Theory of Isotropic/Orthotropic Elasticity 9783031063442, 9783031063459

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K. Bhaskar · T. K. Varadan

Theory of Isotropic/Orthotropic Elasticity

Theory of Isotropic/Orthotropic Elasticity

K. Bhaskar · T. K. Varadan

Theory of Isotropic/ Orthotropic Elasticity

K. Bhaskar Department of Aerospace Engineering Indian Institute of Technology, Madras Chennai, India

T. K. Varadan Department of Aerospace Engineering Indian Institute of Technology, Madras Chennai, India

ISBN 978-3-031-06344-2 ISBN 978-3-031-06345-9 (eBook) https://doi.org/10.1007/978-3-031-06345-9 Jointly published with ANE Books Pvt. Ltd. In addition to this printed edition, there is a local printed edition of this work available via Ane Books in South Asia (India, Pakistan, Sri Lanka, Bangladesh, Nepal and Bhutan) and Africa (all countries in the African subcontinent). ISBN of the Co-Publisher’s edition: 978-9789380156200. © The Author(s) 2023 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publishers, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publishers nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publishers remain neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

Preface

The theory of elasticity is a classical topic and there are indeed many good books dealing with it with reference to isotropic bodies. However, in our experience, we find that many students still consider the subject rather difficult and tiresome, viewing most of the available books as scholarly treatises written chiefly for scholars. Thus, there is a need for a more straightforward and introductory treatment of this important subject, highlighting its essence and illustrating the additional insight it provides regarding structural behaviour. This book is intended to fulfil this need and is written such that a student going through it is motivated towards a more thorough mathematical study of the theory of elasticity, while a practising engineer, with just a rudimentary background in solid mechanics, can use this as self-study material to familiarise himself with the subject. It is often pointed out that modern-day engineers are quite conversant with powerful black-box type structural analysis software and routinely carry out complicated two-dimensional and three-dimensional analysis, but lack a well-nurtured intuitive structural sense. Such a feel for the structure can be generated only by a careful study of the basic theory and its illustrative application to different problems, and thus it is necessary to provide the industrial designer-analyst with sufficiently self-contained, yet easy-to-read, study material. Another motivating factor is the need to provide an introduction to orthotropic elasticity in view of the increasing use of fibre-reinforced composites in all fields of engineering. There are a few books available on this subject as well, but the present book differs from them in that the isotropic and orthotropic cases are juxtaposed here within each problem considered so as to emphasise the similarities and differences. As will be shown at various stages in this book, the conventional engineering theories are often more in error for orthotropic bodies than for corresponding isotropic bodies and hence the elasticity approach is often of greater significance for the former. In keeping with the aforementioned objectives, comprehensive and exhaustive coverage of all kinds of problems and solution methodologies is not attempted herein; instead, attention is focused on a set of fairly simple solutions through which the reader can appreciate the need and nuances of the theory of elasticity and the greater insight that it provides as compared to a preliminary study using the commonly used engineering theories. Further, v

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Preface

only specially orthotropic configurations—those with principal material directions coinciding with body-specific geometric axes—are considered herein. Arbitrary orientations that lead to anisotropic behaviour of the structure characterised by several non-classical coupling effects are not considered; this is not just because the corresponding equations turn out to be quite tedious, but also because such general orientations are not very commonly employed in practice. Though the presentation here is fairly self-contained, appropriate references for further study are included. The contents of the book are conveniently organised into five chapters. The first is an introductory chapter dealing with the field equations of isotropic/orthotropic elasticity. It is followed by a chapter on plane problems in Cartesian coordinates wherein several problems corresponding to beam flexure are studied using Airy stress functions and differences with respect to the corresponding beam theory solutions are highlighted. Then St.Venant’s principle is first put forth as a natural conclusion deduced from a study of two alternative elasticity solutions corresponding to a simply supported beam; this is followed by more direct treatment of the decay of self-equilibrant end stresses and a discussion of the decay lengths in isotropic and orthotropic bodies. Chapter 3 includes a wide variety of plane problems with reference to polar coordinates—dealing with axisymmetric deformation, stress concentration, crack tip stress fields, stresses under concentrated loads, contact stresses and transverse normal stresses in curved beams. Also included herein is a very simple illustrative problem involving the violation of the principle of complementary shear. For most of these problems, full solution details are presented only for the isotropic cases; the corresponding orthotropic solutions, usually involving complex variables, are too cumbersome and hence not discussed here. However, the effect of orthotropy on physical behaviour is clearly pointed out. Chapter 4 presents a fairly comprehensive treatment of torsion of non-circular sections and deals with isotropic and rectilinearly or cylindrically orthotropic shafts. Solid sections of various shapes, thin-walled open and closed sections, laminated orthotropic sections as well as various complicating factors such as grooves, reentrant corners, eccentric holes and warping constraints are considered. Finally, salient features of certain important problems—Boussinesq’s problem, axisymmetric contact, free edge problem and analysis of functionally graded configurations—are explained in Chap. 5. This book is expected to be useful for students undergoing courses on Advanced Structural Mechanics, Elasticity and Composite Structures. The contents of this book may also be used as a guideline for framing the syllabus of a graduate-level elective for students of Aerospace, Civil and Mechanical engineering. Chennai, India

K. Bhaskar T. K. Varadan

Contents

1 Introduction and Mathematical Framework . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Theory of Elasticity versus Conventional Engineering Theories . . . . . . . 1.2 Field Variables of the Theory of Elasticity . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Two-Dimensional Elasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3.1 Plane Stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3.2 Plane Strain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3.3 Field Variables of Two-Dimensional Elasticity . . . . . . . . . . . . . . . 1.4 The Field Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4.1 Constitutive Relations—Generalised Hooke’s Law . . . . . . . . . . . 1.4.2 Equations of Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4.3 Strain–Displacement Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Stress Approach—Compatibility Equations . . . . . . . . . . . . . . . . . . . . . . . . . 1.6 Displacement Approach—Navier Equations . . . . . . . . . . . . . . . . . . . . . . . . 1.7 Stress Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.8 Strain Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.9 Principal Stresses and Strains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.10 Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.11 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1 3 7 7 8 8 8 9 16 17 19 21 22 23 24 25 27

2 Plane Problems in Cartesian Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Airy Stress Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Compatibility Equation in Terms of Stresses and Airy Stress Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.1 Plane Stress—Isotropic Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.2 Plane Strain—Isotropic Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.3 Plane Stress—Specially Orthotropic Case . . . . . . . . . . . . . . . . . . . 2.2.4 Plane Strain—Specially Orthotropic Case . . . . . . . . . . . . . . . . . . . 2.3 Use of Polynomial Stress Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.1 The Simple Case of Uniaxial Tension . . . . . . . . . . . . . . . . . . . . . . 2.3.2 Pure Bending of a Beam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

29 29 30 30 31 32 32 32 33 36 vii

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2.3.3 A Tip-Loaded Cantilever . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3.4 A Simply Supported Beam Under Uniform Load . . . . . . . . . . . . A Fourier Series Solution for the Simply Supported Beam . . . . . . . . . . . Justification of End Conditions Specified in Terms of Integrals . . . . . . . . St.Venant’s Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6.1 Deduction from the Simply Supported Beam Solutions . . . . . . . 2.6.2 Eigensolutions for Isotropic/Orthotropic Rectangular Strip . . . . 2.6.3 Implications of St.Venant’s Principle . . . . . . . . . . . . . . . . . . . . . . . Factors Governing Shear Deformation Effect . . . . . . . . . . . . . . . . . . . . . . . 2.7.1 Clamped Edge Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7.2 Localised Loading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . What is a Long Beam? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

40 45 50 54 56 56 58 63 63 64 66 67 68

3 Plane Problems in Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Field Equations in Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.1 Equilibrium Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.2 Constitutive Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.3 Strain-Displacement Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.4 Compatibility Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.5 Equations for the Axisymmetric Problem . . . . . . . . . . . . . . . . . . . 3.2 Circular Cylinder Under Internal and External Pressure (Lamé’s Problem) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Some Special Cases of Lamé’s Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.1 Solid Cylinder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.2 Externally Pressurised Cylinder with a Pin-Hole . . . . . . . . . . . . . 3.3.3 Pressurised Hole in an Infinite Body . . . . . . . . . . . . . . . . . . . . . . . 3.4 Isotropic Plate with a Circular Hole (Kirsch’s Problem) . . . . . . . . . . . . . . 3.5 Some Similar Stress Concentration Problems . . . . . . . . . . . . . . . . . . . . . . . 3.5.1 Isotropic Plate with an Elliptic Hole (Inglis’ Problem) . . . . . . . . 3.5.2 Orthotropic Plate with a Circular Hole . . . . . . . . . . . . . . . . . . . . . 3.6 Violation of Principle of Complementary Shear . . . . . . . . . . . . . . . . . . . . . 3.7 Linear Crack in an Isotropic Plate (Williams’ Solution) . . . . . . . . . . . . . . 3.8 Stresses Under Concentrated Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.8.1 Concentrated Normal Force on the Boundary of a Half-Plane (Flamant’s Problem) . . . . . . . . . . . . . . . . . . . . . . . 3.8.2 Concentrated Load on an Isotropic Simply Supported Beam (Wilson-Stokes Method) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.9 Frictionless Contact Between Isotropic Cylinders (Hertz Problem) . . . . .

69 69 69 71 71 72 73

2.4 2.5 2.6

2.7

2.8 2.9

74 78 78 79 80 81 84 84 86 87 89 95 96 100 104

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3.10 Bending of a Semicircular Beam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.10.1 Load Case 1: End Moments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.10.2 Load Case 2: End Shear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.10.3 Calculation of the Transverse Normal Stress . . . . . . . . . . . . . . . . 3.11 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

110 110 113 115 117

4 Torsion of Non-circular Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Formulation for Isotropic Shafts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1.1 Displacement Approach (St. Venant’s Warping Function Formulation) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1.2 Stress Approach (Prandtl’s Stress Function Formulation) . . . . . . 4.2 Solutions for Isotropic Simply-Connected Domains . . . . . . . . . . . . . . . . . . 4.2.1 An Elliptical Section . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.2 The Special Case of a Circular Section . . . . . . . . . . . . . . . . . . . . . 4.2.3 An Equilateral Triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.4 A Rectangular Section . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.5 A Comparison of Various Shapes . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Membrane Analogy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 Approximate Analysis of Thin-Walled Open Sections . . . . . . . . . . . . . . . . 4.5 Multiply-Connected Domains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5.1 Formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5.2 Some Simple Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5.3 Approximate Analysis of Thin-Walled Tubes (Bredt-Batho Theory) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5.4 Closed Tube versus Slit Tube . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6 A Brief Discussion of Some Geometrically Complicated Sections . . . . . 4.6.1 Protruding Sharp Corners . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6.2 Re-Entrant Corners . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6.3 Slots and Grooves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6.4 Eccentric Hole in a Circular Section . . . . . . . . . . . . . . . . . . . . . . . 4.6.5 Effect of the Hole Shape . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6.6 Optimum Shape of a Hollow Section . . . . . . . . . . . . . . . . . . . . . . . 4.6.7 Irregular Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.7 Orthotropic Shafts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.7.1 Rectilinear Orthotropy—Simple Solutions . . . . . . . . . . . . . . . . . . 4.7.2 Rectlilinear Orthotropy—Laminated Rectangular Shaft . . . . . . . 4.7.3 Rectilinear Orthotropy—Thin-Walled Open Sections . . . . . . . . . 4.7.4 Shape-Intrinsic Orthotropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.8 Effect of Warping Restraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.9 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

119 119 120 122 127 127 130 131 133 135 136 138 139 139 143 145 147 148 148 149 150 151 152 153 154 155 155 158 164 166 167 171

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5 Some Other Problems of Interest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Concentrated Normal Force on the Boundary of an Isotropic Half-Space (Boussinesq’s Problem) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Frictionless Contact Between Two Isotropic Spheres (Hertz Problem) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Free Edge Phenomenon in Composite Laminates . . . . . . . . . . . . . . . . . . . . 5.4 Functionally Graded Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

173

Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

183

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

195

173 175 176 179

About the Authors

Prof. K. Bhaskar has been with the Department of Aerospace Engineering, Indian Institute of Technology, Madras, since 1992. As part of the Structures Group, he teaches courses related to Solid Mechanics and Elasticity. His research contributions are primarily related to theoretical modelling of thick laminated structures, with about fifty publications in refereed journals. Prof. T. K. Varadan was with the Department of Aerospace Engineering, Indian Institute of Technology, Madras, for about thirty-five years before retiring in 2001. Besides teaching a wide variety of courses related to Structural Mechanics and Aircraft Design, he has made significant research contributions in the areas of Nonlinear Vibrations and Composite Structures, with more than one hundred publications in refereed journals.

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1

Introduction and Mathematical Framework

The theory of elasticity is a mathematical tool that enables the structural analyst to model a structure more realistically than the conventional engineering theories based on the mechanics-of-materials approach. However, the purpose behind the analysis remains the same in either case—to determine the displacements, strains and stresses at every point of the three-dimensional domain of the structure at every instant of time. In this chapter, the difference between the theory of elasticity (TOE) and the conventional engineering theories (CET) is first explained, following which the equations pertaining to the TOE model are presented with an appropriate discussion of their implications. A clear description of material orthotropy is also included with examples.

1.1

Theory of Elasticity versus Conventional Engineering Theories

To understand the difference between these approaches, consider the example problem of a long slender beam. The beam is homogeneous, prismatic with a monosymmetric cross-section, and is made up of an isotropic, linearly elastic material. It undergoes small deformations due to static transverse loading with the resultant at any section passing through the plane of symmetry. The conventional analysis of this problem using the engineering beam theory is based on the following assumptions: (a) All the normal stresses except the longitudinal bending stress are negligible. Further, this non-zero stress does not vary along the width of the beam.

© The Author(s) 2023 K. Bhaskar and T. K. Varadan, Theory of Isotropic/Orthotropic Elasticity, https://doi.org/10.1007/978-3-031-06345-9_1

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1 Introduction and Mathematical Framework

(b) Plane cross-sections remain plane and normal to the deformed axis. (Euler-Bernoulli hypothesis) (c) The transverse displacement at any section does not vary through the thickness (or depth) of the beam. The analysis is actually carried out by reinterpreting the second assumption above as one of linear variation of the longitudinal strain through the depth of the beam and by considering the static equivalence of the corresponding linear bending stress distribution and the resultant bending moment at any cross-section. Further, from purely equilibrium considerations, a transverse shear stress distribution that is consistent with the linear variation of the bending stress is derived and related to the shear force at any section. If one has to analyse the above problem using the TOE approach, all the three assumptions listed above have to be discarded. In general, any assumption of the CET approach is with respect to either neglect of some stress, strain or displacement component compared to others, or the variation of some stress, strain or displacement component with respect to a coordinate direction. Such assumptions are based on pure engineering intuition (for instance, the ones listed above are easily felt to be true for a long slender beam) and are an essential feature of the CET approach, while they have no place in the TOE approach. The TOE approach is based on just a few physically verifiable facts— the constitutive (i.e. stress vs. strain) behaviour of the material, the preservation of the continuum nature of the loaded body, and of course, the basic laws of equilibrium. For example, the following statements, valid for small deformations of any structure made of a Hookean material, form the basis of the linear theory of elasticity: (a) Any stress, when applied alone, causes a corresponding strain which varies in direct proportion to the stress; (b) When the structure is unloaded, it regains its original size and shape; (c) The deformation of the body due to the applied loads is such that no internal voids or cracks can develop. Thus, the displacements are continuous, single-valued functions of the coordinates and one can hence define unique displacement derivatives at any point of the domain. The physical significance of such derivatives is then explained by interpreting them as components of the state of strain; (d) The body as a whole as well as every portion of it is in equilibrium such that the forces, including those due to inertia, in any direction and the moments about any direction sum to zero (essentially Newton’s second law of motion); (e) The body undergoes small deformations such that the laws of equilibrium can be written down with reference to the undeformed geometry and the expressions for the strains in terms of the displacement derivatives do not involve nonlinear terms.

1.2

Field Variables of the Theory of Elasticity

3

The above statements are taken for granted throughout this book. (It should be mentioned that all the above are also an essential component of every linear conventional engineering theory, though they may not always be stated so explicitly.) In place of statement (a) above, one can have a nonlinear constitutive law (referred to as material nonlinearity), and in place of statement (e), one can account for large deformations by employing nonlinear strain-displacement relationships (referred to as geometric nonlinearity); either of these changes results in a nonlinear theory of elasticity which is beyond the scope of the present book. Finally, a brief mention of the need for the TOE approach is in order. As is clear from those of the beam theory put forth as an example, intuitive assumptions are possible only when the structure is nearly one-dimensional or two-dimensional. When all the three dimensions are comparable, one cannot simplify the problem and the TOE approach has to be resorted to; as examples, one can cite the cases of ball and roller bearings. However, even for nearly one- or two-dimensional structures, the TOE approach is called for: (a) when there are sudden discontinuities of geometry like holes, fillets, keyways and notches which lead to stress concentration effects that cannot be captured using the CET approach; (b) when one needs local stresses near supports or points of loading; (c) to establish the limits of applicability of the conventional engineering theories, for instance, to precisely define a “long” beam; (d) to guide the development of non-classical engineering theories based on realistic assumptions regarding displacements and stresses.

1.2

Field Variables of the Theory of Elasticity

As stated earlier, the job of the structural analyst is to determine the deformation and state of stress at every point of the structure. Considering deformation first, the relevant parameters are the displacement vector of any point with respect to its undeformed position and the change of size and shape of any portion of the structure with respect to its undeformed state. While the displacement vector can be completely specified by its three components along with a conveniently chosen set of coordinate axes, say u, v and w with respect to a Cartesian x-y-z system, the deformed shape and size require a more elaborate specification. At this stage, it is sufficient to understand that this state of strain at a point is completely specified in terms of six quantities with reference to any chosen coordinate system—namely, three normal strains εxx , εyy , εzz , and three engineering shear strains γ yz , γ xz , γ xy . These are explained below. At the point of interest P, consider a small element of length Δx oriented along the x-direction in the undeformed state. As the body deforms, the length of the element is

4

1 Introduction and Mathematical Framework

Fig. 1.1 Undeformed (—) and deformed (– –) elements

changed to Δx’. (Its physical position and orientation are also changed (Fig. 1.1), but that is of no interest at the moment.) The normal strain εxx is defined as the change in length per unit length of this element as its original size goes to zero in the limit Δx ' − Δx Δx→0 Δx

εx x = lim

(1.1)

In a similar fashion, εyy and εzz are defined by considering elements oriented along the y- and z-directions, respectively, in the undeformed state. It is important to note these are algebraic quantities, and as per the definition given above, are positive when the element is elongated and negative when compressed. To define the shear strain γ xy at any point of interest, one has to consider two infinitesimally small elements originally parallel to the positive x- and y-directions at that location. As the body deforms, the included angle between the two elements is changed from the original right angle to θ xy radians as shown (Fig. 1.1). Once again, the location and the actual orientation of the deformed elements are not of interest. The engineering shear strain γ xy is defined as γx y =

π − θx y 2

(1.2)

which, once again, is an algebraic quantity. The other two shear strains γ xz and γ yz are defined in a similar fashion, by considering the angle between infinitesimally small elements oriented originally along positive x- and z-directions, and along positive y- and z-directions, respectively.

1.2

Field Variables of the Theory of Elasticity

5

Fig. 1.2 Definition of σ xx , τ xy , τ xz

Regarding the state of stress at a point, it is sufficient to understand that it is completely specified in terms of nine quantities with reference to any chosen coordinate system— namely, three normal stresses σ xx , σ yy , σ zz and six shear stresses τ yz , τ zy , τ xz , τ zx, τ xy and τ yx . All these stem from the rudimentary definition of stress as the internal force per unit area and are explained below. Let P be the point of interest within the loaded body. By passing a plane perpendicular to the x-axis through P, let the body be cut into two free bodies; then, for the equilibrium of the free bodies, a distribution of internal forces should be transmitted through the interface from one free body to the other. Let ΔF be the net force transmitted through a small area ΔA of the interface around P (Fig. 1.2). (Only a net force on the small area is accounted for, but no net moment, because any such moment becomes zero as the area tends to zero in the limit; see Footnote 1.) If the components of this force along the coordinate axes are ΔF x , ΔF y , ΔF z , then one can define the normal stress σ xx and the shear stresses τ xy and τ xz at P as ΔFx ; ΔA→0 ΔA

σx x = lim

ΔFy ; Δ A→0 Δ A

τx y = lim

ΔFz Δ A→0 ΔA

τx z = lim

(1.3)

In the same manner, one can define σ yy , τ yx , τ yz by passing a cutting plane normal to the y-axis through P, and σ zz , τ zx , τ zy when the cutting plane is normal to the z-axis. By now, it will be clear to the reader that the first subscript of any stress component refers to the normal to the plane on which it acts and the second subscript to its direction of action. The state of stress at P is then specified by writing down the components in a convenient matrix form as ⎡ ⎤ σx x τx y τx z ⎢ ⎥ ⎣ τ yx σ yy τ yz ⎦ τzx τzy σzz

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1 Introduction and Mathematical Framework

Fig. 1.3 Positive stress components

All the stress components are algebraic quantities and the accepted sign convention is best understood by considering an infinitesimally small cubic free body at the point of interest. Such a free body subjected to positive stresses is shown in Fig. 1.3. (Such a figure is often referred to as a stress diagram.) Any stress component, acting on a plane whose outward normal is along a positive coordinate direction, is taken to be positive when it acts along a positive coordinate direction; on a plane with the outward normal along a negative coordinate direction, a stress component is positive when it acts along negative coordinate direction. It can be easily seen that this sign convention implies that tensile and compressive normal stresses are positive and negative, respectively; further, when the sign of any shear stress component is reversed, it simply means that the action of a couple of those stresses on two parallel faces of the free body is to rotate it in an opposite sense. Let us consider the equilibrium of the free body in Fig. 1.3. While the forces in any direction identically sum to zero, such is not the case with the moments. For instance, by equating the net moment about the z-axis to zero, one gets τx y = τ yx

(1.4)

1.3 Two-Dimensional Elasticity

7

Thus, these two stress components at any point should have the same values for equilibrium. Similarly, τ xz and τ zx are equal, and τ yz and τ zy are equal. This is often called the principle of complementary shear, which can be stated as1 At any point of a body in equilibrium, a non-zero shear stress component τij always coexists with an equal orthogonal component τji .

Hence, out of the nine stress components mentioned earlier, only six are independent. Thus, the solution to the structural problem involves the determination of fifteen quantities, the so-called field variables, which are all functions of the coordinates x, y, z; they include three displacements, six strain components and six stress components. For convenience, the normal stresses are hereafter denoted with single subscripts—i.e. σ x in place of σ xx , etc.; similarly, εx , etc. are used in place of εxx , etc.

1.3

Two-Dimensional Elasticity

The theory of elasticity is inherently three-dimensional in the sense that all the field variables as discussed above are, in general, functions of all the three coordinates for a loaded structure. However, there are certain situations wherein the deformation of the body is invariant along one coordinate direction leading to a set of final equations in terms of just two coordinates; such problems come under the purview of two-dimensional elasticity. One can easily guess that such a simplification requires that one of the dimensions of the body under consideration be either too small or too large as compared to the other two dimensions. This is indeed so, and the corresponding states, to be described in detail below, are referred to as those of plane stress and plane strain, respectively.2

1.3.1

Plane Stress

This is defined as the two-dimensional state of stress wherein the stress components acting in the out-of-plane direction vanish. More specifically, for a state of plane stress in the x-y plane, the components σ z , τ xz and τ yz are zero, while σ x , σ y and τ xy are functions only of x and y. 1 While deriving the above principle, it has been assumed that there are no body moments, such

as, for example, that produced by a magnetic field on a magnet. In practice, such body moments are very rarely encountered or insignificantly small and are hence not considered anywhere in this book. However, even when body moments are absent, the principle of complementary shear may sometimes be invalid and such an example is presented later in Sec.3.6. 2 It should be noted there are also other possible two-dimensional problems such as those featuring axisymmetric deformation of axisymmetric bodies, or torsion of long prismatic shafts.

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1 Introduction and Mathematical Framework

This state of stress is true for thin plate-like structures, subjected to in-plane forces alone, as the thickness tends to zero in the limit; in practice, it is taken to be applicable as long as the thickness is quite small compared to the in-plane dimensions. A practical example is the case of a thin rotating disc.

1.3.2

Plane Strain

This is defined as the state of two-dimensional strain wherein the strain components specified with reference to the out-of-plane direction vanish. More specifically, for a state of plane strain in the x-y plane, the components εz , γ xz and γ yz are zero, while εx , εy and γ xy are functions only of x and y. This state of strain is applicable for long prismatic structures restrained at the ends and subjected to loads that act normal to the longitudinal axis and are invariant with respect to it. Practical examples include a long pipe carrying pressurised fluid, a roller bearing, and the culvert of a dam (with the longitudinal ends usually bounded by hillocks).

1.3.3

Field Variables of Two-Dimensional Elasticity

For both plane stress and plane strain problems, with respect to the x-y plane, the field variables are taken to be the two in-plane displacements u, v, the three in-plane strains εx , εy , γ xy , and the three in-plane stresses σ x , σ y , τ xy . All these eight quantities are functions of x and y and are the most important non-zero parameters of interest.

1.4

The Field Equations

The equations relating to the field variables are called field equations which form the mathematical TOE model of the physical structure, and which represent the constitutive relationship of the elastic material, the state of equilibrium of every portion of the loaded structure, and the preservation of the continuum nature of the structure during the process of deformation. These field equations are discussed separately in the following sections, wherein, depending on convenience, some of them are first derived for the special case of two-dimensional elasticity and then extended to the more general case.

1.4 The Field Equations

1.4.1

9

Constitutive Relations—Generalised Hooke’s Law

For Isotropic Bodies Let us recollect that an isotropic body is one for which the elastic properties are the same in all directions. The elastic behaviour of an isotropic body is characterised by just two elastic constants—usually the Young’s modulus E and the Poisson’s ratio μ. The other relevant elastic constants, namely, the shear modulus G and the bulk modulus K, can be expressed in terms of E and μ (see Appendix). Starting with the simple Hooke’s law which states that the strain due to any corresponding stress, when applied alone, varies linearly with it, one has εi =

σi for uniaxial loading in the i direction E

(1.5a)

τi j for pure shear in the i- j plane G

(1.5b)

γi j =

Further, due to Poisson’s effect, one has, for uniaxial loading in the i direction εtransverse = −

μσi E

(1.5c)

For the more general state of stress, wherein all the stress components are simultaneously applied, the net deformation is simply the sum of those due to the individual components. Thus, keeping in mind that normal stresses do not produce shear strains and vice versa (i.e. stretch and shear modes are uncoupled), and that a particular shear stress can produce distortion only in its corresponding plane (i.e. all the shear modes are uncoupled with respect to one another), one can write down the net strains as σx − μ(σ y + σz ) E σ y − μ(σz + σx ) εy = E σz − μ(σx + σ y ) εz = E

εx =

τ yz G τx z = G τx y = G

γ yz = γx z γx y

(1.6)

Sometimes, it is necessary to have the stresses in terms of the strains, and this can be done easily by inverting the above relations. The resulting equations are most concisely given as σi = 2Gεi + λe τi j = Gγi j

for i = x, y, z

for i = x, y, z & i / = j

(1.7)

where G and λ (called Lamé’s constants, G being the shear modulus while λ is just mathematically defined) are related to E and μ as

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1 Introduction and Mathematical Framework

G=

E μE ; λ= 2(1 + μ) (1 + μ)(1 − 2μ)

(1.8)

and e = εx + ε y + εz . (It is easy to figure out that e is the volumetric strain, i.e. change in volume per unit volume, by writing down the volume of a cuboid as lbh, taking logarithms and then differentials.) Equations (1.7) are referred to as the generalised Hooke’s law and represent the most general constitutive relations between the stresses and strains for an isotropic linearly elastic body. For Orthotropic Bodies It is first necessary to understand orthotropy. What we mean by an orthotropic material is one for which the elastic properties are direction-dependent but there are three specific orthogonal directions with respect to which stretch and shear modes are uncoupled and also the three shear modes are uncoupled from one another. In other words, if the coordinate axes x, y, z are aligned with these specific directions called the principal material directions, then we have, for general three-dimensional loading, the following stress-strain behaviour: (a) Due to a normal stress (σ x or σ y or σ z ) when acting alone, there is a corresponding linearly proportional normal strain (εx or εy or εz , respectively) and two linearly proportional transverse strains due to Poisson’s effect (εy and εz , or εx and εz , or εx and εy , respectively), but no shear strains; further, all the constants of proportionality (i.e. Young’s moduli and Poisson’s ratios) in the above relationships are not the same with reference to the three directions, but different. (b) Due to a shear stress (τ yz or τ xz or τ xy ) when acting alone, there is a corresponding linearly proportional shear strain (γ yz or γ xz or γ xy, respectively), but no normal strains; further all the constants of proportionality (i.e. the shear moduli) in the above relationships are different in the three different planes. It should be noted that the above features are true only with respect to the principal material directions; with reference to an arbitrary set of reference coordinates, the behaviour turns out to be quite complicated in that the different stretch and shear modes are all coupled. In this book, we shall deal with only such problems where it is convenient to use the principal material directions as the reference axes. The physical implication of this restriction is explained below. Consider some examples of material orthotropy. Wood is a natural orthotropic material with different properties along and across the grains. As far as man-made materials are concerned, one can give several examples—a composite material which is essentially a soft tough plastic reinforced by thin strong fibres running parallel to each other in one or more mutually orthogonal directions, a stiffened plate with closely spaced stiffeners in one

1.4 The Field Equations

11

Fig. 1.4 Examples of specially orthotropic bodies

or two orthogonal directions (idealised as homogeneous but orthotropic), rolled sections which are stiffer in the direction of rolling and so on. In many applications involving such materials, it is common to find that the orthotropic material directions coincide with one or more edges of the structure—for example, a rectangular plate with stiffeners parallel to one set of edges or a circular composite plate with fibres along the circumferential direction (Fig. 1.4); in such cases, the principal material directions are the natural choice for convenient coordinate axes. While such applications are most commonly encountered, it is not totally unusual to have the directions of orthotropy quite different from the natural body coordinates of the structure, for instance, in a composite cylindrical shell with the fibres aligned in a helical direction; however, such cases are rather difficult to analyse because of the different types of coupling as explained earlier and are beyond the scope of the present book. Confining attention to bodies with their material axes coinciding with the chosen x-y-z axes (generally referred to as specially orthotropic configurations), it is first necessary to identify the elastic constants. As already pointed out, the Young’s moduli, shear moduli and Poisson’s ratios are different in the different coordinate directions/planes. While the Young’s moduli E x , E y and E z and the shear moduli Gyz , Gxz , Gxy are easily identified, the Poisson’s ratios need to be specified more carefully. This is because the transverse strain in the y-direction due to uniaxial stress in the x-direction, for instance, is in terms of a different Poisson’s ratio than that in the x-direction due to uniaxial stress in the y-direction. Calling them μxy and μyx , respectively, one has μx y = −

εy when σx alone is applied εx

(1.9a)

μ yx = −

εx when σ y alone is applied εy

(1.9b)

with entirely different values for the two. However, they are not independent of each other; to prove this, let us rewrite the above relations as

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1 Introduction and Mathematical Framework

ε y = −μx y εx = −

μx y σx Ex

i.e., μx y εy =− when σx alone is applied σx Ex

(1.10a)

μ yx εx =− when σ y alone is applied. σy Ey

(1.10b)

and similarly,

By virtue of Maxwell’s reciprocal theorem, the above two left-hand side ratios have to be equal, and hence μ yx μx y = Ex Ey

(1.11)

If E x > E y , μxy is called the major Poisson’s ratio because it is larger than its counterpart μyx , which is called the minor Poisson’s ratio. Similar to Eq. (1.11), one has relations between the Poisson’s ratios in the other two sets of coordinate axes, as given by μx z μzx = ; Ex Ez

μ yz μzy = Ey Ez

(1.12)

Thus, one can conclude that there are nine independent elastic constants for an orthotropic body, which, with reference to the principal material directions x,y,z, can be taken as E x , E y , E z , Gyz , Gxz , Gxy , μyz , μzx and μxy . In terms of these elastic constants, proceeding as was done for the isotropic case, the strain-stress relations for a general triaxial state of stress can be written as μ yx σ y μzx σz σx − − Ex Ey Ez σy μzy σz μx y σx εy = − − Ey Ez Ex μ yz σ y μx z σx σz − − εz = Ez Ex Ey εx =

τ yz G yz τx z = Gxz τx y = Gxy

γ yz = γx z γx y

(1.13)

Inverting the above, one can get the stress-strain relations which are best expressed in terms of the stiffness coefficients C ij of the material as ⎧ ⎫ ⎡ ⎫ ⎡ ⎫ ⎤⎧ ⎫ ⎧ ⎤⎧ ⎪ C11 C12 C13 ⎪ C44 0 0 ⎪ ⎨ εx ⎪ ⎨ γ yz ⎪ ⎨ σx ⎪ ⎨ τ yz ⎪ ⎬ ⎬ ⎪ ⎬ ⎬ ⎢ ⎢ ⎥ ⎥ (1.14a) σy = ⎣ C22 C23 ⎦ ε y ; τx z = ⎣ C55 0 ⎦ γx z ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎩ ⎩ ⎪ ⎩ ⎭ ⎭ ⎪ ⎭ ⎭ σz sym. C33 εz τx y γx y sym. C66

1.4 The Field Equations

13

where C11 =

1 − μx y μ yx 1 − μ yz μzy 1 − μx z μzx ; C22 = ; C33 = E y Ez

Ex Ez

Ex E y

μ yx + μzx μ yz μx y + μx z μzy = C12 = E y Ez

Ex Ez

μzx + μ yx μzy μx z + μx y μ yz = C13 = E y Ez

Ex E y

μzy + μx y μzx μ yz + μ yx μx z C23 = = Ex Ez

Ex E y

(1.14b)

C44 = G yz ; C55 = G x z ; C66 = G x y (1 − μx y μ yx − μ yz μzy − μx z μzx − 2μ yx μzy μx z )

= Ex E y Ez Obviously, this orthotropic constitutive law is more complicated than its isotropic counterpart (Eq. 1.7), but it is very important to realise that the complexity is mathematical rather than physical and that the specially orthotropic body behaves in a qualitatively similar manner as its isotropic counterpart, without any unusual coupling effects; the only difference is that the stiffness coefficients are now direction-dependent. For Plane Stress Considering a state of plane stress in the x-y plane, we need a relationship between the three non-zero stresses σ x , σ y , τ xy and the corresponding strains εx , εy , γ xy . Before proceeding to obtain this relationship, it is important to realise that apart from the above three strain components, the component εz , i.e. that corresponding to the change in thickness of the plate loaded by in-plane forces, is also non-zero because of Poisson’s effect. However, this component is not of great significance (though not negligible) because the thickness itself is very small compared to the in-plane dimensions. The three-dimensional strain-stress relations of an isotropic body (Eq. 1.6), when specialised for this case of plane stress in the x-y plane, yield σ y − μσx τx y σx − μσ y ; εy = ; γx y = E E G −μ(σx + σ y ) εz = ; γ yz = 0; γx z = 0 E

εx =

(1.15)

Inverting the first three relations, one gets E (εx + με y ) (1 − μ2 ) E σy = (ε y + μεx ) (1 − μ2 ) σx =

τx y = Gγx y

(1.16)

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1 Introduction and Mathematical Framework

which is referred to as the plane-stress-reduced constitutive law or simply the plane-stress law. For specially orthotropic bodies, the plane stress law is derived in a similar fashion, as is clear from the following equations: εx =

μ yx σ y σy μx y σx τx y σx − ; εy = − ; γx y = ; Ex Ey Ey Ex Gxy μ yz σ y μx z σx εz = − − Ex Ey   [ ]  σx Q 11 Q 12 εx = ; τx y = Q 66 γx y σy Q 12 Q 22 εy

(1.17)

(1.18a)

where Qij are the plane-stress-reduced stiffness coefficients given by Ey Ex ; Q 22 = ; (1 − μx y μ yx ) (1 − μx y μ yx ) μ yx E x μx y E y = = ; Q 66 = G x y (1 − μx y μ yx ) (1 − μx y μ yx )

Q 11 = Q 12

(1.18b)

The plane stress law (Eq. 1.16) for the isotropic body can also be rewritten in the form of Eq. (1.18a) with Qij as given by Q 11 = Q 22 =

μE E E ; Q 12 = ; Q 66 = G = (1 − μ2 ) (1 − μ2 ) 2(1 + μ)

(1.19)

For Plane Strain The stress-strain law for this case, with z as the out-of-plane direction, is simply obtained from the generalised Hooke’s law (Eq. 1.7) by putting εz = γ xz = γ yz = 0, to yield σx = 2Gεx + λ(εx + ε y ) σ y = 2Gε y + λ(εx + ε y )

(1.20)

τx y = Gγx y i.e., E [(1 − μ)εx + με y ] (1 + μ)(1 − 2μ) E σy = [(1 − μ)ε y + μεx ] (1 + μ)(1 − 2μ)

σx =

τx y = Gγx y

(1.21)

1.4 The Field Equations

15

It is important to note that apart from these stresses, the component σ z is also present here, as given by εz =

σz − μ(σx + σ y ) = 0, i.e. σz = μ(σx + σ y ) E

(1.22)

This component of stress can be easily visualised as that required to prevent the change in length that would occur due to the Poisson effect of the stresses σ x and σ y , and is actually applied as reactions from the restrained ends of the long body. For specially orthotropic bodies, the plane strain constitutive law is simply that of Eqs. (1.14a and 1.14b) for the three-dimensional body, with εz = γ xz = γ yz = 0. Once again, there is a non-zero σ z as given by εz =

μ yz σ y σz μx z σx − − =0 Ez Ex Ey

and hence σz =

E z μ yz σ y E z μx z σx + = μzx σx + μzy σ y Ex Ey

(1.23)

Sometimes, it is necessary to have a strain-stress relationship for this case. This is obtained from the corresponding three-dimensional equation (Eq. 1.6 or 1.13) with the substitution of σ z from Eq. (1.22) or (1.23) above, as   [ ]  εx β11 β12 σx = ; γx y = β66 τx y (1.24a) εy β12 β22 σy where β ij are the plane-strain-reduced compliance coefficients given by ) ( 1 − μ2 −μ(1 + μ) 1 β11 = β22 = ; β12 = ; β66 = E E G

(1.24b)

for the isotropic case, and ) ( 1 − μ yz μzy 1 (1 − μx z μzx ) ; β22 = ; β66 = Ex Ey Gxy ) ) ( ( − μx y + μx z μzy − μ yx + μzx μ yz β12 = = Ey Ex

β11 =

for the orthotropic case.

(1.24c)

16

1.4.2

1 Introduction and Mathematical Framework

Equations of Equilibrium

The second set of field equations accounts for the equilibrium of every point of the structure. These are derived below, first for the simple case of plane stress with reference to x-y coordinates. Consider a thin plate in equilibrium under the action of some applied in-plane tractions (or forces) and support reactions on the boundary contour as well as some in-plane body forces (due to inertia such as weight or centrifugal forces); all these loads do not vary through the small thickness. At any internal point of the plate, consider a small rectangular free body of dimensions Δx, Δy (and of course, the entire thickness, say h) as shown (Fig. 1.5). On each of the four exposed faces of this free body, the normal and shear stresses exerted by the adjoining structure are shown. Since the body is of finite size, the stresses on opposite faces are not exactly equal but have small differences which are denoted by Δσ x , etc. It should be noted that all the stresses, though shown along their positive directions, may actually be either positive or negative; further, the small changes are taken to occur as one moves by the distance Δx or Δy in the positive x- or y-direction. In addition, the free body of Fig. 1.5 is subjected to the body forces which can be generally specified as f x (x, y), f y (x, y) per unit volume along the x-, y-directions, respectively. Now, for the equilibrium of forces in the x-direction, one has Δσx hΔy + Δτ yx hΔx + f x hΔxΔy = 0 i.e., Δτ yx Δσx + + fx = 0 Δx Δy

Fig. 1.5 Derivation of equilibrium equations

1.4 The Field Equations

17

i.e., ∂τ yx ∂σx + + fx = 0 ∂x ∂y

(1.25)

as the size of the free body approaches that of a point in the plan view. Denoting derivatives by a subscript comma, and using the principle of complementary shear (τ xy = τ yx ), the above equation can finally be written as σx,x + τx y,y + f x = 0

(1.26a)

Similarly, by summing forces in the y-direction, one gets τx y,x + σ y,y + f y = 0

(1.26b)

Equations (1.26a and 1.26b) have to be satisfied at every internal point of the x-y domain of the thin plate in equilibrium. In the above derivation, moment equilibrium is not explicitly considered because it is automatically satisfied at every point once the principle of complementary shear is invoked. If one goes through a similar derivation for a thin slice cut out from a long body in plane strain, it is easy to see that the final equilibrium equations turn out to be the same as for plane stress—the only difference is the presence of two equal and opposite forces due to σ z on the two lateral faces of the thin slice. Extending the above derivation for a three-dimensional body and considering the equilibrium of forces acting on a small cuboidal element, one can easily derive the following equations: σx,x + τx y,y + τx z,z + f x = 0 τx y,x + σ y,y + τ yz,z + f y = 0

(1.27)

τx z,x + τ yz,y + σz,z + f z = 0 which should be satisfied at every internal point of the three-dimensional structure in equilibrium. Finally, it should be kept in mind that all the above final equations simply represent a state of equilibrium and are hence valid not just for isotropic and orthotropic elastic structures but for all cases irrespective of the material constitutive law.

1.4.3

Strain–Displacement Relations

These constitute the third set of field equations which are purely kinematic relations that reflect the continuum nature of the deformed body. Starting once again with a simple

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1 Introduction and Mathematical Framework

two-dimensional state—that of plane strain this time, let us focus attention on a small element AB of length Δx originally along the x-direction. After deformation, as AB moves to the position A’B’ (with no out-of-plane motion because of the assumed plane strain conditions), let the displacement components of A be u and v, while those of B be u + Δu and v + Δv, as shown (Fig. 1.6a). Confining attention to small deformations, the final orientation of A’B’ is so close to the original horizontal that its length can be taken as nearly equal to its projected length along the x-direction. Thus, the strain εx is given by εx = lim

AB→0

A' B ' − AB (Δx + u + Δu − u) − Δx ≅ lim Δx→0 AB Δx

i.e. εx = u ,x

(1.28a)

Similarly, by considering the deformation of an initially vertical element, one gets ε y = v,y

(1.28b)

To obtain the expression for the shear strain γ xy , one needs to look at the change in angle between two elements originally along the positive x and y directions. With respect to Fig. 1.6b, this strain is given by γx y = α + β = lim

Δx→0

Δu Δv + lim Δx Δy→0 Δy

i.e. γx y = u ,y + v,x

Fig. 1.6 Derivation of strain-displacement relations

(1.28c)

1.5

Stress Approach—Compatibility Equations

19

Equations (1.28a–1.28c) are valid for plane stress as well as plane strain. The only difference between the two is that there is no out-of-plane displacement for the planestrain problem considered above. i.e. w(x, y, z) = 0

(1.29)

If the problem were to be one of plane stress—i.e. a thin plate subjected to in-plane loading, then, the displacement w would be zero only for the mid-plane, but not for points away from it. However, as the thickness itself is quite small, this variation of w is not of any interest. Coming to the most general case of deformation of a three-dimensional body, it is necessary to invoke the assumption of small deformations once again to argue that all the orientations of deformed elements with respect to their undeformed directions are very small so that one can study the deformation of an element, say, originally in the x-y plane, by simply considering the projection of its deformed length on the x-y plane. Thus, one uses figures identical to those used for the plane strain case above and ends up with the same three strain-displacement relations. Considering the deformations in x-z and y-z planes in an analogous fashion, one gets three more relations, as given by εz = w,z ; γx z = u ,z + w,x ; γ yz = v,z + w,y

(1.30)

One way in which the significance of the strain-displacement relations can be explained is to say that they are merely a statement that unique displacement derivatives exist for a continuum that deforms elastically, and that there is a physical deformation mode associated with each such displacement derivative. Finally, it should be noted that since all the above strain-displacement equations are derived from the considerations of kinematics of deformation alone, they are valid for structures made up of any material, elastic or otherwise, as long as deformations are small.

1.5

Stress Approach—Compatibility Equations

We have a total of fifteen field equations (three equilibrium, six stress-strain and six strain-displacement equations) for the three-dimensional elasticity problem involving fifteen field variables. For plane stress or plane strain, both the field variables and field equations (two equilibrium, three stress-strain and three strain-displacement equations) are eight in number. The above field equations, supplemented by boundary conditions which are a statement of the loading and support conditions of the structure, are sufficient to obtain a solution for the structural problem.

20

1 Introduction and Mathematical Framework

This solution is commonly obtained in two distinct ways: (a) the first, called the stress approach, wherein the stress field is first solved for, and then the strains and displacements are obtained using the constitutive law and by integrating the strain-displacement relations; (b) the second, called the displacement approach, wherein the displacement field is solved for and then strains and stresses are evaluated using strain-displacement relations and constitutive law. While one would expect the equilibrium equations to be the logical starting point for the solution using the stress approach, it is important to know that they are not sufficient to arrive at the correct state of stress in the loaded structure. This is because they form an under-determined system of equations leading to an infinite number of stress solutions out of which only one corresponds to actual reality; all the remaining solutions, though satisfying equilibrium, lead to multi-valued displacement fields which imply loss of the continuum nature of the body after deformation. The additional equations that complement the equilibrium equations and lead to the unique, correct solution of the structural problem are called compatibility equations. The above argument can be put forth in a slightly different fashion. Confining attention to a plane problem, let us note that any state of deformation can be described in terms of two independent single-valued displacement fields u(x, y) and v(x, y). By appropriate differentiation, these then yield the three in-plane strains, which, in turn, are related to the three in-plane stresses. Thus, the three in-plane stresses, or strains, are not independent quantities that can be specified separately, but are related to each other; such a constraining relationship, required to ensure single-valued displacement fields, is called a compatibility equation. A physical explanation of the need for a compatibility equation is provided in Fig. 1.7. If the three in-plane strains of the plane problem are independently specified, then it may

Fig. 1.7 (a) Compatible and (b) incompatible deformations

1.6

Displacement Approach—Navier Equations

21

lead to independent deformations of originally adjacent elements as shown so that they cannot be fitted together as in a jig-saw puzzle without creating an overlap or a void between them. If, on the other hand, the three strains satisfy the compatibility equation, then the deformations of originally adjacent elements are such that they remain adjoining elements with a smooth and continuous interface between them. For the cases of plane stress and plane strain, a single compatibility equation is sufficient. It is obtained by the elimination of displacements from the three strain-displacement equations (which, as we noted earlier, are essentially a statement of preservation of continuum), as given below. ( ) ( ) εx,yy + ε y,x x = u ,x ,yy + v,y ,x x = u ,x yy + v,yx x γx y,x y = (u ,y + v,x ),x y = u ,x yy + v,yx x and hence εx,yy + ε y,x x = γx y,x y

(1.31)

For the general three-dimensional case, there are a total of six compatibility equations, as given by εx,yy + ε y,x x = γx y,x y ε y,zz + εz,yy = γ yz,yz εz,x x + εx,zz = γx z,x z 2εx,yz = (−γ yz,x + γx z,y + γx y,z ),x

(1.32)

2ε y,zx = (−γx z,y + γx y,z + γ yz,x ),y 2εz,x y = (−γx y,z + γ yz,x + γx z,y ),z As with the strain-displacement relations, the above strain compatibility equations are purely kinematic and are hence valid for structures made of any material as long as the deformations are small. It will be necessary to express the compatibility equations in terms of stresses so that they can be solved along with the equilibrium equations for the solution of a structural problem using the stress approach. This, however, will be taken up at an appropriate stage later, because the resulting equations will vary depending on the constitutive law.

1.6

Displacement Approach—Navier Equations

If one wants to attempt a solution for the displacement field first, the logical starting point would be a set of equations in terms of the displacement variables alone. Such a set is

22

1 Introduction and Mathematical Framework

obtained by expressing the equilibrium equations in terms of displacements, which are generally referred to as Navier equations. For the three-dimensional isotropic body, these equations are readily obtained by expressing the stresses in terms of strains using the generalised Hooke’s law (Eq. 1.7) and further in terms of displacements using the strain-displacement relations (Eq. 1.28a– 1.28c, 1.30) and then substituting such expressions in the equilibrium equations (Eq. 1.27). The resulting Navier equations are (G + λ)(u ,x x + v,yx + w,zx ) + G∇ 2 u + f x = 0 (G + λ)(u ,x y + v,yy + w,zy ) + G∇ 2 v + f y = 0

(1.33)

(G + λ)(u ,x z + v,yz + w,zz ) + G∇ w + f z = 0 2

where ∇ 2 ≡ ∂∂x 2 + ∂∂y 2 + ∂∂z 2 is the Laplacian operator. Analogous equations for orthotropic bodies, or the special cases of plane stress or plane strain, can be obtained in a similar manner. Since the displacement approach starts with a solution for the single-valued displacement field, compatibility is automatically taken care of a priori. Thus, the Navier equations form a consistent set and yield a unique solution to the structural problem once the boundary conditions are properly enforced. 2

1.7

2

2

Stress Transformation

It was stated earlier that the state of stress at a particular point is completely specified in terms of six independent components with reference to any convenient set of coordinate axes. The obvious reason for leaving the coordinate system to be chosen arbitrarily and based on one’s convenience is that the stress components with reference to any other coordinate system can then be uniquely determined because there is a definite correspondence between the stress components of the two coordinate systems. The derivation of such a relationship is known as stress transformation. The physical reasoning that results in the transformation equations is essentially a consideration of the equilibrium of a suitably chosen infinitesimally small free body at the point of interest. Without going into the details of such a derivation, which is given in full detail in the Appendix, the final transformation equations are presented below.

1.8

Strain Transformation

23

For two-dimensional transformation from x-y-z system to x’-y’-z system Σ Σ σi j = aik a jl σkl with i , j = x ' , y '

(1.34a)

k=x,y l=x,y

For three-dimensional transformation from x-y-z system to x’-y’-z’ system Σ Σ σi j = aik a jl σkl with i , j = x ' , y ' , z '

(1.34b)

k=x,y,z l=x,y,z

where σ mn is the same as τ mn when m / = n, and amn is the direction cosine of direction m with respect to direction n. (It is customary to use Einstein’s indicial notation and avoid writing the summation signs explicitly in equations such as the above; however, though this is an important aid for concise presentation as well as further manipulation of such equations, it is not used here for the sake of easier understanding and readability.)

1.8

Strain Transformation

The transformation of strains from one set of reference axes to another is quite straightforward, as in the case of stresses, except that one has to consider here the kinematics of deformation of suitably oriented elements. This, once again, is presented in full detail in the Appendix, with the resulting final equations as given below. For two-dimensional transformation from x-y-z system to x’-y’-z system: Σ Σ εi j = aik a jl εkl with i , j = x ' , y '

(1.35a)

k=x,y l=x,y

For three-dimensional transformation from x-y-z system to x’-y’-z’ system: Σ Σ εi j = aik a jl εkl with i , j = x ' , y ' , z '

(1.35b)

k=x,y,z l=x,y,z

where εmn is the not the same as γ mn when m / = n, but defined by εmn =

γmn 2

(1.36)

Thus, one can see that the stress and strain transformation rules are identical provided one uses the mathematically defined εmn instead of the physically defined γ mn ; it should

24

1 Introduction and Mathematical Framework

be kept in mind that εmn , called the tensorial shear strain, is rarely used in any other context.3

1.9

Principal Stresses and Strains

Corresponding to the state of stress at any point, one can identify three mutually orthogonal planes on which the shear stresses vanish—such planes are called the principal planes. The three corresponding normal stresses, called the principal stresses, have the property that they include the largest and the smallest algebraic values of all the normal stresses associated with the infinite number of possible planes passing through that point. Thus, the objective of finding the principal stresses is to determine the largest tensile and compressive stresses acting at a point, so that one can proceed ahead to determine the factor of safety of the loaded structure by comparing it with the corresponding material strength data. From this viewpoint, it should immediately be clear that principal stress analysis is not as meaningful for orthotropic bodies as for isotropic bodies because the material strengths of the former are different in different directions. Often, it is necessary to use the maximum shear stress as the critical design parameter because most of the ductile materials commonly used for structural applications yield considerably before rupturing into two pieces, and this yield behaviour is characterised in terms of shear stresses. Analogous to principal planes for normal stresses, there exist three orthogonal planes, corresponding to the state of stress at a point, on which the net shear stress takes on extreme values. Each of these maximum shear planes is a plane bisecting the right angle between two principal planes; further, the magnitude of the shear stress on each such plane is equal to half that of the difference between the two corresponding principal stresses. It must also be noted that the maximum shear planes are not free of normal stresses. As with stresses, one can also determine the directions in which the normal strain at a point takes on extreme values, or the planes on which the shear strain does so. The equations governing this principal strain analysis are exactly similar to those for stresses.4 The principal axes and the maximum shear planes, for a chosen point, turn out to be the same whether one is interested in stresses or strains, and the extreme values of the tensorial shear strains at a point are related to the principal strains in the same way as specified earlier for stresses. All the equations pertaining to principal stress/strain analysis are developed in the Appendix, and are not presented in the main text because this is often a post-processing

3 Readers familiar with tensors would readily recognise both the state of stress and the state of strain

as second-order tensors, and the transformation rules put forth above as simply that corresponding to such tensors. Towards preserving the introductory, student-friendly nature of this book, we shall not get into such considerations here. 4 Once again, they are general equations applicable to second-order tensors.

1.10

Boundary Conditions

25

step after the stress/strain field for a chosen structural problem has been solved with reference to a convenient set of reference axes. Finally, as mentioned earlier, the analysis for principal stresses or strains is not very meaningful for orthotropic bodies. Instead, one is often concerned about the possibility of structural failure in one of the principal material directions (for instance, due to cracking of the matrix of a unidirectionally reinforced composite parallel to the fibres); hence the stresses and strains have to be determined with reference to these directions and compared with the corresponding allowable values for safe design.

1.10

Boundary Conditions

The field equations, except the constitutive law, are not simple algebraic equations but differential equations that require the specification of boundary conditions to yield a unique solution for a given structural problem. These boundary conditions, which have to be specified for the complete boundary of the domain of the structure, account for the applied external loads and supports. While the field equations account for equilibrium, the continuum nature before and after deformation, and material stress-strain behaviour at every internal point of the loaded body, the boundary conditions serve the same purpose at every point of the bounding surface. In other words, the boundary conditions are simply a statement of the continuous nature of the displacement field between the interior of the body and a supported portion of the bounding surface, or of the continuous nature of the stress field between the interior of the body and a loaded or unrestrained portion of the bounding surface so that infinitesimal free bodies taken on the bounding surface are in equilibrium. Thus, by their very nature, the boundary conditions fall into two distinct categories: (a) essential or geometric boundary conditions, which ensure displacement continuity at the boundary (b) natural or force boundary conditions, which ensure the equilibrium of forces at the boundary. The specification of essential boundary conditions is quite straightforward—for instance, all the displacement field variables have to take on zero values at any rigidly fixed point of the body; the conditions may also be non-homogeneous as in the example case of a long bar fixed at one end and pulled at the other end such that the entire end cross-section moves axially by a specified value. Specification of natural boundary conditions at a point requires the consideration of a suitably chosen infinitesimally small element (with one of its faces on the bounding surface while the other faces are internal surfaces normal to the coordinate axes) and setting up the corresponding equilibrium equations. For instance, with respect to the plane

26

1 Introduction and Mathematical Framework

Fig. 1.8 Specification of natural boundary conditions for (a) a specific load case, (b) general surface tractions

stress problem of a thin triangular plate fixed along one edge, free along the second and subjected to uniform pressure along the third (Fig. 1.8), the natural boundary conditions are to be specified for the second and third edges. For this specific case, a square shape is appropriate for the element to be considered at any point of the free edge, and the corresponding equilibrium equations lead to the following natural boundary conditions: along OB(x = 0, y = 0 to b) :

σx = τx y = 0

(1.37)

For the loaded inclined edge, a triangular element as shown is appropriate, and yields the following conditions: along BA (x = 0 to a, y = b(1-x/a)): σn ' = σx cos2 θ + σ y sin2 θ + 2τx y sin θ cos θ = − p τn ' s ' = (σ y − σx ) sin θ cos θ + τx y (cos2 θ − sin2 θ ) = 0

(1.38)

For a more general case of loading on AB, it is convenient to express it in terms of the x and y components of the surface tractions (or applied loads) per unit area, denoted by T x and T y , respectively. In terms of these, with reference to the same triangular element, one gets

1.11

Summary

27

along AB: σx cos θ + τx y sin θ = Tx τx y cos θ + σ y sin θ = Ty . or, in general lσx + mτx y = Tx lτx y + mσ y = Ty

(1.39)

where l and m are the direction cosines of the outward normal n’ at the chosen boundary point. It can be noted that these general equations automatically yield the simple form of Eq. (1.37) when applied for the free vertical edge of the problem of Fig. 1.8a. Extending the same argument to a three-dimensional body and considering an infinitesimal element of the shape of a tetrahedron at a point of the loaded boundary, one gets lσx + mτx y + nτx z = Tx lτx y + mσ y + nτ yz = Ty

(1.40)

lτx z + mτ yz + nσz = Tz It is quite obvious that the specification of the essential and natural boundary conditions as described above is the same irrespective of the constitutive law of the material. However, it is often necessary to express the natural boundary conditions in terms of the displacement variables and their derivatives while solving the boundary value problem, and then the correct constitutive law as applicable has to be employed.

1.11

Summary

The detailed mathematical formulation of a structural problem within the purview of the theory of elasticity has been presented with reference to Cartesian coordinates. The field equations have been given for both three-dimensional problems and two-dimensional (plane stress/strain) problems, and for isotropic as well as orthotropic bodies. Further, compatibility equations, Navier equations of equilibrium and the specification of boundary conditions have been discussed.

2

Plane Problems in Cartesian Coordinates

This chapter deals with plane problems confined to a rectangular domain, with the focus mainly on problems corresponding to beam flexure. The concept of a stress function is introduced and simple and elegant solutions based on it are discussed. Two alternative elasticity solutions corresponding to the problem of a simply supported beam under uniform load are presented and a comparison of them leads to the enunciation of St.Venant’s principle. The chapter ends with a discussion of the parameters on which the non-classical shear deformation effect depends, and on the range of applicability of the engineering beam theory.

2.1

Airy Stress Function

As stated earlier, a solution of the plane elasticity problem by the stress approach requires that the two equilibrium equations (Eq. 1.26) and the strain compatibility equation (Eq. 1.31) be solved along with appropriate boundary conditions. Obviously, this approach is very appropriate for problems where all or most of the boundary conditions are specified in terms of stresses. The starting point is to employ a two-dimensional function φ, called Airy stress function, defined by φ,yy = σx ;

φ,x x = σ y ;

−φ,x y = τx y

(2.1)

such that the equilibrium equations are satisfied identically when the body forces are absent. Even for a problem with body forces, it is possible to define such a stress function provided the body forces can be derived from a potential function V as

© The Author(s) 2023 K. Bhaskar and T. K. Varadan, Theory of Isotropic/Orthotropic Elasticity, https://doi.org/10.1007/978-3-031-06345-9_2

29

30

2 Plane Problems in Cartesian Coordinates

f x = −V,x ;

f y = −V,y

(2.2)

In that case, the stress function is given by φ,yy = σx − V ;

φ,x x = σ y − V ;

−φ,x y = τx y

(2.3)

With the equilibrium equations satisfied identically, one is left with just one equation to solve—the compatibility equation; this, however, has to be first expressed in terms of stresses/stress function.

2.2

Compatibility Equation in Terms of Stresses and Airy Stress Function

2.2.1

Plane Stress—Isotropic Case

The strain compatibility equation εx,yy + ε y,x x = γx y,x y

(1.31)

along with the strain-stress relations εx =

σ y − μσx τx y σx − μσ y 2(1 + μ)τx y ; εy = ; γx y = = E E G E

(1.15)

yields, (σx − μσ y ),yy + (σ y − μσx ),x x = 2(1 + μ)τx y,x y

(2.4a)

The stress approach requires the solution of the above compatibility equation along with the equilibrium equations. However, it is possible, using the equilibrium equations, to rewrite the above compatibility equation in a more elegant form. To do this, one has to differentiate the first equilibrium equation with respect to x, and the second with respect to y, and sum the two, i.e., ) ) ( ( σx,x + τx y,y + f x ,x + τx y,x + σ y,y + f y,y ,y = 0 to yield 2τx y,x y = −(σx,x x + σ y,yy + f x,x + f y,y ).

2.2

Compatibility Equation in Terms of Stresses …

31

Substitution of the above expression for 2τ xy,xy in Eq. (2.4a) reduces it to ∇ 2 (σx + σ y ) = −(1 + μ)( f x,x + f y,y )

(2.4b)

where ∇ 2 ≡ ∂∂x 2 + ∂∂y 2 is the Laplacian operator. This is the compatibility equation in terms of stresses valid for an isotropic body in plane stress. It can be expressed in terms of φ as 2

2

∇ 4 φ + (1 − μ)∇ 2 V = 0

(2.5)

where ∇ 4 ≡ ∇ 2 .∇ 2 ≡ ∂∂x 4 + 2 ∂ x ∂2 ∂ y 2 + ∂∂y 4 is the biharmonic operator. When the body forces are absent or uniform over the domain, the compatibility equation reduces to the biharmonic form 4

4

4

∇ 4φ = 0

2.2.2

(2.6)

Plane Strain—Isotropic Case

The compatibility equation for this case is derived exactly as done above for plane stress; it is given by ∇ 2 (σx + σ y ) = −

1 ( f x,x + f y,y ) (1 − μ)

(2.7)

i.e. ∇ 4φ +

(1 − 2μ) 2 ∇ V =0 (1 − μ)

(2.8)

Once again, ∇ 4φ = 0

(2.6)

when the body forces are zero or uniform over the domain. Thus, when the body forces are absent or uniform over the domain, the stress function, and hence the stress field, are the same for corresponding problems of plane stress and strain (i.e., with the same two-dimensional geometry and identical traction boundary conditions), though the strain and displacement fields derived thereupon would obviously be different. Further, the stress field is the same irrespective of the elastic constants of the

32

2 Plane Problems in Cartesian Coordinates

material; this feature is the basis on which photoelastic studies carried out with plastic models enable one to find out the stresses in actual metallic structures.1

2.2.3

Plane Stress—Specially Orthotropic Case

The final equation for this case is obtained by writing the strain compatibility equation in terms of stresses using the strain-stress law of Eq. (1.17), and then rewriting it in terms of φ. (No real simplification or elegance of form is achieved by expressing τ xy,xy in terms of the normal stresses as done earlier.) The final equation is ) ( 2μx y (1 − μx y ) (1 − μ yx ) 1 1 1 φ,x x yy + φ,x x x x + − φ,yyyy = − V,yy − V,x x Ey Gxy Ex Ex Ex Ey (2.9)

2.2.4

Plane Strain—Specially Orthotropic Case

For this case, we get the final equation in terms of the plane-strain- reduced compliance coefficients β ij (see Eq. 1.24) as β22 φ,x x x x + (2β12 + β66 )φ,x x yy + β11 φ,yyyy = −(β12 + β22 )V,x x − (β11 + β12 )V,yy (2.10)

2.3

Use of Polynomial Stress Functions

The solution to the plane problem has now been reduced to that of a single compatibility equation in terms of φ. Nevertheless, for most problems, even this single partial differential equation, with appropriate boundary conditions, is not easy to solve if one adopts a rigorous and direct mathematical approach. Instead, it is often convenient to adopt an inverse approach, wherein one examines the stress field corresponding to various simple chosen functions for φ, and, by appropriate superposition of such functions, arrives at a solution for some problem of interest. Confining attention to polynomial functions first, one can immediately see that terms of degree less than two have no significance at all because they lead to zero stress fields everywhere. Further, for problems with zero or constant body forces, the biharmonic equation is identically satisfied by the complete second and third degree polynomials in x and y, and hence each term therein corresponds to a realistic stressed state. For polynomials of a degree more than three, the biharmonic 1 These statements are true only if the body is simply-connected.

2.3

Use of Polynomial Stress Functions

33

equation is satisfied only when the coefficients of the various terms are suitably related to each other. Finally, for obtaining the stress function for some problem of interest, a trial and error approach is necessary to identify the required polynomial terms. However, it has to be emphasised that every structural problem, even when pertaining to a simple rectangular domain, cannot be solved by this approach because all the applied traction boundary conditions cannot be satisfied exactly for most cases. Thus, the trial and error approach is essentially useful to identify the problems that can be solved exactly. Our interest here is not to illustrate the trial and error procedure, but to discuss, in complete detail, some problems amenable to this approach. This is done in the following sections.

2.3.1

The Simple Case of Uniaxial Tension

Let us start with this simple case shown in Fig. 2.1a, with the out-of-plane thickness taken to be very small so that it corresponds to a plane stress problem. The edge conditions are σx = σo , τx y = 0

at x = ±L/2 for all y;

σ y = τx y = 0 at y = ±d/2 for all x

(2.11a)

The chosen stress function should lead to continuity of the above stress values into the interior of the domain, and hence φ,yy = σo , −φ,x y = 0 at x = ±L/2 for all y; φ,x x = −φ,x y = 0 at y = ±d/2 for all x Fig. 2.1 A bar in uniaxial tension

(2.11b)

34

2 Plane Problems in Cartesian Coordinates

Whether one looks at the above conditions together with the biharmonic equation, or takes a cue from the corresponding conventional elementary solution, one is first led to try out the stress function. φ = Ay 2

(2.12)

which indeed turns out to be the correct solution when A = σ o /2. The stress field is obtained as σx = σo , σ y = τx y = 0 for all x, y

(2.13)

The strain field for an isotropic body, for this simple case of uniaxial stress, is εx =

σx σo = ; E E

ε y = −μεx = −μ

σo ; E

γx y =

τx y =0 G

(2.14)

The displacement field is obtained by integration of the strain-displacement equations as σo σo x , and hence, u = + f (y) E E μσo μσo y = εy = − , and hence, v = − + g(x) E E

u ,x = εx = v,y

(2.15)

with the relationship between the unknown functions, f and g obtained from the expression for shear strain as u ,y + v,x = γx y i.e., f ' (y) + g ' (x) = 0. i.e., − f ' (y) = g ' (x)

(2.16)

with the primes denoting appropriate derivatives. The last equation states that the left-hand side which is a function of y alone is equal to the right-hand side which is a function of x alone; this is possible only if each of them is equal to a constant, say k 1 . Proceeding further, one gets f (y) = −k1 y + k2 g(x) = k1 x + k3

(2.17)

2.3

Use of Polynomial Stress Functions

35

where k 2 and k 3 are two more undetermined constants. Thus, the displacement field is obtained in terms of three undetermined constants; as can be seen later, this feature is common for other problems also. The three constants are determined once a set of valid displacement constraints necessary to restrain the three rigid body motions—two translations and one rotation—are imposed; they should be valid in the sense that they should be compatible with the strain field. However, neither the specification of the displacement constraints nor the corresponding solution for k 1 to k 3 is unique. For instance, for the present problem, one can take the two ends x = ± L/2 to be loaded symmetrically by the stress σ o , and then the displacement conditions can be stated as u = 0 at x = 0 for all y v = 0 at y = 0 for all x

(2.18)

since the resulting deformation is symmetric with respect to the axes as shown in Fig. 2.1a. Alternatively, one can take the left end to be held such that u = 0 at x = −L/2 for all y v = 0 at y = 0 for all x

(2.19)

and the right end loaded, so that the deformation is as in Fig. 2.1b. One can easily see that both these specifications are equally valid for the present proboL with lem, though they result in different values for k 1 to k 3 (either all zero, or k2 = σ2E others zero). It should be clear that the difference is only regarding the reference line with respect to which u is measured; apart from this, the two specifications do not really lead to different structural problems with different deformed shapes or states of stress. It is also possible to think of some more valid specifications of displacement conditions; on the other hand, an example of an obviously invalid specification is that corresponding to a rigidly clamped left end, as given by u=v=0

at x = −L/2 for all y

(2.20)

because such a constraint would imply zero εy at this end which is not true for the strain field of Eq. (2.14). Thus, for every stress function, there is a unique stress or strain field, but several possible displacement fields depending on the specification of geometric constraints, with all of them differing from one another only by a rigid body motion. From this viewpoint, it is appropriate to look at any particular φ as uniquely defining a structural problem by

36

2 Plane Problems in Cartesian Coordinates

interpreting it in terms of the corresponding stress field which yields the internal stresses as well as the zero or non-zero tractions on every portion of the boundary; these boundary tractions may be due to actual application of loads or reactions from suitably defined supports. Before proceeding ahead to the next problem, let us note that the extension of the above solution for simple tension to the case of an orthotropic bar is straightforward. Further, for both the isotropic and orthotropic cases, the state of stress or strain is identical to that of the corresponding elementary bar theory solution provided the end tractions are uniformly distributed.

2.3.2

Pure Bending of a Beam

It should be noted that the terms “beam”, “plate”, etc. are applicable only in the CET context wherein intuitive assumptions are made to reduce the problem to one of one or two dimensions. For any such problem, one can think of employing the TOE approach by discarding the intuitive assumptions. However, while doing so, the loads applied on the structure can no longer be specified in terms of integrated quantities or generalised force resultants because such a description is valid only in the CET framework; here, the applied loads have to be specified more carefully as a specific body force or surface traction distribution. Thus, there are an infinite number of elasticity problems that may all correspond to a particular CET problem because there are an infinite number of ways in which a resultant force or moment of specified magnitude and direction may be applied to the body, either as a body force or as a surface traction; for example, corresponding to the CET problem of a uniformly distributed load over the entire length of a prismatic beam, the elasticity problem may be that of a uniform body force in the transverse direction, or a uniform traction applied on the top or bottom surface of the beam or on both surfaces. Of this infinite set of elasticity problems corresponding to a particular CET problem, only a few may be amenable to an exact solution. Such a problem, corresponding to pure bending of a beam, is the topic of this section. The problem is with respect to the rectangular domain shown in Fig. 2.2a, with the out-of-plane thickness being very small and taken as unity without loss of generality; thus it is a case of plane stress. The stress function for the chosen problem is φ = Ay 3

(2.21)

σx = 6Ay; σ y = τx y = 0

(2.22)

which yields the stress field as

On the boundary of the domain, the applied tractions and the corresponding internal stresses have to be in equilibrium; hence the applied loads for this problem are non-zero

2.3

Use of Polynomial Stress Functions

37

Fig. 2.2 Pure bending of a cantilever

normal tractions on the left and right faces alone, and on these faces, they vary linearly from zero at the middle of the depth to a maximum at the top and bottom as shown. Since the resultant of this distribution is a pure moment, the corresponding CET problem is that of pure bending of a beam as shown in Fig. 2.2b; we need a reference line for specification of the displacements, and hence shall take the right end of the beam to be fixed. Before proceeding ahead, it is convenient to express the unknown constant A in terms of the moment M using 

d/2 −d/2

σx |x=0,L ydy = −M 3

(2.23)

3

M , where I = 1 x12d = d12 is the second moment of area of which yields A = − 2M = − 6I d3 the cross-section about the neutral axis. (The minus sign in the above equation is required because the hogging moment M leads to compressive stresses in the positive y region.) Thus the stress field is

σx = −

My ; σ y = τx y = 0 I

(2.24)

The corresponding strain field is εx =

σx −μσx μM y −M y ; εy = = ; = EI E EI E

γx y = 0

(2.25)

Integrating the strain-displacement equations as was done for the earlier problem, one gets u=

−M x y + f (y); EI

and hence γx y = u ,y + v,x =

−M x EI

v=

μM y 2 + g(x) 2E I

+ f ' (y) + g ' (x) = 0.

(2.26)

38

2 Plane Problems in Cartesian Coordinates

i.e., −M x + g ' (x) = − f ' (y) = k1 (say) EI

(2.27)

Thus Mx Mx2 , and hence, g(x) = k1 x + + k2 EI 2E I f ' (y) = −k1 , and hence, f (y) = −k1 y + k3

g ' (x) = k1 +

(2.28)

where k 2 and k 3 are two more undetermined constants. For the present problem, it can easily be verified that the totally restrained edge conditions u=v=0

at x = L for all y

(2.29)

cannot be imposed. Trying out some other such combinations, all of which correspond to a clamped end in the CET context, the appropriate boundary conditions for the present problem can be identified as at x = L :

u = 0 for all y,

v = 0 for y = 0

(2.30)

which serve to restrain the two rigid body translations and also the in-plane rotation. The corresponding values of the undetermined constants are easily obtained as k1 = −

M L2 ML , k2 = , k3 = 0 EI 2E I

(2.31)

A comparison of the present elasticity solution with the CET solution of this problem based on the engineering beam theory is presented in Table 2.1. (For the sake of convenience, the elasticity results are written in such a form that terms common with the CET solution are grouped separately with the additional terms shown in bold font.) From this table, it is seen that the only difference between the two approaches for the present problem is due to the Poisson strain εy which is accounted for in the TOE approach but completely neglected in the CET approach. Due to this so-called thicknessstretch effect (which includes contraction as well), the displacement v actually varies through the depth of the beam; the CET approach, however, leads to a correct prediction of the centroidal deflection curve. In a highly exaggerated form, the qualitative difference between the two approaches is as shown in Fig. 2.3. To get a quantitative feeling of the thickness-stretch effect, let us find out the difference between vtop and vcenter line at the left end as a fraction of the latter. This is given by

2.3

Use of Polynomial Stress Functions

39

Table 2.1 Comparison of results for pure bending of a beam TOE solution σx = −M y/I σ y = τx y = 0

−M y σx = E EI μM y −μσ x = εy = E EI γx y = 0 εx =

−M x y ML + y E I EI ) ( Mx2 M L2 ML x+ + v= − EI 2E I 2E I u=

+

μM y2 2E I

v| y=0 = −

Mx2 M L2 ML x+ + EI 2E I 2E I

Fig. 2.3 Deformation due to pure bending

CET solution σx = −M y/I σ y = 0 (by assumption) VQ =0 τx y = Ib −M y σx = E⎫ EI εy = 0 (by assumption) γx y = 0 εx =

−M x y ML + y EI EI (from u = −yv,x )

u=

Mx2 M L2 ML x+ + EI 2E I 2E I (from E I v,x x = M)

v=−

40

2 Plane Problems in Cartesian Coordinates

( )2 v(0, −d/2) − v(0, 0) d =μ v(0, 0) 2L

(2.32)

which corresponds to a variation of 0.3% for L/d = 5, 0.019% for L/d = 20 and 0.003% for L/d = 50 when μ is taken to be 0.3. Thus, the CET approach is quite accurate for this problem. The above solution and the final conclusions are also valid when the material is orthotropic—the only difference would be that E x and μxy occur in place of E and μ, respectively.

2.3.3

A Tip-Loaded Cantilever

The stress function for a plane stress problem corresponding to a tip-loaded cantilever is ) ( −P 3 (x y 3 − d 2 x y) φ= (2.33) 4 6I where P is the tip force and I = (1)d 3 / 12 with the out-of-plane thickness taken as unity (see Fig. 2.4). The corresponding stress field is given by ( ) −P d 2 −P x y (2.34) σx = ; σ y = 0; τx y = − y2 I 2I 4 which satisfies the following boundary conditions: σ y = τx y = 0 on the top and bottom surfaces (y = ±d/2) ( ) −P d 2 σx = 0, τx y = − y 2 at the left end (x= 0) 4 2I ) ( 2 −P L y −P d σx = , τx y = − y 2 at the right end (x = L) I 2I 4

Fig. 2.4 Tip-loaded cantilever

(2.35)

2.3

Use of Polynomial Stress Functions

41

The resultant of the parabolic shear distribution at the left/right end is a net shear force given by 

d/2

τx y (1)dy = −P

(2.36)

y=−d/2

The minus sign indicates that P is in the negative τ xy direction on both the end-faces, i.e., downward at the left and upward at the right. Similarly, the resultant of the normal stress distribution at the right end is a net moment given by 

d/2 y=−d/2

σx |x=L y(1)dy = − P L

(2.37)

the minus sign indicates that it is a hogging moment. The strain field corresponding to the stress field above, with a single non-zero normal stress and a non-zero shear stress, is given by ) ( −P x y −P d 2 μP x y 2 (2.38) εx = , γx y = −y , εy = EI 2I G 4 EI Integrating εx and εy , one can get u and v displacements as u=

−P x 2 y + f (y); 2E I

v=

μP x y 2 + g(x) 2E I

(2.39)

Putting these in the expression for γ xy , and separating the x and y terms as was done earlier, one gets ( ) −P y 2 Pd 2 μP y 2 Px2 (2.40) + g ' (x) = − + + f ' (y) = a constant k1 − 2E I 2I G 2E I 8I G On integration, these yield Px3 Pd 2 x − + k1 x + k2 6E I 8I G P y3 μP y 3 f (y) = − − k1 y + k3 6I G 6E I

g(x) =

(2.41)

The end conditions, which yield k 1 to k 3 , are generally specified as at x = L, y = 0: u = v = u ,y = 0

(2.42a)

u = v = v,x = 0

(2.42b)

or,

42

2 Plane Problems in Cartesian Coordinates

Fig. 2.5 The two alternative clamped edge conditions

with the first set representing the case where a small element of the vertical length at (L, 0) is restrained from rotating in the x-y plane, while the second set represents such a restraint for a small element of the horizontal length; in either case, u and v displacements at the point (L, 0) are restrained. Thus, the only difference between the two is the rigid body rotation ψ about (L, 0), which represents the shear strain at this point (Fig. 2.5). It is customary to quantify the effect of shear strain in terms of an additional deflection (besides that due to bending which is accounted for in the conventional engineering beam theory), and from this perspective, as is clear from Fig. 2.5, a restraint on the rotation of the vertical element (by holding one or more points of the normal) is appropriate as the definitive specification of a clamped end within the purview of the theory of elasticity. Thus, for the present problem, the clamped end conditions are taken as u = v = u ,y = 0

at x = L, y = 0

(2.43)

which yield, k1 = −

P L2 , 2E I

k2 =

P L3 Pd 2 L + , k3 = 0 3E I 8I G

(2.44)

Table 2.2 presents a comparison of this solution with the CET counterpart. The first and foremost observation is that the CET estimates of the stresses coincide with the TOE estimates, but as far as deflections are concerned, even the centroidal values are not correctly predicted by the CET approach unlike in the earlier problem of pure bending. 2 , which is not captured by Here, there is an additional deflection, of magnitude Pd 8I(L−x) G the CET solution. Very clearly, this is due to the effect of shear strain (as indicated by the presence of G), often referred to as the shear deformation effect; this deflection component varies linearly with respect to x for the present problem because the shear strain occurring due to the shear force is uniform over the beam. The effect of shear deformation is also seen on the displacement u which exhibits a cubic variation with respect to y; this is to be expected because a linear variation as implied by the Euler-Bernoulli hypothesis of plane

2.3

Use of Polynomial Stress Functions

43

Table 2.2 Comparison of results for a cantilever under a tip force TOE solution

CET solution

−P x y I σy = 0 ) ( −P d 2 2 −y τx y = 2I 4

−P x y I σ y = 0 (by assumption) ) ( VQ −P d 2 2 = −y τx y = Ib 2I 4

−P x y EI μPxy εy = EI ( ) − P d2 2 γx y = −y 2I G 4

−P x y σx = E⎫ EI εy = 0 (by assumption) γx y = 0

P y3 P y(L 2 − x 2 ) μ P y3 − + 2E I 6E I 6I G ) 2 3 3 PL PL x Px v= − + 3E I 2E I 6E I

P y(L 2 − x 2 ) (from u = −yv,x ) 2E I P L2x Px3 P L3 − + v= 3E I 2E I 6E I (from E I v,x x = P x)

σx =

σx =

εx =

εx =

u= (

+

P d 2 (L − x) μ P x y2 + 2E I 8I G (

P L3 P L2x Px3 − + v| y=0 = 3E I 2E I 6E I P d 2 (L − x) 8I G P d2 L P L3 + v|x=y=0 = 3E I 8I G

u=

)

+

vti p =

P L3 3E I

cross-sections is true only for zero shear strain. Further, the influence of the thicknessstretch effect (as indicated by a non-zero εy ) is seen on both u and v fields, while for the earlier problem of pure bending, it was seen only on the v field. Since the CET approach underpredicts the deflections, it is of great importance to find out the factors on which the error depends. As pointed out earlier, it is convenient to call the component of the deflection which coincides with the CET solution as the bending deflection vb and the additional deflection as the shear deflection vs . Then vs /vb is a measure of the shear deformation effect and this ratio, at the point (0, 0), is given by

44

2 Plane Problems in Cartesian Coordinates

3E vs = vb 8G

( )2 d L

(2.45)

Thus, the shear deformation effect increases linearly with E/G—the ratio of the material stiffnesses corresponding to the bending and shear modes, but quadratically with d/L—the depth to length ratio. For metallic materials, E/G is given by 2(1 + μ) and is hence nearly 2.6 irrespective of the actual material being used; for this value, vs /vb is 0.975 × 10–4 for L/d = 100, 0.39 × 10–3 for L/d = 50, 0.0024 for L/d = 20, 0.0098 for L/d = 10 and 0.039 for L/d = 5. Thus, for this particular problem, the shear deformation effect is quite small, even for short beams with L/d = 5. The quantitative estimate of the thickness-stretch effect is given by vtop /vcenterline as was done earlier. However, this ratio cannot be calculated at the left end here, because the ( ) term μP x y 2 /2E I becomes zero at that point; in other words, the bending moment is zero at the left end, and thus there is no bending stress or strain or the associated Poisson effect. Hence, by choosing the mid-span location instead, one gets | | | μP x ( −d )2 | vtop − vcentreline || 2E I 2 ) | ( = | | 2 2 3 3 Pd (L−x) P L x P x P L vcentreline | x=L/2 3E I − 2E I + 6E I + 8I G x=L/2 (2.46) ( d )2 μ L = ( d )2 5 3 + 2(1 + μ) L which corresponds to a variation of 0.68% for L/d = 5, 0.045% for L/d = 20 and 0.007% for L/d = 50 when μ is taken to be 0.3. Thus, the thickness-stretch effect is quite small, though slightly more significant than for the earlier problem of pure bending. It should also be noted that shear deformation is always more important than thickness-stretch. If the beam is specially orthotropic, the stress function and the stress field remain unchanged as above, while the strain field becomes ( 2 ) μx y P x y d −P −P x y (2.47) εx = , εy = , γx y = − y2 Ex I Ex I 2I G x y 4 Proceeding as before, one gets | ( ) vs || 3 Ex d 2 = vb |(0,0) 8 Gxy L

(2.48)

This is an important result because it shows that the shear deformation effect in an orthotropic plate depends on the E x /Gxy ratio, the counterpart of the E/G ratio for isotropic plates. When a beam made of metal or plastic has to be reinforced by fibres, the optimum fibre orientation is obviously along the length because the bending stresses occur in

2.3

Use of Polynomial Stress Functions

Table 2.3 Typical properties of unidirectionally reinforced composites

45

Material

E x /Gxy

Boron/aluminium

4–5

Glass/epoxy

5–10

Kevlar/epoxy

33–37

Boron/epoxy

30–37

Graphite/epoxy

25–80

that direction. Due to such reinforcement, E x increases; however, the fibres do not contribute as much to Gxy , which would hence be only slightly higher than G for the metal or plastic, now referred to as the matrix material. Thus, due to reinforcement, the ratio E x /Gxy turns out to be much larger than the original E/G for the isotropic beam; in other words, for the same dimensions d and L, the shear deformation effect is much more significant now than earlier. For instance, if E x /Gxy is 26 (i.e., ten times the common E/G value of 2.6), the shear deflection vs would become about 10% of the bending deflection vb for L/d = 10 instead of the earlier value of just about 1%. To give a quantitative feeling, the E x /Gxy values for some commonly used fibre composite materials are listed in Table 2.3. The thickness-stretch effect for the orthotropic beam is given by | vtop − vcentreline || = | vcentreline x=L/2

2.3.4

μx y 5 3

+

( d )2

L E x ( d )2 Gxy L

(2.49)

A Simply Supported Beam Under Uniform Load

With reference to the problem of Fig. 2.6 (with the out-of-plane thickness taken as small and unity once again), the appropriate stress function is of the form φ = Ax 2 y + By 3 + C x 2 y 3 + Dy 5

Fig. 2.6 Simply supported beam

(2.50)

46

2 Plane Problems in Cartesian Coordinates

If the material is isotropic, φ has to be a biharmonic function with D = −C/5

(2.51)

The stress field is given by σx = 6By + 6C x 2 y − 4C y 3 σ y = 2 Ay + 2C y 3 τx y = −2 Ax − 6C x y

(2.52) 2

Enforcing the conditions at the top and bottom surfaces first, one gets | τx y | y=±d/2 = 0

| qo σ y | y=±d/2 = ± 2

⇒

A=−

3Cd 2 4

qo qo =− 3 d 12I qo qo d 2 and hence, D = , A= 60I 16I ⇒

C =−

(2.53)

If the ends x = ± L/2 are to correspond to simple supports, the common visualisation of a knife-edge suggests the following end conditions: σx = 0 for all y v = 0 for at least one value of y

(2.54)

However, a look at the σ x field tells us immediately that it cannot be made zero at the ends. (It can be verified that this is not possible even if one tries to add some other terms to the chosen φ.) Thus, the end condition for σ x has to be something else instead of the point-wise zero condition, while still corresponding to a simple support. Instead of specifying σ x in a point-wise sense at all, let us choose to leave it as the CET description  M=

d/2 −d/2

σx ydy = 0 at x = ±L/2

(2.55a)

along with the zero net longitudinal force condition 

d/2 −d/2

σx dy = 0 at x = ±L/2

(2.55b)

As will be shown later, this is not a great dilution of the elasticity approach as it appears now.

2.3

Use of Polynomial Stress Functions

47

Table 2.4 Results for a simply supported isotropic beam under uniform load CET solution

TOE solution

qo y σx = 2I

(

) L2 q y3 q d2 y 2 −x − o + o 4 20 I 3I

q y3 q od2 y − o 8I ( 6 I ) −qo x d 2 − y2 τx y = 2I 4

σy =

εx =

(σx − μσ y )

u=

qo x 3 y qo L 2 x y − 8E I 6E I

L2 − x2 4

)

σ y = 0 (by assumption) ) ( VQ −qo x d 2 2 τx y = = −y Ib 2I 4

σx E⎫ εy = 0 γx y = 0

)

(2 +μ)q o x y3 q d 2 (2 + 5μ)x y + − o 40 E I 6) EI ( 5qo L 4 qo L 2 x 2 qo x 4 − + v= 384E I 16E I 24E I ( ) 1 μ q (L 2 − 4x 2 )d 2 1 − − + o 32 I 2G 5E 2E ( ⎡ )⎤ 2 2 2 2 L d q y d −μ − + o 8E I 2 2 5 +

(

εx =

E ) ( σ y − μσ x εy = E τxy γx y = G

(

My qo y σx = = I 2I

( u=

(by assumption)

qo L 2 x y qo x 3 y − 8E I 6E I

)

from u = −yv,x ) ( ) qo L 2 x 2 qo x 4 5qo L 4 − + v= 384E I 16E I 24E I (from E I v,x x =

qo 2 L 2 (x − )) 2 4

(1 + 2μ)q o y4 μq o x 2 y2 − 4E I 24 E I

5qo L 4 384E I ( ) 1 μ q o L2 d2 1 − − + 32 I 2G 5E 2E

v|x=y=0 =

vcentr e =

5qo L 4 384E I

(continued)

48

2 Plane Problems in Cartesian Coordinates

Table 2.4 (continued) CET solution

TOE solution

| ( ) ( ) 6 d 2 4E vs || − 3μ = vb |(0,0) 25 L G | ( ) 4 d 2 σx TOE || = 1 + σx CET |(0,d/2) 15 L | ( )2 3 L 2 σx || = + σ y |(0,d/2) 2 d 5

Since σ x is an odd function of y, the zero net force condition is automatically satisfied; the zero net moment condition yields ( ) ( ) qo L 2 d2 qo L 2 d2 B= 3 − = − (2.56) 2d 2 5 24I 2 5 With all the constants A to D thus fixed up, the stress function is completely defined. Subsequent steps leading to the displacement field are similar to those in the earlier problems; the appropriate displacement boundary conditions are u = 0 at x = 0 for all y (due to symmetry of deformation) v=0

at x = ±L/2, y = 0

(2.57)

A comparison of the stress, strain and displacement fields with their CET counterparts is presented in Table 2.4; the table also includes a comparison of the shear and bending deflections, the maximum bending stresses by the two approaches, and the maximum values of the bending stress σ x and the transverse normal stress σ y . It is possible to extend the above solution in a straightforward manner for the orthotropic case by starting with Eq. (2.9) (with V = 0) in place of the biharmonic equation to obtain ) ( μx y Ex (2.58) − D=C 5 10G x y The other steps are the same as for the isotropic case, and yield A=

qo d 2 μ x y qo d 2 qo L 2 qo d 2 E x , B= − + , 16I 48I 240G x y I 120I ) ( qo qo Ex C =− , D= − 2μx y 12I 120I G x y

(2.59)

2.3

Use of Polynomial Stress Functions

49

The results for this orthotropic case are presented in Table 2.5. From Tables 2.4 and 2.5, it can be seen that both the bending stress σ x and the deflection v are not correctly predicted by the CET approach. The error in either of them Table 2.5 Results for a simply supported orthotropic beam under uniform load TOE solution

CET solution

) ( qo y L 2 − x2 σx = 2I 4 ( )( ) 3 Ex q y q d2 y + o − o − 2μ x y 6I 40 I Gx y

σx =

qo y My = I 2I

(

L2 − x2 4

)

σ y = 0 (by assumption) ( ) VQ −qo x d 2 τx y = = − y2 Ib 2I 4

q y3 q od2 y − o 8I 6I ) ( −qo x d 2 − y2 = 2I 4

σy = τx y

εx =

(σx − μ x y σ y ) (

εy =

σx Ex ⎫ εy = 0 εx =

Ex ) σ y − μ yx σ x Ey

γx y = 0

γx y = τ x y /G x y ) qo x 3 y qo L 2 x y − 8E x I 6E x I ) ) ( ( 3μ x y μx y q x y3 1 1 qo d2 x y + o + − − 40 I Gx y Ex 6I Gx y Ex (

(

u=

u=

v|x=y=0 =

5qo 384E x I

+

L2 d2

qo 80 I

(

1 Gx y

−

3μ x y 4Ex

qo x 3 y qo L 2 x y − 8E x I 6E x I

)

(from u = −yv,x ) ) ( qo L 2 x 2 qo x 4 5qo L 4 − + v= 384E x I 16E x I 24E x I ( ) qo 2 L 2 x − (from E x I v,x x = 2 4

) ( qo L 2 x 2 qo x 4 5qo L 4 − + v= 384E x I 16E x I 24E x I ) ( 3μ x y 1 q o (L 2 − 4x 2 )d 2 − + 80 I Gx y 4Ex ⎡ ⎧ )⎫⎤ ( 2 2 2 μx y qo y 1 d d2 L + − μx y − − 8I 2 E y 2Ex 5 2G x y Ex ( ) 2 2μ x y μx y μ x y q o x 2 y2 q y4 1 − o + − + 24 I E y Gx y Ex 4Ex I

L4

(by assumption)

)

vcentr e =

5qo L 4 384E x I (continued)

50

2 Plane Problems in Cartesian Coordinates

Table 2.5 (continued) CET solution

TOE solution

| ) ( ) ( vs || 6 d 2 4E x = − 3μx y | vb (0,0) 25 L Gxy | ) ( )2 ( Ex 2 d σx TOE || =1+ − 2μx y d | σx CET (0, ) 15 L Gxy 2 | ) ( ) ( σx || 3 L 2 1 Ex = + − 2μx y | σ y (0,d/2) 2 d 5 Gxy

depends on (d/L)2 and a material property parameter involving (E x /Gxy ) and μxy , or their isotropic counterparts; it is important to note that in practice, μxy is usually in the range 0.25–0.35, and hence its contribution to the error expressions is not very significant. Thus, the main parameters with respect to which the CET error varies directly are (d/L)2 and (E x /Gxy ), as discussed earlier with reference to the tip-loaded cantilever. Another observation from Table 2.5 is that the maximum transverse normal stress σ y , which occurs on the lateral surfaces, is much smaller than the maximum bending stress as long as (L/d) is large. One can also quantify the thickness-stretch effect for this case and show that it is less significant than that of shear deformation—this is left as an exercise for the reader. Our immediate interest at this point is in justifying the use of the integral form of the boundary conditions enforced at the ends x = ± L/2, and towards this, an alternative solution for this problem is presented in the following section.

2.4

A Fourier Series Solution for the Simply Supported Beam

Instead of the non-rigorous description of the simply supported ends as in Eqs. (2.55a and 2.55b), let us assume that they are defined here in a rigorous point-wise sense as σx = v = 0 at x = ±L/2 for all y

(2.60)

A solution satisfying these end conditions can be obtained by starting with a Fourier series representation for φ instead of polynomials. The appropriate choice is φ=

∑ m=1,3,..

φm (y) cos

mπ x L

(2.61)

which satisfies the zero σ x conditions at the ends x = ± L/2. Substitution of this in the biharmonic equation yields, for the isotropic case

2.4

A Fourier Series Solution for the Simply Supported Beam 2 '' 4 φm'''' − 2 pm φm + pm φm = 0 f or each m

51

(2.62)

where pm = mπ /L, and the primes denote derivatives with respect to the single variable y. The auxiliary equation of this simple ordinary differential equation has two repeated roots ±pm , and hence the four linearly independent solutions are e pm y , ye pm y , e− pm y , ye− pm y . By expressing exponential functions in terms of hyperbolic functions, one can write the general solution of Eq. (2.62) as φm = A1m cosh pm y + A2m y sinh pm y + A3m sinh pm y + A4m y cosh pm y

(2.63)

where A1m to A4m are undetermined constants. For the present problem with the transverse load distributed equally between the top and bottom surfaces, it is clear that σ y is an odd function of y, and hence φ m is also odd in y. Thus A1m = A2m = 0

(2.64)

with the other constants to be determined by using the conditions at either of the lateral surfaces. Before applying the lateral surface conditions, it is necessary to expand the applied transverse load in a Fourier series. Since the present series methodology is valid for any general load, let us present these equations in general form so as to be useful for any net transverse load q(x) distributed equally between the top and bottom surfaces; let us also assume that the load is symmetric about mid-span. Thus, we have q(x) =

∑ m=1,3,..

qm cos

mπ x L

4qo sin mπ with qm = mπ 2 for uniform load. The lateral surface conditions at y = d/2 then yield qm ⎫ 2 ⎬ − pm φm | y=d/2 = 2 for each m. | φm' | y=d/2 = 0 ⎭

(2.65)

(2.66)

Solving these, one finally gets A3m = where Γ = pm d/2.

−qm cosh Γ qm (cosh Γ + Γ sinh Γ) , A4m = 2 pm (2Γ − sinh 2Γ) pm (2Γ − sinh 2Γ)

(2.67)

52

2 Plane Problems in Cartesian Coordinates

The corresponding stress and displacement fields are given by

σx = σy = τ xy = u=

∑

[ A4 m pm y cosh pm y

∑

−[ A3m sinh pm y + A4 m y cosh pm y ] pm2 cos pm x

∑

[ A4 m pm y sinh pm y

m =1,3,..

m =1,3,..

m =1,3,..

+ ( 2 A4 m + pm A3m )sinh pm y ] pm cos pm x

+ ( A4 m + pm A3m )cosh pm y ] pm sin pm x

[ A4 m (1 + μ ) pm y cosh pm y 1 ∑ E m =1,3,.. + {2 A4 m + (1 + μ ) pm A3m }sinh pm y ]sin pm x

v=−

(2.68)

(2.69)

[ A (1 + μ ) pm y sinh pm y 1 ∑ 4m E m =1,3,.. + {(1 + μ ) pm A3m − (1 − μ ) A4 m }cosh pm y ]cos pm x

with the constants of integration in the displacement field turning out to be zero so as to satisfy the following conditions: u = 0 at x = 0 for all y (due to symmetry of deformation), v = 0 at x = ±L/2 for all y

(2.70)

For the orthotropic beam, the solution methodology is exactly the same except that the use of Eq. (2.9) instead of the biharmonic equation may lead to slightly different forms for φ m . The counterpart of Eq. (2.62) is ) ( 2μx y 1 p 4 φm φm'''' 2 '' pm − − φm + m =0 (2.71) Ex Gxy Ex Ey The roots of the corresponding auxiliary equation are given by ±pm α, ±pm β, where ) /( )2 ( Ex Ex Ex 2 2 − μx y ± − μx y − (2.72) α ,β = 2G x y 2G x y Ey If the material constants are such that the square root term becomes zero, then α and β are equal, and the solution for φ m is similar to that of Eq. (2.63) for the isotropic case. (Such a special case of orthotropy is often referred to as Huber orthotropy.2 ) If α and β are distinct, the four linearly independent solutions are eα pm y , eβ pm y , e−α pm y , e−β pm y

2 see S.P.Timoshenko, S.W.Krieger, Theory of Plates and Shells, McGraw-Hill, 1959.

2.4

A Fourier Series Solution for the Simply Supported Beam

53

which can be expressed in terms of hyperbolic or trigonometric or hyperbolictrigonometric functions, depending on the real or complex nature of α and β. We shall not write down the final solution for each of these cases, but shall confine attention just to the present problem where the x-direction is the preferred direction of reinforcement, and hence E x /Gxy would be quite large leading to positive values of α and β. In that case, φ m is of the form φm = A1m cosh α pm y + A2m cosh β pm y +A3m sinh α pm y + A4m sinh β pm y

(2.73)

with A1m and A2m turning out to be zero once again. The other constants are obtained as −qm β cosh Γ2 cosh Γ2 sinh Γ1 − α cosh Γ1 sinh Γ2 ) qm α cosh Γ1 = 2 (β cosh Γ sinh Γ − α cosh Γ sinh Γ ) 2 pm 2 1 1 2

A3m = A4m

2 (β 2 pm

(2.74)

where Γ1 = α p2m d , Γ2 = β p2m d . The stress and displacement fields are ∑ 2 [A3m α 2 sinh α pm y + A4m β 2 sinh β pm y] pm cos pm x σx = m=1,3,..

σy =

∑

2 −[A3m sinh α pm y + A4m sinh β pm y] pm cos pm x

(2.75)

m=1,3,..

τx y =

∑

2 [A3m α cosh α pm y + A4m β cosh β pm y] pm sin pm x

m=1,3,..

u=

v=−

1 Ex

∑

[ A3m (α 2 + μ xy ) sinh α pm y

m =1,3,..

1 αβ E y

+ A4 m ( β 2 + μ xy ) sinh β pm y ] pm sin pm x

(2.76)

2

∑

m =1,3,..

[ A3m β (1 + α μ yx ) cosh α pm y + A4 mα (1 + β 2 μ yx ) cosh β pm y ] pm cos pm x

Thus, the final solutions for both isotropic and orthotropic cases are in terms of Fourier series; the number of terms considered should be sufficient to yield desired convergence of the displacements, strains and stresses at any point of the domain. Obviously, this solution is more rigorous than the polynomial solution of Sect. 2.3.4 in that the simply supported end conditions are enforced in a point-wise fashion, but this is at the cost of increased complexity and the final solution being in terms of convergent infinite series rather than simple closed-form expressions.

54

2 Plane Problems in Cartesian Coordinates

However, there is an important advantage of the series approach—it enables one to consider, without any major procedural changes, any general transverse load while the polynomial stress function approach is suitable only for very simple cases of loading.

2.5

Justification of End Conditions Specified in Terms of Integrals

As pointed out earlier, the main purpose of deriving the above Fourier series solution is to compare it with the simpler polynomial solution (Sect. 2.3.4) and to examine the legitimacy of the CET-type description of the simple support conditions, viz. in terms of zero longitudinal force and zero bending moment (Eqs. 2.55a and 2.55b), within an elasticity solution. Let us first clearly understand the difference between the two alternative solutions presented for the simply supported beam under uniform load. Both the solutions satisfy the governing equations and the top and bottom boundary conditions exactly and differ only with respect to the end conditions at x = ± L/2. In the polynomial stress function solution, though there is a non-zero distribution of σ x at the ends, it is statically equivalent to a zero longitudinal force and a zero bending moment; thus, in spite of these end stresses, the end sections do rotate resulting in nonzero values of the displacement u at x = ± L/2. At these ends, the displacement v is zero only for the mid-point y = 0, but not for all y, and the nature of the τ xy variation is such as to accommodate the thickness-stretch effect to some extent. In the Fourier series solution, there is no stress σ x at all at the ends, and hence the end sections are freely permitted to rotate. However, the displacement v is completely restrained for all y, by a corresponding distribution of the shear stress τ xy , and thus thickness-stretch is prevented at the ends. In both the solutions, the distribution of the shear stress τ xy at any end should correspond to a resultant upward force of qo L/2 for overall equilibrium. By feeling, one would expect both the solutions to yield more or less the same results when the depth d is quite small as compared to the length L, because, for such long and slender beams, the actual specification of the end conditions cannot be significant as long as it corresponds to simple supports. This will now be verified. Before presenting numerical results, a note about the number of terms required in the Fourier series is appropriate. As a general rule, whenever we differentiate a Fourier series term-by-term, the resulting series is slowly convergent as compared to the original series. Thus, the number of terms required for accurate estimation of strains and stresses would be more than that required for displacements. The numerical results presented below are converged ones obtained by taking terms up to m = 9 for displacements and up to m = 199 for stresses. The orthotropic properties chosen are E x /E y = 25, μx y = 0.25,

G x y /E y = 0.5,

(2.77)

2.5

Justification of End Conditions Specified in Terms of Integrals

55

Table 2.6 Results for the simply supported beam problem Material

L/d

σ x (0,d/2)

v(0,0) Fourier series approach

Polynomial approach

Fourier series approach

Polynomial approach

Isotropic (μ = 0.3) 10

1.022

1.022

1.003

1.003

4

1.141

1.142

1.017

1.017

Orthotropic (see Eq. 2.77)

2

1.552

1.569

1.067

1.067

20

1.119

1.120

1.017

1.017

10

1.476

1.478

1.066

1.066

7

1.964

1.976

1.133

1.135

4

3.890

3.989

1.382

1.412

Note All the results are normalised with respect to CET values

which are typical of a graphite-epoxy composite material with the graphite fibres oriented along x. Table 2.6 presents results for the maximum deflections and the maximum bending stresses by the two solutions for a number of L/d ratios. Very clearly, the two approaches yield almost identical results for isotropic beams with L/d ≥ 4, and for orthotropic beams with L/d ≥ 10, and a difference between the two types of simple support conditions is seen only for deeper beams. Further, such a difference is always very small compared to the discrepancy of the CET counterpart with respect to any of these elasticity solutions. Thus, the approximate integral representation of the end conditions employed in the polynomial solution of Sect. 2.3.4 does not really detract from its value and utility as a standard elasticity solution. As pointed out earlier, the stress approach based on polynomial functions for φ is essentially an inverse trial and error procedure wherein exact point-wise satisfaction of all the boundary conditions is possible only for a very small set of problems. It is in this context that the justification of approximate, CET-type integral representation of the boundary conditions, as proved for the above example case, is valuable because it permits a broader range of problems to be analysed using closed-form stress functions. It should, however, be kept in mind that the approximate boundary conditions have to be on the shorter edges of the domain and the corresponding elasticity solution is not to be used for stress analysis of the region very close to such edges. This is discussed in greater detail below.

56

2 Plane Problems in Cartesian Coordinates

2.6

St.Venant’s Principle

2.6.1

Deduction from the Simply Supported Beam Solutions

The two alternative solutions discussed above have been shown to yield nearly identical results in the central portion of the span as long as the span-to-depth ratio is large. Let us now compare the two solutions with respect to the stresses near the ends x = ± L/2. At these ends, there is no σ x as per the Fourier series solution, but only τ xy acting upward and varying as shown in Fig. 2.7, and completely restraining the v displacement at all y; the corresponding distributions of σ x and τ xy of the polynomial solution are both non-zero as shown and these represent a partial restraining effect on u and v displacements. Fig. 2.7 Restraining stresses at the ends (— polynomial, - - - Fourier series)

Fig. 2.8 Self-equilibrant end-stress distribution

2.6

St.Venant’s Principle

57

However, the two alternative end-stress distributions are statically equivalent because they both correspond to zero longitudinal force, zero bending moment and a vertically upward shear force qo L/ 2 at each end. Thus, if one subtracts one of these solutions from the other, the results would pertain to a rectangle with stress-free lateral surfaces and with self-equilibrating σ x and τ xy stress distributions applied as end tractions at x = ± L/2 as shown in Fig. 2.8. Intuitively. one would expect that the effect of such self-equilibrating loads would be localised near the ends and would not be felt at all in the interior of the strip. And this is exactly what happens as illustrated in Fig. 2.9 which presents plots of σ x at y = 0.2d Fig. 2.9 Decay of stresses from the end (x = L/2 = 10d)

58

2 Plane Problems in Cartesian Coordinates

and τ xy at y = 0 vs. x near the end x = L/2 = 10d for L/d = 20. (Though the stress-free condition of the lateral surfaces for the end-loaded strip of Fig. 2.8 can be satisfied only by taking a very large number of terms in the Fourier series solution that is subtracted from the polynomial solution, convergence of σ x at y = 0.2d and τ xy at y = 0 is obtained much earlier at mmax = 200 itself. These values are used for the plots of Fig. 2.9.) The foregoing example provides a simple illustration of the famous St.Venant’s principle which can be stated thus: If a system of loads acting on a small portion of the boundary of a body is replaced by a statically equivalent system, then such a replacement causes significant changes in stresses and deformation only in the close vicinity of this loaded portion.

Another observation from Fig. 2.9 is that the distance from the end beyond which the effect of the self-equilibrant loading is negligible, referred to as the characteristic decay length, is not the same for the isotropic and orthotropic cases (note the different scales used for the x-axis). This is a very important finding and we shall understand this better by means of a more direct solution for the end-loaded strip problem.

2.6.2

Eigensolutions for Isotropic/Orthotropic Rectangular Strip

Let us consider the plane stress problem of the semi-infinite region x ≥ 0 bounded by the sides y = ± d/2; the out-of-plane thickness is unity. At the end x = 0, a set of selfequilibrant tractions σ x and τ xy are applied while the sides are stress-free. Considering the isotropic case first, we shall seek solutions for the biharmonic equation in terms of exponentially decaying stress functions of the form φ = e−γ x/d Y (y)

(2.78)

which implies that all the stresses decay exponentially from the loaded end. The exponential decay parameter γ should either be a positive real constant or a complex constant with a positive real part; in the former case, the decay is of the simple monotonic exponential nature, while the latter case, with a trigonometric function multiplying the exponential function, implies some oscillation during the exponential decay. The lateral surface conditions, given by σ y = φ,x x = 0 ,

τx y = −φ,x y = 0

at y = ±d/2

can be rewritten as, ) ) ( ( d d = Y' ± =0 Y ± 2 2

(2.79)

2.6

St.Venant’s Principle

59

By virtue of this, the net axial force, bending moment and transverse shear force at any section of the strip become zero as given by  

d/2 −d/2

d/2

 σx dy = 

σx ydy =

−d/2  d/2

−d/2

τx y dy =

d/2

−d/2 d/2

|d/2 φ,yy dy = φ,y |−d/2 = 0

⎡ ⎤d/2 φ,yy ydy = yφ,y − φ −d/2 = 0

−d/2  d/2

−d/2

(2.80)

|d/2 −φ,x y dy = −φ,x |−d/2 = 0

Thus, the solution chosen for φ in Eq. (2.78) is suitable for self-equilibrant end loads. Substitution of this solution in the biharmonic equation reduces it to Y '''' + 2

γ 2 '' γ 4 Y + 4Y =0 d2 d

(2.81)

a homogeneous equation associated with the homogeneous conditions of Eq. (2.79) indicating that it is an eigenvalue problem with some non-trivial solutions for Y possible only for certain distinct values for γ . These are obtained by seeking solutions of the form Y = kemy/d

(2.82)

which yields the auxiliary equation (

m2 + γ 2

)2

=0

(2.83)

with the repeated roots m = ±iγ , ±iγ . Thus, the general solution for the fourth-order equation in Y is given by Y = A cos

γy γy γy γy γy γy +B sin + C sin +D cos d d d d d d

(2.84)

with the constants A to D to be found out from the conditions of Eq. (2.79). At this stage, it is convenient to split the problem into those of symmetric and antisymmetric deformations about the x-axis; the applied end loads can always be decomposed into the corresponding components, the former characterised by σ x being an even function of y and τ xy an odd function of y, and the latter, the converse. The corresponding solutions are discussed below.

60

2 Plane Problems in Cartesian Coordinates

Symmetric Case For this case, the stress function has to be an even function of y and hence C =D=0 The lateral conditions (Eq. 2.79) reduce to ⎡ γ γ cos γ2 2 sin 2 − sin γ2 γ2 cos γ2 + sin

(2.85)

⎤ ⎧ γ 2

⎫ A B

= {0}

(2.86)

A non-trivial solution for A and B is possible only if the determinant of the coefficient matrix is zero, which yields, after some simplification sin γ = −γ

(2.87)

It can easily be verified that this equation has no purely real roots. Seeking complex roots as γ = α + iβ

(2.88)

one can rewrite Eq. (2.87) as sin α cos i β + cos α sin i β = −α − i β i.e., sin α cosh β + i cos α sinh β = −α − i β i.e., sin α cosh β = −α; cos α sinh β = −β i.e., cosh

−1

(

/

⎫ ⎪ α2 ⎪ + cos α − 1 = 0⎪ ⎬ 2 sin α ) ( ⎪ ⎪ ⎪ −1 −α ⎭ β = cosh sin α

−α sin α

)

(2.89)

which can be solved by Newton-Raphson or other techniques to yield an infinite number of roots for α and β, and hence the complex eigenvalues γ. Corresponding to each of these is an associated eigenfunction which is obtained by using any of Eq. (2.86) to get the ratio A/B and then substituting this along with C = D = 0 in the expressions for Y and φ. The solution corresponding to any symmetrical self-equilibrant end loading would then be an appropriate linear combination of the above eigensolutions. Noting that solutions corresponding to higher positive values of α decay faster, an estimate of the decay length can be obtained by considering the solution with the lowest positive α alone,

2.6

St.Venant’s Principle

61

i.e., φ = ( )e−αlowest x/d

(2.90)

where α lowest is referred to as the decay rate parameter. In the increasing order of α, the eigenvalues for the present symmetrically loaded isotropic strip are obtained as 4.2124 + 2.2507i, 10.7125 + 3.1032i, etc. Thus, the decay rate parameter is 4.2124. The characteristic decay length λ is usually defined as the length over which the end stresses decrease to 1% of their value, and is given by e−αlowest λ/d = leading to

λ d

=

ln 100 αlowest

1 100

(2.91)

= 1.093 for the present case.

Antisymmetric Case Proceeding as above, the following equations are obtained for this case. A=B=0 ⎡

sin γ2 cos γ2

γ γ 2 cos 2 γ γ cos 2 − 2 sin γ2

sin γ = γ

(2.92a) ⎤ ⎧

⎫ C D

= {0}

(2.92b) (2.92c)

The root with the lowest real part is 7.4977 + 2.7687i, and hence the characteristic decay length associated with the antisymmetric deformation of an isotropic strip is λ d = 0.614. This conclusion agrees fairly well with the behaviour of the beam of Sect. 2.6.1 (see Fig. 2.9a) which is a case of antisymmetric deformation. Thus, between the cases of symmetric and antisymmetric deformation, St.Venant decay is faster for the latter. For an arbitrary self-equilibrant load on the ends, the decay length is 1.093d; this is often approximated as 1.0d and it is customary to say that replacement of the loading on an isotropic body by a statically equivalent system would cause negligible differences beyond a distance equal to the largest dimension of the loaded area. It should be noted that all the above calculations and the final conclusion are independent of the material properties as long as the body is isotropic. For an orthotropic strip, one has to start with ) ( 2μx y 1 1 1 φ,x x yy + φ,x x x x + − φ,yyyy = 0 (2.93a) Ey Gxy Ex Ex

62

2 Plane Problems in Cartesian Coordinates

However, because of the presence of various material property constants in this equation, further steps would become cumbersome if written out in a general form. For this reason, we shall confine attention to the graphite-epoxy unidirectional material considered earlier (Eq. 2.77). For this case, the above equation reduces to 25φ,x x x x + 49.5φ,x x yy + φ,yyyy = 0

(2.93b)

Assuming a decaying solution as in Eq. (2.78), one gets Y '''' + 49.5

γ4 γ 2 '' Y + 25 4 Y = 0 2 d d

(2.94)

for which the auxiliary equation has four distinct roots given by ±i pγ , ±iqγ with p = 6.999, q = 0.7144. Thus, the general solution for Y can be written in terms of trigonometric functions as Y = A cos

pγ y qγ y pγ y qγ y + B cos + C sin + D sin d d d d

(2.95)

Considering, once again, the symmetric and antisymmetric problems separately, and seeking non-trivial solutions, one obtains the following transcendental equations: pγ qγ pγ qγ sin − p sin cos = 0 (sym.) 2 2 2 2

(2.96a)

pγ qγ pγ qγ sin − q sin cos = 0 (antisym.) 2 2 2 2

(2.96b)

q cos p cos

The first few roots of these equations are purely real and are given by γ =

0.9075, 1.8175, etc. (sym.) 1.2885,2.2155, etc. (antisym.)

Corresponding to the lowest roots, the decay lengths are λ ln 100 = = 5.07 (sym.), 3.57(antisym.) d 0.9075 or 1.2885 The behaviour of the antisymmetrically loaded beam of Fig. 2.9b is seen to agree well with the above result.

2.7

Factors Governing Shear Deformation Effect

63

From the above numerical study, it is clear that because of orthotropy with E x > E y , that the decay length for strongly the St.Venant decay becomes slower. It has been shown3 / orthotropic materials (or large E x /E y ) is proportional to GExxy .

2.6.3

Implications of St.Venant’s Principle

As pointed out earlier, St.Venant’s principle enables one to solve a broader range of problems using the elasticity approach by allowing a CET-type specification of the boundary conditions in terms of appropriate stress resultants; such a specification is itself referred to as a St.Venant-type boundary condition. One will come across more instances of such conditions later on in this book. Another important implication of St.Venant’s principle is with respect to material characterisation tests where it allows ample flexibility in the design of end fixtures without any significant effect on the behaviour of the central length of the test specimen. In this context, it is very important to make sure that the test specimen is long as compared to twice the characteristic decay length of the material, and hence much longer for orthotropic specimens stiffer along the longitudinal direction than for corresponding isotropic specimens. Finally, by virtue of St.Venant’s principle, one can have a physical feel for the local effects of fasteners, joints, sudden changes in geometry, etc. and of how far such effects penetrate into the material. The above discussion of St.Venant’s principle is by no means exhaustive; a more elaborate discussion, including some special problems where the decay is slower than usual, is beyond the scope of this book, but is available elsewhere.4

2.7

Factors Governing Shear Deformation Effect

We have noted that E x /Gxy and (d/L)2 are two parameters with respect to which the shear deformation effect directly increases. In addition, the boundary conditions and the nature of the applied loading also affect the magnitude of the shear deflections as can be seen from the different vs /vb expressions for the problems of Sects. 2.3.3 and 2.3.4. In order to fully understand the influence of these additional factors, let us consider two more illustrative problems. For these problems, we shall present equations as applicable for the orthotropic case; the isotropic counterparts can simply be obtained by replacing E x by E and μxy by μ. 3 K.L.Miller, C.O.Horgan, End effects for plane deformations of an elastic anisotropic semi-infinite

strip, Jl. of Elasticity, 38, 1995, 261–316. 4 see, for example, Y.C.Fung, P.Tong, Classical and Computational Solid Mechanics, World Scien-

tific, 2001.

64

2 Plane Problems in Cartesian Coordinates

Fig. 2.10 Superposed cases for a clamped beam

2.7.1

Clamped Edge Conditions

Let us first consider a problem corresponding to a uniformly loaded beam clamped at both ends. This is solved by superposing two solutions, both corresponding to simply supported ends as shown in Fig. 2.10, with the first being the same as that of Sect. 2.3.4. The second solution (for the problem of Fig. 2.10b) is given by the stress function φ=−

M y3 6I

(2.97)

where M is the net bending moment. The stress and displacement fields are My ; σ y = τx y = 0 I ( ) μx y M y 2 L2 M x2 − + v= 2E x I 4 2E x I

σx = − u=

−M x y ; Ex I

(2.98a)

(2.98b)

which satisfy the following displacement conditions at the simply supported ends: v=0

at (±L/2, 0)

(2.99)

(It should be noted that this problem is the same as that of pure bending of Sect. 2.3.2 because the stress function employed is the same; the two problems are identical in terms of the deformed geometry and traction boundary conditions, the only difference being in the reference chosen for specification of the displacement field.)

2.7

Factors Governing Shear Deformation Effect

65

Fig. 2.11 Deformation due to end moments

For the purpose of superposition to obtain results for clamped ends, the only final result required from this solution is the relation between the central deflection and the rotation of the end cross-sections (Fig. 2.11), which is given by | | 2 | v(0, 0) | | | = M L /8E x I = L (2.100) | u (±L/2, 0) | M L/2E x I 4 ,y The clamped ends are assumed to be such that u = v = u ,y = 0 at (±L/2, 0)

(2.101)

Corresponding to this description, the value of the applied moment M in Fig. 2.10b has to be such as to neutralise the end rotations of the uniformly loaded beam of Fig. 2.10a at the centroidal level (y = 0), given by (see Table 2.5) )| ) ( ( 3μx y || qo x 3 qo d 2 x 1 qo L 2 x u ,y (L/2, 0) = − − + 8E x I 6E x I 40I Gxy E x |x=L/2 (2.102) ) ( 3μx y qo d 2 L 1 qo L 3 − + = 24E x I 80I Gxy Ex Proceeding further, the superposed central deflection for the clamped beam is obtained as v(0, 0) =

=

) ( 3μx y 5qo L 4 qo L 2 d 2 1 + − 384E x I 80I Gxy 4E x )⎤ ⎡ ( 3 2 3μx y L qo L qo d L 1 − − + 4 24E x I 80I Gxy Ex

(2.103)

qo L 4 qo d 2 L 2 + 384E x I 64G x y I

where the first term coincides with the CET solution (easily obtained by superposition once again), and the second term is the shear deflection.

66

2 Plane Problems in Cartesian Coordinates

Thus, for the uniformly loaded beam with clamped ends, one gets | ( )2 Ex vs || d = 6 | vb (0,0) L Gxy

(2.104)

while the counterpart for the simply supported beam is (see Table 2.5) | ( ) ( ) 6 d 2 4E x vs || = − 3μ xy vb |(0,0) 25 L Gxy

(2.105)

Thus, by clamping the ends of the simply supported beam, the relative importance of the shear deflection increases significantly; for materials with μxy much less than E x /Gxy , the increase of the ratio vs /vb for the present problem is by a factor of more than 6. One can deduce from the above result that any stiffening of the boundary constraints, in general, results in a significant decrease of the bending deflections without much effect on the shear deflections, and hence in an increase in the relative importance of the shear deformation effect.

2.7.2

Localised Loading

The other influencing factor is the nature of the applied load; to appreciate this, we compare the solution for a simply supported beam under central localised patch loading with that for uniform loading as obtained in Sect. 2.4. The same methodology is applicable for the central patch load—only the appropriate qm has to be used, as given by qm =

4P N mπ sin mπ L 2N

(2.106)

where the total load P is assumed to be applied on a central patch of width L/N (Fig. 2.12). As before, the total load is taken to be distributed equally between the top and bottom surfaces. A patch size of L/50 is chosen for numerical studies; this is a fairly accurate approximation of a concentrated load, and is in fact more realistic in practical situations because no load can actually be applied through a point or a line contact. (The theoretical case of the concentrated load itself will be discussed later in Chap. 3.) Fig. 2.12 Localised loading

2.8 What is a Long Beam? Table 2.7 Comparison of uniform and patch loads

67

Material

L/d

v(0,0)/vb Uniform load Patch load (patch size = L/50)

Isotropic (μ = 0.3) Orthotropic (Eq. 2.77)

10

1.022

1.027

4

1.141

1.165

20

1.119

1.146

10

1.476

1.571

Note For the patch load, vb is taken to be that for concentrated load, i.e., PL 3 /48EI

A comparison of v/vb for the patch load vs. uniform load is presented in Table 2.7. As can be seen, the ratios are slightly larger for the localised load, with the discrepancy between the ratios of the two loads increasing as L/d decreases. Thus, shear deformation effects are, in general, more important for problems of localised loading, especially when the beam is not very long. It has to be noted that the above comparison is confined to the central deflection alone. The maximum bending stress, right below the patch load, increases rapidly as the patch size decreases, and becomes singular (or infinitely large) for a concentrated load. This phenomenon, which cannot be predicted at all by the engineering beam theory, will be discussed later in Chap. 3.

2.8

What is a Long Beam?

In the light of the above discussion of various problems corresponding to beams, it is appropriate to attempt a clear demarcation of the range of applicability of the conventional engineering beam theory based on the Euler-Bernoulli hypothesis of plane cross-sections. While such a demarcation on the basis of limited studies can never be termed conclusive because it might need to be revised as new elasticity solutions are developed for more severe cases, it can always serve as a very useful thumb rule. From the studies discussed here, it appears that the conventional engineering beam theory is adequate for isotropic ( ) beams with L/d ≥ 20, and for orthotropic beams E x /G x y ≤ 50 with L/d ≥ 100. Thus, a “long beam” has to be viewed as a technical specification and not merely a geometric description because what can be classified as long also depends on the nature of the material used.

68

2.9

2 Plane Problems in Cartesian Coordinates

Summary

The main aim of this chapter has been to illustrate simple solutions of plane elasticity with reference to Cartesian axes and to point out important non-classical effects completely neglected in the engineering beam theory. The greater importance of the shear deformation effect in orthotropic beams than in isotropic beams and the slower St.Venant decay rate have been emphasised. The parameters influencing the shear deformation effect have been highlighted, and the limit of applicability of the conventional engineering beam theory has been demarcated on the basis of the limited studies carried out.

3

Plane Problems in Polar Coordinates

Some problems, by virtue of their geometry, are not amenable to a simple solution when referred to a Cartesian system of axes. The purpose of this chapter is to discuss several such two-dimensional problems which are solved with reference to polar coordinates. Also included herein is an interesting example to show that the principle of complementary shear is not always valid.

3.1

Field Equations in Polar Coordinates

These can be obtained in two ways—either from first principles by referring to a suitable differential element as was done earlier, or by transforming all the earlier equations from Cartesian to polar axes. We shall choose to do this from the first principles because the other approach would amount to a routine mathematical exercise.

3.1.1

Equilibrium Equations

Consider a small element bounded by two radial lines Δθ apart and two circumferential arcs Δr apart as shown (Fig. 3.1). Without loss of generality, the thickness perpendicular to the page is taken as unity. The stresses exerted by the adjoining structure on the faces of the element are as shown, with incremented values as one moves along the r and θ directions; as done earlier, all these stresses are shown along their positive directions. The body force fields are given by f r (r,θ ) and f θ (r,θ ) per unit volume. The equilibrium equations are obtained by summing forces in the radial and circumferential directions taken at the centre of the element. This procedure is similar to what was done earlier

© The Author(s) 2023 K. Bhaskar and T. K. Varadan, Theory of Isotropic/Orthotropic Elasticity, https://doi.org/10.1007/978-3-031-06345-9_3

69

70

3 Plane Problems in Polar Coordinates

Fig. 3.1 Equilibrium of a small element

with respect to Cartesian coordinates, except that one needs to consider the components of the forces acting on the radial faces of the element along and normal to the central r direction, which is Δθ /2 away. Further, since the element is small and reduces to a point in the limiting case, one has cos

Δθ Δθ Δθ ≅ 1, sin ≅ 2 2 2

The equations are (σr + Δσr )(r + Δr )Δθ − σr r Δθ − σθ Δr − (σθ + Δσθ )Δr

Δθ 2

Δθ + (τθr + Δτθr )Δr − τθr Δr + fr r Δθ Δr = 0 2

which reduces to σr ,r +

τr θ,θ (σr − σθ ) + + fr = 0 r r

(3.1a)

(σθ + Δσθ )Δr − σθ Δr + (τr θ + Δτr θ )(r + Δr )Δθ Δθ Δθ + τθr Δr + f θ r Δθ Δr = 0 − τr θ r Δθ + (τθr + Δτθr )Δr 2 2 which reduces to 2τr θ σθ,θ + τr θ,r + + fθ = 0 r r

(3.1b)

3.1

Field Equations in Polar Coordinates

71

It should be noted that the principle of complementary shear has been used above, by taking τ rθ = τ θr .

3.1.2

Constitutive Relations

These need not be explicitly written down because they are of the same form as those with reference to Cartesian coordinates. Once again, for a three-dimensional isotropic or specially orthotropic body with reference to the cylindrical polar coordinates r-θ -z, there is no coupling between the stretch and shear modes or between the shear modes in the three mutually orthogonal planes r-θ , θ -z and r-z. Such a specially orthotropic body is referred to as cylindrically orthotropic—an example would be a circular cylinder reinforced by fibres in the circumferential direction; this has to be contrasted with a body reinforced by a set of parallel fibres, all along a fixed direction in the r-θ plane, and referred to as rectilinearly orthotropic. As before, the idealisations of plane stress and plane strain are applicable when the problem reduces to that of a thin circular disc or a long cylinder with appropriate restraints on the applied loading.

3.1.3

Strain-Displacement Relations

These are derived by considering the changes in the lengths of or the angle between linear elements originally along the radial and circumferential directions, as shown in Fig. 3.2, wherein the radial and circumferential displacements of the points A, B and C, as they move to the positions A’, B’ and C’, are (ur , uθ ), (ur + ur,r Δr, uθ + uθ,r Δr) and (ur + ur,θ Δθ , uθ + uθ ,θ Δθ ), respectively. The derivation is similar to that with reference to Cartesian coordinates, except in the following two respects: (a) the circumferential strain has an additional component dependent on the radial displacement alone, because even for axisymmetric deformation, the length of a circumferential element increases as it moves radially outward; Fig. 3.2 Derivation of strain-displacement relations

72

3 Plane Problems in Polar Coordinates

(b) while calculating the shear strain, it should be noted that even when a small element (of the shape shown in Fig. 3.1) moves as a rigid body in the θ direction (i.e. as it rotates about the z-axis through O), each point of it has a circumferential displacement uθ proportional to its distance r from O. In other words, a gradient of uθ in the radial direction, of value uθ /r, is associated with rigid body motion along the circumferential direction and does not contribute to shear strain. Thus, with reference to Fig. 3.2, the shear strain is given by (α + β). These considerations result in the following relations:

εr = u r ,r εθ = γr θ =

3.1.4

u r ,θ r

ur r

+

u θ,θ r

+ u θ,r −

uθ r

=

u r +u θ,θ r

= u θ,r +

(3.2)

u r ,θ −u θ r

Compatibility Equation

The strain compatibility equation, applicable for plane stress or plane strain, is obtained by eliminating the displacements from the strain-displacement relations (Eq. (3.2)) and is given by εr ,θ θ − r εr ,r + (r 2 εθ,r ),r = (r γr θ ),r θ

(3.3)

Since the further equations are lengthier compared to their Cartesian coordinate counterparts, we shall write them in a generic form applicable for plane stress and plane strain as well as isotropic and cylindrically orthotropic bodies; for this purpose, we define generic two-dimensional compliance coefficients α ij as ⎧ ⎫ ⎡ ⎫ ⎤⎧ ⎪ α11 α12 0 ⎪ ⎨ εr ⎪ ⎬ ⎬ ⎨ σr ⎪ ⎥ ⎢ =⎣ (3.4) α22 0 ⎦ σθ εθ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ ⎭ ⎩ sym. α66 γr θ τr θ with different values for the different cases as can be easily found from the constitutive equations provided in Chap. 1. In terms of these coefficients, the compatibility equation in terms of stresses is (α11 σr + α12 σθ ),θ θ − r (α11 σr + α12 σθ ),r + [r 2 (α12 σr + α22 σθ ),r ],r = α66 (r τr θ ),r θ

(3.5)

3.1

Field Equations in Polar Coordinates

73

The Airy stress function φ is related to the stress components by the relations φ,r φ,θ θ + 2 = σr − V ; φ,rr = σθ − V ; −(φ/r ),r θ = τr θ r r

(3.6)

where V is the body force potential defined by fr = −V,r ;

f θ = −V,θ /r

(3.7)

It can easily be verified that the above definition for φ leads to automatic satisfaction of the equilibrium equations (Eq. (3.1)). The last step is to rewrite the compatibility equation in terms of φ as φ,rr θ θ φ,θ θ θ θ + α11 4 2 r r φ,r θ θ φ,rrr − (2α12 + α66 ) 3 + 2α22 r r φ,θ θ φ,rr φ,r − α11 2 + (2α11 + 2α12 + α66 ) 4 + α11 3 r r r V,θ θ V,r = −(α12 + α22 )V,rr − (α11 + α12 ) 2 + (α11 − 2α22 − α12 ) r r

α22 φ,rrrr + (2α12 + α66 )

(3.8)

For the special case of isotropy, the above equation simplifies to ∇ 4 φ + (1 − μ)∇ 2 V = 0 for plane stress ∇ 4φ +

where ∇ 2 ≡

3.1.5

(

∂2 ∂r 2

+

1 ∂ r ∂r

+

(1 − 2μ) 2 ∇ V = 0 for plane strain (1 − μ)

1 ∂2 r 2 ∂θ 2

)

(3.9a) (3.9b)

and ∇ 4 ≡ ∇ 2 · ∇ 2 .

Equations for the Axisymmetric Problem

When the material properties, geometry, boundary conditions and applied loads are all axisymmetric, the deformation is also axisymmetric, and hence all the above equations can be simplified by deleting the terms involving derivatives with respect to θ. In particular, let us note that the Laplacian operator can be written as ( ) d 1 d r (3.10) ∇2 ≡ r dr dr which facilitates successive integration as will be clear from the following example.

74

3 Plane Problems in Polar Coordinates

3.2

Circular Cylinder Under Internal and External Pressure (Lamé’s Problem)

This solution is very important because it can be used to define the limit of applicability of the conventional theory of thin-walled pressure vessels. Consider a long cylindrical vessel of uniform thickness as shown (Fig. 3.3) and subjected to internal as well as external pressure. Here, attention is confined to the plane strain problem. Taking up the isotropic case first, one has ∇ 4φ = 0 i.e.

[  ( )] d 1 d dφ 1 d r r =0 r dr dr r dr dr

(3.11)

which, on successive integration, yields φ = A ln r + Br 2 ln r + Cr 2 + D

(3.12)

The stress field is given by A φ,r = 2 + B(1 + 2 ln r ) + 2C r r A σθ = φ,rr = − 2 + B(3 + 2 ln r ) + 2C r τr θ = 0 (due to axisymmetry) σr =

which indicates that the r and θ directions are principal stress directions. Fig. 3.3 Lamé’s problem

(3.13)

3.2

Circular Cylinder Under Internal and External Pressure …

75

The lateral surface boundary conditions σr (b) = − po , σr (a) = − pi

(3.14)

are used to fix up A and C, while B is determined by consideration of the displacement ur as follows. We have u r = r εθ = r (β12 σr + β11 σθ ) A = −(β11 − β12 ) + (β11 + β12 )2Cr r + (β12 + 3β11 )Br + (β11 + β12 )2Br ln r and also

 ur =

(3.15a)

 εr dr =

(β11 σr + β12 σθ )dr

A + (β11 + β12 )2Cr r + (β11 + 3β12 )Br + (β11 + β12 )2Br (ln r − 1) + constant A = −(β11 − β12 ) + (β11 + β12 )2Cr r + (β12 − β11 )Br + (β11 + β12 )2Br ln r + constant = −(β11 − β12 )

(3.15b)

where β ij are the plane-strain-reduced compliance coefficients (see Eq. (1.24)) with β 11 = β 22 for the isotropic material. A comparison of the above two expressions for ur leads to zero values for B and the integration constant of Eq. (3.15b). Use of Eq. (3.14) yields A and C, and hence the stress field as ( 2 ( )] [ ) a2 1 2 b 2 1 − p a − 1 + p b σr = − 2 i o (b − a 2 ) r2 r2 ( 2 ( 2 [ ) )] 1 2 b 2 a σθ = 2 p a + 1 − p b + 1 (3.16) i o (b − a 2 ) r2 r2 Since (b2 /r 2 − 1) ≥ 0 and (1 − a2 /r 2 ) ≥ 0, the radial stress is compressive everywhere as can be expected, varying monotonically between the extreme values of pi and po . Further, the circumferential stress due to pi alone is tensile everywhere and that due to po alone is compressive everywhere, with the maximum value due to either of these loads occurring at the inner diameter. These variations are as shown in Fig. 3.4. If the cylinder is orthotropic with the r-θ axes coinciding with the directions of material property symmetry (i.e. cylindrically orthotropic), then the problem remains axisymmetric, and is governed by the following equation (obtained from Eq. (3.8)):

76

3 Plane Problems in Polar Coordinates

Fig. 3.4 Radial and circumferential stresses

β22 φ,rrrr + 2β22

φ,rr φ,r φ,rrr − β11 2 + β11 3 = 0 r r r

(3.17)

where (1 − μθ z μzθ ) (1 − μr z μzr ) ; β22 = Er Eθ −(μθr + μzr μθ z ) −(μr θ + μr z μzθ ) = = Eθ Er

β11 = β12

(3.18)

Equation (3.17) is of the Euler-Cauchy form, and can be converted to an equation with constant coefficients by the coordinate transformation r = es or s = ln r which yields φ,r = φ,s s,r = and similarly,

φ,s r ,

φ,sss − 3φ,ss + 2φ,s φ,ss − φ,s ; φ,rrr = ; 2 r r3 φ,ssss − 6φ,sss + 11φ,ss − 6φ,s φ,rrrr = r4

φ,rr =

(3.19)

3.2

Circular Cylinder Under Internal and External Pressure …

77

Hence, Eq. (3.17) becomes (3.20) where the superscripts indicate derivatives with respect to s. The solution for this equation, after rewriting s in terms of r, is φ = A + Br 2 + Cr 1+α + Dr 1−α

(3.21)

√ √ where α = β11 /β22 , an orthotropy parameter approximately equal to E θ /Er (see Eq. (3.18)). The stresses are given by σr = 2B + C(1 + α)r α−1 + D(1 − α)r −(α+1) σθ = 2B + C(1 + α)αr α−1 − Dα(1 − α)r −(α+1)

(3.22)

As for the isotropic case, one can consider the displacement ur to prove that B = 0, and obtain C and D using the boundary conditions at r = a and b. Finally, the stress field is given by ( 2α ) ( 2α )] [ 1 − r 2α − a 2α α+1 b α+1 r + p p a b σr =− 2α i o (b − a 2α ) r 1+α r 1+α ] ] ) ( 1+α ( ( a )1+α ( ) ) 1 b pi α b2α + r 2α − po α a 2α + r 2α (3.23) σθ = 2α r (b − a 2α ) r which reduces to the isotropic counterpart (Eq. (3.16)) when α = 1. To get a physical feel for the effect of orthotropy, let us consider two values for α, equal to 5 and 1/5; these values are typical of a graphite epoxy unidirectional composite with the fibre reinforcement along the circumferential and radial directions, respectively. Let us also confine attention to a cylinder with b/a = 1.1 and subjected to internal pressure alone. Irrespective of the material, σ r is always smaller than σ θ and always decays monotonically in the radial direction from its maximum value of pi on inner surface. The radial variation of the circumferential stress σ θ is as shown in Fig. 3.5, which also includes results for the isotropic cylinder (α = 1); for all three cases, the maximum stress occurs at the inner diameter. The graph for the radially stiffer cylinder is almost indistinguishable from that for the isotropic cylinder while that for the circumferentially stiffer cylinder shows a steeper stress gradient.

78

3 Plane Problems in Polar Coordinates

Fig. 3.5 Effect of orthotropy on the circumferential stress

Table. 3.1 Maximum circumferential stress due to internal pressure

Rmean t∗

σθ ( pi a/t)

α = 0.2

α = 1 (isotropic)

α=5

50

1.010

1.010

1.013

20

1.025

1.026

1.047

10

1.051

1.055

1.138

5

1.108

1.122

1.456

* t = b − a;

Rmean = a+b 2

Table 3.1 shows the maximum values of σ θ normalised with respect to the CET counterpart based on the membrane shell theory assumption that there is no variation through the thickness at all. These results show that the CET approach is non-conservative because the stresses are under-predicted, but the errors are within 5% as long as the mean radius-to-thickness ratio is 20 or greater; the error rapidly increases for vessels of higher thickness, especially for the circumferentially stiffer case. Thus, the value of 20 for (Rmean /t) can be taken as the demarcation between thin-walled and thick-walled pressure vessels.

3.3

Some Special Cases of Lamé’s Problem

3.3.1

Solid Cylinder

The solution of the solid isotropic cylinder under external pressure (Fig. 3.6a) is obtained from Eq. (3.13) in which the constants A and B have to be taken as zero so that the stresses at the centre (r = 0) are finite, and constant C is obtained from the loaded surface condition. Thus, the state of stress anywhere in the domain is given by

3.3

Some Special Cases of Lamé’s Problem

79

Fig. 3.6 State of stress in a solid cylinder

σr = σθ = − po

(3.24)

This should not be surprising because this problem is no different from that of a rectangular domain with equal biaxial compression (Fig. 3.6b), for which the state of stress everywhere is one of equal biaxial compression and one that remains unaffected by a rotation of the coordinate axes x and y about z. The pressurised cylinder can simply be looked upon as a free body cut out of that domain. We shall not consider the corresponding cylindrically orthotropic case because it does not make sense to specify different properties in the radial and circumferential directions at r = 0 (called the pole of orthotropy), because, at this point, every direction is as much radial as it is circumferential.

3.3.2

Externally Pressurised Cylinder with a Pin-Hole

The solution for this case (Fig. 3.7a) is obtained from Eqs. (3.16) and (3.23) by taking pi = 0. The maximum circumferential stress occurs at the pin-hole and is given by σθ (a) = −

2 po αbα+1 a α−1 b2α − a 2α

(3.25)

The limiting value of this as a → 0 depends on the value of α; it is −2p0 for the isotropic cylinder with α = 1, 0 for a circumferentially stiffer cylinder with α > 1 and Fig. 3.7 A pressurised body with a pin-hole

80

3 Plane Problems in Polar Coordinates

∞ for a radially stiffer cylinder with α < 1. Note that these results are also valid if the domain is subjected to far-field equal biaxial compression (Fig. 3.7b) as explained earlier. Comparing, for the isotropic case, the above maximum stress with the central value for a solid cylinder, one finds that the effect of a pin-hole is quite severe in that the maximum stress is doubled. This phenomenon of a significant increase in stress due to a sudden geometric discontinuity (a stepped configuration or hole or groove or notch) or a metallurgical discontinuity (inclusions of another material) is called stress concentration, and the discontinuity is referred to as a stress-raiser. The factor by which the stress is increased—equal to 2 for the isotropic cylinder with the pin-hole—is called the stress concentration factor. It is very important to note that such stress concentration effects are neither intuitively felt nor captured by the CET approach, but can be analysed only by using the rigorous elasticity approach. It is not possible to specify a stress concentration factor for the orthotropic cylinder with a pin-hole, but very clearly the effect of a pin-hole is very severe in the case of a radially stiffer configuration.

3.3.3

Pressurised Hole in an Infinite Body

This case, besides being of practical relevance, is of interest because one gets to verify, by rigorous analysis, the intuitive feeling that the stresses and deformations would be confined to the local neighbourhood of the hole and would decay as one moves away from it (Fig. 3.8). The solution for this case is obtained from Eq. (3.23) by taking po = 0 and b → ∞, and is given by σr = − pi

( a )1+α r

, σθ = pi α

( a )1+α r

(3.26)

Thus, irrespective of the actual value of α, both the stresses take maximum values at the periphery of the hole and decay monotonically with r. For the isotropic case with α = 1, |σ r |=|σ θ | and both of them decay as given by (1/r 2 ); at a radial distance of 10a, the stresses are only 1% of their maximum value. When the cylinder is circumferentially Fig. 3.8 A pressurised hole

3.4

Isotropic Plate with a Circular Hole (Kirsch’s Problem)

81

stiffer (α > 1), |σ θ |>|σ r | and the decay is faster than for the isotropic case. However, when the cylinder is radially stiffer (α < 1), |σ r |>|σ θ | and the decay is slower than for the isotropic case.

3.4

Isotropic Plate with a Circular Hole (Kirsch’s Problem)

The plate is subjected to far-field uniaxial tension and is thin enough to be treated as a plane stress problem. It is easy to imagine that the effect of the hole on the stress field would be localised and would be negligible at distances large in comparison with the dimensions of the hole in accordance with St.Venant’s principle. Thus, if one considers a free body bounded by a much larger concentric circle (see dotted line in Fig. 3.9), the stress field on this outer boundary would correspond to simple uniaxial tension; this is easily obtained by transformation from Cartesian to polar coordinates. Considering the equilibrium of a triangular element as shown in Fig. 3.9, one gets σo (1 + cos 2θ ) 2 σo = −σo sin θ cos θ = − sin 2θ 2

σr = σo cos2 θ = τr θ

(3.27)

Thus, the problem has now been reduced to the axisymmetric annular domain; it can be decomposed into two sub-cases—the first with axisymmetric radial tension σ o /2 and the second with the non-axisymmetric tractions σr =

σo σo cos 2θ; τr θ = − sin 2θ 2 2

(3.28)

on the outer boundary, say, r = b >> a, the radius of the hole; in either case, the other set of conditions are those for the stress-free hole σr = τr θ = 0 at r = a

(3.29)

The solution for the first sub-case is obtained from the general Lamé’s solution (Eq. 3.16) by putting pi = 0, po = −σ o /2 and a/b → 0 as Fig. 3.9 Kirsch’s problem

82

3 Plane Problems in Polar Coordinates

σr =

) ) ( ( a2 a2 σo σo 1 − 2 ; σθ = 1+ 2 2 r 2 r

(3.30)

For the second sub-case, a look at the relations between Airy’s stress function and the stresses (Eq. (3.6)), and the applied boundary tractions (Eq. (3.28)) suggests the solution for φ as φ = f (r ) cos 2θ

(3.31)

Substitution of this in the biharmonic compatibility equation (Eq. 3.9) reduces it to ( 2 )( 2 ) 4 4f 1 d 1 df d d f =0 − − + + dr 2 r dr r2 dr 2 r dr r2 i.e. f ,rrrr +

9 f ,rr 9 f ,r 2 f ,rrr − 2 + 3 =0 r r r

(3.32)

This equation is of Euler-Cauchy form (as was Eq. (3.17)) and can be solved using the coordinate transformation of Eq. (3.19). The final solution for the second sub-case is thus obtained as ( ) C φ = Ar 2 + Br 4 + 2 + D cos 2θ (3.33) r which yields ) 6C 4D σr = − 2 A + 4 + 2 cos 2θ r r ) ( 6C σθ = 2 A + 12Br 2 + 4 cos 2θ r ) ( 2D 6C τr θ = 2 A + 6Br 2 − 4 − 2 sin 2θ r r (

(3.34)

The constants A to D are obtained using the boundary conditions (Eqs. (3.28) and (3.29)) along with the limiting condition a/b → 0 as A=−

σo σo a 4 σo a 2 , B = 0, C = − , D= 4 4 2

(3.35)

Finally, superposing the solutions of Eqs. (3.30) and (3.34), one obtains the solution to Kirsch’s problem as

3.4

Isotropic Plate with a Circular Hole (Kirsch’s Problem)

) ] [ ( a2 4a 2 3a 4 σo 1 − 2 + 1 − 2 + 4 cos 2θ 2 r r r ) ] [ ( 2 4 a 3a σo σθ = 1 + 2 − 1 + 4 cos 2θ 2 r r ) ( 2 4 2a 3a σo τr θ = − 1 + 2 − 4 sin 2θ 2 r r

83

σr =

(3.36)

which reduces correctly to the far-field conditions of Eq. (3.27) as r → ∞, and correctly satisfies the conditions corresponding to doubly symmetric deformation about the x and y axes centred at the hole. The only non-zero stress at the boundary of the hole is the circumferential stress σ θ which varies as σθ |r =a = σo (1 − 2 cos 2θ )

(3.37)

and has a maximum tensile value of 3σ o at θ = ± π/2 and a maximum compressive value of σ o at θ = 0,π. The stress field converges monotonically to the far-field state of uniaxial tension as one moves away from the hole. Figure 3.10 presents the stress variations near the hole along the vertical and horizontal directions; for the sake of better understanding, the stress components are designated with reference to x and y axes (i.e. along the x-axis (θ = 0), σ r = σ x , σ θ = σ y , and along the y-axis (θ = π/2), σ r = σ y , σ θ = σ x ). These variations should be contrasted with the uniform σ x distribution over the width of the plate that one would obtain by the conventional mechanics of materials approach, with only a slight increase in magnitude at the hole section (x = 0). Another observation from Fig. 3.10 is that the above solution is valid for finite domains also provided the hole diameter is quite small compared to the distance from the nearest edge. From the above discussion, it is seen that the stress concentration factor due to a circular hole is more severe for uniaxial loading as compared to equal biaxial loading for which it has a value of 2 as proved earlier in Sect. 3.3.2; this result for biaxial loading can also be obtained by superposing the results for two orthogonal uniaxial load cases. Similarly, one can also consider the case of far-field pure shear which is equivalent to equal and opposite uniaxial loads, of the same intensity, in two orthogonal directions; the corresponding stress concentration factor is easily obtained as 4. Thus, the above results provide a simple example to show that the stress concentration effect varies significantly with the type of loading even though the geometry remains unchanged.

84

3 Plane Problems in Polar Coordinates

Fig. 3.10 Stresses near the hole

3.5

Some Similar Stress Concentration Problems

There are some other important stress concentration problems for which the complete solutions are too complicated to be presented here; only a summary of results and conclusions are briefly presented below.

3.5.1

Isotropic Plate with an Elliptic Hole (Inglis’ Problem)

When the hole is elliptical instead of circular, one needs curvilinear coordinates and the solution becomes quite complicated.1 For uniaxial tension, the stress concentration is most 1 See S. P. Timoshenko, J. N. Goodier, Theory of Elasticity, McGraw-Hill, 1970.

3.5

Some Similar Stress Concentration Problems

85

Fig. 3.11 Plate with an elliptic hole

severe when the elliptical hole is such that its minor axis is aligned with the direction of loading (Fig. 3.11a); the maximum tensile stress is then at the ends of the major axis and is given by ( a) (3.38) σx max = σo 1 + 2 b where a and b are the semi-major and semi-minor axes, respectively. At the ends of the minor axis, the circumferential stress σ y is compressive and of magnitude σ o , exactly as was the case for the circular hole. If the far-field loading is equal to biaxial tension (Fig. 3.11b), the maximum tensile stress, which occurs at the ends of the major axis, is given by σx

max

a = 2 σo b

(3.39)

If the far-field loading is pure shear with reference to the axes of the elliptical hole (Fig. 3.11c), the maximum circumferential stress on the hole boundary is given by σ max =

(a + b)2 τo ab

(3.40)

86

3 Plane Problems in Polar Coordinates

All the above equations reduce to those for the circular hole when a → b. From the above equations, it is clear that an elliptical hole is a more severe stress-raiser than a circular hole (except when its major axis is along the loaded direction of a plate in uniaxial tension). Further, when the ratio a/b is increased, i.e. as the hole approaches a narrow slit or a crack, all the stress concentration factors increase very rapidly.

3.5.2

Orthotropic Plate with a Circular Hole

For rectilinearly orthotropic plates, Lekhnitskii2 obtained the solution first for an elliptic hole and later specialised it for the circular hole—these solutions are in terms of complex variables. We shall confine attention only to the problem of the circular hole and present some pertinent observations. The main difference in the expressions for the stress concentration factors for the orthotropic plate with respect to their isotropic counterparts is the occurrence of two dimensionless orthotropic parameters: / / ( ) Ex Ex k= ; n = 2 k − μx y + (3.41) Ey Gxy where x, y coincide with the principal material directions. The parameters k and n take the values 1 and 2, respectively, when the material is isotropic. For far-field uniaxial tension in the x-direction (Fig. 3.12), the circumferential stress at the edge of the hole takes the following extreme values: σθ tensile max = σo (1 + n) at A   σθ comp. max  = σo /k at B

(3.42)

A careful look at these equations indicates that the maximum tensile stress is more severe when the loading is along the stiffer direction, i.e. E x > E y , than when E x < E y . Fig. 3.12 Orthotropic plate with a circular hole

2 S. G. Lekhnitskii, Theory of Elasticity of an Anisotropic Body, Mir Publishers, 1981; S. G.

Lekhnitskii, Anisotropic Plates, Gordon and Breach, 1968.

3.6 Violation of Principle of Complementary Shear

87

Table 3.2 Stress concentration factors for circular hole in an orthotropic plate Unidirectional fibre composite

Uniaxial tension in x-direction Fibres along x-direction (E x > E y )

Fibres along y-direction (E x < E y )

Equal biaxial tension with fibres along x-direction

Glass-epoxy σ θ /σ o at A

3.99

2.73

2.26

σ θ /σ o at B

−0.58

−1.73

2.15

σ θ /σ o at A

8.71

2.54

3.71

σ θ /σ o at B

−0.20

−5.00

2.34

Graphite-epoxy

Note Typical properties taken as E L /E T = 3, GLT /E T = 0.5 and μLT = 0.25 for glass-epoxy, and E L /E T = 25, GLT /E T = 0.5 and μLT = 0.25 for graphite-epoxy, where L and T refer to the longitudinal and transverse directions, respectively, with reference to the fibres

It is also possible, in any particular case, for the maximum compressive stress to be of a larger magnitude than the maximum tensile stress. If the far-field loading is equal biaxial tension, then the circumferential stresses are given by σθ = σo (n + 1 − k) at A, ( ) k+n−1 = σo at B k

(3.43)

For better understanding, the values of the stresses at A and B are shown in Table 3.2 for two unidirectionally reinforced composite plates for the two loading cases. In this context, it is very important to keep in mind that the strength also varies with direction for an orthotropic material, and such data is necessary along with the stress concentration factors for failure analysis. For a more comprehensive treatment of stress concentration effects in orthotropic plates, one is referred to the monograph by Savin.3

3.6

Violation of Principle of Complementary Shear

We shall now consider an example4 to show that the principle of complementary shear is not always true. The problem pertains to the two-dimensional rectangular wedge region shown in Fig. 3.13. One face of the wedge is subjected to uniform shear loading while 3 G. N. Savin, Stress Concentration Around Holes, Pergamon, 1961. 4 E. Reissner, Note on the theorem of the symmetry of the stress tensor, Jl. of Mathematics and

Physics, 23, 1944, 192–194.

88

3 Plane Problems in Polar Coordinates

Fig. 3.13 A rectangular wedge

the other is free of load; the applied load is balanced by appropriate support reactions elsewhere. Thus, the boundary conditions are σx = 0, τx y = τo for x = 0 for y = 0 σ y = τ yx = 0

(3.44)

τx y / = τ yx at the corner x = y = 0

(3.45)

with the implication that

To solve this problem, let us assume that τ xy = τ yx everywhere else except at the corner of the wedge and attempt a solution in terms of the Airy stress function. For this purpose, it is necessary to use polar coordinates r-θ centred at O with reference to which the relevant equations are ) ( 2 ) ( 2 ∂ φ ∂ 1 ∂ 1 ∂φ 1 ∂2 1 ∂ 2φ . =0 (3.46a) ∇ 4φ ≡ + + + + ∂r 2 r ∂r r 2 ∂θ 2 ∂r 2 r ∂r r 2 ∂θ 2 σθ = φ,rr = 0, τθr = −(φ/r ),r θ = 0

at θ = 0

σθ = φ,rr = 0, τθr = −(φ/r ),r θ = −τo at θ = π/2

(3.46b)

The solution given by ( φ = τo r 2

π 1 π θ cos 2θ − sin 2θ − + 8 4 8 2

) (3.47)

can be verified to satisfy the above equations identically and is hence the exact solution to the problem.

3.7

Linear Crack in an Isotropic Plate (Williams’ Solution)

89

The stress field corresponding to this solution, after transforming back to the Cartesian axes, is given by ( ) π xy −1 y σ x = τo − + 2 + tan 2 x + y2 x ( ) xy y σ y = τo − 2 + tan−1 x + y2 x ) ( y2 (3.48) τ x y = τo 2 x + y2 which is a well-behaved smoothly varying field at all points of the domain except in the neighbourhood of the corner O. To understand the nature of the stress field near O, let us approach it along the x or y-axis. Along the x-axis, we have. for x > 0, π τ yx = σ y = 0, σx = − τo 2

(3.49a)

π τo 2

(3.49b)

Along the y-axis, we have. for y > 0, τx y = τo , σx = 0, σ y =

Thus, as one approaches O along the x-axis, the value of σ x , constant everywhere for all x > 0, suddenly becomes zero at the corner (see Eq. (3.44)). Similarly, along the y-axis, the value of σ y , constant everywhere for all y > 0, suddenly becomes zero at O. Thus, corresponding to the existence of unequal shear stress components τ xy and τ yx at O, the stress gradients σ x,x and σ y,y are infinitely large. Generalising, one can state that the principle of complementary shear is valid as long as stresses vary smoothly and the corresponding gradients remain finite. However, there are interesting problems wherein, at one or more points of the domain, the complementary shear components may not be equal; when such is the case, this is always accompanied by infinitely large gradients for some stress components at those points.

3.7

Linear Crack in an Isotropic Plate (Williams’ Solution)

In Sect. 3.5.1, it was pointed out that the stress concentration factor due to an elliptical hole depends on the ratio of the major axis to the minor axis; for the limiting case of a straight line crack, the stress concentration factor at the tip of the crack increases to infinity. In this section, we restate the problem as that of the determination of the stresses,

90

3 Plane Problems in Polar Coordinates

Fig. 3.14 Williams’ problem

due to far-field loading, in the neighbourhood of a linear crack on the boundary of a semi-infinite plate (Fig. 3.14a). The solution will be developed in a general form for an elastic wedge of vertex angle 2α with stress-free faces (Fig. 3.14b) and later specialised for α = π to correspond to the cracked plate. With the origin of the polar r-θ coordinates at the tip of the wedge, the relevant equations of this plane stress problem are ) ( 2 ) ( 2 ∂ φ ∂ 1 ∂ 1 ∂φ 1 ∂2 1 ∂ 2φ 4 (3.50a) ∇ φ≡ + + + 2 2 . + 2 2 =0 ∂r 2 r ∂r r ∂θ ∂r 2 r ∂r r ∂θ σθ = φ,rr = 0, τr θ = −(φ/r ),r θ = 0 at θ = ±α

(3.50b)

The appropriate stress function for this5 is of the form φ = r λ+1 f (θ )

(3.51)

where λ is an unknown real parameter. Substitution of this in Eq. (3.50) reduces them to (3.52)

and hence, f = A sin(λ + 1)θ + B cos(λ + 1)θ + C sin(λ − 1)θ + D cos(λ − 1)θ

(3.53)

with

5 M. L. Williams, On the stress distribution at the base of a stationary crack, Jl. of Applied Mechan-

ics, 24, 1957, 109–114.

3.7

Linear Crack in an Isotropic Plate (Williams’ Solution)

A sin(λ + 1)α + B cos(λ + 1)α + C sin(λ − 1)α + D cos(λ − 1)α = 0 −A sin(λ + 1)α + B cos(λ + 1)α − C sin(λ − 1)α + D cos(λ − 1)α = 0 (λ + 1)A cos(λ + 1)α − (λ + 1)B sin(λ + 1)α +(λ − 1)C cos(λ − 1)α − (λ − 1)D sin(λ − 1)α = 0 (λ + 1)A cos(λ + 1)α + (λ + 1)B sin(λ + 1)α +(λ − 1)C cos(λ − 1)α + (λ − 1)D sin(λ − 1)α = 0

91

(3.54)

The first two equations can be added to and subtracted from each other to yield simpler equations having just two terms each; the same can be done with the last two equations. The resulting set of equations form two decoupled sets as given by ] ]    cos(λ + 1)α cos(λ − 1)α B 0 = (3.55a) (λ + 1) sin(λ + 1)α (λ − 1) sin(λ − 1)α D 0 ] ]    sin(λ + 1)α sin(λ − 1)α A 0 = (3.55b) (λ + 1) cos(λ + 1)α (λ − 1) cos(λ − 1)α C 0 The trivial solution (A = B = C = D = 0) corresponds to the unstressed state; for a non-trivial solution, either     cos(λ + 1)α cos(λ − 1)α    =0  (λ + 1) sin(λ + 1)α (λ − 1) sin(λ − 1)α  i.e. λ sin 2α + sin 2αλ = 0 or

(3.56a)

    sin(λ + 1)α sin(λ − 1)α   =0   (λ + 1) cos(λ + 1)α (λ − 1) cos(λ − 1)α 

i.e. λ sin 2α − sin 2αλ = 0

(3.56b)

The above equations can be used to study wedges of any angle α, but our interest here is mainly in the cracked plate problem for which α = π. For this case, both the above equations reduce to sin 2π λ = 0 or λ =

m , m = 1, 2, 3, ... 2

(3.57a)

92

3 Plane Problems in Polar Coordinates

with (m − 2) B =− , D (m + 2) B = −1, D

A = −1 C

for m = 1, 3, 5, . . .

A (m − 2) =− C (m + 2)

for m = 2, 4, 6, . . .

Thus, finally, one can write the solution for φ as ⎡ ⎤ ( ) m m C sin( − 1)θ − sin( + 1)θ m Σ ⎢ ⎥ m 2( 2 )⎥ φ= r 2 +1 ⎢ ⎣ ⎦ m m−2 m +Dm cos( − 1)θ − cos( + 1)θ m=1,3,... 2 m+2 2 ⎡ ( )⎤ m m−2 m Σ ⎢ Cm sin( 2 − 1)θ − m + 2 sin( 2 + 1)θ ⎥ m ⎥ r 2 +1 ⎢ + ( )⎦ ⎣ m m m=2,4,... +Dm cos( − 1)θ − cos( + 1)θ 2 2

(3.57b)

(3.58)

with the stresses given by σr =

( ) φ,θ θ φ,r φ + 2 ; σθ = φ,rr ; τr θ = − r r r ,r θ

(3.59)

One can immediately see that all the stress expressions would have one term proportional to √1r which tends to infinity as r → 0 indicating a square root singularity at the crack tip; the other terms with positive exponents are non-singular, and are relatively negligible in the neighbourhood of the crack tip. Equation (3.58) is a general solution in which terms involving C m are odd functions of θ and those with Dm are even functions of θ —the former vanish when the far-field loading causes a symmetric σ r and σ θ distribution about θ = 0 (e.g. pure tension—crack opening mode or Mode I—Fig. 3.15a), while the latter vanish when it is a case of antisymmetric σ r and σ θ distribution (e.g. pure shear—sliding mode or Mode II—Fig. 3.15b). Confining attention to Mode I alone, one gets, in the neighbourhood of the crack tip, ( ) θ 3θ D1 + negligible non-singular terms σr = √ 5 cos − cos 2 2 4 r ( ) θ 3θ D1 σθ = √ 3 cos + cos + negligible non-singular terms 2 2 4 r ( ) θ 3θ D1 τr θ = √ sin + sin + negligible non-singular terms (3.60) 2 2 4 r

3.7

Linear Crack in an Isotropic Plate (Williams’ Solution)

93

Fig. 3.15 Mode I and Mode II

and hence the crack tip stresses with respect to Cartesian axes as D1 σx | y=0,x≥0 = σr |θ =0 = √ x  D 1 σ y  y=0,x≥0 = σθ |θ=0 = √ x  τx y  = τr θ |θ =0 = 0 y=0,x≥0

Replacing D1 with another undetermined constant defined by √ K I = D1 2π

(3.61)

(3.62)

one can rewrite the above equations as  KI σx | y=0,x≥0 = σ y  y=0,x≥0 = √ 2π x

(3.63)

where K I is called the stress intensity factor corresponding to Mode I, having a definite value for the given geometry and the magnitude of the far-field tension. Some other problems involving cracks are amenable to rigorous analysis, and in all these cases, the variation of the crack tip stress field has been found to be qualitatively similar to that of the above problem involving a square root singularity. In a general form, the variation of any stress component in the neighbourhood of a crack tip can be expressed as K f (θ ) √ 2πr

(3.64)

94

3 Plane Problems in Polar Coordinates

where the stress intensity factor depends on the geometry of the problem and loading. In other words, the stress intensity factor, aptly called so, is simply a measure of the intensity of stresses near the crack tip; it can be determined by using exact or approximate methods and the values are listed for various common geometries and loading in books on fracture mechanics, which is a study aimed at finding out when, why and how a crack initiates and propagates leading to structural failure. For the configuration shown in Fig. 3.14, the values of the stress intensity factor are √ K I = σ∞ πa √ K II = τ∞ πa

(3.65)

where the subscript ∞ refers to far-field values. Thus, along the line of crack, the stresses for the two loads of Fig. 3.15 vary as  σ∞ σ y  y=0,x≥0 = √ for Mode I loading 2x/a  τ∞ for Mode II loading τx y  y=0,x≥0 = √ 2x/a

(3.66)

the equations being valid only in the close neighbourhood of the crack tip (x ≤ a/20); the stress decay for large x, not correctly predicted by the above, is actually to their respective far-field values (σ ∞ ,τ ∞ ) by virtue of St.Venant’s principle. The variation of σ y , for Mode I, is qualitatively shown in Fig. 3.16. In contrast with the above elasticity solution, the CET solution for a plate with a crack fails to predict anything other than a slight increase in stress at the cracked section and is hence totally unacceptable. Before closing the discussion on this problem, it is appropriate to mention that another mode of loading is commonly considered besides Mode I and Mode II; this is called the Fig. 3.16 Plot of σy /σ∞ along y = 0 versus x/a

3.8

Stresses Under Concentrated Loads

95

Fig. 3.17 Mode III

crack tearing mode or Mode III and is caused by anti-plane shear or shear applied so as to cause one face of the crack as shown in Fig. 3.17 to move, relative to the other face, in the direction normal to the plane of the figure. It is also of interest to note that the crack tip stress field displays a square root singularity even if the body is rectilinearly orthotropic6 —only the angular dependence is more complicated than for the isotropic case, and the stress intensity factors are also different. The solution of such crack tip stress fields is rather complicated and beyond the scope of the present book.

3.8

Stresses Under Concentrated Loads

Problems involving (nearly) concentrated loads are of practical significance, and it is important to have an idea of the stress field directly under such loads. In this section, we shall deal with some plane problems involving concentrated loads applied on the boundary—they are either concentrated loads acting on very thin plate-like structures (plane stress problems) or uniform line loads acting along the generators of long prismatic bodies (plane strain problems). Clearly, the CET approach is not acceptable for such problems as can be exemplified by the stress estimates of the engineering beam theory— viz. a linear variation of the bending stress with a finite value under the load and zero values of the transverse normal stress—even at a section of a beam where a concentrated 6 G. C. Sih, P. C. Paris, G.R.Irwin, On cracks in rectilinearly anisotropic bodies, International Jl. of

Fracture, 1(3), 1965, 189–203; A. Hoenig, Near-tip behaviour of a crack in a plane anisotropic elastic body, Engineering Fracture Mechanics, 16, 1982, 393–403.

96

3 Plane Problems in Polar Coordinates

load is applied. We shall start the discussion here with the problem of a concentrated force on a semi-infinite half-plane.

3.8.1

Concentrated Normal Force on the Boundary of a Half-Plane (Flamant’s Problem)

Consider either the plane stress or the plane strain problem confined to the semi-infinite domain −∞ ≤ x ≤ ∞, y ≥ 0 and of unit thickness in the z-direction. The load applied is a concentrated normal force P at the origin as shown (Fig. 3.18a). Considering the isotropic case first, we have, with reference to polar coordinates, ( 2 )( 2 ) ∂ ∂ 1 ∂ 1 ∂ 1 ∂2 1 ∂2 φ=0 (3.67) + + + + ∂r 2 r ∂r r 2 ∂θ 2 ∂r 2 r ∂r r 2 ∂θ 2 along with σθ = φ,rr = 0, τr θ = −(φ/r ),r θ = 0 for θ = ± φ

π (except at r = 0) 2

(3.68a)

φ

The stresses σ θ as well as σ r (given by r,r + r,θ2θ ) are even functions of θ. Further, for a vertical equilibrium of the region bounded by the semicircular arc of arbitrary radius r (Fig. 3.18b), one has 

π/2 −π/2

(τr θ sin θ − σr cos θ )r dθ = P

(3.68b)

which indicates that the stresses σ r and τ rθ are likely to vary inversely with r. The simplest solution for φ suggested by the above considerations is φ = Ar θ sin θ and this satisfies the biharmonic equation and the boundary conditions. Fig. 3.18 Flamant’s problem

(3.69)

3.8

Stresses Under Concentrated Loads

97

From Eq. (3.68b), A can be found to be A=−

P π

(3.70)

The stresses are given by 2P cos θ πr σθ = τr θ = 0 (except at r = 0) σr = −

(3.71)

which is often referred to as a simple radial distribution. Note that σ r is a principal stress, and its value decreases with r for any θ and with θ for any r. In other words, the effect of the concentrated load is most significant along its line of action (θ = 0) and in its close neighbourhood (for small r) as intuitively expected. The associated displacement field can be obtained by integration of the straindisplacement relations as was done in Chap. 2. We shall derive the displacement field for the plane strain case because it will be of use later (in Sect. 3.9). The strain field is given by εr =

σθ − μ(σr + σz ) τr θ σr − μ(σθ + σz ) ; εθ = ; γr θ = E E G

(3.72)

with σz = μ(σr + σθ ). Hence, (1 − μ2 )2P cos θ πr E μ(1 + μ)2P cos θ = πr E = u r ,θ /r + u θ,r − u θ /r = 0

εr = u r ,r = − εθ =

u r + u θ,θ r γr θ

(3.73)

Integrating the first of these equations, one gets ur = −

(1 − μ2 )2P cos θ ln r + f (θ ) πE

(3.74)

Using this and the expression for ε θ , one gets uθ =

(1 − μ2 )2P sin θ μ(1 + μ)2P sin θ + ln r − πE πE

 f (θ )dθ + g(r )

(3.75)

98

3 Plane Problems in Polar Coordinates

Substitution for ur and uθ in the expression for γ rθ yields (1 − 2μ)(1 + μ)Pθ sin θ + C1 sin θ + C2 cos θ πE g = C3r f =−

(3.76)

and hence, (1 − 2μ)(1 + μ)Pθ sin θ (1 − μ2 )2P cos θ ln r − + C1 sin θ + C2 cos θ πE πE (1 − μ2 )2P sin θ (1 − 2μ)(1 + μ)Pθ cos θ μ(1 + μ)2P sin θ uθ = + ln r − πE πE πE (1 − 2μ)(1 + μ)P sin θ + C1 cos θ − C2 sin θ + C3r + πE (1 + μ)P sin θ (1 − μ2 )2P sin θ (1 − 2μ)(1 + μ)Pθ cos θ = + ln r − πE πE πE (3.77) + C1 cos θ − C2 sin θ + C3r ur = −

Due to symmetry of deformation about the y-axis, u θ = 0 at θ = 0 for all r

(3.78a)

and hence, C 1 = C 3 = 0. The remaining constant C 2 , which occurs as an independent term in the expression for ur along the y-axis as seen from Eq. (3.77) with θ = 0, depends on the actual boundary condition corresponding to suppression of this rigid body translation. This can be done in several ways, one of which is to assume that a point directly under the load on the y-axis but at a considerable distance r o below the surface is restrained, i.e. u r = 0 at r = ro , θ = 0

(3.78b)

)2P ln ro . Thus, the vertical displacement directly under the load, which yields C2 = (1−μ πE written conveniently with reference to x–y coordinates, is given by 2

v(0, y) =

(1 − μ2 )2P ro ln πE y

(3.79)

The surface displacements (at θ = ±π/2), which are of particular interest, are given by

3.8

Stresses Under Concentrated Loads

99

(1 − 2μ)(1 + μ)P 2E (1 − μ2 )2P ro (1 + μ)P − ln = πE πE r

u r |θ=±π/2 = − u θ |θ =π/2 = − u θ |θ=−π/2

(3.80)

where it should be noted that all points on either side of the load move towards it by the same radial displacement, consistent with the fact that εr is zero at all these points but is infinitely large under the load. With reference to the Cartesian x–y axes, the surface displacements are given by (1 − 2μ)(1 + μ)P 2E ro (1 + μ)P (1 − μ2 )2P ln − v(±x, 0) = |x| πE πE

u(±x, 0) = ∓

(3.81)

Figure 3.19 presents the variations of the vertical displacement v along the x and y axes, with the former shown such that one can visualise the deformed geometry. The stress field with respect to x–y axes is given by x2 y 2P 2P sin2 θ cos θ = − 2 πr π (x + y 2 )2 y3 2P 2P cos3 θ = − σ y = σr cos2 θ = − 2 πr π (x + y 2 )2 x y2 2P 2P sin θ cos2 θ = − = σr sin θ cos θ = − πr π (x 2 + y 2 )2 σx = σr sin2 θ = −

τx y

(3.82)

If the half-plane is orthotropic with the principal material directions coinciding with the x and y axes, then the solution for the problem is possible only by using complex variables.7 Skipping the mathematical details, we shall just point out some salient features. The most important observation is that the stress field continues to be a simple radial distribution and is given by σr = −

P k f (θ ), σθ = τr θ = 0 r

(3.83)

where k is a constant that depends on the orthotropic material properties and f(θ ) is no more a simple cosine variation. It is thus possible, depending on the actual material properties, for σ r to take on larger values for a particular θ than for θ = 0 (the line of action of the load) for any fixed r. However, for any given θ, the stresses decrease linearly with r as for the isotropic case.

7 S. G. Lekhnitskii, Theory of Elasticity of an Anisotropic Body, Mir Publishers, 1981;

S. G. Lekhnitskii, Anisotropic Plates, Gordon and Breach, 1968.

100

3 Plane Problems in Polar Coordinates

Fig. 3.19 Plots of the vertical displacement

3.8.2

Concentrated Load on an Isotropic Simply Supported Beam (Wilson-Stokes Method)

A direct solution of this problem is possible using the Fourier series approach of Sect. 2.7.2, where instead of a localised patch load one has to consider the limiting case of a point load as the patch size tends to zero. If the load P is applied at mid-span, the Fourier coefficients turn out to be qm =

2P , m = 1, 3, .. L

(3.84)

3.8

Stresses Under Concentrated Loads

101

Instead of this rigorous series approach, we shall discuss a simpler alternative here which is adequate as long as the beam is not very deep; our focus is on the stresses at the loaded section. This approximation is referred to as the Wilson-Stokes method and is based on the superposition of Flamant’s solution involving the concentrated load, and another solution, based on the engineering beam theory, involving only distributed loads. The latter is assumed to be of acceptable accuracy, and attention is confined to not-so-deep beams which fulfil this requirement. The superposition is as shown in Fig. 3.20. The beam of finite length (2 l) and finite depth (2c) is first looked upon as a free body of the infinite domain problem of Flamant, with the forces acting on the sides and the bottom edge, corresponding to the simple radial distribution, in equilibrium with the applied concentrated load. These forces are then reversed and taken as loads for the second case wherein they are balanced by the support reactions. For Flamant’s problem, the stresses along x = 0 are given by σx = τx y = 0 (except at y = 0); σ y = −

Fig. 3.20 (a) The beam under point load and (b) Wilson-Stokes method

2P πy

(3.85)

102

3 Plane Problems in Polar Coordinates

Fig. 3.21 Calculation of bending moment at the central section

For the second case, as per the engineering beam theory, the stresses at this section are given by σx =

H M(y − c) + , σ y = τx y = 0 I 2c

(3.86)

where M is the bending moment about the point (0,c), I = (2c) 12 (assuming unit width for the beam normal to the plane of Fig. 3.20) and H is the total tensile force about the centroidal axis. The bending moment at the central section due to the distributed forces on the bottom and side faces of one half of the beam (Fig. 3.21a) is easily calculated by considering two free bodies as shown in Fig. 3.21b; since the second free body is in equilibrium and the distributed forces on the bottom and right faces are balanced by those on the curved edge, these forces are statically equivalent to those acting on the curved edge of the quadrant-shaped free body. Thus, for the simple radial distribution of tractions given by 3

σr =

2P cos θ π(2c)

(3.87)

the required bending moment about (0,c) is easily calculated by considering the quadrant alone as  π/2 Mdue to distributed loads = − c sin θ σr (2c) dθ 0

 =−

0

π/2

−Pc 2Pc sin θ cos θ dθ = π π

(3.88)

the minus sign indicating that it is a hogging moment. When the moment of the support reaction P/2 is added to this, one obtains the net moment at the central section as

3.8

Stresses Under Concentrated Loads

M=

103

Pc Pl − 2 π

In a similar fashion, the tensile force H is calculated as  π/2 P H= σr sin θ (2c)dθ = π 0

(3.89)

(3.90)

Thus, the net stresses at the loaded section, for the beam with the concentrated load, can finally be written down by combining Eqs. (3.85) and (3.86) and due substitution for M, H and I as ( ) c P 3P l − (y − c) + (except at y = 0) σx = 3 2c 2 π 2π c 2P σy = − πy τx y = 0 (except at y = 0)

(3.91)

all the stresses being singular directly under the load. The CET counterpart of the above solution is σx =

3 Pl (y − c); σ y = τx y = 0 2c3 2

(3.92)

It is easy to imagine that infinite stresses cannot really exist—the mathematical singularity simply means that the material would yield as soon as the concentrated load is applied resulting in a finite loaded area and consequent reduction of stress to a finite (though high) value. This plastic flow is totally out of the purview of the theory of elasticity as much as it is of the CET approach; the former is better only because it provides insight into how (and how fast) the stresses decay as one moves away from the singularity. The other critical point at which a comparison of the two approaches is appropriate is the point (0, 2c) at the bottom of the loaded section. Considering the bending stresses alone, one has, at this point, ( ) 2c 3P l − σx TOE = 2 2c 2 3π 3Pl σx CET = 2 4c 4 (c) σx TOE (3.93) =1− σx CET 3π l which indicates that the conventional engineering beam theory over-predicts the bending stress at the bottom of the section, and the discrepancy is quite small for beams with l/c > 10.

104

3 Plane Problems in Polar Coordinates

There have been more accurate investigations into this problem8 besides the WilsonStokes approach outlined above, but in every case, the beam theory estimate of the bending stress at the bottom of the loaded section has been found to be an over-prediction, and hence erring on the safe side. The corresponding orthotropic problem has been studied using a Fourier series stress function approach9 and it has been shown that. (a) the σ x and τ xy distributions at sections close to a concentrated load are significantly different from the linear and parabolic distributions, respectively, predicted by the engineering beam theory; such discrepancies are more severe than for the isotropic case when E x > E y and increase with the degree of orthotropy E x /E y for any given span-to-depth ratio. (The τ xy distribution is important for orthotropic beams made up of fibre-reinforced layers because excessive values of this interlaminar shear stress can cause debonding or delamination of the layers.) (b) The use of the engineering beam theory for predicting the bending stress at the bottom of the loaded section is not safe, unlike for isotropic beams, because this may be lower than the actual value.

3.9

Frictionless Contact Between Isotropic Cylinders (Hertz Problem)

As another application of Flamant’s solution, let us consider the problem of line contact between two parallel infinitely long isotropic cylinders (Fig. 3.22) and estimate the actual contact width 2a and the corresponding contact pressure distribution p(x) when the cylinders are pressed together by the application of a force. The origin of the coordinate system is taken at the mid-point of the contact width. The surfaces of the cylinders are assumed to be very smooth so that no frictional forces are generated. Further, they are assumed to deform such that the contact width is quite small compared to their radii. Thus, the deformation of either cylinder is confined to a very short arc of its total circumference, and this deformation and the stresses can be captured fairly accurately by treating each body as a half-plane. (In contact mechanics parlance,10 such a contact is referred to as non-conformal.) The only input required for such an idealisation is the amount of overlap h(x) which would occur if the contacting bodies could freely 8 See S. P. Timoshenko, J. N. Goodier, Theory of Elasticity, McGraw-Hill, 1970. 9 J. M. Whitney, Elasticity analysis of orthotropic beams under concentrated loads, Composites

Science and Technology, 22, 1985, 167–184; J. L. Sullivan, H. Van Oene, An elasticity analysis for the generally and specially orthotropic beams subjected to concentrated loads, Composites Science & Technology, 27, 1986, 133–155. 10 K. L. Johnson, Contact Mechanics, Cambridge University Press, 1985.

3.9

Frictionless Contact Between Isotropic Cylinders (Hertz Problem)

105

Fig. 3.22 Two-dimensional contact problem

interpenetrate each other, this being of importance because it represents the sum of the magnitudes of y-displacements of corresponding points on the two surfaces due to the contact pressure. An expression for h(x), most convenient for later use, is given by h(x) = h o −

x2 x2 − 2R1 2R2

(3.94)

which represents a parabolic approximation of the contacting profiles, quite acceptable over the small contact width. If one of the contacting surfaces is concave as shown in Fig. 3.23, the corresponding radius of curvature is taken to be negative. The half-plane problems which yield the surface displacements are as shown in Fig. 3.24 with the contact pressure p(x) unknown; these displacements are obtained by using Flamant’s solution (Eq. (3.81)) for an elemental load p(s)ds acting on an element of length ds at x = s, and integrating over the contact width. Thus, one gets ]  a[ ro (1 + μ) p(s) (1 − μ2 )2 p(s) v(x, 0) = ln ds (3.95) − |x − s| πE πE −a Fig. 3.23 Contact with a concave surface

106

3 Plane Problems in Polar Coordinates

Fig. 3.24 The corresponding half-plane problems

One can rewrite the above equation as. ) a ( 2 1 − μ2 p(s) ln |x − s|ds + a constant independent of x v(x, 0) = − πE −a

(3.96)

Using this result, it is necessary to determine the y-direction displacements v1 and v2 of the two half-planes of Fig. 3.24 by plugging in the corresponding material constants and to finally equate their sum to the overlap h(x) as v1 (x, 0) + v2 (x, 0) = h(x) = h o −

x2 x2 − 2R1 2R2

for − a ≤ x ≤ a

(3.97)

Instead of the above approach, wherein ho is unknown and r o some arbitrarily chosen large value, it is more convenient to enforce the condition obtained by differentiating both sides of Eq. (3.97) with respect to x, i.e. ) ( 1 1 v1,x (x, 0) + v2,x (x, 0) = h ,x = −x (3.98) + R1 R2 The left-hand side derivatives are easily obtained using Eq. (3.96) as v1,x = −

2(1 − μ21 ) π E1



a

−a

2(1 − μ22 ) p(s) ds, v2,x = − x −s π E2



a −a

p(s) ds x −s

by considering the two possibilities x > s and x < s while differentiating ln|x-s|.

(3.99)

3.9

Frictionless Contact Between Isotropic Cylinders (Hertz Problem)

107

Thus, an equation for the contact pressure is finally obtained from Eq. (3.98) as 

a −a

p(s) πE ds = x x −s 2R

(3.100)

where 1 E

=

(1 − μ21 ) (1 − μ22 ) + , E1 E2

1 1 1 + = R R1 R2

(3.101)

the former denoting the reciprocal of a net material stiffness parameter, and the latter, the relative curvature. Equation (3.100) is an integral equation; since the integral involves a point of singularity at x = s, it is often referred to as a singular integral equation. A formal solution of this equation is beyond the scope of this book, but available elsewhere.11 The final result for p(x) is given by p(x) =

E √ 2 a − x2 2R

(3.102)

The unknown contact width can be found out by equating the integral of p(x) to the total normal force P per unit length of the cylinders as  a p(x)d x = P (3.103) −a

which yields / a=2

PR πE

(3.104)

This indicates that the contact width is directly proportional to the square root of the total normal load, and inversely proportional to the square root of the relative curvature/net material stiffness parameter E. (The corresponding relations for a three-dimensional contact are different—see Sect. 5.2.) In terms of P, the contact pressure can be written as p(x) =

2P √ 2 a − x2 πa 2

(3.105)

which indicates an elliptical variation with a maximum value at the centre of the contact width (x = 0) and zero values at the ends (x = ±a). 11 S. G. Mikhlin, Singular Integral Equations, Pergamon, 1957; N.I . Muskhelishvili, Singular Inte-

gral Equations, Noordhoff, 1953.

108

3 Plane Problems in Polar Coordinates

The peak contact pressure is given by / pmax =

2P = πa

PE πR

(3.106)

Finally, the stresses within any solid can be found using Eq. (3.82) as σx = − σy = − τx y

2y π



2y 3 π

2y 2 =− π

a −a a



p(s)(x − s)2 ds ((x − s)2 + y 2 )2

−a a



−a

p(s) ds ((x − s)2 + y 2 )2 p(s)(x − s) ds ((x − s)2 + y 2 )2

(3.107)

with p(s) =

pmax √ 2 2P √ 2 a − s2 = a − s2. 2 πa a

The integrations involved in these equations are not easy in general, but are quite straightforward for the determination of the stresses along the y-axis, i.e. x = 0, as given by " # √  2y pmax a s 2 a 2 − s 2 pmax a 2 + 2y 2 √ σx = − ds = − − 2y 2 2 2 π a a a2 + y2 −a (s + y ) √  2y 3 pmax a pmax a a2 − s2 σy = − ds = − √ 2 2 )2 π a (s + y a2 + y2 −a τx y = 0 (because of symmetry)

(3.108)

Thus, σ x and σ y are principal stresses for points along the y-axis. Both of them are compressive for all y and decay monotonically with y from the surface (y = 0) where both of them have the same magnitude equal to pmax (Fig. 3.25). The maximum in-plane shear stress given by " #   2  σx − σ y  p y max = |τ | =  y − √ (3.109) 2  a a2 + y2 is more important and varies as shown; its maximum value occurs below the surface as given by τmax | y=0.786a = 0.30 pmax

(3.110)

3.9

Frictionless Contact Between Isotropic Cylinders (Hertz Problem)

109

Fig. 3.25 Sub-surface stress variation

To get a better feel for this contact problem, some numerical results are presented in Table 3.3. We shall stop with this simple example of contact mechanics here. Boundary value problems involving contact are of great importance in the study of gears, bearings, rolling contact between railway wheels and rails, dovetail joints in aircraft engines and human hip joints; the corresponding analysis is quite interesting because one has to account for many complicating factors including conforming surfaces, friction, thermal effects and the change in contacting profiles due to plastic deformation or wear.

Table 3.3 Contact width and peak pressure due to a normal load of 200 kN Materials and dimensions

Two cylinders (R2 = R1 ) A cylinder and a flat surface (R2 = ∞)

A cylinder and a concave surface (R2 = − 3R1 )

2a (mm)

pmax (MPa)

2a (mm)

pmax (MPa)

2a (mm)

pmax (MPa)

Steel & steel*, 0.131 R1 = 1.5 cm

966.3

0.186

683.3

0.228

557.9

Steel & steel, R1 = 3 cm

0.186

683.3

0.264

483.1

0.323

394.5

Steel & aluminium*, R1 = 3 cm

0.264

482.9

0.373

341.5

0.457

278.8

*E steel = 3E Al = 200 GPa, μ = 0.3 for both

110

3.10

3 Plane Problems in Polar Coordinates

Bending of a Semicircular Beam

With reference to this problem, we shall focus on one particular aspect of curved beam behaviour as distinct from that of straight beams, namely the presence of significantly high transverse normal stresses. It will be shown that these normal stresses are tensile when the curved beam is unbent towards the straight configuration. While this is not a serious problem as far as isotropic homogeneous beams are concerned, one has to account for them carefully while designing a composite beam; here, the tensile stresses can cause separation of the different fibre-reinforced layers which are usually stacked up and bonded together to form the total thickness of the beam. This debonding of the layers—referred to as delamination—is an important mode of failure of a layered structure because it leads to a significant loss of bending rigidity. Depending on the geometrical proportions and the nature of loading, delamination failure may precede the failure of individual layers and hence laminated composite structures need to be designed against this failure mode. As can be easily appreciated, the estimation of the transverse shear and normal stress at the interface between layers—the so-called interlaminar stresses—is the first step in characterising such delamination failures. The specific problem considered here is a homogeneous, isotropic or cylindrically orthotropic semicircular beam subjected to bending moments or shear forces at its ends (Fig. 3.26); this is a case of plane stress with the out-of-plane thickness taken as unity. The pole of orthotropy is taken to be at the origin of the polar coordinate system. While it is possible to solve this problem using the compatibility equation in terms of Airy’s stress function (Eqs. (3.8) or (3.9a)), we shall present a slightly more elegant formulation. The two load cases of end moment and end shear are dealt with separately.

3.10.1 Load Case 1: End Moments This is a case (Fig. 3.26a) where there is no circumferential variation of the state of stress, because the moment at every cross-section is the same while the shear force is zero. Taking the shear stress τ rθ to be zero, the second equilibrium equation (Eq. (3.1b)) is identically satisfied, while the first (Eq. (3.1a)) reduces to Fig. 3.26 Bending of a semicircular beam

3.10

Bending of a Semicircular Beam

σr ,r +

111

(σr − σθ ) =0 r

(3.111)

This equation is also identically satisfied if the stress components are obtained from a scalar function ψ(r) by the relations σr =

ψ , σθ = ψ,r r

(3.112)

As with Airy’s stress function formulation, the function ψ has to be obtained from the compatibility condition (Eq. (3.5)). For the present case of circumferential invariance, this equation reduces to −r (α11 σr + α12 σθ ),r + [r 2 (α12 σr + α22 σθ ),r ],r = 0

(3.113)

where α ij are the compliance coefficients applicable for plane stress as given by α11 =

1 μr θ μθr 1 , α22 = , α12 = − =− Er Eθ Er Eθ

(3.114)

This compatibility equation can be expressed in terms of ψ as r 3 ψ,rrr + 2r 2 ψ,rr − (E θ /Er )(r ψ,r − ψ) = 0

(3.115)

This equation is of the Euler-Cauchy form and can be converted into an equation with constant coefficients (as was done for Eq. (3.17) in Sect. 3.2) by using the transformation r = es or s = ln r which yields ψ,sss − ψ,ss − (E θ /Er )(ψ,s − ψ) = 0

(3.117)

The solution of this equation is straightforward as given by ψ = Aes + Beλs + Ce−λs = Ar + Br λ + Cr −λ for λ / = 1

(3.117)

where λ2 = E θ /E r . If the body is isotropic, then λ = 1 and the first two terms on the right-hand side do not represent linearly independent solutions. For this case, the solution is given by ψ = Aes + Bses + Ce−s = Ar + Br ln r + C/r

(3.118)

The stresses can then be found from Eq. (3.112). The constants A to C are determined by using the conditions:

112

3 Plane Problems in Polar Coordinates

σr = 0 at r = a, b  b σθ r dr = −M

(3.119)

a

where the latter is the moment condition on the shorter ends enforced in a St.Venant sense. When the above conditions are satisfied, the zero circumferential force condition given by  a

b

σθ dr = 0

(3.120)

can be verified to be automatically satisfied. The stress field is finally obtained as follows. For λ / = 1: ] ] 2λ λ+1 − a λ+1 ) − r λ+1 (b2λ − a 2λ ) 2M(λ2 − 1) r (b σr = 1r λ+1 +a λ bλ (abλ − ba λ ) ] ] 2λ λ+1 − a λ+1 ) − r λ+1 (b2λ − a 2λ ) 2M(λ2 − 1) λr (b σθ = 1r λ+1 −λa λ bλ (abλ − ba λ ) τr θ = 0

(3.121a)

where 1 = (λ + 1)2 (a 2 b2λ + b2 a 2λ ) − (λ − 1)2 (a 2+2λ + b2+2λ ) − 8λa 1+λ b1+λ . For λ = 1: σr = σθ =

4M 2

(

r b a 2 b2 b ln − a 2 ln − b2 ln 2 r a a r

)

( ) 4M 2 a 2 b2 b r b b − a 2 − 2 ln − a 2 ln − b2 ln 2 r a a r

τr θ = 0

(3.121b)

( )2 where 2 = 4a 2 b2 ln ab − (b2 − a 2 )2 . The above solution is an exact elasticity solution for the problem provided the applied normal stresses σ θ corresponding to the moment M on the ends have exactly the same radial variation as in the above equations; otherwise, there will be localised variations in the stress field near the ends which die out as one moves away, as per St.Venant’s principle.

3.10

Bending of a Semicircular Beam

113

A comparison of the present solution with that of the corresponding straight beam problem (Sect. 2.3.2) shows that the most important differences are the presence of nonzero transverse normal stresses σ r here, and that the bending stress σ θ does not vary linearly through the thickness. It is well known that a nonlinear bending stress variation is also predicted by Winkler-Bach conventional curved beam theory12 based on the EulerBernoulli hypothesis; from this theory, it is also known that the neutral axis is not at the mean radius of the beam, but a little shifted towards the centre of curvature, and that the maximum bending stress occurs at the inner diameter. In order to avoid repetition, we shall not present a detailed comparison of the elasticity and CET solutions for curved beam problems as was done for straight beams; we shall merely point out that, for general loading, the non-classical effects of transverse shear deformation and thickness-stretch arise here too and are of essentially the same nature as for straight beams. Our discussion here will be confined to the radial stress which is completely ignored in Winkler’s theory. The maximum radial stress occurs at the critical location r cr given by σr ,r = 0

(3.122)

which yields rcr

]1/2λ [ √ (λ + 1)ab(bλ−1 − a λ−1 ) = ab for λ / = 1 (λ − 1)(bλ+1 − a λ+1 ) / 2 ln(b/a) =ab for λ = 1 b2 − a 2

(3.123)

It is possible to substitute this in the expression for σ r to find out the maximum radial stress. However, since the final expression for this maximum stress turns out to be rather lengthy,13 it is not explicitly presented here; instead, numerical results based on it are presented for some cases later (Sect. 3.10.3), along with those for shear loading.

3.10.2 Load Case 2: End Shear For this case (Fig. 3.26b), the bending moment at any section is proportional to sin θ and the shear force is proportional to cos θ. Thus, the bending stress σ θ would be proportional to sin θ and the shear stress τ rθ to cos θ . A solution corresponding to these variations can be obtained by starting with ) ( ψ ψ ψ sin θ, τr θ = − cos θ (3.124) σr = sin θ, σθ = ψ,r + r r r 12 See E. P. Popov, Engineering Mechanics of Solids, Prentice-Hall, 1998. 13 See W. L. Ko, Delamination stresses in semicircular laminated composite bars, NASA Technical

Memorandum TM 4026, 1988.

114

3 Plane Problems in Polar Coordinates

where ψ(r) is an undetermined function; these can be verified to satisfy the two equilibrium equations (Eq. (3.1)) identically. The function ψ is obtained from the compatibility equation (Eq. (3.5)) which reduces to r 2 ψ,rrr + 3r ψ,rr − αψ,r = 0

(3.125)

12 +α66 , which is equal to 3 if the material is isotropic. Proceeding as where α = α11 +2α α22 with the earlier case, a solution for this equation can be immediately written down as

ψ = A + Br η + Cr −η

(3.126)

√ where η = 1 + α; unlike for the earlier case, the above solution is valid without any change for the isotropic body as well. The boundary conditions required for the determination of the constants A to C are σr = 0 at r = a, b



b

a

τr θ dr = ∓P at θ = 0, π

(3.127)

When the above conditions are satisfied, the zero shear conditions on the two lateral surfaces and the zero moment condition at the ends given by  a

b

σθ r dr = 0 at θ = 0, π

(3.128)

are automatically satisfied. The final solution for the stress field is given by η P(r η − a η )(r η − bη ) sin θ Ωr η+1 η P(r η − a η )(r η − bη ) τr θ = − cos θ Ωr η+1 η P[(r η − a η )(r η − bη ) + η(r 2η − a η bη ) sin θ σθ = Ωr η+1 σr =

(3.129)

where Ω = 2(bη − a η ) − η(a η + bη ) ln ab . Once again, this solution is an exact elasticity solution for the problem only if the applied shear stresses τ rθ corresponding to the force P on the ends have exactly the same radial variation as in the above equation.

3.10

Bending of a Semicircular Beam

115

The critical radius corresponding to maximum σ r is given by ⎡ /[ rcr = ⎣

⎤1/η ] (a η + bη )2 + 4a η bη (η2 − 1) − (a η + bη ) ⎦ 2(η − 1)

(3.130)

It is also important to note from Eq. (3.124) that the radial variation is exactly the same for both σ r and τ rθ , and hence the maximum shear stress also occurs at r = r cr though at a different circumferential location.

3.10.3 Calculation of the Transverse Normal Stress As stated earlier, we shall present numerical values of the maximum transverse normal stress σ r for some chosen cases (Table 3.4). For the sake of comparison, the values of the maximum bending stress σ θ occurring at the inner diameter are also tabulated. Besides the isotropic case, two orthotropic materials are considered, one with E θ /E r = 25 and the other with E θ /E r = 49, with the other constants given by μθ r = 0.25, Grθ = 0.5 E r . These two sets of properties are typical of graphite-epoxy composites with the graphite fibres oriented along the θ-direction. A look at Table 3.4 shows that, for b/a = 1.1, the maximum radial stress is more or less the same for the three materials considered and occurs at nearly the same radial location which is very close to the mean radius. In terms of the non-dimensionalisation employed, the values are almost the same for the two loads considered; the non-dimensionalisation for the shear load case is such that PRm , which is the bending moment at the central Table 3.4 Maximum normal stress in the radial direction for curved beams Details

b/a = 1.5 or Rm /h = 2.5

b/a = 1.1 or Rm /h = 10.5

r cr /a

σ r max

σ θ max at r = a

r cr /a

σ r max

σ θ max at r = a

1.208

1.536

17.33

1.048

1.502

65.07

End moment Isotropic E θ /E r = 25

1.209

1.509

18.13

1.048

1.501

65.29

E θ /E r = 49

1.210

1.485

18.91

1.048

1.499

65.51

End shear Isotropic

1.200

1.522

18.45

1.048

1.501

66.09

E θ /E r = 25

1.193

1.462

21.73

1.048

1.497

66.80

E θ /E r = 49

1.186

1.418

24.60

1.048

1.493

67.51

Note All the stress values are normalised with respect to (M/Rm h) or (P/h) as the case may be; h = b − a; Rm = a+b 2

116

3 Plane Problems in Polar Coordinates

section about its centroidal level, replaces M. The above observations are also true for thinner beams with b/a ≤ 1.1. Thus, for such beams, one can say that the maximum radial stress can be obtained with sufficient accuracy by using the formula σr max = 1.5

Mlocal Rm h

(3.131)

where M local is the local bending moment at any section of the beam due to all the applied loads and calculated about the centroidal level. For thicker beams, the critical radial location as well as the maximum radial stress depends more significantly on the material properties. This critical location is always on the inner side of the mean radius, while the corresponding radial stress may be a little larger or smaller than that given by Eq. (3.131) as can be seen from Table 3.4. As pointed out earlier, the radial variation of the transverse shear stress τ rθ is the same as that of σ r for the case of shear loading, and hence the maximum values of τ rθ are the same as those tabulated for σ r . It is seen that the values of the maximum radial stress are always much smaller than those of the corresponding maximum circumferential bending stress. However, it should be kept in mind that this would not imply that the radial stress is not important, except for the case of an isotropic body with equal strengths in all directions. For the orthotropic cases considered, with reinforcing fibres along the circumferential direction, the strengths in the radial direction are significantly lower and hence the radial stress, though small, can initiate failure. Further, if the total depth is built up by bonding a number of layers together, the transverse strength of the bond, usually referred to as the interlaminar peel strength, is only a few per cent of the longitudinal strength of any layer along the fibre direction. For such cases, as seen in Table 3.4, it is possible for the bond to fail before the individual layers are overstressed, unless the beam is quite thin. Such delamination failures are common not only with curved beams loaded as considered here, but also in curved portions of thin-walled sections like a C-section or cylindrical tubes, and have to be carefully considered during the design process.14 Finally, it is appropriate to point out that the elasticity solutions described here are also useful for the design of composite curved beam test specimens for the determination of the interlaminar peel strength, with appropriate geometric dimensions such that failure occurs in this particular mode; in this context, it is interesting to note that quite a few variations of geometry including elliptical contours and angle sections with a fillet at the junction have been considered for such specimens.15 14 K. T. Kedward, R. S. Wilson, S. K. McLean, Flexure of simply curved composite shapes, Com-

posites, 20, 1989, 527–536. 15 C. C. Hiel, M. Sumich, D. P. Chappell, A curved beam test specimen for determining the interlam-

inar tensile strength of a laminated composite, Jl. of Composite Materials, 25, 1991, 854–68; W. Cui, T. Liu, J. Len, R. Ruo, Interlaminar tensile strength (ILTS) measurement of woven glass/polyester laminates using four-point curved beam specimen, Composites—Part A: Applied Science and Manufacturing, 27,1996, 1097–1105.

3.11

3.11

Summary

117

Summary

A number of polar coordinate solutions have been presented in this chapter, covering a variety of important problems including that of a cracked plate and a contact problem. As in the previous chapter, important features of structural behaviour have been carefully highlighted for each of the problems considered, including some orthotropic cases for which the solutions are too mathematical to be discussed in complete detail here.

4

Torsion of Non-circular Sections

Torsion of non-circular sections is an important problem in the theory of elasticity for which a simple strength of materials approach does not exist, except for some special cases. The fundamental difference between a circular and a non-circular section loaded in torsion is that while the former remains plane with no tendency to undergo out-of-plane warping as it twists, the latter undergoes such warping. When the shaft is prismatic and cross-sectional warping is freely permitted, the problem reduces to one of two dimensions because the rate of twist due to applied end torques is uniform along the length of the shaft and every cross-section undergoes the same warping displacement. Such uniform torsion is referred to as St. Venant torsion and the corresponding two-dimensional analysis is the topic of the present chapter. For the sake of convenience, isotropic shafts are dealt with first, and the effect of orthotropy is examined separately later. The effect of restraining warping displacements is briefly discussed at the end of the chapter.

4.1

Formulation for Isotropic Shafts

The problem of torsion can be solved in two alternative ways—using either the displacement approach or the stress approach. In either case, a semi-inverse method is employed wherein some assumptions regarding the displacement or stress field are put forth a priori such that some of the field equations are automatically satisfied while the others are reduced to a simple form that is amenable to a rigorous solution.

© The Author(s) 2023 K. Bhaskar and T. K. Varadan, Theory of Isotropic/Orthotropic Elasticity, https://doi.org/10.1007/978-3-031-06345-9_4

119

120

4.1.1

4 Torsion of Non-Circular Sections

Displacement Approach (St. Venant’s Warping Function Formulation)

This is based on the following assumptions: (a) Any cross-section of the twisted bar behaves as a rigid disk with respect to deformation in its own plane, i.e. it retains its in-plane shape and size during twisting. (b) Points of any cross-section undergo longitudinal displacements leading to warping of the section from its original flat configuration. Such warping displacements are proportional to the rate of twist. Consider the displacement of a point P(x, y) on a cross-section which rotates through an angle θ when the prismatic shaft is subjected to end torques (Fig. 4.1). The in-plane displacements of this point as it moves to P’ in the cross-sectional view are given by u = −r θ sin α = −θ y v = r θ cos α = θ x

(4.1a)

while its out-of-plane warping displacement is w = ψ(x, y)

dθ dz

(4.1b)

where ψ(x, y) is an undetermined function referred to as St. Venant’s warping function. The rate of twist is hereafter denoted by θ ’. Fig. 4.1 Torsion of a prismatic shaft

4.1

Formulation for Isotropic Shafts

121

The strain field is given by εx = u ,x = 0; εz = w,z = 0;

ε y = v,y = 0 γx y = u ,y + v,x = 0

γx z = u ,z + w,x = θ ' (ψ,x − y)

(4.2)

γ yz = v,z + w,y = θ ' (ψ,y + x) The corresponding stress field has just two non-zero components τx z = Gθ ' (ψ,x − y) τ yz = Gθ ' (ψ,y + x)

(4.3)

These stress components should satisfy the equations of equilibrium (Eq. 1.27) without the body forces f x , etc. Since the rate of twist θ ’ is a constant for a given torque, the first two equilibrium equations are identically satisfied, and the third equation is reduced to the Laplace equation ∇ 2 ψ ≡ ψ,x x + ψ,yy = 0

(4.4)

To supplement these governing equations, the boundary conditions on the ends as well as the lateral surface of the shaft have to be enforced. The lateral surface conditions are enforced using the general boundary condition equations derived in Chap. 1 as. lσx + mτx y + nτx z = Tx lτx y + mσ y + nτ yz = Ty

(1.40)

lτx z + mτ yz + nσz = Tz where l and m are the direction cosines of the outward normal at any point of the crosssectional contour, n being zero since this normal is always orthogonal to the z-axis; T x , T y and T z are zero here since the lateral surface is free of applied tractions. One can immediately see that the first two equations are identically satisfied, and the third is reduced to l(ψ,x − y) + m(ψ,y + x) = 0

(4.5)

at any point (x , y) of the cross-sectional contour. The end conditions are enforced in a St. Venant sense and actually pertain to every cross-section of the shaft as given by (see Fig. 4.2) ¨ ¨ ¨ τzx d x d y = τzy d x d y = 0; (τzy x − τzx y)d x d y = T (4.6) A

A

A

122

4 Torsion of Non-Circular Sections

Fig. 4.2 Stresses on the cross-section

to imply that there is no net shear force at any section but only a net torsional moment T, which is taken to be positive when anticlockwise as seen from the positive z-direction. This displacement formulation is not commonly used because the satisfaction of the lateral surface boundary condition (Eq. 4.5) is not easy. The alternative stress formulation presented below is more convenient and leads to the same final solution; it is also more appropriate for the present problem where all the boundary conditions are specified in terms of stresses.

4.1.2

Stress Approach (Prandtl’s Stress Function Formulation)

This is based on the following assumptions: (a) The applied torsional loading generates only the shear stresses τ zx and τ zy , and all the other stress components are zero. (b) These two non-zero shear stress distributions are the same for all the cross-sections of a prismatic shaft subjected to end torques. It is easy to verify that these lead to automatic satisfaction of the first two equilibrium equations (Eq. 1.27) while the third reduces to τx z,x + τ yz,y = 0

(4.7)

which is satisfied identically if the stress components are obtained from a stress function φ(x, y), referred to as Prandtl stress function, by the relations φ,x = −τ yz , φ,y = τx z

(4.8)

4.1

Formulation for Isotropic Shafts

123

As explained earlier, such a stress formulation requires the satisfaction of certain compatibility equations so as to correspond to a meaningful state of deformation; in a general form, the compatibility conditions are given by Eq. (1.32) in terms of the strain components,. The strain components for the present problem are ε x = ε y = εz = γ x y = 0 φ,y τx z = γx z = G G τ yz −φ,x γ yz = = G G

(4.9)

which satisfy four out of the six compatibility conditions automatically, while the other two are reduced to (−γ yz,x + γx z,y ),x = 0 (−γx z,y + γ yz,x ),y = 0 i.e., γx z,y − γ yz,x = constant

(4.10)

In terms of φ, this can be rewritten as the Poisson equation ∇ 2 φ ≡ φ,x x + φ,yy = constant x G

(4.11)

The lateral surface condition at any point of the cross-sectional contour given by lτx z + mτ yz = lφ,y − mφ,x = 0

(4.12)

can be expressed in a more convenient form by using l=

dy dx , m=− ds ds

(4.13)

where s is the contour coordinate as shown in Fig. 4.3. Thus, φ,x which implies that

∂φ ∂s

=0

dx dy + φ,y =0 ds ds

(4.14)

124

4 Torsion of Non-Circular Sections

Fig. 4.3 Tangential and normal coordinates

i.e., φ = constant along S

(4.15a)

If the cross-section is a simply connected domain (i.e. bounded by a single continuous closed contour line), then the constant in the above equation can be taken as zero because it does not contribute to the derivatives of φ which yield the stresses; thus, we have φ = 0 on the boundary

(4.15b)

If the cross-section is multiply connected, then along each of the closed boundary contour lines, φ is a constant with different values for the different contours; this is a slightly more complicated case and will be discussed separately later (Sect. 4.5). The end conditions are as given by Eq. (4.6). Of these, the zero shear force conditions are automatically satisfied; for example, ) ¨ ¨  (  y τx z d x d y = φ,y d x d y = φ,y dy d x = φ| ytop (4.16) bottom d x = 0 since φ has the same value at the points (x, ytop ) and (x , ybottom ) of the contour line (see Fig. 4.4). The third end condition of Eq. (4.6) becomes ¨ ¨ T =− φ,x xd x dy − φ,y yd x dy ] ]   [   [ (4.17a) xright ytop (yφ)| ybottom − φdy d x =− (xφ)|xleft − φd x dy − leading to, ¨ T =2

φd x d y

(4.17b)

4.1

Formulation for Isotropic Shafts

125

Fig. 4.4 Integration over the cross-section

since (x left or x right , y) and (x, ybottom or ytop ) are points on the contour line where φ vanishes (see Fig. 4.4). (This equation is valid only for simply-connected domains and its counterpart for multiply-connected domains will be derived later in Sect. 4.5). It is important to note that irrespective of the actual shape of the cross-section, half of the total torque is resisted by shear stress τ zx and the other half by τ zy . Before proceeding ahead, let us briefly examine the assumptions of the displacement and stress formulations put forth above. The assumption that all stress components but τ xz and τ yz are zero leads to εx = εy = γ xy = 0 which implies that the cross-sectional shape is preserved during twisting, and to εz = 0 which implies that the warping displacement w is a function of x,y alone. Thus, the two sets of assumptions used in these alternative approaches are only different statements of the same final, expected state of deformation of the prismatic shaft under torsion. The constant in the governing equation (Eq. (4.11)) needs to be determined. This is easily done by comparing Eqs. (4.2) and (4.10), which yields ∇ 2 φ ≡ φ,x x + φ,yy = −2Gθ '

(4.18)

Thus, the torsion problem has been reduced to the solution of this governing equation along with φ = 0 on the boundary contour. If this two-dimensional boundary value problem can be solved exactly, such a solution would be an elasticity solution satisfying all the field equations and specifically corresponding to the state of torsion characterised by free warping and a uniform rate of twist as assumed initially. Since the stress field has been assumed to be invariant along the length of the shaft, the solution would be exact only if the torques at the end sections are actually applied as distributions of τ xz and τ yz given by Eq. (4.8). Otherwise, the solution is not exact but still valid at sections far away from the ends as per St. Venant’s principle. For a given θ’, once φ is determined, the stresses are calculated using Eq. (4.8), the strains using Eq. (4.9), and the warping displacement using Eq. (4.3) which are rewritten as

126

4 Torsion of Non-Circular Sections

w,x =

τx z + θ ' y; G

w,y =

τ yz − θ'x G

(4.19)

The torsional constant J of the cross-section defined as J=

T Gθ '

(4.20)

is obtained by calculating T using Eq. (4.17b). In order to understand the variation of the net shear stress over the cross-section, let us consider a line along which φ is constant (Fig. 4.5). Denoting the tangential and normal coordinates along this line as s and n, respectively, we have τzn = τzx l + τzy m τzs = τzy l − τzx m

(4.21)

where l and m are the direction cosines of n with reference to x–y axes, as given by l=

dy dx or , ds dn

m=−

dx dy or ds dn

Using these relations and Eq. (4.8), Eq. (4.21) can be rewritten as ) ( dx dy = φ,s = 0 τzn = φ,y − φ,x − ds ds dy dx τzs = −φ,x − φ,y = −φ,n dn dn

Fig. 4.5 A shear line

(4.22)

(4.23)

4.2

Solutions for Isotropic Simply-Connected Domains

127

Thus, along the constant φ line, the net shear stress is just τ zs directed along the tangential direction; for this reason, such a line is called a line of shear stress or a shear line. Note, however, that the magnitude of the net shear stress is not constant along this line, but depends on the gradient of φ in the normal direction. (The shear line gives the direction of the net shear stress and is analogous to a stream line of fluid mechanics which gives the direction of the velocity.)

4.2

Solutions for Isotropic Simply-Connected Domains

4.2.1

An Elliptical Section

Considering an elliptical cross-section with its principal axes oriented along the x, y directions (Fig. 4.6), a solution for φ is simply given by the equation of the boundary as (

) x2 y2 φ =k 2 + 2 −1 b a

(4.24)

where k is obtained from the governing equation (Eq. (4.18)) as k=− (

Gθ ' 1 a2

+

1 b2

)

(4.25)

The torque is given by ¨ (

) y2 x2 T = 2k + 2 − 1 dx dy b a2 ) ( 3 πa b πab3 − πab = −kπab + = 2k 4b2 4a 2 Fig. 4.6 An elliptical section

(4.26)

128

4 Torsion of Non-Circular Sections

and hence, k=−

T πab

(4.27)

Thus, in terms of T, the stresses are given by τzx = −

2T y , πab3

τzy =

2T x πa 3 b

(4.28)

which show that the net shear stress variation along any principal axis of the ellipse is linear with respect to the radial distance and this stress acts in a direction perpendicular to the radial line (see Fig. 4.6), exactly as is the case with the circular cross-section; however, this is true only along the two principal axes and not along any general radial direction. The shear lines corresponding to constant φ given by x2 y2 + 2 − 1 = constant 2 a b

(4.29)

is a set of geometrically similar ellipses including the boundary; these are as shown in Fig. 4.7 for a > b and with φ undergoing the same incremental change from one ellipse to the next. Such a figure is useful in two ways—apart from showing the direction of the net shear stress at any point, it also shows the region of maximum shear stress characterised by a very close spacing of the shear lines. From this figure, it is very clear that the shear stress at any point of a radial line acts in the same direction and this direction is the tangential direction to the boundary at its intersection point with the radial line. Further, contrary to preliminary intuition, the maximum shear stress occurs at the ends of the minor axis, which are points closest to the centre, and not at points farthest from it; this maximum stress is given by τmax = |τzx |(0,±b) =

Fig. 4.7 Shear lines

2T πab2

(4.30a)

4.2

Solutions for Isotropic Simply-Connected Domains

129

Fig. 4.8 Shear stress contour lines

while the stress at the ends of the major axis is | | 2T |τzy | = (±a,0) πa 2 b

(4.30b)

which is smaller than τ max by a factor of a/b. In order to get a better feel of the variation of the net shear stress, contour lines corresponding to constant shear stress given by / / x2 2T y2 2 + τ2 = + = constant (4.31) τnet = τzx zy πab b4 a4 are shown in Fig. 4.8, once again for a > b; these are more oblong ellipses compared to the boundary, with the ratio of the axes given by (a/b)2 . Using Eqs. (4.25) and (4.27), one gets θ' =

) ( 2 T 2 + b a πa 3 b3 G

(4.32)

πa 3 b3 a 2 + b2

(4.33)

and hence the torsional constant as J=

Finally, using Eq. (4.19), the warping displacement is obtained as w=

T πa 3 b3 G

) ( 2 b − a2 x y

(4.34)

The corresponding contour lines are hyperbolas as shown in Fig. 4.9; the warping displacement changes sign from one quadrant to the next with zero values on the two axes of the ellipse. The points at which the magnitude of this displacement is maximum can be easily found out by considering the parametric form of the elliptical boundary as given by x = a cos t, y = b sin t

(4.35)

130

4 Torsion of Non-Circular Sections

Fig. 4.9 Warping displacement contour lines

Thus, the warping displacement, proportional to xy, varies with t as w ∝ sin t cos t

(4.36)

and hence takes maximum values for t = ± 45°, ± 135°, i.e. at the points √ √ (±a/ 2, ± b/ 2). This maximum value is given by |w|max =

| 2 | T |b − a 2 | 2 2 2πa b G

(4.37)

To get a physical feel of warping, consider a numerical example. Consider an isotropic shaft with a = 25 mm, b = 15 mm, and twisted through an angle of 4° over its length of 1 m. Using Eq. (4.32), the above expression for wmax can be rewritten as ( ) θ ' ab a 2 − b2 |w|max = (4.38) 2 a 2 + b2 which takes a value for 0.006 mm for this specific example while the curvilinear displacement of a tip of the major axis would be (25 × 4π/180) or 1.75 mm if this cross-section is assumed to rotate while the other end of the shaft is fixed.

4.2.2

The Special Case of a Circular Section

With b = a, the equations of the elliptical cross-section reduce to the following: / Tr 2T x 2 + y2 = τnet = τzs = πa 4 Ip J=

πa 4 = Ip 2 w=0

(4.39)

where I p is the polar moment of inertia about the centroidal axis. Further, the shear lines reduce to a set of concentric circles with s along the circumferential direction. Thus, one can see that the elasticity solution coincides with the simple CET solution for circular

4.2

Solutions for Isotropic Simply-Connected Domains

131

shafts based on the assumption that the cross-sections simply rotate without any in-plane deformation or out-of-plane warping. Let us make note of some important features of the circular section. Rewriting the expression for the torsional constant of the elliptical section (Eq. (4.33)) as J=

πa 3 b3 A4 = a 2 + b2 4π 2 I p

(4.40)

with A and I p denoting the area and the polar moment of inertia about the centroid, one can immediately see that for a given area of cross-section, J is maximum when I p is minimum which is the case corresponding to the circular cross-section. The torque carrying capacity or the torsional strength for a given material depends on the ratio T /τmax . For the elliptical section, this is given by T τmax

=

πab2 Ab = 2 2

(4.41)

which, for a given area A, increases as b increases and is hence maximum when b = a. Thus, it is clear that the circular section is the best from the viewpoints of both torsional strength and rigidity. Though this conclusion has been deduced with respect to the elliptic section, it can be shown that this is true with respect to any other simply-connected shape as well.

4.2.3

An Equilateral Triangle

Similar to the problem of the elliptical section, the equation of the boundary yields the stress function for the equilateral triangle, when the coordinate axes are chosen at the centroid. With reference to Fig. 4.10, this is given by )( ) ( ( √ √ a) 2a 2a x + 3y − φ=k x+ x − 3y − 3 3 3 (4.42) = k(x 3 − 3x y 2 − ax 2 − ay 2 + 4a 3 /27) which can be verified to satisfy the governing Poisson equation when k = Gθ ’/2a. The stresses and the torque are then obtained as (a + 3x)y a 2 + 3y 2 ) (2ax − 3x = Gθ ' 2a

τzx = −Gθ ' τzy

(4.43a)

132

4 Torsion of Non-Circular Sections

Fig. 4.10 Equilateral triangular section

T =

Gθ ' a 4 √ 15 3

(4.43b)

The shear lines and the net shear stress contour lines are as shown in Fig. 4.11, which indicate due symmetry corresponding to the equilateral triangular domain; for the sake of better visualisation, the contour plot is shaded with darker regions indicating higher stress.

Fig. 4.11 (a) Shear lines (b) Stress contour plot

4.2

Solutions for Isotropic Simply-Connected Domains

133

The maximum shear stress occurs at the mid-points of the sides and is given by √ Gθ ' a 15 3T τmax = (4.44) = 2 2a 3 while the shear stress at the centre and the three corners is zero.

4.2.4

A Rectangular Section

A solution for this section can be developed only in terms of infinite series—either a single series or a double series. With reference to the centroidal axes (Fig. 4.12), the required stress function has to be an even function of both x and y because the stress components τ zx (given by φ ,y ) and τ zy (given by φ ,x ), are antisymmetric about the x-axis and y-axis, respectively. A double trigonometric series satisfying this symmetry requirement as well as the zero values at the four edges can be employed for this solution as ∞ Σ

φ=

∞ Σ

Amn cos

m=1,3,.. n=1,3,..

mπ x nπ y cos 2a 2b

(4.45)

and the constants Amn can be found out using the governing equation. However, such a solution is in general slowly convergent, especially with respect to the series in the direction of the longer edge, and is hence not very convenient for numerical calculations. To eliminate this difficulty, a single series solution as described below is commonly employed. Assuming a > b, the solution is sought in terms of a single series in the y-direction as given by ∞ Σ

φ=

X n (x) cos

n=1,3,..

nπ y 2b

(4.46)

which satisfies the zero conditions at y = ± b. This reduces the governing Poisson’s equation to ∞ Σ n=1,3,.

''

(X n −

n2π 2 nπ y X n ) cos = −2Gθ ' 4b2 2b

(4.47)

Expanding the right-hand side also in a cosine series as −2Gθ ' = −2Gθ '

∞ Σ n=1,3,..

4 nπ nπ y sin cos , nπ 2 2b

(4.48)

134

4 Torsion of Non-Circular Sections

Fig. 4.12 A rectangular section

one gets, ''

Xn −

4 n2π 2 nπ X n = −2Gθ ' sin for each n. nπ 2 4b2

(4.49)

This yields X n = An cosh

nπ x nπ nπ x 32Gθ ' b2 sin + + Bn sinh 2b n3π 3 2 2b

(4.50)

with An and Bn to be found out from the zero conditions at x = ± a. The final solution for the stress function is given by φ=

( ∞ x ) cosh nπ nπ 32Gθ ' b2 Σ 1 nπ y 2b sin 1 − cos π3 n3 cosh nπa 2 2b 2b

(4.51)

n=1,3,..

The shear lines and shear stress contour lines are as shown in Fig. 4.13, with zero stress at the centre and the four corners. Fig. 4.13 (a) Shear lines (b) Stress contour plot

4.2

Solutions for Isotropic Simply-Connected Domains

135

The maximum stress, occurring at the mid-points of the longer sides, is given by τmax = |τzx |(0,±b) =

) ( ∞ 1 16Gθ ' b Σ 1 1 − π2 n2 cosh nπa 2b

(4.52a)

n=1,3,..

Since

Σ∞

1 n=1,3,.. n 2

=

π2 8 ,

we get ⎛

τmax

∞ 1 8 Σ = 2Gθ ' b⎝1 − 2 2 π n cosh n=1,3,..

⎞ nπ a 2b

⎠

(4.52b)

The torque is given by ¨ T =2 φ dx dy =

Since

Σ∞

512Gθ ' ab3 π4

1 n=1,3,.. n 4

=

∞ Σ n=1,3,...

π4 96 ,

1 1024Gθ ' b4 − 4 n π5

∞ Σ n=1,3,...

1 nπa tanh 2b n5

(4.53a)

we get ⎛

T = Gθ ' ab3 ⎝

16 1024 b − 5 3 π a

∞ Σ n=1,3,...

⎞ 1 nπa ⎠ tanh 2b n5

(4.53b)

The convergence of the series occurring in the above equations for τ max and T is very rapid for a > b, and only a few terms are required for obtaining results correct up to four significant digits. For the special case of a square of side 2a, the results are T 0.2082(2a)3 J = 0.1406(2a)4

τmax = 0.6753Gθ ' (2a) =

4.2.5

(4.54)

A Comparison of Various Shapes

Because of the practical significance of the torsion problem, numerous studies have been carried out on sections of various shapes; some of them were based on rigorous mathematical methods while the others were carried out using various approximate analytical or numerical techniques. The final results, for some shapes, are summarized in Table 4.1, wherein a comparison of the torsional constants and torsional strengths, for the same total area (or the same weight), is also included; the torsional strength is given by T allowable corresponding to τ max = τ allowable . All the results are normalized with respect to those

136

4 Torsion of Non-Circular Sections

Table 4.1 Torsional properties of various isotropic solid sections Section Circle (radius r) Ellipse (major axis 2a, minor axis 2b)

J

τ max

J Jcircle∗

Tallowable Tcircle∗

πr 4 2

2T πr 3

1

1

2T π ab2

a/b = 2: 0.80 4: 0.47 10: 0.20

0.71 0.50 0.32

0.601T a3

0.88

0.74

at the boundary πa 3 b3 a 2 +b2

at the ends of the minor axis

Square (side 2a)

2.25a4

Rectangle (2a x 2b)

Equation (4.53)

at mid-points of the sides

Equilateral triangle (height a)

At mid-points of long sides a/b = 2: 0.72 (see Eq. (4.52)) 4: 0.44 10: 0.20 √ 15 3T 2a 3

4 a√ 15 3

0.62 0.50 0.35

0.73

0.62

0.66

0.57

0.75

0.71

at mid-points of the sides 17.64T c3

Right-angled triangle 0.0261c4 √ (sides c,c, 2c)

at mid-point of the hypotenuse T 0.395r 3

Semicircle (radius r) 0.294r 4

at mid-point of the curved boundary * circular section of the same area

of the circular section, which, for a given total area, is the best simply-connected shape from the viewpoints of both torsional strength and rigidity.

4.3

Membrane Analogy

This is an analogy between the problem of torsion of an isotropic shaft and the problem of a transversely loaded stretched membrane, put forth by Prandtl. Consider a membrane (a very thin flat plate-like structure without any bending rigidity) stretched in its plane by a uniform tension S per unit length (Fig. 4.14). If the membrane is held at its boundary and subjected to uniform transverse pressure q, then the deflection W is given by the equation ∇ 2 W ≡ W,x x + W,yy = −

q S

(4.55)

4.3

Membrane Analogy

137

Fig. 4.14 The loaded membrane

along with the condition W = 0 on the boundary.

(4.56)

Comparing these with the equations of the shaft, one can see that they are of the same form, and hence corresponding quantities can be related when the shape of the membrane is the same as that of the shaft cross-section. Taking q = 2Gθ ' S

(4.57)

the analogy can be expressed in terms of the following statements: (a) The deflection W of the membrane at any point gives the value of φ at that point. (b) The deflection contour lines (corresponding to constant W ) coincide with the shear lines. (c) At any point of a deflection contour line, the normal slope W ,n gives the magnitude of the net shear stress τ net . (d) Twice the volume enclosed by the membrane and the plane of its boundary, calculated ˜ as 2 W d x dy, gives the torque T. This analogy is very useful for explaining the torsional behaviour of complicated crosssections (as illustrated later), and sometimes to obtain the actual quantitative results by appropriate experiments with transversely loaded membranes.

138

4.4

4 Torsion of Non-Circular Sections

Approximate Analysis of Thin-Walled Open Sections

Let us first discuss the case of a long and narrow rectangular section (Fig. 4.15). The shape of the corresponding loaded membrane would be a cylindrical surface with little change along the length except near the shorter ends. Thus, one can assume that the stress function φ is a function of y alone to get φ,yy = −2Gθ '

(4.58)

τx z = φ,y = −2Gθ ' y

(4.59)

and hence,

without any constant of integration because the variation of τ xz has to be antisymmetric about the x-axis. The further steps yielding the torque are ) ( 2 ' t 2 φ = Gθ (4.60) −y 4  T = 2L φ dy (4.61) Lt 3 ' = Gθ 3 and hence, |τmax | = |τx z

max |

= Gθ ' t =

3T Lt 2

(4.62)

which occurs all along the length of the longer sides, except very near their ends. The other stress component τ yz is taken as zero everywhere because it would be much smaller than τ xz for L > > t. It should be noted that the torque calculation here as given by Fig. 4.15 The deformed membrane corresponding to a narrow rectangular section

4.5

Multiply-Connected Domains

139

Fig. 4.16 An arbitrary thin-walled section

Eq. (4.61) would be fairly accurate because it is based on integration of φ and not on the total moment of the assumed shear stress field; the latter approach would not be acceptable because, as pointed out earlier (see Eq. (4.17)), the contributions of the stress components τ xz and τ yz to the total torque are always equal irrespective of the shape of the cross-section. Before proceeding further, it is appropriate to examine the accuracy of the above approximate analysis. This can be easily done by comparison with the results of the exact solution of Sect. 4.2.4, which would show that the approximate theory is non-conservative in that the torsional rigidity is over-predicted while the maximum stress, due to a given torque, is under-predicted, but the errors are within 3% for L/t > 20. Now, let us consider a more general thin-walled open section consisting of a number of straight or curved segments (Fig. 4.16). The corresponding membrane deformation would now be more or less the same along the length of any segment except near corners and kinks, and thus, one can extend the idea developed above for the narrow rectangle. The resulting equations are T = Gθ '

 Σ Li t3 Gθ ' i t 3 ds or 3 3 τmax = Gθ ' tmax

(4.63)

wherein the possibility of a piecewise or continuous variation of the thickness along the contour has been accounted for.

4.5

Multiply-Connected Domains

4.5.1

Formulation

When the non-circular section has one or more holes or cut-outs, the shear-free boundary condition (corresponding to τ zn , n being normal to the boundary) is given, as in Eq. (4.15a), by

140

4 Torsion of Non-Circular Sections

φ = constant

(4.64)

along each such additional boundary contour. While the value of φ at the outer boundary has been taken as zero, the values on the other boundaries cannot be assumed arbitrarily. Thus, some additional equations arise in the solution of multiply-connected domains; these additional equations simply represent the single-valued nature of the warping displacement w along the hole contours and are as derived below. From Eqs. (4.22) and (4.23), we have −φ,n = τzs = τzy l − τzx m = τzy

dx dy + τzx ds ds

(4.65)

along a constant φ line. Also, from Eq. (4.19), we have τx z = G(w,x − θ ' y) τ yz = G(w,y + θ ' x) Hence, for any closed boundary contour along which φ is a constant, we get ]   [ dy dx −  φ,n ds =  G(w,y + θ ' x) + G(w,x − θ ' y) ds ds ds    = G  w,s ds + Gθ '  xdy − Gθ '  yd x

(4.66)

(4.67)

For w to be a single-valued function, the first integral on the right-hand side should vanish. Observing that each of the next two integrals represents the area enclosed by the closed contour (see Fig. 4.17 and note that the last integral is negative when the contour is traversed in the anticlockwise sense), we have, along each of the hole boundaries, Fig.  4.17 Significance of  xdy and  yd x

4.5

Multiply-Connected Domains

141

 −  φ,n ds = 2Gθ ' Ai

(4.68)

i

where Ai is the area of the ith hole; these are a set of N equations for a cross-section with N holes and are sufficient to determine the N unknown constant values of φ at the different hole boundaries. The implication of Eqs. (4.64) and (4.68) with reference to the loaded membrane would be clear by rewriting them as W = constant = Wi

(4.69a)

  q Ai −  W,n ds = , i.e, −  SW,n ds = q Ai S

(4.69b)

i

i

along each internal closed curve corresponding to a hole of the shaft cross-section. A constant deflection along each such curve can be achieved by attaching a rigid disk, with the same shape as the area included by the curve, to the unloaded, initially horizontal, stretched membrane, and by guiding the disk to move only vertically as the membrane deforms due to transverse pressure. Equation (4.69b) can then be interpreted as the vertical force equilibrium equation of the disk, wherein the upward force due to the transverse pressure is balanced by the vertical component of the membrane tension acting along the boundary of the disk (Fig. 4.18); note that this implies that the rigid disk should be taken as weightless. It should be noted that Eq. (4.69b) is valid for any closed contour on the membrane and thus Eq. (4.68) can be used along any convenient closed contour enclosing a hole and not necessarily coincident with its boundary. The expression for the torque for the multiply-connected domain is different from that given by Eq. (4.17b) for the simply-connected domain. Starting from the third of Eq. (4.6), one gets ¨ ¨ ¨ T = (τzy x − τzx y)d x d y = − φ,x xd x dy − φ,y yd x dy (4.70) A

Fig. 4.18 Equilibrium of the rigid disk

A

A

142

4 Torsion of Non-Circular Sections

Fig. 4.19 Integration over the annular area

where the domain of integration A is now the hollow section. It is convenient to evaluate each of these integrals by splitting the domain1 ; this is as illustrated in Fig. 4.19 for the evaluation of the first integral over a domain with a single hole. For regions 1 and 2, one gets ]  [  ¨ ¨ x (xφ)|xright − φd x dy = φd x d y (4.71) − φ,x xd x dy = − left since φ is zero on the outer boundary. For regions 3 and 4, however, one has ]   [  ¨ ¨ left − x φ,x xd x dy = − (xφ)|xxhole φd x dy = −φ dy+ φd x d y − 1 hole left left 3

 [

¨ φ,x xd x dy = −

−



x

(xφ)|xright hole right −

]  ¨ φd x dy = φ1 xhole right dy+ φd x d y

4

(4.72) where φ 1 is the constant value of φ on the boundary of the hole. Combining the above equations, one has ¨ ¨  − φ,x xd x dy = φ1 (xhole right − xhole left )dy+ φd x d y (3+4)

(3+4)

¨ = φ1 A1 +

φd x d y (3+4)

where A1 is the area of the hole. 1 This can also be done by using Green’s theorem:



( f d x + gdy) =

˜ ∂g ( ∂ x − ∂∂ yf )d xd y.

(4.73)

4.5

Multiply-Connected Domains

143

Thus, for the entire hollow domain including the four regions, one gets ¨ ¨ − φ,x xd x dy = φ1 A1 + φd x d y

(4.74)

˜ and an identical result for − φ,y yd x dy by following similar steps. Putting these in Eq. (4.70) and generalizing for the case of multiple holes, one finally has ¨ Σ T =2 φd x d y + 2 φi Ai (4.75) where the integral on the right-hand side is over the hollow cross-sectional domain, and φ i is the value of φ on the boundary of the ith hole of area Ai . In terms of the membrane analogy, replacing φ by W in the above equation, and noting that W i Ai represents the volume of the prismatic cylindrical portion under the ith rigid disk, it is clear that the torque is given by twice the volume enclosed by the deformed membrane along with its attached rigid disks and the plane of its outer boundary.

4.5.2

Some Simple Solutions

Exact solutions for hollow sections are straightforward when there is an exact solution for the corresponding solid section, and one of the shear lines therein coincides with the boundary of the hole. In such a case, the stress function associated with the solid section yields the solution for the hollow section as well because the additional conditions at the hole boundary, corresponding to a constant φ and a single-valued warping displacement w are automatically satisfied. It is also possible to explain this by visualising the removal of the material corresponding to such a hole from the solid shaft. Even in the twisted solid shaft, the cylindrical surface corresponding to this prismatic hole is a stress-free surface characterised by σn = τns = τnz = 0

(4.76)

where n is the normal direction, since a shear line by definition is one along which the net shear stress is always along the tangential direction; in other words, no load or stress transfer takes place across this cylindrical surface. Thus, for a given rate of twist, the stresses in the outer portion of the solid shaft would be unaffected when the material corresponding to the hole is removed from it. The above conclusions are already known with reference to the case of the concentric annular circular shaft for which the solid shaft equation τ = Gθ ' r

(4.77)

144

4 Torsion of Non-Circular Sections

Fig. 4.20 Torque reduction due to a hole

is applicable without any change and hence yields the same stress distribution for any given θ’; the associated torque, however, is less now, but as is quite well-known, not significantly reduced. This is most conveniently visualised by using the membrane analogy as shown in Fig. 4.20, where the reduction in torque corresponds to the volume of the hatched portion. A more general explanation of this, often employed in structural optimization exercises, is to say that material has been removed only from the least stressed portion of the solid shaft, and hence such removal leads to a more efficient, light-weight design without any significant reduction in the torsional strength or torsional rigidity. Extending the argument to a hollow elliptical cross-section bounded by geometrically similar ellipses x2 y2 + = 1 and a2 b2

x2 y2 + = 1, λ < 1, (λa)2 (λb)2

(4.78)

the stress function of Eq. (4.24) with the constant k defined by Eq. (4.25) is straightaway applicable. The torque of the corresponding solid shaft, as given by Eq. (4.32), is πa 3 b3 Gθ ' ) Tsolid = ( 2 a + b2

(4.79)

Due to the hole, this is reduced by an amount corresponding to the torque carried by 3 (λb)3 Gθ ' . the material which is removed, i.e. by π(λa) ((λa)2 +(λb)2 )

4.5

Multiply-Connected Domains

145

Thus, the torque of the hollow shaft is given by Thollow =

(1 − λ4 )πa 3 b3 Gθ ' ( ) = (1 − λ4 )Tsolid a 2 + b2

(4.80)

In other words, the torsional constants are related by Jhollow = (1 − λ4 )Jsolid

(4.81)

If the same torque were to be applied, then the maximum shear stress and the angle of twist are related by τmax

hollow

=

' θsolid τmax solid ' = ; θ hollow (1 − λ4 ) (1 − λ4 )

(4.82)

Note that these are very small changes compared to the reduction in the weight of the shaft given by π ab − π(λa)(λb) Ahollow = = 1 − λ2 Asolid πab

(4.83)

One can repeat the above steps to obtain results for hollow counterparts of the other solid sections for which solutions are available; however, the inner contour is not a simple scaled-down replica of the outer contour in such cases as can be seen from the shear line plots of the rectangle (Fig. 4.13), and hence such solutions are only of theoretical interest. The more general case where the contour of the hole does not coincide with a shear line is quite complicated and usually requires a numerical method of solution. We shall not discuss such complicated solutions here, but shall try to provide some insight regarding the salient features of such problems, along with some other problems of practical interest, in Sect. 4.6.

4.5.3

Approximate Analysis of Thin-Walled Tubes (Bredt-Batho Theory)

Consider a closed tube whose thickness t, not necessarily uniform along the contour, is small compared to the circumference. Visualising the membrane corresponding to this case, one can immediately see that its normal slope is nearly constant in the portion corresponding to the wall thickness (Fig. 4.21). This merely corresponds to the fact that the shear stress τ zs is more or less the same for the outer and inner contours of the thinwalled tube at any particular circumferential location, and this can be used to develop an approximate theory for thin-walled tubes; a corollary of this is that τ zn is zero everywhere through the wall thickness.

146

4 Torsion of Non-Circular Sections

Fig. 4.21 Membrane analogy for a thin-walled tube

From Fig. 4.21, one has t(−Wn ) = h = constant

(4.84)

τ t = constant

(4.85)

and hence,

along the circumferential direction, where τ is now the net stress. The volume enclosed is given by the area A enclosed by the mean contour of the thin-walled tube multiplied by h, and hence the torque T is given by T = 2τ t i.e., τ=

T 2 At

(4.86)

which is referred to as Bredt-Batho equation. Using Eq. (4.68), and A for the area of the hole without serious error, one gets    T ds 2Gθ ' A = −  φ,n ds =  τ ds =  2A t i.e., θ' =

 ds T  4 A2 G t

(4.87)

4.5

Multiply-Connected Domains

and hence the torsional constant as

147

 ds J = 4 A2 /  t

(4.88)

The accuracy of this theory is easily verified with respect to the exact solution for thin circular tubes; one finds that the value of τ as per Eq. (4.86) is smaller than the actual = 5 and 10, respecmaximum stress at the outer diameter by about 8% and 5% for Rmean t tively, while the torsional rigidity is under-predicted by about 1% and 0.2%, respectively. A similar verification has been carried out for square tubes of uniform thickness2 with the elasticity results obtained using finite element method; this study shows that the thin-wall approximation results in less than 6% error for the maximum shear stress at the midpoints of the sides of the outer boundary, and less than 2% error in the torsional rigidity, for side-to-thickness ratios greater than 25. If a section is of non-uniform thickness, Bredt-Batho equation clearly indicates that thicker portions are under-stressed as compared to thinner ones; thus, a tube of uniform thickness is the optimum from the viewpoint of maximum material utilisation. More accurate alternatives to Bredt-Batho theory have been developed; for instance, a theory based on a linear variation of the shear stress through the wall thickness has been developed for thick-walled rectangular box-beams3 and incorporated into British and European design procedures4 ; such a theory has also been developed for polygonal tubes.5 Finally, it is appropriate to point out that Bredt-Batho theory can straightaway be extended for the study of thin-walled multi-cell tubes as has been explained by Timoshenko.6

4.5.4

Closed Tube versus Slit Tube

At this juncture, it is very informative to discuss the relative advantage, with reference to torsional behaviour, of a closed tubular section over a corresponding open section of the same cross-sectional area. We shall do this with reference to the configurations shown in Fig. 4.22—a closed circular tube and its open counterpart, viz. a slit tube with 2 J. S. Lamancusa, D.A.Saravanos, The torsional analysis of bars with hollow square cross-sections,

Finite Elements in Analysis and Design, 6, 1989, 71–79. 3 J. Marshall, Derivation of torsion formulas for multiply connected thick-walled rectangular sec-

tions, ASME Jl. of Applied Mechanics, 37, 1970, 399–402. 4 see D. J. Ridley-Ellis, J. S. Owen, G. Davies, Torsional behaviour of rectangular hollow sections,

Jl. of Constructional Steel Research, 59, 2003, 641–663. 5 M. R. Hematiyan, A. Doostfatemeh, Torsion of moderately thick hollow tubes with polygonal

shapes, Mechanics Research Communications, 34, 2007, 528–537. 6 S. P. Timoshenko, Strength of Materials, Part 2 Advanced Theory and Problems, CBS Publishers

and Distributors, Delhi, 1986.

148

4 Torsion of Non-Circular Sections

Fig. 4.22 Membrane analogy for closed and slit tubes

a longitudinal cut along a generator. Membrane analogy enables an easy visualisation of the torsional rigidities of the two cases—for the same lateral pressure, one can see that the volume enclosed is much larger for the closed tube. Assuming the walls to be thin, a straightforward application of the formulae developed earlier leads to the following results:

τmax τmax

slit

closed

( )2 Jclosed 2π R 3 t R = =3 Jslit 2π Rt 3 /3 t ( ) 2 3T /2π Rt R for the same torque = =3 2 T /2π R t t

(4.89)

Thus, compared to a thin-walled closed tube, an open counterpart has very low torsional strength and much lower torsional rigidity. Thus, open sections are never used as primary torque-resisting members. In rare instances where they are required to resist torsion, their torsional strength and rigidity can be augmented by providing some warping restraint at the ends—this will be discussed later in Sect. 4.8.

4.6

A Brief Discussion of Some Geometrically Complicated Sections

Only problems amenable to simple solutions have been presented above, but it is possible, on the basis of such experience along with the membrane analogy, to intuitively guess the torsional behaviour of some complicated sections as explained below. We shall not, however, endeavour to discuss the mathematical aspects of the corresponding analytical or numerical solutions.

4.6.1

Protruding Sharp Corners

The behaviour near a sharp corner depends on whether the corner is pointing away from the domain—a protruding corner, or inwards—a re-entrant corner, and hence these are discussed separately. Protruding corners are points of zero stress, as is clear from the results of the triangular and rectangular sections; at any such corner, which is a part of

4.6

A Brief Discussion of Some Geometrically Complicated Sections

149

Fig. 4.23 Protruding corners

two intersecting unloaded straight lines of the boundary contour, both τ zx and τ zy are zero by virtue of the principle of complementary shear. This is also easy to visualise using the membrane analogy—the portion of the membrane, near such a corner on its boundary, would be nearly flat. Thus, the material near a protruding corner is very much under-stressed and hence a section with protruding corners is no better than a lighter counterpart with the sharp corners drastically rounded off, both in terms of torsional rigidity and torsional strength. The situation is exactly the same with respect to corners protruding from the boundary of a hole towards the hollow portion (Fig. 4.23). In this case, the membrane analogy requires such that a similar sharp cut be made in the otherwise smoothly-rounded rigid disk and one can easily visualise that the portion of the flexible membrane within this sharp cut would be nearly flat and in the horizontal plane of the rigid disk, indicating that stress levels are very small.

4.6.2

Re-Entrant Corners

These are corners pointing inwards—into the domain; they may occur at the roots of keyways or gear teeth, or at the junctions of thin-walled sections, or at the intersections of the sides of polygonal holes (Fig. 4.24). Considering the case of a boundary of a simply-connected domain with a single reentrant corner first, the effect of the corner is equivalent to the effect of a similarly shaped sharp knife tip pushing down a small portion of the deformed membrane near its boundary, due to which the portion of the membrane immediately next to the knife tip becomes nearly vertical. The effect of a re-entrant corner on the boundary of a hole is similar as can be visualised by imagining the shape of the membrane close to a sharp corner of the corresponding rigid disk. Thus, a re-entrant corner, unlike its protruding counterpart, is a stress-raiser. If such a corner is sharp, a stress singularity occurs there and the shear stress is theoretically infinitely large; this is an example, similar to that of Sect. 3.6, where the principle of complementary shear does not hold good, and infinitely large values of τ zx and τ zy are associated with zero values of τ xz and τ yz . In practice, these large stresses are relieved by the formation of a permanent set if the material is ductile, or a crack if it is brittle—both these undesirable consequences can be mitigated to some extent by providing a fillet at the corner. It is possible to approximately estimate the local stress

150

4 Torsion of Non-Circular Sections

Fig. 4.24 Re-entrant corners

concentration factor at a filleted corner for thin-walled tubes as well as open sections7 — we shall not present this analysis here, but shall merely point out that this factor is about 1.5 when the fillet radius is equal to the wall thickness (see Fig. 4.24c).

4.6.3

Slots and Grooves

Keyway slots and grooves need to be studied carefully because they are provided on the boundary of shafts which is a high-stress region; here we shall discuss the relative severity of some slotted configurations as shown in Fig. 4.25. Whenever a convex portion of the boundary of a loaded membrane is cut in and made flat or concave, it is easy to visualise that the normal slope of the membrane increases in its neighbourhood. Thus, flat and concave portions of the boundary are high-stress locations in general. This stress concentration has the effect of reducing the torsional Fig. 4.25 Flattened and slotted circular shafts

7 H. D.Conway, A simple formula for the maximum stress in a twisted angle or channel, International

Jl. of Mechanical Sciences, 14, 1972, 615–617.

4.6

A Brief Discussion of Some Geometrically Complicated Sections

Table 4.2 Effect of slots as shown in Fig. 4.25

151

Shaft J/J solid shaft τ max /τ max solid shaft τ max /τ max solid shaft (for same θ’) (for same T) (a)

0.996

1.08

1.08

(b)

0.866

1.28

1.48

(c)

0.898

1.75

1.95

strength of the shaft significantly, while the reduction in torsional rigidity is not likely to be as significant because only a small amount of material is removed. To illustrate this point, the torsional rigidity and the maximum shear stress are tabulated in Table 4.2 for the configurations of Fig. 4.25; with reference to the second configuration shown therein, it is assumed that the root of the rectangular keyway is carefully filleted and hence the corner singularity is not of concern here. The circumferential width of the flat or slotted region is taken to be one-fourth of the diameter of the shaft, as per the common practice while choosing keys. The calculations are based on empirical formulae available for these configurations, along with many others of practical importance, in standard handbooks.8

4.6.4

Eccentric Hole in a Circular Section

A circular shaft with an eccentric circular hole is obviously inferior to a corresponding concentric annular shaft because the former is obtained by removing some highly stressed material from the solid shaft as compared to the latter. It is easy to visualise the membrane analogues of the two shafts for the same lateral pressure, or the same rate of twist (Fig. 4.26). Thus, clearly, the torsional rigidity decreases, and the maximum stress, occurring at the point of the outer circle closest to the hole, increases, as the eccentricity of the hole increases. To provide a feel for the actual variation, results for a shaft with hole Fig. 4.26 Concentric versus eccentric hole

8 R. G. Budynas, A. M. Sadegh, Roark’s Formulas for Stress and Strain, McGraw-Hill, 2020.

152 Table 4.3 Results for an eccentric tube (d/D = 0.5)

4 Torsion of Non-Circular Sections

Eccentricity (e/D)

J/J e=0 *

τ max /τ max(e=0) * (for same θ’)

τ max /τ max(e=0) (for same T)

0.05

0.986

1.08

1.10

0.10

0.941

1.17

1.25

0.15

0.865

1.26

1.46

*J e=0

= π(D 4 − d 4 )/32; τmax (e=0) 16T D/π(D 4 − d 4 )

=

Gθ ' D/2

=

Fig. 4.27 Effect of a small eccentric hole

diameter equal to half of the outer diameter are presented in Table 4.3; for this case, the limiting value of e/D is 0.25 corresponding to zero minimum thickness. The above discussion pertains to the case when the hole diameter is not very small compared to the outer diameter. In case the hole is very small and is located in the outer portion of the domain, the membrane is deformed as shown in Fig. 4.27 indicating that there is a local increase in stress at points A and B and that the stress at B would be the maximum stress. The corresponding stress concentration factor is known to be 2. Obviously, the torsional rigidity is more or less unaffected. The behaviour of a shaft with many circumferentially spaced eccentric holes is qualitatively similar except that torsional rigidity is now significantly reduced. Exhaustive quantitative results for different variants of this case have been tabulated by Naghdi.9

9 A. K. Naghdi, Torsion of multihole circular cylinders, ASME Jl. of Applied Mechanics, 49, 1982,

432–435.

4.6

A Brief Discussion of Some Geometrically Complicated Sections

4.6.5

153

Effect of the Hole Shape

Considering the case of a section with a hole in its central portion, and assuming that the hole is small compared to the size of the outer boundary, membrane analogy suggests that the shape of the deformed membrane would be more or less unaffected when the shape of the hole is changed without a significant increase or decrease in its size. Thus, significant changes are not expected in the torsional rigidity or the maximum shear stress at the outer boundary for a given angle of twist. These intuitive conclusions have in fact been verified with respect to some cases. For example, it has been pointed out10 that the maximum shear stress, in terms of the rate of twist, for a circular section of radius R with a central square hole of side a, is the same as that of a solid circular shaft as long as a/R < 0.5; the corresponding results of the torsional rigidity indicate that it is almost the same when the square hole of side a is replaced by a circular hole of diameter a. A similar comparison is available11 for square shafts with square and circular holes at the centre, again leading to the same conclusions as stated above for hole sizes less than half that of the outer boundary.

4.6.6

Optimum Shape of a Hollow Section

Let us try to answer the following questions: (a) for a given outer boundary and for a given total area, what is the best shape of the hole from the viewpoint of proper material utilisation (so that both torsional strength and rigidity are maximum)? (b) the counterpart question—what is the best shape of the outer boundary for a given hole? The answers depend on whether the hole is small or large. When the hole is large, the section can be approximated as thin-walled, and as concluded earlier using Bredt-Batho theory (Sect. 4.5.3), a uniform thickness configuration is expected to be optimum. This conclusion has been corroborated by a rigorous optimization study12 based on an evolutionary algorithm involving removal of material from low-stress regions and its relocation at high-stress regions—optimum hole shapes were obtained for the example cases of a Greek cross shaft and an octagonal shaft. 10 J. T. Tielking, Torsion of a circular shaft with a square cavity, Mechanics of Structures and

Machines, 15, 1987, 167–176. 11 S. I. Chou, M. Shamas-Ahmadi, Complex variable boundary element method for torsion of hollow

shafts, Nuclear Engineering and Design, 136, 1992, 255–263. 12 Q. Li, G. P. Steven, O. M.Querin, Y .M. Xie, Stress based optimization of torsional shafts using

an evolutionary procedure, International Jl. of Solids and Structures, 38, 2001, 5661–5677.

154

4 Torsion of Non-Circular Sections

When the hole is small and the outer boundary is a reasonably well-rounded one not too different from a circle, the actual shape of the hole is not very significant as pointed out in the last section; to avoid stress concentration effects similar to those at re-entrant corners, a circular hole is often chosen as the best. If, on the other hand, one wants to optimise the shape of a small hole for an oblong outer boundary shape, or the outer boundary for a very oblong hole, the solution of the elliptical section (Sect. 4.2.1) can be used to provide some guidelines. Looking at the corresponding stress level contours (Fig. 4.8) and assuming that the nature of the solution is not altered significantly when a small central portion is removed, it is very clear that material removal should correspond to low stress levels, and thus the hole should be more oblong compared to the outer boundary. This means that, in general, for an optimum hollow section, the thickness should be more at relatively flat regions as compared to regions which are deeply curved; this conclusion agrees with that of more rigorous mathematical optimisation studies.13 Finally, as a general rule for any optimisation exercise for torsion, protruding corners should always be drastically rounded off to reduce material underutilisation, and re-entrant corners should always be avoided or liberally filleted to avoid local overstressing.

4.6.7

Irregular Sections

Many structural members designed for resisting their appropriate primary loads are also subjected to torsion; thus, torsional analysis is often required for very irregular shapes which do not readily admit of simple solutions. As an alternative to brute-force methods of numerical analysis, some thumb-rules and empirical formulae have been suggested for such sections. These are discussed below. For simply-connected shapes, a good estimate of the torsional constant can be obtained by using the equation for the elliptic section with the same area and polar moment of inertia, i.e., J=

A4 4π 2 I p

(4.90)

This was put forth by St. Venant himself based on the study of several complicated shapes using analytical methods. Sometimes, 4π2 in the above equation is replaced by 40.

13 see, for example, N. V. Banichuk, Optimisation of elastic bars in torsion, International Jl. of Solids

and Structures, 12, 1976, 275-286.

4.7

Orthotropic Shafts

155

The maximum shear stress always occurs on the boundary; to determine the actual location, the criterion of the largest inscribed circle has been suggested.14 This says that the maximum stress occurs at or very close to one of the points where the largest inscribed circle touches the boundary, and of all such points, the point where the boundary is most concave or least convex is the most critical one. Separate empirical formulae have been developed for convex (including flat) and concave portions of the boundary—the former in terms of the diameter of the largest inscribed circle, the local radius of curvature of the boundary and the total area of the section, and the latter in terms of these parameters as well as the angle through which the tangent to the boundary rotates in travelling around the concave or re-entrant portion. These formulae are stated to be applicable for fairly compact solid irregular sections as well as elongated straight and curved sections of nonuniform thickness and other common shapes such as I, T, L, Z, etc.

4.7

Orthotropic Shafts

We shall deal with two distinct cases of orthotropy—the first being rectilinear orthotropy with the principal material directions coinciding with the body axes x,y,z, and the second, shape intrinsic orthotropy, quite analogous to cylindrical orthotropy for a circular shaft, with two principal material directions coinciding with the tangential and normal directions of the boundary of the cross-section and the third with the longitudinal axis. We shall not deal with a wide variety of problems as was done for isotropic shafts, but shall restrict our discussion to just a few cases, which will help in pointing out the complications and the change in torsional behaviour due to material orthotropy. We shall continue to assume that all the fundamental hypotheses of St. Venant’s theory (or their equivalents as assumed by Prandtl) are applicable so that the only change that comes about in the governing equations is because of the orthotropic material constitutive law.

4.7.1

Rectilinear Orthotropy—Simple Solutions

Let us first identify the changes in the governing equations due to rectilinear orthotropy with respect to the x–y-z axes; for easy identification with reference to the earlier equations, we shall use the same equation numbers as before with just a prime added. Taking the shear moduli corresponding to τ zx and τ zy as Gxz and Gyz , respectively, the changed equations are τx z = G x z θ ' (ψ,x − y) τ yz = G yz θ ' (ψ,y + x)

(4.3’)

14 see R. G. Budynas, A. M. Sadegh, Roark’s Formulas for Stress and Strain, McGraw-Hill, 2020.

156

4 Torsion of Non-Circular Sections

G x z ψ,x x + G yz ψ,yy = 0

(4.4’)

lG x z (ψ,x − y) + mG yz (ψ,y + x) = 0

(4.5’)

ε x = ε y = εz = γ x y = 0 φ,y τx z = γx z = Gxz Gxz τ yz −φ,x = γ yz = G yz G yz

(4.9’)

φ,yy φ,x x + = constant G yz Gxz

(4.11’)

φ,yy φ,x x + = −2θ ' G yz Gxz

(4.18’)

and,

leading to

with w,x =

τx z + θ ' y; Gxz

w,y =

τ yz − θ'x G yz

(4.19’)

The stress-free boundary condition of the lateral surface reduces to a constant value of φ; for a simply-connected domain, this value can once again be taken as zero. A look at Eq. (4.18’) reveals that it can be reduced to the simple Poisson equation by a change of y-coordinate as defined by / Gxz η=y (4.91) G yz to yield φ,x x + φ,ηη = −2G yz θ '

(4.92)

with φ = 0 on the boundary of the mapped region in the x-η plane corresponding to the actual cross-section in the x–y plane. (Other such transformations are possible including

4.7

Orthotropic Shafts

157

those for cases where the in-plane orthotropic directions are inclined with respect to x,y axes.15 ) The other equations can also be rewritten in terms of η as /

φ,x = −τzy = −τzη

Gxz τzx = τzx or φ,η = √ G yz G x z /G yz ˜ ¨ 2 φ d x dη T =2 φ d x d y = √ G x z /G yz

φ,y = φ,η

φ,y + θ'y Gxz / / Gxz Gxz = φ,η /G x z + θ ' η/ G yz G yz / ( ) φ,η Gxz = + θ 'η / G yz G yz

(4.93)

(4.94)

w,x =

w,y = /

i.e.

i.e.

−φ,x − θ'x G yz

Gxz −φ,x = − θ'x G yz G yz ) / ( Gxz −φ,x ' −θ x / w,η = G yz G yz w,η

(4.95a)

(4.95b)

The implication of these equations is that it is possible to reduce the problem of torsion of a rectilinearly orthotropic section to that of an equivalent isotropic section for which the dimensions in the y-direction alone are scaled as given by Eq. (4.91). The final solution of the orthotropic section for the stresses, warping displacement, etc. can then be found out from the corresponding results of the counterpart isotropic case. From this correspondence, it is possible to infer some important qualitative conclusions. The foremost is that the orthotropic section behaves similar to an isotropic section of a different aspect ratio (the ratio of the vertical to horizontal dimensions); for example, an orthotropic elliptical section with semi-axes a and b such that / Gxz (4.96) a=b G yz 15 see T. Chen, C. J. Wei, St Venant torsion of anisotropic shafts: theoretical frameworks, extremal

bounds and affine transformations, Quarterly Jl. of Mechanics and Applied Mathematics, 58, 2005, 269–287.

158

4 Torsion of Non-Circular Sections

Fig. 4.28 Optimum fibre orientation for an elliptical section

behaves similar to an isotropic circular section in that it is the most efficient shape for a given material stiffness ratio and a given cross-sectional area, and a section which does not warp.16 Arguing the other way, for a given ratio of the semi-axes a/b, it is beneficial to have the principal directions of orthotropy and the stiffness ratio such that Eq. (4.96) is satisfied; assuming, for easy visualisation, the orthotropy to be due to unidirectional fibres aligned in a direction in the plane of the cross-section, the fibres should be parallel to the major axis, i.e. Gxz > Gyz when a > b (Fig. 4.28). This can also be explained more intuitively—a look at Fig. 4.7 shows that the net shear stress acts in a direction close to the major axis over a predominant portion of the domain, including the highly stressed zones near the ends of the minor axis, and thus it is desirable to have the fibre orientation such that the corresponding stiffness Gxz is high. An interesting corollary of the above argument is that an orthotropic circular section does warp under torsion unlike its isotropic counterpart. It is possible to verify these conclusions by carrying out the analysis of the orthotropic elliptical section exactly as for the isotropic case. The orthotropic counterpart of the equilateral triangle of Sect. 4.2.3, amenable to a similar simple closed-form solution, is now an isosceles triangle. For the rectangular section, the Fourier series approach of Sect. 4.2.4 is straightaway applicable for the orthotropic case as well. In this case, again, the optimum fibre orientation, for unidirectional reinforcement in the cross-sectional plane, is along the longer edges. We shall now illustrate such a Fourier series solution for the case of a laminated rectangular bar under torsion, a problem of practical significance.

4.7.2

Rectlilinear Orthotropy—Laminated Rectangular Shaft

Consider a bar of rectangular section made up of, say, two equal thickness layers perfectly bonded to each other (Fig. 4.29); each layer is orthotropic with respect to the x–y-z axes.

16 T. Chen, A homogeneous elliptical shaft may not warp under torsion, Acta Mechanica, 169, 2004,

221–4.

4.7

Orthotropic Shafts

159

Fig. 4.29 A laminated rectangular section

The torsion of this bar, within the purview of St. Venant’s free warping theory, can be analyzed in the same manner as was done in Sec.4.2.4 for the homogenous isotropic case, except for the following additional considerations17 : (a) at the interface of the two layers, the interlaminar shear stress τ yz is non-zero and should be equal for the two layers to satisfy equilibrium; (b) all the displacements should be continuous across the interface; while the continuity of u and v displacements is automatically satisfied by virtue of the assumptions of the theory (see Eq. (4.1a)) that of w needs to be explicitly enforced. If φ 1 (x,y) and φ 2 (x,y) are the Prandtl stress functions for the two layers, then continuity of the stress τ yz across the interface requires that φ1,x = φ2,x at y = 0 for all x i.e., φ2 − φ1 = constant c at y = 0 for all x

(4.97)

The warping displacement for each layer is obtained from the corresponding stress function using the relations w,x =

φ,y + θ ' y; Gxz

w,y =

−φ,x − θ'x G yz

(4.98)

Since the warping displacement is zero for both the layers along the y-axis which is a line of symmetry, the interlaminar continuity of w is automatically satisfied at the origin. Thus, the continuity of w across the entire interface can be satisfied by enforcing the continuity of the derivative w,x as w1,x = w2,x at y = 0 for all x

17 M. Savoia, N. Tullini, Torsional response of inhomogeneous and multilayered composite beams,

Composite Structures, 25, 1993, 587–594; S. R. Swanson, Torsion of laminated rectangular rods, Composite Structures, 42, 1998, 23–31.

160

4 Torsion of Non-Circular Sections

i.e., φ2,y φ1,y = at y = 0 for all x G x z1 G x z2

(4.99)

The stress-free boundary conditions on the outer boundary of the laminate reduce to ] Along y = −b for x = −a to a, φ1 = constant = 0 and along x = ±a for y = −b to 0 (4.100) ] Along y = b for x = −a to a, φ2 = constant = c and along x = ±a for y = 0 to b where one of the constants is taken as zero (as was done earlier for the homogenous hollow section) and the other turns out to be c in view of Eq. (4.97). Noting that both φ1 and φ1 should be even functions of x, solutions for the governing equation (Eq. 4.18’) satisfying the end conditions at x = ± a (Eq. 4.100) can be taken as ∞ Σ

φ1 =

Y1m (y) cos

m=1,3,.. ∞ Σ

φ2 = c +

m=1,3,..

mπ x 2a (4.101)

mπ x Y2m (y) cos 2a

so as to yield ∞ Σ m=1,3,.

''

(

Ykm m 2 π 2 Ykm mπ x − ) cos = −2θ ' , k = 1, 2 G x zk 4a 2 G yzk 2a

(4.102)

Expanding the right-hand-side in a Fourier series as −2θ ' = −2θ '

∞ Σ m=1,3,..

4 mπ mπ x sin cos mπ 2 2a

(4.103)

one gets ''

Ykm 4 m 2 π 2 Ykm mπ − = −2θ ' sin , k = 1, 2 G x zk 4a 2 G yzk mπ 2

(4.104)

for each m, and hence Ykm =

32G yzk θ ' a 2 sin(mπ/2) (1 + Akm cosh λkm y + Bkm sinh λkm y), k = 1, 2 (4.105) π3 m3

4.7

Orthotropic Shafts

161

/ G x zk for each m, where λkm = mπ 2a G yzk , and Akm and Bkm are undetermined constants to be obtained using the lateral surface and interface conditions (Eqs. (4.97), (4.99) and (4.100)), which can be rewritten as ⎫ Y1m (−b) = 0 ⎪ ⎪ ⎪ ⎪ ⎪ Y2m (b) = 0 ⎪ ⎬ (4.106) Y1m (0) = Y2m (0) ⎪ for each m ⎪ ⎪ ' ' ⎪ Y (0) ⎪ Y1m (0) ⎭ = 2m ⎪ G x z1 G x z2 These are four equations which yield the four constants A1m , A2m , B1m and B2m for the two-layered configuration under consideration. Finally, the torque T can be calculated as Σ¨ Σ¨ T = (τzy x − τzx y)d x d y = − (φk,x x + φk,y y)d x d y Ak

Ak

] ]   [  Σ [ xright ytop (yφ)| ybottom − φdy d x =− (xφ)|xleft − φd x dy −

(4.107)

where x left , x right , ybottom and ytop are as shown in Fig. 4.4 but now pertaining to the appropriate individual layer. A simplification of the above equation with substitution from Eqs. (4.97) and (4.100) leads to ¨ ¨ T =2 φ1 d xd y + 2 (φ2 − c)d xd y (4.108) A1

A2

A look at the above equation along with Eq. (4.101) reveals that the constant c does not really affect the value of T, nor does it contribute to the stress field obtained by differentiating φ. Thus, one can assume c to be zero; this is also true for the more general case of a laminated rectangular bar with k layers, where all the stress functions φ K can be taken to be zero on the edges x = ± a, the stress functions for the top and bottom layers to be zero at the top and bottom lateral surfaces, respectively, and the stress functions of any two adjacent layers to be equal at the interface. With this change, the expression for the torque gets reduced to the familiar form Σ¨ T = 2φk d xd y (4.109) k

Ak

We shall now present results for a particular two-layered configuration with G x z1 /G x z2 = 5, G x z1 /G yz1 = G x z2 /G yz2 = 2 and a/b = 2,

162

4 Torsion of Non-Circular Sections

where Gxz is taken to be greater than Gyz for any layer so as to represent a unidirectional composite with reinforcing fibres along the x-direction. (If the fibres were to be along the z-direction, Gxz and Gyz would be equal and hence the analysis would be the same as that of an isotropic bar. The case with Gyz greater than Gxz represents fibres running parallel to each other in the y-direction and will not be considered here because such thickness direction reinforcement is very unlikely in practice.) For these numerical values, one gets √ λ1m = λ2m = λm = mπ/a 2 ) ) ( ( 2 − 5sech λm b 2 + sech λm b ; A2m = A1m = − (4.110) 3 3 2 λm b B1m = B2m = − tanh 3 2 The differentiation of the stress functions φ k to obtain the shear stresses is straightforward; while summing the resulting series to evaluate the numerical values of the stresses, it is often convenient to use the identity (as was done for the isotropic rectangular section in Sect. 4.2.4): ∞ Σ m=1,3,..

1 π2 = m2 8

(4.111)

The integration of the stress functions, as required by Eq. (4.108) is also straightforward and yields the torque and hence the torsional rigidity. Results of the chosen two-layer section are presented in Table 4.4. For each orthotropic layer, both τ zx and τ zy are of interest because the corresponding strengths are not the same. The locations of severe τ zx are the points on the lateral surfaces/interface at x = 0 (see Fig. 4.30). As far as τ zy is concerned, the variation through the thickness of any layer is more complicated now and its maximum value within any layer, occurring at the ends x = ± a, has to be obtained only by considering the complete variation. For the present case, τ zy Table 4.4 Results for a two-layer rectangular laminate under torsion

Parameter τx z aθ ' G yz1

τ yz aθ ' G yz1

Layer 1 At (0, −b) : 1.205

At (0, 0) : −0.1308

At (0, 0) : −0.6539

At (0, b) : −0.4199

At (a, −0.4b) : 0.6442 At (a, −b/2) : 0.6313 At (a, 0) : 0.3411

T a 4 G yz1 θ '

Layer 2

0.3200 for the laminate

At (a, 0) : 0.3411 At (a, b/2) : 0.2114

4.7

Orthotropic Shafts

163

Fig. 4.30 Thickness-wise variation of τ xz at x = 0

within the second layer takes its maximum value at the interface y = 0, while that within the first layer becomes a maximum at about y = − 0.4b as shown in Fig. 4.31. In addition to the possibility of the failure of one of the orthotropic layers, it is also possible that the interfacial bond gives way due to excessive shear; in fact, the latter is more likely to occur first because the interlaminar shear strength of the bond would be much smaller than the strength of either of the layers, and thus the torsional strength

Fig. 4.31 Thickness-wise variation of τ yz at x = a

164

4 Torsion of Non-Circular Sections

Fig. 4.32 Variation of the interlaminar shear stress

calculation is often based on a delamination failure criterion. For the present case, the variation of the interlaminar shear stress τ yz along the interface is as shown in Fig. 4.32. From the above example, it should be clear that laminated orthotropic structures are not only more difficult to analyze than isotropic homogeneous structures but also need a more careful interpretation of the final numerical results.

4.7.3

Rectilinear Orthotropy—Thin-Walled Open Sections

As pointed out in Sect. 4.7.1, the torsional behaviour of a rectilinearly orthotropic section is the same as that of an equivalent scaled isotropic section. Thus, for a thin narrow orthotropic rectangular section (Fig. 4.33a), with dimensions such that the equivalent isotropic section is thin-walled as specified in Sect. 4.4, the analysis can once again be simplified by assuming that the stress function is invariant along the length of the strip. Thus, one has, φ,yy = −2G x z θ '

(4.112)

leading to τx z = −2G x z θ ' y T = |τx z

max |

Lt 3 Gxzθ' 3

= G x z θ 't =

(4.113) 3T Lt 2

which are the same as those for the isotropic case of Sect. 4.4 but for a change of G.

4.7

Orthotropic Shafts

165

Fig. 4.33 Narrow homogeneous and laminated sections

A look at Eq. (4.91) shows that the aspect ratio of the equivalent scaled isotropic section is given by / ( ) L L G yz = (4.114) t equivalent t Gxz and thus, the above approximate formulae lead to less than 3% error as long as this ratio is greater than 20 as proved in Sect. 4.4. It is possible to extend the above analysis to the case of a thin laminate (Fig. 4.33b) by assuming that the stress function for any layer is invariant along the length of the laminate.18 Thus, for the kth layer, one has φk,yy = −2G x zk θ '

(4.115)

φk = G x zk θ ' (Ak + Bk y − y 2 )

(4.116)

leading to,

where the constants Ak and Bk are obtained by enforcing the conditions at the lateral surfaces and the interfaces of the laminate. These conditions are easily identified by noting that the present solution should be identical to that of the rectangular laminate of Sect. 4.7.2 with the aspect ratio tending to a very large value. Thus, the conditions required for the determination of the unknown constants are

18 S. R. Swanson, Torsion of laminated rectangular rods, Composite structures, 42, 1998, 23–31.

166

4 Torsion of Non-Circular Sections

φtop layer = 0 at y = −t/2; φbottom layer = 0 at y = t/2; ⎫ ⎪ φk = φk+1 ⎬ φk+1,y φk,y ⎪ = ⎭ G x zk G x z (k+1)

at the k th interface

(4.117)

where the interface conditions represent the continuity of τ yz and the warping displacement, respectively. Finally, using Eq. (4.111), one can get the torque in terms of θ ’ and hence calculate the effective torsional rigidity. It is important to note that the above analysis is useful for predicting τ xz and the torsional rigidity, but completely ignores τ yz which may actually turn out to be the critical stress, especially at the interfaces of the laminate. However, it provides a useful basis for estimating the torsional stiffness of laminated open sections of various profiles such as angle or channel sections, with the total stiffness simply taken as the sum of those for the individual rectangular portions.

4.7.4

Shape-Intrinsic Orthotropy

We shall simply point out the method of analysis for the two most common instances of shape-intrinsic orthotropy—a circular shaft and a thin-walled tube of arbitrary shape. These are usually filament-wound and exhibit, on a macroscopic scale, orthotropy characterised by material property symmetry with reference to the longitudinal (z), circumferential (θ ) or tangential (s), and the radial (r) or surface normal (n) directions. It is easy to visualise that the physical behaviour of these shafts would be similar to their isotropic counterparts and the only change required is the use of Gθz or Gsz in place of G in the appropriate equations of the simple torsion theory or Bredt-Batho theory. In case of a layered construction, a laminated solid or hollow circular shaft is analyzed exactly in the same way as compound shafts made up of different isotropic materials are treated in the elementary theory—by considering a linear radial variation of the shear strain and a corresponding piecewise linear variation of the shear stress. Similarly, a laminated thin-walled non-circular tube with cylindrically orthotropic layers can be easily accommodated within Bredt-Batho theory19 by assuming a constant shear strain γ sz across the wall thickness and by taking the corresponding thicknessintegrated shear force per unit length (often referred to as the shear flow) to be invariant along the contour, 19 For a more comprehensive discussion, see E. H. Mansfield, A. J. Sobey, The fibre composite

helicopter blade, Aeronautical Quarterly, 30, 1979, 413–449.

4.8

Effect of Warping Restraints

167

i.e., γsz

N Σ

G szk tk = constant along s

(4.118)

k=1

where N denotes the total number of layers and t k denotes the thickness of the kth layer which may vary along s. Proceeding as before, one gets, in place of the equations of Sect. 4.5.3, γsz =

2A

T ΣN k=1

G szk tk

(4.119a)

and hence, G szi T ΣN 2 A k=1 G szk tk  T ds ' θ =  ΣN 4 A2 k=1 G szk tk τszi =

(4.119b) (4.119c)

As stated earlier, the above two examples represent the most common occurrences of shape-intrinsic orthotropy and are also the simplest to analyze. A more general case involving a solid or thick-walled non-circular cylinder is quite complicated and beyond the scope of the present book.20

4.8

Effect of Warping Restraints

St. Venant’s theory of uniform torsion, employed for the different cases considered above, is valid as long as the shaft is prismatic and subjected to equal end torques, and the warping of the non-circular section is freely permitted everywhere along the total length. In practice, this is not always true and additional effects, due to warping restraints, have to be examined. A restraint on the warping displacement can arise either from a rigidly fixed end, or at an intermediate section subjected to a torque where the faces of the two free bodies on either side of the section would have different warping displacements if allowed to separately undergo uniform unrestrained torsion; similarly, a warping restraint is also present whenever the torsional moment is distributed along the shaft or when the shaft itself is not prismatic. These different types are as shown in Fig. 4.34.

20 see, for example, G. A. Kardomateas, Theory of elasticity of filament wound anisotropic ellipsoids

with specialization to torsion of orthotropic bars, ASME Jl. of Applied Mechanics, 55, 1988, 837– 844.

168

4 Torsion of Non-Circular Sections

Fig. 4.34 Examples of warping restraint

For further discussion, let us focus attention on a cantilevered prismatic shaft rigidly fixed at one end and subjected to a torque at the other end. The net stresses in the shaft would be the algebraic sum of the St. Venant shear stresses corresponding to free unrestrained warping and the additional stresses due to suppression of the warping displacements at the fixed end; since the St. Venant shear stress distribution is statically equivalent to the applied torque, the additional stresses at any section have to be selfequilibrant with a zero resultant force and a zero resultant moment. Thus, this additional stress field due to warping restraint has to exhibit decay as one moves away from the fixed end, as per St. Venant’s principle. Though this is true irrespective of the shape of the cross-section, the decay rate depends on the actual shape. It can be proved21 that this decay rate is quite rapid in the case of solid and closed tubular sections, but rather slow in the case of open sections such as a channel or an I-section. An obvious consequence of restraining the warping displacement is to stiffen the shaft locally and hence a cantilevered open section shaft deforms such that the rate of twist is smaller near the restrained end, with this effect being large enough to result in a significant decrease in the angular displacement of the free end from that corresponding to free warping. Thus, while the St. Venant theory of uniform torsion is adequate for most cases of solid and tubular section shafts, one needs a more realistic treatment for the stress and deformation analysis of open sections; this is usually carried out within the purview of an approximate engineering theory—the theory of non-uniform torsion—first put forth by Vlasov and well summarized in the book by Gjelsvik.22 Our interest here is not to provide full details of such a theory, but to simply point out the significance of warping restraint and its effect on deformation and stresses. In order to get a physical feel for the problem, it is instructive to consider a symmetric thin-walled I beam (Fig. 4.35). When twisted without any warping restraints, the flanges of any section warp as shown by the dotted lines while the web remains in the original plane of the section. If such warping displacements are restrained at a section, then the restraining action is equivalent to two bending moments in the planes of the flanges as shown; such a set of self-equilibrant bending moments—which may be two or more for a general cross-section—is referred to as a bimoment. As already stated, there is a decay of 21 J. T. Oden, Mechanics of Elastic Structures, McGraw-Hill, 1967. 22 A. Gjelsvik, The Theory of Thin-walled Bars, John Wiley, 1981.

4.8

Effect of Warping Restraints

169

Fig. 4.35 The bimoment

these restraining moments in the axial direction, and this manifests itself as a horizontal shear force on each flange. Such a twisting couple of equal and opposite horizontal forces serves to resist a fraction of the applied external torque. In other words, the effect of the warping restraint is to change the shear stress distribution, in the neighbourhood of the restrained section, from one of pure closed-loop variation as per St. Venant’s theory; instead, only a fraction of the total torque is now resisted by closed-loop shear stresses as shown in Fig. 4.36a, while the remaining fraction leads to shear stresses in the two flanges, uniform through their wall thickness, as shown in Fig. 4.36b. This latter component of the internal torque is often referred to as the Vlasov torque or warping torque or flange-bending torque; even for a more general thin-walled section, it always corresponds to a uniform distribution of shear stress through the wall thickness and associated bending of different segments of the contour. The relative proportions of these two resisting torques vary along the length of the shaft with the Vlasov component being predominant near the restrained end. This can be demonstrated easily as follows. Consider a cross-section of the cantilevered I-beam at a distance z from the fixed end and let θ be the rotation of this section due to a torque T applied at the free end (Fig. 4.37); let the flange and web dimensions be equal and the thickness be uniform. Let T S and T V be the St. Venant and Vlasov components of the

Fig. 4.36 Shear stresses due to St. Venant and Vlasov torques

170

4 Torsion of Non-Circular Sections

Fig. 4.37 Torsion of a cantilevered shaft

resisting torque at this section; considered independently, these can be related to θ . By the St. Venant theory, one has TS = G J

θ z

(4.120)

with J = (2b + h)t 3 /3 = bt 3 . If the net shear force in each flange due to T V is F, then (hθ/2) corresponds to the tip deflection, due to a tip force F, of a cantilever of rectangular cross-section t x b and length z. Thus, hθ F z3 F z3 = = 2 3E Iflange 3E(tb3 /12)

(4.121)

Etb5 θ Etb3 h 2 θ = 8z 3 8z 3

(4.122)

leading to, TV = Fh = Hence, ( )( )2 ( )2 b b TV Eb4 1 E = = 2 2 TS 8Gz t 8 G z t

(4.123)

where the final expression is in a form suitable for explaining the effect of various parameters. Clearly, the effect of the warping restraint becomes insignificant as b/z decreases, i.e. at distances which are large in comparison with the cross-sectional dimensions. Apart from this, one can also see that the influence of warping restraint becomes more important as E/G and b/t increase; thus, it is more important for fibre-reinforced beams (with fibres running along the longitudinal axis) than for isotropic beams, and for sections with very thin walls.

4.9

Summary

171

In a more general form, the torque-twist relationship of the theory of non-uniform torsion is given by −Eθ ' + G J θ ' = T

(4.124)

where the first term corresponds to the warping torque T V and the latter, to the wellknown St. Venant torque T S . Here, Γ is a cross-sectional property known as the warping constant (with dimensions of length raised to the sixth power) and EΓ is the warping rigidity. The relative importance of the warping restraint, in calculations of the torsional flexibility or the total angle of twist of a shaft of length L, decreases as the parameter /

J L G E increases; this conclusion also applies while evaluating the torsional buckling loads or the natural frequencies of torsional vibration. However, the influence of warping restraints should always be considered for accurate stress analysis in the vicinity of the restrained section. The warping constant is a sectional property, and this can be understood by comparing a thin rectangular strip with the I-section. The free warping displacements of the thin strip would be zero along the two lines of symmetry and distributed antisymmetrically with respect to these lines; thus, it is essentially a case of warping in the thickness direction (the so-called secondary warping). Clearly, the effect of restraining such warping would not be as dramatic as that of the I-section. If the total contour length and the wall thickness are taken to be the same for the two sections, then the St. Venant torsional constant J would be equal for both of them while the warping constant Γ would be considerably different. Formal calculations of the warping constant for different cross-sections are rather straightforward23 ; the final results for commonly encountered shapes are usually tabulated in handbooks. We shall leave this topic here. Interested readers are encouraged to go through the cited references of Gjelsvik and Oden for a more complete discussion of restrained warping on isotropic thin-walled bars. A lucid discussion of warping restraints in orthotropic open and closed sections is also available.24

4.9

Summary

The theory of uniform torsion based on St. Venant’s and Prandtl’s formulations has been put forth in this chapter and its application to various problems has been demonstrated. Prandtl’s membrane analogy has been explained. The special cases of a thin-walled open section and a thin-walled closed tube, which are amenable to a simpler approximate treatment, have been discussed. The effects of various complications such as protruding 23 see, A. Gjelsvik, The Theory of Thin-walled Bars, John Wiley, 1981. 24 J. Loughlan and M. Ata, The behaviour of open and closed section carbon fibre composite beams

subjected to constrained torsion, Composite Structures, 38, 1997, 631–647.

172

4 Torsion of Non-Circular Sections

and re-entrant corners, eccentric holes, etc. have been intuitively discussed. The effect of orthotropy has been explained briefly with reference to some homogenous and laminated sections. Finally, a brief discussion of the effect of warping restraints has been presented.

5

Some Other Problems of Interest

The purpose of this chapter is to present the qualitative aspects of a few problems of practical interest without too many mathematical details.

5.1

Concentrated Normal Force on the Boundary of an Isotropic Half-Space (Boussinesq’s Problem)

A half-space is a semi-infinite body on one side of a bounding plane, say z = 0. When a concentrated normal load P is applied at a point of this plane, which is chosen as the origin of a cylindrical polar coordinate system r-θ-z (Fig. 5.1), the resulting deformation and stresses are axisymmetric about the z-axis. For this problem, which is the three-dimensional counterpart of Flamant’s problem (Sect. 3.8.1), the stresses are given by [ ] R 3r 2 z P σr = (1 − 2μ) − 3 2π R 2 R+z R [ ] R P(1 − 2μ) z (5.1) σθ = − 2π R 2 R R+z σz = −

3P z 3 , 2π R 5

τr z = −

3Pr z 2 2π R 5

© The Author(s) 2023 K. Bhaskar and T. K. Varadan, Theory of Isotropic/Orthotropic Elasticity, https://doi.org/10.1007/978-3-031-06345-9_5

173

174

5 Some Other Problems of Interest

Fig. 5.1 Boussinesq’s problem

P

θ

r z

while the corresponding displacement field is [ ] P r rz − (1 − 2μ) ur = 4π G R 3 R(R + z) ] [ z2 P 2(1 − μ) + 3 uz = 4π G R R

(5.2)

/( ) where R = r 2 + z 2 is the distance of the point under consideration from the origin. To get a better feel for the problem, consider the line of action r = 0 of the load. Along this line, the stresses and displacements are given by σr = σθ =

P(1 − 2μ) 3P , σz = − , τr z = 0 4π z 2 2π z 2 u r = 0, u z =

P(3 − 2μ) 4π Gz

(5.3)

Thus, while the stresses vary inversely with respect to z2 , the vertical displacement varies inversely with respect to z; they become singular right under the concentrated load. While σ z is compressive for all z, σ r is tensile; this can be visualised as due to the surrounding material being pulled in towards the load. For r = 0, σ r is the same as σ θ as it should be.

5.2

Frictionless Contact Between Two Isotropic Spheres …

175

Let us also consider the stresses and displacements on the bounding plane z = 0. The non-zero quantities are P(1 − 2μ) ; σz = ∞ for r = 0, 0 for r > 0 2πr 2 P(1 − μ) P(1 − 2μ) , uz = ur = − 4π Gr 2π Gr

σr = −σθ =

(5.4)

where ur is negative for all r, indicating a radially inward displacement as described above due to which σ r is tensile and σ θ is compressive.

5.2

Frictionless Contact Between Two Isotropic Spheres (Hertz Problem)

Just as Flamant’s solution is the starting point for the contact problem of two cylindrical surfaces, i.e. line contact, Boussinesq’s solution is useful for the analysis of contact between two spherical surfaces, i.e. point contact (Fig. 5.2). Once again, the contact is assumed to be non-conformal and the actual problem is reduced to the study of two halfspaces. Due to axisymmetry, the contact region is bounded by a circle whose radius a is referred to as the radius of contact. As was done earlier, let us define two parameters E and R as 1 E

=

(1 − μ21 ) (1 − μ22 ) + , E1 E2

1 1 1 = + R R1 R2

(5.5)

P

Fig. 5.2 Contact between two spheres

R1

+ r z

+

P

R2

176

5 Some Other Problems of Interest

where R1 and R2 are both algebraic quantities, positive when the corresponding spherical surfaces are convex. In terms of these parameters, the contact radius and the peak pressure are given by ( a=

3P R 4E

)1/3 , pmax

3P = = 2πa 2

(

2

6P E π 3 R2

)1/3 (5.6)

Note that the dependence of pmax on the material and geometric parameters E and R is now different from that of line contact. The contact pressure variation is given by /( ) r2 1− 2 (5.7) p = pmax a where r is the radial coordinate. Regarding the contact stresses, the situation is somewhat similar to the line contact problem in that all the stresses decay monotonically along the z-axis with their maximum values at the surface, but the maximum shear stress is once again at a sub-surface location. Noting that σ r = σ θ along this axis, the maximum values are given by pmax for σ z and (1+2μ) pmax for σ r or σ θ , all of them being compressive. For µ = 0.3, the maximum 2 shear stress is of magnitude 0.31pmax and occurs at a depth of 0.48a. Regarding the variation of the stresses on the surface, while σ z and σ θ are compressive everywhere within the contact circle, σ r is compressive in the central portion of the pmax at circle but becomes tensile in the outer portion taking a maximum value of (1−2μ) 3 the edge of the contact region. Such tensile stresses often lead to ring cracks when the contacting surfaces are of a brittle material such as glass.

5.3

Free Edge Phenomenon in Composite Laminates

A problem that is of considerable practical interest is that of stress analysis of composite laminates with unsupported free edges or internal hole boundaries. In the vicinity of such free edges, the interlaminar stresses—the shearing and peeling stresses at the interfaces between dissimilar layers—grow rapidly to very large values and lead to delamination failures, which often occur much before individual plies fail due to excessive in-plane stresses. This is referred to as the free edge phenomenon, and has been extensively studied using a variety of approaches.1 The most commonly analysed free edge problem is that of a long, thin symmetric laminated plate subjected to uniform axial extension, often referred to as a tensile coupon (Fig. 5.3). Here, we shall restrict our discussion to the case of a 1 See, C. Mittelstedt, W. Becker, Interlaminar stress concentrations in layered structures: Part I—A

selective literature survey on the free-edge effect since 1967, Jl. of Composite Materials, 38, 2004, 1037–1062.

5.3

Free Edge Phenomenon in Composite Laminates

177

P

Fig. 5.3 The tensile coupon specimen

A

B

z y

h P x

symmetric cross-ply lay-up, i.e. one wherein unidirectional fibres are aligned in different layers alternately in the x- and y-directions. The lay-up shown is designated by the fibre angles with respect to the x-axis as (0°/90°/90°/0°), with all the four layers being of the same thickness; one can also look at this as a three-layer configuration with the middle (90°) layer twice as thick as the other two layers. Because of the symmetric lay-up, the plate would remain flat when subjected to the tensile load and simply undergo an elongation along x and a contraction along y (and also z). Since the layers are perfectly bonded to one another, they would deform together and undergo the same strains εx and εy ; for this to happen, each layer has to be subjected to a system of biaxial stresses σ x and σ y , as shown in Fig. 5.4. The resultant of the σ x distribution over all the layers should correspond to the applied load, while that of the σ y distribution should be zero so that these stresses are equal but unlike for the (0°) and (90°) layers. The other stress components—the in-plane shear stress τ xy as well as the transverse components σ z , τ xz, τ yz —are absent. This is true for any element (such as element A) of the plate except near the free edges. The situation is more complicated in the vicinity of a free edge, because the free edge boundary condition requires that σ y be zero. Thus, considering an element adjoining the free edge (element B), each layer has an unbalanced tensile σ y on its left face while the right face is free of stress. The only stress which can balance this σ y is an interfacial shear τ zy in the vicinity of the free edge. However, equilibrium is still not satisfied because there is an unbalanced moment which can be counteracted only by the presence of a corresponding distribution of σ z on the interface. Thus, near the free edge, interlaminar stresses τ yz and σ z are non-zero. The presence of the interlaminar stresses can also be explained using the threedimensional equations of equilibrium. For the present problem where there are no stress

178

5 Some Other Problems of Interest

Fig. 5.4 Stresses in the interior and near the free edge

variations in the x-direction, the equations are τx y,y + τx z,z = 0 σ y,y + τ yz,z = 0

(5.8)

τ yz,y + σz,z = 0 The first equation is identically satisfied by taking τ xy = τ xz = 0 everywhere. The second equation requires that a gradient of σ y in the y-direction has to be accompanied by a thickness-wise variation of τ yz ; both of these are zero in the interior of the plate, but as one approaches the free edge σ y has to drop to zero and hence σ y,y is non-zero and so is τ yz,z . The third equation is satisfied in the interior of the plate by taking τ yz = σ z = 0, but near the free edge, τ yz,y is non-zero and so is σ z,z leading to a thickness-wise variation of σ z . The thickness-wise variations of τ yz and σ z are characterised by zero values on the top and bottom faces of the laminate which are stress-free faces. A closed-form solution is not possible for the above free-edge problem and hence a variety of approximate solutions are available in the literature. We shall not go into the details of any of these solutions, but shall simply point out the salient features of the stress distribution near the free edge. The interlaminar stresses are of primary interest and these, for the 0°–90° interface, are as shown qualitatively in Fig. 5.5. It can be seen that they are zero everywhere except in a small boundary layer region of width of about one laminate thickness near the free edge. In this region, the stress gradients are quite steep; in fact, there is a singularity at the intersection of the interface and the free edge characterised

5.4

Functionally Graded Structures

179

Fig. 5.5 Interlaminar stresses along 0°–90° interface

by infinitely large values of σ z and the derivative τ yz,y . It has been found that the stress distribution in the vicinity of the free edge is not significantly affected by changes in the width of the plate as long as the width is at least twice the laminate thickness. The above example of a cross-ply laminate involves only τ yz and σ z , but for a more general lay-up, all the three interlaminar stresses would be non-zero with both τ xz and σ z tending to infinity at the free edge-interface corner. The study of the free edge effect is very important because it serves to explain the initiation of delamination failures; a delamination failure criterion is often in terms of the average interlaminar stresses in the boundary layer region.2 Such a study is also useful to identify lay-ups which lead to compressive σ z near the free edge and are hence superior from the viewpoint of suppression of delamination. Though most free edge studies have focused on the tensile coupon specimen, other geometries and load conditions have also been considered.3

5.4

Functionally Graded Structures

So far, we have been discussing cases of homogeneous bodies or laminated bodies with piecewise homogeneity. There are many applications wherein a continuous variation of material properties is desirable, and this can be achieved by a continuous variation of the 2 See, for example, J. C. Brewer, P. A. Lagace, Quadratic stress criterion for initiation of delamina-

tion, Jl. of Composite Materials, 22, 1988, 1141–55. 3 See, for example, W. L. Yin, Free-edge effects in anisotropic laminates under extension, bending

and twisting, ASME Journal of Applied Mechanics, 61, 1994, 410–421.

180

5 Some Other Problems of Interest

material composition. A simple example would be the case of a particulate composite where the reinforcement is in the form of small hard particles dispersed in the relatively soft matrix, and the volume fraction of the reinforcement is varied continuously in one or more desired directions. Such materials are referred to as functionally graded materials4 and are of great importance for applications such as the walls of a nuclear reactor where one surface is subjected to high temperatures while the other surface is exposed to the ambient atmosphere. The analysis of functionally graded structures using the theory of elasticity is obviously more complicated than that of homogeneous structures because one ends up with partial differential equations with variable coefficients. While exact solutions are possible for certain special cases, a numerical solution is in general called for. Without going into details of the solution methods, we shall just cite some examples where the benefits of a selective gradation of the reinforcement have been illustrated. With reference to the internally pressurised cylinder, it is known that the hoop stress is maximum at the inner radius and is the most critical stress for design. With reference to this problem, a functionally graded configuration has been studied5 by assuming that Young’s modulus varies radially as per the power law ( r )n E = E1 (5.9) a where a is the inner radius; Poisson’s ratio µ is taken to be a constant. It has been shown that the exponent n can be varied such that the hoop stress at the inner radius decreases, and this is accompanied by an increase in the hoop stress at the outer radius. In some cases, it is also seen that the maximum hoop stress is smaller than the maximum radial stress and hence the latter has to be used as the basis for design. Another interesting example is that of using an orthotropic functionally graded material for a rotating circular disc.6 The total volume of fibres, used for reinforcement in the circumferential direction, is taken to be constant and different spatial distributions of the reinforcement leading to different variations of E θ (r) have been studied. It has been shown that it is possible to design the rotating disc such that the radial displacement at the inner bore is minimum, which is desirable for ensuring a proper fit with the shaft. It has also been pointed out that the fibre reinforcement can be varied to achieve optimal stress patterns, which have to be identified carefully keeping in mind that the fibre reinforcement also changes the strength of the composite from one point to another.

4 V. Birman, L. W. Byrd, Modeling and analysis of functionally graded materials and structures,

Applied Mechanics Reviews, 60, 2007, 195–215. 5 C. O. Horgan, A. M. Chan, The pressurized hollow cylinder or disk problem for functionally

graded isotropic linearly elastic materials, Jl. of Elasticity, 55, 1999, 43–59. 6 J. F. Durodola, O. Attia, Deformation and stresses in functionally graded rotating disks, Compos-

ites Science and Technology, 60, 2000, 987–995.

5.4

Functionally Graded Structures

181

The above two examples are just some samples taken from the literature to illustrate the application of functionally graded materials. It is likely that such applications will be more common in future in the pursuit of better, more efficient designs, and hence there will be many more investigations of such structures using the theory of elasticity.

Appendix

A1 Transformation of Stresses and Strains A1.1 Stress Transformation Rules To derive the stress transformation equations, it is necessary to consider the equilibrium of a suitably chosen infinitesimally small free body at the point of interest. This is easily understood with reference to a two-dimensional transformation. Consider two differently oriented stress diagrams as shown in Fig. A.1, both representing the state of plane stress at the same chosen point. The problem of stress transformation can be stated thus: “Given σ x , σ y , τ xy , and the angle θ between the two coordinate systems, find σ x’ , σ y' , τ x’y’ .” The starting point is to consider an infinitesimally small free body at the point of interest with one face normal to one of the new coordinate axes, say, x’, and the other faces normal to the original x or y or z axis; this body has the shape of a triangular prism which is seen as a triangle in the plan view (Fig. A.2). For convenience, the hypotenuse of the triangle is taken to be of unit length, while the others are related to it in terms of θ. The stresses acting on the different faces of this free-body are as shown. Now, it is necessary to simply write down the equations of statics for this free body in equilibrium. Noting that this involves summing forces and not stresses themselves, and doing this for the x’ direction, we have σx ' (1) = σx (cos θ )( cos θ ) + σ y (sin θ )(sin θ ) +τx y (cosθ )(sin θ ) + τ yx (sin θ )(cos θ ) i.e. σx ' = σx cos2 θ + σ y sin2 θ + 2τx y sin θ cos θ

© The Author(s) 2023 K. Bhaskar and T. K. Varadan, Theory of Isotropic/Orthotropic Elasticity, https://doi.org/10.1007/978-3-031-06345-9

(A.1)

183

184

Appendix

σy

Fig. A.1 Stress transformation

σ y'

τ xy

τ x' y '

σ x'

σx

y

x

y' θ

x'

Fig. A.2 Derivation of σ x’ and τ x’y’

σx

θ

τ x' y '

σ x'

τ xy σy

Similarly, by summing forces in the y’ direction, one gets τx ' y ' = (σ y − σx ) sin θ cos θ + τx y (cos2 θ − sin2 θ )

(A.2)

To obtain σ y' , it is necessary to consider a triangular prism with one face normal to y’; however, the final result can be obtained directly from the equation for σ x’ above by replacing θ by (π/2 + θ ) to yield σ y ' = σx sin2 θ + σ y cos2 θ − 2τx y sin θ cos θ

(A3)

Equations (A.1), (A.2), (A.3) are the stress transformation rules for the plane stress problem and are also applicable for such a two-dimensional transformation (i.e. when the z-axis remains common for the original and the rotated system) for the case of plane strain. Extending the above methodology to a three-dimensional transformation of a general state of stress at a point, it is easy to imagine that the infinitesimal free body to be considered will be a tetrahedron with three faces normal to the original x–y-z axes while the fourth is normal to one of the new axes, x’ or y’ or z’ (Fig. A.3). Summing forces along with one of the new coordinate axes, one can easily derive the transformation rules; they are presented in a concise form as Σ Σ σi j = aik a jl σkl with i , j = x ' , y ' , z ' (A.4) k=x,y,z l=x,y,z

where aik is the direction cosine of direction i with respect to direction k. The two-dimensional transformation rules can also be written in a similar concise form as

Appendix

185 y

Fig. A.3 Three-dimensional transformation

x'

x

z

σi j =

Σ

Σ

aik a jl σkl with i , j = x ' , y '

(A.5)

k=x,y l=x,y

Since stress transformation involves only equilibrium considerations, the equations presented above are applicable for the state of stress at any point of a structure, irrespective of the material constitutive law.

A1.2 Strain Transformation Rules The problem of strain transformation involves the determination of the strain components with reference to a rotated coordinate system when those with reference to the original axes are known. To derive the relations, let us start with the simple case of the plane strain and consider the elongations of elemental lengths oriented along the inclined axes x’ and y’ before deformation, as well as the change in angle between them (Fig. A.4). If the displacement components of A are u,v, then those of B, located at (Δx 1 , Δy1 ) relative to A, are u + u,x Δx 1 + u,y Δy1 , v + v,x Δx 1 + v,y Δy1 .

B'

Fig. A.4 Derivation of εx’

B

A'

y'

y

A

θ

Δy1

θ Δx1

x' x

186

Appendix

Then, we have A' B' − AB AB (A' B' )2 − (AB)2 = AB(A' B' + AB) (A' B' )2 − (AB)2 ≅ for small deformations 2(AB)2

εx ' =

(A.6)

With (A' B' )2 = (Δx1 + u ,x Δx1 + u ,y Δy1 )2 + (Δy1 + v,x Δx1 + v,y Δy1 )2 ≅ Δx12 (1 + 2u ,x ) + 2u ,y Δx1 Δy1 + Δy12 (1 + 2v,y ) + 2v,x Δx1 Δy1 for small u,x , etc. and (AB)2 = Δx12 + Δy12 , one gets εx ' = u ,x (Δx1 /AB)2 + v,y (Δy1 /AB)2 + (u ,y + v,x )(Δx1 /AB)(Δy1 /AB)

(A.7)

= εx cos2 θ + ε y sin2 θ + γx y sin θ cos θ

By means of similar considerations, or by simply replacing θ by (π/2 + θ ) in the above equation, one can get ε y ' = εx sin2 θ + ε y cos2 θ − γx y sin θ cos θ

(A.8)

 ' ' '  To derive the expression for γ x’y’ , the change  ' ' in' angle between B A C and the original  right angle |BAC is required (Fig. A.5). B A C is obtained by considering the direction cosines of A' B' and A' C' . A' B' has the direction cosines given by (Δx1 + u ,x Δx1 + u ,y Δy1 )/A' B' ,(Δy1 + v,x Δx1 + v,y Δy1 )/A' B' Fig. A.5 Derivation of γ x’y’

B'

C'

Δy2

y'

y θ

x'

x

B

A'

C

Δx2

A

θ

Δy1

Δx1

Appendix

i.e.

187

(1+u ,x )Δx1 +u ,y Δy1 v,x Δx1 +(1+v,y )Δy1 , AB(1 + εx ' ) AB(1 + εx ' )

[ ] [ ] i.e. (1 − εx ' ) (1 + u ,x ) cos θ + u ,y sin θ , (1 − εx ' ) v,x cos θ + (1 + v,y ) sin θ for small ε x’ . i.e.(1 − εx ' + εx ) cos θ + u ,y sin θ, (1 − εx ' + ε y ) sin θ + v,x cos θ after neglecting products of εx’ and u,x , etc., and replacing u,x by εx , and so on. Similarly, noting that C is to the left of A, the direction cosines of A' C' can be written as −(Δx2 + u ,x Δx2 − u ,y Δy2 )/A' C' ,(Δy2 − v,x Δx2 + v,y Δy2 )/A' C' i.e. −(1 − ε y ' + εx ) sin θ + u ,y cos θ, (1 − ε y ' + ε y ) cos θ − v,x sin θ The required shear strain can then be found out from ) (π   π  − B' A' C' = cosB' A' C' γx ' y ' = − B' A' C' ≅ sin 2 2 −−'→' −−'→' A B .A C = ' ' ' ' = lA' B' lA' C' + m A' B' m A' C' (A B )(A C )

(A.9)

where l’s and m’s are the direction cosines given above. Substituting for the direction cosines and neglecting products of strains and displacement gradients, one finally gets γx ' y ' = −(εx − ε y )2 sin θ cos θ + γx y (cos2 θ − sin2 θ ) γ

γ

(A.10)

' '

In terms of the tensorial shear strains εxy (= 2x y ) and εx’y’ (= x2y ), these equations can be put in the same form as the 2-D stress transformation equations (Eq. A.5) as Σ Σ εi j = aik a jl εkl with i , j = x ' , y ' (A.11) k=x,y l=x,y

When extended to three dimensions, the equations turn out to be Σ Σ εi j = aik a jl εkl with i , j = x ' , y ' , z '

(A.12)

k=x,y,z l=x,y,z

A1.3 Principal Stresses and Strains The above transformation rules clearly indicate that the stress and strain components take on different values when the reference planes passing through the point under consideration are changed. It is then of interest to know the orientations of the planes corresponding

188

Appendix

to which one or more such components become maximum or minimum or simply vanish. It is in this context that one defines the principal planes; they are planes on which the shear stresses (and shear strains) vanish while the normal stresses take on extreme values. Considering a state of plane stress first, we have the stress σ x’ along an inclined direction as σx ' = σx cos2 θ + σ y sin2 θ + 2τx y sin θ cos θ

(A.1)

For extreme values of this stress, we have dσx ' =0 dθ

(A.13)

to get the principal plane orientations (denoted by subscript P) as tan 2θ P = i.e.

2τx y (σx − σ y )

( ) ( ) 2τx y 2τx y π 1 1 −1 −1 & θ P = tan + tan 2 σx − σ y 2 2 σx − σ y

(A.14)

Thus, there are two principal planes orthogonal to each other. By substituting these values of θ P in the expression for the shear stress τ x’y’ (Eq. A.2), one can immediately see that it vanishes on the principal planes. Thus the principal planes are planes of extreme normal stress as well as those of zero shear. (It is easily verified that there are no other planes of zero shear, unless σ x = σ y and τ xy = 0, which is a special isotropic state of stress for which the normal stress in any direction is the same and every plane is a principal plane free of shear.) The normal stress on the principal planes, or the principal stresses, are then obtained as / ) ( σx − σ y 2 σx + σ y σP = + τx2y (A.15) ± 2 2 which are often denoted by σ P1 and σ P2 and referred to as the major and minor principal stresses, respectively. It should be kept in mind that this terminology is with respect to algebraic quantities, and hence it is always possible that the major principal stress is of smaller magnitude than the minor principal stress and is hence less significant. Proceeding in the same manner as above, one can identify the planes of maximum shear as those with the orientations given by tan 2θτ = −

(σx − σ y ) 2τx y

(A.16)

Appendix

189

i.e. θτ =

( ) ( ) (σx − σ y ) (σx − σ y ) π 1 1 & + tan−1 − tan−1 − 2 2τx y 2 2 2τx y

(A.17)

Since tan 2θ P = − cot 2θτ = tan( π2 + 2θτ ), the maximum shear planes are 45° away from the principal planes, i.e. they are the two orthogonal planes which bisect the angles between the principal planes. The magnitude of the maximum shear stress, which should be the same on the two maximum shear planes because of the principle of complementary shear, is given by  /   ( )2   σx − σ y |τmax | =  (A.18) + τx2y  2   which can also be written as

   σ P1 − σ P2    |τmax | =   2

(A.19)

The actual direction in which this maximum shear stress acts on a maximum shear plane can be found out by looking at the equilibrium of a triangular element bounded by this plane and the two principal planes. It is important to know that the maximum shear planes are not free of normal stress; this stress can be obtained by putting θ = θ τ in the expression for σ x’ (Eq. A.1) as στ =

σx + σ y 2

(A.20)

for both the maximum shear planes. All the foregoing results, developed specifically for the case of plane stress, are valid for the case of plane strain as well. In both cases, the x–y plane is free of shear and is hence another principal plane with the third principal stress σ P3 being zero for plane stress and non-zero (see Eqs. (1.22 and 1.23)) for plane strain. Instead of considering an inclined coordinate system obtained by rotating the axes about the z-direction (or the P3 direction) as done above, if one rotates the axes about the P1 or P2 direction, one can find two more sets of extreme shear planes each 45° away from the corresponding principal planes. The extreme shear stresses so found out are given by        σ P2 − σ P3   &  σ P1 − σ P3  |τmax | =  (A.21)    2 2 Thus, while the maximum tensile and compressive stresses at a point are identified by finding out all the three principal stresses, the maximum shear stress is calculated as

190

Appendix y

Fig. A.6 Identification of the principal planes

σP

x

z

   (  σ P1 − σ P2   σ P2 − σ P3  ,  , |τmax | = max     2 2

 )  σ P1 − σ P3      2

(A.22)

Let us now consider the most general case of three-dimensional stresses at a point. The principal stress analysis for this case is most easily carried out by making use of the fact that the principal planes are shear-free planes on which the net force per unit area acts along the normal direction. Let l, m, n be the direction cosines of a principal plane and σ P be the corresponding principal stress. Consider the equilibrium of an infinitesimal tetrahedron bounded by this principal plane and the planes normal to the x, y and z axes (Fig. A.6). On the face coinciding with the principal plane, the force components per unit area along the coordinate directions are lσ P, mσ P and nσ P , respectively. Summing forces along the three directions x, y and z (as was done earlier in Sect. 1.10 while deriving the natural boundary conditions), one gets lσx + mτx y + nτx z = lσ P lτx y + mσ y + nτ yz = mσ P lτx z + mτ yz + nσz = nσ P i.e. l(σx − σ P ) + mτx y + nτx z = 0 lτx y + m(σ y − σ P ) + nτ yz = 0 lτx z + mτ yz + n(σz − σ P ) = 0 which yield a non-trivial solution for l, m and n only if    (σ − σ )  τx y τx z P  x     =0 τx y (σ y − σ P ) τ yz    τx z τ yz (σz − σ P ) 

(A.23)

Appendix

191

i.e. σ P3 − I1 σ P2 + I2 σ P − I3 = 0

(A.24)

where I1 = σ x + σ y + σz 2 2 I2 = σx σ y + σ y σz + σz σx − τ yz − τzx − τx2y 2 2 I3 = σx σ y σz + 2τ yz τzx τx y − σx τ yz − σ y τzx − σz τx2y

Thus, there are three principal stresses σ P1 , σ P2, σ P3 for the general state of stress; the corresponding principal planes can be found out by substituting the value of σ P in Eqns. (A23) and using the relation l 2 + m 2 + n2 = 1

(A.25)

Since the principal stresses have to be the same irrespective of the choice of the coordinate axes x,y, and z, the constants I 1 to I 3 given above should also be the same; these are known as the stress invariants. The extreme shear stresses once again occur on planes bisecting the angles between two principal planes (we shall not bother to prove this here) and are given by   P2 ) P2  |τmax | =  σ P1 −σ along with a normal stress of (σ P1 +σ 2 2  on planes  bisecting the angle P3  along with a normal between P1 and P2 directions, and similarly |τmax | =  σ P2 −σ 2  σ −σ  (σ P2 +σ P3 ) P1 P3   , and |τmax | = along with a normal stress of (σ P1 +σ P3 ) on stress of 2

2

2

the other planes. Thus, for design purposes, the maximum shear stress is the maximum of the above three quantities, as given in Eq. (A.22) earlier. One can easily verify that all the other earlier relations developed specifically for the two-dimensional states of stress are subsets of the more general equations given above. The principal strain analysis is carried out in an entirely analogous manner as above because the strain transformation equations (Eqs. (A.11–A.12)) are exactly similar to those for stresses (Eqs. (A.4–A.5)). Once again, it should be kept in mind that the tensorial shear strain εij has to be used everywhere instead of γ ij . All the final pertinent equations can be easily written down by analogy; this is not done here, but certainly interesting facts are to be pointed out as given below: (a) The principal planes, for a chosen point of the domain, are the same for stresses and strains. Similarly, the maximum shear stress planes and the maximum shear strain planes are the same at a point. (b) It is often easy to carry out principal stress (or principal strain) analysis for an isotropic body and then calculate the principal strains (or principal stresses) and the maximum shear strain (or stress) directly using the generalized Hooke’s law.

192

Appendix

A2 The Isotropic Elastic Constants The elastic behaviour of an isotropic body is completely described in terms of four physically significant elastic constants—Young’s modulus E, Poisson’s ratio μ, the shear modulus G and the bulk modulus K. However, out of these four constants, only two are independent. Let us now derive the relations which yield G and K in terms of E and μ, the latter being the elastic constants most commonly used for characterisation because they are obtained from a simple tension test.

A2.1 Derivation of G in Terms of E and μ Consider a thin infinitesimal square element under pure shear as shown (Fig. A.7), for which we have τx y = Gγx y

(A.26)

Transforming to the x’-y’ axes which are 45° away with respect to the x–y system, one gets σx ' = −σ y ' = τx y ;

τx ' y ' = 0

(A.27)

εx ' = −ε y ' = εx y or γx y /2; εx ' y ' = 0 These strain and stress fields are related by the plane stress law. Thus, σx ' =

E (εx ' + με y ' ) (1 − μ2 )

τ xy

τ xy

τ xy

y x

Fig. A.7 Derivation of G in terms of E and μ

45o

y'

x'

Appendix

193

i.e. τx y =

γx y E E (1 − μ) = γx y 2 (1 − μ ) 2 2(1 + μ)

(A.28)

Now, from Eqs. (A.26) and (A.28), one gets G=

E 2(1 + μ)

(A.29)

A2.2 Derivation of K in Terms of E and μ The bulk modulus K is defined, with reference to the state of hydrostatic pressure, as the ratio of the pressure to the magnitude of volumetric strain. Considering an infinitesimal cube subjected to hydrostatic pressure p on all its faces, we have ε x = ε y = εz =

− p(1 − 2μ) E

(A.30)

Hence, the volumetric strain is of magnitude 3 p(1 − 2μ) E

(A.31)

E p = |e| 3(1 − 2μ)

(A.32)

|e| = and thus, K =

A2.3 Limiting Values for μ By virtue of their definitions, Young’s modulus E, the shear modulus G and the bulk modulus K should all be positive because they all represent a direct stiffness of the material corresponding to a specific mode of deformation. Using this fact and the above-derived relations between the elastic constants, one has (1+μ) ≥ 0 and (1 − 2μ) ≥ 0 i.e. −1 ≤ μ ≤

1 2

In practice, however, negative values for μ are very uncommon.

(A.33)

Index

B Boundary conditions essential and natural, 25 St.Venant type, 62

C Compatibility equations, 21, 31, 72 Complementary shear principle, 7 violation, 87 Constitutive relations isotropic, 9 orthotropic, 13 plane strain, 14 plane stress, 13

D Displacement approach, 20

E Elastic constants isotropic, 191 orthotropic, 12 Equilibrium equations 2-D Cartesian, 16 3-D Cartesian, 20 plane polar, 69

F Field equations, 8 Field variables, 7, 8

M Membrane analogy, 136

N Navier equations, 22

O Orthotropy basic definition, 10 rectilinear & cylindrical, 71 shape-intrinsic, 166

P Principal stresses & strains, 187 Problems beam flexure, 36, 40, 45, 50, 63, 65 concentrated load, 95, 100, 174 contact, 104, 175 crack, 89 curved beam, 109 free edge phenomenon, 176 functionally graded structure, 180 pressurised cylinder, 74 pressurised hole, 80 simple tension, 36

© The Author(s) 2023 K. Bhaskar and T. K. Varadan, Theory of Isotropic/Orthotropic Elasticity, https://doi.org/10.1007/978-3-031-06345-9

195

196 stress concentration, 80, 84, 86 St.Venant decay, 58 torsion, see Torsion problems

S Strain definition, 4 engineering shear, 4 tensorial shear, 24 transformation, 185 Strain-displacement relations Cartesian, 22 plane polar, 71 Stress definition, 5 invariants, 191 transformation, 183 Stress approach, 20 Stress function Airy, 29, 73 Prandtl, 122

Index St.Venant’s principle, 56, 62 St.Venant’s warping function, 120

T Theory of elasticity assumptions, 1 need, 3 Torsion problems circle, 130 ellipse, 128, 157 equilateral triangle, 131 formulation, 120, 139 hollow sections, 143 irregular shapes, 154 laminated rectangle, 158 thin-walled laminate, 164 thin-walled open section, 138 thin-walled tube, 145 with complications, 148 with warping restraints, 167