Strength Of Materials

Second Edition

STRENGTH OF MATERIALS

A.K. Srivastava • P.C. Gope

Strength of Materials

Strength of Materials SECOND EDITION

A.K. SRIVASTAVA

Manager (Design) Aircraft Upgrade Research and Design Centre Hindustan Aeronautics Limited Nasik

P.C. GOPE

Professor College of Technology G.B. Pant University of Agriculture and Technology Pantnager

New Delhi-110001 2012

` 350.00 STRENGTH OF MATERIALS, Second Edition A.K. Srivastava and P.C. Gope © 2012 by PHI Learning Private Limited, New Delhi. All rights reserved. No part of this book may be reproduced in any form, by mimeograph or any other means, without permission in writing from the publisher. ISBN-978-81-203-4522-5 The export rights of this book are vested solely with the publisher. Third Printing (Second Edition)

¼

¼

¼

Febraury, 2012

Published by Asoke K. Ghosh, PHI Learning Private Limited, M-97, Connaught Circus, New Delhi-110001 and Printed by Meenakshi Art Printers, Delhi-110006.

Dedicated to

MY LATE FATHER

Contents

Foreword Preface Preface to the First Edition

xiii xv xvii

1. SIMPLE AND COMPOUND STRESS

1–48

1.1 Introduction 1 1.2 Stress 1 1.3 Uniformly Distributed Stress 3 1.3.1 Tensile and Compressive Stresses 3 1.3.2 Stress Due to Bending Moment 4 1.3.3 Stress Due to Twisting Moment 5 1.4 Complex Stresses 6 1.4.1 Plane Stress 6 1.4.2 Stresses on an Inclined Plane 7 Exercises 46

2. ANALYSIS OF STRESS AND STRAIN 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11

49–94

Introduction 49 Force Distribution 49 The State of Stress at a Point 50 Stress Notations 50 Stress Tensor at a Point 50 Stress Gradient 51 Differential Equations of Equilibrium 51 Equilibrium Equations for Plane Stress State Generalized Hooke’s Law 53 Direction Cosines 55 Normal and Shear Stresses 56 vii

53

viii

Contents

2.12 Principal Directions 57 2.13 Stress Components on an Arbitrary Plane 57 2.14 Principal Stress 59 2.15 Stress Invariants 60 2.16 Principal Directions 61 2.17 Octahedral Stress 68 2.18 Mean and Deviator Stresses 70 2.19 Strain Analysis 74 2.20 Strain-Displacement Relation 75 2.21 Three-dimensional Strains 79 2.22 Normal and Shearing Strains 80 2.23 Principal Strains 80 2.24 Principal Strain Directions 81 2.25 Concept of Compatibility 81 2.26 St-Venant’s Equations of Compatibility 83 2.27 Solution of Stress Differential Equation 86 2.28 Types of Airy’s Stress Function 88 2.29 Application of Airy’s Stress Function 90 2.30 Mohr’s Circle for the Three-dimensional State of Stress Exercises 94

92

3. THEORY OF FAILURE

95–139

3.1 3.2

Introduction 95 Failure Theory for Ductile Material 96 3.2.1 Maximum Shear Stress Theory 96 3.2.2 Maximum Distortion Energy Theory 102 3.2.3 Strain Energy Density or Total Strain Energy Criterion 105 3.3 Theory of Failure or Yield Criterion for Brittle Materials 106 3.3.1 Maximum Principal Stress Criterion 106 3.3.2 Maximum Principal Strain Criterion 107 3.4 Mohr’s Theory 108 3.5 Experimental Verification of Theory of Failure 111 3.5.1 Comparison of Failure Criteria 113 3.6 Theory of Failure for Cyclic Loads 123 3.6.1 Stress Parameters 123 3.6.2 Strength Parameter 124 Exercises 139

4. ENERGY METHODS 4.1 4.2

Introduction 140 Strain Energy 140 4.2.1 Strain Energy with Simple Loading 143 4.2.2 Strain Energy due to Moment M 143

140–157

Contents

4.2.3 Strain Energy due to Torsional Loading 4.2.4 Strain Energy due to Transverse Shear 4.3 Castigliano’s First Theorem 147 Exercises 156

143 144

5. DEFLECTION OF BEAMS

158–195

5.1 5.2 5.3

Introduction 158 Relation between Slope, Deflection and Radius of Curvature Method for Slope and Deflection 159 5.3.1 Double Integration Method for Slope and Deflection 5.3.2 Macaulay’s Method 172 5.3.3 Moment Area Method 179 5.3.4 Mohr’s Theorems 181 5.4 Indeterminate Structure 185 5.5 Continuous Beam 186 5.6 Clapeyron’s Theorem of Three Moments 186 Exercises 194

158 160

6. CURVED BEAM 6.1 6.2 6.3 6.4

Introduction 196 Stresses in Curved Beam (Winkler–Bach Theory) Position of Neutral Axis 200 Values of h2 200 6.4.1 Rectangular Cross-section 201 6.4.2 Circular Cross-section 201 6.4.3 I-section 202 6.4.4 T-section 202 6.4.5 Trapezoidal Cross-section 203 6.5 Stresses in a Ring 214 6.6 Stresses in a Chain Link 220 Exercises 223

ix

196–224 196

7. THIN CYLINDER AND SPHERE 7.1 Introduction 225 7.2 Classification of Pressure Vessels 225 7.3 Stresses in a Thin Cylindrical Shell due to an Internal Pressure 225 7.4 Circumferential or Hoop Stress 226 7.5 Longitudinal Stress 227 7.6 Effect of Internal Pressure on the Dimensions of a Thin Cylindrical Shell 7.7 The Spherical Shells Subjected to an Internal Pressure 236 7.8 Change in Dimensions of Thin Spherical Shell due to an Internal Pressure Exercises 240

225–240

230 237

x

Contents

8. THICK AND COMPOUND CYLINDER

241–262

8.1 Introduction 241 8.2 Lame’s Theory 241 8.3 Application of Theories of Failure 250 8.4 Compound Cylindrical Shell 252 8.5 Thick Spherical Shells 256 Exercises 262

9. UNSYMMETRICAL BENDING AND SHEAR CENTRE 9.1 9.2 9.3 9.4 9.5

Introduction 263 Definitions 263 Stresses due to Unsymmetrical Bending 264 Deflection of Beam due to Unsymmetrical Bending Shear Centre 275 9.5.1 Shear Centre for Channel Section 275 9.5.2 Shear Centre of Unequal I-section 278 Exercises 280

263–281

266

10. COLUMNS AND STRUTS Introduction 282 Definitions 282 Classification of Column 282 Assumptions Made in the Euler’s Column Theory 283 Expressions for Crippling Load of Different Cases 283 10.5.1 Both the Ends are Hinged or Pinned 283 10.5.2 One End is Fixed and Other is Free 284 10.5.3 Both Ends are Fixed 286 10.5.4 One End is Fixed, Other is Hinged 287 10.6 Effective Length of a Column 288 10.7 Slenderness Ratio 289 10.8 Crippling Stress in Terms of Effective Length and Radius of Gyration 10.9 Limitation of Euler’s Formula 290 10.10 Rankine’s Formula 296 10.11 Eccentric Loading 302 10.12 Johnson’s Formula for Columns 304 10.12.1 Johnson’s Straight Line Formula for Columns 304 10.12.2 Johnson’s Parabolic Formula for Columns 305 Exercises 305

282–307

10.1 10.2 10.3 10.4 10.5

11. SPRING 11.1 Introduction 308 11.2 Definitions 308 11.3 Types of Springs 309

289

308–333

Contents

xi

11.4 Helical Spring 309 11.4.1 Closely-coiled Helical Springs 309 11.4.2 Open-coiled Helical Springs 312 11.5 Strain Energy in the Spring 314 11.6 Springs under Impact Load 315 11.7 Springs in Series 315 11.8 Springs in Parallel 315 11.9 Leaf Springs or Carriage Springs 326 11.9.1 Semi-elliptical Spring 326 11.9.2 Quarter-elliptical Leaf Spring 330 Exercises 333

12. ROTATING DISCS AND CYLINDERS 12.1 Introduction 334 12.2 Rotating Disc 334 12.2.1 Strain Considerations 335 12.3 Hollow Disc (Disc with a Central Hole) 12.4 Solid Disc 338 12.5 Disc of Uniform Strength 338 12.6 Rotating Cylinder 340 12.7 Solid Cylinder 342 12.8 Hollow Cylinder 343 Exercises 348

334–349

337

13. FINITE ELEMENT METHOD AND ITS APPLICATION USING ANSYS SOFTWARE 13.1 Introduction 350 13.2 The Steps 350 13.3 Principle of Minimum Potential Energy 351 13.3.1 Potential Energy 352 13.4 Computer Aided Stress Analysis Technique 353 13.5 Elements Type and Shapes 354 13.6 One-dimensional Problems 358 13.6.1 Natural Coordinate (Intrinsic Coordinate) 358 13.6.2 Isoparametric Element 359 13.6.3 Element Strain Displacement Matrix 359 13.6.4 Element Stiffness Matrix 360 13.6.5 Forces 362 13.7 Application of Finite Element Analysis Using the Ansys Software 13.7.1 Application of Finite Element Analysis Using 1D Element 13.7.2 Application of Finite Element Analysis Using 2D Element 13.7.3 Application of Finite Element Analysis Using 3D Element

INDEX

350–391

367 367 379 386

393–394

Foreword

It is my pleasure to write the Foreword for this comprehensive and practice-oriented text on Strength of Materials by Mr. A.K. Srivastava and Dr. P.C. Gope. The book deals with the basic principles involved in the analysis of structural components. To enhance the scope of this text, it includes a chapter on Finite Element Method (FEM) and its application by FE software, which is a widely used numerical technique for solving various engineering problems. This book is intended to serve as a textbook for undergraduate students of mechanical engineering, production engineering and industrial engineering. Besides, it would be useful to the design personnel working in the industry. Finally, I take this opportunity to congratulate the authors and wish a grand success to the book. Best wishes

P.K. KHANWALKAR General Manager Hindustan Aeronautics Limited Nasik

xiii

Preface

The main objective of the second edition is to enhance the scope of strength of materials in a single volume. Keeping in view the importance of the stress and strain analysis in component design, failure analysis and reliability assessment of a component, this edition includes four new chapters, namely Simple and Compound Stress, Theory of Failure, Energy Methods and Finite Element Method and its Applications Using ANSYS Software. The chapter on Analysis of Stress and Strain has been thoroughly revised. These chapters cover in detail one-dimensional, two-dimensional and threedimensional state of stress. Till now, no book discusses the application of Finite Element Method (FEM) and validation with analytical calculation using the strength of materials approach. This revised edition exposes the reader to the field of Finite Element Analysis (FEA) techniques as practised in the industry. The book additionally covers the fundamentals of finite element analysis and its applications using the software ANSYS. Most of the finite element examples are compared with the examples solved by the strength of materials approach. If readers have any problems with the instructions given in the chapter on Finite Element Method and its Applications Using ANSYS Software, they can email at [email protected] This book is primarily intended for the undergraduate students of mechanical engineering, production engineering, and industrial engineering. In addition, it would be useful to project work, higher studies, and practising professionals. We express our thanks to the readers for sending in many useful comments and suggestions, and hope the interaction would continue in future. Finally, we thank the hard-working staff of PHI Learning for their cooperation at all stages during the printing of the book. A.K. SRIVASTAVA P.C. GOPE

xv

Preface to the First Edition

The subject strength of materials involves analytical methods for determining stress, strength, stiffness and stability of various engineering components and structures. A thorough understanding of the basic principles involved in the analysis of such components and structures is useful to mechanical engineers, civil engineers, architecture, agricultural engineers and design engineers of almost all branches of engineering. Intended as a textbook for the course, Strength of Materials for the undergraduate students of mechanical, production and industrial engineering, this book presents the fundamentals and advanced level topics with illustrations and solutions of the practical problems. Most of the examples are taken from the recent examination of the Indian technical universities as well as professional examining bodies. Each chapter is written in a simple and logical manner, explaining the fundamentals and applications of the theory involved followed by the example. The book is organized in nine chapters. Chapter 1 deals with 3D analysis of stress and strain. Methods for determining the principal stresses graphically and analytically and compatibility concepts have been considered in this chapter. Chapter 2 discusses the determination of slope and deflection of determinate and indeterminate beams. Detailed theory and applications of double integration method, Macaulay’s method and moment area method have been included in this chapter. Chapter 3 covers analysis of curved beam. Methods for determining stresses using Wrinkler–Bach theory has been described in detail. Chapters 4 and 5 explore stress analysis of thin, thick cylindrical and spherical vessels subjected to internal and external pressures. The design and analysis of pressure vessels are illustrated by numerical problems in which solutions have been made simpler and avoided the chance of incomplete understanding of the theory involved and its significance. Chapter 6 presents stress analysis of unsymmetrical bending and determination of shear centre. Buckling analysis of columns and struts using Euler’s and Rankine’s theory is reviewed in Chapter 7. In Chapter 8, analyses of closed, open coiled helical springs and leaf spring are introduced. Finally, Chapter 9 addresses the analysis of rotating disc and cylinder. At the end of each chapter, problems for practice are given. The solved examples are also included, and they have been so selected that almost all the possible types of problems are covered. The contents covered in the book meet the syllabi of most of the Indian universities at degree level.

xvii

xviii

Preface to the First Edition

Despite of best efforts by all concerned, it is possible that some errors might have crept into the book. The authors would appreciate being informed of these errors and will acknowledge them by name and institution in the subsequent editions. Authors are thankful to the publisher for their co-operation at all stages during the printing of this book. A.K. SRIVASTAVA P.C. GOPE

1 1.1

Simple and Compound Stress

INTRODUCTION

Mechanics of materials is the study of the relationship between the external forces applied on a body and the intensity of internal forces or disturbances produced in a deformable body. The external forces applied to any structure are either concentrated loads or distributed forces per unit volume or per unit area. The loads may be axial, tangential or inclined to the surface of the body. The nature of the load may be axial tension, axial compression, bending moment or twisting moment.

1.2

STRESS

Consider a body as shown in Fig. 1.1. The forces acting on the body are F1, F2, and F3, etc. These forces may be moment, couple or distributed over the surface area.

Fig. 1.1

Forces acting on a body.

The effect of these forces on the body can be studied from the internal disturbances produced due to these loads. To study this it is necessary to establish the concept of the state of stress. To develop the concept of the state of stress, consider a section on the body as shown through AB in Fig. 1.1. Consider that the sectioned area consists of a number of small area DA. Now assuming the material to be continuous and cohesive, a finite small force DF acting on the small area DA is shown in Fig. 1.2. The small force DF may have any direction but the force can be resolved 1

2

Strength of Materials

into three components acting parallel to the three mutually perpendicular axes. Let these components be DFx, DFy and DFz in the x, y and z directions respectively. As these forces are parallel to three coordinates, they are normal and tangential to the area. The ratio of force and area defines the stress and describes the intensity of the internal force on the area.

Fig. 1.2 Components of forces in the x, y and z directions.

Now, if DFx acts normal to the area, the force per unit area is known as normal stress and is given as 'Fx T xx lim (1.1) 'A0 'A The other two tangential forces produce tangential stresses and are known as shear stress. txy = lim 'A0

'Fy 'A

(1.2)

txz = lim 'A0

'Fz 'A

(1.3)

In Eqs. (1.1) through (1.3), the double subscript notation is used to denote the direction of the stress and the plane on which the stress is acting. For example, the subscript xy means the stress is acting in the y-direction and is on the x plane. Here y denotes the outward normal direction drawn on the plane. The y plane is nothing but xz plane. Now, if the external force produces uniform deformation, then the normal stress produced is uniform of magnitude s and the each cross-sectional area DA will be subjected to equal force of DF. Hence DF = sDA. The average stress is Ô 'F

=

s=

T Ô 'A F

(1.4)

(1.5) A where F is the average force acting through the centroid of the total area. The units of the stress are N/mm2 (MPa), N/m2 (Pa).

Chapter 1: Simple and Compound Stress

3

Now we can summarize that 1. Stress is defined as the internal resistance developed in the body due to external disturbances over unit area of its cross section. 2. The intensity of force normal to the section is called the normal stress and the intensity of force tangential to the plane is called the tangential or shear stress. The normal stress may be positive or negative depending upon the direction and cause it produces on the element. If the normal stress acting on the surface produces tension, then it is called the tensile stress and if it is producing compression on the element then, it is called the compressive stress. 3. The stress is generally expressed numerically in the unit of N/m2 (Pa) or N/mm2 (MPa). (1 N/mm2 = MPa, 1GPa = 109 Pa, 1 MPa = 106 N/m2 = 106 Pa).

1.3 1.3.1

UNIFORMLY DISTRIBUTED STRESS Tensile and Compressive Stresses

If the stresses set up inside the material due to external loading elongates the fibres of the material, then the stress is said to be tensile. On the other hand, if the stress shortens the fibres, then the stress is said to be compressive. The example of tensile stress (Fig. 1.3) is the tension rod. The normal stress induced in the rod is obtained as

T

F A

(1.6)

where F is the average tensile force and A is the cross section of the tension member.

Fig. 1.3

Tensile stress.

The compressive stresses are obtained when the bar is compressed. A component under the compressive load is shown in Fig. 1.4. A proper sign should be used for tension and compression loading. In case of compression, there is a limitation of length to diameter ratio. The L/D ratio should be less than 2, otherwise buckling may occur and the relations discussed above may not be accurate.

4

Strength of Materials

Fig. 1.4

1.3.2

Compressive stresses.

Stress Due to Bending Moment

When a straight prismatic beam of uniform cross section and made of a homogeneous material is subjected to bending moment, the bending stress can be obtained from the flexure relation given as M

T

E

I

y

R

(1.7)

where M = applied bending moment I = moment of inertia of the beam cross section E = modulus of elasticity R = radius of curvature y = distance from the neutral axis s = stress developed due to application of bending moment. The flexure relation is based on the following assumptions: 1. The beam is straight and subjected to pure bending. It means that the beam is under the action of pure bending only. No axial load either tension or compression and torsional load exists. Hence, shear stress is also absent. 2. The cross section of the beam is uniform. 3. The cross section of the beam remains plane during the bending. It means the plane cross section of the beam initially perpendicular to the axis of the beam remains plane during the deformation and is perpendicular to the deflected neutral plane. 4. The material must be isotropic, homogeneous and follows Hooke’s law. The variation of bending stress and sign convention used for y is shown in Fig. 1.5.

Fig. 1.5

(a) Bending of a beam, (b) cross section of the beam, and (c) stress distribution.

Chapter 1: Simple and Compound Stress

5

Thus, the bending stress is

T

M–y

(1.8)

I

Equation (1.8) shows that 1. The maximum normal stress in the member occurs at a point on the cross section located furthest from the neutral axis. The neutral axis is the locus of all points where the bending stress is zero. For a circular section, the maximum value of y is d/2, d is the diameter of the section. 2. It is applicable to members whose cross section is symmetrical with respect to the axis and the moment M is applied perpendicular to this axis. The moment of inertia and y are obtained as follows for different cross sections. Solid circular cross section of diameter d I=

Qd4

y =

64

d 2

Hollow circular cross section of outer and inner diameter do and di I=

Q (do4  di4 ) 64

y=

do 2

Square cross section with side a I=

a4

y=

12

a 2

Rectangular cross section with sides b ´ d I=

1.3.3

bd 3

y=

12

d 2

Stress Due to Twisting Moment

When a bar is subjected to moment which causes twisting of the bar, the bar is said to be in torsion (Fig. 1.6). When a bar is subjected to twisting moment T, the relation between shear stress developed t, the angle of twist q, length of the bar l, polar moment of inertia J and the modulus of rigidity G is given by T GR U (1.9) J l r

Fig. 1.6

Torsion of circular bar.

6

Strength of Materials

The derivation of the above equation is based on the following assumptions. 1. The bar is straight before the application of torque and of circular cross section. 2. The bar is subjected to pure torque only. 3. The adjacent cross section originally plane and parallel, remains plane and parallel after twisting. 4. The material is homogeneous, isotropic and follows Hooke’s law. 5. The magnitude of the applied torque is not sufficiently high to yield the material. Equation (1.9) shows that the shear stress varies linearly with radius of the bar. It is zero at the centre and maximum at the outer surface or fibre of the bar. If d is the diameter of a circular bar, the maximum shear stress is obtained as Td (1.10) U max 2J The polar moment of inertia for a circular solid cross section is J

Qd4 32

and for the hollow cross section (do = outer diameter and di = inner diameter) J

1.4

Q (do4  di4 ) 32

COMPLEX STRESSES

In the previous section we discussed that stress at a point is the average force acting over an area and it is obtained as the limiting value of the small force acting over a small area passing through the point. Generally, a surface is defined through the direction of outward normal drawn on it. In other words, the outward normal defines the orientation of the area. Now, in the limiting conditions, this small area tends to a point and the state of stress is obtained. It can now be discussed that through a point infinite number of planes can be drawn whose outward normal will be different. Through these surfaces infinite small area can be considered to obtain the state of stress. The totality of all such stresses acting on every possible plane passing through the point is defined to be the state of stress at the point. For understanding the definition of state of stress if the body which we considered in Fig. 1.1, is further sectioned by planes x-y plane (outward normal is in the z-direction) and y-z plane (outward normal is in the x-direction), three faces will contain nine components of stress, three on each face. This is shown in Fig. 1.7. The stress components shown on the faces describe the state of stress at the point for the element.

1.4.1

Plane Stress

A state of stress is said to be plane state of stress when there is no stress in any one direction or in any one plane. If we consider that there is no stress in the z-direction, the plane stress condition exists. So setting the stresses in the z-direction or z-plane zero, the state of stress will become plane stress.

Chapter 1: Simple and Compound Stress

Fig. 1.7

7

Three-dimensional stress system.

This can be represented as shown in Fig. 1.8.

Fig. 1.8

Two-dimensional stress system (plane stress system).

1.4.2 Stresses on an Inclined Plane Consider a bar subjected to biaxial load as shown in Fig. 1.9. Now it is required to find out the stresses on an element which is located at an angle of q to one of the loading axis, say, the x-axis. In other words, it is required to determine the normal and shear stress acting on a plane inclined at an angle q with respect to the x-axis, i.e., outward normal of the plane makes an angle q with the x-axis. The sign convention for angle is as follows. The anticlockwise measurement of angle from reference direction, i.e., x-axis is taken as positive and clockwise measurement form the x-axis is taken as negative. The ABCD is a plane subjected to biaxial stresses as shown in Fig. 1.9. Let EF be an oblique plane inclined at an angle q to the x-axis. The forces acting on the element EFG are shown in Fig. 1.10. snn and tns are normal and tangential stress on the oblique plane EF. Let A be the area of the oblique plane EF. The area of other two planes GF and EG are A sin q and A cos q respectively. The forces acting on each face can be obtained by multiplying the stress with area. Now resolving the forces in x and y directions and applying the force equilibrium conditions, the two unknown stress components acting on the plane EF can be determined.

8

Strength of Materials

Fig. 1.9 Biaxial loading.

Fig. 1.10

Fig. 1.11 Components of forces on oblique plane.

Chapter 1: Simple and Compound Stress

9

Resolving the forces in the x-direction, we can write the equilibrium condition as S Fx = 0

T nn cos R – EF  U ns sin R – EF  T xx – EG  U xy – GF = 0 Substituting the values of area EF, EG and GF,

T nn cos R – A  U ns sin R – A  T xx – A cos R  U xy – A sin R 0 Simplifying, we get

T nn cos R  U ns sin R  T xx cos R  U xy sin R = 0

(i)

Similarly the force equilibrium in the y-direction is obtained as S Fy = 0

T nn sin R – EF  U ns cos R – EF  T yy – GF  U xy – EG = 0 Substituting the values of area EF, EG and GF, we obtain

T nn sin R – A  U ns cos R – A  T yy – A sin R  U xy – A cos R

0

on simplification, we get

T nn sin R  U ns cos R  T yy sin R  U xy

cos R

0

(ii)

From Eqs. (i) and (ii), we get

T nn T xx cos2 R  T yy sin 2 R  U xy sin 2R

(iii)

By substituting, we get cos2 R

1  cos 2R 2

1  cos2R

and sin 2 R

2

Equation (iii) can be simplified as

T nn T x x „

T xx  T yy „

2

Now the normal stress perpendicular to



T xx  T yy 2

cos2R  U xy sin 2R

(1.11)

snn is obtained by substituting q = 90 + q as

T n n = T xx sin 2 R  T yy cos2 R  U xy sin 2R „

„

Tn n = T y y

T xx  T yy

„

„

U ns

„



T xx  T yy 2



T xx  T yy

(1.12) cos 2R  U xy sin 2R 2 2 Similarly, tangential stress component can be obtained by eliminating snn. The corresponding equation after eliminating normal stress is „

sin 2R  U xy cos 2R

The stresses on the inclined plane are shown in Fig. 1.12.

(1.13)

10

Strength of Materials

Fig. 1.12

Stresses on inclined planes.

Maximum shear stress The plane of the maximum shear stress is obtained by differentiating tns with respect to q and equating to zero. dU ns È T xx É Ê

 T yy Ø 2

Ù – Ú

dR

= 0

2 cos 2R  2U xy sin 2R = 0 tan 2R s = 

(T xx  T yy ) 2U xy

(1.14)

Equation (1.14) gives us two values of 2qs, which differs by 180°, i.e., two values of qs differ by 90°. Substituting these values in Eq. (1.13), the maximum shear stress is

U max

“

È T xx É Ê

 T yy Ø Ù Ú

2

2

 U xy 2

(1.15a)

These two values act at the right angle to each other. Substituting the value of sin 2qs and cos 2qs in Eq. (1.11), the normal stress is given as

T nn

Maximum normal stress

T xx  T yy

(1.15b)

2

The plane of maximum normal stress is obtained similarly by differentiation Eq. (1.11) with respect to q and equating to zero. ˜ T nn ˜R

=0

˜ Ë T xx  T yy T xx  T yy Û  cos 2R  U xy sin 2R Ü = 0 Ì ˜R Í 2 2 Ý

Chapter 1: Simple and Compound Stress

2 –

T xx  T yy 2

11

sin 2R  2U xy cos 2R = 0

After simplification the resulting equation is

2U xy

tan 2R p

T xx

(1.16)

 T yy

The multiplication of two Eqs. (1.14) and (1.16) shows that tan 2 R s – tan 2 R p =  1

(1.17)

This shows that these two 2q values differ by 90°. In other words, maximum shear stress planes are ± 45° to the principal stress plane.

EXAMPLE 1.1 A machine member having 50 mm inner diameter and 75 mm outside diameter, 300 mm long and fixed at one end is subjected to tensile load F = 20 kN. Determine the tensile and maximum shear stress.

Fig. 1.13 Example 1.1.

Solution: A=

sxx =

Q 4

F A

(D2  d 2 )

Q 4

20 – 103 2.454 – 10

2.454 – 10 3 m 2

(0.0752  0.052 )

3

8.149 – 10 6 N/m 2 = 8.149 MPa

txy = 0

Maximum normal stress smax = 8.149 MPa Maximum shear stress tmax =

1 2

T max

1 2

– 8.149 4.074 MPa

EXAMPLE 1.2 A circular tension member of diameter 50 mm is to carry an axial load of 360.3 kN. Determine the stresses on a plane: (a) normal to the load applied, (b) inclined at an angle 30° with respect to loading direction and (c) locate the plane of maximum shear stress and magnitude of maximum shear stress.

12

Strength of Materials

Solution: (a) Stress on a plane normal to the load applied AB is normal to the applied load F [Fig. 1.14(a)]. The cross-sectional area of AB plane is

Q

A=

4

sxx =

Q

d2

4

– 502

F

30 – 1000

A

1963.49

1963.49 mm 2

183.499 N/mm 2

183.5 MPa

As there is no force acting on other planes, the stresses on other planes are zero. Hence, the state of stress is written as Ë183.5 Ì Ì0 Ì Ì Í0

T

Fig. 1.14(a)

0

0Û

0

0Ü

0

0 ÜÝ

Ü Ü

Stress distribution of Example 1.2, (b) State of stress.

(b) Stresses on an oblique plane The oblique plane is shown in Fig. 1.14(c), snn and tns are normal and tangential stresses on the oblique plane AB. n is the outward normal drawn on the oblique plane which is at an angle of 30° to the x-axis, the direction of the applied load F.

Fig. 1.14(c)

Stresses on oblique plane of Example 1.2.

From the relation

T nn T x x „

T xx  T yy „

2



T xx  T yy 2

cos2R  U xy sin 2R

Chapter 1: Simple and Compound Stress

13

We can simplify by substituting

syy = txy = 0 snn = T x x = „

„

snn = T x x = „

„

T xx

T xx

+

2

183.5 2

cos 2R

2

+

183.5 2

cos 60’ = 137.625 N/mm 2

Normal stress perpendicular to n(x¢x¢) direction is

T xx

Tn n = Ty y = „

„

„

=

„

2

183.5



2

T xx



2

183.5 2

cos 2R

cos 60°

= 45.874 N/mm2



U ns =

T xx  T yy

= 

sin 2R + U xy cos 2R

2 183.5  0 2

sin 60°

= – 79.456 N/mm2 (c) Plane of maximum shear stress

tan 2Rs = 

T xx  T yy À ‡ 2U xy

2qs = 90°

qs = 45° Hence, the maximum shear stress occurs at 45° to the loading axis. The magnitude of the maximum shear stress is

È T xx  T yy Ø = “ É + U 2xy Ù 2 Ê Ú 2

U max

=

“

È 183.5 É 2 Ê

 0Ø

= ± 91.75 N/mm2

Ù Ú

2

+0

14

Strength of Materials

Principal stresses It is shown that the magnitude of normal and shear stress on any oblique plane depends on the inclination of the plane. The orientation and plane of the maximum normal and shear stresses can be obtained from the results discussed above. The plane on which shear stress is zero is known as principal plane and the associated normal stresses are principal stresses. The normal stress on any oblique plane defined by angle q is given as

T nn =

T xx + T yy 2

+

T xx  T yy 2

cos 2R + U xy sin 2R

(i)

The maximum value of this normal stress can be obtained by substituting the condition for maximum normal stress as given by

tan 2R p =

2U xy T xx



(ii)

T yy

The solution has two roots and angles 2q are differing by 180°. Depending upon the sign (positive or negative) of txy and (sxx – syy) two triangles are drawn in Fig. 1.15.

Fig. 1.15

Computation of principal stresses.

From Fig. 1.15, we can derive one set of angles as

sin 2R p1 =

U xy È T xx É Ê

 T yy Ø 2

Ù Ú

(iii)

2

+ U 2xy

Chapter 1: Simple and Compound Stress

15

and È T xx É Ê

cos 2R p1 =

È T xx  É 2 Ê

 T yy Ø Ù Ú

2

T yy Ø

(iv)

2

U 2xy

+

Ù Ú

and for other set of angles, the equations are U xy

sin 2R p 2 = 

and

cos 2R p 2 =

È T xx É Ê



T yy Ø

2

Ù Ú

È T xx É Ê



2

+ U 2xy

 T yy Ø Ù Ú

2

È T xx  É 2 Ê

T yy Ø 2

+ U 2xy

Ù Ú

Substituting Eq. (iii) and (iv) in Eq. (i), we obtain two principal stresses from two sets of angles defined by equations (iii) and (iv).

T1 =

T2 =

T xx + T yy 2

T xx + T yy 2

+

È T xx É Ê

 T yy Ø

+

È T xx É Ê

 T yy Ø

2

Ù Ú

2

Ù Ú

2

+ (U xy )2

(1.18a)

+ (U xy )2

(1.18b)

2

The sets of equations given by Eq. (1.18) are known as principal stresses and the corresponding planes on which they act are called principal planes. The two principal stresses are right angle to each other. Furthermore, on these planes no shear stress acts. Hence, principal planes are also known as shear less planes.

Mohr’s circle for plane stress Mohr’s circle is a graphical method for analysis of state of stress or determination of principal stresses. For two dimensional state of stress (plane stress), Mohr’s circle is platted on s – t coordinate axes. Let the two dimensional state of stress be given as ËT xx

T = Ì ÌU xy Í

U xy Û Ü

T yy ÜÝ

16

Strength of Materials

The normal and shear stress components on any oblique plane defined by q after rearranging the terms we can express as given below. Consider the normal and shear stress on an oblique plane given as:

T nn = T x x =

T xx + T yy

„ „

U ns =



+

2

T xx  T yy 2

T xx  T yy 2

cos 2R + U xy sin 2R

sin 2R + U xy cos2R

Rearranging the stresses, we can write

T nn 

T xx + T yy 2

=

U ns =

T xx  T yy

cos2R + U xy sin 2R

2 È T xx É Ê

 T yy Ø Ù Ú

2

sin 2R + U xy cos 2R

(1.19)

(1.20)

Squaring and adding Eqs. (1.19) and (1.20) and after simplification, we can write È É T nn  Ê

T xx + T yy Ø 2 2

Ù Ú

2 + U ns =

È T xx É Ê

 T yy Ø 2

Ù Ú

2

+ U 2xy

(1.21)

Now denoting R=

Tm =

È T xx É Ê

 T yy Ø 2

Ù Ú

2 2 + U xy

T xx + T yy 2

(1.22a)

(1.22b)

Equation (1.21) can be written as 2 (T nn  T m )2 + U ns = R2

(1.23)

Equation (1.23) is the equation of a circle having radius R and centre on s-axis at a point with coordinates (sm, 0). This circle is known as Mohr’s circle as it was first studied by German Engineer Otto Mohr. The Mohr’s circle can be drawn on the basis of following sign convention. The normal stress component s and shear stress t are positive when they act on a positive face and in the positive direction. For example s is positive when it acts in the x-direction on the x face and t is positive when it is in up direction or acting in anticlockwise direction measured with reference to the positive x face. This is illustrated in Fig. 1.16. In Fig. 1.16(a) shear stress txy is positive for positive x face and in Fig. 1.16(b) txy is negative.

Chapter 1: Simple and Compound Stress

17

Fig. 1.16 Sign conventions of shear stress.

The steps involved in drawing Mohr’s circle are explained below: Step 1: Plot sxx and syy in the s-axis with a suitable scale according to the magnitude and nature (tensile or compression). Step 2: Since the stress components such as sxx, syy and txy are known, determine the centre of the circle C(sm, 0) using

Tm =

T xx + T yy 2

Step 3: Determine the radius of the circle. To obtain the radius, determine at least one point on the circle. The procedure is discussed below: Consider the case when the normal on the oblique plane coincides with the x-axis [Fig. 1.17(a)]. In this case the inclination angle of the plane with x-axis is zero. Hence, q = 0.

Fig. 1.17

State of stress for q = 0° and q = 90°.

18

Strength of Materials

In this case,

T nn = T x x = T xx „

„

and U ns = T x y = U xy „ „

Now taking the coordinates as A(sxx, txy), locate the point A. This can be considered as a reference point. Now join CA. CA is the radius of the circle. Now rotate the nn axis (presently the nn axis is coincident with the x axis) through 90° anticlockwise as shown in Fig. 1.17(b). In this case q = 90°. From Eqs. (1.11) and (1.13), we get snn = syy = sx¢x¢

and

tns = – txy

These values correspond to the coordinate of point B which is opposite to point A as shown in Mohr’s circle (Fig. 1.18). The radial line CB is 180° anticlockwise from line CA. Hence, 90° anticlockwise rotation in the element is 180° anticlockwise rotation (same direction) in Mohr’s circle.

Fig. 1.18

Mohr’s circle.

Step 4: Two points where this circle cuts the s axis, gives the magnitude and nature of the principal stresses. Step 5: The principal stresses act on planes defined by angles qp1 and qp2 as shown in Fig. 1.18. They are represented by two times of these values in the Mohr’s circle and are measured from the radial reference line CA to line CB and CD, respectively. Step 6: The normal and shear stress components on any arbitrary plane defined by angle q can be obtained from Mohr’s circle by locating the point and its coordinates. To locate the point two times of the angle is measured counterclockwise if q positive or clockwise if negative on the circle from the radial reference line CA. The point P on the circle defines the normal and shear stress on the given plane defined by angle.

Chapter 1: Simple and Compound Stress

19

EXAMPLE 1.3 State of stress is given by (Fig. 1.19)

Fig. 1.19 State of stress of Example 1.3.

(a) Draw Mohr’s circle and determine principal stresses. (b) Determine stresses at a plane having its normal 45° anticlockwise to the direction of y-axis (c) Determine the maximum shear stress. Solution:

Fig. 1.20 Mohr’s circle for Example 1.3.

20

Strength of Materials

(a) The principal stresses are 105 MPa and – 65 MPa (b) Point A is the state of stress which is obtained by rotating at the angle of 2q = 2 ´ 45 = 90° anticlockwise of diameter CD. The normal shear stresses on this plane are 80 MPa and 60 MPa respectively. (c) Maximum shear stress is 85 MPa.

EXAMPLE 1.4 Draw Mohr’s circle for uniaxial tensile load, uniaxial compressive load and pure shear loading corresponding to yielding of the material. The yield strength in tension and compression are different and yield strength in compression is more than yield strength in tension. Solution: Let the yield strength under uniaxial tension be given that

syt and under compression is syc. It is

syc > syt Under the uniaxial tensile loading the principal stresses corresponding to yielding are:

s1 = syt, s2 = s3 = 0 Under the uniaxial compressive loading the principal stresses corresponding to yielding are

s1 = – syc, s2 = s3 = 0 Under the pure shear loading the state of stress is

sxx = syy = szz = 0 txy = t (let)

The principal stresses are

T1 , T 2 =

T xx + T yy

“

2

È T xx É Ê

Substituting, we get s1, s2 = ± t ÿ ÿ s1 = t ÿÿÿÿ

Mohr’s circles are shown in Fig. 1.21.

ÿ

s2 = – t

 T yy Ø 2

Ù Ú

2

+ U 2xy

Chapter 1: Simple and Compound Stress

Fig. 1.21

21

Mohr’s circle under axial tension, compression and pure shear loading.

EXAMPLE 1.5 A force of 1 kN is applied at the end of the beam. Determine and compare the magnitude of the maximum stresses for each cross section. Compare the moment carrying capacity of each section.

1500 mm

1.0 kN

22

Strength of Materials

The cross-section of the beam may be 50

20

100

60 10 0

100

100

100

100

(All dimensions are in mm)

Fig. 1.22

Solution:

The bending stress is given by M

T=

M

y=

I

=

I/ y

M

I

where Z =

Z

y

(a) Circular section

Q

I=

d

y= “

=

y

T=

=

2

I

Z=

Q

d4 =

64

(0.1)4 = 4.909 – 10 6 m 4

64

100

= 50 mm = 0.05 m

2

4.909 – 10 6 0.05

1000 – 1.5 9.818 – 10

5

= 9.818 – 10 5 m3

= 15.278 – 106

N m2

= 15.278 MPa

(b) Square section I=

a4 12

y= “

Z=

I y

T=

=

=

a 2

=

(0.1)4 12 100 2

= 8.333 – 10 6 m 4

= 50 mm = 0.05 m

8.333 – 10 6 0.05

1000 – 1.5 1.67 – 10 4

= 1.67 – 10 4 m3

= 9.0 MPa

20

Chapter 1: Simple and Compound Stress

(c) Rhombus section a4

I=

12

T=

=

2

I y

= 8.333 ´ 10–6 m4

12

a

y= “

Z=

(0.1)4

=

100 2

= 70.711 mm = 0.0707 m

8.333 – 10 6

=

0.0707

1000 – 1.5 1.178 – 10 4

= 1.178 ´ 10–4 m3

= 12.728 MPa

Fig. 1.23(a)

(d) Hollow section I=

Q 64

y= “

Z=

T=

I y

(D 4  d 4 ) = D 2

=

=

100

Q 64

[(0.1)4  (0.05)4 ] = 4.602 ´ 10–6 m4

= 50 mm = 0.05 m

2

4.602 – 10 6 0.05

1000 – 1.5 9.204 – 10 5

= 9.204 ´ 10–5 m3

= 16.298 ´ 106 N/m2

= 16.298 MPa (e) I-section I=

BD3 12



bd 3 12

=

0.1 – 0.13 12



0.08 – 0.063 12

= 6.89 – 10 6 m 4

23

24

Strength of Materials

B

D

d h

b=B–h Fig. 1.23(b)

y= “

Z=

T=

I y

D 2

=

=

100 2

= 50 mm = 0.05 m

6.89 – 10 6 0.05

1000 – 1.5 1.379 – 10  4

= 1.379 ´ 10–4 m3

= 10.88 MPa

If the shaft is made of same material, allowable stress will be same for each section. Hence from bending equation, M = s ´ Z = constant ´ Z. M ‘µ Z The section modulus for hollow section is more as compared to other sections considered in this problem, the moment carrying capacity of hollow section is more as compared to other sections considered in this problem.

Combined stress due to bending moment and torque In practice, bending moment, torque and transverse shear forces are subjected to a machine component. Under the action of moment M, torque T and shear force V, the different elements on a section of the component are subjected to different kinds of stress systems. To explain this aspect, let us consider a part of a solid shaft of diameter d under the action of M, T and V as shown in Fig. 1.24. Consider four points A and C on the extremities along the y-axis and B and D on the extremities along the z-axis. The stresses due to M, T and V acting on A, B, C and D are shown in Fig. 1.25. 1. At the extremities along the z-axis (BD points), bending stresses are zero. 2. At the extremities of the vertical diameter (y-axis), i.e., points A and C are subjected to maximum tensile and compressive stresses, respectively due to bending moment.

Chapter 1: Simple and Compound Stress

25

Fig. 1.24 Shaft under combined loading due to moment, torque and shear force.

Fig. 1.25

Stress distribution due to M, T and V.

3. The shear stress due to shear force V is maximum on the extremities along the z-axis, i.e., at points B and D. 4. Shear stress due to torque is same at all points, i.e., A, B, C and D points are subjected to equal shear stress due to T.

26

Strength of Materials

The value of stresses are calculated as follows:

Tb =

Due to bending moment M

U max =

Shear stress due to torque T

Uf =

Flexural shear stress due to V

32 M

Q d3 16 T Qd

3

3V

=

4A

16V 3Q d 2

EXAMPLE 1.6 A compound shaft made of steel and aluminium transmits 75 kW at 900 rpm. Determine the maximum shear stress and relative angle of twist between surfaces at two extreme ends located at A and C. Neglect any strain concentration effect. Esteel = 207 GPa, EAl = 73 GPa, nAl = 0.334, and nsteel = 0.33. Aluminium

Steel B

C

A 60

75

800

1000 All dimensions are in mm

Fig. 1.26 Example 1.6.

Solution:

For solid circular section, the maximum shear stress is given by U max =

16T Qd

3

The maximum shear stress depends on torque transmitted and minimum diameter of the shaft. The torque transmitted can be obtained from T =

kW – 1000 – 60

=

2Q – rpm

75 – 1000 – 60 2Q – 900

= 795.775 N.m

Shears tress between section A-B will be maximum as the diameter is minimum. U max =

16T Qd

3

=

16 – 795.775 Q

– (60/1000)3

= 18.763 MPa

Chapter 1: Simple and Compound Stress

27

The angle of twist of B relative to A

R B/A =

J=

R B/A =

2(1 + O )

È Tl Ø É Ù Ê J Ú

E

Q

Q

4

32

d =

32

2(1 + 0.334) 73

–

10

9

È 60 Ø É 1000 Ù Ê Ú

4

= 1.27  10–6 m4

È 795.775 É Ê 1.27 –

– 0.8 Ø 10 6

Ù Ú

= 0.0183 rad

The angle of twist of C relative to B

RC/B =

where J =

Q 32

4

d =

Q 32

2(1 + 0.33) 207

È 75 Ø É 1000 Ù Ê Ú

–

10

9

È 795.775 – 1 Ø É Ù Ê 3.106 – 10 6 Ú

= 3.292 ´ 10–3 rad

4

= 3.106 ´ 10–6 m4

The total angle of twist of the shaft is

qC/A = qC/B + qB/A = (3.292 ´ 10–3 + 0.0183) rad = 0.0216 rad

EXAMPLE 1.7 A stepped shaft made of steel is fixed at one end and 1.5 kN.m torque is applied at other end. What additional torque that can be applied at point B if the maximum shear stress is not to be exceeded 105 MPa and the total angle of twist should not exceeds 3°. The modulus of elasticity of steel may be taken as 200 GPa and Poisson’s ratio as 0.3.

Fig. 1.27 Example 1.7.

Solution: JBC = Polar moment of inertia for shaft of BC part =

Q 32

– (0.074  0.0454 ) = 1.955 – 10 6 m 4

T = 1.5 kN.m

28

Strength of Materials

The maximum shear stress will occur at the minimum cross section U max =

Tdo 2J

1.5 – 1000 – 0.07

=

2 – 1.955 – 10 6

= 26.86 MPa

The angle of twist of C relative to B

RC/B =

2(1 + 0.3) 200

–

10

9

È 1500 – 0.6 Ø É Ù Ê 1.955 – 10 6 Ú

= 5.985 ´ 10–3 rad

The two conditions given are maximum shear stress and angle of twist, hence maximum torque that can be taken by AB portion of the shaft is obtained from the following relation.

T=

U max

– J AB – 2 Do

105 – 106 – =

Q 32

(0.094  0.074 ) – 2 0.09

= 9529.50 N.m = 9.53 kN.m

Let qB/A be the rotation of B with respect to A.

qC/A = qB/A + qC/B = rotation of C with respect to A. It is given that the maximum permissible angle of twist is 3° =

3 – Q 180

= 0.0524 rad

qC/A = qB/A + qC/B 0.0524 = qB/A + 5.985 ´ 10–3 Þ qB/A = 0.0464 rad Hence, a torque in the same direction can be applied at B. The magnitude of the torque is obtained as

R B/A =

2(1 + O ) E

T = 0.0464 –

È Tl Ø É JÙ Ê Ú

= 0.0464 =

200 – 10 9 2(1 + 0.3)

–

2(1 + 0.3) 200

–

10

9

È Ø T – 0.75 É 4 4 Ù Ê Q /32{0.09  0.07 } Ú

Q /32 {0.094  0.074} 0.75

= 197973.33 N.m = 197.97 kN.m

Hence, at B 9.53 kN.m torque can be applied to satisfy both the conditions.

EXAMPLE 1.8 A solid circular steel shaft of 25 mm diameter is used for power transmission through a pulley and gear system as shown in Fig. 1.28. The tension in the pulley belts are 1.5 kN and 400 N and acting in the vertical direction. The pitch circle diameter of the pinion and gear is 200 and 400 mm respectively. The pinion is driven by 7 kW motor operating at 550 rpm. Determine the state of stress of the element undergoing greatest stress condition due to bending and torsion, deflection and angle of twist. Take E = 200 GPa, pressure angle f = 20°.

Chapter 1: Simple and Compound Stress

29

Fig. 1.28 Example 1.8.

Solution:

The torque produced at C due to gear is TC =

kW – 60 – 1000 2Q – rpm

=

7 – 1000 – 60 2Q – 550

= 121.54 N.m

Assuming that torque at B is same as torque at C, the torque produced at B due to pulley is TB = TC = (T1 – T2)Rp = (1500 – 400) ´ Rp = 121.54 N.m Þ

Rp = 110.5 mm Dp = 221.0 mm

The tangential component of gear force is obtained from Ft ´ Rg = TC where Rg is the radius of the gear. 221.185 N

607.7 N x y

z

Fig. 1.29(a)

Gear forces.

30

Strength of Materials

Ft =

121.54 0.2

= 607.7 N

The radial component of gear force is Fr = Ft tan f = 607.7 ´ tan 20° = 221.185 N The forces and reactions acting on the shaft are shown in Fig. 1.29(b).

Fig. 1.29(b)

Net effect of the loading on the shaft.

Considering the equilibrium conditions in horizontal and vertical plane, reactions are determined as follows. – RAy – RDy + 1900 – 221.185 = 0

(i)

RAz + RDz – 607.7 = 0

(ii)

and Taking moment of forces in the y-direction about D RAy ´ 500 – 1900 ´ 300 + 221.185 ´ 100 = 0 RAy = 1095.763 N From Eq. (i), RAy + RDy = 1678.815 RDy = 583.052 N Taking the moment of forces in the z-direction about D RAz ´ 500 – 607.7 ´ 100 = 0 RAz = 121.54 N From Eq. (ii), RAz + RDz = 607.7 RDz = 486.16 N

Chapter 1: Simple and Compound Stress

31

To calculate the bending moment in two different planes, i.e., xy plane and xz plane, forces are shown in Fig. 1.29(c), separately for two planes.

Fig. 1.29(c)

Forces and bending moment in the xz plane are shown in Fig. 1.29(d).

Fig. 1.29(d)

Bending moment diagram.

The maximum moment at point B and C is given by MB =

M y2 + M z2

32

Strength of Materials

M B = (  219.15)2 + (  24.31)2 = 220.5 N·m MC = (  58.31)2 + (  48.62)2 = 75.92 N·m The resultant maximum bending moment occurs at point B. In case of circular cross section the resultant bending moment obtained by vector addition of moments My and Mz is located at the angle of b from z-axis. The angle is obtained from

ÑÎ M y ÑÞ ß ÑÐ Mz Ñà

C = tan 1 Ï

(1.24)

It is shown in Fig. 1.30.

Fig. 1.30

Location of resultant bending moment for bending of symmetric beam in two planes.

Chapter 1: Simple and Compound Stress

33

The sign convention used is according to the convention used in stress analysis. If the outward normal is in the positive coordinate direction, all forces and moments are taken as positive. If outward normal is in negative direction, all forces and moments are taken as negative. Moment vectors are shown as positive (according to right hand rule) with a normal in the positive direction and negative with normal in the negative direction.

Î My Ñ Þ Ñ 1 Î 219.15 Þ ß = tan Ï ß = 84.45° Ñ Mz à Ñ Ð  24.31 à Ð

C = tan 1 Ï

Fig. 1.30(a)

The maximum tensile stress occurs at point S1 and maximum compressive stress occurs at S2 point on the surface of the shaft. The stresses are given by At S1

T xx =

M I

ymax =

220.5 – (0.025/2) (Q /64) (0.025)4

= 143.74 MPa

At S2, the maximum compressive stress of magnitude 143.74 MPa occurs. The shear stress due to torque is given as U =

Td 2J

=

121.54 – 0.025 2 – (Q /32) (0.025) 4

= 39.62 MPa

The state of stress in MPa is

T=

Ë14.74 Ì ÍÌ 39.62

39.62 Û 0

Ü ÝÜ

34

Strength of Materials

EXAMPLE 1.9 A shaft is subjected to a force as shown in Fig. 1.31. Determine the state of stress at A, B, C and D.

Fig. 1.31 Figure for Example 1.9.

Solution: The shaft is subjected to axial bending and torsional load. Each load can be analysed separately and added algebraically for the same nature of stresses and by vector method for different nature of stresses. Axial load F causes normal stresses in all four points A, B, C and D as shown in Fig. 1.32. The magnitude of normal stress due to F is obtained as

Ta =

4F

Qd

2

=

4 – 80000

Q – (0.1)2

= 10.19 – 10 6 N/m 2 = 10.19 MPa

Fig. 1.32 Figure for Example 1.9.

Chapter 1: Simple and Compound Stress

35

Stresses due to Bending moment M Force P produces the bending stress as well as transverse shear stress. The bending moment is given as Mz = – PL = –25 ´ 1000 = 25,000 kN.mm = 25 kN·m The bending stress is

Tb =  Iz =

Mz Iz

y

Qd4 64

The bending stresses at different points A, B, C and D are b (T xx )A =

(T bxx ) B =

PL (Q /64) d 4 PL (Q /64) d

4

(0) = 0

( d/2) =

32PL

=

(Q d ) 3

32 – 25000 – 1

Q – (0.1)

3

= 254.65 – 10 6 N/m 2

= 254.65 MPa b (T xx )C =

(T bxx ) D =

PL (Q /64) d 4 PL (Q /64) d

4

(0) = 0 ( d/2) = 

32PL (Q d ) 3

= 

32 – 25000 – 1

Q – (0.1)3

=  254.65 – 10 6 N/m 2

= – 254.65 MPa The shear stress due to transverse load P is given by U =

V Ib

Ay

Hence, (U xy )B = (U xy ) D =

(U xy ) A =

4Vy 3A

= 

=

4Vy 3(Q /4) d

16 – 25000 3Q (0.1) 2

2

=

V Ib V Ib

Ay = 0

Ay = 0

16(  P ) 3Q d

2

= 

16 P 3Q d 2

= – 4.244 ´ 106 N/m2 = – 4.244 MPa

36

Strength of Materials

(txy)C = – 4.244 MPa Shear stress due to torque T The maximum shear stress is given by U xy =

16 T Qd

3

The torsional shear stresses at A, B, C and D are U xy =

16 – 10000 Q (0.1)

3

= 50.93 ´ 106 N/m2 = 50.93 MPa

State of stress at A At point A the shear stress due to torque and transverse load exists. The normal stress due to axial load also exists. The bending stress is zero. By the principle of superposition, we can add equal nature of stresses algebraically. Adding shear stresses algebraically, the shear stress component of the stress at point A is

(U xy ) A =

16 T Qd

3

+

È É Ê

16 P 3Q d

Ø

2Ù Ú

= 50.93 + (– 4.244) = 46.686 MPa

= 46.686 MPa The normal stress at A is

T xx =

4F

Qd2

= 10.19 MPa

Fig. 1.33(a)

State of stress at point B At point B the normal stress due to the bending moment and axial force exists which can be added algebraically to get the net stress. The shear stress that exists at point B is due to torque only. No transverse shear stress develops at point B which is located at the transverse load line.

Chapter 1: Simple and Compound Stress

T xx =

4F

Qd

32 PL

+

2

Qd3

= 10.19 + 254.65 = 264.84 MPa

(U xy ) B =

16 T Qd

3

= 50.93 MPa

Fig. 1.33(b)

State of stress at point C (U xy )C =

16 T Qd

+

3

È 16 P Ø É Ù Ê 3Q d 2 Ú

T xx =

4F

Qd2

= 50.93 + 4.244 = 55.174 MPa

= 10.19 MPa

55.147

10.19

C

Fig. 1.33(c)

State of stress at point D (U xy )D = (T xx ) D =

16 T Qd

3

4F

Qd

2

= 50.93 MPa



32 PL

Q d3

= 10.19 – 254.65 = – 244.46 MPa

37

38

Strength of Materials 50.93

244.46

D

Fig. 1.33(d)

EXAMPLE 1.10 A tube with external diameter 50 mm and 5 mm thickness is subjected to loads as shown in Fig. 1.34. Determine the sate of stress at point A (0, 25, 0) and B (0, –25, 0) and principal stresses. 0.5 kN

y 500 m m

0 20

mm

1.5 kN

z x

Fig. 1.34

Solution: State of stress at A(0, ± 25, 0 mm) Point A will be subjected to both bending moment and twisting moment. The force 0.5 kN will produce bending moment and twisting moment at point A of the following magnitudes. The twisting moment is given by T = 500 –

200 1000

= 100 N·m

The bending moment is M = 500 ´ 0.5 = 250 N·m The stresses are: U xz =

T J

r

Given do = 50 mm = 0.05 m di = do – 2t = 50 – 2 ´ 5 = 40 mm = 0.04 m U xz =

16 Tdo 4 Q (do



di4 )

=

16 – 100 – 0.05 4

Q (0.05

 0.044 )

= 6.9 ´ 106 N/m2 = 6.9 MPa

Chapter 1: Simple and Compound Stress

T xx =

32 Mdo

Q

(do4



di4 )

=

32 – 250 – 0.05

Q (0.054  0.04 4 )

39

= 34.50 ´ 106 N/m2 = 34.50 MPa

Similarly, force 1.5 kN will produce bending moment and axial stress at point A of following magnitude. The bending moment is M = 1500 ´ 0.2 = 300 N·m

T xx =

32 Mdo

Q



(do4

di4 )

=

32 – 300 – 0.05

Q (0.054  0.04 4 )

= 41.41 ´ 106 N/m2 = 41.41 MPa

The direct stress is

T xx =

4F

Q

(do2



di2 )

=

4 – 1500

Q (0.052  0.042 )

= 2.122 ´ 106 N/m2 = 2.122 MPa

The like stresses can be added algebraically, by the principle of superposition method. The total normal stress at point A (0, 25, 0) is

s = 34.50 + 41.41 + 2.122 = 78.032 MPa The shear stress at point A is t = 6.9 MPa

At point B(0, –25, 0) s = –34.50 – 41.41 + 2.122 = –73.788 MPa t = 6.9 MPa

EXAMPLE 1.11 A machine member is subjected to forces as shown in Fig. 1.35. Determine the stresses. 3 kN·m 1.5 kN·m

A

75

20 kN 50

B 300

Fig. 1.35 Example 1.11.

40

Strength of Materials

Solution:

Q

A=

I=

(D 2  d 2 ) =

4

Q

(D4  d 4 ) =

64

Q 4

(0.0752  0.052 ) = 2.454 – 10 3 m 2

Q 64

(0.0754  0.054 ) = 1.246 – 10 6 m 4

J = 2.493 ´ 10–6 m4

Stresses at point A Due to force F,

Ta =

F A

20 – 10 3

=

2.454 – 10 3

= 8.149 ´ 106 N/m2 = 8.149 MPa

Due to moment M = 3000 ´ 0.3 = 900 N·m

Tb =

M I

.y=

900 – 0.0375 1.246 – 10

6

= 27.087 ´ 106 N/m2 = 27.087 MPa

Due to torque T U xy =

Hence,

TD 2J

=

1500 – 0.0375 2.493 – 10 6

= 22.563 ´ 106 N/m2 = 22.563 MPa

sxx = sa + sb = 8.149 + 27.087 = 35.236 MPa (tensile) txy = 22.563 MPa

Stresses at point B sxx = sa – sb = 8.149 – 27.087 = –18.938 MPa (Compression) ÿ ÿ ÿ ÿ ÿ ÿ ÿ ÿ ÿ ÿ ÿ ÿ ÿ txy = 22.563 MPa ÿ

EXAMPLE 1.12 While testing a metallic rod, it is observed that the diameter of the rod is reduced by 0.0025 mm at an axial pull of 20 kN. The original diameter of the rod is 15 mm. Find the Young’s modulus. The rigidity modulus is 50 kN/mm2. Solution:

Given Dd = 0.0025 mm, d = 15 mm, F = 20 kN Stress =

T=

F A

=

4F

Qd2

4 – 20 – 103

Q (15)2

= 113.177 N/mm2

Chapter 1: Simple and Compound Stress

Lateral strain =

'd

0.0025

=

d

41

15

elat = 1.67 ´ 10– 4

\

E=

Longitudinal strain = elong =

stress longitudinal strain stress

113.177

E

E

\

Poisson’s ratio =

O=

F lat F long

1.67 – 10 4 – E 113.177

= 1.476 ´ E ´ 10–6 We have from the relation of E and G as E = 2G(1 + n) = 2 ´ 50 ´ 103 (1 + 1.476 ´ 10–6 E) \

E=

2 – 50 – 103 0.8524

= 117.32 – 103 N/mm 2

EXAMPLE 1.13 The two-dimensional state of stress is given as sxx = 10 MPa, syy = 5 txy MPa and txy = 2.5 MPa. Determine the following on a plane inclined at an angle of 30° from the x-plane in anticlockwise direction. (i) Normal stress, (ii) shear stress, (iii) resultant stress and (iv) principal stresses. Solution:

Given sxx = 10 MPa, txy = 2.5 MPa syy = 5txy = 5 ´ 2.5 = 12.5 MPa

The stresses are shown in Fig. 1.36. AB is the inclined plane at an angle 30° from the x-axis. From the equation the normal stress on the inclined plane is given as (i)

T nn = =

T xx + T yy

+

2 10 + 12.5 2

+

= 12.79 MPa

T xx  T yy 2

10  12.5 2

cos 2R + U xy sin 2R

cos(60°) + 2.5 sin 60°

42

Strength of Materials

Fig. 1.36

(ii) Shear stress U ns =  = 

T xx  T yy

sin 2R + U xy cos 2R

2 10  12.5 2

sin 60° + 2.5 cos 60°

tns = 2.333 MPa 2 2 (iii) Resultant stress U R = T nn + U ns

= (12.79)2 + (2.333)2 = 13.00 MPa (iv) Principal stresses

T1,3 =

=

T xx + T yy 2

10 + 12.5 2

“

“

È T xx É Ê

È 10 É Ê

 T yy Ø Ù Ú

2

 12.5 Ø 2

Ù Ú

2

+ U 2xy

2

+ (2.5)2

= 11.25 ± 2.795 \

s1 = 14.045 MPa, s3 = 8.455 MPa

EXAMPLE 1.14 The state of stress at a critical section is shown in Fig. 1.37. Determine the state of stress on the element inclined at an angle of 30° clockwise from the x-axis. Represent these stresses on the element.

Chapter 1: Simple and Compound Stress

43

50 MPa

100 MPa 25 MPa

Fig. 1.37 State of stress of Example 1.14.

Solution: Given that sxx = 100 MPa, syy = – 50 MPa, txy = –25 MPa, q = –30° or q = 90° – 30° = 60°. The inclined plane is shown in Fig. 1.38.

Fig. 1.38 State of stress and inclined plane.

Consider plane CB. The stress components on the plane CB can be obtained from the relation

Tx x = „

T xx + T yy

„

2

+

È T xx É Ê

 T yy Ø 2

Ù Ú

cos 2R + U xy sin 2R

Here q = 30° (–ve) as it is measured clockwise from reference axis x-axis. \

Tx x =

100  50

„ „

2

+

100 + 50 2

cos (  60°) + (  25) sin (  60°)

= 84.15 MPa Shear stress U x y =  „ „

= 

T xx  T yy 2

100 + 50 2

sin 2R + U xy cos 2R

sin (  60°) + (  25) cos (  60°)

= 52.45 MPa The positive sign indicates that tx¢y¢ acts in the positive direction.

44

Strength of Materials

The stresses on other perpendicular planes such as CD can be obtained by substituting q = 90° – 30° = 60°.

Ty y = „

T xx + T yy

„

=

2 100  50 2

+

+

T xx  T yy 2 100 + 50 2

cos 2R + U xy sin 2R

cos (120°) + (  25) sin (120°)

= – 34.151 MPa Ux y = „ „



100 + 50 2

sin (120°) + (  25) cos (120°)

= – 52.45 MPa The stress components are shown in Fig. 1.39.

Fig. 1.39 Stresses on plane inclined at 30° to the x-axis clockwise.

EXAMPLE 1.15

sxx = –20 MPa, syy = 100 MPa,

The stress components for a plane stress situation are given as

txy = 60 MPa. Determine the principal stresses and orient them to an element.

Solution:

The principal stresses are obtained from the relation

T 1,2 =

=

T xx + T yy 2

20 + 100 2

“

È T xx É Ê

“

È 20 É Ê

s1,2 = 40 ± 84.85 ÿ ÿ s1 = 124.85 MPa ÿ ÿ s2 = – 44.85 MPa

 T yy Ø 2

2

 100 Ø 2

+ U 2xy

Ù Ú

Ù Ú

2

+ (60)2

Chapter 1: Simple and Compound Stress

45

The inclination of the principal stresses can be obtained from the relation

tan 2R p =

2 U xy



T xx

tan 2qp = –1

\ \

ÿ

or

=

T yy

2 – 60 (  20  100)

= 1

2qp = – 45°

qp = – 22.5°.

The difference between two angular positions of two principal stresses is 180°. Hence, 2qp1 = 180° + 2qp2 = 135

qp1 = 67.5°

Two angles are –22.5° and 67.5°. Now substituting any angle in the following equation we can determine the corresponding stress. The normal stress equation is

Tx x = „

„

=

T xx + T yy 2

 20 + 100 2

+

+

T xx  T yy 2

 20  100 2

cos 2R + U xy sin 2R

cos(  45°) + 60 sin (  45°)

= – 44.85 MPa Hence, s2 = – 44.85 MPa acts on the plane defined by qp2 = – 22.5° and s1 = 124.85 MPa acts on the plane defined by qp1 = 67.5°. The results are shown in Fig. 1.40, including the state of stress in terms of elemental stresses.

Fig. 1.40

Elemental and principal stresses on an element.

EXAMPLE 1.16 The state of stresses at a critical point is sxx = –15 MPa, the principal stresses using Mohr’s circle method. Solution:

The stresses are given as

syy = 20 MPa, txy = 10 MPa. Determine

sxx = –15 MPa, syy = 20 MPa, txy = 10 MPa.

46

Strength of Materials

Fig. 1.41 Mohr’s circle values shown as in MPa.

The principal stresses are indicated by points C and D. The point C corresponds to the value of

s1 = –20.156 + 2.5 = –17.656 MPa The point D corresponds to the second principal stress and the value is

s2 = 20.156 + 2.5 = 22.656 MPa

Drawing details

Take s as the x-axis and t as the y-axis. To convert the value of stress to small unit 1 mm = 0.5 MPa is considered. Taking O as reference point sxx = –15 MPa is taken as –ve x-axis and syy = 20 MPa is taken as +ve x-axis. Considering shear stress as +ve downward, 10 MPa is taken downward at A and upward at B. Points E and F thus obtained are joined which is the diameter of the circle. The points C and D thus show the two principal stresses. The orientation of the principal plane is obtained by angle 2qp. tan 2R p =

10 20  2.5

= 0.571

qp = 14.87°

EXERCISES 1.1 Define the stress at a point? 1.2 Draw Mohr’s circle for (a) pure shear, (b) pure biaxial tension, (c) pure uniaxial compression and (d) pure uniaxial tension.

Chapter 1: Simple and Compound Stress

47

1.3 From the given Mohr’s circles (Fig. 1.42) state the nature of the stresses on a differential element in the x and y planes. y-axis

y-axis

y-axis

x-axis x-axis (a)

x-axis

(b)

(c)

Fig. 1.42

1.4 Determine the normal and shear stress on the plane AA as shown in Fig. 1.43. The state of stress is shown below. 75 MPa A 80 MPa 150

150 MPa

80

30° 80 75

A

Fig. 1.43

1.5 A point in a strained material is subjected to tensile stress 65 MPa, compressive stress 45 MPa, acting on two mutually perpendicular planes and a shear stress of 10 MPa on those planes. Determine the normal stress, tangential stress and resultant stress on a plane inclined to 30° with the plane of the compressive stress. 1.6 At a point in a body the normal and shear stresses on two mutually perpendicular planes are given as sxx = –100 MPa, syy = 40 MPa and txy = 50 MPa. Using Mohr’s circle, determine the principal stresses and their planes. 1.7 At a point in a strained material, there are normal stresses of 30 MPa tension, and 20 MPa compression on two planes at right angles to each other, together with shear stress of 15 MPa on the same planes. If the loading on the material is increased so that the stresses reach values of k times those given, find the maximum value of k, if the maximum value of direct stress in the material is not to exceed 80 MPa, and maximum shear stress is not to exceed 50 MPa.

48

Strength of Materials

1.8 If two stress systems as shown in Fig. 1.44 are superposed, determine the principal stresses and the direction. y

y

100 MPa

x

30° x

+

75 MPa 80 MPa

Fig. 1.44

2 2.1

Analysis of Stress and Strain

INTRODUCTION

This chapter presents the three-dimensional theory of stress of a continuous medium. A continuous medium is a material in which each volume of a substance is sufficiently dense so that the concepts (e.g., mass density, stress, etc.) have meaning at every point in the region occupied by the material. The theory of stress depends upon Newton’s law of motion, which is independent of the nature of continuous material. Therefore, the relationships derived here are applicable to all continuous material, whether they behave elastically, plastically, visco-elastically or in any other manner.

2.2 FORCE DISTRIBUTION The forces acting on a body can be of following types: (a) Internal forces (b) External forces (a)

Internal forces: The internal forces are the reactive forces, which are set up due to external applied forces and within elastic limits, the internal forces are numerically equal to the external forces.

(b) External forces: The state of stress and strain in a body arises out of external influences. The external forces acting on a body can be divided into two types. (i)

Surface forces: The forces which are distributed over the surface or boundary of the body and acting from the surrounding medium like atmospheric pressure, hydraulic pressure or contact pressure exerted by one body on another. (ii) Body forces: The forces which act through the body and are not produced by physical contact with other bodies. Body force intensities are designated as forces per unit volume. Examples of this kind of force are the gravitational force, magnetic force and inertia force. 49

50

2.3

Strength of Materials

THE STATE OF STRESS AT A POINT

Since an infinite number of planes can be drawn through a point, we get an infinite number of stress vectors acting at a given point, each stress vector characterized by corresponding plane on which it is acting. The totality of all stress vectors acting on every possible plane passing through the point is defined to be the state of stress at the point. The knowledge of state of stress is important for designer to determine the critical stresses.

2.4

STRESS NOTATIONS

In order to understand clearly the state of stress at a point, double subscript system representing stress components is utilized. The first subscript denotes the direction of the outward drawn normal on the plane on which the stress acts, and the second subscript denotes the direction towards which the stress acts. Therefore, on the positive x face of an elementary parallelopiped, the stresses are txx, txy and txz in the cartesian coordinate system of axes. Out of which txx is the normal stress because the first subscript denotes the direction (i.e., x direction) of the outward drawn normal on the plane x and the second subscript x, denotes the direction (i.e., x direction) along which the stress acts. Whereas txy and txz are the shear stresses because the subscript x gives the direction of the outward drawn normal, i.e., the x direction and the subscripts y and z give the directions (i.e., y and z directions) along which the stresses are measured. The nature of a stress i.e., positive or negative is determined from the direction of the stress and plane on which stress is acting. The product of these two gives the nature of the stress. For example, let if a stress sxx is acting in the negative direction and the outward normal of the plane is also in the negative direction, then the stress sxx is positive. Similarly, if a stress is acting on a negative plane (i.e., outword normal is in the negative direction) and the direction of the stress is in positive direction, the stress is said to be negative.

2.5

STRESS TENSOR AT A POINT

The state of stress at a point in the cartesian co-ordinates can be represented by the following nine component of stress known as the stress tensor:

tij =

ËU xx Ì ÌU yx Ì U Ì Í zx

U xy

U xz Û

U yy

U yz Ü

U zy

U zz ÜÝ

Ü Ü

The first subscript in tij can be considered to identify a row of the array of terms whereas the second subscript can be considered to identify a column of the array. The principle or leading diagonal terms are all normal stress and all off-diagonal terms are shear stresses. The normal stresses can also be written as sxx, syy, szz or sx, sy, sz instead of txx, tyy, tzz or tx, ty, tz. Throughout this book we shall write the normal stresses as sx, sy and sz or sxx, syy and szz. Therefore, the stress tensor may be written as: ËT xx Ì Ì U yx Ì ÌÍ U zx

U xy U xz Û Ü T yy U yz Ü U zy T zz ÜÜÝ

Chapter 2: Analysis of Stress and Strain

51

All the stress components have been shown on a parallelopiped in Fig. 2.1. It can be noted that there are nine components of stress in total for a general state of stress.

Fig. 2.1

2.6

Components of stress tensor acting on the faces of a rectangular parallelopiped.

STRESS GRADIENT

The variation of stress with distance is called stress gradient. If the stress tensor at a point is known, then the stress components in the neighbourhood of the point can be known by expanding the stress components by Taylor series expansion. Thus, if sxx is the stress at a point, then the stress at a point at an infinitely small distance Dx will be given by

Txx 

˜ T 'x + Higher order terms ˜ x xx

Neglecting higher order terms, we get

˜T xx 'x ˜x

Txx  In general, it can be written as:

T ij  2.7

˜ T ij ˜x j

'x j

DIFFERENTIAL EQUATIONS OF EQUILIBRIUM

So far, attention has been focused on the state of stress at a point. One of the important sets of equations used in the analyses of such problems deals with the conditions to be satisfied by the stress components when they vary from point to point. These conditions will be established when the body is in equilibrium. We isolate a small element of the body and derive the equations of equilibrium from its free-body diagram (Fig. 2.2). The inertial or body forces, such as those caused by the weight or the magnetic effect, designated as Bx, By and Bz are associated with the unit volume of the material. Then

+ve

ÇF

x

0

52

Strength of Materials

Fig. 2.2

or

Variation of stresses.

˜U xy Ø È ˜U xz Ø ˜T xx Ø È È ÉÊ T xx  ˜x dx ÙÚ dydz  T xx dydz  É U xy  ˜y dy Ù dxdz  U xy dxdz  ÉÊ U xz  ˜z dz ÙÚ dxdy Ê Ú – txz dxdy + Bx dxdydz = rdxdydz a –x

where Bx = Component of the body force in the x direction, expressed per unit volume a x– = Acceleration in the x direction r = Density of the material of the parallelopipeds Simplifying, we get

˜T xx ˜U xy ˜U xz    Bx ˜x ˜y ˜z



SB x

Similarly, considering the equilibrium of forces in the y and z direction, we get, respectively

˜U yx ˜T yy ˜U yz    By ˜x ˜y ˜z

SB y

˜U zx ˜U zy ˜T zz    Bz ˜x ˜y ˜z

SB z





In the absence of inertia force and body forces, we get equilibrium equations as:

˜T xx ˜U xy ˜U xz   ˜x ˜y ˜z ˜U yx ˜T yy ˜U yz   ˜x ˜y ˜z ˜U zx ˜U zy ˜T zz   ˜x ˜y ˜z

Þ 0Ñ Ñ ÑÑ 0ß Ñ Ñ 0Ñ Ñà

(2.1)

Chapter 2: Analysis of Stress and Strain

53

In tensor notation, Eq. (2.1) can be written as sij, j = 0, where i = x, y and z, j = x, y and z. Comma (,) denotes partial derivative. Hence comma j (i, j) denotes partial derivative with respect to j.

2.8

EQUILIBRIUM EQUATIONS FOR PLANE STRESS STATE

A state of stress is said to be plane state of stress or plane stress state when stresses on one direction vanish. If in a given state of stress, there exists a plane stress in the xy plane, then stresses in the z-direction vanish and

szz = 0,

txz = 0

and

tyz = 0

(2.2)

The state is said to have a plane state of stress parallel to xy plane. This state is also generally known as a two-dimensional state of stress. Thus, differential equations of equilibrium become

˜T xx ˜U xy  ˜x ˜y ˜T yy ˜U yx  ˜y ˜x

2.9

Þ 0Ñ Ñ ß Ñ 0Ñ à

(2.3)

GENERALIZED HOOKE’S LAW

The functional relation of stress and strain, rather load versus deformation was first given by Robert Hooke in 1676. Hooke’s law states that Within elastic limit, stress is proportional to strain. The following are the generalized Hooke's law equations. It can be given a precise expression in terms of stress and strain by stating, in most general form that, if tij ¹ tji, 144444444244444443

sxx = f1(exx, eyy, ezz, gxy, gyz, gzx, gyx, gzy, gxz) syy = f2(exx, eyy, ezz, gxy, gyz, gzx, gyx, gzy, gxz) szz = f3(exx, eyy, ezz, gxy, gyz, gzx, gyx, gzy, gxz) txy = f4(exx, eyy, ezz, gxy, gyz, gzx, gyx, gzy, gxz) tyz = f5(exx, eyy, ezz, gxy, gyz, gzx, gyx, gzy, gxz)

(2.4)

tzx = f6(exx, eyy, ezz, gxy, gyz, gzx, gyx, gzy, gxz) tyx = f7(exx, eyy, ezz, gxy, gyz, gzx, gyx, gzy, gxz) tzy = f8(exx, eyy, ezz, gxy, gyz, gzx, gyx, gzy, gxz) txz = f9(exx, eyy, ezz, gxy, gyz, gzx, gyx, gzy, gxz)

or any stress component sxx, syy, szz, etc. are a function of all the strain components, exx, eyy, etc. These functions f1 ® f9 could be linear or nonlinear. For a small deformation, an elastic material can be considered to be linearly elastic and in that case the functions f1 ® f9 become linear. Hence, we can write in tensor form: s ij = aijkl e kl (2.5)

54

Strength of Materials

This is the Cauchy’s formulation for generalized Hooke’s law. In the most general case, s ij and e kl will have nine components each and aijkl will have eighty-one components, as given herein after: 1444444424444443

sxx = a1,1 exx + a1,2eyy + a1,3ezz + a1,4gxy + a1,5gyz + a1,6gzx + a1,7gyx + a1,8gzy + a1,9gxz syy = a2,1exx + a2,2eyy + a2,3ezz + a2,4gxy + a2,5gyz + a2,6gzx + a2,7gyx + a2,8gzy + a2,9gxz szz = a3,1exx + a3,2eyy + a3,3ezz + a3,4gxy + a3,5gyz + a3,6gzx + a3,7gyx + a3,8gzy + a3,9gxz

– –

– –

– –

– –

– –

– –

– –

– –

– –

– –

(2.6)

txz = a9,1exx + a9,2eyy + a9,3ezz + a9,4gxy + a9,5gyz + a9,6gzx + a9,7gyx + a9,8gzy + a9,9gxz

Special case If the stress and strain tensor are symmetrical, t ij = t ji the above set will consist of only six equations, with thirty-six constants, i.e., from a11 to a66. The stress and strain tensors will have only six components, as: 1444444424444443

(a)

sxx = a1,1exx + a1,2eyy + a1,3ezz + a1,4gxy + a1,5gyz + a1,6gzx syy = a2,1exx + a2,2eyy + a2,3ezz + a2,4gxy + a2,5gyz + a2,6gzx szz = a3,1exx + a3,2eyy + a3,3ezz + a3,4gxy + a3,5gyz + a3,6gzx

– – tzx

– –

(2.7)

– – – – – – – – – – = a6,1exx, + a6,2eyy + a6,3ezz + a6,4gxy + a6,5gyz + a6,6gzx

(b) These equations contain thirty-six elastic constants. These elastic constants are independent of the stress components, at a point for an elastic body to be homogeneous. It should be same at all points within a region. Material is known as isotropic if its elastic constants are the same in all directions at the point. The numbers of independent elastic constant are only two for a perfectly isotropic material. The considerations of homogeneity restrict the total elastic constant within a region to finite value of thirty-six, further the consideration of isotropy reduces these constants to only two.

exx = a11sxx + a12(syy + szz) eyy = a11syy + a12(szz + sxx) ezz = a11szz + a12(sxx + syy) gxy = 2(a11 – a12)txy gyz = 2(a11 – a12)tyz gzx = 2(a11 – a12)tzx

14444442444443

The generalized Hooke’s law equations then reduce to

To calculate the values of a11 and a12, let us take uniaxial stress in the x-direction: syy = szz = 0

This gives a11 =

F xx T xx

1 E

(2.8)

Chapter 2: Analysis of Stress and Strain

55

where E is Young’s modulus of elasticity. Also, eyy = ezz = a12sxx = a12Eexx

a12 =

1 F yy E F xx



O

E

where n = 1/m = Poisson’s ratio Substituting these in Eq. (2.8), we get the final equations as follows:

2.10

exx =

1 [T  O (T yy  T zz )] E xx

(2.9)

eyy =

1 [T  O (T xx  T zz )] E yy

(2.10)

ezz =

1 [T  O (T xx  T yy )] E zz

(2.11)

gxy =

2(1  O ) U xy E

(2.12)

gyz =

2(1  O ) U yz E

(2.13)

gzx =

2(1  O ) U zx E

(2.14)

DIRECTION COSINES

Suppose a plane ABC has some inclination to each of x, y and z axes as shown in Fig. 2.3. The normal to the plane ABC is ON as shown makes angles a, b, g with the x, y and z axes. The cosines of those angles are known as direction cosines and the three direction cosines, cosa = l, cos b = m and cosg = n effectively determine the inclination of the plane ABC, in space. These symbols l, m, n are often used in older books. However, a more comprehensive notation for direction cosines that will be adopted in this book is based on tensor type notation with double subscripts. Thus, any direction cosine is termed aij, which is equal to the cosine of the angle between any two lines i and j. According to this system, l, m, n are anx, any and anz, respectively. It is also convenient to use the direction cosines as nx, ny and nz instead of l, m and n. z g

N

A a b O y

C

B

Fig. 2.3 Direction cosines.

x

56

Strength of Materials

2.11

NORMAL AND SHEAR STRESSES

Consider a rectangular parallelopiped as shown in Fig. 2.4 subjected to the three-dimensional stress system. Let sr be the resultant stress on a plane passing through the point O, and srx, sry and srz its components along three axes of reference. y n

sry sr O tns

srx sn

x

srz z Fig. 2.4

Normal, shear and resultant stress components.

The direction cosines of the resultant stress are: arx =

T ry T T rx , ary = , arz = rz Tr Tr Tr

Let sn and tns be the normal and shear stresses on the plane whose normal n has direction cosines anx, any and anz. Now, sn = sr anr where

anr = anx arx + any ary + anz arz

\

sn = sr (anx arx + any ary + anz arz) = srx anx + sry any + srz anz

Resolving the forces acting on the parallelopiped along the coordinates axes, we can show that

srx = sxxanx + txy any +

txz anz

sry = txy anx + syy any + tyz anz srz = txz anx + tyz any + szz anz

Hence sn = sxx a2nx + syy a2ny + szz a2nz + 2(txy anx any + tyz any anz + txz anx anz)

Now,

sr2 = sn2 + tns2 tns =

where

T r2  T n2

sr2 = srx2 + sry2 + srz2

The direction cosines of the shear stress may be determined as follows: Let asx, asy, asz be the direction cosines of tns.

(2.15) (2.16)

Chapter 2: Analysis of Stress and Strain

srx = snanx +

Now,

asx =

1 U ns

57

tns asx

[T rx  T n anx ]

Similarly, asy = and

2.12

asz =

1 U ns

1 U ns

[T ry  T n any ] [T rz  T n anz ]

PRINCIPAL DIRECTIONS

For the three principal stresses s1, s2 and s3, the principal directions may be determined as follows. For s1 stress, let A1 =

T yy  T 1 U yz U yz T zz  T 1

B1 = 

C1 =

Then

anx1 = any1 = anz1 =

U xy U xz

U xy U xz

U yz T zz T zz

 T1

 T1

U yz

A1 A12

 B12  C12 B1

A12

 B12  C12 C1

A12  B12  C12

A plane which is equally inclined to the three coordinate axes is called the octahedral plane and the stresses acting on this plane are called octahedral stresses. This plane has the direction cosines each equal to “ 1/ 3.

2.13

STRESS COMPONENTS ON AN ARBITRARY PLANE

Now, we can start from small cuboids of material as shown in Fig. 2.5 and cut an elementary tetrahedron OABC, from one corner, as shown. Consider a small tetrahedron at O with three of its faces normal to the coordinate axes. Let the arbitrary plane be identified by its outward drawn normal n whose direction cosines are nx, ny and nz and h be the perpendicular distance from P

58

Strength of Materials y

B O

A

x

C z

Fig. 2.5

Elementary cut from corner of a cuboid.

(point where state of stress is required) to inclined face. If the tetrahedron is isolated from the body and a free-body diagram is drawn, then it will be in equilibrium under the action of the surface force and the body forces. The free-body diagram is shown in Fig. 2.6.

Fig. 2.6 Tetrahedron at point O.

Tzn

Let T n be the resultant stress vector on face ABC. This can be resolved into components Txn, Tyn, parallel to the three axes x, y and z. Area of BOC = Projection of area ABC on the yz plane = Anx Area of COA = Projection of area ABC on the xz plane = Any Area of AOB = Projection of area ABC on the xy plane = Anz

Let the body force components in xy and z direction be Bx, By and Bz per unit volume, respectively. The volume of the tetrahedron is equal to (1/3)Ah. Thus, for equilibrium in the x direction, TxnA – or

sxx Anx –

txy Any – txz Anz +

Txn = sxx nx + txy ny + txz nz –

1 AhBx = 0 3 1 Bh 3 x

Chapter 2: Analysis of Stress and Strain

59

Similarly, for equilibrium in the y and z direction,

and

Tyn = tyx nx + syy ny + tyz nz –

1 B h 3 y

Tzn = tzx nx + tzy ny + szz nz –

1 Bh 3 z

Tyn = tyx nx + syy ny + tyz nz Tzn

= tzx nx + tzy ny + szz nz

1442443

In the limit as h tends to zero, the oblique plane ABC will pass through point O consequently, one gets from above equations: Txn = sxx nx + txy ny + txz nz (2.17)

Equation (2.17) is known as Cauchy stress formula. If T n is resultant stress vector on plane ABC, we have 2

2

2

T n = Txn  Tyn  Tzn

2

If sn and ts are the normal and shear stress components, we have Tn

2

= T n2  U s2

But normal stress can also be expressed as

T n Txn cos B  Tyn cos C  Tzn cos H Txn nx  Tyn ny  Tzn nz Substituting Txn , Tyn and Tzn , we get

T n T xx nx2  T yy n2y  T zz nz2  2 U xy nx ny  2 U yz ny nz  2 U zx nz nx which is same as expression derived earlier.

2.14

PRINCIPAL STRESS

We have seen the normal and shear stress components can be determined on any plane with normal n, using Cauchy’s formula given by Eq. (2.17). Let us assume that there is a plane n with direction cosines nx, ny and nz on which the stress is wholly normal. Let s be the magnitude of this stress vector. Then, we have Tn = sn The components of this along the x, y and z axes are: Txn = s nx, Tyn = s ny, Tzn = s nz Also, from Cauchy’s formula, i.e., Eq. (2.17), Txn = sxx nx + txy ny + txz nz Tyn = tyx nx + syy ny + tyz nz Tzn = tzx nx + tzy ny + szz nz

(2.18)

60

Strength of Materials

Subtracting Eq. (2.18) from the above set of equations, we get (sxx – s) nx + txy ny + txz nz = 0 tyx nx + (syy – s) ny + tyz nz = 0 tzx nx + tzy ny + (szz – s) nz = 0

These set of equations are three simultaneous equations involving the unknown nx, ny and nz. These direction cosines define the plane on which the resultant stress is wholly normal. The determinant of the coefficients of nx, ny and nz must be equal to zero, for non-trivial solution, i.e.,

T xx  T U yz

U xy T yy

U zx



U xz T

U zy

U yz T zz



=0 T

Expanding the above determinant, we get a cubic equation in s as: s 3 – (sxx + syy + szz) s2 + sÿ (sxxsyy + syyszz + szzsxx – txy2 – tyz2 – tzx2)

– (sxxsyyszz – sxxt yz2 – syy tzx2 – szz txy2 + 2txytyztxz) = 0

(2.19a)

T 3  I1T 2  I 2T  I3 0

(2.19b)

or

where I1, I2, and I3 are defined in the next section. Equation (2.19b) is known as Characteristic equation of Principal stresses. The three roots of Eq. (2.19) give three principal stresses.

2.15

STRESS INVARIANTS

The coefficients of s 2, sÿ and the last term in the cubic Eq. (2.19a) can be written as follows: I1 = sxx + syy + szz = Sum of all normal stress component of stress matrix I2 =

T xx U xy T yy U yz T xx U xz   U yx T yy U zy T zz U zx T zz

= Sum of all 2 ´ 2 stress matrix whose diagonal must be from the diagonal of main stress matrix

T xx U xy U xz I3 = U yx T yy U yz U zx U zy T zz = Value of determinant of stress tensor

sij

I1, I2 and I3 are known as the first, second and third invariants of stress, respectively. If the reference coordinate is taken as principal axes (1, 2, 3) instead of (x, y, z), the stress matrix becomes

Chapter 2: Analysis of Stress and Strain

T The stress invariants are I1 = I2 =

ËT1 Ì Ì0 Ì ÌÍ 0

0

0

T2 0

=

2.16

Û Ü 0Ü Ü T 3 ÜÝ

s1 + s2 + s3 T1

0

T2



0

0 T3 T2 = s1s2 + s2s3 + s1s3

I3 =

61

0

T1

0

0

0

T2

0

0

0

T3



T1

0

0

T3

s1 s2 s3

PRINCIPAL DIRECTIONS

Once the principal stresses s1, s2 and s3 are obtained, the principal directions are obtained by seting s = s1, s = s2 and s = s3 in turn, and solving the following equations for anx, any and anz (sxx – s) anx + txy any + txz anz = 0 txy anx + (syy – s) any + tyz anz = 0 txz anx + tyz any + (szz – s) anz = 0 2 2 2 Such that the condition of direction cosines anx  any  anz

1 must satisfy.

EXAMPLE 2.1 In a triaxial stress system, the six components of the stress at a point are: sxx = 6 MPa

txy = 1 MPa

syy = 5 MPa

tyz = 3 MPa

szz = 4 MPa

tzx = 2 MPa

Determine the normal and shearing stress on the plane whose direction cosines are 1/ 3 , 1/ 3 , 1/ 3 . Also calculate the direction of the shear stress. Solution:

Resolving the stresses along the three coordinate axes, we have srx = sxx anx + txy any + txz anz

=

1 3

(6  5  4)

15 3

MPa

62

Strength of Materials sry = txy anx + syy any + tyz anz

=

1 3

9

(1  5  3)

3

MPa

srz = txz anx + tyz any + szz anz

= Resultant stress

sr =

=

1 3

9

(2  3  4)

3

MPa

T rx2  T ry2  T rz2 1 (225  81  81) 3

= 11.35 MPa Normal stress

sn = srx anx + sry any + srz anz

= Q

Shear stress

tns =

=

1 (15 + 9 + 9) = 11 MPa 3

T r2  T n2 128.82  121 = 2.79 MPa

The direction of the shear stress is obtained as follows: asx = = asy = = asz = =

1 U ns

[T rx  T n anx ]

1 2.79

1 U ns

U ns

15 3



11 Û Ü

3Ý

= 0.827

[T ry  T n any ]

1 2.79

1

Ë Ì Í

Ë Ì Í

9 3



11 Û Ü

3Ý

= –0.413

[T rz  T n anz ]

1 2.79

Ë Ì Í

9 3



11 Û Ü

3Ý

= – 0.413

EXAMPLE 2.2 The state of stress at a point is given by sxx = x2y2 + 30,

txy = 3x3y2

syy = xz3 + y2 + 15,

tyz = x2yz

szz = xy2z + 30,

tzx = xz2

Chapter 2: Analysis of Stress and Strain

63

Determine the body force distribution at the point (2, 3, 1) so that the stresses are in equilibrium. All units of stresses are in MPa. Solution:

Using the equilibrium equation: ˜ T xx ˜x



˜ U xy ˜y



˜ U xz ˜z

 Bx = 0

2xy2 + 6x3y + 2xz + Bx = 0 Bx = – (2xy2 + 6x3y + 2xz) At the point (2, 3, 1), Bx becomes Bx = – (2 ´ 2 ´ 9 + 6 ´ 8 ´ 3 + 2 ´ 2 ´ 1) = – (36 + 144 + 4) = –184 N/m3 Similarly, ˜ U xy ˜x



˜ T yy ˜y



˜ U yz ˜z

 By = 0

9x2y2 + 2y + x2y + By = 0 By = – (9x2y2 + 2y + x2y) = – (9 ´ 4 ´ 9 + 2 ´ 3 + 4 ´ 3) = – (324 + 6 + 12) = –342 N/m3 Similarly, using the third equilibrium equations, we get Bz = – (z2 + x2z + xy2) = – (1 + 4 ´ 1 + 2 ´ 9) = – 23 N/m3 The body force distribution required for the equilibrium becomes B = –184 iˆ – 342 ˆj – 23 kˆ

EXAMPLE 2.3 At a point in a stressed material the stress components are:

sxx = –30,

syy = 75,

szz = 110

txy = 72,

tyz = 40,

txz = 30, all are in MPa

Calculate the normal, shear and resultant stresses on a plane whose normal makes an angle of 40° with the x axis and 60° with the y-axis. Solution: Here,

anx = cos 40° = 0.766

and

any = cos 60° = 0.5

Then

anz = =

2 2 1  anx  any

1  (0.766)2  (0.5)2 = 0.404

64

Strength of Materials

\

srx = sxx anx + txy any + txz anz

= –30(0.766) + 72(0.5) + 30(0.404) = 25.14 MPa sry = 72(0.766) + 75(0.5) + 40(0.404) = 108.81 MPa

and Resultant stress

srz = 30(0.766) + 40(0.5) + 110(0.404) = 87.42 MPa sr =

=

(T rx )2  (T ry )2  (T rz )2

(25.14)2  (108.81)2  (87.42)2 = 141.82 MPa

T r2  T n2

Shear stress

tns =

Normal stress

sn = 25.14 (0.766) + 108.81 (0.5) + 87.42(0.404)

= 108.98 MPa \

tns =

(141.82)2  (108.98)2

= 90.75 MPa

EXAMPLE 2.4 In a triaxial stress system, the six components of the stress at a point are: sxx = 6 MPa,

txy = tyx = 1 MPa

syy = 5 MPa,

tyz = tzy = 3 MPa

szz = 4 MPa,

tzx = txz = 2 MPa

Find the magnitude of three principal stresses. Solution: Here, Then

3

(UPTU 2002–03)

2

s – I1s + I2s – I3 = 0

I1 = sxx + syy + szz = 6 + 5 + 4 = 15 MPa 2 2 2 I2 = sxxsyy + syyszz + szzsxx – t xy – t yz – t zx

= 6 ´ 5 + 5 ´ 4 + 4 ´ 6 – 12 – 32 – 22 = 60 MPa and

2 2 2 I3 = sxxsyyszz – sxxt yz – syyt zx – szzt xy + 2txytyztzx

= 6 ´ 5 ´ 4 – 6(3)2 – 5(2)2 – 4(1)2 + 2 ´ 1 ´ 3 ´ 2 = 54 MPa \ Solving

s 3 – 15s 2 + 60s – 54 = 0 s1 = 9 MPa s2 = 4.732 MPa s3 = 1.248 MPa

Chapter 2: Analysis of Stress and Strain

65

EXAMPLE 2.5 At a point P in a body, sxx = 30 kN/cm2, syy = –10 kN/cm2, szz = 10 kN/cm2 and txy = tyz = tzx = 10 kN/cm2. Determine the normal and shearing stress on a plane that is equally inclined to all the three axes. (UPTU 2001–02) Solution:

A plane that is equally inclined to all the three axes will have direction cosines: 1

anx = any = anz = \

3

srx = sxx anx + txy any + txz anz

1

=

3

50

(30 + 10 + 10) =

kN/cm2

3

sry = tyx anx + syy any + tyz anz

1

= and

3

10

[10 + (–10) + 10] =

kN/cm2

3

srz = tzx anx + tzy any + szz anz

1

=

3 È É Ê

30

(10 + 10 + 10) = 50 Ø

2

È É Ê

Ø Ù 3Ú

10

2

3

È É Ê

30 Ø

Resultant stress

sr =

Normal stress

3500 kN/cm2 3 sn = srx anx + sry any + srz anz

Ù

3Ú

kN/cm2 2

Ù

3Ú

=

= Thus,

tns =

=

Ë Ì Í

50 1 3

3



10 1 3

3



30 1 3

3

Û Ü Ý

= 30 kN/cm2

T r2  T n2 3500  (30)2 = 16.33 kN/cm2 3

EXAMPLE 2.6 For the given state of stress, determine the principal stresses and their directions.

[U ij ]

Ë0 Ì Ì1 Ì1 Í

1 1Û Ü

0 1Ü 1 0 ÜÝ

66

Strength of Materials

Solution:

First invariant I1 = 0, Second invariant I2 = –3, Third invariant I3 = 2 f (s ) = s 3 – I1s 2 + I2s – I3

Then or or

s 3 – 3s – 2 = 0 – s 3 + 3s + 2 = 0

or

(– s 3 – 1) + (3s + 3) = – (s + 1)(s 2 + 1 – s) + 3(s + 1) = (s + 1)[–s 2 + sÿ – 1 + 3] = (s + 1)[–s 2 + sÿ + 2] = (s + 1)[–s 2 + 2s – s + 2] = (s + 1)[s (–s + 2) + 1 (–s + 2)] = (s + 1)(s + 1) (–s + 2)

s1 = s2 = –1 and s3 = 2.

Since two of the three principal stresses are equal and s3 is different, the axis of s3 is unique and every direction perpendicular to s3 is a principal direction associated with s1 = s2. For s3 = 2, –2anx + any + anz = 0 anx – 2any + anz = 0 anx + any – 2anz = 0 2 2 2 anx  any  anz =1

and These give

anx = any = anz =

1 3

Standard method of solution: Consider the cubic equation: y3 + py2 + qy + r = 0 where p, q and r are constants. y= x

Substitute

1 p 3

This gives x3 + ax + b = 0 where

a= b=

Substitute

1 (3q – p2) 3

1 (2p3 – 9pq + 27r) 27

cos f = 

b 2( a3 /27)1/2

Chapter 2: Analysis of Stress and Strain

Determine f, and putting g= 2 

a 3

The solutions are: y1 = g È

y2 = g cos É Ê

cos G p  3 3

G 3

ÈG

y3 = g cos É Ê

3

Ø



120 ’Ù



240 ’Ù

Ú

Ø

Ú



p 3



p 3

EXAMPLE 2.7 At a point P, the rectangular stress components are: sxx = 1, syy = –2, szz = 4, txy = 2, tyz = –3 and tzx = 1

all in units of kPa. Find principal stresses and check for invariance. Solution:

Stress matrix is:

[U ij ]

Ë1 Ì Ì2 Ì1 Í

2 2 3

1Û

Ü 3 Ü

4 ÜÝ

I1 = 1 – 2 + 4 = 3 I2 = (1) (–2) + (–2) (4) + 4(1) – 22 – (–3)2 – 12 = –2 + (–8) + 4 – 4 – 9 – 1 = – 20 I3 = 1(–8 – 9) – 2(8 + 3) + 1 (–6 + 2) = – 43 Then

f (s ) = s 3 – I1s 2 + I2s – I3

For this cubic, following the standard method, y = s , p = –3, q = –20, r = 43 a=

1 (–60 – 9) = –23 3

b=

1 (–54 – 540 + 1161) = 21 27

cos f = 

(21/2) (12167/27)1/2

f = –119°.40

67

68

Strength of Materials

The solutions are: s1 = y1 = 4.25 + 1 = 5.25 kPa s2 = y2 = –5.2 + 1 = –4.2 kPa s3 = y3 = 0.95 + 1 = 1.95 kPa

Rearranging such that s1 ³ s2 ³ s3, s1 = 5.25 kPa, s2 = 1.95 kPa, s3 = –4.2 kPa

In terms of the principal stresses, the invariants are: I1 = s1 + s2 + s3 = 3 I2 = s1s2 + s2s3 + s3s1 = –20 I3 = s1s2s3 = –43 These agree with their earlier values.

2.17

OCTAHEDRAL STRESS

Let (1, 2, 3) be the principal stress axes. Now consider the planes whose unit normals satisfy the following relation l2

m2

1

n2

3

with respect to the principal stress axes (1, 2, 3). There are eight such planes as shown in Fig. 2.7.

Fig. 2.7 Octahedral plane for l = m = n = ±

1 3

relative to principal axes (1, 2, 3).

Such planes are equally inclined to three principal axes. A plane which is equally inclinded to three principal axes is known as octahedral plane. The normal and shear stress components associated with these planes are called octahedral normal stress soct and octahedral shear stress toct. The normal stress on any arbitrary plane defined by its direction cosines l, m, n is given as

T n 0 T 1l 2  T 2 m 2  T 3 n 2

(2.20)

Chapter 2: Analysis of Stress and Strain

69

1

Substituting the value of l = m = n =

3 1

T n0

3

in Eq. (2.20), we get

(T 1  T 2  T 3 )

(2.21)

The normal stress given by Eq. (2.21) is known as octahedral normal stress. Hence, soct =

=

1 (s1 + s2 + s3) 3 I1 3

(2.22)

The shear stress on any arbitrary plane is given as U

2

= T r2  T n2 [from Eq. (2.15)]

(2.23)

In terms of principal stresses s1, s2, s3, we can write sn = s1l2 + s2m2 + s3n2

T r = T 12 l2 + T 22 m 2 + T 32 n2

and

(i) (ii)

Now Eq. (2.23) with the help of Eqs. (i) and (ii) becomes, U

2

= T 12 l 2 + T 22 m 2 + T 32 n2  (T 1l2 + T 2 m2 + T 3 n2 )2

(2.24)

Now the octahedral shear stress can be obtained by substituting l = m = n = “ 1/3 in Eq. (2.24) and simplifying, 2

U oct =

or

1

[(T 1  T 2 ) 2 + (T 2  T 3 ) 3 + (T 3  T1 ) 2 ]

9

2 9 U oct = 2(s1 + s2 + s3)2 – 6(s1s2 + s2s3 + s3s1)

(2.25) (2.26)

2 9 U oct = 2I12  6 I2

U oct = U oct =

(2/9) I12  (2/3) I 2 2

(I12  3 I2 )

3

(2.27) (2.28)

So far we have defined the octahedral stress with reference to principal coordinate system and expressed in terms of stress invariants. Hence, these stresses can also be expressed in terms of sxx, syy, szz, txy, etc. associated with arbitrary (x, y, z) axes. Thus,

T oct =

I1 3

=

1 3

(T xx + T yy + T zz )

(2.29)

70

Strength of Materials

U oct =

=

2 3 2 3

(I12  3I 2 )1/2 2 [(T xx + T yy + T zz )2  3(T xx T yy + T yyT zz + T zz T xx  U xy  U yz2  U zx2 )]1/2 (2.30)

Simplifying, we get, U oct =

1 3

2 2 2 1/2 [(T xx  T yy ) 2 + (T yy  T zz )2 + (T zz  T xx )2 + 6(U xy + U yz + U zx )]

(2.31)

The octahedral normal and shear stress are very important in expressions used in yield criterion for ductile material.

2.18

MEAN AND DEVIATOR STRESSES

We have discussed that octahedral normal stress is 1/3 (s1 + s2 + s3) which is same as mean stress of three principal stresses. Hence, the octahedral normal stress also known as mean stress and denoted as sm \

sm = 1/3(s1 + s2 + s3)

= 1/3(sxx + syy + szz)

(2.32)

Experiments indicate that the yielding of metals is independent of the mean stress. Hence, the total state of stress can be written as follows. As the cross shear stresses are equal, hence txy = tyx etc. are taken in the derivation. ËT xx Ì Ì U xy Ì Ì Í U xz

U xy T yy U yz

U xz Û

Ü U yz Ü Ü T zz ÜÝ

=

ËT m Ì Ì 0 Ì Ì 0 Í

0

0

Û Ü 0 Ü Ü T m ÝÜ

Tm

0

or

+

ËT xx  T m Ì U xy Ì Ì U xz Ì Í

U xy T yy  T m U yz

T = Tm + Td

U xz

Û Ü U yz Ü Ü T zz  T m ÜÝ

(2.33)

(2.34)

The stress vector given by Tm is known as mean stress tensor and stress vector Td is known as deviatoric stress tensor. We know that if (x, y, z) coordinates are taken as principal coordinate axes, then s1 = sxx, s2 = syy, s3 = szz and txy = tyz = tzx = 0.

Now Eq. (2.34) yields,

Td =

ËT 1 Ì Ì Ì Ì Í

 Tm 0 0

0

T2



0

0

Tm

Û Ü 0 Ü Ü T 3  T m ÝÜ

The stress invariants for Td are J1 = s1 – sm + s2 – sm + s3 – sm =0

Chapter 2: Analysis of Stress and Strain

J2 =

T1  T m

0

0

T2  T m

+

T2  Tm

0

0

T3  T m

+

T1  T m

0

0

T3  Tm

71

Expanding and simplifying, we get J2 = –1/6[(s1 – s2)2 + (s2 – s3)2 + (s3 – s1)2]

J3 =

T1  T m

0

0

0

T2  Tm

0

0

0

T3  T m

= (s1 – sm) (s2 – sm) (s3 – sm)

State of pure shear A stress vector is said to be a state of pure shear when the first invariant I1 vanishes. Hence, for any coordinate system I1 = 0 indicates that stress matrix is under state of pure shear. For a state of pure shear, the octahedral normal stress is zero. Hence, Td as discussed represents a state of pure shear.

EXAMPLE 2.8 The nonzero stress components are

sxx = –70 MPa, syy = 50 MPa and

(a) Determine the principal stresses (b) Determine the stress invariants (c) Determine the octahedral shear and normal stress. Solution: We have

Given sxx = –70 MPa, syy = 50 MPa and txy = 10 MPa. I1 = sxx + syy + szz = –70 + 50 + 0 = –20 2 I2 = sxxsyy + syyszz + szzsxx  U xy  U 2yz  U zx2

= –70 ´ 50 – 102 = –3600 I3 = 0 \ Principal stress equation is s 3 – I1s 2 + I2s – I3 = 0

or

s 3 + 20s 2 – 3600s = 0

or

s (s 2 + 20s – 3600) = 0

\

s = 0

or

ÿ

s 2 + 20s – 3600 = 0

From the second condition, the roots are

s1,2 =

20 “

202 + 4 – 1 – 3600 2 – 1

txy = 10 MPa.

72

Strength of Materials

s1,2 s1 s2 s3

\ and

= –10 ± 60.827 = –70.827 MPa = 50.827 MPa =0

(c) Octahedral shear stress U oct =

=

2 3 2 3

(I12  3 I2 ) [(  20)2 + 3 – 3600]1/2

2

– 105.83 3 = 49.890 MPa =

EXAMPLE 2.9 Two given principal stresses are s1 and s3. If the order of three principal stresses is s1 > determine the value of s2 for which the octahedral shear stress attains an extreme value. Solution:

s2 > s3,

We have 9t2oct = (s1 – s2)2 + (s2 – s3)2 + (s3 – s1)2

Expanding, we get 2 9U oct = 2T12 + 2T 22 + 2 T 32  2T 1T 2  2T 2T 3  2T 3T 1

For extremum value of toct, differentiating toct with respect to s2, and equating to zero,

˜ U oct =0 ˜T 2

2 ´ 9U oct

= 4s2 – 2s1 – 2s3

˜U oct

˜T 2 \ \

=

1 (2 T 2  T 1  T 3 ) 9

U oct

=0

2s2 – s1 – s3 = 0 s2 =

T1 + T 3 2

EXAMPLE 2.10 Two bodies are subjected to uniaxial tension to s0 value and torsion to t0. If the octahedral shear stress for two loading conditions is the same, determine the ratio of the s0/t0.

Chapter 2: Analysis of Stress and Strain

Solution:

73

Under uniaxial loading condition for first body, the state of stresses is sxx = s0, syy = szz = txy = tyz = tzx = 0

The invariant of stress are I1 = s0, I2 = 0 and I3 = 0 Hence, 2

U oct =

(I12  3I 2 )

3 2

=

T0

3

(i)

For the second body, the state of stresses are sxx = syy = szz = tyz = tzx = 0 and txy = t0. The stress invariants are I1 = 0, I2 = – t02, I3 = 0 2

U oct =

(3 – U 02 )

3 2

=

3

U0

Hence, according to the given statement 2 3

U0 =

2 3

T0

T0 = 3 U0

\

EXAMPLE 2.11 Prove that the state of s stress given below is a state of pure shear.

T=

Ë5 Ì Ì3 Ì Ì Í2

3

2Û

7

4Ü

4

Ü

Ü 12 Ü Ý

Solution: A state of stress is said to be pure state of shear when the first invariant is zero. For the given state of stress I1 = sxx + syy + szz = 5 + 7 – 12 = 0 Hence, the stress is pure shear state of stress.

74

Strength of Materials

2.19

STRAIN ANALYSIS

When a body is subjected to external forces, displacement of a point in the body relative to others occurs. This displacment may be linear or angular one. Also, the deformation occurs in the body is not uniform throughout its volume. These deformations are either characterized by the changes in length of a line element or the changes in the angle between two line element defines the strain. The elongation or contraction of a line element per unit length is known as normal strain. If a line element as shown in Fig. 2.8 (undeformed body) becomes ds¢ after deformation, the strain of the line element is defined as F =

Fig. 2.8

ds  ds „ ds

Line segment PQ in undeformed and deformed body.

Similarly, when any change in angle occurs between two line element originally perpendicular to each other, the deformation is referred to as shear strain. This deformation is denoted by g (gamma) and is measured in radians. This is explained in Fig. 2.9.

(a) Undeformed body Fig. 2.9

(b) Deformed body

Angular deformation of two line element.

Now to develop the concept of state of strain and the strain-displacement relation, we consider the following examples. Let a two demensional element ABCD extended in the x-direction by an amount Du, the strain is said to be occurred in the x-direction and it is normal strain. Similarly, the angular deformation shown in Fig. 2.10 defines the shear strain in the xy plane.

Chapter 2: Analysis of Stress and Strain

Fig. 2.10

2.20

75

Linear strain: (a) in x-direction, (b) y-direction, and (c) shear strain in x-y plane.

STRAIN-DISPLACEMENT RELATION

Consider a body ABCD before deformation. The coordinates of the points A, B, C and D are A(x, y), B(x + Dx, y), C(x + Dx, y + Dy) and D(x, y + Dy). Due to external loads the deformation of the body ABCD occurred and due to deformation all the points A, B, C and D distored to A¢, B¢, C¢ and D¢. Let u, v be horizontal and vertical displacement of line AA¢. Now we can write AM = u, A¢M = v The cordinates of A¢ are A¢(x + u, y + v). Now we can write with reference to Fig. 2.11, BN = u +

˜u ˜x

'x +

1 ˜2 u 2 ˜x 2

('x ) 2 + higher order terms

Fig. 2.11 Direct and shear strain.

76

Strength of Materials

Neglecting the smaller terms, we can write ˜u

BN = u +

'x

˜x

(2.35)

Similarly, we can write the following equations. ˜v

NB „ = v +

˜x

'x

(2.36)

DP = u +

˜u ˜y

'y

(2.37)

D „P = v +

˜v ˜y

'y = DT

(2.38)

Horizontal component of A¢B¢ = A¢R = MN = AN – AM = AB + BN – AM ˜u

= Dx + u + = Dx +

˜u ˜x

˜x

'x – u

'x

(2.39)

This is the final length of AB in the x-direction. \ Strain in the x-direction is denoted as exx

F xx =

F xx =

'x +

˜u ˜x

'x  'x

'x

˜u

(2.40)

˜x

By similar argument, we can derive the strain in the y-direction. From Fig. 2.11, we can write vertical components of A¢D¢. = A¢Q = ST = AT – AS = AD + DT – AS = Dy + v +

˜v Dy – v ˜y

Chapter 2: Analysis of Stress and Strain

= 'y +

˜v ˜y

'y

77 (2.41)

which is the final length of AD in the y-direction. \ Strain in the y-direction is ˜v 'y  'y ˜y 'y

'y + F yy =

F yy =

˜v ˜y

(2.42)

Now, if a and b are the angular displacements of two line element AB and AD, the shear strain can be obtained from the change in angles or from a and b. Consider them as total angular deformation. tan B =

=

B „R A „R

B „N  RN MN

v+ =

˜v ˜x

'x +

'x  v ˜u 'x ˜x

˜v 'x ˜x = ˜u Ø È ÉÊ1 + ˜x ÙÚ 'x

(2.43)

From Eq. (2.42), Eq. (2.43) becomes tan B

˜v = ˜x 1 + F xx

(2.44)

For small deformation theory if exx << 1, Eq. (2.44) reduces to

B

=

˜v ˜x

Similarly, we can derive tan C =

D „Q A „Q

(2.45)

78

Strength of Materials

˜u ˜y = 1 + F yy

(2.46)

Taking for small deformation theory eyy << 1.

C=

˜u ˜y

(2.47)

The shear strain denoted as gxy defined as change in the right angle between two line elements originally at the right angle is expressed as gxy = a + b

˜u ˜v + ˜y ˜x

=

(2.48)

Hence, in two-dimensional state of strain, the strain-displacement relations are ˜u

Þ Ñ Ñ ˜v Ñ F yy = ß ˜y Ñ Ñ ˜u ˜v Ñ H xy = + ˜y ˜x à F xx =

˜x

(2.49)

For simplicity if we define F xy =

=

1 2

1 2

H xy È ˜u É Ê ˜y

+

˜v Ø

(2.50)

Ù ˜x Ú

We can prove F yx =

Hence,

1 2

È ˜u É Ê ˜y

+

˜v Ø Ù ˜x Ú

= exy

(2.51)

exy = eyx

(2.52)

With this concept we can derive the strain-displacement relation for a three-dimensional case too. Let u, v, w be displacements in the x, y and z directions respectively, we can write normal strain as

79

Chapter 2: Analysis of Stress and Strain

˜u Þ ˜x Ñ

F xx =

Ñ

F yy

˜v Ñ = ˜y ß

H xy

=

˜u ˜v + ˜y ˜x

H yz

=

˜v ˜w + ˜z ˜y

(2.53a)

Ñ ˜w Ñ Ñ Fzz = ˜z à and the shear strain as

H zx =

˜w ˜x

+

Þ Ñ Ñ Ñ ß Ñ Ñ Ñ à

˜u ˜z

(2.53b)

Expressing strain as eij where i = x, y or z and j = x, y or z, the strain displacement relations given in Eq. (2.53a, 2.53b) can be written as F ij =

or

2.21

F ij =

1 2 1 2

È ˜ui É Ê ˜x j

+

˜u j Ø Ù ˜x i Ú

(2.54)

(2.55)

(ui, j + u j, i )

THREE-DIMENSIONAL STRAINS

In the case of three-dimensional coordinate system, the components of strain are expressed by the strain matrix: F xx

F xy

F xz

eij = F yx

F yy

F yz

F zx

F zy

F zz

F xx

or

H yx

2

H xy

2 F yy

H zx

H zy

2

2

H xz

2 H yz

2 F zz

Such that gij = 2eij, i, j = x, y, z, where eij are normal strains and gij are the shearing strains.

80

Strength of Materials

2.22

NORMAL AND SHEARING STRAINS

Suppose on a plane the strain components are eij. The direction cosines of the normal to this plane are anx, any and anz. The normal and shearing strains on this plane can be determined as follows:



erx = F xx anx ery = erz =

2 H zx

2

2

any 

H xz

2

anx +

H zy

2

anz

H yz

anx  F yy any +

2 F rx

er =

Resultant strain Normal strain

H yx

H xy

2

anz

any + ez zanz

 F ry2  F rz2

en = erx anx + ery any + erz anz

= exx a2nx + eyy a2ny + ezz a2nz + gxy anx any + gyz any anz + gzx anx anz gns =

Shearing strain

F r2

 F n2

The direction cosines of shearing strain may be determined from: asx = asy = asz =

2.23

2 H ns

2 H ns

2 H ns

[F rx  F n anx ] [F ry  F n any ] [F rz  F n anz ]

PRINCIPAL STRAINS

The principal strains are the roots of the cubic equation: e3 – j¢1e 2 + j¢2e – j¢3 = 0

where is the first invariant of strain.

j¢1 = e xx + e yy + e zz

j¢2 = e xxe yy + e yye zz + e zze xx –

1 2 2 (H  H yz  H zx2 ) 4 xy

is the second invariant of strain. j¢3 = F xx F yy F zz 

F xx H 2yz

4

2



F yyH zx

4

2



F zz H xy

4



H xyH yz H zx

4

is the third invariant of strain. The roots of equation may be determined by a hit-and-trial method or Newton–Raphson method.

Chapter 2: Analysis of Stress and Strain

2.24

81

PRINCIPAL STRAIN DIRECTIONS

The principal strain direction may be determined as follows: For e1, let

A1 =

H yz

 F1

F yy

2

H yz

F zz

2

B1 = 

H xy

H yz

2

2

H zx

2 H xy

C1 =

 F1

2

F zz

F yy

 F1

 F1

H xz

H yz

2

2

Then anx1 =

any1 =

anz1 =

A1 A12

 B12  C12 B1

A12

 B12  C12 C1

A12

 B12  C12

Similarly, direction of e2 and e3 may be determined.

2.25

CONCEPT OF COMPATIBILITY

In simple cases of statistical equilibrium, determinate and indeterminate problems can be distinguished by comparing the number of equations with the number of unknowns. If these are same, a unique solution can be easily obtained and the problem is termed determinate. Here, we shall try to study another aspect of the problem, i.e., the number of equations being more than the number of unknowns. This problem arises if we try to determine the displacement components u, v, w from the strain components, exx, eyy, ezz, gxy, gyz, etc. Here, we have six equations of the type:

u x v u H xy =  x y F xx =

    

(2.56)

82

Strength of Materials

for the three unknowns u, v, w. If the strains are arbitrarily specified, the set of six differential equations given by Eq. (2.56) cannot be integrated to give single valued solutions for u, v and w. Therefore, the strain component must satisfy certain other conditions as well to ensure the uniqueness of the solution. These extra conditions are known as condition of compatibility. This explains the need of compatibility conditions in a mathematical sense. In a physical way, it can be stated that various portions of a deformable body should move, during strains, in such a way that close contact is maintained everywhere along any imaginable interface without separation or overlapping. Thus, on Fig. 2.12(a) , if the line elements AB, BC and AC forming a triangle as shown at (i) deform to the shape given by AB, BC and CA¢ as shown at (ii) and (iii), the condition of compatibility is violated. This means that the components of strains should have a mutual relation so that the apex of the triangle ABC does not open at A, as shown in (ii) or overlap as shown in (iii), after deformations have taken place.

Fig. 2.12

A physical picture for compatibility requirements.

Another example of compatibility requirements is shown in part (b) of the same figure using area elements. In fact, compatibility means accommodation or matching of temperament in ordinary English, and here we want to show the same thing, i.e., how the shape of same type of elements before deformation should remain similar to each other after deformation. Thus, if square elements a b c d, etc., of the 2D body shown at (iv) deform to parallelogram a¢ b¢ c¢ d¢, etc. then each element of the deformed body should have similar shape. Hence, if we isolate an element a² b² c² d² from the deformed body leaving a gap a¢ b¢ c¢ d¢, the shape and size of the isolated element and the gap should be identical for compatibility. In Figure 2.12(v), this is the case and deformed elements a² b²c² d², will completely fill the gap a¢ b¢ c¢ d¢. Hence no small gap will be left and compatibility would be satisfied.

Chapter 2: Analysis of Stress and Strain

2.26

83

ST-VENANT’S EQUATIONS OF COMPATIBILITY

We start from the strain displacement relation and differentiate these to obtain mutual relations between various strain components. First, consider the direct strains. exx =

˜u ˜x

eyy =

˜v ˜y

˜2 F xx

\

˜3u ˜ x ˜ y2

˜y 2 ˜2 F yy

˜x

˜3 v ˜y˜x 2

2

Adding these, we get

˜2 F xx

˜y

\

2

˜2 F xx

˜y 2





˜2 F yy

˜x

2

˜2 F yy

=

˜3 u ˜3 v  2 ˜x ˜y ˜y˜x 2

=

˜2 È ˜u ˜x ˜y É Ê ˜y

=

˜x 2



˜v Ø ˜x Ù Ú

˜ 2H xy ˜ x˜y

Similarly, we can get two more equations. Now, consider the shear strains. gxy =

˜ H xy ˜z

Similarly,

˜ H yz ˜x

˜ u ˜v  ˜y ˜x

=

˜2 u ˜2 v  ˜ y˜ z ˜ x ˜ z

=

˜2 v ˜2 w  ˜x ˜ z ˜y˜ x

˜ H zx ˜2 w ˜2 u =  ˜y ˜x ˜ y ˜z˜ y

˜H xy ˜H ˜H ˜2 u  yz  zx = 2 ˜z ˜x ˜y ˜z˜y ˜ È ˜H xy  ˜x ÉÊ ˜z

˜H yz ˜x



˜H zx Ø ˜y ÙÚ

= 2

˜2 F xx ˜z˜y

Similarly, two more equations can be obtained. The entire set of St. Venant’s equation can be written as:

84

Strength of Materials

˜2 F yy ˜2 F xx  ˜y 2 ˜x 2

˜2H xy ˜x ˜y

˜2 F yy ˜2 F zz  ˜z 2 ˜ y2

˜2H yz ˜ y˜ z

˜2 F zz ˜2 F xx  ˜x 2 ˜z 2 ˜2 F xx 2 ˜ y˜ z

˜2H zx ˜z ˜x ˜ È ˜H yz ˜H zx ˜H xy Ø    ˜x ÉÊ ˜x ˜y ˜z ÙÚ

2

˜2 F yy ˜z˜x

˜ È ˜H zx ˜H xy ˜H yz Ø    ˜y ÉÊ ˜y ˜z ˜x ÙÚ

2

˜2 F zz ˜x ˜y

˜ È ˜H xy ˜H yz ˜H zx Ø    ˜z ÊÉ ˜z ˜x ˜y ÙÚ

Þ Ñ Ñ Ñ Ñ Ñ Ñ Ñ Ñ Ñ ß Ñ Ñ Ñ Ñ Ñ Ñ Ñ Ñ Ñ Ñà

(2.57)

These equations were first found by St. Venant in 1860.

EXAMPLE 2.11 The stress components at a point are given by sxx = 20, syy = 10, szz = 15, txy = 10, tyz = 15, tzx = 20 MPa. Calculate the strain components, taking E = 200 GPa and Poisson’s ratio 0.3. Solution: E = 2G (1 + n) or

G= =

or Now,

E 2(1  O )

200 – 109 2 – 1.3

G = 76.92 GPa exx =

=

1 [s – n(syy + szz)] E xx 1 200 – 109

[20 – 0.3(25)] 106

= 0.0625 ´ 10–3 = 62.5 ´ 10–6 eyy =

=

1 [s – n(szz + sxx)] E yy 1 200 – 109

= –2.5 ´ 10–6

[10 – 0.3(35)]106

Chapter 2: Analysis of Stress and Strain

1

ezz =

and

200 – 109

85

[15 – 0.3(30)] 106

= 30 ´ 10–6 U xy

gxy =

Then

G 10 – 10 6

=

= 130 ´ 10–6

76.92 – 10 9 15 – 10 6

g yz =

76.92 – 10 9

= 195 ´ 10–6 20 – 10 6

g zx =

and

76.92 – 10 9

= 260 ´ 10–6

EXAMPLE 2.12 The strain components at a point are given by exx = 100, eyy = 50, ezz = 40 m strains and gxy = 20, gyz = 10, gzx = 15 m radians. Calculate the normal and shearing strains on a plane whose normal has the direction cosines 1/ 3, 2/3, 0. Solution:

1 1 gxy any + g a 2 2 xz nz

erx = exx anx +

= 100

1 3

ery =

= erz =

= Then resultant strain

er =

=



1 – 20 – 2

2  0 = 65.9 m 3

1 1 gxy anx + eyy any + g a 2 2 yz nz 1 1 – 20 –  50 – 2 3

2  0 = 46.598 m 3

1 1 gxz anx + g a + ezz anz 2 2 yz ny 1 1 1 – 15 –  – 10 – 2 3 2

2 = 8.41 m 3

(F rx ) 2  (F ry ) 2  (F rz )2

 (65.9)  (46.59)  (8.41) m = 81.143 m 2

2

2

86

Strength of Materials en = erx anx + ery any + erz anz

Normal strain

= H ns

Shearing strain

2

gns =

or

2.27

=

È È É 65.9 É Ê Ê

F r2

È

Ø Ù  3Ú

1

46.59 É Ê

2Ø 3 ÙÚ



Ø

8.41(0) Ù m = 76.1 m Ú

 F n2

2 (81.143)



 (76.1)2 m = 56.34 m

2

SOLUTION OF STRESS DIFFERENTIAL EQUATION

We have discussed stress differential equations, stress-strain relations and compatibility equations. Now these equations can be combined together to find the unknown parameters such as stress, strain or displacements. In this section, we will consider plane stress and plane strain problems limited to two-dimension only.

Plane stress case The stress-strain relations for plane stress are F xx =

F yy =

H xy

1 E 1 E =

(T xx 

OT yy )

(2.58)

(T yy 

OT xx )

(2.59)

U xy

(2.60)

2(1 + O ) E

The compatibility equation is given by ˜ 2 F xx ˜y 2

+

˜ 2 F yy ˜x 2

=

˜ 2 H xy

(2.61)

˜x ˜y

Substituting Eqs. (258) to (2.60) in Eq. (2.61), we get 1 E

Ë ˜2 Ì 2 ÍÌ ˜y

(T xx



OT yy ) +

˜

2

˜x

2

(T yy



Û

OT xx ) Ü = ÝÜ

˜

2

˜x ˜y

Ë 2 (1 + Ì E Í

O)

Û

U xy Ü

(2.62)

Ý

From the stress-differential equations given as

˜U xy ˜T xx + =0 ˜x ˜y

   ˜U xy ˜T yy  + =0   ˜x ˜y

(2.63)

Chapter 2: Analysis of Stress and Strain

87

Equation (2.62) can be further simplified as ˜ 2T xx

˜y2

+

˜ 2T yy

˜x 2

2

˜2U xy

=O

˜x ˜y

˜ 2T yy

=O

˜y 2

+O

˜2T xx

˜x 2

+ 2O

2

Ë ˜ 2T xx Ì 2 ÌÍ ˜x

+2

˜ U xy ˜x ˜y

+

˜ 2U xy

˜x ˜y

2 ˜ T yy Û Ü 2 ˜y Ü Ý

(2.64)

Now defining a function f, known as Airy’s stress function as

T yy =

˜ 2G

˜ 2G

˜x

˜y 2

, T xx = 2

and U xy = 

˜2G ˜x ˜y

(2.65)

Equations (2.65) should satisfy the differential equation (2.63). Now substituting Eq. (2.65) in Eq. (2.64) and simplifying, we get

˜ 4G ˜x 4

+2

˜ 4G ˜x 2 ˜y2

+

˜ 4G ˜y 4

=0

Ñ4f = 0

or 2

where

³

=

È ˜2 É 2 Ê ˜x

+

(2.66) ˜

2 Ø

2Ù ˜y Ú

Since solution of Eq. (2.66) satisfy both compatibility and equilibrium equations, any solution of this equation gives a possible stress distribution in an elastic body. If the body force is included in the analysis, the stress differential equations are T xx

U xy

 + Bx = 0  x y   U xy T yy  + + By = 0  x y  +

(2.67)

Now considering a potential function W so that Bx =

 ˜˜:x

Eq. (2.67) becomes

˜T xx ˜x

and By =

˜U  ˜˜:x + ˜yxy

 ˜: ˜y =0

or

˜U xy ˜ (T xx  :) + =0 ˜x ˜y

and

˜U xy ˜ (T yy  :) + =0 ˜y ˜x

(2.68)

88

Strength of Materials

Now assuming a stress function f so that

˜ 2G  ˜y2   ˜ 2G  T yy = : + 2  ˜x   ˜ 2G  txy =   ˜x˜y 

T xx = : +

(2.69)

Equation (2.62) yields 2(1 + O )

˜ 2U xy

˜x ˜y

˜2

=

(T xx  OT yy ) +

˜y 2

˜2

(T yy  OT xx )

˜x 2

(2.70)

Substituting Eq. (2.69) in Eq. (2.70) 2(1 + O )

˜

2

˜x ˜y

2 È ˜ G Ø  É Ù Ê ˜x ˜y Ú

=

˜

˜y

+

Ë Ì: ÌÍ

2 2

˜

Ë Ì: ÌÍ

2

˜x

+

2

˜

2

G

˜y

+

2

˜

2

È

O É: +



G

˜x

2

Ê



È

˜

GØÛ

Ü 2Ù ˜x Ú Ü Ý

O É: + Ê

2

˜

2 ØÛ G

Ü 2Ù ˜y Ú Ü Ý

After simplification, we get ˜

4

G

˜x

4

+2

˜ 2

4

G

˜x ˜y

2

+

˜

4

G

˜y

4

+ (1



O)

È ˜2 : É 2 Ê ˜x

2

+

˜ :Ø 2 Ù ˜y Ú

=0

Ñ4f + (1 – n) Ñ2W = 0

or

(2.71)

For plane strain case ezz = 0 and hence

szz = n(sxx + syy) Now modifying the stress-strain relations, we can derive ³

2.28

4

G+

È 1  2O Ø 2 É Ù ³ : Ê 1  O Ú

=0

TYPES OF AIRY’S STRESS FUNCTION

The following types of stress functions are most commonly used for obtaining the solution: (a) (b) (c) (d)

Algebraic functions Polynomials Fourier series Complex functions

(2.72)

Chapter 2: Analysis of Stress and Strain

(e) Trigonometric series (f) Harmonic functions Other methods for solving two-dimensional problems are: (a) Integral transform method (b) Real potential method (c) Conformal transformation method We shall illustrate some of the stress functions by the use of polynomial only. (i)

Polynomial of first degree.

f1 = a1x + b1y sxx = syy =

˜ 2G1 ˜y2

=0

˜2 G1 =0 ˜x 2

˜2 G1 =0 ˜x ˜ y This stress function gives a stress free body as shown in Fig. 2.13.

txy = –

Fig. 2.13 State of stress for first degree polynomial stress function.

(ii) Polynomial of the second degree.

f2 = sxx = syy =

a2 2 c x  b2 xy  2 y2 2 2 ˜2G2 ˜y 2

= c2

˜2G2 = a2 ˜x 2

txy = 

˜2G2 = –b2 ˜ x˜ y

89

90

Strength of Materials

This state of stress represented by a constant stress yield is shown in Fig. 2.14.

Fig. 2.14 State of stress under constant stress field.

2.29

APPLICATION OF AIRY’S STRESS FUNCTION

Pure tension: function be

Consider a rectangular plate under simple tension as shown in Fig. 2.15. Let Airy’s

f (x, y) =

Fig. 2.15

State of stress under simple tension.

sxx =

Hence

From boundary condition c2 = Similarly,

c2 2 y 2

s0

˜ 2G ˜y 2

= c2

sxx = s0 2 syy = ˜ G2

˜x

=0

Chapter 2: Analysis of Stress and Strain

txy = –

91

˜2 G ˜x ˜ y

=0 Pure shear:

Consider the member under pure shear as shown in Fig. 2.16. Let Airy’s function be f (x, y) = cxy

Fig. 2.16 State of stress under pure shear.

sxx = 0 syy = 0

Hence

txy = c

Pure bending: Consider a member under pure bending. The following stress function is considered. f = ax3 + bx2y + cxy2 + dy3

where a, b, c, and d are constants.

T xx =

Now

T yy = U xy =

˜ 2G ˜y 2

˜2G ˜x2



= 2 cx + 6 dy

= 6 ax + 2 by

˜2 G =  (2 bx + 2 cy) ˜x ˜y

This shows that all stresses vary linearly with respect to x and y. Now if a = b = c = 0, the state of stress would correspond to pure bending as ÿ ÿ ÿ sxx = 6dy syy = 0 txy = 0

Thus, f = dy is the stress function for pure bending which satisfy Ñ4f = 0 condition too. 3

92

Strength of Materials

2.30 MOHR’S CIRCLE FOR THE THREE-DIMENSIONAL STATE OF STRESS We know that in a general state of stress, at every point it is possible to reduce the stresses into three mutually perpendicular normal stresses, i.e., s1, s2 and s3. If Mohr’s circles are drawn for three sets of principal stresses, we get a figure, where all the three circles cluster in the same area [see Fig. 2.17].

Fig. 2.17

Mohr’s circle in three dimensions.

The following procedure may be adopted to obtain the results: 1. Arrange the principal stresses such that algebraically s1 ³ s2 ³ s3 along the sn axis. 2. Locate the centres O1, O2 and O3 at distances 1/2(s2 + s3), 1/2(s1 + s3) and 1/2(s1 + s2) from the origin O, respectively. 3. Draw three circles with centres O1, O2 and O3 having radii 1/2(s2 – s3), 1/2(s1 – s3) and 1/2(s1 – s2), respectively. 4. Draw vertical lines at s1. Set off angle a = cos–1 anx from the vertical line at s1 in the anticlockwise direction and draw the line s1T1T2. 5. With centre O1 draw arc T1T2. 6. Set off angle g = cos–1 anz from the vertical at s3 in the clockwise direction and draw the line s3 N1N2. 7. With centre O3 draw the arc N1N2. 8. Let P3 be the point of intersection of the arc T1T2 and N1N2. 9. From point P3 drop perpendicular P3A and P3P on sn and ts axes, respectively. Then P3A = ts and OA = sn. 10. To check the construction set off angle b = cos –1 any on each side of the vertical at s2. Draw arc P1P2 with O2 as centre. The arc should pass through P3.

EXAMPLE 2.13 The principal stresses at a point are 300 MPa, 50 MPa and –150 MPa. Determine graphically the octahedral normal and shearing stresses and check analytically.

Chapter 2: Analysis of Stress and Strain

Solution:

The direction cosines of the octahedral plane are: anx = any = anz =

a=b=g Normal stress

È

= cos–1 É Ê

sn = =

Resultant stress

sr = =

and shear stress

1 (s + 3 1

Ø Ù 3Ú

1

1 3

= 54° 44¢

s2 + s3)

1 (–150 + 50 + 300) = 66.67 MPa 3 (T 1 )2  (T 2 )2  (T 3 )2 ( 150)2  (50)2  (300)2 = 195.78 MPa

ts = T r2  T n2 =

(195.78)2  (66.67)2 = 184.08 MPa

The graphical construction has been shown in Fig. 2.18. By measurement,

sn = OO2 = 66 MPa ts = ON = 184 MPa

Fig. 2.18

93

94

Strength of Materials

EXERCISES 2.1 Write the short notes on the followings: (i) Stress tensor (ii) Concept of compatibility (iii) Airy’s function 2.2 The cartesian components of stress at a point are: sxx = 15 syy = szz = 8 txy = 6 tyz = 4

txz = 4 MPa

Determine the normal and shear stresses on a plane whose direction cosines are 1/ 3,

1/ 3, 1/ 3 . [Ans. 2.3 The cartesian components of stresses at a point are: sxx = 7, syy = 6, szz = 5, txy = 2, tyz = –2, Determine the values of principal stresses.

19.57 MPa, 3.83 MPa]

txz = 0 MPa

[Ans.

9 MPa, 6 MPa, 3 MPa]

2.4 The state of stress at a point for a given reference xyz is given by the following array of terms: 8 6 Û

Ë 15 Ì Ì 8 Ì 6 Í

Ü

12

5 Ü MPa

8 ÜÝ

5

Determine the principal stresses. [Ans. 19.19 MPa, 10.27 MPa, –10.02 MPa] 2.5 The principal stresses at a point on a plane are: s1 = 50, s2 = 40, s3 = –20 MPa Determine the normal and shear stresses on this plane if its direction cosines are: 1 14

,

2 14

,

3 14

[Ans. 2.14 MPa, 29.8 MPa] 2.6 The strain components at a point are given by exx = 200, eyy = 100, ezz = 50 m strains and gxy = gyz = gzx = 40 m radian. Calculate normal and shearing stains on a plane having direction cosines 1/ 3, 1/ 3, 1/ 3 . [Ans. 156.67 ´ 10–6, 124.72 ´ 10–6] 2.7 Calculate the principal strain for data given in Problem 2.6. 2.8 The principal stresses at a point are 330 MPa, 50 MPa and –120 MPa. Determine graphically the octahedral normal and shearing stresses and check analytically. [Ans. Analytically sn = 86.76 MPa, tns = 185.52 MPa Graphically sn = 86 MPa, tns = 186 MPa]

3 3.1

Theory of Failure

INTRODUCTION

The state of stress at any point on the structure may be simple or complex in nature. Hence, in designing under complex loading or complex state of stress condition, it is difficult to set the maximum limit of the stresses to define the failure condition. All engineering materials are classified as ductile or brittle. A material is said to be ductile when the gross plastic deformation is more than 5% and brittle when it is less than 5%. Hence, the material which shows failure strain ef ³ 5% is termed as ductile material and if ef < 5%, then it is called brittle material. However, under some special conditions a ductile material may fail in a brittle manner. If the material is ductile, the failure is usually specified by the initiation of yielding, whereas if the material is brittle, the failure is characterized by fracture. Hence, yield strength and ultimate strength are generally set as maximum limit stress in ductile and brittle failure respectively. The yield strength and ultimate strength of any material are obtained from tension or compression tests under uniaxial static loading condition. A load is said to be static when the rate of loading is very small or the load is approximately stationary on the member. The test is usually conducted at room temperature with a rate of loading less than 10 mm/min. The material behaviour under uniaxial loading is represented by engineering stress-strain diagram. The stress-strain diagram for ductile material is shown in Fig. 3.1.

Fig. 3.1

Stress-strain behaviour. (in Fig. 3.1(b) line BB¢ is parallel to tangent drawn at A). 95

96

Strength of Materials

The point B in Fig. 3.1(a) and C in Fig. 3.1(b) is the proportional limit. This is the point where the curve first begins to deviate from its linearity or straight line. If the load is removed from this point, there will be no permanent deformation left on the specimen. The stress-strain relation in this linear range according to Hooke’s law is

s = Ee

(3.1)

where s and e are engineering stress and strain respectively. E is the slope of the linear part of the stress-strain curve; E is known as modulus of Elasticity or Young’s modulus. It is the measure of the stiffness of the material. The point D in Fig. 3.1(b) is known as elastic limit. When the specimen is loaded beyond this point, after removal of the load, a permanent deformation or set will be left on the specimen. The points C to D are not straight line but the deformation up to point D is small enough and the relation given above [Eq. (3.1)] holds true. When the load is further increased above the elastic limit, it results in permanent deformation. This behaviour is known as yielding. The stress corresponding to this zone is known as yield strength. In some materials such as low carbon steels, two distinct yield point are seen. The upper yield point occurs first followed by a sudden decrease in strength to a lower yield point. The upper and lower yield points are shown in Fig. 3.1(a) by points B and C respectively. All materials do not have definite yield point. For those the material yield point is obtained by an offset method as shown in Fig. 3.1(b). In offset method a line [BB¢ shown in Fig. 3.1(b)] is drawn parallel to the initial slope or equal to e value with a permanent set equal to 0.2% of the original gauge length, i.e., (e = 0.002). The ultimate strength is the highest strength on stress-strain diagram of a member under uniaxial loading. The points are shown by D in Fig. 3.1(a) and E in Fig. 3.1(b). Several theories of failures based on yielding or fracture have been proposed by many scientists. These theories of failures are sometimes known as yield criteria. A yield criterion is a hypothesis concerning the limit of elasticity under any possible combination of stresses, strains or energy. A yield criterion is expressed mathematically by a yield function f (sij, where sij is the state of stress and Different yield criteria are 1. 2. 3. 4. 5. 6.

sy)

sy is the yield strength in uniaxial tension or compression.

Maximum principal stress theory Maximum shear stress theory Maximum principal strain theory Maximum strain energy theory Maximum distorsion energy theory Mohr’s failure criterion

All these theories are not suitable for both ductile and brittle materials because of their nature of failure. Hence, the theory suitable for ductile material is discussed first.

3.2 3.2.1

FAILURE THEORY FOR DUCTILE MATERIAL Maximum Shear Stress Theory

The maximum shear stress criterion, also known as Tresca criterion, states that yielding begins when the maximum shear stress at a point reaches the maximum shear stress at yield under uniaxial tension

Chapter 3: Theory of Failure

or compression. For multiaxial state of stress, let the three principal stresses be s1, s2 and shear stress is obtained as half of the difference between two principal stresses. Hence, U3

1 |(T 1  T 2 )| 2

U1

1 |(T 2  T 3 )| 2

U2

1 |(T 3  T 1 )| 2

97

s3. The

or (3.2)

or

The maximum shear stress is the largest of (t1, t2, t3). If the order of principal stresses is s1 > s2 > s3, the maximum shear stress can be obtained from the maximum and minimum principal stresses. Hence 1 |(T 1  T 3 )| 2

U max

(3.3)

Now consider on element of the material from uniaxial tensile test subjected to stress equals to the yield strength of the material. The stresses under uniaxial loading condition is

sxx = sy

syy = szz = txy = tyz = tzx = 0

Therefore, using the principal stress equations, these principal stresses are

T 1,2 

T xx 2

 T xx  2    U xy  2  2



Ty

Ty  2     0 2 2   Hence,

2

s1 = sy s2 = 0 and s3 = 0.

(3.4)

The maximum shear stress for the uniaxial state of stress is

tmax = =

1 | (T 1  0) | 2

Ty 2

(3.5)

Now if the principal stresses are unordered, yielding under multiaxial state of stress occurs for any one of the following conditions: |s1 –

s2| = ±sy |s2 – s3| = ±sy |s3 – s1| = ±sy

(3.6)

98

Strength of Materials

For two-dimensional state of stress (biaxial loading condition), let stress theory can be written from Eq. (3.6).

s2 = 0, the maximum shear

|s1| = ±sy |s3| = ±sy |s3 –

s1| = ±sy

(3.7)

The graphical representation of Eq. (3.7) is shown in Fig. 3.2.

Fig. 3.2

Representation of maximum shear stress yield criterion for plane stress.

This shows that for a biaxial state of stress the yield surface predicted by maximum shear stress yield criterion is an elongated hexagon in the s1 – s3 plane when s2 = 0. If three principal stresses s1, s2 and s3 are considered, the yield surface for the maximum shear stress criterion is a regular hexagon in principal stress space. This is shown in Fig. 3.3. To construct Fig. 3.3, let us consider three mutually perpendicular axes Os1, Os2 and Os3; each is considered for a particular principal stress.

Fig. 3.3

Stress space representation of yield criteria.

99

Chapter 3: Theory of Failure

If the principal stresses at a point on the body are s1, in Fig. 3.3, then we can write the coordinates of P as OP1 =

s2 and s3 and the point is denoted by P

s1, P1M = s2 and MP = s3

Now consider a line OH which is equally inclined to three principal axes. Hence, direction cosines of the line are nx =

1 3

, ny =

1 3

and nz =

1 3

PN is perpendicular from the point P on line OH. Hence, we can write ON = =

s1nx + s2ny + s3nz 1 3

(s1 +

s2 + s3)

(3.8)

and PN2 = OP2 – ON2 = (T 12  T 22  T 32 )  =

PN =

1 (T  T 2  T 3 )2 3 1

1 [(T  T 2 )2  (T 2  T 3 )2  (T 3  T1 )2 ] 3 1 1 3

(T 1  T 2 )2  (T 2  T 3 )2  (T 3  T 1 )2

(3.9)

(3.10)

PN is the perpendicular distance from the line OH. Remember Line OH is equally inclined to all three axes. The line PN can now be used to define the yielding. If PN is taken as radius, the every point on the radius define the state of stress as used for point P(s1, s2, s3). Hence, the yield surface is a cylinder of radius equal to PN and the centre of the cylinder is OH. The outer shape of the cylinder depends upon the yield criteria used to determine the yield surface. The plane that contains point P and is normal to the line OH is known as deviatoric plane. The line OH is called hydrostatic axis. This is shown in Fig. 3.4. Similar to point P, point Q differs with respect to hydrostatic stress component. But as we have already discussed that yield criteria are independent of hydrostatic stress. Hence, the deviatoric plane passing through point Q is same as the deviatoric plane passing through P. Hence, PN = QM All such planes when joined produced a solid cylinder, which defines the yield surface. The deviatoric plane for which s1 = s2 = s3 = 0 or hydrostatic stress defined as

sm =

1 (s + 3 1

s2 + s3) = 0

is known as pÿ (pi) plane. Figure 3.5 shows a p plane. The coordinates of point P shown in Fig. 3.3 are the projection lengths in Fig. 3.5(a). These lengths are OP1, P1M and MP. The lengths can be obtained by seting (s1, s2 = s3 = 0) and using Eq. (3.10).

100

Strength of Materials

Fig. 3.4 Deviatoric plane.

Equation (3.10) becomes 2 T 3 1

PN = OP1 = Similarly by setting (s2,

(3.11)

s3 = s1 = 0) and (s3, s1 = s2 = 0), we get 2

P1 M =

and

T3

(3.12)

T2

(3.13)

3 2

MP =

3

Now if the coordinates of point P are (x, y) with respect to OB and Os3 axes OX = x = OP1 cos 30° – MP sin 60° x=

=

2 3

T1 –

3 2



2 3

T2 –

3 2

T1  T 2

(3.14)

2

y = PX = P1M – OP1 sin 30° – PM cos 60° =

2 3

T3 

2 3

T 1 sin 30° 

2 3

cos 60°

Simplifying, we get y=

2T 3 

T 2  T1

(3.15)

6

Now the Tresca (maximum shear stress criterion) yield criterion on p plane is obtained as follows.

Chapter 3:

Theory of Failure

101

Fig. 3.5 Yield surface on p plane.

With reference to Fig. 3.5(b), the yield point along Os1 axis has a state of stress

s2 = s3 = 0 and s1 = sy (yield strength) This point represents the initiation of yielding and the length of the projections on the p plane 2/3 sy. This point is marked as A in Fig. 3.5(b). Let the state of stress in sector AOB be s1 ³ s3 ³ s2 and according to Tresca yield criterion, we have is

s1 – s2 = sy

(3.16)

102

Strength of Materials

Hence, the yield point along OB axes can be located at a distance

Ty

xB =

(3.17)

2

Equation (3.16) represents the yield locus in sector AOB is a line at a distance sy / 2 and parallel to Os3 axis. Similarly, we can draw other points and lines to represent the yield locus on p plane as shown in Fig. 3.5(b).

3.2.2

Maximum Distortion Energy Theory

This theory is also known as von Mises theory. It states that yielding occurs when the distortional strain energy at a point equals the distortional strain energy at yield under uniaxial tension or compression. It has already been explained that the total state of stress at a point given by the principal stresses s1, s2 and s3 can be decomposed into two parts consisting of hydrostatic state of stress and deviatoric state of stress. The strain energy associated with the deviatoric state of stress is known as distortion strain energy. Hence, the distortional strain energy is the energy associated with distortion or change in shape of the body. If s1, s2 and s3 are principal stresses and e1, e2 and e3 are principal strains, then the total strain energy per unit volume is given as UT =

1 2

s1e1 +

1 2

s2e2 +

1 2

s3e3

(3.18)

From the stress-strain relation given as

F1 = F2 = F3 =

and

T1 E

T2 E

T3 E



O E

(T 2 + T 3 )



O (T 1 + T 3 ) E



O (T 1 + T 2 ) E

(3.19)

Equation (3.18) can be simplified to UT =

1 2E

{T 12 + T 22 + T 32  2O (T 1T 2 + T 2T 3 + T 3T1 )}

(3.20)

The energy associated with hydrostatic state of stress s1 = s2 = s3 = p is obtained as  1  2O  2 UH =   (p)  E 

(3.21)

Taking p = 1/3 (s1 + s2 + s3), UH becomes  1  2O  2 UH =   (T 1 + T 2 + T 3 )  E 

(3.22)

Chapter 3:

Theory of Failure

103

The distortional energy per unit volume is the difference between energy from total and hydrostatic state of stress. Hence, UD = UT – UH =

1 2E

[T12 + T 22 + T 32  2O (T 1T 2 + T 2T 3 + T 3T 1 )]

 1  2O  2   (T 1 + T 2 + T 3 )  E 

(3.23)

[T 12 + T 22 + T 32 

(3.24)

Simplifying, we get UD =

1+O 3E

T1T 2  T 2T 3  T 3T1 ]

The state of stress at yield under uniaxial state of stress is s1 = sy, s2 = s3 = 0. Now the distortional energy per unit volume at yield under uniaxial tension or compression is U DO =

1+O 3E

T 2y

(3.25)

Now according to the statement of the theory UD = UDO After simplification of Eqs. (3.24) and (3.25), we get

T 12 + T 22 + T 32  T 1T 2  T 2T 3  T 3T 1 = T y2 or For plane state of stress

(s1 –

s2)2 + (s2 – s3)2 + (s3 – s1)2 = 2 T y2

(3.26) (3.27)

s2 = 0 and Eq. (3.27) becomes

T 12 + T 32  T 1T 3 = T y2

(3.28)

Equation (3.28) shows that von Mises criterion is an ellipse on s1 – s3 axes and the major and minor axes are along the perpendicular bisectors of the angle between s1 and s3. The variation is shown in Fig. 3.6.

Fig. 3.6 Mises yield locus for plane stress.

104

Strength of Materials

s1 = s3 = s (let) s = ±sy s1 = s, s3 = – s (let)

A point B From Eq. (3.28) At point D

T=

From Eq. (3.28)

(3.29)

Ty

(3.30)

3 For this state of stress, the maximum shear stress is U max

T1  T 3

=T 2 Therefore at yield s = tmax = ty, ty is the yield strength in shear. From Eqs. (3.30) and (3.31) ty =

D

=

1 3

(3.31)

sy = 0.577 sy

(3.32)

This value is 15% greater than the prediction made by Tresca criterion. The coordinates of point B in Fig. 3.6 are (sy, sy). Hence, length OB is 2 sy. Therefore, the major axis of the Mises ellipse is 2 2 sy. The minor axis of Mises ellipse lies along CD. The coordinates of point D are ( 1/ 3 sy, – 1/ 3 sy). Hence, minor axis CD is 2 (1/ 3

T y )2

+ (1/ 3

T y )2

= 2 2/3

Ty .

From Eq. (3.10) and (3.27), we can write PN =

2 3

sy

(3.33)

Hence, von Mises yield surface on s1–s2–s3 plane is a solid cylinder of radius 2/3 sy and on 2/3 sy. This is shown in Fig. 3.5(b). The stress space representation of both Tresca and von Mises yield criterion are shown in Fig. 3.7.

p plane it is a circle of radius

Fig. 3.7 Yield surface on principal stress space.

Chapter 3:

3.2.3

Theory of Failure

105

Strain Energy Density or Total Strain Energy Criterion

The strain energy density criterion proposed by Beltrami states that yielding occurs when the strain energy density at a point equals the strain energy density at yield in uniaxial tension or compression. If s1, s2 and s3 are three principal stresses, the total strain energy per unit volume is given as UT =

1 2E

[T12 + T 22 + T 32  2O (T1T 2 + T 2 T 3 + T 3T 1 )]

The state of stress at yield under uniaxial tension test is associated with above state of stress is thus

UTO =

(3.34)

s1 = sy, s2 = s3 = 0. The strain energy

T y2

(3.35)

2E

According to strain energy density criterion UT = UTO

T 12 + T 22 + T 32  2O (T 1T 2 + T 2T 3 + T 3T 1 ) = T y2

or

For two dimensional state of stress (plane stress), let

(3.36)

s2 = 0, (Eq. (3.36) reduces to

T 12 + T 32  2O T 1T 3 = T y2

(3.37)

The yield surface for the strain energy denstiy criterion is an ellipsoid in principal stress space and the shape depends upon the value of Poisson’s ratio n. For n = 0, Eq. (3.37) becomes

T 12 + T 32 = T 2y

(3.38)

and the ellipsoid reduces to a sphere with radius equal to the yield strength sy. The graphical variation of yield locus of strain energy density criterion on s1 – s3 plane is shown in Fig. 3.8 for various values of n.

Fig. 3.8

Strain energy density yield surface for plane stress state.

106

3.3

Strength of Materials

THEORY OF FAILURE OR YIELD CRITERION FOR BRITTLE MATERIALS

3.3.1

Maximum Principal Stress Criterion

The maximum principal stress criterion proposed by Rankine states that yielding occurs at a point when the maximum principal stress reaches the value equal to the maximum stress at yield under uniaxial tension or compression. Hence, according to this theory yielding occurs, when max |s1, If the order of the principal stresses is

s2, s3| = sy

(3.39)

s1 > s2 > s3, then according to the theory s1 = sy

(3.40)

However, brittle materials fail by fracture rather than yielding. Hence, the limiting stress under uniaxial condition should be taken as ultimate strength sut. Replacing sy by sut, the mathematical relation of the maximum principal stress is

or

s1 = sut s3 = sut

  

(3.41)

for plane stress case. The Eq. (3.41) is shown graphically in Fig. 3.9.

Fig. 3.9 Maximum principal stress theory.

The compressive strength and tensile strength of many brittle materials are different and compressive strength is more as compared to tensile strength. If the order of three principal stresses is s1 ³ s2 ³ s3, the maximum principal stress theory predicts the failure as

s1 ³ sut or s3 £ – – suc where sut is the ultimate strength in tension and graphical representation is shown in Fig. 3.10.

(3.42)

suc is the ultimate strength in compression. The

Chapter 3:

Theory of Failure

107

Fig. 3.10 Maximum principal stress theory.

3.3.2

Maximum Principal Strain Criterion

The maximum principal strain criterion known as St. Venant’s Criterion states that yielding occurs when the maximum principal strain equals the maximum principal strain at yield under uniaxial tension or compression. The principal strains in terms of principal stresses are

F1 =

F2 = F3 =

T1 E

T2



E

T3 E

O (T 2 + T 3 ) E





O E

(T 1 + T 3 )

O (T 1 + T 2 ) E

(3.43)

Assuming e1 ³ e2 ³ e3, the maximum principal strain is e1. e1 under uniaxial loading at yield becomes

F10 =

T1 E

=

Ty E

(3.44)

According to the yield criterion, we can write e1 = e10 \

s1 – n(s2 + s3) = sy

(3.45)

For s2 = 0 (biaxial loading), Eq. (3.45) reduces to

s1 – ns3 = sy

(3.46)

For unordered principal strains, the other possibilities of yieldings are

s2 – n (s1 + s3) = sy s3 – n (s1 + s2) = sy

(3.47)

The yield surface for the maximum principal strain criterion for a biaxial loading is shown in Fig. 3.11.

108

Strength of Materials

Fig. 3.11 Maximum principal strain yield locus for biaxial stress state.

3.4

MOHR’S THEORY

Mohr’s theory of failure is based on the Mohr’s circle of three simple test: tension, compression in uniaxial condition and pure shear. Let these tests were conducted on metal whose yield strength or ultimate strength is different in tension and compression and ultimate or yield strength in compression is more than the tensile strength. These three circles on s – t plane are shown in Fig. 3.12.

Fig. 3.12 Three Mohr’s circle for uniaxial compression, uniaxial tension and pure shear syc is the compressive yield strength, syt is the yield strength in tension.

The failure envelop is the line joining the three circles as shown by ABCDE above and below

s axis. In the initial stage, the theory is based on the fact that the line ABCDE may not be straight

one. Later on it was modified and known as Coulomb–Mohr theory or internal friction theory. This theory is based on straight boundary, i.e., line BCD is straight one. The mathematical derivation of the theory can be made assuming the principal stresses such that order of stresses is s1 ³ s2 ³ s3. Let C1, C2 and C3 be centre points of three circles, namely compression circle, tensile circle and general state of stress having largest and smallest stresses as s1 and s3 respectively. Points A1, A2 and A3 are tangent points on the circle. These points are shown in Fig. 3.13.

Chapter 3:

Fig. 3.13

Theory of Failure

109

Mohr’s circle.

Now, consider similar triangles OA1C1, OA2C2 and OA3C3. From the properties of similar triangles, we can write A3C3  A2 C2 OC3  OC2

Now from Fig. 3.13, A3C3 = radius of middle circle =

T1  T 3

2 A2C2 = radius of tensile cirlce

=

T yt

2 A1C1 = radius of compression circle =

T yc 2

OC3 – OC2 = C2C3 = C2S + C3S = C2S + C3s1 – Ss1 = =

T yt 2

+

T yt 2



T1  T 3 2

T1 + T 3 2

OC1 – OC2 = C1C2 = C1S + C2S =

T yc 2

+

T yt 2



T1

=

A1C1  A2C2 OC1  OC2

(3.48)

110

Strength of Materials

Now simplifying Eq. (3.48), we get

T1  T 3 T yt 2

2 

T yt



2 =

T yc 2

T1 + T 3

T yc

2

2

 +

T yt 2

(3.49)

T yt 2

Simplifying Eq. (3.49), we get

T1 T  3 =1 T yt T yc

(3.50)

Equation (3.50) can also be used for ultimate strength. The failure equation in terms of ultimate strength in tension and compression is

T1 T  3 =1 T ut T uc Let now for plane stress case two non zero stresses are

(3.51)

s1 and s3.

(i) If s1 ³ s3 ³ 0, s1 is the maximum principal stress and minimum principal stress is zero. Hence, Eq. (3.51) reduces to following failure condition

s1 = sut

(3.52)

(ii) If s1 ³ 0 ³ s3, the maximum and minimum principal stresses are equation is same as given in Eq. (3.51). (iii) If 0 ³ s1 ³ s3. In this case the failure equation becomes

s1 and s3 and failure

s3 = – suc Figure 3.14(a) shows the Coulomb–Mohr theory of failure in

(3.53)

s1–s3 plane.

Fig. 3.14(a) Coulomb–Mohr theory of failure.

Chapter 3:

Theory of Failure

111

The yield surface for Mohr–Coulomb criterion has the form of an irregular hexagonal pyramid in s1 =ÿs2 =ÿs3 axis. The axis of the pyramid is the hydrostatic axis. This is shown in Fig. 3.14(b) in principal stress space and p plane.

Fig. 3.14(b) Mohr-Coulomb yield surface in (i) principal stress space, (ii) intersection with the p plane.

3.5

EXPERIMENTAL VERIFICATION OF THEORY OF FAILURE

The most common method of experiment is the use of thin walled tube subjected to combined stress as shown in Fig. 3.15.

Fig. 3.15

Thin walled tube under T and F.

The principal stresses are

T1 =

T3 = and

s2 = 0.

T

T  +   + U2 2 2

T 2

2

T  2   +U 2   2



(3.54)

112

Strength of Materials 2 T  T1  T 3 = 2   + U 2 2

Thus, and Tresca criterion becomes

(3.55)

s1 – s3 = sy T  2   + U2 = Ty 2 2

Squaring and simplifying, we get

T   Ty

2

2

  U   +   =1  (T y /2)  

(3.56)

von Mises criterion yields (s1 –

s2)2 + (s2 – s3)2 + (s3 – s1)2 = T 2y

After substituting the values of

s1, s2 and s3 and on simplification, it gives T   Ty

2

2

 U    =1  +   (T y / 3)  

(3.57)

The variation of Eqs. (3.56) and (3.57) is shown in Fig. 3.16.

Fig. 3.16

Experimental verification of Mises and Tresca criterion.

Most of the experimental data falls between the two ellipse shown in Fig. 3.16, but the maximum data points inclined towards the Mises ellipse. Experimental data points from Ni-Cr-Mo steel, 2024

Chapter 3:

Theory of Failure

113

T4 Al alloys, AISI 1023 steel shows that either maximum shear stress theory or von Mises theory are suitable for modelling failure of the material which fails in ductile manner. The material whose behaviour in tension and compression is different and their strengths are different in tension and compression, Mohr’s theory is most suitable. However, Mohr’s theory of failure requires three test, i.e., in tension, compression and shear which are difficult to conduct and expensive, hence modified Mohr’s theroy known as Coulomb–Mohr theory, which requires only tensile and compressive yield strength can be applied with accuracy. Failure of gray cast iron is shown in Fig. 3.17 along with maximum principal stress theory and Coulomb–Mohr theory. Experimental data shows that when the state of stress falls in the first quadrant, i.e., for s1, s3 ³ 0, both theory predict the same failure and data are very close to the prediction. Hence, under such state of stress maximum principal theory is easy to use and hence most suitable. However, when the state of stress falls in the fourth quadrant Coulomb–Mohr theory suits best as compared to maximum principal stress theory.

Fig. 3.17

3.5.1

Comparison of Failure Criteria

The variation of all theories of failure for plane stress situation is shown in Fig. 3.18. Figure 3.18 shows that when the state of stress falls on first quadrant the failure prediction made by different theories is quite close to each other when maximum discrepancy is seen for fourth quadrant. It also indicates that when the minimum and maximum principal stresses are same, i.e., s1 = s3 = s, and s ³ 0, all theories predict the same result. On the other hand, if the material is subjected to pure shear t, then the theories have the largest discrepancy in the prediction of failure. Hence, special care must be taken while predicting the failure loads in pure shear from tension test

114

Strength of Materials

Fig. 3.18

Comparison of different theory of failure.

data. Table 3.1 shows the prediction made by different theories and the relation between yield strength in tension and shear. Table 3.1

Comparison of theory of failure

(a) Yield criterion

Maximum principal stress theory

(b) Predicted maximum utilizable value as obtained from tensile test/comparison test

smax = syt

(c) Predicted maximum utilizable value as obtained from torsion test

smax =

ty

Relation between syt and ty

ty =

syt

Maximum principal strain theory

emax =

syt /E

emax = (5/4) ty /E

ty = (4/5) syt

Maximum shear stress (Tresca) theory

tmax =

syt /2

tmax = ty

ty = 0.5 syt

Distortion energy (von Mises) theory

UD = s 2yt /6G

UD = ty/2G

ty = (1/ 3 ) syt

Chapter 3:

Theory of Failure

115

Design equations Theory of failure discussed in the previous sections is used to determine the limiting condition of component failure. The equations can be modified for design purposes by reducing the strength values associatyed with the failure equation. The commonly used failure theories for design purposes are as follows: Ductile materials 1. Maximum shear stress theory: | s1 –

s3 | =

| s1 –

s2 | =

and | s3 –

s2 | =

or

Ty N

Ty N

if

s1 ³ s2 ³ s3

if

s1 ³ s3 ³ s2

Ty

if s3 ³ s1 ³ s2 N 2. von Mises theory or distortion energy theory

T 12 + T 32  T 1T 3 =

Ty N

3. Strain energy density or total strain energy theory

T 12 + T 32  2O T 1T 3 =

Ty N

4. Mohr’s theory

T1 T 1  3 = T yt T yc N Brittle material 1. Maximum principal stress theory:

s1 = or

s2 =

N

T ut N

if

s1 ³ s2 ³ s3

if

s2 ³ s3 ³ s1

T ut

if s3 ³ s1 ³ s2 N 2. Maximum principal strain theory: or

s3 =

T ut

s1 – ns3 = or

s1 – ns3 =

Ty N

T ut N

if s1 ³

s3 and s2 = 0

116

Strength of Materials

3. Coulomb–Mohr theory

s1 ³ s3 ³ 0

if if

s1 ³ 0 ³ s3 if 0 ³

s1 ³ s3

s1 =

T ut N

T1 T 1  3 = T ut T uc N | s3 | £

T uc N

In all above equations N is the factor of safety.

EXAMPLE 3.1 In a steel member, at a point the major principal stress is 180 MPa and minor principal stress is compressive. If the tensile yield point of the steel is 225 MPa, find the value of minor principal stress using different yield criteria at which yielding will commence. Take Poisson’s ratio = 0.26. Solution:

s1 = 180 MPa sy = 225 MPa

Major principal stress Yield stress Maximum shear stress criteria:

s1 – s2 = sy s2 = s1 – sy = 180 – 225

s2 = – 45 MPa (compressive) Maximum strain energy criteria:

s 12 + s 22 – 2ns1s2 = T y2 (180)2 +

T 22 – 2 ´ 0.26 ´ 180 ´ s2 = (225)2 T 22 – 93.6s2 – 18225 = 0

93.6 “ (93.6) 2  4 – 18225 2 = – 96.08 MPa (compressive)

s2 =

Maximum distortion energy criteria:

T 12  T 22  T 1T 2 = T y2 (180)2 +

T 22 – 180 ´ s2 = (225)2

T 22 – 180s2 – 18225 = 0

Chapter 3:

Theory of Failure

117

2 s2 = 180 ± (180)  4 – 18225

2 = – 72.25 MPa (compressive)

EXAMPLE 3.2 A cylindrical shaft made of steel of yield strength 700 MPa is subjected to static load consisting of bending moment 10 kN·m and a tensional moment 30 kN·m. Determine the diameter of shaft using different theories of failure, and assuming a factor of safety 2. Take E = 210 GPa, Poisson’s ratio = 0.25. Solution:

Given

sy = 700 MPa, M = 10 kN·m, T = 30 kN·m, FS = 2, E = 210 GPa, n = 0.25

Then section modulus of shaft Z =

sb =

Bending stress

=

=

ts =

Shear stress

=

Q 32

d 3 = 0.0982d3 mm3, FS = 2

M Z 10 – 106 0.0982d 3 101.8 – 106 d3

MPa

16T

Qd3 16 – 30 – 106

Q d3

=

152.8 – 10 6 d3

MPa

We know that the maximum principal stress is given by

s1 =

T xx  T yy

È T  T yy Ø 2  É xx ÙÚ  U xy 2 Ê 2

2

sxx = sb and txy = ts and syy = 0 Now,

s1 =

=

101.8 – 10 6 2d 3 50.9 – 10 6 d

3





È 101.8 – 10 6 Ø É Ù 2d 3 Ê Ú

161 – 10 6 d

3

2

È 152.8 – 10 6 Ø É Ù d3 Ê Ú

211.9 – 106 d3

MPa

2

118

Strength of Materials

The minimum principal stress is:

s2 = =

T xx  T yy

È T  T yy Ø 2  É xx ÙÚ  U xy 2 Ê 2

2

50.9 – 106

161 – 106



d3



d3

110.1 – 106

MPa

d3

According to maximum shear stress theory,

Ty 1 (T  T 2 ) = 2 1 2 FS 161 – 10 6

or

d

or

3

700 2–2

=

d = 97.2 mm

According to maximum strain energy theory,

T  T  2OT 1T 2 = 2 1

or

È É Ê

211.9

–

10 6 Ø Ù Ú

d3

2

È  É Ê

110.1 – 10 6 Ø Ù Ú

d3

44902 – 1012

or

2 2

d6



2



2

–

0.25

12122 – 1012 d6

–



ÈTy Ø É FS Ù Ê Ú

211.9

–

2

10 6

d3

11665 – 1012 d6

– 

110.1 – 10 6 d3

È É Ê

700 Ø 2 ÙÚ

2

122500

68689 – 1012

or

= 122500 d6 d = 90.8 mm

or

EXAMPLE 3.3 A shaft of circular cross section having diameter 100 mm is subjected to static bending moment M = 10 kN·m, static torque T = 30 kN·m and axial load of 50 kN. If the yield strength of the shaft material is 750 MPa, determine the design factor of safety using (a) maximum shear stress theory (b) von Mises theory Solution: Given

M = 10 kN·m = 10 ´ 106 N·mm T = 30 kN·m = 30 ´ 106 N·mm P = 50 kN = 50 ´ 103 N d = 100 mm sy = 750 N/mm2

Chapter 3:

Theory of Failure

119

Calculation of elemental stresses due to bending moment M

Tb = =

32 M

Q d3 32 – 10 – 106

Q – (100)3

= 101.859 MPa

Stress due to axial tensile force F = 50 ´ 103 N

Ta = =

4F

Qd2 4 – 50 – 103

Q – (100)2

= 6.366 MPa

Stress due to Torque T = 30 ´ 106 N·mm U =

=

16 T Qd

3

16 – 30 – 10 6

Q – (100)3

= 152.789 MPa

The stresses due to M, T and F on an element is shown in Fig. 3.19.

Fig. 3.19

The maximum normal stress is the addition of stresses due to axial force and bending moment.

s = sxx = sa + sb = 6.366 + 101.859 = 108.225 MPa Shear stress t = 152.789 MPa.

120

Strength of Materials

Calculation of principal stresses

T 1,3 =

=

T 2

ÈT Ø É 2Ù Ê Ú

“

108.225 2

2

+ U2

È 108.225 Ø É Ê 2 ÙÚ

“

2

+ (152.789)2

= 54.1125 ± 162.088

s1 = 216.20 MPa s3 = –107.9755 MPa s2 = 0

\

(a) Maximum shear stress theory gives design factor of safety by

T1  T 3 = \

N=

=

Ty N

Ty T1  T 3 750 216.20 + 107.9755

= 2.314

(b) von Mises theory

T 12 + T 32  T1T 3 = N=

or

=

=

Ty N

Ty T12 + T 32  T 1T 3 750 (216.2)2 + (107.9755)2  216.2 – (  107.9755)

750 285.912

= 2.623

EXAMPLE 3.4 A 30 mm diameter rod made of a ductile material with a yield strength of 200 MPa is subjected to bending moment of 150 N·m. A torque is then gradually applied. Determine the value of the torque when the rod begins to yield using Tresca theory. Solution:

Given M = 150 N·m, d = 30 mm,

sy = 200 MPa, T = ?

Chapter 3:

Theory of Failure

121

Calculation of elemental stresses Bending stress due to moment M = 150 N·m

Tb = =

32M

Qd3 32 – 150 – 1000

Q – (30)

= 56.588 N/mm 2

3

Shear stress due to Torque T U =

=

16 T Qd

3

16 T

Q – (30)3

= 1.8896 – 10  4 T

Calculation of principal stresses

T 1,3 =

T 2

“

56.588

=

2

ÈT Ø É 2Ù Ê Ú

“

2

+ U2

È 56.588 Ø É 2 ÙÚ Ê

2

+ (1.886

–

10  4 T )2

According to maximum shear stress theory

s1 – s3 = sy 2 Ë (28.294)2 + (1.886 ÌÍ

–

10  4 T )2 Û = 200 ÜÝ

Squaring and solving for T T = 508.56 N·m

EXAMPLE 3.5 A part made of cast iron is subjected to following state of stresses. Determine if the component will fail. Assume the ultimate strength in tension as 280 MPa and ultimate strength in compression as 410 MPa. The state of stresses are (i) sxx = 200 MPa, syy = –120 MPa, txy = 100 MPa (ii) sxx = –200 MPa, syy = –120 MPa, txy = 50 MPa (iii) sxx = +200 MPa, syy = +200 MPa, txy = 100 MPa Solution: The component is made of cast iron. Hence, the suitable theory of failure is either maximum principal stress theory or Coulomb-Mohr theory. (i) sxx = 200 MPa, syy = – 120 MPa, txy = 100 MPa

122

Strength of Materials

The principal stresses are

T1,3 =

=

T xx + T yy

“

2 200  120 2

È T xx É Ê

“

 T yy Ø

2 + U xy

Ù Ú

2

È 200 É Ê

2

+ 120 Ø

2

+ (100)2

Ù Ú

2

= 40 ± 188.679

s1 = 228.679 MPa s3 = – 148.679 MPa s2 = 0 (a) According to maximum principal stress theory failure occurs if

s1 (maximum value) ³ sut According to the result we have 228.679 < 280 MPa Hence, the component will not fail. (b) According to Coulomb–Mohr theory: The calculated stresses are s1 > s2 >

s3

Hence, the failure equation is

T1 T  3 =1 T ut T uc LHS =

228.679



280

(  148.679) 410

= 1.179 Hence, 1.179 > 1, Hence, the theory shows that component will fail for the given state of stress. (ii) The principal stresses for given state of stress are

T 1,3 =

200  160 2

“

È  200 + É 2 Ê

= – 160 ± 64.03

s1 = – 95.97 MPa s3 = 224.03 MPa s2 = 0 The order of the stresses is s2 > s1 > s3.

\

120 Ø Ù Ú

2

+ (50)2

Chapter 3:

Theory of Failure

123

According to Coulomb–Mohr theory, failure occurs for the condition of In the present case

s3 = – 224.03

s3 = – suc

T3  224.03 = = 0.546 < 1 T uc  410

or

Hence, the component will not show yielding for the given stresses for given state of stress. (iii) The principal stresses for the given state of stresses are

T 1,3 =

+200 + 200 2

“

È 200 É Ê

 200 Ø 2

Ù Ú

2

+ (100)2

= 200 ± 100

s1 = 300 MPa s3 = 100 MPa s2 = 0 The order of stresses is s1 > s3 > s2. The state of stress falls on the first quadrant. The \

Coulomb–Mohr theory for such state of stress is

s1 (maximum value) = sut As s1 = 300 MPa is more than sut = 280 MPa, failure will occur. 3.6

THEORY OF FAILURE FOR CYCLIC LOADS

Theories of failures discussed so far are applicable for static loads. However, the magnitude of a load on a component may vary during life span. The loads which vary between the two limits with respect to time are known as cyclic loading. The kinds of stresses produced due to fluctuation of values are known as repeated, alternating or fluctuating stresses. Under these conditions a component fails at a load level much below the yield strength of the material. This phenomenon of decreased resistance of material to fluctuating stresses is called fatigue.

3.6.1

Stress Parameters

Under cyclic loads, the magnitude of the stress varies between the two extreme limits. These limit stresses are known as maximum stress and minimum stress. The failure equations under cyclic loads are based on minimum and maximum stress amplitudes. The maximum stress and minimum stress as shown in Fig. 3.20.

smax = greatest algebraic stress in a stress cycle smin = smallest algebraic stress in a stress cycle Stress cycle = smallest portion of the stress history which is repeated periodically and identically Mean stress = algebraic mean of maximum and minimum stresses of a cycle. It is denoted by sm then

Tm =

T max + T min 2

124

Strength of Materials

Fig. 3.20 Variation of stress with time.

Variable stress or stress amplitude: It is the half of the algebraic difference between maximum and minimum stress cycles. It is denoted by sa. Hence,

Ta = 3.6.2

T max  T min 2

Strength Parameter

Endurance limit or fatigue strength is taken as limit stress for cyclic load. Endurance limit or strength is defined as the maximum reversed stress which may be repeated indefinite number of times (>= 107) on a standard polished specimen in reversed bending without failure. It is determined from a S–N tests. S–N test is also known as rotating bending fatigue tests where a specimen of standard size and polished surface finish is subjected to completely reversed bending load till failures. The number of times the load is repeated before failure is known as the number of cycles to failure and denoted as N and the corresponding stress is known as fatigue stress and denoted as S. S–N curve as shown in Fig. 3.21 are obtained from number tests at different stress amplitudes. The strength corresponding to 107 cycles is known as endurance strength or fatigue limit. In the absence of S–N data, endurance strength is obtained from the following empirical relations. Steel material Cast steel Gray cast iron Malleable cast iron Aluminium and magnesium alloys Copper alloy

Se* = = * Se = Se* = Se* = Se* = Se* =

0.5 ´ sut if sut £ 1400 MPa 700 MPa if sut > 1400 MPa 0.4 ´ sut 0.35 ´ sut 0.4 ´ sut (0.25–0.5) sut (0.3–0.4) sut

Chapter 3:

Theory of Failure

125

Fig. 3.21 S–N curve.

Modifying factors for fatigue strength The empirical relations given above are based on standard geometry of circular cross section and diameter 7.6 mm, mirror polished surface finish and completely reversed bending load. However, in actual component these conditions may differ. It is well known that fatigue strength is highly influenced by all these factors. Hence, the endurance strength obtained from these empirical relations must be modified before using in fatigue failure theories. The modifying factor for size, surface finish and loading can be obtained from the following empirical relations.

Surface finish factor, k

a

The surface finish factor is defined as the ratio between the endurance strength obtained with arbitrary surface finish and that obtained with standard mirror polished. The regression equation from the variation of surface finish factor with ultimate tensile strength for different surface conditions is given as ka = a(sut)b where ka is surface finish factor and ultimate strength is in MPa. Table 3.2 shows the values of regression coefficient of equation. Table 3.2 Values of regression coefficients of equation Surface finish

As per Shigley a, (MPa)

Ground Machined or cold rolled Hot rolled As forged

1.58 4.45 56.10 271.0

b – – – –

0.086 0.265 0.719 0.995

a, (MPa) 1.58 4.51 57.7 272

b – – – –

0.085 0.265 0.718 0.995

Size factor, kb The size factor is defined as the ratio of the endurance strength obtained from any arbitrary sized specimen to the endurance strength obtained from standard size specimen (diameter 7.6 mm). Based

126

Strength of Materials

on the experimental observation following data may be used as guide map to select the size factor (Table 3.3). Table 3.3 Size factor Size of the part

Axial loading

Bending

Torsion

£ 7.6 mm > 7.6 mm

1.0 0.9

0.7 to 0.9 0.7 to 0.9

1.0 0.9

Size factor, kb

Based on the experimental data set the size factor under bending and torsion is given as kb = 1.24d–0.107

or

 d     7.62 

 0.107

2.79 £ d £ 51 mm 51 < d £ 254 mm

= 0.859 – 0.000837d

where d is the diameter of circular cross section in mm. For the other cross section d should be replaced with the equivalent diameter de. The equivalent diameter for rectangular section is de = 0.808 hb . For axial load there is no size effect. Hence, kb = 1 for axial load.

Load factor, k

c

The load factor is defined as the ratio of the endurance strength obtained from any arbitrary load to the endurance strength obtained from completely reversed bending load. The following guide map can be used for selection of load factor. For bending load the endurance strength modifying factor is unity. Table 3.4 shows the load factor. Table 3.4

Loading factor

Axial loading

Bending

Torsion

0.8

1.0

0.55

Load factor, kc

The loading factor can also be obtained from the following regression models. For torsional load kc = 0.258 (sut)0.125 For axial load kc = 1.434 (sut)– 0.0783

The endurance strength is now obtained from the relation Se = Se* ´ ka ´ kb ´ kc

Fatigue reduction factor, k

f

Because of any irregularities in the component or specimen stress concentration exits near the discontinuity zone and stresses at this zone are higher than the nominal stress. These irregularities

Chapter 3:

Theory of Failure

127

may be due to the presence of holes, keyways, slots, shoulder, or even any inclusions or impurities in the material. When there is any change in the cross section the distribution of stress does not remain uniform throughout and the basic relations are unable to describe the state of stress at those points. The actual state of stress at a point in a mechanical part is obtained from the product of the nominal stress and the stress concentration factor at that point. The stress concentration factor is defined as the ratio of the maximum stress to the nominal stress. The stress concentration factor kt is thus kt =

Actual stress at a point Nominal stress at a point

Stress concentration is a highly localized effect. The variation in the stress distribution exists only in a very small region in the vicinity of the discontinuity. This region is known as zone or area of stress concentration. The stress raisers have a very little effect on the material with internal irregularities such as gray cast iron, regardless of type of loading. For static loading the stress concentration factor is important for those materials which behave in a brittle manner. In case of ductile material the load applied to the member causes yielding at the discontinuity. The yielding relieves the effect of stress concentration. However, for brittle material the stress relieved is negligible. Engineering materials having some ductility behave in a ductile manner, it is not necessary to use a stress concentration factor. Stress concentration must be considered when the part is made of brittle materials or when they are subjected to fatigue loading. Even under these conditions if a material is insensitive to notch or irregularities, or are not so sensitive to the existence of the notches or discontinuities, full value of the stress concentration factor need not be used. For those materials, it is necessary to use a reduced value of stress concentration factor, kt. The reduced value of stress concentration factor is defined as fatigue stress concentration factor which is the ratio of the endurance limit of the notch free specimen to the endurance of the notched specimen. kf =

Endurance strength of notch free specimen Endurance strength of notched specimen

This is used for brittle materials under static loading also. Sometimes it is known as fatigue strength reduction factor. The fatigue reduction factor and stress concentration factor are related through notch sensitivity and is given as q=

kf  1 kt  1

Based on the experimental results from steel and aluminium material, the Neuber equation for notch sensitivity parameter q is q=

1 a 1+   r 

where r = notch root radius, mm a = material constant which depends on the strength and rigidity.

128

Strength of Materials

Table 3.5 presents material constant a of equation. Table 3.5 Material constant a of equation (a, mm) Material

Axial and bending load

Torsional load

Steel, annealed and normalized Steel quenched and tempered Aluminium, sheet

0.25 0.06 0.5–1.25

0.15 0.036 0.3–0.75

Aluminium, bar

0.08–0.2

0.048–0.12

Note:

‘a’ values for torsional load are obtained by multiplying 0.6 with the values for axial/bending load.

Hence, the fatigue reduction factor is obtained as kf = 1 + q(kt – 1)

Failure theory for fluctuating stresses The following are three theories of failure for cyclic loading. (a) Gerber method (b) Goodman diagram (c) Soderberg diagram

The Gerber Equation The Gerber equation is based on the equation of parabola. Gerber (1874) suggested that the failure curve under fluctuating state of stress is of parabolic nature passing through the endurance strength on the y-axis (variable stress axis) and ultimate strength on mean stress axis (x-axis). The equation of parabola in these two coordinate system is

sa = B – as 2m

Fig. 3.22

Gerber failure locus.

where constants B and a can be obtained from boundary conditions. When When

sm = 0 ÿ ÿ sa

=0

sa = Se sm

=

sut

(3.58) (ultimate strength)

(3.59)

Chapter 3:

From Eq. (3.58)

B = Se

From Eq. (3.59)

0 = B – as 2ut

or

Se – as 2ut = 0

Therefore,

a=

Theory of Failure

129

Se

T ut2

The equation of failure, according to Gerber is

T T a = Se  Se  m  T ut

2

  

 T  T a = Se 1   m    T ut 

2



  

For safe line, all mechanical properties must be divided by factor of safety n.

Ta =

2 Se   nT m   1      n   T ut   

Incorporating fatigue reduction factor we can write the safe design equation as kf 

Ta

  nT S  = e 1   m  T ut n   

   

2

  

The Goodman diagram Goodman states that the failure locus is defined as the straight line joining the endurance limit on the variable stress axis to the ultimate strength on the mean stress axis. Any combinations of mean and variable stress falling below the triangular area bounded by Goodman line allow indefinite or infinite life. Figure 3.23 shows a Goodman line.

Fig. 3.23

Goodman line.

130

Strength of Materials

Let D state of stress with (sm,

sa). DE = sa and CD = OE = sm

From similar triangles ABO and ADE

OA OB or

OA OB

EA

=

DE OA  OE

=

DE

Substituting the values of OA, OB, DE

T ut Se

=

T ut  T m Ta

Simplifying, we get

Ta Se

+

Tm =1 T ut

Now applying stress concentration factor kf and factor of safety n to failure equation safe Goodman equation becomes, kf  Se

Ta

+

Tm 1 = T ut n

The Soderberg criterion The Soderberg line differs from the Goodman line only in that it extends from the endurance strength on the variable stress axis to the yield strength instead of ultimate strength on the mean stress axis. Figure 3.24 represents Soderberg diagram.

Fig. 3.24 Soderberg diagram. Note: An alternate method is discussed below to derive the Soderberg equation. The procedure as explained in the case of Goodman equation can also be applied to derive the Soderberg equation.

Chapter 3:

Theory of Failure

The equation of a straight line passing through endurance strength Se on the strength on sm axis is given as

sa = A + Bsm

131

sa axis and yield (3.60)

where constants A and B are evaluated from the conditions that

sa = Se when sm = 0 sm = sy when sa = 0

(3.61) (3.62)

Substituting Eqs. (3.61) and (3.62) in Eq. (3.60) A = Se

B= 

Se

Ty

Thus, putting values of A and B in Eq. (3.60)

Se

T a = Se 

Ty Ta

or

Se



T m = Se  1  

Tm   T y 

Tm =1 Ty

+

(3.63)

Now if fatigue strength reduction factor kf and factor of safety n are introduced, Eq. (3.63) becomes

kf

Ta Se

+

Tm 1 = Ty n

The above equations can be extended to alternating stress in torsion also. In such a case the stresses are to be replaced by shear stress. Under torsional stress the design equations are Gerber equation

n  k fs  U a Ses Um

Goodman equation

Uu

k fs

Soderberg equation

+ k fs Ua

Ses

+

2

 nU  +  m =1  Uu  Ua

Ses Um Uy

=

=

1 n

1 n

where Um = Ua =

U max + U min

2 U max  U min

2

= mean shear stress = variable shear stress

132

Strength of Materials

As the experimental data for endurance strength in shear, yield strength in shear and ultimate shear strength is limited; it may be estimated from von Mises relation or maximum shear stress criterion. According to von Mises condition of failure Se

Endurance strength in shear Ses = Shear yield strength U y =

3

Ty 3

and ultimate strength in shear U u =

T ut 3

According to Mohr’s circle of failure or maximum shear stress criterion, Endurance strength in shear Ses = Shear yield strength U y =

Se 2

,

Ty 2

and ultimate strength in shear U u =

T ut 2

Fatigue under complex loading (plane stress case) When different nature of loads such as bending and torsion or axial and torsion exist, the fatigue failure equations are modified and used along with static theory of failures. Soderberg Criterion with theory of failure When a component is subjected to bending load or axial load, the Soderberg criterion can be written as

Ty n

= Tm + k f 

Ty    Se 

Ta 

The left-hand side of equation, i.e. sy /n is based on static strength and, hence, is known as “static equivalent normal stress” sen. \

 Ty    Se 

T en = T m + k f  T a 

If the structure or machine component is subjected to variable torsional load, then Soderberg criterion becomes

 Uy  = U m + k fs  U a   n  Ses 

Uy

where ty = 0.577sy from distortion-energy theory and Ses = kc ´ Se

Chapter 3:

Theory of Failure

133

The equivalent static shear stress is

 Uy    Ses 

U es = U m + k fs  U a 

The failure theory according to maximum shear stress theory is

Ty 2n

 T en  2   + U es 2   2

=

Similarly, other criteria can be modified and any theory of failure can be applied considering sen and tes as normal and shear stress component. The results are summarized in Table 3.6. Table 3.6

Equivalent normal and shear stresses

Theory

Equivalent normal stress

Soderberg

T en = T m + kf  T a 

Goodman

T en = T m + kf  T a 

Gerber

T en = T m 2 +

Equivalent shear stress

 Ty    Se 

U es = U m + k fs  U a 

 T ut    Se 

U es = U m + kfs  U a 

 Uy    Ses   Uu    Ses 

2 kf  T a  T ut   Se n 

   

U es =

U m 2

+

k fs  U a  U u2   Ses n 

   

EXAMPLE 3.6 A steel machined shaft of diameter 30 mm is subjected to reversed bending moment. The endurance strength and ultimate strength of steel are 300 MPa and 780 MPa respectively. If the stress concentration factor and notch sensitivity are 1.8 and 0.82 respectively, estimate the maximum bending moment for which the shaft will have indefinite life. Solution: Given Se = 300 MPa, sut = 780 MPa, kt = 1.8, q = 0.82 Due to bending moment, the induced stress on the shaft is

T= The maximum stress is

smax =

32 M max

Q d3

32 M

Qd3

.

The bending moment is completely reversed in nature. Hence, Mmax and Mmin are the same but opposite in nature. Hence, the minimum stress is

T min = 

32 M max

Qd3

134

Strength of Materials

\ Stress amplitude or variable stress is ÿ ÿ ÿ ÿ ÿ sa

sm

(smax –

2

smin

)

32 M max

=

Mean stress

1

=

Q d3

is ÿ sm

1

=

2

(smax +

smin

)

=0 Fatigue reduction factor kf kf = 1 + q(kt – 1) = 1 + 0.82 (1.8 – 1) = 1.656 Using Goodman equation of failure, kf

or

Ta Se

kf 1.656 

M max =

+ sa

Tm =1 T ut = Se

32 M max

Q d3

300  Q  (30) 3 32  1.656

= 300

= 480  10 3 N.mm

Mmax = 480 N·m

EXAMPLE 3.7 Estimate fatigue strength of rotating beam specimen of diameter 30 mm and made of steel having ultimate strength 720 MPa. The surface of the specimen is machined one. Solution:

Given d = 30 mm,

sut

= 720 MPa

From empirical relation the uncorrected endurance strength S e* is S e* = 0.5sut = 0.5 ´ 720 = 360 MPa Modifying factors Surface finish factor ka = asutb For machined surface we have ka = 4.51 (sut)– 0.265

Chapter 3:

Theory of Failure

135

ka = 4.51 ´ (720)– 0.265 = 0.789 kb = 1.24(d)– 0.107

Size factor

= 1.24 ´ (30)– 0.107 = 0.862 Load factor kc = 1 (bending load) \ Endurance strength Se = S e* ´ ka ´ kb ´ kc = 360 ´ 0.789 ´ 0.862 ´ 1 = 244.84 MPa

EXAMPLE 3.8 A transmission shaft made of steel having ultimate strength 500 MPa and yield strength 320 MPa is subjected to fluctuating torque which varies from –75 N·m to 300 N·m. Neglecting any stress concentration effect, determine the diameter of the shaft. The corrected endurance strength is 175 MPa and factor of safety is 2.0. Solution:

sut = 500 MPa, sy = 320 MPa, Se = 175 MPa, Tmin = –75 N.m, Tmax = 300 N.m, n = 2.

Stress components The shear stresses due to torque are U max =

16 Tmax

Qd3

=+

=

U min =

16 Tmin

Qd3

16  300  103

Qd3

48  10 5

Q d3

= 

= 

16  75  103

Q d3 12  10 5

Q d3

The mean shear stress is ÿ ÿ ÿ ÿ ÿ tm

=

=

=

1 2

(tmax +

tmin

)

1  48  10 5 12  10 5   2  Q d 3 Qd3

18  10 5

Q d3

   

136

Strength of Materials

Variable stress ta is 1

=

ÿ ÿ ÿ ÿ ÿ ta

=

=

(tmax –

2

tmin

)

1  48  10 5 12  10 5  + 2  Q d 3 Qd3

   

30  10 5

Q d3

According to Soderberg criterion Um Uy

+

Ua

Se

1

=

n

According to von Mises criterion ty

= 0.577 sy = 0.577 ´ 320 = 184.64 MPa

\

18  10 5

+

Q d  184.64 3

3103.1

or

d

3

+

30  10 5

Q d  175 3

5456.7 d

3

=

=

1 2

1 2

Solving the shaft diameter is d = 25.77 = 26.0 mm

EXAMPLE 3.9 A machine shaft of diameter 30 mm made of steel is subjected to bending moment varies from –100 N·m to 320 N·m and torque + 50 N·m to 100 N·m, if the yield strength of the material is 370 MPa and corrected endurance strength under bending and torsional loads are 230 MPa and 160 MPa respectively, determine (a) the factor of safety guarding against fatigue failure, and (b) factor of safety guarding against static failure. Solution: Given Tmax = 100 N·m, Tmin = +50 N·m, Mmax = 320 N·m, Mmin = –100 N·m, sy = 370 MPa, Se(b) = 230 MPa, Ses = 160 MPa, d = 30 mm Computation of stresses Bending stress due to moment M

T max =

32 M max

Q d3

Chapter 3:

32  320  103

=

Q  (30)3

T min = Mean stress

sm = =

Variable stress

1 2 1 2

sa = =

(smax +

137

= 120.70 MPa

 32  100  103

Q  (30)3

Theory of Failure

=  37.73 MPa

smin)

(120.70 – 37.73) = 41.485 MPa 1 2

(smax –

smin)

1

(120.70 + 37.73) = 79.215 MPa 2 Shear stresses due to torque T Maximum shear stress tmax =

16 Tmax

Q d3

U max =

Minimum shear stress tmin =

16  100  103

Q  (30)3

= 18.863 MPa

16 Tmin

Qd3

U min =

16  50  103

Q  (30)3

= 9.43 MPa

Mean shear stress tm =

=

1 2 1 2

(tmax + tmin) (18.863 + 9.43) = 14.1465 MPa

Variable shear stress ta =

=

1 2 1 2

(tmax – tmin) (18.863 – 9.43) = 4.716 MPa

Factor of safety guarding against fatigue or static failure can be applied by determining equivalent stresses using any one of the fatigue failure theory and applying any theory of failure. Applying Soderberg criterion, the equivalent normal stress is

138

Strength of Materials

T en = T m +

Ty Ta

kf

Se

370  79.215  1

= 41.485 +

230

(k f = 1)

= 168.92 MPa Equivalent shear stress is U es = U m +

U y U a  k fs

Ses

= 14.1465 +

(0.577  370)  4.716  1 160

= 20.439 MPa The state of stress is thus 20.439

168.92

Fig. 3.25

\ The principal stresses are

T1,3 =

168.92 2

2



 168.92  2   + (20.439)  2 

= 84.46 ± 86.898 \

ÿ ÿ s1

= 84.46 + 86.898 = 171.358 MPa

ÿ ÿ s3

= 84.46 – 86.898 = –2.438 MPa

Applying von Mises theory the equivalent von Mises stress is

T von = (171.358)2 + (  2.438)2 + (171.358  2.438) Simplifying, we get svon

= 172.589 MPa

(a) Factor of safety guarding against fatigue failure can be obtained by equating \

nf =

Seb

T von

svon

to Seb.

Chapter 3:

=

230 172.589

Theory of Failure

139

= 1.33

Factor of safety against yielding, we get

Ty T von

ns =

=

370 172.589

= 2.144

EXERCISES 3.1 Define the yield criterion? 3.2 What is the necessity of theory of failure? Explain briefly theories of failure. 3.3 A shaft of diameter 100 mm is subjected to bending moment 10 kN·m and torque 15 kN·m. Find the factor of safety according to (a) maximum shear stress theory (b) distortion energy theory The elastic limit in simple tension is 300 MPa. 3.4 A circular cylindrical shaft is made of steel with a yield strength of 550 MPa. The shaft is subjected to static bending moment 12 kN·m and static torsional moment 25 kN·m. Determine the minimum shaft diameter. Take E = 207 GPa, n = 0.30. 3.5 A component made of brittle material is subjected to stresses sxx = 150 MPa, syy = 0 and txy = 60 MPa. If the ultimate strength of the material is 600 MPa, determine the factor of safety used in design. 3.6 A bar shown in Fig. 3.26 is made from a ductile material. The yield strength of the material is 390 MPa. The load F shown in the figure makes 30° with the positive x-axis. Determine the magnitude of F that will cause the yielding.

Fig. 3.26

4 4.1

Energy Methods

INTRODUCTION

The expressions of strain energy stored in a body are very useful in design and analysis of complex structures. In this chapter the formulations for different energy methods are discussed.

4.2

STRAIN ENERGY

Many practical engineering problems involve complex systems. The deflection analyses by geometric approaches or superposition techniques are not suitable for such complex systems. Energy approaches are sometimes more advantageous and less difficult to get a solution. A mechanical system is shown in Fig. 4.1. A force F is applied gradually at a particulate point Q on the structure. As force is gradually increased, the deflection produced in the direction of the applied force also increases. Also, the deflection produced is directly related to the applied load.

Fig. 4.1

A force system. 140

Chapter 4: Energy Methods

141

The force displacement diagram is shown in Fig. 4.2(a). The diagram depends on the relationship between F and d. The linear and nonlinear F–dÿ relations are shown in Figs. 4.2.

Fig. 4.2 Force displacement diagram.

The total deflection of point Q due to application of force F is the vector sum of vertical and horizontal deflections. The work done is given by E

W=



F dE

(4.1)

0

This indicates that the work done is the area under the force deformation curve. If the relation is linear (Fig. 4.2b), the work done is W =

1 2

FE

(4.2)

Now, consider a cubical element with dimensions (Dx, Dy, Dz) subjected to uniaxial force in the x-direction (Fig. 4.3).

Fig. 4.3

Cube under uniaxial loading.

142

Strength of Materials

The work done is 1 . 1 Fx . E x Fx E x = A'x 2 2 A 'x where A = Dy ´ Dz = cross sectional area W =

W=

1 Fx . E x 2 A 'x

W=

1 2

'y 'z 'x =

1 2

(4.3)

T xx F xx ( 'x 'y 'z )

T xx F xx dV

Now the work done per unit volume is determined by dividing the volume (DxDyDz) and denoting it by u

u=

1 T xx F xx 2

(4.4)

The work done per unit volume is known as strain energy. The strain energy can also be expressed as follows by substituting exx = sxx /E. u=

1 2E

T 2xx

(4.5)

The strain energy per unit volume for three-dimensional state of stress system with only normal forces can be written as u=

1 2

T xx F xx +

1 1 T yy F yy + T zz F zz 2 2

(4.6)

Substituting the stress strain relations, the strain energy per unit volume becomes u=

1 2 [T 2xx + T 2yy + T zz  2O (T xxT yy + T yyT zz + T zzT xx )] 2E

(4.7)

If the shear stress is to be included in the analysis, the strain energy due to shear stress sxy can be written as u=

1 2

T xyH xy

(4.8)

Hence, for general state of stress, the strain energy can be written as

u=

1 (T xx F xx + T yy F yy + T zz F zz + T xyH xy + T yz H yz + T zx H zx ) 2

(4.9)

Substituting stress strain relations u=

1 2 2 [T 2xx + T 2yy + T zz  2O (T xx T yy + T yyT zz + T zzT xx ) + 2(1 + O )(T 2xy + T 2yz + T zx )] (4.10) 2E

143

Chapter 4: Energy Methods

4.2.1

Strain Energy with Simple Loading

The total strain energy in axial loading (Fig. 4.4) is given by

Ua =

1 2

L

2

A, E

1 F   dV E  A

 0

L

As dv = dA . dx

Ua =

Hence,

Ua =

4.2.2

As dV = dA . dx

As

U=

U =



1 F2 dx E A

 0

Fig. 4.4 Uniaxial loading

1 E2 L 2 AE

(4.11)

2



1 M   y  dV 2E  I 



1 M 2   (y dA) dx 2E  I 

M

2

1 M 2 E  I 

2



Fig. 4.5 Bending load.

L

(y 2 dA )



dx

0

y 2 dA = I

U=

4.2.3

L

Strain Energy due to Moment M U=

Hence,

1 2

M2L 2 EI

(4.12)

Strain Energy due to Torsional Loading 2

1 + O T  r dV E  J 

U=



U=

1 + O T    E J

2

L

r 2 dA

 0

1+O T L T L = E J 2 GJ 2

U=



T

dx Fig. 4.6

Torsional load

2

(4.13)

144

Strength of Materials

4.2.4

Strain Energy due to Transverse Shear 1 + O  U= k  EA 

L

0

Vy2 dx =

k 2 AG

L

0

Vy2 dx

(4.14)

The value of k (known as form correction factor for different cross section) is as follows. Form correction factor for shear Beam cross section

k

Rectangular Thin walled circular Circular I section, box section, channel section

1.2 2.0 1.11 1.0

EXAMPLE 4.1 Determine the total strain energy stored in a beam as shown in Fig. 4.7. 12 kN 50

500

1500

25

Fig. 4.7 Example 4.1.

All dimensions are in mm. Take n = 0.3, E = 205 GPa Solution: The beam is of rectangular cross section. Both strain energy due to bending moment and shear force are to be calculated to obtain the total strain energy stored in the beam. The beam reactions are determined from force equilibrium condition. 12 kN

A

B

RA = 9 kN

Fig. 4.8(a)

RB = 3.0 kN

Beam reactions.

Taking the moment about A, the moment equilibrium equation is RB ´ 2.0 = 12 ´ 0.5 Solving RB = 3 kN

Chapter 4: Energy Methods

From the force equilibrium, RA + RB = 12 Þ RA = 9.0 kN Shear force equation 0 < x < 0.5   9 kN Vy =  3 kN 0.5 < x < 2.0 

The Shear force diagram is given in Fig. 4.8(b). 9 kN (–)

0

2000 mm

500 mm (+) 3 kN

Fig. 4.8(b) Shear force diagram.

The bending moment equations are  9 x kN.m Mz =   9 x  12( x  0.5) kN.m

0 < x < 0.5 0.5 < x < 2.0

The bending moment diagram is given in Fig. 4.8(c). 4.5 kN·m

0

500

2000 (mm)

Fig. 4.8(c) Bending moment diagram.

Strain energy due to bending

Ub =

1 2 EI z

0.5

 0

(9000 x )2 dx +

1 2 EI z

2



{9000 x  12000[x  0.5]}2 dx

0.5

Iz =

1  0.025  0.053 = 2.604  10 7 m 4 12

Ub =

1 2 EI z

 (9000)2  0.53 2.0  + {9000 x  12000 [ x  0.5]}2 dx   3   0.5

Ub =

1 2 EI z

 90002  0.53 2.0  + (9000x  12000x + 6000)2 dx   3 0.5  





145

146

Strength of Materials

1 2 EI z

 9000 2  0.53 +  3 

2.0

1 2 EI z

 9000 2  0.53 +  3 

2.0

Ub =

1 2 EI z

  30002 (23  0.53 )   6 3.375 10 + + 6000 2 (2  0.5)  18  10 6  3.75)       3    

Ub =

10 6  13.5 2 EI z

Ub =

Ub =

Ub =



0.5



0.5

 (  3000x + 6000)2 dx    (3000 2 x 2 + 6000 2  2  3000  6000 x ) dx  

106  13.5 2  205  109  260.4  10 9

= 126.5 N.m

Strain energy due to shear force Vy 1 + O  UV =  k  EA 

L

V

2 y

dx

0

 0.5 1 + 0.3   2 =   1.2  (9000) dx + 0  205  10 9  0.025  0.05  



2



0.5

 (3000)2 dx   

UV = 5.07 ´ 10–9 ´ 1.2(81 ´ 106 ´ 0.5 + 9 ´ 106 ´ 1.5) = 0.329 N·m = 329 N·mm

EXAMPLE 4.2 Two circular shafts are made of the same material, one of solid and other of hollow circular have been subjected to torque. The maximum shear stress is to be the same in both the shafts. Find the strain energy stored in both cases if the outer diameters are the same. The ratio of inner diameter to outer diameter in the case of hollow shaft is k. Solution: Solid shaft do = diameter of shaft U max = U s =

where Ts is the applied torque.

16Ts Q  do3

(i)

Chapter 4: Energy Methods

147

Hollow circular shaft do = outer diameter of shaft di = inner diameter of the shaft di = k  di = kdo do U max = U h =

16 Th do Q  (do4  di4 )

=

16 Th

(ii)

Q  do3 (1  k 4 )

Since the maximum shear stresses are the same 16 Th

Q  do3 (1  k 4 )

=

16 Ts

Q  do3

Th = 1  k4 Ts

Strain energy stored in solid shaft

Us =

Ts2  l 2 GJ s

Uh =

Th2  l 2 GJ h

Strain energy stored in hollow shaft

Ratio of strain energy

T  Uh = h Us  Ts 

2

 J  T    s= h  J h   Ts 

2

 1   1  = (1  k 4 )2      4  1  k   1  k4 

Uh = 1  k4 Us

where J s =

4.3

Q 32

do4 and J h =

Q 32

d o4 (1  k 4 )

CASTIGLIANO’S FIRST THEOREM

Consider a series of gradually applied load or forces Fi (i = 1, 2, 3, …, n) to a structure. Let the deflection along the direction of load at each load point be given by di (i = 1, 2, 3, …, n), Fig. 4.9.

Fig. 4.9 Applied forces and deflections.

148

Strength of Materials

The total work is n

W=



Fi dE i

i =1

As energy is conserved, the total work done is equal to the total strain energy of the structure. U=W n

U=

or

U=

1 2

 i =1

Fi dE i

(i)

n

 Fi E i

for linear deformation system

(ii)

i =1

Now, if one of the load let F1 be increased by an infinitesimal amount DF1 keeping the other load the same, the corresponding deflection will also increase by infinitesimal amount Dd1. The additional work done due to increase in load will cause an infinitesimal change in strain energy DU of the system.

Fig. 4.10 Deflection curve for load which is increased by an amount of DF1.

The extra work done at station 1 is (Fig. 4.10)

'W1 = 'U1 =

1 2

1 'F1  'E1 + F1  'E1 =  'F1 + F1   'E1 2



(iii)

The deflection curve for other load/station is given in Fig. 4.11. Extra work done

'W = 'U = F  'E

for i = 2 to nth station

(iv)

Hence, total increase in strain energy = Total extra work done n 1 'Wt = 'U =  (Fi  'E i ) +  'F1 + F1   'E1 i2

2



(v)

Chapter 4: Energy Methods

149

Fig. 4.11 Extra works done.

If the loads F1 + DF1, F2, etc. are applied gradually from zero, the total strain energy would be U + 'U = =

1 1 (F1 + 'F1 ) (E1 + 'E1 ) + F2 (E 2 + 'E 2 ) + ... 2 2 1 1 (F1 + 'F1 ) (E1 + 'E1 ) + 2 2

n



Fi (E i + 'E i )

i =2

Neglecting the product of small terms, we get U + 'U =

=

1 1 F1E1 + 2 2 1 2

'F1  E1 +

n



FiE i +

i 1

U + 'U = U +

1 2

n



1 2

n

 i 1

1 1 F1 'E1 + 2 2

Fi 'E i +

(Fi 'E i ) +

i =1

1 2

n



FiE i +

i =2

1 2

n



(Fi 'E i )

(vi)

i =2

'F1E1

1 'F1  E1 2

(vii)

(Fi 'E i ) + 'F1  E1

(viii)

Hence,

2'U =

n

 i 1

Subtracting Eq. (v) from Eq. (viii), we get 2'U 

n

'U = 'F1  E1 +  (Fi 'E i )  i 1 n

'U = 'F1  E1 +  (Fi 'E i )  i 1

n

 i2 n

 i 1

(Fi 

'E i )   'F1 + F1   'E1 2 

(Fi 'E i ) 

1

1 2

'F1 'E1

150

Strength of Materials

Neglecting the product of smaller terms, we get DU = DF1 ´ d1 or

'U 'F1

= E1

(4.15)

Hence, the partial derivative of strain energy with respect to force gives the deflection under the load in the direction of load. Similarly, we can derive for other loads also. Castigliano’s theorem states that the rate of change of strain energy with respect to statically independent load or force gives the component deflection of the force in the direction of the load. A similar derivation for angle of twist can be derived. This is written as U = Ri M i

(4.16)

Castigliano’s theorem applied to beams In case of beam the internal strain energy is due to bending and shear. For a special case such as if the beam is long and slender, the strain energy due to shear can be neglected compared with that of bending. If a beam is subjected to bending moment M, the strain energy stored is given as

U=

M 2 dx 2 EI



The deflection can be obtained from the Castigliano’s theorem as

E =

 P

L

 0

M 2 dx 2 EI

If E and I are constant, we can write as L

E =

 0

 M  dx M   P  EI

where M is the moment due to external load P and d is the displacement of the point caused by the load P.

EXAMPLE 4.3 Figure 4.12 shows a J-type hanger. A vertical load of 15 kN is acting at a distance of 75 mm from the CG of the hanger as shown. Determine (a) the state of stress at section XX, (b) vertical deflection, (c) horizontal deflection. Consider the deflection due to bending only.

Chapter 4: Energy Methods

151

Fig. 4.12 Example 4.3.

Solution: (a) Moment M = Fl = 15000 ´ 0.075 = 1125 N.m The moment is acting on the section X-X. The Stress due to moment M is

Tb =

M 1125  (0.05/2) y= = 91.67 ´ 106 N/m2 = 91.67 MPa I (Q /64)  0.054

The axial stress due to axial load F is given by

Ta =

F 15000 = = 7.639 ´ 106 N/m2 = 7.639 MPa A (Q /4)  0.052

The total stress (maximum) is

s = sb + sa = 91.67 + 7.639 = 99.31 MPa (b) As the deflection in both the directions is to be obtained by Castigliano’s theorem, apply a dummy load in horizontal direction as shown in Fig. 4.13.

F a Q

Fig. 4.13 Dummy load.

152

Strength of Materials l

U=

 0 l

=

 0

=

(Fa  Qx )2 dx 2EI (F 2 a2 + Q 2 x 2  2FaQx ) dx 2 EI

(F 2 a2 l + Q2 l 3 /3  FaQl 2 ) 2 EI

The vertical deflection is obtained by setting Q = 0 and partially differentiating U with respect to F.

EV =

=

U 1 Fa2 l = (2Fa2 l) = F 2 EI EI 15000  (0.075)2  0.150 200  10 9  (Q /64)  0.054

= 2.063  10 4 m

= 0.206 mm (c)

For horizontal deflection, U should be partially differentiated with respect to dummy load Q and then Q should be set at zero.

EH =

U 1 = 2 EI Q

 2Ql 3   Fal 2     3 

Setting Q = 0

EH =

=

U 1 = (  Fal 2 ) Q 2 EI

 15000  0.075  (0.150)2 200  109  (Q /64)  0.054

= 4.125  10 4 m

= 0.4125 mm The negative sign indicates that the deflection is in the opposite direction of the load.

EXAMPLE 4.4 In Example 4.3, considering bending and axial loads, determine the vertical and horizontal deflection using Castigliano’s theorem.

Chapter 4: Energy Methods A

F a

B

Q

C

Fig. 4.14 Example 4.4.

Solution:

The strain energy equation for part BC is a

U BC =

 0

(Fy)2 dy + 2 EI

a

 0

Q2 dy 2 AE

The strain energy equation for part AB is l

U AB =

 0

(Fa  Qx )2 dx + 2 EI

l

 0

F 2 dx 2 AE

The total strain energy is the sum of all. a

U=

 0

a

U=

 0

U=

(Fy)2 dy + 2 EI (Fy)2 dy + 2 EI

a

 0

a

 0

Q 2 dy + 2 AE Q 2 dy + 2 AE

l

 0 l

 0

(Fa  Qx )2 dx + 2 EI

l

 0

F 2 dx 2 AE

(F 2 a2 + Q2 x 2  2FQax ) dx + 2 EI

F 2 a3 Q2 a F 2 a2 l + Q 2 l 3 /3  FQal 2 F 2l + + + 6 EI 2 AE 2EI 2 AE

The deflection in the direction of F is

EV =

U F

EV =

U Fa3 Fa2 l Fl = + + EI AE F 3EI

EV =

Fa 2  a Fl   + l + EI  3 AE 

Q 0

l

 0

F 2 dx 2 AE

153

154

Strength of Materials

The deflection in horizontal direction is EH =

EH =

U Q

= Q 0

2Qa 2Ql 3 /3  Fal 2 + 2 AE 2 EI

U  Fal 2 = Q Q =0 2 EI

EXAMPLE 4.5 Find out deflection at free end using Castigliano’s theorem for cantilever beam of length l as shown in Fig. 4.15.

Fig. 4.15

Solution: X-X is:

Cantilever beam under end concentrated load and moment.

Consider a section X-X at a distance x from free end. The bending moment at section BM = Px + M

Strain energy due to bending moment, U =

=

=

=

Now,

Slope i =

1 2 EI

1 2 EI

l

Ô

( BM ) 2 dx

0

l

Ô

( Px  M )2 dx

0 l

1 2 EI

Ô

1 2 EI

È É Ê

( P 2 x 2  M 2  2 PMx ) dx

0

P 2 l3 3



M 2l



1 ˜U (2 Ml + Pl 2 ) = 2 EI ˜M =

Ml Pl 2 + EI 2 EI

Ø

PMl 2 Ù Ú

Chapter 4: Energy Methods

155

Deflection at free end, y=

or

y=

1 È 2Pl 3 2EI ÉÊ 3

˜U ˜P



Ø

Ml 2 Ù Ú

Pl 3 Ml 2  3EI 2 EI

EXAMPLE 4.6 Using Castigliano’s theorem, obtain the deflection under a single concentrated load applied to a simply supported beam shown in Fig. 4.16. Given EI = 2 MN/m2. 50 kN X

C A

B 3m

1m

x

Fig. 4.16

X

Simply supported beam under concentrated load.

Solution: Let the load at C be denoted as P. The reaction at A is obtained as RA =

P 4

Consider a section XX at a distance x from A. The moment at section XX is   Mx =   

Now,

Px 0  x  3 4 Px  P (x  3) 3  x  4 4

x 0  x  3  4 M x =  P  x  (x  3) 3  x  4  4

We have

d = U

where

U =

P



M 2 dx 2EI

156

Strength of Materials

d=

\

 P

M 2 dx 2EI



=



1 M 2 M dx 2EI M P

d=



M M dx EI P

Now substituting the values of M and its derivatives we have

d=

1 EI

3

È Ô ÊÉ 0

Px Ø 4 ÚÙ

P x3 = 16 EI 3

= and

È ÊÉ

xØ dx 4 ÚÙ

3

 0



1 EI

4

Ë Px ÔÌ 4  3Í

9P È 16 x 16 EI ÉÊ



x3 3



P( x

4x

2



Ø Ù Ú

Û Ëx  Ì Ý Í4

3) Ü

(x



Û

3) Ü dx Ý

4

3

3P 4EI

d=

3 – 50 – 103 4 – 2 – 106

= 1.875 ´ 10–2 m

= 18.75 mm

EXERCISES 4.1 What do you understand by strain energy absorbed by a system, complimentary strain energy and elastic strain energy? 4.2 State Castigliano’s first and second theorem for strain energy? What are their uses? 4.3 Derive an expression for strain energy in a cantilever beam due to bending and shear under concentrated edge load? 4.4 Derive an expression for strain energy stored in a shaft tapering from a diameter D to a diameter d over a length L and subjected to torque T. 4.5 Mild steel has more toughness than high strength steel. Explain in terms of strain energy. 4.6 A steel rod of dimater 30 mm and length 200 mm is subjected to axial tension of 10 kN. Find the strain energy produced if (a) the load is applied gradually, (b) the load is applied suddenly. 4.7 Find the strain energy stored in the shaft shown in Fig. 4.17.

Fig. 4.17

Chapter 4: Energy Methods

157

4.8 Using Castigliano’s theorem, determine the deflection of point C of the beam shown in Fig. 4.18.

Fig. 4.18

[Ans. (24.772 ´ 103)/EI m] 4.9 A simply supported beam of span 2 m is carrying two loads of 500 kN and 100 kN at 1 m and 1.5 m, respectively from the left hand support. Using Castigliano’s theorem, calculate the deflection at the centre. Take E = 210 GPa, I = 70 ´ 10–6 m4. [Ans. 12 mm]

158

Strength of Materials

5 5.1

Deflection of Beams

INTRODUCTION

It is observed that when a beam or a cantilever is subjected to some type of loading it deflects from its initial/original position. The amount of deflection depends upon its cross-section and bending moment. Strength and stiffness are the two main design criteria for a beam or a cantilever. According to strength criterion of the beam design, the beam should be adequately strong to resist shear force and bending moment. In other words, the beam should be able to resist shear stresses and bending stresses. But according to stiffness (being mathematically calculated as W/d where W is the applied load and d is the maximum deflection or sag) criterion of beam design, the beam should be stiff enough not to deflect more than the permissible limit.

5.2

RELATION BETWEEN SLOPE, DEFLECTION AND RADIUS OF CURVATURE

Figure 5.1 shows a small portion AB of a beam bent into an arc. C y

B

da

dy A dx

a O

Fig. 5.1

a + da x

Section of beam under bending. 158

Chapter 5:

Let

a

Deflection of Beams

159

ds = Length of beam AB C = Centre of the arc a = Angle which the tangent at A makes with xx axis + da = Angle which the tangent at B makes with yy axis

We find from geometry of the figure, that Ð ACB = da and ds = Rda \ or

R=

ds dx = (Assuming ds = dx) dB dB

1 dB = R dx If coordinate of point A are x and y then tana =

or

dy dx

dy dx Differentiating the above equation w.r.t. x, we get

a=

dB d2 y = dx dx 2 1 d2y = R dx 2

(5.1)

Also, we know that

M E = I R M=E´

or

I R

(5.2)

Now, substituting Eq. (5.1) in Eq. (5.2) we get M = EI

5.3

d2 y dx 2

(5.3)

METHOD FOR SLOPE AND DEFLECTION

Though there are many methods to find out the slope and deflection at a section in a loaded beam, yet the following are important from the subject point of view: 1. Double integration method 2. Macaulay’s method 3. Moment area method The first and third methods are suitable for a single load, whereas the second one is suitable for several loads.

160

Strength of Materials

5.3.1

Double Integration Method for Slope and Deflection

We have already discussed in Section 5.2 that the bending moment at a point is: M = EI Integrating Eq. (5.4), we get dy = dx

EI Integrating Eq. (5.5), we have

EIy =

I II

d2 y

(5.4)

dx 2

M

(5.5)

M

It is thus obvious, that after first integration, we get the value of slope at any point. On further integrating, we get the value of deflection at any point. While integrating twice the original differential equation, we will get two constants C1 and C2. The value of these constants may be found by using the end conditions.

Cantilever subjected to concentrated load at free end: A cantilever AB of length L fixed at a point A and free at the point B, and carrying a point load at the free end B is shown in Fig. 5.2. Consider the free end at the origin and measuring distance x from it. X

W

A

B x L X

Fig. 5.2 Cantilever beam under concentrated load at free end.

Referring to Fig. 5.2, and taking hogging moment negative. Mx = –Wx i.e.,

or

EI

d2 y dx 2

EI

= –Wx

dy Wx 2 =– + C1 dx 2

At

x = L,

or

0=–

dy =0 dx

WL2 + C1 2

Chapter 5:

or

C1 =

\

EI

Deflection of Beams

161

WL2 2

Wx 2 WL2 dy =– + dx 2 2

Integrating, EI. y = – The boundary condition is: At

Wx 3 WL2 + x + C2 6 2

x = L, y = 0 0 =–

WL3 WL2 + L + C2 6 2

C2 = –WL3

or

=– \

EI y =

At free end, x = 0.

 1  1   2 6

WL3 3

 Wx 3 6

WL2 WL3 x– 2 3

+

  = WL   2 EI 1   WL   WL = EI  3  3EI 2

1 WL2 dy = 2 dx EI

3

3

and

y =

Cantilever subjected to moment at free end:

Figure 5.3 shows a cantilever beam of span L, flexural rigidity EI subjected to hogging moment M. X

O x

M

L X Fig. 5.3

Cantilever beam under moment at free end.

162

Strength of Materials

Taking origin O at free end, moment at distance x is given by Mx = –M i.e.,

EI

\

d2 y

EI

and

= –M

dx 2

dy = –Mx + C1 dx

EI y =

 Mx 2 2

(5.6)

+ C1x + C2

The boundary conditions available are:

dy =0 dx (B) At x = L, y = 0 (A)

At x = L,

From boundary condition (A), we get 0 = –ML + C1 C1 = ML

or

From boundary condition (B), we get

ML2 + C1L + C2 2 Substituting the value of C1 and rearranging, we have 0=–

C2 =

ML2  ML2 – ML2 = 2 2

From Eqs. (5.6) and (5.7), we get EI and

dy = –Mx + ML = M (L – x) dx EIy = –

Mx 2 ML2 + MLx – 2 2

 !

=M  At free end,

x2 L2  Lx  2 2

"# #$

x = 0. dy 1 ML = ML = dx EI EI

 

 

and

y=

1 L2 M  EI 2

i.e.,

y=

ML2 downward 2 EI

=–

ML2 2 EI

(5.7)

Chapter 5:

Deflection of Beams

163

Cantilever subjected to uniformly distributed load: A cantilever AB of length L fixed at point A and free at point B, carrying a uniformly distributed load throughout the beam span is shown in Fig. 5.4. Consider the free end at the origin and measuring distance x from it. X W/unit length B

A x

L X

Fig. 5.4

Cantilever beam subjected to uniformly distributed load.

Referring to Fig. 5.4 and taking hogging moment negative, Mx = 2

EI \

d y dx 2

EI

=

 Wx 2 2

 Wx 2 2

dy  Wx 3 = + C1 dx 6

At

x = L,

\

0 =–

\ \

C1 = EI

dy =0 dx

WL3 + C1 6

WL3 6

dy Wx 3 WL3 =–  dx 6 6

Integrating Eq. (5.8) again, we get

At

Wx 4 WL3 x  + C2 24 6 x = L, y = 0

\

0 =

or

C2 =

EIy = –

 WL4 24

 WL4 6



WL4 + C2 6



WL4 24

(5.8)

164

Strength of Materials

=– \

EIy =

WL4 8

 Wx 4

+

24

WL3 x WL4 – 6 8

At free end where x = 0, we get

WL3 dy = dx 6 EI

 

 

and

y=

1  WL4 EI 8

or

y=

WL4 downward 8 EI

Simply supported beam with a central point load: Consider a simply supported beam AB of length L and carrying a point load W at the centre of beam C as shown in Fig. 5.5. From the geometry of the figure, we find that the reaction at A and B, RA = RB = W A

W 2 X

C

B

x

yc L/2

L/2 X

Fig. 5.5

Simply supported beam with a point load.

Consider a section XX at a distance x from B. We know that the bending moment at this section, Mx = \

EI

d2 y

W x 2

(plus sign due to sagging)

Wx 2

(5.9)

Wx 2 dy = + C1 dx 4

(5.10)

dx

2

=

Integrating Eq. (5.9), we get EI

where C1 is the first constant of integration. We know when x =

L dy , = 0. Substituting these 2 dx

values in Eq. (5.10), we have 0=

WL2 + C1 16

or

C1 = –

WL2 16

Chapter 5:

Deflection of Beams

165

Substituting this value of C1 in Eq. (5.10), we obtain EI

dy Wx 2 WL2 = – dx 4 16

(5.11)

This is the required equation for the slope, at any section, by which we can get the slope at any point on the beam. A little consideration will show that the maximum slope occurs at A and B. Thus, for maximum slope at B, substituting x = 0 in Eq. (5.11), EI. iB = \ iB =

 WL2 16

 WL2

(Minus sign means that tangent at B makes an angle with AB in the negative or 16 EI anticlockwise direction) or

iB =

 WL2 16 EI

radians

By symmetry, iA =

WL2 radians 16 EI

Integrating Eq. (5.11) once again,

Wx 3 WL2 – x + C2 12 16 where C2 is the constant of integration. We know that when x = 0, y = 0. Substituting these values in Eq. (5.11), we get C2 = 0 EIy =

Wx 3 WL2 – x (5.12) 12 16 This is the required equation for the deflection, at any section, by which we can get the deflection at any point on the beam. A little consideration will show that maximum deflection occurs \

EIy =

at the mid point C. Thus, for the maximum deflection, substituting x =

EIyc =

=

or

yc =

 WL3 48 EI

   

W L 12 2

3

–

WL2 16

L in Eq. (5.12), we get 2

L 2  

WL3  WL3 WL3 – = 96 32 48

(Minus sign means that the deflection is downwards)

166

Strength of Materials

Simply supported beam with a uniformly distributed load: Consider a simply supported beam AB of length L and carrying a uniformly distributed load of W per unit length as shown in Fig. 5.6. From the geometry of the figure, we find that the reaction at A, and B are, WL 2

RA = RB = W/unit length

X

A

C

B

x

L/2

L/2 X

Fig. 5.6 Simply supported beam with a uniformly distributed load.

Consider a section XX at a distance x from B. We know that the bending moment at this section, Mx = \

EI

Wx 2 WLx – 2 2 WLx Wx 2 – 2 2

(5.13)

WLx 2 Wx 3 dy = – + C1 dx 4 6

(5.14)

d2 y dx 2

=

Integrating Eq. (5.13), EI

where C1 is the constant of integration. We know when x = in Eq. (5.14), 0=

or

   

WL L 4 2

C1 = –

2

–

L dy , = 0. Substituting these values 2 dx

   

W L 6 2

3

+ C1

WL3 24

dy WLx 2 Wx 3 WL3 = – – (5.15) dx 4 6 24 This is the required equation for the slope at any section, by which we can get the slope at any point on the beam. We know that the maximum slope occurs at A and B. Thus, for maximum slope, substituting x = 0 in Eq. (5.15), \

EI

EIiB =

 WL3 24

Chapter 5:

Deflection of Beams

167

 WL3

\

iB =

or

iB =

WL3 24 EI

iA =

WL3 24 EI

24 EI

By symmetry,

Integrating Eq. (5.15) once again, EIy =

WLx 3 Wx 4 WL3 x – – + C2 12 24 24

(5.16)

where C2 is the constant of integration. We know when x = 0, y = 0. Substituting these values in Eq. (5.16), we get C2 = 0. \

EIy =

WLx 3 Wx 4 WL3 x – – 12 24 24

(5.17)

This is required equation for the deflection at any section, by which we can get the deflection at any point on the beam. We know that the maximum deflection occurs at the midpoint C. Thus, for the maximum deflection, substituting x = L/2 in Eq. (5.17), EIyc = =

or

yc =

5WL4 384 EI

   

WL L 12 2

   

3

W L 24 2

–

4

–

   

WL3 L 24 2

WL4 WL4 WL4 5WL4 – – =– 384 96 384 48

(Minus sign means that the deflection is downwards) Maximum deflection =

5WL4 384 EI

Simply supported beam with a gradually varying load: Consider a simply supported beam AB of length L, and carrying a gradually varying load from zero at B to W per unit length at A as shown in Fig. 5.7. From the geometry of the figure, we find that the reaction at A, RA =

WL 3

and

WL 6

RB = X

W A

Fig. 5.7

L

X

x

B

Simply supported beam with a gradually varying load.

168

Strength of Materials

Now, consider a section XX at a distance x from B. We know that the bending moment at this section, Mx = RBx –

\

EI

Wx x x ´ ´ L 2 3

=

WLx Wx 3 – 6 6L

=

WLx Wx 3 – 6 6L

(5.18)

WLx 2 Wx 4 dy = – + C1 dx 12 24 L

(5.19)

d2 y dx

2

Integrating Eq. (5.18), we obtain EI

where C1 is the first constant of integration. Integrating Eq. (5.19) once again, EIy =

WLx 3 Wx 5 – + C1x + C2 36 120 L

(5.20)

where C2 is the second constant of integration. We know that when x = 0, y = 0. Therefore, C2 = 0. We also know that when x = L, y = 0. Substituting these values in Eq. (5.20), we get 0= = \

C1 =

WL W ´ L3 – ´ L5 + C1L 36 120 L WL4 WL4 – + C1L 36 120

 WL3

+

36 Now, substituting this value of C1 in Eq. (5.19)

WL3  7WL3 = 360 120

WLx 2 Wx 4 7WL3 dy = – – (5.21) 360 dx 12 24 L This is the required equation for slope at any section, by which we can get the slope at any section on the beam. A little consideration will show that the maximum slope will be either at support A or B. Thus, for slope at A, substituting x = L in Eq. (5.21), we have EI

EI iA = \

WL W 7WL3 WL3 ´ L2 – ´ L4 – = 360 12 24 L 45

WL3 45 EI Now, for slope at B, substituting x = 0 in Eq. (5.21) iA =

EI iB = – \

iB =

7WL3 360

7WL3 radians 360 EI

Chapter 5:

169

Deflection of Beams

Now, substituting the value of C1 in Eq. (5.20), EIy =

\

y =

WL 3 Wx 5 7WL3 x x – – 360 36 120 L

 

1 WLx 3 Wx 5 7WL3 x   36 120 L 360 EI

 

(5.22)

This is the required equation for the deflection at any section, by which we can get deflection at any section on the beam. For deflection at the centre of the beam, substituting x =

     !  

yc =

1 WL L EI 36 2

=

0.00651WL4 EI

3



   

W L 120 L 2

5



 

7WL3 L 360 2

 "#  #$

L in Eq. (5.22), 2

(Minus sign means that the deflection is downwards)

We know that the maximum deflection will occur, where slope of the beam is zero. Therefore, equating the equation to zero,

WLx 2 Wx 4 7WL3 – – =0 360 12 24 L x = 0.519L

\

Now, substituting this value of x in Eq. (5.22), we get ymax =

 1 !

6

1

6

1

1 WL 7WL3 W 3 5 0.519 L  0.519 L  0.519 L 120 L 360 EI 36 =

6"## $

0.00652 WL4 EI

EXAMPLE 5.1 A beam 4 m long, simply supported at its ends is carrying a point load W at its centre. If the slope at the ends of the beam is not to exceed 1°, find the deflection at the centre of the beam. Solution: Given L = 4 m, Central point load = W Then Slope at A iA = 1o =

1–Q = 0.01745 radian 180

We know that the slope at the end, iA =

WL2 16 EI

170

Strength of Materials

and deflection at the centre, yc =

WL3 WL2 L = ´ 3 48 EI 16 EI  WL2   iA =  16 EI  

'

L 3

= iA ´

= 0.01745 ´

4 = 0.02327 m = 2.33 cm 3

EXAMPLE 5.2 A simply supported beam of span 6 m carries a uniformly distributed load (u.d.l.) of 5 kN/m. Find the breadth and depth of the beam if the maximum bending stress is not to exceed 8 N/mm2 and maximum deflection 20 mm. Take E = 104 N/mm2. Solution: Given E = 104 N/mm2 = 104 ´ 106 N/m2 = 104 ´ 106 ´ 10–3 kN/m2 = 107 kN/m2 ymax = 20 mm = (20 ´ 10–3) m Then

or

ymax = (20 ´ 10–3) =

5WL4 384 EI 5 – 5 – (6)4 (384) (10)7 I

I = (4.218 ´ 10–4) m4 = (4.218 ´ 10–4) (1012) mm4

or

= (4.218) ´ 108 mm4

9.

2

Now,

(BM)max =

8

=

5 – (6) 2 8

= 22.5 kN-m = (22.5 ´ 106) N-mm Also

T M = y I 22.5 – 10 6 4.218 – 10

8

=

8 d /2

\

d = 300 mm

Then

I=

or or

4.218 ´ 108 =

1 bd3 12

1 – b – (300) 3 12

b = 187.5 mm

Chapter 5:

Deflection of Beams

EXAMPLE 5.3 Find the deflection at free end in the cantilever beam shown in Fig. 5.8. W

C

a

B

L Fig. 5.8

Solution:

Deflection at free end in the cantilever beam is shown in Fig. 5.9. yB = BD + DE =

Wa 3 Wa 2 + (L – a) 3EI 2 EI

Fig. 5.9

EXAMPLE 5.4 Find the deflection at free end in the cantilever beam, shown in Fig. 5.10. W/unit length C

a L

Fig. 5.10

Solution:

Deflection at free end is shown in Fig. 5.11. yC = CD + DE yC = YB + (L – a) qB =

Wa 4 Wa 3 + (L – a) 8 EI 6 EI

171

172

Strength of Materials

B A

C

qB qB

a

L–a

D E

Fig. 5.11

EXAMPLE 5.5 A cantilever beam carries a u.d.l throughout the span L. The beam is propped at the free end by an unyielding support. Calculate the reaction at free end. Solution:

Deflection of the free end of a cantilever having u.d.l is given by y1 =

Upward deflection of beam,

WL4 8 EI

RL3 3EI y = y1 – y2 y1 = y2 y2 =

Net deflection Unyielding support

R=

 3WL "# !8 $ W

L

R

Fig. 5.12

5.3.2

Macaulay’s Method

The double integration method is convenient if only one symmetrical load is present on the beam. But if there are more than one load then the bending moment equation changes at every load. For such types of problems, Macaulay’s method is very useful.

Steps for Finding deflection using Macaulay’s Method: (1) First the support reactions of simply supported beam are found out by usual method. (2) Considering the LMS at the origin and measuring distance x from it, a single B.M. equation is written indicating the segment lines. (3) While integrating the bracketed expressions are integrated as a whole. The constants of integration are added in the first segment only.

Chapter 5:

173

Deflection of Beams

Deflection of simply supported beam with an eccentric point load:

A simply supported beam AB of length L and carrying a point load W at a distance a from left support and at distance b from right support is shown in Fig. 5.13. The reactions at A and B are given by

Wb Wa and RB = L L Considering the LHS as the origin and measuring distance x from it, the B.M. for all sections of beam can be expressed as: RA =

Mx = RA x ½ –W(x – a) W A

(5.23) X

C a x

RA

B b

L

RB

X

Fig. 5.13 Simply supported beam with an eccentric point load.

The B.M. at any section is also given by Eq. (5.1) as: M = EI

d2 y

(5.24)

dx 2 Hence equating Eqs. (5.23) and (5.24), we get

EI

d2 y dx

2

=

Wb x  W ( x  a) L

(5.25)

Integrating Eq. (5.25), we get EI

dy Wb x 2 W ( x  a)2 =  C1  dx L 2 2

(5.26)

where C1 is a constant of integration. Integrating Eq. (5.26) once again, we get EIy =

W ( x  a) Wb x 3  C1 x  C2 – 3–2 L 6

3

(5.27)

where C2 is another constant of integration. The values of C1 and C2, obtained from boundary conditions, are: (i) At x = 0, y = 0. Substituting these values in Eq. (5.27) up to the first compartment, we get 0 = 0 + 0 + C2 C2 = 0

or

(ii) At x = L and y = 0. Substituting these values in Eq. (5.27), we get (up to last terms) 0=

Wb 3 W ( L  a)3 L + C1L + 0 – 3–2 6L

174

Strength of Materials

WbL2 Wb 3 + C1 L – 6 6

0=

(

' L – a = b)

Wb 2 (L – b2) 6L Substituting the value of C1 in Eq. (5.26), we get C1 = –

EI

(5.28)

Wb x 2  Wb 2 dy  W ( L  b 2 )   ( x  a) 2   = L 2  6L dx  2 =

Wbx 2 Wb 2 W  ( L  b 2 )  ( x  a) 2 2L 6L 2

(5.29)

Equation (5.29) gives the slope at any point in the beam. Slope is maximum at A or B. To find the slope at A, substitute x = 0 in Eq. (5.29) up to first compartment, Wb 2 Wb ´0– (L – b2) 2L 6L

EI iA =

=– or

iA = –

Wb 2 (L – b 2) 6L

Wb (L2 – b2) 6 EIL

Substituting the values of C1 and C2 in Eq. (5.27), we get EIy =

 !

"# $

Wb 3 Wb 2 W x +  (L  b 2 ) x + 0  (x – a)3 6L 6 6L

Deflection at point C under the load, substitute x = a in Eq. (5.30), EI yc = =

or

Wb 3 Wb 2 a – (L – b2)a + 0 6L 6L Wb Wab 2 a(a2 – L2 + b2) = – (L – a2 – b2) 6L 6L

' L = a + b)

=–

Wab [(a + b)2 – a 2 – b2] 6L

=–

Wab 2 [a + b2 + 2ab – a2 – b 2] 6L

=–

Wa 2 b 2 Wab (2ab) = – 6L 3L

yc = –

Wa 2 b 2 3EIL

(

(5.30)

Chapter 5:

Deflection of Beams

175

EXAMPLE 5.6 Find the slope and deflection at C, D and E for the beam shown in Fig. 5.14. Are the slopes at A and B equal? Take E = 200 GPa, I = (450 ´ 106) mm4. 40 kN A

20 kN

X

C

60 kN

D

2m

E

3m

4m

B 1m

x X

Fig. 5.14

Solution: Given E = 200 GPa = (200 ´ 106) kN/m2, I = (450 ´ 106) mm4 = (450 ´ 10–6) m4, RA = 48 kN, RB = 72 kN The general deflection equation is: M = EI

d2 y dx 2

= 48x½–40 (x – 2)½– 20 (x – 5)½–60 (x – 9)

Integrating w.r.t. x, EI

 dy  = 24x ½– 20 (x – 2) ½– 10 (x – 5) ½– 30 (x – 9)  dx  2

2

2

2

+ C1

Integrating w.r.t. x, EI(y) = 8x3 + C1x + C2 + ½

 20

(x – 2)3 ½

 10

(x – 5)3 ½– 10(x – 9)3 3 3 (i) At x = 0, y = 0. Substituting these values in the deflection equation up to the first compartment, we get C2 = 0 (ii) At x = 10, y = 0. Substituting these values in the deflection equation using all the compartments, we get 0 = 8(10)3 + C1 (10) – \

20 10 (10 – 2)3 – (10 – 5)3 – 10 (10 – 9)3 3 3 C1 = – 416

General slope equation is: EI

 dy   dx 

= 24x 2 – 416½–20(x – 2)2½–10(x – 5)2½–30(x – 9)2

The general deflection equation is: EI(y) = 8 Z3  416 Z

 20 3

( Z  2)3

 10 3

( Z  5)3  10( Z  9)3

176

Strength of Materials

Slopes: At A, substitute x = 0 in slope equation in the first compartment, EI

 dy   dx 

= – 416

dy 416 =– = 4.6 ´ 10–3 radians dx EI At C, substitute x = 2 in slope equation in the first compartment, (EI)

dy = 24(2)2 – 416 dx

dy 320 =– = 3.55 ´ 10–3 radians dx EI At D, substitute x = 5 in slope equation in the first and second compartments, EI

 dy  = 24(5)  dx 

2

– 416 – 20(5 – 2)2

= (0.044 ´ 10–3) radians At E, substitute x = 9 in the slope equation in first, second and third compartments, EI

 dy  = 24(9)  dx 

2

– 416 – 20(9 – 2)2 – 10(9 – 5)2

dy 388 = = 4.31 ´ 10–3 radians dx EI At B, substitute x = 10 in the slope equation in all the compartments,

EI

 dy  = 24(10) – 416 – 20(10 – 2)  dx   dy  = 424 = 4.71 ´ 10 radians  dx  EI 2

2

– 10(10 – 5)2 – 30(10 – 9)2

–3

Deflections: At C, substitute x = 2 in deflection equation in the first compartment, EI(y) = 8(2)3 – 416(2) y =

 768

= 8.53 mm EI At D, substituting x = 5 in deflection equation in first and second segments, EI(y) = 8(5)3 – 416(5) – y =

 1260 EI

= 14 mm

20 (5 – 2)3 3

Chapter 5:

Deflection of Beams

177

At E, substitute x = 9 in deflection equation in first, second and third compartments, EI(y) = 8 (9)3 – 416 (9) – y=

 412 EI

20 10 (9 – 2)3 – (9 – 5)3 3 3 = 4.58 mm

EXAMPLE 5.7 A simply supported beam of span 7 m carries loads as shown in Fig. 5.15. Compute deflection at D and E. Point E is the mid point of DB. Take E = 200 GPa and I = 150 ´ 106 mm4. X

24 kN

6 kN/m

A D

C 2m

2m

B

E 1.5 m

1.5 m

X Fig. 5.15

Solution:

Given E = 200 GPa = 200 ´ 106 kN/m2, I = 150 ´ 106 mm4 = 150 ´ 10–6 m4, RA = 21 kN, RB = 21 kN

The general deflection equation is: M = EI

EI Integrating,

( d 2 y) dx

2

( d 2 y) dx

2

 dy  = 21x  dx  2 dy EI   = 10.5x  dx 

2

= 21x  24 ( x  2)

 6 ( x  4) 2 2

= 21x  24 ( x  2 )  3 ( x  4 ) 2

2 3 + C1  24 ( x  2)  3 ( x  4) 2 3

EI

2

+ C1½–12(x – 2)2½–(x – 4)3

Integrating, EI(y) = 3.5x3 + C1x + C2½–4(x – 2)3½

 ( x  4) 4 4

(i) At x = 0, y = 0. Substituting it in the deflection equation in the first compartment, we get C2 = 0

178

Strength of Materials

(ii) At x = 7, y = 0. Substituting it in the deflection equation upto the last segment, 0 = 3.5 (7)3 + 7C1 – 4 (7 – 2)3 – 0.25 (7 – 4)4 C1 = – 97.18

\

General slope equation is: EI

 dy  = 10.5x  dx 

2

– 97.18 – 12 ( x  2 ) 2 – ( x  4) 3

General deflection equation is: EI(y) = 3.5x3 – 97.18x|– 4(x – 2)3 | – 0.25 (x – 4)4 Deflection at D. x = 4 in first and second compartment, \

EI(y) = 3.5(4)3 – 97.18 (4) – 4 (4 – 2)3 y=

 196.72

= 6.557 mm EI Deflection at E. x = 5.5 in the full deflection equation, \

EI(y) = 3.5(5.5)3 – 97.18 (5.5) – 4 (5.5 – 2)3 – 0.25 (5.5 – 4)4 y=

 124.94 EI

= 4.16 mm

EXAMPLE 5.8 A simply supported beam having a span of 9 m is loaded as shown in Fig. 5.16. Calculate the deflection of beam at mid point of CD. Take E = 200 ´ 103 N/mm2 and I = 270 ´ 106 mm4. X

18 kN/m

A

B 2m

C

D

4m

3m

X Fig. 5.16

Solution: Given E = 200 ´ 103 N/mm2 = 200 ´ 106 kN/mm2, I = 270 ´ 106 mm4 = 270 ´ 10–6 m4, RA = 40 kN, RB = 32 kN Note: For Macaulay’s method, the u.d.l must be reached up to the end. So the same load is extended from D to B, and the same magnitude of load is removed from B to D as shown in Fig. 5.17. 18 kN/m A 2m

C

4m

D

B 3m 18 kN/m

Fig. 5.17

Chapter 5:

Deflection of Beams

179

The general deflection equation is:

 d y  dx   d y EI   dx 

 18 ( x  2 ) 2  18 ( x  6)2

2

M = EI

= 40x

2

2

Integrating, EI

2

2

= 40x  9 ( x  2 )2  9 ( x  6)2

 dy  = 20x  dx 

2

+ C1  3 ( x  2 )3  3 ( x  6) 3

Integrating,

20 3 4 4 x + C1x +  0.75( x  2)  0.75( x  6) 3 (i) x = 0, y = 0 gives C2 = 0. (ii) x = 9, y = 0 gives C1 = –346.67. EI(y) =

General slope equation is: EI

 dy  = 20x  dx 

2

– 346.67  3( x  2 )3  3( x  6) 3

General deflection equation is:

3 3 20 3 4 4 x – 346.67x – ( x  2)  ( x  6) 4 4 3 Deflection at mid point of CD, x = 4 m EI(y) =

EI(y) = or

y=

20 3 3 (4) – 346.67(4) – (4 – 2)4 4 3  972 = 17.83 mm EI

Slope at A, x = 0 in slope equation in first compartment,

EI

or

5.3.3

 dy  = 0 – 346.67  dx   dy  =  346.67 = 6.42 ´ 10  dx  EI

–3

radians

Moment Area Method

Consider an element PQ of small length dx at a distance x from B, bent into arc P1Q1 as shown in Fig. 5.18(b).

180

Strength of Materials Area = Mdx

B

x

dy

(a) A

dx

L

B y

M

P Q

A

D dq

P1

Q1

C

(b) R dq

Fig. 5.18 Moment area method.

Let

R dq M P1C Q1D

= = = = =

Radius of curvature of deflected part P1Q1 Angle subtended by the arc P1Q1 at the centre Bending moment between P and Q Tangent at point P1 Tangent at point Q1

For the deflected part P1Q1 of the beam, we have P1Q1 = Rdq P1Q1 @ dx dx = Rdq

But \

dx R

dq =

\

(5.31)

But for beam, we have M E = I R

or

R=

EI M

Substituting the values of R in Eq. (5.31), we get dq =

dx EI M

   

=

Mdx EI

(5.32)

Since the slope at point A is assumed zero, hence total slope at B is obtained by integrating Eq. (5.32) between the limits 0 and L. L

q=

Ô 0

Mdx EI

=

1 EI

L

Ô 0

Mdx

Chapter 5:

Deflection of Beams

181

L

But Mdx represents the area of B.M. diagram of length dx. Hence

Mdx represents the area

0

of B.M. diagram between A and B.

q=

Ô

1 [Area of B.M. diagram between A and B] EI

Slope at B = iB. Therefore,

Area of B.M. diagram between A and B (5.33) EI If the slope at A is not zero then we have total change of slope between B and A is equal to the area of B.M. diagram between B and A divided by the flexural rigidity EI. iB =

iB – iA =

or

Area of B.M. between A and B EI

(5.34)

Now, the deflection, due to bending of the portion P1Q1, is given by dy = x dq Substituting the value dq from Eq. (5.32), we get

Mdx (5.35) EI Since deflection at A is assumed to be zero, hence the total deflection at B is obtained by integrating Eq. (5.35) between the limits 0 and L. dy = x

L

y=

xM dx

Ô 0

EI

=

1 EI

L

Ô

xM dx

0

But xMdx represents the moment of area of the B.M diagram of length dx about a point B. Hence L

Ô

xMdx represents the total area of B.M. diagram between B and A multiplied by the distance of

0

the C.G. of the B.M. diagram area from B. y=

1 Ax ´A´ x = EI EI

(5.36)

where A = Area of B.M. diagram between A and B x = Distance e of C.G. of the area A

5.3.4

Mohr’s Theorems

The results given by equation for slope and for deflection are known as Mohr’s theorems. These are stated as: The change of slope between any two points is equal to the net area of the B.M. diagram between these points divided by EI. The total deflection between any two points is equal to the moment of the area of B.M. diagram between the two points divided by EI.

182

Strength of Materials

Mohr’s theorems are conveniently used for the following cases: 1. Problems on cantilevers (single load) 2. Simply supported beams carrying symmetric loading (single load) 3. Beams fixed at both ends (single load) Expressions for area and centroid are as follows: (a) Area = bh Distance of centroid G =

b 2

G

h

b (a)

(b) Area of triangle =

1 bh 2

Distance of centroid G =

b 3

G

h

b/3

b (b)

(c) Area of ABC =

2 bh 3

Distance of centroid G =

5 b 8 C

D Area A

h

(5/8) b

A

G

b (c)

B

Chapter 5:

(d) Area =

Deflection of Beams

183

1 bh 3

Distance of centroid G =

3b 4

h (3/4) b

G b (d)

(e) Area =

 1  bh  n  1

Distance of centroid G =

 n  1  b  n  2 y = kxn

h

æ n +1ö ç ÷b èn + 2ø

G

b (e)

Fig. 5.19

Determination of area and centroids of simple geometry.

Slope and deflection of a simply supported beam carrying a point load at the centre: Figure 5.20 shows a simply supported beam AB of length L and carrying a point load W at the centre of the beam, i.e., at point C. The B.M. diagram is shown in Fig. 5.20(b). This is a case of symmetrical loading hence slope is zero at the centre, i.e., at point C. Now, using Mohr’s theorem for slope, we get

Area of B. M. diagram between A and C EI Area of B.M. diagram between A and C = Area of triangle ACD Slope at A =

=

1 L WL WL2 – – = 2 2 4 16

184

Strength of Materials W L 2

A

C

W 2

B W 2

L (a) D

A

Fig. 5.20

\

WL 4

B

C

2 L ¥ 3 2 L 2

(b)

(a) Simply supported beam, (b) Bending moment diagram.

Slope at A (or iA) =

WL2 16 EI

Now, using Mohr’s theorem for deflection, we get from Eq. (5.36), Ax y= EI where A = Area of B.M. diagram between A and C

WL2 16 x = Distance of C.G. of area ACD from point A =

=

\

2 L L – = 3 2 3

WL2 L – 3 3 = WL y = 16 EI 48 EI

Slope and deflection of a simply supported beam carrying a uniformly distributed load at the centre: Figure 5.21 shows a simply supported beam AB of length L and carrying a uniformly distributed load of W/unit length over the entire span. The B.M. diagram is shown in Fig. 5.21(b). This is a case of symmetrical loading hence slope is zero at the centre, i.e., at point C. (i) Now, using Mohr’s theorem for slope, we get Slope at A =

Area of B. M. diagram between A and C EI

But Area of B.M. diagram between A and C = Area of parabola ACD =

2 ´ AC ´ CD 3

Chapter 5:

Deflection of Beams

185

W/unit length A

B WL 2

L

WL 2

(a) D WL2 8

A

B

C

5¥L 8 2 L 2

(b)

Fig. 5.21 (a) Beam under uniformly distributed load (b) Bending moment diagram.

= Slope at A =

2 L WL2 WL3 – – = 3 2 8 24 WL3 24 EI

(ii) Now, using Mohr’s theorem for deflection, we get from Eq. (5.36), Ax y= EI where A = Area of B.M. diagram between A and C WL3 24 = Distance of C.G of area ACD from A x

A=

x =

\

5.4

5 5 L 5L ´ AC = – = 8 8 2 16

WL3 5 L – 16 = y = 24 EI

 5  WL  384  EI

4

INDETERMINATE STRUCTURE

Previously, we have dealt the cases of finding out stresses and strain where simple equation of statics were sufficient to solve the problem. But sometimes the equilibrium equations are not sufficient to solve such problems. Such problems are called statically indeterminate problems, and the structure are called statically indeterminate structures. As an example, a cantilever fixed at one end and free at other is statically determinate as the number of reactions is only three [Fig. 5.22(a)], and can be easily found out the three equations for plane structures:

186

Strength of Materials

åFx = 0 åFy = 0 å Mz = 0 where å shows the sum, and Fx, Fy and M are forces in the x, y directions and moment about z axis, respectively. However, if we add a prop, as shown in Fig. 5.22(b), the situation is now altered. Now, we have four unknown reactions, R1 to R4, and only three equations. The problem is hence statically indeterminate. The excess of unknowns over the equations is termed degree of indeterminacy. A

B L (a) P

A

B L (b)

Fig. 5.22

5.5

(a) Fixed beam with concentrated load at free end (b) Indeterminate beam.

CONTINUOUS BEAM

Continuous beam is a beam which is supported on more than two supports. Figure 5.23 shows such a beam, which is subjected to some external loading. The deflection curve is having convexity upwards over the intermediate supports, and concavity upwards over the mid of the span. Hence there will be hogging moments (i.e., negative) over the intermediate supports and sagging moments (i.e., positive) over the mid of the span. The end supports of a simply supported continuous beam will not be subjected to any bending moment. But the end support of fixed continuous beam will be subjected to fixing moments. If the moments over the intermediate supports are known then the B.M. diagram can be drawn. W/unit length

A

B

C

D

E

Fig. 5.23 Continuous beam.

5.6

CLAPEYRON’S THEOREM OF THREE MOMENTS

The moments over the intermediate supports are determined by using Clapeyron’s theorem of three moments which states that

Chapter 5:

Deflection of Beams

187

If BC and CD are any two consecutive span of a continuous beam subjected to an external loading then the moments MB, MC and MD at the supports B, C and D are given by MB L1 + 2MC (L1 + L 2) + MD L 2 = where

L1 = L2 = a1 = a2 =

6 a1 x1 6a2 x 2  L1 L2

(5.37)

Length of span BC Length of span CD Area of B.M. diagram due to vertical loads on span BC Area of B.M. diagram due to vertical loads on span CD

x1 = Distance of C.G. of the B.M. diagram due to vertical loads on BC from B x 2 = Distance of C.G. of the B.M. diagram due to vertical loads on CD from D. Equation (5.37) is known as the equation of three moments or Clapeyron’s equation. B

C

L1

L2

D

(a) x1

Mx

B

C D x dx (b) B.M. diagram due to vertical moment for two spans x¢1 MB

Mx¢

MC

MD

B

C D x dx (c) B.M. diagram due to support moment for each two spans

Fig. 5.24 Beam with three supports.

EXAMPLE 5.9 A continuous beam ABC covers two consecutive spans AB and BC of lengths 5 m and 7 m carrying uniformly distributed loads of 8 kN/m and 10 kN/m, respectively. If the ends A and C are simply supported, find the support moments at A, B and C. Draw also B.M. and S.F. diagrams. Solution: Given L1 = 5 m, L 2 = 7 m, W1 = 8 kN/m, W2 = 10 kN/m Since the ends A and C are simply supported, the support moments at A and C will be zero. \

MA = MC = 0

To find the support moment at B (i.e., MB) Clapeyron’s equation of three moments should be applied. Hence, we get MA L1 + 2MB (L1 + L2) + MC L 2 =

6a1 x1 6 a2 x 2  L1 L2

188

Strength of Materials

0 ´ 5 + 2MB (5 + 7) + 0 ´ 7 =

or

6a1 x1 6 a2 x 2  L1 L2

6 a1 x1 6 a2 x 2  (i) 5 7 Now, a1 is the area of B.M. diagram due to u.d.l. on AB when AB is considered as simply supported beam. Therefore,

or

24MB =

a1 =

W L2 2 2 ´ AB ´ Altitude = ´ AB ´ 1 1 3 3 8 8 – 52 2 –5– = 83.33 8 3

=

The distance of C.G. of this area from one end = or

Span 2

L1 5 = = 2.5 m 2 2 Here, a2 is the area of B.M. diagram due to u.d.l. on BC, therefore,

x1 =

a2 =

W L2 2 2 10 – 7 2 – BC – 2 2 = – 7 – 3 8 3 8

= 285.83 m and

L2 7 = = 3.5 m 2 2 Substituting these values in Eq. (i), we get

x2 =

6 – 83.33 – 2.5 6 – 285.83 – 3.5 + 5 7 = 249.99 + 857.49 = 1107.48 kN-m 24MB = 1107.48 MB = 46.15 kN-m

24MB = \

Now, B.M. diagram due to support moments is drawn as shown in Fig. 5.25 in which MA = 0, MC = 0 and MB = 46.15 kN·m. S.F. Diagram First calculate the reactions RA, RB and RC at A, B and C, respectively. For the span AB, taking moments about B, we get 5 = MB 2 MB = 46.15. Negative sign is taken as the movement at B is hogging.)

RA ´ 5 – 8 ´ 5 ´

= – 46.15

(

'

or or

5RA – 100 = –46.15 RA =

 46.15  100 5

= 10.77 kN

Chapter 5:

Deflection of Beams

189

Similarly, for the span BC, taking moments about B, we get

7 = – 46.15 2 7RC – 245 = – 46.15

RC ´ 7 – 10 ´ 7 ´ or

245  46.15 = 28.41 kN 7 RB = Total load on ABC – (RA + RC) = (8 ´ 5 + 10 ´ 7) – (10.77 + 28.41) = 70.82 kN

or

RC =

Now,

10 kN/m

8 kN/m A

B

5m

C

7m

(a) 46.15

61.25

25 A

B (b) B.M. diagram 41.59

C

10.77 A

B

C

29.23

28.41

(c) S.F. diagram

Fig. 5.25

EXAMPLE 5.10 A continuous beam ABCD of length 15 m rests on four supports covering 3 equal spans, and carries a uniformly distributed load of 2 kN/m length. Calculate the moments and reactions at the supports. Draw the S.F. and B.M. diagram also. Solution: Given L1 = 5 m, L2 = 5 m, L3 = 5 m, W1 = W2 = W3 = 2 kN/m Since ends A and D are simply supported, the support moments at A and D will be zero. \

MA = 0

and

MD = 0

From symmetry MB = MC To find the support moments at B and D Clapeyron’s equation of three moments is applied for ABC and for BCD. For ABC, we get

190

Strength of Materials

or

MAL1 + 2MB (L1 + L 2) + MC L 2 =

6a1 x1 6 a2 x 2  L1 L2

0 ´ 5 + 2MB (5 + 5) + MC ´ 5 =

6a1 x1 6 a2 x 2  5 5

6 (a1 x1  a 2 x 2 ) 5 Now, a1 = Area of B.M. diagram due to u.d.l. on AB when AB is considered as simply supported beam

or

20MB + 5MC =

=

2 ´ AB ´ Altitude 3

=

2 WL ´5´ 1 1 3 8

=

2 2 – 52 ´5´ = 20.833 m2 8 3

2

and

x1 =

L1 5 = = 2.5 m 2 2

Due to symmetry a2 = a1 = 20.833

x 2 = x1 = 2.5

and

Substituting these values in Eq. (i), we get 20MB + 5MC =

6 (20.833 ´ 2.5 + 20.833 ´ 2.5) 5

6 ´ 104.165 = 124.99 5 20MB + 5MB = 124.99 ( MB = MC due to symmetry)

=

or

'

124.99 = 4.99 kN-m 25 MB = MC = 4.99 kN-m

MB = \

Now, the B.M. diagram due to support moments is drawn as shown in Fig. 5.26(b), in which MA = 0, MD = 0, MC = 4.99 kN-m. Let RA, RB, RC and RD are the support reactions at A, B, C and D respectively. Due to symmetry,

and

RA = RD RB = RC For the span AB, taking moments about B, we get MB = RA ´ 5 – 2 ´ 5 ´

5 2

Chapter 5:

or \ Due to symmetry,

– 4.99 = RA ´ 5 – 25 ( RA = 4.002 kN RD = RA = RA + RB + RC + RD = RA + RB + RB + RA = 2(RA + RB) = RA + RB = RB =

Now, or or or or

Deflection of Beams

191

' MB = – 4.99)

4.002 kN Total load on ABCD 2 ´ 15 ( RC = RB, RD = RA) 30 15 10.998 kN

'

Fig. 5.26

EXAMPLE 5.11 A continuous beam ABCD simply supported at A, B, C and D is loaded as shown in Fig. 5.27(a). Find the moments over the beam and draw B.M. and S.F. diagrams. Solution: Given L1 = 6 m, L2 = 5 m, L3 = 4 m, W1 = 10 kN, W2 = 12 kN, W3 = 5 kN/m B.M. diagram due to vertical loads taking each span as simply supported Consider beam AB as simply supported. B.M. at point load at E =

W1 – a – b 10 – 2 – 4 = 6 L1 = 13.33 kN·m

(

' Here a = 2 m, b = 4 m)

Similarly, B.M. at F, considering beam BC as simply supported, =

W2 – a – b 12 – 2 – 3 = 5 L2 = 14.4 kN·m

(

' Here a = 2 m, b = 3 m and L2 = 5)

192

Strength of Materials

The B.M. at the centre of a simply supported beam CD, carrying u.d.l.

W3 – L23 5 – 42 = 8 8 = 10 kN·m

=

B.M. diagram due to support moments Let MA, MB, MC and MD are supports moments at A, B, C and D, respectively. But the end supports of a simply supported beam are not subjected to any bending moment. Hence the support moments at A and D will be zero. \

MA = 0

and

MD = 0

To find the support moments at B and C, Clapeyron’s equation of three moments applied for ABC and for BCD. (a) For spans AB and BC from equation of three moments, we have MA L1 + 2MB (L1 + L2) + MC L2 = or

6 a1 x1 6 a2 x 2  L1 L2

2MB (6 + 5) + 5MC = a1 x1 +

6 a x 5 2 2

(i)

Now a1 x1 = Moment of area of B.M. diagram due to vertical load on AB when AB is considered as simply supported beam about point A =

1 2–2 1 ´ 2 ´ 13.33 ´  ´ 4 ´ 13.33 ´ 2 3 2

 2  1 – 4  3 

= 17.77 + 88.87 = 106.64 and a2 x 2 = Moment of area of B.M. diagram due to vertical load on BC when BC is considered as simply supported beam about point C =

1 2 1 ´ 3 ´ 14.4 ´ ´3+ ´ 2 ´ 14.4 ´ 2 3 2

 3  1 – 2  3 

= 43.2 + 52.8 = 96 Substituting these values in Eq. (i), we get

6 – 96 = 221.84 5 (b) For spans BC and CD from equation of three moments, we have 22MB + 5MC = 106.64 +

MB L 2 + 2MC (L 2 + L3) + MD L3 = or

MB ´ 5 + 2MC (5 + 4) =

or

5MB + 18MC =

(ii)

6 a2 x 2 6 a3 x 3  L2 L3 6 a2 x 2 6 a3 x 3  ( 5 4 6 6 a2 x 2 + a3 x 3 4 5

' MD = 0) (iii)

Chapter 5:

Deflection of Beams

193

where a2 x 2 = Moments of area of B.M. diagram due to vertical load on BC when BC is considered a simply supported beam, about point B =

1 2 1 ´ 2 ´ 14.4 ´ ´2+ ´ 3 ´ 14.4 ´ 2 3 2

 2  1 – 3  3 

= 19.2 + 64.8 = 84 and

a3 x 3 = Moment of area of B.M. diagram due to uniformly distributed load (u.d.l.) on CD, when CD is considered as simply supported beam, about point D =

 2 – Base – Altitude – Base 3  2

2 4 ´ 4 ´ 10 ´ = 53.33 3 2 Substituting these values in Eq. (iii), we get

=

5MB + 18MC =

6 6 ´ 84 + ´ 53.33 = 180.79 5 4

(iv)

Solving Eqs. (ii) and (iv), we get MB = 8.32 kN·m

and

MC = 7.73 kN·m

Now, the B.M. diagram due to support moments is drawn as shown in Fig. 5.27(c), in which MA = 0, MB = 8.32, MC = 7.73 and MD = 0. Support reactions Let RA, RB, RC and RD are the support reactions at A, B, C and D, respectively. For the span AB taking moments about B, we get or or

MB = RA ´ 6 – 10 ´ 4 –8.32 = 6RA – 40

40  8.32 = 5.28 kN 6 For the span CD taking moments about C, we get RA =

MC = R D ´ 4 – 5 ´ 4 ´

4 2

or

–7.73 = 4RD – 40 40  7.73 \ RD = = 8.06 kN 4 Now, taking moments about C for ABC, we get MC = RA(6 + 5) – 10(5 + 4) + RB ´ 5 – 12 ´ 3 or

\ \

–7.73 = 5.28 ´ 11 – 10 ´ 9 + 5RB – 36 5RB = 90 + 36 – 7.73 – 58.08 RB = 12.03 kN

194 Now,

Strength of Materials

RC = Total load on ABCD – (RA + RB + RD) = (10 + 12 + 20) – (5.28 + 12.03 + 8.06) = 16.63 kN

Fig. 5.27

EXERCISES 5.1 What is the Clapeyron’s theorem of three moments? Derive an expression for Clapeyron’s theorem of three moments. 5.2 What is moment area method? Derive an expression for finding out slope and deflection. 5.3 Differentiate first and second Mohr’s theorems. 5.4 Derive an expression for the slope and deflection of a beam subjected to uniform bending moment. Prove that M = EI

d2 y dx 2

where M = Bending moment E = Young’s modulus I = M.O.I. 5.5 What is a Macaulay’s method? Where is it used? Find an expression for deflection at any section of a simply supported beam with an eccentric point load using Macaulay’s method. 5.6 Determine the slope and deflection at the free end of a cantilever of length l carrying a point load at the free end.

Ans. !

R=

wl 2  wl 3 ,E = 2 EI 3 EI

"# $

5.7 A cantilever beam of length 3 m carries a u.d.l. of 2 kN/m spread over the whole length and a point load of 2 kN at the free end. Calculate the deflection at the free end. The crosssection of the beam is 120 mm ´ 240 mm. Take E = 200 GPa. [Ans. y = – 1.38 mm]

Chapter 5:

Deflection of Beams

195

5.8 A simply supported beam of span 6 m carries a u.d.l. throughout. If the maximum deflection is not to exceed 10 mm. Calculate the value of u.d.l. Take E = 200 GPa and I = 40 ´ 10–6 m4. [Ans. W = 4.74 kN/m] 5.9 A continuous beam consists of three successive spans of 8 m, 10 m and 6 m, and carries loads of 6 kN/m, 4 kN/m and 8 kN/m, respectively on the spans. Determine the bending moments and reactions at the supports. [Ans. (i) M A = MD = 0, M C = 32.2 kN·m, M B = 40.16 kN·m (ii) RA = 18.98 kN, RB = 49.82 kN, RC = 48.57 kN, RD = 18.63 kN] 5.10 A continuous beam ABC consists of two consecutive spans AB and BC of length 8 m and 6 m, respectively. The beam carries a uniformly distributed load of 1 kN/m throughout its length. The end A is fixed and the end C is simply supported. Find the support moments and the reactions. Also draw the S.F. and B.M. diagrams. [Ans. (i) MA = 5.75 kN-m, MB = 4.5 kN·m, MC = 0 (ii) RA = 4.15 kN, RB = 7.6 kN, RC = 2.15 kN]

196

Strength of Materials

6 6.1

Curved Beam

INTRODUCTION

T M E = = was derived assuming the beam to be initially straight before the y I R application of a bending moment. However, machine member and structure subjected to bending are not always straight as in the case of chain links and crane hooks, before a bending moment is applied to them. For initially straight beam the simple bending formula is applicable and the neutral axis coincides with the centroidal axis. A simple flexure formula may be used for curved beams for which the radius of curvature is more than five times the beam depth. To deal with such cases Winkler– Bach theory is used. Bending equation

6.2

STRESSES IN CURVED BEAM (WINKLER–BACH THEORY)

The following assumptions are made in this analysis: 1. Plane sections (transverse) remain plane during bending. 2. Radial strain is negligible. 3. The material considered is isotropic and obeys Hooke’s law. 4. The fibres are free to expand or contract without any constraining effect from the adjacent fibres. Consider a portion of a beam ABCD initially in unstrained state and ABC¢D¢ be the strained position of the beam. For simplicity the fibers AB shown on same as before and after the application of bending. Circumferential strain in EF, ec =

=

EF „  EF EF „ = –1 EF EF

( R1  y1 )R T –1= ( R  y)G E

(6.1)

where s is the bending stress in EF.

Fig. 6.1 196

Bending of a curved beam.

Chapter 6:

Curved Beam

197

Let e ¢c is the strain in GH when it increases to GH¢. Then GH „  GH GH „ = 1 GH GH

e ¢c =

R1R –1 RG

=

(1 + e ¢c)Rf = R1q

R R (1  F „c ) = R1 G

(6.2)

Substituting Eq. (6.2) in Eq. (6.1), we have ( R1  y1) R (1  F c )  –1 R1 ( R  y)

ec =

1  y  (1  F „ )  R 1

c

1

=

–1 y R According to assumption (2) radial strain is negligible. Hence

1

y = y1 È É1  Ê

ec =

È É1  Ê

1 =

Adding and subtracting e ¢c

yØ R1 ÙÚ (1  F c„ ) – 1 yØ Ù

RÚ

y y y  Fc„  F c„  1  R1 R1 R y 1 R

y in the numerator and simplifying, we get R

(1  F „c ) e c = e ¢c +

 1  1  y  R R 1

1

(6.3)

y R

The tensile stress in EF¢ becomes

s = ec E =

Ë Ì E ÌF c„  Ì Ì Í

1 1Ø Û  Ù yÜ RÚ Ü Ê R1 Ü y È

(1  F c„ ) É

1

R

Ü Ý

(6.4)

198

Strength of Materials

Total force on the section is: F=

ò s dA

Consider a small strip of elementary area dA, at a distance y from the centroidal layer GH, we have

F=

Ë Ì E Ì Ô F c„ dA  Ô Ì Ì Í

Û 1 1Ø  Ùy Ü RÚ Ê R1 dA Ü Ü y 1 Ü R Ý È

(1  F c„ ) É

È

= Ee c¢ A + E(1 + e c¢) É

1

Ê R1



1Ø R ÙÚ Ô

y dA y 1 R

(6.5)

The total resisting moment is given by M = òs y dA

= Eò e ¢cy dA + E

I

(1  F „c )

 1 R



1

1

1 R

y R

  y dA 2

 1  1  y dA  R R I 1  y 2

= 0 + E(1 + e ¢c)

[

1

 1  1  y dA  R R I 1  y

' ydA = 0]

R

2

M = E(1 + e ¢c)

1

I

R

y

2

dA = Ah2 y 1 R 2 where h is the constant for the cross-section of the bar.

Let

\ Consider,

 1  1   R R

M = E(1 + e c¢)

I

y y 1 R

dA =

I

(6.6)

Ah2

1

Ry dA Ry

I  y  Ry y  dA  y  dA = I y dA – I   R  y  2

=

2

(6.7)

Chapter 6:

1 R

=0–

I

I 

Curved Beam

 dA y 1   R y2

y dA 1 y 2 dA 1 = – = – Ah2 Ô y y R R 1 1 R R

\

199

Hence Eq. (6.5) becomes F = Ee c¢ A – E(1 + e c¢)

 1  1  Ah  R R  R

(6.8)

2

(6.9)

1

Since transverse plane sections remain plane during bending, F=0

 1  1  Ah  R R R  1 1  Ah Ee ¢ A = E(1 + e ¢ )     R R R  1  1  h e ¢ = (1 + e ¢ )   R R R  1 1 F R = (1 + e ¢ )    h  R R

2

0 = Ee c¢ A – E(1 + e c¢)

1

2

or

c

c

1

2

or

c

c

1

„c

(6.10)

c

2

1

Substituting the value of Eq. (6.10) in Eq. (6.7), we get M=E

F „c R

h2 = Ee c¢ RA

M EAR Substituting the value of e c¢ in Eq. (6.4), we get

Ah2

e c¢ =

s=

=

=

 

(6.11)

 

F „c R M y +E y h2 AR 1 R

M +E AR

M y h EAR 1 R

M M  AR AR

y

R

2

 Ry  1  1  y  h  R

2

M Û ' F „ = EAR Ü

Ë Ì Í

c

Ý

200

Strength of Materials

s=

or

 !

M R2 1 2 AR h

 y  "#  y  R  #$

(6.12)

On the other side of GH, y will be negative, and the stress will be compressive.

s=

 !

M R2 1 2 AR h

 y  "#  R  y  #$

(6.13)

If the bending moment is applied in such a manner that it tends to increase the curvature of beam then Eq. (6.12) will give tensile stress and Eq. (6.13) gives compressive stress.

6.3

POSITION OF NEUTRAL AXIS

At the neutral axis,

s = 0.

 !

 

 "#  #$  y   R  y

M R2 y 1 2 AR h Ry

R2 h2

=0

= –1

R2y = – Rh2 – h2y y(R + h2) = – Rh2 2

 Rh   R  h  2

y =–

2

(6.14)

2

Hence neutral axis is located below the centroidal axis.

6.4 Now,

VALUES OF h2 h2 =

=

=

=

h2 =

1 A

I I

y 2 dA y 1 R

[Eq. 6.6]

R y 2 dA R = A A Ry

RË

Ì A ÌÍ Ô

 !

I

%& y( y  R)  R ( y  R)  R ()dA yR ' *

y dA  Ô R dA  Ô

I

R R 2 dA 0  RA  A yR R3 A

I

2

dA – R2 Ry

"# #$

R 2 dA Û Ü

y  R ÜÝ

' Ô y dA = 0 and Ô dA = AÛÝ

Ë Í

(6.15)

Chapter 6:

6.4.1

Rectangular Cross-section

Figure 6.2 shows the rectangular section of width B and depth dy at a distance y from the centroidal layer.

h2 =

R3

dA

dy y

Area of strip dA = Bdy Area of section A = BD

D

D /2

B dy – R2 BD Ô R  y

R

 D /2

B

R3 D/ 2 log e ( R  y)  D / 2  R 2 = D h2 =

6.4.2

201

Curved Beam

 

R3 2R  D log e 2R  D D

 

– R2

(6.16)

O

Fig. 6.2

Rectangular cross-section.

Circular Cross-section

Figure 6.3 shows the circular section of diameter D of a curved bar of radius of curvature R, from the centre of curvature O.

Q

Area of cross section A =

b dy

D2

y

4 Consider a strip of width b and depth dy at a distance y from the centroidal layer. Then

D2  y2 4

b=2 Area of strip dA = bdy

 

2

= 2

Thus,

h2 =

=

R

3 D /2

A

8R

2

Ô

 D/2

3

D /2

Q D2 Ô

 D/2

R

 

D  y 2 dy 4

O

Fig. 6.3

Circular cross-section.

D2  y2 4 dy – R 2 Ry D2  y2 4 dy – R 2 Ry

Expanding the integral by binomial expression and then integrating, we get h2 =

D2 1 D4  +… 16 128 R 2

(6.17)

202

Strength of Materials

6.4.3

I-section

Figure 6.4 shows a I-section. Let Then h2 =

Now,

 !  R A ! R3 A 3

=

I I

R2

R1 R2

R1

R+y =a dy = da

dA  R y b2 da  a

I I I I R3

R2

R3

dA  Ry

t 3 da  a

R2

R4

dA Ry

R3

R4

R3

b1 da a

"# #$ – R

"# #$ – R

2

2

 R2 R R   t3 loge 3  b1 loge 4  – R 2  b2 loge R1 R2 R3   A = b1 t1 + b2 t2 + b3 t3 =

where

R3 A

(6.18)

b1 dy

t1

y

b3

R3 R4

t3

t2

R

b2

R2

R1

O

O

Fig. 6.4

6.4.4

T-section

Figure 6.5 shows a T-section. Let Then Now,

 !  R A !

R3 h = A 2

3

=

I I

R2

R1 R2

R1

R+y =a dy = da

dA  Ry b2 da  a

I I

R3

R2

R3

R2

dA Ry t1 da a

"# #$ – R

"# #$ – R

2

2

 R2 R   t1 loge 3  – R 2 b2 loge R R  1 2 A = b1t1 + b2t2

=

where

I-section.

R3 A

(6.19)

Chapter 6:

Curved Beam

t1 dy y b1

R3

t2

R

b2

R1 R2

O

O

Fig. 6.5 T-section.

6.4.5

Trapezoidal Cross-section

Figure 6.6 shows trapezoidal cross-section.

Fig. 6.6 Trapezoidal cross-section.

Let Then Now,

Z= R +y dZ = dy

 B  C  (R d d 

b=C+

1

Thus,

– Z)

dA = bdy

  B  C  1 R  Z 6"# dZ ! d d  $

= C

Now,

2

2

R3 h = A 2

2

I  I ! R2

[C 

R1

=

R3 A

1

R2

R1

2

BC dZ ( R  Z )] – R2 d1  d2 2 Z

 

CdZ BC  Z d1  d2

 

I

R2

R1

"# #$

( R2  Z ) dZ – R 2 Z

203

204

Strength of Materials

=

=

h2 =

R3 A R3 A R3 A

Ë ÌC Í

log e Z

Ë ÌC Í

log e

È B C Ø R2 | R2  ÉÊ d  d Ù R1 1 2Ú

R2 R1



Û

log e Z  Z |RR2 Ü – R 2

È B C ØÎ Ï R2 ÉÊ d  d Ù 1 2ÚÐ

1

log e

R2 R1



Ý

ÞÛ

( R2  R1 ) ßÜ – R 2 àÝ

Ë È R  d2 Ø È B  C Ø È R  d2 Ø ( R  d2 ) log e É É  (B  ÌC log e É Ù Ù Ê R  d1 Ú Ê d Ú Ê R  d1 Ù Ú Í

where

 B  C  d  2  d  B  2C  =   3  BC 

Û

C ) Ü – R 2 (6.20) Ý

A=

d1

d2 = d – d1

EXAMPLE 6.1 Figure 6.7 shows a circular ring of circular cross-section 8 cm in diameter. The inside diameter of the ring is 16 cm. The load P is 25 kN. Calculate stresses at A and B. P = 25 kN

A

B D = 8 cm

16 cm

Fig. 6.7

Solution:

Area of cross-section A =

Q (8)2 = 50.26 cm2 4

1 D4 D2  +… 16 128 R 2

Then

h2 =

Here

D = 8 cm

and

R = 8 + 4 = 12 cm

\

h2 =

 

64 1 64 – 64  16 128 144

 

= 4.22 cm2

Chapter 6:

Curved Beam

205

Here the bending moment is such that it is tending to increase the curvature of the ring. Therefore Eq. (6.13) will give compressive stress and Eq. (6.12) gives tensile stress at point A. Now,

M = P. R = 25 ´ 103 ´ 12 ´ 10–2 = 3000 N-m

Stress at point A is:

sA = – =–

 !

 

d1 P M R2  1 2 R  d1 A AR h

 "#  #$

= 84.86 MPa (Compressive) Stress at point B is:

sB = – =–

 !

 

 "#  #$

 !

 

 "#  #$

(12 )2 25 – 10 3 3000 – 10 6 4 ´ 104 + 1 50.26 50.26 – 12 4.22 12  4

 !

P M R2  1 2 A AR h

 d  "#  R  d  #$ 2

2

(12 ) 2 25 – 10 3 3000 – 10 6 4 ´ 104 + 1 50.26 50.26 – 12 4.22 12  4

= 43.43 MPa (Tensile)

EXAMPLE 6.2 An open ring having T-section, as shown in Fig. 6.8, is subjected to compressive load of 120 kN. Determine stresses at A and B.

Fig. 6.8

Solution:

y1 =

A1 x1  A2 x 2 A1  A2 100 – 20 – 10  20 – 150 –

=

 150  20 2

100 – 20  20 – 150

= 61 mm

206

Strength of Materials

Area of T-cross-section A = b1t1 + b2t2 = 100 ´ 20 + 20 ´ 150 = 5000 mm2 Now, r1 = 200 mm, r2 = 220 mm, r3 = 370 mm, R = 200 + 61 = 261 mm, y2 = 170 – 61 = 109 mm

 !

R3 r r b2 ln 2  t1 ln 3 A r1 r2

"# – R $

\

h2 =

As

b2 = 100 mm and t1 = 20 mm,

2

12616 100 ln 220   20 ln 370  "# – (261)  200   220  $ 5000 ! 3

\

2

h =

2

= 3555.91 [100 ´ 0.0953 + 20 ´ 0.519] – 68121 = 2677.16 mm2 Here the bending moment is such that it is tending to increase the curvature of the ring. Therefore, Eq. (6.13) will give compressive stress and Eq. (6.12) give tensile stress. Stress at point A is:

sA = –

=–

 !

 

y1 W M R2  1 2 A AR R y1  h

 "#  #$

 !

 

120 – 10 3 120 – 10 3 – 261 (261) 2 61 1  5000 5000 – 261 2677.16 261  61

 "#  #$

= –24 – 162.26 = – 186.25 MPa (Compressive) Stress at point B is:

sB = – =–

 !

W M R2  1 2 A AR h

 y R y 2

2

 "#  #$

 !

 

120 – 10 3 120 – 10 3 – 261 (261) 2 109 1  5000 5000 – 261 2677.16 261  109

 "#  #$

= –179.90 MPa (Tensile)

EXAMPLE 6.3 Figure 6.9 shows a C frame subjected to a load of 100 kN. Determine the stresses at A and B. Solution:

y1 = =

A1 x1  A2 x 2  A3 x3 A1  A2  A3

120 – 20 – 10  20 – 140 – 90  100 – 20 – 170 120 – 20  20 – 140  100 – 20

= 85.55 mm y2 = 180 – 85.55 = 94.45 mm

Chapter 6:

Curved Beam

207

Fig. 6.9

Area of cross-section for I-section = 120 ´ 20 + 20 ´ 140 + 100 ´ 20 = 7200 mm2 As per the notation shown in Figure 6.4. R1 = 150 mm, R2 = 170, R3 = 170 + 140 = 310 mm, R4 = 150 + 180 = 330 mm, R = 150 + 85.55 = 235.55 mm h2 =

Now,

=

 !

R R R R3 b ln 2  t ln 3  b1 ln 4 R3 A 2 R1 3 R2

 !

"# – R $

2

(235.55)3 170 310 330  20 ln  100 ln 120 ln 7200 150 170 310

"# – (235.55) $

2

= 1815.16 [120 ´ 0.125 + 20 ´ 0.6007 + 100 ´ 0.0625] – 55483.80 = 1815.16 [33.264] – 55483.80 = 4895.68 mm2 Here the bending moment is such that it is tending to increase the curvature of the ring. Therefore, Eq. (6.13) will give compressive stress and Eq. (6.12) gives tensile stress.

sA = –

=–

 !

 

y1 W M R2  1 2 A AR h R  y1

 "#  #$

 !

 

 "#  #$

 

 "#  #$

100 – 10 3 100 – 10 3 235.55 (235.55)2 85.55 1  – 7200 7200 235.55 4895.68 235.55  85.55

= –13.88 – 75.88 = –89.77 MPa (Compressive)

sB = –

=–

 !

W M R2  1 2 A AR h

 y R y 2

2

 "#  #$

 !

100 – 10 3 100 – 10 3 235.55 (235.55)2 94.45 1  – 7200 7200 235.55 4895.68 235.55  94.45

= –45.05 MPa (Tensile)

208

Strength of Materials

EXAMPLE 6.4 A curved member shown in Fig. 6.10 has a solid circular cross-section 0.10 m in diameter. If a maximum tensile and compressive stress in the member are not to exceed 150 MPa and 200 MPa, respectively, determine the value of load P that can safely be carried by the member. (AMIE, 1996) P 0.15 m

0.05 m 1

2

D = 0.10 m

R = 0.10 m

Fig. 6.10

Solution: Bending moment M = P (0.15 + 0.1) = 0.25P N-m Area of cross-section is given by A= Then

h2 =

Q 4

´ (0.10)2 = 7.854 ´ 10–3 m2

1 D4 D2  +… 16 128 R 2

(0.10)2 1 (0.10)4  = 7.031 ´ 10–4 m2 16 128 (0.10)2 Bending stress at point 1 is: =

s1 = – 150 ´ 106 = –

 !

 

P M R2 y  1 2 A AR h Ry P 7.854 – 10 3



 "#  #$

P (0.25) 7.854 – 10 3

[\ tension]

1  (0.1)  0.05  "#   – 0.1 ! (7.031 – 10 )  0.10  0.05  #$ 2

4

= –127.32P + 318.31P ´ 5.74 = 1699.78P 150 – 10 6 = 88.25 kN 1699.78 Bending stress at point 2 is:

P=

s2 = –

 !

P M R2  1 2 A AR h

 y  "#  R  y  #$

(i)

[\ compression]

Chapter 6:

200 ´ 106 = –

P 7.854 – 10 3



P (0.25) 7.854 – 10 3

Curved Beam

209

1  (0.1)  0.05  "#   – 0.1 ! (7.031 – 10 )  0.10  0.05  #$ 2

4

= –127.32P – 318.31P ´ 13.22 = –4335.38P

200 – 10 6 = 46.13 kN [compression] 4335.38 Comparing Eqs. (i) and (ii), the safe load P will be lesser of these. P=

(ii)

EXAMPLE 6.5 Determine the ratio of maximum and minimum value of stresses for a curved bar of rectangular section in pure bending. Radius of curvature is 8 cm and depth of beam is 6 cm. Locate the neutral axis. (UPTU – 2001–02) Solution:

Given R = 8 cm, D = 6 cm, d1 = d2 =

Then

h2 =

D = 3 cm 2

  – R    16  6  – 8 = 3.28 cm 8 ln  6  16  6  R3 2R  D ln D 2R  D

2

3

=

2

Location of neutral axis, y=–

=–

smax =

Minimum stress

smin

\

T max T min

R2  h 2

8 – 3.28 = –0.39 cm 64  3.28

   "# !   #$ M  R  d " 1 =  # AR ! h  R  d  #$ 64  3  1   10.707 3.28  8  3  = = 6.32 64  3  1   3.28  11

Maximum stress

Now,

Rh 2

d1 M R2 1 2 AR h R  d1

smax = 1.69smin

2

2

2

2

2

210

Strength of Materials

EXAMPLE 6.6 A curved bar of square section 3 cm, sides and mean radius of curvature 4.5 cm, is initially unstressed. If a bending moment of 300 N-m is applied to the bar to straighten it, find the stresses at inner and outer faces. Solution:

D = 1.5 cm, M = 3 ´ 104 cm 2

Given R = 4.5 cm, d1 = d2 = h2 =

Then

=

smax =

Now,

 

R3 2R  D ln D 2R  D

 

– R2

  – (4.5) = 0.803 cm   M  R  d " 1  # AR ! h  R  d  #$ 3 – 10  (4.5)  1.5  " 1  # 9 – 4.5 ! 0.803  4.5  1.5  #$ (4.5)3 12 ln 3 6

2

2

1

2

1

4

=

2

= –86 MPa

smin =

and

=

2

 !

 

d2 M R2 1 2 AR h R  d2

 !

 

 "#  #$

3 – 10 4 1.5 (4.5)2 1 9 – 4.5 0.803 4.5  1.5

 "#  #$

= 54.10 MPa

EXAMPLE 6.7 A curved beam of rectangular cross-section is subjected to pure bending with couple of 400 N-m. The beam has width 20 mm, depth of 40 mm and is curved in a plane parallel to the depth. The mean radius of curvature is 5 cm. Find the position of neutral axis and the ratio of maximum to the minimum stress. Solution:

Then

Area of cross-section A = 2 ´ 4 = 8 cm2 d1 =

D = 2 cm 2

h2 =

R3 2R  D ln D 2R  D

=

 

 

 

– R2

(5) 3 2 –54 ln 4 2 –54

 

– (5)2 = 1.478 cm2

Chapter 6:

Curved Beam

211

Location of neutral axis, y=– =– Bending stress at the inside face,

s1 = =

 !

= Thus,

R2  h 2

5 – 1.478 = –0.279 cm 25  1.478

 

d1 M R2 1 2 AR h R  d1

 !

 

 "#  #$

25 2 4 – 10 4 1 1.478 5  2 8–5

Bending stress at the outside face,

s2 =

Rh 2

 !

 

d2 M R2 1 2 AR h R  d2

 !

 

 "#  #$

4 – 10 4 25 2 1 8–5 1.478 5  2

T max T = 1 = 1.76 T min T2

 "# = 102.8 MPa  #$  "#  #$

= 58.3 MPa

EXAMPLE 6.8 A curved beam trapezoidal in cross-section as shown in Fig. 6.11 is subjected to pure bending with couple of 400 N-m. The mean radius of curvature is 50 mm. Find the position of the neutral axis and the ratio of the maximum to the minimum stress.

40 mm

30 mm

Solution: Given C = 20 mm, B = 30 mm, d = 40 mm, R = 50 mm Then

d1 =

= and

    40  30  40    3  30  20  d B  2C 3 BC

= 18.67 mm

20 mm Fig. 6.11

d2 = d – d1 = 40 – 18.67 = 21.33 mm Area of cross-section is:

 B  C  ´ d  2   30  20  ´ 40 = 1000 mm =   2 

A=

2

212

Thus,

Strength of Materials

  "#            #$ !   50  21.33    30  20  (50  21.33) ln  50  21.33   (30  20)"# – (50) (50 )  = 20 ln  1000 !  50  18.67   40   50  18.67  #$

h2 =

R  d2 R  d2 R3 BC C ln  ( R  d2 )ln  ( B  C) – R 2 A R  d1 d R  d1 3

2

\

h2 = 140 mm2

Now,

 "#  #$ 400 – 10  50  18.67  " 1 =   # = 77.1 MPa 1000 – 50 ! 140  50  18.67  #$ M  R  d " 1 =  # AR ! h  R  d  #$ 400 – 10  (50)  21.33  " 1 =   # = 50.7 MPa 1000 – 50 ! 140  50  21.33  #$

smax =

 !

 

d1 M R2 1 2 AR h R  d1 3

smin

and

2

2

2

2

3

2

2

T max 77.1 = = 1.52 T min 50.7

\

Location of neutral axis, y=–

=–

Rh 2 R2  h2

50 – 140 = –2.65 mm 2500  140

EXAMPLE 6.9 A central horizontal section of hook shown in Fig. 6.12 is a symmetrical trapezium 50 mm deep, the inner width being 60 mm and outer width being 30 mm. Calculate the extreme intensities of stress, when the hook carries a load of 30 kN, the load line passing 40 mm from the inside edge of section and the centre of curvature being in the load line. (UPTU 2002–03) Solution: Given C = 30 mm, B = 60 mm, d = 50 mm Area of cross-section is given as:

 60  30  ´ 50 = 2250 mm  2  d  B  2C  =   3 BC 

A=

Then

d1

2

Curved Beam

60 mm

Chapter 6:

30 mm

A

B 40 mm

50 mm

30 kN Fig. 6.12

=

 

50 60  60 3 60  30

 

= 22.22 mm

and

d2 = d – d1

\

= 50 – 22.22 = 27.78 Mean radius R = 40 + 22.22 = 62.22

Now,

  "#            #$ !   62.22  27.78    60  30  (62.22  27.78) (62.22 )  = 30 ln  2250 !  62.22  22.22   50   62.22  27.78   (60  30)"# – (62.22) ln   62.22  22.22  #$

h2 =

R  d2 R  d2 R3 BC C ln  ( R  d2 )ln  ( B  C) – R 2 A R  d1 d R  d1 3

2

= 107.055 [30 ´ 0.811 + 54 ´ 0.811 – 30] – 3871.33 = 210.03 mm2 Direct stress

= Bending stress at A,

sA =

or

sA =

sd =

P A

30 – 10 3 = 13.33 MPa 2250

 ! M  R 1 AR ! h

M R2 1 2 AR h 2

2

 y  "#  R  y  #$  d  "#  R  d  #$ 2

2

213

214

Strength of Materials

=

 1 !

6  27.78  "#  62.22  27.78  #$ 2

62.22 30 – 10 3 – 62 .22 1 2250 – 62.22 210.03

= 89.186 (Compressive) So the resultant stress at A is given by (sr)A =

sd + sA

= 13.33 (Tensile) – 89.186(Compressive) (sr)A = 75.86 MPa Bending stress at B,

sB =

=

 !

 

d1 M R2 1 2 AR h R  d1

 !

 "#  #$

 

30 – 10 3 62.22 (62.22 ) 2 22.22 1 – 2250 62.22 210.03 62.22  22.22

 "# = 123.18 MPa  #$

So the resultant stress at B is given as: (sr)B =

sd + sB

= 13.33 + 123.18 = 136.51 MPa (Tensile)

6.5

STRESSES IN A RING

Consider a circular ring loaded as shown in Fig. 6.13. Let M2 be the bending moment at any section x1 – x2 inclined at angle q with the line of action of the applied load W. The portion x1 DFx2 of the ring is in equilibrium under the action of M1 at DF, pull W/2 at DF and the moment M2 at x1 – x2 along with pull T at x1 – x2. We get M2 = M1 +

W R(1 – sinq) 2

Also from Eq. (6.7) M2 = E(1 + e c¢ ) ´

(6.21)

 1  1  Ah  R R

2

(6.22)

1

Comparing Eqs. (6.21) and (6.22), we get E(1 + e c¢ ) ´

 1  1  Ah  R R

2

= M1 +

1

W R(1 – sinq) 2

Multiplying both sides by Rdq and integrating from 0 to p /2, we have Q /2

E

Ô 0

 1 1 (1 + e ¢ )    Ah  R R

2

c

1

R dq =

Q /2

Ô 0

M1R dq +

Q /2

Ô 0

W 2 R (1 – sinq )dq 2

(6.23)

Chapter 6:

Curved Beam

215

W H W1

J K

M2

R A

B

x2 x1

q

C

D

E F M1

R sin q

q W/2

W/2

L M N

W1

W Fig. 6.13 Q /2

or

E

Ô 0

R (1  F c„) 2 Ah dq – E R1

Q /2 Ô

Closed ring.

(1 + e c¢ ) Ah2 dq =

0

Q /2 Ô

M1R dq +

Ô 0

0

W 2 R (1 – sinq )dq 2

R1 = R(1 + e c¢ )

Now, Q /2

\

Ô 0

R (1  F „c ) Q dq = R1 2

  2

Q EAh 2 – Q EAh 2(1 + e ¢ ) = Q M R + WR 2 Q  1 c 1

\

2

2

2

2

or \

–

 !

EA F „c  (1  F c„ )

Q 2

EAh2 e c¢ =

E(1 + e c¢ )

2

M1R +

 

 

WR2 Q 1 2 2

(6.24)

W sinq 2

(6.25)

%& 1  1 () h "# = ' R R* R $

W sinq 2

(6.26)

2

1

From Eq. (6.23), we have

Q

W1 =

Now,

\

Q /2

%& 1  1 () Ah ' R R* R 1

2

=

M1 W  (1 – sinq ) R 2

216

Strength of Materials

Substituting in Eq. (6.26), we get EAe c¢ =

W W M1 M W sin q + 1 (1 – sinq ) =  2 R 2 2 R e c¢ =

or Putting in Eq. (6.24), we get

M1 =

(6.27)

  ! 

 "#  #$

  ! 

 

WR 2 R2 1 2 Q R2  h 2

Substituting in Eq. (6.21), we get M2 =

W M  1 2 EA EAR

WR 2 R2  sin R 2 Q R2  h2

M2 will be maximum at q = 0° and q = 180°. \

Mmax =

(6.28)

"# #$

(6.29)

WR 3

(6.30)

Q ( R2  h 2 )

M2 will be zero, then sinq =

2 R2

(6.31)

Q ( R2  h 2 )

Thus, the bending moment and bending stress are zero at four points, one in each quadrant. Substituting the value of M1 in Eq. (6.27) from (6.28), we get e c¢ =

 !

W R2 AE Q ( R 2  h 2 )

e = e c¢ + (1 + e c¢ )

"# $

(6.32)

 1  1   y   R R   1  y   R 1

s = Ee

and From Eq. (6.22), we have (1 + e c¢ )

 1  1 =  R R  1

\

M1 EAh

2

WR 2 Ah 2 E

(1 – sinq)

%& M  WR (1  sin R )()  Ry  ' EAh 2 Ah E * R  y W % K& R (K)  %K& WR  2  R   1"# – 1  WR (1  sin R )(K)  Ry  AE ' KQ ( R  h ) *K K' 2 ! Q  R  h  #$ EAh 2 Ah E K*  R  y  e = e c¢

2

=



2

1

2

2

2

2

2

2

2

2

Chapter 6:

Now,

s = Ee =

 "# %K& #$ K' ! W R R  A ! Q (R  h ) 2h W R R  A ! Q (R  h ) 2h 2

2

2

2

=

 2 R  1"#  WR (1  sin R )(K)  Ry  K*  R  y  ! Q ( R  h ) #$ 2 Ah "#  %K& 2 R  sin R )(K) –  Ry  "# K*  R  y  #$ #$ !K' Q ( R  h ) K% 2 R  sin R K()  Ry  "# & K' Q (R  h ) K*  R  y  #$

W R2 WR  2 2 A Q (R  h ) 2 Ah 2 2

=

2

2

2

2

2

2

2

2

2

2

2

W sin R 2A Resultant stress sr = sd ± s Direct stress sd =

Then and

217

2

2

2

2

Curved Beam

(6.33)

(6.34)

(i) On a section taken along the line of action of W, q = 0° and the stresses become: (a) At outside of ring,

sr = and is tensile in nature. (b) At inside of ring,

sr =

 

 1  R  y  ! h  R  y

 

  R  y  1"#  ! h  R  y  #$

W R2 QA R 2  h 2

W R2 QA R 2  h 2

2

2

2

2

 "#  #$

2

1

2

1

and is compressive in nature. (ii) On a section perpendicular to the line of action of W, q = 90°, and the stresses become: (a) At outside of ring,

sr =

 !

K%& K'

and is compressive in nature. (b) At inside of ring,

sr =

K()  K* 

y2 2 R2 W R2 R2  1 – 2 2 2 2 2 A Q ( R  h ) 2h Q ( R  h ) R  y2

 !

K%& K'

K()  K* 

 

 "#  W  #$ 2 A "# #$

W R2 R2 W  y1 2 R2 –   1  2 2 2 2 2 A 2h Q (R  h ) R  y1 2A Q (R  h )

and is tensile in nature.

EXAMPLE 6.10 A steel ring of 22 cm mean diameter has a rectangular cross-section 5 cm in the radial direction and 3 cm perpendicular to the radial direction. If the maximum tensile stress is limited to 150 MPa, determine the tensile load the ring can carry.

218 Solution:

Strength of Materials

Area of cross-section is: A = 5 ´ 3 = 15 cm2 h2 =

Then

=

  – R    22  5  – (11) ln   22  5 

R3 2R  D ln D 2R  D

113 5

2

2

= 266.2 ln (1.5882) –121

= 2.144 cm2 The maximum tensile stress occurs at q = 0° on the outside of the ring.

s=

 

W R2 QA R 2  h 2

 1  R  y  "#  ! h  R  y  #$ 2

2

2

2

4

– 10 È 121 ØË Ì1  Q – 15 ÉÊ 121  2.144 ÙÚ Í

W

or

150 ´ 106 =

or or

150 ´ 106 = 2.3889 ´ 103 W W = 62.79 ´ 103 = 62.79 kN

121

È

2.5

ØÛ Ü

2.144 ÉÊ 11  2.5 ÙÚ Ý

EXAMPLE 6.11 A ring made of steel bar of 2.5 cm diameter carries a pull of 15 kN. Calculate the maximum tensile and compressive stresses in the material of ring. The mean radius of the ring is 15 cm. (UPTU 2003) h2 =

Solution:

=

Area of cross-section =

Now,

Q 4

1 D4 D2 +…  16 128 R 2

(2.5)2 1 (2.5)4  = 0.392 cm2 16 128 (15)2

(2.5)2 = 4.909 cm2 = 4.91 cm2

sD =

 

   !

W R2 R2 – 1 2 2 2 QA R  h h

Given R = 15 cm, h2 = 0.392 cm2, y2 = 2.5/2 = 1.25 cm

 

 y  "#  R  y  #$ 2

2

   !

 

=

15 – 10 3 225 225 1.25 – 1 – 10 4 Q – 4.91 225  0.392 0.392 15  1.25

=

15 – 10 7 225 Q – 4.91 225.392

 

 

´ [1 + 44.152]

 "#  #$

Chapter 6:

Curved Beam

219

15 kN D C

Mean radius = 15 cm A

2.5 cm

B

15 kN

Fig. 6.14

=

15 – 10 7 – 225 – 45.152 Q – 4.91 – 225.392

= 43.83 ´ 107 Pa = 438.3 MPa (Tensile)

 

sA = –

=–

=–

 !

%K& 'K

  R  y   1"#  ! h  R  y  #$  225  –  225 –  1.25  1 "# ´ 10  225.392  ! 0.392  13.75  $

W R2 – 2 QA R  h 2

sC =

2

1

2

1

=

15 – 10 3 Q – 4.91

=

15 – 10 7 225 – ´ 51.18 Q – 4.91 225.392

4

= 49.68 ´ 107 Pa = 496.8 MPa (Compressive)

(K)  y   R "#  W *K  R  y  Q ( R  h ) #$ 2 A

W R2 2 R2 1 2 A 2h Q (R2  h 2 ) 15 – 10 3 – 10 4 4.91

2

1

1

2

2

Ë 225 Î 2 – 225 Þ È 1.25 Ø  1ß É Ï Ì Ù  à Ê 13.75 Ú Í 2 – 0.392 Ð Q (225  0.392)

Û Ü Q (225.392) Ý

225

15 – 10 7 2 – 4.91

15 – 10 7 15 – 10 7 [–9.51 – 0.318] + 2 – 4.91 4.91

= (300.24 + 15.27) ´ 106 Pa = 315.51 MPa (Tensile) and

sB =

=

W A

Ë R2 Ì 2 ÌÍ Q ( R 

15 – 10 7 4.91

2

h )



È ØÈ y 2 R2 ØÛ 2 Ü  1Ù É É Ù 2h 2 Ê Q ( R 2  h 2) Ú Ê R  y2 Ú ÜÝ

R2

Ë 225  Ì Q – 225.392 Í

2A

È 2 – 225 Ø È 1.25 Ø Û  1Ù – É ÙÜ  – 0.392 ÉÊ Q – 225.392 Ú Ê 16.25 Ú Ý

225

2

W

15 – 10 7 2 – 4.91

220

Strength of Materials

15 – 10 7 15 – 10 7 [0.31776 – 8.0465] + 2 – 4.91 4.91

=

= (– 236.11 + 15.27) ´ 106 = – 220.84 ´ 106 Pa = 220.84 MPa (Compressive) Maximum tensile stress = 438.3 MPa Maximum compressive stress = 496.8 MPa

6.6

STRESSES IN A CHAIN LINK

Consider a chain link as shown in Fig. 6.15. Let R be the mean radius of the semicircular ends and ‘a’ be the length of the straight sides. Consider the equilibrium of the portion x1 CD x2 of the link. Taking moment about x1x2 section WR (1 – sin q ) 2

M2 = M1 +

Also from Eq. (6.7), M2 = E (1 + e c¢ )

(6.35)

 1  1  Ah  R R 

2

(6.36)

1

Hence by comparing Eqs. (6.35) and (6.36), we get E(1 + e c¢ )

 1  1  Ah  R R 

2

= M1 +

1

E

I

WR (1 – sin q) 2

(6.37)

Multiplying both sides by Rdq and integrate from 0 to p/2, Q /2

0

(1 + e c¢ )Ah 2

I

R dq – E R1

Slope of the tangent at L =

I

Q /2

0

(1 + e c¢ ) ´ Ah2 dq =

I

Q /2

0

I

M1R dq +

Q /2

o

Chain link.

WR 2 (1 – sinq )dq (a) 2

M1 a / 2 EI

(1  F „c ) Q M1a dq =  R1 2 2 EI Hence, from Eqs. (a) and (b), we get \

or

Fig. 6.15

Q /2

R

(b)

0

EAh2

 Q  M a   2 2 EI  1

– E(1 + e c¢ )Ah2

Q 2

 Q R  Aah   2 2 I 

=

2

or

Now, from Eq. (6.27)

M1

=

e c¢ =

Q 2

M1 R +

 





WR 2 Q 1 2 2

 

Q Q WR 2  EAh 2e c¢ 1 2 2 2

 

1 W M1  EA 2 R

 

(6.38)

(6.39)

Chapter 6:

Substituting Eq. (6.39) in (6.38), we get M1

 Q R  Aah  2 2I

2

 = R

 

221

 

WR2 Q Q  Wh 2 1 2 2 4

2 Qh

2

Curved Beam

I = Ak2

Now, where k is radius of gyration.





Q  Q Wh 2 WR 2 1 2 2 4 M1 = 2 Q R  ah  Q h 2 2 2k 2 2 R

Then

R  Q

2

W M1 =

and

R2 h 2  2 2



h2 R 2  R Qk

M2 = W

ah 2

È R2  É Ê Q

R

R2 2

ah



2

Q k2

 

h2 Ø Ù 2Ú 2



h R

(6.40)

+

WR (1 – sinq ) 2

(6.41)

Substituting Eq. (6.40) in Eq. (6.39), we get

 e c¢ =

1 W  EA 2

!

\ From Eq. (6.4),

s = Ee c¢ +

 

W R2 R2 h 2   2 Q 2 2 ah 2 h 2 R 2  R Qk

 !

 "#  # ## #$

WR 2 R2  sin R 2 Q 2 2 Ah ( R  h2)

(6.42)

"#  Ry   W sin R $  R  y  2A

(6.43)

EXAMPLE 6.12 A chain link (see Fig. 6.16) is subjected to a pull of 20 kN. It is composed of steel 2.5 cm diameter and has a mean radius of 3 cm. Its semicircular ends are connected by straight pieces 7.5 cm long. Estimate maximum compressive stress in the link and tensile stress at the same section. Solution:

Here A=

Q 4

´ (2.5)2 = 4.91 cm2

222

Strength of Materials 20 kN

D2 D4  h = 16 128 R 2 2

(2 .5 ) 2 (2 .5 ) 4  16 128 – 9

=

q

=

3 cm radius

R  Q

2

W M1 =

R



R2 h2  2 2

h2 a

h2 R



k 2Q

 

7.5 cm

0.4245 cm2 2.5 cm

Now, radius of gyration is:

k= or \

Qd 4

I = A

64 = d Q d2 4 4

20 kN

Fig. 6.16

k = 0.625 cm k2 = 0.3905 cm2

 9  9  0.4245  – 10 Q 2 2   0.4245 – 7.5    0.4245  3  Q – 0.3905   3 

2

20 – 10 3

Thus,

M1 =

=

200 (2.8648  4.5  0.2122 ) 3  2.59  0.1415

200 ( 1.8474) = – 64.46 N-m 5.7315 From Eq. (6.39), we have =

e0 =

=

=

 

1 W M1  EA 2 R

10 4 E

È 20 – 10 3  É – 4.91 Ê 2

64.46

Ø

Ù 3 – 10 2 Ú

15.99 – 10 6 E

From Eq. (6.42), we have s = Ee 0 +

 

WR 2 Ah

2

 2R  sin R "# –  Ry   W sin R #$  R  y  2A ! Q (R  h ) 2

2

2

Chapter 6:

Curved Beam

223

Compressive stress is maximum at q = 0° on the inside part of the link, \

y = –1.25 cm, q = 0°

and

 !

20 – 10 3 – 3 – 108 2–9 2 – 4.91 – 0.4245 Q – 9.4245

s = 15.99 ´ 106 +

"#   3 – 1.25  $  1.75 

15.99 + (–187.508) = –171.52 MPa Tensile stress at this location (on the outside surface),

 2 – R "#  Ry    2 Ah ! Q ( R  h ) #$  R  y  20 – 10 – 3 – 10  2 – 9 "  3 – 1.25  = 15.99 +  # 2 – 4.91 – 0.4245 ! Q – 9.4245 $  4.25 

s = 15.99 +

2

WR

2

2

2

3

2

2

8

= 93.19 MPa

EXERCISES 6.1 A crane hook is of trapezoidal cross-section having inner side 80 mm, outer side 300 mm and depth 120 mm. The radius of curvature of the inner side is 80 mm. If a load of 100 kN is applied to the hook passing through the centre of curvature, determine the maximum tensile and compressive stresses at the critical cross-section. [Ans. 141.9 MPa, 74.8 MPa] 6.2 Determine the load carrying capacity of a hook of rectangular cross-section. The thickness of the hook is 75 mm, the radius of the inner fibres is 150 mm, while that of the outer fibres is 250 mm. The line of action of the forces passes at a distance of 75 mm from the inner fibres. The allowable stress is 70 MPa. [Ans. 52.51 kN] 6.3 The section of a crane hook is a rectangle 6 cm ´ 4 cm. The centre of curvature of the section is at a distance of 8 cm from the centroid of the section. A load of 15 kN is acting through the centre of curvature. Determine the maximum and minimum bending stresses induced in the hook. [Ans. 6.92 MPa, – 39.51 MPa] 6.4 A circular ring is subjected to a pull of 15 kN. The ring is of T-section as shown in Fig. 6.17 and the internal radius is 10 cm. Determine the maximum and minimum stresses in the ring. [Ans. 7.36 MPa, 5.26 MPa]

Fig. 6.17

224

Strength of Materials

6.5 A ring with a mean radius of curvature of 250 mm is subjected to a load of 2 kN as shown in Fig. 6.18. The ring is made of circular section of 10 mm radius. Calculate the circumferential stress on the inside of the fibre of the ring at A and B. [Ans. sA = 12.30 MPa, sB = 9.71 MPa] 2 kN B

A

Fig. 6.18

6.6 A ring is made of round steel bar 25 mm diameter and the mean radius of the ring is 150 mm. Calculate the maximum tensile and compressive stresses in the material of the ring if it is subjected to a pull of 10 kN. [Ans. 292.2 MN/m2 (Tensile), 331.2 MN/m2 (Comp)] 6.7 A ring is made of round steel bar 24 mm diameter and the mean diameter of the ring is 144 mm. Determine the greatest intensities of tensile and compressive stresses along a diameter XX if the ring is subjected to pull of 12 kN along diameter YY. [Ans. 160.2 MN/m2 (Tensile), 77.36 MN/m2 (Comp)] 6.8 A steel ring 24 cm mean diameter has a rectangular cross-section 6 cm in the radial direction and 3.6 cm perpendicular to the radial direction. If the maximum tensile stress is limited to 144 MN/m2, determine the tensile load that the ring can carry. [Ans. 97.5 kN] 6.9 A chain link is subjected to a pull of 15 kN. It is composed of steel 2 cm diameter and has a mean radius of 2.5 cm. Its semicircular ends are connected by straight pieces 2.5 cm long. Estimate the maximum compressive stress in the link and the tensile stress at the same section. [Ans. – 207.95 MPa, 113.23 MPa]

Chapter 7:

7 7.1

Thin Cylinder and Sphere

225

Thin Cylinder and Sphere

INTRODUCTION

Many problems of practical importance are concerned with solids of revolution which are deformed symmetrically with respect to the axis of revolution. The examples of such solids are circular cylinders subjected to uniform internal and external pressures, spherical shells, rotating disc, etc. In order to meet with several requirements, the fluids are stored under pressure in pressure vessels or shells. The fluid being stored may undergo a change of state inside the pressure vessel as in the case of steam boilers and air compressors.

7.2

CLASSIFICATION OF PRESSURE VESSELS

The pressure vessels may be classified as follows:

According to the dimensions:

The pressure vessels according to their dimensions may be classified as thin shell or thick shell. If the thickness of wall of the shell, t is less than 1/15 of the diameter of the shell, d then it is called a thin shell. On the other hand, if the thickness of wall of the shell is greater than 1/15 of the diameter of the shell then it is said to be a thick shell, i.e., t 1 d £ or > 15 d 15 t

According to the end construction:

The pressure vessels, according to the end construction may be classified as open end or closed end. In case vessels having open ends, the circumferential or hoop stress are induced by the fluid pressure, whereas in case of closed ends, longitudinal stress in addition to circumferential stress are induced.

7.3

STRESSES IN A THIN CYLINDRICAL SHELL DUE TO AN INTERNAL PRESSURE

The analyses of stresses induced in a thin cylindrical shell are made on the following assumptions: 1. The effect of curvature of the cylinder wall is neglected. 2. The tensile stresses are uniformly distributed over the section of the walls. 225

226

Strength of Materials

3. The effect of the restraining action of the heads at the end of the pressure vessel is neglected. When a thin cylindrical shell is subjected to an internal pressure, it is likely to fail in the following two ways: (i) It may fail along the longitudinal section (i.e., circumferentially), as shown in Fig. 7.1(a). (ii) It may fail across the transverse section (i.e., longitudinally) splitting the cylinder into two cylindrical shells, as shown in Fig. 7.1(b). When these cylinders are subjected to internal fluid pressure, the following three types of stresses are developed: (a) Hoop or circumferential stress: Hoop stress acts in a tangential direction to the circumference of the shell. (b) Longitudinal stress: Longitudinal stress acts parallel to the longitudinal axis of the shell. (c) Radial stress: Radial stress in case of thin shell are too small acting in radial direction and can be neglected.

(a) Failure of cylindrical shell along the longitudinal section

Fig. 7.1

7.4

(b) Failure of cylindrical shell along the transverse section

Failure of cylindrical shell.

CIRCUMFERENTIAL OR HOOP STRESS

Consider a thin cylindrical shell subjected to an internal pressure as shown in Figs. 7.2(a) and (b). Tensile stress acting in a direction tangential to the circumference is called circumferential or hoop stress. In other words, it is a tensile stress on longitudinal section. Let

P= d= l= t= sq =

Intensity of internal pressure Internal diameter of the cylindrical shell Length of the cylindrical shell Thickness of the cylindrical shell Circumferential or hoop stress developed in the cylindrical shell.

We know that the total force acting on a longitudinal section (i.e., along the diameter XX) of the shell, = Intensity of pressure ´ Projected area =P´d´l (7.1) and the total resisting force acting on the cylinder walls =

sq ´ 2t ´ l

(7.2)

Chapter 7:

Fig. 7.2

Thin Cylinder and Sphere

227

Circumferential or hoop stress.

From Eqs. (7.1) and (7.2), we have

sq ´ 2t ´ l = P ´ d ´ l Pd 2t P–d t= 2T R

sq =

or or

7.5

LONGITUDINAL STRESS

Consider a closed thin cylindrical shell subjected to an internal pressure as shown in Figs. 7.3(a) and (b). Tensile stress acting in the direction of the axis is called longitudinal stress. In other words, it is tensile stress acting on the transverse or circumferential section YY. Let sL is the longitudinal stress. In this case, the total force acting on the transverse section (i.e., along YY) = Intensity of pressure ´ Cross-sectional area =P´ and

total resisting force =

Q 4

(d) 2

(7.3)

sL ´ p dt

(7.4)

From Eqs. (7.3) and (7.4), we have

sL ´ p dt = P ´

Q 4

\

sL =

P–d 4t

or

t=

P–d 4T L

(d)2

228 If

Strength of Materials

hc is the efficiency of the circumferential joint then t=

P–d 4T LI c

Fig. 7.3 The cylindrical shell.

From above, we see that the longitudinal stress is half of the circumferential or hoop stress. Therefore, the design of a pressure vessel must be based on the maximum stress, i.e., hoop stress. The state of stress on any element of a thin shell subjected to internal pressure is shown in Fig. 7.3(c).

Chapter 7:

Thin Cylinder and Sphere

229

EXAMPLE 7.1 A cylindrical vessel has inside diameter 1.2 m, thickness 20 mm and length 8 m. It is subjected to internal pressure of 4.5 MPa. Compute: (a) Hoop stress, (b) Longitudinal stress, (c) Maximum shear stress. Solution: Here (a) Hoop stress

sq = =

Pd 2t

4.5 – 1200 = 135 MPa 2 – 20

(b) Longitudinal stress

sL =

TR

2 = 67.5 MPa

(c) Maximum shear stress = =

TR  T L 2 135  67.5 = 33.75 MPa 2

EXAMPLE 7.2 A thin cylinder of internal diameter 1.25 m contains a fluid at an internal pressure of 2 MPa. Determine the maximum thickness of cylinder if (a) The longitudinal stress is not to exceed 30 MPa. (b) The circumferential stress is not to exceed 50 MPa. Solution: Given Internal diameter of cylinder d = 1.25 m Internal pressure of fluid P = 2 MPa Longitudinal stress sL = 30 MPa Circumferential stress sq = 50 MPa Circumferential stress or

t=

and Longitudinal stress or

sq =

sL = t=

Pd 2t

2 – 1250 = 25 mm 2 – 50

(i)

Pd 4t

2 – 1250 = 20.8 mm 4 – 30

From Eqs. (i) and (ii), it is clear that t should not be less than 25 mm.

(ii)

230

Strength of Materials

EXAMPLE 7.3 A cylinder of thickness 15 mm has to withstand maximum internal pressure of 2 MPa. If the ultimate tensile stress in the material of the cylinder is 300 MPa, factor of safety 3 and joint efficiency 80%, determine the diameter of the cylinder. Solution:

Allowable stress

sa11 = =

sq =

Hoop stress

or

Diameter d = =

Ultimate tensile stress Factor of safety

300 = 100 MPa 3 Pd = 2 tI j

sa11

2T a11 tI j P

2 – 100 – 15 – 0.8 = 1200 mm 2

EXAMPLE 7.4 A water main 600 mm diameter contains water at a pressure head of 100 m. If the weight density of water is 9810 N/m3. Find the thickness of the metal required for the water main. Given the permissible stress as 40 MPa. Solution:

Pressure of water inside the water main is: P = rgh = 9810 ´ 100 N/m2 = 0.981 N/mm2 Hoop stress

\

7.6

sq =

Thickness t =

Pd 2t

0.981 – 600 = 7.3 mm 2 – 40

EFFECT OF INTERNAL PRESSURE ON THE DIMENSIONS OF A THIN CYLINDRICAL SHELL

We have Hoop stress

sq =

Pd 2t

Longitudinal stress

sL =

Pd 4t

Chapter 7:

Thin Cylinder and Sphere

231

Then hoop or circumferential strain is: ÿ ÿ ÿ ÿ ÿ ÿ eq

Hoop strain

=

TR E

–

O E

TL

=

O Pd Pd – 2 tE 4t E

=

Pd 2tE

eq =

(n = Poisson’s ratio)

O  1   2 

Change in diameter Ed = d Original diameter

(7.5)

(7.6)

(7.7)

Equating the two values of eq given by Eqs. (7.6) and (7.7), we get

Pd Ed = d 2tE

O  1  2   

Change in diameter is:

dd = Longitudinal strain

Longitudinal strain

eL =

Pd 2  O 1   2tE  2

TL E

–

OTR E

=

Pd O Pd – 4tE 2tE

=

Pd 2tE

eL = =

1  2 O  

(7.8) (7.9)

(7.10)

Change in lengh Original length

EL L

(7.11)

Equating the two values of eL given by Eqs. (7.10) and (7.11),

Pd  1 EL  O =  L 2tE  2  Change in length

dL =

Volumetric strain

eV = =

PdL È 1 2tE

Ø É  OÙ Ê2 Ú

(7.12)

(7.13)

Change in volume Original volume

EV V

(7.14)

232

Strength of Materials

Also volumetric strain is given by eV = 2 eq + eL

=

2 Pd 2 Et

È É1  Ê

(7.15)

OØ 2 ÙÚ

+

Pd È 1 2Et ÉÊ 2

Ø  OÙ Ú

Substituting the values of eq and eL, eV =

Pd 2 Et

È5 Ø É  OÙ Ê4 Ú

(7.16)

So change in volume is:

d V = V(2eq + eL) =

Pd È 5 Et

Ø É  OÙ Ê4 Ú

(7.17a) V

(7.17b)

EXAMPLE 7.5 The air vessel of a torpedo is 530 mm external diameter and 10 mm thick, the length being 1830 mm. Find a change in the external diameter and the length which changed with 10.5 MPa internal pressure. Take E = 210 kN/mm2 and Poisson’s ratio n = 0.3. Solution:

Internal diameter = 510 mm

Circumferential stress sq = = Longitudinal stress

sL =

Pd 2t 10.5 – 510

= 267.75 MPa

2 – 10

TR

2 = 133.87 MPa

Circumferential strain eq = = =

TR E 1 E

–

OT L E

(T R  OT L ) 1

210 – 10 3

(267.75 – 0.3 ´ 133.87) = 0.001084

Ed = 0.001084 d Change in external diameter d d = eq d Circumferential strain eq =

= 0.001084 ´ 530 = 0.5744 mm

Chapter 7:

Longitudinal strain eL = = =

TL

–

E 1 E

Thin Cylinder and Sphere

233

OT R E

(T L  OT R )

1

(133.875 – 0.3 ´ 267.75) = 0.000255

210 – 10 3

Change in length d L = eLl = 0.000255 ´ 1830 = 0.4667 mm

EXAMPLE 7.6 A cylindrical shell 900 mm long, 150 mm internal diameter, having thickness of metal as 8 mm, is filled with a fluid at atmospheric pressure. If an additional 20000 mm3 of fluid is pumped into the cylinder, find (i) Pressure exerted by the fluid on the cylinder. (ii) Hoop stress induced. Take E = 200 GPa and Solution:

Let the internal pressure be P. Then

Hoop stress sq = =

Pd 2t

P – 150 = 9.375P MPa 2 –8

Longitudinal stress

sL =

TR

2 = 4.6875P MPa

Circumferential strain eq = =

TR E

–

OT L

P (9.375 – 0.3 ´ 4.6875) E

= 7.96875 Longitudinal strain eL = =

1 E

E

P E

(T L  OT R )

P (4.6875 – 0.3 ´ 9.375) E

= 1.875

P E

n = 0.3.

234

Strength of Materials

Volumetric strain eV = 2eq + eL = (2 ´ 7.96875 + 1.875) = 17.8125

P E

P E

Increase in volume d V = eVV = 20000 mm3 or

17.8125

P 200 – 10

3

Q

´

4

(150)2900 = 20000

(i) P = 14.12 MPa (ii) Hoop stress

sq = =

Pd 2t

14.12 – 150 = 132.4 MPa 2–8

EXAMPLE 7.7 A vertical steam boiler with 2 m internal diameter and 4 m high is constructed with 20 mm thick plates for a working pressure of 10 kg/cm2. The end plates are flat. Calculate: (i) The stress in the circumferential plates due to pressure on the end plates. (ii) Stress in the circumferential plates due to resisting the bursting effect. (iii) Increase in length, diameter and the volume. Take passion’s ratio n = 0.3 and E = 2 ´ 106 kg/cm2. Solution: (i) The stress in the plate due to the pressure to end plates will obviously be the longitudinal stress. \

(ii) Circumferential stress

1 – 2000 Pd = 4 – 20 4t = 25 MPa

sL =

sq = 2sL = 50 MPa

(iii) Longitudinal strain

eL =

= =

=

TL E

TL E

–

OT R E

(1  2O )

25 200 – 10 3

25 – 0.4 200 – 10 3

(1 – 0.3 ´ 2)

=

EL

L

Chapter 7:

dL =

Thus, increase in length

Circumferential strain

eq =

= = Increase in diameter

dd =

TR

25 – 0.4 200 – 10 3



E

O

235

´ 4000 = 0.2 mm

TL

E

1 200 – 10 3

(50 – 25 ´ 0.3)

42.5 200 – 10

Thin Cylinder and Sphere

=

3

42.5

Ed d

´ 2000

200 – 10 3

= 0.425 mm Volumetric strain

eV = 2 eq + eL

=

= Increase in volume

dV =

2 – 42.5 200 – 10

+

3

95 200 – 10

=

3

Q

95

200 – 10 4 3

25 – 0.4 200 – 10 3

EV V (2000)24000

= 5969026.04 mm3 or 5.97 litres

EXAMPLE 7.8 A cylindrical shell 120 mm in diameter and 5 mm metal thickness is 840 mm long. It is filled with incompressible fluid at atmospheric pressure. If an additional 10 cm3 fluid is pumped into the cylinder, calculate the pressure exerted by the fluid on the wall of the cylinder. Take E = 200 GPa, n = 0.3. Solution: Due to additional 10 cm3 incompressible fluid, the volume of the cylinder is increased by 10 cm3. Therefore, d V = 10 cm3 = 104 mm3 Volume of the cylindrical shell is given by V=

Q

=

Q

4 4

d 2l (120)2 ´ 840 = 9500 ´ 103 mm3

Pd È 5 EV = V 2tE ÉÊ 2



Ø

2O Ù Ú

236

Strength of Materials

10 4 9500 – 10

or

7.7

3

=

P – 120 2 – 5 – 200 – 10 3

(2.5 – 2 ´ 0.3)

P = 9.23 MPa

THE SPHERICAL SHELLS SUBJECTED TO AN INTERNAL PRESSURE

Figure 7.4 shows a thin spherical shell of internal diameter d and thickness t and subjected to an internal fluid pressure P. The fluid inside the shell has a tendency to split the shell into two hemispheres along XX axis.

Fig. 7.4

Thin spherical shell.

The force F which has a tendency to split the shell is given as: F=P´

Q 4

d2

The area resisting this force is: A = p dt Hoop or circumferential stress

sq induced in the material of the shell is given by

sq =

Force F Area resisting the force F

P–

4 Q dt

= =

Q

d2

Pd 4t

The stress sq is tensile in nature. The fluid inside the shell is also having tendency to split the shell into two hemispheres along YY axis. Then it can be shown that the tensile hoop stress will also be equal to

Pd . Let this stress be 4t

sL. sL =

\ The stress

sL will be at right angles to sq.

Pd 4t

Chapter 7:

7.8

Thin Cylinder and Sphere

237

CHANGE IN DIMENSIONS OF THIN SPHERICAL SHELL DUE TO AN INTERNAL PRESSURE

The stress

sq and sL are acting at right angles to each other and equal. Strain in any one direction, O TR e= – TL E

= =

TR E Pd

4tE We know that strain in any direction,

E

–

OT R E

' T 

R

=T L =

(1  O )

e =

Ed d

Pd 4t

  (7.18)

(7.19)

Equating the two values of e given by Eqs. (7.18) and (7.19), Pd Ed (1  O ) = 4 tE d EV eV = = 3e V

Volumetric strain:

(7.20)

V = Original volume Substituting the value from Eq. (7.20), we get 3Pd EV (1  O ) = V 4tE

(7.21)

EXAMPLE 7.9 A spherical vessel 1.5 m diameter is subjected to an internal pressure of 2 MPa. Find the thickness of plate required if maximum stress is not to exceed 60 MPa and joint efficiency is 80%. Solution:

Circumferential stress sq = Thickness t =

=

Pd 4tI j

Pd 4T R I j 2 – 1.5 – 10 3 = 15.6 mm 4 – 60 – 0.8

EXAMPLE 7.10 A spherical shell is of 60 cm diameter and of 6 mm wall thickness. If the tensile stress is limited to 600 kg/cm2, calculate the safe working pressure and its volume under the pressure. Take E = 2 ´ 106 kg/cm2, n = 0.25.

238 Solution:

Strength of Materials

Hoop stress

sq = 60 MPa

or

P= = Circumferential strain

eq =

= Volumetric strain

T R – 4t d 60 – 4 – 6 = 2.4 MPa 600

TR E

TR

–

O

TR

E

(1  O )

E

eV = 3eq

= = eV =

dV =

3T R E

(1  O )

3 – 60 200 – 10

3

(1  0.25) = 6.75 ´ 10– 4

EV = 6.75 ´ 10– 4 V

   

4 600 Q 3 2

3

´ 6.75 ´ 10– 4

= 76340.7 mm3

EXAMPLE 7.11 A thin spherical shell 1.5 m diameter with its wall of 1.25 cm thickness is filled with a fluid at atmospheric pressure. What intensity of pressure will develop in it if 160 cu. cm of fluid is pumped into it? Also calculate the hoop stress at that pressure and the increase in diameter. Take E = 2 ´ 106 kg/cm2, n = 0.3. Solution:

Increase in volume = 160 cm3

Now,

V=

4 3 pr 3

4 p (75)3 cm3 3 EV eV = V 160 3eq = V

=

Volumetric strain

eq =

160 Ed = d 3V

Chapter 7:

Increase in diameter

dd =

Thin Cylinder and Sphere

239

160 – 150 4 3 – Q (75) 3 3

= 0.00452 cm Hoop strain

eq =

=

TR E

(1  O )

TR 2 – 10

6

(1  0.3) =

EV

V

7T R 160 = 6 4 20 – 10 Q (75) 3 – 3 3

sq = 86.2 kg/cm2

\ Hoop stress

sq = P= =

Pd 4t 4tT R d

4 – 1.25 – 86.2 = 2.87 kg/cm2 150

EXAMPLE 7.12 The thickness of thin cylindrical shell is reduced by 3 mm because of corrosion. As the effect of this reduction the hoops stress is increased by 12% under the same internal pressure. Find the original thickness of the shell. Solution:

Then

Let t = Original thickness of the shell (t – 3) = Reduced thickness due to corrosion Hoop stress before corrosion =

Pd 2t

Pd 2 (t  3) But given that the hoop stress increases by 12% after corrosion. Therefore, Hoop stress after corrosion =

Pd Pd = 1.12 2 (t  3) 2t t = 1.12(t – 3) t = 28 mm

240

Strength of Materials

EXERCISES 7.1 Derive the expressions for change in volume of a thin spherical shell and cylindrical shells. Ë Ì Ans. Í

Û EV Pd È Pd Ø EV = 3É (1  O )Ù , = (5  4O ) Ü V 4tE Ê 4tE Ú V Ý

7.2 A thin cylindrical shell of diameter 1200 mm is subjected to a fluid pressure of 4 MPa. What should be the thickness of the wall if the maximum stress is not to exceed 100 MPa. Hence what will be the change in volume per metre length. Take, E = 200 GPa, n = 0.3. [Ans. d V = (1.074 ´ 106) mm3] 7.3 A thin spherical shell of diameter 0.5 m and thickness 4 mm is full of fluid under pressure. The change in volume is observed to be 2 ´ 104 mm3. Calculate the internal fluid pressure and hoop stress developed. Take, E = 200 GPa, n = 0.33. [Ans. sq = 30.31 MPa] 7.4 A thin cylindrical shell 3 m long is of 1 m diameter, 10 mm thick and subjected to an internal pressure of 2 MPa. Determine the change in length, diameter and volume. Take, E = 200 GPa, n = 0.25. [Ans. d L = 0.375 mm, dd = 0.4375 mm, d V = 2.356 ´ 106 mm3, P = 5.1 MPa] 7.5 A thin cylindrical shell of length 2 m and having an internal volume of 2.26 m3 is subjected to an internal pressure of 3 N/mm2. If the maximum permissible tensile stress in the material is 180 MPa, evaluate the change in volume of the cylinder. [Ans. d V = 38.7 ´ 105 mm3] 7.6 A cylindrical vessel of 3 m diameter is used for processing rubber and is 10 m long. If the steel plates have the thickness of 24 mm, and vessel operates at 800 kPa internal pressure, determine the total elongation increase in diameter and change in volume. Take, E = 200 GPa, n = 0.3. [Ans. d L = 0.492 mm, dd = 0.62 mm, d V = 32 ´ 106 mm3] 7.7 A thin cylindrical shell with hemispherical ends is subjected to internal pressure of 2 MPa. The length of the cylindrical portion is 1 m and the diameter of both cylindrical and spherical portions is 500 mm. Thickness of the metal is 16 mm. Find the change in volume. Take, E = 210 GPa, n = 0.3. [Ans. d Vl = 58187.5 mm3, d V2 = 10738 mm3, d V = 68925.5 mm3] 7.8 In a thin cylinder the longitudinal stress is 60 MPa, E = 200 GPa and n = 0.25. How much is the volumetric strain?

Ans. !

EV = 1.2 – 10 3 V

"# $

7.9 In a thin cylinder the hoop strain is 3.5 times the longitudinal strain. How much is the Poissons ratio? [Ans. n = 0.25] 7.10 A steel pipe 900 mm diameter has to carry water under a head of 200 m. If the permissible tensile stress is 90 MPa. Determine the minimum value of thickness required for the pipe. [Ans. t = 9.8 mm]

Chapter 8:

8 8.1

Thick and Compound Cylinder

241

Thick and Compound Cylinder

INTRODUCTION

Thick cylinders are cylindrical vessels, containing fluid under pressure whose wall thickness to diameter is not smaller than 1

15

 t • 1  . Radial stress varies along the thickness which is  d 15 

maximum at the inner radius and minimum at the outer radius. The variations of radial as well as circumferential stresses across the thickness are obtained with the help of Lame’s theory.

8.2

LAME’S THEORY

Consider a thick cylinder of length l, internal radius a and external radius b subjected to internal and external uniformly distributed pressure of intensities Pa and Pb, respectively as shown in Fig. 8.1(a). The following assumptions are made in determining the distribution of stresses in the cylinder: (i) (ii) (iii) (iv) Let

The material is homogeneous and isotropic. Plane transverse sections remain plane under the action of internal pressure. The material is stressed within elastic limit as per Hooke’s Law. All the fibres of the material are stressed independently without being constrained by the adjacent fibres.

sr = Radial stress at any radius r sq = Hoop stress at any radius r sL = Longitudinal stress a b r r + dr Pa Pb

= = = = = =

Internal radius of thick cylinder External radius of thick cylinder Internal radius of elemental ring External radius of elemental ring Pressure at internal radius of thick cylinder Pressure at external radius of thick cylinder 241

242

Strength of Materials

1 [sz – v(sq + E and v are all constants, therefore, ez =

We have from Hooke’s law Since

ez, E, sz

sr)]

sq + sr = Constant = 2A (say)

(8.1)

Consider an annular ring of the cylinder between radii r and r + dr as shown in Fig. 5.1(b). Let the radial stress sr and sr + dsr at internal and external radius of annular ring r and r + dr, respectively.

Fig. 8.1

Thick cylinder under pressure.

Applying force equilibrium condition, SFr = 0 i.e., forces in radial direction equated to zero, we get T r   T r + r 

'r  (r + 'r ) R  T r r 'R  2T R 

'R 2

Simplifying we get after neglecting small terms, T T r . 'r 'R + r r 'r R  T R 'r 'R = 0 r

. 'r = 0

Chapter 8:

Tr + r

or

T r  r

TR = T r

Thick and Compound Cylinder

243

TR = 0

T r r

In the limiting case, we have

sq = sr + r

dT r d = (rsr) dr dr

(8.2)

Substituting in Eq. (8.1), we obtain d (rsr) + dr

or

r

sr = 2A

dT r = 2(A – dr

sr)

dT r 2( A  T r ) = dr r

or

dT r

or

A  Tr

Integrating, we get

=

I

2dr r

I

dT r 2dr = ATr r –ln(A – sr) = 2lnr – lnB ln(A – sr) = –2lnr + lnB

or or or

sr) = ln

ln(A –

or

A–

sr =

B r2

B r2

sr = A –

or

B

r2 where A and B are constants. Substituting Eq. (8.3) in Eq. (8.1), we get

sq + A – or

B r2

= 2A

sq = A +

B

r2 Equations (8.3) and (8.4) are known as the Lame’s equations. Constants A and B can be determined from the boundary conditions at r = a and r = b.

(i) At r = a, (ii) At r = b,

sr = –Pa sr = –Pb

(8.3)

(8.4)

244

Strength of Materials

Substituting in Eq. (8.3), we get

B

– Pa = A – and

a2 B

– Pb = A –

b2

 1  1  a b  b  a  = B  a b 

Now,

Pa – Pb = B

2

2

2

2

2 2

\

B=

Then

A=

=

or

a2 b2 (Pa  Pb )

A=

=

B a2

b2  a2

– Pa

b2 (Pa  Pb ) b2  a2

– Pa

b 2 Pa  b 2 Pb  b 2 Pa  a 2 Pa b2  a2

Pa a 2  Pb b 2

b 2  a2 Substituting the values of constants A and B in Eqs. (8.3) and (8.4), we get

sr = sq =

Pa a 2  Pb b 2 b a 2

2

Pa a 2  Pb b 2

–

a2b2 r

2

a2b2

P P   b  a  P P   b  a  a 2

b 2

(8.5)

a b (8.6) 2 2 r2 Now consider the cross-section of a thick cylinder with closed ends subjected to an internal pressure Pa and external pressure Pb. For horizontal equilibrium,

and

b2  a 2

+

Pa pa2 – Pb p b2 = sL p (b2 – a2)

sL =

or

Pa a 2  Pb b 2

Special cases CASE I A cylinder subjected to internal pressure: and (8.6) become

sq =

Pa 2 b2  a 2

b2  a 2

(8.7)

In this case, Pb = 0 and Pa = P. Then Eqs. (8.5)

1  b   r  2

2

(8.8)

Chapter 8:

245

1  b   r 

Pa 2

sr =

and

Thick and Compound Cylinder

2

(8.9) 2 b2  a 2 These equations show that sr is always a compressive stress and sq is a tensile stress. Figure 8.2 shows the variation of radial and circumferential stress across the thickness of the cylinder. Hoop stress is maximum at the inner radius of the cylinder. At

r =a

sq = sr =

and

P (a 2  b 2 )

(Tensile)

b2  a2 P b2  a2

(a2 – b2)

(8.10)

(Compressive)

= –P = Applied pressure r = b,

At

2 Pa 2

sq = sr

and

b2  a2 =0

P sq sr

Fig. 8.2

Cylinder subjected to internal pressure.

CASE II A cylinder subjected to external pressure: In this case, Pa = 0 and Pb = P. Equations (8.5) and (8.6) reduce to

sq = – sr = –

Pb 2 b a 2

2

Pb 2 b2  a 2

1  a  r 1  a  r

2 2 2 2

   

(8.11)

(8.12)

246

Strength of Materials

These equations show that sr and sq are always a compressive stress. The variations of these stresses across the thickness are shown in Fig. 8.3. At

r = a,

sq = – At

2 Pb 2 b2  a2

,

sr = 0

r = b,

sq = – sr = –

P b a 2

2

P b  a2 2

(b2 + a2)

(Compressive)

(b2 – a 2)

(Compressive)

= –P = Applied pressure P

a

b

sr

Fig. 8.3

sq

Cylinder subjected to external pressure.

EXAMPLE 8.1 A thick cylinder with internal radius 10 cm and external radius 16 cm is subjected to an internal fluid pressure of 70 MPa. Draw the variation of hoop and radial stress in the cylinder wall. Also determine the maximum shear stress in the cylinder wall. Solution: Now,

Given a = 10 cm, b = 16 cm, Pa = 70 MPa Hoop stress

sq =

Pa a 2 b2  a2

1  b   r  2

2

Chapter 8:

Thick and Compound Cylinder

1  (160) "# (160)  (100) ! r #$  25600 "# = 44.87 1  ! r $ 70(100) 2

=

2

2

2

2

2

Hoop stresses in thick cylinder are tabulated as follow: r (mm ) T R (MPa )

100 159.73

120 124.63

140 103.47

1  b   r  b a  1  (160) 70(100) = r (160)  (100) !  25600 "# = 44.87 1  ! r $

sr =

Radial stress

Pa a 2 2

160 89.74

2

2

2

2

2

2

2

2

"# #$

2

Radial stresses in thick cylinder are tabulated as follows:

r (mm )

100

120

140

T r (MPa )

 69.99

34.89

13.73

\

Maximum shear stress t max = =

160 0

TR  T r 2

159.73  69.99 = 114.86 MPa 2

200 150

sq , sr(MPa)

sq 100 50 0 20 40 60 80 Fig. 8.4

100

120

140

r (mm) 160

sr

Stress distribution across the cylinder wall.

247

248

Strength of Materials

EXAMPLE 8.2 The cylinder of a hydraulic press has an internal diameter of 30 cm and is designed to withstand a pressure of 10 MPa, without the material being stressed over 20 MPa. Determine the thickness and stress from inner to outer radius of cylinder. Sketch a diagram showing the variation of radial and hoop stress across the thickness of the wall of the cylinder. Solution: Given As we know,

sq = 20 MPa, a = 15 cm, Pa = 10 MPa (As maximum stress is given by sq) Hoop stress

sq =

Pa a 2 b2  a2

1  b   r  2

2

The hoop stress is maximum at inner radius, therefore, 20 =

1  b    b  (150)  150  (150)  b  1  2  150 

10 – (150)2 2

2

2

2

2

[b2 – (150)2] =

or

2

2

b2 – (150)2 = 11250 + 0.5b2

or

0.5b2 = 33750

or or Then

b = 259.8 mm Wall thickness t = b – a = 259.8 – 150 = 109.8 mm

Now,

Hoop stress

sq = =

1  b   r 

Pa a 2

2

b2  a 2

2

10(150) 2 (259.8)  (150) 2

 67500 "# ! r $  67500 "# Radial stress s = 5 1  ! r $ sq = 5 1 

or

and

r

2

1  67500   r  2

2

2

Hoop and radial stresses in cylinder are tabulated as follows: r (mm ) T R (MPa )

T r (MPa )

150 20

170 16.67

190 14.34

210 12.65

230 1138 .

259.8 10

10

6.67

4.34

2.65

138 .

0

Chapter 8:

Thick and Compound Cylinder

249

20 sq

sq, sr(MPa)

15 10 5 0

r (mm) 150

170

190

210

230

250

sr

–5 –10

Fig. 8.5

Stress distribution across the cylinder wall.

EXAMPLE 8.3 Calculate the thickness of metal necessary for a cylindrical shell of internal diameter 160 mm to withstand an internal pressure of 25 MPa, if maximum permissible tensile stress is 125 MPa. Solution: Given Internal radius r = 80 mm, Internal pressure Pa = 25 MPa, Hoop stress sq = 125 MPa. Lame’s equations are:

sq = A + sr = A – At r = 80,

sr = –25 MPa and sq = 125 MPa, –25 = A –

125 =

B r2 B

r2

B (80)2

B 80 2

+A

From Eqs. (iii) and (iv), we have A = 50, B = 480 ´ 103

sr = 0

At

x = b,

\

0 = A – B/b2

or

480 – 10 3 b2 b = 97.98 mm = 98 mm = 50 –

Now,

Thickness of metal = b – a = 98 – 80 = 18 mm

(i) (ii)

(iii)

(iv)

250

8.3

Strength of Materials

APPLICATION OF THEORIES OF FAILURE

The maximum permissible stress in a thick cylinder closed at ends and subjected to internal pressure are:

s1 = sq =

Pa a 2

2

at inner surface.

2

=

Pa (a 2  b 2 )

=

Pa (c 2  1)

b2  a2 \c=

c 1 2

Pa a 2

b a

(from equation (8.7))

b2  a2

Pa

=

For a cylinder with open ends,

2

b2  a 2

s2 = sL =

s3

a b   a 

c2  1 = sr = – Pa

at the inner surface

s2 = 0.

Maximum principal stress theory:

According to this theory,

s1 = syt

Pa (c  1) 2

c2  1

=

as

s1 > s2 > s3

syt

By simplification, we get

T yt  Pa T yt  Pa

c =

t=a

 !

"# #$

T yt  Pa 1 T yt  Pa

(8.13)

where t is the thickness of the wall of the cylinder which is given by t=b–a Equation (8.13) is known as Lame’s formula.

Maximum principal strain theory: According to this theory, s1 – v (s2 + s3) = syt

Pa (c 2  1) c 1 2

v

 P P =s  c  1  a

2

a

yt

Chapter 8:

Thick and Compound Cylinder

 c  1  v  2  c  =  c  1  c  1  2

2

2

=

(c  1) 2

or

T yt

2

c 2 (1  v)  (1  2 v)

Pa

T yt Pa

T yt  Pa (1  2 v) T yt  Pa (1  v)

c=

t=a

 !

251

"# #$

T yt  Pa (1  2 v) 1 T yt  Pa (1  v)

(8.14)

Equation (8.14) is known as the Clavirino’s formula. For a cylinder with open ends, we get t=a

Maximum shear stress theory:

 !

"# #$

T yt  Pa (1  v) 1 T yt  Pa (1  v)

s1 – s3 = syt c2  1

+ Pa =

(2 Pa c 2 )

=

c2  1

syt syt

c=

t=a

Maximum distortion energy theory:

or

 c  ! c

 !

1 1  2 2 1 c 1

T yt

T yt  2 Pa

T yt T yt  2 Pa

"# #$

1

(8.16)

According to this theory,

(s1 –

2

(8.15)

According to this theory, Pa (c 2  1)

P a2

(Birnie’s equation)

s2)2 (s2 – s3)2 + (s3 – s1)2 = 2syt2

   1  1   1  c  1 "#   c  1   c  1 # $ 2

6 c 2 Pa

2

2

2

(c 2  1) 2

2

= 2syt2

3Pa2 c4 = c =

2

syt2 (c2 – 1)2

T yt T yt  3 Pa

2

= 2syt2

252

Strength of Materials

t=a

 !

T yt T yt  3Pa

"# #$

1

(8.17)

EXAMPLE 8.4 A thick cylindrical pipe of inner radius 100 mm is subjected to an internal fluid pressure of 70 MPa. If tensile yield stress for the pipe material is 120 MPa. Calculate the wall thickness of the pipe using maximum principal stress theory. Solution: Given Inner radius r1 = 100 mm, Internal pressure Pa = 70 MPa and Yield stress syt = 120 MPa Then

t = r1

 !

= 100

 !

"# #$

T yt  Pa 1 T yt  Pa

Ë

190

ÍÌ

50

= 100 Ì

8.4

"# #$

120  70 1 120  70 Û  1Ü ÜÝ

= 94.93 mm

COMPOUND CYLINDRICAL SHELL

In a compound cylindrical shell, as shown in Fig. 8.6, the outer cylinder (having inside diameter smaller than the outside diameter of the inner cylinder) is shrunk fit over the inner cylinder by heating and cooling, respectively. On cooling, the contact pressure is developed at the junction of the two cylinders, which induces compressive tangential stress in the material of the inner cylinder and tangential stress in the material of the outer cylinder. Hence inner cylinder subjected to external pressure whereas outer cylinder subjected to internal pressure. This principle is commonly used in the design of gun tube. P Outer cylinder

Inner cylinder P r1 r2

r3

r1 r2

P (b) Inner cylinder

(a) Compound cylinder

Fig. 8.6

r3

P

Stress in compound cylinder.

P r2

P

P P

(c) Outer cylinder

Chapter 8:

Thick and Compound Cylinder

253

Now, consider a compound thick cylindrical shell made up of two tubes as shown in Fig. 8.6(a). Here r1 r2 r3 P

= = = =

Inner radius of inner cylinder Outer radius of inner cylinder Outer radius of outer cylinder Radial pressure at the junction of the two cylinders

For inner tube, Px =

B1 x2

– A1

At x = r1, Px = 0 and x = r2, Px = P. \

0 =

and

P=

B1 r12 B1 r22

– A1

(8.18)

– A1

(8.19)

For outer tube, Px =

B2 x2

– A2

and Hoop stress

(sq)x =

B2 x2

+ A2

At x = r2, Px = P and at x = r3, Px = 0. \

P =

and

0 =

B2 r22 B2 r32

– A2

(8.20)

– A3

(8.21)

The values of A1, B1, A2 and B2 may be found out from these four equations, if the radial pressure P at the junction of the two shells is known. The hoop stress sq may also be obtained with the help of relative expressions. Now, when the fluid under pressure P is admitted inside the compound shell, it will be resisted jointly by both the shells. The hoop stresses may be calculated by the Lame’s formulae as usual. The resultant stresses will be the algebraic sum of the initial stresses and those due to fluid pressure.

EXAMPLE 8.5 A compound cylinder is made by shrinking a tube of 160 mm internal diameter and 20 mm thick over another tube of 160 mm external diameter and 20 mm thick. The radial pressure at the common surface, after shrinking is 80 kg-f/cm2. Find the final stress setup across the section when the compound cylinder is subjected to an internal fluid pressure of 600 kg-f/cm2.

254 Solution:

Strength of Materials

Inner diameter of outer cylinder Thickness Outer diameter of outer cylinder Outer diameter of inner cylinder Inner diameter of inner cylinder

= = = = =

160 mm 20 mm = 2 cm 16 cm + 2 cm + 2 cm = 20 cm 16 cm 16 – 2 – 2 = 12 cm

Thus, Outer radius of outer cylinder r3 = 10 cm Inner radius of outer cylinder r2 = 8 cm = outer radius of inner cylinder Inner radius of inner cylinder r1 = 6 cm First of all, let us apply all the Lame’s equations for the inner and outer cylinder before fluid pressure is admitted. For inner tube,

B1

Px =

x2

– A1

At x = r1, Px = 0, 0 =

B1 – A1 36

(i)

80 =

B1 – A1 64

(ii)

Similarly, at x = r2, Px = 80 kgf/cm2,

Solving Eqs. (i) and (ii) simultaneously, we find that A1 = – 183, B1 = – 6583 Now, using the Lame’s equation, (sq)r1 =

B1

+ A1

r12

=–

6583 6

(sq)r2 = –

and

2

– 183 = – 366 kg/cm2

(iii)

6583

– 183 (8)2 = – 286 kg/cm2

(iv)

For outer tube, Px =

B2 x2

– A2

At x = r2, Px = 80 kg/cm2, 80 =

B2 – A2 64

(v)

0=

B2 – A2 100

(vi)

and at x = r3, Px = 0,

Chapter 8:

Thick and Compound Cylinder

255

Solving Eqs. (v) and (vi), A2 = 142, B2 = 14220 Now,

Hoop stress (sq)r2 = = (sq)r3 =

B2 r22

+ A2

14220 + 142 = 364 kg/cm2 64

(vii)

14220 + 142 100

= 284 kg/cm2

(viii)

Now, let us apply Lame’s equation for the inner cylinder only after the fluid under pressure of 600 kg/cm2 admitted, i.e., Px = or

600 =

and

0=

B3 x2

– A3

B3

– A3

(6 ) 2 B3 (10) 2

(ix)

– A3

(x)

Solving Eqs. (ix) and (x), A3 = 337.5, B3 = 33750 Now,

Hoop stress (sq)r1 = =

and

B3 r12

+ A3

33750 62

(sq)r2 =

33750

(sq)r3 =

33750

82 10 2

+ 337.5 = 1275 kg/cm2 + 337.5 = 865 kg/cm2 + 337.5

= 675 kg/cm2 Resultant hoop stresses across compound cylinder are tabulated as follow:

Hoop stress (kg-f /cm 2 ) x =6 cm

x =8 cm

x =8 cm

x =10 cm

Due to fluid pressure

 366 1275

286 865

 364  865

284 675

Resultant

+909

+579

+1229

+959

Initial

256

Strength of Materials Pb

sq r + dr Pb

Pa

r

a

sr + dsr

sq

Pa b

Pa

Pb

Pa

Pb (a)

sq

(b) Detail view of elemental ring

Fig. 8.7

8.5

sr + dsr

sq

Thick spherical shell under pressure.

THICK SPHERICAL SHELLS

Consider a thick spherical shell of internal radius a and external radius b subjected to internal and external uniformly distributed pressure of intensities Pa and Pb, respectively as shown in Fig. 8.7(a). Let

sr = Radial stress at any radius r sq = Hoop stress at any radius r a b r r + dr Pa Pb

= Internal radius of thick spherical shell = External radius of thick cylinder spherical shell = Internal radius of annular ring spherical shell = External radius of annular ring = Pressure at internal radius of thick spherical shell = Pressure at external radius of thick spherical shell

Consider an annular ring of the shell between radii r and r + dr as shown in Fig. 8.7(b). Let the radial stresses are sr and sr + dsr at internal and external radius of annular ring r and r + dr, respectively. The bursting force on elemental ring acting in radial direction is: = p(r + dr)2 (sr + dsr) – Resisting force = sq(2prdr)

srp r 2

Equating the resisting force and bursting force, we have

p(r + dr)2 (sr + dsr) – sr p r2 = sq(2 prdr) Neglecting small quantities, we get

sq = sr +

rdT r 2 dr

(8.22)

Chapter 8:

Thick and Compound Cylinder

257

Differentiating with respect to r, we get

dT R dT r 1 dT r r d 2T r = + + 2 2 dr dr dr 2 dr 3 dT r r d 2T r + 2 dr 2 2 dr Due to internal pressure, let the radius increases from r to (r + u), then

=

Hoop strain eq = eq =

Also

(8.23)

2 Q (r  u )  2Q r u = 2Q r r

(8.24)

1 [sq – v(sr + E

(8.25)

sq)]

From Eqs. (8.24) and (8.25), we get

or

u 1 = [sq – v(sr + r E

sq)]

r [sq – v(sr + E Differentiating with respect to r, we get

sq)]

u =

   "# + 1 [s – v(s + s )]  $ E ! 1 r  dT dT " (1  v) v + [(1 – v)s – vs ] E! dr dr #$ E  u  du dr  u  dr  du

r dT R dT r dT R du = v  dr E dr dr dr =

Radial strain Also

(8.26)

R

er =

q

r

dr

q

=

q

r

r

dr

(8.27)

(8.28)

er =

1 [sr – v(sq + E

=

1 [sr – 2v(sq)] E

(8.29)

1 du [sr – 2v(sq)] = E dr

(8.30)

sq)]

From Eqs. (8.28) and (8.29), we get

From Eqs. (8.29) and (8.30), we get

 !

"# $

1 r dT R dT 1 [sr – 2v(sq)] = (1  v )  v r  (1  v)T R  vT r E dr dr E E

258

Strength of Materials

Substituting the values of

1 E

T !

r

dT R from Eqs. (8.22) and (8.23), we get dr

sq and

 

 2v T r 

rdr 2 dr

 "# = r (1  v) %K& 3 dT  r d T (K)  v dT "#  $ E ! 'K 2 dr 2 dr K* dr #$ 1 % dT ()  vT "# 1  v6 &T  r + 1 E! ' 2dr * #$ 2

r

r

r

Simplifying, we get

r

r

2

d 2T r dr

+

2

r

4 dT r =0 r r dT r =z dr

Let r

dc + 4z = 0 dr

dz dr = –4 z r

or Integrating both sides, we get

logez = –4loger + logec1 where logec1 is a constant of integration. logez = loge

 c "# !r $

dT r c = 1 or dr 4

1 4

or z =

sr = c1

c1 4

dr r4

Integrating both sides, we get c1 + c2 3r 3 where c2 is another constant of integration. Also we know that

sr = –

rdT r 2 dr

sq = sr +

 !

=  =–

sq =

c1 3r

c1 3r 3 c1

6r 3

3

"# $

 c2  + c2 +

+ c2

(8.31)

rdT r c r c1 = – 13 + c2 + 2 dr 2 r4 3r c1 2r 3

(8.32)

Chapter 8:

Thick and Compound Cylinder

259

Putting c1 = 6B and c2 = A, we get

sq =

B

+A

r3

2B

sr = –

r3

(8.33)

+A

(8.34)

Equations (8.33) and (8.34) are the governing equations for thick shell. Boundary conditions are: At r = a, sr = –Pa and at r = b, sr = –Pb, then

2B

–Pa = –

a3 2B

–Pb = –

b3

+A

(8.35)

+A

(8.36)

Solving Eqs. (8.35) and (8.36), we get A=

Pa a 3  Pb b 3

and

(b 3  a 3 )

a 3b 3 ( Pa  Pb )

B=

2 (b 3  a 3 )

Putting the value of A and B in Eqs. (8.33) and (8.34),

sq = sr =

Pa a 3  Pb b 3 b a 3

+

3

Pa a 3  Pb b 3 b a 3

–

3

   b  a  P P   b  a 

a 3b 3 Pa  Pb 2r

3

a3b 3 r

3

3

3

a 3

b 3

(8.37)

(8.38)

Special cases CASE I A spherical shell subjected to internal pressure: In this case, Pb = 0 and Pa = P. Then Eqs. (8.37) and (8.38) become

sq = sr =

Pa 3

1  b   2r  1  b   r  3

b3  a3 Pa 3

3

3

b3  a 3

3

CASE II A spherical shell subjected to external pressure: Eqs. (8.31) and (8.38) become

1  a   2r  b a  Pb  a  =– 1   b a  r 

sq = – sr

In this case, Pa = 0 and Pb = P. Then

Pb 3

3

3

3

3

3

3

3

3

3

260

Strength of Materials

EXAMPLE 8.6 A thick spherical shell of 40 mm inside diameter is subjected to an internal pressure of 20 MPa. Determine the necessary thickness of the shell, if permissible, stress of the shell material is 40 MPa. Solution: Given Inner diameter = 40 mm, Internal pressure = 20 MPa, Permissible stress sq = 40 MPa Using Lame equation, Px =

2B r3

–A

At r = 20 mm, Px = 20 MPa, \

20 =

or

A=

Now,

sq =

B r3

+A

or

–A=

2B –A 8000

2B – 20 8000

(i)

(Hoop stress is maximum at inner radius.) 40 =

or

2B 20 3

B

+A

20 3

B 8000

A = 40 –

(ii)

Equating Eqs. (i) and (ii), we get 2B B – 20 = 40 – 8000 8000

3B = 60 8000

or or

B = 160000

and

A = 20

Again using Lame equation, Px =

2B r3

–A

At r = b, Px = 0, \

0=

or

b=

2 – 160000 b3 3

320000 = 25.2 mm, therefore, thickness of shell is: 20

t = 25.2 – 20 or

– 20

t = 5.2 mm

Chapter 8:

Thick and Compound Cylinder

261

EXAMPLE 8.7 A thick spherical shell 30 mm internal diameter and 5 mm thick is subjected to an internal pressure of 15 MPa. Determine the variation of radial and hoop stress in the shell. Take E = 200 GPa, v = 0.3. Solution: Given, a = 15 mm, b = 15 + 5 = 20 mm, P = 15 MPa, t = 5 mm Using Lame equation, Px =

2B

–A

r3

At r = 15 mm, Px = 15 MPa, therefore, 15 = or

A=

2B (15)3

–A

2B – 15 3375

(i)

At r = 20 mm, Px = 0 MPa, therefore, 0 =

or

A=

2B (20) 3

–A

2B

(ii)

(20) 3

Equating Eqs. (i) and (ii), we get

2B 2B – 15 = 3375 (20) 3 Solving

B = 43783.8, A = 10.95

sq =

Then Hoop stress

=

sr =

and

=

B

+A

r3

43783.8 r3

2B r3

+ 10.95

–A

2 – 43783.8 r3

– 10.95

Hoop and radial stresses in thick spherical shell are tabulated as follow: r (mm) T R (MPa)

T r (MPa )

15 23.93

16 21.63

17 19.86

18 18.45

19 17.33

14.99

10.42

6.87

4.06

182 .

20 16.42 0

262

Strength of Materials

Fig. 8.8

Stress distribution across the cylinder wall.

EXERCISES 8.1 Prove that the permissible stress induced in the thick cylinder is maximum at inner radius, if the cylinder is subjected to internal pressure. 8.2 A pipe of 400 mm internal diameter and 100 mm thickness contains fluid at a pressure of 8 N/mm2. Find the maximum and minimum hoop stress across the section. Also sketch the radial pressure distribution and hoop stress distribution across the section. [Ans. 208 MPa, 12.8 MPa] 8.3 Find the thickness of metal necessary for cylindrical shell of internal diameter of 80 mm to withstand an internal pressure of 25 N/mm2 and maximum permissible hoop stress is 125 N/mm2. [Ans. t = 9 mm] 8.4 A thick cylindrical shell of 160 mm internal diameter is 45 mm thick. The shell is subjected to an internal pressure of 52.5 N/mm2. Find the maximum and minimum intensities of hoop stress across the section. [Ans. 127.5 N/mm2, 75 N/mm2] 8.5 A thick spherical shell of 400 mm internal diameter is subjected to an internal fluid pressure of 1.5 N/mm2. If the permissible tensile stress in the shell material is 3 N/mm2, find the necessary thickness of the shell. [Ans. 52 mm] 8.6 A steel cylinder of 200 mm external diameter is to be shrunk to another steel cylinder of 100 mm internal diameter. After shrinking the diameter at the junction is 150 mm and radial pressure at the junction is 12.5 N/mm2. Find the original difference in radii at the junction. [Ans. 0.02025 mm] Take E = 2 ´ 105 N/mm2 8.7 A thick spherical shell of 160 mm internal diameter is subjected to an internal pressure of 40 N/mm2. Find the thickness of the shell, if the permissible tensile stress is 80 N/mm2. [Ans. 21 mm] 8.8 A spherical shell of 120 mm internal diameter has to withstand an internal pressure of 30 MPa. If the permissible tensile stress is 80 MPa, calculate the thickness of the shell. [Ans. t = 11.5 mm]

Chapter 9: Unsymmetrical Bending and Shear Centre

9 9.1

263

Unsymmetrical Bending and Shear Centre

INTRODUCTION

If the load line on a beam does not coincide with one of the principal axes of the section, the bending takes place in a plane different from a principal plane. This type of bending is known as unsymmetrical bending. In the case of unsymmetrical bending, the direction of neutral axis is not M T = , it is I y assumed that the neutral axis of the cross-section of the beam is perpendicular to plane of loading. Following are the two reasons of unsymmetrical bending:

perpendicular to the plane of loading. While using the well known bending equation

1. The section is symmetrical, e.g. rectangular, circular, I-section but the load is inclined to both of the principal axes. 2. The section itself may be unsymmetrical, e.g. an angle section or channel section.

9.2

DEFINITIONS

Centroid:

The centroid or centre of area is defined as the point where the whole area is assumed to be concentrated. Let x and y be the distances of the centroid of area from the given axes Y and X, respectively and x1, x2, …, and y1, y2, …, be the distances of the centroid of the areas A1, A2, …, from the given axes Y and X then x =

A1 x1  A2 x2  ž  A1  A2  ž 

y =

A1 y1  A2 y2  ž  A1  A2  ž 

Moment of inertia: The moment of inertia, also called second moment area, about any axis is the product of elemental area and square of the distance of the centroid from the axis. Ix =

I

y2 dA

263

264

Strength of Materials

Iy =

x2dA

I = Ak2

or where k is radius of gyration.

Parallel axis theorem:

I

The moment of inertia of an area about any axis passing through O is

equal to moment of inertia about a centroidal axis passing through the centroid O plus the area of figure multiplied by the square of the distance between the axes as shown in Fig. 9.1. Ix = Ix + A( y )2 Iy = I y + A( x ) 2 Y

Y

X

O

O

X

Fig. 9.1 Coordinate representation to determine moment of inertia.

Product of inertia:

The product of inertia of an area with respect to any two rectangular axes may be defined as the sum of products obtained by multiplying each element of the area by the product of the two coordinates of the element with respect to two rectangular axes, i.e., Ixy =

I

xy dA

where dA is an element of the given area and x, y are the coordinates of the element with respect to the two rectangular axes.

9.3

STRESSES DUE TO UNSYMMETRICAL BENDING

Figure 9.2 shows the cross-section of a beam under the action of bending moment M acting in plane YY. XX, YY = Coordinate axes passing through O UU, VV = Principal axes inclined at an angle q to XX and YY axes, respectively O = Centroid of the section The moment M in the plane YY can be resolved into its components in the plane UU and VV as follows: Moment in the plane UU Moment in the plane VV

MU = M sin q MV = M cos q

Chapter 9: Unsymmetrical Bending and Shear Centre

265

Fig. 9.2 Cross-section of beam with coordinate representations.

The resultant bending stress at the point C(u, v) is given by

sb = =

or

MU u M v + V IVV IUU M sin R u M cos R v  IVV IUU Ë u sin R

sb = M Ì Í

IVV



v cos R Û Ü Ý

IUU

It is more preferable to assign positive sign to term

M v MU u or V , if it produces tension at a IVV IUU

point and negative sign if it produces compression. On any point on the neutral axis, resultant bending stress will be zero.

sb = 0  u sin R v cos R   M  =0 IUU   IVV

or

 u sin R v cos R     =0 IUU   IVV

or

v = –Ì

Ë IUU Í IVV

–

(9.1)

sin R Û Ü

cos R Ý

u

266

Strength of Materials

I  = –  UU tan R  u I  VV  This is an equation of a straight line passing through the centroid O of the section and inclined at an angle a with UU. tan

9.4

Ë IUU

a=–Ì

Í I VV

Û

tan R Ü

(9.2)

Ý

DEFLECTION OF BEAM DUE TO UNSYMMETRICAL BENDING

To determine the deflection of beam due to unsymmetrical bending, the load W may be resolved into components parallel to the principal planes, and deflection caused by these components of the load calculated from the usual equation for deflections of symmetrically loaded beams. The actual deflection of beam is the vector sum of the deflections found from the component of load. The load W can be resolved into the following two components as shown in Fig. 9.3

Fig. 9.3

Components of load and deflection along principal axes.

Let dv = The component W cos q will cause deflection along the line OV du = The component W sin q will cause deflection along the line OU Then

du =

Ë K (W sin R ) l 3 Û Ì Ü E Ivv ÍÌ ÝÜ

(9.3)

and

dv =

Ë K (W cos R ) l 3 Û Ì Ü E Iuu ÌÍ ÜÝ

(9.4)

where l = Length of the beam K = Constant depending on the end conditions of the beam

Chapter 9: Unsymmetrical Bending and Shear Centre

267

The resultant deflection is

d= d=

(E u )2  (E v )2 2

È W sin R Ø È W cos R Ø É Ù É I Iuu ÙÚ Ê Ú Ê vv

K l3 E

2

(9.5)

The inclination g of the direction of d, with the line OV is given by tan g = =

Eu Ev

I uu I vv

tan q

(9.6)

EXAMPLE 9.1 A cantilever, of I-section 2 m long is subjected to a load of 250 N at the free end as shown in Fig. 9.4. Determine the resulting bending stress at corners A and B, on the fixed section of the cantilever. Y1

M

Y, V MV

A

2.5 mm

B

20° X1 45 mm X, U

50 mm

MU O

X1

X, U

2 mm

C 30 mm Y, V

D 1

Y

Fig. 9.4

Solution: Given: Length of the cantilever l = 2 m, Load W = 250 N Since I-section is symmetrical about XX and YY axes, therefore, XX and YY are the principal axes UU and VV. Moment of inertia IUU = IXX =

30 – 50 3 28 – 453  12 12

268

Strength of Materials

2.5 – 30 3 45 – 2 3  12 12 = 1.128 ´ 10–8 m4

IVV = IYY = 2 ´

and

Components of W are: WU = = = WV = = =

and

Wu sin q 250 sin 20 85.50 N Wv cos 20 250 cos 20 234.92 N

Bending moments are given by MU = = MV = =

and

WU l (250 sin 20) ´ 2 = 171 N-m WV l (250 cos 20) ´ 2 = 469.84 N-m

MU will cause a tensile stresses at points A and C and compressive stresses at points B and D. MV will cause a tensile stresses at points A and B and compressive stresses at points C and D. For point A, MU = +171 N-m, MV = +469.84 N-m, uA = 15 mm and vA = 25 mm. \

sA =

M sin R uA M cosR v A  Ivv IUU

or

sA =

MU uA MV v A  Ivv IUU

or

3 ) ( 469.84) (25  10 3 )  sA =  (171)(15  10   8 9.99  10 8  1.128  10 

= (+227.39 + 117.57) ´ 10–6 = 344.96 ´ 10–6 N/m2 = +344.96 MPa For point B, MV = + 469.84 N-m, MU = –171 N-m, UB = 15 mm and VB = 25 mm. MU uB

\

sB =

or

sB = 

I vv



MV vB IUU

 171(15  10  3 ) 8  1.128  10



(  469.84) (25  10  3)   9.99  10  8 

= (–227.39 + 117.57) ´ 10–6 = –109.82 ´ 10–6 N/m2 = –109.82 MPa

Chapter 9: Unsymmetrical Bending and Shear Centre

269

EXAMPLE 9.2 A beam of rectangular section, 80 mm wide and 120 mm deep (see Fig. 9.5) is subjected to a bending moment of 10 kN-m. The trace of the plane of loading is inclined at 45o to the YY axis of the section. Locate the neutral axis of the section, and calculate the maximum bending stress induced in section. (UPTU 2002–03) Y, V C

A MV M

120 mm

X, U

45°

B

MU

X, U

D

80 mm Y, V

(a)

(b) Fig. 9.5

Solution: The components of M along UU and VV are shown in Fig. 9.5. MU will cause tensile stresses at points A and D and compressive stresses at points C and B whereas MV causes a tensile stresses at points C and A and compressive stresses at points B and D. Let the plane of loading be inclined at q with YY axis and the neutral axis be inclined at a with XX axis. Principal moment of inertia is given as: IUU = Ix =

80 – 1203 = 11.52 ´ 106 mm4 12

IVV = Iy =

120 – 803 = 5.12 ´ 106 mm4 12

tan

a=–

IUU tan q IVV

 11.52 – 10  ´ tan 45° = – 2.25  5.12 – 10  6

= – or

a=

6

–66°

This gives the location of the neutral axis NA.

270

Strength of Materials

Maximum stress will occur at the point which is farthest from the principal axis.

sA =

Thus,

MU u MV v  IVV IUU

( 10  106 sin 45°)  40 (+10  106 cos 45°)  60 + 5.12  106 11.52  106 = +55.24 + 52.08 = +107.32 MPa (Tensile) =

sB =

and

=

MU u MV v  IV IU

 10  10 6 sin 45 10  106 cos 45 (40) + (60) 5.12  10 6 11.52  10 6

= –55.24 – 52.08 = –107.32 MPa (Compressive)

EXAMPLE 9.3 A beam of T-section (flange: 100 ´ 20 mm, Web 150 ´ 10 mm) is 3 m in length and is simply supported at the ends. It carries a load of 5 kN inclined at 20o to the vertical and passing through the centroid of the section. If E = 210 GPa, calculate: (i) (ii) (iii) (iv)

Maximum tensile stress Maximum compressive stress Deflection due to the load Position of the neutral axis

Solution: Given Length of beam l = 3 m, Load W = 5 kN Centroid of T-section is given by y =

A1y1  A2 y2  A3 y3  ž A1  A2  A3  ž

100 – 20 – 10  150 – 10 (20  75) 100 – 20  150 – 10 = 46.4 mm =

Since the section is symmetrical about the vertical axis, therefore, the principal axes pass through centroid G and are along UU and VV axes as shown in Fig. 9.6.

=

=

100 – 20 ! 12  20 – 100 ! 12

Ixx = IUU 3

3

"# 10 – 150 $ ! 12

 100 – 20 (46.4  10) 2 

= 9.07 ´ 10–6 m4 Iyy = IVV



150 – 10 3 12

"# = 1.679 ´ 10 $

–6

m4

3

 10 – 150 (123.6  75) 2

"# $

Chapter 9: Unsymmetrical Bending and Shear Centre

271

Fig. 9.6

Components of W are: WU = = WV = =

5 ´ 103 sin 20° 1.71 kN 5 ´ 103 cos 20° 4.69 kN

Bending moments:

WU l 4 1.71 – 3 = = 1.283 kN-m 4 W l MV = V 4 4.69 – 3 = = 3.52 kN-m 4 MU will cause compressive stress at B and D and maximum tensile stress at A and C; whereas MV will cause maximum compressive stress at A and B and maximum tensile stress at C and D. MU =

Maximum tensile stress: M u

M v

U c  V c sC = I IUU VV

 1.283  103  5  10  3 3.52  103  123.6  10  3   =   1.679  10  6 9.07  10  6   = 3.82 ´ 106 + 47.96 ´ 106 = 51.78 ´ 106 N/m2 = 51.78 MPa

272

Strength of Materials

Maximum compressive stress at B:

sB =

MU u B MV v B  Ivv Iuu

 1.283  10 3  50  10  3 3.52  10 3  46.4  10  3    10–6 =  1.679  10  6 9.07  10  6  

= (–38.21 – 18.007) ´ 106 = –56.22 ´ 106 N/m2 = –56.22 MPa (Compressive) Deflection due to load:

d=

KW l 3 E

sin 2 R 2 IVV



cos2 R 2 IUU

where K = 1/48 for a beam with simply supported ends for point load. 3 3  1  5  10 (3) d=    48  210  109

sin 2 20 (1.67  10  6 )



cos2 20 (9.07  10  6 )2

= 1.024 ´ 10–3 m = 1.024 mm Position of neural axis: tan

a=–

IUU tan q Ivv

9.07 – 10  6 ´ tan 20° = – 1.966 1.679 – 10  6 » – 63° =–

a

or

EXAMPLE 9.4 A rectangular beam as shown in Fig. 9.7 is 80 mm wide and 120 mm deep. It is used as simply supported beam on a span of 8 m. Two loads of 4 kN each are applied to the beam, each load being 2 m from a support. The plane of the loads makes an angle 30° with the vertical plane of symmetry. Find the direction of neutral axis and bending stress at the point A. Solution:

Principal moment of inertia:

 80 – (120) "# ! 12 $ 3

IUU =

= 11.52 ´ 106 mm4

120 – (80) "# ! 12 $ 3

and

IVV =

= 5.12 ´ 106 mm4

Chapter 9: Unsymmetrical Bending and Shear Centre

Y, V

A

273

W A

B

120 mm

30° X, U a

X, U 4 kN

C

2m

D

80 mm

2m 8m

N

Y, V

4 kN

Fig. 9.7

\

tan

a=–

IUU tan q IVV

11.52 – 106 ´ tan 30° 512 . – 10 6 = –78.83° =–

MU and MV both produce compressive stress at point A,

\ =

MU = = MV = =

(4 ´ 103 sin 30°) ´ 2 4 ´ 103 N-m (4 ´ 103 cos 30°) ´ 2 6.93 ´ 103 N-m

sA =

MU uA MV v A  IUU IVV

 4 – 10 6 (40)  6.93 – 10 6 (60)  512 . – 10 6 11.52 – 10 6

= –31.25 – 36.09 = – 67.34 MPa

EXAMPLE 9.5 A cantilever beam of I-section is used to support the loads inclined to the V-axis as shown in Fig. 9.8. Calculate the stresses at the corners A, B, C and D. Also locate the neutral axis. Solution:

Bending moments: MU = – (10 ´ 103) sin 30° ´ 1.5 + 5 cos 45° ´ 103 ´ 2.5 = –1.337 kN-m MV = (10 ´ 103) cos 30° ´ 1.5 + (5 ´ 103) cos 45° ´ 2.5 = 21.83 kN-m

274

Strength of Materials

Fig. 9.8

Principal moments of inertia:

15 – 20 3 13 – 16 3  12 12 = 5562.67 cm4

IUU = Ixx =

2 – 2 – 153 16 – 2 3  12 12 = 1135.67 cm4

Ivv = Iyy =

For point A, MU and MV both produce tensile stresses at point A. MU = +1.337 kN-m MV = +21.83 kN-m uA = 7.5 cm

sA =

\

=

vA = 10 cm

MU uA MV v A  Ivv IUU +1.337  10 6 (75) 1135.67  10 4



+ 21.83  10 6 (100) 5562.67  10 4

= +8.83 + 39.24 = +48.07 MPa (Tensile) For point B, MU and MV both produce compressive and tensile stresses at point B. MU = –1.337 kN-m \

MV = +21.83 kN-m uB = 7.5 cm

sB = =

vB = 10 cm

MU uB MV v B  Ivv IUU 1.337  10 6 (75) 1135.67  10 4



 21.83  10 6 (100) 5562.67  10 4

= +30.41 MPa (Tensile)

Chapter 9: Unsymmetrical Bending and Shear Centre

275

For point C, MU and MV both produce tensile and compressive stresses, respectively at point C. MU = +1.337 kN-m, MV = –21.83 kN-m

sC =

\

=

MU uC MV vC  Ivv IUU  1.337  10 6 (75) 1135.67  10 4



21.83  10 6 (100) 5562.67  10 4

= +8.83 – 39.24 = –30.41 MPa (Compressive) For point D, MU and MV both produce compressive stresses at point D. MU = –1.337 kN-m

MV = –21.83 kN-m uD = 7.5 cm

sD =

\

= = Thus,

tan

a= = =

a=

or

9.5

vD = 10 cm

MU uD MV v D  Ivv IUU  1.337  10 6 (75)



 21.83  10 6 (100)

1135.67  10 4 5562.67  10 4 –8.83 – 39.24 = – 48.07 MPa (Compressive) I – uu tan q Iv 5562.67 MU – – 1135.67 MV 5562.67 1.3375 – – 1135.67 21.83 –16.7°

SHEAR CENTRE

The shear centre for any transverse section of the beam is the point of intersection of the bending axis and the plane of the transverse section. In case of beam having two axes of symmetry, the shear centre coincides with the centroid. Shear centre is also known as centre of twist. When the load passes through the shear centre, there will be only bending in the cross-section and no twisting. In case of beam having two axes of symmetry, the shear centre coincides with the centroid. In case of sections having one axis of symmetry, the shear centre does not coincide with the centroid but lies on the axis of symmetry.

9.5.1

Shear Centre for Channel Section

Figure 9.9 shows a channel section (flanges: b ´ t1; web h ´ t2) with XX as the horizontal symmetric axis. Let S = Applied shear force vertical downwards S1 = Shear force in the top flange (there will be equal and opposite shear force in the bottom flange as shown).

276

Strength of Materials

Fig. 9.9 Location of shear centre in channel section.

Now, shear stress t in the flange at a distance x from the right hand edge (of the top flange), t =

Now,

S Ay I xx t

A y = (t1 x)

\

t =

h 2

St1 x h S xh ¹ = I xx t1 2 2 Ixx

(a)

Shear force in elementary area = t dA, But dA = t1dx Hence, total shear force in top flange

I I b

S1 =

t dA

0 b

=

t t1 dx

0

From Eq. (a), we have

I b

S1 =

0

or

S1 =

Sx h Sht1 t1 dx = 2 Ixx 2 Ixx

I b

x dx

0

Sht1 b 2 I xx 4

Let e be the distance of the shear centre SC from the web along the symmetric axis XX. Taking moments about the centre O of the web, we get Se = S1

h h + S1 2 2

Chapter 9: Unsymmetrical Bending and Shear Centre

277

Se = S1 h = \

e=

Now,

Ixx = 2

= or

b – t ! 12

3 1

Sht1 b 2 St h 2 b 2 h= 1 4 Ixx Ixx 4 b 2 h 2 t1 4 I xx

 bt1

 h  2

(9.7) 2

"#  t h #$ 12 2

3

=

bt13 bt1 h2 t2 h3   6 2 12

bt1 h2 t2 h 3 bt 3  (neglecting the term 1 , being negligible in comparison to other terms) 2 12 6

h2 (t2 h + 6bt1) 12 Substituting the value of Ixx in Eq. (9.7), we get Ixx =

e = Let and

12 b 2 h 2 t1 3b 2 t1 – 2 = ( t2 h  6 bt1) 4 h (t2 h  6 bt1)

bt1 = Af (area of the flange) ht2 = Aw (area of the web) 3bA f 3b = e= A Aw  6 A f 6 w Af 3b e= A 6 w Af

Then

i.e.

(9.8)

EXAMPLE 9.6 A channel section has flanges 10 cm ´ 1 cm and web 16 cm ´ 1 cm. Determine the shear centre of the channel. Solution:

Here

b = 10 – 0.5 = 9.5 cm, t1 = 1 cm and t2 = 1 cm, h = 16 + 1 = 17 cm Web area Aw = ht2 = 17 ´ 1 = 17 cm2 Flange area Af = bt1 = 9.5 ´ 1 = 9.5 cm2

We know that e=

=

3b A 6 w Af 3 – 9.5 17 = 3.66 cm 6 9.5

278

Strength of Materials 10 cm 1 cm S 16 cm e 1 cm

1 cm 10 cm

Fig. 9.10

9.5.2

Shear Centre of Unequal I-section

Figure 9.11 shows an unequal I-section which is symmetrical about XX axis. Shear stress t = where

 !

I = Ixx = 2 ( b1  b2 )

I

b1

Shear force S1 =

S Ay It

"# $

t13 h2 h3  (b1  b2 )  t2 12 4 12

t dA

0

As

dA = t1 dx, A y = t1 x

h 2

Fig. 9.11 Location of shear centre in unequal leg I section.

Chapter 9: Unsymmetrical Bending and Shear Centre

I

279

b1

\

S1 =

t dA

0

=

Sxt1 h ´ t1 dx I xx t1 2

I

b1

=

0

=

S ´ h t1 x dx 2 I xx b1

Sht1 x 2 2 I xx 2

= 0

Sht1b12 4 I xx

Similarly, the shear force S2 in the other part of the flange is: Sht1 b22 4 I xx Taking moments of the shear forces about the centre of the web O, we get

S2 =

S2 h = S1 h + Se

(S3 = S for equilibrium)

where e is the distance of shear centre from the centre of the web. or (S2 – S1) h = Se Sh 2 t1 2 (b 2 – b12) = Se 4 I xx

\

e =

t1 h 2 ( b22  b12) 4 I xx

(9.9)

EXAMPLE 9.7 Determine the position of the shear centre of the section of a beam shown in Fig. 9.12. 10 cm

8 cm

4 cm S 40 cm e 2 cm 4 cm 10 cm Fig. 9.12

8 cm

280

Strength of Materials

Solution:

Here t1 = 4 cm,

b1 = 8 cm,

Thus,

Ixx =

b2 = 10 cm,

h = 40 – 4 = 36 cm

18 – 40 3 16 – 363  12 12

= 96000 – 62208 = 33792 cm4 We know

\

e =

t1 h 2 (b22  b12) 4 Ixx

e =

4 – 36 2 (10 2  82 ) = 1.38 cm 4 – 33792

EXERCISES 9.1 If the maximum bending stress allowed in the cross-section of the beam shown in Fig. 9.13 is 15 MPa, determine the value of P. [Ans. P = 5.63 kN] 10 cm 15 cm 60 cm

P

Fig. 9.13

9.2 A simply supported beam of T-section, 2.5 m long carries a central compressive load inclined at 30° to Y-axis as shown in Fig. 9.14. If the maximum compressive and tensile stress in bending are not to exceed 75 MPa and 35 MPa, respectively, find the maximum load the beam can carry. [Ans. 5819.46 N] 10 cm A

30° P

2 cm

N

2 cm

Fig 9.14

15 cm

y

Chapter 9: Unsymmetrical Bending and Shear Centre

281

9.3 An I-beam section is loaded as shown in Fig. 9.15. Determine the stress at A also locate the position of the neutral axis. [Ans. 385.4 MPa, 83° 23¢ clockwise from x-axis]

10 cm

30°

1 cm 1.5 cm

1m

1m

2m

4 cm Fig. 9.15

9.4 A beam of angle section 150 ´ 100 ´ 10 mm is simply supported over a span of 1.6 m with 150 mm leg vertical. A uniformly distributed vertical load of 10 kN/m is applied throughout the span. Determine the maximum bending stress and deflection at the centre. [Ans. 68.3 MPa, 1.55 mm] 9.5 Locate the shear centre of the section shown in Fig. 9.16. [Ans. 5.1 cm] 9.6 Locate the shear centre for the unbalanced I-section shown in Fig. 9.17. [Ans. e = 0.91 cm] 120 mm 20 mm

8 cm

6 cm

4 cm

40 cm

160 mm

2 cm

10 mm

4 cm

20 mm

8 cm

6 cm

120 mm Fig. 9.16

Fig. 9.17

10 10.1

Columns and Struts

INTRODUCTION

Column or strut is defined as a member of a structure, which is subjected to axial compressive load. If the member of the structure is vertical and both of its ends are fixed rigidly while subjected to axial compressive load, the member is known as column, for example, a vertical pillar between the roof and floor. If the member of the structure is not vertical and one or both of its ends are hinged or pin joined, the bar is known as strut. Examples of struts are connecting rods or piston rods.

10.2

DEFINITIONS

Column: A bar or vertical member, subjected to an axial compressive load is called column. Strut: A bar or member of structure in any position other than vertical, subjected to a compressive load is called strut.

Equivalent or effective length:

The distance between adjacent points of inflexion is called equivalent or effective length. The effective length depends on end conditions of the column.

Slenderness ratio:

The ratio of the actual length of a column to the least radius of gyration of the column is known as slenderness ratio.

Safe load: Load under which column will not buckle, known as safe load. Buckling load: The minimum axial compressive load at which column tends

to have lateral

displacement or tend to buckle is called the buckling or crippling load.

10.3

CLASSIFICATION OF COLUMN

Column can be divided into three types based on their slenderness ratio:

Short column:

Column for which slenderness ratio is less than 32 is called short column. When short column of uniform cross-sectional area subjected to axial compressive load, the stress induced 282

Chapter 10:

Columns and Struts

283

in the column corresponding to crushing or direct compressive stress. Short column are designed based on crushing stress, as buckling stresses are very small as compared to crushing stress. l < 32 k

Medium column:

Column for which slenderness ratio is in-between 32 to 120 is called medium column. Medium columns are designed based on crushing as well as buckling stress. 32 £

l £ 120 k

Long column:

Column for which slenderness ratio is more than 120 is called long column. Long column are designed based on buckling stress, as crushing or direct compressive stresses are very small as compared to buckling stress. l > 120 k

10.4

ASSUMPTIONS MADE IN THE EULER’S COLUMN THEORY

The following assumptions are made in the Euler’s column theory: 1. 2. 3. 4. 5. 6. 7.

10.5 10.5.1

The column is initially perfectly straight and the load is applied axially. The cross-section of the column is uniform throughout its length. The column material is perfectly elastic, homogeneous and isotropic and obeys Hooke’s law. The length of the column is very large as compared to its lateral dimensions. The direct stress is very small as compared to the bending stress. The column fails by buckling alone. The self-weight of the column is neglected.

EXPRESSIONS FOR CRIPPLING LOAD OF DIFFERENT CASES Both the Ends are Hinged or Pinned

Figure 10.1 shows a column of length l of uniform cross-section hinged at both the ends. Let P be the crippling load at which the column has just buckled. Consider section XX at a distance x from bottom support. Let y be the deflection at the section. Bending moment M at the section is given by M = EI or or

EI

d2y = –Py d x2

d2y + Py = 0 d x2 d 2 y Py  =0 d x 2 EI

284

Strength of Materials

The solution of above differential equation is:

 

y = C1 cos x

P EI

 + C sin x   2

P EI

 

(10.1)

where C1 and C2 are the constants of integration. At x = 0, y = 0, From Eq. (10.1)

C1 = 0

At x = l, y = 0,

 

From Eq. (10.1)

0 = C2 sin l

or or

P EI

 

C2 = 0

 

sin l

P EI

 =0 

As C1 = 0, and if C2 = 0, then from Eq. (10.1) we will get y = 0. This means the bending will be zero, i.e., column will not bend, which is not true. \ C2 ¹ 0 Now,

  l 

sin l

\

Fig. 10.1 Column with both ends hinged.

 = 0 = sin 0, sin p, sin 2p, sin 3p , ...,  P  = 0, p, 2p, 3p ,..., EI  P EI

Consider the least practical value:

l 

\

10.5.2

P EI

 =p  P=

Q 2 EI l2

(10.2)

One End is Fixed and Other is Free

Consider a column of length l where lower end is fixed and upper being free. Let due to crippling load P the column just buckle. Let deflection at the top end is ‘a’. Consider a section XX at a distance x from the lower end. The bending moment at the section XX is: EI

d2y = P(a – y) dx 2

where y is the deflection at distance x.

Fig. 10.2 Column with one end fixed other free.

Chapter 10:

Now,

EI

d2y + Py = Pa dx 2

The solution of above differential equation is:

 

P EI

y = C1 cos x

 + C sin  x   2

At

x = 0,

\

0 = C1 + a C1 = –a

At

x = 0,

dy = 0, dx

 

 

The slope of the section is: dy =  C1 dx

\

P sin x EI

0 = 0 + C2

P EI

 +a 

P  C2 EI

 

P cos x EI

P EI

It is clear that either C2 = 0

or

\

P =0 EI

l

P ¹ 0. EI

But for the crippling load P the value of l

C2 = 0 Substituting the values of C1 and C2 in Eq. (10.3), we get

 

P EI

 +a 

a = –a cos l

P EI

 

 

 =0 

\

y = –a cos x

At

x = l, y = a,

\

 

P EI

cos l

l

(10.3)

y = 0,

or

or

285

d 2 y Py Pa  = 2 EI EI dx

or

or

Columns and Struts

+a

Q 3Q 5Q P = , , EI 2 2 2

P EI

 

286

Strength of Materials

Consider the least practical value: l

Q P = EI 2

or

P=

10.5.3

Q 2 EI

(10.4)

4l 2

Both Ends are Fixed

P

Consider the column AB of length l fixed at both the ends. Let P be the crippling load and M be the fixed end moment at A and B. Bending moment at section XX = M – Py

EId 2 y = M – Py dx 2

\

d y + Py = M EI dx 2

or or

d2y P M + y= 2 EI EI dx The solution of above differential equation is:

 

P EI

y = C1 cos x

 + C sin  x   2

Slope at the section will be dy = –C1 dx

 

P sin x EI

P EI

 +C 

2

At B,

or At B,

x = 0,

\

or or

 

M cos l P

 

P EI

 

+

P

 

P cos x EI

  = 0   P =1 cos  l  EI   

M 1  cos l P

P EI

X

B

+M  P

dy = 0 = C2 dx C2 = 0 M P

y

X

x

x = 0, y = 0 M 0 = C1 + P M C1 = – P

\

At A, x = l, y = 0= –

P EI

A

l

2

M

P EI

P EI

 

M

Fig. 10.3 Column with both ends fixed.

Chapter 10:

or

l

P = 0, 2p, 4p, 6p EI

l

P = 2p EI

Columns and Struts

287

Consider the least practical value:

\

4Q 2 EI l2

P=

(10.5)

10.5.4 One End is Fixed, Other is Hinged Consider a column AB of length l fixed at lower end and pinned at upper end. Let P be the crippling load. M is the fixed end moment at support. In order to balance the fixed moment M there will be the horizontal reaction at A. Bending moment at the section XX is: EI or

or

EI

d2y = –Py + H(l – x) dx 2

d2y + Py = H(l – x) dx 2

P H d2y + y = (l – x) EI EI dx 2 The solution of the differential equation is:

 

y = C1 cos x

P EI

 + C sin  x   2

The slope at the section is:

 

dy P = –C1 sin x dx EI At x = 0, y = 0

P EI

P EI

P EI

C1 = – At x = 0,

(10.6)

–H  P

0 = C1 +

or

H l P

H l P

P H dy = 0 = C2  dx EI P

or

C2 =

At x = l, y = 0, or

 + H (l – x)  P

 + cos x  

\

Fig. 10.4

0=–

 

H P l cos l EI P

H P

EI P

 + H   P

HI P

 sin  l  

P EI

 

Column with one end fixed other hinged.

288

Strength of Materials

Simplifying, we get

 

P EI

tan l The solution to this equation is: l or or Approximately,

P EI

=l

P = 4.5 radians EI

l2 P = (4.5)2 = 20.25 EI 20.25 EI P= l2 2 20.25 = 2p

or

P=

10.6

 

2Q 2 EI l2

(10.7)

EFFECTIVE LENGTH OF A COLUMN

The effective length of a given column with given end conditions is the length of an equivalent column of the same material and cross-section with hinged ends, and having the value of the crippling load equal to that of the given column. Let

L = Effective length of a column l = Actual length of the column P = Crippling load for the columnm

The crippling load for any type of end condition is given by P=

Q 2 EI

(10.8) L2 The crippling load P in terms of actual length and effective length, and also the relation between effective length and actual length are given in Table 10.1. Table 10.1 End condition and relation between effective length and actual length of Column S.No.

End conditions of column

Crippling load in terms of

Actual length 1.

Both end hinged

2.

One end is fixed and other is free

3.

Both ends fixed

4

One end fixed and other is hinged

Q 2EI l

2

Q 2EI 2

4l 4Q 2EI

l2 2Q 2EI l

2

Relation between effective length and actual length

Effective length

Q 2EI L2

Q 2EI

L2 Q 2EI

L2 Q 2EI 2

L

L =l L = 2l L= L=

l 2

l 2

Chapter 10:

10.7

Columns and Struts

289

SLENDERNESS RATIO

The ratio of the actual length of a column to the least radius of gyration of the column is known as slenderness ratio. Mathematically, slenderness ratio is given by Slenderness ratio = =

Actual length Least radius of gyration

l k

(10.9)

The strength of column depends upon the slenderness ratio and end condition. If the slenderness ratio is increased, the compressive strength of a column decreases as the tendency of buckle is increased.

10.8

CRIPPLING STRESS IN TERMS OF EFFECTIVE LENGTH AND RADIUS OF GYRATION

The moment of inertia I can be expressed in terms of radius of gyration k as: I = Ak2 where A is area of cross-section. Column will tend to bend in direction of least moment of inertia. So column should be designed using the least value of moment of inertia, then k is the least radius of gyration of the column section. Now, crippling load P in terms of effective length is given by P=

Q 2 EI L2

=

Q 2 E – Ak 2

=

Q 2E – A

L2 L2 k2

Crippling stress =

=

Q 2E – A

 L   k

2

Crippling load P = Area A

=

Q 2E – A

=

Q2E

 L A   k

 L   k

2

2

(Substituting the value of P)

(10.10)

290

10.9

Strength of Materials

LIMITATION OF EULER’S FORMULA

From Eq. (10.10), we have Crippling stress =

 

If the slenderness ratio i. e.

l k

 

Q2E

 L   k

2

is small, the crippling stress will be high. But for the column

material, the crippling stress cannot be greater than the crushing stress. Hence when the slenderness ratio is less than a certain limit, Euler’s formula gives a value of crippling stress greater than the crushing stress. In the limiting case, we can find the value of l/k for which the crippling stress is equal to the crushing stress. For a mild stress column with both ends hinged, Crushing stress = 330 N/mm2 Young’s modulus E = 2.1 ´ 105 N/mm2 In the limiting case, equating the crippling stress to the crushing stress, we get Crippling stress = Crushing stress

Q 2E

 l   k

or

or

\ or

2

Q 2 – 2.1 – 105

 l   k

 l   k

2

=

l = k

2

= 330

= 330

Q 2 – 2.1 – 105 330

= 6282

6282 = 79.26, say, 80

Hence if the slenderness ratio is less than 80 for mild steel column with both ends hinged then Euler’s formula will not be valid.

EXAMPLE 10.1 Find the safe load on the column. Both ends are hinged. Length of column = 6 m, Ixx = 8600 cm4, Iyy = 454 cm4, Area = 56 cm2, Factor of safety = 3.0, E = 2 ´ 104 kN/cm2. Solution: Given

Iyy < Ixx

Chapter 10:

Columns and Struts

291

So the column tends to bend in yy direction. I = Iyy = 454 cm4 L = l = 6 m = 600 cm (Both ends are hinged.) A = 56 cm2 E = 2 ´ 104 kN/cm2 = 2 ´ 104 ´ 1000 N/cm2 = 2 ´ 107 N/cm2 Euler’s crippling load P =

=

Q 2 EI L2

Q 2  2  10 7  454 600 2

= 248933.35 N = 249.93 kN Given factor of safety = 3.0 249.93 3 = 82.98 kN

Safe load =

EXAMPLE 10.2 A strut 3 m long is 60 mm in diameter. One end of the strut is hinged while other is fixed. Find the safe compressive load for the member using Euler’s formula. Assume the factor of safety = 3.0 and E = 2 ´ 105 N/mm2. Solution:

Euler’s crippling load =

Q 2 EI

L2 Since one end fixed, other is hinged,

\ Effective length

L =

=

Moment of inertia

I =

= \

P=

l 2 3000 2

= 2121.32 mm

Q d4 64

Q (60) 4 64

= 636.17 ´ 103 mm4

Q 2 – 2 – 105 – 636.17 – 10 3 (2121.32 )2

= 279056.58 N

292

Strength of Materials

Safe compressive load =

279056.58 3

= 93018.86 N = 93.018 kN

EXAMPLE 10.3 A bar of length 6 m when used as a simply supported beam and subjected to uniformly distributed load of 50 kN/m over the whole span deflects 15 mm at the centre. Determine the safe compressive load when it is used as a column with the following end conditions: (i) Both ends fixed (ii) One end fixed and other hinged Take the factor of safety = 2.5 Solution:

For simply supported beam carrying udl, Deflection y =

5 wl 4 384 EI

15 5 50 – 6 4 = ´ 1000 384 EI 5 – 50 – 6 4 – 1000 15 – 384 = 56250 kN-m2

\

EI =

(i) Both ends fixed Effective length L = \

Euler’s crippling load P = =

\

Safe comp. load =

l 6 = =3m 2 2

Q 2E I L2

Q 2 – 56250 32

= 61685 kN

61685 = 24674 kN 2.5

(ii) One end fixed and other hinged Effective length L =

Euler’s crippling load P =

=

l 2

=

6 2

= 4.2426 m

Q 2 EI L2

Q 2 – 56250 (4.24) 2

= 30880.94 kN

Chapter 10:

\

Safe load =

Columns and Struts

293

30880.94 = 12352.37 kN 2.5

EXAMPLE 10.4 A hollow alloy tube 5 m long with external and internal diameters 40 mm and 25 mm, respectively was found to extend 6.4 mm under a tensile load of 60 kN. Find the buckling load for the tube when used as a column with both ends pinned. Also find the safe load for the tube, taking a factor of safety = 4. (AMIE 1989) Solution:

Given L = 5 m = 5000 mm, D = 40 mm, d = 25 mm, y = 6.4 mm, W = 60 kN = 60000 N, Safety factor = 4

Then

and

Cross-sectional area A =

Q

=

Q

Moment of inertia I =

(D 2 – d 2)

4

(402 – 252) = 765.76 mm4

4

Q 64

(D4 – d 4)

Q

(40 4 – 254) = 106488.94 mm4 64 The value of Young’s modulus is obtained as: =

or

d=

PL AE

E=

PL AE

=

60000 – 5000 765.76 – 6.4

= 6.121 ´ 104 N/mm2 Effective length L = 5000 mm Let P is buckling load. Using Euler’s equation, we get P= = And

Q 2  6.121  10 4  106488.94 5000 2

Q 2 EI L2

= 2.573 kN

Safe load = =

Buckling load Factor of safety 2.573 = 0.6432 kN 4

294

Strength of Materials

EXAMPLE 10.5 A solid round bar 5 m long and 5 cm in diameter was found to extend 5 mm under a tensile load of 55 kN. This bar is used as a strut with both ends hinged. Determine the buckling load for the bar and also the safe load taking factor of safety as 3. Solution:

Given L = 5 m = 5000 mm, d = 5 cm = 50 mm, d = 5 mm, W = 55 kN and FS = 3. Area of bar A =

Q

Moment of inertia I =

Q

\

´ 502 4 = 1963.49 mm2

and

=

64

Q

64 The value of Young’s modulus is obtained as:

or

d=

PL AE

E=

PL AE

d4 (50)4 = 306796.16 mm4

55000 – 5000 = 2.801 ´ 104 MPa 1963.49 – 5 Since the strut is hinged at its both ends, therefore, =

Effective length L = l = 5000 mm Using Euler’s equation, we get P= =

Q 2 EI L2

Q 2 – 2.801 – 10 4 – 306796.16 (5000)2

= 3392.52 N and

Safe load = =

Buckling load Factor of safety 3392.52 = 1130.84 N 3

EXAMPLE 10.6 Determine Euler’s crippling load for I-section 40 cm ´ 20 cm ´ 1 cm and 6 m long which is used as a strut with length coefficient of 0.5. Take Young’s modulus as 200 GPa. Solution: Given Dimensions of I-section = 40 cm ´ 20 cm ´ 1 cm Length L = 6 m = 6000 mm

Chapter 10:

295

Columns and Struts

Young’s modulus E = 200 GPa = 200 ´ 103 N/mm2 Moment of inertia of the section about XX axis, IXX = =

1 1 bd 3 – b1d 31 12 12

1 1 ´ 20 ´ 403 – ´ 19 ´ 383 12 12

= 19786 cm4 Similarly, the moment of inertia of the section about YY, IYY =

1 1 ´ 38 ´ 13 + 2 ´ ´ 1 ´ 203 12 12

= 3.166 + 1333.33 = 1336.5 cm4 Fig. 10.5

Least value of the moment of inertia is about YY axis.

Unequal I section.

I = 1336.5 cm4 = 1336.5 ´ 104 mm4

\

l 6000 = = 3000 mm 2 2 Using Euler’s equation, we get

Effective length L =

P= =

Q 2 EI

Q 2 – 200 – 10 3 – 1336.5 – 10 4 (3000)2

L2

= 2931272.51 N = 2931.3 kN

EXAMPLE 10.7 Determine the crippling load for a T-section of dimensions 10 cm ´ 10 cm ´ 2 cm and of length 6 m when it is used as strut with both of its end hinged. Take Young’s modulus E = 2.0 ´ 105 N/mm2.

10 cm

X

Solution: Given:

X

10 cm

Dimension of T-section = 10 cm ´ 10 cm ´ 2 cm Length l = 6 m = 6000 mm Young’s modulus E = 2.0 ´ 105 N/mm2

y

Here Y axis is the axis of symmetry. Therefore, X axis to be located.

y = =

a1 y1  a2 y2 a1  a2

20 – 9  16 – 4 180  64 = = 6.777 cm 20  16 36

2 cm

Fig. 10.6 T section.

296

Strength of Materials

Moment of inertia of the section about the axis XX,

 2  83   10  23   16  2.7772   20  2.2232  +  IXX =     12   12  = (6.667 + 98.834) + (85.333 + 123.387) = 314.221 cm4 Moment of inertia of the section about the axis YY, IYY =

2 – 10 3 8 – 2 3  = 166.67 + 5.33 12 12

= 172 cm4 Least value of moment of inertia is about YY axis, I = 172 cm4 = 172 ´ 104 mm4 Effective length L = l = 6000 mm Using equation, we get P= =

10.10

Q 2 EI L2

Q 2 – 2.0 – 10 5 – 172 – 10 4 6000 2

= 94309.55 N

RANKINE’S FORMULA

As per the limitation of Euler’s formula it is learnt that Euler’s formula gives better result for long column. Rankine established an empirical formula based on results of experiments performed which is applicable to all columns whether they are short or long. Let P be the actual crippling load. Rankine stated an empirical formula as:

1 1 1 =  PC PE P where PC = Crushing load = sC ´ A PE = Crippling load by Euler’s formula =

Q 2 EI

L2 For a given material, the crushing stress sC is a constant. Hence PC will also be constant. So, the value of P will be depending upon the value of PE, and PE depends on effective length L of the column.

(i) If the column is short, i.e., L is small, the value of PE will be large. Hence enough and is negligible as compared to

1 1 . Neglecting , we have PC PE

1 will be small PE

Chapter 10:

Columns and Struts

297

1 1 = PC P \

P = PC

Hence crippling load by Rankine’s formula for a short column is approximately equal to crushing load means short columns fail due to crushing. (ii) If the column is long, i.e., L is large means PE will be small, and compared to

1 will be large as PE

1 1 hence the value of may be neglected. PC PC

\

1 1 = PE P

or

P = PE

Hence the crippling load by Rankine’s formula for long column is approximately crippling load Euler’s formula. Thus the Rankine’s formula (For all lengths of columns ranging from short to long columns) is given as:

1 1 P  PC 1 =  = E PC PE PC PE P or

P=

=

=

PC PE PE  PC PC P 1 C PE

TC A TC A 1  Q 2E I 

[As PC =

sC A and PE =

Q 2 EI L2

]

 2   L 

But

I = Ak2

\

P=

or

P=

TC A TC A = 2 T AL T  L 2 1  2C 1  2C   2 (Q EAk ) Q Ek TC A L 1  a  k

2

(10.11)

298

Strength of Materials

Here a is Rankine’s constant, and is expresed as: a= The values of

TC Q 2E

sC and a for different materials are shown in Table 10.2.

Table 10.2

Rankine’s constant and compressive strength of some material

sC, N/mm2

Material Wrought iron

250

Cast iron

550

Mild steel

320

Strong timber

50

a

1 9000 1 1600 1 7500 1 750

EXAMPLE 10.8 Using Rankine’s formula, find the allowable load on a steel stanchion for the following data: (i) One end fixed and other end is hinged (ii) Length of column is 8 m 1 7500 Yield stress sC = 2500 N/mm2 E = 2 ´ 106 N/mm2 Ixx = 15000 cm4, Iyy = 25000 cm4 Area A = 100 cm2

(iii) Rankine’s constant a = (iv) (vi) (vii) (viii)

Solution:

Allowable load P =

TC A

L 1  a  k Since one end fixed, other hinged,

\

2

Effective length L = =

and

k=

l 2 8000 2

= 5656.85 mm

I A

15000 (I = Imin = Ixx) 100 = 12.25 cm = 122.5 mm

=

Chapter 10:

Now,

2500 – 100 – 100

P= 1

 

1 5656.85 – 7500 122.5

 

Columns and Struts

299

2

= 19465470 N = 19465.470 kN Assume factor of safety = 3.0. \

19465.47 = 6488.49 kN 3.0

Allowable load =

EXAMPLE 10.9 A hollow circular column of length 5 m is having an outer diameter of 15 cm, and an internal diameter of 10 cm. The column is hinged at the ends. Calculate the safe axial load which the column can carry, assuming the safe stress as 100 N/cm2 in compression for the material. Use Rankine’s formula and constant = 1/7500. Solution:

Area of column =

Q

(D2 – d2)

4

A =

Q

Moment of Inertia I =

Q

(152 – 102) 4 = 98.17 cm2

or

64

(D4 – d 4)

Q

(154 – 104) = 1994.175 cm4 64 Effective Length L = l = 5 m (both ends are hinged) =

Safe stress

sC = 100 N/cm2

Rankine constant a = Safe load P =

Radius of gyration k = = Then

1 7500

TC A 1  a ( L/ k ) 2

I A 1994.175 = 4.51 cm 98.17 100  98.17

P = 1

1  500   7500  4.51 

= 3720.25 N The safe load column can carry 3.721 kN.

2

300

Strength of Materials

EXAMPLE 10.10 Compare the safe load obtained in Example 10.9 using the same data with the Euler’s crippling load and state whether Euler’s formula is applicable for this case. Solution:

Safe load P =

Q 2EI l2

From Rankine’s constant a =

1 , the section is of mild steel. 7500

E = 210 GPa P=

Q 2  210  103  19.94  10 6 (5  103 )2

= 1653.119  103 N

= 1653.119 kN 1653.119 Eulers crippling load = 444.27 = 3.721 Rankine’s safe load

EXAMPLE 10.11 A column of effective length 5 m has a sectional area of 4500 mm2. If the least radius of gyration is 60 mm, what is the safe load the column can carry? Using Rankine’s formula. Solution:

Rankine’s critical load P =

TC A L 1  a  k

2

80 – 4500

P=

 

1 5000 – 1500 60 = 63947.368 N 1

 

2

63947.368 2.5 = 25578.947 N

Safe load =

EXAMPLE 10.12 A T-section used as a strut of 4 m length is hinged at both ends. The flange is 150 mm wide and 20 mm thick, and the web is 100 mm deep and 20 mm thick. Determine the safe load the column can carry. Take

sC = 320 MPa, Rankine’s constant =

1 , Factor of safety = 3.0. 7500

Chapter 10:

Fig. 10.7

Solution:

P=

Columns and Struts

301

T section.

TC A L 1  a  k

2

Area of T-section = 150 ´ 20 + 100 ´ 20 = 5000 mm2 Here Y axis is the axis of symmetry. Therefore, X axis to be located.

y =

150 – 20 – 10  100 – 20 – 70 = 34 mm 5000

150 – 20 3 20 – 100 3 + 150 ´ 20 ´ (34 – 10)2 + + 100 ´ 20 ´ (86 – 50)2 12 12 = 100000 + 1728000 + 1666666.7 + 2592000 = 6.08 ´ 106 mm4

\

Ixx =

and

Iyy =

Now,

20 – 150 3 100 – 20 3 + 12 12 = 5.625 ´ 106 + 66.66 ´ 103 = 5.69 ´ 106 mm4 Iyy < Ixx

The column tends to buckle at YY axis, \

k=

I yy A

5.69 – 10 6 = 33.74 mm 5000 Effective length L = l = 4 m = 4000 mm

=

\

320  5000

P= 1

1  4000   7500  33.74 

= 1.575 ´ 106 N

2

302

Strength of Materials

Rankine’s critical load = 1.575 ´ 106 N \

Safe load =

10.11 Let P ex ey k A (a)

1.575 – 10 6 = 5.25 ´ 105 N 3

ECCENTRIC LOADING = = = = =

Eccentric load Eccentricity in x direction Eccentricity in y direction Radius of gyration Area of column

Without buckling:

When there is no buckling of the column,

Resultant stress = Direct stress ± Bending stress sr = sd ± sb

(10.12)

P (Compressive) A If the load is at an eccentricity ey then

where

Direct stress

Bending stress Putting the value of

sd and sb

sd =

P – ey – y My = I Ak 2 in Eq. (10.12), we get

sb =

sr =

 

ey – y P 1“ A k2

 

(10.13)

where y is the distance of the fibre of the column from the natural axis. Consider a rectangular column of width l and thickness h, we get (sr)max =

\

Tensile stress =

    6e  P 1    A h  6e P 1  y A h

y

(10.14)

For zero tensile stress, –1 + or

6e y h

=0

ey =

   

h 1 h = 2 3 6

(10.15)

Thus, if the value of eccentricity ey on either side of X-axis does not exceed h/6, there will be no tensile stress in the column. Similarly, by considering eccentricity ex about Y-axis, we can show

Chapter 10:

Columns and Struts

303

that the eccentricity should not exceed l/6. Thus, for a rectangular section to avoid tension in the column, the load must lie within the middle third of the cross-section of the structure. This is known as the middle third rule. The permissible region has been shown in Fig. 10.8. l

h/6 h/6 l/6

h

l/6

Fig. 10.8 Middle third rule.

Consider a column ABCD subjected to a load P having eccentricity about both the axis as shown in Fig. 10.9. D

Y

C ex

h

ey

X

X

Y

A

B

l

Fig. 10.9

Column having eccentricity about both the axes.

Resultant stress = Direct stress ± Bending stress about XX axis ± Bending stress about YY axis Resultant stress

or

sr = sr =

 "# ! $

P M y “ A I

 P “ Pe !A I

x

yy

 M x "# !I $ Pe " x“ y# I #$ “

XX

Resultant stress at A, (sr)A =

YY

y

(10.16)

xx

 !

6e y 6e P 1 x  A l h

"# $

304

Strength of Materials

(sr)B =

Similarly,

(sr)C = (sr)D =

 ! P  6e 1 A! l P  6e 1 A! l

"# $ 6e "  # h $ 6e "  # h $

6e y P 6e 1 x  A l h

sr = 0

For no tensile stress,

x

x

y

y

6ex 6ey =1  l h

\

e x ey 1  = l h 6

or

This equation represents a straight line, which makes intercepts

l h and on X and Y axes, 6 6

respectively. This rhombus is called core of the column section. (b)

Considering buckling:

Using Rankine’s formula, P=

TC – A

 L 1  a   k

2

Taking eccentricity into account, we get P=

%K1  a  L  K'&  k 

TC – A 2

(K %K1  e x  e y (K )K &K k k )K * *' x 2 y

y 2 x

where x and y are the distances of the fibre from the neutral axis.

10.12

JOHNSON’S FORMULA FOR COLUMNS

Professor Johnson after a series of experiments and observations proposed the following two formulae for columns: 1. Straight line formula 2. Parabolic formula

10.12.1

Johnson’s Straight Line Formula for Columns

This formula was first proposed by Johnson which states   L  P = A T C  n     k  

Chapter 10:

where

Columns and Struts

305

P = Safe load on the column A = Area of the column cross-section sC = Allowable stress in the column material n = A constant, whose value depends upon the column material For mild streel n = 0.57 For structural steel n = 2 For cast iron n = 0.6

L = Slenderness ratio k Hence the stress at critical load for structural streel is

P L = 367.5  2   ; N/mm 2 A k

10.12.2

Johnson’s Parabolic Formula for Columns

Professor Johnson, after proposing the straight line formula, found that the results obtained by this formula are very approximate. He then proposed another formula, which states 2  L  P = A T C  r     k   

where P = Safe load on the column A = Area of the column cross-section sC = Allowable stress in the column material r = A constant, whose value depends upon the column material

L = Slenderness ratio with equivalent column length k

EXERCISES 10.1 Prove that the Euler’s formula is valid for long column. 10.2 Derive an expression for Euler’s crippling load if both the ends are fixed. 10.3 Find the Euler’s crippling load for a hollow cylindrical steel column of 38 mm external diameter and 35 mm thick. The length of column is 23 mm and hinged at its both ends. Take E = 200 GPa. [Ans. 17.25 kN] 10.4 Calculate the safe compressive load on a hollow cast iron column one end rigidly fixed and other pin joined 150 mm outer and 100 mm inner diameter, 10 m long. Use Euler’s formula with a factor of safely of 5 and take E = 95 GN/m2. [Ans. 75 kN] 10.5 Find the Euler’s crushing load for a hollow cast-iron column 15 cm external diameter and 2 cm thick, if it is 6 m long and hinged at both the ends. Take E = 80 GPa. Compare this load with the crushing load as given by the Rankine’s formula, using Rankine’s constant 1/1600. For what length of strut of this cross-section does the Euler’s formula ceases to apply. (Engineering Services) [Ans. 387.4 kN, 393.9 kN, 176.1 cm]

306

Strength of Materials

10.6 Determine the ratio of the strengths of a solid steel column to that of a hollow column of the same material and having the same cross-sectional area. The internal diameter of the hollow column is ½ of its external diameter. Both the columns are of the same length and are pinned at their both ends. [Ans. 5/3] 10.7 A cast iron hollow column, having 8 cm external diameter and 6 cm internal diameter, is used as a column of 2 m length. Using Rankine’s formula, determine the crippling load, when both the ends are fixed. Take sC = 600 MN/m2. [Ans. 660 kN] 10.8 Find the Euler’s crushing load for a hollow cylindrical cast iron column 15 cm external diameter and 2 cm thick, if it is 6 m long and hinged at both ends, E = 80 GN/m2. Compare this load with the crushing load given by Rankine’s formula using sC = 567 MN/m2 and a 1 for what length of column would these two formulae give the same crushing load. 1600 [Ans. 387.8 kN, 406.4 kN, 4.82 m] 10.9 A hollow cast iron column with fixed ends supports an axial load of 1000 kN. If the column is 5 m long and has an external diameter of 250 mm, find thickness of the metal required. Use

=

1 and assume a working stress of 80 MN/m2. 1600 [Ans. 29.4 mm] 10.10 Piston rod of steam engine 80 cm long is subjected to a maximum load of 60 kN. Determine the diameter of rod using Rankine’s formula with permissible compressive stress of 100 MPa.

Rankine’s formula taking a constant of

1 for hinged ends. The rod may be partially fixed 7500 with length coefficient of 0.6. [Ans. 33.23 mm] 10.11 Find the inside diameter of a cast iron column of 200 mm outside diameter which is 5 m long and has to support a safe axial load of 600 kN, one end being rigidly fixed. Use a factor of Take constant in Rankine’s formula as

1 and sC = 567 MN/m2. 1600 [Ans. 179 mm] 10.12 A steel stanchion is built of two rolled steel joists of I- section 30 cm ´ 15 cm ´ 1.25 cm united by plates 2 cm thick and 40 cm wide fastened to flanges. The edges of the plates are flush with the outside edges of the joints. Using Rankine’s formula for a strut, find the safe load for this

safety of 5 in conjunction with Rankine’s formula, a =

1 , FS = 4. [Ans. 1406 kN] 7500 10.13 A hollow cylindrical cast iron column is 4 m long, both ends being fixed. Design the column to carry an axial load of 250 kN. Use Rankine’s formula and adopt a factor of safety (FS) of 5. The internal diameter may be taken as 0.8 times the external diameter. Take sC = 350 MN/m2 stanchion if it is 8 m long, sC = 330 MN/m2 and a =

and a =

1 . 1600

[Ans. D = 136.8 mm, d = 109 mm]

Chapter 10:

Columns and Struts

307

10.14 A built-up beam is shown in Fig. 10.10. It is simply supported at ends. Find its length, given that when it is subjected to a load of 40 kN/m length, it deflects by 1 cm. Find out the safe load if this beam is used in a column with ends fixed. Use Euler’s formula. Take factor of safety 3 and E = 2 ´ 105 MPa. [Ans. 1397.8 cm, 3.037 MN] 30 cm 5cm

2 cm

100 cm X

X

5cm 30 cm

Fig. 10.10 I section.

11 11.1

Spring

INTRODUCTION

Springs are elastic members which distort under load and regain their original shape when load is removed. They are used in railway carriages, motor cars, scooters, motorcycles, rickshaws, governors, etc. According to their uses, the springs perform the following functions: 1. 2. 3. 4. 5.

To To To To To

absorb shock or impact loading as in carriage springs. store energy as in clock springs. apply forces to control motions as in brakes and clutches. measure forces as in spring balances. change the vibrations characteristic of a member as in flexible mounting of motors.

The springs are usually made of either high carbon steel (0.7 to 1.0%) or medium carbon alloy steels. Phosphor bronze, brass, 18/8 stainless steel and monel and other metal alloys are used for corrosion resistance springs.

11.2

DEFINITIONS

Helical springs:

A helical spring is a piece of wire coiled in the form of helix. If the slope of the helix of coil is so small then the bending effects can be neglected, and the spring is called closecoiled spring. In such a spring only torsional shear stress are introduced. On the other hand, if the slope of helix of coil is quite appreciable then both the bending as well as torsional shear stress are introduced in the spring, and a spring of this type is called an open-coiled spring. For all practical purposes, if helix angle is less than 15°, the spring is treated as close coiled helical spring, otherwise it is treated as open coiled helical spring.

Helix angle

It is the angle which the axis of spring wire makes with a horizontal line perpendicular to the axis of the spring. :

Spring index: It is the ratio of the mean coil diameter to the diameter of the spring wire. It is denoted by C. \

C = D/d 308

Chapter 11: Spring

309

Stiffness: It is defined as load per unit deflection, and is denoted by K. Torsional rigidity: Torsional rigidity or torsional stiffness is defined as torque per unit angular twist. 11.3

TYPES OF SPRINGS

Various types of springs are employed for different purposes; some of them are as follows: 1. Helical springs (i) Close-coiled helical springs (ii) Open-coiled helical springs (iii) Tension helical springs (iv) Compression helical springs 2. Leaf springs 3. Torsion springs 4. Circular springs 5. Belleville springs 6. Flat springs

11.4

HELICAL SPRING

11.4.1

Closely-coiled Helical Springs

Consider a closely-coiled helical spring subjected to an axial load neglecting curvature and direct shear effects as shown in Fig. 11.1(a). Let

d D n G W

= = = = =

Diameter of spring wire Mean diameter of the spring Number of turns on coils Modulus of rigidity for the spring material Axial load on spring

w T Fig. 11.1(a) Closely-coiled helical spring under axial load.

Fig. 11.1(b) Section of helical spring with axial force w, producing torque T.

310

Strength of Materials t = Shear stress induced in the wire due to twisting q = Angle of twist in the spring wire d = Deflection of the spring as a result of axial load

Torque on the spring acting about the axis of the spring [Fig. 11.1(b)] WD 2

T =

(11.1a)

We know that the twisting moment is:

Q

T =

16

t1 d 3

(11.1b)

Equating Eqs. (11.1a) and (11.2b), we get t1 =

8WD Qd3

(11.2)

The force W will also induce direct shear stress in the wire. The magnitude can be obtained assuming uniform distribution of stress over the cross section as follows. The direct shear stress is given as t2 =

4W

(11.3a)

Qd2

The distribution of the shear stresses over the cross section is shown in Fig. 11.1(c). Figure 11.1(c) shows that the maximum shear stress develops at the inner side of the coil and the magnitude is obtained as tmax = t1 + t2

=

=

8W

Qd

+

3

4W

Q d2

8WD  d  1+  3  2D  Qd 

(11.4)

Now if a shear factor ks is defined as ks = 1 +

d 2D

0.5   = 1 + C  

(11.5)

The maximum shear stress can be expressed as tmax =

8WD

Q d3

Ks

(11.6)

Now, Length of wire l = Length of one coil ´ No. of coils = 2p R ´ n

where R is the mean coil radius

Chapter 11: Spring

311

Fig. 11.1(c) Shear stress distribution.

We also know, T

=

R or where

GJ l

Tl GJ J = Polar moment of Inertia = Ixx + Iyy

q=

Qd4

or

J=

or

 WD  (2QRn) 2 q=

32

G

(For circular section)

( Q d 4) 32

(11.7)

312

Strength of Materials

Then

Deflection of the springÿ d = Rq

 32WDRn  = 32WDR n  Gd  Gd 2

=R = and

Stiffness of the spring K = =

Wahl’s correction factor:

4

4

8WD3n Gd 4

(11.8)

W E

Gd 4 8 D3n

(11.9)

A.M. Wahl has presented an approximate analysis based upon the theory of the bending of curved beams to determine the stresses in spring wire by taking into account the effect of curvature and direct shear. Equation (11.6) is modified to include these effects by introducing a factor called Wahl’s correction factor. fs = where and,

Wahl’s correction factor Kw = Spring index C =

8WD Kw Q d3

(11.10)

4C  1 0.615  4C  4 C D d

11.4.2 Open-coiled Helical Springs In an open-coiled helical spring, the spring wire is coiled in such a way, that there is a large gap between the two consecutive turns. As a result of this the spring can take compressive load also. An open-coiled helical spring, like a closed-coil helical spring, may be subjected to axial loading or axial twist. In this chapter, we shall discuss only opencoiled helical spring subjected to an axial load as shown in Fig. 11.2. Let

d R P n G W t sb d a

P

d a

= Diameter of the spring wire W = Mean radius of the spring coil D = Pitch of the spring coils Fig. 11.2 Open-coiled helical spring = No. of turns on coils under axial load. = Modulus of rigidity for the spring materials = Axial load on the spring = Maximum shear stress induced in the spring wire due to loading = Bending stress induced in the spring wire = Axial deflection of the spring as a result of axial load = Angle of helix

Chapter 11: Spring

313

A little consideration will show that the load W will cause a moment WR. This moment may be resolved into the following two components: T = WR cos a (It causes twisting of coils) M = WR sin a (It causes bending of coils) Let d = Angle of twist as a result of twisting moment f = Angle of bend as a result of bending moment Length of the spring wire is: l = 2p nR sec a

(11.11a)

and twisting moment is given by WR cos

sb

Now, Bending stress

a = Q ts d 3

(11.11b)

16

My = = I

Qd

' M = T   I y

d 2

WR sin B –

b

4

64 =

q=

and

32 WR sin B Qd3

(11.11c)

' T = GR   J l

Tl WR cos B – l = JG JG

Angle of bend due to bending moment is given as:

f=

Ml WR sin B – l = EI EI

For equilibrium of the spring, External work done = Strain energy stored 1 1 1 Wd = Tq + Mf 2 2 2

\

Wd = Tq + Mf

or

= WR cos a ´

d = WR2l

or

WR cos B – l + WR sin JG

 cos B  sin B "# ! JG EI $ 2

WR sin B – l EI

a´

2

(11.12a)

Now, substituting the values of l = 2p nR sec a, J = we get

d = WR2 ´ 2p nR sec a



cos 2 B

Q

! 32

d 4G

Q 32



(d)4 and I =

sin 2 B E–

Q

64

d4

"# ## $

Q 64

(d)4 in Eq. (11.12a),

314

Strength of Materials

= If we substitute

 !

64 WR3n sec B cos 2 B 2 sin 2 B  G E d4

(11.12b)

a = 0 in Eq. (11.12b), it gives the deflection of a closed-coiled spring, i.e., 64 WR3n Gd 4

d=

8WD3 n

=

11.5

"# $

Gd 4

STRAIN ENERGY IN THE SPRING

We know that the springs are used for storing energy which is equal to work done on it by some external load. Let W = Load applied on the spring d = Deflection in the spring due to load W Energy stored in spring is: U =

1 Wd 2

(11.13)

Maximum shear stress induced in the spring wire,

tmax or

= Kw ´

W =

8WD Q d3

Q d 3U max

(11.14)

8K w D

Deflection of the spring is given by

d=

8  Q d 3U max D3 n 8WD3n  = 8Kw D Gd 4 Gd 4

QU max D 2 n

=

(11.15)

K w dG

Substituting the values of W and d in Eq. (11.13), we have

1 Q d 3U max QU max D2 n  U =  2 8K w D K w dG =

=

U max

2

4 K w2 G U max

(pDn)

Q d  4  2

2

4 K w2 G

V

(11.16)

Chapter 11: Spring

315

where V = Volume of spring wire = Length of spring wire ´ cross-sectional area of spring wire = (pDn)

11.6

Q d  4  2

SPRINGS UNDER IMPACT LOAD

Let a weight W falls onto a spring from a height h measured from the uncompressed state of spring. Let W1 be the equivalent static load and d be the compression of the spring under load W1. Work done by falling weight = W (h + d ) Work done in the spring by W1 =

(11.17)

1 W1d 2

(11.18)

Equating Eqs. (11.17) and (11.18), we get 1 W1 d 2 8W1D3n d= Gd 4

W (h + d ) = where

11.7

(11.19)

SPRINGS IN SERIES

When two springs of different stiffness are joined end to end to carry a common load W, they are said to be connected in series. Figure 11.3 shows spring connected in series. Let k k1, k2 d 1, d 2 d Now,

= = = =

Equivalent or combined stiffness of system Stiffness of springs for 1 and 2, respectively Extension of springs for 1 and 2, respectively Extension of system

k1

Total deflection d = d1 + d 2

W

W W W  = k k1 k2

11.8

k =

k1 k2 k1  k2

2

k2

1 1 1 =  k k1 k2 or

1

W

(11.20)

Fig. 11.3 Springs connected in series.

SPRINGS IN PARALLEL

When two springs are joined in such a way that they have common deflection then they are said to be connected in parallel. Figure 11.4 shows spring connected in parallel. Total load W = W1 + W2

316

Strength of Materials

Deflection d =ÿd 1 = d 2 k d = k1d1 + k2d 2

\ \

k = k1 + k2

where

(11.21)

k1

1

k = Equivalent or combined stiffness of system k1, k2 = Stiffness of springs 1 and 2, respectively d = Deflection of the system

k2

2

W2

W1 W Fig. 11.4

EXAMPLE 11.1

Springs connected in parallel.

A closely coiled helical spring is to carry a load of 500 N. Its mean coil diameter is 10 times than that of the wire diameter. Calculate mean and coil diameter if the maximum shear stress in the material of the spring is 80 MPa. Solution: Given tmax = 80 MPa, W = 500 N, D = 10d

Then

Spring index C =

D d

= 10 Wahl’s stress factor is: Kw =

4C  1 0.615  = 1.14 4C  4 C

Maximum shear stress is: tmax = Kw

8WD Qd 3

80 = 1.14 or

 8 – 500 – 10d "# ! Qd $ 3

d = 13.46 mm Mean diameter is:

D = 10 ´ 13.46 = 134.6 mm

EXAMPLE 11.2 A closed-coiled helical spring of 10 cm mean diameter is made up of 1 cm diameter rod and has 20 turns. The spring carries an axial load of 200 N. Determine the shearing stress. Taking the value of modulus of rigidity = 8.4 ´ 104 N/mm2, determine the deflection when carrying this load. Also calculate the stiffness of the spring and the frequency of free vibration for a mass hanging from it. Solution:

Given D = 10 cm, d = 1 cm, n = 20, W = 200 N, G = 8.4 ´ 104 N/mm2

Chapter 11: Spring

317

D d 10 = = 10 1

Then Spring index C =

Wahl’s stress factor is: Kw =

4C  1 0.615  4C  4 C

=

4 – 10  1 0.615  4 – 10  4 10

=

39 0.615 = 1.14  36 10

Maximum shear stress is given as: tmax = Kw

8WD Qd 3

= 1.14 Deflection of spring is given by d =

=

 8 – 200 – 100 "# = 58.06 MPa ! Q (10) $ 3

8WD3n Gd 4 8 – 200 – (100)3 – 20 = 38.09 mm 8.4 – 10 4 – (10)4

Stiffness of spring is obtained as: W E 200 = = 5.25 N/mm 38.09 Frequency of free vibration is expressed as:

K=

f=

1 2Q

E

g

=

1 2Q

9.81 = 2.55 cycles/s 38.09 – 10  3

EXAMPLE 11.3 Find the maximum shear stress and deflection induced in the helical spring for the following specifications, if it has to absorb 1000 N-m of energy.

318

Strength of Materials

Mean diameter of spring = 100 mm Diameter of steel wire = 20 mm No. of coils = 20 Modulus of rigidity = 80 GPa Solution:

Given U = 1000 N-m, D = 100 mm, d = 20 mm, n = 20, G = 80 GPa D d 100 = =5 20

Then

Spring index C =

Wahl’s stress factor

Kw = =

Volume of spring wire is:

4C  1 0.615  4C  4 C

4 – 5  1 0.615  = 1.31 4–54 5

%&Q (20) () '4 * %Q ( = (p ´ 100 ´ 20) & (20) ) '4 *

V = (p Dn)

2

2

= 1973920.88 mm3 Energy absorbed in the spring is given by 2

U =

U max

4 K w2 G

V

2

1000 ´ 103 =

U max (1973920.88)

4  (1.31)2  80  103

tmax = 460.84 MPa

Deflection in the spring is given as: d =

=

QU max D2 n K w dG

Q (460.84)(100)2 – 20 (1.31)(20)(80 – 10 3)

= 138.15 mm

EXAMPLE 11.4 A closed-coiled helical spring has mean coil diameter 60 mm. It is made up of 6 mm diameter wire and has 16 turns. If its maximum shear stress is not to exceed 110 MPa, determine (i) Maximum value of axial load (ii) Deflection

Chapter 11: Spring

319

(iii) Spring constant (iv) Energy stored under maximum load Take G = 82 GPa. Solution:

Given D = 60 mm, d = 6 mm, n = 16, tmax = 110 MPa

(i)

WR = W ´ 30 =

Q 16

tmaxd3

Q

´ 110 ´ 63 16 W = 155.5 N

or or

d=

(ii)

= = (iii)

K = =

(iv)

U = = =

8WD3n Gd 4 8 – 155.5 – 603 – 16 82 – 103 – 6 4 40.46 mm W 155.5 = E 40.46 3.84 N/mm 1 1 Wd = ´ 155.5 ´ 40.46 2 2 3146 N-mm or 3.146 N-m 3.146 J

EXAMPLE 11.5 A railway wagon weighing 60 kN and moving with a speed of 30 km/hr is to be stopped by 4 buffer springs in which the maximum compression allowed is 30 cm. Calculate the number of turns in each spring in which diameter of wire is 5 cm and mean diameter 20 cm, G = 80 GPa. Solution:

Kinetic Energy of the wagon is given by 1 = mv2 2

 

1 60 – 10 3 30 – 1000 = – – 2 9.81 60 – 60

 

= 212.368 kN-m 212.368 = 53.092 kN-m 4 Energy absorbed = Work done on the spring

Energy to be absorbed by each spring =

53.092 ´ 103 =

1 Wd 2

2

320

Strength of Materials

or

W = Deflection d =

or

0.3 =

Solving we get

2 – 53.092 – 10 3 = 353.95 kN 0.3 8WD3n Gd 4

8  353.95  103  (20  10  2 )3  n 80  109  (5  10  2 )4

n = 6.62  7 turns

EXAMPLE 11.6 A stiff bar of negligible weight transmits a load P to a combination of 3 springs as shown in Fig. 11.5. The three springs are made of the same material with equal rod diameters. They are of the same length before loading. The number of coils in the three springs are 10,12 and 15, respectively, while the mean radii of the coils are in the proportion 1:1.2:1.4, respectively. Find the distance x such that the stiff bar remains horizontal after applying the load. Solution: If the stiff bar is to remain horizontal then the compression of the three springs will be equal. \ Also

d1 = d 2 = d 3 = d G1 = G2 = G3 = G d1 = d2 = d3 = d n1 = 10, n2 = 12, n3 = 15 D1 : D2 : D3 = 1 : 1.2 : 1.4

Let P1, P2 and P3 be the loads on the springs. Then

d1 = d2 =

\

8 P1 D13 n1 80 P1 D13 = Gd 4 Gd 4 8P2 D23 n2 Gd 4

=

96 P2 D23 Gd 4

d3 =

8P3 D33 n3 120 P3 D33 = Gd 4 Gd 4

P1 =

Gd 4E C = 80 D13 10 D13

P2 =

Gd 4E C 3 = 96 D2 12 D23

P x

L

L Fig. 11.5

Chapter 11: Spring

where

P3 =

Gd 4E C 3 = 120 D3 15 D33

C =

Gd 4E 8

321

Taking moments about P, we get P1 ´ x = P2 (L – x) + P3 (2L – x) xC ( L  x ) C ( 2 L  x )C =  10 D13 12 D23 15 D33

or As

D2 D = 1.2, 3 = 1.4 D1 D1

\

x Lx 2L  x = 3  10 12 (1.2) 15 (1.4)3 x 10 20.736 ´ 41.16 ´ x 853.49x (853.49 + 411.6 + 207.36)x 1472.45x x

or or or or or or

= = = = = =

Lx 2L  x  20.736 41160 . 10 ´ 41.16(L – x) + 10 ´ 20.736(2L – x) 411.6(L – x) + 207.36 (2L – x) (411.6 + 414.72)L 826.32L 0.561L

EXAMPLE 11.7 Two close-coiled helical springs wound from the same wire, but with different core radii having equal number of coils, are compressed between rigid plates at their ends. Calculate the maximum shear stress induced in each spring if the wire diameter is 10 mm and the load applied between the rigid plates is 500 N. The core radii of the springs are 100 mm and 75 mm, respectively. Solution: Given n1 = n2, d = 10 mm, W = 500 N R1 = 100 mm R2 = 75 mm Let

W1 = Load shared by outer spring W2 = Load shared by inner spring

Then

d1 = =

Similarly

d2 = =

64W1 R13 n1 Gd 4 64W1 (100)3 – n1 6400 W1 n1 = G G(10)4

(i)

64W2 R23 n2 Gd 4 64 – W2 ( 75)3 – n2 2700 W2 n2 = 4 G (10) G

(ii)

322

Strength of Materials

Since the springs are held between two rigid plates, deflections in both the springs must be equal. Equating Eqs. (i) and (ii) gives 27W2 64 W1 + W2 = 500

W1 =

Also By Eqs. (iii) and (iv),

(iii) (iv)

W2 = 351.6 N W1 = 148.4 N From relation of torque for outer spring, W1 R1 = 148.4 ´ 100 =

Q U1d 3 16

Q

or

´ t1 ´ (10)3 16 t1 = 75.6 N/mm2

Similarly,

t2 =

or

351.6 – 75 – 16 Q (10)3 = 134.3 MPa

EXAMPLE 11.8 A helical spring B is placed inside the coils of a second helical spring A having the same number of coils and free axial length and of same material. The two springs are compressed by an axial load of 200 N which is shared between them. The mean coil diameters of A and B are 80 mm and 60 mm and the wire diameters are 12 mm and 6 mm, respectively. Calculate the load taken and maximum stress in each spring. Solution:

Springs are connected in parallel, PA + PB = P dA = d B

where

Then

or

or

PA PB dA dB

= = = =

(i)

Load shared by spring A Load shared by spring B Deflection of spring A Deflection of spring B

8 PA DA3 nA 8 PB DB3 nB = GA d A4 GB d B4

' G ! n  D   d  =  D  d 

PA DA3 PB DB3 = d A4 d B4

PA PB

= GB A = nB

A

3

4

B

A

A

B

"# $

Chapter 11: Spring

 60   12   80   6  3

=

323

4

= 0.423 ´ 16 = 6.77 PA = 6.77PB

or

Substitute this value of PA in Eq. (i), we get 6.77 PB + PB = 200 PB = 25.74 N PA = 174.26 N

or and

(t)A =

Now,

= (t)B =

and

=

8 PA DA Q d A3

8  (174.26) (80)

Q (12)3

= 20.54 MPa

8 PB DB Q d B3 8 – (25.74 )(60 ) = 18.21 MPa Q (6 )3

EXAMPLE 11.9 A composite spring has two close-coiled springs connected in series, one spring has 10 coils of a mean diameter of 20 mm and wire diameter 3 mm. Find the wire diameter of other spring, if it has 15 coils of mean diameter 35 mm. The stiffness of the composite spring is 2 kN/m. Take G = 80 GPa. Solution:

or

Springs are connected in series, 1 1 1 =  k k1 k2 k1 = =

and

k2 = =

Now, or

Gd14 W = E1 8 D13 n1 80 – 103 – (3) 4 = 10.13 N/mm 8 – (20)3 – 10 Gd 24 8 D23 n2

80 – 10 3 – d24 = 0.0155 d24 N/mm 3 8 – (35) – 15

1 1 1  = k k1 k2 1000 1 1  3 = 10.13 0.0155 d 24 2 – 10

324

Strength of Materials

64.52 d 24 d24 = 160.78 Wire diameter d = 3.56 mm

or

0.5 – 0.0987 =

or Thus,

EXAMPLE 11.10 An open-coiled helical spring consists of 10 coils, each of mean diameter 5 cm, the wire forming the coils being 6 mm diameter, and making a constant angle of 30o with planes perpendicular to the axis of the spring. What load will cause the spring to elongate 1.25 cm and what will be the bending and shearing stresses due to this load? Calculate the value of axial twist which would cause a bending stress of 50 MPa in the coils, E = 210 GPa and G = 84 GPa. Solution: Given n = 10, R = 2.5 cm, d = 0.6 cm, Under axial load,

d=

or

12.5 =

a = 30°, d = 1.25 cm

 !

64 WR 3n sec B cos 2 B 2 sin 2 B  G E d4

"# $

64  W  (2.5)3  10  sec 30  cos2 30 2  sin 2 30     3 (6) 4 210  103   84  10

Solving we get

or Now,

W= = W= Torque T = = Shear stress t = =

124.05 N 0.10076 ´ 10–3 W 124.05 N WR cos a 124.05 ´ 2.5 ´ 10–2 ´ 0.866 = 2.6857 N-m 16 T Qd 3

16 – 2.6857 = 63.325 MPa Q (0.6)3 – 10  6

Bending moment M = WR sin a = 124.05 ´ 2.5 ´ 10–2 ´ 0.5 = 1.5506 N-m Bending stress

sb = =

32 M Q d3 32 – 1.5506 = 73.122 MPa Q (0.6)3 – 10  6

Let Mo is axial torque. Bending moment M = Mo cos a Bending stress sb =

32 Mo – cos B Qd 3

Chapter 11: Spring

50 ´ 106 =

or

Mo =

325

32 Mo – 0.866 Q (0.6)3 – 10  6 50 – Q (0.6)3 = 1.224 N-m 32 – 0.866

EXAMPLE 11.11 In an open-coiled helical spring having 10 coils, the stresses due to bending and twisting are 98 MPa and 105 MPa, respectively, when the spring is axially loaded. Assuming the mean diameter of the coils to be 8 times the diameter of wire, find the maximum permissible load and the diameter of wire for a maximum extension of 2 cm, E = 210 GPa and G = 82 GPa. Solution: Given n = 10, Under axial load,

s b = 98 MPa, t = 105 MPa, D = 8d, R = 4d, d = 2 cm d=

 !

64 WR 3n sec B cos 2 B 2 sin 2 B  G E d4

T = WR cos a

Also

t=

Then

16 T 16 WR cos B = 3 Qd 3 Qd

or

105 ´ 106 =

or

105 ´ 106 =

16 W – 4 d – cos B Qd 3

64 W cos B Qd 2 M = WR sin a

and

sb =

Now,

(i)

32 – M 32 WR sin B = 3 Qd Qd 3

or

98 ´ 106 =

or

98 ´ 106 = Dividing Eq. (ii) by Eq. (i), we get

32 – W – 4 d – sin B Qd 3

128W sin B Qd 2

(ii)

98 = 2 tan a 105 98 tan a = = 0.466 210

or

a=

or Substituting the value of

25.017o

a in Eq. (i), we get

105 ´ 106 = or

"# $

64W cos 25.017

Q d2

=

W = 5.6878 ´ 106 ´ d 2

64 – 0.9062 W Qd 2

326

Strength of Materials

Extension is given by

or or or \

11.9

 !

0.02 =

64 – 5.6878 – 10 6 – d 2 ( 4 d ) 3 10 ( 0.9062 ) 2 2 ( 0.4229) 2 –  82 – 10 9 210 – 10 9 d 4 – 0.9062

0.02 0.02 d W

25.708d [0.100146 + 0.017033] 3.0124 d 6.64 ´ 10–3 m = 6.64 mm 5.6878 (6.64)2 = 250.8 N

= = = =

"# $

LEAF SPRINGS OR CARRIAGE SPRINGS

The carriage springs are widely used in railway wagon coaches and road vehicles these days. These are used to absorb shocks which give an unpleasant feeling to the passengers. The energy absorbed by a laminated spring, during a shock, is released immediately without doing any useful work. A laminated spring, in its simplest form, consists of a number of parallel strips of a metal having different lengths but same width and placed one over the other in laminations as shown in Fig. 11.6. All the plates are initially bent to the same radius and are free to slide one over the other. When the spring is loaded to the designed load, all the plates become flat and the central deflection disappears. The purpose of this type of arrangement of plates is to make the spring of uniform strength throughout. This is achieved by tapering the ends of the laminations. The semi-elliptical type spring rests in the axis of the vehicle and its top plate is pinned at the ends to the chassis of the vehicle. W/2

W/2 l

d

W

Fig. 11.6

Leaf spring.

Laminated springs are of two types: (i) Semi-elliptical (i.e., simply supported at its ends subjected to central load) (ii) Quarter-elliptical (i.e., cantilever types)

11.9.1

Semi-elliptical Spring

Consider a carriage spring pinned at its both ends, and carrying an upward load at its centre as shown in Fig. 11.6. Let l = Span of the spring t = Thickness of plates

Chapter 11: Spring

b n W M d R

= = = = = =

327

Width of the plates Number of plates Load acting on the spring Maximum bending moment developed in the plates Original deflection of the top spring Radius of curvature of the spring

A little consideration will show that the load will be acting on the spring on the lowermost plate, and it will be shared equally on the two ends of the top plate as shown in Fig. 11.6. We know that the bending moment at the centre of the span due to this load, M = And moment resisted by one plate, =

Wl 4

(11.22)

' M = T   I y

Tb I ¹

b

y

bt 3 2 12 = T b ¹ bt = t 6 2 Total moment resisted by n plates,

' I = bt and y t  2  12

Tb –

3

nT b bt 2 (11.23) 6 Since the maximum bending moment due to load is equal to total resisting moment, therefore, equating Eqs. (11.22) and (11.23), M =

Wl nT b bt 2 = 4 6 3W l or sb = 2 nbt 2 From the geometry of the spring figure, we know that the central deflection, l2 8R We also know that in the case of a bending beam,

d=

Tb y or

R=

Ey

Tb

=

E R

=

Et 2T b

(11.24)

(11.25)

' y = t   2

Substituting this value of R in Eq. (11.25),

d=

T l2 l2 = b Et 4 Et 8– 2T b

(11.26)

328

Strength of Materials

Now, substituting the value of

s b from Eq. (11.24) in Eq. (11.26), d=

3W l l2 3Wl 3 – = 2 nbt 2 4 Et 8 Enbt 3

(11.27)

EXAMPLE 11.12 A leaf spring has 12 plates each 50 mm wide and 5 mm thick, the longest plate being 600 mm long. The greatest bending stress in not to exceed 180 N/mm2 and the central deflection is 15 mm. Estimate the magnitude of the greatest central load that can be applied to the spring E = 206 kN/mm2. Solution: (i)

d=

3Wl 3 8E Enbt 3 , W = 8 Enbt 3 3l 3

8 – 15 – 206 – 10 3 – 12 – 50 – 53 3 – (600)3 = 2860 N

or

W=

(ii)

2T nbt 2 s b = 3 Wl2 , W = b 3l

2 nbt

2 – 180 – 12 – 50 – 52 = 3000 N 3 – 600 Allowable load = 2860 N

W=

\

EXAMPLE 11.13 A laminated steel spring, simply supported at the ends and centrally loaded with a span of 75 cm is required to carry a proof load as 7.5 kN, and the central deflection is not to exceed 50 mm. The bending stress must not be greater than 400 N/mm2. Plates are available in multiple of 1 mm thickness and in multiples of 3 mm of width. Determine suitable values for thickness, width and number of plates and the radius to which the plates should be formed. Assume the width to be twelve times the thickness. Take E = 2 ´ 105 N/mm2.

d=

Solution:

nbt 3 = Also

sb =

3 Wl 2 nbt 2

3Wl 3 8 Enbt 3 3Wl 3 8 EE or

(i) nbt 2 =

Dividing Eq. (i) by Eq. (ii), we have t=

l 2T b 3Wl 3 2T b = 8 EE 3Wl 4 EE

3 Wl 2 Tb

(ii)

Chapter 11: Spring

t=

329

750 – 750 – 400 = 5.63 mm 4 – 2 – 10 5 – 50

Let it be 6 mm. \

b = 12 ´ t = 12 ´ 6 = 72 mm Hence from Eq. (ii), n = =

3Wl 2T b bt 2 3 – 7500 – 750 = 8.14 2 – 400 – 72 – 6 2

Number of plates = 9 (say) The modified s b will be

sb = =

\ or or

3 Wl 2 nbt 2 3 7500 – 750 – 2 9 – 72 – 6 2

= 361.7 N/mm2 Tb E = y R E R= ´y

Tb

R= =

Et 2T b

2 – 105 – 3 = 1660 mm 361.7

EXAMPLE 11.14 A leaf spring is to be made of seven steel plates 65 mm wide and 6.5 mm thick. Calculate the length of the spring so that it may carry a central load of 2.75 kN, the stress being limited to 160 N/mm2. Also calculate the deflection at the centre of the spring. Take E = 210 kN/mm2. Solution: Given n = 7, b = 65 mm, t = 6.5 mm, W = 2.75 kN = 2.75 ´ 103 N, E = 210 kN/mm2 = 210 ´ 103 N/mm2 Let l is the length of the spring. Using the relation:

sb = or

160 =

\

l=

3Wl 2 nbt 2

3 – 2.75 – 10 3 – l = 0.215l 2 – 7 – 65 – 6.52 160 = 744.2 mm 0.215

s b = 160 N/mm2,

330

Strength of Materials

Letÿ d is the deflection at the centre of the spring. Now, using the relation: d

= =

T b l2 4 Et

160 – (744.2) 2 = 16.23 mm 4 – 210 – 10 3 – 6.5

EXAMPLE 11.15 A leaf spring 75 cm long is required to carry a central point load of 800 kg. If the central deflection is not to exceed 20 mm, and the bending stress is not greater than 2000 kg/cm2, determine the thickness, width and number of plates. Also compute the radius, to which the plates should be curved. Assume width of the plate equal to 12 times its thickness and E equal to 2.0 ´ 106 kg/cm2. Solution: Given l = 75 cm, W = 800 kg, d = 20 mm = 2 cm, t is the thickness), E = 2.0 ´ 106 kg/cm2. Using the relation:

d=

s b = 2000 kg/cm2, b = 12t (where

T b l2 4 Et

2000 – 752 1.4 = t 4 – 2.0 – 10 6 – t t = 0.7 cm = 7 mm

2 = \

We know that the width of the plates, b = 12 ´ 0.7 = 8.4 cm = 84 mm Let n is the number of plates. Now, using the relation:

s b = 3W l2 2 nbt

2000 = or

11.9.2

n =

3 – 800 – 75 21866 = 2 n 2 – n – 8.4 – 0.7

21866 = 10.9, say, 11 2000

Quarter-elliptical Leaf Spring

The quarter-elliptical type leaf springs are rarely used except as certain parts in some machines. Like a carriage spring quarter, elliptical type leaf spring consists of a number of parallel strips of a metal having different lengths but same width and placed one over the other in laminations as shown in Fig. 11.7. All the plates are initially bent to the same radius and are free to slide one over the other. Now, consider a quarter elliptical type leaf spring subjected to a load at its free end as shown in Fig. 11.7.

Chapter 11: Spring

331 W

Let l t b n W d

= = = = = =

l

Length of the spring Thickness of the plates Width of the plates Number of plates Load acting at the free end of the spring Original deflection of the spring

d

We know that the bending moment at the fixed end of the leaf,

Fig. 11.7

Quarter-elliptical leaf spring.

M = Wl

(11.28)

And moment resisted by one plate is: =

=

T b – bt 3 12

 t  2

TbI y

' I = bt and y = t   12 2 

T b bt 2

=

3

6

Total moment resisted by n plates,

nT b bt 2 (11.29) 6 Since the maximum bending moment due to load is equal to the total resisting moment, therefore, equating Eqs. (11.28) and (11.29), M=

or

Wl =

nT b bt 2 6

sb =

6 Wl nbt 2

(11.30)

From the geometry of the spring, we know that

d (2R – d ) = l . l = l2 d=

\

l2 2R

(Neglecting d 2)

We know that the case of a bending cantilever,

Tb y or

R=

E¹y

Tb

=

E R

=

Et 2T b

' y = t   2

(11.31)

332

Strength of Materials

Substituting this value of R in Eq. (11.31),

d=

l2 2–

Et 2T b

=

T b l2 Et

(11.32)

Now, substituting the value of s b from Eq. (11.30) in Eq. (11.32),

d=

6Wl l 2 6Wl 3 – = nbt 2 Et Enbt 3

(11.33)

EXAMPLE 11.16 A quarter-elliptic leaf spring 80 cm long is subjected to a point load of 1 tonne. If the bending stress and deflection are not to exceed 3.2 t/cm2 and 8 cm, respectively, find the suitable size and number of plates required taking the width as 8 times the thickness. Take E as 2000 t/cm2. Solution:

Given l = 80 cm, W = 1 tonne, s b = 3.2 t/cm2, d = 8 cm, E = 2000 t/cm2, b = 8t (where t is the thickness.)

Let n is number of plates in the spring. Using the relation:

s b = 6Wl2

(with usual notations) nbt 6 – 1 – 80 480 3.2 = = nbt 2 nbt 2

or

(i)

Now, using the relation:

or

d=

6 Wl 3 Enbt 3

8 =

6 – 1 – 803 1536 3 = 2000 nbt nbt 3

(with usual notations)

Dividing Eq. (ii) by Eq. (i),

1536 3 8 3.2 = nbt = 480 3.2 t nbt 2 \

t=

3.2 – 3.2 = 1.28, say, 1.3 cm 8

We know that the width of the plates, b = 8 ´ 1.3 = 10.4 cm Substituting the values of t and b in Eq. (i), we get the required number of plates as: 3.2 = \

n=

480 27.31 = n n – 10.4 – 1.32 27.31 = 8.5, say, 9 3.2

(ii)

Chapter 11: Spring

333

EXERCISES 11.1 A carriage spring 80 cm long is made of 12 plates of 4 cm width. Determine the thickness of the plates, if the bending stress is not to exceed 2000 kg/cm2, and the spring is to carry a load of 600 kg at its centre. Also determine the central deflection. Take E = 2.0 ´ 106 kg/cm2. [Ans. 9 mm, 1.65 cm] 11.2 A carriage spring is built up of 9 plates each 75 mm wide and 6.5 mm thick. Find the length of the spring, so that it may carry a central load of 400 kg, the stress being limited to 1.6 t/cm2. Also find the deflection at the centre of the spring. [Ans. 76.05 cm, 1.78 cm] 11.3 Derive from first principles, making usual assumptions, the formula for the maximum bending stress and for the central deflection of a leaf spring consisting of n leaves and subjected to a central load. 11.4 Derive an equation for the deflection of an open-coiled helical spring. 11.5 A closely-coiled helical spring is to carry a load of 1 kN. Its mean coiled diameter is to be 10 times that of wire diameter. Calculate these diameters if the maximum shear stress in the material of the spring are to be 90 N/mm2. [Ans. 16.82 cm and 1.68 cm] 11.6 A closely-coiled helical spring of round steel wire 8 mm in diameter having 10 complete turns with a mean diameter of 10 cm is subjected to an axial load of 250 N. Determine: (i) the deflection of the spring, (ii) maximum shear stress in the wire and (iii) stiffness of the spring. Take G = 8 ´ 104 N/mm2. [Ans. (i) 6.1 cm, (ii) 124.34 N/mm2, (iii) 4.1 N/mm2] 11.7 A closed-coiled helical spring is to have a stiffness of 70 kN/m and to exert a force of 2.25 kN. If the mean diameter of coils is to be 90 mm, and the shear stress is 230 MPa. Find the required number of coils and diameter of the steel rod from which the spring should be made. Take E = 80 GPa. [Ans. n = 6, d = 13.08 mm] 11.8 A closed-coiled helical spring has to absorb 50 N-m of energy when compressed 50 mm. The coil diameter is 8 times the wire diameter. If there are 10 coils, estimate the diameter of the coil and the wire, and the maximum shear stress. Take G = 85 GPa. [Ans. d = 19.3 mm, D = 154.4 mm, tmax = 109 MPa] 11.9 A leaf spring has 12 plates each 50 mm wide and 5 mm thick, the longest plate being 600 mm long. The greatest bending stress is not to exceed 180 N/mm2 and the central deflection is 15 mm. Estimate the magnitude of greatest central load that can be applied to the spring. Take E = 200 GPa. 11.10 A close-coiled helical spring made of 10 mm diameter steel bar has 8 coils of 150 mm mean diameter. Calculate the elongation, torsional stress and strain energy/unit volume when the spring is subjected to an axial load of 130 N. Take G = 80 GPa. If instead of axial load, an axial torque of 9 Nm is applied, find the axial twist, bending stress and strain energy per unit volume. Take E = 200 GPa.

12 12.1

Rotating Discs and Cylinders

INTRODUCTION

Stresses are set up in circular discs or cylinders on account of rotation about their axis of symmetry. The analysis of the stresses set up in a rotating cylinder or circular disc can be made on the basis of certain simplified assumptions, e.g. density of material is uniform throughout.

12.2

ROTATING DISC

Let a thin circular disc be rotating at an angular speed w rad/sec about its central axis, normal to the plane of rotation. It is assumed that the thickness t of the disc is uniform and very small as compared to the radius. Consider an element ABCD of the disc as shown in Fig. 12.1. Let

sr = Radial stress on face CD at a radius r sr + dsr = Radial stress on face AB at a radius (r + dr) sq = Circumferential stress on face AD

sq = Circumferential stress on face BC Dq = Angle subtended by the element at the center. sr + dsr

t dr A

w

D dq O

q

A C

B

sq tdrdq/2

r

dq/2 sq

B D sr dq

C r

Fig. 12.1

q

(b) Element of the disc

Rotating disc. 334

/2

sq r + dr

(a) Thin circular rotating disc of uniform thickness

dq

sq tdrdq/2

335

Chapter 12: Rotating Discs and Cylinders

Considering the equilibrium of the element and resolving the forces along the radial line, we have rtdqsr + 2sq tdr sin

dR = (sr + dsr) (r + dr)tdq + rtdqdr w 2r2 2

where the last term rw 2r(rdq.dr.t) is centrifugal force arises due to rotation of the disc. Neglecting the product of the two small quantities, and making sin

dR dR » , 2 2 dT r dr

(12.1a)

T R + SX 2 r 2 = 0

(12.1b)

sq = rw 2r 2 + s r + r

We get

d (r T r )  dr

or,

12.2.1 Strain Considerations When the disc starts rotating, due to centrifugal force on each and every element, it will tend to expand. On account of rotation let r becomes r + u and r + dr becomes r + dr + du then radial strain is:

er =

( dr  du)  dr du = dr dr

(12.2)

Hoop strain at radius r, 2Q (r  u)  2Qr u = 2Q r r du 1 er = = (s r – nsq) dr E u 1 eq = = (sq – nsr) r E

eq = Now, and Solving Eqs. (12.4) and (12.5),





sr =

E Ou Odu  dr 1 O2 r

sq =

 u O du     dr  1O r E

2

(12.3) (12.4) (12.5)

(12.6)

(12.7)

Putting Eqs. (12.6) and (12.7) in Eq. (12.1), r

or

d 2u du u X 2 (1 – n2) r2  = –r 2  E dr r dr

d 2 u 1 du u X 2 (1 – n2) r   = – r E dr 2 r dr r 2 Complementary function of differential Eq. (12.8) is: d 2 u 1 du u   =0 dr 2 r dr r 2

(12.8)

336

Strength of Materials

 

d 2u d u  dr 2 dr r

or

=0

This on integration gives du u  = 2A dr r

(12.9)

where 2A is a constant of integration. Equation (12.9) can be rewritten as: r or

du + u = 2Ar dr d (ur) = 2Ar dr

On integration, we get ur = Ar 2 + B u B =A+ 2 r r

or

(12.10)

where B is another constant. From Eqs. (12.9) and (12.10), we get du B =A– 2 dr r For finding the particular integral of Eq. (12.8), we assume

u = cr 3

(12.11)

(12.12)

where c is a constant. u = cr2 r

or By differentiating Eq. (12.12),

du = 3cr2 dr

d 2u = 6cr dr 2

or

(12.13)

Substituting Eq. (12.12) through Eq. (12.13) in Eq. (12.8), we have 6cr + or

1 X 2 (1 – n2) r 3cr2 – cr = – r r E

c =–

SX 2 8E

(1 – n2)

Thus, the complete solution is: u B SX 2 =A+ 2  (1 – n2) r2 r 8E r

(12.14)

Chapter 12: Rotating Discs and Cylinders

or

337

du B 3SX 2 =A– 2  (1 – n 2) r2 (12.15) 8E dr r Putting Eqs. (12.14) and (12.15) in Eqs. (12.6) and (12.7), we get the radial and hoop stresses as:

 ! (1  O ) A  (1  O ) B  (1  3O ) SX 8E )! r

"# $ " (1  O ) r # $

sr =

SX E B (1  O ) A  (1  O ) 2  (3  O ) (1  O 2 ) r 2 2 8E (1  O ) r

sq =

E (1  O 2

2

2

2

2

2

(12.16)

(12.17)

The values of the constants A and B will depend upon the end conditions.

12.3

HOLLOW DISC (DISC WITH A CENTRAL HOLE)

A hollow disc with inner radius r1 and outer radius r2, we have at r = r1, sr = 0 and at r = r2, s r = 0. From Eq. (12.16), (1 + n) A – (1 – n) and

B SX 2 = (3 + n ) (1 – n2) r 22 8E r22

B SX 2 = (3 + n ) (1 – n2) r12 8E r12 Solving Eqs. (12.18) and (12.19), we get (1 + n) A – (1 – n)

A = (3 + n) (1 – n) B = (3 + n) (1 + n)

SX 2 8E

(r12 + r 22)

SX 2

(r 12 r22) 8E Putting these values in Eqs. (12.16) and (12.17), we get

sr =

SX 2 8

 !

(3 + n) r12  r22 

r12 r22  r2 r2

"# $

 3+ O r 2 r 2 1  3O 2  r  SX 2  r22  r12 + 1 2 2    8 3+O r   (i) For maximum value of s r , dT r =0 dr

sq =

or

SX 2 8

(3 + n)

or

 2r r ! r

r= (s r)max =

SX 2 8

2 2 1 2 3

(12.18) (12.19)

(12.20) (12.21)

(12.22)

(12.23)

"# $

 2r = 0

r1r2

(3 + n) (r2 – r1)2

(12.24) (12.25)

338

Strength of Materials

(ii) Maximum value of

sq will occur at r = r1, (sq)max =

 3  O  rw r   1  O  r "#  4  ! 3 O $ 2

2 2

2 1

(12.26)

It can be seen from Eq. (12.25) and (12.26) that (sq)max is greater than (sr)max.

12.4

SOLID DISC

In this case, at r = r2 ,

s r = 0 and at r = 0, u = 0.

By Eq. (12.14), we have u B SX 2 =A+ 2  (1 – n2)r2 8E r r

Equation (12.14) with condition at r = 0, u = 0, B must vanish. Hence B = 0. With the help of first condition (at r = r2, sr = 0), we have from Eq. (12.16) for a solid disc,

A=

SX 2

8E B=0

(3  O )(1  O ) r 22

From Eqs. (12.27), (12.16) and (12.17), we have

sr = sq = Now, s r will be maximum at centre, \

(s r)max =

sq will be maximum at r = 0, \

12.5

(s q)max =

SX 2 8

SX 2 8

SX 2 8

SX 2 8

(K )K *

(12.27)

(3 + n)(r22 – r2)

(12.28)

[(3 + n) r22 – (1 + 3n)r2]

(12.29)

(3 + n) r22

(12.30)

(3 + n) r22

(12.31)

DISC OF UNIFORM STRENGTH

A disc of uniform strength is the one in which the values of radial and circumferential (hoop) stresses are equal in magnitude at all points in the disc, hence

sq = sr = s = constant Consider the equilibrium of the element ABCD of the disc shown in Fig. 12.2. Let t be the thickness of the disc at radius r and t + Dt at radius r + Dr. The outward radial force on face AB. = =

s (t + Dt) (r + Dr)Dq s (tr + rDt + tDr)Dq

Chapter 12: Rotating Discs and Cylinders

Fig. 12.2

Disc of uniform strength.

Centrifugal force acting on the element ABCD is: = r (rDq Drt) w2r Inward radial force acting on face CD =

s trDq

Inward radial force due to component of forces, =

s t DrDq

For equilibrium of the element, Total inward radial force = Total outward radial force or

str Dq +

\

s t Dr Dq = s (tr + rDt + tDr) Dq + r (rDq Drt) w 2r s r Dt Dq + r Dq Drtw2r 2 = 0

't

or

t

X 2 rDr

= –r

T

X 2 rdr dt = –r T t

or Integrating, we get

ln t = – r where A is a constant of integration. or

\ Let

ln

t = –r A

+ ln A

X r   2T 

t = Ae t = t0 at

X 2 r2 2T 2 2

 SX 2 r 2 2T

r = r1 then

339

340

Strength of Materials  X 2 r12    2T 

S

t0 = Ae

 X 2 r2 

S

\

A = t0 e t = t0 e

or



S

2T

1

 

X2 2 2 ( r  r1 ) 2T

(12.32)

This gives the thickness of disc at any radius.

12.6

ROTATING CYLINDER

Stresses are set up in a circular cylinder on account of rotation about its axis of symmetry. Consider a circular cylinder of inner radius r1 and outer radius r2 rotating at speed w (Fig. 12.3). Assume that plane sections of the cylinder remain plane during rotation. r w

r2

r

r2

r1

Z

O

r1

Fig. 12.3

Rotating cylinder.

The axial strain along the Z-axis will be independent of the radius r of the cylinder and will be constant. 1 du Radial strain er = [sr – n(sq + sZ)] = (12.33) E dr 1 u Hoop strain eq = [sq – n(sr + sZ)] = (12.34) E r 1 Axial strain eZ = [sZ – n(sr + sq)] (12.35) E From Eq. (12.34), we have Eu = r[sq – n(sr + sZ)] Differentiating with respect to r, E

du  dT dT Z    dT = sq – n(sr + sZ) + r  R  O  r +  dr dr    dr  dr

= (s r –

sr – n(sq + sZ) (from Eq. (12.33))

 dT  s q) (1 + n) = r  R  O  dT r + dT Z   dr dr dr    

(12.36)

Chapter 12: Rotating Discs and Cylinders

341

From Eq. (12.35), Ee Z = s Z – n(s r + sq) = constant = C1 s Z = C1 + n(s r + sq )

\ Differentiating

sZ with respect to r, dT Z

 dT  dT "# ! dr dr $  dT  O %& dT  O  dT  dT  ()"# – s ) (1 + n) = r ! dr ' dr  dr dr  *$ dT dT "   O (1  O ) = r (1  O ) dr dr #$ !  dT  O dT "# (s – s ) = r (1  O ) dr dr $ ! =n

dr Substituting in Eq. (12.36), we get

(s r

R

r

R

q

r

R

2

r

R

q

r

R

r

r

(12.37)

Also considering the equilibrium of an element of the cylinder between angular positions q and

q + dq and radii r and r + dr, we can get as in the case of rotating disc.

 

dT r  SX 2 r 2 dr Comparing Eqs. (12.37) and (12.38), we have

sr – sq = – r

 !

r (1  O )

 

(12.38)

"# = –r  dT  SX r "# $ $ ! dr dT  dT  SX r  –n =–  dr  dr

dT R dT O r dr dr

r

2

dT R 2 r r dr dT dT (1 – n) r + (1 – n) R = – rw 2r dr dr dT r dT R  (1 – n) = – rw 2 r dr dr dT r dT R S =– w 2r  (1  O ) dr dr

(1 – n)

 !

"# $

d S (sr + sq) = – ( 1  O) dr

w 2r

Integrating, we get

sr + sq = – where C2 is a constant of integration. Adding Eqs. (12.38) and (12.39),

 !

2s r = – r

S

(1  O )

"# $

X2 r2 2

+ C2

dT r S X 2r2  SX 2 r 2  + C2 dr (1  O ) 2

(12.39)

342 or

Strength of Materials

 

dT r 3  2O = – rw 2r 2 2 (1  O ) dr Multiplying both sides by r,

2s r + r

2rsr + r2

2

  + rC    3  2O  + rC  1O 

dT r X 2 r 3 3  2O = –r 2 1O dr

d 2 X 2r3 (r sr) = – r 2 dr

or

 + C 

Integrating, r2sr = – r where C3 is constant of integration.

2

2

X 2 r 4  3  2O   r 2 C + C 3 8  1O  2 2

   1O 

2 2 C C sr = – r X r 3  2O  2  23

8

Substituting the value of

sr in Eq. (12.39),

2

   1O 

r

2 2 C C 1  2O sq = – r X r  2  23

8

2

r

Equations (12.40) and (12.41) are the governing equations for a rotating cylinder.

12.7

SOLID CYLINDER

From Eqs. (12.40) and (12.41), respectively

sr =

C2 C3 X 2 r 2  3  2O  + 2 S   2 8  1O  r

 

 

C2 C3 X 2 r 2 1  2O  2 S 2 8 1O r Constant C3 must be zero, since the stress remains finite at r = 0.

sq =

\

sr =

 

 

C2 1 3  2O rw2r 2  2 8 1O

C2 1  1  2O  2 2    rw r 2 8 1 O  For a solid cylinder with a free surface at r = r2

sq =

sr = 0

\

 

 

C2 1 3  2O = rw2r 22 8 1O 2

(12.40)

(12.41)

Chapter 12: Rotating Discs and Cylinders

   1O  1  3  2O  =   rw !r   13  22OO  r "#$ 8 1O 

\

s r = 1 3  2O rw2[r22 – r2]

and

sq

(12.42)

8

2

2 2

343

2

(12.43)

The maximum stresses occur at the centre of the cylinder. At centre, r = 0.

 

12.8

 

1 3  2O rw2r 22 8 1O

(sr)max = (sq )max =

\

HOLLOW CYLINDER

From Eq. (12.40),

 

sr \

Solving for C2 and C3,

 

C2 C3 1 3  2O  2  rw2r 2 2 8 1O r = 0 at r = r1 and r = r2 ,

sr =

     3  2O  rw r  1 8 1O 

0 =

C2 C3 1 3  2O  2  rw2r12 2 8 1O r1

0 =

C2 C3  2 2 r2

2 2 2

 

 rw (r + r )  1  3  2O  C = –   rw r r 8 1O  1  3  2O  =   rw  r  r  r r r  r  8 1O   1  3  2O  =   rw r  r  r r r   13  22OO  r 8 1O  C2 1 3  2O = 8 1O 2

2

sr

\

sq

sq is maximum at r = r1.

If

(sq)max =

 

2 1

2 2

2 2 2 1 2

3

\

(12.44)

2

2 1

2 2

2 2 1 2 2

2

2 1

2 2

2 2 1 2 2

 

1 3  2O rw 2 r22 4 1O

r1 r is very small 1 » 0 then r2 r2 (sq)max =

 

2

  1  2O  r12  1    2   3  2O  r 2 

 

1 3  2O rw 2 r 22 4 1O

(12.45)

2

  

(12.46)

(12.47)

(12.48)

344

Strength of Materials

Comparing Eqs. (12.47) and (12.48), we can observe that (sq)max in a cylinder with a small hole at the centre is twice that of (sq)max in a solid cylinder. For s r to be maximum, dT r =0 dr

or

–2r +

2r12 r22 r3

=0

or

r =

\

(sr)max =

(12.49)

r1r2

 

 

1 3  2O rw 2(r2 – r1)2 8 1O

(12.50)

EXAMPLE 12.1 A hollow cylinder of 40 cm external diameter and 20 cm internal diameter is rotating at 3000 rpm. Determine the distribution of radial and hoop stresses in the cylinder. Density of the cylinder material = 7800 kg/m3, n = 0.3. Solution:

For hollow cylinder,

        1  3  0.6  2Q – 3000   =  ´ 7800 ´     !10 60 8  1  0.3   40000  r  = 0.329927 ´ 10 500    r

r2 r 2 3  2O sr = 1 rw 2 r12  r22  1 2 2  r 2 8 1O r

2

2

 20 2 

"# $

100 – 400  r 2 ´ 10–4 r2

2

–4

2

(sr)max at r = or

and

r=

r1 r2 ,

10 – 20 = 14.142 cm

r (cm )

10

14.142

T r (MPa )

0

3.299

 !

   1O 

15

20

3.207

0

  "#  3  2O  $ r 500  40000  0.6667r "# r ! $

r2 r2 sq = 1 3  2O rw 2 r12  r22  1 2 2  1  2O r 2

8

= 0.329927 ´ 10–4

2

2

r (cm )

10

15

20

T R (MPa)

27.493

17.41

10.997

The variation of stresses is shown in Fig. 12.4:

Chapter 12: Rotating Discs and Cylinders

345

sq , sr(MPa)

30

sq

20

10 sr 10 12 14 16 18 20

r (mm)

Fig. 12.4 Stress distribution.

EXAMPLE 12.2 A turbine rotor, 0.4 m external diameter and 0.2 m internal diameter is revolving at 1200 rpm. Taking the weight of rotor as 7800 kg/m3 and Poisson’s ratio 0.3, find the maximum hoop and radial stresses assuming (i) (ii) (iii) (iv)

Rotor Rotor Rotor Rotor

Solution:

to to to to

be be be be

a a a a

thin disc long cylinder solid disc solid cylinder

(i) For thin disc

w=

2Q N 2Q – 1200 = = 125.66 rad/s 60 60

(sr)max =

SX 2 8

(3 + n)(r2 – r1)2

7800 – (125.66)2 (3.3) (0.1)2 N/m2 8 = 0.5081 MPa =

(sq )max =

SX 2 4

[(3 + n)r 22 + (1 – n)r 21]

7800 – (125.66)2 [3.3(0.2)2 + 0.7(0.1)2] N/m2 4 = 4.28 MPa =

(ii) For long cylinder (sr)max =

 

 

SX 2 3  2O (r2 – r1)2 8 1O

   

7800 – (125.66)2 2.4 (0.1)2 8 0.7 = 0.528 MPa =

346

Strength of Materials

(sq)max = =

SX 2 [(3 – 2n)r22 + (1 – 2n)r 12] 4 (1  O ) 7800 – (125.66 )2 [2.4(0.2)2 + 0.4(0.1)2] N/m2 4 – 0.7

= 4.39 MPa (iii) For thin solid disc (sr)max = (sq )max =

SX 2

(3 + n)r 22

8

7800 – (125.66)2 (3.3) (0.2)2 N/m2 8 = 2.03 MPa =

(iv) For long solid cylinder (sr)max = (sq)max =

 

 

SX 2 3  2O 2 r2 8 1O

   

7800 – (125.66 )2 2.4 (0.2)2 N/m2 8 0.7 = 2.111 MPa

=

EXAMPLE 12.3 A disc of 50 cm diameter and uniform thickness is rotating at 2000 rpm. Determine the maximum stress induced in the disc. If a hole of 10 cm diameter is drilled at the centre of the disc, determine the maximum intensities of radial and hoop stresses induced. Take n = 0.28, r = 7800 kg/m3. Solution:

For the solid disc,

 3  O  rw r  8   3  0.28 ´ 7800 ´  2Q – 2000  ´ 25 =  8   60  (sr)max = (sq)max =

2 2 2

2

2

´ 10–4

= 8.76 MPa For the hollow disc,

 3  O  rw (r – r )  8   3  0.28 ´ 7800 ´  2 – Q – 2000  =  8   60 

(sr)max =

= 5.61 MPa

2

2

1

2

2

(25 – 5)2 ´ 10–4

347

Chapter 12: Rotating Discs and Cylinders

(sq)max =

 3  O  rw  1  O  r  4  ! 3  O  2

 3  0.28  =   ´ 7800 ´ 4  

2 1

"# $

 r22

 2 – Q – 2000   60 

2

´

= 17.69 ´ 106 N/m2 = 17.69 MPa

 1  0.28  25  625"# ´ 10 ! 3  0.28  $

–4

EXAMPLE 12.4 A solid rotor of a turbine is 0.6 m diameter at the blade ring. It is keyed to a 50 mm diameter shaft. If the minimum thickness is 9.5 mm, what should be the thickness at the shaft for a uniform stress of 200 N/mm2 at 10000 rpm? Take density = 7700 kg/m3. Solution: At



S r 2X 2 2T



SX 2 – 0 .09 2T



SX 2 – 0 .0006 2T

t = Ae r = 0.3 m,

Here

t = 9.5 = A e At r = 0.025 m, t = Ae



= 9.5 e

SX 2 – 0. 0894 2T

7700  

1000 Q  0 .0894 30  2 – 200 – 10 6 2

= 9.5 e = 9.5e1.89 = 63 mm

EXAMPLE 12.5 A turbine rotor is 15 cm diameter below the blade ring and 2 cm thick. The turbine is running at 36000 rpm. The allowable stress is 150 MPa. What is the thickness of the rotor at a radius of 5 cm and at the centre? Assume uniform strength r = 7800 kg/m3. Solution:

t = t0 e

Here

At



SX 2 2 2 r2  r1 2T





r = 5, 7800 2 Q – 36000  – ( 25  56.25) –10  4  60 2 –150 –10 6  2

t = 2e At

= 2e1.15474 = 2 ´ 3.1732 = 6.3464 cm r = 0, 7800 2 Q – 36000  (  56.25 ) – 10  4 –  60 2 – 150 –10 6  2

t = 2e

= 2e2.07854 = 2 ´ 7.9928 = 15.985 cm

348

Strength of Materials

EXAMPLE 12.6 A grinding wheel is 300 mm diameter with the bore at the centre of 25 mm diameter. If the thickness of the wheel at the outer radius is 25 mm, what should be the thickness at the bore diameter for a uniform allowable stress of 10 MPa at 2800 rpm? Take density of the wheel material as 2700 kg/m3. Solution: Then

Now,

Given

s = 107 N/m2, r = 2700 kg/m3, r1 = 0.15 m, r2 = 0.0125, t1 = 0.025 m. w=

2Q ´ 2800 = 293.215 rad/s 60

t1 = A e



SX 2 r12 2T

 2700 – ( 293.215) 2 – ( 0 .15 )2    2 –10 7  

or or

0.025 = A e A = 0.025e0.26115 m = 32.46 mm

and

t2 = A e



SX 2 r22 2T

 2700 – ( 293.215)2 ( 0 .0125 )2 2 –10 7

= 32.46 e = 32.46 e–0.0018135 = 32.40 mm

EXERCISES 12.1 For a rotating disc with a central hole, show that the maximum value of radial stress is found at a distance r1 r2 from the centre of disc where r1 and r2 are the inner and outer radii of the disc, respectively. 12.2 Starting from the basic principles, derive an expression for the thickness of a solid rotor of uniform strength. 

SX 2 2 2 3 r2  r1 8 2T ]

[Ans. t = t0 e 12.3 Prove that the maximum circumferential stress in a rotating disc with a central pin hole is twice the value for a solid disc of the same dimension. 12.4 The rotor of a steam turbine is a solid disc of uniform strength and is 20 cm diameter at the blade ring and 2.5 cm thick at the centre. It is running at a constant speed of 30000 rpm. Calculate the thickness of the rotor at a radius of 5 cm. The material density is 7470 kg/m3, and the maximum allowable stress in the rotor is 145 MPa. [Ans. 16.83 cm] 12.5 A thick uniform disc of 25 cm diameter with a central hole of 5 cm diameter runs at 10000 rpm. Calculate the maximum principal stresses and the maximum shearing stress in [Ans. 33.79, 106.49, 36.35 MPa] the disc. n = 0.3 and r = 7470 kg/m3. 12.6 A thin solid disc of 75 cm diameter is to rotate at 3000 rpm. The material density is 7600 kg/m3 and Poisson’s ratio is 0.28. Plot the variation of radial and hoop stresses in the disc. [Ans. 43.25 to zero MPa, 43.25 to 18.98 MPa]

Chapter 12: Rotating Discs and Cylinders

349

12.7 A long hollow cylinder is of 20 cm external diameter and is 5 cm thick. It is revolving at a constant speed of 2400 rpm. Calculate the maximum radial and hoop stresses induced in the cylinder, r = 7600 kg/m3, n = 0.30. [Ans. 0.514 MPa, 4.286 MPa] 12.8 A solid cylinder of 25 cm diameter is rotating at 1500 rpm. Determine the maximum hoop stress induced in the cylinder if its material density is 7800 kg/m3, Poisson’s ratio is 0.28. Also draw the variations of radial and hoop stresses in the cylinder. [Ans. 1.274 to zero MPa, 1.274 to 0.459 MPa] 12.9 Determine the greatest values of radial and hoop stress for a rotating disc in which the outer and inner radii are 0.3 m and 0.15 m. The angular speed is 150 rad/s. Take Poisson’s ratio [Ans. 1.6 MPa, 13.6 MPa] as 0.304 and density 7700 kg/m3.

350

Strength of Materials

13 13.1

Finite Element Method and Its Application Using ANSYS Software

INTRODUCTION

This chapter presents a brief discussion of finite element approach as followed in the analysis packages, e.g., ANSYS. In this chapter, step-by-step instructions are given to model and analyse the different types of problems using 1D, 2D and 3D elements. The students are advised to solve the problems given in the book using the analytical approach presented in Chapters 1 to 12 and validate the results in the same way as stated in this chapter. This chapter discusses the GUI (Graphic User Interface) steps used for solving a few examples of the preceding chapters by the finite element approach. The finite element method approximates the numerical techniques used in design analysis to find the value of the field variable/unknown. This method can be extensively used to obtain solutions to a large class of engineering problems involving stress analysis, heat transfer, electromagnetism, vibration, impact, crash, analysis, etc.

13.2

THE STEPS

Finite element method (FEM) consists of six steps as follows: 1. 2. 3. 4. 5. 6.

Discretization Selecting the solution approximation Forming element matrices Assembly Finding the unknown Interpretation

Discretization Discretization is the first step in the finite element method and hence it is an important step. In this step continuum is divided into finite number of elements. The elements are connected to each other through a node. The discretization of continuum in element and node is called the mesh generation. Dividing the continuum into more number of elements is always preferable as more accurate solution is obtained. More elements of continuum lead to expensive solution in the discretization process. 350

Chapter 13: Finite Element Method and Its Application Using ANSYS Software

351

Selecting the solution approximation To find out the value of field variables within the element or over the element domain, certain expressions are assumed. This expression is useful to find out the value of field variables over the element. Generally, the expression is assumed as polynomial because it is easier to integrate or differentiate. Assumption of expression is one of the important step in the finite element method.

Forming element matrices The total analysis of continuum is first carried out by considering only one element out of the total number of elements. Thus, the single element is analysed for equilibrium. The equation of equilibrium is generally written in matrix from and therefore element matrix is formed. The element matrix is stiffness matrix, displacement matrix, force matrix, etc. Element matrix is formed by using four methods as follows: (a) (b) (c) (d)

Direct method Variational approach method Weighted residual approach Energy method

Generally, the variational approach method and weighted residual approach out of these four methods are mostly used in mechanical engineering problems. The variational approach is used for stress analysis problem whereas the weighted residual method is used for thermo fluid problems.

Assembly The element matrix developed for all elements is assembled together or summed up together to form a global matrix. Thus, the global matrix is the representation of whole continuum. Algebraic equations based on global matrix are then solved, before solving equation, it is necessary to apply the boundary condition.

Finding the unknown The algebraic equation developed by assembly matrices are solved using Gauss elimination method to find out the value of field variables or unknowns. Generally, the values of field variables are found out at the nodal point.

Interpretation The results obtained by using total analysis are then interpreted by the design analyst in order to modify the design to yield better design product. This is important and critical stage for taking a decision for modification of design.

13.3

PRINCIPLE OF MINIMUM POTENTIAL ENERGY

This is an important principle used for finite element analysis. It states that the summation of strain energy and work potential for any elastic body is minimum, if a body is in equilibrium. Work potential is the potential energy due to external force.

352

Strength of Materials

Consider an elastic body subjected to force P. Let u be the deformation of spring due to external force and K be the stiffness of spring (Fig. 13.1). The spring stiffness is expressed as K = P/u. P u K

Fig. 13.1(a) Spring with point force.

The strain energy stored in the spring is U = 1/2 (force in the spring) (displacement) = 1/2 (Ku) u = 1/2 Ku2 The potential energy of the external load P is WP = (load) (displacement from zero potential state) = – Pu Let P represent the summation of strain energy and work potential. P is called the total potential energy. Total potential energy, P = Total strain energy + Work potential P = U + WP = 1/2 Ku2 – Pu for minimum of Total Potential energy, ¶P/¶u = 0 Ku – P = 0 or

13.3.1

Ku = P

(13.1)

Potential Energy

The total potential energy P of an elastic body is defined as the sum of total strain energy (U) and the work potential Fig. 13.1(b). P = Strain energy + Work potential (U) (WP) For linear elastic materials, the strain energy per unit volume in the body is 1/2 s Te. transpose of stress matrix s . For an elastic body, the total strain energy is given by U =

1 2

T V

T

F dV

s T is the

Chapter 13: Finite Element Method and Its Application Using ANSYS Software

Fig. 13.1(b)

353

Three-dimensional body.

The work potential WP is given by

WP = 

u

T

f dV 

 u T dS   u P T

T i i

i

V

S

where f = distributed force per unit volume T = surface traction P = concentrated force The total potential energy for the general elastic body is

3=

1 2

T

T

F dV 

V

u V

T

f dV 

 u T dS   u P T

T i i

(13.2)

i

S

The expression for total potential energy for whole body is

3=

1 2

QT KQ  QT F

where Q = Global displacement matrix K = Global stiffness matrix which is obtained by assembling element stiffness matrix F = Global force matrix which is obtained by assembling body force, traction and point force

13.4

COMPUTER AIDED STRESS ANALYSIS TECHNIQUE

The following steps are used to do the analysis using Computer Aided Engineering (CAE) packages:

Preprocessing (i) Changing jobname (ii) Defining element type

354

Strength of Materials

(iii) (iv) (v) (vi) (vii) (viii)

Defining material properties Defining section or real constant Creating geometrical model Defining mesh attributes Creating mesh Defining loading and boundary conditions

Solution (i) Solve

Postprocessing (i) (ii) (iii) (iv)

13.5

Plot the deformed shape Plot the contour plot of deflection and stresses List the reactions Exit the ANSYS program

ELEMENTS TYPE AND SHAPES

There are basically three types of elements used in finite element analysis. • 1D element • 2D element • 3D element

1D element The 1D element is used for geometries having one dimensions very large as compared to the other two dimensions (Fig. 13.2). The geometrical model of 1D elements represents 1 out of three required dimensions and can be modeled by a line (length of the part). When the element is created by connecting two nodes or meshing the line, only one out of three dimensions is defined. Remaining two dimensions, i.e., area of cross-section (width and height) must be defined with additional input data and assigned to respective elements.

Fig. 13.2 1D model for rectangular cross-section.

Different types of 1D element are used in finite element analysis, e.g., rod, beam, pipe, spring, damper and gap element, etc. (Fig. 13.3). The 1D element is easy to generate and can be used for specific application such as membrane and bending. It can be used for analysis of long shaft, beam, tapered beam, etc.

Chapter 13: Finite Element Method and Its Application Using ANSYS Software

355

J

x Y I X

Fig. 13.3

1D model for circular cross-section.

2D elements The 2D element is used when two of the dimensions are very large as compared to the third dimension (Fig. 13.4). It can be used for meshing of all sheet metal parts, panels and control surfaces of aircraft structures, etc. In general, 2D meshing is used for parts having width/thickness ratio > 20.

Fig. 13.4 2D model.

2D meshing is carried out on mid surface of the part. Geometrical model of 2D elements represent 2 out of 3 required dimensions (Fig. 13.5). The third dimension, i.e., thickness has to be provided by user during meshing stage. Each element in the mesh can have different thickness and material properties.

Fig. 13.5 2D model showing mid surface.

356

Strength of Materials

Mathematically, element thickness is specified half on element’s top and half on bottom side. Hence, in order to represent the element geometry appropriately, it is necessary to extract mid surface and mesh on the mid surface. There are basically two types of 2D elements used in finite element analysis, e.g., tria and quad (Fig. 13.6). Each node of 2D element has got 6 DOF.

Fig. 13.6

2D elements.

The elements considered so far is linear type of element, i.e. joint between two node is a line. This element is useful for discretizing the body which is having the straight boundary. The body having irregular or curved, quadratic or cubic nature elements are used. Numbers of nodes/element are increased for quadratic or cubic nature of element (Fig. 13.7).

Fig. 13.7 Quadratic elements.

3D elements 3D element is used when all the three dimensions are comparable (Fig. 13.8). Solid elements could not be twisted or bent and have very high bending/torsional stiffness and hence each node of 3D element has got 3 DOF (translation). 3D elements are used for analysis of thick body-like parts made of casting, stamping, brackets, etc. There are three types of 3D solid elements used in finite element analysis (Fig. 13.9). (i) Tetrahedron (ii) Penta or wedge (iii) Hex or brick

Chapter 13: Finite Element Method and Its Application Using ANSYS Software

357

Fig. 13.8 3D model.

Fig. 13.9

3D elements.

Hex or brick elements are preferable as compared to the tetra element. Hex meshing is difficult to generate for curved body with small radius fillet. Most of the time hex meshing is generated in combination with penta mesh, whereas tet meshing is quite easy to generate. The majority of engineering analysis is carried out considering tetra element as it is ease to generate.

358

13.6 13.6.1

Strength of Materials

ONE-DIMENSIONAL PROBLEMS Natural Coordinate (Intrinsic Coordinate)

Suppose node 1 and 2 of e element are located at x1 and x2 distance from the absolute datum. Consider a typical finite element e in the local coordinate system (Fig. 13.10). We define a natural coordinate system, denoted by

Y=

2 x2  x1

(x  x1 )  1

(13.3)

Fig. 13.10 Typical element in x- and x-coordinate.

Natural coordinate is used to define the shape function for 1D element and can be defined as N1 (Y ) = N 2 (Y ) =

1  Y 2 1+Y 2

(13.4)

For 1D element there are two shape functions, i.e., N1 and N2. These shape functions are used for linear approximation, which helps to find displacement of any point over the element in terms of nodal displacement. Let q1 and q2 be the displacement of nodes 1 and 2, then displacement of any point u within the element e can be expressed as (Fig. 13.11) u = N1q1 + N2q2

Fig. 13.11 Intorpolation of displacement field within an element.

The variation of shape function over the element is shown in Fig. 13.12.

(13.5)

Chapter 13: Finite Element Method and Its Application Using ANSYS Software

359

Fig. 13.12 Variation of shape function over the element.

13.6.2

Isoparametric Element

We know the displacement of point over element is expressed as u = N1q1 + N2q2

(13.6)

It is observed that geometric position of any point can also be expressed in terms of shape function. x = N1x1 + N2x2

(13.7)

From Eqs. (13.6) and (13.7), it is clear that displacement as well as geometry of element is expressed by the same parameter (shape function) of the same order; such elements are called isoparametric elements.

13.6.3

Element Strain Displacement Matrix

The equation of strain for the element can be given as F =

du dx

where du is small displacement of element having length dx. Using the chain rule of differentiation, we have F =

du d Y

(13.8)

d Y dx

We have from Eq. (13.3)

Y=

Therefore,

dY dx

=

2(x  x1 ) ( x2  x1 ) 2 x2  x1

 1

(13.9)

360

Strength of Materials

We know

u = N1q1 + N2q2 1  Y  1 + Y  =   q1 +   q2 2    2  du

 q1 + q2

=

dY

2

(13.10)

Substituting Eqs. (13.9) and (13.10) in Eq. (13.8), we have q1 + q2  2    2  x2  x1 

F =

=

=

1

(  q1 + q2 )

x2  x1 1 le

(  q1 + q 2 )

(13.11)

where le is the length of element. Equation (13.11) can be written in matrix form as e = Bq

where

B=

1 le

q1  [–1 1] and q =   q2 

Here B is called the element strain displacement matrix. The stress over the element can be given as s = EBq

where E is modulus of elasticity.

13.6.4

Element Stiffness Matrix

Expression for strain energy for element can be given as Ue =

1 2

 T F dV T

e

or

Ue =

1 2

 T F A dx T

e

where A is the area of element, and Ue the strain energy.

(13.12)

Chapter 13: Finite Element Method and Its Application Using ANSYS Software

Substituting

361

s = EBq and e = Bq in expression of strain energy Ue =

 (EBq) BqA dx

1

T

2

e

=

1 2

 q B E BqA dx T

T

e

   T EAB B dx  q = q   2   e  1



T

X2     T = q  AEB B dx  q 2   X1  

1



T

(13.13)

We know Y=

2(x  x1 )

 1

(x 2  x1 )

Therefore, dY dx

2

=

dx =

x2  x1 le 2

dY

(13.14)

Substituting Eq. (13.14) in (13.13) Ue

 = q  AEBT B  2   1

=

=

=

=

1 2

1 2 1 2 1 2

T

1



1

l  q T  AEBT B e 2 

qT AE

qT

 dY  q  2  

le

 2 q 

1 1   [  1 1] le q le2 +1

AE  1  le  1

qT K e q

1   q 1  (13.15)

362

Strength of Materials

where element stiffness matrix

Ke =

AE  1  le  1

1   1 

(13.16)

It is clear from the expression of Ke that element stiffness matrix depends upon material parameter (E) and geometrical property (A and le).

13.6.5

Forces

The forces acting on body are classified into three types (Fig. 13.13): (i) Body force (ii) Traction force (iii) Point force Body force: The body force is the force acting on the body which is uniformly distributed. The body force is always expressed as force/volume, e.g., weight. Traction force: The traction force acts all over the surface of the body. The traction force is expressed as force/area. Frictional force, viscous force and surface force are example of the traction force. In case of 1D element traction force is expressed as force/length. Point force: The point force is the external force acting on the body. It is always expressed in absolute unit of force.

Fig. 13.13 One-dimensional bar loaded by traction, body and point loads.

Chapter 13: Finite Element Method and Its Application Using ANSYS Software

363

Elemental body force matrix The potential energy due to body force is given by

u

T

f A dx

le

where, f is the body force/volume. Substituting u = N1q1 + N2q2, we have potential energy due to body force =

 (N q

1 1

+ N 2 q2 ) fA dx

le

=



 N q fA dx

N1q1 fA dx +

2 2

le

le

Writing the equation in matrix form, we get  f A  = qT   f A  

We have



N1 dx =

le

 le

N

 dx     N2 dx    1

le

 le

(13.17)

1  Y  le    dY 2 2 1

=

le 2

1 Y



1

2

dY

1

=

le  1 1 2  Y  Y  2 2 4  1

 N dx = 2

(13.18)

 N dx = 2

(13.19)

le

1

le

Similarly, we have

le

2

le

364

Strength of Materials

Substituting Eqs. (13.18) and (13.19) in Eq. (13.17), we have potential energy due to body force.   fA The potential energy due to body force = q T    fA 

le   2 le   2

= qT f e

(13.20)

where elemental body force is fe =

fA le 1   2 1

(13.21)

Elemental traction force The potential energy due to traction force is given by



uT T dx

le

where T is the traction force/length. Substituting u = N1q1 + N2q2, we have potential energy due to body force =



(N1q1 + N 2 q2 ) T dx



N1q1 T dx +

le

=

le

 N q T dx 2 2

le

Writing the equation in matrix form



   T N1 dx   le   = qT    T N 2 dx     le 



(13.22)

we have



N1 dx =



N 2 dx =

le

le

le 2 le 2

(13.23)

(13.24)

Chapter 13: Finite Element Method and Its Application Using ANSYS Software

365

Substituting Eqs. (13.23) and (13.24) in Eq. (13.22), we have potential energy due to body force

= qT

 le  T   2  = qTT e  le  T   2

where

Te =

Tle 1   2 1

(13.25)

where, Te is the elemental traction force matrix.

EXAMPLE 13.1 Consider the stepped bar as shown in Fig. 13.14. Each element is the cross section and length A1, A2, A3, l1, l2 and l3 respectively and body force f per unit volume. Traction forces on each element are T1, T2 and T3 per unit length. The modulus of elasticity for stepped bar is E. A concentrated load P is applied at node 2. Explain the complete procedure for finding the stresses and reactions.

Fig. 13.14

Stepped bar.

Solution: Global force matrix can be written by considering the body force, traction force and point force. These are written for each node 1, 2, 3 and 4. fA1l1 Tl   + 11   2 2    fA1l1  fA2 l2 T1l1 T2 l2  + + + + P 2 2 2  2  F=   fA3 l3 T2 l2 T3 l3   fA2 l2 + + +  2 2 2 2      fA3 l3 T3 l3 +     2 2

366

Strength of Materials

The potential energy of whole body can be written as

3=

1

QT KQ  QT F

2

By applying the minimum potential energy, the matrix equation obtained will be KQ = F

 K11   K 21   K31   K 41

K12

K13

K22

K23

K32

K33

K 42

K 43

K14   Q1   F1      K24  Q2   F2    =   K34  Q3   F3      K 44  Q4   F4 

The element of global stiffness matrix K known as K11, K12, etc. are already evaluated. Similarly, F1, F2, F3 and F4 represent the force acting on node 1, 2, 3 and 4. The value of force is already evaluated. In this problem Node 1 is fixed, therefore the first element of global force matrix, i.e., F1 the reaction force R1 is to be added. The equation can be solved using the elimination approach. Before applying the elimination approach it is necessary to apply the boundary condition. The elimination approach state, element of displacement matrix if boundary condition applies then eliminate the corresponding row and column from matrix equation for solving the algebraic equation. In this problem the boundary condition is applied to first element of global displacement matrix. Therefore, delete the first row and the first column from matrix equation ○

○

○

K12 ○

○

○

K13 ○

○

K 22

K23

K32

K33

K 42

K 43

○

○

○

○

○

○

○

○

 K11   K21   K31   K 41

○

K14   Q1  F1 + R1      K24  Q2   F2     =  K34  Q3   F3      K 44  Q4   F4  ○

○

○

○

○

○

○

○

○

○

○

The matrix equation reduces to

 K 22   K32   K 42

K24  Q2   F2      K34  Q3  =  F3      K 44  Q4   F4 

K 23 K33 K 43

The matrix equation leads to three algebraic equations with three unknowns, i.e., Q2, Q3 and Q4, which can be evaluated to get displacement at the node point. From displacement at nodal point the strain, stress over the element can be evaluated by the following equations. Strain e1 = B1q F =

 q1  [  1 1]   le q2  1

Chapter 13: Finite Element Method and Its Application Using ANSYS Software

367

Stress

s1 = Ee Similarly, the strain and stress for element 2 and 3 can be evaluated. The reaction force at Node 1 can also be find out considering the algebraic equation formed using the first row of matrix. Thus, we have K11Q1 + K12Q2 + K13Q3 + K14Q4 = F1 + R1

13.7

APPLICATION OF FINITE ELEMENT ANALYSIS USING THE ANSYS SOFTWARE

13.7.1 Application of Finite Element Analysis Using 1D Element EXAMPLE 13.2 The stepped bar in Fig. 13.15 of 600 mm length, carries point force, P = 300 kN. Using the elimination approach for handling the boundary conditions, determine the stress and reaction on bar. Verify the results using the ANSYS software. E = 200 GPa, n = 0.3.

Fig. 13.15

Stepped bar.

Solution: Dividing the complete body into 3 elements and 4 nodes, the element stiffness matrix for element 1, 2 and 3 is

K1 =

=

A1 E  1  l1  1

1  1

250  200  103  1  150 1 1

 1 = 333.33 ´ 103   1

1  1

2 1 1  1 2

368

Strength of Materials

K2 =

A2 E  1  l2 1

1  1

2 1  = 333.33 ´ 103   1

K3 =

A3 E  1  l3  1

3 1 2  1 3

1  1

3  1 = 266.66 ´ 103   1

4 1 3  1 4

Global stiffness matrix 1

 333.34  333.34 3  K = 10   0   0  333.34  333.34 3 = 10   0   0

2

3

333.33

0

333.33  333.34

333.34

333.34

333.34  266.6

0

266.6

333.33

0

666.67

333.34

333.34

599.94

0

266.6

4

  0   266.6   266.6  0

1 2 3 4

  0   266.6   266.6  0

Neglect the body force and traction force. The global force matrix contains only the point force P act at Node 2, while at Nodes 1 and 4 reaction force acts represented by R1 and R4.

R1    3 300  10  F=   0     R4  

369

Chapter 13: Finite Element Method and Its Application Using ANSYS Software

Using the principle of minimum potential energy the matrix equation can be written as F = KQ ○

○

○

○

○

○

○

○

○

○

○

○

○

○

○

○

○

0

○

○

○

○

○

○

○

○

266.6 ○

○

○

0

○

○

○

○

○

○ ○

○ ○

○

333.34 + 266.6

○

○

○

333.34

○

○

0

333.34

○ ○

○

333.33 + 333.34

○

○

  0   266.6   266.6  ○

○

○

○

○

○

○

○

○

○

○

○

○

○

○

○

○

○

○

○

333.33

○

 333.34  333.34 3  10   0   0

○

R1  Q1       3 Q2  300  10   =   0 Q3        R4  Q4   ○

○

○

○

○

○

○

○

○

○

○

○

○

○

○

○

○

○

○

○

○

○

Using the elimination approach, deleting the first and fourth rows and column as the beam is fixed at node 1 and 4. The matrix equation reduces to 333.34   599.9 

 666.68 10 3   333.34

Q2  300  10 3    =   0 Q3   

The matrix equation leads to two algebraic equations as 103(666.66Q2 – 333.34Q3) = 300 ´ 103

(13.26)

103(–333.3Q2 – 599.9Q3) = 0

(13.27)

Solving Eqs. (13.26) and (13.27), we get Q3 = 0.346 mm Q4 = 0.623 mm Strain on element 1 F1 = B1q =

F1 =

Q1  [  1 1]   le Q2  1

 0  [  1 1]   = 0.0041 150 0.623 1

Stress on element 1

s1 = Ee1 = 200 ´ 109 ´ 0.0041 = 0.82 ´ 109 N/m2 Strain on element 2 F 2 = B2 q =

F2 =

Q2  [  1 1]   le Q3  1

0.623  [  1 1]   =  0.0018 150 0.396  1

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Stress on element 2

s2 = Ee2 = 200 ´ 109 ´ (– 0.0018) = – 0.36 ´ 109 N/m2 Strain on element 3 F 3 = B3 q =

Q3  [  1 1]   le Q4  1

0.396  [  1 1]   =  0.0011 150  0  1

F3 =

Stress on element 3

s3 = Ee3 = 200 ´ 109 ´ (– 0.0011) = – 0.22 ´ 109 N/m2 Reaction force at node 1 103(333.34Q1 – 333.34Q2) = R1 R1 = –207.67 ´ 103 N Reaction force at node 4 103(–266.6Q3 + 266.6Q4) = R4 R4 = –92.32 ´ 103 N Solution using ANSYS Software

Preprocessing Step 1: Define the type of element Preprocessor>Element type>Add/Edit/Delete>Add>Link>2D spar1>OK> Close. Step 2: Define the material model Preprocessor>Material props>Material 200E3 and PRXY=0.3>OK.

models>Structural>Linear>Elastic>Isotropic>EX=

Step 3: Define the real constant Preprocessor>Real constants>Add/edit/delete>Add>OK>Cross-sectional area>Enter 250>OK. Preprocessor>Real constants>Add/edit/delete>Add>OK>Cross-sectional area>Enter 400>OK. Step 4: Build the model Preprocessor>Modeling>Create>Node>In active CS>Key point number=1 & X, Y, Z locations in active CS=0, 0, 0>Apply>Node number=2 & X, Y, Z locations in active CS=150,0,0>Apply> Node number=3 & X, Y, Z locations in active CS=300,0,0>Apply>Node number=4 & X, Y, Z locations in active CS=600,0,0>Apply Step 5: Meshing (a) Mesh attributes for element 1 and 2 Preprocessor>Modeling>Create>Element>Element attribute>Element type number=1 link 1, material number 1, real constant set number 1.

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(b) Mesh generation for element 1 and 2 Preprocessor>Modeling>Create>Element>Pick node 1&2>Apply>Again pick node 2&3>OK (c) Mesh attributes for element 3 Preprocessor>Modeling>Create>Element>Element attribute>Element type number=1 link 1, material number 1, real constant set number 2. (d) Mesh generation for element 3 Preprocessor>Modeling>Create>Element>Pick node 3 & 4>OK. Step 6: Loading Preprocessor>Loads>Define loads>Apply>Structural>Force/moment>On nodes>Pick node 2> Apply force in Fx direction (300e3 N). Step 7: Boundary conditions Preprocessor>Loads>Define loads>Apply>Structural>Displacement>On nodes>Pick node at point 1&4>All DOF. Step 8: Solution Solution>Solve>Current LS.

Postprocessing Step 1: Contour plot of displacement General postproc>Plot results>Contour plot>Nodal solu>DOF solution>Displacement vector sum. Step 2: Contour plot of stress (a) Define additional elemental data General postproc>Elemental table>add>Define additional elemental data appears, in the scroll box on the left click once on by sequence num>highlight LS on the right scroll box>Enter 1 after LS, in the white box below the right scroll box>OK. (b) Plot elemental table General postproc>Elemental table>Plot elem table>OK. (c) Reaction General postproc>List results>Highlight all struc forc F>OK. (d) Model save File>save as>Stepped_beam.db (e) Exit ANSYS Quit from ANSYS toolbar. Table 13.1 presents the comparison of analysis results. Table 13.1 Stress/reactions Stress at Stress at Stress at Reaction Reaction

element 1(N/mm2) element 2(N/mm2) element 3(N/mm2) on node 1(kN) on node 4(kN)

Comparison of Analysis Results

Analytical Approach (FEM) 820 –360 –220 –207.67 –92.32

FEM (ANSYS) Using

% difference

830.769 –369.231 –235.897 –207.69 –92.31

1.3 2.5 6.8 0 0

372

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EXAMPLE 13.3 A cantilever rectangular beam (200 mm ´ 150 mm) of 3 m long carries vertical point load of 4 kN at free end. Find the deflection of beam at free end. Take E = 210 GPa and Poisson’s ratio = 0.3. Solution:

Preprocessing Step 1: Define the type of element Preprocessor>Element type>Add/Edit/Delete>Add>Beam (2 node 188)>OK>Close. Step 2: Define the material model Preprocessor>Material Props>Material models>Structural>Linear>Elastic>Isotropic>EX=210E3 & PRXY=0.3 >OK. Step 3: Define the section of beam Preprocessor>Sections>Beam>Common sections, sub type pick rectangular section (define ID=1, Name=Rectangular beam, B=200 & H=150)>Apply to see the sectional property>close. Step 4: Build the model Preprocessor>Modeling>Create>Keypoints>In active CS>Key point number=1 & X, Y, Z locations in active CS=0, 0, 0>Apply>Keypoints number=2 & X, Y, Z locations in active CS=3000, 0, 0>OK. Preprocessor>Modeling>Create>Lines>Lines>Straight line>Pick key point 1 & 2 > OK. y z x B

A

Fig. 13.16 Model.

Step 5: Meshing (a) Mesh attributes Preprocessor>Meshing>Mesh tool>Element attributes>Set (Element type number = 1 Beam 188, Material Number=1 & Section number=1 rectangular beam), press button OK. (b) Element size Preprocessor>Meshing>Mesh tool>Size controls>Global set>Element edge length (Define 25). (c) Mesh generation Preprocessor>Meshing>Mesh tool>Mesh: picked line>Press button mesh>picked the line from graphics>OK. Step 6: Display of meshed model in 3D Plot control>Style>Size & shape>Display of element, Press button on. Step 7: Loading Preprocessor>Loads>Define loads>Apply>Structural>Force/Moment>On nodes>Pick node at free end (Point A)>Apply force in FY direction (– 4000 N). Step 8: Boundary conditions Preprocessor>Loads>Define loads>Apply>Structural>Displacement>On nodes>Pick node at fixed end (Point B)>All DOF.

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Solution: Solution>Solve>Current LS.

Postprocessing Step 1: Contour plot General postproc>Plot results>Contour plot>Nodal solu>DOF solution>Displacement vector sum. Step 2: 3D display of results Put power graph button on ANSYS toolbar menu to see the results in 3D. Step 3: Model save File>Save as>Rectangular_beam.db. Step 4: Exit ANSYS Quit from ANSYS toolbar. Table 13.2 shows the comparison of analysis results. Table 13.2

Comparison of Analysis Results Deflection of beam (mm)

Analytical approach

3.047

FEM(ANSYS)

3.056

% difference 0.3

EXAMPLE 13.4 A cantilever I-section beam of 2 m long carries vertical point load of 5 kN (Fig. 13.17). Find the deflection of beam at free end. Take E = 210 GPa and Poisson’s ratio = 0.3. Solution:

Preprocessing Step 1: Define type of element. Preprocessor>Element type>Add/Edit/Delete>Add beam (2 node 188)>OK>Close. Step 2: Define material model Preprocessor>Material props>Material models>Structural>Linear>Elastic>Isotropic, define EX= 210E3 & PRXY=0.3>OK Step 3: Define section of beam Preprocessor>Sections>Beam>Common>Sections, Sub type pick eye section (define ID=1, Name= Eye section, W1=200 mm, W2=200 mm, W3=420 mm, t1=t2=t3=10 mm)>OK. Step 4: Build the model Preprocessor>Modeling>Create>Keypoints>In active CS>Key point number =1 & X, Y, Z locations in active CS=0, 0, 0 Apply>Keypoints number =2 & X, Y, Z locations in active CS=2000, 0, 0 > OK. Preprocessor>Modeling>Create>lines>lines>Straight line>pick key point 1 & 2 > OK.

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Fig. 13.17 Sectional details. Y B

Z

X

A

Fig. 13.18 Model.

Step 5: Meshing (a) Mesh attributes Preprocessor>Meshing>Mesh tool>Element attributes>Set (Element type number = 1 Beam 188, Material number=1 & Section number= 1 Eye section), press button OK. (b) Element size Preprocessor>Meshing>Mesh tool>Size controls>Global set>Element edge length (Define 20). (c) Mesh generation Preprocessor>Meshing>Mesh tool>Mesh: picked line>Press button mesh>picked the line from graphics>OK. Step 6: Display of meshed model in 3D Plot control>Style>Size & Shape>Display of element, Press button on. Step 7: Loading Preprocessor>Loads>Define loads>Apply>Structural>Force/Moment>On nodes>Pick node at point A>Apply force in FY direction (–5000 N). Step 8: Boundary conditions Preprocessor>Loads>Define loads>Apply>Structural>Displacement>On nodes>Pick node at point B>All DOF.

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Solution: Solution>Solve>Current LS.

Postprocessing Step 1: Contour plot General postproc>Plot results>Contour plot>Nodal solu>DOF solution>Displacement vector sum. Step 2: 3D Display of Results Put power graph button on ANSYS toolbar menu to see the results in 3D. Step 3: Model save File>save as>beam.db. Step 4: Exit ANSYS Quit from ANSYS toolbar. Table 13.3 presents the comparison of analysis results. Table 13.3

Comparison of Analysis Results Deflection of beam (mm)

Analytical approach

0.3

FEM(ANSYS)

0.32

% difference 6.6

EXAMPLE 13.5 A cantilever tapered I-section beam of 2 m long carries vertical point load of 5 kN. The tapered beam with fixed end and free end sectional details are shown in Fig. 13.19. Find the deflection of beam at free end. Take E = 210 GPa and Poisson’s ratio = 0.3. Solution:

Preprocessing Step 1: Define the type of element Preprocessor>Element type>Add/Edit/Delete>Add beam (2 node 188)>OK>Close. Step 2: Define material model Preprocessor>Material props>Material models>Structural>Linear>Elastic>Isotropic> EX= 210E3 & PRXY=0.3>OK. Step 3: Define the starting and end section Preprocessor>Sections>Beam>Common sections, sub type pick eye section (ID=1, Name= Starting section, W1=200 mm, W2=200 mm, W3=416 mm, t1=t2=t3=8 mm)>OK. Preprocessor > Sections > Beam > Common sections sub type pick eye section (ID=2, Name= End section, W1=100 mm, W2=100 mm, W3=308 mm, t1=t2=t3=4 mm)>OK.

376

Strength of Materials

Fig. 13.19

Sectional details for Example 13.5.

Step 4: Define the taper section Preprocessor>Sections>Taper sections>By XYZ locations>New taper section ID (3), New section>name (Tapered beam), Beginning section ID (1, Starting section), XYZ locations of beginning set (0, 0, 0), Ending section ID (2, End section), XYZ locations of ending section (2000, 0, 0)>OK Step 5: Build the model Preprocessor>Modeling>Create>Keypoints>In active CS>Key point number =1 & X,Y,Z locations in active CS=0,0,0>Apply>Keypoint number =2 & X,Y,Z locations in active CS=2000, 0,0>OK. Preprocessor>Modeling>Create>lines>Lines>Straight line>Pick key point 1 & 2>OK. Y B

Z

X

A

Fig. 13.20 Model.

Step 6: Meshing (a) Mesh the attributes Preprocessor>Meshing>Picked lines>pick the line from graphics>press button OK>Define material number (1), Element type number (1 Beam 188) and Element section (3 taper)>OK. (b) Element size Preprocessor>Meshing>Mesh tool>Size controls>Global set>Element edge length (Define 20).

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(c) Mesh the generation Preprocessor>Meshing>Mesh tool>Mesh: picked line>Press button mesh>picked the line from graphics>OK. Step 7: Display of meshed model in 3D Plot control>Style>Size & shape>Display of element, Press button on. Step 8: Loading Preprocessor>Loads>Define loads>Apply>Structural>Force/Moment>On Nodes > Pick node at point A > Apply Force in FY direction (–5000 N). Step 9: Boundary conditions Preprocessor>Loads>Define loads>Apply>Structural>Displacement>On nodes>Pick node at point B>All DOF. Solution: Solution>Solve>Current LS.

Post processing Step 1: Contour plot General postproc>Plot results>Contour plot>Nodal solu>DOF solution>Displacement vector sum. Step 2: 3D display of results Put power graph button on ANSYS toolbar menu to see the results in 3D Step 3: Model save File>Save as>Tapered beam.db Step 4: Exit ANSYS Quit from ANSYS toolbar.

EXAMPLE 13.6 Refer to Example 13.4, find out the displacement of beam for axial compressive/tensile load of 5 kN Take E = 210 GPa and Poisson’s ratio = 0.3. Solution:

Preprocessing Step 1: Define the type of element Preprocessor>Element type>Add/Edit/Delete>Link (2D spar)>OK>close. Step 2: Define the real constant Preprocessor>Real constant>Add/Edit/Delete>Add>OK>Cross-sectional area>800>Close. Step 3: Define the material model Preprocessor>Material props>Material models>Structural>Linear>Elastic>Isotropic, define EX=210E3 & PRXY=0.3>OK.

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Step 4: Build the model Preprocessor>Modeling>Create>Keypoints>In active CS>Key point number =1 & X, Y, Z locations in active CS=0, 0, 0 apply>Key point number =2 & X,Y, Z locations in active CS=2000,0,0>OK. Preprocessor>Modeling>Create>Lines>Lines>Straight line>Pick key point 1 & 2>OK Y B

Z

X

A

Fig. 13.21 Model.

Step 5: Meshing (a) Mesh the attributes Preprocessor>Meshing>Mesh tool>Element attributes>Set (Element type number = 1 link, material number=1 & Real constant set number =1), press button OK. (b) Element size Preprocessor>Meshing>Mesh tool>Size controls>Global set>Element edge length (Define 20). (c) Mesh generation Preprocessor>Meshing>Mesh tool>Mesh: picked line>Press button mesh>picked the line from graphics>OK. Step 6: Display of meshed model in 3D. Plot control>Style>Size & shape>Display of element, press button on. Step 7: Loading Preprocessor>Loads>Define loads>Apply>Structural>Force/Moment>On nodes>Pick node at point A>Apply force in FX direction (–5000 N). Step 8: Boundary conditions Preprocessor>Loads>Define loads>Apply>Structural>Displacement>On nodes>Pick node at point B>All DOF. Solution: Solution>Solve>Current LS.

Postprocessing Step 1: Contour plot General postproc>Plot results>Contour plot>Nodal solu>DOF solution>Displacement vector sum. Step 2: 3D display of Results Put power graph button on ANSYS toolbar menu to see the results in 3D. Step 3: Exit ANSYS Quit from ANSYS toolbar Table 13.4 presents the comparison of analysis results.

Chapter 13: Finite Element Method and Its Application Using ANSYS Software Table 13.4

Comparison of Analysis Results Deflection of beam (mm)

13.7.2

379

Analytical approach

0.059

FEM (ANSYS)

0.059

% difference 0

Application of Finite Element Analysis Using 2D Element

EXAMPLE 13.7 Refer to example 9.1. Determine the resulting bending stress at corners A and B, on the fixed section of the cantilever. Take E = 210 GPa and Poisson’s ratio = 0.3. Solution:

Preprocessing Step 1: Define the type of element Preprocessor>Element type>Add/Edit/Delete>Shell)>Elastic 4 node 63>OK>Close. Step 2: Define the real constant Preprocessor>Real constant>Add/Edit/Delete>Add>Shell thickness at node I TK (1)>2>Close> Add>Shell thickness at node I TK (1)>2.5>Close. Step 3: Define the material model Preprocessor>Material props>Material models>Structural>Linear>Elastic>Isotropic, define EX= 210E3 & PRXY=0.3>OK. Step 4: Build the model Preprocessor>Modeling>Create>Keypoints>In active CS>Key point number =1 & X, Y, Z locations in active CS=0, 0, 0 Apply>Key point number=2 & X,Y, Z locations in active CS=2000,0,0 >Apply>Key point number =3 & X,Y, Z locations in active CS=2000,47.5,0>Apply>>Key point number = 4 & X,Y, Z locations in active CS=0, 47.5,0>Close. Preprocessor>Modeling>Create>Area>Arbitrary>Through KPS>Pick1, 2, 3, 4 Keypoints>OK. Preprocessor>Modeling>Create>Keypoints>In active CS>Key point number =5 & X, Y, Z locations in active CS=0, 0, 15>OK. Preprocessor>Modeling>Create>Lines>Straight line>Pick keypoints 1 and 5>OK. Preprocessor>Modeling>Operate>Extrude>Lines>Area>Along lines>Pick lines (L1 & L3)> OK>Pick line L5>OK. Preprocessor>Modeling>Reflect>Areas>Pick area A2 & A3>OK>Pick X-Y Plane>Item to be reflected>Pick area>Existing areas will be>Copied>OK.

380

Strength of Materials

Fig. 13.22 Model.

Step 5: Meshing (a) Mesh the attributes Preprocessor>Meshing>Mesh tool>Element attributes>Set (Element type number = 1 Shell 63, Material number=1 & Real constant set number =1), press button OK. (b) Element size Preprocessor>Meshing>Mesh tool>Size controls>Global set>Element edge length (Define 10). (c) Mesh the generation Preprocessor>Meshing>Mesh tool>Mesh: Area>Free>Press button mesh>Picked the area A1 (web of beam) from graphics>OK. Preprocessor>Meshing>Mesh tool>Element attributes>Set (Element type number = 1 Shell 63, Material number=1 & Real constant set number =2), press button OK. Preprocessor>Meshing>Mesh tool>Mesh: Area>Free>Press button mesh>Picked the area A2, A3, A4 &A5 (top & bottom flange) from graphics>OK. (d) Merge the node Preprocessor>Numbering ctrls>Merge items>Range of coincidence>0.001>OK. Step 6: Display of meshed model in 3D Plot control>Style>Size & Shape>Display of element, Press button on Step 7: Loading Preprocessor>Loads>Define loads>Apply>Structural>Force/Moment>On nodes>Pick all the sectional nodes at free end>Apply>Force in FY direction = –235/No. of nodes>OK. Preprocessor>Loads>Define loads>Apply>Structural>Force/Moment>On nodes>Pick all the sectional nodes at free end>Apply>Force in FZ direction = 85.5/No. of nodes>OK. Step 8: Boundary conditions Preprocessor>Loads>Define loads>Apply>Structural>Displacement>On nodes>Pick all sectional node at fixed end of beam>All DOF. Solution: Solution>Solve>Current LS.

Postprocessing Step 1: Contour the plot. General postproc>Plot results>Contour plot>Nodal solu>Nodal solution>Stress>X component stress.

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Step 2: Query of results General postproc>Query results>Subgrid solu>Stress X-direction stress SX>Pick the nodes at fixed end (Point A&B) to see the magnitude of stress>OK. Exit ANSYS Quit from ANSYS toolbar. Table 13.5 shows the comparison of analysis results. Table 13.5

Comparison of Analysis Results

Resultant bending stress (N/mm2) at point A 2

Resultant bending stress (N/mm ) at point B

Analytical approach

FEM (ANSYS)

% difference

344.96

351.8

1.9%

–109.82

–116.7

6.3%

EXAMPLE 13.8 Refer to Example 9.3. Determine the resulting deflection and bending stress at corners B and C. Take E = 210 GPa and Poisson’s ratio = 0.3. Solution:

Preprocessing Step 1: Define type of element Preprocessor>Element type>Add/Edit/Delete>Shell)>Elastic 4 node 63>OK>Close. Step 2: Define real constant Preprocessor>Real constant>Add/Edit/Delete>Add>Shell thickness at node I TK (1)>10>Close> Add>Shell thickness at node I TK (1)>20>Close. Step 3: Define the material model Preprocessor>Material props>Material models>Structural>Linear>Elastic>Isotropic, define EX=210E3 & PRXY=0.3>OK. Step 4: Build the model Preprocessor>Modeling>Create>Keypoints>In active CS>Key point number =1 & X, Y, Z locations in active CS=0, 0, 0 apply>Key point number =2 & X,Y, Z locations in active CS=1500,0,0 >Apply>Key point number =3 & X,Y, Z locations in active CS=3000,0,0>Apply>Key point number =4 & X,Y, Z locations in active CS=3000,160,0>Apply>Key point number =5 & X,Y, Z locations in active CS=1500,160,0>Apply>Key point number = 6 & X,Y, Z locations in active CS=0,160,0> Close. Preprocessor>Modeling>Create>Area>Arbitrary>Through KPS>Pick 1, 2, 5 & 6 Keypoints >Apply>Pick 2, 3, 4 &5 Keypoints>OK. Preprocessor>Modeling>Create>Keypoints>In active CS>Key point number =7 & X, Y, Z locations in active CS=0, 160, 50>OK. Preprocessor>Modeling>Create>Lines>Straight line>Pick keypoints 6 and 7>OK. Preprocessor>Modeling>Operate>Extrude>Lines>Area>Along lines>Pick the top web line of both area>OK>Pick line L8>OK.

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Strength of Materials

Preprocessor>Modeling>Reflect>Areas>Pick area A3 & A4>OK>Pick X-Y Plane>Item to be reflected>Pick area>Existing areas will be>Copied>OK.

Fig. 13.23 Model.

Step 5: Meshing (a) Mesh attributes Preprocessor>Meshing>Mesh tool>Element attributes>Set (Element type number = 1 shell 63, Material number=1 & Real constant set number =1), press button OK. (b) Element size Preprocessor>Meshing>Mesh tool>Size controls>Global set>Element edge length (Define 10). (c) Mesh generation Preprocessor>Meshing>Mesh tool>Mesh: Area>Free>Press button mesh>picked the area A1 & A2 (web of beam) from graphics>OK. Preprocessor>Meshing>Mesh tool>Element attributes>Set (Element type number = 1 Shell 63, Material number=1 & Real constant set number =2), press button OK. Preprocessor>Meshing>Mesh tool>Mesh: Area>Free>Press button mesh>Picked the area A3, A4, A5 &A6 (top flange of beam) from graphics>OK. (d) Merge Node Preprocessor>Numbering ctrls> Merge items>Range of coincidence>0.001>OK. Step 6: Display of Meshed Model in 3D Plot control>Style>Size & Shape>Display of element, Press button on. Step 7: Loading Preprocessor>Loads>Define loads>Apply>Structural>Force/Moment>On nodes>Pick all the sectional nodes at the edge of Area A1 & A2>Apply>Force in FY direction = – 4690/No. of nodes>OK. Preprocessor>Loads>Define loads>Apply>Structural>Force/Moment>On nodes>Pick all the sectional nodes at the edge of Area A1 & A2>Apply>Force in FZ direction = –1710/No. of nodes>OK. Step 8: Boundary conditions Preprocessor>Loads>Define loads>Apply>Structural>Displacement>On nodes>Pick all sectional node at simply supported end of beam>Pick UY and Uz>OK.

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Solution: Solution>Solve>Current LS.

Postprocessing Step 1: Contour the plot General postproc>Plot results>Contour plot>Nodal solu>Nodal solution>Stress>X Component stress. Step 2: Resultant displacement General postproc>Plot results>Contour plot>Nodal solu>DOF solution>Displacement vector sum. Step 3: Query of results General postproc>Query results>Subgrid solu>Stress X-direction stress SX>Pick the nodes at fixed end (Point A&B) to see the magnitude of stress>OK. Step 4: Exit ANSYS Quit from ANSYS toolbar. Table 13.6 presents the comparison of analysis results. Table 13.6

Comparison of Analysis Results Analytical approach

Resultant deflection(mm)

1.024 2

–56.22

2

51.78

Resultant bending stress (N/mm ) at point B Resultant bending stress (N/mm ) at point C

FEM (ANSYS) % difference using 2D (shell 63) 1.1 –60 54.7

7% 6.7% 5.6%

EXAMPLE 13.9 A cantilever I-section beam of 2 m long carries vertical point load of 4 kN (Fig. 13.24). Find the deflection of beam at free end considering 1D element for top and bottom flange and web using 2D element. Take E = 210 GPa and Poisson’s ratio = 0.3. Solution:

Preprocessing Step 1: Define the type of element Preprocessor>Element type>Add/Edit/Delete>Shell>Elastic 4 node 63>Apply>Preprocessor> Element type>Add/Edit/Delete>Add>Beam (2 node 188)>OK>Close. Step 2: Define the real constant Preprocessor>Real constant>Add/Edit/Delete>Add>Pick type 1 shell 63>OK>Shell thickness at node I TK(1)>5>Close. Step 3: Define the material model Preprocessor>Material props>Material models>Structural>Linear>Elastic>Isotropic, define EX=210E3 & PRXY=0.3>OK.

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Strength of Materials

Fig. 13.24 Example 13.9.

Step 4: Define the section of beam Preprocessor>Sections>Beam>Common sections, Sub type pick rectangular section (define ID=1, Name=Rectangular beam, B=200 & H=10)>OK. Step 5: Build the model Preprocessor>Modeling>Create>Keypoints>In active CS>Key point number=1 & X, Y, Z locations in active CS=0, 0, 0 Apply>Key point number=2 & X, Y, Z locations in active CS=2000,0,0>Apply>Key point number =3 & X, Y, Z locations in active CS=2000,390,0>Apply> Key point number = 4 & X, Y, Z locations in active CS=0,390,0>OK. Preprocessor>Modeling>Create>Area>Arbitrary>Through KPS>Pick 1, 2, 3 & 4 >OK.

Fig. 13.25 Model.

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Step 6: Meshing (a) Mesh the attributes for 2D Preprocessor>Meshing>Mesh tool>Element attributes>Set (Element type number = 1 Shell 63, Material number=1 & Real constant set number =1), press button OK. (b) Element size Preprocessor>Meshing>Mesh tool>Size controls>Global set>Element edge length (Define 20). (c) Mesh generation for 2D Preprocessor>Meshing>Mesh tool>Mesh: Area>Free>Press button mesh>Picked the area A1 from graphics>OK. (d) Mesh attributes for 1D Preprocessor>Meshing>Mesh attribute>Picked the top web line>OK>Material number 1, Element type number 2 beam 188, Element section 1 rectangular, Pick orientation keypoints-click yes>Pick the node number 240 to define the orientation or any node along vertical axis at fixed end of beam (Refer to Fig. 10.25)>OK. Preprocessor>Meshing>Mesh attribute>Bottom web line>OK>Material number 1, Element type number 2 beam 188, Element section 1 rectangular, Pick orientation keypoints-click yes>Pick the node number 240 to define the orientation or any node along vertical axis at fixed end of beam (Refer to Fig. 10.25)>OK. (e) Mesh the generation for 1D Preprocessor>Meshing>Mesh tool>Mesh: line>Press button mesh>Picked the top web line>OK Preprocessor>Meshing>Mesh tool>Mesh: line>Press button mesh>Picked the bottom web line>OK. Step 7: Display of meshed model in 3D Plot control>Style>Size & Shape>Display of element, Press button on. Step 8: Loading Preprocessor>Loads>Define loads>Apply>Structural>Force/Moment>On nodes>Pick all the sectional nodes at the free edge of Area A1, Point B>OK>Force in FY direction = – 4000/No. of nodes>OK. Step 9: Boundary conditions Preprocessor>Loads>Define loads>Apply>Structural>Displacement>On nodes>Pick all sectional node at fixed end of beam, point A>Pick UY and Uz>OK. Solution: Solution>Solve>Current LS.

Postprocessing Step 1: General postproc>Plot results>Contour plot>Nodal solu>Nodal solution>Stress>X Component stress. Step 2: General postproc>Plot results>Contour plot>Nodal solu>DOF solution>Displacement vector sum. Step 3: Exit ANSYS Quit from ANSYS toolbar.

386

13.7.3

Strength of Materials

Application of Finite Element Analysis Using 3D Element

EXAMPLE 13.10 Refer to Example 9.3. Determine the resulting deflection and bending stress at corners B and C using 3D brick elements. Take E = 210 GPa and Poisson’s ratio = 0.3. Solution:

Preprocessing Step 1: Define the type of element Preprocessor> Element type>Add/Edit/Delete>Solid>Brick 8 Node 45>OK>Close. Step 2: Define the material model Preprocessor>Material props>Material models>Structural>Linear>Elastic>Isotropic, define EX=210E3 & PRXY=0.3>OK. Step 3: Build the model Preprocessor>Modeling>Create>Keypoints>In active CS>Key point number =1 & X, Y, Z locations in active CS=0, 0, 0 Apply>Key point number=2 & X, Y, Z locations in active CS=100,0,0>Apply>Key point number=3 & X, Y, Z locations in active CS=100, –20, 0>Apply>Key point number=4 & X, Y, Z locations in active CS=0, –20, 0>Apply>Key point number=5 & X, Y, Z locations in active CS=45, –20, 0>Apply>Key point number=6 & X, Y, Z locations in active CS=55, –20, 0>Key point number=7 & X, Y, Z locations in active CS=45, –170, 0>Apply>Key point number=8 & X, Y, Z locations in active CS=55, –170, 0>Close. Preprocessor>Modeling>Create>Area>Arbitrary>Through KPS>Pick 1, 2, 3, 6, 8, 7, 5, 4 & 1 keypoints>Apply. Preprocessor>Modeling>Operate>Extrude>Area>Along normal>Pick the area from graphics> Press OK>Length of extrusion>1500>OK. Preprocessor>Modeling>Operate>Extrude>Area>Along normal>Pick the cross-sectional area from end of beam from graphics>Press OK>Length of extrusion>1500>OK.

Fig. 13.26 Model.

Step 4: Meshing (a) Mesh the attributes Preprocessor>Meshing>Mesh tool>Element attributes>Set (Element type number = 1 Solid 45 & Material number=1), Press button OK.

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(b) Element size Preprocessor>Meshing>Mesh tool>Size controls>Global set>Element edge length (Define 10). (c) Mesh generation Preprocessor>Meshing>Mesh tool>Mesh: Volume>Hex/Wedge>Sweep>Button sweep>Pick both the volume from graphics>OK. (d) Merge node Preprocessor>Numbering Ctrls>Merge items>Range of coincidence>0.001>OK. Step 5: Loading Preprocessor>Loads>Define loads>Apply>Structural>Force/Moment>On nodes>Pick all the sectional nodes at the edge of volume V1 & V2>Apply>Force in FY direction = – 4690/No. of nodes>OK. Preprocessor>Loads>Define loads>Apply>Structural>Force/Moment>On nodes>Pick all the sectional nodes at the edge of volume V1 & V2>Apply>Force in FX direction = –1710/No. of nodes>OK. Step 6: Boundary conditions Preprocessor>Loads>Define loads>Apply>Structural>Displacement>On nodes>Pick all sectional node at simply supported end of beam>Pick UY and UX>OK Solution: Solution>Solve>Current LS.

Postprocessing Step 1: General postproc>Plot results>Contour plot>Nodal solu>Nodal solution>Stress>Z Component stress. Step 2: Exit ANSYS Quit from ANSYS toolbar. Table 13.7 presents the comparison of analysis results. Table 13.7

Comparison of Analysis Results

Resultant bending stress (N/mm2) at point B 2

Resultant bending stress (N/mm ) at point C

Analytical approach

FEM (ANSYS) using 3D (Solid 45)

–56.22

–60

51.78

56.4

% difference 6.7% 8.9%

EXAMPLE 13.11 Refer to Example 9.4. Determine the resulting bending stress at point A using 3D elements. Take E = 210 GPa and Poisson’s ratio = 0.3.

388

Strength of Materials

Solution:

Preprocessing Step 1: Define the type of element Preprocessor>Element type>Add/Edit/Delete>Solid>Brick 8 node 45>OK>Close. Step 2: Define the material model Preprocessor>Material props>Material models>Structural>Linear>Elastic>Isotropic, define EX=210E3 & PRXY=0.3>OK. Step 3: Build the model Preprocessor>Modeling>Create>Keypoints>In active CS>Key point number=1 & X, Y, Z locations in active CS=0, 0, 0 Apply>Key point number=2 & X, Y, Z locations in active CS=80,0,0> Apply>Key point number=3 & X, Y, Z locations in active CS=80,120,0>Apply>Key point number =4 & X, Y, Z locations in active CS=0,120,0>OK. Preprocessor>Modeling>Create>Area>Arbitrary>Through KPS>Pick 1, 2, 3 & 4 Keypoints >Apply. Preprocessor>Modeling>Operate>Extrude>Area>Along normal>Pick the area from graphics> Press OK>Length of extrusion>2000>OK. Preprocessor>Modeling>Operate>Extrude>Area>Along normal>Pick the cross-sectional area from end of beam from graphics>Press OK>Length of extrusion>6000>OK. Preprocessor>Modeling>Operate>Extrude>Area>Along normal>Pick the cross-sectional area from end of beam from graphics>Press OK>Length of extrusion>2000>OK.

Fig. 13.27 Model.

Step 4: Meshing (a) Mesh attributes Preprocessor>Meshing>Mesh tool>Element attributes>Set (Element type number = 1 Solid 45& Material Number=1), Press button OK. (b) Element size Preprocessor>Meshing>Mesh tool>Size controls>Global set>Element edge length (Define 10). (c) Mesh generation Preprocessor>Meshing>Mesh tool>Mesh: Volume>Hex/Wedge>Sweep>Button sweep>Pick volume V1, V2 & V3 from graphics>OK. (d) Merge the node Preprocessor>Numbering ctrls>Merge items>Range of coincidence>0.001>OK.

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Step 5: Loading Preprocessor>Loads>Define loads>Apply>Structural>Force/Moment>On nodes>Pick all the sectional nodes at the edge of volume V1 & V2>Apply>Force in FY direction = –3464/No. of nodes>OK. Preprocessor>Loads>Define loads>Apply>Structural>Force/Moment>On nodes>Pick all the sectional nodes at the edge of volume V1 & V2>Apply>Force in FX direction = – 2000/No. of nodes>OK. Preprocessor>Loads>Define loads>Apply>Structural>Force/Moment>On nodes>Pick all the sectional nodes at the edge of volume V2 & V3>Apply>Force in FY direction = –3464/No. of nodes>OK. Preprocessor>Loads>Define loads>Apply>Structural>Force/Moment>On nodes>Pick all the sectional nodes at the edge of volume V2 & V3>Apply>Force in FX direction = – 2000/No. of nodes>OK. Step 6: Boundary conditions Preprocessor>Loads>Define loads>Apply>Structural>Displacement>On nodes>Pick all sectional node at the end of simply supported beam>All DOF>OK. Solution: Solution>Solve>Current LS.

Post processing Step 1: General postproc>Plot results>Contour plot>Nodal solu>Nodal solution>Stress>Z Component stress. Step 2: Exit ANSYS Quit from ANSYS toolbar. Table 13.8 shows the comparison of analysis results. Table 13.8

Comparison of Analysis Results Analytical approach

Resultant bending stress (N/mm2) at point A

–67.34

FEM (ANSYS) using 3D (Solid 45) –59.5

% difference 11.6%

EXAMPLE 13.12 Refer to Example 9.5. Determine the resulting deflection and bending stress at point A using 3D elements. Take E = 210 GPa and Poisson’s ratio = 0.3. Solution:

Preprocessing Step 1: Define the type of element Preprocessor>Element type>Add/Edit/Delete>Solid>Brick 8 node 45>OK>Close.

390

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Step 2: Define the material model Preprocessor>Material props>Material models>Structural>Linear>Elastic>Isotropic, define EX= 210E3 & PRXY=0.3>OK. Step 3: Build the model Preprocessor>Modeling>Create>Keypoints>In active CS>Key point number =1 & X, Y, Z locations in active CS=0, 0, 0 Apply>Key point number =2 & X, Y, Z locations in active CS=150,0,0>Apply> Key point number=3 & X, Y, Z locations in active CS=150, –20,0>Apply>Key point number = 4 & X, Y, Z locations in active CS=85, –20,0>Apply>Key point number=5 & X, Y, Z locations in active CS=85, –180,0>Apply>Key point number=6 & X, Y, Z locations in active CS=150, –180,0> Apply> Key point number=7 & X, Y, Z locations in active CS=150, –200,0>Apply>Key point number=8 & X, Y, Z locations in active CS=0, –200,0>Apply>Key point number=9 & X, Y, Z locations in active CS=0, –180,0>Apply>Key point number=10 & X, Y, Z locations in active CS=65, –180,0>Apply> Key point number=11 & X, Y, Z locations in active CS=65, –20,0>Apply>Key point number=12 & X, Y, Z locations in active CS=0, –20,0>OK. Preprocessor>Modeling>Create>Area>Arbitrary>Through KPS>Pick 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 & 12 keypoints>OK. Preprocessor>Modeling>Operate>Extrude>Area>Along normal>Pick the area from graphics> Press OK>Length of extrusion>1500>OK. Preprocessor>Modeling>Operate>Extrude>Area>Along normal>Pick the cross-sectional area from end of beam from graphics>Press OK>Length of extrusion>1000>OK.

Fig. 13.28 Model.

Step 4: Meshing (a) Mesh attributes Preprocessor>Meshing>Mesh tool>Element attributes>Set (Element type number = 1 Solid 45& Material Number=1), press button OK. (b) Element size Preprocessor>Meshing>Mesh tool>Size controls>Global set>Element edge length (Define 10). (c) Mesh generation Preprocessor>Meshing>Mesh tool>Mesh: Volume>Hex/Wedge>Sweep>Button sweep>Pick volume V1 & V2 from graphics>OK.

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(d) Merge the node Preprocessor>Numbering ctrls>Merge items>Range of coincidence>0.001>OK. Step 5: Loading Preprocessor>Loads>Define loads>Apply>Structural>Force/Moment>On nodes>Pick all the sectional nodes at the edge of volume V1 & V2>Apply>Force in FY direction = –8660.3/No. of nodes>OK. Preprocessor>Loads>Define loads>Apply>Structural>Force/Moment>On nodes>Pick all the sectional nodes at the edge of volume V1 & V2>Apply>Force in FX direction = –5000/No. of nodes>OK. Preprocessor>Loads>Define loads>Apply>Structural>Force/Moment>On nodes>Pick all the sectional nodes at the free edge of volume V2>Apply>Force in FY direction = –3535.5/No. of nodes>OK. Preprocessor>Loads>Define loads>Apply>Structural>Force/Moment>On nodes>Pick all the sectional nodes at the edge of volume V2 & V3>Apply>Force in FX direction = 3535.5/No. of nodes>OK. Step 6: Boundary conditions Preprocessor>Loads>Define loads>Apply>Structural>Displacement>On nodes>Pick all sectional node at the fixed end of beam>All DOF>OK. Solution: Solution>Solve>Current LS.

Postprocessing Step 1: General postproc>Plot results>Contour plot>Nodal solu>DOF solution>Displacement Vector sum. Step 2: General postproc>Plot results>Contour plot>Nodal solu>Nodal solution>Stress>Z component stress. Step 3: Exit ANSYS Quit from ANSYS toolbar. Table 13.9 shows the comparison of analysis results. Table 13.9

Comparison of Analysis Results Analytical approach

Resultant Resultant Resultant Resultant Resultant

displacement (mm) bending stress (N/mm2) bending stress (N/mm2) bending stress (N/mm2) bending stress (N/mm2)

at at at at

point point point point

A B C D

— 48.07 30.41 –30.41 –48.07

FEM (ANSYS) using 3D (Solid 45) 4.5 47.3 30.06 –30.06 –47.3

% difference

1.6% 1% 1% 1.6%

Index

Airy’s stress function, 87–88 ANSYS, 350–354

Bending stress, 4 Body forces, 49 Brittle material, 95 Buckling, 282

Castigliano’s theorem, 150 Cauchy stress formula, 59 Circumferential or hoop stress, 225 Close-coiled spring, 308 Combined stress, 111 Compatibility, 81 Condition of compatibility, 82 Crippling load, 282

Degree of indeterminacy, 186 Determinate, 81 Deviatoric plane, 99 Double integration method, 159–160 Ductile material, 115

Endurance limit, 124 Euler’s formula, 290 Experimental verification of theory of failure, 111

Fatigue, 123, 127 reduction factor, 127 Finite element method, 350

Generalized Hooke’s law, 53 Gerber equation, 128 Goodman diagram, 128–129

Helical spring, 308 Hooke’s law, 53 Hoop or circumferential stress, 226, 236

Indeterminate structure, 185

Lame’s equations, 243 theory, 241 Load factor, 126 Long column, 283 Longitudinal stress, 225–226

Macaulay’s method, 159, 172 Maximum distortion energy theory, 102 normal stress, 10

393

394

Index

principal strain criterion, 107 principal stress criterion, 106 shear stress, 10 shear stress theory, 96 Mean and deviator stresses, 70 stress, 123 Medium column, 283 Mohr’s circle, 15 theory, 108 Mohr–Coulomb criterion, 111 Moment area method, 159, 179

Neutral axis, 196, 200 Normal stress, 2

Octahedral plane, 68 stress, 68 Open-coiled spring, 308

PI plane, 99 Plane stress, 6 Principal stress, 14

Radial stress, 226 Radius of gyration, 289 Rankine’s formula, 296

Safe load, 282 Second moment area, 263 Shear stress, 5, 8 Short column, 282 Size factor, 125 Slenderness ratio, 228–283 Soderberg criterion, 132, 136 Solution of stress differential equation, 86

Spring index, 312 St. Venant’s equations, 83 State of pure shear, 71 State of stress, 71 Statically indeterminate problems, 185 structures, 185 Stiffness, 158, 315 Strain energy density, 105 Strain-displacement relation, 74–75 Stress, 1–2 gradient, 51 invariants, 60 on an inclined plane, 7 tensor, 60 Strut, 282 Surface finish factor, 125 forces, 125

Tensile and compressive stresses, 3 Theory of failure, 132 Thick shell, 225 Thin shell, 225 Torsional rigidity, 309 Tresca criterion, 96 Two-dimensional state of stress, 41

Unsymmetrical bending, 263

Variable stress, 124 von Mises theory, 102

Wahl’s correction factor, 312 Winkler-Bach theory, 196

Yield criterion, 96

Second Edition

STRENGTH OF MATERIALS A.K. Srivastava • P.C. Gope The book, now in the Second Edition, presents the fundamental principles of strength of materials and focuses on 3D analysis of stress and strain, double integration method, Macaulay’s method, moment area method and method for determining stresses using Winkler–Bach theory. It also covers the analyses of helical springs and leaf spring, and buckling analysis of columns and struts using Euler’s and Rankine’s theory. This edition includes four new chapters, namely Simple and Compound Stress, Theory of Failure, Energy Methods and Finite Element Method and its Applications Using ANSYS Software. The chapter on Analysis of Stress and Strain has been thoroughly revised. The text is primarily designed for the undergraduate students of mechanical engineering, production engineering, and industrial engineering. Besides students, practising engineers would also find the book useful. KEY FEATURES ◆ A large number of numerical problems ◆ Open-ended or synthesis-type examples wherever required ◆ Chapter-end exercises THE AUTHORS A.K. SRIVASTAVA (M.Tech., G.B. Pant University of Agriculture and Technology, Pantnagar) is Manager (Design) in Aircraft Upgrade Research and Design Centre, Hindustan Aeronautics Limited (Ministry of Defence), Nasik. Earlier, he was with RITES Ltd. (under the Ministry of Railways), and Dehradun Institute of Technology, Dehradun. Mr. Srivastava has also published ten technical research papers in reputed national and international journals. P.C. GOPE (Ph.D., National Institute of Technology Jamshedpur) is Professor in Mechanical Engineering at the College of Technology, G.B. Pant University of Agriculture and Technology, Pantnagar. With over two decades of teaching experience, he has published/presented 63 research papers in reputed national and international journals/conferences. A recipient of the Sir Rajendra Nath Mukherjee Memorial Award for his best research paper by the Institution of Engineers (India) in 2004, Dr. Gope has worked on five industry and CSIR-sponsored research and investigative projects.

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