advanced-mechanics-of-solids-by-l-s-srinath-.pdf

About the Author LS Srinath received his PhD from Illinois Institute of Technology, Chicago, and has served as Professor of Mechanics and Aerospace Engineering at University of Kansas, Lawrence. He is a distinguished alumnus of IIT, Chicago. He has also served as Professor of Mechanical Engineering at the Indian Institute of Technology, Kanpur, and also the Indian Institute of Science, Bangalore. Besides these, he was the Director of Indian Institute of Technology, Madras, Chennai. Professor Srinath has authored several books and papers and has been on Advisory Boards of quite a few national bodies dealing with engineering education. Advanced Mechanics of SOLIDS Third Edition L S Srinath Former Director Indian Institute of Technology Madras Chennai Tata McGraw-Hill Publishing Company Limited NEW DELHI McGraw-Hill Offices New Delhi New York St Louis San Francisco Auckland Bogotá Caracas Kuala Lumpur Lisbon London Madrid Mexico City Milan Montreal San Juan Santiago Singapore Sydney Tokyo Toronto Tata McGraw-Hill Published by the Tata McGraw-Hill Publishing Company Limited, 7 West Patel Nagar, New Delhi 110 008. Copyright © 2009 by Tata McGraw-Hill Publishing Company Limited. No part of this publication may be reproduced or distributed in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise or stored in a database or retrieval system without the prior written permission of the publishers. The program listings (if any) may be entered, stored and executed in a computer system, but they may not be reproduced for publication. This edition can be exported from India only by the publishers, Tata McGraw-Hill Publishing Company Limited. ISBN 13: 978-0-07-13988-6 ISBN 10: 0-07-13988-1 Managing Director: Ajay Shukla General Manager: Publishing—SEM & Tech Ed: Vibha Mahajan Sponsoring Editor: Shukti Mukherjee Jr. Editorial Executive: Surabhi Shukla Executive—Editorial Services: Sohini Mukherjee Senior Production Manager: P L Pandita General Manager: Marketing—Higher Education & School: Michael J Cruz Product Manager: SEM & Tech Ed: Biju Ganesan Controller—Production: Rajender P Ghansela Asst. General Manager—Production: B L Dogra Information contained in this work has been obtained by Tata McGraw-Hill, from sources believed to be reliable. However, neither Tata McGraw-Hill nor its authors guarantee the accuracy or completeness of any information published herein, and neither Tata McGraw-Hill nor its authors shall be responsible for any errors, omissions, or damages arising out of use of this information. This work is published with the understanding that Tata McGraw-Hill and its authors are supplying information but are not attempting to render engineering or other professional services. If such services are required, the assistance of an appropriate professional should be sought. Typeset at Astral Pre Media Pvt. Ltd., A-39, Sector-58, Noida (UP) and printed at Rashtriya Printers, M-135, Panchsheel Garden, Naveen Shahdara, Delhi 110 032 Cover: Rashtriya Printers RCXDCRQXRYDXL The McGraw-Hill Companies Contents Preface List of Symbols SI Units (Systeme International d’Unit’es) Typical Physical Constants (As an Aid to Solving Problems) 1. Analysis of Stress 1.1 Introduction 1 1.2 Body Force, Surface Force and Stress Vector 2 1.3 The State of Stress at a Point 4 1.4 Normal and Shear Stress Components 4 1.5 Rectangular Stress Components 4 1.6 Stress Components on an Arbitrary Plane 6 1.7 Digression on Ideal Fluid 11 1.8 Eqality of Cross Shears 11 1.9 A More General Theorem 13 1.10 Principal Stresses 14 1.11 Stress Invariants 16 1.12 Principal Planes are Orthogonal 17 1.13 Cubic Equation has Three Real Roots 17 1.14 Particular Cases 19 1.15 Recapitulation 19 1.16 The State of Stress Referred to Principal Axes 24 1.17 Mohr’s Circles for the Three-Dimensional State of Stress 25 1.18 Mohr’s Stress Plane 26 1.19 Planes of Maximum Shear 28 1.20 Octahedral Stresses 29 1.21 The State of Pure Shear 31 1.22 Decomposition into Hydrostatic and Pure Shear States 31 1.23 Cauchy’s Stress Quadric 34 1.24 Lame’s Ellipsoid 36 1.25 The Plane State of Stress 38 1.26 Differential Equations of Equilibrium 40 1.27 Equilibrium Equations for Plane Stress State 42 1.28 Boundary Conditions 45 1.29 Equations of Equilibrium in Cylindrical Coordinates 45 v xiii xv xvi 1 vi Contents 1.30 Axisymmetric Case and Plane Stress Case 48 Problems 49 Appendix 1 Mohr’s Circles 54 Appendix 2 The State of Pure Shear 56 Appendix 3 Stress Quadric of Cauchy 60 2. Analysis of Strain 2.1 Introduction 63 2.2 Deformations 64 2.3 Deformation in the Neighbourhood of a Point 65 2.4 Change in Length of a Linear Element 67 2.5 Change in Length of a Linear Element—Linear Components 69 2.6 Rectangular Strain Components 70 2.7 The State of Strain at a Point 70 2.8 Interpretation of gxy, gyz, gxz as Shear Strain Components 71 2.9 Change in Direction of a Linear Element 73 2.10 Cubical Dilatation 74 2.11 Change in the Angle between Two Line Elements 77 2.12 Principal Axes of Strain and Principal Strains 78 2.13 Plane State of Strain 83 2.14 The Principal Axes of Strain Remain Orthogonal after Strain 84 2.15 Plane Strains in Polar Coordinates 85 2.16 Compatibility Conditions 86 2.17 Strain Deviator and its Invariants 90 Problems 91 Appendix on Compatibility Conditions 94 63 3. Stress–Strain Relations for Linearly Elastic Solids 3.1 Introduction 97 3.2 Generalised Statement of Hooke’s Law 97 3.3 Stress–Strain Relations for Isotropic Materials 98 3.4 Modulus of Rigidity 99 3.5 Bulk Modulus 101 3.6 Young’s Modulus and Poisson’s Ratio 102 3.7 Relations between the Elastic Constants 102 3.8 Displacement Equations of Equilibrium 104 Problems 107 97 4. Theories of Failure or Yield Criteria and Introduction to Ideally Plastic Solid 4.1 Introduction 109 4.2 Theories of Failure 110 4.3 Significance of the Theories of Failure 117 4.4 Use of Factor of Safety in Design 121 4.5 A Note on the use of Factor of Safety 124 4.6 Mohr’s Theory of Failure 129 4.7 Ideally Plastic Solid 132 4.8 Stress Space and Strain Space 134 109 vii Contents 4.9 4.10 4.11 4.12 4.13 General Nature of the Yield Locus 135 Yield Surfaces of Tresca and Von Mises 136 Stress–Strain Relations (Plastic Flow) 137 Prandtl–Reuss Equations 139 Saint Venant–Von Mises Equations 140 Problems 140 5. Energy Methods 5.1 Introduction 143 5.2 Hooke’s Law and the Principle of Superposition 143 5.3 Corresponding Force and Displacement or Work-Absorbing Component of Displacement 145 5.4 Work Done by Forces and Elastic Strain Energy Stored 146 5.5 Reciprocal Relation 147 5.6 Maxwell–Betti–Rayleigh Reciprocal Theorem 148 5.7 Generalised Forces and Displacements 149 5.8 Begg’s Deformeter 152 5.9 First Theorem of Castigliano 153 5.10 Expressions for Strain Energy 155 5.11 Fictitious Load Method 161 5.12 Superposition of Elastic Energies 163 5.13 Statically Indeterminate Structures 164 5.14 Theorem of Virtual Work 166 5.15 Kirchhoff ’s Theorem 169 5.16 Second Theorem of Castigliano or Menabrea’s Theorem 170 5.17 Generalisation of Castigliano's Theorem or Engesser’s Theorem 173 5.18 Maxwell–Mohr Integrals 176 Problems 181 143 6. Bending of Beams 6.1 Introduction 189 6.2 Straight Beams and Asymmetrical Bending 190 6.3 Regarding Euler–Bernoulli Hypothesis 198 6.4 Shear Centre or Centre of Flexure 201 6.5 Shear Stresses in Thin-Walled Open Sections: Shear Centre 6.6 Shear Centres for a Few Other Sections 208 6.7 Bending of Curved Beams (Winkler-Bach Formula) 209 6.8 Deflections of Thick Curved Bars 216 Problems 223 189 7. Torsion 7.1 Introduction 230 7.2 Torsion of General Prismatic Bars–Solid Sections 7.3 Alternative Approach 236 7.4 Torsion of Circular and Elliptical Bars 240 7.5 Torsion of Equilateral Triangular Bar 243 7.6 Torsion of Rectangular Bars 245 202 230 232 viii Contents 7.7 7.8 7.9 7.10 7.11 7.12 7.13 Membrane Analogy 248 Torsion of Thin-Walled Tubes 249 Torsion of Thin-Walled Multiple-Cell Closed Sections Torsion of Bars with Thin Rectangular Sections 255 Torsion of Rolled Sections 256 Multiply Connected Sections 259 Centre of Twist and Flexural Centre 264 Problems 265 251 8. Axisymmetric Problems 8.1 Introduction 269 8.2 Thick-Walled Cylinder Subjected to Internal and External Pressures—Lame’s Problem 271 8.3 Stresses in Composite Tubes—Shrink Fits 280 8.4 Sphere with Purely Radial Displacements 287 8.5 Stresses Due to Gravitation 292 8.6 Rotating Disks of Uniform Thickness 294 8.7 Disks of Variable Thickness 298 8.8 Rotating Shafts and Cylinders 300 8.9 Summary of Results for use in Problems 303 Problems 305 269 9. Thermal Stresses 310 9.1 Introduction 310 9.2 Thermoelastic Stress–Strain Relations 311 9.3 Equations of Equilibrium 311 9.4 Strain–Displacement Relations 312 9.5 Some General Results 312 9.6 Thin Circular Disk: Temperature Symmetrical about Centre 314 9.7 Long Circular Cylinder 316 9.8 The Problem of a Sphere 320 9.9 Normal Stresses in Straight Beams due to Thermal Loading 323 9.10 Stresses in Curved Beams due to Thermal Loading 325 Problems 328 10. Elastic Stability 10.1 Euler’s Buckling Load 331 I. Beam Columns 335 10.2 Beam Column 335 10.3 Beam Column Equations 335 10.4 Beam Column with a Concentrated Load 336 10.5 Beam Column with Several Concentrated Loads 339 10.6 Continuous Lateral Load 340 10.7 Beam-Column with End Couple 342 II. General Treatment of Column Stability Problems (As an Eigenvalue Problem) 344 10.8 General Differential Equation and Specific Examples 344 331 Contents 10.9 10.10 III. 10.11 10.12 10.13 10.14 10.15 10.16 10.17 10.18 ix Buckling Problem as a Characteristic Value (Eigenvalue) Problem 350 The Orthogonality Relations 352 Energy Methods for Buckling Problems 355 Theorem of Stationary Potential Energy 355 Comparison with the Principle of Conservation of Energy 357 Energy and Stability Considerations 358 Application to Buckling Problems 359 The Rayleigh–Ritz Method 360 Timoshenko’s Concept of Solving Buckling Problems 364 Columns with Variable Cross-Sections 366 Use of Trigonometric Series 368 Problems 371 11. Introduction to Composite Materials 11.1 Introduction 374 11.2 Stress–Strain Relations 375 11.3 Basic Cases of Elastic Symmetry 377 11.4 Laminates 381 11.5 Ply Stress and Ply Strain 404 11.6 Failure Criteria of Composite Materials 406 11.7 Micromechanics of Composites 411 11.8 Pressure Vessels 421 11.9 Transverse Stresses 422 Problems 424 374 12. Introduction to Stress Concentration and Fracture Mechanics 428 I. 12.1 12.2 12.3 12.4 12.5 12.6 II. 12.7 12.8 12.9 12.10 12.11 12.12 12.13 12.14 12.15 12.16 Stress Concentration 428 Introduction 428 Members under Tension 429 Members under Torsion 439 Members under Bending 443 Notch Sensitivity 445 Contact Stresses 446 Fracture Mechanics 457 Brittle Fracture 457 Stress Intensity Factor 458 Fracture Toughness 460 Fracture Conditions 462 Fracture Modes 464 Plane Stress and Plane Strain 468 Plastic Collapse at a Notch 471 Experimental Determination of KIc 475 Strain-Energy Release Rate 476 Meaning of Energy Criterion 479 x 12.17 12.18 12.19 12.20 12.21 12.22 12.23 12.24 12.25 Contents Design Consideration 482 Elasto-Plastic Fracture Mechanics (EPFM) Plane Body 485 Green’s Thorem 486 The J-Integral 486 Path Independence of the J-Integral 487 J-Integral as a Fracture Criterion 489 ASTM-standard Test for JIC 490 Relationships of KC, GC, and J 491 Problems 491 482 Appendix 494 Index 500 Kirchhoff. . and Maxwell–Mohr integrals. and Stress– Strain relations for isotropic solids. The discussions in this chapter are important because of their applicability to a wide variety of problems. The contents of these chapters provide a firm foundation to the mechanics of deformable solids which will enable the student to analyse and solve a variety of strength-related design problems encountered in practice. Chapter 5 deals with energy methods. a chapter on composite materials was also added and this addition was well received. Mohr’s theory of failure has been considerably enlarged because of its practical application. Castigliano. still. The treatment starts with Analysis of stress. many suggestions were being received to add several specialized topics. Chapter 4 dealing with Theories of Failure or Yield Criteria is a general departure from older texts. is highly desirable. based on many suggestions received. the first five chapters deal with the general analysis of mechanics of deformable solids. Several worked examples illustrate the applications of these theorems. Engesser. Instances are many where equations presented in handbooks are used to solve practical problems without examining whether the conditions under which those equations were obtained are satisfied or not. Also. Analysis of strain. which is one of the important topics and hence. Since this is a second-level course addressed to senior level students. depending on the material properties.Preface The present edition of the book is a completely revised version of the earlier two editions. The coverage is exhaustive and discusses the theorems of Virtual Work. Menabria. is discussed in great detail. a few topics due to their importance needed to be included and a new edition became necessary. This treatment is brought earlier because. The second reason is to bring into focus the assumptions made in obtaining several basic equations. The second edition provided an opportunity to correct several typographical errors and wrong answers to some problems. While it was difficult to accommodate all suggestions in a book of this type. These chapters are quite exhaustive and include materials not usually found in standard books. an understanding of the possible factors for failure. in addition. As in the earlier editions. in applying any design equation in strength related problems. elliptical. Torsion is covered in great detail in Chapter 7. their equivalents in SI units are also given. are covered with adequate number of worked examples. failure criteria for composites. off-axis loading. Using the theory-of-elasticity approach. a few can be found with kgf. Stresses and deformations caused in bodies due to thermal gradients need special attention because of their frequent occurrences. Stress concentration and fracture are important considerations in engineering design. Chapter 12 provides a fairly good coverage with a sufficient number of worked examples. Analysis of axisymmetric problems like composite tubes under internal and external pressures. In addition to topics on Beam Columns. However. Elastic instability problems are covered in Chapter 10. equilateral triangular bars. Timoshenko. are discussed. problems in these aspects are discussed in books solely devoted to these. use of trigonometric series. angle-ply and cross-ply laminates. This chapter provides a good foundation to this topic. Usually. these problems are treated in books on Thermoelasticity. is not found in standard texts.mhhe. in spite of its importance. Modern– day engineering practices and manufacturing industries make use of a variety of composites. kgf is used for force and weight measurements.. While SI units are used in most of numerical examples and problems. Chapter 9 in this book covers thermal stress problems. This chapter also discusses the validity of Euler–Bernoulli hypothesis in the derivations of beam equations. Centre of flexure. The subject material is a natural extension from isotropic solids to anisotropic solids. this chapter exposes the student to the instability problem as an eigenvalue problem. etc. meter and second units. The analysis of thermal stress problems are not any more complicated than the traditional problems discussed in books on Advanced Mechanics of Solids. Introduction to the mechanics of composites is found in Chapter 11.. Curved Beams. rotating disks. Orthotropic materials.com/srinath/ams3e and it contains the following material: For Instructors Solution Manual PowerPoint Lecture Slides . effects of Poisson’s ratio. to solve buckling problems find their place in this chapter. Energy methods as those of Rayleigh–Ritz. thin-walled multiple cell sections. shafts and cylinders can be found in Chapter 8. This is done deliberately to make the student conversant with the use of both sets of units since in daily life.. Torsion of circular.xii Preface Bending of beams. etc. Another notable inclusion in this chapter is the torsion of bars with multiply connected sections which. etc. In those problems where kgf units are used. are covered in Chapter 6.. The web supplements can be accessed at http://www. Several practical problems can be solved with confidence based on the treatment provided. etc. This is an important concept that a student has to appreciate. a good introduction to these important topics can be provided in a book of the present type. SRM Institute of Science and Technology. L S SRINATH . Ms Shukti Mukherjee. NIT Jamshedpur. Their names are given below. Chennai. I wish to thank my family members for their patience. Ms Sohini Mukherjee and Mr P L Pandita. Guwahati. Assam P K Sarkar Department of Mechanical Engineering. Tamil Nadu E V M Sargunar Sree Sastha Institute of Engineering and Technology. Mumbai. SSVPS’s B S Deora College of Engineering. New Delhi P Venkitanarayanan Department of Applied Mechanics. I am also thankful to the staff at McGraw-Hill Education India. K S R K Murthy Department of Mechanical Engineering. Indian Institute of Technology. Bihar S R Pandey Department of Applied Mechanics. Lastly. especially Ms Vibha Mahajan. National Institute of Technology. Indian Institute of Technology. Uttar Pradesh G Rajesh Kumar Department of Mechanical Engineering. Dhule. Patna. Feedback and suggestions are always welcome at srinath_ls@sify. Ms Surabhi Shukla. Rizvi College of Engineering. Indian Institute of Technology. Tamil Nadu In addition to this. Bihar M K Singha Department of Applied Mechanics. Maharashtra D Prasanna Venkatesh Department of Mechanical Engineering.com. Chennai. Jharkhand Dr Amit Kumar Department of Mechanical Engineering. Kanpur. Dhanbad. Indian School of Mines. for their cooperation during the different stages of this project.xiii Preface For Students MCQ’s (interactive) Model Question Papers I am thankful to all the reviewers who took out time to review this book and gave me their suggestions. support and love given to me during the preparation of this manuscript. Maharashtra C A Akhadkar Department of Mechanical Engineering. yz. y. gzx wx. wz D = exx+ eyy+ ezz e 1 . I2 . z-plane shear stress on x-plane in y-direction. Eyz. z directions principal stresses at a point first. Ezz exx. axial (polar) direction spherical coordinates shear stresses in polar coordinates displacements in x. n z s 1. J 2. circumferential. z t xy. z directions linear strains in x-direction. J 3 normal stress force force vector on a plane with normal n components of force vector in x. y. y. t qz u x. s 2. zx n x. ny. second. t gz. third invariants of strain . eyy. I3 s oct t oct s r.List of Symbols (In the order they appear in the text) s Fn T n T x. z axes cubical dilatation principal strains at a point first. y. second. Eyy. third invariants of stress normal stress on octahedral plane shear stress on octahedral plane normal stresses in radial. e2 . gyz. e 3 J 1. wy. uy. y-plane. s 3 I1 . shear stress on y-plane in z-direction. y. s q. u z Exx. y. z A A t s x. j t gq. s z g. y-direction. z-direction (with non-linear terms) linear strains (with linear terms only) shear strain components (with non-linear terms) shear strain components (with linear terms only) rigid body rotations about x. shear stress on z-plane in x-direction direction cosines of n in x. ezz Exy. q. z directions area of section normal to the section shear stress normal stress on x-plane. Ezx gxy. circumferential. w t Kt N f r D. y. m G=m m E K P u sy U U* s ut sct a ij b ij Mx. eq. Iy . z axes linear deflection. I z Ir Ixy. axial directions Lame’s constants rigidity modulus engineering Poisson’s ratio modulus of elasticity bulk modulus. z axes polar moment of inertia products of inertia about xy and yz coordinates torque. Iyz T Y a Q R V n ij b. temperature warping function coefficient of thermal expansion lateral load axial load elastic potential Poisson’s ratio in i-direction due to stress in j-direction width thickness theoretical stress concentration factor normal force stream function fillet radius radii notch sensitivity fracture toughness in mode I offset yield stress angular velocity fracture resistance fracture stress boundary J-integral . y. d q Kc. My. ez l. complementary energy ultimate stress in uniaxial tension ultimate stress in uniaxial compression influence coefficient.List of Symbols eg. generalized deflection moments of inertia about x. KIc Sy w R sfr G J xv strains in radial. material constant compliance component moments about x. stress intensity factor pressure Poisson’s ratio yield point stress elastic energy distortion energy. Mz d I x . 000 001 0.066 0.010197 0.746 1 .8066 9.SI Units (Systeme International d’Unit’es) (a) Base Units Quantity Unit (Symbol) length mass time force pressure meter (m) kilogram (kg) second (s) newton (N) pascal (Pa) force is a derived unit: kgm/s2 pressure is force per unit area: N/m2: kg/ms2 kilo-watt is work done per second: kNm/s (b) Multiples giga (G) mega (M) kilo (k) milli (m) micro (m) nano (n) 1 000 000 000 1 000 000 1 000 0.10197 1 0.001 0.8066 ¥ 104 98.000 000 001 (c) Conversion Factors To Convert to Multiply by kgf kgf/cm2 kgf/cm2 newton Pa kPa HP HP kW newton Pa kPa kgf N/m2 kgf/cm2 kW kNm/s kNm/s 9.746 0. Typical Physical Constants (As an Aid to Solving Problems) Material Ultimate Strength Yield Strength (MPa) (MPa) Elastic Poisson’s Coeff.7 17. gray Carbon steel Stainless steel 414 210 690 568 414 825 690 568 221 — 552 — 300 — 415 276 170 — 250 — 73 90 200 207 28 41 83 90 per °C ¥ 10–6 0.0 . Comp Shear Tens or Shear Tens Shear Comp Aluminium alloy Cast iron.292 0.2 10.334 0. Modulus Ratio Therm (GPa) Expans. Tens.4 11.211 0.291 For more accurate values refer to hand-books on material properties 23. In books on the theory of elasticity. In reality. in the limit we shall not come across a void. all materials are composed of many discrete particles. however. is subjected to a tensile force F along the axis of the bar. For example. which are often microscopic. say a circular rod of diameter d.1 1 INTRODUCTION In this book we shall deal with the mechanics of deformable solids. Such a body . The concept of stress has already been introduced in the elementry strength of materials. which deals with the geometry of deformation without considering the forces that cause the deformation. an automobile crankshaft or a piston inside an engine cylinder or an aircraft wing are subject to loadings that are both complex as well as dynamic in nature. and the loads acting on it will also be complex. in the limit one may end up in a void. a structural member or a machine element will not possess uniform geometry of shape or size. When a bar of uniform cross-section. one is more familiar with forces. In such cases. so that if we consider a small volume element of the matter surrounding a point and shrink this volume. It will be assumed that the matter of the body that is being considered is continuously distributed over its volume.CHAPTER Analysis of Stress 1. and when an arbitrarily selected volume element is shrunk. though the measurement of force is usually done through the measurement of deformations caused by the force. one usually starts with the analysis of strain. The starting point for discussion can be either the analysis of stress or the analysis of strain. one will have to introduce the concept of the state of stress at a point and its analysis. But in our analysis. the average stress induced across any transverse section perpendicular to the axis of the bar and away from the region of loading is given by σ = F = 4 F2 Area π d It is assumed that the reader is familiar with the elementary flexural stress and torsional stress concepts. However. Books on the strength of materials. which will be the subject of discussion in this chapter. In general. we assume that the matter is continuously distributed. begin with the analysis of stress. we shall not deal with forces that vary with time. However. 1. are concentrated forces. the corresponding area 1 1 1 DT ¢ at P¢ is D A¢. 1.1. then p F2 each part is in equilibrium under the action of the externally applied forces and the internally distributed forces C across the interface.1 Body subjected to forces support reactions R 1. y. the surface force distribution o x is often termed surface traction.2 Advanced Mechanics of Solids is called a continuous medium and the mechanics of such a body or bodies is called continuum mechanics. the inertia force and the mag1 netic force. Let the body be cut into two parts C and D by a plane 1-1 passing through point P. These two areas are distinn¢ 1 DT guished by their outward drawn normals n DA P D 1 1 R1 R2 F3 Fig. 1 the ratio ∆T tends to a definite limit. z). F 3 . F r.2 BODY FORCE. Let P be a point inside the body with coordinates (x. The surface forces act P on the surface or area elements of 1 the body. the body will be subjected to two types of forces— p body forces and surface forces. Examples of this kind of force are the gravitational R3 force. . The F1 y F2 body forces act on each volume element of the body. In Fig. and ∆A 7/3/2008. We assume that as DA tends to zero. the body is in equilibrium. In general. 1. It is explicitly assumed that under the action of both body forces and surface forces. as shown in Fig.pmd 1 n and n ¢. 1. F 2. The action of part C on DA at point P can be represented by the force 2 vector DT and the action of part D on DA¢ at P¢ can be represented by the force vector 1 DT ¢. the surface forces F 1. When the area considered R1 lies on the actual boundary of the R2 F3 body. . 5:25 AM .2. In part C. R2 and R 3 are also surface forces. The Fig. In part D. 1. 1. as shown in Fig. z while p is a distributed force.2 Free-body diagram of a body cut into two parts Chapter_01. If we consider the F1 free-body diagrams of C and D. SURFACE FORCE AND STRESS VECTOR Consider a body B occupying a region of space referred to a rectangular coordinate system Oxyz.1. let DA DA¢ R3 be a small area surrounding the point P¢ P. We thus have 1 1 T = −T′ (1.Analysis of Stress 3 further. nz of the outward drawn normal). z) but also on the plane passing through the point (identified by direction cosines nx. ny.1 is cut by a different C2 2 plane 2-2 with outward drawn normals n and 2 P¢ 2 DT ¢ 2 n¢ 2 n 2 DT P D2 n ¢ passing through the same point P. then the stress vector representing the action 2 of C2 on D 2 will be represented by T (Fig. The limiting vector is written as 1 1 ∆T d T 1 = =T lim dA ∆ A→ 0 ∆ A (1. at point P¢. 2 ∆T T = lim = ∆A ∆ A→ 0 2 1 In general. y. Hence the stress at a point depends not only on the location of the point (identified by coordinates x. i. the moment of the forces acting on area D A about any point within the area vanishes in the limit.2) 1 Vectors T and T ′ are called the stress vectors and they represent forces per unit area acting respectively at P and P ¢ on planes with outward drawn normals 1 1 n and n ¢. . which is applicable to particles.3)). 1 We further assume that stress vector T representing the action of C on D 1 at P is equal in magnitude and opposite in direction to stress vector T ¢ representing the action of D on C at corresponding point P¢. This assumption is similar to Newton’s third law. (1.3 Body cut by another plane P on a plane with outward drawn normal n 2 will be different from stress vector T acting 2 at the same point P. stress vector T acting at point 1 Fig. 1.e.3) If the body in Fig.1) Similarly. but on a plane with outward drawn normal n . the action of part D on C as D A¢ tends to zero. can be represented by a vector 1 1 ∆T ′ d T ′ 1 = = T′ dA′ ∆ A′→0 ∆ A′ 0 lim 1 (1. 1. 1.4 Resultant stress vector. We n have. y. 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. This can be resolved into two components. we have n 2 n n n T = Tx2 + Ty2 + Tz2 (1. T z . y.Advanced Mechanics of Solids 4 1.3 THE STATE OF STRESS AT A POINT Since an infinite number of planes can be drawn through a point. each stress vector characterised by the corresponding plane on which it is acting. be cut by a plane parallel to the yz plane. being normal to the plane.5). therefore.5 n n n components are denoted by T x .5) RECTANGULAR STRESS COMPONENTS Let the body B. shown in Fig. This vector can be resolved into three components parallel to the x. will be denoted by sx (instead of by x s ).6 that if the stress vectors acting on three mutually perpendicular planes passing through the point are known. 1.4) T = σ2 + τ2 n n where T is the magnitude of the resultant stress. n Stress vector T can also be resolved into three components parallel to the x. z axes. The normal to this plane is parallel to the x axis and hence. . If these Fig.4). we can determine the stress vector acting on any other arbitrary plane at that point. z axes. It is the knowledge of this state of stress that is of importance to a designer in determining the critical planes and the respective critical stresses. the plane is called the x x plane. normal and shear stress components 1.1.1.4 NORMAL AND SHEAR STRESS COMPONENTS n Let T be the resultant stress vector at point P acting on a plane whose outward drawn normal is n (Fig. we get an infinite number of stress vectors acting at a given point. the relation: n Tx n 2 Tz tn n n (1. 1. one along the normal n and the other perpendicular to n. 1. 1. The n component parallel to n is called the normal stress n n s Ty and is generally denoted by σ . It will be shown in Sec. The components parallel to the y and z axes are shear stress components and are denoted by txy and txz respectively (Fig. The resultant stress vector at P acting on this will be T . T y . The component pern T pendicular to n is known as the tangential stress or n shear stress component and is denoted by τ . The component parallel to the x axis. 6. one should have denoted the normal stress component as txx. the y plane and the z plane. to distinguish between a normal stress and a shear stress. the x plane. tyz on y plane sz. To maintain x o consistency. tyx. 1. 1. the normal stress is denoted by s and the shear stress by t. However. Similarly. y sy tyx tyz txy tzy sx sx tzx txz sz tyx sy x z Fig. txy. The three hidden faces have their outward drawn normals . Consequently. For exn sx txz ample. At any point P. Following the notation mentioned above.6 Rectangular stress components One should observe that the three visible faces of the rectangular element have their outward drawn normals along the positive x. txz is the stress component on the x plane in z direction. the positive stress components on these faces will also be directed along the positive axes. txy is the stress component on the x plane in y direction.Analysis of Stress 5 y In the above designation. txz on x plane sy. the normal and shear stress components on these planes are sx.5 Stress components on x plane x plane in the x direction. 1. the first subscript x indicates the plane on which the stresses are acting and the x txy second subscript (y or z) indicates the T direction of the component. y and z axes respectively. tzx. one can draw three mutually perpendicular planes. This z would be the stress component on the Fig. tzy on z plane These components are shown acting on a small rectangular element surrounding the point P in Fig. e. 1.7 Tetrahedron at point P in Fig. The volume of the tetrahedron is equal to Ah where h is the 3 perpendicular distance from P to the inclined face. we shall speak in terms of the average stresses n over the faces. For equilibrium of the . Let the three mutually perpendicular planes be the x. 1 per unit volume. The positive stress components on these faces will. y and z. the bottom face has its outward drawn normal along the negative y axis. y Consider a small tetrahedron at P with three of its faces n normal to the coordinate Ty n axes. x and z axes. 1. The free-body diagram is shown Fig. then it will be in equilibrium under the C action of the surface forces sy z and the body forces.. Let h be the perpendicun n lar distance from P to the Tz Tx inclined face.6 STRESS COMPONENTS ON AN ARBITRARY PLANE It was stated in Section 1. For example. If A is the area of the inclined face then Area of BPC = projection of area ABC on the yz plane = Anx Area of CPA = projection of area ABC on the xz plane = Any Area of APB = projection of area ABC on the xy plane = Anz Let the body force components in x. therefore. txy. and the inclined face B having its normal parallel to sz n.7. txz. sy. y and z planes and let the arbitrary plane be identified by its outward drawn normal n whose direction cosines are nx.3 that a knowledge of stress components acting on three mutually perpendicular planes passing through a point will enable one to determine the stress components acting on any plane passing through that point. n y and nz. sz. i. tyz. be directed along the negative axes. the rectangular stress components are sx. 1. T y . On the three faces. This can be n n n resolved into components T x . Since the size of the tetrahedron considered is very small and in the limit as we are going to make h tend to zero. Hence.6 Advanced Mechanics of Solids in the negative x. T z . tyx and tyz are directed respectively along the negative y. tzx and tzy. sy. gy and gz respectively. parallel to the three axes x. y and z axes. If the tetrahedron is isolated from the P sx x body and a free-body diaA gram is drawn. tyx. y and z directions be gx. Let T be the resultant stress vector on face ABC. the positive stress components on this face. and the average stress components acting on the faces will tend to their respective values at point P acting on their corresponding planes.Analysis of Stress 7 tetrahedron. one can write Eq.9) n Tz = nx txz + ny tyz + nz sz Equation (1.11b) n Since the normal stress is equal to the projection of T along the normal. Consequently. tzy = tyz and tzx = txz. n Tx = sx nx + tyx ny + tzx nz – 1 hg 3 x (1. sy = tyy and sz = tzz.8 that among these nine rectangular stress components only six are independent.9) as n Ti = nx tix + ny tiy + nz tiz = ∑ nj tij (1. and sx = txx. This is because txy = tyx. for equilibrium in x direction n Tx A – sx Anx – tyx Any – tzx Anz + 1 Ahgx = 0 3 Cancelling A. for equilibrium in y and z directions n T y = txynx + sy ny + tzynz – n 1 hg (1.6) 1 hg 3 y (1. the oblique plane ABC will pass through point P.8) 3 z In the limit as h tends to zero. n n n sn = nx Tx + ny Ty + nz Tz (1. Ty and Tz along n.10) j where i and j can stand for x or y or z. This equation shows that the nine rectangular stress components at P will enable one to determine the stress components on any arbitrary plane passing through point P. This is known as the equality of cross shears. 1. n If T is the resultant stress vector on plane ABC. one gets from equations (1. (1.9) is known as Cauchy’s stress formula.8) Tz = txznx + tyzny + sznz – and n Tx = nx sx + ny tyx + nz tzx n Ty = nx txy + ny sy + nz tzy (1.11a) =Tx +Ty + Tz If sn and tn are the normal and shear stress components.12a) .6)–(1. we have n 2 T n2 n2 n2 (1. y and z directions must individually vanish. In anticipation of this result. it is n n n also equal to the sum of the projections of its components Tx . It will be shown in Sec. the sum of the forces in x. Thus.7) Similarly. we have n 2 T = σ n2 + τ n2 (1. Hence. 9) s n = nx2 sx + ny2 sy + nz2 sz + 2nxny txy + 2nynz tyz + 2nznx tzx (1.1 A rectangular steel bar having a cross-section 2 cm ¥ 3 cm is subjected to a tensile force of 6000 N (612.11b) and (1. 1. If the axes are chosen as shown in Fig. This is the normal stress s y.8 Example 1.n =0 2 z (ii) nx = 0. 000 N 2 /cm 4 τ n = 500 N/cm 2 (51 kgf/cm 2 ) . The other stress components are zero.12a) n n T y = 1000 .1 Solution Area of section = 2 ¥ 3 = 6 cm2.8 Advanced Mechanics of Solids n n n Substituting for Tx . ny = nx = (iii) nx = ny = nx = y 1 2 1 3 n T n F x (a) n T (b) (c) F z Fig. Ty and Tz from Eq.8.12b) Equation (1. 1. (i) Using Eqs (1. The average stress on this plane is 6000/6 = 1000 N/cm2.11) can then be used to obtain the value of tn Example 1.9). determine the normal and shear stresses on a plane whose normal has the following direction cosines: (i) nx = ny = 1 . n Tz = 0 σ n = 1000 = 500 N/cm 2 2 n 2 τ n2 = T − σ n2 = 250. (1. 2 T x = 0.2 kgf ). (1. 9 shows a cantilever beam in the form of a trapezium of uniform thickness loaded by a force P at the end. sy = –5. sx = 10. 2 Ty = 9 Tz = 0 σ n = 500 N/cm 2 . Determine the normal and shearing stresses on a plane that is equally inclined to all the three axes.000 N/cm2.4 kgf/cm 2 ) Example 1. txy = tyz = tzx = 10.000 N/cm2.3 Figure 1.000 N/cm2 (–510 kgf/cm2). and τ n = 500 N/cm2 (51 kgf/cm 2 ) n n n T y = 1000 .2 At a point P in a body.8) n Tx = n Ty = n Tz = n 2 \ T 1 3 1 3 1 3 3 N/cm2 (10000 + 10000 + 10000) = 10000 (10000 – 5000 + 10000) = –5000 3 N/cm2 (10000 – 10000 – 5000) = –5000 3 N/cm2 = 3 [(108) + (25 ¥ 106) + (25 ¥ 106)] N2/cm4 = 450 ¥ 106 N2/cm4 \ t n2 = 450 ¥ 106 – 400 ¥ 106 = 50 ¥ 106 N2/cm4 or tn = 7000 N/cm2 (approximately) Example 1. If it is assumed that the bending stress on any vertical section of the beam is distributed according to the elementary . Tz = 0 σ n = 1000 N/cm 2 3 τ n = 817 N/cm2 (83. n 1000 .Analysis of Stress n n (ii) T x = 0. Solution A plane that is equally inclined to all the three axes will have n x = n y = nz = 1 3 since nx2 + n 2y + nz2 = 1 From Eq. 3 (iii) T x = 0.6)–(1. sz = –5.12) 1 [10000 – 5000 – 5000 + 20000 + 20000 + 20000] 3 = 20000 N/cm2 sn = From Eqs (1.000 N/cm2 (1020 kgf/cm2). (1. But on plane AB at point A there can exist a shear stress. let axes x and y be chosen along and perpendicular to the edge. the plane perpendicular to edge EF. the resultant stress is along the normal (i. On the x plane. x axis).9(b).9(c) and (d). . 1. i.9 Example 1.. These are shown in Fig.10 Advanced Mechanics of Solids flexure formula.9). The normal to plane AB makes an angle q with the x axis.e. Let the normal and shearing stresses on this plane be s1 and t1. 1. τ xy = τ yz = τ zx = 0 The direction cosines of the normal to plane AB are nx = cos θ . where s 1 is the flexural stress Mc . 1. n y = sin θ .3 Solution At point A. as shown in Fig. We have σx = σ. 1.e. There is no shear stress on this plane since the top edge is a free surface (see Sec. nz = 0 The components of the stress vector acting on plane AB are n Tx = s1 = nx sx + ny tyx + nz tzy = s cosq n Ty = nx txy + ny sy + nz tzy = 0 n Tz = nx txz + ny tyz + nz s z = 0 n n n Therefore. σ y = σ z = 0. I y s1 A s x E P A A F q c s1 B B (a) s s1 c (b) t1 A s s1 t 1 A s1 (d) (c) Fig. show that the normal stress s on a section perpendicular to the top edge of the beam at point A is σ1 cos 2 θ . the normal stress on plane AB = σ n = nx T x + n y T y + nz T z = s cos2 q. . This aspect will be discussed again in Sec. the resultant stress on plane AB is n 2 T Hence n n n = T 2x + T 2y + T 2z = σ 2 cos 2 θ t 2 = s 2 cos2q – sn2 = s 2 cos2q – s 2 cos4q or 1 s sin 2q 2 t= 1. tzx and tzy. only six are independent.14. (1. txy. tyx. sz.7 DIGRESSION ON IDEAL FLUID By definition. T y = − pn y . one concludes that the resultant stress vector on any plane is normal and is equal to –p. n T z = − pnz (1. one has from Eqs. tyz. Dy and D z (Fig. then n n T x = − pnx . an ideal fluid cannot sustain any shearing forces and the normal force on any surface is compressive in nature. 1. p ≥ 0 n n The rectangular components of T are obtained by taking the projections of T along the x. The word isotropy means ‘independent of orientation’ or ‘same in all directions’. ny and nz are the direction cosines of n. This is the type of stress that a small sphere would experience when immersed in a liquid.9). This is because txy =tyx. This can be represented by n T = − pn.14) Comparing Eqs (1. Consider an infinitesimal rectangular parallelpiped surrounding point P. the state of stress at a point where the resultant stress vector on any plane is normal to the plane and has the same magnitude is known as a hydrostatic or an isotropic state of stress.14) sx = sy = sz = –p Since plane n was chosen arbitrarily. Hence. T y = ny σy . 1. n T z = nz σ z (1.13) and (1. sy.Analysis of Stress 11 Since sn = s1 s = σ1 cos 2 θ = Mc I cos 2 θ Further. txz.13) Since all shear stress components are zero.10). 1.8 EQUALITY OF CROSS SHEARS We shall now show that of the nine rectangular stress components sx. These are known as cross-shears. n n T x = nx σ x . tyz = tzy and tzx = txz. If nx. Let the dimensions of the sides be D x. y and z axes. 12 Advanced Mechanics of Solids tyy + Dtyy Y Dz tyx + Dtyx txy + Dtxy txx Dy tzx + Dtzx txx + Dtxx txz + Dtxz X txy Dx tyy Z Fig. 1.10 Stress components on a rectangular element Since the element considered is small, we shall speak in terms of average stresses over the faces. The stress vectors acting on the faces are shown in the figure. On the left x plane, the stress vectors are txx, txy and txz. On the right face, the stresses are txx + Dtxx, txy + Dtxy and txz + Dtxz. These changes are because the right face is at a distance D x from the left face. To the first order of approximation we have ∆τ xx = ∂τ xx ∆ x, ∂x ∆τ xy = ∂τ xy ∆x , ∂x ∆τ xz = ∂τ xz ∆x ∂x Similarly, the stress vectors on the top face are tyy + Dtyy, tyx + Dtyx and tyz + Dtyz, where ∆τ yy = ∂τ yy ∆y, ∂y ∆τ yx = ∂τ yx ∆y, ∂y ∆τ yz = ∂τ yz ∆y ∂y On the rear and front faces, the components of stress vectors are respectively tzz, tzx, tzy tzz + Dtzz, tzx + Dtzx, tzy + Dtzy where ∆τ zz = ∂τ zz ∆ z, ∂z ∆τ zx = ∂τ zx ∆ z, ∂z ∆τ zy = ∂τ zy ∆z ∂z For equilibrium, the moments of the forces about the x, y and z axes must vanish individually. Taking moments about the z axis, one gets txx Dy D z ∆y ∆y – (txx + Dtxx) Dy Dz + 2 2 (txy + Dtxy) Dy D z D x – tyy D x D z (tyy + Dtyy) D x D z ∆x + 2 ∆x – (tyx + Dtyx) D x D z Dy + 2 Analysis of Stress tzy Dx Dy 13 ∆x ∆x ∆y – tzx Dx Dy – (tzy + Dtzy) Dx Dy + 2 2 2 (tzx + Dtzx) D x Dy ∆y =0 2 Substituting for Dtxx, Dtxy etc., and dividing by Dx Dy Dz − ∂τ xy ∂τ yy ∆y ∂τ xx ∆y + τ xy + ∆x + − ∂x 2 ∂x ∂y 2 τ yx − ∂τ yx ∂τ zy ∆x ∂τ zx ∆y ∆y − + =0 ∂y ∂z 2 ∂z 2 In the limit as D x, Dy and Dz tend to zero, the above equation gives txy = tyx. Similarly, taking moments about the other two axes, we get tyz = tzy and tzx = txz. Thus, the cross shears are equal, and of the nine rectangular components, only six are independent. The six independent rectangular stress components are sx, sy, sz, txy, tyz and tzx. 1.9 A MORE GENERAL THEOREM The fact that cross shears are equal can be used to prove a more general theorem which states that if n and n¢ define two planes (not necessarily orthogonal but in the limit n¢ passing through the same point) n′ with corresponding stress vectors T n n T and T , then the projection of n n′ T along n¢ is equal to the pro- T ⋅n n′ n′ n′ T ⋅ n′ n′ jection of T along n, i.e. T ◊ n = T ◊ n (see Fig. 1.11). n The proof is straightforward. If T n nx′ , n y′ and nz′ are the direction cosines of n¢, then n T ◊ n n n n¢ = Tx nx′ + Ty n y′ + Tz nz′ From Eq. (1.9), substituting for Fig. 1.11 Planes with normals n and n¢ n n n Tx , Ty and Tz and regroup- ing normal and shear stresses n T ◊ n¢ = sx nx n¢x + sy ny n¢y + sz nz n¢z + txy nx n¢y + tyx ny n¢x + tyz ny n¢z + tzy nz n¢y + tzx nz n¢x + txz nx n¢z 14 Advanced Mechanics of Solids Using the result txy = tyx, tyz = tzy and tzx = tzx n T ⋅ n¢ = sx nx n¢x + sy ny n¢y + sz nz n¢z + txy (nx n¢y + ny n¢x) + tyz (ny n¢z + nz n¢y) + tzx (nz n¢x + nx n¢z) Similarly, n′ T ⋅ n = sx nx n¢x + sy ny n¢y + sz nz n¢z + txy (nx n¢y + ny n¢x) + tyz (ny n¢z + nz n¢y) + tzx (nz n¢x + nx n¢z) Comparing the above two expressions, we observe n n′ T ⋅ n′ = T ⋅ n (1.15) Note: An important fact is that cross shears are equal. This can be used to prove that a shear cannot cross a free boundary. For example, consider a beam of rectangular cross-section as shown in Fig. 1.12. y tyx = 0 A txy = 0 A x (a) Fig. 1.12 (b) (a) Element with free surface; (b) Cross shears being zero If the top surface is a free boundary, then at point A, the vertical shear stres component t xy = 0 because if t xy were not zero, it would call for a complementary shear t y x on the top surface. But as the top surface is an unloaded or a free surface, t yx is zero and hence, t xy is also zero (refer Example 1.3). 1.10 PRINCIPAL STRESSES We have seen that the normal and shear stress components can be determined on any plane with normal n, using Cauchy's formula given by Eqs (1.9). From the strength or failure considerations of materials, answers to the following questions are important: (i) Are there any planes passing through the given point on which the resultant stresses are wholly normal (in other words, the resultant stress vector is along the normal)? (ii) What is the plane on which the normal stress is a maximum and what is its magnitude? (iii) What is the plane on which the tangential or shear stress is a maximum and what it is its magnitude? Answers to these questions are very important in the analysis of stress, and the next few sections will deal with these. Let us assume that there is a plane n with Chapter_01.pmd 14 7/3/2008, 5:27 AM Analysis of Stress 15 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 n T = sn The components of this along the x, y and z axes are n T x = σ nx , n (1.16) n T y = σ ny , T z = σ nz (1.17) Also, from Cauchy’s formula, i.e. Eqs (1.9), n T x = sx nx + txy ny + txz nz n T y = txy nx + sy ny + tyz nz n T z = txz nx + tyz ny + sz nz Subtracting Eq. (1.17) from the above set of equations we get (sx – s) nx + txy ny + txz nz = 0 txy nx + (sy – s) ny + tyz nz = 0 (1.18) txz nx + tyz ny + (sz – s ) nz = 0 We can view the above set of equations as three simultaneous equations involving the unknowns nx, ny and nz. These direction cosines define the plane on which the resultant stress is wholly normal. Equation (1.18) is a set of homogeneous equations. The trivial solution is nx = ny = nz = 0. For the existence of a non-trivial solution, the determinant of the coefficients of nx, ny and nz must be equal to zero, i.e. (σ x − σ ) τ xy τ xz (σ τ xy y −σ τ yz ) τ xz τ yz =0 (σ z − σ ) (1.19) Expanding the above determinant, one gets a cubic equation in s as s3 – (sx + sy + sz)s 2 + (sx sy + sy sz +sz sx – τ x2 y − τ 2y z − τ z2 x ) s – 2 2 2 (sx sy sz + 2txy tyz tzx – sx τ yz − σ y τ xz − σ z τ xy ) = 0 (1.20) The three roots of the cubic equation can be designated as s1, s2 and s3. It will be shown subsequently that all these three roots are real. We shall later give a method (Example 4) to solve the above cubic equation. Substituting any one of these three solutions in Eqs (1.18), we can solve for the corresponding nx, ny and nz. In order to avoid the trivial solution, the condition. nx2 + n 2y + nz2 = 1 (1.21) is used along with any two equations from the set of Eqs (1.18). Hence, with each s there will be an associated plane. These planes on each of which the stress vector is wholly normal are called the principal planes, and the corresponding 16 Advanced Mechanics of Solids stresses, the principal stresses. Since the resultant stress is along the normal, the tangential stress component on a principal plane is zero, and consequently, the principal plane is also known as the shearless plane. The normal to a principal plane is called the principal stress axis. 1.11 STRESS INVARIANTS The coefficients of s 2, s and the last term in the cubic Eq. (1.20) can be written as follows: l 1 = sx + sy + sz (1.22) 2 2 − τ 2yz − τ zx l 2 = sxsy + sysz + szsx – τ xy = σx τ xy τ xy σy + σy τ yz τ yz σz + σ x τ xz τ xz σ z (1.23) 2 2 l 3 = sx sy sz + 2txy tyz tzx – sx τ 2yz – sy τ zx – sz τ xy σx τ xy τ zx = τ xy σy τ yz τ zx τ yz σz (1.24) Equation (1.20) can then be written as s3 – l1s 2 + l2s – l3 = 0 The quantities l 1, l 2 and l 3 are known as the first, second and third invariants of stress respectively. An invariant is one whose value does not change when the frame of reference is changed. In other words if x¢, y¢, z¢, is another frame of reference at the same point and with respect to this frame of reference, the rectangular stress competence are σ x′ , σ y ′ , σ z ′ , τ x′ y ′ , τ y ′z ′ and τ z ′x ′ , then the values of l 1, l 2 and l 3, calculated as in Eqs (1.22) – (1.24), will show that sx + sy + sz = sx¢ + sy¢ + sz¢ i.e. and similarly, l1 = l′1 l2 = l2′ and l3 = l3′ The reason for this can be explained as follows. The principal stresses at a point depend only on the state of stress at that point and not on the frame of reference describing the rectangular stress components. Hence, if xyz and x¢y¢z¢ are two orthogonal frames of reference at the point, then the following cubic equations s 3 – l1 s 2 + l 2 s – l 3 = 0 and s 3 – l¢1 s 2 + l¢2 s – l¢3 = 0 Analysis of Stress 17 must give the same solutions for s. Since the two systems of axes were arbitrary, the coefficients of s 2, and s and the constant terms in the two equations must be equal, i.e. l1 = l1′, l2 = l2′ and l3 = l3′ In terms of the principal stresses, the invariants are l 1 = s1 + s2 + s3 l 2 = s 1s 2 + s 2 s 3 + s 3 s 1 l 3 = s1 s 2 s 3 1.12 PRINCIPAL PLANES ARE ORTHOGONAL The principal planes corresponding to a given state of stress at a point can be shown to be mutually orthogonal. To prove this, we make use of the general theorem in Sec. 1.9. Let n and n¢ be the two principal planes and s1 and s2, the corresponding principal stresses. Then the projection of s1 in direction n¢ is equal to the projection of s2 in direction n, i.e. s1n¢ ◊ n = s2n ◊ n¢ (1.25) If nx , ny and nz are the direction cosines of n, and n¢x, n¢y and n¢z those of n¢, then expanding Eq. (1.25) ( ) ( σ1 nx nx′ + n y n ′y + nz nz′ = σ 2 nx nx′ + n y n ′y + nz nz′ ) Since in general, s1 and s2 are not equal, the only way the above equation can hold is nx nx′ + n y n y′ + nz nz′ = 0 i.e. n and n¢ are perpendicular to each other. Similarly, considering two other planes n¢ and n¢¢ on which the principal stresses s2 and s3 are acting, and following the same argument as above, one finds that n¢ and n¢¢ are perpendicular to each other. Similarly, n and n¢¢ are perpendicular to each other. Consequently, the principal planes are mutually perpendicular. 1.13 CUBIC EQUATION HAS THREE REAL ROOTS In Sec. 1.10, it was stated that Eq. (1.20) has three real roots. The proof is as follows. Dividing Eq. (1.20) by s 2, l l σ − l1 + 2 − 32 = 0 σ σ For appropriate values of s, the quantity on the left-hand side will be equal to zero. For other values, the quantity will not be equal to zero and one can write the above function as l l 2 (1.26) − 3 = f (σ ) σ σ2 Since l1, l2 and l3 are finite, f (s ) can be made positive for large positive values of s. Similarly, f (s ) can be made negative for large negative values of s. Hence, if one σ − l1 + 18 Advanced Mechanics of Solids f (s) plots f (s ) for different values of s as shown in Fig. 1.13, the curve must cut the s axis at least once as shown by the dotted curve and for s this value of s, f (s ) will be equal to A C B zero. Therefore, there is at least one real root. Let s3 be this root and n the asFig.1.13 Plot of f(s ) versus s sociated plane. Since the state of stress at the point can be characterised by the six rectangular components referred to any orthogonal frame of reference, let us choose a particular one, x¢y¢z¢, where the z¢ axis is along n and the other two axes, x¢ and y¢, are arbitrary. With reference to this system, the stress matrix has the form. ⎡ σ x ′ τ x ′y ′ 0 ⎤ ⎢ ⎥ 0⎥ (1.27) ⎢τ x ′y ′ σ y ′ ⎢ 0 ⎥ 0 σ 3⎦ ⎣ z¢ sz ¢ tx ¢y ¢ sy ¢ o y¢ x¢ Fig 1.14 Rectangular element with faces normal to x¢, y¢, z¢ axes Expanding Figure 1.14 shows these stress vectors on a rectangular element. The shear stress components tx¢z¢ and ty¢z¢ are zero since the z¢ plane is chosen to be the principal plane. With reference to this system, Eq. (1.19) becomes (σ x ′ − σ ) τ x ′y ′ (σ 0 τ x ′y ′ y′ −σ 0 0 ) 0 = 0 (1.28) (σ 3 − σ ) (s3 – s ) [s 2 – (sx¢ + sy¢) σ + σ x ′σ y ′ − τ x2′y ′ ] = 0 This is a cubic in s. One of the solutions is s = s3. The two other solutions are obtained by solving the quadratic inside the brakets. The two solutions are 1 σ1,2 = σ x′ + σ y ′ 2 ⎡⎛ σ − σ ⎞ 2 ⎤2 x′ y′ 2 ⎥ ± ⎢⎜ + τ x ′y ′ ⎟⎠ 2 ⎢⎝ ⎥ ⎣ ⎦ ( ) (1.29) The quantity under the square root power 1 is never negative and hence, 2 s1 and s2 are also real. This means that the curve for f (s ) in Fig. 1.13 will cut the s axis at three points A, B and C in general. In the next section we shall study a few particular cases. Analysis of Stress 1.14 19 PARTICULAR CASES (i) If s1, s2 and s3 are distinct, i.e. s1, s2 and s3 have different values, then the three associated principal axes n1, n2 and n3 are unique and mutually perpendicular. This follows from Eq. (1.25) of Sec. 1.12. Since s1, s2 and s3 are distinct, we get three distinct axes n1, n2 and n3 from Eqs (1.18), and being mutually perpendicular they are unique. (ii) If s1 = s2 and s3 is distinct, the axis of n3 is unique and every n3 direction perpendicular to n3 is a principal direction associated with s1 = s2. This is shown in Fig. 1.15. 0 To prove this, let us choose a Fig. 1.15 Case with s1 = s2 frame of reference Ox¢y¢z¢ such and s3 distinct that the z¢ axis is along n3 and the x¢ and y¢ axes are arbitrary. From Eq. (1.29), if s1 = s2, then the quantity under the radical must be zero. Since this is the sum of two squared quantities, this can happen only if sx¢ = sy¢ and tx¢y¢ = 0 But we have chosen x¢ and y¢ axes arbitrarily, and consequently the above condition must be true for any frame of reference with the z¢ axis along n3. Hence, the x¢ and y¢ planes are shearless planes, i.e. principal planes. Therefore, every direction perpendicular to n3 is a principal direction associated with s1 = s2. (iii) If s1 = s2 = s3, then every direction is a principal direction. This is the hydrostatic or the isotropic state of stress and was discussed in Sec. 1.7. For proof, we can repeat the argument given in (ii). Choose a coordinate system Ox¢y¢z¢ with the z¢ axis along n3 corresponding to s3. Since s1 = s2 every direction perpendicular to n3 is a principal direction. Next, choose the z¢ axis parallel to n2 corresponding to s2. Then every direction perpendicular to n2 is a principal direction since s1 = s3. Similarly, if we choose the z¢ axis parallel to n1 corresponding to s1, every direction perpendicular to n1 is also a principal direction. Consequently, every direction is a principal direction. Another proof could be in the manner described in Sec. 1.7. Choosing Oxyz coinciding with n1, n2 and n3, the stress vector on any arbitrary plane n has value s, the direction of s coinciding with n. Hence, every plane is a principal plane. Such a state of stress is equivalent to a hydrostatic state of stress or an isotropic state of stress. 1.15 RECAPITULATION The material discussed in the last few sections is very important and it is worthwhile to put it in the form of definitions and theorems. 20 Advanced Mechanics of Solids Definition n For a given state of stress at point P, if the resultant stress vector T on any plane n is along n having a magnitude s, then s is a principal stress at P, n is the principal direction associated with s, the axis of s is a principal axis, and the plane is a principal plane at P. Theorem In every state of stress there exist at least three mutually perpendicular principal axes and at most three distinct principal stresses. The principal stresses s1, s2 and s3 are the roots of the cubic equation s 3 – l1s 2 + l2s – l3 = 0 where l1, l2 and l3 are the first, second and third invariants of stress. The principal directions associated with s1, s2 and s3 are obtained by substituting si (i = 1, 2, 3) in the following equations and solving for nx, ny and nz: (sx – si) nx + txy ny + txz nz = 0 txy nx + (sy – si) ny + tyz nz = 0 nx2 + n 2y + nz2 = 1 If s1, s2 and s3 are distinct, then the axes of n1, n2 and n3 are unique and mutually perpendicular. If, say s1 = s2 π s3, then the axis of n3 is unique and every direction perpendicular to n3 is a principal direction associated with s1 = s2. If s1 = s2 = s3, then every direction is a principal direction. Standard Method of Solution Consider the cubic equation y3 + py2 + qy + r = 0, where p, q and r are constants. 1 Substitute y=x– p 3 This gives x3 + ax + b = 0 1 where a = 1 3q − p 2 , b= 2 p3 − 9 pq + 27r 3 27 b Put cos f = − ( ) ( 1/ 2 ⎛ 3⎞ 2⎜ − a ⎟ ⎝ 27 ⎠ Determine f, and putting g = 2 − a/3 , the solutions are φ p − 3 3 ⎛φ ⎞ p y 2 = g cos ⎜ + 120°⎟ − ⎝3 ⎠ 3 y 1 = g cos ⎛φ ⎞ p y 3 = g cos ⎜ + 240°⎟ − ⎝3 ⎠ 3 ) Analysis of Stress 21 Example 1.4 At a point P, the rectangular stress components are σ x = 1, σ y = −2, σ z = 4, τ xy = 2, τ yz = − 3, and τ xz = 1 all in units of kPa. Find the principal stresses and check for invariance. Solution The given stress matrix is ⎡1 2 1 ⎤ ⎡τ ij ⎤ = ⎢ 2 −2 −3⎥ ⎣ ⎦ ⎢ ⎥ ⎢⎣ 1 −3 4 ⎥⎦ From Eqs (1.22)–(1.24), l1 = 1 – 2 + 4 = 3 l 2 = (–2 – 4) + (–8 – 9) + (4 – 1) = –20 l 3 = 1(–8 – 9) –2(8 + 3) + 1 (–6 + 2) = – 43 \ f (s) = s 3 – 3s 2 – 20s + 43 = 0 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 21 ( 2) cos f = − (12167 27 ) 1/ 2 \ f = –119° 40 The solutions are s 1 = y1 = 4.25 + 1 = 5.25 kPa s 2 = y2 = –5.2 + 1 = – 4.2 kPa s 3 = y3 = 0.95 + 1 = 1.95 kPa Renaming such that s1 ≥ s2 ≥ s3 we have, s 1 = 5.25 kPa, s2 = 1.95 kPa, s3 = – 4.2 kPa The stress invariants are l 1 = 5.25 + 1.95 – 4.2 = 3.0 l 2 = (5.25 ¥ 1.95) – (1.95 ¥ 4.2) – (4.2 ¥ 5.25) = –20 l 3 = – (5.25 ¥ 1.95 ¥ 4.2) = – 43 These agree with their earlier values. Example 1.5 With respect to the frame of reference Oxyz, the following state of stress exists. Determine the principal stresses and their associated directions. Also, check on the invariances of l1, l2, l3. s3 = 2 – 3 Check on the invariance: With the set of axes chosen along the principal axes. 0 .e. s1 = –1 is in the direction +1/ 2.. which are s = 2 ± 3 . ny and nz represent the same line. nx = ± 1/ 2 . ny = ± 1/ 2 .e. nz = 0. s2 = 2 + 3. from Eqs (1. l 1 = –1 + 2 + 3 +2– 3 =3 l 2 = (–2 – 3 ) + (4 – 3) + (–2 + l 3 = –1 (4 – 3) = –1 Directions of principal axes: (i) For s1 = –1.22 Advanced Mechanics of Solids ⎡1 2 1⎤ [tij] = ⎢ 2 1 1⎥ ⎢ ⎥ ⎣⎢1 1 1⎥⎦ Solution For this state l1 = 1 + 1 + 1 = 3 l 2 = (1 – 4) + (1 – 1) + (1 – 1) = –3 l 3 = 1(1 – 1) – 2(2 – 1) + 1(2 – 1) = –1 f (s) = s 3 – l1 s 2 + l2 s – l3 = 0 3 i. − 1/ 2. one solution is s = –1. The other two solutions are obtained from the solution of the quadratic equation. Using this in the third ( ) ( ) and fourth equations and solving. ( ) Hence. It should be noted that the plus and minus signs associated with nx.18) and (1. . \ s1 = –1. s – 3s 2 – 3s + 1 = 0 or (s 3 + 1) – 3s (s + 1) = 0 i.21) (1 + 1)nx + 2ny + nz = 0 2nx + (1 + 1)ny + nz = 0 nx + ny + (1 + 1)nz = 0 together with 3 ) = –3 nx2 + n2y + nz2 = 1 From the second and third equations above.. the stress matrix will have the form 0 ⎡ −1 0 ⎤ ⎢ ⎥ [t ij] = ⎢ 0 2 + 3 0 ⎥ ⎢ 0 0 2 − 3 ⎥⎦ ⎣ Hence. (s + 1) (s 2 – s + 1) – 3s (s + 1) = 0 or (s + 1) (s 2 – 4s + 1) = 0 Hence. n2 and n3 form a rightangled triad. ny and nz in a manner similar to the preceeding one or get the solution from the condition that n1. the axis of s3 is unique and every direction perpendicular to s3 is a principal direction associated with s1 = s2.6 directions. i. nz = 1 2 ⎛ 1 ⎞ ⎜⎝ 1 + ⎟ 3⎠ 1/ 2 For the given state of stress. we get ⎛ 1 ⎞ nx = n y = ⎜ 1 + ⎟ ⎝ 3⎠ 1/ 2 nz = 1 (3 + 3 ) 1/ 2 (iii) For s3 = 2 – 3 We can solve for nx. n3 = n1 ¥ n2.Analysis of Stress (ii) For s2 = 2 + 23 3 (–1 – 3 )nx + 2ny + nz = 0 2nx + (–1 – 3 )ny + nz = 0 nx + ny (–1 – 3 )nz = 0 together with nx2 + n y2 + nz2 = 1 Solving. 1⎛ 1 ⎞ 1− ⎜ ⎟ 2⎝ 3⎠ 1/ 2 . l3 = 2 f (s) = –s 3 + 3s + 2 = 0 = (–s 3 – 1) + (3s + 3) = –(s + 1) (s 2 – s + 1) + 3(s + 1) = (s + 1) (s – 2) (s + 1) = 0 s 1 = s2 = –1 and s3 = 2 Since two of the three principal stresses are equal. l2 = –3.e. and s3 is different. For s3 = 2 –2nx + ny + nz = 0 nx – 2ny + nz = 0 . determine the principal stresses and their ⎡0 1 1 ⎤ [tij] = ⎢1 0 1 ⎥ ⎢ ⎥ ⎢⎣1 1 0 ⎥⎦ Solution \ l 1 = 0. The solution is nx = n y = − Example 1. If one denotes this loading axis by z¢. l2 = 0.e. on a plane that is equally inclined to xyz axes. l1 = 3r. the axis of loading making equal angles with the given xyz axes. i.30) . and the planes normal to these. This is the case of a uniaxial tension. there is a tensile stress of magnitue 3r.24 Advanced Mechanics of Solids nx + ny – 2nz = 0 nx2 + n2y + nz2 = 1 These give nx = ny = nz = 1 3 Example 1. We then have for the stress matrix ⎡σ 1 0 [τ ij ] = ⎢⎢ 0 σ 2 0 ⎣⎢ 0 0⎤ 0 ⎥⎥ σ 3 ⎦⎥ (1. are stress free. s 2 = s3 = 0. Thus. we can choose the principal axes as the coordinate axes and refer the rectangular stress components accordingly. can be chosen arbitrarily. one gets nx = n y = n z = 1/ 3 . the other two axes. x¢ plane and y¢ plane. 1.16 THE STATE OF STRESS REFERRED TO PRINCIPAL AXES In expressing the state of stress at a point by the six rectangular stress components. For s1 = 3r ( r – 3r)nx + rny + rnz = 0 rnx + (r – 3p) ny + rnz = 0 rnx + rny + (r – 3r) nz = 0 or –2nx + ny + nz = 0 nx – 2ny + nz = 0 nx + ny – 2nz = 0 The above equations give n x = ny = nz With nx2 + n 2y + nz2 = 1 . x¢ and y¢. the solutions are s 1 = 3r. l3 = 0 Therefore the cubic is s 3 – 3rs 2 = 0.7 The state of stress at a point is such that sx = sy = sz = txy = tyz = tzx = r Determine the principal stresses and their directions Solution For the given state. Ty = σ 2 n y . the components of the stress vector are. It will be shown in Sec.35) MOHR’S CIRCLES FOR THE THREE-DIMENSIONAL STATE OF STRESS We shall now describe a geometrical construction that brings out some important results. we get different values of s and t. s2 and s3 along the s axis and construct three circles with diameters (s1 – s2).18 that the point Q(s. (s2 – s3) and (s1 – s3) as shown in Fig. In this plane we can mark a point Q with co-ordinates (s. The plane with the s axis and the t axis is called the stress plane p. (1. then s = σ1nx2 + σ 2 n y2 + σ 3 nz2 n 2 and t2 = T (1.34) = nx2 n 2y (σ1 − σ 2 ) 2 + n 2y nz2 (σ 2 − σ 3 ) 2 + nz2 nx2 (σ 3 − σ 1 ) 2 The stress invariants assume the form l 1 = s1 + s2 + s3 l 2 = s 1s 2 + s 2 s 3 + s 3 s 1 l 3 = s1 s 2 s 3 1. 1. (No numerical value is associated with this symbol). let the frame of reference Pxyz be chosen along the principal stress axes. Corresponding to each plane n. n Tz = σ 3 n z (1. From Fig. a point Q can be located with coordinates (s. For different planes passing through point P. Arrange the principal stresses such that algebraically s1 ≥ s2 ≥ s3 Mark off s1.17 19/8/14 (1.Analysis of Stress 25 On any plane with normal n.9).16. 1. B and C represent the three principal stresses and the associated shear stresses are zero. . the following points can be observed: (i) Points A. 1. n n T x = σ 1 nx . t ).16. from Eq. This region is called Mohr’s stress plane p and the three circles are known as Mohr’s circles. t) for all possible n will lie within the shaded area. Consider a plane with normal n at point P. Take another set of axes s and t. t) for all possible directions n. t ) representing the values of the normal and shearing stress on the plane n.32) If s is the normal and t the shearing stress on this plane. The problem now is to determine the bounds for Q (s.33) − σ2 (1. At a given point P.31) The resultant stress has a magnitutde n 2 T = σ12 nx2 + σ 22 n 2y + σ 32 nz2 (1. Let s be the normal stress and t the shearing stress on this plane. 16. The normal stresses σ + σ3 σ2 + σ3 associated with the principal shears are respectively 1 . 2 (iii) Just as there are three extremum values s1. t2 and t3 where 2τ 3 = (σ1 − σ 2 ). section we shall prove this. . 2 (v) When s1 = s2 = s3. t) * E C s3 B s2 0 Fig. It will be 2 shown in Sec. 1.18 MOHR’S STRESS PLANE It was stated in the previous section that when points with coordinates (s.36) (iv) When s1 = s2 π s3 or s1 π s2 = s3. 2 2 σ + σ2 and 1 . 1. 1. the three circles reduce to only one 1 circle and the shear stress on any plane will not exceed (s1 – s3) or 2 1 (s1 – s2) according as s1 = s2 or s2 = s3. as in Fig. The principal shears are denoted by t1. 2τ1 = (σ 2 − σ 3 ) (1. 2τ 2 = (σ 1 − σ 3 ). While the planes on which the principal normal stresses act are free of shear stresses. and 1 . s2 and s3 for the normal stresses. 1.19 that the principal shear planes are at 45° to the principal normal planes. the three circles collapse to a single point on the s axis and every plane is a shearless plane. t ) for all possible planes passing through a point are marked on the s – t plane. these σ − σ3 σ 2 − σ3 σ − σ2 . These are indicated by points D.16 F A s1 s Mohr's stress plane (ii) The maximum shear stress is equal to 1 (s1 – s3) and the associated nor2 mal stress is 1 (s1 + s3). the principal shear planes are not free from normal stresses. This is indicated by point D on the outer circle.26 Advanced Mechanics of Solids t D p Q(s. E and F in Fig. 1. The planes on which these shear being 1 2 2 2 stresses act are called the principal shear planes. In this.16. the points are bounded by the three Mohr’s circles. there are three extremum values for the shear stresses. the numerator must also be positive.32) and (1. the numerator must also be negative. (1. the right-hand side expressions in the above equations must all be positive.39) The above three equations can be used to solve for nx2 .33) n 2 T = s 2 + t 2 = σ12 nx2 + σ 22 n 2y + σ 32 nz2 and also (1.42) must be positive. the numerator in Eq. (1. Case (i) s1 > s2 > s3 Case (ii) s1 = s2 > s3 Case (iii) s1 = s2 = s3 We shall consider these cases individually. Similarly. (1. Recall that we have arranged the principal stresses such that s1 ≥ s2 ≥ s3. Case (i) s1 > s2 > s3 For this case. the denominator being negative. There are three cases one can consider.38) 1 = nx2 + n 2y + nz2 (1.40) is positive and hence. the resultant stress vector T and the normal stress s are such that from Eqs (1.40) (σ − σ 3 ) (σ − σ1 ) + τ 2 (σ 2 − σ 3 ) (σ 2 − σ1 ) (σ − σ1 )(σ − σ 2 ) + τ 2 nz2 = (σ 3 − σ1 )(σ 3 − σ 2 ) n 2y = (1. n 2y and nz2 are all positive.37) s = σ1nx2 + σ 2 n y2 + σ 3 nz2 (1.42) Since nx2 . the denominator in Eq. n 2y and nz2 yielding nx2 = (σ − σ 2 ) (σ − σ 3 ) + τ 2 (σ 1 − σ 2 ) (σ 1 − σ 3 ) (1. In Eq.Analysis of Stress 27 Choose the coordinate frame of reference Pxyz such that the axes arenalong the principal axes. (s – s2)(s – s3) + t 2 ≥ 0 (s – s3)(s – s1) + t 2 £ 0 (s – s1)(s – s2) + t 2 ≥ 0 The above three inequalities can be rewritten as ⎛ τ 2 + ⎜σ − ⎝ σ 2 + σ3 ⎞ 2 ⎛ σ 2 − σ3 ⎞ ⎟⎠ ≥ ⎜⎝ 2 ⎟⎠ 2 2 . Therefore.41) (1.41). On any plane with normal n. 17 and also from Fig. t) should lie inside the shaded area of Fig. in this case. using Mohr’s circles. t) must lie on or outside the circle AFB with radius equal to (s1 – s2) 2 1 and centre at (s1 + s2). for this case. The last equation 2 2 indicates that the point must lie outside a circle of zero radius (since s1 = s2). Case (iii) s1 = s2 = s3 This is a trivial case since this is the isotropic or the hydrostatic state of stress. t) must lie inside or on the circle ADC with radius Case (ii) s1 = s2 > s3 Following arguments similar to the ones given above. one has for this case from Eqs (1. the maximum 1 1 shear stress is (s1 – s3) = t2 and the associated normal stress is (s1 + s3). t ) must lie on or 1 1 outside a circle of radius (s2 – s3) with its centre at (s2 + s3) along the s axis 2 2 (Fig. since s1 = s2.28 Advanced Mechanics of Solids σ + σ1 ⎞ ⎛ ⎛ σ − σ1 ⎞ τ + ⎜σ − 3 ≤⎜ 3 2 ⎟⎠ 2 ⎟⎠ ⎝ ⎝ 2 2 2 ⎛ ⎝ τ 2 + ⎜σ − σ1 + σ 2 ⎞ 2 ⎛ σ1 − σ 2 ⎞ ⎟⎠ ≥ ⎜⎝ 2 ⎟⎠ 2 2 According to the first of the above equations. the point (s. Similarly. 2 Hence. 1. The point Q lies on the circle BEC. 1.16. point (s. 2 2 . See Appendix 1 for the graphical determination of the normal and shear stresses on an arbitrary plane. the point Q(s. 1. Mohr’s circles collapse to a single point on the s axis.40)–(1. that the point (s.16 for the case s1 > s2 > s3. 1.16). The second equation indicates 1 (s – s3) 2 1 1 and centre at (s1 + s3) on the s axis.42) σ + σ3 ⎞ ⎛ ⎛ σ − σ3 ⎞ τ + ⎜σ − 2 =⎜ 2 2 ⎟⎠ 2 ⎟⎠ ⎝ ⎝ 2 2 2 ⎛ τ 2 + ⎜σ − ⎝ ⎛ τ 2 + ⎜σ − ⎝ σ 3 + σ1 ⎞ 2 ⎛ σ 3 − σ1 ⎞ ⎟⎠ = ⎜⎝ 2 ⎟⎠ σ1 + σ 2 ⎞ 2 2 ⎛ σ1 − σ 2 ⎞ ⎟⎠ ≥ ⎜⎝ 2 ⎟⎠ 2 2 2 From the first two of these equations. the Mohr’s circles will reduce to a circle BC and a point circle B. Hence.19 PLANES OF MAXIMUM SHEAR From Sec. t ) must lie on the 1 1 circle with radius (s1 – s3) with its centre at (s1 + s3). the last equation indicates that 2 1 the point (s. This is the circle with BC as diameter. 1. There are eight such planes. The principal shear plane will.e. 1. n y = 0 and n z = ±1 2. we get nx = 0/0.Analysis of Stress 29 Substituting these values in Eqs. This is so because. Since n x + n y + n z = 1 .37)–(1.17 t2 1/2 (s 1 + s 3) (b) (a) Principal planes (b) Planes of maximum shear If s1 = s2 > s3. make a fixed angle with s 3 axis (45° or 135°) but will have different values depending upon the selection of s 1 and s 3 axes.18. an octahedral s3 plane will be defined by nx = ny = nz = ± 1/ 3 . The normal and shearing stresses on these s3 planes are called the octahedral normal stress s2 and octahedral shearing stress respectively. since s 1 = s 2 π s 3 .14). Such a plane will have nx = 2 2 2 n y = n z .1.39) in Sec. every direction perpendicular to s 3 is a principal direction associated with s 1 = s 2 (Sec. whereas. make angles of 45° and 135° with the s1 and s2 planes as shown in Fig. Substituting these values in Eqs (1.1. . as shown in s1 s1 Fig. one gets nx = ± 1/2 . n x and n y are indeterminate.37)–(1. therefore.1. with the associated normal 2 1 stress (s2 + s3). 2 n y = 0/0 and n z = ±1 2 i.(1.18.17. the axis of s 3 is unique. A plane that is equally inclined to these three axes is called s2 an octahedral plane. 1.16) 1 and the maximum shear stress will be (s2 – s3) = t1. This means that the planes on which t 1 is acting makes angles of 45° and 135° with the s 3 axis but remains indeterminate with respect to s 1 and s 2 axes. s2 s2 s1 (s 1 1/2 s3 ) + s3 (a) Fig.18 Octahedral planes and (1. 1. then the three Mohr’s circles reduce to one circle BC (Fig. Substituting nx = ny = nz = ±1/ 3 in Eqs (1.20 OCTAHEDRAL STRESSES Let the frame of reference be again chosen along s1. s2 and s3 axes.34).39).33) Fig. 1. 1. This means that the planes (there are two of them) on which the shear stress takes on an extremum value. then the normal stresses on the octahedral planes will be zero and only the shear stresses will act. their associated normal stresses. it may be interpreted as the mean normal stress at a given point in a body. txy. s2 = 80 MPa.23). Using Eqs (1. This is important from the point of view of the strength and failure of some materials (see Chapter 4). the octahedral shear stress and its associated normal stress. (1. or toct = soct = 1 (sx + sy + sz) 3 (1.46) The octahedral normal stress being equal to 1/3 l1.44c) 2 3 1 It is important to remember that the octahedral planes are defined with respect to the principal axes and not with reference to an arbitrary frame of reference.36). one can express these in terms of sx.8 The state of stress at a point is characterised by the components sx = 100 MPa. sy = –40 MPa. t1 = t2 = σ2 −σ3 2 σ 3 − σ1 2 = 80 + 40 = 60 MPa 2 = −40 − 100 = − 70 MPa 2 σ1 − σ 2 100 − 80 = = 10 MPa 2 2 The associated normal stresses are t3 = σ 1* = σ2 +σ3 2 = 80 − 40 = 20 MPa 2 . Example 1. sz. sy.44b) 9 2 (l 2 − 3l )1/ 2 (1.43) 2 τ oct = 1 [(s1 – s2)2 + (s2 – s3)2 + (s3 – s1)2] (1. txy = tyz = tzx = 0 Determine the extremum values of the shear stresses. Solution The given stress components are the principal stresses.44a) 2 9τ oct = 2(s1 + s2 + s3)2 – 6(s1s2 + s2s3 + s3s1) (1. since the shears are zero.30 Advanced Mechanics of Solids soct = 1 (s1 + s2 + s3) = 1 l1 3 3 and or (1. s3 = – 40 MPa Hence from Eq. If in a state of stress. Since soct and toct have been expressed in terms of the stress invariants. s1 = 100 MPa. the first invariant (s1 + s2 + s3) is zero. tyz and tzx also.45) ( 2 2 2 2 + τ yz + τ zx 9τ oct = (sx – sy)2 + (sy – sz)2 + (sz – sx)2 + 6 τ xy ) (1.22) and (1. Arranging the terms such that s1 ≥ s2 ≥ s3. sz = 80 MPa. For such a state. Hence. the necessary condition for a state of pure shear to exist is that l1 = 0.47) . Hence.21 THE STATE OF PURE SHEAR The state of stress at a point can be characterised by the six rectangular stress components referred to a coordinate frame of reference.7 MPa 3 3 1. The magnitudes of these components depend on the choice of the coordinate system. l1 = sx + sy + sz = 0.Analysis of Stress σ 2* = σ 3 + σ1 2 = 31 −40 + 100 = 30 MPa 2 σ 3* = σ1 + σ 2 = 100 + 80 = 90 MPa 2 2 2 2 1/ 2 toct = 1 ⎡(σ1 − σ 2 )2 + (σ 2 − σ 3 ) + (σ 3 − σ1 ) ⎤ = 61.8 MPa ⎦ 3⎣ 1 140 soct = (σ1 + σ 2 + σ 3 ) = = 46. for a pure shear stress state. an octahedral plane is subjected to pure shear with no normal stress. we find that sx = sy = sz = 0. Since l1 is an invariant. It can be shown (Appendix 2) that this is also a sufficient condition.22 DECOMPOSITION INTO HYDROSTATIC AND PURE SHEAR STATES It will be shown in the present section that an arbitrary state of stress can be resolved into a hydrostatic state and a state of pure shear. It was remarked in the previous section that when l1 = 0. 1. If. then a state of pure shear is said to exist at point P. the stress matrix will be ⎡ 0 τ xy τ xz ⎤ ⎢ ⎥ ⎡⎣τ ij ⎤⎦ = ⎢τ xy 0 τ yz ⎥ ⎢ ⎥ ⎣⎢τ xz τ yz 0 ⎥⎦ For this coordinate system. with that particular choice of coordinate system. the octahedral plane (remember that this plane is defined with respect to the principal axes and not with respect to an arbitrary set of axes) is free from normal stress. this must be true for any choice of coordinate system selected at P. for at least one particular choice of the frame of reference. Let the given state referred to a coordinate system be ⎡σ x τ xy τ xz ⎤ ⎢ ⎥ [tij] = ⎢τ xy σ y τ yz ⎥ ⎢ ⎥ ⎣⎢τ xz τ yz σ z ⎥⎦ Let p = 1/3(sx + sy + sz) = 1/3l1 (1. ⎡6 0 0 ⎤ ⎡ 4 4 6 ⎤ [tij] = ⎢0 6 0 ⎥ + ⎢ 4 − 4 8 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣0 0 6 ⎥⎦ ⎢⎣ 6 8 0 ⎥⎦ . [Refer Sec. Determine the normal and shearing stresses on an octahedral plane. Resolve the given state into a hydrostatic state and a pure shear state. the hydrostatic state and the pure shear state the same or are they different? Explain why.47) If the given state is referred to the principal axes. p = 1/3(s1 + s2 + s3) = 1/3l1. 1. the decomposition into a hydrostatic state and a pure shear state can once again be done as above. as before. Compare these with the soct and toct calculated for the hydrostatic and the pure shear states.e. as shown: ⎡σ x τ xy τ xz ⎤ ⎡ p ⎢ ⎥ ⎢ ⎢τ xy σ y τ yz ⎥ = ⎢ 0 ⎢ ⎥ ⎢0 ⎣⎢τ xz τ yz σ z ⎥⎦ ⎣ 0 p 0 τ xy τ xz 0 ⎤ ⎡σ x − p ⎢ ⎥ σy − p τ yz 0 ⎥ + ⎢ τ xy ⎢ p ⎦⎥ ⎢ τ xz τ yz σz − ⎣ ⎤ ⎥ ⎥ ⎥ p ⎦⎥ (1. (1.47).14(iii). Are the octahedral planes for the given state.49) where. Eq.32 Advanced Mechanics of Solids The given state can be resolved into two different states.9 The state of stress characterised by t ij is given below. The pure shear state of stress is also known as the deviatoric state of stress or simply as stress deviator.] The second state is a state of pure shear since the first invariant for this state is l¢1 = (sx – p) + (sy – p) + (sx – p) = sx + sy + sz – 3p = 0 from Eq. Solution l1 = 10 + 2 + 6 = 18. Example 1. 0 ⎤ ⎡p ⎡σ1 0 ⎢0 σ 0 ⎥⎥ = ⎢⎢ 0 2 ⎢ ⎢⎣ 0 0 σ 3 ⎥⎦ ⎢⎣ 0 0 p 0 0 ⎤ ⎡σ1 − p 0 0 0 ⎥⎥ + ⎢⎢ 0 σ2 − p 0 0 σ3 − p ⎥⎦ ⎢⎣ 0 ⎤ ⎥ ⎥ p ⎥⎦ (1. ⎡10 4 6 ⎤ [tij] = ⎢ 4 2 8 ⎥ ⎢ ⎥ ⎢⎣ 6 8 6 ⎥⎦ 1l =6 3 1 Resolving into hydrostatic and pure shear state.48) The first state on the right-hand side of the above equation is a hydrostatic state. (1. i. 10 p Solution From elementary strength of materials. The octahedral planes for the given state (which are identified after determining the principal stress directions). and hence. the octahedral planes corresponding to the given state and the pure shear state are identical. toct = 0. and toct for the given state is equal to toct for the pure shear state. The cirpd cumferential or the hoop stress is . . For the hydrostatic state. the value of toct for the pure shear state is 2 396 1/ 2 = 2 22 ( ) 3 Hence.10 A cylindrical boiler. d 4t the diameter and t the thickness.8 cm thick. soct = 0 since the first invariant of stress for the pure shear state is zero. the value of soct for the given state is equal to the value of soct for the hydrostatic state. Therefore. 1. soct = 6. is made of plates 1. the axial stress in the plate is pd where p is the internal pressure. every direction is a principal direction. (1. 180 cm in diameter. Determine the maximum shearing stress in the plate at point P and the plane on which it acts.44) toct = ( 2 l 2 − 3l 2 2 3 1 ) 1/ 2 1/ 2 = 2 ⎡⎣182 − 3 ( 20 − 16 + 12 − 64 + 60 − 36 ) ⎤⎦ 3 1/ 2 = 2 ( 396 ) = 2 22 3 For the hydrostatic state. 1. pd/4t P pd/2t Fig. and is subjected to an internal pressure 1400 kPa. The value of the second invariant of stress for the pure shear state is l¢2 = (–16 – 16 + 0 – 64 + 0 – 36 ) = –132 Hence.19. the octahedral normal and shear stresses are: s oct = 1 l1 = 6 3 From Eq. the hydrostatic state and the pure shear state are all identical. since every plane is a principal plane with s = 6 and consequently. toct = Example 1. the principal stress directions for the given state and the pure shear state are identical. 2t The state of stress acting on an element is as shown in Fig.Analysis of Stress 33 For the given state.19 Example 1. For the pure shear state. 34 Advanced Mechanics of Solids The principal stresses when arranged such that s1 ≥ s2 ≥ s3 are pd pd σ1 = . 1. 1.33). as shown in Fig. then we get a surface S. This surface is known as the stress surface of Cauchy. 700 kPa 2 ⎝ 2 × 1. 1. s = σ 1 nx2 + σ 2 n y2 + σ 3 nz2 Along the normal n to the plane. By the previous definiQ tion. Let s1.20(b). σ3 = – p 2t 4t The maximum shear stress is therefore. then this normal m is parallel to the n resultant stress vector T at P. Fig.8 ⎠ τ max = 1400 ⎜ 1. This surface determines the normal component of stress on every plane passing through P. Let m S Q be a point on the surface.20 (a) Cauchy’s stress quadric (b) Resultant stress vector and normal stress component (1. This has a very interesting property.23 CAUCHY’S STRESS QUADRIC We shall now describe a geometrical description of the state of stress at a point P. σ2 = . The normal stress on this plane is from Eq. the n resultant stress vector T can be easily determined.50) As different planes n are chosen at P. Since the direction of the result- n ant vector T is known. Choose a frame of reference whose axes are along the principal axes.8 × 100 ⎞ + 1⎟ = 35. . ( 2t ) τ max = 1 (σ 1 − σ 3 ) = 1 p d + 1 2 2 Substituting the values ⎛ 1. s2 and s3 be the principal stresses. If such Qs are marked for every plane passing through P. Consider a plane with normal n.51) R If m is a normal to the tangent plane to the surface S at point Q. choose a point Q such that PQ = R = 1 σ (1. and its component s along the normal is known. the length PQ = R is such that R n the normal stress on the plane whose n T n normal is along PQ is given by T σ = 12 P (b) (a) Fig. we get different values for the normal stress s and correspondingly different PQs.20(a). (1. 52) n T x = 1 σ 1 x. then y Q (x. 1. Hence. n Ty = σ1n y . where a is a constant of proportionality. n Tz = σ 3 nz Substituting for nx. if mx. the resultant stress vector. is σ = σ 1nx2 + σ 2 n 2y + σ 3 nz2 If the coordinates of the point Q are (x. mz = 2as3z n n n (1. (1. (1. m is the normal perpendicular ∂x ∂y ∂z to the tangent plane to S at Q. then mx = α ∂ F . ∂F and ∂F .21). Let Pxyz be the principal axes at P (Fig.31). the surface S has the equations (a surface of second degree) z Fig.53b) when s is compressive We know from calculus that for a surface with equation F(x.51).20). Hence. are n Tx = σ1nx . T y . and mz are the direction cosines of m. 1. The plus sign is used when s is tensile and the minus sign is used when s is compressive. y. R n T z = 1 σ3z R . (1.53b) mx = 2as1x. z) and the length PQ = R. my.54) n T is the resultant stress vector on plane n and its components T x . as before. (1. R n T y = 1 σ 2 y.52) Substituting these in the above equation for s n x P s R2 = s1x2 + s2 y2 + s3z2 From Eq. R ny = y . (1. R nz = z R (1. y. ∂x my = α ∂ F .53a) or Eq. we have sR2 = ±1. y. From Fig. and T z according to Eq. n is the normal to a particular plane at P. the normal to the tangent plane at a point Q on the surface has direction cosines proportional to ∂F . (1. z) = 0. ny and nz from Eq.53a) s1x2 + s2 y2 + s3z2 = –1 (1.Analysis of Stress 35 n We shall now show that the normal m to the surface S is parallel to T . The normal stress on this plane. z) R nx = x . my = 2as2 y.21 Principal axes at P and n to a plane when s is tensile s1x2 + s2 y2 + s3z2 = +1 (1. ∂y mz = α ∂ F ∂z From Eq. has the following properties: (i) If Q is a point on the stress surface. T y and T z . It would be of interest to know the shape of the stress surface for different states of stress. n T y = σ 2 ny . The coordinates (x. and the third inbetween (Fig. 2 x2 + y + z 2 = 1 2 2 2 σ1 σ2 σ3 y n Q x n z Q P Fig. we get from the above two equations. mx. according to Eq. my and mz are proportional to T x .e. n (ii) The normal to the surface at Q is parallel to the resultant stress vector T on the plane with normal PQ. On a plane passing through P with normal n. y. 1. z) of the point Q are then n x = Tx. This aspect will be discussed in Appendix 3. One of its three semiaxes is the longest. i.36 Advanced Mechanics of Solids n n σ1 x = R T x . Therefore. 1. n Hence. parallel to the principal axes at P.1. then PQ = 1/ σ where s is the normal stress on a plane whose normal is PQ. (1.24 LAME’S ELLIPSOID Let Pxyz be a coordinate frame of reference at point P. σ3z = RT z Substituting these in Eq. for instance . PQ = | T | . This ellipsoid is called the stress ellipsoid or Lame's ellipsoid. n m y = 2α R T y .22). The stress surface of Cauchy. are n n T x = σ 1n x . (1.e. z =Tz Since nx2 + n 2y + nz2 = 1. T z = σ 3 nz Let PQ be along the resultant stress vector and its length be equal to its n magnitude. therefore. n n y = T y. n mz = 2α R T z n n i.22 Lame’s ellipsoid n T (1. the stress surface of Cauchy completely defines the state of stress at P. These are the extermum values.31). If two of the principal stresses are equal. the resultant stress vector n is T and its components.54) n n mx = 2α R T x . the other the shortest.55) This is the equation of an ellipsoid referred to the principal axes. or n σ2 y = RT y . m and T are parallel. at a point (x0. Consequently. s1 = s2 = s3.11 Show that Lame’s ellipsoid and the stress-director surface together completely define the state of stress at a point. z0 ) has direction cosines proportional to ∂F . if m is the normal to the tangent plane (Fig. Lame's ellipsoid becomes a sphere. y. the equation of the ellipsoid referred to principal axes is given by 2 x2 + y + z 2 = 1 2 2 2 σ1 σ2 σ3 The stress-director surface has the equation 2 x2 + y + z 2 = 1 σ1 σ2 σ3 It is known from analytical geometry that for a surface defined by F(x. mz = α z0 σ3 Ellipsoid Surface lle para n m l to m L (x0. y0 . Each radius vector PQ of the stress ellipsoid represents to a certain scale.23 Stress director surface and ellipsoid surface . 1. s2 and s3 are the principal stresses at a point P. Lame’s ellipsoid and the stress-director surface together completely define the state of stress at point P. evaluated at (x0. Hence.23). Solution If s1. z0). the resultant stress on one of the planes through the centre of the ellipsoid. the normal to the tangent at a point (x0. y0. then mx = α x0 σ1 . y0.11) that the stress represented by a radius vector of the stress ellipsoid acts on the plane parallel to tangent plane to the surface called the stress-director surface. ∂F . Example 1. It can be shown (Example 1. defined by 2 x2 + y + z 2 = 1 σ1 σ2 (1.56) σ3 The tangent plane to the stress-director surface is drawn at the point of intersection of the surface with the radius vector. my = α y0 σ2 . ∂F .Analysis of Stress 37 s1 = s2. y0. z) = 0. z0) ∂x ∂y ∂z on the stress ellipsoid. Lame’s ellipsoid is an ellipsoid of revolution and the state of stress at a given point is symmetrical with respect to the third principal axis Pz. z ) 0 P Q Stress-director Surface Fig. If all the principal stresses are equal.1. 57). there exists a coordinate system Oxyz such that for this system sz = 0. z0). On this plane. my and mz m T x = α x0 . m m T y = α y0 . the resultant stress m vector is T with components proportional to x0. 1. 1. m is the normal to the tangent plane at L.e. (1. From Eq. n PQ being an extension of PL and equal to T in magnitude. tyz = 0 (1. the resultant n stress vector will be T with components given by m T x = σ 1mx . sy sy txy sx txy sx sx sx txy txy sy (a) sy (b) Fig.25). m m T y = σ 2my . 1.24 (a) Plane state of stress (b) Conventional representation Consider a plane with the normal lying in the x y plane. T z = α z0 i. the plane having this resultant stress will have m as its normal. L(x0. y0 and z0. txz = 0. 1.38 Advanced Mechanics of Solids Consider a plane through P with normal parallel to m. the components of stress on the plane with normal m are proportional to the coordinates (x0. ny and nz are the direction cosines of the normal.25 THE PLANE STATE OF STRESS If in a given state of stress.57) then the state is said to have a ‘plane state of stress’ parallel to the x y plane. T y and T z . ny = sin q and nz = 0 (Fig. T z = σ 3 mz Substituting for mx. This state is also generally known as a two-dimensional state of stress. we have nx = cos q. On a plane through P with normal m. z0) is a point on the stress-director surface. y0. The state of stress is shown in Fig. Hence the stress-director surface has the following property. (1.24. If nx. y0.9) . This means that the m m m components of PL are proportional to T x . All the foregoing discussions can be applied and the equations reduce to simpler forms as a result of Eq. 25 Normal and shear stress components on an oblique plane n T x = sx cos q + txy sin q n T y = sy sin q + txy cos q (1.62) The above equation gives two planes at right angles to each other.29) as s1.Analysis of Stress 39 sy n txy t n sx s sx q q sx txy txy sy sy (a) (b) Fig.59) n 2 2 2 t 2 = Tx + Ty − σ t= σx −σy sin 2q + tx y cos 2q 2 The principal stresses are given by Eq.11b) s = sx cos2 q + sy sin2 q + 2txy sin q cos q = σx + σy 2 n and or + σx − σy 2 cos 2q + tx y sin 2q (1.11a) and (1. If the principal stresses s1. s2 = σx + σy 2 (1. s2 and s3 are arranged such that s1 ≥ s2 ≥ s3.63a) .58) n Tz =0 The normal and shear stress components on this plane are from Eqs (1. (1. 1.60) 1/ 2 ⎡⎛ σ − σ ⎞ 2 ⎤ x y 2 ± ⎢⎜ ⎟ + τ xy ⎥ 2 ⎢⎝ ⎥ ⎠ ⎣ ⎦ (1.61) s3 = 0 The principal planes are given by (i) the z plane on which s3 = sz = 0 and (ii) two planes with normals in the x y plane such that tan 2φ = 2τ xy σx −σy (1. the maximum shear stress at the point will be τ max = σ1 − σ 3 2 (1. ∂x τ xy + ∂τ xy ∆ x. the average stress components are σx + ∂σ x ∆ x. face 3 No.26 ⎡⎛ σ − σ ⎞ 2 ⎤ x y 2 = ⎢⎜ ⎟ + τ xy ⎥ 2 ⎢⎝ ⎥ ⎠ ⎣ ⎦ (1.e. 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. attention has been focussed on the state of stress at a point. we shall deal with sz average values of the stress 5 components on each face. (1. face Fig.26).27. These conditions will be established when the body (and. y and Dz 4 tend to zero. txy and txz. In general. 1. 3 x sz +Dsz etc. 1.27 Variation of stresses No.e.64) DIFFERENTIAL EQUATIONS OF EQUILIBRIUM So far. i. ∂x τ xz + ∂τ xz ∆x ∂x .63b) 1/ 2 τ max 1. we y + Dsy are going to make Dx. Since in the limit. One of the fundamental problems in a book of this kind is the determination of the state of stress at every point or at any desired point in a body. Dy and Dz isolated from its pary ent body. 1. 1. 2. 1. the average stress components are sx. the state of stress in a body varies from point to point. every part of it) is in equillibrium. On the left hand face.26 Isolated cubical element equality of cross shears. i. 1.40 Advanced Mechanics of Solids In the xy plane. A similar procedure was adopted in (a) Sec. 2. in equilibrium Consider a small rectangular element with sides Dx. the maximum shear stress will be τ max = 1 (σ 1 − σ 2 ) 2 and from Eq. On z sy the right hand face. therefore.61) (1. 6 The faces are marked as 1. We isolate a small element of the body and derive the equations of equilibrium from its free(b) body diagram (Fig.8 for establishing the Fig. 1 These stress components are sx sx + Dsx 2 shown in Fig. Analysis of Stress 41 This is because the right hand face is Dx distance away from the left hand face. Following a similar procedure, the stress components on the six faces of the element are as follows: σ x , τ xy , τ xz Face 1 Face 2 σx + ∂σ x ∆ x, ∂x ∂τ xy ∆ x, ∂x τ xz + ∂τ xz ∆x ∂x τ yz + ∂τ yz ∆y ∂y τ zy + ∂τ zy ∆z ∂z σ y , τ yx , τ yz Face 3 Face 4 σy + ∂σ y ∆y, ∂y τ yx + ∂τ yx ∆y, ∂y σ z , τ zx , τ zy Face 5 Face 6 τ xy + σz + ∂σ z ∆ z, ∂z τ zx + ∂τ zx ∆ z, ∂z Let the body force components per unit volume in the x, y and z directions be g x , g y, and g z . For equilibrium in x direction ∂τ yx ⎛ ⎞ ∂σ x ⎛ ⎞ ⎜ σ x + ∂ x ∆ x ⎟ ∆y ∆ z − σ x ∆y ∆ z + ⎜τ yx + ∂ y ∆y ⎟ ∆ z ∆x − τ yx ∆ z ∆ x + ⎝ ⎠ ⎝ ⎠ ∂τ z x ⎛ ⎞ ∆ z ⎟ ∆ x ∆y − τ zx ∆ x ∆y + γ x ∆ x ∆y ∆ z = 0 ⎜τ zx + ∂ z ⎝ ⎠ Cancelling terms, dividing by Dx, Dy, Dz and going to the limit, we get ∂σ x ∂τ yx ∂τ zx + + + γx = 0 ∂x ∂y ∂z Similarly, equating forces in the y and z directions respectively to zero, we get two more equations. On the basis of the fact that the cross shears are equal, i.e. t xy = tyx, tyz = tzy, t xz = t zx, we obtain the three differential equations of equilibrium as ∂σ x ∂τ xy ∂τ zx + + + γx = 0 ∂x ∂y ∂z ∂σ y ∂τ xy ∂τ yz + + +γy = 0 ∂y ∂x ∂z (1.65) ∂σ z ∂τ xz ∂τ yz + + + γz = 0 ∂z ∂x ∂y Equations (1.65) must be satisfied at all points throughout the volume of the body. It must be recalled that the moment equilibrium conditions established the equality of cross shears in Sec.1.8. 42 Advanced Mechanics of Solids 1.27 EQUILIBRIUM EQUATIONS FOR PLANE STRESS STATE The plane stress has already been defined. If there exists a plane stress state in the xy plane, then sz = tzx = tyz = gz = 0 and only sx, sy, txy, gx and gy exist. The differnetial equations of equilibrium become ∂σ x ∂τ xy + + γx = 0 ∂x ∂y ∂σ y ∂τ xy + +γy =0 ∂y ∂x (1.66) Example 1.12 The cross-section of the wall of a dam is shown in Fig. 1.28. The pressure of water on face OB is also shown. With the axes Ox and Oy, as shown in Fig. 1.28, the stresses at any point (x, y) are given by (g = specific weight of water and r = specific weight of dam material) sx = –g y o x ⎛ ρ 2γ − sy = ⎜⎜ tan β tan 3 β ⎝ b txy = τ yx = − B A y Fig 1.28 Example 1.12 γ tan 2 β tyz = 0, tzx = 0, ⎞ ⎛ γ ⎞ − ρ ⎟⎟ y ⎟⎟ x + ⎜⎜ 2 ⎠ ⎝ tan β ⎠ x sz = 0 Check if these stress components satisfy the differential equations of equilibrium. Also, verify if the boundary conditions are satisfied on face OB. Solution The equations of equilibrium are ∂σ x ∂τ xy + + γx = 0 ∂x ∂y and ∂σ y ∂τ xy + + γy = 0 ∂y ∂x Substituting and noting that gx = 0 and gy = r, the first equation is satisifed. For the second equation also γ −ρ− γ +ρ =0 tan β tan 2 β On face OB, at any y, the stress components are sx = –g y and txy = 0. Hence the boundary conditions are also satisfied. 2 Example 1.13 Consider a function f (x, y), which is called the stress function. If the values of s x , s y and txy are as given below, show that these satisfy the differential equations of equilibrium in the absence of body forces. Analysis of Stress σx = ∂ 2φ , ∂ y2 σy = ∂ 2φ , ∂ x2 τ xy = − 43 ∂ 2φ ∂x∂ y Solution Substituting in the differential equations of equilibrium ∂ 3φ ∂ 3φ − =0 ∂ y2 ∂ x ∂ y2 ∂ x ∂ 3φ ∂ 3φ − =0 ∂ x2 ∂ y ∂ x2 ∂ y Example 1.14 Consider the rectangular beam shown in Fig. 1.29. According to the elementary theory of bending, the ‘fibre stress’ in the elastic range due to bending is given by σx = − 12 My My =− l bh3 y y h/2 x z h/2 b Fig. 1.29 Exmaple 1.14 where M is the bending moment which is a function of x. Assume that sz = tzx = tzy = 0 and also that txy = 0 at the top and bottom, and further, that sy = 0 at the bottom. Using the differential equations of equilibrium, determine txy and sy. Compare these with the values given in the elementary strength of materials. Solution From Eq. (1.65) ∂σ x ∂τ xy ∂τ xz + + =0 ∂x ∂y ∂z Since txz = 0 and M is a function of x − or Integrating 12 y ∂ M ∂τ xy + =0 ∂y bh3 ∂ x ∂τ xy 12 ∂ M y = ∂y bh3 ∂ x txy = 6 ∂ M y 2 + c f ( x) + c 1 2 bh3 ∂ x 44 Advanced Mechanics of Solids where f (x) is a function of x alone and c 1 , c 2 are constants. It is given that t xy = 0 at y = ± h 2 6 h2 ∂ M \ = –c1 f (x) – c2 bh3 4 ∂ x c1 f (x) + c2 = − 3 ∂ M 2bh ∂ x or 2 3 ∂ M ⎛ 4 y − 1⎞ ⎜⎜ 2 ⎟⎟ 2bh ∂ x ⎝ h ⎠ From elementary strength of materials, we have txy = \ txy = V lb h/2 ∫y y ′ dA where V = − ∂Μ is the shear force. Simplifying the above expression ∂x ∂ M 12 ⎛ h 2 − y 2 ⎞ b txy = − ⎟2 ∂ x b 2 h3 ⎜⎝ 4 ⎠ or txy = 2 3 ∂ M ⎛ 4 y − 1⎞ ⎜⎜ 3 ⎟⎟ 2bh ∂ x ⎝ h ⎠ i.e. the same as the expression obtained above. From the next equilibrium equation, i.e. from ∂σ y ∂ τ xy ∂ τ yz + + =0 ∂y ∂x ∂z 2 ∂σ y 3 ⎛ 4 y − 1⎞ ∂ 2 M = − ⎜⎜ 2 ⎟⎟ 2 2bh ⎝ h ∂y ⎠ ∂x we get 3 ∂ 2M sy = − 2bh ∂ x 2 \ ⎛ 4 y3 ⎞ ⎜⎜ 2 − y ⎟⎟ + c3 F ( x) + c4 ⎝ 3h ⎠ where F(x) is a function of x alone. It is given that sy = 0 at y = − h . 2 Hence, c3F(x) + c4 = 3 ∂ 2M h 2bh ∂ x 2 3 2 = 1 ∂ M 2b ∂ x 2 Substituting 3 ∂ 2M sy = − 2bh ∂ x 2 ⎛ 4 y3 h⎞ ⎜⎜ 2 − y − ⎟⎟ 3⎠ ⎝ 3h Analysis of Stress 45 At y = +h/2, the value of sy is 1 ∂ 2M = w b ∂ x2 b where w is the intensity of loading. Since b is the width of the beam, the stress will be w/b as obtained above. sy = 1.28 BOUNDARY CONDITIONS Equation (1.66) must be satisfied throughout the volume of the body. When the stresses vary over the plate (i.e. the body having the plane stress state), the stress components sx, sy and txy must be consistent with the externally applied forces at a boundary point. Consider the two-dimensional body shown in Fig.1.30. At a boundary point P, the outward drawn normal is n. Let Fx and Fy be the components of the surface forces per unit area at this point. y Fy P Fy n sx Fx n Fx txy P1 o x sy (b) (a) Fig. 1.30 (a) Element near a boundary point (b) Free body diagram Fx and Fy must be continuations of the stresses sx, sy and txy at the boundary. Hence, using Cauchy’s equations n T x = Fx = sx nx + txy ny n T y = Fy = sy ny + txy nx If the boundary of the plate happens to be parallel to y axis, as at point P1, the boundary conditions become Fx = sx and Fy = txy 1.29 EQUATIONS OF EQUILIBRIUM IN CYLINDRICAL COORDINATES Till this section, we have been using a rectangular or the Cartesian frame of reference for analyses. Such a frame of reference is useful if the body under analysis happens to possess rectangular or straight boundaries. Numerous problems 46 Advanced Mechanics of Solids exist where the bodies under discussion possess radial symmetry; for example, a thick cylinder subjected to internal or external pressure. For the analysis of such problems, it is more convenient to use polar or cylindrical coordinates. In this section, we shall develop some equations in cylindrical coordinates. Consider an axisymmetric body as shown in Fig. 1.31(a). The axis of the body is usually taken as the z axis. The two other coordinates are r and q, where q is measured counter-clockwise. The rectangular stress components at a point P(r, q, z) are sr , sq , sz , tq r , tq z and tzr These are shown acting on the faces of a radial element at point P in Fig.1.31(b). s r , sq and sz are called the r radial, circumferential and P axial stresses respectively. If z the stresses vary from point to point, one can derive the sz q appropriate differential equations of equilibrium, as in tzq Sec. 1.26. For this purpose, consider a cylindrical eletrq trz sq ment having a radial length tqz sr Dr with an included angle Dq and a height Dz, isolated from (b) the body. The free-body diagram of the element is shown in Fig.1.32(b). Since the element is very small, we work with the average stresses acting on each face. The area of the face aa¢d¢d is r Dq Dz and the area of face (a) bb¢c¢c is (r + Dr) Dq Dz. The Fig. 1.31 (a) Cylindrical coordinates of a point areas of faces dcc¢d¢ and abb¢e¢ (b) Stresses on an element are each equal to Dr Dz. The faces abcd and a¢b¢c¢d¢ have each an area r + ∆r Dq Dr. The average 2 stresses on these faces (which are assumed to be acting at the mid point of eace face) are On face aa¢d¢d normal stress sr tangential stresses trz and trq On face bb¢c¢c z ( normal stress sr + ∂σ r ∆r ∂r ) Analysis of Stress 47 z r a d Dq d¢ Dr b c y a¢ b¢ c¢ y σ θ′ r τr′ θ r x σ r′ /2 (a) (b) Fig. 1.32 (a) Geometry of cylindrical element (b) Variation of stresses across faces tangential stresses τr z + ∂τ rz ∆r ∂r and τ rθ + ∂τθr ∆r ∂r The changes are because the face bb¢c¢c is Dr distance away from the face aa¢d¢d. On face dcc¢d¢ normal stress sq tangential stresses trq and tq z On face abb¢a ∂σθ ∆θ ∂θ ∂τ rθ ∂τ θ z tangential stresses trq + Dq and tq z + Dq ∂θ ∂θ normal stress sq + The changes in the above components are because the face abb¢a is separated by an angle Dq from the face dcc¢d¢. On face a¢b¢c¢d¢ normal stress sz tangential stresses trz and tq z On face abcd normal stress sz + ∂σ z ∆z ∂z tangential stresses trz + ∂τ rz D z and t + ∂ τθ z ∆z qz ∂z ∂z Let gr , gq and gz be the body force components per unit volume. If the element is in equilibrium, the sum of forces in r, q and z directions must vanish individually, Equating the forces in r direction to zero, ∂σ r ⎞ ∂τ rz ⎞ ⎛ ∆r ⎞ ⎛ ⎛ ⎜⎝ σ r + ∂ r ∆r ⎟⎠ ( r + ∆ r ) ∆θ ∆ z + ⎜⎝ τ rz + ∂ z ∆ z⎟⎠ ⎜⎝ r + 2 ⎟⎠ ∆θ ∆ r 48 Advanced Mechanics of Solids ( ) − σ r r ∆θ ∆z − τ rz r + ∆r ∆θ ∆r − σ θ sin ∆θ ∆r ∆z 2 2 ∂σ ⎛ ⎞ − τ rθ cos ∆θ ∆r ∆z − ⎜ σ θ + θ ∆θ ⎟ sin ∆θ ∆r ∆z 2 2 ∂θ ⎝ ⎠ ( ) ∂τ ⎛ ⎞ + ⎜τ rθ + rθ ∆θ ⎟ cos ∆θ ∆r ∆z + γ r r + ∆r ∆θ ∆r ∆z = 0 ∂θ 2 2 ⎝ ⎠ Cancelling terms, dividing by Dq Dr Dz and going to the limit with Dq, Dr and Dz, all tending to zero r or ∂τ ∂σ r ∂τ + r rz + rθ + σ r − σ θ + r γ r = 0 ∂r ∂z ∂θ ∂σ r ∂τ rz 1 ∂τ rθ σ r − σθ + + + + γr = 0 r ∂r ∂ z r ∂θ (1.67) Similarly, for equilibrium in z and q directions, we get and ∂τ rz ∂σ z 1 ∂τθ z τ rz + + + + γz = 0 r ∂r ∂ z r ∂θ (1.68) ∂τ rθ ∂τθ z 1 ∂σ θ 2 τ rθ + + + + γθ = 0 r ∂r ∂ z r ∂θ (1.69) Equations (1.67)–(1.69) are the differential equations of equilibrium expressed in polar coordinates. 1.30 AXISYMMETRIC CASE AND PLANE STRESS CASE If an axisymmetric body is loaded symmetrically, the stress components do not depend on q. Since the deformations are symmetric, trq and tq z do not exist and consequently the above set of equations in the absence of body forces are reduced to ∂σ r ∂τ rz σ r − σ θ + + =0 r ∂r ∂z ∂τ r z ∂σ z τ rz + + =0 ∂r ∂z r A sphere under diametral compression or a cone under a load at the apex are examples to which the above set of equations can be applied. If the state of stress is two-dimensional in nature, i.e. plane stress state, then only sr, sq , trq , gr, and gq exist. The other stress components vanish.These nonvanishing stress components depend only on q and r and are independent of z in the absence of body forces. The equations of equilibrium reduce to ∂σ r 1 ∂τ rθ σ r − σ θ + + =0 r ∂ r r ∂θ ∂τ rθ 1 ∂σθ 2τ rθ + + =0 r ∂θ r ∂r (1.70) Analysis of Stress 49 Example 1.15 Consider a function f(r, q ), which is called the stress function. If the values of sr , sq , and trq are as given below, show that in the absence of body forces, these satisfy the differential equations of equilibrium. sr = 2 1 ∂φ + 1 ∂ φ 2 r ∂ r r ∂θ 2 sq = ∂ 2φ ∂ r2 ∂ 2φ ∂φ t rq = − 1 + 1 r ∂ r ∂θ r 2 ∂θ Solution The equations of equilibrium are ∂σr 1 ∂τ rθ σ r − σ θ + + =0 r ∂ r r ∂θ ∂τ rθ 1 ∂σ θ 2τ rθ + + =0 r ∂θ r ∂r Substituting the stress function in the first equation of equilibrium, ⎛ ∂φ 1 ∂ 2φ 2 ∂ 2φ 1 ∂ 3φ ∂ 3φ ∂ 2φ ⎞ − 12 + − 3 + 2 + 1 ⎜⎜ − 1 + 12 ⎟ 2 2 2 2 r ∂r r ∂r r ∂θ r ∂θ ∂ r r ⎝ r ∂θ ∂ r r ∂θ 2 ⎟⎠ ∂φ 1 ∂ 2φ 1 ∂ 2φ + 12 + − =0 r ∂ r r 3 ∂θ 2 r ∂ r 2 Hence, the first equation is satisfied. Similarly, it can easily be verified that the second condition also holds good. 1.1 It was assumed in Sec.1.2 that across any infinitesimal surface element in a solid, the action of the exterior material upon the interior is equipollent (i.e. equal in strength or effect) to only a force. It is also possible to assume that in addition to a force, there is also a couple, i.e. at any point across any plane n, n n M n there is a stress vector T and a couple-stress n T P Fig. 1.33 Problem 1.1 Chapter_01a.pmd 49 n vector M , as shown in Fig. 1.33. Show that a set of equations similar to Cauchy’s equations can be derived, i.e. if we know the couple-stress vectors on three mutually perpendicular planes passing through the point P, then we can determine the couple-stress vector on any plane n 7/3/2008, 5:29 AM 50 Advanced Mechanics of Solids passing through the point. The equations are n M x = Mxx nx + Myx ny + Mzx nz n M y = Mxy nx + Myy ny + Mzy nz n M z = Mxz nx + Myz ny + Mzz nz n n n n Mx , My , Mz are the x, y and z components of the vector M acting on plane n. 1.2 A rectangular beam is subjected to a pure bending moment M. The crosssection of the beam is shown in Fig. 1.34. Using the elementary flexure formula, determine the normal and shearing stresses at a point (x, y) on the plane AB shown. y A M M (x, y) 45° h x b B Fig. 1.34 Problem 1.2 6My ⎤ ⎡ ⎢ Ans. σ n = τ n = 3 ⎥ bh ⎦ ⎣ 1.3 Consider a sphere of radius R subjected to diametral compression (Fig. 1.35). Let sr, sq and sf be the normal stresses and trq , tqf and tfr the shear stresses at a point. At point P(o, y, z) on the surface and lying in the yz plane, determine the rectangular normal stress components sx, sy and sz in terms of the spherical stress components. [Ans. sx = sq ; sy = sf cos2 f ; sz = sf sin2 f] z P f q x Fig. 1.35 Problem 1.3 y The pressure of water on face OB is also shown. 1.1. nx = ± ⎜ ⎟ ⎝b −a⎠ ⎝a −b⎠ ⎢⎣ ⎥⎦ 1. n = ± 2 ⎤ 11 x ⎢⎣ 3 y 3 z 3 ⎥⎦ 1. The stresses at any point (x. . ⎡T11 2 1 ⎤ ⎡⎣τ ij ⎤⎦ = ⎢ 2 0 2⎥ ⎢ ⎥ ⎣⎢ 1 2 0 ⎥⎦ ⎡ Ans. Consider an element OCD and show that this element is in equilibrium under the action of the external forces (water pressure and gravity force) and the internally distributed forces across the section CD. on which the resultant stress vector is tangential to the plane ⎡a 0 d ⎤ ⎡⎣τ ij ⎤⎦ = ⎢ 0 b e ⎥ ⎢ ⎥ ⎣⎢ d e c ⎥⎦ 1/ 2 1/ 2 ⎡ ⎤ ⎛ b ⎞ ⎛ ⎞ . Determine the direction cosines of the normal to that plane.36 Problem 1.6 A cross-section of the wall of a dam is shown in Fig. determine the unit normal of a plane parallel to the z axis. nz = 0. n = ± 2 . y) are given by the following expressions sx = –g y ⎛ ρ 2γ − sy = ⎜⎜ tan β tan 3 β ⎝ γx txy = tyx = − tan 2 β tyz = tzx = sz = 0 o x ⎞ ⎛ γ ⎞ − ρ ⎟⎟ y ⎟⎟ x + ⎜⎜ 2 ⎠ ⎝ tan β ⎠ o ␤ C B A y Fig.6 D where g is the specific weight of water and r the specific weight of the dam material.e. T = 2. n = ± 1 .Analysis of Stress 51 1.5 If the rectangular components of stress at a point are as in the matrix below. n y = ± ⎜ a ⎟ . i. nz = 0 ⎥ ⎢ Ans.36. Determine T11 such that there is at least one plane passing through the point in such a way that the resultant stress on that plane is zero.4 The state of stress at a point is characterised by the matrix shown. 48 ⎤ ⎢ t2 = 7. sy = 8.616. sz = 0.627. nx = nz = 0. sn = 12. nz = 0 and q = 9.455 ⎥ ⎥ ⎢ s3 = 0. nx = 0. ny = 0.6. 18° with x axis.65. nx = ny = 0. nz = 0 ⎤ ⎢ s2 = 10. greatest shearing stress and octahedral stresses. nx = 0. ⎡ Ans. σ z′ = 2c ⎥ ⎢ ⎥ ⎢ s oct = –c. nx = 0.788. nx = 0. stress deviators.288 ⎣ 1. ny = 0. sz = 10.643. sy = c.52 ⎦ ⎣ 1. ny = 0.31.394.941.8. ny = 0. determine the principal shears and the associated normal stresses. txy = 3. nx = 0. –72° with x axis. s1 = 13.27.709. s1 = 23.94. tyz = tzx = 0. nz = 0.8. ny = 0. ⎡ Ans.52 Advanced Mechanics of Solids 1. sn = 8. sz = 4.53 ⎥ ⎥ ⎢ t1 = 3.8 The state of stress at a point is characterised by the components ⎤ ⎥ ⎥ ⎦ sx = 12.304 ⎢ s3 = – 47.10 For the state of stress at a point characterised by the components (in 1000 kPa) sx = 12. where c = 1000 kPa. t3 = 3.88. s1 = 16. sy = 4. sn = 4. nz = 1 ⎥ ⎥ ⎢ s3 = 3.188. nz = 0. sz = c.912. nz = 0.55.153.798 ⎦ ⎣ 1. nz = 0. tyz = tzx = 0 determine the principal stresses and their directions. nz = 1 ⎥ ⎢ ⎢ s3 = ( –2 – 10 )c. ny = 0. ny = 1 ⎦ ⎣ 0⎤ ⎡ 3 −10 ⎢ (ii) ⎡⎣τ ij ⎤⎦ = −10 0 30 ⎥⎥ ⎢ ⎣⎢ 0 30 −27 ⎦⎥ ⎡ Ans.41.322 ⎤ ⎢ s2 = 8. s1 = ( –2 + 10 )c. nz = 0. nx = 0. nz = 0. toct = 78 c ⎥ ⎢ 3 ⎥⎦ ⎢⎣ . σ x′ = – 4c.9 For Problem 1.84 Find the values of the principal stresses and their directions ⎡ Ans.11 Let sx = –5c. nz = 0. tyx = 5.7 Determine the principal stresses and their axes for the states of stress characterised by the following stress matrices (units are 1000 kPa).34 txy = 4. ny = 0.20. σ ′y = 2c. txy = –c.583.2° with x axis ⎥ ⎥ ⎢ ⎥ ⎢ tmax = 10 c. nx = 0.96. nz = 0 and q = 9. s1 = 50.6 ⎥ ⎥ ⎢ s 3 = –50.473 ⎢ s2 = 0. nz = 0. Determine the principal stresses.274. 0 24 ⎤ ⎡ 18 ⎢ (i) ⎡⎣τ ij ⎤⎦ = 0 −50 0 ⎥ ⎢ ⎥ ⎢⎣ 24 0 32 ⎥⎦ ⎡ Ans. ny = 0. nx = 0.2° with y axis ⎤ ⎥ ⎢ s2 = c. nz = 0 ⎦ ⎣ 1. ny = 0. principal axes.8 ⎤ ⎢ s2 = 0. nx = 0. ⎡ Ans.95. 000 N and a torque T = 5000 N cm. ⎡ Ans.13 Check whether these satisfy the equations of equilibrium. t = t = 1 Gq x 2 + y 2 .14 For the state of stress given in Problem 1. the octahedral shearing stress and the maximum shearing stress. txz = tzx = –Gq y. At any point P of the cross-section LN. At point A on the surface. Also show that the lateral surface is free of load. σ 1. 1.37) is subjected to a tensile force P = 10. determine the principal shears. show that n n n Tx =Ty =Tz = 0 1. i e. 1. t = –Gq 1 3 2 ⎢ 2 ⎢ ⎢ toct = 6 Gq x 2 + y 2 .38) is subjected to a torque T.13 A cylindrical rod (Fig.Analysis of Stress 53 1.2 = 2000 1 ± 13 / 5 Pa ⎤ ⎥ π ⎢ ⎥ ⎢ ⎥ 2000 13 ⎢ Pa τ max = ⎥ 5 π ⎢ ⎥ ⎢ ⎥ ⎢ τ oct = 4000 22 Pa ⎥ ⎢ 3π 5 ⎦ ⎣ 1.12 A solid shaft of diameter d = 10 cm (Fig. the following stresses occur sx = sy = sz = txy = tyx = 0. soct = 0 ⎢ 3 ⎢⎣ ( ) x2 + y2 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥⎦ . octahedral shear stress and its associated normal stress.38 Problem 1. 1.37 Problem 1.13. determine the principal stresses. tyz = tzy = Gq x x n L T y P z N Fig. 1. T P A Fig.12 ( ) ⎡ Ans. y and z. then n x = cos a.1) n 2y = (σ − σ 3 ) (σ − σ 1 ) + τ 2 (σ 2 − σ 3 ) (σ 2 − σ 1 ) (A1.3) For the above equations.18 are nx2 = (σ − σ 2 ) (σ − σ 3 ) + τ 2 (σ1 − σ 2 ) (σ1 − σ 3 ) (A1. nz = cos g. If nx. t z P3 n (s.A1.2) nz2 = (σ − σ 1 ) (σ − σ 2 ) + τ 2 (σ 3 − σ 1 ) (σ 3 − σ 2 ) (A1. y and z axes.Appendix 1 Mohr’s Circles It was stated in Sec. the principal axes coincide with the coordinate axes x. respectively with the x. 1. A plane through P with PN as normal will be parallel to a tangent plane at N to the unit sphere.16.40)–(1. the points are bounded by the three Mohr’s circles.1). A1. t ) for all possible planes passing through a point are marked on the s –t plane. P 2 and P 3 are the poles of this sphere (Fig.1. The same equations can be used to determine graphically the normal and shearing stresses on any plane with normal n. Consider a point N on the surface of the sphere. n y and nz are the direction cosines of the normal n to such a plane through P.17 that when points with coordinates (s. 1. Equations (1.1 P2 y s3 0 σ1 + σ 3 2 R2 R3 s 2 σ 1 + σ 2 s1 2 Mohr’s circles for three-dimensional state of stress s . t) R1 N x P P1 Fig. as in Fig.42) of Sec. The radius vector PN makes angles a. P1. Construct a sphere of unit radius with P as the centre. ny = cos b. b and g. The values of s and t for different positions of N moving along this circle can be obtained again from (Eq. one gets a family of circles. all with (σ1 + σ 3 ) centres at . The third circle with centre at and 2 radius R1 is not an independent circle since among the three direction cosines nx. The intersection point of σ2 + σ3 these two circles locates (s. the point N on the unit sphere moves on a circle parallel to the circle P1P3. we get another circle with centre at and 2 radius R1.2) as (s – s3)(s – s1) + t 2 = n2y (s2 – s3) (s2 – s1) 2 or σ1 + σ 3 ⎞ ⎛ (σ − σ 3 ) 2 + t 2 = n2y (s2 – s3)(s2 – s1) + 1 = R22 ⎜σ − ⎟ 2 ⎠ 4 ⎝ (σ1 + σ 3 ) and 2 radius equal to R2. nz). t ). If n = 0 we get a of nz. From Eq. This gives a circle parallel to the equatorial circle P1P2.3) (s – s1)(s – s2) + t 2 = nz2 (s3–s1)(s3–s2) 2 or σ1 + σ 2 ⎞ ⎛ (σ − σ 2 ) 2 + t 2 = nz2 (s3 – s1)(s3 – s2) + 1 = R32 ⎜σ − ⎟ 2 4 ⎝ ⎠ Since nz = cos g is a constant. the above equation describes a circle in the s – t σ + σ 2 on the s axis and radius equal to R . with 2 (σ 2 + σ 3 ) nx = cos a kept constant. (A1. all with centres at 1 z 2 Mohr’s circle. Similarly. if ny = cos b is kept constant. For different values of ny. we get the outermost circle. With ny = 0. A1. This describes a circle in the s – t plane with the centre at . only two are independent. This circle plane with the centre at 1 3 2 gives the values of s and t as N moves with g constant. ny.Analysis of Stress 55 Let point N move in such a manner that g remains constant. In order to determine the normal stress s and shear stress t on a plane with normal n = (nx. Similarly. For different values σ + σ 2 . we get a family of circles. ny and nz. we describe two circles with centres and radii as σ + σ3 and radius equal to R2 centre at 1 2 centre at σ1 + σ 2 and radius equal to R3 2 where R2 and R3 are as given in the above equation. This is therefore a necessary condition. If s1.3) The problem is to find the locus of Q.1 Normal n to a plane through P y′ ′ ′ nx′ = x .2) We have to show that there exist at least three mutually perpendicular planes on which the normal stresses are zero. T is a state of pure shear at P.Appendix 2 The State of Pure Shear n Theorem: A necessary and sufficient condition for T to be a state of pure shear is that the first invariant should be equal to zero. then. y¢. A2. i. y ¢. z¢) be a point on this normal (Fig. . n y′ = . Let n be the normal to one such plane. ⎡ 0 ⎢ ⎡⎣τ ij ⎤⎦ = ⎢τxy ⎢ ⎢⎣τxz τxy τxz ⎤ 0 τ yz ⎥ τ yz ⎥ ⎥ 0 ⎥⎦ n Therefore. If PQ = R.1) From Cauchy’s formula.e. A2. must be equal to zero. we proceed as follows: Given l1 = sx + sy + sz = 0. such that with respect to that frame. Let Q(x¢. (A2. Let Px¢y¢z¢ be the principal axes at P. y¢ Q (x ¢. if the state of stress T is a pure shear state. To prove that l1 = 0 is also a sufficient condition. the normal stress sn on a plane n with direction cosines nx′ . if there exists at least one frame of reference Pxyz. an invariant. l1 = 0. Since l1 = 0. s2 and s3. are the principal stresses then l1 = s1 + s 2 + s3 = 0 (A2. from Eq. n Proof: By definition. nz ′ = z R R R Since PQ is a pure shear normal. then l1. two cases are possible. z ¢) n P x¢ z¢ Fig.1). n y′ nz′ is sn = s1 nx2′ + s2 n 2y ′ + s3 nz2′ (A2.2) s1x¢2 + s2 y¢2 + s3z¢2 = R2sn = 0 (A2. Case (ii) One of the principal stresses (say s3) is zero. σ3 = 0 The above two cases cover all posibilities.2(a).Analysis of Stress 57 Case (i) If two of the principal stresses (say s 1 and s2) are positive. A2. as shown in Fig. an axis lying in the plane x¢ = y¢ and another lying the plane x¢ = –y¢. lies on the z¢ axis.3) gives s1x¢2 + s2 y¢2 – (s1 + s3)z¢2 = 0 (A2. A2.2 (a) Planes at 45° (b) Principal stress on an element under plane state of stress The above solutions show that for case (ii). the point Q. i. Let us consider case (ii) first since it is the easier one. This represents the z¢ axis. σ 1 > 0.e when s3 = 0 and s1 = –s2. as shown in Fig. This is the elementary case usually discussed in a plane state of stress. σ1 > 0. These are the z¢ axis.e. (ii) x¢ = +y¢ or x¢ = –y¢. Case (i) Since s3 = – (s1 + s2).4) This is the equation of an elliptic cone with vertex at P and axis along PZ¢ (Fig. and the other is negative. Case (ii) From Eq. there are three pure shear normals. A2. Eq.3) s1x¢ 2 – s1 y¢ 2 = 0 or x¢ 2 – y¢ 2 = 0 The solutions are (i) x¢ = 0 and y¢ = 0.e. . z¢ P x¢ p /4 y¢ (a) s3 = 0 s2 s1 = s2 (b) Fig. σ 2 = −σ 1 < 0. the point Q can lie in either of these two planes. z¢) can be anywhere on the surface of the cone. as the result will show. i. (A2.3).2(b). A2. y¢. i. (A2. σ 2 > 0. the third principal stress s3 is negative. so that one of the remaining principal stress s1 is positive. These represent two mutually perpendicular planes.e.e. The point Q(x¢. σ 3 = − (σ1 + σ 2 ) < 0 The case that s1 and s2 are negative and s3 is positive is similar to the above case. i. i. we get (σ 2 x1′2 + σ 1 y1′2 ) y ′2 + 2σ1 y ′y1′ + ⎡⎣σ 1 − (σ 2 + σ 1 ) x1′2 ⎤⎦ = 0 (A2. y1′ . Then.8) If Q(x¢.e.5).6) or equivalently Eqs (A2. it must satisfy Eq. Let Q1 ( x1′ .8). (A2. Let ( x2′ . y1′ . s1x¢2 + s2 y¢2 – (s1 + s2) = 0 (A 2.9) ⎡σ1 − (σ 2 + σ1 ) x1′2 ⎤ ⎣ ⎦ ′ ′ y2 y3 = 2 2 ⎡ ′ ′ ⎤ ⎣σ 2 x1 + σ1 y1 ⎦ (A 2. i. (A2.5) If Q lies on the elliptic cone also. Let Q(x¢.7) for x¢ and Eq. S being perpendicular to PQ 1. y3′ ) be the solutions. y2′ ) and ( x3′ . Clearly ⎡σ 2 − (σ1 + σ 2 ) y1′2 ⎤ ⎦ x2′ x3′ = ⎣ ⎡σ 2 x1′2 + σ1 y1′2 ⎤ ⎣ ⎦ (A 2.7) and (A2.6) by x1′2 and substituting for x¢x¢ 1 from Eq.5) and (A2. A2. y¢. (A2.8) for y¢. 1) be a point on the cone and let S be a plane passing through P and perpendicular to PQ 1.7) Similarly. (A2. we get two solutions for each. (A2. i. This gives σ 1 x ′2 y1′2 + σ 2 ( x1′ x′ + 1) 2 − (σ1 + σ 2 ) y1′2 = 0 or (σ 1 y1′2 + σ 2 x1′2 ) x′2 + 2σ 2 x1′ x′ + ⎡⎣σ 2 − (σ 1 + σ 2 ) y1′2 ⎤⎦ = 0 (A 2.10) Adding the above two equations x2′ x3′ + y2′ y3′ = σ 1 + σ 2 − σ 1 y1′2 − σ 2 y1′2 − σ 2 x1′2 − σ 1 x1′2 σ 2 x1′2 + σ 1 y1′2 . then it must satisfy Eqs (A2. multiplying Eq. x1′ x′ + y1′ y ′ + 1 = 0 (A2.3 Cone with vertex at P and axis along PZ ¢ Now one has to show that there are at least three mutually perpendicular generators of the above cone.6) by 2 y1′2 and substitute for y ′y1′ from Eq.58 Advanced Mechanics of Solids Q2 z Q1 Q3 P y x Fig. We have to show that the plane S intersects the cone along PQ 2 and PQ 3 and that these two are perpendicular to each other.5). (A2. 1) is a point lying in S as well as on the cone. PQ is perpendicular to PQ1.e.4). 1) be a point in S.6) Multiply Eq. Since these are quadratic. (A2. One can solve Eq. x2′ x3′ + y2′ y3′ + 1 = 0 Consequently. y3′ are real.7) and (A2.e. since s1 > 0 and s2 > 0. the solutions of Eqs (A2. the righthand side is equal to –1.e. . i. y3′ . If x2′ . y2′ . 1) i. x2′ . 1) be specifically Q1(1.8). 1) are real. x3′ and y2′ . 1) is on the cone and recalling that s1 + s2 = –s3. x3′ and y2′ . 1.Analysis of Stress 59 Since Q1 ( x1′ . y1′ . y1′ .7) and (A2. choose x1′ = y1′ = 1. PQ 2 and PQ 3 are perpendicular to each other if Q 2 = ( x2′ . let Q1 ( x1′ . For this. Therefore.8) then are 4(σ12 + σ1σ 2 + σ 22 ) The above quantity is greater than zero. y3′ . then the descriminants must be greater than zero. 1) and Q3 ( x3′ . Both the descriminants of Eqs (A2. are to be real. The St and S are right second-order cylinders whose generators are parallel to nz and whose cross-sections have axes along nx and ny. then we get a surface S and this surface determines the normal component of stress on any plane through P. then S = St is an elliptic cylinder. then St is a hyperbolic cylinder whose cross-section has vertices on the nx axis and Sc is a hyperbolic cylinder. St and Sc are each a central quadric surface about P with axes along nx. . then St is a hyperboloid with one sheet and Sc is a double sheeted hyperboloid. In particular. then St and Sc are hyperboloids of revolution with a polar axis along nz. A3. s2 > 0 and s3 < 0. (i) If s1. if s1 = s2. then S = St is a spheroid with polar axis along nz (b) If s1 = s2 = s3. (ii) If s1. The stress surface S associated with a given state of stress T is defined as the locus of all points Q. each perpendicular to n x and equidistant from P. s2 and s3 are not all of the same sign. The surface consists of St and Sc— the tensile and the compressive branches of the surface. There are two cases (a) If s1 = s2 π s3. In this case. s2 π 0 and s3 = 0 (i. s2 > 0. the vertices of which are along the nz axis. (ii) If s1 > 0 and s2 < 0. s3 > 0 then S = St is an ellipsoid with axes along nx. say s1 > 0.e. Thus. Case (ii) Let s1 π 0. such that PQ = Rn 1 where R = PQ = 1/ 2 σ ( n) ( ) and s (n) is the normal stress component on the plane n. If s1 = s2 then S = St is a circular cylinder. In particular. s2 > 0. two possibilities can be considered. The following cases are possible. If such Qs are marked for every plane passing through P. (i) If s1 > 0. n The normal to the surface S at Q(n) is parallel to T . s3 π 0. then S = St is a sphere.Appendix 3 Stress Quadric of Cauchy n Let T be the resultant stress vector at point P (see Fig. s2 and s3 all have the same sign. say s1 > 0. ny and nx at P. ny and nz. Case (iii) If s 1 π 0 and s 2 = s 3 = 0 (uniaxial state) and say s 1 > 0 then S = S t consists of two parallel planes.1) on a plane with unit n normal n. This means that a point Q is chosen along n such that R = 1/ σ . plane state). s2 π 0. S completely determines the state of stress at P. Case (i) s1 π 0. of which an equation in that plane is expressed by Eq. s2 π 0.53) of Sec. the surface of which is made of straight lines parallel to the z-axis. s3 < 0 St : (1/ x2 σ1 + ) (1/ 2 y2 σ2 − ) (1/ 2 z2 σ3 ) 2 =1 sc Sc : − Fig. the above equation becomes s1x2 + s2 y2 + 1 2 or x2 + y =1 (1/σ 1 )2 (1/σ 2 )2 .3) This is obviously a second-order cylinder. s3 > 0 Equation (A3. 1. s2 > 0.3). s2 > 0.1 Ellipsoidal surface st 1/ σ 3 (1/ x2 (1/ x2 σ1 σ1 ) (1/ 2 y2 + ) (1/ 2 y2 + σ2 σ2 + ) (1/ 2 + ) (1/ 2 z2 σ3 z2 σ3 ) 2 ) 2 =1 = −1 (A3. s3 π 0 (i) s1> 0.23. This equation represents an ellipsoid. Case (ii) Let s1 π 0. (i) If s1 > 0 and s2 > 0. s2 π 0 and s3 = 0.Analysis of Stress 61 One can prove the above statements directly from Eqs (1. passing through every point of the curve in the xy plane.1) shows that Sc is an imaginary surface and hence.2. This is shown in Fig. S = St.1) Case (i) s1 π 0. These equations are z S t : s 1x 2 + s 2 y 2 + s 3z 3 = 1 Sc : s1x2 + s2 y2 + s3z2 = –1 These can be rewritten as y St : P x Sc : Fig. Then Eq.2 One-sheeted and two sheeted hyperboloids (1/ x2 σ1 − ) (1/ 2 y2 σ2 + ) (1/ 2 z2 σ3 ) 2 =1 (A3. A3. (a) If s1 = s2 π s3 the central section is a circle (b) If s1 = s2 = s3 the surface is a sphere (ii) If s1 > 0.53) reduces to s1x2 + s2 y2 = ±1 (A3. A3. (A3. A3. (1. St is a one-sheeted hyperboloid and Sc is a two-sheeted hyperbloid.2) Hence. St is given by a hyperbolic cylinder. the elliptic cylinder becomes a circular cylinder. this becomes x 2 = 1/s1 or x = ±1/ σ 1 This represents two straight lines parallel to the y axis and equidistant from it. In particular. Eq. each perpendicular to nx and equidistant from P. Hence.62 Advanced Mechanics of Solids This is the equation of an ellipse in xy plane. s2 = s3 = 0. S = St is an elliptic cylinder. (ii) If s1 > 0 and s2 < 0. the cross-sectional vertices of which lie on the nx axis and Sc is given by a hyperbolic cylinder with its cross-sectional vertices lying on the n axis. Case (iii) If s1 π 0. S = St is given by two parallel planes. (1. then the equation becomes s1x2 – |s2| y2 = ±1 or x2/(1/s1)2 – y2 (1/σ 22 ) = ± 1 This describes conjugate hyperbolas in the xy plane. Hence.53) reduces to s1 x2 = ±1 When s1 > 0. . if s1 = s2. CHAPTER 2 Analysis of Strain 2. Dux Duy q2 Dy Dy q1 Dx Duy Dx Dux (b) (a) (c) Fig.1(c) shows the shear strain g xy in the xy plane. the element undergoes an extension Dux in x direction. In each case. These are two-dimensional strains. the initial or undeformed position of the element is indicated by full lines and the changed position by dotted lines. In elementary strength of materials two types of strains were introduced: (i) the extensional strain (in x or y direction) and (ii) the shear strain in the xy plane. 2.1(a). Figure 2. 2. If ex denotes the linear strain in x direction.1(b)] is ex = (2. Shear strain g xy is defined as the change in the initial right angle between two line elements originally . 2. the linear strain in y direction [Fig.1 INTRODUCTION In this chapter the state of strain at a point will be analysed.2) Figure 2.1 (a) Linear strain in x direction (b) linear strain in y direction (c) shear strain in xy plane In Fig.1) ∆u y ey = ∆y (2. then ∆u x ∆x Similarly.1 illustrates these two simple cases of strain. The extensional or linear strain is defined as the change in length per unit initial length. Since P is displaced to P¢. uy and uz along the x. x O y and z axes respectively.2 Displacement of point P to P¢ r¢ = r + u u = r¢ – r (2.2). the total change in the angle is q1 + q2.64 Advanced Mechanics of Solids parallel to the x and y axes. y¢. and r¢ that of point P¢.3) Therefore. we will study strains in three dimensions and we will begin with the study of deformations. y ¢.4) Reduction in the initial right angle is considered to be a positive shear strain. z¢) (Fig. then one can put q1 (in radians) + q2 (in radians) = tan q1 + tan q2 From Fig. Let a point P belonging to the body and having coordinates y (x. In the figure. If q1 and q2 are very small. This is change in volume per unit original volume. In this chapter. z) components ux. In addition to these two types of strains.6) . 2. called the volumetric strain. If r is the position vector of point P.2 DEFORMATIONS r¢ In order to study deformation or change in the shape of a body.1(c) tan q1 = ∆u y . 2. y. then Fig.e. We assume that the displacement is defined throughout the volume of the body. the vector segment PP¢ is called the P¢(x ¢. y. 2. the shear strain g xy is g xy = q1 + q2 = ∆u y ∆u x + ∆y ∆x (2. ∆x tan q2 = ∆u x ∆y (2. Then we can say that a displacement vector field has been defined throughout the volume of the body. the displacement vector u (both in magnitude and direction) of any point P belonging to the body is known once its coordinates are known. 2. z) be displaced after deformations to P¢ with coordinates (x¢. was also introduced in elementary strength of materials. and one can write u = iux + juy + kuz z (2. since positive shear stress components txy and tyx cause a decrease in the right angle. we compare the positions of material points before and after deformation. u The displacement vector u has r P(x.5) The displacement undergone by any point is a function of its initial coordinates. z ¢) displacement vector and is denoted by u. i. a third type of strain. y + Dy. 1. 2. Assuming that the displacement field given in Example 2. Hence. uz = 0. what is the distance between points P and Q after deformation? Solution The displacement vector at point P is u(P) = (0 + 0)i + (3 + 1) j + (0 + 0)k = 4j The displacement components at P are ux = 0. 1. 3). uz = 4 and the coordinates of Q¢ are (6. uz) (Fig. ux = 4. The distance P¢Q¢ is therefore d¢ = (62 + 22 + 22)1/2 = 2 11 2. –2) is u = (32 + 1)i + (3 – 2) j + (32 + 2)k = 10i + j + 11k The initial position vector r of point P is r = 3i + j – 2k The final position vector r¢ of point P¢ is r¢ = r + u = 13i + 2 j + 9k Example 2. –2)? Solution The displacement vector u at (3.3 DEFORMATION IN THE NEIGHBOURHOOD OF A POINT Let P be a point in the body with coordinates (x.1 65 The displacement field for a body is given by u = (x2 + y)i + (3 + z) j + (x2 + 2y)k What is the deformed position of a point originally at (3. 2. y. –1) respectively. Consider a small region surrounding the point P. Let Q be a point in this region with coordinates (x + Dx. uy = 2.1 has been imposed on the body. Let the displacement vector u at P have components (ux. . the points P and Q move to P¢ and Q¢. the displacement components at point Q are. uy.3). 1) and (2. 0.Analysis of Strain Example 2. uy = 4. When the body undergoes deformation. 4. the final coordinates of P after deformation are P¢ : x + ux = 0 + 0 = 0 y + uy = 0 + 4 = 4 z + uz = 1 + 0 = 1 or P¢: (0. 1) Similarly. z). z + Dz). 2 Two points P and Q in the undeformed body have coordinates (0. 0. the segment P¢Q¢ has components D x + Dux. Dy and D z along the three axes. the displacements at Q are. z) P¢: (x + ux. z + uz) Q: (x + Dx. One can represent these in the form of a matrix called the displacement-gradient matrix as ⎡ ∂ ux ⎢ ∂x ⎢ ⎡ ∂ ui ⎤ ⎢∂ uy ⎢∂ x ⎥ = ⎢ ∂ x ⎣ j⎦ ⎢ ⎢ ∂ uz ⎢ ∂x ⎣ ∂ ux ∂y ∂ uy ∂y ∂ uz ∂y ∂ ux ⎤ ∂z ⎥ ⎥ ∂ uy ⎥ ∂z ⎥ ⎥ ∂ uz ⎥ ∂ z ⎥⎦ . Terms like. These are the gradients of the displacement components (at a point P ) in x. we can also interpret the third term. Dy + Duy. the coordinates of Q¢ are. After deformation. Q¢ P¢ P Q x z Fig.3 Displacements of two neighbouring points P and Q ∆u x = ux + Dux. i. Dz + Duz along the three axes. Dy and Dz. P¢ and Q are P: (x. uz + Duz. For Duy and Duz too. y + Dy. we have ∆u y = ∂ uy ∂ uy ∂ uy ∆x + ∆y + ∆z ∂x ∂y ∂z (2. the segment PQ had components D x.7b) ∆u z = ∂ uz ∂u ∂u ∆ x + z ∆y + z ∆ z ∂x ∂y ∂z (2. z + Dz) The displacement components at Q differ slightly from those at P since Q is away from P by Dx. y and z directions. Similarly. Dy. Dx. y + Dy + uy + Duy. ∂x ∂y ∂z are important in the analysis of strain. Q¢ = (x + D x + ux + Dux. .7a) The first term on the right-hand side is the rate of increase of ux in x direction multiplied by the distance traversed. z + Dz + uz + Duz) (2. Consequently. uy + Duy. 2.e.8) Before deformation. then to first-order approximation ∂ ux ∂ ux ∂ ux ∆x + ∆y + ∆z ∂x ∂y ∂z (2. y + uy. ∂ ux ∂ ux ∂ ux .7c) Therefore. etc.66 Advanced Mechanics of Solids y The coordinates of P. y. The second term is the rate of increase of ux in y direction multiplied by the distance traversed in y direction. If Q is very close to P. . 2 Ds Dy = ny Ds = 0. y + Dy + uy + Duy. y + Dy. nz = 0.3 67 The following displacement field is imposed on a body u = (xyi + 3x2z j + 4k)10–2 Consider a point P and a neighbouring point Q where PQ has the following direction cosines nx = 0.072 + 0.555 Ds Using Eqs (2. y. uy and uz are the displacement components. Dy and Dz.01 Dx + 0. 3) therefore.800. 1.7a)–(2.218 Ds Dy + Duy = (0.555 Ds Hence. ux = pxy uy = 3px2z uz = 4p ∂ ux = py ∂x ∂ uy = 6 pxz ∂x ∂ uz =0 ∂x ∂ ux = px ∂y ∂ uy =0 ∂y ∂ uz =0 ∂y ∂ ux =0 ∂z ∂ uy = 3 px 2 ∂z ∂ uz =0 ∂z At point P(2.12 Dz) = (0.02 Dy = (0. the new vector P¢Q¢can be written as P¢Q¢ = (0. We are using p = 10–2.4 CHANGE IN LENGTH OF A LINEAR ELEMENT Deformation causes a point P(x. find the components of P¢¢ Q¢¢ after deformation.36 Dx + Dy + 0.067) Ds = 0. z + Dz + uz +Duz). z + Dz) gets displaced to Q¢ with new coordinates (x + Dx + ux + Dux. If PQ = D s.016) Ds = 0. z) in the solid body under consideration to be displaced to a new position P¢ with coordinates (x + ux.8 + 0. Hence. Solution Before deformation. A neighbouring point Q with coordinates (x + Dx.Analysis of Strain Example 2.939j + 0. the components of PQ are Dx = nx Ds = 0.555k)Ds 2. the values of Dux.202 + 0. it is possible to determine the . ny = 0. z + uz) where ux.8 Ds Dz = nz Ds = 0.218i + 0.939 Ds Dz + Duz = Dz = 0. 1. 3). Dux = (yDx + xDy)p = (Dx + 2Dy)p Duy = (6xzDx + 3x2Dz)p = (36Dx + 12Dz)p Duz = 0 Substituting for Dx.200. the components of Ds¢ = |P¢Q¢| are Dx + Dux = 1. Duy and Duz can be calculated. y + uy.555 Point P has coordinates (2.7c). Dz¢ = Dz + Duz) \ Ds¢ 2: (P¢Q¢ )2 = (Dx + Dux)2 + (Dy + Duy)2 + (Dz + Duz)2 From Eqs (2.10) 2 where 2 ⎣ 2 2 ⎡ ∂ uz 1 ⎛ ∂ ux ⎞ ⎛ ∂ u y ⎞ ⎛ ∂ u z + ⎢ + + E zz = ∂ z 2 ⎢⎝⎜ ∂ z ⎠⎟ ⎝⎜ ∂ z ⎠⎟ ⎜⎝ ∂ z ⎣ ∂ ux ∂ u y ∂ ux ∂ u x ∂ u y ∂ u y Exy = + + + ∂y ∂x ∂x ∂y ∂x ∂y ∂ u y ∂ uz ∂ ux ∂ ux ∂ u y + + + ∂z ∂y ∂y ∂z ∂y ∂ uy ∂ ux ∂ uz ∂ u x ∂ ux Exz = + + + ∂z ∂x ∂x ∂z ∂x Eyz = It is observed that Exy = Eyx. ⎛ ∂u ∆ x′ = ⎜ 1 + x ∂x ⎝ ∂ ux ∂ ux ⎞ ⎟ ∆ x + ∂ y ∆y + ∂ z ∆z ⎠ ∆y ′ = ∂ uy ∂ uy ⎞ ∂ uy ⎛ ∆ x + ⎜1 + ⎟ ∆y + ∂ z ∆ z ∂x ∂ y ⎝ ⎠ ∆z ′ = ∂ uz ∂u ⎛ ∂u ⎞ ∆ x + z ∆y + ⎜ 1 + z ⎟ ∆ z ∂x ∂y ∂z ⎠ ⎝ (2. Its components are Ds¢: (D x¢ = Dx + Dux.11) ∂ uz ∂ u z ∂x ∂y ∂ u y ∂ uz + ∂z ∂y ∂ u y ∂ uz + ∂z ∂x ∂ uz ∂z ∂ uz ∂z Exz = Ezx (2. Dz) \ Ds2: (PQ)2 = Dx2 + Dy2 + Dz2 Let Ds¢ be the length of P¢Q¢.68 Advanced Mechanics of Solids change in the length of the line element PQ caused by deformation. Dy¢ = Dy + Duy.7a)–(2. Its components are Ds: (Dx. Dy. ⎦ 2⎤ ⎞ ⎥ ⎟ ⎠ ⎥⎦ + (2.9) We take the difference between Ds¢2 and Ds2 (P¢Q¢ )2 – (PQ)2 = Ds¢ 2 – Ds2 = (D x¢ 2 + Dy¢ 2 + Dz¢ 2) – (D x2 + Dy2 + Dz2) = 2(Exx D x2 + Eyy Dy2 + Ezz Dz2 + Exy D x Dy + Eyz Dy Dz + Exz Dx Dz) Exx = 2 2 ∂ u x 1 ⎡⎛ ∂ u x ⎞ ⎛ ∂ u y ⎞ ⎛ ∂ u z ⎞ ⎤ ⎥ + ⎢⎜ + + ∂ x 2 ⎢⎝ ∂ x ⎠⎟ ⎝⎜ ∂ x ⎟⎠ ⎜⎝ ∂ x ⎟⎠ ⎥ ⎣ ⎦ Eyy = ∂ u y 1 ⎡⎛ ∂ u x ⎞ 2 ⎛ ∂ u y ⎞ ⎛ ∂ u z ⎞ 2 ⎤ ⎥ + ⎢ + + ∂ y 2 ⎢⎜⎝ ∂ y ⎟⎠ ⎜⎝ ∂ y ⎟⎠ ⎜⎝ ∂ y ⎟⎠ ⎥ (2.7c). Let Ds be the length of the line element PQ. We introduce the notation E PQ = ∆s ′ − ∆s ∆s Eyz = Ezy.12) . ⎛ ∆s ′ − ∆ s ( ∆ s ′ − ∆ s ) 2 ⎞ 2 ∆s′2 − ∆s 2 + = ⎜ ⎟ Ds 2 2 ∆s 2 ⎝ ∆s ⎠ ( ) ) Ds 1 2 = EPQ + 2 EPQ Ds2 ( 1E 2 PQ 2 From Eq.. .13) 2 ) EPQ 1 + 1 E PQ Ds2 = Exx Dx2 + Eyy Dy2 + Ezz Dz2 2 + Exy Dx Dy + Eyz Dy Dz + Exz Dx Dz If nx. then nx = ∆x . substituting for (Ds¢ – Ds2) = EPQ 1+ ( (2. etc. ∆s ∆s Substituting these in the above expression ( EPQ 1 + 1 E PQ 2 ) =E 2 xx nx nz = ∆z ∆s 2 + Eyy n y + Ezz nz2 + Exy nx ny + Eyznynz + Exznxnz (2.18) .14) Equation (2. etc.10).5 CHANGE IN LENGTH OF A LINEAR ELEMENT—LINEAR COMPONENTS Equation (2. . ∂uy / ∂y. ny and nz are the direction cosines of PQ. ∂uy /∂y. Ez = [1 + 2Ezz]1/2 – 1 (2. (2.16) This gives the relative extension of a line segment originally parallel to the x-axis. .14) gives the value of the relative extension at point P in the direction PQ with direction cosines nx. …. etc. ny = ∆y . By analogy. as well as non-linear terms.11) in the previous section contains linear quantities like ∂ ux /∂ x. If the line segment PQ is parallel to the x axis before deformation. Now. Retaining only linear terms. (∂ ux /∂x◊ ∂ux /∂y). then nx = 1. are extremely small so that their squares and products can be neglected. If the deformation imposed on the body is small. This quantity will be called the relative extension at point P in the direction of point Q.17) 2. Ex = [1 + 2Exx]1/2 – 1 (2. ∂ux /∂y. ∂y ezz = ∂ uz ∂z (2. ny and nz .15) 2 Hence. the following equations are obtained exx = ∂ ux . the quantities like ∂ux /∂x. ∂x eyy = ∂ uy ..Analysis of Strain 69 EPQ is the ratio of the increase in distance between the points P and Q caused by the deformation to their initial distance. like (∂ ux /∂x)2. ny = nz = 0 and ( ) Ex 1 + 1 E x = Exx (2. we get Ey = [1 + 2Eyy]1/2 – 1. The relations expressed by Eqs (2. When nx = 1. gyz and gxz represent shear strains in xy. 2. we could have written exy = gxy. 1/2gxy is written as exy. the line element PQ is parallel to the x axis and the linear strain is Similarly. (2.20) Equation 2. Since all that is required to determine the state of strain are the six rectangular strain components. y and z directions. ny = nz = 0. ny. It will be shown later that gxy.19) (2. eyy = eyy. these six components are said to define the state of strain at a point. i. we have retained this notation. 2. + ∂y ∂x gyz = ∂ u y ∂ uz . This definition is similar to that of the state of stress at a point.20 directly gives the linear strain at point P in the direction PQ with direction cosines nx.6 RECTANGULAR STRAIN COMPONENTS exx. nz. eyz = gyz.e. we will use only the linear terms in strain components and neglect squares and products of strain components.20). + ∂z ∂y gxz = ∂ ux ∂ uz + ∂z ∂x 2 EPQ ª ePQ = exx nx2 + eyy n y + ezz nz2 + exy nx ny + eyz ny nz + exz nx nz (2. In the subsequent analyses. 1 g = 1 ⎛ ∂ ux + ∂ u y ⎞ = e xy ∂ x ⎟⎠ 2 ⎜⎝ ∂ y 2 xy If we follow the above notation and use exx = exx. yz and xz planes respectively.70 Advanced Mechanics of Solids gxy = ∂ ux ∂ u y . using Eq. exz = gxz but as it is customary to represent the shear strain by g.18) and (2.22) . In the theory of elasticity.19) are known as the strain displacement relations of Cauchy. ezz = ezz (2.7 THE STATE OF STRAIN AT A POINT Knowing the six rectangular strain components at a point P. one can calculate the linear strain in any direction PQ.21) To maintain consistency. Ex ª exx = ∂ ux ∂x Ey ª eyy = ∂ uy ∂y and Ez ª ezz = ∂ uz ∂z are the linear strains in y and z directions respectively. We can write this as ⎡ε xx γ xy ⎢ [eij] = ⎢γ xy ε yy ⎢ ⎢⎣γ xz γ yz γ xz ⎤ ⎥ γ yz ⎥ ⎥ ε zz ⎥⎦ (2. eyy and ezz are the linear strains at a point in x. Analogous to the rectangular stress components. these six strain components are called the rectangular strain components at a point. The totality of all linear strains in every possible direction PQ defines the state of strain at point P. y) are ∂ uy ∂ ux ux + D x and uy + Dx ∂x ∂x . uy be the displacements of point P. gxz = + + + ∂y ∂x ∂z ∂y ∂z ∂x represent the shear strains in the xy. ∂x ∂ uy . gyz = . y and z. yz and xz planes respectively. Consider two line elements. point P moves to P¢. ∂y ∂ uz ∂z represent the linear strains of line elements parallel to the x.4 (a) Change in orientations of line segments PQ and PR-shear strain Let the coordinates of P be (x. 2.4a). Since point Q is Dx distance away from P.Analysis of Strain 71 then Eq. so that the coordinates of P¢ are (x + ux.8 INTERPRETATION OF g xy . g yz . exx = eyy = ezz = ∂ ux y ∂y y R¢ R q2 Q¢ Dy q1 ∂ uy ∂x Dx P¢ uy P Q ux x O Dx Fig. 2. g xz AS SHEAR STRAIN COMPONENTS It was shown in the previous section that ∂ ux . It was also stated that ∂ u y ∂ uz ∂ ux ∂ u y ∂ ux ∂ uz gxy = . y) before deformation and let the lengths of PQ and PR be Dx and Dy respectively. y and z axes respectively. After deformation. Let ux. (2. y + uy). PQ and PR. the displacement components of Q(x + Dx. This can be shown as follows. Note that eij = eji 2.20) can be written in a very short form as e PQ = ∑ ∑ eij ninj i j where i and j are summed over x. point Q to Q¢ and point R to R¢. originally perpendicular to each other and parallel to the x and y axes respectively (Fig. 4(a). Similarly.27) wx = .72 Advanced Mechanics of Solids Similarly. If the element PQR undergoes a pure rigid body rotation through a small angular displacement. the shear strains gyz and gzx can be interpreted appropriately. then q1 ª tan q1 = ∂ uy ∂x ∂ ux ∂y so that the total change in the original right angle is q2 ª tan q2 = q1 + q 2 = ∂ ux ∂ u y + = gxy ∂y ∂x (2.24) This is the shear strain in the xy plane at point P. No strain occurs during this rigid body displacement. 2. then from Fig.4 ∂u y taking the counter-clockwise rotation as positive.26) ∂u ⎞ ⎛∂u wy = 1 ⎜ x − z ⎟ = wxz 2⎝ ∂z ∂x ⎠ (2. 2.4(a). for rotations about the x and y axes. The negative sign in (–∂ux /∂ y) comes since a positive ∂ ux/∂ y will give a movement from the y to the x axis as shown in Fig. it is seen that if q1 and q2 are small. We define (b) Change in orientations of line segments PQ and PRrigid body rotation wz = 1 ⎛ ∂ u y − ∂ ux ⎞ = wyx (2. we get 1 ⎛ ∂ uz − ∂ u y ⎞ = wzy ∂ z ⎟⎠ 2 ⎜⎝ ∂ y (2. Similarly. the displacement components of R(x.23) (2. y +Dy) are ∂ uy ∂ ux ux + Dy and uy + Dy ∂y ∂y From Fig.25) ∂ y ⎟⎠ 2 ⎜⎝ ∂ x This represents the average of the sum of rotations of the x and y elements and is called the rotational component.4(b) we note ω zo = ∂uy ∂u =− x ∂x ∂y y R¢ wzo = ∂u x ∂y Q¢ wzo = P¢ ∂x x O Fig. 2. 2. with direction cosines nx. Ds¢ = Ds (1 + ePQ) (2. 0. uz + Duz where Dux. After deformation.12). ny and nz.4 Consider the displacement field u = [y2i + 3yz j + (4 + 6x2)k]10–2 What are the rectangular strain components at the point P(1. y. ∂x eyy = ∂ uy = 6 ¥ 10–2. then the displacement components of point Q are. If ux. P¢ and Q¢ are as follows: P: (x.7c). with direction cosines n'x. From Eq. uz are the displacement components of point P. z + D z) P¢: (x + ux. remembering that in the linear range EPQ = ePQ.Analysis of Strain 73 Example 2. (2. 2)? Use only linear terms. ux = y2 ◊ 10–2 uy = 3yz ◊ 10–2 uz = (4 + 6x2) ◊ 10–2 ∂ ux =0 ∂x ∂ uy =0 ∂x ∂ uz = 12x ◊ 10–2 ∂x ∂ ux = 2y ◊ 10–2 ∂y ∂ uy = 3z ◊ 10–2 ∂y ∂ uy = 3y ◊ 10–2 ∂z ∂ uz =0 ∂y Solution ∂ ux =0 ∂z ∂ uz =0 ∂z The linear strains at (1.7a)–(2. Let PQ be the element of length Ds. 0. y + Dy. (2. 2) are exx = ∂ ux = 0. y + uy. the element becomes P¢Q¢ of length Ds¢. ux + Dux. 0.28) The coordinates of P. uy. uy + Duy. Q. Duy and Duz are given by Eq. z) Q: (x + D x.9 CHANGE IN DIRECTION OF A LINEAR ELEMENT It is easy to calculate the change in the orientation of a linear element resulting from the deformation of the solid body. z + uz) Q¢: (x + D x + ux + Dux. 2) are gxy = ∂ ux ∂ u y =0+0=0 + ∂y ∂x ∂ u y ∂ uz =0+0=0 + ∂z ∂y ∂ ux ∂ uz gxz = = 0 + 12 ¥ 10–2 = 12 ¥ 10–2 + ∂z ∂x gyz = 2. y + Dy + uy + Duy. n'y and n′z. ∂y e zz = ∂ uz =0 ∂z The shear strains at (1. z + D z + uz + Duz) . 7c) nx′ = n′y = ⎡⎛ ∂ ux ⎞ ∂ ux ∂ ux ⎤ 1 1+ n + n + n 1 + ε PQ ⎢⎣⎜⎝ ∂ x ⎟⎠ x ∂ y y ∂ z z ⎥⎦ 1 1 + ε PQ ⎡∂ uy ∂ uy ⎞ ∂ uy ⎤ ⎛ nx + ⎜ 1 + n ⎢ ⎟ ny + ∂y ⎠ ∂ z z ⎥⎥⎦ ⎝ ⎣⎢ ∂ x (2. z + Dz).10 CUBICAL DILATATION Consider a point A with coordinates (x.7c) Dux = ∂ ux ∂ ux ∂ ux ∆x + ∆y + ∆z ∂x ∂y ∂z Duy = ∂ uy ∂ uy ∂ uy ∆x + ∆y + ∆z ∂x ∂y ∂z ∂ uz ∂ uz ∂ uz Dz ∆x + ∆y + ∂x ∂y ∂z The displaced segement A¢B¢ will have the following components along the x. ny = ∆y . y. (2. After deformation.30) .7a)–(2. uy and uz are the components of diplacements of point A. ∆x . nz = ∆s ′ ∆s ′ ∆s ′ nx = Substituting for Ds¢ from Eq. z + Dz + uz + Duz) where ux.74 Advanced Mechanics of Solids Hence. Duy. and from Eqs (2. y + Dy + uy + Duy. (2. (2. y + uy. y + Dy. nz = ∆z ∆s ∆s ∆s ∆y + ∆u y ∆ x + ∆u x ∆ z + ∆u z n′x = . Duz from Eq. z + uz) B¢ : (x + Dx + ux + Dux. n′y = . y and z axes: Duz = x axis: Dx + Dux = ⎛1 + ∂ u x ⎞ ∆x + ∂ u x ∆y + ∂ u x ∆z ⎜ ∂ x ⎟⎠ ∂y ∂z ⎝ y axis: Dy + Duy = ∂uy ∂uy ⎞ ∂uy ⎛ ∆x + ⎜ 1 + ∆y + ∆z ⎟ ∂x ∂ y ∂z ⎝ ⎠ z axis: Dz + Duz = ∂ u z ∆x + ∂ u z ∆y + ⎛ 1 + ∂ u z ⎜ ∂x ∂y ∂z ⎝ ⎞ ⎟ ∆z ⎠ (2. z) and a neighbouring point B with coordinates (x + Dx.7a)–(2.20). nz′ = 2. the points A and B move to A¢ and B¢ with coordinates A¢ : (x + ux.29) ⎡ ∂ uz ∂ uz ∂ uz ⎞ ⎤ ⎛ 1 n + n + 1+ n 1 + ε PQ ⎣⎢ ∂ x x ∂ y y ⎜⎝ ∂ z ⎟⎠ z ⎥⎦ The value of ePQ is obtained using Eq.28) and for Dux. y and z axes as P¢Q¢ P¢R¢ P¢S¢ x axis: ∂ ux ⎞ ⎛ ⎜1 + ∂ x ⎟ Dx ⎝ ⎠ ∂ ux Dy ∂y ∂ ux Dz ∂z y axis: ∂ uy Dx ∂x ∂ uy ⎞ ⎛ Dy ⎜1 + ∂ y ⎟⎠ ⎝ ∂ uy Dz ∂z z axis: ∂ u z Dx ∂x ∂ uz Dy ∂y ∂ uz ⎞ ⎛ ⎜1 + ∂ z ⎟ Dz ⎝ ⎠ The volume of the right parallelepiped before deformation is equal to V = Dx Dy Dz.5).30) the projections of P¢Q¢ will be ∂ ux ⎞ ⎛ along x axis: ⎜1 + Dx ∂ x ⎟⎠ ⎝ along y axis: ∂ uy Dx ∂x along z axis: ∂ uz Dx ∂x Hence. the right parallelepiped PQRS becomes an oblique parallelepiped P¢Q¢R¢S¢.5 Deformation of right parallelepiped Consider now an infinitesimal rectangular parallelepiped with sides Dx. one can successively identify AB with PQ (Dy = Dz = 0).Analysis of Strain 75 y R¢ R S¢ Q¢ P¢ S Q P O x z Fig. P¢R¢ and P¢S¢ along the x. 2. Dy and Dz (Fig. 2. PS ( Dx = Dy = 0) and get the components of P¢Q¢. The volume of the deformed parallelepiped is obtained from the well-known formula from analytic geometry as V ¢ = V + DV = D ◊ D x Dy Dz . When the body undergoes deformation. PR (Dx = Dz = 0). from Eqs (2.5 with AB of Eqs (2.30). Then. one has Dy = Dz = 0. Identifying PQ of Fig. 2. 64) – 0. the above determinant becomes Dª1+ ∂ ux ∂ uy ∂ uz + + ∂x ∂y ∂z = 1 + exx + eyy + ezz Hence.06 + 0 = 0.20) ePQ = 0. Example 2.02⎥⎥ ⎢ ⎢⎣ 0 − 0.76 Advanced Mechanics of Solids where D is the following determinant: ∂u x ⎞ ∂u x ∂u x ⎛ ⎜1 + ∂x ⎟ y ∂ ∂z ⎝ ⎠ ∂u y ∂u y ⎞ ∂u y ⎛ D= ⎜1 + ⎟ ∂x ∂ y ∂z ⎝ ⎠ ∂u z ∂u z ∂u z ⎞ ⎛ ⎜1 + ∂z ⎟ ∂x ∂y ⎝ ⎠ (2. determine ePQ .04 0 ⎤ [eij] = ⎢ − 0.48) = 0.5.08 . the new volume according to the linear strain theory will be V ¢ = V + DV = (1 + exx + eyy + ezz) D x Dy Dz (2. is equal to the sum of three linear strains. according to the linear theory.31) If we assume that the strains are small quantities such that their squares and products can be negelected.36) + 0. ny = 0 and nz = 0.5 The following state of strain exists at a point P ⎡ 0.8.34) Therefore.6 In Example 2. the volumetric strain.06 0. Solution From Eq.33) The volumetric strain is defined as D = ∆V = exx + eyy + ezz V (2.02 0 ⎥⎦ In the direction PQ having direction cosines nx = 0.02 − 0.02 (0) + 0 (0.007 Example 2.6.34) D = exx + eyy + ezz = 0.06 (0) + 0 (0.04 0.04 (0) – 0. (2. what is the cubical dilatation at point P? Solution From Eq.32) (2.02 + 0. also known as cubical dilatation.02 (0. (2. then q ¢ = 90∞ – a (2.36a) Now (90∞ – q ¢) represents the change in the initial right angle. then after strain. Therefore Eq. (2. 2. n ′y 2 and nz′ 2 can be substituted from Eq. nz1 and PR be another line element with direction cosines nx 2. (Fig. ny1. nz2 = 0 2 3 Determine the angle between the two segments before and after deformation. (2. 2.36a) gives the shear strain a between PQ and PR. Neglecting squares and products of small strain components.11 77 CHANGE IN THE ANGLE BETWEEN TWO LINE ELEMENTS Let PQ be a line element with direction cosines nx1. ny2.6).36c) Example 2. (2.7 The displacement field for a body is given by u = k(x2 + y)i + k(y +z)j + k(x2 + 2z2)k where k = 10–3 At a point P(2. 2. nz′1 .6 Change in angle between two line segments cos q¢ = cos q ¢= nx′1 nx′ 2 + n ′y1 n y′ 2 + nz′1 nz′ 2 The values of nx′1 n ′y1 .Analysis of Strain 2. R¢ y After deformation. nx′ 2 . Let q be the angle between the two line elements before deformation.36c) since a is small. if the two line segments PQ and PR are at right angles to each other before strain. q P¢ From analytical geometry Q P O x z Fig. nz2.35) In particular. 1 [(1 + 2exx) nx1nx2 + (1 + 2eyy) ny1ny2 (1 + ε PQ ) (1 + ε PR ) + (1 + 2ezz) nz1nz2 + gxy(nx1ny2 + nx2ny1) + gyz(ny1nz2 + ny2nz1) + gzx(nx1nz2 + nx2 nz1)] (2. PQ : nx1 = ny1 = nz1 = .36a). We can determine q ¢ easily from the Q¢ results obtained in Sec.29). consider two line segments PQ and PR having the following direction cosines before deformation 1 1 . cos q¢ = 1 [2exx nx1 nx2 + 2eyy ny1 ny2 + 2ezz nz1 nz2 (1 + ε PQ ) (1 + ε PR ) + gxy (nx1ny2 + nx2ny1) + gyz(ny1nz2 + ny2nz1) + gzx(nx1 nz2 + nx2 nz1)] (2.36b) and (2.36b) or cos q ¢ = cos (90∞ – a) = sin a ª a (2. 2. then gx¢y¢ at P = cos q ¢ = expression given in Eqs (2.9. If we represent the directions of PQ and PR at P by x¢ and y¢ axes. 3). the line segments beR come P¢Q¢ and P¢R¢ with an included angle q¢ q ¢. PR: nx2 = ny2 = . If this is denoted by a. nz) along which the strain is an extremum (i. (2. ∂ePQ /∂nz. (2.20) ( ) ( ) ( ) ( ) ( ) = k ⎡⎢( 4 × 1 ) + (1 × 1 ) + (12 × 0 ) + (1 × 1 ) + 0 + 0 ⎤⎥ = 3k 2 2 2 ⎣ ⎦ 1 1 1 1 1 1 23 k ePQ = k ⎡⎢ 4 × + + 12 × + 1 × + 1 × + 4 × ⎤⎥ = 3 3 3 3 3 3 ⎦ 3 ⎣ ePR After deformation.e. ny. we would have to equate. ny and nz are not all independent since they are related by the condition nx2 + n 2y + nz2 = 1 (2. Now we ask ourselves the following questions: What is the direction (nx. ny and nz change. ezz = ∂y ∂x ∂z gxy = ∂ u y ∂ uz ∂ ux ∂ u y ∂ uz ∂ u x = k. ny and nz were all independent. ∂ePQ /∂nx .37) . nx.78 Advanced Mechanics of Solids Solution Before deformation. we get different values of strain ePQ.20) as ePQ = exx nx2 + eyy n2y + ezz nz2 + gxynxny + gyznynz + gxznxnz As the values of nx. to zero. gyz = = k. the relative extension (i. the angle q between PQ and PR is cos q = nx1 nx2 + ny1ny2 + nz1nz2 = 1 + 1 = 0. 2. ∂ePQ /∂ny . eyy = = k. the angle beteween P¢Q¢ and P¢R¢ is from Eq.12 PRINCIPAL AXES OF STRAIN AND PRINCIPAL STRAINS It was shown in Sec.35) cos q ¢ = 1 ⎡ (1 + 23/3 k ) (1 + 3k ) ⎢⎣ (1 + 8k) 1 + (1 + 2k ) 1 + 0 6 6 ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎤ + ⎜ 1 + 1 ⎟ k + ⎜ 0 + 1 ⎟ k + ⎜ 0 + 1 ⎟ 4k ⎥ 6⎠ 6⎠ 6⎠ ⎦ ⎝ 6 ⎝ ⎝ = 0. gzx = = 4k + + + ∂y ∂x ∂z ∂y ∂x ∂z The linear strains in directions PQ and PR are from Eq. strain) at P in the direction PQ is given by Eq.8144 and q = 35. in order to find the maximum or the minimum.5 that when a displacement field is defined at a point P.5° 2. if nx.3° \ The strain components at P after deformation are exx = ∂ uy ∂ ux ∂ u z = 4kz = 12k = 2kx = 4k.8165 6 6 q = 35. maximum or minimum) and what is the corresponding extremum value? According to calculus. (2.e. However. which determine the direction along which the relative extension is an extremum. 2exxnx + gxyny + gxznz – 2enx = 0 (2. From Eq. 2nx ε xx + n y γ xy + nz γ zx nxγzx + n y γzy + 2nz ε zz = nx nz nx γzx + n y γzy + 2nz ε zz 2n y ε yy + nx γxy + nz γ yz = nz ny Denoting the right-hand side expression in the above two equations by 2e and rearranging.39c) the values of nx.38). the right-hand side becomes equal to 2e. Equations (2.39c) to get the values of nx. the value of e is equal to this extremum. we get 2 2(exx nx2 + eyy n y + ezz nz2 + gxynxny + gyznynz + gzxnznx) = 2e ( nx2 + n 2y + nz2 ) If we impose the condition nx2 + n 2y + nz2 = 1. Multiplying the first equation by nx.37) with respect to nx and ny we get 2nx + 2nz ∂ nz =0 ∂ nx ∂n 2ny + 2nz z = 0 ∂ ny (2. (2.39a) (2.39c) can be written as (exx – e)nx + 1 gxyny + 1 gxznz = 0 2 2 1 g n + (e – e)n + 1 g n = 0 (2. ny and nz.39b) gxynx + 2eyyny + gyznz – 2 eny = 0 and gzxnx + gzyny + 2ezznz – 2enz = 0 (2. ny and nz determine the direction along which the relative extension is an extremum and further. Therefore ePQ = e This means that in Eqs (2.38) Differentiating ePQ with respect to nx and ny and equating them to zero for extremum 0 = 2nxexx + nygxy + nzgzx + ∂ nz (n g + nygzy + 2nzezz) ∂ nx x zx 0 = 2nyeyy + nxgxy + nzgyz + ∂ nz (n g + nygzy + 2nzezz) ∂ n y x zx Substituting for ∂ nz/∂ nx and ∂ nz/∂ ny from Eqs (2.20). second by ny and the third by nz and adding them.40a) yy y 2 yx x 2 yz z 1 g n + 1 g n + (e – e)n = 0 zz z 2 zx x 2 zy y .39c) One can solve Eqs (2.39a)–(2.39a)–(2. (2. the left-hand side is the expression for 2ePQ. Let us assume that this direction has been determined.39a)–(2.Analysis of Strain 79 Taking nx and ny as independent and differentiating Eq. e 3 – J1e 2 + J2e – J3 = 0 (2.80 Advanced Mechanics of Solids If we adopt the notation given in Eq. (2. gyz and gzx. then e is the principal strain at P and n is the principal strain direction associated with e.9).45) are all similar to Eqs (1.40a) can be written as 1g = e zx 2 zx (exx – e)nx + exyny + exznz = 0 (2.13) and (1. The principal directions associated with e1. (exx – ei)nx + exy ny + exznz = 0 exynx + (eyy – ei)ny + eyznz = 0 (2. the determinant of its coefficient must be equal to zero. eyz and ezx not gxy. J3 are the first. (ε xx − ε ) exy exz e yx (ε yy − ε ) e yz ezx ezy exy exz J 3 = e yx ezx ε yy e yz ezy ε zz (2. we get e 3 – J 1e 2 + J 2 e – J 3 = 0 where J 1 = exx + eyy + ezz ε xx exy ε yy e yz ε xx + + J2 = ezx e yx ε yy ezy ε zz ε xx =0 (2. In every state of strain there exist at least three mutually perpendicular principal axes and at most three distinct principal strains. 2.12). (1.15 can be applied to the case of strain. 1.e. ny and nz. J2.8). 1g =e .42) (2. i.10–1.14). The results of Sec.43) exz ε zz (2. ny and nz. i.40b) eyxnx + (eyy – e)ny + eyznz = 0 ezxnx + ezyny + (ezz – e)nz = 0 The above set of equations is homogeneous in nx. second and third invariants of strain. are the roots of the cubic equation. 3) in the following equations and solving for nx. if the relative extension (i.e. The principal strains e1.44) (2.22). e2 and e3. (1.46) where J1. For the existence of a non-trivial solution.41)–(2. e2 and e3 are obtained by substituting ei (i = 1. put 1g =e . xy 2 xy 2 yz yz then Eqs (2. The problem posed and its analysis are similar to the analysis of principal stress axes and principal stresses.e. For a given state of strain at point P. (1. Equations (2. strain) e is an extremum in a direction n.45) It is important to observe that J2 and J3 involve exy.47) n 2x + n 2y + n 2z = 1 .41) (ε zz − ε ) Expanding the determinant. e2 and e3 are distinct.15 ( ) a = 1 − 69 − 36 = − 47 3 2 2 b = 1 ( − 432 − 621 + 243) = − 30 27 ezx = 1 γ zx 2 . 1. γ zx = z + x = 2 x ∂y ∂x ∂z ∂y ∂x ∂z At point (3.43) – (2.Analysis of Strain 81 If e1. 1. ε zz = z = 0 ∂x ∂y ∂z ∂ u y ∂ uz ∂ ux ∂ u y ∂u ∂u gxy = + = 1. e yz = 1 γ yz .46) is ε 3 − 6ε 2 − 23 ε + 9 = 0 2 Following the standard method suggested in Sec.8 The displacement field in micro units for a body is given by u = (x2 + y)i + (3 + z) j + (x2 + 2 y)k Determine the principal strains at (3. gzx = 6 The strain invariants from Eqs (2. (2. say e1 = e2 π e3. uz = x2 + 2y. The rectangular strain components are ∂ uy ∂ ux ∂u exx = = 2 x. then every direction is a principal direction. ε yy = = 0. gyz = 3. 2 2 1 6 3 2 3 = −9 J3 = 1 0 2 2 3 3 0 2 The cubic from Eq. then the axis of n3 is unique and every direction perpendicular to n3 is a principal direction associated with e1 = e2. ux = x2 + y. If e1 = e2 = e3 . ezz = 0 gxy = 1. n2 and n3 are unique and mutually peprendicular. Example 2. –2) and the direction of the minimum principal strain. then the axes of n1. If. uy = 3 + z. γ yz = + = 3. Solution The displacement components in micro units are. eyy = 0. –2) the strain components are therefore. 1.45) are J 1 = exx + eyy + ezz = 6 6 J2 = 1 2 1 0 2 + 3 0 2 3 6 2 + 3 0 3 0 = − 23 2 Note that J2 and J3 involve exy = 1 γ xy . exx = 6. Solution From the strain matrix given. The minimum principal strain is –2.02 − 0.01 − 0. ny = 0.02 0.01 0 The cubic equation is e3 – 0.06 + 0 = 0. the invariants are J 1 = exx + eyy + ezz = 0.39 ¥ 2 ¥ 0.01 = 0. determine the principal strains and the directions of the maximum and minimum principal strains.22 + 4.06 − 0. (2. the first invariant J1 is exx + eyy + e zz = e1 + e2 + e3 = 7.08e 2 + 0.6 The principal strains in micro units are e 1 = g cos φ /3 + 2 = + 7.9 For the state of strain given in Example 2.0001) + 0 + 0 = − 0.02 0 0 0 = (0.684 f = 46°48¢ \ g = 2 − a /3 = 5.49 The third invariant J3 is e1e2e3 = 7.02 − 0.39 – 2 + 0.06 + 0.02 −0.06 − 0.02 0 J 3 = − 0.78 – 1.47) (6 + 2) nx + 1 n y + 3nz = 0 2 1 n + 2n + 3 n = 0 y 2 x 2 z 2 2 nx + n y + nz2 = 1 The solutions are nx = 0.0001) + 0 = 0. Example 2.02 + 0. from Eq.801.267.39 e 2 = g cos (φ /3 + 120°) + 2 = − 2 e 3 = g cos (φ /3 + 240°) + 2 = + 0.51 = –11.0004) + (– 0.01 0 + 0.534 and nz = – 0.02 (−0.0012 – 0.000002 = 0 .61 As a check. For this.000002 0 −0.0007 0.61 = 6 The second invariant J2 is e1e2 + e2e3 + e3e1 = –14.61 = – 9 These agree with the earlier values.08 J2 = 0.02 0.82 Advanced Mechanics of Solids cos f = − −30 2 × − a 3 /27 = 0.0007e + 0.5. observing that the constant J3 in the cubic is very small.176 and nz = 0.08 J 2 = e1e2 + e2e2 + e3e1 = (0.4 ± 0.02nx + 0.20) as ε PQ = ε xx nx2 + ε yy n 2y + γ xy nx n y or if PQ makes an angle q with the x axis.47) 0.07 + 0.88. To determine the direction of e1 = 0. The non-vanishing strain components are exx.Analysis of Strain 83 Following the standard procedure described in Sec.06 – 0. ny. 2. such that for this system ezz = 0.005 Rearranging such that e1 ≥ e2 ≥ e3. For the other two solutions (e not equal to zero).02nx – 0. the principal strains are e1 = 0. 1. Similarly. eyy and gxy.13 PLANE STATE OF STRAIN If.0007 = 0 The solutions of this quadratic equation are e = 0.01nz = 0 nx2 + n 2y + nz2 = 1 The solutions are nx = 0.07 ¥ 0.02ny = 0 – 0. gzx = 0 (2.07. there exists a coordinate system Oxyz. gyz = 0.035.08 e + 0.07) ny – 0. e3 = 0 As a check: J 1 = e1 + e2 + e3 = 0. i. e2 = 0. from Eqs (2. with direction cosines nx.15. for e3 = 0. (2.01. one can determine the principal strains.07. then the relative extension or the strain ePQ is obtained from Eq.02nx + (0.0007e = 0 One of the solutions obviously is e = 0. one can ignore it and write the cubic as e 2 – 0.0007 J 3 = e 1 e2 e 3 = 0 (This was assumed as zero) Hence.08e 2 + 0.49) . If PQ is a line element in this xy plane. these values agree with their previous values.01 = 0. in a given state of strain.44.02 – 0.01nz = 0 nx2 + n 2y + nz2 = 1 The solutions are nx = ny = 0.47) (0.48) then the state is said to have a plane state of strain parallel to the xy plane.02ny = 0 –0. However.236 and nz = 0. then ε PQ = ε xx cos 2 θ + ε yy sin 2 θ + 1 γ xy sin 2θ 2 (2. from Eqs (2.944.06ny – 0.075 and 0.01) = 0. dividing by e e 2 – 0.e.07) nx – 0. 0. ny = – 0. given above. it remains perpendicular after strain also. (2. Since PR is an arbitrary perpendicular line to the principal axis PQ. exxnx1nx2 + eyyny1ny2 + ezznz1nz2 + exy(nx1ny2 + ny1nx2) + eyz (ny1nz2 + ny2nz1) + ezx (nx1nz2 + nx2nz1) = 0 Multiplying by 2 and putting 2exy = gxy. and exy = 1 gxy .50) ⎟ +⎜ ⎟ ⎥ 2 2 2 ⎢⎝ ⎠ ⎝ ⎠ ⎥⎦ ⎣ Note that e3 = ezz is also a principal strain.40b). .36a). 2ezx = gzx we get 2exxnx1nx2 + 2eyyny1ny2 + 2ezznz1nz2 + gxy (nx1ny2 + ny1nx2) + gyz (ny1nz2 + ny2nz1) + gzx (nx1nz2 + nz1nx2) = 0 Comparing the above with Eq. 2eyz = gyz. Since ePQ and ePR are quite general. and ezz in place of sx. at point P. ny2 and nz2 respectively and adding. we get cos q ¢ (1 + ePQ) (1 + ePR) = 0 where q ¢ is the new angle between PQ and PR after strain.40b) (exx – e1)nx1 + exyny1 + exznz1 = 0 exynx1 + (eyy – e1)ny1 + eyznz1 = 0 exznx1 + eyzny1 + (ezz – e1)nz1 = 0 Let nx2. ny2 and nz2 be the direction cosines of a line PR. 2 eyz = 1 gyz. Therefore. the line segments remain perpendicular after strain also. 2 2 2. perpendicular to PQ before strain. In particular. eyy. every line perpendicular to PQ before strain remains perpendicular after strain.e. Therefore. (2. ny1 and nz1.84 Advanced Mechanics of Solids If e1 and e2 are the principal strains. PR can be the second principal axis of strain. tyz and tzx respectively. if PS is the third principal axis of strain perpendicular to PQ and PR. i. such that γ xy tan 2 φ = (2. then 1/ 2 ⎡⎛ ε − ε ⎞ 2 ⎛ γ ⎞ 2 ⎤ xx yy xy ± ⎢⎜ ε1 . Then according to Eqs (2. The principal strain axes make angles f and f + 90∞ with the x axis. ezx = 1 gzx in place of txy. sy and sz respectively. = (2. we get. by nx2. q ¢ = 90∞. Repeating the above steps. to satisfy the equation. nx1nx2 + ny1ny2 + nz1nz2 = 0 Multiplying Eq.51) ε xx − ε yy ε xx + ε yy The discussions and conclusions will be identical with the analysis of stress if we use exx.14 THE PRINCIPAL AXES OF STRAIN REMAIN ORTHOGONAL AFTER STRAIN Let PQ be one of the principal extensions or strain axes with direction cosines nx1. ε 2 . Point a with coordinates d ¢ c b¢ q2 (r. 2. q ) g e t s moved to b¢ with coordinates Fig. The strain due to this part is ∂θ ∂ uθ ∆θ 1 ∂ uθ = ∂θ r ∆θ r ∂θ The total circumferential strain is therefore ∂u u εθ = r + 1 θ r r ∂θ Chapter_02. 2. 5:32 AM .53) 7/3/2008.52) The circumferential strain eq is caused in two ways.15 PLANE STRAINS IN POLAR COORDINATES We now consider displacements and deformations of a two-dimensional radial element in polar coordinates. The strain due to this radial movement alone is ur ∆θ ur = r ∆θ r In addition to this. The radial and cirx O cumferential displacements u q α= θ u u ∂ u are denoted by u r and u q .Analysis of Strain 85 we can identify a small rectangular element. c¢ q + a). The polar coordinates of a point a are r and q.7 Displacement components of a radial element ∂ ur ⎛ ⎞ ∂α ⎜ r + ∆r + ur + ∂ r ∆r . The neighbouring y p o i n t b( r + D r . then the length ad = r Dq changes to (r + ur)Dq. q ) gets displaced after q1 deformation to position a¢ with coordinates (r + u r . the change in ad is ∂uθ Dq. + r r r a r ∂r ∆ r Consider an elementary rab dial element abcd. 2. that will remain rectangular after strain also. as shown a¢ d in Fig.pmd 85 (2. If the element abcd undergoes a purely radial displacement.7. θ + α + ∂ r ∆r ⎟ ⎝ ⎠ The length of a¢b¢ is therefore ∆r + The radial strain er is therefore ∂ ur ∆r ∂r εr = ∂ ur ∂r (2. the point d moves circumferentially to d ¢ through the distance ∂u uθ + θ ∆θ ∂θ Since point a moves circumferentially through uq . with faces normal to the principal axes of strain. q1 = ∂ ur Dq.19). ezz. r u ∂u ⎛ ⎞ q 2 = ⎜ uθ + θ ∆r − uθ − θ ∆r ⎟ 1 r ∂r ⎝ ⎠ ∆r ∂u u = θ − θ ∂r r a= Similarly. uy. The magnitude of q2 is uq + ∂r ⎡⎛ ∂ uθ ⎤ 1 ⎞ ⎢⎜ uθ + ∂ r ∆r ⎟ − α ( r + ∆r ) ⎥ ∆r ⎠ ⎣⎝ ⎦ But Hence. gyz and gzx.54) COMPATIBILITY CONDITIONS It was observed that the displacement of a point in a solid body can be represented by a displacement vector u.16 (2. However. uz. the shear strain grq is ∂ ur ∂ uθ uθ + − r ∂θ ∂r r γ rθ = θ1 + θ 2 = 1 2.86 Advanced Mechanics of Solids To determine the shear strain we observe the following: The circumferential displacement of a is uq . given by Eqs (2. ux. determination of the three displacement functions from the six strain components is more complicated since it involves integrating six equations to obtain three functions. The total number of these relations are six and they fall into two groups. whereas that of d is ur + Hence. the reverse operation. the radial displacement of a is ur. whereas that of b is ∂ uθ Dr. uθ .e. i. The deformation at a point is specified by the six strain components. along the three axes x. The determination of the six strain components from the three displacement functions is straightforward since it involves only differentiation. . that all the six strain components cannot be prescribed arbitrarily and there must exist certain relations among these. y and z respectively. gxy. ∂θ 1 ⎡ ⎛ u + ∂ ur ∆ ⎞ − u ⎤ r ∆θ ⎢⎣⎜⎝ r ∂θ θ ⎟⎠ r ⎥⎦ ∂ ur =1 r ∂θ Hence. exx. therefore. One may expect. eyy.18) and (2. which has components. The three displacement components and the six rectangular strain components are related by the six strain displacement relations of Cauchy. exx and gzx. and ezz. we get two more conditions. We have ∂ ux ∂ u y gxy = + ∂y ∂x ∂ u y ∂ uz gyz = + ∂z ∂y ∂ uz ∂ u x gxz = + ∂x ∂z Differentiating ∂γ xy ∂ 2u y ∂ 2ux = + ∂ z∂ y ∂ z∂ x ∂z ∂γ yz ∂ 2u y ∂ 2uz = + ∂ x∂ z ∂ x∂ y ∂x . 2 2 ∂ 2ε xx ∂ ε yy ∂ γ xy + = ∂ x∂ y ∂ y2 ∂ x2 ∂ 2ε yy ∂ z2 + 2 ∂ 2ε zz ∂ γ yz = ∂ y 2 ∂ y∂ z (2. ∂y γ xy = ∂ ux ∂ u y + ∂y ∂x Differentiate the first two of the above equations as follows: 2 ∂ 2ε xx ∂ 3u x ⎛ ∂ ux ⎞ = = ∂ ⎜∂y ⎟ 2 2 ∂ x ∂ y ∂y ∂ x∂ y ⎝ ⎠ ∂ 2ε yy ∂ x2 = ∂ 3u y ∂ 2 ⎛ ∂ uy ⎞ = ⎜ ⎟ ∂ y ∂ x2 ∂ x ∂ y ⎝ ∂ x ⎠ Adding these two. we get 2 ∂ 2 ⎛ ∂ u x + ∂ u y ⎞ = ∂ γ xy ⎜ ⎟ ∂ x∂ y ⎝ ∂ y ∂ x ⎠ ∂ x∂ y i. by considering eyy.55) ∂ 2ε zz ∂ 2ε xx ∂ 2γ zx + = ∂ z∂ x ∂ x2 ∂ z2 Second group: This group establishes the conditions among the shear strains. 2 2 ∂ 2ε xx ∂ ε yy ∂ γxy + = ∂ x∂ y ∂ y2 ∂ x2 Similarly. This leads us to the first group of conditions. ∂x ε yy = ∂ uy .Analysis of Strain 87 First group: We have ε xx = ∂ ux .e. ezz and gyz. 56e) 2 ∂ ⎛ ∂γ xy + ∂γ yz − ∂γ zx ⎞ = ∂ ε yy ⎜ ⎟ 2 ∂y⎝ ∂z ∂x ∂y ⎠ ∂x∂z (2. these equations are also known as continuity equations. the six strain compatibility relations are 2 ∂ 2γ xy ∂ 2 ε xx ∂ ε yy = + ∂x ∂ y ∂ y2 ∂ x2 ∂ 2 ε yy (2. Collecting all equations.56c) ∂ z2 + ∂ ⎛ ∂γ yz + ∂γ zx − ∂γ xy ⎞ = ∂ 2ε zz 2 ∂ z ⎜⎝ ∂ x ∂y ∂ z ⎟⎠ ∂x∂ y (2.88 Advanced Mechanics of Solids ∂γ zx ∂ 2uz ∂ 2ux = + ∂y ∂ x∂ y ∂ y∂ z Adding the last two equations and subtracting the first ∂γ yz ∂γ zx ∂γ xy ∂ 2uz + − =2 ∂x ∂y ∂z ∂x∂ y Differentiating the above equation once more with respect to z and observing that ∂ 3u z ∂ 2ε zz = ∂ x∂ y∂ z ∂x∂ y we get. We can give a geometrical interpretation to the above equations. It is easy to conceive that if the components of strain are not connected by certain relations.56b) ∂ 2ε zz ∂ 2ε xx ∂ 2γ zx = + ∂z∂x ∂ x2 ∂ z2 (2. . By a cyclic change of the letters we get the other two equations. imagine an elastic body cut into small parallelepipeds and give each of them the deformation defined by the six strain components. ∂ 3u z ∂ 2ε zz ∂ ⎛ ∂γ yz + ∂γ zx − ∂γ xy ⎞ = =2 ⎜ ⎟ 2 ∂z ⎝ ∂x ∂y ∂z ⎠ ∂x∂ y∂z ∂x∂ y This is one of the required relations of the second group.56d) ∂ ⎛ ∂γ zx + ∂γ xy − ∂γ yz ⎞ = ∂ 2ε xx 2 ∂ x ⎜⎝ ∂ y ∂z ∂ x ⎟⎠ ∂y∂z (2. For this purpose. SaintVenant’s compatibility relations furnish these conditions. Hence. it is impossible to make a continuous deformed solid from individual deformed parallelepipeds.56a) ∂ 2γ yz ∂ 2 ε zz = ∂ y ∂z ∂ y2 (2.56f) The above six equations are called Saint-Venant’s equations of compatibility. e = 0 ∂z ∂ ux at ‘o’ is b and hence. ∆ z and ∆z ∂z ∂z ∂z Similarly. according to condition (b) ∂ uy ∂ ux ∆ z = 0 and. y Solution (i) Since point ‘o’ does not have any displacement ux = c = 0. f. (ii) Determine the strain components. ∆z = 0 ∂z ∂z and according to condition (c) ∂ ux ∆y = 0 ∂y Applying these requirements ∂ uy at ‘o’ is e and hence. f and k are constants. b. b. y. and t is the shear Fig. z) are obtained as o z x ux = –t yz + ay + bz + c uy = t xz – ax + ez + f uz = – bx – ey + k where a. a = 0 ∂y Consequently.Analysis of Strain 89 Example 2.8 stress. the displacement components at any point (x.8 Example 2. c. uz = k = 0 The displacements of a point Dz from ‘o’ are ∂ uy ∂ ux ∂ uz ∆ z. (iii) Verify whether these strain components satisfy the compatibility conditions. 2. (i) Select the constants a.10 For a circular rod subjected to a torque (Fig. uy = f = 0. (b) The element Dz of the axis rotates neither in the plane xoz nor in the plane yoz (c) The element Dy of the axis does not rotate in the plane xoy. c. e. the displacements of a point Dy from ‘o’ are ∂ uy ∂ ux ∂ uz ∆y. 2. ∆y and ∆y ∂y ∂y ∂y Hence. b = 0 ∂z ∂ ux at ‘o’ is a and hence.8). e. the displacement components are ux = –t yz. uy = t xz and uz = 0 . k such that the end section z = 0 is fixed in the following manner: (a) Point o has no displacement. (2. ∂y ∂ uy = −τ z + τ z = 0 ∂x ∂u + z =τx ∂x ∂ ux + = −τ y ∂z + (iii) Since the strain components are linear in x. ε yy = ∂ uy = 0.e. 1. (2.59) =0 This means that an element subjected to deviatoric strain undergoes pure deformation without a change in volume.57) is the spherical part of the strain matrix.17 STRAIN DEVIATOR AND ITS INVARIANTS Similar to the analysis of stress.57) is an isotropic state of strain.22. (2.34). i. the second matrix on the righthand side of Eq. Hence.90 Advanced Mechanics of Solids (ii) The strain components are ∂ ux ∂x ∂ ux gxy = ∂y ∂ uy gyz = ∂z ∂ uz gzx = ∂x exx = = 0. then according to Eq.57) ) (2. The spherical part of the strain matrix. we can resolve the eij matrix into a spherical (i. It is called isotropic because . the Saint-Venant’s compatibility requirements are automatically satisfied. This discussion is analogous to that made in Sec.e. The eij matrix is ⎡ε xx ⎢ ⎡⎣eij ⎤⎦ = ⎢ exy ⎢ ⎣⎢ exz exy ε yy e yz exz ⎤ ⎥ e yz ⎥ ⎥ ε zz ⎦⎥ This can be resolved into two parts as ⎡ε xx − e exy exz ⎤ ⎢ ⎥ ⎡eij ⎤ = ⎢ exy ε yy − e e yz ⎥ + ⎣ ⎦ ⎢ ⎥ e yz ε zz − e ⎦⎥ ⎣⎢ exz ( ⎡e ⎢ ⎢0 ⎢0 ⎣⎢ 0 0⎤ ⎥ e 0⎥ 0 e ⎥⎥ ⎦ (2. ε zz = 0. this part is also known as the pure shear part of the strain matrix. 2. y and z. If an isolated element of the body is subjected to the strain deviator only. isotoropic) and a deviatoric part.58) e = 1 ε xx + ε yy + ε zz 3 represents the mean elongation at a given point. the volumetric strain is equal to where ∆V = (e – e) + (e – e) + (e – e) xx yy zz V = exx + eyy + ezz – 3e (2. The first matrix represents the deviatoric part or the strain deviator. The second matrix on the righthand side of Eq. . 2. (i) Linear invariant is zero since J1′ = (exx – e) + (eyy – e) + (ezz – e) = 0 (ii) Quadratic invariant is (2. with a strain of magnitude e. 3)? [Ans. Use only linear terms.61) 2⎤ ⎥⎦ (iii) Cubic invariant is ⎡ε xx − e exy exz ⎤ ⎢ ⎥ J 3′ = ⎢ exy ε yy − e e yz ⎥ ⎢ ⎥ ezy ε zz − e ⎦⎥ ⎣⎢ exz (2. (1.0007.62) The second and third invariants of the deviatoric strain matrix describe the two types of distortions that an isolated element undergoes when subjected to the given strain matrix eij.0014)] 2.3 For the displacement field given in Problem 2. 3. 1). ⎡ ⎢ ⎢ Ans.20). then every direction is a principal strain direction. (2.2 The displacement field for a body is given by u = [(x 2 + y 2 + 2)i + (3x + 4y 2) j + (2x 3 + 4z)k]10–4 What is the displaced position of a point originally at (1. A sphere subjected to this state of strain will uniformally expand or contract and remain spherical. 2. ⎢⎣ ⎡4 1 0⎤ ⎤ ⎢0 0 1⎥ ⎥ ⎢ ⎥⎥ ⎢⎣ 4 2 0 ⎥⎦ ⎥⎦ 2. 3).60) exy ⎤ ⎡ε yy − e e yz ⎤ ⎡ε xx − e ⎡ε xx − e exz ⎤ J 2′ = ⎢ ⎥+⎢ ⎥+⎢ ⎥ e ε − − e ε e e ε e ⎢⎣ xy ⎥⎦ ⎢⎣ yz ⎥⎦ ⎣ xz zz − e ⎦ yy zz ( = − 1 ⎡ ε xx − ε yy 6 ⎢⎣ ( 2 2 ) + (ε yy − ε zz ) + (ε zz − ε xx )2 + 6 exy + e yx + ezx ) (2.1 The displacement field for a body is given by u = (x2 + y)i + (3 + z)j + (x2 + 2y)k Write down the displacement gradient matrix at point (2.2. what are the strain components at (1. 2.Analysis of Strain 91 when a body is subjected to this particular state of strain. according to Eq. Consider the invariants of the strain deviator.0019. 3. 2. These are constructed in the same way as the invariants of the stress and strain matrices with an appropriate replacement of notations. uy = kxy.4 What are the strain acomponents for Problem 2.3. 3) having direction cosines before deformation as (a) PQ: PR: (b) PQ: PR: nx1 = 0.7∞ = 4. Eyy = 16p + 136p . (a) ⎣ε ij ⎦ = k ⎢ x + y ⎢ ⎢⎣ 2 z ⎢ ⎢ (b) ε PQ = 4k ( x + y + 3 ⎣ x + y 2z ⎤⎤ ⎥⎥ x 2z ⎥⎥ 2z 2( x + y ) ⎥⎦ ⎥ ⎥ ⎥ z) ⎦ 2. with k = 0. 2. ny1 = nz1 = 1 2 nx2 = 1. Determine the principal strains and the direction of the maximum unit strain (i.6.0016. uz = 2k(x + y)z where k is a constant small enough to ensure applicability of the small deformation theory.8∞ = 0.5∞ – 50. exx = 0. if non-linear terms are also included? 2 2 2 ⎤ ⎡ Ans.8∞ ⎥⎦ 2. uz = 4kz2 k is a very small constant. (b) 4k . emax). Ezx = 6p + 24p where p = 10–4 ⎥⎦ 2. (c) 17. gzx = 0.5p .8 ⎡ Ans. The rectangular components of a small strain at a point is given by the following matrix.0004 ⎤ ⎢⎣ ⎥⎦ gxy = 0. eyy = 0.5 If the displacement field is given by ux = kxy. nz = 0. uy = k(4x + 2y2 + z). Ezz = 4p + 8p 2 2 ⎢⎣ Exy = 7p + 56p . ny = 1/ 2 . 0 ⎤ ⎡1 0 ⎢ ⎡⎣ε ij ⎤⎦ = p 0 0 − 4⎥⎥ where p = 10–4 ⎢ ⎢⎣0 − 4 3 ⎥⎦ . (a) 90∞ – 89. 3) in directions (a) nx = 0.6.001.8. determine the change in angle between two line segments PQ and PR at P(2. What are the strains at (2.0006 2.76k ⎤ ⎢⎣ ⎥⎦ 2 2. 2. (a) write down the strain matrix (b) what is the strain in the direction nx = ny = nz = 1/ 3 ? ⎡ ⎡y ⎢ ⎢ ⎡ ⎤ ⎢ Ans. gyz = 0. ny = 0.0002.2∞ ⎤ ⎢⎣ (b) 55.0007.7 For the displacement field given in Problem 2. ny1 = nz1 = ⎡ Ans.e. ezz = 0. Eyz = 0. nz2 = 0. nz = 1/ 2 (b) nx = 1. Exx = 2p + 24. ny2 = nz2 = 0 1 2 nx2 = 0. (a) 33 k .8 nx1 = 0. ny2 = 0. ny = nz = 0 (c) nx = 0.92 Advanced Mechanics of Solids ⎡ Ans.6 The displacement field is given by ux = k(x2 + 2z).6. nz = 0. gxy = A + Bxy (x2 + y2 – c 2). eyy = px.13 For the state of strain given in Problem 2. ezz = 2p(x + y) eyz = 2pz.10 2. B = 4. e1 = 4p.12. State the conditions under which the following is a possible system of strains: exx = a + b(x2 + y2) x4 + y4. ezz = 0 [Ans. b + b + 2c2 = 0] Given the following system of strains exx = 5 + x2 + y2 + x4 + y4 eyy = 6 + 3x2 + 3y2 + x4 + y4 gxy = 10 + 4xy (x2 + y2 + 2) ezz = gyz = gzx = 0 determine whether the above strain field is possible. Not satisfied] Verify whether the following strain field satisfies the equations of compatibility.894 ⎥ ⎢ for e : n = 1.894. gxy = 2xy + 2x3 [Ans. verify whether the compatibility condition is satisfied: exx = 3x2y.12 93 ⎡ Ans. ⎡ Ans. ezx = 2pz [Ans. gyz = 0 2 2 4 4 gzx = 0 eyy = a + b (x + y ) + x + y . If it is possible.447. n = n = 0 ⎥ 2 x y z ⎢ ⎥ ⎣ for e3 : nx = 0. ⎤ u x = 5 x + 1 x3 + xy 2 + 1 x5 + xy 4 + cy ⎥ ⎢ 3 5 ⎢ ⎥ ⎢ u y = 6 y + 3 x 2 y + y 3 + x 4 y + 1 y 5 + cx ⎥ 5 ⎣ ⎦ 2. Yes] gxy = p(x + y).9 2. e2 = p. p is a constant: exx = py. It is possible. e3 = –p ⎤ ⎢ for e1 : nx = 0. ⎡ Ans. Components of spherical part are ⎢ ⎢ e = 1 [11 + 4(x2 + y2) + 2(x4 + y4)] ⎢ 3 ⎢ Volumetric strain = 11 + 4(x2 +y2) + 2(x4 + y4) ⎢⎣ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥⎦ . ny = 0. ny = 0. nz = 0. eyy = 4y2x + 10–2.447 ⎦ For the following plane strain distribution.11 2.Analysis of Strain 2. write down the spherical part and the deviatoric part and determine the volumetric strain. determine the displacement components in terms of x and y. assuming that ux = uy = 0 and wxy = 0 at the origin. 2. We try to determine the displacements (ux.1) connected body. these conditions reduce to ∂ 2 exx ∂y 2 + ∂ 2 e yy ∂x 2 =2 ∂ 2 exy ∂ x ∂y Generally. eyy = eyy. dux = ∂u x ∂u dx + x dy ∂x ∂y Q ux(x2. and these were called compatibility conditions. x Consider the displacement ux Fig. continuous displacement functions. uy) at another point Q in terms of the known functions exx. 1 1 1 ezz = ezz. y2) joining the points P and Q. exx = exx. A. y2) = u°x + ∂u Q ∂u x x ∫ ∂x dx + ∫ ∂y dy P P Q Q ∂u x e d + = u°x + ∫ xx x ∫ ∂y dy P P . exy. their requirement for the existence of single-value displacement is not shown. w xy by means of a line integral over a simple continuous curve C P(x1. u°y) are known. In a two-dimensional case. P Since. and showing their equivalence in the above manner. y2) simply connected region at which the displacement (u°x. y2) = u°x + ∫ du x (A. eyy.y1) be some point in a Q(x2. eyy. ezx = γ zx ) should satisfy certain necessary 2 2 2 conditions for the existence of single-valued. In this section. this aspect will be treated y for the plane case. exy.16 that the six strain components eij (i.Appendix On Compatibility Conditions It was stated in Sec. eyz = γ zy . However. Let P(x1 .1 Continuous curve connecting Q P and Q in a simply ux(x2. these equations are obtained by differentiating the expressions for exx..e. exy = γ xy . 6) .2) becomes Q Q Q Q ⎛ ∂ω ∂ω yx ⎞ yx dx + dy ⎟ ux(x2.Analysis of Strain 95 ∂u x 1 ⎛ ∂u x ∂u y ⎞ 1 ⎛ ∂u x ∂u y ⎞ = ⎜ + + − ∂y ∂x ⎟⎠ 2 ⎜⎝ ∂y ∂x ⎟⎠ 2 ⎝ ∂y Now. ⎝ ⎠ Since the order of differentiation is immaterial.3) Substituting.22) and (2. ∂ω yx 1 ∂ ⎛ ∂ux ∂ux ⎞ 1 ∂ ⎛ ∂ux ∂u y ⎞ = + − + ∂x ∂x ⎟⎠ 2 ∂x ⎜⎝ ∂y ∂x ⎟⎠ 2 ∂y ⎜⎝ ∂x ∂ ∂ = exx − exy ∂y ∂x Similarly. y2) = u°x + ∫ exx dx + ∫ exy dx − ( yω yx ) − ∫ y ⎜ ∂x ∂y ⎠ P P ⎝ P P (A. ∂ω yx 1 ∂ ⎛ ∂u x ∂u y ⎞ = − ∂x ∂x ⎟⎠ 2 ∂x ⎜⎝ ∂y = 1 ∂ ⎛ ∂u x ∂u y ⎞ 1 ∂ ⎛ ∂u x ∂u x ⎞ − + − ∂x ⎟⎠ 2 ∂y ⎜⎝ ∂x ∂x ⎟⎠ 2 ∂x ⎜⎝ ∂y 1 ∂ ⎛ ∂ux ⎞ adding and subtracting 2 ∂y ⎜ ∂x ⎟ .25).2) Integrating by parts. Q Q Q P P P ux(x2. ∂ω xy 1 ∂ ⎛ ∂u x ∂u y ⎞ = − ∂y ∂x ⎟⎠ 2 ∂y ⎜⎝ ∂y 1 ∂ ⎛ ∂u x ∂u y ⎞ 1 ∂ ⎛ ∂u y ∂u y ⎞ = − + − ∂x ⎟⎠ 2 ∂x ⎜⎝ ∂y ∂y ⎟⎠ 2 ∂y ⎜⎝ ∂y 1 ∂ ⎛ ∂u x ∂u y ⎞ 1 ∂ ⎛ ∂u y ∂u y ⎞ = + − + ∂x ⎟⎠ 2 ∂x ⎜⎝ ∂y ∂y ⎟⎠ 2 ∂y ⎜⎝ ∂y = ∂ exy − ∂ e yy ∂y ∂x Chapter_02.pmd 95 7/3/2008. (A. = exy . 5:33 AM (A.5) (A. Eq. y2) = u°x + ∫ exx dx + ∫ exy dy − ∫ ω yx dy \ (A.4) Now consider the terms in the last integral on the right-hand side. the last integral on the right-hand side Q Q Q P P P ∫ ω yx dy = ( yω yx ) ∫ − ∫ yd (ω yx ) Q Q ⎛ ∂ω ∂ω yx ⎞ yx = ( yω yx ) − ∫ y ⎜ dx + dy⎟ ∂x ∂y ⎠ P P ⎝ (A.w yx from equations (2. This means ∂exy ⎤ ∂ ⎡ ∂e yx ∂e yy ⎤ ∂e ∂ ⎡ e − y xx + y = e −y +y ∂y ⎢⎣ xx ∂y ∂x ⎥⎦ ∂x ⎢⎣ xy ∂y ∂x ⎥⎦ i. 7/3/2008.6) in (A. the integral should be independent of the path of integration. This means that the integral is a perfect differential.. 5:34 AM .4) Q Q Q P P P ux(x2. Q Q⎡ ∂exy ⎤ ∂e dx ux(x2. the above equation reduces to ∂ 2 exx ∂y 2 + ∂ 2 e yy ∂x 2 =2 ∂ 2 exy ∂x∂y Q P Fig. Hence. ∂ 2 exy ∂exy ∂ 2 exy ∂ 2 e yy ∂exx ∂exx ∂ 2 exx ∂exy − −y + + y = − y + y ∂y ∂y ∂x ∂ x ∂y ∂x ∂x∂y ∂y 2 ∂x 2 Since exy = eyx.8) An identical expression is obtained while considering the displacement uy (x2. A multiply connected body can be made simply connected by a suitable cut.2 Chapter_02. The displacement functions will then become singlevalued when the path of integration does not pass through the cut.pmd Continuous curve connecting P and Q but not passing through the cut of multiply connected body 96 (A.(yw yx) + ∫ exx dx + ∫ ex y dy ⎡⎛ ⎞ ⎛ ⎞ ⎤ − ∫ y ⎢⎜ ∂ exx − ∂ exy ⎟ dx + ⎜ ∂ e yx − ∂ e yy ⎟ dy ⎥ y x y x ⎝ ∂ ∂ ⎠ ⎝ ∂ ∂ ⎠ ⎣ ⎦ Regrouping. the compatibility condition is a necessary and sufficient condition for the existence of single-valued displacement functions in simply connected bodies. y2) = u°x . y2) = u°x .e.(yw yx) + ∫ ⎢exx − y xx + y ∂y ∂x ⎥⎦ P P⎣ + ∂e ⎡ Q ∂e ⎤ yx yy ∫ ⎢ exy − y ∂y + y ∂x ⎥ dy P⎣ ⎦ (A. it is a necessary but not a sufficient condition. For a multiply connected body. A.7) Since the displacement is single–valued.96 Advanced Mechanics of Solids Substituting (A.5) and (A. y2). The constant of proportionality is the Young’s Modulus E. i. 3. its lateral dimensions. The strain components were related to the displacement components through six of Cauchy’s strain-displacement relationships.1 CHAPTER 3 INTRODUCTION In the preceding two chapters we dealt with the state of stress at a point and the state of strain at a point. In this chapter.1) It is also well known that when the uniform rod elongates.e. They depend on the manner in which the material resists deformation. they are approximations of the true behavioural pattern.Stress–Strain Relations for Linearly Elastic Solids 3. Consequently. i. the equations relate the state of stress at a point to the state of strain at the point. when P is gradually increased from zero to some positive value.e. The constitutive equations describe the behaviour of a material. We now extend this information or knowledge to relate the six rectangular components of stress to the six rectangular components of strain. decreases. the length of the rod also increases.2 GENERALISED STATEMENT OF HOOKE’S LAW Consider a uniform cylindrical rod of diameter d subjected to a tensile force P. not the behaviour of a body. In elementary strength of materials. its diameter. the relationships between the stress and strain components will be established. it is postulated in elementary strength of materials that the axial stress s is proportional to the axial strain e up to a limit called the proportionality limit. the ratio of lateral strain to longitudinal strain was termed as Poisson’s ratio n. Therefore. The constitutive equations are mathematical descriptions of the physical phenomena based on experimental observations and established principles. since an accurate mathematical representation of the physical phenomena would be too complicated and unworkable. Based on experimental observations. As is well known from experimental observations. Such equations are termed constitutive equations. We assume that each of the six independent . σ e = E or s = Ee (3. and vice versa. These planes were termed the principal planes and the stresses on these planes the principal stresses. The mathematical expressions of this statement are the six stress–strain equations: sx = a11ε xx + a12 ε yy + a13ε zz + a14γ xy + a15γ yz + a16 γ zx sy = a21ε xx + a22 ε yy + a23ε zz + a24γ xy + a25γ yz + a26γ zx s z = a31ε xx + a32 ε yy + a33ε zz + a34γ xy + a35γ yz + a36γ zx (3. b11.e.2) or that given by Eq. it was shown that at any point one can identify before strain. .14. . Since in an isotropic material. For homogeneous. i.98 Advanced Mechanics of Solids components of stress may be expressed as a linear function of the six components of strain and vice versa. whatever they may be. Solving Eq. (3. (3. Assuming that the material is isotropic.2) as six simultaneous equations. a small rectangular parallelepiped or a box which remains rectangular after strain. Whether we use the set given by Eq. 3. are constants for a given material. will remain rectangular . it has no directional property. linearly elastic material. In Chapter 1. This means that the material we are dealing with is isotropic.3). six strain-stress equations of the type: exx = b11σ x + b12σ y + b13σ z + b14τ xy + b15τ yz + b16τ zx (3. b12.3). For example.3) are known as Generalised Hooke’s Law. and that of homogeneity. the six Eqs (3.e. . Care must be taken to distinguish between the assumption of isotropy. one can show that only two independent elastic constants are involved in the generalised statement of Hooke’s law.2) txy = a41ε xx + a42 ε yy + a43ε zz + a44γ xy + a45γ yz + a46γ zx tyz = a51ε xx + a52 ε yy + a53ε zz + a54γ xy + a55γ yz + a56γ zx tzx = a61ε xx + a62 ε yy + a63ε zz + a64γ xy + a65γ yz + a66γ zx Or conversely. 36 elastic constants are apparently involved. . a small rectangular box the faces of which are subjected to pure normal stresses. one can get Eq. In Sec. there are no shear stresses on these planes. a12.3) eyy = . which is a statement that the stress-strain properties. . i. timber of regular grain is homogeneous but not isotropic. . it was shown that at any point there are three faces (or planes) on which the resultant stresses are wholly normal. 2. are the same at all points. (3. (3. The normals to the faces of this box were called the principal axes of strain.2) or (3.3 STRESS–STRAIN RELATIONS FOR ISOTROPIC MATERIALS We now make a further assumption that the ideal material we are dealing with has the same properties in all directions so far as the stress-strain relations are concerned. etc where a11. which is a particular statement regarding the stress-strain properties at a given point. As the next sections show. Since Ox¢ and Oy¢ are at right angles to each other. For the principal stress s1 the equation becomes s1 = ae1 + be2 + ce3 where a.e. x′ T z = nz1σ 3 These are the components in x. y and z coincide with the principal stress and principal strain directions. for s1 the equation becomes s 1 = ae1 + b(e2 + e3) = (a – b)e1 + b(e1 + e2 + e3) by adding and subtracting be1.9).6a) Similarly. the principal strain axes will also be along Ox.6b) . Oz¢. In other words. this can be extended to the relations between rectangular stress and strain components also. such that the direction cosines of Ox¢ are nx1. which are both at right angles to s1. 2 2 2 sn = σ x ' = nx1 σ 1 + n y1 σ 2 + nz1 σ 3 (3. ny2.4b) s 3 = lD + 2me3 (3. nz1 and those of Oy¢ are nx2. Oy.Stress–Strain Relations for Linearly Elastic Solids 99 after deformation (no asymmetrical deformation). x′ T y = n y1σ 2 . Denoting b by l and (a – b) by 2m. the normal to these faces coincide with the principal strain axes. 3. i. The resultant stress vector on the x¢ plane will have components as x' T x = nx1σ 1 . s3 with the three principal strains e1. for an isotropic material. nx1nx2 + ny1ny2 + nz1nz2 = 0 (3. For an isotropic body. Oz coincide with the principal stress axes. Oz. b and c are constants.4a) Similarly. Oy. the shear stress component on this x¢ plane in y¢ direction is obtained as the sum of the projections of the components in y¢ direction. Thus tx¢y¢ = nx1nx 2σ1 + n y1n y 2σ 2 + nz1nz 2σ 3 (3. for s2 and s3 we get s 2 = lD + 2me2 (3. ny1. Hence. But (e1 + e2 + e3) is the first invariant of strain J1 or the cubical dilatation D. Eqs. nz2. nz2. the equation for s1 becomes s 1 = lD + 2me1 (3. s2. ny2. Oy¢. The normal stress on this x¢ plane is obtained as the sum of the projections of the components along the normal. one can relate the principal stresses s1. Hence. must be the same for an isotropic material. y and z directions. there are only two elastic constants involved in the relations between the principal stresses and principal strains for an isotropic material.4 MODULUS OF RIGIDITY Let the co-ordinate axes Ox. Thus. e2 and e3 through suitable elastic constants.4c) The constants l and m are called Lame’s coefficients. (1. the effect of s1 in any direction transverse to it is the same in an isotropic material. Let the axes x. But we observe that b and c should be equal since the effect of s1 in the directions of e2 and e3. Consider another frame of reference Ox¢. which has direction cosines nx2.5) The normal stress sx¢ and the shear stress tx¢y¢ are obtained from Cauchy’s formula. 100 Advanced Mechanics of Solids On the same lines, if e1, e2 and e3 are the principal strains, which are also along x, y, z directions, the normal strain in x¢ direction, from Eq. (2.20), is 2 2 2 ex¢x¢ = nx1 ε1 + n y1 ε 2 + nz1 ε 3 (3.7a) The shear strain gx¢y¢ is obtained from Eq. (2.36c) as gx¢y¢ = 1 ( ⎡ 2 nx1nx 2ε1 + n y1n y 2ε 2 + nz1nz 2ε 3 (1 + ε x′ ) (1 + ε y′ ) ⎣ ) + nx1nx 2 + n y1n y 2 + nz1nz 2 ⎤⎦ Using Eq. (3.5), and observing that ex¢ and ey¢ are small compared to unity in the denominator, gx¢y¢ = 2 (nx1nx 2ε1 + n y1n y 2ε 2 + nz1nz 2ε 3 ) (3.7b) Substituting the values of s 1 , s 2 and s 3 from Eqs (3.4a)–(3.4c) into Eq. (3.6b) tx¢y¢ = nx1nx 2 (λ ∆ + 2 µ ε1 ) + n y1n y 2 (λ ∆ + 2 µ ε 2 ) + nz1nz 2 (λ ∆ + 2 µ ε 3 ) = λ ∆(nx1nx 2 + n y1n y 2 + nz1nz 2 ) + 2 µ (nx1nx 2 ε1 + n y1 n y 2 ε 2 + nz1 nz 2 ε 3 ) Hence, from Eqs (3.5) and (3.7b) (3.8) tx¢y¢ = mgx¢y¢ Equation (3.8) relates the rectangular shear stress component tx¢y¢ with the rectangular shear strain component gx¢y¢. Comparing this with the relation used in elementary strength of materials, one observes that m is the modulus of rigidity, usually denoted by G. By taking another axis Oz¢ with direction cosines nx3, ny3 and nz3 and at right angles to Ox¢ and Oy¢ (so that Ox¢y¢z¢ forms an orthogonal set of axes), one can get equations similar to (3.6a) and (3.6b) for the other rectangular stress components. Thus, 2 2 2 sy¢ = nx 2σ 1 + n y 2σ 2 + nz 2 σ 3 sz¢ = nx 32σ 1 + n y 32σ 2 + nz 32σ 3 (3.9a) (3.9b) ty¢z¢ = nx 2 nx 3σ1 + n y 2 n y 3σ 2 + nz 2 nz 3σ 3 (3.9c) tz¢x¢ = nx 3 nx1σ1 + n y 3 n y1σ 2 + nz 3 nz1σ 3 (3.9d) Similarly, following Eqs (3.7a) and (3.7b) for the other rectangular strain components, one gets 2 2 2 ey¢y¢ = nx 2ε1 + n y 2ε 2 + nz 2ε 3 (3.10a) 2 2 2 ez¢z¢ = nx 3 ε1 + n y 3 ε 2 + nz 3 ε 3 (3.10b) gy¢z¢ = 2(nx 2 nx 3ε1 + n y 2 n y 3ε 2 + nz 2 nz 3ε 3 ) (3.10c) gz¢x¢ = 2(nx 3 nx1ε1 + n y 3n y1 ε 2 + nz 3 nz1ε 3 ) (3.10d) From Eqs (3.6a), (3.4a)–(3.4c) and (3.7a) 2 2 2 sx¢ = nx1 σ 1 + n y1 σ 2 + nz1 σ 3 Stress–Strain Relations for Linearly Elastic Solids ( ) ( 2 2 2 2 2 2 = λ ∆ nx1 + n y1 + nz1 + 2µ ε1nx1 + ε 2 n y1 + ε 3 nz1 = l D + 2m ex¢x¢ Similarly, one gets sy¢ = l D + 2m ey¢y¢ sz¢ = l D + 2m ez¢z¢ Similar to Eq. (3.8), ty¢z¢ = m gy¢z¢ tx¢z¢ = m gz¢x¢ 101 ) (3.11a) (3.11b) (3.11c) (3.12a) (3.12b) Equations (3.11a)–(3.11c), (3.8) and (3.12a) and (3.12b) relate the six rectangular stress components to six rectangular strain components and in these only two elastic constants are involved. Therefore, the Hooke’s law for an isotropic material will involve two independent elastic constants l and m (or G). 3.5 BULK MODULUS Adding equations (3.11a)–(3.11c) σ x′ + σ y ′ + σ z ′ = 3λ∆ + 2µ (ε x′ x′ + ε y ′ y′ + ε z ′ z′ ) (3.13a) Observing that σ x′ + σ y′ + σ z ′ = l1 = σ1 + σ 2 + σ 3 (first invariant of stress), ε x′x′ + ε y ′ y ′ + ε z′ z ′ = J1 = ε1 + ε 2 + ε 3 (first invariant of strain), and Eq. (3.13a) can be written in several alternative forms as s1 + s2 + s3 = (3l + 2m)D sx¢ + sy¢ + sz¢ = (3l + 2m)D l 1 = (3l + 2m)J1 (3.13b) (3.13c) (3.13d) Noting from Eq. (2.34) that D is the volumetric strain, the definition of bulk modulus K is K= pressure p = volumetric strain ∆ (3.14a) If s1 = s2 = s3 = p, then from Eq. (3.13b) 3p = (3l + 2m)D or 3 p = (3l + 2m) ∆ and from Eq. (3.14a) K = 1 (3l + 2m) (3.14b) 3 Thus, the bulk modulus for an isotropic solid is related to Lame’s constants through Eq. (3.14b). 102 Advanced Mechanics of Solids 3.6 YOUNG’S MODULUS AND POISSON’S RATIO From Eq. (3.13b), we have D= σ1 + σ 2 + σ 3 ( 3λ + 2µ ) Substituting this in Eq. (3.4a) s1 = or e1 = λ ( 3λ + 2µ ) (σ 1 + σ 2 + σ 3 ) + 2 µ ε1 λ+µ µ ( 3λ + 2µ ) ⎡ ⎤ λ (σ 2 + σ 3 ) ⎥ ⎢σ 1 − 2 (λ + µ ) ⎣ ⎦ (3.15) From elementary strength of materials e 1 = 1 [σ 1 − v (σ 2 + σ 3 )] E where E is Young’s modulus, and n is Poisson’s ratio. Comparing this with Eq. (3.15), E= 3.7 µ ( 3λ + 2 µ ) ; (λ + µ ) ν= λ 2 (λ + µ ) (3.16) RELATIONS BETWEEN THE ELASTIC CONSTANTS In elementary strength of materials, we are familiar with Young’s modulus E, Poisson’s ratio n, shear modulus or modulus of rigidity G and bulk modulus K. Among these, only two are independent, and E and n are generally taken as the independent constants. The other two, namely, G and K, are expressed as G= E , 2 (1 + ν ) K= E 3 (1 − 2ν ) (3.17) It has been shown in this chapter, that for an isotropic material, the 36 elastic constants involved in the Generalised Hooke’s law, can be reduced to two independent elastic constants. These two elastic constants are Lame’s coefficients l and m. The second coefficient m is the same as the rigidity modulus G. In terms of these, the other elastic constants can be expressed as E= µ ( 3λ + 2 µ ) , (λ + µ ) λ 2 (λ + µ ) + 2µ ) νE , G ∫ m, l= (3.18) , 3 (1 + ν ) (1 − 2ν ) It should be observed from Eq. (3.17) that for the bulk modulus to be positive, the value of Poisson’s ratio n cannot exceed 1/2. This is the upper limit for n. For n = 1/2, 3G = E and K = • K= (3 λ n= Stress–Strain Relations for Linearly Elastic Solids 103 A material having Poisson’s ratio equal to 1/2 is known as an incompressible material, since the volumetric strain for such an isotropic material is zero. For easy reference one can collect the equations relating stresses and strains that have been obtained so far. (i) In terms of principal stresses and principal strains: s 1 = l D + 2me1 s 2 = l D + 2me2 (3.19) s 3 = lD + 2me3 where D = e1 + e2 + e3 = J1. e1 = ⎡ ⎤ λ+µ λ σ − (σ + σ 3 ) ⎥ µ ( 3λ + 2µ ) ⎢⎣ 1 2 ( λ + µ ) 2 ⎦ e2 = λ+µ µ ( 3λ + 2µ ) ⎡ ⎤ λ (σ 3 + σ 1 ) ⎥ ⎢σ 2 − + λ µ 2 ( ) ⎣ ⎦ (3.20) ⎡ ⎤ λ (σ 1 + σ 2 ) ⎥ ⎢σ 3 − + λ µ 2 ( ) ⎣ ⎦ (ii) In terms of rectangular stress and strain components referred to an orthogonal coordinate system Oxyz: sx = l D + 2mexx sy = l D + 2meyy s z = l D + 2mezz (3.21a) where D = exx + eyy + ezz = J1. e3 = λ+µ µ ( 3λ + 2µ ) txy = mgxy, tyz = mgyz, tzx = mgzx (3.21b) ⎡ ⎤ λ σy + σz ⎥ ⎢σ x − 2 (λ + µ ) ⎣ ⎦ exx = λ+µ µ ( 3λ + 2µ ) eyy = ⎡ ⎤ λ+µ λ (σ z + σ x ) ⎥ ⎢σ y − µ ( 3λ + 2µ ) ⎣ 2 (λ + µ ) ⎦ ezz = ⎡ ⎤ λ+µ λ σ x + σ y )⎥ ( ⎢σ z − µ ( 3λ + 2µ ) ⎣ 2 (λ + µ ) ⎦ gxy = 1 txy, µ ( gyz = 1 tyz, µ ) gzx = 1 tzx µ (3.22a) (3.22b) In the preceeding sets of equations, l and m are Lame's constants. In terms of the more familiar elastic constants E and n, the stress-strain relations are: (iii) with exx + eyy + ezz = J1 = D, ⎤ E ⎡ ν ∆ + ε xx ⎥ ⎢ + − 1 ν 1 2 ν ( ) ⎣( ) ⎦ = l J1 + 2Gexx sx = 104 Advanced Mechanics of Solids ⎤ E ⎡ ν ∆ + ε yy ⎥ ⎢ (1 + ν ) ⎣ (1 − 2ν ) ⎦ = l J1 + 2Geyy sy = sz = (3.23a) ⎤ E ⎡ ν ∆ + ε zz ⎥ ⎢ + − 1 ν 1 2 ν ( ) ⎣( ) ⎦ = l J1 + 2Gezz txy = Ggxy, tyz = Ggyz, ( txx = Ggzx (3.23b) ) exx = 1 ⎡σ x − ν σ y + σ z ⎤ ⎦ E⎣ eyy = 1 ⎡⎣σ y − ν (σ z + σ x ) ⎤⎦ E ( (3.24a) ) ezz = 1 ⎡σ z − ν σ x + σ y ⎤ ⎦ E ⎣ gxy = 1 txy, G 3.8 gyz = 1 tyz, G gzx = 1 tzx G (3.24b) DISPLACEMENT EQUATIONS OF EQUILIBRIUM In Chapter 1, it was shown that if a solid body is in equilibrium, the six rectangular stress components have to satisfy the three equations of equilibrium. In this chapter, we have shown how to relate the stress components to the strain components using the stress-strain relations. Hence, stress equations of equilibrium can be converted to strain equations of equilibrium. Further, in Chapter 2, the strain components were related to the displacement components. Therefore, the strain equations of equilibrium can be converted to displacement equations of equilibrium. In this section, this result will be derived. The first equation from Eq. (1.65) is ∂σ x ∂τ xy ∂τ zx =0 + + ∂x ∂y ∂z For an isotropic material sx = λ ∆ + 2 µ ε xx ; τ xy = µ γ xy ; τ xz = µ γ xz Hence, the above equation becomes ∂γ xy ∂γ xz ⎞ ⎛ ∂ε λ ∂∆ + µ ⎜ 2 xx + + =0 ∂x ∂y ∂ z ⎟⎠ ⎝ ∂x From Cauchy’s strain-displacement relations exx = ∂ ux , ∂x gxy = ∂ ux ∂ u y , + ∂y ∂x gzx = ∂ ux ∂ uz + ∂z ∂x Stress–Strain Relations for Linearly Elastic Solids 105 Substituting these 2 ⎛ ∂ 2u ∂ 2ux ∂ u y ∂ 2ux ∂ 2uz ⎞ x ⎟ =0 + + + + λ ∂∆ + µ ⎜ 2 ⎜ ∂ x2 ∂x ∂ y 2 ∂ x∂ y ∂ z 2 ∂ x∂ z ⎟⎠ ⎝ or ⎛ ∂ 2u ⎛ ∂ 2ux ∂ 2ux ∂ 2ux ⎞ ∂ 2u y ∂ 2uz ⎞ x ⎜ ⎟ =0 + + + + + λ ∂∆ + µ ⎜⎜ µ ⎟ 2 ⎜ ∂ x 2 ∂ x∂ y ∂ x∂ z ⎟ ∂x ∂ y2 ∂ z 2 ⎟⎠ ⎝ ∂x ⎝ ⎠ or ⎛ ∂ 2u ∂ 2ux ∂ 2ux ⎞ ∂ ⎛ ∂ ux + ∂ u y + ∂ uz ⎞ + λ ∂∆ + µ ⎜⎜ 2x + =0 ⎟⎟ + µ 2 2 ∂x ∂ x ⎜⎝ ∂ x ∂y ∂ z ⎟⎠ ∂y ∂z ⎠ ⎝ ∂x Observing that D = ε xx + ε yy + ε zz = ⎛∂u ( λ + µ ) ∂∂x ⎜ ∂ xx ⎝ + ∂ ux ∂ u y ∂ uz + + ∂x ∂y ∂z ∂ u y ∂ uz ⎞ ⎛ ∂ 2 u x ∂ 2 u x ∂ 2u x ⎞ + + + ⎟=0 ⎟ + µ ⎜⎜ 2 ∂y ∂z ⎠ ∂ y2 ∂ z 2 ⎟⎠ ⎝ ∂x This is one of the displacement equations of equilibrium. Using the notation —2 = ∂2 + ∂2 + ∂2 ∂ x2 ∂ y 2 ∂ z 2 the displacement equation of equilibrium becomes ( λ + µ ) ∂∂∆x + µ ∇ 2 u x =0 (3.25a) Similarly, from the second and third equations of equilibrium, one gets ( λ + µ ) ∂∂∆y + µ ∇2 u y =0 ( λ + µ ) ∂∂∆z + µ ∇ 2 u z =0 (3.25b) These are known as Lame’s displacement equations of equilibrium. They involve a synthesis of the analysis of stress, analysis of strain and the relations between stresses and strains. These equations represent the mechanical, geometrical and physical characteristics of an elastic solid. Consequently, Lame’s equations play a very prominent role in the solutions of problems. Example 3.1 A rubber cube is inserted in a cavity of the same form and size in a steel block and the top of the cube is pressed by a steel block with a pressure of p pascals. Considering the steel to be absolutely hard and assuming that there is no friction between steel and rubber, find (i) the pressure of rubber against the box walls, and (ii) the extremum shear stresses in rubber. 106 Advanced Mechanics of Solids p z l y x Fig. 3.1 Example 3.1 Solution (i) Let l be the dimension of the cube. Since the cube is constrained in x and y directions exx = 0 and eyy = 0 and sz = –p Therefore ( ) exx = 1 ⎡σ x − ν σ y + σ z ⎤ = 0 ⎦ E⎣ eyy = 1 ⎡⎣σ y − ν (σ x + σ z ) ⎤⎦ = 0 E Solving ν ν s =– p 1− ν z 1 −ν If Poisson’s ratio = 0.5, then sx = sy = sz = –p (ii) The extremum shear stresses are sx = sy = σ − σ3 σ − σ2 , , τ3 = 1 τ1 = 2 2 2 2 If n £ 0.5, then sx and sy are numerically less than or equal to sz. Since sx, sy and sz are all compressive t2 = σ1 − σ 3 s1 = sx = – ν p 1 −ν s2 = sy = – ν p 1 −ν s 3 = sz = –p \ ( ) 1 − 2ν 1 − 2ν t1 = p 1 − ν = p, τ 2 = p, τ 3 = 0 1 −ν 1 −ν 1 −ν If n = 0.5, the shear stresses are zero. Example 3.2 A cubical element is subjected to the following state of stress. sx = 100 MPa, sy = –20 MPa, sz = – 40 Mpa, txy = tyz = tzx = 0 Stress–Strain Relations for Linearly Elastic Solids 107 Assuming the material to be homogeneous and isotropic, determine the principal shear strains and the octahedral shear strain, if E = 2 ¥ 10 5 MPa and n = 0.25. Solution Since the shear stresses on x, y and z planes are zero, the given stresses are _ s 2 _> s 3 principal stresses. Arranging such that s 1 > s 1 = 100 MPa, s2 = –20 MPa, The extremal shear stresses are s3 = – 40 MPa t1 = 1 (s2 – s3) = 1 (–20 + 40) = 10 Mpa 2 2 t2 = 1 (s3 – s1) = 1 (–40 – 100) = –70 Mpa 2 2 t 3 = 1 (s1 – s2) = 1 (100 + 20) = 60 Mpa 2 2 The modulus of rigidity G is G= 2 × 105 E = 8 ¥ 104 MPa = 2 (1 + ν ) 2 × 1.25 The principal shear strains are therefore τ1 g1 = g2 = g3 = = 10 = 1.25 ¥ 10–4 8 × 104 =− 70 = − 8.75 × 10−4 8 × 104 G τ2 G τ3 G = 60 = 7.5 ¥ 10–4 8 × 104 From Eq. (1.44a), the octahedral shear stress is t0 = 1 [(s1 – s2)2 + (s2 – s3)2 + (s3 – s1)2]1/2 3 = 1 [1202 + 202 + 1402]1/2 = 61.8 MPa 3 The octahedral shear strain is therefore g0 = τ0 G = 61.8 4 = 7.73 ¥ 10–4 8 × 10 3.1 Compute Lame’s coefficients l and m for (a) steel having E = 207 ¥ 106 kPa (2.1 ¥ 106 kgf/cm2) and n = 0.3. (b) concrete having E = 28 ¥ 106 kPa (2.85 ¥ 105 kgf/cm2) and n = 0.2. Chapter_03.pmd 107 7/3/2008, 5:39 AM 108 Advanced Mechanics of Solids ⎡ Ans. (a) 120 ¥ 106 kPa (1.22 ¥ 106 kgf/cm2), 80 ¥ 106 kPa ⎡ ⎢ (8.1680 ¥ 105 kgf/cm2) ⎢ ⎢ 6 4 (b) 7.8 ¥10 kPa (7.96 ¥ 10 kgf/cm2), 11.7 ¥ 106 kPa ⎢ ⎢ ⎢ (1.2 ¥ 105 kgf/cm2) ⎣ ⎣ 3.2 For steel, the following data is applicable: E = 207 ¥ 106 kPa (2.1 ¥ 106 kgf/cm2), and G = 80 ¥ 106 kPa (0.82 ¥ 106 kgf/cm2) For the given strain matrix at a point, determine the stress matrix. 0 −0.002 ⎤ ⎡ 0.001 ⎢ [eij] = 0 −0.003 0.0003⎥⎥ ⎢ ⎢⎣ −0.002 0.003 0 ⎥⎦ 0 −160⎤ ⎡ −68.4 ⎡ ⎡ ⎢ ⎥ ⎢ ⎢ 0 − 708.4 24 3 ⎥ ¥ 10 kPa ⎢ ⎢ Ans. [tij] = ⎢ ⎢ ⎢ 24 −228.4⎦⎥ ⎣⎢ −160 ⎣ ⎣ 3.3 A thin rubber sheet is enclosed between two fixed hard steel plates (see Fig. 3.2). Friction between the rubber and steel faces is negligible. If the rubber plate is subjected to stresses sx and sy as shown, determine the strains exx and eyy, and also the stress ezz z sy x sx Fig. 3.2 Example 3.2 y ⎤ Ans. sz = +n (sx + sy) ⎥ ⎥ exx = + 1 + ν [(1 – n)sx – nsy] E ⎥ ⎥ ⎥ e = + 1 + ν [(1 – n)s ns ] y x ⎥⎦ yy E ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦ Theories of Failure or Yield Criteria and Introduction to Ideally Plastic Solid 4.1 CHAPTER 4 INTRODUCTION It is known from the results of material testing that when bars of ductile materials are subjected to uniform tension, the stress-strain curves show a linear range within which the materials behave in an elastic manner and a definite yield zone where the materials undergo permanent deformation. In the case of the so-called brittle materials, there is no yield zone. However, a brittle material, under suitable conditions, can be brought to a plastic state before fracture occurs. In general, the results of material testing reveal that the behaviour of various materials under similar test conditions, e.g. under simple tension, compression or torsion, varies considerably. In the process of designing a machine element or a structural member, the designer has to take precautions to see that the member under consideration does not fail under service conditions. The word ‘failure’ used in this context may mean either fracture or permanent deformation beyond the operational range due to the yielding of the member. In Chapter 1, it was stated that the state of stress at any point can be characterised by the six rectangular stress components—three normal stresses and three shear stresses. Similarly, in Chapter 2, it was shown that the state of strain at a point can be characterised by the six rectangular strain components. When failure occurs, the question that arises is: what causes the failure? Is it a particular state of stress, or a particular state of strain or some other quantity associated with stress and strain? Further, the cause of failure of a ductile material need not be the same as that for a brittle material. Consider, for example, a uniform rod made of a ductile material subject to tension. When yielding occurs, (i) The principal stress s at a point will have reached a definite value, usually denoted by sy; (ii) The maximum shearing stress at the point will have reached a value equal to t = 1 s y; 2 (iii) The principal extension will have become e = sy /E; (iv) The octahedral shearing stress will have attained a value equal to ( 2 /3) sy; and so on. Chapter_04.pmd 109 7/3/2008, 5:42 AM 110 Advanced Mechanics of Solids Any one of the above or some other factors might have caused the yielding. Further noteworthy observations on the bursting action of a liquid which is used to transmit pressure were made by Bridgman who found that cylinders of hardened chrome-nickel steel were not able to withstand an internal pressure well if the liquid transmitting the pressure was mercury instead of viscous oil. is slightly different in its approach and will be discussed separately. There are six main theories of failure and these are discussed in the next section. It is necessary to remember that failure may mean fracture or yielding. metals and other crystalline solids (including consolidated natural rocks) which are impervious. Further. In less compact solid materials. Another theory. however. the maximum principal stress in the material determines failure regardless of what the other two principal stresses are. so long as they are algebraically smaller.On the other hand. high pressure gradients are caused on the surface of the material which tend to burst or explode the glass. Most solid materials can withstand very high hydrostatic pressures without fracture or without much permanent deformation if the pressure acts uniformly from all sides as is the case when a solid material is subjected to high fluid pressure. there will be many criteria or theories of failure. are elastically compressed and can withstand very high hydrostatic pressures. called Mohr’s theory. The critical value obtained from this test will have to be applied for the stress or strain at a point in a general machine or a structural member so as not to initiate failure at that point. This is so because the state of stress or strain which causes the failure of the material concerned can easily be calculated. but fail either during the period the pressure is being reduced or later when the pressure is rapidly released. undergo considerable permanent deformation when subjected to high hydrostatic pressures. a marked evidence of failure has been observed when these solids are subjected to hydrostatic pressures. like that of a uniaxial tension or a pure torsion test. It is stated that the liquid could have penentrated through the fine invisible surface cracks and when the pressure was released. whereas the large molecules of oil are not able to penentrate so easily. Whatever may be the theory adopted. this penentration and the consequent failure of the material can be prevented if the latter is covered by a thin flexible metal foil and then subjected to high hydrostatic pressures.2 THEORIES OF FAILURE Maximum Principal Stress Theory This theory is generally associated with the name of Rankine. 5:42 AM . Consequently. like glass bulbs. As Karman pointed out. the entrapped liquid may not have been able to escape fast enough. Chapter_04. the factor that causes a ductile material to yield might be quite different from the factor that causes fracture in a brittle material under the same loading conditions.pmd 110 7/3/2008. the information regarding it will have to be obtained from a simple test. it has been observed that even brittle materials. which are subject to high hydrostatic pressure do not fail when the pressure is acting. Materials with a loose or porous structure such as wood. as pointed out earlier. This theory is not much supported by experimental results. 4. Consequently. According to this theory. It appears that small atoms of mercury are able to penentrate the cracks. Further. 1. Assuming s1 > s2 > s3.1 Rectangular element with 45° plane If s1 is equal to s2 in magnitude. for ductile materials subjected to pure torsion. yielding. we say that failure occurs when the maximum principal stress at any point reaches a value equal to the tensile (or compressive) elastic limit or yield strength of the material obtained from the uniaxial test. the maximum principal stress theory. Notwithstanding all these. then on a 45° plane. Further evidence to show that the maximum principal stress theory cannot be a good criterion for failure can be demonstrated in the following manner: Consider the block shown in Fig. occurs when the maximum shearing stress Chapter_04. experiments reveal that the shear stress limit causing yield is much less than s1 in magnitude. Using information from a uniaxial tension (or compression) test. Such a state of stress occurs in σmaximum 1 − σ3 a cylindrical bar subjected to pure torsion.63b). if s1 > s2 > s3 are the principal stresses at a point and sy the yield stress or tensile elastic limit for the material under a uniaxial test. according to this theory.τ If the principal stress theory max = 2 was valid.Theories of Failure or Yield Criteria and Introduction to Ideally Plastic Solid 111 From these observations. subjected to stress s1 and s2. s1 t t s2 s2 45° t s1 Fig. is considered to be reasonably satisfactory for brittle materials which do not fail by yielding. Thus. the shearing stress will have a magnitude equal to s1. 4. s1 would have been the limiting value. then failure occurs when s1 ≥ sy (4. we draw the conclusion that a pure state of hydrostatic pressure [s1 = s2 = s3 = –p ( p > 0)] cannot produce permanent deformation in compact crystalline or amorphous solid materials but produces only a small elastic contraction. from Eq. (1. 4.pmd 111 7/3/2008. 5:42 AM . However. This contradicts the maximum principal stress theory. where s1 is tensile and s2 is compressive. because of its simplicity. provided the liquid is prevented from entering the fine surface cracks or crevices of the solid.1) Maximum Shearing Stress Theory Observations made in the course of extrusion tests on the flow of soft metals through orifices lend support to the assumption that the plastic state in such metals is created when the maximum shearing stress just reaches the value of the resistance of the metal against shear. 2 Biaxial state of stress less than sy. However. differs considerably from the planes of maximum shear. Tension tests conducted on mild steel bars show that at the time of yielding. On the other hand. according to this theory. failure occurs at a point in a body when the maximum strain at that point exceeds the value of the maximum strain in a uniaxial test of the material at yield point. which usually develop along these planes. for ductile load carrying members where large shears occur and which are subject to unequal triaxial tensions. in a block subjected to a biaxial tension. since equal triaxial tensions cannot produce any shear. Further. Therefore. as remarked earlier. this is not supported by experiments. thus contradicting this theory. If s1 > s2 > s3 are the three principal stresses at a point. While the maximum strain theory is an improvement over the maximum stress theory. it is not a good theory for ductile materials. whereas. the angle of the slip planes or of the shear fracture surfaces. In a bar subject to uniaxial tension or compression. Thus. 4.112 Advanced Mechanics of Solids reaches a critical value. Also. the values of the maximum shear in tension and compression are not equal. for brittle crystalline materials which cannot be brought into the plastic state under tension but which may yield a little before fracture under compression. The maximum shearing stress theory is accepted to be fairly well justified for ductile materials. if s1. if s2 were compressive. as shown in Fig. 4. (4. However. failure occurs when σy ε1 = 1 ⎡⎣σ1 − ν (σ 2 + σ 3 )⎤⎦ ≥ E s1 s2 s2 (4. the maximum shearing stress theory is used because of its simplicity. the principal strain e1 is ε1 = 1 (σ1 − νσ 2 ) E and is smaller than s1/E because of s2.pmd 112 7/3/2008. failure occurs when τ max = σ1 − σ 3 ≥ σy 2 2 where sy /2 is the shear stress at yield point in a uniaxial test. s1 can be increased more than sy without causing failure. the s1 magnitude of s1 to cause failure would be Fig. For materials which fail by Chapter_04. the maximum shear stress occurs on a plane at 45° to the load axis. thus supporting the theory. the so-called slip lines occur approximately at 45°. 5:42 AM . in these brittle materials. s2 and s3 are the principal stresses at a point.3) E We have observed that a material subjected to triaxial compression does not suffer failure.2.2) Maximum Elastic Strain Theory According to this theory. Failure of material under triaxial tension (of equal magnitude) also does not support this theory. Further. which will be discussed subsequently.44a) and (1.Theories of Failure or Yield Criteria and Introduction to Ideally Plastic Solid 113 brittle fracture. s2 and s 3 are the principal stresses at a point. and toct is equal to zero. This theory is equivalent to the maximum distortion energy theory. y and z directions respectively.4b) This theory is supported quite well by experimental evidences. from Fig. failure cannot occur and this.44c) 1/ 2 2 2 toct = 1 ⎡(σ1 − σ 2 )2 + (σ 2 − σ 3 ) + (σ 3 − σ 1 ) ⎤ ⎦ 3⎣ = ( 2 l 2 − 3l 2 3 1 ) 1/ 2 In a uniaxial test. is ∆W = 1 σ 1 ∆y ∆ z (δ∆ x ) + 1 σ 2 ∆ x∆z ( δ∆y ) + 1 σ 3 ∆ x ∆y (δ∆ z ) 2 2 2 where dD x.3(b).pmd 113 7/3/2008. then the work done by the forces. If e1. The tangential stress on this plane is the octahedral shearing stress. failure occurs at a point where the values of principal stresses are such that 1/ 2 2 2 2 τ oct = 1 ⎡(σ1 − σ 2 ) + (σ 2 − σ 3 ) + (σ 3 − σ1 ) ⎤ 3⎣ ⎦ (l or 2 1 ≥ 2σ 3 y ) − 3l2 ≥ σ y2 (4. failure at any point in a body subject to a state of stress begins only when the energy per unit volume absorbed at the point is equal to the energy absorbed per unit volume by the material when subjected to the elastic limit under a uniaxial state of stress. according to this theory. s1 = s2 = s3 = –p. one may prefer the maximum strain theory to the maximum stress theory. e2 and e3 are the corresponding principal strains. s2 and s3 be the principal stresses and let their magnitudes increase uniformly from zero to their final magnitudes. Octahedral Shearing Stress Theory According to this theory. ny and nz = 1/ 3 . The normal to this plane has direction cosines nx. then from Eqs (1. If s1. at yield point. According to this theory. as stated earlier. the octahedral stress ( 2 /3) sy = 0. Chapter_04. 5:42 AM . Oy and Oz is called the octahedral plane. Consequently.47sy. 4. the critical quantity is the shearing stress on the octahedral plane. Hence. is supported by experimental results. To calculate the energy absorbed per unit volume we proceed as follows: Let s1. The plane which is equally inclined to all the three principal axes Ox. according to the present theory. Maximum Elastic Energy Theory This theory is associated with the names of Beltrami and Haigh.4a) (4. dDy and dDz are extensions in x. when a material is subjected to hydrostatic pressure. failure occurs when σ 12 + σ 22 + σ 32 − 2ν (σ 1 σ 2 + σ 2 σ 3 + σ 3 σ 1 ) ≥ σ y2 (4.3 (a) Principal stresses on a rectangular block (b) Area representing work done From Hooke’ s law δ∆ x = ε1 ∆ x = 1 ⎡⎣σ 1 − ν (σ 2 + σ 3 ) ⎤⎦ ∆ x E δ∆y = ε 2 ∆y = 1 ⎡⎣σ 2 − ν (σ 1 + σ 3 ) ⎤⎦ ∆y E δ∆ z = ε 3 ∆ z = 1 ⎡⎣σ 3 − ν (σ1 + σ 2 ) ⎤⎦ ∆ z E Substituting these ∆W = 1 ⎡⎣σ 12 + σ 22 + σ 32 − 2ν (σ1 σ 2 + σ 2 σ 3 + σ 3 σ 1 ) ⎤⎦ ∆ x ∆y ∆ z 2E The above work is stored as internal energy if the rate of deformation is small. Consequently. von Mises and Hencky. According to this theory.114 Advanced Mechanics of Solids s3 s2 s Work Done = Area Under Triangle Dx s1 e O Dz Dy s1 (b) s3 s2 (a) Fig. it is not the total energy which is the criterion for failure.5) In a uniaxial test. 5:42 AM . Energy of Distortion Theory This theory is based on the work of Huber. the energy stored per unit volume at yield point or elastic limit is 1/2 E σ y2 . the energy U per unit volume is 1 ⎡σ 2 + σ 2 + σ 2 − 2ν σ σ + σ σ + σ σ ⎤ ( 1 2 2 3 3 1 )⎦ 2 3 2E ⎣ 1 (4. in fact the Chapter_04. Hence.6) This theory does not have much significance since it is possible for a material to absorb considerable amount of energy without failure or permanent deformation when it is subjected to hydrostatic pressure. 4.pmd 114 7/3/2008. 7) where p = 1 (s1 + s2 + s3). it was shown that any given state of strain can be resolved uniquely into an isotropic and a deviatoric state of strain.pmd 115 (4. If s1.9) 7/3/2008.8) that.Theories of Failure or Yield Criteria and Introduction to Ideally Plastic Solid 115 energy absorbed during the distortion of an element is responsible for failure. we have ⎡ε1 0 ⎢0 ε 2 ⎢ ⎢⎣0 0 0 ⎤ ⎡e 0 0⎤ ⎡ ε1 − e 0 0 ⎤ 0 ⎥⎥ = ⎢⎢0 e 0⎥⎥ = ⎢⎢ 0 ε2 − e 0 ⎥⎥ ε 3 ⎥⎦ ⎢⎣0 0 e ⎥⎦ ⎢⎣ 0 0 ε3 − e ⎥⎦ (4.22) that any given state of stress can be uniquely resolved into an isotropic state and a pure shear (or deviatoric) state. 5:42 AM . It is easy to see from Eqs (4.17). e2 and e3 are the principal strains at the point. recall that the necessary and sufficient condition for a state to be a pure shear state is that its first invariant must be equal to zero. 3 It was also shown that the volumetric strain corresponding to the deviatoric state of strain is zero since its first invariant is zero. 1. Similarly. the isotropic state of strain is related to the isotropic state of stress because ε1 = 1 ⎡⎣σ 1 − ν (σ 2 + σ 3 ) ⎤⎦ E ε 2 = 1 ⎡⎣σ 2 − ν (σ 3 + σ 1 ) ⎤⎦ E ε 3 = 1 ⎡⎣σ 3 − ν (σ 2 + σ 1 )⎤⎦ E Adding and taking the mean 1 ε + ε +ε ( 2 3) = e 3 1 = 1 ⎡⎣(σ 1 + σ 2 + σ 3 ) − 2ν (σ 1 + σ 2 + σ 3 ) ⎤⎦ 3E e = 1 ⎡⎣(1 − 2ν ) p ⎤⎦ E or Chapter_04. by Hooke’s law. The energy of distortion can be obtained by subtracting the energy of volumetric expansion from the total energy. It was shown in the Analysis of Stress (Sec. If e1.7) and (4. Also.8) where e = 1 (e1 + e2 + e3). 3 The first matrix on the right-hand side represents the isotropic state and the second matrix the pure shear state. in the Analysis of Strain (Section 2. s2 and s3 are the principal stresses at a point then 0 ⎤ ⎡ p 0 0 ⎤ ⎡σ1 − p 0 0 ⎤ ⎡σ1 0 ⎢0 σ 0 ⎥ = ⎢0 p 0 ⎥ = ⎢ 0 σ − p 0 ⎥⎥ 2 2 ⎢ ⎥ ⎢ ⎥ ⎢ ⎢⎣0 0 σ 3 ⎥⎦ ⎢⎣0 0 P ⎥⎦ ⎢⎣ 0 0 σ 3 − p ⎥⎦ (4. the bulk modulus. This is obtained by simply putting s1 = sy and 6G (1 + ν ) 2 s2 = s3 = 0 in Eq.11a) 2 (1 + ν ) 2 σ 1 + σ 22 + σ 32 − σ1 σ 2 − σ 2 σ 3 − σ 3 σ 1 6E (1 + ν ) ⎡ = (σ − σ 2 )2 + (σ 2 − σ 3 )2 + (σ 3 − σ1 )2 ⎤⎦ 6E ⎣ 1 ( = Substituting G = (4. failure occurs at that point where s1. (1. 3E Hence.9) U¢ = 3 (1 − 2ν ) p 2 2E (4. the proportionality constant being equal to 3 (1 – 2n) = K.pmd 116 ⎦ 7/3/2008.11b) (4.13) But we notice that the expression for the octahedral shearing stress from Eq.116 Advanced Mechanics of Solids i. 5:42 AM . according to the distortion energy theory. In a uniaxial test. E Consequently. (4. subtracting U’ from U U * = 1 σ 12 + σ 22 + σ 32 − ν (σ 1 σ 2 + σ 2 σ 3 + σ 3 σ 1 ) 2E E = ( − ) 1 − 2ν (σ1 + σ2 + σ3 )2 6E (4. This states that the volumetric strain 3e is proportional to the pressure p. e is connected to p by Hooke’s law.e. Hence. (4.11c) E for the shear modulus. This is also equal to σ y from Eq.12b) This is the expression for the energy of distortion.10) 1 − 2ν (σ 1 + σ 2 + σ 3 ) 2 6E The total elastic strain energy density is given by Eq.5).22) is 1/ 2 2 2 2 τ oct = 1 ⎡(σ1 − σ 2 ) + (σ 2 − σ 3 ) + (σ 3 − σ1 ) ⎤ 3⎣ Chapter_04.11c). the energy of distortion is equal to 1 σ y2 . (3. (4.14). the work done or the energy stored during volumetric change is U¢ = 1 pe + 1 pe + 1 pe = 3 pe 2 2 2 2 Substituting for e from Eq. 2(1 + ν ) ( or ) ) U * = 1 σ 12 + σ 22 + σ 32 − σ 1 σ 2 − σ 2 σ 3 − σ 3 σ 1 6G 2 2 2 U * = 1 ⎡ (σ 1 − σ 2 ) + (σ 2 − σ 3 ) + ( σ 3 − σ 1 ) ⎤ ⎦ 12G ⎣ (4.s2 and s3 are such that (σ1 − σ 2 )2 + (σ 2 − σ 3 )2 + (σ 3 − σ1 )2 ≥ 2σ y2 (4.12). (4. Eq.12a) (4. for example. Further. in complex loading conditions. subjecting it to all the possible conditions of loading that the member would be subjected to during operation. shape and temperature of the member. 2 and in the case of pure shear. a brittle material may be made to yield before failure. Eq. strain and energy) associated with it.pmd 117 7/3/2008. toct = ≥ or 4. and the factor (such as stress. fracture without permanent deformation is the characteristic feature of brittle materials. These are shown in Fig. (4. In discussing the various theories of failure. the distortion energy theory states that failure occurs when 2 9τ oct = ≥ 2σ y2 2σ (4. the normal stresses on a 45° element are s and –s. Chapter_04.4). as remarked earlier. 5:42 AM . Consequently. While yielding or permanent deformation is the characteristic feature of ductile materials. the maximum shear stress t at a point is equal to 1 s. then one could determine the maximum loading condition that does not cause failure. Consequently. we have expressed the critical value associated with each theory in terms of the yield point stress sy obtained from a uniaxial tensile stress. 4. Further. if the loading conditions are suitably altered. Even ductile materials fail in a different manner when subjected to repeated loadings (such as fatigue) than when subjected to static loadings.14) or equivalently. using sy or ty is equivalent because during a uniaxial tension.3 SIGNIFICANCE OF THE THEORIES OF FAILURE The mode of failure of a member and the factor that is responsible for failure depend on a large number of factors such as the nature and properties of the material. All these factors indicate that any rational procedure of design of a member requires the determination of the mode of failure (either yielding or fracture).14) 3 y This is identical to Eq. one can also express the critical value associated with each theory of failure in terms of the yield point shear stress ty. type of loading. If tests could be performed on the actual member. the octahedral shearing stress theory and the distortion energy theory are identical. one has to identify the factor associated with the failure of a member and take precautions to see that this factor does not exceed the maximum allowable value. But this may not be possible except in very simple cases.13) expresses well the condition under which the ductile metals at normal temperatures start to yield.Theories of Failure or Yield Criteria and Introduction to Ideally Plastic Solid 117 Hence. Experiments made on the flow of ductile metals under biaxial states of stress have shown that Eq. In a sense. It is equally easy to perform a pure torsion test on a round specimen and obtain the value of the maximum shear stress ty at the point of yielding.4. This was done since it is easy to perform a uniaxial tensile stress and obtain the yield point stress value. (4. that the mode of failure of a ductile material differs from that of a brittle material. Therefore. We have observed. where s is numerically equivalent to t. the purely elastic deformation of a body under hydrostatic pressure (toct = 0) is also supported by this theory. This information is obtained by performing a suitable test (uniform tension or torsion) on the material in the laboratory. (4. etc. 4 Uniaxial and pure shear state of stress If one uses the yield point shear stress ty obtained from a pure torsion test. failure occurs when the shear stress t at a point in the member reaches a value t ≥ ty (iii) Maximum Strain Theory According to this theory. failure occurs when the maximum strain at any point in the member reaches a value ε = 1 [σ 1 − ν (σ 2 + σ 3 )] E From Fig. from Eq. s2 = 0. s3 = –s = –t t oct = 1 (σ 2 + σ 2 + 4σ 2 )1/ 2 3 = 6σ= 3 Chapter_04. the octahedral shear stress is 1/ 2 toct = 1 ⎡(σ 1 − σ 2 ) 2 + (σ 2 − σ 3 ) 2 + (σ 3 − σ 1 ) 2 ⎤ ⎣ ⎦ 3 Observing that s1 = s = t.pmd 118 2τ 3 7/3/2008. in the case of pure shear s1 = s = t. the maximum and minimum normal stresses at a point are s and –s (each numerically equal to the shear stress t ). 5:42 AM . failure occurs when the strain e at any point in the member reaches a value ε = 1 (τ y + ντ y ) = 1 (1 + ν )τ y E E (iv) Octahedral Shear Stress Theory When an element is subjected to pure shear. the maximum normal stress s is numerically equivalent to ty.44a). Corresponding to this. failure occurs when the normal stress s at any point in the stressed member reaches a value s ≥ ty This is because. 4. 4. (1. then the critical value associated with each theory of failure is as follows: (i) Maximum Normal Stress Theory According to this theory.4. s2 = 0. 4. s3 = –s = –t Hence. in a pure torsion test when yielding occurs.4. as shown in Fig. (ii) Maximum Shear Stress Theory According to this theory.118 Advanced Mechanics of Solids ty s sy sy t s ty s = ty t = sy /2 Fig. Theories of Failure or Yield Criteria and Introduction to Ideally Plastic Solid 119 So. σ 2 = 0. failure occurs when the octahedral shear stress at any point is t oct = 2τ 3 y (v) Maximum Elastic Energy Theory The elastic energy per unit volume stored at a point in a stressed body is. For example. from Fig.12). then the values of sy and ty obtained from tests conducted on the material should be related by the corresponding expressions.5). (4.pmd 119 7/3/2008. 5:42 AM . (4. U = 1 ⎡⎣σ 12 + σ 22 + σ 32 − 2ν (σ 1σ 2 + σ 2 σ 3 + σ 3 σ 1 ) ⎤⎦ E In the case of pure shear. σ 3 = −τ 1 ⎡τ 2 + τ 2 + 4τ 2 ⎤ ⎦ 12G ⎣ = 1 τ2 2G So. σ1 = τ . So. failure occurs when the distortion energy density at any point is equal to 2 (1 + ν ) 2 U * = 1 τ y2 = 1 ⋅ τy 2G 2 E = (1 + ν ) E τ y2 The foregoing results show that one can express the critical value associated with each theory of failure either in terms of sy or in terms of ty. U * = 1 ⎡⎣ (σ 1 − σ 2 ) 2 + (σ 2 − σ 3 ) 2 + (σ 3 − σ 1 )2 ⎤⎦ 12G Once again. σ 3 = −τ ( ) U = 1 ⎡τ 2 + τ 2 − 2ν −τ 2 ⎤ ⎦ 2E ⎣ 1 2 = (1 + ν ) τ E Hence. Assuming that a particular theory of failure is correct for a given material. by observing that in the case of pure shear σ1 = τ . from Eq. 4. if the distortion energy is a valid theory for a Chapter_04. failure occurs when the elastic energy density at any point in a stressed body is such that U = 1 (1 + ν ) τ y2 E (vi) Distortion Energy Theory The distortion energy density at a point in a stressed body is. from Eq.4. U* = σ 2 = 0. 577 times the value of sy obtained from a uniaxial tension test conducted on the same material.60 of the tensile yield strength sy. 4⎠ ⎝ U= U* = 120 Relationship sy = t y Max. The first column lists the six theories of failure. Table 4.8 σ y τ = σy 1⎞ ⎛ Max.57.5 σ y 5 τy 4 E τ y = 0. according to the octahedral shear stress theory. The second column lists the critical value associated with each theory in terms of sy. then the value of the energy in terms of sy and that in terms of ty should be equal. strain ⎜ν = ⎟ 4⎠ ⎝ ε= Octahedral shear τ oct = 1⎞ ⎛ Max. Thus.50 and 0. failure occurs when the octahedral shear stress at a point assumes a value equal to 2 /3 σ y . then the value of ty obtained from pure torsion test should be equal to 0. U* = (1 + ν ) τ y2 = E (1 + ν ) 3E σ y2 1 σ = 0. normal stress Chapter_04. failure occurs at a point when the octahedral shear stress equals a value 2/3 τ y . energy ⎜ν = ⎟ .632 σ y τ oct = U= U * = (1 + ν ) τ 2y E τ y = 0.577 σ y y 3 This means that the value of t y obtained from pure torsion test should be equal to 0.pmd ty = s y 1 2 ty τ y = 0. Tests conducted on many ductile materials reveal that the values of ty lie between 0.1 Failure theory Tension Max. For example. 5:42 AM . The fourth column gives the relationship that should exist betweenty and sy in each case if each theory is valid.1 summarizes these theories and the corresponding expressions. This result agrees well with the octahedral shear stress theory and the ty = or Table 4. the yield point shear stress value in pure torsion.577 times the yield point stress sy obtained from a uniaxial tension test. For example. the average value being about 0. The third column lists the critical value associated with each theory in terms of ty. Assuming octahedral shear stress theory is correct.577 σ y 7/3/2008. the yield point stress in uniaxial tension test.120 Advanced Mechanics of Solids material. shear stress Distortion energy Shear sy 1 σy E 2 σy 3 1 2 σy 2E 2 1 +ν σ y 3 E ε= 2 τy 3 τ y = 0. according to octahedral shear stress theory.577 σ y 5 1 2 τy 4 E τ y = 0. Use a factor of safety N. The maximum shear stress theory gives values for design on the safe side. The maximum shear stress theory predicts that shear yield value ty is 0. Consequently. proportional to F 2. because of its simplicity. but rather the load-carrying capacity is increased by N.3 = 1 σ ± 1 2 Chapter_04.1 Determine the diameter d of a circular shaft subjected to a bending moment M and a torque T.Theories of Failure or Yield Criteria and Introduction to Ideally Plastic Solid 121 distortion energy theory. 5:42 AM (4. If d is the diameter.4 USE OF FACTOR OF SAFETY IN DESIGN In designing a member to carry a given load without failure. Example 4. according to the several theories of failure.16) = 16T3 πd Therefore. 4. then designing the member to safely carry a load equal to NF is equivalent to designing the member for a critical factor equal to X/N. the normal stress s at P on a plane normal to the axis of the shaft is. This is about 15% less than the value predicted by the distortion energy (or the octahedral shear) theory. say. σ2 = 0 7/3/2008. Solution Consider a point P on the periphery of the shaft. this theory is widely used in machine design dealing with ductile materials.pmd 121 2 (σ 2 ) + 4τ 2 .15) = 32M3 πd The shearing stress on a transverse plane at P due to torsion T is Td ⋅ 32 τ = Td = 2I P 2π d 4 (4. Also. from elementary strength of materials. in using the factor of safety. then designing the member to safely carry a load to equal to NF is equivalent to limiting the critical factor to X / N . then owing to the bending moment M. However.5 times the tensile yield value. Hence. The purpose is to design the member in such a way that it can carry N times the actual working load without failure. but is. one can use a factor appropriately reduced during the design process. the principal stresses at P are σ 1. care must be taken to see that the critical factor associated with failure is not reduced by N. σ= My d 64 =M 2 πd4 I (4. If X is directly proportional to F. usually a factor of safety N is used. Let X be a factor associated with failure and let F be the load. This point will be made clear in the following example.17) . if X is not directly proportional to F. It has been observed that one can associate different factors for failure according to the particular theory of failure adopted. we have σ 1 − ν σ 3 = Chapter_04. the shear stress becomes Nt. (4. 2 + 4τ 2 1/ 2 1/ 2 σy 2 σy N Substituting for s and t 32 M 2 + T 2 π d3 32 M 2 + T 2 or. Equating the maximum normal stress to sy. With a factor of safety N. tmax = ( ) = ) = ( ) = ( ) = Nτ max = 1 N σ 2 + 4τ 2 2 (σ or.pmd 122 σy N σy N 7/3/2008. when the load is increased N times.. Hence. (4. 16M + 16 M 2 + T 2 1/ 2 1/ 2 1/ 2 π d3 σ y N From this. (iii) Maximum Strain Theory tor of safety N is 1/ 2 1/ 2 σy N π d 3σ y N The maximum elastic strain at point P with a fac- ε max = N ⎡⎣σ 1 − ν (σ 2 + σ 3 ) ⎤⎦ E From Eq.. the normal and shearing stresses are Ns and Nt. 5:42 AM . the maximum shearing stress ( ) 1 σ − σ = 1 σ 2 + 4τ 2 1/ 2 ( 3) 2 1 2 When the load is increased N times. the maximum normal stress should not exceed sy.17) is At point P. the value of d can be determined with the known values of M.e.e. the yield point stress in tension.3) σ 1 − ν (σ 2 + σ 3 ) = Since s2 = 0. 32M + 1 × 32 M 2 + T 2 π d3 πd3 i.122 Advanced Mechanics of Solids (i) Maximum Normal Stress Theory At point P. (ii) Maximum Shear Stress Theory from Eq. T and sy. 2( σ max = σ1 = N ⎡⎢ σ + 1 σ 2 + 4τ 2 ⎣2 ) 1/ 2 ⎤ ( ) = 2σ y N ( ) = 2σy N ( ) = σ + σ 2 + 4τ 2 or ⎥⎦ = sy i. and using a factor of safety N. 5:42 AM . Ns2 and Ns3.4a). the factor N 2 appears in the expression for U. is 1/ 2 Ntoct = N ⎡(σ1 − σ 2 ) 2 + (σ 2 − σ 3 )2 + (σ 3 − σ 1 ) 2 ⎤ = 2 σ y ⎦ 3 3 ⎣ 1/ 2 ⎡(σ 1 − σ 2 ) 2 + (σ 2 − σ 3 )2 + (σ 3 − σ 1 )2 ⎤ ⎣ ⎦ or = 2σ N y With s2 = 0 1/ 2 ⎡ 2σ 12 + 2σ 32 − 2σ 1 σ 3 ⎤ ⎣ ⎦ 1/ 2 ⎡σ 12 + σ 32 − σ 1 σ 3 ⎤ ⎣ ⎦ or 2σ N y σ = y N = Substituting for s1 and s3 ( ) ( ⎡ 1 σ 2 + 1 σ 2 + 4τ 2 + 1 σ σ 2 + 4τ 2 ⎢⎣ 4 4 2 ( − 1 σ σ 2 + 4τ 2 2 (σ or 2 ) 1/ 2 + 3τ 2 ) 1/ 2 ( + 1 σ 2 + 1 σ 2 + 4τ 2 4 4 ( ) ) 1/ 2 − 1 σ 2 + 1 σ 2 + 4τ 2 ⎤⎥ 4 4 ⎦ ) = ) = 1/ 2 = σy N σy N Substituting for s and t ( 16 4M 2 + 3T 2 π d3 1/ 2 16 (4M 2 + 3T 2 )1/ 2 = or σy N π d3 σ y N (v) Maximum Energy Theory The maximum elastic energy at P from Eq. 2 Chapter_04. (4. In the previous four cases.pmd 123 7/3/2008.6) and with a factor of safety N is 2 σy U = N ⎡⎣σ12 + σ 22 + σ 32 − 2ν (σ1 σ 2 + σ 2 σ 3 + σ 3 σ1 ) ⎤⎦ = 2E 2E Note: Since the stresses for design are Ns1. only N appeared because of the particular form of the expression. (4.Theories of Failure or Yield Criteria and Introduction to Ideally Plastic Solid or 2( σ + 1 σ 2 + 4τ 2 ) 1/ 2 2 Substituting for s and t ( (1 − ν ) 16M3 + (1 + ν ) πd or ) − ν σ + ν σ 2 + 4τ 2 2 2 1/ 2 ( 16 M 2 + T 2 π d3 (1 − ν ) 16M + (1 + ν ) 16 ( M 2 + T 2 ) 1/ 2 = = 123 σy N ) 1/ 2 = σy N π d 3σ y N (iv) Octahedral Shear Stress Theory The octahedral shearing stress at point P from Eq. 5 A NOTE ON THE USE OF FACTOR OF SAFETY As remarked earlier. (2σ12 + 2σ 32 − 2σ1 σ 3 ) = or (σ12 + σ 32 − σ1 σ 3 )1/ 2 = 2 σ y2 N2 σy N This yields the same result as the octahedral shear stress theory.11c). 4. (σ 12 + σ 32 σ y2 − 2ν σ 1 σ 3 ) = N2 Substituting for s1 and s3 ( ) ( ⎡ 1 σ 2 + 1 σ 2 + 4τ 2 + 1 σ σ 2 + 4τ 2 ⎢⎣ 4 4 2 ( ) ( ) 1/ 2 + 1 σ2 4 + 1 σ 2 + 4τ 2 − 1 σ σ 2 + 4τ 2 4 2 σ 2 + ( 2 + 2ν ) τ 2 = or 1/ 2 ⎡σ 2 + ( 2 + 2ν ) τ 2 ⎤ ⎣ ⎦ i. when a factor of safety N is prescribed. or = ) 1/ 2 σ y2 + 2ντ 2 ⎤⎥ = 2 ⎦ N σ y2 N2 σy N 16 ⎡ 4M 2 + 2 + 2ν T 2 ⎤1/ 2 = σ y ( ) ⎦ N π d3 ⎣ 1/ 2 ⎡ 4M 2 + 2 (1 + ν ) T 2 ⎤ ⎣ ⎦ = π d 3σ y 16 N (vi) Maximum Distortion Energy Theory The distortion energy associated with Ns1. then see that this factor at any point in the member does not exceed X/N. i. Ns2 and Ns3 at P is given by Eq. Equating this to distortion energy in terms of sy Ud = = N 2 (1 + ν ) ⎡ (σ − σ 2 )2 + (σ 2 − σ 3 )2 + (σ 3 − σ1 )2 ⎤⎦ ⎣ 1 6E 1 +ν 2 σ 3E y With s2 = 0.e. Chapter_04.pmd 124 7/3/2008. (4.e.124 Advanced Mechanics of Solids With s2 = 0. (ii) If the factor associated with failure is X. 5:42 AM . we may consider two ways of introducing it in design: (i) Design the member so that it safely carries a load NF. 25. using the octahedral shear stress theory and the maximum shear stress theory. But method (i) gives 2 E 2 2 σy U = N ⎡⎣σ12 + σ 22 + σ 32 − 2ν (σ1 σ 2 + σ 2 σ 3 + σ 3 σ1 ) ⎤⎦ = 2E 2E 2 The result obtained from method (i) is correct. 4. As one an see. the result will be identical. So long as X is directly proportional to F. method (ii) may give wrong results.5 Example 4. 4. Ns2 and Ns3 are the principal stresses corresponding to the load NF. n = 0.15 = 6750 Nm The normal stress due to M at A is 64Md = − 32M s= − 2π d 4 π d3 Chapter_04. both in tension and compression.2 = 9000 Nm and the torque on the shaft is T = 45. if we adopt method (ii) with the maximum energy theory. determine the diameter of the shaft. With E = 207 ¥ 106 kPa. is 1 y .000 kPa. Consider a point on the periphery at section A for analysis.pmd 125 7/3/2008.000 ¥ 0. For example. the member is to be designed for N times the load. Example 4. A 20 cm A A 15 cm F Fig. the result will be σ y2 U = 1 ⎡⎣σ12 + σ 22 + σ 32 − 2ν (σ1 σ 2 + σ 2 σ 3 + σ 3 σ1 ) ⎤⎦ = 1 N 2E 2E σ where X. If X is not directly proportional to F. since Ns1.5 at uniform speed. 5:42 AM . the results are not the same.2 Solution The moment at section A is M = 45. the factor associated with failure. The result given by method (ii) is not the right one.2 A force F = 45. Use a factor of safety N = 2. The crank shaft is made of ductile steel whose elastic limit is 207. whether one uses NF or X/N for design analysis.Theories of Failure or Yield Criteria and Introduction to Ideally Plastic Solid 125 But the second method of using N is not correct.000 N is necessary to rotate the shaft shown in Fig. since by the definition of the factor of safety.000 ¥ 0. 3 = 1 σ ± 1 σ 2 + 4τ 2 2 2 (i) Maximum Shear Stress Theory ) 1/ 2 .126 Advanced Mechanics of Solids and the maximum shear stress due to T at A is 32Td = 16T 2π d 4 π d 3 The shear stress due to the shear force F is zero at A. Thus. Ntmax = 1 114591.6 Pa d3 This should not exceed the maximum shear stress value at yielding in uniaxial tension test.61) are t= ( σ 1.pmd 126 ) 1/ 2 7/3/2008. the value of tmax becomes 114591.8 d3 πd With a factor of safety N = 2.4 cm (ii) Octahedral Shear Stress Theory 1/ 2 τ oct = 1 ⎡⎣ (σ 1 − σ 2 ) 2 + (σ 2 − σ 3 ) 2 + (σ 3 − σ 1 )2 ⎤⎦ 3 With s2 = 0. The principal stresses from Eq.6 = σ y = 207 × 106 ( ) 2 2 d3 \ d 3 = 1107 × 10− 6 m3 or d = 10. 1/ 2 τ oct = 1 ⎡⎣ 2σ 12 + 2σ 32 − 2σ 1 σ 3 ⎤⎦ 3 Substituting for s1 and s3 and simplifying ( 2 σ 2 + 3τ 2 3 t oct = = ) 1/ 2 2 ⎡ 32 M 2 + 3 16T 2 ⎤1/ 2 ( ) ( ) ⎦ 3π d 3 ⎣ ( = 16 23 4 M 2 + 3T 2 3π d Chapter_04. σ2 = 0 tmax = 1 (σ 1 − σ 3 ) 2 = 1 (σ 2 + 4τ 2 )1/ 2 2 = 1 323 ( M 2 + T 2 )1/ 2 2 πd ( ) 1/ 2 Pa = 163 90002 + 67502 = 57295.35 × 10−2 m = 10. 5:42 AM . (1. n = 0. 5:42 AM .25) (10802 + 34002 )1/ 2 = π d × 207 × 106 3 3 Chapter_04. determine the maximum bending moment it can support in addition to the torque.75 × 1080) + (16 × 1.18 cm Example 4.3 A cylindrical bar of 7 cm diameter is subjected to a torque equal to 3400 Nm. = 2 × 2 × 343418 = 2 σ 3 y 3π d 3 or 2 × 343418 = π d 3 σ y = π d 3 × 207 × 106 \ d 3 = 1.e. and using a factor of safety N = 2.056 × 10−3 or d = 0.4 In Example 4.25.pmd 127 7/3/2008. if failure is governed by the maximum strain theory.1018 m = 10. If the bar is at the point of failing in accordance with the maximum principal stress theory.1(iii) 16(1 − ν ) M + 16(1 + ν ) ( M 2 + T 2 )1/ 2 = π d σ y N 3 (16 × 0. and a bending moment M. factor of safety N = 3 and sy = 207 MPa. Solution From Example 4. The tensile elastic limit for the material is 207 MPa. and the factor of safety to be used is 3. determine the diameter of the bar if it is subjected to a torque T = 3400 Nm and a bending moment M = 1080 Nm.Theories of Failure or Yield Criteria and Introduction to Ideally Plastic Solid 127 2 2 1/ 2 = 16 23 ⎡ 4 ( 9000 ) + 3 ( 6750 ) ⎤ ⎦ 3π d ⎣ 2 × 343418 3π d 3 Equating this to octahedral shear stress at yielding of a uniaxial tension bar.1(i) 3 16 M + 16 ( M 2 + T 2 )1/ 2 = π d σ y N i. or or \ 3 −6 6 16 M + 16 ( M 2 + 34002 )1/ 2 = π × 7 × 10 × 207 × 10 3 ( M 2 + 34002 )1/ 2 = 4647 − M M 2 + 34002 = 4647 2 + M 2 − 9294 M M = 1080 Nm Example 4.3. Solution According to the maximum strain theory and Example 4. The elastic modulus for the material is E = 103 ¥ 106 kPa. 77 ¥ 106 d 3 or d 3 = 389 ¥ 10–6 or d = 7. 4.6. (Refer to Fig.5 Solution The lower end of the rope is subjected to a triaxial state of stress. σ1′ = W + (γ r − γ ) H .3 ¥ 10–2 m = 7.e. The rope has a cross-sectional area A.6 Example 4. according to the maximum shear stress theory. near the hoist. where σ1 = W . The rope attached to the instrument has a specific weight gr and the water has a specific weight g.. When the equipment is above the surface of the water. σ 2′ = σ 3′ = 0 A Therefore.5 An equipment used in deep sea investigation is immersed at a depth H.128 Advanced Mechanics of Solids i. at lower section τ max = σ1 − σ 3 2 and at the upper section τ max = σ1′ − σ 3′ 2 = ( = 1 W +γH 2 A ( ) 1 W − γH + γrH 2 A ) If the specific weight of the rope is more than twice that of water.) s1¢ s1¢ = H s1 p W + Wr A p=gH p p=gH s1 = W A Fig. then the upper section is the critical section. 4. 5:42 AM . The weight of the equipment in water is W.pmd 128 and σ2 = σ3 = 0 7/3/2008. the stress is σ1 = W ′ A Chapter_04. A σ 2 = σ 3 = − γ H (compression) At the upper section there is only a uniaxial tension σ 1′ due to the weight of the equipment and rope immersed in water. There is a tensile stress s1 due to the weight of the equipment and two hydrostatic compressions each equal to p. Analyse the strength of the rope. 12960 + 71348 = 216.3 cm Example 4. Theories of Failure or Yield Criteria and Introduction to Ideally Plastic Solid 129 τ max = 1 W ′ 2 A W ¢ is the weight of equipment in air and is more than W. 4. as far as failure is concerned.7. then the plane having this normal stress and a maximum shear stress accompanying it. we would get two different values for the critical shear stress. 4.7 Mohr’s circles s1.pmd 129 7/3/2008. This was that the criterion of failure is unaltered by a reversal of sign of the stress. Chapter_04. 5:42 AM . with diameter (s 1 – s 3 ). for a general state of stress. will be the critical plane. the critical point for the normal stress sn will be the point A. represented by point A or A¢. From Mohr’s circle diagram. this is not true for a brittle material. 4.6 MOHR’S THEORY OF FAILURE In the previous discussions on failure. Consider the line ABB¢A¢. according to the maximum shear stress theory. While the yield point stress sy for a ductile material is more or less the same in tension and compression. In such a case. t A B s3 s2 O sn s1 sn B¢ A¢ Fig. all the theories had one common feature. Hence. The maximum shear stress associated with this normal stress value is t. the planes having maximum shear stresses for given normal stresses. The fundamental assumption is that if failure is associated with a given normal stress value. It is also necessary to check the strength of the rope for this stress. The points lying on BA and B¢A¢ represent a series of planes on which the normal stresses have the same magnitude sn but different shear stresses. consider Mohr’s circles. the critical circle is the outermost circle in Mohr’s circle diagram. To explain the basis of Mohr’s theory. have their representative points on the outer circle. Mohr’s theory is an attempt to extend the maximum shear stress theory (also known as the stress-difference theory) so as to avoid this objection. shown in Fig. s2 and s3 are the principal stresses at the point. Consequently. In simple tension. In simple compression. Similarly.8 shows this limiting state of stress. causing failure.8. we can perform many more tests (like combined tension and torsion) until failure occurs in each case. we can draw an envelope. If one uses a factor of safety N. a large number of tests will have to be performed on a single material to determine the envelope for it.8. In the limit. relating to simple tension. For practical application of this theory. 4. s2 = s3 = 0. In general. we can construct the outermost circle. s1 = s2 = 0 and s3 = –syc. If the yield point stress in simple tension is small.8 Diagram representing Mohr’s failure theory In addition to the three simple tests. representing a finite limit for ‘hydrostatic tension’. on the left-hand side. The outermost circle in the circle diagram for this case is represented by C. then the envelope will cut the horizontal axis at point L. For pure shear. tangents to the circles as shown in Fig. The outermost circle for this state is indicated by S. The point of contact of the outermost circle for a given state with this envelope determines the combination of s and t. In each case. When a member is subjected to a general state of stress. as shown in Fig. on a given material. For all these circles.130 Advanced Mechanics of Solids Now. 4. the envelope rises indefinitely. the Mohr’s circle with (s1 – s3) as diameter should lie within the envelope. for no failure to take place. tys = s1 = –s3 and s2 = 0. we conduct three experiments in the laboratory. syc will be greater than syt numerically. one assumes the envelopes to be straight lines. Chapter_04.8. t A B C s yc C H G O D E s3* sy t s1* L F s T S Fig. pure shear and simple compression. compared to the yield point stress in simple compression. 5:42 AM . Figure 4. indicating no elastic limit under hydrostatic compression. the circle can touch the envelope. The outermost circle in the circle diagram (there is only one circle) corresponding to this state is shown as T in Fig. 4.pmd 130 7/3/2008. the test is conducted until failure occurs. where σ1* = N σ1 and σ 3* = Nσ 3 . s1 = syt.e. 4. Obviously. i. then the circle with N(s1 – s3) as diameter can touch the envelopes. and correspondingly for each state of stress. The plane on which failure occurs will have its representative point on this outer circle. for a brittle material. 6 Consider the problem discussed in Example 4. For a brittle material with no yield stress value. k will become equal to 1 and Eq. Example 4. k= σ ut σ uc (4. and after simplification.18a) becomes identical to the maximum shear stress theory.18b) Equation (4. determine the allowable force F according to Mohr’s theory of failure. Let the crankshaft material have s yt = 150 MPa and s yc = 330 MPa. (4. triangles LCF. and k is the ratio of syt to syc for the material.e. to avoid failure according to Mohr’s theory. LBE and LAD are similar. (4.Theories of Failure or Yield Criteria and Introduction to Ideally Plastic Solid 131 The envelopes being common tangents to the circles. Eq. Then.pmd 131 7/3/2008. 5:42 AM . Solution Bending moment at section A = (20 × 10−2 F ) Nm Torque = (15 × 10−2 F ) Nm Chapter_04. σ yt = σ 1* − σ yt * σ σ yc 3 = N (σ 1 − kσ 3) where k= (4.2). BG = AH (a) CG CH Now. When syt = syc. Consider a point on the surface of the shaft where the stress due to bending is maximum. the condition is σ 1 − kσ 3 ≤ σ yt = σ eq N where N is the factor of safety used for design.18c) syt /N is sometimes called the equivalent stress seq in uniaxial tension corresponding to Mohr’s theory of failure.2. i. If the diameter of the shaft is 10 cm. making CBG and CAH similar. (a). k is the ratio of s ultimate in tension to s ultimate in compression. Draw CH parallel to LO (the s axis).18a) states that for a general state of stress where s1 and s3 are the maximum and minimum principal stresses. BG = BE – GE = BE – CF = 1 σ yt − 1 (σ 1* − σ 3*) 2 2 CG = FE = FO – EO = 1 (σ 1* + σ 3*) − 1 σ yt 2 2 AH = AD – HD = AD – CF = 1 σ yc − 1 (σ 1* − σ 3*) 2 2 CH = FD = FO + OD = 1 (σ1* + σ 3*) + 1 σ yc 2 2 Substituting these in Eq.18a) σ yt σ yc (4. Let the factor of safety be 2. As can be observed.4545 σ yc 330 N (σ 1 − kσ 3) = 2 F(2106 + 31.5 F = σ yt = 150 × 106 Pa F = 34092N or 4. 2 2 σ2 = 0 σ1. the stress–strain diagram would be like the one shown in Fig. π k= − 68.9(a).pmd Stress Stress B O Strain (b) Stress–strain diagram for (a) Ductile material (b) Brittle material 132 7/3/2008.132 Advanced Mechanics of Solids σ (bending) = 64 Md4 = 32M3 Pa πd 2π d \ τ (torsion) = 32Td4 = 16T3 Pa πd 2π d σ 1. Part OA is linear.7) = 2106 F . (4. If a specimen is loaded within this limit and gradually unloaded.9 Chapter_04. it returns to its original length A O C D Strain (a) Fig. 4. 4274. such as mild steel. 4.75 F σ yt 150 = = 0.5 F \ From Eq.18a). 5:42 AM .7 IDEALLY PLASTIC SOLID If a rod of a ductile metal.3 = 1 σ ± 1 (σ 2 + 4τ 2 )1/ 2 . signifying that in this region.3 = 16M3 ± 8 4 (4 M 2 + T 2 )1/ 2 πd πd = 8 F ⎡ 2 (20 × 10−2 ) ± 10−2 (1600 + 22 F )1/ 2 ⎤ ⎦ π × 10−3 ⎣ = 80 F (40 ± 42. the curve has several distinct regions. the strain is proportional to the stress. is tested under a simple uniaxial tension.25) = 4274. pmd 133 7/3/2008.e. the elastic region is shown exaggerated for clarity. If the specimen is strained further.10 Ideal stress-strain diagram for a material that is (a) Linearly elastic (b) Rigidperfectly plastic (c) Rigid-linear work hardening (d) Linearly elastic-perfectly plastic (e) Linearly elastic-linear work hardening Chapter_04. Beyond A. Point B.10(b) represents a material which is rigid (i. while Fig. This is the linear elastic region and point A denotes the limit of proportionality. Figure 4. perfectly plastic and linearly elastic–linear work hardening. represents the elastic limit. 4.10(d) and (e) represent respectively linearly elastic. Brittle materials. indicating work hardening. 5:42 AM . Fig. 4.9(b). After D. 4. However. the strain upto point B is still elastic. 4. In order to develop stress-strain relations during plastic deformation.10(c) shows the behaviour of a material which is rigid for stresses below sy and for stress levels above sy a linear work hardening characteristics is exhibited. In these. In the figure.10(a) represents a linearly elastic material. Most metals and alloys do not have a distinct yield point. Strain (d) Strain (e) Fig. the curve becomes slightly nonlinear. Figure 4. A material exhibiting this characteristic behaviour is designated as rigid linear work hardening. further straining is accompanied by increased stress.10. allow very little plastic deformation before reaching the breaking point.Theories of Failure or Yield Criteria and Introduction to Ideally Plastic Solid 133 Strain (c) Stress Strain (b) Stress Strain (a) Stress Stress Stress without any permanent deformation. therefore. 4. The stress-strain diagram for such a material would look like the one shown in Fig. the actual stress-strain diagrams are replaced by less complicated ones. such as cast iron. Such a material is called a rigid perfectly plastic material. These are shown in Fig. the stress drops suddenly (represented by point C) and thereafter the material yields at constant stress. has no deformation) for stresses below sy and yields without limit when the stress level reaches the value sy. The change from the purely elastic to the elastic-plastic state is gradual. titanium carbide or rock material. dimensional strain space. several yield criteria have been considered. In particular. In the previous sections of this chapter. 0 ⎤ ⎡ p 0 0 ⎤ ⎡σ1 − p 0 0 ⎡σ1 0 ⎢0 σ 0 ⎥ = ⎢0 p 0 ⎥ = ⎢ 0 σ2 − p 0 2 ⎢ ⎥ ⎢ ⎥ ⎢ 0 σ3 − ⎣⎢0 0 σ 3 ⎦⎥ ⎣⎢0 0 P ⎥⎦ ⎢⎣ 0 or Chapter_04. possible to represent the yield surface in a three-dimensional space with coordinate axes s 1. The function f characterises the yielding of the material as follows: As long as f < 0 no plastic deformation or yielding takes place. i. e2. one may express the function as f (s1. s2. A basic assumption that is now made is that there exists a scalar function called a stress function or loading function. Similarly. The function f = 0 represents a closed surface in the stress space. K). l2 and l3 are the stress invariants. pure shear) state. which depends on the states of stress and strain.19) 7/3/2008. The Deviatoric Plane or the p Plane In Section 4. f > 0 has no meaning. One can imagine a six-dimensional space called the stress space. Yielding occurs when f = 0.8 STRESS SPACE AND STRAIN SPACE The state of stress at a point can be represented by the six rectangular stress components tij (i. 134 ⎤ ⎥ ⎥ p ⎦⎥ (i = 1. e3). the state of strain at a point can be represented by a point in a six. It is. it was stated that most metals can withstand considerable hydrostatic pressure without any permanent deformation.pmd [σ i ] = [ p] + [σ i* ]. 3). and the history of loading.e. s3) and the principal strains (e1. l 3) where l 1. We have also observed that a material is said to be isotropic if the material properties do not depend on the particular coordinate axes chosen. j = 1. 4. It has also been observed that a given state of stress can be uniquely resolved into a hydrostatic (or isotropic) state and a deviatoric (i. eij.2(a). s2 and s3. For materials with no work hardening characteristics.134 Advanced Mechanics of Solids In the following sections. Similarly. 2. therefore. we shall very briefly discuss certain elementary aspects of the stress-strain relations for an ideally plastic solid.e. a state of plastic strain eij(p) can be so represented. the yield function can be expressed as f (l1. 3) (4. the parameter K = 0. 5:42 AM . represented by f (tij. in which the state of stress can be represented by a point. s2. the plastic characteristics of the material are said to be isotropic if the yield function f depends only on the invariants of stress. Equivalently. It is assumed that the material behaviour in tension or compression is identical. strain and strain history. The isotropic stress theory of plasticity gives function f as an isotropic function of stresses alone. l2. 2. s3). A history of loading can be represented by a path in the stress space and the corresponding deformation or strain history as a path in the strain space. For such theories. These criteria were expressed in terms of the principal stresses (s1. we can discuss its characteristics with reference to its trace on the p plane.e. 6) also should cause yielding. This plane (s1. 3) Consequently.20) s2 represents a plane passing through the origin. (0. σ 2* . s 2*. s2 and s3 axes on this plane as s ¢1.11 The p Plane moved parallel to OD. These are (a. 0) lies on the yield surface. f ( l2* . c. 2. the yield function will represent a cylinder perpendicular to the p plane. this will generate 11 other (or a total of 12) points on the surface. Therefore. According to the isotropic stress theory. 4. For the deviatoric state. the yield function will be independent of the hydrostatic state. Thus. b. Figure 4. b. 0.Theories of Failure or Yield Criteria and Introduction to Ideally Plastic Solid 135 p = 1 (σ 1 + σ 2 + σ 3 ) 3 is the mean normal stress. i. appealing to isotropy and the property of the material in tension and compression. the yield function will be a function of the second and third invariants of the devatoric state. l3* ). 0). one point on the yield surface locates five other points. l1* = 0. (b. as we have assumed that the material behaviour in tension is identical to that in compression. Further. 0. b). s2 = 0.e. – 6) also cause yielding. O This is shown by point P in Fig. 0) and (0. b. c) and the remaining six are obtained by multiplying these by –1. 5:42 AM . – 6. The equation σ1* + σ 2* + σ 3* = 0 (4. σ 3*) causes yielding. c) (c. (i = 1. the yield locus is a symmetrical curve. c) on the yield surface. s2. a). b).9 GENERAL NATURE OF THE YIELD LOCUS Since the yield surface is a cylinder perpendicular to the p plane. i. a. If the stress state (σ1* . s 3*) Since the addition or subtraction of an isotropic state does not affect the yielding process. 0) and (0. 0. If we choose a general point (a. s3) is called the deviatoric plane or the p p plane plane. s2 and s3.11. (c. a. causes yielding. Since we have assumed isotropy. 0. Hence. 6. s3 P (s *1. whose normal OD is equally inclined (with direction cosines 1/ 3) D to the axes s1.12 shows the p plane and the projections of the s1. and where σ i* = σ i − p. i. therefore. the states (0. the states (– 6. These projections make an angle of 120° with each other.e. 4. c. s ¢2 and s ¢3. 4. s3 = 0. a) (b.pmd 135 7/3/2008. with reference to the yield locus. point P can be Fig. Let us assume that the state (6. Chapter_04. (a. the state s1 = 6. The trace of this surface on the p plane is the yield locus. the point represents1 ing this state will lie in the p plane. the point s1 = sy will have its projection on the p plane as sy cos φ = 2 / 3 σ y along the s ¢1 axis.2 was stated by the maximum shear stress theory. therefore.13. If s1. l2 . 4. s ¢2. According to this criterion. yielding occurs when s1 = sy. Similarly.136 Advanced Mechanics of Solids s 2¢ s 2¢ D f s 1¢ O O s 1¢ s 3¢ q p plane (b) (a) Fig. one can * deal with the deviatoric state (for which l1 = 0 ) and write the above condition as f (l2* .13.21) Since a hydrostatic state of stress does not have any effect on yielding. l3) = f (l12 − 3l2 ) = σ y2 (4. [Fig. Consequently. s ¢3 axes in Fig. be written as f = l2* + 1 σ y2 = l2* + s 2 3 Chapter_04.23) 7/3/2008. along s ¢1.4b). 4. Eq. s2 and s3 are all acting (with s1 > s2 > s3).10 (a) The yield locus (b) Projection of p plane YIELD SURFACES OF TRESCA AND VON MISES One of the yield conditions studied in Section 4.e. yielding occurs when s1 – s3 = sy. The s1 axis is inclined at an angle of φ to its projection s ¢1 axis on the p plane. In other words.12(b)]. (4. Another yield criterion discussed in Section 4. This condition is generally named after Tresca. the yielding starts 1 when the maximum shear (s1 – s3) becomes equal to the maximum shear sy /2 in 2 uniaxial tension yielding. other points on the p plane will be at distances of ± 2 / 3 σ y along the projections of s1.. Let us assume that only s1 is acting. l3*) = f (l2* ) = −3 l2* = σ y2 (4. Then.pmd 136 (4. This defines a straight line joining points at a distance of sy along s1 and –s3 axes. the yield locus is a hexagon. s2 and s3 axes on the p plane.2 was the octahedral shearing stress or the distortion energy theory. The projection of this line on the p plane will be a straight line joining points at a distance of 2 / 3 σ y along the s ¢1 and –s ¢3 axes. yielding begins when s1 – s3 = sy. as shown by AB in Fig.22) The yield function can. Hence. if s1 > s2 > s3. i. According to this theory. yielding occurs when f (l1 . and sin f = cos q = 1/ 3 . 4. 5:42 AM .12 4. 4. s = 1 σy 3 (4.27) D s2 s 2¢ 2/3 B sy s1 A s 3¢ s 1¢ s3 (a) (b) Fig. 5:42 AM .Theories of Failure or Yield Criteria and Introduction to Ideally Plastic Solid 137 where s is a constant. a right circular cylinder.13.11 STRESS–STRAIN RELATIONS (PLASTIC FLOW) The yield locus that has been discussed so far defines the boundary of the elastic zone in the stress space. 3 or.24) (σ1 − p) 2 + (σ 2 − p)2 + (σ 3 − p) 2 = 2s 2 (4. therefore.26) σ1* + σ 2* + σ 3* = 2s 2 (4. 4. The yield surface is defined by l2* + s 2 = 0 σ1σ 2 + σ 2σ 3 + σ 3σ1 − 3 p 2 = − s 2 or The other alternative forms of the above expression are (4. 4. This yield criterion is known as the von Mises condition for yielding. plastic deformation takes place. Consequently.23) s 2 = 1 σ y2 .25) can also be written as (4. one can speak of only the change in the plastic strain rather than the total plastic strain because the latter is the sum total of all plastic strains that have taken place during the previous strain history of the specimen. The yield surface according to the von Mises criterion is.e.13 Yield surfaces of Tresca and von Mises This is the curve of intersection between the sphere σ12 + σ 22 + σ 32 = 2s 2 and the p plane defined by σ1* + σ 2* + σ 3* = 0 .25) (σ1 − σ 2 ) 2 + (σ 2 − σ 3)2 + (σ 3 − σ1) 2 = 6s 2 Equation (4. (4. the cylinder of von Mises circumscribes Tresca’s hexagonal cylinder. In this context.28) Hence. From Eq. the radius of the cylinder is 2 / 3 σ y i. a circle with radius 2s in the p plane. therefore. When a stress point reaches this boundary. This is shown in Fig.pmd 137 7/3/2008. 4. This curve is. the stress–strain relations for plastic flow relate the Chapter_04. 29) (iii) The total strain increments deij are made up of the elastic strain increments d ε ije and plastic strain increaments d ε ijp d ε ij = d ε ije + d ε ijp (4. The following assumptions are made: (i) The body is isotropic (ii) The volumetric strain is an elastic strain and is proportional to the mean pressure (sm = p = s) ε = 3kσ or d ε = 3kdσ (4. the equations governing plastic deformation cannot. 5:42 AM . but must be differential relations. Consequently.pmd 138 (4. Another way of explaining this is to realise that the process of plastic flow is irreversible. be finite relations concerning stress and strain components as in the case of Hooke’s law.138 Advanced Mechanics of Solids strain increments. the volumetric strain is purely elastic and hence (4. in principle.33) 7/3/2008. ε xxp + ε yyp + ε zzp = 0 Chapter_04.30) (iv) The elastic strain increments are related to stress components s ij through Hooke’s law e d ε xx = 1 [σ x − ν (σ y + σ z )] E 1 e d ε yy = [σ y − ν (σ x + σ z )] E e (4.32) e e ε = ε xx + ε eyy + ε zz But e e e p ε = ε xx + ε yy + ε zz + (ε xxp + ε yy + ε zzp ) Hence. From (ii). that most of the deformation work is transformed into heat and that the stresses in the final state depend on the strain path.31) d ε zz = 1 [σ z − ν (σ x + σ y )] E 1 e d ε exy = d γ xy = τ xy G e e d ε yz = d γ yz = 1 τ yz G e e d ε zx = d γ zx = 1 τ zx G (v) The deviatoric components of the plastic strain increments are proportional to the components of the deviatoric state of stress p + ε zzp ) ⎤ = ⎡σ x − 1 (σ x + σ y + σ z ) ⎤ dλ d ⎡ε xxp − 1 (ε xxp + ε yy ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ 3 3 where dl is the instantaneous constant of proportionality. from Eqs (4.Theories of Failure or Yield Criteria and Introduction to Ideally Plastic Solid 139 Using this in Eq.37) we have d λ ≥ 0 If the von Mises condition is applied. (4.12 (4. It is also observed that the principal axes of stress and plastic strain increments coincide.35) PRANDTL–REUSS EQUATIONS Combining Eqs (4.35) d ε ij = d ε ij( e ) + d λ sij (4. dW p = d λT 2 Since dW p ≥ 0 2 ) + (σ z − p )2 + τ xy2 + τ yz2 + τ zx2 ⎤⎦⎥ (4.31) and (4.e.30). the above equations and the remaining ones are d ε xxp = dλ sxx p d ε yy = d λ s yy d ε zzp = dλ szz d γ xyp (4. as given in Eq. consider the work done during the plastic strain increment p dW p = σ x d ε xxp + σ y d ε yy + σ z d ε zzp + τ xy d γ xyp + τ yz d γ yzp + τzx d γ zxp ( = dλ σ x sxx + σ y s yy + σ z szz + τ xy sxy + τ yz s yz + τ zx szx ( ) ) 2 2 2 ⎤ = dλ ⎡σ x (σ x − p ) + σ y σ y − p + σ z (σ z − p ) + τ xy + τ yz + τ zx ⎣ ⎦ ( or 2 dW p = d λ ⎡(σ x − p ) + σ y − p ⎣⎢ i.30).36) where d ε ij( e) is related to stress components through Hooke’s law. (4. 5:42 AM .pmd 139 7/3/2008.31).34) = d λ sxy d γ yzp = d λ s yz d γ zxp = dλ szx Equivalently d ε ijp = d λ sij 4. Equations (4. It is easy to show that dl is non-negative.35) dWp = dl2s2 Chapter_04.31) and (4.35) constitute the Prandtl–Reuss equations.32) d ε xxp = dλ ⎡σ x − 1 (σ x + σ y + σ z ) ⎤ ⎢⎣ ⎥⎦ 3 Denoting the components of stress deviator by sij. For this.23) and (4. (4. (4. 14 shows three elements a. In this case.e dl is proportional to the increment of plastic work. it should be observed that the principal axes of strain increments coincide with the axes of the principal stresses.39) Expanding this ( ) dexx = 2 dλ ⎡σ x − 1 σ y + σ z ⎤ ⎢⎣ 3 2 ⎦⎥ deyy = 2 dλ ⎡σ y − 1 (σ z + σ x ) ⎤ ⎢⎣ 3 2 ⎦⎥ ( ) de zz = 2 dλ ⎡σ z − 1 σ x + σ y ⎤ ⎢⎣ 3 2 ⎦⎥ dgxy = dltxy dgyz = dltyz (4.1 Figure 4. b. the elastic components of strain are very small compared to plastic components. Which one of these three. c subjected to different states of stress.1 Chapter_04.000 a b c Fig. 5:42 AM 21. In such a case d ε ij ≈ d ε ijp and this gives the equations of the Saint Venant–von Mises theory of plasticity in the form de ij = dl sij (4. 4.40) dgzx = dltzx The above equations are also called Levy–Mises equations. will yield first according to (i) the maximum stress theory? (ii) the maximum strain theory? 7500 28. 4. 4.14 Problem 4. do you think.000 30.000 .pmd 140 7/3/2008.140 Advanced Mechanics of Solids dl = or dW p (4.38) 2s 2 i.500 21.13 SAINT VENANT–VON MISES EQUATIONS In a fully developed plastic deformation. and (b) according Fig. The tensile elastic limit is 413. when yields s ing starts. a = 0.000 N and T = 1800 Nm. The factor of safety N = 1. The tensile yield limit is 280 ¥ 103 kPa (2.85 kPa (2.6 A torque T is transmitted by means of a system of gears to the shaft shown in Fig. Hint: The forces F and 0.000 kPa [Ans. σ yc = 310 MPa . Note that when a torque is being transmitted. d = 4.510 kgf cm).1 × 10 6 kgf /cm 2 ) and s yp is 207.15) if the tensile load F is 35.17(b).3.15 Problem 4.4F acting on the gear A are shown in Fig.5. d = 4. where F is the tangential force. There are two bending moments—one in the vertical plane and the other in the (0.16. 5:42 AM .17. used to transmit 50 hp at 600 rpm. This is shown in Fig. σ y = 290000 kPa . in addition to the tangential force. R = 0. The shaft experiences bending owing to its own weight also.08 m. 4.1 kgf/cm2)] 4. Use a factor of safety N = 1.5. F T Fig.86 ¥ 103 kgf/cm2) and the shear yield limit is 140 ¥ 103 kPa (1.6 cm] 4. (iii) c] 4. F = 35. The factor of safety is 2. 4. ( 2.85 kPa. (ii) a. (b) 206.4F. 0.6 m long. (a) zero. If T = 2500Nm (25. [Ans. what is the tensile stress s at the point (a) according to the maximum t shearing stress theory. Use the maximum shear stress theory.000 N and the torsional moment T is 1800 Nm.17(b).5 to the octahedral shearing stress theory? [Ans. the state of t stress is as shown in Fig. (i) b. horizontal plane.1 cm] Use the maximum shear stress theory. the maximum moment is (a + b) Chapter_04.4 Fab) . 4. In the vertical plane.4 For the problem discussed in Problem 4.2 Determine the diameter of a cold-rolled steel shaft.3 4. Use a factor of safety 2. determine the diameter of the shaft. using the maximum shear stress theory.8 m and b = 0. determine the diameter according to Mohr’s theory if σ yt = 207 MPa.pmd 141 7/3/2008. d = 3.43 ¥ 103 kgf/cm2).Theories of Failure or Yield Criteria and Introduction to Ideally Plastic Solid 141 (iii) the maximum shear stress theory? Poisson’s ratio n = 0. If the shearing stress at the point is 206. E = 207 ¥ 10 6 kPa (2100 kgf/cm2). The shaft is simply supported at its ends in bearings. 4. there occurs a radial force equal to 0. 4.2 cm] 4.7 kPa.5 At a point in a steel member. 4. [Ans.16 Problem 4.1 m. The reactions at the bearings are also shown.25 [Ans.3 Determine the diameter of a ductile steel bar (Fig 4. The resultant bending is 1/ 2 ⎡⎛ 0. The other conditions are as given in Problem 4.4 Fab ⎞ 2 ⎛ Fab ⎞2 ⎤ (M)max = ⎢⎜ ⎟ +⎜ ⎟ ⎥ ⎢⎣⎝ a + b ⎠ ⎝ a + b ⎠ ⎥⎦ = 1.4.08F ab a+b The critical point to be considered is the circumferential point on the shaft subjected to this maximum moment. [Ans.pmd 142 7/3/2008.4 F C T a a+b F = T/R 0.4F b a+b (b) Fig.4F 0. [Ans.6 in the horizontal plane the maximum moment is ( Fab) .4 has syt = 207 ¥ 106 Pa and σ yc = 517 × 106 Pa determine the diameter of the bar according to Mohr's theory of failure.7 If the material of the bar in Problem 4. both these maxi( a + b) mums occur at the gear section A.142 Advanced Mechanics of Solids a b + A C D + + + F (a) F b a+b A a a+b D 0. d = 4. d ª 65 mm] 4. 4. 5:42 AM .17 Problem 4.6 cm] Chapter_04. The law asserts that ‘deflections are proportional to the forces which produce them’. the stresses and strains at a point were related through the material or the constitutive equations. In this chapter. relate the complete system of forces acting on the body to the deformation of the body as a whole. This is a very general assertion without any restriction as to the shape or size of the loaded body. Except for the condition that the material we considered was a continuum.e. 5.2 HOOKE’S LAW AND THE PRINCIPLE OF SUPERPOSITION We have observed in Chapter 3 that the rectangular stress components at a point can be related to the rectangular strain components at the same point through a set of linear equations that were designated as the generalised Hooke’s Law.1 5 INTRODUCTION In Chapters 1 and 2. we shall consider the entire body or structural member or machine element. we shall state Hooke’s law as applicable to the elastic body as a whole. the work done by these forces is stored as strain energy inside the body. however. . Many of the theorems developed in this chapter can be used with great advantage to solve displacement problems and statically indeterminate structures and frameworks. the shape or size of the body as a whole was not considered. the material properties rather than the behaviour of the body as such was not considered. i. Here too. The strain energy methods are extremely important for the solution of many problems in the mechanics of solids and in structural analysis.CHAPTER Energy Methods 5. When the body deforms under the action of the externally applied forces. It is assumed that the forces are applied gradually. which can be recovered when the latter is elastic in nature. In Chapter 3. also discussed the critical conditions to impend failure at a point. attention was focussed on the analysis of stress and strain at a point. on the theory of failure. Hooke’s law will relate the force acting on the body to the displacement. Chapter 4. along with the forces acting on it. In this chapter. 2) where a23 is the influence coefficient for vertical deflection at point 2 due to a force applied in the specified direction (that of F3) at point 3.144 Advanced Mechanics of Solids F1 1 3 q F3 2 D2 d2 Fig. then Hooke's law asserts that d 2 = a21F1 (5. Therefore. This is the principle of superposition. then a21 is the actual value of the vertical deflection at 2. according to Hooke’s Law. all having the direction of F1. Let us consider the vertical component of the deflection of point 2. (5. we have D2 = k21F1 where k 21 is some proportionality constant. are applied simultaneously at 1. is proportionate to F1. Let F1 be applied first. point 2 undergoes a displacement or a def lection. one can consider the displacement of point 2 in any specified direction and apply Hooke’s Law. and in consequence. The question that we now examine is whether the principle of superposition holds true to two or more forces. a force F1 is applied at point 1. When F 1 is increased. Consider a force F3 acting alone at point 3. then d 2 = a21F1 where a21 is a constant. 5. the resultant vertical deflection which they produce at 2 will be the resultant of the deflections which they would have produced if applied separately. and then F3. The vertical deflection at 2 is d 2 = a21F1 + a 23 ¢ F3 (5. then a deflection equal and opposite to the earlier deflection takes place. 5.3) .1) where a21 is a constant called the ‘influence coefficient’ for vertical deflection at point 2 due to a force applied in the specified direction (that of F1) at point 1. This def lection of point 2 may take place in a direction which is quite different from that of F1. If D2 is the actual def lection. If d2 is the vertical component. Let d 2 be the component of D 2 in a specified direction. such as F 1 and F 3. If several forces. if we fix our attention on the deflection in a specified direction. If F1 is a unit force. which act in different directions and at different points.1 Elastic solid and Hooke's law In Fig.1. which according to Hooke's law.e.1) d 2¢ = a23F3 (5. as stated by Eq. If q is the angle between D 2 and d 2 d 2 = D2 cosq = k21 cosq F1 If we keep q constant. If a force equal and opposite to F1 is applied at 1. i. D 2 also increases proportionately. Then. and let d2¢ be the vertical component of the deflection of 2. Only F3 is a21 acting now. including force F2 at 2 is d 2 = a21F1 + a22F2 + a23F3 + . valid for two different forces acting at two different points. if it exists. therefore.6) where k is a constant independent of F1 and F3.a23 F3 = 0 ′ F3 .a23 ′ )F3 (a21 . the The difference a21 . The forces of reaction at the points of support will also be considered as applied forces. is due to the presence of F1 when F3 is applied. (5.5) becomes ′ ′ a − a23 a21 − a21 = 23 =k F1 F3 d 2′′ (5. . left-hand side is a function of F3 alone.4) Since the elastic body is not subjected to any force now. . . This means that the deflection at 2 due to any number of forces. unless k vanishes. (5. If we apply -F3.a21 or ′ ′ a − a23 a21 − a21 = 23 F1 F3 (5.kF1 a23 Substituting this in Eq.5) ′ .3 CORRESPONDING FORCE AND DISPLACEMENT OR WORK-ABSORBING COMPONENT OF DISPLACEMENT Consider an elastic body which is in equilibrium under the action of forces F1. the deflection finally becomes ′ F1 . ′ F1 . the final deflection given by Eq.a21 ′ exists. . (5. Then ′ F1 ′ F3 . the right-hand side must be a function F1 alone. therefore.Energy Methods 145 where a'23 may be different from a23. Hence.7) 5. Hence.3) d 2 = a21F1 + a23F3 . Hence ′ = a23 .a21 d 2′′ = a21F1 + a23 (5. if it exists. (5. This difference. .4) must be zero. must be due to the action of F3.kF1F3 The last term on the right-hand side in the above equation is non-linear.2. and ′ a 23 = a23 and ′ a21 = a21 The principle of superposition is. Hence. F3. Consequently. if the difference a23 . 5. which is contradictory to Hooke's law.a21 = a21F1 + a23 ′ may be different from a21. Similarly. This is shown in Fig.a21 a21F1 + a23 i. F2. This can be extended by induction to include a third or any number of other forces.a23 it must be due to the action of F1 and. k = 0.a23 F3 ′ F3 .e. Eq. since F3 is acting when -F1 is applied. Now apply -F1. . ′ )F1 = (a23 . If the points of support a. For example.etc. Assuming that the forces are increased in constant proportion and the increase is gradual. . . . The corresponding displacement is also called the work-absorbing component of the displacement. (5. . when the forces F1. . . . .. when F1. . . F2. . d2. One can apply Hooke’s law to these corresponding displacements and obtain from Eq. . etc. . . 5.4 WORK DONE BY FORCES AND ELASTIC STRAIN ENERGY STORED Equations (5. .. a13. depend on all the forces F1. If the actual displacement is D1 and takes place in a direction as shown in Fig. ..7) F1 1 D1 F4 1 c F3 a 2 F2 b F5 Fig.. (5. . the deflections also increase similarly. (5.. the deflections reached are also equal to half their full magnitudes. are the influence coefficients of the kind discussed earlier. . etc. . the work done by F1 at its point of application will be W1 = 1 Fd 2 1 1 1 F (a F + a12F2 + a13F3 + . etc. 2 1 2 2 1 d = a21 1 F1 + a22 1 F2 + .. . therefore.8) where a11. F2. This corresponding displacement is denoted by d1. . . etc. b and c do not yield. then the component of this displacement in the direction of force F1 is called the corresponding displacement at point 1. The total work done by external forces is. are one-half of their full magnitudes. reach two-thirds of their full magnitudes. .. (5. a corresponding displacement can be identified. . .146 Advanced Mechanics of Solids d The displacement d1 in a specified direction at point 1 is given by Eq. . a12. . Similarly.9) 2 1 11 1 Similar expressions hold good for other forces also.7). etc. given by = W1 + W2 + W3 + . At every loaded point. If we slowly increase the forces F1. the deflections are ( ) ( ) ( ) ( ) 1 d = a11 1 F1 + a12 1 F2 + . . from zero to their full magnitudes. F2. 5.2).) 2 1 1 . d2 = a21F1 + a22F2 + a23F3 + .e.8) show that the displacements d1.2 Corresponding forces and displacements d1 = a11F1 + a12F2 + a13F3 + .) (5.. . then at these points the corresponding displacements are zero. = 1 (F d + F2d2 + F3d3 + . the deflections reached are also equal to two-thirds of their full magnitudes. . . F2. 2 2 2 2 i. etc. .. a 12 = a21 (5. are increased in constant conservation of energy principle and the superposition principle expression for U must hold without restriction on the manner application of these forces. apply force F2 at point 2. the dictate that this or order of the RECIPROCAL RELATION It is easy to show that the influence coefficient a12 in Eq. negative work will be done and the total work will be recovered. If it were not so. The potential energy that is stored as a consequence of the deformation of any elastic body is termed elastic strain energy. The corresponding deflection at point 2 is a22F2 and that at point 1 is a12F2. The energy stored is 1 1 F d = a F2 2 1 1 2 11 1 d1 = a11F1 U1 = since Next. keeping their ratios constant. This shows that the work done is stored as potential energy and its magnitude should be independent of the order in which the forces are applied. the additional energy stored is 1 F (a F ) + F1(a12F2) 2 2 22 2 The total elastic energy stored is therefore U2 = 1 1 a F 2 + a 22 F22 + a12 F1F2 2 11 1 2 Now. In general.5 (5. d3 are the corresponding displacements then the elastic strain energy stored is 1 (F d + F2d2 + F3d3 + .11b) U¢ = . force F1 is fully acting and hence. . aij = aji. it would be possible to store or extract energy by merely changing the order of loading and unloading. F2. .) 2 1 1 It must be noted that though this expression has been obtained tion that the forces F1. If F1.10) on the assumpproportion. consider a force F1 applied at point 1. d2. .. This would be contradictory to the principle of conservation of energy.11a) or in general a ij = aji (5. U and U¢ must be equal. F3 are the forces in a particular configuration and d1. When the forces are gradually reduced to zero. U= 5. (5. F2. if F2 is applied before F1. the elastic energy stored is U = U1 + U2 = 1 1 a F 2 + a11 F12 + a21F1F2 2 22 2 2 Since the elastic energy stored is independent of the order of application of F1 and F2. During this displacement.Energy Methods 147 If the supports are rigid. F3 .8) is equal to the influence coefficient a21. Consequently. then no work is done by the support reactions. To show this. . and let d1 be the corresponding displacement. .. . + a1nFn) + F2′ (a21F1 + a22F2 + . + ann Fn2 ) 2 11 1 + = 5. + Fn δ n′ i. + a1nFn) 2 1 11 1 +. as shown in the next section. + Fn′ dn = F1′(a11F1 + a12F2 + . .. . + a2nFn) + .6 1 F (a F + an2F2 + . .10) U= = 1 (F d + F2d2 + . . (5.. One can obtain an expression for the elastic strain energy in terms of the applied forces. . . . etc. F2′ . F1δ1′ + F2δ 2′ + . + annFn) = a11F1 F1′ + a22F2 F2′ + annFn Fn′ + a12( F1′F2 + F2′ F1) + a13( F1′F3 + F3′ F1) + . .. + Fndn) 2 1 1 1 F (a F + a12F2 + . + a1n F1Fn + .11b) has great importance in the mechanics of solids. .) do the same . . and δ 1′ . F2′ . . + Fn′ (an1F1 + an2F2 + . (5. . . using the above reciprocal relationship. .. ..e. F2′ . . . . . Let d 1. . . and F1′.. . etc. . + a1n( F1′Fn + Fn′ F1) (5. . δ 2′ . .148 Advanced Mechanics of Solids The result expressed in Eq. . . . . F2. . . . . From Eq. Then. . . .) 2 12 1 2 1 S(a11 F12 ) + S(a12F1F2) 2 (5. (5. . . .14) In words: ‘The forces of the first system (F1. . . . F2. F2. ..12) MAXWELL–BETTI–RAYLEIGH RECIPROCAL THEOREM Consider two systems of forces F1. d 2. . + annFn) 2 n n1 1 1 (a F F + a13F1F3 + .. .13) The symmetry of the expressions between the primed and unprimed quantities in the above expression shows that it is equal to F1 δ 1′ + F2 δ 2′ + . . .11) we have F1′d 1 + F2′ d 2 + . . both systems having the same points of application and the same directions. . . = F1δ′ 1 + F2′δ 2 + . . (5. U= 1 (a F 2 + a22 F22 + . making use of the reciprocal relation given by Eq.) acting through the corresponding displacements produced by any second system ( F1′. . the corresponding displacements caused by F1′. be the corresponding displacements caused by F1. . Betti and Rayleigh can also be given wider meaning with these extended definitions. . 5. Similarly. . subjected to a concentrated force F1 at point 1 and a couple F2 = M at point 2. a12 stands for the corresponding linear displacement of point 1 caused by a unit couple F2 applied at point 2. . as shown in Figs. the term 'displacement' may mean linear or angular displacement. is equal to the angular rotation at point 2 in the direction of the moment F2 caused by a unit load acting alone at point 1.3 Generalised forces W = 1 ( F1δ1 + F2δ 2 + .1 F2 = 1 .Energy Methods 149 amount of work as that done by the second system of forces acting through the corresponding displacements produced by the first system of forces’. This fact will be demonstrated in the next few examples. Check the deflections at points 1 and 2. It is possible to extend the term 'force' to include not only a concentrated force but also a bending moment or a torque. 5. etc. . represented concentrated forces and d1. If F1 is a unit force acting alone. For example.4 Example 5. . the corresponding linear displacements. and displacements Example 5.3. . Consider. d2. The reciprocal relation a12 = a21 can also be F1 interpreted appropriately. F1 F2. 2/3 L F1 = 1 2 2 a 12 a 21 1 (a) (b) Fig. 2 With the above generalised definitions for forces and displacements. a21 gives the corresponding angular rotation of point 2 caused by a unit concentrated force F1 at point 1.4(a) and (b).1 Consider a cantilever loaded by unit concentrated forces. . Betti and Rayleigh.3. d1 will now stand for the corresponding linear displacement of point 1 and d2 for the corresponding angular rotation of point 2. 5. the above relation reveals 1 that the linear displacement at point 1 in the direction of F1 caused by a unit couple acting alone at point 2. the elastic body shown in Fig. making reference to Fig. for example. Similarly. .7 GENERALISED FORCES AND DISPLACEMENTS In the above discussions. This is the reciprocal theorem of Maxwell. . then a11. etc. the work done when the forces are gradually increased from zero to their full magnitudes is given by Fig. gives the linear displacement of point 1 corresponding to the direction of F1. . + Fnδ n ) 2 The reciprocal theorem of Maxwell. 5. the influence coefficient. 5. 5. etc.5). 5. Similarly. one observes the deflections at 1 as F is moved along the beam to get the required information.6(a) is subjected to a bending moment M = F1 at point 1. a31 = a13. 5. 2/3 L d2 = Ma21 P 2 2 1 q1 = Pa12 1 M = F1 (a) (b) Fig.4(a).3 . errors would creep in.e.4(b) the unit load F2 acts at point 2 and as a result. the deflection of point 1 is a12. we have from elementary strength of materials 3 d2 due to F1= deflection at 1 due to F1 + deflection due to slope 3 3 = 8L + 4L 81 EI 54 EI d1 due to F2 = deflection at 1 due to a unit load at 1 + deflection at 1 due to a moment (L/3) at 1 3 3 = 8L + 4L 81 EI 54 EI Example 5. In Fig. 5. Example 5. On the other hand. Let us determine the curve of deflection for the beam. The deflection at 2 due to F at 1 is the same as the deflection at 1 due to F at 2.6(b). F 4 3 2 1 Fig.2 Solution One obvious method would be to use a travelling microscope and take readings at points 2. 5. Point 2 is 2/3 L from the fixed end. The reciprocal relation conveys that these two deflections are equal. These readings would be very small and consequently. and in Fig. a21 = a12.e. i. 4. i. 5.3 The cantilever beam shown in Fig. the deflection of point 2 is a21. the deflection at 3 due to F at 1 is the same as the deflection at 1 due to F at 3.150 Advanced Mechanics of Solids Solution In Fig. Hence. it is subjected to a concentrated load P = F2 at point 2.5 Example 5. the reciprocal relation can be used to obtain this curve of deflection more accurately. Verify the reciprocal theorem.6 Example 5. As a result. 5. 3.2 Consider a cantilever beam subjected to a concentrated force F at point 1 (Fig 5. If L is the length of the cantilever and if point 1 is at a distance of 2 L from the fixed end. the unit load F1 acts at point 1. 7(b) is e = σ − 2ν σ = (1 − 2ν ) σ E E E The two points of application A and B. the solution of which is apparently difficult. therefore. Consequently. (Fig. P s B h P s s A (a) Solution This is a very general (b) Fig. 5. as shown. the unit contraction in any direction from Fig.7 Example 5.Energy Methods 151 Solution From elementary strength of materials the deflection at point 2 due to the moment M at point 1 is ( ) 21EI = 29MLEI 2 d2 = M 2 L 3 2 The slope (angular displacement) at point 1 due to the concentrated force P at point 2 is ( ) 21EI = 29PLEI q1 = P 2 L 3 2 2 Hence. Example 5. we can get a solution very easily by applying the reciprocal theorem.4 Determine the change in volume of an elastic body subjected to two equal and opposite forces. 5. Dh = h (1 − 2ν ) σ E Now we have two systems of forces: System 1 Force Volume change P DV . Every volume element will be in a state of hydrostatic (isotropic) stress. 5. the work done by M through the displacement (angular displacement) produced by P is equal to 2 Mq1 = 2 MPL 9 EI This is equal to the work done by P acting through the displacement produced by the moment M. However. The distance between the points of application is h and the elastic constants for the material are E and n. Let the elastic body be subjected to a hydrostatic pressure of value s. move towards each other by a distance.4 problem.7). H ¢. we shall demonstrate the application of the reciprocal theorem to a problem in experimental mechanics. A theoretical analysis is quite difficult for an odd structure like the one shown. C. 0 at B. The two systems involved in the reciprocal theorem are as follows: System 1 Specified Forces V. 0 at E (i. M ¢ at B (unknown) and other reactive forces at A. d1 (unknown) at E. 0. M at B (unknown) and other reactive forces at A. horizontal and angular) at B equal to zero. displacement in the direction of P) is measured. = 5. Let this be δ 1′ . To determine V at B we proceed as follows. D fixed and preventing angular rotation and horizontal displacement at B. 0 at A. M ¢ and H ¢ that are induced at B are not measured. D (also unknown). 5. The reactions at the other supports also are such that the displacement at these supports are zero. System 2 Experimental Forces V ¢. 0. C.8 Reactions due to force P A known vertical displacement δ 2′ is imposed at B. P at E (known) Corresponding displacements 0. The reaction forces are V. It is required to determine the forces of reaction at point B. The corresponding displacement at E (i. 0.8 shows a structural member subjected to a force P at point E. P E A H M B D d¢2 C V Fig.8 BEGG¢¢ S DEFORMETER In this section. the change in volume is zero. keeping A. Figure 5. H. D (all unknown). the forces V ¢.e. point E not loaded) . C. During the vertical displacement of B. C and D.5. H and M and these make the displacements (vertical.e.152 Advanced Mechanics of Solids System 2 Force Distance change From the reciprocal theorem s Dh P Dh = s DV DV = P ∆h or σ Ph (1 − 2ν ) E If n is equal to 0. The corresponding displacement δ1′ at E is measured.8). A known horizontal displacement δ 2′ is imposed at B. 0. + a1n F1 Fn ) + . the value of V can be determined. we proceed as above. .15) Since δ 2′ is the known displacement imposed at B and δ1′ is the corresponding displacement at E that is experimentally measured. with all other displacements being kept zero. The rate at which U increases with F1 is given by ∂ F1 above expression for U. etc. . Eq. Hence. The reciprocal theorem again gives M = −P 5.12). . The result is H = -P δ1′ δ 2′ To determine M at B. The corresponding displacement δ1′ resulting at E is measured. .9 δ1′ θ′ FIRST THEOREM OF CASTIGLIANO From Eq. δ 1′ at E Applying the reciprocal theorem (V ⋅ δ 2′ ) + ( H ⋅ 0 ) + ( M ⋅ 0 ) + 0 + ( P ⋅ δ1′ ) = (V ′ ⋅ 0 ) + ( H ′ ⋅ 0 ) + ( M ′ ⋅ 0 ) + 0 + ( 0 ⋅ δ1 ) i. i. . 0.16) . moments or torques. then ∂U = d 1 ∂ F1 (5. + a F 11 1 12 2 13 3 1n n ∂ F1 This is nothing but the corresponding displacement at F1. if d1 stands for the generalised displacement (linear or angular) corresponding to the generalised force F1. . keeping all other displacements zero. It is necessary to remember that the corresponding displacement δ1′ at E is positive when it is in the direction of P. From the coefficients.- V= −P δ1′ δ 2′ (5. 0 at B. 0. a11. the expression for the elastic strain energy is ( U = 1 a11 F12 + a22 F22 + . etc. a12. are the corresponding influence ∂ U . concentrated loads.e. are the generalised forces. + ann Fn2 2 ) + ( a12 F1 F2 + a13 F1 F3 + . . . F1. ∂U = a F + a F + a F + . F2. a known amount of small rotation q ¢ is imposed at B.Energy Methods 153 Corresponding displacements δ 2′ . In the above expression. (5. C and D. To determine H at B. (5. 0 at A.. ..e. . Fn Let the strain energy stored be U. Let the elastic system be free of all forces. . where under forces F1.16). etc. Some of these are concentrated F3 loads and some are couples and torques.154 Advanced Mechanics of Solids In exactly the same way.9). That is to say. Now F4 F5 increase one of the forces. etc. F1 . ∂ F3 = δ 3 . Let D Fn be applied first. Several examples will illustrate these subseF1 quently. when DFn Æ 0 ∂U = δ n ∂ Fn . This theorem is extremely useful in deF2 termining the displacements of structures as well as in the solutions of many statically indeterminate structures.. the elementary force DFn is acting all the time with full magnitude at the point n which is undergoing a displacement dn. 5. one can show that ∂U ∂U ∂ F2 = δ 2 .9 Elastic body in equilibrium creases to U + DU. by DFn and as a result the strain energy inFig. 5. . The total energy stored is therefore 1 ∆ F ∆δ 2 n n Equating this to the previous expression. say Fn.etc.e. . Fn. The energy stored is 1 ∆ F ∆δ 2 n n where Ddn is the elementary displacement corresponding to D Fn. we get U + ∆ Fn δ n + U+ ∆U 1 ∆ Fn = U + ∆ Fn δ n + ∆ Fn ∆δ n ∆ Fn 2 In the limit. But while these displacements are taking place. We can give an alternative proof for this theorem as follows: Consider an elastic system in equilibrium under the force F1. i. This is a quantity of the second order which can be neglected since D Fn will be made to tend to zero in the limit. . In the form derived in Eq. Hence. This is Castigliano's first theorem. 5. bodies satisfying Hooke's Law (see Sec. . . These forces by themselves do an amount of work equal to U. this elementary force does work equal to D Fn dn. . ∆U ∆F DU = ∆ Fn n Now we calculate the strain energy in a different manner. F2. we put all the other forces. etc. (Fig. . F2. ‘the partial differential coefficient of the strain energy function with respect to Fr gives the displacement corresponding with Fr’.15). F2 . the theorem is applicable to only linearly elastic bodies. . . Next. (5. we must express the strain energy in terms of the forces (including moments and couples) since it is the partial derivative with respect to a particular force that gives the corresponding displacement. this section will be subjected to three forces Fx. bending moment and torsion. we can assume that these forces and moments remain constant over Ds. Similarly. the forces and moments have opposite signs. At the left-hand section of this elementary member. 5.10 Reactive forces at a general cross-section . during the twist caused by the torque T = Mx. shear force.Energy Methods 155 it is important to note that dn is a linear displacement if Fn is a concentrated load. We shall make use of the formulas available from elementary strength of materials. the remaining forces and moments do no work.10(a) shows an elastic member subjected to several forces. Further. Fy and Fz and three moments Mx. Figure 5. no work is assumed to be done (since the deformations are extremely small) by the other forces and moments.10 EXPRESSIONS FOR STRAIN ENERGY In this section we shall develop expressions for strain energy when an elastic member is subjected to axial force. During the deformation caused by the axial force Fx alone. My and Mz (Fig. Let Ds be an elementary length of the member. Moment M x is the torque T and moments My and Mz are the bending moments about the y and z axes respectively. In the next section. then when Ds is very small. In general.10(b)). Consequently. The force Fx is the axial force and forces Fy and Fz are the shear forces across the section. expressions for strain energies in terms of forces will be obtained. Consider a section of the member at C. B A My C F3 F1 C Fy DS Fx F2 Mx Fz Mz (a) (b) Fig. 5. 5. the work done by each of these forces and moments can be determined individually and added together to determine the total elastic strain energy stored by Ds while it undergoes deformation. or an angular displacement if Fn is a couple or a torque. If we assume that the shear force is distributed uniformly across the section (which is not strictly correct).11) D s Dg and the work done by Fy will be \ DU = Dg Ds Dg DU = 1 Fy ∆s ∆γ 2 From Hooke's law. to take into account the different cross-sections and nonuniform distribution. the work done is DU = DU = 1 M z ∆φ 2 .18) ∆s 2 AG A similar expression is obtained for the shear force Fz. (ii) Elastic energy due to shear force: The shear force Fy (or Fz) is distributed across the section in a complicated manner depending on the shape of the cross-section.156 Advanced Mechanics of Solids (i) Elastic energy due to axial force: If dx is the axial extension due to Fx. the error caused in assuming uniform distribution of the shear force across the section will be very small.12. then DU = 1 Fxδ x 2 F = 1 Fx ⋅ x ∆ s AE 2 using Hooke's law. the shear displacement will be (from Fig. Fx2 (5. Hence. 5. With this or DU = k Fy2 (5. 5. Fy AG where A is the cross-sectional area and G is the shear modulus. which is often ignored. if Df is the angle of rotation due to the moment Mz(or My). Substituting this Dg = Ds Fig. However.17) ∆s 2 AE A is the cross-sectional area and E is Young's modulus. 5. a factor k is introduced. (iii) Elastic energy due to bending moment: Making reference to Fig.11 Displacement due to shear force F DU = 1 Fy ∆ s y 2 AG Fy2 ∆s 2 AG It will be shown that the strain energy due to shear deformation is extremely small. (iv) Elastic energy due to torque : Because of the torque T.23) M y2 ∫ 2 EI ds y 0 (5. we have Df Mz = E Iz R R Mz Ds Mz Fig.20) give important expressions for the strain energy stored in the elementary length Ds of the elastic member. The elastic energy for the entire member is therefore (i) Due to axial force U1 = S F2 x ∫ 2 AE ds (5. 5. Hence Df = ∆ s Mz = ∆s R EI z Substituting this DU = M z2 ∆s 2 EI z (5.Energy Methods 157 From the elementary flexure formula.20) Equations (5.19) A similar expression can be obtained for the moment My. DU = 1 T ∆θ 2 = T 2 ∆s 2GI p (5. The work done due to this twist is.e.22) k z Fz2 ∫ 2 AG ds 0 (5. Dq = T ∆s GIp lp is the polar moment of inertia.21) 0 (ii) Due to shear force U2 = U3 = (iii) Due to bending moment U4 = S k F2 y y ∫ 2 AG ds 0 (5.17)-(5. the elementary member rotates through an angle Dq according to the formula for a circular section T = G ∆θ ∆s Ip i.24) S S .12 Displacement due to bending moment Mz 1 = R EI z or where R is the radius of curvature and lz is the area moment of inertia about the z axis. If the beam is of a rectangular section and A = bd .22) is (putting k1 = 1) U2 = L 2 P2 L P ∫ 2 AG dx = 2 AG 0 One can now show that U2 is small as compared to U1. I = 1 bd 3 12 2G ª E Substituting these 3 P2 L U2 ⋅ 6bd2 3 ⋅ 2G = 2bdG 12 P L U1 = d2 2 L2 For a member to be designated as beam. the length must be fairly large compared to the cross-sectional dimension.26) p Example 5. L > d and the above ratio is extremely small.13 Example 5. (5. (5.25) 2 T ∫ 2GI ds 0 (5. Consequently.24) L Px dx P 2 L3 U1 = ∫ ( ) = 2 EI 6 EI 0 2 The elastic energy due to shear from Eq. .13. from Eq. one can neglect shear energy as compared to bending energy. therefore.5 Determine the deflection at end A of the cantilever beam shown in Fig. 5. With U= P 2 L3 6 EI we get ∂ U P L3 = = δA ∂ P 3EI which agrees with the solution from elementary strength of materials. Hence.158 Advanced Mechanics of Solids U5 = (iv) Due to torque U6 = M z2 ∫ 2EI ds z 0 S S (5. 5.5 M = Px The elastic energy due to bending moment is. Solution The bending moment x at any section x is A B L P Fig. 15 Example 5. L2 L1 I2 I1 P A Solution The bending energy is U= ( Px )2 L ( Px )2 ∫ 2 EI dx + ∫ 2 EI dx 1 2 L 0 1 = Fig.Energy Methods 159 Example 5. R= Pa 2 3b + 2a 2 ( b + a )3 Remembering that a + b = L. 5.7 L1 2 L31 ( P + P L3 − L31 6 EI1 6 EI 2 2 ) ) Determine the support reaction for the propped cantilever (Fig. Neglect shear energy. the length of cantilever.) P a b A R Fig. The energy is U= b ( − Rx )2 ∫ 2 EI 0 2 ⎡ − R ( b + x ) + Px ⎤⎦ dx + ∫ ⎣ dx 2 EI 0 a ⎛ R 2 b3 R 2 b 2 a R 2 a 3 R 2 ba 2 U= 1 ⎜ + + + 2 6 2 EI ⎝ 6 + ∂U = 1 EI ∂R P 2 a3 PRb a 2 2 PRa 3 ⎞ − − ⎟ 6 2 6 ⎠ ⎛ Rb3 Ra 3 Pba 2 Pa 3 ⎞ 2 2 ⎜ 3 + Rb a + 3 + Rba − 2 − 3 ⎟ ⎝ ⎠ Equating this to zero and solving for R. 5.7 Solution The reaction R at A is such that the deflection there is zero.6 P L31 dA = ∂ U = + P L3 − L31 ∂ P 3EI1 3EI 2 ( Example 5. ( ) ( 32 − 2aL ) R= P a L 2 .15. 5. determine the deflection at end A.14 Example 5.6 For the cantilever of total length L shown in Fig. 5.14. 8 ( ) 2 \ dA = ∂ U = 3 π r + 4 L Pr 2 EI ∂P Example 5.pmd 160 7/3/2008. 5. The energy due to torque is.Advanced Mechanics of Solids 160 Example 5. is subjected to torque T.8 end A? For the structure shown in Fig. What is the twist of end A? The member is circular in section. Solution The torque is a mo- y O T A q r x T A q T co T q sin O sq ment in the xy plane and can be represented by vector T. 5. 5. Then π P2 r 3 2P2 r 2 L U= 3 + EI 4 EI Fig.16 Example 5.9 The end of the semi-circular member shown in Fig.24).17 Example 5. from Eq. this vector can be resolved into two components T cosq and T sinq. The component T cosq acts as torque and the component T sinq as a moment. what is the vertical deflection at Solution The moment at any section q of the curved part is Pr (1 . At any section q. π T cos θ ) U1 = ∫ ( r dθ 2GI P 0 2 (a) (b) Plan View Fig.cos q ).9 2 = π rT 4GI P The energy due to bending is. π U2 = ∫ (T sin θ )2 2 EI 0 r dθ = π rT 4 EI 2 Chapter_05. (5. The bending energy therefore is r A q L ( 2 Pr )2 ⎡⎣ Pr (1 − cos θ )⎤⎦ r dθ + ∫ dx ∫ 2 EI 2 EI 0 0 2 π P L We neglect the energy due to the axial force.17. from Eq. 5. The bending moment for the vertical part of the structure is a constant equal to 2Pr. 7:55 AM . (5.26). as shown.16. i.10 Determine the slope at end A of the cantilever in Fig. Situations arise where it may be desirable to determine the displacement (either linear or angular) at a point where there is no force (concentrated load or a couple) acting.18 Example 5. we Fig. and in the final result.10 assume a fictitious moment M to be acting at A and determine the slope caused by P and M. L U = ∫ 0 ( Px + M )2 2 EI dx 2 3 2 2 = P L + M L + MPL 6 EI 2 EI 2 EI .Energy Methods 161 Ip is the polar moment of inertia. the fictitious load is put equal to zero. The following example will describe the technique. the total energy is 2 ⎛ ⎞ U = U1 + U 2 = π rT ⎜ 1 + 1 ⎟ 4 ⎝ GI p EI ⎠ Hence. So. the twist is ( = 1 ( 1 + 1 )T r 2G E ) q = ∂ U = π rT 1 + 1 2 4 ∂T 2 2G E π r 3 5. But no moment is acting at A. Example 5. For a circular member 4 Ip = 2 I = π r 2 Substituting. M is equated to zero. 5. 5. P Solution To determine the slope by Castigliano's method we have to determine U and L A M take its partial derivative with respect to the corresponding force. Since the magnitude of M is actually zero.e. in the final result. a moment. we assume a small fictitious or dummy load to be acting at the point where the displacement is required.18 which is subjected to load P. The energy due to P and M is. Castigliano's theorem is then applied.11 FICTITIOUS LOAD METHOD Castigliano's first theorem described above helps us to determine the displacement at a point corresponding to the force acting there. In such situations. If M is zero. 5.162 Advanced Mechanics of Solids 2 q = ∂ U = ML + PL ∂ M EI 2 EI This gives the slope when M and P are both acting. The bending moment at section q of the semi-circular part is M 1 = Pr (1 . determine the ratio of L to r if the horizontal and vertical deflections of the loaded end A are equal.cos q ) . the slope due to P alone is PL2 2 EI If on the other hand.11 For the member shown in Fig. apply a horizontal fictitious force F to the right. P is zero and M alone is acting the slope is q= q= ML EI Example 5.Fr sin q ) At any section x in the vertical part. Solution In addition to the vertical for P at A. U= 1 2 EI π L 2 1 ∫ ⎣⎡ Pr (1 − cos θ ) − Fr sin θ ⎦⎤ r dθ + 2 EI ∫ ( 2Pr + Fx ) dx 0 0 2 ∂ U = − r 2 π ⎡ Pr (1 − cos θ ) − Fr sin θ ⎤ sin θ dθ + 1 L ( 2 Pr + Fx ) x dx ⎦ EI 0∫ ⎣ EI 0∫ ∂F \ and ∂U ∂F 2 π F =0 = δh = − r EI L 1 ∫ ⎡⎣ Pr (1 − cos θ ) sin θ ⎤⎦ dθ + EI ∫ 2 Pr xdx 0 0 ( 2 Pr 3 PrL2 Pr + = −2r 2 + L2 EI EI EI From Example 5.16.8. Example 5. P is the only force acting.8 =− ( 2 dv = Pr 3 π r + 4 L EI 2 Equating dv to dh ( ) ) ) ( Pr 2 3 π r + 4 L = Pr −2 r 2 + L2 EI 2 EI ( ) L2 − 4 Lr − r 2 3π + 2 = 0 2 Dividing by r2 and putting L = r r 2 ρ − 4 ρ − 3π + 2 = 0 2 or ( ) ) . the moment is M 2 = 2Pr + Fx Hence. However. The reason for this is simple. U2 is the work energy caused by F 2 alone. 5. ρ= or ρ =2+ 6+ 3π 5.e M is not a fictitious force as was assumed in that example.10. Hence. Example 5.e. Another way of observing this is to note that the strain energy functions are not linear functions. As a specific example. When F 2 is applied next (with F1 continuing to act).Energy Methods 4 ± ⎡⎣16 + 4 ( 3π /2 + 2 ) ⎤⎦ 2 Solving. the total energy is given by U1 + U2 + work done by P during the displacement caused by M.18. one cannot obtain the total elastic energy by adding the energies caused by individual forces. plus the work done by F1 during the displacement caused by F 2. Consider an elastic body subjected to two forces F 1 and F 2. i. the total energy stored when both F1 and F2 are acting is equal to (U1 + U 2 + U3). the energy due to M alone is U2 = 1 2 EI L 2 M L 2 ∫ M dx = 2 EI 0 Obviously. U1 + U2 is not equal to U. i. if P is applied first and then M. When F1 is applied first. where U1 is the work energy caused by F1 alone. and U3 is the energy due to the work done by F1 during the displacement caused by F 2. Let P and M be actual forces acting on the cantilever. let the energy stored be U 1. In other words.12 163 2 SUPERPOSITION OF ELASTIC ENERGIES When an elastic body is subjected to several forces. individual energies cannot be added to get the total energy. the additional energy stored is equal to U 2 due to F 2 alone. The deflection at the end of the cantilever (where P is acting with full magnitude) caused by M is 2 d *A = ML 2 EI During this deflection. Hence. The elastic energy stored due to P and M is given by (a). 2 3 2 2 U = P L + M L + MPL 6 EI 2 EI 2 EI The energy due to P alone is U1 = 1 L ( Px)2 dx = P 2 L3 2 EI 0∫ 6 EI Similarly. consider the cantilever shown in Fig. the work done by P is ⎛ ML2 ⎞ U3 = P ⎜ ⎟ ⎝ 2 EI ⎠ . the sum of individual energies is not equal to the total energy of the system. The order of loading is immaterial. It is immaterial whether P is applied first or M is applied. Hence. This is true in the case of small deformation as we have been assuming throughout our discussions. For example. 5. Example 5. one should be careful in applying the superposition principle to the energies. as shown in Fig. we get 2b M0 b U¢ = ∫ 0 M 02 dx + 2 EI a ∫ 0 ( M 0 − P /2 x )2 2 EI dx 2 ⎛ ⎞ = 1 ⎜ M 02 b + M 02 a − M 0 P a + 1 P 2 a 3 ⎟ 2 EI ⎝ 2 12 ⎠ Because of symmetry.13 STATICALLY INDETERMINATE STRUCTURE Many statically indeterminate structural problems can be conveniently solved. an axial force causing linear deformation will not do work during an angular deformation (or twist) caused by a torque. 5.164 Advanced Mechanics of Solids If this additional energy is added to U1 + U2. 5. as shown in (b). using Castigliano's theorem. then one gets the previous expression for U. The technique is to determine the forces and moments to produce the required displacement. However. b Solution The symmetry conditions indicate that the top and bottom members C C deform in such a manner that the tangents at the points of loading remain horiP/2 zontal.12 the frame.19.19 Example 5. Hence ∂U ′ = 1 ⎡ 2M 0 (a + b) − 1 Pa 2 ⎤ ⎥⎦ ∂ M0 2 EI ⎢⎣ 2 .7 was one such problem. the individual energies caused by axial force.12 A rectangular frame with all four sides of equal cross section is subjected to forces P. Similarly. bending moment and torsion can be added since the force causing one kind of deformation will not do any work during a different kind of deformation caused by another force. one can consider only a quarter part of Fig. Thus. Also. a bending moment will not do any work during axial or linear displacement caused by an axial force. the change in slope at section C is zero. Determine the moment at section C and also the increase in the disP tance between the two points a of application of force P. there is no a a P change in slopes at sections (a) (b) C-C. The following example will further illustrate this method. Example 5. Considering only the bending energy and neglecting the energies due to direct tension and shear force. one can consider only a quarter of the ring for calculation as shown in Fig. determine the change in the vertical diameter. we determine the partial derivative of 4U¢ with respect to P (assuming that the bottom loaded point is held fixed).13 A thin circular ring of radius r is subjected to two diametrically opposite loads P in its own plane as shown in Fig. M0 = Pa 2 4 ( a + b) To determine the increase in distance between the two load points. ⎡ 2 4 2 4 2 3⎤ U = 4U ′ = 4 ⎢ P a 2 ( a + b ) − P a + P a ⎥ 2 EI ⎢16 ( a + b ) 12 ⎥ 8( a + b) ⎣ ⎦ ∂ U = Pa 3 ( a + 4b ) ∂ P 12 EI ( a + b ) \ Example 5. The value of M0 is such as to cause no change in slope at B.20 Example 5. Also. So. P A r A A q B M0 M0 B P/2 M0 q B P/2 P/2 (b) (c) P (a) Fig. Section at A can be considered as built-in.20(a). 5. Moment at q = M = P r (1 − cos θ ) − M 0 2 U= π/ 2 2 1 ⎡ P r (1 − cos θ ) − M ⎤ r dθ 0⎥ 2 EI 0∫ ⎢⎣ 2 ⎦ Since there is no change in slope at B π/ 2 ∂U = − r 2 ⎡ P r (1 − cos θ ) − M 0 ⎤ dθ = 0 2 EI 0∫ ⎢⎣ 2 ∂ M0 ⎦⎥ .13 Solution Because of symmetry. during deformation there is no change in the slopes at A and B.Energy Methods 165 Equating this to zero. 5.20(c). 5. Obtain an expression for the bending moment at any section. . . Let one of the displacements d1 be increased by a small quantity Dd1. which is (P/2). Observe that in order to determine the deflection at B. U* = U* = = \ π/ 2 2 1 ⎡Qr 2 − cos θ ⎤ r dθ ⎥⎦ π 2 EI 0∫ ⎢⎣ ( Q2 r3 π / 2 2 − cos θ 2 EI 0∫ π ) 2 dθ 3 π/ 2 ⎛ 4 ⎞ ∂U* Qr = + cos2 θ − 4 cos θ ⎟ dθ ∂Q π EI 0∫ ⎜⎝ π 2 ⎠ = Qr 3 ⎛ 4 π π 4 ⎞ + − EI ⎜⎝ π 2 2 4 π ⎟⎠ ( ) ( ) ( ) 3 Qr 3 π 2 − = Pr π − 2 EI 4 π 2 EI 4 π As this gives only the increase in the radius. etc. F2. .e. 5. the elastic energy is \ π/ 2 ( ) ( ) 2 1 ⎡ Pr 2 ⎤ ∫ 2 EI ⎢⎣ 2 π − cos θ ⎥⎦ r dθ 0 The differential of this with respect to (P/2) will give the vertical deflection of the end B with reference to A. Remember that these are the work absorbing components (linear and angular displacements) in the corresponding directions of the forces (Fig. Fig. d2. . one has to take the differential with respect to the particular load that is acting at that point.e. . During this additional displacement. 5. etc.21). or ( ) P r π −1 − M π = 0 0 2 2 2 ( ) M 0 = Pr 1 − 2 2 π ( ) ( ) M at q = P r (1 − cos θ ) − P r 1 − 2 = Pr 2 − cos θ 2 2 2 π π To determine the increase in the diameter along the loads.. all other displacements where forces are acting are . If one considers the quarter ring. π/ 2 ⎡P ⎤ ∫ ⎢⎣ 2 r (1 − cos θ ) − M 0 ⎥⎦ dθ = 0 0 i. . be the corresponding displacements. = 3 dv = Pr π − 2 EI 4 π 5. the increase in the diameter is twice this quantity.20(c).166 Advanced Mechanics of Solids i.14 THEOREM OF VIRTUAL WORK Consider an elastic system subjected to a number of forces (including moments) F1. Let d1. one has to determine the elastic energy and take the differential.e. . Putting (P/2) = Q. i. (except F1) do no work at all because Fig. i. the strain energy must be expressed in terms of d1. 5.27) This is the theorem of virtual work.Energy Methods 167 held fixed. then according to the principle of virtual work. ∂ U = W. .21 Generalised forces and their points of application do not move displacements (at least in the work-absorbing direction). etc. Further.16). All the members have the same cross-sectional areas and are of the same material. linear or nonlinear. caused by the change in F 1. . For d2 F3 example. 5. If d1 and d2 are the vertical and horizontal displacements. . as derived in Eq.e. . . Hence DU = F1 Dd1 + k DF1 Dd1 or and ∆U = F + k ∆ F 1 1 ∆δ1 Lt ∆U = ∂ U = F1 ∆δ1 → 0 ∆δ1 ∂δ1 (5. F2. In applying this virtual displacement. It is important to observe that in obtaining the above equation. whereas in the application of Castigliano's theorem U had to be expressed in terms of F1. (5. This aspect will be discussed further in Sec.14 Three elastic members AD. then Dd1 must be consistent with such a constraint. . .. . if point I is constrained in such a manner that it can move only in a particuF1 lar direction. This additional work is stored as strain energy DU. 5. it is possible for D to move vertically and horizontally.22. What is the deflection of D under load W? Solution Under the action of load W.15. A hypothetical d1 displacement of such a kind is called a virtual displacement. is strictly applicable to linear elastic or Hookean materials. as shown in Fig. whereas Castigliano's first theorem. . .. etc. the small d 3 F2 displacement Dd1 that is imposed must be consistent with the constraints acting. BD is 100 cm long and members AD and CD are each 200 cm long. Note that in this case.. ∂ U = 0 ∂δ 2 ∂δ1 where U is the total strain energy of the system. d2. that it obeys Hooke's law. Example 5. The theorem is applicable to any elastic body. which means that additional forces may be necessary to maintain such a condition. BD and CD are connected by smooth pins. etc. we have not assumed that the material is linearly elastic. The only force doing work is F1 by an amount F 1 Dd 1 plus a fraction of DF 1 Dd 1. F 2. the forces F1. 2 2 Hence. BD will not undergo any changes in length but AD will extend by d1 cosq and CD will contract by the same amount. the forces in the members are (with d as corresponding extensions) in AD: aEδ = aE 1 3 δ1 + δ 2 1 2 200 L 1 in BD : aEδ = aEδ2 L 100 ( ) ( ) ( ) ( ) in CD : aEδ = aE 1 − 3 δ1 + δ 2 1 2 200 L The total elastic strain energy taking only axial forces into account is P 2 L = aE ⎡ 1 U= Σ 2aE 2 ⎢⎣ 800 + = aE \ W= ( ( 3 δ1 + δ 2 1 − 3δ +δ 1 2 800 ) ⎤⎥⎦ ( 8003 δ ) 2 1 + 1 δ 22 160 ∂ U = 3aE δ ∂δ1 400 1 2 ) 2 + 1 δ 22 100 . 5. assuming Hooke's law. BD will extend by d2 and AD and CD each will extend by 1 d. For the present example.14 Becacse of d1. and Example 5. (a).17. the total extension of each member is AD extends by 1 3 δ1 + δ 2 cm 2 BD extends by d2 cm CD extends by 1 − 3 δ1 + δ 2 cm 2 To calculate the strain energy. From Fig.22 Example 5. one needs to know the force-deformation equation for the non-Hookean members.168 Advanced Mechanics of Solids A A d2 sin q q q d2 D B D d1 d1 W d1 cos q C (a) (b) (c) (d) Fig. 3 2 cos q = Because of d2.17. This aspect will be taken up in Sec. 5. d2. . Let the specified displacements be d1. dr and the specified forces be Fs. the corresponding forces are F1′. This amounts to the statement that U is an essentially positive quantity. may be negative but the sum of these products taken as a whole is positive. (δt′ − δt′′ ). dr have the values F1′′ . . . Ft. dr displacements Second System Forces Fs. 0. d2.Fn. º. it would have been possible to extract energy by applying an appropriate system of forces. Fn have the values δ s′′. Ft. . . Fr′ . . . If this system is not unique. δ n′ Corresponding d1. If this were not so. we shall prove an important theorem dealing with the uniqueness of solution. . . (δ n′ − δ n′′ ) . by the principle of superposition the difference between these two systems must also be an equilibrium configuration.. This means that some of the products Fn dn etc. Fn δ s′ . d2. On the basis of this and the superposition principle. the corresponding displacement are δ s′ . . . Fr′ and at those points where forces are prescribed.. . . . Fr′′ Corresponding displacements d1.. will have a unique equilibrium configuration. When the body is elastic. ( Fr′ − Fr′′ ) . δ t′ . . this work is stored as elastic strain energy. ( F2′ − F2′′ ) . .. . . . 0 º. . . . . δ t′ . every portion of the body must store positive energy or no energy at all. . F2′′.. . δ n′ . Let this be the equilibrium configuration. . Hence. δ n′′ We have assumed that these are possible equilibrium configurations.. . Ft... º.Energy Methods and 0= 169 ∂ U = aE δ ∂δ 2 80 2 Hence. . U will vanish only when every part of the body is undeformed. Forces F1′. which means that D moves only vertically under W and the value of this vertical deflection d1 is d1 = 400 W 3aE 5. . . δ t′′. which states the following: An elastic body for which displacements are specified at some points and forces at others. Fs. . δ n′′.. we get the third equilibrium configuration as First System Forces Corresponding displacements ( F1′ − F1′′ ) . . Subtracting the second system from the first.. . . F2′ . F2′′. . . . . .. . Accordingly.. . . we can prove Kirchhoff's uniqueness theorem. Fr′′ and the displacements corresponding to the forces Fs. . we observe that the applied forces taken as a whole work on the body upon which they act.15 KIRCHHOFF'S THEOREM In this section. It is necessary to observe that it is not possible to prescribe simultaneously both force and displacement for one and the same point. . Consequently. 0 (δ s′ − δ s′′ ). . Fn F1′′. 0. dr δ s′′. We therefore have two distinct systems.. d2. Hence. δ t′′. . . then there should be another equilibrium configuration in which the forces corresponding to the displacements d1. 0 0. d2 is zero. ... F2′ . at those points where displacements are prescribed. Ft. . . . . . First. since all the external forces by themselves are in equilibrium and. For each joint. Thus. For a plane framework.170 Advanced Mechanics of Solids The strain energy corresponding to the third system is U = 0. Consequently the body remains completely undeformed. the framework is termed a plane frame. F2. satisfy the three force equilibrium equations and the three moment equilibrium equations. If the skeleton diagram lies wholly in one plane. Fr are known. the frame is redundant. there is a unique equilibrium configuration. the framework will have become statically determinate and the elastic strain energy of the remaining members can be determined. Hence.2j + 3 Castigliano's second theorem (also known as Menabrea's theorem) can be stated as follows: The forces developed in a redundant framework are such that the total elastic strain energy is a minimum.e. the number of independent equations are 3j . thus giving a total of 3j number of equations. This means that the first and second systems are identical. then ∂ U = 0. . . Then.28) . . F2 and Fr are the forces in the redundant members of a framework and U is the elastic strain energy.. we can write three force equilibrium equations (in a general three-dimensional case). therefore.16 SECOND THEOREM OF CASTIGLIANO OR MENABREA'S THEOREM This theorem is of great importance in the solution of redundant structures or frames.6 and if the number of members exceed 3j . i. Let a framework consist of m number of members and j number of joints. 5.23(b). . Then by Castigliano's first theorem. ∂ U = 0. The number N = m . 5. . the 'increase' in the distance between the joints a and b is given as δ ab ′ = − ∂ Us ∂ Fi (5. all these equation are not independent. the degree of redundancy is given by the number N = m . However. Let Us be the strain energy of these members.6. . if M > 3j . ∂ U = 0 ∂ F2 ∂ Fr ∂ F1 This is also called the principle of least work and can be proven as follows: Let r be the number of redundant members. as shown in Fig. if F1. The reason is as follows.6 the frame is termed a redundant frame. . Assuming that the values of these redundant forces F1.3j + 6 is termed the order of redundancy of the framework. Remove the latter and replace their actions by their respective forces. 5. its length will increase by dab = Fi l i Ai E i (5. Substituting this in Eq. given by Eq. The increase in the distance given by Eq. . .29).29) where li is the length and Ai is the sectional area of the member. . U2 = 2 2 . will vanish when differentiated with respect to Fi. Ur = \ Fl 2 l1 F2 l F2 l + 2 2 + .. . The reactive force on the redundant members ab being Fi. (5. (5. . .28) must be equal to the increase in the length of the member ab. . . Hence − ∂ Us F l = i i Ai Ei ∂ Fi (5. If U is this total energy ∂U = 0 ∂ Fi . + U r ) = 0 ∂ Fi ∂ Fi 1 ∂ U + U + . (5.. Ur = r r 2 A1 E1 2 A2 E2 2 Ar Er Hence.30) The elastic strain energies of the redundant members are U1 = Fl 2 l 1 F2 l F2 l . + U = Fi li ( 2 r) Ai Ei ∂ Fi 1 since all terms.23 (a) Redundant structure (b) Structure with redundant member removed The negative appears because of the direction of Fi. + r r 2 A1 E1 2 A 2 E2 2 Ar Er ∂ U + U + .30) − or ∂ Us = ∂ (U + U 2 + ...Energy Methods Pi Pi b 171 b Fi Fi a a (a) (b) Fig. other than the ith term on the right-hand side. . . the total elastic energy of all redundant members is U1 + U2 + . + U + U = 0 ( 2 r s) ∂ Fi 1 The sum of the terms inside the parentheses is the total energy of the entire framework including the redundant members. BF 2h −2T 3h CF. 5. we get ∂ U = 0. . Solution Let T be the tension in the mem- 2P E ber AB. . ∂ U = 0. Determine the force in the horizontal redundant member. DE The total strain energy is 3−P 3 − 2P 3+0 . All the members are of the same section and material.172 Advanced Mechanics of Solids Similarly. BD h T AF.31) This is the principle of least work.15 h 2 EA U= 2h 2T FE h − 2T ⎡ ⎛ 2 T 2 2 PT 2 ⎢2 3 T + 2 ⎜ P + 3 − 3 ⎝ ⎣ ( + 2 3 T 2 + 3P 2 − 2 PT 3 ⎞ 16T 2 ⎟+ 3 ⎠ ) 2⎤ ⎛ 2 ⎞ + 16 ⎜ T + P 2 − 2 PT ⎟ + 4T ⎥ 3 ⎦ 3 ⎠ ⎝ 3 The condition for minimum strain energy or least work is ∂U = 0 = h 2 EA ∂T ⎡ 4T 4 P + 32T + 4 3 T ⎢ 4 3T + 3 − 3 3 ⎣ ⎤ -12 P + 32T − 32 P + 8T ⎥ 3 3 3 ⎦ \ or ( ) ⎛ ⎞ T 4 3 + 4 + 32 + 4 3 + 32 + 8 = P ⎜ 4 + 12 + 32 ⎟ 3 3 3 3 3⎠ ⎝ 3 T= 9 ( )P 3 +1 6 3 + 19 3+0 −T + P 3 CE. . ∂ U = 0 ∂ F2 ∂ Fr ∂ F1 (5.24 contains a redundant bar. 5. The forces in the members are 60° 60° C D F A P h 60° 60° B P Members Length Force AB 2 3h +T AC. .15 The framework shown in Fig. Example 5.24 Example 5. by considering the redundant members one-by-one. DF Fig. 26(b). 5. 5.16 U1 = 1 Rδ = 1 R R = R 2 2 2 k 2k where d is the deflection of the spring.e. It is represented by U = x ∫ F dx 0 (5.e. Consider the area of OBC which is the strain energy. The strain energy in the spring is Fig. However. whose load-displacement curve is as given in Fig.26(a). Consider the spring showns in Fig.Energy Methods 173 Example 5. though the body may be elastic. i. the body is linearly elastic. situations exist where the deformation is not proportional to load. we have explicitly assumed that the elastic body satisfies Hooke's law. A Solution Let R be the unknown reaction at A.25).32) . The strain energy in the beam is U2 = L M 2 dx ∫ 2EI 0 L =∫ ( Rx − wx 2) 2 0 ( 2 dx 2 EI = 1 1 R 2 L3 + 1 w2 L5 − 1 RwL4 40 8 EI 6 ) Hence. The spring is a non-linear spring. 5. the total strain energy for the system is ( 2 U = U1 + U 2 = R + 1 1 R 2 L3 + 1 w2 L5 − 1 RwL4 2k EI 6 40 8 ) From Castigliano's second theorem ( ) ∂ U = R + 1 1 RL3 − 1 wL4 = 0 k EI 3 8 ∂R R= \ 5.16 A cantilever is supported at the free end by an elastic spring of spring constant k. i. 5. The cantilever beam is uniformly loaded.17 3kwL4 8 3EI + kL3 ( ) GENERALISATION OF CASTIGLIANO'S THEOREM OR ENGESSER'S THEOREM It is necessary to observe that in developing the first and second theorems of Castigliano. The intensity of W loading is W. Determine the reaction at A (Fig. R is the force on the spring.25 Example 5. expressing U in terms of F. It is represented by F B A F DF O X C Dx (b) (a) Fig. Differentiating the complementary energy with respect to F yields dU * = x (5. we must take the derivative of the complementary energy and not that of the strain energy.26 Hence F U* = ∫ x dF (5. we notice that to obtain the corresponding deflection.16). The expression given by Eq. the curve OB is a straight line and consequently. we get From these U= 1 ⋅ a ⎡ 1 ⋅ F 1/ n ⎤ ⎥⎦ n + 1 ⎢⎣ a1/ n n +1 1 ax n +1 n +1 . discussed in Sec.14. Consider as an example an elastic spring the force deflection characteristic of which is represented by F = axn where a and n are constants. and is applicable whether the elastic member is linear or non-linear. The strain energy is x x 0 0 U = ∫ F dx = ∫ a ( x ′ ) dx ′ = The complimentary strain energy is U* = F F n () F ∫ x dF = ∫ a 0 0 1/ n dF = 1/1n ⋅ n F(1 + 1/ n) n +1 a dU = axn = F dx dU * = 1 ⋅ F1/ n = x dF a1/ n Further. Now consider the area OAB.34) represents Engesser's theorem. 5. (5.174 Advanced Mechanics of Solids dU = F dx This is the principle of virtual work. the strain energy and the complementary strain energy are equal and it becomes immaterial which one we use in Castigliano's first theorem. When a material obeys Hooke's law. 5.34) dF This gives the deflection in the direction of F. (b) Nonlinear load-displacement curve This is termed as a complementary energy. 5. If we compare with Castigliano's first theorem (Eq.33) 0 (a) Non-linear spring. \ Example 5. the deflection is not linearly related to the load.27. The strain energy is δ lW 4/3 U = ∫ Wd δ = 0 ( aE )1/3 . Hence the principle of virtual work is valid both for linear and non-linear elastic material. then F= W ≈ Wl 2 sin α 2δ l2 + δ 2 − l 1 δ 2 ≈ l 2 l2 and e= Also e = F = Wl aE 2δ aE Equating the two strains 2 Wl = δ 2δ aE 2l 2 or ( ) d =l W Ea 1/3 i. Check Castigliano's first theorem. 5. l l C A B d a C1 W Fig. whereas to obtain deflection using Castigliano's first theorem.17 Solution When C has displacement CC1 = d.e. it is immaterial wheather we use U or U*.Energy Methods 175 () 1/ n 1 dU = 1 F = x n a n dF and this does not agree with the correct result. we have to use the complementary energy U* if the material is non-linear. a the cross-sectional area and e the strain.17 Consider Fig. 5. which shows two identical bars hinged together.27 Example 5. If it is linearly elastic. since both are equal. tan α ≈ sin α ≈ δ /l If F is the force in each member. using the elastic and complementary strain energy. carrying a load W. we have from the figure for small a. and in the final result. we apply a fictitious force Q in the corresponding direction at A.11.28.18 MAXWELL–MOHR INTEGRALS Castigliano's first theorem gives the displacement of points in the directions of the external forces where they are acting. we shall deA velop certain integrals. At any section. Castigliano's first theorem applied to the strain energy. For example. 5. a general section M X = MxP + MxQ . a fictitious force in the direction of the required displacement is assumed at the point. 5. the moments and forces of reaction are caused by the actual external forces plus the fictitious Fx Mz load Q. does not yield the deflection d. which are based on the fictitious load techniques. 5.29 Moments and forces across F X = FxP + FxQ. In this section. 5. the value of the fictitious load is considered equal to zero.176 Advanced Mechanics of Solids 1/3 \ ∂ U = 4lW 1/3 ∂W 3 ( aE ) Hence. Engesser's theorem gives the correct result. about the x axis we Fz Mx have Fig. When a displacement is required at a point where no external force is acting.29. This is so because the load defection equation is not linearly related. w U* = ∫ δ dW = 0 = w l 1/3 ∫ W dW ( Ea)1/3 0 3lW 4/3 1/3 4 ( Ea) ( ) W ∂ U* = l Ea ∂W 1/3 =δ Hence. 5. If we consider the complementary energy. This technique was discussed in Sec. Since no external force load P is acting at A in the corresponding direction. 5. we need to determine the moments and forces across a general Fy section. In order to calculate the strain energy in the elastic My member. Consider the determination of the verQ tical displacement of point A of a strucP ture which is loaded by a force P. This is shown in Fig.28 A general structure under shown in Fig. as Fig. These force factors vary from section to section.35) Note that in Fig. The above method is sometimes known as the unit load method.Energy Methods 177 where FxP is caused by the actual external forces. where Fx1. It is essential to observe that the additional force factors. the force factors due to the actual loads and fictitious force are Fx = FxP + Fx1Q. are directly proportional to Q. Fy = FyP + Fy1Q. These sets of integrals are known as Maxwell-Mohr integrals. 5. etc. Mz = MzP + Mz1Q (5. are the force factors caused by a unit fictitious generalised force. etc. Consequently. such as P. Mx1.37) . M y and M z act as bending moments. Mx = MxP + Mx1Q My = MyP + My1Q Fz = FzP + Fz1Q. Hence. If Q is doubled. Mx1Q. and FxQ is due to the fictitious load Q.36) If the fictitious force Q is replaced by a fictitious moment or torque. These integrals can be used to solve not only problems of finding displacements but also to solve problems connected with plane thin-walled rings. The above set of equations is generally written as dA = M M M My y x x ∫ GI ds + ∫ EI x y l l ds + ∫ l Mz Mz ds EIz k F F + ∫ Fx Fx ds + ∫ y y y ds + ∫ k z Fz Fz ds EA GA GA l l l (5.29 while M x acts as a torque. such as FxQ. etc. The total elastic energy is U= ∫ ( ( MxP 2 MyP + M y1 Q + Mx1 Q ) ds +∫ 2GIx 2 EI y l l +∫ ( MzP + Mz1 Q )2 ds 2 Elz l +∫ l ( k y FyP + Fy1 Q ) 2 ds 2GA l +∫ ( FxP ) 2 ds + Fx1 Q ) ds 2 EA 2 k ( F + Fz1 Q ) ds + ∫ z zP 2GA l 2 Differentiating the above expression with respect to Q and putting Q = 0 δA = ∂U = ∂ Q Q = 0 ∫l M yP M y1 ds M xP M x1 ds +∫ GIx EI y l +∫ MzP M z1 ds F F ds + ∫ xP x1 EIz EA l + k y FyP Fy1 ds k F F ds + ∫ z zP z1 GA GA l l ∫ l (5. MxQ. one can write these as Fx1Q. these factors also get doubled. we get the corresponding deflection qA. 178 Advanced Mechanics of Solids where Mx. To determine the rotation. M z are the force factors caused by a generalised unit fictitious force applied where the appropriate displacement is needed. Example 5.74 ) (1 − cos φ ) dφ EI ⎡ aq ( a + x ) − 1 qx 2 − 0.M1 M at any section in quadrant = -1 M at any section in the top horizontal member = aq (a + x) .cos f) . My . The unknown moment M 1 is the redundant unknown generalised force. M at any section in quadrant = aq . we assume a unit moment in the same direction as M1. To determine the vertical displacements of the midpoints of the horizontal members.qx 2/2 . . . 5. the vertically upward displacement of point A is π/2 dA = ∫ − 0 a 4 q (1 − cos φ − 0.M1 M at any section in the top horizontal member = -1 a a q ( a + x ) − qx 2 /2 − M − a 2 q (1 − cos φ ) + M 1 1 ad φ − ∫ dx EI EI 0 ( ) ( ) EIqA = − a 3 q π + 1 + M 1a π + 1 = 0 2 3 2 M1 = a 2 q 3π + 2 ≈ 0. 5..(a + x) Hence. The moment due to this fictitious unit moment at any section is M . y a a q x A a B x q A M1 aq f Fig.18 π/ 2 \ qA = ∫ 0 or \ Solution Consider one quarter of the ring.74 a 2 q 3 (π + 2 ) This is the value of the redundant unknown moment.18 Determine by what amount the straight portions of the ring are bought closer together when it is loaded.30 consider only the bending energy. a (1 .74a 2 q ⎤ ( a + x ) ⎢ ⎥⎦ 2 + ∫− ⎣ dx EI 0 a . . Owing to symmetry.30 Example 5. we apply a fictitious force Pf = 1 in an upward direction at point A of the quarter ring. as shown in Fig.cos f) M at any section in top horizontal part = . Because of this M at any section in quadrant = -a (1 . there is no rotation of the section at point A. A y r B (a) A q qr dq F1 M1 q f q B (b) (c) Fig. The reactive forces at section A are F1 and M1.19 A thin walled circular ring is loaded as shown in Fig. we may consider one half of the ring.1) + M1 + F1r (1 .86 a 4 q EI Hence.31 Example 5. we assume a fictitious moment and a fictitious horizontal force.cosf) .72 EI EI Example 5. Determine the vertical displacement of point A. section A does not rotate and also does not have a horizontal displacement. Hence in addition to M1 and F1. Take only the bending energy.31. The moment at any section f due to the distributed loading q is φ M q = ∫ qr dθ r (sin φ − sin θ ) = qr 2 (φ sin φ + cos φ − 1) 0 M at any section f with distributed loading F1 and M1 is M = qr2 (f sinf + cosf .Energy Methods 179 0. 5.19 Solution Because of symmetry. 5. Because of symmetry. each of unit magnitude at section A. the two horizontal members approach each other by a distance equal to = 2 (0.86) a 4 q a4q = 1. we have 1 2 M1 + 3 F1r = qr 4 2 M at any section f due to unit fictitious moment is M =1 \ qA = (5.32 Example 5.180 Advanced Mechanics of Solids M at any section f due to the unit fictitious horizontal force is M = r (1 . M at any section f due to Pf = 1 is r sinf M1 = − π \ dv = 2 2 ∫ r ⎡⎣ qr (φ sin φ + cos φ − 1) + M1 + F1r (1 − cos φ )⎤⎦ sin φ dφ 0 ⎛ 2 ⎞ qr 4 qr 4 = ⎜ π − 2⎟ ≈ 0.39) qr 2 qr and F1 = 2 2 To determine the vertical displacement of A we apply a fictitious unit force Pf = 1 at A in the downward direction. The radius of the member is R and the member subtends an angle a at the centre. .38) and (5.20 Figure 5.467 EI ⎝ 4 ⎠ EI a O D q A Vertical Load W C x y Fig. Taking only bendng and torsional energies into account. determine the vertical deflection of the loaded end A.20 Example 5. 5.32 shows a circular member in its plan view. we have M1 + F 1 r = 0 (5.38) I π r ⎡qr 2 φ sin φ + cos φ − 1 + M + F r 1 − cos φ ⎤ dφ ( ) 1 1( )⎦ EI 0∫ ⎣ = r ( π M 1 + F1rπ ) EI Since this is also equal to zero. It carries a vertical load W at A perpendicular to the plane of the paper.cos f) π dA = I EI \ 2 2 ∫ r ⎡⎣ qr (φ sin φ + cos φ − 1) 0 + M1 + F1r (1 − cos φ ) ⎤⎦ (1 − cos φ ) d φ = ( r2 π 3π − qr 2 + π M 1 + F1r EI 4 2 ) Since this is equal to zero.39) Solving Eqs (5. and C of the beam as dA = 3 cm dB = 8 cm dC = 5 cm Find the deflection of point R when the beam is loaded at points.000 N at A.34 Problem 5. 5.2 Chapter_05.34. [Ans. A.2 For the horizontal beam shown in Fig. 12. PB = 3500 N and PC = 5000 N.6 cm of support B causes a reaction Ra = 10. 5:51 AM . M = W ¥ AD = WR sinq T = W ¥ DC = WR (1 . 5.1 5.)] P A R B C Fig. or π. Thus.33 produces vertical deflections at three points A. B and C by PA = 7500 N. Rb = 13.33 Problem 5.Energy Methods 181 Solution At section C. [Ans. 2 ⎡ ⎛ 3π − 8 ⎞ ⎤ δV = WR3 ⎢ π − 1 ⎜ 4 EI GJ 4 ⎟⎥ ⎣ ⎝ ( EI δV = WR 3 π 1 − 3 GJ ⎠⎦ ) 5. Hence.1 A load P = 6000 N acting at point R of a beam shown in Fig. if a= if a = π. Determine the reaction Rb at B due to a vertical displacement of 0. 5. 2 α 1 WR sin θ 2 R dθ + α 1 ⎡WR 1 − cos θ ⎤ 2 R dθ 1 Wd = U = ( ) )⎦ ∫ ∫ 2GJ ⎣ ( V 2 EI 2 0 0 ( ) ( ) 1 α − 1 sin 2α + 1 3 α + 1 sin 2α − 2 sin α ⎤ d V = WR3 ⎡⎢ ⎥⎦ 2 GJ 2 4 ⎣ 2 EI This is the same as ∂U/∂W.8 cm at support A. B. the work done by W during its vertical deflection is 1 W dV and this is stored as the elastic energy U.333 N] B A 30 cm D C 30 cm 60 cm Fig. a vertical displacement of 0. 5. the moment of the force about x axis acts as bendng moment M and the moment about y axis acts as torque T.pmd 181 7/3/2008.cosq) α \ U= α 2 1 1 ∫ 2 EI (WR sin θ ) R dθ + ∫ 2GJ ⎡⎣WR (1 − cos θ ) ⎤⎦ R dθ 0 0 2 When the load W is gradually applied.6 cm (approx. 5.36.3 A closed circular ring made of inextensible material is subjected to an arbitrary system of forces in its plane.5 Determine the rotation of point C of the beam under the action of a couple M applied at its centre (Fig.3 5. 5. no deformation occurs. Since⎡ ⎢ the material is inextensible.5 B .4 5.35 Problem 5. ⎢⎣ ⎣ F2 F1 F4 F3 Fig. ⎡ Hint: Subject the ring to uniform internal pressure.37 Problem 5.35).36 Problem 5. Show that the area enclosed by the frame does not change under this loading. Ml ⎤ ⎡ ⎢⎣ Ans. Assume small displacements (Fig. ⎡ Ans. All members have the same cross-section and the same rigidity EA. δ = Wl 7 + 4 2 ⎤ A ⎢⎣ ⎥⎦ EA ( l l ) A W l Fig.182 Advanced Mechanics of Solids 5.37). 5. 5. 5. θ = 12 EI ⎥⎦ M C A l /2 Fig.4 Determine the vertical displacement of point A for the structure shown in Fig. 5.⎢ ⎢ Now apply the reciprocal theorem. ⎡ PR 3 ⎤ ⎢⎣ Ans. 5. The member is of uniform circular crosssection (Fig.7 What is the relative displacement of points A and B when subjected to forces P. Consider only bending energy (Fig. Take all energies into account. 5.7 5.Energy Methods 183 5.8 Determine the vertical displacement of the point of application of force P.39).8 a .40).38).39 Problem 5.38 Problem 5. 5. δAB = 3π EI ⎥⎦ r P A B P Fig.6 5. 5. 5. 5. δ A = 2 P ⎜⎝ 3EI + 2 EI + 2GI + AG + 2 AE ⎟⎠ ⎥ P ⎣ ⎦ a P b Fig. P ⎡ Pa 3 Pa 2b ⎤ ⎢⎣ Ans.6 What is the relative displacement of points A and B in the framework shown? Consider only bending energy (Fig.40 Problem 5. δAB = 6 EI + 2 EI ⎥⎦ a/2 A B P b a Fig. ⎡ ⎛ a3 a 2b a3 ka b ⎞⎤ ⎢ Ans. 5. 5.11 Two conditions must be satisfied by an ideal piston ring. and (b) it should exert a uniform pressure all around.43 determine the stiffness of the system.9 What are the horizontal and vertical displacements of point A in Fig. δ = 17 Ph . Each of the springs has a length l and moment of inertia I for bending in the plane of the moment.e. if the ring is closed inside the cylinder. End B is free to rotate but can move only in a vertical direction (Fig. M ≈ 8EI ⎤ θ l ⎦⎥ ⎣⎢ . show that the initial gap width must be 3p pr4/EI.42 Problem 5. the torque per unit angle of twist of the shaft. ⎡ Ans.184 Advanced Mechanics of Solids 5.12 For the torque measuring device shown in Fig.42).9 5. 5. i. δ = 1. 5. p is the uniform pressure per centimetre of circumference.10 Determine the vertical displacement of point B under the action of W. Assume AB to be rigid.10 5.41. 5. ⎡ Wa 3 = Ans .73Ph ⎤ V H EA EA ⎦⎥ ⎣⎢ 30∞ 30∞ h A B a a P a h Fig.41 Problem 5. (a) It should be truly circular when in the cylinder. Assuming that these conditions are satisfied by specifying the initial shape and the cross-section. EI is kept constant. δ B ⎢ EI ⎣ ⎛ 3π 1 ⎞⎤ ⎜ 4 − 9π + 8 ⎟ ⎥ ⎝ ⎠⎦ D a C A B W Fig. ⎡ Ans. 5. The included angle between any two forces is 2a and A is the crosssectional area of the member. 5. 5.45. Show that the vertical deflection caused by angular velocity w is given by 2ρ ω 2 r 5 E t2 where r is the hoop radius. 5.44). Consider both bending and axial energies.12 5. 5. Determine the changes in the radius of the ring along the line of action of the forces. t the thickness of the section and r the weight density of the material.13 A circular steel hoop of square cross-section is used as the controlling element of a high speed governor (Fig. d= w t d Free to Slide Fig. ⎡ ⎞ ⎛ ⎞⎤ PR 3 ⎛ cot α + α − 1 ⎟ + PR ⎜ cot α + α2 ⎟ ⎥ ⎢ Ans.45 Problem 5.13 5.44 Problem 5.Energy Methods 185 M l Fig. ⎜ 2 2 EI ⎝ 2 2 sin α α ⎠ 4 EA ⎝ sin α ⎠ ⎥⎦ ⎣⎢ P R 2a P Fig.14 A thin circular ring is loaded by three forces P as shown in Fig.43 Problem 5.14 P . 5. 16 In the previous problem determine the force in the member OC by Castigliano's second theorem.186 Advanced Mechanics of Solids 5. Beam ACB has Young's modulus E and member CD has a value E '.47 Problem 5. ⎡ 5 wl 4 aE ′ ⎢ Ans.15 5.17 5. 5.15 For the system shown (Fig. 2W/3] 5. W = 3aE δ ⎤ 2l ⎥⎦ ⎣⎢ B O C Fig.47). 4 6 EIh + qE ′ l 3 ⎢⎣ ( ) ⎤ ⎥ ⎥⎦ l w/unit length l A C B h D Fig. 5. 5. ⎥ A1 a 3 + h 3 + A2 l 3 ⎥ ⎢⎣ ⎦ ( ) . ⎡ ⎤ WA1 lh 2 ⎢ Ans.46) determine the load W necessary to cause a displacement d in the vertical direction of point O. [Ans.18 A pin jointed framework is supported at A and D and it carries equal loads W at E and F.46 Problem 5. and of all the other members are A2 each.17 Using Castigliano's second theorem. determine the reaction of the vertical support C of the structure shown (Fig. a is the cross-sectional area of each member and l is the length of each member. The cross-sectional area of CD is a. 5. A W ⎡ Ans. Determine the tensions in BF and CE. Use the principle of virtual work. The lengths of the members are as follows: AE = EF = FD = BC = a BE = CF = h BF = CE = AB = CD = l = (a2 + h2)1/2 The cross-sectional areas of BF and CE are A1 each. 49.20 5.19 A ring is made up of two semi-circles of radius a and of two straight lines of length 2a.19 Fig. M = Pa .18 5.711 Pa ⎥ ⎢ ⎣ EI ⎦ a a a q A a a B a a p Fig.48 Problem 5.49 Problem 5.50 Problem 5.51 is subjected to a torque T at A. EI ⎥⎦ 12 ( 2 + π ) ⎣ 5.20 Determine reaction forces and moments at the fixed ends and also the vertical deflection of the point of loading. determine the change in distance between A and B. T = 0. 5.21 A semi-circular member shown in Fig. 5.50). 5. 5.Energy Methods B A C D F E 187 W W Fig 5.387 Pa⎤ ⎢ ⎥ 2 3⎥ ⎢ δ = 0. Also determine the vertical deflection of A. ⎡ 6 − 17π − 6π 2 qa 4 ⎤ ⋅ ⎢ Ans. Assume G = 0. as shown in Fig. 5. M = T . ⎡ Ans. When loaded as shown. Torque = − T ⎤ ⎢ 2 9π ⎥ 2 ⎢ R T 9π + 1 − 5 ⎥ ⎢ δV = ⎥ 8EI 4 π ⎣ ⎦ ( ) . Determine the reactive moments at the built-in ends B and C. Consider only bending energy. ⎡ Ans.4E (Fig. 188 Advanced Mechanics of Solids C R T A B Fig.51 Problem 5. 5.12 determine the change in the horizontal diameter ( ) ⎡ Pr 3 2 1 ⎤ ⎢⎣ Ans. δ h = − EI π − 2 ⎥⎦ .21 5.22 In Example 5. The effects of transverse shear stresses will be discussed in Sec.CHAPTER 6 Bending of Beams 6.1 INTRODUCTION In this chapter we shall consider the stresses in and deflections of beams having a general cross-section subjected to bending.1 shows an initially straight beam having a vertical section of symmetry and subjected to a bending moment acting in this plane of symmetry.4-6. These lateral forces.1 Beam with a vertical section of symmetry subjected to bending The plane of symmetry is the xy plane and the bending moment Mz acts in this plane. Assuming that the sections that are plane before bending remain so after bending. also induce shear stresses. Because of pure bending moments. In general. the flexural stress sx is obtained in elementary strength of materials as sx = - Mz y Iz (6.1) The origin of the co-ordinates coincides with the centroid of the cross-section and the z axis coincides with the neutral axis. The minus sign is to take care of the .6. y Mz x Mz z Fig. Figure 6. 6. in addition to causing bending or flexural stresses in transverse sections of the beams. In elementary strength of materials only beams having an axis of symmetry are usually considered. only normal stresses are induced. Flexural stresses are normal to the section. Owing to symmetry the beam bends in the xy plane. the moments causing bending are due to lateral forces acting on the beams. 6. For this. The moment is shown vectorially. σ Mz =− x = E Iz y R (6. The origin O is taken at the centroid of the cross-section. We shall consider each one separately. It is once again assumed that sections that are plane before bending remain plane after bending. 6. The x axis is along the axis of the beam and the z axis is chosen to coincide with the moment vector. This is usually known as the Euler-Bernoulli hypothesis. There are three general methods of solving this problem. the stretch or contraction of any fibre will be proportional to the perpendicular distance from BB. Since the cross-section rotates about BB during bending. 6. the strain in any fibre is ex = k¢y¢ y y DA x O Mz (a) B y¢ b z Mz B (b) z Fig. consider a small area DA lying at a distance y¢ from BB. the beam is said to be under symmetrical bending.2 STRAIGHT BEAMS AND ASYMMETRICAL BENDING Now we shall consider the bending of initially straight beams having a uniform cross-section.2 shows a beam subjected to a pure bending moment Mz lying in the xy plane. Otherwise it is said to be under asymmetrical bending.2 Beam with a general section subjected to bending .190 Advanced Mechanics of Solids sign of the stress. the equations from elementary strength of materials give. 6. Method 1 Figure 6.2) The above set of equation is usually called Euler-Bernoulli equations or NavierBernoulli equations. if E is the Young's modulus of the beam material and R the radius of curvature of the bent beam.2(b) this is represented by BB and it can be shown that it passes through the centroid O. as shown. produces a compressive stress at a point with the positive y co-ordinate. When the bending moment acts in the plane of symmetry. In Fig. Further. Hence. Points lying on this axis will not experience any stress and consequently this axis is the neutral axis. Iz is the area moment of inertia about the neutral axis which passes through the centroid. A positive bending moment Mz. This means that the cross-section will rotate about an axis such that one part of the section will be subjected to tensile stresses and the other part above this axis will be subjected to compression. 3) where k is an appropriate constant. integrating the above equation over the area of the section.6a) and k (Iyz cos b . which means that BB is a centroidal axis. we take moments of the normal stress distribution about the y and z axes. 6.3 Location of neutral axis and distance y ¢ of point C from it k ∫∫ ( yz sin b . Iyz sin b .Iz sin b ) = Mz (6. (6.4) The above equation shows that the first moment of the area about BB is zero.Bending of Beams 191 where k¢ is some constant.DF = y sin b .z cos b Substituting this in Eqs (6.e.6b) From the first equation tanb = Iy I yz This gives the location of the neutral axis BB. DFx = k y¢ D A For equilibrium. It is important to observe that the beam in general will not bend in the plane of the bending moment and the neutral axis BB will not be along the applied moment vector Mz. sx = k¢Ey¢ = k y¢ (6. The minus sign is because a positive stress at a positive (y. from Hooke's law. (6. The neutral axis BB in general will be inclined at an angle b to the y axis. The moment about the y axis must vanish and the moment about the z axis should be equal to -Mz. Hence y C z y¢ B D (6. k ∫∫ y ′ dA = 0 (6.3) as y¢= CF .Iy cos b = 0 (6. Assuming only sx to be acting and sy = sz = 0.5b) y¢ can now be expressed in terms of y and z coordinates (Fig. The force acting on DA is therefore.5) y b F z ∫∫ σx z dA = ∫∫ ky′ z dA = 0 z cos b B Fig.yz cos b ) dA = -Mz i. z) point produces a moment vector in the negative z direction.6b) in Eq. 6. the resultant normal force acting over the cross-section must be equal to zero. Hence. Next. Substituting for k from Eq.7) .5a) ∫∫ sx y dA = ∫∫ ky¢y dA = -Mz (6.3) sx = = M z ( y sin β − z cos β ) I yz cos β − I z sin β y tan β − z M I yz − I z tan β z (6.z2 cos b ) dA = 0 and k ∫∫ ( y2 sin b . next. (6. The stress at any point (y. The principal axes Oy¢ and Oz¢ are inclined such that tan 2q = 2 I yz Iz − I y . Method 2 we observe from Eq. an alternative method of solving the problem is to determine the principal axes of the section.192 Advanced Mechanics of Solids Substituting for tan b from Eq.4. The bending moment M acts in the xy plane with the moment vector along the z axis. and then to apply the elementary flexure formula.7) that b = 90° when Iyz = 0. the beam bends in a plane which in general does not coincide with the plane of the moment. Equation (6. (6. we conclude that when a beam with a general cross-section is subjected to a pure bending moment Mz. to resolve the bending moment into components along these axes. 6. sx = yI y − zI yz 2 I yz − IyIz (6.7). y y¢ Mz q y q y¢ b¢ q z sin O Mz z Mz z¢ Mz cos q z¢ Fig.8). This procedure is shown in Fig. if the y and z axes happen to be the principal axes of the cross-section.e. z) is given by Eq. In summary. This is so because for a symmetrical section.8) Mz The above equation helps us to calculate the normal stress due to bending. This means that if the y and z axes are the principal axes and the bending moment acts in the xy plane (i. the moment vector Mz is along one of the principal axes).4 Resolution of bending moment vector along principal axes y and z axes are a set of arbitrary centroidal axes in the section. The neutral axis is inclined at an angle b to the y axis such that tan b = Iy /Iyz.8) then reduces to sx = - Mz y Iz This is similar to the elementary flexure formula which is valid for symmetrical bending. 6. (6. the principal axes coincide with the axes of symmetry.e. So. the beam bends in the plane of the moment with the neutral axis coinciding with the z axis. i. 5 Resolution of bendthe strain at any point in the cross-section will ing moment vector be proportional to the distance from the neualong two arbitrary tral axis BB.e. The neutral axis is determined by equating sx to zero. Choose a convenient set of centroidal axes Oyz about which the moments and product of inertia can be calculated easily. this means that the neutral axis passes through the centroid O. such as BB. the elementary flexure formula can be used.10) The angle b ¢ is with respect to the y¢ axis.9) sx = − I y′ I z′ It is important to observe that with the positive axes chosen as in Fig. according to which the sections z Mz that were plane before bending remain plane B after bending.Bending of Beams 193 The moment resolved along the principal axes Oy¢ and Oz¢ are M y¢ = Mz sin q and Mz¢ = Mz cos q. a minus sign is used in the equation. For equilibirum.4. ∫∫ sx dA = k ∫∫ y¢ dA = 0 As before. Let M be the My applied moment vector (Fig. y¢ = y sin b . sx = Ek¢y¢ = ky¢ (a) where k is some constant. the cross-section will rotate about an axis. a point with a positive y coordinate will be under compressive stress for positive Mz¢ = Mz cosq. M Resolve the moment vector M into two comB y¢ b ponents My and Mz along the y and z axes respectively. orthogonal axes ex = k¢y¢ Assuming that only sx is non-zero. Fig. Hence. i. M y′ z ′ M z ′ y′ =0 − I y′ I z′ or z′ = tan b¢ = M z ′ I y ′ M y′ I z′ y′ (6. Hence. since only a moment is acting. For each of these moments.5).3). 6.z cos b (b) . 6. With the principle of superposition. the total force over the cross-section should be equal to zero. 6. 6. M y′ z ′ M z′ y ′ (6. From geometry (Fig. y Method 3 This is the most general method. Let b be the angle between the neutral axis and the y axis. We assume the Euler-Bernoulli hypothesis. Consequently. 13) sx = ( ) ( M z yI y − zI yz + M y yI yz − zI z 2 I yz − I y Iz ) (6.12) The above two equations can be solved for k and b.7). Choose a centroidal set of axes Oyz such that the z axis is along the M vector.e.13) This gives the location of the neutral axis BB. substituting for k from Eq. Next. In recapitulation we have the following three methods to solve unsymmetrical bending.11) (6. The neutral axis is given by Eq.14) When My = 0 the above equation for sx becomes equivalent to Eq.8). (6. the moments of the forces about the axes should yield ∫∫ sx z dA = ∫∫ ky¢ z dA = My ∫∫ sx y dA = ∫∫ ky¢y dA = -Mz Substituting for y' ( k ∫∫ ( y ) sin β − yz cos β ) dA = -M k ∫∫ yz sin β − z 2 cos β dA = My i. (6.11) into equations (a) and (b) sx = = M y ( y sin β − z cos β ) I yz sin β − I y cos β M y ( y tan β − z ) I yz tan β − I y Substituting for tan b from Eq. Dividing one by the other My I yz sin β − I y cos β =Mz I z sin β − I yz cos β My I yz tan β − I y = Mz I z tan β − I yz or i.e. z) is then given by Eq. (6.Mz k I yz sin β − I y cos β = My sin β − I yz (6. tan b = I y M z + I yz M y I yz M z + I z M y (6. . and 2 ( k ( Iz z ) cos β ) = . The stress sx at any point (y.8).194 Advanced Mechanics of Solids For equilibrium. (6. (6. Method 1 Let M be the applied moment vector. 6) near the built-in-end? y 30° 1000 N y 0 1000 N x 6 cm z 400 cm (a) 4 cm (b) D Fig. such that the y¢ and z¢ axes are the principal axes.000 ⎜ 2 sin 30° − 3 cos 30° ⎟ ⎝ Iy Iz ⎠ . such that My = . (6.400.Bending of Beams 195 Method 2 Let M be the applied moment vector. Choose a centroidal set of axes Oy¢z¢. (6. z¢) is given by Eq.(1000 cos 30°) 400 = .9) is sx = M y z Mz y − Iy Iz ( = − 400. M z = .000 sin 30° N cm The coordinates of point D are (y. Example 6.000 cos 30 ) ( −I 3) ° y z ⎛ ⎞ = 400. Resolve the moment into components My¢ and Mz¢ along the principal axes. The normal stress sx and the orientation of the neutral axis are given by Eqs (6. Resolve the applied moment M into components My and Mz.14) and (6.000 cos 30° N cm The horizontal component also produces a negative moment about the y axis. What is the stress due to bending at point D (Fig.400.6 Example 6.10).1 A cantilever beam of rectangular section is subjected to a load of 1000 N (102 kgf ) which is inclined at an angle of 30° to the vertical. Then the normal stress s x at any point (y¢.9) and the orientation of the neutral axis is given by Eq. y and z axes are symmetrical axes and hence these are also the principal axes. 6. The force along the vertical axis produces a negative moment Mz (moment vector in negative z direction). The force can be resolved into two components 1000 cos 30° along the vertical axis and 1000 sin 30° along the z axis. 6. -2). z) = (-3.000 sin 30° ) ( −I 2) − (−400.13) respectively.(1000 sin 30°) 400 = . the normal stress at D from Eq.1 Solution For the section. Method 3 Choose a convenient set of centroidal axes Oyz about which the product and moments of inertia can easily be calculated. Hence. (6. 48 N/cm (= 0.(28. Since the member has equal legs. 6.1 cm4. My¢ = Mz cos 45° = .wL 2 1.000 N cm 2 The centroid of the section is located at = (100 × 10 × 50) + (90 × 10 × 5) (100 × 10) + (90 × 10) = 28. 000 = .48 × 90.13 cm z = .87 cm . shown in Fig.66.7 mm from the outer side of the vertical leg.7) = -71.47. is subjected to its own weight. Determine the stress at point A near the built-in section. y y¢ 3m 28.151 kgf/cm). 000 ⎜ 2 − 3 3 ⎟ 2 × 32 2 × 72 ⎝ ⎠ = -1934 N/cm2 = -19340 kPa (= -197 kgf/cm2) Example 6.196 Advanced Mechanics of Solids Iy = \ 6 × 43 = 32 cm 4 . It is given that the beam weighs 1.2 Solution The bending moment at the built-in end is 2 M z = .7 . The principal axes are the y¢ and z¢ axes.10) = -18. 12 Iz = 4 × 63 = 72 cm 4 12 ⎛ ⎞ sx = 400.3 mm = -7.7 mm = -1. The components of Mz along y¢ and z¢ axes are.7 mm z 100 mm A z¢ 10 mm Fig.100 N cm Mz¢ = Mz cos 45° = . the z¢ axis is at 45° to the z axis.28. 6.2 A beam of equal-leg angle section.(100 .100 N cm \ sx = M y′ z ′ M z′ y ′ − I y′ I z′ For point A and y = .47. therefore.7 Example 6.7. The principal moments of inertia are 284 cm4 and 74. 3 S= 17.y sin 45° = -13.1055 = -3419 N/cm2 = -341.22 = . Member Area y¢ z¢ Ay¢ Az¢ y z A B C D 6.3 cm ΣA 17.7 -26.42 = +37.100 × 3.5 z* = ΣAz′ = 460 = 26.6.8 Example 6.6 mm = . 6.22 + 50.8 shows a unsymmetrical one cell box beam with fourcorner flange members A.63.5 265 460 Therefore.1 -15. 2500 kgf = Py y 125 cm 640 kgf y¢ A B 30 cm 20 cm z C z¢ D 40 cm Fig. C and D.1 cm ΣA 17.1 13.Bending of Beams 197 Hence. Assume that the sheet-metal connecting the flange members does not carry any flexual loads.2 mm = 3.5 .100 × 6. the coordinates of the centroid from D are y* = ΣAy′ 265 = = 15.72 cm 47.1 284 = -2364 . y¢ = y cos 45° + z sin 45° = -50.3 Solution The front face ABCD is assumed built-in.5 3.72 47.3 Figure 6. Determine the stresses in the flange members A and D.13.3 13.9 4.36 − 74.900 kPa \ sx = - Example 6.36 cm and z¢ = z cos 45° .42 . B.5 30 20 0 0 40 0 40 0 195 70 0 0 260 0 200 0 14.0 2.9 -15.7 -26. Loads Px and Py are acting at a distance of 125 cm from the section ABCD.5 5. 09z (sx)A = .1 1326.3 13.7 -26.57y . we shall discuss here the condition necessary for a plane section to remain plane.14) sx = \ 6.2 cm4 \ One should be careful to observe that the loads P y and P z are acting at x = -125 cm \ Moment about z axis = Mz = -312500 kgf cm = -30646 Nm Moment about y axis = My = +80000 kgf cm = +7845.7) = -1440 kgf.1) .3) = +1460 kgf.2 y − 3237 z ) (834.3 −312500 ( 6308.9 4.4 1729.0 2.(.09 ¥ 26. We have from Hooke's law ( ) ex = 1 ⎡σ x − ν σ y + σ z ⎤ ⎦ E⎣ ey = 1 ⎡⎣σ y − ν (σ z + σ x ) ⎤⎦ E ( ) e z = 1 ⎡σ z − ν σ x + σ y ⎤ ⎦ E⎣ Solving the above equations for the stress sx we get sx = νE (1 + ν ) (1 − 2ν ) (ε x + ε y + ε z ) + E ε x 1 +ν (c) .9 y − 834.2 z ) + 80000 (834.(-96.(96.3 y2 z2 Ay2 222 24 228 228 187.5 5.9 -15. In order to determine the actual deformation of an intially plane section of a beam subjected to a general loading.5 3.2)2 − (3237 × 6308.3 992.7 1443 84 1140 570 Az2 Ayz 1220.1 13.5 y z 14.(0. It is natural to question how far this assumption is valid.7 691.57 ¥ 14.7 -26. (6. we will have to use the methods of the theory of elasticity.9 cm4 Iyz = SAyz = +834.cm2 = -141227 kPa (sx)D = .198 Advanced Mechanics of Solids Member Area A B C D 6.09 ¥ 13.7 187.1 -15.8 2421 -451 938.8 Iz = SAy2 = 3237 cm4 Iy = SAz2 = 6308.7 691. Since this is beyond the scope of this book.57 ¥ 15.cm2 = 143233 kPa REGARDING EULER–BERNOULLI HYPOTHESIS We were able to solve the flexure problem because of the nature of the crosssection which remained plane after bending.0.3 Nm From Eq.9) = -96.9) .0.5 -1034. 15) sx = l J1 + 2Gex (6. from Eq. (6. ∂ 2u x ∂σ x = E ∂x ∂ x2 (6.Bending of Beams 199 or from Eq.G ⎜⎜ 2 ∂ z2 ⎝∂y ⎞ ∂ ε +ε ⎟⎟ − G z ∂ x y ⎠ ( ) (6.e.15) where l is a constant and G is the shear modulus. sx = Eε x = E ∂ ux ∂x (6. (c). we have sy = sz = 0 Hence.16a) Differentiating.G ⎜⎜ 2 ∂x ∂ z2 ⎝∂y i. According to the Euler-Bernoulli hypothesis.ν s x E Hence. (6.16b) From equilibrium equation and stress-strain relations ∂τ ∂τ ∂σ x = − xy − xz ∂y ∂z ∂x = −G ∂ ⎛ ∂ ux + ∂ u y ⎞ − G ∂ ⎛ ∂ ux + ∂ uz ⎞ ∂ y ⎜⎝ ∂ y ∂ x ⎟⎠ ∂ z ⎜⎝ ∂ z ∂ x ⎟⎠ ⎛ ∂ 2ux ∂ 2u x + = −G ⎜⎜ 2 ∂ z2 ⎝∂y ⎞ ∂ ⎛ ∂ u y + ∂ uz ⎞ ⎟⎟ − G ∂ x ⎜⎝ ∂ y ∂ z ⎟⎠ ⎠ ⎛ ∂ 2 u x ∂ 2u x + = .17b) . Eq. E 2 2 ∂ 2u x GE ⎛ ∂ u x + ∂ u x + ⎜ ∂ x 2 E − 2ν G ⎜⎝ ∂ y 2 ∂ z2 ⎞ ⎟⎟ = 0 ⎠ (6. ( ∂σ x 1 − 2ν G E ∂x ) ⎛ ∂ 2 u x ∂ 2u x + = .16b). (3.17a) Since sy = sz = 0.17a) becomes ⎛ ∂ 2u x ∂ 2 u x ∂σ x + = .G ⎜⎜ 2 ∂ z2 ⎝∂y ⎞ 2ν G ∂σ x ⎟⎟ + E ∂x ⎠ ⎞ ⎟⎟ ⎠ 2 2 ∂σ x GE ⎛ ∂ u x + ∂ u x =⎜⎜ E − 2ν G ⎝ ∂ y 2 ∂x ∂ z2 or ⎞ ⎟⎟ ⎠ Substituting in Eq. ey = e z = . Iz is the moment of inertia about the neutral axis which is taken as the z axis. In or K1 My (which is equivalent to Iz plane sections remaining plane) will be valid only in those cases where the bending moment is a constant or varies linearly along the axis of the beam. z ) = ∂ y2 ∂ y2 E ∂ 2u x ∂ 2 φ ( y.e. i. Hence. z ) ∂ 2 φ ( y. Differentiating the above expression and E ∂ 2u x ∂ 2 φ ( y.e.200 i. the Euler-Bernoulli hypothesis that sx = . ∂ M ( x) = a constant ∂x or M (x) = K3x + K5 This means that M(x) can only be due to a concentrated load or a pure moment. (6. (d) Eux = y Iz ∫ M dx + φ ( y. z ) ∂ 2 φ ( y. z ) where f is a function of y and z only.16a) A or ⎛ ∂ 2ux ∂ 2ux ∂ 2ux +G⎜ + 2 ⎜ ∂ y2 ∂x ∂ z2 ⎝ sx = ∂ ux My =E Iz ∂x (6.18). z ) ⎤ ∂ M ( x) + = K2 ⎢ ⎥ 2 ∂x ∂ z2 ⎦ ⎣ ∂y The left-hand side quantity is a function of x alone or a constant and the righthand side quantity is a function of y and z alone or a constant. M is a function of x only and y is the distance measured from the neutral axis. z ) = ∂ z2 ∂ z2 Substituting these in Eq. both these quantities must be equal to a constant. (6. Advanced Mechanics of Solids ( E − 2ν G ) ⎛ ∂ 2u x ∂ 2 u x ∂ 2u x + + G ⎜⎜ 2 ∂ x2 ∂ z2 ⎝∂y ⎞ ⎟⎟ = 0 ⎠ ⎞ ⎟⎟ = 0 ⎠ where A is a constant. Ay ∂ M ( x) G ⎡ ∂ 2 φ ( y. Then E ∂ 2u x y ∂M = 2 I ∂z z ∂x Integrating Eq. z ) ⎤ + ⎢ + ⎥ =0 EI z ∂ x E ⎣ ∂ y2 ∂ z2 ⎦ ⎡ ∂ 2 φ ( y. From flexure formula and Eq. other words.18) (d) In the above equation. The values of txy and t xz are to be determined with the equations of equilibrium and compatibility conditions. When the force P is parallel to the z-axis. But P can be applied at such a distance from the centroid that twisting does not occur. 6. for a beam with a general cross-section. . If a transverse force is applied at this point. Iz sy = sz = tyz = 0 (6. the load P in addition to causing bending will also twist the beam. the beam carries loads which are transverse to the axis of the beam and which cause not only normal stresses due to flexure but also transverse shear stresses in any section. We shall not try to determine these. we assume that sx = - P ( L − x) y . the load P will have to be applied at a distance e from the centroid O.4 SHEAR CENTRE OR CENTRE OF FLEXURE In the previous sections we considered the bending of beams subjected to pure bending moments.Bending of Beams 201 6.10). Consider the cantilever beam shown in Fig.9 e x z o Cantilever beam loaded by force P Let Ox be the centroidal axis and Oy. a position can once again be established for which no rotation of the centroidal elements of the cross-sections occur. one important point should be noted. the load has to be along the axis of symmetry to avoid twisting. Let the load be parallel to one of the principal axes (any general load can be resolved into components along the principal axes and each load can be treated separately). This load in general. 6. In general.19) This is known as St. we can resolve it into two components parallel to the y and z-axes and note from the above discussion that these components do not produce rotation of centroidal elements of the cross-sections of the beam. This point is called the shear centre of flexure or flexural centre (Fig. y y P P (L – x) L Fig. As said above. cause (i) Normal stress sx due to flexure. this will cause both bending and twisting. However. Venant's assumption. For a section with symmetry. Oz the principal axes of the section. For the same reason. The determination of txy and t xz for a general cross-section can be quite complex. will at any section. The value of s x as given above is derived according to the flexure formula of the previous section. 6. (ii) Shear stresses t xy and t xz due to the transverse nature of the loading and (iii) Shear stresses t xy and t xz due to torsion In obtaining a solution. In practice.9 carrying a load at the free end. The point of intersection of these two lines of the bending forces is of significance. can be quite complex. 6. to locate the shear centre of the section. as shown in Fig. i. As mentioned in the previous section. For a section of a general shape. In other words. the shear stresses near the boundary lines of the section are parallel to the boundary. Our object in this section is to locate that point through which the load V y should act so as to cause no twist. as discussed in the next section. Consider an element of length Dx of the beam at section x. Consequently. the shear stress can be taken to be parallel to the centre line of the section at every point as shown in Fig. approximate locations of the shear-centres can be determined by an elementary analysis. Since the section of the beam is thin.5 It is important to observe that the location of the shear centre depends only on the geometry. bending. 6. 6. Then there will be normal stress distribution due to bending and shear stress distribution due to vertical load. as shown in Fig. SHEAR STRESSES IN THIN-WALLED OPEN SECTIONS: SHEAR CENTRE Consider a beam having a thin-walled open section subjected to a load V y. the shape of the section. the boundary of the section is an unloaded boundary.10 Load P passing through shear centre 6. 6. . as mentioned earlier. twisting and shear in the beam.12.11(b). However. i.e. The thickness of the wall is allowed to vary.11(a). which.11 ts (b) Thin-walled open section subjected to shear force The surface of the beam is not subjected to any tangential stress and hence. for thin-walled beams with open sections. the shear stresses near the boundary cannot have a component perpendicular to the boundary. 6.e. y txs s=0 Vy z x x s Dx (a) Fig. There will be no shear stress due to torsion. the load V y produces in general. the location of the shear centre depends on the distribution of txy and txz.202 Advanced Mechanics of Solids y P P1 o z P2 Fig. Let us assume that load V y is applied at the shear centre. 20) s Vy 1 ( I y − I yz z ) t ds tsx = ∫ 2 ts I yz − I y I z 0 y or tsx = − ( Vy ts I y I z − 2 I yz ) ⎡ ⎢I y ⎣ s s ⎤ 0 0 ⎦ ∫ yt ds − I yz ∫ zt ds ⎥ (6. (6. the unbalanced normal force is balanced by the shear stress tsx distributed along the length Dx. yI y − zI yz ∂ M z ∂σ x = 2 ∂x I yz − I y I z ∂ x (6. and substituting in Recalling from elementary strength of materials y ∂x Eq. (6. its value at any s is also given by Eq. 6. the normal stress sx is given by Eq.8) as yI y − zI yz Mz sx = 2 I yz − I y I z Hence. therefore.22) The first integral on the right-hand side represents the first moment of the area between s = 0 and s about the z axis.e. Since txs is the complementary shear stress.8). For equilibrium.12 ∂σ x ∆x ∂x ∂M z ∆x ∂x Free-body diagram of an elementary length of beam Let Mz be bending moment at section x and Mz + ∂ M z Dx the bending moment at ∂x ∂σ x ∆x.21) ∂ M z = -V . are corresponding flexural stresses at these ∂x two sections. the sign of the stress will be correctly given by Eq. . Considering a length s of the section. (6. (6. However. (6. Observing that My = 0.22). The second integral is the first moment of the same area between s = 0 and s about the y axis.Bending of Beams 203 txs sx σx + tsx x Mz + Mz Fig.20) ts is the wall thickness at s. section x + Dx◊ sx and sx + ⎛ s s 0 0⎝ τ sx ts ∆x − ∫ σ x t ds + ∫ ⎜ σ x + ∂σ x ⎞ ∆x ⎟ t ds = 0 ∂x ⎠ s ∂σ τ sx = − 1 ∫ x t ds ts 0 ∂ x i. It is important to observe that for the moments shown the normal stresses should be compressive and not as shown in the figure. i.14. . Vy ez = moment of txs about O y Vy ez y Vy V Vz Shear Centre O z z ey ez O txs Fig. is called the shear centre or the centre of flexure. 6. the shear stress distribution txs must be statically equivalent to the load Vy. Vy ⎡⎣ I y Qz − I yz Q y ⎤⎦ τ sx = τ xs = − (6. 6.e.25). Hence. 6. 6. Then.22) gives the shear stress distribution at section x due to the vertical load Vy acting under the explicit assumption that no twisting is caused. through which Vz and Vy should act to prevent the beam from twisting. That is. then the moment of Vz about the centroid O must be equal to the moment of txs about the same point.25) )⎣ If the above shear stress distribution is due to the shear force alone and not due to twisting also.4.23) 2 ts I y I z − I yz ( ) Equation (6.204 Advanced Mechanics of Solids Let Qz be the first moment of the area between s = 0 and s about the z axis and Qy the first moment of the same area about the y axis.13 Location of shear centre and flow of shear stress Fig.24) ⎡ I z Qy − I yz Qz ⎤⎦ (6. The point with coordinates (ey. we can determine the shear stress txs at any s as txs = - or txs = - ( Vz 2 ts I y I z − I yz ( Vz 2 ts I y I z − I yz ) ⎡ ⎢Iz ⎣ s s ⎤ 0 0 ⎦ ∫ zt ds − I yz ∫ yt ds ⎥ (6. 6. (ii) The moment of txs about the centroid (or any other convenient point) must be equal to the moment of Vy about the same point. This means the following: (i) The resultant of txs integrated over the section area must be equal to Vy. Vy ez = moment of txs about O where ez is the eccentricity or the distance of Vy from O to avoid twisting (Fig. This is shown in Fig. If a force Vz is acting instead of Vy. ez). as mentioned in Sec.13).22) and (6.14 Location of shear centre for a general shear force Any arbitrary load V can be resolved into two components Vy and Vz and the resulting shear stress distribution t xs at any s is given by superposing Eqs (6. 15). the shear stress increases linearly from s = 0 to s = b. Also. txs = F ( ) Qz = bt1 h + 0 + ( s − b − h ) t1 − h 2 2 2 ⎛ ⎞ = ⎜ bh + h − h s ⎟ t1 2 2 ⎝ ⎠ . ts = t1 and the statical moment is Hence.e. so that Iyz = 0. and the statical moment is Hence.23) then.4 Solution Let Oyz be the principal axes. (6. and the statical moment is the moment of the shaded area in Fig.15) about the z axis. (6. the shear varies parabolically from s = b to s = b + h.15 Example 6. 6. From Eq. Qz = t1 sh 2 txs = Fsh 2I z for 0 ≤ s < b (6. For s in the vertical web. y t1 F F y z o s = 0 h t2 b Fig. 6. For s in the horizontal flange. ( ) ( ) h h 1 h ⎡ ⎤ Qz = bt1 2 + 2 − y t2 ⎢ y + 2 2 − y ⎥ ⎣ ⎦ = ⎛ h2 ⎞ ⎤ 1⎡ bt1 h + ⎜ − y 2 ⎟ t2 ⎥ ⎢ 2⎣ 4 ⎝ ⎠ ⎦ ⎡ ⎛ 2 ⎞ ⎤ (6.27) bt1 h + ⎜ h − y 2 ⎟ t2 ⎥ for − h < y < + h ⎢ 2t2 I z ⎣ 2 2 ⎝ 4 ⎠ ⎦ i. noting that F is negative. i.4 Determine the shear stress distribution in a channel section of a cantilever beam subjected to a load F.e.26) i.e. as shown. or ( txs = F I Q ts I y I z y z txs = FQz ts I z ) where Qz is the statical moment of the area from s = 0 to s about z axis. Considering the top flange. ts = t1. locate the shear centre of the section (Fig.Bending of Beams 205 Example 6. ts = t2. i.29) +h/ 2 Hence.16. the resultant of txs over the area is equal to F.15(b)] M = ( resultant of τ xs in top flange ) × h 2 + ( resultant of τ xs in bottom flange ) × h 2 = 2 ( resultant of τ xs in top flange ) × h 2 ( = 2 average of τ xs in top flange × area × h 2 ) . and therefore. 6. When s = 2b + h. the resultant of the stress in the top and bottom flange cancel each other.206 Advanced Mechanics of Solids 2 ⎛ ⎞ txs = F ⎜ bh + h − h s ⎟ 2I z ⎝ 2 2 ⎠ Hence. we have ⎤ ⎛h F 2⎞ ∫ τ xs t2 dy = 2 I ⎢ ∫ bt1h dy + ∫ ⎜ 4 − y ⎟ t2 dy ⎥ z ⎣ ⎝ ⎠ −h / 2 ⎦ 2 3 3 ⎡ ⎤ = F ⎢bt1h 2 + h t2 − h t2 ⎥ 2I z ⎣ 4 12 ⎦ ⎡ t h3 ⎤ = F ⎢bt1h 2 + 2 ⎥ 2 I z ⎣⎢ 6 ⎥⎦ Now for the section 2 2 3 Iz = bt1 h + bt1 h + t2 h 4 4 12 2 3 = bt1 h + t2 h 2 12 (6. ∫ τ xs ts dy = F −h / 2 Hence.e. 6. there is no horizontal resultant. the right tip of the bottom flange.4—Shear stress distribution diagrams +h/ 2 ⎡ for 2b + h ≥ s > b + h (6.e. ez C F (a) (b) Fig. Integrating txs over the vertical web.16 Example 6. the shear varies linearly. Taking moment about the midpoint of the vertical web [(Fig. The variation of txs is shown in Fig. In addition. This shear stress distribution should be statically equivalent to applied shear force F. It is easy to see that this is equal to F in magnitude. the shear is zero. On integrating txs over the area of the section.28) i. 6. it has a moment. 6.17). Hence. the shear centre is located at a distance ez from C [Fig.Bending of Beams 207 ⎛ ⎞ = 2 ⎜ Fbh × bt 1 × h ⎟ 2⎠ ⎝ 4I z = Fb 2 h 2t1 4I z This must be equal to the moment of F about the same point. Locate the shear centre (Fig. (6.5 Determine the shear stress distribution for a circular open section under bending caused by a shear force.17 Example 6. Example 6. 6. y F q R df O 2F π Rt 2R (a) O (b) Fig.16(b)].5 Solution The static moment of the crossed section is Qz = θ ∫ ( R dφ t ) R sin φ 0 = R 2t (1 − cos θ ) . F must act at a distance ez from C such that Fez = or ez = Fb 2 h 2t1 4I z b 2 h 2 t1 4I z Substituting for Iz from Eq.29) ez = or ez = 3b2 h 2t1 6bt 1 h2 + t2 h3 3b 2 t1 6bt1 + t2 h Hence. 6. the moment about O is negative since it is directed from + z to + y.6 SHEAR CENTRES FOR A FEW OTHER SECTIONS In a thin-walled inverted T section. and for a vertically upward shear force F. The moment of this distributed stress about C is obviously zero.18(a).17(b).23). 6. 6. from Eq. txs = − F (1 − cos θ ) π Rt For q = 180° txs = − 2F π Rt The distribution is shown in Fig. Hence the.18 Shear Centre C (b) Location of shear centres for inverted T section and angle section For the angle section. the distribution of shear stress due to transverse shear will be as shown in Fig. FQz txs = − = − F R 2t (1 − cos θ ) tI z tI z Iz = π R 3 t Hence. the shear centre for this section is C. For F positive. 6. . The moment of this distribution about O is. 6. Shear Centre C (a) Fig. the moment of the shear stresses about C is zero and hence. force F must be applied at the shear centre C. But M= 2π ∫ τ xs ( R dθ t ) R 0 2π = − F ∫ R 2 t (1 − cos θ ) dθ π Rt 0 = -2FR This should be equal to the moment of the applied transverse force F about O. C is the shear centre.208 Advanced Mechanics of Solids Hence. noting that Iyz = 0.19 shows how the beams will twist if the loads are applied through the centroids of the respective sections and not through the shear centres. (6. Figure 6. which is at a distance of 2R from O. Hence. This is possible when the beam section is symmetrical about the plane of curvature and the bending moment M acts in this plane. The unstretched length before bending is (r 0 . Let r 0 be the initial radius of curvature of the centroidal surface. Fibres above this surface get compressed and fibres below this surface get stretched. (c) (b) (a) Fig.20 Geometry of bending of curved beam (b) . S.19 Twisting effect on some cross-sections if load is not applied through shear centre 6. SN is the trace of the neutral surface with radius of curvature r0.y) Df. let the section AB rotate through dDf and occupy the position A¢B¢. Fibres lying in the neutral surface remain unaltered. the neutral axis.20. 6. 6. We consider the case where bending takes place in the plane of curvature. it is again assumed that sections which are plane before bending remain plane after bending.7 BENDING OF CURVED BEAMS ( WINKLER–BACH FORMULA) So far we have been discussing the bending of beams which are initially straight. Consider a fibre at a distance y from the neutral surface. Now we shall study the bending of beams which are initially curved. Hence. Owing to the moment M.C.Bending of Beams 209 S. 6.C. a transverse section rotates about an axis called the neutral axis.C. S. The section rotates about NN. Consider an elementary length of the curved beam enclosing an angle Df. The change in y Df dDf r0 r 0 B¢ r M S G y M B r0 r0 N C A¢ C A N –y N e C x sxdA F (a) Fig. As in the case of straight beams. as shown in Fig. 32b) r0 − y 0 ⎜⎝ r r0 ⎟⎠ The above expression brings out the main distinguishing feature of a curved beam. ex = − Ey ⎛ ⎞ r 1− 1 (6. For equilibrium. r0 is very large compared to y. the beam must have a large curvature in which the dimensions of the cross-sections of the beam are comparable with the radius of curvature r0. the resultant of s x over the area should be equal to zero and the moment about NN should be equal to the applied moment M. the denominator (r0 .e. Also SN = r0 Df Hence. The value of (d Df)/Df can be obtained from Fig. strain ≡ ε x = − y (δ ∆φ ) (r0 − y ) ∆φ (6.32a). (6. the above equation reduces to that of the straight beam. for a fibre above the neutral axis.32a) r0 − y 0 ⎜⎝ r r0 ⎟⎠ Now we shall assume that only s x is acting and that s y = sz = 0. (∆φ + δ ∆φ )r =1 ∆φ r0 i.31) y ⎛ ⎞ r 1− 1 (6.e. It is seen that SN = (Df + d Df) r where r is the radius of curvature of the neutral surface after bending. On this assumption.20(a).y) will be less than that for a negative y. (6.32b) becomes σx = − ⎛ ⎞ σ x = −Ey ⎜ 1 − 1 ⎟ ⎝r r ⎠ 0 With r 0 Æ •. On the other hand.210 Advanced Mechanics of Solids length due to bending is y(d Df). It should be observed that the strains in fibres above the neutral axis will be numerically greater than the stains in fibres below the neutral axis. then Eq. This is similar to the Bernoulli-Euler hypothesis for the bending of straight beams. Noting that for the moment as shown.e. i. the strain is negative. if the curvature (i. Since the resultant normal force . (6.30) r δ ∆φ = 0 −1 ∆φ r ⎛1 1⎞ = r0 ⎜ − ⎟ ⎝ r r0 ⎠ (6. since for positive y. The value of y must be comparable with that of r0. 1/r0) is very samll. i. 6.e. Substituting in Eq. i. This is evident from Eq.30) It is assumed here that the quantity y remains unaltered during the process of bending.e. The second integral is zero according to Eq.33). (6. (6. the normal stress varies non-linearly across the depth. The maximum stress may occur either at the top or at the bottom of the section.e and r0 = r0 . the neutral axis gets shifted towards the centre of the curvature. 6. ⎛ ⎞ Er0 ⎜ 1 − 1 ⎟ Ae = M r r 0 ⎠ ⎝ (6. we have. Eq.35) becomes y′ − e sx = − M Ae ρ 0 − e − y ′ + e .34) − or σ x ( r0 − y ) y Ae = M y M sx = − Ae ( r0 − y ) (6. The distribution is hyperbolic and one of its asymptotes coincides with the line passing through the centre of curvature.32) ⎛ ⎞ σ (r − y ) Er0 ⎜ 1 − 1 ⎟ = − x 0 r r y 0 ⎠ ⎝ Substituting this in Eq. Equation (6.Bending of Beams 211 is zero. (6.e.21(a). Thus. If the origin is taken at the centroid and y¢ is the distance of any fibre from this origin.35) is often referred to as the Winkler-Bach formula. (6. (6.35) shows. For equilibrium. In some texts.34) But from Eq. then putting y = y¢ . depending on its shapes. as shown in Fig. the origin of the coordinate system is taken at the centroid of the section instead of at the point of intersection of the neutral axis and the y axis. where e is the distance of the centroid from the neutral axis NN and this moment is negative.35) As Eq.33) The second equation can be written as ⎛ + Er0 ⎜ 1 − 1 ⎝ r r0 ⎞⎡ ⎟ ⎢ − ∫ y dA + r0 ⎠⎣ A y dA ⎤ ∫ r − y⎥ = M ⎦ A 0 The first integral represents the static moment of the section with respect to the neutral axis and is equal to (-Ae). ⎛ 1 1 ⎞ y dA ∫ σ x dA = − Er0 ⎜ r − r ⎟ ∫ r − y = 0 ⎝ A 0 ⎠ A 0 ⎛ ⎞ y 2 dA − ∫ σ x y dA = + Er0 ⎜ 1 − 1 ⎟ ∫ =M ⎝ r r0 ⎠ A r0 − y A From the first equation above and y dA ∫ r − y =0 A 0 (6. 33).36) Ae ρ0 − y ′ To use Eq. We shall calculate these for a few of the commonly used sections.y the equation becomes r0 − u ∫ u dA = 0 A A (6.35). dA = b du and u = r 0 . Introducing the new variable u u = r0 . r0 = ∫ dA/ u or A The integral in the denominator represents a geometrical characteristic of the section. In other words. consider Eq. one requires the value of r0.212 Advanced Mechanics of Solids r0 r0 Neutral Axis e N (a) C N (b) Fig.y¢.39b) . For this.22. Hence. the values of r0 and e are independent of the moment within elastic limit. (6. Rectangular Section From Fig.38) (6. r0 = h ⎛ρ + ⎜ 0 log n ⎜ ⎜ ρ0 − ⎝ h⎞ 2⎟ h ⎟⎟ 2⎠ = h log n (r2 / r1 ) The shift of the neutral axis from the centroid is h e = ρ0 − ⎛ρ + h⎞ ⎜ 0 2⎟ log n ⎜ ⎟ ⎜ ρ0 − h ⎟ ⎝ 2⎠ h or e = ρ0 − ⎛r ⎞ log n ⎜ 2 ⎟ ⎝ r1 ⎠ (6. ρ + h/2 ρ0 + h b du dA = 0 2 = b log n ∫ ∫A u u h ρ0 − h / 2 ρ0 − 2 Hence. 6.21 Distribution of normal stress and location of neutral axis y′ − e sx = − M (6. (6.37) Hence.39a) (6. 6. the above equation reduces to that of the previous case. (6.y h1 − e ⎡ dA ∫ u = ∫ ⎢ −h − e ⎣ \ 2 b2 + (b1 − b2 ) ( h2 + e + y )/ h ⎤ ⎥ dy ρ0 − e − y ⎦ r2 − (b1 − b2 ) r1 When b1 = b2.41) I-Section For the I-section shown in Fig. 6. we obtain for r r dA 3 2 ∫ u = b1 log r + b 2 log r 1 3 (6. r r r dA 3 4 2 ∫ u = b1 log r + b2 log r + b 3 r 1 3 4 (6. 6. 6.23 Parameters for a trapezoidal section to calculate r0 according to Eq. The variable width of the (b1 − b 2 ) (h 2 + e + y) h dA = dy [b 2 + (b1 − b2 ) (h 2 + e + y )/ h ] and u = r0 . 6.e . following the same procedure as in the preceding case.31) Trapezoidal Section (see Fig.31) Let h1 + h2 = h. (6.24) the section 1 ⎫ − (b1 − b2 ) ⎬ ⎭ (6.40) Proceeding as in the previous case.42) . = [b2 + r2 (b1 − b2 )/ h ] log r0 = ( b1 + b2 ) h ⎧ r b2 + r2 (b1 − b2 )/ h ] log 2 [ ⎨ 2 r ⎩ T-section (see Fig. 6.25.Bending of Beams y r0 y r1 r0 r0 u r1 r0 r2 u r2 b1 h1 h C du b Fig.23) section is b = b2 + Fig.22 213 h2 b2 Parameters for a rectangular section to calculate r0 according to Eq. 6 Determine the maximum tensile and maximum compressive stresses across the Sec. 6. 27. .214 Advanced Mechanics of Solids y r3 r1 r3 r1 b1 b1 r2 r4 r2 b2 b2 b3 Fig. Load P = 2000 kgf (19620 N).31) Fig.(a cos q . AA of the member loaded. q a dA ∫ u = 2aπ A Fig.26) u = r0 . 6.24 Parameters for T-section to calculate r0 according to Eq. where b = 0 a θ b − cos 0 2 Adding and substracting (b cos q + b ) to the numerator. 6.y = (r0 .a cos q du = a sin q dq dA = 2a sin q du = 2a2 sin2 q dq y r0 r0 π 2 2 dA ∫ u = ∫ 2a sin θ / ( ρ0 − a cos θ ) dθ 0 A u π 1 − cos 2 θ ρ = 2a ∫ dθ .31) Circular Section (see Fig. 6.e) .25 Parameters for I-section to calculate r0 according to Eq.31) and r0 = ( ) ⎡b − b 2 − 1 1/ 2 ⎤ ⎢⎣ ⎥⎦ ( = 2π ⎡⎢ ρ0 − ρ02 − a 2 ⎣ ) 1/ 2 ⎤ ⎥⎦ a2 2 ⎡⎣ ρ0 − ( ρ02 − α 2 )1/ 2 ⎤⎦ Example 6. 6. as shown in Fig.26 Parameters for a circular section to calculate r0 according to Eq. (6.e) = r0 . (6. (6. h = 6 cm.27 Example 6. y = -(e + h/2) = -3.35).27 and. 6.73. (σ x′ )C = − and (σ x′ ) D = − 19 P 2. \ log ρ0 + h /2 = log 7 = 0.27 (10. (6.001 P 24 × 0.0417 P 24 A Hence the combined stresses are (sx)C = (0.27) Hence.Bending of Beams a a A 4 cm P a a = 8 cm M C A 6 P D P cm Fig. owing to bending moment M r0 = y Ae ( r0 − y ) σ x′ = .5596 ρ0 − h /2 4 From equations (6.27 0.5596 From Eq. y = h − e = 2.M y M 24 × 0.38) and (6.73) The stress due to direct loading is σ x′′ = − P = − P = − 0.73 − 2.73 2 (−3.0. b = 4 cm.73 = − 1.73 − y ) =− For the problem M = P (a + a + h/2) = 19P At C.73 + 3. e = 11 − 10.6 Solution For the section r0 = 11 cm.6848 .27 (10.39) 6 = 10.73 = 0.0417) P = 0.6431P = 129 kgf/cm2 (12642 kPa) P 215 .6848 P 24 × 0.27) 19 P × = 0.27 (10. at D. 0427 P = -209 kgf/cm2 (20482 kPa) Example 6. r2 = 10 cm.216 Advanced Mechanics of Solids and (sx)D = (-1.204 × 17. 6. 400 2.001 .28 Example 6.5 The total normal stress is therefore. across any section. there is a tangential force P and a moment (Px . The energy method will be used. Consider the member shown in Fig.0.907 kPa 6.363 cm \ A = 1 (b1 + b2 )h = 35 = 17.204 − 2.5 cm 2 2 2 r 0 = A/3.5/3. Solution For the section P dA 10 ∫ u = [1 + 10(4 − 1)/7] log 3 − (4 − 1) = 3. force P = 3000 kgf (29400 N). r1 = 3 cm.363 = 5. In the straight part of the U-ring.400 kgf cm (1705 Nm) The normal stress due to bending is therefore b2 Fig.8 DEFLECTIONS OF THICK CURVED BARS In Chapter 5. we shall discuss a few problems involving thick rings.596 The moment across section D is M = -3000 r0 = -17.r0 = 0.0417) P = -1. h = 7 cm.7 (σ x′ )D = y −M Ae r0 − y 17. In the curved part of the member. the problems of thin rings and thin curved members were analyzed using energy methods.596 5. or 136. there will . b2 = 1 cm.M 0). 6.148 kPa) =+ The normal stress due to axial loading is (σ x′′ ) D = 3000 = 3000 = 171 kgf /cm 2 A 17. 6. In this section.363 = 17. (sx)D = 1397 kgf/cm2.80 cm 3 (b1 + b2 ) 5 e = r0 .28) having a trapezoidal section with the following dimensions: b1 = 4 cm.204 cm r0 = 3 + D \ P b1 r1 r2 h (b1 + 2 b2 )h = 3 + 14 = 5.5 × 0.7 Determine the stress at point D of a hook (Fig.2 = 1226 kgf/cm2 (120.29(a). In general. which for a straight beam passes through the centroid of the section.45) .43) ∆UV = αV ∆s 2 AG where a is a numerical factor depending on the shape of the cross section. Because of the bending moment M. the strain energy due to V is small as compared to that due to M. (6. Their values are V = P cos q N = P sin q M = M0 . the energy stored is 2 ∆U M = M ∆s (6. Ds = r0 Df. 6.29 Geometry of deflection of a curved bar be a tangential force V. enclosing an angle Df.(d + r0 sin q ) P To calculate the strain energy stored we proceed as follows (we make use of the expressions developed in Chapter 5): (i) In the straight part of the member: Owing to the shear force V. is 2 ∆UV = αV ∆s 2 AG If r0 is the radius of curvature of the centroidal fibre. (ii) In the curved part of the member: Owing to the shear force V.44) 2 EI where I is the moment of inertia about the neutral axis. the strain energy stored in a small sectoral element. a normal force N and a bending moment M. the strain energy stored in a small length Ds is 2 (6.Bending of Beams d N V M0 M q P P r0 217 M0 (b) N C (a) M d Df Df r0 (c) Fig. A is the area of the section and G is the shear modulus. 48) The same result is obtained if M is applied first and then N. (6. the centroid C [Fig.20).46) ∆U N = N ∆s 2 AE Owing to bending moment M. . where e0 is the strain at C and consequently. DUM = Putting Df = DUM = M 2 ∆φ 2 AeE ∆s ρ0 M 2 ∆s 2 AeE ρ0 (6. This is according to the principle of superposition. M e0 = + Aρ0 E Hence the work done by N is DMN = MN ∆ s Aρ 0 E (6.34).31). the energy stored is equal to the work done. y0 is the distance of the centroidal fibre from the neutral axis and is equal to -e. the force N does additional work equal to DUMN = N ε 0 ∆ s e0 from Eq.29 (c)] DUM = 1 M (δ ∆φ ) 2 From Eq.29(c)] moves through a distance e0 Ds. 6. (6. M is positive. 2 (6.35) is e0 = σx y0 =− M E AeE (r0 − y0 ) In the above equation. This can be seen by referring to Fig. If d Df is the change in the angle due to bending [Fig. 6. owing to the rotation of the section. 6. With these. which is valid for small deformations. Also. (6.30. 6. substituting for the right-hand part in the above equation d Df = ∆φ M AeE Hence.47) If N is applied first and then M. according to the convention followed (Fig. r0 = r0 + e.218 Advanced Mechanics of Solids Because of the normal force N. ⎛1 1 ⎞ d Df = ∆φ r0 ⎜ − ⎟ ⎝ r r0 ⎠ From Eq. which is assumed to be acting at the centroid of the cross-section. With the strain energy calculated as above and using Castigliano's theorem. 6. since the lengths of the fibres are different. 6.31.8 A ring with a rectangular section is subjected to diametral compression. Solution We observe that the deformation of the ring will be symmetrical about the horizontal and vertical axes. one can solve for the unknown—either the deflection or the indeterminate reaction. the work done by N when M is applied is zero since the section rotates about the neutral axis which passes through the centroid. the moment M does work equal to Fig. the total strain energy is. We can. consider only a quadrant of the circle for the analysis. Determine the bending moment and stress at point A of the inner radius across a section q. This is shown in Fig.31(c). M0 is the unknown internal moment. Combining all the energies detailed above. Its value can be determined from . the last expression will be zero and the third expression (which becomes indeterminate since e = 0 and r0 = •) is replaced by M 2/2EI. We shall illustrate this through an example. as shown in Fig. The angle enclosed between AB and A¢B¢ is therefore A M B¢ B dq δθ = Df r2 219 r1 ε n ∆φ ( r2 − r1 ) ( r2 − r1 ) = ε n ∆φ Owing to this rotation of A¢B¢. U = ∫ (∆UV + ∆U N + ∆U M + ∆U MN ) s 2 2 2 ⎛ ⎞ (6.Bending of Beams N A¢ The normal force N acting across the section produces uniform strain en.49) = ∫ ⎜ αV + N + M + MN ⎟ ds 2 2 2 AG AE AeE ρ AE ρ 0 0 ⎠ s⎝ For the straight part of the beam. The extension of the fibre at b will be en r1 Df. 6. Example 6. therefore. 6.30 Deformation of a section of curved bar DUNM = M ε n ∆φ en = N AE Since DUNM = MN ∆φ AE = MN ∆s AE ρ0 For a straight beam.31(b)]. Consequently. face AB will not shift parallel to itself. r 1 and r 2 are the inner and external radii respectively. This is also confirmed in the above expression where r0 = • for a straight beam and therefore DUMN = 0. there will be no changes in the slopes of the vertical and horizontal faces of the ring [Fig. as shown in Fig. 6. Their values are N = − P ρ0 cos φ and V = − P sin φ 2 2 The total strain energy for the quadrant from Eq.49) is π /2 U= ∫ α P 2 sin 2 φ 8 AG 0 π /2 ρ0 d φ + ∫ 0 P 2 cos 2 φ ρ0 d φ 8 AE 2 + π /2 ∫ 0 ⎡ M − P ρ (1 − cos φ )⎤ ⎣⎢ 0 2 0 ⎦⎥ d φ 2 AeE ⎡ M − P ρ (1 − cos φ ) ⎤ P cos φ ⎢ 0 2 0 ⎥⎦ − ∫ ⎣ dφ 2 AE 0 π /2 2 ⎞ ⎛ α P2 =⎜ + P ⎟ π ρ0 ⎝ 8 AG 8 AE ⎠ 4 ( ) (6. We shall use Castigliano’s theorem to determine this moment.220 Advanced Mechanics of Solids P P r1 r0 r0 A r2 f q M0 P/2 M0 M0 P/2 P/2 (c) (b) b V = P (a) P sin φ 2 M P N= cos φ 2 P/2 (d) Fig. there is a normal force N and a shear force V.P r0 (1 .31 Example 6.50a) ( ) ⎡ 2 π P2 2 π π ⎤ π ⎢ M 0 2 + 4 ρ0 2 + 4 − 2 − M 0 ρ0 P 2 − 1 ⎥ ⎣ ⎦ Pρ Pρ π ⎞ P ⎛ (6.cos f) 2 In addition.50b) M0 − 0 + 0 ⎟ − ⎜ AE 2 ⎝ 2 2 4⎠ + 1 2 AeE . 6. Across any section f.8 the condition that the change in the slope of this section is zero. (6. the moment is M = M0 .31(d). As the change in slope at the section where M is applied is zero. M0 is still an unknown quantity.39b) =− y = h − e.51) ⎜ 1 − π + πρ ⎟ 0 ⎠ ⎝ If we ignore the initial curvature of the member while calculating the strain energy.e. from Eqs (6. same as given in Eq. this moment is the same as in Example 5. ∂U = 1 ∂ M0 2 AeE ( ) ⎡M π − ρ P π − 1 ⎤ − P = 0 0 ⎢⎣ 0 ⎥⎦ 2 AE 2 P ρ0 2 ⎛ 2 2e ⎞ (6. (6. With the value of M0 known.38) and (6. (6. \ ⎡ M − P ρ (1 − cos φ )⎤ ⎣⎢ 0 2 0 ⎦⎥ d φ 2 EI π /2 ∂U * = 1 ⎡ M − P ρ (1 − cos φ ) ⎤ ρ dφ = 0 ∂ M 0 EI 0∫ ⎢⎣ 0 2 0 ⎦⎥ 0 M 0 π − P ρ0 π + P ρ0 = 0 2 2 2 2 M0 = ( ) P ρ0 1− 2 π 2 i.12.P r0 (1 .Bending of Beams 221 In the above expression. log (r2 / r1 ) e = ρ0 − r2 − r1 = ρ0 − r0 log (r2 / r1 ) . Also. 2 r0 = r2 − r1 .e. the bending moment at any section q is obtained as M = M0 . i.35) and adding additional stress due to the normal force N.51) with e Æ 0 and r0 Æ •. y M ⋅ +N sA = − Ae ( r0 − y ) A P ρ0 2 Ae P cos θ ⎛ 2e 2⎞ y ⎜ cos θ + πρ − π ⎟ r − y − 2 A 0 ⎝ ⎠ 0 For point A.e.cos q ) 2 = Pρ0 ⎛ 2e 2 ⎞ − cos θ + 2 ⎜⎝ πρ0 π ⎟⎠ The normal stress at A can be calculated using Eq. then \ M0 = π /2 U* = ∫ α P 2 sin 2 φ 8 AG 0 π /2 ρ0 d φ + ∫ 0 P 2 cos2 φ ρ0 d φ 8 AE 2 + π /2 ∫ 0 and i. that of a thin ring. then the chage in the vertical diameter is obtained as δv = ( P ρ03 π 2 − EI 4 π ) This can be seen from Eq.50b). Solution From Eq. Determine the change in the vertical diameter. and simplifying ⎧ δ v = 4 P ρ 0 ⎨ απ + 1 ⎩16 AG 2 AE 2 ρ0 ⎛2 π 2e ⎞ ⎜ π − 8 − πρ ⎟ + 2 AEe ρ 0 0 ⎠ ⎝ ⎛π 1 e2 ⎞ ⎪⎫ ⎜⎜ − + 2 ⎟⎟ ⎬ ⎝ 8 π πρ 0 ⎠ ⎪⎭ If e is small compared to r0.15 AE AEe If we assume that the ring is thin and the effect of the strain energies due to the direct force and shear force are negligible. then dv ª = α π ρ0 P 4 AG α π P ρ0 4 AG + ( ) ( 2 P ρ0 2 π 2 P ρ03 π 1 − + − AE π 8 AEe ρ 0 8 π + 0. (6.13.488 ) P ρ0 P ρ02 + 0.9 A circular ring of rectangular section. (6. the total energy for the complete ring is ⎧⎪ α P 2 π π P2 + 1 + U = 4 ρ0 ⎨ ρ0 32 AG 32 AE 2 AeE ⎪⎩ − M 0 ρ0 P where M0 = ( ⎡ π M 02 ρ02 P 2 3π + −2 ⎢ 4 4 ⎣⎢ 2 (π ) 2 ) ( ) Pρ0 π ⎤⎫ P ⎡ ⎤ M0 + −1 ⎥− − 1 ⎥⎬ ⎢ A E ρ 2 2 4 ⎦ 0 ⎣ ⎦⎭ ρ0 P ⎛ ⎞ 1 − 2 + 2e ⎟ 2 ⎜⎝ π πρ0 ⎠ dv = ∂ U ∂P Using the above expression for U (remembering that M is also a function of P). is subjected to diametral compression.31. Also.35).Advanced Mechanics of Solids 222 Using these ⎪⎧ ρ (π cos θ − 2 ) + 2e ( h − 2e ) ⎪⎫ + cos θ ⎬ σA = − P ⎨ 0 2 A ⎩⎪ πe − 2 h ρ ( 0 ) ⎭⎪ Example 6. When r0 is large compared to y and e Æ 0. shown in Fig. 6. . check with Example 5. Aer0 becomes equal to I according to flexure formula. 32) with a 10 cm ¥ 15 cm section is used as a simply supported beam of 3 m span. carries an end load of 45 kgf (441 N). sA = 73 kgf/cm2 = 7126 kPa ⎡ ⎢⎣ N. is at 36. If M = 1. 6.1 A rectangular wooden beam (Fig.1550 kgf cm (1133 Nm). Calculate the maximum flexural stress at the built-in end and also locate the neutral axis.2 A cantilever beam with a rectangular cross section. Calculate the maximum flexural stress at midspan and also locate the neutral axis for the same section.34). find the absolute maximum flexural stress in the section.3 A bar of angle section is bent by a couple M acting in the plane of the larger side (Fig. 6.2 2 ⎡ Ans. Find the centroidal principal axes Oy¢z¢ and the principal moments of inertia.33.8° to the longerside 6.5 kgf/cm = 10052 kPa ⎢⎣ N. 5 cm ¥10 cm which is built-in in a tilted position.Bending of Beams 223 6. It carries a uniformly distributed load of 150 kgf (1470 N) per meter. 6. y 30° W C D 15 cm z A 3m B 10 cm Fig.0 cm 6.32 Problem 6. ⎡ ⎣⎢ .33 Problem 6. s = ±102. as shown in Fig.A cuts side AD such that DN = 1. 6. 6. ⎡ ⎣⎢ 5 cm 10 cm 1 3 Fig.1 Ans.A. The load acts in a plane making 30° with the vertical. The length of the cantilever is 1.2 m. f = ±14° 32¢ ⎢ Iy¢ = 41. principal moments of inertia 116 cm4.5 ⎡ ⎣⎢ 2 ⎡ Ans.36 Problem 6.9 cm4 .1° with vertical . s = 914 kgf/cm (89640 kPa) ⎢⎣ f = 60° w.3 cm4.2 cm4.4 ⎡ ⎣⎢ 2 ⎡ Ans.35.3 ⎡ ⎢ ⎢ ⎣ 12.4 Determine the maximum absolute value of the normal stress due to bending and the position of the neutral axis in the dangerous section of the beam shown in Fig.5 m and P = 200 kgf (1960 N).7 mm z¢ 6. centroid at 2. 30° 20 mm P = 1000 kgf 140 mm 2m 20 m 4m m 120 mm Fig.27 cm from the base.7 mm f 127 mm z o ⎡ Ans. Given a = 0.35 Problem 6.7 mm a z P P 80 mm 8 mm CG P a 2a (b) (a) Fig.6. Iz = 73. 6. 30. (Fig 6. Iz¢ = 391 cm4 ⎢ s = 33600 kPa max ⎣ 64 mm Fig. 6. y axis 6. 1454 kgf/cm (142588 kPa) ⎢⎣ f = 60. y 22.34 Problem 6.36). 6. Section properties: equal legs 80 mm.224 Advanced Mechanics of Solids y y¢ 12.5 Determine the maximum absolute value of the normal stress due to bending and the position of the neutral axis in the dangerous section of the beam.t.r. P = 500 kgf (4900 N). P 20 cm P 0. Calculate the stresses in members A and D. 6. C and D.5 kgf/cm (11032 kPa) ⎢⎣ f = -25°36¢ with vertical ⎡ ⎣⎢ 1. each of equal area connected by a thin web.Bending of Beams 225 6.5 m 12 cm Fig. A P L h b Fig. B.5 m 6. determine the maximum absolute value of the flexural stress and also locate the neutral axis at the section where this maximum stress occurs. 6. 6. (sx)A = -464 kgf/cm (-45503 kPa) 2 ⎢⎣ (sx)D = 448 kgf/cm (43934 kPa) .6 cm 2. as indicated.7 [Ans. The area of each stringer is 0. ⎡ ⎣⎢ 2 ⎡ Ans. 2133 kgf/cm2 ( 209175 kPa)] 6.39 shows an unsymmetrical beam section composed of four stringers A.38 Problem 6. It is assumed that the web will not carry any bending stress. Solve for the stress at A near the built-in end. b = 10 cm and L = 150 cm.6 2 ⎡ Ans.38) of length L has right triangular section and is loaded by P at the end.37. P = 200 kgf (1960 N). 112.7 A cantilever beam (Fig. The beam section is subjected to the bending moments M y and M z.37 Problem 6.6 For the cantilever shown in Fig.8 Figure 6. h = 15 cm. 6. Advanced Mechanics of Solids Mz = 10. 6. Calculate the forces F A = (s x)¢ A A¢ A = (s x )¢ A A A and FD = (sx)¢D A¢D. Now.40 is at e = 4R/p measured from point 0. what will be the bending stresses in stringers A and D? Est st = 2 ¥ 106 kgf/cm2 (196 ¥106 kPa) Emg alloy = 0. using the original areas calculate the stress as (sx)A = (sx)¢A A¢A/AA = (sx)¢A (sx)D = (sx)¢D A¢D /AD 2 ⎡ Ans.4 ¥ 106 kgf/cm2 (39. to produce the same strain. Solve the problem in the usual manner. 6. and Est Est The areas of A and B remain unaltered.39 Problem 6. ⎡ ⎣⎢ e t R o Fig. 6.8 6. if stringers C and D are made of magnesium alloy and stringers A and B of stainless steel.9 In the above problem. the areas of stringers C and D in equivalent steel will be Emag Emag AC′ = AC × AD′ = AD × .10 .40 Problem 6. Determine the stresses (sx)¢A and (s x )¢ D . (sx)A = -480 kgf/cm (-47072 kPa) ⎢⎣ (sx)D = 425.000 kgf cm 226 A 12 cm B My = 500 kgf cm 20 m C 8 cm D Fig. For example. using all equivalent steel stringers.2 ¥106 kPa) Hint: Assume once again that sections that are plane before bending remain plane after bending. Convert all the stringer areas into equivalent areas of one material.10 Show that the shear centre for the section shown in Fig.6 kgf/cm2 (41737 kPa) 6. Hence. the stress will be proportional to E. 11 6.41 Problem 6.13 For the section given in Fig. 6. In Fig. and (c).2 a.Bending of Beams 227 6.13 .41 show that the shear centre is at a distance e= R 4(sin α − α cos α ) 2α − sin 2α from the centre of curvature O of the section.12 Locate the shear centres from C.43 Problem 6.11 For the section shown in Fig.12 [Ans. 6. 6. (b) 0. (b). 6. a 2t a t t 2a 2a t a a (a) (c) a/2 (b) Fig.42(a).43. 6.42 Problem 6.705 a (c) 0.Gs for the sections shown in Fig. 6.76 a] 6. show that the shear centre is located at a distance e from O such that b1 t e x b R O t Fig. (a) 1. 6. t o p – a p – a R e Fig.42(b) the included angle is p /2. (sx)A = 3591kgf/cm2 (352310 kPa) ⎡ ⎢ ⎢ (sx)B = -1796 kgf/cm2 (-176147 kPa) ⎢ ⎢ ⎣ ⎣ 6.44 Is loaded by forces P.228 Advanced Mechanics of Solids where e= A B A = 12 + 6p ( ) b + b1 +6 b R R 2 2 3 b ⎛b ⎞ ⎛b ⎞ + 12 b 1 + 3π ⎜ 1 ⎟ − 4 ⎜ 1 ⎟ b ⎝R⎠ ⎝R⎠ R R R 2 b + b1 b ⎞ ⎛b ⎞ ⎛ + 3⎜ 1 ⎟ ⎜4 + 1 ⎟ R ⎝R⎠ ⎝ R⎠ Note: one can particularise this to the more familiar sections by putting b or b1 or both equal to zero. Develop the expression for r0 in terms of the altitude h of the triangle and R the radius of curvature of the centroidal axis. 6. ⎡ ⎤ 3h2 ⎢ Ans.700 N). 6. The web thickness is 1 cm. r0 = ⎥ 3 R 2 h + ⎡ ⎤ ⎢ 2 ⎢( 3R + 2h ) log − 3h ⎥ ⎥ 3R − h ⎣ ⎦ ⎦⎥ ⎣⎢ 6. 6. [Ans.44 Problem 6.14 The open link shown in Fig.45. Find the maximum tensile and compressive stresses in the curved end at section AB. each of which is equal to 1500 kgf (14.14 ⎡ Ans. and B = 3p + 12 P A B 3 cm P 3 cm Fig.16 Find the maximum tensile stress in the curved part of the hook shown in Fig. 3299 kgf/cm2 (328680 kPa)] . 6.15 A curved beam has an isosceles triangular section with the base of the triangle in the concave face. 18 Determine the ratio of the numerical value of smax and smin for a curved bar of rectangular cross-section in pure bending if r0 = 5 cm and h = r2 r1 = 4 cm.Bending of Beams 229 4 cm 8 cm 8 cm 2700 kgf (26. [Ans. [Ans.600 N) 45° 8c m Fig.46 Problem 6.25. Apply Castigliano’s theorem.1. [Ans. r3 = 4 cm. 6. Calculate the total strain energy.16 6. 6.17 6. 6.20 Determine the dimensions b 1 and b 3 of an I-section shown in Fig. 6.460 N) 4 cm 1 cm Fig.21 For the ring shown in Fig.19 Solve the previous problem if the bar is made of circular crosssection.89] 6. and b1 + b3 = 5 cm.76] 6. b1 = 3.31 determine the changes in the horizontal diameter. r4 = 6 cm.17 Find the maximum tensile stress in the curved part of the hook shown in Fig. r2 = 7 cm. ( ) ( ) ⎡ P ρ0 ⎧ α 1 4 1 1 ⎡ 2 1 ⎤ ⎫⎤ 2 2 ⎢ Ans. sx = 2277 kgf/cm2 (223300 kPa)] 6 cm 2000 kgf (19. 1. [Ans.67 cm. b3 = 1.45 Problem 6. δ H = A ⎨ − 2G + E π − 2 − Eeρ ⎢ 2e − ρ 0 π − 2 ⎥ ⎬ ⎥ ⎣ ⎦ ⎭⎦ 0 ⎩ ⎣ . The other dimensions are r1 = 3 cm. b2 = 1 cm. 6.33 cm] 6.46. to make smax and smin numerically equal in pure bending. Hint: Apply two horizontal fictitious forces Q along the diameter. the top face BD in Fig. To observe this. Owing to this strain.2) 7/3/2008.1(a). 7.pmd 230 lγ r (7. 7. 7. 7.1(c). 7.1) G where G is the shear modulus.1 Deformation of a thin sheet under shear stress and the resulting tube If the deformed sheet is now folded to form a tube. 7. There. then from Hooke’s law g = τ (7.CHAPTER 7 7. point D moves to D¢ [Fig. When the sheet is folded into a tube. subject to shear stress t. we were able to obtain a solution to this problem under the assumption that the cross-sections of the bar under torsion remain plane and rotate without any distortion during twist. the sides AB and CD can be joined without any discontinuity and this joined face will assume the form of a flat helix. If g is the shear strain.1 Torsion INTRODUCTION The torsion of circular shafts has been discussed in elementary strength of materials. 8:09 AM .1(b)].1(b).1(c). rotates with respect to the bottom face through an angle q* = Chapter_07. as shown in Fig. The sheet deforms through an angle g. consider the sheet shown in Fig. D t lg B D t D¢ B¢ D B t l t g A C A C (a) A C t (c) (b) Fig. as shown in Fig. such that DD¢ = lg. the shear stress at the corner of a rectangular section must be zero. . the shear stress must be tangential to the boundary.. at the boundary. We can see that both these cannot be valid for a non-circular shaft. From the above analysis we observe that for circular shafts. and there is no distortion of the section. i. Hence. it would have a component perpendicular to the boundary. if T1 is the torque on the first tube with polar moment of inertia Ip1. both boundaries are unloaded boundaries [Fig. According to this method.2(a)]. displacements ux.t varies linearly with r. if the shear stress were always perpendicular to the radius OB [Fig. This is obviously inadmissible since the surface of the shaft is unloaded and a shear stress cannot cross an unloaded boundary.4). + Tn T1 T T = = 2 = . In order to solve the torsion problem in general..1) q* = θ* or τ ⋅l G r τ = Gr l Also. Now one can stack a series of tubes.2(b)]. the cross-sections remain plane before and after. give T τ Gθ * = = Ip r l (7. Eq.5) the familiar equations from elementary strength of materials. In the case of circular shafts. the ratio r is the same for each tube. For. since the shear stresses on both the vertical faces are zero. we shall adopt St. 7. These stacked tubes can form the section of a solid (or a hollow) shaft if the top Gθ * is the same for each face of each tube has the same rotation Gq *. (7. the shear stresses are perpendicular to a radial line and vary linearly with the radius.4) where lp is the second polar moment of area. = n = 1 I pn I p1 + I p 2 + .e .5) would be valid. T= (7. Therefore. etc.e. i. But. . uy and uz are . . by the same argument. or in other words. T2 the torque on the second tube with polar moment of inertia Ip2.Torsion 231 where r is the radius of the tube. if l τ tube. then T T + T2 + . . this will no longer be valid. + I pn Ip Ip1 Ip 2 where T is the total torque on the solid (or hollow) shaft and Ip is its polar moment of inertia. Substituting for g from Eq. Further.e. 7.. for a non-circular section. one inside the other and for each tube.3) and (7. Further. the moment about the centre of the tube is T = r ¥ 2prtt or i. therefore. (7. Venant’s semi-inverse method.3) 2π r 3tτ τ I p = r r T τ = Ip r (7. Equations (7. 19)].12. (b) shear stress at the corner of a rectangular section being zero as shown in (c). Fig. we try to identify the problem for which the assumed displacements and the associated stresses are solutions. On the basis of the solution of circular shafts.18) and (2. A point P(x. Using Hooke’s law. The strains are then determined from strain-displacement relations [Eqs (2. It is assumed here that the shaft does not contain any holes parallel to the axis. The components of this displacement are ux = . the twist per unit length being q. as shown in Fig.rq z sin b uy = rq z cos b y y x qz P z O z T y (a) P¢ uy R b r qz rqz P(x. therefore. Applying the equations of equilibrium and the appropriate boundary conditions. 7. rotate through q z.3.232 Advanced Mechanics of Solids O O B B = 0 = 0 0 = (c) (b) (a) (a) Figure to show that shear stress must be tangential to boundary.2 assumed. y) in this section will undergo a displacement rq z. In Sec. 7. y) (b) b ux x (c) Fig.2 TORSION OF GENERAL PRISMATIC BARS–SOLID SECTIONS We shall now consider the torsion of prismatic bars of any cross-section twisted by couples at the ends. 7. we assume that the crosssections rotate about an axis. 7.3 Prismatic bar under torsion and geometry of deformation x . A section at distance z from the fixed end will. the stresses are then determined. multiply-connected sections will be discussed. 7. Venant’s displacement components are ux = – qyz (7. t yz = Gg yz . we can calculate the associated strain components. Substituting for sin b and cos b. t zx = Gg zx sz = where D = exx + eyy + ezz Substituting Eq. We have. This is called warping.18) and (2. exx = ∂ uy ∂ ux ∂ uz .7) u z = qy(x. γ yz = + .Torsion 233 From Fig. from Eqs (2.9) . eyy = . (7. the point P may undergo a displacement uz in z direction. y) and is independent of z. This means that warping is the same for all normal cross-sections. St.6) uy = qxz (7. 7.3(c ) y and cos b = x r r In addition to these x and y displacements.19).7) exx = eyy = ezz = gxy = 0 ⎛ ∂ψ ⎞ gyz = q ⎜ + x⎟ y ∂ ⎝ ⎠ ⎛ ∂ψ gzx = q ⎜ − ⎝∂x From Hooke’s law we have (7. ezz = ∂x ∂y ∂z gxy = ∂ u y ∂ uz ∂ ux ∂ u y ∂u ∂u + . y) sin b = y (x. γ zx = x + z ∂y ∂x ∂z ∂y ∂z ∂x From Eqs (7. y) is called the warping function.8) in the above set s x = s y = s z = t xy = 0 ⎛ ∂ψ ⎞ t yz = Gq ⎜ + x⎟ y ∂ ⎝ ⎠ (7. we assume that the z displacement is a function of only (x. From these displacement components.8) ⎞ y⎟ ⎠ sx = νE ∆ + E ε xx (1+ ν ) (1 − 2ν ) 1 +ν sy = νE ∆ + E ε yy (1 + ν ) (1 − 2ν ) 1 +ν νE ∆ + E ε zz (1 + ν ) (1 − 2ν ) 1 +ν t xy = Gg xy .6) and (7. ∂σ x ∂τ xy ∂τ zx + + =0 ∂x ∂y ∂z ∂τ xy ∂σ y ∂τ yz + + =0 ∂x ∂y ∂z (7. ny. These stress components should satisfy the equations of equilibrium given by Eq.4 Cross-section of the bar and the boundary conditions . nz) at a point on the surface [Fig. 7.e.e.12) nx t xz + ny t yz + nz s z = Fz y n(nx. Now let us consider the boundary conditions.3(b)]. then from Eq. (1.10) ∂τ zx ∂τ yz ∂σ z + + =0 ∂x ∂y ∂z Substituting the stress components. o) y Ds y s R (a) x Dy * –Dx x Ds x n * (b) Fig.e. (7.4(a)]. i. If Fx. 7. it satisfies the Laplace equation) everywhere in region R [Fig.9) nxs x + ny t xy + nz t xz = Fx nx t xy + ny s y + nz t yz = Fy (7. 7. ny. the first two equations are satisfied identically. the warping function y is harmonic (i. From the third equation. Fy and Fz are the components of the stress on a plane with outward normal n (nx.234 Advanced Mechanics of Solids ⎛ ∂ψ ⎞ t zx = Gq ⎜ − y⎟ ⎝∂x ⎠ The above stress components are the ones corresponding to the assumed displacement components.11) Hence.65). (1. we obtain ⎛ ∂ 2ψ ∂ 2ψ + Gθ ⎜⎜ 2 ∂ y2 ⎝ ∂x ⎞ ⎟⎟ = 0 ⎠ ∂ 2ψ ∂ 2ψ + = ∇2 ψ = 0 ∂ x2 ∂ y2 i. 4(b) nx = cos (n. (7. nz ∫ 0. (7. and satisfies Eq. Further. ⎛∂ψ ⎞ ⎛∂ψ ⎞ ⎜ ∂ x − y⎟ = ⎜ ∂ x − y⎟ + ⎝ ⎠ ⎝ ⎠ ⎛ ∂ 2ψ ∂ 2ψ + x ⎜⎜ 2 ∂ y2 ⎝∂x Hence.Torsion 235 In this case.9) must be equivalent to the applied torque. (7. In addition. Next. i. ds ⎛ ∂ψ ⎞ ⎜ ∂ y + x ⎟ ny = 0 ⎝ ⎠ ny = cos (n. 7. y) = – dx ds (7.e. we find that the first two equations in the boundary conditions are identically satisfied. Thus. The resultant force in x direction is ∂ψ ⎛ ⎞ ∫∫ tzx dx dy = Gq ∫∫ ⎜ ∂ x − y ⎟ dx dy (7.e. Eq. (7. each problem of torsion is reduced to the problem of finding a function y which is harmonic.14) Therefore. according to Eq. the above surface integral can be converted into a line integral. Using the stress components from Eq. x) = dy . The third equation yields ⎛ ∂ψ ⎞ Gθ ⎜ − y ⎟ nx + Gθ ⎝∂x ⎠ From Fig.11) in region R.13) Substituting ⎛∂ψ ⎞ dy ⎛ ∂ ψ ⎞ dx ⎜ ∂ x − y ⎟ ds − ⎜ ∂ y + x ⎟ ds = 0 ⎝ ⎠ ⎝ ⎠ (7.9). there are no forces acting on the boundary and the normal n to the surface is perpendicular to the z-axis.15) ⎝ ⎠ The right-hand side integrand can be written by adding — 2y as R R ⎛ ∂ 2ψ ∂ 2ψ + x ⎜⎜ 2 ∂ y2 ⎝ ∂x since — 2y = 0. satisfies Eq. (7. i. on the two end faces. the stresses as given by Eq. ⎡ ∂ψ ∂ψ ⎤ ⎛ ⎞ ⎛ ⎞ ∫∫ τ zx dx dy = Gθ Ú ⎢ x ⎜⎝ ∂ x − y ⎟⎠ nx + x ⎜⎝ ∂ y + x⎟⎠ n y ⎥ ds S ⎣ ⎦ R ⎡ ∂ψ ⎞ dy ⎛ ∂ψ ⎞ dx ⎤ = Gq Ú x ⎢ ⎛⎜ − y⎟ +⎜ + x⎟ ds ⎠ ds ⎝ ∂ y ⎠ ds ⎥⎦ S ⎣⎝ ∂ x =0 .15) becomes ⎛ ∂ψ ⎞ ⎜ ∂ x − y⎟ + ⎝ ⎠ ⎧ ⎡ ⎞ ∂ ⎟⎟ = ⎠ ∂x ∂ψ ⎞ ⎟⎟ ⎠ ⎡ ⎛ ∂ψ ⎞⎤ ⎞⎤ ∂ ⎡ ⎛ ∂ψ ⎢ x ⎜ ∂ x − y ⎟ ⎥ + ∂ y ⎢ x ⎜⎝ ∂ y + x⎟⎠ ⎥ ⎠⎦ ⎣ ⎝ ⎣ ⎦ ⎤ ⎡ ∂ψ ⎤⎫ ⎛ ⎞ ⎛ ⎞ ∂ ∂ ∫∫ τ zx dx dy = Gθ ∫∫ ⎨ ∂ x ⎢ x ⎜ ∂ x − y ⎟ ⎥ + ∂ y ⎢ x ⎜ ∂ y + x ⎟ ⎥ ⎬ dx dy ⎝ ⎠ ⎝ ⎠⎦ ⎭ ⎣ ⎦ ⎣ R R⎩ Using Gauss’ theorem. the resultant forces in x and y directions should vanish.14) on boundary s. (7.11). have to satisfy the compatibility conditions. i. For non-circular shafts.19) With this stress function (called Prandtl’s torsion stress function). The remaining stress components sx. independent of z.56). y) called the stress function. the quantity J reduces to the familiar polar moment of inertia. ezz = 0 (7. the product GJ is retained as the torsional rigidity. Based on the result of the torsion of the circular-shaft.17) The above equation shows that the torque T is proportional to the angle of twist per unit length with a proportionality constant GJ. The assumed stress components. The strain components from Hooke's law are exx = 0.236 Advanced Mechanics of Solids according to the boundary condition Eq.20) . the third condition is also satisfied. which is usually called the torsional rigidity of the shaft. In order to satisfy the equations of equilibrium we should have ∂τ yz ∂τ zx ∂τ yz ∂τ zx = 0. sz and txy are assumed to be zero. + =0 ∂z ∂x ∂y ∂z (7. we assume a function f (x. (2. the stresses are the same in every normal cross-section. sy. ∂y τ yz = − ∂φ ∂x (7. We can substitute these directly into the stress equations of compatibility. Alternatively. let the nonvanishing stress components be tzx and tyz.9) T = ∫∫ (tyz x – tzx y) dx dy R = Gθ ∂ψ ∂ψ ⎛ 2 ⎞ 2 ∫∫ ⎜ x + y + x ∂ y − y ∂ x ⎟ dx dy ⎠ R ⎝ Writing J for the integral ∂ψ ∂ψ ⎞ ⎛ −y dx dy J = ∫∫ ⎜ x 2 + y 2 + x ∂ ∂ x ⎟⎠ y R ⎝ (7. eyy = 0. if they are to be proper elasticity solutions. In order to satisfy the third condition.14). (7.e.3 ALTERNATIVE APPROACH An alternative approach proposed by Prandtl leads to a simpler boundary condition as compared to Eq. referring to Fig. the principal unknowns are the stress components rather than the displacement components as in the previous approach. 7.16) we have T = GJq (7.4(a) and Eq. = 0.18) If it is assumed that in the case of pure torsion. Similarly. (7. For a circular cross-section. In this method. such that tzx = ∂φ . then the first two conditions above are automatically satisfied. (7. 7. we can show that ∫∫ t yz dx dy = 0 R Now coming to the moment.14). we can determine the strains corresponding to the assumed stresses and then apply the strain compatibility conditions given by Eq. The first two of these are identically satisfied. a shaft having holes).13) ∂φ dy ∂φ dx + =0 ∂ y ds ∂ x ds dφ =0 (7. (7. and the moment about O should be equal to the applied torque T.21) The stress function. the resultants in x and y directions should vanish.19) ∂φ gyz = – 1 . (7. Since the stress components depend only on the differentials of f.23) For a multi-connected region R (i.e.e.e. ∂φ ∫∫ tzx dx dy = ∫∫ ∂ y dx dy R R . Next. certain additional conditions of compatibility are imposed. ∂ 2φ ∂ 2φ + = —2f = a constant F ∂ x2 ∂ y2 ∂ ∂y ⎛ ∂ 2φ ∂ 2φ ⎜⎜ 2 + ∂ y2 ⎝∂ x ⎞ ⎟⎟ = 0 ⎠ (7. we consider the boundary conditions [Eq. 7. The constant F is yet unknown. 2 2 ∂ ⎛ ∂ φ + ∂ φ ⎞ = 0. G 237 gzx = 1 tzx G Substituting from Eq. The resultant in x direction is i. On the two end faces. gyz = 1 tyz. (7. G ∂x and gzx = 1 ∂φ G ∂y From Eq. The third equation gives nx ∂φ ∂φ − ny =0 ∂y ∂x Substituting for nx and ny from Eq. ⎜⎜ 2 ⎟ ∂ x ⎝∂ x ∂ y 2 ⎟⎠ Hence. no loss of generality is involved in assuming f = 0 on s (7.56).9. the non-vanishing strain compatibility conditions are (observe that f is independent of z) ∂ ⎛ − ∂γ yz + ∂γ zx ⎞ = 0 ∂ x ⎜⎝ ∂ x ∂ y ⎟⎠ ∂ ∂y ⎛ ∂γ yz ∂γ zx ⎞ − =0 ⎜ ∂ y ⎟⎠ ⎝ ∂x i.Torsion gxy = 0. f is constant around the boundary. This will be discussed in Sec. should satisfy Poisson's equation. (2. for a simply connected region.22) ds Therefore.12)]. therefore. (7.25). (∂/∂z) (wz) is the rotation per unit length.8].21).24) Hence. ∂ 2φ ∂ 2φ + = —2f = –2Gq ∂ x2 ∂ y2 (7. all differential equations and boundary conditions are satisfied if the stress function f obeys Eqs (7. the resultant in y direction also vanishes. (7. To determine this. Similarly. (7. we have termed it as twist per unit length and denoted it by q.24).238 Advanced Mechanics of Solids = ∫ dx ∫ ∂φ dy ∂y =0 since f is constant around the boundary. we find that each integral gives ∫∫ f dx dy Thus T = 2 ∫∫ f dx dy (7. 7. y) about the z-axis [Eq. In this chapter. from Fig.4(a) T= ∫∫ (xtzy – ytzx) dx dy R ∂φ ⎞ ⎛ ∂φ +y = – ∫∫ ⎜ x dx dy ∂ ∂ x y ⎟⎠ ⎝ R =– ∂φ ∂φ ∫∫ x ∂ x dx dy − ∫∫ y ∂ y dx dy R R Integrating by parts and observing that f = 0 of the boundary. Hence. 2. we observe that half the torque is due to tzx and the other half to tyz. Thus.21). Sec. we observe from Eq.19) ∂ 2φ ∂ 2φ ∂τ zx ∂τ yz + 2 2 = ∂ y − ∂x ∂x ∂y ∂γ yz ⎞ ⎛ ∂γ = G ⎜ zx − ∂ ∂ x ⎟⎠ y ⎝ ⎡ ∂ ⎛ ∂ ux ∂ uz = G ⎢∂ y ⎜ ∂ z + ∂ x ⎝ ⎢⎣ ⎞ ∂ ⎛ ∂ u y ∂ uz ⎟−∂x⎜ ∂z + ∂y ⎠ ⎝ ⎞⎤ ⎟⎥ ⎠ ⎥⎦ ∂ ⎛ ∂ ux ∂ u y ⎞ =G ∂z ⎜ ∂y − ∂x ⎟ ⎝ ⎠ =G ∂ (–2wz) ∂z where wz is the rotation of the element at (x. But there remains an indeterminate constant in Eq.25) . Regarding the moment. (2.23) and (7. Torsion 239 According to Eq. tyz = – ∂y ∂x That is. Let s be the contour line of f = constant [Fig. (7.5(a)] along this contour tzx = y n Dy contour line f = const. the shear acting in the x direction is equal to the slope of the stress function f (x. tzn = 0 (7. x) .19).e. (7. Hence.5(b) – dy dx = cos (n. This condition may be generalised to determine the shear stress in any direction. ∂φ ∂φ . y) – tzx sin (n.26b) dy –t yz dx + τ zx =0 ds ds (7. Therefore. The resultant shearing stress is therefore t zs = tyz sin (n. tyz s n s (n. dφ =0 ds (7. 7. Consider a line of constant f in the cross-section of the bar. y) + tzx cos (n.26a) ∂φ dx ∂φ dy + =0 (7. x) = ds dn where n is the outward drawn normal. 7. 7. 7.27a) From Fig. The shear stress acting in the y direction is equal to the negative of the slope of the stress function in the x direction.27b) This means that the resultant shear at any point is along the contour line of f = constant at that point. Eq. These contour lines are called lines of shearing stress.5 Cross-section of the bar and contour lines of f i. the expression on the left-hand side is equal to tzn. y) in the y direction. as follows.26c) dx ds or dy ds From Fig.5(c). y) = dn ds dy dx = cos (n. the component of resultant shear in the direction n. x) *D s Dn * x tzx –Dx (b) (c) (a) Fig.26c) becomes tyz cos (n. x) = 0 and – (7. (7. The shear stresses are given by Eq.e. Since the fixed end has zero uz at least at one point. the boundary is a circle. From Eq. uz = q c.7). y) dy = tyz dx − τ zx dn dn =– or t zs = – ∂φ dx ∂φ dy − ∂x dn ∂y dn (7. (7. the crosssection does not warp.9) as Tx tyz = Gq x = Ip tzx = – Gq y = – Ty Ip . uz = q c = Tc GIp which is a constant.16) J = ∫∫ (x2 + y2) dx dy = Ip R the polar moment of inertia for the section. the boundary condition given by Eq. from Eq. y) are the coordinates of any point on the boundary. 7.17) T = GIpq T GI p or q= Therefore. normal to the contour line at the concerned point.29) dy − x dx = 0 ds ds 2 2 d x + y =0 ds 2 x2 + y2 = constant i.11) is y = constant = c With y = c. uz is zero at every cross-section (other than rigid body displacement).4 TORSION OF CIRCULAR AND ELLIPTICAL BARS (i) The simplest solution to the Laplace equation (Eq. 7.28) ∂φ ∂n Thus. (7. i. x) – tzx cos (n. (7. 7.e. the magnitude of the shearing stress at a point is given by the magnitude of the slope of f (x.14) becomes –y or (7. Hence. (7. Thus. discussed in Sec. From Eq. y) measured normal to the tangent line. The above points are very important in the analysis of a torsion problem by membrane analogy.240 Advanced Mechanics of Solids = tyz cos (n. Hence. where (x.7. 30) where A is a constant. Eq. the direction of the resultant shear t is such that. yields (1 + A) x2 (1 – A) y2 = constant This is of the form 2 x2 + y = 1 2 2 a b (7.e. Further 2 2 + τ zx t 2 = τ yz = or t= T 2 ( x2 + y 2 ) I p2 Tr Ip where r is the radial distance of the point (x. (ii) The next case in the order of simplicity is to assume that y = Axy (7. from Fig.6 tan a = τ zy = − Gθ x = − x τ zx Gθ y y y a r y x x Fig. all the results of the elementary analysis are justified.6 a Torsion of a circular bar Hence. This also satisfies the Laplace equation. The boundary condition. 7.Torsion 241 Therefore.14) gives.31) . Thus. (7. y). 7. the resultant shear is perpendicular to the radius. (Ay – y) dy – (Ax + x) dx = 0 ds ds dy – x (A + 1) dx = 0 ds ds or y (A – 1) i. dy (A + 1) 2x dx – (A – 1) 2y =0 ds ds or d ⎡(A + 1) x2 – (A – 1) y2 ⎤ = 0 ⎦ ds ⎣ which on integration. 33a) 2Ty π ab3 (7. one gets 4 4 J= π a 3b 3 a 2 + b2 Hence.17) T = GJq = Gq or π a 3b 3 a 2 + b2 a 2 + b2 q= T G π a3b3 (7.9) as ⎛ ∂ψ ⎞ t yz = Gq ⎜ + x⎟ ⎝ dy ⎠ =T ⎞ a 2 + b2 ⎛ b2 − a2 + 1 ⎟⎟ x 3 3 ⎜ 2 2 π a b ⎜⎝ b + a ⎠ = 2Tx π a 3b (7. is J= 2 2 2 2 ∫∫ (x + y + Ax – Ay ) dx dy R = (A + 1) ∫∫ x2 dx dy + (1 – A) 2 ∫∫ y dx dy = (A + 1) Iy + (1 – A) Ix 3 3 Substituting Ix = π ab and Iy = π a b .33b) and similarly.32) The shearing stresses are given by Eq. as given in Eq. the function y= b2 − a 2 b2 + a2 xy represents the warping function for an elliptic cylinder with semi-axes a and b under torsion. (7. from Eq. (7. (7. t zx = .242 Advanced Mechanics of Solids These two are identical if a2 = 1 − A b2 1 + A or b2 − a 2 A= b2 + a2 Therefore.16). The value of J. where uz is positive.35) This satisfies the Laplace equation. the warping is prevented at that end and consequently.7) t max = 2T π a 3b 3 uz = qy = T (b 2 − a 2 ) π a 3b 3G xy depressed uz negative The contour lines giving uz = constant are the hyperbolas shown in Fig. are shown by dotbar and contour lines ted lines. the displacement uz can easily be determined. the maximum shear stress occurs when y is maximum. i. y) is 1/ 2 2T ⎡b 4 x 2 + a 4 y 2 ⎤1/ 2 (7. However. 2 2 a b giving t= 2T π a 3b 3 or = ⎛ y2 ⎞ x2 = a2 ⎜⎜1 − 2 ⎟⎟ b ⎠ ⎝ [a2b4 + a2 (a2 – b2) y2]1/2 Since all terms under the radical (power 1/2) are positive.14) yields (–6Axy – y) dy – (3Ay2 – 3Ax2 + x) dx = 0 ds ds .7 Cross-section of an elliptical face is depressed. there of uz are no normal stresses. therefore. we substitute for x from 2 2 ⎤ + τ zx t = ⎡⎣τ yz ⎦ 2 x 2 + y = 1.5 TORSION OF EQUILATERAL TRIANGULAR BAR Consider the warping function y = A(y3 – 3x2y) (7. are uz positive a indicated by solid lines. when y = b. and the concave portions or where the surFig.Torsion 243 The resultant shearing stress at any point (x. Thus. 7.34) π ab With the warping function known. which can easily be verified. We have from Eq. (7. tmax occurs at the ends of the minor axis and its value is (a 4 b 2 )1/ 2 = 2T 2 (7. 7. i. The boundary condition given by Eq. normal stresses are induced which are positive in one quadrant and negative in another. For a torque T as shown. b 7. if one end is built-in.e. (7.e.7. If the ends are free.33c) ⎦ π a 3b 3 ⎣ 2 To determine where the maximum shear stress occurs. These are similar to bending stresses and are. called the bending stresses induced because of torsion. the convex portions of the crosselevated section. (7. 7. 7. where b is a constant.244 Advanced Mechanics of Solids dy + (3Ay2 – 3Ax2 + x) dx = 0 ds ds or y(6Ax + 1) i.36) – 1 (3xy2 – x3) + 1 (x2 + y2) – 2 a2 = 0 2 6a 3 (x – 3 y + 2a) (x + or 3 y + 2a) (x – a) = 0 (7. The equations of the boundary lines are C y 3a B x o 2a a D Fig.36) becomes A(3xy2 – x3) + (7.16) J= 2 2 2 2 ∫∫ ⎡⎣ x + y + Ax ( 3 y − 3x ) − Ay ( −6 xy ) ⎤⎦ dx dy R = ∫0 3a a dy ∫ − a + ∫− 3 y − 2a a 3a dy ∫ − = 9 3 a 4 = 3 Ip 5 5 ( ) ⎡ x 2 + y 2 + Ax 3 y 2 − 3 x 2 − Ay ( −6 xy )⎤ dx ⎣ ⎦ 3 y − 2a ( ) ⎡ x 2 + y 2 + Ax 3 y 2 − 3x 2 − Ay ( −6 xy ) ⎤ dx ⎣ ⎦ (7. 1 2 1 2 x + y =b 2 2 2a 2 . (7.e.37) is the product of the three equations of the sides of the triangle shown in Fig. d 3 Axy 2 − Ax3 + 1 x 2 + 1 y 2 = 0 2 2 ds ( ) Therefore.8.8 Cross-section of a triangular bar and plot of tyz along x-axis x–a=0 on CD x– 3 y + 2a = 0 on BC x+ 3 y + 2a = 0 on BD From Eq. If we put A = – 1 and b = + 3 6a Eq.38) .37) Equation (7. 42) 2 At the corners of the triangle. y C y T B convex 2b x o 2a x concave D 2a (a) A 2a T (b) Fig. as shown in Fig. Along the x-axis.9(a).6 TORSION OF RECTANGULAR BARS The torsion problem of rectangular bars is a bit more involved compared to those of elliptical and triangular bars. tzx = 0 and the variation of tyz is shown in Fig.41) ( x − a) a The largest shear stress occurs at the middle of the sides of the triangle. the shear stresses are zero. 7. The stress components are ⎛ ∂ψ ⎞ t yz = Gθ ⎜ + x⎟ y ∂ ⎝ ⎠ ( = Gθ 3 Ay 2 − 3 Ax 2 + x ( = Gθ x 2 − y 2 + 2ax 2a ) ) (7. tyz is also zero at the origin 0. with a value = 3Gθ a (7. 7.9 (a) Cross-section of a rectangular bar (b) Warping of a square section .8. Let the sides of the rectangular cross-section be 2a and 2b with the origin at the centre. tmax = 7. We shall indicate only the method of approach without going into the details.Torsion 245 Therefore.40) ⎛ ∂ψ ⎞ t zx = Gθ ⎜ − y⎟ ⎝∂ y ⎠ and Gθ y (7. q= T =5 T GJ 3 GI P (7. 7.39) Ip is the polar moment of inertia about 0. the governing equation is ∂ 2ψ 1 ∂ 2ψ 1 + =0 ∂ x2 ∂ y2 over region R. Assuming that b > a.246 Advanced Mechanics of Solids Our equations are.1. such that ψ = xy − ψ 1 In terms of y1. The final results which are important are as follows: The function J is given by J = Ka3b For various b/a ratios. Substitution into the Laplace equation for y1 yields two linear ordinary differential equations with constant coefficients. we have nx = 0 and ny = ±1. we have nx = ±1 and ny = 0. and the boundary conditions become ∂ψ 1 = 0 ∂x ∂ψ 1 = 2x ∂y on on x = ±a y = ±b It is assumed that the solution is expressed in the form of infinite series ∞ y = ∑ X n ( x ) Yn ( y ) n=0 where Xn and Yn are respectively functions of x alone and y alone. Hence. ∂ 2ψ ∂ 2ψ + =0 ∂ x2 ∂ y2 over the whole region R of the rectangle. Now on the boundary lines x = ±a or AB and CD. as before. the boundary conditions become ∂ψ = y ∂x ∂ψ = –x ∂y on x = ±a on y = ±b These boundary conditions can be transformed into more convenient forms if we introduce a new function y1. the corresponding values of K are given in Table 7. and ⎛ ∂ψ ⎞ ⎛ ∂ψ ⎞ ⎜ ∂ x − y ⎟ nx + ⎜ ∂ y + x ⎟ n y = 0 ⎝ ⎠ ⎝ ⎠ on the boundary. On the boundary lines BC and AD. Further details of the solution can be obtained by referring to books on theory of elasticity. it is shown in the detailed analysis that the maximum . This formula is applicable to a large number of squatty sections with an error not exceeding 10%.860 1.0 5. as given in Table 7.0 2.000 0.401 0.0 • 2.936 1.136 3.994 1.970 1. The zones where uz is positive are shown by solid lines and the zones where uz is negative are shown by dotted lines. Hence. . 7.664 3.250 2. If 4p 2 is replaced by 40. On these sides tmax = K1 Ta J The values of K1 for various values of b/a are given in Table 7. the warping is as shown in Fig. which is applicable to an elliptical section.656 4.484 0.696 1.430 0.1.32).9 (b).496 4.1. can be written as T = π a 3b3 G = 1 GA4 θ a 2 + b2 4 π 2 Ip (a 2 + b 2 ) A is the polar moment 4 of inertia.350 1.443 0. Empirical Formula for Squatty Sections Equation (7.11).656 3.2 1.468 0.5 2.000 2. and Ip = T = GA4 θ 40 I p is an approximate formula that can be applied to many sections other than elongated or narrow sections (see Secs 7.518 1.5 3.998 2.375 shearing stress is at the mid-points of the long sides x = ±a of the rectangle.e.508 0.0 10.992 5. the mean error becomes less than 8% for many sections.541 0.1 b/a K K1 K2 1 1.Torsion 247 Table 7.984 4. Substituting for J.571 0. i.328 1. where A = pab is the area of the ellipse. the above expression can be written as tzx = 0 tmax = K2 and Ta a 2b where K2 is another numerical factor.600 0. b/a = 1.208 4.0 4. For a square section.10 and 7. 7. Let F be the uniform tension per unit length of the membrane.10 Stretching of a membrane When the membrane is subjected to a uniform lateral pressure p. it undergoes a small displacement z where z is a function of x and y. the component of FDy in z ( ) direction is − F ∆ y ∂z since sin b ª tan ª b for small values of b.10). The force on face ∂x BC is also F Dy but is inclined at an angle (b + Db) to the x-axis.248 7. the force acting is FDy. In such situations. analytical solutions tend to become more involved and difficult. Its slope is therefore ( ) ∂z + ∂ ∂z ∆ x ∂x ∂x ∂x and the component of the force in z direction is ( ) ∂ z + ∂ ∂ z ∆ x⎤ FDy ⎡⎢ ⎥⎦ ⎣∂ x ∂ x ∂ x Similarly. On face AD. 7. The value of the initial tension F is large enough to ignore its change when the membrane is blown up by the small pressure p. Hence.7 Advanced Mechanics of Solids MEMBRANE ANALOGY From the examples worked out in the previous sections. Consider the equilibrium of an infinitesimal element ABCD of the membrane after deformation. it becomes evident that for bars with more complicated cross-sectional shapes. tan b is the slope of the face AB and is equal to ∂ z/∂ x. the components of the forces FDy acting on faces AB and CD are – FD x ∂ z and F ∆ x ∂y ⎡∂z ⎤ ∂ ⎛ ∂z⎞ ⎢ ∂ y + ∂ y ⎜⎝ ∂ y ⎟⎠ ∆ y ⎥ ⎣ ⎦ . Let a thin homogeneous membrane like a thin rubber sheet be stretched with uniform tension and fixed at its edge. This is inclined at an angle b to the x-axis. it is desirable to resort to other techniques—experimental or otherwise. which is a given curve (the cross-section of the shaft) in the xy-plane (Fig. The membrane analogy introduced by Prandtl has proved very valuable in this regard. z F C D x F Dx B A y F F b + Db F Dy x F B b A Z + z (a) ∂z ∆x ∂x (b) Fig. as shown. assuming that the membrane deflection is small. The twisting moment is numerically equivalent to twice the volume under the membrane [Eq. 7. (7.43) of the membrane becomes identical to Eq. the resultant force in z direction due to tension F is 2 ⎡ ⎤ − F ∆ y ∂ z + F ∆y ⎢ ∂ z + ∂ z2 ∆ x ⎥ − F ∆ x ∂ z ∂x ∂ ∂y x ∂ x ⎣ ⎦ ⎡∂ z ∂2z ⎤ + ∆y ⎥ + F ∆x ⎢ 2 ∂ y ∂y ⎣ ⎦ ⎛ ∂2z ∂2z ⎞ = F ⎜ 2 + 2 ⎟ ∆ x ∆y ∂y ⎠ ⎝∂x The force p acting upward on the membrane element ABCD is p D x D y. then Eq. the shear stresses will be essentially parallel to the boundary. (7.23)].8 TORSION OF THIN-WALLED TUBES Consider a thin-walled tube subjected to torsion. if we adjust the membrane tension F or the air pressure p such that p/F becomes numerically equal to 2Gq. The slopes of the membrane are then equal to the shear stresses and these are in a direction perpendicular to that of the slope.11 Torsion of a thin-walled tube D t2 .Torsion 249 Therefore.43) Now. The shear stresses (complementary shears) are t1 and t2. 7. Since the thickness is small and the boundaries are free. (7. Consider the equilibrium of an element of length Dl. then the height z of the membrane becomes numerically equal to the torsion stress function [Eq. For equilibrium in z direction we should have −τ1 t1 ∆l + τ 2 t2 ∆l = 0 C Dl B A D t1 A B t1 Z C Fig.25) of the torsion stress function f.24)]. Further. The areas of cut faces AB and CD are respectively t1 Dl and t2 Dl. 7. if the membrane height z remains zero at the boundary contour of the section. Let t be the magnitude of the shear stress and t the thickness. therefore 2 ⎞ ⎛ 2 F ⎜⎜ ∂ 2z + ∂ z2 ⎟⎟ = − p ∂y ⎠ ⎝∂ x or ∂ 2z + ∂ 2z = − p F ∂ x2 ∂ y 2 (7. (7. The thickness of the tube need not be uniform (Fig.11). For equilibrium. 47) . 7.45) is generally known as the Bredt–Batho formula. This is called the shear flow q.250 or Advanced Mechanics of Solids τ1 t1 = τ 2 t2 = q.12(b). 7. DF = q Ds h t d Ds Dl O (b) (a) Fig. we should have V1A1 = V2A2. the total torque is T = Σ 2q ∆ A = 2qA (7. Because of shear strain g . the torque is DT = q ∆ sh = 2q ∆ A where DA is the area of the triangle enclosed at O by the base s. To determine the twist of the tube. Equation (7.46) = T 2 ∆l ∆ s 8 A2 G t (7. since the equation is similar to the flow of an incompressible liquid in a tube of varying area. Consider next the torque of the shear about point O [Fig. the shear force on the element is t t Ds = q Ds. where A is the area and V the corresponding velocity of the fluid there. Hence. For continuity.44) Hence. 7. the force does work equal to DU = 1 (τ t ∆ s ) δ 2 = 1 (τ t ∆ s ) γ ∆l 2 = 1 τ (τ t ∆ s ) ∆ l 2 G = q 2 ∆l ∆ s 2G t (7. Referring to Fig. a constant (7. we make use of Castigliano’s theorem.12(a)].45) Where A is the area enclosed by the centre line of the tube. the quantity t t is a constant.12 Cross-section of a thin-walled tube and torque due to shear The force acting on an elementary length Ds of the tube is DF = τ t ∆ s = q ∆ s The moment arm about O is h and hence. 14 Section of a thin-walled multiple cell beam and moment axis Chapter_07.13 Torsion of a thin-walled multiple cell closed section Consider the equilibrium of an element at the junction. (7. (7.51) This is again equivalent to a fluid flow dividing itself into two streams.Torsion 251 using Eq.13(b). the twist or the rotation per unit length (Dl = 1) is ∂U ds = T ∂ T 4 A2G ∫ t Using once again Eq.. 7.50) TORSION OF THIN-WALLED MULTIPLE-CELL CLOSED SECTIONS We can extend the analysis of the previous section to torsion of multiple-cell sections. In the direction of the axis of the tube − τ1 t1 ∆l + τ 2t2 ∆l + τ 3t3 ∆l = 0 or q2 Cell 2 A2 q2 Cell 1 q1 A1 q1 O Fig. 7. Choose any moment axis. Consider the two-cell section shown in Fig. Dl Dl q1 = t1t1 A1 q3 A2 q2 = t2t2 t1 t 2t 1 t3 t2 t3 (a) (b) Fig.45) q= q= 7.9 q 2 AG (7. 7.45). (7.13. The moment about O due to q1 flowing in cell 1 (with web included) is [Eq. 7. as shown in Fig.e. q1 = q2 + q3 (7. 7.45)] T1 = 2q1 A1 7/3/2008.pmd 251 t 1t1 = τ 2 t2 + τ 3t3 i.48) Hence. since q3 = q1 – q2.14). The shear flow in the web can be considered to be made up of q1 and –q2.49) ds ∫ t (7. The total elastic strain energy is therefore U= T 2 ∆l 8 A2G ds ∫ t (7. such as point O (Fig. 5:56 AM . 252 Advanced Mechanics of Solids where A1 is the area of cell 1. Hence. with A1* as the area enclosed at O outside cell 2.50). 5:57 AM .15 Example 7. 7. the twist of each cell should be the same. Next. determine the shear flows and the angle of twist of the member per unit length.1 Figure 7.52) A1 and A2 are the areas of cells 1 and 2 respectively. the moment about O due to q2 flowing in cell 2 (with web included). the twist of each cell is given by q ds 2Gq = 1 Ú t A Let a 1 = Ú ds for cell 1 including the web t a 2 = Ú ds for cell 2 including the web t a 12 = Ú ds for the web t Then. For continuity. t a 2t t t /2 a t /2 2 1 t a Fig. Similarly. If the member is subjected to a torque T. Example 7.52)–(7.53) For cell 2 2Gq = 1 (a q − a q ) A2 2 2 12 1 (7.15 shows a two-cell tubular section whose wall thicknesses are as shown. we shall consider the twist. the total torque is T = T1 + T2 = 2q1 A1 + 2q2 A2 (7.54) Equations (7.pmd 252 7/3/2008.1 Chapter_07. for cell 1 2Gq = 1 (a q − a12 q2 ) A1 1 1 (7. is T2 = 2q2 ( A2 + A1* ) − 2q2 A1* The second term with the negative sign on the right-hand side is the moment due to the shear flow q2 in the middle web. (7.54) are sufficient to solve for q1. According to Eq. q2 and q. ds a a a 2a 5a Ú t = t + t + t + t = t = a2 For web. ) 1 (a q − a12 q1 ) A2 2 2 ( ) 1 5a a 1 q2 − q1 = (5q2 − q1 ) 2 t t at a 7 q − q = 5q − q 2 2 1 2 1 or From Eq. 7.16 shows a two-cell tubular section as formed by a conventional airfoil shape. 2Gq = = Equating. 2Gq = 1 (a q − a12 q2 ) A1 1 1 ( ) ( = 12 7a q1 − a q2 = 1 7 q1 − q2 t at 2 a 2t For cell 2.2 Figure 7.Torsion 253 Solution For cell 1. and having one interior web. An external torque of 10000 Nm (102040 kgf cm) is acting in a clockwise direction. ds a a a a 7a Ú t = t + 2t + t + t = 2t = a1 For cell 2. .53) For cell 1. Determine the internal shear flow distribution.52) ( q2 = 3 q1 4 ) T = 2q1 A1 + 2q2 A2 = 2a 2 q1 + 3 q1 = 7 a 2 q1 4 2 \ q1 = 2Gq = = or 2T 7a 2 and q2 = 3T 2 14a 1 (5q2 − q1 ) at ( ) 1 15 − 1 q = 11 q 1 at 4 4at 1 ( ) 1 11 ⎛ 2T ⎞ 2G 4at ⎜⎝ 7a 2 ⎟⎠ 11 T = 28 a3tG q= Example 7. ds a Ú t = t = a12 From Eq. (7. (7.16. The cell areas are as follows: A1 = 680 cm2 A2 = 2000 cm2 The peripheral lengths are indicated in Fig. 254 Advanced Mechanics of Solids 0.36 q1 The torque due to shear flows should be equal to the applied torque.09 For cell 2.20 q2 – 0.18 q1 Hence.08 33 = 366 For web.54) 1 (a q − a12 q2 ) For cell 1.09 cm Cell 1 s = 33 cm 3 cm s = 6 s = 67 cm q2 0.20 q2 – 0. a 2 = 33 + 63 + 48 + 67 = 2409 0.09 cm 0. Hence.16 Example 7. a 1 = 67 + 33 = 1483 0.18 q1 or.19 q1 – 0.189 q1 − 0. 1 (2409 q − 366 q ) 2 1 2000 = 1.09 From Eqs (7.2 Solution Let us calculate the line integrals Ú ds/t. 7.09 0. 2. 2Gq = A1 1 1 For cell 1. equating the above two values 2.e. 1 (1483q − 366 q ) 1 2 680 = 2.09 0. (7.74 q2 = 0 i.54 q2 1 (a q − a q ) 2Gq = 12 1 A2 2 2 = For cell 2.53) and (7. from Eq. a 12 = 0.52) T = 2q1A1 + 2q2A2 = or 10000 ¥ 100 = 2q1 ¥ 680 + 2q2 ¥ 2000 = 1360 q1 + 4000 q2 Substituting for q2 106 = 1360q1 + 5440q1 = 6800q1 .09 0.06 cm q1 Cell 2 s = 48 cm cm 67 = s 0.37 q1 – 1.54 q2 = 1.06 0.09 cm 0.08 cm Fig. q 2 = 1. one has f = 0 at y = ± t/2.56a) ∂y These shears are shown in Fig. the above equation becomes Fig. B A Let f = 0.Torsion kgf ⎞ ⎛ q 1 = 147 N ⎜ 15. Substituting these a2 = a 1 = 0. the above equations are not valid near the ends. and (tzx)max = ± Gq t (7. we can take f (x. y) = f (y). Obviously. x t From Eq.17 shows the section of a rectangular bar subjected to a torque T. Hence.55) From Eq. 7. 7. and Gθ t 2 4 f = Gq ⎛ t − y2 ⎞ ⎜ ⎟ 2 ⎝4 ⎠ (7. (7. (7.17 Torsion of a thin rectangular bar ∂ 2φ = –2Gq ∂ y2 Integrating f = −Gθ y 2 + a1 y + a2 Since f = 0 around the boundary.24).01 cm ⎝ cm ⎟⎠ 7. Therefore.25) D C b ∂ 2φ ∂ 2φ + = –2Gq ∂ x2 ∂ y2 Excepting at the ends AD and BC. Let the thickness t be small compared to the width b. (7. the stress function is fairly uniform and is independent of x.17.4 ⎜ cm ⎝ cm ⎟⎠ TORSION OF BARS WITH THIN RECTANGULAR SECTIONS Figure 7. The maximum shearing stresses are at the surfaces y = ± t/2.19) ∂φ = 0 ∂x ∂φ and tzx = = –2Gq y (7.10 q2 = 200 255 N ⎛ kgf ⎞ 20. T = 2 ∫∫ φ dx dy 2 ⎞ b/2 t/2 ⎛ t = 2Gθ ∫− b / 2 dx ∫− t / 2 ⎜ − y 2 ⎟ dy ⎝4 ⎠ . The section consists of only one boundary and the value of y the stress function f around this boundary is constant.56b) tyz = – From Eq. so that the section becomes an angle. This section has only one boundary with f = constant = 0.56) can be applied to any narrow cross-section which has a relatively small curvature. If the thickness changes. G bt 3 τ zx = − 6T3 y.57) become ⎛ t2 ⎞ f = Gq ⎜ i − y 2 ⎟ ⎝4 ⎠ (i = 1. Eqs (7.55) and (7. the length b = b1 + b2 if the thickness is uniform.17.256 Advanced Mechanics of Solids T = 1 bt 3 Gθ 3 or (7.18(b). provided b is taken as the total length of both legs of the angle concerned and y is the rectangular coordinate in the direction of the local thickness. for the thin rectangular section. we imagine a 90∞ bend in the middle of the rectangle shown in Fig. b1 t t b t b b2 t t t (a) (b) (c) b1 t1 b2 t2 t3 b3 (d) Fig. 7. as shown in Figs 7.17. 7. Hence.18(a)–(d). the shape across the thickness will be similar to that shown in Fig.57) The results are q= 7.11 1 3T .18 Torsion of rolled sections In the case of a T-section shown in Fig. Excepting for the local effects near the corner.17(d). 7. To see this. as shown in Fig. 7. Eqs (7. 2 or 3) .58) TORSION OF ROLLED SECTIONS The argument leading to the approximations given by Eqs (7.57) can be applied.55) and (7.55) and (7. (τ zx ) max = ± 3T2 bt bt (7. 7. 3—Torsion of a closed tubular section and a slit tubular section Solution For the closed tube. the force on the rivets will be Tl F = τ tl = 2π r 2 . 7.3 Analyze the torsion of a closed tubular section and the torsion of a tube of the same radius and thickness but with a longitudinal slit. as shown in Fig. 7. If the slit tube is riveted along the length to form a closed tube of length l. we have from elementary analysis T = (2π rτ t ). as shown in Figs 7. following the same analysis as for a thin rectangular section tmax = ∓ Gθ t The shear stress directions in the slit tube are shown in Fig.Torsion T = 1 Gθ (b1t13 + b2 t23 + b3t33 ) 3 This is obtained by adding the effect of each rectangular piece. 7.59) Example 7. and 257 (7. According to Eq.19 Example 7. The ratio of the torsional rigidities is T= T1 = (2π r 3tGθ ) 1 (2π rt 3Gθ ) 3 T2 () = 3 r t 2 For a thin tube with r/t = 10. there is only one boundary and on this f = 0.19(b). y T y n r x x (a) (b) (c) Fig.57) 1 bt 3Gθ = 1 2π rt 3Gθ 3 3 Further. if t is the shear stress. tube (a) is 300 times as stiff as tube (b). (7. r = 2π r 2tτ and θ= τ Gr 3 Therefore.19(c).19(a) and (b). T = 2π r tGθ For the slit tube. 75 + 28.54). the section becomes a two-cell structure for which we can apply Eqs (7.20 Example 7. (7.75 1.25 cm thick. 1.5t ) .25 cm thick flat 30 cm plates are welded onto the sides of the section.75 + 28. as shown by dotted lines.58) (∑ b t ) 2 i i tmax = 3T = 3 × 50 000 30 × (5/4) + 30 × (5/4) 2 + (30 − 2.4 Solution (i) Using Eq.52)–(7.258 Advanced Mechanics of Solids where for t we have put the value T 2π r 2 t as for a non-slit tube.5) × (5/4)3 ×1 G = 878/G radians per cm length (ii) When the two side plates are welded.00 T = 2q1 A1 + 2q2 A2 or = 4q1 A1 = 4τ × 1 × 28.25 2 T = 1322. 7. For each cell ( a 1 = a2 = Ú ds = 1 28. is subjected to a torque T = 50000 kgfcm (4900 Nm).20). Find the maximum shear stress and the angle of twist per 30 cm unit length. with flanges and with a web 1. t= Example 7.75 × 28.75 t 1.cm I beam (Fig. 1. Fig.25 2 2 = 69. Find the maximum shear stress and the angle of twist.75 + 28. then the average force on each rivet is F/n.4 (i) A 30.25 cm (ii) In order to reduce the stress and the angle of twist. If there are n rivets in a length l. 7.5) × (5/4)2 2 = 1097 kgf/cm2 (107512 kPa) q= = 1 3T G ∑ b t3 i i ( ) 3 × 50 000 30 × (5/4)3 + 30 × (5/4)3 + (30 − 2. pmd ∂φ gyz = 1 τ yz = − 1 G G ∂x 259 7/3/2008.2 and 7. ∂z ∂τ zx ∂τ y z =0 + ∂x ∂y ∂τ z x = 0. 7. Figure 7.75 28. It is easy to extend the same analysis for the solution of shafts. 7. The equations of equilibrium yield ∂τ y z = 0.21 Torsion of multiply-connected sections Once again.5 = 37. as in Sec.06/G radians per cm length 7.3. t = 50000/1322. 5:58 AM . The holes have boundaries C1 and C2. we considered the torsion of shafts with sections which do not have holes. such that tzx = ∂φ . ∂y tyz = – ∂φ ∂x The non-vanishing strain components are ∂φ gzx = 1 τ zx = 1 G G ∂y and Chapter_07. n C1 A1 C0 C2 n T Fig. the cross-sections of which contain one or more holes. we assume that tyz and tzx are the only non-vanishing stress components.21 shows the section of a shaft subjected to a torque T.81 kgf /cm2 (3705 kPa) 2Gq = = 1 (a q − a12 q2 ) A1 1 1 1 q (a − a12 ) A1 1 1 Therefore ( q= 1 × 2 × 1 69 − 28.Torsion 259 Therefore.12 MULTIPLY CONNECTED SECTIONS In Sec.25 ) = 0.75 2G 28. y) be a stress function. ∂z Let f (x.75 1. 7.3. i.12).22(a).61) i.e.7).e. say on C0. (2. We can assume that f = 0 on one boundary. dφ =0 ds f for Ci = Ki (7.60) So far. and then determine the corresponding values of Ki on each of the remaining boundaries Ci. the value of dy integrated around any closed contour Ci should be equal to zero. Ú Ci y dx = ∫ GKH y dx + ∫HLG y dx = area G′GKHH ′ − area H ′ HLGG′ = area enclosed by Ci = Ai . (7. on each boundary f is a constant.3. 7.63) Hence.e. To do this. y). we observe that the displacement of the section in z direction.e. uz = qy (x.2 should be satisfied.13) ∂φ dy ∂φ dx =0 + ∂ y ds ∂ x ds or i.260 Advanced Mechanics of Solids The compatibility conditions given by Eq. we should have nx ∂φ ∂φ =0 − ny ∂y ∂x Substituting for nx and ny from Eq. we observe that there are several boundaries and on each boundary the conditions given by Eq. Sec. (7.62) From Eq. the analysis is identical to that given in Sec.9). (7. Considering the boundary conditions. Since each boundary is a free boundary. i. for the single valuedness of uz ∂φ ⎞ ⎛ ∂φ 1 dx − dy + ( y dx − x dy ) = 0 Ú Ci ⎜ ∂ x ⎟⎠ Ú Ci ⎝∂y Gθ (7.56) yield ∂ 2φ ∂ 2φ + = —2f = a constant F 2 2 ∂x ∂y (7. ∂ψ ⎞ ⎛ ∂ψ Ú Ci dψ = Ú Ci ⎜⎝ ∂ x dx + ∂ y dy⎟⎠ = 0 (7. must be single valued. Consequently. 7. and using the stress function ∂φ ∂ψ = 1 τ zx + y = 1 +y θ θ ∂ G G y ∂x and ∂ψ ∂φ = 1 τ yz − x = − 1 −x ∂y Gθ Gθ ∂ x (7. Unlike the case where the section did not contain holes. from Eq. we cannot assume that f = 0 on each boundary. This can be seen from Fig. (7. 7.64) The second integral on the left-hand side is equal to twice the area enclosed by the contour Ci. 66b) where n is the outward drawn normal to the boundary Ci.65) ∂φ ⎞ ⎛ ∂φ ⎛ ∂φ dx ∂φ dy ⎞ Ú Ci ⎜⎝ ∂ y dx − ∂ x dy⎟⎠ = Ú Ci ⎜⎝ ∂ y ds − ∂ x ds ⎟⎠ ds (7.22(b) ⎛ ∂φ dy ∂φ dx ⎞ = – Ú Ci ⎜ ds + ⎝ ∂ y dn ∂ x dn ⎟⎠ = –Ú C i ∂φ ds ∂n (7. Ú C ( y dx − x dy ) = 2Ai i The first integral in Eq. (7.22 Evaluation of the integral around contour Ci Similarly. 7. Using Eq. Therefore. (7.3 remain unaltered. Ú x dy = ∫LGK x dy + ∫KHL x dy = area L′ LGKK ′ − area K ′ KHLL′ = –Ai Therefore.16) can be obtained for a multiply-connected body in terms of the stress function f. as follows.24) and (7.63) J= = ∂ψ ∂ψ ⎛ ⎞ ∫∫ ⎜ x 2 + y 2 + x ∂ y − y ∂ x ⎟ dx dy ⎠ R ⎝ ∂φ ∂φ ⎞ x − x2 − 1 y − y 2 ⎟ dx dy ∫∫ ⎛⎜ x 2 + y 2 − 1 R ⎝ Gθ ∂ x Gθ ∂ y ⎠ .67) on each boundary Ci.66a) and from Fig. Eqs (7. (7. 7. (7. 7.64) becomes Ú Ci ∂φ ds = 2GqAi ∂n (7. i.e. Eq.64) can be written as (7. Ai is the area enclosed by Ci .25) are Torque T = 2 ∫∫ φ dx dy R 2 and 2 ∂ φ ∂ φ + = ∇ 2φ = − 2Gθ 2 2 ∂x ∂y The value of J defined in Eq.Torsion y K K¢ Ci G n Dn H L¢ L o 261 H¢ G¢ Dy Dy Ds –Dx x (b) (a) Fig. The remaining equations of Sec. 1.. K2. (7. the above equation reduces to Eq. (7.61). .68) T = GJθ = 2 ( ∫∫ φ dx dy + ∑ K i Ai ) (7. Example 7. . . Since we have chosen f to be zero over the boundary C0 J = 2 φ dx dy + 1 1 v∫ C1 ( y dx − x dy ) K1 + Gθ v∫ C2 ( y dx − x dy ) K 2 + . Hence. assumes the form J= (7. . And from Eq. . (7.69) For a solid shaft with no holes.) Gθ R Gθ where Ai is the area enclosed by curve Ci. 2. C2 .17).21.28).5 . are the values of f on C1. 7.) shown in Fig. . therefore.69). . .. . 2 ⎡ φ dx dy + ∑ K A ⎤ i i⎦ Gθ ⎣ ∫∫ Equation (7. using equations (7.5 Analyse the torsion problem of a thin-walled. 7. . Put contour S1 n n contour S2 C2 n cell 2 K ∂φ =− 2 t ∂n cell 1 t f = K2 C0 C1 f = K1 t f=0 f=0 Fig.. (7. Assume uniform thickness t.262 Advanced Mechanics of Solids =– 1 Gθ = 1 Gθ ∂φ ∂φ ⎛ ∫∫ ⎜ ∂ x x + ∂ y R⎝ ∂ ⎡ R⎣ ⎞ y ⎟ dx dy ⎠ ∂ ⎤ ⎦ ∫∫ ⎢ 2φ − ∂ x ( xφ ) − ∂ y ( yφ )⎥ dx dy = 2 ∫∫ φ dx dy + 1 v∫ C φ ( y dx − x dy ) Gθ R Gθ where we have made use of Gauss’ theorem and the subscript C on the line integral means that the integration is to be performed in appropriate directions over all the contours Ci (i = 0.23.24). Solution Consider the two-celled section shown in Fig. . Gθ ∫∫ G θ R where K1. the stress function f is constant around each boundary. 7. According to Eq.9) J = 2 ∫∫ φ dx dy + 1 ( 2 K1 A1 + 2 K 2 A2 + .56) and (7. . multiple-cell closed section.23 Example 7. e. 7. K2 and q.67).6 Using equations (7. prove that the shear flow is constant for a thin-walled tube (shown in Fig. (7. From Eq. Solution Let S be the contour of the centre line and t s the thickness at any section. Since the thickness t is small.72) = = τ2 ∂n t t From Eq.73)–(7. Example 7. for cell 2 t2(S2 – S12) – t12S12 = 2GqA2 or K2 K −K ( S2 − S12 ) + 2 t 1 S12 = 2GqA2 t i.28) and (7. K2 S2 S − K1 12 = 2GqA2 t t (7. the ∂φ resultant shear stress is given by tzs = – where n is the normal to the contour ∂n of f . (7. i. Further. According to Eq.Torsion 263 f = 0 on boundary C0. for cell 1 t1(S1 – S12) + t12S12 = 2GqA1 where S1 is the peripheral length of cell 1 including the web.24) subjected to torsion.75) will enable us to solve for K1. the integral on the right-hand side can be omitted. − ∂φ 0 − K1 K1 = − = = τ1 ∂n t t (7.70) For the web contour S12 K − K1 ∂φ = − 2 = τ12 ∂n t and for contour S2 of cell 2 − (7.28).69) T = 2 ( K1 A1 + K 2 A2 ) (7. Substituting for t1 and t12 − K1 K − K1 S1 − S12 ) − 2 S12 = 2GqA1 ( t t K1 or S1 S − K 2 12 = 2GqA1 t t (7. the stress function f is constant around . With this From Eq. (7.71) 0 − K 2 K2 ∂φ = − (7. (7. f = K1 on C1 and f = K2 on C2.73) Similarly.61). and S12 the length of the web. the line of shear stress.74) T = 2 ( ∫∫ φ dx dy + ∑ i K i Ai ) Compared to Ai.75) Equations (7. since f = 0 on C0 and f = K1 on C1. the area of the solid part of the tube section is very small and hence.e.61). we have for contour S1 of cell 1. the lines of shear stress follow the contours of the cells. 5 that there exists a point in the crosssection.e. is zero since the force P2 is acting through the flexural centre. But a21 is the deflection (i. the section as a whole will rotate and only one point will remain at rest. the beam bends without the section rotating. 6.e. the only point which does not undergo rotation. Consider a cylindrical rod with one end firmly fixed so that no deformation occurs at the built-in section (Fig. Similarly. P2 = F Fig. from Eq.25 Centre of twist and flexural centre P1 = T For such a built-in cylinder.6 7. This point is termed the centre of twist. the rotation. t st s = K1 a constant i. q is constant. it was stated in Sec. This point is called shear centre or flexural centre. Since this is equal to zero. i. To see this. through point 2.e. the flexural centre. CENTRE OF TWIST AND FLEXURAL CENTRE We have assumed in all the previous analyses in this chapter that when a twisting moment or a torque is applied to the end of a shaft. at any section ts − n C1 C0 Fig 7. Let f = 0 on C 0 and f = K 1 on C 1 . displacement) of point 2 due to force P1 (= T). 7. and since during twisting. such that when a transverse force is applied passing through this point.13 ∂φ = 0 − K1 K1 − = = τs ts ts ∂n Therefore. it can be shown that the centre of twist and the flexural centre coincide. a12 = 0. displacement) of the flexural centre due to torque. Let d1 be the rotation caused at point 1 due to force P2 (= F) and let d2 be the deflection (i. That is.28). deflection. from the reciprocal theorem.e. . let the twisting couple be T = P1 and the bending force be F = P2.Advanced Mechanics of Solids 264 each boundary. 7. It is important to note that for this analysis to be valid it is necessary for the end to be firmly built-in. Then.25). But d1. a21 = 0. is the centre of twist. the flexural centre and the centre of twist coincide. which is the centre of twist and P2. It is assumed that P1 is applied at point 1.24 Example 7. Consequently. (7. 7. y=– Gθ 2 2 1 Fig.1 stress in a shaft with a groove where b tends to be very small.26 Problem 7. Neglecting stress concentration. (a) t = p a/64 ⎡ ⎢ 3 πa ⎢ (b) t = ⎢ 4 64 ⎢⎣ ⎣ .2 ⎡ Ans. 7.28).Torsion 265 7. 7.28 Problem 7. is the Saint-Venant warping function (also Prandtl stress function) for the torsion of a round shaft with a a semi-circular keyway.3 A thin-walled box section of dimensions 2a ¥ a ¥ t is to be compared with a solid section of diameter a (Fig. 7. (b) tmax = Gq (2a – b) ⎡ ⎢⎣ (c) Ratio Æ 2 as b Æ 0 ⎢⎣ 7. Find the thickness t so that the two sections have 2a a a t Fig. find the ratio of the shear stresses for (a) equal twisting moments in the two cases and t a (b) equal angles of twist in the t two cases. (a) 1 : 4 /p ⎡ ⎢⎣ (b) 1 : p /4 ⎢⎣ 7.27 Problem 7. ⎡ Ans.26 y and C is a constant.3 (a) the same maximum stress for the same torque and (b) the same stiffness. (c) What is the ratio of the maximum stress in a shaft without a groove to the maximum Fig. 7. 7.2 The two tubular sections shown in Fig.1 (a) Verify that ⎛ 2 2b 2 ax − b2 ⎞ 2 ⎜⎜ x + y − 2ax + 2 ⎟⎟ x + y2 ⎝ ⎠ where a and b are as shown in Fig. ⎡ Ans. x (b) Obtain an expression for the maximum b stress in the section.27 have the same wall thickness t and same circumference. 7.31 Problem 7. 1.5 cm (c) What is the shear stress in the middle web? ⎡ Ans.Advanced Mechanics of Solids 266 7. The wall thickness is uniformly 1. q1 = 2 2 4 π + 18 1 a π + 12π + 16 ⎢⎣ ( ) θ= Fig.30 Problem 7.06∞ ⎡ ⎢⎣ (b) 2352 Nm.6 cm 0. 0.4 ⎡ Ans. Determine the shear stresses in the walls and the angle of twist per unit length of the box. Find the twisting moment that can be taken up by the section and the angle of twist if the length of the member is 3 m.837∞⎢⎣ 7.5 ¥ 106 kPa.4 A hollow aluminium section is designed. Take G = 157.30. q2 = ⎢ Ans.31 is subjected to a torque T. (a) What is the maximum allowable 1.6 cm 10 cm 10 cm (b) (a) Fig 7.29 Problem 7. find the allowable twisting moment and the angle of twist.6 ( 2π + 3) T ⎤ ⎥ 2Ga t π + 12π + 16 ⎥ ⎦ 3 ( 2 ) . as shown in Fig.5 A steel girder has the cross-section shown in Fig.5 2a a 2a t t ⎡ (π + 2 ) T 5π + 8 q .44 kNm ⎡ ⎢ ⎢ 1 8 ⎢ (b) (4.6 cm 5. (a) 2352 Nm. The stress due to twisting should not exceed 350000 kPa (3570 kgf/cm2).6 cm 5.25 cm.25 cm torque? (b) What is the twist per metre length under that torque? 12. 25 cm Fig. 0.6 A thin-walled box shown in Fig. 7.29(a). 7. If the member is redesigned.29(b). Neglect stress concentrations. as shown in Fig.2 ¥ 10 ) radians ⎢ G ⎢ ⎢ ⎢⎣ (c) zero ⎢⎣ 7. 7.6 cm 0. 7. 7. (a) 273.9 cm 0. for a maximum shear stress of 35000 kPa (357 kgf/cm2). neglecting stress concentrations. t4 = 0. t3 = 0. t6 = 0. t4 = 0.33 is subjected to a torque T = 113000 Nm (115455 kgf cm).08 cm. The dimensions are as indicated.13 cm t1 t2 t1 t5 a t5 2. t2 = 0. t 1 = 394649 kPa ⎡ ⎢ t 2 = 518388 kPa ⎢ ⎢ t 3 = 460472 kPa ⎢ ⎢ ⎢ t 4 = –136862 kPa⎢ ⎢ t 5 = 57920 kPa ⎢ ⎢ ⎢⎣ t 6 = 368377 kPa ⎢⎣ 7. 7.Torsion 267 7.06 cm.06 cm.1 MPa⎢ ⎢ = –211 MPa ⎢ = 140 MPa ⎣ .7 Figure 7.8 A thin tubular bar shown in Fig.06 cm.8 ⎡ Ans. t1 = 0.32 shows a tubular section with three cells.7 cm.33 Problem 7.6 MPa⎢ = 393.08 cm.4a 4a t2 t3 t4 t2 t4 t3 t3 t3 t2 Fig. Given a = 12.10 cm. The thin-walled tube is subjected to a torque T = 113000 Nm (115455 kgf cm).2 MPa ⎡ = 558. t3 = 0. t1 = 0.10 cm ⎡ Ans. t 5 = 0.08 cm. 7. 7. ⎢ ⎢ ⎢ ⎢ ⎣ t1 t2 t3 t4 t5 = 441.7 cm.32 Problem 7. Determine the shear stresses in the walls.08 cm. Determine the shear stresses in the walls of the section.13 cm. t5 = 0. T 2a 2a t2 t2 t4 t1 t3 t5 t3 t6 a t1 2a t5 t4 t3 t2 T t6 t3 Fig. t2 = 0.7 a = 12. 4 cm 25.8 cm 1. 7. neglecting stress concentrations. the torque per unit radian of twist? [Hint: Treat cell 1 as a closed box and cell 2 as made of two narrow rectangular members. T = 112.95 kgf/cm2) ⎢ ⎢ 2 t B = 34475 kPa (351.10 ⎡ Ans. The shear flow near the junction is shown in Fig. Determine the allowable twisting moment for a maximum shear stress of 68950 kPa (703. i.34.4 cm C 0. 7.6 kgf/cm2).268 Advanced Mechanics of Solids 7. τ = ≈ T2 1 2 2 ⎢ ⎥ θ 2t (a + t ) 2a t 1 ⎣ ⎦ 7. What is the shear stress for a given torque and what is the stiffness.4 cm 50. 7. as shown in Fig.] 2a a 1 2 (b) t (a) Fig. 7.34 Problem 7.9 A thin-walled box section has two compartments.63 cm Fig.35 Problem 7. Calculate the shear stresses in the different parts of the section.8 kgf/cm ) ⎢ ⎢ tC = 68950 kPa (703.33(b).e.126 Nm (1.27 cm A 0.63 cm B 25.6 kgf/cm2) ⎣ ⎣ .10 A section which is subjected to twisting is as shown in Fig.35. 25. It has a constant wall thickness t.13 ¥ 106 kgf cm) ⎡ ⎢ ⎢ t A = 2151 kPa (21.9 T T = Gat (a 2 + t 2 ) ≈ Ga3t ⎤ ⎡ Ans.7. pmd 269 sr + Dsr D F E A Fig. etc. The stress vectors acting on its faces are as shown. Since the deformation is symmetrical about the axis. it is convenient to use cylindrical coordinates.1 8 INTRODUCTION Many problems of practical importance are concerned with solids of revolution which are deformed symmetrically with respect to the axis of revolution. 8. z sz + Dsz H Dz o sr r Chapter_08. Examples of such solids are circular cylinders subjected to uniform internal and external pressures. the stress components do not depend on q.1 G sz Dr C B sq An axisymmetric body 6/18/2008. Further. 2:22 PM .69)] can be reduced to our special case. it is instructive to derive the relevant equations applicable to axisymmetric problems from first principles. Consequently. However. Let an elementary radial element be isolated. rotating circular disks.CHAPTER Axisymmetric Problems 8. Consider an axisymmetric body shown in Fig.1. spherical shells subjected to uniform internal and external pressures. the differential equations of equilibrium [Eqs (1. a few of these problems will be investigated. In this chapter. Let the axis of revolution be the z-axis.67)–(1. trq and tq z do not exist. The deformation being symmetrical with respect to the z-axis. 8. On face ABFE.3) ∂ ur ∂r (8. Hence. the stresses are sz and tzr. the equilibrium equation for q direction is identically satisfied. the circumferential strain is eq = ( r + ur ) ∆θ − r ∆θ = ur (8. 2:25 PM . the length of the arc becomes (r + ur) Dq.1) Similarly. the normal stresses are sq and there are no shear stresses. On face CDHG. consider an arc AE at distance r. for equilibrium in r direction we get ∂σ r ∂τ rz σ r − σθ + gr = 0 + + r ∂r ∂z (8. On face BCGF. we need expressions for the circumferential strain eq and the radial strain er. Since the stress components are independent of q.2) where gr is the body force per unit volume in r direction. Consequently. subtending an angle Dq at the centre. The radial displacement is ur. 8. er = Chapter_08.4) r ∆θ The radial strain is. Hence. the normal and shear stresses are sz + Dsz = sz + ∂σ z Dz ∂z trz + Dtrz = trz + ∂τ rz Dz ∂z On face AEHD. ∂σ z ⎛ ∂τ rz ∆r ⎞ (r + Dr) Dr Dq Dz ⎜⎝ r + ⎟⎠ ∆r ∆θ ∆z + ∂z 2 ∂r ⎛ ∆r ⎞ + trz Dr Dq Dz + gz ⎜ r + ⎟ Dr Dq Dz = 0 ⎝ 2⎠ Cancelling Dr Dq Dz and going to limits ∂σ z ∂τ rz τ rz + gz = 0 + + r ∂z ∂r (8. ∂σ r ∂τ rz the stresses are sr + Dr and trz + Dr ∂r ∂r For equilibrium in z direction ∂σ z ⎞ ⎛ ∂τ rz ⎞ ∆r ⎞ ⎛ ⎛ ⎜⎝ σ z + ∂ z ∆z⎟⎠ ⎜⎝ r + 2 ⎟⎠ ∆θ ∆r + ⎜⎝ τ rz + ∂ r ∆r ⎟⎠ (r + Dr) Dq Dz ⎛ ⎝ – trzr Dq Dz – sz ⎜ r + ∆r ⎞ ⎟ Dq Dr + gz 2⎠ ∆r ⎞ ⎛ ⎜⎝ r + ⎟ Dq Dr Dz = 0 2⎠ where gz is the body force per unit volume in z direction.2(b). For the problems that we are going to discuss in this chapter. 8. from Fig.pmd 270 r 6/18/2008.270 Advanced Mechanics of Solids On faces ABCD and EFGH. the normal and shear stresses are sr and trz. The arc length is rDq.2(a). Referring to Fig. pb pb b pa a pa (b) (a) Fig. Appropriate solutions will be obtained for each case. 8.2 THICK-WALLED CYLINDER SUBJECTED TO INTERNAL AND EXTERNAL PRESSURES—LAME'S PROBLEM Consider a cylinder of inner radius a and outer radius b (Fig. In subsequent sections we shall consider the following problems: Circular cylinder subjected to internal or external pressure Sphere subjected to internal or external pressure Sphere subjected to mutual gravitational attraction Rotating disk of uniform thickness Rotating disk of variable thickness Rotating shaft and cylinder 8. It is possible to treat this problem either as a plane stress case (sz = 0) or as a plane strain case (ez = 0).pmd 271 6/18/2008.3 Thick-walled cylinder under internal and external pressures Chapter_08.3).Axisymmetric Problems Dq E A r Dr B ur + ur A A¢ Dr B 271 ∂ur ∆r ∂r B¢ (b) (a) Fig. 8. 8.5a) where uz is the axial displacement.2 Displacements along a radius The axial strain is ez = ∂ uz ∂z (8. 2:22 PM . Let the cylinder be subjected to an internal pressure pa and an external pressure pb. Neglecting body forces.2) reduces to ∂σ r σ r − σ θ + =0 ∂r r (8. (8.272 Advanced Mechanics of Solids Case (a) Plane Stress Let the ends of the cylinder be free to expand.4) σr = σr = u ⎞ E ⎛ dur +ν r⎟ ⎜ r⎠ 1 − ν 2 ⎝ dr (8.7) (u r )⎤ = 0 dr ⎢⎣ r dr r ⎦⎥ If the function ur is found from this equation. 2 + d ⎛ dur ur ⎞ + ⎟ =0 dr ⎜⎝ dr r⎠ d ⎡1 d (8.6b).6b) dur ⎞ ⎛ dur ⎞ ⎛ ur ⎜⎝ r dr + ν ur ⎟⎠ − ⎜⎝ r + ν dr ⎟⎠ = 0 dur d 2 ur du u du +r +ν r − r −ν r =0 2 dr dr r dr dr d 2 ur 1 dur ur − 2 =0 r dr dr r This can be reduced to i.8) 6/18/2008.3) and (8. trz = 0.5c) εθ = 1 (σ θ − νσ r ) E E ( ε r + νεθ ) σθ = E 2 ( εθ + νε r ) 1− ν2 1− ν Substituting for er and eq from Eqs (8. From Hooke’s law ε r = 1 (σ r − νσ θ ) .6a) dur ⎞ E ⎛ ur ⎜ + ν dr ⎟⎠ 1− ν 2 ⎝ r Substituting these in the equation of equilibrium given by Eq. (8.1) is identically satisfied.5b) Since r is the only independent variable. the above equation can be written as d ( rσ r ) − σ θ = 0 dr Equation (8.pmd 272 C2 r (8.5c) σθ = d dr or (8.7) is or ur = C1 r + Chapter_08. (8. The solution to Eq. Owing to uniform radial deformation. 2:22 PM . the stresses are then determined from Eqs. We shall assume that sz = 0 and our results will justify this assumption.6a) and (8.e. (8. E or the stresses in terms of strains are (8. Eq. and cross-sections perpendicular to the axis of the cylinder remain plane.10)–(8. the elements can be considered to be in a state of plane stress. sr = –pa when r = b. i.e. Substituting this function in Eqs. Chapter_08. (8. When r = a.9a) σθ = E 1− ν2 ⎡ 1⎤ ⎢⎣C1 (1 + ν ) + C2 (1 − ν ) r 2 ⎦⎥ (8. Hence. the deformation undergone by the element does not interfere with the deformation of the neighbouring element. Hence. the stresses sr and sq produce a uniform extension or contraction in z direction.pmd 273 6/18/2008.10) (8.8) and (8. independent of r.12).11) b2 − a 2 (8.12) b2 − a 2 It is interesting to observe that the sum sr + sq is constant through the thickness of the wall of the cylinder.9) we get ur = σr = σθ = 1 − ν pa a 2 − pb b 2 1 + ν a 2 b2 pa − pb r+ 2 2 E E r b2 − a2 b −a pa a 2 − pb b 2 2 b −a 2 pa a 2 − pb b 2 2 b −a 2 − a 2 b 2 pa − pb + a 2 b 2 pa − pb r2 r2 (8.9b) The constants C1 and C2 are determined from the boundary conditions. 2:22 PM .Axisymmetric Problems 273 where C1 and C2 are constants of integration. It is important to note that in Eqs (8. whence. sz = 0. as we assumed at the beginning of the discussion. according to Hooke’s Law.e. If we consider two adjacent cross-sections. i.6a) and (8. pa and pb are the numerical values of the compressive pressures applied.6b) σr = E 1− ν 2 ⎡ ⎤ C1 (1 + ν ) − C2 (1 − ν ) 12 ⎥ r ⎦ ⎣⎢ (8. sr = –pb Hence. E 1− ν2 ⎡ 1⎤ ⎢⎣C1 (1 + ν ) − C2 (1 − ν ) a 2 ⎥⎦ = − pa E 1− ν 2 ⎡ 1⎤ ⎢C1 (1 + ν ) − C2 (1 − ν ) b2 ⎥ = − pb ⎣ ⎦ C1 = 1 − ν pa a 2 − pb b 2 E b2 − a2 C2 = 1 + ν a 2b2 ( p − pb ) E b2 − a2 a Substituting these in Eqs (8. 16) a – – pb 2 σr = − 274 pb 2 2 2 (8. 2:22 PM . 8.5 Cylinder subjected to external pressure Chapter_08.4 shows the variation of radial and circumferential stresses across the thickness of the cylinder under internal pressure.11) and (8. Figure 8.13) σθ = pa 2 ⎛ b2 ⎞ + 1 ⎜ ⎟ b2 − a 2 ⎝ r2 ⎠ (8. pa = 0 and pb = p.12) become In this case pb = 0 and pa = p. (sq)max is always greater than the internal pressure and approaches this value as b increases so that it can never be reduced below pa irrespective of the amount of material added on the outside. 6/18/2008.4 (b) Cylinder subjected to internal pressure Hence.14) These equations show that sr is always a compressive stress and sq a tensile stress.15) 2 p sq sr p – p + b 2 + a2 b 2 − a2 p (a) Fig. where (σ θ )max = ( p a2 + b2 2 b −a ) (8. Equations p b sr sq Fig.12) reduce to In this case. i. the stresses are uniformly distributed in the cylinder with s r = s q = –p. The circumferential stress is greatest at the inner surface of the cylinder. Cylinder Subjected to External Pressure (8.11) and (8.17) The variations of these stresses across the thickness are shown in Fig.274 Advanced Mechanics of Solids Cylinder Subjected to Internal Pressure Eqs (8.5. 8. 8. If there is no inner hole.e.pmd ⎛ a2 ⎞ − 1 ⎜ ⎟ b2 − a2 ⎝ r2 ⎠ σθ = − ⎛ a2 ⎞ ⎜⎝ 1 + 2 ⎟⎠ b −a r (8. if a = 0. Then σr = pa 2 ⎛ b2 ⎞ − 1 ⎜ ⎟ b2 − a 2 ⎝ r2 ⎠ (8. 07 kPa). Assume that the ends of the cylinder are closed. The maximum and minimum principal stresses are s1 = sq and s3 = sr.25. where σθ = p s r = − p. σz = p a2 b − a2 2 (assumed) In the above expressions. Solution The critical point lies on the inner surface of the cylinder. Hence. it is assumed that away from the ends. 2:22 PM . b2 + a 2 2 b −a 2 . The yield point for the material (in tension as well as in compression) is syp = 5000 kgf/cm2 (490000 kPa).45 cm 3 b= Example 8.pmd p σy b2 =1 2 2 N b −a 2 275 6/18/2008. sz caused by p is uniformly distributed across the thickness. The yield stress in tension for the material is 350 MPa. v = 0. 1 σ − σ = p b2 ( 3) 2 1 b2 − a2 tmax = Substituting the numerical values (1 atm = 98. determine the maximum working pressure p according to the major theories of failure. The inner radius is 5 cm. E = 207 ¥ 106 kPa.Axisymmetric Problems 275 Example 8.5 100 + 25 (ii) Maximum shear stress theory Maximum shear stress = 1 (σ θ − σ r ) at r = a 2 2 ⎞ ⎛ = 1 p ⎜ 22b 2 ⎟ 2 ⎝b − a ⎠ \ Chapter_08.1 Select the outer radius b for a cylinder subjected to an internal pressure p = 500 atm with a factor of safety 2. Solution (i) Maximum normal stress theory Maximum normal stress = sq at r = a = p \ p (b 2 + a 2 ) 2 = 2 (b − a ) or p= (b 2 + a 2 ) (b 2 − a 2 ) σy N 350 × 106 100 − 25 × = 140 × 103 kPa 1.2 A thick-walled steel cylinder with radii a = 5 cm and b = 10 cm is subjected to an internal pressure p. Using a factor of safety of 1. 5 a = 6.5. i.5 \ p = 100 ¥ 103 kPa (v) Energy of distortion theory This will give a value identical to that obtained based on octahedral shear stress theory. 2:22 PM .5 \ 350 × 106 × 75 = 121. Example 8.3 A pipe made of steel has a tensile elastic limit sy = 275 MPa and E = 207 ¥ 106 kPa.25 × 75)⎤⎦ = (100 − 25) 1. p = 100 ¥ 103 kPa.276 Advanced Mechanics of Solids 350 × 106 100 − 25 × = 87.e.7 × 103 kPa 1.5 × 143. 3 Chapter_08.75 p= (iv) Octahedral shear stress theory 1/ 2 t oct = 1 ⎡σ θ2 + σ r2 + (σ r − σ θ )2 ⎤ at r = a ⎦ 3⎣ 1/ 2 1⎡ 2 (σ r − σ θ ) 2 + 2σ r σ θ ⎤⎦ 3⎣ = 1/ 2 = 2 3 2 ⎧⎡ 2 2 ⎫ p (b 2 + a 2 ) ⎤ ⎪ 2 (b + a ) ⎪ − p ⎨⎢ − p − ⎬ ⎥ (b 2 − a 2 ) ⎦ (b 2 − a 2 ) ⎪ ⎪⎩ ⎣ ⎭ 2 σy 3 N = 1/ 2 \ or (b 2 + a 2 ) ⎤ 2 ⎡ 4b 4 − p⎢ 2 ⎥ 2 2 3 (b 2 − a 2 ) ⎦ ⎣ (b − a ) p ( 40000 125 − 5625 75 ) 1/ 2 = = 2 σy 3 N 350 1. determine the proper thickness for the pipe wall 4 according to the major theories of failure. Use a factor of safety N = .5 × 103 kPa 3 100 (iii) Maximum strain theory Maximum strain = eq = 1 (σ θ − ν σ r ) at r = a E or p= = p a2 2 E (b − a 2 ) ⎡⎛ ⎛ b2 ⎞ b2 ⎞ ⎤ ⎢⎜⎝1 + 2 ⎟⎠ − ν ⎜⎝ 1 − 2 ⎟⎠ ⎥ a a ⎦ ⎣ \ σ p ⎡ ( a 2 + b 2 ) − ν (a 2 − b 2 ) ⎤ = y ⎦ NE E (b2 − a 2 ) ⎣ or 350 × 106 p ⎡⎣125 + ( 0.pmd 276 6/18/2008. If the pipe has an internal radius a = 5 cm and is subjected to an internal pressure p = 70 ¥ 103 kPa. 75 × 25 × 10 −4 ) [b 2 − (25 × 10 −4 )] ⎣ 3 × 275 × 106 4 1312.13 b2 – 2578.38 cm \ Wall thickness t = 2. (4.38 ¥ 10–2 m = 7. s2 = 0.Axisymmetric Problems Solution (i) Maximum principal stress theory Maximum principal stress = sq at r = a =p (b 2 + a 2 ) 2 2 (b − a ) = σy N \ 70 × 106 ⎡⎣(25 × 10 − 4 ) + b 2 ⎤⎦ 275 × 106 × 3 = 4 ⎡b 2 − (25 × 10−4 ) ⎤ ⎣ ⎦ or 1750 × 10−4 + 70b 2 = 825b 2 − 20625 × 10−4 4 or 136.82 ¥ 10–2 m = 8. 2:22 PM 277 .38 cm (iv) Maximum distortion energy theory From Eq.25 × b 2 ) ⎤⎦ = or \ b = 7.25b2 – 5156.12 cm \ Wall thickness t = 2.82 cm (iii) Maximum strain theory (with n = 0.12 cm (ii) Maximum shear stress theory tmax = 1 (σ θ − σ r ) at r = a 2 = \ σy pb 2 = (b 2 − a 2 ) N 70 × 106 b 2 [b 2 − (25 × 10− 4 )] = 3 × 275 × 106 8 or 70b2 = 103.pmd 277 6/18/2008.1 ¥ 10–4 \ b = 8.82 cm \ Wall thickness t = 3.5 ¥ 10–4 + 87. s3 = sr = –p Chapter_08.25b2 = 6906.5b2 = 206.25) emax = 1 (σ θ − νσ r ) at r = a E = \ σ p ⎡(a 2 + b 2 ) − ν (a 2 − b 2 ) ⎤ = y 2 ⎣ ⎦ NE E (b − a ) 2 70 × 106 ⎡ (0.12) with s 1 = sq .25 ¥ 10–4 \ b = 7.12 ¥ 10–2 m = 7.25 ¥ 10–4 + (1. 4) 10–4 or b = 7.pmd 278 6/18/2008.946 \ b 2 = (63 or 13. sections that are far from the ends can be considered to be in a state of plane strain and we can assume that sz does not vary along the z-axis. 1 [σ 2 + σ r2 + (σ r − σ θ )2 ] 12G θ σ r2 − σθ σ r = σ 2y N2 2 ⎡ (b 2 + a 2 ) 2 (b 2 − a 2 ) ⎤ σ y +1+ 2 p ⎢ 2 ⎥= 2 2 (b + a 2 ) ⎦ N 2 ⎣ (b − a ) 2 ⎛ σy ⎞ Putting ⎜ = fy and simplifying one gets ⎝ p N ⎟⎠ (3 − f y2 ) b 4 + 2a 2 f y2 b 2 + (1 − f y2 ) a 4 = 0 \ b2 = = With −2a 2 f y2 ± ⎡ 4a 4 f y4 − 4a 4 (1 − f y2 ) (3 − f y2 ) ⎤ ⎣ ⎦ 2 (3 − f y2 ) a 2 ⎡ − f y2 ± (4 f y4 − 3) ⎤ ⎣⎢ ⎦⎥ 2 (3 − f y2 ) a = 5 ¥ 10–2 fy = 275 × 106 × 3 70 × 106 × 4 = 2.278 Advanced Mechanics of Solids U* = = (1 + ν ) (2σθ2 + 2 σ r2 − 2σ r σθ ) 6E = 2 (1 + ν ) 2 1+ ν σy (σ θ + σ r2 − σ θ σ r ) = 3E E N2 2 σθ + \ i.e. the equation is d (rσ r ) − σ θ = 0 dr From Hooke's law e r = 1 [σ r − ν (σ θ + σ z ) ] E e q = 1 [σ θ − ν (σ r + σ z ) ] E e z = 1 [σ z − ν (σ r + σ θ ) ] E Chapter_08.9 cm Case (b) Plane Strain When the cylinder is fairly long. As in the case of plane stress.9 cm Wall thickness t = 2.9 ¥ 10–2 m = 7. 2:22 PM . (8. (8.4). (8. The solution is the same as in Eq. C2 r where C1 and C2 are constants of integration.20) and (8.22c) 6/18/2008.19b) On substituting for er and eq from Eqs (8.18) is sz = − Chapter_08.20) sr = E (1 − 2ν ) (1 + ν ) dur ur ⎤ ⎡ ⎢(1 − ν ) dr + ν r ⎥ ⎣ ⎦ (8. Further.7) for the plane stress case.8).5c) d dr dur dur ur ⎡ ⎤ ⎢(1 − ν ) r dr + ν u r ⎥ − ν dr − (1 − ν ) r = 0 ⎣ ⎦ or dur d 2 ur ur +r − =0 dr r dr 2 i.pmd 2ν E C (1 − 2ν ) (1 + ν ) 1 279 (8. one has from the last equation s z = n (sr + sq) er = 1+ ν [(1 − ν ) σ r − νσ θ ] E (8.21) Substituting these in the equation of equilibrium. we observe that sr + sq is a constant independent of r .19a) sr = E [(1 − ν ) ε r + νεθ ] (1 − 2ν ) (1 + ν ) (8. Eq.21) u r = C1 r + sq = E (1 − 2ν ) (1 + ν ) C2 ⎤ ⎡ ⎢⎣C1 + (1 − 2ν ) r 2 ⎥⎦ (8. 2:22 PM . d ⎛ dur ur ⎞ + ⎟ =0 dr ⎜⎝ dr r⎠ This is the same as Eq. From Eqs (8. the axial stress from Eq.3) and (8.22b) Once again.e. the above equations become sq = E (1 − 2ν ) (1 + ν ) ur ⎤ ⎡ dur ⎢ν dr + (1 − ν ) r ⎥ ⎣ ⎦ (8.22a) sr = E (1 − 2ν ) (1 + ν ) C2 ⎤ ⎡ ⎢C1 − (1 − 2ν ) r 2 ⎥ ⎣ ⎦ (8.Axisymmetric Problems 279 Since ez = 0 in this case.18) 1+ ν [ (1 − ν ) σθ − νσ r ] E Solving for sq and sr eq = sq = E [νε r + (1 − ν ) εθ ] (1 − 2ν ) (1 + ν ) (8. (8. 23) r2 pa − pb a 2 b 2 b2 − a 2 (8. involves two cylinders made of two different materials and fitted one inside the other. whereas in the plane strain case. 8. 2:22 PM . On the other hand.24) r2 pb a 2 − pa b 2 (8. Since no such high-strength material exists. When the cylinders come to room temperature. The internal radius of the outer cylinder is less than c by D. C1 = (1 − 2ν ) (1 + ν ) pb b 2 − pa a 2 E a 2 − b2 and C2 = 1 + ν ( pb − pa ) a 2 b2 E a2 − b2 Substituting these.sz has a constant value given by Eq. the inner cylinder has an internal radius a and an external radius c. i. If the inner cylinder is cooled and the outer cylinder is heated. The sum of these two quantities.pmd 280 6/18/2008. one fitting inside the other. For example. the yield point of the material must be at least 30000 kgf/cm2 (2940000 kPa). if we need a vessel to withstand a pressure of say 15000 atm.e. The contact pressure pc acting on the outer surface of the inner cylinder reduces its outer radius by u1. But in the plane stress case. sz = 0.25). the same contact pressure increases the inner radius of the outer cylinder by u 2 . Before assembling. the stress components become sr = sq= pa a 2 − pb b 2 b2 − a 2 pa a 2 − pb b 2 b2 − a 2 s z = 2ν − + pa − pb a 2 b 2 b2 − a 2 (8.25) b2 − a2 It is observed that the values of sr and sq are identical to those of the plane stress case. The problem lies in determining the contact pressure pc between the two cylinders. (8. s r = – pb when r = b E (1 − 2ν ) (1 + ν ) C2 ⎤ ⎡ ⎢C1 − (1 − 2ν ) a 2 ⎥ = − pa ⎣ ⎦ E (1 − 2ν ) (1 + ν ) C2 ⎤ ⎡ ⎢C1 − (1 − 2ν ) b 2 ⎥ = − pb ⎣ ⎦ Solving. The above construction is often used to obtain thick-walled vessels to withstand high pressures. its internal radius is c – D. Its external radius is b.3 STRESSES IN COMPOSITE TUBES—SHRINK FITS The problem which will be considered now. shrink-fitted composite tubes are designed.280 Advanced Mechanics of Solids Applying the boundary conditions s r = – pa when r = a. a shrink fit is obtained. Chapter_08. then the two cylinders can be assembled. 26) and (8.17). For the inner tube u1 = or 1 − ν1 E1 ⎛ c 2 ⎞ c + 1 + ν1 a 2 c 2 ⎜ − pc 2 ⎟ E1 c c − a2 ⎠ ⎝ pc ⎞ ⎛ ⎜⎝ − 2 ⎟ c − a2 ⎠ cpc ⎡(1 − ν1 ) c 2 + (1 + ν1 ) a 2 ⎤ ⎦ E1 (c 2 − a 2 ) ⎣ u1 = − For the outer tube u2 = − or 1 − ν2 ⎛ 1 + ν 2 c 2b2 c2 ⎞ + p c c ⎟ E2 ⎜⎝ E2 c b2 − c 2 ⎠ ⎛ pc ⎞ ⎜⎝ 2 ⎟ b − c2 ⎠ cpc ⎡(1 − ν 2 ) c 2 + (1 + ν 2 )b 2 ⎤ ⎦ E2 (b 2 − c 2 ) ⎣ u2 = − In calculating u2. To determine u1 and u2. we should have –u1 + u2 = D cpc ⎡ (1 − ν1 ) c 2 + (1 + ν1 ) a 2 ⎤ ⎦ E1 ( c 2 − a 2 ) ⎣ i. Eq. the internal pressure p causes a tensile tangential stress sq . therefore. but.13) and (8.27).Axisymmetric Problems 281 i. At the inner surface of the inner cylinder. Equation (8. the contact pressure pc is given by pc = ∆ /c 1 ⎡c + a − ν ⎤ + 1 1⎥ E1 ⎢⎣ c 2 − a 2 ⎦ E2 2 2 ⎡ b2 + c2 ⎤ + ν2 ⎥ ⎢ 2 2 ⎣b − c ⎦ (8.e. 8. Hence. Because of shrink fitting. we make use of Eq. Eq (8.26) will then reduce to pc = 2 2 2 2 E ∆ (c − a ) (b − c ) 2c 3 (b 2 − a 2 ) (8. The stress distribution in the assembled cylinders is shown in Fig.14).pmd 281 6/18/2008. (–u 1 + u2) must be equal to D. assuming a plane stress case. a composite cylinder can Chapter_08. the difference in the radii of the cylinders. we have neglected D since it is very small as compared to c. D is the difference in radii between the inner cylinder and the outer jacket.27) It is important to note that in Eqs (8.e. the contact pressure pc causes at the same points a compressive tangential stress.10). + cpc ⎡(1 − ν 2 )c2 + (1 + ν 2 )b2 ⎤ = ∆ ⎦ E2 (b2 − c 2 ) ⎣ (8.26a) Regrouping. the inner cylinder is under external pressure pc. (8. then E1 = E2 and n1 = n2. If the composite cylinder made up of the same material is now subjected to an internal pressure p. then the two parts will act as a single unit and the additional stresses induced in the composite can be determined from Eqs (8.14).6. Noting that u1 is negative and u2 is positive.26b) If the two cylinders are made of the same material. 2:26 PM . (8. 28) At point B of the outer cylinder. 8. due to the pressure p. which are composite tube respectively equal to sq and sr.e. from Eqs (8.pmd 282 6/18/2008.282 Advanced Mechanics of Solids sq a b c c Fig. s1 and s3 are the Fig. from Eqs (8.6 – sr D pc Streses in composite tubes support greater internal pressure than an ordinary one. at the inner points of the jacket or the outer cylinder. However.7 Equal strength maximum and minimum normal stresses.13) and (8. (s1 – s3) at point A of the inner cylinder should be equal to (s1 – s3) at point B of the outer cylinder. circumferential) stresses sq . (sq – sr)A = − 2 pc c2 c − a2 2 Hence. 8. (sq – sr)A = p b2 + a 2 b2 − a 2 = 2p − (− p) b2 b − a2 2 Because of shrink-fitting pressure pc.16) and (8.14). due to internal pressure p. the resultant value of (sq – sr) at A is (sq – sr)A = 2 p b2 − 2 p c2 c b2 − a2 c2 − a2 (8. at the same point. 8.7) c is the common radius of the two cylinders at the contact surface when the composite cylinder is sq a + experiencing an internal pressure p and the shrinkA B sr – fit pressure pc.14). To determine this value of D. At point A. and c and b the radii of the jacket (see Fig. then according to the maximum shear stress theory.13) and (8. Chapter_08. and observing that r = c in these equations. the internal pressure p and the contact pressure pc both will induce tensile tangential (i. 2:26 PM .17). For design purposes. from Eqs (8. If the strengths of the two cylinc b ders are the same. one can proceed as follows. one can choose the shrink-fit allowance D such that the strengths of the two cylinders are equal. since the composite involves the same material. Let a and c be the radii of the inner cylinder. equating Eqs (8. substituting for pc from Eq.31a) Also.27).Axisymmetric Problems 283 ⎡ a 2 (c 2 + b 2 ) a 2 (c 2 − b 2 ) ⎤ (sq – sr)B = p ⎢ 2 2 − 2 2 2 2 ⎥ ⎣ c (b − a ) c (b − a ) ⎦ = 2p a2b2 c (b 2 − a 2 ) 2 At the same point B.30) becomes or 2 2 2 2 ⎤ ⎡ b2 ⎤ ∆E (c − a ) (b − c ) ⎡ b 2 c2 a 2b2 + = p − ⎢ ⎥ ⎢ 2 2 2 2 2 ⎥ 3 2 2 2 2 2 2 2c (b − a ) ⎣ (b − c ) (c − a ) ⎦ ⎣ (b − a ) c (b − a ) ⎦ or 2 2 2 2 4 pb 2 (c 2 − a 2 ) ∆E (2b c − b a − c ) = 2c 3 (b 2 − a 2 ) c 2 (b2 − a 2 ) D= or b 2 c (c 2 − a 2 ) 2p E b2 (c 2 − a 2 ) − c 2 (b 2 − c 2 ) (8. from Eq.31b) .14).pmd 2b 2 (b 2 − a 2 ) 283 ⎡ ⎤ (c 2 − a 2 ) (b 2 − c 2 ) ⎢1 − 2 2 2 2 2 2 ⎥ ⎣ b ( c − a ) + c (b − c ) ⎦ 6/18/2008. is pc = p sq – sr = p Chapter_08. (8.27) and it is this value of D that is required now for equal strength.29) 2 For equal strength.31). 2:26 PM (8. with internal radius equal to c and external radius b.13) and (8. b 2 (c 2 − a 2 )2 (b 2 − c 2 ) c 2 (b 2 − a 2 ) ⎡⎣ b 2 (c 2 − a 2 ) + c 2 (b 2 − c 2 ) ⎤⎦ The value of (sq – sr) either at A or at B. (8. Eq. due to the contact pressure pc. (8.28) and (8.29) 2p b2 c2 a 2b2 b2 p p p − 2 = 2 + 2 c c c 2 (b 2 − a 2 ) (b 2 − a 2 ) (c 2 − a 2 ) (b 2 − c 2 ) ⎡ b2 ⎤ ⎡ b2 ⎤ c2 a 2b2 (8. from Eqs (8.30) pc ⎢ 2 p + = − ⎥ ⎢ 2 2 2 2 2 2 2 2 ⎥ (c − a ) ⎦ ⎣b − c ⎣ (b − a ) c (b − a ) ⎦ The shrink-fitting pressure pc is related to the negative allowance D through Eq. from Eqs (8. Hence.30).28) and (8. ⎡ c2 + b2 c2 − b2 ⎤ (sq – sr)B = pc ⎢ 2 − 2 ⎥ 2 b − c2 ⎦ ⎣b − c b2 b − c2 = 2 pc 2 The resultant value of (sq – sr) at B is therefore (sq – sr)B = 2 p a 2b2 b2 + 2 pc 2 2 2 (b − c 2 ) c (b − a ) (8. (8. D depends upon the difference between the external radius of the inner cylinder and the internal radius of the jacket. where b2 + c2 b2 − c 2 c 2 − a 2 Differentiating with respect to c and equating the differential to zero.31a). one can determine the optimum value of c for minimum (sq – sr) at A and B.pmd 284 6/18/2008.34) ab Example 8. i.4 Determine the diameters 2c and 2b and the negative allowance D for a two-layer barrel of inner diameter 2a = 100 mm.284 Advanced Mechanics of Solids or ⎡ ⎢ 1 sq – sr = p 22b 2 ⎢1 − 2 2 b (b − a ) ⎢ + 2c 2 ⎢ 2 2 b −c c −a ⎣ 2 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ (8. (8. With a. for composites made of the same material. The material is steel with E = 2(10)6 kgf/cm2 (196 ¥ 105 kPa).33) Also. In other words. Further. (8. and this depends on the internal pressure p. from Eq. From Eq. is ⎡ ⎤ ⎢ ⎥ 1 ⎢1 − ⎥ (sq – sr)min = p 2b (b 2 − a 2 ) ⎢ b2 ab ⎥ + ⎢ b(b − a) a(b − a) ⎥⎦ ⎣ 2 = p or (sq – sr)min = p 2b2 (b − a 2 ) 2 ⎡ (b − a ) ⎤ ⎢1 − 2b ⎥ ⎣ ⎦ b (b − a) (8. 2:26 PM .32). this depends on c(+) and c(–). Chapter_08. (8. the minimum value of (sq – sr) is obtained when the denominator of the second expression within the square brackets is a maximum. The maximum pressure the barrel is to withstand is pmax = 2000 kgf/cm2 (196000 kPa).32) Therefore.32). D opt = p 1 pc = E E (8. D= 2c (c 2 − a 2 ) − 2c3 2cb2 dD = + =0 dc (b2 − c 2 )2 (c 2 − a 2 )2 Simplifying. when D is a maximum. syp in tension or compression is 6000 kgf/cm2 (588 ¥ 103 kPa). The factor of safety is 2.e. the shrink-fit allowance D that is necessary is given by Eq. the optimum value of D is from Eq. one gets c = ab The corresponding value of (sq – sr). in order to have equal strength according to the shear stress theory. b and p fixed. (8.31a). from Eqs (8.25 − 25) × ( 0. (8. determine the contact pressure pc and the values of sr and sq at the inner and external points of the inner cylinder and of the jacket.5 A steel shaft of 10 cm diameter is shrunk inside a bronze cylinder of 25 cm outer diameter. p 2000 ab = (50 × 150) = 0.3 × 156. 2b = 300 mm. a = 0.18 × 10 × 25 1. and E steel = 2.005 cm difference between the radii).34 6/18/2008. D = 0.4 kgf/cm2 (22775 kPa) The shaft experiences equal sr and sq at every point. The shrink allowance is 1 part per 1000 (i. the value of D is. The numerical values are therefore.3 Substituting in Eq. Solution In Eq.14).33). 0.25 − 25) ⎜⎝ r2 ⎠ When r = 5 cm and r = 12. 2:26 PM . (8.26). Find the circumferential stresses in the bronze cylinder at the inside and outer radii and the stress in the shaft.34). c = 5. 5 pc 5 pc (0.005. 2c = 173 mm.17). pc = 610 kgf/cm (59780 kPa) For the bronze tube. Ecu = 1 ¥ 106 kgf/cm2 (98 ¥ 106 kPa). If the compound cylinder is subjected to an internal pressure p = 1500 kgf/cm2 (147009 kPa). With c = ab.26a).Axisymmetric Problems 285 Solution From Eq.09 × 10 × (156. 6000 = 2000 b b−a 2 b = 3a Since c = ab .25 ⎞ 1+ sq = ⎟ (156.4 kgf/cm2 (82555 kPa) sq = 232.5. s r = sq = –610 kgf/cm2 (59780 kPa) Example 8.3.7 × 25) + 6 6 2. c = 3 a.6 A compound cylinder made of copper inner tube of radii a = 10 cm and c = 20 cm is snug fitted (D = 0) inside a steel jacket of external radius b = 40 cm. Hence.3 for both metals. from Eq. n1 = n2 = 0. Therefore. (8.pmd 285 ncu = 0.005 2 or. from Eq. (8. 610 × 25 ⎛ 156.09 ¥ 106 kgf/cm2 (107 ¥ 106 kPa) n = 0.16) and (8. b = 12.5 cm sq = 842.866 mm D= E 2 × 106 Example 8. Use the following data: E st = 2 ¥ 106 kgf/cm2 (196 ¥ 106 kPa). (8. Chapter_08. the circumferential stress is. 2a = 100 mm. nst = 0.7 × 25 + 1.e.25) = 0.18 ¥ 106 kgf/cm2 (214 ¥ 106 kPa) E bronze = 1. nst = 0. b = 40.10). from Eq. 2:26 PM . ⎡ (ur ) p + (ur ) p at r = c ⎤ = ⎡(u r ) p at r = c ⎤ c c ⎣ ⎦ cu ⎣ ⎦ st For copper cylinder. 1 − ν cu pa 2 c 1 + ν cu a 2 c 2 p + (ur)p = E 2 2 Ecu c (c 2 − a 2 ) (c − a ) cu (u ) pc = − (ur)total = 1 − ν cu pc c3 1 + ν cu a 2 c 2 pc − 2 2 2 Ecu (c − a ) Ecu c (c − a 2 ) pc c 2 pa 2 c ⎡ c 2 (1 − ν cu ) + a 2 (1 + ν cu ) ⎤ − 2 2 ⎦ Ecu (c − a ) Ecu (c 2 − a 2 ) ⎣ For steel jacket. i. the increase in the external radius of the copper cylinder under p and contact pressure pc should be equal to the increase in the internal radius of the jacket under the contact pressure pc. c = 20.pmd pc = 433 kgf/cm2 (42453 kPa) 286 6/18/2008. ncu = 0.3 ⎤ 3000 × 100 pc ⎢ + ⎥= 300 × 106 2 × 1200 × 106 ⎦ 300 × 106 ⎣ \ Chapter_08. When the compound cylinder is subjected to an internal pressure p. from Eq. (8.e. the initial contact pressure is zero.286 Advanced Mechanics of Solids Solution Since the initial shrink-fit allowance D is zero. ⎡ 500 − 300 × 0. (u ) pc = = 1 − ν st pc c3 1 + ν st c 2 b 2 pc + Est (b 2 − c 2 ) Est c (b 2 − c 2 ) pc c ⎡ c 2 (1 − ν st ) + b 2 (1 + ν st ) ⎤ ⎦ Est (b 2 − c 2 ) ⎣ Equating the (ur)s pc c 2 pa 2 c ⎡c 2 (1 − ν cu ) + a 2 (1 + ν cu ) ⎤ − 2 2 ⎦ Ecu (c − a ) Ecu (c 2 − a 2 ) ⎣ = or pc c ⎡ c 2 (1 − ν st ) + b 2 (1 + ν st ) ⎤ ⎦ Est (b 2 − c 2 ) ⎣ ⎡ (c 2 + a 2 ) − ν cu (c 2 − a 2 ) (b 2 + c 2 ) + ν st (b 2 − c 2 ) ⎤ + pc ⎢ ⎥ Ecu (c 2 − a 2 ) Est (b 2 − c 2 ) ⎣ ⎦ = p Ecu 2a 2 (c 2 − a 2 ) 2 With p = 1500 kgf/cm .34.10). (8.34 2000 + 1200 × 0.3. a = 10. 11) and (8.pmd ∂ ur ∂r 287 6/18/2008. sq at r = a is = (1500 × 100) − (433 × 400) 100 × 400 (1500 − 433) + × (400 − 100) 100 (400 − 100) = 1357 kgf/cm2 (compressive) (ii) Outer surface: sr at r = c is – 433 kgf/cm2 and. Consider a particle situated at radius r before deformation. another particle at distance (r + Dr) along the same radial line will undergo a ∂u ⎛ ⎞ displacement ⎜ ur + r ∆r ⎟ . ∂r ⎝ ⎠ Hence. For copper tube. from Eqs (8.12). (i) Inner surface:. 2:26 PM . sq at r = b is = (433 × 400) 400 × 1600 433 + × (1600 − 400) 1600 (1600 − 400) = 289 kgf/cm2 8. from Eqs (8. (i) Inner surface: sr at r = c is – 433 kgf/cm2 and.Axisymmetric Problems 287 Now. sq at r = c is = (433 × 400) 400 × 1600 433 = × (1600 − 400) 400 (1600 − 400) = 722 kgf/cm2 (ii) Outer surface: sr at r = b is zero and. Similarly.4 SPHERE WITH PURELY RADIAL DISPLACEMENTS Consider a uniform sphere or spherical shell subjected to radial forces only.11) and (8. sr at r = a is –1500 kgf/cm2 and. the radial strain is er = Chapter_08. The sphere or the spherical shell will then undergo radial displacements only. After deformation. the spherical surface of radius r becomes a surface of radius (r + ur) and the particle undergoes a displacement ur. sq at r = c is = (1500 × 100) − (433 × 400) 100 × 400 (1500 − 433) + (400 − 100) 400 (400 − 100) = 279 kgf/cm2 For steel jacket. such as internal or external pressures. pc will act as an external pressure on the copper tube and as an internal pressure on the steel jacket.12). the three extensional strains along the three axes are ∂ ur . s q = s f. the circumferential strain is 2π ( r + ur ) − 2π r ur = 2π r r This is the strain in every direction perpendicular to the radius r.35) . ef = sf sr + Dsr q sr f (ef) r (c) r (er) sf sq 2q q (eq) s r + Dsr sr (a) sf 2q sq sf (b) Fig. 2:26 PM . er = εθ = –sr (2q r)(2q r) + (sr + Dsr) (r + Dr)2q (r + Dr)2q ( −2 r + ( ) Putting sq = sf and Dsr = r2 ) ∆r ∆r 2θ∆rσ φ sin θ − 2 r + 2θ ∆rσ θ sin θ + γ r 4θ 2 r 2 ∆r = 0 2 2 σr r ∆r . subtending a small angle 2q at the centre. the above equation reduces in the limit to ∂σ r + 2rσ r − 2rσ φ + r 2γ r = 0 ∂r Since r is the only independent variable. Consider a spherical element of thickness Dr at distance r. there are no shear stresses and shear strains. the radius becomes (r + ur) and the circumference of the great circle is 2p (r + ur) Hence. Because of complete symmetry.8 Sphere with purely radial displacement Thus. 8.pmd 288 (8. we can choose a frame of reference. the above equation can be rewritten as d 2 (r σ r ) − 2rσ φ + r 2γ r = 0 dr Chapter_08. Let gr be the body force per unit volume in the radial direction. For equilibrium in the radial direction.288 Advanced Mechanics of Solids Before deformation.36) 6/18/2008. ∂r ur u (8. 8. the circumference of any great circle on the surface of radius r is 2p r. as shown in Fig. The stress equations of equilibrium can also be derived easily.8. After deformation. Because of spherical symmetry. εφ = r r r Because of symmetry. 40) d (rσ ) = d ⎛ y ⎞ = 1 dy − 1 y r dr dr ⎜⎝ r ⎟⎠ r dr r 2 Therefore. (8.39) can be solved. 1 d (r 2σ ) = 2 σ r r φ r 2 dr From Hooke’s law (8. 2:26 PM .40) becomes d 2 y ν dy ν y y ν dy 1 1 − ν) 2 − + 2 − 2 + =0 ( 2 r dr r dr dr r r d2y y −2 2 =0 2 dr r This is a homogeneous linear equation with the solution or (8.38) and e f = 1 ⎡⎣σ φ − ν (σ φ + σ r ) ⎤⎦ E or ur = 1 ⎡⎣(1 − ν ) σ φ − νσ r ⎤⎦ E r Equations (8.39) u r = 1 ⎡⎣(1 − ν )r σ φ − ν rσ r ⎤⎦ E Differentiating with respect to r dur 1 ⎡ (1 − ν ) d (rσ φ ) − ν d (rσ r ) ⎤ = E ⎢⎣ dr dr ⎥⎦ dr Subtracting the above equation from Eq.37) e r = 1 ⎡⎣σ r − ν (σ θ + σ φ ) ⎤⎦ E or dur = 1 (σ r − 2νσ φ ) E dr (8. (8.39) (8. From Eq.38) d ( rσ φ ) d ( rσ r ) +ν + σ r − 2νσ φ dr dr Substituting for sf from Eq. (8.Axisymmetric Problems 289 If body force is ignored. Eq.36) 0 = − (1 − ν ) 2 d 2 (r 2σ r ) d (rσ r ) 1 ν d (r σ r ) = 0 (1 − ν ) ν σ − − + r 2 dr r dr dr 2 2 If r sr = y.37)–(8.pmd 289 6/18/2008.41) y = Ar 2 + B r Chapter_08. (8. (8. 290 Advanced Mechanics of Solids where A and B are constants.48) 6/18/2008. sr = A + B r3 (8.43) Problem of Thick Hollow Sphere Consider a spherical body formed by the boundaries of two spherical surfaces of radii a and b respectively. (8. and sr = –pb when r = b From Eq.45) (8. pb = 0. B= 3 ⎡ 3 ⎤ a 3b 3 3 ⎢ −b pb + a pa + 3 ( pb − pa )⎥ r ⎣ ⎦ ⎡ 3 1 a 3b 3 p − p ⎤ 3 − + − b p a p ( b a )⎥ b a ⎢ b3 − a 3 ⎣ 2r 3 ⎦ If the sphere is subjected to internal pressure only.47) ⎛ 1 ⎞ a3 − 13 ⎟ 3 3 ⎜ 3 ⎝ 1 − (a / b ) b r ⎠ sf = σ θ = pa ⎛ 1 a3 1 ⎞ ⎜ + ⎟ 1 − ( a 3 / b3 ) ⎝ b3 2r 3 ⎠ In the case of a cavity inside an infinite or a large medium.44) (8.46) a3 ⎛ b3 ⎞ 1 + ⎜ ⎟ b3 − a 3 ⎝ 2r 3 ⎠ The above two equations can also be written as sf = σ θ = pa s r = pa (8. Hence.pmd 290 (8.37) ( ) sf = 1 d Ar 2 + B = A − B3 2r dr r 2r The constants A and B are determined from the boundary conditions. b3 pb − a 3 pa a 3b 3 ( pb − pa ) b −a b − a3 Thus. and sf = σ θ = s r = pa a3 ⎛ b3 ⎞ 1 − ⎜ ⎟ b3 − a 3 ⎝ r3 ⎠ (8. (8. The boundary conditions are therefore s r = –pa when r = a. 2:26 PM .42) B B –p a = A + 3 and − pb = A + 3 a b Solving. (8.42) And from Eq. b Æ • and the above equations reduce to a3 s r = − pa 3 r Chapter_08. Let the hollow sphere be subjected to an internal pressure pa and an external pressure pb . the general expressions for sr and sf are A= − sr = 3 1 b − a3 3 3 . 7 Calculate the thickness of the shell of a bomb calorimeter of spherical form of 10 cm inside diameter if the working stress is s kgf/cm2 (98 s kPa) and the internal pressure is s/2 kgf/cm2 (49 s kPa).51) Using the strain–displacement relations er = Chapter_08.47).46) and (8. Solution From equations (8. 1 d (r 2σ ) − 2 σ = 0 r r φ r 2 dr given by Eq. Example 8. sr = E ⎡ε (1 − ν ) + 2νεφ ⎤ ⎦ (1 + ν )(1 − 2ν ) ⎣ r (8. in terms of the displacement component ur. i.50) sf = E (νε r + εφ ) (1 + ν )(1 − 2ν ) (8. the thickness of the shell is 1. using Hooke's law and strain–displacement relations. (8. provided the outer surface of the body is free from pressure and provided that every point of this outer surface is at a distance greater than four or five times the diameter of the hole from its centre.Axisymmetric Problems 291 a3 (8. s f = σ θ = + pa Example 8. Solution We have er = 1 ⎡⎣(σ r − 2νσ φ ) ⎤⎦ E e f = 1 ⎡⎣(1 − ν ) σ φ − νσ r ⎤⎦ E Solving for sr and sf.3 cm Hence.49) 2r 3 The above equations can also be used to calculate stresses in a body of any shape with a spherical hole under an internal pressure pa. 2:26 PM . 3 3 ⎞ ⎛ sf = σ 3 5 3 ⎜ 1 + b 3 ⎟ 2 b −5 ⎝ 2×5 ⎠ Equating this to the working stress s 53 b − 53 3 ⎛ b3 ⎞ ⎜1 + ⎟=2 2 × 53 ⎠ ⎝ \ b ª 6. the maximum tensile stress is due to sf.8 Express the stress equation of equilibrium.e. Hence.3 cm. which occurs at r = a.pmd dur dr 291 and εφ = ur r 6/18/2008.37). The problem of a sphere strained by the mutual gravitation of its parts will now be considered .54) The complementary solution is u r = Cr + C1 r2 and the particular solution is (1 + ν ) (1 − 2ν ) ur = 1 ρ gr 3 10 E (1 − ν ) a Hence. we get E (1 − ν ) (1 + ν )(1 − 2ν ) u ⎞ E (1 − ν ) d ⎛ dur +2 r⎟ =0 r⎠ (1 + ν )(1 − 2ν ) dr ⎜⎝ dr or 8. From the general solution dur 3(1 + ν ) (1 − 2ν ) = C+ ρ gr 2 .52)]. (8. (8. 2:26 PM . Expressing the equations of equilibrium in terms of displacement ur [Eq. from Eq.50) sr = Chapter_08.52) STRESSES DUE TO GRAVITATION When body forces are operative. 10 E (1 − ν ) a dr and from Eq. r is the radius of any point from the centre and g is the acceleration due to gravity.pmd ur (1 + ν ) (1 − 2ν ) = C+ ρ gr 2 10 E (1 − ν ) a r E ⎡ε (1 − ν ) + 2νεφ ⎤ ⎦ (1 + ν ) (1 − 2ν ) ⎣ r 292 6/18/2008. C1 should be equal to zero as otherwise the displacement will become infinite at r = 0.53) r r r φ r 2 dr where gr is the body force per unit volume.36). the stress equation of equilibrium is. we have u ⎞ E (1 − ν ) d ⎛ dur r + 2 r ⎟ − ρg = 0 r⎠ a (1 + ν ) (1 − 2ν ) dr ⎜⎝ dr (8. r is the mass density. It is known from the theory of attractions ( ) gr = − ρ g r a where a is the radius of the sphere. The remaining constant is determined from the boundary condition sr = 0 at r = a. (8.292 Advanced Mechanics of Solids and substituting in the equilibrium equation.5 ⎡ d 2 ur 2 dur 2 ⎤ ⎢ 2 + r dr − 2 ur ⎥ = 0 r ⎢⎣ dr ⎥⎦ (8. 1 d r 2σ − 2 σ + γ = 0 (8. the complete solution is u r = Cr + C1 r 2 + 1 (1 + ν ) (1 − 2ν) ρ gr 3 10 E (1 − ν )a For a solid sphere. 3(1 + ν ) (1 − 2ν ) dur = C+ 1 ρ gr 2 10 E (1 − ν )a dr (3 − ν ) (1 − 2ν ) 3(1 + ν ) (1 − 2ν ) =− 1 ρ ga + 1 ρ gr 2 10 10 E (1 − ν ) E (1 − ν )a (1 + ν ) (1 − 2ν ) ρ g ⎡ 2 (3 − ν ) 2 ⎤ = 1 3r − a 10 E (1 − ν ) a ⎢⎣ (1 + ν ) ⎥⎦ The above value is zero when r2 = (3 − ν ) 2 a 3(1 + ν ) Hence. there is a definite surface outside which the radial strain is an extension. i. At the centre (r = 0).pmd 293 6/18/2008.e. if n is positive (which is true for all known materials). which is compressive. they are equal and have a magnitude 3−ν sr = σ φ = σ θ = 1 ρ ga (compressive) 10 1 − ν Further. of course. This result is due. 2:26 PM . = E (1 + ν ) (1 − 2ν ) C= − Hence.55) and from Eq.53) (3 − ν )a 2 − (1 + 3ν ) r 2 ρ g sf = σ θ = − 1 10 (1 − ν ) a (8. to the ‘Poisson effect’ of the large circumferential stress. Chapter_08. hoop stress. (8.Axisymmetric Problems = u ⎤ ⎡ dur E (1 − ν ) + 2ν r ⎥ r ⎦ (1 + ν ) (1 − 2ν ) ⎢⎣ dr = E (1 + ν ) (1 − 2ν ) 293 3(1 + ν ) (1 − 2ν ) ⎡ ρ gr 2 + 2νC ⎢⎣C (1 − ν ) + 10 Ea + ν(1 + ν ) (1 − 2ν ) ⎤ ρ gr 2 ⎥ 5E (1 − ν )a ⎦ (3 − ν ) (1 + ν ) (1 − 2ν ) ⎡ ⎤ ρ gr 2 ⎥ ⎢⎣C (1 + ν ) + 10 Ea (1 − ν ) ⎦ From the boundary condition sr = 0 at r = a. In other words. (3 − ν ) (1 − 2ν ) ρ ga 10 E (1 − ν ) (3 − ν ) 2 ρg sr = − 1 (a − r 2 ) 10 (1 − ν ) a (8.56) It will be observed that both stress components sr and sq are compressive at every point. for 1/ 2 ⎡ (1 + ν ) ⎤ r > a⎢ ⎣ 3(1 + ν ) ⎥⎦ the radial strain er is positive though the radial stress sr is compressive everywhere. 57b) dur u and εθ = r dr r From Hooke’s law. (8. Then.6 ROTATING DISKS OF UNIFORM THICKNESS We shall now consider the stress distribution in rotating circular disks which are thin. (8.294 Advanced Mechanics of Solids 8.57a) d rσ − σ + ρω 2 r 2 = 0 ( r) θ dr The strain components are.59) and rearranging (8.5b) can be used.59) Let rsr = y Then. or (8.60b) σθ = d2y dy +r − y + (3 + ν ) ρω 2 r3 = 0 2 dr dr The solution of the above differential equation is r2 (8. as before.62) (3 + ν ) s r = C + C1 12 − ρω 2 r3 8 r (8.e.pmd 294 6/18/2008. i.63) From Eq. The equation of equilibrium given by Eq.58) er = e r = 1 (σ r − νσ θ ) E e q = 1 (σ θ − νσ r ) E From Eq. (8. (8.2)] we put the body force term equal to the inertia term rw 2r. The z-axis is the axis of rotation. 2:26 PM . where w is the angular velocity of the rotating disk and r is the density of the disk material. provided we add the inertia force term rw 2r. (8. ∂σ r σ r − σ θ + + ρω 2 r = 0 r ∂r (8. We assume that over the thickness.61) (3 + ν ) y = Cr + C1 1 − ρω 2 r3 r 8 (8.60) Chapter_08. (8. from Eq.60a) dy + ρω 2 r 2 dr Substituting these in Eq. with sz = 0. the radial and circumferential stresses do not vary and that the stress sz in the axial direction is zero. (8. in the general equation of equilibrium [Eq.57b) (8.58) e r = d ( r εθ ) dr From Hooke’s law 1 (σ − νσ ) = e = d (r ε ) = 1 d (rσ − ν rσ ) r θ θ θ r E dr dr E r (8. (sr)r=b = 0 from which we find that 3+ν ρω 2 b 2 + a 2 .69) (8. since otherwise the stresses sr and sq become infinite at the centre. (sr)max = Chapter_08. If there are no forces applied there.67) 2 2 3+ν ⎛ 1 + 3ν 2 ⎞ r ⎟ ρω 2 ⎜ b 2 + a 2 + a b2 − 8 3+ν ⎝ ⎠ r The radial stress sr reaches its maximum at r = ab where 3+ν ρω 2 (b − a) 2 8 The maximum circumferential stress is at the inner boundary.64) ( C= sr = sq = ) C1 = − 3+ν ρω 2a 2b 2 8 2 2 3+ν ⎛ ⎞ ρω 2 ⎜ b2 + a 2 − a b2 − r 2 ⎟ 8 ⎝ ⎠ r (8.70) .pmd 295 (8.Axisymmetric Problems (1 + 3ν ) sq = C − C1 12 − ρω 2 r3 8 r The integration constants are determined from boundary conditions.65b) ρω 2b 2 − ρω 2r 2 8 8 These stresses attain their maximum values at the centre of the disk. where sq = sr = σ θ = 3+ν ρω 2b 2 8 (8. 3+ν ρω 2b 2 8 and the stress components become C= sr = 3+ν ρω 2 (b 2 − r 2 ) 8 (8. then (sr)r=a = 0. 8 Substituting these in Eqs (8.64) Solid Disk For a solid disk. we must take C1 = 0. 295 (8. The constant C is determined from the condition at the periphery (r = b) of the disk. then.63) and (8. (sr)r = b = C − 3+ν ρω 2b 2 = 0 8 Hence.68) 6/18/2008. 2:26 PM (8.65a) 3+ν 1 + 3ν (8. where 3+ν 1 − ν 2⎞ ⎛ (sq)max = ρω 2 ⎜ b 2 + a 4 3 + ν ⎟⎠ ⎝ It can be seen that (sq)max is greater than (sr)max.66) Circular Disk with a Hole of Radius a If there are no forces applied at the boundaries a and b. (8.0075 cm in radius). the radial displacement of the disk at the hole will be greater than the radial displacement of the shaft at its boundary. When the shaft and the disk are rotating.68) u disk = r (σ θ − νσ r ) E 2 2 3+ν ⎡ 1 + 3ν 2 = r ρω 2 ⎢b 2 + a 2 + a b2 − r E 8 3+ν r ⎣ 2 2 ⎛ ⎞⎤ −ν ⎜ b2 + a 2 − a b2 − r 2 ⎟ ⎥ ⎝ ⎠⎦ r Chapter_08.52 − 7. (i) What are the stresses due to shrink-fit? (ii) At what rpm will the shrink-fit loosen up as a result of rotation? (iii) What is the circumferential stress in the disk when spinning at the above speed? Assume that the same equations as for the disk are applicable to the solid rotating shaft also.52 − 0) or pc = 1044 kgf/cm2 (102312 kPa) The tangential stress at the hole will be the largest stress in the system and from Eq. there will be no radial pressure on any boundary.0075 (7.296 Advanced Mechanics of Solids When the radius a of the hole approaches zero.18 ¥ 106 kgf/cm2 (214 ¥ 106 kPa).52 ⎛ 37.27) pc = 2. In other words.67) and (8. the maximum circumferential stress approaches a value twice as great as that for a solid disk [Eq. by making a small circular hole at the centre of a solid rotating disk. From Eqs (8. Solution (i) To calculate the shrink-fit pressure. we have from Eq.e. (8.52 − 7.5 cm radius. The shrink-fit allowance is 1 part in 1000 (i.71) E Example 8.24) sq = 1044 × 7. 2:28 PM . i. (8.53 (37.pmd 296 6/18/2008.9 A flat steel disk of 75 cm outside diameter with a 15 cm diameter hole is shrunk around a solid steel shaft. u r = r εθ = r (σ θ − νσ r ) (8. (8. an allowance of 0.52 − 0) (37. (8.52 ⎞ 1 + ⎜ ⎟ (37.0075 cm at 7.71).58).52 ⎠ = 1131 kgf/cm2 (110838 kPa) (ii) When the shrink-fit loosens up as a result of rotation.66)].e. The displacement ur for all the cases considered above can be calculated from Eq. we double the maximum stress.52 ) ⎝ 7. The difference between these two radial displacements should equal D = 0. E = 2.52 ) × 2 × 7.18 × 106 × 0. 18 × 10 2 2 ⎛ 1.7 3. (8.18 × 106 = 34 ¥ 106 rw 2 Therefore.3 × 7.52 + ⎟ 0.Axisymmetric Problems 297 ( 3 + ν ) (1 − ν ) ρω 2 ⎛ b 2 + a 2 + 1 + ν b 2 a 2 − 1 + ν r 2 ⎞ = r ⎜⎝ 1 − ν r2 3 + ν ⎟⎠ E 8 3.0081 = 226066 (rad/s)2 Therefore.3 × 37.65a) and (8. (8.0075 or w 2 = 0.0075 × 106 × 1 × 981 4018 0.0081 × 226066 981 = 2184 kgf/cm2 (214024 kPa) Example 8. From Eq.52 + 37. w = 475 rad/s or 4536 rpm (iii) The stresses in the disk can be calculated from Eq. The maximum stress at this speed is to be 1164 kgf/cm2 (114072 kPa).3 × 7.3 × 7.pmd 3+ν 1 − ν 2⎞ ⎛ ρω 2 ⎜ b 2 + c 4 3 + ν ⎟⎠ ⎝ 297 6/18/2008. the maximum circumferential stress due to rotation alone is (sq)1 = Chapter_08.52 ) 8 × 2.52 − 1.3 × 0.52 8 3.7 ρω 2 × 7.65b) ushaft = r (σ θ − νσ r ) E = 1− ν ρω 2 r ⎡⎣(3 + ν ) b 2 − (1 + ν ) r 2 ⎤⎦ 8E = 0.7 7. (4052 – 34) ¥ 10–6 rw 2 = 0.9 × 7. Solution Let c be the radius of the shaft and b that of the disk.52 ⎞ ¥ ⎜ 37.3 7.52 + 7.52 − 1.10 A flat steel turbine disk of 75 cm outside diameter and 15 cm inside diameter rotates at 3000 rpm.5 − 1.5 × 7.3 ) = 1170 rw 2 = 1170 × 0.70).52 + 7.3 ρω 2 37.68) ( s r = 3.5 ρω 2 = × 6 8 2. at which speed the blades and shrouding cause a tensile rim loading of 44kgf/cm2 (4312 kPa). Find the maximum shrinkage allowance on the diameter when the disk and the shift are rotating. 2:28 PM .71). (8.52 ⎝ ⎠ = 4052 ¥ 10–6 rw 2 From equations (8.5 (3. Let h be the thickness of the disk.0008 cm 0. Hence.52 37.52 + 0.52 ) = 0.7 = 118.064 × 2.52 37. from Eq.9.27).52 = 1.53 37.52 ⎠ 37. and the tensile rim loading pb.52 − 7.064 ED or D= 110 × 10−6 = 0. the combined stress at 7.e.0081 − 91.5 + 1 ⎜ ⎟ 37.12) (sq)2 = 2 pc c 2 ⎛ b2 ⎞ + 2 pb b + 1 ⎜ ⎟ b2 − c2 ⎝ c2 ⎠ b2 − c2 = pc 2 ⎛ 7. pc = 110 kgf/cm2 The corresponding shrink-fit allowance is obtained from Eq. The equation of equilibrium can be obtained by referring to Fig. (8.7 2 = 1164 − 1170 × (100π ) × 0.7 This should be equal to 1164 kgf/cm2.8 Hence.3 ρω 2 37. the method of analysis developed in the previous section for thin disks of constant thickness can be extended also to disks of variable thickness. 1.08pc + 91.52 E ∆ × 110 = 2 × 7.5 cm radius is sq = 1170rw 2 + 1. i. (8.298 Advanced Mechanics of Solids ( = 3. ( 7.08pc + 91.18 8.7 DISKS OF VARIABLE THICKNESS Assuming that the stresses do not vary over the thickness of the disk.7 981 = 1164 – 953.pmd 298 6/18/2008.52 − 7.52 4 3.52 − 7. 8.08 pc = 1164 – 1170rw 2 – 91. 2:28 PM ) .52 ⎞ + 2 44 × 37.52 ⎝ 7.7 × 7.3 ) = 1170 rw 2 Owing to shrinkage pressure pc. For equilibrium in the radial direction ( ) ( ) ( ⎛ ∂ ( hσ r ) ⎞ ∆r ∆h 2 ∆r ⎜⎝ hσ r + ∂ r ∆r ⎟⎠ ( r + ∆r ) ∆θ + ρ r + 2 ∆θ h + 2 ω r + 2 ( ) − hσ r r ∆θ − 2σ θ h + ∆h ∆r sin ∆θ = 0 2 2 Chapter_08.7 Hence.5 – 91. varying with radius r. 59) hs q = (8. 2:28 PM . dur u er = and εθ = r dr r d Hence. Chapter_08.74) can easily be integrated.pmd 299 6/18/2008. Eq. er = (r εθ ) dr From Hooke’s law and Eq. (8.58).72) (8.e. i.73a) dy + hρω 2 r 2 dr The strain components remain as in Eq.9 Rotating disk of variable thickness Simplifying and going to the limit r d (hσ r ) + hσ r + ρω 2 r 2 h − σ θ h = 0 dr d (r hσ ) − σ h + ρω 2 r 2 h = 0 r θ dr Putting y = rhsr or (8.Axisymmetric Problems 299 Dr r Dq Fig 8.73a) and (8. The general solution has the form y = mrn + 2 + Ara + Brb r2 in which m= − (3 + ν )ρω 2 c (ν n + 3n + 8) and a and b are the roots of the quadratic equation x2 – nx + rn – 1 = 0 A and B are constants which are determined from the boundary conditions.73b) d2y dy ⎛ dy ⎞ +r − y + (3 + ν ) ρω 2 hr 3 − r dh ⎜ r − ν y⎟ = 0 (8.73b) 1 (σ − νσ ) = 1 d (rσ − ν rσ ) θ θ r E r E dr Substituting for sr and sq from Eqs (8. (8.75) in which C is a constant and n any number. (8.74) ⎠ dr h dr ⎝ dr dr 2 In the particular case where the thickness varies according to the equation y = Crn (8. pmd dur . the above equation gives we get i.e. dεr =0 dr e r = constant Hence. i.11 Determine the shape for a disk with uniform stress.e. Equilibrium Eq.8 ρω 2 2 r + C1 2σ 2 ⎞ ⎡ ρω 2 2 ⎤ ⎛ r + C1 ⎥ = C exp ⎜ − ρω 2 r ⎟ h = exp ⎢ − 2σ ⎠ ⎝ ⎣ 2σ ⎦ ROTATING SHAFTS AND CYLINDERS In Sec. εθ = 1 (σ θ − νσ r ) E E if sr = sq then er = eq .76) εz = ∂uz =0 ∂z (8. we assumed that the disk was thin and that it was in a state of plane stress with sz = 0. sr = sq . dr 300 εθ = ur . Solution From Hooke's law e r = 1 (σ r − νσ θ ) . as before.73) gives hs = d (rhσ ) + hρω 2 r 2 dr = hσ + rσ dh + ρω 2 hr 2 dr 1 dh 1 2 or = − ρω r h dr σ which upon integration gives log h = − or 8. r (8. It is also possible to treat the problem as a plane strain problem as in the case of a uniformly rotating long circular shaft or a cylinder.77) 6/18/2008.57): d (rσ ) − σ + ρω 2 r 2 = 0 r θ dr The strain components are. from Hooke's law.300 Advanced Mechanics of Solids Example 8. Let sr = sq = s. (8. er = Chapter_08. 2:28 PM . The equation of equilibrium is the same as in Eq. 8. Let the z-axis be the axis of rotation. (8.6. From strain–displacement relations er = dur dr and εθ = ur r e r = d ( r εθ ) = d ( r ε r ) dr dr Since er = eq .5 and 8. sr and sq are not only equal but also constant throughout the disk. 79c) 301 6/18/2008. (8. s z = ν (σ r + σ θ ) and hence.77) eq = e r = d (r εθ ) dr and using the above expressions for er and eq . (8. Eq.Axisymmetric Problems 301 From Hooke's law e r = 1 [σ r − ν (σθ + σ z )] E e q = 1 [σθ − ν (σ r + σ z )] E e z = 1 [σ z − ν (σ r + σ θ ) ] E Since ez = 0 (plane strain).79a) (1 + 2ν ) sq = C − C1 12 − ρω 2 r 2 8(1 − ) ν r (8.78) sq = (1 − ν ) y dy −ν − νρω 2 r 2 = d r dr dr ⎡ ⎤ ⎛ dy 2 3⎞ ⎢(1 − ν ) ⎜⎝ r dr + ρω r ⎟⎠ − ν y ⎥ ⎣ ⎦ 3 − 2ν d2y dy +r − y+ ρω 2 r 3 = 0 2 1 − ν dr dr The solution for this differential equation is or r2 (3 − 2ν ) y = Cr + C1 1 − ρω 2 r 3 r 8(1 − ν ) Hence. 2:28 PM . substituting in equations for er and eq er = 1+ ν (1 − ν ) σ r − νσ θ ] E [ 1+ ν (1 − ν ) σ θ − νσ r ] E [ From strain–displacement relations given in Eq.78) dy + ρω 2 r 2 dr Substituting for sr and sq in Eq.79b) ⎡ ⎤ 1 s z = ν ⎢ 2C − ρω 2 r 2 ⎥ 2(1 − ν ) ⎣ ⎦ (8. and Chapter_08. (8. we get d [(1 − ν )r σθ − ν r σ r ] dr With rsr = y.pmd (3 − 2ν ) s r = C + C1 12 − ρω 2 r 2 8(1 − ) ν r (8.76) gives for sq (1 − ν ) σ r − νσ θ = (8. 87) Comparing Eq. Chapter_08.82) sq assumes a maximum value at r = a and its value is (sq)max = (3 − 2ν ) ⎛ 2 1 + 2ν 2 ⎞ 2b + a 2 − a ρω 2 8 (1 − ν ) ⎜⎝ 3 − 2ν ⎟⎠ If a2/b2 is very small. from Eqs (8.84) sq = (3 − 2ν ) ⎛ 2 1 + 2ν 2 ⎞ b − r ρω 2 8 (1 − ν ) ⎜⎝ 3 − 2ν ⎟⎠ (8.80) sq = (3 − 2ν ) 8 (1 − ν ) ⎡ 2 a 2b 2 1 + 2ν 2 ⎤ 2 2 ⎢ ( a + b ) + 2 − 3 − 2ν r ⎥ ρω r ⎣ ⎦ (8. the maximum circumferential stress is doubled in its magnitude. the stresses are sr = (3 − 2ν ) 2 (b − r 2 ) ρω 2 8 (1 − ν ) (8. since otherwise the stresses would become infinite at r = 0. the constant C1 must be equal to zero.83). we find that (sq)max ≈ (3 − 2ν ) 2 2 b ρω 4 (1 − ν ) (8. From these C = K (a 2 + b 2 ) and C1 = − Ka 2b 2 where K= (3 − 2ν ) ρω 2 8 (1 − ν ) Hence. 2:28 PM . we find that by drilling a small hole along the axis in a solid shaft.83) (ii) For a long solid shaft.79a–c) sr = (3 − 2ν ) 8 (1 − ν ) ⎡ 2 a 2b 2 − r 2 ⎤ ρω 2 2 ( a + b ) − ⎢ ⎥ r2 ⎣ ⎦ (8.87) with Eq. (8. (8. sr = 0 at r = a and r = b which are the inner and outer radii.85) sz = ν ⎡b 2 (3 − 2ν ) − 2r 2 ⎤ ρω 2 ⎦ 4 (1 − ν ) ⎣ (8.302 Advanced Mechanics of Solids (i) For a hollow shaft or a long cylinder.86) The value of sq at r = 0 is (sq)max = (3 − 2ν ) 2 2 b ρω 8 (1 − ν ) (8. we find that C= (3 − 2ν ) ρω 2b 2 8 (1 − ν ) Hence.pmd 302 6/18/2008. the radius of the shaft. Using the other boundary condition that sr = 0 when r = b.81) sz = ν ⎡ (a 2 + b 2 ) (3 − 2ν ) − 2r 2 ⎤ ρω 2 ⎦ 4 (1 − ν ) ⎣ (8. 3 × π × 30 4 × 0.86).0081 × 300 2 × π 42 2 981 602 = 3120 kgf (31576 N) Tensile force Fz = 8.0081 kgf/cm3 (0.07938 N/cm3).Axisymmetric Problems 303 Example 8.9 SUMMARY OF RESULTS FOR USE IN PROBLEMS (i) For a tube of internal radius a and external radius b subjected to an internal pressure pa and an external pressure pb. sq is greater than zero if pa > pb ⎛ b 2 ⎞ + 1⎟ 2 ⎜⎝ a 2 ⎠ The maximum and minimum stresses are (sr)max = σ r (at r = b) = − pb (sr)min = σ r (at r = a ) = − pa (sq)max = σ θ (at r = a) = Chapter_08. 2:28 PM . the radial and circumferential stresses are given by (according to plane stress theory) sr = sq = pa a 2 − pbb 2 2 b −a 2 pa a 2 − pbb 2 2 b −a 2 2 2 p − p a b − a b2 r b2 − a 2 2 2 p − p a b + a b2 r b2 − a 2 sz = 0 The stress sr < 0 for all values of pa and pb. If the shaft is constrained at its ends so that it cannot expand or contract longitudinally. is rotating at a speed of 300 rpm.12 A solid steel propeller shaft.3. The weight of steel may be taken as 0. Fz = ν ⎡b2 (3 − 2ν ) π b 2 − π b4 ⎤ ρω 2 ⎦ 4(1 − ν ) ⎣ = ν π b4 ρω 2 2 Substituting the numerical values 0. Solution The total axial force is b Fz = ∫ σ z 2π r dr 0 and from Eq. calculate the total longitudinal thrust over a cross-section due to rotational stresses. 60 cm in diameter. substituting for sz. Poisson’s ratio may be taken as 0. (8.pmd 303 pa (a 2 + b 2 ) − 2 pb b 2 b2 − a 2 6/18/2008. whereas sq can be greater or less than zero depending on the values of pa and pb. the contact pressure pc due to difference D between the outer radius of the inner tube and the inner radius of the outer tube is given by pc = ∆ /c ⎡ 1 ⎛c + a ⎞ − ν1 ⎟ + 1 ⎢ ⎜ 2 2 ⎠ E2 ⎢⎣ E1 ⎝ c − a 2 2 ⎛ b2 + c2 ⎞⎤ ⎜ b2 − c 2 − ν 2 ⎟ ⎥ ⎝ ⎠ ⎦⎥ where E1. 2:28 PM . the stresses are given by sq = σ φ = sr = 1 b − a3 3 3+ν ρω 2 (b2 − r 2 ) 8 3+ν 1 + 3ν ρω 2b 2 − ρω 2 r 2 8 8 These stresses attain their maximum values at the centre r = 0. c and b are the corresponding values for the outer tube. Poisson’s ratio. n1. where sq = sq = σ r = Chapter_08.n2. the radial and circumferential stresses are given by sr = 1 b − a3 3 ⎡ 3 ⎤ a 3b 3 3 ⎢ −b pb + a pa + 3 − ( pb − pa ) ⎥ r ⎣ ⎦ ⎡ 3 a 3b 3 − ( p − p ) ⎤ 3 b a ⎥ ⎢ −b pb + a pa − 2r 3 ⎣ ⎦ (iv) For a thin solid disk of radius b rotating with an angular velocity w. inner radius and outer radius respectively. pa = 2 pa a 2 − pb (a 2 + b 2 ) b2 − a 2 1 p ⎛ b 2 + 1⎞ ⎟⎠ 2 b ⎜⎝ a 2 (sq)min = 0 2 ⎡ ⎛ b2 + a 2 ⎤ ⎞ (ur)r = a = a ⎢ pa ⎜ 2 + ν ⎟ − 2 pb 2 b 2 ⎥ 2 E ⎢ ⎝b − a b − a ⎦⎥ ⎠ ⎣ 2 ⎡ ⎛ b2 + a 2 ⎞⎤ (ur)r = b = b ⎢ 2 pa 2 a 2 − pb ⎜ 2 − ν⎟ ⎥ 2 E⎢ b −a ⎝b − a ⎠ ⎥⎦ ⎣ (ii) Built-up cylinders: When the cylinders are of equal length.304 Advanced Mechanics of Solids (sq)min = σ θ (at r = b) = If then. E2. a and c refer to the inner tube’s modulus.pmd 304 3+ν ρω 2b 2 8 6/18/2008. If E1 = E2 and n1 = n2. then (b 2 − c 2 )(c 2 − a 2 ) pc = ∆E3 b2 − a 2 2c (iii) For a sphere subjected to an internal pressure pa and an external pressure pb. p = 17000 kPa (174 kgf/cm2)] 8.2 In the above problem. determine the changes in the radii.1. If E = 1. assume s-tensile strength = 30000 kPa (307. Dr1 = 0.3 In Example 8.1 A thick-walled tube has an internal radius of 4 cm and an external radius of 8 cm.pmd 305 6/18/2008. if one uses the energy of distortion theory. which says that (σ )max − n (σ ) min ≤ σ tenslie strength where n is the ratio of s-tensile strength to s-compressive strength. For the present problem.Axisymmetric Problems 305 The radial outward displacement at r = b is 1− ν ρω 2b3 4E (v) For a thin disk with a hole of radius a.2 ¥ 108 kPa (1. [Ans.8 kgf/cm2).23 ¥ 106 kgf/cm2) and n = 0. ⎡ Ans. It is subjected to an external pressure of 1000 kPa (10. rotating with an angular velocity w.01 mm ⎤ ⎢⎣ Dr2 = 0. determine the internal pressure according to Mohr’s theory of failure.24.4 A thick-walled tube with an internal radius of 10 cm is subjected to an internal pressure of 2000 kgf/cm2 (196000 kPa). what will be the external radius of the cylinder? The rest of the data remain the same.007 mm ⎥⎦ 8. = 6.2 kgf/cm2) and s-compressive strength = 120000 kPa (1228. E = 2 ¥ 106 kgf/cm2 Chapter_08. [Ans. the stresses are (ur)r = b = sr = 2 2 3+ν ⎛ ⎞ ρω 2 ⎜ b 2 + a 2 − a b2 − r 2 ⎟ 8 ⎝ ⎠ r sq = 3+ν ρω 2 8 ⎛ 2 a 2b 2 1 + 3ν 2 ⎞ 2 ⎜⎝ b + a + 2 − 3 + ν r ⎟⎠ r ( ) (sr)max = σ r at r = ab = (sq)max = σ θ (at r = a) = and 3+ν ρω 2 (b − a )2 8 3+ν ρω 2 4 ⎛ 2 1 + ν 2⎞ ⎜⎝ b + 3 + ν a ⎟⎠ (sq)max > (sr)max The radial displacements are (ur)r = a = 3+ν 1 − ν 2⎞ ⎛ ρω 2 a ⎜ b 2 + a 4E 3 + ν ⎟⎠ ⎝ (ur)r = b = 3+ν 1 − ν 2⎞ ⎛ ρω 2b ⎜ a 2 + b 4E 3 + ν ⎟⎠ ⎝ 8.05 cm] 8. 2:28 PM .24 kgf/cm2). Calculate the change in the internal radius due to the pressure.7 Determine the dimensions of a two-piece composite tube of optimum dimensions if the internal pressure is 2000 kgf /cm 2 (196000 kPa). Est = 2 ¥ 106 kgf/cm2 (196 ¥106 kPa) and Ecu = 106 kgf/cm2 (98 ¥ 106 kPa).3.023 cm ⎥⎦ 8. 8. Determine the value of the external radius if the maximum shear stress developed is limited to 3000 kgf/cm2 (294 ¥ 106 kPa).10).pmd 306 6/18/2008. Chapter_08. Its internal and external radii are 10 cm and 15 cm respectively.5 A thick-walled tube is subjected to an external pressure p2. Calculate the stresses at the inner and outer radius points of each tube.3 and E = 200000 MPa (2041 ¥ 103 kgf/cm2). 2:28 PM .3 cm ⎤ ⎢⎣ Dr1 = 0. If the maximum shear stress is limited to 200000 kPa (2041 kgf/cm2). internal radius r1 = 8 cm and E = 2 ¥ 106 kgf/cm2 (196 ¥ 106 kPa). 8.000 kPa) in a composite tube consisting of an inner copper tube of radii 10 cm and 20 cm and an outer steel tube of external radius 40 cm.014 cm ⎥ ⎢ pc = 500 kgf/cm2 ⎥ ⎢ (49030 kPa) ⎦ ⎣ 8.6 Determine the pressure p0 between the concrete tube and the perfectly rigid core. determine the value of p2 and also the change in the external radius.10 Problem 8. ⎡ Ans. r2 = 17.306 Advanced Mechanics of Solids (196 ¥ 106 kPa) and n = 0. ⎡ Ans.3. Take r1/r2 = 0.16. Determine the contact pressure also. external pressure p2 = 0.nst = 0.34.6 8.19 mm 8. p 2 = 111 MPa (1133 kgf/cm2) ⎤ ⎢⎣ ⎥⎦ Dr2 = –0. [Ans. r2 ª 14 cm. n = 0.4 kgf/cm2] r1 p = 12 kgf/cm 2 (1126 kPa) r2 Fig.5 (Fig.8 Determine the radial and circumferential stresses due to the internal pressure p = 2000 kgf/cm2 (196. ncu = 0. rc = 0. Assume Ec = 2 ¥ 106 kgf/cm2. ⎡ Ans. p0 = 17. The maximum shear stress is to be limited to 1500 kgf/cm2 (147 ¥ 10 6 kPa). Check the strength according to the maximum shear theory. r3 = 24 cm ⎤ ⎥ ⎢ D = 0. 5 ¥ 10– 6. ⎡ Ans. E = 2. r3 = 140 mm.11): (a) Stresses due to the heavy-force fits with interferences of D1 = 0.06 mm and D2 = 0.2 ¥ 106 kgf/cm2. For inner tube: ⎤ ⎢ s r = –2000 kgf/cm2 (–196000 kPa) ⎥ ⎢ s r = –248 kgf/cm2 (–24304 kPa) ⎥⎥ ⎢ s t = 2672 kgf/cm2 (262032 kPa) ⎥ ⎢ ⎢ ⎥ s t = 920 kgf/cm2 (90221 kPa) ⎢ ⎥ For outer tube: ⎢ ⎥ 2 s r = –248 kgf/cm (–24304 kPa) ⎥ ⎢ sr = 0 ⎢ ⎥ ⎢ ⎥ s t = 413 kgf/cm2 (40474 kPa) 2 ⎢ ⎥ s t = 165 kgf/cm (16170 kPa) ⎢ ⎥ 2 pc = 248 kgf/cm (24304 kPa) ⎣ ⎦ 8.11 Determine for the composite three-piece tube (Fig.10 A composite tube is made of an inner copper tube of radii 10 cm and 20 cm and an outer steel tube of external radius 40 cm. Est.7. ⎡ Ans. If the temperature of the assembly is raised by 100°C.pmd 307 6/18/2008.9 In problem 8. 8. if the inner tube is made of steel (radii 10 cm and 20 cm) and the outer tube is of copper (outer radius 40 cm). Ecu and ncu are as in Problem 8. Chapter_08. For inner tube: ⎤ ⎢ 2 s r = –2000 kgf/cm (–196000 kPa) ⎥⎥ ⎢ s r = –577 kgf/cm2 (–56546 kPa) ⎥ ⎢ ⎢ s t = 1800 kgf/cm2 (176400 kPa) ⎥ ⎢ ⎥ s t = 371 kgf/cm2 (36358 kPa) ⎢ ⎥ For outer tube: ⎢ ⎥ s r = –577 kgf/cm2 (–56546 kPa) ⎥ ⎢ ⎢ ⎥ sr = 0 2 ⎢ ⎥ s t = 962 kgf/cm (94276 kPa) ⎢ ⎥ 2 s t = 385 kgf /cm (37730 kPa) ⎢ ⎥ ⎣ ⎦ pc = 577 kgf/cm2 (56546) 8. determine the radial and tangential stresses at the inner and outer radius points of each tube.12 mm in diameters (b) Stresses due to the internal pressure p = 2400 kgf/cm2 r1 = 80 mm. r2 = 100 mm. determine the circumferential and radial stresses at the inner and outer radii points of each tube. nst. ast = 12. r4 = 200 mm. For inner tube: ⎤ ⎢ ⎥ sr = 0 ⎢ 2 s r = –173 kgf/cm (–16954 kPa) ⎥⎥ ⎢ s t = – 461 kgf/cm2 (– 45080 kPa) ⎥ ⎢ ⎢ s = –288 kgf/cm2 (–28243 kPa) ⎥ ⎢ ⎥ For outer tube: ⎢ ⎥ 2 s r = –173 kgf/cm (–16954 kPa) ⎥ ⎢ sr = 0 ⎢ ⎥ ⎢ ⎥ s t = 288 kgf/cm2 (28243 kPa) ⎢⎣ ⎥⎦ s t = 115 kgf/cm2 (11270 kPa) 8. 2:28 PM . acu = 16.Axisymmetric Problems 307 ⎡ Ans.5 ¥ 10– 6. ⎡ Ans. find the inward radial displacement at the inside of a thick cylinder subjected to external pressure. u = pR 2 (1 − ν )/(2 Eh) ⎤ r ⎥ ⎢ ⎥ ⎢ pR ⎥ sq = σ φ = ⎢ 2h ⎥ ⎢ ⎥ ⎢ ⎥⎦ (sr) average = 1 p ⎢⎣ 2 8. the circumferential stress and the radial displacement. Determine the radial and circumferential stresses at point r. ⎡ ⎛ R2 ⎞ ⎢ Ans.11 Problem 8.308 Advanced Mechanics of Solids r4 r3 r1 r2 p Fig 8.14 An infinite elastic medium with a spherical cavity of radius R is subjected to hydrostatic compression p at the outside. 2:28 PM ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ .13 A thin spherical shell of thickness h and radius R is subjected to an internal pressure p. Determine the mean radial stress.12 The radial displacement at the outside of a thick cylinder subjected to an internal pressure pa is (2 p r r ) E (r − r ) 2 a b a 2 b 2 a By Maxwell's reciprocal theorem.11 8. Show that the circumferential stress at the surface of the cavity exceeds the pressure at infinity.pmd 308 6/18/2008. ur = ⎥ E (b 2 − a 2 ) ⎦ ⎣ 8. s r = − p ⎜1 − 3 ⎟ ⎝ r ⎠ ⎢ ⎢ ⎡ ⎢ R3 ⎤ sq = sf = − p ⎢1 + 3 ⎥ ⎢ 2r ⎦ ⎣ ⎢ ⎢ ⎢ (sq) at cavity = − 3 p 2 ⎣ Chapter_08. ⎡ 2 pb ab 2 ⎤ ⎢ Ans. nal = 0. where dimensions are in inches and pounds. (b) Find the rotation speed at which the contact pressure is zero. (c) What is the maximum tangential stress at the above speed. g = 2. (a) 4013 rpm ⎤ ⎢⎣ (b) 4028 rpm.pmd 309 6/18/2008. show that (a) the maximum horsepower is transmitted when w = w 0 / 3 and (b) this maximum horsepower is equal to 0. (a) Assuming that the shaft does not change its dimensions because of its own centrifugal force.0045 cm (a) Find the maximum tangential stress in the disk when it is at a standstill. producing a radial interface pressure p in the non-rotating condition.18 A disk of thickness t and outside diameter 2b is shrunk on to a shaft of diameter 2a. It is then rotated with an angular velocity w rad/s. (a) 647 kgf/cm2 (6349 kPa) ⎥ ⎢ (b) 4990 rpm ⎥ ⎢ 2 (c) 2622 kgf/cm (257129 kPa) ⎦ ⎣ 8.17 A steel disk of 75 cm diameter is shrunk on a steel shaft of 7. (b) Solve the problem without making assumption (a). [Ans. 13420 rpm] Chapter_08. Neglect the expansion of the shaft caused by rotation.0081/g kgm/cm3. Est = 7.001 cm/cm.76 10–3 kgf/cm3.19 A steel shaft of 7.5 cm diameter has an aluminium disk of 25 cm outside diameter shrunk on it. 2:28 PM . ⎡ Ans.3 ¥ 105 kgf/cm2 (175 ¥ 105 kPa). The interference on the diameter is 0.000366 a2t fpw0. find the speed at which the disk is just free on the shaft.15 A perfectly rigid spherical body of radius a is surrounded by a thick spherical shell of thickness h. ⎡ ⎤ 3(1 − ν )b3 p ⎢ Ans. Take r = 0. ⎤ ⎡ Ans.16 A steel disk of 50 cm outside diameter and 10 cm inside diameter is shrunk on a steel shaft so that the pressure between the shaft and disk at standstill is 364 kgf/cm2 (3562 kPa). 8. Calculate the rpm of rotation at which the shrink-fit loosens up.3. σ r = 3 3⎥ 2(1 − 2ν )a + (1 + ν )b ⎦ ⎣ 8. ⎥⎦ 8. determine the radial and circumferential stresses at the inner surface of the shell (b = a + h). If the shell is subjected to an external pressure p.5 cm diameter. If f is the coefficient of friction between disk and shaft and w0 is the value of the angular velocity for which the interface pressure falls to zero.Axisymmetric Problems 309 8. The shrink allowance is 0. If the bar is prevented from completely expanding in the axial direction. The increase in the length of a uniform bar of length L. y. In this chapter. for complete restraint. x.CHAPTER 9 9.e. y. the thermal stress needed is s = –aE (T – T0) where the negative sign indicates the compressive nature of the stress. (ii) Long circular cylinders—hollow and solid. then the average compressive stress induced is ∆L L where E is the modulus of elasticity. z) The body under consideration may be restrained from expansion or movement in some regions and external tractions may be applied to other regions.1 Thermal Stresses INTRODUCTION It is well known that changes in temperature cause bodies to expand or contract. we shall restrict ourselves to the analysis of the following problems: (i) Thin circular disks with symmetrical temperature variation. It is assumed in the above analysis that E and a are independent of temperature. i. in an elastic continuum. The determination of stresses under such situations may be quite complex. is D L = a L (T – T0) where a is the coefficient of thermal expansion. It is a function of time and the space coordinates (x. T = T(t. z). when its temperature is raised from T0 to T. If the expansion is prevented only partially.pmd 310 7/3/2008. the temperature change is not uniform throughout. 6:04 AM . (iv) Straight beams of arbitrary cross-section. Thus. s=E Chapter_09. In general. (iii) Spheres with purely radial temperature variation—hollow and solid. (v) Curved beams. then the stress induced is s = –kaE (T – T0) where k represents a restraint coefficient. the temperature rise is not uniform. therefore. The total strains at each point of a body are thus made up of two parts. ∂σ x ∂τ xy ∂τ xz + + +γx=0 ∂x ∂y ∂z . These are repeated for convenience. (9.2a) s y = le + 2mey – (3l + 2m) a T s z = le + 2mez – (3l + 2m) a T txy = mgxy. (1. The total strains at each point can. γ zx = 1 τ zx G G G The stresses can be expressed explicitly in terms of strains by solving Eq. If.1a) ) e z = 1 ⎡σ z − ν σ x + σ y ⎤ + αT ⎦ E⎣ gxy = 1 τ xy . These are s x = le + 2mex – (3l + 2m) a T (9. be written as ( ) ex = 1 ⎡σ x − ν σ y + σ z ⎤ + αT ⎦ E⎣ ey = 1 ⎡⎣σ y − ν (σ x + σ z )⎤⎦ + αT E ( (9.3) EQUATIONS OF EQUILIBRIUM The equations of equilibrium are the same as those of isothermal elasticity since they are based on purely mechanical considerations. tyz = mgyz.2 THERMOELASTIC STRESS–STRAIN RELATIONS Consider a body to be made up of a large number of small cubical elements. this normal strain in any direction is equal to a T.1b) γ yz = 1 τ yz . then distortions of the elements and consequently stresses should occur in the body. If the coefficient of linear thermal expansion is a. this expansion is the same in all directions and in this manner only normal strains and no shearing strains occur. (9. In rectangular coordinates these are given by Eq. The first part is a uniform expansion proportional to the temperature rise T. 9. The second part of the strains at each point is due to the stress components. we shall develop the general thermoelastic stress–strain relations and discuss two important general results.1a).Thermal Stresses 311 Before these specific problems are analysed. each element will tend to expand by a different amount and if these elements have to fit together to form a continuous body.2b) (9. then all the cubical elements will expand uniformly and all will fit together to form a continuous body.3 νE . If the temperatures of all these elements are uniformly raised and if the boundary of the body is unconstrained. however. (1 + ν ) (1 − 2ν ) µ=G= E 2(1 + ν ) (9. For any elementary cubical element of an isotropic body. tzx = mgzx The Lame constants l and m (= G) are given by l= 9.65). txy.e.e. if T(x. gzx 3 displacement components ux. the equations are the same as in isothermal elasticity. the problem of thermoelasticity consists in determining the following 15 functions: 6 stress components sx. gyz. y. on a part of the boundary. Displacement Boundary Conditions Here. (9. the surface tractions are prescribed and on the remaining part.5 ex = ∂ ux . 9.6) . displacements are prescribed. t) = a(t) + b(t)x + c(t)y + d(t)z (9. ∂x gxy = ∂ ux ∂ u y + . To repeat. Hence. Eq. (9. the prescribed boundary conditions may be a combination of the above two. uz so as to satisfy the following 15 equations throughout the body 3 equilibrium equations. (i) The method of arriving at a solution depends in general on the specific nature of the problem. In some cases. tyz. In most problems. the boundary conditions belong to one of the following two cases: Traction Boundary Conditions In this case. ∂z ∂y (9. uy. ∂y ∂x εy = ∂uy .5) and the prescribed boundary conditions. gy and gz are body force components.312 Advanced Mechanics of Solids ∂τ xy ∂σ y ∂τ yz + + + γy = 0 ∂x ∂y ∂z (9.19).1) 6 strain–displacement relations.5a) γ zx = ∂ uz ∂ ux + ∂x ∂z (9. ∂y γ yz = εz = ∂ uz ∂z ∂ u y ∂ uz + .4 STRAIN–DISPLACEMENT RELATIONS Only geometrical considerations are involved in deriving strain–displacement relations. these are given by Eqs (2. i. sz. the displacement components determined should agree with the prescribed displacements at the boundary.5b) SOME GENERAL RESULTS When the temperature distribution is known. It is shown in books on thermoelasticity that if the temperature distribution in a body is a linear function of the rectangular Cartesian space coordinates. Eq.4) 3 stress–strain relations. the stress components determined must agree with the prescribed surface traction at the boundary. these are 9. ez. sy. z. Eq.4) ∂τ xz ∂τ yz ∂σ z + + + γz = 0 ∂x ∂y ∂z where gx. i. In rectangular coordinates.18) and (2. tzx 6 strain components ex. gxy. ey. (9. (9. However. We shall first allow free expansion of the body due to temperature rise T0 with no restraint whatsoever.7a)] corresponding to free thermal expansion. uz = 0 ex = ey = ez = 0. Conversely. uy and uz together with the values of ux.2. Consequently. sx = sy = sz = –p and txy = tyz = tzx = 0. uz = a T 0 z (9. e x = e y = e z = a T0 Therefore. Since all cubical elements of the body expand freely. then the solution of the corresponding thermoelastic problem is ux = 0. as is obvious. provided that all external restraints. the strain components are ex = ε y = ε z = 1 [ − p − ν (− p − p) ] E 1 (1 − 2ν ) p E gxy = gyz = gzx = 0 =− Therefore. This has been discussed in Sec.Thermal Stresses 313 where t represents time. the above values of ux. If the body is subject to a hydrostatic state of stress. follows from the above statement and from the linearity of the boundary-value problem as formulated through Eqs (9. gxy = gyz = gzx = 0 txy = τ yz = τ zx = 0. this is the only temperature distribution for which all stress components are identically zero. i. then all the stress components are identically zero throughout the body. sx = sy = sz = 0. txy = tyz = tzx = 0 gxy = gyz = gzx = 0. σ x = σ y = σ z = − Eα T0 1 − 2ν To show this. under those provisions. ux = a T0 x. uy = a T0 y. body forces and displacement discontinuities are absent. therefore.4) and (9. we shall apply the principle of superposition. E u z = − 1 (1 − 2ν ) pz (9. no stresses develop and all elements expand in an identical manner. It. .7a) Now we apply boundary tractions to prevent this displacement. E uy = − 1 (1 − 2ν ) py. ux = − 1 (1 − 2ν ) px . Hence. With this state of stress. These results are obtained immediately by considering the stress compatibility relations.7b) E To get the original problem. the strains and displacements are altered.5) that a linear function may be added to or subtracted from a given temperature distribution without affecting the resulting stress distribution. then all elements of the body will experience the same state of stress (–p). should give zero displacements. uy and uz [Eq. (ii) We shall now show that if a body is subjected to a uniform temperature rise T = T0(t) and if the boundary of the body is prevented from having any displacements. uy = 0.1). (9.e. Corresponding to this state of stress. the equations of equilibrium are identically satisfied. 9. trq = 0. it is taken that the stresses and displacements also do not vary over the thickness. εθ = r dr r Substituting in Eq. from Eqs (8. we find sr = E [ε r + νεθ − (1 + ν ) α T ] 1− ν2 sq = E [εθ + νε r − (1 + ν ) α T ] 1− ν2 (9.1a) take the form. the stress–strain relations given by Eq. (9. With sz = 0.8) as stated earlier.11) Substituting these in the equation of equilibrium d dT (9.12) (ε + νεθ ) + (1 − ν ) (ε r − εθ ) = (1 + ν )α r dr r dr The strain-displacement relation for a symmetrically strained body. satisfy the equilibrium equation dσ r σ r − σθ =0 (9.314 Advanced Mechanics of Solids αT0 x − 1 (1 − 2ν ) px = 0 E or. because of symmetry.10) e q = 1 (σ − νσ ) + α T r E θ Solving the above equations for sr and sq . It is assumed further that it does not vary over the thickness and consequently.13) . Also.4). are r dur u .12) er = d 2 ur dr 2 + dT 1 dur ur − 2 = (1 + ν )α r dr dr r (9. (9.6 THIN CIRCULAR DISK: TEMPERATURE SYMMETRICAL ABOUT CENTRE Consider a thin disk subjected to a temperature distribution which varies only with r and is independent of q. The stresses sr and sq . (9. therefore.9) + dr r Body forces are ignored. e r = 1 (σ r − νσ θ ) + α T E (9. in polar coordinates. p= Eα T 1 − 2ν 0 sx = σ y = σ z = − p = − Eα T 1 − 2ν Hence.3) and (8. 9. therefore. (9. (8. a = 0 and in Eq.15). the outer radius of the disk. (9.5 that the stresses in a body are zero if the temperature distribution is linear with respect to a Cartesian frame of reference and if the body is free from external restraints and body forces.Thermal Stresses 315 This may be written as d ⎡1 d dT (rur ) ⎤ = (1 + ν )α dr ⎢⎣ r dr dr ⎦⎥ Integration of the above equation yields. The results are r s r = −α E 1 ∫ Tr dr + E 2 r a 1− ν2 sq = α E ⎡ ⎤ C1 (1 + ν ) − C2 (1 − ν ) 12 ⎥ ⎣⎢ r ⎦ 1 r E Tr dr − α ET + 2 ∫ r a 1− ν2 ⎡ 1⎤ ⎢⎣C1 (1 + ν ) + C2 (1 − ν ) r 2 ⎦⎥ (9. For a disk with a hole. a is zero. (9.14) u r = (1 + ν )α 1 ∫ Tr dr + C1r + 2 ra r It can be observed that the above expression becomes identical to Eq. the constant C2 should be equal to zero.13) and using the results in Eq.15) (9. We shall now consider two specific cases. y and z-axes. as otherwise ur would become infinite at r = 0.17) (9.8) if T is put equal to zero. a is the inner radius and for a solid disk. It should be observed that a linear variation of temperature with r will also induce stresses. A linear radial variation will not give a linear variation with respect to the x. The remaining constant C1 is determined from the condition that sr = 0 at r = b. (9.18) . In fact T = kr = k x2 + y2 + z 2 Solid Disk of Radius b In the case of a solid circular disk.11). From Eq. The lower limit a in the integral above depends on the disk. the stresses are b r ⎛ ⎞ s r = α E 1 Tr dr − 1 Tr dr 2 ∫ ⎜ b2 ∫ ⎟ r 0 ⎝ ⎠ 0 b r ⎛ ⎞ 1 1 sq = α E ⎜ −T + 2 ∫ Tr dr + 2 ∫ Tr dr ⎟ b r ⎝ ⎠ 0 0 (9.14) it is observed from L¢ Hospital’s rule that 1r ∫ Tr dr = 0 r →0 r 0 Hence.16) The constants C1 and C2 are determined by the boundary conditions. This does not contradict the statement made in Sec. The stress components are determined by substituting the value of ur in Eq. r C (9. we get lim b C 1 = (1 − ν ) α2 ∫ Tr dr b 0 Substituting this. 9. 16) C 1 = α (1 − ν ) s r = α 2E r r ⎡ r 2 − a2 b ⎤ Tr dr − ∫ Tr dr ⎥ ⎢ 2 2 ∫ a ⎣b − a a ⎦ (9. sr = 0 at r = a and r = b. we get b 1 a 2 b Tr dr = α − ν Tr dr .15) C1 (1 + ν ) − C2 (1 − ν ) 12 = 0 a b −α E 12 ∫ Tr dr + E 2 b a 1− ν ⎡ 1⎤ ⎢C1 (1 + ν ) − C2 (1 − v ) b 2 ⎥ = 0 ⎣ ⎦ Solving. then T is a function of r alone and is independent of z.19) Disk with a Hole of Radius a For a disk with a hole and traction free surfaces.21) and from Eq. 9.22) Tr dr ⎥ ∫ 2 2 r ⎣ b −a a a ⎦ If the temperature T is constant. all the shear stress components are zero and there are now three normal stress components sr.7 LONG CIRCULAR CYLINDER We shall now consider the nature of the thermal stresses induced in a long circular cylinder when the temperature is symmetrical about the axis and does not vary along the axis. owing to symmetry. sections far from the ends can be considered to be in a state of plane strain and we can analyse this problem with uz. then all the stress components are zero and the radial displacement is ur = a rT. (9. If the z-axis is the axis of the cylinder and r the radius. Once again. The stress-strain relations are e r = 1 [σ r − ν (σ θ + σ z ) ] + α T E . assumed to be zero.14) r ⎡ ⎤ (1 − ν )r 2 + (1 + ν ) a 2 b u r = α ⎢(1 + ν ) ∫ Tr dr + (9. Substituting these in Eq.316 Advanced Mechanics of Solids From L¢ Hospital’s rule 1 1 r Tr dr = T0 ∫ 2 r →0 r 2 0 where T0 is the temperature at the centre of disk. Since the cylinder is long. the axial displacement. Hence. sq and sz. C (1 ) ∫ ∫ 2 b2 − a 2 a b2 − a 2 a Substituting in Eqs (9. at r = 0 lim ⎛ 1 r 1 ⎞ sr (0) = σ θ (0) = α E ⎜ 2 ∫ Tr dr − T0 ⎟ 2 ⎠ ⎝b 0 (9. (9.20) sq = α 2E r r ⎡ r 2 + a2 b ⎤ Tr dr + ∫ Tr dr − Tr 2 ⎥ ⎢ 2 2 ∫ a ⎣b − a a ⎦ (9.15) and (9. Hence.29) sq = (9. be written from equations (9.30) 1 − ν (1 + ν ) (1 − 2ν ) When T = 0.16) as eq = ur = C 1+ ν 1 r Tr dr + C1r + 2 α 1 − ν r a∫ r r C ⎞ ⎛ C1 s r = − α E 12 ∫ Tr dr + E ⎜ − 2 1− ν r a 1 + ν ⎝ 1 − 2ν r 2 ⎟⎠ α ET + E ⎛ C1 + C2 ⎞ E 1 r Tr dr − ∫ 2 1− ν r a 1 − ν 1 + ν ⎜⎝ 1 − 2ν r 2 ⎟⎠ and from Eqs.24) (9. sr and sz can.24) can be written as Let er = (9.29) 2ν EC1 s z = − α ET + (9.27) (9. (9. in place of E. (9. therefore. (9. Solid Cylinder of Radius b As before. given by Eq. The constants C1 and C2 are determined from the boundary conditions. n1 and a1. from the last equation we get s z = ν (σ r + σ θ ) − ET α Substituting this in the expressions for ez and eq er = 1− ν2 E ⎛ ⎞ ν ⎜⎝ σ r − 1 − ν σ θ ⎟⎠ + (1 + ν ) α T 1− ν2 ⎛ ⎞ eq = σ − ν σ + (1 + ν ) α T E ⎝⎜ θ 1 − ν r ⎠⎟ ν =ν . The expressions for ur.23) (1 + ν ) α = α1 (9. (9. (9. Since the equation of equilibrium is the same as in the plane stress case. n and a respectively.22a–c).28). we have ez = 0. 1 1 1− ν 1− ν2 Then.28) (9.23).14) and (9. Normal force given by Eq.26) E1 θ Comparing the above expressions with Eq.Thermal Stresses 317 eq = 1 [σ θ − ν (σ r + σ z )] + α T E e z = 1 [σ z − ν (σ r + σ θ ) ] + α T E Since uz = 0.25). from L¢ Hospital’s rule lim r →0 1r Tr dr = 0 r ∫0 . the equations become identical to Eqs (8. E =E. Eqs (9. further analysis is identical to that in the plane stress case. and (9.30) is necessary to keep uz = 0 throughout. it is immediately observed that the expressions for er and eq in the plane strain case is similar to those in the plane stress case if we use E1.10).25) 1 (σ − ν1σ θ ) + α1T E1 r 1 (σ − ν1σ r ) + α1T (9. We shall now consider two particular cases. 28) with C2 = 0 and a = 0 in the lower limit of the integration C1 α 1 b Tr dr = 1 − ν b 2 0∫ (1 + ν ) (1 − 2ν ) or C1 = (1 + ν ) (1 − 2ν ) α b ∫ Tr dr (1 − ν ) b2 0 Comparing this with the value of C1 obtained for the plane stress case. accordingly. (9.34) The radial displacement is given by b b ⎤ 1+ ν ⎡ (9.23) sz = α E ⎛ 2ν b Tr dr − T ⎞ ⎟⎠ (1 − ν ) ⎜⎝ b2 0∫ (9.25).27) we observe that C2 should be equal to zero.35) α ⎢(1 − 2ν ) r2 ∫ Tr dr + 1 ∫ Tr dr ⎥ 1− ν ⎣ r0 b 0 ⎦ Note: In obtaining Eq. The values of sr and sq are.31) sq = α E ⎛ −T + 1 b Tr dr + 1 r Tr dr ⎞ ∫ ∫ ⎟⎠ (1 − ν ) ⎜⎝ b2 0 r2 0 (9. (9. we get from Eq. Consequently. from Eq. n and a to E1. and then converting these according to Eq. however. we observe that C1 for the plane strain case can be obtained from C1 for the plane stress case. For the solid cylinder. (9. This condition can be achieved by superposing a uniform stress distribution σ z′ = C3 so that the resultant force is zero. The value of C3 to make the total force zero in z direction is. from Eq. the ends of the cylinder are free. therefore. If. sr = α E ⎛ 1 b Tr dr − 1 r Tr dr ⎞ ∫ ⎟⎠ (1 − ν ) ⎜⎝ b 2 0∫ r2 0 (9.34). given by b 2ν EC1 C3pb2 = 2πα E ∫ Tr dr − π b2 (1 − ν ) 0 (1 + ν ) (1 − 2ν ) . as otherwise ur would be infinite at r = 0. (9. then the resultant force in z direction should be equal to zero. (9. we have assumed a plane strain condition with ez = 0. n1 and a1. Further. a stress distribution sz as given by Eq. the resultant of sz from Eq.318 Advanced Mechanics of Solids Hence.30) was necessary to maintain uz = 0. merely by changing E.32) and at r = 0 sr(0) = σ θ ( 0) = α E ⎛ 1 b Tr dr − 1 T ⎞ (1 − ν ) ⎜⎝ b 2 ∫0 2 0 ⎟⎠ (9. (9. Since sr = 0 at r = b.33) where T0 is the temperature at r = 0.30) is ur = 2ν EC 2πα E ∫ 2π rσ z dr = − (1 − ν ) ∫ Tr dr + (1 + ν ) (1 −1 2ν ) π b 2 0 0 b b The resultant of the superimposed uniform stress σ z′ = C3 is p b2C3. (9. s and then converting these according to Eq. calculate the stresses. the temperature at any distance r from the centre is given by the expression T= Ti log (b /r ) log (b /a ) Substituting this in Eqs (9. n.35) plus (–n C3r/E).36) sq = α E 1 ⎡ r + a b Tr dr + r Tr dr − Tr 2 ⎤ ⎥ ∫ (1 − ν ) r 2 ⎢⎣ b 2 − a 2 ∫a a ⎦ (9.36)–(9. we get sr = α E 1 ⎡ r − a b Tr dr − r Tr dr ⎤ ⎥ ∫ (1 − ν ) r 2 ⎢⎣ b 2 − a 2 ∫a a ⎦ (9. s1 in place of E.39) 2 2 2 2 Example 9.25).Thermal Stresses 319 The resultant sz distribution is given by sz = α E ⎛ 2 b Tr dr − T ⎞ ⎟⎠ (1 − ν ) ⎜⎝ b 2 ∫0 The ur displacement is then given by Eq. Eqs (9.22). (9. Assuming steady-state conditions. Accordingly.37) and ur = ⎤ (1 + ν )α ⎡ r (1 − 2ν ) r 2 + a 2 b Tr dr + ⎢ ∫ ∫ Tr dr ⎥ 2 2 (1 − ν )r ⎣ a (b − a ) a ⎦ (9. n1.39) ⎡ b a2 ⎢ − log r − 2 b − a2 ⎣ ⎛ b2 ⎞ b⎤ ⎜⎝1 − 2 ⎟⎠ log a ⎥ r ⎦ sr = α ETi 2 (1 − ν ) log (b / a ) sq = 2 ⎡ α ETi 1 − log b − 2 a 2 ⎢ r b −a 2 (1 − ν ) log (b / a ) ⎣ sz = α ETi 2 (1 − ν ) log (b / a ) ⎛ b 2 ⎞ log b ⎤ + 1 ⎜⎝ ⎟ a ⎦⎥ r2 ⎠ ⎡ b 2a 2 log b ⎤ ⎢1 − 2 log r − 2 a ⎦⎥ b − a2 ⎣ . by putting E1. What are the values of sq and sz near the inner and outer surfaces? Solution Under steady heat flow conditions. (9.1 The inner surface of a hollow tube is at temperature Ti and the outer surface at zero temperature.38) Also sz = b ⎞ αE ⎛ 2 Tr dr − T ⎟ ⎜ (1 − ν ) ⎝ (b 2 − a 2 ) ∫a ⎠ (9.20)–(9. Hollow Cylinder with Inner Radius a We shall write the solutions for this from the plane stress case results. These values are (sq)r = a = (σ z ) r = a = (sq)r = b = (σ z ) r = b = α ETi ⎛ 2b2 b⎞ 1− 2 log ⎟ ⎜ a⎠ b ⎝ b − a2 2(1 − ν ) log a α ETi ⎛ b⎞ 2a2 1− 2 log ⎟ ⎜ a⎠ b ⎝ b − a2 2(1 − ν ) log a If Ti is positive.37). whereas sq and sz are compressive at the inner surface and tensile at the outer surface. 9. we get d 2 ur dr 2 + 2 dur 2ur 1 + ν dT − 2 = α r dr 1 − ν dr r .e. the shear stresses are all zero and the normal stresses are such that sq = sf.42) Substituting these in the expressions for sr and sq and then substituting these in the equilibrium equation.320 Advanced Mechanics of Solids sr = 0 at r = 0 and r = b. i. The stress components sq and sz attain their maximum positive and negative values at r = a and r = b. These tensile stresses cause cracks in brittle materials such as stone. T is a function of r alone. (8.40) sf = E ⎡ε + νε r − (1 + ν ) α T ⎤ ⎦ (1 + ν ) (1 − 2ν ) ⎣ φ (9.41) From strain-displacement relations. brick and concrete. 1 d 2 2 (r σ r ) − σ φ = 0 2 dr r r d 2 (r σ r ) − 2rσ φ = 0 dr or The stress–strain relations are e r = 1 (σ r − 2νσ φ ) + α T E e f = εθ = 1 ⎡⎣σ φ − ν (σ r + σ φ ) ⎤⎦ + α T E Solving the above equations for sr and sf sr = E ⎡ (1 − ν ) ε r + 2νεφ − (1 + ν ) α T ⎤ ⎦ (1 + ν ) (1 − 2ν ) ⎣ (9. from Eq. Because of symmetry. The equation of equilibrium in the radial direction is. the radial stress is compressive at all points.8 THE PROBLEM OF A SPHERE We shall now consider the problem of a sphere subjected to purely radial temperature variation. we have er = dur dr and εφ = εθ = ur r (9. Solid Sphere In this case. Hence.44) (9. (9. The remaining constant C1 is determined from the condition that sr = 0 at r = b. In Eq. as otherwise.40).46) (9. EC1 2 EC2 1 − ⋅ =0 1 − 2ν 1 + ν a 3 . Hence. if the sphere is hollow.43). (9. from Eq. the strain components er and ef can be determined from Eq.44). the limit ⎛ 1 r ⎞ lim ⎜ − 2 ∫ Tr 2 dr ⎟ = 0 r →0 ⎝ r 0 ⎠ according to L¢ Hospital’s rule. Consequently.44) and (9. the displacement ur would become infinite at r = 0.47) Hollow Sphere Let a be the radius of the inner cavity and b the outer radius of the sphere. the lower limit a in the integrals may be taken as zero. The boundary conditions are sr = 0 at r = a and r = b. From the expression for ur.45) we shall consider two specific cases.45) r ⎛ b ⎞ s r = 2α E ⎜ 13 ∫ Tr 2 dr − 13 ∫ Tr 2 dr ⎟ (1 − ν ) ⎝ b 0 ⎠ r 0 sq = σ φ = α E ⎛ 2 b Tr 2 dr + 1 r Tr 2 dr − T ⎞ ∫ ⎟⎠ (1 − ν ) ⎜⎝ b3 ∫0 r3 0 (9. (9. from Eq. b EC1 − 2α E 13 ∫ Tr 2 dr + =0 (1 − ν ) b 0 1 − 2ν or C1 = 2α (1 − 2ν ) 1 b 2 Tr dr (1 − ν ) b3 ∫0 Substituting this in Eqs (9.44). (9. the constant C2 should be equal to zero. (9.43) α 12 ∫ Tr 2 dr + C1r + 22 1− ν r a r where C1 and C2 are constants to be determined from boundary conditions. the inner radius. The lower limit of the integral in the above equation is zero if the sphere is solid or is equal to a.42) and substituted in Eq.Thermal Stresses 321 This can also be written as d ⎡ 1 d (r 2u ) ⎤ = 1 + ν α dT r ⎥ dr ⎢⎣ r 2 dr ⎦ 1 − ν dr The solution is r C 1+ ν (9. The results are ur = sr = − 2α E 1 r Tr 2 dr + EC1 − 2 EC2 1 − ν r 3 a∫ 1 − 2ν (1 + ν )r 3 s q = sf = α E 1 r Tr 2 dr + EC1 + EC2 − ET α 1 − ν r 3 a∫ 1 − 2ν (1 + ν )r 3 1 − ν (9. 2 Let the inner surface of a hollow sphere be at temperature Ti and the outer surface at temperature zero. sf = 2α E 1− ν Example 9. according to the boundary conditions. Let the system be in a steady heat flow condition.48) and (9. sr = 0 at r = a and r = b. and (srf ) r = a = − α ETi b(b − a ) (a + 2b) 2(1 − ν ) b3 − a 3 (sf ) r = b = − α ETi a (b − a ) (2a + b) 2(1 − ν ) b3 − a 3 . Solution Substituting the above expression for T in Eqs (9.45). it is observed that sr is a maximum or a minimum when r2 = 3a 2 b 2 b 2 + ab + a 2 The expression for sf shows that its value increases with r for Ti positive.44) and (9. The temperature distribution is then given by T= Ti a b−a ⎛b ⎞ ⎜⎝ − 1⎟⎠ r Determine the stress distribution. the stress components can be calculated if the distribution of temperature is known.49) Tr dr + 2 ∫ Tr 2 dr − T ⎥ ⎢ 3 3 3 ∫ 2 ⎦ 2r a ⎣ 2(b − a )r a Therefore. we get sr = α ETi ab 1 − ν b3 − a 3 sq = σ φ = ⎡ a 2b2 ⎤ 1 2 2 ⎢ a + b − r (b + ab + a ) + 3 ⎥ r ⎦ ⎣ α ETi ab 1 − ν b3 − a 3 ⎡ a 2b2 ⎤ 1 2 2 ⎢ a + b − 2r (b + ab + a ) − ⎥ 2r 3 ⎦ ⎣ As can be seen. Differentiating the expression for sr with respect to r and equating the resulting expression to zero.49). The result is s r = 2α E 1− ν r ⎡ r 3 − a3 b 2 ⎤ Tr dr − 13 ∫ Tr 2 dr ⎥ ⎢ 3 3 3 ∫ r a ⎣ (b − a ) r a ⎦ (9.322 Advanced Mechanics of Solids b EC1 2 EC2 1 − 2α E 13 ∫ Tr 2 dr + − ⋅ =0 1−ν b a 1 − 2ν 1 + ν b3 The above equations can be solved for C1 and C2 and substituted in Eqs (9.48) ⎡ 2r 3 + a 3 b 2 1 r 1 ⎤ (9. Let this be f0 (x). The section that was plane before bending will. From Hooke’s law. the total axial displacement. ∂ ux ex = (9. we shall develop an elementary formula for normal stresses in free beams subjected to thermal loadings. therefore. then the displacement of any section in the axial direction due to temperature rise will be a function of the axial coordinate x. Let the y and z-axes lie in the plane of the section and let the x-axis be the axis of the beam (Fig. which is inclined at b to the y-axis. The axial strain ex is. The beam is assumed to be statically determinate and free of external loads. If now.1 Beam subjected to thermal loading The analysis is similar to the one used in Chapter 6 for the bending of beams. This is what was done in Chapter 6. x. 9. Hence.1(b). therefore. 9.9 NORMAL STRESSES IN STRAIGHT BEAMS DUE TO THERMAL LOADING In this section. 9. remain plane after bending and axial displacement. since sy and sz are assumed to be zero. such as BB in Fig. then the displacement in x direction of any point (y. According to this assumption. z) in a plane will be a linear function of the coordinates y and z. We shall make use of the Bernoulli–Euler assumption mentioned in Chapter 6. y and z-axes form a set of centroidal axes. the beam is allowed to undergo bending with the plane section remaining plane. This is equivalent to saying that the cross-section rotates about an axis. If the beam is prevented from bending and if warping is not allowed.1). The unknowns k and b are now replaced by f1′( x) and f 2′( x) .50) = f 0′( x) + yf1′( x) + zf 2′( x) ∂x The strain represented by the last two terms on the right-hand side is similar to the one expressed in Chapter 6. The temperature variation is arbitrary and the cross-section of the beam is also arbitrary. according to the Euler–Bernoulli hypothesis. sections which are plane and perpendicular to the axis before loading remain so after loading and the effect of lateral contraction (due to Poisson effect) may be neglected. and write the strain as ex = f 0′( x ) + ky ¢. We can also assume that the section rotates about an axis. sx = E(ex – a T ) . where y¢ is the perpendicular distance of a point from BB. will be ux = f0(x) + yf1(x) + zf2(x) where f1 and f2 are functions of x alone.Thermal Stresses 323 9. y y o z (a) x B z b o B (b) Fig. f1′ and f 2′ are then given by − I y M zt − I yz M yt I z M yt + I yz M zt p f1′ = f = . E ∫∫ αTy dA = − M zt .14) since the analyses in both cases have proceeded on similar lines. the above conditions become f 0′ ∫∫ dA + f1′ ∫∫ y dA + f 2′ ∫∫ z dA = ∫∫ αT dA f 0′ ∫∫ y dA + f1′ ∫∫ y 2 dA + f 2′ ∫∫ yz dA = ∫∫ α Ty dA (9. from Eq. 2 ∫∫ z dA = I y .55) reduces to M yt P M z sx = − α ET + t − zt y + (9. ∫∫ σ x z dA = M y = 0 i. (9. The expressions ∫∫ y dA = ∫∫ z dA = 0 because of the selection of the centroidal axes.324 Advanced Mechanics of Solids Substituting for ex sx = E[ f 0′( x) + yf1′( x ) + zf 2′( x) − α T ] (9. (9. The solutions for f 0′ .52) f 0′ ∫∫ z dA + f1′ ∫∫ yz dA + f 2′ ∫∫ z 2 dA = ∫∫ αTz dA The integrations extend over the entire cross-section. (9. f 0′ = t . then Iyz = 0 and Eq. ∫∫ dA = A.55) bears a very close resemblance to Eq. Further.53) E ∫∫ αTz dA = M yt A minus sign is used in the second expression in order to make the final result similar to the result of Chapter 6.e.55) Equation (9. (9. Substituting the expression for sx. or sx = −α ET + ( Iz M yt + I yz M zt ) pt ( I y M zt + I yz M yt ) − y+ z 2 A ( I y Iz − I yz ) ( I y Iz − I 2yz ) sx = −α ET + pt M zt ( yI y − zI yz ) + M yt ( yI yz − zIz ) + A I 2yz − I y Iz (9.56) A Iz Iy . the conditions to be satisfied at any section are ∫∫ σ x dA = 0. the axial stress sx is. If the axes chosen happen to be the principal axes of the section.51).52) can be written as Af 0′( x) = Let ∫∫ αT dA f1′I z + f 2′ I yz = ∫∫ αTy dA f1′I yz + f 2′ I y = ∫∫ αTz dA E ∫∫ αT dA = pt . ∫∫ σ x y dA = M z = 0. the resultant force over the section is zero and the moments about the y and z axes should individually vanish. (6. 2 ∫∫ y dA = I z . Eq.51) Since a free beam without external loading is considered. ′ (9.54) 2 EA E ( I y I z − I 2 yz ) E ( I y I z − I 2 yz ) Substituting these. ∫∫ yz dA = I yz Substituting these. 3 Deformation of a curved beam . T(r. 6. (6.2 Curved beam subjected to thermal loading Fig. i. Consider a free curved beam of arbitrary constant cross-section.2). the length of the same fibre becomes y ⎡ ⎤ (9. q ). the element deforms and it is assumed that sections which were plane before.33)] such that y dA ∫∫ r − y = 0 0 (9. remain plane after deformation.Thermal Stresses 325 9. the radius of curvature of the neutral surface is given by r0 [Eq. After deformation.e. 9. It is assumed that this arc lies in one of the principal planes of the beam. the origin 0 lies on the neutral axis and y is measured towards the centre of curvature. We shall follow the notations used in Chapter 6. the centre line of which is an arc of a circle (Fig. 9. In the isothermal case. Let the elementary length of an undeformed element enclose an angle Dq. The change in the length of the fibre is therefore y ⎡ ⎤ ⎢ r0′ − y − ∫ αT dy ⎥ (∆θ + δ∆θ ) − (r0 − y ) ∆θ 0 ⎣ ⎦ y y ⎡ ⎤ ⎡ ⎤ = ⎢ r0′ − r0 − ∫ αT dy ⎥ ∆θ + ⎢ r0′ − y − ∫ αT dy ⎥ (δ ∆θ ) 0 0 ⎣ ⎦ ⎣ ⎦ Y r0 o q r0 c Dq Y Dq + dDq ′ r0 q dD O Centroidal line Fig.7 STRESSES IN CURVED BEAMS DUE TO THERMAL LOADING An elementary analysis of the stresses developed in curved beams may be developed on the same basic assumptions as in the case of straight beams.3.7. 9. 9. Let the temperature vary as a function of r and q.57) As in Sec. A view of the deformed element is given in Fig.58) − − r y ′ ⎢0 ∫ αT dy ⎥ (∆θ + δ∆θ ) 0 ⎣ ⎦ The third term in the first bracket above represents the thermal expansion in y direction. Because of thermal loading. A fibre at a distance y from the chosen origin has a length (r0 – y) Dq before deformation. then we get a free curved beam subjected to thermal loading only.4(c). i. on this curved beam. ∫∫ σθ y dA = 0 Beam with Rectangular Section A somewhat more accurate result can be obtained for a curved beam with a rectangular cross-section and temperature independent of q.3 and 9.59) sq = E(eq – a T ) Therefore. If. This is obtained by superposing the result for a thin circular disk subjected to radial thermal loading with the result for the bending of a curved beam subjected to pure bending moment. Since the beam is free of external loadings.4(b). we apply an equal and opposite moment Fm.e. the ends of the element will be found (Examples 9. 9. as shown in Fig. y ⎡ ⎤ 1 r0′ − r0 − ∫ αT dy ⎥ + δ ∆θ (r0′ − y ) ⎢ (r0 − y ) ⎣ 0 ⎦ ∆θ From Hooke’s law. 9.4) to be subjected to zero resultant circumferential force and some moment Fm. . If a sectoral element is isolated from a disk.326 Advanced Mechanics of Solids Hence. the strain is y y ⎡⎛ ⎞ δ ∆θ ⎛ ⎞⎤ 1 − − r y αT dy⎟ ⎥ ′ ⎢⎜ r0′ − r0 − ∫ αT dy ⎟ + ∫ 0 ⎜ ∆ θ ( r0 − y ) ⎢⎣⎝ ⎠ ⎝ ⎠ ⎥⎦ 0 0 We observe that εθ = y ∫ αT dy << y 0 Hence. we should have ∫∫ σθ dA = 0. ⎧⎪ σθ = E ⎨ 1 y ⎫⎪ ⎡⎛ ⎤ ⎞ δ ∆θ (r0′ − y ) ⎥ − α T ⎬ ⎢⎜ r0′ − r0 − ∫ αT dy ⎟ + ∆θ ⎠ 0 ⎪⎩ r0 − y ⎣⎢⎝ ⎦⎥ ⎭⎪ δ ∆θ The two unknowns r0′ and are determined from the boundary conditions of ∆θ the beam. taking only sq into account εθ = (9. ∫ σθ dA = Fq = 0 A ∫ σθ y dA = Fm A where Fm is the moment about the median line. as shown in Fig. 4 Curved beam with rectangular section Example 9. 9.21).Thermal Stresses A C A Fm B B D = r0 C Fm D Fq = 0 Fq = 0 (b) A C Fm (a) 327 B D Fm (c) Fig. Fθ = αE ⎡ 1⎞ ⎤ 2 ⎛1 ⎢ β (b − a ) − β a ⎜⎝ b − a ⎟⎠ ⎥ b −a ⎣ ⎦ 2 2 ⎡⎛ ⎞ 1r + α E ⎢⎜ − ∫ Tr ′ dr ′⎟ r ⎢⎝ ⎠ a ⎣ b a ⎤ b b + ∫ T dr − ∫ T dr ⎥ ⎥ a a ⎦ In the above expression. The resultant circumferential force across any section is r b Fθ = ∫ σ θ dr = a b 2 b ⎡b b ⎤ a + dr Tr dr dr Tr dr ⎥ ∫ ∫ ∫ ∫ ⎢ 2 2 2 b − a ⎣a a ar a ⎦ αE r b ⎡b 1 ⎤ + α E ⎢ ∫ 2 dr ∫ Tr ′ dr ′ − ∫ T dr ⎥ a a ⎣a r ⎦ Let b ∫ Tr dr = β a Then. (9. the value of the circumferential stress sq is ⎡ r 2 + a2 σθ = α E 2 ⎢ 2 2 b r 2⎤ ∫ Tr dr + ∫ Tr ′ dr ′ − Tr ⎥ a ⎣b − a a ⎦ Let the disk be of unit thickness perpendicular to the plane of the paper. Solution From Eq. we have made use of the formula .3 Show that the resultant circumferential force across any radial section of a hollow disk subjected to thermal loading is zero. Example 9. the moment becomes Fm = b b 2 b ⎡b ⎤ a ⎢ ∫ r dr ∫ Tr dr + ∫ r dr ∫ Tr dr ⎥ b − a ⎣a a a a ⎦ αE 2 2 b ⎡b 1 r ⎤ + α E ⎢ ∫ dr ∫ Tr ′ dr ′ − ∫ Tr dr ⎥ r a a ⎣a ⎦ b Putting ∫ Tr dr = β .Advanced Mechanics of Solids 328 d dα V (α ) V (α ) dF (α . α ) d d dα α α U (α ) ∫ F (α . Solution If r0 is the radius of the median line and sq the circumferential stress on a fibre at r from the centre of curvature (Fig. 6:05 AM . calculate the stresses. from Example 9.pmd 328 7/3/2008. x) dx = ∫ U (α ) Substituting the limits.4 Determine the bending moment due to the circumferential stress across a section of a thin hollow disk subjected to radial thermal variation.21). Chapter_09. Using Eq.1 A thin hollow tube has its inner surface at temperature Ti and its outer surface at zero temperature. (9. The inner radius is a and the thickness of the tube is t.4b). the above expression becomes a Fm = ⎡β 2 ⎤ (b − a 2 ) + a 2 β log b ⎥ a⎦ b − a ⎣⎢ 2 αE 2 2 r ⎡ ⎤ + α E ⎢ log r ∫ Tr ′ dr ′ ⎥ a ⎣ ⎦ = b b − α E ∫ (log r ) Tr dr − α E β a a b ⎡ ⎤ b ⎞⎤ ⎡ 2 2 ⎛ b a α E β b 2 log 1 (log 1) (log r) Tr dr ⎥ + − + − − ∫ ⎢ ⎜ ⎟ ⎢ ⎥ 2 2 ⎝ ⎠ a 2(b − a ) ⎣ ⎦ a ⎣ ⎦ α Eβ 9. then the moment about the median line is b F m = ∫ σ θ (r − ρ0 ) dr a b b a a = ∫ σ θ r dr − ρ0 ∫ σ θ dr The second integral on the right-hand side is zero. x) dV dU dx + F(V . Assuming steady-state conditions. 9. α ) − F (U .3. it is observed that Fq = 0 There is no resultant circumferential force across any section. Thermal Stresses 329 ⎡ α ETi ⎛ t ⎞ ⎢ Ans. show that the circumferential stress at any point is equal to 2α E multiplied by the following expression: 3(1 − ν ) ⎡⎣ (mean temperature of the whole sphere) + (1/2 the mean temperature within the sphere of radius r) – 3 T ⎤⎦ 2 9. T = T(r). Calculate the resultant moment due to sq about the median line of the section across any radial section. The faces of the disk are kept at temperature Ti and the circumference is kept at temperature T0. Show that the radial stress sr at any radius r is proportional to the difference between the mean temperature of the whole sphere and the mean temperature of a sphere of radius r. ⎡ ⎡ α Eβ ⎡ 2 b ⎛ ⎞⎤ b + a 2 ⎜ 2 log − 1⎟ ⎥ + α E [ β (log b − 1) ⎢ ⎢ Ans. The temperature variation along r from the centre is given by r2 b2 The heavy ring is at temperature T0 and its strain is assumed to be negligible. (σ θ ) r = a = (σ z ) r = a = − ⎜⎝ 1 + ⎟ ν 2(1 ) 3 − a⎠ ⎢ ⎢ α ETi ⎛ t ⎞ ⎢ (σ θ ) r = b = (σ z ) r = b = ⎜1 − ⎟ ⎢⎣ 2(1 − ν ) ⎝ 3a ⎠ ⎤ ⎥ ⎥ ⎥ ⎥ ⎥⎦ 9. Consider a sectoral element.5 The temperature distribution in a long cylindrical conductor due to the passage of current is given by T = l(b2 – r2) . Show that the radial compressive stress in the disk at radius r is T = Ti − (Ti − T0 ) ⎛ 3 − ν r2 ⎞ σ r = 1 Eα (Ti − T0 ) ⎜ − 4 ⎝ 1 − ν b 2 ⎟⎠ 9.4. 9.4 A thin. uniform disk of radius b is surrounded by a heavy ring of the same material. i. Fm = 2 2 ⎢ ⎝ ⎠ a − 2( b a ) ⎣ ⎦ ⎢ ⎢ b ⎢ ⎤⎢ − ∫ log r Tr dr ⎥ ⎢ ⎢ a ⎦⎢ ⎢ ⎢ ⎢ b ⎢ ⎢ where β = ∫ Tr dr ⎢ ⎢⎣ a ⎢⎣ 9.e.2 A solid sphere of radius b is subjected to thermal loading T = T(r).3 A thin disk of inner radius a and outer radius b is subjected to a temperature variation which is symmetrical about the axis. The assembly just fits when the disk and the ring are at a uniform temperature. Also. as shown in Fig. 9. Eα λ ⎡ 2 2 ⎡ ⎢ Ans.6 A beam of rectangular section (see Fig.5) is subjected to a temperature distribution of the form ⎛ y2 ⎞ T = T0 ⎜ 1 − 2 ⎟ ⎝ c ⎠ Show that the normal stress induced is given by ⎛ y2 ⎞ σ x = 2 α T0 E − α T0 E ⎜ 1 − 2 ⎟ 3 ⎝ c ⎠ y x c z c l Fig. s r = − 4 (1 − ν ) (b − r ) ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎢ Eα λ (3r 2 − b 2 ) ⎢⎢ 4 (1 − ν) sq = − ⎢ ⎢ Eα λ sz = (2r 2 − b 2 ) ⎢⎢ 2 (1 − ν ) ⎣ 9.5 Problem 9. Determine the stresses due to thermal loading only. 9.330 Advanced Mechanics of Solids where l is a constant.6 . If the column is perfectly straight and is ideal in every respect. 10. the member will return to its straight equilibrium position. It may be observed that in order to keep point A in line with B. there exists a critical axial load Pcr. Q if the column is given a small lateral deflection by a force Q and the lateral force is removed. 10. following this approach.pmd 331 7/3/2008. such that under the action of Pcr.2 are as follows: 4p 2 EI p 2 EI p 2 EI for case (a). 6:08 AM . then it will remain straight and will be in equilibrium. 10. Fig. Consider a centrally loaded column with the lower end built-in and the upper end hinged (Fig. known as the Euler’s critical load or the buckling load can. be obtained by considerP ing the equilibrium of a slightly buckled column. When Q is removed. the column will continue to remain in the slightly buckled form and will be in equilibrium. a lateral reaction R will be necessary.1 Column Euler’s critical loads for the columns shown in Fig. The value of Pcr. therefore. which is slightly more complicated than the above cases. and for case (c) 4 L2 L2 L2 The method followed in elementary strength of materials to derive the above formulas will be applied to the following problem. eral load The critical loads for the first modes shown in Fig. The critical value of the compressive load is that value of Pcr which can keep the strut in a slightly buckled shape.3). In elementary strength of materials. However. 10. for case (b).CHAPTER Elastic Stability 10 10. Chapter_10A.2 with lathave been obtained. If now a small lateral force Q is P applied in addition to the axial force P (Fig. 10.1 EULER’S BUCKLING LOAD Consider a long slender column subjected to an axial force P.1). the member will act as a beam and will assume a deflected form and will remain deflected as long as the lateral force Q is acting. pmd 332 7/3/2008.x) EI dx 2 (10.Advanced Mechanics of Solids 332 P P P L /4 L /2 L L /4 P (c) P (b) (a) Fig. we obtain the following equations: y=0 at x = 0 and at x = L. 10. at x = 0 C1 + R L = 0 P C1 cos kL + C2 sin kL = 0 kC2 . These are dy =0 dx Substituting these. 10.R = 0 P Chapter_10A. 8:11 AM .1) The general solution of this equation is y = C1 cos kx + C2 sin kx + R (L .2 (a) Column with one end fixed and the other end free. (c) Column with both ends fixed The bending moment at any section x is M = Py .3 Column with one end fixed and the other end hinged d 2 y = .x) Using the expression P A R M = -EI L x d 2y dx 2 d 2y = -Py + R(L .x) P The constants C1 and C2 and the reaction R will have to be determined from the boundary conditions.R(L .k2y + R (L .x) dx 2 Let k2 = P EI The differential equation then becomes EI y B Fig. (b) Column with both ends hinged. Elastic Stability 333 The trivial solution is C1 = C2 = R = 0. The smallest value of kL satisfying this equation is kL = 4. a fairly general problem of a column with a varying cross-section and with end as well as intermediate loading. EI1 Chapter_10A. Example 10. C2 = P k P Substituting into the second equation.699 L)2 As another example. then to have zero moments at the hinged ends A and B.y2) P1 + P2 = k32 .x) . 6:08 AM .P2(d .4. 10. which means Pcr = k2EI = 20. A R Solution If the equilibrium position of the buckI1 led column is as shown in Fig. d x R I2 y B M = P1y1 + R(L . it is I2. (a). which means that the column remains straight. the moment is P2 L2 R = P2d /L 333 δ P2 L P2 = k22 .19 EI L2 ≈ π 2 EI (0. we obtain the transcendental equation C1 = - tan kL = kL (10.2) A solution to this can be obtained from a graphical plot.493.4) is compressed by two forces P1 and P2. For the L1 portion. 10.pmd (a) Let y1 be the deflection at any section of the L1 portion and y2 the deflection at any section of the L2 portion.y2) or − EI 2 d 2 y2 dx Using the notations 2 = P 1y 2 + P1 = k12 .x) and for the L2 portion.1 A column AB with hinged ends (Fig. EI 2 P2 = k42 EI1 7/3/2008.P2(d .x) Using Eq.1 − EI1 d 2 y1 dx 2 = P1 y 1 + δ P2 L (L . it is necessary to have a horizontal reaction R such that L1 RL = P2d or. The non-trivial solution is R 1 R L. EI 2 (L . The moment of inertia for the length L1 of the column is I1 and for the remaining length L2.x) . Determine the critical P1 value of the force P1 + P2. P1 + P 2 Fig.4 Example 10. 10. we shall analyse by the elementary method. the moment is M = P1y2 + R(L . i.334 Advanced Mechanics of Solids the differential equations become d 2 y1 = − k12 y1 − dx 2 d 2 y2 δ k 2 (L − x) 4 L δ = − k32 y2 − k22 x L dx 2 The solutions of the above equations are . as an example. 2 δ k4 L k12 δ k22 L k32 at x = L2. Chapter_10A. (ii) Stability problem as an eigenvalue problem and (iii) Energy methods to obtain approximate solutions to buckling problems.pmd 334 7/3/2008.e.87 L) 2 In this chapter. we get (P1 + P2)cr = π 2 EI (0. y 1 = C1 sin k1x + C2 cos k2x - y 2 = C3 sin k3x + C4 cos k4x + The boundary conditions are y1 = 0 at x = L. at x = L2 The first four conditions yield C1 = δ (k12 L + k42 L1 ) k12 L (sin k1 L2 − tan k1 L cos k1 L2 ) C 2 = −C1 tan k1 L. y 1= d y2 = 0 at x = 0. (L . L1 = L2. I1 = I2 = I and P1 = P2 are taken. C4 = 0 Substituting the values of these constants into the continuity condition. C3 = δ (k32 L − k22 L2 ) k32 L sin k3 L2 . If. ⎛ dy2 ⎞ ⎛ dy1 ⎞ ⎜⎝ dx ⎟⎠ = ⎜⎝ dx ⎟⎠ at x = L2 the following transcendental equation is obtained: k42 k12 − k12 L + k42 L1 k22 k32 L − k22 L2 = 2+ k1 tan k1 L1 k3 k3 tan k3 L2 For any particular case. we shall discuss three specific topics: (i) Beam columns. 6:08 AM .x) x y2 = d ⎛ dy1 ⎞ ⎛ dy2 ⎞ ⎜⎝ dx ⎟⎠ = ⎜⎝ dx ⎟⎠ at x = L2. the above equation can be solved to give the critical load. we could apply the principle of superposition because of the linear relationship between the load acting and the deflection produced. The lateral load q will be assumed to be positive when it is acting downward. In such cases.pmd 335 7/3/2008.Elastic Stability 335 I.6(a).6(b). The beams that are subjected to axial loads in addition to lateral loads are known as beam columns. We shall restrict our analysis to beam columns having symmetrical cross-sections. it was found that stresses were directly proportional to the applied loads.3 BEAM COLUMN EQUATIONS Consider the beam shown in Fig. in the case of a beam which is subjected to lateral forces Q1 and Q2 as well as to axial forces P as shown in Fig. the principle of superposition cannot be applied without certain modifications. 10.10.10. then the deflection d due to the combined action of Q1 and Q2 is d 1 + d 2. If d1 is the deflection caused at point S due to Q1 alone. 6:08 AM V + DV .5 (a) Cantilever with loads Q1 and Q2. 10. An elementary length D x of the beam before deflection is considered.6 Beam column with varying lateral load Chapter_10A. Similarly. 10.5(a) shows a cantilever loaded by forces Q1 and Q2. (b) Beam column with axial and lateral loads In arriving at this result it is assumed that the deflection d1 caused by Q1 does not affect the action of Q2. In addition. The free body diagram of length D x is shown in Fig. it can be seen that the bending moment caused by P depends on the deflection y produced by the lateral forces Q1 and Q2. The beam carries a distributed lateral load of intensity q. and d2 the deflection at the same point S due to Q2 alone.2 BEAM COLUMN In the theory of bending discussed in Chapter 6. which is a function of x. q( x ) P q P M P V S dy/dx M + DM P Dx y Dx (a) (b) Fig. In these cases. Figure 10. the beam is subjected to an axial compressive force P. 10. it is assumed that the deformations produced by one load do not affect the action of the other loads.5(b). However. Q2 S Q1 Q2 Q1 P P y (b) (a) Fig. BEAM COLUMNS 10. in determining the deflections. pmd 336 7/3/2008. (10. 10.5). and 10. Q The bending moment at any seca c tion x is due to Q as well as P.3)-(10. The relations among the load.7) are the basic differential equations for the bending of beam columns. Chapter_10A. Using Eq.dV (10. the bending moment due to P cannot be calculated until the deflection is determined. in the limit as D x Æ 0 q = . L y The beam column is therefore Fig. x P P However. we ignore the effects of shearing deformation and assume that the curvature of the beam axis is given by d 2y = -M (10.7) simply supported and carrying a load Q at distance a from the right hand support. Eqs (10. In the limit.7) dx 4 dx 2 Equations (10. 10. The beam is subjected to an axial force P.4) and (10.7 Beam column with concentrated load statically indeterminate.3) can be written as EI EI d3y dx3 d4y +P +P dy = -V dx (10.6) d2y =q (10. Summing forces in Y directions -V + q D x + (V + DV ) = 0 or.336 Advanced Mechanics of Solids The shearing force V and bending moment M acting on the sides of the element are assumed to be positive in the directions shown.3) dx Taking moments about point S and assuming that the angle between the axis of the deformed beam and the horizontal is small. we get ∆x dy + (V + ∆V ) ∆ x − ( M + ∆M ) + P ∆x =0 2 dx ∆x dy + V + ∆V − ∆M + P = 0 q ∆x dx 2 M +q∆x or.5) dx 2 where E is the Young’s modulus of the beam material and I is the moment of inertia about the neutral axis.4 EI BEAM COLUMN WITH A CONCENTRATED LOAD Consider a uniform beam of span L (Fig. as D x Æ 0 dy V = dM − P (10. These equations reduce to the familiar beam bending equations when P is equal to zero. 6:08 AM .4) dx dx As in the case of the bending of beams. shearing force V and bending moment M are obtained from the equilibrium considerations of the element. a) Q( L − a) (L .x) . (10.9) P EI the above equations become Putting k2 = d2y dx 2 + k2 y = - Qa x EIL Q ( L − a) ( L − x) EIL dx The general solutions of these equations are d2y 2 + k2 y = - y = A cos kx + B sin kx − Qa x PL x £ (L .a) (10.Py L for for x ≥ (L . The conditions are (i) y = 0 at x = 0 and at x = L (ii) y at x = (L .a) x £ (L .5) M= EI EI d 2y dx 2 d 2y dx 2 =- Qa x . from Eq.pmd Qa ( L − a) PL Qa PL 337 7/3/2008.a) would be the same according to both solutions.Elastic Stability 337 The bending moment at any x is Qa x + Py L M= for x £ (L .a) should be the same according to both solutions.8) for x ≥ (L . From condition (i) A = 0 and C = -D tan kL Conditions (ii) and (iii) give y = C cos kx + D sin kx − B sin k ( L − a) − Qa ( L − a) PL = D [sin k ( L − a ) − tan k L cos k ( L − a) ] − Bk cos k ( L − a ) − Chapter_10A.a) (10.x) + Py L Hence.a) PL The constants A. 6:08 AM .a) Q( L − a) ( L − x) x ≥ (L . (iii) The tangent at x = (L .Py L = − Q (L − a) (L . B. C and D are to be determined from the conditions of the beam. Q sin ka Qa dy cos kx − = P sin kL PL dx 0£x£L-a Q sin k ( L − a) Q( L − a) dy cos k ( L − x) + = − x≥L-a£L P sin kL PL dx d2y dx 2 = − 2 d y dx 2 = Qk sin ka sin kx 0 £ x £ L .e. we obtain the following formulae.13) QL3 3(tan u − u ) (10. when d= p/2 = or Chapter_10A. 6:08 AM .10b).11) (10.a P sin kL Qk sin k ( L − a) sin k ( L − a) P sin kL (10.14) 48EI u3 It is observed from the above equation that d becomes infinite when u = p /2. Pk sin kL D=− Q sin k ( L − a) Pk tan kL Substituting these constants.e. if a = L/2.10a) for x ≥ ( L − a ) (10. then d=y Putting u= at L Q = 2 2 Pk kL L = 2 2 ⎛ P⎞ ⎜⎝ ⎟⎠ EI kL kL ⎤ ⎡ ⎢⎣ tan 2 − 2 ⎥⎦ 1/2 (10. the solutions are y= Q sin ka Qa sin kx − x Pk sin kL PL y= Q sin k ( L − a) Q ( L − a) sin k ( L − x) − ( L − x) Pk sin kL PL for x ≤ ( L − a) (10.10a) and (10. which are useful.pmd P= L 2 P EI π 2 EI L2 338 = Pcr 7/3/2008. i. the load acts at midspan.12) x≥L-a£L As a particular case.10b) By the differentiation of Eqs (10.Advanced Mechanics of Solids 338 = Dk [ cos k ( L − a ) + tan kL sin k ( L − a )] + Q ( L − a) PL From the above two equations we get B= Q sin ka . i. 6:08 AM . the lateral deflections becomes very large. however small Q is. a2 and a3 respectively from the right hand side support B. Because of the linear relationship between deflection y and load Q. then the bending moment at section x is M= Chapter_10A. (10. 10.15) If Q1.Elastic Stability 339 So. let y1 be the deflection due to Q1 alone with P.8. From Eq.8 a2 P x a1 Beam-column with several lateral loads At some section left of Q3. Q3 A P Q2 Q1 B y a3 Fig. (10. We should recall that Pcr given above is the Euler buckling load for a slender column with hinged ends.8)]. provided the same axial force acts on the bar. y2 the deflection at the same point due to Q2 alone with P. 10. the principle of superposition in a modified form can be used for the effect of the lateral load. then the deflection also is doubled. Q2 and Q3 are acting together with P. Consider the beam shown in Fig. whereas the relation between the deflection and axial force P is more complicated. we get the following: EI d 2 y1 EI d 2 y2 EI d 2 y3 dx dx dx 2 2 2 =- Q1a1 x − Py1 L =- Q2 a2 x − Py2 L =- Q3 a3 x − Py3 L By adding these equations EI d 2 ( y1 + y2 + y3 ) dx 2 = − Q1a1 Qa Qa x − 2 2 x − 3 3 − P ( y1 + y2 + y3 ) L L L (10. which is subjected to an axial force P and three lateral loads Q1.5) with each Q and P [(similar to Eq. 10.pmd Q1a1 Qa Qa x + 2 2 x + 3 3 x + P( y1 + y2 + y3 ) L L L 339 7/3/2008.5 BEAM COLUMN WITH SEVERAL CONCENTRATED LOADS Equations (10. Q2 and Q3 acting at distances a1.10a and b) show that deflection y is proportional to lateral load Q. if Q is doubled (with P remaining unaltered). when P becomes equal to Pcr. and y3 the deflection due to Q3 alone with P. Hence. the resultant deflection can be obtained by the superposition of the deflection produced by each lateral load acting in combination with axial force P. the P x load on an elementary length qDa Da is q Da.. an respectively from the right hand support. therefore.. am. Q2. a2. Let a beam be acted upon by n lateral loads Q1. when there are several loads acting on a bar with an axial force P.10) as EI y= sin kx m x m ∑ Qi sin kai − ∑ Qi ai Pk sin kL i =1 PL i =1 + sin k ( L − x) m L−x m ∑ Qi sin k ( L − ai ) − ∑ Qi ( L − ai ) Pk sin kL i = m +1 PL i = m +1 (10... 10. (10.5).. we get d 2 ( y1 + y2 + y3 ) Qa Qa Qa = . y At distance x from A.10a) is Dy = q ∆ a sin ka qa∆a sin kx − x Pk sin kL PL Assuming a to vary from 0 to (L .e.9 Beam-column with continuous deflection due to the load lateral load q Da from Eq. Therefore. 10...x) from the right hand support.. the deflection at x due to the loading to the left of x. the deflection due to this part of the load. Qm+1.6 CONTINUOUS LATERAL LOAD The result obtained for a single load and the method of superposition can be used to solve the problem of a beam subjected to a continuously distributed load and an axial force. using Eq.. Qn at distances a1. (10. which lies between Qm.1 1 x − 2 2 x − 3 3 x − P ( y1 + y2 + y3 ) L L L dx 2 This is identical to Eq. is y1 = sin kx Pk sin kL L− x ∫0 q sin ka da − x PL L− x ∫0 qa da Similarly...10a) and the principle of superposition. and Qm+1. (10. the Fig. (10. i.. Q2. At distance a from B A q P the right hand support. Qm+2.. Let q be the intensity of loading. for a varying from (L .9. (10. Da x a 10... Qm lying to the right of x and Eq...x) to L. Qm.10b) for loads Qm+1. from Eq. (10.Qn lying to the left of x.. . (10..340 Advanced Mechanics of Solids From Eq. .10b) and the principle of superposition. the total deflection is obtained from Eq. we have made use of Eq.15). (10.. am+1.16) In the above equation.. At some point x.pmd sin k ( L − x) Pk sin kL 340 L−x L L ∫L − x q sin k ( L − a) da − PL ∫L − x q ( L − a) da 7/3/2008. 6:08 AM . shown in Fig.. is y2 = Chapter_10A.10a) for loads Q1. using the general differential equation for a beam-column given by Eq.7). (10. (10. we have put f = 2ux/L. (10. y= Chapter_10A. we obtain = qL4 4 ⎡ cos (u − f ) ⎤ qL2 1 ( L − x) x − − ⎢ ⎥ 2 16 EIu 4 ⎣ cos u ⎦ 8EIu as in Eq. y = A sin kx + B cos kx + Cx + D + y=0 d2y =0 dx 2 and at x = 0 and at x = L These give B = -D = q k2P . Summing the above two quantities and integrating with q as constant. A= q 1 − cos kL .17). for a beam uniformly loaded laterally and subjected to an axial force P. sin kL k2P C=− qL 2P Therefore.Elastic Stability 341 The total deflection y due to complete loading is y = y 1 + y 2. C and D are constants. 6:08 AM . y= q 1 − cos kL q qL qx 2 + − − + sin kx (cos kx 1) x 2P 2P k 2 P sin kL k2P Putting kL 4u 2 EI and P = 2 L2 the above equation can be written as u= y= ⎡1 − cos 2u qL2 2ux 2ux ⎤ sin cos 1 ( L − x) x + − − ⎢ ⎥ 2 L L 16 EIu 4 ⎣ sin 2u ⎦ 8EIu qL4 ⎡ sin f − cos 2u sin f + sin 2u cos f ⎤ qL2 ( L − x) x − 1⎥ − ⎢ 2 sin 2u 16 EIu ⎣ ⎦ 8EIu where.2 Determine the deflection y.e.7) is d4y d2y +P 2 =q 4 dx dx where q is a constant. B. Simplifying. i. The boundary conditions are that the deflection and bending moment are zero at x = 0 and x = L. 1 with u = kL as 2 y= qL2 ⎡ cos (u − 2ux L) ⎤ 1 − − ⎥⎦ 8EI u 2 ( L − x) x cos u 16 EIu 4 ⎢⎣ qL4 (10. Solution The general differential Eq.pmd qL4 341 7/3/2008. we obtain the result.17) Example 10. The general solution of this equation is EI qx 2 2P where A. x) for x.e. From Eq. 10. and θb = − ⎜ ⎟ at x = L ⎝ dx ⎠ ⎝ dx ⎠ Chapter_10A.19) − + − P ⎜⎝ sin kL L ⎟⎠ P ⎢⎣ sin kL L ⎥⎦ The slopes qa and qb at A and B can be obtained by differentiating the above expression and putting x = 0 and x = L.18) gives the deflection produced by Mb. In this equation. keeping the product Qa = Mb constant. as shown in Fig. (10. M b ⎛ sin kx x ⎞ M a ⎡ sin k ( L − x) L − x ⎤ (10. 6:08 AM .Advanced Mechanics of Solids 342 10.11. . 10.7 BEAM-COLUMN WITH END COUPLE Consider the beam shown in Fig.10a) y= Q sin ka Qa sin kx − x Pk sin kL PL Now. i.pmd 342 7/3/2008.10. ⎟ = k(Qa) = Mbk 3! ⎝ ⎠ Hence. Thus. 10.11 Beam-column with two end couples curve for Ma and Mb acting together.10 Beam-column with one end couple load Q is applied is made to approach zero. For this purpose.18) If two couples Ma and Mb are applied at the ends A and B of the bar. we get the deflection Fig. if we substitute Mb Ma B A Ma for Mb and (L . however. . we assume y that the distance a where the Fig. the limit of Q sin ka as a Æ 0 and Q Æ•. the equation for the deflection curve can be obtained by applying the modified principle of superposition. y= ⎛ dy ⎞ ⎛ dy ⎞ qa = ⎜ ⎟ at x = 0. where a moment Mb is applied at support B. 10. y= Mbk M sin kx − b x Pk sin kL PL or y= M b ⎛ sin kx x ⎞ − P ⎜⎝ sin kL L ⎟⎠ (10. Equation (10. we obtain moment Mb acting at support B. The solution to this can be Mb obtained from the equation for A B the deflection curve due to a P P single concentrated load. Adding these results. qa qb P P we obtain the deflection produced by Ma. so that Qa = Mb remains constant is ⎛ ⎞ k 3a3 Lt Q ⎜ ka − + . In this manner. dy k cos kx L − x qa sin kx q ( L − x) = sin ka da − sin k ( L − x) ∫ dx Pk sin kL 0 L Pk sin kL L − Chapter_10A. 10. V ) − F (α . The slopes are where qa = Ma L M L ψ (u ) + b φ (u ) 3EI 6 EI (10. x) dx = ∫ U (α ) Then. Replacing P a x L Fig.12 Example 10. U ) dα dα dα U (α ) ∫ F (α .20a) qb = Mb L M L ψ (u) + a φ (u) 3EI 6 EI (10. The loading Qi will q A B q L now be equal to ada acting P at distance a from B.Elastic Stability 343 The negative sign in qb expression is because of the sign convention adopted [Fig. 6:08 AM . (10. as shown in Fig.16).3 y= the summation by integration. we make use of the formula d dα V (α ) V (α ) dF (α . x) dV dU dx + F (α .12. Solution We make use of Eq.pmd L−x 2 k cos k ( L − x) qa 2 1 x q( L − x) da + − × ∫ PL 0 L PL L Pk sin kL 343 7/3/2008.20b) f (u) = 3⎛ 1 1⎞ − u ⎜⎝ sin 2u 2u ⎟⎠ y (u) = 3 ⎛ 1 1 ⎞ − ⎜ 2u ⎝ 2u tan 2u ⎟⎠ u= and 1 1 ⎛ P⎞ kL = L ⎜ ⎟ 2 2 ⎝ EI ⎠ 1/ 2 Example 10. Find the slopes at the ends of the column.3 A beam-column carries a triangular load. the equation becomes sin kx L − x qa x L − x qa 2 sin ka da − da ∫ ∫ Pk sin kL 0 L PL 0 L + sin k ( L − x) Pk sin kL L qa L−x sin k ( L − a) da − ∫ PL L−x L L qa ( L − a ) da ∫ L−x L Now. (10. 10.11)]. ⎧ dy ⎫ q ⎨ ⎬ = (6 sin kL − 6kL − k 2 L2 sin kL) ⎩ dx ⎭ x = L 6 Pk 2 L sin kL II GENERAL TREATMENT OF COLUMN STABILITY PROBLEMS (As an Eigenvalue Problem) 10. given by Eq.22) Equations (10. For the present. Hence. then the general equation becomes.344 Advanced Mechanics of Solids L sin k ( L − x) q ( L − x) sin kx L qa ∫ L sin k ( L − a)da + Pk sin kL ⋅ L−x + 1 PL L L − x q ( L − x) x L qa ( L − a) da − ∫ PL L−x L L q q L 2 ⎧ dy ⎫ a ka da − sin = ⎨ ⎬ ∫ ∫ a da PL2 0 ⎩ dx ⎭ x = 0 PL sin kL 0 \ = q 2 3Pk L tan kL (3 tan kL − 3kL − k 2 L2 tan kL) Similarly.21) and (10.22) are the equilibrium equations of a slightly buckled beam subjected to axial load only. Eq. i.e.21). 6:08 AM .8 GENERAL DIFFERENTIAL EQUATION AND SPECIFIC EXAMPLES The general differential equation derived for a beam column. then the general equation can be derived on the same lines as in Sec. 10. The boundary conditions. can be quite general. Hence. with q = 0.23) 7/3/2008.3. If the column is not subjected to lateral loads. On using the notation k2 = P . the axial load will represent the critical load. d4y d2y (10.21) + P =0 dx 4 dx 2 If EI varies along x. can be used as a general equation to determine the critical loads of buckled bars. the end conditions.21) becomes EI d4y dx 4 Chapter_10A. these equations represent the general differential equations for a column.7). (10. giving EI d2y⎞ d2y d2 ⎛ EI + P =0 ⎜ ⎟ dx 2 ⎝ dx 2 ⎠ dx 2 (10. (10.pmd 344 + k2 d2y dx 2 =0 (10. we shall assume that EI is constant along x and use Eq. (10. 13. and sin kL = 0 Hence. the deflection and bending moments are zero at the two ends.e. and d2y dx 2 = 0 at x = 0 and x = L These conditions give B = 0. i. 10. C = 0. D = 0.13 L/3 (c) (b) Various modes of buckling of a column with two ends hinged k= nπ ⎛ P ⎞ =⎜ ⎟ L ⎝ EI ⎠ where A is undetermined. i. 6:08 AM .25) Pcr = n2π 2 EI L2 Column with One End Fixed and the Other End Free The end conditions are at fixed end (i.6) d3y dy +P =0 dx dx3 From these. Chapter_10A. the shapes of the buckled bar are as shown in Fig.pmd A=C=0 345 and C=0 cos kL = 0 7/3/2008. i.11). at x = 0).e. moment and shear force are zero. the amplitude of the deflection remains undetermined. Hence. d2y/dx2 = 0 and from Eq.e.Elastic Stability 345 The general solution of this equation is y = A sin kx + B cos kx + Cx + D (10. C = -Ak. y = 0 and dy/dx = 0 at free end (at x = L). (10.e. The corresponding loads are obtained from the equation 1/ 2 or nπ x L (10. For n = 1. n = 2 and n = 3. 10. We can consider the following particular cases: Column with Hinged Ends (Fundamental Case) In the case of a bar with hinged ends (Fig. the constants are determined as B = -D. EI A sin kL + B cos kL = 0. in determining the load that keeps the column in a slightly buckled form. 10. kL = np The deflection curve is then obtained as y = A sin kx = A sin A L/2 x (a) Fig.24) The constants are determined from the end conditions of the bar. y = 0. When n = 2. . 3. etc. 10.26) 2 or. we obtain or (10. L x L/3 L/5 (a) (b) (d) (c) Fig. 6:08 AM . Pcr = π EI 2 4 L2 This is the smallest load that can keep the column in a slightly buckled shape. 4L The corresponding deflection curves are shown in Fig. 10. .14 Various modes of buckling of a column with one end fixed and the other end free Column with One End Fixed and Other End Pinned This case is shown in Fig. 10. 10. R L x d2y = 0 at x = L dx 2 With these.pmd P 346 Fig. The end conditions are dy y = 0.346 Advanced Mechanics of Solids kL = (2n − 1) π 2 The deflection curve is therefore y = B(1 . = 0 at x = 0 dx y = 0..15 Column with one end fixed and other end pinned 7/3/2008. Since the top end is pinned. we get the other critical loads as kL = π 2 Pcr = 9π EI .1. . a lateral force R is necessary to keep the column in that position.14. etc. 10.cos kx) or y = B ⎡1 − cos (2n − 1) π x ⎤ ⎢⎣ 2 L ⎥⎦ With n = 1..15 and was discussed in Sec. 4 L2 2 Pcr = 25π 2EI . the general solution yields the following equations: B+D=0 Ak + C = 0 Chapter_10A. dy = 0 for x = 0 dx dy = 0 for x = L dx Substituting these in Eq. Chapter_10A. i.e.24). -sin kL + kL cos kL = 0 or tan kL = kL (10. (10. B+D=0 Ak + C = 0 A sin kL + B cos kL + CL + D = 0 Ak cos kL . For the existence of a non-trivial solution. satisfy the above transcendental equation.27) The load which keeps the column in a slightly buckled form should. the boundary conditions are y = 0.e. which means that the deflection curve is a straight line (i. The determinant is 0 1 0 1 k 0 1 0 = − sin kL + kL cos kL 0 0 L 1 sin kL cos kL 0 0 For the existence of a non-trivial solution. we get y = 0. therefore. 6:08 AM .pmd 2(cos kL − 1) + kL sin kL = 0 347 7/3/2008.699 L)2 Column with Ends Fixed For a column with both ends fixed.19 EI L2 = π 2 EI (0.493 and the corresponding critical load is Pcr = 20. the above quantity should be equal to zero.Elastic Stability 347 CL + D = 0 A sin kL + B cos kL = 0 A trivial solution for the above set of equations is A = B = C = D = 0.Bk sin kL + C = 0 For the existence of a non-trivial solution. The smallest root of this equation is kL = 4. y = 0). the following determinant should be equal to zero 0 1 0 1 k 0 1 0 =0 sin kL cos kL L 1 k cos kL − k sin kL 1 0 i. the determinant of the coefficients should be equal to zero.e. 10. C = 0.28) is kL kL kL cos − sin =0 2 2 2 kL kL = 2 2 The lowest root of this transcendental equation is kL/2 = 4.6) EI Chapter_10A. because the force P becomes inclined.18π 2 EI L2 Fig. 6:08 AM .16 Modes of buckling of a The deflection curves corresponding to column with both these two critical loads are shown in ends fixed Fig. B = -D and the deflection curve is therefore 2nπ x ⎞ ⎛ y = B ⎜ cos − 1⎟ ⎝ ⎠ L (10. (10. (10. During buckling. The load may be applied through the tension of a cable passing through the fixed point O.17). a shearing force is developed at the top end.28) One solution is sin kL =0 2 kL = 2nπ and hence Pcr = i. Since the deflection is assumed to be very small. whenever sin kL/2 = 0.e. 10.493 and hence or L/2 tan Pcr = 8.pmd d3y dx 3 +P dy δ =P dx c 348 7/3/2008. the vertical component will be almost equal to P and the horizontal component is V =−P δ c From Eq. 4n 2π 2 EI L2 Noting that sin kL = 0 and cos kL = 1.16. the constants are found as A = 0.348 Advanced Mechanics of Solids or sin ( ) kL kL kL kL cos − sin =0 2 2 2 2 (10. 10. This shearing force is equal to the horizontal component of the inclined force P.29) A second solution to Eq. Column with Load Passing Through a Fixed Point Consider a column with one end fixed and the other end loaded in such a manner that the load passes through a fixed point (Fig. 30) 7/3/2008.17 or (d) o (b) o (a) d3y dy k 2δ + k2 = 3 dx c dx This is one of the boundary conditions.Elastic Stability d 349 P P P P P δ c d L c c c o C x o (c) Column with load passing through a fixed point Fig. kc kc Substituting these. 10. 6:08 AM . The other conditions are dy = 0 at x = 0 dx y = 0 and d2y = 0 at x = L dx 2 Substituting these in the general solution given by Eq. (10. the deflection curve is obtained as y= δ kc δ c [ tan kL (cos kL − 1) + kL − sin kL ] c⎞ ⎛ tan kL = kL ⎜1 − ⎟ ⎝ L⎠ or Chapter_10A.24) B+D =0 Ak + C = 0 C= δ c A sin kL + B cos kL = 0 Solving these equations. the constants are obtained as A=− δ δ . B= tan kL = − D.pmd C= 349 (10. pmd 350 (10.9 BUCKLING PROBLEM AS A CHARACTERISTIC VALUE (EIGENVALUE) PROBLEM In Sec. When c approaches infinity. i. the values of kL and c =0 L L2 Pcr π2 E I are as follows: π p = 2. the critical load is obtained as kL = π and Pcr = π 2 EI L2 which is the same as the one obtained for case (i). the column acts as a hinged-end column.350 Advanced Mechanics of Solids The above equation gives the value of the critical load for any given ratio of c/L. 10. The moment at the top end is also zero and consequently.e. 10. The differential equation then becomes d2 ⎡ d2y⎤ d2y + =0 EIp ( x ) P ⎢ ⎥ dx 2 ⎣ dx 2 ⎦ dx 2 Chapter_10A. This can be explained by the fact that when the line of action of P passes through the base point. For three specific values of c/L. When c = L. the column behaves like it did in case (ii). We can write I x = Ip( x) where I is a constant moment of inertia and p(x) is a dimensionless function of x.25 When c = 0. starting with the general differential equation of equilibrium of a slightly buckled column subjected to axial load only.493 L2 Pcr 2 1 0. which is the case discussed in (iii). we get the case of a column pinned at the top and fixed at the bottom. The specific examples analysed. the fundamental case. which will be discussed now.22) as d2 ⎛ d2 y⎞ d2y + =0 EI P ⎜ x ⎟ dx 2 ⎝ dx 2 ⎠ dx 2 (10. These features give us a better insight into the problem and provide a basis for the application of energy methods to buckling problems. where the load is aways vertical.31b) 7/3/2008. 6:08 AM .31a) It is assumed that the moment of inertia Ix can vary along the axis of the column.05 π 2 EI • 1 kL = 4. bring out some general characteristic features of the differential equation. the buckling problem was discussed.8. The general differential equation of equilibrium of a slightly buckled column with general boundary conditions was given by Eq. the moment at the base is zero and the end behaves like a hinged end. (10. (10. x) + C3 + C4 L where C1. be referred to as homogeneous boundary conditions. a non-trivial solution. in which we are interested. we get a set of four linear homogeneous equations with the following general form α11C1 + α 21C2 + α 31C3 + α 41C4 = 0 α12 C1 + α 22 C2 + α 32C3 + α 42 C4 = 0 α13C1 + α 23C2 + α 33C3 + α 43C4 = 0 α14 C1 + α 24 C2 + α 34C3 + α 44 C4 = 0 (10. 10. When EIx is constant. therefore.32) This is a homogeneous differential equation with the following general solution: x (10.35) Some of these coefficients a s are transcendental functions of the parameter k while others are constants. However.Elastic Stability 351 Dividing by EI and using the notation k2 = P . the above equation can be EI written as 2 d2 ⎡ d2y⎤ 2 d y + =0 p ( x ) k ⎢ ⎥ dy 2 ⎣ dx 2 ⎦ dx 2 (10. (10.35) is that C1 = C2 = C3 = C4 = 0. x) + C2φ 2 (k . The most frequently encountered boundary conditions are For freely supported end y=0 For a fixed end y = 0 and For a free end d2y dx 2 and d2y dx 2 =0 dy =0 dx (10.36) 7/3/2008. exists if the determinant of the coefficients is zero. α11 α12 ∆= α13 α14 Chapter_10A.6) as d ⎡ d2y⎤ 2 dy =0 ⎢ p( x) 2 ⎥ + k dx ⎣ dx dx ⎦ These boundary conditions are linear and homogeneous equations and will. (10. x) and f2(k. A trivial solution of Eq.33) y = C1φ1 (k . We have come across these kinds of equations in Sec. The constants have the dimensions of length and are determined from the boundary conditions. representing a straight undeflected column (y = 0). x) are transcendental functions of x and k. C3 and C4 are constants and f1(k. the general solution has the form given by Eq. Substituting these homogeneous boundary conditions for a specific case in Eq.34) = 0 and shear = 0 The zero shear force condition is represented by an equation similar to Eq. i. C2. (10.8.e.pmd 351 α 21 α 22 α 23 α 24 α 31 α 32 α 33 α 34 α 41 α 42 =0 α 43 α 44 (10.24).33). 6:08 AM . the magnitude) of deflection however remains indeterminate..32) associated with the set of particular boundary conditions of the case under consideration. (10. C2i. Let ki 2 = C2 i = C2 i . we get four equations to determine the four constants C1i. Chapter_10A. (10.33). (10..e. However.). as these equations are homogeneous. x) + C 2i φ 2 (ki .e. For each value of the parameter k a corresponding critical load to keep the column in that buckled shope is obtained from Eq. C1i C4 i = C4i C1i Substituting these in Eq. The relationship between the two groups of problems is as follows: Problem Equation D = 0 Buckling Vibrations Stability criterion Frequency equation 10. Eq. (10.32) shows that there exists a set of values of the parameter k for which a deflected configuration of the column is possible.36) is a transcendental equation providing an infinite number of roots ki (i = 1. C3i and C4i.37) The constant C1i remains undetermined. 2. x) + C 4i φ4 (ki . we get the deflection curve corresponding to the load Pi as yi = C1i ⎡⎣φ1 (ki ... (10. In general. 6:08 AM . There is a close similarity between the analysis of a buckling problem and the analysis of a vibration problem connected with small oscillations. The amplitude (i. only the ratios of these constants can be determined. These are called the characteristic roots or values of the parameter k and for each ki .35). there is a corresponding critical load Pi given by Pi EI Introducing one of the characteristic values ki into the system of Eq.pmd 352 7/3/2008.10 Characteristic values Buckling loads Frequencies Characteristic functions Buckling modes Principal modes of vibration THE ORTHOGONALITY RELATIONS The characteristic functions yi satisfying the homogeneous differential equation have an important property which play an important role in the energy methods. Consider the homogeneous differential equation 2 d2 ⎡ d2y⎤ 2 d y =0 p ( x ) + k ⎢ ⎥ dy 2 ⎣ dx 2 ⎦ dx 2 This is satisfied by any characteristic function yi and the associated characteristic value of the parameter ki. i. x) + C 3i φ3 (ki .37). which is the only unknown in this equation.352 Advanced Mechanics of Solids The expansion of the determinant is D and equating it to zero yields an equation for the parameter k. The above analysis of the homogeneous differential Eq. x ) ⎤⎦ (10. yi is called characteristic function of the homogeneous differential Eq. (10. C1i C3i = C3i . 38) yields L d2 y i 2 0 dx ki2 ∫ dy yk dx = ki2 dxi yk L 0 dy dy − ki2 ∫ i k dx dx dx Substituting these in Eq. Consequently.34). L ∫ p( x) 0 d 2 yi d 2 yk dx 2 dx 2 L dyi dyk dx = 0 0 dx dx dx − ki2 ∫ (10. we can interchange yi and yk and obtain L ∫ p( x) d 2 yk d 2 yi 2 2 dx dx 0 Subtracting one from the other Chapter_10A. we obtain d 2 yi ⎤ d2 ⎡ p ( x ) ⎢ ⎥ yk dx 2 dx2 ⎦⎥ ⎢ 0 dy ⎣ L ∫ d = dy ⎡ d 2 yi ⎤ p ( x ) ⎢ ⎥ yk dx 2 ⎦⎥ ⎣⎢ L 0 d 2 yi ⎤ dyk d ⎡ − ∫ ⎢ p( x) 2 ⎥ dx dy ⎢⎣ dx ⎥⎦ dx 0 L Integrating once again by parts d = dy L ⎡ d 2 yi ⎤ p ( x ) ⎢ ⎥ yk dx 2 ⎦⎥ ⎣⎢ 0 ⎡ d 2 yi ⎤ dyk − ⎢ p ( x) ⎥ dx 2 ⎦⎥ dx ⎣⎢ L 0 L⎡ d 2 yi ⎤ d 2 yk + ∫ ⎢ p( x) dx ⎥ dx 2 ⎦⎥ dx 2 ⎢ 0⎣ Similarly.38). we obtain ⎡ ⎧⎪ d d 2 yi dy ⎫⎪ ⎤ p ( x) + ki2 i ⎬ yk ⎥ ⎢⎨ 2 dx ⎭⎪ ⎦⎥ dx ⎣⎢ ⎩⎪ dy L 0 ⎡ d 2 yi dyk ⎤ − ⎢ p( x ) ⎥ dx 2 dx ⎦⎥ ⎢⎣ L 0 L⎡ L d 2 y d 2 yk ⎤ 2 dyi dyk + ∫ ⎢ p( x) 2i − dx k ⎥ ∫ dx dx dx = 0 i dx dx2 ⎦⎥ ⎢ 0⎣ 0 The first term within the brackets vanish for any combination of the homogeneous boundary conditions given by Eq.38) Integrating the first term by parts.Elastic Stability 353 2 ⎡ d 2 yi ⎤ 2 d yi =0 ⎢ p ( x) 2 ⎥ + ki dx ⎥⎦ dx 2 ⎢⎣ Multiply this equation by any other characteristic functions yk and integrate over the length L of the column. obtaining d2 dy 2 L 2 d 2 yi ⎤ d2 ⎡ 2 d yi p ( x ) y dx + k yk dx = 0 ⎥ k ∫ 2⎢ i ∫ 2 dx2 ⎦⎥ ⎢ 0 dy ⎣ 0 dx L (10. 6:08 AM .pmd 353 L dyk dyi dx = 0 0 dx dx dx − kk2 ∫ 7/3/2008. the second term in Eq. (10.39) Since the above equation is valid for each combination of two characteristic functions. (10. (10. 41) and (10.. we observe that L ∫ p( x) 0 d 2 yi d 2 yk dx 2 dx 2 dx = 0 (10.32) and these are known as the orthogonality relations of the characteristic functions yi. Chapter_10A.32) p(x) = 1 and consequently. (10. If Ix is independent of x. •) L which are the terms of a Fourier expansion.pmd 354 7/3/2008. (10.19. 10. (10.41) and (10.42) If i = k. L 2 i − kk2 i dy dy i k ∫ dx dx dx = 0 0 (10.44). The advantage of representing a deflection curve by a series like πx 2π x 3π x y = a1 sin +. 6:08 AM .41) Using Eq. y = 2 = 0 for x = 0 and x = L⎟ . (10..e. + a2 sin + a3 sin L L L will be demonstrated in Sec.354 Advanced Mechanics of Solids (k ) ∫ dydx L dyk (10.39). the functions ⎝ dx ⎠ nπ x yn = sin L are the characteristic solutions which satisfy the orthogonality conditions.43) ∫ p ( x ) ⎜ 2 ⎟ dx ≠ 0 and ∫ ⎜⎝ dx ⎟⎠ dx ≠ 0 dx ⎝ ⎠ 0 0 Equations (10.41) in Eq.44) For example.42) express the fundamental properties of the characteristic solutions of the differential Eq. ⎜ i. 2.39). (10. as k i2 is also different from zero.32) with prescribed boundary conditions is said to constitute a complete system of orthogonal functions. then in Eq. we observe that 2 2 L ⎛ d 2 yi ⎞ ⎛ dyi ⎞ (10.42) reduce to L L L dy dy d2 y d2y i k k dx = 0 ∫ dx dx dx = ∫ 2i dx 2 0 0 dx (10. the sequence of functions yi = sin i πx (i = 1. A family of functions consisting of all of Eq. and consequently. We may recall that for a ⎛ ⎞ d2y hinged column. (10..40) dx = 0 dx 0 2 2 If i is different from k then in general ki − kk will be different from zero.. (10. Eqs (10. form a complete system of orthogonal functions satisfying conditions given by Eq.. the integral 2 ⎛ dyi ⎞ ∫ ⎜⎝ dx ⎟⎠ dx 0 L cannot be equal to zero since the integrand is always a positive quantity and in Eq.. 48) is also stated as that in which the quantity U + V assumes a stationary value.d W = 0 Using Eq. (10.46). the body deforms and consequently.48) The expression U + V is known as the total potential of the system. If a point of the body is constrained to lie on the surface of another body. Equation (10.47) In the above equation. if a point of the body is fixed. The work done by external surface and body forces Pi during these virtual displacements is δW = ∑ Piδ ∆ i + higher order terms (10. For example.46) where dDi are the work absorbing components of the virtual displacements. These virtual displacements being very small. a decrease in potential energy in the form of an equation − δV = ∑ Pi δ ∆ i = δW (10. Consequently. Let the body be in equilibrium. From Eq.11 THEOREM OF STATIONARY POTENTIAL ENERGY The energy method of analysing the problems of elastic stability is based on an extremum principle of mechanics. First-order change means a change in which only those terms that contain first-order terms are considered.46). (10. i. W=U (10. Eq.e. the changes necessary in the external forces to bring about these virtual displacements will also be very small and will vanish in the limit. as in Eq. These are small displacements that are consistent with the constraints imposed on the body.49) Chapter_10A. (10. i.Elastic Stability 355 III ENERGY METHODS FOR BUCKLING PROBLEMS 10. During the application of these forces. i. the work done by the external forces should be equal to the internal elastic energy U. If a part of the body is subjected to distributed external forces. the summation must be replaced by a surface integral. are ignored. U + V = stationary (10. the above equation can be written as d (U + V ) = 0 (10. then over that part.47) -d V . The internal forces which are set up inside the elastic body also do work during the deformation process and this is stored as elastic strain energy. we have neglected the higher order terms of Eq. these forces do a certain amount of work W. then the virtual displacement there should be tangential to the surface of the contacting body. (10.pmd 355 7/3/2008. then the virtual displacement there is zero. When external forces are applied gradually and no dissipation of energy takes place due to friction etc.45). 6:08 AM . It is convenient to define a potential V of the external forces in such a manner that the work done during virtual displacements is equal to -d V. Consider an elastic body subjected to external surface and body forces.e.48) can be stated as follows: The first-order change in the total potential energy must vanish for every virtual displacement when an elastic body is in equilibrium. Terms of higher order.45) Let portions of the body be given small virtual displacements. (10.e. Example 10. the strain energies are U= Chapter_10A.356 Advanced Mechanics of Solids It is shown in books on elasticity that for stable equilibrium. The total extension of each member is 12 5 q1 + q2 13 13 member BD q1 4 3 member CD q1 − q2 5 5 If A is the cross-sectional area and E is the Young’s modulus. 10. any virtual displacement will cause a positive change in the total potential energy of the system. the point D of which is subjected to P units of force. which means that for a system in stable equilibrium. the total potential energy is minimum. Hence. A 90 50 A A C B 5 13 12 120 5 q2 13 D D q2 12 q1 13 D q1 P q2 q1 Fig. Applying the principal of minimum potential energy.18 shows a three-bar truss. then the elastic strain energy stored in a member of length L is member AD EA 2 δ 2L where d is the elongation.4 Figure 10.18 Example 10. The members have equal cross-sectional areas.4 Solution Let the point D have a vertical displacement q1 and a horizontal displacement q2. This is shown in the figure for member AD. determine the vertical and horizontal displacements of D due to the load. The elongation caused in each member due to these displacements can be calculated geometrically.pmd for AD U1 = EA ⎛ 12 5 ⎞ q2 ⎟ ⎜⎝ q1 + 2 (130) 13 13 ⎠ for BD U2 = EA q12 2 (120) for CD U3 = EA ⎛ 4 3 ⎞ ⎜⎝ q1 − q2 ⎟⎠ 2 (150) 5 5 356 2 2 7/3/2008. 6:08 AM . for the three members. pmd 357 7/3/2008. d (U + V ) = EA [ 958(2q1 dq1) . ( ) U = EA 958q12 − 47 q1q2 + 177 q22 × 10−5 Taking the undeformed position as the datum. Thus. The latter principle states that in an equilibrium condition. i. the work done is equal to 1 Pi yi 2 where yi is the work absorbing component of the deflection at Pi.e.12 COMPARISON WITH THE PRINCIPLE OF CONSERVATION OF ENERGY It is important to realise that the principle of the minimum total potential is different from the law of conservation of energy. i.Elastic Stability 357 The total elastic strain energy is the sum of the above three quantities. (10. the work done by all external forces during the loading process is equal to the internal elastic strain energy stored.95 EA EA It should be observed that we have not made use of any equation of statics in solving the problem.47(q1d q2 + q2d q1) or ( + 177(2q2dq2) ) [ ¥ 10-5 . the virtual work done is dW = SPi Dyi W= ∑ Chapter_10A.pq1 For equilibrium position. On the other hand.49) becomes U-W =0 If the loading is done gradually. the total potential energy is ( ) U + V = EA 958q12 − 47q1q2 + 177q22 ¥ 10-5 . the first-order variation of the above quantity should be equal to zero. the quantities inside the parentheses should vanish individually.36 10. we obtain P P and q2 = 6. Eq.Pd q1 = 0 1916q1 − 47q2 − P × 105 δ q1 + ( −47q1 + 354q2 ) δ q2 = 0 EA Since d q1 and d q2 are arbitrary virtual displacements. the potential energy in the deformed configuration is V = -Pq1 Hence. 1916q1 − 47q2 = P × 105 EA − 47q1 + 354q2 = 0 Solving these two equations. Hence.e. q1 = 52. 6:08 AM . aL P P B B L a A a A Fig.e. as shown in Fig. we have demonstrated the use of the theorem of stationary potential energy in solving a statically indeterminate problem. It carries a centrally applied load P.pmd 358 7/3/2008. If the decrease in the potential 2 energy is greater than the energy stored in the spring. It is assumed that the bar is infinitely rigid. how energy considerations can be used to analyse stability problems. 10. if 1 1 PL α 2 < Sα 2 L2 2 2 then the system is stable.358 Advanced Mechanics of Solids There is no 1/2 factor in this case since the forces pis are acting with full magnitude during the virtual displacements Dyi s. i. if 1 1 PL α 2 > Sα 2 L2 2 2 then the system is unstable. 10. with reference to a specific problem. Now we shall show. Because of this displacement.19 Vertical bar hinged at one end and supported by spring at the other end Let the bar be displaced through a small angle a.13 ENERGY AND STABILITY CONSIDERATIONS In Example 10. Pcr = SL then we get the value of the critical load which keeps the column in equilibrium in a slightly displaced configuration.19. At 2 the same time. Chapter_10A. On the other hand. 6:08 AM . 1 PL α 2 .e.4. Consider a vertical bar hinged at one end and supported at the other end by a spring. the spring elongates by an amount a L and the energy stored due to this is 1 S (α L )2 where S is the spring constant. If 1 1 PLα 2 = Sα 2 L2 2 2 i. the load P is lowered by the amount L − L cos α = L(1 − cos α ) ≈ L α 2 2 The decrease in potential energy is equal to the work done by P. i.e. 10. 51) The total potential energy is. 2 2 ⎡ ⎤ L ⎛ d2 y⎞ 1 1 L ⎛ dy ⎞ d (U + V) = δ ⎢ E ∫ I x ⎜ 2 ⎟ dx − P ∫ ⎜ ⎟ dx ⎥ = 0 2 0 ⎝ dx ⎠ ⎢ 2 0 ⎝ dx ⎠ ⎥ ⎣ ⎦ Chapter_10B.pmd 359 7/3/2008. For equilibrium in the displaced position. we shall consider only the bending energy.Elastic Stability 359 The same conclusion can be obtained by applying the principles of statics.20 Column with varying moment of Inertia L DL = ∫ (∆s − ∆x) 0 2 DS = (∆ x 2 + ∆y 2 )1/ 2 ≈ ∆ x + But 1 ⎛ ∆y ⎞ ∆x 2 ⎜⎝ ∆ x ⎟⎠ 2 Hence. therefore. Pa L = Sa L2 or Pcr = SL An analysis of stability problems in column buckling. i.e. 10.20(b)) Fig. D x Let the buckled form be expressed by D s Dx Ds x y = f (x). given by 2 U+V= 2 1 L ⎛ d2 y⎞ 1 ⎛ dy ⎞ E ∫ I x ⎜ 2 ⎟ dx − P ⎜ ⎟ dx 2 0 ⎝ dx ⎠ 2 ⎝ dx ⎠ (10. In DL calculating the strain energy. The end B of the column is subjected to a vertical load P and a horizontal force Sa L. From the straight equilibrium configuration. (10. 10. we observe from Fig.14 APPLICATION TO BUCKLING PROBLEMS We shall now discuss the application of the minimum total energy principle to column buckling problems.52) For equilibrium. will be taken up again in Sec. PDL = 1 L ⎛ dy ⎞ P ∫ ⎜ ⎟ dx 2 0 ⎝ dx ⎠ (10. Let the moment of inertia Ix be variable.20. the variation of the above quantity should vanish. using the above concept. The elastic strain energy is (b) U= P (a) 1 L E ∫ Ix 2 0 2 ⎛ d2 y⎞ ⎜ 2 ⎟ dx ⎝ dx ⎠ (10. let L the column be moved to a neighbouring Dy y bent configuration. Consider the column shown in P Fig.17. the moment about A should be zero. 10. 10. 6:09 AM (10.53) . the potential energy in the buckled form is V = –PDL To calculate DL. carrying an axial load P. For moment equilibrium about A.50) Taking the undeflected position as datum. . . The coefficients a correspond to a set of parameters. These are called coordinate functions.49) is obtained by the Rayleigh–Ritz method.31a). + anfn The f terms are a set of arbitrarily chosen functions of x. it may be mentioned that the final result agrees with the differential equation given by Eq. .54) y = a1f1 + a2f2 + .58) Chapter_10B. an ) − PF2 (a1 . the conditions become ∂ (U + V ) = 0. These lead to an expression involving the n parameters a. 6:09 AM . revealing the importance of the energy criterion. the determinant of the coefficients should be equal to zero. Hence.57) ∂ F1 ∂F − P 2 =0 ∂ an ∂ an The above set of equations involve only linear functions since these are derivatives of the quadratic expressions involved in U + V. While the mathematical procedure will not be discussed here.360 Advanced Mechanics of Solids The above equation permits the determination of the function y = f (x) by applying the technique of the calculus of variations. The deflection of the buckled column is expressed in the form of a finite series: (10. (10. is to be a solution of the problem.50) and (10.31a). i.. been reduced to the familiar maximum–minimum problem involving the parameters a1. (10.54). the elastic strain energy and the potential energy can be calculated. If y.. . We shall demonstrate the method with respect to the buckling problem discussed in the previous section. an ) (10.57) is a set of homogeneous equations. (10. 10.e.15 THE RAYLEIGH–RITZ METHOD A direct solution to the extremum problem stated by Eq. With the value of y as given by Eq. (10. n) ∂ ai i. the merit of the energy criterion for the solution of stability problems becomes evident in the Rayleigh–Ritz method discussed in the next section.pmd 360 7/3/2008. for the existence of a non-trivial solution... 2.. (10. a2. and having the form U + V = F1 (a1 .e.. D =0 (10. The problem has. using Eqs (10. therefore.51). . a2 ..... a2 . as yet undetermined. . . an. (10. (10.54). as given by Eq. then the parameters a must be chosen so as to make the total potential energy an extremum [Eq. While this differential equation can be derived from static equilibrium considerations as was done in the derivation of Eq.56) ∂ F1 ∂F − P 2 =0 ∂ a1 ∂ a1 ∂ F1 ∂F − P 2 =0 ∂ a2 ∂ a2 (10. Since Eq.55) in which F1 and F2 are quadratic forms of the parameters a. such that each term satisfies the prescribed boundary conditions of the column. (10.55)]. (i = 1. . involves less labour than is involved in solving the differential equation and the associated eigenvalue (i.54) as y = a1 (φ1α1 + φ 2α 2 + .10.5 Consider a pin-ended column subjected to an axial compressive load P.59) The importance of the Rayleigh–Ritz method lies in the fact that it offers a method of obtaining an approximate solution to the buckling problem. a set of n linear homogeneous equations is obtained from which the ratios of the parameters a can be determined. 6:09 AM . (10. a few terms of the series in Eq. This is the reason why Fourier Series play such an important role in the applications of the Rayleigh–Ritz method. The smallest of the roots gives the critical load Pcr. a1 an = αn a1 the buckling mode is obtained from Eq.(10.57). . 10. In the majority of cases.20. The method. + φ nα n ) (10. . a1 a3 = α 3 . The coordinate function chosen satisfies the boundary conditions which are y = 0 at x = 0 and at x = L d2y Solution = 0 at x = 0 and at x = L dx 2 From Eq. 10. (10. satisfactory results can be obtained only when the coordinate functions chosen form a system of orthogonal functions discussed in Sec.50). as shown in Fig. the strain energy is obtained as 2 L⎛ 2 ⎞ U = 1 EI ∫ ⎜ d y ⎟ dx 2 2 0 ⎝ dx ⎠ () L = 1 EI ∫ a 2 π 2 L 0 = Chapter_10B.. Calling a2 = α2. . Example 10. (10. .54) give a sufficiently accurate result.e. . Introducing P = Pcr in Eq.Elastic Stability 361 This is an equation of degree n in the unknown P and is the stability condition from which P can be determined. in many cases. In the majority of cases. Success or failure in applying the Rayleigh–Ritz method to any problem depends largely on the proper choice of the coordinate functions. characteristic value problem) . Assume that the buckled shape is given by y = a sin πx L where a is an unknown parameter.pmd 4 sin 2 π x dx L 1 4 2 ⎛ EI ⎞ π a ⎜ 3⎟ ⎝L ⎠ 4 361 7/3/2008. The strain energy is equal to Chapter_10B. The first two conditions are satisfied by the coordinate function.6 at x = 0. It is subjected to a compressive load P at the free end.21 Example 10.21). 6:09 AM . dy = 0 at x = 0 dx d2y dx 2 =0 at x = L (i) Let us ignore the last condition for the time being. we must have U+V= 1 4 EI 1 a π a 3 − Pπ 2 = 0 2 2 L L 1 2 a ⎛ 2 EI ⎞ π ⎜ π 2 − P⎟⎠ = 0 2 L⎝ L The non-trivial solution is obtained when or P = Pcr = π 2 EI L2 We have been able to obtain an exact solution since the coordinate function we used happens to give the exact deflected shape for the column.6 Consider a column fixed at one end and free at the other end (Fig.362 Advanced Mechanics of Solids From Eq. (10. 10.pmd 362 7/3/2008. Determine the approximate critical load assuming the deflection curve as d P 2 ⎛ x⎞ ⎛ x⎞ y = a1 ⎜ ⎟ + a2 ⎜ ⎟ ⎝ L⎠ ⎝ L⎠ 3 Solution The boundary conditions are y=0 x and Fig. 10. Example 10. the potential energy is obtained as 2 L dy ⎛ ⎞ V = − 1 P ∫ ⎜ ⎟ dx 2 0 ⎝ dx ⎠ () L = – 1 P ∫ a2 π 2 0 L =– 2 cos2 π x dx L ⎛ a2 ⎞ 1 Pπ 2 ⎜ ⎟ 4 ⎝ L⎠ Thus.51). the total potential energy is 1 4 2 EI 1 a2 π a 3 − Pπ 2 4 4 L L For this to be an extremum. the determinant D of the coefficients should be equal to zero.18 EI L2 The smaller value is Pcr = 2.pmd 363 EI L2 π 2 EI 4 L2 . 6:09 AM . Hence. 2 ⎛ 4 EI 4 P ⎞ ⎛ 12 EI 9 P ⎞ ⎛ 6 EI 3P ⎞ − − − =0 D= ⎜ 3 − ⎝ L 3L ⎟⎠ ⎜⎝ L3 5L ⎟⎠ ⎜⎝ L3 2 L ⎟⎠ or 3P2L4 – 104PL2EI + 240E2I2 = 0 Therefore.92%.Elastic Stability 363 2 L⎛ d2 y⎞ 1 EI ∫ ⎜ 2 ⎟ dx 2 0 ⎝ dx ⎠ U= 2 L 6a x ⎞ 1 ⎛ 2a = EI ∫ ⎜ 21 + 32 ⎟ dx ⎝ 2 L ⎠ 0 L ( = 2 EI a12 + 3a1a2 + 3a22 L3 The potential energy is to equal to ) 2 1 L ⎛ dy⎞ V =− P∫ ⎜ ⎟ dx 2 0 ⎝ dx⎠ 2 =− 2 1 L ⎛ 2a1 x 3a2 x ⎞ P ∫⎜ 2 + ⎟ dx 2 0⎝ L L3 ⎠ ( P 20a12 + 45a1a2 + 27a22 30 L Hence. 7/3/2008. the error is only +0.49 Compared to the exact value Chapter_10B. P = 2.49 EI 2 L or 32. the total potential energy is =− ) ⎛ 6 EI 3P ⎞ ⎛ 6 EI 9 P ⎞ 2P ⎞ 2 ⎛ 2 EI + a1a2 ⎜ 3 − + a22 ⎜ 3 − U + V = a1 ⎜ 3 − 3L ⎟⎠ 2 L ⎟⎠ 10 L ⎟⎠ ⎝ L ⎝ L ⎝ L For an extremum we should have ∂ (U + V ) ⎛ 4 EI 4 P ⎞ ⎛ 6 EI 3P ⎞ = ⎜ 3 − ⎟ a1 + ⎜⎝ 3 − ⎟ a2 = 0 ⎝ L ∂ a1 3L ⎠ 2L ⎠ L ∂ (U + V ) ⎛ 6 EI 3P ⎞ ⎛ 12 EI 9 P ⎞ = ⎜ 3 − ⎟ a1 + ⎜⎝ 3 − ⎟ a2 = 0 ⎝ L 2L ⎠ 5L ⎠ ∂ a2 L For the existence of a non-trivial solution. 364 Advanced Mechanics of Solids (ii) In the above analysis. in the slightly bent configuration.e. this means that when P is less than Pcr. i. In terms of energy. If we use this condition d2 y dx2 (at x = L) = 2a1 2 L + 6a2 L L3 =0 a 1 = –3a2 or Using this 2 1 ⎛ x⎞ ⎛ x⎞ y = a1 ⎜ ⎟ − a1 ⎜ ⎟ ⎝ L⎠ 3 ⎝ L⎠ 3 Substituting a1 = –3a2 in the expressions for U and V U= a12 ⎞ 2 EI 2 2 EI ⎛ 2 2 − + = a a a1 1 ⎜ 1 3 ⎟⎠ 3 L3 L2 ⎝ V=– and P L 4 P 2 ⎛ 2 2 1 2 1 2⎞ a1 ⎟ = − a1 ⎜⎝ a1 − a1 + ⎠ 3 2 10 15 L Therefore. then the column is in stable equilibrium. U–W>0 (10. 10. the quantity inside the parentheses should be equal to zero.60) Chapter_10B. 4 P 3 EP = 15 L 4 L3 or Pcr = 2.5 EI L2 which is almost identical with the previous solution but the solution has been obtained with comparative ease.16 TIMOSHENKO’S CONCEPT OF SOLVING BUCKLING PROBLEMS Consider a straight column subjected to an axial load P. it will return to its vertical position as soon as the disturbing force is removed. at x = L. we have ignored the third boundary condition. d2y dx 2 = 0. the elastic strain energy stored in the bent column is greater than the work done by the axial load in moving through a distance DL. If P is less than the critical load. which means that if the column is slightly displaced from its straight equilibrium position by any transverse disturbing force.pmd 364 7/3/2008. 6:09 AM . i.e.e. i. ⎛ 2 EI 4 P ⎞ 2 U+V= ⎜ − a ⎝ 3 L3 15 L ⎟⎠ 1 For an extremum. (10. define the amplitudes of the terms.51) 2 W= 1 L ⎛ dy ⎞ P ∫ ⎜ ⎟ dx = PF2 (a1 . The strain energy is given by 2 L⎛ 2 ⎞ 1 d y U = EI ∫ ⎜ 2 ⎟ dx = F1 (a1 . + an φn The f terms are functions of x so that each term satisfies the boundary conditions of the column.61) L 2 2 2 ∫ (d y / dx ) dx P = EI 0 L = 2 ∫ (dy / dx) dx F1 (a1 . . an ) F2 (a1 . . . . we can assume that the buckled column curve can be expressed by Eq. . On the other hand.Elastic Stability 365 where U is the strain energy and W = P DL. if the column is slightly displaced. the work done by the external load P will exceed the strain energy in bending and the equilibrium becomes unstable.61) characterises the state when the equilibrium configuration changes from stable to unstable. a2 . (10. an ) 2 0 ⎝ dx ⎠ The work done by the external force during deformation is from Eq.54) as y = a1φ1 + a2φ 2 + .W=0 (10.62) or Eq. This requires that Chapter_10B. (10. . a2 .63) is made a minimum. when P exceeds Pcr. . .. The constants a1. . the critical load is obtained when the expression in Eq. . . . . the condition U.62) can also be written as L EI ∫ (dy / dx)2 dx P= 0 L (10. . . Consequently. (10. a2 . (10. Following the same procedure as in the Rayleigh–Ritz method. . an ) 2 0 ⎝ dx ⎠ Using Eq. an ) (10.62) 0 Observing that for a pin-ended column or a column with one end free d2y M = – EI dx 2 M = Py and that 2 and ⎛ d 2 y⎞ P2 EI ∫ ⎜ 2 ⎟ dx = EI ⎝ dx ⎠ 2 ∫ y dx Eq. a2 . 6:09 AM . (10... . .. a2.pmd 365 7/3/2008.63) ∫ y dx 2 0 Since we need the minimum value for the load P. 62) and (10. (i = 1.15.63) gives a slightly more accurate result. (10. in the examples considered. . . Equation (10. There is a fundamental difference between Eqs (10. (10. 2.57). (10.65) or M = Py where P is the axial force acting on a pin-ended column or a column with one end free end the other end fixed. Example 10. 6:09 AM . . then M could be expressed in two ways U= M = – EI d2y dx 2 (10. The elastic strain energy is obtained from the expression 1 L 2 ∫ M dx 2 EI 0 If we take the deflection curve as y = y (x).63) with respect to each coefficient ai must vanish. This discussion is identical to that given in Sec..63) as the Timoshenko formula. (10. a non-trivial solution exists when the determinant of the coefficients is equal to zero. This yields ∂ F1 ∂F − P 2 = 0. (10.65). Assume that the deflection curve can be represented by the series nπ x L Since the deflection curve must be symmetrical with respect to the middle point of the column (because the moment of inertia is symmetrical about the middle point). 10. If we use the first expression in Eq. the even parameters in the above series vanish. we get the strain energy for any assumed form of the column.7 Consider a column with the moment of inertia of the cross-sectional area varying according to the equation π x⎞ ⎛ I = I 0 ⎜ 1 + sin ⎟ ⎝ L⎠ Solution The column is hinged at both ends. 10. The energy method will be found to be very suitable to obtain fairly good solutions. we take the external force also into account and consequently.62) is generally referred to as the Rayleigh–Ritz formula and Eq. If we use the second expression.pmd 366 πx L + a3 sin 3π x 5π x + a5 sin + . we have treated the moment of inertia I as independent of x. the final result obtained for Pcr using Eq.63) though they appear to be equivalent. The deflection equation then becomes y = ∑ an sin y = a1 sin Chapter_10B.62) or Eq.17 COLUMNS WITH VARIABLE CROSS-SECTIONS So far.366 Advanced Mechanics of Solids the derivatives of Eq. Since there are n homogeneous equations.) ∂ ai ∂ ai (10.64) These are identical to Eq.. (10. L L 7/3/2008. We shall consider a few problems where I varies with x. 48 ⎛ 5832 ⎞ + 81π − 9 P* ⎟ a3 = 0 a1 + 2 ⎜ ⎝ 35 ⎠ 5 and − where P* = PL2 π EI 0 For the existence of a non-trivial solution. (10.62) P = EI 0 π (8 / 3 + π )a12 − (48 / 5) a1a3 + (5832 / 35) + 81 π )a32 2 a12 L + 9a32 = F1 F2 For minimum P.746 or P = 18. This gives D = 9P*2 − ( ) 5832 + 81π + 24 + 9π P* 35 2 ⎛8 ⎞ ⎛ 5832 ⎞ ⎛ 24 ⎞ + ⎜ + π⎟ ⎜ + 81π ⎟ − ⎜ ⎟ = 0 ⎝3 ⎠ ⎝ 35 ⎠ ⎝ 5⎠ Solving. P* = 5. the determinant of the coefficients should be equal to zero.05 Chapter_10B. we should have from Eq. (10. 6:09 AM . Thus πx y = a1 sin L + a3 sin 3π x L 2 ⎛ d2y⎞ 1 U = ∫ EI ⎜ 2 ⎟ dx 2 0 ⎝ dx ⎠ L = 2 2 2 1L π x⎞ ⎡ πx 3π x ⎤ ⎛ ⎛π⎞ ⎛ 3π ⎞ + − − 1 sin sin sin EI a a dx ⎢ ∫ 0 ⎜⎝ ⎟ 1⎜ ⎟ 3 ⎜ ⎝ L⎠ ⎝ L ⎟⎠ 20 L ⎠ ⎣⎢ L L ⎥⎦ = 1 24 ⎛ π ⎞ ⎡⎛ 4 π ⎞ ⎛ 2916 81π ⎞ 2 ⎤ + EI 0 ⎜ ⎟ ⎢⎜ + ⎟ a12 − a1a3 + ⎜ ⎟ a3 ⎝ ⎠ ⎝ ⎠ ⎝ 35 2 5 2 ⎠ ⎥⎦ L ⎣ 3 2 3 and 1 L ⎛ dy ⎞ 1 ⎛π⎞ ⎛ L 2 L 2⎞ ∫ ⎜ ⎟ dx = 2 ⎜⎝ L ⎟⎠ ⎜⎝ 2 a1 + 2 × 9a3 ⎟⎠ 2 0 ⎝ dx ⎠ 2 DL = 2 Substituting in Eq.64) ∂ (F – PF2) = 0 and ∂ a1 1 ∂ (F1 – PF2) = 0 ∂ a2 48 ⎛8 ⎞ a3 = 0 2 ⎜ + π − P* ⎟ a1 − ⎝3 ⎠ 5 Thus.Elastic Stability 367 We shall consider only two terms of the series.pmd 367 EI 0 L2 7/3/2008. 368 Advanced Mechanics of Solids 10. . 6:09 AM .. a2..11 that the functions satisfying Eqs (10. 0 for n π m These are the orthogonality relations expressed by Eqs (10. terms containing products like am an vanish and only terms like an2 remain.. The strain energy is given by y = a1 sin πx 2 L⎛ 2 ⎞ 1 d y U = EI ∫ ⎜ 2 ⎟ dx 2 0 ⎝ dx ⎠ 2π x 32 π 2 3π x a sin − –. the square of the above expression will involve terms of two kinds Now.42) and (10.31b) also satisfy orthogonality conditions. it will be useful to represent the deflection curve in the form of a trigonometric series. Let the deflection curve be represented by 2π x nπ x + a2 sin + .67) 4 L n =1 Similarly. Then U= = = ⎡ π4 L ⎤ 1 24 π 4 L 34 π 4 L EI ⎢ a12 4 + a22 4 + a32 4 + ..pmd 1 L ⎛ dy ⎞ ∫ ⎜ ⎟ dx 2 0 ⎝ dx ⎠ 368 7/3/2008. the above series can be made to represent any deflection curve.31a) and (10. Hence. + an sin +. ∞ ∑ n 4 an2 (10.. d2y = – a1 π2 an2 sin n 4π 4 4 L πx − a2 sin 2 22 π 2 sin nπ x L nπ x mπ x sin L L L By direct integration it can be seen that and n2 m2 π4 2 am an L ∫ sin 0 2 4 L nπ x dx = . These coefficients may be calculated by a consideration of the strain energy of the beam or the column. and 2 L sin L nπ x mπ x ∫ sin L sin L dx = 0. . in the expression for strain energy. . . 3 2 2 2 2 L L L L L L dx Hence. The trigonometric series which we shall consider now is made up of such functions. ...18 USE OF TRIGONOMETRIC SERIES In many instances. We have discussed in Sec. . 10. if we consider the expression 2 DL = Chapter_10B..66) L L L By properly determining the coefficients a1.⎥ 2 L 2 L 2 ⎣ L 2 ⎦ EI π 4 4 L3 EI π 4 3 (1 a 2 1 ) + 24 a22 + 34 a32 + .43). (10. .22).. Determine the deflection curve using the energy method. This virtual displacement is obtained by changing one of the terms an sin (np x/L) to (an + d an) sin (npx/L). 10. DL = π2 ∞ ∑ n2an2 (10.68) 4L n=1 With these expressions for U and DL.68) y = a1 sin πx + a2 sin dW = Q δ an sin nπ c π2 2 +P 2n anδ n 4L L The increase in strain energy is ∂U dU = ∂ a δ an n Chapter_10B.8 A beam column is subjected to an axial force P and a lateral force Q at x = c (Fig. 6:09 AM . Example 10.22 Example 10. we can consider the following example.pmd 369 7/3/2008. L L Let a virtual displacement d yn be given. the deflection curve d an sin (np x/L) is superimposed on the original deflection curve. The work done by the external forces Q and P is nπ c dW = Q d an sin + Pd (D L) L Using Eq. A Q c a P B P x L y Fig. (10.8 Solution Let the deflection curve be 2π x + . 10. In other words.Elastic Stability 369 we find the integrand to consist of two kinds of terms an2 n 2π 2 2 L nπ x L cos 2 mπ x nπ x cos L L L By direct integration it can be shown that and 2 am an L ∫ cos 2 0 L and ∫ cos2 0 mnπ 2 2 cos L nπ x = 2 L nπ x mπ x cos dx = 0 L L for n π m Consequently. 67) EI π 4 n 4 an δ an 2 L3 Since the increase in strain energy should be equal to the work done. 10.9 Solution In the solution of the previous example consider c to be very small. we have dU = EI π 4 4 nπ c π2 2 n an δ an δ an + P n an δ an = L 2L 2 L3 Q sin from which. M P P L Fig. an = EI π 2 2QL3 1 EI π n (n − β ) 4 2 2 sin nπ c L The deflection curve is. 1 nπ c sin L EI π 4 ⎛ 2 PL2 ⎞ ⎜⎝ n − 2⎟ EI π ⎠ If we use the notation 2QL3 an = PL2 b= then. such that Qc = M = constant. given by 2QL3 y= EI π 4 ∞ 1 n =1 n 2 (n 2 − β ) ∑ sin nπ c nπ x sin L L Example 10.pmd y= 2ML2 EI π 3 ∞ 1 n =1 n (n − β ) ∑ 370 2 sin nπ x L 7/3/2008. determine the deflection curve for the beam column shown in Fig. 10.370 Advanced Mechanics of Solids From Eq. nπ c nπ c ª L L Then sin and y= ∞ 2 L3 π n nπ x Qc sin ∑ 4 2 2 L n =1 n ( n − β ) EI π L Let c Æ 0 and Q Æ•.23 Example 10.23. Then. 6:09 AM . Chapter_10B. (10.9 Using an infinite series. therefore. Determine the moment Mb at the built-in end (Fig. It is subjected to an axial force P.1 ⎡ ⎡ qL2 β (u ) ⎢ ⎢ Ans.1 A uniformly loaded beam is built-in at one end and simply supported at the other end. y = − ⎥ ⎢ ⎥+ sin kL ⎦ P ⎣ L P ⎢⎣ sin kL L ⎥⎦ ⎥ ⎢ ⎢ δπ 2 πx⎥ + 2 sin ⎢ ⎥ 2 2 L ⎥⎦ π −k L ⎢⎣ 10.2 A beam of uniform cross-section has an initial curvature given by the equation y0 = δ sin πx L It is subjected to end couples Ma and Mb and to an axial force P (Fig. 10. L L If the bar is simply supported and subjected to axial force P only.24 Problem 10. Mb = + ⎢ ⎢ 8 φ (u ) ⎢ ⎢ ⎢ ⎢ 3 ⎢ ⎢ where. .. b (u) = 3 (tan u – u) u ⎢ ⎢ ⎢ ⎢ ⎢ 3 ⎛ 1 1 ⎞ ⎢ f (u) = − ⎢ ⎢ 2u ⎜⎝ 2u tan 2u ⎟⎠ ⎢ ⎢⎣ ⎣ 10.Elastic Stability 371 10. 10.2 ⎡ M a ⎡ L − x sin k ( L − x ) ⎤ M b ⎡ sin kx x ⎤ ⎤ − − ⎢ Ans.3 The initial shape of a bar can be approximated by the series 2π x + δ 2 sin + .pmd 371 7/3/2008. 10. A P Ma Mb B P Fig. .. 10. q P A P B Mb Fig. where α = 2 π L L 2 −α ⎝1 − α ⎠ Chapter_10B.25). Determine the deflection curve. 6:09 AM .⎟ .25 Problem 10. show that the deflection curve due to P is given by y = δ1 sin πx k 2 L2 ⎛ δ ⎞ δ πx 2π x y 1 = α ⎜ 1 sin + 2 2 sin + .24). 10. the equation y = δ sin πx L ⎡ π 2 EI qL ⎤ ⎢ Ans.26 Problem 10. Find the critical value of P by assuming. Pcr = 2. 10.27 Problem 10.6 A prismatic bar with hinged ends (Fig.27. The column has a moment of inertia I1 for half the length and moment of inertia I2 for the other half.26) is subjected to the action of a uniformly distributed axial load of intensity q and an axial compressive force P. EI ⎤ ⎡ ⎢⎣ Ans. Chapter_10B. it is assumed that the deflection curve has the form y= δ x2 L2 where L is the length of the column and x is measured from the fixed end. 6:09 AM . Pcr = 2 − ⎥ 2 ⎦ L ⎣ P P1 R B L1 I1 d C L d L2 I2 x P2 q R x A P1 + P 2 Fig. 10.372 Advanced Mechanics of Solids 10. 10.88 L2 ⎥⎦ 10. EI ⎤ ⎡ ⎢⎣ Ans.7 10. for the deflection curve.7 Determine the critical load (P1 + P2) by the energy method for the case shown in Fig.5 The deflection curve for a pin-ended column is represented by a polynomial as y = ax4 + bx3 + cx2 + dx + e Determine the critical load by the energy method. determine the critical load. Using the energy method.5 L2 ⎥⎦ 10.pmd 372 7/3/2008.6 Fig. Pcr = 9.4 For a column with one end built-in and the other end free and carrying an axial load P. Elastic Stability 373 Assume the deflection curve in the form y = δ sin πx L ⎡ ⎡ Ans. 6:09 AM .pmd 373 7/3/2008. (P + P ) = 1 2 cr ⎢ ⎢ ⎢ ⎢ 2 2 ( π EI / L ) ( m + 1) 2 ⎢ ⎢ 2 2 ⎢ ⎡ ⎤ m m − 1⎞ 8 1 m ⎛ m − 1⎞ 8 m −1 ⎢ ⎢ m + ⎜⎛ − 2 (m − 1) + n ⎢ + ⎜ + 2 ⎥⎢ ⎟ ⎟ ⎝ m ⎠ 6 ⎝ m ⎠ π m ⎦⎥ ⎢ π ⎢ ⎣⎢ m 6 ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ where m = P1 + P2 and n = I 2 ⎢⎣ ⎢⎣ P1 I1 Chapter_10B. drained and allowed to harden. Generally speaking. when the same bundle of filaments or fibres is dipped into a bath of resin. bending and torsional forces. A structure built up of bricks is good to carry compressive loads. composites are produced when two or more materials are joined to give a combination of properties that cannot be attained in the original materials. Generally speaking. 7:09 AM . one would like to use materials that are suitable to specific applications. are both homogeneous and isotropic. then it is said to be homogeneous. It has a definite shape. a durable surface and it can be machined. Materials like steel. If the properties are the same at every point in the body. This is an example of a composite material that is designed for a specific operation. Materials that are not isotropic are called anisotropic materials. These qualifications are with respect to their elastic properties. Concrete.CHAPTER 11 11. aluminium. It has no shape and no defined hard surface for machining purposes. etc. fibre and laminar—based on the shapes of the constituent materials. it behaves as a new material possessing properties that are comparable to those of steel or other metals. The bundle can resist tensile forces. which is useless when used as an engineering structure. Such a material is called a composite material. Composites can be placed into three categories—particulate. consider a bundle of glass fibres or carbon filaments.1 Introduction to Composite Materials INTRODUCTION Till now. the elastic properties are different for different directions. we have been considering materials that are homogeneous and isotropic. copper. A reinforced concrete beam with steel reinforcement at the bottom is good to carry bending loads which will induce compressive stresses in concrete and tensile stresses in the steel reinforcement. For example. But. At any point of such a body. Directions for which the elastic properties are the same are said to be elastically equivalent. but it is useless for compressive. and can resist forces in tension. However. a mixture of Chapter_11A. wood is homogeneous but is not isotropic since its strength along the fibres is greater than its strength in a direction transverse to the fibres.pmd 374 7/3/2008. a cable wire or rope that is used for hauling purposes needs to have the required tensile strength in the direction of the cable. compression and bending. Isotropy implies that the properties are independent of directions. On the other hand. one obtains 2 2 ∂σ y ∂τ xy ∂σ x ∂σ x = ∂ V = . from Eq. = ∂ V = . tzx. having alternating layers of wood veneer. . . (11.4) ∂ε x∂ε y ∂ε x ∂γ xy ∂ε x ∂γ xy ∂ε x ∂ε y From Eqs. . is a laminar composite. a31 = a13. a65 = a56 Chapter_11A. the stresses are linearly related to the strains and are given by sx = sy = a21 ε x + a22 ε y + a23 ε z + a24 γ xy + a25 γ yz + a26 γ zx s z = a31 ε x + a32 ε y + a33 ε z + a34 γ xy + a35 γ yz + a36 γ zx txy = a41 ε x + a42 ε y + a43 ε z + a44 γ xy + a45 γ yz + a46 γ zx (11. sy. In this chapter. . (11. we shall focus our attention mainly on fibre-reinforced composites and laminates.gzx in terms of sx. Let us assume that there exists an elastic potential V such that sx = ∂V .1). is a fibre-reinforced composite. it was stated that for a linearly elastic body..2) gyz = b51 σ x + b52 σ y + b53 σ z + b54 τ xy + b55 τ yz + b56 τ zx gzx = b61 σ x + b62 σ y + b63 σ z + b64 τ xy + b65 τ yz + b66 τ zx where the coefficients bijs are related to aijs.. .2 STRESS–STRAIN RELATIONS In Chapter 3.. (11.1) tyz = a51 ε x + a52 ε y + a53 ε z + a54 γ xy + a55 γ yz + a56 γ zx tzx = a61 ε x + a62 ε y + a63 ε z + a64 γ xy + a65 γ yz + a66 γ zx Assuming that the sixth-order determinant of the coefficients aijs in Eq.. The expressions for the strain components will then be b15a12 τ yzε y+ +b16a13τ zxε z + a14 γ xy + a15 γ yz + a16 γ zx ex = b11 σ x + b12 σ y + b13 σ z + b14aτ11xyε+x + ey = b21 σ x + b22 σ y + b23 σ z + b24 τ xy + b25 τ yz + b26 τ zx e z = b31 σ x + b32 σ y + b33 σ z + b34 τ xy + b35 τ yz + b36 τ zx gxy = b41 σ x + b42 σ y + b43 σ z + b44 τ xy + b45 τ yz + b46 τ zx (11. ∂ εy . (11. etc. ey. . τ zx = ∂V ∂ γ zx (11. .. this number can be reduced based on the material properties. fiberglass. (11.3) The physical meaning of the potential V will become clear soon.. However. ∂ εx σy = ∂V . .3).Introduction to Composite Materials 375 cement and gravel. and plywood.pmd 375 7/3/2008. 7:09 AM .1) is not zero.4) and (11. Assuming the existence of such a potential. containing glass fibres embedded in a polymer. . one immediately gets a 12 = a21.1) contain altogether 36 elastic constants. one can solve for ex. 11. The stress–strain relations given by Eq. is a particulate composite. .6) gives the expression for V which can be verified by differentiating with respect to ex. ey. (11.e without dynamic effects.pmd 376 7/3/2008. based on the existence of such a potential. ey. .5b) As a result of this. a13 = a31. .2) are valid and let a uniform stress s x = p be applied along the x-axis.1). in general. (11. (11. .1) or the bijs in Eq. . However. i. ez. Thus.2) get reduced to 21.. + a36γ zx ) ε z 2 1 + (a14 ε x + a24 ε y + a34 ε z + a44γ xy + .6) 1 + a44γ 2xy + a45γ xy γ yz + a46γ xy γ zx 2 1 + a55γ 2yz + a56γ yz γ zx 2 1 + a66 γ 2 zx 2 By differentiating Eq.8) This is nothing but the elastic energy per unit volume at any point of the body. 2 Using Eq.e. it is claimed by material scientists that the number of independent elastic constants does not exceed 18. b16 = b61. . . (11. 7:09 AM . Consider a body for which Eqs (11. since a12 = a21. Equation (11. + a16γ zx ) ε y 2 (11. etc.) γ xy + . . when forces are applied uniformly. .2) (11. (11.7) 1 + (a13ε x + a23ε y + a33ε z + . one gets expressions for sx. 6) Consequently. . the number of independent elastic constants is 21. thus verifying Eq. .6) can be grouped to give for V an equivalent expression as V = 1 (a11ε x + a12 ε y + a13 ε z + a14 γ xy + a15γ yz + a16γ zx ) ε x 2 1 + (a12 ε x + a22 ε y + a23ε z + . V = 1 a11 ε 2x + a12 ε x ε y + a13 ε x ε z + a14 ε xγ xy + a15 ε xγ yz + a16 ε xγ zx 2 1 + a22 ε 2y + a23 ε y ε z + a24 ε yγ xy + a25 ε y γ yz + a26ε y γ zx 2 1 + a33 ε 2z + a34 ε z γ xy + a35 ε z γ yz + a36 ε z γ zx 2 (11. for a general case of anisotropy. in Eq. a ij = aji (i. .6) with respect to ex..3).. 3. . b13 = b31. (11. and comparing with Eq. sz. ez. . j = 1.. resulting in 21 constants. 2. the number of independent elastic constants in Eqs (11..1) and (11. aij = aji and bij = bji. the elastic potential assumes the form V = 1 (σ x ε x + σ y ε y + σ z ε z + τ xy γ xy + τ yz γ yz + τ xz γ zx ) 2 (11. 6) (11. . . (11.5a) b ij = bji (i. (11. The remaining stress components are Chapter_11A. a16 = a61 and similarly. .2) gets reduced by15. etc. i.376 Advanced Mechanics of Solids And. sy. . In other words. So.. because of the symmetry of the material properties. The terms of Eq.1). .. the 36 elastic constants aijs in Eq. 3.1) and (11. b12 = b21. etc. 2. j = 1. . E the Young’s modulus and m the Poisson’s ratio. 7:09 AM . but the position of the figure as a whole is not changed. 11. For an isotropic body. In other words. In general. Eq.9) γ yz = b51 p γ zx = b61 p gxy = b41 p It becomes obvious that small segments passing through different points and parallel to x-axis get extended by the same amount. but not on the axis.2) is 36. When a surface of revolution is rotated through any angle about the axis of revolution. all segments parallel to a given direction n and drawn through different points. it has been shown earlier that the number of independent elastic constants is only 2.3 BASIC CASES OF ELASTIC SYMMETRY The number of elastic constants involved either in Eq. Fig. no matter where they are located. is said to possess an axis of symmetry. rectangular parallelepipeds deform into oblique parallelepipeds having no right angles between the faces. material constants or components of modulus are used to determine stresses from strains. it should be noted that.9) show that elements of the same size in the form of rectangular parallelepipeds with respective parallel faces deform identically. Transversely Isotropic Consider a fibre-reinforced body in which the filaments are fairly long and are all oriented in the same direction. The constants aijs appearing in Eq. A body which after rotation can be brought into coincidence with itself by reflection in a plane. then it is obvious that the elastic properties at any point in the x–y plane.4). For such a body.e ε y = b21 p ε z = b31 p ex = b11 p (11. i. the surface can be made to coincide with itself after an operation which changes the positions of some of its points. Any geometrical figure which can be brought into coincidence with itself after an operation which changes the position of any of its points is said to possess symmetry.2) are constant for every point. the strain components as given by Eq. all parallel directions are elastically equivalent. A body which can be brought into coincidence with itself by a rotation about an axis. (11. is said to possess a plane of symmetry. However. The constants bijs appearing in Eq. in general. Then.1) or Eq. gets changed. the position of every point on the surface. For an anisotropic body. (11. Let the z-axis be parallel to the fibre elements and let the x and y axes lie in a plane perpendicular to the element orientation.1) are sometimes called material constants or components of modulus. (11. or the engineering constants.Introduction to Composite Materials 377 zero.pmd 377 7/3/2008.1. and components of compliance are used to get strains from stresses. If the fibres are uniformly distributed in the matrix.2) are called components of compliance. the number of independent elastic constants (according to material scientists) does not exceed 18. (3. (11. Equations (11. the number of independent elastic constants get reduced depending on the type of symmetry that exists. (11. It was also stated that even in the most general case. undergo equal elongations. and these get reduced to 21 independent elastic constants under the assumption that an elastic potential V (equivalent to strain energy density function) exists. In other words. which is the Chapter_11A. A homogeneous anisotropic body exhibiting this property is said to be rectilinearly anisotropic. these being the Lame’s constants l and m. 11. Equations (11. are independent of the directions in that plane. γ yz = b55τ yz . The strain components will be exx = ε yy = ε zz = 0.2) can therefore be written as (using double subscripts for ex. Now consider the shearing stresses. γ yz = γ zx = 0 exx = ε yy = ε zz = 0. the deformation of a rectangular parallelepiped will only be in the xy plane. In general. It is assumed that no debonding between the fibres and matrix occurs when the body is stressed. all strain components other than gzx = b66 tzx will be zero. Now consider with t yz alone. ε yy = b21σ x . γ xy = γ yz = γ zx = 0 One should observe that the transverse strains eyy and ezz will not be equal. ε yy = b23σ z = b13σ z . b13 = b23 (i. with tzx alone. and other stress components being zero. If a uniform force p is applied in the z direction. 7:09 AM . for txy alone (with other stress components being zero). γ xy = γ yz = γ zx = 0 Because of transverse isotropy.1 Transversely isotropic composite plane of symmetry.e. ε zz = b31σ x . e y and ez). γ xy = b44τ xy . Thus. with all stress components sij and tij acting.10) gyz = b55τ yz gzx = b66τ zx Chapter_11A. such that sz = p and all other stress components are zero. the strain components eij can be obtained by superposition of all the above expressions. transverse strains are equal).378 Advanced Mechanics of Solids y y x x z Fig. 11. exx = b11σ x + b12σ y + b13σ z eyy = b12σ x + b11σ y + b13σ z e zz = b31σ x + b31σ y + b33σ z gxy = b44τ xy (11. ε xx = b13σ z . A similar set of equations can be written with stress sy. Thus. γ zx = γ xy = 0 Similarly. the strain components will be exx = b11σ x . then any rectangular parallelepiped with faces parallel to x. ezz = b33σ z . with the remaining stress components being zero. For a uniform stress sx. The axis normal to the plane of elastic symmetry is some times called the principal direction.pmd 378 7/3/2008. A body of this nature is said to be transversely isotropic. y and z planes will deform into a rectangular parallelepiped with equal lateral contraction (or extension) in x and y directions. For a shearing stress txy in the xy plane. This contraction is indicated by a negative sign. i. Also.14) G= E 2 (1 + ν ) (11. ν ν exx = 1 σ x − xy σ y − xz σ zz E xx E yy E zz eyy = – e zz = ν yx E xx ν zx E xx + ν yz 1 σy − σ E yy E zz zz σx − gxy = 1 τ Gxy xy gyz = 1 τ G yz yz gzx = 1 τ G yz zx ν zy E yy σ y + 1 σ zz E zz (11. in the plane of isotropy. nxy is the Poisson’s ratio in x direction due to a stress in y direction. b12 and b44 are also related. Since xy plane is a plane of isotropy.11) In these equations. Also. there are eight constants. the elastic constants in the plane of isotropy. i. Also. b13 = b31. However.e. and as a result. one has from Eq.e. n yx = nxy. What is important is the reciprocal identity. n zx = nzy. because of the reciprocal relations.e bij = bji. because of reciprocal identity.12a) Exx Further.e the ratio of lateral contraction in x direction to axial extension in y direction. Exx = Eyy. i.10) are written using the familiar engineering constants E.pmd Exx Gxy = = 2(1 + ν xy ) 379 1 ν xy ⎞ ⎛ 2⎜ 1 + ⎝ Exx Exx ⎟⎠ 7/3/2008.12b) i. b11. Gyz = Gzx. ν yz E zz = ν zy E yy and ν xz E zz = ν zx (11. as a result of the plane of symmetry. Further. 7:09 AM . as follows. Slight variations exist in the subscript representation of Poisson’s ratio from book to book. One can see this if Eq. b55 = b66. (11. the number gets reduced to seven elastic constants. (3. Chapter_11A. the strain–stress equations are written in different ways.Introduction to Composite Materials 379 In the above set of equations.v and G. 11) Exx = 1 . nxy. (11. Ezz. one gets 1 1 1 = ⋅ b44 2 (b11 − b12 ) i. One can rewrite Eq. Consequently.e. y and z axes are planes of symmetry (by reflection) and the body is said to be orthogonally anisotropic or orthotropic. Fig.2. the shear stresses cause deformations only in their respective planes.380 Advanced Mechanics of Solids In Eq. 7:09 AM . y and z are the principal directions. (11. retaining the difference. for an axial force in z direction. Gyz. But.12b). (11.11) are only five. the fibres parallel to x-axis may be different from the fibres parallel to y-axis. Similarly. The axes x. 11. There are thus altogether nine independent elastic constants. the lateral contractions (or extensions) in x and y directions will be different. the strain–stress relations will be (recalling that bij = bji): exx = b11σ x + b12σ y + b13σ z eyy = b12σ x + b22σ y + b23σ z e zz = b13σ x + b23σ y + b33σ z (11. z-axis is taken normal to this plane. 11. Gxy = 1 b44 Substituting these in Eq. the number of independent elastic constants in Eq. the elasticity moduli corresponding to these two directions will also be different. b11 − ν xy E xx = b12 . However. nxz.14) using the familiar engineering constants (nij is the Poisson’s effect in i direction due to a force in j direction): exx = Chapter_11A. these being Exx. y y x x z Fig. b 44 = 2(b11 – b12) (11.13) As a result of this.2 Orthotropic composite To be quite general. the properties in x and y directions tend to become equal. Orthotropic Body Let the fibres in a composite be aligned along the x and y axes and let these be uniformly distributed. (11.pmd 380 1 σ − ν xy σ − ν xz σ E xx x E yy y E zz z 7/3/2008. when the thickness in z direction is large.14) gxy = b44txy gyz = b55tyz gzx = b66tzx Observe that because of orthogonal symmetries. The planes normal to x. which are essentially thick sheets. the rectangular stress components will be sx.pmd 381 7/3/2008.15) 1 τ Gzx zx In the above equations. or wrapping one sheet after another about a mandrel).16) E yy ν yz = Ezz ν zy It should be observed that nyx π nxy. composites are manufactured keeping in view certain specific applications.Introduction to Composite Materials eyy = − e zz = – ν yx ν yz σx + 1 σ y − σz Exx ν zx E xx 381 E yy σx − gxy = 1 τ Gxy xy gyz = 1 τ Gyz yz ν zy E yy E zz σy + 1 σz Ezz (11. At any point. sy and txy. an analysis of composite laminates becomes important. For example. b 11 = b22 .4 LAMINATES So far. 11. Let xy plane represent the midplane of a composite laminate. Chapter_11A. 7:09 AM . Exx vxz = E zz ν zx . in the sense that the composite element that was being considered was an element in a three-dimensional or a bulk material having special properties.17) Consequently. (11. Unidirectional Laminates Let the composite consist of fibres all aligned parallel to x-axis. our discussion has been quite general. b55 = b66 (11. b13 = b23 .e Exx vxy = Eyy vyx and. We assume that the laminate will be subjected to a plane state of stress. the following additional relations hold good among the elastic constants. the following conditions hold good because of the reciprocal identity: gzx = ν xy E yy = ν yx E xx i. are produced not only to be used as such or in a moulded form as corrugated sheets but also to produce bulk materials (by cementing one sheet on top of another. with z-axis normal to the plane. the number of elastic constants gets reduced to six. unlike in the case of an isotropic body. Such a composite will obviously be stronger in the x direction than in the y direction. So. a plywood with veneers oriented in different directions is essentially a laminate designed to meet specific requirements. If the fibres in the x and y directions are identical in their elastic properties and assuming that they are uniformly interwoven. Laminates. n zx π nzx and nzy π nyz. However. 34 0. for a plane state of stress with szz = 0.19) σ yy 1 τ Gxy xy nxy is the Poisson’s effect (ratio) in x direction due to a stress in y direction.pmd 382 7/3/2008. Poisson’s ratio.18). There are six independent elastic constants called compliance coefficients. Thus. A negative sign is used to indicate lateral contraction for an extensional stress at right angles. nzx is the Poisson’s ratio in z direction due to a stress in x direction.0 1.3 18. ez and gxy. one can express the stress components in terms of strain components.6 76 10.46 Similar to Eq. shear modulus.6 2. Using engineering constants E and v.18) e zz = b31σ xx + b32 σ yy gxy = b44txy We shall be using double subscripts for stresses and strains in order to be consistent with aijs.26 0. volume fraction of fibre Vf . ν xy exx = 1 σ xx − σ E xx E yy yy eyy = − e zz = − gxy = ν yx Exx ν zx E xx σ xx + 1 σ yy E yy σ xx − ν zy E yy (11.70 0.5 8.60 Sp.3 0. Table 11.14 2.5 7. etc.1 4.382 Advanced Mechanics of Solids The strain components will be ex.3 n yx Vf 0. and specific gravity for selected unidirectional (along x-axis) composites.23 0.59 7. As in Eq. ν xy E yy = ν yx (11.1 gives typical values of Young’s moduli in x and y directions. their values being (remembering that b12 = b21): exx = b11σ xx + b12σ yy eyy = b12σ xx + b22σ yy (11. Chapter_11A.8 1. 7:09 AM .20) E xx Table 11.96 8.6 1. ey.1 Fibres Along xx-axis Material Graphite + Epoxy Boron + Epoxy Graphite + Epoxy Glass + Epoxy Kevlar + Epoxy Exx (GPa) E yy (GPa) G xy (GPa) 181 204 138 38. bijs and tijs.27 5. (11. because of reciprocal identity. (11.28 0.17 5.5 0.16).66 0.45 0. gravity 1. ex¢x¢.22) txy = Gxygxy The reciprocal identity given by Eq. say x¢ and y¢ directions that are oriented at an angle q to x and y axes. bij θ ⎤⎦ Chapter_11A.e. one can solve Eq. sy¢y¢ and tx¢y¢ and let the corresponding strain components be.) along the principal directions can be determined. (11.60) e ij = f2(sij. then one can directly use Eq. Let the stresses applied be sx¢x¢.59) and (1.19) to obtain expressions for sxx. the axes x and y were chosen along the principal directions. (11. However. syy and txy in terms of the engineering constants and the strain components. and gx¢y¢. q ) according to Eqs (1. etc. If the laminate is stressed such that the rectangular stress components for these axes can easily be determined.3 Off-axis loading angle is positive when it is measured counter-clockwise. Consider Fig. q ) \ according to Eqs (2. E yy n yx. then. The procedure to get ei¢j¢. along and perpendicular to the fibre directions. ey¢y¢. The axes x¢ and y¢ are rotated through an angle q counter-clockwise. bij ). (11.Introduction to Composite Materials 383 sxx = a11 ε xx + a12 ε yy syy = a12ε xx + a22ε yy (11. θ ⎤ ⎣ ⎦ { } = f3 f 2 ⎡⎣ f1 ( σ i ' j ' . if the loading is in an arbitrary direction.11.3. (11.20) and (2. and in practice. θ ). in terms of si¢j¢ is as follows: sij = f1(si¢j¢. 7:09 AM . using the elastic constants y y¢ transformed to these new axes.e. bij) according to Eq. q x The positive x¢-axis makes an angle q with fibre direction (i.pmd 383 7/3/2008. ey¢y¢. the elastic constants (like E xx.36a) ei¢j¢ = f3 ⎡ f2 (σij .18) ei¢j¢ = f3(eij. it is desirable to get e x¢x¢.19).20) holds good Off-axis Loading in writing the strain–stress equations. through experiments or otherwise. 11. These come out as E xx sxx = (ε + ν xy ε yy ) 1 − ν xy ν yx xx syy = E yy 1 − ν xy ν yx (ν yx ε xx + ε yy ) (11. x-axis) and the Fig. etc.21) txy = a44gxy Or. i. where the principal directions are x and y and the arbitrary x¢ loading directions are x¢ and y¢. quantities of higher order compared to unity. one ignores in the denominator. ny = sin q. For x¢-axis. and the normal to this plane which is the x-axis.20) and (2. ex¢x¢ = ε xx cos 2 θ + ε yy sin 2 θ + γ xy sin θ cos θ ey¢y¢ = ε xx sin 2 θ + ε yy cos 2 θ + γ xy sin θ cos θ (11.24). Similarly.59) and (1. (11. and for y¢-axis. To transform si¢j¢ into sij. nx¢ = cos (90 – q ) = sin q and ny¢ = cos q. (11.23) txy = + 1 (σ x′ x ′ − σ y ′ y ′ ) sin 2θ + τ x ′ y ′ cos 2θ 2 From Eq. n x ¢ = cos q and ny¢ = –sin q.11. makes angles –q and (90 + q) with x¢ and y¢ axes.Advanced Mechanics of Solids 384 y¢ sy ¢y ¢ y x¢ tx ¢y ¢ syy y¢ e y¢ eyy txy exx x x¢ e x¢ gx ¢y ¢ gxy sxx sx ¢x ¢ Fig. we make use of Eqs (1. nx = cos q.36). ex¢x¢ = cos 2 θ ⎡⎣b11 (σ x ′ x′ cos 2 θ + σ y ′ y ′ sin 2 θ − τ x′ y ′ sin 2θ ) + b12 (σ x′ x ′ sin 2 θ + σ y ′ y ′ cos 2 θ + τ x ′ y ′ sin 2θ ) ⎤⎦ + sin 2 θ ⎡⎣b12 (σ x ′ x′ cos 2 θ + σ y ′ y ′ sin 2 θ − τ x′ y ′ sin 2θ ) Chapter_11A. (2. eyy and gxy from Eq. Hence. nx = cos (90 + q ) = – sin q. for the y-axis.4 Off-axis stresses and strains The above transformations are illustrated in Fig.60). 2 2 sxx = σ x′x ′ cos θ + σ y ′y ′ sin θ − 2τ x ′ y ′ sin θ cos θ syy = σ x′ x ′ sin 2 θ + σ y ′ y ′ cos 2 θ + 2τ x ′ y ′ sin θ cos θ (11. 11.24) e zz = b31 (σ x ′ x ′ cos θ + σ y ′ y ′ sin θ − τ x′ y ′ sin 2θ ) 2 2 + b32 (σ x ′x′ sin 2 θ + σ y ′y ′ cos 2 θ + τ x′y ′ sin 2θ ) gxy = + 1 b44 ⎡⎣(σ x ′ x′ − σ y ′ y ′ ) sin 2θ + 2τ x ′ y ′ cos 2θ ⎤⎦ 2 To obtain ei¢j¢ in terms of eij we make use of Eqs (2.25) gx¢y¢ = – 2ε xx cos θ sin θ + 2ε yy sin θ cos θ + γ xy (cos 2 θ − sin 2 θ ) Substituting for exx. ny = cos q.18) 2 2 exx = b11 (σ x′ x′ cos θ + σ y′ y′ sin θ − τ x′ y′ sin 2θ ) + b12 (σ x ′ x′ sin 2 θ + σ y ′ y ′ cos 2 θ + τ x ′ y ′ sin 2θ ) eyy = b12 (σ x ′ x′ cos 2 θ + σ y ′ y ′ sin 2 θ − τ x′ y ′ sin 2θ ) + b22 (σ x ′ x ′ sin 2 θ + σ y ′ y ′ cos 2 θ + τ x′ y ′ sin 2θ ) (11.36a) for the shear strain. sxx is the normal stress on the x plane. 7:09 AM . In using Eq.4.pmd 384 7/3/2008. 26) .18) and (11. the expression for ex¢x¢ can be simplified as ⎡ n4 2ν xy m4 ⎛ 1 + +⎜ − ex¢x¢ = ⎢ ⎜ E ⎣⎢ xx E yy ⎝ Gxy E yy ⎡⎛ 1 2ν xy 1 1 + ⎢⎜ + − + ⎜ E yy Gxy E yy ⎢⎣⎝ Exx Chapter_11A.19). ⎡ 1 ⎛ 2ν xy 1 1 − n4 + m4 − ⎜ ex¢x¢ = ⎢ ⎜ E E E G ⎢ xx yy xy ⎝ yy ⎣ ⎡⎛ 1 1 1 + ⎢⎜ + − ⎢⎣⎜⎝ Exx E yy Gxy ⎞ 2 2⎤ ⎟ n m ⎥ σx′ x′ ⎟ ⎥⎦ ⎠ ⎞ 2 2 ν xy ( n4 + m4 ⎟n m − ⎟ E yy ⎠ ⎤ )⎥ σ y ′ y ′ ⎥⎦ ⎡ ⎛ 1 ⎞ ⎛ 2ν xy 1 1 + ⎢− 2 ⎜ − n2 − m 2 ⎟ nm − ( n 2 − m 2 ) nm ⎜ ⎜ ⎟ ⎜ E yy E yy G ⎢⎣ ⎝ Exx xy ⎠ ⎝ ⎞⎤ ⎟⎥ τ x ′ y ′ ⎟⎥ ⎠⎦ Observing that n4 + m4 = n4 + m4 + 2n 2 m2 − 2n2 m2 = (n 2 + m 2 )2 − 2n 2 m 2 = 1 − 2n2 m 2 . b12 and b44 from equations (11.pmd 385 ⎞ 2 2⎤ ⎟n m ⎥ σx′ x′ ⎟ ⎠ ⎦⎥ ⎞ 2 2 ν xy ⎟n m − ⎟ E yy ⎠ ⎤ ⎥ σ y′ y′ ⎥⎦ 7/3/2008. one can rewrite the expression for ex¢x¢ as ( ) ex¢x¢ = b11n4 + 2b12 n 2 m2 + b22 m 4 + b44 n 2 m 2 σ x ′ x′ ( ) + b11n m + b12 n + b12 m + b22 n m − b44 n 2 m2 σ y ′ y ′ 2 2 4 4 2 2 + ⎡⎣ −2b11n3 m + 2b12 n3 m − 2b12 nm3 + 2b22 nm3 + b44 nm (n2 − m 2 ) ⎤⎦ τ x ′ y ′ Substituting for b11.Introduction to Composite Materials 385 + b22 (σ x ′x′ sin 2 θ + σ y ′y ′ cos 2 θ + τ x′y ′ sin 2θ ) ⎤⎦ 1 sin 2θ ⎡⎣b44 (σ x′ x ′ − σ y ′ y ′ ) sin 2θ + 2b44 τ x ′ y ′ cos 2θ ⎤⎦ 4 Grouping the coefficients of sx¢x¢. we get + ( ex¢x¢ = σ x′ x ′ b11 cos 4 θ + 2b12 cos 2 θ sin 2 θ + b22 sin 4 θ + 1 b44 sin 2 2θ 4 ( ) + σ y ′y ′ b11 cos 2 θ sin 2 θ + b12 cos 4 θ + b12 sin 4 θ + b22 cos2 θ sin 2 θ − ( 1 b sin 2 2θ 4 44 ) + τ x ′ y ′ − b11 cos2 θ sin 2 θ + b12 cos2 θ sin 2θ − b12 sin 2 θ sin 2θ ) + b22 sin 2θ sin 2 θ + 1 b44 sin 2θ cos 2θ 2 Using the notation n = cos q and m = sin q. 7:09 AM (11. sy¢y¢ and tx¢y¢. b22. Es and n s) along the principal directions are known.29) can be written in a compact from using the notations of Eqs (11.28) and (11.19). if the material constants (i. Equations (11. The compliance coefficients for these will come out as 4 4 2ν xy ⎞ ⎛ ′ = m + n +⎜ 1 − b22 ⎟ n2 m2 E xx E yy ⎜ Gxy ⎝ E yy ⎟⎠ ⎡ ⎛ 4 2 ⎞ ⎛ 2ν xy ′ = ⎢2 ⎜ n − m ⎟ − ⎜ 1 − b24 ⎜E Exx ⎟⎠ ⎜⎝ Gxy E yy ⎣⎢ ⎝ yy ⎤ ⎞ 2 ⎟ ( n − m 2 ) ⎥ nm ⎟ ⎥⎦ ⎠ (11. the compliance coefficients for any arbitrarily oriented axes x¢ and y¢. Hence. 7:09 AM .29) 2ν ⎛ ⎞ ′ = 4 ⎜ 1 + 1 + xy − 1 ⎟ n 2 m 2 + 1 b44 ⎜ ⎟ ⎝ E xx E yy E yy Gxy ⎠ Gxy Since changes in thickness of the laminate are not of much concern. expressions for ey¢y¢ and ex¢y¢.28) ⎤ ) ⎥ nm ⎥⎦ Proceeding on the lines for getting ex¢x¢.e.27) 4 4 2ν xy ⎞ ⎛ b11′ = n + m + ⎜ 1 − ⎟ n2 m2 Exx E yy ⎜ Gxy ⎝ E yy ⎟⎠ 2ν xy ⎛ b12′ = ⎜ 1 + 1 − 1 + ⎜ ⎝ Exx E yy Gxy E yy ⎞ 2 2 ν xy ⎟n m − ⎟ E yy ⎠ ⎡ ⎛ m2 2ν xy n2 ⎞ ⎛ 1 − − ⎟+⎜ b14′ = ⎢ 2 ⎜⎜ ⎟ ⎜ Exx ⎠ ⎝ Gxy E yy ⎢⎣ ⎝ E yy ⎞ 2 ⎟ ( n − m2 ⎟ ⎠ (11.386 Advanced Mechanics of Solids ⎡ ⎛ m2 n2 + ⎢2 ⎜ − ⎜E Exx ⎣⎢ ⎝ yy 2ν xy ⎞ ⎛ 1 − ⎟+⎜ ⎟ ⎜ Gxy E yy ⎠ ⎝ ⎤ ⎞ 2 ⎟ ( n − m 2 ) ⎥ nmτ x ′ y ′ ⎟ ⎥⎦ ⎠ Similarly. it is preferable to use compliance coefficients while expressing strains in terms of stresses. can be obtained. for a unidirectionally reinforced laminate. (11. +x axis.28) and (11.pmd 386 7/3/2008. one can write ′ σ x ′x′ + b12 ′ σ y ′y ′ + b14 ′ τ x ′y ′ ex¢x¢ = b11 ′ σ x′x ′ + b22 ′ σ y ′ y ′ + b24 ′ τ x′ y ′ ey¢y¢ = b12 ′ σ x′ x ′ + b32 ′ σ y ′ y ′ + b34 ′ τ x′ y ′ ez¢z¢ = b31 b σ + b σ + b ′ ′ ′ τ x′ y ′ gx¢y¢ = 14 x ′ x ′ 24 y ′ y ′ 44 From Eq. have not been written.29). b11′ = b11n 4 + b22 m 4 + 2b12 n 2 m2 + b44 n2 m 2 b12′ = b11n 2 m2 + b22 n 2 m2 + b12 (n 4 + m 4 ) − b44 n2 m2 Chapter_11A. Thus.26). one can get from Eqs (11. i. For an anisotropic body.18) and (11. (11. Thus.e.. One should remember that positive x¢-axis makes an angle +q counter-clockwise with the fibre axis. the compliance coefficients for ez¢z¢. one can get expressions for ey¢y¢ and gx¢y¢. 31c) Substituting these for b11′ in Eq. 7:09 AM . E xx b22 = 1 .30a) can be further modified for ease of applications when we take up for analysis multidirectional composites. b44 = 1 Gxy (11.31a) (11. (11.30b) Equations (11.32a) 1 (b + b22 − 2b12 − b44 ) cos 4θ 8 11 = P1 + P2 cos 2θ + P3 cos 4θ Similarly. b12′ = 1 (b11 + b22 + 6b12 − b44 ) − 1 (b11 + b22 − 2b12 − b44 ) cos 4θ 8 8 = P4 − P3 cos 4θ b14′ = − P2 sin 2θ − 2 P3 sin 4θ ′ = P1 − P2 cos 2θ + P3 cos 4θ b22 (11. E yy b12 = − ν xy E yy =− ν yx Exx .30a) ′ = −2b11nm3 + 2b22 n3 m − 2b12 (n 2 − m2 ) nm − b44 (n 2 − m2 ) nm b24 ′ = 4b11n2 m 2 + 4b22 n 2 m2 − 8b12 n2 m2 + b44 (n 2 − m2 )2 b44 where b 11 = 1 . 8 n2m2 = 1 (1 − cos 4θ ).31b) (11.Introduction to Composite Materials b14′ = – 2b11n3 m + 2b22 nm3 + 2b12 (n2 − m2 ) nm + b44 (n2 − m 2 ) nm ′ = b11m 4 + b22 n 4 + 2b12 n 2 m2 + b44 n2 m 2 b22 387 (11. Towards this.30a). consider the following trigonometric identities: n4 = 1 (3 + 4 cos 2θ + cos 4θ ).pmd 387 7/3/2008. substituting for other components in Eq.30a) 1 1 b11′ = 8 (3 + 4 cos 2θ + cos 4θ ) b11 + 8 (3 − 4 cos 2θ + cos 4θ ) b22 + = 1 (1 − cos 4θ ) (2b12 + b44 ) 8 1 1 (3b + 3b22 + 2b12 + b44 ) + (b11 − b22 ) cos 2θ 8 11 2 + (11.32b) ′ = − P2 sin 2θ + 2 P3 sin 4θ b24 ′ = 1 (b11 + b22 − 2b12 + b44 ) − 1 (b11 + b22 − 2b12 + b44 ) cos 4θ b44 2 2 = P5 − 4 P3 cos 4θ Chapter_11A. (11. 8 n3 m = nm3 = 1 (2 sin 2θ + sin 4θ ) 8 1 (2 sin 2θ − sin 4θ ) 8 m4 = 1 (3 − 4 cos 2θ + cos 4θ ) 8 (11. 28). Example 11.3 − 4.25) ] ¥ 0.094) × 0.61 + (139.pmd 388 7/3/2008.33) 8 P 4 = 1 (b11 + b22 + 6b12 − b44 ) 8 P 5 = 1 (b11 + b22 − 2b12 − b44 ) 2 These equations are useful in two ways.53 × 10−3 (GPa)−1 ′ = [ 2(72.28.09 − 139.47 − 3.25)] ¥ 0. Chapter_11A.47 − 3. n = 0. n = cos 30° = 0.15 × 10−3 (GPa)−1 (b) q = 90°.47 ] × 10−12 b44 = 114.75) × 10−3 (GPa)−1 b12′ = [ (5. when we take up multi-direction composites.1875] × 10−12 = (34.3 + 68. Gxy = 7. they can be used for any off-axis loading direction. m = 1.1875] × 10−12 b22 = 80.068 + (139.433 ¥ 10–12 b24 −12 −3 −1 = (142.094) (0.345 + 54. ν yx = 0. Exx = 181 GPa.47 − 3.28) and (11.5 From Eqs.47 + 3.09 + 3.75 − 0.17 GPa.29).433 × 10−12 = 46.1875 − 1.3 GPa. 7:09 AM .01594 Obtain the compliance coefficients appropriate to x¢y¢ axes which are at (a) +30° (counter-clockwise) to xy axes and (b) +90° to xy axes Solution: (a) q = 30°. Secondly.094) (0.525 + 97.98 × 10−3 (GPa)−1 ′ = [ 0.27) and (11. Once these quantities are determined.88 × 10−3 ) (GPa)−1 b14′ = [ 2(24.33 × 10 (GPa) ′ = [ 4(5.094) × 0.188) × 0.094 − 139.1875 + 139.866. Firstly.388 Advanced Mechanics of Solids In the above expressions for b¢ij P1 = 1 (3b11 + 3b22 + 2b12 + b44 ) 8 P 2 = 1 (b11 − b22 ) 2 1 P 3 = (b11 + b22 − 2b12 − b44 ) (11. the quantities Pis are material properties of the composite.75 − 0.094) × 0. (11.86 − 68. b11′ = [ 3.433 × 10 = 32.381) − (139.5 − 3. these equations become useful.525 + 97.1 Consider a graphite-epoxy laminate whose elastic constants along and perpendicular to the fibres are as follows. E yy = 10.143) + (139.188) × 0.47) × 0. m = sin 30° = 0.107 + 6. as mentioned before.433 ¥ 10–12 = (40.547] × 10−12 = − (7.81 − 1. From Eqs (11. vxy = 0. 5 Example 11. The elastic constants along the principal directions of the laminate are Exx = 100 GPa. = Ey¢y¢ = Exx and Gx¢y¢ = Gxy as the results show.25 Gxy = 5 GPa.Introduction to Composite Materials 389 b11′ = 1 = 97. x¢ τ x′ y ′ = 30 MPa The laminate is unidirectionally reinforced and the fibre orientation is 30° to x¢-axis. 11. and the y¢-axis will be along the x-axis (but in the opposite direction). x sy ¢y ¢ tx ¢y ¢ sy ¢y ¢ Fig. y¢ y σ y ′y ′ = 30 MPa.525 × 10−3 (GPa) −1 E xx ′ = 0 b24 ′ = 1 = 139.2 2 ⎡⎛ σ x ′x ′ − σ y ′ y ′ ⎞ 1 ⎢ = (σ x ′x ′ + σ y ′ y ′ ) ± ⎜ ⎟ +τ 2 2 ⎢⎝ ⎠ ⎣ ⎤ 2 ⎥ x′y′ ⎥ ⎦ = 1 (130) ± ⎡⎣(35)2 + 302 ⎤⎦ = 111 or 19 2 \ Chapter_11A.2 At a point in a laminate the following stress state exists: sx¢x¢ = 100 MPa.61). 30° sx ¢x ¢ tx ¢y ¢ sx ¢x ¢ E yy = 10 GPa.5. the values of the elastic constants get switched since the x¢-axis will be along the y-axis.09 × 10 −3 (GPa) −1 E yy b12′ = − ν xy Exx = − 1. (1.2 Determine the principal stresses. the principal stresses are s1. ν yx = 0. principal strains and their orientations in the plane of the laminate. Ex¢x¢ = Eyy.5 × 10 −3 (GPa) −1 b44 Gxy It should be observed that x¢y¢ frame is obtained through rotation of the xy frame by 90° counter-clockwise.pmd s 1 = 111 MPa and s2 = 19 MPa (check: σ x ′x ′ + σ y ′ y ′ = σ1 + σ 2 ) 389 7/3/2008. 7:09 AM . as shown in Fig. 11. Thus. ′ = b22 Example 11.55 × 10−3 (GPa)−1 b14′ = 0 1 = 5. Consequently. Solution From Eq. (11. Let us transform the given stress components to xy axes.25 × 73. and then use Eq.62). (11. and s2 = 19 MPa makes an angle of 110.20). from Eq. we need the corresponding compliance coefficients.52 − 0. (1.207) × 10−3 ± ⎡⎣(3.3° and φ 2′ = 110. (2.19) and using the reciprocal identity. To obtain the strain components with respect to xy axes.25) + (30 × 0. From Eq.433) = 73.48 × 10−3 = 0.062 × 10−3 5 The principal strains corresponding to these are. Thus.52 MPa syy = (100. the algebraically maximum principal stress.25) − (2 × 30 × 0.413) 2 + (4. (11. To determine the principal strains.48 MPa (check σ xx + σ yy = σ x ′ x ′ + σ y ′ y ′ ) txy = 1 (70) × 0.8571 70 \ φ1′ = 20.3° with x¢-axis. (11.531) 2 ⎤⎦ × 10 −3 2 = (3. 7:09 AM .3° with x¢-axis (counter clockwise).2 = ε xx + ε yy 2 ± ⎡⎛ ε − ε yy ⎢⎜ xx ⎢⎜ 2 ⎣⎝ 2 ⎞ ⎛ γ xy ⎟ +⎜ ⎟ ⎜ 2 ⎠ ⎝ ⎞ ⎟ ⎟ ⎠ 2 ⎤ ⎥ ⎥ ⎦ = 1 (0. Eq. lies within the principal 45° angle.866 + (30 × 0.4668 ¥ 10–3 or – 1. e1.433) = 56.6726) = 9.3815 + 7.75) + (2 × 30 × 0.25 × 56.0 × 0.31 MPa 2 From Eq.19).3815 × 10−3 100 100 ) eyy = − 0.31 × 10−3 = 9.390 Advanced Mechanics of Solids From Eq. which in our present case is s1 = 111 MPa.75) + (30 × 0. tan 2f¢ = 2τ x′y ′ σ x′x′ − σ y′y ′ = 60 = 0. To determine the components with respect to x¢y¢ axes.pmd 390 7/3/2008. we need to transform the given stress components to these axes.8784 ¥ 10–3 (check: exx + eyy = e1 + e2) Chapter_11A.3° From Strength of Materials. sxx = (100 × 0. the required rectangular components can be obtained either with respect to x¢y¢ or with respect to xy axes.23). s1 = 111 MPa makes an angle of 20.52 + 73.207 × 10−3 100 10 gxy = 45.7942 ¥ 10–3) ± (5.5) = 45. ( ( ) exx = 56.50).48 × 10−3 = 7. 4275 × 10−12 b′24 = [ 2(74.2. Example 11.6° and 130. n cos 30° = 0.3277 ε xx − ε yy 0.433 × 10−12 = 20.4275 × 30) + (20. the directions of the principal strains were obtained by transforming the applied stresses to the principal direction axes. the principal stress axes do not coincide with the principal strain axes in an anisotropic body.5° These angles are with respect to the x-axis.pmd εx ′ x ′ + ε y ′ y ′ 2 ± 391 ⎡⎛ ε − εy′ y′ ⎢⎜ x ′ x ′ ⎢⎜ 2 ⎣⎝ 2 ⎞ ⎛ γ x′ y′ ⎟ +⎜ ⎟ ⎜ 2 ⎠ ⎝ ⎞ ⎟ ⎟ ⎠ 2 ⎤ ⎥ ⎥ ⎦ 7/3/2008.3728 × 30)] × 10− 6 = 0.5° and 123.1875 − 2.625 + 56.Introduction to Composite Materials 391 From Eq.5°) to x¢-axis. Solution: q = 30°.5° and f2* = +63.4375 × 30) + (57.996 − 2.25 × 30)] × 10− 6 = 0.28) and (11. in general. (11.5) − (200 − 5) × 0.25 + (200 − 5) × 0. By subtracting 30°.5] × 10−12 = − 18. we get the orientations of the principal strain axes with respect to x¢-axis. Show that the same results can be obtained by using the loading or the stress axes as reference and obtaining the corresponding compliance coefficients. Thus.3728 × 100) + (20.564 × 30) + (136.564 × 30) ] × 10 − 6 = 0.25). (2.4375 × 100) + (93.2 = Chapter_11A. In this example.4996) + (200 − 5) × 0. while the principal strain axes are at 33.1875 + 200 ] × 10−12 = 136. the directions of these principal strains are γ xy 9.207 tan 2f* = \ f 1* = –26.564 × 10−12 b′44 = [ 4(10 + 100 + 5 − 200) × 0.001576 (check: ε x ′ x ′ + ε y ′ y ′ = ε xx + ε yy ) gx¢y¢ = [ (57.3728 × 10−12 b′22 = [ 0.29) b′11 = [ 5.51).5 From Equations (11.5] × 0.010434 The principal strains are e 1.3 In Example 11.062 = = − 1.6243 + 6.006012 ey¢y¢ = [ −(18. – 56. Unlike an isotropic case.5° (i.866. the principal stress axes are at 40.4368 × 100) − (18.1875] × 10 −12 = 93.5° and f2* = 33.24 + (200 − 5) × 0.25 × 10−12 From Eq.1875] × 10−12 = 48.433 × 10 −12 = 57.3815 − 7.5°.4375 × 10−12 b′14 = [ 2(25 − 7. f 1* = – 56. ex¢x¢ = [ (48.5] × 0.e.4368 × 10−12 b′12 = [ (10 + 100 − 200 + 5) × 0.6°. m = sin 30° = 0. 7:09 AM . To get this. ′ σ x ′ x′ + b12 ′ σ y ′ y ′ + b14 ′ τ x′ y′ ex¢x¢ = b11 ′ σ x′x′ + b22 ′ σ y ′y′ + b24 ′ τ x ′y ′ ey¢y¢ = b12 ′ σ x′x′ + b42 ′ σ y ′y′ + b44 ′ τ x ′y ′ gx¢y¢ = b41 In Eq. On the other hand.576 ⎞ 2 ⎤ −3 2 = 1 (6.36a) s ij = f 2 (εij . aij ]. we need expressions for ex¢x¢. (11. e ij = f1 (ε i ′ j ′ . (11.217) ⎥ × 10 ⎝ 2 2 ⎢⎣ ⎥⎦ = (3.669) × 10–3 = 0 . composites are used to comply with situations where stresses or loads are prescribed which usually are not along principal directions.012 + 1. To estimate the deformations using Eq. the solutions for sx¢x¢. are sx¢x¢ = 1 [(b22 ′ b44 ′ − b24 ′ 2 ) ε x′x′ − (b12 ′ b44 ′ − b14 ′ b24 ′ ) ε y ′y′ ∆ ′ b24 ′ − b14 ′ b22 ′ ) γ x′y′ ] + (b12 ′ b44 ′ − b24 ′ b14 ′ ) ε x′x′ − (b11 ′ b44 ′ − b14 ′ 2 ) ε y′y ′ sy¢y¢ = 1 [(b12 ∆ ′ b24 ′ − b14 ′ b12 ′ ) γ x′y′ ] + (b11 tx¢y¢ = 1 [(b12 ′ b42 ′ − b22 ′ b14 ′ ) ε x ′x′ − (b11 ′ b24 ′ − b12 ′ b14 ′ ) ε y ′y ′ ∆ Chapter_11A.00946. aij ) from Eq.29).27). For this. – 0. (a).28) and (11. from Eq.012 − 1. aij ). (a). (11.59) and (1. one needs compliance coefficients from known elastic constants along principal directions. 7:09 AM .392 Advanced Mechanics of Solids ⎡⎛ 6. θ ) from Eqs (2. θ Alternatively. ey¢y¢ and gx¢y¢.27).794 ± 5.e obtain si¢j¢ in terms of ei¢j¢s.21) si¢j¢ = f3 (σ ij . The determinant of the coefficients in Eq.60) = f3 ⎡⎣ f 2 (ε ij .576) × 10−3 ± ⎢⎜ ⎟⎠ + (5. θ ⎤⎦ { } = f3 f2 [ f1 (ε i ' j ' . sy¢y¢ and tx¢y¢.20) and (2. sy¢y¢ and tx¢y¢. The motivation for considering this first is that in practice. one can solve for sx¢x¢. i.pmd 392 7/3/2008. (a) is ′ ( b22 ′ b44 ′ − b24 ′ 2 ) − b12′ ( b12′ b44 ′ − b24 ′ b14′ ) + b14′ ( b12′ b24 ′ − b22 ′ b14′ ) D = b11 ′ b22 ′ b44 ′ + 2b12 ′ b24 ′ b14 ′ − b22 ′ b14 ′ 2 − b11 ′ b24 ′ 2 − b44 ′ b12 ′2 = b11 Hence. b¢41 = b¢14 and b¢42 = b¢24. θ ). when we need to analyse multidirectional fibre composites we need to know the stress values for given off-axis strain values. θ ) from Eqs (1. from Eq. one can follow a similar procedure as was adopted earlier.00187 Off-axis Components of Modulus In the previous discussions we obtained the off-axis components of compliances b¢ijs from Eqs (11. 11. 11.35) ′ ε x ′x′ + a24 ′ ε ′y ′y ′ + a44 ′ γ x′y ′ tx¢y¢ = a14 Comparing the coefficients ei¢j¢ in Eqs (11.6.pmd 393 7/3/2008. one can follow the procedure adopted earlier. ′ ε x′x ′ + a12 ′ ε ′y ′y ′ + a14 ′ γ x ′y ′ sx¢x¢ = a11 ′ ε x ′x′ + a22 ′ ε ′y ′y ′ + a24 ′ γ x′y ′ sy¢y¢ = a12 (11.36) a′22 = − 1 (b11 ′ b44 ′ − b14 ′ 2) ∆ a′24 = 1 (b11 ′ b24 ′ − b14 ′ b12 ′ ) ∆ a′44 = 1 (b11′ b22 ′ − b12 ′ 2) ∆ Application of Eq.34) and (11. (11. similar to Eq. 7:09 AM (11.35). Instead. for a laminate under plane state of stress with an off-axis coordinate system.36) to get a¢ij s involves calculations of b¢ijs. (11. This is shown schematically in Fig. one has. one gets a′11 = 1 (b22 ′ b44 ′ − b24 ′ 2) ∆ a′12 = − 1 (b12 ′ b44 ′ − b14 ′ b24 ′ ) ∆ a′14 = 1 (b12 ′ b24 ′ − b14 ′ b22 ′ ) ∆ (11.6 Off-axis components of stresses and strains 4 4 2 2 2 2 a′11 = a11 n + a 22 m + 2 a12 n m + 4 a 44 n m 2 2 2 2 4 4 2 2 a′12 = a11 n m + a 22 n m + a12 ( n + m ) − 4 a 44 n m 3 3 2 2 2 2 a′14 = − a11 n m + a 22 nm + a12 ( n − m ) nm + 2 a 44 ( n − m ) nm 3 3 2 2 2 2 a′24 = − a 11 nm + a 22 n m − a12 ( n − m ) nm − 2 a 44 ( n − m ) nm 4 4 2 2 2 2 a′22 = a 11 m + a 22 n + 2 a12 n m + 4 a 44 n m Chapter_11A.Introduction to Composite Materials + (b11 ′ b22 ′ − b12 ′ 2 ) γ x ′y ′ ] 393 (11.34) From the general stress–strain equations. The results are the following: x¢ y y¢ q ex ¢x ¢ txy sy ¢y ¢ sx ¢x ¢ exx x gx ¢y ¢ tx ¢y ¢ syy eyy gxy sxx ey ¢y ¢ Fig.27).37a) . e. 7:09 AM .38a) a′22 = Q1 − Q2 cos 2θ + Q3 cos 4θ a′12 = Q4 − Q3 cos 4θ a14′ = − 1 Q2 sin 2θ − Q3 sin 4θ 2 (11.39) Q4 = 1 (a11 + a22 + 6a12 − 4a44 ) 8 Q5 = 1 (a11 + a22 − 2a12 + 4a44 ) 8 The Qis involve only material properties and once they are determined.31). a12 = + . Chapter_11A. i. the compliance coefficients for any off-axis direction can be determined using Eq. (11. Q1 = 1 (3a11 + 3a22 + 2a12 + 4a44 ) 8 Q2 = 1 (a11 − a22 ) 2 Q3 = 1 (a11 + a22 − 2a12 − 4a44 ) 8 (11. (11.37b) Equations (11.38b) a′24 = − 1 Q2 sin 2θ + Q3 sin 4θ 2 ′ = Q5 – Q3 cos 4q a44 where.21) and (11. from Eqs (11. for other components we get.32) and (11. Using the trigonometric identities given by Eq.37a) can be recast as was done in the case of compliance coefficients.pmd 394 7/3/2008. Eqs (11.394 Advanced Mechanics of Solids 2 2 2 2 2 2 2 2 2 a′44 = a 11 n m + a 22 n m − 2 a12 n m + a 44 ( n − m ) where. a44 = Gxy 1 − ν yx ν xy 1 − ν yx ν xy 1 − ν yx ν xy (11.33).38b). a′11 = 1 (3 + 4 cos 2θ + cos 4θ ) a11 + 1 (3 − 4 cos 2θ + cos 4θ ) a22 8 8 + = 1 (1 − cos 4θ ) (2a12 + 4a44 ) 8 1 1 (3a11 + 3a22 + 2a12 + 4a44 ) + (a11 − a22 ) cos 2θ 8 2 + 1 (a + a22 − 2a12 − 4a44 ) cos 4θ 8 11 = Q1 + Q2 cos 2θ + Q3 cos 4θ Similarly. (11. E yy ν yx Exx E xx a11 = . a22 = .22). 866 and m = sin q = 0.891 × 0. a24 ′ = − 42.78 GPa.562) + (10.17 GPa Further. 7:09 AM .37b).1083.34 × 0. Substituting in Eq.6245) − (4 × 7.8 × 0.05 GPa.87 GPa.3248) + (10.64 GPa.433) + (2 × 7.1875) = 32. (11.59 GPa Multi-directional Laminates Multi-directional laminates can be formed by cementing plies with different fibre orientations.87 GPa. Similarly.433) = –54. ν xy = 0.0045 = 10. a 11 = E xx = 181 × 1.2 GPa a′12 = (181.40) 7/3/2008. a12 ′ = 42.34 × 0. m = 0.1875) = 109.45 GPa a′14 = − (181. n = 0.562.59 GPa.891 × 0.17 × 0. a14 ′ = − 42. The effective in-plane modulus of laminate plies is found to be simply the arithmetic mean of the moduli of the constituent plies. m 2 = 0. nm3 = 0.5.1875 a′11 = (181. n = 0.433 n = 0.19 Gpa ′ = 23. A multi-directional composite laminate is defined by a code which describes the stacking sequence of the ply groups.pmd 395 (11.0045 = 2.0159 × 181 × 1.1083) + (2.5 × 0. (11.6 GPa.625) + (2 × 2.3 × 1. Gxy = 7.750. a44 ′ = 46.0625 n 2 m2 = 0.17 × 0. Laminates with midplane symmetry will behave like homogeneous anisotropic plates.866.0045 Estimate the components of moduli for an off-axis orientation of (a) q = +30° and (b) q = +45°.25.17 × 0.3 GPa . n = cos q = 0. m 4 = 0.5 × 0.32 GPa. (1 − ν xy ν yx ) −1 = 1.8 × 0. nm = 0. the code [02 /902 /45/ − 453 ]S Chapter_11A.891) (0.8 × 0. Soluion: (a) for q = +30°.0159.17 GPa . for (b) with q = + 45° a11 ′ = 56.5.34 × 0.4 A unidirectionally reinforced composite of ‘Toray’ filament and ‘Nameo’ resin has the following moduli and Poisson’s ratio.Introduction to Composite Materials 395 Example 11. For example.1875) + (10.0045 = 181.1875) + (2. a22 ′ = 46.8 GPa 1 − ν yx ν xy a22 = E yy = 10.1875) + (4 × 7.37a). From Eq. Eyy = 10.891 GPa 1 − ν yxν xy a 44 = Gxy = 7. Exx = 181 GPa . 2 4 n3m = 0. a44 ′ = 36.3248.34 GPa 1 − ν yx ν xy a 12 = ν yx E xx = 0. a24 a22 ′ = − 20. the laminate described by the code given in Eq. (ii) The strain is uniformly the same across the thickness of the laminate.e z = 0. at z = – h . the components of moduli for any given direction are not the same for each ply. the first ply group has two plies of 0° 2 orientation.7 shows the laminate schematically.43) γ *xy γ xy ( z ) = The above assumption is fairly reasonable when the total laminate thickness is small and bonding between plies is good.40) can be written with the following code also: (11. both the ply orientation and the ply material modulus are symmetric with respect to the midplane of the laminate.e.42b) Hence. the stresses vary from ply to ply. The subscript S denotes that the 2 laminate is symmetric with respect to the midplane. followed by one 45° ply.e z = + h . 11. i. Hence.e.7 Multidirectional laminate—Schematic representation means the following: The thickness of the laminate is h. Starting from the bottom of the laminate. and finally the last group with three – 45° plies. Thus. i. the ascending order from the bottom face is identical to the descending order from the top face. (11. 7:09 AM . A subscript T is used to describe the total laminate without resorting to describe symmetry or otherwise. i.396 Advanced Mechanics of Solids z y h/2 x Fig. ε xx ( z ) = ε *xx ε yy ( z ) = ε *yy (11.pmd 396 7/3/2008. and it is useful to discuss in terms of average stresses across the thickness of the laminate.41b) [02 /902 / 45/ − 456 /45/902 /02 ]T where the middle six ply groups with the same orientations have been grouped together. the following assumptions are made: (i) The laminate is symmetric.41a) [02 /902 /45/ − 453 / − 453 / 45/902 /02 ]T or (11. x and y axes are arbitrary axes with reference to which the strains are prescribed. These axes may not in general coincide with any fibre axes. Figure 11. q (z) = q (–z) (11. Because of different orientations of the plies. Chapter_11A. followed by the next group with two 90° plies. i. for a given uniform strain. For a symmetric laminate. The upper half of the laminate is the same as the lower half except that the stacking sequence is reversed to maintain midplane symmetry. For example.42a) and aij(z) = aij (–z) (11. Inplane Stress–Strain Relations In deriving the stress–strain relations for a multi-directional laminate. 45a) where.45b) −h/2 Similarly.45c) τ xy = 1 ⎡⎣ A14 ε *xx + A24 ε *yy + A44 γ *xy ⎤⎦ (11. −h / 2 h/2 A12 = ∫ a12 dz.45e) −h / 2 σ xx .e.pmd 397 7/3/2008.45d) h h where. are the average stresses across the thickness of the laminate. σ yy and τ xy Nxx = A11 ε *xx + A12 ε *xy + A14 γ *xy Nyy = A12 ε *xx + A22 ε *yy + A24 γ *xy (11. for a laminate of thickness h. the stress resultants are Nxx = hσ xx .Introduction to Composite Materials 397 h/ 2 σ yy = 1 ∫ σ yy dz h (11. σ yy = 1 ⎡⎣ A12 ε *xx + A22 ε *yy + A24 γ *xy ⎤⎦ (11. −h / 2 h/2 A14 = ∫ a14 dz (11. (11. 7:09 AM . Hence. h/2 σ xx = 1 ∫ ⎡⎣ a11 ε *xx + a12 ε *yy + a14 γ *xy ⎤⎦ dz h −h / 2 1 h 1 = h = ⎡ε *xx ∫ a11 dz + ε *yy ∫ a12 dz + γ *xy ∫ a14 dz ⎤ ⎣ ⎦ ⎡ A11 ε *xx + A12 ε *yy + A14γ *xy ⎤ ⎣ ⎦ (11.44) −h/2 h/2 τ xy = 1 ∫ τ xy dz h −h / 2 Now. −h / 2 h/2 A44 = ∫ a44 dz (11. σ xx = a11 ε xx + a12 ε yy + a14 γ xy σ yy = a12 ε xx + a22 ε yy + a24 γ xy τ xy = a14 ε xx + a24 ε yy + a44 γ xy Since the strains are the same in the plies. from Eq. i. h/2 A 11 = ∫ a11 dz.35) for any ply.46) Nxy = A14 ε *xx + A24 ε *yy + A44 γ *xy Chapter_11A. stresses per unit thickness. N yy = hσ yy . remembering that x and y are arbitrary axes. N xy = hτ xy Substituting for σ xx . −h / 2 h/2 A24 = ∫ a24 dz. σ yy and τ xy . h/2 A 22 = ∫ a22 dz. So. However. A 22 = Q1h − Q2V1 + Q3V2 A 12 = Q4 h − Q3V2 A 14 = − 1 Q2V3 − Q3V4 2 (11. 7:09 AM . Hence the limits for the integrals can be changed to z = 0 and z = + h . etc. continuing the integration sign as used earlier.47d) It was assumed that the laminate consists of even number of plies and are symmetrically positioned. the values of aijs change in discrete steps from ply to ply and they are not continuous functions of z as indicated by Eqs (11.e. In practice. i. from Eqs (11. i.47a) where V1 = h/2 h/2 −h / 2 −h / 2 ∫ cos 2θ dz. are the integrated values (across the thickness) of the off-axis components of moduli of the laminate. Evaluation of In-plane Moduli The resultant values of the moduli components are obtained by integrating the moduli component values across the thickness. from z = – h to z = 0 is the same as from z = + h to 2 2 z = 0. Chapter_11A. i. the integration sign can be changed to summation sign. but the Qs are the same for each ply. from ply to ply.pmd 398 7/3/2008.398 Advanced Mechanics of Solids The quantities A11.e. but in the bonding process.45 b and e).47b) Similarly. A 11 = Q1h + Q2V1 + Q3V2 (11. i. and V2 = ∫ cos 4θ dz (11. q changes from ply to ply.47c) A 24 = − 1 Q2V3 − Q3V4 2 A 44 = Q5 h − Q3V2 where h/2 h/2 −h / 2 −h / 2 V3 = ∫ sin 2θ dz . Also. A22.45b and e). and the quantities multiplied by 2. when different plies of finite thicknesses are bonded to get the laminate. it was explicitly stated that the 2 orientations q change in finite steps. the mid-plane. the plies are put with their fibre axes oriented differently. and V4 = ∫ sin 4θ dz (11. the positioning sequence from the bottom of the laminate. Thus.e. h/2 A 11 = ∫ a11 dz −h/2 = ∫ (Q1 + Q2 cos 2θ + Q3 cos 4θ ) dz = Q1 ∫ dz + Q2 ∫ cos 2θ dz + Q3 ∫ cos 4θ dz The Qs for a laminate are constant because of our assumption that the laminate consists of plies of the same kind having identical material constants.e. . Equation (11.) (11. Eqs (11. = 1 where. .48) i =1 th where qi is the fibre orientation of the i ply whose thickness is hi. Thus. .pmd 399 7/3/2008. . V1 = 2k1h1 cos 2θ1 + 2k 2 h2 cos 2θ 2 + .. The summation in Eq. . ) V 2 = t (2k1 cos 4θ1 + 2k2 cos 4θ2 + . so that the number of plies from z = 0 to z = h is n. = h h1 h h + 2k2 2 + ..50) V *1 = 1 = v1 cos 2θ1 + v2 cos 2θ2 + . as mentioned earlier. and the total number of plies in the laminate is 2n. Then. then Eq.47b and d) can be rewritten as or 2k1 V1 = v1 cos 2θ1 + v2 cos 2θ2 + . 2k 2 number of plies with fibre orientation q2. 7:09 AM . i..e. = 1 h h h It is easily seen that (2kihi/h) is the volume fraction of plies with fibre orientation i in the laminate. h V V3* = 3 = v1 sin 2θ1 + v2 sin 2θ 2 + . Let the laminate be composed of 2k1 number of plies with 2 fibre orientation q1... (11. n..) where. Chapter_11A. + 2 ki hi cos 2θ i + . 2k1 is the number of plies in the laminate with q1 orientation of fibres. .. ..49).51) h V4* = V4 = v1 sin 4θ1 + v2 sin 4θ 2 + . + 2 ki + . ... + vi + .. say t.. h V V2* = 2 = v1 cos 4θ1 + v2 cos 4θ2 + . (11.. (11.48) is over all the plies from 2 h z = 0 to z = . h v1 + v2 + .50) is simply the rule mixtures equation which will be discussed later..52) V 3 = t (2k1 sin 2θ1 + 2k 2 sin 2θ 2 + .. then on the basis of Eq. + 2ki i + . etc.. h If the thickness of each ply is the same. and 2ki number with qi orientation. and 2k1 + 2 k2 + . . (11. ..49) Also... . . + vi cos 2θi + . 2k 2 is the number of plies in the laminate with q2 orientations. ... . If the volume fractions are indicated by vis.Introduction to Composite Materials 399 n V1 = 2∑ cos 2θi hi (11. ..) V 4 = t (2k1 sin 4θ1 + 2k 2 sin 4θ 2 + . (11.. = 2 n 2k1h1 + 2k2 h2 + . one can write V1* = V 1 = t (2k1 cos 2θ1 + 2 k2 cos 2θ 2 + .. . + 2ki hi + .49) can be written as V (11. 39) are obtained from Eq.51) or Eq. the values of Qis are the same for each ply and these can be evaluated from Eq. sin 2θ = sin 4θ = 0 q 2 = 90°. (11. In general. (11. The modulus data for this ply was given in Example 11. Let v0 be the volume fraction of plies with q = 0° orientation. Further. both being zero. Eq. v0 and v90 are not equal.47) gives the values of Aijs. (11.51) or (11.52). cos 2θ = − 1. sin 2θ = sin 4θ = 0 q 1 = 0°.400 Advanced Mechanics of Solids Using either Eq. These are being reflected in A14 and A24.51). (11.5 Estimate the in-plane moduli and compliances for a cross-ply laminate formed by using unidirectional composites with ‘Toray’ filament and ‘Namco’ resin.4 and is repeated here: Chapter_11B. When v0 = v90. Further. a shear stress txy produces only shear strain in the xy plane. the values of Vi*s (or Vi s) can be determined. From Eq. The values of aijs needed in Eq. Exx and Eyy can be different. (11. It can easily be seen from the figure that x because of the difference in fibre densities in x and y directions. and v90 be the volume fraction of plies with fibres at q = 90° orientation. When v0 or v90. A threez y dimensional body formed by cross-ply laminates was said to be orthogonally anisotropic or orthotropic. V1 = v0 h − v90 h = (v0 − v90 )h V2 = (v0 + v90 )h = h V 3 = V4 ∫ 0 From Eqs (11.pmd 400 7/3/2008. cos 4θ = + 1. A 22 A 44 A 12 A 14 Example 11.53) = [Q4 – Q3]h = A24 ∫ 0 A laminate of this type is called an orthotropic laminate. Since the plies are identical. Using Eqs (11.8 Cross-ply laminate tion or in y direction. The units of Aijs are Pa m or Nm–1. (11. As an illustration of the steps involved consider the following case: Cross-ply composites are commonly used in practice when uniform strength is required in both x and y directions. The information needed is orientation and volume fraction (or the number of plies) of each ply group.8. Fig. 11.39). the volume fractions are equal and A11 = A22. The laminate consists of plies with fibres oriented at q = 0° and q = 90° The laminate is symmetric. A 11 = [Q1 + Q2 (v0 − v90 ) + Q3 ] h = [Q1 − Q2 (v0 − v90 ) + Q3 ] h = [Q5 – Q3]h (11.52). the moduli components A11 and A22 vary linearly with (v0 – v90). we get unidirectional composite laminates.37b). and does not cause any linear strain either in x direcFig. Finally. 11.47a and c). 7:11 AM . is zero. one can easily compute the in-plane moduli of multi-directional laminates with any ply orientation. cos 2θ = cos 4θ = 1. (11. 8 + 10.17 GPa Q1 = 1 (3a11 + 3a22 + 2a12 + 4a44 ) 8 = 1 [ (3 × 181.89 h 12 A 14 = A24 = 0 The average values of the compliance coefficients are obtained by the inversion of Eq.34) = 85.8) + (3 × 10.891) − (4 × 7. −1 (1 − ν yx νxy ) = 1. (11. a11 = 181. 7:11 AM . (11.17) ] 8 = 76.pmd 401 7/3/2008.34) + (2 × 2. (11.891) + (4 × 7.17 GPa.34 GPa.17)] = 22. (11.88 GPa 8 Substituting these in the expressions for Aijs from Eq. E yy = 10.71 0 90 h 11 1 A = 76.0159.73 GPa 2 2 Q3 = 1 (a11 + a22 − 2a12 − 4a44 ) 8 = 1 [181.71 0 90 h 22 (c) 1 A = 26.17 h 44 1 A = 22.46).0045 Solution From Example 11.3 GPa.891 GPa.73 + 19. 401 Gxy = 7.71 = 7.8 GPa.39).88 − 19. then D = A11 ( A22 A44 − A224 ) − A12 ( A12 A44 − A24 A14 ) + A14 ( A12 A24 − A22 A14 ) Corresponding to Eq. a12 = 2. From Eq. 1 A = 76. one can write for εij* Chapter_11B.36 − (v − v ) 85.46).36 GPa Q2 = 1 (a11 − a22 ) = 1 (181. If D is the determinant of the Aijs in Eq.53).4.8 − 10. νxy = 0.34 − (2 × 2.36 + (v − v ) 85. (11.6 GPa 8 Q5 = 1 (a11 + a22 − 2a12 + 4a44 ) = 26.8 + 10.71 GPa 8 Q4 = 1 (a11 + a22 + 6a12 − 4a44 ) 8 (b) = 1 [181.60 − 19.34 + (6 × 2.17)] = 19. a44 = 7. the values of aijs for any ply are a22 = 10.73 + 19.Introduction to Composite Materials Exx = 181 GPa.891) − (4 × 7.71 = 2.46). one gets b11 = 1 2 ( A22 A44 − A24 ) ∆ b12 = − 1 ( A12 A44 − A14 A24 ) ∆ b14 = 1 ( A12 A24 − A14 A22 ) ∆ (11.89h.17 h.07 (96.07 × 7.0151 × 10−30 (96.17) × 1018 = − 0.46) for εij* s and comparing with the coefficients in Eqs (11.17) − 2.pmd 402 7/3/2008. In these laminates.0151 × 10−30 (96. one has.0151 × 10−30 h −3 ∆ \ b11h = 0.07 × 7.55) 2 b22 = 1 ( A11 A44 − A14 ) ∆ b24 = 1 ( A11 A24 − A14 A12 ) ∆ 2 b44 = 1 ( A11 A22 − A12 ) ∆ To find the values of Aijs we need the values of v0 and v90.89 (2. from Eqs (11. A 11 = 96.40 × 10−12 ( Pa) −1 b12 h = − 0. Assuming v0 = v90 = 0.07h.89 × 7. Hence.17) = 66.17) × 1018 = 10.07h.42a and b).12 × 10 h ( Pa) 1 = 0. A22 = 96. (11.313 × 10−12 ( Pa) −1 (d) b22 h = 0.402 Advanced Mechanics of Solids ε *xx = b11 N xx + b12 N yy + b14 N xy ε *yy = b12 N xx + b22 N yy + b24 N xy γ *xy = b41 N xx + b42 N yy + b44 N xy (11. the following: Chapter_11B. A14 = A24 = 0 Substituting these.892 ) × 1018 = 138.07 × 96. A44 = 7.17) × 1018 = 10.54).5.07 × 7.3 × 10−12 ( Pa)−1 Angle-ply Laminates Another class of laminates that are commonly used in practice are the angle-ply laminates.54) Solving Eq. for a balanced angle-ply laminate assuming complete symmetry.40 × 10−12 ( Pa) −1 b44 h = 0.89 × 7.0151 × 10−30 (2. The laminate is said to be balanced when there are equal number of plies with positive and negative orientations.0151 × 10−30 (96. 30 3 3 D = 96.07 − 2. there are only two ply orientations which have the same magnitudes but are of opposite signs. A12 = 2. 7:11 AM . 55) b11h = b 22 h = 39.31 h GPa A 44 = (26. (11. Q1 = 76. (11.88 GPa For a laminate formed from these composites.71)h = 42.71 GPa.36 − 19.56) Substituting these in Eq.5. A 14 = A24 = 0 For the composite considered in Example 11. For such a laminate. A12 = (Q4 + Q3 ) h.59 h GPa A 14 = A24 = 0 The components of compliance are obtained from Eq.05 GPa b11h E yy = 1 = 25. consider a balanced symmetric angle-ply laminate with b = 45°. 7:11 AM .59 GPa b 44 h 403 7/3/2008. Q2 = 85. Q4 = 22. V1* = 1 (cos 2β + cos 2β ) = cos 2β 2 V2* = cos 4b (11. A 11 = A22 = (76. θ2 = − β . v1 = v2 = 1 2 403 (11.pmd E xx = 1 = 25.71) h = 46.81 × 10−12 ( Pa )−1 b44 h = 21. the values of Qis are. Q3 = 19.6 + 19.46 × 10−12 ( Pa)−1 b14 = b 24 = 0 The corresponding Engineering constants are Chapter_11B.05 GPa b 22 h Gxy = 1 = 46.65 h GPa A 12 = (22.51).71)h = 56.36 GPa.73 GPa. A 11 = [Q1 + Q2 cos 2β + Q3 cos 4β ] h A 22 = [Q1 − Q2 cos 2 β + Q3 cos 4β ] h A 12 = [Q4 − Q3 cos 4β ] h (11.88 + 19.91 × 10−12 ( Pa)−1 b12 h = − 29.47a and c).58) A 44 = [Q5 − Q3 cos 4 β ] h A 14 = A24 = 0 As an example. A44 = (Q5 + Q3 ) h A 11 = A22 = (Q1 − Q3 ) h. Q5 = 26.6 GPa.57) V3* = V4* = 0 From Eqs (11.Introduction to Composite Materials q1 = + β . Let the thickness of each ply be 130 ¥ 10–6 m. \ a21 = 2.313 × 10−12 = − 1 × 0.4. Let the plies be of the same composite material that we have been discussing so far. (d).746.15 × 106 ) = 908.891 GPa sxx = (181.891 × 0.08 ¥10–3 m The compliance coefficients for the laminate from Eq.40 × 10−9 = 5 × 10−9 (N/m)−1 h 2. This is a symmetric cross-ply laminate having a total of 16 plies. for a given strain.08 b12 = − 1 × 0.9 MPa txy = 0 Chapter_11B.15 × 106 ) = 51. 7:11 AM .3 MPa 6 syy = (2. a 11 = 181. Also.8 × 0. consider a laminate having the code [04/904]s. \ a21 = 2.15 × 106 ) = −12. the strains are ε *xx = b11 N xx = 5 × 10−9 × 106 = 5 × 10−3 ε *yy = b12 N xx = − 0.5 PLY STRESS AND PLY STRAIN The stress analysis of symmetrical laminates discussed in the previous section was based on the fundamental assumption that all the plies in the laminate experienced uniform strains. a given load or stress gets distributed according to the stiffness of each ply.54). and let the laminate be subjected to a uniaxial stress resultant Nxx = 1 MN/m.891 × 0. the stress–strain equations are sxx = a11ε xx + a12ε yy syy = a21ε xx + a22ε yy From Example 11. In this particular case.34 × 0.15 × 10 ) = 12. a22 = 10.746 The reason for working out the values of the Engineering constants is to show that in the case of a composite. As the fibre orientations in the plies are different. the Poisson’s ratio in the x direction is 0.891 × 5 × 106 ) − (181.pmd 404 7/3/2008.15 × 10−9 × 106 = − 0.8 × 5 × 10 ) − (2. As an example.6 MPa.34 × 5 × 10 ) − (2.8 GPa.313 × 10 −9 = − 0. 11.e. Thickness of laminate = h = 16 ¥ 130 ¥ 10–6 = 2. the values of hbij s are as given in Eq.34 GPa.891 GPa sxx = (10.34 GPa. i.9 MPa txy = 0 For the 90° fibre orientation. the stresses induced in individual plies will be different. (11.15 × 10−3 These are the strains experienced by each ply.8 GPa. under the same assumption of uniform strain. (d) are b11 = 1 × 10. a 11 = 10. For the ply. 6 6 6 syy = (2.08 b14 = 0 From Eq.5.15 × 10 −9 (N/m)−1 h 2.Advanced Mechanics of Solids 404 vxy = −b12 E xx h = 0. a22 = 181.891 × 5 × 10 ) − (10.40 × 10−12 = 1 × 10. the value of Poisson’s ratio can be greater than 0. for 0° fibre orientation. 59) * t 12 = τ12 * The stresses sij are referred to the principal directions 1 and 2.08 × 10−3 m) = 998 × 103 N/m  1 MN/m This checks with the resultant applied stress Nxx. based on the maximum strain criterion. whereas for 90° ply group. In this section. ε xx = 5 × 10 is the strain perpendicular to the fibres and this is greater than 4.5 × 10−3 * −3 then.5 ¥ 10–3.pmd 405 7/3/2008. sxx and syy are the stresses along and perpendicular to the fibres. which is the limit. Hence.Introduction to Composite Materials 405 It should be observed that for 0° fibre orientation ply group. Maximum Stress Theory The maximum stress theory assumes that failure occurs when any of the stresses in the principal material axes reach a critical value.6 + 51. σ 22 is the ultimate tensile or Chapter_11B.9 MN/m 5 11. sxx and syy are stresses perpendicular and parallel to the fibres.95 × 106 Pa) × (2. failure would have occurred in the 90° ply group.3) = 479. the average stresses in x and y directions are σ xx = 1 (908. and the conditions for these are * s 11 = σ11 * s 22 = σ 22 (11. For example. From these results. We shall restrict our discussion to orthotropic materials in general and to laminates in particular.6 FAILURE CRITERIA OF COMPOSITE MATERIALS It is obvious from the discussions so far that the failure theories for composite materials would be quite different and more complex compared to theories of failure for an isotropic solid. for the 90° ply group. we shall briefly consider some of the failure criteria found suitable for composites. One of the reasons in estimating the stresses and strains in individual plies is to check whether they meet the failure criteria. Failure criteria will be discussed in the next section.95 MPa 2 σ y = 0. 7:11 AM . There are three possible modes of failure. in the present case if the maximum strain criterion is applied with the limit that ε max along fibre ≤ 10 × 10−3 ε max perpendicular to fibre ≤ 4. when the resultant applied stress reached a value N xx (max) = 4.5 = 0. τ xy = 0 The resultant stress in x direction is σ xx h = (479. σ11 is the * ultimate tensile or compressive stress in direction 1. the maximum stress theory and the maximum strain theory are the basic theories that are considered first. As in the case of isotropic materials. ε 22 is the * maximum tensile or compressive strain in direction 2. i. (11. E22 and G12 are the material constants.406 Advanced Mechanics of Solids * compressive stress in direction 2 and τ12 is the ultimate shear stress acting on * * plane 1 (i. then from Eq.60) t 12 = + σ x′ x′ sin θ cos θ Combining Eqs (11.59) and (11.62) t 12 = 1 (σ x′ x′ − σ y′ y′ ) sin 2θ + τ x′ y ′ cos 2θ 2 If the applied stresses sx¢x¢ . 11. the stresses in the principal directions 1 and 2 are s 11 = σ x′ x′ cos 2 θ + σ y′ y ′ sin 2 θ − τ x′ y′ sin 2θ s 22 = σ x′ x′ sin 2 θ + σ y ′ y′ cos 2 θ + τ x′ y′ sin 2θ (11.23). sy¢y¢ and tx¢y¢. and if x¢. according to the maximum stress theory. If the load were to be applied at an angle q to the fibre axis direction 1. failure occurs when any of the following conditions hold: * ε11 Chapter_11B. If E11. failure occurs when the strain along any principal direction assumes a critical value. then.e when * e 11 = ε11 * e 22 = ε 22 (11. then.pmd 406 7/3/2008. Maximum Strain Theory According to this theory. plane with normal in direction 1) in direction 2.e. failure occurs when sx¢x¢ assumes the smallest of the following three values sx¢x¢ = * σ 11 cos 2 θ sx¢x¢ = * σ 22 sin 2 θ sx¢x¢ = * τ12 sin θ cos θ (11.61) Instead of sx¢x¢ alone. The values of σ11 . and γ 12 is the maximum shear strain in plane 1-2.5. according to the maximum strain theory. 7:11 AM . σ 22 * and τ12 are obtained experimentally for a given composite. from the transformation equations s 11 = σ x′x′ cos 2 θ s 22 = σ x′ x′ sin 2 θ (11.60). sy¢y¢ and tx¢y¢ either individually or in combination cause s11 or s22 or s12 to exceed their maximum allowable values. Fig. failure occurs. y¢ are the corresponding frame of reference for the applied stresses. if the stresses acting are sx¢x¢.63) * g 12 = γ 12 * where is the maximum tensile or compressive strain in direction 1. (11.65) represents a six-dimensional surface. σ 22 .Introduction to Composite Materials e 11 = σ11 E11 − ν12 e 22 = −ν12 g 12 = τ11 G12 σ11 E11 σ 22 E22 + 407 * ≥ ε11 σ 22 E22 * ≥ ε 22 (11. t12 are obtained from Eq.66a) *2 σ11 Similarly. Eq. with σ11 alone. Let σ11 be the normal or * yield strengths in the directions of anisotropic symmetry. (4. M and N are parameters characterising the anisotropy of the material. For an isotropic solid. s22 and. Another theory which is commonly used in design processes is the energy of distortion theory. The values of the parameters are obtained from * * * . Eq. If the applied stresses lie within the surface.pmd 407 1 (11. In the stress-space.65) where 1. Distortion Energy Theory While the maximum stress and maximum strain theories are easy to apply. 7:11 AM . this expression is generalised and written as U* = 2f (sij) = F (σ11 − σ 22 )2 + G (σ 22 − σ 33 ) 2 + H (σ 33 − σ11 ) 2 2 2 2 + 2M τ 23 + 2 N τ 31 =1 + 2 L τ12 (11. 2 and 3 are the principal directions of symmetry and F. then the values of s11. which can then be substituted into Eq. σ 33 tests conducted on a sample of the composite. For an orthotropic solid. Eq.66c) *2 σ 33 7/3/2008. L.12) gives the distortion energy as 1 ⎡ (σ − σ )2 + (σ − σ )2 + (σ − σ )2 ⎤ 2 2 3 3 1 ⎦ 12G ⎣ 1 where s1.23).65) gives *2 *2 Fσ11 + H σ11 =1 F+H= 1 (11. with σ 2* and σ 3* individually applied. (11. which sometimes is called the Tsai–Hill theory. This theory is similar to the distortion energy or the deviatoric stress theory applied to isotropic solids.66b) *2 σ 22 1 (11. G. (11. H. The critical values of sijs and tijs will give the limits to this yield surface. then no failure occurs. they have limitations since experiments do not completely support them. sy¢y¢ and tx¢y¢ are the stresses applied. Then. (11. one gets F+G= G+H= Chapter_11B.64) * ≥ γ 12 If sx¢x¢. s2 and s3 are the principal stresses and G is the shear modulus.62). G.68) then reduces to 2 2 2 ⎛ σ 1 ⎞ ⎛ σ11 ⎞ ⎛ σ 22 ⎞ ⎛ σ 22 ⎞ ⎛ τ 12 ⎞ ⎜⎜ * ⎟⎟ − ⎜⎜ * ⎟⎟ ⎜⎜ * ⎟⎟ + ⎜⎜ * ⎟⎟ + ⎜⎜ * ⎟⎟ = 1 ⎝ σ 11 ⎠ ⎝ σ11 ⎠ ⎝ σ 11 ⎠ ⎝ σ 22 ⎠ ⎝ τ 12 ⎠ (11. Hence.67) ⎞ ⎟⎟ ⎠ Equation (11. the plane yz will be a transverse plane of isotropy.60) and (11.67) describes a failure surface in a six-dimensional space. τ12 are equal. G.68) In the case of a laminate with unidirectional reinforcements.69) describes a failure envelope and so long as the point (σ11 . *2 τ12 2M = 1*2 .1. for t12. H. then from Eqs (11.pmd 408 7/3/2008. for this body. From Eqs (11. one can solve for F.65) becomes H= 1 *2 2σ11 1⎛ 1 − 1 + 1 ⎜ *2 *2 *2 2 ⎜⎝ σ11 σ 22 σ 33 ⎡(σ11 − σ 22 )2 − (σ 22 − σ 33 ) 2 + (σ 33 − σ11 ) 2 ⎤ + 1 (σ 22 − σ 33 ) 2 ⎣ ⎦ σ *2 22 ( ) 2 2 + 1*2 τ12 + τ 31 + 1*2 (τ 23 )2 = 1 τ12 τ 23 (11. So long as the point (σ11 .408 Advanced Mechanics of Solids On the same lines. (11. In our * * * * and σ 33 and τ13 notation. If the unidirectional laminate is subjected to a stress s x¢x¢ at an angle q to x-axis. Also. Equation (11. G and H. Let x-axis be along the fibre direction instead of z as shown in that figure. and for this plane. substituting the present values for F.66d) τ 31 where τij* s are the yield strengths in shear.. τ 23 . if the state of stress is a plane state. The solutions are F= 1⎛ 1 + 1 − 1 ⎜ *2 *2 *2 2 ⎜⎝ σ11 σ 22 σ 33 ⎞ ⎟⎟ ⎠ G= 1 ⎛− 1 + 1 + 1 ⎞ ⎜ *2 *2 *2 ⎟ ⎟ 2 ⎜⎝ σ11 σ 22 σ 33 ⎠ (11. Eq.69) failure occurs when Chapter_11B. no failure occurs. Then. H. M and N substituted from Eqs (11. σ 23 .e. L = N. one gets 2L = 1 .66a–c). If the point happens to fall either on the surface or outside the surface. the transverse yield strengths σ *yy and σ *zz will be equal to each other. i. 11.69) Equation (11. τ12 . L. failure occurs. σ 22 .t23 and t31.65) with values for F. then one has s33 = t31 = t23 ∫ 0. τ 31 ) lies within this surface. for an orthotropic body. this means that σ 22 are equal. σ 22 .66d) and (11. Consider now a unidirectionally reinforced composite as shown in Fig. 7:11 AM . etc. τ12 ) lies within the surface no failure occurs. τ 23 2 N = 1*2 (11. 25 σ *11 (tens) σ*22 (tens) τ*12 E22 = 17.Introduction to Composite Materials σ x′ x ′ ⎡ 4 ⎛ = ⎢ cos*2θ + ⎜ 1*2 − 1*2 ⎜τ ⎢⎣ σ11 ⎝ 12 σ 11 ⎞ 2 sin 4 θ ⎤ 2 ⎟⎟ sin θ cos θ + *2 ⎥ σ 22 ⎥⎦ ⎠ 409 −1 2 (11.433 Failure occurs when sx¢x¢ ≥ 127. (e) 7/3/2008.74 sx¢x¢ = * σ 22 = 27.5 MPa (Compression) (b) Maximum Strain Theory: From Eq.25 sx¢x¢ = * τ12 = 55. Chapter_11B. (b) maximum strain theory (tension) and (c) distortion energy theory (tension).6 For a class of E-glass-epoxy composite with unidirectional reinforcement.7 MPa 0.4 MPa (tension).9 GPa = 1304 MPa σ*11 (comp) = 1034 MPa = 27.5 MPa sin θ cos θ 0.2 = 127.5 MPa sin θ cos θ 0.7 MPa 0.pmd 409 σ 22 = σ x′ x′ sin 2 θ .433 Failure occurs when sx¢x¢ ≥ 110.4 MPa 2 sin θ 0.64 MPa σ*22 (comp)= 138 MPa = 55. G12 = 8. (11. (ii) Compression: From Eq. sx¢x¢ = * σ 11 cos 2 θ = 1304 = 1378.2 = 127.2 MPa Determine the minimum value of sx¢x¢ applied at an angle of 30° to the fibre axis to cause failure according to (a) maximum stress theory (tension and compression).25 sx¢x¢ = * τ12 = 55.70) Example 11.8 GPa. (11. the following data apply: E 11 = 53.6 = 110.60) s 11 = σ x′ x′ cos2 θ .61) sx¢x¢ = * σ 11 cos 2 θ = 1034 = 1378. (11.61). 7:11 AM . Solution (a) Maximum Stress Theory (i) Tension: From Eq.75 sx¢x¢ = * σ 22 = 138 = 552 MPa sin 2 θ 0.6 GPa v 12 = 0. the shear strain at the time of yielding is * γ 12 = * τ12 = 0.01922 (0.64) and (e) e 11 = σ x′ x ′ E11 cos 2 θ − ν12 σ x′ x′ E22 sin 2 θ 2 ⎞ ⎛ cos 2 θ − ν 12 sin θ ⎟ = σ x′ x ′ ⎜ E22 ⎠ ⎝ E11 2 2 ⎞ ⎛ e 22 = σ x′x′ ⎜ −ν 21 cos θ + sin θ ⎟ E11 E22 ⎠ ⎝ ⎛ 1 sin 2θ ⎞ g 12 = σ x′x′ ⎜ ⎟ ⎝ 2 G12 ⎠ ⎞ Substituting the critical values for the strains and solving for sx¢x¢ ⎟ with the ⎠ ν12 ν 21 ⎞ reciprocal identity E = E ⎟ 22 11 ⎠ 2 sin 2 θ ⎤ * ⎡ cos θ − ν σ x′x′ = ε11 ⎢ E 12 E22 ⎥⎦ ⎣ 11 = 0.01397)−1 = 0. the maximum tensile strain in direction 1 when yield * (or failure) stress σ 11 is applied is * ε11 = * σ 11 = 1. (11.01922 53.034 = 0. from Eq.64).01047 + 0. the maximum tensile strain in direction 2 when yielding (or failure) occurs is * ε 22 = * σ 22 = 0.8 E11 Similarly.0552 = 0.01394 − 0.001542 17. 7:11 AM .003492) −1 = 1.pmd 410 7/3/2008.0276 = 0.00642 8.4406 GPa = 441 MPa Chapter_11B.001542 (−0.6 G12 From Eqs (11.410 Advanced Mechanics of Solids t 12 = 1 σ x′ x′ sin 2θ 2 Further.8395 GPa = 1839 MPa cos 2 θ + sin 2 θ ⎤ * ⎡ σ x′x′ = ε 22 ⎢ −ν 12 ⎥ ⎣ E22 −1 E22 ⎦ = 0.9 E22 Further. 70) σ x′x′ = 108 × [ 0. m and f refer to composite. unidirectional composite rod.71) The subscripts c. (11. This means that the strain in the matrix and the strains in the fibres Chapter_11B.04511) −1 = 0.pmd 411 7/3/2008.82047] = 83 MPa −1/ 2 11. Dividing Eq. (11.0052612 + (3. 7:11 AM . respectively. (11.Introduction to Composite Materials ⎛ sin 2θ ⎞ 1 * σ x′x′ = γ 12 ⎜2 G ⎟ ⎝ 12 ⎠ 411 −1 = 0.72) where vv is the volume of voids that the composite element may contain.1275 GPa = 142 MPa Based on the minimum of the three values. Assume that the rod extends uniformly with no delaminations between the matrix and the fibres. Consider a composite of mass mc and volume vc.9 Unidirectional composite rod Assume that transverse sections that were plane before loading remain plane after loading.74) where the Ms and Vs stand for mass and volume fractions.e. i.73) Vm + V f + Vv = 1 (11. The total mass of the composites is the sum of the matrix mass mm and the reinforcing fibre mass mf. 11. subjected to a force Pc in the direction of the fibres. Consider a rectangular. matrix and fibre.1875 + 0. one gets and Mm + M f = 1 (11.9. the micromechanical aspects of fibrous composites based on the rule of mixtures. 11.00935) × 0. (c) Distortion Energy Theory: From Eq.7 MICROMECHANICS OF COMPOSITES In this section. Fig. the critical stress value is 142 MPa. The volume vc of the composite is given by vc = vm + v f + vv (11. Pc Pc L Fig.2819 − 0.71) by mc and Eq. where the sharing of the loads by the matrix and the fibre is dependent on the volume-weighted averages of the component properties.00642 (0.72) by vc. mc = mm + m f (11. will be examined. (11. Thus. and the matrix as one group. This is called the rule of mixtures for the Young’s modulus in the fibre direction. (11. The situation depicted by Eq.76a) and (11.e.80) If the composite is loaded in the transverse direction. 11. σ f = E f ε cl If Ac is the total cross-sectional area of the composite. From Eqs (11. 7:11 AM . v f = A f L.10). the matrix and the fibres are all equal. one can obtain an expression for the composite strength in the fibre direction as σ cl = σ m Vm + σ f V f (11. It is further assumed that the Poisson’s ratios of the matrix and the fibres are equal. ε cl = ε m = ε f = ∆L (11. then Pc = Pm + Pf i. If Em and Ef are the Young’s moduli for the matrix and the fibre.76a) = ( Am Em + Af E f ) ε cl or (11. and if it is assumed once again that there is no separation between the fibres and the matrix.75) is known as the isostrain situation.10 Chapter_11B. tf tc tm Fig. then the stresses are sm = Em ε cl .75) L where ecl indicates the strain in the composite in the longitudinal direction. (11. v m = Am L.pmd 412 Two phases of unidirectional composite rod 7/3/2008. Fig.78) Eq.78). and v Am = m = Vm .412 Advanced Mechanics of Solids are the same.77) becomes Ecl = Em Vm + E f V f = E11 (11.76b) Af A σc = Ecl = Em m + E f Ac Ac ε cl (11. then one can group the fibres together as one phase material that is continuous. sc Ac = σ m Am + σ f A f (11.77) Since the lengths of the composite. vc Ac vc = Ac L Af v f = = Vf Ac vc (11.79) E11 is the Young’s modulus for the composite in the fibre direction. If L is the length of the member as shown in Fig. (11.81). Some aspects of this will be discussed subsequently. = σ ct Em Vm + σ ct Ef Vf V V 1 = m + f Ect Em E f (11. and the linear strain in the rod are given by ecl = Chapter_11B. when each is considered as one phase material.85) for the values of the Young’s moduli in the axial direction (i.85) gives the Young’s modulus for the composite in a direction transverse to the fibre direction according to the rules of mixtures. then Lt f = v f . ect = σ ct Ect εm = . ect = ε m Vm + ε f V f Now.84) σm Em . then the transverse stresses in the two phases are equal. (11.9.Introduction to Composite Materials 413 If the applied load is uniformly distributed across the transverse faces. Eq.81) The total transverse displacement is ∆tc = ∆tm + ∆t f (11.85) Equation (11.83) Dividing Eq. 11. i. (11. in the direction of the fibres) and the transverse direction are obtained under the assumption that the Poisson’s ratios for the matrix and the fibres are equal. If the Poisson’s ratios are different. (11. Ltm = vm .82) tm and tf are the equivalent gauge lengths of the matrix and the fibre respectively. When a composite cylindrical rod of uniform cross-section is subjected to a force Pc.82) by tc ect = = ∆tc ∆tm ∆t f = + tc tc tc ∆tm tm ∆t f t f ⋅ + ⋅ tm tc t f tc = εm vf vm +εf vc vc i.e. assuming that cross-sections remain plane.79) and (11. It should be observed that equations (11.e. 7:11 AM .84) becomes σ ct Ect or.pmd σf Ef = 413 σm Em 7/3/2008. Ltc = vc (11.e. εf = σf Ef Using Eq. then the analysis becomes complicated. σ ct = σ m = σ f (11. the stresses in the fibre and matrix. under some simple assumptions.e. Assume. if the shear moduli for the matrix and the composite are not equal. an expression can be obtained as indicated next.11. 11.11(b). Chapter_11B. the shear stresses on the complementary faces are equal. (11. as shown in Fig.414 Advanced Mechanics of Solids = Pc Ac Ecl = ⎤ Pc ⎡ 1 . t tc df t t tf t b dm (b) l (a) tm t (c) Fig.11(c). there will be some discontinuity in the shear strains as shown in Fig. 11.86) σ m = Em ε cl (11. Ignoring this discontinuity. As shown in Fig. (11.87) The determination of the shear modulus Gc for the composite in terms of shear moduli for the fibre and matrix is not simple. that the composite can be considered to be a combination of two continuous phase materials. However.pmd df = 414 τ (δ f / t f ) τt f Gf and and Gm = δm = τ (δ m / tm ) τ tm Gm 7/3/2008.79) ⎢ Ac ⎣ EmVm + E f V f ⎥⎦ s f = E f ε cl . 11. Consequently. 7:11 AM . from Eq. then Gf = i.11 Assumed shear deformation ⎛δ f ⎞ the shear strain in the fibre is ⎜ ⎟ ⎝ tf ⎠ ⎛δ ⎞ the shear strain in the matrix is ⎜ m ⎟ ⎝ tm ⎠ ⎛ δ f + δm ⎞ and the total shear strain in the composite is ⎜ ⎟ ⎝ tc ⎠ The shear modulus for the composite is Gc = τ (δ f + δ m ) tc = τ tc δ f + δm (11. one that of fibre and the other that of matrix.88) If Gf and Gm are the shear moduli for the fibre and matrix. 11. For this. Hence. (11. Thus. (11. one can get the composite Poisson’s ratio in terms of the matrix and fibre ratios under some simple assumptions. Among the several important properties of composites.89) gives the composite shear modulus in terms of the constituent shear moduli. In obtaining an expression for the composite elastic modulus E11 in the fibre direction. r is the density and E is the modulus of elasticity. the Poisson’s ratio for the composite is nc = ε ct = Vmν m + V f ν f ε cl (11. The transverse strain is therefore ect = δ tc tc = ε cl tc (tmν m + t f ν f ) = ε cl (Vmν m + V f ν f ) (11. In the metric system the yield strength s will be expressed in kgf/cm2 and the density in kgf/cm3. is positive. If the ratios happen to be different. consider Fig. One should be careful about the units involved in Eq. the longitudinal strain. 11.79). the specific strength will be expressed in Chapter_11B.90a) using Eq. (11. the specific strength and specific modulus are the special characteristics.90b) It should be observed that the transverse strain ecl as given by Eq.83). These are defined as follows: Specific strength = σ (11.Introduction to Composite Materials 415 Substituting these in Eq. Properties of some typical fibres are given in Table 11.90a) is negative when ecl. (11. τ tc (τ t f / G f ) + (τ tm / Gm ) Gc = tc (t f / G f ) + (tm / Gm ) = G f Gm (11. it was assumed that the Poisson’s ratios for the fibre and matrix were equal. 7:11 AM . The highest specific modulus is usually found in materials having a low atomic number and covalent bonding.88).pmd 415 7/3/2008. then the change in the transverse dimension tc is δ tc = tmν mε cl + t f ν f ε cl This is under the assumption that there are no transverse stresses when the bar is subjected to uniaxial tension.9 and Eq.89) V f Gm + Vm G f Equation (11. (11.91) ρ Specific modulus = E ρ where s is the yield or tensile strength.2.91). such as carbon and boron. The longitudinal strain for the composite is or Gc = ecl = σ cl Ecl = σ cl EmVm + E f V f If vf and vm are the Poisson’s ratios for the constituents. 9 85 1. Estimate the density.76 3450 4480 4137 72.416 Advanced Mechanics of Solids 3 kg/cm cm3 cm 2 ⋅ cm : ⋅ : 2 2 kg kg cm s cm s2 In SI units also.80). modulus of elasticity and tensile strength parallel to the fibre axis. (11. (11.71).36 1275 3450 303 379 2.50 2.4 m3 of fibres and 0.85).7 One of the important light weight composites used for high temperature applications is borasic-reinforced aluminium containing 40% by volume of fibres. kgf 2 Table 11.70 × 103 ) + 0. The following data is given: Material Density (kg/m3) E (GPa) 2. scl = 35(0. the specific strength or the specific modulus will be expressed in m2/s2.4(2. 8:18 AM .6 m3 of aluminium.14 4480 3300 124 172 1. Chapter_11B.6) + 2760 (0.36 ¥ 103 2. (11.56 × 103 kg/m3 From Eq. Hence. from Eq.4 86.75 1.6(2. (11.90 5650 1860 276 531 Example 11.4) = 1125 MPa In a direction perpendicular to the fibre axis.36 × 103 ) = 2.44 1. Also estimate the modulus of elasticity perpendicular to the fibres. from Eq.pmd 416 7/3/2008. the density rc (kg/m3) of the composite is rc = 0.2 Material Density (Mg/m3) Polymers Kevlar Polyethelene Metals Be Boron Glass E-glass S-glass R-glass Carbon High strength High modulus Tensile strength (MPa) Elasticity modulus (GPa) 1.79). Ecl = 70(0.6) + 380(0.70 ¥ 103 380 70 Fibres Aluminium Tensile strength (MPa) 2760 35 Solution A cubic metre of composite consists of 0.4) = 194 GPa From Eq.83 2.55 2. 3 0. Calculate the percentage of load carried by the fibres when the composite is loaded.6 + 0. = Example 11. \ i.3 Hence. the fibres carry 92% of the applied load. The direction of loading in the composite will be in the fibre direction. 7:11 AM .9 GPa \ Example 11.4 = 9. Solution Assuming isostrain condition.7(σ m /σ f ) + 0.8 A glass fibre reinforced nylon composite contains E-glass ibres 30% by volume. ecl = ε m = ε f But.7(1/25.3) = 0. the specific modulus of the composite should not be lower than that of the current material. It is desired that while weight reduction is important.3) σ m (0.3 0. E (nylon) = 2.9 An important part of a structure which currently is being made of an aluminium alloy having a modulus of elasticity of 60 GPa is to be replaced by a composite material consisting of E-glass fibres in nylon matrix. The moduli of elasticity of the constituents are E (glass) = 72 GPa. (11.71) + 0.8 GPa.3 = 0.7) + σ f (0.e em = σm Em = σm Em and ε f = σf Ef σf Ef σf Ef = == 72 = 25.624 × 10−3 Ect 70 380 Ect = 103.71 Em 2. the fraction of the load carried by the fibre is Ff σ f Af = Fm + F f σ m Am + σ f A f = σ f vf .8 σm The load carried by the composite is Fc = F m + F f Hence. The density of aluminium alloy used is 2. using Eq.8 ¥ 103 kgf / m3 Chapter_11B.78).92 0. σ m vm + σ f v f = σ f (0.Introduction to Composite Materials 417 1 = 0.pmd 417 7/3/2008. 14 ¥ 103 kg/m3 and 2. If we use 60% by volume of glass fibres in the composite. The volume fraction of Kevlar fibres in these intermediate epoxy sheets is 55%.4 mm thick aluminium and four sheets of 0. the fibre content is 55%. the modulus along the fibre according to the rule of mixtures is 3 1 [(2 × 70) + (0. and its modulus is 72 GPa.2 ¥ 0.55) = 0.09 mm3 Since there are four such fibre reinforced epoxy sheets.44 mm3.2 mm thickness.8 GPa. Ec = (0.4) × 1. The aluminium sheets being isotropic.87 GPa To evaluate the modulus perpendicular to the fibre orientation.6) × 2. then the density and modulus of the composite will be rc = (0.4) × 2.2) = 2. For nylon. the corresponding values are 1. Hence.9 × 106 m 2 sec −2 While the specific modulus is marginally increased by 10%.55 × 103 + (0. 7:11 AM . its modulus will be direction Ecl = Chapter_11B.36 mm3.32 × 109 Nm −2 1. and that of fibres is 0.36 × 3) + (0. Solution In each epoxy sheet of 0. the fibre content is (0. the pure epoxy content is 0.986 × 103 kg m −3 = 22.2.4) + (4 × 0.8 × 103 kg m−3 = 60 × 109 Nm3 2.88 mm Out of this.36 mm3.10 A microlaminate is produced using five sheets of 0. the aluminium content is 2 mm3.55 ¥ 103 kg/m3. A microlaminate of size 1 mm ¥ 1 mm has a total volume equal to Vc = (5 × 0.9 × 106 Nkg −1m = 22. we have to do it in two steps.8 = 44.11 mm3 and that of pure epoxy content is (0.8 × 103 kg m2 = 20.418 Advanced Mechanics of Solids Solution The specific modulus of the aluminium alloy is 60 GPa 2. Thus. and that of pure epoxy is 0. the density is reduced by 29% of the original values. Calculate the modulus of elasticity of the microlaminate parallel and perpendicular to the fibre alignment.45) = 0.8 = 69.pmd 418 7/3/2008.986 × 103 kg/m3.44 × 124)] 2. the density of E-glass is 2.6) × 72 + (0.2 mm epoxy which is reinforced with unidirectionally oriented Kevlar fibres.32 GPa \ specific modulus of the composite is 44. Example 11. in a 1 mm ¥ 1 mm sheet size. the total fibre content is 0.44 mm3.14 × 103 = 1.69 × 106 m2 sec–2 From Table 11.2 ¥ 0. So.88 × 10−3 3000 mm smax = ε max × E = 0. (11. one can use the isostrain condition and obtain a modulus value as Ect = 1 [(2 × 70) + (0.5 mm. and fibre reinforced epoxy (volume content = 0. This is to make sure that if the fibres break. we have to use Eq.477 GPa Now. Example 11. Assume a modulus of 3.8 GPa The actual value will however be in between these two values. For each of the reinforced epoxy sheets. if the bonding is good between the aluminium sheets and the reinforced epoxy sheets. (11. The structure is to carry a load of 2. Solution If the member is made entirely of epoxy without any fibres. and this gives a low modulus value.92 × 106 Nm −2 Chapter_11B.83 × 10−3 × 3.85). However.8 ⎤ 1 = + = 0. for the laminate. we have used Eq. if E¢cl is the modulus in a direction transverse to fibre orientation. then Vf V 1 = m + E′cl Em E f = 1 ⎛ 0.2 kN and is to be 3 m long having a circular cross-section.Introduction to Composite Materials 419 independent.8 × 6. It is also required that the composite member should not stretch more than 2.5 GPa for the epoxy.pmd 419 7/3/2008. However.11⎞ −1 + ⎜ ⎟ = 0.85) assuming isostress conditions. then emax = 25 mm = 0.1544 (GPa) 0.8 ⎢⎣ 70 6. the laminate consists of aluminium (volume content = 2 mm 3 . E¢ct = 6.2 ⎝ 3 124 ⎠ \ Ect = 6. for the reinforced epoxy.8 mm 3.477 GPa). the epoxy will be able to carry the load without any catastrophic failure.11 It is desired to design a tensile member made of a uni-directional composite material. The matrix is to be epoxy with a yield strength of 80 MPa.477)] 2. the modulus will be 1 ⎡2 0.477 ⎥⎦ \ Ect = 18. E al = 70 GPa). a modulus in the transverse direction has already been determined as E¢ct. The yield strength of the composite should not exceed the yield strength of the epoxy. Hence. for the reinforced epoxy. 7:11 AM .09 0.0543 Ect 2.4 GPa In getting the above answer.8 = 51.5 × 109 = 2. 420 Advanced Mechanics of Solids Area of section = 2.2 × 103 2.92 × 106 Nm −2 = 0.753 × 10−3 m2 \ Diameter of the member = d = 31 mm Assuming a specific weight of 1.25 ¥ 103 kgf m–3, the weight of the tensile member will be W(epoxy) = (1.25 × 103 ) (0.753 × 10−3 ) (3) = 2.83 kgf For the composite, the maximum strain permitted is still 0.83 ¥ 10–3. The maximum yield strength for the composite is 80 MPa. Hence, the minimum modulus for the composite will be σ = 80 × 10 Nm ε 0.83 × 10−3 6 Ec (minm) = −2 = 96.4 × 109 Nm −2 From Table 11.2, the moduli of glass fibres are less than the minimum required. So one has to look for a fibre having a higher modulus. High-modulus carbon having a modulus of 531 GPa, and a density of 1.90 Mgm–3 meets our requirement. If Vf is the volume fraction of the carbon fibre in the composite, the modulus of the composite will be Ec = V f (531) + (1 − V f ) × 3.5 = 96.4 This should be equal to or greater than 96.4. Thus, V f (531) + (1 − V f ) × 3.5 = 96.4 Vf = 0.176 or The volume fraction of the carbon is 0.176 and that of epoxy is 0.824. A composite of this nature will have a modulus not less than 96.4 GPa. If the structure is made of such a composite, and if the fibres break when a load of 2.2 kN is applied, then the epoxy alone should be able to carry the load. If Ac is the total area of section, then 0.824 Ac is the area of epoxy and the stress on this should not exceed 80 MPa. Thus, 0.824 Ac × 80 × 106 Nm −2 = 2.2 × 103 N or Ac = 2.2 × 103 N 0.824 × 80 × 106 Nm −2 = 0.0333 ¥ 10–3 m2 = 33.4 mm2 The diameter of the composite is 6.5 mm. Volume = 33.4 ¥ 10–6 m2 ¥ 3 m = 10 ¥ 10–5 m3 Weight = [(1.9 ¥ 0.176) + (1.25 ¥ 0.824)] ¥ 10 ¥ 10–5 = 0.137 kgf Therefore, the carbon fibre reinforced structure is less than one-quarter the diameter of pure epoxy structure, and one-twentieth the weight of pure epoxy. Chapter_11B.pmd 420 7/3/2008, 7:11 AM Introduction to Composite Materials 11.8 421 PRESSURE VESSELS Let the thickness of the vessel be small compared to the radius of the vessel, so that the curvature effects on the fibres can be neglected. The problem concerned is with the orientation of the fibres for optimum strength. In the netting theory, it is assumed that only the fibres take the load and that too in the directions of the fibres only. The strength of the fibre in its transverse direction is taken as zero. The contribution of the matrix to the strength is ignored. Consider a cylindrical pressure vessel with closed ends, as shown in Fig. 11.12(a), subjected to an internal pressure p. The longitudinal and hoop stresses are pa pa (11.92) , σθ = h 2h where a is the radius of the vessel and h is the thickness of the vessel. Assume a helical winding as shown in Fig. 11.12, and let us consider the stresses along the fibre orientation. If s is the stress along the fibre orientation, then the stress in the z direction is s cos2 f and that in the hoop direction is s sin2 f. For equilibrium, σz = σ cos 2 φ = pa 2h and σ sin 2 φ = pa h (11.93) From these two, tan 2 φ = 2 or φ ≅ 55° (11.94) Hence, the optimum orientation of the fibre does not coincide with the principal stress direction. The shear stress shown in Fig. 11.12(c) is balanced by the shear stress caused by the fibre in the –f direction. In practice, the fibres are not made to run in the optimum directions as given by Eq. (11.94), because such a pattern cannot be used to form the end domes. Generally, a small winding angle f is used to form both the cylindrical portion and the end domes, and then an overlay of fibres in the circumferential direction is put to resist the hoop stresses. Thus, in practice, the fibres run approximately in the principal stress directions. p z z (a) s sin2 f s sin f cos f s cos2 f f s (b) (c) Fig. 11.12 Composite pressure vessel Chapter_11B.pmd 421 7/3/2008, 8:20 AM Advanced Mechanics of Solids 422 11.9 TRANSVERSE STRESSES In the previous sections, it was assumed that the Poisson’s ratios of the matrix material and of the fibres were equal. When the ratios are different, one can expect forces between the surfaces of contact because of different contractile tendencies. To see this, consider a cylindrical composite member having a a single fibre as a core, Fig. 11.13. Let a be the b radius of the fibre and b the outer radius of the matrix. Let the composite cylinder be subjected matrix to a uniaxial load in the z direction. Ef E m, v m , s m vf Let er , eq , ez be the strains and sr , sq , sz the sf stress components in the polar coordinate system. Then, the general Hooke’s law with Young’s modulus in the longitudinal direction as E and Poisson’s ratio as n, is Fig. 11.13 Cylinder with a single fibre εr 0 0 εθ 0 0 0 1 +ν 0 = E εz σr 0 0 σθ 0 0 0 0 − σz (11.95) 1 0 0 − ν (σ r + σ θ + σ z ) 0 1 0 E 0 0 1 The equation of equilibrium is dσ r σ r − σ θ + =0 dr r This is the only equilibrium equation for the case under consideration. Let the strain in the z direction be constant. The strain-displacement equations are dur u (11.96) , εθ = r , ε z = constant dr r where ur is the radial displacement. Equation (11.95) can be solved for sr and sq in terms of er, eq and ez. The results are er = s r = K [(1 − ν ) ε r + ν (εθ + ε z )] sq = K [(1 − ν ) ε θ + ν (ε r + ε z )] (11.97a) E (1 + ν ) (1 − 2ν ) Substituting for er and eq from Eq. (11.96), one gets where K= (11.97b) du u ⎡ ⎤ s r = K ⎢(1 − ν ) r + ν r + νε z ⎥ dr r ⎣ ⎦ Chapter_11B.pmd 422 (11.98) 7/3/2008, 7:11 AM Introduction to Composite Materials 423 u ⎡ du ⎤ sq = K ⎢ν r + (1 − ν ) r + νε z ⎥ r ⎣ dr ⎦ Substituting into the equilibrium equation, the result appears as d 2ur dr 2 + 1 dur ur − 2 =0 r dr r (11.99) The solution of the above differential equation is C2 (11.100) r where C1 and C2 are constants to be determined from the boundary conditions. Equation (11.100) is valid for both matrix and fibre. Representing the fibre equation by subscript f, and the matrix equation by m, Eq. (11.100) becomes ur = C1r + u rf = C1 f r + C2 f r (11.101) C2 m (11.102) r The boundary conditions to determine the constants are (i) At the free surface r = b, srm = 0 (ii) At the interface r = a, because of continuity, urf = urm and srf = srm (iii) At r = 0, urf = 0 This gives C2f = 0 Applying the above boundary conditions, the following equations are obtained: u rm = C1m r + C ⎛ (i) (1 − ν m ) ⎜ C1m − 22m b ⎝ C2 m ⎞ ⎛ ⎟ + ν m ⎜ C1m − 2 b ⎠ ⎝ C1m + or, C ⎛ (ii) (1 − ν m ) ⎜ C1m − 22m a ⎝ ⎞ ⎟ +νm ⎠ ⎞ ⎟ + ν mε z = 0 ⎠ 1 + 2ν m C2 m ⎛ ⎜ C1m + 2 a ⎝ b2 C2 m = −ν m ε z (f) ⎞ ⎟ +νm εz ⎠ = (1 − ν f ) C1 f + ν f C1 f + ν f ε z or, C1m + i.e. C1m + 1 + 2ν m a2 1 + 2ν m C2 m − C1 f = (ν f − ν m ) ε z (g) C2 m (h) − Cif a = 0 a2 Equations (f) – (h) can be solved for the constants. The stress (–p), at the interface is obtained as Also, Chapter_11B.pmd C1m a + a2 C2 m + ν m ε z = C1 f + ν f ε z 423 7/3/2008, 7:11 AM 424 Advanced Mechanics of Solids p= 2ε z (ν m − ν f )Vm (V f /2 K m ) + (Vm /2 K f ) + (1/ Gm ) where K is given by Eq. (11.97b). The stress components in the fibre are s rf = σθ f = − p s zf = E f ε z − 2ν f p The stress components in the matrix are ⎛ a2 ⎞ ⎛ b2 ⎞ 1 − s rm = p ⎜⎜ 2 ⎟ ⎜ ⎟ 2 ⎟ r2 ⎠ ⎝b − a ⎠⎝ ⎛ a2 ⎞ ⎛ b2 ⎞ 1 + s qm = p ⎜⎜ 2 ⎟ ⎜ ⎟ 2 ⎟ r2 ⎠ ⎝b − a ⎠⎝ (11.103) 2 ⎛ ⎞ s zm = E m ε z + 2ν m p ⎜ 2 a 2 ⎟ ⎝b − a ⎠ 11.1 A particular laminate has the following elastic constants along the principal axes x-y: Exx = 200 GPa , E yy = 20 GPa , Gxy = 10 GPa , ν yx = 0.25 At a point in the laminate, the following state of stress exists: sx¢x¢ = 200 MPa , σ y ′ y ′ = 20 MPa , τ x ′ y ′ = 20 MPa The x¢-axis makes an angle of 30° with the fibre axis, counter-clockwise. Calculate the principal stresses, the principal strains and their orientations. ⎡ Ans. s1,2 = ⎢ f¢ = ⎢ ⎢ ⎢ e1,2 = ⎢ ⎢⎣ f* = ⎡ ⎢ 6.25° and 96.25° ⎢ ⎢ 7.207 × 10−3 ; − 2.255 × 10 −3⎢ ⎢ ⎢⎣ − 33.6° and 56.4° 11.2 For a graphite-epoxy laminate having uniaxial reinforcements (parallel to x-axis), the following elastic constants apply: Exx = 181 GPa, E yy = 10.3 GPa, 202.2 MPa; 7.8 MPa Gxy = 7.17 GPa, ν yx = 0.28 Obtain the off-axis compliance coefficients when the axes are rotated by (a) +45° and (b) +60°. Express the results in (1012 Pa)–1 units. Chapter_11B.pmd 424 7/3/2008, 7:11 AM Introduction to Composite Materials Ans. 425 ′ = 59.75; b22 ′ = 59.75; b12 ′ = − 9.99 (a) b11 ′ = 105.7; b14 ′ = − 45.78; b24 ′ = − 45.78 b44 ′ = 80.53; b22 ′ = 34.75; b12 ′ = − 7.88 (b) b11 ′ = 114.1 b14 b44 ′ = − 32.34; b24 ′ = − 46.96 11.3 Estimate the components of moduli and compliances for a cross-ply laminate formed from composites consisting of Toray filament and Namco resin. The modulus data are Exx = 181 GPa, E yy = 10.3 GPa, ν yx = 0.159, Gxy = 7.17 GPa, (1 − ν yx ν xy ) −1 = 1.0045 The laminate code is (a) [02/90]s (b) [04/90]. Assume that the composites are of uniform thicknesses. Express Ans. (a) A11 = 124.65h; A22 = 67.49h; A44 = 7.17h; A12 = 2.89h; b11h = 8.03; b22 h = 14.82; b44h = 139.47; b12 h = − 0.344; (b) A11 = 147.51h; A22 = 44.63h; A44 = 7.17 h; A12 = 2.89h b11h = 6.78; b22 h = 22.43; b44h = 139.47; b12 h = − 0.440 11.4 A laminate is formed from angle-ply composite plies having elastic constants given in Example 11.5. Estimate the components of moduli and compliances for the laminate described by the following codes: (a) f = ±30° and (b)f = ±60°. Ans. (a) A11 = 109.3h; A22 = 23.6 h; A12 = 32.46 h A44 = 36.73 h; A14 = A24 = 0 hb11 = 15.42; hb 22 = 71.36; hb12 = –21.8; hb44 = 27.22; hb14 = b 24 = 0 (b) A11 = 23.6 h; A22 = 109.3 h; Chapter_11B.pmd 425 7/3/2008, 7:11 AM 426 Advanced Mechanics of Solids ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦ (b) A11 = 23.6 h; A22 = 109.3 h; A12 = 32.46 h ⎤ ⎥ ⎥ A44 = 36.73 h; A14 = A24 = 0 ⎥ ⎥ hb11 = 71.36; hb22 = 15.42; ⎥⎥ ⎥ hb12 = −21.18; hb 44 = 27.22; ⎥ ⎥ ⎥⎦ b14 = b 24 = 0 11.5 For the laminates of Problem 11.4, estimate the average values of the engineering constants (Es and n s) corresponding to x and y axes. ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣ Ans. (a) E xx = 64.9 GPa; E yy = 14 GPa; ⎤ ⎥ G xy = 36.7 GPa; ν xy = 1.376; ⎥ ⎥ (b) E xx = 14 GPa; E yy = 64.9 GPa; ⎥ ⎥ ⎥⎦ Gxy = 36.7 GPa; νxy = 0.297 11.6 For the laminate described in Example 11.6, determine the minimum failure stresses sx¢x¢ applied at q = 45° and q = 60° to the fibre axis according to (a) maxmium stress theory in tension and compression; (b) maximum strain theory in tension only; (c) distortion energy theory in tension and compression. Use the data given in Example 11.6. θ = 45°, σ x′x′ = 55.2 MPa Ans. (a) Tension: ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣ θ = 60°, σ x′x′ = 36.8 MPa Compression: θ = 45°, σ x′x′ = 110.4 MPa (b) Tension: θ = 60°, σ x′x′ = 127.5 MPa θ = 45°, σ x′x′ = 73.6 MPa θ = 60°, σ x′x′ = 40 MPa (c) Tension: θ = 45°, σ x′x′ = 49.4 MPa θ = 60°, σ x′x′ = 35.3 MPa Compression: θ = 45°, σ x′x′ = 102 MPa ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣ θ = 60°, σ x′x′ = 105 MPa 11.7 A cemented carbide cutting tool used for machining contains 75% by weight tungsten carbide (WC), 15% by weight titanium carbide (TiC), 5% by weight TaC, and 5% by weight cobalt (Co). Estimate the density of the composite, given the following densities for the constituents: Chapter_11B.pmd ρ wc = 15.77 Mgm−3 , ρTic = 4.94 Mgm−3 , ρTac = 14.5 Mgm−3 , ρco = 8.90 Mgm−3 , 426 7/3/2008, 8:22 AM Introduction to Composite Materials 427 11.8 An electrical contact material is produced by infiltrating copper into a porous tungsten-carbide (WC) compact. The density of WC is 15.77 Mgm –3 and that of the final composite is 12.3 Mgm –3 . Assuming that all of the pores are filled with copper, and given rc = 8.94 Mgm–3, calculate (i) the volume fraction of copper in the composite, (ii) the volume fraction of pores in WC compact before infiltration and (iii) the original density of WC compact. [Ans. (a) 0.507; (b) 0.507; (c) 7.775] 11.9 An epoxy matrix is reinforced with 40% by volume E-glass fibres to produce a 20 mm diametre composite to carry a load of 25 kN. Calculate the stress acting on the fibre elements. The modulus of epoxy is 3 Gpa and that of glass fibre is 72.4 Gpa. [Ans. 187.3 MPa] 11.10 In the design problem of Example 11.11, if one uses high strength carbon instead of the high modulus carbon, what will be the changes as compared to the pure epoxy member? [Ans. Diameter = 7.3 mm; Weight = 0.179 kgf] 11.11 If Kevlar fibres are used instead of carbon fibres in Example 11.11, show that the volume fraction of fibre needed would be 0.8, and the diameter of the member would be 13.1 mm, and the weight 0.57 kgf. Chapter_11B.pmd 427 7/3/2008, 7:11 AM Introduction to Stress Concentration and Fracture Mechanics CHAPTER 12 I STRESS CONCENTRATION 12.1 INTRODUCTION While analysing the stresses induced in members subjected to tension, compression, torsion, and bending, it is generally assumed that members do not have abrupt changes in their cross-sections. In the case of a tapered member under tension or compression, the cross-section changes uniformly. But, abrupt changes in the cross-sections of load-bearing members cannot be avoided. Shafts subjected to torsion will have shoulders to take up thrusts, and key-ways for pulleys and gears. Oil grooves, holes, notches, etc., are common. In such cases, the analysis of stresses and strains become complicated. Elementary equations derived under the assumption of no abrupt changes in the geometry of the section are no longer valid. Sectional discontinuities are called stress raisers, and the distribution of stresses in the neighbourshood of such regions are higher than in other regions. They are called regions of stress concentration. Generally, stress concentration is a highly localized effect. Figures 12.1(a) and (b) show members with stepped cross-sections under tension and torsion respectively. Let the members be circular in their cross-sections. In the case of the member under tension, let A1, A2, and A3 be respectively the cross-sectional areas of the parts A, B, and C. If P is the axial tensile force, the stresses in the parts according to P P P elementary analysis are A1 , A 2 , and A3 . However, these values are valid in regions for removed from sectional discontinuities including the region where the load P is applied. The corners where the discontinuities occur are regions of stress concentration. These are shown by dots. Similarly, in the case of the torTr Tr sion member, the shear stresses by elementary analysis are I and I , where Ia b a and I b are the polar moments of inertia of the parts A and B. As before, these average stress values are valid in regions far removed from geometrical discontinuities. At points of discontinuities and nearabout, the stress values are high. Chapter_12a.pmd 428 7/6/2008, 2:13 PM Introduction to Stress Concentration and Fracture Mechanics A 429 BB BA BB BT (b) C P (a) Fig. 12.1 Stepped cross-sections 12.2 MEMBERS UNDER TENSION Figure 12.2 shows a two-dimensional member having two semi-circular grooves and subjected to tensile loading. The distribution of normal stresses across the section mn is shown qualitatively in the figure. At points m and n, the stress magnitudes are high and they fall rapidly to a uniform value as shown. Ignoring stress concentration, the average or the nominal stress across the section mn is σ0 = s σ bt = σb ( b − 2r ) t ( b − 2r ) where b is the width and t, the thickness of the plate. At points m and n, the stresses are maximum, and let their values be s max .The ratio of smax to the nominal or average stress s0 is called the stress-concentration factor Kt; i.e., m n 2r b Kt = σ max σ max ( b − 2r ) . = σ0 σb The subscript t in Kt represents that this stress concentration factor is obtained theoretically or experimentally and does not depend on the mechanical properties (within the elastic limit) of the plate material. Sometimes, instead of using the area across mn, the area away from discontinuity is used to calculate the nominal stress. In the present case, this will be s Fig. 12.2 Plate with semicircular grooves σ 0′ = σ bt = σ bt Chapter_12a.pmd 429 7/6/2008, 5:59 PM 9 1.0 2r 2. where d r is the radius of the groove or the fillet and d is the width of the plate near the groove or the fillet. Poisson’s ratio is taken as 0.2 − 0.3.1 2r where d is the width of the plate at the grooves.3 in the foregoing equation.1 0.0 r d Fig. 12.4 0. the maximum stress occurs again at the bottom of the grooves.1(a) and (b). 2:13 PM .4 (b) fillets 1. In the case of a circular member of large diameter with hyperbolic grooves and subjected to tension.7 0. and r is the radius of curvature at the bottom of the groove. 12.4 dispalys qualitatively how the stress concentration factor in plates Fig. one should be careful about the meaning of the stress concentration factor. σ so.5 0.2 K t 1.3 0. these regions are r smoothened by fillets as r shown in Fig. while referring to design tables. 3.8 0.8 7/6/2008. The majority of data for design purposes are obtained experimentally. Determination of stress concentration factors purely from theoretical analysis for sectional discontinuities of several shapes is difficult and complicated. Figure 12.6 r d d (a) Semicircular grooves 2. it was said that the regions where the cross-sections abruptly change are zones of high stress concentration.8 (b) Fillets (a) grooves 1.2 0.0 0 0. The stress concentration factor is given by Kt = Chapter_12a. With reference to Figures 12.6 0.4 Stress concentration factor for grooves and fillets The case of a very wide plate with hyperbolic grooves has been solved theoretically and the solution shows that the stress concentration factor near the roots of the grooves can be represented approximately by the formula d (a) + 1. 12.Advanced Mechanics of Solids 430 Kt ′ = and σ max . To reduce stresses.3 Members with fillets r varies with the ratio .pmd 430 0. The detailed solution. Kt = 2.5 shows a plate of width w and thickness t with a small circular hole of radius c. (a) with Eq.5 (b). 12. the radius of the hole.93 in the 2r case of the plate. 2:13 PM . The width w is assumed to be large compared to c. Chapter_12a. which is a function of j.85 + 0.5 (c). with in both cases.5 in the case of the cylinder. it can be assumed that the stress condition around the circumference of the circle is not affected by the presence of the hole. Fig. For example. the stress across PQ is sin ϕ σ'= σ wt = σ sin ϕ . σ A σ B A B Q w σ sinϕ = σ’ D D D ϕ ϕ ϕ m σ` = σsin ø P m (c) c b (b) σ (a) Fig. The radius vector makes an angle j with mm. The plate is subjected to a tensile stress s at a distance far removed from the hole. This problem has an exact solution given by the theory of elasticity. An approximate solution can also be obtained using the energy method and curved beam theory discussed in chapters 5 and 6. ( wt / sin ϕ ) This stress distribution. consider a large circle drawn concentric with the hole and having a radius b.08 (b) 2r Comparing Eq. and Kt = 2. Kt = 0.pmd 431 7/6/2008.5 (a) Plate with a Circular Hole Figure 12. consider a tangential plane PQ at point D.Introduction to Stress Concentration and Fracture Mechanics 431 d + 0. it is seen that the stress concentration factor in the case of a cylinder under tension is smaller than the stress concentration factor for d = 10 a plate under tension. (b). 12.5 Plate with a circular hole To determine the stress distribution around the circumference. is shown in Fig. which is fairly simple. Since this circle is far removed from the hole. is given in the Appendix at the end of this chapter. For this. 12. Hence. The area of the section across PQ is wt . 6(b). 12.6 Thick ring subjected to periferal loading (c) Consider a qudrant mn of the ring across the section mm.6 .5 (b) and 12. The reactive forces across mn consist of a longitudinal force N0 and a moment M0 which maintains the slope there as zero. 12. Cosider a section of the quadrant at angle q. 2:13 PM . outer radius b. The value of N0 is obtained by integrating s sin j from 0 to p/2. Fig.8 ) is used to obtain M0.432 Advanced Mechanics of Solids The problem is now reduced to a thick circular ring of thickness t with inner radius c. N0 = π/2 ∫ σ sin ϕ bt d ϕ = σ bt 0 The strain energy method ( similar to the method in Example 6. 12. S K θ ϕ m M0 n n m m n K M ρ0 N=N′ cos θ N′ M0 σ′ N0 (b) N0 θ M0 N0 (a) Fig.pmd 432 7/6/2008.6 (c). and subjected to loading s sin j around the periphery as shown in Figures 12.. The face of this section is subjected to the following moments: moment due to M0 = M0 moment due to N 0 = − N 0 ( ) b+c b+c − ρ 0 cos θ = − N0 (1 − cos θ) 2 2 ( θ moment due to distributed forces = ∫ σ t sin φ bd φ bcoφ − 0 θ = σ bt ∫ sin φ(b cos φ − 0 b+c cos θ 2 ) b+c cos θ)dφ 2 θ 1 b+c = σ bt ⎡ b sin 2 φ + cos θ cos φ ⎤ 2 ⎣⎢ 2 ⎦⎥ 0 1 b+c = σ bt ⎡ b sin 2 θ + cos θ(cos θ − 1) ⎤ 2 ⎣⎢ 2 ⎦⎥ Chapter_12a.e. Fig. i. U = π/2⎛ 2 2 ⎞ 2 αV N M MN ∫ ⎜ 2 AG + 2 AE + 2 AeE ρ − AE ρ ⎟ ρ0 dθ ⎝ 0 0⎠ 0 Here. θ N ′ + ∫ σ sin φt bd φ = N 0 0 N ′ = − σ bt ( cos θ − 1) + σ bt = σ bt cos θ or The face at section q is subjected to moment M. Observing that the direction of M is opposite to the one shown in Fig. Since there is no change of slope across mn. the total strain energy V for the quadrant from Eq. A is the cross-sectional area. ( ) π/2 M N ∂M ∂U dθ =0= ∫ − AEe AE ∂M 0 ∂M 0 0 Substituting for M and N.49) is..e.e. and shear force V = N ′ sin θ = σ bt cos θ sin θ. ∂M 0 ) ⎡1 1 2 2 ⎤ ∫ ⎢ e M 0 − 2 σ btc sin θ − σ bt cos θ ⎥ d θ = 0 ⎦ 0 ⎣ i. and E is young’s modulus. 2:13 PM . (6. and observing that π/2 ( ∂M = 0.. normal force N = N ′ cos θ = σ bt cos 2 θ.e. 6.Introduction to Stress Concentration and Fracture Mechanics 433 The total moment M = M 0 − σ bt = M 0 − σ bt b+c (1 − cos θ) + 12 σ b2t sin 2 θ + σ bt b +2 c cos θ ( cos θ − 1) 2 b+c (1 − cos θ) (1 + cos θ) + 12 σ b2t sin 2 θ 2 1 M = M 0 − σ btc sin 2 θ. 2 The vertical force N ′ on the face at q is obtained from statics. i. 1 π 1 π π M − σ btc − σ bt = 0 e 02 2 4 4 ∴ M0 = = Chapter_12a.. ⎧1 ⎨ ⎩e ( ) ( ) 2 ) i.30.e.. i.pmd ( π ⎫ 1 1 ⎡ θ 1 ⎤ θ ⎢⎣ M 0 θ − 2 σ b tc 2 − 4 s in 2 θ ⎥⎦ − σ b t 2 + 4 s in 2 θ ⎬⎭ = 0 0 ( 2 1 σ btc π + eσ bt π π 2 4 4 ) ( ) σ bt c + e 2 2 433 7/6/2008. r0 is the radius of the centre line. e is the distance of the neutral axis of the curved member from the centroid of the section. M0 = ( ) σ bt c + e 2 2 A = (b − c ) t y= b−c −e 2 b−c b+c +c−e = −e 2 2 Substituting these. ⎡ b ⎤ c ⎥ ⎢ b − 1⎥ ⎣ c ⎦ σ2 = ⎢ e = ρ0 − (b − c ) ( c) log b = b + c (b − c ) − 2 log b c ( ) b + 1 b −1 e = c − c c 2 log b c . (6.5147 c 2 log 5 σ1 = σ ⎡⎢ 5 ( )( ) 1 1.35).pmd 434 ⎣⎢ 5 − 1 ⎦⎥ = σ1 + σ 2 = (1.25σ .20. 2:13 PM . 7/6/2008. 6. ( ) or Let b/c = 5 as an example. From Eq.08σ . simplifying. σ max ∴ Chapter_12a. e 5 + 1 5 −1 = − = 0.83 + 1.83σ 4 ⎣ 4 1. Then. one gets r0 = ρ0 − e = ⎡ 2e ⎞ ⎤ ⎛ b c c ⎟⎥ ⎢ ⎜ σ1 = σ +1 1− b ⎢ 4c 2e ⎜⎝ ⎟⎥ c − 1⎠ ⎦ ⎣ ( ) Similarly.0294 ⎤ + 1 1− ⎥⎦ = 1.434 Advanced Mechanics of Solids The normal stress at the point n of the section mn is σ 1 due to the moment M0. plus σ 2 due to the longitudinal force N0. 6. making reference to Fig. and expressing in the form of ratios.25) σ = 3. σ1 = M0 y Ae ( r0 − y ) In this equation.20.0294 σ 2 = σ ⎡ 5 ⎤ = 1. and since M0 is opposite to one in Fig. It is also seen from the table. when a glass plate is subjected to compressive stress s at the boundary. Instead of the stress s at the boundary being tensile. 12.8(b). at the points s-s.09 1.. the hole cannot be considered as small.22σ 2 4 16 The exact theory also tells that at the point s i. When r = c.e.11 2. Fig.Introduction to Stress Concentration and Fracture Mechanics 435 Table 12. the results of the approximate calculation agree closely with the exact solution. i.50 2. At point r = 2c.33 1. From the exact theory. and the point s will experience a tensile stress of magnitude s. Figure 12. the stress is compressive and is equal to s. The reason for this is that the stress calculated for the curved beam according to the elementary theory is not accurate enough.1 gives the values of σ1 . the stresses at points n-n are -sx each.e. when b/c > 8. 2:13 PM . Brittle materials are strong in compression and weak in tension. 12. 12. If the hole is small Chapter_12a.1 b/c= 3 4 5 6 7 8 s1= s2= smax = 1. 12. 12.08 1. the approximate value deviates substantially from the exact value.14 1. the distribution of stress on the outer perifery of the bigger circle is no longer what was assumed. due to tensile stress. cracks develop at points s. The combined stresses at points n-n are each (-sx+3sy).33 3. the stress is ( ) σ r = 2c = 1 σ 2 + 1 + 3 = 1. and those to sy are 3sy each.. though the hole is small.. if it is compressive as shown in Fig.pmd 435 7/6/2008. where both stresses are tensile. σ 2 .03 1. Consequently. as shown in Fig. i.e.25 1.20 1. when j in Fig.sy). the point n at the hole experiences a tensile stress of magnitude 3s.95 3. Table 12. 12. and σ max for several values of b/c. This means that when the plate is subjected to uniform tensile stress s. A thin tube with a hole and subjected to torsion is shown in Fig. the combined stresses due to sx and sy are each (3sx. Similarly. it can be seen that for b/c between 5 and 8. the stress sr at a distance r from the centre across the section mm is given by 2 4⎞ ⎛ σ r = 1 σ ⎜ 2 + c 2 + 3c4 ⎟ 2 ⎝ r r ⎠ where s is the uniform tensile stress across the ends of the plate.30 Comparing the values in the table with the exact solution smax = 3s for a very small hole.7(a).83 1. Due to sx.7 (b) .83 3. When b/c < 5. This is important if the material is brittle like glass.26 1. the stress falls down rapidly as shown in Fig. the tensile stress sr = 3s as stated earlier. and the point s at the hole experiences a compressive stress of magnitude s.7 is equal to π2 .83 3.19 3. at the boundary. the sign of the stresses around the hole become reversed.8 (a) shows a plate subjected to a biaxial state of stress sx and sy. When r increases.5 (a). Hence.7 (b).93 3. at the point n of the hole. the point n will experience a stress of magnitude –3s. the stresses at n and s are respectively –3sy and +sy. Due to sx.pmd 436 7/6/2008. The net stresses are therefore: at n: -sx .Advanced Mechanics of Solids 436 σ σ s m n m n s (a) s m m s σ (b) σ Fig.8 (a) Sheet subjected to biaxial stress state. 12. the stresses at n and s are respectively -sx and +3sx. since ΩsxΩ=ΩsyΩ= s. Due to sy. 12. 2:13 PM .7 (a) Plate under tensile stress. at s: 3sx+ sy = + 4s. These are equal in magnitude.3sy = -4s. (b) Thin tube subjected to pure torsion Chapter_12a. y σy y σx 16 s n σx s n σy σx x s n n s x σy (b) (a) Fig. the area around the hole can be considered to be subjected to a biaxial state of stress with + sx and .sy. (b) Plate under compressive stress compared to the radius of the tube and is far removed from the ends. The exact analysis of the problem gives the magnitude of the stress at point n of the major axis of the ellipse as ( σ∗ = σ 1+ 2 a b ) y n σ a s b x σ Fig. is subjected to pure torsion in the direction shown in Fig.3s sq at m = 0.e. 12. 12. there will be tensile stresses which are four times the shear stress in the tube..75s. b if the radius c of the hole is equal to 2 . where b is half-width of the plate. The problem of a hole in a plate of finite width has also been solved theoretically. and the plate is subjected to uniform tension s at the ends in the direction of the minor axis of the ellipse. the width 2b Æ •) with a hole.Introduction to Stress Concentration and Fracture Mechanics 437 Hence.8(b). and the plate is subjected to a uniform tensile stress s at the ends.9 (b) Plate with an Elliptical Hole Figure 12. In the previous discussions. Referring to Fig.9. it was assumed that the hole was small compared to the width of the plate and was far from the loaded ends. when a thin tube with a hole. 12. at points such as s.10 Plate with an elliptical hole under tension Chapter_12a. then σ θ at points n and m are sq at n = 4. m σθ σθ σ n σ σ c b Finite plate with a hole Fig. the stress sq at n is more than that for a large plate (theoretically. 12. Hence. i. for a finite plate with a hole.pmd 437 7/6/2008.10 shows a plate of large width (theoretically infinite) with a hole which is elliptical in shape. distance of the straight edge from the centre of the hole. 2:13 PM . tends to infinity. when the ellipse becomes very narrow. 12.11. Fig. When the uniaxial tension s is along the major axis of the ellipse. As the equation shows. and sq = 3. The value of the maximum stress is σ ∗ = σ2 ( a + b )2 ab and the minimum stress is σ ∗ = −σ ( a + b) .438 Advanced Mechanics of Solids where a is the semi-major axis and b is the semi-minor axis of the ellipse. i. i. the stress sq at the ends of the major axis keeps increasing as the ellipse becomes more and more slender. point s and its value is –s. The solution from the theory of elasticity shows that the stresses at the tips of both major and minor axes. it is eqvivalent to subjecting the plate to a tensile stress s = t at p/4 and a compressive stress -s at 3p/4 to the x-axis. the ellips degenerates into a circle. When the plate with an elliptical hole is subjected to pure shear t parallel to the x and y axes.pmd 438 Plate with an elliptical hole under shear 7/6/2008..11 Chapter_12a.. points n and s respectively are both zero. and the narrow slit is along the direction of the external loading. the value of sq tends to s. b → 0. the maximum value of the stress sq occurs at the tips s of the minor axis. In the case of the elliptical hole. When a = b. In the limit. ab y τ t=r s x n ⫺σ τ Fig. and its value is ( σ∗ = σ 1+ 2 b a ) In this case. the least value of stress occurs at the ends of the minor axis.e.e. 12. when b tends to zero. 7:34 PM . This agrees with the previous discussion of a hole in a wide plate. 2. This analogy compares the torsional stresses in a bar of uniform cross-section with that of the motion of a frictionless fluid circulating in a shell having the same cross-section as that of the torsion member. At point A in Fig.Introduction to Stress Concentration and Fracture Mechanics These occur at points whose location depends on the ratio b a 439 . let Vx and Vy be the components in the x and y directions respectively of the velocity of the circulating fluid. notches and grooves do not affect the ultimate strength of the material. though at the bottom of fine scratches. the distribution of stresses depend on the ductility of the material. where the curve tends to 3 as dr tends to zero. which is a brittle material. 2. ∂u y ∂u z ∂u ∆ = ∆V = x + + V ∂x ∂y ∂z Chapter_12a. holes. in the case of glass.. and the points where they occur are close to the tips of the major axis. the stress magnitudes should be quite high. Plates with semicircular grooves subjected to tension as shown in Fig. change in volume per unit volume is given by Eq.1) 7/6/2008. This is seen in Fig.34. geometrical discontinuities or irregularities in members under torsion act as stress raisers. Figure 12. 12. 12. As an explanation to this. In the case of brittle materials however. are nearly three times the stress at the ends of the plate as the radius r of the groove is very small in comparison with the width d of the minimum section.pmd 439 (12. An ideal fluid is characterized by two qualities. This is the reason why in the case of ductile materials.12. small holes are drilled at the ends of a crack. 12. Beyond the elastic limit. It becomes clear why cracks perpendicular to the direction of tensile loading tend to spread. 12.e. But.e. the volumetric strain. the value of s* becomes very high. the hydrodynamical analogy is useful. fine surface scratches do not produce any noticeable weakening effect. also experience stress concentration as stated earlier. In the case of deformable solids.4. Experiments reveal that the stresses at points m and n. A ductile material can be stretched considerably beyond the elastic limit without a great increase in stress. There are no redistribution of stresses. since the stresses tend to get distributed more and more uniformly as the member gets stretched.. i. This is the reason why brittle members with grooves or fillets show a lower ultimate strength compared to members with no geometrical changes.When the ellipse becomes very narrow. the stress concentration caused by grooves and fillets remain up to the point of breaking. to prevent spreading of cracks. i.3 MEMBERS UNDER TORSION Similar to members in tension or compression. Since the maximum stress in the case of a circular hole is finite (stress concentration factor being 3). and (b) frictionlessness. 2:13 PM . it is stated that common glass. in its natural state has many microscopic cracks and defects and that a few additional ones deliberately caused do not substantially affect the strength. In discussing torsion problems.12 shows the cross-section of a shell in which an ideal fluid is circulating. (a) incompressibility. All of the foregoing conclusions regarding stress distribution assume that the maximum stresses are within the elastic limits of the materials under test. are the displacements at a point in the x. and z directions. the vorticity or rotation is given by the expression ω yx = ∂v y ∂vx − ∂x ∂y (12. and the conservation of mass equation becomes ∂v x ∂v y + =0. ∂x ∂y (12. the density r is independent of time. this becomes y Vy A Fig. ω yx = i. ∂x ∂y (12. For a two-dimensional body.8) Hence. and uz.. for a two-dimensional flow field as in Fig.440 Advanced Mechanics of Solids where ux.pmd 440 (12. y. ∂x ∂y (12.5) Then the fluid is said to be incompressible.4) where r is the density of the fluid. 12. the continuity equation is the mathematical expression of the conservation of mass. 12.12. uy. ∂v y ∂v x − = constant.25 gave wyx = wz as rigid body rotation about the z-axis without strain or deformation. In Chapter 2. The terms inside the brackets represent the volumetric strain. If the rigid body rotation is uniform every where. dealing with the analysis of strain. 1 ⎛ ∂u y ∂u x ⎞ − = ω z = constant 2 ⎜⎝ ∂x ∂y ⎟⎠ then. in the case of a fluid. 2.2) If the body under consideration is incompressible. an incompressible fluid circulating with uniform vorticity in a shell is expressed by ∂vx ∂v y + =0 ∂x ∂y Chapter_12a.6) Similarly.12 ∆= ∂u x ∂u y + ∂x ∂y (12.3) In the case of a fluid in motion. If r is the density. ∂u y ∂u x − = constant. then Vx ∆= x ∂u x ∂u y + =0 ∂x ∂y (12.e. 2:13 PM .9a) 7/6/2008.7) The condition of uniform vorticity is therefore. If the flow is steady. Eq. the conservation of mass gives Circulating ideal fluid in a shell ∂v y ⎞ ⎛ ∂v dρ +ρ⎜ x + =0 ∂y ⎟⎠ dt ⎝ ∂x (12. 12) From the stream function and Eq.9 (a).9b) Now define a stream function f such that ∂φ ∂φ . Consider Fig. or Prandtl’s torsion stress function.13 which shows a shaft with a small eccentric hole. 2:13 PM . τ zx = ∂φ ∂φ .13) This means that the velocity components vx and vy correspond to shear stress components τ zx and τ zy respectively. From torsion stress function and Eq. Such a cylinder obviously alters the velocity distribution in the neighbourhood of the obstruction. we should have ∂2φ ∂x 2 + ∂2φ ∂y 2 = constant. and v y = − ∂y ∂x vx = (12.21 for the forsion stress function. 12. 12. the velocities are doubled. The effect of this hole on the stress distribution is similar to the velocity distribution of a circulating fluid in a shell with a solid cylinder of the same diameter as the hole. while at points m and n. 7.10) Such a function satisfies Eq. 12.11) This stream function coincides with Eq. 7. when the shaft with a small circular hole is subjected to torsion. According to hydrodynamic analysis.13 n m (c) (b) (a) Shaft with a circular hole. (c) Shaft with a key way 441 7/6/2008. m m Chapter_12a. vx = ∂φ ∂φ . and τ zy = − ∂y ∂x (12.Introduction to Stress Concentration and Fracture Mechanics ∂v y ∂v x − = constant ∂x ∂y and 441 (12. (b) Shaft with a semicircular groove. In order to satisfy Eq. the velocities of the circulating fluid in the front and rear points of the solid cylinder are zero.19.pmd n m n (a) Fig.10. (12. Let the shaft be subjected to torsion. and v y = − ∂y ∂x (12. 12.9 (b). Analogously therefore. 12. the shear stresses in the immediate neighbourhood of the hole will be twice of what it would be in the absence of the hole. 12. 12. the maximum stress induced ⎣ ⎦ depends on the ratio a . Fig.13 (c). and vertices a of reentrant corners. point m. When a and b become equal. Figures 12. and protruding corners experiences zero stresses. If the diameter changes gradually.pmd (b) (a) Shaft with an elliptical hole. the shearing stresses at these corresponding points in the torsion problem are zero. f ) corners dial direction of the shaft. Figure 12. reentrant corners are points of high stress-concentration. This explains why shafts with radial cracks are weak in torsion. is about twice the shearing stress at the surface of the shaft far away from the groove. (b) shaft with a radial crack 442 7/6/2008. Some of these are b c sharp and some are rounded corners. i. a hole in the shaft.15 (a) and (b) illustrate these.14 Reentrant (b. Circular shafts with abrupt changes in diameters are subjected to high stress concentrations under torsion. then one may use the elementary analysis to get the values of the stresses.14 illustrates protruding corners or projecting corners. the velocities of the circulating fluid are theoretically infinite.13 (b) shows a shaft with a semicircular groove at the periphery. Based on the hydrodynamic analogy. Thus. points n-n.. This means that even a small torque will induce permanent set at these points. At these points. 12. When b becomes very small. then the shearing stresses at the ends of the major axis () ⎡ a ⎤ a are increased in the proportion ⎢1 + b ⎥ :1. Hence. the ellipse tends to become b a circle. Generally speaking. the shear stress at the bottom of the groove.15 Chapter_12a. and the shearing stresses at the tips of this crack become very high. To reduce the a (a) Fig. Let the principal axes be a and Fig.e. e) and b. If the principal axis a is along the raprotruding (a. In the case of a key way with sharp corners. In the corresponding torsion problem. The corners m-m are called reentrant corners. the shearing stresses at these points are also very high.442 Advanced Mechanics of Solids Figure 12. the ellipse resembles a crack in the radial direction. the hydrodynamic analogy indicates a zero velocity of the circulating fluid at the corners protruding or projecting outwards. c. The stress concentration can however be reduced by rounding the corners n-n. and the discussion can be applied. 2:13 PM . The hydrodynamic analogy explains d f the effects of a small hole of elliptical corss-section or of a groove with a semie elliptic cross-section in a shaft under torsion. However.4 MEMBERS UNDER BENDING Equations obtained for normal and shearing stresses in the case of prismatic beams are very often applied to cases of beams of variable cross-section.1 1. The value of t0 is given by τ 0 = Td = 16T3 2J πd where T is the torque applied and J is the polar moment of inertia of the smaller shaft. r Vd VD Fig. 7:47 PM . 12. 12.8 D/d = 2 1.15 r/d Fig. when these structural members or machine components are subjected to fluctuating loads. i. 16T τ0 These localized high stresses may not be dangerous for ductile materials subjected to static loading. Fig.. 12. 0 0 0. and d and D are the two diameters of the circular shaft. If the Chapter_12a.17 illustrates the stress concentration factors Kt as a function of r/d for two values of D/d. d.16. 2 Kt 1.Introduction to Stress Concentration and Fracture Mechanics 443 occurrence of high stresses. 0 2.2 0. as in the case of turbine rotors and crankshafts.05 0. 4 D/d = 1. these stress concentrations will have pronounced effects. Kt = 3 τ max = τ max π d . fillets or shoulders are provided in stepped shafts.e. 6 2. Thus.16 Shaft with variable diameter Figure 12.pmd 443 7/7/2008. The magnitude of the maximum stress depends on the ratios r/d and D/d.25 12. The stress concentration factor Kt is equal to the ratio of the maximum shear stress τ max occurring at the fillet to the stress t0 occurring in the shaft with the smaller diameter. 3.17 Variation of stress concentration factor 0. where r is the radius of the fillet.10 0. 18(b).18(a). Kt = 0. Only in limited number of cases. (12. Fig. If the changes are abrupt. m d d n (b) (a) Fig. 12.pmd 444 7/6/2008. For example.14b) where d is the diameter of the minimum cross-section and r is the smallest radius of curvature at the bottom of the groove.Advanced Mechanics of Solids 444 changes in the sections of the beam are not abrupt and are gradual.14a) can be replaced with sufficient accuracy by the following approximate equation 3 d (12. the radius of curvature at the bottom of the groove. can approximately represented by Kt = d + 0.08 + 0. then.12. a circular shaft with hyperbolic notches subjected to bending has a smaller stress concentration factor at the roots than a wide plate with hyperbolic notches under bending.355 Chapter_12a. Fig. 2:13 PM . a circular shaft with a hyperbolic groove. a large plate with hyperbolic notches subjected to pure bending has also been rigorously analysed for stress distribution near the notches. the values of Kt have been obtained using the equations of the theory of elasticity. 12.14d) r where d is the minimum width of the plate and r.14a) where d d + 1 + (1 + 4v ) +1 + 2r 2r 1+ v 1+ d + 1 2r (12. the maximum stress values will be greater than those obtained from elementary formulas. the stress concentration factor in the case of pure bending is obtained as ⎡ ⎤ ⎡ 3d ⎤ d d ⎢1 + 2r + 1 ⎥ ⎢ 2r + 4 + v − (1 − 2v ) 2r + 1 ⎥ ⎣ ⎦⎣ ⎦ Kt = 3 4N N =3 ( ) (12. As in the case of tension. 12. m and n. v is the Poisson’s ratio for the material.2. The stress concentration factor near the roots fo the notch.18 Shaft and plate with hyperbolic grooves under bending When d 2r is fairly large. Sec.85 (12. as in the previous cases of tension and torsion. The maximum stress can be expressed as σ max = Kt σ in which s is the stress at the point under consideration as obtained from the prismatic beam formula. the solutions obtained by the application of elementary analysis are fairly satisfactory.14c) 4 2r similar to the circular shaft. and Kt is the stress concentration factor. Eq. On the other hand. aluminium alloy and steel whose sult = 0. there are some materials that are not very sensitive to notches. 2:13 PM . Next. keyways.6 q 0.7 GPa.19 Variation of q with notch radius 445 4.Introduction to Stress Concentration and Fracture Mechanics 12.5 Notch radius r (mm) Fig.0 7/6/2008.4 Aluminium alloy 0. for these materials. The value smax of stress at these highly stressed zones was obtained by multiplying the nominal stress value s0 by a factor Kt called the stress concentration factor.0 1. For design purposes.1) (12. the factor Kt is obtained first for a given geometry either from theoretical considerations or experimental results.0 2. i. grooves.16) Figure 12. for the material under consideration. at these zones.15) where q is usually between zero and unity. stresses higher than the nominal stress values are induced. This factor Kt is independent of the material.(a).. For such materials. fillets.pmd 0.5 445 NOTCH SENSITIVITY It was stated earlier in this chapter that when the sectional geometry of a member under stress has geometrical discontinuities like grooves.7GPa 0. if q = 1.0 2.2 0 0 Chapter_12a. the maximum stress value is σ max = K f σ 0 (b) where Kf is a reduced value of Kt and s0 is the nominal stress value.5 1. a lower stress concentration factor can be used for design purpose.0 Steel.5 3. Notch sensitivity q is defined by the equation q= Kf −1 Kt − 1 (12.. and the material under consideration has no sensitivity to notches at all. σ max = Kt σ 0 (a) However.15) shows that if q = 0.8 0. With these. then Kf = 1. holes. find q from design charts. The notch 1.19 shows how the notch sensitivity factor varies with the notch radius for two materials. In line with Eq. then Kf = Kt and the material has full notch sensitivity. s ult = 0. 12. etc.5 3. the value of Kf is obtained from the equation Kf = 1 + q (Kt . Equation (12.e. etc. When pressure is applied between the contacting bodies.446 Advanced Mechanics of Solids sensitivity factor curves involve considerable scatter and because of this many design calculations involve only the stress concentration factor Kt. In the present discussion. the area of contact will be a point as the expression reveals. a circular area of contact is developed due to axial symmetry. The stresses at all points within the area of contact in the two spheres are not uniform..6 CONTACT STRESSES Stresses developed during the pressing actions of two bodies need careful attention since the occurrence of such cases are very frequent. v 2 be the respective elastic constants of the two spheres. the radius a of the contact surface is given by a 3F 8 1 v12 E1 1 v22 E2 1/ d1 1/ d 2 (12. i. (a) Two Spheres in Contact Consider two spheres of diameters d1 and d2 brought into contact. Figure 12. pits or flaking in the surface material. Gears. the spheres experience point contacts. The general analysis of contact stresses involve bodies having double radius of curvature. named after the scientist who developed the theory. When a force F is applied. 12. are familiar examples.20 Body having a double radius of curvature Chapter_12a. ball-and-roller bearings.21(a) shows two spheres in contact and the frame of reference xyz. the stresses developed will be high.e. They have a semi-elliptical distribution. 2:13 PM . when no pressure is applied. etc.pmd 446 7/6/2008. When bodies with curved surfaces come into contact without any pressure or forces between them. only two special cases will be considered.20 illstrates a body having a double radius of curvature.. the point or line contact become area contacts. The stresses developed are generally referred to as Hertzian stresses. v 1 and F E2. Since the areas of contact are small. because of their importance. that is.17) If the spheres are extremely rigid with E1Æ• and E2Æ•. the geometry of contact is in general either a point or a line. 12. Let this contact area have a radius a. According to Hertzian analysis. contacting spheres and contacting cylinders. when the radius in the plane of rolling is different from the radius in a perpendicular plane. and let E 1. The axis of z is downword and the force F acts along Fig. Figure 12. wheel on rails. Typical failures due to these high contact stresses are seen as cracks. Initially. and this is shown in Fig. (c) Enlarged sketch The maximum pressure pmax occurs at the centre of the contact area. 12. (b) Sphere inside another spherical surface All points within the spheres experience stresses.21(b) shows the stress distribution in the spheres and in the area of contact. d2 is negative. For a sphere d1 in contact within another internal spherical surface. but the stresses along the zaxis are maximum. in any diametrical plane at z the stresses at point ( 0.18) are general expressions in the sense that they are valid for a sphere in contact with a plane surface. These cases are shown in Fig. or a sphere inside another spherical surface. (b) Stress distribution within the area of contact.. VF VF Vx Vx Vx Vd1 Vy Vy Vy Va Vd2 a Vz (c) VF VF Vz (a) Vz (b) Fig. Figure 12. For a sphere of diameter d1 in contacte with a plane.17) and (12.22 (a) Sphere in contact with a plane. 12. 1 d (negative) 2 2 d1 d1 d2 = • (a) (b) Fig.e. separately.Introduction to Stress Concentration and Fracture Mechanics 447 the z-axis.21(c). and its magnitude is given as pmax = 3F 2 (12. i. 2:13 PM . z ) Chapter_12a.18) 2π a Equations (12. 12. 12.22. d2 = •. 0.pmd 447 7/6/2008.21 (a) Geometry of spheres. 20) consist of a point and circle. is pmax = 2 ⎡ 3 F ⎢ 3 FE 2 ( r1 + r2 ) 0. The plot of txz = tyz .22) Equations (12. ⎢⎣ E ⎥⎦ (12.19) and (12. These are the principal stresses at any point z along the z-axis. pmax is taken as the unit.23 shows graphically the plots of sz and sx = sy. Figure 12. and the maximum pressure are ⎡ FE 2 ⎤ pmax = 0.109 ⎢ 3 1 ⎥ . (12.pmd 448 7/6/2008. need to be used. Hence. Eq. the maximum pres- sure pmax as given by Eq.388 = 2 πa 2 ⎢ r12 r22 ⎣ ⎤ ⎥ ⎥ ⎦ (12. the radius of the contact area. and taking v = 0. For the stresses.18). but appropriate value for Poisson’s ratio corresponding to the sphere considered.448 Advanced Mechanics of Solids are maximum. τ xy = 0. the maximum pressure which is compressive.23) 2 One can plot the values sx and sz along z to display their variations as a functions of the distance from z = 0.18) which occurs at the centre of the contact area. the radius a of the surface of contact is taken as the unit. 2:13 PM .3. Putting r2 = •.23). is also shown.20) where pmax is the numerical value as given by Eq.3.22) are valid for both spheres.19) (12. and τ xz = τ yz = σx −σz 2 = σy −σz (12. In measuring the distances along the z-axis. and v is taken as equal to 0. All the normal stresses are compressive in nature. Further as σ x = σ y . Assuming both the spheres have the same elastic properties.388 ⎢ 3 2 ⎥ ⎣⎢ r1 ⎥⎦ ⎡ Fr ⎤ a = 1. Let the sphere with radius r1 be pressed on to a plane surface which has the same elastic properties as that of the sphere. (12. Their values are ⎧ ⎫ ⎪⎡ ⎪ ⎛ ⎞⎤ 1 σ x = σ y = − pmax ⎨ ⎢1 − z tan −1 ⎜ 1 ⎟ ⎥ (1 + v ) − 2 2 ⎬ a z / a ⎝ ⎠ 2 1 z / a + ⎦ ⎪⎩ ⎣ ⎪ ( )⎭ ⎡ ⎤ 1 2 2⎥ ⎣1 + z / a ⎦ σ z = − pmax ⎢ (12. Chapter_12a.21) where r1 and r2 are the radii of the two spheres. The Mohr’s circles for the state of stress described by equations (12.17) to (12. ( ) 2 The average stress at the area of contact is F / πa . is 1 1 2 times the average stress. (12. the maximum shear stress occurs inside the sphere at approximately half the distance of the radius of the contact area.1 pmax (12. τ xz and τ yz At the central point of contact. i.5 3.0 Fig.0 0 449 s/Pmax sx = sy 0.23 Plot of σ max.24c) This being too small.σ x .3. 2:13 PM .6 0.31pmax. The lubricant.4 0. (12.24) is ( ) τ xz = τ yz = 1 (σ x − σ z ) = 1 − 1 + 2v + 1 pmax = 1 − 2v pmax 2 2 With 2 v= 0. The maximum shearing stress at this point.0 z/a 2. for v = 0. This point must be considered as the weakest point in such materials as steel.20). 12. which depend on shear stresses for yielding.5 1. (12.5 2. Chapter_12a. σ x = σ y = − pmax ⎡(1 + v ) − 1 ⎤ = − 1 + 2v pmax ⎣⎢ 2 ⎦⎥ (12. at x = y = z = 0.0 txz = tyz sz 1. which is under pressure. In fact. σ z = − pmax The maximum shear stress at the point from Eq.24b) 4 τ xz = τ yz = 0.19). (12.8 1. the values of sx = sy are from Eq.3 is about 0..pmd 449 7/6/2008. (12. enters the fine crack and wedges the chip loose.24a) 2 From Eq.e. σ y . It is suggested that cracks originate at this point below the surface and progresses to the surface.2 0. it dose not cause any yielding of materials such as steel.Introduction to Stress Concentration and Fracture Mechanics 0 0. 12. the bodies deform and the line of contact becomes a narrow rectangle of width 2b and length l. and E2. then d2 is negative.25) and (12.25). pmax = Chapter_12a.. Fig. (12. then the half-width b of the rectangle area of contact is given by b= ⎡ 2 ⎢ 2F 1 − v1 ⎢ πl ⎢ ⎣ ( ) ( E 1+ 1 − v22 1+ 1 d1 d2 ) ⎤ E2 ⎥ ⎥ ⎥ ⎦ (12. 2:13 PM .e. v2 are the respective elastic constants. pressing against each other.24. The pressure distribution within the area of contact is once again semi-elliptical as in the case of the two spheres.26) are general and are applicable to both cylinders.24 (a) Two cylinders in contact. i. A wheel pressing on a rail is a case where d2 = •. then d2 becomes infinite in Eq. compressive stress σ max which occurs along the middle line of the contact area is given by 2F (12. F F x x d1 l y y d2 z b F F z (b) (a) Fig. 12. and if E1. (b) Pressure distribution in the contact area If d1 and d2 are the diameters of the cylinders. there will be a line of contract l. If the cylinder is in contact with a hollow cylinder of diameter d2. v1.450 Advanced Mechanics of Solids (b) Two Cylinders in Contact Now consider two cylinders.pmd 450 7/6/2008.26) πbl Equations (12. Before the application of force F.25) The maximum pressure. If a cylinder of diameter d1 presses on a plate. After the application of the pressing force. 27) ⎥⎦ ⎤ ⎞ 1 1 + z2 b2 − 2 z ⎥ 2 2⎟ b 1 + z /b ⎠ ⎥⎦ 1 (12.128 ⎢ .pmd 451 1 − v12 1 − v2 2 .28) (12. 2 ⎡ ⎤ ⎡ EF d1 + d 2 ⎤ F (1 − v) d1d 2 ⎥ b = 1.5 sx 1.25 Plots of sx .564 ⎢ ⎥ l (1 − v 2 )d1d 2 ⎥ (12.30) ⎢⎣ El (d1 + d 2 ) ⎥⎦ ⎣⎢ ⎦ Figure 12.8 1. 0 0.6 tyz 2. then equations (12.0 sz 1. and let k1 = Chapter_12a.2 0. and k2 = E1 E2 (12. The unit of depth is b.5 3.29) 1 + z 2 / b2 If the elastic properties of the two cylinders are identical.Introduction to Stress Concentration and Fracture Mechanics 451 The state of stress along the z-axis for two cylinders is given by 2⎞ ⎡ ⎛ ⎤ σ x = −2vpmax ⎢ ⎜1 + z 2 ⎟ − z ⎥ ⎝ ⎠ b ⎢⎣ ⎡⎛ b σ y = − p m ax ⎢ ⎜ 2 − ⎣⎢ ⎝ σ z = − pmax (12. the half-width of the contract area. pmax = 0.0 s/Pmax sy 0.26) reduce to the following.3) Let the force applied per unit length of cylinders be F* = F/l.5 z/a 0. and sz as a function of depth form the centre of the contact area.0 Fig.31) 7/6/2008. 12. sy . and the unit of stress is pmax.0 2. sz and tyz (v = 0.25) and (12.4 0. sy.25 is a plot of sx. 2:13 PM . 3 pmax = 0.36) π 2 (k1 + k2 )r1r2 If the materials of both cylinders are the same and v = 0. Chapter_12a. b = 1. Instead of maximum shear stress theory for failure of materials. and its magnitude is 0. pmax = 0.34) For the case of contact of cylinder with a plane surface.35) Substituting for b from Eq.418 F * E (r1 + r2 ) r1r2 (12.32) where r1 and r2 are the radii of the cylinders. (12.52 F *r E (12.78b. sometimes the octahedral shear stress theory is used. b = 1.3.38) Based on the plot of tyz. then b = 1.301 pmax .pmd 452 7/6/2008. one gets pmax = F * (r1 + r2 ) (12. From Eq.33) into Eq. 2:13 PM .26). the maximum shearing stress occurs at a depth z = 0.418 F*E r (12.452 Advanced Mechanics of Solids Then b= ⎡ 4 F * (k1 + k2 ) r1r2 ⎤ ⎢ ⎥ r1 + r2 ⎣⎢ ⎦⎥ (12.33) In the case of two equal radii r1 = r2 = r.52 F * r1r2 E (r1 + r2 ) (12.08 F *r E (12.44a). (1.37) In case of contact of a cylinder with a plane surface. (12. If both cylinders have the same elastic constants and v = 0. τ oct = 2 ⎡ − 1 + 2v pmax + pmax ⎤ 3 ⎣⎢ 2 ⎦⎥ 2 (1 − 2v) p max 6 v = 0. substituting from equations (12. σ1 = σ x . arranging them algebraically. σ 2 = σ 3 = σ z ∴ 1/ 2 2 2 τ oct = 1 ⎡(σ x − σ z ) + (σ z − σ x ) ⎤ 3⎣ = ⎦ 2 (σ − σ ) = 2 (1 − 2v) p x z max 3 3 (12. = With (12. For carbon steel. stresses.19) and (12.1 Two carbon steel balls. and v = 0. σ x = −2vpmax . E = 207 GPa. σ1 = σ 2 = σ x = σ y .28). Since these are compressive.094 pmax (ii) Two Cylinders in Contact and (12.3.20). s2 and s3 are principal stresses. and σ 3 = σ z ∴ 1/ 2 2 2 τ oct = 1 ⎡(σ x − σ z ) + (σ z − σ x ) ⎤ 3⎣ = = ⎦ 1/ 2 1⎡ 2σ 2 + 2σ z 2 − 4σ xσ z ⎤⎦ 3⎣ x 2 σ −σ ( x z) 3 (12. the maximum shear stress.Introduction to Stress Concentration and Fracture Mechanics 453 1/ 2 2 τ oct = 1 ⎡(σ1 − σ 2 ) + (σ 2 − σ 3 ) 2 + (σ 3 − σ1 )2 ⎤ 3⎣ where s1. 2:13 PM . At the centre of the area of contact. (12. σ z = − pmax Arranging algebraically.39) At z = 0. determine the values of the principal stresses.292. each 25 mm in diameter are pressed together by a force F = 18N.40) τ oct = 0.pmd 453 7/6/2008. Chapter_12a.41) Example 12. σ y = − pmax . and sz are the principal stresses at the centre of the area of contact. At z = 0. and the octahedral shear stress. from equations (12.29). ⎦ (i) Two Spheres in Contact sx = sy .27). 091 mm From Eq. (12.17) ⎡ 2 (1 − v 2) / E ⎢ 2/d ⎢⎣ a= 3 3F 8 a= 3 2 3 F ⎡ (1 − v ) d ⎤ ⎢ ⎥ 8 ⎣⎢ E ⎥⎦ ⎤ ⎥ ⎥⎦ Substituting the given values.07 2 σ x = σ y = − 1 + 2v pmax 2 = . 3 × 18 × 1010 = 1045 MPa 2 × π × 9.18).24).828 MPa sz = .0. pmax = 3F 2 2π a = From Eq. −3 ⎤ ⎡ a = 3 ⎢ 3 × 18 × 0.7459 = 9. s1 = s2 = -828 MPa.pmax= . s3 = .454 Advanced Mechanics of Solids Solution From Eq.5 MPa 2 Octahedral shearing stress is 1/ 2 2 2 2 τ oct = 1 ⎡(σ1 − σ 2 ) + (σ 2 − σ 3 ) + (σ 3 − σ1 ) ⎤ 3⎣ ⎦ 1/ 2 2 2 = 1 ⎡(σ 2 − σ 3 ) + (σ 3 − σ 2 ) ⎤ ⎣ ⎦ 3 = Chapter_12a.915 × 259× 10 ⎥ 8 × 207 × 10 ⎣ ⎦ = 10−4 3 0. (12.1045 MPa Arranging algebraically.792 pmax = .1045 MPa Maximum shear stress is τ max = σ1 − σ 3 2 1 = (−828 + 1045) = 108.07 × 10−5 m = 0. (12.pmd 454 2 (σ 2 − σ 3 ) 3 7/6/2008. 2:13 PM . 17).1143 mm From Eq.43 From Eq.2 In Example 12.792 pmax = −521× 106 Pa σ z = − pmax = −658 ×106 Pa . the maximum shearing stress. Solution From Eq.1. and the octahedral shearing stress.5 MPa.18). Maximum shear stress is τ max = Chapter_12a. from Eq. a= 3 ⎡ 3 × 18 × 2 × 0. determine the principal stresses. For F = 18N. (12.584) × 104. (12. one of the steel balls is replaced by a flat carbon plate. Arranging algebraically.(12.1. 2:13 PM . = Example 12. (12. pmax = 3F 2 2 a 10 = 3 × 18 × 10 2 = 658 × 106 Pa .3 MPa 3 Also.492 = 11. 2 × × 11.915 × 25 × 10 −3 ⎤ ⎢ ⎥ 8 × 207 × 10 9 ⎣ ⎦ = 10 4 3 1.y) τ oct = 2 (1 − 2v) Pmax 6 2 (1 − 0. at the centre of the contact area. σ1 = σ 2 = −521 × 106 Pa. with d2 = •.24).5 × 107 6 = 102.43 ×10 − 5 m = 0.pmd 455 σ1 − σ 3 2 7/6/2008. σ 3 = −658 × 106 Pa. σ x = σ y = − 1 + 2v pmax 2 = − 0.Introduction to Stress Concentration and Fracture Mechanics 455 = 2 (− 828 + 1045) = 102. a= 3 3F 8 ⎡ 2(1 − v) 2 d1 ⎤ ⎢ ⎥ E ⎣⎢ ⎦⎥ Substituting the given values from Example 12. 3 In Example 12. 2 From Eq.456 Advanced Mechanics of Solids 1 = (−521 + 658)106 = 68.y). At what distance from the contact surface do they occur? Solution The maximum shear stress and octahedral shear stress occur approxi- mately at half the radius of the contact area. (12.4 ⎣⎢ 2 ⎦⎥ = − 0.292) − ⎬ ⎢ ⎥⎦ ⎪⎣ 2 2× 5⎪ 4⎭ ⎩ { } = − pmax ⎡1 − 1 × 1. 2:13 PM . i.292) − 0. At this 2 point.8 pmax = −526 MPa .2.5 MPa .19) and (12. = Example 12.107 ⎤ (1. the octahedral shear stress is τ oct = 2 (1 − 2v) pmax 6 2 × 0.e.177 pmax = −116 MPa and ⎡ ⎤ 1 ⎥ ⎢ σ z = − pmax ⎢ ⎥ 1+ 1 ⎣⎢ 4 ⎦⎥ = −0. Hence. the maximum shear stress is τ max = 1 (σ x − σ z ) 2 = 1 (−116 + 526) = 205 MPa .pmd 456 2 (−116 + 526) = 193 MPa .416 × 658 × 106 6 = 64. determine the maximum shear stress and the maximum octahedral shear stress.5 MPa. 3 7/6/2008. from equations (12.20) ⎧ ⎫ ⎪⎡ 1 1 ⎪ −1 ⎤ σ x = σ y = − pmax ⎨ 1 − tan (2) (1. at z = 1 a = 5.7 × 10−5 m.. 2 The Octahedral shear stress is τ oct = 2 (σ x − σ z ) 3 = Chapter_12a. Chapter_12a. Most of the factors used in design calculations are based on the results of experimental investigations. and renders the stress concentration factor useless. the energy required to produce a spontaneous crack extension thus creating a new fracture surface. In order to use these factors in practice. the designer has to know precisely the geometrical parameter present in his structural or machine member. notch depth. When there exists a crack. Consequently. the elastic stress concentration factor approaches infinity as the root radius approaches zero. This knowledge will help in calculating the average stress necessary to initiate a crack. or an inclusion.7 BRITTLE FRACTURE It is generally known that materials always show a strength that is much smaller than what might be expected from the analysis of molecular forces. or a flaw. It was shown theoretically. the theoretical strength in tension based on the analysis of molecular forces is about 11 GPa. in the case of ductile materials. etc. fillet radius. that the stress s required to extend the crack spontaneously is inversely proportional to the square root of the length of the crack. based on the strain energy principle.Introduction to Stress Concentration and Fracture Mechanics II 457 FRACTURE MECHANICS 12. This discrepancy between theory and experiments was attributed to the fact that glass in its natural state contains a large number microscopic crackproducing regions of high stress concentration. a designer is interested in two factors associated with the problem of a crack in a specimen. (a) The state of stress field in the close vicinity of the crack tip (b) If a crack already exists. For example. a new approach is required while dealing with cracks in structural or machine members. which may not be easy. Let a uniform ten2a sion s be applied at the two ends of the plate. Further. In this context. Figure 12.Considering the crack as a microscopic alliptical hole.26 shows a glass plate with a narrow crack of length 2a. The specimens needed for experimental investigations have to be prepared very carefully since they involve factors like root radius. 12. Hence. Experimental investigations made on glass sheets in which s cracks of known length were made with a glass Fig.26 Glass plate with a cutter’s diamond showed a very satisfactory crack of length 2a agreement. the theoretical strength would be much higher than the experimental results. 2:13 PM . But tensile tests conducted on glass rods reveal s a strength of only 180 MPa. for a glass specimen. Previous discussions on stress concentrations revealed that very few problems involving regular geometrical irregularities could be solved theoretically to determine stress concentration factors. zones of high stresses make the material yield with the stresses getting redistributed.pmd 457 7/6/2008. τ xy = Chapter_12a. and c) show that the elastic normal and elastic shear stresses in the vicinity of the crack tip depend on the radial distance r from the tip. The fracture modes will be discussed separately. The plate is subjected to a uniform tensile stress s applied at the ends. and are with reference to Fig. the configuration of the stressed body (i.42a) σy = K 2π r ⎡ 3 ⎤ ⎢⎣ cos 2 1 + sin 2 sin 2 ⎥⎦ (12. 12.458 Advanced Mechanics of Solids 12. the location of crack in the plate. y s h r q x 2a 2b Fig. The crack of length 2a is a through crack in the plate of thickness t.8 STRESS INTENSITY FACTOR Consider a plate of uniform thickness having a centrally located crack. and the mode of crack opening. ratio of crack length to the width of the plate. However.42a. This means that the state of stress at a given point in the vicinity depends completely on the factor K called the stress intensity factor. the orientation q of the point of interest.27.e..pmd 458 7/6/2008. These are expressed by the following equations.27 h s Plate with a through crack of length 2a ( ( ) ) σx = K 2π r ⎡ 3 ⎤ ⎢⎣ cos 2 1 − sin 2 sin 2 ⎥⎦ (12.). 2:13 PM . b.42c) 2 2 ⎦⎥ 2π r ⎣⎢ 2 Equations (12. 12. etc. and the factor K. this factor K depends on the nature of loading. The stress field in the vicinity of the crack tip has been obtained theoretically.42b) K ⎡sin cos cos 3 ⎤ (12. 28(a) and (b) show graphically the values of K1/Ko .0 1.2 0. and a/b = 0. 12.43) where K is in (N/mm2) mm or MPa m .8 s 0 0. Chapter_12a.44) Figures 12.6 a ratio b (a) 7. Values of K have been determined for a variety of situations employing both theory of elasticity approach and numerical techniques. 2:13 PM . (b) Edge crack of length a. and the geometry of the specimen.4 0. (12. in an infinite plate subjected to a uniform tensile stress s as shown in Fig.pmd 459 7/6/2008.0 KI/K O b h 5. the stress intensity factor K is given by K0 = σ π a (12. (a) Central crack of length 2a. 12.Introduction to Stress Concentration and Fracture Mechanics 459 For a central crack of length 2a. and the meaning of this will be discussed subsequently. For example.43) is modified as K I = ασ π a (12.0 KI/K 0 h = 0. the value of K depends on the type of loading. The subscript I in K indicates that it is mode I fracture.5 b 3.27.0 s 6.6 a ratio b Fig. s 4.0 0 0.4 b h h 2a 2.4 0.0 s 2. for several values of h/b and a/b.5.28 Values of KI/Ko as a function of a/b for different bh values.32σ π a In order to take care of this dependence of Ko on the type of loading and the geometry.0 1.7 b h = l b 0. If h/b = I.0 h = 0. the magnitude of Ko gets modified and becomes K I = 1. where Ko is taken as the base unit. As mentioned earlier.2 (b) 0.0 2b 3.0 4. Eq.0 h a h = 0. Although the fracture toughness KIc is a basic material property in the same sense as yield strength. These tests are usually conducted on single edge-notch specimens subjected to mode I. 12. For a given flaw size. Through carefully controlled testing of the specimen of a given material. This critical crack length gives the critical value of Kc by the equation. Figure 12. Chapter_12a. (12.e.460 12. also called fracture toughness.11). If KIc is known.pmd 460 7/6/2008.45) Allowable stress s KIc is a basic material parameter called fracture toughness.29 illustrates graphically the relationship between crack length a and maximum allowable stress s to prevent sudden extension of crack for two materials.. 2:13 PM . and under plane strain conditions. and for a given operating stress the maximum allowable crack size is proportional to the square of KIC.29 s1 High KIc s2 Low KIc a1 Crack length a Crack length and allowable stress for high and low KIc materials As the figure illustrates. K Ic = ασ π ac (12. it varies as a function of strain rate and temperature. This aspect deals with the strength of the material.This is designated by the symbol Kc . the maximum allowable stress s1 is higher for a material with high KIc than the allowable stress s2 for a material with low KIc. i. one with high KIc and the other with lower KIc. Fig. Therefore. the allowable stress is directly proportional to KIc. This dependence on the strain rate and temperature decreases as the temperature decreases. for a given crack length a1. Closely associated with this aspect is the inherent behavioural property of the material of the specimen.45). the maximum allowable stress to prevent brittle fracture for a given flaw size. This is characterized by the critical stress intensity factor. the opening mode (discussed in section 12.9 Advanced Mechanics of Solids FRACTURE TOUGHNESS The previous discussion dwelt on the stresses induced in a specimen with a central crack subjected to external loading. then it is possible to compute from Eq. increasing the value of KIc has a much larger influence on allowable crack size than on allowable stress. for a known applied stress. the critical crack length ac which suddenly propagates is noted. Eq. It was suspected that an existing crack.44) can be used with a = 1. The sheet has a 60 mm long central transverse crack . Eq. 2:13 PM .8 MPa This is the stress value at which catastrophic failure will occur. −3 Hence.8 ¥ 106 Pa = 92.005. (12.Introduction to Stress Concentration and Fracture Mechanics 461 Example 12. Solution Assuming that the plate was long and the length of the crack was small compared to the width. 12-m wide. one gets σ= K Ic πa 28. The ratio of this stress value to yield strength is σ yp 240 = = 2.44) can be used with a = 1. If the yield strength for the material is 240 MPa. and 20-m long . Tests have revealed that under the conditions. b 6 h 10 Further. the value σ = KI / π a Since fracture occurs when K1 = K1C . b = 6 =1. For this case. 2a = 60 mm. and centally located. how does the failure stress compare with it? Solution Making reference to Fig. 2b = 12 m. 12. the material has a fracture toughness factor KIc= 28. the crack may be considered to be present in a very long plate. Hence. Example 12. or σ = 0. σ = KIc πa with s = 90 MPa and KIc = 20 MPa Chapter_12a. Since ab is very small.386 σ 92. Calculate the tensile stress for catastrophic failure. presumably in the middle went undetected.8 σyp Thus.67 .5 MPa m . The ambient temperature was cold. The steel sheet is subjected to a tensile stress in the direction of its length. Determine the maximum length of the crack that could have escaped the crack-detector’s attention.5 × 106 = ⎡π × 30 × 10−3 ⎤ ⎣ ⎦ = 92.59. the ratio of crack length to width of the plate is a = 30 × 10 = 0. and 2h = 20 m .27. 7/6/2008.5 A 10-m wide plate used in a heavy-machine-shop construction operation had a catastrophic failure during assembly when the sheet was subject to a stress of 90 MPa. The operating temperature is below its ductile-to-brittle transition temperature.4 An off-shore drilling platform has a steel sheet 35-mm thick . catastrophic failure will occur at 0. (12. The critical stress intensity factor for the material was 20 MPa m . This gives for s.pmd 461 m .386 syp. This is the point at which the curve begins to deviate from a straight line. This is shown in Fig.5 per cent of the original gauge length. Figure 12. and ultimate stress. Hooke’s law. Point P in this figure is called the proportional limit. 2:13 PM .5 per cent are used. yield strength syp is often defined by an offset method. 12.. although 0.30.30 Tensile test diagram for a ductile material Chapter_12a. These qualifications or distinctiveness are based on the stress–strain curves obtained usually during a tensile test. It is generally known that there are materials like copper.7 mentioned brittle fracture without explaining what we mean by brittle fracture. This is usually 0.222 or 90 a = 0. maximum stress. In this method.2 or 0. 0.01. This point is called the yield point. The other points U and F correspond to ultimate strength and the fracture or breaking stress.30 is a typical curve obtained from standard tensile test of a ductile material.462 Advanced Mechanics of Solids 90 × 106 = 20 × 106 πa π a = 20 = 0. and \ Length of the centrally located crack = 2a = 142 mm.1 and 0. 12. the specimen will regain its original length without any permanent set. if the load is gradually removed.071m or 71 mm. 12. the yield strength correspond to a definite amount of permanent set. which states that stress is proportional to strain. U F Y E P Stresss o A ey eu eF Strain e Fig.pmd 462 7/6/2008. Many materials reach a point at which the strain begins to increase very rapidly without a corresponding increase in stress. At this point of stressing. Point E is called the elastic limit. Not all materials have an obvious yield point. mild-steel etc.10 FRACTURE CONDITIONS Section 12. For this reason. which clearly have a defined yield points stress. applies only up to the proportional limit P. 5 ⎜ Ic ⎟ ⎝ σ yp ⎠ 2 (12. This term means fracture without yielding occurring throughout the fractured cross section. The test requirements for measuring KIc call for plane strain values..Introduction to Stress Concentration and Fracture Mechanics 463 Materials which exhibit definite yield zones are called ductile materials. These materials fail catastrophically after reaching a finite stress state. Thin specimens do not exhibit flat fractured surfaces. are example. 12.46) Fracture stress where sypis 0. or ductile-to-brittle temperature.31 exhibits this phenomenon. It was also stated that the values of stress intensity factors are valid under plane strain conditions. increasing the thickness of a part leads to a decrease in KIc . Cold temperature is definitely an influencing facture for brittle fracture. conditions in which no yielding occurs. They reveal ductile-brittle mode (mixed mode) of failures.31 Effect of thickness on fracture stress In general. But Equations. The fracture mechanic concept is correct only for linear elastic materials i. As Fig. the stresses become very high and yielding occurs. 6:21 PM . operations below room temperature is an indicator of possible brittle fracture. Before fracture occurs. The minimum thickness to obtain plane strain conditions and valid K Ic measurement is ⎛K ⎞ t = 2. the yield strength is temperature dependent. Chapter_12b. Tables of transition temperatures for various materials are not available. There exists a temperature for a given material. However.1). and c ) show that as r approaches zero near the crack tip. The term relatively brittle is used in test procedures.42 a. This means that the thickness of the specimen is critical. wherein below that temperature the material suddenly exhibits a brittle nature without any yield characteristic.e.2 per cent offset yield strength (see Sec.12. and therefore the published values invariably refer to plane strain values. if the yield zone is very small compared to the crack width (generally of the order of 0. cast iron. This minimum value is called the plane strain critical stressintensity factor. Glass. and fracture stress is a function of the thickness of the specimen.31 shows. Materials which do not have yield points are called brittle materials. Mixed mode Plane strain fracture Thickness t Fig. Figure 12.e. As the thickness increases. the value of KIc becomes asymptotic to a minimum value with increasing thickness. the elastic solutions i. b. This temperature is called the transition temperature. So. In the case of ductile materials.. 12. the value of fracture stress becomes constant. they exhibit strong yield characteristics. possibly because of great variations on their values even for a single material.14). ( 12. Liner Elastic Fracture Mechanics ( LEFM ) solutions obtained for stress intensity factors cabe used.pmd 463 7/6/2008. Generally. and are designated as mode II. 12. 6:21 PM . and for n. Figure 12. the first mode. and are called mode III. This was the reason for putting the subscript Ic to the critical stress intensity factor K.32(a). For its locations. Mode III is called the tearing mode. and steel-n has yield strength of 1500 MPa.11 FRACTURE MODES In our discussion so far. A factor of safety of 1.6 A plate of 1.pmd 464 7/6/2008. KIc =100 MPa m .32(a) is commonly called the opening mode and is designated by I. the designer has to assume the worst conceivable locations. Solution (a) We shall first determine the thickness of each sheet based on the yield strengths of the materials. because considerable amount of analysis and experimental investigations have been done on this mode.464 Advanced Mechanics of Solids When a crack is visible and its length can be measured. Chapter_12b.32 Fracture modes (a) Opening mode. this data can be used along with its location in the member.32(c). Steel-m has yield strength of 850 MPa.7 mm. In Fig. In our discussion here. Experimental methods to detect through thickness edge cracks are valid only for cracks longer than 2. the attention has been mainly on mode I. The corresponding critical stress intensity factors for the two materials are: for m. since more than one location for the crack may be critical. (c) Tearing/Shearing mode Example 12. The expected load in the longitudinal direction is 4 MN. These are shown in Figures 12.5-m width and 3-m length is required for construction operations. 12. 12. (b) and (c). It has an edge crack and the forces attempt to extend the crack. Which of the two materials should be selected? Inspection did not reveal any apparent cracks in the two sheets. 12. In Fig. (b) Shearing mode .32(b) and (c) are called the shearing modes. (a) (b) (c) Fig. Two steel plates m and n are being considered for this purpose . three ways of separating a plate are considered in facture mechanics. KIc = 60 MPa m . the displacements are out of plane. the designer has to assume for 2a the longest length that goes undetected by any of the crack detection techniques. Figures 12. When a crack is not visible. attention has been focussed on opening mode or the first mode. Minimum weight is important.32(b) the displacements stay within the plane of the plate.5 is to be used. the thickness required is therefore With a factor of safety = 1.7 mm. which is t = 4.5 × 4 = 4.76mm 1. (12.1 π a 60 1.5 For these values.7mm.5 MPa.44).1 σ π a σ = 100 1.8 × 10−3 = 0.67 × 10−3 m or 2. 100 = 1. the curve in Fig.5.75 × 10−3 m = 6.5 × 850 1. t= Steel-n : a 2.5 × 4 = 2. i.27.7 = 1.28 gives a value for K1/Ko as 1. we shall assume a crack in each sheet whose maximum length goes undetected.5 × 1. = b 1. Steel-n: With K Ic = 60 MPa m.5 × 103 b 1. 60 = 1. This value is lower than sy /1.1 π × 2.7 × 10−3 = 987 MPa This crictical stress for crack extension is greater than the yield stress for the material. for both materials.5 × 1500 (b) We shall next determine the thickness based on the critical stress that each sheet can bear without crack growth.67 mm 1.1 π a or = 100 1.5 × t = F 1.44). Based on Fig.1.5F 1. Steel-m : With KIc=100 MPa m .7 mm 1. and Eq.. the allowable critical stress for steel-n is t= Chapter_12b. and Eq.e.5 × σ yp t= or 465 = 1.Introduction to Stress Concentration and Fracture Mechanics σ yp Steel-m : 1. (12.7 × 10−3 = 592 MPa 592 = 395 1. 12. To carry a load of 4 MN.pmd 465 4 = 6.0018 h 3/ 2 = = 1. Hence the thickness based on syp prevails. 2. Since inspection did not reveal any apparent cracks.1 σ π a σ= or = 60 1.5 = 1000 MPa.1 π × 2. 12.7 × 10−3 m or 4. 6:21 PM .5 × 395 7/6/2008. the m-steel with a higher KIc and lower yield strength requires a thickness of only 4.5 × 106 ⎞ 35 × 10 −3 = 2. whereas. Example 12. (12. So.46)? Solution t = 35mm.76 mm.5 ⎜ ⎟⎠ σ0 ⎝ 2 0.7 In Example 12. (a) gives for w2 the value b ω1 = c a Fig. Eq. Substituting.pmd 2 466 a 7/6/2008. the thickness of the steel sheet was 35mm and the value of KIc was given as 28.68).4 dealing with off-shore platform. (8. Solution From Eq.466 Advanced Mechanics of Solids Hence. This stress reaches its maximum value at the inner radius c. Determine the critical speed for the disk based on (i) the yield stress syp .70).8 A rotating disk with a bore radius c and an outer radius b has a small radial crack of length a at the bore.5 MPa m . the n-steel with a lower a KIc and a higher yield strength requires a thickness of 6. and is equal to () ⎡ ⎤ σ max = 3 + v ρω 2 b 2 ⎢1 + 1 − v c ⎥ 4 3+ v b Nc (a) ⎣ ⎦ (i) With smax= syp. for a disk rotating with an angular velocity of w rad/sec. 12. KIc = 28. the m-steel is recommended for the task.014 σ 0 = 28. (8. (12.7 mm. Example 12. ⎛ 28.5 × 106 or or σ 0 = 241 MPa This agrees well with the value given in the example. What should be the 0. Eq.2 per cent yield strength to ensure plane-strain condition according to Eq.5 MPa m .33 Rotating disk with a radial crack at the bore 2 σ yp ( 3 + v) ρ ⎡⎢⎣b2 + 13 −+ vv c 2 ⎤⎥⎦ or ω1 = 2 σ yp ρ ⎡⎣( 3 + v ) b 2 + (1 − v ) c 2 ⎤⎦ (ii) From Eq.44) K Ic = ασ Chapter_12b. the circumferential stress sq at a radial distance r from the centre is σθ = 2 2 ⎡ ⎤ 3+ v ρω 2 ⎢b 2 + c 2 + b c2 − 1 + 3v r 2 ⎥ + 8 3 v r ⎣ ⎦ where r is the mass density of the material. 6:21 PM . and (ii) the critical stress intensity factor KIc. (ii) The design is to be based on the critical stress intensity factor KIc.8 what is the ratio of the critical speed N1 based on the yield stress to the critical speed N2 Based on the critical stress intensity factor? ω N N1 = 30 ω1 . one can get the corresponding critical 60 speeds N1 and N2. N1 = N2 ( × 2. The inner radius is fixed and the outer radius is to be determined.9 In Example 12. a = 2.83 = 1. A factor of safety n is involved. where N is the rpm. and KIc= 50 MPa m . K Ic α a = () 2⎤ ⎡ 3+ v ρω 2 2 b 2 ⎢1 + 1 − v c ⎥ 4 ⎣ 3+ v b ⎦ 1 = ρω 2 2 ⎡⎣(3 + v ) b 2 + (1 − v ) c 2 ⎤⎦ 4 or ω2 = 2 K Ic (π a ) 1 4 αρ ⎡⎣( 3 + v ) b 2 + (1 − v ) c 2 ⎤⎦ (c) 2π N .12. Chapter_12b. there is the danger of catastrophic failure when the speed reaches 0.54 mm.Introduction to Stress Concentration and Fracture Mechanics σ= or K Ic α 467 (c) a Substituting this into Eq.57N1. 6:21 PM . (a). Putting ω = Example 12.75 This example shows that if N1 is the rpm decided by the yield strength criterion. Example 12.pmd 467 7/6/2008. 1 ασ y N1 = (π a ) 4 N2 K Ic If a = 1. and N 2 = 30 ω 2 giving 1 = 1 π π N2 ω2 Solution From Equations (b) and (c). The cylinder has a small radial crack of length a at the bore.12 × 1515 50 (b) = 0.10 A cylinder subjected to internal pressure p has an inner radius c and an outer radius b. sy = 1515 MPa.54 × 10 −3 ) 1 4 1. (i) The value of the outer radius b is to be determined according to the maximum shear stress theory ignoring the crack.3 × 5. 35(a). it is critical to consider sq. Fig. In a pipe of uniform cross-section. σ θ = p b 2 + c 2 . b2 = nK Ic + pα π a nK Ic − pα π a c2 12. τ max = 1 (σ1 − σ 3 ) = p 2 1 σ . 12. K Ic = ασ π a σ= or Equating this to K Ic σ πa 1 σ. in the (case of a body of uniform section with no discontinuities. n θ K Ic 1 b2 + c2 p = n b2 − c 2 α π a or. (12. and equations (8. the load lines will) be uniform and all parallel. 6:21 PM .13) and (8. a point at the inner radius is considered. with and without geometrical discontinuities. 12. These are similar to stream lines in fluid flow. Similarly. If there is Chapter_12b.1. for an edge crack of length c. 2n yp Equating this to p c p a b Fig.12 PLANE STRESS AND PLANE STRAIN While discussing stress concentration. At this point. based on Example 8. when the body is loaded longitudinally. If some sort of obstruction to the flow exists then the streamlines get crowded near the obstruction and the velocity of flow near the obstruction will no longer be uniform. σ z = p b −c c2 b − c2 2 The maximum and minimum pressures are σ1 = σ and σ 3 = σ r .34 Cylinder with an internal crack under pressure or b2 b − c2 2 b2 = 1 σ yp b − c 2 2n 2 b2 = σ yp σ y − 2np c2 (ii) To prevent crack extension.468 Advanced Mechanics of Solids Solution (i) To apply the maximum shear stress theory. the steady flow of a fluid can be represented by streamlines which are all parallel to the flow direction. 2 2 σ r = − p. From Eq. it is helpful to consider load-path or load-flow lines in a body such as a wide plate.pmd 468 7/6/2008.14).44). Hence. Introduction to Stress Concentration and Fracture Mechanics 469 a geometrical discontinuity. As seen in Fig. 12. At the root radius of the notch sx = 0. (b) Bar with a notch.e.35 (a) Load lines in a uniform bar. i. The load lines also indicate the direction of the load or the stress. sx increase. Fig. Figure12.35(b). the situation is a plane stress case.46) 7/6/2008. This means that though the member is subjected to uniaxial y A A x (a) (c) (b) Fig. However. 6:21 PM . the load lines bend near the notch and the tangents to the lines give the directions of the resultant stresses. 12. 12.35(b). 12.36 shows qualitatively the bi-axial nature of the stress distribution near the notch section of a uniaxially loaded member.36 The bi-axial state of stress near the notch The two faces of the plate are stress-free. (c) Load lines indicating bi-axial state of stress loading. The tangent at A to one of the lines has two components. the load lines are all straight indicating the uniaxial state of stress. in Fig. reaches a maximum and at a far distance from the notch tip becomes zero. At the notch tip. the load lines get crowded near the notch tip and the stresses near that region will no longer be uniform. the state of stress is bi-axial. one in x-direction and another in y-direction. However. 12. such as a notch. y sy sx sx = 0 x Fig. 12. In Fig. Hence. sy is maximum and becomes uniform at the far end of x..pmd 469 σx E −v σy E =− v (σ + σ y ) E x (12.35(c). but ez is not equal to zero. sz = 0. as x increase. ε z = −v Chapter_12b. Very close to the notch tip. since the surface of the notch is stress-free.35(a). at point A. 12. as shown in Fig. (b) Stresses preventing contraction Chapter_12b. it builds up rapidly inside..47). Consequently. A sufficiently thick plate with a crack will therefore be in a state plane strain. Inside the plate near the crack tip. the two faces of the plate are stress-free and the value of σ z is zero at the two faces.38 (a) Cylindrical material surrounding the plane stress crack tip. free contraction.37 = Contraction near the notch tip s s s s crack plane s s B (thickness) s (a) s (b) Fig. A small cylindrical material surrounding the crack tip will therefore experience σ z in the z direction. (12.pmd 470 7/6/2008. there will be a triaxial state of stress taking into account sx.39. The stress σ z in the thickness direction that is required to completely prevent ezwill be σz E ( v σ +σy E x ( σz = v σx +σy i. ) ) (12. Hence. in the z direction) to contract.39 Thin plate. the material near the crack tip is heavily constrained in the thickness direction (i. Since the faces of the plate are stressfree. there is not enough material surrounding the notch tip to constrain or prevent contraction in the zdirection. this is a case of plane stress. However. Tries to contract Fig. this is a as localized effect in a wide plate. ez is not zero. 12.e. 12.470 Advanced Mechanics of Solids and contraction in the plate thickness occurs. Fig 12. But.38. Fig.37. 12.36 and sz as per Eq. sy from Fig. In the case of a thin plate. 12. So.e. 6:21 PM . but sz is zero. as shown in Fig.47) As mentioned. In the case of a thick specimen. 12. σ z will be zero at these faces. a small dimple appears near the crack tip in the two faces. 34σ y = 2σ yp . In Chapter 4. 7:27 PM . yielding occurs according to the maximum shear stress theory when 1 1 (σ − 2vσ y ) = σ yp 2 y 2 0. near the crack tip s1 = s2 = sy . several theories of yielding were discussed. The maximum shear is Chapter_12b.33).34σ y = σ yp.txy = 0 according to equations (12. In the case of a thick plate.42) and sz = v (sx + sy) according to Eq. or σ y = 3σ yp . As 2 1 2 471 7/6/2008. As load increases. sz = s3 = 0 and therefore it is a plane stress case. According to octahedral shearing stress theory (also called the distortion energy theory).Introduction to Stress Concentration and Fracture Mechanics 471 12.13 PLASTIC COLLAPSE AT A NOTCH The presence of a high state of stress near the notch tip suggests the occurrence of plastic yielding near the tip. In the case of a thin plate. Plastic yielding occurs when sy = syp. At low loads. According to the octahedral shearing theory yielding occurs when (with v = 0.49) τ oct = ⎡( σ1 − σ 2 ) + ( σ 2 − σ 3 ) + ( σ 3 − σ1 ) ⎤ 2 = σ ⎦ 3⎣ 3 yp where s1.33. (12. yielding occurs when sy = 3syp. ( ) 2 1/ 2 + (2vσ y − σ y ) 2 ⎤ ⎥⎦ = 2 σ 3 yp ) 2 0. = s3 = sz = v (sx + sy) = 2vsy (12. (12. s2. the local stress is less than 3syp and hence the material remains elastic. ( 1⎡ σ − 2vσ y 3 ⎢⎣ y i. and syp the yield point stress for the material.e.50) Assuming v = 0.51) This means that in the case of plane strain (thick plate). the maximum shear stress theory and the octahedral shear stress theory are applicable to a large number of materials.. the yield point stress for the material. Thus. i. sy becomes equal 3syp and yielding occurs. near the crack tip. Among these.pmd 1 1 (σ − σ 3 ) = σ y . Figure 12. sx = sy . or σ y = 3σ yp i.e.b) These are the principal stresses also. and s3 are the principal stresses arranged algebraically.e.. since txy = 0 in the plane of symmetry. yielding will occur at a point when 1 2 2 2 1 2 (12. This is the same as the maximum shearing stress theory. According to the maximum shear stress theory yielding will occur at a point when 1 (σ − σ 3 ) = 12 σ yp (12. along the x-axis. and syp..40 (a) represents the situation.48) 2 1 where σ 1 and s3 are respectively the maximum and the minimum principal stresses at the point. 51) becomes applicable. the nominal stress keeps decreasing linearly with increasing crack length.52) where B is the thickness of the plate. the maximum load carrying capacity (for a plate with single edge notch) is Pmax = B(W − a )σ yp (12. 12. Such a situation is called plastic collapse.40(a) and (b) can be reached before fracture. Thus. In plane stress cases they are limited to syp. The stress peak is local and the average stress across the section cannot become much higher than in the case of plane stress. The maximum stress that the cross section across the notch can carry will be limited to the yield point stress. the plastic zone keeps spreading until the entire remaining section yields unless fracture occurs earlier. = Chapter_12b.472 Advanced Mechanics of Solids loading increases. and a is the crack length. Figure 12. under the limit load is σ nom = Pmax WB (W − a ) σ yp (12.pmd 472 7/6/2008. In plane strain case. From the foregoing discussions it becomes clear that if the stress distributions as shown in figures 12. (12.5(b). The nominal stress in the section where there is no crack. and Eq.53) W As can be seen. 12. W is the width. Consider the case where there is no strain hardening. then plastic collapse can occur.40 (a) Yielding at notch tip under plane strain (b) Yielding at notch tip under plane stress It is important to observe that in plane strain cases the notch tip stresses are much higher than syp when yielding occurs. 6:21 PM . in plane stress where the stress in the entire cross section is equal to yield strength at the time of collapse. the stresses in most of the ligament are still close to syp. when tearing or plastic collapse commences at the notch tip. the yield area across the section keeps enlarging until the entire area cannot have stress greater than syp. or in general non-plane stress case. In the case of a work-hardening material. Thus. Fig. the stress distribution after the onset of yielding is not uniform. This failure load is called the collapse load or the limit load. plane strain condition is more severe and can more easily lead to fracture and cracks. As the load increases. sy sy 3syp Increasing applied load 2syp syp Elastic-plastic Elastic-plastic syp Elastic Elastic x (a) (b) x Fig.40(b) depicts the situation. 3 × 106 ) = 874875 N  875 kN Chapter_12b. The width of plate is 30 cm. Thus. Kt = 1. and thickness is 1. K 't = 1 + 2 ( )σ 5 = 11 .25 × 10−2 )(233. Here.12 For Exercise 12. Solution σ nom = (W − 2a ) σ yp W (30 − 10) = × 350 = 233. Calculate the collapse load. 6:21 PM . but will be less than the ultimate stress sult. Hence. and hence * ε * = σ . a = 5 cm and b = 1 cm.Introduction to Stress Concentration and Fracture Mechanics 473 because the strain gradient is very steep. calculate the nominal stress in the full section at the time of collapse if the yield stress is 350 MPa. and semi-minor axis of 1cm.pmd 473 7/6/2008. for a work-hardening material Eq. the entire section across the notch is experiencing uniform stress and there is no stress concentration. In general. and ε nom = E σ nom E ( Kt )'ε = ( Kt )'σ = 11 When the notched section has fully yielded with no work hardening. (12.25 cm. Hence. What is the strain concentration factor in the elastic case? What is the stress concentration factor after the notched section has fully yielded in plane stress assuming no work hardening? Solution The theoretical stress concentration factor Kt' for an elliptical notch is ( K 't = σ * = 1 + 2 a σ nom b ) where a and b are respectively the semi-major and semi-minor axes of the ellipse.11 Calculate the theoretical stress concentration factor of an elliptical notch with semi-major axis of 5cm perpendicular to the applied load.54) It should be noted that the foregoing discussion is strictly for uniform applied loading. = (W − a ) σ nom col 1 W When the member is still in an elstic state.11.53) changes into (12. This average stress across the ligament is called the collapse strength σ col . stress is proportional to strain. Example 12. \ Example 12. the average stress can be higher than syp.3 MPa 30 Collapse load = Wtσ nom = (30 × 10−2 )(1. to bring down the stress concentration factor from • to a finite value of 5.5 cm long. 6:21 PM . if the stress concentration factor is 5. This can be recast in terms of the radius of curvature r of the ellipse at the end of the major axis. the theoretical stress concentration factor is given by a b where a and b are the semi-major and semi-minor axes of the elliptical hole. before the drilling of the stop hole.13 As mentioned earlier in this chapter. the diameter of the stop hole should be Chapter_12b. 7/6/2008. Assume that the crack with the stop hole is an ellipse. if a crack appears in practice one sometimes drills a stop hole at the crack tip as a temporary repair. Calculate the theoretical stress concentration factor before and after drilling the stop hole. The crack tip radius is almost zero. and its length is a.pmd 474 a 2 . The radius of curvature is given by K't = 1 + 2 ρ=b 2 a Using this and substituting for b ⎛ 1 ⎞ K ' t = 1 + 2a ⎜ ⎟ ⎝ ap ⎠ a K' t = 1 + 2 or ρ In the case of a circle. If the crack is 2. determine the diameter of the hole to be drilled to give a theoretical stress concentration factor of 5. Solution For an elliptical hole in an infinite plate. A hole of diameter d is drilled with its centere coinciding with the crack tip.e. If d is the diameter of the hole drilled 2a d For the present example.. d= i.474 Advanced Mechanics of Solids Example 12. Suppose a crack has started at the edge of a strip. then K't = 1 + 2 5 = 1+ 2 or 2a d 2a =4 d a 2 Hence. K't = 1 + 2 a =∞ 0 After drilling the hole. r = R the radius of the circle. 5 ⎜ Ic ⎟ ⎝ σ yp ⎠ 2 ⎛K ⎞ B ≥ 2. To ensure that cracking occurs within a certain envelope and to reduce scatter. starter notches are generally used.0 ¥ 102 54 98 76 m) (continued) Chapter_12b. (12. The standard recommendations according to ASTM are: ⎛K ⎞ a ≥ 2. To ensure plane strain conditions. or use of large thickness specimens is recommended.55).pmd 475 7/7/2008. Among the several standard specimens recommended.6 ¥ 102 14.14 EXPERIMENTAL DETERMINATION OF KIc The American Society for Testing and Materials (ASTM) has set standard test methods to determine the values of plane strain fracture toughness of metallic materials.55) 2 yp Since the value of KIc is not known prior to testing.1W (min) 4W Fig.5 ⎜ Ic ⎟ ⎝ σ yp ⎠ 2 ⎛K ⎞ W ≥ 5.1 Modulus (MPa) Material Steels Medium carbon High strength alloys Maraging steel 2. Table 12.41 ASTM specimen for three-point bend test Table 12. the value of KIc obtained is used to validate the dimensions of the specimen according to Eq. 12. 8:27 PM . the specimen dimensions must be large enough.41. one of them. 12.1 ¥ 105 Yield stress syp (MPa) Toughess KIc (MPa 2.1 gives the representative values of plane strain fracture toughness for selected engineering alloys.6 ¥ 102 18. At the end tests. namely the three-point bending specimen is shown in Fig.Introduction to Stress Concentration and Fracture Mechanics 475 12.1W (min) W a B =W 2 2.0 ⎜ Ic ⎟ ⎝σ ⎠ (12. some estimate based on other experiments is used. Fatigue crack Load Machined notch 2. The specimens are then fatigure pre-cracked prior to testing to simulate an ideal plane crack with essentially zero tip radius to agree with the assumptions of LEFM. 6:21 PM .0 ¥ 105 10. i.57) M Chapter_12b.42 Single edge-crack extention As the load P on the body is gradually increased.56) 2 where d is the displacement of the point of application. displacement of the point of application occurs. Consider a body with a crack of length a and subjected to a load P as shown in Fig.0 ¥ 102 73 38 12.42(a).2 ¥ 102 5. In terms of stiffness.15 STRAIN−ENERGY RELEASE RATE It is obvious that a body with a crack or a void is less stiff than a similar body without a void. stiffness M of a given member is defined as the force or load necessary to cause a unit deflection under the load or in the direction of loading.6 ¥ 102 11. Let the body be of unit thickness.42(b). Under uniaxial loading. the load-displacement line OP1 will be as shown in Fig. 1 Pδ (12. 12. 12.pmd 476 7/6/2008. The elastic strain energy U stored is equal to the work done by the load P.e. P P P1 dd da P2 M1 a P'1 M2 o d dd P (b) (a) Fig. since U= δ= P (12.476 Advanced Mechanics of Solids Aluminium alloys 2024 T3 2024 T8 7075 T6 Titanium alloys Ti-6A-4V (high strength) 70 ¥ 104 3.4 ¥ 102 44 27 30 1.. 12.45 ¥ 102 4. and for a linearly elastic body. 58) ( ) ∂u ∂a δ ( ) ⎤⎥⎥ ⎡ ∂ 1 ⎢ 2 P ∂P M 1 2 = ⎢ +P 2 M ∂a ∂a ⎢ ⎣ ( ) ⎥ ⎦ (12. 6:21 PM . the stiffness gets reduced and the load moves down. Fig 12.43 Load-extension plots for crack extension (a) Fixed-grip ( fixed displacement). but the original displacement d1 remains unchanged. (i) In the case of fixed grip. (b) Constant load Chapter_12b.e. 12. P1 P = δ2 = 2 M1 M2 δ1 = (12. Let the crack length a be increased by an amount da.59) Further.58) where M is the stiffness of the body with a crack of length a. Crack begins to increase from a ) ) da (a + da + Crack is longer by da (a P2 Crack is longer by da (a) Crack begins to P increase from a (a) P1 d1 d d1 d2 d (a) (b) Fig.. There are now two cases to consider. the stiffness gets reduced from M1to M2. The quantity ∂a is called the strain-energy release rate. the strain energy gets reduced such that from Eq.. (ii) The load P1 is held constant and the crack is extended.e. Due to this. after the initial displacement d under the load P1) and the crack is extended. (i) The loading grips are held fixed (i. (12.60) ∂u where the subscript d indicate that it is fixed-grips. the load P1 gets reduced to P2 corresponding to the reduced stiffness M 2. i.Introduction to Stress Concentration and Fracture Mechanics 477 the energy is 2 U=1P 2M (12.pmd 477 7/6/2008. As a result of this.43(a) and (b). with the additional cut da. due to additional crack length da. Chapter_12b. 6:21 PM .57) 1 ∂P ∂δ =0= +P ∂a M ∂a ∂ 1 M ∂P = − PM ∂a ∂a ( ) or ( ) ∂ 1 M ∂a Substituting in Eq. the change in strain–energy due to the extension of the crack is dU = U 2 − U1 = Since ( ) δ = P . from Fig.Advanced Mechanics of Solids 478 Differentiating Eq.61) is the strain–energy release rate under fixedgrip condition.12.62) 7/6/2008.61) The expression given by Eq. (ii) In the case of constant load P. =P ∂a P ∂δ1 da ∂a ∂δ ⎤ ⎞ 1 ⎡⎛ P ⎢⎜ δ1 + 1 da⎟ − δ1 ⎥ ∂a ⎠ 2 ⎣⎝ ⎦ ∂ δ 1 1 = P da 2 ∂a ⎡ ∂ 1 ⎤ M ⎥ 1 ⎢ = P ⎢P da ∂a ⎥ 2 ⎢ ⎥ ⎣ ⎦ dU = ( ) i.pmd ( ) ∂U ∂a 478 p 1 = P2 2 ( ) ∂ 1 M ∂a (12. ∂δ M ∂a δ 2 = δ1 + and ∴ 1 P(δ 2 − δ1 ) 2 ( ) ∂ 1 M .. (12. (12.37(b).60) ( ) ∂U ∂a δ ( ) ( ) ⎡ ∂ 1 ∂ 1 ⎤⎥ M M 1 ⎢ 2P = ⎢− PM + P2 2 M ∂a ∂a ⎥ ⎢ ⎥ ⎣ ⎦ 1 ∂ M 1 = − P2 2 ∂a ( ) (12. (12.e. It is necessary that the plastic deformation at the load tip is kept to a minimum.. If the loading is uniform tension. The procedure is repeated until the slit length is greater than the longest crack to be used in the test specimen.e. and the slit is extended by small increments between each pair of consecutive measurements. Compliance C is the deflection per unit load on the specimen. the crack size a should appear as a2 in U2* . Hence. The slit length for each measurement must be measured accurately. B is the specimen thickness. width W and thickness B with a small central crack of length 2a. the strain–energy release rate is independent of the type of load application (e.g. fixed-grip. Gc can be obtained by noting the load at fracture. Hence.64) 2E The stress due to a crack depends upon the crack tip stresses.16 MEANING OF ENERGY CRITERION The change in strain energy due to extension of crack can be interpreted as the energy necessary to create a fracture over da. Thus the strain energy due to U*l (body with no crack) = σ 2 . The energy will also be proportional to the E thickness B (thicker the plate greater will be the energy). the critical strain–energy release rate Gc is dU 1 = Gc = P 2 da 2 The factor C= 1 M ( ) ∂ 1 M 1 ∂C = P2 2 ∂a ∂a (12. 6:21 PM . constant load. From figures 12.e.).43(a) and (b). A set of compliance measurements is made on a specimen. The compliance coefficients are generally expressed in the dimensionless form EBC.Ul*and U2*should have the same dimensions of energy. Hence. the energy due to crack dependes on the crack size a E also. the compliance at a given slit length a.(12.Introduction to Stress Concentration and Fracture Mechanics 479 This is identical to Eq. etc. Consequently. 12. which depends on the crack size. 2 U 2* = C σ Ba2 E Chapter_12b. U2* will be proporcrack will be proportional to tional to σ2 B Further. i. where E is Young’s modulus.pmd 479 (12. then the elastic strain energy of the uncracked body is 1σ LBW (12.. and C.65) 7/6/2008. one can write U* = U*1(body with no crack) + U*2 (due to crack) Consider a large plate of length L. combinations of load change and displacements. at the instant that the crack length is about to get extended. i.63) is called the compliance of the cracked plate.61) excepting for the sign. These crack tip stresses are proportional to the applied stress s. Once the compliance versus crack length relationship is established for a given specimen configuration. at instability. 480 Advanced Mechanics of Solids where C is a dimensionless constant of proportionality. per unit thickness.44). determine the value of fracture resistance R. The plate is pulled to fracture and the fracture load is 800 kN.5 MPa m .68) 2 Gc = R and R = πσ a E 12. (12. Thus.14 A 75-cm wide steel plate has a central crack of length 2a = 10 cm. ER.66) dU = 2πσ 2 a (12. Determine the stress intensity factor assuming a W as small. Hence. 6:21 PM . Also. a = 1 in Eq. This is fracture toughness. namely. fracture if : K1c = ER = toughness (12. The factor πσ 2 a is equal to the square of the stress intensity factor K1. E for the material is 207 GPa. Eq. The plate is 5 mm thick. A detailed analysis shows that C = p.69) shows that fracture occurs when (πσ 2 a) reaches a certain value.3 MPa K Ic = 213.69) Equation (12. Thus.e. i. 000 (75 × 10−2 )(5 × 10−3 ) = 2133 × 105 N / m2 = 213. while the energy release rate is denoted by Gc. In other words. ER. (12.pmd 480 7/6/2008. The total strain energy of a plate of unit thickness. we have from equations (12. The fracture energy per unit crack extension is called fracture resistance and is denoted by R.3 × 106 ⎡⎣ π × 5 × 10−2 ⎤⎦ = 84. Since all considerations are for one crack tip.70) fracture resistance R = K Ic 2 E (12. having a centre crack of length 2a is therefore 2 2 U = U1 + U 2 = 1 σ LW + πσ a 2 2 E E (12.63) and (12. Solution Since Wa is small .61) tells that fracture occurs when K12 reaches a certain values. K I = σ πa The nominal stress s of the uncracked specimen at the time of fracture is σ= 800.67) da E This is for a crack with two tips. Chapter_12b.71) Example 12. ∴ dU πσ 2 a = (12.68) da E per crack tip. (12.5.e.5 − 5) × 480 = 288 MPa 15. K I c2 E R= (84. the nominal stress at which failure takes place (or the remaining strength due to the presence of crack).Introduction to Stress Concentration and Fracture Mechanics 481 From Eq. Use a =2.5 = = 101.15 Using the result of the previous example.1.5) 2 × 1012 = 207 ×109 = 34.53) is σ nom = (W − a ) σ yp W (12. from Eq.. K Ic = ασ fr πa \ σ fr = K Ic α πa Toughness = = α πa 84. calculate the residual strength of a plate with an edge crack of length a = 5 cm. plastic collapse does not occur. = Chapter_12b.5 cm. i.3)2 × 1012 × 5 × 10−2 207 × 109 = = 34. Solution The residual strength sfr is σ fr = K IC α πa 84.2 Since 288 >101.1 × π × 5 × 10−2 The nominal stress at the time of collapse.5 MPa 2. (12.pmd 481 7/6/2008. 6:21 PM . the fracture resistance is R= πσ 2 a E π × (213.5 1 × π × 5 × 10−2 = 213 MPa Example 12. Check for collapse.5 × 103 N/m Also.61).5 × 103 N/m The residual strength is the fracture stress sfr. The width of the plate W = 12. Thus. and syp = 480 MPa. a certain aluminium alloy is selected because of its high strength and light weight. Under the action of external forces and prescribed displacements. the value of Kc is fixed. then the allowable flaw size is given by K2c/(ps2).e. if the design stress s is set at a high level to increase the payload capacity of the aircraft. let the stress–strain curve be as shown in Fig. As a consequence of this.44 (a). a catastrophic fracture may occur. In the case of a linearly elastic body.τ *zx . Is it the material selection because of the environment. design stress. (a) lies in the fact that it is essential to decide what is most important in the design of a component. σ *y . stress intensity factor Kc is given by (a) K = σ πa c where s is the design stress and a is the flaw size.γ *zx are the final or terminal values reached at the end of gradual loading.18 ELASTO-PLASTIC FRACTURE MECHANICS (EPFM) Consider a body B having linear or non-linear elastic properties containing a crack or a void. For instance.8). with the design stress and crack size. ε *y . ε*x . Fig. For example. etc.72) where σ *x . Added to this. The significance of Eq.17 DESIGN CONSIDERATION The conditions for fracture in a component depend on the interaction of material properties. the selection of the material generally depends on the environmental conditions in which the designed product will be functioning. 12. or the flaw size that must be tolerated for safe functioning of the component? Once any two combinations to the three variables (fracture toughess. In the case of a non-linearly elastic body. availability. 12. the body will undergo deformation and store strain energy. 6:21 PM . such as the toughness. the elastic energy per unit volume at any point of the body is given by Eq. and prescribed displacements D on the boundary SD.482 Advanced Mechanics of Solids 12. The energy stored is equal to the work done by the internal stresses during the deformation process. In the process of using this equation in the design process. the value of the third variable is fixed.44 (b). In addition. Once a selection like this is made. the value of the critical stress intensity factor Kc is essentially fixed. loaded by surface traction F on the boundary SF. assume that for the wing skin of a military aircraft. This flaw may get covered up by a rivet head.. i.pmd 482 7/6/2008. 12.. and cost consideration. the conditions may be such as to require a corrosion resistant material. W= ( 1 * * σ ε + σ *y ε *y + σ *z ε *z + τ *xy γ *xy + τ *yz γ *yz + τ *zx γ *zx 2 x x ) (12. If this flaw size goes undetected due to the limitations of the inspection process. For a large plate with a central crack. size.. (11. and the flaw size) is identified. or the high level of design stress because of weight. Let the body have a volume V. and the crack may get extended from the rivet hole and cause failure. Consider an elementary rectangular volume of Chapter_12b.. if the situation allows for the presence of a relatively large crack—one that can be readibly detected and repaired—the design stress is fixed and must be less than K c / π a ... the energy stored is equal to the work done by it and is equal to ∫ (σ x∆y∆z ) ∆ ε x ∆x = Lt.5).e. can be written. 4. ∫ (σ x ∆ ε x ) ∆x∆y∆z = ∫ (σ x d ε x ) dx dy dz Lt.75) V Now consider the body B with the cavity. 12. sy* . The stresses acting on the rectangular faces are shown in the figure. 12. the work done by other forces sy DxDz. Due to sx acting on the area DyDz. Similarly. (12. The limits of the integration are from zero to final values at the end of loading. and the elementary volume are shown enlarged in the figure. and that of DB be drawn as in Fig. (12. ε * ∫ o (σ x dε x ) + σ y dε y + σ z dε z + τ xy γ xy + τ yz γ yz + τ zx γ zx dxdydz (12.73). where W = ∫ (∑ σ xdε x ) (12. Let DB be a small elementary volume adjacent to the cavity.Introduction to Stress Concentration and Fracture Mechanics 483 the body with sides Dx. Dy and Dz. U = ∫ WdV = ∫ ⎡⎣ ∫ ∑ σ x d ε x ⎤⎦ dV V (12. also called strain energy density.45(b). 8:00 PM . szDxDy. the elastic strain energy stored in the elementary volume is External loads SF sy s* B Dy s sx sz Dx Prescribed displacements SD e* e De (a) Dz (b) Fig. the strain energy density given by Eq.2. Fig.72).. (12.73) reduces to that given by Eq. sx*. 12.. The total strain energy stored in the body is therefore. etc. and the surface Chapter_12b.pmd 483 7/7/2008. (b) Non-linear elastic curve σ *.73) The quantity in the parenthesis under the integral sign is the strain energy per unit volume.t*zx. Let the elementary volume of the body DB be isolated from B and let free-body diagrams of the newly created void..44 (a) Elastic body with cavity. The strain energy density at any point is denoted by W. i.74) The summation sign under the integral stands for the expanded version given in Eq. at the point considered.In the case of a linearly elastic solid.45(a). Only a part of the cavity. Assuming that deformations are small and that superposition principle is applicable. where ∆ ε x ∆ x is the elementary extension in the x direction (refer sec. The elementary part DB will be having surface fractions T *. This is similar to action and reaction discussed in reference to Fig. Tx* = nxσ *x + nyτ *yx + nz τ *zx Ty* = nxτ *xy + n yσ *y + nz τ *zy Chapter_12b.e.σ *z .1. y. T and U are proportional to each other. and let σ *x.τ *yz .484 Advanced Mechanics of Solids of the newly created cavity (i.. u y . 6:21 PM . During the loading process.τ *xy . In the case of a lineraly elastic body. and u*z in x. Then. The traction vector T * and the displacement vector u are the final or the terminal values at the end of loading the body. 12.45 (a) Body with cavity .pmd 484 (12. y. and τ *zx.76) ∆B Consider a point P and an elementary area dS surrounding it on the surface of the newly created cavity.σ *y .2.77) 7/6/2008.e. F P B C P * C T DC DB T* T* n C DC (a) (b) Fig.45 (b). Let n be the normal to this area and T * * Fig.9). Let n x . the point P will have undergone * displacements u*x . 12. In the case of a non-linearly elastic body. U = ∫ Wd V = ∫ W d V + ∫ Wd V V −∆B V (12. The traction vector T will have components Tx* . the rectangular stress components at P. (b) Newly created cavity and small volume removed The total elastic strain energy of the original body is now equal to the strain energy of the body with the newly created cavity. i.. n y . they are not proportional . and n z be the direction cosines of the normal n. the space that was occupied by DB) will be having equal and opposite surface fraction T * . B. plus the strain energy stored in the elementary volume DB that is removed. and z directions. and z directions. Ty* . the traction and Tz* in x. from equations (1. Fig. and the surface of the elementary cavity DC had surface traction equal and opposite to T *.79) Now we try to make the newly created cavity traction free so that it becomes a virtual extension of the original cavity. 6:21 PM . It is important to recollect what has been done so far. work is done on the body and the energy stored due to this is given by Eq.pmd 485 (12. Finally. having a cavity C and loaded by surface traction F on SF and prescribed displacements D on SD. we apply gradually.79).80) ∆S Hence. and (b) that of the isolated body DB. the forces T applied will do work on the body and this is equal to Eq. During this process. so that we have now a body with an extended cavity C + DC.e. (12. We started with a body B (linearly or non-linearly elastic). Equation (12.Introduction to Stress Concentration and Fracture Mechanics 485 Tz* = nxτ *zx + nyτ *zy + nzσ *z During the loading process. The total work done by traction forces acting on the entire surface area of the new cavity during the loading process is ⎡ T * . Next.44(a). in order to make the surface of DC. Free-body diagrams of the body with the old cavity C and the newly created cavity DC.u* w = ∫ ∆ w ds = ∫ ⎢ ∫ Tx du x + T y du y + T z du z ∆S ∆S ⎢ ⎣ 0 ( ⎤ )⎥⎥ds ⎦ (12. traction free. During the process of applying T to the surface of DC. The elastic strain energy of the original body B was decomposed into two parts: (a) that of the body B with the newly created cavity DC.78) This expression is valid for both linear and non-linear elastic bodies. This is easily achieved by applying equal and opposite traction forces T at the surface of the new cavity. the work done by the traction T acting on the area ds is equal to [similar to Eq. i.19 PLANE BODY Let the body B considered be a plane body.81) ∆S 12.79).81) can then be written as −∆U = ∫ WdA − ∫ ∆Wds ∆A Chapter_12b. the elastic body stored strain energy U given by Eq. T *. u * ∆w = ∫ 0 (T du x x ) + Ty du y + Tz du z ds (12. the decrease in energy in the process of creating a void or a cavity is −∆ U = ∫ W d V − ∫ ∆ W ds ∆B (12. equal and opposite forces T. an elementary volume of body DB adjacent to the cavity was identified and this was isolated from the parent body. 12.82) ∆S 7/6/2008.74)].75). (12. The elementary body DB was acted upon by surface traction T *. The strain energy stored now in the body B–DB. (12. During the deformation process.. in the body with extended cavity is U ' = ∫ WdV − V ∫ WdV + ∫ ∆Wds ∆B (12. (12. and the elementary volume DB were drawn. 12.83) GREEN’S THOREM Γ be the closed boundary of a domain A. Boundary Γ Domain A P (x. Now in the limit. Γ contour 12. and let P(x. 6:21 PM .84) Integration path Fig.46.y) and Q(x.y) be two ∂q ∂p functions that are continuous together with their partial derivatives ∂x and ∂y in the domain A and the boundary .486 Advanced Mechanics of Solids where DA is the area of the material removed in forming a void and DS represents the newly created traction-free boundary surface. 12. let the cavity or the void considered become a crack of length a.47 such that the origin is at the crack-tip. and Eq.12. Fig. da=dx.46. the direction in which the is traversed is chosen so that the domain A remains to the left. Fig. (12.20 Let (12.83) can be written as y Crack x da a Fig. the rate of change of energy with crack growth can be expressed as − ∂U = ∫ ∫ ∂W dxdy − ∫ ∂wds ∂a ∂a ∂a A Γ 12. Then. 12.y) ∂Q ∂p ⎛ ⎞ ∫ ∫ ⎜⎝ ∂x − ∂y ⎟⎠ dxdy = ∫ ( Pdx + Qdy ) A Γ (12. For an infinitesimal crack extension. Then.y) Q (x.46 Boundary G enclosing the domain A In this case. 12.47 Body with extended crack Chapter_12b.21 THE J-INTEGRAL Let the co-ordinate system be as shown in Fig.pmd 486 7/6/2008. i.pmd 487 7/6/2008.48 (a). 12.Introduction to Stress Concentration and Fracture Mechanics − ∂u = ∫ ∫ ∂W dxdy − ∫ ∂w ds ∂a ∂x ∂x A Γ 487 (12. the area integral in Eq. J = − ∂U = ∫ Wdy − ∫ ∂w ds (unit: Nm–1) ∂a Γ ∂x Γ (12.e. The main restriction is that the body must be subjected to monotonically increasing loading and must not experience any unloding.89) ⎟ ∂x ∂x ⎠ ∂y ⎝ ∂x ∂x ⎟⎠ ⎦⎥ A⎢ ⎣ ∂x ⎝ Chapter_12b.86) ( ) ∂u The quantity − ∂ a is called the J-integral. (12.85) Using Green’s theorm.87) is its applicability to plastic behaviour under certain restrictions.58) and (12.59) ∂ γ xy ∂ε y ∂ε ∂W = σ x x + τ xy + σy ∂x ∂x ∂x σx (12. (12. J is thus a measure of the input work to the system and not the amount of work recoverable on unloading. An important consequence of Eq.. γ xy = x + .87) J is thus the drop in potential energy per unit virtual extension of crack. i. ∂x ∂y ∂y ∂x Equation (12.88) Since. εx = ∂u y ∂u y ∂u x ∂u . ∂W ∫ ∫ ∂x dxdy A becomes for a plane body from equations (12.e. The first term on the right-hand side of the expression.85) with reference to the closed path Γ shown in Fig.88) becomes ∂u y ⎞ ∂W ∂ ⎛ ∂u ⎞ ∂ ⎛ ∂u ∂ ⎛ ∂u y ⎞ = σ x ⎜ x ⎟ + τ xy ⎜ x + + σy ⎜ ∂x ∂x ⎝ ∂x ⎠ ∂x ⎝ ∂y ∂x ⎟⎠ ∂x ⎝ ∂y ⎟⎠ ⎡ ∂W ∂ ⎛ ∂u x ⎞ ∂ ⎛ ∂u x ∂u y ⎞ ∂ ⎛ ∂u y ⎞ ⎤ dxdy = ∫ ∫ ⎢σ x ⎜⎝ ∂x ⎟⎠ + τ xy ∂x ⎜ ∂y + ∂x ⎟ + σ y ∂x ⎜ ∂y ⎟ ⎥ dxdy ∂ x ∂ x ⎝ ⎠ ⎥⎦ ⎝ ⎠ A A⎢ ⎣ ∴∫ ∫ ⎡ ∂ ⎛ ∂u ∂u y ⎞ ∂ ⎛ ∂u y ⎞ ⎤ ∂u = ∫ ∫ ⎢ ⎜ σ x x + τ xy + ⎜ τ xy x + σ y ⎥ dxdy (12.85) can be converted to line integral giving − ∂u ∂w = W dy − ∫ ds ∂a Γ∫ ∂x Γ (12..22 PATH INDEPENDENCE OF THE J-INTEGRAL Consider Eq. 12. εy = . (12. 6:27 PM . 12. the J-integral is zero. n x ds = ds cos θ = dy . 12. (12. ∂x Γ From Eq. (12. Hence.89)..77) ∂u ⎡ ∂u ⎤ y ∂w x ∫ ∂x ds = ∫ ⎢ ∂x ( nxσ x + n y τ yx ) + ∂x ( nxτ xy + n yσ y )⎥ ds Γ Γ⎣ ⎦ (12. i.pmd 488 Γ ∂w ds = 0 ∂x (12. ⎡ ∂u ⎤ ∂u y ∂w x ∫ ∂x ds = ∫ ⎢Tx ∂x + Ty ∂x ⎥ ds Γ Γ⎣ ⎦ From Eq.48(b).85)..48(a). J = ∫ Wdy − ∫ Γ Chapter_12b.90) ⎛ ∂u ⎞ ∂u ⎛ ∂u ⎞ ∂u y y ∂w x x ∫ ∂x ds = ∫ ⎜ σ x ∂x + τ xy ∂x ⎟ dy − ∫ ⎜ τ xy ∂x + σ y ∂x ⎟ dx ⎠ ⎠ Γ Γ⎝ Γ⎝ Using Green’s thorem. (12. (12.48 (a) Closed contour G surrounding the domain A. ∫ ∂w ds. for a closed contour G as in Fig. the above expression can be written as ⎛ ∂u ∂u ⎞ ⎛ ∂u ⎞ ∂u y y ∂w ∂ ∂ x x ∫ ∂s ds = ∫ ∫ ∂x ⎜ σ x ∂x + τ xy ∂x ⎟ + ∂y ⎜ τ xy ∂x + σ y ∂x ⎟ dxdy ⎝ ⎠ ⎝ ⎠ A Γ y n Path G dy q q ds ds O A –dx x (b) (a) Fig. n y ds = ds sin θ = − dx. (b) segment ds of the path with normal n This expression is identical to Eq.91) 7/6/2008. Substituting these in Eq.90) From Fig.488 Advanced Mechanics of Solids Now consider the second term on the right-hand side of Eq (12.e. i. 6:35 PM .e. 12.79) for a plane body. 50(a) depicts these. Accordingly. A closed contour ABCDEFA. i.93) where B is the specimen thickness. for the flanks. (12.e. 12. C A Crack E D G2 B C X F A G2 E D B F G1 G1 (First path) (a) (b) Fig.e. the J-integral according to Eqs.87). 12. CD and AF is shown. Figure 12. (b) Two paths in the same direction The two paths G1 and G2 of the contour are in opposite direction to each other. Hence.Introduction to Stress Concentration and Fracture Mechanics 489 Consider Fig. we have J = J ABC + J CD + J DEF + J FA = 0 Along the flanks CD and FA of the crack.87) and (12.49(a) which shows a body with a crack. From Eq. etc.pmd 489 7/6/2008. Let the crack-lengths be a1. Since the J-integral for a closed contour is zero. J is the drop in potential energy per unit virtual extension of crack. This contour consists of two paths G1 and G2.78).49 Closed contours for a body with a crack (a) Two paths in different directions. y = dy = 0 and the traction force T is also zero. and the two flanks. First. 12. load displacement diagrams are obtained for a number of pre-cracked specimens.93) is as follows. (12. 7:23 PM . Fig. (12. U1 is Chapter_12b. the domain or the area being to the left of the traversing direction).49 (b). The energy per unit thickness u1 delivered to the specimen at a given level of displacement d is obtained as the area under the load-displacement curve. J =− ( ) 1 ∂u B ∂a (12. is zero. a3. which includes the two flanks of the crack. If the path DEF is in the same direction (i. J Γ1 = J Γ 2 (12. J ABC = − J DEF The second path DEF is opposite in direction to the first path ABC..23 J-INTEGRAL AS A FRACTURE CRITERION The path independency of the J-integral can be used as a fracture criterion in the same manner that the stress intensity factor is used.. 12.92) This implies that the J-integral is path independent when applied around a crack tip from one crack surface to the other. The procedure indicated by Eq. a2. d2. (b) Energyvs-slitlength. The negative slopes of U1-a curves are plotted against displacement for any desired crack length between the shortest and the longest used in testing. then d¢1. Fig.50 (a) Load-vs-displacement for different slit lengths a1.51 Notched three-point bend specimen 2.50 (b).. and d3.25 W(min) 7/6/2008.490 Advanced Mechanics of Solids then plotted as a function of crack-length for several constant values of displacement d1. Displacement measurement point a W W 2 =B Fig.pmd 490 2. The recommended specimens are the notched bend and compact tension. (c) J-vs-displacement 12.d calibration curve for each initial crack length. 12. if JC is an appropriate criterion of crack extension. 6:21 PM .24 ASTM-STANDARD TEST FOR JIC The American society for Testing and Materials has standardized a test method to determine J1C as a measure of toughness.25 W(min) Chapter_12b. a3. a2 a 3 d1 d2 d3 Displacement d Crack length a Fig. The objective is to determine the value of J at the initiation of crack growth. etc. It is not intended to characterize crack growth beyond the initiation stage. du1 da a1 d1 d2 d3 Displacement d a2 a3 d3 d2 d1 J a1 J= a1 a2 a3 Energy per unit thickness U1 Load per unit thickness P B The slopes represent − ∂Ua at a given value of displacement which is obvi∂a ously J.51 shows the sketch of the notched bend specimen. 12. 12. d2. a2. Fig. d¢2 and d¢3 are the displacements on the onset of crack extensions for the respective crack length d1. Figure 12. A knowledge of the displacement δ on the onset of crack extension enables the determination of JC from the J . (12. d3.50(c ). Alternatively. For the three-point bend specimen. W is the width of the specimen.7 GPa.25 RELATIONSHIPS OF KC.211. 12. For the three point bend specimen. the maximum pressure at the centre of the contact area. This means that the critical stress intensity factor K1c. The elastic constants for the materials are: cast iron–E = 41. In order to obtain a valid J1cvalue.326. copper–E = 44. Determine the width of the contact area.94 b) 12. 8:01 PM .1 A 20-mm long cast iron rod of 25-mm diameter is pressed on to a thick copper plate with a force of 20 N. B > 25 ⎜ lc ⎟ ⎝σ f ⎠ (a) In order to ensure that the crack tip stress/strain field is characterized by pathindependent integrals. v = 0. and the J-integral are related. the initial uncracked ligament dimension (W-a)=b. and ao is the original crack size. and the width B must satisfy the condition ⎛J ⎞ b. ⎛a ⎞ f ⎜ 0⎟ = 2 ⎝W ⎠ The initial values of J1c obtained from the measurements of the area under the load versus displacement curve is validated by the condition (a). AND J It is obvious that the changes involved in the process of extension of a crack in a loaded body is intimately connected with the stress field existing in the neighborhood of the crack tip. B is the specimen thickness.pmd 491 7/7/2008. v = 0. GC .Introduction to Stress Concentration and Fracture Mechanics ( 491 ) a The specimen has a deep initial crack w ≥ 0. the critical strain energy release rate Gc.5 . Evaluations of the J-integral are made from load versus load–displacement curves using the area under the load displacement curve. and the octahedral stress at the centre of the contact area. Chapter_12b. b is the initial uncracked ligament (W − ao ) . the crack length a.94 a) for plane stress (12. the J-integral is given by J= A ⋅ Bb ⎛a ⎞ f ⎜ 0⎟ ⎝W ⎠ where A is the area under the load versus load-point displacement diagram. The relationships are as follows: J = Gc = J = Gc = K Ic 2 1 − v2 E ( K Ic 2 E ) for plane strain (12.4 GPa. A factor of safety 2 is used. Using a factor of safety of 1. where b is the halfwidth of the contact area.15MPa] 12. Weight is a major consideration.35 Tube with an external crack under internal pressure Chapter_12b.7 mm.5. p max = 930 GPa. [Ans: 3. Determine the maximum longitudinal stress the sheet can withstand without the danger of catastrophic failure.301 pmax. ⎥ ⎣ oct ⎦ 12.4 An aluminium plate of 1. c p b Fig.1 mm ⎥⎦ 12. 6:21 PM .85 × 10-7 mm.3 For two cylinders pressed together. Consider points at the inner and outer boundaries.2 For two spheres in contact under pressure. ⎤ ⎢⎣ t = 6.⎤ ⎢τ = 253 GPa. show that the maximum shear occurs at a depth of z = 0. Determine the maximum internal pressure that can be applied without yielding or fracture occurring.5-m width and 3-m length is required to support a force of 2 MN in the 3-m direction. Inspection procedures can detect a through-thickness edge cracks longer than 2. and Sy = 490 MPa. The two ends of the cylinder are closed. select the proper sheet and its thickness. 12. 12. show that the maximum shear stress on the z-axis occurs very nearly at half the distance of the radius of the contact area and its value is 0.31pmax. Al-2024 and Al–7178 are the materials under consideration. KIc = 33 MPa m .pmd 492 7/6/2008. ⎡ Ans: Use Al-7178.6 A cylinder with an internal radius of 5 cm and external radius of 6. 12. The material has a yield strength of 490MPa.5 cm has a radial crack of 2-mm length on the outer periphery. Al-2024 has a value of 26 MPa m for KIc.5 A steel sheet that is 16 m long and 8 m wide is found to have a central transverse crack of 40-mm length. For Al-7178.492 Advanced Mechanics of Solids ⎡ Ans: 6.78b and its magnitude is 0. and a yield stress Sy = 455 MPa. The material of the sheet has a fracture toughness factor KIc = 25 MPa m . σ nom =355 MPa ] 12. ⎥ ⎣ ⎦ Chapter_12b. thickness B = 1.7.pmd 493 7/6/2008.7 cm.9 Calculate the fracture toughness of a material for which a plate test with central crack gives the following information: Width W = 50 cm. ⎤ ⎢ No collapse. 6:21 PM .86. a = 1. No collapse. ⎡ Ans : σ fr = 222 MPa. ⎥ ⎣ ⎦ 12. Loading is perpendicular to the major axis.8 For Problem 12.9 cm. Check for collapse.031 cm ⎤ ⎣ ⎦ 12. ⎡ Ans : K t' = 23. and an yield strength of syp = 520 MPa. and thickness t = 0. [ Ans : Pmax =133. The yield strength is syp= 480MPa.7 MPa m ⎥ Ic ⎣ Ic ⎦ 12. ρ = 0. failure load P = 1335 kN. and a minor axis equal to 0.15cm.5 cm. What is the radius of curvature at the ends of the major axis? Assume that there is no yielding. ⎡ Ans : Toughness = 39.10 Given a toughness of K = 77 MPa m .9 kN.943 ⎤ ⎢ K if K < 25.2 MPa m. calcute the nominal stress in the full section at the time of collapse if the yield strength is 525 MPa.01. ⎤ ⎢ Yes.7 Calculate the theoretical stress concentration factor for an elliptical notch with a major axis equal to 8 cm. determine the residual strength of a centre cracked plate of 45 cm width and crack length 2a = 7. crack length 2a = 5 cm. What is the fracture load? Width W = 25 cm. Is this plane Strain? Check for collapse.Introduction to Stress Concentration and Fracture Mechanics 493 ⎡ Ans : p = 50 MPa or p =1. ∂ 2σ x ∂y 2 + ∂ 2σ y ∂x 2 ⎛ ∂ 2σ y ∂ 2σ ⎞ ⎛ ∂ 2σ ∂ 2σ y ⎞ x x − v⎜ + = − + + (1 v ) ⎟ ⎜ ⎟ 2 2 ∂x 2 ⎠ ∂y 2 ⎠ ⎝ ∂y ⎝ ∂x . (d).56 a) ∂ 2 ε xx ∂y 2 + ∂ε yy ∂x 2 = ∂ 2γ xy ∂x∂y (a) This can be converted to stress compatibility equation for the plane stress case using the stress–strain relations: ε xx = 1 (σ x − vσ y ).494 Advanced Mechanics of Solids APPENDIX The strain compatibility condition for the two-dimensional case is. (a). (f ) with respect to y and adding. ε yy = 1 (σ y − vσ x ) E vxy = E 2(1 + v) 1 τ = τ xy G xy E (b) (c) Substituting in Eq. one gets 2 ∂ 2τ xy ∂ 2σ x ∂ σ y 2 =− − ∂ x ∂y ∂x 2 ∂y 2 Substituting the above in Eq. (e) with respect to x. and Eq. ∂2 ∂2 ∂2 ⎡ (σ x − vσ y ) + 2 (σ y − vσ x ) = 2(1 + v)τ xy ⎤⎦ 2 ∂x ∂y ⎣ ∂y ∂x (d) The equations of equilibrium in the absence of body forces from equations (1.65) are ∂σ x ∂τ xy + =0 ∂x ∂y (e) ∂σ y ∂τ xy + =0 ∂y ∂x (f ) Differentiating Eq. from Eq.(2. It can easily be converted into polar co-ordinates. The usual method of solving an elasticity problem is by introducing a function f of x. q) so defined by Eq. 495 (g) Equation (g) is the stress equation of compatibility. τ xy = − 2 ∂ 2φ ∂x∂y (h) As can be checked. we should have 2 2 ⎛ ∂2 ∂2 ⎞ ⎛ ∂ φ + ∂ φ ⎞ = + ⎟ 0 ⎜ 2 ⎟⎜ ⎝ ∂x ∂y 2 ⎠ ⎝ ∂x 2 ∂Y 2 ⎠ i. In order that it may satisfy the compatibility condition (g)..e. ∂ 4φ ∂x 4 +2 ∂ 4φ ∂y 4 + ∂ 4φ ∂y 4 (j) =0 (k) A function f (x. We have.e. (m) satisfies the equations of equilibrium. Eq. and let ∂φ 1 ∂2φ + r ∂r r 2 ∂θ 2 σr = 1 σθ = ∂ 2φ (m) ∂r 2 2 ∂φ ∂ ⎛ 1 ∂φ ⎞ 1 ∂ φ τ rθ = 12 − = − ⎜ θ r r θ r ⎝ r ∂θ ⎟⎠ ∂ ∂ ∂ ∂ r The function f (r. We shall transform Eq.Appendix ⎛ ∂2 ∂2 ⎞ ⎜ 2 + 2 ⎟ (σ x + σ y ) = 0 ⎝ ∂x ∂y ⎠ i. that satisfies the equations of equilibrium (e) and (f ). in polar coordinates. this function satisfies the equations of equilibrium.70). If it satisfies in addition. and the appropriate boundary conditions. (k) into polar coordinates to solve axi-symmetric problems. (1. (k) satisfies the equations of equilibrium and the compatibility condition. Equation (g) is the stress equation of compatibility expressed in Cartesian co-ordinates. Let a function f (x. q). boundary conditions of a given problem. the compatibility condition (g). Let the stress function in polar coordinates be f(r.. then such a function is the proper function for that problem. y) which satisfies Eq. y) be choosen such that σx = ∂ 2φ ∂ 2φ ∂y ∂x . = = sin θ ∂x ∂y r r . σy = 2 . r2 = x2 + y2 and q = arctan y x from which ∂r ∂r y x = = cosθ . and y. q) satisfying this equation will satisfy equilibrium equations and compatibility condition. For this function ∂φ ∂φ ∂r ∂φ ∂θ ∂φ 1 ∂φ = + = cosθ − . then the stress distribution around the circumference of the circle is the one caused just by s. = = ∂x r ∂y r 2 r r Let the stress function in polar coordinates be f(r. without being affected by the hole. q). then it is the proper stress function for that problem. Fig. If the function in addition. sinθ ∂x ∂r ∂x ∂θ ∂x ∂r r ∂θ Symbolically. satisfies the boundary conditions for a given problem. If a large circle of radius b is drawn concentric with the hole. A-1. since the hole is very small and the boundary . (j) can be written as 2 2 2 2 2 ⎛ ∂2 1 ∂ 1 ∂ 2 ⎞ ⎛ ∂ φ + 1 ∂φ + 1 ∂ φ ⎞ = 0 ∂2 ⎞ ⎛ ∂ φ ∂ φ ⎞ ⎛ ∂ ⎜ 2 + 2 ⎟ ⎜ 2 + 2 ⎟ = ⎜⎝ 2 + r ∂r + 2 ⎜ 2⎟ 2 2 r ∂r r ∂θ 2 ⎟⎠ ⎝ ∂x ∂y ⎠ ⎝ ∂ x ∂y ⎠ r ∂ θ ⎠ ⎝ ∂r ∂r (q) Any function f (r. WIDE PLATE WITH A SMALL CIRCULAR HOLE Consider a wide plate with a small circular hole of radius a. subjected to a uniform tensile stress s.496 Advanced Mechanics of Solids y ∂θ sin θ ∂θ x cosθ =− 2 =− . Eq. ( ∂ ∂ 1 ∂ = cos θ − sin θ ∂x ∂r r ∂θ ∴ ∂ 2φ ∂x 2 = = ) ( ) ∂ 1 ∂ ⎛ ∂φ cos θ − 1 sin θ ∂ ⎞ ∂ ⎡ ∂φ ⎤ cos θ − sin θ = ⎢ ⎥ ∂θ ⎟⎠ r ∂ r r ∂ θ ⎜⎝ ∂r ∂ x ⎣ ∂x ⎦ ∂2φ ∂r 2 cos 2 θ − 2 ∂2 φ ∂φ sin 2 θ ∂φ sin θ cos θ + + 2 + ∂r ∂θ ∂r r ∂θ r2 ∂ 2 φ sin 2 θ ∂θ 2 r In the same manner one gets ∂ 2φ ∂y 2 = ∂ 2φ ∂r 2 (n) sin 2 θ + 2 ∂ 2 φ sin θ cos θ ∂φ cos 2 θ + r r ∂r ∂θ ∂r ∂φ sin θ cos θ ∂ 2φ cos 2 θ + 2 ∂θ ∂θ 2 r 2 r Adding together equations (n) and (p) −2 ∂ 2φ ∂x 2 + ∂ 2φ ∂y 2 = ∂ 2φ ∂r 2 + (p) 2 1 ∂φ 1 ∂ φ + 2 r ∂r r ∂θ 2 Using this. 16) and (8. on a small area bdq. Substituting ⎛ ∂2 1 ∂ 1 ∂2 ⎞ ⎜⎝ 2 + r ∂r + 2 ⎟ r ∂θ 2 ⎠ ∂r ⎧ ∂2 ⎫ 1 ∂ 1 ∂2 × ⎨ 2 [ f ( r ) cos 2θ ] + f ( r ) cos 2θ ] + 2 [ [ f (r ) cos 2θ ]⎬ = 0 2 ∂ r r r ∂θ ⎩ ∂r ⎭ i. At an angle q.e. the stress is s cosq. and the other tangential:.2(a). A wide plate with a small hole subjected to a tensile stress s. The stress distribution around the big circle can be determined from statics as was done in Section 12.s cosq sinq. (p).s cosq sinq n s Fig. A-1. The ring is now subjected to the following stresses: radial: σ cos 2θ = 1 σ (1 + cos 2θ ) 2 tangential: -s cosq sinq = − 1 s sin2q 2 (r) 1 σ on the ring can be solved using equations 2 (8.Appendix s 497 scos 2q scosq m q . The remaining parts consisting of the varying radial stress The case of uniform radial stress 1 σ cos2θ .17). of the circle is far removed from the hole. and the tangential stress 1 σ sin 2θ can be analysed through the stress 2 2 function method. ⎛ ∂2 1 ∂ 1 ∂2 ⎞ + + ⎜⎝ 2 ⎟ r ∂r r 2 ∂θ 2 ⎠ ∂r ⎡ ∂ 2 f (r ) ⎤ 4 f (r ) 1 ∂f ( r ) ×⎢ cos 2θ + 2 cos 2θ − cos 2θ ⎥ = 0 2 2 ∂ r r r ⎣⎢ ∂r ⎦⎥ . This can be resolved into two components. This large circular thick plate with a small hole can be isolated and analysed as equivalent to the original problem. This has to satisfy the compatibility condition given by Eq. Let the stress function be of the form f = f (r) cos2q. one radial : s cos2q. one obtains 4 2 A = − σ . (m) are 2 ∂φ ⎛ 1 ∂ φ 6C 4D ⎞ σr = 1 + 2 = − ⎜ 2 A + 4 + 2 ⎟ cos 2θ 2 ⎝ r ∂r r ∂θ r r ⎠ σθ = ∂ 2φ ⎛ 6C ⎞ = ⎜ 2 A + 12Br 2 + 4 ⎟ cos 2θ ⎝ ∂r r ⎠ (s) 2 ⎛ ∂φ ⎞ ⎛ ⎞ − 2D sin 2θ τ rθ = − ∂ ⎜ 1 ⎟ = ⎜ 2 A + 6 Br 2 − 6C 4 2 ⎟ ∂r ⎝ r ∂θ ⎠ ⎝ r r ⎠ The constants of integration are now determined from the conditions: (i) that the edge of the inner hole is free from external forces. These conditions give the following equations. and (ii) the outer boundary is subjected to stresses given Eq. B = 0. D = a σ.498 Advanced Mechanics of Solids Cancelling cos 2q. and observing that the differential equation involves only f (r). + 4D = − 1σ 2 A + 6C 4 2 2 b b 2 A + 6C + 4D = 0 4 a a2 − 2D = − 1σ 2 A + 6 Bb2 − 6C 4 2 2 b b 2 A + 6 Ba 2 − 6C4 − 2 D =0 a a2 Solving these and putting a b 0 because of a very wide plate. (r). the partial differential equation becomes an ordinary differential equation. 4 4 2 . C = − a σ. which is 2 ⎛ d2 4f ⎞ 1 d 4 ⎞⎛d f 1 df + − + − 2⎟ =0 ⎜⎝ 2 ⎟ ⎜ 2 r dr r dr dr r ⎠ ⎝ dr r ⎠ The general solution is f (r ) = Ar 2 + Br 4 + C 12 + D r The stress function is therefore ⎛ ⎞ φ = f ( r )cos2θ = ⎜ Ar 2 + Br 4 + C 12 + D⎟ cos 2θ ⎝ ⎠ r The corresponding stresses from Eq. 17). .. Remembering that in addition to these. the final solutions are uniform radial stress 2⎞ 4 2⎞ ⎛ ⎛ σ r = σ ⎜1 − a2 ⎟ + σ ⎜1 + 3a4 − 4a2 ⎟ cos 2θ 2⎝ 2⎝ r ⎠ r r ⎠ 2 ⎞ 4⎞ ⎛ ⎛ σ θ = σ ⎜1 + a2 ⎟ − σ ⎜ 1 + 3a4 ⎟ cos 2θ 2⎝ r ⎠ 2⎝ r ⎠ (t) 4 2⎞ ⎛ τ rθ = − σ ⎜ 1 − 3 a4 + 2 a2 ⎟ sin 2θ 2⎝ r r ⎠ When r is very large. and σ θ = σ (1 − 2 cos 2θ ) (u) It can be seen that sq is greatest when θ = π or θ = 3π . When θ = 0 or θ = π . At these points. at the ends m and n of 2 2 the diameter perpendicular to the direction of s. the stresses approach the values given by Eq.Appendix 499 Substituting these in Eq. σ θ = −σ . (s) we get the stresses in the large disc (equivalently in the plate) due to the varying radial stresses 1 σ cos 2φ and the tangential stresses 2 1 σ sin 2φ .e.16) and (8. (r). the disc is subjected to the 2 1 σ on the outer boundary. sq = 3s. At the edge of the small hole σ r = τ rθ = 0. whose solution is given by 2 equations (8. i. equations 335 with concentrated lateral load 336 with continuous lateral load 340 with end couple 342–344 with several loads 339 Beam with hyperbolic groove 444 Begg’s deformeter 152 Biaxial stress state 470 Body force 2 Boundary conditions 45. 94 Complementary energy 174 Compliance 480 Components of compliance 377 of modulus 377 off axis 386 Composite cross-ply 400 cylinder of equal strength 282 failure (see Failure) in-plane moduli of 398 micro-mechanics of 411 . 312 Bredt-Batho formula 250 Brittle fracture 457 material 463 Buckling as an eigenvalue problem 350–352 energy methods 355 of columns with variable cross-sections 366 Rayleigh-Ritz method 360 Timoshenko’s concepts 364 use of trigonometric series 368 Bulk modulus 101 Castigliano first theorem of 153 generalized theorem of 173 second theorem of 170 (see also Menabrea’s theorem) Cauchy’s strain-displacement relations 70 stress formula 7 stress quadric 34. 60–62 Centre of flexure 201 of twist 264 shear 201 Change in angle of a linear segment 73 in direction of a linear segment 73 in length of a linear segment 69. 67 Circular disk Thermal stresses in 314 (see also rotating disks) Collapse load 473 strength 473 Column stability general treatment of 344 with both ends fixed 347 with hinged ends 345 with load passing through a fixed point 348 with one end fixed and other end free 345 with one end fixed and other end pinned 346 with variable cross-section 366 Compatibility conditions 86–88. 476 Axis of symmetry 377 Axisymmetric body 269 Axisymmetric problems 269–306 Beam bending 181 Beam column.INDEX Anisotrophy orthogonal 401 rectilinear 377 Asymmetrical bending 190 ASTM standards 490. 481 Free boundary 14 Eigenvalue problem 350 Elastic constants. 274 under internal pressure 271. 407 of composites 406 significance of 117 Fictitious load method 161 Flexural centre 201. 274 Cylinders in contact 450. 71 Decomposition 31 Deformation in neighborhood of point 64 of right parallelepiped 75 Deviatoric part of strain matrix 90 (see strain deviator) Deviatoric plane 134 (see also pi-plane) Deviatoric state of stress 32 Disks of variable thickness 298 Displacement along a radius 271 equations of equilibrium 104 generalized 149 gradient matrix 66 vector 64 vector field 64 work absorbing component of 145 Distortion energy 119. 119. 476. 431 maximum shearing stress 111. relations of 102 energy 376. 407 methods 144. 270 in cylindrical coordinates 45–48 in plane-stress case 42 in rectangular coordinates 40–42 Eqvivalent stress 131 Euler-Bernoulli equations 190 hypothesis 190. 118 maximum principal stress 113. 264 Fluid vorticity 440 Fracture conditions 463 elasto-plastic 483 first mode 464 modes 464 resistance 481 toughness 460. 124 Failure theory distortion energy 118.Index off-axis loading of 383 pressure vessel 421 rod. 407 maximum elastic energy 113 maximum elastic strain 112. 483 Generalized displacement 149 force 149 Green’s theorem 486 . 198 Extensional strain 63. 118 Mohr’s 129 octahedral shearing stress 113. 118. 355 criterion 480 Engesser’s theorem 173 Equality of cross-shears 11 Equilibrium equations differential 40-41 displacement 104 for axisymmetric case 42. unidirectional 412 stress-strain relations 375 transverse stresses in 422 tube 280 Constitutive equations 97 Contact stresses 445 Continuity conditions (see compatibility conditions) Continuity equations 440 Continuum mechanics 2 Corresponding displacements 145 forces 145 Critical stress intensity factor 460 Cross-shears 11 Cubic equation real roots 17 standard solution 20 Cubical dilatation 74 Curved beams bending of 209 deflections of 216 thermal stresses in 325 Winkler-Bach formula 209 Cylinder under external pressure 271. maximum 118 energy stored 146 energy superposition of 163 potential 375 stability 331 symmetry 377 Ellipsoid surface 61 Energy complementary 174 distortion 119. 453 501 energy. 407 Factor of safety 121. 42 strain 460. 471. 102 off-axis components of 392 of rigidity 99 specific 416 Mohr’s circles 25. 54–55 stress plane 26 theory of failure 129 Multidirectonal laminates 395 148 Pi-plane 134 Plane Octahedral 28 of maximum shear 28 of symmetry 377 shearless 16. 7 Notch sensitivity 445 Octahedral normal stress 29 planes 29 shear stress 29 Off-axis components of compliance 386 of modulus 392 Off-axis loading 383 Opening mode 464 Order of redundancy 170 Orthogonal symmetry 380 Orthogonality relations 352 Orthotropic body 380 laminate 401 J-fracture criterion 490 –integral 487 –path independence 487 Kirchhoff’s theorem 169 Lame’s coefficients 102. 375 Hydrostatic state of stress 11. 472 Plastic collapse 471. 31 Ideal fluid 439 Incompressible material 103 Influence coefficient 144 Isostrain 412 Isotropic state of strain 90 state of stress 11. 19. 469. 28 state of strain 83 state of stress 38. 90 stress 16 Netting theory 421 Neutral axis location 191.502 Index Hertzian stresses 446 Hooke’s law 144 generalized statement 97. 99 ellipsoid 36 displacement equations 105 problem 271 Laminate angle-ply 408 cross-ply 400 multidirectional 395 off-axis loading 383 orthotropic 401 unidirectional 381 Laplace equation 234 Least work principle 170 Levy-Mises equations 140 Limit load 473 Linear strain 71 Lines of shearing stress 239 Load lines 469 Material anisotropic 374 composite 374 constants 377 isotropic 98 Maximum shear plane 28 Maxwell-Betti-Rayleigh theorem Maxwell-Mohr integrals 176 Membrane analogy 248 Menabrea’s theorem 170 Micromechanics 411 Modulus in-plane components 398 of elasticity 97. 473 Plastic solid. 40. 212 Normal stress components 4. 19 state function 134 transversely 377 Ideal fluid 11 Ideally plastic solid 109. ideal 132 Plate with elliptical hole 437 with circular hole 431 Ply strain 404 stress 404 Poission equation 237 Poission’s ratio 99. 379 Potential energy 355 . 471 stress 469. 132 Invariants of strain 80. 90 isotropic state of 90. 411 Young’s modulus. 17 planes are orthogonal 17 shear planes 28 strains 78 strain axes 78. 39 stress axes 15. 440 Rigidity modulus 99 Rotating disks of uniform thickness 294–298 disks of variable thickness 298 shafts and cylinders 300–302 Rule of mixtures 400. 453 Spherical part of strain matrix 90 Stability elastic 331 (see also columns) energy. 55 Quadric. 84 stress 16. and 355 State of plane stress 38 pure shear 31. for 413 Saint Venant compatibility equations 88 Saint Venant’s assumption 201 Saint Venant-von Mises equations 140 Shaft with eccentric hole 441 wiht key way 442 with semicircular groove 442 Shear centre 201. maximum 28 strain components 64. 376 extensional 63.Index Prandtl-Reuss equations 139 Prandtl-torsion stress function Pressure vessels 421 Principal axes of bending 192 direction 378 planes 15. 264 flow 250 in thin-walled sections 202 plane. 39 Principle of least work 170 of superposition 143 Pure shear state 31. 28 503 Shrink fits 280 Specific modulus 416 strength 416 Sphere hollow 290 radial displacement 287–290 stresses due to gravitation 292 thermal stress 320 Spheres in contact 446. 55 (see also decomposition) pure shear strain 90 strain at a point 70 stress at a point 4 stress referred to principal axes 24 Statically indeterminate structures 164 Steady flow 440 Stiffness 477 Strain deviator 90 displacement relations 70 energy expression 155. 71 stress components 4. 95 linear 71 off-axis 384 plane state of 83 principal 78 pure shear part 90 rectangular components 70 shear 71 space 134 spherical part of 90 two dimensional 63 Strain energy density 484 release rate 477. 7 Shear mode 465 Shearless plane 16. 60–62 Rayleigh-Ritz method 360 Reciprocal identity 379 relation 147 theorem 148 Rectangular strain components 70 stress components 4 Relative extension 69 Residual strength 482 Resultant stress vector 7 Rigid body rotation 72. Cauchy’s stress 236 34. 479 Stream function 441 Stress components on arbitrary plane 6 concentration factor 429 under bending 444 under tension 429 under torsion 439 . 71 function 134 invariants 80. 408 strain relation. 38 quadric 34. 102 . surface 2 Transverse stress 422 Transversely isotropic 377 Tresca yield surface 136 Tsai-Hill theory 407 Unit load method 177 Unsymmetrical bending 190 Virtual work theorem 166 Volumetric strain 76 (See also cubical dilatation) Vorticity 440 Warping function 233 Winkler-Bach formula 209 Work absorbing component 145 Yield criteria 109 Yield locus 135 Yield surface 136 Young’s modulus 97. 375 strain relations. 60–62 raiser 428 rectangular components 4 space 134.504 Index Stress (continued) deviator 32 difference theory 129 due to gravitation 292 ellipsoid 36 equivalent 131 field near a crack 458 function 42 function. linear 97. Mohr’s 25 plane state of 38 principal 14. isotropic materials 98 strain relations. equilibrium of 6 Theorem of Kirchoff 169 of stationary potential energy 355 of virtual work 166 Theories of failure (see failure) Thermal stress in curved beam 325–328 disk with hole 316 long circular cylinder 316–319 solid disk 315 sphere 320–322 straight beam 322 thin circular disk 314 Thermoelastic strain-displacement relations 312 stress-strain relations 311 Thick-walled cylinder plane strain 278–280 plane stress 272–273 Torsion of circular bar 230. torsion 236 in composite tubes 280 intensity factor 458 invariants 16 isotropic state of 11. plastic flow 137 strain relations. 240 elliptical bar 240. 19 normal components 4 octahedral 29 off-axis 384 plane. thermoelastic 311 transverse 422 vector 3 Stresses due to gravitation 292 Superposition of energies 163 Superposition principle 143 Surface force 2 traction 2 Tangential stress 4 Tetrahedron. 243 equilateral triangular bar 243 multiple-cell closed section 251 multiply connected sections 259 prismatiac bars 232 rectangular bars 245 rolled sections 256 slit tube 257 squatty sections 247 thin rectangular bars 255 thin-walled tubes 249 Torsional rigidity 236 Torsion stress-function 236 Traction.