
x
Spandrel parabola
A1 x3
Tangent at A A
vB
D
B
B A1
bh 3EI
A2
3
3a 4
2
a
3
wa 2 EI
A3
wa 2 EI
x1 a
1 3
wa 6 EI
a(wa )
1 2
x2
7a 4
(App. A.3) 2
1 wa 2 EI
5a 2
x3
3
8a 3
( a ) B A B A A1 A 2 A 3 B
3
wa 12 EI
(2 6 6)
7 wa 6 EI
3
( b ) t B A v B A1 x 1 A 2 x 2 A 3 x 3 vB
4
wa 24 EI
(7 30 32)
69 w a 24 EI
4
23 w a 8 EI
4
SOLUTION (4.32) P
4EI (a)
EI
A L/2
B
L/2
C
P 2
P 2
M/EI PL/4EI
Px/2E I
PL/16EI A
A2
A1
C
x B
D x B
A v m ax
B t BD
Tangent at B
1 PL 2 16 EI
(2)
A2
1 PL 2 4 EI
(2)
L
L
2
1 PL 64 EI 2
1 PL 16 EI
t A B A1 ( 3 ) A 2 ( L
D
t AB
A1
Tangent at D
61
B
1 L
t AB
2L 3
)
3
3 PL 64 EI
2
3 PL 64 EI
(CONT.)
4.32 (CONT.)
( b ) BD B D B 0 .
B
Also
1 Px 2 2 EI
2
(x)
Px 4 EI
Hence 2
3 PL 64 EI
2
Px 4 EI
x
,
3 4
L
2x 3
)
Thus, v m ax t B D
2
1 Px 2 2 EI
3 3L P 6 EI 64
3
(
3
Px 6 EI
3 P L3 128 EI
SOLUTION (4.33)
M
Tangent at A
A
2a
A
P Let R B be redundant.
a
B
RA
A 0
C
RB
A1
M M
1 2
M A (2a ) M Aa
A2
1 2
Pa(2a ) Pa
A3
1 2
Pa
2
2
4a/3 v B t BA 0
A
B
A1
A
A2
or
C x
A3
Pa
M
A
Statics: R B
2a/3
1 EI 1 2
7 4
4a 3
[ A1 (
) A2 (
2a 3
)]
Pa
P
RA
3 4
P
SOLUTION (4.34) Assume R c as redundant. w A
tCB
Tangent at B
RA
M wL
2
B L/2
L
t AB
C RC
RB
L/2 L/3
8
A1
A1 A2
2L/3
x
A3
R C L/2
2 wL 3 8
2
(L)
A2
1 2
Rc L
A3
1 2
Rc L
2
2
wL 12
3
(App. A.3)
(L)
Rc L
(2)
Rc L
L
2
4 2
8
(CONT.)
62
4.34 (CONT.) E I t A B A1 ( 2 ) A 2 ( L
2L 3
Rc L
E It CB A3 ( 3 ) L
4
)
Rc L
wL 24
3
6
3
24
Since
2 t CB t AB :
or
Rc
1 6
wL
Statics: R B
3 4
wL
RcL
3
12
wL 24
RA
4
RcL
3
6
wL
5 12
SOLUTION (4.35) S
m ax
y
n
2
2
W g
v E AL
gS y AL
W a ll
,
2
2
n v E
Substitute given data: 6
2
9 .8 1 ( 2 5 0 1 0 ) [
W a ll
2
( 0 .0 2 ) ( 2 )]
1 6 .6 N
4 2
2
9
( 3 ) ( 3 .5 ) ( 2 1 0 1 0 )
SOLUTION (4.36) P=mg 1.25=a
0 .7 5 2
x
B
P
st
Table A.8 ( Case 6 ).
b=0.75
A
1. 2 5 2
L=2 (L b 2
Pbx 6 LEI
20 ( 9 . 81 )( 0 . 75 )( 1 . 25 )
x )
2
P
2
6 ( 2 )( 210 10
9
)( 0 . 06 0 . 08
3
12 )
(2
2
0 . 75
2
1 . 25 ) 0 . 0535 2
mm
Max. moment is under load. Thus
M
st
m ax
c
We have K 1
( 0 .6 2 5 0 .7 5 2 0 9 .8 1 ) ( 0 .0 4 )
I
0 .0 6 0 .0 8
1
3
12
1 [1
2h
st
1 .4 3 7 M P a 1
2 ( 0 .5 ) 0 . 0535 10
3
] 2 137 . 7
( a ) m a x s t K 7 .3 7 m m ( b ) m ax st K 1 9 8 M P a SOLUTION (4.37)
A
4
max
( 2 5 ) 1 5 6 .2 5 m m 2
W A
[1
1
from which
2h
W
Solving W 2(
hE L
[
st
2 2 m ax A 2
2 m ax
]; 2
Since s t W L A E , thus:
2
m ax A
W
m ax
W
A
1]
2
1 1
1 2 hAE WL
A
6
)
2
( 2 5 0 1 0 ) (1 5 6 .2 5 1 0
6
)
9
2[
st
(1) (P4.37)
m ax
Substituting given data: W
2h
(1 .1 )( 2 0 0 1 0 )
3 1 9 .0 7 N
6
2 5 0 1 0 ]
4 .6
63
SOLUTION (4.38) Refer to Solution of Prob. 4.37. From Eq.(1), we obtain A
2W
[
2 m ax
hE L
6
(P4.38)
]
m ax
9
2 (90 )
(1 2 5 1 0 )
2
[
1 .2 ( 2 0 0 1 0 )
1 2 5 1 0 ] 1 .8 4 5 1 0 6
1 .5
3
m
2
Thus d
2
1 .8 4 5 1 0
4
3
d 0 .0 4 8 5 m = 4 8 .5 m m
,
SOLUTION (4.39) A (10 )
314 . 2 mm
2
From Eq. (P4.37): L
Solving, h
2W (
m ax ) m ax A 2
hE L
( A m ax 2W )
m ax
2W E
3 ( 350 10
2
6
)
2 ( 500 )( 170 10
9
)
(P4.39)
[ 314 . 2 350 2 500 ] 673
mm
SOLUTION (4.40) We have W 2 4 9 .8 1 2 3 5 .4 N and M I
1 12
6
( 0 .0 4 )( 0 .0 6 ) 0 .7 2 1 0 3
m
m ax
1 4
WL
4
W h
C
A L/2
B
k
L/2
W=196.2 N
R st
R
R k
Using Table A.9: st
or
(W R ) L
3
3
R
48EI
650 10
6
R(
1
L
R
k
) 10
6
R[
48EI
7 8 .2 180 10
k
3
6 5 0 R ( 5 3 .3 0 7 ) ,
st
RL
48EI
k
or Hence
3
0 .6 5 1 0
3
1 0 .2
R 7 8 .2 N
(2)
3
]
4 8 ( 0 .0 7 )( 0 .7 2 )
and
W R 1 5 7 .2 N
0 .4 3 4 m m
Maximum stress occurs at midspan:
st
Mc I
1 5 7 .2 ( 2 ) ( 0 .0 3 ) 4 ( 0 .7 2 1 0
6
3 .2 8 M P a
)
(CONT.)
64
4.40 (CONT.)
K 1
2 ( 0 .0 5 )
1
0 .4 3 4 1 0
1 6 .2
3
( a ) v m a x 1 6 .2 ( 0 .4 3 4 ) 7 .0 m m ( b ) m a x 1 6 .2 (3 .2 8 ) 5 3 .1 M P a SOLUTION (4.41) Refer to Example 4.14. EkG
m ax 2
or
AL
4 1 6 .6 7 9 1 0
2 1 0 1 0 2[ 6
;
44100 10
12
4 .4 1 8 1 0
3
9
L m in
]
12
4 ( 7 4 4 9L.3 9 1 0 ) m im
or L m in 0 .6 7 6 m
Thus, we obtain 2 Ek L
m ax
2 ( 4 1 6 .6 )( 0 .6 7 6 )
GJ
0 .0 4 8 ra d = 2 .7 5
3
7 9 1 0 ( 3 .1 1 )
o
SOLUTION (4.42) The E k in wheel B must be absorbed by the shaft. We have J
32
(1 2 5 ) 2 .4 (1 0 4
5
) m
4
Substitute Eqs.(4.42a) and (4.42b) into (4.41): Ek
1 4
b t 4
2 Ek L
(a)
m ax
(b)
m ax 2
GJ
EkG AL
2
[
( 0 . 075 ) ( 0 . 025 )( 1800 )( 150060 2 ) 4
4
1
2 ( 2 7 .5 9 )( 0 .3 ) 9
(1 9 1 0 )( 2 .4 1 0
2[
5
)
9
( 2 7 .5 9 )(1 9 1 0 )
] 2 0 .0 0 6 0 2 5
27 . 59 N m
2
ra d 0 .3 5
o
1
] 2 1 1 9 .3 3 M P a
2
( 0 .0 2 5 ) ( 0 .3 )
4
SOLUTION (4.43) The E k in wheel A must be absorbed by the shaft. Refer to Solution of Prob.4.42. We have Ek
1 4
b t
(a)
m ax
(b)
m ax 2
4
2 Ek L GJ
EkG AL
2
[
( 0 . 0625 ) ( 0 . 025 )( 1800 )( 120060 2 ) 4
4
1
2 ( 8 .5 2 )( 0 .3 ) 9
(1 9 1 0 )( 2 .4 1 0
5
)
9
( 8 .5 2 )(1 9 1 0 )
2[
] 2 0 .0 0 3 3 5 ra d 0 .1 9
1
] 2 6 6 .3 1 M P a
2
( 0 .0 2 5 ) ( 0 .3 )
4
65
o
2
8 . 52 N m
SOLUTION (4.44) Let r r y , then r x r xy , Hence, Eq. (4.46) for z
max
Et 2 ( 1
2
t 2
1 ) r
M M
,
y
M
M
x
xy
0.
:
;
126 (10
6
200 10
)
9
( 3 . 2 10
3
) 1 r
2
2 (1 0 .3 )
or r 2 .7 9 1 m and D 5 .5 8 2 m
Equations (4.50) lead to
6M
max
t
m ax
;
2
6
126 (10
6M
)
m ax
( 3 . 2 10
3
)
2
or M
2 1 5 .0 4
m ax
N m m
SOLUTION (4.45) (a) y
Dw ' ' ' ' p 0 sin
Dw ' ' ' ( b ) p 0 cos
b y
Dw ' ' ( ) p 0 sin 2
b
c1 y c 2
b
y b
c1
Dw ' ( b ) p 0 cos 3
y
b
1 2
c1 y
2
c2 y c3
and y
D w ( ) p 0 s in 4
b
b
1 6
3
c1 y
1 2
c2 y
2
c3 y c4
(1)
Boundary conditions: w ' ( 0 ) 0:
c 3 ( ) P0 ,
w ( b ) 0:
c2
1 3
w ( 0 ) 0:
3
b
c1 b
2
p0b ,
3
c4 0
w ' ( b ) 0:
2
c1 0 ,
c2
2
3
p0b
2
Equation (1) becomes p0b
w
D
4 4
( s in
y
b
b
2
y
2
b
(2)
y)
( b ) At y=0: m ax
(c) D
Et
3
1 2 (1
2
)
6M t
m ax 2
9
7 0 1 0 (1 0 1 0
2
6 t
3
)
d w
[D
2
3
dy
2
]y0
6 t
2
[0 0
2 p0b
3
2
]
12 p0
3
b
(t)
2
6 .4 1 k N m
2
1 2 (1 0 .3 )
Thus w m a x w ( b2 )
p0b 4
4
D
(1
) 4
3
3 5 1 0 ( 0 .5 )
4
3
( 6 .4 1 1 0 )
and
m ax
3
1 2 ( 3 5 1 0 )
3
( 00.0.51 ) 3 3 .8 6 2
4
M Pa
66
(1
4
) 0 .7 5 2 1 0
3
m = 0 .7 5 2 m m
SOLUTION (4.46) ( a ) Dw ' ' ' ' p 0 , Dw'
Dw ' ' ' p 0 y c 1 ,
p0 y 3
1 6
1 2
2
c2 y c3
4
c1 y
Dw ' '
1 2
p0 y
2
c1 y c 2
and Dw
1 24
p0 y
c1 y 3
1 6
1 2
c2 y
2
c3 y c4
(1)
Boundary Conditions: w (0) 0 : c4 0,
w ' ' (0) 0 : c2 0
w ( b ) 0 a n d w ' ( b ) 0:
c1
3 8
p0 b,
c3
1 48
p0b
3
Equation (1) is therefore p0b
w 2
(b)
d w dy
2
p0 2D
(y
4
48 D
2
y
y
y
[ ( b ) 3( b ) 2 ( b ) ]
3 yb 4
3
4
(2)
)
At y=b: M
m ax
2
D
p0
d w dy
2
2
b 8
m ax
,
6M t
m ax 2
0 .7 5 p 0 ( t ) b
2
( c ) From Solution of Prob. 4.45, D 6 4 .1 k N m Thus w m ax w ( 2 ) b
p0b
4
192 D
p0b
4
[ ( 2 ) 3( 2 ) 2 ( 2 ) ] 1
48 D
1
3
( 3 5 1 0 )( 0 .5 ) 3
4
1 9 2 ( 6 .4 1 1 0 )
3
1
4
0 .0 0 1 8 m = 1 .8 m m
and
0 .7 5 p 0 ( bt ) 0 .7 5 (3 5 1 0 )( 51000 ) 2
m ax
3
2
6 5 .6 3 M P a
End of Chapter 4
67
CHAPTER 5
ENERGY METHODS AND STABILITY
SOLUTION (5.1) Axial strain energy
Use Eq. (5.11), circular Part:
2
U
P L
C
2
2
P L
2 AE
d
2(
2P L
2
d E 2
)E
4
Square Part: 2
U
P L
S
2
2
P (L 2)
2
2 AE
P L
2a E
2
4a E
Requirement:: 2
U
US;
C
d
2
2P L
P L
d E 2
2
4a E
8
a
SOLUTION (5.2) Refer to solution 5.1: U
2
P L 2
4a E
2
2P L 2
d E
2
P L E
(
1 4a
2
2
d
(1)
)
2
Using Eq. (5.27), U
1 2
P
(2)
Equating Eqs. (1) and (2), we obtain PEL ( 2 4 ) a
2
d
2
SOLUTION (5.3) T A B 2 .5 k N m
T B C 1 .5 k N m
Segment AB J AB
( 4 5 ) 4 0 2 .5 8 1 0 4
AB
m
4
( 2 .5 1 0 ) ( 0 .5 4 )
2
U
9
32
T L
3
2G J
2
2 ( 4 0 1 0 )( 4 0 2 .5 8 1 0 9
9
1 0 4 .8 J
)
Segment BC J BC
(3 0 ) 7 9 .5 2 1 0 4
BC
m
4
32 2
U
9
T L 2G J
(1 .5 1 0 ) ( 0 .3 6 ) 3
2
2 ( 4 0 1 0 )( 7 9 .5 2 1 0 9
9
7 5 .8 3 J
)
(CONT.)
68
5.3 (CONT.) Total strain energy U 1 0 4 .8 1 2 7 .3 2 3 2 .1 J
SOLUTION (5.4) T AB 3 k N m
TBC 5 k N m
30 mm
20 mm
A
B TB=2 kN m
45 mm
C
TC=5 kN m
0.36 m
0.54 m
Segment AB J AB
( 4 5 2 0 ) 3 8 6 .9 1 0 4
4
AB
m
4
(3 1 0 ) ( 0 .5 4 )
2
U
9
32 3
T L
2
2 ( 4 0 1 0 )(3 8 6 .9 1 0 9
2G J
9
1 5 7 .0 2 J )
Segment BC J BC
(3 0 ) 7 9 .5 2 1 0 4
BC
m
4
32
(5 1 0 ) ( 0 .3 6 ) 3
U
9
2
2 ( 4 0 1 0 )( 7 9 .5 2 1 0 9
9
1, 4 1 5 J )
Total strain energy U 1 5 7 .0 2 1 4 1 5 1 5 7 2 J
SOLUTION (5.5) See solution of Prob. 5.3: J A B 4 0 2 .5 8 1 0
J B C 7 9 .5 2 1 0
9
9
m
m
4
4
Therefore 2
2
U
C
T L
2G J TL
TC
[
0 .5 4
2 ( 2 8 ) 4 0 2 .5 8
GJ
TC
0 .5 4
[
2 8 4 0 2 .5 8
0 .3 6
] 1 0 4 .8 1 0
6
7 9 .5 2
0 .3 6
] 2 0 9 .6 1 0
7 9 .5 2
6
2
TC
TC
or C (1 0 ) 6
TC
2 0 9 .6
Then U
1 0 4 .8 1 0
6
( C 1 0
( 2 0 9 .6 )
2
2
12
)
2 3 8 5 C 2 3 8 5 ( 0 .0 6 ) 2
69
2
8 .5 8 6 J
SOLUTION (5.6) See solution of Prob. 5.3: J A B 4 0 2 .5 8 1 0 TC
U
Thus,
2
2G
L AB
(
J
3
(1 .4 1 0 )
LBC
2
2 (80 )
m
J B C 7 9 .5 2 1 0
4
9
m
4
)
J BC
AB
9
( 4 00 .52 .54 8
0 .3 6 7 9 .5 2
)
7 1 .8 9 J
Equation (5.29): U
1 2
TC C ;
1 4 0 0 C
7 1 .8 9
2
from which C 0 .1 0 2 7 r a d 5 .8 8
o
SOLUTION (5.7) P
B
A
C
a
V AB P ,
6 5
V BC
Pa L
L Pa/L
V
x P
U
s
L
2
Vx
dx
2 AG
0
3 5 AG
[
P dx 2
0
L
2
P a L
0
2
2
dx ]
2
3P a 5 AGL
(L a)
SOLUTION (5.8) Vx U
s
wx
wL 2 L 0
2
Vx
2 AG 2
2
3w 5 AG
L 4
2
dx
x
3w 5 AG 2
x L 2
6 5
L
( 0
L
3
x 3
x ) dx 2
L 2
2
3
3w L 5 AG 12
0
2
1 w L 20 AG
3
SOLUTION (5.9) I bh
P A
B
C
P
U
a
L
(a) U U
b
t
1 2 EI
1 2E
[
a
M 0
2 AB
dx
2
dx
dA
L
M 0
1 2E
2 CB
(
P A
M 2
dx' ]
My I
12
2
P (La ) 2 AE
2
P (La ) 2 Ebh
We have
Pa/L
x’
x
a
3
2
P a 6 EI
2
) dA
AB
(a L )
Px 2
2P a Ebh
2
3
M
CB
Pa L
x'
(a L )
(1) (CONT.)
70
5.9 (CONT.) But
Thus
U
2
P (La )
t
bh
U
3
12
0 ,m ax
U b.
a
2
a
6 Pa bh
U
t
[1 4 ( h ) ]
2 Ebh
Pa ( h 2 )
( b ) b ,m ax Hence
y d A 0 , and Eq. (1) becomes U
2
a ,m ax
,
2 m ax
2E
2
P
2
2 Eb h
2
(1
P bh
6a h
)
m ax
,
P bh
(1
6a h
)
2
SOLUTION (5.10) M
( Lx x ) 2
w 2
A x
wL 2
U
0 ,m ax
U
b
w B
1 2 EI
2 m ax
2E
2
9w L
2
m ax
I bh
1 8
3
m ax
wL
2
c h 2
12 , M
m ax c
I
3wL 4 bh
2 2
4
2
32 Eb h
M dx
h
wL 2
L
M
(1)
4
1 2 EI
L w 2
( 0
) ( Lx x ) dx 2
2
2
2
w L
5
20 Ebh
(2)
3
It is required to obtain C : U 0 , m a x C U V , or C U
V 0 ,m ax U
(3)
Substituting Eqs.(1) and (2) into (3): C 4 5 8 . Thus
U
0 ,m ax
45 U 8 V
SOLUTION (5.11) From Solutions of Probs. 5.9 and 5.7: U
b
2
2
(a L )
P a 6 EI
U
Thus or
2
p a
2
6 EI
U
2
s
(a L )
3P a 5 AGL 2
(a L )
(a L )
3P a 5 AGL
v A 2 P a ( a L )[ 6 E I a
3 5 AGL
1 2
Pv A
]
SOLUTION (5.12) x
w
x A 2
U
M dx 2EI
1 2EI
L
L 0
(
1 2
B 2
w x ) dx 2
2
1 w L 40
71
EI
5
M
x
1 2
wx
2
SOLUTION (5.13) P
P A
a
2a
C
B
D a
MAC=Px MCD=Pa
P
P Pa
M
x Segment AC U
AB
a
2
M dx
0
P
2EI
2
2EI
2
a
x dx 2
P a
0
3
6EI
Segment CD U
CD
2
P a
3a 0
2
2
dx
2EI
By symmetry: U
3
P a EI
U
AC
BD
.
Total strain energy : Pa
U 2
3
2
P a
6EI
3
2
4 P a
EI
3
3
EI
SOLUTION (5.14) x
w
b h
A
L
6
We have
Vx wx
5 U
s
L 0
V
2
6
dx
2 AG
3 w
2
x
5 AG
B
1
5 2 AG 3
L
2
3
1 w L
L
( w x) dx 2
0
3
5 AG
0
SOLUTION (5.15) We have V AC V BD P
6 5
VCD 0
Thus U
s
V
2
2 AG
a
dx 2 0
2
P dx
2(
2 AG
2
6 P a 5 AG
72
6 5
2
)
P a 2 AG
SOLUTION (5.16) See solution of Probs. 5.13 and 5.15: 2
U
b
3
3
4 P a 3
s
5 AG
2
6 P a
EI
vC 2 P a (
or
U
6 P a
(due to symmetry) 2
U
2
EI
vC v D
Thus
3
4 P a
1
5 AG 2a
2
3
3EI
1
P (vCb vCs )
2
P vC
)
5G A
SOLUTION (5.17) ( a ) Axial strain energy in bolt. 2
U
b
2
P L
T L
2 AE
2 AE 2
T ( 0 .0 3 )
2[
2 .6 5 2 6 (1 0
9
)T
2
(6 ) ( 2 0 0 1 0 )] 2
3
4
2 .6 5 2 6 (1 0 Ub
Thus
1 2
T ;
9
)( 6 3 0 ) 1 .0 5 2 8 J 2
1 .0 5 2 8 N m
1 2
( 6 3 0 )
3 .3 4 2 m m
( b ) Bending strain energy in link. I bh U
b
L
3
12 216 m m
2
0
2EI
2EI
{[
0 .0 2 5
(420 x) dx 2
0
0 .0 5
2
( 2 1 0 x ') d x '} 0
A
EI
B
Substituting the data, U
b
50 mm
25 mm
1 .3 8
4
M dx
1
1 2 (1 2 )( 6 )
3
3
6
630
420
( 2 0 0 1 0 )( 2 1 6 )
3 1 9 (1 0
x’
x
1 .3 8
210
) J
SOLUTION (5.18) a x Q
P B
A
C
M
AC
Qx
M
CB
Qx P(x a )
L (CONT.)
73
5.18 (CONT.) Thus
M
vA
1 EI
1 EI
[ ( Q x )( x )d x
M
i
dx
Q
i
a
L
[Q x
0
P ( x a ) ]( x ) d x
a
Set Q 0 , and integrate: vA
1 EI
L
P ( x a )xdx
( 2 L 3a L a ) 3
P 6 EI
a
2
3
SOLUTION (5.19) M
BC
Px
M
B
Thus
1 EI
[
P L P R s in
CA
L
M 0
M
BC
P
BC
dx
2
M
M
Rd ]
CA
P
CA
0
( 4 L 6 R L 2 4 R L 3 R ) 3
P 12 EI
2
2
3
SOLUTION (5.20)
A a
P a
a
Pa/2
C
P/2
M
B
D P/2
Pa/2
M
+
AD
P 2
vD
1 EI
vD
P 2 EI
( x a ), M
[
M i
2a
i
P (xa ) 2
0
M
BD
P 2
dx 2
dx
a 0
2
x 2
dx
x
Pa/2 Integrating, 3
vD
Pa 4 EI
SOLUTION (5.21) B
C
M
A RA
L/2
AB
Consider R A as redundant.
L RB
x
M
0
RAx
M
CB
RC
x’ (
M
0
L
RA
2
M
0
L
RA 2
) x ' M 0 ,
vA 0
1 2
Thus L
vA
2
0
( R A x )xdx
L
( 0
M
0
L
RA 2
) x ' M 0 ]
x' 2
After integrating RA
Then
RC
2 3
M
4 3
M
0
0
L
L
For the entire beam,
F y 0:
RB 2
74
M L
0
dx 0
M
M
i
i R A
dx
x
SOLUTION (5.22) U
a
D
2
P (2L) 2 AE
U P
U
a
2 PL AEa
PL 3EsI
2 PL AEa
or
Solving P
3
3
wL 8
1 EsI
2
M 2Es
0
L
(Px
0 4
0
a
)
wL 8 EsI
3 AE
(
Ls
s
2
AE a L 6 E s I
M Px
dx wx 2
2
wx 2
2
U U
a
U
s
)xdx 0
SOLUTION (5.23) w
M
AB
1 2
M
BC
3x 5
1 EI
M
Q x
x A
B
A
4
3
C
1 EI
L AB 2 a
2
wx
( 2 wa ) 2 wa
{
M Q
i
i
4x 5
Q
2
L BC 5 a
dx
2a
2
1
( 2 w x )( 0 )d x
0
2
5a
0
( 65x wa 2 wa
4x 5
Q )(
4x 5
) dx }
Set Q 0 and integrate:
A
60
wa EI
4
SOLUTION (5.24)
C
M
P
x
A
AB
Px
M
BC
Pa
a
B
x L
A
1 EI
M
M i
dx
i
P
a
1 EI
{ ( P x ) xd x 0
L
( P a )( a )d x} 0
Integrating, A
2
(a 3L )
Pa 3EI
SOLUTION (5.25) Introduce a fictitious horizontal force Q at end B. Vertical reactions at supports are P/2. We have M Thus
BC
R (1 c o s ) 2
B
2 EI
2 EI
M 0
2
M BC
BC
Q
P 2
( R sin ) Q
P 2
M
BC
( R s in ) Q ] [ R s in ] R d
Setting Q=0 and integrating: B
CA
Rd
[ R (1 c o s )
0
M
3
PR 2 EI
75
SOLUTION (5.26) (a)
Let fictitious couple C=0:
x C
B a
2a
a
x
A
P
M
AB
Px
M
BC
Pa
M
DC
Px Pa
C x
D
P Pa+C A
Thus
M
M
i
dx
P
i
a
P x dx Pa 2
2
0
4 3
2 Pa EI
3
P
0
2a
(x
2ax a )dx
2
2
0
Pa ( 3 4 2)
3
Pa
a
dx
3
8
( b ) We now have M
AB
A
Px C
[ M
1 EI
M
M
i
M
DC
Px Pa C
dx ]
C
i
Pa C
BC
Hence, after setting C=0: A
a
P EI
a
[ xdx a dx 0
0
2a
(x
a )dx ]
0
3 Pa 2 EI
2
SOLUTION (5.27) M
(a)
A
AB
Pz,
M
a
1 EI
[ M
P EI
(
0
3
a 3
M
P
AB 3
L 3
AB
)
Px,
BC
dz
L
0
1 GJ
(
T0 L 2
T B C T 0 (1
M
M
BC
P
BC
dx ]
1 GJ
x L
L
0
) Pa
[ T 0 (1
x L
) Pa ]( a ) dx
Pa L) 2
( b ) Introduce a fictitious couple C about x axis at point B. T B C C T 0 (1
x L
) Pa
Hence B
1 EI
[
a
M 0
0 0
M
1 GJ
AB
C
AB L 0
dz
L
M 0
[ C T 0 (1
x L
M BC
1 GJ
(
T0 L 2
dx ]
) P a ] (1 ) d x
Setting C=0 and integrating: B
BC
C
PaL )
76
1 GJ
L 0
TBC
TBC C
dx
SOLUTION (5.28) Due to Symmetry: F A B F C D , Statics: R A x 6 0 k N
F AC FBD
,
R A y 1 2 .5
Method of joints: Joint A
kN
,
R D y 1 2 .5
Joint B
F AC 30 kN ( T )
F BC 12 . 5 kN ( C )
F AB 32 . 5 kN ( T )
Total strain energy: U
2
Fi L i
6
2 AE
[ 3 2 .5 (1.3 ) 2 3 0 (1.2 ) 2 1 2 .5 ( 0 .5 ) ] 2
10 2 AE
2
2
This gives, substituting the given data, U 4 7 .4 7 N m
Hence U W;
4 7 .4 7
mm
1 2
( 6 0 1 0 ) D 3
or D
1 . 582
SOLUTION (5.29)
2P
B
Introduce a fictitious force Q at C.
P
Statics:
0.9 m A
R Ax P Q
C
R Ax
R A y 0 .2 4 P
Q R Ay
1.2 m
0.675 m
R C 1.7 6 P
RC
Apply method of joints at C and B: F B C 2 .2 P
(C )
F A B 0 .4 P
(C )
F A C 1.3 2 P Q
Thus C
1 AE
Fi
Fi Q
Li
Differentiating and setting Q=0: C
1 . 32 P ( 1 ) AE
(1 . 875 )
2 . 475 P AE
77
(T )
kN
SOLUTION (5.30) A
P
B
Introduce Q at C. By method of joints:
L
F AB 0,
L
F AC ( P Q )
P
C
D
FBC P
FC D Q
2,
Q
We write ( C ) v
1 AE
Fi
Fi Q
Li
[ 0 ( P )( 0 ) ( P Q )
1 AE
2(
2)
2 ( Q )( 1 )] L
This yields, for Q 0: ( C ) v
2
( C ) h
1 AE
( C ) h
1 AE
[0 ( P ) ( 1 ) ( P )
2PL AE
2 .8 2 8
PL AE
Similarly
or
1 2 2 AE
Fi
Fi P
Li
P L 3.8 2 8
PL AE
2(
2)
2 ]L
SOLUTION (5.31) Introduce Q at point C. F.b.d. - Entire truss ( from
M 0,
F 0 ):
R A y 1.2 5 P 0 .9 3 7 5 Q
R Ax Q
R B y 2 .2 5 P 0 .9 3 7 5 Q
Joint A:
F AC
13 Q
12
F A C 3 .2 5 P 2 .4 3 7 5 Q
5
F A B 3 P 1. 2 5 Q
F AB
A
(T )
(C )
R Ay F B C 3 .7 5 P 1.5 6 2 5 Q
Joint B:
(C )
Thus, we have ( C ) h
1 AE
Fi L i
( C ) v
1 AE
Fi Li
Fi Q
P AE
[1 2 6 1.7 9 1 2 9 .2 9 6 9 ] 1 0 3.0 8 8
P AE
[ 3 ( 3 . 2 )( 3 ) 3 . 25 ( 7 . 8 )( 3 . 25 )
P AE
and Fi P
( 3 . 75 )( 3 . 75 )( 5 )] 181 . 5
78
P AE
SOLUTION (5.32) F BD 0 .
C
32
Q D
B 32
Introduce Q at point D. Reactions, as found by statics, are shown in the figure. We shall apply the method of joints, as needed. 32 kN C Joint C
15
FBC
42-Q/2
Joint E
17
E
A
8
42+Q/2
E
4
FDE 7 0
5 6
F AE 5 6
Q
Thus
D
(C )
F AD
Q
A
3
56+2Q/3
42-Q/2
(C ) F AD 3 0
(T )
1 AE
10 AE
2 8 6 .1 3 3 1 0 AE
3
5 4
32
42+Q/2
2 3
(32 ) 68 kN
F BC 60 kN ( T )
60
5 3
17 8
Joint A
FDE
F AE
FC D
FC D
F jL j
Q
(C )
F j Q
[0 0 0 0 2 ( 7 0 ) ( 3
5 6
) 3 .2 ( 5 6 ) ( 3 ) 2 ( 3 0 ) (
5 6
2
5 6
)]
SOLUTION (5.33) Joint B FBD
FBC 3
FBA
4
FBD
5 4
FBA
(1)
B
FBC P
3 4
FBA
P A 0
1 AE
1 AE
[( P
FjL j 3 4
F j FBA
F B A )(
3L 4
)( 4 ) ( 3
0 .5 6 2 5 P L 3 .3 7 5 F B A L 0 ,
5 4
F B A )(
5 24
P
FBC
7 8
)(
5 4
) F B A ( L )(1 )]
F B A 0 .1 6 6 P P 6
Then Eqs.(1) give FBD
5L 4
P
79
SOLUTION (5.34) M F R s in P R (1 c o s ) ,
v 0
1 EI
M
M F
dx
Therefore v
1 EI
[ FR sin PR ( 1 cos )]( R sin ) Rd
0
3
FR
2 EI
2 PR EI
3
0,
F
4P
4P
SOLUTION (5.35)
P
C
M
Px 2
c o s Q x s in
L x
A
Q B
P/2
P/2 Line of symmetry
( a ) With Q 0: B
2 EI 3
L
c o s ) ( x s in ) d x
Px 2
( 0
s in c o s
PL 3EI
3
PL 6 EI
s in 2
( b ) With Q R B h : B 0
2 EI
L
c o s R B h x s in ) ( x s in ) d x
Px 2
( 0
or R Bh
P cot
1 2
SOLUTION (5.36) Statics: F B C
2P
FC D
,
3
P
F AB
,
3
P 3
,
M
AB
Px
Thus U
B A
2
FAB 2 AE
1 2 AE
dx
2L 0
( 6 AE 11L
2
P 3
M
)P
2 AB
2 EI
A
dx
3
4L 3EI
B
1 2 RI
dx
2L
C D
2
FBC 2 AE
P x dx 2
2
0
2
We have C
U P
PL 3E
( 11A
8L I
2
dx
)
80
1 2 AE
D
2
FC D
dx
2 AE
C
2L 0
4 3
P dx 2
1 2 AE
L 0
2
P 3
dx
SOLUTION (5.37) Let the load at B be designated by Q. Locate origin of coordinates at A: V AB P
M
Px
AB
V AB
1,
P
M
AB
P
x
Locate origin of coordinates at B: V BC P Q
M
P(x
BC
L 2
) Qx
V BC
1,
P
M
BC
P
x
L 2
( a ) Equation(5.38), with Q=P, gives vA
M
1 EI
M
i
P
i
dx
L
L 2
0 3
7 PL 16 EI
( x )dx
Px EI
V
1 AG
P(2xL 2)
2
EI
0
Vi
dx
P
L
(x
L 2
)dx
L 2
0
1.2 P AG
(1 ) d x
2
0
1.2 ( 2 P ) AG
1 .8 P L AG
(1 ) d x
(P5.37)
( b ) From Table B.1: E=200 GPa and G=79 GPa. Given L/h=5. We have A=bh and I b h 1 2 . Eq.(P5.25) is rewritten as 3
Ebv A P
3
7 (1 2 ) L
7 (1 2 )( 5 )
16 h
3
3
1 .8 E L Gh
16
1.8 ( 2 0 0 ) ( 5 ) 79
656 . 25 22 . 78 679 . 03
Error:
22 . 78 679 . 03
(100 ) 3 . 35 %
SOLUTION (5.38) Inasmuch as horizontal displacement at C, h is zero, Eq.(5.41) gives h
U H
0
1 EI
[ HR sin FR ( 1 cos )] R (sin ) Rd
0 2R
1 EI
[( H
P ) x 2 F R ]x d x
0
H ( 2
8 3
) 2F
8P 3
or 4 . 2375 H 2 F 2 . 6667 P 0
(1)
Similarly, vertical displacement at C is zero: v
U F
0
1 EI
[ HR sin FR ( 1 cos )][ R ( 1 cos ) Rd
0 2R
1 EI
[( H
P ) x 2 F R ]2 R d x
0
2H F(
3 2
8) 4P
or 2 H 1 2 .7 1 2 4 F 4 P 0
(2)
Solving Eqs.(1) and (2): H 0 . 5193 P
F 0 . 2329 P
81
SOLUTION (5.39) The structure is statically indeterminate to the first degree. Select R, the reaction at B, as redundant. From the equilibrium of forces at joint D, R with F A D F D C F : F F 4 3
F
D
5 8
(W R )
(1)
W Substituting Eq.(1), into Eq.(5.45), together with Eq. (5.38) we have B 0
L
F R
[2 F
1 AE
0
1 AE
h
(W R ) L
25 32 AE
R R
R 0
dx ]
(W R )( 85 ) L 0 . 8 RL ]
5 8
[( 2 )
dx
0 .8 R L AE
Solving, R 0 .4 9 4 W Thus F A D F C D 0 .3 1 6 W
F B D 0 .4 9 4 W
and
SOLUTION (5.40) We write M M Thus
U R A
vA
U M A
A
6.
M R A
M
dx
1 EI
L
( M
0
RAx
A
1 6
kx )xdx 0
(1)
k x )( 1) d x 0
(2)
3
0
1 EI
3
0
1 EI
R A x kx
A
M M A
M
dx
1 EI
L
( M
0
A
RAx
1 6
3
Integrating and simplifying Eqs.(1) and (2) we obtain, respectively:
1 2
M
A
1 3
RAL
M
A
1 2
RAL
1 30
kL 1 24
3
kL
3
Solving RA
3 20
kL
2
M
A
1 30
kL
3
SOLUTION (5.41) U W
Thus
EI 2
L
(d
2
dx
0
v 2
) dx 2
L 0
EI 2
L
0
L
w v d x w 0 a s in 0
U W :
E I 2L
3
4
a
( L ) a sin 4
2 x L
w0L 2
dx
w0 EI
L
4
( ) s in
w0L 2
a
,
We have v
2 x L
2
x L
82
dx
a
w0 EI
L
( )
4
4
EI 4L
3
a
2
SOLUTION (5.42) L A D L C D 3 .4 6 m U
AL i E
1 2
2
1 2
AL i E ( v cos / L i )
2
Vertical load of the joint, by Eq.(5.46): 3
U v
W
E jAj Lj
v cos 2
1
or
W E A v [
2
cos 30 L AD
o
2
o
E A v [ c o3s .4360
2
cos 30 LC D 2
o
cos 30 3 .4 6
o
1 LBD
]
13 ] 0 .7 6 6 9 E A v
Substitute the given data W 0 .7 6 6 9 ( 0 .0 0 6 )( 2 0 0 1 0 )( 6 2 5 1 0 9
6
) 5 7 5 .2 k N m
SOLUTION (5.43) v
(a)
2
ax
2L
U
3
EI 2
(3L x )
L
v" 3
(v" ) dx 2
0
9 EI 2L
6
a
2
a L
3
(L x)
L
( L x ) dx 2
0
3EI 2L
3
a
2
(1)
We have W P v A
(2)
Virtual work principle, U W , is thus P a
and
3EI 2L
3
(2a a )
a PL 3EI 3
v
2
Px 6 EI
(3L x )
(3)
( b ) At x=L, Eq.(3) gives v m ax
3
PL 3EI
Using Eq.(3): and at x=L: m ax
(4) dv dx
Px 2 EI
(2 L x )
2
PL 2 EI
(5)
SOLUTION (5.44) We have v ax ( L x ) axL ax ,
v' aL 2ax,
2
So, Also
U
EI 2
L
( v " ) d x 2 a E IL 2
2
0
W P v A P (acL ac ) 2
From u W , it follows that E IL ( 4 a a ) ( P c L P c ) a , 2
a
Pc ( L c ) 4 E IL
Hence, at x=c: vA
2
Pc ( Lc )
2
4 E IL
83
v" 2a
SOLUTION (5.45) I
d
4
64
2
d
A
,
r
,
4
I A d 4 , Le L
We have
cr
2
E
P A
4P
;
2
( Le r )
d
2
2
2
Ed
16 L
2
d
,
4
64 PL
2
3
E
Substituting given data: d [
Check:
L r
3
64 ( 50 10 )( 1 . 2 ) 3
( 210 10
4 ( 1. 2 )
4L d
9
0 .0 2 9
1
2
] 4 29 mm
)
1 6 5 .5 . Also
P A
3
4 ( 50 10 )
( 0 . 029 )
2
75 . 7 MPa
600
MPa.
OK.
SOLUTION (5.46) Substituting the given data: ( L e r ) c
E S y 45 . 7
Try Johnson’s formula Pc r
S
A
S
y
2 y 2
4 E
(
Le r
)
2
or 3
2 2 0 (1 0 )
d
Check
Le
2
d 4
5 2 0 (1 0 )
6
( 5 2 0 1 0 )
6
4
4
2
2
2
( 0 .2 5 ) 1 6
9
(1 1 0 1 0 )
d 2 5 .7 m m
;
2
d
38 . 9 OK.
250 ( 4 ) 25 . 7
Try Euler formula: d [
and
4 Le
d
2
6 4 P Le 3
E
1
]4 [
3
6 4 ( 2 2 0 1 0 )( 0 .2 5 ) 3
2
1
] 4 2 2 .5
9
(1 1 0 1 0 )
44 . 4 45 . 7
4 ( 250 ) 22 . 5
mm
does not apply.
SOLUTION (5.47) ( a ) By Eq.(5.61) with A d d [
64
2 Pc r L e 2
E
]
4 and r d 4 :
2
1 4
( b ) Equation (5.61) with A b h , b
I bh
3
12 ,
r
2
h
2
1 2:
2
1 2 Pc r L e 2
Eh
3
SOLUTION (5.48) ( a ) Same area:
(D
4
2
d ) ao ai 2
ai ao 2
2
4
2
(D
2
2
d ) 50 2
2
(5 0 3 5 ) 2
2
4
(CONT.)
84
5.48 (CONT.) a i 3 8 .7 m m ,
or
1
t
2
( a o a i ) 1 1 .3 m m
( b ) Circular bar I
4
(D
d ) 4
64
( 5 0 3 5 ) 2 3 3 .1 1 0 4
4
9
m
4
64
EI
( 7 2 ) ( 2 3 3 .1)
2
Pc r
2
2
Le
3 4 .2 k N
2
( 2 .2 )
Square bar I
1
4
12
1
(ao ai ) 4
( 5 0 3 8 .7 ) 3 3 3 .9 1 0 4
9
m
4
12
EI
( 7 2 ) ( 3 3 3 .9 )
2
Pc r
4
2
2
Le
2
( 2 .2 )
49 kN
SOLUTION (5.49) I
d
(8 )
4
64 A
d
4
2 0 1 .0 6 m m
4
64 2
(8 )
4
2
5 0 .2 7 m m
2
4
EI 2
P A Pc r
L
2
( 2 0 0 1 0 )( 2 0 1 .0 6 1 0 2
9
( 0 .4 )
2
12
)
2 .4 8 k N
The corresponding stress is
cr
Pc r
A
2480 5 0 .2 7 1 0
6
4 9 .3 M P a 2 5 0 M P a OK .
We have
M
0:
C
( a b ) Q b Pc r
1 8 0 Q 3 0 ( 2 4 8 0 ), or Therefore
Q a ll
Q
n
4 1 3 .3
Q 413 N
295 N
1 .4
SOLUTION (5.50) Pc r n P 2 .6 ( 2 2 ) 5 7 .2 k N , I
d
A
4
d
64
2
r
4
L e 0 .7 L 0 .7 (1) 0 .7 m d 4
Equation (5.61) gives 2
d
4
6 4 Pc r L e
E 3
6 4 (5 7 .2 1 0 )( 0 .7 ) 3
(200 10 ) 3
9
2
,
d 0 .0 2 3 m = 2 3 m m
(CONT.)
85
5.50 (CONT.) Hence Le
0 .7
r
1 2 1 .7
0 .0 2 3 4
Equation (5.62): Le
(
r
E
)c
Sy
200 10
9
250 10
6
8 8 .8 6 1 2 1 .7
Euler formula is valid, Therefore d 23 m m
SOLUTION (5.51) Refer to Solution of Prob. 5.50. Now we have L e 0 .7 ( 6 2 5 ) 4 3 7 .5 m m and Pc r 2 .6 (1 2 5 ) 3 2 5 k N . Equation (5.61): 6 4 (3 2 5 1 0 )( 0 .4 3 7 5 )
2
d
4
6 4 Pc r L e
E 3
3
2
3
d 0 .0 2 8 m
,
(200 10 ) 9
and Le
r
4 3 7 .5
6 2 .5 1 2 1 .7
28 4
Euler formula does not apply. Apply Johnson formula, Eq. (5.66): d 2(
or
2
Pc
Sy
S y Le
E 2
1 2
)
325 10
2[
3
250 10
2 5 0 1 0 ( 0 .4 3 7 5 ) 6
6
(200 10 ) 2
9
2
1
]
2
d 0 .0 4 1 9 m = 4 1 .9 m m
SOLUTION (5.52)
L B C 0 .6 5 m
F AB
5 12
FBC
P,
13 12
P
P B
F AB
12 13 5 FBC
Bar AB ( F A B ) cr
2
EI 2
Le
2
9
( 2 1 0 1 0 )[
( 5 1 0
3
( 0 .4 )
4
) ]
4 2
6 .3 5 9 k N
5 12
Pc r ,
Pc r 1 5 .2 6 k N
Bar BC ( F B C ) cr
2
9
( 2 1 0 1 0 )[
( 7 .5 1 0
4 ( 0 .6 5 )
2
3
4
) ]
1 2 .1 9 k N
Choose the small value, Pc r 1 1.2 5 with n 2 .5 . Thus Pall
Pcr n
11 . 25 2 .5
4 . 5 kN
86
13 12
Pc r ,
Pc r 1 1 .2 5 k N
SOLUTION (5.53) ( a ) Applying the method of joints at A: F AB 40 kN ( C ) and F B C 2 2 0 k N ( C ) L A B 2 .5 m .
cr
FAB
We have r
cr
3
4 0 (1 0 )
2 5 0 (1 0 ) 6
;
A
d
2
d 1 4 .3 m m
,
4
I A d 4 and Euler’s formula: 2
E
(L r)
2
; 250 (10
)
6
2
( 210 10
9
( 2 . 5 0 . 25 d )
)
, d 109 . 8 mm
2
Use, a commercial size of : d 110 mm diameter ( b ) L B C 1 .8 7 5 m
FBC
cr
3
2 2 0 (1 0 )
2 5 0 (1 0 ) 6
;
A
d
2
d 3 3 .5 m m
,
4
Euler formula:
cr
2
E
(L r)
2
6
; 250 (10
)
2
( 210 10
9
)
( 1 . 875 0 . 25 d )
2
, d 82 . 4 mm
Use 8 3 m m diameter
SOLUTION (5.54) I r
64
( 0 . 04 ) I A
d 4
9
125 . 66 10
4
10 mm
L r
m 1600 10
4
160
Euler’s formula: 2
P cr
EI
F a ll
1 .0 0 .5
nL
2
2
( 200 125 . 66 ) 1 .5 (1 .6 )
2
64 . 6 kN
Thus Pc r 2 ( 6 4 .6 ) 1 2 9 .2 k N
SOLUTION (5.55) Two angles: I x 2 I x 2 ( 4 2 6 1 0 ) 8 5 2 (1 0 ) m m 3
3
4
I y 2 ( I y A x ) 2[1 5 3 (1 0 ) 7 4 4 (1 0 .5 ) ] 4 7 0 , 0 5 2 m m 2
I m in 4 7 0 , 0 5 2
Use smaller I : 2
Pc r
Pa ll
3
EI y 2
Le
Pc r n
2
1 2 2 .7 2 .5
mm
9
4
( 2 0 0 1 0 )( 4 7 0 , 0 5 2 1 0 ( 2 .7 5 )
2
We have L e 2 .7 5 m Thus
12
)
2
4 9 .1 k N
87
1 2 2 .7 k N
4
SOLUTION (5.56) A d I
2
4 ( 60 )
(60 )
4
d 64
P cr
2
L
4
64
EI 2
2
2
4 2 ,827
6 3 6 ,1 7 3 m m ( 210 10
9
1
)( 636 10
12
)
mm
4
r
1319
2
2
I A
d 4
60 4
15 mm
kN
Using Eq.(5.72): Py
350 ( 10 ) 6
3
2 . 827 10
0 . 002 ( 2 . 827 10
[1
636 ,173 ( 10
12
3
)
1
) 0 . 03 1
]
Py 3
1319 ( 10 )
or P y 2 6 6 0 1 0 4 .4 P y 1.3 0 5 1 0 2
Solving, this quadratic gives: P y 649
12
0
kN . Thus, Pa ll 6 4 9 3 2 1 6 .3 k N
SOLUTION (5.57) I m in
1
(1 6 0 8 0 1 3 0 5 0 ) 5 .4 7 2 5 1 0 3
3
A 1 6 0 8 0 1 3 0 5 0 6 .3 1 0 rm in
6
mm
4
12
A 2 9 .5 m m
I m in
3
mm
3
L e 0 .5 L 2 .7 5 m
L e r 2 7 5 0 2 9 .5 9 3 .2
Hence, E
(7 2 1 0 )
2
cr
( Le r )
2
2
9
( 9 3 .2 )
2
8 1 .8 M P a
SOLUTION (5.58) L e 0 .7 L 3 .8 5 m . From solution of Prob.5.57: rm in 2 9 .5 m m .
We now have L e r 3 8 5 0 2 9 .5 1 3 0 .5
Hence, E 2
cr
(7 2 1 0 ) 2
( Le r )
9
(1 3 0 .5 )
2
4 1 .7 M P a
SOLUTION (5.59) A (5 0 4 4 ) 1, 7 7 2 2
2
mm
2
Equation (5.68): r
1 4
D
2
d
2
1 4
100
2
88
2
33 . 3 mm ,
L r
2000 33 . 3
60 . 1
(CONT.)
88
5.59 (CONT.)
r
12(50 )
ec
(a)
2
( 3 3 .3 )
Py
Refer to Fig.5.23b:
r
2
( 3 3 .3 )
Refer to Fig.5.23b:
160 M Pa,
A
9( 50 )
ec
(b)
0 .5 4 1
2
P y 1 6 0 (1, 7 7 2 ) 2 8 3 .5
kN
0 .4 0 6
2
Py
175 M Pa,
A
P y 1 7 5 (1, 7 7 2 ) 3 1 0 .1 k N
SOLUTION (5.60) We have
I
(D
d ) and A
4
4
64
(D
2
d ); 2
r
1
I A
4
D
2
d
2
4
A
P n ( 4 4 0 ) 1 .5 ( 4 4 0 ) 6 6 0 k N ,
(200 175 ) 7363 m m 2
2
2
4 I
( 2 0 0 1 7 5 ) 3 2 .5 (1 0 ) m m 4
4
6
4
64 r
1
200 175 2
2
6 6 .4 4 m m
4
Hence,
P
A ec r
2
660 10 7363 10
3 6
r
0 .1 e ( 6 6 .4 4 1 0
L
8 9 .6 M P a
3
)
2
3 .6 6 6 6 .4 4 1 0
5 5 .1
3
2 2 .6 5 e
Substitute these into Eq. (5.74a): 2 1 0 8 9 .6 1 2 2 .6 5 e s e c 2 7 .5 5
8 9 .6 1 0 190 10
6
9
from which e 0 .0 4 9 0 2 m = 4 9 .0 2 m m
SOLUTION (5.61) From solution of Prob. 5.60:
P
A 7363 m m
2
I 3 2 .5 (1 0 ) m m 6
4
Hence EI 2
Pc r
v m ax
L
2
( 2 0 0 1 0 )(3 2 .5 1 0 2
9
( 4 .6 )
e
6
)
2
3 .0 2 3 M N
Figure S5.61
(CONT.)
P
89
5.61 (CONT.) ( a ) Using Eq. (5.73): 1 .2 5 e s e c 2
45 10
3
3 .0 3 2 1 0
6
o 1 e s e c (1 0 .9 6 ) 1 ,
e 6 7 .2 8 m m
( b ) Referring to Fig S5.61: M P ( v m a x e ) 4 5 (1 .2 5 6 7 .2 8 ) 3 0 8 0 N m
Hence,
P
m ax
Mc
A
I
45 10
3
7363 10
6
3 0 8 0 ( 0 .1) 3 2 .5 1 0
6
1 5 .5 9 M P a
SOLUTION (5.62) L e 2 ( 4 .6 ) 9 .2 m . Refer to solution of Prob. 5.61
EI 2
Pc r
( 2 0 0 1 0 )(3 2 .5 1 0 2
2
9
Le
(9 .2 )
2
6
)
7 5 7 .9 k N
( a ) Equation (5.73): 1 .2 5 e s e c 2
o 1 e [s e c ( 2 1 .9 3 ) 1] , 7 5 7 .9 45
e 1 6 .0 2 m m
( b ) M P ( v m a x e ) 4 5 (1 .2 5 1 6 .0 2 ) 7 7 7 .2 N m Therefore,
m ax
P A
Mc I
45 10
3
7363 10
6
7 7 7 .2 ( 0 .1) 3 2 .5 1 0
6
6 .1 1 2 2 .3 9 1 8 .5 M P a
SOLUTION (5.63) A
L e 2 L 3 .6 m
75 mm C 15 mm
A 150 75 120 45
5 .8 5 1 0
P
150 mm D
B
1
I
3
mm
2
(1 5 0 7 5 1 2 0 4 5 ) 3
3
12
P
4 .3 6 2 1 0 r
L
6
mm
4
I A 2 7 .3 m m
e c 3 7 .5 m m
Thus, ec r
2
3 7 .5 3 7 .5 ( 2 7 .3 )
2
1 .8 8 7
Le
6 5 .9 3
2r
(CONT.)
90
5.63 (CONT.) Use Eq.(5.74b) with L L e :
m ax
160 10 5 .8 5 1 0
1 1 .8 8 7 s e c 6 5 .9 3
3 3
9 9 .3 M P a 9 3 2 0 0 1 0 (5 .8 5 1 0 ) 160 10
3
SOLUTION (5.64) 150 mm
C
L e 2 L 3 .6 m
D
P
A 150 75 120 45
75 mm B
A P
5 .8 5 1 0
3
mm
2
15 mm 1
I
(7 5 1 5 0 4 5 1 2 0 ) 3
3
12
1 4 .6 1 1 0
6
mm
4
e c 75 m m
r
I A 4 9 .9 7 m m
Therefore, ec r
2
75 75
( 4 9 .9 7 )
2 .2 5
2
Le
3 6 .0 2
2r
Apply Eq.(5.74b) with L L e :
m ax
160 10 5 .8 5 1 0
3 3
1 2 .2 5 s e c 3 6 .0 2
9 4 .8 M P a 9 3 2 0 0 1 0 (5 .8 5 1 0 ) 160 10
3
SOLUTION (5.65) We have A ao ai 120 100 2
1
I
2
(ao ai ) 4
12
ao
3
2
(1 2 0 1 0 0 ) 8 .9 5 (1 0 ) m m 4
4
6
4
12
8 .9 5 (1 0 )
A c
1
4
4 .4 (1 0 ) m m
2
3
I
r
2
4 5 .1 m m
4 .4
60 m m
2
Le 2 L 2 (2 ) 4 m
Hence ec r
2
(5 5 )(6 0 ) ( 4 5 .1)
EI
2
2
Pc r
2
Le
3
A
( 7 0 1 0 )8 .9 5 (1 0 2
P
1 .6 2 2 , 9
(4)
2
2 0 0 (1 0 ) 4 .4 (1 0 6
)
6
4 5 .5 M P a
)
3 8 6 .5 k N
(CONT.)
91
5.65 (CONT.) P
200
Pc r
0 .5 1 7
3 8 6 .5
( a) Apply Eq. (5.73): v m ax e[ s e c (
P
2
Pc r
) 1]
(5 5 )[se c (
0 .5 1 7 ) 1] 7 3 .7 6 m m
2
( b ) Use Eq.(5.74a)
m ax
P
[1
A
ec r
sec(
2
P
2
Pc r
)]
( 4 5 .5 ) [1 (1 .6 2 2 ) s e c (
0 .5 1 7 ) ] 2 1 8 .3 M P a
2
SOLUTION (5.66) From solution of Prob. 5.65: A a o b i 4 .4 (1 0 ) m m 2
1
I
2
3
( a o a i ) 8 .9 5 (1 0 ) m m 4
12
2
4
6
L e 2 ( L ) 2 (1 .9 ) 3 .8 m
EI 2
Pc r
P
2
Pc r
9
Le
( 3 .8 )
300
c 60 m m
( 2 0 0 1 0 ) (8 .9 5 1 0 2
4
6
)
2
1, 2 2 3 k N
0 .2 4 5
1223
( a ) Equation (5.73): v m ax e[ s e c (
P
2
Pc r
) 1]
or 1 5 e[ s e c (
0 .2 4 5 ) 1] ,
e 3 7 .2 m m
2
(b) M
m ax
m ax
P ( e v m a x ) 3 0 0 (3 7 .2 1 5 ) 1 5 .6 6 k N m
P A
M
m ax
I
c
300 10 4 .4 (1 0
3
3
)
1 5 .6 6 1 0 ( 0 .0 6 ) 3
4 .9 5 (1 0
1 7 3 .2 M P a
92
6
)
SOLUTION (5.67) Figure 5.17a: L e 2 L 2 ( 2 ) 4 m . ( a ) Cylindrical tube: A 500 m m
2
4
2
(D
d ) 2
or d
D
2
4A
40 2
4 (500 )
31 m m
Thus I
64
(D
d )
4
4
[ 4 0 3 1 ] 8 .0 3 3 (1 0 ) m m 4
64
4
4
4
and r
4
( 8 .0 3 3 )(1 0 )
I A
1 2 .6 8 m m
500
It follows that L e r 4 0 0 0 1 2 .6 8 3 1 5 .5
Since L e r 2 0 0 , the Euler formula applies. Hence Pc r
2
EI 2
Le
2
9
(1 0 5 1 0 )( 8 .0 3 3 1 0 (4)
8
)
5 .2 0 3 k N
2
( b ) Square tube. The cross-sectional area: A a o a i . Inner diameter is 2
ai
ao A 2
2
4 0 5 0 0 3 3 .1 7 m m 2
Then I
1 12
( a o bi ) 4
1 12
r
I A
10
1 1 .2 5 500
15 m m ,
4
( 4 0 3 3 .1 7 ) 1 1 .2 5 1 0 4
4
4
mm
4
and 2
Le r 4 0 0 1 5 2 6 7
Since L e r 2 0 0 , the Euler formula is valid. Therefore Pc r
2
EI 2
Le
2
9
(1 0 5 1 0 )( 0 .1 1 2 5 1 0 (4)
2
6
)
7287 N
Comment: Hollow square has a critical load that is 1.4 times more than for a hollow circular section.
SOLUTION (5.68) From Table B.1: E 2 0 0 G P a
S y 250 M Pa
The properties of area are A b h (3 5 )(1 0 ) 3 5 0 m m , 2
I
1 12
bh 3
1
(3 5 )(1 0 ) 2 9 1 7 m m 3
4
12
(CONT.)
93
5.68 (CONT.) I
r
2917
A
2 .8 8 7 m m ,
(
Le r
350
E
)c
200 10
Sy
3
8 8 .9
250
( a ) From Fig. 5.17c, L e 0 .7 L 0 .7 (1 8 0 ) 1 2 6 m m . Hence Le
126
r
4 3 .6
2 .8 8 7
Since 4 3 .6 8 8 .9 , Johnson Formula should be used. Thus: S y ( Le r )
Pc r A S y [1
2
]
4 E 2
2
2 5 0 ( 4 3 .6 )
( 3 5 0 ) ( 2 5 0 ) [1
4
200 10
2
3
] 8 2 .2 3 k N
( b ) Now we have Le
0 .7 ( 5 0 0 )
r
1 2 1 .2 8 8 .9
2 .8 8 7
Euler formula applies. So EI
( 2 0 0 1 0 )( 2 9 1 7 1 0
2
Pc r
2
2
9
Le
( 0 .3 5 )
12
2
)
47 kN
SOLUTION (5.69) ( a ) Cross-sectional area: A
4
2
(D
d ) 2
( 6 2 .5 6 0 ) 2 4 0 .5 m m 2
4
2
2
Moment of inertia:
I
64
4
(D
d ) 4
64
( 6 2 .5 6 0 ) 1 1 2 , 8 4 1 .5 m m 4
4
4
and r Le
I A
r
750 2 1 .7
2 1 .7 m m
1 1 2 ,8 4 1 .5 2 4 0 .5
3 4 .5 6
Also (
Le r
2
E
)c
Sy
2
9
( 7 0 1 0 )
2 7 0 1 0
6
5 0 .5 8
Since L e r 5 0 .9 6 , the Johnson formula applies. Thus Pc r A S y [1
Sy 2
4 E
(
Le r
) ] 2 4 0 .5 1 0 2
6
( 2 7 0 1 0 )[1 6
6
2 7 0 1 0 ( 3 4 .5 6 ) 4
2
9
( 7 0 1 0 )
2
]
5 7 .3 6 k N
( b ) We have C D 2 and: ce r
2
3125 (3) ( 2 1 .7 )
2
0 .2
A 2 4 0 .5 m m
2
P 16 kN
(CONT.)
94
5.69 (CONT.)
Equation (5.74a) gives then
m ax
[1
P A
2
3
2 4 0 .5 1 0
P Pc r
)]
[1 0 .2 s e c ( 2
1 6 (1 0 )
s e c ( 2
ec r
6
16 5 7 .3 6
)]
7 9 .8 M P a
SOLUTION (5.70) From Prob.5.55 for both angles: I m in 4 7 0 , 0 5 2 m m r
1,4 8 8 2
2 E Sy
Cc
Le
1 7 .7 7 m m ,
4 7 0 ,0 5 2
2
2
( 200 10
240 10
9
)
6
r
4
and A 7 4 4 2 1 4 8 8 m m
2 .7 5 1 0 1 7 .7 7
3
1 5 4 .8
128
Since C c 1 5 4 .8, use Eqs. (5.77b):
a ll
2
9
( 2 0 0 1 0 )
1 .9 2 (1 5 4 .8 )
4 2 .9 M P a
2
Hence Pa ll
a ll
A 4 2 .9 (1 4 8 8 ) 6 3 .8 4 k N
SOLUTION (5.71) Cc
2
2 E Sy
[
2
2
1
3
( 200 10 )
Le
] 2 106 ,
350
r
0 . 65 ( 3 ) d 4
7 .8 d
(Eq.c of Sec. 6.2)
Equation (5.77b):
Le
Check:
3
5 0 (1 0 )
a ll
d
r
2
4
6 0 . 0441
2
9
( 2 0 0 1 0 )
1 .9 2 ( 7 .8 d )
2
d 4 4 .1 m m
,
136 C c
OK.
SOLUTION (5.72) L e 0 . 5 L 2 m . Assume 1 1 L e d 2 6 and use Eq.(5.80b). Thus
a ll
P A
3
1 0 0 (1 0 ) d
2
8 .2 7 [1
1 3
(
Le d 26
2
) ]1 0
6
This gives 1 0 0 (1 0 ) 8 .2 7 d 3
2
0 .0 2 4 5,
d 109 m m
Then
P A
3
1 0 0 (1 0 ) ( 0 .1 0 9 )
Check: 8 .4 2 1 0
2
8 .4 2 M P a
O K .,
L e d 2 0 .1 0 9 1 8 .3
95
OK.
2
Thus
SOLUTION (5.73) A
(3 5 0 3 0 0 ) 2 5 , 5 2 5 m m , 2
4
2
L e 0 .7 L 0 .7 ( 6 .1) 4 .2 7 m .
2
Equation (5.68): r
1 4
d
2
D
2
350 300 2
1 4
Le
1 1 5 .2 4 m m ,
2
r
4 .2 7 1 0 1 1 5 .2 4
3
3 7 .0 5
Using Eq.(5.79b): a ll [1 4 0 0 .8 7 (3 7 .0 5 )] 1 0 7 .7 7 M P a Hence Pa ll 1 0 7 .7 7 1 0 ( 2 5 , 5 2 5 1 0 6
6
) 2751 kPa
SOLUTION (5.74) 0 .1 2 0 .0 8 12
I m in
3
6
5 .1 2 (1 0
A 0 .0 0 9 6 m ,
rm in
2
Cc
4
) m , I m in
2
2 E Sy
A 2 3 .0 9 m m ,
1 2 1 .7 Le r
3,5 0 0 2 3.0 9
1 5 1.6
By Eq.(5.77b):
a ll
2
E
1 .9 2 ( L e r )
2
2
9
( 2 1 0 1 0 )
1 .9 2 (1 5 1 .6 )
2
4 6 .9 7 M P a
We have 600 9 .6
62 . 5 MPa
280
MPa
SOLUTION (5.75) Table A.6: A 27 . 5 10
3
mm
r z 101 . 1 mm
2
Buckling in xy plane: C c [ 2
2
( 200 10
1
3)
280 ] 2 119 , L e 0 . 7 L 4 . 2 m
and L e r z 4 .2 0 .1 0 1 1 4 1.5 C c . Apply Eq.(5.77a):
n
5 3
all
3 8
( 119 )
1 8
( 1 1 9 ) 1.7 9
1 2
.5 ( 41 ) ] 146 . 9 MPa 119
4 1.5
280 10 1 . 79
6
[1
3
4 1 .5
2
and Pall 146 . 9 ( 27 . 5 ) 4 , 040
kN
Buckling in xz plane: L e 0 . 5 (12 ) 6 m n
5 3
a ll
3 8
( 3171.59 )
2 8 0 1 0 1 .7 7
6
[1
r y 161
mm ;
1 8
( 31 71 .39 ) 1 .7 7
1 2
( 31 71 .39 ) ] 1 5 0 .4 M P a ,
L r y 37 . 3 C c
2
2
96
Pa ll 1 5 0 .4 ( 2 7 .5 ) 4 ,1 3 6 k N
SOLUTION (5.76) A a
r
2
a
I A 2
a
I a L
a
4
r
12 0 .5
3
a (2
1 .7 3 2 a
3)
Assume: 9 .5 L e r 6 6 . Using Eq. (5.79b):
a ll
250 10
a
3
[1 4 0 0 .8 7 (
2
1 .7 3 2
) ]1 0
6
a
or 3
a 1 0 .7 6 1 0 2
a 1 .7 8 6 1 0
3
0,
a 48 m m .
So, r a 2
3 48 2
3 1 3 .9 m m
L r 5 0 0 1 3 .9 3 6 6 6
Our assumption was correct. Use a 4 8 m m
SOLUTION (5.77) A 200 100 160 60 10, 400 m m I m in rm in
1
2
( 2 0 0 1 0 0 1 6 0 6 0 ) 1 3, 7 8 6 , 6 6 6 .6 7 m m 3
3
4
12 A 3 6 .4 1 m m ,
I m in
L e rm in 4 .3 1 0
3
3 6 .4 1 1 1 8 .1
Use Eq. (5.79c):
a ll
350 10 (1 1 8 .1)
3
2 5 .0 9 M P a
2
Pa ll 2 5 .0 9 (1 0 , 4 0 0 ) 2 6 0 .9 4 k N
SOLUTION (5.78) For the situation described L e 2 L (see Fig. 5.17a) and d=62 mm So, Le
2 (1 .2 2 1 0 ) 3
d
3 9 .4
62
Since 39.4 > 26, use Eq. (5.80c). 0 .3 (1 2 1 0 ) 9
a ll
(3 9 .4 )
2
2 .3 2 M P a
and Pa ll
a ll
A ( 2 .3 2 ) (8 8 6 2 ) 1 2 .6 6 k N
97
SOLUTION (5.79) We now have L e 2 L 2 ( 7 5 0 ) 1 5 0 0 m m and d=62 mm Le
1500
d
2 4 .2
62
Since 24 .2< 26, apply Eq. (5.80b).
1 2 4 .2 2 ( ) ] 5 .8 8 M P a 3 26
8 .2 7[1
a ll
Thus Pa ll
a ll
A ( 5 .8 8 ) (8 8 6 2 ) 3 2 .0 8 k N
SOLUTION (5.80) Boundary conditions: v ( 0 ) 0 , v ( L ) 0 , With w=0, the solution of Eq.(5.82b):
M (0) M 0 ,
M (L) M
0
v A sin k x B c o s k x C x D
(1)
v " A k s in k x B k
(2)
2
2
cos kx
Substituting boundary conditions into these equations, we have v (0 ) B D 0,
v ( L ) A s in k L B c o s k L C L D 0
and since E Iv " M , M ( 0 ) B E Ik
Solving, and setting k M
A
M 0,
M ( L ) A E Ik sin k L B E Ik c o s k L M 2
2
P E I as needed,
2
1 c o s k L s in k L
0
P
2
B D
,
M
0
P
C 0
,
Equation (1) is thus M
v
0
1 c o s k L s in k L
(
P
s in k x c o s k x 1 )
SOLUTION (5.81) Apply Eq.(4.14) 2
EI
dx
M Pv
M
d v 2
2
dx
2
(x
2
Lx )
P
or d v
w 2
k x 2
w 2
(x
2
wL 2
Lx )
A
B
x
wL 2
Figure S5.81 We have, the deflection: v vh vP
(1)
v h A s in k x B c o s k x
(2)
Here It can be shown that, for the case of uniform loading w (Fig. S5.81): vP
w 2 E Ik
2
(x
2
Lx
2 k
2
(3)
)
Boundary conditions v h ( 0 ) 0 and v h ( L ) 0 give A and B. In doing so, Eq.(1) results in: v
w E Ik
4
[(1 c o s k L )
s in k x s in k L
cos kx
98
2
k 2
(x
2
L x ) 1]
0
SOLUTION (5.82) Refer to Example 5.23.
v
m x L
a m sin
(1)
m 1
Hence U W
EI 2
1 2
L
(d
0
2
dx
v 2
L
4
EI
) dx 2
P ( v ') d x 2
0
4L
L
3
4
w vdx
0
2
m am
2
P 4L
m am 2
2
2wL
am
1 m
Applying U W , we obtain am
4 wL 5
4
EI
1 3
2
m (m b)
where b
PL
2
2
EI
Substitution of this into Eq.(1) gives the solution.
End of Chapter 5
99
Section II FAILURE PREVENTATION CHAPTER 6
STATIC FAILURE CRITERIA AND RELIBILITY
SOLUTION (6.1) Table 6.2: K c 5 9 1 0 0 0 M P a
mm
S y 1503 M Pa
and
1 . 01
Case A of Table 6.1: with a w 0 . 1 , From Eq.(6.3), with n 1 :
Kc
a
59
2 0 8 .4
1000
(1 .0 1 )
( 25 )
M Pa
Thus, we have P ( 2 w t ) 2 0 8 .4 ( 0 .5 0 .0 2 5 )1 0 2 , 6 0 5 6
kN
The nominal stress at fracture
6
2 .6 0 5 (1 0 )
2 3 1 .6
0 .0 2 5 ( 0 .5 0 .0 5 )
M Pa
This is well below the yield strength of 1503 MPa.
SOLUTION (6.2) Py
all
A net
( 350 15 ) 2 . 844
650 1 .2
Case B of Table 6.1: with a w 0 .0 7 ,
By Eq.(6.3),
Kc n
a
MN
1.1 2
100 1000 1 .2 (1 .1 2 )
( 25 )
2 6 5 .5 M P a
Since 265.5 < 650, fracture controls. Thus P f 265 . 5 ( 350 15 ) 1 , 394
kN
SOLUTION (6.3) From Table 6.2: K c 23 1000
MPa
Case B of Table 6.1: a w 0 .1 6 ,
mm
and S
1.1 2
By Eq.(6.3), with n 1 :
Kc
a
8 1 .9 3 M P a
23 1000 1 .1 2
( 20 )
It follows that P ( wt ) 81 . 93 (125 25 ) 256
Then
P ( w a )t
3
256 ( 10 ) ( 0 . 125 0 . 02 ) 0 . 025
97 . 5 MPa
S
97 . 5 MPa
y
100
kN
y
444
MPa
SOLUTION (6.4) Table 6.1, Case A: 1 .0 3 . Table 6.2, K c 5 9 M P a
S y 1503 M Pa .
m,
We have
Sy n
6 0 1 .2 M P a
1503 2 .5
Equation (6.1) gives K
a (1 .0 3 )( 6 0 1 .2 ) ( 2 1 0
3
) 4 9 .0 8 M P a
Using Eq. (6.2), we find n
Kc
K
59 4 9 .0 8
1 .2
SOLUTION (6.5) Table 6.1, Case B: 1 .3 7 . Table 6.2, K c 3 1 M P a
m,
S y 392 M Pa .
We have
Sy n
140 M Pa
392 2 .8
Equation (6.1): K
a 1 .3 7 (1 4 0 ) ( 0 .0 0 5 ) 2 4 .0 4 M P a
Applying Eq. (6.2), n
Kc
K
31 2 4 .0 5
1 .2 9
SOLUTION (6.6) Table 6.1: a w 5 0 5 0 0 0 .1, Table 6.2: K c 1 1 1 M P a
m,
1 .0 1 . S y 798 M Pa .
( a ) Equation (6.1), K
a 1 .0 1(1 5 0 ) ( 0 .0 5 ) 6 0 M P a
Equation (6.2) gives the safety factor for fracture as, n
Kc
K
111 60
1 .8 5
Safety factor for yielding is n
Sy
798 150
5 .3 2
( b ) Using Eq. (6.3) with n=1, fracture stress is
f
Kc
a
111 (1 .0 1 )
( 0 .0 5 )
2 7 7 .3 M P a
101
m
m
m
SOLUTION (6.7) Case A of Table 6.1 and Table 6.2: K
c
59
1000
1 . 01
MPa
S y 1503 M Pa
mm
(assumed)
By Eq.(6.3):
Kc n
(a)
pfr
(b)
pfr
2t
t
a
59
3 1 0 .7
1000
(5)
(1 .5 )(1 .0 1 )
M Pa S y
,
p f 2 ( 4 9 .7 1) 9 9 .4 2 M P a
,
pf
t r
3 1 0 .7 ( 4 )
25
4 9 .7 1 M P a
SOLUTION (6.8) Table 6.2: K c 31 MPa Case D of Table 6.1:
m
y
392
MPa
1.3 2
0 .4
a w
S
Using Eq.(6.3) with n 1 and 6 M tw : 2
Kc
a ;
6M tw
2
3 1(1 0 ) 1.3 2 6
6M 0 . 0 2 5 ( 0 .1 )
2
( 0 .0 4 )
Solving M 2 . 76 kN m
SOLUTION (6.9) ( a ) K c 59 M Pa
m
S y 1 5 0 3 M P a (Table 6.2).
and
Case B of Table 6.1: with a=12.5 mm and w=125 mm:
a w
0 . 1 1 . 12
Eq.(6.3) with n 1 : 5 9 (1 .1 2 )
(1 2 .5 1 0
3
),
2 6 5 .8 M P a
Therefore Pa ll ( w t ) (1 2 5 2 5 )( 2 6 5 .8 ) 8 3 0 .6 k N
Note that the nominal stress at fracture
8 3 0 .6 2 5 (1 2 5 1 2 .5 )
2 9 5 .3 M P a S y
( b ) Table 6.2: K c 6 6 M P a
m
and S y 1149 M P a
Thus 66 10
6
(1 . 12 )
a (1 . 12 )(
Solving a 0 .0 1 5 m = 1 5 .6 m m
Comment: This value of a satisfies Table 6.2
102
830 . 6 10 125 25 10
3 6
) a
SOLUTION (6.10) a w
Refer to Example 6.3. We now have a 2 .1 1
0 .4 :
15 37 . 5
b 1.3 2
and
Equation (6.5) is therefore ( 2 . 11 )
Table 6.2: K c 7 7 M P a
m
6 ( 0 . 175 P )
(1 . 32 )
P 0 . 0375 ( 0 . 0125 )
0 . 0125 ( 0 . 0375 )
2
83 , 349 P
and S y 690 M P a
Note that both a and t satisfy Table 6.2. Using Eq.(6.3),
Kc
a
n
7 7 1 0
8 3, 3 4 9 P
:
2
3
( 0 .0 1 5 )
Solving P 2 .1 2 8 k N
The nominal stress at fracture, ( 2 .1 2 8 1 0 ) [ 0 .0 1 2 5 ( 0 .0 3 7 5 0 .0 1 5 )] 7 .5 7 M P a S y . 3
SOLUTION (6.11) By Table 6.2, K c 5 9 M P a
m and S y 1503
M P a . Note that values of a and t satisfy
Table 6.2. a w
From Table 6.1:
a ll
Kc
n
a
1.1 2 . Therefore,
0 .1
5 9 1 0
(1 .4 )(1 .1 2 )
6
( 0 .0 0 5 )
3 0 0 .2 2 M P a
and Pa ll
a ll
( w t ) 3 0 0 .2 2 ( 0 .0 5 0 .0 2 5 ) 0 .3 7 5 M N = 3 7 5 k N
Comment: The nominal stress at the fracture 375 kN t(2w2a)
3
3 7 5 1 0 ( 0 .0 2 5 )( 0 .1 0 .0 1 )
1 6 6 .7 M P a S y
SOLUTION (6.12)
(a) (b)
x
xy
S
y
n
S
y
n
32 M 3
D
16 T
D
3
(
x
(
x
2
2
4P
D
2
3
32 ( 5 10 )
( 0 .1 ) 3
16 ( 8 10 )
( 0 .1 )
3
2
3
4 ( 50 10 )
( 0 .1 )
2
57 . 3 MPa
1
260 n
[ 5 7 .3 3 ( 4 0 .7 4 ) ] 2 ,
260 n
[ 5 7 .3 4 ( 4 0 .7 4 ) ] 2 ,
1
4 xy ) 2 ;
40 . 74 MPa
1
3 x y ) 2 ; 2
3
2
2
103
2
2
1
n 2 .8 6
n 2 .6 1
SOLUTION (6.13) Table B.1: S u 2 4 0 M P a and S u ' 6 5 0 M P a From the solution of Prob. 6.12, we have
5 7 .3 M P a and
x
xy
4 0 .7 5 M P a .
Thus
x
2
1 ,2
(
x
2
) 2
2 xy
2 8 .6 5
( 2 8 .6 5 ) ( 4 0 .7 5 ) 2
2
or 1 7 8 .4 6 M P a
2 2 1 .1 6 M P a
Thus 1
u n
1;
2
u n
7 8 .4 6 240
2 1 .1 6 650
1 n
or n 2 .7 8
SOLUTION (6.14) y A
F
B
0.8 m
A
P T
We have
T
D
M
P=20F,
T=0.4F,
P
x
M=0.8F
Stresses at fixed end:
b
32 M
a
D
3
32 ( 0 . 8 F )
( 0 . 04 )
20 F
( 0 .0 4 )
2
4
3
127 , 324 F
1 5 , 9 1 5 .5 F
x
16 T
D
3
16 ( 0 . 4 F )
( 0 . 04 )
3
31 , 831 F
a b 1 4 3 , 2 3 9 .5 F
Apply Eq.(6.16): S
y
n
1
[ x 3 ] 2 ; 2
2
6
2 5 0 (1 0 ) 1.4
2
or F 1 . 163
kN
SOLUTION (6.15) Refer to solution of Prob. 6.14. We have m ax
(
[(
x
2
) 2
1 4 3 , 2 3 9 .5 F 2
2
) (3 1, 8 3 1 F ) ] 2
2
Hence, m ax
Sy 2n
;
7 8 , 3 7 4 .7 F
1
F [(1 4 3 , 2 3 9 .5 ) 3 ( 3 1, 8 3 1 ) ] 2
6
2 5 0 (1 0 ) 2 (1 .4 )
or F 1 .1 3 9 k N
104
1 2
7 8 , 3 7 4 .7 F
2
x
SOLUTION (6.16) T 0 .4 F
At the fixed end A:
M
0 .8 F
z
Vy F
Thus,
x
32M
D
A 1 6T
D
( 0 .0 6 )
( 0 .0 6 )
3 7 .7 2 6 F
3
1 6 ( 0 .4 F )
3
3 2 ( 0 .8 F )
3
9 .4 3 1 F
3
Then 1, 2
3 7 .7 2 6 F
3 7 .7 2 6 F
) ( 9 .4 3 1 F ) 2
2
2
1 3 9 .9 5 1 0 F 140 10
(
2
3
(a)
m a x 2 1 .0 9 1 0 F
2 .2 3 1 0 F 3
2
3
6
2 1 .0 9 1 0 F ,
F 4 .1 5 k N
3
1 .6
( b ) 1 1 2 2 ( S y n ) 2
2
2
[3 9 .9 5 3 9 .9 5 ( 2 .2 3 ) ( 2 .2 3 ) ] 2
2
1 2
(1 0 ) F ( 2 6 0 1 .6 )1 0 3
6
or F 3 .9 5 k N
SOLUTION (6.17) Q B 102 (10 . 3 )( 74 . 85 ) 78 . 6 10
3
3
mm
Q C Q B 6 . 6 ( 69 . 7 )( 34 . 85 ) 94 . 6 10
3
mm
3
We have
A
Mc I
VQB
3
24 ( 10 )( 0 . 08 ) 13 . 4 ( 10
6
)
143 . 3 MPa ,
B
(143 . 3 ) 124 . 9 MPa
69 . 7 80
and B
Ib
3
6 0 (1 0 )( 7 8 .6 1 0 1 3 .4 (1 0
6
6
)
)( 0 .0 0 6 6 )
5 3 .3 M P a ,
c
VQc Ib
6 4 .2 M P a
Thus ( 1, 2 ) B
1 2 4 .9 2
[(
1 2 4 .9 2
1 B 1 4 4 .6 M P a
or
(
max
1
) 5 3.3 ] 2 6 2 .5 8 2 .1 2
2
2 B 1 9 .6 M P a
) B 82 . 1 MPa
Hence ( m ax ) B
S
y
2(2 )
320 4
8 2 .1 8 0
;
F a ils
Alternatively, using Eq.(6.11), we have [1 2 4 .9
2
1
4 ( 5 3.3 ) ] 2 2
320 2
;
1 6 4 .2 1 6 0
105
F a ils
SOLUTION (6.18) From Solution of Prob.6.17, at point B: 1 1 4 4 .6 M P a 2 1 9 .6 M P a Thus 1
S
[ 1 1 2 2 ] 2 2
2
y
n
320 2
160 1
or
2
[( 144 . 6 )
(144 . 6 )( 19 . 6 ) (19 . 6 ) ] 2 155 . 3 160 2
No failure
Alternatively, by Eq.(6.16): 1
3 ( 53 . 3 ) ] 2
2
[124 . 9
2
; 155 . 3 160
320 2
No failure
SOLUTION (6.19) 1
(a)
1
(b)
1 1
3
pr
t
Sy n
2
2
3 .5 ( 0 .2 5 )
0 .8 7 5 t
;
t
250 1 .5
S
(
2 2
0 .8 7 5 t
,
2
0 .4 3 7 5 t
t 5 .2 5 1 0
,
3 0
, 3
m = 5 .2 5 m m 1
2
y
) ; [(
n
0 . 875 t
)
2
( 0 . 875 20 . 4375 ) ( 0 . 4375 ) ]2 t 2
t
or 1
1 t
[ 0 .7 5 8 ] 2
250 1 .5
t 4 .5 4 8 1 0
,
3
m = 4 .5 4 8 m m
SOLUTION (6.20) From Solution of Prob.6.19: 1 0 . 875 , 2 t 1
(a)
Su
;
n
0 .8 7 5 t
0 . 4375 t
350 1 .5
,
3
0
t 3 .7 5 1 0
,
3
3 .7 5 m m
( b ) Using Eq.(6.26): 1
Su
Su
0 .8 7 5 t
;
n
350 1 .5
t 3 .7 5 1 0
,
3
m = 3 .7 5 m m
and 2
n
0 .4 7 5 t
;
350 1 .5
,
t 1 .8 7 5 1 0
3
SOLUTION (6.21) From Solution of Prob.6.17, at point B: 1 1 4 4 .6 M P a 2 1 9 .6 5 M P a (a)
1
( b ) By Eq.(6.25),
Su n
,
n
n
Su
1
280 144 . 6
280 144 . 6 19 . 65 ( 280 620 )
1 . 94
1 . 82
106
m = 1 .8 7 5 m m
250 1 .5
SOLUTION (6.22)
M
A
0:
Fy 0 :
1 7 5 ( 2 .4 6 )( 2 .4 6 2 ) R B (1 .7 ) 0 ,
R B 3 1 1 .5 k N
R A 119 kN
y
175 kN/m z
A 1.7 m
119 kN
b
311.5 kN 133
V, kN
2b
0.76 m C
B
3 ft
119
x -178.5
M, kN m x -50.54 At B at the upper outermost fiber:
m ax
M
m ax
c
5 0 .5 4 1 0 b 3
1
I
b (2b )
7 5 .8 1 1 0 b
3
3
3
12
Thus,
m ax
7 5 .8 1 1 0
; a ll
b
3
140 10 , 6
3
b 0 .0 8 2 m = 8 2 m m
At B at the neutral axis axis: m ax
3V
2A
3 1 7 8 .5 1 0 2
2b
3
2
1 3 3 .8 7 5 1 0 b
3
2
And m ax 1 2
Thus, m ax
; a ll
1 3 3 .8 7 5 1 0 b
2
3
140 10 , 6
Use a 8 2 m m by 1 6 4 m m rectangular beam.
107
b 0 .0 3 1 m = 3 1 m m
SOLUTION (6.23) We now have
N.A.
A 0 .3 0 .1 2 0 .0 3 6 m
B
120 mm
A
I
1 12
( 0 .1 2 )( 0 .3 ) 0 .2 7 1 0 3
M 0 .4 0 .1 5 0 .5 5 P
300 mm
c A c B 0 .1 5 m
3 ft Referring to Example 6.9:
P A
McA
A
P A
M cB
B
2 7 .7 7 8 P 3 0 5 .5 5 6 P
I
2 7 .7 7 8 P 3 0 5 .5 5 6 P
I
Thus 1 3 3 3 .3 3 4 P
2 0
2 2 7 7 .7 7 8 P
1 0
It follows that 3 3 3 .3 3 4 P
6
1 7 0 (1 0 ) 2 .5
P 204 kN
, 6
6 5 0 (1 0 )
2 7 7 .7 7 8 P
P 936 kN
,
2 .5
SOLUTION (6.24) S u 170
S uc 650
MPa
(Table B.1).
MPa
Thus 1, 2
and (a)
100 50 2
[(
1 127 . 6 MPa 127 . 6 , n 1 . 33
170 n
7 7 .6 ,
1
) (70 ) ]2 2
2
170 n
n
(b)
100 50 2
2
77 . 6 MPa
n 2 .1 9
170 1 2 7 .6 7 7 .6 (1 7 0 6 5 0 )
1.1 5
SOLUTION (6.25) Table B.1: S u 170 1, 2
or (a)
(b)
120 60 2
1 140
S uc 650
MPa [(
MPa ,
120 60 2
) 40 ] 2
2
140 ,
n 1 . 21
170 n
80 ,
n 2 . 13
170 1 4 0 8 0 (1 7 0 6 5 0 )
MPa 1 2
40 MPa ,
170 n
n
2
1.0 6
108
2
3
80 MPa
3
m
3
SOLUTION (6.26) The circumferential, axial, and radial stresses are given by 1
pr
24 p
t
2
pr
2t
3 0
12 p
Insertion of these expression into Eqs. (6.6) and (6.14) provide the critical pressures. ( a ) For the maximum shearing stress theory: 24 p 0
(250 10 ) 6
1 1 .4
p 7 .4 4 M P a
( b ) For the maximum energy of distortion theory: p (24 24 12 12 ) 2
2
1 2
1 1 .2
(250 10 ) 6
p 1 0 .0 2 M P a
Comment: The permissible value of the internal pressure is conservatively limited to 7.44 MPa.
SOLUTION (6.27) Maximum shear stress criterion, substituting the given expressions for n
Sy
1
2
S yt ( 3 . 56 1 . 70 ) a
0 . 19
2
1
and
2
into Eq.(6.6):
S yt
a
2
Maximum distortion energy criterion, using Eq.(6.14) together with the given expressions for 1 and 2 : Sy
n 2
[ 1 1
2
2 2
S yt
1
0 . 215
1 2
2
[ 3 . 56 ( 3 . 56 )( 1 . 70 ) ( 1 . 70 ) ] 2 a
]2
2
S yt
a
2
The factor of safety based on energy of distortion theory is therefore 11.6 % larger than that based on the maximum shear stress theory. This indicates that the maximum energy of distortion criterion is less conservative, as expected.
SOLUTION (6.28) Principle stress criterion, carrying the given expressions for n
Su
1
Su
3 . 56 a
2
t
0 . 281
1
and
2
into Eq.(6.22):
Sut
a
2
Coulomb-Mohr criterion, applying Eq.(6.25) together with the given expressions for
1
and 2 : n
Su
1
2
S u S uc
Sut [ 3 . 56 ( 1 . 7 )( 1 5 )] a
2
0 . 256
Sut
a
2
The n according to the principle stress criterion is thus 8.9 % larger than that on the basis of The Coulomb-Mohr criterion. This indicates that the Coulomb-Mohr theory is more conservative, particularly when S u c S u .
109
SOLUTION (6.29) Table B.1; S y 2 5 0 M P a
x
4P
,
2
D
xy
16T 3
D
Equation (6.11) results in xy
1 2
Sy
[(
1
) x ]2 2
n
2
6
{ ( 2 5 01 .51 0 ) [
1 2
2
3
4 ( 4 5 1 0 )
( 0 .0 5 )
2
1
] } 2 8 2 .5 M P a 2
Thus T
D
3
( 0 .0 5 )
xy
16
3
16
(8 2 .5 1 0 ) 2 .0 2 5 k N m 6
SOLUTION (6.30) Refer to solution of Prob.6.29, Equation (6.16) gives xy
1
[(
3
Sy n
) x] 2
2
1 3
6
{ ( 2 5 01 .51 0 ) [ 2
3
4 ( 4 5 1 0 )
( 0 .0 5 )
2
1
] } 2 9 5 .3 1 M P a 2
and T
D
3
xy
16
(2)
3
16
(9 5 .3 1 1 0 ) 2 .3 3 9 k N m 6
SOLUTION (6.31) We have 1 ,
2 .
Table B.2: S u 1 5 0 M P a ,
S uc 5 7 5 M P a
( a ) Equation (6.22):
Su
n
107 . 1 MPa
150 1 .4
( b ) Equation (6.24) is thus 150
575
1 1 .4
or 84 . 98 MPa
SOLUTION (6.32) We have n=2, Table B.1: S u 340 MPa ,
1, 2
1 2
( 102 0 )
( 10 0 )
2
S uc 620
Equation (6.24): 1 Su
2
S uc
1 n
Thus 5
25 340
2
5
25 620
2
1 2
Solving 111 . 1 MPa
110
2
5
MPa 25
2
SOLUTION (6.33) We have n=2. Table B.1: S y 3 4 5 M P a ( a ) Equation (6.11): S
(
y
n 345 2
or
1
4 xy ) 2
2
2
x
1
8 6 .1 M P a
1
9 9 .4 3 M P a
[( 1 0 ) 4 ] 2 , 2
2
( b ) Equation (6.16): S
(
y
n 345 2
or
1
3 x y ) 2
2
2
x
[ ( 1 0 ) 3 ] 2 , 2
2
SOLUTION (6.34)
We have
y
xy 4 5 M P a
30 M Pa
x
0.
Thus
1, 2
y
y
(
2
) 2
2
1 3 2 .4 M P a
2 xy
30
2
(
30
) (45) 2
2
2
2 6 2 .4 M P a
( a ) Maximum principal stress theory: 1 1;
3 2 .4 5 5
1;
6 2 .4 5 5
Su
But
2
(failure occurs)
Su
( b ) Coulomb-Mohr Theory: 1 2
Su
32
1;
S uc
55
6 2 .4
1
160
gives 0 .5 8 0 .3 9 0 .9 7 1
(no fracture)
SOLUTION (6.35)
x
32M
D
3
xy
1 6T
D
3
M 7 5 0 ( 0 .3 ) 2 2 5 N m
(CONT.)
111
6.35 (CONT.) 350
( S u ) a ll ( S u c ) a ll
Also
140 M Pa
2 .5 630
252 M Pa
2 .5
1, 2
x
(
2
x
) 2
2
2 xy
16
D
3
(M
M
T
2
2
)
Substituting the given data 16
1,2
or
D
(225
3
2
3
D
2
D
4 7 9 3 .7 9
1
1
225 680 )
2
3
(1 1 4 5 .9 2 3 6 4 7 .8 7 )
2 5 0 1 .9 5 D
3
( a ) Maximum Principal Stress Theory 4 7 9 3 .7 9
1 4 0 (1 0 ) 6
3
D
( b ) Coulomb-Mohr Theory: 1 2
( S u ) a ll
or
or
D 0 .0 3 2 5 m = 3 2 .5 m m
4 7 9 3 .7 9
1;
( S u c ) a ll
140 D
3
2 5 0 1 .9 5 252 D
3
10
6
D 0 .0 3 5 3 m = 3 4 .3 m m
SOLUTION (6.36) M (W F ) L ( 9 2 ) 0 .2 5 2 .7 5 k N m
T W a 9 0 .3 2 .7 k N m
An element at point A:
x
32M
t
D 1 6T
D
3
( 0 .0 5 ) 1 6 ( 2 .7 )
3
3 2 ( 2 .7 5 )
( 0 .0 5 )
3
224 M Pa
3
y A T
110 M Pa
z
So m ax
(
x
2 (
224
) t 2
2
Sy
) (1 1 0 ) 2
C
B
W+F
n 2
157 M Pa
y
2
210
x
n
A
Solving, n 1 .3 4
B x
t
t
An element at point B:
z
x
Figure (a) (CONT.)
112
6.36 (CONT.) 4 (W F )
m ax d t
1 6T
D
3A
4 (1 1 1 0 )
3
3
3 ( 0 .0 2 5 )
2
1 1 0 (1 0 ) 1 1 7 .4 7 M P a 6
210 n
or n 1 .7 9
SOLUTION (6.37) Table B.1: S y 250
and 7 . 86 Mg
MPa
m , n 2 .1 3
State of stress, at a point C at bottom surface on midspan:
x
C
32 M
D
3
16T
D
3
We have w 7 . 86 ( 9 . 81 )( D M
Thus
wL
m ax
2
32( w L
x
D
2
D
8)
2 .7 8 1 0 D
6
7 .7 2 8 1 0
6
2
8D
2
3
3
6
2 .7 8 (1 0 ) D
3
D
3
2
) 3(
2
) ( 6 )1 0
2 .0 4 1 0 D
2
3
3
(S ) ( 2
2
y
n) :
2 5 0 1 0 2 .1
6
or D
2
1 4 1.7 1 0
1 2 .4 8 4 8 D
6
8
Solving, by trial and error: D 34 . 34 mm
Use a 35-mm diameter shaft.
SOLUTION (6.38) Applying Eq.(6.34), we have z
From Fig.6.15:
s l
2 s
2 l
kN
2 . 0 4 (1 0 )
x 3
Equation (6.16), (
3 2 ( 6 0 .6 D
3
2
8
3
16( 400 )
4 ) 60 . 6 D
2
400 250 2
30 35
2
3 . 25
R 9 9 .9 4 % .
113
)
2
m
SOLUTION (6.39) The state of stress is
d
d
4P
. Thus
2
d
l
4 ( 200 )
2
l
4 (30 )
2
4
4
d
d
2
2
2 5 4 .6 d 3 8 .2 d
kPa
2
2
kPa
Figure 6.16, for R=99.7 %: z 2 . 75 Equation (6.34), in rearranged form: z (
2 s
2
1
)2 s 1
2 .7 5[ ( 3 5 1 0 ) ( 3 8 2.2 ) ] 2 3 5 0 1 0 3
2
2
3
2 5 4 .6
d
d
2
from which (3 5 1 0 ) ( 3 8 2.2 ) 3
2
2
d
3
(1 2 7 .2 7 3 1 0 3
9 2 .5 8 2 d
2
)
2
6
or d 1 .5 7 4 (1 0 ) d 0 .4 7 5 (1 0 ) 0 Solving, d 0 .0 3 4 2 m and d 3 4 .2 m m 4
2
SOLUTION (6.40) Apply Eqs.(6.30a) and (6.30b): 12
x
1 12
12
xi
912 12
y
76,
i 1
1 12
yi
936 12
78
i 1
12
x [
1 11
1
1
( x i x ) ] 2 [ 111 ( 2 2 0 4 )] 2 1 4 .1 5 5 2
i 1
12
[ 11 1
y
1
1
( y i y ) ] 2 [ 1 1 (1, 3 5 1.3 4 )] 2 1 1.0 8 4 2
1
i 1
SOLUTION (6.41) (a) x
n( xi )
2
n
nx
520
7
3640
7( -44.78 )
2
460
2
920
2(-104.78 )
2
430
5
2150
5(-134.78 )
2
545
5
2725
5( -19.78 )
2
570
10
5700
10(
5.22 )
2
575
18
10350
18( 10.22 )
2
595
8
4760
8( 30.22 )
2
600
3
1800
3( 35.22 )
2
620
6
3720
6( 55.22 )
2
660 -----
4 ----68
2640 4( 95.22 ) ------------ ---------------38405 196521.69
2
(CONT.)
114
6.41 (CONT.) Thus, Eq.(6.30a):
5 6 4 .7 8 M P a
38 ,405 68
Eq.(6.30b): n
[ 617
1
( x i 5 6 4 .7 8 ) ] 2 5 4 .1 5 9 M P a 2
i 1
( b ) Eq.(6.34), z
Figure 6.16:
s l
5 2 5 5 6 4 .7 8 5 4 .1 5 9
m
0 .7 3 5
R 76 %
SOLUTION (6.42) Maximum load: l 2 5 k N ,
l 3 kN
Strength of part: s 3 0 k N ,
s 2 kN
Equation (6.33a): m s l 3 0 2 5 5 k N Equation (6.33b):
m
2 s
2 l
2 3 2
2
3 .6 k N
Thus, failure impends at m
z
Figure 6.16:
m
5 3 .6
1 .3 8 9
R 92 %
SOLUTION (6.43) ( a ) We have
nom
Pn o m
A
4 (35 )
( 0 .0 1 2 5 )
2
35 1 .2 2 7 2 1 0
4
2 8 5 .2 M P a
So n
350 2 8 5 .2
1 .2 3
( b ) Mean stress value equals l n o m 2 8 5 .2 M P a . Estimated standard deviation equals 2 .5 l 2 0 .3 7 2 M P a 1 .2 2 7 2 1 0
4
The margin of safety, Eq. (6.34), is thus z
s l
2 s
2
l
3 5 0 2 8 5 .2 2
2 8 2 0 .3 7 2
2
1 .8 7
(CONT.)
115
6.43 (CONT.) From Fig. 6.16, reliability corresponding to z=1.87 is R 97 %
Hence, failure percentage equals 100-97=3%. Comment: In foregoing calculations, statistical variability of dimensions is omitted.
SOLUTION (6.44) Equation (6.34): Figure 6.16:
z
25 20 2
2 .5 3
2
1 . 28
R 90 %
Thus failure percentage is 10 %. End of Chapter 6
116
CHAPTER 7
FATIGUE FAILURE CRITERIA
SOLUTION (7.1) Use Figure 7.5 for steel (1020). (a)
At infinite life, S e ' 2 3 0 M P a The maximum stress in the beam is
m ax
M
m ax c
I
32 M
D
m ax 3
from which D
3
32 M
(1)
m ax
m ax
Letting m a x S e ' , D
(b)
3
3
3 2 ( 4 1 0 ) 6
( 2 3 0 1 0 )
5 6 .2 m m
5
At 1 0 cycles, S 3 1 0 M P a . Equation (1) gives Then D
3
3
3 2 ( 4 1 0 ) 6
( 3 1 0 1 0 )
5 0 .8 m m
SOLUTION (7.2) Use Figure 7.5 for aluminum alloy (2024). (a)
Endurance strength, S n ' 1 3 5 M P a The maximum stress in the beam:
m ax
32 M
D
or D
m ax 3
3
32 M
(1)
m ax
m ax
With m a x S n ' 1 3 5 M P a , D
(b)
3
3
3 2 (1 .5 1 0 ) 6
(1 3 5 1 0 )
4 8 .4 m m
7
At 1 0 cycles, S 1 6 5 M P a . Equation (1) results in D
3
3
3 2 (1 .5 1 0 ) 6
(1 6 5 1 0 )
4 5 .2 m m
SOLUTION (7.3) To determine the K t , we use Fig.C.1. Structural steel: S u 400
MPa
(Table B.1).
At section C: D d
38 30
1.2 6 7
(a)
r d
4 30
0 .1 3 3
( b ) Figure 7.9a:
max
1 .7
3
15 ( 10 ) 0 . 03 ( 0 . 01 )
85 MPa
(CONT.)
117
7.3 (CONT.) K
t
1. 7
r 4 mm : q 0 . 78
K
1 0 .7 8 (1.7 1 ) 1.5 5
f
(Eq.7.13b)
Similarly, at D: D d
r d
38 34 2 34
K
t
1.1 1 8
(a)
0 .0 5 9
( b ) Figure 7.9a:
1.8
3
15 ( 10 )
1 .8
max
79 . 41 MPa
0 . 034 ( 0 . 01 )
r 2 mm : q 0 . 72
K
1 0 .7 2 (1.8 1 ) 1.5 8
f
SOLUTION (7.4) S u 400
Table B.1:
MPa ,
S e C f C rC sC t
where
K
f
S Se
1.5 8 at D (from Solution of Prob.7.3)
S e 0 .4 5 S u 1 8 0
M Pa
A S u 272( 400 ) b
f
MPa
'
1 K f
'
C
250
y
0 .9 9 5
0 .7
C r 0 .8 7 (Table 7.3) C s 1 (axial loading)
C t 1 0 .0 0 5 8 ( 4 7 5 4 5 0 ) 0 .8 5 5
Thus S e ( 0 .7 )( 0 .8 7 )(1)( 0 .8 5 5 )
1 1 .5 8
(1 8 0 ) 5 9 .3 M P a
SOLUTION (7.5) S e C f C r C s C t (1 K f ) S e '
We have
r d
t
2 .5 (Fig.C.3)
K
2 26
(1)
0 .0 7 7 ,
1.1 5 4
D d
Table B.4: S u 655 MPa , S e 0 .4 5 S u 2 9 4 .8 '
H
B
197
M Pa
Table 7.3: C r 0 .8 9 Fig.7.9a: q 0 . 8 ,
K
f
1 0 .8 ( 2 .5 1) 2 .2
Table 7.2: C f A S u 4 .5 1( 6 5 5 ) b
Use C s 1 (axial loading)
0 .2 6 5
0 .8 0 9
Ct 1
Equation (1) is therefore S e ( 0 .8 0 9 )( 0 .8 9 )(1)(1)( 21.2 )( 2 9 4 .8 ) 9 6 .4 8
118
M Pa
SOLUTION (7.6) S u 626
Table B.4:
MPa
H
B
179
From Eq.(7.1): S e 0 . 5 S u 313 '
MPa
(Note: by Eq.(2.22): S u 3500 (179 ) 626 . 5 MPa. ) C s 0 .8 5
From Eq.(7.9): By Eq.(7.7):
Ct 1
C r 0 .8 9
Using Table 7.3:
C
A S u 4 .5 1( 6 2 6 b
f
0 .2 6 5
) 0 .8 1 9
For Fillet: r d
4 25
0 .1 6
D d
35 25
1.4
K t 1.4 5 Hence, from Fig.C.9: From Fig.7.9a, q=0.82 Equation (7.13b): K f 1 0 .8 2 (1.4 5 1 ) 1.3 7
Thus S e C f C r C s C t ( K1 ) S e ( 0 . 819 )( 0 . 89 )( 0 . 85 )( 1 )( 1 .137 ) 313 141 . 6 MPa '
f
SOLUTION (7.7) Table B.3: S u 4 7 0 M P a . We apply S e C f C r C sC t (
1 K f
'
)Se
where K
f
2 .5 ,
S e 0 .5 S u 2 3 5 '
C s 0 .7 ,
C r 0 .8 4 (Table 7.3),
C
f
AS
b u
M Pa
57 . 7 ( 470 )
0 . 718
Ct 1
0 . 696
Thus S e ( 0 .6 9 6 )( 0 .8 4 )( 0 .7 )(1)( 21.5 )( 2 3 5 ) 3 8 .5 M P a
SOLUTION (7.8) S e C f C r C s C t (1 K f ) S e '
(1)
We have D d
30 25
1 .2
1 d
2 25
0 .0 8
Hence, from Fig. C.12, K t 1 .9 5 Table B.4: S u 6 5 8 M P a ,
H
B
192
Equation (7.1): S e ' 0 .5 S u 3 2 9 M P a Figure 7.9a: r 2 m m ; Therefore, C
A S u 4 .5 1( 6 5 8 b
f
q 0 .8 2 0 .2 6 5
) 0 .8 0 8
(CONT.)
119
7.8 (CONT.) Table 7.3: C r 0 .8 7 K
1 q ( K t 1) 1 0 .8 2 (1 .9 5 1) 1 .7 8
f
Equation (7.9): C s 0 .8 5 Ct 1
(T 4 5 0 C ) o
Equation (1) results in then S e ( 0 .8 0 8 )( 0 .8 7 )( 0 .8 5 )(1)( 1 .71 8 )(3 2 9 ) 1 1 0 .4 M P a
SOLUTION (7.9) Refer to solution of Prob. 7.8, Now, Fig. C.11 give 1 .2 ,
D d
0 .0 8,
r d
Table B.3: S u 7 7 0 M P a ; r 2 mm;
Figure 7.9b:
K t 1 .5 H
B
229
q 0 .9 8
K
f
1 q ( K t 1) 1 0 .9 8 (1 .5 1) 1 .4 9
C
f
A S u 5 7 .7 ( 7 7 0
Also b
C r 0 .8 7 ,
0 .7 1 8
C t 0 .5 6 5,
) 0 .4 8 8
C s 0 .8 5
S e ' 0 .5 S u 3 8 5 M P a
Thus S e C f C r C s C t (1 K f ) S e '
( 0 .4 8 8 )( 0 .8 7 )( 0 .8 5 )( 0 .5 6 5 )( 1 .41 9 )(3 8 5 ) 5 2 .7 M P a
SOLUTION (7.10) Refer to definitions given by Eqs. (7.14). ( a ) m 12 ( m a x m in ) 12 (8 4 8 4 ) 0 a
1 2
(
R
Thus,
m ax
m in
m ax
m in
)
84 84
1 2
(8 4 8 4 ) 8 4 M P a
1
and
A
a m
12 0
(b) m
1 2
(8 4 1 4 ) 3 5 M P a
a
1 2
(8 4 1 4 ) 4 9 M P a
R
14 84
(c) m a R
0 84
0
A
42 42
1
1 6
1 2
,
A
7 5
(8 4 0 ) 4 2 M P a
120
SOLUTION (7.11) M
max
M
M ,
min
m
0,
a
651 , 898 M
32 M
( 0 . 025 )
3
Equation (7.7): C
f
AS
b u
4 . 51 ( 700 )
0 . 265
0 . 794
Also Ct 1
C r 0 .8 7
Table 7.3:
Equation (7.9):
C s 0 .8 5
Equation (7.1):
S e 0 .5 ( 7 0 0 ) 3 5 0 '
M Pa
From Fig. C.9, with D d 1 . 5 , r d 0 . 05 : K t 2 .1 q 0 .7 7
By Fig.7.9a: and K Hence
1 0 . 77 ( 2 . 1 1 ) 1 . 85
f
S e C f C r C sC t (
1 K f
'
)Se
( 0 .7 9 4 )( 0 .8 7 )( 0 .8 5 )(1)( 1 .81 5 )(3 5 0 ) 1 1 1 .1 M P a
Thus, Eq.(7.24): n
Se
6
1 1 1 .1 1 0 6 5 1 ,8 9 8 .6 M
1 .5
;
a
or M 1 1 3 .6 N m
SOLUTION (7.12) Table B.3: S u 4 7 0 M P a
H
B
131
S e 0 .4 5 S u 2 1 1 .5 '
M Pa
Tensile area through the hole: 2 ( R r ) t 2 (10 4 )( 2 . 5 ) 30 mm
m a
and
F 2A
F 2( 30 )
2
F 60
(1)
We have C r 0 .7 0 (Table 7.3)
C
A S u 4 .5 1( 4 7 0 ) b
f
Ct 1 0 .2 6 5
0 .8 8
C s 1 (axial loading)
From Fig.C.5: d D
0 .4 , q 0 .8
By Fig.7.9a: Hence
K
f
K t 2 .8
1 0 . 8 ( 2 . 8 1 ) 2 . 44
Therefore, S e C f C r C s C t (1 K f ) S e '
( 0 .8 8 )( 0 .7 )(1)(1)(1 2 .4 4 )( 2 1 1 .5 ) 5 3 .4
M Pa
(CONT.)
121
7.12 (CONT.) By Eq.(7.20):
m
4 7 0 1 .4
470
(1 )(
3 4 .2 5
(2)
M Pa
) 1
5 3 .4
From Eqs.(1) & (2): 34 . 25
F 60
or F 2 .0 6
kN
SOLUTION (7.13) Refer to solution of Prob. 7.12. We have, m a F 6 0 and C r 0 .8 7
(Table 7.3)
C t 1 0 .0 0 5 8 ( T 4 5 0 )
(Eq. 7.11)
1 0 .0 0 5 8 ( 5 4 5 4 5 0 ) 0 .4 5
Endurance limit becomes S e 5 3 .4 ( 00.8.77 )( 0 .4 5 ) 2 9 .8 7 M P a
The SAE criterion from Eq. (7.20) with S u S
m
S
a m
n
f S
f
Se
1
4 1 5 1 .2 (1)
415 2 9 .8 7
1
f
is
2 3 .2 2 M P a
Thus F 6 0 2 3 .2 2 ,
F 1 .3 9 k N
SOLUTION (7.14) Table B.3: S y 3 9 0 M P a
Refer to solution of Prob. 7.12. We have, m a F 6 0 and C r 0 .8 9
(Table 7.3)
C t 1 0 .0 0 5 8 ( T 4 5 0 )
(Eq. 7.11)
1 0 .0 0 5 8 ( 5 4 0 4 5 0 ) 0 .4 8
Endurance limit is then S e 5 3 .4 ( 00.8.79 )( 0 .4 8 ) 3 2 .5 9 M P a
The Soderberg criterion from Eq. (7.20) with S u S f :
m
Sy n
a
Sy
m
Se
1
3 9 0 2 .2 (1)
390 3 2 .5 9
1
1 3 .6 7 M P a
Thus F 60
1 3 .6 7 ,
F 8 2 0 .2 N
122
SOLUTION (7.15) A 10 ( 25 5 ) 200
Pm
1 2
mm
( 5 25 ) 15 kN ,
Pa 10 kN
( a ) Stress concentration factor is neglected for ductile materials under static loading. Thus
max
Pmax
3
25 ( 10 )
A
S
n
6
200 ( 10
y
m ax
)
580 125
125
MPa
4 .6 4
( b ) We now have 0 .2 ,
d D
S u 690
K t 2 .4 5 (Fig.C.5)
MPa ,
S
y
580
MPa ,
H
197
B
(Table B.3)
q 0 .8 3 (Fig 7.9a)
K
1 0 .8 3 ( 2 .4 5 1 ) 2 .2
f
Ct 1
C r 0 .8 4 (Table 7.3)
C s 1 (axial loading)
A S u 4 .5 1( 6 9 0 ) b
C
f
0 .2 6 5
0 .7 9 8
S e 0 .4 5 S u 3 1 0 .5 M P a '
Hence S e ( 0 .7 9 8 )( 0 .8 4 )(1)(1)( 21.2 )(3 1 0 .5 ) 9 4 .6 1 M P a
We have
m
Pm A
3
15 ( 10 )
200 ( 10
6
)
75 MPa ,
a
50 MPa
Equation (7.22) gives n 75
1 . 57
690 690
( 50 )
94 . 61
SOLUTION (7.16) Refer to Solution of Prob.7.15 (a)
n
S
y
m ax
580 125
4 .6 4
( b ) We now have Pm
1 2
[ 25 ( 5 )] 10 kN ,
Pa 15 kN
Hence
a
75 MPa ,
m
50 MPa
Thus n 50
690 690
1 . 16 ( 75 )
94 . 61
123
SOLUTION (7.17) D
A
4
2
( 0 .0 5 3 1 2 5 )
2
2 .2 1 7 1 0
4
( a ) Su 670 M Pa
H
3
m
1 9 7 (Table B.4)
B
P S u A 6 7 0 ( 2 .2 1 7 1 0
and
2
3
) 1 .4 8 5 M N
( b ) S e C f C r C s C t (1 K f ) S e '
where K
f
1
C r 0 .8 7 (Table 7.3)
C s 1 (axial load)
C
f
AS
4 . 51 ( 670 )
b u
0 . 265
0 . 804
C t 1 0 . 0058 ( 480 450 ) 0 . 826 S e ' 0 .4 5 S u 0 .4 5 ( 6 7 0 ) 3 0 1 .5
M Pa
S e ( 0 .8 0 4 )( 0 .8 7 )(1)( 0 .8 2 6 )(1 / 1)(3 0 1 .5 ) 1 7 4 .2
and
M Pa
Thus P A S e ( 2 .2 1 7 1 0
3
)(1 7 4 .2 ) 3 8 6 .2 k N
SOLUTION (7.18) Refer to Solution of Prob.7.17. We now have A d
2
4 ( 0 .0 5 )
2
4 1 .9 6 3 1 0
3
m
2
( a ) For a static fracture of a ductile material, the groove has little effect. Hence, P S u A ( 6 7 0 1 0 ) (1 .9 6 3 1 0 6
(b)
r d
0 .0 2 5,
D d
1 .0 6 2 5
3
) 1 .3 1 5 M N
K t 2 .6 (Fig.C.10)
From Fig.7.9a, with S u 6 7 0 M P a and r 1 .2 5 m m and
K
f
q 0 .7 5
1 q ( K t 1 ) 1 0 . 75 ( 2 . 6 1 ) 2 . 2
We now have S e 1 7 4 .2 2 .2 7 9 .1 8
M Pa
Thus P A S e (1 .9 6 3 1 0
3
)( 7 9 .1 8 1 0 ) 1 5 5 .4 3 k N 6
SOLUTION (7.19) From Table B.4: S u 519
MPa ,
S
y
353
MPa ,
H
B
149
By Eq.(6.20), S ys 0 . 577 S y 203 . 7 MPa ( a ) Thus, S y s 1 6 T d : 3
T
3
( 0 . 025 ) ( 203 . 7 10 16
6
)
624 . 9 N m
(CONT.)
124
7.19 (CONT.) ( b ) S es C f C r C s C t (1 K f ) S es '
where Ct 1
C r 0 .8 4 (Table 7.3)
C s 0 .8 5 (Eq.7.9)
C
f
AS
1 . 58 ( 519 )
b u
S e 0 .2 9 S u 1 5 0 .5 '
From Fig.C.8, with
From Fig.7.9b: q 0 .9 ,
K
f
0 . 929
D d
(Eq.7.7)
(Eq.7.4)
M Pa
0 . 05 and
r d
0 . 085
2
K t 1 . 72
1 q ( K t 1 ) 1.6 5
Hence S e s ( 0 .9 2 9 )( 0 .8 4 )( 0 .8 5 )(1)( 1 .61 5 )(1 5 0 .5 ) 6 0 .5
M Pa
Refer to Eq.(7.24): S es 16 T d . 3
Thus T
3
6
( 0 .0 2 5 ) ( 6 0 .5 1 0 ) 16
1 8 5 .6
N m
SOLUTION (7.20) Refer to Solution of Prob.7.19. ( a ) A d
2
4 ( 25 )
4 490 . 874
2
mm
2
, and S
y
353
MPa . Thus
P S y A 490 . 874 ( 353 ) 173 . 3 kN
( b ) We now have with r d
0 .0 5 ,
D d
2
From Fig. C.7: K t 2 .5 2 Figure 7.9a: q 0 .7 and K
f
1 q ( K t 1 ) 1 0 .7 ( 2 .5 2 1 ) 2 .0 6
By Eq.(7.3): S e 0 .4 5 S u 2 3 3 .6 M P a Hence S e ( 0 .9 2 9 )( 0 .8 4 )( 0 .8 5 )(1)( 2 .01 6 )( 2 3 3 .6 ) 7 5 .2 2
M Pa
Thus P S e A ( 7 5 .2 2 )( 4 9 0 .8 7 4 ) 3 6 .9 2
kN
SOLUTION (7.21) From Table B.3: S u 830
MPa ,
S
y
460
MPa ,
H
B
248
By Eq.(6.20), S
ys
0 . 577 S
y
265 . 4 MPa
Refer to Solution of Prob.7.19. (CONT.)
125
7.21 (CONT.) 3
d S ys
(a) T
3
6
( 0 .0 2 5 ) ( 2 6 5 .4 1 0 )
16
16
( b ) C f A S u 4 .5 1( 8 3 0 ) b
8 1 4 .2 N m
0 .2 6 5
0 .7 6
S e s 0 .2 9 S u 2 4 0 .7 M P a '
S es C f C r C s C t (1 K
and
'
f
) S es
( 0 .7 6 )( 0 .8 4 )( 0 .8 5 )(1)( 1 .61 5 )( 2 4 0 .7 ) 7 9 .1 6 M P a
Therefore T
3
d S es
3
6
( 0 .0 2 5 ) ( 7 9 .1 6 1 0 )
16
16
2 4 2 .9 k N m
SOLUTION (7.22) Su 658 M Pa
H
1 9 2 (Table B.4)
B
S e C f C r C s C t (1 K f ) S e '
where C r 0 .8 1 (Table 7.3)
C
AS
f
b u
C s 0 .8 5 (Eq.7.9)
1 . 58 ( 658 )
0 . 085
0 . 91
(Table 7.2)
S e 0 .5 S u 3 2 9 M P a '
0 . 125 ,
d D
and
K t 2 . 18
From Fig.7.9a: q 0 . 79 ,
K
f
Fig.C.13) 1 0 . 79 ( 2 . 18 1 ) 1 . 93
Thus S e ( 0 .9 1)( 0 .8 1)( 0 .8 5 )(1)( 1 .91 3 )(3 2 9 ) 1 0 6 .8 M P a
We have M
m
m
1 2
(1 5 8 .2 5 6 .5 ) 1 0 7 .3 5 N m , M
( D
3
m
32 ) ( dD
2
6)
M
1 0 7 .3 5
( 0 .0 2 5 )
3
3 2 [ ( 0 .0 0 3 1 2 5 ) ( 0 .0 2 5 )
2
6]
a
5 0 .8 5 N m
8 8 .8 3 M P a
(Fig.C.13)
a 8 8 .8 3( 15007.8.355 ) 4 2 .0 8 M P a
Equation (7.22): n 8 8 .8 3
658 658
1 .8 9 ( 4 2 .0 8 )
1 0 6 .8
SOLUTION (7.23) Refer to Solution of Prob.7.22. We now have Su 680 M Pa,
H
B
2 0 1 (Table B.3)
q 0 . 8 (Fig.7.9a)
Also
d D
0 .1 2 5
K
f
K t 2 .6 5 (Fig.C.13)
1 0 . 8 ( 2 . 65 1 ) 2 . 32
S e 0 .4 5 S u 0 .4 5 ( 6 8 0 ) 3 0 6 M P a '
(CONT.)
126
7.23 (CONT.) S e ( 0 .9 1)( 0 .8 1)( 0 .8 5 )(1)( 2 .31 2 )(3 0 6 ) 8 2 .6 4 M P a
and
We write Pm
1 2
(60 20) 40 kN ,
Pa 2 0 k N
A D
2
4 Dd
(Fig.C.13)
and
m
9 6 .9 1 M P a ,
40
( 0 .0 2 5 )
2
4 [( 0 .0 2 5 )( 0 .0 0 3 1 2 5 )]
Equation (7.22) is therefore n
1 .3 7
680 680
9 6 .9 1
( 4 8 .4 6 )
8 2 .6 4
SOLUTION (7.24) From Table B.2: S u 5 2 0 M P a ,
H
B
149
C r 0 .7
Table 7.3:
Equation (7.7) and Table 7.2: C
f
4 .5 1(5 2 0
0 .2 6 5
) 0 .8 6
S e ' 0 .5 S u 2 6 0 M P a
Thus
S e C f C r C s C t (1 K f ) S e '
( 0 .8 6 )( 0 .7 )(1)(1)(1 1)( 2 6 0 ) 1 5 6 .5 M P a
We have
32 M
a
Pm
m
d
( 0 .0 2 5 ) 3
4 (1 5 1 0 )
A
3 2 (1 5 0 )
3
( 0 .0 2 5 )
2
3
9 7 .7 8 M P a 3 0 .5 6 M P a
Goodman criterion
a
Se
m
Su
1 n
9 7 .7 8 1 5 6 .5
;
3 0 .5 6 520
1 n
Solving, n 1 .4 6
SOLUTION (7.25) Refer to solution of Prob. 7.22. We have S y 380 M Pa (Table B.4) C r 0 .8 9
(Table 7.3)
C t 1 0 .0 0 5 8 ( T 4 5 0 )
(Eq. 7.11)
1 0 .0 0 5 8 ( 4 5 5 4 5 0 ) 0 .9 7
Endurance limit is therefore S e 1 0 6 .8 ( 00 .8.8 91 )( 0 .9 7 ) 1 1 3 .8 3 M P a
The Soderberg criterion (by Table 7.4)
a
Se
m
Sy
1 n
Inserting the data, we have 4 2 .0 8 1 1 3 .8 3
8 8 .8 3 380
1 n
,
n 1 .6 6
127
a
9 6 .9 1
20 40
4 8 .4 6 M P a
SOLUTION (7.26)
1
pr t
, a
pa
m
pr
2
pm
2 3
,
2t
0.
3
pm 1260 kPa,
pa 840 kPa
.
Replace S u with S y in Eq.(7.30): S
[1
y
n
1e
or
1
1 2
] 2 1 e 0 .8 6 6 1 e
1 4
Sy
(1)
0 .8 6 6 n
Similarly, Eq.(7.25):
1e
S
1m
y
Se
(1260
1 t
1a
280 210
r t
S
( pm
840 )
y
pa )
Se
2380 t
(2)
By Eq.(1) and (2): 280 ( 0 .8 6 6 ) 2 .5
2380 t
t 0 .0 1 8 4 m = 1 8 .4 m m
,
SOLUTION (7.27) 1
pr t
,
2
pr 2t
,
3
0.
p m 1 . 7 MPa ,
p a 1 . 1 MPa
Refer to Solution of Prob.7.26: 1e
Su
(1)
0 .8 6 6 n
Use Eq.(7.25): Su
1e 1m
Se
1a
r t
( pm
Su Se
p a ) 3 0[1 .7
From Eqs.(1) and (2): 128,
350 0 .8 6 6 n
n 3.1 6
SOLUTION (7.28) b 0 .0 1 m ,
M
max
M
a
1 4
M
M PL 4.
( 20 )( 0 . 1) 0 . 5 N m ,
0 . 25 N m ,
m
6M
m
We have,
L 0 .1 m ,
m 2
bh
6 ( 0 .2 5 ) 0 .0 1 h
2
m
150 h
2
Equation (7.20) gives
m
980 4 (1 )
98
7 1 .0 1 M P a
1
40
Thus h
150 71 . 01 10
6
1 . 45 mm
128
M a
min
0
35 15
(1 .1)] 1 2 8 M P a
(2)
SOLUTION (7.29) From Solution of Prob.7.28, we have a m
150 n
.
2
(1)
Equation (7.20) by replacing S u with S y :
m
S
a
m
n
y
S
y
1
Se
Substituting the given data gives
m
620 4
62
(1 )(
6 0 .7 8 M P a
(2)
) 1
40
By Eqs.(1) and (2), 6 0 .7 8 ( 1 0 ) 6
150 h
2
or h 1 . 57 mm
SOLUTION (7.30) At fixed end M
M M
m
P L 2 .2 5 ( 0 .0 3 7 5 ) 0 .0 8 4 4 N m Hence
m ax
M
m
a
bh
2
Equation (7.20):
m ax
M
m in
2 6 ( 0 .0 4 2 2 )
6M
M
a
( 0 .0 0 3 1 2 5 ) h
m
2
1 0 5 0 1 .2 (1 )
1050
0 .0 4 2 2 N m
81 h
2
2 8 3 .7 8 M P a
1
504
Thus 2 8 3 .7 8 1 0
81 h
2
6
Pa,
h 0 .5 3 4 1 0
3
m = 0 .5 3 4 m m
SOLUTION (7.31) Refer to solution of Prob. 7.30. Equation (7.11): C t 1 0 .0 0 5 8 ( T 4 5 0 ) 1 0 .0 0 5 8 ( 4 7 0 4 5 0 ) 0 .8 8
and S e 5 0 4 ( 0 .8 8 ) 4 4 3 .5 2 M P a
Equation (7.20), replacing S u by S f , becomes
m
S
a
m
n
f S
f
Se
1
Inserting numerical values:
m
6 9 0 1 .5 (1)( 4 4639.50 2 ) 1
1 7 9 .9 9 M P a
Therefore, 81 h
2
1 7 9 .9 9 1 0 , 6
h 0 .6 7 1 1 0
129
3
m = 0 .6 7 1 m m
SOLUTION (7.32) P0
Pm Pa
Eq.(7.20):
,
2
m
M 850 4
85
(1 )
0 . 2 Pm 0 . 1 P0 ,
m
Pa Pm 1
a
3 6 .2 8 M P a
1
1 7 .5 6M
m
Also
bh
6 ( 0 .1 ) P0
m 2
0 . 0 0 5 1 0 0 (1 0
1.2 (1 0 ) P0 6
6
)
Thus 1 . 2 P0 36 . 28 ,
P0 30 . 23 N
SOLUTION (7.33) M
M
0 .2 P0 ,
m ax
m
P0
M
m in
0 .2 ( 0 .5 P0 ) 0 .1 P0
( 0 .2 0 .1 ) 0 .0 5 P0 ,
2
Equation (7.20): m
850 2 1 .5 ( 3 )
85
M
3 5 .6 2 9
a
P0 2
( 0 .2 0 .1 ) 0 .1 5 P0
M Pa
1
35 6M
m
Also
bh
m 2
6 ( 0 . 0 5 P0 )
0 . 0 0 5 1 0 0 (1 0
0 .6 (1 0 ) P0 6
6
)
Thus 0 . 6 P0 35 . 629 ,
P0 59 . 38 N
SOLUTION (7.34) I
bh 12
3
24 ( 4 )
3
128
12
mm
4
. Table B.3: S u 690
MPa
For a wide cantilever beam (see Secs 4.4 and 4.10, and Case 1 of Table A.8): 3
(1 ) 2
This gives Pmin
3 EI 0 . 91 L
0 .9 1
PL 3EI
3
min
3
PL 3EI
9
3 ( 200 10 )( 128 10 0 . 91 ( 0 . 3 )
3
12
)
( 0 . 01 )
and hence Pmax 62 . 52 N
31 . 26 N
Thus Pm 46 . 89 N ,
Pa 15 . 63 N and
46 . 89 ( 0 . 3 )( 2 10
3
m
a
. 63 219 . 8 ( 15 ) 73 . 27 MPa 46 . 89
0 . 128 ( 10
9
)
)
219 . 8 MPa
Eq. (7.22): Su
n
m
Su Se
a
219 . 8
690 690
1 . 63 ( 73 . 27 )
250
130
m
1
SOLUTION (7.35) Table B.4: S
500
y
MPa
Refer to Solution of Prob.7.34. Replacing S u by S y in Eq.(7.22): S
n
m
y
S
y
Se
219 . 8
a
500 500
1 . 36 ( 73 . 27 )
250
SOLUTION (7.36) Refer to solution of Prob. 7.34. We now have (Table 7.3) C r 0 .7 5 C t 1 0 .0 0 5 8 ( T 4 5 0 )
(Eq. 7.11)
1 0 .0 0 5 8 ( 4 9 0 4 5 0 ) 0 .7 7
Endurance limit becomes S e 2 5 0 ( 0 .7 5 )( 0 .7 7 ) 1 4 4 .4 M P a
The Gerber criterion (Table 7.4):
(
a
Se
) 1 2
m
Su
Substitution of the numerical values gives ( 2 1699.80 n ) 1 2
7 3 .2 7 n 1 4 4 .4
or 0 .5 0 7 4 n 0 .3 1 8 6 n 1 0 2
Solving this quadratic equation: 0 .5 0 7 4
n
2
( 0 .5 0 7 4 ) 4 ( 0 .3 1 8 6 )( 1 ) 2 ( 0 .3 1 8 6 )
or n 1 .1 5
SOLUTION (7.37) I
4
( 0 . 03 )
4
0 . 636 10
m
a
23 . 6 MPa
M
m
c
I
6
m
2000 ( 0 . 5 )( 0 . 03 )
0 . 636 10
6
4
Pm 4 kN ,
,
Pa 2 kN
47 . 2 MPa
Eq. (7.16), with replacing S u by S y : S
y
n
m
S
y
Se
a;
280 2 .5
4 7 .2
280 150 K
f
( 2 3 .6 )
(CONT.)
131
7.37 (CONT.) Solving K
f
1.4 7 .
From Fig.7.9a: q=0.8 and
K
f
1 q ( K t 1 );
1.4 7 1 0 .8 ( K t 1 ),
K t 1.5 9
Then from Fig. C.9: K
t
r d
1.5 9 9 60
0 .1 5
D
3
d
and D 3 d 3 ( 60 ) 180
mm
SOLUTION (7.38) Table B.4: S u 634 I
bh 12
3
MPa ,
0 . 02 ( 0 . 06 )
3
1 . 8 kN m
D d
120 60
r d
4 60
192
B
0 . 36 10
12
m
M
H
M
6
m
4
1 . 2 kN m
a
Figure C.2: 2
0 .0 6 7
K t 2 .1
Figure 7.9a: q 0 . 82 and K
f
1 0 .8 2 ( 2 .1 1) 1 .9 0 2 .
Se
400 1 .9 0 2
2 1 0 .3 M P a
We have
m
a
150 ( 11 .. 28 ) 100
M
m
c
I
1800 ( 0 . 03 ) 0 . 36 10
6
150
MPa
MPa
Equation (7.22): Su
n
m
Su Se
a
150
634 634
1 .4 ( 100 )
210 . 3
SOLUTION (7.39) Refer to solution of Prob. 7.38. We now have M
m
1 2
(2200 200) 1200 N m
M
a
1 2
(2200 200) 1000 N m
Thus, m 1 5 0 ( 11 28 00 00 ) 1 0 0 M P a ,
a 1 0 0 ( 11 02 00 00 ) 8 3 .3 M P a
(CONT.)
132
7.39 (CONT.) The Gerber criterion (by Table 7.4).
(
a
Se
m
Su
) 2
1 n
Inserting the numerical values, results in 8 3 .3 n 2 1 0 .3
( 160304n ) 1,
0 .3 9 6 1 n 0 .0 2 4 9 n 1 0
2
2
Solving this quadratic equation: 0 .0 2 4 9
n
2
( 0 .0 2 4 9 ) 4 ( 0 .3 9 6 1 )( 1 ) 2 ( 0 .3 9 6 1 )
or n 1 .5 6
SOLUTION (7.40) Refer to solution of Prob. 7.38. We now have S
434 M Pa
f
(from Table B.4)
C t 1 0 .0 0 5 8 ( 4 7 5 4 5 0 ) 0 .8 6
(by Eq. 7.11))
1 0 .0 0 5 8 ( 4 9 0 4 5 0 ) 0 .7 7
C r 0 .8 7
(Table 7.3)
and S e 2 1 0 .3( 0 .8 7 )( 0 .8 6 ) 1 5 7 .3 M P a
Also M
m
1 .5 k N m
M
a
0 .5 k N m
and m 1 5 0 ( 11 58 00 00 ) 1 2 5 M P a ,
a 1 0 0 ( 1520000 ) 4 1 .6 7 M P a
The SAE criterion from Table 7.4:
a
Se
m
S
f
1 n
Introducing the data, we obtain 4 1 .6 7 1 5 7 .3
125 434
1 n
Solving, n 1 .8 1
SOLUTION (7.41) See Tables 7.5, 6.2, and 6.1. We have A 5 .6 1 0
12
n 3 .2 5
K c 77 M Pa
m
S y 690 M Pa
1 .1 4 for a w 0 .4 (Case B, Table 6.1)
(CONT.)
133
7.41 (CONT.) Note that values of a and t satisfy Table 6.2. The stresses are:
m ax
Pm a x
2wt
3
9 5 0 (1 0 ) 2 ( 0 .0 8 )( 0 .0 3 4 )
1 7 4 .6
m in
7 9 .4 M P a
The range stress is then 1 7 4 .6 7 9 .4 9 5 .2 M P a The final crack length at fracture, by Eq. (7.39): af
1
Kc
(
) 2
m ax
( 1 .1 4 7 17 7 4 .6 ) 0 .0 4 7 6 m 4 7 .6 m m 2
1
Equation (7.41) results in then N
0 .0 4 7 6 5 .6 (1 0
12
0 .6 2 5
0 .0 3 2
0 .6 2 5
) ( 0 .6 2 5 ) [ (1 .7 7 ) (1 .1 4 ) ( 9 5 .2 ) ]
3 .2 5
2 0 , 4 5 4 cycles
For a period of 20 second, approximate fatigue life L is L
20 ,454 ( 20 ) 60 (60 )
1 1 3 .6 h
End of Chapter 7
134
CHAPTER 8
SURFACE FAILURE
SOLUTION (8.1) From Table 8.2, we have K 10
2
(steel on steel) 5
K 2 10 K 10
(lead on steel)
3
(brass on steel)
7
K 10 (polyurethylene on steel) The Brinell hardness numbers are in MPa: Steel
Su 58 10 H
s
3
58 10
(Table B.1)
psi 3
500 116 Bhn
(Eq. 2.22)
1 1 6 ( 9 .8 1) 1 1 3 8 M P a
Lead H l 3(9 .8 1) 2 9 .4 M P a
Brass H b 8 (9 .8 1) 7 8 .5 M P a
Polyurethylene H
p
7 (9 .8 1) 6 8 .7 M P a
Thus referring to Eq. (a) of Sec. 8.5: Lead
H K 2 9 .4 (1 0 ) 2 (1 0
Brass
H K 7 8 .5 (1 0 ) 1 0
6
6
3
5
) 1470 G Pa
7 8 .5 G P a
Polyurenthylene H K 6 8 .7 (1 0 ) 1 0 6
7
687, 000 G Pa
2
Steel H K 1 1 3 8 (1 0 ) 1 0 1 1 3 .8 G P a Comment: Polyurethylene gives much longer life than any other material. 6
SOLUTION (8.2) Through the use of Eq. (8.2), wear volume is V K
WL H
K 10
4
where W 40 N
(by Table 8.3),
H b 9 .8 1( 6 0 ) 5 8 9 M P a , L 80 (2
mm c y c le
) (1 5 0 0
c y c le s s ix m o n th s
H
s
) 240, 000 m m
Therefore Vb
(1 1 0
Vs
(1 1 0
4
)( 4 0 )( 2 4 0 ) 6
5 8 9 (1 0 ) 4
)( 4 0 )( 2 4 0 ) 6
1 0 3 0 (1 0 )
9 .8 1(1 0 5 ) 1 0 3 0 M P a
1 .6 3 m m
3
0 .9 3 m m
3
135
SOLUTION (8.3) From Eq. (8.2), we have V K
WL H
Here K 3 10
H
s
H
f
5
W 45 N
(by Table 8.3),
9 .8 1(1 6 0 ) 1 5 6 9 .6 M P a 9 .8 1( 4 5 0 ) 4 4 1 4 .5 M P a
L (3 7 .5 2
mm c yc le
)( 6 0 0 0
c yc le s m o n th
)(1 2 m o n th s ) 5 , 4 0 0 , 0 0 0 m m
Hence, V s (3 1 0
V f ( 4 .6 )
5
)
( 4 5 )( 5 , 4 0 0 ) 6
1 5 6 9 .6 (1 0 )
1 5 6 9 .6 4 4 1 4 .5
4 .6 m m
1 .6 4 m m
3
3
SOLUTION (8.4) Follow procedure of Example 8.1. We have Copper disk: 110 Vickers hardness, V c 0 .9 8 m m Aluminum pin: 95 Brinell hardness, V a 4 .1 m m
3
3
Contact force: W 2 5 N at a radius R 2 4 m m Test duration: t 1 8 0 m in at a speed n 1 2 0 r p m Total length of sliding is then L 2 R n t
2 ( 2 4 )(1 2 0 )(1 8 0 ) 3 .2 6 1 0
6
mm
The hardness of pin and disk are H a 9 .8 1(9 5 ) 9 3 2 M P a H c 9 .8 1(1 1 0 ) 1 0 7 9 M P a
From Eq. (8.2); K V H W L . Thus 4 .1 ( 9 3 2 )
Ka
2 5 ( 3 .2 6 1 0 )
Kc
2 5 ( 3 .2 6 1 0 )
6
0 .9 8 (1 0 7 9 ) 6
4 .9 6 1 0 1 .3 1 0
5
5
SOLUTION (8.5) Follow procedure of Example 8.1. Given data: Steel disk: 248 Brinell hardness, V s 0 .9 8 m m Copper pin: 85 Vickers hardness, V c 4 .1 m m
3
3
Contact force: P 3 5 N at a radius R 2 4 m m Test duration: t 1 8 0 m in at a speed n 1 2 0 r p m Total length of sliding equals L 2 R n t
2 ( 2 4 )(1 2 0 )(1 8 0 ) 3 .2 6 1 0
6
mm
(CONT.)
136
8.5 (CONT.) Hardness of pin and disks are H c 9 .8 1(8 5 ) 8 3 4 M P a ,
H s 9 .8 1( 2 4 8 ) 2 4 3 3 M P a
Using Eq. (8.2); K V H W L . Therefore 4 .1 ( 8 3 4 )
Kc
3 5 ( 3 .2 6 1 0 )
Ks
3 5 ( 3 .2 6 1 0 )
3 10
6
0 .9 8 ( 2 4 3 3 )
5
2 .1 1 0
6
5
SOLUTION (8.6) Case B ( 1st column ), Table 8.4 with r1 r2 , (a)
1 E
1 E
a 0 . 88
3
( b ) p 0 1 .5
0 . 88
r E
F
1 .5
P
a
m
2 E
2
1 r
210 ( 10
500 2
( 0 . 624 ) ( 10
2 r
0 . 624
0 . 15
500
3
1 r
6
)
E1 E 2 .
9
)
mm
613 . 1 MPa
( c ) Equations (8.4) at z=0:
x
z
p 0 613 . 1 MPa
yz
p 0 [( 1 )
y
xz
1 2
(
1 2
]
z)
x
p0 2
1 2 2
p 0 0 . 8 ( 613 . 1 ) 490 . 5 MPa
( 0 . 8 1 ) 61 . 31 MPa
SOLUTION (8.7) Refer to Table 8.4 (Case C, Column 2). n
( a ) a 0 .8 8 3 F where
2 E 1(1 0 n
1 r1
1 r2
11
1 0 .0 0 6
F 2 .2 k N
)
1 0 .0 0 6 0 5
)
1
1 .3 7 7 4
Thus
and
a 0 .8 8[
2 2 0 0 (1 0
p o 1 .5
F
11
1 .3 7 7 4
a
2
1 .5
]
3
2 .2 1 6 2 m m 3
2 .2 (1 0 ) 2
( 2 .2 1 6 2 ) (1 0
Since 2 1 4 M P a 4 3 6 M P a
6
214 M Pa
)
OK
( b ) 0 .7 7 5 F n 3
2
2
0 .7 7 5[( 2 2 0 0 ) (1 0 2
11
2
1
) (1 .3 7 7 4 )]
3
0 .0 0 6 7 7 m m
( c ) From Fig. 8.9a, m a x is at about z 0 .4 a and m a x p o 0 .3 2 . Therefore z 0 .4 ( 2 .2 1 6 2 ) 0 .8 8 6 m m
m a x 0 .3 2 ( 2 1 4 ) 6 8 .4 8 M P a
137
SOLUTION (8.8) See Table 8.4 (Case B, Column 3). We have 11
2 E 1(1 0
( a ) a 1 .0 7 6
m
),
1 0 .0 2 5
1 0 .0 7 5
5 3 .3 3 3 3
Lm
F
2 2 0 (1 0
11
)
1 .0 7 6[ ( 0 .0 2 5 )( 5 3 .3 3 3 3 ) ]
1
4 .3 7 0 7 (1 0
2
5
) m 0 .0 4 3 7 m m
Then po
2
F aL
220
2
5
( 4 .3 7 0 7 1 0
)( 0 .0 2 5 )
1 2 8 .2 M P a 3 2 0 M P a
( b ) From Fig. 8.9b,
OK
is at z 0 .7 5 a and
yz ,m ax
p o 0 .3 .
yz ,m ax
Hence, z 0 .7 5 ( 0 .0 4 3 7 ) 0 .0 3 2 8 m m
yz ,m ax
0 .3 (1 2 8 .2 ) 3 8 .4 6 M P a
SOLUTION (8.9) Refer 2nd column of C, Table 8.4. E 1 E 2 E 210
GPa
1
r1 7 mm
r 2 45 mm
F 1 200
kN
m
Hence
( a ) a 1 . 076 [ ( b ) p0
2
F aL
2 E
2 9
2 1 0 (1 0 )
105 ( 10
)( 120 . 635 )
3
0 . 135 ( 10
n
,
] 2 0 . 135
200 ( 10 )
2
9
1 0 5 (1 0 )
1 0 .0 0 7
1 0 .0 4 5
1 2 0 .6 3 5
1
3
200 ( 10 ) 9
3
mm
943 . 1 MPa
)
SOLUTION (8.10) We have r1 ' . Substitution of the numerical values into Eqs.(8.6) through (8.10)gives m A
4 2 0 .2 5
2 0 .0 3 4 9
n
0 .0 3 4 9
1 0 .0 0 9 3 7 5
5 7 .3 0 6 6 ,
B
1 2
9
4 ( 2 0 0 1 0 ) 3 ( 0 .9 1 )
2 9 3 .0 4 (1 0 ) 9
1
[( 0 .0 019 3 7 5 0 ) ( 0 ) ( 0 ) ] 2 5 3 .3 3 3 3 2
2
2
From Eq.(8.9) cos
5 3 .3 3 3 3 5 7 .3 0 6 6
2 1 .4 6
0 .9 3 0 7 ,
o
Then, interpolating in Table 8.5, we have c a 3 .6 2 5 1 and c b 0 .4 2 0 4 . Through the use of Eq.(8.7): 3
a 3 .6 2 5 1[
2 .2 5 1 0 ( 0 .0 3 4 9 )
b 0 .4 2 0 4[
2 .2 5 1 0 ( 0 .0 3 4 9 )
9
2 9 3 .0 4 (1 0 ) 3
9
2 9 3 .0 4 (1 0 )
1
] 3 2 .3 3 7 1 m m 1
] 3 0 .2 7 1 0 m m
(CONT.)
138
8.10 (CONT.) The maximum contact pressure, by Eq.(8.6), is thus 2 .2 5 1 0
p 0 1 .5
3
1 6 9 6 .2 M P a
( 2 .3 3 7 1 )( 0 .2 7 1 0 )
Thus stress may be satisfactory for the steel used. SOLUTION (8.11) In this case: r2 r2 ' as well as r1 ' . Substituting w=L=6.25 mm. and given data into the equations on the second column of Case A of Table 8.4: a 1 .0 7 6 p0
Thus
2
F aw
3
2 .2 5 1 0 0 .0 0 6 2 5
r1 1 .0 7 6
F w
3
2 ( 2 .2 5 1 0 )
(1 .9 7 6 7 1 0
4
)( 0 .0 0 6 2 5 )
( 0 .0 0 9 3 7 5 )(
2 2 0 0 1 0
9
) 1 .9 7 6 7 1 0
4
m
1 1 5 9 .4 2 3 M P a
Comments: With the flat-faced follower inaccuracies in machining, misalignment, and shaft deflection may cause edge contact and hence higher actual stress than 1159.423. This cannot occur with a spherical face follower, however.
SOLUTION (8.12) Refer to Table 8.4 (Case A, Column 2). r2 and 2 E a 0 .8 8 3 F r1
Thus
0 .8 8[3 6 0 ( 0 .0 5 6 2 5 )(
(a)
p o 1 .5 1 .5
1
2 2 0 0 1 0
)]
9
3
5 .1 6 7 6 1 0
4
m = 0 .5 1 6 7 6 m m
F
a
2
360
( 5 .1 6 7 6 1 0
4
)
2
6 4 3 .7 M P a
2 2 r1
( b ) 0 .7 7 5 3 F
0 .7 7 5[(3 6 0 ) ( 2
4 .7 5 1 1 1 0
6
1
2
2 2 0 0 1 0
9
) ( 0 .0 516 2 5 )]
m = 4 .7 5 1 1 0
3
3
mm
SOLUTION (8.13) See Table 8.4 (Case C, Column 3). 2 E and n 1 r1 1 r2 1 0 .0 1 5 1 0 .0 1 6 2 5 5 .1 2 8 2 ( a ) a 1 .0 7 6
F Ln
9 .0 1 5 1 0 po
2F aL
1 .0 7 6[ 4
3
1 3 .5 1 0 ( 2 )
1
9
( 2 0 0 1 0 )( 0 .0 3 7 5 )( 5 .1 2 8 2 )
]
2
m = 0 .9 0 1 5 m m 3
2 (1 3 .5 1 0 )
( 0 .9 0 1 5 )( 0 .0 3 7 5 )
2 5 4 .2 M P a
2 5 4 .2 M P a 3 1 7 M P a
OK. (CONT.)
139
8.13 (CONT.) ( b ) Equations (8.5) at z=0: x 2 p o 2 ( 0 .3)( 2 5 4 .2 ) 1 5 2 .5 M P a
y
p o 2 5 4 .2 M P a
z
p o 2 5 4 .2 M P a
xy
1 2
( 1 5 2 .5 2 5 4 .2 ) 5 0 .9 M P a
xz
1 2
(
x
z ) 5 0 .9 M P a
1 2
(
y
z) 0
yz
( c ) From Fig. 8.9b,
yz ,m ax
is at z 0 .7 5 a and
p o 0 .3 .
yz ,m ax
So, z 0 .7 5 (9 .0 1 5 5 1 0
4
) 6 .7 6 2 1 0
4
m = 0 .6 7 2 m m
0 .3 ( 2 5 4 .2 ) M P a 7 6 .2 6 M P a
yz ,m ax
SOLUTION (8.14) Refer to Table 8.4 (Case C, Column 2). 2 E 2 (200 10 ) 1 10 9
n
( a ) a 0 .8 8 3 F
1 r1 n
F
a
2
1 0 .0 5
1 0 .0 5 5
0 .8 8[5 6 0
1 .2 8 0 3 6 1 0
( b ) p o 1 .5
1 r2
3
1 .5
11
1 .8 1 8 2 11
1 1 0 1 .8 1 8 2
1
]
3
m = 1 .2 8 0 3 6 m m 560
(1 .2 8 0 3 6 1 0
3
)
2
1 6 3 .1 M P a
( c ) 0 .7 7 5 F n 0 .7 7 5[(5 6 0 ) (1 1 0 3
2 .9 8 3 1 0
2
2
2
6
m = 2 .9 8 3 1 0
3
11
1
2
mm 3
( d ) At z 0 .3 7 5 a 0 .3 7 5 (1 .2 8 0 3 6 1 0 ) 0 .4 8 0 1 1 0
yz ,m ax
3
) (1 .8 1 8 2 )]
3
m = 0 .4 8 0 1 m m (Fig. 8.9a):
0 .3 2 p o 0 .3 2 (1 6 3 .1) 5 2 .1 9 2 M P a
SOLUTION (8.15) See Table 8.4 (Column 2). We have 2 E 2 ( 2 0 0 1 0 ) 1 1 0 9
11
( a ) Case A: r2 . Thus a 0 .8 8 3 F r1 0 .8 8[(5 6 0 )( 0 .0 5 )(1 1 0
5 .7 5 7 1 1 0 p o 1 .5
F
a
2
4
1 .5
11
1
)]
3
m 560
( 5 .7 5 7 1 1 0
4
)
2
8 0 6 .7 1 8 M P a
(CONT.)
140
8.15 (CONT.) ( b ) Case B: r1 r2 1 0 0 m m , m
a 0 .8 8 3 F p o 1 .5
0 .8 8[(5 6 0 )
1 .5
F
a
2
m 2 r 2 0 .1 2 0
560
( 5 .7 5 7 1 0
4
)
11
1 1 0 20
2
1
]
3
5 .7 5 7 1 0
4
m = 0 .5 7 5 7 m m
8 0 6 .7 M P a
SOLUTION (8.16) Use Table 8.4 (Case A, Column 3). 2 E 2 (200 10 ) 1 10 9
a 1 .0 7 6
11
3
r 1 .0 7 6[ 1 .80.11 0 ( 0 .0 1 2 5 )(1 1 0
F L
5 .1 0 3 9 1 0
5
11
1
)]
2
m = 0 .0 5 1 0 4 m m 4
2 a 1 .0 2 0 7 8 1 0
m = 0 .1 0 2 1 m m
Therefore
2 r1
( 13 ln
0 .5 7 9 F EL
3
0 .5 7 9 (1 .8 1 0 ) 9
( 2 0 0 1 0 )( 0 .1 )
a
) 2 ( 0 .0 1 2 5 )
[ 13 ln
3 .4 0 1 4 1 0
7
5
( 5 .1 0 3 9 1 0
)
]
m = 3 .4 0 1 4 1 0
-4
mm
SOLUTION (8.17) We have r1 r2 r . Thus, Eqs. (8.8) and (8.9) become m
2 r 2 ( 0 .2 2 ) 0 .4 4
4 (1 r ) (1 r )
(1 r ) (1 r )
cos
(1 r ) (1 r )
90
0,
o
From Table 8.5 it can be concluded that surface of contact has a circular boundary: c a c b 1 . Then n
9
4 ( 2 0 6 1 0 ) 2
3 (1 0 .2 5 )
2 .9 2 9 7 8 (1 0 ) 11
Eqs. (8.7) gives, a b 1[
3
2 (1 0 ) ( 0 .4 4 ) 2 .9 2 9 7 8 1 0
11
1 3
]
1 .4 4 3 m m
Hence, Eq. (8.6): p o 1 .5
6
2 (1 0 )
(1 .4 4 3 )(1 0
3
)
6 6 1 .8 M P a
SOLUTION (8.18) We have r1 0 .5 m , r2 0 .3 5 m , r1 ' , r2 ' , and 90 . Thus, using Eqs.(8.6) through (8.10): o
m A
4 1 0 .5 2 m
,
1 0 . 35
n
0 . 824
B
1 2
( r1 1
1 r2
9
4 ( 206 10 ) 3 ( 1 0 . 09 )
3 . 0183 (10
11
)
)
(CONT.)
141
8.18 (CONT.) cos
B A
1 0 .5 1 0 .3 5 1 0 .5 1 0 .3 5
7 9 .8 6
0 .1 7 6 ,
Interpolating from Table 8.5: c a 1.1 2 8 , 3
5 ( 10 )( 0 . 824 )
a 1 . 128 [
3 . 0183 ( 10
p 0 1 .5
)
3
5 ( 10 )( 0 . 824 )
b 0 . 893 [
Thus
11
3 . 0183 ( 10
11
c b 0 .8 9 3 . Hence
1
] 3 2 . 6958
mm
1
] 3 2 . 1342
mm
3
5 ( 10 )
1 .5
F
ab
)
o
( 2 . 6958 2 . 1342 10
6
414 . 9 MPa
)
SOLUTION (8.19) We now have F
300
600 2
S
r2 5 .2 m m ,
and r2 ' 3 0 m m .
r1 r1 ' 5 m m ,
and
1500
N for each row,
y
MPa
We proceed as in Example 8.3. (a) m
4
2 0 . 005
1 0 . 0052
1 0 . 030
0 . 0229 ,
n 293 . 0403 10
9
and A
87 . 3362
2 0 . 0229
B
1 2
[( 0 )
(
2
1 0 . 0052
1
1 0 . 03
)
2
2 ( 0 ) ] 2 79 . 4872 2
Using Eq.(8.9), cos
Table 8.5: c a 3.3 3 3 0 Hence
b 0 . 4441 [
o
c b 0 .4 4 4 1 1
300 0 . 0229
a 3 . 3330 [
24 . 48
0 . 9101
79 . 4872 87 . 3362
293 . 0403 10
9
mm
1
300 0 . 0229 293 . 0403 10
] 3 0 . 9539
9
] 3 0 . 1271
mm
1 ,181
MPa
Then p 0 1 .5
(b) n
300
( 0 . 9539 0 . 1271 10
6
)
Fy
(1)
F
It is required that S y 1.5 F y a b S
y
1 .5
or
Fy
3
cacb (m n ) Substituting the data given
2 3
1 .5
1500 ( 10 ) 6
3
( 3 . 3330 0 . 4441 )(
Fy
Solving F y 614
N
Equation (1): n
614 300
2
0 . 0229 293 . 0403 10
2 .0 5
142
9
)
3
SOLUTION (8.20) Refer to Table 8.4 (Case A, Column 3). a 1 .0 7 6
r1
F L
where F 5 kN ,
2 E
L 25 m m ,
2 0 6 1 0
r1 5 0 0 m m
0 .9 7 1(1 0
2 9
11
)
Therefore 3
( 5 1 0 )
a 1 .0 7 6[
( 0 .5 ) ( 0 .9 7 1 1 0
0 .0 2 5
11
1
]
1 .0 6 0 3 m m
2
Hence, po
2
3
2 ( 5 1 0 )
F aL
(1 .0 6 0 3 1 0
3
120 M Pa
)( 0 .0 2 5 )
SOLUTION (8.21) See Table 8.4 (Case C, Column 3).
2
F aL
F Ln
( a ) a 1 .0 7 6 ( b ) po
2 E
1
1 .0 7 6[
1 0 .0 1
3 0 0 1 0
3
9
1 0 0 (1 0 )( 8 4 )
1 0 .0 6 2 5
1
0 .2 0 3 3 m m
]
2
84,
0 .2 0 3 3
( c ) From Fig. 8.9b, z 0 .7 5 a 0 .1 5 2 5 m m
F L 300 kN
9 3 9 .4 M P a
300
2
n
,
9
1 0 0 (1 0 )
yz ,m ax
and
0 .3 p o 0 .3 (9 3 9 .4 ) 2 8 2 M P a
SOLUTION (8.22) Given: r1 r1 ' 0 .0 2 m , Equation (8.8), m
4 1 0 .0 2
1 0 .0 2
1 0 .0 2 5
1 0 .1
r2 0 .0 2 5 m , 0 .0 8 ,
n
0 .3,
r2 ' 0 .1 m , 9
4 ( 2 0 0 1 0 ) 2
3 (1 0 .3 )
E 200 G Pa
2 .9 3 (1 0 ) 11
Also A
2 0 .0 8
cos
1 15 25
2 m
2 5,
B
5 3 .1 3 0 1
1 2
[( 0 ) ( 2
1 0 .0 2 5
1 0 .1
2
o
From Table 8.5, we find c a 1 .6 6 4 5
c b 0 .6 6 4 2
The semiaxes are then a 1 .6 6 4 5[
(1 2 0 0 ) ( 0 .0 8 )
b 0 .6 6 4 2[
(1 2 0 0 ) ( 0 .0 8 )
2 .9 3 (1 0
11
2 .9 3 (1 0
11
)
)
1
]
3
0 .0 0 1 1 4 7 m 1 .1 5 m m
3
0 .0 0 0 4 6 m 0 .4 6 m m
1
]
Thus p o 1 .5
F
ab
1 .5
1200
(1 .1 5 0 .4 6 )(1 0
6
)
1083 M Pa
143
1
) 2 ( 0 )] 2 1 5
SOLUTION (8.23) We have r1 r1 ' 0 .0 1 8 m ,
r2 0 .0 2 m ,
r2 ' 0 .1 m ,
0 .3,
E 200 G Pa
Equations (8.8), m
4 1 0 .0 1 8
1 0 .0 1 8
1 0 .0 2
1 0 .1
0 .0 7 8 3,
n
9
4 ( 2 0 0 1 0 ) 2
3 (1 0 .3 )
2 .9 3 (1 0 ) 11
Equations(8.10) A
2 m
B
2 5 .5 4 2 8 1 2
cos
[( 1
1 0 .0 2
20 2 5 .5 4 2 8
1 0 .1
2
3 8 .4 6
From Table 8.5, c a 2 .2 1 6 4 Equations (8.7): a 2 .2 1 6 4[
b 0 .5 5 5 6[
1
) ]2 20
c b 0 .5 5 5 6
( 9 0 0 ) ( 0 .0 7 8 4 ) 2 .9 3 (1 0
11
)
( 9 0 0 ) ( 0 .0 7 8 3 ) 2 .9 3 (1 0
11
o
)
1
]
3
0 .0 0 1 3 7 9 m 1 .3 8 m m
3
0 .0 0 0 3 5 m 0 .3 5 m m
1
]
It follows that p o 1 .5
F
ab
1 .5
900
(1 .3 8 0 .3 5 )(1 0
6
)
890 M Pa
End of Chapter 8
144
Section III
APPLICATIONS
CHAPTER 9
SHAFTS AND ASSOCIATED PARTS
SOLUTION (9.1) T
9549 kW n
9 5 4 9 (1 0 )
3 8 .1 9 6 N m
2500
and
Also
c 3
2
a ll L
T
a ll
T GJ
3 8 .1 9 6 6
1 5 0 1 0
3
c 7 .8 6 4
,
2 ( 1 8 0 )
;
3 8 .1 9 6 ( 2 ) 9
8 0 (1 0 ) c
4
mm c 9 .6 6 m m
,
A 20-mm dia. shaft would probably be selected.
SOLUTION (9.2) Refer to Example 9.1. We now have S u 3 6 5 M P a (T a b le B .2 ), n 2 .5 , and S y is replaced by S u . Equation (9.9) gives D [ 1 6Sn ( M u
c
M
2 c
Tc ] 2
1 3
Substituting the given data, D [
1 6 ( 2 .5 ) 6
( 3 6 5 1 0 )
( 2 0 2 .5
( 2 0 2 .5 ) (1 6 2 ) ] 2
2
1 3
0 .0 2 5 2 5 6 m = 2 5 .2 6 m m
SOLUTION (9.3) T AC
(a)
9 5 4 9 ( 7 .5 )
9549 kW n
1200
5 9 .6 8 N m
and 2
c 3
T
a ll
5 9 .6 8 6
2 1 0 (1 0 ) 2
c 7 .1 2 6
,
D A C 1 4 .2 5
mm
Similarly TC B 2
c
9 5 4 9 ( 2 2 .5 ) 1200
3
179
179
c 1 0 .2 8
,
6
2 1 0 (1 0 ) 2
N m
( b ) We have T L G J , with J c Thus AC
5 9 .6 8 ( 2 )
4
mm
D C B 2 0 .5 6
2.
0 .3 5 9 6 ra d 2 0 .6
4
9
( 0 .0 0 7 1 2 5 ) ( 8 2 1 0 )
2
BC
179 ( 4 )
4
0 .4 9 8 ra d 2 8 .5 3 9
( 0 .0 1 0 2 8 ) ( 8 2 1 0 )
2
Hence A B B C A C 7 .9 3
o
145
o
o
mm
mm
SOLUTION (9.4) G a 28 GPa ,
Table B.1:
S u 520
Table B.3:
MPa ,
We have a s but L a L s , Js Ja
Ga
0 . 354
Gs
G s 79 GPa S
y
440
MPa
Ta Ts
where J
D
32
4
Hence Ds Da
Wa
Also
Ws
0 .7 7 1
or
Va a
( D a 4 ) a
V s
s
D a 1. 2 9 6 D s
4
2
(Ds
4 )
s
2
Da a 2
Ds
s
a 2 .8 M g m , 3
Table B.1:
s 7 .8 6 M g m
3
Thus Wa Ws
2
(1. 2 9 6 D s ) ( 2 . 8 )
2 Ds
0 .5 9 8
( 7 .8 6 )
SOLUTION (9.5) ( a ) Use Eq.(9.5): 6
2 6 0 (1 0 ) n
3
4 (1 0 )
( 0 .1 )
3
1
[( 8 5 5 0 0 .1 ) ( 8 8 ) ] 2 2
2
n 2 .6 1
or
( b ) Apply Eq.(9.6): 6
2 6 0 (1 0 ) n
3
4 (1 0 )
( 0 .1 )
3
1
[ ( 8 5 5 0 0 .1 ) 4 8 ( 8 ) ] 2 2
2
or n 2 .8 6
SOLUTION (9.6)
F y 0 : 0 and
M
A
0 yield
960
Mz (N m)
480 x A
My (N m) A
R By 8 0 0 N
R Az 5 0 0 N
R Bz 1 0 0 0 N
M
960 300
2
C
1006 N m
M
600 480
2
D
7 6 8 .4 k N m
B
C 300
R Ay 1 6 0 0 N
600
2
T 1 .2 5 k N m
x D
2
B
Critical section is at C.
(CONT.)
146
9.6 (CONT.) We have 1 6T
(a)
n
32M
m ax
D
m ax
Sy
( b ) m ax n
D
(
1 6 (1, 2 5 0 )
( 0 .0 4 5 )
( 0 .0 4 5 )
3
3
6 9 .9 M P a
1 1 2 .5 M P a
3 .0 7
345 1 1 2 .5
3 2 (1, 0 0 6 )
3
3
) 2
2
(5 6 .2 5 ) ( 6 9 .9 ) 2
8 9 .7 M P a
2
2
m ax
S ys
2 .3 4
210 8 9 .7
SOLUTION (9.7) y
0 . 6 kN m
0 . 6 kN m
x
A
B
C z
6 kN
3.75 kN
Critical point is just to the right of C.
2.25 kN
1 . 125
kN m
M x 0 . 6 kN m
T
x S
(a)
y
n
32
D
3
2
M
T
2
6
2 5 0 (1 0 )
;
1 .5
3
3 2 (1 0 )
D
3
[1.1 2 5
2
1
0 .6 ] 2 2
or D 42 . 71 mm 250 ( 10
(b)
1 .5
6
)
3
32 ( 10 )
D
3
[1 . 125
2
3 4
2
1
( 0 .6 ) ] 2
Solving D 4 2 .3 m m SOLUTION (9.8) We have S y 3 4 0 M P a and n 1 .5 . The reactions and the torques, as found by the equations of statics, are marked in Fig. S9.8a. Moment and torque diagrams are shown in Fig. S9.89b. Critical sections is at C. (CONT.)
147
9.8 (CONT.) y A
2m
90 N
C
7.86N m
2m z
360 N
D
7.86N m.
450 N
360 N
1m
720 N
B
x
360 N
(a)
720
Mz (N m) A
C
My (N m)
x
D
Figure S9.8
360
180
B
x
T (N m) -7.86
C
x
B
(b) Eq. (9.7) is thus D
32n 3
Sy
M
M
2 z
2 y
T
2
Inserting the numerical values: [
or
3 2 (1 .5 )
(3 4 0 1 0 ) 6
(
7 2 0 1 8 0 ( 7 .8 6 ) ) ] 2
2
2
1 3
D 0 .0 3 2 2 m = 3 2 .2 m m
SOLUTION (9.9) Refer to solution of Prob. 9.8. Now apply Eq. (9.8): D
[
32n 3
Sy
M
3 2 (1 .2 )
(3 4 0 1 0 ) 6
M
2 z
(
2 y
3
T
2
4
720 180 2
2
3 4
or D 0 .0 2 9 9 m = 2 9 .9 m m
148
( 7 .8 6 ) ) ] 2
1 3
SOLUTION (9.10) Table B.1: S
520
y
xy
We have M 0 ,
MPa
16T
D
3
By Eq.(1.15), T
9549 kW n
9549 (70 ) 110
6 .0 7 7 k N m
( a ) Equation (6.11) Sy n
2
xy
,
D
3
32Tn Sy
32 (6077 ) 4
6
( 5 2 0 1 0 )
D 7 8 .1 m m
,
( b ) Equation (6.16): Sy n
3
xy
,
D
3
16
3T n
Sy
16
3 (6077 ) 4 6
( 5 2 0 1 0 )
D 7 4 .4 m m
or
SOLUTION (9.11) We have S u 5 9 0 M P a and n 1 .4 Ay B y 8 kN
From statics:
8 kN
y
8 kN
0.5 m
A z
1m
0.5 m
C 0.48 kN m
Ay
D 0.48 kN m
B
x
(a)
By
8 0.5=4 kN m
Mz
x T
0.48 kN m
x
(b)
(c)
Figure S9.11 The critical section is between C and D. Thus M
2 z
T
2
4 0 .4 8 2
2
4 .0 2 9 k N m
Apply Eq. (9.9): 1 6 1 0 (1 .4 ) 3
D
3
(5 9 0 1 0 ) 6
( 4 4 .0 2 9 ) 0 .0 4 6 m
Use a 4 6 m m diameter shaft. SOLUTION (9.12) Now Fig. S9.11a (see: Solution of Prob. 9.11) becomes as shown below. From statics: (CONT.)
149
9.12 (CONT.) Ay 2 kN
B y 6 kN
Az 6 k N
Bz 2 kN
y A
1m
0.5 m
D 0.48 kN m
C 0.48 kN m
Az Ay
z
8 kN
8 kN 0.5 m
B
x
(a)
x
(b)
x
(c)
x
(d)
By
Bz
Mz 2 0.5=1 kN m
2 1.5=3 kN in.
3 kN m My
1 kN m
T
0.48 kN m
The critical section is just to the right of C (or just to the left of D). Thus M
2 z
M
2 y
T
2
3 1 0 .4 8 2
2
2
3 .1 9 9 k N m
Applying Eq. (9.9): 1 6 1 0 (1 .5 ) 3
D
3
(5 9 0 1 0 ) 6
( 3 3 .1 9 9 ) 0 .0 4 3 1 m
Use a 4 4 m m diameter shaft. SOLUTION (9.13) y A Az
0 . 6 kN m
0.3 m
C 5.2 kN
3 kN
Ay
z
x
0 . 6 kN m
0.5 m
Bz
30o
975
M
B y 1 . 95 kN ,
B
B z 1 . 125
A y 3 . 25 kN
kN , A z 1 . 875
kN
By
N m
z
x M
563
Critical point is just to the right of C.
N m
M
y
x T
600
1
[ 9 7 5 5 6 3 ] 2 1 .1 2 6 k N m 2
C
2
N m
x
150
(CONT.)
9.13 (CONT.) Thus
M
0
m
M
1 . 126
a
kN m
T m 600
N m
From Sec. 8.6: C t 1 0 .0 0 5 8 ( 5 0 0 4 5 0 ) 0 .7 1 S e 0 . 5 S u 260 '
MPa
C
f
AS
b u
4 . 51 ( 520
0 . 265
) 0 . 86
C r 0 .8 9
Assume D 51 mm and C s 0 .7 0 . Thus
S e C f C r C s C t ( K1 ) S e ( 0 . 86 )( 0 . 89 )( 0 . 70 )( 0 . 71 )( 11. 2 )( 260 ) 82 . 42 MPa '
f
Substitute the data and K s b K s t 1.5 (Table 9.1) into Eq.(9.13): 6
520 (10 ) 1 .5
[ 1.5 ( 8 2 . 4 2 1 ,1 2 6 )
32
D
520
3
2
1
1.5 ( 6 0 0 ) ] 2 2
Solving, D 63 . 5 mm SOLUTION (9.14) Statics: R A 9 .4 5 k N ,
R B 6 .7 5
kN ,
M
D
2 .5 3 k N m M
m ax
Refer to Secs. 7.6 and 7.7: C s 0 .7 K
C r 0 .8 1 (Table 7.3)
C
AS
f
b u
0 . 085
1 . 58 (1260
1, S e 0 . 5 S u 630 '
f
) 0 . 861
S e ( 0 .8 6 1)( 0 .8 1)( 0 .7 )(1)( 6 3 0 ) 3 0 7 .5 6 M P a
and
We have M
m
0
M
a
2 .5 3 k N m
T m 2 .2 6 k N m
T a 0 .2 2 6 k N m
Using Eq.(9.12): 6
1 2 6 0 (1 0 ) n
3
3 2 (1 0 )
( 6 2 .5 1 0
3
[( 310276.506 2 .5 3 ) 2
)
3
3 4
( 2 .2 6
1260 3 0 7 .5 6
1
0 .2 2 6 ) ] 2 2
n 2 .8 2
or
SOLUTION (9.15) Statics: R A 5 .2 2 k N , M
m
0,
R B 8 .2 8 k N ,
T m 3 .4
kN m ,
From Secs. 7.6 and 7.7: C r 0 .8 9
M
2 .6 1 k N m M
C
a
. We have
T a 0 .6 8 k N m
C s 0 .7
K
f
1
C t 1 0 . 0058 ( 510 450 ) 0 . 652 S e 0 .5 S u 7 0 0 '
M Pa
C
A S u 5 7 .7 (1 4 0 0 b
f
0 .7 1 8
S e ( 0 .3 1 8 )( 0 .8 9 )( 0 .7 )( 0 .6 5 2 )(1)( 7 0 0 ) 9 0 .4 M P a
and
Apply Eq.(9.13), with replacing S u by S y and K s t 2 (Table 9.1): 910 ( 10 n
Solving,
6
)
3
32 ( 10 )
( 87 . 5 10
3
)
3
[(
910 90 . 4
2 . 61 )
2
n 1 .9 9
151
2 (3 .4
1
910 90 . 4
0 . 68 ) ] 2 2
) 0 .3 1 8
SOLUTION (9.16) From Fig. C.13, for d D 0 .2 : K t 1 .5 (torsion)
K t 2 .2 (bending)
Figures (7.9) : q 0 .9 5 (torsion) q 0 .8 (bending) The endurance limit is From Eq. 7.13b we have K
fb
1 q ( K t 1) 1 0 .8 ( 2 .2 1) 1 .9 6
K
ft
1 0 .9 5 (1 .5 1) 1 .4 8
Endurance limit: S e C f C r C s C t (1 K t ) S e '
where C
4 .5 1(5 2 0
f
0 .2 6 5
) 0 .8 6
C r 0 .8 1
(Table 7.3)
C s 0 .8 5
(Eq. 7.9)
C t 1 0 .0 0 5 8 ( 4 9 0 4 5 0 ) 0 .7 7 S e ' 0 .5 S u 2 6 0 M P a
(Eq. 7.11)
(Eq. 7.1)
Thus, S e ( 0 .8 6 )( 0 .8 1)( 0 .8 5 )( 0 .7 7 )(1 1 .9 6 )( 2 6 0 ) 6 0 .5 M P a
Also
M
a
120 N m ,
M
0,
m
Tm 6 0 0 N m ,
Ta 0
Equation (9.16a): D
3
32 n
Su
S
[( S u M a ) 2
3 4
e
2
1 2
Tm ]
Substituting the given data, we have 3
( 4 0 ) (1 0
9
)
[ ( 6502.50 1 2 0 ) 2
32 n 6
( 5 2 0 1 0 )
2
1
2
1
3 4
(6 0 0 ) ]
3 4
(6 0 0 ) ]
2
Solving, n 2 .8 3
SOLUTION (9.17) Refer to solution of Prob.9.16. Apply Eq. (9.16b): D
3
32 n
Sy
[(
Sy Se
M a) 2
3 4
2
Tm ]
1 2
Inserting the given numerical values, 3
( 4 0 ) (1 0
9
)
[ ( 6404.50 1 2 0 ) 2
32 n 6
( 4 4 0 1 0 )
From which n 2 .7 2
152
2
SOLUTION (9.18) M S
200
a
N m,
290
y
T m 500
N m,
S u 455
MPa ,
M
0,
m
Ta 0,
K
st
2
MPa
Refer to Secs. 7.6 and 7.7: C
4 . 51 ( 455
f
0 . 265
Assume C s 0 . 7 ,
) 0 . 89 ,
C r 0 . 87 ,
S e ( 0 . 89 )( 0 . 87 )( 0 . 7 )( 1)(
S e 0 . 5 S u 227 . 5 MPa '
1 2 .2
)( 227 . 5 ) 56 . 05 MPa
Through the use of Eq.(9.14), we have 6
4 5 5 (1 0 ) 1 .5
32
D
[1( 5 6 . 0 5 2 0 0 ) 2
455
3
D 38 . 8 mm 51 mm
or
3 4
1
2
( 2 )(5 0 0 ) ] 2
Assumption is incorrect.
Assume C s 0 .8 5 : Se
0 . 85 0 .7
( 56 . 05 ) 68 . 06 MPa
Equation (9.14): 6
4 5 5 (1 0 ) 1.5
32
D
[1( 6 8 . 0 6 2 0 0 ) 2
455
3
D 36 . 7 mm
or
3 4
1
2
( 2 )(5 0 0 ) ] 2
Assumption is correct.
SOLUTION (9.19) Refer to Secs. 7.6 and 7.7: q 0 .9 2 K
1 q ( K t 1 ) 1 0 .9 2 (1.8 1 ) 1.7 4
f
S e 0 . 5 S u 500 '
C
Thus
AS
f
b u
MPa ,
57 . 7 (1000
0 . 718
C s 0 . 85 ,
C r 0 . 84
) 0 . 405
S e ( 0 .4 0 5 )( 0 .8 4 )( 0 .8 5 )(1)( 1 .71 4 )(5 0 0 ) 8 3 .0 9 M P a
We have: M
0,
m
M
a
500
N m,
T m 600
N m,
T a 90 N m
Apply Eq.(9.11), with replacing S u by S y : 6
600 ( 10 ) n
32
( 0 . 05 )
3
[(
600 83 . 09
500 ) ( 600 2
600 83 . 09
1
90 ) ] 2 2
or n 1 . 93
SOLUTION (9.20) The endurance limit is expressed as S e C f C r C s C t (1 K f ) S e '
Here C
f
4 .5 1( 6 9 0
0 .2 6 5
) 0 .7 9 8
(Eq. 7.7 and Table 7.2)
C r 0 .8 7
(Table 7.3),
C t 1 (Eq. 7.11)
C s 0 .7 0
(Assuming D > 51 mm, Eq. 7.9) (CONT.)
153
9.20 (CONT.) S e ' 0 .5 S u 0 .5 ( 6 9 0 ) 3 4 5 M P a
(Eq. 7.1)
Therefore S e ( 0 .7 9 8 )( 0 .8 7 )( 0 .7 )(1)(1 1 .2 )(3 4 5 ) 1 3 9 .7 M P a
Equation (9.13) with T a 0 becomes
3
D
32 n
Sy
[K s (M
m
Su Se
M a ) K sT m ] 2
2
1 2
Introducing the given numerical values, we have 3 2 ( 3 .5 )
3
D
6
( 6 9 0 1 0 )
6
( 6 9 0 1 0 )
{1 .5[ 2 0 0
6
(1 3 9 .7 1 0 )
( 6 0 0 )] 1 .5 (3 6 0 ) } 2
2
1 2
Solving, D 0 .0 5 8 6 m 5 8 .6 m m
Since D 5 1 m m , our assumption is correct. SOLUTION (9.21) We now have K s 2 (from Table 9.1) and S u replaced by S y . See solution of Prob. 9.20. Equation (9.14) becomes
3
D
32 n
Sy
[K s (M
m
Sy Se
M a ) K s ( 34 T m )] 2
2
1 2
or 3 2 ( 3 .5 )
3
D
6
( 5 8 0 1 0 )
{ 2[ 2 0 0
580 1 3 9 .7
(6 0 0 )] 2
3 2
1
2
(3 6 0 ) }
2
Solution is D 0 .0 6 1 7 4 m 6 1 .7 m m
Since D 5 1 m m , our assumption is correct. SOLUTION (9.22) 2 kN m
M
0 . 8 kN m
y
Critical section is just to the left of point D
x C
A M
D
B
z
600
T
2 .5 7 k N m
M
D
[ 2 . 57
x
M
a
2 . 692
T m 600
N m
1
2
0 . 8 ] 2 2 . 692 2
kN m ,
N m,
M
0
m
T a 30 N m
x Refer to Secs. 7.6 and 7.7: S e 0 . 5 S u 330 '
S e C f C rC sC t (
MPa , 1 K f
C
f
AS
b u
4 . 51 ( 660
0 . 265
) 0 . 807
) S e ( 0 . 807 )( 1 )( 0 . 7 )( 1 )( 1 )( 330 ) 186 . 4 MPa '
Equation (9.14): 6
6 6 0 (1 0 ) n
32
( 0 .0 7 5 )
[1.5 ( 1 8 6 . 4 2 , 6 9 2 ) 2
660
3
Solving n 2 .3 4
154
3 4
(1.5 ) ( 6 0 0
660 1 8 6 .4
kN m
1
30 ) ]2 2
SOLUTION (9.23) From Solution of Prob. 9.22 M
0,
m
M
2 . 692
a
kN m ,
T a 30 N m ,
T m 600
N m
S e 1 8 6 .4 M P a
Now we have C r 0 .8 9 . Thus, S e 1 8 6 .4 ( 0 .8 9 ) 1 6 5 .9 M P a Through the use of Eq.(9.13), with replacing S u by S y : 6
550 ( 10 ) n
550 [1 . 5 ( 165 2 , 692 ) 1 . 5 ( 600 .9 2
32
( 0 . 075 )
3
550 165 . 9
1
30 ) ] 2 2
n 2 . 08
or
SOLUTION (9.24) R B y 1 .0 3 8 k N ,
0.45 kN
y
R A y 6 .1 6 2 k N R B z 3 .7 3 8 k N ,
1.35 kN
z x
A C
R Ay M
D
R A z 0 .1 3 8 k N
B
M
( kN m)
B 6.3 kN R B z
2.7 kN
9 kN x
D
C
R Az
R By
z
x
A
y
(kN m)
0.208
x 0.021
0.748
0.924
We have T 0 .9 k N m and 1
M
D
[ 0 .2 0 8 0 .7 4 8 ] 2 0 .7 7 6 k N m
M
C
[ 0 .9 2 4 0 .0 2 1 ] 2 0 .9 2 4 k N m
M
m
0,
2
2
2
2
1
Hence M
a
0 .9 2 4 k N m ,
Ta 0 ,
T m 0 .9 k N m
From Table B.3: S u 5 9 0 M P a . Refer to Secs. 7.6 and 7.7: C r 0 .8 7 ,
C
and
f
AS
C s 0 .7 (assumed D>50 mm)
4 . 51 ( 590 )
b u
0 . 265
0 . 83
S e ( 0 .8 3)( 0 .8 7 )( 0 .7 )(1)( 11.8 )( 0 .5 5 9 0 ) 8 2 .8 4 M P a
Applying Eq.(9.12): 5 9 0 1 0 1 .6
6
[ ( 8529.804 9 2 4 ) 2
32
D
3
3 4
2
1
(9 0 0 ) ] 2
D 0 .0 5 6 8 m = 5 6 .8 m m
155
SOLUTION (9.25) R B y 2 .0 8 k N ,
R A y 6 .9 2 k N
9 kN
y
x C
6.92 kN
M
x
( kN m )
We have T 0 .9 k N m and M
C
M
x 0.021
D
1 2
[ (1 .0 3 8 ) ( 0 .0 2 1) ] 1 .0 3 8 k N m M 2
3.74 kN
y
(kN m )
0.416
B
D 0.748
2.08 kN
1.038
C
C
0.14 kN
z
M
x
A
B
D
R A z 0 .1 4 k N
6.3 kN
2.7 kN
z
A
M
R B z 3 .7 4 k N ,
2
Refer to Solution of Prob. 9.24: C f 0 .8 3 ,
C s 0 .7
a
From Table B-3: S y 4 9 0 M P a
From Sec. 7.6: S e C f C r C s C t ( K1 ) S e ( 0 .8 3 )( 0 .7 5 )( 0 .7 )(1)( 11.2 )( 0 .5 5 9 0 ) 1 0 7 .1 2 M P a '
f
Using Eq.(9.11), with S u replaced by S y : D
or
3
32 n
Sy
[(
Sy Se
1
32 ( 2 )
M a ) Tm ] 2 2
2
D 0 .0 5 8 6 m = 5 8 .5 m m
SOLUTION (9.26) Refer to Case 6 of Table A.8:
We have I
Pbx 6 LEI
4
(L b x ) 2
(12 . 5 )
2
4
2
19 . 175 10
3
mm
4
W D W E 15 9 . 81 147 . 2 N
Deflection at D: D '
D ''
2
2
2
1 4 7 .2 (1 )( 0 .4 )(1 .4 1 0 .4 ) 6 (1 4 0 0 )( 2 1 0 1 9 .1 7 5 ) 2
1 .3 9 3 m m
2
2
1 4 7 .2 ( 0 .4 )( 0 .4 )(1 .4 0 .4 0 .4 ) 6 (1 4 0 0 )( 2 1 0 1 9 .1 7 5 )
1 .1 4 2 m m
and D E 1 .3 9 3 1 .1 4 2 2 .5 3 5 m m
Equation (9.18) results in n cr
1 2
594
2 gW 2W
2
1 2
g
1 2
9 . 81 2 . 535 10
rpm
156
1
[( 1 0479.10 2 1, 0 3 8 ) 9 0 0 ] 2 2
6
( 4 9 0 1 0 )
3
9 . 901 cps
2
SOLUTION (9.27) Refer to Case 6 of Table A.8:
Pbx 6 LEI
(L b x ) 2
2
2
We have W D W E 15 9 . 81 147 . 2 N At midspan C: C
0 . 5 ( 10
3
)
2
147 . 2 ( 0 . 4 )( 0 . 7 )
(1 . 4
6 ( 1 . 4 ) EI
2
0 .4
2
0 .7 ) 2
or EI 25 . 711 (10 ) N m 3
2
Deflection at D: D ' D ''
2
2
2
1 4 7 .2 (1 )( 0 .4 )(1 .4 1 0 .4 )
0 .2 1 8
3
6 (1 .4 )( 2 5 .7 1 1 1 0 ) 2
2
2
1 4 7 .2 ( 0 .4 )( 0 .4 )(1 .4 0 .4 0 .4 ) 3
6 (1 .4 )( 2 5 .7 1 1 1 0 )
mm
0 .1 7 9
mm
and
D
E
0 . 218 0 . 179 0 . 397
mm
Equation (9.18) gives n cr
g
1 2
9 .8 1
1 2
0 .3 9 7 1 0
3
2 5 .0 2
cps
SOLUTION (9.28) We have I
4
(15 )
39 . 76 10
4
3
mm
4
Refer to Case10 of Table A.8: a
P C
b
A B L 2
C
Pb L 3 EI
C
150 ( 0 . 4 )( 1000 )
Thus 2
3 ( 210 39 . 76 )
0 . 958
mm
Equation (9.18) results in n cr
1 2
1 2
0 .9 5 8 1 0
g
9 .8 1 3
1 6 .1 1 c p s 9 6 7
157
rp m
6 . 428 EI
SOLUTION (9.29) See Table A.8 (Case 6) and Example 9.5. We have I
4
d
( 0 .0 5 6 2 5 )
64
4
0 .4 9 1 4 (1 0
64
6
) m
4
( a ) Deflection at C owing to 450 N: C '
4 5 0 ( 2 .2 5 )( 0 .6 2 5 )
( 2 .8 7 5 0 .6 2 5 2 .2 5 ) 1 .0 5 m m 2
9
6 ( 2 .8 7 5 )( 2 0 0 1 0 )( 0 .4 9 1 4 1 0
6
)
2
2
Deflection at C due to 270 N: C "
2 7 0 (1 )( 0 .6 2 5 )
( 2 .8 7 5 0 .6 2 5 1 ) 0 .6 8 4 m m 2
9
6 ( 2 .8 7 5 )( 2 0 0 1 0 )( 0 .4 9 1 4 1 0
6
)
2
2
Total deflection at C: C 1 .0 5 0 .6 8 4 1 .7 3 4 m m Deflection at D owing to 450 N: 4 5 0 ( 0 .6 2 5 )( 2 .8 7 5 1 .8 7 5 )
D '
9
6 ( 2 .8 7 5 )( 2 0 0 1 0 )( 0 .4 9 1 4 1 0
[ 2 ( 2 .8 7 5 )(1 .8 7 5 ) 0 .6 2 5 1 .8 7 5 ] 1 .1 4 1 m m 2
6
)
2
Deflection at D due to 270 N: D "
2 7 0 (1 )(1 .8 7 5 )
( 2 .8 7 5 1 .8 7 5 1 ) 1 .1 2 m m 2
9
6 ( 2 .8 7 5 )( 2 0 0 1 0 )( 0 .4 9 1 4 1 0
6
)
2
2
Total deflection at D: D 1 .1 4 1 1 .1 2 2 .2 6 1 m m Applying Eq. (9.18), we obtain n cr
1 2
[
9 .8 1[( 4 5 0 )(1 .7 3 4 1 0 4 5 0 (1 .7 3 4 1 0
3
3
) ( 2 7 0 )( 2 .2 6 1 1 0
2
) 2 7 0 ( 2 .2 6 1 1 0
3
)
3
)]
2
1
]
2
1 1 .2 4 c p s 6 7 4 .4 r p m
( b ) Apply Eq. (9.21): n c r ,C n c r , D
n cr
where
Hence
2
2
n c r ,C n c r , D
g
n cr ,C
1 2
C '
n cr , D
1 2
D "
n cr
g
( 9 2 3 )( 8 9 4 ) 2
1 2
(923) (894 )
2
1 2
9 .8 1 1 .0 5 1 0
3
9 .8 1 1 .1 2 1 0
3
1 5 .3 8 c p s 9 2 3 rp m 1 4 .9 c p s 8 9 4 rp m
6 4 2 rp m
SOLUTION (9.30) Refer to Table A.8 (Case 6) and Solution of Prob. 9.29. We now have I D
4
6 4 ( 0 .0 8 7 5 )
4
6 4 2 .8 7 7 4 1 0
6
m
4
( a ) Deflection at C owing to 450 N: C '
4 5 0 ( 2 .2 5 )( 0 .6 2 5 ) 9
( 2 .8 7 5 0 .6 2 5 2 .2 5 ) = 0 .1 7 9 3 m m 2
6 ( 2 .8 7 5 )( 2 0 0 1 0 )( 2 .8 7 7 4 4 1 0
6
)
2
2
Deflection at C due to 270 N: C "
2 7 0 (1 )( 0 .6 2 5 ) 9
6 ( 2 .8 7 5 )( 2 0 0 1 0 )( 2 .8 7 7 4 1 0
( 2 .8 7 5 0 .6 2 5 1 ) 0 .1 1 6 9 m m 2
6
)
2
2
Total deflection at C: C 0 .1 7 9 3 0 .1 1 6 9 0 .2 9 6 2 m m (CONT.)
158
9.30 (CONT.) Deflection at D owing to 450 N: D '
4 5 0 ( 0 .6 2 5 )( 2 .8 7 5 1 .8 7 5 )
[ 2 ( 2 .8 7 5 )(1 .8 7 5 ) 0 .6 2 5 1 .8 7 5 ] 0 .1 9 4 8 m m 2
9
6 ( 2 .8 7 5 )( 2 0 0 1 0 )( 2 .8 7 7 4 4 1 0
6
)
2
Deflection at D due to 270 N: D "
2 7 0 (1 )(1 .8 7 5 )
( 2 .8 7 5 1 .8 7 5 1 ) 0 .1 9 1 2 m m 2
9
6 ( 2 .8 7 5 )( 2 0 0 1 0 )( 2 .8 7 7 4 4 1 0
6
)
2
2
Total deflection at D: D 0 .1 9 4 8 0 .1 9 1 2 0 .3 8 6 m m Applying Eq. (9.18), we find n cr
1 2
9 .8 1[( 4 5 0 )( 0 .2 9 6 2 1 0
[
4 5 0 ( 0 .2 9 6 2 1 0
3
3
) ( 2 7 0 )( 0 .3 8 6 1 0 3
2
) 2 7 0 ( 0 .3 8 6 1 0
)
3
)]
2
1
]
2
2 7 .2 1 c p s 1 6 3 3 r p m
( b ) Apply Eq. (9.21): n c r ,C n c r , D
n cr
where
Hence
2
2
n c r ,C n c r , D
g
n c r ,C
1 2
C '
n cr , D
1 2
D "
g
2
( 2234 ) ( 2163)
0 .1 7 9 3 1 0
2
3
9 .8 1
1 2
( 2 2 3 4 )( 2 1 6 3 )
n cr
3 7 .2 3 c p s 2 2 3 4 rp m
9 .8 1
1 2
0 .1 9 1 2 1 0
3
3 6 .0 5 c p s 2 1 6 3 rp m
1 5 5 4 rp m
SOLUTION (9.31) Refer to Table A.8 (Case 6) and Example 9.5. We have I
4
d
64
( 0 .0 4 6 9 )
4
64
0 .2 3 7 5 1 0
6
m
4
( a ) Deflection at C owing to 340 N: C '
3 4 0 (1 .1 2 5 )( 2 5 0 .6 2 5 )
(1 .7 5 1 .1 2 5 0 .6 2 5 ) 1 .2 8 3 9 m m 2
9
6 (1 .7 5 )(1 0 5 1 0 )( 0 .2 3 7 5 1 0
6
)
2
2
Deflection at C due to 500 N: C "
5 0 0 ( 0 .3 7 5 )( 0 .6 2 5 )
(1 .7 5 0 .3 7 5 0 .6 2 5 ) 1 .1 3 2 9 m m 2
9
6 (1 .7 5 )(1 0 5 1 0 )( 0 .2 3 7 5 1 0
6
)
2
2
Total deflection at C: C 1 .2 8 3 9 1 .1 3 2 9 2 .4 1 6 8 m m Deflection at D owing to 340 N: D '
3 4 0 ( 0 .6 2 5 )(1 .7 5 1 .3 7 5 ) 9
6 (1 .7 5 )(1 0 5 1 0 )( 0 .2 3 7 5 1 0
[ 2 (1 .7 5 )(1 .3 7 5 ) 0 .6 2 5 1 .3 7 5 ] 0 .7 7 0 3 m m 2
6
)
2
Deflection at D due to 500 N: D "
5 0 0 ( 0 .3 7 5 )(1 .3 7 5 ) 9
(1 .7 5 1 .3 7 5 0 .3 7 5 ) 1 .0 1 5 4 m m 2
6 (1 .7 5 )(1 0 5 1 0 )( 0 .2 3 7 5 1 0
6
)
2
2
Total deflection at D: D 0 .7 7 0 3 1 .0 1 5 4 1 .7 8 5 7 m m Applying Eq. (9.18), we obtain n cr
1 2
[
9 .8 1 ( 3 4 0 2 .4 1 6 8 1 0 3 4 0 ( 2 .4 1 6 8 1 0
3
3
5 0 0 1 .0 1 5 4 1 0
2
) 1 1 0 (1 .0 1 5 4 1 0
3
3
)
) 2
1
]
2
1 1 .4 9 c p s 6 9 0 r p m
(CONT.)
159
9.31 (CONT.) ( b ) From Eq. (9.21), we have n c r ,C n c r , D
n cr
where
2
2
n c r ,C n c r , D
g
n c r ,C
1 2
C '
n cr , D
1 2
D "
g
9 .8 1
1 2
1 .2 8 3 9 1 0
3
9 .8 1
1 2
1 .0 1 5 4 1 0
3
1 5 .9 1 c p s 8 3 5 rp m 1 5 .6 5 c p s 9 3 9 rp m
It follows that ( 8 3 5 )( 9 3 9 )
n cr
2
6 2 4 rp m
2
(835 ) (939 )
SOLUTION (9.32) Shaft
Sy
a ll
Key
Sy
a ll
a ll
n
n S ys n
580 2 .3
2 5 2 .2 M P a
3 0 .8 2 .3
154 2 .3
1 3 3 .9 1 M P a 6 6 .9 5 M P a
Torque: 9549 kW
T
9549 (90 )
n
9 5 4 .9 N m
900
Force at the shaft surface is then F
(a) L (b) L (c) L
F (w 2)
a ll
F (w 2)
a ll
F
a ll ( w )
T r
9 5 4 .9 1 8 .7 5
5 0 .9 3 k N
5 0 .9 3 1 0
3
2 5 2 .2 ( 9 .3 5 1 0
3
5 0 .9 3 1 0
2) 3
6
1 3 3 .9 1 1 0 ( 9 .3 5 1 0 5 0 .9 3 1 0
3
6
6 6 .9 5 1 0 ( 9 .3 5 1 0
0 .0 4 3 2 m = 4 3 .2 m m
3
)
3
2)
0 .0 8 1 4 m = 8 1 .4 m m
0 .0 8 1 4 m = 8 1 .4 m m
SOLUTION (9.33) From Table B.3: For shaft: S y S y c 4 4 0 M P a For key: S y S y c 3 9 0 M P a n 2 . Shaft:
a ll T
and
F
0 .5 8 S y n
a ll J r
1 0 .5 7 0 .0 7 5 2
0 .5 8 ( 4 4 0 ) 2
a ll ( D
4
D 2
32 )
1 2 7 .6 M P a
(1 2 7 .6 ) ( 0 .0 7 5 )
3
16
1 0 .5 7
kN m
2 8 1 .9 k N
Key length: Based on bearing on shaft: L
2 8 1 .9 1 0
F ( S y n )( h 2 )
(
4 4 0 1 0
6
)(
3
9 .3 7 5 1 0
2
Based on bearing on key: L
F ( S y n )( h 2 )
2 8 1 .9 1 0
(
3 9 0 1 0
6
Based on shear in key: L
)(
(
)w
n
(
0 .5 8 3 9 0 1 0 2
160
0 .3 0 8 4 m = 3 0 8 .4 m m
3
)
2 2 8 1 .9 1 0
)
3
9 .3 7 5 1 0
2 F 0 .5 8 S y
0 .2 7 3 4 m = 2 7 3 .4 m m
3
2
3
6
)(1 8 .7 5 1 0
0 .1 3 3 m = 1 3 3 m m 3
)
SOLUTION (9.34) We have F
T r
0 .4 1 0 3 7 .5 1 0
a ll w L
By Eq.(9.23): n
3
3
6
( 7 0 1 0 ) ( 9 .3 7 5 1 0
2F
2 1 .3 3 k N
2
3
) ( 7 5 1 0
3
)
3
2 ( 2 1 .3 3 1 0 )
1 .1 5
SOLUTION (9.35)
Shaft
Sy
a ll
460 4
460 2
340 4
n S ys
a ll
n
115 M Pa 5 7 .5 M P a
4
Key
Sy
a ll
n Sy 2
a ll
85 M Pa
4 2 .5 M P a
n
Torque in shaft. We have J D
3 2 (6 0 )
4
4
3 2 1 .2 7 2 1 0
6
mm
4
Hence, a ll J
T
r
5 7 .5 (1 .2 7 2 )
2 .4 4 k N m
0 .0 3
Force at the shaft surface: F
(a) L (b) L (c) L
F (w 2)
a ll
F (w 2)
a ll
F
a ll ( w 2 )
T r
2 .4 4 0 .0 3
2 .4 4 k N m 7 0 .7 m m
81300 6
1 1 5 (1 0 )( 0 .0 1 )
9 5 .6 m m
81300 6
8 5 (1 0 )( 0 .0 1 ) 81300 6
4 2 .5 (1 0 )( 0 .0 2 )
9 5 .6 m m
SOLUTION (9.36) Shear force in each bolt: F
T RN
Average shear stress in each bolt, a ll
Sy n
F A
T RN
db 4
4T 2
db RN
Solving db
4Tn
(P9.36)
RNS y
Introducing the given data: db [
3
4 ( 5 1 0 )(1 .2 )
1 6
( 0 .0 8 )( 6 )( 2 6 0 )(1 0 )
]
2
7 .8 2 m m
161
SOLUTION (9.37) From Eq. (1.15), 9549 kW
T
n
9549 (30 )
2 .3 8 7 k N m
120
Force at shaft surface F
T r
F L(w 2)
7 9 .6 k N
2 .3 8 7 0 .0 3
( a ) Key
7 5 ( 5 )(1 0
6
3
7 9 .6 (1 0 )
F L(w)
3
7 9 .6 (1 0 )
7 5 (1 0 )(1 0
6
)
212 M Pa;
)
n
106 M Pa;
n
S ys
Sy
1 .9 8
420 212
210 106
1 .9 8
210 3 1 .3
6 .7 1
420 2 4 .6
1 7 .1
( b ) Bolts Force at bolt circle F
T Db 2
2 .3 8 7 0 .0 7 2
3 3 .1 5 k N
Sear Stress in bolts
F 2
6 ( d b
4)
3 1 .3 M P a ;
n
S ys
2 4 .6 M P a ;
n
Sy
3 3 .1 5 ( 4 ) 6 ( 0 .0 1 5 )
2
( c ) Bearing on bolts in flange
F 6 d b Ft
3 3 .1 5 6 (1 5 )(1 5 )1 0
6
SOLUTION (9.38)
(a)
T
9549 kW n
F
T r
9549 ( 45 )
200
2148 0 .0 2 5
2 .1 4 8 k N m
8 5 .9 2 k N 3
3
Area in shear in key (1 4 .0 6 2 5 1 0 )(8 7 .5 1 0 ) 1 .2 3 1 0 3
3
3
m
2
Area in bearing for key [(1 4 .0 6 2 5 2 ) 1 0 ](8 7 .5 1 0 ) 6 .1 5 1 0
4
m
2
Shear stress in key:
8 5 .9 2 1 0 1 .2 3 1 0
3
3
6 9 .8 5 M P a
Bearing stress in key: (b)
8 5 .9 2 1 0 6 .1 5 1 0
3
4
1 3 9 .7 M P a
Area in shear for bolts 6 ( 4 )(9 .3 7 5 ) 4 1 4 .1 7 m m 2
2
Force at bolt circle: F
2 .1 4 8 0 .0 7 5
2 8 .6 k N
Shear stress in bolts:
28600 4 1 4 .1 7
6 9 .0 5 M P a
(CONT.)
162
9.38 (CONT.) ( c ) Area in bearing for bolts 6 (9 .3 7 5 )( 2 1 .8 7 5 ) 1 2 3 0 .5 m m
2
Bearing stress in bolts:
28600 1 2 3 0 .5 (1 0
6
)
2 3 .2 M P a
( d ) Area in shear at edge of hub D h t f (1 0 0 )( 2 1 .8 7 5 ) 6 8 7 2 .2 m m Force at edge of hub: F Shear stress in web:
T Dh 2
4 2 .9 6 0 1 0 6 8 7 2 .2 (1 0
2148 0 .0 5
3
6
)
2
4 2 .9 6 k N
6 .2 5 M P a
End of Chapter 9
163
CHAPTER 10
BEARINGS AND LUBRICATION
SOLUTION (10.1) We have H 9 .8 1( 6 5 ) 6 3 8 M P a
Equation (8.3') will be used.
KWl HAp
(1)
where A p D L 2 4 (1 2 ) 2 8 8 m m
2
l n D t 1 8 ( 2 4 )( 6 0 1 2 0 0 ) 9 7 .7 1 0
( a ) Good Lub: K 2 1 0
( 2 1 0
6
6
6
mm
(Table 8.3), Equation (1) gives: 6
)(1 5 0 )( 9 7 .7 1 0 ) 6
6
( 6 3 8 1 0 )( 2 8 8 1 0
)
0 .1 6 m m
( b ) Excellent Lub: K 1 1 0
(1 1 0
7
7
(Table 8.3), Equation (1) is then 6
)(1 5 0 )( 9 7 .7 1 0 ) 6
( 6 3 8 1 0 )( 2 8 8 1 0
6
)
0 .0 0 8 m m
SOLUTION (10.2) Refer to solution of Prob. 10.1. H 9 .8 1(1 0 5 ) 1 0 3 0 M P a
(Table B.6 and Sec. 8.5)
Use Eq. (8.3'),
KWl HAp
(1)
Here A p D L ( 2 5 )( 2 5 ) 6 2 5 m m
2
l n D t 1 8 ( 2 5 )( 6 0 1 5 0 0 ) 1 2 7 .2 1 0
( a ) Good Lub: K 2 1 0
( 2 1 0
6
6
6
(Table 8.3), Then, Eq. (1) result in 6
)(1 1 5 )(1 2 7 .2 1 0 ) 6
(1 0 3 0 1 0 )( 6 2 5 1 0
6
)
0 .0 4 5 m m
( b ) Excellent Lub: K 1 1 0
(1 1 0
7
7
(Table 8.3), Equation (1) gives 6
)(1 1 5 )(1 2 7 .2 1 0 ) 6
(1 0 3 0 1 0 )( 6 2 5 1 0
6
)
0 .0 0 2 m m
164
mm
SOLUTION (10.3) We have H 9 .8 1( 6 0 ) 5 8 8 .6 M P a
Equation (8.3) will be used, KH WA l
(1)
a
where 0 .1 5 m m , 7
K 1 10
W 450 N
(Table 8.3)
l n D t n ( 2 5 )( 6 0 2 4 3 6 5 1 .2 ) 4 9 .5 4 1 0 n m m 6
A p ( 2 5 )( 2 5 ) 6 2 5 m m
2
Equation (1) is thus (1 1 0
0 .1 5
7
6
)( 4 5 0 )( 4 9 .5 4 1 0 n ) 6
( 5 8 8 .6 1 0 )( 6 2 5 1 0
6
)
Solving, n 2 4 .8 r p m
SOLUTION (10.4) Equation (10.10): Tf
159 kW n
1 5 9 (1 .5 ) 1200 60
1 1 .9 N m
Equation (10.7): Tfc
4
2
3
Lr n
11 . 9 ( 0 . 09 ) 10 4
t 80
From Fig.10.8:
o
2
3 3
( 0 . 22 )( 0 . 08 ) ( 20 )
12 . 04 mPa s
C
SOLUTION (10.5) (a) Tf
2
f
c
Tf n
( b ) kW (c)
3
4 Lr n
159
Tf
Wr
4
2
( 4 .1 4 1 0
3
3
)( 0 .1 )( 0 .0 3 7 5 ) ( 2 4 0 0 0 6 0 ) 0 .0 7 5 (1 0
4 .6 ( 4 0 0 ) 159
4 .6 2 2 5 0 ( 0 .0 3 7 5 )
3
)
4 .6 4 N m
1 1 .6
0 .0 5 5
SOLUTION (10.6) T
f
(a)
fWr 0 . 01 ( 8000 )( 0 . 06 ) 4 . 8 N m Tfc 4
2
3
Lr n
4 . 8 ( 0 . 05 ) 10 4
2
3 3
( 0 . 12 )( 0 . 06 ) ( 10 )
23 . 45 mPa s
( b ) Figure 10.8: SAE 40 oil
165
SOLUTION (10.7) P
W DL
12 ( 10
3
)
0 . 768
( 0 . 125 )( 0 . 125 )
f
MPa,
F W
80 12 10
3
0 . 667 (10
2
)
Equation (10.9):
fP 2
2
0 . 667 ( 10
( cr )
n
2
)( 0 . 768 )( 10 2
2
(4)
6
)
( 0 . 0004 ) 25 . 95 mPa s
SOLUTION (10.8) From Fig. 10.8: 1 1 m P a s (a) P
W DL
400
2250 ( 0 .1 5 )( 0 .0 3 7 5 )
n
kPa
1500 60
25
rp s
Equation (10.9): 2 n r P c
f 2
2
2 11 ( 10
3
)( 25 )
400 10
3
1 0 . 001
0 . 014
( b ) T f fW D 2 ( 0 .0 1 4 )( 2 .2 5 )( 7 5 ) 2 .3 6 Tf n
kW
2 .3 6 ( 2 5 )
159
159
N m
0 .3 7
SOLUTION (10.9) ( a ) From Eq. (10.10): Tf
159 kW n
1 5 9 (1 .4 ) 2100 60
6 .3 6 N m
Equation (10.7) gives then
Tfc 2
3
4 Lr n
( 6 .3 6 )( 0 .1 7 5 ) 4
2
3
( 0 .2 )( 0 .0 7 5 ) ( 3 5 )
9 .5 5 m P a s
Using Fig. 10.8: t 90 C o
( b ) Equation (10.8) results in f
Tf
Wr
6 .3 6 2 (75 )
0 .0 4 2
SOLUTION (10.10) Figure 10.8: 8 .4 m P a s , Figure 10.14:
h0 c
0 . 025 0 . 05
0 .5
10
6 8 .4 (1 0
c 0 .0 0 1 r 0 .0 0 1(5 0 ) 0 .0 0 5 m m ,
S 0 . 52
Equation (10.17): S ( cr )
2 n P
3
P
)( 3 0 )
0 .5 2 ,
and W P D L 4 8 4 .6 ( 0 .1) ( 0 .0 5 ) 2 .4 2 k N
166
P 4 8 4 .6 k P a
L D 1 2
SOLUTION (10.11) P
W DL
3
1 2 (1 0 )
0 .7 6 8
( 0 .1 2 5 )( 0 .1 2 5 )
f
M Pa,
F W
80 3
1 2 (1 0 )
0 .6 6 7 (1 0
Thus f
r c
( 0 .6 6 7 1 0
1 0 .0 0 0 4
2
) 1 6 .7
From Fig. 10.15: S 0 .8 . The viscosity is therefore
S P n ( r c )2
0 .8 ( 0 .7 6 8 )1 0 4 (1 0 .0 0 0 4 )
6
2
0 .0 2 4 5 8
P a s 2 4 .5
mPa s
SOLUTION (10.12) ( a ) From Solution of Prob.10.8: 1 1 m P a s , P 4 0 0 k P a , n 2 5 rp s , L D 1 4 We thus have S ( cr )
2 n P
1 ( 0 . 001 )
2 11 ( 10
3
)( 25 )
0 . 69
3
400 ( 10 )
From Fig. 10.15: f 20,
r c
( b ) Thus
fW D
Tf
2 Tfn
kW
159
f 2 0 ( r ) 2 0 ( 0 .0 0 1 ) 0 .0 2 c
0 .0 2 ( 2 .2 5 )( 0 .0 7 5 ) 3 .3 7 5 N m
3 . 375 ( 25 ) 159
0 . 531
SOLUTION (10.13) ( a ) We have P
W DL
1 .4 5 ( 0 .0 3 )( 0 .0 3 )
1 .6 1 M P a
L D 30 30 1
Somerfield number: S ( cr ) ( 2
n P
) ( 01.05 3 ) [ 2
2 0 .7 (1 0
3
)( 4 0 ) 6
1 .6 1 (1 0 )
] 0 .1 2 9
Then, Fig. 10.14 gives, h o c 0 .4 2 and 1 h o c 0 .5 8 . It follows that e c 0 .5 8 ( 0 .0 3 ) 0 .0 1 7 4 m m
( b ) Figure 10.15. With S 0 .1 2 9 , L D 1;
( r c ) f 3 .5 .
So, f 3 .5 ( c r ) 3 .5 ( 0 .0 3 ) (1 5 ) 0 .0 0 7 N m Friction torque, T fW ( D 2 ) 0 .0 0 7 (1 .4 5 )(1 5 ) 0 .1 5 2 N m
Thus,
kW
Tn 159
0 .1 5 2 ( 4 0 ) 159
0 .0 3 8 2
( c ) Figure 10.16 ( S 0 .1 2 9 with L D 1 ) gives P P
0 .4 2
m ax
or Pm a x P 0 .4 2 1 .6 1(1 0 ) 0 .4 2 3 .8 3 M P a 6
167
2
),
L D 1
SOLUTION (10.14) P
W DL
3
1 5 (1 0 )
2 .0 8 3 M P a
6
(1 2 0 )( 6 0 )1 0
Figure 10.8: 1 6 m P a s Apply Eq. (10.17): n
S ( cr ) ( 2
P
o
(for SAE 40 oil at 8 0 C )
)
or 0 .1 5 (
120 2 c
)
2 (1 6 1 0
3
)(1 5 0 0 6 0 ) 6
2 .0 8 3 (1 0 )
Solving, c 0 .0 6 7 9 m m
It follows that c r 0 .0 6 7 9 6 0 0 .0 0 1 1
OK (See Sec. 10.8)
Figure 10.14: (for S 0 .1 5 and L D 1 2 )
h o c 0 .2 8
from which h o 0 .2 8 ( 0 .0 6 7 9 ) 0 .0 1 9 m m
SOLUTION (10.15) h0 c
S ( cr )
Figure 10.14: S 0 . 13 ,
0 .4 ,
0 . 0025 0 . 00625 2 n P
( 0 .05006 2 5 )
( a ) Fig.10.8:
t 80 C
( b ) Fig.10.15:
r c
2 (900 60 ) 3
7 0 0 (1 0 )
,
L D 1
9 .4 5 m P a s
o
f 3 , f 3 ( 0 . 00625 ) 0 . 00375 50
( c ) T f fW r 0 .0 0 3 7 5 ( 7 0 0 0 .1 0 .1)( 0 .5 ) 1 3 .1 N m kW
Tfn 1050
13 . 1 ( 15 ) 159
1 . 24
SOLUTION (10.16) P
W DL
S ( cr )
( a ) Figure 10.14: ( b ) Figure 10.15: T
f
1500 0 .0 2 5 0 .0 2 5
2 n P
h0 c
r c
2 .4 M P a ,
( 0 .010 0 8 )
2 5 0 (1 0
3
6
Tf n 159
f 1 1,
0 .1 6 5 (1 6 . 6 7 ) 159
0 .5 4 3,
0 . 75 ; h 0 0 . 75 ( 0 . 01 ) 0 . 008
2 ) 0 . 165
0 .0 1 7
168
L D 1
c 0 .0 1 m m
mm
f 1 1( 0 .0 0 0 8 ) 0 .0 0 8 8
fWr ( 0 . 0088 )( 1500 )( 0 . 025
kW
)(1 6 .6 7 )
2 .4 (1 0 )
n 1 6 .6 7 rp s ,
N m
SOLUTION (10.17) P
W DL
1 .2 5 M P a ,
4000 0 .0 8 0 .0 4
S (c) r
( a ) Figure 10.14: ( b ) Figure 10.16:
h0 c
c 0 .0 8
rp s ,
mm,
L D 1 2
30 mPa s
Fig.10.8: Thus
n 10
2 n P
( 0 .0 0 2 ) 1
0 .1 6 ,
3
)(1 0 ) 6
1 . 2 5 (1 0 )
0 .0 6
h 0 0 .1 6 ( 0 .0 8 ) 0 .0 1 3 m m
0 .2 6 ,
P p m ax
2 3 0 (1 0
p m ax
1 .2 5 0 .2 6
4 .8 0 8 M P a
SOLUTION (10.18) 8 mPa s
Figure 10.8: T
119 hp
n
Try S=0.03: Fig.10.15:
1 1 9 ( 0 .1 6 ) 2 7 .2
0 .7 N m ,
f 1 . 45 ,
r c
n 1 6 3 0 6 0 2 7 .2 F
T r
0 .7 0 .0 2 5
rp s ,
L D 1
28 N
f 1 . 45 ( 0 . 001 ) 0 . 00145
From Eq.(10.17): P ( cr )
Thus
2 n S
2 8 1 0
( 0 .01 0 1 )
3
( 2 7 .2 )
0 .0 3
7 .2 5 M P a
W P D L 7 .2 5 (1 0 )( 0 .0 5 )( 0 .0 5 ) 1 8 .1 3 k N 6
F fW 0 .0 0 1 4 5 (1 8 .1 3 1 0 ) 2 6 .3 N 3
Since
28 > 26.3, assumption S=0.03 is incorrect.
Try S=0.025: Fig.10.15: P ( 0 .01 0 1 )
r c
f 1.3 ,
2 ( 8 1 0
3
)( 2 7 .2 )
0 .0 2 5
f 1.3 ( 0 .0 0 1 ) 0 .0 0 1 3 8 .7 M P a
W P D L 8 .7 (1 0 )( 0 .0 5 )( 0 .0 5 ) 2 1 .8 6
kN
F fW 0 .0 0 1 3( 2 1 .8 1 0 ) 2 8 .3 N , 3
Assumption is correct.
( a ) W 2 1 .8 k N ( b ) Fig.10.14:
h0 c
0 . 13 , h 0 0 . 13 ( 0 . 001 25 ) 0 . 0033
( c ) Equation (10.20): 1
h0 c
mm
1 0 . 13 0 . 87
SOLUTION (10.19) h0
c 0 .0 0 1 5 (5 0 ) 0 .0 7 5 m m ,
Figure 10.14: P
c
0 .0 2 5 0 .0 7 5
1 3
,
L D 1 2
S=0.022 W DL
8000 ( 0 . 1 )( 0 . 05 )
1 . 6 MPa
n 900
60 15 rps
(CONT.)
169
10.19 (CONT.) ( a ) Equation (10.17):
SP n
0 . 22 ( 1 . 6 10
6
)
15
f 6 .5 ,
r c
( b ) Figure 10.15:
2
( cr )
( 0 . 0015 )
kW
( 80 0 . 050 ) 15
159
52 . 8 mPa s
f 6 .5 ( 0 .0 0 1 5 ) 0 .0 1
F fW 0 . 01 ( 8000 ) 80 Tfn
2
N
0 . 377
159
SOLUTION (10.20) P
W DL
1 .2 M P a ,
6000 0 .1 0 .0 5
r 50 m m ,
From Sec. 10.10: A 1 2 .5 D L 1 2 .5 (1 0 0 5 0 ) 1 0 Table 10.3: C 7 . 4 watts Equation (10.17): S ( cr )
2 n P
1 0 .0 0 1
)
L D 1 2
0 .0 6 2 5
2
m ,
C o
2 ( 2 0 1 0
3
)( 5 )
6
1.2 (1 0 )
f 3.2 ,
r c
Figure 10.15:
(
m
2
6
0 .0 8 3
f 3 .2 ( 0 .0 0 1 ) 0 . 0 0 3 2
Through the use of Eq.(10.24), we have H fW ( 2 r n ) 0 .0 0 3 2 ( 6 , 0 0 0 ) ( 2 0 .0 5 5 ) 3 0 .1 6 w a tts
Thus, by Eq.(10.23): t0 ta
20
H CA
30 . 16 7 . 4 ( 0 . 0625 )
20 65 . 2 85 . 2
o
C
SOLUTION (10.21) Refer to Solution of Prob.10.20. We have C 8 . 5 watts
m
Using Eq.(10.23): t 0 t a
2
C (Table 10.3) o
30
H CA
30 . 16 8 . 5 ( 0 . 0625 )
30 56 . 8 86 . 8
o
SOLUTION (10.22) C 14 kN
Table 10.5:
Fa
Table 10.7:
V Fr
3 1.2 ( 2 )
C s 6 . 95 kN Fa
1.2 5 e
X 0 .5 6
Cs
n 1500 3 6 .9 5
0 .4 3 2
Y 1.0 3 7 (by interpolation)
Thus, we obtain P XVF
r
YF a 0 . 56 (1 . 2 )( 2 ) 1 . 037 ( 3 ) 4 . 455
P VF
1 . 2 ( 2 ) 2 . 4 kN
or r
Hence L1 0
6
10 60 n
( CP )
a
6
10 6 0 (1 5 0 0 )
( 4 .41 45 5 ) 3 4 4 .8 h 3
170
kN
C
n 5 rp s
SOLUTION (10.23) Table 10.5:
C 1 4 .8 k N
Table 10.8:
V Fr
Fa
C s 7 .6 5
1.2 5 e 0 .9 5 ;
X 0 .3 7 ,
Y 0 .6 6
We have then P XVF
r
YF a 0 . 37 (1 . 2 )( 2 ) 0 . 66 ( 3 ) 2 . 868
P VF
1 . 2 ( 2 ) 2 . 4 kN
kN
or r
Thus L1 0
6
10 60 n
( CP )
a
6
10 6 0 (1 5 0 0 )
( 21.84 .86 8 ) 1 5 2 6 .9 3
h
SOLUTION (10.24) Refer to Solution of Prob. 10.22: C 14 kN ,
C s 6 .9 5
Fa
kN ,
Cs
0 .4 3 2
Table 10.7: Fa VF
r
3 1( 2 )
1 .5 e ;
X 0 . 56
Y 1 . 037
(by interpolation)
It follows that P K s [ X V F r Y F a ] 1 .5[( 0 .5 6 )1( 2 ) 1 .0 3 7 (3)] 6 .3 4 7 k N
or P K s VF r (1 . 5 )1 ( 2 ) 3 kN
Thus L1 0
6
10 60 n
( CP )
a
6
10 6 0 (1 5 0 0 )
( 6 .31 44 7 ) 1 1 9 .2 3
h
SOLUTION (10.25) Table 10.5: C 5 5 .9 k N
C s 3 5 .5 k N
Table 10.8: Fa V Fr
6 .7 5 (1 )( 2 2 .5 )
iF a
0 .3 e ,
Cs
1 ( 6 .7 5 ) 3 5 .5
0 .1 9
and X 1,
Y 0 .9 2
Thus P X V F r Y F a 1(1)( 2 2 .5 ) ( 0 .9 2 )( 6 .7 5 ) 2 9 k N
or P V F r (1)( 2 2 .5 ) 2 2 .5 k N
Hence L1 0
6
10 60 n
( CP ) 3
6
10 60 (700 )
( 5259.9 ) 1 7 0 .5 h 3
171
SOLUTION (10.26) Table 10.5: C 1 4 k N
C s 6 .9 5 k N
Table 10.7: Fa Cs
Fa
0 .2 9 e ,
2 6 .9 5
V Fr
2 (1 )( 4 )
0 .5 e
and X 0 .5 6 ,
Y 1 .1 4 2
(by interpolation)
Hence P X V F r Y F a 0 .5 6 (1)( 4 ) (1 .1 4 2 )( 2 ) 4 .5 2 k N
or P V F r (1)( 4 ) 4 k N
Thus L1 0
6
10 60 n
( CP )
a
6
10 60 (3500 )
( 41.542 ) 1 4 1 .5 h 3
From Fig. 10.26: K r 0 .6 2 . So L 5 K r L1 0 0 .6 2 (1 4 1 .5 ) 8 7 .7 h
SOLUTION (10.27) Variation factor: V 1 (Eq. 10.25) C 1 9 .5 k N
Table 10.5:
C s 10 kN
Table 10.7: Fa Cs
1 .7 10
0 .1 7 ,
Fa
e 0 .3 4 ;
V Fr
1 .7 (1 )( 4 .5 )
0 .3 8 e
and X 0 .5 6 ,
Y 1 .3 1
Thus P X V F r Y F a 0 .5 6 (1)( 4 .5 ) (1 .3 1)(1 .7 ) 4 .7 4 7 k N
P V F r (1)( 4 .5 ) 4 .5 k N
or
It follows that L1 0
6
10 60 n
( CP )
a
6
10 60 (600 )
( 41.79 .54 7 ) 1 9 2 6 h 3
SOLUTION (10.28) K s 2 .5
Table 10.9:
V 1 .2
and Fa V Fr
1 .7 (1 .2 )( 4 .5 )
0 .3 1 5 e ;
X 0 .5 6 ,
Y 1 .3 1
Therefore, Eq. (10.27): P K s ( X V F r Y F a ) 2 .5[( 0 .5 6 )(1 .2 )( 4 .5 ) (1 .3 1)(1 .7 )] 1 3 .1 3 k N
(CONT.)
172
10.28 (CONT.) or P K sV F r 2 .5 (1 .2 )( 4 .5 ) 1 3 .5 k N
Hence L1 0
6
10 60 n
a
( CP )
6
10 60 (600 )
( 11 39 .5.5 ) 8 3 .7 h 3
From Fig. 10.26: K r 0 .6 2 . So, L 5 K r L1 0 0 .7 (8 3 .7 ) 5 8 .6 h
SOLUTION (10.29) P2 3
3
L '10 P1 2 L " 10
0 . 5 P1 ,
P 2 0 . 794 P1
3
20 . 6 %
SOLUTION (10.30) 10
10 3
P2
L '10 P1 3 2 L " 10
10
0 . 5 P1 3 ,
P 2 0 . 812 P1
18 . 8 %
SOLUTION (10.31) Table 10.5:
C 5 5 .9 k N ,
Table 10.8:
VF
Fa
1 .5 1 .2 ( 5 )
r
C s 3 5 .5 k N iF a
0 . 25 e ,
Cs
2 ( 1 .5 ) 35 . 5
0 . 085
and X 1,
Y 1.3 8 6 (by interpolation)
Then we obtain P XVF
r
YF a 1(1 . 2 )( 5 ) 1 . 386 (1 . 5 ) 8 . 079
or P V F r 1 .2 (5 ) 6 k N
Thus L1 0
6
10 60 n
( CP )
a
6
10 6 0 (1 0 0 0 )
( 85.05 7.99 ) 5 5 2 1 h 3
and 5 L1 0 5 (5 5 2 1) 2 7 , 6 0 5 h
SOLUTION (10.32) P K s [V X F r 0 ] 1 .7[1 .2 (1)5 ] 1 0 .2
35 mm-02-series (Table 10.6): L1 0
6
10 60 n
( CP ) a
6
10 60 ( 2400 )
10
[ 13 01 .2.9 ] 3 3 1 0 .6 5 h
35 mm-03-series (Table 10.6): L1 0
6
10 60 ( 2400 )
10
[ 14 04 .2.6 ] 3 9 4 9 .3 h
173
kN
kN
SOLUTION (10.33) P K sV F r 1(1)( 2 5 ) 2 5
(Eq.10.26)
kN
75 mm-02-series (Table 10.6): L1 0
6
10 60 n
( CP )
a
6
10
10 60 ( 2000 )
[ 9215.3 ] 3 6 2 5 .1 h
and 5 L1 0 3,1 2 5 h r 75 mm-03-series (Table 10.6): L1 0
10
6
10 60 ( 2000 )
[ 12853 ] 3 6 3 4 6 h
and 5 L1 0 3 1, 7 3 0 h
SOLUTION (10.34) P K sV F r 2 (1 1 1 2 .5 ) 2 5
(Eq.10.26)
kN
We have L1 0
6
10 60 n
6
24
a
( CP ) :
02-series (Table 10.6): 03-series (Table 10.6):
10 60 ( 2400 )
3
[ 2C4 ] ,
C 3 6 .2 8 6 k N
40-mm-bore bearing 30-mm-bore bearing
SOLUTION (10.35) Fa
Table 10.7:
Cs
0 , use 0.014: X 1 ,
Y 0
We have, by Eq.(10.27), P K s XVF
r
2 . 5 (1)( 1 . 2 ) 8 24 kN
Thus L 10
6
10 60 n
( CP )
From Table 10.5:
a
40 5
:
6
10 60 ( 900 )
C [ 24 ] , C 18 . 1 kN 3
30-mm-bore bearing
SOLUTION (10.36) Refer to Solution of Prob.10.22: C 14 kN , Figure 10.26: K r 0 .6 2 .
P 4 . 455
We now have n 1200
rpm .
Hence L5 K r
6
10 60 n
( CP ) 0 .6 2 3
6
10 6 0 (1 2 0 0 )
( 4 .41 45 5 ) 2 6 7 h 3
174
kN
SOLUTION (10.37) Figure 10.26:
K
r
0 .3 2 ,
Refer to Solution of Prob.10.32:
35 mm-02-series: L 5 K r L1 0 0 .3 2 (3 1 0 .6 5 ) 9 9 .4 1 h
35 mm-03-series: L 5 K r L1 0 0 .3 2 (9 4 9 .3) 3 0 3 .8 h
End of Chapter 10
175
CHAPTER 11
SPUR GEARS
SOLUTION (11.1) Equation (11.4): m
d N
d 6 (3 2 ) 1 9 2 m m ,
,
r
d 2
96 m m
Table 11.1: a m 6 mm ,
h 2 .2 5 ( 6 ) 1 3 .5 m m ,
hk 2 ( 6 ) 1 2 m m
Equation (11.9) of Sec.11.4: rb r c o s 9 6 c o s 2 0
9 0 .2 1 1 m m ,
o
r0 r a 9 6 .2 1 m m
SOLUTION (11.2) N1
Equation (11.8): r s
N
2
or N 1
1 3
N
2
3
Equations (11.6) and (11.5b): N1 N 2
c r1 r2
2P
m 2
(N1 N 2)
3 2
(
N
2
3
N2)
3 2
(
4 3
N 2 ) 360
or N
1 8 0 teeth.
2
Thus N 1
180 3
6 0 teeth
SOLUTION (11.3) ( a ) Pitch circle is P m 1 0 3 1 .4 2 m m
Pitch diameters are d
N
p
p
1 8 (1 0 ) 1 8 0 m m ,
d g 4 2 (1 0 ) 4 2 0 m m
The center distance: d
c
p
dg 2
1 2
(1 8 0 4 2 0 ) 3 0 0 m m
Base circle are Pb p r p c o s 9 0 c o s 2 0 Pb g 2 1 0 c o s 2 0
(b) c
d
p
dg
o
o
8 4 .5 7 m m
1 9 7 .3 4 m m
300 8 308 m m
2
(1)
The velocity ratio does not change. Hence d
p
dg
18 42
(2)
Solving Eqs. (1) and (2), d
p
1 8 4 .8 2 m m
d g 4 3 1 .1 9 m m
Since rb r c o s , the new pressure angle is new c o s
1
rb p d
p
2
cos
1
8 4 .5 7 1 8 4 .8 2 2
2 3 .7 7
176
o
SOLUTION (11.4) Addendum and circular pitch: a 1 m 1 (4) 4 m m
(Table 11.1)
p m ( 4 ) 1 2 .5 5 6 m m ; (Eq. 11.5a)
Tooth thickness, t 1 .1 5 7 1 m 1 .5 7 1( 4 ) 6 .2 8 4 m m
(Table 11.1)
SOLUTION (11.5) Refer to Table 11.1;
m 1 P :
with
Dedendum,
b d 1 .2 5 m 1 .2 5 (1 2 ) 1 5 m m
Clearance,
f 0 .2 5 m 0 .2 5 (1 2 ) 3 m m
h k 2 m 2 (1 2 ) 2 4 m m
Working depth,
t 1 .5 7 1 m 1 .5 7 1(1 2 ) 1 8 .8 5 2 0 m m
Thickness,
Circular pitch, by Eq. (11.3): p m (1 2 ) 3 7 .6 9 9 1 m m
SOLUTION (11.6) d g 4d
p
c
p
(Eq. 11.8)
We have 1 2
(d
d g ) 2 0 0;
5d
p
400
or
and m d
p
N
4N
p
p
,
N
p
d
p
m 80 5 16
Thus, N
g
64
SOLUTION (11.7) Equation (11.8), rs
N
p
N
g
1200 2400
1 2
,
N
g
2N
p
2 (1 8 ) 3 6
Equation (11.4), d
p
N p m 1 8 (3 ) 5 4 m m
and c
1 2
(d
p
dg)
1 2
(5 4 3 6 ) 4 5 m m
Equation (11.5a) gives p m ( 3 ) 9 .4 2 4 8 m m
177
d
p
80 m m
SOLUTION (11.8) rs
Equation (11.8): m d
p
p
0 . 5234
mN
p
N
p
N
g
1 4
N
p
84
,
N
p
2 1 teeth
4
13 . 1942
4 ( 2 1) 8 4 m m ,
dg mN
p
4 (8 4 ) 3 3 6 m m
Therefore c
1 2
(d g d p )
1 2
(3 3 6 8 4 ) 2 1 0 m m
SOLUTION (11.9) ( a ) From Eq. (11.4), pitch diameters are d g N g m 6 0 (8 ) 4 8 0 m m d
p
2 4 (8 ) 1 9 2 m m
The center distance c
1 2
(d
p
dg)
1 2
(1 9 2 4 8 0 ) 3 3 6 m m
The base circles are rb p
rb g
d
p
2
480 2
cos
cos 20
192 2
o
cos 20
o
9 0 .2 1 m m
2 2 5 .5 2 m m
( b ) When center distance is increased by 6 mm, we have c=342. This corresponds to a 1.79% increased in center distance. However, d
p
d g 2 (3 4 2 ) 6 8 4 m m
(1)
and N
p
d
p
N
g
dg
;
24 dp
60 dg
(2)
Solving Eqs. (1) and (2): d g 4 8 8 .6 m m
d
p
1 9 5 .4 m m
We have m
d
p
N
p
1 9 5 .4 24
8 .1 4 2 m m
Circular pithch p m 8 .1 4 2 2 5 .5 8 m m
Changing center distance does not affect the base circle. Thus, new c o s
1
(
rb p rp
) cos
1
( 1 99 50 .4.2 12 ) 2 2 .6
178
o
SOLUTION (11.10) Equation (11.8): rs
N
p
N
g
1 4
N
4N
g
p
4 ( 22 ) 88 teeth
Equation (11.4): d g N g m 88(4) 352
d
p
mm
N p m 22 ( 4 ) 88 mm
Hence, c
(3 5 2 8 8 ) 2 2 0
1 2
mm
SOLUTION (11.11)
From Eq. (11.5b), the modulus is 1 m
N
p
d
p
or d p N p m 2 5 (3 ) 7 5 m m
The speed at the contact point, V p V g V : V w p rb p w p
d
p
2
cos 20
o
( 3 46 00 0 ) 2 ( 725 ) c o s 2 0
o
1 2 , 5 4 7 m m s 1 2 .5 5 m s
SOLUTION (11.12)
From Eq. (11.5a), the circular pitch, p m ( 2 ) 6 .2 8 m m
By Eqs. (11.5b) and (11.6), the center distance: c
1 2
(N
p
N g )m
1 2
(3 0 1 0 0 )( 2 ) 1 3 0 m m
The pitch circular radii are rp
1 2
N pm
1 2
(3 0 )( 2 ) 3 0 m m
rg
1 2
N gm
1 2
(1 0 0 )( 2 ) 1 0 0 m m
Using Eq. (11.9), The base radii: rb p r p c o s 3 0 c o s 2 0
o
2 8 .1 9 m m
rb g rg c o s 1 0 0 c o s 2 0 9 3 .9 7 m m o
The addendum is equal to a p
1 p
m 2 mm
(CONT.)
179
11.12 (CONT.) The order radii are therefore ro p r p a 3 0 2 3 2 m m
ro g 1 0 0 2 1 0 2 m m
and
ro p ro g 1 3 4
SOLUTION (11.13)
Refer to Solution of Prob. 11.9. We have p m (8 ) 2 5 .1 3 3 m m , rg 4 8 0 2 2 4 0 m m , rb p 9 0 .2 1 m m ,
20
o
rp 1 9 2 2 9 6 m m
rb g 2 2 5 , 5 2 m m ,
c 336 m m
From Table 11.1, the addendum for gears are a 1 p 8 mm
The outside radii are then ro p r p a 9 6 8 1 0 4 m m ro g 2 4 0 8 2 4 8 m m
Substituting the data into Eq. (11.14), the contact ratio: Cr
[ 1 0 4 9 0 .2 1 2
1 2 5 .1 3 3 c o s 2 0
o
2
1 .6 9 4
SOLUTION (11.14)
We have e (
N1 N3
N
)( N 3 ) 4
24 96
0 .2 5
Substituting into Eq. (11.18) 0 .2 5
0 n2 120 n2
,
n 2 2 4 rp m
Direction is the same as that of the sun gear.
180
2 4 8 2 2 5 .5 2 ] 2
2
3 3 6 ta n 2 0 2 5 .1 3 3
o
SOLUTION (11.15) kW
Tn 9549
T1
,
22 (9549 )
d 1 2 7 (8 ) 2 1 6 m m ,
( a ) Ft1
T1 r1
5 2 .5 2 0 .1 0 8
d 2 4 8 (8 ) 3 8 4 m m ,
d 3 3 6 (8 ) 2 8 8 m m
(Eq.11.2)
486 N
Equation (11.19): F r F t ta n ( b ) RC
5 2 .5 2 N m
4000
486 227 2
2
o
4 8 6 ta n 2 5
o
2 2 7 lb
T C 4 8 6 ( 0 .1 4 4 ) 7 0 N m
536 N ,
486 227
A
486
B
486
2 2 .9
o
C RC
486 TC
SOLUTION (11.16) T1
9549 kW n
22 (9549 ) 4000
5 2 .5 2 N m
d 1 2 7 (5 ) 1 3 5 m m ,
( a ) Ft 1
T1 r1
5 2 .5 2 0 .0 6 7 5
d 2 4 8 (5 ) 2 4 0 m m ,
d 3 3 6 (5 ) 1 8 0 m m
778 N
Equation (11.19): F r 1 F t 1 ta n 2 0 2 8 3 N o
( b ) RC
778 283 2
2
T C 7 7 8 ( 0 .0 9 ) 7 0 N m
828 N ,
778 B
A
45
283
o
778
778
20
778
o
C RC TC
181
(Eq.11.4)
SOLUTION (11.17) Equation (11.4): d 1 2 0 ( 6 ) 1 2 0 m m ,
d 2 40(6) 240 m m ,
d 3 20(6) 120 m m ,
( a ) T1
9549 kW n
Ft 1
9549 (37 )
2 9 4 .4 N m
1200
F r 1 4 .9 1 ta n 2 0
4 .9 1 k N ,
2 9 4 .4 0 .0 6
d 4 60(6) 360 m m
We have 4 .9 1( 0 .1 2 ) 0 .0 6 F t 3 ,
o
1 .7 9 k N
F t 3 9 .8 2 k N ,
(Eq.11.19)
F r 3 9 .8 2 ta n 2 0 3 .5 7 k N o
2 ( b ) F x 0 : F x 4 .9 1 3 .5 7 1 .3 4 k N
B
3
3.57
9.82 o
9 .5
RB
4.91 RB
F y 0 : F y 9 .8 2 1 .7 9 8 .0 3 k N 8 .0 3 1 .3 4 2
8 .1 4 k N
2
TB 0
1.79
SOLUTION (11.18) ( a ) T1
9549 kW n T1
F t1
r1
9549 ( 80 )
1600
477 . 5 0 . 054
8 . 843
T1
2
1
F r 1 8 . 843 tan 20
kN ,
3
B
A
mN
r1
2
8843
1
477 . 5 N m ,
TC
8843
20
2 o
54 mm
3 . 219
( b ) r3
C
3219
6 ( 18 )
(Eq.11.4) (Eq.11.19)
kN
mN 2
3
6 ( 50 )
150
2
mm
T C 8 . 843 ( 0 . 15 )
o
1 .3 2 6 k N m
RC
RC
8 . 843
9 . 411
2
3 . 219
2
kN
SOLUTION (11.19) 9549 kW
( a ) T1
n
9549 ( 80 )
1600
Equation (11.4): r1 F r 1 6632 tan 25
( b ) r3
mN3
RC
2
8 (50 ) 2
o
477 . 5 N m mN
1
2
8 ( 18 ) 2
3 . 093
200
6632 3093 2
72 mm .
F t1
r1
477 . 5 0 . 072
6 . 632
kN
kN
mm 2
7 .3 1 8
T C 6 6 3 2 ( 0 .2 0 0 ) 1 .3 2 6 k N m
kN ,
6632 B
A
T1
C
T1
TC
3093
25
o
6632 RC
182
SOLUTION (11.20) Equation (11.4): d 1 N 1 m 1 5 (5 .2 ) 7 8 m m , d 3 2 0 (5 .2 ) 1 0 4 m m ,
( a ) T1
9549 kW n
Ft 1
9 5 4 9 ( 7 .5 ) 1500
d 4 4 0 (5 .2 ) 2 0 8 m m
4 7 .5 7 N m F r 1 1 .2 2 ta n 2 5
1 .2 2 k N F t 2 ,
4 7 .5 7 0 .0 3 9
d 2 3 5 (5 .2 ) 1 8 2 m m ,
o
0 .5 6 k N F r 2
T 2 1 .2 2 ( 0 .0 9 1) 1 1 1 N m Ft 3
2 .1 3 k N ,
111 0 .0 5 2
F r 3 2 .1 3 ta n 2 5
o
0 .9 9 k N
(b) 4
2 3
2.13
RC TC
B
65
0.99
0.56
C
0.99
1.22
o
T C 2 .1 3( 0 .1 0 4 ) 2 2 1 .5 N m
RC
SOLUTION (11.21)
d 3 18(4) 72 m m ,
( a ) T1
9549 kW n
Ft 1 Ft 2
9 5 4 9 (1 5 ) 1800
7 9 .5 8 0 .0 4 8
7 9 .5 8
d 2 36(4) 144 m m , d 4 42(4) 168 m m
N m
1 .6 6 k N
F r 1 1 .6 6 ta n 2 5
o
0 .7 7
kN
T 2 T 3 1 .6 6 ( 0 .0 7 2 ) 1 1 9 .5 N m Ft 3
3 .3 2
1 1 9 .5 0 .0 3 6
kN ,
F r 3 3 .3 2 ta n 2 5
( b ) R x 1 . 66 1 . 55 0 . 11 kN RB
R x R y 2 .5 5 2
2
0.77
RB
1.66
2 2 .5
o
B
o
1 .5 5 k N
R y 3 . 32 0 . 77 2 . 55 kN
kN
1.55
3 3.32
183
2
2 .3 5 k N
2.13
Equation (11.4): d 1 2 4 ( 4 ) 9 6 m m ,
0 .9 9 2 .1 3
2
SOLUTION (11.22) Equations (11.2) and (11.8): d 1 1 5 (5 .2 ) 7 8 m m , n 2 n1
N1
1 5 0 0 ( 13 55 ) 6 4 2 .9
N2
0b Ym K
1
f
rp m
Table 11.4: 0 1 7 2 M P a and 2 5 0 B h n
( a ) Table 11.3: Y=0.443, Fb
d 2 3 5 (5 .2 ) 1 8 2 m m ,
6
1 7 2 1 0 ( 0 .0 1 2 ) 0 .4 4 3 ( 0 .0 0 5 2 ) 1 .6
1
( b ) Table 11.10: K 1 . 117 ,
Q
2 .9 7 2
2 ( 35 )
1 .4
15 35
(Eq.11.33)
kN
(Eq.11.40)
F w d p b Q K 7 8 (1 2 )(1 .4 )(1 .1 1 7 ) 1 .4 6 4 k N
( c ) V2
dn
12
( 0 .1 8 2 )( 6 4 2 .9 ) 60
6 .1 2 m s ,
Fd
Thus
2 .9 7 2 3 .0 0 7 F t ,
Ft 9 8 8
and
1 .4 6 4 3 .0 0 7 F t ,
Ft 4 8 7 N
(Eq.11.38) 3 .0 5 6 .1 2 3 .0 5
F t 3 .0 0 7 F t
(Eq.11.24a)
N
SOLUTION (11.23) Equations (11.4) and (11.8): d 3 mN
V3
3
dn
5 ( 20 ) 100
( 0 .1)( 1 16 20 5 ) 5 .8 9
60
( a ) Table 11.3: Y 0 .3 2 0 , Fb
mm ,
0 bYm K
f
1 1 .5
( c ) Fd
3 . 0 5 5 .8 9 3.0 5
1500 ( 34 ) 1125
N3
rpm
m s
Table 11.4:
0
172
and 250 Bhn
MPa
(172 )( 15 )( 0 . 320 )( 5 ) 2 . 75 kN
( b ) Table 11.10: K 0 .9 0 3 M P a , F w d 3 bQK
N1
n 3 n1
Q
2N4 N3 N4
(Eq.11.33)
100 (15 )( 43 )( 0 . 903 ) 1 . 81 kN
2 ( 40 ) 20 40
4 3
(Eq.11.40)
(Eq.11.38)
F t 2 .9 3 1 F t
Thus
2 . 75 10
and
1 . 81 10
3
3
2 . 931 F t ,
F t 938
2 . 931 F t ,
F t 617 . 5 N
N
SOLUTION (11.24) Refer to Solution of Prob. 11.23. Equation (11.4) and (11.8): d 1 m N 1 5 (1 5 ) 7 5 m m ,
V1
dn 60
n 2 n1
N1 N2
1 5 0 0 ( 13 55 ) 6 4 2 .9 rp m
( 0 .0 7 5 )( 1 56 00 0 ) 5 .8 9 m s
(CONT.)
184
11.24 (CONT.) ( a ) Table 11.3: Y 0 .2 8 9 , Fb
obYm K
1 1 .5
f
Table 11.4: 1 7 2 M P a at 2 5 0 B h n (1 7 2 ) (1 5 ) ( 0 .2 8 9 ) ( 5 ) 2 .4 9 k N
( b ) Table 11.10: K 0 .9 0 3 M P a ,
2N2
(Eq. 11.40)
F w d 1 b Q K 7 5 (1 5 )(1 .4 )( 0 .9 0 3) 1 .4 2 k N
(Eq. 11.38)
N1 N 2
2 (35 )
1 .4
3 .0 5 5 .8 9 3 .0 5
( c ) Fd
Q
(Eq. 11.33)
15 35
F t 2 .9 3 1 F t
Thus 2 .4 9 1 0 2 .9 3 1 F t ,
F t 8 4 9 .5 N
1 .4 2 1 0 2 .9 3 1 F t ,
F t 4 8 4 .5 N
3
and 3
SOLUTION (11.25)
We have d N m 2 2 ( 2 .5 ) 5 5 m m ( a ) o 2 2 1 M P a (Table 11.4), Y 0 .3 3 (Table 11.3) Equation (11.33): Fb
(b) V
dn 60
obYm K
2 2 1 ( 3 0 ) ( 0 .3 3 ) ( 2 .5 )
1 .5
f
( 0 .0 5 5 )(1 5 0 0 )
4 .3 2 m p s
60
Fd
Equation (11.24a):
3 .6 5 k N
3 .0 5 4 .3 2 3 .0 5
F t 2 .4 2 F t
Hence 3 .6 5 2 .4 2 F t ,
F t 1 .5 0 8 k N
Equation (11.22) is then kW
Ft V 1000
1 5 0 8 ( 4 .3 2 ) 1000
6 .5 1
SOLUTION (11.26) Equations (11.4), (11.8), and (11.20'): d 2 36(4) 144 m m , n 2 1 8 0 0 ( 23 ) 1 2 0 0
rpm ,
d 3 18(4) 72 V
dn2 60
mm
( 0 .1 4 4 )(1 2 0 0 ) 60
9 .0 5
m ps
(CONT.)
185
11.26 (CONT.) ( a ) Table 11.3: Y 0 .4 4 6 (by interpolation), Table 11.4: 0 1 7 2 M P a (for steel of 200 Bhn) 0b Ym
Fb
K
1
f
1 7 2 (1 0 ) ( 0 .4 4 6 ) ( 4 ) 1 .4
2 .1 9 2 k N (Eq.11.33)
( b ) Table 11.10: K 0 .6 7 6 M P a ,
2 ( 42 )
Q
18 42
F w d 3 b Q K 7 2 (1 0 )( 2 1 1 5 )( 0 .6 7 6 ) 6 8 1 .4
( c ) Fd
3 . 05 9 . 05 3 . 05
F t 3 . 97 F t
(Eq.11.40) (Eq.11.38)
N
( Eq.11.24a)
2 .1 9 2 3 .9 7 F t ,
Thus
21 15
Ft 5 5 2
N
SOLUTION (11.27) Equation (11.4): d 1 24 (10 ) 240 ( a ) Table 11.3: Y 0 .3 9 3 (interpolated). 0 bYm
Fb
K
f
( b ) Table 11.10: F w d 1 bQK
1 1 .5
mm
Table 11.4:
0
172
[172 (15 )( 0 . 393 )( 10 )] 6 . 76 kN
K 0 . 545
Q
MPa,
2 ( 36 ) 24 36
(for steel of 200 Bhn)
MPa
(Eq.11.33)
6 5
(Eq.11.40)
240 (15 )( 65 )( 0 . 545 ) 2 . 35 kN
(Eq.11.38)
SOLUTION (11.28)
Refer to Solution of Prob. 11.27. Equation (11.4): d 1 2 4 (1 0 ) 2 4 0 m m , ( a ) Table 11.3: Y 0 .3 3 7 , Table 11.4: Fb
obYm K
f
1 1 .5
d 3 1 8 (1 0 ) 1 8 0 m m
o 1 7 2 M P a (for steel of 200 Bhn)
[1 7 2 (1 5 ) ( 0 .3 3 7 ) (1 0 ) ] 5 .8 k N (Eq. 11.33)
( b ) Table 11.10: K 0 .5 4 5 M P a ,
Q
2 N3 N3 N4
2 (1 8 ) 18 42
0 .6
(Eq. 11.40)
F w d 3 b Q K 1 8 0 (1 5 )( 0 .5 4 5 ) 8 8 3 N
(Eq. 11.38)
SOLUTION (11.29) d
p
N p m 24(2) 48
V
d pnp
Kv
60
6 . 1 4 . 02 6 .1
( 0 .0 4 8 )(1 6 0 0 ) 60
1 . 659
mm,
d g 60(2) 120
4 .0 2 m p s
mm
(Eq.11.4)
(Eq.11.20')
(Fig.11.15) (CONT.)
186
11.29 (CONT.) Ft
1000 kW V
1 0 0 0 ( 0 .9 )
2 2 3 .9
4 .0 2
N
J 0 .2 6
Pinion: (Fig.11.16):
Table 11.7: S t 1 9 7 .5 M P a
(mid-point).
Equation (11.35'): Ft K 0 K v
KsKm
1 bm
2 2 3 .9 (1) (1 .6 5 9 )
J
1 (1 .6 ) 1 0 .0 1 5 ( 0 .0 0 2 ) 0 .2 6
7 6 .2 M P a
Equation (11.36):
a ll
St K L
KT K R
1 9 7 .5 (1 )
1 (1 .2 5 )
158
J 0 .3 0 ,
Gear: (Fig.11.16):
2 2 3 .9 (1)(1 .6 5 9 )
M Pa
1 0 .0 1 5 ( 0 .0 0 2 )
0 .3
,
Yes.
S t 58 . 6 ksi
Table 11.7: 1 (1 .6 )
a ll
66 M Pa
and
a ll
5 8 .6 (1 ) 1 (1 .2 5 )
4 6 .9
M Pa
a ll
,
No.
SOLUTION (11.30) From Solution of Prob.11.29: d p 4 8 m m , F t 2 2 3 .9 N ,
C H 1 . 001
mG d g d
p
d g 120 m m ,
2 .5 ,
ng np
N
p
N
g
V 4 .0 2 m p s , 24 1600 ( 60 ) 640
(from Eq.11.46)
Pinion: c C p ( Ft K 0 K v
where
C
p
I
166
Ks
KmC
bd
I
f
1
)
(Eq.11.42)
2
M P a (Table 11.11)
sin cos
mG
2
m G 1
0 . 161
60 84
0 . 115
Thus c 1 6 6 1 0 [ 2 2 3 .9 (1)(1 .6 5 9 ) 3
S c 620 . 5 MPa
c , a ll
ScC LC H
KT K R
1 .6 (1 ) 1 ( 0 .0 1 5 )( 0 .0 4 8 ) 0 .1 1 5
(Table 11.12)
6 2 0 .5 (1 )(1 .0 0 1 ) 1 (1 .2 5 )
1
] 2 4 4 4 .7 M P a
(mid-point)
4 9 6 .9 M P a
(Eq.11.44)
Hence c , a ll c
Gear: Table 11.12:
Yes.
S c 4 8 5 .5
(mid-point)
M Pa
c 1 6 6 1 0 [ 2 2 3 .9 (1)(1 .6 5 9 ) 3
c , a ll
4 8 2 .5 (1 )(1 ) 1 (1 .2 5 )
1 .6 (1 ) 1 ( 0 .0 1 5 )( 0 .1 2 ) 0 .1 1 5
386 M Pa
Hence c , a ll c
Yes.
187
1
] 2 2 8 1 .3 M P a
rpm
SOLUTION (11.31) d
p
N p m 2 0 (3 ) 6 0 m m ,
V
d pnp
( 0 .0 6 )(1 2 0 0 )
60
Ft
1000 kW V
(Eq. 11.4)
3 .7 7 m p s
60
K 1 .6 1 8
d g 5 0 (3 ) 1 5 0 m m
(Fig. 11.15)
1 0 0 0 (1 .0 )
2 6 5 .3 N
3 .7 7
J 0 .2 4 5 (mid-point)
Pinion, (Fig. 11.16):
Table 11.7: S t 2 8 6 M P a
Equation (11.35'): Ft K o K v
KsKm
1 bm
2 6 5 .3 (1)(1 .6 1 8 )
J
1 (1 .7 ) 1 ( 0 .0 1 5 )( 0 .0 0 3 ) 0 .2 4 5
6 6 .2 M P a
Equation (11.36):
a ll
St K L
KT K R
2 8 6 (1 ) 1 (1 .3 )
Table 11.7: S t 8 9 .6 M P a
2 6 5 .3 (1)(1 .6 1 8 )
8 9 .6 (1 )
a ll
1 (1 .3 )
, Yes. a ll
Gear, (Fig. 11.16): J 0 .3 ,
and
220 M Pa
1 (1 .7 ) 1 ( 0 .0 1 5 )( 0 .0 0 3 ) 0 .3
6 8 .9 M P a
a ll
5 4 .1 M P a
, Yes.
SOLUTION (11.32) Refer to Solution of Prob. 11.31 d
p
60 m m ,
ng n p
N
p
N
g
d g 150 m m ,
V 3 .7 7 m p s ,
1 2 0 0 ( 52 00 ) 4 8 0 rp m ,
C H 1 .0 0 8
F t 2 9 6 .7 N ,
mG d g d
p
2 .5
K 4 .2
(from Eq. 11.46).
Pinion:
where
C
p
I
Then
c
Ks
C p [ Ft K o K v
166 10
3
s in c o s
mG
2
m G 1
1 .6 1
50 70
KT K R
7 8 9 (1 )(1 .0 0 8 ) 1 (1 .3 )
1
]
2
(Eq. 11.42)
1 .1 5
S c 8 7 9 M P a (Table 11.12) ScC LC H
f
(Table 11.11)
1 6 6 1 0 [ 2 9 6 .7 (1) ( 4 .2 )
c , a ll
I
M Pa
3
c
K mC
bd
1 .7 (1 ) 1 ( 0 .0 1 5 )( 0 .0 6 ) 1 .1 5
1
]
2
2 3 7 .5 M P a
(mid-point)
6 8 1 .6 M P a
(Eq.11.44) (CONT.)
188
11.32 (CONT.) Hence, c , a ll c
Yes.
Gear: Table 11.12: S c 5 5 1 .5 M P a
1 6 6 1 0 [ 2 9 6 .7 (1) ( 4 .2 ) 3
c
c , a ll
So
(mid-point)
5 5 1 .5 (1 )(1 )
1
]
2
1 5 0 .2 M P a
4 2 4 .2 M P a
1 (1 .3 )
c , a ll
1 .7 (1 ) 1 ( 0 .0 1 5 )( 0 .1 5 ) 1 .1 5
Yes.
c
SOLUTION (11.33) N
p
20, 2( 40 )
Q
20 40
Thus
N
4 3
40,
g
d
2 0 (5 ) 1 0 0
p
Table 11.10: S e 6 2 1 k s i,
,
b 50 m m
mm,
K 1 .8 2 1 M P a
F w d p b Q K 1 0 0 (5 0 )( 43 )(1 .8 2 1) 1 2 .1 4
(Eq.11.38)
kN
Equations (11.20') and (11.24a): dn
V
60
( 0 .1 )( 9 0 )
60
0 .4 7 m p s ,
Then, we obtain 1 2 .1 4 1 .1 5 4 F t ,
Fd
3 .0 5 0 .4 7 3 .0 5
F t 1 .1 5 4 F t
F t 1 0 .5 2 k N
Hence kW
Ft V 1000
1 0 ,5 2 0 ( 0 .4 7 )
1000
4 .9 5
SOLUTION (11.34) From Solution of Prob.11.33: N
p
20, N
g
40, d ScCLC H
c , a ll
KT K
1 0 0 m m , b 5 0 m m , V 0 .4 7 m p s , m G N
p
where S c 5 5 1 .5 M P a (Table 11.12)
,
R
C
L
Then
5 5 1 .5 (1 .1 )(1 )
6
F t ( 6 0 6 C.7 1 0 )
2
1 K0Kv
p
where
C
p
149
Kv
M Pa
5 . 56 0 . 47 5 . 56
K 0 1,
I
6 0 6 .7
(1 )1
K
1,
sin cos
mG
2
m G 1
I K mC
p
2
(mid-point)
K T 1,
K
R
1
M Pa
(Eq.11.42) f
(Table 11.11)
1 . 08 s
bd Ks
N
1.1 (Fig.11.19)
C H 1,
c , a ll
g
(Curve B, Fig.11.15) K m 1 .3,
C
f
1
0 . 107
(CONT.)
189
11.34 (CONT.) 6
F t ( 6 0 6 .7 1 03 )
Thus
0 .0 5 ( 0 .1 ) 0 .1 0 7 1 1 (1 .0 8 ) 1 1 .3 (1 )
2
1 4 9 1 0
6 .3 1 8 k N
and Ft V
kW
1000
6 3 1 8 ( 0 .4 7 )
2 .9 7
1000
SOLUTION (11.35)
N
2 5,
p
2 (50 )
Q
25 50
50,
g
4 3
, Table 11.10: S e 6 2 1 M P a ,
d
N m 2 5 (5 ) 1 2 5 m m ,
N
p
b 40 m m
K 1 .8 2 1 M P a
Thus, F w d p b Q K 0 .1 2 5 ( 0 .0 4 )( 43 )(1 .8 2 1 1 0 ) 1 2 .1 4 k N 6
(Eq. 11.38)
Equations (11.20') and (11.24a) with 6 0 0 fp m 6 0 0 1 9 6 .8 3 .0 5 m s : dn
V
60
( 0 .1 2 5 )(1 2 0 )
0 .7 8 5 m s ,
60
Fd
3 .0 5 0 .7 8 5 3 .0 5
F t 1 .2 5 7 F t
Then, we have 1 2 .1 4 1 .2 5 7 F t ,
F t 9 .6 6 k N
Hence, kW
Ft V 1000
9 6 6 0 ( 0 .7 8 5 ) 1000
7 .5 8
SOLUTION (11.36) N
N
p
ng g
60
np
Table 11.4:
240 600
24,
8 7 .2
0g
Y p 0 .3 3 7 ,
M Pa,
0 p
Y g 0 .4 2 1
172
M Pa
Hence Y p
0 p
0 .3 3 7 (1 7 2 ) 5 8
Yg
0g
0 .4 2 1(8 2 .7 ) 3 4 .8 2
Gear is weaker
Thus Fb
0bYm K
8 2 .7 ( 9 0 ) ( 0 .4 2 1 ) ( 4 ) 1 .4
f
8 .9 5 3
kN
We have d g 60(4) 240 m m ,
V
dn 60
( 0 .2 4 )( 2 4 0 ) 60
3 .0 2 m p s
Hence Fd
3 .0 5 3 .0 2 3 .0 5
F t 1 .9 9 F t
8 .9 5 3 1 .9 9 F t ,
and kW
Ft V 1000
4 5 0 0 ( 3 .0 2 ) 1000
1 3 .5 9
190
F t 4 .5 k N
SOLUTION (11.37) N
N
p
ng
60
np
g
Table 11.4:
24,
240 600
Y p 0 .3 3 7 ,
82 . 7 MPa ,
0g
Y p 0 p Yg 0 g
Y g 0 .4 2 1
172
0 p
MPa
the gear is weaker
Thus 0 bYm
Fb
K
82 . 7 ( 80 )( 0 . 421 )( 4 )
7 . 958
1 .4
f
kN
We have the quantities: d
g
N g m 60 ( 4 ) 240
mm
V d n 6 0 ( 0 .2 4 )( 2 4 0 6 0 ) 3 .0 2 m p s 3 .0 5 3 .0 2 3 .0 5
Fd
F t 1 .9 9 F t
(from Eq.11.24a)
and 7 , 9 5 8 1 .9 9 F t ,
Ft 4 k N
Equation (1.15) results in Ft V
kW
1000
4 0 0 0 ( 3 .0 2 ) 1000
1 2 .1
SOLUTION (11.38) Gear is weaker and can transmit lower hp. d g 60(4) 240 m m , K
KT K
L
s
1,
Table 11.7: S t 3 9 .3 M P a , V
d g ng 60
( 0 .2 4 ) 2 4 0 60
3 .0 2 m p s
We have
St K L
a ll
KtKR
3 9 .3 (1 ) 0 .8 5 (1 )
4 6 .2
M Pa
Also K
v
3 . 56
3 . 02
3 . 56
1 . 49 (Curve C, Fig.11.15)
m 4 m m , b 9 0 m m . (given)
J 0 .3 0 (from Fig.11.16, N K
0
1.2 5 (Table 11.5)
K
s
1 (standard gear),
K
m
g
60 ) 1.7 (Table 11.6)
Thus, Eq.(11.35'): Ft
4 6 .2 ( 9 0 )( 4 )( 0 .3 0 ) 1 .2 5 (1 .0 )(1 .7 )(1 .4 9 )
1 .5 7 6 k N
Hence, kW
Ft V 1000
1 5 7 6 ( 3 .0 2 ) 1000
4 .7 6
191
Table 11.9: K
R
0 .8 5
SOLUTION (11.39) N
N
g
Q
V
p
rs
2N N
p
dn 60
g
N
3 5,
28 4 5
70 28 35
g
d
70 63
( 0 .1 6 8 )( 6 0 0 )
N p m 2 8(6 ) 1 6 8 m m ,
p
K 1 .8 6 2
,
b 50 m m
M P a (Table 11.10)
5 .2 8 m p s
60
We calculate that F w d p b Q K 1 6 8 (5 0 )( 76 03 )(1 .8 6 2 ) 1 7 .3 8
kN
and 3 .0 5 5 .2 8 3 .0 5
Fd
F t 2 .7 3 1 F t ,
1 7 .3 8 2 .7 3 1 F t
F t 6 .3 6 4 k N
Hence Ft V
kW
1000
6 3 6 4 ( 5 .2 8 )
1000
3 3 .6
SOLUTION (11.40) N
g
35 , 2 ( 35 )
Q
63
N
70 63
28 ,
p
p
K 1 . 862
,
F w d p bQK
d
28 ( 6 ) 168
168 ( 60 )( 1 . 862 )(
b 60 mm
(Table 11.10)
MPa 70 63
mm ,
) 20 . 85 kN
We obtain V d n 6 0 ( 0 .1 6 8 )( 6 0 0 6 0 ) 5 .2 8 m p s Using Eq.(11.24a), 3 .0 5 5 .2 8 3 .0 5
Fd
F t 2 .7 3 F t
2 0 .8 5 1 0 2 .7 3 F t
F t 7 .6 3 7 k N
3
Equation (1.15) is thus Ft V
kW
1000
7 6 3 7 ( 5 .2 8 )
1000
4 0 .3
SOLUTION (11.41) From Solution of Prob. 11.39: N
g
3 5,
V 5 .2 8
N
p
2 8,
d
p
168 m m ,
b 50 m m ,
mG N
g
N
p
1 .2 5
m ps
Apply Eq. (11.44): c , a ll
ScCLC H KT K
R
Here S c 1 0 2 0 M P a (mid-point, Table 11.12)
K
v
5 . 56
5 . 28
5 . 56
1 . 19
K 0 K s K T 1,
CL C
f
1,
C H 1 (Eq.11.46)
(Curve A, Fig.11.15) K
R
1. 2 5
(Table 11.9) (CONT.)
192
11.41 (CONT.) Thus c , a ll
1 0 2 0 (1 )(1 )
816 M Pa
(1 )(1 .2 5 )
(Eq.11.44)
We find that C
p
191 M Pa
C
f
1, K m 1 . 3
I
sin cos
mG
2
m G 1
(from Table 11.11) (from Table 11.6)
0 . 161
35 28 35
0 . 089
Hence, from Eq.(11.45): Ft (
c , a ll
C
p
)
2
6
( 8 1 6 1 03 ) 1 9 1 1 0
1 KsKv
2
bd Ks
I KmC
f
( 0 .0 5 )( 0 .1 6 8 ) 0 .0 8 9 1 1 (1 .1 9 ) 1 1 .3 (1 )
8 .8 2 k N
and kW
Ft V 1000
8 8 2 0 ( 5 .2 8 ) 1000
4 6 .6
End of Chapter 11
193
CHAPTER 12
HELICAL, BEVEL, AND WORM GEARS
SOLUTION (12.1) mn
( a ) Use Eqs.(12.1) and (12.2'). m
c o s
p m ( 4 .6 1 9 ) 1 4 .5 1 1 m m ,
4 cos 30
o
p a p c o t 1 4 .5 1 1 c o t 3 0
ta n
( b ) P 1 m 1 4 .6 1 9 0 .2 1 6 m m 5 .4 8 6 in , -1
(c) dp
Nmn
c o s
20 ( 4 ) cos 30
9 2 .4 m m ,
o
p n m n 1 2 .5 6 6 m m
4 .6 1 9 m m ,
40 ( 4 )
dg
cos 30
o
1 ta n n c o s
o
ta n
2 5 .1 3 4 m m 1 ta n 2 5 o cos 30
o
2 8 .3
o
1 8 4 .7 5 m m
Gear (d) 30
Thrust
o
Pinion, R.H. SOLUTION (12.2) ( a ) Apply Eqs.(12.1) through (12.2'). m
mn c o s
3 .1 7 5 cos 30
o
3 .6 6 6 m m , p n m n 9 .9 7 5 m m ,
p m (3 .6 6 6 ) 1 1 .5 1 7 m m , p a p c o t 1 1 .5 1 7 c o t 2 0
( b ) P 1 m 1 3 .6 6 6 0 .2 7 3 m m 6 .9 3 4 in . , ta n -1
(c) dp
Nmn c o s
1 8 ( 3 .1 7 5 ) cos 20
o
6 0 .8 1 8 m m ,
dg
5 5 ( 3 .1 7 5 ) cos 20
o
1 ta n n c o s
o
ta n
3 1 .6 4 m m 1 ta n 1 4 .5 o cos 20
o
1 5 .3 9
o
1 8 5 .8 3 2 m m
Gear (d)
Pinion, L.H. SOLUTION (12.3) ( a ) d N m 3 0 (3 .1 7 5 ) 9 5 .2 5 m m , m n m c o s 3 0 2 .7 5 m m , o
p m 9 .9 7 5 m m
p n p c o s 9 .9 7 5 c o s 4 0
o
7 .6 4 m m
p a p c o t 9 .9 7 5 c o t 4 0
o
1 1 .8 9 m m
(CONT.)
194
12.3 (CONT.) ( b ) Pn 1 m n 1 2 .7 5 0 .3 6 4 m m 9 .2 5 in . ta n
1
ta n n
(
) ta n
c o s
1
o
( ta n 1 4 .5o ) 1 8 .6 5
-1
o
cos 40
SOLUTION (12.4) rs
c
1 4
Pn
N
2n
N
p
N
g
N
p
g
c o s
18 Ng
;
N
g
1 5 .6 2 5 1 8 7 2 2 c o s
250
;
72
Solving, c o s 0 .8 9 5,
2 6 .5
Comment: The helix angle of 2 6 .5
o
o o
o
and
P N d . Thus
is in usual range of 1 5 to 3 0 .
SOLUTION (12.5) cos P Pn where m n 1 Pn
Equation (12.2): Nmn
cos
d
32( 6 ) 260
4 2 .4
0 .7 3 8 5 ,
o
F t F n c o s n c o s 1 0 (c o s 2 0 )(c o s 4 2 .4 ) 6 .9 4 N o
kW
Thus
( dn ) F t 60
o
( 0 . 26 )( 800 )( 6 . 94 ) 60
(Eq.12.10)
75 . 58
SOLUTION (12.6) 1
( a ) n ta n
(ta n c o s ) ta n
1
o
o
(ta n 2 0 c o s 3 0 );
n 1 7 .5
o
N ' N c o s 3 5 c o s 3 0 5 3 .8 9 3
3
o
( b ) A super gear of equal strength would have 53.89 teeth and 1 7 .5
o
SOLUTION (12.7) 1 3
N
p
N
g
40 Ng
,
N
Equation (12.5) c
Pn 2
g
3(4 0 ) 1 2 0
P Pn c o s :
with
N
p
N
c o s
g
14 2
40 120 cos15
o
3 6 9 .1 m m
SOLUTION (12.8) ( a ) n ta n
1
(ta n c o s ) ta n
1
o
o
(ta n 2 0 c o s 2 2 );
n 1 8 .6
N ' N c o s 3 5 c o s 2 2 4 3 .9 1 teeth 3
3
o
( b ) A super gear of equal strength would have 43.91 teeth and 1 8 .6
195
o
o
SOLUTION (12.9) a p a g a 1 P m 1 .5 m m
( a ) Addendum:
p m 1 .5 4 .7 1 2 m m
Circular Pitch: c
1 2
rp
(d 1 2
dg)
p
N Pm
1 2
N
1 2
p
N
g
P
1 2
(N
p
N g )m
( 2 0 )(1 .5 ) 1 5 m m ,
rg
1 2
1 2
( 2 0 1 2 0 )(1 .5 ) 1 0 5 m m
(1 2 0 )(1 .5 ) 9 0 m m
The radii of base circles: rb p r p c o s 1 5 c o s 2 0
1 4 .0 9 5 m m
o
rb g rg c o s 9 0 c o s 2 0 8 4 .5 7 2 m m o
Outside radii: ro p r p a r p m 1 5 1 .5 1 6 .5 m m ro g r g a 9 0 1 .5 9 1 .5 m m
From Eq. (11.14) the contact ratio is therefore Cr
1 p cos
[
( rp a ) ( rp c o s ) 2
1 ( 4 .7 1 2 ) c o s 2 0
[
o
2
(1 6 .5 ) (1 4 .0 9 5 ) 2
2
( rg a ) ( rg c o s ) ] 2
2
(9 1 .5 ) (8 4 .5 7 2 ) ] 2
2
c ta n p
1 0 5 ta n 2 0 4 .7 1 2
o
1 .7 1 5
The total contact ratio is C r t C r rr a 1 .7 1 5
4 0 ta n 3 0 4 .7 1 2
o
6 .6 1 6
( b ) To obtain a total contact ratio of 4.0, the helix angle has to be C ra
b ta n p
ta n
4 .0 1 .7 1 5 2 .2 8 5
2 .2 8 5 ( 4 .7 1 2 ) 40
1 5 .0 7
;
o
SOLUTION (12.10) 299
(a)
153.9
tan
299
1 tan n cos
tan
1 tan 20 o cos 45
27 . 24
o
o
(Eq. 12.3)
Ft F n c o s n c o s 4 5 0 c o s 2 0 c o s 4 5 o
F a F t ta n 2 9 9 ta n 4 5
o
o
299 N
F r F t ta n 2 9 9 ta n 2 7 .2 4
o
(Eqs.12.10)
1 5 3 .9 N
( b ) m m n c o s 2 .5 c o s 4 5 3 .5 3 6 m m o
d g N g m 6 0 (3 .5 3 6 ) 2 1 2 .1 6 m m , d
Thus
299 N
T p 2 9 9 ( 0 .1 123 1 5 ) 1 6 .9 2 N m
T g 2 9 9 ( 0 .2 122 1 6 ) 3 1 .7 2 N m
196
p
(Eq.12.2') 3 2 (3 .5 3 6 ) 1 1 3 .1 5 m m
(Eq.12.3)
SOLUTION (12.11) T2
2
(a)
1178 T3 B
549
C 1
549
1178
429 3
2
T1
429 T1
9549 kW n1
9 5 4 9 (1 5 )
9 5 .4 9 N m
1500
m m n c o s 6 .3 5 c o s 2 0
o
6 .7 5 8 m m ,
d 1 N 1 m 2 4 ( 6 .7 5 8 ) 1 6 2 .1 9 m m
(Eq.12.2)
We have ( rs ) 1 2
24 36
d1
( rs ) 1 3 0 .5 Ft1
T1
d1 2
d1
2 3
d3
d2
d2
,
d3
,
1 6 2 .1 9 2 3
1 6 2 .1 9 0 .5
2 4 3 .3 m m
3 2 4 .4 m m
1 .1 7 8 k N F t 2 F t 3
9 5 .4 9 0 .1 6 2 .1 9 2
F r 1 F t 1 ta n 1 .1 7 8 ta n 2 5
F a 1 F t 1 ta n 1 .1 7 8 ta n 2 0
( b ) T1 9 5 .4 9 N m ,
5 4 9 N Fr 2 Fr 3
o
o
429 N
T2 0 ,
(Eqs.12.10)
T 3 9 5 .4 9 ( 12 ) 1 9 1 N m
SOLUTION (12.12) Equations (12.7b) and (12.2'): N ' d
p
N
p 3
cos
2 9 .6 , m m n c o s 3 .1 2 c o s 2 5
22 3
cos 25
o
2 2 ( 3 .4 4 )
N pm
3
cos 25
o
o
3 .4 4 m m
1 0 1 .6 6 m m
Table 11.4: o 1 7 2 M P a
Table 11.3: Y 0 .3 5 7 , We have 0b Ym n
Fb
K
1
f
d pnp
V
60
1 7 2 ( 5 0 ) 0 .3 5 7 ( 3 .1 2 ) 1 .5
1
( 0 .1 0 1 6 6 )(1 8 0 0 ) 60
6 .3 8 6
kN
(Eq.11.33, modified)
9 .5 8 m p s
Also Fd
or and
5 .5 6
9 .5 8
5 .5 6
Fb F d : Ft
1000 kW V
F t 1 .5 5 7 F t
(1)
6 .3 8 6 1 .5 5 7 F t ,
1000 ( 22 ) 9 .5 8
2 .2 9 6 k N
Hence, by Eqs.(2) and (3): n
4 .1 0 1 2 .2 9 6
1 .7 9
197
F t 4 .1 0 1 k N
(2) (3)
SOLUTION (12.13) ( a ) Speed ratio: 37 05
5
50 25
Center distance is c m(N
p
mn
Ng)
c o s
(N
p
Ng)
(1)
Thus, for equal center distance: (3 0 7 5 )
4 c o s 2 9 .8
(b)
4 c o s 2 9 .8
o
(3 0 7 5 )
5 .5 c o s
3 1 .5
( 2 5 5 0 );
5 .5 c o s
5 3 .8
( 2 0 3 2 );
o
o
SOLUTION (12.14) ( a ) ta n
1 ta n n
ta n
c o s
1 ta n 2 5 o cos 30
m m n co s 4 co s 3 0 d
p
2 8 .3
o
o
o
4 .6 2 m m
N p m 2 2 ( 4 .6 2 ) 1 0 1 .6 m m
( b ) V d n ( 0 .1 0 1 6 )( 2 4 0 0 6 0 ) 1 2 .8 m p s Ft
1000 kW V
1 0 0 0 (1 .5 )
117 N
1 2 .8
( c ) F r F t ta n 1 1 7 ta n 2 8 .3 6 3 N o
F a F t ta n 1 1 7 ta n 3 0
Fn
Ft c o s n c o s
117 o
cos 25 cos 30
o
o
6 7 .5 N
1 4 9 .1 N
SOLUTION (12.15) Refer to Solution of Prob.12.12. We now have Equation (11.40): Q and Fw
d pbQ K 2
cos
2( 40 ) 22 40
40 31
, Table 11.10: K 0 .9 0 3 M P a
(1 0 1 .6 6 )(5 0 )( 43 01 )( 0 .9 0 3 )(
Equation (1) for F w F d :
) 7 .2 1 k N
1 2
cos 25
o
7 .2 1 1 .5 5 7 F t ;
(Eq.11.38, modified)
F t 4 .6 3 1 k N
(4)
Thus, by Eqs.(3) and (4): n
4 .6 3 1 2 .2 9 6
2 .0 2
SOLUTION (12.16) Equation (12.2'), (12.4) and (12.7b): m m n c o s 4 .2 c o s 3 5
o
5 .1 2 7 m m ,
d 2 6 5 ( 5 .1 2 7 ) 3 3 3 .3 m m ,
N1
N1
'
3
cos
d 1 N 1 m 3 0 (5 .1 2 7 ) 1 5 3 .8 m m
30 3
cos 35
o
5 4 .5 8
(CONT.)
198
12.16 (CONT.) and Table 11.4: 0 1 2 4 M P a
Table 11.3: Y 0 .4 8 3 , Thus Ym n
Fb 0 b
0 .4 8 3 ( 4 .2 )
1 2 4 (3 8 )
1
9 .5 6 k N
1
(Eq.11.33, modified)
From Table 11.10: K 0 . 3 5 2 M P a We have the quantities: 2N
Q
N
5 . 56
Fd
2
cos 35
5 . 56
4 .1 9 k N
o
(Eq.11.38, modified)
1 9 .3 3 m p s
60 19 . 33
F t 1 . 791 F t
4 .1 9 1 .7 9 1 F t ,
Hence
(Eq.11.40)
1 5 3 .8 ( 3 8 ) (1 .3 6 8 ) ( 0 .3 5 2 )
( 0 .1 5 3 8 )( 2 4 0 0 )
60
1.3 6 8
30 65
2
cos
d 1 n1
V
g
d 1b Q K
Fw
and
2( 65 )
g
N
p
(Eq.11.24c, modified)
F t 2 .3 4 k N
Therefore Ft V
kW
2 ,3 4 0 (1 9 .3 3 )
1000
1000
4 5 .2
SOLUTION (12.17) ( a ) Equations (12.2'), (12.5), and (12.7b): rs
1 3
N1
N
d1
c
,
d2
2
(N1 N 2 )
m 2
or
N 1 44
and
d 1 4 4 (1 .7 ) 7 4 .8 m m ,
N1 '
N1 cos
3
and N
3
30
1 .7 2
(N1 3N1)
132
2
d 2 2 2 4 .4 m m
74 . 8 mm , m n m cos (1 . 7 ) cos 30
44 cos
150
o
Using 69.28 teeth, Y 0 .4 2 8 (interpolated, Table 11.3). Fb
0b Ym n K
1
f
1 7 2 ( 6 4 ) 0 .4 2 8 (1 .4 7 ) 1 .4
1
4 .9 5 k N
o
1 . 47 mm
By Table 11.4: 0 1 7 2 M P a (Eq.11.33, modified)
Similarly, Table 11.10: K 0 .5 4 5 M P a and Q
2N N
g
N
g
d 1b Q K
Fw
V
p
2
cos
d 1 n1 60
5 .5 6
2 ( 132 ) 44 132
1 .5
( 7 4 .8 ) ( 6 4 ) (1 .5 ) ( 0 .5 4 5 ) 2
cos 30
( 0 .0 7 4 8 )( 9 0 0 ) 60
Also
Fd
and
4 .9 5 1 .3 3 7 F t ,
3 .5 2
5 .5 6
o
(Eq.11.40) 5 .2 2 k N
(Eq.11.38, modified)
3 .5 2 m p s
F t 1 .3 3 7 F t
(Eq.11.24c, modified)
F t 3 .7 k N
Thus kW
Ft V 1000
3 7 0 0 ( 3 .5 2 ) 1000
1 3 .0 2
(CONT.)
199
12.17 (CONT.) ( b ) Equation (11.36): St K
a ll
L
KT KR
We have S t 2 1 0 M P a (by interpolation of mid-point values, Table 11.7) K
L
1
K
R
0 . 85
KT 1
(indefinite life) (Table 11.9)
and
a ll
2 1 0 (1 )
247 M Pa
(1 )( 0 .8 5 )
Equation (11.35): bJm K0K sK mKv
Ft
(where a ll )
where K 0 1.2 5
Ks 1
3 .5 6
Kv
3 .5 2
3 .5 6
K m 1.4
1 .2 4 (Fig.11.15)
(Table 11.6)
J 0 .4 8 (Fig.12.6a) J multiplier 1 . 01
(Fig.12.6b)
and J 0 .4 8 ( 1.0 1 ) 0 .4 8 5
Hence
Ft
2 4 7 ( 6 4 )( 0 .4 8 5 )(1 .7 ) (1 .2 5 )(1 )(1 .4 )(1 .2 4 )
6 kN
Then kW
Ft V 1000
6 0 0 0 ( 3 .5 2 ) 1000
2 1 .1
SOLUTION (12.18) ( a ) d p N p m 2 0 (3 ) 6 0 m m , ( b ) p ta n
1 20 42
2 5 .4 6 , o
g ta n
d g 4 2 (3 ) 1 2 6 m m 1 42 20
6 4 .5 4
o
(Eq.12.12a)
(Eq.12.12b)
(c) 2
60
1
L [ 6 0 1 2 6 ] 2 2 1 3 9 .6 m m
L
2
Equation ( a ) of Sec.12.6:
p
g
L 3
46 . 5 mm ;
Thus
b 30 m m
126 ( d ) Equation ( b ) of Sec.12.6: c 0 .1 8 8 m 0 .0 0 5 0 .1 8 8 ( 3 ) 0 .0 0 5 0 .5 6 9 m m
200
10 m 30
SOLUTION (12.19) Equation (12.13): d p 2 0 0 m m , ta n
g
N
g
N
p
2;
1 rs
d
dg
g 6 3 .4 3 ,
o
p
rs
p
200 0 .5
400 m m
9 0 6 3 .4 3 2 6 .5 7
o
Therefore rp ,avg rp
b 2
s in
r g , a v g rg
b 2
s in g 2 0 0 3 2 .5 s in 6 3 .4 3 1 7 0 .9 m m
Vp
d
p ,a v g
np
o
8 5 .4 6 m m
o
60
1 0 0 3 2 .5 s in 2 6 .5 7
p
( 0 .1 7 0 9 )( 5 0 0 ) 60
4 .4 7 m p s
Hence Ft
1 0 0 0 (1 1 )
1000 kW Vp
4 .4 7
2 .4 6 k N
Equations (12.17): F a p F t ta n s in
p
2 .4 6 (ta n 2 0 )(s in 2 6 .5 7 ) 4 0 0 N F r g
F r p F t ta n c o s
p
2 .4 6 (ta n 2 0 )(c o s 2 6 .5 7 ) 8 0 0 .8 N F a g
o
o
2460
800.8
o
o
Gear
400 800.8
Pinion SOLUTION (12.20)
From Solution of Prob.12.19: V p 2 .4 7 m p s ,
F t 2 .4 6 k N ,
b 65 m m
Thus Fd N
p
3 .0 5 4 .4 7 3 .0 5
d
( 2 .4 6 ) 6 .0 7 k N
m 2 0 0 3 .5 5 7 ,
p
Hence, from Table 11.3:
(Eq.11.24a) N
p
'
N
p
cos
p
57 c o s 2 6 .5 7
(using N p ' 6 3 .7 , interpolated)
Y 0 .4 2 4
0 124 M Pa
Also, from Table 11.4: Hence, Fb
Since
0b Ym K
f
Fb Fd ,
1
1 2 4 ( 6 5 ) 0 .4 2 4 ( 3 .5 ) 1 .4
1
o
8 .5 4 k N
the gears are safe.
201
(Eq.11.33)
6 3 .7
SOLUTION (12.21) From Solution 12.20:
F d 6 .0 7
kN ,
114,
N
N
p
57,
N
p
' 6 3 .7
Hence N
g
N
p
rs 2N
Q '
N
p
g
57 0 .5
'
' N
2 ( 2 5 4 .9 )
'
g
1 .6
6 3 .7 2 5 4 .9
g
'
N
g
cos
114 c o s 6 3 .4 3
g
o
2 5 4 .9
(Eq.11.40, modified)
From Table 11.10: K 0 .6 9 M P a Thus 0 .7 5 d p b K Q '
Fw
Since
cos
0 .7 5 ( 2 0 0 )( 6 5 )( 0 .6 9 )(1 .6 ) c o s 2 6 .5 7
p
Fw Fd ,
o
1 2 .0 4 k N
(Eq.11.38, modified)
gears are safe.
SOLUTION (12.22) Table 11.10: K 0 .5 4 5 M P a , Table 11.4: 0 1 7 2 M P a ,
N
p
' N
c o s 1 3 0 c o s 2 0 3 1 .9 3 o
p
Table 11.3: Y 0 . 364 (using N p ' 31 . 93 , interpolated)
We have r1
N 1m
2
1 tan
30 (8 )
1 120 240
r1 , a v g 1 2 0 V1
d 1 , a v g n1
N2
26 . 6 ,
2
60 (8 ) 2
60
60 cos 63 . 4
o
240 m m
90 26 . 6 63 . 4
s in 1 1 0 4 .3 m m ,
( 0 .2 0 8 7 )( 7 2 0 )
cos 2
r2
o
70 2
60
N2'
120 m m ,
2
o
(Eq.12.12b)
r2 , a v g 2 4 0
70 2
s in 2 2 0 8 .7 m m
7 .8 7 m p s
134 ,
N 1'
30 cos 26 . 6
o
33 . 6
(Eq.12.15)
Hence Fb
Q'
0b Ym K
1 7 2 ( 7 0 ) 0 .3 6 4 ( 8 )
1
f
2N2' N 1 ' N 2 '
1 .5
1 .6
1
2 3 .4 k N
(Eq.11.33)
(Eq.11.40, modified)
It follows that Fw
0 .7 5 d 1 b K Q ' cos 1
0 .7 5 ( 2 4 0 )( 7 0 )( 0 .5 4 5 )(1 .6 ) c o s 2 6 .6
o
1 2 .2 9 k N
Since F w F b , power capacity depends on F w . We find that Fd
and
3 . 05 7 . 87 3 . 05
F t 3 . 58 F t
1 2 .2 9 3 .5 8 F t ,
(Eq.11.24a)
F t 3 .4 3 k N
Therefore kW
Ft V 1000
3 4 3 0 ( 7 .8 7 ) 1000
27
202
(Eq.11.38, modified)
SOLUTION (12.23) ( a ) d p N p m 30(4) 120 m m , d
g
ta n
1
1
dg
( d ) ta n p
p ,a v
d
p
V av d
Ft
b s in
o
1 2 0 4 5 s in 1 8 .4
p
7 4 5 .7 ( 4 0 )
Vav
( 13 26 00 ) 7 1 .6 ,
1500 500
360 m m
9 0 7 1 .6 o
p
o
o
1 8 .4
o
1 0 5 .8 m m
n ( 0 .1 0 5 8 )(1 5 0 0 6 0 ) 8 .3 1 m p s
p ,av
7 4 5 .7 h p
d g (1 2 0 )
3 .5 9 k N
8 .3 1
( b ) F a F t ta n s in
p
3 .5 9 ta n 2 0 s in 1 8 .4
F r F t ta n c o s
p
3 .5 9 ta n 2 0 c o s 1 8 .4
o
o
0 .4 1 2 k N
o
o
1 .2 4 k N
Equation (1.16): Tp
9549 kW
Tg
9549 kW ng
np
9549 (30 )
9549 (30 )
1500
500
191 N m
573 N m
SOLUTION (12.24) r1
1 2
r2
1 2
mN mN
1 ta n
1
2
( 8 . 5 )( 30 ) 127 . 5 mm
1 2
1 1 2 7 .5 255
o
Table 11.4:
70 2
70 2
sin 26 . 6
sin 63 . 4
172
0
o
(Eq.11.4)
mm
2 6 3.4
2 6 .6 ,
r1 , av 127 . 5
r 2 , av 255
( 8 . 5 )( 60 ) 255
1 2
o
(Eq.12.12b)
111 . 8 mm
o
223 . 7 mm
Table 11.10: K 0 . 545
MPa ,
MPa
Note that number of teeth is as same as in Prob.12.14. Hence, from Solution of Prob.12.22: Q ' 1.6 , Y 0 . 364 Therefore, we have: Fb
0b K
Ym
0 . 75 d 1 bKQ '
Fw
6
1 7 2 (1 0 )( 0 .0 7 ) 1 .5
f
cos 1
( 0 .3 6 4 )( 0 .0 0 8 5 ) 2 4 .8 3 k N
0 . 75 ( 0 . 255 )( 0 . 07 )( 0 . 545 10 cos 26 . 6
o
6
)( 1 . 6 )
13 . 06 kN
(Eq.11.33) (Eq.11.38, modified)
Also V 1 d 1 n1 6 0 ( 0 .2 2 3 6 )( 7 2 0 6 0 ) 8 .4 3 m s Ft
3 .0 5 8 .4 3 3 .0 5
F t 3 .7 6 4 F t ;
1 3 .0 6 3 .7 6 4 F t ,
Hence kW
F V1 1000
3 , 4 7 0 ( 8 .4 3 ) 1000
2 9 .2 5
203
F t 3 .4 7
kN
(Eq.11.24a)
SOLUTION (12.25) Worm: Vg
ta n
d
tan w
g
c
1 2
( d g 2 )wg
Vw
(d
2 )w
p
dw
tan 35
rs
dg dw
o
(Eq.12.21)
rs dw
28 d w
0 . 025
( d w d g ) 150 ,
29 d w 300,
and
p
dw d
300
g
d w 1 0 .3 4 5 m m
d g 2 8 (1 0 .3 4 5 ) 2 8 9 .7 m m
Then L d w ta n w (1 0 .3 4 5 )(ta n 3 5 ) 2 2 .7 6 m m o
Gear: pg
p ng
c o s
1 1 .4 8 m m (Eq.12.1)
9 .4 cos 35
o
But p g p w . Hence
Nw
L pw
2 2 .7 6 1 1 .4 8
1 .9 8 or 2 teeth
(Eq.12.20)
SOLUTION (12.26) (T w ) i
9549 kW n
9549 (30 )
1800
1 5 9 .2 N m
( T w ) o 1 5 9 .2 0 .9 0 1 4 3 .3 N m ng nw
N
w
N
g
1800 ( 802 ) 45 rpm
(T g ) o (T w ) o
nw
(Eq.12.18)
1 4 3 .3 ( 1 84 05 0 ) 5 .7 3 2
ng
kN m
Thus (Tg )0
Ft
d 2
5 .7 3 2 0 .2 5 2
4 5 .8 6 k N
SOLUTION (12.27) n 3 ( 22 00 )( 550 )( 2 5 0 ) 2 5 rp m
Gear 3 rotates
counterclockwise (negative) at 25 rpm.
SOLUTION (12.28) ( a ) N g N w 40 2 20,
d g 4 0 (3 .2 ) 1 2 8 m m ,
L 2 (1 0 .0 5 ) 2 0 .1 m m ,
(b)
ta n
c
1 2
L
dw
d w 3 .5 (1 0 .0 5 ) 3 5 .1 8 m m
2 0 .1
( 3 5 .1 8 )
(d w d g )
1 2
p (3 .2 ) 1 0 .0 5 m m
,
1 0 .3
o
(3 5 .1 8 1 2 8 ) 8 1 .5 9 m m
204
SOLUTION (12.29) ( a ) dw c
2 210
0 .8 7 5
ta n
L
dw
0 .8 7 5
m ( 5 3 .8 )
2 5 3 .8 m m ,
6 5 3 .8
6 .3 6
;
d g 2 c d w 4 2 0 5 3 .8 3 6 6 .2 m m
o
( b ) V w d w n w ( 0 .0 5 3 8 )(1 2 0 0 6 0 ) 3 .3 8 m s where d g 3 6 6 .2 m m ,
b 25 m m
K w 1 5 0 p s i 1 5 0 ( 6 8 9 5 ) 1 .0 3 4 M P a
(Table 12.2)
Thus, F w (3 6 6 .2 )( 2 5 )(1 .0 3 4 ) 9 .4 7 k N
SOLUTION (12.30) p g m ( 4 ) 1 2 .5 7 m m p w
( a ) L N w p w 4 (1 2 .5 7 ) 5 0 .2 8 m m ( b ) ta n
1
L
dw
ta n
1 5 0 .2 8 (60 )
1 4 .9 4
(Eq.12.20)
o
(Eq.12.21)
( c ) d g N g m 90(4) 360 m m c
(d w d g )
1 2
1 2
(6 0 3 6 0 ) 2 1 0 m m
SOLUTION (12.31) (a) rs c
1 20 1 2
pg
N
w
N
g
,
N
g
3( 2 0 ) 6 0
(d g d w ) 175 d g N
( 275 )
(d g 75)
d g 275 m m
14 . 4 mm p w
60
g
1 2
Then, Eqs.(12.20) and (12.21): L p w N w 1 4 .4 (3) 4 3 .2 m m
tan
1
L
d w
tan
1
43 . 2
( 75 )
10 . 39
o
(b) Vw
d w nw 60
Fw t F ga
( 0 .0 7 5 )(1 0 0 0 ) 60 1000 kW Vw
3 .9 3 m p s
1 0 0 0 ( 7 .5 ) 3 .9 3
1 .9 1 k N
(CONT.)
205
12.31 (CONT.) (c) Vs
Vw
cos
4 m ps
3 .9 3 c o s 1 0 .3 9
o
f 0 . 024
Hence
(Eq.12.28)
(from Table 12.3)
Thus, Eq.(12.27): c o s n f ta n
e
o
o
o
o
c o s 2 0 0 . 0 2 4 ( ta n 1 0 . 3 9 )
cos n f cot
c o s 2 0 0 . 0 2 4 ( c o t 1 0 .3 9 )
0 .8 7 4
or
8 7 .4 %
Power delivered to machine ( k W ) m 0 .8 7 4 ( 7 .5 ) 6 .5 6 k W
SOLUTION (12.32) p w 1 4 .4 m m ,
From Solution of Prob.12.31:
pn
We have Table 11.4:
Fb
p w c o s
1 4 .4 c o s 1 0 .3 9
0 172 M Pa 0b Ym n K
1
f
o
0 .2 2 2
o
mn
d g 275 m m
4 .5 1 m m
1 pn
Table 12.1: Y 0 .3 9 2
1 7 2 ( 3 8 ) 0 .3 9 2 ( 4 .5 1 ) 1 .4
1 0 .3 9 ,
1
8 .2 5 k N
(Eq.11.33, modified)
Also Vg
T Ft
Hence
Fd
d g ng 60
9540 kW ng
T dg 2
6 .1 V g 6 .1
( 0 .2 7 5 )( 5 0 )
9 5 4 0 ( 7 .5 )
60
50
1 .4 3 0 .2 7 5 2
0 .7 2 m p s
1 .4 3 k N m
1 0 .4 k N
6 .1 0 .7 2 6 .1
F t 1 .1 2 (1 0 .4 ) 1 1 .6 5 k N
(Eq.11.24b)
Since Fb Fd
gears are not safe.
SOLUTION (12.33) From Solution of Prob.12.32: Table 12.2:
K
w
F d 1 1 .6 5 k N
1 . 295 MPa
Therefore F w d g b K w 2 7 5 (3 8 )(1 .2 9 5 ) 1 3 .5 3 k N
Since
Fw Fd
gears are safe.
206
(Eq.12.22)
SOLUTION (12.34) We have c 1 7 5 m m 7 in . and p o w e r 7 .5 k W 1 0 h p Equation (12.24): A 0 .3( 7 )
1 .7
8 .2 ft
2
e 8 7 .4 % , (from Solution of Prob.12.31)
By Fig.12.17:
o C 5 6 ft lb m in ft F 2
Hence, Eqs. (12.25) and (12.26):
Since
(hp )d
C At 3 3 ,0 0 0
( hp ) i
( hp ) d 1 e
5 6 ( 8 .2 )(1 0 0 ) 3 3 ,0 0 0
1 . 392 1 0 . 874
1 .3 9 2 h p = 1 .0 4 4 k W
11 . 05 8 . 288 kW
( hp ) i 10 hp or ( k W ) i 7 .5 k W , overheating will not be a problem.
End of Chapter 12
207
CHAPTER 13
BELTS, CHAINS, CLUTCHES, AND BRAKES
SOLUTION (13.1) ( a ) w (5 )(1 0 0 )(1 0 , 8 0 0 ) 5 .4 N m V
dn
( 0 .1 2 5 )(1 5 0 0 )
60
60
F1 F 2
Fc
w g
2
V
5 .4 9 .8 1
[
7 6 3 .7 N
9 .8 2
( 0 .1 2 5 )(1 5 0 0 ) 60
2 1 8 0 2 s in o
e Thus
f
e
0 .3 ( 2 .9 7 8 )
1500
4 7 .7 5 N m
9 .8 2 m p s
1 0 0 0 ( 7 .5 )
1000 kW V
9 5 4 9 ( 7 .5 )
T1
] 5 3 .0 5 N
1
2
T1 r1
(Eq.13.13)
( 1 8 71.55 2 56 2 .5 ) 1 7 0 .6
2 .4 4 3
F1 F c ( 1 )
(Eq.13.3)
o
2 .9 7 8 ra d
(Eq.13.21)
5 3 .0 5 ( 12 .4.4 44 33 )
4 7 .7 5 0 .0 6 2 5
1 .3 4 7 k N
(Eq.13.20)
F 2 1, 3 4 7 7 6 3 .7 5 8 3 .3 N
( b ) L 2 c ( r1 r2 )
( r2 r1 )
2
(Eq.13.9)
c
2 (1, 5 2 4 ) ( 6 2 .5 1 8 7 .5 )
(1 2 5 )
2
1525
3 8 4 5 .6 m m .
SOLUTION (13.2) ( a ) T1
9549 kW
9549 ( 10 )
n1
2800
( b ) T1 ( F1 F 2 ) r1 , r1 r2
n1 n2
34 . 1 N m
F1 F 2
T1 r1
F 2 2 2 7 .3 F 2
3 4 .1 0 .1 5
n
r2 r1 ( n 1 ) 0 .1 5 ( 12 68 00 00 ) 0 .2 6 2 5
,
(Eq.13.19)
m
2
r2 r1
s in
c
0 . 2 6 2 5 0 .1 5 0 .7
2 2 . 819
9 .2 5
0 .1 6 1 ,
o
(Eq.13.6)
rad
( c ) V d n 6 0 ( 0 .3)( 2 68 00 0 ) 4 3 .9 8 m p s w 25 , 000 ( 0 . 06 0 . 0005 ) 0 . 75 N m
Fc
w g
V
2
0 . 75 9 . 81
T1
F1 F c ( 1 )
r1
( 43 . 98 )
2
147 . 9 N ,
1 4 7 .9 ( 10 .7.7 55 77 )
3 4 .1 0 .1 5
e
0 . 2 ( 2 . 819 )
6 7 5 .5 N
1 . 757
(Eq.13.20)
F 2 6 7 5 .5 2 2 7 .3 4 4 8 .2 N
We have K s 1.5 (Table 13.5). Therefore F max 1 . 5 ( 675 . 5 ) 1 . 013
max
1 , 013 60 0 . 5
kN
33 . 77 MPa
208
(Eq.13.22)
SOLUTION (13.3) We have r2 2 0 0 m m and f 0 .2 5 . The remaining data are the same. Equation (13.6) gives 1
s in [
r2 r1 c
] s in
1
[ 2 01 09 506 2 ] 4 .0 5 8
o
Then 2 1 8 0 2 ( 4 .0 5 8 ) 1 7 1 .9 o
e
f
e
o
( 0 .2 5 ) (1 7 1 .9 ) ( 1 8 0 )
o
2 .1 1 7
and F1 2 1 4 F2 2 1 4
2 .1 1 7
or F1 2 .1 1 7 F 2 2 3 9
(1)
Also F1 F 2 6 2 5 N
Solving, F1 1, 3 9 8 .5 N ,
F 2 7 7 3 .5 N
Length of the belt, by Eq. (13.9): L 2 c ( r2 r1 )
1 C
( r2 r1 )
2
2 (1 .9 5 ) ( 0 .2 0 0 0 .0 6 2 )
1 1 .9 5
( 0 .2 0 0 0 .0 6 2 )
2
4 .7 3 3 m
SOLUTION (13.4) Pulley A Using Eq. (13.16), with F1 2 .5 k N , F1 F2
e
f
,
2 .5 F2
e
0 .1 5 (1 2 0 )( 1 8 0 )
f 0 .1 5 ,
1 .3 6 9 ;
Fc 0 ,
120 :
F 2 1 .8 2 6 k N
Pulley B Then, equilibrium condition
T
0 applied to free-body diagram of
pulley B gives
T
B
0;
T B 2 .5 ( 0 .1 5 ) 1 .8 2 6 ( 0 .1 5 ) 0
or TB 1 0 1 N m
B TB
F1 2 .5 k N
0.15 m F 2 1 .8 2 6 k N
Similarly, we have
T
A
0; T A 2 .5 ( 0 .0 2 ) 1 .8 2 6 ( 0 .0 2 ) 0 ,
209
TA 13 N m
o
SOLUTION (13.5) ( a ) V1
2 r1 n1
60
2 ( 0 .0 3 7 5 )( 2 5 0 0 )
9 .8 2 m p s
60
From Eq. (1.15); with F F1 F 2 : kW
FV 1000
1 .5
;
( F1 F 2 ) 9 .8 2 1000
F1 F 2 1 5 2 .7 5 N
(1)
( b ) Evaluate V 2 as, V1
or
2 r2 n 2 60
9 .8 2
;
r2 9 3 .8 in .,
2 r2 (1 0 0 0 ) 60
d 2 1 8 7 .6 m m
and 1
2 2 s in [ 2 s in
1
r2 r1 c
]
[ 9 3 .86 2 53 7 .5 ] 2 .9 6 ra d
( c ) Centrifugal force, using Eq. (13.13), Fc
w g
V
2
1 .7 5 9 .8 1
(9 .8 2 ) 1 7 .2 N 2
Then, Eq. (13.16): F1 1 7 .2
e
F 2 1 7 .2
0 .3 5 ( 2 .9 6 )
2 .8 1 8
(2)
Solving Eqs. (1) and (2), we have F1 2 5 4 N
F 2 1 0 1 .2 N
SOLUTION (13.6) We have w g 8 .8 9 .8 1 0 .8 9 7
and P V ( F1
w g
V ) V ( 2 5 0 0 0 .8 9 7 V ) 2
2
Thus P V
0 5 0 0 0 .8 9 7 (3 )V ; 2
So V 2 r n ; Solving
V 3 0 .4 8 m p s
3 0 .4 8 2 ( r )( 4 2 0 0 6 0 )
r 6 9 .3 m m
SOLUTION (13.7) By Eq. (13.13): Fc
w g
V
2
1 .4 9 .8 1
[ 2 ( 0 .0 8 )(3 0 0 0 6 0 )] 9 0 .1 N 2
From Eq. (13.16), 1 1 0 0 9 0 .1 F 2 9 0 .1
e
o
( 0 .2 5 ) ( s in 1 8 ) (1 6 0 1 8 0 )
1 .2 4 1
Solving F 2 9 0 3 .9 N
(CONT.)
210
13.7 (CONT.) Then T A ( F1 F 2 ) r1 (1 1 0 0 9 0 3 .9 )( 0 .0 8 ) 1 5 .7 N m
and P
3 0 0 0 (1 5 .7 )
VT 9549
9549
4 .9 3 k W
SOLUTION (13.8) Fc
w g
2
V
F1 F c F2 Fc
8 9 .8 1
[ ( 0 .2 )( 1 66 00 0 )] 2 2 8 .9 N 2
f s in
e
3 , 0 0 0 2 2 8 .9
;
F2 2 2 8 .9
e
1 7 0 2 .9 6 7 ra d o
0 .1 5 ( 2 . 9 6 7 ) s in 1 9
o
3.9 2 4
Solving F 2 935 . 1 N Thus T ( F1 F 2 ) r (3 0 .9 3 1) 1 0 ( 0 .1) 2 0 6 .9
N m
3
kW
Tn 9549
2 0 6 .9 (1 6 0 0 )
3 4 .7
9549
SOLUTION (13.9) ( a ) Fc
w g
F1 F c
2
V
3 9 . 81
f s in
e
8 0 0 Fc
[ ( 0 . 3 )(
e
3000 60
)]
2
0 . 2 5 ( 2 . 7 9 3 ) s in 1 9
o
679 . 1 N ,
160
o
2 . 793
rad
8 .5 4
or F1 679 . 1
8 . 54 , F 1 1, 711 . 6 N
800 679 . 1
Also T ( F 1 F 2 ) r1 (1711 . 6 800 )( 0 . 15 ) 136 . 7 N m kW
Tn 9549
136 . 7 ( 3000 ) 9549
42 . 95
(b)
F1
max
A
1 , 711 . 6 6
145 ( 10
)
11 . 8 MPa
SOLUTION (13.10) r2 r1
n1 n2
2700 , r 2 0 . 1 ( 1800 ) 0 . 15 m
sin
T1
9549 kW
e Fc
0 . 15 0 . 1 0 .5
2700
w g
V
o
9 5 4 9 (1 5 ) 2700
0 . 2 ( 2 . 9 4 1 ) s in 1 7
2
5 . 74 ;
,
2 .5 9 . 81
o
180
5 3 .0 5 N m
o
2 ( 5 . 74 ) 2 . 941
rad
V d n ( 0 .2 )( 2 67 00 0 ) 2 8 .2 7 m p s
7 .4 7 7
( 28 . 27 )
2
203 . 7 N
(CONT.)
211
13.10 (CONT.)
Thus F 1 F c ( 1 )
T1 r1
203 . 7 ( 76 .. 477 ) 477
53 . 05 0 .1
816 . 1 N
Table 13.5: K s 1.4 and F max K s F 1 1 . 4 ( 816 . 1) 1 . 143
kN
SOLUTION (13.11) N 1 n1
(a) N2
2 2 (1 4 0 0 )
n2
4 4 teeth
700
( b ) H d H r K 1K 2 where H r 2 6 .6 h p = 1 9 .8 k N K 1 1 .5
N 1 P n1
1000 H
60 1 0 0 0 ( 5 0 .6 )
d
V1
So,
2 2 ( 0 .0 1 9 0 5 )(1 4 0 0 )
60
F1
K 2 1 .7 by Table 13.10)
(from Table 13.9,
H d 2 6 .6 (1 .5 )(1 .7 ) 6 7 .8 h p = 5 0 .6 k W
Hence ( c ) V1
(by Table 13.8)
9 .7 8
n
F a ll
n
6 2 .6 5 .1 7
9 .7 8 m p s
5 .1 7 k N
F a ll 3 1 .3 ( 2 ) 6 2 .6 k N
F1
(by Table 13.7)
and 1 2 .1
SOLUTION (13.12) N 1 n1
(a) N2
n2
1 8 (1 6 0 0 ) 640
4 5 teeth
( b ) H d H r K 1K 2 where H r 1 6 .1 h p = 1 2 k W K 1 1 .2
(from Table 13.8)
(by Table 13.9),
K 2 3 .3 (from Table 13.10)
and H d 1 6 .1(1 .2 )(3 .3) 6 3 .8 h p = 4 7 .6 k W
( c ) V1 N 1 P n1 1 8 ( 0 .0 1 9 0 5 )(1 6 0 0 6 0 ) 9 .1 4 m p s 7 4 5 .7 H
F1
(d) n
d
V1
F a ll F1
,
7 4 5 .7 ( 6 3 .8 ) 9 .1 4
5 .2 1 k N
where F a ll 2 (3 1 .3) 6 2 .6 k N
Thus, r2
6 2 .6 5 .2 1
12
212
(by Table 13.7)
SOLUTION (13.13) n1 n2
4 1
3
Use c 2 ( r2 r1 )
(Sec.13.6)
We have r1
N1P
2
2 2 (1 6 ) 2
5 6 .0 2 m m ,
r2
n1 n2
r1
4 1
(5 6 .0 2 ) 2 2 4 .0 8 m m
Thus c 2 ( 2 2 4 .0 8 5 6 .0 2 ) 3 3 6 .1 2 m m
r1 r2 2 8 0 .1 m m
Since
sprocket will clear.
SOLUTION (13.14) n1 n2
r1
2 . 19 3 Use c 2 ( r1 r2 )
4600 2100 N1P
2
1 4 (1 4 ) 2
3 1 .1 9 m m ,
n1
r2
n2
(Sec. 13.6)
r1 2 .1 9 (3 1 .1 9 ) 6 3 .3 1 m m
c ( r1 r2 ) 3 1 .1 9 6 3 .3 1 9 4 .5 m m c 2 ( 9 4 .5 ) 1 8 9 m m
Use c 1 9 0 m m
SOLUTION (13.15) ( a ) Speed ratio is ; 3:1. Thus N 2 3( 2 3) 6 9 teeth
( b ) H d H r K 1K 2 where H r 1 9 .5 h p
(from Table 13.8)
K 1 1 .3
(from Table 13.9),
K 2 1 .7 (by Table 13.10)
H d 1 9 .5 (1 .3)(1 .7 ) 4 3 .1 h p
and
( c ) V 1 N 1 P n1 2 3( 0 .0 1 9 )(1 8 0 0 6 0 ) 1 .3 1 m p s 7 4 5 .7 H
F1
(d) n
d
V1
F a ll F1
7 4 5 .7 ( 4 3 .1 ) 1 3 .1
2 .4 5 k N
where F a ll 4 (3 1 .3) 1 2 5 .2 k N
,
(Table 13.7)
So, n
1 2 5 .2 2 .4 5
5 1 .1
SOLUTION (13.16) ( a ) N 2 3(3 5 ) 1 0 5 teeth ( b ) H d H r K 1K 2 where
H r 3 6 .6 h p = 2 7 .1 k W
(by Table 13.8) (CONT.)
213
13.16 (CONT.) K 1 1 .3 ,
K 2 1 .7
(by Table 13.9 & 13.10)
and H d 3 5 (1 .3)(1 .7 ) 7 7 .3 5 h p = 5 7 .6 8 k W
( c ) V 1 N 1 P n1 3 0 ( 0 .0 1 9 0 5 )(9 0 0 6 0 ) 8 .5 7 m p s F1
(d) n
7 4 5 .7 H
F a ll F1
7 4 5 .7 ( 7 7 .3 )
d
V1
6 .7 3 k N
8 .5 2
where F a ll 2 (3 1 .3) 6 2 .6 k N
,
Hence n
9 .3
6 2 .6 6 .7 3
SOLUTION (13.17) (a)
p m ax
T
2 Fa
2(6)
d (D d )
( 0 .1 5 )( 0 .2 5 0 .1 5 )
Fa f ( D d )
1 4
1 4
2 5 4 .6 k P a
(Eq.13.28)
( 6 , 000 )( 0 . 3 )( 0 . 25 0 . 15 ) 180
N m
(Eq.13.30)
(b) p max
T
1 3
4 Fa
(D
2
d
2
3
Fa f
D d D
2
)
d
3
4(6) 2
( 0 . 25 0 . 15
2
2
)
191
3
1 3
( 6 , 000 )( 0 . 3 )
(Eq.13.32)
kPa
0 . 25 0 . 15 0 . 25
2
0 . 15
3 2
183 . 8 N m
SOLUTION (13.18) (a) T
9549 kW n
9549 (30 )
500
5 7 2 .9 N m
N 2
and T
1 12
fp
max
3
(D
577 , 267 . 7 d
d ) 3
3
( 0 . 25 )( 140 10 )[ 64 d 3
12
3
d ] 3
(Eq.13.33)
572 . 9 2
Solving d 7 9 .2 m m and D 4 d 3 1 6 .8 m m ( b ) Fa
1 4
p m ax ( D d ) 2
2
4
(1 4 0 1 0 )[1 6 d 3
2
d ] 1 .6 4 9 (1 0 ) d 2
6
2
1 0 .3 4
kN
SOLUTION (13.19) ( a ) T 5 7 2 .9 N m We now have T
1 8
N 2 (from solution of Prob. 13.18)
fp m a x d ( D d )
206 ,167 d
2
3
2
8
( 0 .2 5 )(1 4 0 1 0 ) d (1 6 d 3
2
d ) 2
(Eq.13.29)
572 . 9 2
Solving, d 1 1 1 .6 m m and D 4 d 4 4 6 .4 m m (CONT.)
214
13.19 (CONT.) ( b ) Fa
1 2
p max d ( D d )
( 0 .1 4 )(1 1 1 .6 )[ 4 4 6 .4 1 1 1 .6 ] 8 .2 2 k N
2
(Eq.13.28)
SOLUTION (13.20) ( a ) From Eq. (13.29), with a factor of safety n : d (D
d )
2
2
8 nT
;
fPm a x
2
0 .0 5 ( D
0 .0 0 2 5 )
8 (1 .6 )(1 3 5 .6 ) 6
( 0 .3 )(1 .6 1 0 )
Solving D 1 5 9 .8 m m ( b ) From Eq. (13.30), we have Fa
4 nT f (Dd )
4 (1 .6 )(1 3 5 .6 )
( 0 .3 )(1 5 9 .8 5 0 )
1 3 .8 k N
SOLUTION (13.21) (a) T
9549 kW n
9 5 4 9 ( 3 7 .5 )
8 9 5 .2 N m
400
and T
fp
1 8
0 .8 (1 0
Fa
2
( b ) p avg
d (D
max 3
Fa 2
(D d
2
)
d ) 2
) p m ax
p m ax d ( D d )
2
2
8
( 0 . 2 ) p max 0 . 15 [ 0 . 3
2
0 . 15
2
2
( 0 .3 0 .1 5 )
1 8 6 .6 k P a
4
SOLUTION (13.22) ( a ) Use Eq. (13.29) by multiplying 3 6 0 : T
360
[ 81 fPm a x d ( D
2
d )] 2
from which p m ax
8 (360 )T 2
fd ( D d
2
)
8 ( 3 6 0 8 0 ) (1 8 0 0 )
2
2
( 0 .3 ) ( 0 .2 ) ( 0 .2 8 0 .2 )
8 9 5 2 .5 k P a
( b ) From Eq. (13.28) by multiplying 3 6 0 : Fa
360
[ 12 p m a x d ( D d )]
80 360
[ 2 (8 9 5 2 .5 )( 0 .2 )( 0 .4 8 )]
300 kN
( c ) Each cylinder supplies a force F a 2 . Thus p hyd
Fa 2
d
2
4
3 0 0 ,0 0 0 2
( 0 .2 )
2
4
]
(Eq.13.29)
p m a x 2 7 9 .8 k P a
,
( 0 .2 7 9 8 )(1 5 0 )(1 5 0 ) 9 .8 9 k N
9890 ( 4 ) 2
8 9 5 .2 4
4 .7 7 M P a
215
(Eq.13.28)
SOLUTION (13.23) ( a ) Use Eq. (13.33) by multiplying 3 6 0 :
T
[ 12 fPm a x ( D d )] 3
360
3
from which 12 (360 )T
p m ax
3
3
f (D d )
1 2 ( 3 6 0 8 0 ) (1 8 0 0 ) 3
3
( 0 .3 ) ( 0 .2 ) ( 0 .2 8 0 .2 )
7 .3 9 M P a
( b ) From Eq. (13.32), by multiplying by 3 6 0 :
Fa
(c)
p hyd
Fa 2
d
2
360
4
[ 4 p m a x ( D
2 ( 49500 )
( 0 .0 4 )
d )]
2
2
80 360
[ 4 ( 7 , 3 9 0 )( 0 .2 8 0 .2 )] 4 9 .5 k N 2
2
1 9 .7 M P a
2
SOLUTION (13.24) T
9549 (35 ) 800
and T
4 1 7 .8 N m
fp m a x 1 2 s in
d )
3
(D
3
R is e 3
( 0 .3 ) 4 2 0 (1 0 ) 1 2 s in 8
o
1 2
( D d ) w s in
[ 0 .2 5 d ] 4 1 7 .8 3
3
(Eq.13.41)
Solving d 2 4 0 .2 m m Therefore w
1 D d 2 s in
2 5 0 2 4 0 .2 2 s in 8
3 5 .2 1 m m
o
SOLUTION (13.25) T 4 1 7 .8 N m (from Solution of Prob. 13.24)
Now we have T
fp m a x d 8 s in
(D
Solving 0 .0 6 2 5 d d
2
d ) 2
0 .0 0 1 2
3
3
( 0 .3 ) 4 2 0 (1 0 ) 8 s in 8
o
[ 0 .0 6 2 5 d d ] 4 1 7 .8 3
or d 2 4 0 m m (by trial and error)
Thus, we have w
1 D d 2 s in
250 240 2 s in 8
o
3 5 .9 m m
SOLUTION (13.26) ( a ) Rise w sin 80 sin 10
o
13 . 89 mm
D 5 0 0 1 3 .8 9 5 1 3 .8 9 m m d 5 0 0 1 3 .8 9 4 8 6 .1 1 m m
Equation (13.38): T
( 0 . 2 )( 500 )( 0 . 48611 ) 8 sin 10
o
[ 0 . 51389
2
0 . 48611
2
] 3 . 05 kN m
Equation (13.37): F a 12 (5 0 0 )( 0 .4 8 6 1 1)[ 0 .5 1 3 8 9 0 .4 8 6 1 1] 1 0 .6 1 k N ( b ) kW
Tn 9549
3 ,0 5 0 ( 5 0 0 ) 9549
1 5 9 .7
216
(Eq.13.38)
SOLUTION (13.27) Fa
4
2
p max ( D
d ), 2
D
2
d
4 Fa
2
p max
0 . 016
4(5)
( 400 )
We have 1 2
( D d ) 0 .2 5
D d 0 .5
or
(1)
Thus d
2
D
2
( D d )( D d ) 0 .5 ( D d ) 0 .0 1 6
or D d 0 .0 3 2
(2) D 266
From Equation (1) & (2):
d 234
mm ,
mm
Equation (13.41): T
5000 ( 0 . 2 ) 0 . 266
3
0 . 234
3
o
2
0 . 234
2
3 sin 12
0 . 266
602
N m
SOLUTION (13.28) d Fn
Fn
d/2
D/2
r
D 2
d 2
p max
F h 0:
2 rdr sin
F a F n s in
D 2
d 2
p max (
2 rdr sin
) sin
Fa
This, after integration, yields Eq.(13.40).
T fF n r
Similarly,
D 2
d 2
fp
max
(
2 rdr sin
)r
Integrating gives Eq.(13.41).
SOLUTION (13.29) F 1 wrp
max
( 0 . 1)( 0 . 2 )( 700 ) 14 kN
f 0 .3 ( 2 4 0
2 360
(Eq.13.45)
) 1.2 5 7
We have F2
F1 e
f
14 e
1 .2 5 7
3 .9 8 3 k N
Thus T ( F 1 F 2 ) r (14 3 . 983 )( 0 . 2 ) 2 . 003
and
kW
Tn 9549
2 , 0 0 3(1 5 0 ) 9549
3 1.4 6
217
kN m
SOLUTION (13.30) 2 1 0 3 .6 6 5 ra d
T I 2 .3( 2 0 0 ) 4 6 0 N m
o
We have F1 F 2 F1 F2
e
f
T r
e
460 0 .1 2 5
3 .6 8 k N
0 .3 ( 3 . 6 6 5 )
(Eq.13.42)
3 .0 0 3
Solving F 1 3 . 003 F 2 3 . 003 ( F 1 3 . 68 )
or F1 5 .5 2 k N
F 2 1 .8 4 k N
Thus Fa F2
1, 8 4 0 ( 13 20 50 ) 7 6 7 N
r a
(Eq.13.43)
SOLUTION (13.31) T
9549 kW n
9549 ( 40 )
636 . 6 N m
600
(1)
Also F1 F 2 e
f
F2 e
5 .7 2 7 F 2
0 . 4 ( 4 .3 6 3 )
and T r ( F1 F 2 ) 0 .2 5 ( 5 .7 2 7 F 2 F 2 ) 1.1 8 2 F 2
Equation (1) and (2) give F 2 538 . 6 N
F 1 3 , 085
N
SOLUTION (13.32) 240
o
4 . 189
rad
(a) F 1 p max wr 600 ( 0 . 075 )( 0 . 15 ) 6 . 75 kN
Also F1 F2
e
f
e
0 . 4 ( 4 .1 8 9 )
5 .3 4 2
Thus F1 5 .3 4 2 F 2
Fa F2
r a
and
F 2 1 . 264
(1 . 264 10 ) 3
150 400
474
( b ) T r ( F 1 F 2 ) 0 . 15 ( 6 . 75 1 . 264 ) 10 kW
Tn 9549
823( 200 ) 9549
kN
1 7 .2 4
218
3
N
823
N m
(2)
SOLUTION (13.33) (a) F 1 p max wr 800 ( 0 . 06 )( 0 . 15 ) 7 . 2 kN T
9549 ( 10 )
9549 kW n
434
220
T ( F1 F 2 ) r ,
N m
F 2 F1
7 .2
T r
0 . 434 0 . 15
4 . 307
kN
We have e
f
F1 F2
f ln 1.6 7 2 0 .5 1 4
1.6 7 2 ,
7 .2 4 .3 0 7
Thus
0 .5 1 4 0 .1 4
(b)
3 .6 7 1 ra d 2 1 0 .3
o
129.4 From triangle ABC: 3 0 .4
s 2 0 6 .6 c o s 3 0 .4 178 . 2 mm
o
o
O 150
75.9
200 124.1 3 0 .4
o
C
A s B 206.6
72.8
SOLUTION (13.34) e
f
e
F1 e
0 . 12 ( 3 . 665 )
f
1 . 552
210
o
3 . 665
rad
F 2 1.5 5 2 F 2
F2
A
M
A
0:
F1 s F 2 c F a a
F1
or 1 . 552 F 2 ( 80 ) 50 F 2 1 . 5 ( 300 )
from which F 2 6 .0 7 k N , and F1 9 .4 2 k N
Thus kW
( F1 F 2 ) r n 9549
( 9 .4 2 6 .0 7 )( 4 )(1 0 0 ) 9549
1 0 .5 2
219
S
Fa
c a
SOLUTION (13.35) T
F1 F 2
F2
900
T r
F1
Also
9 5 4 9 (1 5 )
9549 kW n
f
o
ra d
7 9 5 .7 5 N
1 5 9 .1 5 0 .2
e
2 1 0 3 .6 6 5
1 5 9 .1 5 N m
e
0 .4 ( 3.6 6 5 )
(1)
4 .3 3 2
or F1 4 .3 3 2 F 2
(2)
From Eqs.(1) and (2): F 2 2 3 8 .8 N
F1 1 0 3 4 .6
N
We therefore obtain Fa
( cF 2 sF 1 )
1 a
1 250
111 . 4 N
[100 ( 238 . 8 ) 50 (1034 . 6 )]
Yes. Self-locking
SOLUTION (13.36) From Solution of Prob.13.35: T 159 . 15 ,
F 2 238 . 8 N ,
F 1 1034 . 5 N
Now we have
A F2
F1
c
Fa
M
A
0:
a
( a ) From Eq. (13.45): F1 p m a x w r 5 0 0 (1 0 )( 0 .0 2 )( 0 .1) 1 k N 3
Equation (13.44), 1(1 0 ) e 3
0 .2 5 ( 2 6 5 1 8 0 )
315 N
Equation (13.42): T r ( F1 F 2 ) ( 0 .1)(1 0 0 0 3 1 5 ) 6 8 .5 N m
( b ) Fa
c F 2 s F1 a
4 0 ( 3 1 5 ) 1 0 (1 0 0 0 ) 200
13 N
If F a 0 , the brake will self-lock: s
c F2 F1
[ c F1 s F 2 ]
No. Not self-locking
SOLUTION (13.37)
f
1 a
366 . 04
S
F 2 F1 e
Fa
40 (315 ) 1000
1 2 .6 m m
220
N
SOLUTION (13.38) ( a ) From Eq. (13.45): F1 w r p m a x 7 5 ( 2 5 0 )( 0 .4 9 ) 9 .1 9 k N
Equation (13.44), F 2 F1 e
0 .2 5 ( 4 .5 3 8 )
9 .1 9 ( 0 .3 2 2 ) 2 .9 6 k N
Using Eq. (13.42): T ( 9 .1 9 2 .9 6 ) ( 2 5 0 ) 1 .5 5 8 N m
( b ) Using Eq. (13.46): 2 .9 6 (1 5 0 ) 9 .1 9 ( 3 5 )
Fa
1 9 5 .8 N
0 .6 2 5
From Eq.(13.46): F a 0 for s 2 .9 6 (1 5 0 ) 9 .1 9 4 8 .3 m m The brake is self-locking (for f 0 .2 5 ), if: s 4 8 .3 m m
SOLUTION (13.39) (a) T
9549 kN n
Fn
T fr
9549 ( 25 )
800
298 . 4 0 . 25 ( 0 . 3 )
Hence F a
Fn
298 . 4 N m
3979
N
( b fc )
a
3979 1
( 0 . 4 0 . 25 0 . 05 ) 1 . 542
kN
No. Not self-locking ( b ) R Ax fF n 0 . 25 ( 3979 ) 994 . 8 N R Ay F n F a 3979 1542 R A [ 994 . 8
2437
2
2437
N
1
2
] 2 2 . 632
kN
SOLUTION (13.40) ( a ) From Solution of Prob.13.39: F n 3979
N
We now have Fa
Fn a
( b fc )
3979 1
( 0 . 4 0 . 25 0 . 05 ) 1, 641
N
No. Not self-locking ( b ) From Solution of Prob.13.39: R A 2 . 632
kN
SOLUTION (13.41)
M
A
0:
Fn
4 ( 0 .4 5 ) 0 .2
9 kN
We now use p avg
Fn 2 ( r s in
)w
9 ,0 0 0 o
2 ( 0 .1 5 )(s in 4 5 )( 0 .0 7 5 )
0 .5 6 6 M P a
2
(CONT.)
221
13.41 (CONT.) T fF n r ( 0 .3 5 )(9 1 0 )( 0 .1 5 ) 4 7 2 .5 N m 3
Thus and
kW
Tn 9549
12 . 37
472 . 5 ( 250 ) 9549
Comment: Short-Shoe analysis overestimates pressure, torque, and power capacity of the brake.
SOLUTION (13.42) ( a ) From Eq.(13.47): F n Pm a x [ 2 ( r s in
2
)] w (8 0 0 )(1 0 )[ 2 ( 0 .1 5 s in 3
and
fF n 0 .2 5 ( 4 4 9 5 ) 1 1 2 4 N
So
M
0:
A
44 2
o
)]( 0 .0 5 ) 4 4 9 5 N
0 .5 F a 0 .0 2 5 (1 1 2 4 ) 4 4 9 5 ( 0 .2 ) 0 ,
F a 1 .7 4 2 k N
and T fF n r 1 1 2 4 ( 0 .1 5 ) 1 6 8 .6 N
(b) R A [( 4 4 9 5 1 7 4 2 ) 1 1 2 4 ] 2
2
1
2 .9 7 4 k N
2
Fa
300
200 A
25
fF n
Fn
SOLUTION (13.43) ( a ) From Eq. (13.48): T fF n r ,
(b)
or
M
Fn
T fr
250 ( 0 .4 )( 0 .3 5 )
1 .7 9 k N
( 0 .3 5 )[ 0 .4 1 .7 9 (1 0 )] ( 0 .9 F a ) 0 .3 2[1 .7 9 (1 0 )] 0 3
o
3
Fa 9 1 5 N
SOLUTION (13.44) Equation (13.47): F n p m a x [ 2 ( r s in
2
)] w 7 0 0[ 2 ( 0 .1 s in
30 2
o
1 .4 5 k N
Then, Eg.(13.49): Fa
Fn a
( b fc )
3
1 .4 5 (1 0 ) 0 .2
[ 0 .1 2 0 .2 ( 0 .0 3 ) ]
827 N
222
)]( 0 .0 4 )
SOLUTION (13.45) ( a ) From Eq. (13.57), T ( s in ) m
p m ax
(1)
2
fw r ( c o s 1 c o s 2 )
Substituting Eq. (13.52) and the data: p m ax
1 3 6 s in 9 0
o
2
o
2 .7 6 M P a
o
( 0 .3 5 ) ( 0 .0 2 5 ) ( 0 .0 7 5 ) ( c o s 0 c o s 9 0 )
( b ) Introducing Eq. (13.52) and the data into Eq. (1); we have p m ax
1 3 6 s in 6 5
o
2
o
o
( 0 .3 5 ) ( 0 .0 2 5 ) ( 0 .0 7 5 ) ( c o s 2 0 c o s 6 5 )
4 .8 4 M P a
SOLUTION (13.46) c (cos 2 2 cos 2 1 ) 4 r (cos 2 cos 1 )
Equation (13.54): (1)
where 1 0 , 2 9 0 Equation (1) thus o
AO c
o
c ( 1 1) 4 r ( 1)
c 2r
or
and b c cos 45
o
2 r cos 45
o
1 . 414 r
SOLUTION (13.47) Refer to Figs. P13.47, 13.22, and 13.23. 1
c [ 2 5 0 1 7 5 ] 2 3 0 5 .2 m m 2
tan
2
1 250 175
1 10 . 01 , 2 100 . 01
55 . 01 ; o
We have 2 9 0
o
o
o
hence ( s in ) m 1.
( a ) Equation (13.53): M
n
6
( 0 .0 5 )( 0 .2 )( 0 .3 0 5 2 )( 0 .9 1 0 )
[ 2 ( 2 ) s in 2 0 0 .0 2 s in 2 0 .0 2 ] 2 6 2 7 .5 N m o
4 (1 )
o
Equation (13.54): M
f
6
( 0 . 3 )( 0 . 05 )( 0 . 2 )( 0 . 9 10 ) 4 (1 )
[ 0 . 3052 (cos 200 . 02
o
cos 20 . 02 ) o
4 ( 0 .2 )(c o s 1 0 0 .0 1 c o s 1 0 .0 1 )] 2 3 8 .5 N m o
o
Equation (13.55): Fa
1 a
(M
n
M
f
)
1 0 .5 5
[ 2 3 8 9 ] 4 .3 4 4 k N
( b ) Equation (13.57): T
2
6
( 0 .3 )( 0 .0 5 )( 0 .2 )( 0 .9 1 0 ) 1
[ c o s 1 0 .0 1 c o s 1 0 0 .0 1 ] 6 2 5 .6 N m o
Thus kW
Tn 9549
625 . 6 ( 600 ) 9549
39 . 31
223
o
SOLUTION (13.48) Refer to Figs.P13.48 and 13.22. 1 1 5 ,
2 120 90
o
c
150 s in 3 0
o
o
300
o
and (s in ) m 1,
105
a 300 cos 30 250 510 o
m,
o
mm
( a ) Equations (13.53), (13.54), and (13.55): 6
M
( 0 .0 6 )( 0 .2 )( 0 .3 )( 0 .8 1 0 )
n
M
( 0 .3 )( 0 .0 6 )( 0 .2 )( 0 .8 1 0 )
f
4 (1 ) 6
4 (1 )
[ 2 (1 .8 3 3 ) s in 2 4 0 s in 3 0 ] 3, 6 2 3 N m o
o
[ 0 .3 ( 0 .5 0 .8 6 6 ) 4 0 .2 ( 0 .5 0 .9 6 6 )] 5 4 9 N m
Thus Fa
(b) T
1 a
(M
n
M
2
6
( 0 .3 )( 0 .0 6 )( 0 .2 )( 0 .8 1 0 ) 1
f
)
1 0 .5 1
[3, 6 2 3 5 4 9 ] 6 .0 2 7 k N
[ 0 .9 6 6 0 .5 ] 8 4 4 N m
(Eq.13.57)
Hence kW
Tn 9549
( 844 )( 500 ) 9549
44 . 19
SOLUTION (13.49) ( a ) We have 2 4 5 , Apply Eq. (13.58) to obtain o
a
4 r s in
2
2 2 s in 2
4 (1 2 5 ) s in 4 5
o
2 ( 4 5 )( 1 8 0 ) s in 4 5
o
1 5 5 .2 m m
( b ) Applying Eqs. (13.59) and (13.60): R Ax
2wr a
2
Pm a x s in 2
2 ( 0 .0 5 )( 0 .1 2 5 ) 0 .1 5 5 2
2
(1 .5 5 1 0 ) s in 4 5 6
o
1 1 .0 3 k N
R A y fR A x 0 .3 (1 1 .0 3 ) 3 .3 1 k N
( c ) From Eq. (13.61b): T R A y a (3 .3 1)( 0 .1 5 5 2 ) 0 .5 1 4 k N m
End of Chapter 13
224
CHAPTER 14
MECHANICAL SPRINGS
SOLUTION (14.1)
(a) J
32
TL GJ
(8 )
4
; 80
402 . 124
mm
1 . 396
rad
o
2
T ( 1 . 25 ) 79 ( 10
9
)( 402 . 124 10
12
)
or T 35 . 48 N m
(b)
16T
d
3
1 6 ( 3 5 .4 8 )
( 8 1 0
3
)
353 M Pa
3
SOLUTION (14.2) (a) d
3
3
1 6 ( 2 1 0 )( 0 .1 5 )
16 PR
a ll
( 3 5 0 1 0
6
d 1 8 .7 1 m m
,
1 .5 )
( b ) Equation (14.2): L
4
d G
4
(1 8 .7 1 ) ( 7 9 )( 4 0 )
2
32 PR
3
3 2 ( 2 1 0 )(1 5 0 )
0 .8 4 5 m 8 4 5 m m
2
SOLUTION (14.3) Angle of twist, T L J G ;
T J G L , where
2 0 ( 1 8 0 ) 0 .3 4 9 ra d o
J d
3 2 (1 2 )
4
3 2 2 .0 3 6 (1 0 ) m m
4
3
4
2 .0 3 6 (1 0
Thus, 0 .3 4 9 ( 2 .0 3 6 1 0
T
9
9
)( 7 9 1 0 )
4 6 .8 N m
1 .2
From Eq.(14.1): 1 6T d
3
1 6 ( 4 6 .8 )
( 0 .0 1 2 )
3
1 3 7 .9 M P a
SOLUTION (14.4) D Cd 5 ( 7 ) 35 mm
a ll K s
8 PC
d
2
450
;
8 P2 ( 5 )
(7 )
2
(1
0 .6 1 5 5
),
Thus k
P2 P1
2 1
1 5 4 2 1 0 0 0 20
2 7 .1 N m m
and N
a
dG 3
8C k
3
7 ( 29 10 ) 3
8 ( 5 ) ( 27 . 1 )
7 . 49 coils
225
P2 1 .5 4 2 k N
9
) m
4
SOLUTION (14.5) Apply Eq. (14.10): P
4
d G 3
8D Na
where N a p d 1 2 .5 9 .5 3 m m
So,
P
4
9
( 0 .0 0 9 5 ) ( 7 9 1 0 )( 0 .1 2 5 ) 8 ( 0 .0 5 )
1 .9 3 k N
3
By Eq. (14.6), 8 PD K s d
For
3
C D d 5 0 9 .5 5 .2 6 3
Ks 1
1 .1 1 7
0 .6 1 5 5 .2 6 3
(Eq. 14.7)
It follows that
8 (1 .9 3 )( 0 .0 5 )
( 0 .0 0 9 5 )
(1 .1 1 7 ) 3 2 0 .1 M P a
3
From Table (14.3), S y s 0 .4 S u By Eq. (14.12), Su Ad
1 6 1 0 (9 .5 )
b
0 .1 9 3
1 0 4 2 .6 M P a
Hence S y s 0 .4 5 (1 0 4 2 .6 ) 4 6 9 .2 M P a
Since 3 2 0 .1 4 6 9 .2
Yes.
SOLUTION (14.6) From Eq. (14.11), k
4
P
d G 3
8D Na
As d =constant and k proportional to d
4
3
D , The largest active coil will have the smallest
value of k . That is, the bottom coil will deflect to zero pitch first. Using Eq. (14.10), with N a p 6 m m : P
4
d G
3
8D Na
4
9
( 0 .0 4 ) ( 7 9 1 0 )( 0 .0 0 6 ) 3
8 ( 0 .0 6 ) (1 )
7 0 .2 N
The total deflection is P N a 6 (5 ) 3 0 m m SOLUTION (14.7) ( a ) From Eq. (14.11), k
4
d G 3
8D Na
Outer spring: ko
4
(7 ) (79000 ) 3
8 ( 40 ) ( 4 )
9 2 .6 2 N m m
Inner spring: ki
4
( 4 .5 ) ( 7 9 0 0 0 ) 3
8 ( 22 ) (8 )
4 7 .5 4 N m m
(CONT.)
226
14.7 (CONT.) By Eq. (14.11), k W or W k . Hence
1 4 .2 7 m m
2000 ( 4 7 .5 4 9 2 .6 2 )
( b ) Force on each spring: W o k o (9 2 .6 2 )(1 4 .2 7 ) 1 3 2 2 N W i k i ( 4 7 .5 4 )(1 4 .2 7 ) 6 7 8 N
For outer spring: C 4 0 7 5 .7 1, K s 1
0 .6 1 5 5 .7 1
For inner spring: C 2 2 4 .5 4 .8 9 , K s 1
1 .1 1
0 .6 1 5 4 .8 9
1 .1 3
(8 w D d ) K s 3
Apply Eq. (14.6): i
8 (1 3 2 2 )( 4 0 )
i
8 ( 6 7 8 )( 2 2 )
(7 )
3
( 4 .5 )
(1 .1 1) 4 3 5 M P a (1 .1 3 ) 4 7 1 M P a
3
SOLUTION (14.8) C 15 3 5
S u Ad
b
1510 ( 3
Table 14.3: ( a)
k
3
Nt N
S
0 . 42 S u 509
MPa
3
A 1510
23 . 7 N mm
3
2 12 , h s ( N t 1) d 13 ( 3 ) 39 mm 8 Pmax C
( b ) S ys K s
d
2
Pm a x
Thus
MPa
8 ( 5 ) ( 10 )
a
a
ys
0 . 201
) 1 , 211
3 ( 79 10 )
dG 8C N
b 0 .2 0 1 ,
Table 14.2:
Pmax
,
( 5 0 9 )( 3 ) 8 ( 5 )(1
2
0 .6 1 5
2
S ys d 8 CK
s
3 2 0 .4 N )
5
SOLUTION (14.9) (a) d
2
(1
8 PCn S ys
0 . 615 C
3
8 ( 2 10 )( 5 )( 1 . 3 )
)
6
( 500 10 )
(1
0 . 615 5
)
or d 8 . 62 mm
( b ) D 8 .6 2 ( 5 ) 4 3 .1 m m , N
a
dG
hs (N
(c)
c hf
3
8C K
a
2 4 .4 4 106
3
( 8 .6 2 )( 7 9 1 0 ) 3
8(5 ) (90 )
c 1 .1 0
hf D
90
2 4 .4 4 m m
7 .5 7
2 ) d 82 . 49 mm,
0 .2 3
3
2 (1 0 )
1 0 6 .9 4 3 .1
h f 24 . 44 82 . 49 106 . 9 mm
2 .4 8
Spring is safe (Curve A, Fig.14.10)
227
SOLUTION (14.10) (a) Nt
hs
d
12
2 1.6 1.8
C
N a 12 2 10 ( S ys n ) d
P k
2
8 K sC
3
( 9 0 0 2 )(1 .8 )
( 1 . 8 )( 79 10 )
1 .1 5
P k
1.0 7 4
0 .6 1 5 8 .3 3
64 N
3 . 075
3
8 ( 8 . 33 ) ( 10 )
a
2
8 (1 .0 7 4 )( 8 .3 3 )
s 1 .1 5
Thus
Ks 1
3
dG 8C N
8 .3 3
15 1 .8
N mm
64 3 .0 7 5
2 3 .9 3 m m
h f h s s 2 1 .6 2 3 .9 3 4 5 .5 3
(b)
s h
h
0 . 53
3 . 04
f
D
f
mm
Spring is safe (Curve B, Fig.14.10)
SOLUTION (14.11) d
2
8CP
S ys n
(1
0 .6 1 5 C
)
8 ( 8 )( 2 0 0 ) 6
( 4 2 0 1 0 ) 2 .5
(1
0 .6 1 5 8
),
d 5 .1 1 m m
D 5 .1 1(8 ) 4 0 .8 8 m m
N
a
5 . 11 ( 79 10
dG 3
8C k
3
8 ( 8 ) ( 9 10
6 3
) )
10 . 95
N t 1 0 .9 5 2 1 2 .9 5,
s 1 .2 0
h s (1 2 .9 5 )(5 .1 1) 6 6 .2 m m
2 6 .7 m m
200 9000
h f 6 6 .2 2 6 .7 9 2 .9 m m
Therefore s h
h
0 . 287 ,
f
D
f
2 . 23 Spring is safe (Curve A, Fig.14.10)
SOLUTION (14.12) ( a ) a ll
8 PD
d
3
,
d
3
8 PD
a ll
8 ( 2 )75 6
( 5 2 5 1 0 )
,
d 8 .9 9 m m
( b ) C D d 7 5 8 .9 9 8 .3 4 Na
dG 3
8C k
s 1 .1 0
(c)
s hf
0 .5 3,
P k
3
8 .9 9 ( 7 9 1 0 )
7 .2 9 ,
3
8 ( 8 .3 4 ) ( 2 1 )
1 .1 0 hf D
2 21
h s ( N a 2 1)8 .9 9 9 2 .5 1 m m
1 0 4 .8 m m ,
2 .6 3
h f h s s 1 9 7 .3 m m
Spring is safe (Curve B, Fig.14.10)
228
SOLUTION (14.13) We have C D d 2 0 2 .5 8 .0 ( a ) From Table 14.3: S y s 0 .4 0 S u where Su Ad
b
2 0 6 0 ( 2 .5 )
0 .1 6 3
1774 M Pa
and m a x 0 .4 0 (1 7 7 4 ) 7 0 9 .6 M P a
By Eq. (14.8): 2
d m ax
Pm a x
8CK
where K w 1
w
2
( 2 .5 ) ( 7 0 9 .6 )
0 .6 1 5 8
1 .0 7 7
(Eq. 14.7)
2 0 2 .1
8 ( 8 )(1 .0 7 7 )
Use Eq. (14.11), k
4
d G 3
8D Na
4
( 2 .5 ) ( 7 9 0 0 0 ) 3
8 ( 2 0 ) (1 1 )
4 .3 8 N m
( b ) Pm a x k 2 0 2 .1 4 .3 8 4 6 .1 4 m m By Fig. 14.8c: h s N t d h s ( N a 2 ) d (1 1 2 ) 3 2 .5 m m
Thus, h f 3 2 .5 4 6 .1 4 7 8 .6 4 m m
( c ) h f D 7 8 .6 4 2 0 3 .9 3 2 h f 4 6 .1 4 7 8 .6 4 0 .5 8 7
From Fig. 14.10, for Case B:
Yes, Buckling occurs.
SOLUTION (14.14) ( a ) Use Eq. (14.12) and Table 14.2. Su Ad
b
1 6 1 0 ( 0 .9 )
0 .1 9 3
1643 M Pa
Table (14.3): S y s 0 .4 5 S u 0 .4 5 (1 6 4 3 ) 7 3 9 .4 M P a
( b ) Spring index, with C D d 1 0 0 .9 1 1 .1 : Ks 1
0 .6 1 5 C
1
0 .6 1 5 1 1 .1
1 .0 5 5
(Eq. 14.7)
Rearrange Eq. (14.6), let S y s , and solve for P . 3
P
d S ys 8KsD
3
( 0 .9 ) ( 7 3 9 .9 ) 8 (1 .0 5 5 )(1 0 )
2 0 .1 N
From Fig. 14.7, N a 1 4 .5 2 1 2 .5 . Equation (14.11) is then k
Gd 3
4
8D Na
( 7 9 , 0 0 0 )( 0 .9 ) 3
8 (1 0 ) (1 2 .5 )
4
0 .5 1 8 N m m
(CONT.)
229
14.14 (CONT.) (c) s
P k
3 8 .8 m m
2 0 .1 0 .5 1 8
From Fig. 14.7c: h s ( N a 3) d (1 2 .5 3)( 0 .9 ) 1 3 .9 5 m m h f s h s 3 8 .8 1 3 .9 5 5 2 .7 5 m m
Figure 14.7: p
(d)
s hf
3 8 .8 1 3 .9 5
hf 2d
5 2 .7 5 2 ( 0 .9 )
Na
1 2 .5
hf
2 .7 8 ,
D
5 2 .7 5 10
4 .0 7 6 m m
5 .2 7 5
Figure 14.10, Curve A: No buckling failure.
SOLUTION (14.15) ( a ) Equation (14.4), C D d 1 4 1 .5 9 .3 3 3 Equation (14.7): Ks 1
0 .6 1 5 9 .3 3 3
1 .0 6 6
From Fig. 14.7: N t N a 2 1 6 2 1 8,
h s d N t 1 .5 (1 8 ) 2 7 m m
The pitch equals P ( h f 2 d ) N a [3 5 2 (1 .5 )] 1 6 2 m m
Solid deflection s h f hs 3 5 2 7 8 m m
( b ) Equation (14.12) and Table 14.2: S u 1 6 1 0 (1 .5 )
0 .1 9 3
1498 M Pa
From Table (14.3): S y s 0 .4 5 S u 0 .4 5 (1 4 8 9 ) 6 7 0 M P a
From Eq. (14.11): P
Gds 3
8C N a
( 2 9 , 0 0 0 ) (1 .5 ) ( 8 )
3 .3 4 4 N
3
8 ( 9 .3 3 3 ) (1 6 )
and m ax
8DKsP
8 (1 4 ) (1 .0 6 6 ) ( 3 .3 4 4 )
d
3
670 3 7 .6 5
(1 .5 )
3
3 7 .6 5 M P a
Then n
(c)
s hf
8 3 .5
a ll m ax
0 .2 2 9 ,
hf D
1 7 .8
35 14
2 .5
From Fig. 14.10 No buckling.
230
SOLUTION (14.16) ( a ) Pm
42 2
4 C 1 4C4
Kw a
m
3 kN
Kw
Pa
Ks
Pm
42 2
Pa
0 .6 1 5 C
1.3 1 1
1 . 311 1 . 123
1 3
1 kN
C
Ks 1
15 3
5
1.1 2 3
0 .6 1 5 C
0 . 389
Equation (14.24), with replacing S u s by S y s : m
5 0 0 1 .3 ( 0 .3 8 9 )( 2 5 0 0 2 8 0 )
1 9 2 .3 M P a 1
280
and d
2
8 Pm C
Ks
3
8 ( 3 1 0 )( 5 )
1 .1 2 3
m
6
(1 9 2 .3 1 0 )
s 1 . 10
( b ) D 14 . 94 ( 5 ) 74 . 7 mm N
a
dG
hs (N
(c)
s h
f
3
8C k
a
3
1 4 .9 4 ( 7 9 1 0 )
h
f
D
mm
48 . 89 mm
4000 90
1 3.1 1
3
8( 5 ) ( 90 )
2 ) d 225 . 7 mm
0 .1 8 ,
d 1 4 .9 4
,
3 .6 8
h f 274 . 6 mm
Spring is safe (Curve A, Fig.14.10)
Also fn
356000 d 2
D N
356 , 620 ( 14 . 94 )
2
( 74 . 7 ) ( 13 . 11 )
a
72 . 8 cps 4 , 370
cpm
SOLUTION (14.17) From Eq. (14.20): S e s ' 4 6 5 M P a ,
P (400 0) 2 200 N
k P 200 10 20 N m m
By Eq. (14.8), m a x
8 Pm a x D
d
Kw;
3
465
8 ( 4 0 0 )( 4 0 )
d
3
(1 .3 ) .
Solving d 4 .8 5 m m
Equation (14.11): k d G 8 D N a or 4
Na
4
d G 3
8D k
3
4
( 4 .8 5 ) ( 7 9 , 0 0 0 ) 3
8 ( 40 ) ( 20 )
4 .2 7
Crash allowance =15 % s 1 .1 5 42000 2 3 m m h s ( N a 3) d ( 4 .2 7 3)( 4 .8 5 ) 3 5 .2 6 m m h f 3 5 .2 6 2 3 5 8 .2 6 m m
231
SOLUTION (14.18) From Eq. (14.20): S e s ' 3 1 0 M P a k P 200 10 20 N m m
From Eq. (14.8): 8 Pm a x D
m ax
310
Kw;
3
d
8 ( 4 0 0 )( 4 0 )
d
3
(1 .3 )
Solving d 5 .5 5 m m
Equation (14.11): k d G 8 D N a or 4
4
Na
4
( 5 .5 5 ) ( 7 9 , 0 0 0 )
d G 3
8D k
3
3
8 ( 40 ) ( 20 )
7 .3 2
Crash allowance =8 % s 1 .0 8 42000 2 1 .6 m m h s ( N a 3) d ( 7 .3 2 3)(5 .5 5 ) 5 7 .2 8 m m h f 5 7 .2 8 2 1 .6 7 8 .8 8 m m
SOLUTION (14.19) Refer to Solution of Prob. 14.15. ( a ) We now have 4 C 1 4C 4
Kw
0 .6 1 5 C
Pm a x Pm in
Pa
14 4 2
2
4 ( 9 .3 3 3 ) 1
4 ( 9 .3 3 3 ) 4
5 N,
0 .6 1 5 9 .3 3 3
Pm
1 .1 5 6 14 4 2
9 N
From Eq. (14.8): 8CK
a Pm
m a(
(b) n
Pa
S ys
Pa
2
8 ( 9 .3 3 3 ) (1 .1 5 6 ) ( 5 )
(1 .5 )
2
6 1 .0 5 M P a
) 6 1 .0 5 ( 95 ) 1 0 9 .9 M P a
a m
w
d
3 .9 2
670 6 1 .0 5 1 0 9 .9
( c ) From Table 14.3 the modified endurance limit is S e s ' 0 .2 2 S u 0 .2 2 (1 4 8 9 ) 3 2 7 .6 M P a
Thus S es '
n
a
3 2 7 .6 6 1 .0 5
5 .3 7
Pm
180 30 2
SOLUTION (14.20) ( a ) C 15 3 5 Kw a m
4 C 1 4C 4
Kw
Pa
Ks
Pm
0 . 615 C
1 . 311
105
Pa
N
Ks 1
0 . 615 C
180 30 2
75 N
1 . 123
0 . 834
1 . 311 75 1 . 123 105
Thus
m
K
8 Pm C s
d
2
1 . 123
8 ( 105 )( 5 )
( 3 10
3
)
2
166 . 8 MPa
a 0 . 834 (166 . 8 ) 139 . 1 MPa
(CONT.)
232
14.20 (CONT.) Table 14.2 and Eq.(14.12): S u Ad
b
2060 ( 3
0 . 163
Table 14.3: S es 0 . 23 S u 396 '
) 1 , 722
MPa
Equation (7.5a): S us 0 . 67 S u 1,154
MPa
Equation (14.25) results in 1 ,1 5 4 ( 3 9 6 )
n
( b ) hs ( N a 2 ) d 7 2 m m , s 1 .1 0
(c)
fn
(d)
h
s
1.3 8
1 3 9 .1 ( 2 1 ,1 5 4 3 9 6 ) 1 6 6 . 8 3 9 6
k
1 8 .3 8 m m ,
180 1 0 .7 7
1 356 , 620 d 2 2 D Na
0 .2 ,
356 , 620 ( 3 )
h
6 . 03
f
D
f
108
2
2 ( 15 ) ( 22 )
dG 3
8C N a
3
3 ( 7 9 1 0 ) 3
8 (5 ) ( 22 )
1 0 .7 7 N m m
h f 7 2 1 8 .3 8 9 0 .4
mm
cps 6 , 480
(for fixed-free ends)
cpm
Buckling will occur (Curve B, Fig.14.10)
SOLUTION (14.21) D C d 8 (5 ) 4 0 m m Kw
4 C 1 4C4
m Ks
0 .6 1 5 C
8 Pm C
d
0 .6 1 5 8
1 .0 7 7
1.1 8 4 8 ( 5 0 0 )( 8 )
1 .0 7 7
2
Ks 1
(5)
2
4 3 8 .8 M P a
Table 14.3 and Eq.(14.12): Su Ad
b
2 0 6 0 (5
0 .1 6 3
) 1585 M Pa
Equation (7.5a): S u s 0 .6 7 (1 5 8 5 ) 1 0 6 2 M P a Table 14.3: S e s 0 .2 3 (1 5 8 5 ) 3 6 4 .6 M P a '
Equation (14.23) gives a
1 2
( 3 6 4 .6 )(1 0 6 2 1 .2 4 3 8 .8 ) 1062 (
1 2
)( 3 6 4 .6 )
9 2 .4 7 M P a
Thus Pa
Ks
a
Kw m
Pm
1 .0 7 7 9 2 .4 7 1 .1 8 4 4 3 8 .8
(5 0 0 ) 9 5 .8 N
and Pm in 5 0 0 9 5 .8 4 0 4 .2 N Pm a x 5 0 0 9 5 .8 5 9 5 .8 N
233
MPa
SOLUTION (14.22) ( a ) C 24 5 4 . 8 4 C 1 4C4
Kw
Pm
320
640 2
1.3 2 5
0 .6 1 5 C
Pa
N
Ks 1
0 .6 1 5 C
160 2
80 N
1.1 2 8
Table 14.2 and Eq.(14.12): S u Ad
0 . 163
2060 ( 5
b
S es 0 . 23 S u 365 '
Table 14.3:
) 1 , 585
MPa
MPa
Equation (7.5a): S us 0 . 67 S u 1, 062
MPa
Therefore we obtain
m
K
a
8 Pm C s
K
w
Pa
K
s
Pm
8 ( 320 )( 4 . 8 )
1 . 128
2
d
m
1 . 325 1 . 128
( 0 . 005 ) 80 320
176 . 5 MPa
2
(176 . 5 ) 51 . 83 MPa
Equation (14.25): n
1 , 062 ( 365 ) 51 . 83 ( 2 1 , 062 365 ) 176 . 5 ( 365 )
2 . 49
( b ) Load of 160 N causes deflection of 72-65=7 mm. N
G d
a
8 PC
3
7 ( 7 9 , 0 0 0 )( 5 ) 8 (1 6 0 )( 5 )
1 7 .3
3
SOLUTION (14.23) 90 20 2
( a ) Pa
4 C 1 4C4
Kw a m
35 N
K
w
Pa
K
s
Pm
0 .6 1 5 C
1 . 253 1 . 103
90 20 2
Pm 1.2 5 3
55 N
Ks 1
0 .6 1 5 C
1.1 0 3
0 . 723
35 55
Use Eq.(14.24) with replacing S u s by S y s . m
Thus
5 6 0 1 .6 0 .7 2 3 ( 2 5 6 0 3 1 5 )
1 2 2 .9 M P a 1
315
d
2
K
8 Pm C
m
s
1 .1 0 3
8 ( 5 5 )( 6 ) 6
(1 2 2 .9 1 0 )
,
d 2 .7 5 m m
( b ) D d C 2 .7 5 ( 6 ) 1 6 .5 m m k
P
Na
90 20 0 .0 7 5 0 .0 6 5
dG 3
8C k
7 kN m 3
2 .7 5 ( 2 9 1 0 ) 3
8(6) (7 )
6 .6 ,
h s 8 .6 ( 2 .7 5 ) 2 3 .7 m m ,
N t 6 .6 2 8 .6 h f 0 .0 7 5 1 .1 0 ( 7 02 00 0 ) 7 8 .1 4 m m
s h f h s 7 8 .1 4 2 3 .7 5 4 .4 m m
(c)
(d)
fn s hf
3 5 6 ,6 2 0 d 2
D Na
0 .7 ,
3 5 6 , 6 2 0 ( 2 .7 5 ) 2
(1 6 .5 ) ( 6 .6 )
hf D
4 .7
5 4 5 .8
cps 32, 748
cpm
Spring is safe (Curve A, Fig.14.10)
234
SOLUTION (14.24) 4 C 1 4C4
(a) Kw
1.2 5 3
0 .6 1 5 C
Pa
5 0 1 0 2
20 N
a
K
m
w Pa
K s Pm
1 . 253 1 . 103
m
Thus
Ks 1 5 0 1 0 2
Pm
0 .6 1 5 C
1.1 0 3
30 N
0 . 757
20 30
5 6 0 1 .6
1 1 9 .3 M P a
0 .7 5 7 ( 2 5 6 0 3 1 5 )
1
315
Hence 2
d
8 Pm C
K
1 .1 0 3
m
s
8 ( 3 0 )( 6 ) 6
(1 1 9 .3 1 0 )
d 2 .0 6 m m
or
( b ) D 2 .0 6 ( 6 ) 1 2 .3 6 m m 5 0 1 0 0 .1 2 5 0 .1 0 5
k
P
N
dG
a
3
2 . 06 ( 29 10 )
3
8C k
2 kN m 17 . 29
3
8(6) (2)
N t 17 . 29 2 19 . 29 h s 1 9 .2 9 ( 2 .0 6 ) 3 9 .7 4 m m h f 1 2 5 1 .1 0 ( 2 10 00 0 ) 1 2 5 .0 0 6 m m
s h f h s 1 2 5 .0 0 6 3 9 .7 4 8 5 .2 7 m m
(c)
(d)
fn s
3 5 6 , 6 2 0 ( 2 .0 6 )
hf
0 .6 8 ,
hf
2 7 8 .1 c p s 1 6 , 6 8 6 c p m
2
(1 2 .3 6 ) (1 7 .2 9 )
1 0 .1
D
Spring will buckle (Curve A, Fig.14.10)
SOLUTION (14.25) ( a ) Pm 400 Kw
Pa 100
N
4 C 1 4C4
a
K
1
s
0 . 615 C
1.2 5 3
0 .6 1 5 C
m K s
N
8 Pm C
d
K
w
Pa
K
s
Pm
2
1.1 0 3
m
1.2 5 3 1 .1 0 3
8 ( 4 0 0 )( 6 )
1
( 4 )(
6,741 d
2
6,741
2
(d )
d
)
2
1,9 1 4 d
2
Equation (14.23) leads 1,9 1 4 d
2
(720
330 2
)
330 2
720 1. 6
(
6,741 d
2
)
or 6 , 438 d
2
450
6 , 741 d
2
,
d 5 . 41 mm
( b ) D 5 . 41 ( 6 ) 32 . 46 mm N
a
dG 8 PC
3
5 .4 1( 7 9 , 0 0 0 )( 8 ) 8 ( 5 0 0 3 0 0 )( 6 )
3
9 .8 9
235
1 . 103
SOLUTION (14.26) ( a ) Pm 470
Pa 130
N,
C 30 6 5
N,
Table 14.2 and Eq.(14.12): S u Ad
b
0 . 201
1510 ( 6
) 1053
MPa
Table 14.3: S
0 . 42 (1053 ) 442
ys
Ks 1 P2 P1
We have k N
a
1G d 8 P1 C
K
m
a
3
2 1
d
Kw
Pa
Ks
Pm
600 340 13
1 7 ( 2 9 0 0 0 )( 6 )
8 ( 3 4 0 )( 5 )
8 Pm C s
1.1 2 3 ,
0 .6 1 5 C
3
m
1 .3 1 1 1 .1 2 3
'
Kw
4 C 1 4C4
0 .6 1 5 C
1
20 N mm ,
340 20
MPa
1.3 1 1 17 mm
8 .7 8 ( 470 )( 5 )
1 . 123
2
S es 0 . 21 (1053 ) 221
MPa ,
(6)
2
186 . 7 MPa
( 14 37 )(1 8 6 .7 ) 6 0 .2 9
M Pa
It follows that n
(b)
hs ( N k
s
3
a
1 . 20
fn
(d)
hf
6 ( 79000 )
(Eq.14.25)
54 . 48 N mm
3
8 ( 5 ) ( 8 .7 )
13 . 22 mm ,
600 54 . 48
h
356 , 620 ( 6 )
1 356 , 620 d 2 2 D Na
0 .1 7 ,
1.2
2 ) d (10 . 7 ) 6 64 . 2 mm
dG 8C N
(c)
s
a
442( 221) 6 0 .2 9 ( 2 4 4 2 2 2 1 ) 1 8 6 .7 ( 2 2 1 )
2
2 ( 30 ) ( 8 . 7 )
2 .6
f
D
h f 64 . 2 13 . 22 77 . 42 mm
136 . 6 cps 8 ,196
cpm
(fixed-free ends)
Spring is safe (Curve B, Fig.14.10)
SOLUTION (14.27) Refer to Example 14.6. The critical torsional shear stress in the hook is from Eq. (14.34b): B (
rm ri
)
8 PD
d
3
S ys
where S y s 0 .4 0 S u 0 .4 0 (1 2 5 6 ) 5 0 2 .4 M P a
rm 3 .7 5 m m D 7 .5 m m
ri 3 .7 5 ( 2 .5 2 ) 2 .5 m m d 2 .5 m m
Thus, 5 0 2 .4 ( 32.7.55 )
8 P (1 2 .5 )
( 2 .5 )
3
,
P 1 6 4 .4 N
The larger load is PA 2 3 2 .1 N , as found in Part (b) of Example 14.6. This shows that failure will occur first by shear stress in the hook.
236
SOLUTION (14.28) k 4 d1 3 D1
Gd
4
Since k 1 k 2 :
3
8D Na
d
4 2
3
D2
d 2 d1 4
,
3 D2
4
3
D1
3 ( 81 ) , 4
d 2 14 . 27 mm
3
SOLUTION (14.29) C
4 C 1 4C 4
Kw a m
10
5 0 .5
K
w
Pa
K
s
Pm
K
1
s
0 .6 1 5 C
1 . 0615
0 . 615 C
1 .1 4 5,
Pm 3 N ,
m Ks
0 .7 1 9
8 Pm C
d
2
Pa 2 N
1 .0 6 1 5
8 ( 3 ) (1 0 )
( 0 .5 )
2
3 0 5 .6 M P a
a 0 .7 1 9 (3 0 5 .6 ) 2 1 9 .7 M P a
Table 14.2 and Eq.(14.12): Su Ad
2 0 6 0 ( 0 .5
b
0 .1 6 3
) 2306
M Pa
Table 14.3: S e s 0 .2 3 S u 5 3 0 .4 M P a
Equation (7.5a): S u s 0 .6 7 S u 1 5 4 5 M P a
'
Equation (14.25) is therefore n
1545 ( 530 . 4 )
1 . 13
219 . 7 ( 2 1545 530 . 4 ) 305 . 6 ( 530 . 4 )
SOLUTION (14.30) ( a ) We have C D d 2 .4 0 .6 4 . So, 4 C 1 4C 4
Kw
0 .6 1 5 C
4 ( 4 ) 1
4(4)4
0 .6 1 5 4
1 .2 5 0 .1 5 3 7 5 1 .4 0 3 7 5
From Eq. (14.8): m ax K w
8 PD
d
3
8 ( 6 )( 2 .4 )(1 0
1 .4 0 3 7 5
( 0 .6 )(1 0
3
9
)
)
2 3 8 .3 M P a
( b ) Use Eq.(14.34). Here ( ri ) A ( rm ) A d 2 1 .2 0 .3 0 .9 m m . Hence (
A
)c
3
1 .2 1 6 ( 6 )( 2 .4 1 0 ) 9 0 .9 ( 0 .6 )(1 0 )
4 (5) 2
( 0 .6 ) (1 0
6
)
4 5 2 .7 1 7 .6 8 4 7 0 .3 8 M P a
From Eq. (14.12) and Table 14.2 for music wire: Su Ad
2 0 6 0 ( 0 .6 )
b
0 .1 6 3
2239 M Pa
We also have S y S y s 0 .5 7 7 ( 0 .4 0 S u 0 .5 7 7 0 .6 9 3 S u 0 .6 9 3 ( 2 2 3 9 ) 1, 5 5 2 M P a
Thus n
Sy (
A
)c
1552 4 7 0 .3 8
3 .3
SOLUTION (14.31) ( a ) I bh
ML EI
Hence
3
12 (1 . 2 ) L
,
N
a
L
D
EI M
4
12 0 . 1728 ( 16 2 )( 207 10
14 ,983
(1 8 )
240
mm 3
)( 0 . 1728 )
4
,
C D d 18 1 . 2 15
14 , 983
mm
265
(CONT.)
237
14.31 (CONT.) ( b ) P a 1 5 (1 6 ) 2 4 0 N m m Ki
2
3 C C 0 .8 3 C ( C 1)
i Ki
1 .0 4 6 ,
6 Pa bh
2
1 .0 4 6
6 ( 240 ) (1 .2 )
3
8 7 1 .7 M P a
SOLUTION (14.32) The spring index is C D d 1 5 1 .5 1 0 . From Eq. (14.36): 2
4 C C 1 4 C ( C 1 )
Ki
2
4 (1 0 ) 1 0 1
1 .0 8 1
4 (1 0 )(1 0 1 )
Equation (14.12) and Table 14.2 for hard-drawn wire: Su Ad
1 5 1 0 (1 .5 )
b
0 .2 0 1
1392 M Pa
S y S y S y s 0 .5 7 7 ( 0 .4 2 0 .5 7 7 ) S u 1 0 1 3 M P a 1 0 1 3 N m m
From Eq. (14.39): 3
M
d Sy
3
(1 .5 ) (1 0 1 3 )
32 K i
3 2 (1 .0 8 1 )
3 1 0 .5 N m m
We have I d
6 4 (1 .5 )
4
4
6 4 0 .2 4 9 m m
4
Equation (14.41), with r a d 1 .2 , is thus M lw
1 .2
EI
3 1 0 .5 [ (1 5 ) N a ] 3
( 2 1 0 1 0 ) ( 0 .2 4 9 )
N a 4 .2 8 8
,
SOLUTION (14.33) From Eq.(14.12) and Table 14.2 for oil-tempered wire: Su Ad
b
1610(2)
0 .1 9 3
1 4 0 8 .4 M P a
( a ) Equation (7.5b) and Table 14.3: S y S y s 0 .5 7 7 0 .4 5 S u 0 .5 7 7 0 .7 8 S u 0 .7 8 (1 4 0 8 .4 ) 1 0 9 8 .6 M P a
Spring index: C D d 1 2 .5 2 6 .2 5 . Equation (14.36) : 4 ( 6 .2 5 ) 6 .2 5 1
Ki
4 ( 6 .2 5 )( 6 .2 6 1 )
Equation (14.39) with i
32 Pa
d
3
a ll
0 .1 3 5 S y n 1 0 9 8 .6 1 .8 6 1 0 .3 M P a :
6 1 0 .3 (1 0 ) 6
K i;
3 2 P ( 0 .0 2 8 )
( 0 .0 0 2 )
3
( 0 .1 3 5 )
Solving, P 1 2 6 .8 N ( b ) L w D N a (1 2 .5 )(3 .5 ) 1 3 7 .4 m m I d
4
64 (2)
4
6 4 0 .7 8 5 4 m m
4
Equation (14.41), with M P a 1 2 6 .8 ( 0 .0 2 8 ) 3 .5 5 N m Hence rad
M lw EI
( 3 .5 5 ) ( 0 .1 3 7 4 ) 9
2 0 0 1 0 ( 0 .7 8 5 4 1 0
12
)
3 .1 1 r a d
238
2
SOLUTION (14.34) ( a ) We have a ll m a x n 8 0 0 2 .5 3 2 0 M P a From Eq. (14.42):
a ll
6 PL nbh
3 2 0 (1 0 ) 6
;
2
3
6 ( 4 0 1 0 )( 0 .7 ) 8 ( 6 )( 0 .0 2 2 )
b 136 m m
,
2
( b ) Eq. (14.43): (1 ) 2
( hl ) (1 0 .3 ) 3
6P Enb
3
6 ( 4 0 1 0 )
2
9
2 0 0 (1 0 )( 8 )( 0 .1 3 6 )
( 0 0.0.72 2 )
3
0 .0 3 2 3 m 3 2 .3 m m
SOLUTION (14.35) ( a ) C 0 .6 0 .0 8 7 .5 Table 14.2 and Eq.(14.12): Su Ad
S ys
Table 14.3: S y
0 .1 9 3
1610(2
b
0 .4 5 S u
0 .5 7 7
1 0 9 8 .4 M P a
0 .5 7 7 2
4 C C 1 4 C ( C 1 )
Equation (14.36): K i
) 1 4 0 8 .4 M P a
1 . 11
Equation (14.39) is therefore 3
( 2 ) (1 .0 9 8 4 ) 1 .4
M Pa
3 2 (1 .1 1 )
( b ) L (1 5 ) 6 2 8 2 .7 m m ,
ML EI
0 .5 5 5 ( 0 .2 8 2 7 )
9
( 2 0 0 1 0 )( 0 .7 8 5 4 1 0
12
)
0 .5 5 5 N m
I
4
d 64
0 .7 8 5 4 m m
0 .9 9 9 ra d 5 7 .3
4
o
SOLUTION (14.36) ( a ) P m 700 m
6 Pm L bh
6 ( 7 0 0 )( 0 . 4 )
2
Also
P a 400
N
m
2
40 h ( h )
h
1
'
m
( Se K
f
a m
Pa Pm
4 7
3
1 4 0 0 1 .2
Su
a
42
Su n
N
(
4 7
)
1400
)(
3 6 0 .1 M P a ) 1
5 0 0 1 .4
Thus we obtain 360 . 1 (10
)
42 h
3
,
h 4 . 89 mm
b 195 . 6 mm
Hence (b) k
6
Ebh 3
3
3 L ( 1
2
)
3
( 207 10 )( 195 . 6 )( 4 . 89 ) 3
3 ( 400 ) ( 0 . 91 )
3
27 . 1 N mm
End of Chapter 14
239
CHAPTER 15
POWER SCREWS, FASTENERS, AND CONNECTIONS
SOLUTION (15.1) ( a ) By Fig.15.4b: Thus
thread depth=7.5 mm,
thread width=7.5 mm
d m 75 7 . 5 67 . 5 mm ,
( b ) By Fig.15.4a: Hence
d r 75 15 60 mm ,
thread depth=7.5 mm,
d m 67 . 5 mm ,
L p 15 mm
thread width at pitch line=7.5 mm
d r 60 mm ,
L p 15 mm
SOLUTION (15.2) 4 , p 6 . 35
25 . 4 p
Table 15.3:
( a ) L n p 2 ( 6 .3 5 ) 1 2 .7 m m p
dm d ta n
2
Fig.15.4a: 1 4 .5
3 8 .1 3 .1 7 5 3 4 .9 2 5 m m
L
dm
1 2 .7
( 3 4 .9 2 5 )
6 .6
;
o
ta n n c o s ta n c o s 6 .6 (ta n 1 4 .5 );
n 1 4 .4 1
o
( b ) For starting, we have f Tu Td
(c) e (d) F
Wd
f cos n tan
m
cos n f tan
2
Wf
4 3
( 0 .1 ) 0 .1 3 ,
o
o
o
(c o s 1 4 .4 1 ) ( 0 .1 3 ) ta n 6 .6
7 .5 ( 3 4 .9 2 5 )
0 .1 3 (c o s 1 4 .4 1 ) ta n 6 .6
a
5 4 .2 0 .1 9
7 .5 ( 0 .1 1 )( 5 0 .8 )
7 .5 ( 0 .1 1 ) ( 5 0 .8 )
o
o
(c o s 1 4 .4 1 ) ( 0 .1 3 ) ta n 6 .6
cos n f cot
Tu
o
o
c o s n f ta n
o
o
o
o
o
c o s 1 4 .4 1 ( 0 .1 3 ) ta n 6 .6
4 3
( 0 .0 8 ) 0 .1 1
2
2
2
fc
o
cd c
0 .1 3 (c o s 1 4 .4 1 ) ta n 6 .6
7 .5 ( 3 4 .9 2 5 )
o
c o s 1 4 .4 1 ( 0 .1 3 ) c o t 6 .6
2
2
5 4 .2 N m 2 3 .3 N m
0 .4 6 4 6 %
2 8 5 .3 N
SOLUTION (15.3) f cos n tan
T o 0 d m [ cos
Solving, ta n 1 2 .5 or Then, Eq.(15.12):
n
f tan
] d c fc
f fc cos n ( d c d m ) cos n f fc ( d c d m )
dm [
o
0 .1 0 .0 8 (c o s 1 4 .4 1 )(1 .4 5 4 5 ) o
c o s 1 4 .4 1 ( 0 .1 )( 0 .0 8 )(1 .4 5 4 5 )
o
d m ta n
e
(Eq.15.11)
f c o s n ta n c o s n f ta n
o
3 4 .9 2 5 (ta n 1 2 .5 )
] d c fc
o
3 4 .9 2 5 [
o
o
c o s 1 4 .4 1 0 .1 (ta n 1 2 .5 )
240
0 .4 9 4 4 9 .4 %
o
0 .1 (c o s 1 4 .4 1 )(ta n 1 2 .5 )
] 5 0 .8 ( 0 .0 8 )
SOLUTION (15.4) d m 32 2 30 mm ,
ta n
1
1
ta n
L
d m
Wd
Tu
4 .8 5
8
( 30 )
f tan
m
1 f tan
2
n 0
L 2 ( 4 ) 8 mm ,
Wf
cdc
2
o
6 ( 30 )
2
[
0 . 1 tan 4 . 85 1 0 . 1 tan 4 . 85
o o
]
6 ( 0 . 15 ) 50 2
39 . 28 N m
We have n
V L
40 8
5 rps 300
rpm
Thus kW
Tn 9549
1 . 23
39 . 28 ( 300 ) 9549
SOLUTION (15.5) n 0,
c o s n 1,
L 25 m m ,
V 0 .1 5 6 0 9 m m in T
9549 kW n
9549 ( 4 ) 360
n
V L
9 0 .0 2 5
ta n
360
L
dm
25
( 45 )
rp m
1 0 6 .1 N m
Hence, by Eq.(15.9): Tu
Wd
f tan
m
1 f tan
2
; 106 . 1
10 ( 45 )
f 0 . 1768
2
1 f ( 0 . 1768 )
Solving f 0 . 272
SOLUTION (15.6) N 2,
( a ) Table 15.3:
p
2 5 .4 2
1 2 .7
m m th re a d ,
V 0 .0 1 m p s 0 .6 m m in ,
( b ) We have
fc 0,
ta n
dm d
1
p
ta n
dm
1
p 2
n
70
1 2 .7 2
[ 1( 623.7.6 ) ] 3 .6 4
0 .6 0 .0 1 2 7
n 0
4 7 .2 rp m
6 3 .6 m m o
Then, we obtain Tu
d mW
kW
2
f c o s n ta n
[ cos
Tu n 9549
n
f ta n
]
1 7 1 5 ( 4 7 .2 ) 9549
6 3 .6 ( 2 5 0 ) 2
[
0 .1 5 (1 ) ta n 3 .6 4
o
1 ( 0 .1 5 ) ta n 3 .6 4
o
8 .4 8
Hence ( k W ) req
8 .4 8 0 .8 5
9 .9 8
241
] 1 .7 1 5 k N m
0 .1 7 6 8
SOLUTION (15.7) ( a ) From Fig. 15.4b, d dm
P 4
24
2 5 .5 m m
6 4
Equation (15.6), with n 0 , Wdm
T
f dm L
(
2
d m fL
100 ( 24 )
2
)
c o s n 1, ta n L d m ,
l p , becomes:
W fc d c 2
( 0 .0 1 ) ( 2 4 ) 6
[ ( 2 4 ) ( 0 .1 1 )( 6 ) ]
1 0 0 ( 0 .1 1 )( 3 6 ) 2
2 2 9 .5 1 9 8 4 2 7 .5 N m
( b ) The screw is self locking, if: f
L
dm
6
( 24 )
0 .0 8
SOLUTION (15.8) ( a ) Because of the triple-threaded screw, L 3 p 3 (8 ) 2 4 m m
From Fig. 15.4a: thread depth 0 .5 p 0 .5 (8 ) 4 m m d m d 0 .5 p 5 0 4 4 6 m m
From Eq. (15.2), ta n
1
L
dm
ta n
1
( 4264 ) 9 .4 3
o
( b ) For starting, we have fc
( 0 .1 2 ) 0 .1 6
4 3
f
4 3
( 0 .1 3) 0 .1 7
From Eq. (15.8): n ta n
1
(ta n c o s ) ta n
1
(ta n 1 4 .5 c o s 9 .4 3 ) 1 4 .3 1 o
o
o
By Eq. (15.7): Td
Wdm
f c o s n ta n
2
c o s n f ta n
15 ( 46 ) 2
W fc d c
2
o
0 .1 7 c o s 1 4 .3 1 t a n 9 .4 3
o
o
o
c o s 1 4 .3 1 ( 0 .1 7 ) ta n 9 .4 3
1 5 ( 0 .1 6 ) ( 6 8 ) 2
3 4 5 ( 0 .0 0 9 1) 8 1 .6 8 4 .7 N m
SOLUTION (15.9) ( a ) We have n
0 .0 2 ( 6 0 )( m m in )
5 0 rp m
0 .0 2 4 ( m r e v )
From Eq. (15.6), with f c 0 .1 2 and f 0 .1 3 : Tu
Wdm
f c o s n ta n
2
c o s n f ta n
W fc d c
2 o
o
1 5 ( 4 6 ) ( 0 .1 2 ) c o s 1 4 .3 1 ta n 9 .4 3 2
o
c o s 1 4 .3 1 ( 0 .1 3 ) ta n 9 .4 3
o
1 5 ( 0 .1 2 )( 6 8 ) 2
3 4 5 ( 0 .2 9 6 5 ) 6 1 .2
1 6 3 .5 N m
(CONT.)
242
15.9 (CONT.) Power: ( h p ) out ( h p ) in
FV 7 4 5 .7 Tn 7121
1 5 0 0 0 ( 0 .0 2 )
0 .4 0 2 h p
7 4 5 .7
1 6 9 .4 ( 5 0 ) 7121
1 .1 8 9 h p
Efficiency is thus e
0 .3 4 3 4 %
0 .4 0 2 1 .1 8 9
( b ) From Eq. (15.10): f c o s n ta n c o s 1 4 .3 1 ta n 9 .4 3 o
o
0 .1 6 1
Since f 0 .1 3 0 .1 6 1 the screw is not self-locking and overhauling
SOLUTION (15.10) n 0, e
or
ta n
c o s n 1.
1 f ta n
1 0 .1 2 ta n 1 0 .1 2 c o t
0 .7 0
;
1 f c o t
[
1 0 .1 2 ta n ta n 0 .1 2
] ta n
2 .5 ta n 0 .7 0 0
2
Solving tan 0 . 3213 or 2 . 1783 ; 17 . 81
o
or 65 . 34
o
Thus Wd
T0
m
2
f tan
[ 1 f
tan
]
50 ( 30 )
o
[ 0 . 12 tan 17 . 81o ] 145 . 3 N m 1 0 . 12 tan 17 . 8
2
SOLUTION (15.11) n 0, ta n
p
dm
cos n 1
p d m ta n ( 4 6 .8 ) ta n
,
V 0 .2 ( 6 0 ) 1 2 m m in n p
(n in rp m )
(1) (2)
From Eqs.(1) and (2): n
V p
12
( 0 . 0468 ) tan
We have the torque expressed as
and
Tu
9549 kW n
Tu
Wd 2
m
9549 ( 3 . 75 ) ( 0 . 0468 ) tan
f tan 1 f tan
12
438 . 7 tan
438 . 7 tan
Thus 438 . 7 tan 12 ( 23 . 4 )[ 10. 015. 15 tan ] tan
or
438 . 7 tan 65 . 8 tan
2
42 . 12 280 . 8 tan
or
157 . 9 tan 65 . 8 tan
2
42 . 12
or
tan
2
2 . 4 tan 0 . 64 0
(CONT.)
243
15.11 (CONT.) Solving ta n 1 , 2
2
b
b 4 ac
2a
2 .4 1 .7 9 2
Use ta n 0 .3 0 5 . Then, Eq.(1) gives p 4 6 .8 ( 0 .3 0 5 ) 4 4 .8 4 m m
SOLUTION (15.12) Table 15.1 and Fig.15.3: A t 3 5 3 m m , 2
h d d r 3 .6 8 m m , b
(a) (b) b
P A
2 h ta n 3 0
3 8
3
6 0 (1 0 )
3 5 3 (1 0
P
d m hne
6
170
)
ne
;
p 3 mm,
dm
d dr 2
2 4 2 0 .3 2 2
0 .3 7 5 2 (3 .6 8 ) ta n 3 0
o
d r d - 1 .2 2 7 p = 2 0 .3 2 m m
o
2 2 .1 6 m m
4 .6 2 m m
M Pa
P
d m h
b
3
6 0 (1 0 )
( 2 2 .1 6 ) ( 3 .6 8 ) 7 0
3 .3 4 6
th r e a d engaged
Minimum nut length is thus L n p n e (3 )(3 .3 4 6 ) 1 0 .0 4 m m e
( c ) Screw:
Nut:
3P 2 d r neb
3P 2 d neb
p
8 2
3
3 ( 6 0 1 0 )
2 ( 2 2 .1 6 )( 4 .6 2 )(1 0
6
)( 3 .3 4 6 )
3
3 ( 6 0 1 0 ) 2 ( 2 4 )( 4 .6 2 )(1 0
6
8 3 .6 M P a
7 7 .2 M P a
)( 3 .3 4 6 )
SOLUTION (15.13) d m 46 m m ,
(a) (b) b
W A
h
2
W 2
dr
W
d m h ne
;
4
4 mm, 3
4 (1 5 1 0 )
( 0 .0 4 2 )
ne
2
d r d p 50 8 42 m m ,
1 0 .8 M P a
W
b d m h
3
1 5 (1 0 ) 6
1 0 (1 0 ) ( 0 .0 4 6 )( 0 .0 0 4 )
2 .6 th re a d engaged
L n p n e 8 ( 2 .6 ) 2 0 .8
Thus, minimum length of nut:
e
( c ) Screw: Nut:
3W 2 d r neb
3W 2 d neb
b
3
3 (1 5 1 0 ) 2 ( 0 .0 4 2 ) ( 2 .6 ) ( 0 .0 0 4 ) 3
3 (1 5 1 0 ) 2 ( 0 .0 5 ) ( 2 .6 ) ( 0 .0 0 4 )
1 6 .4 M P a
1 3 .8 M P a
244
mm
p 2
8 2
4 mm
SOLUTION (15.14) kb
kp
2
d E s
0 . 58 50 0 . 5 15 0 . 58 50 2 . 5 15
kb
kb k
3.5 3 4 (1 0
4 ( 0 .0 5 )
0 . 58 ( E s 3 )( 0 . 015 ) 2 ln[ 5
C
2
( 0 .0 1 5 ) E s
4L
4 . 512 (10
3
)E s
(Eq.15.31a)
)E s
(Eq.15.34)
]
0 . 439
3 . 534 3 . 534 4 . 512
p
3
We have the preload: Fi
T 0 .2 d
72 0 . 2 ( 0 . 015 )
24 kN
Thus F b C P F i 0;
0 .4 3 9 P 2 4 ,
P 5 4 .6 7
kN
SOLUTION (15.15) At 3 5 3 m m , 2
Table 15.1:
Table 15.4: S u 5 2 0 M P a We have Pm 6 0 k N ,
C r 0 .8 7
Table 7.3: Table 15.6: K
f
Pa 2 0 k N ,
3 n 1 .4
Equation (15.39) gives S e ( 0 .8 7 )( 13 )( 0 .4 5 5 2 0 ) 6 7 .8 6
( a ) kb
2
d Es
2
( 0 .0 2 4 ) E s
0 .0 0 9 E s
4 ( 0 .0 5 )
0 .5 8 ( E s 3 )( 0 .0 2 4 )
kp
2 ln [ 5
C
4L
0 .5 8 ( 0 .0 5 ) 0 .5 ( 0 .0 2 4 ) 0 .5 8 ( 0 .0 5 ) 2 .5 ( 0 .0 2 4 )
kb
kb k p
0 .0 0 9 0 .0 0 9 0 .0 0 8 7
M Pa
0 .0 0 8 7 E s ]
0 .5 0 8
F b a C Pa 0 .5 0 8 ( 2 0 ) 1 0 .1 6
ba
kN ,
1 0 ,1 7 0 353
Therefore we obtain Su n
or
520 1 .4
bm
bm
Su Se
520 67 . 86
ba
( 28 . 8 ),
bm
150 . 74 MPa
F b m b m A t 1 5 0 .7 4 (3 5 3) 5 3 .2 1 k N
Also
F bm CP m F i ;
53 . 21 0 . 509 ( 60 ) F i
or F i 2 2 .7 k N
( b ) T K d F i 0 .2 ( 2 0 )( 2 2 .7 ) 9 0 .8 N m
245
2 8 .8 M P a
SOLUTION (15.16) Table 15.1: A t 245 mm
2
,
K
3 . 8 (Table 15.6),
f
C r 0 .8 9 (Table 7.3),
Ct 1
Using Eq.(15.39): S e ( 0 . 89 )( 1)(
1 3 .8
)( 0 . 45 750 ) 79 . 05 MPa
( a ) F bm 160 ( 245 ) 39 . 2 kN kb k
p
2
d Es 4L
2
( 0 . 02 ) E s
0 . 58 ( 50 ) 0 . 5 ( 20 )
2 ln[ 5
0 . 58 ( 50 ) 2 . 5 ( 20 )
kb
kb k p
6 .2 8 3 6 .2 8 3 6 .7 2 2
3
6 . 283 (10
4 ( 0 . 05 )
0 . 58 ( E s 3 )( 0 . 02 )
C
6 . 722 ( 10
)E s
3
)E s
]
0 .4 8 3
Hence, we obtain F b m C Pm F i 0 . 4 8 3 P m 2 5
or
39 . 2 0 . 483 Pm 25 , Pm 29 . 4 kN
and S
y
n
bm
or
Also
F ba CP a ;
ba
S
y
Se
ba ;
620 2 .2
15 . 53 MPa
and
160
620 7 9 .0 5
ba
F ba 15 . 53 ( 245 ) 3 . 8 kN
3 . 8 0 . 483 Pa ,
Pa 7 . 867
kN
Then Pmax Pm Pa 29 . 4 7 . 867 37 . 27 kN Pmin Pm Pa 29 . 4 7 . 867 21 . 53 kN
( b ) T 0 . 15 ( 20 )( 25 ) 75 N m SOLUTION (15.17) ( a ) From Eq. (15.23): Fb C P Fi ( k
kb b
4 kb
)5 4 .2 5 .2 k N
By Eq. (15.24): 4k
F P (1 C ) P F i ( 5 k b )5 4 .2 0 .2 k N b
( b ) There is a compression (of- 0 .2 k N ) in parts. Thus they do not separate under 5 k N load. SOLUTION (15.18) ( a ) Each bolt supports P 1 1 .6 2 5 .8 k N . From Eqs. (15.23) and (15.24) Fb C P Fi ( 3 k
kb b
kp
) ( 5 .8 ) F i 1 4 .5 F i 2 kb
F p (1 C ) P F i ( 3 k
b
kp
) ( 5 .8 ) F i 2 .9 F i
(1) (2) (CONT.)
246
15.18 (CONT.) Joint separates when F p 0 . Equation (2) is then F p 2 .9 F i 0 ,
F i 2 .9 k N
( b ) Using Eq. (1): F b 1 .4 5 2 .9 4 .3 5 k N Fb
From Table 15.4: S p
3 1 0 (1 0 ) 6
;
At
4 ,3 5 0 At
,
A t 1 4 .0 3 m m
2
So, by Table 15.1, select: I S O 5 0 .8 steel bolt (with tensile stress area closest to A t 1 4 .0 3 m m .) 2
SOLUTION (15.19) ( a ) kb
2
d Es 4L
3
4 ( 0 .0 3 2 )
0 .5 8 E c d
kp
2 ln [ 5
C
2
(1 8 ) ( 2 0 0 1 0 )
0 .5 8 L 0 .5 d 0 .5 8 L 2 .5 d
kb kb k
p
N m
0 .5 8 (1 6 5 1 0 )(1 8 ) 2 ln [ 5
1 . 59 1 . 59 3 . 49
9
6
]
1 .5 9 1 0
0 .5 8 ( 0 .0 3 2 ) 0 .5 ( 0 .0 1 8 ) 0 .5 8 ( 0 .0 3 2 ) 2 .5 ( 0 .0 1 8 )
3 .4 9 1 0
9
N m
]
0 . 313
At 2 1 6 m m , 2
Tables 15.1 and 15.4:
S p 600 M Pa
F i 0 .9 S p A t 0 .9 ( 6 0 0 )( 2 1 6 ) 1 1 6 .6 4 k N
Equation (15.20): Then
F b C P F i 0 .3 1 3( 860 ) 1 1 6 .6 4 1 2 0 .8 k N
b
(b)
Fb At
120800 216
5 5 9 .3 M P a
T 0 .2 F i d 0 .2 (1 1 6 .6 4 )(1 8 ) 4 1 9 .9 N m
SOLUTION (15.20) Pm Pa 1 0 k N ,
Su 830 M Pa ,
Table 15.6: K
f
3,
Equation (7.11): C t 1 0 .0 0 3 2 ( 9 0 0 8 4 0 ) 0 .8 1 Refer to Solution of Prob.15.19:
C 0 .3 1 3
At 2 1 6 m m
2
Equation (15.39): S e ( 0 .8 4 )( 0 .8 1)( 13 )( 0 .4 5 8 3 0 ) 8 4 .7 M P a Thus, Eq.(15.40) results in n
( 8 3 0 )( 2 1 6 ) 1 1 6 , 6 4 0 ( 0 .3 1 3 )(1 0 , 0 0 0 )[(
830
1 .8 5
) 1]
8 4 .7
247
C r 0 .8 4 (Table 7.3)
SOLUTION (15.21) ( a ) Compression of the parts is lost when F p 0 . Thus, from Eq. (15.24): F i (1 C ) P F P ( k
(k
4 kb b
kp kp
b
) P FP
)(3 5 ) 0 2 8 k N
4 kb
( b ) Minimum force in parts occurs when fluctuating load is maximum. Using Eq. (15.24): Fp ( k
kp b
) P Fi
kp
4 5
(3 5 ) 3 8 1 0 k N
SOLUTION (15.22) ( a ) Compression of the parts is lost when F p 0 . Therefore, by Eq. (15.24): Fi ( k
(k
kp b
)P Fp
kp
2 kb b
)(3 5 ) 0 1 7 .5 k N
3 kb
( b ) Minimum force in parts takes place when fluctuating load is maximum. From Eq. (15.24): Fp ( k
kp b
kp
)P Fp
1 2
(3 5 ) 3 8 2 0 .5 k N
SOLUTION (15.23) ( a ) From Eq. (15.24): F p (1 C ) P F i ( k
2 .5 ( 3 3 1 ) P 1 3,
kp kp
b
) P Fi
P 2 0 .6 7 k N
Fb Fi 1 3 k N
( b ) Load off: Load on: F b
1 4
( 2 0 .6 7 ) 1 3 1 8 .1 7 k N
Hence, Pm
1 3 1 8 .1 7 2
1 5 .5 9 k N
Pa
1 8 .1 7 1 3 2
2 .5 8 5 k N
SOLUTION (15.24) ( a ) From Eq. (15.24): F p (1 C ) P F i ( k 600 ( 2k
kp p
kp
kp p
) P 8000,
kb
) P Fi P 2 5, 8 0 0 N
(CONT.)
248
15.24 (CONT.) Fb Fi 8 0 0 0 N
( b ) Load off: Load on: F b
1 3
( 2 5, 8 0 0 ) 8 0 0 0 1 6 , 6 0 0 N
So, 8 0 0 0 1 6 ,6 0 0
Pm
1 2 .3 k N
2 1 6 ,6 0 0 8 0 0 0
Pa
4 .3 k N
2
SOLUTION (15.25) ( a ) By Eq. (15.24): F p (1 C ) P F i ( k
1800
4 5
P 6000,
kp p
kb
) P Fi
P 9700 N
Fb Fi 6 0 0 0 N
( b ) Load off:
Load on: F b 6 0 0 0
(9 7 5 0 ) 7 9 5 0 N
1 5
Thus Fm
6000 7950 2
Fa
7950 6000 2
6975 N 975 N
SOLUTION (15.26) Table 15.4:
d 20 mm ,
A t 245
2
mm
( a ) We have F b 0 . 9 S y A t 0 . 9 ( 630 )( 245 ) 138 . 915 and
kb
kp
AE
s
L
2
d Es
4L
2
( 20 ) E s 4 ( 60 )
0 . 58 ( E s 2 )( 20 ) 2 ln[ 5
0 . 58 ( 60 ) 0 . 5 ( 20 ) 0 . 58 ( 60 ) 2 . 5 ( 20 )
]
F i 0 . 75 F b 104 . 2 kN
kN
5 . 236 E s
9 . 379 E s
C
5 . 236 5 . 236 9 . 379
0 . 358
Thus Pb C P F i 0 .3 5 8 ( 4 0 ) 1 0 4 .2 1 1 8 .5
(b)
T KdF
i
kN
0 . 15 ( 20 )( 104 . 2 ) 312 . 6 N m
SOLUTION (15.27) Tables 15.1, 15.4, and 15.6:
A t 8 4 .3 m m ,
K
S p 380 M Pa,
S y 420
2
f
2 .2 M Pa ,
Su 520 M Pa
Table 7.3: C r 0 .8 9 Equation (15.39): S e ( 0 .8 9 )(1)( 21.2 )( 0 .4 5 5 2 0 ) 9 4 .7 M P a (CONT.)
249
15.27 (CONT.) 20 4 2
( a ) We have Pa
8 kN ,
Pm
20 4 2
12 kN
Thus a ( MPa )
m
9 4 .9 M P a ,
8 8 4 .3
12 8 4 .3
1 4 2 .3 M P a
Not safe. (Fig. S15.27)
a
(a) No preload
140
Soderberg Line
a
Se
105
Modified Goodman Line
70
ba
Figure S15.27
35
0
(b) With preload m
210 280
140
70
45o
0
S
420 490 520
350 bm
Su
m
( MPa )
y
( b ) F i 0 .7 5 (3 8 0 8 4 .3) 2 4 k N kb
2
d Es
2
( 0 .0 1 2 ) E s
4L
0 .0 0 2 3 E s
4 ( 0 .0 5 )
0 .5 8 ( E s 2 )( 0 .0 1 2 )
kp
2 ln [ 5
0 .5 8 ( 0 .0 5 ) 0 .5 ( 0 .0 1 2 ) 0 .5 8 ( 0 .0 5 ) 2 .5 ( 0 .0 1 2 )
0 .0 0 5 E s ,
C
]
0 .0 0 2 3 0 .0 0 2 3 0 .0 0 5
We have F b m C Pm F i 0 .3 1 5 (1 2 ) 2 4 2 7 .8 k N , ba
F b a 0 .3 1 5 (8 ) 2 .5 2 k N ,
2520 8 4 .3
0 .3 1 5
bm
2 7 ,8 0 0 8 4 .3
3 2 9 .8 M P a
2 9 .9 M P a
It is seen from Fig. S15.27 that the joint fails according to the Soderberg theory, while it is safe on the basis of Goodman criteria. ( c ) Use Eq.(15.37): 5 2 0 ( 8 4 .3 ) 2 4 , 0 0 0
n
520
( 0 .3 1 5 )[ 8 0 0 0 (
1 .1 3
) 1 2 ,0 0 0 ]
9 4 .7
( d ) Applying Eq.(15.28), we have ns
1 . 75
24 20 ( 1 0 . 315 )
SOLUTION (15.28) We have S
p
600
A t 115
i
mm
0 . 75 S
Pa Pm
a
m
(Table 15.5)
MPa
(Table 15.2)
0 . 75 ( 600 ) 450
p
MPa
5 kN
10 2
2
3
5 ( 10 ) 115 ( 10
6
)
43 . 48 MPa
Equation (15.39): Se CrCt (
1 K f
) ( 0 .4 5 S u )
(CONT.)
250
15.28 (CONT.) where
C r 0 .8 7
(Table 7.3)
C t 1 0 .0 0 5 8 ( 4 9 0 4 5 0 ) 0 .7 7
K
3 .8
f
(Table 15.6)
S u 830
Hence
(Table 15.5)
MPa
S e ( 0 . 87 )( 0 . 77 )(
We have
a
S u
n C
a
[(
(Eq.7.11)
1 3 .8
)( 0 . 45 830 ) 65 . 84 MPa
. With preload, Eq.(15.38) gives
m
i
Su
) 1]
Sy 830 450
( 43 . 48 )( 0 . 31 )(
2 . 07
830
1)
65 . 84
Without preload, C=1 and i 0 : n
1 . 40
830 830
1)
( 43 . 48 )(
65 . 84
Comment: The presence of preload is beneficial.
SOLUTION (15.29) Refer to Solution of Prob.15.28. We have A t 1 1 5 m m , 2
Hence,
C 0 .3 1,
S p 600 M Pa.
F i 0 . 75 ( A t S p ) 0 . 75 (115 600 ) 51 . 75 kN
Apply Eq.(15.27), S p At Fi
P
Cn
600 ( 115 ) 51 . 75 10 ( 0 . 31 )( 2 )
3
27 . 82 kN
Using Eq.(15.28b): P
Fi n s (1 C )
37 . 5 kN
51 . 75 2 ( 1 0 . 31 )
Comment: Failure owing to separation will not take place before bolt failure. SOLUTION (15.30) ( a ) We have 380
MPa
At 1 5 7
mm
Pa Pm
12 2
S
p
S u 520 2
MPa
Se 100 M Pa
(Table 15.1),
6
(Table 15.4)
k N b o lt
F i 0 .9 S p A t 0 .9 (3 8 0 )(1 5 7 ) 5 3 .6 9 k N
a
i
0 .9 S
m
6 1 0
1 5 7 (1 0
p
3 6
)
3 8 .2 2
M Pa
0 . 9 ( 380 ) 342
MPa
(CONT.)
251
15.30 (CONT.) kb
C
We have
S u C
a
m
. Eq.(15.38), with preload:
i
Su
[(
0 . 167
kb 5kb
a
n
kb
kb kb
) 1]
Sy 520 342
( 0 . 167 )( 38 . 22 )(
520
4 .5 1)
100
Without preload, C=1 and i 0 : n
2 . 19
520 520
1)
( 38 . 22 )(
100
( b ) Equation (15.28b): Fi
ns
P (1 C )
5 . 37
53 . 69 12 ( 1 0 . 167 )
Comments: The presence of preload is very beneficial. Joint will separate before bolts fail.
SOLUTION (15.31) Repeating section L P
P
We have L=w=50 mm and total number of rivets in the joint n=2. Therefore
P n (d
2
4)
b
P ndt
t
P ( w d e )t
3
4 ( 32 10 )
2 ( )( 0 . 019 ) 3
32 ( 10 ) 2 ( 0 . 019 )( 0 . 01 )
56 . 43 MPa
2
84 . 21 MPa 3
32 ( 10 )
[ 50 ( 19 3 )] 10 ( 10
6
)
114 . 3 MPa
SOLUTION (15.32) Sketch is the same as that given in Solution of Prob.15.31 and n=2. The allowable loads are: F s a ll n d Fb
b , a ll
Ft
t , a ll
2
4 1 0 5 ( 2 )( )(1 8 )
4 5 3 .4 4 k N
n d t 3 3 0 ( 2 )(1 8 )(1 0 ) 1 1 8 .8
kN
( w d e ) t 1 5 0[ 6 0 (1 8 1 .5 )]1 0 6 0 .7 5
Thus e
2
6 0 .7 5 (1 5 0 )( 6 0 )(1 0 )
6 7 .5 %
252
kN
SOLUTION (15.33) For members, by Table B.3: S y 4 6 0 M P a ,
S y s 0 .5 7 7 ( 4 6 0 ) 2 6 5 .4
M Pa
For bolts, from Table 15.4: S y 9 4 0 M P a ,
S y s 0 .5 7 7 (9 4 0 ) 5 4 2 .4
M Pa
from Table 15.1: d r 1 8 1 .2 2 7 ( 2 .5 ) 1 4 .9 3 m m (1 4 .9 3 )
As 2
Shear on bolts:
As S ys
Fs
3 5 0 .1 4 ( 5 4 2 .4 )
n
2
Ab S y
360 (940 )
n
3 5 0 .1 4 m m
2
95 kN
A b 2 (1 8 )(1 0 ) 3 6 0 m m
Bearing on bolts: Fb
2
4
2
1 1 2 .8 k N
3
Bearing on members: 360 ( 460 )
Fb
6 6 .2
2 .5
A t ( 6 0 1 8 )1 0 4 2 0 m m
Tension in members: 420 ( 460 )
Ft
kN 2
5 5 .2 k N Pa ll
3 .5
SOLUTION (15.34) For members, using Table B.3:
S
210
y
MPa
For bolts, from Table 15.1: d r 1 4 1 .2 2 7 ( 2 ) 1 1 .5 5 m m
2 (1 1 .5 5 )
As
Shear of bolt:
P As
2 0 ,0 0 0 2 0 9 .5
2 0 9 .5 m m
119 M Pa,
2 0 ,0 0 0 168
2
9 5 .4 7 M P a ,
n
A b 2 ( 6 )( 14 ) 168
Bearing on bolt: b
2
4
mm
n
640 119
3 .8 8
370 9 5 .4 7
2
5 .3 8
Bearing on members: n
1 . 76
210 119
A t ( 60 2 14 ) 6 192
Tension on members:
t
20 , 000 192
104 . 2 MPa ,
n
210 104 . 2
mm
2
2 . 02
SOLUTION (15.35) Table 15.1: A t 8 4 .3 m m ,
d r 1 2 1 .2 2 7 (1 .7 5 ) 9 .8 5 3 m m
2
Hence, A s
4
(9 .8 5 3 ) 7 6 .2 5 m m 2
2
P 4 As
P 4 ( 7 6 .2 5 )
P 305
Pivots about point A. Row 1 is the highest loaded. Apply Eq.(15.43) with j=2 and e=250 mm: F1
Thus
t ,m ax
M r1 2
2
r1 r2
2
t
250 P ( 225 ) 2
2
2 ( 225 ) 2 (75 )
(
2
t
) 2
t
0 .5 P , 2
P 3 3 7 .2
0 .5 P 8 4 .3
( 3 3P7 .2 ) ( 3 P0 5 ) 2
( 0 .0 0 3 0 .0 0 4 4 ) P 0 .0 0 7 4 P
Hence, 0 .0 0 7 4 P 1 4 0; or
m a x 0 .0 0 4 4 P 8 4 :
Pa ll 1 8 .9 2 k N P 1 9 .0 9
253
P 1 6 8 .6
kN
2
SOLUTION (15.36) The maximum design load is Pm a x n P ( 2 .3)( 2 5 ) 5 7 .5 k N . From geometry: F 2 ( 4 1 5 ) F1 . Here F1 and F 2 are tensile forces in bolts 1 and 2, respectively. We have
M
0:
A
1 0 0 0 (5 7 .5 ) 1 0 0 ( 145 F1 ) 3 7 5 F1 ;
F1 1 4 3 .2 k N
The required tensile stress area is then At
143,200 600
239 m m
2
From Table 15.1: The required thread size is about M 2 0 2 .5 C . SOLUTION (15.37) ( a ) Using Table 15.4, S p 3 1 0 M P a ,
S y 340 M Pa
S y s 0 .5 7 7 S y 1 9 6 .2 M P a
The tensile stress area: ( F o r c e )( n )
At
S
2 ,5 0 0 ( 3 ) 310
p
2 4 .2 m m
2
From Table 15.1, select M 7 1 C thread with A t 2 8 .9 m m
2
( b ) Table 15.1 and Fig. 15.3: p 1 mm, h
1 2
d 7 mm,
d r d 1 .2 2 7 p 5 .7 7 m m ,
( d d r ) 0 .6 1 5 m m
d m d 0 .6 5 p 6 .3 5 m m ,
b
p 8
2 h ta n 3 0
o
0 .8 3 5 m m
Apply Eq. (15.19a): 3K tPp 2 dbLn
S ys
3 ( 4 )( 2 5 0 0 )(1 )
n
2 ( 7 )( 0 .8 3 5 ) L n
1 9 6 .2 3
Solving, L n 1 2 .5 m m
SOLUTION (15.38)
(a)
Refer to Solution of Prob. 15.37. Using Table 15.4: S p 310 M Pa,
S y 340 M Pa ,
F o rce ( n )
6 4 .5 2 m m
At
Sp
4000 (5 ) 310
S y s 0 .5 7 7 S y 1 9 6 .2 M P a
2
From Table 15.1: Select M 1 2 1 .7 5 C thread with A t 8 4 .3 m m (b)
2
Table 15.1 and Fig. 15.3: d 12 m m ,
p 1 .7 5 m m
d m d 0 .6 5 p 1 0 .8 6 m m d r d 1 .2 2 7 p 9 .8 5 m m h
1 2
( d d r ) 1 .0 7 5 m m ,
b 1 .7 5 8 2 h ta n 3 0
o
1 .4 6 m m
(CONT.)
254
15.38 (CONT.) Using Eq. (15.19a): 3 ( 4 ) ( 4 0 0 0 ) (1 .7 5 )
2 (1 2 ) (1 .4 6 ) L n
1 9 6 .2 5
from which L n 1 9 .4 m m
SOLUTION (15.39) Table 15.1:
At 2 4 5 m m ,
Shear area:
As
d r 2 0 1 .2 2 7 ( 2 .5 ) 1 6 .9 3 m m
2
P 3
As
(1 6 .9 3 ) 2 2 5 2
4
2
1 , 481 P
P 3 ( 225 ) 10
mm
6
Pivots about point A. Bolt 1 is the highest loaded. M r1
F1
r1 r2 r3
t
2 4 5 (1 0
2
2
)
From Mohr’s circle:
2
2
140 90 40
2
0 .5 9 7 P
2 ,4 3 7 P
0 .5 9 7 P 6
1 2 5 P (1 4 0 )
2
t , max
t
2
(
t
2
) 2
2
2 , 437 P 2
(
2 , 437 P 2
) ( 1 , 481 P ) 2
2
( 1 , 219 1 , 918 ) P 3 ,137 P
Hence 3,1 3 7 P 1 4 5 1 0 ,
P 4 6 .2 2
6
or
1, 9 1 8 P 8 0 1 0 ,
kN
Pa ll 4 1 .7 1 k N
6
SOLUTION (15.40) Vertical and horizontal components of the force P=10 kN are 8 kN and 6 kN at B, respectively. Rivet B is most heavily loaded. The centroid of the group of rivets is at C. We have FB
M rB
8 3 0 (1 5 0 )
2
rj
2
2
2
2[ 3 0 9 0 1 5 0 ]
0 .5 7 1 k N
6/6=1 kN B
1
V B [1 1 . 904 2
2
] 2 2 . 15 kN
(Fig.a) 8/6+0.571 =1.904 kN
Therefore B
B
VB
d
2
VB
4
dt
2 ,1 5 0
( 20 )
2 ,1 5 0 2 0 1 5
2
4
6 .8 4 4
7 .1 6 7
M Pa
VB
M Pa
Figure S15.40
SOLUTION (15.41) Rivet A is the most heavily loaded. As
Vd
P 5
4
(15 )
50 5
2
176 . 715
10
kN
mm
2
V a ll 1 0 0 (1 7 6 .7 1 5 ) 1 7 .6 7 2
kN
Centroid C of the line AB, with respect to A, is determined from: 5 0 x 7 0 (1 0 ) 2 5 0 (1 0 ) 3 2 0 (1 0 ) 3 9 0 (1 0 ) ,
x
A
x 206
mm
C d
e
B (CONT.)
P
255
15.41 (CONT.) Thus Mr
FA
A 2
rj
50 , 000 e ( 206 ) 206
2
2
136
2
44 114
2
2
184
93 . 875 e
V A 1 0 , 0 0 0 9 3 .8 7 5 e 1 7 , 6 7 2 ,
e 8 1 .7 m m
and d 54 . 3 mm
SOLUTION (15.42) The A is the highest loaded point. A s (15 )
4 176 . 715
2
Thus
FA
A 2
rj
46 P ( 206 ) 206
2
136
2
2
x 206
Refer to Solution of Prob. 15.41: Per
mm
2
44 114
2
2
184
F
,
mm,
P 5
0 .2 P
e 206 70 90 46 mm
0 . 086 P
Hence V A ( 0 .2 0 .0 8 6 ) P 0 .2 8 6 P and
A
100 10 ;
P 6 1 .7 9 k N
6
0 .2 8 6 P 1 7 6 .7 1 5 (1 0
6
)
SOLUTION (15.43) A 0 . 707 hL 0 . 707 ( 7 )( 60 ) 297 P
S
ys
A
n
200 ( 297 ) 2 .5
mm
2
23 . 76 kN
Cross-sectional area of one plate A p 40 (10 ) 400
mm
2
. Then
P ( S y n ) A p ( 2 5 0 2 .5 ) 4 0 0 4 0 k N
Hence the capacity of the plate significantly exceeds that of the weld.
SOLUTION (15.44) Table B.4:
for plates,
Table 15.8:
for weld,
S y 427 M Pa . S y 345 M Pa .
Since 3 4 5 4 2 7 , the weld should yield first. The maximum load that can be applied equals P
SyA n
3 4 5 (1 5 8 7 .5 ) 4
1 1 3 .2 k N
SOLUTION (15.45) Refer to Solution of Prob. 15.44. S y 0 .5 S y 0 .5 ( 4 2 7 ) 2 1 3 .5 M P a From Eq. (15.44):
(for plate)
S y s 0 .5 S y 0 .5 ( 3 4 5 ) 1 7 2 .5 M P a
(for weld) (CONT.)
256
15.45 (CONT.) Since 1 7 2 .5 2 1 3 .5
The weld should yield first.
Hence, S ys A
P
1 7 2 .5 (1 5 8 7 .5 )
n
3
7 5 .4 7 k N
SOLUTION (15.46) For plates: S y 4 6 0 M P a
(Table B.3), S y s 0 .5 S y 2 3 0 M P a
For weld: S y 3 7 9 M P a
(Table 15.8), S y s 0 .5 S y 1 8 9 .5 M P a
(Eq. 15.44) (Eq. 15.44)
Since 1 8 9 .5 2 3 0
weld would yield first.
Thus, Fig. 15.27a: S ys
P
n
1 8 9 .5 ( 0 .7 0 7 )( 8 )( 2 7 0 )
4 2 .8 7 k N
3 .5
SOLUTION (15.47) For plates: S y 5 3 0 M P a
(Table B.3),
S y s 0 .5 S y 2 7 5 M P a
For weld: S y 4 1 4 M P a
(Table 15.8),
S y s 0 .5 S y 2 1 2 M P a
(Eq. 15.44) (Eq. 15.44)
Since 212 275
weld would yield first.
Thus, Fig. 15.26a: S ys
P
n
2 1 2 ( 0 .7 0 7 )( 5 )( 2 6 0 ) 4
2 2 .5 k N
SOLUTION (15.48) A p 75 10 750
mm
2
P A
,
all
750 (140 ) 105
kN
R 1 21 kN
(along AB)
M
D
0:
R 1 ( 75 ) 105 (15 ) 0 ;
M
A
0:
R 2 ( 75 ) 105 ( 75 15 ) 0 ;
Check: Hence
R 1 R 2 105 L1
R1 Sw
21 1 .2
R 2 84 kN
(along DE)
kN
17 . 5 mm ,
L2
R2 Sw
84 1 .2
70 mm
SOLUTION (15.49) I x 2 (1 0 0 t )( 6 2 .5 ) 2
3
2 (1 2 5 ) t 12
1 .1 0 7 (1 0 ) t
A 2 (1 0 0 t 1 2 5 t ) 4 5 0 t m m
We have
6
1 2
[( 1455000t0 ) ( 3 .71 .15 0672t .5 ) ]
Thus
a ll ;
and
h
0 .1 3 4 0 .7 0 7
2
2 1 4 .3 t
5 5,
a ll 5 5 M P a
M 1 5 ( 2 5 0 ) 3 .7 5 M N m m
2
2
4
mm ,
2 1 4 .3 t
t 3 .9 m m
5 .5 2 m m
257
SOLUTION (15.50) I x 1 .1 0 7 (1 0 ) t m m , 6
From Solution of Prob.15.49: We have Pm 1 5 k N ,
Pa 5 k N ,
K
Refer to Example 15.12. We obtain C
f
A 450t m m
4
1.5
(Table 15.9)
A Su 272(420) b
f
2
0 .9 9 5
0 .6 6 7 .
Equation (7.5a): S u s 0 .6 7 ( S u ) 0 .6 7 ( 4 2 0 ) 2 8 1 .4 M P a S e C s C f (1 K f ) S e ( 0 .7 )( 0 .6 6 7 )( 11.5 )( 0 .5 4 2 0 ) 6 5 .3 7 M P a '
Then, from Eq.(7.21): S us n
m (
a
)
m
Su
2 8 1 .4 2 .5
1
3 5 .8 3 M P a
1
420 ) 1 3 6 5 .3 7
(
Se
Also, from Solution of Prob.15.49: m Thus
2 1 4 .3 t
3 5 .8 3
and
h
5 .9 8 0 .7 0 7
1 . 072 t
t 5 .9 8 m m
or
8 .4 6 m m
SOLUTION (15.51) Table 15.8: S y 4 1 4 M P a
S y s 0 .5 S y 2 0 7 M P a
We have t 0 . 707 (12 ) 8 . 484
mm
A one weld 8 . 484 L
T 100 ( 60 ) 6 MN mm
Total
3
J 2
2
tL 12
8 .4 8 4 L 12
3
1.4 1 4 L
3
At point A: A S ys
and
P A
n
207 2 .5
Tr J
100 ,000
1 6 .9 6 8 L
82 . 8
6
6 (1 0 )( L 2 ) 1. 4 1 4 L
3
5 ,893 L
2 ,1 2 2 , 0 0 0
L
(1)
2
(2)
A
From Eqs.(1) and (2): 82 . 8 L 5 ,893 L 2 ,122 , 000 0 ; 2
L 71 . 17 L 25 , 628 0 2
Solving L
1 2
[ 7 1 .1 7
7 1 .1 7 4 ( 2 5 , 6 2 8 ) ] 1 9 9 .6 2
mm
SOLUTION (15.52) Pm 100
Pa 20 kN
kN ,
Table 15.9: K
f
1.5
Refer to Solution of Prob.15.51: m
Also
5 ,893
L
S u 496
C
f
S
ys
AS
2 ,1 2 2 , 0 0 0 L
MPa , b u
(1)
2
S
y
272 ( 496 )
0 . 5 ( 414 ) 207
414 0 . 995
MPa
MPa
(Table 15.8),
0 . 566
Refer to Example 15.12:
S e (0 .7 )(0 .5 6 6 )( 11.5 )(0 .5 4 9 6 ) 6 5 .5 1 M P a
(CONT.)
258
15.52 (CONT.) We have, from Eq.(7.21) with S y s and S y replacing S u s and S u : S ys n
m (
a m
)
Sy
2 0 7 2 .5
1
(
Se
3 6 .5 7 M P a
1
414 ) 1 5 6 5 .5 1
(2)
Equations (1) and (2) are therefore L 161 . 1 L 58 , 025 . 7 0 2
L
or
1 2
[1 6 1 .1
1 6 1 .1 4 (5 8 , 0 2 5 .7 ) ] 3 3 4 .5 2
mm
SOLUTION (15.53) S u s 0 .6 7 S u 2 8 6 .1 M P a ;
Table 15.8: S u 4 2 7 M P a; We have C
f
1 .5
A b o th 2[ 0 .7 0 7 ( 6 .5 ) 2 5 0 ] 2 2 9 7 .8
mm
A Su 272(427) b
f
0 .9 9 5
K
(Table 15.9)
0 .6 5 7
Refer to Example 15.12: S e ( 0 .7 )( 0 .6 5 7 )( 11.5 )( 0 .5 4 2 7 ) 6 5 .5 M P a
We have Pm Pa 3
J 2
tL 12
Pm a x , AL 12
2
2 , 2 9 7 .8 ( 2 5 0 )
2
12
1 2 (1 0 ) 6
We have Pm 0 .5 Pm a x ,
At point A: m
and
1 2
Pm
A
Tm r J
Pm a x
4
T m 7 5 Pm 3 7 .5 Pm a x
3 7 .5 Pm a x (1 2 5 )
2 ( 2 2 9 7 .8 )
mm
2
6
1 2 (1 0 )
0 .0 0 0 6 Pm a x
Also, from Eq.(7.21): S us n
m (
a m
)
Su
2 8 6 .1 2
1
(1 )
427
1 9 .0 2 M P a
1
6 5 .5
Se
Thus 0 .0 0 0 6 Pm a x 1 9 .0 2 ,
Pm a x 3 1 .7 k N
SOLUTION (15.54) 150 mm
y A
B 8
200 mm
C
6
x
T
30 ( 10 A
D
E
40 ( 10 A
3
)
3
)
T ( 200 )
Figure S15.54
J
T ( 75 ) J
Inspection of Fig. S15.54 shows that point E has the highest stress. We write T 4 0 (1 7 5 ) 3 0 (1 0 0 ) 4 k N m
A b o th 2 (1 5 0 t ) 3 0 0 t m m
2
(CONT.)
259
15.54 (CONT.) t (1 5 0 )
I x 2[
3
1 5 0 t (1 0 0 ) ] 2 ( 0 .2 8 1 t 1 .5 t )(1 0 ) 3 .5 6 2 (1 0 ) t m m 2
12
6
I y 2[ 0 1 5 0 t ( 7 5 ) ] 1 .6 8 7 (1 0 ) t m m , 2
6
6
J 5 .2 4 9 (1 0 ) t m m
4
6
4
4
We have 3
v
and
4 0 (1 0 ) 300 t
4 (75 )
6
5 .2 4 9 (1 0 ) t 1 2
E [ v h ] 2
2
1 3 3 .3 t
1 6 6 .6 t
h
,
3
3 0 (1 0 ) 300 t
4 (1 0 0 ) 6
5 .2 4 9 (1 0 ) t
100 t
M Pa
Therefore, by Eq.(15.44), n E 0 .5 S y ;
3(
1 6 6 ,6 t
) 0 .5 (3 5 0 ),
t 2 .8 6 m m
and h
4 .0 5 m m
2 .8 6 0 .7 0 7
SOLUTION (15.55) J 5 .2 4 9 (1 0 ) t m m , 6
From Solution of Prob.15.54: From Table 15.9: C
K
2 .7 ;
f
A Su 272(427) b
f
A b o th 3 0 0 t m m
4
2
S u s 0 .6 7 S u 0 .6 7 ( 4 2 7 ) 2 8 6 M P a
0 .9 9 5
0 .6 5 7
Refer Example 15.12: S e ( 0 .7 )( 0 .6 5 7 )( 21.7 )( 0 .5 4 2 7 ) 3 6 .4 M P a We have Pm 3 0 k N ,
Pa 2 0
a m 2 3
kN ,
Thus, by Eq.(7.21): S us n
m (
a m
)
Su
2 8 6 1 .5
1
(
Se
2
427 ) 1 3 3 6 .4
2 1 .6 M P a
(1)
At point E: Use the method of Solution of Prob.15.54, with P Pm 3 0 k N Then Pm v 2 4 k N
Pm h 1 8 k N
T m 2 4 (1 7 5 ) 1 8 (1 0 0 ) 2 .4
kN m
Therefore
and
3
vm
2 4 (1 0 )
hm
1 8 (1 0 )
300 t 3
300 t
2 .4 ( 7 5 )
6
5 .2 4 9 (1 0 ) t 2 .4 (1 0 0 )
6
5 .2 4 9 (1 0 ) t 1 2
m ( v m h m ) 2
2
80 t
60 t
100 t
(2)
Equations (1) and (2) give 100 t
2 1 .6 ,
t 4 .6 m m
and h
4 .6 0 .7 0 7
6 .5 1 m m
End of Chapter 15
260
CHAPTER 16
MISSELLANEOUS MACHINE COMPONENTS
SOLUTION (16.1) Equation (16.17) and Eq.(16.16a) at r=a: 1 , max p i
(a) 1
Sy;
2
2
2
2
2
a b b a
5 4
pi
2
p i ( p i ) 260 ,
5 4
pi
r , max
p i 115 . 6 MPa
( b ) 1 1 2 2 S y 2
2
2
1
p i [( 54 )
( 54 )( 1 ) ( 1 ) ] 2 260 ,
2
p i 133 . 2 MPa
2
SOLUTION (16.2) T
b
3
2
( 100 ) b
3
157 . 08 b
2
3
Also
T 2 b fp l 2 b fp ( 3b ) 6 fb p
and
6 fb p 157 . 08 b ,
2
2
3
Hence
p
3
h
We have E h E s E ,
3
2 bpc
2
2
2
2 bp ( 4 b )
2
E (c b )
2
0 . 706 (10
8 ( 55 . 56 ) b 3
3 ( 210 10 )
(Eq.16.27b)
55 . 56 MPa
, and a=0. Equation (16.25) becomes
s
8 pb
2
E (4b b )
157 . 08 6 ( 0 . 15 )
3E 3
)b
SOLUTION (16.3) ( a ) From Table B.1: S y 2 5 0 M P a ,
0 .3 .
E 200 G Pa,
Applying Eq. (16.17), p i ,m ax
2
2
2
2
b a b a
250
2
2
2
2
0 .1 8 0 .1 2 0 .1 8 0 .1 2
9 6 .1 5 M P a
Equation (16.16c) at r a : u m ax
2
Pi a
2
( b2 a2 ) b a
E
6
9 6 .1 5 (1 0 ) 9
2 0 0 (1 0 )
2
2
( 0 .1 2 )( 0 .1 8 2 0 .1 2 2 0 .3 ) 0 .1 8 0 .1 2
0 .1 6 7 m m
( b ) Equation (16.19): 2
Po , m a x
b a 2b
2
2
250
2
0 .1 8 0 .1 2 2
2 ( 0 .1 8 )
2
6 9 .4 4 M P a
SOLUTION (16.4) ,m ax p i
(a) Su and
5 3
pi ,
2
2
2
2
b a b a
pi
Su pi ,
pi 1,
5 3
3 5
r ,m ax p i 2
( 350 ) 210 p i 350
MPa
(governs)
MPa
(CONT.)
261
16.4 (CONT.) (b)
1 Su
2
S uc
5 pi
1;
pi
3( 350 )
1
650
p i 158 . 7 MPa
or
SOLUTION (16.5) ( a ) Use Eq.(16.17) with p i p , ,m ax p
2
2
2
2
c b c b
a b,
b c:
6 2 .5 p 1 0 0 1 0
9
[
6 0 1 0 p
2
2
c b
By Eq.(16.25), with a 0 , 0 .0 5
2
c b
60 M Pa,
6
2
(Fig. 16.6b) 6
6 0 (1 0 )
0 . 05 mm :
b 62 . 5 ,
0 .3 ]
6 2 .5 p 2 0 0 1 0
(1)
p
[1 0 .3 ]
9
or 50 10
6
37 . 5 10
6
0 . 188 p 0 . 219 p ,
p 30 . 71 MPa
( b ) Equation (1) becomes 2
2
2
2
c 0 .0 6 2 5
60 3 0 .7 1
c 0 .0 6 2 5
c 110 m m
,
2c 220 m m
SOLUTION (16.6) ( a ) a=0, b=12.5 mm, c=50 mm F
T b
12 kN
150 0 . 0125
Thus
1 2 ( 1 0 ) 2 b fp l 2 ( 0 .0 1 2 5 )( 0 .1 5 )( 0 .0 5 ) p
or
p 20 . 37 MPa
3
(Eq.16.27a)
Equation (16.25) gives then
2
3
1 2 .5 ( 2 0 .3 7 )
5 0 1 2 .5
1 0 0 (1 0 )
( b ) ,m ax p
2
[ 5 0 2 1 2 .5 2 0 .3 ]
1 2 .5 ( 2 0 .3 7 )
2
2
2
2
c b c b
2 0 0 1 0
2
3
(1 0 .3 ) ( 3 .6 5 0 .8 9 1)1 0
3
0 .0 0 5
mm
2
2 0 .3 7[ 5 0 2 1 2 .5 2 ] 2 3 .0 9
M Pa
5 0 1 2 .5
SOLUTION (16.7) We have a=15 mm, b=25 mm, c=50 mm. Equation (16.25): 0 . 025
2
25 p 210 10
9
2
[ 50 2 25 2 0 . 3 ] 50 25
2
25 p 105 10
9
2
[ 25 2 15 2 0 . 3 ], 25 15
Steel, Eq.(16.17): ,m ax p
2
2
2
2
c b c b
3 7 .3 9
2
2
2
2
50 25 50 25
6 2 .3 2
M Pa
Bronze, Eq.(16.19): ,m ax 2 p
b 2
2
b a
2
2 (3 7 .3 9 )
25 2
2
2 5 1 5
262
2
1 1 6 .8
M Pa
p 37 . 39 MPa
SOLUTION (16.8) Equation (16.25) with a=0, b=50 mm, c=150 mm.
bp Ec
2
2
b c
bp
[ c2 b2 c ] 2
150 50
50 p
0 .0 3
12010
9
Es
[1 s ]
2
50 p
[ 1 5 0 2 5 0 2 0 .2 5 ]
21010
9
( 0 .7 )
Solving p 3 7 .8 9 M P a Shaft: r p 3 7 .8 9 M P a Cylinder: ,m ax p
2
2
2
2
c b c b
2
150 50
p 3 7 .8 9
r ,m ax
2
3 7 .8 9[ 1 5 0 2 5 0 2 ] 4 7 .3 6
M Pa
M Pa
SOLUTION (16.9) Equation (16.23):
(u d ) rb
2
bp E
2
b c
2
bp
[ c2 b2 ]
E
( 3 0 .3 ) 1.9 6 7 5
bp E
or p
E 3 .9 3 4 b
Then, Eq.(16.24) with a=0, us
bp E
(1 )
0 .7 3 .9 3 4
0 .1 7 8
Therefore, d s 0 .3 5 6
SOLUTION (16.10) Equation (16.31a) with ( r ) p 0 and b c :
r
3 8
(a b 2
3 0 .3 4 8
2
2
a b r
2
2
r ) 2
[( 0 .0 2 ) ( 0 .0 4 ) 2
2
2
p 2
( 0 .0 2 ) ( 0 .0 4 ) ( 0 .0 3 )
2
2
]8 5 0 0
2
9 0 (1 0 ) 6
Solving, 8 , 0 7 5 .5 tp s 4 8 , 4 5 3 r p m
SOLUTION (16.11) m ax
3600 60
2 1 2 0 ra d s ,
m in 1 1 4
ra d s ,
p 0
( a ) Equation (16.32), with p 0 ; ,m ax
4
2
[ (1 ) a
2
( 3 )b ] 2
(CONT.)
263
16.11 (CONT.) Let a=b and b=c=4b: 75 10
7 , 800 ( 120 )
6
2
3 .3( 4 b ) ]
2
[ 0 .7 b
4
2
Solving, b 7 1 .1 2 m m ,
c 2 8 4 .4 m m
( b ) Equation (16.37) of Sec.16.5: I
l
7 .8 ( 5 0 )
(c b ) 4
2
4
( 0 .2 8 4 4 0 .0 7 1 1 2 ) 3 .9 9 2 1 N m s 4
2
4
2
Thus Ek
I ( m a x m in ) 2
1 2
2
(3 .9 9 2 1)(1 2 0 1 1 4 ) 2
1 2
2
2
2 7 .6 5 9
kN m
SOLUTION (16.12) 2 4 0 0 ( 2 6 0 ) 2 5 1 .3 ra d s
( a ) Equation (16.35) with a=b and b=c:
bp E
2
b
2
[ b2 c 2 1] c b
20 ( 10
20 E
3
2
4E
2
[ 0 .7 b
2
[ 100 2 20 2 1 ]
)p
E
2
20 ( 7 . 8 )
20
100
3 .3 c ]
2
[ 2 . 083 p 64 . 896 ]( 10 2
For p 3 . 2 MPa 1 .0 2 5 1 0
[ 0 . 7 ( 0 . 02 ) 3
2
3 .3( 0 .1) ] 2
(1)
)
2 5 1 .3 ra d s , Eq.(1) gives.
and 3
2
4E
mm
We have at 0: 3
1 .0 2 5 1 0
20 2 1 0 1 0
p 5 .1 6 7 M P a
( 2 .0 8 3 p ) ,
9
( b ) Thus p
2
2
2
2
b c c b
5 .1 6 7 (1 .0 8 3 ) 5 .5 9 6 M P a
SOLUTION (16.13) ( a ) Equation (16.26) with a=0: p
2
E b
c b
2
2
2c
200 ( 10
9
2
)( 0 . 02 ) 300
50
50
2 ( 300 )
2
2
38 . 89 MPa
Then
, max
p
2
2
2
2
c b c b
38 . 89
300 300
2
50
2
2
50
2
41 . 11 MPa
( b ) Equation (16.35) with b=0.05 m, c=0.3 m, p=0, 7 .8 k N m , and E 2 0 0 G P a : 0 . 02 ( 10
3
)
b 4E
2
[ 0 .7 b
2
3 . 3 c ] 145 . 64 10 2
Solving, 3 7 0 .6
ra d s . Thus
n 3 7 0 .6 ( 6 0 2 ) 3 5 3 9 rp m
264
2
12
SOLUTION (16.14) ( a ) From Eq. (16.17), we have p ,m ax
2
2
2
2
c b b c
30
2
2
2
2
( 0 .1 2 ) ( 0 .0 6 ) ( 0 .0 6 ) ( 0 .1 2 )
18 M Pa
( b ) Using Eq. (16.27a): F 2 b p fl 2 ( 0 .0 6 )(1 8 1 0 )( 0 .1 8 )( 0 .2 ) 6
2 4 4 .3 k N
( c ) By Eq. (16.27b): T F b 2 4 4 .3 ( 0 .0 6 ) 1 4 .6 6 k N m
SOLUTION (16.15) 4 I
2
m a x 2 4 0 0 ( 2 6 0 ) 2 5 1 .3 ra d s
(b a )l 4
4
2
m in 1 2 5 .7 ra d s
[ 0 .2 0 .0 5 ]( 6 0 )( 7 .8 ) 1 .1 7 2 N m s 4
4
2
Therefore T
I ( m a x m in ) 2
1 2
2
or T
2
2
1 .1 7 2 ( 2 5 1 .3 1 2 5 .7 ) 2 (4 )
2 .2 0 8 k N m
SOLUTION (16.16) (a) I
2
(b a )l
m ax
4
4
3000 60
[ 0 .2 5 0 .0 2 5 ]( 7 .8 )( 6 0 ) 2 .8 7 1 N m s 4
2
4
( 2 ) 1 0 0 ,
2
m in 0 .9 (1 0 0 ) 9 0 ra d s
Equation (16.32) with p=0: , max
( b ) E k
1 2
4
2
[( 1 ) a
7 ,8 0 0 (1 0 0 )
2
2
(3 )b ] 2
[ 0 .7 ( 0 .0 2 5 ) 3 .3 ( 0 .2 5 ) ] 3 9 .7 8 M P a 2
4
I ( max min ) 2
2
1 2
2
( 2 . 871 )( 100
2
90
2
)
26 . 919
2
kN m
SOLUTION (16.17)
Dimensions are in millimeters.
A2
25 50
B
A
25
A1
r A ri 2 1 5 m m , A 12 , 500 r
150
50
A1 r1 2 A 2 r2 A1 2 A 2
mm
rB 4 1 5 m m
2
( 5 0 1 0 0 )( 2 1 5 2 5 ) 2 ( 2 5 1 5 0 )( 4 1 5 7 5 ) 5 0 1 0 0 2 ( 2 5 1 5 0 )
300 m m
r
(CONT.)
265
16.17 (CONT.) Equation (16.50): R
A
12, 500
dA r
265 dr r
100 215
2 8 8 .4 3 9 6 m m
415
25 265
dr r
e r R 3 0 0 2 8 8 .4 3 9 6 1 1 .6 0 4 m m
Equations (16.55a): ( ) A
100
P A
[1
P 1 2 ,5 0 0
r ( R rA ) e rA
[1
]
3 0 0 ( 2 8 8 .4 3 9 6 2 1 5 ) (1 1 .5 6 0 4 )( 2 1 5 )
]
or P 1 2 6 .7 k N
( ) B
100
P A
[1
P 1 2 ,5 0 0
r ( R rB )
[1
e rB
]
3 0 0 ( 2 8 8 .4 3 9 6 4 1 5 ) (1 1 .5 6 0 4 )( 4 1 5 )
]
or P 1 8 0 .8 k N
SOLUTION (16.18) ro ri h 5 0 4 0 9 0 m m A b h ( 2 0 )( 4 0 ) 8 0 0 m m
2
We have h
R
ln
ro
40
ln
1
6 8 .0 5 1 9
50
ri
r
90
( ri r ) 7 0 m m
2
e r R 1 .9 4 8 1 m m
( a ) Using Eq.(16.52): i
(b)
7 0 0 ( 6 8 .5 1 9 5 0 ) 8 0 0 (1 0
6
7 0 0 ( 6 8 .0 5 1 9 9 0 )
o
o
i
8 0 0 (1 0
6
Mc I
1 6 2 .2 M P a
) (1 .9 4 8 1) ( 0 .0 5 )
1 0 9 .5 M P a
) (1 .9 4 8 1) ( 0 .0 9 )
7 0 0 ( 0 .0 2 ) 1
( 0 .0 2 ) ( 0 .0 4 )
1 3 1 .3 M P a 3
12
266
SOLUTION (16.19) h
ri r
2 0 0 3 2 .5 1 6 7 .5 m m
2 h
ro r
2 0 0 3 2 .5 2 3 2 .5 m m
2
A b h ( 4 5 )( 6 5 ) 2 9 2 5 m m
2
Therefore h
R
ro
ln
65
ln
2 3 2 .5
1 9 8 .2 2 7 m m
1 6 7 .5
ri
e r R 2 0 0 1 9 8 .2 2 7 1 .7 7 3 m m
( a ) Applying Eq.(16.52): i
M ( R ri )
1 .5 1 0 (1 9 8 .2 2 7 1 6 7 .5 ) 3
A e ri
5 3 .0 6 M P a
( 2 9 2 5 ) (1 .7 7 3 ) (1 6 7 .5 )
( b ) Use Eq. (16.52):
o
M ( R ro )
1 .5 1 0 (1 9 8 .2 2 7 2 3 2 .5 ) 3
A e ro
4 2 .6 4 M P a
( 2 9 2 5 ) (1 .7 7 3 ) ( 2 3 2 .5 )
SOLUTION (16.20)
45 mm
rA 1 2 5 m m ,
5 mm
5 mm
A 325
A 25 mm
B
r
A1 A 2
2
r
Equation (16.50):
A1 r1 A 2 r2
mm
A1
A2
R
rB 1 7 0 m m
A
325
dA r
130
25
125
dr r
2 5 ( 5 )(1 2 7 .5 ) ( 4 0 5 )(1 5 0 ) 125 200
1 3 9 .9 7 5 3 m m
170
5
130
dr r
1 4 1 .3 4 6 2 m m
e r R 1 .3 7 9 8 m m
Equations (16.53), with M P ( r 2 5 ) 1 6 6 .3 4 6 2 P and P P : ( ) B
P A
120
P 325
[1
[1
( r 2 5 )( R rB ) e rB
]
1 6 6 .3 4 6 2 (1 3 9 .9 7 5 3 1 7 0 ) (1 .3 7 9 8 )(1 7 0 )
]
or P 1 .9 2 2 k N
267
SOLUTION (16.21) h
R
ln
60
ro
ln
1 2
2 4 8 .7 9 5 4 m m
220
ri
r
280
1
( ri ro )
(220 280) 250 m m
2
A ( 6 0 )( 6 0 ) 3 6 0 0 m m
2
e r R 2 5 0 2 4 8 .7 9 5 4 1 .2 0 4 6 m m
Using Eq.(16.52): i
M ( R ri )
M ( 2 4 8 .7 9 5 4 2 2 0 )
1 5 0 (1 0 ) 6
;
A e ri
3 6 0 0 (1 0
6
, M 4 .9 7 k N m
)(1 .2 0 4 6 )( 0 .2 2 )
Similarly,
o
M ( R ro )
M ( 2 4 8 .7 9 5 4 2 8 0 )
; 1 5 0 (1 0 ) 6
A e ro
3 6 0 0 (1 0
6
,
M 5 .8 4 k N m
)(1 .2 0 4 6 )( 0 .2 8 )
Therefore M
a ll
4 .9 7 k N m
SOLUTION (16.22) A 5 0 b 2 (1 5 0 2 5 ) 5 0 b 7 5 0 0
We have rA 1 5 0 m m and rB 3 5 0 m m . Applying Eq. (16.52):
A
B
( M )( R r A )
( M )( R rB )
A e rA
A e rB
from which rB ( R r A ) r A ( R rB ) ,
or
R 210 m m
Then
R
A
or
350(R 150) 150(R 350)
A 210
dA r
200 b d r 1 5 0 r
350 200
2(25)dr 210 r
6 0 .4 1 3 2 b 5 8 7 5 .9 6 5 8
Hence or
5 0 b 7 5 0 0 6 0 .4 1 3 2 b 5 8 7 5 .9 6 5 8 b 156 m m
268
4 7 b ln 4 0 ln 3 4
SOLUTION (16.23) r 150 m m
A 5000 m m
2
M ( R d )P
Table 16.1, Case A: R
h ln
ro r2
100 ln
2 1
1 4 4 .2 6 9 5 m m
e r R 1 5 0 1 4 4 .2 6 9 5 5 .7 3 0 5 m m
Equation (16.53): ( ) A
P A
( r d ) P ( R rA ) A e rA
]
P A
( r d )( R r A )
[1
e rA
]
or 3
2 5 (1 0 )
80
5000
[1
(1 5 0 d )(1 4 4 .2 6 9 5 1 0 0 ) ( 5 .7 3 0 5 )(1 0 0 )
]
1 5 ( 5 7 3 .0 5 ) (1 5 0 d ) ( 4 4 .2 6 9 5 ) 6 , 6 4 0 .4 2 5 4 4 .2 6 9 5 d
Solving, d 4 4 .1 7 m m
SOLUTION (16.24) Locate centroid : 120 mm 50 mm
A2
r
C A1
O
75 mm
80 mm
A1 r1 A 2 r2 A1 A 2 4 5 0 0 (1 2 0 ) ( 3 0 0 0 ) (1 6 0 ) 4500 3000
136 m m
r
1 2
R
h ( b1 b 2 ) 2
( b1 ro b 2 ri ) ln
ro ri
h ( b1 b 2 )
( 0 .5 ) (1 2 0 ) ( 7 5 5 0 ) 2
[ ( 7 5 ) ( 2 0 0 ) ( 5 0 ) (8 0 ) ] ln
200
1 2 7 .1 3 3 0 m m
P
(1 2 0 ) ( 7 5 5 0 )
80 e r R 8 .8 6 7 m m
( a ) Use Eq.(16.55a),
A
P
[1
r ( R rA )
A
A
B ]
O
M=P r
e rA
P
r
(1 3 6 ) (1 2 7 .1 3 3 0 8 0 ) 1 1 0 0 .4 M P a 6 7 5 0 0 (1 0 ) (8 .8 6 7 ) (8 0 ) 75000
(CONT.)
269
16.24 (CONT.) ( b ) From Eq.(16.55b):
P
B
r ( R rB )
[1
A 75000
]
e rB
7 5 0 0 (1 0
6
(1 3 6 ) (1 2 7 .1 3 3 0 2 0 0 ) 1 4 5 .9 M P a ) (8 .8 6 7 ) ( 2 0 0 )
SOLUTION (16.25) A d
r 100 m m
4 7854 m m
2
2
From Table 16.1, Case B: R
A 2 (r
r
2
c
2
9 3 .3 0 1 5
7854 2 (1 0 0
)
2
100 50
2
)
e r R 6 .6 9 8 5 m m
(a)
Equations (16.55a): ( ) A
P A
[1
r ( R rA ) e rA
]
or 150
P 7854
( ) B
(b)
P A
1 0 0 ( 9 3 .3 0 1 5 5 0 )
[1
( 6 .6 9 8 5 )( 5 0 )
[1
r ( R rB ) e rB
]
P 8 4 .5 8 k N
], 3
8 4 .5 8 (1 0 ) 7 8 5 4 (1 0
6
)
[1
1 0 0 ( 9 3 .3 0 1 5 1 5 0 ) ( 6 .6 9 8 5 )(1 5 0 )
]
50 M Pa
SOLUTION (16.26) rA 1 2 5 m m , r
1
rB 2 2 9 m m , 1
( ro ri )
2
a 100 m m ,
b 50 m m
(229 125) 177 m m
2
From Table 16.1 A a b (1 0 0 )(5 0 ) 5 0 0 0 m m
dA
r
2 b
[r
r
2
2
P
a ] 2
a
2 (5 0 )
[1 7 7
1 7 7 1 0 0 ] 9 7 .2 4 9 5 m m 2
2
r
A
100
Hence A
R
dA A
5 0 0 0
1 6 1 .5 2 1 5 m m
9 7 .2 5
P M=-P r B
r
(CONT.)
270
16.26 (CONT.) e r R 1 5 .4 7 8 5 m m
Equation (16.55a) with M P r :
A
P
[1
r ( R rA )
A
3
1 7 7 (1 6 1 .5 2 1 5 1 2 5 ) 1 ) (1 5 .4 7 8 5 ) (1 2 5 )
1 2 5 (1 0 )
]
e rA
5 0 0 0 (1 0
6
3 4 .5 4 M P a
B
P
[1
r ( R rB )
A
3
1 7 7 (1 6 1 .5 2 1 5 2 2 9 ) 1 ) (1 5 .4 7 8 5 ) ( 2 2 9 )
1 2 5 (1 0 )
]
e rB
5 0 0 0 (1 0
6
1 8 .8 6 M P a
SOLUTION (16.27) rA 1 3 0 m m r
1
rB 2 0 0 m m
( ro ri ) 1 6 5 m m
2
A a b ( 7 0 )(3 5 ) 2 4 5 0 m m
dA
2 b
r
[r
r
2
2
a ] 2
P
a
2 (3 5 )
[1 6 5
1 6 5 7 0 ] 4 8 .9 6 0 1 m m 2
2
70
A
R
dA A
r
2 4 5 0
A
1 5 7 .2 0 7 6 m m
4 8 .9 6 0 1
P
r
e r R 7 .7 9 2 4 m m
M=-P r B
Using Eq.(16.55) with M P r :
B
P
[1
A
r ( R rB )
]
e rB
9 0 (1 0 ) 6
Pa ll 2 4 5 0 (1 0
6
(1 6 5 ) (1 5 7 .2 0 7 6 2 0 0 ) 1 ) ( 7 .7 9 2 4 ) ( 2 0 0 )
Solving Pa ll 1 9 6 .2 k N
271
SOLUTION (16.28) 1
A
bh
2
The section width w varies linearly with r. Thus (1) w c 0 c1 r Since w b
( a t r ri )
w 0
dr (2)
( a t r ro )
r
Substituting Eq.(1) into Eq.(2); c1
b
c0
h
w
O
b
b ro h
ri
h r
Then
dA A
r
ro ri
w
c 0 c1 r
dr
r
dr
r
Inserting c1 and c 0 into this, after integrating and rearranging, we have dA
b(
ro
r
ln
h
ro
1)
ri
Therefore 1
A
R
dA r
2
ro
ln
h
h ro
1
ri
SOLUTION (16.29) r A c
2
r
Through use of the polar coordinates we write: w 2 c s in
r r c cos
d r c s in d
C
d A w d r 2 c s in d 2
2
w
O
c
dA r
2 c s in
0
r c cos
2
2
2
d
c (1 c o s )
0
r c cos
2
ccos
2
2
c cos (r c ) 2
2
2
2
r c cos
0
2 (r c cos )d 2(r 0
2
r
dr
2
c )
2
0
d r c cos
(CONT.)
272
16.29 (CONT.)
r c 2
2r
0
2 c s in
2(r
0
2
2
c ) 2
ta n
r c 2
2
ta n
2
1
r c
2
0
This gives
dA
2 ( r
r c ) 2
2
r
SOLUTION (16.30) A
1 2
( b1 b 2 ) h
ri
h w
The section width w varies linearly with r as
b2
w c 0 c1 r
(1)
( a t r ri )
w b2
( a t r ro )
O
r
dr
We have w b1
b1
r
(2)
Introduce Eq.(1) into Eq.(2), then solve for c 0 and c1 c0
ro b1 ri b 2
c1
h
b1 b 2
(3)
h
Then, we write
dA r
ro
w
ri
dr
r
ro
c 0 c1 r
ri
r
d r c 0 ln
ro ri
c 1 ( ro ri )
This gives, substituting Eqs.(3):
dA
ro b1 ri b 2
r
h
ln
ro ri
( b1 b 2 )
Hence 1 A
R
dA r
2
h ( b1 b 2 ) 2
( ro b1 ri b 2 ) ln
ro ri
( b1 b 2 )
SOLUTION (16.31) As in Example 16.7, we take c 1 c 2 0 in Eq.(16.60). Then, substituting Eq.(16.60) into (16.56a) we obtain an expression for the radial moment M r . (CONT.)
273
16.31 (CONT.) Boundary conditions: w 0
M
0
r
(r a )
now give p0a
c4
2
c3a
4
0
64 D
p0a
c3
2
32 D
3 1
from which c 4 p 0 a ( 5 ) 6 4 (1 ) D 4
The plate deflection is thus p0a
w
4
r
p0a
w m ax
At r=0:
4
3 1
( a4 2
64 D
4
r
2 2
a
5 1
(1)
)
5 1
64 D
Substituting Eq.(1) into Eq.(16.56b): p0
M
[( 3 ) a
16
2
( 1 3 ) r ] 2
At r=a: 6 M
,m ax
t
3 (1 ) p 0
2
4
( at )
2
SOLUTION (16.32) Refer to Example 16.7. At r=a:
r
6M t
r
2
S
( a ) 1 2 or
y
n
1 2
p0a
3 4
2
2,
2
t
p0 ( t ) (
a
S
2
2
2
p 0 ( t ) [ 16 4
y
1
S
y
)
n
1 4
3 4
6M t
S
) S
n 2
,
n
( b ) 1 1 2 2 ( or a
2
a
;
p0 ( t )
2
y
4t
2
2
a
2
p0a
1
y
n
t
(a)
P0
2
2
2
3 16
t
2
9 16
] (
Sy
2
7 4
) ;
n
p0 ( t )
Sy n
Solving, n 1.5 1 2
S
y
p0
(a)
SOLUTION (16.33) Table 16.2: w m ax 1 .2 5
Also
or
m ax
3 16
p0a
(1 ) ( 5 )
Et
2
3
6
3 16
( 0 .7 )(5 .3 )
Sy
3 8
3 .5 (1 0 )( 4 0 ) 9
2 0 0 (1 0 ) t
(3 ) p 0
a t
2
3
2
t 0 .2 5 m m
,
5 6 0 1 .2 3 8
;
t 0 .5 6 m m
274
3 .5 ( 4 0 ) t
2
SOLUTION (16.34) From Example 16.7 1
p0
a
(t)
4
2
2
3 4
a
p0 ( t ) S
We have 1 1 2 2 ( 2
2
p 0 ( at ) [ 161 2
or
4
4 ( 10
or
10
4
)a
( 10 ) ( 10
4
12
3 16
y
)
n
2 6
] ( 150 2 10 ) ;
9 16
12 , 857 . 14 (10
)
2
12
p 0 ( at ) ( 167 ) 5 , 625 ( 10
2
2
4
12
)
), a 238 . 1 mm
SOLUTION (16.35) r 1 7 5 1 1 7 4
r 7 5 1 7 4
F
mm,
mm,
F ri p z ,
p pz,
2
t 2 mm
ri 7 3 m m
90 , o
p 100 kPa
N
Equation (16.64b): N
and
2
( 0 .0 7 3 ) ( 1 0 0 )
F 2 r s in
2 ( 0 .0 7 4 ) (1 )
3 .6
kN
1 . 8 MPa
3 , 600 0 . 002
Then, Eq.(16.64a): r
or
r
6
pz
1 .8 ( 1 0 )
;
t
0 .1 7 4
0 .0 7 4
3
1 0 0 (1 0 ) 0 .0 0 2
2 . 93 MPa
SOLUTION (16.36) N is maximum at A ( 9 0 ): o
,m ax
pa ( b a 2 ) ( b a )t
S
y
n
240 1. 2
;
2 . 2 ( 5 0 )( 2 5 0 2 5 )
( 25050 )t
or t 0 . 619
Similarly
n
mm pr
240 1 .2
;
2t
2 . 2 ( 50 )
, t 0 . 275
2t
mm
SOLUTION (16.37) 1
pa
2 x
t
Pa 2t
( a ) Since 1 and 2 are of the same sign ( 3 0 ) : 1 0
S
y
n
;
pa t
S
y
n
,
t
pan S
y
(CONT.)
275
16.37 (CONT.) S
( b ) 1 1 2 2 ( 2
(
(c)
2
pa
) [1 2
t
1
Su
2
S uc
1 2
1 n
1 4
S
] (
y
n
pa
;
y
)
n
) , t 2
pa
tS u
2
3 2
4 tS u
1 n
pan S
y
3 pan
t
,
4 Su
SOLUTION (16.38) (a)
p 200 x 200 15 (16 ) 440
all
pa t
pr
; t
3
440 ( 10 )( 5 )
150 ( 10
all
6
)
14 . 67 mm
( b ) p 200 15 [ 3 (16 4 )] 380 t
pr
3
380 ( 10 )( 5 )
all
150 ( 10
6
kPa
12 . 67 mm
)
( c ) Top-end plate, with p 200 kPa m ax
3 4
t 158
or
a
(Fig.16.18):
1 5 0 (1 0 )
2
6
p( t ) ;
3 4
( 2 0 0 1 0 )( t )
2
3 4
( 4 4 0 1 0 )( t )
2
3
5
mm
Bottom-end plate, with p 440 m ax
kPa
3 4
a t
kPa :
1 5 0 (1 0 )
2
6
p( ) ;
3
5
or t 235
mm
SOLUTION(16.39) We have r a . The loading is p p z a ( 1 c o s ) . The resultant force F for the part intercepted by : F 2 a
2
a ( 1 c o s ) s in c o s d
0
2a [ 6 3
1
1 2
c o s (1 2
2 3
c o s )]
Substitution of this into Eqs.(16.65) and rearranging yield the equations quoted in this problem.
SOLUTION (16.40) ( a ) L w 1 .7 2 a t 1 .7 2 7 5 0 1 2 1 6 3 .2 m m (b)
cr
0 . 605
Et a
0 . 605
210 10
9
750
( 12 )
2 , 033
MPa
No failure in buckling, since c r S y and material would yield.
276
SOLUTION (16.41)
cr
0 .6 0 5
Et a
0 .6 0 5
9
2 0 0 1 0 (1 0 ) 600
2, 017 M P a
Thus Pc r 2 a t c r
2 ( 6 0 0 ) (1 0 ) ( 2 0 1 7 ) 7 6 , 0 3 9 k N
End of Chapter 16
277
CHAPTER 17
FINITE ELEMENT ANALYSIS IN DESIGN
SOLUTION (17.1) We have (
AE L
)1,2
A(2 E )
6 AE
L 3
(
L
AE L
)3
2 A(E )
L 3
6 AE L
There are four displacement components ( u 1 , u 2 , u 3 , u 4 ) and so the order of the system matrix is 4x4. Using Eq. (17.1): [ k ]1 [ k ] 2 [ k ] 3
1 1
6 AE 1 L 1
( a ) System matrix, [ K ] [ k ]1 [ k ] 2 [ k ] 3 , by superposition is thus u1
u2
1 6 AE 1 [K ] L 0 0
u3
1
0
(1 1)
1
1
(1 1)
0
1
u4
u1
u2
u3
u4
0 0 6 AE 1 L 1
1 1 0 0
1
0
2
1
1
2
0
1
0 0 1 1
( b ) Refer to Eq. (17.7b): F1 x F2 x F 3x F 4x
AE L
1 1 0 0
1
0
2
1
1
2
0
1
0 0 1 1
u1 u2 u 3 u 4
(1)
Boundary conditions are u 1 u 4 0 and F x 3 P . Equation (1) is then 0 AE 2 L 1 P
Solving u 2
PL 9 AE
1 u 2 2 u3
u3
PL 18 AE
( c ) Equations (1) result in F1 x F2 x F 3x F 4x
6 AE L
1 1 0 0
1
0
2
1
1
2
0
1
0 0 1 1
The reactions are R1
2 3
P
R4
1
P
3
278
PL PL
2 P 3 9 AE L P 18 AE AE 0 0 P 3 0
SOLUTION (17.2) We have (
AE L
)1
AE
(
AE
L
L
[ k ]1 [ k ] 2
AE 1 L 1
AE
)2
(
AE
L
L
)3
4 AE
0 .8
5L
AE L
Equation (17.1): 1 1
[ k ]3
A E 0 .8 L 0 .8
0 .8 0 .8
( a ) System matrix, [ K ] [ k ]1 [ k ] 2 [ k ] 3 , by superposition is then 1 AE 1 [K ] L 0 0
1
0
(1 1)
1
1
(1 0 .8 )
0
0 .8
0 0 .8 0 .8 0
( b ) Refer to Eq. (17.7b): F1 x F2 x F 3x F 4x
AE L
u1
u2
u3
1 1 0 0
1
0
2
1
1
1 .8
0
0 .8
u4 0 0 .8 0 .8 0
u1 u2 u 3 u 4
Boundary conditions are u 1 u 4 0 , We have F x 3 P . Thus 0 AE 2 L 1 P
1 u2 1 .8 u 3
Solving PL
u2
2 .6 A E
u3
PL 1 .3 A E
( c ) Equations (1) yield F1 x F2 x F 3x F 4x
AE L
1 1 0 0
1
0
2
1
1
1 .8
0
0 .8
0 0 .8 0 .8 0
The reactions are R1
1 2 .6
P
R4
0 .8
P
1 .3
279
P P
P 2 .6 2 .6 L 0 1 .3 A E P 0 1 .6 P 2 .6
0
(1)
* SOLUTION (17.3) We have A E L the same for all elements. Thus, Eq. (17.1): 1 1
AE 1 L 1
[ k ]1 [ k ] 2 [ k ] 3 [ k ] 4
( a ) By superposition, we have 1 1 AE 0 [K ] L 0 0
1
0
0
2
1
0
1
2
1
0
1
2
0
0
1
0 0 0 1 1
(b) F1 x F 2x F3 x F 4x F 5 x
AE L
1 1 0 0 0
1
0
0
2
1
0
1
2
1
0
1
2
0
0
1
0 0 0 1 1
u1 u 2 u3
(1)
u 4 u 5
Boundary conditions: u 1 0 and u 5 . We have F 2 x F 3 x F 4 x 0 . Hence, 0 AE 20 L 0
1 0 0
2
1
0
1
2
1
0
1
2
0 0 1
0 u 2 u3 u 4
This equation may be rewritten, after transposing the product of the appropriate stiffness coefficients by the known displacement ( ) to the left side. In so doing, 0 0 AE L
AE L
2 1 0
1 2 1
0 1 2
u2 u3 u4
Solving u2
4
u3
2
u4
3 4
(CONT.)
280
17.3 (CONT.) (c)
Equations (1) give then F1 x F 2x F3 x
AE L
F 4x F 5 x
1 1 0 0 0
1
0
0
2
1
0
1
2
1
0
1
2
0
0
1
0 4 4 0 AE 2 0 L 3 4 0 4
0 0 0 1 1
The reaction is R1
1
AE
4L
SOLUTION (17.4) We have AE L
1
(1 2 0 0 1 0
6
)(7 2 1 0 ) 5 4 1 0 9
6
N m
1 .6
c cos 30
o
s s in 3 0
3 2
o
1 2
3 4
( a ) Through the use of Eq. (17.14), we have 3 4 3 4 6 [ k ]1 5 4 (1 0 ) 3 4 3 4
3 4
1 4
3
3 4 3 4
4
3
1 4
4
14 3 4 1 4 3
4
or 0 .7 5 0 .4 3 3 6 [ k ]1 5 4 (1 0 ) 0 .7 5 0 .4 3 3
0 .4 3 3
0 .7 5
0 .2 5
0 .4 3 3
0 .4 3 3
0 .7 5
0 .2 5
0 .4 3 3
0 .4 3 3 0 .2 5 N m 0 .4 3 3 0 .2 5
( b ) Equation (17.11b), { } e [ T ]{ } e : u1 0 .8 6 6 0 .5 v1 0 u 3 v 0 3
0 .5
0
0 .8 6 6
0
0
0 .8 6 6
0
0 .5
0 0 .5 0 .8 6 6 0
1 .5 1 .8 9 9 1 .2 0 .2 8 9 mm 2 .2 1 .9 0 5 0 1 .1 0 0
(CONT.)
281
17.4 (CONT.) ( c ) Inserting the given data into (17.15) with i=1 and j=3, we obtain the axial force 6 F1 F1 3 5 4 (1 0 )
3
1 2
2
2 .2 1 .5 3 (1 0 ) 2 0 5 .4 k N 0 1 .2
So, axial stress in the element equals 1 F1 A 1 7 1 .2 M P a Comment: The negative sign denotes compression.
SOLUTION (17.5) We have E 2 0 0 G P a (by Table B.1). Refer to Solution of Prob. 17.4. AE L
1
(1 2 0 0 1 0
6
)( 2 0 0 1 0 ) 1 5 0 1 0 9
6
N m
1 .6
c cos 60
o
1 2
s s in 6 0
o
3 2
( a ) From Eq. (17.14): 1 4 3 4 6 [ k ]1 1 5 0 (1 0 ) 1 4 3 4
3 4 3 4 3 4 3 4
0 .2 5 0 .4 3 3 6 1 5 0 (1 0 ) 0 .2 5 0 .4 3 3
1 4
43 3 4 3 4
3 4 1 4 3 4
3
4
0 .4 3 3
0 .2 5
0 .7 5
0 .4 3 3
0 .4 3 3
0 .2 5
0 .7 5
0 .4 3 3
0 .4 3 3 0 .7 5 0 .4 3 3 0 .7 5
( b ) Equation (17.11b), { } e [ T ]{ } e : u1 0 .5 0 .8 6 6 v1 u 0 3 v 0 3
( c ) F1 F1 3 1 5 (1 0 ) 12 6
3 2
0 .8 6 6
0
0 .5
0
0
0 .5
0
0 .8 6 6
0 0 .8 6 6 0 .5 0
1 .5 1 .7 8 9 1 .2 0 .6 9 9 mm 2 .2 1 .1 0 1 .9 0 5
2 .2 1 .5 3 (1 0 ) 4 3 3 .4 k N 0 1 .2
1 F1 A 3 6 1 .2 M P a
282
SOLUTION (17.6)
Table 17.6 Data for truss of Fig. P17.6. E le m e n t
o
1
45
2
315 o
3
0
4
90
5
90
o
o
o
c
s
2 2
2 2
2 2
c
2 2
2
s
2
cs
1 2
1 2
1 2
1 2
1 2
1 2
1
0
1
0
0
0
1
0
1
0
0
1
0
1
0
Equation (17.14) u1
[ k ]1
1 1 2 AE 2L 1 1
v1
u2
2
1 2
1 2
2
1 2
1 2
2
1 2
1 2
2
1 2
1 2
v2 1 2 1 2 1 2 1 2
Similarly, for elements 2, 3, 4, and 5, we obtain
u1
[ k ]2
1 2 AE 1 2L 1 1
u1 1 0 AE [ k ]3 L 1 0
v1
v1
u3
2
1 2
1 2
2
1 2
1 2
2
1 2
1 2
2
1 2
1 2
u4
v4
0
1
0
0
0
1
0
0
0 0 0 0
u4
v4
u2
v2
0 AE 0 [ k ]4 L 0 0
0
0
1
0
0
0
1
0
0 1 0 1
v3 1 2 1 2 1 2 1 2
u3 0 AE 0 [ k ]5 L 0 0
283
v3
u4
0
0
1
0
0
0
1
0
v4 0 1 0 1
SOLUTION (17.7) Table P17.7 Data for the truss of Fig P17.7 E le m e n t
L e n g th ( m )
1
7 .5
3 6 .9
2
6
0
3
4 .5
90
4 .5
0
4 .5
2
s
0 .8
0 .6
1
0
o
4 5
o
c
o
135
2
2
cs
s
0 .6 3 9
0 .4 8
0 .3 6
1
0
0
0
1
0
0
1
1
0
1
0
0
0 .7 0 7
0 .7 0 7
0 .5
0 .5
0 .5
o
o
c
Apply Eq.(17.14): AE [ k ]1 7 .5
AE [ k ]2 6 .0
AE [ k ]4 4 .5
u1
v1
u2
v2
0 . 639
0 . 48
0 . 639
0 . 48
0 . 361
0 . 48
0 . 639
0 . 48
0 . 639
0 . 48
0 . 361
0 . 48
u1
v1
u3
1
0
1
0
0
0
1
0
1
0
0
0
u3
v3
1
0
1
0
0
0
1
0
1
0
0
0
u4
0 . 48 u 1 0 . 361 v 1 0 . 48 u 2 0 . 361 v 2
v3
u2
0 0 0 0
0 AE 0 [ k ]3 4 .5 0 0
u1 v1 u3 v3
v4
v2
u3
v3
0
0
1
0
0
0
1
0
0 1 0 1
u2
0 0 0 0
AE [ k ]5 4 .5 2
u3 v3 u4 v4
v2
u2 v2 u3 v3
u4
0 .5
0 .5
0 .5
0 .5
0 .5
0 .5
0 .5
0 .5
0 .5
0 .5
0 .5
0 .5
v4 0 .5 u 2 0 .5 v 2 0 .5 u 4 0 .5 v 4
SOLUTION (17.8) Table P17.8 Data for the truss of Fig.P17.8 E le m e n t
1
0
2
o
60
o
3
120
4
0
5
o
o
60
o
c
s
c
1
0
0 .5
2
2
cs
s
1
0
0
0 .8 6 6
0 .2 5
0 .4 3 3
0 .7 5
0 .5
0 .8 6 6
0 .2 5
0 .4 3 3
0 .7 5
1
0
1
0
0
0 .5
0 .8 6 6
0 .2 5
0 .4 3 3
0 .7 5
(CONT.)
284
17.8 (CONT.) ( a ) Use Eq.(17.14):
AE [ k ]1 L
u1
v1
u4
1
0
1
0
0
0
1
0
1
0
0
0
u1
AE [ k ]2 L
AE [ k ]3 L
AE [ k ]4 L
v4
0 0 0 0
u1 v1 u4 v4
v1
u2
v2
0 .2 5
0 .4 3 3
0 .2 5
0 .4 3 3
0 .7 5
0 .4 3 3
0 .2 5
0 .4 3 3
0 .2 5
0 .4 3 3
0 .7 5
0 .4 3 3
0 .4 3 3 0 .7 5 0 .4 3 3 0 .7 5
u4
v4
u2
v2
0 .2 5
0 .4 3 3
0 .2 5
0 .4 3 3
0 .7 5
0 .4 3 3
0 .2 5
0 .4 3 3
0 .2 5
0 .7 5
0 .4 3 3
u2
v2
u3
1
0
1
0
0
0
1
0
1
0
0
0
v1 u2 v2
0 .4 3 3 0 .7 5 0 .4 3 3 0 .7 5
0 .4 3 3
v3
0 0 0 0
u1
u2 v2 u4 v4
u3
u2 v2 u3 v3
AE [ k ]5 L
v3
u4
v4
0 .2 5
0 .4 3 3
0 .2 5
0 .4 3 3
0 .7 5
0 .4 3 3
0 .2 5
0 .4 3 3
0 .2 5
0 .4 3 3
0 .7 5
0 .4 3 3
0 .4 3 3 0 .7 5 0 .4 3 3 0 .7 5
u3 v3 u4 v4
( b ) Global Stiffness Matrix [ K ] u1 AE L
v1
u2
v2
u3
v3
u4
1. 2 5
0 .4 3 3
0 .2 5
0 .4 3 3
0
0
1
0 .4 3 3
0 .7 5
0 .4 3 3
0 .7 5
0
0
0
0 .2 5
0 .4 3 3
1.5
0
1
0
0 .4 3 3
0 .7 5
0
1.5
0
0
0 .2 5 0 .4 3 3
0
0
1
0
1. 2 5
0 .4 3 3
0 .2 5
0
0
0
0
0 .4 3 3
0 .7 5
0 .4 3 3
1
0
0 .2 5
0 .4 3 3
1.5
0
0
0 .4 3 3
0 .7 5
0
0 .2 5 0 .4 3 3
0 .4 3 3 0 .7 5
v4 0 0 .4 3 3 0 .7 5 0 .4 3 3 0 .7 5 0 1.5 0
u1 v1 u2 v2 u3 v3 u4 v4
(CONT.)
285
17.8 (CONT.) R1 x R 1y 0 P Q 0 R 4x R 4y
{ F } [ K ]{ };
0 0 u2 v2 K u 3 v 3 0 0
SOLUTION (17.9) Table P17.9 Data for the truss of Fig.P17.9 E le m e n t
L e n g th ( in . )
1
7 .2 1 1
5 6 .3 1
2
5
3
o
o
3
90
5
3 6 .8 7
o
7 .2 1 1
5 6 .3 1
o
4 5
3 6 .8 7
o
c
s
0 .5 5 5
0 .8 3 2
0 .8
c
2
2
cs
s
0 .3 0 8
0 .4 6 2
0 .6 9 2
0 .6
0 .6 4
0 .4 8
0 .3 6
0
1
0
0
1
0 .8
0 .6
0 .6 4
0 .4 8
0 .3 6
0 .5 5 5
0 .8 3 2
0 .3 0 8
0 .4 6 2
0 .6 9 2
( a ) Use Eq. (17.14): u1
v1
0 .3 0 8
0 .4 6 2
0 .3 0 8
0 .4 6 2
0 .6 9 2
0 .4 6 2
0 .3 0 8
0 .4 6 2
0 .3 0 8
0 .4 6 2
0 .6 9 2
0 .4 6 2
u1
v1
0 .6 4
0 .4 8
0 .6 4
0 .4 8
0 .3 6
0 .4 8
0 .6 4
0 .4 8
0 .6 4
0 .4 8
0 .3 6
0 .4 8
AE [ k ]1 7 .2 1 1
AE [ k ]2 5 u2
0 AE 0 [ k ]3 3 0 0
u2
u3
v2
u3
v3
0
0
1
0
0
0
1
0
0 1 0 1
v2
0 .4 6 2 0 .6 9 2 0 .4 6 2 0 .6 9 2
u1 v1 u2 v2
v3
0 .4 8 0 .3 6 0 .4 8 0 .3 6
u1 v1 u3 v3
u2 v2 u3 v3
(CONT.)
286
17.9 (CONT.) u3
[k ]4
AE 5
v3
u4
0 . 64
0 . 48
0 . 64
0 . 48
0 . 36
0 . 48
0 . 64
0 . 48
0 . 64
0 . 48
0 . 36
0 . 48
u2
AE [ k ]5 7 .2 1 1
v4
0 . 48 u 3 0 . 36 v 3 0 . 48 u 4 0 . 36 v 4
v2
u4
0 .3 0 8
0 .4 6 2
0 .3 0 8
0 .4 6 2
0 .6 9 2
0 .4 6 2
0 .3 0 8
0 .4 6 2
0 .3 0 8
0 .4 6 2
0 .6 9 2
0 .4 6 2
v4
0 .4 6 2 0 .6 9 2 0 .4 6 2 0 .6 9 2
u2 v2 u4 v4
( b ) Global Stiffness Matrix [ K ]
AE
u1
v1
u2
v2
0 .1 7 1
0 .1 6
0 .0 4 3
0 .0 6 4
0 .1 2 8
0 .0 9 6
0
0 .1 6
0 .1 6 8
0 .0 6 4
0 .0 9 6
0 .0 9 6
0 .0 7 2
0
0 .0 4 3
0 .0 6 4
0 .0 8 5
0
0
0 .0 6 4
0 .0 9 6
0
0 .5 2 5
0
0 .1 2 8
0 .0 9 6
0
0
0 .2 5 6
0
0 .0 9 6
0 .0 7 2
0
0
0 .4 7 7
0 .0 9 6 0 .1 7 1
0 .3 3 3
u3
v3
0 0 .3 3 3
0
0
0 .0 4 3
0 .0 6 4
0 .1 2 8
0 .0 9 6
0
0
0 .0 6 4
0 .0 9 6
0 .0 9 6
0 .0 7 2
{ F } [ K ]{ };
R1 x R 1y P 0 0 Q R 4x R 4y
0 0 u2 v2 K u 3 v 3 0 0
287
u4
0 .0 4 3 0 .0 6 4 0 .1 2 8
0 .1 6
v4
0 0 .0 6 4 0 .0 9 6 0 .0 9 6 0 .0 7 2 0 .1 6 0 .1 6 8 0
u1 v1 u2 v2 u3 v3 u4 v4
SOLUTION (17.10)
Table P17.10 Data for the truss of Fig.P17.10 E le m e n t
L e n g th
1
1. 2
0
2 3
c
s
c
1
0
0 .9 2 3 0
o
1.3
2 2 .6 2
0 .5
90
o
o
o
4
1. 2
0
5
1.3
2 2 .6 2
o
2
2
cs
s
1
0
0
0 .3 8 5
0 .8 5 2
0 .3 5 5
0 .1 4 8
1
0
0
1
1
0
1
0
0
0 .9 2 3
0 .3 8 5
0 .8 5 2
0 .3 5 5
0 .1 4 8
( a ) Use Eq.(17.14); AE [ k ]1 1.2
uA
vA
uC
1
0
1
0
0
0
1
0
1
0
0
0
uA
AE [ k ]2 1.3
0 0 0 0
uA vA uC vC
vA
uB
vB
0 .3 5 5 0 .1 4 8 0 .3 5 5 0 .1 4 8
0 .8 5 2
0 .3 5 5
0 .8 5 2
0 .3 5 5
0 .1 4 8
0 .3 5 5
0 .8 5 2
0 .3 5 5
0 .8 5 2
0 .3 5 5
0 .1 4 8
0 .3 5 5
uB
0 AE 0 [ k ]3 0 .5 0 0 uB
AE [ k ]4 1.2
vC
vB
uC
vC
0
0
1
0
0
0
1
0
0 1 0 1
vB
uD
1
0
1
0
0
0
1
0
1
0
0
0
uA vA uB vB
uB vB uC vC
vD
0 0 0 0
uB vB uD vD
(CONT.)
288
17.10 (CONT.)
AE [ k ]5 1.3
uC
vC
uD
vD
0 .8 5 2
0 .3 5 5
0 .8 5 2
0 .3 5 5
0 .1 4 8
0 .3 5 5
0 .8 5 2
0 .3 5 5
0 .8 5 2
0 .3 5 5
0 .1 4 8
0 .3 5 5
0 .3 5 5 0 .1 4 8 0 .3 5 5 0 .1 4 8
uC vC uD vD
( b ) Global Stiffness Matrix [ K ] AE
uA
vA
uB
vB
uC
vC
uD
1. 4 8 9
0 .2 7 3
0 .6 5 5
0 .2 7 3
0
0
0 .2 7 3
0 .1 1 4
0 .2 7 3
0 .1 1 4
0
0
0
0 .6 5 5
0 .2 7 3
1. 4 8 9
0 .2 7 3
0
0
0 .2 7 3
0 .1 1 4
0 .2 7 3
2 .1 1 4
0
2
0 .8 3 3
0 .8 3 3 0
0 .8 3 3
0
0
0
1. 4 8 9
0 .2 7 3
0 .6 5 5
0
0
0
2
0 .2 7 3
2 .1 1 4
0 .2 7 3
0
0
0
0 .6 5 5
0 .2 7 3
1. 4 8 9
0
0
0
0 .2 7 3
0 .1 1 4
0 .2 7 3
0 .8 3 3 0
R Ax R Ay 0 0 0 0 P R Dy
{ F } [ K ]{ };
0 0 uB vB K u C v C uD 0
vD
0 0 0 0 .2 7 3 0 .1 1 4 0 .2 7 3 0 .1 1 4 0
SOLUTION (17.11) We have AE=30 MN. Table P17.11 Data for the truss of Fig.P17.11 E le m e n t
L e n g th
1
5
5 3 .1 3
2
4
0
o
o
c
s
0 .6
0 .8
1
0
c
2
2
cs
s
0 .3 6
0 .4 8
0 .6 4
1
0
0
( a ) Use Eq.(17.14): (CONT.)
289
uA vA uB vB uC vC uD vD
17.11 (CONT.) AE [ k ]1 5
AE [ k ]2 4
u1
v1
0 .3 6
0 .4 8
0 .3 6
0 .4 8
0 .6 4
0 .4 8
0 .3 6
0 .4 8
0 .3 6
0 .4 8
0 .6 4
0 .4 8
u2
v2
u3
1
0
1
0
0
0
1
0
1
0
0
0
u2
v2
0 .4 8 0 .6 4 0 .4 8 0 .6 4
u1 v1 u2 v2
v3
0 0 0 0
u2 v2 u3 v3
( b ) Global Stiffness Matrix [K ] AE
u1
v1
u2
v2
u3
0 .0 7 2
0 .0 9 6
0 .0 7 2
0 .0 9 6
0
0 .0 9 6
0 .1 2 8
0 .0 9 6
0 .1 2 8
0
0 .0 7 2
0 .0 9 6
0 .3 2 2
0 .0 9 6
0 .0 9 6
0 .1 2 8
0 .0 9 6
0 .1 2 8
0
0
0 .2 5
0
0
0
0
0
0
0 .2 5 0 0
v3
0 .2 5
( c ) Boundary Conditions: u 1 v 1 u 3 v 3 0 . F2 x AE F 2y
0 .3 2 2 0 .0 9 6
u2 1 4 AE 3 v2
F1 x F1 y (d) F 3x F3 y
AE
0 .0 9 6 u 2 0 .1 2 8 v 2 3
0 0 .0 0 1 0 m 1 0 .0 6 3 1 0 , 0 0 0 0 .0 0 3 4
0 .0 7 2 0 .0 9 6 0 .2 5 0 0
0 .0 9 6 7 .6 3 2 0 .1 2 8 0 .0 0 1 0 1 0 .1 7 6 kN 0 .0 0 3 4 0 7 .5 0 0 0 0
( e ) Use Eq.(17.16) F1 2
F23
AE 5 AE 4
0 .6 1
0 .0 0 1 0 .8 1 2 .7 2 k N 0 .0 0 3 4 0 .0 0 1 0 7 .5 k N 0 .0 0 3 4
(C )
290
(C )
0 0 0 0 0 0
u1 v1 u2 v2 u3 v3
SOLUTION (17.12) We have E 2 0 0 G P a
A 2 (1 0 ) m m 3
2
Table P17.12 Data for the truss of Fig.P17.12 E le m e n t
L e n g th ( m m )
1
3000
90
2
3000
3
2
45
3000
0
c
s
c
0
1
0 .7 0 7 1
o
o
o
2
2
cs
s
0
0
1
0 .7 0 7
0 .5
0 .5
0 .5
0
1
0
0
( a ) Use Eq.(17.14) u1
[ k ]1
v1
0 0 0 0
AE 3000
u2
0
0
1
0
0
0
1
0
u1
2
AE
[k ]2
3000
u1
[k ]3
AE 3000
v2
0 u1 1 v1 0 u2 1 v2
v1
0 .5
0 .5
0 .5
0 .5
0 .5
0 .5
0 .5
0 .5
0 .5
0 .5
0 .5
u4
0
1
0
0
0
1
0
1
0
0
0
(10
11
0 u1 0 v1 0 u4 0 v4
) . Adding zero’s in proper locations and adding :
u1
[K ]
4 3
(10
11
1 . 354 0 . 354 0 0 ) 0 . 354 0 . 354 1 0
0 .5 u 1 0 .5 v 1 0 .5 u 3 0 .5 v 3
v4
1
4 3
v3
0 .5
v1
( b ) The common factor
u3
v1
u2
v2
u3
v3
u4
0 . 354
0
0
0 . 354
0 . 354
1
1 . 354
0
1
0 . 354
0 . 354
0
0
0
0
0
0
0
1
0
1
0
0
0
0 . 354
0
0
0 . 354
0 . 354
0
0 . 354
0
0
0 . 354
0 . 354
0
0
0
0
0
0
1
0
0
0
0
0
0
v4
0 u1 0 v1 0 u2 0 v2 0 u3 0 v3 0 u4 0 v 4
(CONT.)
291
17.12 (CONT.)
(c)
u1 25000 v 1 F2x 0 F2 y 0 K F3x 0 0 F3 y F4x 0 0 F4 y 0
Eliminate rows and columns in [K] corresponding to zero displacements: 0 4 (10 3 25000
11
u1 1 11 ( 4 3 ) (1 0 ) v1
0 . 354 u 1 1 . 354 v 1
1 . 354 ) 0 . 354
0 .2 0 7 0 0 .0 3 8 9 6 (1 0 ) m m 0 .7 9 3 2 5 0 0 0 0 .1 4 8 6
0 .7 9 3 0 .2 0 7
(d) R2 y R3x R 3y R 4x
4 11 (1 0 ) 3
1
0 0 .3 5 4 0 .3 5 4 1
1 9 .8 1 3 0 .3 5 4 u 1 5 .1 7 8 kN 0 .3 5 4 v 1 5 .1 7 8 0 5 .1 8 7
( e ) Use Eq.(17.16): AE
F1 2
L1
F1 3
u1 2 (1 0 ) 2 0 0 (1 0 ) 1 0 3000 v1 3
0
AE L2
2 (1 0 ) 2 0 0 (1 0 )
F1 4
AE L3
0 .0 3 8 9 6 0 .7 0 7 (1 0 ) 0 .1 4 8 6
9
3000
0 .7 0 7
2
u1 2 (1 0 ) 2 0 0 (1 0 ) 0 1 3000 v1 3
1
0 .0 3 8 9 6 1 (1 0 ) 1 9 .8 1 3 k N ( T ) 0 .1 4 8 6
u1 0 .7 0 7 v1
0 .7 0 7 3
9
7 .3 1 2 2 k N
(T )
0 .0 3 8 9 6 0 (1 0 ) 5 .1 8 6 7 k N ( C ) 0 .1 4 8 6
9
SOLUTION (17.13) Table P17.13 Data for the truss of Fig.P17.13 E le m e n t
L e n g th
1
5
5 3 .1 3
2
4
180
o
o
c
s
0 .6
0 .8
1
0
c
2
2
cs
s
0 .3 6
0 .4 8
0 .6 4
1
0
0
(CONT.)
292
17.13 (CONT.) ( a ) Use Eq.(17.14): AE [ k ]1 5
AE [ k ]2 4
u1
v1
0 .3 6
0 .4 8
0 .3 6
0 .4 8
0 .6 4
0 .4 8
0 .3 6
0 .4 8
0 .3 6
0 .4 8
0 .6 4
0 .4 8
u2
v2
1
0
1
0
0
0
1
0
1
0
0
0
u2
u3
v2
0 .4 8 0 .6 4 0 .4 8 0 .6 4
u1 v1 u2 v2
v3
0 0 0 0
u2 v2 u3 v3
( b ) Global Stiffness Matrix [K ] AE
u1
v1
u2
v2
u3
0 .0 7 2
0 .0 9 6
0 .0 7 2
0 .0 9 6
0
0 .0 9 6
0 .1 2 8
0 .0 9 6
0 .1 2 8
0
0 .0 7 2
0 .0 9 6
0 .3 2 2
0 .0 9 6
0 .0 9 6
0 .1 2 8
0 .0 9 6
0 .1 2 8
0
0
0 .2 5
0
0
0
0
0
0
0 .2 5 0
0 .2 5
v3
0 0 0 0 0 0
u1 v1 u2 v2 u3 v3
(c) 0 . 096 u 2 0 . 128 v 2
0 0 . 322 AE 60000 0 . 096 u2 1 6 v 2 1 0 (1 0 )
4 3
3
0 18 3 (1 0 ) 1 0 .0 6 3 6 0 0 0 0 6 0 .4
m
(d) R1 x R1 y R 3x
AE
0 .0 7 2 0 .0 9 6 0 .2 5
0 .0 9 6 4 5 .0 2 4 u2 0 .1 2 8 6 0 .0 3 2 v 2 0 45.
kN
( e ) Use Eq.(17.16) F1 2 A E 0 .6
u2 6 0 .8 1 0 (1 0 ) 0 .6 v 2
0 .0 1 8 0 .8 3 7 5 .2 0 .0 6 0 4
kN
(C )
(CONT.)
293
17.13 (CONT.) u2 6) 0 1 0 (1 0 1 v 2
F23 A E 1
0 .0 1 8 0 180 0 .0 6 0 4
kN
(T )
SOLUTION (17.14) We have AE 125
MN .
Table P17.14 Data for the truss of Fig.P17.14 E le m e n t
L e n g th
1
5
1 4 3 .1 3
2
5
3
o
3 6 .8 7
5
1 4 3 .1 3
o
o
c
s
c
0 .8
0 .6
0 .8 0 .8
2
2
cs
s
0 .6 4
0 .4 8
0 .3 6
0 .6
0 .6 4
0 .4 8
0 .3 6
0 .6
0 .6 4
0 .4 8
0 .3 6
( a ) Apply Eq.(17.14): AE [ k ]1 25L
AE [ k ]2 25L
u1
v1
u2
16
12
16
12
9
12
16
12
16
12
9
12
u2
v2
u3
16
12
16
12
9
12
16
12
16
12
9
12
v2
12 9 12 9
u1 v1 u2 v2
v3 12 9 12 9
AE [ k ]3 25 L
u2 v2 u3 v3
u2
v2
u4
16
12
16
12
9
12
16
12
16
12
9
12
v4
12 9 12 9
u2 v2 u4 v4
( b ) Global Stiffness Matrix AE [K ] 25L
u1
v1
u2
v2
u3
v3
u4
v4
16
12
16
12
0
0
0
9
12
9
0
0
0
16
12
48
12
16
12
16
12
9
12
27
12
9
12
0
0
16
12
16
12
0
0
0
12
9
12
9
0
0
0
16
12
0
0
16
0
0
12
9
0
0
12
0 0 12 9 0 0 12 9
12
25000 AE 48 25L 12 40000
(c)
u1 v1 u2 v2 u3 v3 u4 v4
12 u2 27 v2
(CONT.)
294
17.14 (CONT.) 12 25000 1 . 0026 (10 48 40000 1 . 927
u 2 27 25 L AE (1152 ) 12 v2
3
) m
(d) F1 x F 1y F 3 x F 3y F4x F 4 y
1 1152
16 12 16 12 16 12
( e ) Use Eq.(17.16): We have u 1 v 1 0 ;
F 12
F 32
F 42
AE
L
AE
u 2 0 . 6 v2
0 . 8
L
AE
u3 v 3 0;
u 2 0 . 6 v2
0 .8
L
1 1152
1 1152
kN
u 4 v 4 0.
27 0 . 6 12
0 .8
27 0 . 6 12
0 .8
12 25000 8 . 854 48 40000
12 25000 48 . 958 48 40000
27 0 . 6 12
0 . 8
1 1152
u 2 0 . 6 v2
0 .8
7 . 083 5 . 313 39 . 167 12 25000 48 40000 29 . 375 7 . 083 5 . 313
12 9 12 27 9 12 12 9
kN
12 25000 8 . 854 48 40000
kN
(C )
(C )
kN
(C )
SOLUTION (17.15) We have E 210 GPa
4
A 5 10
m
2
Table P17.15 Data for the truss of Fig.P17.15 E le m e n t
L e n g th
1
5
5 3 .1 3
2
4
90
o
c
s
0 .6
0 .8
0
1
o
c
2
2
cs
s
0 .3 6
0 .4 8
0 .6 4
0
0
1
( a ) Apply Eq.(17.14): AE [ k ]1 5
u1
v1
u2
0 .3 6
0 .4 8
0 .3 6
0 .4 8
0 .6 4
0 .4 8
0 .3 6
0 .4 8
0 .3 6
0 .4 8
0 .6 4
0 .4 8
v2
0 .4 8 0 .6 4 0 .4 8 0 .6 4
u1 v1 u2 v2
(CONT.)
295
17.15 (CONT.) u1
v1
u3
v3
0 AE 0 [ k ]2 4 0 0
0
0
1
0
0
0
1
0
0 1 0 1
u1 v1 u3 v3
( b ) Global Stiffness Matrix u1
v1
0 .7 5 6 1.0 0 8 0 .7 5 6 7 [K ] 10 1.0 0 8 0 0
u2
v2
u3
1.0 0 8
0 .7 5 6
1.0 0 8
0
3 .9 6 9
1. 0 0 8
1.3 4 4
0
1.0 0 8
0 .7 5 6
1.0 0 8
0
1.3 4 4
1. 0 0 8
1.3 4 4
0
0
0
0
0
0
0
0 2 .6 2 5
v3
2 .6 2 5 0 0 0 2 .6 2 5 0
u1 v1 u2 v2 u3 v3
(c) F1 x 100000
10
100 10
3
1 . 008 0 . 025 v1 3 . 969
0 . 756 1 . 008
7
(1 . 008 10 )( 0 . 025 ) 3 . 696 10 v 1 7
F 1 x ( 0 . 756 10
7
7
)( 0 . 025 ) 1 . 008 10
7
v1
or
or
v 1 0 . 00887
m
F 1 x 99 . 6 kN
( d ) Support reactions: F2x F2 y F 3x F3 y
10
7
0 . 756 1 . 008 0 0
1 . 008 99 . 59 1 . 344 0 . 025 132 . 79 0 . 00887 0 0 232 . 84 2 . 625
kN
( e ) Use Eq.(17.16): We have u 2 v 2 0 ,
F 12
F 13
AE
0 . 6
L1
AE L2
0
u3 v 3 0.
u1 105 10 0 . 8 5 v1 u1 105 10 1 4 v1
6
0
6
0 . 6
0 . 025 0 . 8 166 0 . 00887
0 . 025 1 232 . 8 kN 0 . 00887
296
kN
(C )
(T )
SOLUTION (17.16) We have AE 20 MN . Table P17.16 Data for the truss of Fig.P17.16 E le m e n t
L e n g th
1
5
3 6 .8 7
2 3
8
0
5
c
s
0 .8
0 .6
1 0 .8
o
o
1 4 3 .1 3
o
c
2
2
cs
s
0 .6 4
0 .4 8
0 .3 6
0
1
0
0
0 .6
0 .6 4
0 .4 8
0 .3 6
( a ) Apply Eq.(17.14): AE [ k ]1 5
AE [ k ]2 8
u1
v1
0 .6 4
0 .4 8
0 .6 4
0 .4 8
0 .3 6
0 .4 8
0 .6 4
0 .4 8
0 .6 4
0 .4 8
0 .3 6
0 .4 8
u1
v1
1
0
1
0
0
0
1
0
1
0
0
0
u2
AE [ k ]3 5
u3
u2
v2
0 .4 8 0 .3 6 0 .4 8 0 .3 6
u1 v1 u2 v2
v3
v2
0 0 0 0
u1 v1 u3 v3 u3
v3
0 .6 4
0 .4 8
0 .6 4
0 .4 8
0 .3 6
0 .4 8
0 .6 4
0 .4 8
0 .6 4
0 .4 8
0 .3 6
0 .4 8
0 .4 8 0 .3 6 0 .4 8 0 .3 6
u2
v2
u3
0 .1 2 5
u2 v2 u3 v3
( b ) Global Stiffness Matrix [K ] AE
u1
v1
0 .2 5 3
0 .0 9 6
0 .1 2 8
0 .0 9 6
0 .0 9 6
0 .0 7 2
0 .0 9 6
0 .0 7 2
0 .1 2 8
0 .0 9 6
0 .2 5 6
0
0 .0 9 6
0 .0 7 2
0
0 .1 1 4
0 .0 9 6
0 0 .1 2 8
0 .1 2 5
0
0 .1 2 8
0 .0 9 6
0 .2 5 3
0
0
0 .0 9 6
0 .0 7 2
0 .0 9 6
v3
0 0 .0 9 6 0 .0 7 2 0 .0 9 6 0 .0 7 2 0
u1 v1 u2 v2 u3 v3
(CONT.)
297
17.16 (CONT.)
(c)
F2 x F2 y F3 x
AE
0 0 .1 1 4 0 .0 9 6
0 .1 2 8 u 2 0 .0 9 6 v 2 0 .2 5 3 u 3
1
K F
1 AE
u2 v2 u 3
0 .2 5 6 0 0 .1 2 8
1 2 0 1 0
6
F1 x ( d ) F1 y A E F 3x
3 .8 9 2 9
6 .2 1 7 7 3 .8 9 2 7 4 .6 2 2 9 0 .1 2 8 0 .0 9 6 0 .0 9 6
1 5 .3 2 8 5 7 .7 8 5 9 0 .0 9 6 0 .0 7 2 0 .0 7 2
4 .6 2 2 9 7 .7 8 5 9 9 .2 4 5 2
0 .1 2 5 0. 0 .0 9 6
40000 80000 0
0 .0 2 8 0 0 .0 6 9 1 m 0 .0 4 0 4
0 .0 2 8 0 4 0 .0 0 8 0 .0 6 9 1 4 5 .7 4 4 0 .0 4 0 4 7 5 .6 9 6
kN
( e ) Use Eq.(17.16): We have u 1 v 1 v 3 0 . F1 2
F1 3
F3 2
AE
0 .0 2 8 0 0 .6 7 6 .2 4 0 .0 6 9 1
0 .8
5 AE
1
8 AE 5
0 .0 4 0 4 0 1 0 1 .0 0
0 .8
kN
kN
(C )
(T )
0 .0 2 8 0 0 .0 4 0 4 0 .6 1 2 6 .1 6 0 0 .0 6 9 1
kN
(C )
SOLUTION (17.17) Due to symmetry, only one-half of the beam need be considered. 12 6L EI [ k ]1 3 L 12 6L
P/2
L
2
6L 4L
2
12 6L
6 L 2L
12
2
6L
0
0
0
0
0
0
0
0
6L 2 2L 6 L 2 4L
1 1 2
k/2
0 3 E I kL E I [ k ]2 3 0 L 0
0 0 0 0
(CONT.)
298
17.17 (CONT.) Therefore 12 6L EI [K ] 3 L 12 6 L
12
6L 4L
6L 2 2L 6L 2 4 L
6 L
2
6L
kL
12
3
EI 2L
6 L
2
( a ) Boundary conditions are v1 0 and 2 0 . Equation (17.19a) with F 2 y P 2 and M 1 0 : 4L 0 EI 3 L 6L P 2
6L 1 3 kL 12 v2 E I
2
Introduce the data and solve: 1 5 .1(1 0
3
v 2 1 3 .5 (1 0
) ra d
3
) m
(b) 12 F1 y 6L 0 EI 3 F L 12 2 y M 2 6 L
12
6L 4L
6 L
2
6L
12
kL
3
EI 2L
6 L
2
6L 2 2L 6L 2 4 L
0 5 .1 3 (1 0 ) 1 3 .5 0
or F1 y 7 .4 2 5 k N F 2 y 9 .8 5 5 k N M 3 0 .1 5 0 k N m 2 F s p r in g 1 8 0 (1 3 .5 ) 2 .4 3 k N
(C )
From symmetry: F1 y F 3 y 7 .4 2 5 k N
SOLUTION (17.18) L 6 .7 m ,
P = 9 kN , v1
12 6L EI [ k ]1 3 L 12 6L
1 2
6 L 2L
5
u2
6L 4L
E I = 6 5 (1 0 ) N m ,
2
12 6L 12 6L
4
2
k 210 kN m v2
6L 2 2L 6 L 2 4L
0 3 E I kL E I [ k ]2 3 0 L 0
2
u3 3
0
0
0
0
0
0
0
0
0 0 0 0
(CONT.)
299
17.18 (CONT.) Thus 12 6L EI [K ] 3 L 12 6 L
12
6L 4L
6L
6L 2 2L 6L 2 4 L
6 L
2
kL
12
3
EI 2L
6 L
2
(1)
( a ) Boundary conditions are v1 0 and 1 0 . Equation (17.19a), with F 2 y P and M
2
0: kL P E I 1 2 EI 3 L 0 6 L
6 L v2 2 2 4 L
3
Substituting the given numerical values and solving, we have v 2 0 .0 3 2 7 m
2 0 .0 0 7 3 3 ra d
(b) 12 F1 y 6L EI M1 3 F L 12 2 y M 2 6 L
12
6L 4L
6L
2
6L
12
kL
3
EI 2L
6L
2
6L 2 2L 6 L 2 4 L
0 0 0 .0 3 2 7 0 .0 0 7 3 3
Introducing the data and multiplying: 2 .1 3 8 k N F1 y M 1 1 4 .2 4 5 k N m F2 y 9 .0 0 5 k N M 0 .0 8 1 k N m 2
The spring force is Ps p r in g 2 1 0 ( 0 .0 3 2 7 ) 6 .8 6 7 k N
(C )
SOLUTION (17.19) We have E I 7 0 1 0 N m , ( a ) Use Eqs.(17.19): v1 1 v2 4
12 18 EI [ k ]1 3 L 12 18
2
18
12
36
18
18
12
18
18
L 3 m,
P 50 kN
2 18 18 18 36
v1
1 v2
2
(CONT.)
300
17.19 (CONT.) v2
2
v3
12 18 EI [ k ]2 3 L 12 18
18
12
36
18
18
12
18
18
( b ) Global Stiffness Matrix v1 1 12 18 EI 12 [K ] 3 L 18 0 0
3 18 18 18 36
v2
2 v3
3
v2
2
v3
3
18
12
18
0
36
18
18
0
18
24
0
12
18
0
72
18
0
12
18
12
0
18
18
18
0 0 18 18 18 36
v1
1 v2
2 v3
3
Use Eq.(17.20a): F2 y 24 EI M 2 3 0 L M 18 3
3
L EI
18 v 2 18 2 36 3
0 72 18
1
K F
v2 3 L 2 EI 3
0 .0 7 2 9 1 7 0 .0 1 0 4 1 7 0 .0 4 1 6 6 7
0 .0 1 0 4 1 7 0 .0 1 7 3 6 1 0 .0 1 3 8 8 9
0 .0 4 1 6 6 7 0 .0 1 3 8 8 9 0 .0 5 5 5 5 6
50 10 0 0
3
0 .1 4 0 6 m 0 .0 2 0 1 r a d 0 .0 8 0 4 r a d
(c) F1 y EI M1 3 L F 3 y
12 18 1 2
18 18 18
0 0 1 8
v 2 3 4 .3 6 2 k N 2 5 6 .2 3 3 k N m 3 1 5 .6 0 2 k N
From a free-body diagram of element 2: ( M 2 ) 2 4 6 .8 0 6 k N m and ( F 2 ) 2 1 5 .6 0 2 k N
(CONT.)
301
17.19 (CONT.)
(d) 50 kN 2
1 1
1.5 m
3
1.5 m
2
34.362
V
+
(kN )
x 15.602
M
46.806
(kN m )
+
x
56.28
SOLUTION (17.20) ( a ) The element stiffness matrices are, from Eq. (17.19a): v1
1
v2
2
v3
12 6L EI 12 [ k ]1 3 L 6L 0 0
6L
12
6L
0
6 L
2L
12
6 L
6 L
4L
0
0
0
0
0
0
0
0
0 0 0 EI [ k ]2 3 0 L 0 0
4L
2
6 L 2L
2
2
0
2
0
0
0
0
0
0
0
0
0
12
6L
12
0
6L
4L
12
12
kL
3
EI 0
6L
2L
6 L
2
0 6L 2 2L 6 L 2 4 L 0
6 L
2
6 L
0 0 0 0 0 0
0
0
0
3
(CONT.)
302
17.20 (CONT.) (b) 12 6L 12 [K ] 6L 0 0
12
6L 4L
2
6L
2L
24
0
0
8L
6L 2L
2
12
0
6L
0
2
0 12 6L
2
6 L
12
kL
3
EI 0
6L
2L
2
0 6L 2 2L 6 L 2 4 L 0
6 L
( c ) Boundary conditions are: v1 0 , 1 0 , v 2 0 System governing equations, by Eqs. (17.19a), after rearrangement: 8 L2 0 EI 2 0 3 2L L P 6L
2L 4L
2
2
6L
6L 6L 3 kL 12 EI
2 3 v 3
Solving, v3
2
where k 1 k L
7PL
3
(
EI 3PL
3
EI
1 1 2 7 k1
2
(
1 1 2 7 k1
)
),
3 3 2
EI
SOLUTION (17.21) Boundary conditions are v1 1 v 3 3
v 2 (given)
Equation (17.19): 24 EI 12 L [ k ]1 3 L 24 1 2 L
12 L 8L
2
12 L 4L
2
24 12 L 24 12 L
12 L 2 4L 12 L 2 8L
12 6L EI [ k ]2 3 L 12 6L
6L 4L
2
6L 2L
2
12 6L 12 6L
6L 2 2L 6L 2 4L
After assembling [ K ] and considering the boundary conditions, the pertinent equations are found as (CONT.)
303
17.21 (CONT.) P EI 3 L 0
24 12 1 2 L 6 L
12 L 6 L 2 2 8L 4L 2
Multiplying, EI
P
( 3 6 6 L 2 )
(1)
(6 L 12 L 2 )
(2)
L 0
EI L
3
3
2
( a ) Equation (1) is then P 33
EI L
3
( b ) Equation (2) gives 2
2L
SOLUTION (17.22) through (17.26) It is important to take into account any conditions of symmetry which may exist. Use a 2-D finite element program such as ANSYS. End of Chapter 17
304
CHAPTER 18
CASE STUDIES IN MACHINE DESIGN
SOLUTION (18.1) a2=0.16 m a1=2.6 m
B
P=15 kN
L=2.5 m
C
H
D 40o
a3=1.0 m
80o FBG
A
FC F
60
o
o
50
Figure S18.1 Free Body diagram of loader arm FAE
( a ) Member forces Dismember the arm ABD. It is assumed that the links and hydraulic cylinder are all in tension. The conditions of moment equilibrium are applied to Fig. S18.1:
M
0:
B
F C F c o s 1 0 ( 0 .1 6 ) F A E s in 5 0 (1 .0 ) 1 5 ( 2 .7 6 ) 0 o
o
F A E 0 .2 0 6 F C F 5 4 .0
Fx 0 :
(1)
F A E c o s 6 0 F B G c o s 4 0 F C F s in 1 0 0 o
o
o
F B G 3 5 .2 4 6 0 .3 6 1 F C F
Fy 0 :
(2)
F A E s in 6 0 F B G c o s 4 0 F C F c o s 1 0 1 5 0 o
o
o
FC F 4 2 k N ( C )
Substitution of this into Eqs. (1) and (2) result in F A E 4 5 .3 5 k N ( T )
F B G 2 0 .0 8 k N ( C )
Comments: Since the result obtained for F A E is positive. A negative sign means that the sense of the force is opposite to that taken originally. ( b ) Diameters of pins at A, B, and C in double shear. We have S ys n
F 2
d
2
4
d [ 2 SF n ]
,
1 2
ys
Thus, dA [
2 ( 4 5 .3 5 ) ( 2 .4 )
dB [
2 ( 2 0 .0 8 ) ( 2 .4 )
dC [
3
(1 5 0 1 0 )
3
(1 5 0 1 0 ) 2 ( 4 2 ) ( 2 .4 ) 3
(1 5 0 1 0 )
1
]
2 1 .5 m m
2
1 4 .3 m m
1
] 1
]
2
2
2 0 .7 m m
305
SOLUTION (18.2) The location of the critical point is at K, where the maximum moment and the shear force are M P L 1 5 ( 2 .5 ) 3 7 .5 N m
V P 15 kN
The cross-sectional area properties: A ( c 2 c 1 ) ( 7 5 5 0 ) 9 .8 1 7 5 1 0 2
I
4
2
2
( c 2 c1 ) 4
4
2
3
( 7 5 5 0 ) 1 9 .9 4 1 8 1 0 4
4
4
mm 6
2
mm
4
The maximum tensile stress due to the bending occurs at point K. Therefore,
m ax
M c2
I
3
3 7 .5 (1 0 )( 0 .0 7 5 ) 1 9 .9 4 1 8 (1 0
6
)
141 M Pa
The shearing stress is zero, 0 , at point K. The maximum shearing stress is at the neutral axis z and parallel to y axis. From third case, Table 3.2: m ax 2
V A
3
1 5 (1 0 )
2
9 .8 1 7 5 (1 0
3
)
3 .0 6 M P a
This is a very low stress for the specified material. The bending stress vanishes at the neutral axis, H 0 . The factor of safety is thus n
3 .4
480 141
SOLUTION (18.3) A sketch of Mohr's circle is shown in Fig. S18.3 constructed by obtaining the position of point C at ( x y ) 2 4 0 0 μ on the horizontal axis and of point at ( x ,
xy
2 ) (1 0 0 0 μ , 3 5 0 μ ) from the origin O.
The principal strains are represented by points A1 and B1 . Thus, referring to the figure:
( 1 0 0 02 2 0 0 ) 3 5 0
400
2
1 ,2
2
'=400 D
y B(-200, 350) B1 O
”s C
’p
A1
A(1000, -350) x
E Figure S18.3 or 1 1 0 9 4 .6 μ ,
2 2 9 4 .6 μ
(CONT.)
306
18.3 (CONT.) As a check, note that x y 1 2 8 0 0 . The planes of principal strains are 2
1 350 600
' ta n
p
3 0 .3
o
2 p " 3 0 .3 1 8 0 2 1 0 .3
and
o
and p ' 1 5 .1 5
o
p " 1 0 5 .1 5
and
o
(1)
From Mohr's circle, p ' locates the 1 direction. The maximum shearing strains are given by points D and E: m ax 2
( 1 0 0 02 2 0 0 ) 3 5 0 2
1389
2
Alternatively, 1 2 1 0 9 4 .6 2 9 4 .6 1 3 8 9 μ . SOLUTION (18.4) From Prob. 18.3, we have 1 1 0 9 4 .6 μ , 2 2 9 4 .6 μ , m a x 1 3 8 9 μ The first two of Eqs. (2.7) together with (1) give 1
2
2 1 0 1 0
3
1 ( 0 .2 8 )
2
2 1 0 1 0
3
1 ( 0 .2 8 )
(1)
[1 0 9 4 .6 0 .2 8 ( 2 9 4 .6 ) ] 2 3 1 M P a
2
[ 2 9 4 .6 0 .2 8 (1 0 9 4 .6 ) ] 2 .7 1 M P a
From the last of Eqs. (2.7): m ax
E 2 (1 )
m ax
3
2 1 0 1 0 2 (1 0 .2 8 )
p ' 1 5 .1 5
From Prob. 18.3:
o
(1 3 8 9 ) 1 1 4 M P a
s 6 0 .1 5
and
o
Using Eq. (3.35), ' 12 ( 2 3 1 2 .7 1) 1 1 6 .9 M P a SOLUTION (18.5) Use Eq. (3.40): a x c o s a y s in a 2
1 1 0 4 (1 0
6
2
xy
s in a c o s a
) x c o s 0 y s in 0 2
o
2
o
o
xy
x 1104 μ
o
s in 0 c o s 0 ,
Similarly, b x c o s b y s in b 2
4 3 2 (1 0
4 3 2 (1 0
6
6
2
xy
s in b c o s b
) x c o s ( 6 0 ) y s in ( 6 0 ) 2
o
) 0 .2 5 x 0 .7 5
2
y
0 .4 3 3
o
xy
s in ( 6 0 ) c o s ( 6 0 ) o
xy
o
(1)
and c x c o s c y s in c 2
9 6 (1 0
6
2
) 0 .2 5 x 0 .7 5
y
xy
s in c c o s c
0 .4 3 3
Subtract Eq. (2) from Eq. (1): 5 2 8 0 .8 6 6
Thus
xy
610 μ ,
xy
y
144 μ
307
xy
(2)
SOLUTION (18.6) Equation (5.78a): 2 E Sy
Cc
2
2
3
( 2 1 0 1 0 )
1 2 8 .8
250
For the 1.6 m link column, L B G r 1 6 0 0 1 0 .6 9 1 4 9 .7 C c
and Eq. (5.77b) is used. Hence,
a ll
2
E 1 .9 2 ( L B G r )
2
9
( 2 1 0 1 0 )
1 .9 2 (1 4 9 .7 )
2
4 8 .2 M P a
Comment: This stress is very low compared to 250 MPa; The link will not yield.
SOLUTION (18.7) From Prob. 18.6: A 3 1 8 m m ,
E 200 G Pa ,
2
r 1 0 .6 9 m m .
The required value of a ll :
a ll
FB G
A
3
1 1 .3 4 (1 0 ) 3 1 8 (1 0
6
3 5 .7 M P a
)
(1)
Assume L m r C c . Equation (5.77b):
a ll
2
E
1 .9 2 ( L m r )
2
2
9
( 2 0 0 1 0 )
1 .9 2 ( L m r )
Lm
Equating Eqs. (1) and (2), we obtain
(2)
2
r
1 6 9 .7
Since L m r C c , our assumption is OK. Thus Lm r
Lm 1 0 .6 9
1 6 9 .7
or L m 1 .8 1 4 m
Comment: If this link is more than 1.814 m in length, it will buckle.
SOLUTION (18.8) ( a ) Central cross brace. Stain energy due to bending. Equation (5.18) with M W x 2 : U 2[
c 2 M 2 EI
0
dx]
1 EI
c 2 0
W 4
2
x dx 2
2
3
W c 96 EI
Due to shear, Eq. (5.23) with V W 2 : U
c 0
2
3V 5GA
dx
2
3W c 20 G A
Side supports. Strain energy owing to bending, with M W x 2 : U 4[
b 2 0
2
M 2 EI
dx]
2 EI
b 2 0
2
W 16
x dx 2
2
3
W b 384 EI
(CONT.)
308
18.8 (CONT.) Due to shear with V W 4 : U 2
b
2
2
dx
3V 5GA
0
3W b 40 G A
Total strain energy is then Ut
2
3
W c 96 EI
2
3
W b 384 EI
2
2
3W c 20 G A
2W b 40 G A
or 3
3
U t W [ E1I ( 9c 6 2
b 384
)
(c
3 20 G A
b 2
Q.E.D.
)]
(P18.8a)
( b ) Introducing Eq. (P18.8a) into Eq. (5.31), we obtain st
U t
3
W [ 4 81E I ( c 3
W
b 4
)
(c
3 10 G A
b 2
(P18.8b)
)]
SOLUTION (18.9) ( a ) Substituting the given data into Eq. (P18.8b) result in s t W [ 4 81E I ( c 3
W{
)
3
4 8 ( 2 0 0 1 0 0 .2 7 8 )
6
6
(c
3 10 G A
[( 0 .8 ) 3
1
W (1 0 10
3
b 4
b 2
)]
(1 .2 )
3
4
]
3 3
1 0 ( 7 9 1 0 8 1 9 )
[( 0 .8 )
1 .2 2
]}
){0 .3 7 4 7[ 0 .5 1 2 0 0 .4 3 2 ] [ 0 .0 0 4 6 4[1 .4 ]}
W {0 .3 5 3 7 0 .0 0 6 5} 1 0
6
W ( 0 .3 6 0 2 )
or s t 0 .3 6 0 2 (1 0 5 .4 0 3(1 0
3
6
) W ( 0 .3 6 0 2 1 0
6
)(1 5 1 0 ) 3
) m
( b ) The impact factor, by Eq. (4.32): K 1
1
2h
st
1
2 ( 250 )
1
1 0 .6 7
5 .4 0 3
Equation (4.35) is therefore m a x K s t 1 0 .6 7 (5 .4 0 3) 5 7 .7 m m
SOLUTION (18.10) ( a ) Thin-walled cylinder is in biaxial stress ( 3 0 ) with 1 and a 2 . From Eqs. (3.6) at r a : 1
pa t
20 (350 )
t
7000 t
,
Sy
2
2
1 2
3500 t
(1)
Equation (6.14): 1 1 2
2
2 2
(
n
)
(2)
Substituting Eqs. (1) into (2), we have 6
[( 7 ) ( 7 )(3 .5 ) (3 .5 ) ] 1 02 ( 5 55 2 ) , 2
2
2
t
t 55 m m
Hence, b a t 405 m m
( b ) We have a t 6 .3 6 : the thin-walled analysis does not apply. So, the solution is not valid.
309
SOLUTION (18.11) Refer to Prob. 18.10. For a closed-ended thick-walled cylinder under internal pressure, the critical section where the maximum stresses occur is at r a (see Fig. 16.3). The stresses, from Eqs. (16.16a), 16.16b), and (16.15) with p p i and p o 0 : r p p
p
a
2
2
2
2
b a b a a
(1)
2
2
b a
2
We observe that 1 , a 2 , and r 3 , where algebraically 1 2 3 . With a safety factor included, Eq. (6.13) appears as [ ( 1 2 ) ( 2
3 ) ( 2
2
1) ] 2( 2
3
Sy n
(2)
)
Substituting Eq. (1) into (2), we have [(b a a ) ( a b a ) ( b a b a ) ]( 2
or
4
3b p
2
2
2
2
2
(b a ) ( 2
2
2
Sy n
2
)
2
2
2
2
2
2
p
2
2
b a
) 2( 2
2
2
Sy n
)
2
(3)
By introducing the given data ( a 0 .3 5 m , S y 5 5 2 M P a , p 2 0 M P a , n 5 ) and simplifying Eq. (3) becomes: 9 .1 5 7 b 2 .4 8 8 b 0 .1 5 2 0 4
2
Let x b and solve the resulting quadratic equation to find x 0 .1 7 7 8 . The outer radius is then 2
b 0 .4 2 1 7 m = 4 2 1 .7 m m
Hence, t b a 4 2 1 .7 3 5 0 7 1 .7 m m Comment: The outer diameter equals 2 b 8 4 4 m m . A standard cylinder with about 8 4 5 m m outer diameter and 7 0 0 m m inner diameter should be selected.
SOLUTION (18.12) ( a ) Element Stiffness Matrix. Referring to Fig. P18.12 we sketch Fig. S18.12. F1 y , v 1
L
1 F1 x , u 1
45o
F3 x , u 3
2
F2 x , u 2
F2 y , v 2
3 F3 y , v 3
Figure S18.12 Finite element model (CONT.)
310
18.12 (CONT.) Using Eq. (17.4), Fig. S18.12, Table P18.1, and the given data: u1
v1
c cs 7 [ k ]1 4 (1 0 ) c2 c s 2
u1 0 0 7 [ k ] 2 4 (1 0 ) 0 0
u2 c
cs s
cs
c
s
cs
2
u3
0
0
1
0
0
0
1
0
u1
cs 2 s cs 2 s
2
cs
2
v1
v2
2
1 v1 0 7 4 (1 0 ) 1 u2 v2 0 u1
v3
v1
u2
v2
0
1
0
0
0
1
0
0
0 0 0 0
u2
0 1 0 1
0 .5 0 .5 7 2 (1 0 ) 0 .5 0 .5
u1 v1
[ k ]3 2
u3 v3
u1 v1 u2 v2
v2
u3
v3
0 .5
0 .5
0 .5
0 .5
0 .5
0 .5
0 .5
0 .5
0 .5 0 .5 0 .5 0 .5
u2 v2 u3 v3
In the foregoing, the column and row of each stiffness matrix are labeled according to the nodal displacements associated with them. Observe that displacements u 3 and v 3 are not involved in element 1; u 2 and v 2 are not involved in element 2; u 1 and v1 are not involved in element 3. Thus, before adding [ k ]1 , [ k ] 2 , and [ k ] 3 to form the system matrix, two row and columns of zero must be added to each of the element matrices to account for the absence of these displacements. 7
In so doing, and using a common factor 1 0 , the element stiffness matrices become: u1
v1
u2
v2
u3
4 0 4 7 [ k ]1 (1 0 ) 0 0 0
0
4
0
0
0
0
0
0
0
4
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0 0 0 0 0 0
u3
v3
u1
v1
u2
v2
0 0 0 7 [ k ] 2 (1 0 ) 0 0 0
0
0
0
0
4
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
4
0
0
0
v3
0 4 0 0 0 4
u1 v1 u2 v2 u3 v3
u1 v1 u2 v2 u3 v3
(CONT.)
311
18.12 (CONT.) u1 0 0 0 7 [ k ] 3 (1 0 ) 0 0 0
v1
u2
v2
u3
0
0
0
0
0
0
0
0
0
0
0
2
2
v3
2
2
0
2
2
2
2
0
2
2
2
2
0
2
2
2
2
u1 v1 u2 v2 u3 v3
System Stiffness Matrix. There are a total of six components of displacement for the truss before boundary constraints are imposed. Hence, the order of the truss stiffness matrix must be 6 6 . Before addition of the terms from each element stiffness matrix into their corresponding locations in [ K ] , we obtain the global matrix for the truss: u1 4 0 4 7 [ K ] (1 0 ) 0 0 0
v1
u2
v2
u3
0
4
0
0
4
0
0
0
4
0
2
2
2
0
2
2
2
4
2
2
2
4 2 2 2 4 2 0
0
2
v3
2
2
u1 v1 u2
(1)
v2 u3 v3
System Force-Displacement Relationship. From Fig. P18.12, the boundary conditions are u 1 0 , v1 0 , and u 3 0 . In addition F 3 y 0 . Equation (17.17 is therefore R1 x R 1y P W R 3x 0
[K
0 0 u 2 ] v 2 0 v 3
(2)
where [ K ] is given by Eq. (1). ( b ) Displacements. To find u 2 , v 2 , and v 3 only part of Eq. (2) relating to these displacements is considered: (CONT.)
312
18.12 (CONT.) (4 2 ) 24, 000 7 3 6 , 0 0 0 (1 0 ) 2 0 2
P W 0
2 2
2 (4 2 ) 2
2
u2 v2 v 3
(3)
Inverting, u2 8 v 2 2 .5 (1 0 ) v 3
1
1 1 0
0 1 1
4 .8 2 8 1
24 1 .5 3 3 6 (1 0 ) 4 .9 4 m m 0 0 .9
Reactions. Inserting of the preceding values of u 2 , v 2 , and v 3 into Eq. (2) gives the reactional forces: R1 x R1 y R 3x
4 0 2
7 10
0 0 2
0 1 .5 60 3 4 4 .9 4 (1 0 ) 3 6 k N 36 2 0 .9
The results may be verified by applying the equations of equilibrium to the free-body diagram of the entire truss, Fig. SP18.12. Axial Forces in Bars. Using Eqs. (17.16), (3) and Table P18.12, we obtain F1 F1 2
AE L
[1
F 2 F1 3
AE L
[0
F3 F 2 3
AE 2L
u2 7 0 ] 4 (1 0 ) [1 v2
1 .5 3 0] (1 0 ) 6 0 k N 4 .9 4
0 7 1] 4 (1 0 ) [ 0 v 3 u2 1] 2 v3 v2
[1
0 3 1] (1 0 ) 3 6 k N 0 .9 2 (1 0 ) [ 1 7
1 .5 3 1] (1 0 ) 0 .9 4 .0 4
7 1 .8 k N
Stresses in Bars. Driving the element forces by the cross-sectional area results in 1
3
6 0 (1 0 ) 4 8 0 (1 0
6
)
125 M Pa
2
1 2 5 ( 36 60 ) 7 5 M P a
3 1 2 5 ( 7610.8 ) 1 4 9 .6 M P a
The negative sign means a compressive stress. Factor of safety against yielding. Dividing the yield strength of S y 2 5 0 M P a (from Table B.1) by each stress, we obtain n1
250 125
2
n2
250 75
3 .3 3
n3
250 1 4 9 .6
1 .6 7
Comments: The bar axial stresses found are relatively low for the well known material considered (see Sec. 1.6). End of Chapter 18
313
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