Continuum mechanics and linear elasticity 9789402417692, 9789402417715

Table of contents :
Preface......Page 7
Contents......Page 10
Part I Elements of Continuum Mechanics......Page 15
1.1 Matrices......Page 16
1.2 Vector Spaces......Page 21
1.3 Euclidean Vector Spaces......Page 24
1.4 Euclidean Point Spaces......Page 31
1.5 Second-Order Tensors: Fundamentals......Page 33
1.6 Examples of Tensors: Elementary Projections......Page 38
1.7 Basic Properties of Tensors......Page 39
1.8 Linear Mappings as Geometric Transformations......Page 45
1.9 Transformation Rules for a Change of Basis......Page 48
1.10 Higher Order Tensors......Page 51
1.11 A Special Property: Isotropy......Page 53
1.12 Pseudo-scalars/vectors/tensors......Page 54
1.13 Other Uses of Vector and Tensor Products......Page 56
1.14 Fourth-Order Tensors......Page 63
1.15 Finite Rotations......Page 69
1.16 On the Relationship Between Skw and mathscrV......Page 72
1.17 More General Bases......Page 74
1.17.1 Cylindrical Polar Coordinates......Page 76
1.17.3 Physical Components for Vectors and Tensors......Page 78
1.18 Invariants......Page 80
1.19 Eigenvalues......Page 84
1.20 Some Important Theorems......Page 89
1.21 Projections in Lin......Page 94
1.22 Vector and Tensor Calculus......Page 95
1.22.1 Differential Calculus in Finite-Dimensional Euclidean Vector Spaces......Page 98
1.22.2 Differential Operators: DIV, GRAD, CURL......Page 104
1.22.3 Vector and Tensor Identities......Page 109
1.22.4 Non-Cartesian Coordinates......Page 112
1.22.5 Some Important Integral Theorems......Page 115
1.23 Differentiation of Tensor Functions......Page 118
1.24 Exercises......Page 125
Bibliography......Page 133
2.1 Deformable Bodies: Definition and Generalities......Page 134
2.2 Examples of Deformations/Motions......Page 136
2.3 Velocity and Acceleration Fields. Material Time Derivatives......Page 140
2.4 The Deformation Gradient......Page 145
2.5 Changes in Area and Volume......Page 149
2.6 Strain Tensors......Page 152
2.7 Examples of Particular Strain Tensors......Page 156
2.9 The Spatial Gradient of Velocity......Page 160
2.10 Transport Formulae......Page 164
2.11 Exercises......Page 167
Bibliography......Page 172
3 Balance Laws......Page 174
3.1 The Principle of Mass Conservation......Page 175
3.2 Body and Surface Forces......Page 179
3.3 Global Form......Page 183
3.4 Local Form......Page 184
3.5 Some Technical Proofs......Page 186
3.6 Further Properties of the Cauchy Stress Tensor......Page 191
3.7 Particular States of Stress......Page 197
3.8 The Piola–Kirchhoff Stress Tensors......Page 199
3.9 Exercises......Page 202
Bibliography......Page 209
4 Constitutive Relationships......Page 210
4.1 The Principle of Material Frame-Indifference......Page 211
4.2 Other Important Constitutive Principles......Page 219
4.3 Cauchy-Elastic Materials......Page 222
4.4 Material Symmetry......Page 223
4.5 Hyperelastic Solids......Page 230
4.6 Constitutive Representations for Isotropic Hyperelastic Solids......Page 236
4.7 Internal Constraints......Page 240
4.8 Particular Forms of the Strain-Energy Function......Page 244
4.9 Simple Fluids......Page 245
4.10 Exercises......Page 250
Bibliography......Page 253
Part II Topics in Linear Elasticity......Page 254
5.1 Introduction......Page 255
5.2 Linearised Kinematics......Page 256
5.3 Distortional and Spherical Strain......Page 257
5.4 Linearised Constitutive Behaviour......Page 258
5.5 Linearised Field Equations......Page 263
5.6 Restrictions on the Elastic Constants......Page 267
5.7 The Navier–Lamé Equations......Page 270
5.8 Principle of Superposition......Page 272
5.9 Saint-Venant's Principle......Page 274
5.10 Worked Examples......Page 275
5.11 Standard Simplifications......Page 283
5.11.1 Plane Strain......Page 284
5.11.2 Plane Stress......Page 285
5.11.3 Antiplane Strain/Stress......Page 286
5.12 Exercises......Page 287
Bibliography......Page 292
6.1 Introduction......Page 293
6.2 Simply and Multiply Connected Domains......Page 295
6.3 The `Incompatibility' Operator......Page 297
6.4 Conservative Fields......Page 300
6.5 Cesàro–Volterra Formula......Page 301
6.6 Alternative Forms of the Compatibility Equation......Page 308
6.7 Beltrami–Michell Equations......Page 311
6.8 Explicit Calculations and Examples......Page 313
6.9 Weingarten-Volterra Dislocations......Page 319
6.10 Exercises......Page 323
Bibliography......Page 330
7.1 Introduction......Page 331
7.2 Some Auxiliary Notation......Page 333
7.3 Governing Equations......Page 335
7.4 Circular Cylinder......Page 336
7.5 Non-circular Cylinder......Page 339
7.6 A Closer Look at the Torsional Rigidity......Page 346
7.7 Prandtl Stress Function......Page 351
7.8 Modified Stress Function......Page 354
7.9 Multiply Connected Domains......Page 357
7.10 The Shear Stress......Page 359
7.11 Complex Variables Formulation......Page 362
7.12 Worked Examples......Page 366
7.13 Exercises......Page 377
Bibliography......Page 381
8.1 Introduction......Page 383
8.2 Plane Strain......Page 385
8.3 Plane Stress......Page 387
8.4 Generalised Plane Stress......Page 388
8.5.2 The Governing Equation for Φ......Page 391
8.5.3 Physical Interpretation of the Boundary Conditions......Page 393
8.5.4 The Displacement Field......Page 396
8.6 Worked Examples......Page 397
8.7 Volterra Distortions......Page 416
8.8 Exercises......Page 420
Bibliography......Page 430
9.1 The Bi-harmonic Equation via Fourier Transforms......Page 431
9.3 The Direct Approach......Page 441
9.4 A Modification of the Method......Page 446
9.5 The Elastic Quarter-Plane......Page 449
9.6 Displacement Boundary-Value Problems......Page 454
9.6.1 The Papkovitch–Neuber Representation......Page 455
9.6.2 The Stress Tensor in Terms of ψ and B......Page 458
9.6.3 Particular Case: Plane Elasticity......Page 459
9.7 Exercises......Page 464
Bibliography......Page 470
Appendix A Vector and Tensor Identities......Page 471
B.1 Cylindrical Coordinates......Page 473
B.2 Spherical Coordinates......Page 476
Appendix C Geometry of Areas......Page 479
D.1 Definitions and General Properties......Page 486
D.2 Fourier Transforms for Multivariate Functions......Page 488
D.3 The Dirac Delta Function......Page 490
D.4 The Use of Complex Variables......Page 494
D.5 Some Calculations for Chap. 9......Page 498
D.6 Useful Sine and Cosine Fourier Transforms......Page 499
E.1 Cartesian Coordinates......Page 501
E.2 Complex Variables......Page 505
E.3 Polar Coordinates......Page 514
Appendix References......Page 522
Index......Page 524

Citation preview

Solid Mechanics and Its Applications

Ciprian D. Coman

Continuum Mechanics and Linear Elasticity An Applied Mathematics Introduction

Solid Mechanics and Its Applications Volume 238

Founding Editor G. M. L. Gladwell, University of Waterloo, Waterloo, ON, Canada Series Editors J. R. Barber, Department of Mechanical Engineering, University of Michigan, Ann Arbor, MI, USA Anders Klarbring, Mechanical Engineering, Linköping University, Linköping, Sweden

The fundamental questions arising in mechanics are: Why?, How?, and How much? The aim of this series is to provide lucid accounts written by authoritative researchers giving vision and insight in answering these questions on the subject of mechanics as it relates to solids. The scope of the series covers the entire spectrum of solid mechanics. Thus it includes the foundation of mechanics; variational formulations; computational mechanics; statics, kinematics and dynamics of rigid and elastic bodies; vibrations of solids and structures; dynamical systems and chaos; the theories of elasticity, plasticity and viscoelasticity; composite materials; rods, beams, shells and membranes; structural control and stability; soils, rocks and geomechanics; fracture; tribology; experimental mechanics; biomechanics and machine design. The median level of presentation is the first year graduate student. Some texts are monographs defining the current state of the field; others are accessible to final year undergraduates; but essentially the emphasis is on readability and clarity. Springer and Professors Barber and Klarbring welcome book ideas from authors. Potential authors who wish to submit a book proposal should contact Dr. Mayra Castro, Senior Editor, Springer Heidelberg, Germany, e-mail: [email protected] Indexed by SCOPUS, Ei Compendex, EBSCO Discovery Service, OCLC, ProQuest Summon, Google Scholar and SpringerLink.

More information about this series at http://www.springer.com/series/6557

Ciprian D. Coman

Continuum Mechanics and Linear Elasticity An Applied Mathematics Introduction

123

Ciprian D. Coman School of Computing and Engineering University of Huddersfield Huddersfield, UK

ISSN 0925-0042 ISSN 2214-7764 (electronic) Solid Mechanics and Its Applications ISBN 978-94-024-1769-2 ISBN 978-94-024-1771-5 (eBook) https://doi.org/10.1007/978-94-024-1771-5 © Springer Nature B.V. 2020 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, 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 publisher, 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 publisher 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 publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature B.V. The registered company address is: Van Godewijckstraat 30, 3311 GX Dordrecht, The Netherlands

To the loving memory of my father and to my mother

Preface

Continuum Mechanics represents a logical evolutionary step in the development of several engineering disciplines involved with the mechanics of deformable objects. It provides a unified high-level description for the behaviour of liquids, solids and gases, by dealing only with the observed, macroscopic effects experienced by these substances in response to various external agents. What distinguishes Continuum Mechanics from other physical theories like, for example, Quantum Mechanics or Statistical Physics, is that all the fine microscopic details are disregarded; in particular, it is assumed that the highly discontinuous atomic structure of matter can be replaced by a smoothed hypothetical body, usually referred to as a ‘continuum’.1 The upshot of introducing this simplifying hypothesis is intimately connected with the powerful tools of Vector Calculus involved in formulating the classical mathematical models of deformable continua. This book represents an outgrowth of more than 15 years of teaching various courses related to mathematical modelling at several universities in the UK. My audiences have largely been final-year undergraduates on Mathematics Honours programmes; on several occasions, I also delivered similar material to first-year (applied maths) Ph.D. students through a couple of regional Doctoral Training Centres (in Glasgow and Nottingham). An embryonic collection of supplementary classroom notes was first written around 2002/04, and covered most of the material in the first four chapters. As time went by, the scope of the narrative evolved from a mere outline to a more detailed treatment of the subject, which included additional topics related to Linear Elasticity and several other adjacent areas. The final result, the printed material within these pages, represents a scaled-down attempt to produce the kind of introductory textbook I wish I had as an undergraduate student. A question likely to arise in the prospective reader’s mind is, probably, why another book on these particular topics? It is also reasonable to ask in what way this text is different from others available. The answers lie partly in the subtitle: one 1 The word continuous comes from the Latin root con (‘together with’) and tenere (‘hold’), meaning ‘uninterrupted’ or ‘holding together’. ‘Continuum Mechanics’ is the mechanics of continuous media.

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Preface

of the intended goals was to produce a reasonably self-contained text that promotes the elementary use of coordinate-free equations and capitalises on two staple courses in undergraduate Mathematics curricula: Linear Algebra and Vector Calculus. In the UK, courses that cover these topics are taken by most students in the first 2 years of their studies. In a sense, both Continuum Mechanics and Linear Elasticity can be regarded as natural extensions of the aforementioned core subjects, and allow a mix of calculational work (both routine and advanced) with more theoretical material. Continuum Mechanics helps students develop physical intuition, while Linear Elasticity builds confidence in the mathematical techniques studied in other courses. Because the two main topics of the book are well-established academic disciplines and have so many separate branches, I chose to concentrate on those chapters that best reveal the spirit of Rational Mechanics and are still within the grasp of the intended undergraduate audience. By restricting the selection of topics discussed, an attempt was made to develop the ideas clearly, logically and without gaps. Many of the derivations included in the next chapters give deliberately more details than strictly necessary or than most other similar books at this level give. In particular, ample amount of support is provided for many of the formal vector and tensor manipulations, so that beginning students can develop a confident understanding and an appreciation for a range of calculations in different formats. There is also less emphasis on stereotyped problems, which I believe do not facilitate much personal discovery during the learning process. To this end, a generous supply of exercises of varying degrees of difficulty is included at the end of each chapter (53 pages in total); these are regarded an integral part of the text and form one of its unique features. The student who wants to achieve a thorough grasp of the material discussed in this book must invest some non-trivial amount of time in solving at least some of them before moving on to the next chapter. The exercises included have been gathered over a period of years for various classes I have given; some are taken from others’ textbooks and papers, while some are modified or original. It is no longer possible to properly acknowledge the original sources. My debt to earlier writers is certainly substantial. As already mentioned, the text assumes familiarity with elementary Linear Algebra and standard Vector Calculus; prior exposure to Vector Mechanics in three dimensions is desirable, but not absolutely essential. Since it is my experience that almost no student retains much detailed information from one semester to the next, the first chapter includes a convenient outline of matrix algebra, linear spaces, vectors and Euclidean point spaces. Complex Analysis makes several ‘guest appearances’ in the second part of the book. It is first mentioned briefly in the chapter on torsion, but it plays only a marginal role and can be avoided (if one so desires). However, the last chapter deals with the application of Fourier integral transforms to a particular class of Linear Elasticity problems. Although Complex Analysis does have a stronger presence in that chapter, most of our manipulations will be formal, and the inverse transforms will be carried out with the help of a small number of standard integrals. One of the appendices at the end of the text summarises all the information that is needed for making that chapter reasonably

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self-contained. There is also another appendix in which the solution of the biharmonic equation is discussed from a number of different perspectives, one of which hinges on the use of more advanced Complex Analysis; again, that part can be skipped without any interference with the rest of the text. I should perhaps emphasise that this is not a text on ‘Applied Analysis’ or ‘PDEs’, and I permit a certain degree of carelessness in restrictions and conditions. Although some results are stated as ‘theorems’ and ‘propositions’, I omit rigorous proofs and offer instead informal arguments and examples that either shed light or amplify the formal statements. That is partly because I intended this book to be no more than an introduction, one that could act as a bridge for the gap between the simple-minded accounts and the really advanced treatments of Continuum Mechanics and Linear Elasticity available in the literature. I hope it will be read in conjunction with the classics in these areas, and that it will encourage further exploration. The only consolation I can offer to readers who feel frustrated about my lack of rigorousness, or the number of times I seem to be cutting corners, is that the book is not addressed to them. The manuscript was developed on and off over a period of years, mostly by compiling a series of handouts and hand-written notes used in my past classes. A first draft of the book was produced around 2010/12. Due to objective circumstances, I was unable to continue with this project for many years afterwards; it was eventually finalised during the academic year 2018/19 (in short bursts of activity, whenever my day job permitted). Given the prolonged gestation period of the manuscript and the long hiatus, it would be rather difficult to accurately track down all sources that I consulted at various times. For this reason, I have refrained from adding many bibliographical references in the main text, which would only have served as a distraction in a book of this type; a list of general references that have helped me clarify various points regarding the material presented is mentioned at the end of each chapter. Of course, I make no claim for originality in the basic theory, other than in the organisation and details of presentation. I would like to express my sincere thanks to the publisher, Springer Media, for not giving up on this project. Their patience and understanding in waiting for the delivery of a long-overdue manuscript is deeply appreciated. As I have learned through my own experience, the best lectures are a two-way conversation between the lecturer and his class. By its very nature, a book limits the dialogue and some of that elusive quality is lost. Nevertheless, I would be very pleased to receive questions, comments and criticism at [email protected]. An errata list will be available on my personal website, www.cipriancoman.net, and will be updated regularly. Clifton Grove, Nottingham, UK June 2019

Ciprian D. Coman

Contents

Part I

Elements of Continuum Mechanics

1 Vector, Tensors, and Related Matters . . . . . . . . . . . . . . . . . 1.1 Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Euclidean Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . 1.4 Euclidean Point Spaces . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Second-Order Tensors: Fundamentals . . . . . . . . . . . . . . 1.6 Examples of Tensors: Elementary Projections . . . . . . . . 1.7 Basic Properties of Tensors . . . . . . . . . . . . . . . . . . . . . 1.8 Linear Mappings as Geometric Transformations . . . . . . 1.9 Transformation Rules for a Change of Basis . . . . . . . . 1.10 Higher Order Tensors . . . . . . . . . . . . . . . . . . . . . . . . . 1.11 A Special Property: Isotropy . . . . . . . . . . . . . . . . . . . . 1.12 Pseudo-scalars/vectors/tensors . . . . . . . . . . . . . . . . . . . 1.13 Other Uses of Vector and Tensor Products . . . . . . . . . . 1.14 Fourth-Order Tensors . . . . . . . . . . . . . . . . . . . . . . . . . 1.15 Finite Rotations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.16 On the Relationship Between Skw and V . . . . . . . . . . 1.17 More General Bases . . . . . . . . . . . . . . . . . . . . . . . . . . 1.17.1 Cylindrical Polar Coordinates . . . . . . . . . . . . . 1.17.2 Spherical Polar Coordinates . . . . . . . . . . . . . . 1.17.3 Physical Components for Vectors and Tensors . 1.18 Invariants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.19 Eigenvalues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.20 Some Important Theorems . . . . . . . . . . . . . . . . . . . . . 1.21 Projections in Lin . . . . . . . . . . . . . . . . . . . . . . . . . . .

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1.22 Vector and Tensor Calculus . . . . . . . . . . . . . . . . . . . . . . . . . 1.22.1 Differential Calculus in Finite-Dimensional Euclidean Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.22.2 Differential Operators: DIV, GRAD, CURL . . . . . . . . 1.22.3 Vector and Tensor Identities . . . . . . . . . . . . . . . . . . . 1.22.4 Non-Cartesian Coordinates . . . . . . . . . . . . . . . . . . . . 1.22.5 Some Important Integral Theorems . . . . . . . . . . . . . . 1.23 Differentiation of Tensor Functions . . . . . . . . . . . . . . . . . . . . 1.24 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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85 91 96 99 102 105 112 120

2 Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Deformable Bodies: Definition and Generalities . . . . . . . . . . 2.2 Examples of Deformations/Motions . . . . . . . . . . . . . . . . . . . 2.3 Velocity and Acceleration Fields. Material Time Derivatives . 2.4 The Deformation Gradient . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Changes in Area and Volume . . . . . . . . . . . . . . . . . . . . . . . 2.6 Strain Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7 Examples of Particular Strain Tensors . . . . . . . . . . . . . . . . . 2.8 The Interpretation of the Polar Decomposition Theorem . . . . 2.9 The Spatial Gradient of Velocity . . . . . . . . . . . . . . . . . . . . . 2.10 Transport Formulae . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.11 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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3 Balance Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 The Principle of Mass Conservation . . . . . . . . 3.2 Body and Surface Forces . . . . . . . . . . . . . . . . 3.3 Global Form . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Local Form . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5 Some Technical Proofs . . . . . . . . . . . . . . . . . . 3.6 Further Properties of the Cauchy Stress Tensor 3.7 Particular States of Stress . . . . . . . . . . . . . . . . 3.8 The Piola–Kirchhoff Stress Tensors . . . . . . . . . 3.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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4 Constitutive Relationships . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 The Principle of Material Frame-Indifference . . . . . . . 4.2 Other Important Constitutive Principles . . . . . . . . . . . 4.3 Cauchy-Elastic Materials . . . . . . . . . . . . . . . . . . . . . . 4.4 Material Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5 Hyperelastic Solids . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6 Constitutive Representations for Isotropic Hyperelastic Solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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4.7 Internal Constraints . . . . . . . . . . . . . . . . . . . . . 4.8 Particular Forms of the Strain-Energy Function 4.9 Simple Fluids . . . . . . . . . . . . . . . . . . . . . . . . . 4.10 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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5 Linear Elasticity: General Considerations and Boundary-Value Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Linearised Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Distortional and Spherical Strain . . . . . . . . . . . . . . . . . . . . . 5.4 Linearised Constitutive Behaviour . . . . . . . . . . . . . . . . . . . . 5.5 Linearised Field Equations . . . . . . . . . . . . . . . . . . . . . . . . . 5.6 Restrictions on the Elastic Constants . . . . . . . . . . . . . . . . . . 5.7 The Navier–Lamé Equations . . . . . . . . . . . . . . . . . . . . . . . . 5.8 Principle of Superposition . . . . . . . . . . . . . . . . . . . . . . . . . . 5.9 Saint-Venant’s Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.10 Worked Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.11 Standard Simplifications . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.11.1 Plane Strain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.11.2 Plane Stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.11.3 Antiplane Strain/Stress . . . . . . . . . . . . . . . . . . . . . . 5.12 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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243 243 244 245 246 251 255 258 260 262 263 271 272 273 274 275 280

6 Compatibility of the Infinitesimal Deformation Tensor . 6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Simply and Multiply Connected Domains . . . . . . . 6.3 The ‘Incompatibility’ Operator . . . . . . . . . . . . . . . 6.4 Conservative Fields . . . . . . . . . . . . . . . . . . . . . . . . 6.5 Cesàro–Volterra Formula . . . . . . . . . . . . . . . . . . . 6.6 Alternative Forms of the Compatibility Equation . . 6.7 Beltrami–Michell Equations . . . . . . . . . . . . . . . . . 6.8 Explicit Calculations and Examples . . . . . . . . . . . . 6.9 Weingarten-Volterra Dislocations . . . . . . . . . . . . . . 6.10 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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281 281 283 285 288 289 296 299 301 307 311 318

7 Torsion . . . . . . . . . . . . . . . . . . 7.1 Introduction . . . . . . . . . . 7.2 Some Auxiliary Notation . 7.3 Governing Equations . . .

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Part II

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xiv

Contents

7.4 Circular Cylinder . . . . . . . . . . . . . . . . . 7.5 Non-circular Cylinder . . . . . . . . . . . . . . 7.6 A Closer Look at the Torsional Rigidity 7.7 Prandtl Stress Function . . . . . . . . . . . . . 7.8 Modified Stress Function . . . . . . . . . . . . 7.9 Multiply Connected Domains . . . . . . . . 7.10 The Shear Stress . . . . . . . . . . . . . . . . . . 7.11 Complex Variables Formulation . . . . . . 7.12 Worked Examples . . . . . . . . . . . . . . . . . 7.13 Exercises . . . . . . . . . . . . . . . . . . . . . . . Bibliography . . . . . . . . . . . . . . . . . . . . . . . . .

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324 327 334 339 342 345 347 350 354 365 369

8 Two-Dimensional Approximations . . . . . . . . . . . . . . . . . . . . . . 8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Plane Strain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3 Plane Stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4 Generalised Plane Stress . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5 The Airy Stress Function . . . . . . . . . . . . . . . . . . . . . . . . . 8.5.1 Main Definitions . . . . . . . . . . . . . . . . . . . . . . . . . 8.5.2 The Governing Equation for U . . . . . . . . . . . . . . . 8.5.3 Physical Interpretation of the Boundary Conditions 8.5.4 The Displacement Field . . . . . . . . . . . . . . . . . . . . 8.6 Worked Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.7 Volterra Distortions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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371 371 373 375 376 379 379 379 381 384 385 404 408 418

9 Special Two-Dimensional Problems: Unbounded Domains 9.1 The Bi-harmonic Equation via Fourier Transforms . . . 9.2 Remarks on the Displacement Field . . . . . . . . . . . . . . 9.3 The Direct Approach . . . . . . . . . . . . . . . . . . . . . . . . 9.4 A Modification of the Method . . . . . . . . . . . . . . . . . . 9.5 The Elastic Quarter-Plane . . . . . . . . . . . . . . . . . . . . . 9.6 Displacement Boundary-Value Problems . . . . . . . . . . 9.6.1 The Papkovitch–Neuber Representation . . . . . 9.6.2 The Stress Tensor in Terms of w and B . . . . . 9.6.3 Particular Case: Plane Elasticity . . . . . . . . . . 9.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Appendix A: Vector and Tensor Identities . . . . . . . . . . . . . . . . . . . . . . . . 459 Appendix B: Cylindrical and Spherical Coordinates . . . . . . . . . . . . . . . . 461

Contents

xv

Appendix C: Geometry of Areas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 467 Appendix D: Fourier Transforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 475 Appendix E: The Bi-harmonic Equation . . . . . . . . . . . . . . . . . . . . . . . . . . 491 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 513 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 515

Part I

Elements of Continuum Mechanics

Chapter 1

Vector, Tensors, and Related Matters

Abstract All mathematical models of Continuum Mechanics are described by using vectorial and tensorial quantities. For this reason, a certain fluency in vector and tensor manipulations is imperative for much of the rest of the book. The purpose of the current chapter is largely twofold. While the first few sections can serve as a brief review of several key topics in Linear Algebra, for the most part the attention will be directed towards the concept of second-order tensor and its higher order generalisations. The last couple of sections indicate how the usual vector calculus can be extended to tensor fields and functions whose arguments are tensors.

1.1 Matrices A matrix is an m × n array of scalars, where m, n ∈ N. These numbers can be real or complex, but in this book we shall work mostly with matrices involving real numbers. If the matrix is A, then ⎡ ⎤ a11 a12 . . . a1n ⎢ a21 a22 . . . a2n ⎥ ⎢ ⎥ A=⎢ . . . ⎥. ⎣ .. .. . . . .. ⎦ am1 am2 . . . amn

The numbers ai j , i = 1, 2, . . . , m and j = 1, 2, . . . , n are called the elements of the matrix. The first index (i) identifies the row, while the second index ( j) gives the column in which the element ai j is situated. A matrix is denoted for short by A = [ai j ]. The set of all real matrices is denoted by Mm×n (R). If m = n then A is called a square matrix. An element of M1×n (R), that is, a matrix of the form  a11 a12 . . . a1n

© Springer Nature B.V. 2020 C. D. Coman, Continuum Mechanics and Linear Elasticity, Solid Mechanics and Its Applications 238, https://doi.org/10.1007/978-94-024-1771-5_1

3

4

1 Vector, Tensors, and Related Matters

is called a row vector, and an element of Mm×1 (R), i.e. a matrix of the form ⎡

a11 ⎢ a21 ⎢ ⎢ .. ⎣ .

⎤ ⎥ ⎥ ⎥ ⎦

am1 is called a column vector. The transpose of A ∈ Mm×n (R) is defined to be the n × m matrix A T obtained by interchanging rows and columns in A. More precisely, if A = [ai j ] then [A T ]i j = a ji . It should be evident that for all matrices (A T )T = A. Consider the case of a square matrix A = [ai j ]. The set of numbers a11 , a22 , …, ann is called the principal diagonal of A. The remaining elements ai j are said to be ‘above’ (respectively, ‘below’) the diagonal depending on whether i < j or i > j. A square matrix is called diagonal if ai j = 0 for i = j and each i, j = 1, 2, . . . , n. The particular case of a diagonal matrix in which all the elements along the principal diagonal are equal to unity is called the identity matrix. It is denoted by In , ⎡

1 ⎢0 ⎢ In = ⎢ . ⎣ ..

0 1 .. .

0

0

... ... ... ...

⎤ 0 0⎥ ⎥ .. ⎥ , .⎦ 1

and when there is no risk of confusion we drop the subscript n on the left-hand side. We can write this for short as In = [δi j ], where δi j is the Kronecker delta symbol, or simply the Kronecker delta, defined by

δi j =

1 0

if i = j . if i = j

(1.1)

Basic operations: Matrix addition is defined entry-wise for arrays of the same dimensions. More precisely, if A, B ∈ Mm×n (R) the sum of A and B is defined to be the matrix A + B ∈ Mm×n (R), whose entries are defined by [A + B]i j = [A]i j + [B]i j . The symbol ‘+’ is used in two different ways—it denotes addition between scalars and addition between matrices at other places. Although these are two distinct algebraic operations, no ambiguities will arise if the context in which the symbol ‘+’ appears is observed. Also, note that the requirement that A and B have the same size prevents adding a row vector to a column vector, even though the two may contain the same number of entries.

1.1 Matrices

5

The matrix (−A), called the additive inverse of A, is defined to be the matrix obtained by negating each entry of A. That is, if A = [ai j ] then (−A) = [−ai j ]. This process allows the subtraction of matrices to be defined in a natural way. By definition, the difference of A, B ∈ Mm×n (R), denoted by A − B, is obtained by adding to A the additive inverse of B, or A − B := A + (−B). The addition and subtraction of matrices has the well-known properties listed below. If A, B, C ∈ Mm×n (R) then closure property: associativity: commutativity:

A + B is again an element of Mm×n (R); (A + B) + C = A + (B + C); A + B = B + A.

These first two properties hold for the subtraction of matrices as well. A matrix in Mm×n (R) whose all elements are zero is denoted by Omn and is called the zero matrix. For every A ∈ Mm×n (R) we have A + (−A) = Omn . When there is no risk of confusion, we shall simply write O for Omn . Another basic operation is scalar multiplication. The product of a scalar α ∈ R with a matrix A ∈ Mm×n (R) is defined to be the matrix obtained by multiplying each entry of A by α. That is, [α A]i j = α[A]i j for each i = 1, 2, . . . , m and j = 1, 2, . . . , n. The rules for combining addition and scalar multiplication are what one would naturally expect. For the sake of completeness these properties are briefly recalled below closure property: associativity:

α A is again an element of Mm×n (R); α(A + B) = α A + α B; (α + β)A = α A + β A, α, β ∈ R.

Multiplication of matrices: Let A ∈ Mm× p (R) and B ∈ M p×n (R). The product of A and B is a matrix C ∈ Mm×n (R), C = [ci j ], whose elements are defined by ci j =

p 

aik bk j = ai1 b1 j + ai2 b2 j + · · · + ai p b pj ,

k=1

for each i = 1, 2, . . . , m and j = 1, 2, . . . , n. We set AB := C. The product of two matrices is defined only if the number of columns in A coincides with the number of rows of B. In general, the product of two matrices is not commutative, i.e. AB = B A, but a number of intuitive properties hold A(BC) = (AB)C, (A + B)C = AC + BC, A(B + C) = AB + AC, (α A)B = A(α B) = α(AB), (AB)T = B T A T , AIn = In A = A

if A ∈ Mn×n (R),

6

1 Vector, Tensors, and Related Matters

provided that all operations indicated are well defined. Determinants: A permutation p = ( p1 , p2 , . . . , pn ) of the numbers (1, 2, . . . , n) is simply any re-arrangement. For example, the set {(1, 2, 3); (1, 3, 2); (2, 1, 3); (2, 3, 1); (3, 1, 2); (3, 2, 1)} contains the six distinct permutations of (1, 2, 3). In general, the sequence (1, 2, . . . , n) has n! = n · (n − 1) · (n − 2) . . . 2 · 1 different permutations. In relation to this concept, there is the problem of restoring the natural order by a sequence of pairwise interchanges, and the sign of a permutation is defined to be

σ ( p) :=

⎧ +1 ⎪ ⎪ ⎪ ⎨ ⎪ −1 ⎪ ⎪ ⎩

if p can be restored to natural order by an even number of interchanges, if p can be restored to natural order by an odd number of interchanges.

(1.2)

For example, if p = (1, 4, 3, 2) then σ ( p) = −1 and if p = (4, 3, 2, 1) then σ ( p) = +1. The sign of the natural order p = (1, 2, 3, 4) is evidently σ ( p) = +1. A determinant is a function of square matrices: it takes a square matrix as input and gives a number as output. More precisely, for a matrix A ∈ Mn×n (R), A = [ai j ] the determinant of A is defined to be the scalar  σ ( p)a1 p1 a2 p2 . . . anpn , (1.3) det(A) = p

where the sum is taken over the n! permutations p = ( p1 , p2 , . . . , pn ) of (1, 2, . . . , n). Another notation commonly used is  a11  a21  det(A) =  .  ..  an1

 a12 . . . a1n  a22 . . . a2n  .. . . . . . . ..  an2 . . . ann 

It must be observed that each term that enters the sum in (1.3) contains exactly one entry from each row and each column of A. A minor determinant (or simply a minor) of A ∈ Mn×n (R) is defined to be the determinant of any k × k sub-matrix of A (for any 1 ≤ k ≤ n). An individual entry of A can be regarded as a 1 × 1 minor, while det(A) itself is considered to be an n × n minor of A. The cofactor of the element ai j of a matrix A ∈ Mm×n (R) is defined as Γi j = (−1)i+ j Δi j ,

1.1 Matrices

7

where Δi j is the (n − 1) × (n − 1) minor obtained by deleting the ith row and the jth column of A. The matrix of cofactors of A T is called the adjugate of A and is denoted by A∗ or adj(A). Thus, the element found at the intersection between row i and column j in A∗ is Γ ji . In this book, we shall also use the matrix of cofactors of A, which will be identified by the notation cof (A) such that [cof (A)]i j = Γi j . The cofactors of a square matrix appear naturally in the expression of det(A). The following important expansion formulae hold true: det A = =

n  k=1 n 

aik Γik ≡ ai1 Γi1 + ai2 Γi2 + · · · + ain Γin

(1.4a)

a k j Γk j ≡ a 1 j Γ1 j + a 2 j Γ2 j + · · · + a n j Γn j .

(1.4b)

k=1

Through the application of either of these two formulae, the original n × n determinant is reduced to calculating several (n − 1) × (n − 1) minors. Formula (1.4a) represents the cofactor expansion by row i of the determinant, while (1.4b) is the expansion by column j. More generally, (1.4) can be generalised to δi j det(A) =

n 

aik Γ jk =

k=1

n 

aki Γk j .

k=1

For i = j, we recover the earlier formulae. Some fundamental properties of determinants appear recorded below det(α A) = α n det(A),

det(AB) = det(A) det(B),

det(A ) = det(A), T

for all A, B ∈ Mn×n and α ∈ R; the second property above holds also for a product with an arbitrary finite number of terms. Inverse matrix: A square matrix A ∈ Mn×n (R) is called singular if det(A) = 0 and non-singular otherwise. For a matrix that has the latter property it is always possible to find another n × n matrix, called the inverse of A and denoted by A−1 , such that A−1 A = A A−1 = In . It can be shown that this inverse, A−1 = [di j ], is given by A−1 =

1 A∗ , det(A)

di j =

Γ ji , det(A)

The evident property det(In ) = 1 implies that det(A−1 ) = 1/ det(A); we also have (A−1 )−1 = A,

(A−1 )T = (A T )−1 ,

(AB)−1 = B −1 A−1 ,

8

1 Vector, Tensors, and Related Matters

provided that the matrices A, B ∈ Mn×n (R) are invertible (i.e. they have inverses). Finally, the cofactor matrix can be expressed as cof (A) = (det A)A−T ,

(1.5)

i.e. this is the adjugate of A T or, what is the same, the transpose of the adjugate.

1.2 Vector Spaces A linear space is a set in which elements can be added together and multiplied by real numbers to form another element from that set. Definition 1.1 A real linear vector space (vector space for short) consists of a set V , a mapping (u, v) → u + v of V × V into V and a mapping (λ, v) → λv of R × V into V , such that the following conditions are satisfied: (C1) V is a commutative group with respect to the operation ‘+’; (C2) the associative law λ(μv) = (λμ)v holds, (∀) λ, μ ∈ R, (∀) v ∈ V ; (C3) the distributive laws (λ + μ)v = λv + μv and λ(u + v) = λu + λv hold, (∀) λ, μ ∈ R and (∀) u, v ∈ V ; (C4) 1v=v, (∀) v ∈ V . The first condition above (C1) means the following: (C1-1) u + (v + w) = (u + v) + w, (∀) u, v, w ∈ V ; (C1-2) u + v = v + u, (∀) u, v ∈ V ; (C1-3) there exists an element 0 ∈ V (the zero element) such that u + 0 = u, (∀) u ∈ V ; (C1-4) for every u ∈ V there is an element −u (the inverse of u) such that u + (−u) = 0. The elements of a vector space are called vectors. Their individual properties are irrelevant; what matters is that the addition and scalar multiplication in linear spaces satisfy the axioms mentioned above. Definition 1.2 (a) The elements v1 , v2 , . . . , vn of V are called linearly independent if there exist no real numbers λ1 , λ2 , . . . , λn such that λ1 v1 + λ2 v2 + · · · + λn vn = 0

and

n 

(λ j )2 > 0.

(1.6)

j=1

(b) If there exist real numbers λ1 , λ2 , . . . , λn such that (1.6) holds, then we say that v1 , v2 , . . . , vn are linearly dependent. We can then express any one of the v j , with λ j = 0, as a linear combination of the others

1.2 Vector Spaces

vj = −

9

λ j−1 λ j+1 λ1 λn v1 − · · · − v j−1 − v j+1 − · · · − vn . λj λj λj λj

The vectors v1 , v2 , . . . , vn are said to span a subset M ⊂ V if and only if every vector in M is a linear combination of this set of elements. In this case, we write M = span(v1 , v2 , . . . , vn ). Definition 1.3 A finite-dimensional vector space V is one for which there exists a v2 , . . . ,  vn ∈ V , with the following properties: finite number of vectors, say,  v1 ,  v2 , . . . ,  vn are linearly independent, (a)  v1 ,  (b) (∀) v ∈ V, (∃) λ1 , λ2 , . . . , λn ∈ R such that v =

n 

λ j vj.

(1.7)

j=1

In this case we say that the dimension of V is equal to n and write dim(V ) = n, if n ∈ N is the least with this property. A finite-dimensional linear space ‘looks’ a lot like Rn , for some suitable value of n ∈ N, and for all practical purposes we shall tacitly assume this identification henceforth. In other words, the elements of such a space will appear as ordered n-tuples of the form (x1 , x2 , . . . , xn ) where xi ∈ R.1 In connection to Definition 1.3 some observations are in order. First, the system of v2 , . . . ,  vn } is said to form a basis in V . Thus, in a finite-dimensional elements { v1 ,  vector space each element can be generated using a system of n linearly independent vectors. Second, the representation (1.7) is unique, and the numbers λ1 , λ2 , . . . , λn are called the components of v relative to the basis { v1 ,  v2 , . . . ,  vn }. The standard basis in an n-dimensional space is defined by n-tuples e j with n entries, the jth entry being equal to 1 and the others 0. For instance, in the case of a three-dimensional linear space we have e1 := (1, 0, 0),

e2 := (0, 1, 0),

e3 := (0, 0, 1).

Definition 1.4 Let V1 and V2 be two linear spaces. A linear transformation T : V1 → V2 is a mapping such that T (u + v) = T (u) + T (v)

and

T (αu) = αT (u),

for all scalars α ∈ R and all u, v ∈ V1 . Two linear transformation T , S : V1 → V2 are said to be equal if S(v) = T (v) for all v ∈ V1 . In the case when both the domain and the co-domain of these vi ) for transformations are finite-dimensional, it suffices to demand that S( vi ) = T ( v2 , . . . ,  vn } ⊂ V1 corresponds to a basis. What this i = 1, 2, . . . , n, where { v1 ,  observation is saying is that a linear transformation is completely determined if we must be emphasised that an n-tuple is not a 1 × n (row) matrix; the elements of a vector space can be represented as either row or column matrices, depending on the situation at hand.

1 It

10

1 Vector, Tensors, and Related Matters

know its effect on a full set of basis elements. Furthermore, it can be shown that a linear transformation may be defined by assigning arbitrary images for such a set of basis elements. Linear transformations between finite-dimensional linear spaces are equivalent to v2 , . . . ,  vn } ⊂ V be a fixed basis of V and for sima matrix. Indeed, let B ≡ { v1 ,  plicity let T : V → V be a linear mapping (we take the domain and co-domain to be the same, an assumption that can be relaxed at little extra cost). Then, for each i = 1, 2, . . . , n, T ( vi ) is a linear combination of elements in B, and its corresponding coefficients can be used to define the rows or columns of an n × n matrix, which together with the basis B completely determine the linear transformation T . The question of whether we should let T ( vi ) give the rows or columns of the matrix corresponding to T is settled by requiring that the matrix of a product of two transformations be the product of their corresponding matrices. Details of this process for the general case when V ≡ Rn are presented below. If v ∈ V is an arbitrary element, then this will admit a unique representation of n λi ei (with respect to the standard basis). By the linearity of T , the form v = i=1  T (v) = T

n 

 λi ei

=

i=1

n 

λi T (ei ).

(1.8)

i=1

But, since T (ei ) ∈ V , it can also be represented with respect to the basis of V , T (ei ) =

n 

α ji e j ,

i = 1, 2, , . . . , n,

(1.9)

j=1

for some scalars α ji ∈ R. Combining (1.8) and (1.9) we arrive at T (v) =

n n  

λi α ji e j ,

i=1 j=1

and it must be clear that the n × n matrix A = [α ji ] completely characterises the action of T on every vector v. By changing v ∈ V , the only modification on the right-hand side in the above equation corresponds to the set of λi ’s, that is, the components of that vector. As pointed out above A := [col(T (e1 )), col(T (e2 )), . . . , col(T (en )], i.e. it is the matrix whose columns are the vectors T (ei ) for i = 1, 2, . . . , n. We can write further T (v) = Av, with the understanding that the matrix A depends on the choice of bases in the domain and the co-domain of T . In fact, A depends not only on the set of basis vectors, but also on the order in which they are given as well. The short discussion above suggests that matrices are a convenient way to encode linear transformations between finite-dimensional linear spaces. This is a powerful

1.2 Vector Spaces

11

idea which forms the foundation for most of the tensor algebraic manipulations in the later sections of this chapter. Sometimes one encounters linear spaces within larger linear spaces, the precise idea being made clear next. Definition 1.5 Let V be a linear space. A non-empty subset M of V is a subspace if it is closed under both vector addition and scalar multiplication; that is, M must satisfy (∀) u, v ∈ M, (∀) α ∈ R ⇒ u + v ∈ M and αu ∈ M.

1.3 Euclidean Vector Spaces From now on, we shall agree to denote the elements of vector spaces by boldface letters, and V rather than V will be used to denote a three-dimensional Euclidean vector space. Definition 1.6 Let V be a three-dimensional real vector space. We say that V is a Euclidean vector space if we can define an additional operation (u, v) ∈ V × V −→ u · v ∈ R,

(1.10)

such that for all u, v, w ∈ V and α, β ∈ R the following properties hold: (P1) u · v = v · u; (P2) (αu + βv) · w = α(u · w) + β(v · w); (P3) u · u ≥ 0; u · u = 0 if and only if u = 0. Formula (1.10) defines the so-called scalar (or ‘dot’ or ‘inner’) product of u and v. Strictly speaking, Euclidean vector spaces can have any dimension; a typical example is Rn (with n ≥ 1). For any two vectors u = (u1 , u2 , . . . , un ) ∈ Rn and v = (v1 , v2 , . . . , vn ) ∈ Rn , one can set u · v :=

n 

ui vi ,

i=1

which is easily shown to satisfy the conditions (P1) through (P3) above. The magnitude of a vector u, denoted by |u|, is a positive number defined by |u| := (u · u)1/2 . A vector of unit magnitude is called a unit vector; any u = 0 can be normalised, i.e., replaced by a proportional vector u˜ of unit magnitude, by merely setting u˜ := u/|u|. Consider any two non-zero vectors u, v ∈ V ; the angle between them, θ ≡ θ (u, v), is defined by the trigonometric equation

12

1 Vector, Tensors, and Related Matters

cos θ :=

u·v , |u||v|

(0 ≤ θ ≤ π );

(1.11)

in particular, if u · v = 0 then θ = 90◦ , and the two vectors are said to be orthogonal. Of course, Eq. (1.11) will have no solution unless the right-hand side belongs to the interval [−1, 1]. That this condition is satisfied can be checked by using (P3) written for u → u + αv, with α ∈ R an arbitrary parameter. Expanding the corresponding scalar product, we find |u|2 + 2α(u · v) + α 2 |v|2 ≥ 0 for all α ∈ R, which can be regarded as a quadratic in α. Since its sign does not change, irrespective of what α is, we conclude that the discriminant D (say) of this quadratic must be negative, i.e., D = 4 (u · v)2 − 4|u|2 |v|2 ≤ 0. This confirms that (1.11) has a solution for any two non-zero vectors u, v ∈ V . The structure of a three-dimensional Euclidean vector space makes it possible to introduce another operation on V , the so-called vector product, (u, v) ∈ V × V −→ u ∧ v ∈ V .

(1.12)

By definition, according to this new operation any two elements of V can be combined to yield a new vector with the following properties, (P4) (P5) (P6) (P7)

u ∧ v = −v ∧ u; (αu + βv) ∧ w = α(u ∧ w) + β(v ∧ w); u · (u ∧ v) = 0; v · (u ∧ v) = 0; (u ∧ v) · (u ∧ v) = (u · u)(v · v) − (u · v)2 .

It is worthwhile emphasising that the conditions (P4) through (P7) can be derived from the Euclidean structure introduced in Definition 1.6. Here, we shall take (1.12) at face value and regard it as a new operation that enjoys the properties listed above, while postponing the exact details of how it is introduced until later in Sect. 1.16. The vector product of u and v is orthogonal to both vectors (cf. (P6)), and if the vectors are parallel then their vector product will be the zero vector (cf. (P4)). The magnitude of u ∧ v can be found from (1.11) and (P7) by taking |u| = 1 = |v|, thus leading to |u ∧ v| = |u||v| sin θ . In conclusion, we can write down the following expressions for the two products: u · v = |u||v| cos θ

and

u ∧ v = |u||v| sin θ k,

(1.13)

where k is a unit vector orthogonal to both u and v. Definition 1.7 The scalar triple product of three vectors u, v, w ∈ V , denoted by [u, v, w], is the real number defined by [u, v, w] := u · (v ∧ w).

(1.14)

This is an important operation that will be used extensively in this book. The axioms satisfied by the earlier two products can be combined to deduce a number of

1.3 Euclidean Vector Spaces

13

volume= [u, v, w] w

u∧v v

2 area= |u ∧ v|

v

θ O

u

O

u

Fig. 1.1 The vector product (left) of two vectors u and v is another vector perpendicular to both factors of the product; its magnitude is equal to the area of the parallelogram having u and v as two adjacent sides. The scalar triple product (right) of three vectors u, v and w is a number whose absolute value represents the volume of the parallelepiped having these vectors as three adjacent sides

key properties of the expression defined in (1.14). In the interest of brevity we only list some of these, and leave their routine justification to the reader. Proposition 1.1 For ui ∈ V (i = 1, . . . , 4) and α i ∈ R (i = 1, 2) the following are true, 1. the triple scalar product remains unchanged by a cyclic permutation of the three vectors u1 , u2 , u3 ; by swapping the place of only two vectors its sign changes, but the absolute value of the product remains the same; for example, [u1 , u2 , u3 ] = [u2 , u3 , u1 ] = [u3 , u1 , u2 ] = −[u2 , u1 , u3 ] = −[u3 , u2 , u1 ] = −[u1 , u3 , u2 ]; 2. the mapping (u1 , u2 , u3 ) → [u1 , u2 , u3 ] is linear in each argument, e.g. [α1 u1 + α2 u2 , u3 , u4 ] = α1 [u1 , u3 , u4 ] + α2 [u2 , u3 , u4 ]; 3. [u1 , u2 , u3 ] = 0 ⇐⇒ u1 , u2 , u3 are linearly dependent. We recall in passing the interpretation given in elementary treatments of vector geometry for the vector and scalar triple products. As the left sketch in Fig. 1.1 illustrates, the magnitude of the vector product represents the area of a certain parallelogram; a triple scalar product, on the other hand, measures a volume (right sketch, same figure). In the next section, we shall make contact with the geometrical aspects of abstract vector spaces. The above developments have made no use of any preferred basis in V and have emphasised the absolute or intrinsic character of vector manipulations. We illustrate now a different aspect that corresponds to representing vectors relative to a chosen basis in V .

14

1 Vector, Tensors, and Related Matters

As already mentioned, in a three-dimensional vector space a basis has exactly three elements that will generally be denoted by e1 , e2 , e3 . This set of vectors will often be abbreviated as {ei }, with the range of the subscript ‘i’ being implicitly understood (from 1 to 3). Relative to this generic basis a vector u ∈ V may be written as u = u1 e1 + u2 e2 + u3 e3 ≡

3 

ui ei ,

(1.15)

i=1

where (u1 , u2 , u3 ) are the components of u with respect to the given basis. Here, we may take the basis vectors to be orthonormal (i.e. mutually orthogonal unit vectors), a condition that can then be expressed more compactly as ei · e j = δi j ,

(1.16)

where δi j is the Kronecker delta already defined in (1.1). The existence of an orthonormal basis in the general case of a Euclidean vector space follows from the Gram– Schmidt orthogonalisation process. Going back to (1.15), we can write it more succinctly by using the so-called summation convention (1.17) u = ui ei ; this convention says that whenever an index is repeated (only once) in the same term, a summation over the range of this index is implied unless otherwise indicated. The index ‘i’ that is summed over is said to be a dummy index because it can arbitrarily be replaced by any other letter. Any index over which there is no summation is called a free index. We now look at some basic aspects of this new notation system. To this end, let a, b, c ∈ V ; the components of these vectors will be denoted by ai , bi , ci (i = 1, 2, 3). Note how the summation convention is employed in the following examples: • ai ai = a j a j = a12 + a22 + a32 . • a p b p = a1 b1 + a2 b2 + a3 b3 . • More complicated examples can be readily imagined. For instance, ai ai bi bi = ai2 bi2 = a12 b12 + a22 b22 + a32 b32 , but we must be careful to distinguish between this and the product of ai ai and bi bi , namely (a12 + a22 + a32 )(b12 + b22 + b32 ),

1.3 Euclidean Vector Spaces

15

which should not be written as ai ai bi bi since it involves two separate summations. To avoid confusion, we represent the latter as ai ai b j b j = (ai ai )(b j b j ) = (a12 + a22 + a32 )(b12 + b22 + b32 ), which requires separate summation over i and j from 1 to 3. Note that this is the same as ai b j ai b j or ai b j b j ai , for example, as may be checked by carrying out the summation over i and j in turn. • The summation index cannot be repeated more than once. For instance, ai bi ci is meaningless, and is not equal to a1 b1 c1 + a2 b2 c2 + a3 b3 c3 . Another symbol that is particularly useful in many of our subsequent calculations is the so-called alternator or alternating symbol (also known as Ricci’s permutation symbol), defined by2 ⎧ ⎨ +1 if (i jk) is a cyclic permutation of (123), ∈i jk = −1 if (i jk) is a non-cyclic permutation of (123), ⎩ 0 otherwise.

(1.18)

This is just a particular notation for the sign of a permutation (1.2) when n = 3, which is better suited for the algebra of subscripts. Each i, j, k can only take on values 1, 2 or 3, so ∈i jk represents a collection of 33 = 27 numbers, ∈123 = ∈231 = ∈312 = +1,

∈132 = ∈321 = ∈213 = −1,

and all the other ∈i jk = 0 (e.g. ∈112 = ∈121 = 0, and so on). We can generalise these observations as indicated below • ∈i jk = ∈ki j = ∈ jki , hence a cyclic permutation of the indices on the permutation symbol does not alter its value; • ∈i jk is anti-symmetric in any pair of indices, i.e., an interchange in any two indices on the permutation symbol introduces a minus sign. This corresponds to an anticyclic permutation of indexes, ∈i jk = − ∈ik j = − ∈ jik = − ∈k ji . The alternator permits us to write the determinant of a 3 × 3 matrix in a compact form. If T = [Ti j ] represents such a matrix, it can be shown by direct calculations that det T = ∈ pqr T p1 Tq2 Tr 3 = ∈ pqr T1 p T2q T3r , ∈i jk det T = ∈ pqr T pi Tq j Tr k .

(1.19a) (1.19b)

2 The cyclic permutations of the ordered triple (1, 2, 3) are (1, 2, 3), (2, 3, 1) and (3, 1, 2); note that

an even number of transpositions is needed to restore the original order in each of these cases.

16

1 Vector, Tensors, and Related Matters

One of the most important uses of the permutation symbol in Continuum Mechanics is for expressing vector products in terms of the components of the constituent factors; we first need an auxiliary result. Proposition 1.2 If B V ≡ {ei } is an orthonormal basis of the Euclidean vector space V , then e3 ∧ e1 = ±e2 , e1 ∧ e2 = ±e3 , (1.20) e2 ∧ e3 = ±e1 , the ‘+’ or ‘–’ signs holding together. From (1.18), it can be seen that (1.20) can be expressed more compactly as ei ∧ e j = ± ∈i jk ek .

(1.21)

To justify (1.20), we start with the following observation. Let a ∈ V and consider the representation of this vector in B V , a = ai ei (say). Multiplying both sides of this relation by an arbitrary e p ∈ B V results in a · e p = (ai ei ) · e p = ai (ei · e p ) = ai δi p = a p . That is, a p = a · e p , which provides an expression for the p-component of a with respect to B V , here denoted by a p . Hence, we have shown that a = (a · ei )ei , (∀) a ∈ V .

(1.22)

Let us go back now to our original task. Set a := e2 ∧ e3 in (1.22) and then expand the resulting right-hand side e2 ∧ e3 = [(e2 ∧ e3 ) · ei ]ei = [e2 , e3 , ei ]ei = [e1 , e2 , e3 ]e1 ; that is, e2 ∧ e3 = [e1 , e2 , e3 ] e1 . Using exactly the same arguments (but different choices for a in (1.22)) another two similar relations are easily established, namely, e3 ∧ e1 = [e1 , e2 , e3 ] e2 and e1 ∧ e2 = [e1 , e2 , e3 ] e3 . A quick inspection of these last three formulae confirms that we are done if we manage to check that [e1 , e2 , e3 ] = ±1. This turns out to be true, as can be easily seen by using (P7) in Definition 1.6 with u := ei , v := e j , and i = j. The choice of sign in (1.20) is a matter of convention. If we choose this value to be +1 then the basis is said to be right-handed, otherwise it is left-handed. From now on, all bases in V will be assumed to be right-handed unless the contrary is explicitly mentioned. Irrespective of the orientation of the basis, (1.21) can be expressed as ∈i jk = Δ[ei , e j , ek ],

for i, j, k ∈ {1, 2, 3},

(1.23)

where Δ = +1 or Δ = −1 depending on whether the {ei } is right- or left-handed, respectively. More generally, three vectors a, b, c ∈ V form a basis if [a, b, c] = 0. The basis is right-handed if [a, b, c] > 0 and left-handed if [a, b, c] < 0. If one or all vectors of such a right-handed system have their directions reversed the resulting system is left-handed. However, if the direction of any two vectors in this set is reversed then the right-handedness is preserved.

1.3 Euclidean Vector Spaces

17

The vector product of two vectors u and v can have one of the two perpendicular directions on the linear subspace spanned by the two terms in the product. The convention adopted here will be that the ordered system of vectors {u, v, u ∧ v} has the same ‘handedness’ as the underlying basis {ei }.3 Returning to the calculation of u ∧ v, we have u ∧ v = (ui ei ) ∧ (v j e j ) = ui v j (ei ∧ e j ) = ∈i jk ui v j ek = ∈i j1 ui v j e1 + ∈i j2 ui v j e2 + ∈i j3 ui v j e3 = (u2 v3 − u3 v2 )e1 + (u3 v1 − u1 v3 )e2 + (u1 v2 − u2 v1 )e3 , and this last result may be written symbolically in the well-known form   e1  u ∧ v = u1 v1

e2 u2 v2

 e3  u3  . v3 

(1.24)

The determinant can be expanded with the help of (1.19a) to give u ∧ v = ∈i jk ui v j ek .

(1.25)

Similar calculations can be carried out for the triple scalar product u · (v ∧ w), u · (v ∧ w) = uk (v ∧ w)k = uk (∈ki j vi w j ) = ∈ki j uk vi w j .

(1.26)

In light of the formula (1.19), an alternative to (1.26) is   u1   v1  w1

u2 v2 w2

 u3  v3  w3 

or

 u1  u2  u3

v1 v2 v3

 w1  w2  w3 

(1.27)

from which the symmetries that appear stated in Proposition 1.1, u · (v ∧ w) = w · (u ∧ v) = v · (w ∧ u), may be deduced. We also remark that the expressions (1.27) will hold if the vectors u, v, w are expressed with respect to any orthonormal system (not necessarily the standard basis). The vector triple product of three vectors u, v, w ∈ V is by definition the expression u ∧ (v ∧ w); the result of these operations turns out to be a linear combination of v and w as the next formula indicates 3 In

the usual three-dimensional Euclidean point space (to be introduced later), this convention corresponds to the ‘right-hand rule’: if you put the index of your right hand on u and the middle finger on v, then your thumb should point in the direction of u ∧ v.

18

1 Vector, Tensors, and Related Matters

n

a

(a · n)n n∧a∧n

Fig. 1.2 Decomposition of a vector a along a given direction specified by the unit vector n and another direction perpendicular to it. Note that the latter is contained within the plane passing through the tail of a and perpendicular to n

u ∧ (v ∧ w) = (u · w)v − (u · v)w,

(∀) u, v, w ∈ V .

(1.28)

If n ∈ V is a given unit vector (i.e., |n| = 1), we can write any arbitrary vector a ∈ V in the form4 a = (a · n)n + n ∧ (a ∧ n), (1.29) thus expressing a as the sum of its components along n (first term) and another component orthogonal to n (second term); in elementary treatments of vector geometry the situation described above corresponds to that seen in Fig. 1.2.

1.4 Euclidean Point Spaces Definition 1.8 A (three-dimensional) Euclidean point space is a set E3 , whose elements are called points, together with a (three-dimensional) Euclidean vector space V that enjoy the following properties: (E1) For every ordered pair of points (P, Q) ∈ E3 × E3 , there corresponds a unique vector denoted by f (P, Q) ∈ V ; in other words, there exists a mapping (1.30) (P, Q) ∈ E3 × E3 −→ f (P, Q) ∈ V , and it is further postulated that the correspondence (1.30) satisfies: (E2) f (P, Q) = − f (Q, P) for all P, Q ∈ E3 . (E3) f (P, Q) = f (P, R) + f (R, Q) for all P, Q, R ∈ E3 . (E4) (∀) P ∈ E3 and (∀) u ∈ V there exists a unique point Q ∈ E3 such that f (P, Q) = u. general, the vector triple product is not associative: a ∧ (b ∧ c) = (a ∧ b) ∧ c. If the first and the last terms in such a product are equal, then it does not matter the order in which the products are carried out, e.g. n ∧ (a ∧ n) = n ∧ a ∧ n = (n ∧ a) ∧ n.

4 In

1.4 Euclidean Point Spaces

19

−−→ To make things more intuitive, it is convenient to adopt the notation P Q for f (P, Q). The vector space V is the translation space of E3 and its elements are called translations of E3 . The dimension of a Euclidean point space is that of its translation space, while the notions of distance and angle are derived from V . In principle, higher dimensional Euclidean point spaces can be defined in a similar way, but in this book they will play only a minor role. The distance between two points P, Q ∈ E3 , denoted by d(P, Q) is defined in −→ the obvious manner as being the magnitude of the vector P Q, −→ −→1/2 −−→ d(P, Q) := | P Q | = P Q · P Q .

(1.31)

It can be checked that d : E3 × E3 −→ R satisfies the conditions for a metric, i.e. d(P, Q) = d(Q, P), d(P, Q) ≥ 0, with equality if only if P = Q, d(P, Q) ≤ d(P, R) + d(R, Q), for all P, Q, R ∈ E3 . The first two properties are trivial by the definition of our concept of ‘distance’. As for the last one, we can write −−→ −−→ −−→ −−→ −−→ −−→ d 2 (P, Q) = P Q · P Q = P R + R Q · P R + R Q −−→ −−→ −−→ −−→ = | P R | 2 + | R Q |2 + 2 P R · R Q −−→ −−→ −−→ −−→ ≤ | P R |2 + | R Q |2 + 2| P R || R Q |  −−→ 2 −−→ 2  = | P R | + | R Q | = d(P, R) + d(R, Q) , and hence the last property is checked. Property (E4) above makes it clear that the vectors of V induce a mapping of E3 onto itself: for a given u ∈ V this mapping associates to each point P ∈ E3 another −→ point Q ∈ E3 such that P Q = u, i.e. the point P gets translated into Q. Note that according to (E2), the vector (−u) translates Q into P. These observations suggest that points can be added to vectors to produce points, and we can define the addition between a point in E3 and a vector in V through P + u = Q. The difference between two points can also be defined, and it turns out to be a vector. Given P, Q ∈ E3 , axiom (E1) ensures the existence of a unique v ∈ V associated with these points; we can simply define Q − P := v. The operations introduced above allow us to define simple geometrical concepts such as lines or planes, in E3 . For instance, if U ⊂ V is a subspace with dim(U) = 1, then the line in the Euclidean point space passing through a point P and parallel to U is just the set {P + tu | t ∈ R} ⊂ E3 , where u ∈ U; other geometrical concepts can be defined just as easily.

20

1 Vector, Tensors, and Related Matters

In a Euclidean point space, there is no canonical choice of where the origin should be placed because it can be translated anywhere within that space. When a certain point is chosen, it can be regarded as the origin and subsequent calculations may ignore the distinction between a point and a vector. This is made more precise next. Definition 1.9 An arbitrary point O ∈ E3 together with an orthonormal basis {ei } in V define a system of (rectangular) Cartesian coordinates in E3 . The point O is known as the origin of the system of coordinates. −−→ If P ∈ E3 is an arbitrary point, then the vector O P = x ∈ V is known as the −−→ position vector of P. The components xi (i = 1, 2, 3) of the vector O P relative to the aforementioned basis are called the Cartesian coordinates of P relative to {O; ei }; we shall also refer to {O; ei } as the reference triad. It is now clear that by fixing a preferred point as the origin one can identify the Euclidean point space E3 with R3 . More precisely, we shall identify geometrical points in E3 with their position vectors in R3 , the distinction between these two different mathematical entities being clear from the context. On a few occasions, we shall work directly with geometrical vectors in order to avoid the risk of cluttering the presentation with too many notational conventions. The distinction between points and vectors is an important one, especially in a text on Continuum Mechanics. As we shall see in the next chapter, deformable bodies consist of ‘material particles’ that are identified with geometrical points in E3 . These particles have associated to them a number of vectorial quantities such as velocity, acceleration and so on. The use of E3 and R3 (or V ) in that particular context will help us avoid serious ambiguities in the later chapters. Elementary accounts of Linear Algebra that emphasise Euclidean point spaces can be found in [1–3], while Noll’s text [4] contains higher calibre material.

1.5 Second-Order Tensors: Fundamentals The standard setting for our discussion will be a three-dimensional Euclidean vector space V (as described above). Definition 1.10 A second-order tensor A is simply a linear transformation of V onto itself. The meaning of this definition is the following: to each vector u ∈ V there corresponds a new vector, denoted by Au ∈ V such that A(αu1 + βu2 ) = α( Au1 ) + β( Au2 ),

(∀) u1 , u2 ∈ V , (∀) α, β ∈ R.

If A and B are two second-order tensors, we can define the sum A + B, the difference A − B, and the scalar multiplication α A by the following rules:

1.5 Second-Order Tensors: Fundamentals

21

( A ± B)u := Au ± Bu,

(α A)u := α( Au),

(1.32)

where u ∈ V denotes an arbitrary vector. There are two particularly important second-order tensors, namely the secondorder identity tensor I and the second-order zero tensor O, Iu = u

and

Ou = 0,

(∀) u ∈ V .

(1.33)

By using the definitions (1.32), the set of all second-order tensors becomes a linear vector space (see Definition 1.1); we agree to call this space Lin(V , V ) or Lin for short. An important question is whether or not Lin is a finite-dimensional vector space, because if that is the case then we can expect every second-order tensor to be represented as a linear combination of some simpler linear mappings. An equally important aspect here is how we can actually construct such simpler mappings. The next concept turns out to be of critical importance in answering these questions. Definition 1.11 The tensor product of the vectors u and v is denoted by5 u ⊗ v, and represents the (simple) linear transformation defined according to (u ⊗ v)a := (a · v)u,

(∀) a ∈ V .

(1.34)

The tensor product of two vectors is called a dyad; a sum of such products is referred to as a dyadic. By its very definition, the tensor product is a linear mapping in its constituent factors; more precisely, for all u, v, w ∈ V and α, β ∈ R, (αu + βv) ⊗ w = α(u ⊗ w) + β(v ⊗ w), u ⊗ (αv + βw) = α(u ⊗ v) + β(u ⊗ w). In general, the tensor product is not commutative (u ⊗ v = v ⊗ u) as suggested by the geometrical interpretation of this operation included in Fig. 1.3. If a, b, c ∈ V are such that a ⊗ c = b ⊗ c, this does not necessarily imply that a = b. However, if a ⊗ c = b ⊗ c for all c ∈ V then a = b. The justification of this fact is straightforward. Let i ∈ {1, 2, 3} be fixed, but otherwise arbitrary. Set c := ei in the above equality to find a ⊗ ei = b ⊗ ei ⇒ (a ⊗ ei )ek = (b ⊗ ei )ek ⇒ a(ei · ek ) = b(ei · ek ), whence a = b if we choose k = i (no summation). The next result confirms that Lin is finite-dimensional and shows that, given an orthonormal basis of V , one can immediately construct a basis for Lin.

this notation, u ⊗ v stands for the ordered pair of u and v, while the multiplication sign is just a reminder of what to expect.

5 In

22

1 Vector, Tensors, and Related Matters

Fig. 1.3 Geometrical interpretation of Definition 1.11. The tensor product of two vectors acts as an oblique ‘projection’ on one of them (depending on the order of the vectors in the product). The sketch also makes it clear that u ⊗ v and v ⊗ u are distinct geometrical operations

Proposition 1.3 If B V ≡ {ei } is a basis of V , then the set   B⊗ := ei ⊗ e j | 1 ≤ i, j ≤ 3

(1.35)

forms a basis of Lin. In particular, this is a nine-dimensional vector space, i.e., dim (Lin) = 9. Hence, any A ∈ Lin admits the representation A = A i j ei ⊗ e j ,

(1.36)

where the scalars Ai j ∈ R represent the components of A relative to B⊗ . By using (1.36), it can be further shown that Ae j = Ak j ek

and

Ai j = ei · ( Ae j ).

(1.37)

Some general remarks about this result are listed below, 1. Note that B⊗ consists of the following nine elements: e1 ⊗ e1 , e1 ⊗ e2 , e1 ⊗ e3 , e2 ⊗ e1 , e2 ⊗ e2 , e2 ⊗ e3 , e3 ⊗ e1 , e3 ⊗ e2 , e3 ⊗ e3 , while the components Ai j are the entries of the matrix ⎡

A11 [ A] := ⎣ A21 A31

A12 A22 A32

⎤ A13 A23 ⎦ ; A33

the components of the vector Aei are found in the ith column (i = 1, 2, 3). It is important to distinguish between the tensor A and its component representation

1.5 Second-Order Tensors: Fundamentals

23

[ A]. For second-order tensors, this matrix representation is useful for manipulative purposes, but it does not generalise conveniently to the more general cases discussed in Sect. 1.10. In order to avoid overloading the notation, in what follows we shall not always distinguish explicitly between A and [ A] as the meaning should always be clear from the context. 2. Proposition 1.3 assures us (albeit indirectly) that what is true for the simple tensor product (1.34) it is also going to be true for second-order tensors, in general (owing to the fact that both obey the distributive law of multiplication). 3. If u, v ∈ V , u ⊗ v = (ui ei ) ⊗ (v j e j ) = ui v j ei ⊗ e j , so that the components of u ⊗ v may be displayed in matrix form ⎡

u1 v1 [u ⊗ v] = ⎣u2 v1 u3 v1

u1 v2 u2 v2 u3 v2

⎤ u1 v3 u2 v3 ⎦ . u3 v3

4. Relative to B⊗ , the identity tensor defined in (1.33) admits the following representation: (1.38) I = e1 ⊗ e1 + e2 ⊗ e2 + e3 ⊗ e3 ≡ e p ⊗ e p . This follows immediately by using Definition 1.11 and Eq. (1.22) Iu = (e p ⊗ e p )u = (u · e p )e p = u,

(∀) u ∈ V .

Note also that the Cartesian components of the identity tensor are Ii j ≡ δi j , for i, j = 1, 2, 3, i.e. ⎡ ⎤ 1 0 0 [I] = ⎣0 1 0⎦ . 0 0 1 5. The equality of two tensors is independent of the nature of B⊗ , in the sense that it does not matter what basis of V is used to construct the set (1.35). The key point here is that when trying to establish the equality of two given tensors we have the freedom to choose the basis that simplifies their component representation the most. We start the proof of Proposition 1.3 by first checking that B⊗ represents a system of linearly independent elements in Lin. This amounts to showing that an assumed relation of the form (αi j ∈ R), (1.39) αi j ei ⊗ e j = O, implies that all αi j = 0. To this end, let p ∈ {1, 2, 3} be fixed, but arbitrary. Then, 0 = αi j (ei ⊗ e j ) e p = αi j (e j · e p ) ei = αi j δ j p ei = αi p ei ,

24

1 Vector, Tensors, and Related Matters

where we have used (1.34) and the definition of the Kronecker delta symbol (1.1). Since the elements of B V are linearly independent, the last relation implies that αi p = 0 for i = 1, 2, 3; but p was arbitrary, hence αi p = 0 for i, p = 1, 2, 3, as required. Now, all that remains to be checked is that every tensor in Lin can be represented as a linear combination of elements from B⊗ . A by-product of this simple exercise will be the explicit coefficients formulae in (1.37). Let A ∈ Lin and write A = A i j ei ⊗ e j ,

(1.40)

our task being that of finding Ai j ∈ R. Let p, q ∈ {1, 2, 3} be fixed, but arbitrary. Combining (1.40) with the definition of the tensor product (1.34) leads to Aeq = Ai j (ei ⊗ e j ) eq = Ai j δ jq ei = Aiq ei =⇒ e p · Aeq = Aiq ei · e p = Aiq δi p = A pq . This shows that the scalars Ai j in (1.40) are uniquely determined by A, in agreement with the second relation in (1.37). The proof of Proposition 1.3 is now complete. The result we have just finished proving can be phrased in an equivalent way. It basically says that given any A ∈ Lin there exist three vectors u1 , u2 , u3 ∈ V such that (1.41) A = e1 ⊗ u1 + e2 ⊗ u2 + e3 ⊗ u3 . To justify this representation, it is enough to notice that uk = Ak j e j for k = 1, 2, 3; the components of the vector uk are the entries found on row k in the component representation of A. In the same way, any A ∈ Lin admits a representation of the form A = v k ⊗ ek , where v k = A jk e j . The choice of basis in V is not unique, so a question arises in relation to representing the linear mapping A : V → V relative to two different bases. Let B V(1) = {E i } and B V(2) = {e j } be two different bases in the same linear space V . Formally, we can indicate the choice of different bases by writing A : {V , B V(1) } −→ {V , B V(2) }.

(1.42)

The set of all such mappings can be shown to be a linear space and the set  B⊗12 := {ei ⊗ E j  1 ≤ i, j ≤ 3}

(1.43)

forms a basis in this space; hence, A in (1.42) admits the following representation, A = A i j ei ⊗ E j ,

Ai j := ei · ( AE j ).

1.6 Examples of Tensors: Elementary Projections

25

1.6 Examples of Tensors: Elementary Projections An elementary example of second-order tensor emerges in connection with the projection of an arbitrary vector onto a given direction specified by a (unit) vector. Our setting is that of the Euclidean vector space V , but we shall tacitly assume the geometrical correlation between elements of this space and geometrical vectors. With this in mind, let n be the given unit vector that defines a line/direction (d) in space. Then, by definition, (1.44) Proj n a := (n · a)n represents the parallel projection of a ∈ V along the unit vector n. Geometrically, this projection represents the oriented line segment cut from (d) by the planes drawn through the end points of a, and perpendicular to (d), taken with the plus sign if the direction of the projection of the starting point of a coincides with the positive direction of n, and with the minus sign otherwise. It must be clear that |(a · n)| always gives the length of the segment of the axis (d) between the planes drawn as mentioned above. If n = 0 and |n| = 1, Eq. (1.44) is modified according to Proj n a :=



 a·n n. |n|2

(1.45)

Formula (1.44) represents a linear operation; it associates to each vector a ∈ V a new vector Proj n a ∈ V whose magnitude is equal to |n · a| and is parallel to n. Indeed, for all u, v ∈ V and α, β ∈ R, Proj n (αu + βv) = [(αu + βv) · n] n = α(u · n)n + β(v · n)n = α Proj n u + β Proj n v. In light of (1.34), it is easily seen that Proj n = n ⊗ n

or

Proj n =

n⊗n , |n|2

(1.46)

depending, respectively, on whether |n| = 1 or |n| = 1. Observing further that   I = I − Proj n + Projn

⇒

  a = I − Proj n a + Projn a,

by comparing this result with (1.29) it transpires that    Proj ⊥ n a := I − Proj n a ≡ n ∧ a ∧ n

(1.47)

26

1 Vector, Tensors, and Related Matters

represents the projection of the vector a onto a plane of unit normal n. We shall refer to this as the normal projection associated with the unit vector n ∈ V .6 Note that the earlier formula (1.29) now admits the intuitive representation a = Proj n a + Proj ⊥ n a,

(∀) a ∈ V .

The normal projection onto a plane defined by two orthogonal vectors m and k (not necessarily of unit magnitude) can also be expressed as k⊗k m⊗m + ; |m|2 |k|2

(1.48)

if n is another vector such that {n, m, k} form a right-handed basis in V , then the projection along a direction perpendicular to this plane will be given by (1.46).

1.7 Basic Properties of Tensors Since the elements of Lin are mappings, we can consider the composition of two or more tensors. To fix ideas, let A, B, C ∈ Lin and suppose that we subject an arbitrary vector u to the linear transformation B, obtaining a vector v := Bu, and afterwards subject v to the linear transformation A, obtaining a third vector w := Av. Then w can be regarded as a function of the argument u since w = Av = A(Bu) =: Cu. The transformation u → Cu ≡ w is in fact linear, C(u1 + u2 ) = A(B(u1 + u2 )) = A(Bu1 + Bu2 ) = A(Bu1 ) + A(Bu2 ) = Cu1 + Cu2 and C(αu) = A(B(αu)) = A(α Bu) = α A(Bu) = α(Cu). The linear transformation C is called the product of the transformations A and B, written C = AB, where the factors are written from right to left in the order in which the corresponding transformations are carried out. Multiplication of elements in Lin is associative, i.e., A(BC) = ( AB)C,

(∀) A, B, C ∈ Lin,

but it is not commutative; in general AB = B A. 6 CAVEAT:

in the literature this is usually called the ‘parallel’ projection, while (1.44) is referred to as the ‘orthogonal’ projection.

1.7 Basic Properties of Tensors

27

If A, B ∈ Lin have component representations [ A] and [B] relative to some orthonormal basis B V = {ei }, and the product AB has a matrix [C] in the same basis, then [C] = [ A][B]. This is a crucial result because it shows that multiplication of second-order tensors is likely to be intimately connected to the multiplication of 3 × 3 matrices. A number of important concepts that will help us to better appreciate the similarity between tensors and matrices are introduced next. Definition 1.12 Let A ∈ Lin. • The transpose of A is the unique tensor AT ∈ Lin such that   u · AT v = v · ( Au),

(∀) u, v ∈ V .

(1.49)

• A is said to be symmetric if AT = A. If AT = − A then A is called a skew-symmetric tensor. • A is said to be invertible if there exists a tensor A−1 ∈ Lin, called the inverse of A such that (1.50) A A−1 = A−1 A = I. • A is said to be 1. positive semi-definite if u · ( Au) ≥ 0, (∀) u ∈ V . 2. positive definite if u · ( Au) > 0, (∀) u ∈ V , u = 0. • A is said to be orthogonal if it preserves length, i.e., | Au| = |u|,

(∀) u ∈ V .

(1.51)

• There exists a unique tensor cof ( A) ∈ Lin, called the cofactor of A, such that cof ( A)(u ∧ v) = ( Au) ∧ ( Av),

(∀) u, v ∈ V .

(1.52)

Orthogonal transformations play a central role in Continuum Mechanics, so it is useful to have some alternative characterisations. The proof of the next result is straightforward and is left as an exercise for the reader. Proposition 1.4 The following statements concerning A ∈ Lin are equivalent: 1. A is orthogonal; 2. ( Au) · ( Av) = u · v for all u, v ∈ V ; 3. For an arbitrary orthonormal basis B V ≡ {ei }, the set { Aei } also forms an orthonormal set of vectors; 4. AT A = A AT = I. The set of all orthogonal tensors will be denoted by Ort. In addition to this, two of its important subsets are

28

1 Vector, Tensors, and Related Matters

  Ort+ := A ∈ Ort | [ Ae1 , Ae2 , Ae3 ] > 0 ,   Ort− := A ∈ Ort | [ Ae1 , Ae2 , Ae3 ] < 0 ;

(1.53a) (1.53b)

Ort+ represents the set of proper orthogonal tensors, while the elements of Ort− will be referred to as improper orthogonal tensors. As we shall see later the set (1.53a) is closed under tensor multiplication and forms a subspace of Ort. Note that Ort itself is closed under tensor multiplication, i.e., the product of any number of orthogonal tensors is yet another element of Ort. This can be seen easily by taking A1 , A2 ∈ Ort, so that (1.51) holds for both tensors. Then |( A1 A2 )u|2 = ( A1 A2 u) · ( A1 A2 u) = A1 ( A2 u) · A1 ( A2 u) = ( A2 u) · ( A2 u) = u · u = |u|2 , and hence (1.51) holds for A1 A2 as well. Products of more than two factors are handled similarly. On many occasions, we have to be able to manipulate powers of A−1 or AT . It is convenient to introduce the notation A−T := ( A−1 )T ≡ ( AT )−1 . It can also be shown that ( AT )n = ( An )T for any n ∈ Z. Next, we look at a number of rules that need to be well understood before one can take advantage of Definition 1.12. For instance, the transpose of a second-order tensor can be shown to satisfy ( AB)T = B T AT ,

( AT )T = A,

(α A + β B)T = α AT + β B T , for all A, B ∈ Lin and α, β ∈ R. The first relation can be extended to an arbitrary number of terms, e.g. ( ABC)T = C T B T AT , and so on. The method of proof for these results is straightforward; we are going to explain it in the context of a somewhat simpler example, (u ⊗ v)T = v ⊗ u,

(∀) u, v ∈ V .

(1.54)

To establish (1.54), we need to use the definition of the transpose—see Eq. (1.49). The transpose of A := u ⊗ v ∈ Lin, denoted by AT := (u ⊗ v)T , is a second-order tensor which must satisfy x · ( AT y) = y · ( Ax),

(∀) x, y ∈ V .

(1.55)

The right-hand side of this last equation can be written as y · ( Ax) = y · [(u ⊗ v)x] = y · [(v · x)u] = (v · x)( y · u) = x · [(u · y)v] = x · [(v ⊗ u) y].

(1.56)

1.7 Basic Properties of Tensors

29

Comparing (1.55) and (1.56), we get  x · ( AT y) = x · (v ⊗ u) y ,

(∀) x, y ∈ V ,

which is the same with AT = v ⊗ u, and this completes the proof. An immediate consequence of (1.54) is the following relationship between the component representations of a tensor and its transpose: A = A i j ei ⊗ e j

=⇒

AT = A ji ei ⊗ e j .

Proposition 1.5 (useful properties for ⊗) Let a, b, c, d ∈ V and A ∈ Lin . Then, (a ⊗ b)(c ⊗ d) = (b · c)(a ⊗ d), A(a ⊗ b) = ( Aa) ⊗ b, (a ⊗ b) A = a ⊗ ( AT b).

(1.57a) (1.57b) (1.57c)

We start with the justification of (1.57a); to this end, set T := (a ⊗ b)(c ⊗ d)

and

S := (b · c)(a ⊗ d).

The following observations can be made: (i) T , S ∈ Lin; (ii) Equation (1.57a) asserts that T = S, i.e. the two tensors T and S are equal. This is true if and only if the images of every vector u ∈ V under the two tensors are identical. We are thus left to prove that T u = Su for all u ∈ V . This amounts to calculating T u and Su, and then comparing the results T u = [(a ⊗ b)(c ⊗ d)]u = (a ⊗ b)[(c ⊗ d)u] = (a ⊗ b)[(u · d)c] = (u · d)[(a ⊗ b)c] = (u · d)(c · b)a,

(1.58)

and Su = (b · c)(a ⊗ d)u = (b · c)[(a ⊗ d)u] = (b · c)[(d · u)a] = (b · c)(d · u)a.

(1.59)

Comparing (1.58) and (1.59) confirms the expected equality. In a similar way, we can prove (1.57b). The definitions for T and S must now be amended to and S := ( Aa) ⊗ b. T := A(a ⊗ b) Next, evaluate T u and Su, and then compare the results T u = [ A(a ⊗ b)]u = A[(a ⊗ b)u] = A[(b · u)a] = (b · u) Aa,

(1.60)

30

1 Vector, Tensors, and Related Matters

and Su = [( Aa) ⊗ b]u = (b · u) Aa.

(1.61)

A quick glance at (1.60) and (1.61) shows that T u = Su, and we are done. The property (1.57c) can be checked following similar arguments. As an application, let us find the components of the second-order tensor C := AB, where A, B ∈ Lin. If B V ≡ {ei } is an orthonormal basis of V , then we can write A = A i j ei ⊗ e j

and

B = B pq e p ⊗ eq ,

so, with the help of (1.57a), C = AB = (Ai j ei ⊗ e j )(B pq e p ⊗ eq ) = Ai j B pq (ei ⊗ e j )(e p ⊗ eq ) = Ai j B pq (e j · e p )(ei ⊗ eq ) = Ai j B pq δ j p (ei ⊗ eq ) = Ai p B pq ei ⊗ eq = Ai p B pj ei ⊗ e j . This sequence of calculations shows that C = C i j ei ⊗ e j ,

with

Ci j = Ai p B pj .

(1.62)

In a similar way, we find that if A, B, C ∈ Lin, then D := ABC is a second-order tensor that has the representation D = Di j ei ⊗ e j , with Di j = Ai p B pq Cq j .

(1.63)

Proposition 1.3 in conjunction with Definition 1.12 and Eq. (1.57) allow a simple characterisation of symmetric and skew-symmetric tensors in terms of their component representations. More precisely, if T is a symmetric tensor and has components [T ] = [Ti j ], then Ti j = T ji for all i, j = 1, 2, 3; for a skew-symmetric tensor W with [W ] = [Wi j ] the components satisfy W ji = −Wi j . A simple counting argument suggests that a symmetric second-order tensor has six independent components, whereas a skew-symmetric tensor has only three. Any arbitrary tensor A can be written as the sum of a symmetric and a skewsymmetric part 1 1 (1.64) A = ( A + AT ) + ( A − AT ) . 2 2       symmetric

skew-symmetric

As a short-hand notation, we shall sometimes write sym( A) and skew( A) to indicate, respectively, the symmetric and the skew-symmetric parts defined above. In connection to what has been said above we also introduce the following sets,   Sym := A ∈ Lin | AT = A ,

  Skw := A ∈ Lin | AT = − A ,

(1.65)

1.7 Basic Properties of Tensors

31

which are linear subspaces of Lin, with dim (Sym) = 6 and dim (Skw) = 3. Since a skew-symmetric tensor has only three distinct components, one might suspect that, in some sense, such a tensor could be ‘identified’ with a certain vector. The important issue here concerns the nature of this relationship that is made clear in the following result. Proposition 1.6 For any W ∈ Skw there exists a unique vector w ∈ V (called the ‘axial vector’ of W ) such that W a = w ∧ a,

(∀) a ∈ V .

(1.66)

To check (1.66) we first note that Proposition 1.3 gives W = W23 (e2 ⊗ e3 − e3 ⊗ e2 ) + W13 (e1 ⊗ e3 − e3 ⊗ e1 ) +W12 (e1 ⊗ e2 − e2 ⊗ e1 ), where we have taken into account that W ji = −Wi j . Next, by using the definition of the tensor product (1.34), a direct evaluation of the left-hand side in (1.66) leads to W a = (W13 a3 + W12 a2 ) e1 + (−W12 a1 + W23 a3 ) e2 +(−W23 a2 − W13 a1 ) e3 ,

(1.67)

which clearly mimics a vector product expression. This is made apparent by introducing the vector w := wi ei with w1 := −W23 ,

w2 := W13 ,

w3 := −W12

or, equivalently, wi =

1 ∈i jk Wk j 2

(i = 1, 2, 3).

(1.68)

Finally, calculating w ∧ a with the help of (1.24) and then comparing the result with (1.67) confirms that W a = w ∧ a; this completes the proof. Note that the components of W can be related to those of w; according to (1.37) Wi j = ei · (W e j ) = ei · (w ∧ e j )   = ei · (wk ek ) ∧ e j = wk ei · (ek ∧ e j  = wk ei , ek , e j = −[ei , e j , ek ]wk = − ∈i jk wk . In conclusion, Wi j = − ∈i jk wk , where wk are the components of the axial vector of the skew-symmetric tensor of components Wi j .

32

1 Vector, Tensors, and Related Matters

Definition 1.13 The vector of a tensor A ∈ Lin is defined to be minus twice the axial vector of its skew-symmetric part; this vector is indicated as either A× or vec( A). If A = Ai j ei ⊗ e j then A× may be obtained also by replacing ‘⊗’ with ‘∧’ in each term of this double sum A× = Ai j ei ∧ e j = (a23 − a32 )e1 + (a31 − a13 )e2 + (a12 − a21 )e3 = Ai j ∈i jk ek . Alternatively, the justification of this formula follows from the fact that the operation A → A× is linear and (a ⊗ b)× = a ∧ b for all a, b ∈ V . From the above definition one can easily recover the so-called Stokes’ transformation, Aa = AT a − A× ∧ a,

(∀) a ∈ V .

1.8 Linear Mappings as Geometric Transformations Any A ∈ Lin can be interpreted as a geometric operation or transformation that acts between certain subsets of the usual three-dimensional Euclidean point space. To avoid the proliferation of unnecessary symbols, points in this space will be identified with their position vectors; this means that the coordinates of the points correspond to the components of their position vectors, the origin being fixed. The positive powers of A correspond to repetitions of the transformation, while negative powers represent inverse transformations. The identity tensor I ∈ Lin corresponds to the identical transformation, that is, no transformation at all. Consider u ∈ V as drawn from a fixed origin, and another vector u ∈ V as drawn from the same origin. If we let u = Au, then this equation may be regarded as defining a transformation of the point P, whose position vector is u, into the point P  , represented by the position vector u . Note that the origin remains fixed. Any point on the line (d) : x = a + t b (here t ∈ R is the line parameter, a, b fixed vectors) becomes a point on the line (d  ) : y = a + t b , where a = Aa and b = Ab. Hence, straight lines are transformed into straight lines, and lines parallel to b go into lines parallel to Ab. In a similar way, planes go into planes and the notion of parallelism is preserved. By way of example, consider the particular tensor A = −e1 ⊗ e1 + e2 ⊗ e2 + e3 ⊗ e3 .

(1.69)

Geometrically the transformation u → Au is a spatial reflection with respect to the (23)-plane: if u = ui ei then Au = −u1 e1 + u2 e2 + u3 e3 . That is, P(u1 , u2 , u3 ) → P  (−u1 , u2 , u3 ), which confirms our assertion (see Fig. 1.4). This transformation

1.8 Linear Mappings as Geometric Transformations

33

Fig. 1.4 Geometrical illustration of simple linear mappings; the sketch on the left corresponds to the mapping (1.69), while the one on the right is related to (1.70)

replaces each three-dimensional set of points by a congruent copy situated on the opposite side of this plane. The repeated application of this transformation brings us back where we started from. Indeed, taking into account that (ei ⊗ ei )(e j ⊗ e j ) = (ei ⊗ e j )(ei · e j ) = δi j (ei ⊗ e j ) (no sum) for any i, j = 1, 2, 3, it is a simple exercise in algebra to check that A2 = A A = I. Since A2 = I, this tensor A can be regarded as some sort of square root of the identity tensor; this type of square root will be made precise in Sect. 1.20. Note that the tensor B = e1 ⊗ e1 − e2 ⊗ e2 − e3 ⊗ e3 ≡ − A

(1.70)

has the same property, B 2 = I, but geometrically it represents a reflection in the 1-axis. The effect of this transformation on a set of points is equivalent to a rotation about the 1-axis through an angle of 180◦ in a plane parallel to the (23)-plane; see Fig. 1.4. Theorem 1.1 The necessary and sufficient condition that a tensor A ∈ Lin represents a rotation about some axis is that it be reducible to the form A = e1 ⊗ e1 + e2 ⊗ e2 + e3 ⊗ e3 ,

(1.71)

where {O; ei } and {O; ei } are two right-handed reference triads in the usual threedimensional Euclidean point space. We give an informal justification of this result. The geometrical interpretation of the meaning of (1.71) is best understood by considering an arbitrary parallelepiped whose three adjacent sides are the non-coplanar vectors a = ai ei , b = b j e j , c = ck ek , and by investigating the action of A on this set of points.

34

1 Vector, Tensors, and Related Matters

Let us start by noticing that Aa = (ep ⊗ e p )(ai ei ) = ai δ pi ep = ai ei and, similarly, Ab = b j ej and Ac = ck ek . Hence, the volume of the transformed parallelepiped is V  := [ Aa, Ab, Ac] = [a, b, c] =: V, by using (1.27) and taking into account that the coordinates of the three vectors are unchanged. Furthermore, the angles between the vectors a, b, c remain the same because, for instance, ( Aa) · ( Ab) = (ai ei ) · (b j ej ) = ai bi = a · b, etc., and the lengths are preserved, too, | Aa|2 = ( Aa) · ( Aa) = ai ai = |a|2 , and so on. These observations show that the original parallelepiped (which was arbitrary) is transformed into a new parallelepiped congruent with the old one. This change is akin to the effect of a force acting upon a rigid-body: while the body may move as a result of such action, its shape remains unaffected. Note that the reference triad {O; ei } is changed into {O; ei } and any vector u is transformed from its position relative to the original reference triad to the same position relative to the second triad. The converse can be checked just as easily by using standard properties of orthogonal transformations. As we shall see later on, a rotation is a special type of orthogonal tensor (more precisely, such a tensor is in Ort+ ). Note that A−1 = AT if A ∈ Ort; in particular, the inverse of (1.71) corresponds to A−1 = e1 ⊗ e1 + e2 ⊗ e2 + e3 ⊗ e3 . Next, we glance at the interpretation of another tensor, B ∈ Ort, defined by B := e1 ⊗ e1 + e2 ⊗ e2 − e3 ⊗ e3 = (e1 ⊗ e1 + e2 ⊗ e2 − e3 ⊗ e3 )(e1 ⊗ e1 + e2 ⊗ e2 + e3 ⊗ e3 ) = B2 B1. Our writing above indicates that B can be viewed as the superposition of two linear transformations: first a rotation due to B 1 followed by a reflection in the (1 2 )-plane, which is accomplished by B 2 . The characterisation of rotations using formula (1.71) is not entirely satisfactory because it requires too much information (knowledge of the six vectors that form the two orthonormal triads). If the axis of rotation is parallel to one of the vectors ei , then the analytical expression of the tensor describing such a rotation is relatively simple; the general case of an arbitrary axis of rotation will be dealt with in Sect. 1.15. For now let us take e1 as the axis of rotation and let θ be the angle of rotation measured positive in the counter-clockwise direction. The rotation takes {O; ei } into {O; ei } leaving e1 unchanged, so we can write e1 = e1 , e2 = cos θ e2 + sin θ e3 , e3 = − sin θ e2 + cos θ e3 .

1.8 Linear Mappings as Geometric Transformations

35

Replacing these vectors in (1.71) gives A = e1 ⊗ e1 + e2 ⊗ e2 + e3 ⊗ e3 = e1 ⊗ e1 + (cos θ e2 + sin θ e3 ) ⊗ e2 + (− sin θ e2 + cos θ e3 ) ⊗ e3 = e1 ⊗ e1 + (e2 ⊗ e2 + e3 ⊗ e3 ) cos θ + (e3 ⊗ e2 − e2 ⊗ e3 ) sin θ or, A = e1 ⊗ e1 + (I − e1 ⊗ e1 ) cos θ + (e3 ⊗ e2 − e2 ⊗ e3 ) sin θ,

(1.72)

where I = ei ⊗ ei ∈ Lin is the usual identity tensor. While this last form of A is more in line with the absolute character of rotations and their independence of component representations, the presence of the skewsymmetric tensor e3 ⊗ e2 − e2 ⊗ e3 is still unsatisfactory. It is clear that this tensor can be reduced to its axial vector, so there is still some redundancy in (1.72). In Sect. 1.15, we shall see how this formula can be brought into a form that requires only the axis and the angle of rotation, the two elements that fully determine a spatial rigid rotation.

1.9 Transformation Rules for a Change of Basis Up to this point our description of vectors and tensors has been purely abstract; vectors were elements of a Euclidean vector space V , while tensors represented linear mappings from that set onto itself. In Physics, interest in these concepts arises mainly in connection with the invariance of equations expressed in vector and tensor form, as physical phenomena cannot depend on the particular system of coordinates one chooses for their description. This aspect is not obvious and needs a separate discussion. It must be emphasised that vectors and tensors themselves remain invariant upon a change of basis; however, their respective components do depend on the choice of basis in V (which is arbitrary, in principle). The way the components of vectors and tensors transform when the standard basis is changed hinges entirely on how the basis vectors change. This is what we want to explore in what follows. Let V denote the usual three-dimensional Euclidean vector space, and select the right-handed orthonormal basis B V(1) ≡ {ei } ⊂ V so that ei · e j = δ i j . With respect to the chosen basis an arbitrary vector u ∈ V can be written as u = ui ei .

(1.73)

ei } (see Fig. 1.5). Again, we have Now, choose a second orthonormal basis B V(2) ≡ { the connections  ej = δij, (1.74) ei ·

36

1 Vector, Tensors, and Related Matters

e3

e3 e2

e1

O e2

e1

Fig. 1.5 Change of basis in the Euclidean point space V . The two bases {ei } and { e j } share the same origin. Note that Q i j ≡ ei · e j represents the cosine of the angle between the unit vectors ei and  e j (the angles between ei and  ei for i = 1, 2, 3 are indicated on the above sketch)

and we may also decompose u (the same vector as above) with respect to this new basis, ei , (1.75)  u = ui  where we have indicated by  u the expression of the vector u in the new basis.7 Since V is three-dimensional,  ei must be expressible as a linear combination of e1 , e2 , e3 for each i = 1, 2, 3. We write this in the form ⎧ ⎪ e1 = Q 11 e1 + Q 21 e2 + Q 31 e3 , ⎨  e2 = Q 12 e1 + Q 22 e2 + Q 32 e3 , ⎪ ⎩  e3 = Q 13 e1 + Q 23 e2 + Q 33 e3 or, by using the summation convention,  ei = Q ji e j

(i = 1, 2, 3).

(1.76)

It follows from (1.76) that ej, Q i j = ei ·

(1.77)

ej. which represent the direction cosines of ei relative to  From (1.74) and (1.76), we have ei · e j = (Q ki ek ) · e j = Q ki ( e j · ek ) = Q ki Q k j , δij = and, similarly, Q ik Q jk = δ i j . In conclusion, Q ik Q jk = δ i j = Q ki Q k j ,

(1.78)

7 We must emphasise that the vector itself does not change, it is just its components that are modified;

the use of different labels for our vector is employed here solely for emphasising this distinction.

1.9 Transformation Rules for a Change of Basis

37

so the set of numbers [Q i j ] can be regarded as being the components of an orthogonal tensor Q which satisfies [ Q][ Q]T = [ Q]T [ Q] = [I].

(1.79)

Note that ei = Q ki Q ji e j = δ k j e j = ek , Q ki and, by combining this with (1.76), we can write ei = Q i j ej.

 ei = Q ji e j ,

(1.80)

Also, since ei , u = u j e j = u j Q ji

ei ,  u = ui

we obtain  ui = Q ji u j

and

ui = Q i j uj

(1.81)

and

[u] = [ Q][ u].

(1.82)

or [ u] = [ Q]T [u]

Thus, the components of a vector transform in the same way as the basis vectors under a change of basis (compare with (1.80)). Next, we investigate the tensorial transformation law, i.e., the way the components of an arbitrary element of Lin change when we switch to a different basis of V . To this end, let us write A = A i j ei ⊗ e j

and

 i j ei ⊗ A= A ej,

where, according to (1.37), we have Ai j = ei · ( Ae j ) in B V(1) ,

(1.83a)

 Ai j =  ei · ( A e j ) in

(1.83b)

B V(2) .

We emphasise again that  A is just the expression of A in the new tensorial basis formed with the ‘hat’ vectors. Combining (1.83a) with (1.76) and making use of (1.83b), we find successively i j =  ei · ( A e j ) = (Q pi e p ) · A(Q q j eq ) A = Q pi Q q j (e p · Aeq ) = Q pi Q q j A pq = Q pi A pq Q q j .

38

Thus,

1 Vector, Tensors, and Related Matters

 Ai j = Q pi Q q j A pq

and

Ai j A pq = Q pi Q q j 

(1.84)

and

[ A] = [ Q][ A][ Q]T .

(1.85)

or, expressed more compactly, [ A] = [ Q]T [ A][ Q]

The transformation rules in (1.85) relate the different components of the same tensor in the two different representations, A and  A, respectively. In many books, the transformation rules (1.84) represent the starting point for the definition of the concept of tensor. According to that point of view, a Cartesian tensor of order n ≥ 0 is a mathematical entity characterised by the components Ai1 i2 ...in (n indices) relative to an orthonormal basis B V(1) , which transform like j1 j2 ... jn = Q i1 j1 Q i2 j2 . . . Q in jn Ai1 i2 ...in A

(1.86)

e j = Q i j ei ; here [Q i j ] is taken to be under a change of basis B V(1) → B V(2) given by a proper orthogonal 3 × 3 matrix. The set of all such tensors of order n can be shown to be a linear vector space that herein after will be denoted Lin n . Clearly, with this point of view in mind, scalars are zeroth-order tensors, while vectors correspond to first-order tensors. We shall keep the notation Lin for Lin 2 , and note in passing that the dimension of Lin n is equal to 3n . In Sect. 1.12 we return to a more in-depth examination of (1.86), after first discussing an alternative way to introduce tensors of order n.

1.10 Higher Order Tensors As already announced, we are now going to see how the mathematical framework used for introducing second-order tensors in Sect. 1.5 can be easily adapted to cover the general case of a tensor of order n > 2. Let us first recall that second-order tensors were defined as linear mappings acting between a three-dimensional linear space V ; the set of all such linear mapping was denoted by Lin. Definition 1.14 A third-order tensor is a linear mapping from V to Lin. The set of all such linear mappings is denoted by Lin 3 . Taken at face value this new concept does not seem immediately of much use. To unlock its power we need to generalise the tensor product of two vectors. Definition 1.15 The tensor product of u1 , u2 , u3 ∈ V , denoted by u1 ⊗ u2 ⊗ u3 , represents the linear mapping from V to Lin defined according to (u1 ⊗ u2 ⊗ u3 )a := (a · u3 )u1 ⊗ u2 ,

(∀) a ∈ V .

1.10 Higher Order Tensors

39

It is a routine exercise to show that Lin 3 forms a linear space and, by following the same line of reasoning as in the proof of Proposition 1.3, it can be further demon-  strated that if {ei } is a basis in V then the set ei1 ⊗ ei2 ⊗ ei3 | 1 ≤ i 1 , i 2 , i 3 ≤ 3 forms a basis in Lin 3 . This observation is important because it means that given any A ∈ Lin 3 one can find a set of 33 numbers Ai1 i2 i3 (1 ≤ i 1 , i 2 , i 3 ≤ 3) such that A = Ai1 i2 i3 ei1 ⊗ ei2 ⊗ ei3 . At this juncture, it is worth asking ourselves whether this new way of introducing the third-order tensors has anything in common with the method mentioned at the ei }, linked end of Sect. 1.9. Assuming that the basis {ei } is replaced by a new basis { to the old one by the transformation formulae (1.76), it is a trivial matter to show that the components of the original tensor in the two bases are related by j1 j2 j3 = Q i1 j1 Q i2 j2 Q i3 j3 Ai1 i2 i3 , A which is exactly of the same form as (1.86) for n = 3. The extension of these concepts to the general case of tensors of order n can now be done inductively, and it is made explicit below. Definition 1.16 A tensor of order n > 2 is a linear mapping from V to Lin n−1 . The set of all such linear mappings is denoted by Lin n . To fully appreciate this definition, we need a further extension of the tensor product. Definition 1.17 The tensor product of n > 2 vectors u1 , u2 , . . . , un ∈ V is denoted by u1 ⊗ u2 ⊗ · · · ⊗ un and represents the linear mapping from V to Lin n−1 defined according to the rule (u1 ⊗ u2 ⊗ · · · ⊗ un )a := (a · un )u1 ⊗ u2 ⊗ · · · ⊗ un−1 ,

(∀) a ∈ V .

By analogy with Proposition 1.3, a basis in the linear space Lin n is given by 

 ei 1 ⊗ e i 2 ⊗ · · · ⊗ ei n | 1 ≤ i 1 , i 2 , . . . , i n ≤ n ,

and every A ∈ Lin n can be represented as a linear combination of elements from this set, i.e., A = Ai1 i2 ...in ei1 ⊗ ei2 ⊗ · · · ⊗ ein ; in the above definition ei1 , ei2 , . . . , ein are elements of the set {e1 , e2 , e3 }.

40

1 Vector, Tensors, and Related Matters

1.11 A Special Property: Isotropy As seen in Sect. 1.9, the underlying property of a tensor is a certain law of transformation for its components, dictated essentially by arbitrary linear changes of coordinates. It is of interest to know whether there are any cases in which the components of a general tensor of rank n ≥ 0 remain the same under fairly general classes of transformations for Cartesian systems of coordinates. The discussion included next shows that such tensors are remarkably simple. Definition 1.18 If the components of A ∈ Lin n are unchanged under arbitrary rotations of orthonormal bases, then A is said to be isotropic. We illustrate this definition with a number of simple examples: • Lin 0 : all scalars are isotropic. • Lin 1 : 0 (the zero vector) is the only isotropic vector. This may be proved as u j = u j for all follows. To satisfy the definition of isotropy, we require Q i j ui ≡  rotations [Q i j ]. We are free to choose [Q i j ] to be any rotation. The choice ⎡

0 [Q i j ] = ⎣ −1 0

1 0 0

⎤ 0 0⎦ 1

(representing a rotation of 90◦ about e3 ) leads to u1 = u2 = 0. Any other choice leads immediately to u3 = 0, and hence u = 0. • Lin: scalar multiples of δi j are the only isotropic second-order Cartesian tensors. pq = A pq for all rotations [Q i j ]. Indeed, for isotropy we require Q i p Q jq Ai j ≡ A The same choice as in the previous example gives A11 = A22 and A13 = A23 = 0. Choosing now ⎡ ⎤ 1 0 0 0 1⎦ [Q i j ] = ⎣ 0 0 −1 0 (a rotation through 90◦ about e1 ), gives A22 = A33 and A21 = A31 = 0. Thus, Ai j = A11 δi j , where A11 is an arbitrary scalar. • Lin 3 : scalar multiples of ∈i jk are the only isotropic third-order Cartesian tensors. The details leading to this result are rather tedious, and are thus omitted. • Lin 4 : the only (independent) isotropic fourth-order Cartesian tensors, with free indices i, j, k, l, are scalar multiples of δi j δkl , δik δ jl , δil δ jk , and the most general isotropic element of Lin 4 has components of the form α δi j δkl + β δik δ jl + γ δil δ jk , where α, β, γ ∈ R are arbitrary scalars.

1.11 A Special Property: Isotropy

41

• For tensors of order higher than 4, we note here that 1. if n = even, then the only isotropic tensors in Lin n are linear combinations of products of Kronecker delta symbols; 2. if n = odd, then the only isotropic tensors in Lin n are linear combinations of the products of one permutation symbol and an appropriate number of Kronecker delta symbols.

1.12 Pseudo-scalars/vectors/tensors Let us return to Eq. (1.86). Without restricting [Q i j ] to proper orthogonal matrices that equation corresponds to the definition of tensor adopted in Physics. Since n ≥ 0 in Eq. (1.86), scalars and vectors can also be viewed as particular tensors of order 0 and 1, respectively. A scalar would correspond to a quantity that remains invariant under any change of coordinates (1.76), while a vector is described by a set of three numbers that under the same circumstances follow the pattern (1.81). Writing out these rules by using (1.86) and including second-order tensors as well, we have  = φ, φ  u j1 = Q i1 j1 ui1 , j1 j2 = Q i1 j1 Q i2 j2 Ai1 i2 , A

(1.87a) (1.87b) (1.87c)

which must be satisfied for any orthogonal matrix [Q i j ]; the ‘hats’ denote transformed quantities. It must be stressed that, even though our original departure point was quite different, our vectors (elements of a vector space) and tensors (linear mappings) satisfy the above requirements by default. However, there is a subtle aspect that forced us to impose the restriction to proper orthogonal matrices in (1.86), and this deserves further explanations. The scalar triple product defined in (1.14) represents a number. If we consider a change of basis by letting  ei = −ei (i = 1, 2, 3) then the matrix [Q i j ] corresponds to [−δi j ], which is obviously orthogonal. Under this transformation the scalar triple product changes its sign, so it is no longer invariant (i.e., the rule (1.87a) is violated). Mathematically, the triple scalar product is still a scalar, but from the point of view of Physics it is not. To enlarge the scope of the rules (1.87) the concept of pseudo-tensor is introduced by modifying them according to  = (det Q)φ, φ

(1.88a)

 u j1 = (det Q)Q i1 j1 ui1 , j1 j2 = (det Q)Q i1 j1 Q i2 j2 Ai1 i2 ; A

(1.88b) (1.88c)

these new rules define pseudo-scalars, pseudo-vectors and second-order pseudotensors.

42

1 Vector, Tensors, and Related Matters

We look at three representative examples that illustrate the points made above. The first one concerns the so-called permutation pseudo-tensor that is determined by the permutation symbol introduced in (1.18), ∈ = ∈i jk ei ⊗ e j ⊗ ek .

(1.89)

∈, with A change of basis from {ei } leads to a different representation of (1.89),  = ∈ pqr pqr (Q i p ei ) ⊗ (Q jq e j ) ⊗ (Q kr ek ) ∈ ep ⊗  eq ⊗  er = ∈ pqr Q i p Q jq Q kr ei ⊗ e j ⊗ ek . =∈

(1.90)

Comparing (1.89) and (1.90) leads to pqr = ∈i jk Q pi Q q j Q r k = (det Q) ∈ pqr , ∈

(1.91)

where the last equality follows from formula (1.19b). Although ∈ is a tensor by definition, the set of numbers ∈i jk behave like a pseudo-scalar according to (1.91). pqr = ∈ pqr , i.e. ∈ ei } have the same ‘handedness’ then Q ∈ Ort+ and ∈ If {ei } and { is an isotropic tensor. If the two bases have different handedness then det Q = −1 pqr = − ∈ pqr . and ∈ The next example is the vector product of two vectors. We ask ourselves if the result of this operation is indeed a vector in the sense of definition (1.87b). If a, b ∈ V are arbitray, this will require calculating the components of a ∧ b in the two bases. To this end , let Δ j ∈ {+1, −1} ( j = 1, 2) be associated with the orientation of the two bases as explained when discussing formula (1.23). Then,    (a ∧ b) p = (a ∧ b) ·  e p = (Q im ai em ) ∧ (Q jn b j en ·  ep mnp ai b j = Q im Q jn ai b j [ em , en , e p ] = Q im Q jn Δ2 ∈ = Q im Q jn Δ2 (det Q) ∈mnp ai b j = (det Q)Q kp Δ1∈i jk ai b j = (det Q)Q kp ai b j [ei , e j , ek ] = (det Q)Q kp (a ∧ b)k ,

(1.92)

where we have taken advantage that the matrix [Q i j ] is orthogonal, and hence Q im Q jn ∈mnp = ∈i jk Q kp (det Q); also, use has been made here of the evident relation Δ1 = (det Q)Δ2 . The result obtained in (1.92) confirms that the components of a ∧ b transform like those of a pseudo-vector. This is a general result: vector products are always pseudo-vectors. The final example pertains to the axial vector of a skew-symmetric tensor W —see Proposition 1.6. If w is the axial vector of such a tensor, by analogy with the work in (1.92), we have

1.12 Pseudo-scalars/vectors/tensors

43

mn ( mn Δ2 ∈ pmn −2 wp = W em ∧ en ) ·  ep = W = Q im Q jn Wi j Δ2 (det Q) ∈ pmn = Δ1 Q im Q jn ∈mnp Wi j = Δ1 ∈i jk Q kp (det Q)Wi j = (det Q)Q kp Wi j (ei ∧ e j ) · ek = (det Q)Q kp (−2wk ),

(1.93)

which indicates that the axial vector of a skew-symmetric tensor is a pseudo-vector. In Physics, pseudo-vectors are usually referred to as axial vectors; this class consists of vectorial quantities whose direction depends on the ‘orientation’ of the reference triad (left- or right-handed); examples include the moment of a force, angular momentum, and angular velocity. On the other hand, vectors such as force, velocity and linear momentum, whose direction depends only on their physical meaning and remains unaffected by the ‘handedness’ of the reference triad, are called polar vectors. Starting from the definition (1.88), one can easily check what is the outcome of combining polar and axial vectors. As we have already seen above, the vector product of two polar vectors is an axial vector, but if one of the vectors is polar and the other axial, the result will be a polar vector. In this latter case, the dot product yields a pseudo-scalar. Also, vector products between two axial vectors produce another axial vector. For obvious reasons an axial vector can never be equated to a polar vector (strictly speaking).

1.13 Other Uses of Vector and Tensor Products The importance of the ‘dot’ and ‘cross’ products in Mechanics requires no justification. Up to now, we have seen that the tensor product is one of the major devices as far as tensor manipulations are concerned. In this section, we shall introduce a few more standard ways for combining vectors and tensors. A first class of new operations that we wish to introduce concerns the so-called contracted products. In general, the process of contraction in a tensor of order n consists in setting two indexes equal and performing the summation indicated by that process. Contracting a tensor yields another tensor whose order is less by two than that of the original one. For example, consider a fourth-order tensor A which according to the remarks made previously admits a representation of the form A = Ai jkl ei ⊗ e j ⊗ ek ⊗ el , where the numbers Ai jkl represent its components relative to the usual standard basis {ei } of V . Each of the following expressions represents a different contraction of this tensor, Ai jil e j ⊗ el , Ai jkk ei ⊗ e j ; Aiikl ek ⊗ el , note how this operation maps Lin 4 on to Lin 2 ≡ Lin.

44

1 Vector, Tensors, and Related Matters

It should be pointed out that the contraction of a second-order tensor A results in a scalar, the so-called trace of the tensor, denoted by tr ( A) or | A|—this will be formally defined later on. Also, the scalar product of two vectors a, b ∈ V can be regarded as the contraction of a ⊗ b ∈ Lin. Definition 1.19 Let m, n ∈ N. The tensor product of A ∈ Lin n and B ∈ Lin m is an element of Lin m+n , denoted by A ⊗ B, and defined according to the rule     A ⊗ B ≡ Ai1 i2 ...in ei1 ⊗ ei2 ⊗ · · · ⊗ ein ⊗ B j1 j2 ... jm e j1 ⊗ e j2 ⊗ · · · ⊗ e jm := Ai1 i2 ...in B j1 j2 ... jm ei1 ⊗ ei2 ⊗ · · · ⊗ ein ⊗ e j1 ⊗ e j2 ⊗ · · · ⊗ e jm . (1.94)     There is no need to distinguish between, say, ei ⊗ e j ⊗ ek and ei ⊗ e j ⊗ ek , and we therefore write ei ⊗ e j ⊗ ek , omitting the parentheses. It is possible to apply the contraction operation more than once, thus obtaining multiple contractions. For tensor products, depending on the order of the tensors involved, this procedure can be carried out in a number of different ways. In the case of a tensor product between second-order tensors there are only two possible double contractions which we shall call, respectively, the vertical and the horizontal double-dot products, (a ⊗ b) : (c ⊗ d) := (a · c)(b · d), (a ⊗ b) · ·(c ⊗ d) := (a · d)(b · c),

(1.95a) (1.95b)

for all a, b, c, d ∈ V . For second-order tensors represented in the standard basis, with the help of Proposition 1.3, we deduce A : B = Ai j Bi j

and

A · ·B = Ai j B ji ;

(1.96)

the magnitude of A ∈ Lin is taken to be  A := ( A : A)1/2 .

(1.97)

Some important properties of the double-dot contraction are listed below | A| ≡ tr( A) = I : A = A : I,

(1.98a)

A : (BC) = (B A) : C = ( AC ) : B, A : (u ⊗ v) = u · ( Av) = (u ⊗ v) : A,

(1.98b) (1.98c)

A : B = B : A and A : B T = AT : B,

(1.98d)

T

T

for all u, v ∈ V and A, B, C ∈ Lin. Equation (1.96) can also be written in a more suggestive way as A : B = tr( AB T )

and

A · ·B = tr( AB),

(1.99)

1.13 Other Uses of Vector and Tensor Products

45

whence the relation between the two double-dot products becomes apparent, A : B = A · ·B T = AT · ·B. . The vertical triple-dot product .. is sometimes useful when dealing with three consecutive contractions of tensors of order greater than or equal to three . (a1 ⊗ a2 ⊗ a3 ) .. (b1 ⊗ b2 ⊗ b3 ) ≡ (a1 · b1 )(a2 · b2 )(a3 · b3 ),

(1.100)

for all a j , b j ∈ V ( j = 1, 2, 3). Extension to products between elements of Lin 3 and Lin n is done similarly. The next item on the agenda is the extension of the scalar and vector products between elements of both V and Lin n (n > 1). Definition 1.20 Given a vector a ∈ V and a tensor A ∈ Lin n , we define the left and right inner products between a and A by a · A = Ai1 i2 ...in (a · ei1 )ei2 ⊗ ei3 ⊗ · · · ⊗ ein = ai1 Ai1 i2 ...in ei2 ⊗ ei3 ⊗ · · · ⊗ ein

(1.101)

and, respectively, A · a = Ai1 i2 ...in ei1 ⊗ ei2 ⊗ · · · ⊗ ein−1 (ein · a) = Ai1 i2 ...in ain ei1 ⊗ ei2 ⊗ · · · ⊗ ein−1 .

(1.102)

As a particular case, we note that if a ∈ V and A ∈ Lin, with this definition in mind, we have and a · A = ai Ai j e j , A · a = Ai j a j ei that is, A · a ≡ Aa

and

a · A ≡ AT a.

(1.103)

This property becomes particularly useful when dealing with certain differential operations that will be introduced in Sect. 1.22. We shall make use of the equivalence of the operations stated in (1.103) throughout the book (usually without a warning). In view of this notation, for A ∈ Lin, the relation (1.37) can be written as Ai j = ei · A · e j , and the vector of this tensor will then be given by A× = ∈i jk (ei · A · e j )ek . Since the permutation symbol is anti-symmetric in any pair of indexes, from (1.49) it follows that A× = 0 if the tensor is symmetric—as naturally expected. The inner product between a vector and a tensor in Lin n is always an element of Lin n−1 . It can be shown that Definition 1.20 is independent of the choice of basis. e j } (say), related to each For instance, if we replace the basis {e j } by a new one, { other through (1.76), then we find successively

46

1 Vector, Tensors, and Related Matters

a · A = ai1 Ai1 i2 ...in ei2 ⊗ ei3 ⊗ · · · ⊗ ein = Q i1 j1 a j1 Ai1 i2 ...in (Q i2 j2 e j2 ) ⊗ (Q i3 j3 e j3 ) ⊗ · · · ⊗ (Q in jn e jn ) = (Q i1 j1 Q i2 j2 . . . Q in jn Ai1 i2 ...in ) a j1 e j3 ⊗ · · · ⊗ e jn e j2 ⊗  e j2 ⊗ = a j1 A j1 j2 ... jn e j3 ⊗ · · · ⊗ e jn , the last result being of the same form as (1.101). The independence of this operation on the choice of basis employed in the representation of the terms that enter into the contracted product holds in general (e.g. the double-dot products (1.95a) and (1.95b) enjoy a similar property). Definition 1.20 can easily be extended to deal with the more general inner product of two tensors of arbitrary order, as the next definition indicates. Definition 1.21 Let m, n ∈ N. The inner product of A ∈ Lin n and B ∈ Lin m is an element of Lin m+n−2 , denoted by A · B and defined according to the rule     A · B ≡ Ai1 i2 ...in ei1 ⊗ ei2 ⊗ · · · ⊗ ein · B j1 j2 ... jm e j1 ⊗ e j2 ⊗ · · · ⊗ e jm := Ai1 i2 ...in B j1 j2 ... jm (ein · e j1 )ei1 ⊗ ei2 ⊗ · · · ⊗ ein−1 ⊗ e j2 ⊗ · · · ⊗ e jm , = Ai1 i2 ...in−1 p B pj2 ... jm ei1 ⊗ ei2 ⊗ · · · ⊗ ein−1 ⊗ e j2 ⊗ · · · ⊗ e jm . (1.104) If A, B ∈ Lin then

A · B = AB,

(1.105)

where the right-hand side corresponds to the result already found in (1.62). These equivalent notations for the multiplication of two tensors will be used interchangeably without further ado. An interesting application of the tensorial inner product is found in writing a skew-symmetric tensor W with axial vector w in the compact form W =∈ · w, with ∈ being the third-order permutation pseudo-tensor introduced in (1.89). Definition 1.22 Given a vector a ∈ V and a tensor A ∈ Lin n , we define the left and right vector products between a and A by a ∧ A = Ai1 i2 ...in (a ∧ ei1 ) ⊗ ei2 ⊗ ei3 ⊗ · · · ⊗ ein = ∈i1 pq a p Aqi2 ...in ei1 ⊗ ei2 ⊗ ei3 ⊗ · · · ⊗ ein

(1.106)

and A ∧ a = Ai1 i2 ...in ei1 ⊗ ei2 ⊗ · · · ⊗ ein−1 ⊗ (ein ∧ a) = ∈ pqin Ai1 i2 ...in−1 p aq ei1 ⊗ ei2 ⊗ · · · ⊗ ein ,

(1.107)

respectively. In particular, for a second-order tensor A, we get a ∧ A = ∈i jk a j Akp ei ⊗ e p

and

A ∧ a = ∈i jk A pi a j e p ⊗ ek ,

(1.108)

1.13 Other Uses of Vector and Tensor Products

47

whence, by inspection, ( A ∧ a)T = −a ∧ AT ,

(a ∧ A)T = − AT ∧ a.

(1.109)

Another particularisation of this definition is obtained by letting A → b ⊗ c, for two arbitrary vectors b, c ∈ V . Then, according to (1.106), a ∧ (b ⊗ c) = (a ∧ b) ⊗ c.

(1.110)

There are situations when one encounters products of three factors (vectors and tensors) containing both the vector and inner products. In such cases, it is important to know whether these operations are associative. It turns out that the multiplication between second-order tensors and vectors, whether the multiplication is performed with a ‘wedge’ or a ‘dot’, is associative in the following cases: (a ∧ A) · B = a ∧ ( A · B), (a ∧ A) · b = a ∧ ( A · b), a ∧ ( A ∧ b) = (a ∧ A) ∧ b,

( A · B) ∧ a = A · (B ∧ a), a · ( A ∧ b) = (a · A) ∧ b,

where A, B ∈ Lin and a, b ∈ V . However, note also the following: A · (a ∧ b) = ( A · a) ∧ b, ( A · a) ∧ B = A ∧ (a · B),

(a ∧ b) · A = a ∧ (b · A), A · (a ∧ B) = ( A · a) ∧ B.

These examples can be summarised by the observation that in the case in which the vector occurs at any other place than the end, the product is not associative. When considering the scalar triple product (1.14) it was pointed out that a · (b ∧ c) = (a ∧ b) · c for any vectors a, b, c. A similar property is found when we replace some of these vectors by tensors, namely, a · (b ∧ A) = (a ∧ b) · A, ( A ∧ a) · b = A · (a ∧ b), A · (a ∧ B) = ( A ∧ a) · B. We remark in passing that if W ∈ Skw with axial vector w, then W = w ∧ I = I ∧ w;

(1.111)

in particular, b ⊗ a − a ⊗ b = I ∧ (a ∧ b) for a, b ∈ V . There is an alternative way to introduce the vector product between a vector and a tensor that will be illustrated now for the particular case of second-order tensors. Definition 1.23 Given a ∈ V and A ∈ Lin, a ∧ A ∈ Lin is defined by (a ∧ A) · b := a ∧ ( A · b),

(∀) b ∈ V .

(1.112)

48

1 Vector, Tensors, and Related Matters

Similarly, A ∧ a is the element of Lin that satisfies ( A ∧ a) · b := A · (a ∧ b),

(∀) b ∈ V .

It should be clear that the tensor a ∧ A thus defined makes sense since the righthand side of (1.112) involves the vector product of two elements of V and is linear in b. To find the component representation of the ‘new’ tensor, simply set b → e j in (1.112) and then take the dot product with ei , (a ∧ A)i j = ei · (a ∧ A) · e j = [ei , a, ( A · e j )] = [( A · e j ), ei , a] = ( A · e j ) · (ei ∧ a) = (Ak j ek ) · (ei ∧ a) = Ak j [ek , ei , a] = Ak j [a, ek , ei ] = Ak j a · (ek ∧ ei ) = ∈ki p a p Ak j = ∈i pk a p Ak j . This result coincides with the components of a ∧ A recorded in (1.108) by making the substitutions p → j and j → p in the former expression. Many of the properties listed after Definition 1.22 can be checked very quickly by using (1.112). We illustrate the strategy for a somewhat more complicated example that provides an expansion formula for a ∧ (b ∧ A). To this end, let c ∈ V be arbitrary and set B := b ∧ A. Then, by using (1.112) repeatedly in conjunction with the vector triple product result (1.28), we find 

a ∧ (b ∧ A) · c = (a ∧ B) · c = a ∧ (B · c)   = a ∧ (b ∧ A) · c = a ∧ b ∧ ( A · c)  = a · ( A · c) b − (a · b)( A · c)  = ( AT · a) · c b − (a · b)( A · c)   = b ⊗ ( AT · a) · c − (a · b) A · c   = b ⊗ a) · A · c − (a · b) A · c,

whence a ∧ (b ∧ A) = (b ⊗ a) · A − (a · b) A.

(1.113)

In deriving this formula, we have also used the definition of the transpose (1.49) and Eq. (1.57c). An interesting particular case of (1.113) is obtained by choosing a = b and A = I, a ∧ (a ∧ I) = a ⊗ a − (a · a)I. (1.114) Definition 1.24 Let m, n ∈ N. The vector product of A ∈ Lin n and B ∈ Lin m is an element of Lin m+n−1 , denoted by A ∧ B, and defined according to the rule

1.13 Other Uses of Vector and Tensor Products

49

    A ∧ B ≡ Ai1 i2 ...in ei1 ⊗ ei2 ⊗ · · · ⊗ ein ∧ B j1 j2 ... jm e j1 ⊗ e j2 ⊗ · · · ⊗ e jm := Ai1 i2 ...in B j1 j2 ... jm ei1 ⊗ ei2 · · · ⊗ ein−1 ⊗ (ein ∧ e j1 ) ⊗ e j2 · · · ⊗ e jm . As a simpler example consider the case when A → a ⊗ b and B → c ⊗ d, (a ⊗ b) ∧ (c ⊗ d) = a ⊗ (b ∧ c) ⊗ d ∈ Lin 3 . A very useful operation, the so-called double-vector product of two tensors, is defined first for dyads according to (a ⊗ b) × × (c ⊗ d) ≡ (a ∧ c) ⊗ (b ∧ d),

(1.115)

for all a, b, c, d ∈ V . If A, B ∈ Lin, with A = Ai j ei ⊗ e j and B = Bkl ek ⊗ el , then A× × B = Ai j Bkl (ei ∧ ek ) ⊗ (e j ∧ el ); while this operation is commutative, it does not enjoy any associative features, i.e. × A× × B = B × A,

but

× × × (A× × B) × C = A × (B × C).

It can be further shown that × × ( AB) × × ( AB) = ( A × A)(B × B), and

(∀) A, B ∈ Lin,

I× × I = 2I,

where I is the usual identity element in Lin. Checking these formulae is fairly straightforward. We illustrate the corresponding manipulations on a somewhat similar example, namely, the identity T I× × A = (I : A)I − A .

Assuming that A = A jk e j ⊗ ek we have successively, × I× × A = (ei ⊗ ei ) × (A jk e j ⊗ ek ) = A jk (ei ∧ e j ) ⊗ (ei ∧ ek ) = A jk (∈i j p e p ) ⊗ (∈ikq eq ) = ∈i j p ∈ikq A jk e p ⊗ eq = (δ jk δ pq − δ jq δ pk )A jk e p ⊗ eq = A j j e p ⊗ e p − Aq p e p ⊗ eq = (I : A)I − AT .

(1.116)

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1 Vector, Tensors, and Related Matters

There are two further products that can be defined for dyads; these involve both an inner and a vector product, • (c ⊗ d) ≡ (a · c)b ∧ d, (a ⊗ b) ×

(1.117a)

(a ⊗ b) × • (c ⊗ d) ≡ (b · d)a ∧ c,

(1.117b)

for all a, b, c, d ∈ V . If A = Ai j ei ⊗ e j and B = Bkl ek ⊗ el then • B = A B (e · e )(e ∧ e ) = δ A B (e ∧ e ) A× i j kl i k j l ik i j kl j l = Ak j Bkl (e j ∧ el ). It is also immediately seen that the vector of any A ∈ Lin assumes the representations • A = A × I. A× = I × •

1.14 Fourth-Order Tensors Let us return now to the double-dot (contracted) product introduced earlier in (1.95a). This is an important concept which facilitates the routine simplification of notation in many complicated calculations, as we shall discover in some of the later chapters. Definition 1.25 Let a, b, c, d, u, v ∈ V be arbitrary vectors. The right and left double-dot products between the tensor products a ⊗ b ⊗ c ⊗ d and u ⊗ v are defined by the following intuitive rules: (a ⊗ b ⊗ c ⊗ d) : (u ⊗ v) = (a ⊗ b)(c · u)(d · v), (u ⊗ v) : (a ⊗ b ⊗ c ⊗ d) = (u · a)(v · b)(c ⊗ d).

(1.118a) (1.118b)

If C ∈ Lin 4 and A ∈ Lin then C = Ci jkl ei ⊗ e j ⊗ ek ⊗ el and A = Amn em ⊗ en ; according to (1.118a) and (1.118b), we can, therefore, write C : A = Ci jkl Amn ei ⊗ e j (ek · em )(el · en ) = Ci jkl Amn δkm δln ei ⊗ e j = Ci jkl Akl ei ⊗ e j and, similarly, A : C = Ai j Ci jkl ek ⊗ el . Definition 1.26 Let a, b, c, d, u, e, f , g, h ∈ V be arbitrary vectors. The double-dot product of a ⊗ b ⊗ c ⊗ d and e ⊗ f ⊗ g ⊗ h is by definition an element of Lin 4 calculated by the rule

1.14 Fourth-Order Tensors

51

 (a ⊗ b ⊗ c ⊗ d) : (e ⊗ f ⊗ g ⊗ h) = (a ⊗ b ⊗ g ⊗ h) (c ⊗ d) : (e ⊗ f ) = (a ⊗ b ⊗ g ⊗ h)(c · e)(d · f ).

(1.119)

Hence, if C = Ci jmn ei ⊗ e j ⊗ em ⊗ en and S = S pqkl e p ⊗ eq ⊗ ek ⊗ el , we can introduce the (double-dot) product of two fourth-order tensors according to C : S = Ci j pq S pqkl ei ⊗ e j ⊗ ek ⊗ el ,

(1.120)

or more succinctly, (C : S)i jkl = Ci j pq S pqkl ; evidently, this is an element of Lin 4 . Definition 1.27 The identity tensor in Lin 4 is defined by I := ei ⊗ e j ⊗ ei ⊗ e j ≡ δik δ jl ei ⊗ e j ⊗ ek ⊗ el .

(1.121)

The components of I are (I)i jkl = δik δ jl , and it is a routine matter to check that I : C = C : I = C,

(∀) C ∈ Lin 4 .

In Definition 1.12, we defined the transpose of a tensor A ∈ Lin as being the unique element of Lin with the property u · AT · v = v · A · u for all u, v ∈ V . By analogy, the transpose of C ∈ Lin 4 , can be introduced as the element CT ∈ Lin 4 that satisfies (a ⊗ b) : CT : (c ⊗ d) = (c ⊗ d) : C : (a ⊗ b)

(∀) a, b, c, d ∈ V . (1.122)

It is clear that the transpose will also satisfy A : CT : B = B : C : A,

(∀) A, B ∈ Lin.

(1.123)

Definition 1.28 A tensor C ∈ Lin 4 is called symmetric if CT = C, where CT represents its transpose in the sense of Eq. (1.122). A skew-symmetric element of Lin 4 is characterised by CT = −C. In relation to this last definition, we introduce the set   Sym 4 := C ∈ Lin 4 | CT = C . Routine calculations indicate that this set forms a linear subspace of Lin 4 . Furthermore, since the components of a symmetric tensor C satisfy the symmetry properties Cklmn = Cmnkl , it can be further shown that dim (Sym 4 ) = 45. In Linear Elasticity, an important role is played by a subset of these symmetric tensors. This new class is characterised by those elements C of Sym 4 which in addition also satisfy

52

1 Vector, Tensors, and Related Matters

C : (a ⊗ b) = C : (b ⊗ a),

(∀) a, b ∈ V .

(1.124)

Such tensors are called super-symmetric and the set of all such tensors forms a linear subspace of Sym 4 ; for notational convenience we introduce   SSym 4 := C ∈ Lin 4 | CT = C and satisfies (1.124) . In terms of components, super-symmetry translates into the following restrictions: Cklmn = Cmnkl = Clkmn = Cklnm , and it can be shown without difficulty that dim (SSym 4 ) = 21. The tensor  I= Iklmn ek ⊗ el ⊗ em ⊗ en , where

(1.125)

1  Iklmn = (δkm δln + δkn δlm ), 2

belongs to SSym 4 and has the property that  I : C = C : I = C,

(∀) C ∈ SSym 4 ,

i.e.,  I plays the role of the identity element in SSym 4 . Definition 1.29 An element C ∈ SSym 4 is called invertible if there exists another tensor C−1 ∈ SSym 4 , known as the inverse of C, such that I. C : C−1 = C−1 : C =  The inverse of such a tensor, if it exists, is unique. In the remaining of this section, we shall discuss some additional concepts related to the elements of Lin 4 . We start with an alternative notation for the vertical double dot product, designed specifically for the repeated application of such products. This is introduced next through a few examples. For C ∈ Lin 4 and A ∈ Lin, the right contracted product of these two tensors will be denoted by C[ A] ≡ C : A = Ci jkl Akl ei ⊗ e j . If B ∈ Lin, the contracted product of C with A and B is the real number C[ A, B] ≡ (C : B) : A = Ci jkl Ai j Bkl .

(1.126)

1.14 Fourth-Order Tensors

53

For C ∈ Lin 6 and A, B, C ∈ Lin, we can define the following contracted products: C[ A] = Ci jklmn Amn ei ⊗ e j ⊗ ek ⊗ el , C[ A, B] = Ci jklmn Akl Bmn ei ⊗ e j ,

(1.127a) (1.127b)

C[ A, B, C] = Ci jklmn Ai j Bkl Cmn .

(1.127c)

In general, if C ∈ Lin n and A j ∈ Lin n j , j = 1, 2, . . . , p, with m = n − (n 1 + n 2 + · · · + n p ) ≥ 0 then, following the same pattern as above, we can construct the contracted product C[ A1 , A2 , , . . . , A p ], which will be a tensor of order m, if m ≥ 2; for m = 1 the result is a vector, while if m = 0 it will be a scalar. With the notation just introduced the symmetry condition in Definition 1.28 becomes C[ A, B] = C[B, A] for all A and B ∈ Lin, which has a more intuitive form. The components of an arbitrary fourth-order tensor C can also be calculated from i, j, k, l ∈ {1, 2, 3}. (1.128) Ci jkl = C[ei ⊗ e j , ek ⊗ el ], Any tensor in Lin 4 can be regarded as a linear mapping from Lin back to itself, an observation that suggests a natural extension of the tensor product of two vectors to the elements of Lin; for A, B ∈ Lin define ( A ⊗ B)[C] := (B : C) A,

(∀) C ∈ Lin.

(1.129)

The linear mapping A ⊗ B is in Lin 4 and its components are easily calculated by using (1.128), namely, (1.130) ( A ⊗ B)i jkl = Ai j Bkl . The transpose of this mapping can be obtained directly from (1.123); since ( A ⊗ B)T [C, D] = ( A ⊗ B)[ D, C] = (( A ⊗ B) : C) : D) = (B : C)( A : D) = ((B ⊗ A) : D) : C) = (B ⊗ A)[C, D], for all C, D ∈ Lin, we conclude that ( A ⊗ B)T = B ⊗ A—compare with (1.54). Two other properties, which mirror closely those recorded in (1.57b) and (1.57c), are the following: C : ( A ⊗ B) = (C : A) ⊗ B, ( A ⊗ B) : C = A ⊗ (CT : B), where C ∈ Lin 4 and A, B ∈ Lin.

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1 Vector, Tensors, and Related Matters

In addition to the generalised tensor product (1.129), there are some other similar operations that are worth mentioning. Relative to the standard basis, and for any A, B ∈ Lin, we define the new operations A⊗B ∈ Lin 4 ,

( A⊗B)i jkl := Aik B jl

(1.131)

A⊗B ∈ Lin 4 ,

( A⊗B)i jkl := Ail B jk .

(1.132)

and

These operations are related to the identity tensors I and  I introduced earlier, in the sense that  1 I⊗I + I⊗I , I = I⊗I and  I= (1.133) 2 with I being the usual second-order identity tensor. Simple calculations based on the component representation (1.125) indicate that 1  I[ A] = ( A + AT ), 2

(∀) A ∈ Lin,

suggesting that the identity in SSym 4 plays the role of a symmetrising operator  I

A ∈ Lin −−−−→

1 ( A + AT ) ∈ Sym. 2

We also note that T := I⊗I is the transposition operator, i.e. T[ A] = AT for every A ∈ Lin, T

A ∈ Lin −−−−→ AT ∈ Lin. The linear mapping W :=

 1 I⊗I − I⊗I 2

(1.134)

transforms every second-order tensor into its skew-symmetric part and is usually referred to as the anti-symmetriser; we note here the evident relations W=

1 (I − T) 2

and

1  I = (I + T). 2

In the literature, the notation in (1.131) is usually replaced by the symbol ‘’, and is referred to as the square tensor product (also known as the conjugation product); given A, B ∈ Lin, then A  B ∈ Lin 4 is defined by the rule ( A  B)[C] := A · C · B T ,

(∀) C ∈ Lin.

(1.135)

1.14 Fourth-Order Tensors

55

The components of this tensor can be found from (1.128) ( A  B)i jkl = ( A  B)[ei ⊗ e j , ek ⊗ el ] = (ei ⊗ e j ) : ( A  B)[ek ⊗ el ]   = (ei ⊗ e j ) : A · (ek ⊗ el ) · B T  = ei ⊗ e j : ( A · ek ) ⊗ (B · el ) = (ei · A · ek )(e j · B · el ) = Aik B jl , which matches the formula provided in (1.131). The conjugation product can be combined with the transposer T to express the operation in (1.132) in the forms A⊗B = ( A  B) : T,

B⊗ A = T : ( A  B).

In addition, a more unified representation for the sum of the two operations in (1.131) and (1.132) can be given by introducing a new product; more precisely, for A, B ∈ Lin let A ⊗ B ∈ Lin 4 , with 1 (∀) C ∈ Lin, ( A  B)[C + C T ], 2   1 A ⊗ B i jkl = (Aik B jl + Ail B jk ). 2

( A ⊗ B)[C] :=

(1.136a) (1.136b)

It can be seen by inspection that  I= I ⊗ I

and

I = I  I.

We conclude our discussion about fourth-order tensors with a list of properties of the conjugation product ( A  B)T = AT  B T ,

(1.137a)

( A  B) : (C  D) = ( A · C)  (B · D),

(1.137b)

( A ⊗ B) : (C  D) = A ⊗ (C · B · D),

(1.137c)

( A  B) : (C ⊗ D) = ( A · C · B ) ⊗ D,

(1.137d)

T

T

for any A, B, C, D ∈ Lin. The multitude of tensor operations reviewed above might seem daunting on a first encounter. One should keep in mind that tensors are a means to an end, and not all of the above operations will be needed at once. In the case of fourth-order tensors, these operations are simply introduced to take care of the transposition of the various vector terms in tetrads like a ⊗ b ⊗ c ⊗ d. The transposition operation introduced in (1.123) is hardly effective for this purpose as it can only swap the first and the last two-vector blocks in such expressions.

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1 Vector, Tensors, and Related Matters

1.15 Finite Rotations We are now in a position to complete the discussion started in Sect. 1.8 in relation to characterising finite rotations without the need of specifying a reference triad in the usual Euclidean point space. To keep this section reasonably short, we shall rely heavily on purely geometric arguments. Consider a fixed point O in the three-dimensional Euclidean point space and a line (d) passing through it and parallel to the direction given by the unit vector a ∈ V . We want to describe geometrically the transformation that maps an arbitrary point −−→ −−→ P to P1 in a plane perpendicular to a and such that | O P | = | O P1 |—for details see Fig. 1.6. If the position vectors of the two points with respect to the origin O are x and x  , respectively, our aim will be to find the latter quantity as a function of the former. −−→ −−→ Let us choose M ∈ P P1 such that | M P | = | M P1 | and let N be the foot of the −−→ −−→ perpendicular from M to (d). Then, P P1 being perpendicular to both O N and −−→ −−→ O M will also be perpendicular to the plane (N O M) and, in particular, to N M . As a consequence ∠P N M ≡ ∠M N P1 , and we denote by θ/2 their common value. Elementary vector geometry indicates that −−→ P P1 = x  − x,

−−→ 1 O M = (x + x  ). 2

−−→ −−→ −−→ −−→ As P M is perpendicular to both O M and O N , it will be parallel to a ∧ O M , i.e., 1 −−→ (1.138) P M  a ∧ (x + x  ). 2 −−→ −−→ Note that in M N P, | P M | = | N M | tan(θ/2) with     1 −−→ −−→ | N M | = | O M | sin ∠N O M =  a ∧ (x + x  ), 2

Fig. 1.6 The geometry of finite rotations: P1 is obtained by a counter-clockwise rotation of P by an angle θ about the axis (d). This rotation takes place in a plane perpendicular to (d) (shown shaded)

( )

1

/2

1.15 Finite Rotations

57

    θ 1 −−→    | P M | =  a ∧ (x + x ) tan . 2 2

and hence

(1.139)

Comparing (1.138) and (1.139), we can thus write θ −−→  −−→ P M = a ∧ O M tan . 2

(1.140)

This last equation can be cast in a more suggestive way by taking advantage of the −−→ −−→ obvious relation P P1 = 2 P M ; more precisely, (1.140) becomes ! " 1 θ 1  (x − x) = a ∧ (x + x  ) tan 2 2 2   θ ∧ (x + x  ). ⇒ x  − x = a tan 2

(1.141)

Although (1.141) fully capture the nature of our geometrical transformation, it is still not explicit as x  appears on both sides of the equality sign. With a little bit more effort this feature can be eliminated. To this end, the equation is rewritten in the equivalent form x  − (a ∧ x  ) tan

θ θ = x + (a ∧ x) tan . 2 2

(1.142)

Taking the vector product of (1.142) with a, and using of the expansion formula (1.28), yields  θ θ a ∧ x  − (a · x  )a − (a · a)x  tan = a ∧ x + [(a · x)a − (a · a)x] tan . 2 2 Since a · x  = a · x from (1.141), and a · a = 1, this last equation can be reduced to a ∧ x  + x  tan

θ θ θ = a ∧ x − x tan + 2a(a · x) tan . 2 2 2

(1.143)

Finally, to eliminate a ∧ x  between (1.142) and (1.143), the latter equation is multiplied by tan θ/2 and the result is added to (1.142), with the outcome x  sec2

  θ θ θ θ = x 1 − tan2 + 2(a ∧ x) tan + 2a(a · x) tan2 2 2 2 2

or, after some further simplifications, x  = x cos θ + (a ∧ x) sin θ + a(a · x)(1 − cos θ ).

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1 Vector, Tensors, and Related Matters

This explicit relation can be put in tensor form by using two properties already encountered, a ∧ x = (I ∧ a) · x and a(a · x) = (a ⊗ a) · x, so that x  = R(a; θ ) · x,

(1.144a)

R(a; θ ) ≡ a ⊗ a + cos θ (I − a ⊗ a)+ sin θ (I ∧ a).

(1.144b)

This equation is known as Rodrigues’ formula and is a classic result in vector geometry. To reiterate, Eqs. (1.144) give an analytical description of the rotation through angle θ in a plane whose unit normal is a (which defines the rotation axis). Some particular cases are examined next. If θ = 90◦ then R(a; 90◦ ) = a ⊗ a + I ∧ a. A vector which is perpendicular to a is reduced to zero by the first term on the righthand side. The action of I ∧ a ≡ a ∧ I on vectors situated in planes whose normal is a represents a rotation through an angle of 90◦ . If θ = 180◦ , then (1.144b) becomes R(a; 180◦ ) = 2a ⊗ a − I.

(1.145)

This linear transformation leaves vectors parallel to a unchanged, but reverses in direction any vector perpendicular to a. For this reason, (1.145) is usually called a planar reversal. It is intuitively clear that any two finite rotations about the same axis can be replaced by an equivalent single rotation. The question is whether the same is true for finite rotations that have different axes. Unexpectedly, this happens to be true, although the justification of this result is best handled by using (1.141), which shall be written in the form x  − x = Θ ∧ (x  + x),

θ Θ := a tan . 2

(1.146)

Consider now a second rotation that rotates P1 through an angle θ  about a new axis defined by the unit vector a . If P2 represents the image of P1 as a result of this geometric transformation, then x  − x  = Θ  ∧ (x  + x  ),

Θ  := a tan

θ , 2

(1.147)

where x  denotes the position vector of P2 relative to O. Adding up (1.146) and (1.147) results in x  − x = Θ ∧ (x + x  ) + Θ  ∧ (x  + x  )

= (Θ + Θ  ) ∧ (x + x  ) + Θ ∧ (x  − x  ) + Θ  ∧ (x  − x).

(1.148)

1.15 Finite Rotations

59

The x-differences on the right-hand side of (1.148) can be replaced by the expressions provided by (1.146) and (1.147), which introduces two vector triple products. Expanding them using (1.28) gives x  − x = (Θ + Θ  ) ∧ (x + x  ) − [Θ · (x  + x  )]Θ  + (Θ · Θ  )(x  + x  ) +[Θ  · (x + x  )]Θ − (Θ  · Θ)(x + x  ). (1.149) By taking the dot product of (1.146) with Θ and the dot product of (1.147) with Θ  shows that and Θ  · x  = Θ  · x  . (1.150) Θ · x = Θ · x Substituting (1.150) into the right-hand side of (1.149) allows for some significant simplifications, x  − x = (Θ + Θ  ) ∧ (x + x  ) − Θ  · [Θ · (x + x  )] +Θ[Θ  · (x + x  )] + (Θ · Θ  )(x  − x) = (Θ + Θ  ) ∧ (x + x  ) + (x + x  ) ∧ (Θ ∧ Θ  ) + (Θ · Θ  )(x  − x), from where we finally deduce 



x −x=

Θ +Θ−Θ ∧Θ 1−Θ ·Θ



∧ (x  + x).

(1.151)

Comparison with (1.146) suggest that (1.151) represents a rotation through an angle θ  about an axis a , with a tan

Θ +Θ−Θ ∧Θ θ  = . 2 1−Θ ·Θ

(1.152)

Both the rotation angle and the vector a can be found by using (1.152). In conclusion, the action of two consecutive rotations about arbitrary axes can be replaced by a single equivalent rotation; the axis and the angle of this resultant rotation are determined as mentioned above. Furthermore, since Θ ∧ Θ  = Θ  ∧ Θ, formula (1.152) confirms that the order of the rotations is important; in particular, a rotation is not a vectorial quantity. If Θ ∧ Θ  = 0, i.e. the rotation axes are parallel, then the order in which the rotations are performed is unimportant.

1.16 On the Relationship Between Skw and V This section involves a slight detour to tie up a loose end. At the beginning of this chapter, in our discussion about Euclidean linear spaces (see Sect. 1.3), the vector product was dealt with expediently by first postulating the existence of a

60

1 Vector, Tensors, and Related Matters

new operation that satisfied a certain set of axioms. Later, it was confirmed that the geometrical interpretation of the new operation corresponded to the familiar vector product of directed line segments. In this section, we revisit briefly the aforementioned aspect in order to illustrate an alternative (non-rigorous) way of introducing a vector product operation on a threedimensional Euclidean space. As a collateral outcome, we shall also gain further insight into the nature of the linear subspace Skw ⊂ Lin. For the purposes of this section only, the vector product will be indicated by the usual cross ‘×’ (we are thus free to re-define ‘∧’ as we please). We start by introducing a new way to combine ordered pairs of vectors from V × V ; the so-called exterior product of any two vectors a, b ∈ V , denoted by a ∧ b, is a skew-symmetric second-order tensor defined by ∧ : V × V −→ Skw,

a ∧ b := a ⊗ b − b ⊗ a.

(1.153)

Recalling our earlier discussion of such tensors it should be clear that any W ∈ Skw can be represented as W=



Wi j ei ∧ e j =

i< j

1 Wi j ei ∧ e j , 2

(1.154)

where in the last expression the summation subscripts are unrestricted. The linear mapping τ , τ : Skw −→ V ,

τ (a ∧ b) =

1 ∈ : (a ∧ b), 2

(1.155)

where ∈ is the permutation (pseudo)tensor, establishes a bijective correspondence between Skw and V . If {ei } is an orthonormal basis, a simple exercise shows that τ (ei ∧ e j ) = ∈ki j ek ,

i, j ∈ {1, 2, 3},

(1.156)

so we have the correspondences τ

e1 ∧ e2 −→ e3 ,

τ

e2 ∧ e3 −→ e1 ,

τ

e3 ∧ e1 −→ e2 .

(1.157)

In light of (1.154) and (1.156), to every W ∈ Skw there corresponds a unique vector ω ≡ ω i ei (say), with ω :=

1 Wi j τ (ei ∧ e j ), 2 i< j

ωi =

1 ∈i jk W jk ; 2

(1.158)

we write ω = W  to indicate this correspondence. Routine algebraic manipulations lead to the component expressions recorded in (1.158) and indicate that ω = −w,

1.16 On the Relationship Between Skw and V

61

where w represents the axial vector mentioned in Proposition 1.6—see Eq. (1.68) in particular. The vector product of two elements a, b ∈ V , denoted by a × b, can now be defined be setting a × b := a ∧ b. (1.159) Clearly, a × b represents another element of V and this operation satisfies the axioms listed in Sect. 1.3. From (1.156) and (1.159) it transpires that a × b = ∈i jk a j bk ei , which also agrees with our earlier formula (1.25). We close this section with an important observation: the bijective mapping τ defined in (1.155) may be used to identify the exterior product with the vector product. In this case the arrows in (1.157) may be replaced by equality signs, leading again to a familiar result from Sect. 1.3. Furthermore, skew-symmetric tensors can also be identified with axial vectors (through the same mapping τ ).

1.17 More General Bases At the end of Sect. 1.4, we declared our intention to stick to the standard basis {ei } for all of our calculations using component representations of vectors and tensors (but most of what has been said so far remains true if we use a more general orthonormal basis). There are two particular system of coordinates that deserve a closer look, and this will form the main objective of the present section. Many important problems in Continuum Mechanics have cylindrical or spherical symmetry, so one would normally expect that their solutions can be simplified by adopting a system of coordinates that is particularly tailored to their particular geometry. A general system of coordinates in a region D of E3 , consists of three scalar functions, which uniquely identify the position of a generic point P ∈ D with an ordered triple of real numbers. To make things more precise, starting from the usual rectangular Cartesian coordinates (x1 , x2 , x3 ), we introduce a curvilinear system by simply specifying a oneto-one correspondence xk = xk (θ1 , θ2 , θ3 ),

(k = 1, 2, 3),

(1.160)

between a domain Ω1 ⊂ R3 and its image Ω2 ≡ x(Ω1 ). Here, the function x ≡ (x1 , x2 , x3 ) : Ω1 → Ω2 ⊂ R3 is assumed to be continuously differentiable, with non-vanishing Jacobian  J (θ ) ≡ det

∂ xk ∂θ j

 = 0, (∀) θ ≡ (θ1 , θ2 , θ3 ) ∈ Ω1 .

(1.161)

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1 Vector, Tensors, and Related Matters

It can be shown that (1.160) is then globally invertible and the inverse mapping θk = θk (x1 , x2 , x3 ),

(k = 1, 2, 3),

(1.162)

is also continuously differentiable on Ω2 . Hence, the position of a generic point P in D can be described by using either the Cartesian coordinates (x1 , x2 , x3 ) or the curvilinear coordinates (θ1 , θ2 , θ3 ). For a fixed j ∈ {1, 2, 3} the condition θ j = const. in Eq. (1.160) defines a surface in E3 called the θ j -coordinate surface. Through each point of D passes exactly one of each family of the coordinate surfaces. Furthermore, two coordinate surfaces, θ1 = const. and θ2 = const. (say) will intersect along a curve which is parametrised by θ3 , and this is known as the θ3 -coordinate curve. The θ2 - and θ1 -coordinate curves are introduced in a similar way. Through any point of D passes exactly one of each family of coordinate curves. Definition 1.30 The curvilinear coordinates θ j ( j = 1, 2, 3) are called orthogonal if the coordinate curves are mutually orthogonal (i.e., the θk - and θ j -coordinate curves are orthogonal for k = j with k, j ∈ {1, 2, 3}). Since angles between curves at an intersection point are described by the angles between their tangents at the point in question, it is natural to introduce the vectors G j :=

∂x ∂ xk = ek , ∂θ j ∂θ j

( j = 1, 2, 3)

(1.163)

at an arbitrary point P ∈ D. These are tangent vectors to the θ j -coordinate curves passing through the point P (where the partial derivatives in (1.163) are evaluated). According to what has been said above, these vectors are mutually perpendicular; it should be also noted that they are linearly independent because of the Jacobian condition (1.161). Therefore, the system of vectors 

G 1 (P), G 2 (P), G 3 (P)



forms a so-called natural basis at P, corresponding to the curvilinear system of coordinates θ j ( j = 1, 2, 3). Unlike the standard basis {ei }, which is typically fixed in space, the natural basis changes from point to point; in this sense it is a local basis (Fig. 1.7). The disadvantage of a natural basis is that, if θ j have different physical dimensions, then the components of any vector relative to the G j ( j = 1, 2, 3) will share the same feature. This shortcoming can be easily rectified by normalising the basis vectors. With this observation in mind, we introduce 1 g j := G j, hj

# h j := |G j | ≡

∂ xk ∂ xk ∂θ j ∂θ j

$1/2 (no sum over j),

(1.164)

1.17 More General Bases

63

x3

Fig. 1.7 Curvilinear system of coordinates at an arbitrary point P in space. The unit vectors g j (P) are obtained from G j (P) ( j = 1, 2, 3) by the usual normalisation procedure (i.e., dividing the vectors by their corresponding magnitudes)

θ3 -curve

G1

x

e1

O

g3

g1

θ1 -curve

e3

G3 g2

P

G2

θ2 -curve

e2

x2

x1 where the h j ’s are known as the Lamé parameters. As a consequence, the basis 

g 1 (P), g 2 (P), g 3 (P)



is now orthonormal, i.e. for any point P ∈ D, g i (P) · g j (P) = δi j ,

(∀) i, j = 1, 2, 3.

From (1.80) we have g j = Q k j ek

and

e j = Q jk g k ,

where Q i j = g j · ei . Combining (1.163) and (1.164) it transpires that Qk j =

1 ∂ xk h j ∂θ j

(no sum over j).

1.17.1 Cylindrical Polar Coordinates In this case θ1 ≡ r , θ2 ≡ θ , θ3 ≡ z and x1 = r cos θ,

x2 = r sin θ,

where r ∈ [0, ∞), θ ∈ [0, 2π ), z ∈ R.

x3 = z,

(1.165)

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1 Vector, Tensors, and Related Matters

x3

Fig. 1.8 Cylindrical polar   coordinates: er , eθ , ez form an orthonormal set of vectors attached to the current point P; its position vector can be resolved in terms of these vectors

ez eθ P e3 O

e1

er

e2 θ

x2

r

x1

It can immediately be checked that the Jacobian J = r ≥ 0, hence the mapping (1.165) satisfies (1.161) except on the z-axis. The inverse of this mapping is r = (x12 + x22 )1/2 ,

θ = tan−1 (x2 /x1 ),

z = x3 .

(1.166)

Since ∂x = cos θ e1 + sin θ e2 , ∂r ∂x = −r sin θ e1 + r cos θ e2 , Gθ = ∂θ ∂x = e3 , Gz = ∂z Gr =

it follows that h r = |G r | = 1,

h θ = |G θ | = r,

h z = |G z | = 1,

and then we can formally write ⎡ ⎤ ⎡ er cos θ ⎣eθ ⎦ = ⎣ − sin θ 0 ez

sin θ cos θ 0

⎤⎡ ⎤ 0 e1 0 ⎦ ⎣ e2 ⎦ , 1 e3

(1.167)

where g 1 , g 2 , g 3 have been replaced by the more standard notation er , eθ , ez .

1.17 More General Bases

65

1.17.2 Spherical Polar Coordinates Let8 θ1 ≡ r , θ2 ≡ θ , θ3 = ϕ and x1 = r sin θ cos ϕ,

x2 = r sin θ sin ϕ,

x3 = r cos θ,

(1.168)

where r ∈ [0, ∞), θ ∈ [0, π ], ϕ ∈ [0, 2π ) (see Fig. 1.9). The Jacobian J = r 2 sin θ , so the mapping (1.168) is singular only on the z-axis. Partial differentiation applied to the position vector x = r sin θ cos ϕe1 + r sin θ sin ϕe2 + r cos θ e3 , in conjunction with (1.163), leads to ∂x = sin θ cos ϕe1 + sin θ sin ϕe2 + cos θ e3 , ∂r ∂x = r cos θ cos ϕe1 + r cos θ sin ϕe2 − r sin θ e3 , Gθ = ∂θ ∂x = −r sin θ sin ϕe1 + r sin θ cos ϕe2 . Gϕ = ∂ϕ Gr =

Finally h r = |G r | = 1,

h θ = |G θ | = r,

h ϕ = |G ϕ | = r sin θ,

and formally ⎡ ⎤ ⎡ er sin θ cos ϕ ⎣ eθ ⎦ = ⎣ cos θ cos ϕ − sin ϕ eϕ

sin θ sin ϕ cos θ sin ϕ cos ϕ

⎤⎡ ⎤ cos θ e1 − sin θ ⎦ ⎣e2 ⎦ , 0 e3

(1.169)

where, again, the vectors g 1 , g 2 , g 3 have been replaced by the more standard notation er , eθ , eϕ .

1.17.3 Physical Components for Vectors and Tensors The components of vectors and tensors expressed in the bases {er , eθ , ez } or {er , eθ , eϕ } of the previous section are known as physical components; the terminology comes from the fact that they have the same dimensions as the corresponding vectors and tensors. Following the developments of Sect. 1.9, it seems logical to inquire about the relationship between the components relative to the standard basis the definition of the angle θ for spherical polar coordinates is different from that in the previous section; refer to Figs. 1.8 and 1.9 for the corresponding differences.

8 CAVEAT:

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1 Vector, Tensors, and Related Matters

x3

Fig. 1.9 Spherical polar  coordinates: er , eθ , eϕ represents an orthonormal set of vectors attached to the current point

er

e3 O

e1

θ

r

eϕ P

eθ

e2

x2

ϕ

x1

{ei } and the physical components. This sort of information will be needed on several occasions later in the book. We have already done most of the work in this direction— it is just a matter of making an appropriate choice for the orthogonal matrix in (1.82) and (1.85). In the case of cylindrical polar coordinates, we choose the 3 × 3 matrix [Q i j ] that appears in (1.167), while for spherical coordinates the appropriate choice will involve the matrix found in (1.169). For example, in the former case, if u ∈ V then u = u1 e1 + u2 e2 + u3 e3 = ur er + uθ eθ + uz ez , with

⎡ ⎤ ⎡ ur cos θ ⎣uθ ⎦ = ⎣ − sin θ 0 uz

sin θ cos θ 0

⎤⎡ ⎤ 0 u1 0 ⎦ ⎣u2 ⎦ 1 u3

(1.170)

or ur = u1 cos θ + u2 sin θ,

(1.171a)

uθ = −u1 cos θ + u2 cos θ, uz = u3 .

(1.171b) (1.171c)

In the case of a tensor A ∈ Lin things are slightly more complicated, but the general principle is the same. We write A = A11 e1 ⊗ e1 + A12 e1 ⊗ e2 + A13 e1 ⊗ e3 + · · · = Arr er ⊗ er + Ar θ er ⊗ eθ + Ar z er ⊗ ez + · · · ,

1.17 More General Bases

67

where the dots stand for the obvious remaining (six) terms. By (1.85), the correspondence between the two sets of components is obtained from the transformation formula   T   A C A RT Q , (1.172) A P O L AR = Q where the labels ‘POLAR’ and ‘CART(ESIAN)’ have the obvious interpretations, and [ Q]T corresponds to the square matrix in (1.170). Carrying out the calculations on the right-hand side produces some fairly lengthy (and difficult to remember) relations of the form included below Arr = A11 cos2 θ + A22 sin2 θ + (A12 + A21 ) sin θ cos θ, Ar θ = (A22 − A11 ) sin θ cos θ + A12 cos2 θ − A21 sin2 θ, Aθr = (A22 − A11 ) sin θ cos θ + A21 cos2 θ − A12 sin2 θ, Aθθ = A11 sin2 θ + A22 cos2 θ − (A12 + A21 ) sin θ cos θ, (1.173) Ar z = A13 cos θ + A23 sin θ, Aθ z = −A13 sin θ + A23 cos θ, A zr = A31 cos θ + A32 sin θ, A zz = A33 .

A zθ = −A31 sin θ + A32 cos θ,

For A ∈ Sym some simplifications can be made, eventually leading to Arr = Ar θ = Aθθ = Ar z = Aθ z =

1 1 (A11 + A22 ) + (A11 − A22 ) cos 2θ + A12 sin 2θ, 2 2 1 Aθr = (A22 − A11 ) sin 2θ + A12 cos 2θ, 2 1 1 (A11 + A22 ) − (A11 − A22 ) cos 2θ − A12 sin 2θ, 2 2 A zr = A13 cos θ + A23 sin θ, A zθ = −A13 sin θ + A23 cos θ, A zz = A33 .

(1.174)

Our next item on the agenda will be a discussion of several concepts and results from Linear Algebra that play a fundamental role in Continuum Mechanics.

1.18 Invariants There are several ways to introduce the concept of principal invariant for a given second-order tensor. The route taken here is somewhat indirect, but it does have the merit of highlighting a number of important properties of the scalar triple product (see Definition 1.7).

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1 Vector, Tensors, and Related Matters

Proposition 1.7 Corresponding to an arbitrary tensor A ∈ Lin, there are three scalars, I A , I I A , I I I A , such that [ A · a, b, c] + [a, A · b, c] + [a, b, A · c] = I A [a, b, c],

(1.175a)

[ A · a, A · b, c]+[ A · a, b, A · c] + [a, A · b, A · c] = I I A [a, b, c], (1.175b) [ A · a, A · b, A · c] = I I I A [a, b, c], (1.175c) for all a, b, c ∈ V . These scalars are called the principal invariants of A. We shall sometimes use the alternative notation I1 ( A) ≡ I A ,

I2 ( A) ≡ I I A ,

I3 ( A) ≡ I I I A

(1.176)

or, if there is no risk of confusion, simply I1 , I2 , I3 . Given a basis of V , say B V ≡ {ei }, our next objective will be to find an explicit representation for I A , I I A and I I I A in terms of the components of A relative to B V . Let us start with (1.175a) in which we set a → e1 , b → e2 , c → e3 . The tensor A is assumed to have components Ai j relative to the given basis. Provided that B V is right-handed (i.e., [e1 , e2 , e3 ] = 1), with the help of the first relation in (1.37) and the usual properties of the triple scalar product, we get [ A · e1 , e2 , e3 ] + [e1 , A · e2 , e3 ] + [e1 , e2 , A · e3 ] = [Ai1 ei , e2 , e3 ] + [e1 , A j2 e j , e3 ] + [e1 , e2 , Ak3 ek ] = Ai1 [ei , e2 , e3 ] + A j2 [e1 , e j , e3 ] + Ak3 [e1 , e2 , ek ] = (A11 + A22 + A33 ) [e1 , e2 , e3 ] = A11 + A22 + A33 . In conclusion, I A = A11 + A22 + A33 =: tr A,

(1.177)

where the trace of A ∈ Lin is defined by the middle expression in (1.177). As already mentioned in (1.98a), an alternative notation for this scalar is | A|. It is important to observe that the trace is a linear mapping from Lin into R, i.e., tr(α A + B) = α tr( A) + tr(B),

(∀) A, B ∈ Lin, (∀) α ∈ R;

other useful properties are recorded below tr( A · B) = tr(B · A),

tr( AT ) = tr( A), (∀) A, B ∈ Lin.

(1.178)

The representation of I I A is found in the same way as above; evaluating the right-hand side of (1.175b) for the same choices of a, b, c, we obtain

1.18 Invariants

69

[ A · e1 , A · e2 , e3 ] + [e1 , A · e2 , A · e3 ] + [ A · e1 , e2 , A · e3 ] = [Ai1 ei , A j2 e j , e3 ] + [e1 , Ak2 ek , A p3 e p ] + [Aq1 eq , e2 , Ar 3 er ] = Ai1 A j2 [ei , e j , e3 ] + Ak2 A p3 [e1 , ek , e p ] + Aq1 Ar 3 [eq , e2 , er ] = (A11 A22 − A12 A21 ) + (A22 A33 − A23 A32 ) + (A11 A33 − A13 A31 ),

where use was made of [ei , e j , ek ] = ∈i jk . In a more compact form,        A11 A12   A22 A23   A11 A13  1      =  + + I IA =  (tr A)2 − tr( A2 ) . A21 A22   A32 A33   A31 A33  2

(1.179)

Finally, for the expression for I I I A the process is identical. By using (1.175c) this time, it turns out that I I I A = ∈ pqr A p1 Aq2 Ar 3 =: det A,

(1.180)

with the determinant of A ∈ Lin being defined by the expression in the middle part of this last equation; this is nothing else other than the usual determinant of the component matrix [ A]. Unsurprisingly, the determinant of a second-order tensor enjoys the same properties as that of a square matrix; for future reference we list below some of those important properties det (α A) = α 3 det ( A), det ( AT ) = det ( A),

det ( A · B) = det ( A) det (B), det ( A−1 ) = 1/ det ( A), (if A is invertible),

where A, B ∈ Lin; in addition, det(I) = 1 and det(O) = 0. In Sect. 1.7, we introduced two subsets of the set of orthogonal tensors that were defined in terms of the sign of a certain scalar triple product. A quick glance at (1.175c) indicates that the definitions of those sets can be cast more conveniently in terms of determinants. For an orthogonal tensor Q, we recall that Q · Q T = I; hence, by taking the determinant of this relation and using the properties listed above, we discover that det Q = ±1. The case det Q = +1 corresponds to proper orthogonal tensors, while the other, det Q = −1, characterises the improper ones; that is, the definitions (1.53) become Ort+ := { Q ∈ Ort | det Q = +1} , Ort− := { Q ∈ Ort | det Q = −1} .

(1.181a) (1.181b)

Proper orthogonal tensors correspond to rotations and are characterised by an axis (defined by an axial vector) and a rotation angle—see Sect. 1.15. An example of improper orthogonal tensor is (−I) and geometrically represents a central inversion. It has the property of turning a right-handed basis of V into a left-handed one. The elements of Ort are either rotations or products between a rotation and (−I), i.e.,

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1 Vector, Tensors, and Related Matters

given any orthogonal second-order tensor Q, either Q or (− Q) represents a rotation. We shall revisit proper orthogonal tensors in the next section. Before leaving this section, we call attention to some useful properties of the double-cross product vis-à-vis the cofactor of an invertible tensor A ∈ Lin—see Definition 1.12. By letting a, b, c ∈ V and using the aforementioned definition 

 cof( AT ) · (a ∧ b) · c = [ AT · a, AT · b, AT · ( A−1 )T · c]   = (det AT )(a ∧ b) · ( A−1 )T · c   = (det AT ) A−1 · (a ∧ b) · c,

where successive use was made of (1.49), (1.50), (1.52) and (1.175c). Since a, b, c are arbitrary vectors it follows that (det AT ) A−1 = cof( AT ), and letting A → AT we finally obtain (1.182) cof( A) = (det A) A−T . The cofactor satisfies the following properties: A·(cof( A))T = (det A)I = (cof( A))T · A, cof( A · B) = cof( A) · cof(B),

(1.183a) (1.183b)

cof( AT ) = (cof( A))T .

(1.183c)

The formula (1.182) can also be expressed in terms of the double-vector product introduced in (1.115), namely, cof( A) =

1 × A A, 2 ×

and then, by direct calculations,   I I A = tr cof( A) ≡ |cof( A)|. The determinant of A is also characterised by I I I A ≡ det A =

1 1 × A × A : A ≡ A : cof( A). 6 3

Up to this point nothing has been said about the invariance of the expressions defined in Proposition 1.7. In other words, the principal invariants of A are meant to be combinations of the components of A that are insensitive to the choice of bases in V . This feature is really apparent in (1.175) because each of the scalars I A , I I A , and I I I A is independent of the choice of vectors a, b, c ∈ V , and thus should remain the same irrespective of the choice of basis in V .

1.18 Invariants

71

A more direct justification of what has been just said requires the lengthier calculations included below. Consider the representation of an arbitrary tensor in two different bases, and let us identify them as [ A] and [ A], respectively. If we let I and  I A] = [ Q]T [ A][ Q], represent the values of I A in two cases then, according to (1.85), [ and  I = tr([ Q]T [ A][ Q]) = ([ Q]T [ A][ Q])ii = Q pi A pq Q qi = (Q pi Q qi )A pq = δ pq A pq = A pp = I. The invariance of I I A and I I I A can be established in the same way and is omitted in the interest of brevity.

1.19 Eigenvalues We are now in a position to introduce the concept of eigenvalue and explore its ramifications in relation to second-order tensors. Definition 1.31 A number λ ∈ C is an eigenvalue of a tensor A ∈ Lin if there exists a non-zero vector p ∈ V such that A · p = λp

or ( A − λI) · p = 0,

(1.184)

in which case p is called an eigenvector. The set of all eigenvalues of a second-order tensor is called its spectrum, and we shall use the notation Sp ( A) to refer to it. If we adopt a geometrical point of view, the eigenvectors of a tensor A define directions in the Euclidean point space that remain unaffected by the action of A; the eigenvalue λ in (1.184) represents the amount of ‘stretch’ or ‘shrink’ to which the eigenvector p is subjected when transformed by A. As it clearly follows from this definition, the eigenvalues of a tensor do not depend on the choice of basis in V (i.e., λ is a true scalar). This suggests that other tensor invariants (besides those already mentioned) can be constructed by considering various algebraic combinations of the eigenvalues. One may also suspect that the converse statement is true, i.e., λ = λ(I A , I I A , I I I A ) or, put differently, the eigenvalues are functions of the principal invariants. The validity of this last supposition is confirmed below. Let us start by first noticing that the second equation in (1.184), when expressed in a given orthonormal basis, represents a system of (three) linear homogeneous equations that has non-trivial solution if and only if PA (λ) ≡ det( A − λI) = 0.

(1.185)

This algebraic equation in λ represents the so-called characteristic equation of A, while PA is known as the characteristic polynomial. As (1.185) is in fact a cubic, we

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1 Vector, Tensors, and Related Matters

expect that any second-order tensor A will have exactly three eigenvalues that can be real or complex. Next, let us take a closer look at the characteristic equation by recalling (1.175c) and (1.180). Setting A → A − λI in the former equation and using (1.185) yields 0 = [ A · a − λa, A · b − λb, A · c − λc].

(1.186)

Expanding the right-hand side of (1.186) and then using the relations (1.175), the arbitrary factor [a, b, c] can be removed, and we end up with the following explicit form for (1.185): PA (λ) ≡ λ3 − I A λ2 + I I A λ − I I I A = 0. Definition 1.32 If λ0 ∈ Sp ( A) such that (λ − λ0 )r | PA (λ) and (λ − λ0 )r +1  PA (λ), then r ∈ N is called the algebraic multiplicity of the eigenvalue λ0 . The spectrum of A will be regarded as the set of its eigenvalues with each eigenvalue repeated a number of times equal to its algebraic multiplicity. Note that since PA is a third-order polynomial, we have 1 ≤ r ≤ 3. By way of example, consider the tensor A = 4I + e1 ⊗ e2 . Then, PA (λ) = (λ − 4)(λ − 2)2 , so Sp ( A) = {2, 2, 4}; the eigenvalue λ1 = λ2 = 2 has algebraic multiplicity equal to 2, while λ3 = 4 has algebraic multiplicity 1. Next, let A ∈ Lin and set E λ := {u ∈ V | ( A − λI) · u = 0} , where λ ∈ Sp ( A). It is an easy exercise to check that E λ is a subspace of C3 ; this is called the eigenspace (or the characteristic space) of A ∈ Lin corresponding to the eigenvalue λ. Definition 1.33 The geometric multiplicity of the eigenvalue λ is by definition the dimension of the eigenspace corresponding to λ. These two concepts of multiplicity for the eigenvalues of a tensor are quite distinct and convey different information. For example, it is known that if the algebraic multiplicity of λ ∈ Sp ( A) is greater than 1, then one cannot predict the geometric multiplicity of that eigenvalue without additional information. Let us look at these two concepts in the particular case of an A ∈ Lin that admits the representation A = −e1 ⊗ e1 − 3e1 ⊗e2 − 9e1 ⊗ e3 + 5e2 ⊗ e2 +18e2 ⊗ e3 − 2e3 ⊗ e2 − 7e3 ⊗ e3 .

1.19 Eigenvalues

73

Routine calculations show that PA (λ) = −(λ + 1)3 , so λ1 = λ2 = λ3 = −1 is an eigenvalue of algebraic multiplicity 3. To find the eigenspace E −1 , we assume u = ui ei and compute ( A + I) · u = 0, which produces 2 equations in 3 unknowns (the components of u). Two linearly independent solutions are a = [0, −3, 1]T and b = [1, −3, 1]T , from where we conclude that dim(E −1 ) = 2; hence, the geometric multiplicity of the eigenvalue λ = −1 is equal to 2. Proposition 1.8 A rotation tensor R ∈ Ort+ has an eigenvalue λ = 1. The corresponding eigenspace, E 1 := {u ∈ V | R · u = u} is one-dimensional and is called the axis of R. For the justification of the first part, R − I = R − R · R T = R · (I − R)T . Thus, by taking the determinant of this relation we find  det(R − I) = det(R) det (I − R)T = det(I − R) = − det(R − I), whence det(R − I) = 0. In other words, the characteristic equation of R has a root λ = 1; let u0 be the corresponding unit eigenvector. Regarding the dimension of the corresponding eigenspace, we start with the evident observation that dim(E 1 ) ≤ 3. If dim(E 1 ) = 3 then Ort+ = {I}, which is clearly not true because we can easily find a proper orthogonal tensor different from I (e.g. see the tensor in Eq. (1.72)). Assume by contradiction that dim(E 1 ) = 2. This means that there exists another unit vector v 0 ∈ E 1 such that u0 and v 0 are linearly independent and span E 1 . If w ∈ V is a vector with w · u0 = w · v 0 = 0 then, since orthogonal tensors preserve the scalar product, we have (R · w) · u0 = (R · w) · v 0 = 0. Hence, R · w must be a scalar multiple of w, i.e., R · w = αw for some α ∈ R. We note that w lies in the subspace E 1⊥ , the orthogonal complement of E 1 , and recall that V = E 1 ⊕ E 1⊥ . The upshot here is that the set of vectors {u0 , v 0 , w} forms a basis in V , so we can use these vectors to find a representation of the tensor R. Since the coefficient matrix associated with this representation is diagonal, det R = α. But our R is in Ort+ , which demands α = 1 and then R · w = w or w ∈ E 1 . We have, therefore, reached a contradiction, so it remains that dim(E 1 ) = 1, as required. Proposition 1.9 1. The eigenvalues of a positive-definite tensor are strictly positive. 2. If A ∈ Sym then its characteristic equation has three real eigenvalues. 3. If A ∈ Sym then the eigenvectors corresponding to two distinct eigenvalues are orthogonal. A brief justification of these properties follows next. Let A ∈ Lin be a positivedefinite tensor, and consider λ ∈ Sp ( A) with the corresponding eigenvector p ∈ V . By definition, A · p = λ p; multiplying this equality by p gives p · ( A · p) = λ ( p · p)

=⇒

λ=

p · ( A · p) . | p|2

Since the tensor is positive definite, p · A · p > 0, and the expression for λ found above indicates that this eigenvalue is clearly strictly positive.

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1 Vector, Tensors, and Related Matters

(b) Assume that A has a complex eigenvalue λ = α + iβ √ (α, β ∈ R), and let p = a + ib (a, b ∈ V ) be the corresponding eigenvector (i ≡ −1). Both the real and imaginary parts of A · p − λ p must be equal to zero. These conditions translate into two separate relations, A · a = α a − β b,

A · b = β a + α b.

(1.187)

The first equation in (1.187) is multiplied by b and the second by a; taking the difference of the results thus obtained, we discover a · ( A · b) − b · ( A · a) = β(|b|2 + |a|2 ).

(1.188)

As A ∈ Sym, the left-hand side of (1.188) is identically zero (see Definition 1.31), but its right-hand side does not vanish unless β = 0. We conclude that the imaginary part of λ being zero, the eigenvalue must be real, as required. (c) Let λ1 and λ2 be two distinct eigenvalues of A ∈ Sym , with p1 and p2 the corresponding eigenvectors: A · p1 = λ1 p1 and A · p2 = λ2 p2 . Multiplying the former relation by p2 and the latter by p1 , followed by the subtraction of the corresponding results, gives p2 · ( A · p1 ) − p1 · ( A · p2 ) = (λ1 − λ2 ) p1 · p2 . The symmetry of A renders the left-hand side zero; since λ1 = λ2 , we are left with p1 · p2 = 0, and this completes the proof of the proposition. Theorem 1.2 (Cayley–Hamilton) Every tensor A ∈ Lin satisfies its own characteristic equation, (1.189) A3 − I A A2 + I I A A − I I I A I = O. This is called the Cayley-Hamilton equation. For particular tensors, in addition to (1.189), some simpler equations can be shown to hold. In the case of a tensor A ∈ Lin for which the eigenvalues satisfy λ1 = λ2 = λ and λ3 = λ, it can be demonstrated that there exists scalars a, b ∈ R such that A2 + a A + b I = O. Furthermore, for a spherical tensor, i.e. a tensor for which λ1 = λ2 = λ3 , it turns out that it is always possible to find a scalar c ∈ R with the property that A + c I = O. By using Theorem 1.2, any integral powers higher than the second of any A ∈ Lin can be expressed as tensor polynomials of the second degree; the key observation is to use (1.189) to obtain An = I1 An−1 − I2 An−2 + I3 An−3 ,

n ≥ 3,

which can then be applied recursively. The next example illustrates this calculation strategy

1.19 Eigenvalues

75

A5 = (A3 )(A2 ) = (I1 A2 − I2 A + I3 I) A2 = I1 ( A3 ) A − I2 A3 + I3 A2 = (I1 A − I2 I)(I1 A2 − I2 A + I3 I) + I3 A2 = · · · = α A2 + β A + γ I, where α, β, γ are functions of the principal invariants, α := I13 − 2I1 I2 + I3 ,

β := I22 + I1 I3 − I12 I2 ,

γ := I3 I12 − I2 I3 ,

and the dots stand for routine calculations that are omitted. This observation holds for arbitrary tensor functions f ( A), not necessarily some integral power like f ( A) = A5 . For simplicity we shall assume that the eigenvalues of A are distinct—the general case requiring only minor modifications. The secondorder polynomial P(Z ) = α0 + α1 Z + α2 Z 2 with the property that P(λ j ) = f (λ j ), j = 1, 2, 3 is uniquely determined by this assumption. Moreover, f (λ) = P(λ) + (λ − λ1 )(λ − λ2 )(λ − λ3 )Q(λ), where Q(λ) is a function such that Q(λ j ) = 0 for j = 1, 2, 3. Letting λ → A and using Theorem 1.2 for the last term on the right-hand side above, we discover that B := f ( A) = P( A); hence, f ( A) = α0 I + α1 A + α2 A2 ,

(1.190)

where α0 , α1 , α2 depend on λ j ( j = 1, 2, 3) as can easily be seen from the Lagrange interpolation polynomial P(Z ) =

(Z − λ3 )(Z − λ1 ) (Z − λ2 )(Z − λ3 ) f (λ1 ) + f (λ2 ) (λ1 − λ2 )(λ1 − λ3 ) (λ2 − λ3 )(λ2 − λ1 ) +

(Z − λ1 )(Z − λ2 ) f (λ3 ). (λ3 − λ1 )(λ3 − λ2 )

In conclusion, letting Z → A and using (1.190), f ( A) =

( A − λ2 I) · ( A − λ3 I) ( A − λ3 I) · ( A − λ1 I) f (λ1 ) + f (λ2 ) (λ1 − λ2 )(λ1 − λ3 ) (λ2 − λ3 )(λ2 − λ1 ) +

( A − λ1 I) · ( A − λ2 I) f (λ3 ). (1.191) (λ3 − λ1 )(λ3 − λ2 )

Equation (1.191) is known as the Lagrange–Sylvester equation; it makes clear the remarkable fact that the dependence of B ≡ f ( A) on f is only through the linear dependence on f (λ j ), j = 1, 2, 3. With a bit of more effort (1.191) can be extended to the case when not all of the λ j ’s are distinct.

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1 Vector, Tensors, and Related Matters

If λ1 = λ2 , say, then by using a limiting process it is found that the right-hand side of (1.191) will depend linearly on f (λ1 ), f  (λ1 ) and f (λ3 ). If λ1 = λ2 = λ3 , similar limiting arguments show that the dependence will be on f (λ1 ), f  (λ1 ), and f  (λ1 ). These calculations show that the determination of non-linear functional relations between tensors is reduced to the determination of the values of the function under consideration at the points which correspond to the roots of the characteristic equation of the tensors in question. If the roots are multiple, the values of derivatives of the function considered also have to be known at the points corresponding to the multiple roots. An obvious, but noteworthy consequence of (1.190), is that if the characteristic equation PA (λ) = 0 has multiple roots then PB (λ) = 0 also has multiple roots; if the tensor A is spherical, the B must be also spherical. If A ∈ Skw then, in general, B∈ / Skw, but ‘symmetry’ is obviously a tensorial property that is preserved.

1.20 Some Important Theorems Earlier in this chapter it was shown that, relative to a given basis in V , a secondorder tensor tensor is characterised by a 3 × 3 matrix, whose entries represent its components. The judicious choice of a basis often can simplify the form of that matrix and reduce the complexity of various algebraic manipulations. Thus, it is desirable to find a basis for V such that the component matrix of A with respect to it is as simple as possible. Fortunately, this is entirely analogous to similar situations in Linear Algebra that provides powerful theorems in this direction. Here, we confine ourselves to stating a particular case that is most relevant for our immediate needs, and refer to specialised Linear Algebra texts for more general theory. Theorem 1.3 (Spectral Representation Theorem) If A ∈ Sym then there is an orthonormal basis for V consisting entirely of eigenvectors of A. Moreover, for any such basis { pi }, the corresponding eigenvalues λi (i = 1, 2, 3) form the entire spectrum of A, and A = λ1 p 1 ⊗ p 1 + λ 2 p 2 ⊗ p 2 + λ 3 p 3 ⊗ p 3 .

(1.192)

Conversely, if A has the form (1.192) with { pi } an orthonormal system of vectors then λi (i = 1, 2, 3) correspond to the eigenvalues of A with pi (i = 1, 2, 3) being a set of associated eigenvectors. The vectors pi (i = 1, 2, 3) are called the principal directions or axes of A, corresponding to the eigenvalues (or principal values) λi (i = 1, 2, 3). Let us note that the decomposition (1.192) can be written as A = A1 · A2 · A3 ,

(1.193)

1.20 Some Important Theorems

77

where Ak := λk ek ⊗ ek +



ei ⊗ ei

(k = 1, 2, 3),

i=k

the order of the factors in (1.193) being immaterial. Geometrically, the above decomposition means that a symmetric linear transformation is described by three independent expansions (or contractions) along the three perpendicular axes determined by the principal directions { pi } (i = 1, 2, 3). Consider, for instance, A3 ≡ p1 ⊗ p1 + p2 ⊗ p2 + λ3 e3 ⊗ e3 . If u ∈ V , u = ui pi and it can be seen that A3 · u is such that the first two components of u remain unaltered. The third component, u3 , is either stretched or compressed by a factor equal to λ3 ; for λ3 > 1 there is elongation, while λ3 < 1 corresponds to compression. If λ1 = λ2 = λ3 then the set of orthonormal vectors p j ( j = 1, 2, 3) is uniquely determined (to within a reversal of directions and relabelling of vectors), so in this case A has exactly three principal directions. If exactly two eigenvalues coincide, for instance, λ2 = λ3 = λ and λ1 = λ2 , then every vector in the plane π determined by p2 and p3 is an eigenvector of A. In other words, there is one principal direction corresponding to λ1 and infinitely many principal directions perpendicular to the direction associated to λ1 . To see this, let a be a vector in this plane, so (∃) α, β ∈ R such that a = α p2 + β p3 . Then A · a = α( A · p2 ) + β( A · p3 ) = λ(α p2 + β p3 ) = λa, which proves our assertion. Therefore, any pair of orthogonal vectors lying in π can be chosen as p2 and p3 . In this case, A represents a homothetic transformation with coefficient λ in a plane perpendicular to p1 , combined with an expansion with coefficient λ1 along the p1 -axis. If λ1 = λ2 = λ3 = λ then A = λI, i.e. A is a homothetic transformation with coefficient λ in the whole space. In this case, any three orthonormal vectors can be chosen as the basis { pi } (i.e., any direction in space is a principal direction of such a tensor). Many tensors in Continuum Mechanics are symmetric. One of the merits of the Spectral Representation Theorem is that it affords significant simplifications in manipulating such tensors. For instance, the expressions of the principal invariants for a symmetric tensor are particularly easy to remember. If λ1 , λ2 , λ3 denote its eigenvalues then, from (1.177), (1.179) and (1.180), we obtain I 1 = λ1 + λ2 + λ3 ,

I 2 = λ1 λ2 + λ2 λ3 + λ3 λ1 ,

I 3 = λ1 λ2 λ3 .

Another example is provided by the inverse of a tensor A ∈ Sym that admits the spectral representation 3  A= λi pi ⊗ pi . (1.194) i=1

The inverse of a tensor represented in an arbitrary basis is difficult to calculate in general, but if the tensor happens to be invertible and we know (1.194), then this inverse is simply given by

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1 Vector, Tensors, and Related Matters

A−1 =

3 

λi−1 pi ⊗ pi ;

i=1

the conditions A−1 · A = A · A−1 = I are easily checked with the help of (1.57a). Definition 1.34 Two symmetric second-order Cartesian tensors are said to be coaxial if their eigenvectors coincide in some order. The following result characterises when this happens. Proposition 1.10 A, B ∈ Sym are co-axial if and only if A · B = B · A. For the direct statement: if the two tensors are co-axial then by Theorem 1.3 we can write 3 3   A= λi pi ⊗ pi and B= μj pj ⊗ pj, i=1

j=1

where { pi } is the common set of eigenvectors of A and 3B. Invoking the properties λi μi ( pi ⊗ pi ). of the tensor product ‘ ⊗ ’, we find A · B = B · A = i=1 The converse implication is not difficult either. We assume the commutativity of the two tensors, A · B = B · A, and aim to show that they have the same eigenvectors (not necessarily in the same order). To this end, let p ∈ V be an arbitrary eigenvector of B with the corresponding eigenvalue μ, i.e. B · p = μ p. In view of the assumed commutativity, ( A · B) · p = (B · A) · p, or A · (B · p) = B · ( A · p). This implies that μ( A · p) = B · ( A · p), and therefore A · p must be proportional to p, i.e. (∃) λ ∈ R such that A · p = λ p. In other words, p must be an eigenvector of A. The proof is completed by noticing that p was arbitray, so we can repeat the above arguments for the remaining eigenvectors of B. Theorem 1.4 (Square-Root Theorem) Let S ∈ Sym be a positive-definite tensor. Then, there is a unique positive-definite tensor U ∈ Sym such that U 2 = S. We formally write S1/2 or

√

S for the tensor U.9

The justification of this important result hinges on Theorem 1.3. Since S is symmetric, we may write it in the spectral form S=

3 

λi pi ⊗ pi ,

i=1

where λi (i = 1, 2, 3) are the (real) eigenvalues of S and { pi } represent the corresponding unit eigenvectors. The tensor S is also positive definite and, according to 9 As

the notation suggests, U represents the square root of S.

1.20 Some Important Theorems

79

Proposition 1.9, λi > 0 (i = 1, 2, 3). We can therefore define a second-order tensor U such that 3  % λi pi ⊗ pi . U := i=1

Our claim is that U defined above is positive definite, symmetric, and U 2 = S. The symmetry and the positive-definite nature of U are obvious; for instance,  U = T

3  %

T λi pi ⊗ pi

=

i=1

3  %

λi ( pi ⊗ pi ) = T

i=1

3  %

λi pi ⊗ pi = U,

i=1

which confirms the former property. A direct calculation also gives U 2 = U · U, 3 3   % U = λi λ j ( pi ⊗ pi ) · ( p j ⊗ p j ) 2

i=1 j=1 3  3  3 3   % % λi λ j ( pi · p j )( pi ⊗ p j ) = λi λ j δi j ( pi ⊗ p j ) = i=1 j=1

=

3 

i=1 j=1

λi ( pi ⊗ pi ) = S.

(1.195)

i=1

To prove the uniqueness we assume that we can find another tensor, U ∗ (say), with the properties mentioned in the statement of this theorem. Both U and U ∗ commute with S; for example, U ∗ · S = U ∗ · U 2∗ = U 3∗ = U 2∗ · U ∗ = S · U ∗ , and so on. By Proposition 1.10, these two pairs of co-axial tensors have the same eigenvectors. If μi (i = 1, 2, 3) are the eigenvalues of U ∗ then U ∗ · pi = μi pi (no sum) and leftmultiplying this equation by U ∗ we find λi pi = S · pi = U 2∗ · pi = U ∗ · (U ∗ · pi ) = μi2 pi , √ whence μi = λi for i = 1, 2, 3. Thus, the tensor U ∗ has the same eigenvalues and eigenvectors as U, which concludes the uniqueness of the square-root tensor. It is important to realise that in order to calculate explicitly the square root of a given tensor, we first need to bring it to its diagonal form; in other words, we must find its spectral representation. Theorem 1.5 (Polar Decomposition Theorem) An arbitrary invertible tensor A ∈ Lin can be decomposed into the following two unique forms A = R · U = V · R,

(1.196)

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1 Vector, Tensors, and Related Matters

where R is an orthogonal tensor and U, V ∈ Sym are positive-definite tensors. The products R · U and V · R are, respectively, called the right polar decomposition and the left polar decomposition of A. Before we sketch the proof of this remarkable theorem, a word about terminology is in order. Every complex number can be written in polar form as z = r eiθ = a + ib, (a, b ∈ R), where r = |z| is a non-negative real number and θ is its argument (0 ≤ θ < 2π ), with |eiθ | = 1. Thus, z is decomposed into a stretching factor r and a rotation factor eiθ . The Polar Decomposition Theorem A = V · R is an analogous decomposition for invertible elements of Lin. The price we pay for reversing the order of the two factors is that V must be replaced by another ‘stretching’ transformation, U, because tensor multiplication is not commutative. The tensors A · AT and AT · A are symmetric and positive definite. Hence, by Theorem 1.4, there exist unique positive-definite tensors U , V ∈ Sym such that V 2 = A · AT and U 2 = AT · A. Now define R := A · U −1 . This new tensor is orthogonal (see Definition 1.12), R T · R = ( A · U −1 )T · ( A · U −1 ) = U −1 · ( AT · A) · U −1 = U −1 · U 2 · U −1 = I. One can check in exactly the same way, R · R T = I. Evaluating the determinant det R = det( A · U −1 ) = (det A) (det U)−1 , we discover that R is a proper or improper orthogonal tensor, depending on whether det A > 0 or det A < 0, respectively. Since U is unique, R must also be unique, and hence A = R · U constitutes the unique representation of A as the product between an orthogonal tensor and a symmetric positive-definite one. Following exactly the same arguments, we also find that A = V · P has the same uniqueness property as above, where P := V −1 · A is also an orthogonal tensor. In other words, both the right and the left polar decompositions are unique; it only remains to show that P = R. Since P is an orthogonal tensor and V has a square root, we can write A= R·U

    = I · A = P · P T · (V · P) = P · P T · V · P = P · {(V 1/2 · P)T · (V 1/2 · P)}.

This equation suggests that both R · U and P · ( P T · V · P) represent the right polar decompositions of the same tensor. Thus, R = P and U = R T · V · R, in view of the uniqueness property of the right polar decomposition. The next result complements the statement of this theorem by providing more detailed information about the relationship between U and V . Proposition 1.11 Within the context of the Polar Decomposition Theorem, if U has eigenvalues λi and eigenvectors pi (i = 1, 2, 3), then λi > 0 and these values are also the eigenvalues of V with the corresponding eigenvectors q i := R · pi .

1.20 Some Important Theorems

81

The property λi > 0 follows from the symmetry and positive definiteness of U. In addition, V · (R · pi ) = V · R · pi = R · U · pi = R · (U · pi ) = R · (λi pi ) = λi (R · pi ), that is, V · q i = λi q i , which confirms that q i (i = 1, 2, 3) are the eigenvectors of V .

1.21 Projections in Lin Before continuing with introducing more fundamental concepts about tensors, we take time out to revisit the projection tensors mentioned briefly in Sect. 1.6. Their effect on vectors has already been clarified; our interest now is in showing how they can be used to extract lower dimensional tensors from the elements of Lin. The second-order tensors introduced earlier in this chapter can justly be regarded as three-dimensional since dim V = 3, and this is reflected by the size of their component matrix. By contrast, two-dimensional second-order tensors correspond to the case in which dim V = 2; their component matrix would then be 2 × 2. Most of the concepts introduced in the previous sections carry over to such tensors and will not be pursued here any further. Our interest is in a somewhat different matter: given a three-dimensional tensor A, how can we ‘extract’ a two-dimensional tensor from it? For instance, if this tensor has the component matrix [Ai j ] in the standard basis {ei }, ⎡

⎤ A11 A12 A13 ⎢ ⎥ [ A] = ⎣ A21 A22 A23 ⎦ , A31 A32 A33 picking up the 2 × 2 block in the upper-left corner would presumably give us the component matrix of a two-dimensional tensor. However, this is somewhat unsatisfactory as it depends on a choice of basis in V . It is possible to achieve the same result without involving any specific basis, as shown next. To this end, a particular case will be considered first. Let N(e3 ) ≡ Proj ⊥ e3 = I − e3 ⊗ e3 be the normal projection (1.47) corresponding to n → e3 , and note its effect on A, ⎡ ⎡ ⎤ ⎤ A11 A12 A13 A11 A12 0 [N(e3 ) · A] = ⎣ A21 A22 A23 ⎦ , [ A · N(e3 )] = ⎣ A21 A22 0⎦ . 0 0 0 A31 A32 0 Left multiplication (i.e., N(e3 ) · A) appears to ‘clear’ the last row, while right multiplication (i.e., A · N(e3 )) ‘clears’ the last column. This is not accidental; in fact, if

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1 Vector, Tensors, and Related Matters

N(ei ) ≡ I − ei ⊗ ei (no summation), then left multiplication by this projection tensor will ‘clear’ the ith row, whereas right multiplication does the same to the ith column. It is, therefore, to be expected that if we consider the concerted action of these two multiplication operations the outcome will be precisely the two-dimensional block in the upper-left corner of [ A]. For the general case, assume that n is a given unit vector and take N ≡ I − n ⊗ n; the tensor A ∈ Lin can then be represented in the form A = Aint + s(1) ⊗ n + n ⊗ s(2) + κ(n ⊗ n),

(1.197)

where Aint := N · A · N, s

(1)

:= N · A · n,

κ := A : (n ⊗ n), s

(2)

(1.198a)

:= n · A · N.

(1.198b)

In the representation (1.197), Aint can be regarded as a two-dimensional tensor and will be referred to as the interior part of A (relative to n). Likewise, s(i) (i = 1, 2) will be regarded as vectors, and κ as a scalar. If m and k are unit vectors orthogonal to n and such that {m, k, n} form a right( j) ( j) handed basis in V , then [ Aint ] := [A†αβ ] (α, β = 1, 2) and s( j) = s1 m + s2 k ( j = 1, 2) or, putting everything together ⎡ [ A] = ⎣

(1)

[ Aint ]

[s ]

[s(2) ] T

k

⎤

⎡

A†11

⎢ † ⎦≡⎢ ⎢ A21 ⎣ s1(2)

A†12 A†22 s2(2)

s1(1)

⎤

⎥ s2(1) ⎥ ⎥ . ⎦ κ

1.22 Vector and Tensor Calculus With a few exceptions, the vectors and tensors considered so far were assumed to be constant; however, in Continuum Mechanics this is seldom the case. For example, the position vector of a moving particle will be a vector which depends on time; the deformation experienced by an elastic solid subjected to external loads is a tensor that, in general, assumes different values at different locations within such a deformable solid. In this section, we want to review briefly some relevant material from standard courses on Vector Calculus, in order to introduce a handful of new definitions and concepts related partly to Tensor Calculus.

1.22 Vector and Tensor Calculus

83

We shall start with the simplest case, in which vectors or tensors are allowed to depend on just one independent variable (e.g. a time-like variable ‘ t’). Let I ⊂ R be an open interval, and consider a mapping u : I → V , which associates to any t ∈ I a vector u(t) ∈ V . We shall agree to refer to such a mapping as a vector function. If {ei } denotes a fixed orthonormal basis in V , then such a function is described by three real-valued functions u j (t), j = 1, 2, 3—the components of u relative to the given basis, i.e., u(t) = u j (t)e j . The continuity or differentiability of such a vector function is equivalent to its components sharing the same property. For instance, u(t) is continuous at t0 ∈ I ⇐⇒ u j (t) are continuous at t0 ∈ I for j = 1, 2, 3.

(1.199)

The product rule for the differentiation of real-valued functions admits a natural extension when the product is of a more general nature. If u ≡ u(t) and v ≡ v(t) d (u • v) = dt



du dt



 •v+u•

 dv , dt

(1.200)

where the ‘bullet’ can be any of the products introduced so far. It is possible to have an expression involving several terms and different types of products. As a simple example, consider the product rule for the differentiation of a scalar triple product containing three vectors u, v, w depending on some parameter t, ! " ! " ! " du dv dw d [u, v, w] = , v, w + u, , w + u, v, . dt dt dt dt The above formula is obtained by the repeated application of (1.200). Recalling (1.26) and (1.27), this also leads immediately to the rule for differentiation of a 3 × 3 determinant. The case of a tensor function, that is, a mapping A which assigns to each t ∈ I a tensor A(t) ∈ Lin n is entirely similar to what has been said above. For the sake of brevity, let us consider n = 2; such a tensor will then admit the representation A(t) = Ai j (t)ei ⊗ e j with respect to the fixed orthonormal basis {ei }. By analogy with (1.199) the differentiability (say) of this tensor function is equivalent to the differentiability of each of its components Ai j (t) (1 ≤ i, j ≤ 3). The product rule continues to hold in a number of different variations like, for example, d dA du ( A ⊗ u) = ⊗u+ A⊗ , dt dt dt

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1 Vector, Tensors, and Related Matters

where here A ≡ A(t) ∈ Lin and u ≡ u(t) ∈ V for all t ∈ I . It also remains true for contracted products; if C(t) ∈ Lin 4 and A(t), B(t) ∈ Lin, then   ! " d dA d C[ A] = C [ A] + C , dt dt dt   ! " ! " d d dA dB C[ A, B] = C [ A, B] + C , B + C A, . dt dt dt dt In Continuum Mechanics, various physical quantities such as the stress and strain tensors depend on position as well as on time. The latter dependence can be ignored in a large class of problems known as static, but the dependence on the spatial coordinates is the very essence of the mathematical models studied in the following chapters. Definition 1.35 Let Ω ⊂ E3 be an open10 region; • a scalar field Φ is a mapping that assigns a scalar Φ(x) ∈ R to each point x ∈ Ω, Φ : x ∈ Ω → Φ(x) ∈ R; • a vector field u on Ω corresponds to the mapping u : x ∈ Ω → u(x) ∈ V ; • a second-order tensor field A on Ω represents a mapping A : x ∈ Ω → A(x) ∈ Lin. More general tensor fields can easily be imagined (e.g. fourth-order tensor fields). Relative to a fixed orthonormal basis vector and tensor fields have components that depend on position (x ∈ Ω). The continuity and differentiability properties of such fields can then be phrased in terms of their components with respect to such a fixed basis. Although not immediately obvious, it can be shown that ‘continuity’ and ‘differentiability’ are properties independent of the choice of basis. The partial derivatives with respect to xk (k = 1, 2, 3) of u(x) and A(x) above are defined in the natural way, e.g. ∂u ∂ui = ei , ∂ xk ∂ xk

∂ Ai j ∂A = ei ⊗ e j . ∂ xk ∂ xk

(1.201)

The product rule and everything else that was said about vector and tensor functions remain valid for the corresponding fields as well. P ∈ E3 and ε > 0, we say that an ε-neighbourhood of P is the set of all points that are less than ε in distance away from P. An interior point of a given region Ω ⊂ E3 is one which has an ε-neighbourhood that lies completely in that region; Ω is said to be open if every point in that region is an interior point.

10 If

1.22 Vector and Tensor Calculus

85

The classical notation for partial derivatives employed in (1.201) is poorly suited for performing long calculations—that is why a different notation is commonly used in Continuum Mechanics. Not only is this new notation simpler, but it can also be incorporated naturally into the index notation. Henceforth, we shall adhere to the following rule: differentiation with respect to a coordinate xi is shown by an index ‘ i’ preceded by a comma. Note that the comma does not interfere with other rules of index notation; in particular, the summation convention applies across the comma as well. For example, if φ ≡ φ(x) is a scalar field, then φ, i

stands for

∂φ , i = 1, 2, 3 ∂ xi

(any of the partial derivatives indicated), and φ, ii

corresponds to

∂ 2φ ∂ 2φ ∂ 2φ ∂ 2φ ≡ + + . ∂ xi ∂ xi ∂ x12 ∂ x22 ∂ x32

If A ≡ A(x) = Ai j (x) ei ⊗ e j then Ai j, i

is the same as

Ai j, 1 + A2 j, 2 + A3 j, 3 or

∂ A1 j ∂ A2 j ∂ A3 j + + . ∂ x1 ∂ x2 ∂ x3

1.22.1 Differential Calculus in Finite-Dimensional Euclidean Vector Spaces In preparation for the remaining sections of this chapter, several fundamental ideas from Vector Calculus will be reviewed at this juncture. For full generality, the discussion will be placed within the framework of finite-dimensional Euclidean point spaces and their associated vector spaces (e.g. see [5, 6]). As already pointed out previously, Euclidean vector spaces can be identified with Rn , for some suitable n ∈ N. It is also recalled that the elements of En can be viewed either as points in this space or as position vectors of such points, i.e. vectors in Rn . In the interest of keeping the presentation at a low level, we are not going to distinguish explicitly between En and Rn ; we take the latter as being a set whose elements can be viewed as either points or vectors. With these caveats in mind, we introduce the notations Z1 ≡ Rn ,

Z2 ≡ Rm ,

Z3 ≡ R p

(n, m, p ∈ N),

and let  · 1 and  · 2 denote the magnitudes of elements in Z1 and Z2 , respectively. In this section the set of all linear mappings between two of the above spaces, Z1 and Z2 (say), will be denoted by L(Z1 , Z2 ).

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1 Vector, Tensors, and Related Matters

Consider a mapping f : Ω ⊂ Z1 → Z2 , with Ω an open set, and let x 0 ∈ Ω. If there exists a linear mapping A ∈ L (Z1 , Z2 ) such that lim

h→0

 f (x 0 + h) − f (x 0 ) − Ah2 = 0, h1

(1.202)

then we say that f is differentiable at x 0 , and we write f  (x 0 ) := A. In this formula, x 0 is regarded as a point in Z1 , while h represents a vector; adding a vector (h) to a point (x 0 ) we get another point, x 0 + h, which is in the close proximity of the original one since h → 0 (see Sect. 1.4). The expression f (x 0 + h) − f (x 0 ) − Ah represents a vector in Z2 , irrespective of whether the images of f are regarded as points in Z2 or as vectors (because the difference of two points in a Euclidean point space is a vector). If f is differentiable at every x ∈ Ω, we say that f is differentiable on Ω. The derivative defined in (1.202) is sometimes called the differential of f at x 0 , and Ah is often written11 d f (x 0 )[h] or f  (x 0 )[h]. It is important to note that if f is differentiable on Ω then, for each fixed x ∈ Ω, f  (x) represents a linear mapping from Z1 to Z2 , i.e. f  (x) ∈ L (Z1 , Z2 ); but f  is also a mapping (non-linear, in general) that maps Ω into L (Z1 , Z2 ). If this latter mapping is continuous, then we say that f is continuously differentiable on Ω. The relation (1.202) can be written in the equivalent form f (x 0 + h) = f (x 0 ) + f  (x 0 )h + R(h),

(1.203)

where lim h→0 R(h)2 /h1 = 0 (i.e., R(h) is infinitesimally small in Z2 when 0 < h1  1). Equation (1.203) may be interpreted as saying that the derivative f  (x 0 ) provides a linear approximation for the function f at points in the vicinity of x 0 . From this relation, it also follows at once that f is continuous at any points at which it is differentiable. The second-order derivative is obtained as the derivative of the derivative in (1.202). Assume that f  (x) ∈ L(Z1 , Z2 ) exists at all points x in a close vicinity of x 0 ∈ Ω; if a bilinear mapping B : Z1 × Z1 → Z2 can be found such that  f  (x 0 + h1 )[h2 ] − f  (x 0 )[h2 ] − B[h1 , h2 ]2 = 0, h1 →0 h1 1 lim

(1.204)

for all h2 ∈ Z1 , then f is twice differentiable at x 0 and we write f  (x 0 ) := B or d2 f (x 0 ) := B. Higher order derivatives are defined easily by the same iterative procedure: the derivative of order p is the derivative of the derivative of order ( p − 1), and this will be a p-linear mapping from Z1 × Z1 × · · · × Z1 to Z2 . 11 We use square brackets to indicate that h f  (x0 ) is a non-linear mapping in general.

→ f  (x 0 )[h] is a linear mapping; by contrast, x 0 →

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The structure of the linear mapping A ∈ L(Z1 , Z2 ) in (1.202) is intimately connected to the so-called partial derivatives of the original function. For f : Ω ⊂ Z1 → R, the partial derivative with respect to the jth variable, and evaluated at x 0 ∈ Ω, corresponds to the limit f (x 0 + t e j ) − f (x 0 ) ∂f (x 0 ) := lim , t→0 ∂x j t

j = 1, 2, . . . , n,

(1.205)

provided that the right-hand side exists. Here, e j is the element of the standard basis in Z1 , whose jth entry is 1 and all the others are zero. This definition makes sense for vector-valued functions as well. If f (x) = f i (x)ei , then ∂f ∂ fi (x 0 ) := (x 0 )ei ≡ f i, j (x 0 )ei , ∂x j ∂x j

j = 1, 2, . . . , n,

i.e., the partial derivative of f with respect to x j is obtained by taking the partial derivatives with respect to x j of the components of that function. The Jacobian matrix of such a vector-valued function of several variables is the matrix whose columns are the vectors f, j for j = 1, 2, . . . , n, ⎡

⎤ f 1, 1 (x 0 ) f 1, 2 (x 0 ) . . . f 1, n (x 0 ) ⎢ f 2, 1 (x 0 ) f 2, 2 (x 0 ) . . . f 2, n (x 0 ) ⎥ ⎢ ⎥ J ( f )(x 0 ) = ⎢ ⎥. .. .. .. ⎣ ⎦ . . . f m, 1 (x 0 ) f m, 2 (x 0 ) . . . f m, n (x 0 )

Theorem 1.6 If there is a linear transformation A such that (1.202) is satisfied then all partial derivatives of f at x 0 exists, and the matrix representing A is the Jacobian matrix defined above. In particular, such a linear transformation is unique. It is possible for all partial derivatives of a vector field to exist and yet the Jacobian matrix not to be its derivative. The next result clarifies when we can rely on partial derivatives to find the derivative. Theorem 1.7 If Ω is an open subset of Z1 and f : Z1 → Z2 is a mapping such that all partial derivatives of f exist and are continuous on Ω, then f is differentiable on Ω and its derivative is given by the Jacobian matrix J ( f ). Rules for manipulating derivatives: Unless otherwise specified, the functions that are mentioned below are supposed to be differentiable at x 0 ∈ Ω ⊂ Z1 . 1. If f : Z1 → Z2 is linear, then it is differentiable everywhere, and its derivative at all points x 0 is f , i.e. d f (x 0 )[h] = f (h),

(∀) h ∈ Z1 .

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2. If φ : Ω → R and f : Ω → Z2 , then φ f is differentiable at x 0 ; if in addition φ(x 0 ) = 0, then f /φ is also differentiable at x 0 . In either case, for all h ∈ Z1 d(φ f )(x 0 )[h] = φ(x 0 )d f (x 0 )[h] + (dφ(x 0 )[h]) f (x 0 ), !   " f 1 1 d (x 0 ) [h] = d f (x 0 )[h] − 2 (dφ(x 0 )[h]) f (x 0 ). φ φ(x 0 ) φ (x 0 ) 3. Let Ω1 ⊂ Z1 and Ω2 ⊂ Z2 be open sets, let g : Ω1 → Ω2 , f : Ω2 → Z3 be mappings, and let x 0 ∈ Ω1 . If g is differentiable at x 0 and f is differentiable at g(x 0 ) then the composition f ◦ g : Ω1 ⊂ Z1 → Z3 is differentiable at x 0 and d( f ◦ g)(x 0 )[h] = d f (g(x 0 ))dg(x 0 )[h],

(∀) h ∈ Z1 .

4. If f , g : Ω ⊂ Z1 → Z2 are both differentiable at x 0 , then so is the inner product f · g : Ω → R, and d( f · g)(x 0 )[h] = d f (x 0 )[h] · g(x 0 ) + f (x 0 ) · dg(x 0 )[h],

(∀) h ∈ Z1 .

The calculation of the derivatives of various functions that depend on tensors and vectors is most easily carried out by using a link between (1.202) and the so-called directional derivative. This is explained briefly next. Let h ∈ Z1 be a fixed unit vector (|h| = 1) and consider x 0 ∈ Ω ⊂ Z1 , with Ω the open set already mentioned before. The directional derivative of f at x 0 in the direction h is defined as δ f (x 0 ; h) := lim

t→0

f (x 0 + t h) − f (x 0 ) , t

(1.206)

the limit being taken in the usual sense of convergence in Z2 . It is easily seen that δ f represents a linear operation on the function f , that is, δ( f 1 + f 2 )(x 0 ; h) = δ f 1 (x 0 ; h) + δ f 2 (x 0 ; h), and it is also a homogeneous function in h, in the sense that δ f (x 0 ; αh) = αδ f (x 0 ; h) for all α ∈ R. Higher order directional derivatives can be defined as well. For instance, the second-order directional derivative of a function f : Ω → R corresponds to # $  ∂2 f (x 0 + sh1 + t h2 )  ∂s∂t (s, t)=(0,0) # # $ $   d d  δ f (x 0 + t h1 ; h2 )  ≡ δ f (x 0 + t h2 ; h1 )  . = dt dt t=0 t=0

δ 2 f (x 0 ; h1 , h2 ) :=

A key result that will play an important role in the calculation of derivatives in what follows is stated below.

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Proposition 1.12 Let f be a function as in (1.202). If it is differentiable at x 0 then the directional derivative at that point exists in every direction and   d d f (x 0 )[h] = δ f (x 0 ; h) ≡ f (x 0 + t h) , dt t=0

(1.207)

for all h ∈ Z1 . The converse of this statement is in general false, even if the partial derivatives (1.205) are assumed to exist at each x 0 ∈ Ω. A similar result to (1.207) holds for higher order derivatives, when they exist, d2 f (x 0 )[h1 , h2 ] = δ 2 f (x 0 ; h1 , h2 ),

(∀) h1 , h2 ∈ Z1 ,

and the pattern can be easily extended to third- and higher order derivatives. Consider now the case Z2 ≡ R and let f : Ω ⊂ Z1 → R be a differentiable function in the sense of (1.202). By invoking a classical result in Linear Algebra12 , we deduce the existence of ∇ f (x 0 ) ∈ Z1 such that d f (x 0 )[h] = h · ∇ f (x 0 ) = δ f (x 0 ; h).

(1.208)

Definition 1.36 The vector ∇ f (x 0 ) is called the gradient of f at x 0 , and is denoted also grad f (x 0 ). It should be clear that ∇ f does not depend on the choice of basis in Z1 (since neither (1.202) nor (1.206) rely on that concept). Nevertheless, it is sometimes useful to have a more explicit representation for the gradient in a particular basis. If we use the standard basis {ei } then ∇ f (x 0 ) :=

n  ∂f (x 0 )ei , ∂ xi i=1

as can be easily seen from a direct application of (1.207). If the previous mapping f is differentiable and ∇ f : Ω ⊂ Z1 → Z1 is continuously differentiable in Ω, then for each x 0 ∈ Ω, (∇ f ) (x 0 ) is a symmetric element of L(Z1 , Z1 ) and the bilinear form B in (1.204) satisfies B[h1 , h2 ] = h1 · (∇ f ) (x 0 )[h2 ],

(∀) h1 , h2 ∈ Z1 .

(1.209)

Definition 1.37 The symmetric linear mapping (∇ f ) (x 0 ) is called the Hessian of f at x 0 and in this text will be denoted by (∇ ⊗ ∇) f (x 0 ).

12 This

is known as the Riesz Representation theorem and asserts that for every linear functional T ∈ L (Z1 , R) there exists a unique a ∈ Z1 such that T [x] = a · x for all x ∈ Z1 .

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In light of (1.209) and this new definition, the second-order derivative of a twice continuously differentiable function f : Ω ⊂ Z1 → R can be cast in the form d2 f (x 0 ; h, h) = (h ⊗ h) : (∇ ⊗ ∇ f )(x 0 ) = δ 2 f (x 0 ; h, h),

(1.210)

where, relative to the standard basis in Z1 , the Hessian admits the representation (∇ ⊗ ∇ f )(x 0 ) :=

n n   ∂2 f (x 0 )ei ⊗ e j . ∂ xi ∂ x j i=1 j=1

We close this section by rewriting Taylor’s expansion theorem using the notations introduced above. Theorem 1.8 1. (Taylor’s second-order formula) Let f : Ω ⊂ Rn → R have continuous partial derivatives of third order. Assume that x 0 ∈ Ω and h ∈ Rn is such that h  1, where  ·  denotes the magnitude of elements in Rn . Then, f (x 0 + h) = f (x 0 ) + h · ∇ f (x 0 ) +

1 (h ⊗ h) : (∇ ⊗ ∇) f (x 0 ) + R2 (x 0 ; h), 2!

where R2 (x 0 ; h)/h2 → 0 as h → 0. 2. (Taylor’s third-order formula) Everything as above, except that f has continuous partial derivatives of fourth-order. Then, 1 (h ⊗ h) : (∇ ⊗ ∇) f (x 0 ) 2! . 1 + (h ⊗ h ⊗ h) ..(∇ ⊗ ∇ ⊗ ∇) f (x 0 ) + R3 (x 0 ; h), 3! (1.211)

f (x 0 + h) = f (x 0 ) + h · ∇ f (x 0 ) +

where R3 (x 0 ; h)/h3 → 0 as h → 0. In Cartesian coordinates, the component version of (1.211) is equivalent to f (x 0 + h) = f (x 0 ) +

n  i=1

hi

n ∂f ∂2 f 1  (x 0 )+ hi h j (x 0 ) ∂ xi 2 i, j=1 ∂ xi ∂ x j

+

n 1  ∂3 f hi h j hk (x 0 ) + R3 (x 0 ; h). 6 i, j,k=1 ∂ xi ∂ x j ∂ xk

The proof of (1.211) follows immediately by applying the classical Taylor’s Theorem to g : (−ε, ε) → R, g(t) := f (x 0 + t h) about t0 = 0; here, 0 < ε  1 is small enough so that x 0 + t h ∈ Ω for all t ∈ (−ε, ε).

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1.22.2 Differential Operators: DIV, GRAD, CURL We return now to our usual three-dimensional setting. The scalar, vector and tensor fields that appear below are assumed to be differentiable as many times as we like. According to the discussion in Sect. 1.22.1, the gradient of a scalar field φ(x) is a vector field that associated to each point x ∈ E3 a vector ∇φ(x) ∈ V . Relative to the standard basis, it was noted that (grad φ)(x) =

∂φ ∂φ ∂φ ∂φ(x) e i ≡ φ, i e i = e1 + e2 + e3 . ∂ xi ∂ x1 ∂ x2 ∂ x3

(1.212)

Motivated by the above representation we introduce the so-called ‘nabla’ or ‘del’ operator, ∂ ∂ ∂ ∂ ∇ := e j ≡ e1 + e2 + e3 , (1.213) ∂x j ∂ x1 ∂ x2 ∂ x3 which will be regarded as a symbolic vector operator, the result of its action on a scalar function being a vector function. Thus, the notation for the gradient ∇φ can be taken to literally mean the multiplication between the ‘vector’ ∇ and the scalar φ. Recall that the gradient of a scalar field does not depend on the chosen system of coordinates. The independence of (1.212) on the particular coordinates used in writing it can also be checked directly by using the developments of Sect. 1.9. Let { ei } be a new orthonormal basis in V . According to (1.82) the coordinates xi ) of a generic point in E3 relative to the two systems of (xi ) and, respectively, ( x p /∂ xi = Q i p . Use of this fact in coordinates are linked by  x p = Q i p xi , and hence ∂ conjunction with the chain rule allow us to write xp ∂φ ∂φ ∂ ∂φ ej) ej = (Q i j = Q i j ∂ xi ∂ xi ∂ x p ∂ xi ∂φ ∂φ ∂φ ∂φ ej ej = Q i j Q i p = (Q i p Q i j ) ej = δ pj = ep , ∂ xp ∂ xp ∂ xp ∂ xp

grad φ = ei

which establishes the independence of grad φ ≡ ∇φ on the chosen coordinates. Definition 1.38 The gradient of a vector field u is an element of Lin, denoted by grad u, and which satisfies  grad u(x) : (a ⊗ b) = b · grad [a · u(x)],

(∀) a, b ∈ V .

(1.214)

To find the component representation relative to the usual rectangular Cartesian system of coordinates, we start with the observation that the left-hand side in (1.214) is a · [grad u(x) · b. Setting a → ei and b → e j in (1.214) gives (grad u)i j = ei · (grad u) · e j = e j · grad (u · ei ) = e j · (ui, k ek ) = δ jk ui, k = ui, j .

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Thus, we have the following component representation for grad u ∈ Lin: grad u =

∂ui ei ⊗ e j ≡ ui, j ei ⊗ e j ; ∂x j

(1.215)

in the interest of brevity, we have left out the dependence on x ∈ E3 , a rule that will be applied whenever there is no risk of confusion. An alternative way to introduce the gradient of a vector field is by replacing (1.214) with   d [grad u(x)] · h = u(x + t h) , (∀) h ∈ V , dt t=0 which has the advantage of yielding the intuitive formula grad u =

∂u ⊗ ej. ∂x j

It is also instructive to look at the gradient of a vector field from the point of view of the ‘del’ operator. For instance, the result in (1.215) can be expressed as grad u = u ⊗ ∇ ≡ (∇ ⊗ u)T . One should keep in mind that the symbolic expression u ⊗ ∇ is just a notation for the transpose of ∇ ⊗ u, the latter being calculated as explained below 

∂ ∇ ⊗ u = ej ∂x j = e j ⊗ ei

 ⊗ (ui ei ) = e j ⊗

∂ (ui ei ) ∂x j

∂ui = ui, j e j ⊗ ei , ∂x j

and use of the rule (1.54) gives u ⊗ ∇ ≡ (∇ ⊗ u)T = ui, j (e j ⊗ ei )T = ui, j ei ⊗ e j . Note that ∂/∂ x j being scalar operators pass by ⊗ just as normal scalars do. We have also used the fact that the basis vectors e j did not depend on position so they, too, behaved like constants. Definition 1.39 The divergence of a vector field u is a scalar field, denoted by div u, and represents the trace of its gradient; more precisely, div u = tr(grad u) = tr((grad u)T ) ≡ (∇ ⊗ u) : I, where I is the second-order identity tensor.

(1.216)

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The representation of the divergence in Cartesian coordinates follows immediately from (1.215) (1.217) div u = ui, i or, by using the ‘del-dot’ operator  ∂ ∂ · (ui ei ) = e j · (ui ei ) ∇ · u = ej ∂x j ∂x j ∂ui = δi j ui, j = ui, i . = e j · ei ∂x j 

Definition 1.40 The curl or rotation of a vector field u is a new vector field, denoted by curl u, and represents the vector of its gradient; more precisely,  (curl u) ∧ a = grad u − (grad u)T · a,

(∀) a ∈ V .

(1.218)

The Cartesian-component representation of the curl is easily extracted from (1.218). To this end, we write curl u = (curl u) p e p and set a → e j in the definition, with the outcome (curl u) p (e p ∧ e j ) = (grad u) · e j − (grad u)T · e j . Taking the dot product of this equation with ek and remembering that [ek , e p , e j ] = [e p , e j , ek ] = ∈ pjk , with the help of (1.215) we obtain (curl u) p ∈ pjk = uk, j − u j, k .

(1.219)

Finally, to eliminate the ‘rogue’ permutation symbol on the left-hand side, this equation is multiplied by ∈i jk keeping in mind that ∈ pjk ∈i jk = 2δi p ; noting further that ∈i jk u j, k = − ∈i jk uk, j , Eq. (1.219) yields (curl u)i = ∈i jk uk, j , and thus curl u = ∈i jk

∂uk ei ≡ ∈i jk uk, j ei ; ∂x j

the same conclusion is reached by employing the ‘del-cross’ operator, 

∂ ∇ ∧ u = ej ∂x j = e j ∧ ek

 ∧ (uk ek ) = e j ∧

∂ (uk ek ) ∂x j

∂uk = ∈ jki ei uk, j = ∈i jk uk, j ei . ∂x j

(1.220)

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Definition 1.41 If φ is a scalar field, the Laplacian ∇ 2 φ is defined as ∇ 2 φ = ∇ · (∇φ).

(1.221)

In Cartesian coordinates, this scalar field becomes ∇ 2 φ = φ, ii =

∂ 2φ ∂ 2φ ∂ 2φ + + . ∂ x12 ∂ x22 ∂ x32

If u is a vector field, its Laplacian (∇ 2 u) will be the new vector field ∇ 2 u = ∇ · (∇ ⊗ u),

(1.222)

with the following Cartesian coordinate representation: ∇ 2 u = (uk ek ), j j = uk, j j ek ≡ (∇ 2 uk )ek . The operations of gradient, divergence and curl can be introduced for general tensors of order n. Here we restrict our exposition to some more modest results involving only second-order tensors. Definition 1.42 The divergence of a second-order tensor field A is a vector field, denoted by div A, and defined according to (div A) · a = div ( A · a),

(∀) a ∈ V .

(1.223)

Note that the right-hand side involves the divergence of a vector field, A · a, which we already know how to calculate (cf. Definition 1.39). By choosing a → e j in (1.223) it follows immediately that (div A) j = Ai j, i , and hence we find the explicit representation of the divergence div A =

∂ Ai j e j ≡ Ai j, i e j . ∂ xi

(1.224)

The same result can be obtained with the formalism of the del-dot operator   ∂ · (Ai j ei ⊗ e j ) ∇ · A = ek ∂ xk ∂ (Ai j ei ⊗ e j ) = ek · (ei ⊗ e j )Ai j, k = ek · ∂ xk = Ai j, k (ek · ei )e j = Ai j, k δik e j = Ai j, i e j . It is important to draw the attention to a different definition of div A that is commonly used in the literature, and which corresponds to

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A · ∇ :=

∂ Ai j ei ≡ Ai j, j ei . ∂x j

(1.225)

Definition 1.43 The gradient of a second-order tensor field A is a third-order tensor field, denoted by grad A, and which is defined by   d (grad A) · h = A(x + t h) , dt t=0

(∀) h ∈ V .

(1.226)

The condition (1.226) is adopted to preserve the consistency with the gradient of a vector field. Some authors swap the place of the two terms on the left-hand side, i.e., they consider h · (grad A) while leaving the right-hand side unchanged. It is a simple matter to show that in Cartesian coordinates the condition (1.226) leads to A⊗∇ ≡

∂A ⊗ ek = Ai j, k ei ⊗ e j ⊗ ek , ∂ xk

(1.227)

whereas the modified one produces ∇ ⊗ A ≡ ek ⊗

∂A = Ai j, k ek ⊗ ei ⊗ e j . ∂ xk

(1.228)

Definition 1.44 The Laplacian of a second-order tensor field A is a new tensor field, denoted by ∇ 2 A, and assumed to satisfy  (∇ 2 A) : (a ⊗ b) = ∇ 2 A : (a ⊗ b) ,

(∀) a, b ∈ V .

(1.229)

In Cartesian coordinates, this tensor field admits the representation ∇ 2 A = Ai j, kk ei ⊗ e j .

(1.230)

Definition 1.45 The curl of a second-order tensor field A is another second-order tensor field, denoted by ∇ ∧ A, and which is defined by (∇ ∧ A) · a = ∇ ∧ ( A · a),

(∀) a ∈ V .

(1.231)

Since this ‘curl’ is a second-order tensor field, ∇ ∧ A = (∇ ∧ A)i j ei ⊗ e j and the components are given by (∇ ∧ A)i j = ei · (∇ ∧ A) · e j . By using (1.231) we find  (∇ ∧ A) · e j = ∇ ∧ ( A · e j ) = ∇ ∧ (An j en ) = An j, m em ∧ en , and then (∇ ∧ A)i j = ei · (∇ ∧ A) · e j = ∈imn An j, m .

(1.232)

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Another curl-type operation for tensors is A ∧ ∇, which is defined by13 T  A ∧ ∇ := − ∇ ∧ AT .

(1.233)

We shall return to the curl of a tensor field later in the book (Chap. 6).

1.22.3 Vector and Tensor Identities In Continuum Mechanics, it is frequently necessary to apply the ‘grad’, ‘div’ and ‘curl’ operators to combinations of scalar, vector and tensor fields. Most of the time these expressions are in the form of products. They need to be expanded so that the aforementioned differential operations can be ‘transferred’ to the individual terms in the original product. We first illustrate these remarks with a basic example. Let us assume that we want to expand div (φu), where φ ≡ φ(x) and u ≡ u(x) are differentiable scalar and vector fields, respectively. The result of multiplying together scalar and vector fields is a vector field, so we are prompted to use (1.217) followed by the standard product rule: ∂ ∂φ ∂ui ∂ (φu)i = (φui ) = ui + φ ∂ xi ∂ xi ∂ xi ∂ xi = (grad φ) · u + φ div u.

div (φu) =

It is clear now that in this last expression the initial ‘div ’ operator contributes to the presence of both grad φ and div u (i.e. it was ‘transferred’ to those terms). There are a couple of important remarks vis-à-vis this simple example. First, note that we have used Cartesian coordinates to justify the identity

or

div (φu) = (grad φ) · u + φ div u ∇ · (φu) = (∇φ) · u + φ(∇ · u).

(1.234)

Second, all the operations involved on the right-hand side of (1.234) are independent of the choice of coordinates, as we have already pointed out on several occasions. In consequence, this result will hold for other types of coordinates, although a direct proof might be a lot harder to establish in those cases (e.g. if the basis vectors are not constant). More generally, to prove an identity like (1.234) it will always suffice to restrict ourselves to selecting rectangular Cartesian coordinates; the general statement will follow from the invariance of the various operations (scalar, vectorial or tensorial products) and the coordinate-free nature of the nabla operator and its ‘siblings’ (del-dot, del-cross and del-tensor). But it is important that the expanded

13 Sometimes

u ∧ ∇ is defined for vector fields as well, but in that case u ∧ ∇ := −∇ ∧ u.

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expression is re-arranged so that it features only invariant (differential) operations. A few additional examples will consolidate these points. Example 1.1 Let φ ≡ φ(x) and A ≡ A(x) be differentiable scalar and tensor fields, respectively. Our objective is to expand div (φ A). This time multiplying together scalar and tensor fields yields a new tensor field, so the divergence operator is the one defined (1.224). Use of that formula and the product rule results in, ∂ Ai j ∂ ∂φ ∂ (φ A)i j e j = (φ Ai j )e j = Ai j e j + φ ej ∂ xi ∂ xi ∂ xi ∂ xi   ∂ Ai j ∂φ = (A T ) ji ej + φ e j = AT · grad φ + φ div A. ∂ xi ∂ xi

div (φ A) =

We have thus proved the identity div (φ A) = (grad φ) · A + φdiv A or

∇ · (φ A) = (∇φ) · A + φ(∇ · A).

(1.235)

Example 1.2 Let u ≡ u(x) be a vector field. We want to check the identity ∇ ∧ (∇ ∧ u) = ∇(∇ · u) − ∇ 2 u.

(1.236)

Expressions involving the repeated application of the cross product invariably require the use of the ‘ε − δ’ identity (see Exercise 1.9), ∈ki j ∈kpq = δi p δ jq − δiq δ j p . Using the definition of the curl (1.220) we have successively ∇ ∧ (∇ ∧ u) = ∈i jk (∇ ∧ u)k, j ei = ∈i jk (∈kpq uq, p ), j ei = ∈ki j ∈kpq uq, pj ei = (δi p δ jq − δiq δ j p )uq, pj ei = δi p δ jq uq, pj − δiq δ j p uq, pj = u j, i j ei − ui, j j ei = (u j, j ), i ei − (ui ei ), j j ≡ ∇(∇ · u) − ∇ 2 u.

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When checking a given identity, as in this last example, sometimes it is more convenient to express the left- and right-hand sides separately in terms of their Cartesian-coordinate representations. By comparing the two sets of results, it is usually straightforward to identify whether they are equal or not (invariably, this requires re-labelling the summation subscripts in one of the two expressions found). We follow this approach in the next two examples. Example 1.3 For two arbitrary tensor fields A ≡ A(x) and B ≡ B(x), the following identity is true ∇ · ( A · B) = (∇ · A) · B + A : (∇ ⊗ B).

(1.237)

On the left-hand side (L H S), we have the divergence of a tensor, hence L H S = (Ai p B pj ), i e j = Ai p, i B pj e j + Ai p B pj, i e j = (∇ · A) p B pj e j + Ai p B pj, i e j .

(1.238)

For evaluating the right-hand side (R H S), we remember (1.228) and the definition of the double-dot product (1.96); with these in mind, R H S = (∇ · A) p B pj e j + (Amn em ⊗ en ) : (Bi j, k ek ⊗ ei ⊗ e j ) = (∇ · A) p B pj e j + Amn Bi j, k δmk δni e j = (∇ · A) p B pj e j + Aki Bi j, k e j .

(1.239)

If in the second term on the last line of (1.239) the summation subscripts are relabelled according to k → i and i → p, then a quick comparison between (1.238) and (1.239) indicates that the identity (1.237) must be true. Example 1.4 If A ≡ A(x) and u ≡ u(x) are tensor and vector fields, respectively, then • (∇ ⊗ u). ∇ · ( A ∧ u) = (∇ · A) ∧ u + A × (1.240)

The justification of this identity will follow the same approach as in the previous example. Let us start to evaluate the left-hand side. Recalling Eq. (1.108), we can write L H S = (∈i jk A pi u j ), p ek = ∈i jk A pi, p u j ek + ∈i jk A pi u j, p ek = ∈i jk (∇ · A)i u j ek + ∈i jk A pi u j, p ek .

(1.241)

1.22 Vector and Tensor Calculus

99

The first term on the second line in (1.241) is easily recognised as being (∇ · A) ∧ u; as for the second term, we suspect it matches the remaining bit on the right-hand side in the identity that we need to prove. Next, we evaluate that particular term • (∇ ⊗ u) = (A e ⊗ e ) • (u e ⊗ e ) A× mn m n × q, p p q = Amn uq, p (em · e p )(en ∧ eq ) = Amn uq, p δmp ∈nqk ek = A pn uq, p ∈nqk ek .

(1.242)

(1.243)

Changing the summation subscripts on the last line, n → i and q → j, confirms our expectation. From (1.241) and (1.242) we then conclude that (1.240) is true.

1.22.4 Non-Cartesian Coordinates It is important to have an understanding of how the differential operators introduced above change when we switch to cylindrical or spherical coordinates. The matter is not trivial because the elements of the local bases will then be vector fields themselves. In this book, we take a pragmatic approach that has the merit of being entirely elementary and eminently suitable for mechanistic manipulations. This is explained below, first for the case of cylindrical polar coordinates. Seen as a vector, the ‘del’ operator will have components that change according to the rules consistent with Sect. 1.9. In the usual Cartesian system of coordinates, the components of this symbolic vector are ∂/∂ x j =: ∇ j ( j = 1, 2, 3), while in the local basis {er , eθ , ez } they will appear as ∇r , ∇θ , ∇z (which are yet to be found). We can then write ∂ ∂ ∂ + e2 + e3 ∂ x1 ∂ x2 ∂ x3 = er ∇r + eθ ∇θ + ez ∇z ,

∇ = e1

our immediate aim being to find the latter set of components for the ‘del’ vector. According to (1.82) and (1.167), the link between the two sets of components is provided by the symbolic relation ⎤ ⎡ cos θ ∇r ⎣∇θ ⎦ = ⎣ − sin θ 0 ∇z ⎡

sin θ cos θ 0

⎤⎡ ⎤ 0 ∇1 0 ⎦ ⎣∇2 ⎦ . 1 ∇3

(1.244)

The quantities ∇ j ( j = 1, 2, 3) can be expressed in terms of (r, θ, z) by using the chain rule in connection with (1.166), and the following self-evident results:

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1 Vector, Tensors, and Related Matters

∂r x1 = , ∂ x1 r ∂θ x2 = − 2, ∂ x1 r ∂z = 0, ∂ x1

∂r x2 = , ∂ x2 r ∂θ x1 = 2, ∂ x2 r ∂z = 0, ∂ x2

∂r x3 = , ∂ x3 r ∂θ = 0, ∂ x3 ∂z = 1. ∂ x3

Then, ∇1 ≡

∂ ∂ ∂r ∂ ∂θ ∂ ∂z = + + ∂ x1 ∂r ∂ x1 ∂θ ∂ x1 ∂z ∂ x1 sin θ ∂ ∂ − = cos θ ∂r r ∂θ

and, similarly, ∇2 = sin θ

cos θ ∂ ∂ + , ∂r r ∂θ

∇3 =

∂ . ∂z

(1.245)

(1.246)

On substituting (1.245) and (1.246) into (1.244), it is found that ∇r =

∂ , ∂r

and hence ∇ = er

∇θ =

1 ∂ , r ∂θ

∇z =

∂ , ∂z

∂ 1 ∂ ∂ + eθ + ez ∂r r ∂θ ∂z

(1.247)

represents the ‘del’ operator in cylindrical polar coordinates. The expressions for the gradient, divergence and curl of any vector or tensor field can be found routinely by using (1.247). The only other piece of information we need is the derivatives of the polar basis elements with respect to r , θ , z. With the help of (1.167), it is found that ∇r er = 0, 1 ∇θ er = eθ , r ∇z er = 0,

∇r eθ = 0,

∇r ez = 0,

(1.248a)

1 ∇θ eθ = − er , r ∇z eθ = 0,

∇θ ez = 0,

(1.248b)

∇z ez = 0.

(1.248c)

Appendix B contains an extensive list of differential operators expressed in both cylindrical and spherical coordinates. By way of example, we calculate below the divergence of a vector field in cylindrical polar coordinates. Simple manipulations give ∇ · u = (er ∇r + eθ ∇θ + ez ∇z ) · (ur er + uθ eθ + uz ez ) = er · ∇r (ur er + uθ eθ + uz ez ) + eθ · ∇θ (ur er + uθ eθ + uz ez ) + ez · ∇z (ur er + uθ eθ + uz ez ). (1.249)

1.22 Vector and Tensor Calculus

101

The last three scalar products can be simplified considerably by applying the chain rule in conjunction with (1.248). Hence, " ! ∂ur ∂uθ ∂uz er + eθ + ez ∇ · u = er · ∂r ∂r ∂r " ! 1 ∂ur ur 1 ∂uθ uθ 1 ∂uz er + eθ + eθ − er + ez + eθ · r ∂θ r r ∂θ r r ∂θ " ! ∂ur ∂uθ ∂uz + ez · er + eθ + ez . ∂z ∂z ∂z Finally, ei · e j = δi j for i, j ∈ {r, θ, z}, and we conclude that ∇·u=

ur 1 ∂uθ ∂uz 1 ∂ ∂uz 1 ∂uθ ∂ur + + + = (r ur ) + + . ∂r r r ∂θ ∂z r ∂r r ∂θ ∂z

A second example that we consider to illustrate the above points involves the expression of the divergence of a second-order tensor field A in cylindrical polar coordinates. Relative to the basis {er , eθ , ez } this tensor field has the representation A = Ai j ei ⊗ e j , where i, j ∈ {r, θ, z}, so ∇ · A = (er ∇r + eθ ∇θ + ez ∇z ) · (Arr er ⊗ er + Ar θ er ⊗ eθ + Ar z er ⊗ ez ) + (er ∇r + eθ ∇θ + ez ∇z ) · (Aθr eθ ⊗ er + Aθθ eθ ⊗ eθ + Aθ z eθ ⊗ ez ) + (er ∇r + eθ ∇θ + ez ∇z ) · (A zr ez ⊗ er + A zθ ez ⊗ eθ + A zz ez ⊗ ez ). (1.250) The dot-products on each line in (1.250) can be treated in exactly the same way as Eq. (1.249); many terms will not contribute much, either because the dot products of different basis vectors are zero or ∇i (e j ⊗ ek ) = 0 for some of the i, j, k ∈ {r, θ, z}. More specifically, the consecutive evaluation of the expressions on the three lines in (1.250) gives, respectively,       ∂ Arr Arr Ar θ Ar z ∂ Ar θ ∂ Ar z + er + + eθ + + ez , ∂r r ∂r r r   ∂r Aθr 1 ∂ Aθr 1 ∂ Aθθ 1 ∂ Aθ z − Aθθ er + + eθ + ez , r ∂θ r ∂θ r r ∂θ ∂ A zr ∂ A zθ ∂ A zz er + eθ + ez . and ∂z ∂z ∂z Putting these results together, we obtain the following components of ∇ · A: 1 ∂ Aθr ∂ A zr 1 ∂ Arr + + + (Arr − Aθθ ), ∂r r ∂θ ∂z r 1 ∂ Aθθ ∂ A zθ 1 ∂ Ar θ + + + (Ar θ + Aθr ), (∇ · A)θ = ∂r r ∂θ ∂z r 1 ∂ Aθ z ∂ A zz 1 ∂ Ar z + + + Ar z . (∇ · A)z = ∂r r ∂θ ∂z r (∇ · A)r =

102

1 Vector, Tensors, and Related Matters

In spherical polar coordinates, the components of the ‘del’ operator can be shown to be given by ∇ = er ∇r + eθ ∇θ + eϕ ∇ϕ , with ∇r =

∂ , ∂r

or ∇ = er

∇θ =

1 ∂ , r ∂θ

∇ϕ =

∂ 1 r sin θ ∂ϕ

∂ ∂ 1 ∂ 1 + eθ + eϕ , ∂r r ∂θ r sin θ ∂ϕ

(1.251)

where use was made of the following relations: ∇r er = 0, 1 ∇θ er = eθ , r 1 ∇ϕ er = eϕ , r

∇r eθ = 0,

∇r eϕ = 0,

(1.252a)

1 ∇θ eθ = − er , r cot θ ∇ϕ eθ = eϕ , r

∇θ eϕ = 0,

(1.252b)

1 cot θ ∇ϕ eϕ = − er − eθ . r r

(1.252c)

Several useful formulae expressed in spherical polar coordinates can be found in Appendix B.

1.22.5 Some Important Integral Theorems We start with Ω ⊂ E3 , a bounded domain and assume that the surface bounding this set, ∂Ω,14 is sufficiently regular. By regular we mean that the outward unit normal n ≡ n i ei to this surface depends continuously on its position (examples of acceptable domains include cylinders, spheres, cubes, etc., as well as combinations of such geometries). A key result in multivariate integral calculus is a theorem that establishes a remarkable connection between surface and volume integrals. It is called the Divergence Theorem (also sometimes referred to as Gauss’ Theorem). In its most primitive form, this theorem asserts that if φ is a scalar field over the region Ω, then &

& ∂Ω

φn i d A =

Ω

∂φ dV (i = 1, 2, 3), ∂ xi

(1.253)

where n i is the ith component of the unit outer normal to the surface ∂Ω. In all the theorems that appear below, we shall assume that the integrals in question exist. If u = ui ei is the Cartesian representation of a vector field, by setting φ → ui in (1.253) and using the summation convention, we discover the Divergence Theorem in its more traditional form, boundary of a subset Ω ⊂ E3 , denoted ∂Ω, consists of those points P ∈ E3 such that every ε-neighbourhood of P (ε > 0) intersects both Ω and the complement of Ω. This definition applies to subsets of E2 as well.

14 The

1.22 Vector and Tensor Calculus

103

&

& ∂Ω

n · u dA =

Ω

∇ · u dV.

(1.254)

This form of the theorem is independent of the chosen system of coordinates because of the invariance properties of the del-dot operator and the dot product. A number of other useful formulae are derivable immediately from (1.253). For instance, on multiplying that equation by the constant basis element ei (i = 1, 2, 3) and adding up the results, we end up with &

& ∂Ω

φn d A =

Ω

∇φ dV,

(1.255)

which asserts that the volume integral of the gradient of a scalar field may be expressed in terms of the values assumed by the function on the bounding surface ∂Ω of that region. Another identity can be obtained from (1.254) by letting u → u ∧ a, where a ∈ V is a constant vector. Since ∇ · (u ∧ a) = (∇ ∧ u) · a, the Divergence Theorem gives #&

& ∂Ω

(u ∧ a) · n d A = a ·

$ Ω

(∇ ∧ u) dV .

The integrand on the left-hand side is easily recognised as being the scalar triple product [n, u, a]. We know that its value does not change when the elements are permuted cyclically, whence &

#& $ a · (n ∧ u) d A = a · (∇ ∧ u) dV ∂Ω $Ω #& $ #& (n ∧ u) d A = a · (∇ ∧ u) dV , ⇒a· ∂Ω

Ω

(∀) a ∈ V .

Since a was arbitrary, the last equality yields the interesting integral identity &

& ∂Ω

n ∧ u dA =

Ω

∇ ∧ u dV.

(1.256)

Similar results involving tensors (instead of vector integrands) are readily obtained by following an identical route; a couple of such identities are stated below. If φ → u j in (1.253), & & ∂Ω

n ⊗ u dA =

Ω

∇ ⊗ u dV.

(1.257)

The Divergence Theorem for tensor fields states that &

& ∂Ω

AT · n d A =

Ω

∇ · A dV.

(1.258)

104

1 Vector, Tensors, and Related Matters

All of the above integral formulae can be found as particular cases of the more general identity recorded below. Theorem 1.9 (Generalised Divergence Theorem) Let u be a vector field and A a tensor field, each continuous in Ω ∪ ∂Ω and continuously differentiable in Ω. Then, &

& ∂Ω

u ⊗ ( AT · n) d A =

Ω

 u ⊗ div A + (grad u) · A dV.

(1.259)

For instance, if u is a constant vector then we recover (1.258), while if A ≡ I then (1.257) is obtained. Finally, by taking the trace of (1.257) the classical form (1.254) emerges. The next theorem is named after the British scientist G. Stokes, and relates a line integral around a closed path to a certain surface integral &

' C

u · dx =

Σ

n · (∇ ∧ u) d A.

(1.260)

In the above statement, Σ is any open surface whose boundary is a fixed closed curve C ≡ ∂Σ, n represents the outward unit normal to Σ, and the curve C is traversed in the direction such that Σ appears to the left of an observer moving along C with n at points near C pointing from the observer’s feet to his head (see Fig. 1.10). The theorem asserts that the line integral of the tangential component of a vector field over this closed path C is equal to the surface integral of the normal component of the curl of that function integrated over any surface bounded by C . Stokes’ Theorem admits a relatively straightforward extensions to second-order tensor fields that can be obtained directly from (1.260). If A = A(x) is such a tensor field, then A = v k ⊗ ek with v k = A jk e j (k = 1, 2, 3); see the comments made immediately after Eq. (1.41). By writing (1.260) for u → v k and then taking the product of the resulting relation with ek , gives

Σ Σ C

C

Fig. 1.10 The geometrical setting for Stokes’ theorem: examples of two different surfaces Σ ‘spanning’ the same boundary curve C . A small sample of outward unit normals to Σ is included as well

1.22 Vector and Tensor Calculus

105

&  ek (v k · dx) = ek (∇ ∧ v k ) · n d A C Σ ' &  =⇒ dx · (v k ⊗ ek ) = n · (∇ ∧ v k ) ⊗ ek d A.

'

C

Σ

  But, ∇ ∧ A = ∇ ∧ v k ⊗ ek , so this last equation can be cast in the equivalent form &

' C

dx · A =

Σ

n · (∇ ∧ A) d A,

(1.261)

which represents the tensorial version of Stokes’ Theorem. Both (1.260) and (1.261) require the fields u and A to be at least C 1 -differentiable in a vicinity of Σ ∪ C .

1.23 Differentiation of Tensor Functions The calculus of functions having tensors as their arguments, and being possibly tensor valued, can be reduced to that of mappings f : Rn → Rm for some suitable m, n ∈ N—see Sect. 1.22.1. For the most part, our discussion below will follow this route, although a more elementary point of view will be introduced as well. Definition 1.46 Consider a scalar-valued tensor function ϕ : Lin → R, with ϕ = ϕ( A), and assume that ϕ(Ai j ) has continuous partial derivatives with respect to all its variables. By definition the gradient of ϕ is the second-order tensor ϕ A that admits the representation ∂ϕ ∂ϕ ei ⊗ e j , (1.262) ϕA ≡ := ∂A ∂ Ai j relative to the standard basis. Alternatively, the gradient of ϕ with respect to A is the unique element ϕ A ∈ Lin that satisfies ϕ A [H] ≡ ϕ A : H =

 d  ϕ( A + s H) , s=0 ds

(∀) H ∈ Lin.

(1.263)

Both formulae in the above definition lead to the same results, but it should be clear that (1.263) is more general as it does not rely on any specific component representation. We will use this formula to compute the gradient of a few simple functions. Example 1.5 Calculate the gradients of the following scalar tensor functions: (i) tr( A);

(ii) tr( A2 );

(iii) tr( A3 ).

Let ϕ( A) ≡ tr( A) = I : A. According to (1.263),

106

1 Vector, Tensors, and Related Matters

ϕ A [H] =

  d  d  I : ( A + s H)  I : A + s(I : H)  = =I:H s=0 s=0 ds ds

or ϕ A : H = I : H for all H ∈ Lin, whence ϕ A = I. Recalling that the trace of a tensor is in fact its first principal invariant I1 , the result we have just obtained can be stated in the form ∂ I1 = I. (1.264) ∂A For (ii) above, take ϕ( A) ≡ tr( A2 ) = I : A2 and use again (1.263). Thus,  d  I : ( A + s H)2  s=0 ds  d  2 I : [ A + s( A · H + H · A) + s 2 H 2 ]  = s=0 ds = I : (H · A) + I : ( A · H) = 2 AT : H,

ϕ A [H] =

and we conclude that

∂ tr( A2 ) = 2 AT . ∂A

(1.265)

For (iii) we note that we need to calculate the derivative with respect to s ∈ R of ( A + s H)3 , then set s = 0, and take the double-dot product with the identity tensor. The only terms that will ‘survive’ all these steps will be those linear in the parameter ‘s’. Since tensor multiplication is not commutative, we end up with I : ( A · H · A) + I : ( A · A · H) + I : (H · A · A)  = 3( AT · AT ) : H = 3 A2 )T : H, where use has been made of the properties (1.98b). Keeping in mind (1.263) we can now state the final result, ∂ tr( A3 ) = 3 ( A2 )T . (1.266) ∂A

Example 1.6 Calculate the gradient of the principal invariants (i) I2 =

1 (tr( A))2 − tr( A2 ) ; 2

(ii) I3 = det A ( A invertible).

The gradient defined in (1.263) is a linear operation, in the sense that the gradient of a sum of functions is the sum of the gradients of the individual terms in the sum. This observation can be used to calculate the gradient of I2 . Furthermore, the first term in the expression of I2 is the composition of t → t 2 /2 and A → tr( A) ≡ I1 . Since the chain rule is applicable here, in light of (1.264), the gradient of the first term will be I1 I, a result that can be combined with (1.265) to give

1.23 Differentiation of Tensor Functions

107

∂ I2 = I1 I − AT . ∂A

(1.267)

For (ii) we use the Cayley–Hamilton theorem. Taking the trace of Eq. (1.189) in that result leads to 1 (1.268) I3 = tr( A3 ) − I1 tr( A2 ) + I2 tr( A) . 3 The gradients of all the terms in this expression have already been calculated; using the product rule on the last two terms, in conjunction with (1.264)–(1.267), results in ∂ I3 1   T = 3 A2 − tr( A2 )I − 2I1 AT + I2 I + I1 (I1 I − AT ) ∂A 3 1   T = 3 A2 + 3I2 I − 3I1 AT , 3

(1.269)

where the last line was obtained by replacing tr( A2 ) = I12 − 2 I2 (which follows from the definition of I2 ). To further simplify the gradient of I3 , we need to resort again to the Cayley–Hamilton equation (1.189). Multiplying it by A−1 and taking the transpose of the corresponding result shows that the expression enclosed between the square brackets in (1.269) is equal to 3I3 A−T , from which ∂ I3 = I3 A−T . ∂A

(1.270)

There is a subtle point in connection with (1.263) that needs further discussion. To keep things simple, the tensor function considered in Definition 1.46 was defined on the entire Lin. This is not always the case; in many instances ϕ is defined only on some subset of it like, for example, Sym. Since A + s H enters in the argument of ϕ it is clear that we need to restrict H to Sym as well. The upshot of this observation is that now (1.263) must hold only for symmetric tensors H. If ϕ A were not in Sym then its skew-symmetric part would not be uniquely determined by Eq. (1.263). Consequently, the gradient of a scalar function that is defined only on Sym must be an element of the same subspace. To achieve this outcome, the procedure outlined in the above examples must be amended as we now explain. The domain of ϕ is first extended to Lin by setting  ϕ( A) := ϕ

A + AT 2

 ,

(∀) A ∈ Lin,

and then the gradient of ϕ( A) is computed exactly as before. After this has been done A must be restricted again to Sym. The next example illustrates this point. Assume u, v ∈ V (fixed vectors), and consider ϕ : Sym → R defined by ϕ( A) = u · A · v,

(∀) A ∈ Sym.

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1 Vector, Tensors, and Related Matters

According to our earlier discussion, the extension of this function to Lin is ϕ( A) =

1 1 u · A · v + u · AT · v, 2 2

and its gradient is calculated easily  d  ϕ( A + s H) s=0 ds   1 d  1 d  u · ( A + s H) · v  + u · ( A + s H)T · v  = s=0 s=0 2 ds 2 ds 1 1 1 = u · H · v + u · H T · v = (u ⊗ v + v ⊗ u) : H, 2 2 2

ϕ A [H] =

whence ϕA =

1 (u ⊗ v + v ⊗ u), 2

which is clearly a symmetric tensor. Definition 1.47 Let T : Lin → Lin, be a tensor-valued function of tensorial argument, T = T ( A), and assume that the components Ti j (Akl ) relative to the standard basis have continuous partial derivatives ∂ Ti j /∂ Akl . By definition, the gradient of T with respect to A is the fourth-order tensor TA ≡

∂ Ti j ∂T := ei ⊗ e j ⊗ ek ⊗ el . ∂A ∂ Akl

(1.271)

Note that (1.271) asserts that for each fixed A ∈ Lin, relative to the standard basis, the tensor T A ( A) ∈ Lin 4 has components (T A )i jkl =

∂ Ti j . ∂ Akl

(1.272)

The gradient may also be viewed as the derivative of the mapping A → T ( A) between the linear space Lin and itself. It is defined by the condition T A [H] ≡ T A : H =

 d  T ( A + s H) , s=0 ds

(∀) H ∈ Lin.

Example 1.7 If A ∈ Lin is an invertible tensor then ∂ A−1 [H] = − A−1 · H · A−1 , ∂A

(∀) H ∈ Lin.

(1.273)

1.23 Differentiation of Tensor Functions

109

Applying (1.273) leads to  ∂ A−1 d  [H] = ( A + s H)−1  , s=0 ∂A ds

(∀) H ∈ Lin.

(1.274)

This suggests that we must evaluate the right-hand side in (1.274). To this end, differentiate with respect to s the obvious identity ( A + s H)−1 · ( A + s H) = I, to obtain ! " " ! d d ( A + s H)−1 · ( A + s H) + ( A + s H)−1 · ( A + s H) = O. ds ds By setting s = 0 in this relation, !

" " !   d d   · A + A−1 · =O ( A + s H)−1  ( A + s H) s=0 s=0 ds ds  d  ( A + s H)−1  · A + A−1 · H = O ⇒ s=0 ds  d −1  ( A + s H)  ⇒ = − A−1 · H · A−1 , s=0 ds

(1.275)

which completes the solution of this example. Example 1.8 Find the gradient of the following tensor-valued function G( A) = A · AT ,

(∀) A ∈ Lin,

and then show that 1 ∂  ( A · AT ) : ( A · AT ) = A · AT · A. 4∂A To solve the first part of this question we just use the definition (1.273)  d  ∂ ( AT · A)[H] = ( A + s H)T · ( A + s H)  s=0 ∂A ds  d  T A · A + s( AT · H + H T · A) + s 2 H T · H  = s=0 ds = AT · H + H T · A.

110

In conclusion,

1 Vector, Tensors, and Related Matters

∂ ( AT · A)[H] = AT · H + H T · A. ∂A

The second part can be solved routinely as above, by using again the definition. However, we choose a different route in order to illustrate a simple chain-rule strategy. Owing to the symmetry of the relation to be proved, let us consider the scalar-valued tensor function ϕ(B) = B : B for B ∈ Sym . It is easy to check that the gradient of this function is ϕ B = 2B. On the other hand, we can think of B as being a function of A and we write B = B( A) ≡ A · AT to indicate this dependence. Then,  ∂  d  ( A · AT ) : ( A · AT ) = ϕ(B( A + s H)) s=0 ∂A ds  d  T = ϕ(B + s( A · H + H · AT ) + s 2 H · H T ) s=0 ds = ϕ B [ A · H T + H · AT ] = (2 A · AT ) : ( A · H T ) + (2 A · AT ) : (H · AT ) = (2 AT · A · AT ) : H T + (2 A · AT · A) : H = (4 A · AT · A) : H, the last line being a consequence of C T : D = C : D T for any C, D ∈ Lin. The required result is now proved. Although one can always calculate gradients of tensor functions from first principles (as already illustrated above), it is hardly the best way for anything but very simple expressions. In many instances, the tensor functions of interest involve the composition of simpler mappings or various types of multiplication operations. It is, therefore, helpful to be aware of how tensor gradients behave in relation to the product and chain rules for differentiation. The simplest case is that of two scalar-valued tensor functions, ϕ ≡ ϕ( A) and ψ ≡ ψ( A); in this case,   (1.276) ϕψ A = ϕψ A + ψϕ A , a result that follows immediately from first principles. If the function to be differentiated is a composition of the form ϕ(G( A)), with G : Lin → Lin, then ∂ϕ ∂G kl ∂ϕ = ∂ Ai j ∂G kl ∂ Ai j

⇒

(ϕ A )i j = (ϕ G )kl (G A )kli j

or ϕ A = (G A )T : ϕ G .

(1.277)

A related case is that of G(K ( A)), with both G and K tensor-valued functions; a similar reasoning as above suggests that

1.23 Differentiation of Tensor Functions

111

G A = G K : K A,

(1.278)

both factors on the right-hand side being elements of Lin 4 . Differentiation with respect to a tensor transpose is a common scenario  ∂A ∂( AT )T d T T [H] = [H] = + s H) = H T = T : H, ( A  s=0 ds ∂ AT ∂ AT  d ∂ AT  [H] = ( A + s H)T  = H T = T : H, s=0 ∂A ds Therefore, we can summarise these calculations ∂A ∂ AT = T, = ∂A ∂ AT

(1.279)

where T was introduced in Sect. 1.14. By the same token, ∂A ∂ AT = I, = ∂A ∂ AT ∂G ∂ϕ ∂G ∂ϕ , : T, =T: = T T ∂A ∂A ∂A ∂A

(1.280a) (1.280b)

where ϕ and G are scalar and tensor-valued functions, respectively, while I represents the identity tensor in Lin 4 (already encountered in Sect. 1.14). The next formulae are less intuitive and are recorded separately for easy reference. Proposition 1.13 If ϕ ≡ ϕ( A) ∈ R, G ≡ G( A) ∈ Lin, K ≡ K ( A) ∈ Lin are functions that depend on A ∈ Lin, then the following product rule formulae are true: (ϕG) A = G ⊗ ϕ A + ϕ G A ,  T  T (G : K ) A = G A : K + K A : G.

(1.281a) (1.281b)

These formulae are easily justified by using the relevant definitions and some of the properties of fourth-order tensors discussed in Sect. 1.14. In the interest of brevity we check only (1.281b),  d  ∂  G : K )[H] = G( A + s H) : K ( A + s H)  s=0 ∂A ds = G A [H] : K + G : K A [H] = (G A : H) : K + G : (K A : H) = G A [K , H] + K A [G, H]  T  T = G A [H, K ] + K A [H, G]     = (G A )T : K : H + (K A )T : G : H.

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1 Vector, Tensors, and Related Matters

Since H is an arbitrary second-order tensor the formula is thus proved. Proposition 1.14 If G ≡ G( A) ∈ Lin and K ≡ K ( A) ∈ Lin are functions that depend on A ∈ Lin, then the following product rule formula holds: (G · K ) A = (G  I) : K A + (I  K T ) : G A .

(1.282)

Using again the definition (1.273), we have  d  ∂  G · K )[H] = G( A + s H) · K ( A + s H)  s=0 ∂A ds = G A [H] · K + G · K A [H],

(1.283)

which is almost identical to what we have found above save for the inner product that now replaces the double-dot multiplication. Formula (1.282) follows by applying the identities from Exercise 1.60, with C → G A , A → H, K → B for the first term in (1.283), and C → K A , B → G, A → H for the second one.

1.24 Exercises 1. If a, b ∈ V are orthogonal, then solve the vector equation X ∧ a = b. 2. Given a, b, c, d ∈ V such that a · c = 0, solve X ∧ a + (X · b)c = d. 3. Show that the general solution of the vector equation α X + X ∧ a = b, where α ∈ R and a, b ∈ V , is provided by the formula X=

α 2 b + (b · a)a + α(a ∧ b) . α(α 2 + |a|2 )

4. Show that the equation of the plane (Π ) through three given points A, B, C, with position vectors a, b, and c, respectively, can be cast in the form x · n = p, where n = b ∧ c + c ∧ a + a ∧ b is a normal vector to (Π ) and p = [a, b, c] represents the distance from the origin to the same plane. 5. Show that the length of the common perpendicular to the skew (non-parallel) lines (d1 ) : r = a1 + t1 b1 and (d2 ) : r = a2 + t2 b2 where t1 , t2 ∈ R and r is the position vector of a generic point on the lines, is given given by ±

(b1 ∧ b2 ) · (a1 − a2 ) , |b1 ∧ b2 |

where the sign is chosen so as to make the expression positive. 6. Let r = αa + β b + γ c, where r is any given vector and a, b, c are known non-coplanar vectors. Show that the scalars α, β, γ are given by the formulae

1.24 Exercises

113

α=

[r, c, a] [r, a, b] [r, b, c] , β= , γ = . [a, b, c] [a, b, c] [a, b, c]

7. Prove that n is orthogonal to (a − Proj n a) for all vector n, a ∈ V , where n = 0. 8. Show that   (i) Proj n Proj n a = Proj n (a);

  (ii) Proj n a − Proj n a = 0,

for all n, a ∈ V . Explain the geometrical meaning of these results. 9. Use basic properties of triple scalar products to show that  a · c (a ∧ b) · (c ∧ d) =  b·c

 a · d  . b · d

Hence, or otherwise, establish the so-called ‘ε − δ’ identity, ∈i jk ∈i pq = δ j p δkq − δ jq δkp . 10.

(1.284)

a. This is a generalisation of the preceding question; if a, b, c, u, v, w ∈ V ,  a · u  [a, b, c][u, v, w] =  b · u c · u b. Hence, or otherwise, establish the formula   δi j δim  ∈ikl ∈ jmn = δk j δkm  δl j δlm

a·v b·v c·v

 a · w b · w  . c · w

 δin  δkn  . δln 

c. Check that ∈ikl ∈ikn = 2δln and ∈ikl ∈ikl = 6. 11. Show that (b ∧ c) ∧ (c ∧ a) = [a, b, c]c, and then deduce the following identity [b ∧ c, c ∧ a, a ∧ b] = [a, b, c] 2 . 12. If a, b, c, d ∈ V prove that (b ∧ c) ∧ (a ∧ d) + (c ∧ a) ∧ (b ∧ d) + (a ∧ b) ∧ (c ∧ d) = −2[a, b, c]d.     13. Consider Ai jkl := 21 δik δ jl + δil δ jk and Bi jkl := 21 δik δ jl − δil δ jk . If Ti j and Si j are the components of, respectively, symmetric and skew-symmetric elements of Lin, show that

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1 Vector, Tensors, and Related Matters

Ai jkk = δi j , Bi jkl Tkl = 0,

Bi jkk = 0,

Ai jkl Tkl = Ti j ,

Ai jkl Skl = 0,

Bi jkl Skl = Si j .

14. Let T , E ∈ Lin such that, relative to the standard basis, [T ] = [Ti j ] and [E] = [E i j ]. The expressions E i j = E i j − 13 E kk δi j and T i j = Ti j − 13 Tkk δi j define the so-called deviatoric parts of the foregoing tensors. If Ti j = 2μE i j + λ(E kk )δi j for some λ, μ ∈ R, then show that (i) E kk = T kk = 0, T i j = 2μE i j ; 1 (ii) E i j E i j = E i j E i j = E i j E i j − (E kk )2 ; 3 1 1 1 (iii) Ti j E i j = T i j E i j + Tkk Ell . 2 2 6 15. Let {ei } be the usual standard basis in the Euclidean vector space V , and consider T ∈ Lin defined by T · e1 = e1 + e2 , T· e2 = e2 + e3 , T · e3 = e3 + e1 . (i) Find the representation of T relative to ei ⊗ e j ; (ii) calculate T · v, if v := e1 + e2 + e3 ; (iii) determine the additive decomposition of T into its symmetric and skew-symmetric parts. Find the axial vector of its skew-symmetric part. 16. If l, m, n are orthonormal vectors in V , then check that Q = m⊗n+n⊗l +l ⊗m is in Ort. Show further that Q 2 = Q T and Q ∈ Ort+ . 17. Let u1 and u2 be arbitrary vectors. Show that u2 ⊗ u1 − u1 ⊗ u2 is a skewsymmetric tensor, and then check that u1 ∧ u2 represents its axial vector. 18. Show that symmetry and anti-symmetry of the components of a second-order tensor are properties independent of the choice of orthonormal basis. 19. Let a, b, c ∈ V . It is possible to construct two partial transposition operations for elements of Lin 3 by defining the following ‘rules’, (a ⊗ b ⊗ c) t · u := a ⊗ c(b · u),

t

(a ⊗ b ⊗ c) · u := b ⊗ a(c · u),

for all u ∈ V , where t (. . . ) and (. . . )t are the two new operations mentioned above. Show that these operations are independent of the basis chosen for representing the elements of Lin 3 . 20. Let u1 and u2 be arbitrary vectors. By using Proposition 1.7 show that tr (u1 ⊗ u2 ) = u1 · u2 and det (u1 ⊗ u2 ) = 0. 21. Show that if W ∈ Skw, then W : (a ⊗ a) = 0 for all a ∈ V . Geometrically, this means that skew-symmetric tensors map any vector onto planes perpendicular to the original vector. Also, convince yourself that W · w = 0, where w represents the axial vector of W . 22. If S ∈ Sym and W ∈ Skw, then S : W = 0. Is it true that S · ·W = 0 as well?

1.24 Exercises

115

23. Let p be a non-zero vector and S ∈ Sym. Prove that p is an eigenvector of S if and only if p ⊗ p commutes with S. 24. Find the most general A ∈ Lin such that a ∧ A = A ∧ a for every a ∈ V . 25. Let a, b, c ∈ V arbitrary vectors. Show that a ⊗ (b ∧ c) + b ⊗ (c ∧ a) + c ⊗ (a ∧ b) = [a, b, c]I. Explain why the right-hand side does not change if we swap the order of the terms in the tensor products on the left-hand side. 26. If A ∈ Lin and a, b ∈ V , then check the formula det( A + a ⊗ b) = det A + a · cof( A) · b. 27. Show that det( A + B) = det A + det B + cof( A) : B + A : cof(B).

(1.285)

28. If a j , b j ∈ V ( j = 1, 2, 3), check the identity (a1 ⊗ b1 ) × × (a2 ⊗ b2 ) : (a3 ⊗ b3 ) = [a1 , a2 , a3 ][b1 , b2 , b3 ]. × 29. If A, B, C ∈ Lin, then ( A × × B) : C = A : (B × C).

30. Let A, B ∈ Lin and a, b ∈ V . Show that

× ( A ∧ a) × × (B ∧ a) = ( A × B) · (a ⊗ a), (a ∧ I) × × (b ∧ I) = a ⊗ b + b ⊗ a. 31. Let A, B ∈ Lin. Prove the following formula:  A× × B = tr( A)tr(B) − tr( A · B) I −tr( A)B T − tr(B) AT + ( A · B + B · A)T . 32. Let I ∈ Lin be the usual identity tensor and consider T := I ∧ I ∈ Lin 3 . Show that: (i) u · T · v = u ∧ v for all u, v ∈ V ; (ii) ∈= −I ∧ I, where ∈ was defined in (1.89) and its components are Ricci’s permutation symbols. 33. By using (1.144) show that two finite rotations through angles θ1 and θ2 about the same axis, specified by the unit vector a, are equivalent to a single rotation through an angle θ1 + θ2 about a. 34. Given two unit vectors a, b ∈ V , consider Q 1 = 2a ⊗ a − I and Q 2 = 2b ⊗ b − I, two planar reversals in planes having these vectors as unit normals. Then Q 1 · Q 2 is a rotation whose axis is parallel to the common per-

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1 Vector, Tensors, and Related Matters

pendicular of a and b, and the angle of which is twice the angle between these two vectors. 35. a. Show that if b ∈ V then (I − I ∧ b)−1 =

1 (I + I ∧ b + b ⊗ b) . 1 + |b|2

b. The tensor R ∈ Lin in (1.144) can be written as R = (I − I ∧ b)−1 (I + I ∧ b),

b :=

θ 1 a tan . 2 2

36. Let W 1 , W 2 and W 3 be skew-symmetric tensors with axial vectors w1 , w2 and w3 , respectively. Show that (ii) W 1 : W 2 = 2w1 · w2 ; (i) W 1 · W 2 =w2 ⊗ w1 − (w1 · w2 )I; (iii) W 1 : (W 2 · W 3 ) = w1 · (w2 ∧ w3 ). 37. If a, b, c, d ∈ V , then 1 (a ∧ b) ⊗ (c ∧ d) = T · S − (T · ·S)I, 2 38.

where S := b ⊗ a − a ⊗ b and T := d ⊗ c − c ⊗ d. a. For a, b, c and d ∈ V , show that cof (a ⊗ b + c ⊗ d) = (a ∧ c) ⊗ (b ∧ d);

(1.286)

in particular, if a, b, c are orthonormal, then b ⊗ a = −cof (a ⊗ b + c ⊗ c). b. Using (1.286) show that if u ∈ V is a unit vector then cof (I − u ⊗ u) = u ⊗ u. Assuming now that u ∈ V is an arbitrary vector (i.e. |u| = 1), find how the right-hand side of this formula must be modified. 39. Given two non-parallel unit vectors u, v ∈ V , the rotation (1.144) that sends u into v about an axis perpendicular to both u and v is given by R = I +v⊗u−u⊗v+

! " u⊗u+v⊗v u·v (u ⊗ v + v ⊗ u) − . 1+u·v u·v

40. Let W ∈ Skw associated with the unit axial vector w, and let ε > 0. Show that

1.24 Exercises

117

det(I + εW ) =1 + ε2 , (I + εW )−1 = (1 + ε2 )−1 (I + ε A + ε2 B), where A, B ∈ Lin are tensors that you must specify. 41. Let α, β ∈ R and consider m and n two orthogonal unit vectors. Determine the eigenvalues, the eigenvectors and a spectral representation for the following tensors A = α I + βm ⊗ m, B = m ⊗ n + n ⊗ m. 42. Let A ≡ A(t) be an invertible tensor which depends on t ∈ R. Assuming that the derivative d A/dt exists, prove that d dA : A−T . (det A) = (det A) dt dt

(1.287)

43. Let P ∈ Lin and consider P S = sym( P), P A = skew( P). If w ∈ V represents the axial vector of P A , then P T · P = P 2S + I(w · w) − w ⊗ w − w ∧ P S − (w ∧ P S ) T ,  P· P T − P T · P = 2 w ∧ P S + (w ∧ P S ) T . 44. Let a ≡ a(t), where t ∈ R. Show that d (i) dt

(

d a d2a , a, dt dt 2

)

(

) $ & # d2a da d a d3a a ∧ 2 dt = a ∧ , + c, = a, ; (ii) dt dt 3 dt dt

where c is a constant vector. 45. Let x ≡ x(t) and y ≡ y(t) be differentiable vector functions from the interval (a, b) ⊂ R into V . If the derivatives x  (t) and y (t) satisfy x  = ax + b y and y = cx − a y, where a, b, c ∈ R, show that x(t) ∧ y(t) is a constant vector. 46. Show that a given tensor S ∈ Lin commutes with every W ∈ Skw if and only if S is a spherical tensor, i.e. S = α I for some α ∈ R. 47. a. Let f be a real polynomial, A ∈ Lin, and λ an eigenvalue of A. Show that f (λ) is an eigenvalue of f ( A) and that an eigenvector of A associated with λ is also an eigenvector of f ( A) associated with f (λ). b. Are there any non-trivial elements in Lin possessing the three principal invariants all equal to zero? Justify your answer. 48. Relative to the standard basis {ei } in V , a tensor A ∈ Lin has components ⎡

2 [ A] = ⎣2 1

1 3 2

⎤ 3 4⎦ . 1

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1 Vector, Tensors, and Related Matters

Find the representation of the same tensor in the basis { ei } defined by  e1 := e1 ,  e3 := e1 + e2 + e3 . Repeat the question if the basis is now defined e2 := e1 + e2 , e2 := e2 + e3 ,  e3 := 2e3 . by  e1 := −e1 + e3 ,  49. If φ, v and S ∈ Lin are scalar, vector and tensor fields, respectively, such that div v = 0 and S = −φ I + 2β sym (grad u) for β ∈ R, then show the following: (i) div S = −grad φ + β ∇ 2 v; (ii) grad (|S|) = −3 grad φ. 50. If I j ≡ I j ( A) ( j = 1, 2, 3) are the principal invariants of A ∈ Sym, then check the following properties: | A3 | = 3(I3 − I1 I2 ) + | A|3 , 2(I12 − 3I2 ) = (λ2 − λ3 )2 + (λ3 − λ1 )2 + (λ1 − λ2 )2 , where λ j are the eigenvalues of A, and | . . . | stands for the usual ‘trace’ operation. 51. Show from first principles that a ∧ b = (I ∧ a) · b = (a ∧ I) · b,

b ∧ a = b · (I ∧ a) = b · (a ∧ I).

52. If a ∈ V and A := I ∧ a + a ⊗ a, then A4 = I. 53. Let a, b, c ∈ V be three mutually perpendicular vectors. Show that a. I ∧ a = c ⊗ b − b ⊗ c, I ∧ b = a ⊗ c − c ⊗ a, and I ∧ c = b ⊗ a − a ⊗ b. b. (I ∧ a)2 = −a ⊗ a − b ⊗ b, (I ∧ a)3 = −b ⊗ a + a ⊗ b, and (I ∧ a)4 = a ⊗ a + b ⊗ b. 54. Derive the following formulae, (i) ∇ ⊗ (T · u) = (∇ ⊗ T ) · u + (∇ ⊗ u) · T T , (ii) ∇ · (T · u) = (∇ · T ) · u + T : (∇ ⊗ u), (iii) ∇ ∧ (u ∧ v) = u(∇ · v) + v · (∇ ⊗ u) − u · (∇ ⊗ v) − v(∇ · u), where u, v are vector fields and T is a second-order tensor field. 55. Show that if u is a vector field and I the usual second-order identity tensor, then I ∧ (∇ ∧ u) = u ⊗ ∇ − ∇ ⊗ u, ∇ ∧ (I ∧ u) = u ⊗ ∇ − I(∇ · u),

∇ · (∇ ∧ u) = 0, ∇ ∧ (∇ ⊗ u) = 0.

56. Consider C, S ∈ Lin 4 , with components given by Cklmn = λ δkl δmn + μ(δkm δln + δkn δlm ), λ 1 δkl δmn + (δkm δln + δkn δlm ), Sklmn = − 2μ(3λ + 2μ) 4μ where λ, μ are real constants, (3λ + 2μ)μ = 0. Show that C is the inverse of S in the sense of the Definition 1.29.

1.24 Exercises

57. Show that

119

&



∂Ω

&



x ∧ T · n dA = T

Ω



 x ∧ div T + T × dV,

where x is the position vector of a generic point in Ω ⊂ E3 , while T × represents the vector of the second-order tensor field T (see Definition 1.13). 58. If A, B, C ∈ Lin, then × A× × (B × C) = (B ⊗ C)[ A] + (C ⊗ B)[ A] −(B  C T )[ AT ] − (C  B T )[ AT ]. Hence, deduce that × I× × ( A × B) = (B ⊗ I)[ A] + ( A ⊗ I)[B] − ( A · B + B · A). 59. Let A be a second-order tensor. Show that for any positive integer k,   ∂  tr( Ak ) = k Ak−1 )T . ∂A 60. Let A, B ∈ Lin and C ∈ Lin 4 . Show that   I  B T : C [ A],    B · C[ A] = (B  I : C [ A].

C[ A] · B =



(1.288a) (1.288b)

61. Show that ∂  4 A = A3  I + I  B 3 + A2  B + A  B 2 , ∂A where A ∈ Lin and B ≡ AT . 62. If A ∈ Lin, then derive the following formulae:    ∂  I1 A2 = A2 ⊗ I + I1 A  I + I  AT , ∂A    ∂  I2 A = A ⊗ I1 I − AT + I2 (I  I), ∂A where I j ≡ I j ( A) ( j = 1, 2) are the principal invariants of A.

(1.289a) (1.289b)

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1 Vector, Tensors, and Related Matters

Bibliography 1. Dodson CTJ, Poston T (1991) Tensor geometry. Springer, Berlin 2. Farin G, Hansford D (2005) Practical linear algebra: a geometry toolbox. A K Peters, Wellesley, MA 3. Schneider PJ, Eberly DH (2003) Geometric tools for computer graphics. Morgan Kaufmann Publishers, Amsterdam 4. Noll W (1987) Finite-dimensional spaces. Martinus Nijhoff Publishers, Dordrecht 5. Jost J (1998) Postmodern analysis. Springer, New York 6. Lang S (1997) Undergraduate analysis, 2nd edn. Springer, New York 7. Akivis MA, Goldberg VV (1977) An introduction to linear algebra and tensors. Dover Publications Inc, New York 8. Block HD (1978) Introduction to tensor analysis. Charles E. Merrill Books Inc, Columbus, Ohio 9. Borisenko AI, Tarapov IE (1979) Vector and tensor analysis. Dover Publications Inc, New York 10. Chen HC (1985) Theory of electromagnetic waves: a coordinate-free approach. McGraw-Hill Book Company, New York 11. Chou PC, Pagano NJ (1967) Elasticity: tensor, dyadic, and engineering approaches. D. Van Nostrand Company Inc, Princeton, New Jersey 12. Curtis CW (1984) Linear algebra: an introductory approach. Springer, New York 13. Danielson DA (1997) Vectors and tensors in engineering and physics, 2nd edn. Addison-Wesley, Reading, Massachusetts 14. Drew TB (1961) Handbook of vector and polyadic analysis. Reinhold Publishing Corporation, New York 15. Gibbs JW, Wilson EB (1901) Vector analysis. Yale University Press, New Haven, Connecticut 16. Gurtin ME (1981) An introduction to continuum mechanics. Academic Press, New York 17. Halmos P (1987) Finite-dimensional spaces. Springer, Berlin 18. Hjelmstad KD (1997) Fundamentals of structural mechanics. Prentice-Hall Inc, Upper Saddle River, New Jersey 19. Soos E, Beju I, Teodorescu PP (1983) Euclidean tensor calculus with applications. Abacus Press, Tunbridge Wells, Kent (UK) 20. Lewis PE, Ward PJ (1989) Vector analysis for engineers and scientists. Addison-Wesley, Wokingham (UK) 21. Lichnerowicz A (1962) Elements of tensor calculus. Methuen & Co, Ltd., London 22. Lohr E (1950) Vektor- und Dyadenrechnung für Physiker und Techniker, 2nd edn. Walter de Gruyter & Co, Berlin (in German) 23. Lurie AI (2005) Theory of elasticity. Springer, Berlin 24. Marsden JE, Hughes TJR (1983) Mathematical foundations of elasticity. Prentice-Hall Inc, Englewood Cliffs, New Jersey 25. Meyer CD (2000) Matrix analysis and applied linear algebra. SIAM (Society for Industrial and Applied Mathematics), Philadelphia 26. Nadeau G (1964) Introduction to elasticity. Holt, Rinehart and Winston Inc, New York 27. Pach K, Frey T (1964) Vector and tensor analysis. Terra, Budapest, Hungary 28. Podio-Guidugli P (2000) A primer in elasticity. Kluwer Academic Publishers, Dordercht, The Netherlands 29. Schumann W, Dubas M (1979) Holographic interferometry. Springer, Berlin 30. Schumann W, Zürcher J-P, Cuche D (1985) Holography and deformation analysis. Springer, Berlin 31. Simmonds JG (1995) A brief on tensor analysis, 2nd edn. Springer, New York 32. Weatherburn CE (1960) Advanced vector analysis. G. Bell and Sons Ltd., London

Chapter 2

Kinematics

Abstract Kinematics represents the branch of Mechanics that studies the motion of a mechanical body or a system of such bodies without consideration given to its mass or the forces acting on it. In this chapter, we make the first few steps towards describing the geometrical features of the mathematical model for a deformable body. In contrast to the kinematics of a rigid body, here the description will be largely local in nature because the changes undergone by a deformable body under the action of external forces differ from place to place.

2.1 Deformable Bodies: Definition and Generalities Definition 2.1 A body B is a set whose elements can be put into one-to-one correspondence with points of a region in the three-dimensional Euclidean point space. We shall refer to such a region as a configuration of the body. The elements of B are called particles or material points (see Fig. 2.1). According to this definition, we have an invertible mapping κ : B × [0, ∞) → E3 which associates to each element P ∈ B at time t ≥ 0, a point in E3 , say, κ(P, t). The dependence on time (here denoted by t) is included because during a motion the body changes its configuration with time. In order to provide a natural description of the motions undergone by B, a fixed (but otherwise arbitrary) configuration is chosen as reference configuration, and this will be denoted by Br . For the sake of simplicity, we choose Br := κ(B, 0). On the other hand, the configuration of B at time t > 0 will be denoted by Bc := κ(B, t), © Springer Nature B.V. 2020 C. D. Coman, Continuum Mechanics and Linear Elasticity, Solid Mechanics and Its Applications 238, https://doi.org/10.1007/978-94-024-1771-5_2

121

122

2 Kinematics

. B

B

.

. B

Fig. 2.1 The geometry of the motion for deformable bodies: the same material particle P in the body is characterised by a position vector X in the reference configuration (Br ), while in the current configuration its position is given by x. These position vectors are linked by the invertible mapping χ through the equation x = χ(X, t)

and this represents the current configuration of the body; to make the dependence on ‘time’ more conspicuous, sometimes we shall also use the alternative notation Bt for Bc . We note that by letting X := κ(P, 0)

and

x := κ(P, t),

the point P can be eliminated from these two relations, x = κ(κ −1 (X, 0), t) ≡ χ (X, t), where χ (· , ·) is defined by the right-hand side of the above equality. This mapping is evidently bijective, and we shall assume that it has all the differentiability properties required in what follows. In Continuum Mechanics, χ is known as the motion of the body B. At fixed time, χ defines the deformation of the body from Br to Bc , while at a fixed (generic) point X, χ describes the trajectory of the point as time passes. In the latter case, t → χ (X, t) ∈ E3 represents a three-dimensional curve. The understanding of the kinematics of B is facilitated by the introduction of an appropriate system of coordinates. For maximum clarity, we shall consider two Cartesian systems of coordinates associated with the reference and current configurations, respectively. Fields in Br are referred to a system of coordinates defined by the origin O ∈ E3 and the orthonormal set of vectors {E 1 , E 2 , E 3 }. Thus, the position

2.1 Deformable Bodies: Definition and Generalities

123

of a point P ∈ B in the reference configuration is given by the position vector1 of X ≡ κ(P, 0) situated in Br . Similarly, we introduce another fixed orthonormal Cartesian system of coordinates with origin o ∈ E3 and basis vectors {e1 , e2 , e3 }. The position of the same point P in the current configuration is given by the position vector of x ≡ κ(P, t) which can be resolved in terms of the vectors e1 , e2 , e3 . For the remaining of this chapter, the two systems of coordinates are taken to be coincident, but we shall retain the notational distinction between the vectors E α and ei in order to better highlight the nature of the various tensors introduced in the subsequent sections. Problems in Continuum Mechanics may be formulated by employing either (X, t) as independent variables—in the so-called material or Lagrangian description—or (x, t) as independent variables—in the so-called spatial or Eulerian description. To switch from one formulation to another, it is necessary to use χ (X, t) and its inverse χ −1 (x, t). The referential coordinates of a (generic) point are denoted by X α (α = 1, 2, 3) and its spatial coordinates by xi (i = 1, 2, 3). In general, if f s (x, t) is the spatial description of a scalar, vector or tensor field, the corresponding material description f m (X, t), is defined by χ (X, t), t) ≡ f s (x, t) , f m (X, t) := f s (χ and vice versa

χ −1 (x, t), t) ≡ f m (X, t) . f s (x, t) := f m (χ

The presence of X and x throughout this chapter (and later) requires the adoption of a notation that would allow us to distinguish easily between the reference and current configurations. To this end, we agree to use small Greek letters for indices associated with material fields, and small Roman letter indices for spatial fields. As far as possible, we shall denote scalar, vector and tensor quantities evaluated in the reference configuration by capital letters, while the corresponding quantities evaluated in the current configuration will be identified by lower case letters. Another ambiguity generated by the use of Lagrangian and Eulerian descriptions involves various differential operators defined in terms of partial derivatives. In this respect, we shall denote the ‘gradient’, ‘divergence’ and ‘curl’ operators with regard to X ∈ Br by Grad, Div and Curl, respectively; similarly, the corresponding operators with reference to x ∈ Bc will be written as grad, div and curl.

2.2 Examples of Deformations/Motions Definition 2.2 A motion of a deformable body is said to be rigid if the distance between any two of its material points remains unchanged. Such motions have the 1 Recall that, according to the discussion in Sect. 1.4 of Chap. 1, points are identified with their position vectors.

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2 Kinematics

B B

Fig. 2.2 A particular case of rigid deformation corresponding to (2.3)

general expression x = χ (X, t) ≡ Q(t) · X + c(t) ,

(2.1)

where Q(t) represents a proper orthogonal tensor (which may depend on t) and c(t) is a given vector (which may also depend on t). In the expression (2.1), c ≡ c(t) represents a translation and Q ≡ Q(t) a rotation. Note also that the inverse motion is easily obtained by left-multiplication of (2.1) by Q T , in conjunction with Q T · Q = I, so that X = χ −1 (x, t) ≡ Q T (t) · (x − c(t)) .

(2.2)

To illustrate geometrically the effect of a rigid deformation, consider the following particular case of (2.1) given by x1 = c1 + X 1 cos θ − X 2 sin θ , x2 = c2 + X 1 sin θ + X 2 cos θ , x3 = X 3 . (2.3) Figure 2.2 shows what happens to a 3-by-3 square in the reference configuration under the action of the mapping given by (2.3): every point in the square is shifted c1 units horizontally, c2 units vertically, and then all the points in the square are rotated by the same amount (which is controlled by the angle θ ). To convince ourselves that the motion (2.1) has indeed the property asserted above, we evaluate the distance between two arbitrary points, x, y ∈ Bc —the images of X, Y ∈ Br under the deformation χ (· , t) defined in (2.1). We have successively,

2.2 Examples of Deformations/Motions

simple extension

125

simple shear

B

B

B

Fig. 2.3 Simple extension (left) and simple shear (right)

|x − y|2 = (x − y) · (x − y) = [ Q · (X − Y )] · [ Q · (X − Y )] = [( Q T · Q) · (X − Y )] · (X − Y ) = (X − Y ) · (X − Y ) = |X − Y |2 or, by taking square roots, |x − y| = |X − Y |, which confirms our foregoing expectation. In the remaining of this section, we record several basic examples of typical deformations encountered in Solid Mechanics. Simple extension: Consider the linear mapping χ : E3 → E3 defined by x1 = (1 + α)X 1 ,

x2 = X 2 ,

x3 = X 3 ,

(2.4)

where α > 0 is a given constant. A simple two-dimensional illustration of this deformation is shown in the left sketch of Fig. 2.3. Any straight line in Br is mapped to another straight line in Bc . More precisely, lines parallel to the 2-direction are simply translated horizontally and do not undergo any change of length. On the other hand, lines parallel to the 1-direction will be stretched (i.e. their length will increase), while still remaining parallel to that direction; any other lines will undergo both an extension and a rotation (relative to their original slope). Simple shear: This deformation is characterised by a relative displacement of parallel planes by an amount proportional to the distance between the planes, and in a direction parallel to the planes. A prototypical example of mapping that reproduces this type of behaviour is given by x1 = X 1 + α X 2 ,

x2 = X 2 ,

x3 = X 3 ,

(2.5)

where α > 0 is a given constant (referred to as the amount of shear). The right sketch in Fig. 2.3 illustrates what happens to a 2-by-2 square under the action of (2.5). The material particles move (or deform) only in the 1-direction, their

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2 Kinematics

Fig. 2.4 Geometrical illustration of simple deformations. The left sketch corresponds to the dilatation deformation defined in (2.6); a unit cube in Br is transformed into a similar cube of side α > 1 in Bc . On the right, the mapping (2.7) is applied to a right circular cylindrical deformable body, whose lower end is fixed; the straight generator aa  in Br is mapped to the curve aa  in the current configuration Bc

displacement being proportional to their 2-coordinate. Also, the planes perpendicular to the 1-direction are rotated about an axis parallel to the 3-axis through an angle tan−1 (α), known as the angle of shear. The planes X 2 = constant are known as the shearing planes and the positive X 1 -direction is the shear direction; furthermore, the planes X 3 = constant represent the planes of shear, while the lines parallel to the X 3 -axis correspond to the axes of shear. Dilatation: x1 = α X 1 ,

x2 = α X 2 ,

x3 = α X 3 ,

(2.6)

where α ∈ R. If α > 1 then we have the situation seen on the left in Fig. 2.4; a unit cube is changed into another larger cube (of side α). However, for 0 < α < 1, the cube will be compressed into a smaller cube (not shown in the foregoing Figure). We can represent (2.6) as x = α I · X. Simple torsion: The last two deformations we include here are more complicated. The first one, described by the Eq. (2.7) below, corresponds to a torsional mode of deformation for right circular cylindrical bodies ⎧ ⎪ ⎨x1 = X 1 cos(τ X 3 ) − X 2 sin(τ X 3 ) , x2 = X 1 sin(τ X 3 ) + X 2 cos(τ X 3 ) , ⎪ ⎩ x3 = X 3 .

(2.7)

In these equations τ > 0 is a given constant (commonly referred to as the twist per unit length); see the right sketch in Fig. 2.4. The planar transverse cross sections of the cylinder remain plane, but they are rotated through an angle τ X 3 about the X 3 -axis.

2.2 Examples of Deformations/Motions

127

B

B

Fig. 2.5 Pure bending: a deformable cuboid (Br ) is transformed into a sector of a circular cylindrical tube (Bc ). Shown on the right is an illustration of the two-dimensional effect the mapping (2.8) has on the transverse cross sections of the original cuboid (the original rectangular shapes are mapped to annular sectors)

Pure bending: The description of this mode of deformation involves Cartesian coordinates X β (β = 1, 2, 3) in Br , and cylindrical polar coordinates (r, θ, z) in Bc . Its corresponding equations are r = f (X 1 ) ,

θ = g(X 2 ) ,

z = λX 3 ,

(2.8)

where f = f (x) and g = g(x) are given functions of one variable. Figure 2.5 provides an intuitive picture of the main features of this mapping: a deformable cuboid is bent into a sector of a thick circular cylindrical shell. The material planes perpendicular to the X 1 -axis become circular cylindrical surfaces in Bc , while the planes perpendicular to the X 2 -axis are deformed into radial planes containing the z-axis in the current configuration.

2.3 Velocity and Acceleration Fields. Material Time Derivatives The velocity and acceleration fields of a continuum body are the primary kinematic fields used in describing its motion. Since every deformable body is endowed with two different types of configurations (material and spatial), the velocity and acceleration fields will also admit two distinct representations.

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2 Kinematics

Definition 2.3 (a) The velocity V ≡ V (X, t) of a particle P ∈ B is defined as V (X, t) :=

χ (X, t) ∂χ , ∂t

(2.9)

where X is the position of P in Br . This is the rate of change of the position of P in the Lagrangian description. The corresponding vector field in the Eulerian description is  χ (X, t)  ∂χ χ −1 (x, t), t) . ≡ V (χ (2.10) v(x, t) :=  ∂t −1 χ X=χ (x, t) (b) The acceleration A ≡ A(X, t) of a particle P ∈ B is A(X, t) :=

∂ 2χ (X, t) . ∂t 2

(2.11)

Note that this represents a Lagrangian field; its Eulerian counterpart is  ∂ 2χ (X, t)  χ −1 (x, t), t) . a(x, t) := ≡ A(χ  ∂t 2 χ −1 (x, t) X=χ

(2.12)

It is worth emphasising that X ∈ Br and x ∈ Bc above represent the position vectors of the (same) particle P ∈ B in the reference and the current configurations, respectively. As we shall see next, it turns out that the velocity field plays an important role in calculating the rate of change of other fields given in the Eulerian description. This is related to the concept of material time derivative. Let us imagine the following situation: suppose that a certain field (scalar, vectorial or tensorial) is defined over the body and we wish to know its rate of change as would be recorded at a given particle X during the motion. There are two situations to consider, as explained below. In the material description the independent variables are X and t, so all we have to do is take the partial derivative of the given field with respect to t. For example, if φ ≡ φ(X, t) is a scalar field, then φ˙ ≡

∂φ(X, t) Dφ := , Dt ∂t

where the first two equalities indicate the standard notation for the material time derivative. We can similarly calculate material time derivatives of any vector and tensor fields given in the Lagrangian description. For an Eulerian field, things are a bit more complicated: we must calculate the total derivative with respect to time of the material description of the field under consideration, keeping X fixed. This process is carried out with the help of the chain rule for multivariate functions. By way of example, let us consider in detail the case of a scalar field φ ≡ φ(x, t). According to what we have just said, the material time

2.3 Velocity and Acceleration Fields. Material Time Derivatives

129

derivative of this Eulerian field is (by definition) φ˙ ≡

 χ (X, t), t]  Dφ dφ[χ := ,  Dt dt χ −1 (x, t) X=χ

(2.13)

and we note that the same rule applies for vector and tensor fields as well (with some obvious changes, of course). In practice, the material time derivative can be calculated as follows. The first step is to convert the field in question into its material representation by using the equation x = χ (X, t), χ (X, t), t] ≡ φ[χ1 (X, t), χ2 (X, t), χ2 (X, t), t]. φ(x, t) = φ[χ Differentiating with respect to t this equation, while keeping X fixed (constant), leads to χ (X, t), t] ∂χi (X, t) ∂φ[χ χ (X, t), t] χ (X, t), t] ∂φ[χ dφ[χ = + . dt ∂ xi ∂t ∂t

(2.14)

The last step consists in returning to the spatial description by making the substitution X = χ −1 (x, t) in (2.14). Thus, we eventually find  χ (X, t), t]  dφ[χ ∂φ(x, t) ∂φ(x, t) , = vi (x, t) +  dt ∂ x ∂t −1 i χ (x, t) X=χ where vi (x, t) are the components of the Eulerian velocity field defined in (2.10). Note that the formula we have proved can also be stated in intrinsic form, φ˙ ≡

Dφ ∂φ(x, t) = v(x, t) · grad φ(x, t) + ; Dt ∂t

(2.15)

it is customary to leave out the dependence on x and t, in which case φ˙ ≡

Dφ ∂φ = (gradφ) · v + . Dt ∂t

For an Eulerian vector field, these calculations can be easily adapted as we indicate below. According to the definition (2.13), the material time derivative of such a vector field u ≡ u(x, t) is  χ (X, t), t]  Du du[χ u˙ ≡ := . (2.16)  Dt dt χ −1 (x, t) X=χ If u(x, t) = ui (x, t)ei , we find

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2 Kinematics

 χ (X, t), t]  ∂ui (x, t) ∂ui (x, t) dui [χ = v j (x, t) +  dt ∂ x ∂t −1 j χ (x, t) X=χ = (grad u(x, t))i j v j (x, t) + and thus, u˙ ≡

∂ui (x, t) (i = 1, 2, 3) , ∂t

Du ∂u(x, t) = (grad u(x, t)) · v(x, t) + Dt ∂t

or u˙ ≡

(2.17)

∂u Du = (gradu) · v + , Dt ∂t

where in the last equation the dependence on x and t has been suppressed. A common use of the dot-notation for material time derivatives appears in the representation of the velocity and acceleration fields, namely, x˙ ≡ v(x, t) , x¨ ≡ a(x, t) , χ˙ ≡ V (X, t) , χ¨ ≡ A(X, t) . We close this section with a number of simple examples that illustrate the calculations required to obtain concrete expressions for the concepts introduced above. Example 2.1 The motion of a deformable body is described by x1 = X 1 ,

x2 = X 2 cosh t + X 3 sinh t ,

x3 = X 2 sinh t + X 3 cosh t . (2.18) It is required to find the components of the velocity field in both their material (Lagrangian) and spatial (Eulerian) forms. The Lagrangian form of the required components of the velocity field V = (V1 , V2 , V3 ) are obtained by simply taking the partial derivatives of the expressions in (2.18). This gives     V (X, t) = X 2 sinh t + X 3 cosh t e2 + X 2 cosh t + X 3 sinh t e3 .

(2.19)

According to (2.10), the Eulerian description of this velocity field is obtained by making use of the inverse motion corresponding to (2.18). Note that



−1



cosh t sinh t x2 cosh t − sinh t x2 X2 = = , X3 x3 x3 sinh t cosh t − sinh t cosh t

(2.20)

with the last relation providing X 2 and X 3 as functions of x2 and x3 . Consequently,

2.3 Velocity and Acceleration Fields. Material Time Derivatives

131

χ −1 (x, t), t) v2 = V2 (χ = (x2 cosh t − x3 sinh t) sinh t + (−x2 sinh t + x3 cosh t) cosh t = x3 (cosh2 t − sinh2 t) = x3 , and, similarly, χ −1 (x, t), t) v3 = V3 (χ = (x2 cosh t − x3 sinh t) cosh t + (−x2 sinh t + x3 cosh t) sinh t = · · · = x2 . Hence, the spatial form of the velocity field is v = x3 e2 + x2 e3 . Example 2.2 The Eulerian description of the velocity field of a deformable body is x1 2x2 3x3 , v2 = , v3 = . (2.21) v1 = 1+t 1+t 1+t Determine the motion of the body by integrating these equations and using the consistency condition x(t = 0) = X. Also, find the spatial representation of the acceleration field for this deformable body. The first part of the question is easily answered by observing that v j = d x j /dt ( j = 1, 2, 3), and hence the particular expressions in (2.21) lead to separable ordinary differential equations. For example, for the first equation we have x1 d x1 = dt 1+t

⇒

d x1 dt = x1 1+t

⇒ ln x1 = ln K (1 + t) ,

whence x1 = K (1 + t), for some K ∈ R. Since x1 (t = 0) = X 1 we immediately find x1 = X 1 (1 + t). For the second equation, 2x2 d x2 = dt 1+t

⇒

d x2 2dt = x2 1+t

⇒ ln x2 = ln K 2 (1 + t)2 ,

so that x2 = K 2 (1 + t)2 . Applying further the initial condition x2 (t = 0) = X 2 it transpires that x2 = X 2 (1 + t)2 . Finally, the last equation is handled in a similar way, eventually leading to x3 = X 3 (1 + t)3 . In conclusion, the motion χ (say) corresponding to the Eulerian velocity field (2.21) is given by the equations x1 = X 1 (1 + t) ,

x2 = X 2 (1 + t)2 ,

x3 = X 3 (1 + t)3 ,

(2.22)

x3 , (1 + t)3

(2.23)

while the inverse motion χ −1 is X1 =

x1 , (1 + t)

X2 =

x2 , (1 + t)2

X3 =

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2 Kinematics

The Lagrangian description of the acceleration field is obtained by calculating the second-order partial derivatives of the expressions in (2.22). According to (2.12), the Eulerian form of the same field is then obtained by replacing the X j ( j = 1, 2, 3) in those results with the expressions in (2.23); the outcome of these trivial manipulations is recorded below a1 = 0 ,

a2 =

2x2 , (1 + t)2

a3 =

6x3 . (1 + t)2

The same result could be obtained by taking into account that a = v, ˙ and then making use of (2.17) with u → v; see Exercise 2.3(a) at the end of this chapter. Example 2.3 The motion of a deformable body is given by x1 = X 1 + X 22 t 2 ,

x2 = X 2 + X 2 t 3 ,

x3 = X 3 .

(2.24)

(a) Find the acceleration at t = 2 of the particle that was initially at (1, 3, 2); (b) Find the acceleration of a particle which is at (1, 3, 2) at t = 2. To answer the first part, let X 0 := (1, 3, 2) and t0 := 2, and note that we are asked to calculate A(X 0 , t0 ). The general expression of the Lagrangian description of the acceleration field is calculated directly from (2.11), giving A1 (X, t) = 2X 22 ,

A2 (X, t) = 6t X 2 ,

A3 (X, t) = 0 .

(2.25)

Evaluating these expressions at X = X 0 and t = t0 leads to A(X 0 , t0 ) = (18, 36, 0). For the second part of the question, let x 0 := (1, 3, 2) and t0 := 2; we are asked to calculate a(x 0 , t0 ). Recalling formula (2.12), and taking into account (2.25), the general expression of the Eulerian form of the acceleration field turns out to be a1 (x, t) =

2x23 , (1 + t 3 )2

a2 (x, t) =

6x2 t , 1 + t3

a3 (x, t) = 0 .

(2.26)

Hence, letting x → x 0 and t → t0 in these formulae we find the required answer, a(x 0 , t0 ) = (2/9, 4, 0).

2.4 The Deformation Gradient The mapping χ introduced in the first section of this chapter provides a global description of the motion experienced by a deformable body. Its local behaviour, however, is dictated by the gradient of χ —a quantity that will play a prominent role for the rest of this book.

2.4 The Deformation Gradient

133

. B

B

Fig. 2.6 The deformation gradient F provides a local description of the motion. More precisely, d x = F · dX, where d X and d x are infinitesimal vectors based at X ∈ Br and x ∈ Bc , respectively, with x = χ (X, t); see text for full details

To motivate the introduction of the next concept, let us consider two neighbouring particles in B, whose position vectors in Br are X and X + dX; in Fig. 2.6 they are −−→ identified as P and Q, respectively. We note in passing that the vector P Q is equal to dX. Under the action of the mapping χ , these two points will be located at P  and Q  , whose position vectors are χ (X, t) and χ (X + dX, t), respectively. The −−−→ difference of these two quantities will give us the vector P  Q  . Furthermore, since χ is continuous, it is natural to expect that the length of this vector in Bc will be an infinitesimal quantity dx, and we can also write    χ (X, t) · dX + O |dX|2 . dx = χ (X + dX, t) − χ (X, t) = Gradχ

(2.27)

If the separation between the points P and Q in Br is very small, then the higher χ plays a key order terms in (2.27) can be ignored, and it becomes clear that Gradχ role in describing the local geometry of the motion for all points near X ∈ Br . Definition 2.4 The deformation gradient tensor F ≡ F(X, t) is defined by F(X, t) := Grad x ≡ Grad χ (X, t) .

(2.28)

The component representation of this second-order tensor field is F = F pα e p ⊗ E α ,

F pα = x p, α ≡

∂xp . ∂ Xα

(2.29)

χ −1 (· , t) : Bc → χ (· , t) : Br → Bc is bijective for all t > 0, it has an inverse,χ Sinceχ Br (for each fixed t > 0). Thus, we can define the second-order tensor field F −1 (x, t) := grad X ≡ grad χ −1 (x, t) , with the component representation

(2.30)

134

2 Kinematics −1 F −1 = Fαp Eα ⊗ e p ,

−1 Fαp = X α, p ≡

∂ Xα . ∂xp

(2.31)

We can then immediately show that this F −1 defined in (2.30) is in fact the inverse of the deformation gradient F. To this end, we need to check that F · F −1 = F −1 · F = I, where I is the usual second-order identity tensor. Indeed, −1 E β ⊗ eq ) F · F −1 = (F pα e p ⊗ E α ) · (Fβq −1 −1 = F pα Fβq (E α · E β )(e p ⊗ eq ) = F pα Fβq δαβ (e p ⊗ eq ) −1 = F pα Fαq (e p ⊗ eq ) = x p, α X α, q (e p ⊗ eq )

= x p, q (e p ⊗ eq ) = δ pq e p ⊗ eq = e p ⊗ e p = I, where use has been made of the chain rule, x p,α X α,q = x p,q . It can also be shown that, similarly, F −1 · F = I, and hence the desired conclusion. The deformation gradient is an example of two-point tensor: it maps infinitesimal vectors between the reference and the current configurations. We refer to our earlier discussion at the end of Sect. 1.5 which sheds further light on the representation given in Eq. (2.29). Note that since F is invertible, J ≡ J (X, t) := det F(X, t) = 0 .

(2.32)

Taking into account that F = I in the reference configuration (at time t = 0), and det F is a continuous real-valued mapping which does not vanish, we deduce that (∀) X ∈ Br , (∀) t ≥ 0 ;

J (X, t) > 0,

J is known as the Jacobian of the motion (or, the Jacobian determinant, for short). Proposition 2.1 If φ and u are, respectively, scalar and vector fields representing properties of a deformable body, then Grad φ = F T · grad φ , Grad u = (grad u) · F . For (2.33a) we have successively, Grad φ − F T · grad φ = φ, α E α − (x p, α e p ⊗ E α )T · (φ, q eq ) = φ, α E α − (x p, α E α ⊗ e p ) · (φ, q eq ) = φ, α E α − x p, α φ, q E α (e p · eq ) = φ, α E α − x p, α φ, q δ pq E α = φ, α E α − φ, p x p, α E α = 0 , since φ, p x p, α = φ, α . In a similar way, for the second relation (2.33b),

(2.33a) (2.33b)

2.4 The Deformation Gradient

135

Grad u − (grad u) · F = uα, β E α ⊗ E β − (u p, q e p ⊗ eq ) · (xr, ρ er ⊗ E ρ ) = uα, β E α ⊗ E β − u p, q xr, ρ (eq · er )e p ⊗ E ρ = uα, β E α ⊗ E β − u p, q xr, ρ δqr e p ⊗ E ρ = uα, β E α ⊗ E β − u p, q xq, ρ e p ⊗ E ρ = uα, β E α ⊗ E β − u p, β e p ⊗ E β = (uα E α − u p e p ), β ⊗ E β = O , since u = uα E α = u p e p . Proposition 2.2 If u is a vector field and T is a tensor field associated with a deformable body, then Div u = J div (J −1 F · u) , Div T = J div (J

−1

F · T) ,

(2.34a) (2.34b)

where J represents the Jacobian determinant defined in (2.32). The justification of these two formulae rests on the following auxiliary result (known as the Piola identity—see Exercise 2.8) div (J −1 F) ≡ (J −1 F pα ), p E α = 0 .

(2.35)

With the formula (2.35) in hand, we have successively ∂uα ∂ −J (J −1 F pβ uβ ) ∂ Xα ∂xp

∂uβ ∂ ∂uα −1 −1 −J (J F pβ ) uβ + J F pβ = ∂ Xα ∂xp ∂xp

 

Div u − J div(J −1 F · u) =

(2.35)

=

∂uβ ∂ x p ∂uα ∂uα ∂uβ − J J −1 = − = 0. ∂ Xα ∂ x p ∂ Xβ ∂ Xα ∂ Xβ

  uβ, β

Similarly, for (2.34b) we obtain ∂ Tαβ ∂ Eβ − J (J −1 F pα Tαq )eq ∂ Xα ∂xp

∂ Tαβ ∂ Tαq ∂ = Eβ − J (J −1 F pα ) Tαq eq + J −1 F pα eq ∂ Xα ∂xp ∂xp

 

Div T − J div(J −1 F · T ) =

(2.35)

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2 Kinematics

=

∂ Tαβ ∂ Tαq ∂ x p E β − J J −1 eq ∂ Xα ∂ x p ∂ Xα

  Tαq, α

=

∂ Tαβ ∂ Tαq ∂ Eβ − eq = (Tαβ E β − Tαq eq ) = 0 . ∂ Xα ∂ Xα ∂ Xα

The last equality is justified by the following argument: T = Tαβ E α ⊗ E β = Tαq E α ⊗ eq =⇒ E α ⊗ [Tαβ E β − Tαq eq ] = O =⇒ Tαβ E β − Tαq eq = 0 . The mapping χ introduced in Sect. 2.1 provides a means of describing the degree of ‘departure’ of Bc from Br . This feature can be made more conspicuous by introducing the concept of displacement field associated with a deformable body. Definition 2.5 The displacement u of a particle is defined as u := x − X ≡ χ (X, t) − X ≡ x − χ −1 (x, t) .

(2.36)

Note that the displacement can be regarded either as a Lagrangian field (independent variables: X and t) or as an Eulerian field (independent variables: x and t). The definition (2.36) yields x = X +u

=⇒

Grad x = Grad X + Grad u ,

where the last relation can also be cast in the equivalent form F = I+H,

H := Grad u .

(2.37)

The tensor H is called the displacement gradient and represents an important kinematic quantity in the linearised theories of Continuum Mechanics (e.g. Linear Elasticity).

2.5 Changes in Area and Volume One of the principal aims of kinematics in relation to deformable bodies is to provide concrete formulae for the changes in length, area, and volume experienced by various material sets as they move between Br and Bc . Changes in length will be postponed until we reach Sect. 2.6 as they are best captured by the so-called strain and deformation tensors. Our next immediate task is to find out how the element of area and the element of volume change when moving from the reference to the current configuration.

2.5 Changes in Area and Volume

137

B

B

. Fig. 2.7 Transformation of surface areas under the action of χ (· , t) for a fixed t > 0

Consider a surface Sr ⊂ Br which deforms into another surface Sc ⊂ Bc (that is, Sc = χ (Sr , t)). Assume that X ∈ Sr , and let x ∈ Bc be its image under the mapping χ (· , t) at some fixed instant of time. Let dX and dY be material line elements based at X, and let dx and d y be their counterparts in the current configuration (see Fig. 2.7). If F denotes the deformation gradient, then dx = F · dX,

d y = F · dY .

(2.38)

By definition, the elements of area in the reference and current configurations, respectively, are given by d A := |dX ∧ dY |

and

da := |dx ∧ d y| .

(2.39)

If we denote by N and n the unit normals to the two surfaces, at X ∈ Br and x ∈ Bc , respectively, then dX ∧ dY ⇒ |dX ∧ dY | dx ∧ d y n= ⇒ |dx ∧ d y|

N=

Nd A = dX ∧ dY ,

(2.40a)

nda = dx ∧ d y.

(2.40b)

From (2.40b) to (2.38), it follows that nda = dx ∧ d y = (F · dX) ∧ (F · dY ) = cof(F) · (dX ∧ dY ),

(2.41)

the last equality being obtained with the help of Definition 1.12. To go further, we need the formula (1.182), according to which cof(F) = (det F)F −T . Going back into (2.41) with this result, we find nda = (det F)F −T · (dX ∧ dY ) = (det F)F −T · Nd A . In conclusion,

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2 Kinematics

B

B

. Fig. 2.8 Change in the volume elements associated with Br and Bc

nda = J F −T · Nd A ,

(2.42)

where J ≡ det F. This is a very important result, known as Nanson’s formula, describing how an oriented element of surface area deforms under a given motion. Changes in the volume elements associated with the reference and current configurations are tackled next. The referential element of volume dV ≡ dV (X) (at X ∈ Br ) is by definition dV := dX · (dY ∧ dZ) ≡ [dX, dY , dZ] ,

(2.43)

where dX, dY , dZ denote mutually perpendicular infinitesimal vectors based at X. During the motion, dV is carried into an element of volume dv ≡ dv(x) (at x ∈ Bt — see Fig. 2.8), whose definition is dv := dx · (d y ∧ dz) ≡ [dx, d y, dz] .

(2.44)

Remembering that dx = F · dX ,

d y = F · dY ,

dz = F · dZ ,

we can relate the two volume elements with the help of (2.43) and (2.44), dv = [F · dX, F · dY , F · dZ] = (det F)[dX, dY , dZ] , the last equality being a consequence of Proposition 1.7. In conclusion, dv = J dV ,

(2.45)

where J ≡ det F is the Jacobian of the motion. It is clear from (2.45) that J measures the change in volume during a motion; later, in Sect. 2.10, we are going to look at the rate of change of this quantity. Definition 2.6 If a motion χ is such that there is no change in volume, then the motion is said to be isochoric. In that case,

2.5 Changes in Area and Volume

139

J (X, t) ≡ det F(X, t) = 1,

(∀) (X, t) ∈ Br × (0, ∞) .

(2.46)

In order to check that a given motion is isochoric we must first calculate the deformation gradient F by using (2.29), and then evaluate its determinant. If this determinant is identically equal to unity, then the motion is isochoric.

2.6 Strain Tensors Let us investigate now how material line elements are stretched or contracted during χ is the fundamental measure an arbitrary deformation χ (· , t). The tensor F ≡ Gradχ of deformation that plays a key role in capturing such information. Assuming that M and m are unit vectors along dX and dx, respectively, dX = M |dX|

and

dx = m |dx| .

By using these formulae in dx = F · dX, we get m|dx| = (F · M)|dX| and hence,     (m|dx|) · (m|dx|) = (F · M)|dX| · (F · M)|dX| ⇒ (m · m)|dx|2 = (F · M) · (F · M)|dX|2 .

(2.47)

But, the left-hand side in (2.47) is just |dx|2 (since m · m = |m|2 = 1), while the right-hand side can be cast in the alternative form (F · M) · (F · M)|dX|2 = [M · (F T · F) · M]|dX|2 .

(2.48)

Combining together (2.47) and (2.48) leads to |dx| = |dX|

 M · (F T · F) · M = |F · M| .

(2.49)

The right Cauchy–Green deformation tensor, C ≡ C(X, t), is defined as C := F T · F ,

(2.50)

with the component representation Cαβ = Cαβ E α ⊗ E β ,

Cαβ = x p, α x p, β ≡ F pα F pβ .

(2.51)

Note that C is a Lagrangian measure of deformation that operates on the reference configuration Br . Definition 2.7 The stretch in the direction M at X ∈ Br is defined as

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2 Kinematics

B

. B Fig. 2.9 Change of angle between material line elements as a result of the motion χ . The angle Θ in Br becomes θ in Bc , with the difference Θ − θ being known as the ‘angle of shear’

λ(M) ≡

√ |dx| = M·C·M, |dX|

(2.52)

where the last equality follows from (2.49) and (2.50). We note in passing that the stretch is a non-dimensional quantity that satisfies 0 < λ(M) < ∞. Let us see now how angles between arbitrary material lines are locally changed during the deformation χ (· , t). To this end, consider a pair of material line elements dX 1 , dX 2 based at X. During the motion they are mapped to dx 1 and dx 2 , respectively, and will be based at x = χ (X, t); in consequence, dx 1 = F · dX 1 and dx 2 = F · dX 2 (see Fig. 2.9). As before, let M 1 , M 2 and m1 , m2 be unit vectors along dX 1 , dX 2 and dx 1 , dx 2 , respectively. If we set Θ := ∠ (M 1 , M 2 ) ,

θ := ∠ (m1 , m2 ) ,

then cos Θ = M 1 · M 2 and cos θ = m1 · m2 . Our immediate objective is to work out an expression for cos θ in terms of quantities that we know or have already been defined. We have successively, dx 2 (F · dX 1 ) · (F · dX 2 ) dx 1 · = |dx 1 | |dx 2 | |dx 1 ||dx 2 | (F · M 1 ) · (F · M 2 ) [F · (dX 1 /|dX 1 |)] · [F · (dX 2 /|dX 2 |)] = = (|dx 1 |/|dX 1 |)(|dx 2 |/|dX 2 |) λ(M 1 )λ(M 2 ) T M 1 · (F · F) · M 2 = , λ(M 1 )λ(M 2 )

cos θ = m1 · m2 =

and hence cos θ =

M1 · C · M2 . λ(M 1 )λ(M 2 )

(2.53)

2.6 Strain Tensors

141

The decrease Θ − θ in the angle between the pairs of line elements is called the angle of shear of the directions M 1 and M 2 , in the plane of shear defined by M 1 and M 2 . The components of the right Cauchy–Green deformation tensor admit a simple physical interpretation. For instance, by choosing M := E α in (2.52), gives λ(E α ) =



Eα · C · Eα =

 Cαα

or Cαα = λ2 (E α ) ,

(α = 1, 2, 3, no sum) .

(2.54)

According to (2.54), Cαα (α = 1, 2, 3, no sum) is the square of the stretch undergone by a material line element which in Br is situated at X and aligned in the direction defined by the basis vector E α . Similarly, we can find the interpretation of Cαβ (α = β). It suffices to take M 1 := E α and M 2 := E β in (2.53), cos θ =

Cαβ Eα · C · Eβ  =√ , λ(E α )λ(E β ) Cαα Cββ

(α = β).

Hence, Cαβ (α = β) relates to a pair of material line elements which in Br meet at X and are in the orthogonal directions specified by E α and E β ; more specifically, Cαβ is the product of the stretches undergone by these elements and the cosine of the angle between them in Bc . There are many other measures of strain. The direct counterpart of C defined above is the so-called left Cauchy–Green deformation tensor B ≡ B(X, t), B := F · F T . Note that this is still a Lagrangian tensor, but it operates on the current configuration, its component representation being given by B = Bi j ei ⊗ e j ,

Bi j = xi, α x j, α ≡ Fiα F jα .

Recall that the second-order tensor field C was essentially introduced by measuring strain as the change in the ratio |dx|/|dX|. We can also associate strain with the change in the squared lengths of infinitesimal line elements in the two configurations, i.e. |dx|2 − |dX|2 . A simple calculation indicates that |dx|2 − |dX|2 = dX · (C − I) · dX ,

(2.55)

which leads to the Lagrangian strain tensor E ≡ E(X, t), E :=

 1 C−I , 2

E αβ =

 1 Cαβ − δαβ . 2

(2.56)

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2 Kinematics

The 1/2 pre-factor in the above definition is introduced for historical reasons (so that the linearised version of the expression in (2.56) coincides with the infinitesimal strain tensor in Linear Elasticity). Two other important deformation tensors, especially in Solid Mechanics, come from the deformation gradient. Since this is an invertible tensor field we can apply the Polar Decomposition Theorem to write F = R·U = V · R,

(2.57)

where R is a proper orthogonal tensor (R · R T = R T · R = I, det R = 1), while U, V are symmetric (U T = U, V T = V ) and positive definite. From the proof of the Polar Decomposition Theorem in Chap. 1, it follows that U and V are the square roots of C and B, respectively; that is, U 2 = F T · F = C and V 2 = F · F T = B .

(2.58)

The tensors U and V in (2.57) are known as the right and left stretch tensors, respectively; R is called the rotation tensor. The main difference between the two stretch tensors U and V is that the former operates on Br , while V is associated with Bc . Their mechanical significance will be discussed shortly in Sect. 2.8. The Spectral Representation Theorem applied to the symmetric and positive  definite right-stretch tensor establishes the existence of an orthonormal basis N (α) (α = 1, 2, 3) consisting entirely of the eigenvectors of U, i.e. U · N (α) = λα N (α) (α = 1, 2, 3; no sum) , so that U=

3 

λα N (α) ⊗ N (α) ,

α=1

where λα > 0 are known as the principal stretches of the deformation,   the eigenvalues and N (α) (α = 1, 2, 3) represent the Lagrangian principal directions or axes. This terminology is in agreement with the physics of motion; to see this, we choose M := N (α) in the definition (2.52) and then take into account (2.58), 1/2  (α) 1/2  λ(N (α) ) = N (α) · C · N (α) = N · U 2 · N (α) 1/2  1/2  = (U · N (α) ) · (U · N (α) ) = λα N (α) · λα N (α) = λα (N (α) · N (α) )1/2 = λα ,

(no sum) ,

that is, λα = λ(N (α) ), which justifies the adopted terminology. By invoking the same result as above,  left stretch tensor V also admits a diag the onal representation relative to a basis n(i) consisting entirely of its eigenvectors,

2.6 Strain Tensors

143

V · n(i) = λi n(i) (i = 1, 2, 3; no sum) , and V =

3 

λi n(i) ⊗ n(i) .

i=1

We further note that the eigenvalues of V coincide with those of U, but its eigenvectors (known as the Eulerian principal directions or axes) are obtained by a simple rigid rotation of the Lagrangian principal axes n(1) = R · N (1) ,

n(2) = R · N (2) ,

n(3) = R · N (3) .

(2.59)

These formulae follow from the proof of the Polar Decomposition Theorem, where it was shown that V = R · U · R T (see Chap. 1).

2.7 Examples of Particular Strain Tensors Here, we revisit some of the examples in Sect. 2.2 to provide further details in the description of those deformations. It is useful to distinguish between various deformations according to whether their corresponding F (deformation gradient) is constant or not. Definition 2.8 A motion of the form x = χ (X, t) ≡ F · X + c

(2.60)

in which F is independent of X and t, while c is a constant vector, is known as a homogeneous motion. For such motions, all measures of strain introduced above will be constant tensors. This makes it particularly easy to calculate the Lagrangian (or Eulerian) principal axes because they will be constant in space (i.e. they will not change from point to point, as it is typically the case). For the dilatation (2.6), it can be checked easily that F = αI ,

C = B = α2 I .

The fact that C is simply a diagonal tensor confirms that the dilatation does not change angles. Furthermore, since J ≡ det F = α 3 , we must require α > 0. If α > 1 the deformation is a uniform extension, while for α < 1 it represents a uniform contraction. The deformation is isochoric if α = 1 (a trivially degenerate case). One can consider a more realistic modification of (2.6) by assuming that

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2 Kinematics

x1 = α X 1 ,

x2 = β X 2 ,

x3 = β X 3 ,

(2.61)

where α, β ∈ R. Loosely speaking this corresponds to a simple extension in the 1-direction and a lateral contraction in the 2- and 3-directions. The deformation gradient follows immediately: F = α e1 ⊗ E 1 + β(e2 ⊗ E 2 + e3 ⊗ E 3 ) . Note that J = αβ 2 , so we must require α > 0. If α > 1 the deformation is a uniform extension in the 1-direction, while for α < 1 it corresponds to a contraction in the same direction. Since for β > 0 the diagonal terms of F are all positive, it follows by inspection that U = F and R = I. The parameter β measures the lateral extension (β > 1) or contraction (β < 1), in the (2, 3)-plane. The case β < 0 is slightly more complicated. Now, ⎡

⎤ ⎡ ⎤ α 0 0 1 0 0 [U] = ⎣ 0 −β 0 ⎦ and [R] = ⎣ 0 −1 0 ⎦ , 0 0 −β 0 0 −1 and we note that (−β) characterises the amount of lateral extension or contraction; a rotation through an angle π about the 1-axis is also included in this deformation. The deformation (2.61) is isochoric if αβ 2 = 1. This implies that if α > 1 then |β| < 1, or vice versa. In other words, an extension in the 1-direction produces a contraction in the lateral direction, or an extension in the lateral direction produces a contraction in the 1-direction. The Cauchy–Green deformation tensors are easily found to be C = α 2 E 1 ⊗ E 1 + β 2 (E 2 ⊗ E 2 + E 3 ⊗ E 3 ) , B = α 2 e1 ⊗ e1 + β 2 (e2 ⊗ e2 + e3 ⊗ e3 ) or, alternatively (using matrices), ⎡

⎤ α2 0 0 [B] = [C] = ⎣ 0 β 2 0 ⎦ . 0 0 β2 Again, the lack of nonzero off-diagonal components in these matrices confirms that the right-angles between directions parallel to the vectors E α (α = 1, 2, 3) remain unchanged under the action of (2.61). Unlike the examples discussed above, the simple shear deformation (2.5) changes both lengths and angles. This is already apparent in the expression of the deformation gradient, which can be written as F = I + α e1 ⊗ E 2 ≡ e1 ⊗ E 1 + e2 ⊗ E 2 + e3 ⊗ E 3 + α e1 ⊗ E 2

2.7 Examples of Particular Strain Tensors

or, simply,

145

⎡

1 [F] = ⎣0 0

α 1 0

⎤ 0 0⎦ , 1

whence the Cauchy–Green tensors are immediately found to be ⎡

1 [C] = ⎣α 0

⎤ 0 0⎦ , 1

α 1 + α2 0

⎡

1 + α2 ⎣ α [B] = 0

α 1 0

⎤ 0 0⎦ . 1

As U = C 1/2 , some further calculations reveal that the right-stretch tensor assumes the following component representation ⎡

2K [U] = ⎣α K 0

αK (2 + α 2 )K 0

⎤ 0 0⎦ , 1

1/2  . This matrix has the eigenvalues with K := 4 + α 2 λ1 :=

 1  −1 K +α , 2

λ2 :=

 1  −1 K −α , 2

λ3 := 1

(2.62)

associated with the eigenvectors (Lagrangian principal axes) N (1,2) := −

 1 α ∓ K −1 E 1 + E 2 , 2

N (3) := E 3 ,

(2.63)

where the superscripts ‘1’ and ‘2’ correspond to the minus and plus signs (in this particular order). It can be checked that λ1 > 1 and λ2 < 1, which means that the material line elements are stretched along N (1) and compressed along N (2) . The rotation tensor can be found from R = F · U −1 and has the component representation ⎡ ⎤ 2K α K 0 [R] = ⎣−α K 2K 0⎦ . 0 0 1 If we let θ := tan−1 (α/2), this matrix can be cast in the form ⎡

cos θ [R] = ⎣− sin θ 0

sin θ cos θ 0

⎤ 0 0⎦ , 1

which represents a clockwise rotation about the 3-axis by the angle θ .

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2 Kinematics

The torsion and pure bending deformations mentioned in Sect. 2.2 are inhomogeneous, a feature that complicates considerably their analysis. In the interest of brevity, below we discuss briefly only the latter deformation. We start with the observation that (2.8) can also be expressed as x1 = f (X 1 ) cos g(X 2 ) , x2 = f (X 1 ) sin g(X 2 ) , x3 = λX 3 .

(2.64)

  with These equations hold in a Cartesian system of coordinates o; ei associated  Bc , and which is related to the cylindrical polar coordinates o; er , eθ , ez by er = cos θ e1 + sin θ e2 , eθ = − sin θ e1 + cos θ e2 , ez = e3 .

(2.65)

The deformation gradient of (2.64) will be F = xi, α ei ⊗ E α , and by taking into account (2.65), we discover that F = f  (X 1 )er ⊗ E 1 + f (X 1 )g  (X 2 )eθ ⊗ E 2 + λez ⊗ E 3 ,

(2.66)

where the ‘dash’ represents the derivative with respect to either X 1 or X 2 . The rotation tensor R is given by R = er ⊗ E 1 + eθ ⊗ E 2 + ez ⊗ E 3 , while the left stretch tensor V assumes the representation V = f  (X 1 )er ⊗ er + f (X 1 )g  (X 2 )eθ ⊗ eθ + λez ⊗ ez .

(2.67)

  This simple form of V is due to the fact that er , eθ , ez are the Eulerian principal axes; we can also read off the principal stretches from (2.67), λ1 = f  (X 1 ) ,

λ2 = f (X 1 )g  (X 2 ) ,

λ3 = λ .

If the pure bending deformation is isochoric, J ≡ det F = det V = λ1 λ2 λ3 = 1 (identically). This gives the separable ordinary differential equation λ f  f g  = 1, which admits the general solution  r=

2X 1 +β α

1/2 ,

θ=

α X2 , λ

(α, β ∈ R) .

The arbitrary constants α and β in these expressions can be determined in terms of the initial dimensions of the cuboid, but we do not pursue the matter further.

2.8 The Interpretation of the Polar Decomposition Theorem

147

2.8 The Interpretation of the Polar Decomposition Theorem The right and left polar decompositions for the deformation gradient F have already been stated in (2.57). Their expressions involve the rigid rotation R, as well as the stretch tensors U and V that were defined in terms of the Cauchy–Green deformation tensors—cf. (2.58). Next, we explore the implications of this important result in interpreting the local kinematics of a deformable body. Since dx = F · dX, we can write dx = F · dX = (R · U) · dX = R · (U · dX) , dx = F · dX = (V · R) · dX = V · (R · dX) .

(2.68a) (2.68b)

Recall that, as explained in Sect. 2.4, dX must be understood as the separation between two nearby particles in Br , while dx represents its counterpart in Bc . Equation (2.68a) admits the following interpretation: the deformation undergone by our deformable body in the close vicinity of a given particle X ∈ Br can be regarded as a two-step process (see Fig. 2.10). • STEP 1a: Impose the stretches λα in the directions in Br defined by N (α) (α = 1, 2, 3). This represents U · dX and the result is going to still be in Br because U operates on the reference configuration. • STEP 2a: Next, perform a rigid rotation which maps the output of the previous step to Bc ; this is achieved by the orthogonal tensor R (which is a two-point tensor acting between Br and Bc —in this order). Because we also have the alternative decomposition (2.68b), the deformation of the points nearby the same particle X mentioned above, can also be interpreted in reverse order as indicated below. • STEP 1b: Perform a rigid rotation which maps dX from Br to a new position in Bc . This corresponds to R · dX (remember that R is a two-point tensor). • STEP 2b: Next, impose the stretches λi in the directions in Bc defined by n(i) (i = 1, 2, 3). This is effected by the application of V on the output of the previous step, and the result is still going to be in Bc since V operates on the current configuration.

2.9 The Spatial Gradient of Velocity In this section, we shall introduce some of the main kinematical variables that describe the instantaneous time rates of deformation. These variables, referred to as kinematical rate tensors, are not, in general, the time rates of the deformation tensors introduced previously. The reason for this is that deformation tensors are functions of two

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2 Kinematics

stretching + rotaƟon

B

B

rotaƟon + stretching

B

B

Fig. 2.10 The geometric interpretation of the Polar Decomposition Theorem

configurations (a reference and a current configuration, respectively), whereas rate tensors are, by definition, functions of the current configuration alone. A fundamental rate tensor is the so-called velocity gradient. Definition 2.9 The velocity gradient tensor L is defined as the spatial gradient of velocity, L(x, t) := grad v ≡ grad v(x, t) , (2.69) and has the component representation L = L i j ei ⊗ e j ,

L i j = vi, j ≡

∂vi . ∂x j

(2.70)

Note that L corresponds to an Eulerian tensor field and is defined without recourse to any reference configuration. This is in contrast to the deformation gradient F, which is a two-point tensor (operating between Br and Bc ). Next, we derive some important formulae involving the tensor introduced in (2.69) above. Using (2.33b) for the velocity vector field v, we find Grad v = (grad v) · F = L · F. The left-hand side of this equation can be written successively χ (X, t), t) ∂ V p (X, t) ∂v p (χ e p ⊗ Eα = e p ⊗ Eα ∂X ∂ Xα α



∂ ∂χ p (X, t) ∂ ∂χ p (X, t) e p ⊗ Eα e p ⊗ Eα = = ∂ Xα ∂t ∂t ∂ Xα

Grad v =

(2.71)

2.9 The Spatial Gradient of Velocity

=

149

D ˙ (F pα e p ⊗ E α ) = F. Dt

On the second line above, we have taken into account that the material time derivative of a Lagrangian field commutes with the material gradient, i.e.



∂ ∂(. . . ) ∂ ∂(. . . ) . = ∂ Xα ∂t ∂t ∂ X α

(2.72)

In general, the material time derivative of an Eulerian field does not commute with the spatial gradient, i.e. if we replaced X α with xi in (2.72) the equality would no longer be true. Going back to (2.71), we can now record the important formula ˙ = L · F or L = F ˙ · F −1 . F

(2.73)

Another useful relation involving L can be obtained with the help of (1.287) at the end of Chap. 1: replacing A → F and τ → t in that result, gives   D ˙ · F −1 = (det F) tr (L) , (det F) = (det F) tr F Dt where use has been made of (2.73). Alternatively, by remembering that J ≡ det F and tr (L) = L ii = ∂vi /∂ xi = div v, the above result can be re-arranged in the form J˙ = J tr(L) = J div v .

(2.74)

According to (2.45) and (2.74), div v measures the rate at which volume changes during the motion. In the case of an isochoric motion, J ≡ 1, so J˙ = 0 and hence div v = 0 .

(2.75)

As seen earlier, in Chap. 1, any second-order Cartesian tensor admits an additive decomposition into a symmetric and a skew-symmetric part. Applied to L this observation gives L = D+W, (2.76) where D :=

1 (L + L T ) 2

and

W :=

1 (L − L T ) 2

(2.77)

are known as the stretching tensor or the strain-rate tensor, and the spin tensor, respectively. The terminology used for D and W is justified by the physical interpretation of these tensors. To this end, let us consider a material line element dX in Br which is deformed into dx (situated in Bc ); the ‘arclengths’ of these material line elements are

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2 Kinematics

ds 2 := dx · dx ≡ |dx|2

and

dS 2 := dX · dX ≡ |dX|2 .

We also assume that M and m are unit vectors along dX and dx, hence M = dX/dS and m = dx/ds. Then, ds 2 − dS 2 = dx · dx − dX · dX = (F · dX) · (F · dX) − dX · dX = dX · (F T · F) · dX − dX · I · dX = dX · (F T · F − I) · dX . By taking the material time derivative of the above result yields, in conjunction with the formula (2.73), D D ˙ +F ˙ T · F) · dX (ds 2 − dS 2 ) = dX · (F T · F − I) · dX = dX · (F T · F Dt Dt = dX · (F T · L · F + F T · L T · F) · dX = (F · dX) · (L + L T ) · (F · dX) = 2 dx · D · dx or, equivalently

˙ ds = m· D · m. ds

If we choose m := ei (i = 1, 2, 3) in this formula we arrive at the physical interpretation of the diagonal components of D: the diagonal component Dii (no sum) is the rate of extension per unit length of a material line element which, in the current configuration Bc , is situated at x and momentarily aligned with the direction defined by the base vector ei (i = 1, 2, 3). If m j := dx j /|dx j | ( j = 1, 2) and θ := ∠(dx 1 , dx 2 ), then it can be shown that θ˙ =

(m1 · L · m1 + m2 · L · m2 )(m1 · m2 ) − 2m1 · D · m2 . |m1 ∧ m2 |

Assuming that m1 and m2 are orthogonal, this last relation becomes 1 − θ˙ = m1 · D · m2 2 and provides an interpretation for the non-diagonal components of the stretching tensor D: the off-diagonal component Di j (i = j) is half the rate of decrease of the angle between a pair of material line elements which, in Bc , intersect at x and are instantaneously in the direction represented by the basis vectors ei and e j . Since D ˙ · dX = (L · F) · dX = L · dx = ( D + W ) · dx, (dx) = F Dt

2.9 The Spatial Gradient of Velocity

151

and we have already an interpretation of D (as discussed above), it remains to interpret W . We do this by setting D ≡ O (the second-order zero tensor), so that D (dx) = W · dx = w ∧ dx, Dt where w is the axial vector of W . This shows that the motion is locally a rigid rotation and W represents a measure of the rate of rotation of line elements. In a broad sense, the spin tensor W can be interpreted as the mean angular velocity in the continuum body. The combination of D and W shows that the motion consists of stretching and rotation (analogous to the interpretation of U and R in the polar decomposition of F).

2.10 Transport Formulae We are going to conclude this chapter with a number of results known as the transport formulae that arise in connection with calculating the rates of change of integrals over material curves, surfaces, and volumes in the current configuration. Most of these formulae are of major significance in Fluid Mechanics, but they also play a key role in establishing the main equations of motion for deformable bodies (as we shall see in the next chapter). To motivate the subsequent three-dimensional calculations, let us first take a quick look at the much simpler case of a one-dimensional real-valued function f = f (x, t), where a ≤ x ≤ b and t ∈ R. Under certain technical conditions (e.g. see [1, 2]), the rate of change with respect to this latter variable of the integral of f (x, t) with respect to ‘x’ is simply d dt



b

 f (x, t) dx =

a

a

b

∂ f (x, t) dx . ∂t

(2.78)

However, if we now want to address the same issue, but the integration limits are allowed to depend on ‘t’ as well, that is, a = a(t) and b = b(t), then d dt



b(t)

a(t)

 f (x, t) dx =

b(t)

a(t)

∂ f (x, t) dx + b (t) f (b(t), t) − a  (t) f (a(t), t) , ∂t

where the ‘dash’ stands for differentiation with respect to ‘t’ (e.g. see [1, 2] for a proof). This suggests that the required rate of change cannot be simply obtained by interchanging the t-derivative and the integration sign—as was done in (2.78). We now turn to the fully three-dimensional case. In what follows Ct , St and Rt will denote, in turn, a material curve, a material surface and a material volume in the current configuration Bc of a deformable body B. If φ ≡ φ(x, t) is a continuously differentiable scalar field representing some property of B, and u ≡ u(x, t) corresponds to a similar vector field, then the following formulae hold true:

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2 Kinematics

    D φ˙ dx + φ L · dx , φ dx = Dt Ct C t     D φn da = φ˙ + φ tr L n − φ L T · n da, Dt St S  t    D φ˙ + φ tr L dv, φ dv = Dt Rt Rt

(2.79a) (2.79b) (2.79c)

and     D u˙ + L T · u · dx, u · dx = Dt Ct Ct   D u · n da = (u˙ + u tr L − L · u) · n da, Dt St St   D u dv = (u˙ + u tr L) dv , Dt Rt Rt

(2.80a) (2.80b) (2.80c)

where n ≡ n(x, t) denotes the outward unit normal to St . In trying to justify these formulae, the main complication is that we have time dependence not only in the integrands, but also in Ct , St and Rt . Nevertheless, (2.79) and (2.80) can be checked without any significant difficulties by following the simple three-step strategy outlined below. The first step is to transfer to Br the integrals on the left-hand side in each of the above transport formulae. This is accomplished by using the important formula (2.73), together with 1. d x = F · d X for line integrals; 2. nda = J F −T · Nd A for surface integrals; 3. dv = J d V for volume integrals. Once in the reference configuration, the differentiation with respect to ‘t’ and the integrals commute with each other (in other words, we can differentiate under the integral sign); this constitutes the second step. Finally, the result from the previous step is transferred back to the current configuration. This is similar to the first step above, except that all calculations should be performed in reverse order. To re-iterate the comments made above we can state the general situation, symbolically, in the form D Dt

 Pt

 (. . . ) =

Pt

D(. . . ) , Dt

D Dt

 Pr

(. . . ) =

 Pr

D(. . . ) ; Dt

(2.81)

here, Pr ⊂ Br is the counterpart of Pt ⊂ Bc , i.e. Pt = χ (Pr , t), Pr can be a curve or a two-dimensional surface or a three-dimensional region in Bc , while the ‘dots’ in (2.81) are meant to be replaced by any suitable (scalar, vector or tensor) fields.

2.10 Transport Formulae

153

As an illustration of the strategy outlined above, the complete proof of (2.80b) is included below. The remaining transport formulae can be established by following the same approach: D Dt



D Dt

u · n da = St



u · (J F −T · N) d A Sr

(note the integral is now over Sr )  D (J F −1 · u) · N d A = Dt Sr (using the definition of transpose)  D (J F −1 · u) · N d A = Sr Dt

 D J F −1 · u˙ + J˙ F −1 · u + J = (F −1 ) · u · N d A . Dt Sr ˙ From (2.74), we have J˙ = J tr (L). It can also be checked that F −1 = −F −1 · L, −1 by differentiating with respect to time the obvious identity F · F = I and then making use of (2.73). Thus, D Dt





[J F −1 · u˙ + J tr (L)F −1 · u − J F −1 L · u] · N d A

u · n da = St



Sr



Sr



Sr

=



 J F −1 · [u˙ + tr (L)u − L · u] · N d A

[u˙ + tr (L)u − L · u] · (J F −T · N) d A

=

[u˙ + tr (L)u − L · u] · n da .

= St

We close this chapter with a result that has many applications in Fluid Mechanics. Theorem 2.1 (Reynolds’ Transport Formula) Let B be a deformable body and assume that Rt ⊂ Bc is a closed region bounded by the material surface ∂ Rt , with outward unit normal n. If u ≡ u(x, t) is a regular vector field associated with Bc , D Dt



 u dv = Rt

Rt

∂u dv + ∂t

 ∂ Rt

(u ⊗ v) · n da ,

(2.82)

where v ≡ v(x, t) represents the Eulerian description of the velocity field in the deformable body. The above formula is directly related to the result stated in (2.80c), and its justification is relatively straightforward. By using the general strategy outlined in connection with the transport formulae, it transpires that

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2 Kinematics

D Dt

 u dv = Rt

D Dt 

=





˙ + u J˙) dV (uJ  ˙ + uJ div v) dV = (uJ (u˙ + u div v) dv , (uJ ) dV =

R0

R0

R0

(2.83)

Rt

with R0 ⊂ Br such that Rt = χ (R0 , t). The integrand of (2.83) can be written in expanded form with the help of the earlier formula (2.17). Taking further into account that div (v ⊗ u) = (grad u) · v + u(div v), the above calculations become D Dt



 u dv = Rt

Rt

∂u dv + ∂t

 ∂ Rt

div (v ⊗ u) dv .

(2.84)

Finally, the original formula (2.82) is recovered by applying the Divergence Theorem (1.258) to the second integral on the right-hand side of (2.84). On the right-hand side in Reynolds’ Transport Formula, the second integral can be regarded as the net ‘flow’ of u across the (time-dependent) surface ∂ Rt , while the first term corresponds to a rate of change as if the the region Rt were ‘frozen’ in Bc . We mention in passing that the formula (2.79c) represents the counterpart of (2.82) for a scalar field, and can also be re-stated as    ∂φ D dv + φ dv = φ(v · n) da . (2.85) Dt Rt Rt ∂t ∂ Rt

2.11 Exercises 1. Consider the deformation x = χ (X) defined by x1 = AX 1 + B X 2 ,

x2 = B X 1 + AX 2 ,

x3 = C X 3 ,

where A, B, C ∈ R are given constants. Find the inverse deformation χ −1 (x). 2. Obtain the Lagrangian descriptions of velocity and acceleration for the motions   a. x1 = X 1 1 + t 2 , x2 = X 1 t + X 2 , x3 = X 3 . x2 = X 2 e−ct , x3 = X 3 e−ct , (c ∈ R) . b. x1 = X 1 e−ct , x2 = X 2 /(1 + t), x3 = X 3 + X 1 t . c. x1 = X 1 + X 2 t 2 , 3.

Also, find the spatial (Eulerian) descriptions v(x, t) and a(x, t). a. If v and a denote the Eulerian descriptions of the velocity and acceleration fields, respectively, show that a = (grad v) · v +

∂v , ∂t

this equation being set in the spatial description.

2.11 Exercises

155

b. A motion is said to be steady if the velocity at any given point is independent of time, so that v = v(x). The velocity at a point x ∈ Bc in a body of fluid in steady flow is given by a 2 x1 x2 a 2 (x 2 − x22 ) e + 2U v(x) = U  1 1   2 e2 + V e3 , 2 x12 + x22 x12 + x22 where U, V and a are constants. Show that div v = 0 and find the acceleration of the particle at x, at some time t > 0. 4. An electromagnetic fluid is subjected to a decaying electric field of magnitude φ(x, t) = e−At (x12 + x22 + x32 )−1/2 , where A ∈ R is a constant and x = (x1 , x2 , x3 ). The velocity of the fluid is v(x, t) = x1 x3 e1 + x22 t e2 + x2 x3 t e3 . a. Determine the rate of change of φ at t = 1 of the particle which occupies the point with coordinates (2, −2, 1) in the current configuration. b. The acceleration of the same particle at the same time. 5. In the current configuration, a certain physical property of a continuum body is given by φ(x, t) = |x|2 . Let an observer O move with the velocity v(x, t) = e1 + 2e2 + 3e3 . What is the rate of change of φ found by O at (a) x = (1, 2, 1); (b) x = (0, 1, 1); (c) x = (1, 2, 3)? 6. A motion of a deformable body is defined by the velocity components v1 =

3x1 , 1+t

v2 =

x2 , 1+t

v3 =

5x32 . 1+t

Assume that Br is at t = 0, so that x(t = 0) = X. a. Derive the particle path, i.e. the motion x = χ (X, t). b. Compute the Lagrangian representation of the velocity field. c. Compute the associated acceleration in the material and spatial descriptions. 7. Consider the motion x = χ (X, t) given by x1 = X 1 + X 3 t 3 , x2 = X 2 + X 1 t 3 , x3 = X 3 + X 2 t 3 . Find the velocity of a. the particle that was at the point X 0 = (4, 4, 4) at the reference time t = 0; b. the particle that occupies the point x 0 = (4, 4, 4) at time t > 0. 8. If J = det F is the Jacobian of the motion, then justify the formulae Grad J = J div F , div (J

−1

F) = 0 ,

−1

Div (J F ) = 0 , −1

grad J = −J Div (F ) .

(2.86a) (2.86b) (2.86c) (2.86d)

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2 Kinematics

9.

a. For a rigid motion x = c(t) + Q(t) · X of a material body B, show that the velocity v ≡ v(x, t) and the acceleration a ≡ a(x, t) can be written as v = c˙ + W 0 · (x − c) , ˙ 0 + W 20 ) · (x − c) , a = c¨ + ( W

(2.87a) (2.87b)

where W 0 is a skew-symmetric tensor which you should identify. b. If w is the axial vector of W 0 , show that (2.87) can be cast in the form v = c˙ + w ∧ (x − c) ,

 a = c¨ + w ˙ ∧ (x − c) + w ∧ w ∧ (x − c) .

(2.88a) (2.88b)

(Interpretation: c˙ and c¨ represent, respectively, the translational velocity and acceleration of B, while w is its angular velocity.) 10. Consider the deformation x1 = α1 (X 1 + β X 2 ), x2 = α2 X 2 , x3 = α3 X 3 , where α j ( j = 1, 2, 3) and β > 0 are given constants. a. Under what conditions is this deformation isochoric? b. A unit cube in Br has its edges parallel to the coordinate axes and is subject to the above deformation. Find the length of its edges in Bc . c. Determine the changes in the angles of the lower and upper bases of the cube mentioned above. 11. Consider an arbitrary motion x = χ (X, t). Show that the deformation gradient expressed in the Lagrangian and Eulerian principal axes assumes the form F = λ1 n(1) ⊗ N (1) + λ2 n(2) ⊗ N (2) + λ3 n(3) ⊗ N (3) , where λ j ( j = 1, 2, 3) are the principal stretches. 12. Consider the deformation x1 = X 1 +

3 9 X2 + , 2 2

x2 =

3 7 X2 + , 2 2

x3 = X 3 .

A square of side 2 in Br is situated in the plane X 3 = 3 and has vertices at the points A1 (−1, −1, 3), A2 (1, −1, 3), A3 (1, 1, 3), A4 (−1, 1, 3). If the square is subjected to the above deformation, then a. find the deformed lengths of A1 A2 and A2 A3 in Bc ; b. what is the deformed angle ∠ (A4 A1 A2 ) in Bc ? 13. Consider the deformation x1 =

X 12 − X 22 , (X 12 + X 22 )1/2

x2 =

2X 1 X 2 , (X 12 + X 22 )1/2

x3 = X 3 .

2.11 Exercises

157

a. Find the components of the right Cauchy–Green deformation tenor C. b. Determine the Lagrangian principal axes and the corresponding principal stretches. 14. Use Nanson’s formula relating area elements to check the following formulae, D (da) = (tr(L) − n · L · n)da, Dt n˙ = (n · L · n)n − L T · n, relating to a material surface element of area da with unit normal n, L being the velocity gradient at the current location of the element. 15. A motion of a deformable body is given by ⎧ −at ⎪ ⎨x1 = e (X 1 cos Θ − X 2 sin Θ) , x2 = e−at (X 1 sin Θ + X 2 cos Θ) , ⎪ ⎩ x3 = φ(t)X 3 , where Θ ≡ Θ(X 3 , t) := bX 3 t and a, b > 0 are given constants. a. Find the form of φ(t) for which the motion is isochoric. b. Determine, for the isochoric motion, the components of velocity and acceleration in the Eulerian description. 16. Let Ω ⊂ E3 and E ∗ := sym(H), where H is the displacement gradient. Define E :=

1 vol(Ω)

 Ω

E ∗ dV ,

to be the average of E ∗ over the region Ω. Show that 1 E= vol(Ω)

 ∂Ω

(u ⊗ n + n ⊗ u) d A ,

so that E depends only on the boundary values of the displacement u. 17. Let E ∗ := sym(H) and W ∗ := skw(H), where H is the material gradient of the displacement field in a deformable body. Show that E ∗ 2 + W ∗ 2 = H2

and

E ∗ 2 − W ∗ 2 = H · H T .

18. For the homogeneous deformation x = χ(X, t) defined by x1 = α X 1 + β X 2 , x2 = −α X 1 + β X 2 , x3 = μX 3 , where α, β, μ > 0, determine the component representation of the right Cauchy–Green tensor and the principal stretches. Find R and U for the polar decomposition of the original deformation. 19. Let F denote the deformation gradient and define  F := J F −1 . Show that

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2 Kinematics

D  ( F) =  F · [(div v)I − L] , Dt where L is the velocity gradient.   20. Consider a unit cube Ωcube ≡ X = (X 1 , X 2 , X 3 ) : 0 < X j < 1, j = 1, 2, 3 , and let χ be the deformation x1 = X 1 + ε X 2 X 32 ,

x2 = X 2 + ε X 3 X 12 ,

x3 = X 3 + ε X 1 X 22 ,

where 0 < ε  1. Show that χ (Ωcube )) = 1 − vol(χ

ε3 3ε2 + . 4 3

  21. Consider the cylinder Ωcyl ≡ X=(X 1 , X 2 , X 3 ) : X 12 + X 22 = 1, 0≤X 3 < 4 , and let χ be the deformation x = (1 + ε|X|2 )X, with 0 < ε  1. a. Show that the deformation gradient is given by   F = I + ε |X|2 I + 2X ⊗ X , and then deduce that det F = 0 for sufficiently small ε. b. Hence, or otherwise, check that the volume of the deformed cylinder is   28 χ (Ωcyl )) = 4 + 10ε + ε2 + 3ε3 π . vol(χ 3 22. A deformation is described by the equations x1 = X 1 + ε X 2 ,

x2 = X 2 + ε X 1 ,

x3 = X 3 ,

where 0 < ε  1. Show that under this deformation the lateral surface of the cylinder considered in the previous question has an area given by  4

2π

 1/2 1 + ε2 − 2ε sin 2θ dθ .

0

Evaluate numerically this integral for ε = 10−2 . 23. In the pure strain deformation defined by x j = λ j X j (no sum, j = 1, 2, 3), show that the angle between a pair of line elements in the (1, 2)–plane which were initially orthogonal is changed by at most sin−1 by the deformation.

 2    λ1 − λ22 / λ21 + λ22

2.11 Exercises

24.

159

a. Show that the length of a material line element, initially in the direction of the unit vector M, is unchanged by the deformation provided that |F · M| = 1, where F is the deformation gradient. b. A body is subjected to a homogeneous pure strain in which the basis vectors are stretched in the ratios λ−1/2 α, λ−1/2 α −1 , λ. If no stretch occurs in either of the two directions (cos θ, ± sin θ, 0), 0 ≤ θ ≤ π/2, show that  1/2 α 2 = [λ ± λ2 − sin2 2θ ]/2 cos2 θ.

25.

a. In its reference configuration, a body contains a spherical cavity with centre O and radius A, filled with explosive. The explosive is detonated at t = 0 and produces a spherically symmetric motion of the body given by x = (r/R)X, with r = f (R, t) and R = |X|, r = |x|; the referential (X) and current (x) positions are both taken relative to O as origin. i. Show that the deformation gradient is given by 1 F= 2 R



∂f f − ∂R R

 X⊗X+

f I, R

where I is the usual second-order identity tensor. ii. If the motion is isochoric, then show that f must satisfy the differential equation ∂f = R2. f2 (2.89) ∂R Using appropriate initial conditions solve this equation and find an explicit formula for f . Determine the velocity and acceleration fields.

Bibliography 1. Adams RA (1999) Calculus: a complete course. Addison-Wesley, Longman Ltd., Ontario, Canada 2. Salas SL, Hille E (1995) Calculus: one and several variables. Wiley, New York 3. Atkin RJ, Fox N (1980) An introduction to the theory of elasticity. Longman, London 4. Chadwick P (1999) Continuum mechanics: concise theory and problems. Dover Publications Inc, Mineola, New York 5. Ciarlet PG (1993) Mathematical elasticity. North-Holland, Amsterdam 6. Gurtin ME (1981) An introduction to continuum mechanics. Academic Press, New York 7. Hjelmstad KD (1997) Fundamentals of structural mechanics. Prentice-Hall Inc, Upper Saddle River, New Jersey 8. Gill AJ, Bonet J, Wood RD (2016) Nonlinear solid mechanics for finite element analysis: statics. Cambridge University Press, Cambridge (UK) 9. Jaunzemis W (1967) Continuum Mechanics. The McMillan Company, New York 10. Shih Liu I (2002) Continuum mechanics. Springer, Berlin 11. Malvern LE (1969) Introduction to the mechanics of a continuum medium. Prentice-Hall Inc., Englewood Cliffs, New Jersey

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2 Kinematics

12. 13. 14. 15.

Narasimhan MNL (1993) Principles of continuum mechanics. Wiley Inc, New York Ogden RW (1984) Non-linear elastic deformations. Ellis Horwood Ltd., Chichester Spencer AJM (1980) Continuum mechanics. Longman Ltd., Essex (UK) Timoshenko SP, Goodier JN (1970) Theory of elasticity. McGraw-Hill Book Company, Auckland, international edition Truesdell C (1966) The elements of continuum mechanics. Springer, Berlin Truesdell C (1977) A first course in rational continuum mechanics, vol 1. Academic Press, New York Truesdell C, Noll W (2004) The non-linear field theories of continuum mechanics. Springer, Berlin Rubin D, Lai WM, Krempl E (2010) Introduction to continuum mechanics, 4th edn. Elsevier, Amsterdam

16. 17. 18. 19.

Chapter 3

Balance Laws

Abstract In our foregoing discussion of the changes in shape and size of deformable bodies, no reference was made to the mechanical interactions between either various parts of the bodies, or parts of the bodies and the surrounding environment. For all practical purposes such mechanical interactions can be represented by suitably defined macroscopic forces. As we shall see shortly, the familiar description of the concept of force from Newtonian Mechanics will require some re-evaluation before we can provide a satisfactory mechanical framework for deformable continua. As a prelude to our discussion in the later sections of this chapter, we review below some of the key concepts from the mechanics of rigid bodies.

The linear momentum P of a rigid material particle P is defined as the product between its mass (m) and velocity (v), that is P := mv. If the particle is subjected to a system of forces whose resultant is F tot , then d d P = (mv) = F tot , dt dt

(3.1)

assuming that m = constant and using Newton’s Second Law (F tot = ma, with a representing the acceleration of the particle). Since the linear momentum is a vector, one can define the moment of this vector with respect to a fixed point O. Denoting by L O this moment of the (linear) momen−→ tum, we have L O := r ∧ P ≡ r ∧ (mv), where r ≡ O P represents the position vector of P relative to O. Note that  d d LO = r ∧ (mv) = v ∧ (mv) + r ∧ (ma) dt dt = r ∧ F tot =: M tot ,

(3.2)

where M tot denotes the resultant moment (or torque) with respect to the same point O of all the forces acting on the particle; L O defined above is usually referred to as the angular momentum.

© Springer Nature B.V. 2020 C. D. Coman, Continuum Mechanics and Linear Elasticity, Solid Mechanics and Its Applications 238, https://doi.org/10.1007/978-94-024-1771-5_3

161

162

3 Balance Laws

These concepts are easily extended for systems that contain a finite number of system material particles  {Pi } having masses m i . The linear momentum of such a  ri ∧ is then P := m i v i , while its angular momentum corresponds to L O := (m i v i ), with r i being the position vector of Pi relative to O. By considering the rate of change of these quantities, it follows immediately that  d P= F tot dt

and

 d LO = M tot , dt

(3.3)

  M tot where F tot is the resultant force acting on the system of particles, and stands for the resultant moment with respect to O of all the forces acting on the particles {Pi }. The equations (3.3) are known as the principles of conservation of linear and angular momentum, respectively, and they play a central role in the dynamics of systems of rigid particles. By letting the number of particles increase without a bound, the various sums in the foregoing definitions will become integrals, and one can then formulate the counterpart of the two equations in (3.3) for rigid bodies; these are also known as Euler’s first and second laws of motion. In this chapter, we are going to be concerned with extending these fundamental results to deformable bodies.

3.1 The Principle of Mass Conservation In Continuum Mechanics, each deformable body B possesses an intrinsic unchanging property, called mass, which is characterised by a positive dimensional scalar denoted by m or m(B). It is typically defined to be a measure of the material content of a body. Like a small number of other concepts in Physics (e.g. ‘length’, ‘time’, ‘force’ and so on), mass is considered a primitive concept whose interpretation is left to our intuition and experience. We express the invariance of mass by stipulating that D m(B) = 0, Dt

(3.4)

where D/Dt denotes the material time derivative introduced in the previous chapter (i.e. the derivative with respect to time, while keeping fixed X ∈ Br ). In this section, we are going to be concerned with writing (3.4) in a way that takes advantage of the mathematical formalism introduced in the previous chapter. To this end, the mass of a body is postulated to obey the following two intuitive properties: 1. m(P1 ∪ P2 ) = m(P1 ) + m(P2 ), for all P1 , P2 ⊂ B with P1 ∩ P2 = ∅. 2. m(P) → 0 as vol(P) → 0, where vol(P) stands for the volume of P ⊂ B. Since our deformable bodies are a collection of material points, we associate to each material particle P ∈ B a positive number m(P) that measures, relative to some standard reference mass, its own invariant mass content. If B degenerates into

3.1 The Principle of Mass Conservation

163

a discrete set of points {P I } (I = 1, 2, . . . , k), then the first property above allows us to calculate the total mass of our degenerate body by simply summing up the mass of all of its individual particles: m(B) ≡ m({P I }) =

k 

m(P I ),

(3.5)

I =1

and we note that if k → ∞ then, intuitively, the sum in (3.5) will have to be replaced by an appropriate integral. In Continuum Mechanics, we consider bodies whose mass is distributed continuously, so we need to be able to describe the mass of any portion of the body, however small it might be. As our primary interest is in deformable bodies, it is also sensible to assume that the mass of each particle will depend on the configuration we are looking at. Let us recall that a generic material point P ∈ B has position vectors X and x in Br and Bc , respectively. Thus, in what follows dm(X) will be taken to represent the infinitesimal element of mass at X ∈ Br in the reference configuration, while by dm(x, t) we shall denote its value at x ∈ Bc in the current configuration. By integrating the infinitesimal element of mass over an entire region (situated either in Br or Bc ) we find the total mass of that region (in the corresponding configuration). For example, if P ⊂ B is an arbitrary region, then   m(P) = dm(X) = dm(x, t) , (3.6) P0 Pt  



 mass of P in Br mass of P in Bc where P0 and Pt are the regions occupied by P at time t = 0 and t > 0, respectively. Although Eq. (3.6) captures the essence of the mass conservation property for deformable bodies, the integration with respect to the infinitesimal mass elements (i.e., dm) is unsatisfactory. This is where the foregoing second postulate comes in1 : it allows us to replace the integrals in (3.6) by standard volume integrals via certain scalar fields known as mass densities. These are defined next. Let X ∈ Br and x ≡ χ(X, t) ∈ Bc , where t > 0 is fixed, and consider two families of nested regions in Br and Bc , respectively, with the following properties: 1.

P(n) 0 ⊂ Br ,

X ∈ P(n+1) ⊂ P(n) 0 0 ,

(∀) n ≥ 0,

(n) with P(n) 0 → {X} as n → ∞, i.e. the sets P0 ‘collapse’ to the point X in the limit n → ∞.

1 In

the usual three-dimensional Euclidean space, the mass can be regarded as a Borel measure that is assumed to be absolutely continuous with respect to the volume measure; the existence of a mass density is then assured by the Radon–Nikodym Theorem in Measure Theory.

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3 Balance Laws

2.

P(n) t ⊂ Bc ,

x ∈ P(n+1) ⊂ P(n) t t ,

(∀) n ≥ 0.

Again, we require P(n) t → {x} as n → ∞, with the same interpretation as above. Definition 3.1 (a) The reference and the current mass densities of B correspond to the scalar fields ρ0 ≡ ρ0 (X) and ρ ≡ ρ(x, t) defined by ρ0 (X) := lim

n→∞

m(P(n) 0 )

vol(P(n) 0 )

,

ρ(x, t) := lim

n→∞

m(P(n) t ) vol(P(n) t )

,

(3.7)

where ‘vol’ represents the usual volume of a three-dimensional region in space. The important point here is that the mass density depends on the configuration, but the mass of any part of B is preserved during a given motion; the latter part of this statement follows directly from (3.6). In order to make use of the definitions just introduced, we shall write (3.7) in the more intuitive form dm(X) ⇒ dm(X) = ρ0 (X) dV, dV (X) dm(x, t) ⇒ dm(x, t) = ρ(x, t) dv, ρ(x, t) = dv(x, t) ρ0 (X) =

(3.8a) (3.8b)

where dV ≡ dV (X) and dv ≡ dv(x, t) represent the usual infinitesimal volume elements defined in the reference and current configurations, respectively. As already hinted at earlier, the formulae (3.8) allow us to convert the integrals in (3.6) to standard volume integrals, so that (3.6) can now be re-stated as   m(P) = ρ0 (X) dV = ρ(x, t) dv , P P  0 



 t  mass of P in Br mass of P in Bc whence a first version of the Principle of Mass Conservation (PMC) emerges, 

 P0

ρ0 (X) dV =

Pt

ρ(x, t) dv,

(∀) P0 ⊂ Br , Pt ≡ χ(P0 , t).

(3.9)

This is a global (or integral) principle because it applies to arbitrary regions rather than to arbitrary points of a deformable body. There is also the added inconvenience of having integrals calculated on both configurations (Br and Bc ). To cast (3.9) in a more sensible way we need the following key result.

3.1 The Principle of Mass Conservation

165

Proposition 3.1 (Localisation Principle) Let Φ be a continuous scalar, vector, or tensor field defined on an open set P ⊂ E3 . Then, given any ζ 0 ∈ P, we have 1 δ→0 vol(Ωδ )



Φ(ζ 0 ) = lim

Ωδ

Φ dV,

where Ωδ ⊂ P is the closed ball of radius δ > 0 centred at ζ 0 , while vol(Ωδ ) denotes the volume of Ωδ in E3 . From the previous chapter we know that dv = J dV , where J ≡ det F is the Jacobian of the motion. Using this result in (3.9), with the help of the Localisation Principle we find the following local form of the P MC in Lagrangian description, ρ0 (X) = J (X, t)ρ(χ(X, t), t), (∀) X ∈ Br , t > 0,

(3.10)

also known as the referential equation of continuity. It is customary to state this result in the more succinct way, ρ0 = ρ J. A physical principle expressed by an equation that holds at every point of a deformable body and for all times, for example, Eq. (3.10), is referred to as the local form of that principle (here, ‘local’ means ‘pointwise’). There is also a local form of the P MC corresponding to the Eulerian description. This is obtained by writing (3.4) on the current configuration, D Dt

 Pt

ρ(x, t) dv = 0,

and by taking into account formula (2.79c) from the previous chapter. The desired result then follows by a simple application of the Localisation Principle, ρ(x, ˙ t) + ρ(x, t) div v(x, t) = 0,

(∀) x ∈ Bt , t > 0

(3.11)

or, more compactly, ρ˙ + ρ div v = 0,

(3.12)

bearing in mind the dependence on x ∈ Bc and t > 0 of the scalar and vector fields involved in (3.12). This represents the differential (or local) form of (3.4), also known as the spatial equation of continuity. A very important consequence of this principle is included next. Proposition 3.2 If Φ is the Eulerian description of an arbitrary scalar, vector, or tensor field, then

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3 Balance Laws

D Dt



 Pt

ρ Φ dv =

Pt

ρ

DΦ dv, Dt

(3.13)

for all Pt ⊂ Bc . Basically, this result asserts that whenever we want to calculate the rate of change of an integral in Bc and its integrand involves the product between the current mass density (ρ) and another field (Φ), the material time derivative is simply transferred to that other part of the integrand (Φ). We justify (3.13) for the case of a scalar field Φ (the proof for vector and tensor fields is entirely similar). To this end, we recall again the formula (2.79c) from Chap. 2, according to which,    D ρ Φ dv = (ρ ˙Φ) + (ρ Φ) div v dv. (3.14) Dt Pt Pt The integrand on the right-hand side of (3.14) can be written successively, ˙ + ρΦ div v (ρ ˙Φ) + (ρ Φ) div v = (ρ˙ Φ + ρ Φ) = (ρ˙ + ρ div v)Φ + ρ Φ˙ = ρ

DΦ , Dt

(3.15)

where on the second line we have made use of (3.12). Putting together (3.14) and (3.15), we recover (3.13), and this completes the proof of Proposition 3.2.

3.2 Body and Surface Forces In Newtonian Mechanics forces are treated as primitive concepts, in the sense that they cannot be defined in terms of simpler concepts. Within the realm of this province of Mechanics, one of the prevalent practices is to assign quantitative measures to forces by relying on various other properties of rigid bodies (e.g. mass) and changes in their associated kinematical variables (e.g. acceleration). For instance, according to Newton’s Second Law, a force may be measured by the product of mass and acceleration (but this does not mean that a force is defined by such a product). In Continuum Mechanics, forces are derived from different considerations, so that Newton’s Second Law can be used to calculate accelerations from forces, rather than the other way around. The change of shape and volume of deformable bodies under the action of external mechanical loads is resisted by internal contact forces that develop inside such bodies. A basic assumption underlying all Continuum Mechanics is that forces are continuously distributed, that is, at every point of the continuum there is a system of such forces. Consequently, in a state of static deformation every part of our deformable

3.2 Body and Surface Forces Fig. 3.1 Arbitrary deformable body subjected to a system of forces F j ( j = 1, 2, 3) in static equilibrium (left-hand side). If the body is cut through a plane in two different parts (labelled ‘A’ and ‘B’) then, if the body is to continue to remain in equilibrium, F1 and F2 must be balanced by F AB . This last force represents the cumulative effect of all the forces exerted on the plane π A by the other half (‘B’) in the uncut body

167

F1 F1 A A

F2

FBA

πA πB

B

F3

F2

FAB

B

F3

bodies must be in equilibrium under the joint action of the applied external forces and the internal contact forces. This is easily understood by considering the following thought experiment. Consider a deformable solid body acted upon by a system of external forces F1 , F2 , F3 (say) in static equilibrium, as seen in the left sketch included in Fig. 3.1. If this body were cut along an arbitrary plane, as shown in the right sketch of the same figure, the separate halves would no longer be in equilibrium and would fly apart. However, before cutting, the portion of the body labelled ‘A’ was in equilibrium, so an internal resultant force F AB must be exerted on the material in the plane π A by the material in the plane π B such that F AB , F1 , and F2 form a system of forces in equilibrium (the two halves are shown separately for the sake of clarity). In Fig. 3.1 the plane π has a particular orientation, but the same conclusion would follow whatever orientation was chosen. Thus, if any plane is drawn in a deformable body subjected to a system of external forces, the material in this plane will be acted upon by an internal force exerted by adjacent material. The mechanical stress is the quantity which enables this internal force to be determined. With these preliminary remarks in mind, in the remaining of this section we shall outline a more quantitative approach for the mathematical description of the concept of force needed in what follows. Traditionally, it is assumed that during an arbitrary motion (or deformation) of a body, the mechanical interactions between any of its parts, or its parts and the surrounding medium, are described by two general classes of forces: (i) body forces, and (ii) contact forces. The former are of secondary interest in Continuum Mechanics and correspond to forces exerted on the interior points of a body by the surrounding medium (i.e. they are caused by action at a distance); gravitational and magnetic forces are of this type. By contrast, contact forces—which have already been briefly mentioned above, are typical for deformable bodies. They

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3 Balance Laws

can be further subdivided into: (a) contact forces exerted on the external bounding surface of Bc by its environment, and which are transmitted indirectly to the interior of the body through its constituent particles (e.g. a heavyweight placed on top of a deformable body); (b) contact forces between separate (internal) parts of the current configuration Bc . For any sub-part P ⊂ B, we define the total applied force in a given motion (or deformation) χ as F (P, χ ) := F B (P, χ ) + FC (P, χ ),

(3.16)

where the subscripts ‘B’ and ‘C’ are used for identifying the contributions due to the body and contact forces, respectively. Both terms on the right-hand side of (3.16) are expressed through appropriate vectorial densities,  F B (P, χ ) ≡

Pt

 ρb dv,

FC (P, χ ) ≡

t da,

(3.17)

∂Pt

with Pt = χ (P, t) (for any t > 0). Here, b ≡ b(x, t) is the body force density (or body force, for short), which represents distributed force defined per unit mass in the current configuration due to an external agency (e.g. gravitational or electromagnetic forces). The integrand in the second equation in (3.17), that is, t ≡ t(x, t, ∂Pt ), represents the contact force density or the Cauchy traction field or traction vector (force measured per unit current surface area of ∂Pt ). The indicated dependence on x and ∂Pt is intended to convey that t depends not only on position, but also on the material surface upon which it acts. The Cauchy traction vector t is intimately connected to the orientation of ∂Pt . Let x be an arbitrary point on this surface, and let us choose arbitrarily the positive sense of the normal n to ∂Pt at x. This defines for our surface a positive side, ∂P+ t (say), the one containing n; the other side of the surface will be the negative side, ∂P− t (say). The French scientist A. L. Cauchy introduced the simplifying assumption that the traction on all like-oriented surfaces with a common tangent plane at x is the same; more specifically, t(x, t, ∂Pt ) = t(x, t, n),

(3.18)

where n ≡ n(x, t) represents the oriented unit normal to ∂Pt at x and time t > 0; a pictorial illustration of Cauchy’s postulate is included in Fig. 3.2. If da is the area element based at x, then t ≡ t(x, n) is defined as a vector such that t da represents the surface force on the positive side of ∂Pt experienced at the location x on this surface. That is, the resultant of all forces exerted through da by the material on ∂P+ t upon the material on ∂P− t . Since traction vectors are defined in conjunction with the surfaces that transmit them, it is meaningless to refer to the stress vector t(x) at x ∈ Bc —the surface on which t acts (or, equivalently, the unit normal to that surface) must be also specified. It is also necessary to emphasise that Cauchy’s simplifying hypothesis assumes not

3.2 Body and Surface Forces

169

Fig. 3.2 Two surfaces ∂P( j) ( j = 1, 2) that share the same outward unit normal n at a common (1) (2) (1) (2) point x ∈ ∂Pt ∩ ∂Pt . Cauchy’s postulate asserts that t(x, t, ∂Pt ) = t(x, t, ∂Pt )

just one traction vector at each point, it assumes a traction vector state t(x, t, n) such that, at one and the same point x at a given instant of time t > 0, there is a distinct traction vector associated with each direction n. Strictly speaking, we have an infinite number of such stress vectors at any given point (x, t). Definition 3.2 The totality of (Cauchy) traction vectors t ≡ t(x, t, n) at a fixed (x, t) ∈ Bc × (0, ∞) and for all directions n (with |n| = 1) defines the state of stress at the point x and time t > 0. As we shall see later in this chapter, the state of stress at any point (in space and time) is completely determined if the traction vectors associated with three mutually perpendicular planes, passing through x at time t > 0, are known, and these traction fields are continuous in the vicinity of x. We close our discussion of contact forces in deformable bodies by pointing out that the interpretation of FC depends on the relative position of ∂Pt ; as already mentioned above, at points x ∈ ∂Pt interior to Bc , the stress vector t(x, t, n) represents force per unit area exerted on Pt by the material outside this region. On the other hand, at points x ∈ ∂Pt ∩ ∂Bc (i.e., on the external surface of the body), the stress vector t(x, t, n) represents force per unit area applied to the surface of the body by an external agency, and it is usually referred to as surface traction. We state below the final form of (3.16) that will prove to be suitable for our subsequent purposes. Definition 3.3 The resultant force acting on the material occupying a sub-region Pt ⊂ Bc is defined by  F (Pt ) :=

 Pt

ρb dv +

t(n) da,

(3.19)

∂Pt

Finally, the generalisation of the concept of resultant moment (or torque) relative to a fixed point follows naturally from (3.19) and is recorded below for easy reference.

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3 Balance Laws

Definition 3.4 The resultant moment (or torque) with respect to the fixed point o acting on the material occupying Pt ⊂ Bc is defined by  M(Pt ; o) :=

Pt

 x ∧ (ρb) dv +

∂Pt

x ∧ t(n) da,

(3.20)

where x is a position vector relative to o.

3.3 Global Form Having given a precise meaning to the resultant force and the resultant moment acting on any sub-part Pt ⊂ Bc of a deformable body B, we can now contemplate the extension of the classical conservation principles mentioned at the beginning of this chapter. To this end, we first generalise the concepts of linear and angular momentum as indicated below. Definition 3.5 (a) The linear momentum of the sub-part Pt ⊂ Bc is defined by  P(Pt ) :=

Pt

ρv dv.

(3.21)

(b) If x ∈ Pt represents the position vector of a representative point of Pt ⊂ Bc , relative to an origin o, then the angular momentum of the sub-part Pt with respect to o is defined by  L(Pt ; o) := x ∧ (ρv) dv. (3.22) Pt

Although in classical rigid-body mechanics the conservation of angular momentum is (loosely speaking) a consequence of the conservation of linear momentum, this is no longer true for deformable bodies. The main equilibrium equations of Continuum Mechanics turn out to be a direct consequence of the following two conservation principles (regarded as being independent of each other). Conservation of Linear Momentum (CLM): D P(Pt ) = F (Pt ), (∀) Pt ⊂ Bc . Dt

(3.23)

Conservation of Angular Momentum (CAM): D L(Pt , o) = M(Pt , o), (∀) Pt ⊂ Bc . Dt

(3.24)

3.3 Global Form

171

These two conservation principles hold independently of the choice of origin, although both L and M in (3.23) do depend on such a choice. By taking into account (3.19) and (3.20), Eqs. (3.23) and (3.24) can be written in expanded form and, thus, we arrive at the global forms of the above conservation principles, D Dt D Dt

 Pt



 Pt

ρ v dv = 

x ∧ (ρ v) dv =

Pt

 Pt

ρ b dv +

t(n) da,

(3.25a)

x ∧ t(n) da.

(3.25b)

∂Pt



x ∧ (ρ b) dv +

∂Pt

We re-iterate that (3.25a) and (3.25b) must hold for any Pt ⊂ Bc and t > 0 (they are also known as the global equilibrium equations for deformable bodies). In the next section, we make the assumption that the integrands of the above integral relations are continuous and take advantage of the Localisation Principle to derive a more useful local form of the above two equilibrium equations.

3.4 Local Form We start with the observation that the material time derivative of the integrals on the left-hand side in Eqs. (3.25a) and (3.25b) can be taken care of by making recourse to Proposition 3.2; e.g. take Φ → v for the former equation, and Φ → x ∧ v for the latter. Hence (3.25) can be re-cast in the simpler form 

  Pt

Pt

ρ(a − b) dv =

ρ x ∧ (a − b) dv =



t(n) da,

(3.26a)

x ∧ t(n) da,

(3.26b)

∂Pt

∂Pt

where we recall that a ≡ v˙ is the Eulerian description of the acceleration field. Note that, although all the integrals in (3.26) are taken in Bc , we have a mixture of both volume and surface integrals. Thus, the Localisation Principle cannot be applied in a straightforward manner. For that we still need a couple of additional new results that are stated below, but their proof will be postponed until the next section. Proposition 3.3 (Principle of Action and Reaction/Cauchy’s Lemma) The traction vectors acting at the same point x ∈ Bc (at time t > 0), but on opposite sides of a material surface, are equal in magnitude and opposite in direction (see Fig. 3.3). Mathematically, this is expressed by the following property of the traction vector: t(x, t, −n) = −t(x, t, n).

(3.27)

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3 Balance Laws

Fig. 3.3 Traction vectors associated with both faces of a flat surface element, at a given point x ∈ Bc at time t > 0; see Proposition 3.3

n(x)

t(n)

da x −t(n)

−n(x)

This result, together with the CLM, can be used to show that the correspondence n → t(n) is a linear mapping, and hence defines a second-order tensor. Proposition 3.4 (Cauchy’s Stress Theorem) There exists a unique second-order Eulerian tensor field σ ≡ σ (x, t), with (x, t) ∈ Bc × (0, ∞), such that t(x, t, n) = σ T (x, t) · n ≡ n · σ (x, t),

(∀) x ∈ Bc , t > 0.

(3.28)

It is common practice to write (3.28) in the simpler form t(n) = σ T · n,

(3.29)

keeping in mind that both σ and n depend on position and time (to lighten up the notation, in what follows we shall sometimes suppress the dependence of σ and t on x ∈ Bc and t > 0, whenever there is no imperative need for it). Next, we use (3.29) to obtain the local form of the CLM. On substituting that formula into (3.26a), with the help of the Divergence Theorem, we get 

 Pt

Therefore,

 Pt

ρ(a − b) dv =

∂Pt

 σ T · n da =

  div σ − ρ(a − b) dv = 0,

Pt

div σ dv.

(∀) Pt ⊂ Bc ,

and a simple application of the Localisation Principle yields div σ + ρb = ρa;

(3.30)

this is the local form of the C L M, also known as Cauchy’s first equation of motion. It can be shown that the CAM equation is equivalent to the symmetry of the Cauchy stress tensor σ , that is, (3.31) σT = σ.

3.4 Local Form

173

This is sometimes referred to as Cauchy’s second equation of motion. A proof of (3.31) is given in the next section (see also Exercise 3.1).

3.5 Some Technical Proofs For the remainder of this section, we shall return to the justification of (3.27) and (3.28). We start with the simpler proof of the former result. Proof of Proposition 3.3 We pick up an arbitrary point P ∈ Bc whose position vector is x. Next, consider an imaginary small volume Ω ⊂ Bc surrounding P. Through this point, let us draw an imaginary surface Σ that divides the volume Ω into two parts, Ω1 and Ω2 (say), both of which are assumed to be situated entirely within Bc ; a two-dimensional sketch of this scenario is included in Fig. 3.4. We are aiming to apply the CLM in Ω = Ω1 ∪ Ω2 , but first let us make an important observation. The boundaries of Ω1 and Ω2 will have Σ in common, although the outward unit normals of Ω j ( j = 1, 2) along Σ, denoted by m j ≡ m j ( y), with y ∈ Σ, will have opposite directions; that is, m1 + m2 = 0 at each point of Σ. Letting n be the outward unit normal on S1 ∪ S2 and observing that ∂Ω j = S j ∪ Σ ( j = 1, 2)

and

S = S1 ∪ S2 ,

(3.32)

the consecutive application of the CLM to the volumes Ω, Ω1 , and Ω2 gives 

 Ω

ρ(a − b) dv =

t(x, n) da,

(3.33)

∂Ω

together with 

 Ωj



ρ(a − b) dv =

t(x, n) da + Sj

Σ

t( y, m j ) da

Fig. 3.4 Cross-sectional view of the volume Ω and the surface Σ used in the proof of Proposition 3.3 (see text for details)

( j = 1, 2).

(3.34)

S1

m −m n ∂Ω2 = S2 ∪ Σ

Ω2

∂Ω1 = S1 ∪ Σ Ω1

n

S2 Σ

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3 Balance Laws

C m e3 P

e2

x B

e1 A −e1 −e2

S2

S1 S3 −e3

Fig. 3.5 Cauchy’s tetrahedron and details of the notation defined in the text

Subtracting (3.34) from (3.33), and using (3.32), it transpires that  Σ



 t( y, m1 ) + t( y, m2 ) da = 0,

(∀) Σ ⊂ Bc , with x ∈ Σ.

But m1 + m2 = 0, so if m := m2 and we allow Σ to shrink to the point P (hence y → x), the Localisation Principle in conjunction with the continuity of the corresponding integrands lead to t(x, −m) = −t(x, m) for all unit vectors m; changing m to n Eq. (3.27) is immediately recovered, and this completes the proof of Proposition 3.3. Proof of Proposition 3.4 The structure of the stress tensor in terms of three linearly independent vectors, which is the essence of (3.28), requires more effort to justify, although the general idea is very similar to what we have just done. We start by considering a point P ∈ Bc whose position vector with respect to the basis B = {e1 , e2 , e3 } is x, and let m be a unit vector (i.e., |m| = 1) that defines an arbitrary direction in space. We then consider a tetrahedron Pt ≡ P ABC ⊂ Bc with the following properties: (a) P is one of its vertices; (b) the faces which share P are parallel to the coordinate planes; (c) m is the outward normal to the face ABC. Thus, S1 := B PC is parallel to the O x2 x3 -plane, S2 := A PC is parallel to the O x1 x3 -plane, and S3 := B P A is parallel to the O x1 x2 -plane. If the distance from P to S := ABC is denoted by δ and we introduce the additional notations (Fig. 3.5), A(δ) := area(S) ( j = 1, 2, 3), A j := area(S j ),

3.5 Some Technical Proofs

175

then it follows by elementary geometrical arguments that (∃) C1 , C2 ∈ R such that A(δ) = C1 δ 2 , and hence vol(Pt ) = C2 δ 3 ; Ai = (ei · m)A(δ) for i = 1, 2, 3.

(3.35a) (3.35b)

Note that  CLM ⇒

 ∂Pt



⇒

t(n) da = ρ(a − b) dv Pt

t(n) da

≤ max |ρ(a − b)| · vol(Pt ). Pt

∂Pt

(3.36)

Since all fields appearing in (3.36) are continuous, 0 ≤ C := maxPt |ρ(a − b)| < ∞. Taking into account (3.35a), from (3.36) it follows that



C × (C2 δ 3 ) 1

≤ t(n) da = const × δ → 0,

A(δ) ∂Pt C1 δ 2

as δ → 0.

(3.37)

Next, we exploit the fact that the lateral surface of the tetrahedron Pt is the union of S j ( j = 1, 2, 3) and S, so the left-hand side of (3.37) can be simplified further. First, by using the additivity of the integral,  ∂Pt

 t(n) da =

t(m) da + S

3   i=1

t(−ei ) da,

(3.38)

Si

where n is the generic label for the outward unit normals to S j and S ( j = 1, 2, 3). At the same time, the Localisation Principle gives  1 t(m) da → t(x, m), A(δ) S  1 Ai t(−ei ) da → t(x, −ei ), A(δ) S A(δ)  

as δ → 0,

(3.39a)

as δ → 0.

(3.39b)

ei ·m

In light of these results, by taking the limit δ → 0 in (3.38), we arrive at t(x, m) = −

3 3   (ei · m)t(x, −ei ) = (ei · m)t(x, ei ), i=1

(3.40)

i=1

the last equality being a consequence of (3.27). We note in passing that, more generally, if a, b, c are three linearly independent vectors and α, β, γ ∈ R, then t(x, αa + β b + γ c) = αt(x, a) + β t(x, b) + γ t(x, c).

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3 Balance Laws

(

22

2

21

23

(

2)

( 1)

12

3)

1 3

32 11

31

13

33

Fig. 3.6 The physical interpretation of the components of the Cauchy stress tensor σ

Finally, we are now in position to define the Cauchy stress tensor by setting σ T (x) :=

3 

t(x, ei ) ⊗ ei .

(3.41)

i=1

The notation for the vector m in (3.40) is irrelevant, and we shall stick to the more familiar n henceforth. It is a simple exercise to check the assertion recorded in (3.28), particularly, σ ·n= T

 3 

 t(x, ei ) ⊗ ei · n =

i=1

3 

t(x, ei )(ei · n) = t(x, n),

i=1

where use has been made of (3.40). To elucidate further the meaning of the i jcomponent of σ , we write (cf. Proposition 1.3)   σi j = ei · σ · e j = e j · σ T · ei  3     3  = ej · t(ek ) ⊗ ek · ei = e j · t(ek )δik = e j · t(ei ) k=1

k=1

or σi j = e j · t(ei ). Thus, σi j (x, t) represents the component in the direction e j of the stress vector which acts on a plane normal to ei at x, at time t (see Fig. 3.6); more specifically, t(x, t, ei ) = σi j (x, t)e j = σi1 (x, t)e1 + σi2 (x, t)e2 + σi3 (x, t)e3 .

(3.42)

3.5 Some Technical Proofs

177

A direct consequence of (3.42) is the following: if t = ti ei and n = n j e j , then ti (x, t, n) = σ ji (x, t)n j

(i = 1, 2, 3);

the nine quantities σi j correspond to the components of a second-order Eulerian tensor field σ (x, t). Its component representation relative to the basis {ei } is ⎡

⎤ σ11 σ12 σ13 [σ ] = ⎣σ21 σ22 σ23 ⎦ . σ31 σ32 σ33

σ (x, t) = σi j (x, t) ei ⊗ e j ,

Proof of Eq. (3.31) Replacing the traction vector t(n) in (3.26b) with its expression given by formula (3.29), and then using (3.30), it turns out that the CAM can be re-arranged as  Pt

  x ∧ div σ dv =

 ∂Pt

  x ∧ σ T · n da, (∀) Pt ⊂ Bc .

Let us multiply this equation by an arbitrary constant vector h. Then, standard properties of the scalar triple product allow us to write  Pt





h · x ∧ divσ ) dv =  =

∂Pt





h · x ∧ (σ · n) = T

∂Pt

∂Pt

∂Pt

[h, x, (σ T · n)] da

 [(σ · n), h, x] da = T

 =



  n · σ · (h ∧ x) da =

∂Pt

(σ T · n) · (h ∧ x) da



Pt

  div σ · (h ∧ x) dv,

where use has also been made of the Divergence Theorem and the definition of the transposition operation. Therefore, we have reduced the CAM to the equality of two volume integrals  Pt

 h · x ∧ divσ ) dv =

 Pt

  div σ · (h ∧ x) dv.

(3.43)

The integrand on the right-hand side in this equation can be expanded by using formula (A.2d) from Appendix A, so that    div σ · (h ∧ x) = (div σ ) · (h ∧ x) + tr σ · grad(h ∧ x)).

(3.44)

Replacing (3.44) into (3.43) and invoking again the properties of the scalar triple product, it follows that

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3 Balance Laws

 Pt

 tr σ · grad(h ∧ x)) dv = 0,

(∀) Pt ⊂ Bc ,

 whence tr σ · grad(h ∧ x)) for all x ∈ Bc . Since grad(h ∧ x) = − ∈i j p h p ei ⊗ e j , this last equation can be written in expanded form as indicated below     tr σ · grad(h ∧ x) = σ ji grad(h ∧ x i j = − ∈i j p σ ji h p = − ∈i j1 σ ji h 1 − ∈i j2 σ ji h 2 − ∈i j3 σ ji h 3 = −(σ32 − σ23 )h 1 − (σ13 − σ31 )h 2 − (σ21 − σ12 )h 3 . The vector h in this last equation was arbitrary, so we can take h 1 = 1, h 2 = h 3 = 0 to get σ32 = σ23 . By making similar particular choices for h, we discover that σ13 = σ31 and σ12 = σ21 . In conclusion, the Cauchy stress tensor σ is symmetric. An alternative justification of the result we have just proved is suggested in Exercise 3.1. Before we start exploring further some other simple properties of σ in the next section, it is important to emphasise that it is the (Cauchy) traction vector that should be regarded as a primitive concept, a physical object which can be understood intuitively. By contrast, the Cauchy stress tensor σ ≡ σ (x, t) is merely a mathematical tool for keeping track of the states of stress corresponding to the position x ∈ Bc at time t > 0; only the elements of that state, i.e., the stress vectors, are open to direct mechanical interpretation. This observation is also reflected by the many different types of stress tensors used in the literature, some of which will be mentioned briefly in the last section of this chapter (see Sect. 3.8).

3.6 Further Properties of the Cauchy Stress Tensor We now return to the newly defined stress tensor σ in order to highlight some of its key properties and introduce additional terminology. Definition 3.6 Let σ ≡ σ (x, t) be the Cauchy stress tensor at a particular point x ∈ Bc and time t > 0. If σ · n = λ n, then λ ∈ R is known as a principal stress and n represents a principal (stress) direction. The symmetry of the Cauchy stress tensor assures us that there are three principal stresses at each point x ∈ B (and for each time t > 0); cf. Proposition 1.9. Associated with these three numbers there are also three principal directions (or vectors) that can be used to construct an orthonormal basis (cf. the Spectral Representation Theorem); σ has a diagonal representation when written in this basis, a feature that is particularly attractive in the formulation of various constitutive laws.

3.6 Further Properties of the Cauchy Stress Tensor Fig. 3.7 The decomposition of the Cauchy traction vector into normal and shear components

179

t(n) ≡ σ · n (traction vector)

normal force n

shearing force

x da

Consider an arbitrary planar material surface passing through x ∈ Bc , and which has a unit normal n (see Fig. 3.7). The traction vector t(x, n) can be expressed as the sum of two vector components: one parallel to the plane, t  (say), and the other one normal to the plane, t ⊥ (say); more specifically, t = t ⊥ + t , where t ⊥ := (n ⊗ n) · t ≡ (t · n)n,   t  := I − n ⊗ n · t ≡ t − (t · n)n.

(3.45a) (3.45b)

These are called the normal and, respectively, the shear force (per unit area) on the plane with unit normal n. According to (3.45a), the magnitude (N ) of the normal force is given by N := n · t(n) = n · (σ · n) ≡ σi j n i n j ,

(3.46)

and is usually referred to as the normal stress at (x, t); it is said to be tensile when positive (t · n > 0), and compressive when negative (t · n < 0). Similarly, the magnitude (T ) of the shearing force is easily obtained from (3.45b), T := |t − t ⊥ | =

 |t|2 − N 2 ,

(3.47)

and is known as the tangential or shear stress. Note that if m is a unit vector that specifies the direction of the orthogonal projection of t onto the plane of normal vector n, then t = N n + T m. (3.48) If the components of the Cauchy stress tensor σ are known, computing the normal and shear stresses corresponding to any given plane is a routine matter that requires a straightforward application of the formulae (3.46) and (3.47). In such cases, the tangential direction m can be found from (3.48), namely,

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3 Balance Laws

m=

1 (σ · n − N n); T

(3.49)

it is clear that m = 0 if n coincides with one of the principal (stress) directions. The Cauchy stress vector satisfies an interesting symmetry property, m · t(x, n) = n · t(x, m),

(∀) n, m ∈ V , |n| = |m| = 1,

whose origin lies in the symmetry of the Cauchy stress tensor σ (more precisely, this is a consequence of the C AM). In words, this says that the projection along the direction m of the stress vector associated with an element of surface of normal n is equal to the projection along the direction n of the stress vector corresponding to an element of surface of normal m. It is of practical interest to have an understanding of the directions n which lead to extremal values for the normal and tangential stresses. The method of Lagrange multipliers provides a relatively straightforward answer to these questions. Thus, for finding the maximum or minimum values of N defined in (3.46) subject to the constraint ϕ := 1 − n i n i = 0, we must extremise N + λϕ, where λ ∈ R is the usual Lagrange multiplier. Hence ∂ (N + λϕ) = 0 ∂n i

⇒

2(σi j n j − λn i ) = 0

⇒

σ · n = λn,

so the extremal values for the normal stress coincide with the principal stress directions. Next, let us assume that (x, t) is fixed. We want to find the directions n for which the tangential stress (3.47) at (x, t) is either maximum or minimum. To this end, we shall consider the basis { f 1 , f 2 , f 3 } consisting of the eigenvectors of σ (x, t); the corresponding eigenvalues will be denoted by σ1 , σ2 , σ3 . Taking advantage of the diagonal representation of σ in the aforementioned basis, t(x, n) = σ1 n 1 f 1 + σ2 n 2 f 2 + σ3 n 3 f 3 , and Eq. (3.47) shows that T 2 = |t(x, n)|2 − N 2 = σ12 n 21 + σ22 n 22 + σ3 n 23 − (σ1 n 21 + σ2 n 22 + σ3 n 23 )2 = (σ1 − σ2 )2 n 21 n 22 + (σ2 − σ3 )2 n 22 n 23 + (σ3 − σ1 )2 n 23 n 21 . The extremal values of T 2 subject to ϕ ≡ 1 − n i n i = 0 will depend on the principal stresses, as we now confirm. If σ1 = σ2 = σ3 then T 2 = 0 for all n ∈ V . If only two principal stresses are equal, σ1 = σ2 (say), then T 2 = (σ3 − σ1 )2 n 23 (1 − n 23 ).

3.6 Further Properties of the Cauchy Stress Tensor

181

In this case T 2 is maximum or minimum for n 3 = 0 or n 3 = ±1. The other two components of n will be found from the normalisation condition, i.e. n 21 + n 22 = 1 − n 23 . The last case to consider is when all the principal stresses are distinct; without loss of generality, we shall assume that σ1 > σ2 > σ3 . On introducing the Lagrange multiplier μ ∈ R, we are led to extremising T 2 + μϕ with respect to n i (i = 1, 2, 3). Hence,   n 1 σ12 − 2σ1 (σ1 n 21 + σ2 n 22 + σ3 n 23 ) + μ = 0,   n 2 σ22 − 2σ2 (σ1 n 21 + σ2 n 22 + σ3 n 23 ) + μ = 0,   n 3 σ32 − 2σ3 (σ1 n 21 + σ2 n 22 + σ3 n 23 ) + μ = 0.

(3.50a) (3.50b) (3.50c)

The option n 1 = n 2 = n 3 = 0 is excluded because the constraint ϕ = 0 would be violated in that case. If n i = 0 for i = 1, 2, 3, then σ12 + μ σ2 + μ σ2 + μ = 2 = 3 , σ1 σ2 σ3

(3.51)

which cannot be satisfied simultaneously. There are two distinct possibilities that need to be evaluated: (i) one component of n is zero (e.g. n 1 = 0) and the other two are different from zero (e.g. n 2 , n 3 = 0); (ii) two components of n are zero (e.g. n 1 = n 2 = 0) and the third is different from zero (e.g. n 3 = 0). In the case (i) Eq. (3.50a) is automatically satisfied, while (3.50b) and (3.50c) yield the last equality in (3.51) whence μ = σ2 σ3 . It remains that we must solve the system n 22 + n 23 = 1, 2(σ2 n 22 + σ3 n 23 ) = σ2 + σ3 , √ √ which has solutions n 2 = ±1/ 2 and n 3 = ±1/ 2. To summarise, after considering the other two very similar cases that we omitted for the sake of brevity, we have the following solutions for the maximum of T : n 1 = 0,

√ n 1 = ±1/ 2, √ n 1 = ±1/ 2,

√ n 2 = ±1/ 2, n 2 = 0,

√ n 2 = ±1/ 2,

√ n 3 = ±1/ 2, √ n 3 = ±1/ 2,

4T 2 = (λ2 − λ3 )2 ,

n 3 = 0,

4T 2 = (λ1 − λ2 )2 .

4T 2 = (λ1 − λ3 )2 ,

In the case (ii), n 1 = n 2 = 0, and from the constraint ϕ = 0 we get n 3 = ±1. Equations (3.50) give μ = σ32 . Hence, the general solutions in this second case can be shown to be n 1 = 0,

n 2 = 0,

n 3 = ±1,

T = 0,

n 1 = ±1, n 1 = 0,

n 2 = 0, n 2 = ±1,

n 3 = 0, n 3 = 0,

T = 0, T = 0,

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3 Balance Laws

and correspond to the minimum of T . This completes our discussion about finding the directions n in space for which the tangential stress T assumes extremal values. A geometric solution of this problem is also available, and is due to the German engineer C. O. Mohr. This is reviewed briefly next. Let us start by first finding the components of an arbitrary direction n in terms of T , N and σ j ( j = 1, 2, 3). The three equations that will help us to accomplish this are recorded below σ1 n 21 + σ2 n 22 + σ3 n 23 = N , σ12 n 21

+ σ22 n 22 + n 21 + n 22

σ32 n 23 + n 23

(3.52a)

= N +T ,

(3.52b)

= 1,

(3.52c)

2

2

the first two being taken straight from (3.46) and (3.47), while the last is just the normalisation condition for unit vectors. It must be noticed that these relations constitute a linear system of three algebraic equations in n i2 (i = 1, 2, 3), whose solutions is readily worked out to be

n 21 =

T 2 + (N − σ2 )(N − σ3 ) , (σ1 − σ2 )(σ1 − σ3 ) n 23 =

n 22 =

T 2 + (N − σ3 )(N − σ1 ) , (σ2 − σ3 )(σ2 − σ1 )

T 2 + (N − σ1 )(N − σ2 ) . (σ3 − σ1 )(σ3 − σ2 )

Taking into consideration that σ1 > σ2 > σ3 (as explained earlier), these formulae indicate that the denominators of the first and the last fractions must be non-negative, while that of the second fraction must be non-positive. This observation leads to the following set of simultaneous inequalities  2  1 σ2 − σ3 2 , T + N − (σ2 + σ3 ) ≥ 2 2   2  1 σ1 − σ3 2 , T 2 + N − (σ1 + σ3 ) ≤ 2 2   2  1 σ1 − σ2 2 2 . T + N − (σ1 + σ2 ) ≥ 2 2 

2

(3.53a) (3.53b) (3.53c)

In the (N , T ) plane (see Fig. 3.8), the first inequality (3.53a) is satisfied in the exterior of the circle C1 with centre on the horizontal axis and whose intersection with this line are the points of abscissae σ2 and σ3 . The second inequality (3.53b) holds true inside the circle C2 centred at a point on the O N axis and which cuts it in the points of abscissae σ1 and σ3 . Finally, the third inequality (3.53c) is satisfied for points outside the circle C3 whose centre lies again on O N and which intersects this horizontal axis in points of abscissae σ2 and σ1 . The set of points in the (N , T ) plane which

3.6 Further Properties of the Cauchy Stress Tensor Fig. 3.8 Geometrical interpretation of the extremal tangential stresses (see text for details)

183

T A

C2

C1 O

σ3

C3 σ2

O

N σ1

satisfies all three inequalities (3.53) will be situated in the shaded region shown in Fig. 3.8. It can be seen that |T |max is equal to the length of the line segment O  A and is equal to the radius of the circle C2 , i.e. 2|T |max = σ1 − σ3 and corresponds √ to 2N = σ1 + σ3 . Substituting these values in (3.52) we recover n 1 = n 3 = ±1/ 2 and n 2 = 0. Some of the above calculations are summarised in the following main result. Proposition 3.5 Let x ∈ Bc and t > 0 be fixed. If the Cauchy stress tensor σ ≡ σ (x, t) admits the distinct eigenvalues σ1 > σ2 > σ3 corresponding to the orthonormal eigenvectors f 1 , f 2 , and f 3 , i.e. σ · f i = σi f i (i = 1, 2, 3, no sum), then 1. the mapping n → n · t(n),

|n| = 1,

has a maximum value equal to σ1 (corresponding to n = f 1 ) and a minimum value equal to σ3 (attained for n = f 3 ); 2. the symmetric mapping (n, m) → m · t(n),

m · n = 0,

|m| = |n| = 1,

admits a maximum value equal to 1 (σ1 − σ3 ), 2 corresponding to 1 n = √ ( f 3 + f 1) 2

and

1 m = √ ( f 3 − f 1 ). 2

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3 Balance Laws

3.7 Particular States of Stress Some particular expressions of the Cauchy stress tensor are worth spelling out separately as they provide further insight into the mechanical interpretation of its various components. Definition 3.7 If in some configuration the shear stress is identically zero and the normal stress is independent of n, the stress is said to be spherical or hydrostatic (see Fig. 3.9). In this case, there is a scalar field p ≡ p(x, t), called pressure, such that t(n) = − p n

or

σ = − p I.

(3.54)

This kind of state of stress is typically associated with inviscid fluids, but such states can also be encountered in certain idealised situations involving solid bodies. There are several other special states of stress that deserve mentioning. Definition 3.8 Let x ∈ Bc and t > 0. 1. A state of pure normal stress in the direction of the unit vector n corresponds to a Cauchy stress tensor of the form σ = σ (n ⊗ n),

(3.55)

where σ ≡ σ (x, t). 2. A pure shear stress state in the plane spanned by the unit vectors n and m is defined by the Cauchy stress tensor σ = τ (n ⊗ m + m ⊗ n),

(3.56)

where τ ≡ τ (x, t) > 0. It is a simple exercise to check that the state of stress (3.55) is characterised by a normal stress σ , while the corresponding shear stress is zero; pure tension corresponds to σ > 0, while pure compression is associated with σ < 0. In light of the remarks made immediately after Definition 3.6, at any point x ∈ Bc (and time t > 0) σ can be represented in the basis of its principal axes, and therefore one may regard the state of stress at that point as being completely described by a superposition

Fig. 3.9 Hydrostatic (or spherical) state of stress: σ = −pI

n Bc

x

t(n) = −pn

3.7 Particular States of Stress

185

of three states of pure normal stress acting on the surface elements perpendicular to the principal axes of the Cauchy stress tensor. The state of stress defined in (3.56) is characterised by shear stress equal to τ , while the normal stress is zero—see Eqs. (3.46) and (3.47). In the general representation of σ with respect to an arbitrary orthonormal basis, in addition to the terms of the form σii ei ⊗ ei (i = 1, 2, 3, no sum)—that were already touched upon in the previous paragraph, there are also three sets of terms σi j (ei ⊗ e j + e j ⊗ ei ) (i, j = 1, 2, 3, i = j, no sum) associated with pure shear stress states in the coordinate planes. Consequently, relative to an arbitrary orthonormal system of coordinates, the state of stress at (x, t) ∈ Bc × (0, ∞) can be seen as the superposition of three states of pure normal stress and three other states of pure shear stress. This point of view mirrors our kinematic interpretation of the right Cauchy–Green deformation tensor C in Sect. 2.6 (see Chap. 2). The importance of the expressions (3.54)–(3.56) stems also from the fact that for constant p, σ , τ they represent actual equilibrium states of stress that can be maintained solely by surface loads applied on the external boundary of deformable solids, in the absence of body forces. Given a vector field ˚t on ∂Bc , the equilibrium of Bc subject to the tractions ˚t is expressed by the equations div σ = 0 in Bc ,

t(n) = ˚t on ∂Bc ,

(3.57)

where n represents the usual outward unit normal on ∂Bc . If Bc assumes a spherical geometry and ˚t = − pn, then the stress tensor σ given by (3.54) trivially satisfies the differential equations in (3.57), and the boundary condition holds because t(n) = n · σ = n · (− p I) = − pn = ˚t . In a similar way, we can let Bc be a right circular cylinder, whose axis is parallel to the unit constant vector n, and consider ˚t = ±σ n on the end faces of the cylinder and zero on its lateral surface (here, σ > 0 is a constant). In this case, the state of stress in the cylinder is described by the particular Cauchy stress tensor recorded in (3.55). Finally, by letting Bc be a right cuboid subjected to uniformly distributed tangential loads on two pairs of opposite faces, then the stress distribution inside the cube will correspond to (3.56). An intuitive representation of these last two particular situations is included in Figs. 3.10 and 3.11. Another important special state of stress, that will also be discussed at length in Chap. 8, is particularly relevant to plate-like deformable bodies subjected to in-plane forces, and is defined by the requirements stated below.

m

Fig. 3.10 Pure normal stress: σ = σ (n ⊗ n), σ > 0. Note that t(±m) = 0

Bc n t(−n) = −σn

t(n) = +σn

186 Fig. 3.11 Pure shear stress: σ = τ (n ⊗ m + m ⊗ n), with τ > 0

3 Balance Laws

m

t(m) = +τ n n t(n) = +τ m

t(−n) = −τ m

Bc

t(−m) = −τ n

Definition 3.9 A plane stress state at x ∈ Bc and t > 0 is defined by a Cauchy stress tensor of components σi j that satisfy σ13 = σ23 = σ33 ≡ 0, and σ11 , σ12 , σ22 are functions of the coordinates x1 and x2 only.

3.8 The Piola–Kirchhoff Stress Tensors We return now to a further examination of the concept of stress for a deformable body. Traditionally, the mechanical stress at a point P in such a body can be broadly defined as ΔF , (3.58) lim ΔS→P ΔA where ΔS is an oriented material surface passing through P, ΔA represents its area, and ΔF stands for the resultant distribution of forces acting on a given side of this surface. The main idea encapsulated in (3.58) is that ‘stress = force divided by area on which the force acts’. Since Continuum Mechanics operates with two distinct types of configurations, the reference and the current configurations, respectively, it seems natural to inquire as to where the force ΔF and the area ΔA are being measured. Either configuration will do equally well for recording ΔA, but each choice comes with its own pros and cons. However, because of its very nature, ΔF will always represent force acting on the current configuration. At this juncture, it is important to be aware that each one of the above two choices has important consequences when we set up a particular boundary value problem. The previous discussions about the Cauchy stress vector t has involved the current configuration Bc . It can be easily seen that the Cauchy stress tensor σ measures the contact force per unit area in the deformed (current) configuration. In problems of Solid Mechanics, it is not always convenient to work with σ since the deformed

3.8 The Piola–Kirchhoff Stress Tensors

187

configuration is not known in advance. For this reason we want to introduce a stress tensor that gives force measured per unit area in the undeformed (or reference) configuration. This is the so-called first Piola–Kirchhoff stress tensor, S ≡ S(X, t), defined by S := J F −1 · σ ,

(3.59)

where J ≡ det F is the usual Jacobian. In contrast to the Cauchy stress tensor, which deals entirely with Bc , the first Piola– Kirchhoff tensor is a two-point tensor (note the presence of the deformation gradient in (3.59) which provides a link with the reference configuration Br ). The component form of S relative to the usual two rectangular Cartesian systems associated with the orhonormal sets {E α } and {ei } is S = Sαp E α ⊗ e p ,

with

Sαp = J Fαi−1 σi p .

The rationale behind the introduction of (3.59) can be traced back to the transformation formulae in Proposition 2.2. Writing (2.34b) for an arbitrary tensor S, it follows immediately that Div S = J div (J −1 F · S).

(3.60)

If we want to write Cauchy’s first equation of motion (3.30) on Br , it seems sensible to choose S such that the expression inside the spatial divergence in (3.60) is equal to the Cauchy stress tensor σ . This leads precisely to the expression recorded in (3.59), and for this particular choice Div S = J div σ . Multiplying (3.30) by J and using this last result together with the P MC (ρ0 = ρ J ) leads to the material version of Cauchy’s first equation of motion, Div S + ρ0 b0 = ρ0 A,

(3.61)

where b0 ≡ b0 (X, t) ≡ b(χ (X, t), t) and A ≡ A(X, t) ≡ a(χ (X, t), t) represent the Lagrangian versions of the body force density and the acceleration field, respectively. Despite the close similarity between (3.61) and its spatial counterpart, it is important to notice that the symmetry of the stress tensor has been lost. Indeed, using (3.59) to re-write σ T = σ in terms of S, we end up with ST · F T = F · S,

(3.62)

which represents the local version of the CAM in material description. In this case, one would have to consider (3.62) as additional equations that have to be added to the three scalar equations of motion in (3.61). One way to circumvent the above difficulty is to augment the expression of S so that it becomes symmetric, while preserving the Lagrangian nature of the original

188

3 Balance Laws

tensor. This leads to the second Piola–Kirchhoff stress tensor Π ≡ Π(X, t), Π := S · F −T ≡ J F −1 · σ · F −T ,

(3.63)

which admits the following component representation: Π = Παβ E α ⊗ E β ,

−1 Παβ = J Fαi−1 σi p Fβp .

(3.64)

Since S = Π · F T , Eq. (3.61) becomes   Div Π · F T + ρ0 b0 = ρ0 A,

(3.65)

while the symmetry of σ translates into Π T = Π. The interpretation of the first Piola–Kirchhoff tensor can be given along the following lines. Consider a material volume P in B. In the reference configuration, this occupies a region P0 ⊂ Br , while in the current configuration the same material volume is found in Pt ⊂ Bc . The total surface force on P at time t can be calculated using the Cauchy stress tensor    t(n) da ≡ σ T · n da = σ · n da. (3.66) ∂Pt

∂Pt

∂Pt

By using Nanson’s formula, n da = J F −T · N d A, we can write the surface integral (3.66) on the reference configuration 

 ∂Pt

σ T · n da =

∂P0



 J σ · F −T · N d A =

 ∂P0

ST · N d A,

where n and N, respectively, are the outward unit normal fields on ∂Pt and ∂P0 . Thus, we can formally write σ T · n da = ST · N d A

⇒

ST · N =

t da . dA

(3.67)

The expression ST · N is sometimes referred to as the Piola–Kirchhoff stress vector; it should be clear that it has the same direction as the original Cauchy traction vector t ≡ σ T · n (but its magnitude will be different). According to (3.67), the first Piola–Kirchhoff tensor specifies the contact force in Bc per referential area with respect to the orientation in Br of the surface involved. We can go one step further with the physical interpretation of this stress tensor, by looking more closely at a generic component Sαp of S,

3.8 The Piola–Kirchhoff Stress Tensors

189

Sαp (X, t) = E α · S(X, t) · e p = (ST (X, t) · E α ) · e p = or, Sαp (X, t) =

e p · t(x, t, E χα ) da dA

e p · t(x, t, E χα ) da , dA

(3.68)

where E χα is the image of E α through χ (as it follows from Nanson’s formula). Equation (3.68) provides the following information: Sαp represents the component in the p-direction of the force acting on an element of surface in Bc which corresponds to an element of surface in Br with normal along E α , the force being expressed per unit area of the element in Br . A caveat needs to be issued here regarding the terminology and notation used in this section. In the literature S is often called the nominal stress tensor, the name of first Piola–Kirchhoff stress tensor being reserved for ST . This has no implications on the relationship between σ and Π, but it does affect the equation of motion (3.65). Furthermore, there are many other measures of stress that we have not touched upon because their applicability is application-specific and they will not be needed later in this book. For example, the expression τ ≡ J σ is known as the Kirchhoff stress tensor, while the non-symmetric material tensor T B ≡ R T · ST is associated with name of the American scientist M. A. Biot (here R is the orthogonal rotation tensor in the polar decomposition of the deformation gradient F).

3.9 Exercises 1.

a. Let A be a second-order tensor field and a a vector field. Show that x ∧ ( A · a) = a · (x ∧ A)T ,

(3.69a)

div(x ∧ A) = 2ω + x ∧ div A , T

T

(3.69b)

where ω is the axial vector of skw( A). b. Use (3.69) in conjunction with the global equation for the conservation of angular momentum—see (3.26b), to establish that skw(σ ) = O, where σ represents the Cauchy stress tensor. Hence, deduce that σ must be symmetric. 2. Let Pt ⊂ Bc be a material region that undergoes a rigid motion, x = c(t) + Q(t) · X, with Q ≡ Q(t) and c ≡ c(t) as in Definition 2.2. Define

x :=

1 m(Pt )



 Pt

ρ x dv,

J (Pt ; o) :=

Pt

  ρ (x · x)I − x ⊗ x dv, (3.70)

190

3 Balance Laws

which represent, respectively, the position of the mass centre o and the (mechanical) inertia tensor relative to the origin o of the material occupying Pt . Use Definitions (3.21) and (3.22) to show that P(Pt ) = m(Pt )v,

L(Pt , o) = m(Pt )x ∧ v + J (Pt ; o)w,

(3.71)

where v = c˙ + w ∧ (x − c) represents the velocity of the mass centre, w ˙ · Q T , and J (P; o) denotes the axial vector of the skew-symmetric tensor Q corresponds to the (mechanical) inertia tensor relative to the mass centre o (see Exercise 2.9 in Chap. 2). 3. Consider a Cauchy stress field whose component representation in a fixed Cartesian system of coordinates is given by ⎡

6x1 x32

⎢ [σ ] = ⎣ 0

−2x33

0 1 2

−2x33

⎤

⎥ 2 ⎦. 3x12

a. Assuming that body forces are negligible, does the stress field satisfy the equilibrium equations? b. Determine the Cauchy traction vector acting at the point x 0 := (2, 3, 2) on the plane 2x1 + x2 − x3 = 5. c. Determine the normal and projected shear tractions acting at x 0 on this plane. 4. The Cauchy stress tensor in a material which is at rest in a certain frame of reference has the Cartesian component representation ⎡ 2 ⎤ 0 x1 2x1 x2 1 ⎢ ⎥ 0 [σ ] = α 2 ⎣2x1 x2 x22 ⎦, 4 2 2 0 0 2(x1 + x2 ) where α ∈ R is a constant. Find the body force per unit volume which must be acting in that frame of reference. 5. The distribution of (Cauchy) stress in a square slab which occupies the region

 Ω := x = (x1 , x2 , x3 ) |x1 | ≤ a, |x2 | ≤ a, |x3 | ≤ h is given by σ (x) = σ11 e1 ⊗ e1 + σ22 e2 ⊗ e2 + σ12 (e1 ⊗ e2 + e2 ⊗ e1 ), where σ11 = − p(x12 − x22 )/a 2 , σ22 = p(x12 − x22 )/a 2 , σ12 = 2 px1 x2 /a 2 ( p ∈ R).

a. Assuming that body forces are absent, verify that the slab is in equilibrium and calculate the components of the surface traction on each of its faces. b. Show that the resultant force which acts on each of the faces x1 = ±a, x2 = ±a has magnitude 8 pah/3.

3.9 Exercises

191

c. Find the resultant torque with respect to O (the origin of the system of coordinates) acting on each of the faces x1 = ±a and x2 = ±a. 6. The Cauchy stress field in an elastic body has the Cartesian component representation ⎡ ⎤ 1 0 2x2 ⎢ ⎥ 1 4x1 ⎦ . [σ ] = ⎣ 0 2x2

4x1

1

a. Assuming that body forces are absent, does this stress field satisfy the equilibrium equations? b. Determine the traction vector acting at the point x 0 := (1, 2, 3) on the plane x1 + x2 + x3 = 6. c. Determine the normal and projected shear tractions acting at x 0 on this plane. d. Determine the principal stresses and directions of σ at the same point. 7.

a. The components of the Cauchy stress tensor in a rectangular Cartesian coordinate system O x1 x2 x3 at a point P are given (in appropriate units) by ⎡

3

2

2

⎤

⎢ [σ ] = ⎣2

4

⎥ 0⎦ .

2

0

2

i. Find the traction at P on the plane: (I) normal to the x1 -axis; (II) whose normal has direction ratios 1 : −3 : 2; (III) parallel to the plane x1 + 2x2 + 3x3 = 1. ii Find the principal directions at P and the directions of the principal axes of stress at P. Verify that the principal axes are mutually orthogonal. iii. Consider now a change of coordinates according to the transformation x1 =

1 1 (x1 − 2x2 + 2x3 ), x 2 = (−2x1 + x2 + 2x3 ), 3 3 1 x 3 = (−2x1 − 2x2 − x3 ). 3

Verify that these equations define an orthogonal transformation, and then find the components of the above stress tensor in the new system of coordinates. Use this answer to check the results obtained at (ii) above. 8. A cantilever beam with rectangular cross section occupies a region

 Ω := x = (x1 , x2 , x3 ) |x1 | ≤ a, |x2 | ≤ h, 0 ≤ x3 ≤  . The end x3 =  is built-in and the beam is bent by a vertical force P applied at the free end x3 = 0 and acting in the x2 -direction. The Cauchy stress tensor is

192

3 Balance Laws

σ = (A + Bx22 )(e2 ⊗ e3 + e3 ⊗ e2 ) + C x2 x3 e3 ⊗ e3 , where A, B, C ∈ R. a. Show that this stress field satisfies the equations of equilibrium with no body forces, provided that 2B + C = 0. b. Determine the relation between A and B if no traction acts on the sides x2 = ±h. c. Express the resultant force on the free end x3 = 0 in terms of A, B, and C. d. Using (a) and (b) above, show that C = −3P/4ah 3 . 9. Relative to a rectangular system of coordinates O x1 x2 x3 , the Cauchy stress tensor in a deformable solid assumes the matrix representation ⎡

α

0

⎤

0

⎢ [σ ] = ⎣ 0

x2 + α x3

⎥ Φ(x2 , x3 )⎦ ,

0

Φ(x2 , x3 )

x2 + β x3

where α, β ∈ R. a. If Φ(0, 0) = 0, find Φ(x2 , x3 ) so that the given Eulerian stress field satisfies the equilibrium equations with zero body forces. b. Considering the expression of Φ(x2 , x3 ) determined above, find the Cauchy stress vector t acting on the plane x1 + x2 + x3 = 3. 10. Consider the following stress distribution for a circular cylindrical bar of length L > 0, having the 1-direction as its axis, ⎡

0

⎢ [σ ] = ⎣−αx3 αx2

−αx3 0 0

αx2

⎤

⎥ 0 ⎦. 0

a. If the lateral surface of the cylinder is given by the equation x22 + x32 = 4, find the distribution of the traction vector on this surface. b. Find also the expression of the traction vector on the ends x1 = 0 and x1 = L, and then calculate the corresponding resultant forces and moments. Discuss your results. 11. An elliptical bar of finite length has a lateral surface defined by the equation x22 + 2x32 = 1, and experiences a Cauchy stress distribution described by ⎡

0

⎢ [σ ] = ⎣−2x3 x2

−2x3

x2

⎤

0

⎥ 0⎦.

0

0

3.9 Exercises

193

Show that the lateral surface of the cylinder is traction free, and then find the resultant force acting on the face x1 = 0. Determine also the resultant moment about the origin O of the tractions acting on the same face of the cylinder. 12. The Eulerian description of the velocity field in a liquid of constant density ρ is given by the following components: v1 = −

α x2 , + x22

x12

v2 =

α x1 , + x22

x12

v3 = 0,

where x12 + x22 = 0 and α ∈ R. Show that this velocity field satisfies the equation of mass continuity. 13. The stress distribution in a beam of length L of circular cross section and radius r = a, with its axis aligned along the x1 -axis, is given by a Cauchy stress tensor with components 4W W (1 + 2ν) x2 x3 , (x1 − L)x3 , σ12 = − 4 4 πa πa (1 + ν)      3 ν 2W 1 ν 2 2 (a − x3 ) − x2 , = + − πa 4 (1 + ν) 4 2 4 2 2 = σ23 = σ33 = 0,

σ11 = σ13 σ12

where W and ν are constants. a. Verify that the lateral surface of the cylinder, x22 + x32 = a 2 , is traction free, and that the resultant of the forces on the end x3 = L is equal to (0, 0, W ). b. Show that the resultant torque about the origin is (0, −W L , 0). 14. Establish the relation ρa =

∂ (ρv) + div (ρv ⊗ v), ∂t

connecting the mass density (ρ), the Eulerian description of the velocity (v), and the Eulerian acceleration field (a) of a moving deformable body. 15. Assume that (x, t) is fixed, and consider the orthonormal system of vectors { f 1 , f 2 , f 3 } consisting of the principal directions of σ (x, t); the corresponding eigenvalues (principal stresses) are denoted by σ1 , σ2 , σ3 . If we consider the principal directions as coordinate axes, then the plane whose normal vector forms equal angles with the coordinate system axes is called an octahedral plane. There are eight such planes forming a regular octahedron as illustrated in Fig. 3.12a. a. Show that the normal stress N on any of the eight octahedral planes is given by 1 N = (σ1 + σ2 + σ3 ). 3 b. Show that the shearing stress T on any of the eight octahedral planes assumes the form

194

3 Balance Laws 3

(a)

2

(b) 1 2

2 1

1

2

1

Fig. 3.12 a Regular octahedron in the space of principal stresses; b biaxial tension for a deformable body in a state of plane stress

T2 =

 1 (σ1 − σ2 )2 + (σ2 − σ3 )2 + (σ3 − σ1 )2 , 9

and then express this in terms of the principal invariants Iσ and II σ . 16. Consider a deformable body B in a state of plane stress, subjected to biaxial tension characterised by the Cauchy stress tensor σ = σ1 e1 ⊗ e1 + σ2 e2 ⊗ e2

(σ1 = σ2 ),

(3.72)

and let n = cos θ e1 + sin θ e2 (0 ≤ θ ≤ π ) be the normal to an arbitrary transverse plane passing through Bc (see Fig. 3.12b). a. Explain why e1 and e2 are the principal stress directions of this biaxial state of stress. b. Show that the normal stress N on the plane considered above is given by N=

1 1 (σ1 + σ2 ) + (σ1 − σ2 ) cos 2θ. 2 2

(3.73)

c. Show further that the shear stress T on the same plane is  2 T 2 = σ12 cos2 θ + σ22 sin2 θ − σ1 cos2 θ + σ2 sin2 θ ,

(3.74)

and then deduce that the maximum value of T is Tmax =

1 |σ1 − σ2 |, 2

corresponding to the plane that bisects the angle between the two principal directions.

3.9 Exercises

195

17. Show that if B is a body in equilibrium, then 

 Br

F dV =

  Br

Br



∂Br



∂Br

S dV =

F · S dV =

x ⊗ N d A,  X ⊗ (ST · N) d A + ρ0 X ⊗ b0 dV, Br  x ⊗ (ST · N) d A + ρ0 x ⊗ b0 dV,

∂Br

Br

where F and S are the deformation gradient and the first Piola–Kirchhoff stress tensor, respectively, while Br is the reference configuration occupied by B. 18. (Bernoulli’s Theorem) Consider a flow with hydrostatic pressure σ = − p I, where p = p(x, t) is a scalar field. The body force is assumed to be conservative, i.e. b = −grad φ for some scalar field φ = φ(x, t). a. If the flow is steady (i.e. ∂v/∂t = 0), then  v · grad

 1 2 1 |v| + φ + v · grad p = 0. 2 ρ

b. If in addition the flow is also irrotational (i.e. curl v = 0), then  grad

 1 2 1 |v| + φ + grad p = 0. 2 ρ

19. For a body held in equilibrium in its current configuration Bc , show that 

 Bc

σ dv =

 ∂Bc

(σ · n) ⊗ x da +

Bc

ρ(b ⊗ x) dv,

where ρ is the mass density of Bc , b represents the body force density, σ denotes the Cauchy stress tensor and n is the outward unit normal on ∂Bc . 20. The Cauchy stress tensor σ at some location inside a deformable body has components ⎡ ⎤ 0 0 −2β ⎢ ⎥ 0 β ⎦, [σ ] = ⎣ 0 −2β

β

0

√ where β ∈ R is a constant. Show that the principal stresses are 0 √ and ±β 5. Find the corresponding direction for which the principal stress is β 5. 21. In the absence of body forces, a deformable body is in equilibrium under a state of stress given by σ (x) = ψ(x) a ⊗ a,

196

3 Balance Laws

where ψ(x) represents a scalar field and a is a constant unit vector. a. Show that grad ψ must be everywhere perpendicular to a. b. Prove that the stress vector acting across an area with arbitrary unit normal n is parallel to a. c. Evaluate the stress vector when: (i) n is parallel to a; (ii) n is perpendicular to a. 22. The homogeneous Cauchy stress field σ in a deformable body is characterised by the following component matrix, ⎡

σ11

2

1

⎤

⎢ [σ ] = ⎣ 2

0

⎥ 2⎦ .

1

2

0

Determine a value of σ11 such that there will be traction-free planes in the body, and then determine the unit normal n to these traction-free planes.

Bibliography 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13.

Atkin RJ, Fox N (1980) An introduction to the theory of elasticity. Longman, London Gurtin ME (1981) An introduction to continuum mechanics. Academic Press, New York Hunter SC (1976) Mechanics of continuous media. Ellis Horwood Ltd., Chichester Jaunzemis W (1967) Continuum mechanics. The McMillan Company, New York Malvern LE (1969) Introduction to the mechanics of a continuum medium. Prentice-Hall Inc, Englewood Cliffs, New Jersey Narasimhan MNL (1993) Principles of continuum mechanics. Wiley Inc, New York Ogden RW (1984) Non-linear elastic deformations. Ellis Horwood Ltd., Chichester Segel LA (1987) Mathematics applied to continuum mechanics. Dover Publications Inc, New York Spencer AJM (1980) Continuum mechanics. Longman Ltd., Essex (UK) Truesdell C (1966) The elements of continuum mechanics. Springer, Berlin Truesdell C (1977) A first course in rational continuum mechanics, vol 1. Academic Press, New York Truesdell C, Noll W (2004) The non-linear field theories of continuum mechanics. Springer, Berlin Rubin D, Lai WM, Krempl E (2010) Introduction to continuum mechanics, 4th edn. Elsevier, Amsterdam

Chapter 4

Constitutive Relationships

Abstract The balance laws discussed in the previous chapter represent general principles that all deformable bodies must satisfy. They do not distinguish between fluids and solids, and are equally applicable to all bodies. It also turns out that the number of equations found so far is insufficient for determining the deformations and stresses in an arbitrary deformable body.

A typical situation of interest in the mechanics of deformable solids involves a body that occupies a known reference configuration Br (at t = t0 ). External loads applied to this configuration will usually result in a sequence of gradual changes in the shape and size of the original body, which at the current instant of time (t > t0 ) will be found in a new configuration Bc . The change from Br to Bc is achieved through the motion χ discussed in Chap. 2, and which is one of the main unknowns (note that the Eulerian velocity field v and the displacement field u are both kinematic quantities equivalent to χ ). A complete description of the current configuration Bc would also require knowledge of the stress distribution and the current mass density. However, the equations found so far are ⎧ ⎨ ρ˙ + ρ div v = 0 , div σ + ρ b = ρ v˙ , ⎩ T σ =σ,

(mass conservation) , (conservation of linear momentum) ,

(4.1a) (4.1b)

(conservation of angular momentum) ;

(4.1c)

equations (4.1a) and (4.1b) provide 4 scalar equations for 10 unknowns: ρ (scalar: 1 component), σ (symmetric tensor: 6 components) and v (vector: 3 components). This indicates that the system (4.1) is under-determined (i.e. there are more unknowns than equations), and it must be augmented by 6 additional equations in order to make the problem solvable in general. This role is fulfilled by equations that relate the internal stresses to the kinematics. Broadly speaking, a constitutive law is any relationship between one of the stress tensors introduced in the previous chapter and a kinematic measure of deformation. For example, σ = G2 (C) , S = G3 (H) , σ = G1 (F) , © Springer Nature B.V. 2020 C. D. Coman, Continuum Mechanics and Linear Elasticity, Solid Mechanics and Its Applications 238, https://doi.org/10.1007/978-94-024-1771-5_4

197

198

4 Constitutive Relationships

where G j ( j = 1, 2, 3) are given tensor-valued functions defined on suitable subspaces of Lin; here, F is the deformation gradient, C ≡ F T · F denotes the right Cauchy–Green tensor, while H ≡ Grad u stands for the displacement gradient. Of course, not just any choice of such functions will lead to the description of a real material. As a rule, a number of further physical requirements need to be imposed on a constitutive law, and it is this aspect that we want to explore in the rest of this chapter.

4.1 The Principle of Material Frame-Indifference Physical phenomena occur as events that have both spatial and temporal dimensions. In order to mathematically describe them quantitatively, a frame of reference, or observer, must be introduced, so that relative distances and time-intervals could be measured and recorded. An event in the physical world can be identified with its location, which represents a point x in the usual three-dimensional Euclidean point space E3 , and a particular time that corresponds to a number t on the real line R. We can, therefore, describe an event as an ordered pair {x, t}, where x ∈ E3 and t ∈ R. The totality of all events corresponds to the Cartesian product of these two sets, i.e. E3 × R, which is usually referred to as space-time. Definition 4.1 A change of observer, from O to O∗ (say), is a one-to-one mapping of E3 × R onto itself such that distances, time-intervals, and temporal order are preserved. Let us consider two arbitrary events E 1 , E 2 ∈ E3 × R which are recorded as {x, t} and {x 0 , t0 } by an observer O; under a change of observer, O → O∗ , the same events are also perceived as {x ∗ , t ∗ } and {x ∗0 , t0∗ }. Note that the above definition demands that the following conditions must be satisfied: |x − x 0 | = |x ∗ − x ∗0 |

and

|t − t0 | = |t ∗ − t0∗ | .

(4.2)

Since x, x 0 ∈ E3 are arbitrary, it can be deduced from the first condition in (4.2) that x ∗ − x ∗0 = Q(t) · (x − x 0 ) ,

(4.3)

where Q(t) is a second-order orthogonal tensor. Likewise, as t and t0 are arbitrary, the second requirement in (4.3) can only be satisfied if t∗ = t + a ,

(4.4)

for some constant a ∈ R. The equality (4.3) involves vectors in V (as the difference between the points) and as such it is independent of a specific choice of origin in E3 .

4.1 The Principle of Material Frame-Indifference

199

We can, of course, choose an arbitrary origin in space and then identify points with their position vectors; by setting c(t) := x ∗0 − Q(t) · x 0 , the change of frame formulae becomes t∗ = t + a . (4.5) x ∗ = c(t) + Q(t) · x , The orthogonal tensor Q(t) is uniquely determined by the transformation O → O∗ , but it should be clear that the point c(t) depends on the choice of origin. The first equation in (4.5) represents a Euclidean transformation, since it preserves the Euclidean geometry of the original space; the second part of (4.5) simply expresses the fact that the time scale is independent of the reference system. Definition 4.2 A Galilean transformation is a transformation between frames of reference related to each other by a translational motion with constant velocity. Such transformations are a particular case of (4.5) and correspond to Q(t) = Q 0 = constant ,

c(t) = c0 + c1 t ,

(∀) t ∈ R ,

where c0 and c1 are constant vectors. We recall briefly that the importance of this particular subset of the Euclidean transformations stems from their status in Classical Mechanics: inertial frames are preserved by such transformations. Put differently, the basic equations of Classical Mechanics are invariant with respect to Galilean transformations. We shall revisit this aspect later in this chapter. Returning now to (4.5), there is a sharp distinction between a change of observer and a coordinate transformation. While the former affects the points in E3 , the latter does not; it is only the coordinates of the points that are altered when we perform a change of coordinates. Up to now, our discussion has been quite general and without direct reference to deformable bodies. Next, we consider a deformable body that, relative to some observer O, has a motion described by x = χ (X, t). A second observer O∗ in relative motion to the first will describe the same motion using the transformation formulae (4.5), x ∗ = χ ∗ (X, t ∗ ) := c(t) + Q(t) · χ (X, t) and t ∗ = t + a. Both motions, χ and χ ∗ , describe the same event as perceived by two different observers, O and O∗ , respectively. However, from a strictly mathematical point of view, the latter motion is obtained from the former by superimposing a Euclidean transformation and shifting the original time scale; as a result, instead of having two observers that record the same event, we can think of the same situation as involving only one observer that superimposes arbitrary Euclidean transformations on what is being recorded. This is a much more convenient point of view for our subsequent purposes. Definition 4.3 Two motions associated with a moving deformably body, χ (X, t) and χ ∗ (X, t ∗ ), are called objectively equivalent if they are related by χ ∗ (X, t ∗ ) = c(t) + Q(t) · χ (X, t) ,

t∗ = t + a ,

(4.6)

200

4 Constitutive Relationships

where Q ≡ Q(t) ∈ Ort+ , a ∈ R is a constant, and we recall that     Ort+ ≡ A ∈ Lin  AT · A = A · AT = I, det A = +1 represents the set of all proper second-order orthogonal tensors. We have restricted the rotations in (4.6) to proper orthogonal tensors in order to ensure that right-handed systems of coordinates are preserved under a change of observer. Two (objectively) equivalent motions differ by a rigid motion, and henceforth the relations (4.6) will be taken to represent the mathematical conditions for a change of observer. In trying to describe the behaviour of a deforming body, different observers make measurements of various quantities associated with its current configuration. If these observers are in relative motion to each other, their measurements will, in general, be different. A meaningful description of the material behaviour would require the observers to make the same measurement of various kinematical or mechanical fields despite their relative movement. Before we are in a position to explore in some depth the implications of this fundamental requirement, we need to assign a meaning to what is meant for a spatial field to be objective (i.e. ‘unchanged’ by a change of observer). Definition 4.4 (a) Suppose that φ(x, t) is a spatial scalar field corresponding to a motion χ (X, t); under the change of observer (4.6) this field becomes φ ∗ (x ∗ , t ∗ ). Then, φ is said to be an objective scalar field if φ ∗ (x ∗ , t ∗ ) = φ(x, t) .

(4.7)

(b) Suppose that u(x, t) is a spatial vector field corresponding to a motion χ (X, t); under the change of observer (4.6) this field becomes u∗ (x ∗ , t ∗ ). Then, u is said to be an objective vector field if u∗ (x ∗ , t ∗ ) = Q(t) · u(x, t) .

(4.8)

(c) Suppose that T (x, t) is a spatial tensor field corresponding to a motion χ (X, t); under the change of observer (4.6) this field becomes T ∗ (x ∗ , t ∗ ). Then, T is said to be an objective tensor field if T ∗ (x ∗ , t ∗ ) = Q(t) · T (x, t) · Q T (t) .

(4.9)

Throughout this chapter, ‘objectivity’ will be understood in the sense of these definitions. It will also be tacitly assumed that all observers perceive the reference configuration to be the same (since the notion of objective tensor is independent of the chosen reference configuration). For example, this means that if a material tensor is observed as T and T ∗ by two different observers, then T = T ∗ (which should be contrasted with (4.9)); the same holds true for a material vector that is observed

4.1 The Principle of Material Frame-Indifference

201

B

fixed

Ɵme:

moving Ɵme:

Fig. 4.1 Change of observer and a particular choice of the coordinate systems for the two observers: ei∗ (t) = Q(t) · ei (i = 1, 2, 3)

as u and u∗ : the above assumption requires that u = u∗ . Sometimes, the definition of objectivity is broadened to accommodate these situations, the aforementioned vector and tensor fields being called objective material fields. Within this extended framework for objectivity, there is the issue regarding the hybrid nature of two-point tensors. These are sometimes referred to as being objective if they map an objective material vector into an objective spatial vector (or vice versa, depending on their definition). If such an objective two-point tensor is observed as T and T ∗ , then v = T · u and v ∗ = T · u∗ with u = u∗ and v ∗ = Q(t) · v. Thus, T ∗ = Q(t) · T , i.e. ‘objective’ two-point tensors transform like objective spatial vectors. Another important point that we want to make is that, according to Definition 4.4, the components of the vector field u and the tensor field T are unchanged by an observer transformation. To see this more clearly, let {ei } be an orthonormal basis associated with the original observer O; as a consequence of the change of observer, O → O∗ , this system of basis vectors is transformed into {ei∗ (t)}, where ei∗ (t) := Q(t) · ei ,

(i = 1, 2, 3) .

(4.10)

Furthermore, we shall assume that the observer O∗ uses this system of vectors to record the position of events in space— see Fig. 4.1. Put differently, under these particular assumptions, the orthogonal tensor Q(t) can be regarded as a two-point tensor that admits the representation Q(t) = ei∗ (t) ⊗ ei . The i-component of the vector field u, as seen by the observer O∗ , is u∗ · ei∗ . Taking into account (4.8) and (4.10) we can then write    u∗ (x ∗ , t ∗ ) · ei∗ (t) = Q(t) · u(x, t) · ei∗ (t) = Q(t) · u(x, t) · Q(t) · ei   = Q T (t) · Q(t) · u(x, t) · ei = u(x, t) · ei ,

202

4 Constitutive Relationships

since Q(t) is an orthogonal tensor. The last expression in the above sequence of calculations is the i-component of the same vector field u in the original frame. Similarly, it is a routine matter to check that ei∗ (t) · T ∗ (x ∗ , t ∗ ) · e∗j (t) = ei · T (x, t) · e j ,

(∀) i, j = 1, 2, 3 .

The Principle of Material Frame-Indifference (PMFI): of a deformable body is independent of the observer.

The material response

The implications of this fundamental result will be explored throughout this chapter. Since the field equations of Continuum Mechanics contain a number of kinematic quantities derived directly from χ , one of our main tasks is to find out which fields are objective (in the sense of Definition 4.4). A second item on the agenda is to establish how objective fields are modified by the application of the spatial gradient and divergence differential operators. It turns out that these operations do not interfere with the objectivity of the corresponding fields, and the next formal result addresses this aspect in some detail. Proposition 4.1 The following are true: 1. the spatial gradient of an objective scalar field is an objective vector field; 2. the spatial divergence of an objective vector field is an objective scalar field; 3. the spatial divergence of an objective tensor field is an objective vector field. The justification of these properties is included below, and we start by making a simple   observation. Let x = (x1 , x2 , x3 ) be the coordinates of a point in Bc relative to ei , while x ∗ = (x1∗ , x2∗ , x3∗ ) represent the coordinates of the same point recorded after a change of observer. With this notation in mind, (4.6) allows us to write (x ∗ − c) · ei∗ = ( Q · x) · ei∗

 = x · ( Q T · ei∗ ) = x · ( Q T · Q) · ei = x · ei , ⇒

and thus,

xi∗

= xi + c ·

ei∗

,

[see (4.10)]

(i = 1, 2, 3) ,

∂ ∂ ∂ x ∗p ∂ ∂ = ∗ = ∗ δ pi = ∗ . ∂ xi ∂ x p ∂ xi ∂xp ∂ xi

From this result, it follows immediately that ∂φ ∗ ∗ ∗ ∂φ (x, t) , ∗ (x , t ) = ∂ xi ∂ xi ∂ T jk∗ ∂ xi∗

∂u∗j

(x ∂ xi∗

(x ∗ , t ∗ ) =

∗

, t ∗) =

∂u j (x, t) , ∂ xi

∂ T jk (x, t) , ∂ xi

where φ, u, T and φ ∗ , u∗ , T ∗ are as in (4.7)–(4.9).

(4.11a)

(4.11b)

4.1 The Principle of Material Frame-Indifference

203

To prove the first part of the proposition, we recall that (by definition) (grad φ)(x, t) = and (grad∗ φ ∗ )(x ∗ , t ∗ ) =

∂φ (x, t) ei ∂ xi

(4.12)

∂φ ∗ ∗ ∗ ∗ (x , t ) ei . ∂ xi∗

(4.13)

Using Eqs. (4.11a) and (4.10) in (4.13), we get 

 ∂φ ∂φ (x, t) Q(t) · ei = Q(t) · (x, t) ei . (grad φ )(x , t ) = ∂ xi ∂ xi ∗

∗

∗

∗

(4.14)

Comparing (4.12) and (4.14) leads to (grad∗ φ ∗ )(x ∗ , t ∗ ) = Q(t) · (grad φ)(x, t) ,

(4.15)

which confirms the objectivity of grad φ. For the justification of the second part, we recall that the divergence of a vector field is a scalar field. Assuming that u is objective, we get with the help of (4.11a), (div ∗ u∗ )(x ∗ , t ∗ ) =

∂ui∗ ∗ ∗ ∂ui (x , t ) = (x, t) = (div u)(x, t) . ∂ xi∗ ∂ xi

Finally, for checking the last part of Proposition 4.1, we take into account that the divergence of a tensor field is a vector field. The condition (4.8) is satisfied because (div ∗ T ∗ )(x ∗ , t ∗ ) =

∂ Ti∗j ∂ xi∗

∂ Ti j (x, t) ( Q(t) · e j ) ∂ xi 

∂ Ti j (x, t) e j = Q(t) · ∂ xi  = Q(t) · (div T )(x, t) ,

(x ∗ , t ∗ ) e∗j =

and this completes our proof. The results checked above can be handled more expediently by using a coordinatefree approach. For the purposes of illustrating this strategy, we denote the differential operators ‘grad ’ and ‘grad ∗ ’ by ∇ and ∇ ∗ , respectively. In light of (4.5), these symbolic vectors are related by ∇ ∗ = Q(t) · ∇ and we have ∇ ∗ · u∗ = ( Q · ∇) · ( Q · u) = ( Q T · Q) · (∇ · u) = ∇ · u ,

204

4 Constitutive Relationships

and ∇ ∗ · T ∗ = ( Q · ∇) · ( Q · T · Q T ) = ( Q T · Q) · (∇ · T · Q T ) = ∇ · (T · Q T ) = (∇ · T ) · Q T = Q · (∇ · T ) . In the above calculations, we have relied on Q = Q(t) being independent of the spatial coordinates of both x and x ∗ . Proposition 4.2 1. The deformation gradient F and the velocity gradient L are not objective. 2. The left Cauchy–Green deformation tensor B is objective. 3. The stretching tensor D = sym(L) is objective. We start the justification of these basic properties by taking the material gradient of Eq. (4.6) F ∗ (X, t ∗ ) ≡ Grad χ ∗ (X, t ∗ ) = Q(t) · Grad χ (X, t) = Q(t) · F(X, t) , where use was made use of (4.11a). For future reference, we record this important result separately, (4.16) F∗ = Q · F . A simple glancing at (4.9) and (4.16) reveals that the deformation gradient is not an objective tensor field in the sense of Definition 4.4 since the post-factor Q T is missing from (4.16). This is to be expected because F is a two-point tensor and acts between Br and Bc . ˙ · F −1 , so we can write According to Eq. (2.73) in Sect. 2.9, L = F ˙ · F + Q · F) ˙ ∗ · (F ∗ )−1 = ( Q ˙· F) · ( Q · F)−1 = ( Q ˙ · (F −1 · Q −1 ) L∗ = F ˙ · QT + Q · ( F ˙ · F −1 ) · Q T = Q ˙ · QT + Q · L · QT . = Q

(4.17)

˙ · Q T , and taking the material time derivative of Q · Q T = I, it Letting Ω Q := Q transpires that Ω Q is skew-symmetric, i.e. Ω Q + Ω TQ = O. A quick comparison between (4.17) and (4.9) suggests that L is not an objective tensor field (owing to the presence of Ω Q ). The objectivity of the left Cauchy–Green deformation tensor B is fairly straightforward. Under an observer transformation, this tensor becomes B ∗ = F ∗ · (F ∗ )T = ( Q · F) · ( Q · F)T = Q · (F · F T ) · Q T = Q · B · Q T , where use was made of formula (4.16).

4.1 The Principle of Material Frame-Indifference

205

Since the Cauchy–Green deformation tensor C operates on Br (i.e. it is not a spatial tensor), it is unsurprising that it does not enjoy the transformation rule for objective tensors, cf. (4.9). However, the tensor is invariant, in the sense that C ∗ = (F ∗ )T · F ∗ = ( Q · F)T · ( Q · F) = F T · ( Q T · Q) · F = F T · F = C . The objectivity of D is a consequence of Eq. (4.17) derived above, as the following calculations confirm

1 1 ∗ 1 L + (L ∗ )T = ( Q · L · Q T + Q · L T · Q T ) + (Ω Q + Ω TQ ) 2 2 2 1 1 T T T = Q · (L + L ) · Q + (Ω Q + Ω Q ) 2 2 (4.18) = Q · D · QT ;

D∗ =

the proof of Proposition 4.2 is now complete. Testing for the material frame-indifference of any kinematical quantity mirrors closely the strategy illustrated in the above calculations. One invariably starts with an equivalent motion (in the sense of Definition 4.3) and then calculates the ‘starred’ version of the quantity of interest. By comparing the result obtained with the transformation rules recorded in Definition 4.4, it is then a straightforward matter to identify whether or not the original quantity is objective. For forces and moments, which are regarded as primitive concepts, it is postulated a priori that they are independent of the observer. In particular, we shall assume in what follows that the body force (per unit mass) b and the traction vector t are both frame-indifferent. That is, if b∗ and t ∗ represent their starred versions as a result of the change of observer (4.6), then b∗ = Q · b

and

t∗ = Q · t .

(4.19)

The objectivity of the Cauchy stress tensor σ follows now as a corollary of the second equation in (4.19). To this end, let us note that the outward unit normal n to a surface f (x, t) = 0 is given by n = grad f /|grad f |. The scalar function f is clearly objective, while the frame-indifference of grad f is an immediate consequence of Proposition 4.1, whence grad ∗ f ∗ = Q · (grad f ). Taking the dot product of this relation with itself leads to |grad ∗ f ∗ | = |grad f |, and we conclude that n∗ = Q · n, i.e. the unit normal to a surface in Bc is a frame-indifferent (or objective) vector. If the Cauchy stress tensor is observed as σ and σ ∗ by two observers, then

and

t ∗ = σ ∗ · n∗ = σ ∗ · ( Q · n) = (σ ∗ · Q) · n ,

(4.20)

t ∗ = Q · t = Q · (σ · n) = ( Q · σ ) · n .

(4.21)

206

4 Constitutive Relationships

Equating the last terms in (4.20) and (4.21) we obtain (σ ∗ · Q) · n = ( Q · σ ) · n. Since this must hold for every unit vector n, it implies that σ ∗ · Q = Q · σ or, after right-multiplication by Q T , σ ∗ = Q · σ · QT . (4.22)

4.2 Other Important Constitutive Principles Both fluid and solid materials are assumed to obey the PMFI discussed in the previous section, as well as two other principles. Before we can state them, we need to introduce some new vocabulary. Assume that θ (X, t) is a typical Lagrangian scalar field representing some property of a deformable body B; to fix ideas, let us consider this field as representing the temperature of a moving body. The history up to time ‘t’ of the function θ at a fixed X ∈ Br is the totality of the values θ (X, t − s),

0≤s ≤t.

We shall denote these collective values by θ t (X, ·) or, if there is no risk of confusion, simply by θ t (X). If only one particle at a time is considered, then it is customary to drop the dependence of θ on X. Thus, a fixed history is a family of functions that depends on one parameter, the present time ‘t’. The functions that describe histories of entire temperature fields will be denoted by θ t (·), with the value of θ t (·) at X ∈ Br being the temperature history at X, i.e. the function θ t = θ t (X) defined by θ t (X)(s) := θ (X, t − s),

0≤s ≤t;

(4.23)

note that the only quantity that is being varied in Eq. (4.23) is ‘s’ (the independent variable). If our θ is a smooth function (differentiable as many times as we like), then one can write down a Taylor expansion for θ t (·), θ t (Y ) = θ t (X)+(Y − X) · ∇θ t (X) 1 + (Y − X) ⊗ (Y − X) : (∇ ⊗ ∇)θ t (X) + . . . , 2

(4.24)

valid for all Y ∈ Br ∩ Nε (X), where Nε (X) := {Y ∈ E3 : |Y − X| < ε } is a sufficiently small neighbourhood around X. In principle, the history θ t (Y ) can be approximated by retaining a finite number of terms in (4.24); an approximation of grade N corresponds to keeping N ≥ 1 terms in the above Taylor expansion.

4.2 Other Important Constitutive Principles

207

Principle of Determinism for the Stress (PDS): The stress in a continuum body is determined by the history of the motion of that body. Taking the Cauchy stress σ as a typical dependent variable, this principle translates into the following mathematical statement: χ t (·); X, t) , σ (x, t) = F (χ

x = χ (X, t) ,

(4.25)

where F represents the constitutive functional, and the explicit dependence of this functional on ‘X’ allows for the deformable material under consideration to be inhomogeneous, whereas the explicit dependence upon ‘t’ accounts for possible degradation of the material properties in time. Equation (4.25) states that the value of the Cauchy stress tensor at x ∈ Bc and time t > 0 is uniquely determined by the entire history of the motion up to time ‘t’, and this process may vary from particle to particle, as well as from one instant of time to the next. The form in which we stated this principle is quite general and not very useful for practical purposes. In order to arrive at a more tractable material model, we need to narrow down the dependence of F on the history of the motion, and that is the purpose of the next axiom. Principle of Local Action (PLA): In determining the stress at point x = χ (X, t), the history of the motion outside an arbitrary neighbourhood of X ∈ Br may be disregarded. This postulate is adopted in order to exclude long-range interactions (actions ata-distance) and stipulates that the stress response at a particle x = χ (X, t) is determined by what happens in an arbitrarily small neighbourhood of X ∈ Br ; everything else is irrelevant. Translated into mathematical language, the PLA asserts that for any two motions χ and χ with the same history up to time t > 0 in some neighbourhood Nε (X), the value of (4.25) is the same, i.e. χ t (·); X, t) , χ t (·); X, t) = F (χ F (χ

(4.26)

provided that χ (Y , τ ) = χ (Y , τ ).

(∀) Y ∈ Nε (X) , (∀) τ ≤ t ,

where 0 < ε 1 is arbitrary. Most materials of practical interest can be modelled within the class of the socalled simple materials; in this case the response at a point depends on the history of the deformation gradient, relative to an arbitrary choice of reference configuration, at that point. For such materials (4.25) is modified according to σ (x, t) = F (F t (X); X, t) ,

x = χ (X, t) ,

(4.27)

208

4 Constitutive Relationships

where we have kept the same notation for the constitutive functional (although F in (4.25) and (4.27) are understood as being distinct). The motivation behind adopting (4.27) rather than (4.25) is a direct consequence of the PLA, as we now explain. Writing (4.24) for χ i (X, t) (i = 1, 2, 3) and truncating the corresponding result after the first two terms leads to χ it (Y ) = χ it (X) + (Yα − X α )

∂χ it (X) + . . . , ∂ Xα

(4.28)

for all Y ∈ Nε (X) (and for sufficiently small ε > 0). According to the PLA, we know that the value of the functional F at X ∈ Br and time t > 0 is determined by the history up to time t > 0 of any other motion, χ (say), which has identical history to χ in a sufficiently small vicinity of X. For such a motion Eq. (4.28) shows that χ t (Y ) χ t (X) + F t (X) · (Y − X) ,

(∀) Y ∈ Nε (X) ,

(4.29)

which confirms that the history of the deformation gradient at a given particle X characterises, in a first approximation, the history of the motions at ‘nearby points’. Before we can take full advantage of (4.29) we need to take a look at the restrictions imposed by the PMFI on the constitutive law (4.25). If χ ∗ ∼ χ as in Definition 4.3, then the constitutive law for the former motion can be expressed as χ ∗ ) t ; X, t) , σ ∗ (x ∗ , t) = F ((χ

x ∗ = χ ∗ (X, t ∗ ) .

(4.30)

χ ∗ ) t = c t + Q t · χ t , and we are free to chose as we please the histories c t But (χ t and Q . We make the special choice a := 0 and χ (Y , t − s) , c(t − s) := −χ

Q(t − s) := I , s ≤ t ,

(4.31)

so that χ ∗ (X, t − s) = χ (X, t − s) − χ (Y , t − s)

F(X, t − s) · (X − Y ) .

(4.32)

In Eq. (4.32), the term X − Y depends on Nε (X), while F t (X) is independent of it. Since by our choice in (4.31) it also transpires that σ ∗ = σ , we deduce from (4.30) and (4.32) that, in a first approximation, we can work with the simple-material constitutive law (4.27) instead of (4.25). Including further terms in the approximation (4.28) results in more complicated constitutive relationships and forms the basis of higher order continuum theories, which are beyond the scope of this introductory text.

4.3 Cauchy-Elastic Materials

209

4.3 Cauchy-Elastic Materials For an elastic material, the stress in the current configuration depends only on the current state of deformation of that configuration and is independent of the previous time history. According to our foregoing discussion vis-à-vis the PLA and simple materials, this information is encapsulated in the deformation gradient calculated with respect to a fixed (but otherwise arbitrary) reference configuration. Thus, for such elastic materials the constitutive law can be expressed as σ (x, t) = Gr (F(X, t), X),

x = χ (X, t) ,

(4.33)

where Gr denotes the response functional of the elastic material relative to a chosen reference configuration Br . Note that this symmetric tensor-valued functional is defined on the space of all deformation gradients F (invertible linear transformations). To avoid cluttering the notation, in what follows Eq. (4.33) will be often written in the more compact form σ = G (F) ,

(4.34)

leaving out the explicit dependence on X as well. Sometimes the appellative ‘Cauchyelastic’ is added to the ‘materials’ that obey the constitutive relationship (4.34); as we shall see later in this chapter this distinction is important when there is a risk of confusion, as there are other types of elastic materials in Continuum Mechanics. Under the change of observer (4.6) the response functional remains unchanged, but σ and F in (4.34) will assume the new values σ ∗ and F ∗ , respectively; these are linked by (4.35) σ ∗ = G (F ∗ ) . From Eq. (4.16), we already know that F ∗ = Q · F; we have also seen in (4.22) that the Cauchy stress tensor is an objective spatial tensor (in the sense of Definition 4.4), and hence σ ∗ = Q · σ · Q T . Using this information in combination with (4.34) and (4.35) yields G ( Q · F) = Q · G (F) · Q T ,

(∀) Q ∈ Ort+ , F ∈ Invlin ,

(4.36)

where we have introduced the notation,     Invlin := A ∈ Lin  det A = 0 for the set of all invertible second-order tensors. Equation (4.36) expresses the restriction imposed by the PMFI on the general constitutive law (4.34) for Cauchy-elastic materials. It is a requirement that must be satisfied by any potential constitutive functional G in order to ensure that the material frame-independence for stresses holds true.

210

4 Constitutive Relationships

4.4 Material Symmetry In addition to the conditions imposed by material frame-indifference, constitutive equations are also subject to restrictions given by intrinsic symmetries that a material may display in a particular reference configuration. In order to motivate the next definition, let us investigate first the effect that changing the reference configuration has on the material response. It will prove useful to introduce two new sets of second-order tensors, Invlin+ := { A ∈ Lin | det A > 0 } , Unim+ := { A ∈ Invlin+ | det A = 1 } ; the first one is known as the special linear group, while the second is called the special unimodular group. As their names suggest, each one of these two sets forms a group under tensor multiplication, with the identity element given by the usual identity tensor I. We let σ be the Cauchy stress in the current configuration Bc , and consider F (1) , (2) F to be the deformation gradients relative to two distinct reference configurations, Br(1) and Br(2) , respectively; see Fig. 4.2. Also, we agree to denote by G1 and G2 the response functionals relative to these configurations, so that σ = G1 (F (1) ) = G2 (F (2) ) .

(4.37)

( j)

Since Br (for j = 1, 2) are configurations of the considered body, we can define the deformation gradient of the transformation Br(1) → Br(2) (at this point, we still regard the first as a reference configuration, but the second is temporarily regarded as a current configuration); calling this material gradient H ∈ Invlin+ , we claim

B

B

B

Fig. 4.2 Change of reference configuration and its effect on the material response

4.4 Material Symmetry

211

that

F (1) = F (2) · H .

(4.38)

Equation (4.38) is just a consequence of the chain-rule differentiation formula; labelling the position vectors of generic particles in Br(1) and Br(2) by X (1) and X (2) , respectively, use of the aforementioned Calculus result gives (1) = F pα

∂xp ∂ X α(1)

=

(2) ∂ x p ∂ Xβ

∂ X β(2)

∂ X α(1)

(2) = F pβ Hβα ,

which confirms our claim. Substituting (4.38) into (4.37), we then discover G1 (F (2) · H) = G2 (F (2) ) . This last relation provides a definition for G2 in terms of G1 and the deformation gradient H. In general, the response of the material relative to Br(2) differs from that relative to Br(1) , i.e. G1 = G2 . However, it is quite natural to expect that for some particular choice of H we may have G1 = G2 , in which case G2 (F (2) · H) = G2 (F (2) ),

(∀) F (2) ∈ Invlin+ .

(4.39)

According to this formula, in order to calculate σ it is not necessary to distinguish between the two reference configurations. To make things more precise let us introduce some new terminology. Definition 4.5 A material symmetry is represented by a deformation after which the response of the material is indistinguishable from that relative to the former configuration. More precisely, if H ∈ Invlin+ is the deformation gradient of a symmetry transformation, in terms of the response functional we have G (F · H) = G (F), (∀) F ∈ Invlin+ .

(4.40)

The set of tensors H ∈ Invlin+ for which (4.40) holds defines the symmetry of the material relative to Br (remember that the response functional was defined with respect to some reference configuration, Br in this case). Intuitively, the larger this set is the more symmetry the material has, and the easier it is to describe it mathematically. The concept of material symmetry is illustrated by an intuitive example corresponding to the sketch included in Fig. 4.3. A prismatic solid made of a soft rubberlike material is reinforced by metallic fibres aligned to the geometry of the solid as indicated on the left-hand side of that Figure. The fibres are assumed to be strong in tension, but have little bending stiffness. Also, let us assume that this configuration is subjected to a hypothetical mechanical test that consists in producing uniform displacements (equal to 0 < δ 1) along the top and the bottom sides of our sample; the force required to produce this deformation is then recorded and represents the

212

4 Constitutive Relationships

Fig. 4.3 Mechanical testing of an anisotropic solid (fibre-reinforced) by using two different reference configurations that differ by a rotation of 90◦

result of the test. If we rotate the original solid as indicated in the right sketch in Fig. 4.3 and then repeat the test, in light of our assumptions, it is intuitively clear that the force recorded will be smaller. The only difference between these two scenarios is the original orientation of the solid (i.e. its reference configurations). We can, therefore, conclude that in this example the mechanical response corresponding to the two particular reference configurations is different. Motivated by the above discussion we define the set  S := {H ∈ Unim+  G (F · H) = G (F), (∀) F ∈ Invlin+ } ,

(4.41)

where G is calculated relative to a fixed reference configuration. As shown below, S defines a multiplicative group called the symmetry group of the material relative to the fixed reference configuration just mentioned. Taken at face value our definition of S seems inconsistent with Eq. (4.40), in that we have replaced Invlin+ with Unim+ —the reason for this will be explained shortly. To show that the set (4.41) defines a multiplicative group we have to check three formal properties: ‘closure’, ‘inverse’ and ‘identity’; for the time being we shall allow the elements of S to be in Invlin+ . If H 1 , H 2 ∈ Invlin+ and satisfy (4.40), then det(H 1 · H 2 ) = (det H 1 )(det H 2 ) > 0 and G (F · (H 1 · H 2 )) = G ((F · H 1 ) · H 2 ) = G (F · H 1 ) = G (F) , which confirms that H 1 · H 2 belongs to the symmetry group; this completes the checking of the closure property. We also note that if H ∈ Invlin+ then this tensor admits an inverse H −1 that will satisfy G (F · H −1 ) = G ((F · H −1 ) · H) = G (F · (H −1 · H)) = G (F) ,

4.4 Material Symmetry

213

so H −1 is still in the symmetry group and the ‘inverse’ property is thus checked. Finally, the ‘identity’ condition is trivially satisfied since G (F · I) = G (F), and hence I belongs to the symmetry group as well. In general, the symmetry group changes with a change in the reference configuration according to Noll’s rule: if P represents the deformation gradient associated with the change of reference configuration Br(1) → Br(2) , then S2 = P · S1 · P −1 ,

(4.42)

where S1 is the symmetry group of the material relative to Br(1) and S2 is that relative to Br(2) . Formula (4.42) represents an equality between two sets and has the following interpretation: if H 2 ∈ S2 then H 2 = P · H 1 · P −1 for some H 1 ∈ S1 and, conversely, if H 1 ∈ S1 then H 1 = P −1 · H 2 · P for some H 2 ∈ S2 . Noll’s rule (4.42) is true even if we regard the elements of the symmetry group as being in Invlin+ . Note also that if P is a spherical tensor then S1 = S2 , i.e. a dilatation has no effect on the symmetry group. Since any H ∈ Invlin+ can be written as the product of a spherical tensor and an element from Unim+ (e.g. H = {(det H)1/3 I} · {(det H)−1/3 H}), we can restrict ourselves to elements of the special unimodular group in our discussion of the symmetry group. From the definition of S we have the following estimate of the size of this group: {I} ⊂ S ⊂ Unim+ .

(4.43)

The smallest symmetry group is the trivial one, {I}, and such materials are called triclinic. Noll’s rule implies that this symmetry group remains unaffected by any change of reference configuration, i.e. new symmetries cannot be created by any deformation. On the other hand, the largest S is Unim+ , and Noll’s rule also shows that the symmetry of this material cannot be reduced by any deformation. We also mention the obvious set ordering, {I} ⊂ Ort+ ⊂ Unim+ ,

(4.44)

which, in connection to (4.43), raises a question about the relative ‘position’ of S and Ort+ . This turns out to be an important point that will be discussed shortly. For now, we note that the latter set is a maximal subgroup of Unim+ in the sense that if G is any tensor group satisfying Ort+ ⊂ G ⊂ Unim+ , then either G = Ort+ or G = Unim+ . The following result is a simplified version of one of Noll’s more general theorems, which identifies an important property of the constitutive functional when S contains a rotation. Proposition 4.3 Assume a material with a constitutive law of the form σ = G (F), where G is calculated relative to a fixed reference configuration. A rotation tensor Q ∈ Ort+ is in the symmetry group of this material if and only if

214

4 Constitutive Relationships

G ( Q · F · Q T ) = Q · G (F) · Q T ,

(∀) F ∈ Invlin+ .

(4.45)

The justification of this elegant formula is sketched next. As F in (4.41) is arbitrary, we can take F → Q · F to find G ( Q · F · H) = G ( Q · F) ,

(4.46)

which must hold for every F ∈ Invlin+ and every H ∈ S. If Q ∈ S, then Q T ≡ Q −1 ∈ S. Letting H → Q T in (4.46) results in G ( Q · F · Q T ) = G ( Q · F) ,

(∀) F ∈ Invlin+ .

(4.47)

At this stage, we recall Eq. (4.36) found when discussing the restriction imposed on G by the PMFI; taken in conjunction with (4.47) it yields (4.45). Conversely, we must show that if this latter equation is satisfied, then Q ∈ S. Indeed, from (4.36) and (4.45) we find that (4.47) must be satisfied as well; by taking F → Q T · F in this last equation we then discover that Q −1 ≡ Q T ∈ S. Since the inverse of this element must also be in S we conclude that Q = ( Q −1 )−1 ∈ S, as required. The concept of symmetry group is crucial in defining different classes of materials. The discussion in the remaining of this section is directed towards Cauchy-elastic materials. Definition 4.6 1. An isotropic Cauchy-elastic material is one for which Ort+ ⊂ S ,

(4.48)

for at least one reference configuration. Such a configuration is an undistorted state of the material. A stress-free undistorted configuration is called a natural configuration. 2. A Cauchy-elastic material is called a solid if there exists a reference configuration such that S is the full proper orthogonal group or a subgroup of it, S ⊂ Ort+ or S ≡ Ort+ . The preferred configuration is again described as an undistorted state of the material. 3. A Cauchy-elastic solid is called anisotropic if its symmetry group relative to an undistorted state is a non-trivial subgroup of Ort+ , i.e. S ⊂ Ort+ and S ≡ Ort+ . Recalling the earlier observation regarding the maximality of Ort+ as a subgroup of Unim+ , the condition (4.48) results in two possible outcomes: either S = Ort+ (in which case we are dealing with an isotropic solid) or S = Unim+ (this corresponds to a simple fluid, as we shall see later in Sect. 4.9). We also note that, from a Physics

4.4 Material Symmetry

215

point of view, one can interpret the second part of the above definition as saying that a solid has a preferred configuration such that any non-rigid deformation from it alters its material response. For isotropic materials, the symmetry condition (4.40) becomes (∀) Q ∈ Ort+ .

σ = G (F) = G (F · Q),

(4.49)

According to the Polar Decomposition Theorem, F = R · U, and setting Q → R T in (4.49) we obtain σ = G (F) = G (F · R T ) = G (V · (R · R T )) = G (V ) or, since B = V 2 ,

σ = G (B) ,

(4.50)

(4.51)

for some new constitutive functional G . Equation (4.50) shows that the rotation tensor R plays no role in the determination of the stress σ for an isotropic material because only the left-stretch deformation tensor V appears in that reduced constitutive equation. It is convenient to consider the left Cauchy–Green deformation tensor B as the main independent variable in the constitutive response functional for isotropic materials. In relation to (4.51), note that (∀) F ∈ Invlin+ ,

G (F · F T ) := G (F),

(4.52)

which allows (4.45) to be re-cast in the equivalent form G ( Q · B · Q T ) = Q · G (B) · Q T ,

(∀) Q ∈ Ort+ .

(4.53)

This formula can be obtained as a particular case of Proposition 4.3, but here we present a justification based on (4.52), G ( Q · B · Q T ) = G ( Q · B 1/2 · ( Q · B 1/2 )T ) = G ( Q · B 1/2 ) = G (Q · V)

[using (4.52)]

= Q · G (V ) · Q T

[using (4.36)]

= Q · G (B) · Q . T

The condition identified in (4.53), and which characterises isotropic elastic materials, places strong restrictions on the potential forms of G that make it possible to actually solve the functional Eq. (4.53). Some further terminology is needed before we can make this statement more precise.

216

4 Constitutive Relationships

Definition 4.7 (a) Let φ( A) be a scalar function of A ∈ Lin. Then φ is said to be an isotropic scalar function if for every tensor A, (∀) Q ∈ Ort+ .

φ( Q · A · Q T ) = φ( A) ,

(b) Let G ( A) be a symmetric second-order tensor function of A ∈ Lin. Then G is said to be an isotropic tensor function if for every tensor A, G ( Q · A · Q T ) = Q · G ( A) · Q T ,

(∀) Q ∈ Ort+ .

For such isotropic scalar and tensor functions powerful representations formulae are available, which allow us to express G in (4.53) as a quadratic polynomial in B. A few of these theorems are listed below without proof. Theorem 4.1 A real-valued function ϕ : Sym → R is isotropic if and only if there exists a function  ϕ : R 3 → R such that ϕ( A) =  ϕ (I1 , I2 , I3 ) ,

(∀) A ∈ Sym ,

(4.54)

where I j ≡ I j ( A) ( j = 1, 2, 3) are the principal invariants of A. Theorem 4.2 A function G : Sym → Sym is isotropic if and only if there exists three isotropic scalar functions α0 , α1 , α2 : R 3 → R such that G ( A) = α0 I + α1 A + α2 A2 ,

(∀) A ∈ Sym .

(4.55)

According to the Cayley–Hamilton theorem mentioned in Chap. 1 every secondorder tensor satisfies its own characteristic equation, that is, A3 − I1 A2 + I2 A − I3 I = O. If the tensor A is invertible, then multiplying this relation by its inverse we can express A2 as a linear combination of I, A and A−1 . Hence, for tensor functions whose natural domain includes symmetric and invertible tensors, Theorem 4.2 can be re-stated in the alternative form recorded below. Theorem 4.3 Let D be a set of elements in Sym that are invertible. A function G : D → Sym is isotropic if and only if there exists three isotropic scalar functions β0 , β1 , β−1 : R 3 → R such that G ( A) = β0 I + β1 A + β−1 A−1 ,

(∀) A ∈ D .

(4.56)

The presence of A2 or A−1 in the above representations, as well as the dependence of the coefficients on the principal invariants, makes these isotropic tensor functions fairly nonlinear. The situation is much simpler if we drop the nonlinearity; in this case we are left with the simple expression recorded in (4.57). Theorem 4.4 A linear function G : Sym → Sym is isotropic if and only if there exist λ, μ ∈ R such that

4.4 Material Symmetry

217

G ( A) = 2μ A + λ | A| I ,

(∀) A ∈ Sym ,

(4.57)

where | A| ≡ tr( A) is the trace of A. The representation theorems listed above show that for an elastic isotropic material the constitutive relation can be given in one of the two equivalent forms σ = G (B) ≡ α0 I + α1 B + α2 B 2 , σ = G (B) ≡ β0 I + β1 B + β−1 B

−1

(4.58a) .

(4.58b)

In these formulae, the real-valued coefficients α j ≡ α j (I1 , I2 , I3 ) for j = 0, 1, 2, and β j ≡ β j (I1 , I2 , I3 ) for j = 0, 1, −1, are called the response or material coefficients. They depend on the particular isotropic material under consideration; in practice they can be determined by performing simple experiments on samples of material having regular geometries (e.g. biaxial tension of a rectangular sheet). The elements of each set of material coefficients in (4.58) are typically independent, but knowledge of one set leads to the other via the Cayley–Hamilton Theorem.

4.5 Hyperelastic Solids Before we touch upon the main topic of this section, we start with a brief review of certain basic notions from Classical Mechanics associated with the fundamental concepts of work and power. For simplicity, we consider only a single rigid material particle of mass m and having velocity v, which is subjected to a system of forces whose resultant is F tot . The elementary work dW of this net force during the elementary displacement d r is defined as dW := F tot · d r.1 The power P supplied to the particle represents the amount of work dW per unit time, i.e. P :=

dW dr = F tot · = F tot · v . dt dt

(4.59)

By Newton’s Second Law, F tot = ma = m v, ˙ and then Eq. (4.59) becomes P=

d dW = dt dt



1 m|v|2 2

 ,

(4.60)

where the expression that is differentiated on the right-hand side is easily recognised as being the kinetic energy (K ) of the particle. Equation (4.60) is usually expressed in the equivalent form dW = d K , whence 1 Generally,

d W is not the total differential of any function. This is true only when the forces are conservative, i.e. F tot = −∇U for some U = U (r). In this case d W = −dU , and the right-hand side of this last equation is a true total differential.

218

4 Constitutive Relationships

W AB = ΔK ≡ K B − K A .

(4.61)

Here, the expression on the left-hand side is the total work performed in changing the state of our original particle from a place ‘A’ to another location ‘B’. In Classical Mechanics, Eq. (4.61) is usually referred to as the work–energy theorem. It states that the work done by the external forces on a rigid particle is balanced by the change in kinetic energy of the particle. We are interested to extend Eq. (4.61) to deformable bodies. It is clear that the presence of internal forces in this case will make the task non-trivial; to prepare the ground, some new terminology is introduced next. Definition 4.8 Let Ωt ⊂ Bc be an arbitrary sub-region in the current configuration of a deformable body. (a) The kinetic energy of Ωt is defined by K (Ωt ) :=

1 2

 Ωt

ρ |v|2 dv .

(b) The stress power supplied to Ωt is defined by  Pdef (Ωt ) :=

Ωt

σ : D dv ,

(4.62)

where D ≡ sym(L) represents the symmetric part of the velocity gradient L. (c) The external power supplied to Ωt corresponds to  Pext (Ωt ) :=

Ωt

 ρb · v dv +

∂Ωt

t(n) · v da ,

(4.63)

where t(n) ≡ n · σ represents the Cauchy traction vector, and n denotes the usual outward unit normal on ∂Ωt . The starting point for finding a relationship between these new quantities is Cauchy’s (first) equation of motion. Taking the dot product of this equation with the Eulerian velocity field v, followed by integration over Ωt , leads to 

 Ωt

v · div σ dv +

Ωt

 ρ b · v dv =

Ωt

ρ v˙ · v dv .

(4.64)

We aim to transform the first integral in (4.64) with the help of the Divergence Theorem. To this end, we shall use the identity v · div σ = div (σ · v) − σ : D , whose justification is recorded below

(4.65)

4.5 Hyperelastic Solids

219

v · div σ = v j σi j, i = (v j σi j ), i − v j, i σi j = (σi j v j ), i − σi j v j, i = div (σ · v) − σ : D . Integrating (4.65) over Ωt , 

 v · div σ dv =

Ωt



Ωt

div (σ · v) dv −

 =

∂Ωt



Ωt

σ : D dv =



v · (σ T · n) da −



∂Ωt

(σ · v) · n da −



Ωt

σ : D dv =

∂Ωt

Ωt

σ : D dv

 v · t(n) da −

Ωt

σ : D dv .

Finally, substituting this result into (4.64) yields  Ωt

 ρ b · v dv +

∂Ωt

t(n) · v da =

D Dt

 Ωt

  1 ρ |v|2 dv + σ : D d v (4.66) 2 Ωt

or, using the nomenclature of Definition 4.8, Pext (Ωt ) =

D K (Ωt ) + Pdef (Ωt ) , Dt

(∀) Ωt ⊂ Bc .

(4.67)

Equation (4.67) is known as the mechanical energy-balance equation. The term Pdef represents the deformation work done in Ωt per unit time, while Pext is the cumulative work per unit time done by the contact forces acting on ∂Ωt , together with the body forces acting inside the same region. Thus, Eq. (4.67) shows that the work done by the body and surface forces is converted into kinetic energy and deformation work. In general, the energy is not conserved during the motion and the stress power will incorporate both dissipative and conservative contributions; for example, the deformation work mentioned above might result in an increase in the temperature of the body, as well as heat that is dissipated in the surrounding environment. To keep things simple, thermal effects are not considered in this book. In connection to (4.67), we are interested in the special case in which the energy-balance equation represents conservation of energy. This arises when the material response is governed by an elastic potential function. Motivated by the foregoing observations, we introduce the elastic stored energy W ≡ W (F) per unit volume in Br such that D Dt

 Ωt

ρ W (F) dv = ρ0

 Ωt

σ : D dv ,

(4.68)

for all Ωt ⊂ Bc . The presence of (ρ/ρ0 ) in the above equation is due to the fact that W is measured per unit volume in the reference configuration, whereas the integrals in (4.68) are written on the current configuration. Assuming the continuity of the corresponding integrands, the local version of (4.68) is

220

4 Constitutive Relationships

J −1

D W = σ : D, Dt

(4.69)

since ρ0 = ρ J ; the real-valued function W (F) is also referred to as the elastic or strain energy (per unit volume in Br ). We mention in passing that the expression on the right-hand side in (4.69) represents the deformation or stress power per unit volume in Bc . Definition 4.9 An elastic material that possesses a strain-energy function is said to be a hyperelastic or Green-elastic material. Substituting (4.68) into (4.66), the energy-balance equation for such a hyperelastic material can now be cast in the form Pext (Ωt ) =

 D KINETIC energy + STRAIN energy , Dt

(4.70)

which states that (if thermal effects are entirely neglected), the power expended on Ωt must equal the rate at which the total energy (kinetic + internal) of Ωt is changing. Next, we want to study the implications of material frame-indifference on the possible expressions of W (F). According to that principle, W must be independent of the motion of the observer, which amounts to W (F) = W (F ∗ ) = W ( Q · F),

(∀) Q ∈ Ort+ ,

(4.71)

where F and F ∗ are the deformation gradients associated with the equivalent motions that appear in Definition 4.3. Using the Polar Decomposition Theorem F = R · U and setting Q → R T , we find W (F) = W (R T · R · U) = W (U) ,

(4.72)

i.e. the strain-energy function does not depend on the rotation tensor R. We can actually take this result one step further. Since U = C 1/2 , instead of writing W as a function of U, we notice that it may also be regarded as a new function of C, i.e. W (F) = W (C) ,

(4.73)

where W (C) := W (C 1/2 ) for all C ∈ Sym. It is also easy to show that (4.73) is a sufficient condition for (4.71) to hold. We illustrate the duality of the above representation by taking a closer look at some specific strain energy functions used in the literature. The first example is the Hadamard–Green-elastic material which corresponds to W (F) =

 b a F4 − (F · F T ) : (F · F T ) + ψ(det F) , F2 + 2 4

4.5 Hyperelastic Solids

221

for every F ∈ Invlin+ ; a, b ∈ R and ψ : R+ → R is a twice continuously differentiable function. Note the following properties: F2 ≡ F : F = I : (F T · F) = I : C ≡ |C|, (F · F T ) : (F · F T ) = (F T · F) : (F T · F) = C : C ≡ C2 ,   2 det F = (det F) = det (F T · F) = (det C)1/2 , which are easily checked by using the definition of the double-dot contracted product. With this in mind, the above stored energy function can immediately be transformed into W (C) :=

 b 2 a |C| + |C| − C2 + ψ((det C)1/2 ) , (∀) C ∈ Sym . 2 4

Another similar example involves the Blatz-Ko hyperelastic material, with the stored energy function W (F) =

1 1 F2 + (det F)−m , 2 m

(∀) F ∈ Invlin+ ,

for some m > 0. In this case, we can re-arrange the above expression in the form W (C) =

1 1 |C| + (det C)−m/2 , (∀) C ∈ Sym . 2 m

We were able to quickly switch between the F- and C-representations because the former expressions satisfied (4.72). Let us now return to Eq. (4.69) with the aim of obtaining an explicit representation for the Cauchy stress tensor σ in terms of the stored energy function W = W (C). Using the chain rule for partial differentiation, D ∂W ˙ W = Cαβ = W C : C˙ , Dt ∂Cαβ

(4.74)

where the dot on C indicates a material time derivative and W C is the derivative of W with respect to the right Cauchy–Green tensor. To simplify (4.74), we need to ˙ take a closer look at C, ˙ ˙ T · F + FT · F ˙ = (L · F)T · F + F T · (L · F) C˙ = F T · F = F = F T · (L T + L) · F = 2F T · D · F ,

(4.75)

222

4 Constitutive Relationships

where use has been made of (2.73) and D represents the usual stretching tensor. Replacing (4.75) back into (4.74) and using the standard properties of the double-dot product it transpires that (σ − 2J −1 F · W C · F T ) : D = 0 .

(4.76)

If we now let P be the symmetric second-order tensor that appears in the first term on the left-hand side of (4.76), then P : D = 0; this equation holds true for every D. But, P does not depend on D and, as this latter tensor can be chosen independently of F, we are eventually led to P ≡ 0 or, what is the same  σ =2J

−1

F·

∂W ∂C

 · FT .

(4.77)

The important formula (4.77) suggests that F · W C · F T is a symmetric tensor, which in turn is equivalent to the symmetry of the middle term in this product. For this property to hold, we need to regard W as a symmetric function of both Cαβ and Cβα , even though these components are equal to each other. The following example clarifies the point just made. Let us assume that W has the form 2 2 2 W (C) := C11 + C22 + C13 ,

so that

(4.78)

∂W ∂W = 2 C13 = 0 = , ∂C13 ∂C31

which would suggest that W C is not symmetric. However, if Cαβ enters in the expression of W (C) symmetrically, then ∂W ∂W = , ∂Cαβ ∂Cβα and the expected symmetry of the tensorial derivative is recovered. Indeed, for our particular example it suffices to express (4.78) as 1 2 2 W (C) = C11 + C22 + (C13 + C31 )2 , 4 and then the previous difficulty is eliminated. We refer back to Chap. 1, where a more in-depth discussion of this issue was provided (see the discussion immediately after Eq. (1.270)).

4.6 Constitutive Representations for Isotropic Hyperelastic Solids

223

4.6 Constitutive Representations for Isotropic Hyperelastic Solids We have already seen that for a hyperelastic material that is isotropic relative to some fixed Br , the strain-energy function W (F) is unaffected by rotations in Br (prior to deformation)—see Eq. (4.72). Thus, W (F · Q T ) = W (F) , for all proper rotations Q. In terms of the right Cauchy–Green deformation tensor C (see (4.73)), this equation becomes W ( C) = W (C) ,

(4.79)

where  C is the right Cauchy–Green deformation tensor corresponding to the deformation gradient F · Q T , i.e.  C ≡ (F · Q T )T · (F · Q T ) = Q · (F T · F) · Q T = Q · C · Q T .

(4.80)

Substituting (4.80) back into (4.79), we find that for every C ∈ Sym, W (C) = W ( Q · C · Q T ),

(∀) Q ∈ Ort+ .

(4.81)

The property found in (4.81) is immediately recognised as the isotropy of the scalar function W : Sym → R. Recalling the representation available for such functions (see Theorem 4.1), this feature ensures the existence of a new scalar function  : R3 → R such that W  (I1 , I2 , I3 ) , W (C) = W (4.82) where I j ≡ I j (C), ( j = 1, 2, 3) are the principal invariants of C; if λk (k = 1, 2, 3) are the principal stretches (the eigenvalues of either U or V ), then I1 = λ21 + λ22 + λ23 ,

I2 = λ21 λ22 + λ22 λ23 + λ23 λ21 ,

I3 = λ21 λ22 λ23 .

(4.83)

A convenient representation of the constitutive equation for a hyperelastic material can be obtained directly by combining (4.77) and (4.82), as we show next. The idea is to calculate σ in terms of W C , by taking advantage of the chain rule and the derivatives of the principal invariants already found in Chap. 1. Thus, we start with the obvious chain-rule formula  ∂W  ∂Ij ∂W = ∂C ∂ Ij ∂C j=1 3

(4.84)

224

4 Constitutive Relationships

 /∂ I j =: W  j are known (since W  is given a priori) and the only thing in which ∂ W left to do is to evaluate the terms ∂ I j /∂ C. At this juncture, two observations are in order. First, note that the principal invariants of C are the same as those of B (the left Cauchy–Green deformation tensor). Second, we aim to have the final expression for σ that follows from (4.82) and (4.84) in terms of B rather than C. Particularising the general derivative formulae (1.264), (1.267), and (1.270) for A → C, gives ∂ I1 (C) = I, ∂C ∂ I2 (C) = (I1 (C)) I − C T ≡ (I1 (B)) I − C , ∂C ∂ I3 (C) = (I3 (C)) C −T ≡ (I3 (B)) C −1 . ∂C

(4.85a) (4.85b) (4.85c)

We can now evaluate the individual contribution to (4.77) of each of the three terms that make up the sum on the right-hand side of (4.84). For the first term 1 (F · F T ) = 2J −1 W 1 B. 2J −1 W

(4.86)

The contribution of the second term is slightly more complex, 2 I1 − 2J −1 (F · F T ) · (F · F T )W 2 2 F · (I1 I − C) · F T = 2J −1 (F · F T )W 2J −1 W 2 B − 2J −1 W 2 B 2 . = 2J −1 I1 W (4.87) Finally, 2J −1 F · (I3 C −1 ) · F T = 2J −1 I3 (F · F −1 ) · (F −T · F T ) = 2J −1 I3 I .

(4.88)

Going back to (4.84) with the information gathered from (4.86)–(4.88), and taking into account (4.77), we arrive at σ = α0 I + α1 B + α2 B 2 ,

(4.89)

where α0 ≡ 2



3 , I3 W

 2  2 , W1 + I 1 W α1 ≡ √ I3

2  α2 ≡ − √ W 2, I3

(4.90)

that is, α j = α j (I1 , I2 , I3 ) with I j√= I j (B) for j = 1, 2, 3; note that we have also used I3 = (det F)2 = J 2 or J = I3 . An alternative to (4.89) can be obtained by applying the Cayley–Hamilton Theorem to B and then multiplying the result by B −1 , with the final outcome

4.6 Constitutive Representations for Isotropic Hyperelastic Solids

225

B 2 = I1 B − I2 I + I3 B −1 . If the expression of B 2 is substituted back in (4.89), then the following equivalent representation is obtained: σ = β0 I + β1 B + β−1 B −1 ,

(4.91)

where  2   3 , I 2 W2 + I 3 W β0 ≡ √ I3

2  β1 ≡ √ W 1, I3

 2 , β−1 ≡ −2 I3 W

(4.92)

i.e. β j = β j (I1 , I2 , I3 ) with I j = I j (B) for j = 0, 1, −1. We remark in passing that in contrast to the response coefficients defined for Cauchy-elastic materials, in the above two cases (4.90) and (4.92), these coefficients are not independent. Also, note that knowledge of the strain-energy function  (I1 (B), I2 (B), I3 (B)) completely determines the constitutive behaviour of the W (Cauchy) stress tensor through either (4.89) or (4.91). For an isotropic elastic material, the Cauchy stress tensor σ and the left-stretch tensor V are co-axial (see Exercise 4.16), which means that they  the same  share eigenvectors. In particular, referred to the Eulerian principal axes n( p) , the spectral representation of σ is 3  σ = σ p n( p) ⊗ n( p) , (4.93) p=1

with σ p ( p = 1, 2, 3) being the principal (Cauchy) stresses. One can take advantage of this particular feature in order to express the constitutive behaviour of isotropic elastic materials in terms of the principal stretches λi > 0 (i = 1, 2, 3); more precisely, σ p = σ p (λ1 , λ2 , λ3 ) for p = 1, 2, 3. The corresponding calculations are outlined below. Equation (4.82) and the particular form of (4.83) suggest that W (C) can be equivalently represented in terms of the principal stretches λi > 0, which correspond to the positive square roots of the eigenvalues of C. Thus,  (λ1 , λ2 , λ3 ) , W (C) = W

(4.94)

 , i.e. for some suitably defined symmetric function W  (λ2 , λ3 , λ1 ) = W  (λ3 , λ1 , λ2 ) ,  (λ1 , λ2 , λ3 ) = W W

(∀) λ1 , λ2 , λ3 > 0 .

Using the definition of the strain-energy function in (4.69) and the representation (4.93), we can write 3  D W =J σ p D pp , (4.95) Dt p=1

226

4 Constitutive Relationships

where D pp ( p = 1, 2, 3; no sum) denote the normal components of D referred to the Eulerian principal axes. At the same time, Eq. (4.94) leads to the alternative result 3   ∂W D W = λ˙ p . Dt ∂λ p p=1

(4.96)

Our next objective is to calculate the expressions of D pp (no sum) in (4.95). Since ˙ · F −1 and F = R · U, D is the symmetric part of L = F 1 1 (4.97) R · U˙ · U −1 · R T + R · U −1 · U˙ · R T . 2 2   Referred to the Lagrangian principal axes N (q) , the right stretch tensor admits the spectral representation 3  U= λq N (q) ⊗ N (q) , (4.98) D=

q=1

and we recall that the Eulerian and Lagrangian principal axes are related by the transformation formulae n( p) = R · N ( p) or, equivalently, R T · n( p) = N ( p) for p = 1, 2, 3. We can now write   D pp ≡ n( p) · D · n( p) = n( p) · R · U˙ · U −1 · R T · n( p)     = R T · n( p) · U˙ · U −1 · R T · n( p)  ( p) = N ( p) · (U˙ · U −1 ) · N ( p) = λ−1 · U˙ · N ( p) (no sum) , p N

(4.99)

where use has been made of the symmetry of the two terms in the sum (4.97) and the aforementioned transformation relations. To complete the above calculations, the last bracketed term in (4.99) must be evaluated. Taking the material time derivative of (4.98) gives U˙ =

3

 ˙ (q) , ˙ (q) ⊗ N (q) + λq N (q) ⊗ N λ˙ q N (q) ⊗ N (q) + λq N q=1

whence N ( p) · U˙ · N ( p) =

3  



 ˙ (q) (no sum) . λ˙ q δ pq δ pq + 2λq δ pq N ( p) · N

(4.100)

q=1

As N ( p) are unit vectors, N ( p) · N ( p) = 1 (no sum) and differentiation with respect ˙ ( p) = 0 (no sum). It is now clear from (4.100) to ‘t’ of this equation leads to N ( p) · N that N ( p) · U˙ · N ( p) = λ˙ p , and further use of (4.99) allows us to conclude that

4.6 Constitutive Representations for Isotropic Hyperelastic Solids

˙ D pp = λ−1 p λ p , for p = 1, 2, 3 (no sum) .

227

(4.101)

Substituting this result back into (4.95) and comparing with (4.96), eventually leads to the required formulae σ p = J −1 λ p

 ∂W , ∂λ p

for p = 1, 2, 3 (no sum) ,

(4.102)

where J = λ1 λ2 λ3 . Equation (4.102) is applicable to any isotropic hyperelastic solids, and can be regarded as a particular version of their constitutive relationships (i.e. referred to the Eulerian principal axes). It a simple matter to adapt (4.102) to other stress tensors. Consider, for example, the case of the first Piola–Kirchhoff stress tensor S ≡ J F −1 · σ . Taking into account that ( p) , F −1 = (R · U)−1 = U −1 · R T , so that F −1 · n( p) = U −1 · R T · n( p) = λ−1 p N use of (4.93) yields the representation S=

3 

t p N ( p) ⊗ n( p) ,

tp ≡

p=1

 ∂W ( p = 1, 2, 3) . ∂λ p

(4.103)

4.7 Internal Constraints Deformable bodies are sometimes subjected to kinematic constraints, i.e. conditions that place restrictions on the class of admissible motions they can undergo. For example, a simple material is said to be incompressible if no changes in volume take place during its possible motions. Thus, as already mentioned in Chap. 2, if the material is incompressible the deformation gradient must satisfy J (X, t) ≡ det F(X, t) = 1,

(∀) X ∈ Br , (∀) t > 0 .

By their very nature, kinematic constraints exist independently of motions or their histories, and they need to be maintained by suitable internal forces. These forces are not derived from a typical constitutive law like, for example, one that gives the Cauchy stress tensor. Instead, it is postulated a priori that the mechanical effect of such constraints is to give rise to a reaction stress which does no work in any motion compatible with the constraints. We shall elaborate shortly on this last statement, but not before we make a few observations about the general form of the constraints. An internal constraint at a point in a deformable body is generally defined by an algebraic relation of the form ζ (F) = 0 , (4.104) where ζ : Invlin → R is a given function and F represents the deformation gradient at that point. The constraint functional ζ depends on the choice of reference

228

4 Constitutive Relationships

configuration and, since it is a particular form of constitutive equation, it must satisfy the PMFI, i.e. (4.105) ζ ( Q · F) = ζ (F), (∀) Q ∈ Ort+ . According to the Polar Decomposition Theorem F = R · U, so we can take Q → R T in (4.105) above to find that ζ is unaffected by rigid-body rotations and depends on F only through U = C 1/2 . Hence, ζ can be regarded as a function of the right Cauchy–Green deformation tensor C, and we shall replace (4.105) by its equivalent representation γ (C) = 0 , (4.106) for some scalar function γ : Sym → R. The following result plays a key role in dealing with internal constraints. Constitutive Principle for Elastic Materials with Constraints: For a simple elastic material with the kinematic constraint (4.106), the Cauchy stress σ (x, t) at time t > 0 is determined by the deformation gradient only to within a stress N(x, t) that does no work in any motion satisfying the constraint. If we employ an Eulerian description, this principle stipulates the following: σE = N +σ , N : D = 0, (∀) χ (X, t) such that γ (C) = 0 ;

(4.107a) (4.107b)

here, σ E represents the so-called extra stress which depends on F in the usual fashion (e.g. see Eq. (4.34)), N is the constraint stress which is not specified by any constitutive relation, but is restricted according to (4.107b), and σ denotes the (Cauchy) stress tensor. As it stands, Eq. (4.107b) is rather opaque—some further work is needed to bring it to a useful form. To this end, let us take the material time derivative of the constraint equation (4.106); the outcome is γ˙ (C) =

∂γ ˙ Cαβ = γC : C˙ = 0 , ∂Cαβ

(4.108)

where γC stands for the gradient of γ with respect to C. Using the expression of C˙ found in (4.75) we can go back into (4.108) to get 0 = γC : C˙ = 2 γC : (F T · D · F) = (2 F · γC · F T ) : D or, (F · γC · F T ) : D = 0 ,

(4.109)

which holds for all stretching tensors D that satisfy (4.107b). Since the double contraction ‘ : ’ defines an inner product in the six-dimensional space Sym, we can inter-

4.7 Internal Constraints

229

pret the workless constraint on N as follows: both N and F · γC · F T are orthogonal to D and, therefore, N and F · γC · F T must be ‘parallel’ to each other, i.e.2  N = q F · γC · F ≡ q F · T

∂γ ∂C

 · FT ,

(4.110)

for some scalar field q. This important formula provides an explicit representation of the constraint stress. It shows that this stress is determined to within a scalar multiplier by the form of the constraint and the deformation gradient F. The scalar field q is part of the unknowns and has to be determined at the same time with them. Increasing the number of constraints introduces only additional minor complications. For instance, it can be shown by following the same route as above that, in the case of several distinct constraints given by γ j (C) = 0

j = 1, 2, . . . , m ,

the reaction stress will have to be of the form N=

m  j=1

 qj F ·

∂γ j ∂C

 · FT ,

(4.111)

where the scalar fields q j ( j = 1, 2, . . . , m) are determined at the same time with the other unknowns of the problem. The general theory sketched above is illustrated by two particular cases commonly encountered in Elasticity. (a) Incompressibility: As already mentioned at the beginning of this section, a material is incompressible if it is susceptible only to isochoric motions. Thus, at any place in the body and at any time, det F − 1 = 0 must be satisfied. This condition can be cast in terms of the right Cauchy–Green strain tensor C ≡ F T · F by simply taking γ (C) ≡ det C − 1. The gradient of γ with respect to C is easily calculated by using (1.270) ∂ ∂γ = (det C) = I3 (C) C −T ∂C ∂C = (det F)2 (F T · F)−T = F −1 · F −T , and then the constraint stress N is immediately found by plugging this γC back into formula (4.110),  N =qF·

∂γ ∂C



· F T = q F · (F −1 · F −T ) · F T = q I .

result can be proved more rigorously: if A, B ∈ Sym such that A : C = 0 and B : C = 0 for all C ∈ Sym, then A = α B, for some α ∈ R.

2 This

230

4 Constitutive Relationships

Hence, in the absence of any other constraints, the extra stress for an incompressible elastic body reduces to σE = σ +qI

=⇒

σ = −q I + σ E .

(4.112)

In this formula, σ corresponds to the stress in the incompressible body, σ E is given by a constitutive law, as seen in the previous sections, and q is typically determined in an ad hoc way from the equations of the particular problem one wants to solve. Equation (4.112) confirms that for incompressible elastic materials the stress is determined by the deformation gradient only to within a hydrostatic pressure q. It is important to draw a clear distinction between incompressible elastic materials and unconstrained materials subjected to isochoric motions. The stress in the latter case is simply determined by the usual constitutive law adopted for that material (and hence by the isochoric motion whose influence is ‘felt’ by the deformation gradient or other similar kinematic quantity). As seen above, this is no longer true for incompressible materials since in this scenario there is always an indeterminate spherical tensor that is not directly linked to the motion/deformation experienced by that material. (b) Inextensibility: We turn now to the case when the material in a deformable body is constrained against deformations along a fixed direction specified by the unit vector M in Br . Recalling the interpretation of C from Chap. 2 such a constraint is defined by the equation γ (C) := M · C · M − 1 = 0 , (4.113) which identifies the material as being inextensible in the direction of M. We can also understand the nature of the constraint in (4.113) by using the following rationale. If X is the position vector of an arbitrary point along the direction defined by M and dX is an infinitesimal vector based at that point, then we can write M := dX/|dX|. Hence,  |dx|2 = dx · dx = (F · dX) · (F · dX) = dX · (F T · F) · dX

    dX dX ·C· = |dX|2 |dX| |dX| = |dX|2 (M · C · M) = |dX|2 or, |dx| = |dX|, i.e. the material behaves like a rigid body along the direction specified by M. A routine calculation shows that γC = M ⊗ M; used in conjunction with the formula (4.110) this yields  N =qF·

∂γ ∂C

 · F T = q F · (M ⊗ M) · F T = q(F · M) ⊗ (F · M) .

4.7 Internal Constraints

231

Noting that m := F · M is a unit vector specifying the orientation of M in the current configuration, the constraint stress can be cast in the form N = q (m ⊗ m), i.e. N is a uniaxial stress in the direction taken up by the inextensible direction after deformation; the arbitrary scalar field q represents the tension in the inextensible direction.

4.8 Particular Forms of the Strain-Energy Function We have found that the stress response of an isotropic hyperelastic material is derived  which is expressible in terms of the principal from the given strain-energy function W invariants of the left Cauchy–Green strain tensor B ≡ F · F T . In this section, some well-established such functions are recorded. =W  (I1 , I2 ). For incompressible materials det F = 1, so that I3 (B) = 1, and W Assuming that the reference configuration is unstressed, the strain-energy function must vanish for C = I, i.e. when I1 = I2 = 3. Thus, it is convenient to assume that  is a function of I1 − 3 and I2 − 3, and vanishes in the reference configuration. W 1. Neo-Hookean material: This constitutive law was proposed by L. R. G. Treloar in 1948, and has the following form:  (I1 , I2 ) = c(I1 − 3) , W

(4.114)

where c ∈ R is a material constant. 2. Mooney–Rivlin material:  (I1 , I2 ) = c1 (I1 − 3) + c2 (I2 − 3) , W

(4.115)

where c1 , c2 ∈ R are material constants. This was first proposed by Mooney in 1940. 3. Ogden material: This includes the Neo-Hookean and Mooney–Rivlin forms as special cases. Its advantage is that can it be calibrated to account for realistic behaviour observed in rubber-like materials. The constitutive law is phrased in  rather than W  terms of W  (λ1 , λ2 , λ3 ) = W

M  μm βm β β (λ1 + λ2 m + λ3 m − 3) , β m m=1

(4.116)

in which the μm are constants, and the βm are not necessarily integers and may be positive or negative; M is a positive integer which may be taken as large or as small as desired. Typical values for the case M = 3 are recorded below

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4 Constitutive Relationships

β1 = 1.3 ,

β2 = 5.0 ,

β3 = −2.0 ,

μ1 = 6.3 ,

μ2 = 0.012 ,

μ3 = −0.1 .

To obtain the incompressible form of (4.116), it suffices to impose λ1 λ2 λ3 = 1, which gives  (λ1 , λ2 ) = W

M  μm βm β (λ1 + λ2 m + (λ1 λ2 )−βm − 3) . β m m=1

(4.117)

Note that the Mooney–Rivlin material corresponds to M = 2, β1 = 2, β2 = −2 and the Neo-Hookean material is recovered by choosing M = 1, β1 = 2 in (4.117).

4.9 Simple Fluids One of the main properties of fluids is their ability to take the shape of the container which they are poured in. This indicates the indifference of these materials with respect to their initial configuration and suggests that their symmetry group is maximal. Definition 4.10 A simple fluid is a simple material for which S ≡ Unim+ . We start our discussion with a particular case, the so-called elastic fluid which corresponds to a constitutive law of the form σ = G (F) (leaving out the dependence on ‘x’ and ‘t’). It is of interest to explore the restrictions placed on G by the symmetry group. According to Sect. 4.4 and the above definition, the condition G (F · H) = G (F) ,

(∀) F ∈ Invlin+

(4.118)

must be satisfied for all H ∈ Unim+ . The tensor H 0 := J 1/3 F −1 , with J ≡ det F, is a proper unimodular tensor, so we can take H → H 0 in (4.118) to find G (F) = G (F · H 0 ) = G (F · (J 1/3 F −1 )) = G (J 1/3 I)

(4.119)

or G (F) = G (J 1/3 I). This result confirms that G depends on F only through a spherical tensor that involves its third principal invariant. The implication of this simplification vis-à-vis the objectivity requirement (4.36) is addressed next, Q · G (J 1/3 I) · Q T = Q · G (F) · Q T = G (F · Q) = G (det(F · Q)1/3 I) = G (J 1/3 I)

(4.120)

4.9 Simple Fluids

233

for every Q ∈ Ort+ . It is easily recognised that (4.120) expresses the isotropy of the tensor-valued function G (J 1/3 I). Thus, we conclude (see Exercise 4.14) that this function must be a spherical tensor, i.e. σ = f (J )I for some continuous function f : (0, ∞) → R. Recalling the principle of mass conservation, ρ0 = ρ J , we can finally state the full constitutive law for an elastic fluid, σ (x, t) = − p(x, t; ρ)I ,

(4.121)

where the scalar field p ≡ p(x, t; ρ) is known as the pressure of the fluid. The presence of the current mass density in (4.121) makes this fluid compressible. By contrast, an incompressible fluid corresponds to a constitutive law like (4.121), but with the mass density assumed to be constant, i.e. p ≡ p(x, t). In both cases, the state of stress is a hydrostatic one, so such fluids cannot sustain shear stresses (see the discussion in Sects. 3.7 and 3.8). The incompressible elastic fluid discussed above is known as the ideal fluid and provides a good approximation for materials such as water under certain conditions. The fundamental physical property of fluids is their inability to sustain shear stresses when in equilibrium; ideal fluids retain this property when in motion as well. On the other hand, in a real fluid undergoing a simple shear motion (say), adjacent layers of fluid will encounter some resistance in sliding over each other. Such frictional forces are related to the rate of the deformation gradient which would have to be included in the corresponding expression of the constitutive law for a fluid, if such effects are to be captured by a general mathematical formula. This aspect is dealt with next. Consider a fluid with a constitutive law of the type ˙ , σ = L (F, F)

(4.122)

for some material response functional L : Invlin+ × Lin → Sym. The form of the expression on the right-hand side in (4.122) still falls within the class of simple materials. Our aim is to provide a reduced form of this constitutive relation that would allow us to generalise (4.121) to real fluids. ˙ = L · F, where L ≡ gradv represents the spatial First, recall from Chap. 2 that F velocity gradient. Thus, Eq. (4.122) can be re-cast as σ = H (F, L) ,

(4.123)

with H (F, L) := L (F, L · F) ,

(∀) F ∈ Invlin+ , L ∈ Lin .

(4.124)

A simple extension of the objectivity requirement (4.36) derived in Sect. 4.3, together with the calculation of L ∗ in Proposition 4.2 shows that H must satisfy ˙ · Q T ) = Q · H (F, L) · Q T , H ( Q · F, Q · L · Q T + Q

(4.125)

234

4 Constitutive Relationships

for every Q ∈ Ort+ , every F ∈ Invlin+ and every L ∈ Lin. By breaking up L into its symmetric ( D) and skew-symmetric (W ) parts, the term involving the spatial velocity gradient on the left-hand side becomes ˙ · QT , Q · D · QT + Q · W · QT + Q and we are going to choose Q ≡ Q(t) such that the underlined terms vanish. To this end, let t0 > 0 be fixed (but otherwise arbitrary); for any given skew-symmetric tensor W , we can define an orthogonal tensor Q(t) (see Exercise 4.13) such that Q(t0 ) = I ,

˙ 0 ) = −W , Q(t

and hence the evaluation of (4.125) at t = t0 leads to H (F, L) = H (F, D) , which implies that H cannot depend on the spin tensor W . In conclusion, for the constitutive law (4.122) to be frame-indifferent, it must reduce to σ = H (F, D) .

(4.126)

Further restrictions on (4.126) follow from the symmetry group. Let H ∈ Unim+ be fixed (time independent). We make the observation that D corresponding to the deformation gradient F · H is the same as the stretching tensor associated with F. In light of this property, we are left with the following condition that replaces (4.40): H (F · H, D) = H (F, D) ,

(∀) H ∈ Unim+ ,

(4.127)

and (∀) D ∈ Sym, F ∈ Invlin+ . The result just found suggests that the symmetry restrictions on the two independent variables in the constitutive functional can be handled separately. First, by repeating the arguments developed at the beginning of this section we obtain H (F, D) = H ((det F)1/3 I, D) ,

(∀) F ∈ Invlin+ , D ∈ Sym ,

and then the original Eq. (4.122) can be re-stated once again in the form σ = h(ρ, D) ,

(4.128)

for some suitable tensor function h : R × Sym −→ Sym. This is already isotropic in the first argument because the mass densities associated with deformation gradients F and Q · F · Q T , respectively, are the same. The isotropy in the second argument (which is a consequence of Proposition 4.3 and the fact that Ort+ ⊂ Unim+ ), requires h(ρ, Q · D · Q T ) = Q · h(ρ, D) · Q T for every Q ∈ Ort+ and D ∈ Sym.

4.9 Simple Fluids

235

That is, h(ρ, D) is an isotropic tensor function in its second argument and Theorem 4.2 provides an explicit representation for such functions. We summarise the work done above in the next key result. Proposition 4.4 The constitutive relation for a viscous fluid characterised by a con˙ admits the reduced representation stitutive law of the form σ = L (F, F) σ = α0 (ρ, ‫ ג‬D )I + α1 (ρ, ‫ ג‬D ) D + α2 (ρ, ‫ ג‬D ) D2 ,

(4.129)

for suitable functions α j ( j = 0, 1, 2) that depend on density and the principal invariants of the stretching tensor, ‫ ג‬D ≡ {I1 ( D), I2 ( D), I3 ( D)}. Materials with the constitutive law (4.129) are known as Reiner–Rivlin fluids. A Newtonian visous fluid is a particular case of this class of materials, which corresponds to α0 := − p(ρ) + λ(ρ)| D| ,

α1 := 2μ(ρ) ,

α2 ≡ 0 ,

in which case (4.129) reduces to σ = (− p + λ| D|)I + 2μ D .

(4.130)

The functions p ≡ p(ρ), λ ≡ λ(ρ), and μ ≡ μ(ρ) are material coefficients; μ and λ are, respectively, the first and second coefficients of viscosity (or shear viscosity and dilatational viscosity coefficients, respectively). Fluids like water and air are well described by this class of Newtonian viscous fluids, but there are many ‘fluids’ whose behaviour cannot be explained by (4.130); examples include pastes and glues, blood, clay, emulsions, paints and so on. The study of these non-Newtonian fluids is the object of a branch of Continuum Mechanics known as Rheology. If a Reiner–Rivlin fluid is incompressible, then I1 ( D) = | D| = div v = 0, and in this case Eq. (4.129) is replaced by σ = −π(x, t)I + α1 (ρ, ‫ ג‬D ) D + α2 (ρ, ‫ ג‬D ) D2 ,

(4.131)

where α0 has been absorbed into the indeterminate pressure π ≡ π(x, t) which corresponds to the reaction stress associated with the incompressibility constraint. An incompressible linearly viscous fluid also admits a slightly simpler constitutive equation than (4.130), namely, σ = − p(x, t)I + 2μ D .

(4.132)

The compressible version of this fluid requires that μ be a function of density, while for incompressible fluids μ = μ0 (a constant). A Navier–Stokes fluid is usually taken to be an incompressible linearly viscous fluid with constant viscosity and constant density. The so-called Navier–Stokes equations, used extensively in Fluid Mechanics,

236

4 Constitutive Relationships

are obtained from Cauchy’s first equation of motion under the assumption that the Cauchy stress tensor is given by Eq. (4.132). We sketch below the derivation of these equations. The first step is to take the spatial divergence of (4.132). Note that for the hydrostatic pressure, div ( p I) = (grad p) · I = grad p, and then div D =

1 1 div(grad v) + div ((gradv)T ) . 2 2

The first term on the right-hand side of this equation will be zero because of the incompressibility condition (i.e. div v = 0), div (gradv) = (grad v)i j, i e j = (vi, j ), i e j = (vi,i ), j e j = (divv), j e j = 0 , while for the second term we have div((grad v)T ) = (grad v) ji, i e j = (v j, i ), i e j = (v j ), ii e j = (v), ii = ∇ 2 v . Summarising these calculations div σ = −grad p + μ∇ 2 v ,

(4.133)

and we recall that, from our discussion in Sect. 2.3, a = v˙ can be expressed as a=

∂v + (grad v) · v . ∂t

(4.134)

Substituting (4.133) and (4.134) into divσ + ρb = ρa (Cauchy’s first equation of motion), we eventually find the Navier–Stokes equation

 ∂v μ∇ v − grad p + ρb = ρ + (grad v) · v . ∂t 2

(4.135)

Each term in this equation admits a distinct physical interpretation in terms of forces (computed per unit volume of the fluid) acting on each material particle. For instance, the two terms on the right-hand side correspond to inertia forces arising because of the local rate and the convective rate of change of linear momentum. As for the terms on the left-hand side: the first one represents the viscous frictional force acting on a typical fluid particle, the second term is a pressure gradient force and the last one is the usual body force (per unit volume). The Navier–Stokes equation expresses the balance of these forces for each fluid particle. The vector Eq. (4.135) is equivalent to a system of three scalar partial differential equations in which the unknowns are the components of the Eulerian velocity field v;

4.9 Simple Fluids

237

even though the constitutive behaviour described by (4.132) is linear, in general, the Navier–Stokes equations have strong kinematic nonlinearities due to the presence of the term (grad v) · v.

4.10 Exercises 4.1. Let A, B be two objective (second-order) tensor fields, and consider two objective vector fields u1 and u2 . Indicate which of the following expressions is objective, u1 ⊗ u2 ; A· B;

u1 ∧ u2 ;

u1 · u2 ;

A: B ;

A · u1 .

4.2. Show that the stress power per unit volume in Bc is an objective scalar field. 4.3. For each of the following constitutive equations decide whether or not the principle of objectivity is satisfied  (α, β ∈ R and p a scalar-valued function of time): (i) σ = − p(t)I; (ii) σ = α F + F T ; (iii) σ˙ = (α tr D) I + β D3 . 4.4. Show that the current mass density is an objective scalar field. 4.5. If u and K are, respectively, spatial vector and tensor fields associated with the current configuration Bc of a deformable body, then the convected rates of change of u and K are defined according to u˚ = u˙ + L T · u

and

K˚ = K˙ + L T · K + K · L .

(4.136)

Assuming that u and K are objective, show that u˚ and K˚ share the same property. ˙ · Q T is a skew4.6. a. If Q = Q(t) is an orthogonal tensor, show that W 0 := Q T 2 ¨ ˙ symmetric tensor and Q · Q = W 0 + W 0 . b. Assuming that v and a are the Eulerian descriptions of the velocity and the acceleration fields of a moving body, show that neither of these is an objective vector field. 4.7. The stress response of a certain type of simple material which exhibits both elastic and viscous properties is described by the constitutive equation ˙ σ = f (F, F),

(4.137)

f being a symmetric tensor-valued function. Investigate the restrictions imposed on f by the principle of objectivity and hence show that the most general objective form of (4.137) is ˙ · RT σ = R · f (U, U)

238

4 Constitutive Relationships

with R being the rotation tensor in the polar decomposition of the deformation gradient F = R · U = V · R. 4.8. The Rivlin–Ericksen tensors A1 , A2 , . . . are defined recursively by (n)

C := F T · An · F ,

n ≥ 1,

where the superscript ‘(n)’ denotes the nth material time derivative and F is the usual deformation gradient associated with the motion of a deformable body. Show that A1 = 2 D and An = A˚ n−1 , n ≥ 2, where the superimposed circle denotes the convected rate of change defined in (4.136b). Deduce that each of the Rivlin–Ericksen tensors is symmetric and objective. 4.9. Let σ be the Cauchy stress tensor associated to a deformable body, and let D and W be the usual stretching and spin tensors, respectively. Show that the Jaumann  stress rate σ , defined by 

σ := σ˙ − W · σ + σ · W , 

is an objective tensor. Deduce that σ = α(tr D) I + β D is a constitutive equation that obeys the principle of material frame-indifference (α, β ∈ R). 4.10. a. The current mass density of a deformable body can be regarded as a function of the deformation gradient corresponding to the motion it undergoes; that is, ρ = ρ(F). Is this function isotropic in the sense of Definition 4.7? b. The stretching tensor D can be regarded as a tensor-valued function of the deformation gradient F, according to the formula D(F) :=

4.11.

  1 ˙ ˙ · F −1 T , F · F −1 + F 2

(∀) F ∈ Invlin+ .

Check that the isotropy of this function is equivalent to the frame-indifference of D. a. Show that the constitutive law for an elastic fluid can also be expressed as σ = K (B) = K ((det B)1/3 B) ,

(∀) B ∈ Sym ,

(4.138)

where K is a suitably defined tensor-valued constitutive functional. b. Show that K (B) is isotropic and then deduce that (4.138) can be stated in the more explicit form σ = γ 0 I + γ 1 B + γ 2 B2 ,

(4.139)

4.10 Exercises

239

with γ j ≡ γ j (‫ ג‬B ) for j = 1, 2, 3, and ‫ ג‬B ≡ {I1 (B), I2 (B), I3 (B)}. c. For every B ∈ Sym one can define a new tensor  B :=



1 det B 2

1/3 (e1 ⊗ e1 + I) .

Explain why K (B) = K ( B), and then use this observation to show that γ 1 = γ 2 ≡ 0 in (4.139). Hence, recover the reduced constitutive law (4.121) for elastic fluids. 4.12. If G : Lin → Lin is a tensor function and A ∈ Lin, then the eigenvalues of G ( A) are isotropic scalar functions of A. Deduce that the eigenvalues of A are isotropic scalar functions. 4.13. The exponential of a second-order tensor A is defined by the infinite series exp( A) = I + A +

A3 A2 + + ... . 2! 3!

Show that: a. if A, B ∈ Lin, then exp( A) · exp(B) = exp( A + B); b. if AT = − A, then exp( A) is an orthogonal tensor; c. for any tensor W such that W T = −W , the tensor function Q(t) := exp(−(t − t0 )W ) satisfies the following initial-value problem, ˙ + Q · W = 0, Q

Q(t0 ) = I .

4.14. If A ∈ Sym and Q · A · Q T = A for every Q ∈ Ort+ , show that A is a spherical tensor, i.e. A = α I for some α ∈ R. 4.15. Consider a non-Newtonian viscous fluid characterised by the constitutive law σ = K (ρ, L, x˙ , x, t) ,

(4.140)

depending on the current mass density ρ, spatial velocity gradient L, velocity x˙ , present place x and time t > 0. If this constitutive relation satisfies the principle of material frame-indifference, show that (4.140) reduces to σ = K# (ρ, D) , where D represents the stretching tensor and K# is a suitably defined tensorvalued functional. 4.16. a. It was shown in (4.50) that the constitutive law for an isotropic elastic solid can be expressed in the form σ = G (V ). Show that

240

4 Constitutive Relationships

(∀) Q ∈ Ort+ ,

G ( Q · V · Q T ) = Q · G (V ) · Q T ,

and deduce that for such materials σ is co-axial with the left-stretch tensor V. b. If σi (i = 1, 2, 3) are the principal Cauchy stresses, then σi = φ0 + φ1 λi + φ2 λi2 ,

4.17.

(i = 1, 2, 3) ,

where λi > 0 (i = 1, 2, 3) represent the principal stretches and φi (i = 0, 1, 2) are isotropic scalar functions that depend on V . a. Take the derivative with respect to F (deformation gradient) in (4.73) and use the chain rule to show that ST =

∂ W (F) , ∂F

where S is the first Piola–Kirchhoff stress tensor defined in (3.59). b. The strain-energy function W (C) in (4.73) can be expressed as a function of the Lagrangian strain tensor E by writing W (C) = W (I + 2E) =: W # (E). If Π represents the second Piola–Kirchhoff stress tensor, show that Π=

∂ W # (E) . ∂E

Bibliography 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

Gurtin ME (1981) An introduction to continuum mechanics. Academic Press, New York Holzapfel GA (2000) Nonlinear solid mechanics. Wiley Ltd., Chichester Jaunzemis W (1967) Continuum mechanics. The McMillan Company, New York Shih Liu I (2002) Continuum mechanics. Springer, Berlin Marsden JE, Hughes TJR (1983) Mathematical foundations of elasticity. Prentice-Hall Inc, Englewood Cliffs, New Jersey Narasimhan MNL (1993) Principles of continuum mechanics. Wiley Inc, New York Spencer AJM (1980) Continuum mechanics. Longman Ltd., Essex (UK) Truesdell C (1966) The elements of continuum mechanics. Springer, Berlin Truesdell C (1977) A first course in rational continuum mechanics, vol 1. Academic Press, New York Truesdell C, Noll W (2004) The non-linear field theories of continuum mechanics. Springer, Berlin

Part II

Topics in Linear Elasticity

Chapter 5

Linear Elasticity: General Considerations and Boundary-Value Problems

Abstract In this chapter, we introduce the mathematical model for a linearly elastic solid and its associated boundary-value problems. This is achieved by a judicious particularisation of the general theory developed in the previous chapters; broadly speaking, the reference and current configurations will be assumed to be very close to each other (in a sense that will be made clear shortly). The upshot of this simplification is the linearity of the aforementioned boundary-value problems, which can then be solved by a number of indirect strategies involving: superposition, semi-inverse approaches, and the Saint-Venant’s Principle. The last section touches upon some well-established approximations whereby a three-dimensional situation is reduced to a two-dimensional problem. These specific approximations are taken up in much greater detail in some of the subsequent chapters.

5.1 Introduction In the second half of the book, we are going to look in more depth at a particular theory of Continuum Mechanics known as Linear Elasticity. The geometrical setting is the same as before—see Fig. 5.1. We consider a deformable solid body B that occupies a region Ω0 ⊂ E3 in the reference configuration; as a result of the applied loads the body deforms, and the current configuration fills up another region Ωc ⊂ E3 . The displacement of the material points between the two configurations are described by the vector field u = x − X, where x and X denote the position vectors of the same particle in the reference and current configurations, respectively. The key assumption underlying the linear theory that we want to develop here is that both u (displacement) and H (displacement gradient) are infinitesimally small. To make this statement more precise, let us recall from Chap. 1 that  1/2 |u| ≡ u · u

and

 1/2 H ≡ H : H = (Hi j Hi j )1/2

characterise, respectively, the magnitudes of the vector u and the tensor H. In Linear Elasticity, the equations of Continuum Mechanics are systematically linearised © Springer Nature B.V. 2020 C. D. Coman, Continuum Mechanics and Linear Elasticity, Solid Mechanics and Its Applications 238, https://doi.org/10.1007/978-94-024-1771-5_5

243

244

5 Linear Elasticity: General Considerations and Boundary-Value Problems

Fig. 5.1 The geometry of deformation in Linear Elasticity; the reference and current configurations are assumed to be close to each other in the sense made clear in the main text

applied loads

around the reference configuration, while terms of order |u|2 and H2 are neglected. The details of the linearisation process are explained next.

5.2 Linearised Kinematics Let us start with the observation that1 if H  1,   F −1 = I − H + O H2 .

(5.1)

This is easily justified since F = I + H (from u = x − X ⇒ H = F − I) and a straightforward calculation quickly shows that   F −1 · F = F · F −1 = I + O H2 , i.e. F −1 · F = F · F −1  I is correct to first order if H  1. In Linear Elasticity, there is no need to distinguish between Eulerian and Lagrangian derivatives. Indeed, from Chap. 2 we know that Grad u = (grad u) · F, so (5.1) gives, grad u = (Grad u) · F −1 = H · (I − H)  H. Thus, to first order, we can replace (Grad u) by (grad u); henceforth we use the ‘nabla’ notation (∇) to indicate gradients. For H  1 the Green–Lagrange strain tensor becomes E=

1 1 1 T (F · F − I) = (H + H T + H T · H)  (H + H T ) . 2 2 2

Within the context outlined above, the symmetric part of the displacement gradient is known as the infinitesimal strain tensor, and represents the main strain measure.

f (x) and g(x) are two functions of x, then f (x) = O (g(x)) as x → x0 if there exist δ, M > 0 such that | f (x)| ≤ M|g(x)| if 0 < |x − x0 | < δ.

1 The ‘big Oh’ notation is widely used in Mathematics: if

5.2 Linearised Kinematics

245

Note that the displacement gradient admits the additive decomposition H = E+Ω,

(5.2)

where the tensors E and Ω are formally defined below,  1 u⊗∇+∇⊗u , 2  1 Ω = u⊗∇−∇⊗u , 2

1 (ui, j + u j, i ) , 2 1 Ωi j = (ui, j − u j, i ) . 2 Ei j =

E=

(5.3a) (5.3b)

Clearly, E ∈ Sym and Ω ∈ Skw, the latter being called the infinitesimal rotation tensor.

5.3 Distortional and Spherical Strain As discussed in Chap. 2, the Jacobian J ≡ det F represents a measure of how local volume elements are affected by the motion of a deformable body. By using Eq. (1.285) from Chap. 1, it follows directly that   J ≡ det F = det(I + H) = 1 + |H| + O H2

(5.4)

and thus, since dv = J d V , dv − d V = J − 1  |H| = |E| = ∇ · u . dV

(5.5)

This shows that the trace of the infinitesimal strain tensor E is a measure of the (infinitesimal) volume change (also known as the volumetric dilatation). Next, let us introduce the new tensors 1 E d := E − |E|I 3

and

E s :=

1 |E|I , 3

(5.6)

which are usually referred to as the distortional and volumetric/spherical strain tensors, respectively; note that E = E d + E s . The distortional strain tensor, having a vanishing trace, describes only changes in shape, without any change in magnitude of the volume element. By contrast, the volumetric strain represents only a change in size, without any change in shape of the volume element.

246

5 Linear Elasticity: General Considerations and Boundary-Value Problems

5.4 Linearised Constitutive Behaviour Let us recall (from Chap. 4) that an elastic material is typically defined by a constitutive law of the form σ = g (F) , (5.7) for some suitably defined tensor-valued function  g : Invlin+ → Sym. We also know that, by using the Polar Decomposition Theorem for the deformation gradient (F = R · U = V · R) and the Principle of Material Frame-Indifference, Eq. (5.7) can be expressed in the reduced form σ = R · g(U) · R T ,

(5.8)

where g : Sym → Sym is defined in terms of the previous  g (its precise form will be irrelevant for our immediate purposes). We are interested in linearising (5.8) around the reference configuration. This corresponds to considering a perturbation of the right-stretch tensor U around I and, to make things simple, we are also going to assume that there are no initial stresses in the reference configuration; that is,   g(I) ≡ g(U)

U=I

= O.

(5.9)

It is a routine matter to check that within the framework of Linear Elasticity the following approximations hold true: U  I + E,

U −1  I − E ,

R  I +Ω,

(5.10)

where E and Ω are the tensors that appear in (5.3). On the other hand, by Taylor-expanding g(U), g(I + E) = g(I) +

  ∂g (I)[E] + O E2 ; ∂U

(5.11)

since E is just the symmetric part  it should be clear that the ignored terms in  of H, the above expansion are also O H2 . Thus, Eq. (5.11) can be rewritten as   g(I + E) = C[E] + O H2 ,

(5.12)

where we have used (5.9) and introduced the new notation C≡

∂g  .  ∂U U=I

This is known as the (fourth-order) stiffness tensor and has components

(5.13)

5.4 Linearised Constitutive Behaviour

247

Ci jkl =

 ∂gi j (U)  ; ∂Ukl U=I

the quantities Ci jkl are called the elastic moduli of the linear elastic solid, relative to the initial stress-free reference configuration. If (5.10) and (5.12) are substituted back into (5.8), then   (5.14) σ = C[E] + O H2 or, to first order, Ti j = Ci jkl E kl ,

T = C[E] ,

(5.15)

after relabelling σ → T . The reason for this change of notation is because, to first order, there is no distinction between the Cauchy (σ ) and the first Piola–Kirchhoff (S) stress tensors, S = J F −1 · σ  (1 + |E|)(I − H) · σ  σ ,

(5.16)

where use has been made of(5.1) together with (5.5), and the ignored terms in the  above equation are O H2 . The stress tensor in what follows will be denoted by T as this seems a more neutral terminology, with no reference to either σ or S. For the remainder of this book, a linear elastic solid will be one for which the components of the stress tensor (Ti j ) are linear combinations of the infinitesimal strain tensor (E i j ), as in (5.15)—these constitutive equations are known as the generalised Hooke’s law. The elastic stiffness tensor enjoys some straightforward symmetry features that follow directly from (5.15). Since E i j = E ji and Ti j = T ji the elastic moduli will satisfy the so-called minor symmetries, Ci jkl = C jikl = Ci jlk ; typically, the major symmetries Ci jkl = Ckli j also hold. In this case the elastic material is said to be hyperelastic and there exists a strain-energy density function W = W (E), with W ≡

1 1 E : C : E = Ci jkl E i j E kl , 2 2

and T=

∂W ∂E

or

Ti j =

∂W . ∂ Ei j

(5.17)

(5.18)

248

5 Linear Elasticity: General Considerations and Boundary-Value Problems

For such a hyperelastic linear elastic material (which essentially covers all linear models used in applications) there are, in general, 21 elastic moduli. A general treatment of a linear elastic theory involving so many constitutive parameters is beyond the scope of this elementary textbook. Instead, here we shall focus on an easier scenario, known as Isotropic Linear Elasticity. In this case the elastic moduli Ci jkl depend on just two constitutive real constants, λ and μ, according to the formula2 Ci jkl = λδi j δkl + μ(δik δ jl + δil δ jk ) .

(5.19)

With these expressions, the generalised Hooke’s law (5.15) becomes T = 2μE + λ|E|I ,

Ti j = 2μE i j + λ(E kk )δi j .

(5.20)

The two constitutive parameters in these equations are called the Lamé constants. It is customary to express (5.20) in terms of certain combinations of λ and μ that are very useful in the interpretation of simple elastic deformations. For example, E and ν defined by λ μ(3λ + 2μ) , ν := , (5.21) E := λ+μ 2(λ + μ) are widely used as an alternative to λ and μ; they are known as the Young’s modulus (E) and the Poisson’s ratio (ν). To shed light on the expressions that feature in (5.21), let us consider the (infinitesimal) uniform uniaxial tension (or compression) of a circular bar (here regarded as a long, slender cylindrical body with the property that any characteristic dimension of a transverse cross section is much smaller than the length of this body). Without loss of generality we shall assume that the bar is extended or compressed in the 1-direction; see Fig. 5.2. For this configuration, it is reasonable to hypothesise that T = T11 (e1 ⊗ e1 ) (since |Ti j |  |T11 | for (i, j) = (1, 1)). By specialising the generalised Hooke’s law to our setting it follows at once that E 22 = E 33 (by symmetry) and E i j = 0 for i = j. In fact, the only non-trivial equations in (5.20) are T11 = 2μE 11 + λ(E 11 + 2E 22 ) whence T11 =

μ(3λ + 2μ) E 11 λ+μ

and

0 = 2μE 22 + λ(E 11 + 2E 22 ) ,

and

E 22 = −

λ E 11 . 2(λ + μ)

According to the notation introduced in (5.21), these equations can be stated in the equivalent forms E 22 T11 and ν=− , E= E 11 E 11 2 The expression on the right-hand side of (5.19) is a particular type of fourth-order isotropic tensor;

see the discussion in Sect. 1.11.

5.4 Linearised Constitutive Behaviour

249 reference configuraƟon

Poisson eīect

deformed configuraƟon

Fig. 5.2 Uniaxial tension test illustrating the Poisson effect for a linearly elastic isotropic material (the displacements/deformations are exaggerated) Fig. 5.3 Simple shear test that illustrates the role of G := μ (shear modulus). The elastic material is assumed to extend indefinitely in both the X 1 - and X 3 -directions

2 22

reference configuraƟon

deformed configuraƟon

≡0 21 12

11

≡0

1

3

33

≡0

which provide a natural interpretation for E and ν. The Young’s modulus is the ratio between the axial stress and strain in a uniaxial test, while the Poisson’s ratio represents the negative of the ratio between the lateral and axial strains in a similar context. The Lamé parameter μ is also known as the shear modulus and it is sometimes denoted by G. Its physical interpretation follows by considering a particular deformation described by the displacement field u = γ X 2 e1 ; see Fig. 5.3. In this case u1, 2 = γ and ui, j = 0 otherwise; furthermore, E = (γ /2)(e1 ⊗ e2 + e2 ⊗ e1 ) and, in particular, |E| = 0. Thus, Hooke’s law gives the only non-trivial equation T12 = μγ ,

(5.22)

which confirms that μ =: G determines the response of the body in shear. Next, we look at a typical situation which further illustrates the usefulness of the two new material constants E and ν. Noting that the constitutive equation (5.20) provides T as a function of E, the question we want to address next is whether or

250

5 Linear Elasticity: General Considerations and Boundary-Value Problems

not we can ‘invert’ this relationship (the question is not trivial because the right-hand side of (5.20) depends not only on E, but also on its trace, |E|). To this end, we start off by taking the trace of both sides in the constitutive equation, to get |T | = 2μ|E| + 3λ|E| 1 1 − 2ν =⇒ |E| = |T | = |T | 3λ + 2μ E

(since |I| = 3) (by Exercise 5.1) .

Solving for E in (5.20) we also have λ 1+ν ν 1 T− |E|I = T − |T |I , 2μ 2μ E E

E=

where we have used again the results of Exercise 5.1 and the expression of |E| found above. In conclusion, the inverted form of the generalised Hooke’s law for an isotropic linear elastic solid may be stated as  E=

1+ν E

 T−

ν E

 |T |I ,

Ei j =

1+ν E

 Ti j −

ν (Tkk )δi j . E

(5.23)

The result we have just proved tells us that the elastic stiffness tensor C is invertible as an element of SSym 4 . It is possible to write (5.23) in the form E = S[T ] ,

E i j = Si jkl Tkl ,

(5.24)

where the fourth-order tensor S ≡ C−1 is known as the compliance tensor. It is instructive to point out that the constitutive equation (5.20) can also be expressed in terms of the distortional and volumetric strain tensors introduced in (5.6). First, let us note that any second-order tensor T can be written as  1 1 T = T − |T |I + |T |I 3 3 

or T = T d + T s , where 1 T d := T − |T |I 3

and

T s :=

1 |T |I . 3

(5.25)

If T is the infinitesimal stress tensor, then T d and T s are referred to as the distortional and spherical stress, respectively. By elementary direct calculations one can check that, for a linearly elastic isotropic solid, distortional stresses generate distortional strains, while spherical stresses induce only volumetric strains; more precisely T s = 3K E s ,

T d = 2μE d ,

(5.26)

5.4 Linearised Constitutive Behaviour

251

where the new material constant K := λ +

2μ 3

(5.27)

is known as the bulk modulus (sometimes referred to as the modulus of compression). According to Exercise 5.1, an alternative form for (5.27) is K =

E . 3(1 − 2ν)

(5.28)

Note that K → ∞ as ν → 0.5, which is consistent with the characterisation of such materials as being incompressible. For the perfectly compressible material, ν = 0, so the bulk modulus K = E/3.

5.5 Linearised Field Equations At this juncture, we are almost ready to state the relevant equations needed to formulate the fundamental problems of (isotropic) Linear Elasticity. One of the key missing ingredients is Cauchy’s equation of motion. From Chap. 3 we know that this equation can be presented either in an Eulerian or a Lagrangian description, div σ + ρb = ρa

or

Div S + ρ0 b0 = ρ0 A ,

(5.29)

¨ where the respectively. In the first equation above the Eulerian acceleration a = u, ‘dot’ stands for the material derivative. From Eq. (2.17) in Chap. 2, u˙ =

 ∂u ∂u  + grad u · v  ∂t ∂t

=⇒

u¨ 

∂ 2u ; ∂t 2

here, v is the Eulerian velocity field, and all the neglected terms are of second order in H  1. This simple exercise confirms that in Linear Elasticity material derivatives can be replaced by ordinary partial derivatives with respect to the temporal variable ‘t’ (at a fixed point x). The reference (ρ0 ) and current mass densities (ρ) are related by the equation of continuity

 ρ0 = Jρ  1 + (∇ · u) ρ  ρ , where for the first approximation we have used (5.4), and the second one is required for the consistent linearisation of the Eulerian equation of motion (if we replace ρ → ρ0 /(1 + ∇ · u) in that equation the result will contain nonlinear terms, which is unacceptable in the current context).

252

5 Linear Elasticity: General Considerations and Boundary-Value Problems

Finally, by recalling Eq. (2.34) from Chap. 2, it is clear that div  Div. Together with (5.16), we can then replace (5.29) with just one form of those equations, ∇ · T + ρ0 b0 = ρ0

∂ 2u , ∂t 2

in Ω0 .

(5.30)

This represents the equation of motion for a linearly elastic solid; if the inertial effects are negligible, its right-hand side is ignored and (5.30) becomes the static equilibrium equation. It is worth emphasising that (5.30) is written on the original (undeformed) domain, Ω0 . However, it is also important to keep in mind that we still distinguish between the reference and current configurations, in particular x = X since otherwise u ≡ x − X will be rendered meaningless. Once the displacement field has been found, the deformed configuration at a fixed time t > 0 corresponds to the image of Ω0 under the mapping x = X + u(X, t). It is customary to omit the ‘zero’ subscript on ρ and b in (5.30), and this convention will be adopted here as well; we stress that, in general, ρ = ρ(X) and b = b(X), for X ∈ Ω0 (although, more often than not, especially in the subsequent chapters, ρ will be a constant and b ≡ 0). The key equations of Linear Elasticity can be subdivided into three sets: (a) the equation of motion (5.30); (b) the kinematic/geometric relations (5.3a) that define E (deformation) in terms of u (displacement); (c) the constitutive law (5.15) connecting T (stress) to E (deformation). It is possible to reduce these equations to a more manageable size. In the next chapter we shall re-formulate them so that the main unknowns are the components of the infinitesimal stress tensor T ; later in this chapter, an equivalent displacement equation will also be given. This displacement formulation is not tied to isotropy, so it is instructive to see how it works in the general case first. To this end, let us note that due to the symmetries of C and E, Hooke’s generalised law is equivalent to Ti j = Ci jkl uk, l ,

i, j ∈ {1, 2, 3} .

(5.31)

Once we replace (5.31) in (5.30), it transpires that   ∂ 2u j Ci jkl uk, l , i + ρb j = ρ 2 , ∂t

j = 1, 2, 3 .

(5.32)

A homogeneous elastic solid is one for which the elastic moduli Ci jkl are constant. If these quantities depend on position, that is, Ci jkl = Ci jkl (X), with X ∈ Ω0 , then the material is called inhomogeneous; thus, the properties of such a material change from place to place inside such an elastic solid. In trying to further simplify (5.32) in this latter case, one must take extra care to properly apply the product rule for derivatives to the two terms enclosed between the parentheses. We shall not elaborate on this as the matter is only tangentially relevant to our immediate goals; instead, we point out

5.5 Linearised Field Equations

253

that in the homogeneous case the Eq. (5.32) becomes Ci jkl uk, li + ρb j = ρ

∂ 2u j , ∂t 2

( j = 1, 2, 3) ;

(5.33)

this represents a system of three coupled second-order partial differential equations in the components of the displacement field u j ( j = 1, 2, 3). Going back to the general system of the Linear Elasticity, note that the number of unknowns (the components of u, E and T ) matches that of the available equations (15 in total). Unfortunately, taken on their own those equations are insufficient; one still needs to supplement them with appropriate boundary and initial conditions. Broadly speaking, the former correspond to various particular types of mechanical loading applied to the surface of the elastic body. To make this statement more precise we shall assume that ∂Ω0 = ∂Ω0u ∪ ∂Ω0t ,

∂Ω0d ∩ ∂Ω0t = ∅ .

(5.34)

On ∂Ω0u the displacements are known, while the tractions are prescribed on ∂Ω0t ,   ˚ u u = u(X, t) , ∂Ω0    n · T  t = ˚t (X, t) , ∂Ω0

X ∈ ∂Ω0u , t ≥ 0 ,

(5.35a)

X ∈ ∂Ω0t , t ≥ 0 ,

(5.35b)

where u˚ and ˚t are known vector fields. In addition to (5.35), the presence of the second-order temporal derivative in (5.30) demands two initial conditions, u(X, 0) = f init (X) , ∂u (X, 0) = g init (X) , ∂t

X ∈ Ω0 ,

(5.36a)

X ∈ Ω0 ,

(5.36b)

where, again, f init (X) and g init (X) are given vector fields. If ∂Ω0t = ∅ in (5.34) then the condition (5.35a) applies on the entire boundary and it is referred to as a displacement boundary condition, whereas the case when ∂Ω0u = ∅ is known as a traction boundary condition—see Fig. 5.4. The general case corresponding to both ∂Ω0t = ∅ and ∂Ω0u = ∅ is called a mixed boundary condition; that is, displacements are prescribed on ∂Ω0u , while tractions are given on the remainder of ∂Ω0 . It is not possible to prescribe both the displacement and the traction at the same place (which is the reason for the second condition in (5.34)). Displacement boundary conditions are usually associated with fixed (immovable) parts of the boundary; in that case u = 0 along the part that is fixed. Traction boundary conditions, on the other hand, correspond to forces that are applied to the external bounding surface of the body. Note that free surfaces of an elastic body require traction-free boundary conditions, i.e. the traction vector is prescribed to be zero

254

5 Linear Elasticity: General Considerations and Boundary-Value Problems

Ω0

Ω0 ≡ Ω0

Ω0

tracƟon boundary condiƟons

Ω0 ≡ Ω0

Ω0

Ω0

displacement boundary condiƟons

Ω0

mixed boundary condiƟons

Fig. 5.4 Types of boundary conditions corresponding to (5.34) and (5.35)

there. Strictly speaking, a traction boundary condition cannot impose constraints on all the components of the stress tensor T at the point on ∂Ω0t where it is applied. In this sense, it is not a true stress boundary condition as the next observation clarifies. For instance, if ∂Ω0t ≡ {X 2 = const.} then the outward unit normal to that surface is n = (0, ±1, 0) and t(n) ≡ n · T = ±(T12 , T22 , T23 ). These are the only components of the stress tensor that could be prescribed in this particular situation. The values of the other components of T on ∂Ω0t will be dictated by the specifics of the particular solution for the problem at hand—in other words, they are found as part of the solution. For future reference we state below the main equations comprising the system of Linear Elasticity for an isotropic solid, ⎧ ∂ 2u ⎪ ⎪ ∇ · T + ρb = ρ , ⎪ ⎪ ∂t 2 ⎪ ⎨ 1 ⎪ E = (u ⊗ ∇ + ∇ ⊗ u) , ⎪ ⎪ 2 ⎪ ⎪ ⎩ T = 2μE + λ|E|I ,

(5.37a) (5.37b) (5.37c)

and we recall that (5.37) must be solved subject to the boundary constraints (5.35) and the initial conditions (5.36). If the inertial term on the right-hand side of (5.37a) is present, we speak of linear elastodynamics. Most of the remaining chapters will be on linear elastostatics. By contrast, in this case the aforementioned inertial effect is discarded, so that (5.37a) is replaced by ∇ · T + ρb = 0 ;

(5.38)

the time dependence of the relevant fields will play no role, and there is no need to impose the initial conditions (5.36).

5.5 Linearised Field Equations

255

˚ ˚t , f init and g init (i.e. ‘given data’) the above system is For suitably chosen ρ, f , u, expected to have a solution that consists of the displacement field u, the (infinitesimal) deformation tensor E, and the (infinitesimal) stress tensorT . We shall  agree to call this solution an elastic state, and will typically write it as u, E, T . For the mathematical model of the isotropic linearly elastic solid to be a useful description of real-world phenomena, it must enjoy several basic properties: (a) existence, (b) uniqueness and (c) well-posedness. ‘Existence’ refers to whether a solution can be found for a given set of data; another aspect here is about how the differentiability properties of the elastic state are connected to similar properties of the data. These are difficult questions that require a sophisticated mathematical apparatus and will not be touched upon in this book. We shall always assume that a solution exists and will aim to produce a concrete expression for the elastic states. The second aspect mentioned above, ‘uniqueness’, is related to the number of possible elastic states for a given data. Of course, if one set of data generates more than one solution it is quite doubtful that our model does what it is supposed to. In the next section, we shall revisit briefly the topic of uniqueness in order to understand its impact on the constitutive parameters in Hooke’s law. Finally, the third property, ‘well-posedness’ deals with the continuous dependence of the elastic states on the given data; to put it differently, small variations in the data must lead to small changes in the elastic states.

5.6 Restrictions on the Elastic Constants As already mentioned at the end of the previous section, the solution of the boundaryvalue problem of Linear Elasticity must be unique. In this section, we are going to explore the restrictions imposed by that requirement on the constitutive parameters that appear in (5.37c). To make things reasonably simple, we shall ignore inertial effects (i.e. the acceleration term will be missing from our equations). We begin by considering an elastic body occupying a domain Ω0 ⊂ E3 and bounded by a surface ∂Ω0 ; the latter is assumed to consists of two disjoint parts, ∂Ω0u and ∂Ω0t , such that ∂Ω0 = ∂Ω0u ∪ ∂Ω0t . On ∂Ω0u the displacements are known ˚ while on the remaining part of ∂Ω0 the elastic body is subjected to a and equal to u, given surface traction distribution ˚t . We ask: under what condition does the elastic state generated by this loading is unique? To answer this question, we shall assume that there is more than just one elastic state. Let {u( j) , E ( j) , T ( j) } ( j = 1, 2) be two elastic states corresponding to the scenario outlined above. According to what was said in the previous section, these elastic states will be obtained by solving the following equations in Ω0 ( j = 1, 2): ⎧ ∇ · T ( j) + ρb = 0 , ⎪ ⎪ ⎨  1 ( j) u ⊗ ∇ + ∇ ⊗ u( j) , E ( j) = ⎪ 2 ⎪ ⎩ ( j) T = 2μE ( j) + λ|E ( j) |I ,

(5.39a) (5.39b) (5.39c)

256

5 Linear Elasticity: General Considerations and Boundary-Value Problems

subject to the boundary conditions ( j = 1, 2)   u( j) 

∂Ω0u

= u˚

  (n · T ( j) )

and

∂Ω0t

= ˚t .

(5.40)

The difference of the two elastic states will be yet another elastic state due to the linearity of the above equations and their boundary constraints; thus, we are motivated to introduce the new fields u := u(1) − u(2) ,

E := E (1) − E (2) ,

T := T (1) − T (2) .

By subtracting the equations and the boundary conditions satisfied by the two elastic states {u(1) , E ( j) , T ( j) } ( j = 1, 2), we get ⎧ ∇ · T = 0, ⎪ ⎨ 1 E = (u ⊗ ∇ + ∇ ⊗ u) , ⎪ 2 ⎩ T = 2μE + λ|E|I ,

(5.41a) (5.41b) (5.41c)

which must be solved subject to   u

∂Ω0u

=0

  (n · T )

and

∂Ω0t

= 0.

(5.42)

Right-multiplying (5.41a) by u and then integrating the result over Ω0 leads to 

 Ω0

∇ · (T · u) dV −

Ω0

T : (∇ ⊗ u) dV = 0 ,

(5.43)

since ∇ · (T · u) = (∇ · T ) · u + T : (∇ ⊗ u). The first term on the left-hand side of (5.43) can be transformed into a surface integral by using the Divergence Theorem (see (1.254) in Chap. 1), and then the resulting integral can be split up into two contributions because of (5.34). Thus,    n · (T · u) d A + n · (T · u) d A − T : (∇ ⊗ u) dV = 0 , ∂Ω0u

∂Ω0t

Ω0

and further use of (5.42) indicates that the first two integrals in this last equation will actually vanish, leaving us with  Ω0

T : (∇ ⊗ u) dV = 0 .

Recalling (5.41c), the integrand in the above equation can be shown to be

5.6 Restrictions on the Elastic Constants

257

T : (∇ ⊗ u) = μ(E : E) +

λ |E|2 , 2

  λ 2 μE : E + |E| dV = 0 . 2 Ω0



and therefore

(5.44)

Our original task has been reduced to finding conditions under which Eq. (5.44) implies E ≡ O. Before attacking this question we digress briefly by reviewing some basic mathematical facts relevant to the above issue. A quadratic form involving n variables z j ( j = 1, 2, . . . , n), and associated with a square matrix A = [ai j ] ∈ Mn×n (R), is the scalar expression Q ≡ Q(z) =

n  n 

ai j z i z j = z T Az ,

z = [z 1 , z 2 , . . . , z n ]T .

(5.45)

i=1 j=1

Such a quadratic form is said to be positive definite if Q(z) > 0 for all z = 0. An important result from Linear Algebra (known as Sylvester’s criterion) states that the quadratic form (5.45), with A symmetric, will be positive definite if all principal minors in the top left-hand corner of A are positive, so that Δ1 := a11 > 0 ,

  a11 a12   > 0,  Δ2 :=  a12 a22 

  a11 a12 a13    Δ3 := a12 a22 a23  > 0 , . . . . a13 a23 a33 

Going back to (5.44) it should be clear that the integrand can be regarded as a quadratic form in which z 1 ≡ E 11 , z 2 ≡ E 22 , z 3 ≡ E 33 , z 4 ≡ E 12 , z 5 ≡ E 13 , z 6 ≡ E 23 , and ⎤ ⎡ λ + 2μ λ λ 0 0 0 ⎢ λ λ + 2μ λ 0 0 0⎥ ⎥ ⎢ ⎢ λ λ λ + 2μ 0 0 0 ⎥ ⎥. ⎢ A=⎢ 0 0 4μ 0 0 ⎥ ⎥ ⎢ 0 ⎣ 0 0 0 0 4μ 0 ⎦ 0 0 0 0 0 4μ Thus, Eq. (5.44) can be cast in the form  Ω0

z T Az dV = 0 .

If Q = z T Az is positive definite then the above condition requires that z T Az = 0 and hence z = 0, i.e. E ≡ O; the constitutive equation (5.41c) shows that T ≡ O as well. Of course, under these conditions u need not be identically zero, but as we are going to see later, in Chap. 6, the displacement field associated with such elastic states corresponds to a rigid-body deformation at most.

258

5 Linear Elasticity: General Considerations and Boundary-Value Problems

The principal minors of the above matrix A are as follows: Δ1 =λ + 2μ ,

Δ2 = 4μ(λ + μ) ,

Δ4 = 4μΔ3 ,

Δ5 = 16μ2 Δ4 ,

Δ3 = 4μ2 (3λ + 2μ) , Δ6 = 64μ3 Δ5 ,

so (by Sylvester’s criterion) all these expressions must be positive. A simple exercise in elementary algebra shows that this requirement is satisfied if μ>0

and

3λ + 2μ > 0

(5.46)

or, in terms of the Young’s modulus and the Poisson’s ratio, E >0

and

−1 0 since we would expect that in the uniaxial tensile test mentioned in Sect. 5.4 the elongation of the bar will be accompanied by a decrease in the area of its normal cross sections. On the other hand, it is reassuring that both conditions in (5.46) have clearly defined physical interpretations. Since a positive shearing strain (γ > 0) must produce a positive shearing stress (T12 > 0), Eq. (5.22) leads to μ > 0. As for the second inequality in (5.46), it simply asserts that the bulk modulus is strictly positive. That this is to be expected can be verified by considering a pure pressure loading, i.e. T = − p I, acting on an elastic body. Equation (5.26a) specialised to this situation gives   d V − dv . p=K dV If p > 0 (i.e. pure compression) then the volume occupied by the body is expected to decrease (d V > dv) and the above equation yields K > 0; for p < 0 (i.e. pure expansion) one would expect that d V < dv, and then again K > 0.

5.7 The Navier–Lamé Equations This section explores an equivalent formulation for the main system of Linear Elasticity (5.37). As already pointed out earlier in Sect. 5.5, by substituting (5.37b) in (5.37c), and then using the resulting expression of T in (5.37a), one can obtain a system of three equations in the components of the displacement field u. However, without any additional simplifications, the outcome of these straightforward manipulations will assume a rather awkward form. Here, we are interested in a more optimal formulation of that result.

5.7 The Navier–Lamé Equations

259

We start by taking the divergence of both sides in the constitutive equation (5.37c), and find that   ∇ · T = 2μ(∇ · E) + λ∇ · |E|I . (5.48) Our immediate task is to rewrite the two terms on the right-hand side of (5.48) as expressions of the displacement field u only. From Chap. 1, we recall that ∇ · (φ I) = ∇ φ for any scalar field φ so, by taking φ → |E|, we find     ∇ · |E|I = ∇ |E| = ∇ ∇ · u ,

(5.49)

since |E| = ∇ · u. On the other hand, by taking the divergence of both sides in Eq. (5.37b) it turns out that   2 ∇ · E = ∇ · (u ⊗ ∇) + ∇ · (∇ ⊗ u)  = ∇ ∇ · u) + ∇ 2 u ,

(5.50)

where the last line will be justified shortly. By plugging (5.49) and (5.50) back into (5.48), we discover that   ∇ · T = (λ + μ)∇ ∇ · u + μ∇ 2 u , and then Eq. (5.37a) leads to the so-called Navier–Lamé system,   ∂ 2u μ∇ 2 u + (λ + μ)∇ ∇ · u + ρb = ρ 2 . ∂t

(5.51)

Unlike the original system (5.37) that involves both stresses and displacements, the Navier–Lamé system is formulated only in terms of displacements. Before we address the issue of boundary conditions for these new equations, we return briefly to the justification of (5.50). To this end, note that       ∇ · u ⊗ ∇ = u ⊗ ∇ i j, i e j = ui, j , i e j = ui, ji e j = ui, i j e j       = ui, i , j e j = ∇ · u , j e j = ∇ ∇ · u , and       ∇ · ∇ ⊗ u = ∇ ⊗ u i j, i e j = u j, i , i e j

  = u j, ii e j = u j e j , ii = ∇ 2 u ,

which confirm the validity of (5.50).   With the help of the identity ∇ ∧ (∇ ∧ u) = ∇ ∇ · u − ∇ 2 u from Chap. 1, the Navier–Lamé system can also be arranged in a slightly different form, namely,

260

5 Linear Elasticity: General Considerations and Boundary-Value Problems

    ∂ 2u (λ + 2μ)∇ ∇ · u − μ ∇ ∧ (∇ ∧ u + ρb = ρ 2 . ∂t

(5.52)

Of course, neither (5.51) nor (5.52) will be very useful unless the traction boundary conditions associated with (5.37) are expressed in terms of the displacement field as well. This is our next item on the agenda; let us start with a preliminary result. Proposition 5.1 If f and g are vector fields, then      f ∧ ∇ ∧ g = f · g ⊗ ∇ − g ⊗ ∇) · f .

(5.53)

As usual, to check this identity we evaluate the two sides of (5.53) separately, and then compare the results. For the left-hand side, we get successively     f ∧ ∇ ∧ g =∈i jk f j ∇ ∧ g k ei =∈i jk ∈kpq f j gq, p ei   = δi p δ jq − δiq δ j p f j gq, p ei = f j g j,i ei − f j gi, j ei , where use has been made of the ‘ε − δ’ identity. As for the right-hand side,        f · g ⊗ ∇ − g ⊗ ∇) · f = ( f p e p ) · gs,t es ⊗ et − gm,n em ⊗ en · ( f q eq ) = f p gs,t δ ps et − f q gm,n δnq em = f p g p,t et − f q gm,q em . If on the last line the summation (dummy) indexes are changed according to p → j, t → i, q → j, and m → i, we recover the expression obtained for the left-hand side. Returning to the aforementioned issue regarding the boundary conditions, we want to show that the stress vector t ≡ n · T can be expressed as a function of u. Simply using the constitutive law (5.37c),

     t = μ n · u ⊗ ∇ + (n · ∇)u + λ ∇ · u n

     = μ ∇ ⊗ u · n + (n · ∇)u + λ ∇ · u n .

(5.54)

Although this already shows that t ≡ t(u), which confirms our earlier assertion, by taking f → n and g → u in (5.53), we can express (5.54) in the equivalent form     t = λ ∇ · u n + 2μ(n · ∇)u + μ n ∧ (∇ ∧ u) .

(5.55)

5.8 Principle of Superposition Linear Elasticity is characterised by the obvious linearity of its governing equations and their associated boundary conditions. One of the key consequences is the additivity of the elastic states—a feature that makes it possible to solve complex elastostatics problems by adding up the elastic states associated with some simpler situations. This is made precise by the following formal result.

5.8 Principle of Superposition

261

Proposition 5.2 (The Superposition Principle) Let B be a linearly elastic body occupying a domain Ω0 ⊂ E3 in the reference configuration. Consider the elastic states u( j) , E ( j) , T ( j) ( j = 1, 2, . . . , m) corresponding to the elastostatic problems (5.37b)–(5.37c), (5.38), with body forces b → b( j) , and subject to   u( j) 

∂Ω0u

= u˚ ( j)

and

  [n · T ( j) ]

∂Ω0t

( j) = ˚t ,

( j = 1, 2, . . . , m) .

  Then, the elastic state u, E, T , where u≡

m 

u( j) ,

E≡

j=1

m 

E ( j) ,

T≡

j=1

∂Ω0u

=

m 

u˚ ( j)

and

j=1



 n·T 

∂Ω0t

T ( j) ,

j=1

is the solution of the same differential system, with b → boundary conditions   u

m 

=

m 

˚t ( j) ,

m j=1

b( j) and the modified

( j = 1, 2, . . . , m) .

j=1

A graphical illustration of this principle for m = 2 (in the case of a traction boundary-value problem) is shown in Fig. 5.5. A thin elastic plate is subjected to uniaxial stretching in two opposite directions—these correspond to the problems labelled (2)  and (3); note that the unloaded sides are supposed to be traction-free. If we let u( j) , E ( j) , T ( j) ( j = 2, 3) be the elastic states associated with these two   problems and u(1) , E (1) , T (1) represents the corresponding state for the problem labelled (1) then, by the above superposition result, we have u(1) = u(2) + u(3) ,

E (1) = E (2) + E (3) ,

T (1) = T (2) + T (3) .

1

Fig. 5.5 Principle of superposition at work: the elastic state of the bi-axial stretching problem (1) is the sum of the elastic states corresponding to the two simpler problems (2) and (3)

262

5 Linear Elasticity: General Considerations and Boundary-Value Problems

5.9 Saint-Venant’s Principle Our earlier discussion regarding the boundary-value problems of linear elastostatics— see Sect. 5.5, has revealed that at each point on the external surface bounding the reference configuration one must prescribe either the traction or the displacement (but not both). It is often the case that in many practical problems involving slender mechanical configurations, such as bars and plates, it is very difficult to determine the actual pointwise distribution of traction acting on any parts of the boundary, although the resultant of the applied traction can be measured accurately. This suggests that it would be desirable to replace a pointwise traction boundary condition with a weaker requirement involving only the traction resultant. But this raises the question as to whether the solution of the new problem is in any way close to the original situation. Fortunately, the answer is affirmative under certain general conditions that are spelled out below. First, we need to introduce some new terminology.   Definition 5.1 Let B be an elastic body. Consider two systems of forces f (i) i∈I   and g ( j) j∈J acting at the points Oi ∈ B (i ∈ I ) and O j  ∈ B ( j ∈ J ), respectively. These systems of forces are said to be statically equivalent (or equipollent) if the following two conditions are satisfied: 

(i)

f (i) =

i∈I



g ( j) ,

(5.56a)

j∈J

 −−→   −−−→  O Oi ∧ f (i) = O O j ∧ g ( j) ,

(ii)

i∈I

(5.56b)

j∈J

for some point O ∈ B. Condition (5.56a) expresses the fact that the two systems of forces have the same force resultants, while the (5.56b) stipulates that their resultant moments about the point O ∈ B are also identical. Proposition 5.3 (Saint-Venant’s Principle) For an elastic body B occupying a region Ω0 ⊂ E3 consider the two separate boundary-value problems of linear elastostatics consisting of the differential equations (5.37b)–(5.37c) and (5.38) subject to   u (1)

∂Ω0u

= u˚

(1)

and

where ˚t ≡ ˚t (X) and ˚t distributions, i.e. 

(2)

∂Ω0t



∂Ω0t

( j) = ˚t ,

( j = 1, 2) .

(2) ≡ ˚t (X), X ∈ ∂Ω0t , are statically equivalent traction

˚t (1) (X) d A = (1)

∂Ω0t

  (n · T )

X ∧ ˚t (X) d A =

 ∂Ω0t



∂Ω0t

˚t (2) (X) d A , (2)

X ∧ ˚t (X) d A .

(5.57a) (5.57b)

5.9 Saint-Venant’s Principle

263

    If area ∂Ω0t  area ∂Ω0 , then  (1) (1) (1)   (2) (2) (2)  u , E ,T  u , E ,T at points in Ω0 away from ∂Ω0t . In the vicinity of ∂Ω0t the two elastic states are in general different. In other words, by changing the traction boundary conditions over a small area, with a statically equivalent system of forces, will have a local effect on the corresponding elastic states in the body. Away from that area, at distances that usually exceed its linear characteristic dimension, the elastic states in the two cases will be almost indistinguishable. This very useful principle was first postulated by the French scientist B. de Saint-Venant in 1855 in his definitive studies on the torsion and pure bending of elastic bars (cylindrical bodies that are long compared to their cross-sectional linear dimensions). He introduced it as a hypothesis, stating that the difference between stresses and displacements produced by two different statically equivalent distributions of end tractions will be only a small perturbation everywhere except near the ends of the elastic bars in question.

5.10 Worked Examples Below, we illustrate one the most elementary solution strategies for certain classes of linear elastostatics problems, namely, the semi-inverse method. It consists of partially specifying the form of the displacements or stresses, and then using the equations of the linear theory of Elasticity to fix the arbitrary elements in those assumed solutions. If one succeeds in completely eliminating any degree of arbitrariness, the resulting functions will constitute the complete solution of the original problem—according to the uniqueness result discussed in Sect. 5.6. The simple examples included in this section provide a detailed look at how the semi-inverse method works. Example 5.1 We are interested to determine the elastic state in a heavy circular cylinder suspended from its upper base (see Fig. 5.6).

In the interest of simplicity it will be assumed that the length of the cylinder is > 0 and its cross-sectional area is equal to unity. Thus,    Ω0 ≡ (X 1 , X 2 , X 3 ) ∈ E3  0 < X 3 < , (X 1 )2 + (X 2 )2 < π −1 , and ∂Ω0 = ∂Ω − ∪ ∂Ω0lat ∪ ∂Ω + , where ∂Ω0− ≡ {X 3 = 0} , ∂Ω0lat ≡ {0 < X 3 < , (X 1 )2 + (X 2 )2 = π −1 } , ∂Ω0+ ≡ {X 3 = } .

264

5 Linear Elasticity: General Considerations and Boundary-Value Problems

Fig. 5.6 Heavy elastic cylinder hanging under its own weight

reference configuration

deformed configuration

According to Saint-Venant’s Principle the reaction loads acting on ∂Ω0+ are equivalent with a resultant traction R0 = ρg e3 (the weight of the cylinder), where g represents the gravity acceleration constant. Since the applied forces act along the cylinder axis (O X 3 ), we expect that at each point in Ω0 the traction vector t(e3 ) = (T31 , T32 , T33 ) will dominate over both t(e2 ) and t(e1 ). Furthermore, it is plausible to assume that T31 = T32  0 as well, so the only ‘active’ component of T is T33 . In conclusion, it will be assumed right from the outset that T33 = 0 , Ti j ≡ 0 ,

for (i, j) = (3, 3) .

(5.58)

Since the body force (self-weight in this case) is equal to (−ρge3 ), the equilibrium equation (5.38) becomes ∇ · T − ρge3 = 0 and, in light of (5.58), it is trivially reduced to just one equation that is readily integrated; more specifically, ∂ T33 − ρg = 0 ∂ X3

=⇒

T33 = ρg X 3 + C ,

(5.59)

for some arbitrary constant C ∈ R. Since ∂Ω0− is traction free, i.e. t(−e3 ) = 0 on ∂Ω0− , by making use of (5.59) it transpires that C = 0. Although we have now fully determined T (by a semi-inverse approach), we still need to check that the remaining boundary conditions are satisfied. To this end, note that the unit normal to ∂Ω0lat is a two-dimensional vector n = (n 1 , n 2 , 0), where the precise values of n j ( j = 1, 2) are not required here. Since the lateral surface is traction free, we need to check that t(n) ≡ n · T = 0 or T11 n 1 + T21 n 2 = 0 , T12 n 1 + T22 n 2 = 0 , T13 n 1 + T23 n 2 = 0

on ∂Ω0lat ,

5.10 Worked Examples

265

which are trivially satisfied due to (5.58). Finally, integrating T33 in (5.59) over ∂Ω0+ we get that t(e3 ) = R0 on the upper face of the cylinder. By invoking the uniqueness result of Sect. 5.6 we can therefore conclude that T = ρg X 3 e3 ⊗ e3 represents the complete stress solution for the proposed problem. The deformation of the cylinder is immediately found by using the inverted form of Hooke’s law (5.23), E 11 = E 22 = −

νρg X3 , E

E 33 =

ρg X3 , E

E i j = 0 , (i = j) .

We can now retrieve the displacement components u j = u j (X) ( j = 1, 2, 3) by integrating the expressions of E i j and remembering the definition (5.3a). This leads to the following partial differential equations: νρg ∂u1 X3 , =− ∂ X1 E ∂u1 ∂u2 + = 0, ∂ X2 ∂ X1

∂u2 νρg X3 , =− ∂ X2 E ∂u2 ∂u3 + = 0, ∂ X3 ∂ X2

∂u3 νρg X3 , = ∂ X3 E ∂u3 ∂u1 + = 0, ∂ X1 ∂ X3

whence3 u1 = −

 νρg νρg ρg 2 X 1 X 3 , u2 = − X 2 X 3 , u3 = X − 2 + ν(X 12 + X 22 ) . E E 2E 3

In obtaining these expressions we have left out a possible rigid-body deformation of the bar. This can be justified by assuming that both u (displacement) and Ω (infinitesimal rotation) are zero at the point (0, 0, )—this is the same as saying that the cylinder is rigidly fixed at the centre of its top face. The deformation undergone by the cylinder is x = X + u or, in component form   νρg νρg X3 , X3 , x2 = X 2 1 − x1 =X 1 1 − E E

 ρg (X 32 − 2 ) + ν(X 12 + X 22 ) . x3 = X 3 + 2E Some observations regarding these last expressions are in order. Points that were situated on the axis of the cylinder before deformation (X 1 = X 2 = 0) remain on the same axis after the deformation has taken place, being shifted by an amount equal to u3 = ρg(X 32 − 2 )/(2E) < 0 (i.e. downward). On the other hand, points that before the deformation were situated in a normal cross section X 3 = k, (0 < k < ) are no longer in a plane after; they are found on a paraboloid of equation

3 The

general problem of determining the displacement field from the strain tensor will be taken up in greater detail in the next chapter. In this example the integration of the kinematic equations is elementary; for details of a similar calculation see Example 6.3.

266

5 Linear Elasticity: General Considerations and Boundary-Value Problems

Fig. 5.7 Pure bending of a cylindrical bar subjected to applied terminal couples; the generic transverse cross section Σ is normal to the axis of the bar, but need not be circular

x3 = k +

νρg −2 2 ρg 2 νρg  (k − 2 ) + 1− k (x1 + x22 ) . 2E 2E E

Example 5.2 The end faces of a long cylindrical elastic bar of length 2L (L > 0) are acted upon by systems of forces that are statically equivalent to two couples whose plane of action passes through the axis of the bar; the moments of the couples are equal in magnitude and have opposite signs. Assuming that the lateral surface of this cylindrical body is traction free, let us determine the elastic state in the bar. We start by choosing a Cartesian system of coordinates O X 1 X 2 X 3 such that O X 3 coincides with the axis of the bar and the origin is symmetrically positioned with respect to the end faces (for simplicity we shall also assume that the end planes are normal to the bar axis). The O X 1 and O X 2 axes are taken to be the principal centroidal axes of inertia of a cross section; see Fig. 5.7 for a sketch of this situation. By the results in Appendix C, 

 Σ

X1 d A =

 Σ

X 2 d A = 0 and

Σ

X1 X2 d A = 0 ,

(5.60)

where Σ ⊂ E2 denotes a typical transverse cross section, and the coordinates X 1 , X 2 are measured with respect to the centroid of Σ. We make use of the semi-inverse method and take T to be a tensor with Ti j = 0 , (i, j) = (3, 3) ,

T33 = α X 1 ,

(5.61)

5.10 Worked Examples

267

where α ∈ R is a constant that we aim to determine, and X 1 is measured in the cross section in which T33 is calculated (relative to the corresponding centroid). To formulate the boundary conditions note that the domain occupied by the bar corresponds to    Ω0 ≡ (X 1 , X 2 , X 3 ) ∈ E3  (X 1 , X 2 ) ∈ Σ, −L ≤ X 3 ≤ L ; its bounding surface is ∂Ω0 = ∂Ω0− ∪ ∂Ω0lat ∪ ∂Ω0+ , where ∂Ω0lat ≡ ∂Σ × (−L , L) ,

  ∂Ω0± ≡ X 3 = ±L .

The boundary conditions can then be cast in the form  

∂Ω0±

∂Ω0±

ζ ± ∧ t(±e3 ) d A = ±M e2 ,

(5.62a)

t(±e3 ) d A = 0 ,

(5.62b)

t(n) = 0 ,

on ∂Ω0lat ,

(5.62c)

where ζ ± ≡ X 1 e1 + X 2 e2 ∈ ∂Ω0± and n is the outward unit normal on ∂Ω0lat . Condition (5.62c) is automatically satisfied because of (5.61), while (5.62b) follows from our choice of the X β -axes (β = 1, 2). It remains to see what new information transpires from (5.62a); expanding the vector product we find       X 12 d A + e1 α X 1 X 2 d A = M e2 , −e2 α Σ

Σ

and then use of the last equation in (5.60) leads to M α=− , I

 where I ≡

Σ

X 12 d A

denotes the second moment of area of the cross section with respect to the X 2 -axis. By invoking the uniqueness result of Sect. 5.6 we can therefore conclude that T = −(M/I )X 1 e3 ⊗ e3

(5.63)

represents the complete stress solution for the proposed bending problem. The deformation of the cylinder is immediately found by using the inverted form of Hooke’s law (5.23), E 11 = E 22 = −

ν νM T33 = X1 , E EI

E 33 = −

M X1 , EI

E i j = 0 , (i = j) .

268

5 Linear Elasticity: General Considerations and Boundary-Value Problems

To find the displacements we follow the same route as in the previous example. The kinematic conditions that require integration are νM ∂u2 νM ∂u3 M ∂u1 X1 , X1 , X1 , = = =− ∂ X1 EI ∂ X2 EI ∂ X3 EI ∂u1 ∂u2 ∂u2 ∂u3 ∂u3 ∂u1 + = 0, + = 0, + = 0, ∂ X2 ∂ X1 ∂ X3 ∂ X2 ∂ X1 ∂ X3 whence u1 =

 M 2 νM M X + ν(X 12 − X 22 ) , u2 = X 1 X 2 , u3 = − X1 X3 , 2E I 3 EI EI

after discarding the rigid-body deformation terms. This completes the solution of the second example and we can now make some remarks about the results obtained. First, notice that T33 = 0 along the axis of the bar (X 1 = X 2 = 0), which is also known as the neutral axis. Material lines parallel to the neutral axis4 and situated above it will be subject to compressive stresses (T33 < 0), while those below it will experience stretching (T33 > 0). The neutral axis neither lengthens nor shortens as a result of the pure bending deformation, hence its name. In fact, all material lines parallel to O X 3 and situated in the X 2 X 3 -plane will share the same property; O X 2 X 3 is called the neutral plane. For a point on the neutral axis, the displacement is u0 = (u10 , u20 , u30 ), where u10 =

M 2 X , 2E I 3

u20 = u30 ≡ 0 ,

(5.64)

showing that it deflects in the O X 1 X 3 plane, known as the bending plane. Notice that in the particular bending scenario discussed here the line of centroids of the bar deflects in the plane of the applied couple. However, this is a rather special result and, in general, the plane of bending and the plane of the couple will be different. For a curve given in Cartesian coordinates as y = y(x), the curvature κ and the radius of curvature R at a point (x0 , y0 ) can be calculated from the well-known formula   

1 y  (x0 ) κ (x0 ,y0 ) ≡  = = y  (x0 ) 1 + O(y  (x0 ))2 ,  2 3/2 R (x0 ,y0 ) [1 + (y (x0 )) ] where the dash stands for differentiation with respect to ‘x’ and the last equality above is valid if |y  (x0 )|  1. If we make the substitutions y → u10 and x → X 3 in (5.64) then, to first-order, we get M 1 = R EI 4 These

=⇒

are sometimes referred to as fibres.

M=

EI R

or M = E I κ ,

(5.65)

5.10 Worked Examples

269

which is known as the Euler–Bernoulli bending law. This approximate result (widely used in engineering approximations) links the magnitude of the applied terminal bending moments (M) to the curvature of the neutral axis (κ), as well as the mechanical (E) and geometrical properties of the cross section (I ); the quantity E I is called the flexural rigidity of the bar. Also, note that for Euler–Bernoulli bending the neutral axis deforms into an arc of circle of radius R ≡ E I /M. Finally, we remark in passing that Eq. (5.64) shows that any cross section X 3 = constant transforms into a plane section after deformation.

Example 5.3 Let us consider the classical problem of finding the stress distribution in a thick spherical shell subjected to surface forces resulting from the presence of a fluid at pressure p1 > 0 inside and pressure p2 > 0 outside the shell (for simplicity it will be assumed that there are no body forces).

A sketch of the situation we are interested in is included in Fig. 5.8, where the bounding surfaces of the shell are shown as the thick concentric circles. Due to the spherical symmetry of the problem it is advantageous to use sphericalcoordinates  whose origin is at the centre of the shell; the usual basis vectors are er , eθ , eϕ . With these conventions in mind, the domain occupied by the shell is    Ω0 ≡ (r, θ, ϕ)  R1 < r < R2 , 0 ≤ θ ≤ π, 0 ≤ ϕ < 2π , while the two bounding surfaces can be collectively represented as     ∂Ω0 ≡ r = R1 ∪ r = R2 . By using the semi-inverse method, the displacement field will be taken to be spherically symmetric, i.e. (5.66) u = f (r )er , for some function f = f (r ) that is unknown at this stage. The boundary-value problem for the Navier–Lamé system in this particular geometry consists of     (λ + 2μ)∇ ∇ · u − μ ∇ ∧ (∇ ∧ u) = 0

in Ω0 ,

(5.67)

subject to the boundary conditions        − p1 er if r = R1 , λer ∇ · u + 2μ er · ∇ u + μ er ∧ (∇ ∧ u) = − p2 er if r = R2 . In spherical coordinates—see Eq. (1.251),

(5.68)

270

5 Linear Elasticity: General Considerations and Boundary-Value Problems

outward unit normal

elasƟc material

O

Fig. 5.8 Thick spherical shell subjected to hydrostatic pressure, both internally and externally. The geometry of the shell is defined in the left sketch, while the loading is illustrated on the right-hand side. Note that the loading on both spherical boundaries has opposite directions to the outward unit normals (hence the minuses in front of p1 and p2 )

    df 2f + ∇ · u = ∇ · ( f er ) = ∇ f · er + f ∇ · er = dr r

(5.69)

and  df  ∇ ∧ u = (∇ f ) ∧ er + f ∇ ∧ er = (5.70) er ∧ er + f (er ∧ er ) = 0 . dr   Also, (er · ∇)u = er · ∇ ⊗ u , and to simplify this expression we still need to compute ∇ ⊗ u in spherical coordinates (but only for the special choice (5.66)); this is done below,       ∇ ⊗ u = ∇ ⊗ f er = ∇ f ⊗ er + f ∇ ⊗ er     ∂ df ∂ 1 ∂ 1 er ⊗ er + f er + eθ + eϕ ⊗ er = dr ∂r r ∂θ r sin θ ∂ϕ    df f  er ⊗ er + eθ ⊗ eθ + eϕ ⊗ eϕ = dr r     df f f I+ − er ⊗ er . (5.71) = r dr r On making use of (5.69) and (5.70), the Navier–Lamé system reduces to r2

d2 f df −2f = 0, + 2r 2 dr dr

5.10 Worked Examples

271

which is easily recognised as being an Euler-type ordinary differential equation; its solution is readily found, f (r ) = Ar +

B , r2

A, B ∈ R (arbitrary) .

(5.72)

To fix the arbitrary constants in the expression of f (r ) we need to bring in the boundary conditions. Use of (5.69), (5.70) and (5.71) in (5.68) leads to  λer

df 2f + dr r

 + 2μer ·

     f df f − I+ er ⊗ er = − p j er r dr r

for r = R j ( j = 1, 2), or df (λ + 2μ) + 2λ dr

   f − p1 if r = R1 , = r − p2 if r = R2 .

Finally, replacing (5.72) in this last equation yields a linear system for A and B,  A(3λ + 2μ) −

4μ R13



 B = − p1 ,

A(3λ + 2μ) −

4μ R23

 B = − p2 ,

whence 1 A= 3λ + 2μ



p1 R13 − p2 R23 R23 − R13

 and

R3 R3 B= 1 2 4μ



p1 − p2 R23 − R13

 .

In conclusion, the displacement in the pressurised shell is u=

1 3 R2 − R13



   p1 R13 − p2 R23 R 3 R 3 p1 − p2 r + 12 2 er ; 3λ + 2μ r 4μ

(5.73)

the strains and stresses can be routinely found by using the formulae (5.37b) and (5.37c).

5.11 Standard Simplifications All the general theory developed thus far in this chapter has been of a general nature and is applicable to any problem concerning linearly elastic isotropic bodies. However, without introducing additional simplifying assumptions the number of theoretical solutions of the general boundary-value problem (5.35), (5.37b), (5.37c) and (5.38) is fairly limited (or the mathematical apparatus required is well beyond the scope of an introductory text). Fortunately, there are several main classes of prob-

272

5 Linear Elasticity: General Considerations and Boundary-Value Problems

Fig. 5.9 Long prismatic body: the geometry of the plane strain approximation typical cross-secƟon

loading independent of

lems amenable to a well-established dimensional reduction that makes it possible to formulate general solution strategies for the corresponding simplified boundaryvalue problems. We outline these simplifications below, while the next chapters will amplify this succinct treatment with more analytical details and a range of examples. Since in Linear Elasticity there is no need to distinguish between the reference and current configurations, henceforth (until further notice) we shall use the Greek indices to range from 1 to 2, while Latin indices will still range from 1 to 3. For instance, X α eα = X 1 e1 + X 2 e2 , while X i ei = X 1 e1 + X 2 e2 + X 3 e3 .

5.11.1 Plane Strain This approximation is applicable to long prismatic or cylindrical bodies bounded by plane ends normal to their generators; the applied loading and boundary conditions are also required to be uniform in the direction of the generators. To make this statement more precise, consider the situation of the elongated prismatic solid in Fig. 5.9, in which the generators are parallel to a given vector a ∈ V . By choosing a Cartesian system of coordinates O X 1 X 2 X 3 for which the X 3 -axis is parallel to a, the assumption of plane strain requires that the deformation be independent of X 3 . It is thus postulated that u3 ≡ 0 , and therefore

uα = uα (X 1 , X 2 ) , ⎡

E 11 [E] = ⎣ E 21 0

E 12 E 22 0

(α = 1, 2) , ⎤ 0 0⎦ , 0

(5.74)

(5.75)

with E αβ = E αβ (X 1 , X 2 ) for α, β = 1, 2. It should be clear that the deformations experienced by this elastic body are confined to the planes X 3 =constant, like the shaded transverse cross section Σ ⊂ E2 in the aforementioned figure.

5.11 Standard Simplifications

273

loading independent of

middle cross-secƟon (shaded)

Fig. 5.10 A plate-like elastic solid occupying a thin domain in E3 , bounded by the flat surfaces X 3 = ±h/2. The deformation of the midplane (X 3 = 0) can be described by a plane stress approximation

There is one caveat to what was said above. The conditions in (5.74) have an approximate character: to satisfy the constraint u3 = 0 on the end faces would require appropriate tractions to be imposed on those planes, and in a practical situation this may be difficult to arrange. If the length of the body is large compared to the linear characteristic dimensions of the cross section, Saint-Venant’s Principle assures us that a true state of plane strain does develop sufficiently far away from the end faces.

5.11.2 Plane Stress This approximation deals with exactly the opposite situation described above. It is applicable to thin plates, which can be regarded as very short cylindrical bodies (see Fig. 5.10). The flat end faces are traction free and the plate is assumed to be loaded only in its middle plane (a plane that is symmetrically placed with respect to the end faces; in our sketch the distance between the middle plane and the end faces is constant and equal to h/2). Choosing a Cartesian system of coordinates in which the X 3 -axis is normal to the midplane, the underlying assumptions of the plane stress approximation can be stated in the form Ti3 = 0 , (i = 1, 2, 3)

Tαβ = Tαβ (X 1 , X 2 ) , (α, β = 1, 2) , (5.76)

and

were T denotes the infinitesimal stress tensor. Written in matrix form, the component representation of T is qualitatively identical to that seen earlier in (5.75), ⎡

T11 [T ] = ⎣T21 0

T12 T22 0

⎤ 0 0⎦ . 0

(5.77)

274

5 Linear Elasticity: General Considerations and Boundary-Value Problems

The first condition in (5.76) is justified by the fact that the components T13 , T23 and T33 are typically small compared to the remaining components of the stress tensor. Noting that the outward unit normals on the end faces of the plate are n = (0, 0, ±1) and t(n) ≡ n · T must vanish there, it follows immediately that T13 = T23 = T33 = 0 at those points. If the plate is thin it is a reasonable assumption to take these stress components as being small throughout the plate.

5.11.3 Antiplane Strain/Stress In the first case (antiplane strain) the main assumption is that uα ≡ 0 , (α = 1, 2)

and

u3 = u3 (X 1 , X 2 ) ,

(5.78)

which gives the following deformation and stress tensors ⎡

⎤ 0 0 E 13 [E] = ⎣ 0 0 E 23 ⎦ E 31 E 32 0

⎡

and

⎤ 0 0 T13 [T ] = ⎣ 0 0 T23 ⎦ , T31 T32 0

(5.79)

with E 3α = E α3 = E α3 (X 1 , X 2 ) and T3α = Tα3 = Tα3 (X 1 , X 2 ) for α = 1, 2. Antiplane stress represents a slightly more general assumption, in the sense that T is postulated to have the form Tαβ = 0 , (α, β = 1, 2) ,

Ti3 = Ti3 (X) , (i = 1, 2, 3) .

(5.80)

In fact, the dependence of the non-zero stress components on X turns out to be much simpler, as can be easily inferred from the static equilibrium equations. In the absence of body forces, ∇ · T = 0 becomes T31, 3 = 0 , T32, 3 = 0 , T13, 1 + T23, 2 + T33, 3 = 0 ,

(5.81)

so T31 and T32 are functions of X α (α = 1, 2), while T33 is a linear function of X 3 (and which may depend on X 1 and X 2 as well). The antiplane stress approximation is particularly relevant in problems involving the deformation of bars under terminal loads, a typical configuration being that discussed in Example 5.2 of the previous section. Below, we give a brief classification of such antiplane problems. The main task is to find the elastic state in a cylindrical bar of arbitrary cross section, assuming that its lateral surface is traction free and the bar is loaded by surface forces applied on its end sections. All these problems are subject to Saint-Venant’s Principle, in the sense that the applied loads are taken to be statically equivalent to a resultant force R and a resultant moment M O (with respect to some point O). Although, in general, these vectors are allowed to have arbitrary orientations relative to the normal cross sections of the bar, they can be resolved into normal and tangential

5.11 Standard Simplifications

275

(b)

(a)

(c)

(d)

Fig. 5.11 Typical situations where the anti-plane stress hypothesis is applicable: a uniaxial tension, b torsion, c pure bending, and d vertical bending (or shear)

components to those cross sections. Then, by considering the superposition principle, the particular solutions corresponding to the simplified loads can be added together to build the solution for the general loading scenario. Historically, the particular solutions mentioned above are divided into four broad classes, (a) Uniaxial tension :

MO = 0

and n ∧ R = 0 ,

(5.82a)

(b) Torsion : (c) Pure bending :

R=0 R=0

and n ∧ M O = 0 , and n · M O = 0 ,

(5.82b) (5.82c)

(d) Vertical bending :

MO = 0

and n · R = 0 ;

(5.82d)

here, n is the unit normal to a typical transverse cross section. A pictorial summary of these basic cases is included in Fig. 5.11. We have already touched upon the uniaxial tension case, as well as a particular case of pure bending. The torsion problem will be taken up in greater detail in Chap. 7.

5.12 Exercises 5.1. Show that 2μ =

E , 1+ν

and then deduce that

λ=

Eν , (1 + ν)(1 − 2ν) λ ν = . 2μ 1 − 2ν

3λ + 2μ =

E , 1 − 2ν

276

5 Linear Elasticity: General Considerations and Boundary-Value Problems

5.2. Prove that the components of the compliance stress tensor S in (5.24) can be written as ν 1+ν (δik δ jl + δil δ jk ) . Si jkl = − δi j δkl + E 2E 5.3. Use Exercise 5.1 together with (5.28) to check the following relations, λ= 5.4.

3ν K , 1+ν

K =

GE 3K E , G= , 3(3G − E) 9K − E

E=

G(3λ + 2G) . λ+G

a. Within the approximations of Linear Elasticity show that J=



det(I + 2E) ,

where E is the infinitesimal deformation tensor defined in (5.3a). b. Show that det (I + 2E) = 1 + 2I E + 4I I E + 8I I I E , where I E , I I E and I I I E are the principal invariants of E. Hence establish that the volumetric dilation is given by the trace of E. 5.5. A fourth-order Cartesian tensor C is called strongly elliptic if (a ⊗ b) : C[a ⊗ b] > 0 ,

for all a = 0 , b = 0 .

(5.83)

Show that the elasticity tensor for isotropic solids (5.19) is strongly elliptic if and only if μ>0 and λ + 2μ > 0 . (5.84) 5.6. A fourth-order Cartesian tensor C is said to be positive definite if A : C[ A] > 0 ,

for all A ∈ Lin ,

AT = A = O .

(5.85)

In Sect. 5.6 it was established by using Sylvester’s criterion that the linear elasticity tensor for isotropic solids must be positive definite in order to guarantee the uniqueness of the linear elastostatics boundary-value problem. Prove the positive definiteness of the same tensor directly by using the definition (5.85). 5.7. Recall from Chap. 4 that the constitutive law for an isotropic elastic solid can be represented as (5.86) T = β0 I + β1 B + β−1 B −1 , where B ≡ F · F T is the left Cauchy–Green deformation tensor and β j = β j (I B , I I B , I I I B ) (for j = 0, 1, −1) are scalar functions of the principal invariants of B. a. Show that for a stress-free reference configuration the counterpart of (5.10) is

5.12 Exercises

277

β0 (3, 3, 1) + β1 (3, 3, 1) + β−1 (3, 3, 1) = 0 . b. If H  1 then I B = 3 + 2|E| + . . . , I I B = 3 + 4|E| + . . . , I I I B = 1 + 2|E| + . . . , where E is the infinitesimal strain tensor (5.3a) and the dots stand for the  ignored O H2 terms. c. Taking into account that, to first-order, B  I + 2E and B −1  I − 2E ,   use the Taylor expansion of (5.86), up to and including O H terms, to show that one can recover Hooke’s law (5.20), with    ∂Ψ  ∂Ψ  ∂Ψ  λ := 2 +4 +2 , ∂ I B 0 ∂ I I B 0 ∂ I I I B 0  Ψ := β0 + β1 + β−1 , μ := (β1 − β−1 )0 , where the above functions and derivatives are evaluated for the stress-free state (this is indicated by the vertical bar and the zero subscript). 5.8. Consider again the situation in Example 5.3. Assuming that there is only an internal pressure and the outer surface of the shell is traction free, show that (5.73) becomes u = u(r )er , with u(r ) =

  p1 R13 R23 1 (1 + ν) . (1 − 2ν)r + 2 r2 E(R23 − R13 )

(5.87)

a. Show that in this case the stresses in the shell are given by Trr =

p1 R13 3 R2 − R13

 1−

R23 r3

 ,

Tθθ = Tϕϕ =

p1 R13 3 R2 − R13

  R3 1 + 23 . 2r

b. From the above expressions deduce the ‘thin-shell’ formulae Trr  p1 (r − R2 )/ h , Tθθ = Tϕϕ 

1 1 p1 R1 / h , u(r )  (1 − ν) p1 R12 /Eh . 2 2

5.9. Consider again the thick spherical shell from Example 5.3, and assume that now there is an internal pressure on the inner surface r = R1 , while on the outer surface r = R2 the deformation is arrested by a surrounding spherical rigid wall. a. Use the same spherical symmetry assumption to show that

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5 Linear Elasticity: General Considerations and Boundary-Value Problems

u(R1 ) =

R2 p1 R1 (1 − 2ν)(η3 − 1) , η := . E{1 + 2η3 [(1 − 2ν)/(1 + ν)]} R1

b. By finding the stresses in the shell, check the formulae 3 p1 (1 − ν) , (1 + ν) + 2η3 (1 − 2ν) νTrr (R2 ) . Tθθ (R2 ) = Tϕϕ (R2 ) = 1−ν

Trr (R2 ) = −

5.10. An isotropic elastic material with Lamé coefficients λ1 , μ1 occupies a spherical region 0 ≤ r ≤ R1 , and another elastic material with Lame coefficients λ2 , μ2 occupies the region R1 ≤ r ≤ R2 (for some 0 < R1 < R2 ). The spherical surface r = R2 is subject to a uniform pressure p > 0. If the body force is absent, assuming that the stress and displacement fields are continuous across the spherical surface r = R1 , find the equilibrium displacement in both regions. 5.11. Consider the same situation as in Example 5.2, but this time assume that the resultant moment of the applied couple is M = M1 e1 + M2 e2 (see Fig. 5.12), where e1 and e2 define the principal axes of inertia of a cross section. a. Use the superposition principle to obtain the elastic state in the bar. b. Find the angle between the direction of M and the plane of bending. 5.12.

a. In Example 5.2, show that the bar experiences a constant bending moment M on every cross section. b. (Anticlastic curvature) The solution given in the example mentioned above is valid for cylindrical or prismatic beams having arbitrary cross sections; in particular, the solution given in Sect. 5.10 applies to the rectangular geometry in Fig. 5.13. Show that the material line elements (‘fibres’) parallel to the X 2 -axis acquire

Fig. 5.12 Pure bending of a cylindrical elastic bar subjected to applied terminal couples having a moment resultant M = M1 e1 + M2 e2 (the cross sections need not be circular)

5.12 Exercises

279

bending plane neutral plane

Fig. 5.13 Pure bending for a rectangular beam

Fig. 5.14 Deformed configuration of the axial section (left) and the transverse cross section (right) for the rectangular beam seen in Fig. 5.13. The material line elements (‘fibres’) parallel to the X 2 -axis acquire an anticlastic curvature

a curvature in the opposite sense of the X 3 -axis—as seen in Fig. 5.14. Determine the relationship between the corresponding radii of curvature. Furthermore, convince yourself that the ‘fibres’ parallel to the X 1 -axis are rotated, but remain straight. 5.13. Show that the strain-energy density function (5.17) can be expressed in terms of the first two principal invariants of the infinitesimal strain tensor E, namely   λ 2 I − 2μI I E . W = W# (I E , I I E ) := μ + 2 E

(5.88)

Calculate ∂ I E /∂ E and ∂ I I E /∂ E, and then deduce from (5.88) the generalised Hooke’s law (5.20).

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5 Linear Elasticity: General Considerations and Boundary-Value Problems

Bibliography 1. Barber JR (2002) Elasticity. Kluwer Academic Publishers, Dordrecht, The Netherlands 2. Biezeno CB, Grammel R (1955) Engineering dynamics, vol 1 (Theory of Elasticity). Blackie & Son Ltd., London and Glasgow 3. Biezeno CB, Grammel R (1956) Engineering dynamics, vol 2 (Elastic Problems of Single Machine Elements). Blackie & Lon Ltd., London and Glasgow 4. Chou PC, Pagano NJ (1967) Elasticity: tensor, dyadic, and engineering approaches. D. Van Nostrand Company Inc, Princeton, New Jersey 5. Filonenko-Borodich M (1963) Theory of elasticity. Mir Publishers, Moscow 6. Leipholz H (1974) Theory of elasticity. Noordhoff International Publishing, Leyden 7. Lekhnitskii SG (1981) Theory of elasticity of an anisotropic body. Mir Publishers, Moscow 8. Little RW (1973) Elasticity. Prentice-Hall Inc, Englewood Cliffs, New Jersey 9. Love AEH (2011) A treatise of the mathematical theory of elasticity. Dover Publications, Mineola, New York 10. Malvern LE (1969) Introduction to the mechanics of a continuum medium. Prentice-Hall Inc, Englewood Cliffs, New Jersey 11. Nadeau G (1964) Introduction to elasticity. Holt, Rinehart and Winston Inc, New York 12. Teodorescu PP (2013) Treatise on classical elasticity. Springer, Dordrecht, The Netherlands 13. Timoshenko SP, Goodier JN (1970) Theory of elasticity, International edn. McGraw-Hill Book Company, Auckland

Chapter 6

Compatibility of the Infinitesimal Deformation Tensor

Abstract Our next main task will be to investigate how the displacement field can be recovered if the infinitesimal stress or strain tensors are known. Although the scope of this question turns out to be rather broad, the gist of what we are about to do is contained in the following basic scenario. Assume that D ⊂ E2 is an open set and g j ∈ C 1 (D) are given scalar fields ( j = 1, 2); we are interested in finding a new scalar field φ ∈ C 2 (D) such that φ , 1 = g1 and φ , 2 = g2 in D. Of course, if such a field exists then φ , 12 = φ , 21 or g1, 2 = g2, 1 . This latter condition is necessary for the existence of a φ having the foregoing stated properties. To put it differently, the obtained condition ensures the compatibility (or consistency) of the two equations satisfied by φ. It is fairly straightforward to show that under certain circumstances the condition is also sufficient. We note in passing that if g := (g1 , g2 , 0), the compatibility condition in this basic case can be cast as ∇ ∧ g = 0. The extension of these naive calculations to vector and tensor fields (as explained in the next sections) leads naturally to a discussion of the Beltrami–Michell equations and the concept of Weingarten-Volterra dislocation in multiply connected linearly elastic bodies.

6.1 Introduction As we have already seen in the previous chapter (e.g. Examples 5.1 and 5.2), in certain Linear Elasticity problems the infinitesimal deformation tensor E is obtained before the displacements are known. In such situations finding the associated displacement field u = ui ei requires the integration of the following system of first-order PDEs ∂u1 = E 11 , ∂ X1 ∂u2 = E 22 , ∂ X2 ∂u3 = E 33 , ∂ X3

  1 ∂u1 ∂u2 = E 12 , + 2 ∂ X2 ∂ X1   1 ∂u2 ∂u3 = E 23 , + 2 ∂ X3 ∂ X2   1 ∂u3 ∂u1 = E 31 , + 2 ∂ X1 ∂ X3

© Springer Nature B.V. 2020 C. D. Coman, Continuum Mechanics and Linear Elasticity, Solid Mechanics and Its Applications 238, https://doi.org/10.1007/978-94-024-1771-5_6

(6.1a) (6.1b) (6.1c)

281

282

6 Compatibility of the Infinitesimal Deformation Tensor

Fig. 6.1 Geometrical interpretation of compatible (top) and incompatible (bottom) deformation fields in a linearly elastic solid (the original square is assumed to be infinitesimal); see text for explanations

which follow immediately from the definition of E. The number of unknowns (3) is less than the number of equations available (6), so it must be intuitively clear that a solution of such a system will not always be possible unless some additional conditions are satisfied; by analogy with similar situations encountered in Linear Algebra, we say that the system (6.1) is overdetermined. It is the aim of this chapter to establish necessary and sufficient conditions under which such an over-determined system is solvable, and to provide a closed-form expression for the vector field u in terms of the tensor field E. Before we embark on more technical issues, we illustrate heuristically the idea of strain compatibility in Linear Elasticity with the help of the following thought experiment. Let us consider an imaginary (infinitesimally small) two-dimensional square sheet of metal on which we draw a grid as seen in the leftmost part of Fig. 6.1. This configuration is then subjected to an infinitesimal deformation characterised by a strain tensor E. It is further assumed that, as a result of this process, the grid lines remain straight and parallel with each other (but their lengths and the angles between them may change); one such possibility can be seen in the upper part of Fig. 6.1, the first sketch.1 Note that we have chosen a simple shear deformation for the sake of simplicity, but this is not essential. If we now cut the metal sheet along the deformed grid lines and separate the pieces apart (i.e. we superimpose a rigid-body transformation/displacement on our deformed pieces) then, provided that the cutting process did not introduce any additional strains, we end up with the situation seen in the upper part of Fig. 6.1, second configuration. It should be obvious that starting from this state, we can recover the previous one, in the sense that the individual pieces will fit together perfectly (no gaps will be present—see last sketch, upper part of the same Figure). The strain field of this configuration is an example of compatible strain

1 The

deformation is exaggerated here in order to help make the point.

6.1 Introduction

283

field, in the sense that it has its origin in a continuous and single-valued displacement field (modulo the aforementioned rigid-body transformation). On the other hand, consider the different situation seen in the lower part of Fig. 6.1. The same square sheet of metal is first cut along the grid lines and subjected to a rigid-body transformation. Following that, the individual pieces undergo independent deformations. If we subsequently try ‘putting them together’, in general, they will not match perfectly, i.e. there will be a mismatch between adjacent sides as shown in the last sketch in the lower part of Fig. 6.1. In this scenario, the strain field experienced by the ensemble of individual pieces is not compatible; that is, there is no continuous and single-valued displacement field that can be traced back to the origin of the global deformation experienced by this ensemble.

6.2 Simply and Multiply Connected Domains Although the approach taken in this text vis-à-vis mathematical technicalities is a pragmatic one, some of the results discussed in this chapter require extra care. In particular, certain topological features of our initial configuration, Ω0 , will have to be spelled out explicitly. For the rest of this section Ω0 will be regarded as a spatial region of E3 , but everything that is said about such domains remains valid for planar regions as well (subject to obvious modifications). Let Γ j ( j = 1, 2) be two (smooth) closed curves in Ω0 . If Γ1 can be deformed continuously to coincide with Γ2 (without leaving the confines of Ω0 ), then the two curves are called homotopic and are considered to be equivalent (written Γ1 ∼ Γ2 ). The concept of homotopy defines an equivalence relation2 on the set of all closed curves in our spatial domain. A point in Ω0 can be regarded as a trivial closed curve, and we can thus define a contractible curve as being a closed curve homotopic to a point. Any two contractible curves are homotopic and therefore belong to the same (homotopy) equivalence class. A closed curve that is not contractible in the sense defined above will be referred to as irreducible. A domain Ω0 is said to be simply connected provided every closed curve in Ω0 is contractible. We remark in passing that such a domain has just one equivalence class (i.e. any two closed curves in Ω0 are homotopic). In E2 typical examples of simply connected domains are regions of the plane without ‘holes’ (e.g. disks, rectangles, polygons). This is, however, no longer true in E3 ; a sphere with an internal cavity (or a set of such disjoint cavities) is still simply connected, but a hollow cylinder or a torus is not. By way of example, let us consider a three-dimensional region with two cylindrical holes drilled across, corresponding to the sketch seen in Fig. 6.2. The closed Mathematics, an equivalence relation ‘∼’ on a set X is a binary relation that satisfies three axioms: (∀) A, B, C ∈ X, (i) A ∼ A; (ii) if A ∼ B then B ∼ A; (iii) if A ∼ B and B ∼ C then A ∼ C. An equivalence relation provides a partition of the underlying set X into disjoint subsets called equivalence classes. Any two elements in one of such equivalence classes are equivalent to each other; they are indistinguishable from the point of view of ‘∼’.

2 In

284

6 Compatibility of the Infinitesimal Deformation Tensor

Fig. 6.2 Cross-sectional view of a triply connected domain: a three-dimensional domain with two cylindrical holes drilled across (these are identified by the labels H1 and H2 )

irreducible curves Γ j ( j = 1, 2) encircling the first cylindrical hole (H1 ) are homotopic and belong to the same equivalence class. Note that, similarly, the irreducible curves γ j ( j = 1, 2) that run around the second hole (H2 ) belong to a different equivalence class. A third distinct class of equivalence consists of closed curves that do not encircle either hole, like C j ( j = 1, 2) in the foregoing Figure (these last ones are contractible). Such a domain Ω0 is said to be triply connected (note that a closed curve that encircles both H1 and H2 is homotopic to a pair of curves like Γ1 and γ1 , and is not regarded as an element of a new equivalence class). We can also remark that if one of the holes H j ( j = 1, 2) is missing, we can only define two distinct types of homotopy equivalence classes. The corresponding domains will then be called doubly connected (e.g. a hollow cylinder or a solid torus in E3 ; an annulus would also be an archetypal example of such doubly connected domain in E2 ). A multiply connected region in E3 can be transformed into a simply connected one by inserting suitable surface barriers, both sides of which are taken as forming part of the boundary of the new region thus defined. For instance, the domain occupied by a thick hollow cylinder can be rendered simply connected by choosing a barrier in the shape of a half-plane passing through the axis of the cylinder and having that axis as part of its finite boundary. In our discussion above we have distinguished between simply, doubly, and triply connected domains based on the number of distinct equivalence classes that can be formed in relation to the homotopy concept for closed curves situated in those domains. An alternative, more practical, classification can also be given according to the least number of barriers that must be introduced to render the domain in question a simply connected one. In light of this remark, a region Ω0 is said to be n-tuply connected (n ∈ N, n ≥ 2) if there exist a set of non-intersecting barriers S j ( j = 1, 2, . . . , n − 1) such that n−1  Sj Ω0 \ j=1

6.2 Simply and Multiply Connected Domains

285

S

S

S S

Fig. 6.3 The triply connected region Ω0 is rendered simply connected by introducing the (arbitrary) surface barriers S j ( j = 1, 2). The resulting domain Ω0 \ (S1 ∪ S2 ) is simply connected because closed curves in this new region can no longer encircle the two holes, H1 and H2 . Both faces of the barriers participate in the boundary of the simply connected region: this boundary is a surface supported by the closed curve abcde f e d gb a ha. Although, for illustrative purposes, ab and a b are shown as geometrically distinct, a ≡ b and a ≡ b (but these two pairs of points belong to different sides of S1 ). A similar observation is valid for ed and e d (in relation to S2 )

is simply connected. Figure 6.3 illustrates this transformation for the previous example included in Fig. 6.2; the two barriers S j ( j = 1, 2) are entirely arbitrary smooth surfaces that connect the outer boundary of Ω0 to its inner components (i.e. the cylindrical surfaces of H1 and H2 ). Throughout this chapter, we shall work mostly with a linearly elastic solid occupying a simply connected domain Ω0 ⊂ E3 . Multiply connected regions will appear in Sect. 6.9 and in several other places in the book. To minimise technicalities, it will be tacitly assumed, unless otherwise specified, that all fields of interest (strains, stresses, etc) are continuously differentiable (smooth) as many times as we like.

6.3 The ‘Incompatibility’ Operator We start by introducing a new tensorial operator called the incompatibility, and which is formally defined by T  (6.2) inc A := −∇ ∧ (∇ ∧ AT ) , where A ≡ A(X), with X ∈ Ω0 ⊂ E3 , is an arbitrary second-order tensor field. In the usual orthonormal Cartesian system of coordinates determined by the vectors {ei } the incompatibility operator admits the following representation inc A = − ∈ikl ∈ jmn Aln, km ei ⊗ e j .

(6.3)

286

6 Compatibility of the Infinitesimal Deformation Tensor

Indeed, if A = Aln el ⊗ en , then   ∂ ∇ ∧ A T = ek ∧ (Aln en ⊗ el ) = Aln, k (ek ∧ en ) ⊗ el . ∂ Xk By taking the curl of the transpose of this tensor we further get    T   ∂ ∧ Aln, k el ⊗ (ek ∧ en ) ∇ ∧ (∇ ∧ AT ) = em ∂ Xm = Aln, km (em ∧ el ) ⊗ (ek ∧ en ) . The last result can be brought into a form identical to (6.3) by noticing the simplifications, em ∧ el = ∈mli ei = ∈iml ei , ek ∧ en = ∈kn j e j = ∈ jkn e j , and by taking into account that the order of the second-order partial differentiation is immaterial if the tensor field in question is at least C 2 (Ω0 ) (i.e. A, km = A, mk for such fields). Before we continue, a word about notation is necessary. To simplify the repeated writing of multiply transposed ‘curls’ of tensor fields we recall the earlier definition (see Sect. 1.22.2), T  (6.4) A ∧ ∇ := − ∇ ∧ AT , which allows the alternative representation for the incompatibility operator, inc A = ∇ ∧ A ∧ ∇ .

(6.5)

Taken at face value the right-hand side of (6.5) seems to be ambiguous because we can think of it as (∇ ∧ A) ∧ ∇

or

∇ ∧ ( A ∧ ∇) ,

(6.6)

where the parentheses in these expressions indicate the priority of the vector-product operations involved. However, if we adopt the definition (6.4) to interpret the corresponding expressions in (6.6), it turns out that the final results will be the same in both cases. This can be easily checked by using a Cartesian coordinates representation of both A and ∇, but the corresponding calculations are omitted here. To keep things consistent, in what follows we shall agree that an expression of the form ∇ ∧ (. . . ) ∧ ∇, where the dots stand for a second-order tensorial expression, is to be interpreted by using the second rule in (6.6) in conjunction with formula (6.4). According to the definition (6.2) and the component representation (6.3) the incompatibility of an arbitrary second-order tensor field will have nine components in general. But, this operator is usually associated with symmetric tensor fields (as we shall see shortly), and in that case the incompatibility itself is a symmetric tensor field. This useful property is implicit in the next result.

6.3 The ‘Incompatibility’ Operator

287

Proposition 6.1 For any second-order tensor field A, (inc A)T = inc( AT ) .

(6.7)

If A is symmetric, then A = AT and hence inc( A) = inc ( AT ). Since inc ( AT ) = (inc A)T by (6.7), we conclude that inc( A) = (inc A)T , i.e. the tensor field inc ( A) is symmetric. The justification of (6.7) is immediate. Indeed, use of (6.3) gives (inc A)T = − ∈ikl ∈ jmn Aln, km e j ⊗ ei ,

(6.8)

while inc( AT ) = − ∈ikl ∈ jmn Anl, km ei ⊗ e j = − ∈ jkn ∈iml Anl, km e j ⊗ ei , where the last equality in the second equation above was obtained by relabelling the summation indexes, i → j, j → i, n → l, l → n. Invoking again the commutativity of the second-order partial derivatives of A, we can also interchange k → m and m → k in the same last equation. The result obtained in this way coincides with (6.8), and hence the proof of Proposition 6.1 is completed. Proposition 6.2 If A and u are tensor and vector fields, respectively, then inc(grad u) = O   inc (grad u)T = O div(inc A) = 0

i.e.

∇ ∧ (u ⊗ ∇) ∧ ∇ = O ,

(6.9a)

i.e. i.e.

∇ ∧ (∇ ⊗ u) ∧ ∇ = O , ∇ · (∇ ∧ A ∧ ∇) = 0 .

(6.9b) (6.9c)

To convince ourselves that these identities are true we employ the componentrepresentation formula (6.3). For example, inc (∇ ⊗ u) = − ∈ikl ∈ jmn (∇ ⊗ u)ln, km ei ⊗ e j

  = − ∈ikl ∈ jmn un, lkm ei ⊗ e j = − ∈ikl ∈ jmn un, m , kl ei ⊗ e j = − ∈ikl (∇ ∧ u) j, kl ei ⊗ e j .

If the indices i and j are fixed we know that ∈ikl is antisymmetric in k and l, whereas (∇ ∧ u) j, kl is symmetric. When summed, the product of these quantities will give a zero contribution to the total sum, and hence (6.9b) is proved. The proof of the identity (6.9a) is very similar to what has been said above and is left to the reader to check. As for (6.9c), this is just a particular case of the more general result ∇ · (∇ ∧ B) = 0, which holds for any second-order tensor fields B. It must be clear from (6.2) that inc A = ∇ ∧ B, for a suitably defined second-order tensor field B. With this in mind, since ∇ ∧ B =∈ikm Bm j,k ei ⊗ e j , we immediately find ∇ · (∇ ∧ B) = (∇ ∧ B)i j, i e j =∈ikm Bm j, ki e j .

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6 Compatibility of the Infinitesimal Deformation Tensor

Clearly, this sum must be zero because, when m and j are fixed, ∈ikm is antisymmetric in the indexes i and k, but Bm j, ki is symmetric in the same pair of subscripts. The justification of the identities (6.9) is now completed.

6.4 Conservative Fields In this section we recall briefly some useful concepts from Vector Calculus and indicate their corresponding extension to tensor fields. We start by considering a vector field w ∈ C 1 (Ω0 ), where Ω0 ⊂ E3 is a simply connected domain. This w is said to be conservative if there can be found a scalar field φ ∈ C 2 (Ω0 ) such that w = ∇φ. If this is possible, then φ is called a potential function or simply a potential of w. Proposition 6.3 A necessary and sufficient condition for the vector field w to be conservative is ∇ ∧ w = 0 in Ω0 . In this case the scalar potential is given by φ(X) = φ(X 0 ) +

w · dX ,

(∀) X ∈ Ω0 ,

(6.10)

P0 P

where the path integral is calculated along an arbitrary curve in Ω0 , starting at the point P0 (X 0 ) and ending at P(X). The proof of this standard result relies on Stokes’ Theorem (1.260) and the fact that the irrotational nature of the vector field w ensures that the path integral in (6.10) depends on the start and end points only. We mention in passing that if Ω0 is not simply connected, additional conditions will have to be imposed in order for φ to be single-valued. We are interested in the extension of the above result when the original vector field is replaced by a tensor field. To this end, within the setting already outlined above, let us consider a second-order tensor field A ∈ C 1 (Ω0 ). If there exists a vector field v ∈ C 2 (Ω0 ) such that ∇ ⊗ v = A in Ω0 , then A is said to be conservative; in this case v represents a potential vector field for A. Proposition 6.4 A necessary and sufficient condition for the tensor field A to be conservative is ∇ ∧ A = O in Ω0 . In this case the vector potential is given by v(X) = v(X 0 ) +

dX · A ,

(∀) X ∈ Ω0 ,

(6.11)

P0 P

where the path integral is calculated along an arbitrary curve in Ω0 , starting at the point P0 (X 0 ) and ending at P(X). The justification of this result is entirely similar to that of Proposition 6.3, with the only difference that use must be made of the tensorial version of Stokes’ Theorem

6.4 Conservative Fields

289

(1.261). Loosely speaking, finding the displacement field associated with a given infinitesimal deformation tensor is just an extension of Proposition 6.4 (albeit a nontrivial one). This aspect is taken up in the next section.

6.5 Cesàro–Volterra Formula The concept of compatibility hinted at in Sect. 6.1 will be made precise now by introducing the following formal definition. Definition 6.1 Assume that Ω0 ⊂ E3 is simply connected. A compatible infinitesimal strain field E ∈ C 2 (Ω0 ) is one for which the equation 1 (u ⊗ ∇ + ∇ ⊗ u) = E 2

(6.12)

admits a single-valued solution u ∈ C 3 (Ω0 ). Theorem 6.1 A necessary and sufficient condition for the symmetric tensor field E ∈ C 2 (Ω0 ) to be a compatible infinitesimal strain field in a simply connected domain Ω0 ⊂ E3 is to satisfy the so-called compatibility equation, ∇∧ E∧∇ = O

in Ω0 .

(6.13)

The Cartesian-coordinate version of the condition (6.13) was first obtained by Saint-Venant in 1860—the corresponding equations are known as the Saint-Venant’s compatibility conditions. If Ω0 is multiply connected, then (6.13) is still necessary for the existence of a single-valued displacement field u; however, this is no longer sufficient. At the end of this chapter, we shall elaborate on the global conditions that need to be imposed in addition to the local (i.e. pointwise) requirement (6.13). Before giving a proof of this important theorem it is instructive to write explicitly (6.13) in Cartesian coordinates. According to (6.3) we have ∈ikl ∈ jmn Eln, km = 0

for all i, j ∈ {1, 2, 3} ,

and Proposition 6.1 assure us that there are only six such distinct equations. These correspond to setting the pair of indexes (i, j) equal to (1, 1), (2, 2), (3, 3), (1, 2), (1, 3), and (2, 3). For example, if (i, j) = (1, 1), then 0 = ∈1kl ∈1mn Eln, km = ∈123 ∈123 E 33, 22 + ∈123 ∈132 E 32, 23 + ∈132 ∈123 E 23, 32 + ∈132 ∈132 E 22, 33 = E 33, 22 − E 23, 23 − E 23,23 + E 22, 33 .

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6 Compatibility of the Infinitesimal Deformation Tensor

A similar set of calculations can be carried out to obtain the remaining compatibility equations. The final results are listed below for future reference, 2E 12, 12 = E 11, 22 + E 22, 11 ,

(6.14a)

2E 13, 13 = E 11, 33 + E 33, 11 , 2E 23, 23 = E 33, 22 + E 22, 33 ,

(6.14b) (6.14c)

E 11, 23 = (E 12, 3 − E 23, 1 + E 31, 2 ), 1 , E 22, 13 = (E 21, 3 − E 13, 2 + E 32, 1 ), 2 , E 33, 12 = (E 31, 2 − E 12, 3 + E 23, 1 ), 3 .

(6.14d) (6.14e) (6.14f)

We return now to the proof of Theorem 6.1. The first part (the ‘necessity’) is an immediate consequence of the properties of the ‘curl’ operator. Let us start by assuming that E is compatible, so that (6.12) admits a solution; in other words, E represents the symmetric part of the gradient of a sufficiently smooth vector field. Taking the ‘curl’ of both sides in (6.12), and then transposing the result, leads to (∇ ∧ E)T =

T T 1 1 ∇ ∧ (∇ ⊗ u) + ∇ ∧ (u ⊗ ∇) . 2 2

By using the convention explained in (6.4) and (6.5), we can then write  T  T 1 1 ∇ ∧ E ∧ ∇ = − ∇ ∧ ∇ ∧ (∇ ⊗ u) − ∇ ∧ ∇ ∧ (u ⊗ ∇) 2 2 1 1 = ∇ ∧ (u ⊗ ∇) ∧ ∇ + ∇ ∧ (∇ ⊗ u) ∧ ∇ = O ; 2 2 the vanishing of the two terms on the last line of this equation is guaranteed by (6.9a) and (6.9b) in Proposition 6.2. For the second part (the ‘sufficiency’) we are going to prove that if the compatibility condition (6.13) is satisfied, then we can construct a displacement field u such that E is the symmetric part of its gradient. Note that the system of PDEs (6.1) requires initial conditions in order to find a unique solution (Theorem 6.1 deals only with existence, but says nothing about the uniqueness of the displacement field). For reasons that will become transparent as we go along, it will be assumed that both the displacement and the infinitesimal rotation vector are known at an arbitrary (but momentarily fixed) point P0 (X 0 ), with X 0 ∈ Ω0 , which actually amounts to knowing both u(X 0 ) and Ω(X 0 ). The aim is to find the value of the displacement field u at an arbitrary point P(X), with X ∈ Ω0 . Since du = dX · (∇ ⊗ u), and E and Ω represent, respectively, the symmetric and antisymmetric parts of u ⊗ ∇ ≡ (∇ ⊗ u)T , it follows immediately that du = E · dX + Ω · dX .

6.5 Cesàro–Volterra Formula

291

P

P P

P

Fig. 6.4 Integration paths between P0 and P in the elastic solid occupying the domain Ω0 ⊂ E3 . If this domain is simply connected (left), provided that the compatibility condition (6.13) is satisfied, the choice of the path going between the two points mentioned is irrelevant. For multiply connected domains (right) this is no longer true—note that in the particular case depicted here Ω0 has two cylindrical cavities shown in white

By integrating this equation along an arbitrary path in Ω0 , going from P0 (X 0 ) to P(X)—see Fig. 6.4, results in



u(X) = u(X 0 ) +

E(Y ) · dY + P0 P

Ω(Y ) · dY .

(6.15)

P0 P

Noting the product rule for differentials3 d [Ω(Y ) · (Y − X)] = dΩ(Y ) · (Y − X) + Ω(Y ) · d(Y − X) , the integral containing Ω in (6.15) is amenable to the following manipulations

Ω(Y ) · dY = P0 P

Ω(Y ) · d(Y − X)

P0 P

=

d [Ω(Y ) · (Y − X)] −

P0 P

X

= Ω(Y ) · (Y − X)

− X0

dΩ(Y ) · (Y − X) P0 P

dΩ(Y ) · (Y − X) .

(6.16)

P0 P

This can be further simplified by taking into account that dΩ(Y ) = dY · (∇ Y ⊗ Ω), so (6.15) can eventually be cast in the form E(Y ) · dY u(X) = u(X 0 ) + Ω(X 0 ) · (X − X 0 ) + P0 P − dY · [(∇ Y ⊗ Ω) · (Y − X)] .

(6.17)

P0 P

3 Here

and subsequently, X is kept fixed and the differentials are taken with respect to Y . Also, the notation ∇ Y will be used to indicate that the gradient is calculated relative to Y .

292

6 Compatibility of the Infinitesimal Deformation Tensor

Thus, we have managed to find the displacement field in terms of E and Ω. Because the latter field is not given, we have yet to express the integrand in the last term of (6.17) as a function of the infinitesimal deformation tensor only. To this end, we need the following auxiliary result. Proposition 6.5 Let u be a vector field. If E=

1 (u ⊗ ∇ + ∇ ⊗ u) 2

and

Ω=

1 (u ⊗ ∇ − ∇ ⊗ u) , 2

then (∇ ⊗ Ω) · a = −(E ∧ ∇) ∧ a ,

(6.18)

for all vector fields a. The justification for this formula is straightforward. For example, evaluating the component form of the expressions on either side of (6.18) it can easily be checked that they are identical. We leave this as an exercise to the reader and, instead, here we give a proof that makes minimal recourse to the aforementioned component representations. As ∇ ∧ (∇ ⊗ u) = 0,    T    ∂ ∂ ∧ ej ⊗u 2 ∇ ∧ E = ∇ ∧ (u ⊗ ∇) = ek ∂ Xk ∂Xj = (ek ∧ u, k j ) ⊗ e j = (ek ∧ u,k ), j ⊗ e j  T = (∇ ∧ u), j ⊗ e j = ∇ ⊗ (∇ ∧ u) , whence

 T 1 ∇ ∧ E = ∇ ⊗ (∇ ∧ u) . 2

(6.19)

Let us recall that 21 ∇ ∧ u ≡ ω is the axial vector of the skew-symmetric tensor Ω so, according to (6.19), we expect the ‘curl’ of E to be related to the gradient of Ω. To uncover the link between the two we are going to use (A.1d), in which we set T → Ω and u → a, where this a can be either a vector field or simply a constant vector. Therefore, (∇ ⊗ Ω) · a = ∇ ⊗ (Ω · a) − (∇ ⊗ a) · Ω T = ∇ ⊗ (ω ∧ a) + (∇ ⊗ a) · Ω = (∇ ⊗ ω) ∧ a − (∇ ⊗ a) ∧ ω + (∇ ⊗ a) · Ω  1 = ∇ ⊗ (∇ ∧ u) ∧ a − (∇ ⊗ a) ∧ ω + (∇ ⊗ a) ∧ ω 2  1 = ∇ ⊗ (∇ ∧ u) ∧ a , (6.20) 2

6.5 Cesàro–Volterra Formula

293

because (∇ ⊗ a) · Ω = (∇ ⊗ a) · (I ∧ ω) = [(∇ ⊗ a) · I] ∧ ω = (∇ ⊗ a) ∧ ω. Comparing (6.19) and (6.20) establishes the validity of (6.18), and the proof of Proposition 6.5 is now completed. If in (6.18) we take a → Y − X and remember that, due to the symmetry of the infinitesimal deformation tensor, E · dY = dY · E, Eq. (6.17) becomes u(X) = u(X 0 ) + ω(X 0 ) ∧ (X − X 0 ) 

  + dY · E(Y ) + E(Y ) ∧ ∇ Y ∧ (Y − X) .

(6.21)

P0 P

Next, let us define the second-order tensor   K (Y ; X) := E(Y ) + E(Y ) ∧ ∇ Y ∧ (Y − X) ,

(6.22)

so that the formula for the displacement can be written in the more compact form u(X) = u(X 0 ) + ω(X 0 ) ∧ (X − X 0 ) +

dY · K (Y ; X) ;

(6.23)

P0 P

Eq. (6.23) represents the Cesàro–Volterra formula. So far, we have shown that given an infinitesimal deformation tensor E(X), with X ∈ Ω0 ⊂ E3 , it is possible to construct a displacement field u(X) in Ω0 such that the Eqs. (6.1) are satisfied but, taken at face value, the formula we have obtained depends on the integration path between X 0 and X. According to the discussion of Sect. 6.4, this integral is path-independent if the ‘curl’ of the integrand vanishes identically. To establish if this is indeed the case for (6.22), requires some extra effort. Proposition 6.6 If A ≡ A(X) is a second-order tensor field then ∇ ∧ ( A ∧ X) = (∇ ∧ A) ∧ X + AT − I| A| .

(6.24)

The validity of this formula is easily confirmed with the help of the identity (A.3h) from Appendix 1. Letting u → X, we find ∇ ∧ ( A ∧ X) = (∇ ∧ A) ∧ X − (∇ ⊗ X) × × A.

(6.25)

As ∇ ⊗ X = I, by using the formula (1.116) from Chap. 1 the last term on the T × right-hand side of (6.25) can be rewritten as (∇ ⊗ X) × × A = I × A = | A|I − A . Finally, by substituting this last result back into (6.25) gives (6.24).

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6 Compatibility of the Infinitesimal Deformation Tensor

Proposition 6.7 For K defined in (6.22) the following identity holds true   ∇ Y ∧ K (Y ; X) = ∇ Y ∧ E(Y ) ∧ ∇ Y ∧ (Y − X) .

(6.26)

The justification of (6.26) is included below. By taking the ‘curl’ of both sides in the definition (6.22), we get ∇ Y ∧ K (Y ; X) = ∇ Y ∧ E(Y ) + ∇ Y ∧



  E(Y ) ∧ ∇ Y ∧ (Y − X) .

(6.27)

The first term on the right-hand side does not require any further simplification, but for the second term we can apply Proposition 6.6, with A → E(Y ) ∧ ∇ Y and X → Y − X. Thus, 

    ∇ Y ∧ E(Y ) ∧ ∇ Y ∧ (Y −X) = ∇ Y ∧ E(Y ) ∧ ∇ Y ∧ (Y − X)  T   + E(Y ) ∧ ∇ Y − tr E(Y ) ∧ ∇ Y I . (6.28) T T Since  E = ET  and by definition E ∧ ∇ := −(∇ ∧ E) , we have tr(E ∧ ∇) = −tr (∇ ∧ E) = −tr(∇ ∧ E). Recall that according to (1.232), (∇ ∧ E)i j = ∈kmi E m j, k , so that the trace of the ‘curl’ of E will be obtained by setting i = j. In this case, tr(∇ ∧ E) = ∈kmi E mi, k ; for fixed k ∈ {1, 2, 3}, the first term on the right-hand side is antisymmetric in the indexes ‘m’ and ‘i’, whereas the second term in that product is symmetric in the same pair of indexes. Because those are summation subscripts, the net result will be zero for each k, and hence tr(∇ ∧ E) = 0. Going back into (6.27) with (6.28), we discover that

T  ∇ Y ∧ K (Y ; X) = ∇ Y ∧ E(Y ) + E(Y ) ∧ ∇ Y

  + ∇ Y ∧ E(Y ) ∧ ∇ Y ∧ (Y − X) .

(6.29)

The proof of Proposition 6.7 is completed once we notice that the sum of the first two terms on the right-hand side of this last equation is actually zero by (6.4). The order of the terms under the path integral that appears in the Cesàro–Volterra formula (6.23) can be reversed by taking into account that dY · K = K T · dY . This leaves us with the task of simplifying K T . To this end, let us notice that the difference K − E is a tensor of the form A ∧ a, where A ≡ E(Y ) ∧ ∇ Y and a ≡ Y − X. From Chap. 1 we also know that ( A ∧ a)T = −(a ∧ AT ). Thus,



T  E(Y ) ∧ ∇ Y ∧ (Y − X)  T = E(Y ) − (Y − X) ∧ E(Y ) ∧ ∇ Y   = E(Y ) + (Y − X) ∧ ∇ Y ∧ E(Y ) .

K T = E T (Y ) +

6.5 Cesàro–Volterra Formula

295

In conclusion, formula (6.23) can be cast in the alternative form  u(X) = u(X 0 ) + ω(X 0 ) ∧ (X − X 0 ) + K (Y ; X) · dY ,

(6.30)

P0 P

where

   K (Y ; X) := E(Y ) + (Y − X) ∧ ∇ Y ∧ E(Y ) .

The expression of u found above consists of two parts. The terms outside the path integral in (6.30) represent a rigid-body transformation; strictly speaking, such terms are arbitrary because the deformation tensor is insensitive to linear functions of X. This suggests that in order to obtain a unique displacement field, additional information about u0 := u(X 0 ) and ω0 := ω(X 0 ) must be available at the point chosen to play the role of X 0 . One possibility would be to choose X 0 be a point of Ω ∪ ∂Ω which is fixed (i.e. u0 = ω0 = 0); other choices may be available depending on the particular problem that is to be solved. We conclude this section with a simple result about the form of the incompatibility operator in a case of practical interest. Proposition 6.8 The compatibility condition (6.13) is identically satisfied for infinitesimal deformation fields of the form E(X) = φ(X)a ⊗ a , where a is a constant vector and φ ≡ φ(X) represents a scalar field that depends linearly on the coordinates X i (i = 1, 2, 3) of X. Furthermore, for such deformation fields (not necessarily linear in X i ), ∇ ∧ E ∧ ∇ = −(∇ ⊗ ∇φ) × × (a ⊗ a) . The first part of this result is trivial because according to (6.14) the compatibility conditions involve exclusively second-order derivatives of E. Because E depends linearly on X i (i = 1, 2, 3), each term in (6.14) will be identically zero. In fact the infinitesimal deformation field E need not have the particular form considered in the above proposition, as long as it is symmetric and its components display at most linear dependence on X. The second part of Proposition 6.8 is a simple application of the vector identities from Appendix A. Note that   ∇ ∧ E = ∇ ∧ φ(a ⊗ a) = (∇φ) ∧ (a ⊗ a) + φ∇ ∧ (a ⊗ a) = (∇φ) ∧ (a ⊗ a) ,

296

6 Compatibility of the Infinitesimal Deformation Tensor

according to the identity (A.3a). By the conventions adopted earlier in this chapter, E ∧ ∇ = −(∇ ∧ E)T = (a ⊗ a) ∧ ∇φ. Hence, with the further help of the identity (A.3h), ∇ ∧ (E ∧ ∇) = ∇ ∧ [(a ⊗ a) ∧ ∇φ] = [∇ ∧ (a ⊗ a)] ∧ ∇φ − (∇ ⊗ ∇φ) × × (a ⊗ a) = −(∇ ⊗ ∇φ) × × (a ⊗ a) , which completes the proof of Proposition 6.8.

6.6 Alternative Forms of the Compatibility Equation We are now going to derive an equivalent form of the compatibility equation (6.13) that is still in intrinsic notation, but the iterated ‘curl’ will be eliminated. The equation we are going to derive works for any symmetric second-order tensor field, not just the symmetric part of the displacement gradient. Proposition 6.9 If A is a symmetric second-order tensor field for which

then

∇∧ A∧∇ = O,

(6.31)

  ∇ ⊗ (∇ · A) + (∇ · A) ⊗ ∇ − ∇ ⊗ ∇| A| − ∇ 2 A = O .

(6.32)

The converse is also true. The justification of this result is not immediate and requires several intermediate steps. The main idea is to take the double-cross product of Eq. (6.31) with the identity tensor I, that is, we shall aim to expand the expression I × × (inc A). Let us observe that inc A = ∇ ∧ B ,

with

B := A ∧ ∇ ≡ −(∇ ∧ A)T ,

so the expression we need to find is I × × (∇ ∧ B). Using Cartesian component representation for I and B gives successively,

6.6 Alternative Forms of the Compatibility Equation

297

  × B (e ∧ e ) ⊗ e I× (∇ ∧ B) = (e ⊗ e ) p p i j, k k i j × ×   = Bi j, k e p ∧ (ek ∧ ei ) ⊗ (e p ∧ e j ) = (Bi j, k δ pi ek − Bi j, k δ pk ei ) ⊗ (e p ∧ e j ) = Bi j, k ek ⊗ (ei ∧ e j ) − Bi j, k ei ⊗ (ek ∧ e j )   ∂ ⊗ (Bi j ei ∧ e j ) + Bi j, k ei ⊗ (e j ∧ ek ) = ek ∂ Xk = ∇ ⊗ B× + B ∧ ∇ , where B × is the vector of B. To summarise, we have just shown that I× × (∇ ∧ B) = ∇ ⊗ B × + B ∧ ∇ .

(6.33)

It remains to evaluate the expressions on the right-hand side of (6.33). The vector of the second-order tensor B ≡ −(∇ ∧ A)T is calculated first. Since ∇ ∧ A = Amn, k (ek ∧ em ) ⊗ en , its transpose will be obtained by swapping the places of the two vectors in the tensor product, and hence B × = −Amn, k en ∧ (ek ∧ em ) = −Amn, k (δmn ek − δnk em ) = −Amm, k ek + Amk, k em = −| A|, k ek + (∇ · A)m em or,4 B × = −∇| A| + ∇ · A .

(6.34)

As for the second term on the right-hand side of Eq. (6.33), this is in fact the transpose of ∇ ∧ (∇ ∧ A). Again, the quickest way to evaluate this expression is by recourse to its Cartesian component representation,      ∂ ∧ Ai j, p (e p ∧ ei ) ⊗ e j = Ai j, pk ek ∧ (e p ∧ ei ) ⊗ e j ∇ ∧ (∇ ∧ A) = ek ∂ Xk = Ai j, pk δki e p ⊗ e j − Ai j, pk δkp ei ⊗ e j 

= (Ai j, i ), p e p ⊗ e j − Ai j, kk ei ⊗ e j = e p ⊗ (Ai j, i e j ), p − (Ai j ei ⊗ e j ),kk = ∇ ⊗ (∇ · A) − ∇ 2 A . Transposing the result just obtained leads us to  T ∇ ∧ (∇ ∧ A) = (∇ · A) ⊗ ∇ − ∇ 2 A .

(6.35)

that, in general, (∇ ∧ A)× = −∇| A| + ∇ · AT for any second-order tensor A, not necessarily symmetric.

4 Note

298

6 Compatibility of the Infinitesimal Deformation Tensor

Finally, taking (6.34) and (6.35) into (6.33) gives the equation recorded in (6.32). This completes the proof of Proposition 6.9. In Cartesian coordinates, (6.32) takes on a relatively simple form that we are going to encounter later in this text, ∇ 2 Ai j −

∂ 2 A jk ∂ 2 Aik ∂ 2 Akk − + = 0, ∂ Xi ∂ Xk ∂ X j ∂ Xk ∂ Xi ∂ X j

(i, j = 1, 2, 3) .

(6.36)

Not all of the six equations obtained from (6.36) are independent. The tensor inc A is also divergence-free according to Proposition 6.2, which amounts to three additional scalar equations. Thus, there are only three independent equations in (6.36). Earlier in this chapter, we wrote down the detailed expression of the compatibility conditions—see Eqs. (6.14). It is possible to reduce their number at the expense of increasing the order of the derivatives. For example, it is a routine exercise to show that (6.14) can be replaced by just three equations that feature only fourth-order partial derivatives,   (no sum over i, j or k) , (6.37) E ii, j jkk = − E jk, i + E ik, j + E i j,k , 123 where i = j = k, with i, j, k ∈ {1, 2, 3}. These equations are justified by differentiating (6.14d) with respect to X 2 , X 3 , (6.14e) with respect to X 1 , X 3 , and (6.14f) with respect to X 1 , X 2 . The resulting expressions thus obtained match (6.37). As we have used only three out of the six equation in (6.14), it is not immediately clear whether the remaining equations can provide additional information. To investigate the matter further, let us differentiate twice: (6.14a) with respect to X 3 , (6.14b) with respect to X 2 and, finally, (6.14c) with respect to X 1 . The corresponding outcomes are 2E 12, 1233 = E 11, 2233 + E 22, 1133 ,

2E 13, 1322 = E 11, 3322 + E 33, 1122 ,

2E 23, 2311 = E 33, 2211 + E 22, 3311 .

(6.38)

Consider now the expressions that follow from (6.37) when written for i = 1, j = 2, k = 3 and, respectively, i = 2, j = 1, k = 3. Adding up the two corresponding equations, after cancelling the like terms of opposite signs, gives E 11, 2233 + E 22, 1133 = 2(E 12, 3 ), 123 . The right-hand side of this last equation can also be written as E 12, 3123 = E 12, 1233 , so it is seen at once that we have in fact recovered the first equation in (6.38). By the same token, the other equations in (6.38) are found to be a consequence of (6.37). Whether (6.37) represent a more advantageous form of the compatibility conditions (6.14) recorded earlier, is an arguable issue. Even though the number of equations has dropped from six to three, the order of differentiation has increased from two to four. This increase in the order of differentiation introduces a higher degree of arbitrariness in the solutions of (6.37), and demands that additional restrictions would have to be imposed in order to guarantee the same solution that follows from the integration of the original system of Eqs. (6.14).

6.7 Beltrami–Michell Equations

299

6.7 Beltrami–Michell Equations In a linearly elastic solid there is a one-to-one correspondence between strains and stresses, so it is natural to expect that the Saint-Venant compatibility equation (6.13) admits an equivalent formulation in terms of the infinitesimal stress tensor T . This result forms the object of our next investigation. Theorem 6.2 Let E and T denote, respectively, the infinitesimal strain and stress tensors in a linearly elastic isotropic solid that occupies a simply connected domain Ω0 ⊂ E3 . If E is a compatible strain field in the sense of (6.13), then ∇2T +

     ν  1 ∇ ⊗ ∇|T | = − ∇· f I − ∇⊗ f + f ⊗∇ 1+ν 1−ν

(6.39)

or, in component form, Ti j, kk +

1 ν Tkk, i j = − ( f k, k )δi j − ( f i, j + f j, i ) , 1+ν 1−ν

for all i, j ∈ {1, 2, 3}. Here, f ≡ ρb denotes the usual body force (per unit volume). The starting point in the justification of (6.39) is the alternative form of the SaintVenant compatibility equation (6.32); written for A ≡ E it reads   ∇ ⊗ (∇ · E) + (∇ · E) ⊗ ∇ − ∇ ⊗ ∇|E| − ∇ 2 E = O .

(6.40)

Since the new form of the compatibility equation (6.40) contains the divergence and the trace of E (i.e. ∇ · E and |E|, respectively), we begin the proof of Theorem 6.2 by expressing these two quantities in terms of the infinitesimal stress tensor. Recall that E and T are related to each other by the inverse of Hooke’s Law, E=

ν 1+ν T − |T |I , E E

so that |E| ≡ E : I =

  ν 1+ν T : I − |T | I : I E E

or, |E| =

1 − 2ν |T | , E

because I : I = 3 and T : I = |T |. By taking the divergence of both sides in (6.41), we see that ∇·E=

(6.41)

  1+ν ν ∇ · T − ∇ · |T |I . E E

(6.42)

300

6 Compatibility of the Infinitesimal Deformation Tensor

Since T satisfies the equilibrium equations ∇ · T + f = 0 and ∇ · (φ I) = ∇φ for any scalar field φ, this last equation yields ∇·E=−

ν 1+ν f − ∇|T | . E E

(6.43)

With this in mind, the first two terms in (6.40) amount to  2ν   1+ν ∇⊗ f + f ⊗∇ − ∇ ⊗ ∇|T | . E E (6.44) The third term in (6.40) can be rewritten with the help of (6.42) as ∇ ⊗ (∇ · E) + (∇ · E) ⊗ ∇ = −

    1 − 2ν ∇ ⊗ ∇|T | . ∇ ⊗ ∇|E| = E

(6.45)

Finally, in light of the constitutive law (6.41), the last term in (6.40) becomes ∇2 E =

1+ν 2 ν 2  ∇ T− ∇ |T | I . E E

(6.46)

This would suffice for the purpose of writing the Saint-Venant compatibility equation (6.40) in terms of T , but the last term in (6.46) admits a further simplification; we pursue this next. By taking the double-dot product of (6.40) with the identity tensor I, 2∇ · (∇ · E) − 2∇ 2 |E| = 0 .

(6.47)

The first term in this equation can be obtained from (6.43) by taking the divergence of both sides, ∇ · (∇ · E) = −

ν 1+ν ∇ · f − ∇ 2 |T | , E E

while, on account of (6.42), the second term in (6.47) can be cast as ∇ 2 |E| =

1 − 2ν 2 ∇ |T | . E

Substituting the newly found expressions into (6.47) it transpires that ∇ 2 |T | = −

1+ν (∇ · f ) , 1−ν

and then (6.46) can be re-arranged to give the more convenient form    1+ν 2 ν 1+ν  2 ∇· f I. ∇ E= ∇ T+ E E 1−ν

(6.48)

6.7 Beltrami–Michell Equations

301

With (6.44), (6.45) and (6.48) back into (6.40), the result stated in Theorem 6.2 follows immediately now. Equations (6.39) are named after the two scientists who discovered them; E. Beltrami (1892) derived the particular case of these equations corresponding to uniform body forces (i.e. ∇ · f = 0, ∇ ⊗ f = f ⊗ ∇ = O), while J. H. Michell (1900) obtained the general form stated in Theorem 6.2. Because T is a symmetric tensor, and (6.39) is in fact a tensorial equation, it means that we have six scalar equations for the components Ti j (i, j ∈ {1, 2, 3}, Ti j = T ji ). However, not all of these are independent since, if equilibrium equations are to be satisfied, then ∇ · T + f = 0, which represents an additional set of three constraints involving the components of T .

6.8 Explicit Calculations and Examples Provided that the compatibility condition (6.13) is satisfied, in evaluating the path integrals in (6.30) we have the freedom to choose the integration path so that the corresponding integrands simplify as much as possible. It turns out that, in practice, a convenient choice is a path that is parallel to the Cartesian coordinate axes; thus, each path will consist of three straight lines on which two of the coordinates will be fixed and the other will be variable. Such a choice suggests that the path integral will be reduced to the sum of three standard integrals on the real line. These statements are made more precise below. Let L j(i) ( j = 1, 2, 3) denote the straight lines that make up the path extending from the initial point P0 (X 0 ), X 0 = X 10 e1 + X 20 e2 + X 30 e3 , to the current point P(X), X = X 1 e1 + X 2 e2 + X 3 e3 . There are three such paths (i = 1, 2, 3) that we shall consider, each corresponding to a different component (ui ) of the displacement field u = u1 e1 + u2 e2 + u3 e3 , path from P0 to P := L1(i) ∪ L2(i) ∪ L3(i)

(i = 1, 2, 3) ;

Figure 6.5 gives a pictorial representation of these paths in the particular case when X 0 = 0. Analytically, we can represent these paths in parametric vector form as indicated below. For u1 , ⎧ ⎪ X 0 e + ζ e2 + X 30 e3 , X 20 ≤ ζ ≤ X 2 : L2(1) ⎪ ⎨ 1 1 : L3(1) Y (ζ ) = X 10 e1 + X 2 e2 + ζ e3 , X 30 ≤ ζ ≤ X 3 ⎪ ⎪ ⎩ζ e 1 + X 2 e 2 + X 3 e 3 , X 0 ≤ ζ ≤ X 1 : L (1) . 1

1

302

6 Compatibility of the Infinitesimal Deformation Tensor

u1

u2

X3 P2

P1

X3

P1

P P0

u3

X3

X2

X1

P

P2

P0

P1

X2

X1

P

P0

P2

X1

X2

Fig. 6.5 Possible choice of integration paths in the Cesàro–Volterra integral. For P0 (0, 0, 0) → P1 (0, X 2 , 0) → P2 (0, X 2 , X 3 ) → P(X 1 , X 2 , X 3 ); for

u2

:

u1 :

P0 (0, 0, 0) →

P1 (0, 0, X 3 ) → P2 (X 1 , 0, X 3 ) → P(X 1 , X 2 , X 3 ); and for u3 : P0 (0, 0, 0) → P1 (X 1 , 0, 0) → P2 (X 1 , X 2 , 0) → P(X 1 , X 2 , X 3 )

For u2 ,

For u3 ,

⎧ ⎪ X 0 e + X 20 e2 + ζ e3 , X 30 ≤ ζ ≤ X 3 ⎪ ⎨ 1 1 Y (ζ ) = ζ e1 + X 20 e2 + X 3 e3 , X 10 ≤ ζ ≤ X 1 ⎪ ⎪ ⎩ X 1 e1 + ζ e2 + X 3 e3 , X 0 ≤ ζ ≤ X 2 2 ⎧ ⎪ ζ e + X 20 e2 + X 30 e3 , X 10 ≤ ζ ≤ X 1 ⎪ ⎨ 1 Y (ζ ) = X 1 e1 + ζ e2 + X 30 e3 , X 20 ≤ ζ ≤ X 2 ⎪ ⎪ ⎩ X 1 e1 + X 2 e2 + ζ e3 , X 0 ≤ ζ ≤ X 3 3

: L3(2) : L1(2) : L2(2) . : L1(3) : L2(3) : L3(3) .

Note that dY = e p dζ on L p(i) for all p, i ∈ {1, 2, 3}. In consequence, the original path integrals (6.30) are reduced to ordinary integrals, as already pointed out above. For each component ui (i = 1, 2, 3) of the displacement field u we can write (ignoring the rigid-body transformation), ui =

3  p=1

=

L 1(i)

 · e p )i dζ = (K P0 P

i1 dζ + K

3  p=1

L 2(i)

L p(i)

i2 dζ + K

i p dζ K



L 3(i)

i3 dζ K

(no sum over i) .

For the sake of completeness, we record below the full expressions of the components of the displacement field (again, leaving out the purely rigid-body transformation)

6.8 Explicit Calculations and Examples

u1 (X) =



 E 12 + (X 2 − ζ )(E 12, 2 − E 22, 1 ) + (X 3 − X 30 )(E 12, 3 − E 32, 1 ) dζ   E 11 dζ , E 13 + (X 3 − ζ )(E 13, 3 − E 33, 1 ) dζ + +

L 2(1)

L 3(1)



L 1(1)



 E 23 + (X 3 − ζ )(E 23, 3 − E 33, 2 ) + (X 1 − X 10 )(E 23, 1 − E 13, 2 ) dζ L 3(2)   E 21 + (X 1 − ζ )(E 21, 1 − E 11, 2 ) dζ + + E 22 dζ ,

u2 (X) =

u3 (X) =

303

L 1(2)

L 2(2)



 E 31 + (X 1 − ζ )(E 31, 1 − E 11, 3 ) + (X 2 − X 20 )(E 31, 2 − E 21, 3 ) dζ   E 32 + (X 2 − ζ )(E 32, 2 − E 22, 3 ) dζ + E 33 dζ . +

L 1(3)

L 2(3)

L 3(3)

These formulae provide elementary closed-form solutions for the components ui of displacement field u associated with a compatible strain described by the infinitesimal deformation tensor E. The Cesàro–Volterra formula leads to more manageable results when E = E 11 (X)e1 ⊗ e1 + E 22 (X)e2 ⊗ e2 + E 33 (X)e3 ⊗ e3 ,

(6.49)

i.e. E is a diagonal tensor. In this case the above general expressions become u1 (X 1 , X 2 , X 3 ) =

X1 X 10

E 11 (ζ, X 2 , X 3 ) dζ − −

u2 (X 1 , X 2 , X 3 ) =

X2 X 20

X 30



X3 X 30

X 20



X1 X 10





X2 X 20

X3 X 30

(X 3 − ζ )E 33, 2 (X 10 , X 20 , ζ ) dζ

(X 1 − ζ )E 11, 2 (ζ, X 20 , X 3 ) dζ , (6.50b)

E 33 (X 1 , X 2 , ζ ) dζ − −

(X 2 − ζ )E 22, 1 (X 10 , ζ, X 30 ) dζ

(X 3 − ζ )E 33, 1 (X 10 , X 2 , ζ ) dζ , (6.50a)

E 22 (X 1 , ζ, X 3 ) dζ − −

u3 (X 1 , X 2 , X 3 ) =

X3

X2

X1 X 10

(X 1 − ζ )E 11, 3 (ζ, X 20 , X 30 ) dζ

(X 2 − ζ )E 22, 3 (X 1 , ζ, X 30 ) dζ . (6.50c)

If the components E ii (X) (no sum) in (6.49) are linear functions of X j (i, j = 1, 2, 3), then the derivatives E ii, k (k = 1, 2, 3, no sum) are constants, so the integrals in (6.50)

304

6 Compatibility of the Infinitesimal Deformation Tensor

X2

X2 −Me2

Me2 X1

O

X3

X3

Fig. 6.6 Normal bending of a cylindrical rod

are particularly easy to evaluate. Even when these functions are polynomials (which is usually the case) the above integrals pose no significant challenges. In closing this section we are going to look at several examples in which the infinitesimal deformation tensor is given a priori, and our task will be to find the corresponding displacement fields that generate states of deformation consistent with the given expressions. Example 6.1 The infinitesimal deformation tensor for the normal bending of a cylindrical rod (see Fig. 6.6) has the form   E(X) = −AX 2 e1 ⊗ e1 − B(e2 ⊗ e2 + e3 ⊗ e3 ) ,

(6.51)

where A, B ∈ R are two constants that depend on the mechanical and geometrical properties of the rod, and X = X i ei . We aim to find the displacement field associated with (6.51). According to Proposition 6.8, the compatibility condition is trivially satisfied, so we are assured that a continuous single-valued displacement vector field does exist. All that is left to do is apply the formulae (6.50) for this particular case. Note that E 11, 2 = −A , E 11, 3 = 0 ,

E 22, 1 = 0 , E 22, 3 = 0 ,

E 33, 1 = 0 , E 33, 2 = AB .

Hence

X1

u1 (X) = 0

AX 2 dζ = −AX 1 X 2 ,

(6.52a)

6.8 Explicit Calculations and Examples Fig. 6.7 Off-axis bending of a rectangular rod (its dimensions are exaggerated in the interest of clarity). In this example, the applied bending moments are directed along the axis defined by the unit vector a parallel to the X 2 X 3 -plane

305

X2

X2

−Ma

Ma X1

O

X3

X3



X2

u2 (X) =



X3

ABζ dζ −

0



X1

AB(X 3 − ζ ) dζ +

0

 1  2 A X 1 + B(X 22 − X 32 ) , 2 X3 AB X 2 dζ = AB X 2 X 3 . u3 (X) =

A(X 1 − ζ ) dζ

0

=

(6.52b) (6.52c)

0

Example 6.2 The infinitesimal deformation tensor for the off-axis bending of a rectangular rod (see Fig. 6.7) has the form   E(X) = (C X 3 − AX 2 ) e1 ⊗ e1 − B(e2 ⊗ e2 + e3 ⊗ e3 ) ,

(6.53)

where A, B, C ∈ R and X = X i ei . Again, the question is to find the infinitesimal displacements experienced by the rod.

By inspection, E 11, 2 = −A , E 11, 3 = C ,

E 22, 1 = 0 , E 22, 3 = −BC ,

E 33, 1 = 0 , E 33, 2 = AB .

Then, a direct application of (6.50) yields u1 (X) =

X1

(C X 3 − AX 2 ) dζ = (C X 3 − AX 2 )X 1 , X2 X3 B(C X 3 − Aζ ) dζ − AB(X 3 − ζ ) dζ + u2 (X) = − 0

0

=

0

 1  2 A X 1 + B(X 22 − X 32 ) − BC X 3 X 2 , 2

0

X1

A(X 1 − ζ ) dζ

306

6 Compatibility of the Infinitesimal Deformation Tensor



X3

u3 (X) = −

B(C X 3 − AX 2 ) dζ −

0

=

X1



X2

C(X 1 − ζ ) dζ +

0

 1  C − X 12 + B(X 22 − X 32 ) + AB X 2 X 3 . 2

BC(X 2 − ζ ) dζ

0

Example 6.3 Let us re-do the first example without making recourse to the Cesàro–Volterra formula. As already pointed out, because E in (6.51) is compatible, the direct integration of the system of PDEs (6.1) must lead to a solution. For our particular situation, these equations become ∂u2 ∂u3 = = AB X 2 , ∂ X2 ∂ X3 ∂u1 ∂u2 + = 0, ∂ X2 ∂ X1

∂u1 = −AX 2 , ∂ X1

∂u1 ∂u3 + = 0, ∂ X3 ∂ X1

∂u3 ∂u2 + = 0. ∂ X2 ∂ X3

(6.54)

(6.55)

Direct integration of (6.54) gives 1 AB X 22 + f 2 (X 1 , X 3 ) , 2 u3 = AB X 3 X 2 + f 3 (X 1 , X 2 ) ,

u1 = −AX 1 X 2 + f 1 (X 2 , X 3 ) ,

u2 =

(6.56a) (6.56b)

where the functions f j ( j = 1, 2, 3) are arbitrary at this stage. To fix them, we need to enforce the constraints (6.55) on these expressions, that is, −AX 1 +

∂ f1 ∂ f2 ∂ f1 ∂ f3 + = 0, + = 0, ∂ X2 ∂ X1 ∂ X3 ∂ X1 ∂ f3 ∂ f2 AB X 3 + + = 0. ∂ X2 ∂ X3

Next, we make the change of (dependent) variable  1   f 2 := f 2 + A B X 32 − X 12 , 2 so that now ∂ f1 ∂ f2 + = 0, ∂ X2 ∂ X1

∂ f1 ∂ f3 + = 0, ∂ X3 ∂ X1

∂ f3 ∂ f2 + = 0. ∂ X2 ∂ X3

(6.57)

6.8 Explicit Calculations and Examples

307

It can be shown (see Exercise 6.4 at the end of this chapter) that the functions f 1 ,  f 2 , and f 3 satisfying (6.57) must be the components of a rigid-body transformation. Leaving out this part of the solution it is seen that the formulae (6.56) are identical to the result found previously in (6.52).

6.9 Weingarten-Volterra Dislocations All of our developments in the previous sections were carried out within the context of simply connected domains. If this assumption is relaxed, that is, if we allow for Ω0 ⊂ E3 to be multiply connected, then the displacement field associated with an infinitesimal deformation tensor E ∈ C 2 (Ω0 ) need not be continuous. This counterintuitive fact is also related to the concepts of self-stresses and Volterra dislocation that we shall briefly touch upon in this last section. Self-stresses can be generated inside an elastic body when two (or more) parts of the body can be joined together, so that each constraints the other to an unnatural size of shape (e.g., see Fig. 6.10). In a simply connected body such operations will generally produce discontinuous internal stresses/strains at certain locations within the body. However, this restriction disappears if the body is multiply connected; recall from Sect. 6.2 that the form of these bodies is such that some of their sections can be cut completely across without separating them in two distinct pieces. By way of example, consider an infinitely long elastic cylinder at rest. To avoid unnecessary complications we shall assume a state of plane strain in what follows; this means that we need only deal with its annular cross section. Assume that this annulus is cut along the radial lines aa and bb , and the wedge aa b b of angle θ0 is removed (see Fig. 6.8). If the exposed cut aa is kept fixed and we join bb to it, the new annular configuration is in a state of self-stress. This operation produces multi-valued displacements, as explained next.

Fig. 6.8 A simple example of 2D dislocation in an annular domain. The annulus is cut along the radial lines aa and bb (left); the wedge aa b b is then removed (middle); finally, the exposed cut bb is displaced so that it matches the cut aa , and the external forces that have acted on the annulus during these operations are removed. The annulus thus deformed will generally be in a state of self-stress

308

6 Compatibility of the Infinitesimal Deformation Tensor

If the position of an arbitrary point P in the strained annulus is expressed in the polar coordinates r = O P, θ = ∠ a O P, then the displacement uθ (P) = r θ θ0 /2π . Clearly uθ (Q 1 ) = 0 = r θ0 = uθ (Q 2 ), where Q j ( j = 1, 2) are geometrically coincident points situated on the two joined cuts, as seen in Fig. 6.8. To avoid the inconvenience of multi-valued functions, one can view the displacement as being discontinuous across aa . This change of tack is justified by the fact that the elastic stress state in any part of the body does not depend directly on the displacement field, but only on its derivatives. If instead of cutting out the wedge aa b b we cut the wedge cc d d, the resulting self-strained annulus would be identical to the original one as far as the stress and strain distributions are concerned. After the above preliminary considerations we are now ready to flesh out these ideas by exploiting the earlier work on the Cesàro–Volterra formula. Definition 6.2 A Volterra dislocation on a domain Ω0 ⊂ E3 is a vector field u with the following properties: 1. u ∈ C 2 (Ω0 \ S), where S is a smooth surface in Ω0 ; 2. u is not continuous across S; 3. the infinitesimal strain tensor E associated with u is continuous across S and the extension (by continuity) of E to Ω0 is of class C 2 . Let S be a two-sided surface barrier that reduces Ω0 to a simply connected region Ω# := Ω0 \ S, and consider a point Q ≡ Q(X) ∈ S. In what follows the notation u+ (X) stands for the limit of u(P) as P approaches Q from one side of S (which is arbitrarily chosen to be the positive side), while u− (X) represents the limit of u(P) as P approaches Q from the other side of S (identified as the negative side). It should be clear that the quantity Δu(X) := u+ (X) − u− (X)

(6.58)

provides a measure of the discontinuity in u at the point X ∈ S. (i.e. Δu(X) = 0 implies that u is continuous at X). In the simply connected domain Ω# defined above, Theorem 6.1 is still applicable and yields a single-valued vector field u, except that now this might be discontinuous across S. It follows from (6.23) that   dY · K (Y ; X) , Δu(X ) = dY · K (Y ; X ) , (6.59) Δu(X) = Γ

Γ

where Γ and Γ are two closed paths in Ω0 that intersect S at X and X , respectively; the orientation of Γ and Γ is such that the integrals in (6.59) are calculated by moving from the negative to the positive side of S (see Fig. 6.9). By the same token, from Exercise 6.9 it also follows that    T  T Δω(X) = dY · ∇ Y ∧ E(Y ) , Δω(X ) = dY · ∇ Y ∧ E(Y ) . Γ

Γ

6.9 Weingarten-Volterra Dislocations

309

It is a direct consequence of the assumption E ∈ C 2 (Ω0 ) in conjunction with Stokes’ Theorem (1.261) that the following equalities hold,  Γ

and

 Γ

dY · K (Y ; X ) =

 T dY · ∇ Y ∧ E(Y ) =

 Γ

 Γ

dY · K (Y ; X )

(6.60)

 T dY · ∇ Y ∧ E(Y ) .

(6.61)

Note that this last equation confirms that Δω(X) = Δω(X ) for all X, X ∈ S, so we can simply use Δω for the common value of these expressions. By subtracting the two equations in (6.59) we finally find   T  ∧ (X − X ) , Δu(X ) = Δu(X) − dY · ∇ Y ∧ E(Y ) whence

Δu(X ) = Δu(X) + Δω ∧ (X − X) .

(6.62)

The geometrical interpretation of the Volterra dislocation is now clear: it corresponds to an infinitesimal rigid-body deformation of S+ relative to S− (here, regarded as two distinct material surfaces); see the right sketch in Fig. 6.9. In the most general case, a Volterra dislocation is completely specified by prescribing the six kinematic quantities Δu and Δω at an arbitrary point on S. By using a very similar approach as above, it can be proved that in the case of a multiply connected domain the following general result holds.

Fig. 6.9 Left: an illustration of the notation used in Eq. (6.59). Right: displacement fields for Volterra dislocations—see (6.62)

310

6 Compatibility of the Infinitesimal Deformation Tensor

Theorem 6.3 (J. Weingarten) Let Ω0 ⊂ E3 be a multiply connected and   p domain let S j be non-intersecting barriers in Ω0 such that Ω# := Ω0 \ ∪ j=1 S j is simply connected. If the infinitesimal strain tensor E ∈ C 2 (Ω0 ) is single-valued, then



Δu(X)

Sj

= b j + c j ∧ X S j ,

(6.63)

where  c j :=

Γj

 b j :=

Γj

 T dY · ∇ Y ∧ E(Y ) ,

(6.64a)

  T dY · E(Y ) − ∇ Y ∧ E(Y ) ∧ Y ,

(6.64b)

for j = 1, 2, . . . , p; the closed paths Γ j in (6.64) are arbitrary irreducible curves intersecting only the barrier S j , and oriented from S−j to S+j . The constants b j and c j defined in (6.64) are independent of the choice of barriers and the curves Γ j . In fact if Γ j∗ is another curve in Ω0 such that Γ j∗ ∼ Γ j then b j (Γ j∗ ) = b j (Γ j ) and c j (Γ j∗ ) = c j (Γ j ) (no summation). Since the position of the barriers is arbitrary, instead of regarding u discontinuous across S j , one can regard this displacement field as being multi-valued and of class C 3 (Ω0 ). The multi-valued nature of this vector field can be traced back to the Cesàro-integral representation (6.23), for which the condition (6.13) is no longer sufficient to ensure the pathindependence of the integral term when Ω0 is multiply connected. With these observations in mind, the implications of Weingarten’s result vis-à-vis the compatibility issues discussed in this chapter become clear and are spelled out below. Theorem 6.4 Assume that Ω0 and S j ( j = 1, 2, . . . , p) are as in the above theorem. A necessary and sufficient condition for the infinitesimal strain tensor E to be compatible in Ω0 is to satisfy the following conditions ∇∧ E∧∇ = O in Ω0 ,   T dY · ∇ Y ∧ E(Y ) = 0 , 

Γj

Γj

  T dY · E(Y ) − ∇ Y ∧ E(Y ) ∧ Y = 0 ,

(6.65a) (6.65b) (6.65c)

where Γ j are arbitrary irreducible curves intersecting only the barrier S j , and oriented from S−j to S+j ( j = 1, 2, . . . , p). In contrast to the Saint-Venant’s compatibility condition (6.65a) which may be regarded as local (or pointwise), the additional requirements (6.65b) and (6.65c) represent global conditions. We close this chapter with two more examples of configurations that can accommodate self-stresses (see Fig. 6.10). Both examples have a historical status as they

6.9 Weingarten-Volterra Dislocations

311

Fig. 6.10 Two classical examples of Volterra dislocations in an infinitely long hollow elastic cylinder: edge dislocation (left) and screw dislocation (right); see main text for details

have been used extensively as elastic models for certain types of imperfections in real crystals. Consider again an infinitely long hollow cylinder, which is a doubly connected domain in E3 . A plane cut is made along a half-plane passing through the axis of the cylinder; we recall that this process renders the original configuration simply connected. The two faces thus exposed by the cut are displaced relative to each other within the foregoing half-plane (we take Δω = 0 in (6.62)); for the left sketch in Fig. 6.10 we have Δu1 = b, while in the right one Δu3 = b. The two displaced faces of the cuts are rejoined together, so that the new configurations are once again doubly connected. Both configurations are now in a state of self-stress. The elastic states of the two situations in Fig. 6.10 can be calculated by adopting the simplified strategies outlined in Sect. 5.11. We postpone the solution for the edge dislocation until the end of Chap. 7, where we also include a couple of other examples of incompatible stress fields. The determination of the elastic state for the screw dislocation is based on the anti-plane strain approximation, and is outlined in Exercise 6.20 below.

6.10 Exercises 1. Let a and b be two constant vectors, and consider u(X) := a + b ∧ X. Without using the index notation, show that the infinitesimal strain tensor E associated with this u is identically zero. 2. For a second-order tensor field A defined on Ω0 ⊂ E3 , we set A · ∇ := ∇ · AT

and

∇ · A · ∇ := ∇ · ( A · ∇) .

(6.66)

Assuming that A ∈ C 4 (Ω0 ), prove the identity     inc (inc A) = ∇ ⊗ ∇ (∇ · A · ∇) − ∇ ⊗ ∇ · (∇ 2 A) − (∇ 2 A) · ∇ ⊗ ∇ + ∇ 4 A ,

 where ∇ 4 (. . . ) = ∇ 2 ∇ 2 (. . . )) is the bi-Laplacian operator (see Appendix E).

312

6 Compatibility of the Infinitesimal Deformation Tensor

3. By using Cartesian coordinates show that  T T  ∇ ∧ (∇ ∧ A)T = ∇ ∧ ∇ ∧ AT . 4. Consider a vector field h = h j e j , where h 1 ≡ h 1 (X 2 , X 3 ), h 2 ≡ h 2 (X 1 , X 3 ), and h 3 ≡ h 3 (X 1 , X 2 ). If these functions satisfy the system of PDEs ∂h 1 ∂h 2 + = 0, ∂ X2 ∂ X1

∂h 2 ∂h 3 + = 0, ∂ X3 ∂ X2

∂h 3 ∂h 1 + = 0, ∂ X1 ∂ X3

in some domain Ω0 ⊂ E3 , then show that h(X) = u0 + ω0 ∧ X, where u0 and ω0 are arbitrary constant vectors, and X = X j e j . 5. Define the formal matrix differential operator ⎡

0 ⎢ ∂ ⎢ M∇ := ⎣ ∂ X 3 − ∂ ∂X 2

− ∂ ∂X 0 ∂ ∂ X1

3

∂ ∂ X2 − ∂ ∂X

⎤ 1

⎥ ⎥. ⎦

(6.67)

0

The action of M∇ on second-order Cartesian tensor fields is done by following the rules of matrix multiplication, with the convention that this operator acts only from left to right. In an expression such as M∇ · A, with A a second-order tensor field, the result of the multiplication between an entry of (6.67) and one of this tensor, results in either zero or the corresponding partial derivative of the latter. Show that the compatibility condition inc E = O can be formally written as  T M ∇ · M∇ · E = O . 6. Consider the deformation of an elastic body such that its displacement field u does not depend on X 3 and the displacement component in the e3 −direction vanishes identically; more specifically, u = u(X 1 , X 2 ) and u · e3 = 0. a. Show that the compatibility condition reduces to just one single equation, ∂ 2 E 11 ∂ 2 E 22 ∂ 2 E 12 + = 2 . ∂ X 1∂ X 2 ∂ X 22 ∂ X 12 b. Within the same context, the deformation of an elastic body is such that the infinitesimal strain tensor assumes the form E = −κ(X 1 )X 2 e1 ⊗ e1 , for some given function κ(X 1 ). Show that this E is compatible and the associated displacements have the expressions recorded below,

6.10 Exercises

313

1

a a

X1

P X2 Fig. 6.11 Bending of a two-dimensional cantilever beam due to a vertical tip load



X1

u1 (X 1 , X 2 ) = −X 2 κ(ζ ) dζ + · · · , 0 X1 κ(ζ )(ζ − X 1 ) dζ + · · · , u2 (X 1 , X 2 ) = − 0

where the dots stand for rigid-body deformation terms. 7. The deformation of a cantilever beam of rectangular cross section and acted upon by a downward force can be approximated by the two-dimensional situation seen in Fig. 6.11. This results in an infinitesimal deformation tensor field of the form E(X) = E 11 e1 ⊗ e1 + E 22 e2 ⊗ e2 + E 12 (e1 ⊗ e2 + e2 ⊗ e1 ) ,

(6.68)

where E 11 = −α X 1 X 2 ,

E 22 = αβ X 1 X 2 ,

1 E 12 = − α(β + 1)(a 2 − X 22 ) , 2

and α, β ∈ R are given constants. Assuming further that the displacement field u ≡ u(X 1 , X 2 ) (i.e. there is no dependence on X 3 ) and u3 ≡ 0, show that a. E ≡ E(X 1 , X 2 ) given by (6.68) is compatible in the sense of Definition 6.1. b. The displacements in the beam corresponding to (6.68) are given by   1 α X 2 (β + 2)X 22 − 3X 12 − α(β + 1)a 2 X 2 + AX 2 + B , 6   1 u2 (X 1 , X 2 ) = α X 1 X 12 + 3β X 22 − AX 1 + C , 6 u1 (X 1 , X 2 ) =

where A, B, C ∈ R are arbitrary constants. Indicate how these constants can be determined.

314

6 Compatibility of the Infinitesimal Deformation Tensor

8. Let ω = ωi ei be the axial vector of the skew-symmetric part of the displacement gradient tensor. Verify by direct calculations that ∂ E 12 ∂ E 11 ∂ω2 ∂ E 11 ∂ E 13 ∂ω3 = − , = − , ∂ X1 ∂ X1 ∂ X2 ∂ X1 ∂ X3 ∂ X1 ∂ω1 ∂ E 13 ∂ E 12 = − . ∂ X1 ∂ X2 ∂ X3 Find the derivatives of ωi (i = 1, 2, 3) with respect to X 2 and X 3 as well. How can these formulae be written in a coordinate-free form? 9. Consider an elastic body that occupies the simply connected domain Ω0 ⊂ E3 . Let X 0 , X ∈ Ω0 and assume that ω(X 0 ) is known. Show that ω(X) = ω(X 0 ) +

X X0

  ∇ Y ∧ E(Y ) · dY .

(This result says that the symmetric part of the displacement gradient determines the skew-symmetric part if the value of the latter is known at a given point in Ω0 .) 10. Find the displacement field for the off-axis bending of the rectangular rod of Example 6.2 by direct integration of Eqs. (6.1). 11. If X = X j e j ∈ R3 , let g ≡ g(X 3 ) be a given function and β ∈ R. Define the infinitesimal strain tensor field   1 1 E = ∇ ⊗ g(X 3 )e3 − X 32 X + e3 ⊗ ∇ (β X 32 + X 3 |X|2 ) , 2 2 where |X|2 = X · X. a. Show that   1 2 1 β X 3 + X 3 |X|2 e3 − X 32 X u(X) = g(X 3 ) + 2 2 is a particular solution of the equation u ⊗ ∇ + ∇ ⊗ u = 2E.   12. Consider the infinitesimal strain tensor E = −K X 1 (1 + α)e3 ⊗ e3 − α I , where α, K ∈ R are given constants. a. Show that this tensor field can be cast in the form   E = K e1 ⊗ ∇φ − ∇ ⊗ a ,

(6.69)

where a ≡ a(X) and φ ≡ φ(X) are, respectively, vector and scalar fields that you should identify.

6.10 Exercises

315

b. Check that  u(X) = K

 1 2 X 3 + α(X 12 − X 22 ) e1 + α X 1 X 2 e2 − X 1 X 3 e3 2



satisfies the equation u ⊗ ∇ + ∇ ⊗ u = 2E. c. By using (6.69) deduce that the particular solution of this equation stated above could have been obtained by inspection. The same assertion applies to the previous question. 13. An isotropic elastic body occupies the simply connected domain Ω0 ⊂ E3 and is subjected to a state of deformation such that the corresponding displacement field, u = ui ei , satisfies u1 = u2 ≡ 0,

u3 = u3 (X 1 , X 2 ) .

a. Note that in this case E 13 ≡ E 13 (X 1 , X 2 ) and E 23 ≡ E 23 (X 1 , X 2 ) are the only non-zero components of the infinitesimal strain deformation tensor E, and then show that the Saint-Venant compatibility conditions are reduced to ∂ 2 E 13 ∂ 2 E 23 = , 2 ∂ X 1∂ X 2 ∂ X1

∂ 2 E 23 ∂ 2 E 13 = . ∂ X 1∂ X 2 ∂ X 22

(6.70)

Hence deduce that these equations are equivalent to ∂ E 13 ∂ E 23 = . ∂ X2 ∂ X1

(6.71)

b. Because there is no dependence on X 3 in E, a direct application of the formulae derived in Sect. 6.8 is not immediately possible. Convince yourself that in this case one may choose the integration path in (6.49) as shown in Fig. 6.12, and then deduce that the expression of the non-trivial component of u becomes   E 31 + (X 1 −ζ )E 31, 1 + (X 2 − X 20 )E 31, 2 dζ u3 (X 1 , X 2 ) = (3) L1   E 32 + (X 2 − ζ )E 32, 2 dζ , + L 2(3)

where L1(3) : Y = ζ e1 + X 20 e2 ,

X 10 ≤ ζ ≤ X 1 ,

L2(3) : Y = X 1 e1 + ζ e2 ,

X 20 ≤ ζ ≤ X 2 .

316

6 Compatibility of the Infinitesimal Deformation Tensor

X3

Fig. 6.12 Integration path for the Cesàro–Volterra formula in the particular case of Exercise 6.13

u3

X20

X10 X1

X2

P0 P1

P

14. Consider an infinitesimal deformation tensor E ≡ E(X) that admits the following representation in a Cartesian system of coordinates ⎡

⎤ 0 0 E 13 [E] = ⎣ 0 0 E 23 ⎦ . E 13 E 23 0

(6.72)

a. Show that  ×   S S inc E = −2 ∇ ⊗ ∇ E 13 × × (e1 ⊗ e3 ) − 2 ∇ ⊗ ∇ E 23 × (e2 ⊗ e3 ) , where (. . . ) S represents the symmetric part of the tensor enclosed between the parentheses. b. If E 13 ≡ E 13 (X 1 , X 2 ) and E 23 ≡ E 23 (X 1 , X 2 ), then use the result obtained at (a) to establish the following formula, ∇ ∧ E ∧ ∇ = 2(B12 − A22 )(e1 ⊗ e3 ) S + 2(A12 − B11 )(e2 ⊗ e3 ) S , where A := −∇ ⊗ ∇ E 13 , B := −∇ ⊗ ∇ E 23 , and Ai j , Bi j are the components of these tensors. Deduce the two equations in (6.70) by using the above results. 15. Show that Saint-Venant’s compatibility relations (6.14) can be obtained as a direct consequence of the assumption u, Ω ∈ C 2 (Ω0 ), where Ω is the skewsymmetric part of gradu ≡ u ⊗ ∇. In particular, show that E ik, jl − E jk, il = E il, jk − E jl, ik , 16.

(∀) i, j, k, l ∈ {1, 2, 3} .

a. If φ ≡ φ(X) is a smooth scalar field, then the following identity holds true, ∇ ∧ (φ I) ∧ ∇ = ∇ ⊗ ∇φ − (∇ 2 φ)I . b. Show that ∇ ∧ (∇ ⊗ ∇φ) = O.

6.10 Exercises

317

c. In a linearly elastic solid occupying a simply connected domain Ω0 ⊂ E3 , consider a state of deformation described by an infinitesimal deformation tensor of the form   E = ∇ 2 φ I − (1 + α)∇ ⊗ ∇φ , where α ∈ R and φ ≡ φ(X 1 , X 2 ) is a smooth scalar field. Show that if E ≡ E(X 1 , X 2 ) defined above satisfies the compatibility equation (6.13), then ∇ 4 φ = 0 in Ω0 . 17.

a. Let w be a vector field defined on Ω0 ⊂ E3 . If w ∈ C 3 (Ω0 ) show that     ∇2 ∇ ⊗ w = ∇ ⊗ ∇2w .

(6.73)

b. Consider an isotropic elastic body in equilibrium for which the body force per unit volume is derived from a harmonic potential, i.e. f ≡ ρ0 b := −∇φ(X) and ∇ 2 φ(X) = 0. Show that in this case ∇ 2 |T | = 0 and ∇ 4 T = 0 , where T is the usual infinitesimal stress tensor, and |T | ≡ T : I. 18. A linearly elastic isotropic body is in equilibrium in the absence of body forces. Show that if u = u(X) denotes the displacement field of this body, then ∇ 2 (∇ · u) = 0 , ∇ 2 (∇ ∧ u) = 0 . Furthermore, deduce that u is a bi-harmonic function, i.e. ∇ 4 u = 0. 19. Consider the screw dislocation seen in the sketch on the right in Fig. 6.10. This is a typical situation amenable to the anti-plane strain assumption (5.78). a. Check that the only non-zero components of the strain and stress tensors are E α3 = E α3 (X 1 , X 2 ) and Tα3 = Tα3 (X 1 , X 2 ), respectively, where E α3 =

1 u3, α , 2

Tα3 = μu3, α ,

(α = 1, 2).

(6.74)

Verify also that in the absence of body forces the equilibrium equation is ∇ 2 u3 = 0 .

(6.75)

b. In cylindrical polar coordinates the anti-plane strain assumption requires ur = uθ ≡ 0 and uz = uz (r, θ ). Show that in this case (6.74) are changed to Eθ z =

1 ∂uz , 2r ∂θ

Er z =

1 ∂uz , 2 ∂r

318

6 Compatibility of the Infinitesimal Deformation Tensor

and Tθ z =

μ ∂uz , r ∂θ

Tr z = μ

∂uz , ∂r

while all the other components of E and T are identically zero. How is the equilibrium equation in this case different from what has been found in the first part of this problem? c. For a cylinder of inner radius R1 and outer radius R2 , its inner (r = R1 ) and outer (r = R2 ) lateral surfaces are assumed to be traction free, i.e. Trr = Tr θ = Tr z = 0 ,

r = R1 , R2 .

In making the screw dislocation, the infinitely long cylinder is cut on the half-plane X 2 = 0, X 1 > 0, the top face of the cut (X 2 = 0+ ) being held fixed and the bottom face (X 2 = 0− ) is displaced by b in the z−direction. The displacement boundary conditions are uz (X 1 , 0+ ) − uz (X 1 , 0− ) = b ,

ur (X 1 , 0± ) = uθ (X 1 , 0± ) = 0 .

(i) Determine the elastic state of the screw dislocation by using the antiplane strain approximation together with uz = U (θ ), for some function U that you must specify. (ii) Write down the expressions of the Cartesian components of the stress tensor, and then check that the tractions on the two faces of the cut are equal and opposite. (ii) If the cylinder is of finite length, show that the traction distributions on the end surfaces that arise from the above solution are statically equivalent to a torque applied to the ends of the cylinder (hence, if the end surfaces are not constrained, the cylinder will twist when the screw dislocation is created).

Bibliography 1. Adams RA (1999) Calculus: a complete course. Addison-Wesley, Longman Ltd., Ontario 2. Salas SL, Hille E (1995) Calculus: one and several variables. Wiley, New York 3. Chou PC, Pagano NJ (1967) Elasticity: tensor, dyadic, and engineering approaches. D. Van Nostrand Company Inc, Princeton, New Jersey 4. Fraeijs de Veubeke BM (1979) A course in elasticity. Springer, New York 5. Gurtin ME (1972) Theory linear theory of elasticity. In: Truesdell C (ed) Handbuch der Physik. Springer, Berlin, pp 1–273 6. Love AEH (2011) A treatise of the mathematical theory of elasticity. Dover Publications, Mineola, New York 7. Malvern LE (1969) Introduction to the mechanics of a continuum medium. Prentice-Hall Inc, Englewood Cliffs, New Jersey 8. Nadeau G (1964) Introduction to elasticity. Holt, Rinehart and Winston Inc, New York 9. Teodosiu C (1982) Elastic models of crystal defects. Springer, Berlin

Chapter 7

Torsion

Abstract A simple example of anti-plane problem has already been discussed in some detail earlier in Chap. 5 (Example 5.2, as well as Exercises 5.11 and 5.12). We now consider a more elaborate anti-plane situation related to the torsional deformation of long and slender cylinders. Our treatment relies on the displacement formulation of the problem (based on the Navier–Lamé system) together with a semi-inverse approach. It is possible to give a more direct solution to the torsion problem by circumventing the need for ‘guessing’ the form of the displacement field. Such a treatment was first reported in the literature by the Italian engineer R. Baldacci (1957), who used the Beltrami–Michell equations as the starting point in his analysis; an expanded version of his original solution can be found in Baldacci (Scienza delle Costruzioni, 1983) [1] (pp. 200–238). L. Solomon (Élasticité Lineaire, 1968) [2] has also pursued a similar direct route, albeit his work was partly based on the use of complex variables (see pp. 140–183 of his book).

7.1 Introduction Let us consider a right circular cylinder C of length 2L > 0 as shown in Fig. 7.1a. The lateral surface of the cylinder, (Σ ), is assumed to be traction free, while its lower (Σ − ) and upper faces (Σ + ) are subjected to load distributions having a null resultant force and corresponding resultant moments equal to (+M a) and (−M a), respectively. Here, a represents a unit vector parallel to the symmetry axis of C (the latter being shown in that figure as the vertical dashed line) and M > 0 is a known number. As a result of these opposite axial moments the cylinder will be twisted: initially straight generators of the cylinder in the undeformed configuration become helical curves after the deformation has taken place. The question here is to quantify the twisting deformation in terms of the applied loading, M, and calculate the stresses, strains and displacements in the deformed cylinder; we shall make this statement more precise in Sect. 7.4. This problem was initially solved by the French scientist C. A. Coulomb in 1784, who postulated that under the twisting action mentioned above, cross sections perpendicular to the axis of the cylinder are merely being rotated in their own plane, © Springer Nature B.V. 2020 C. D. Coman, Continuum Mechanics and Linear Elasticity, Solid Mechanics and Its Applications 238, https://doi.org/10.1007/978-94-024-1771-5_7

319

320

7 Torsion

(a)

+Ma

(b)

+Ma

a

L

L

L -Ma

-Ma Fig. 7.1 Torsion of circular cylinders. In (a) the cylinder is assumed to be free and is deformed under the action of opposite twisting moments acting on its upper and lower faces; as explained in the text, transverse cross sections experience rigid rotations in their own plane, with the exception of the one shown shaded. The cylinder in (b) is fixed at the lower end and, again, subjected to a twisting moment on its upper face. Note that mechanical equilibrium requires the presence of an equal but opposite twisting moment on the constrained face

but no warping (i.e. out-of-plane deformation) takes place. He also assumed that the rotation undergone by each cross section depends only on the distance measured along the axis of the cylinder. It must be intuitively clear that the rotations undergone by Σ + and Σ − have opposite signs, and that the rotation of any generic cross section Σ is a continuous function of the distance measured along the axis of the cylinder. Thus, by the Intermediate Value Theorem in elementary Calculus there is at least one cross section that remains unaffected by the twisting deformation of C ; symmetry considerations indicate that this is the cross section equally spaced from the two flat ends of the cylinder. In conclusion, we can assume without loss of generality that the cylinder has its lower base fixed as in Fig. 7.1b. In Sect. 7.4 we are going to look at this problem following Coulomb’s original hypotheses. As it happens, his set of assumptions is applicable only to cylindrical solids of circular transverse cross sections. In the more general situation of noncircular cylinders, it turns out that the normal cross sections experience both in-plane and out-of-plane deformations that must be considered simultaneously. Although at first sight, it might seem that in this case we are dealing with a three-dimensional problem, in reality, the two modes of deformation can be uncoupled by using an extension of the Coulomb’s method that was first proposed by B. de Saint-Venant in 1855. His approach falls within the class of semi-inverse methods of solution on

7.1 Introduction

321

which we touched briefly in Chap. 5; the details of Saint-Venant’s solution for the torsion of non-circular cylinders will be presented in Sect. 7.5. The rest of the chapter contains a number of alternative formulations of this particular solution, as well as several examples that illustrate the theory introduced herein. In keeping with the spirit of this book, the derivations will be presented in a coordinate-free language whenever possible, since many of the results related to this problem are independent of the particular choice of coordinate systems (as long as the axis along which twisting takes place is parallel to the generators of the cylinder).

7.2 Some Auxiliary Notation As already mentioned in the previous section, one of the key features of the torsion problem for non-circular cylinders consists of the possibility of decoupling the deformation experienced by transverse cross sections in their own plane, and the effects related to the changes undergone by these originally flat surfaces in the axial direction. This particularity motivates the introduction of some new notations that will facilitate the application of vector and tensor methods in the subsequent sections. Let V be the usual three-dimensional Euclidean vector space, and consider φ : Ω ⊂ E3 → R. If a ∈ V is a unit vector (|a| = 1) and Π represents a plane normal to a, we know from Chap. 1 that a vector1 X ∈ V can be decomposed into normal and tangential components, X || and X ⊥ , respectively, according to the formulae X || := (a ⊗ a) · X and X ⊥ := (I − a ⊗ a) · X = a ∧ X ∧ a. By analogy with this representation, we introduce the formal decomposition of ∇φ into parallel and orthogonal parts with respect to the unit vector a, ∇ || φ(X) := (a ⊗ a) · ∇φ(X), ∇ ⊥ φ(X) := (I − a ⊗ a) · ∇φ(X),

(7.1a) (7.1b)

so that, formally, ∇φ = ∇ || φ + ∇ ⊥ φ. If we choose a Cartesian system of coordinates with a ≡ e3 and let eγ (γ = 1, 2) denote the unit vectors along the other two mutually orthogonal axes, then X = X i ei and we can write ∇ || φ ≡ e3

∂φ , ∂ X3

∇ ⊥ φ ≡ e1

∂φ ∂φ + e2 . ∂ X1 ∂ X2

(7.2)

Note that in this case X ⊥ ≡ X 1 e1 + X 2 e2 and X || ≡ X 3 e3 . As ∇φ is independent of the choice of coordinate system, and the projectors a ⊗ a and I − a ⊗ a enjoy the same property, the parallel and orthogonal gradients defined above describe intrinsic operations. The calculus of ∇ || and ∇ ⊥ is subject to similar rules and properties as the original ‘del’ operator. For example, formally, ∇⊥2 = ∇ ⊥ · ∇ ⊥ and all of the 1 Recall

that points in E3 are identified with their position vectors—see Sect. 1.4.

322

7 Torsion

Fig. 7.2 Decomposition of the position vector X into parallel (X || ) and orthogonal (X ⊥ ) components with respect to a given unit vector a

X3 Σ

O

X⊥

X2

P

a ≡ e3 X1

X||

X X2

O X1

Σ−

identities recorded in (A.1)–(A.3) in Appendix A hold true with obvious minor modifications. In addition, these operators benefit from a number of other properties that lead to simplifications in many circumstances. For example, a ∧ X = a ∧ X ⊥ since a ∧ X || = 0; similarly, a ∧ (∇φ) = a ∧ (∇ ⊥ φ). Let us also further note that a · (∇ ⊥ φ) = 0 and ∇φ(X ⊥ ) = ∇ ⊥ φ(X ⊥ ),

∇φ(X || ) = ∇ || φ(X || ).

The meaning of X ⊥ and X || within the context of the torsion problems discussed in this chapter is illustrated in Fig. 7.2, which shows a typical cylindrical segment (not necessarily circular), with Σ − being the lower base of the cylinder, while Σ represents an arbitrary transverse cross section. Included in that sketch is also a Cartesian system of coordinates in which the X 3 -axis is directed along the symmetry axis of the cylinder and is parallel to the unit vector a that defines the direction of the twisting/torsion axis. The origin of this system of coordinates is on the lower flat face, i.e. O ∈ Σ − , and coincides with the centroid of this two-dimensional domain; the orientation of the X 1 - and X 2 -axes is of little importance, and is subject only to the requirement that O X 1 X 2 X 3 forms a right-handed Cartesian system of coordinates in the sense made clear in Chap. 1. The X 3 -axis meets Σ at a point O , which is likewise the centroid of Σ, and one can immediately set up another system of coordinates O X 1 X 2 X 3 in which the X 1 - and X 2 -axes are parallel to their corresponding −−→ −−→ counterparts in Σ − . If P ∈ Σ is an arbitrary point, then X ⊥ ≡ O P and X || ≡ O O ; in particular, note that X ⊥ is the position vector of P with respect to the origin O of the system of coordinates defined in Σ. As we shall see shortly, it is X ⊥ rather than X that features in most of the calculations for the torsion problem, and this is a direct consequence of the decoupling between the two modes of deformation mentioned earlier on. With the notations introduced above, the cylinder in Fig. 7.1b can be identified with Σ × (0, L). Equivalently, one can think of the cylinder as the set of points P ∈ E3 whose position vectors X have the special form indicated below,

7.2 Some Auxiliary Notation

323

X = X ⊥ + X || ∈ V , a · X ⊥ = 0,

X || = X 3 a, (0 < X 3 < L).

(7.3)

Implicit in this description is the fact that the cylinders in question are geometrically uniform from one end to the other, which is the case with all cylindrical bodies considered henceforth. Another point worth emphasising is that Σ ± (upper/lower flat ends) are congruent with the typical cross section Σ. As shown later on in this chapter, a consequence of this feature is the possibility of reducing the torsion problem to what is, essentially, a two-dimensional boundary-value problem.

7.3 Governing Equations For easy reference, we record below the appropriate equations and boundary conditions that define the torsion problem in Linear Elasticity. As this is a static problem and body forces are ignored, the equilibrium equation (7.4a) is particularly simple. In addition, we need the geometrical relations (7.4b) that define the linearised deformation tensor, the compatibility condition (7.4c), and the constitutive law for linearly isotropic materials (7.4d). Equations (7.5) define the various conditions that must be enforced on the surface of the cylinder. Assuming that n is the outward unit normal to the lateral surface of the cylinder Σ , the condition (7.5a) is the requirement for this lateral surface to be traction free. On the flat ends of the cylinder (Σ ± ) we choose to impose averaged conditions corresponding to a zero resultant force (7.5c), and a resultant moment that is statically equivalent to a twisting moment perpendicular to the end faces (7.5d).2 To have equilibrium it is necessary that if (+M a) is the twisting moment on Σ + , then the corresponding value on Σ − must by (−M a). Finally, Eq. (7.5b) represents the requirement that the normal tractions on the flat end faces of the cylinder are null, and it is supposed to be satisfied pointwise on Σ ± ; this condition says that material line elements parallel to the longitudinal axis of the cylinder will not be subjected to any tensile or compressive forces along their length. ∇ · T = 0, (7.4a) 1 E = (∇ ⊗ u + u ⊗ ∇) , (7.4b) 2 ∇∧ E∧∇ = O, (7.4c) T = 2μE + λ|E|I ,

(7.4d)

t(n) = 0 on Σ , (7.5a) ± T : (a ⊗ a) = 0 on Σ , (7.5b)  t(±a) d A = 0 , (7.5c) ± Σ X ⊥ ∧ t(±a) d A = ±M a . Σ±

(7.5d)

2 It is tacitly assumed in this equation that an origin has been chosen in Σ, relative to which we measure X ⊥ ; this choice and the effect it has on the various quantities of interest (stresses, displacements, and so on) will be taken up at length in Sect. 7.6.

324

7 Torsion

7.4 Circular Cylinder As already mentioned in Sect. 7.1, the key assumption in the solution of the elastic torsion problem for circular cylinders is that cross sections normal to the axis of the cylinder undergo rigid rotations in their own plane and preserve their planarity (so there is no displacement in the a-direction). This suggests that the displacement field in the cylinder will be of the form u(X) = R(a; θ ) · X − X,

(7.6)

where R(a; θ ) ≡ a ⊗ a + cos θ (I − a ⊗ a) + sin θ (I ∧ a) represents the rotation through an angle equal to θ about an axis defined by the unit vector a. Because we are interested in small rotations, θ  0, and then cos θ  1, sin θ  θ . With these simplifications in mind, the rotation tensor becomes R(a; θ ) = I + θ (I ∧ a). The angle θ is not constant, but depends on the distance from the fixed end of the cylinder measured along its axis, i.e. θ = θ (X || ). Thus, from (7.6) we then get u(X) = (a ∧ X)θ (X || ).

(7.7)

At this stage the dependence of θ on X || is still unknown. However, if (7.7) is to represent the displacement field of the torsion problem, then it must satisfy (7.4a) or, equivalently, the Navier–Lamé equation μ∇ 2 u + (λ + μ)∇(∇ · u) = 0.

(7.8)

As we are about to see, enforcing this requirement will lead to the conclusion that θ ≡ θ (X || ) is a linear function. To this end, let us start by evaluating the differential operators that feature in (7.8) for u given by (7.7),   ∇ · u = ∇ · θ (a ∧ X) = (∇ || θ ) · (a ∧ X) + θ ∇ · (a ∧ X)   = (∇ || θ ) · (a ∧ X) + θ (∇ ∧ a) · X − a · (∇ ∧ X) dθ a · (a ∧ X) = 0, = (∇ || θ ) · (a ∧ X) = d X3 since ∇ ∧ X = 0 = ∇ ∧ a and ∇ || θ = a(dθ /d X 3 ). In conclusion, ∇ · u = 0 and hence ∇(∇ · u) = 0. Next, let us deal with the only other term left in (7.8), that is, ∇ 2 u ≡ ∇ · (∇ ⊗ u). Simple calculations show that   ∇ ⊗ u = ∇ ⊗ θ (a ∧ X) = (∇ || θ ) ⊗ (a ∧ X) + θ ∇ ⊗ (a ∧ X) = (∇ || θ ) ⊗ (a ∧ X) − θ (I ∧ a),

(7.9)

7.4 Circular Cylinder

325

where use has been made of the relation ∇ ⊗ (a ∧ X) = −I ∧ a. Taking the divergence of (7.9) yields successively,     ∇ 2 u = ∇ · (∇ || θ) ⊗ (a ∧ X) − ∇ · θ (I ∧ a)     = (∇||2 θ )(a ∧ X) + (∇ || θ ) · ∇ ⊗ (a ∧ X) − (∇ || θ) · (I ∧ a) + θ∇ · (I ∧ a) = (∇||2 θ )(a ∧ X) − 2(∇ || θ ) · (I ∧ a).

(7.10)

This is amenable to a further simplification because (∇ || θ ) · (I ∧ a) = (I ∧ a)T · ∇ || θ = −(a ∧ I) · ∇ || θ = −(I ∧ a) · ∇ || θ = −a ∧ (∇ || θ ).

(7.11)

Replacing (7.11) into (7.10), and remembering that ∇ || θ is parallel to a, allows us to conclude that ∇ 2 u = (∇||2 θ )(a ∧ X). If Eq. (7.8) is to be satisfied for all points in C , then d 2θ or = 0. ∇||2 θ = 0 d X 32 Integrating twice this last equation gives θ = α X 3 + β, for some α, β ∈ R. Remembering that the end face X 3 = 0 was assumed to be fixed, θ (X 3 = 0) = 0, and hence θ = α X 3 . The constant α is known as the twist per unit length of the cylinder; it measures the angle of rotation of the cross section per unit length, that is, the rate of twist of the cylinder.3 At this stage, α is not yet determined and will have to be found later in terms of the applied twisting moment, by using (7.5d). In conclusion, we have shown that the displacement field for the torsion of circular cylinders has the expression u(X) = α(a · X)(a ∧ X) ≡ α(a · X || )(a ∧ X ⊥ ).

(7.12)

Our next task will be to check that (7.12) is compatible with the boundary conditions stated in Eqs. (7.5a)–(7.5c). For this we require an explicit expression for the stress vector. As we have already calculated ∇ ⊗ u in (7.9), we deduce immediately that u ⊗ ∇ ≡ (∇ ⊗ u)T = α(a ∧ X) ⊗ a + α X 3 (I ∧ a) and, since I ∧ a is a skewsymmetric tensor, then E=

 1  α (a ⊗ a) ∧ X − X ∧ (a ⊗ a) . 2

(7.13)

The trace of this tensor is simply obtained by replacing ‘⊗’ with ‘·’, after taking into account the associativity of the operations represented by ‘⊗’ and ‘∧’. Because that α = dθ/dX 3 , so it represents the rate of change of θ along the X 3 -axis, i.e. along the cylinder axis.

3 Note

326

7 Torsion

the resulting scalar triple products have two identical terms, they will be identically zero, and hence |E| = 0 (i.e. the volumetric dilatation is zero). The constitutive law (7.4d) becomes T = 2μE or, with the help of (7.13),   T = αμ (a ⊗ a) ∧ X − X ∧ (a ⊗ a) .

(7.14)

Note that the boundary condition (7.5b) is satisfied by this expression. The stress vector corresponding to the normal cross sections is      t(a) = a · T = αμ a · (a ⊗ a) ∧ X − a · X ∧ (a ⊗ a)   = αμ a ∧ X − [a, X, a]a = αμ(a ∧ X), so that t(a) = αμ(a ∧ X ⊥ ).

(7.15)

The boundary condition (7.5c) is equivalent to  (αμ) a ∧

Σ±

 X ⊥ d A = 0.

Observing that the integral in the above equation is just the static moment vector of Σ ± (see Appendix C), this condition will be automatically satisfied, provided that the axis of the cylinder passes through the centroid of the cross section. As for condition (7.5a), it is obviously satisfied; indeed,  

t(n) = n · T = αμ (n · a)(a ∧ X ⊥ ) − n · (X ⊥ ∧ a) a ,

X ⊥ ∈ ∂Σ

which is identically zero because n ⊥ a and n · (X ⊥ ∧ a) = [n, X ⊥ , a] = [a, n, X ⊥ ] = a · (n ∧ X ⊥ ) = 0 (due to n ∧ X ⊥ = 0, if ∂Σ is a circle). Finally, the boundary condition (7.5d) will provide the link between M and α. In light of (7.15), this constraint can be written as 

 Σ+

X ⊥ ∧ (a ∧ X ⊥ ) d A =

 M a. αμ

Expanding4 further the integrand on the left-hand side according to X ⊥ ∧ (a ∧ X ⊥ ) = |X ⊥ |2 a − (X ⊥ · a)X ⊥ = |X ⊥ |2 a,

this equation is linear in a and the outward unit normal on Σ − is parallel to (−a), the other equation in (7.5d) does not give any new information.

4 As

7.4 Circular Cylinder

327



we discover that M = μα J0 ,

J0 :=

Σ+

|X ⊥ |2 d A,

(7.16)

where J0 is the polar moment of inertia of Σ + . It can be seen without difficulty that Σ + in the expression of J0 can be replaced by a generic cross section Σ without altering the result. This completes the solution of the elastic torsion problem for circular cylinders, since Eq. (7.16) provides an explicit dependence for the twist per unit length (α) in terms of the applied twisting moment (M). Before we move on to solving the same problem for non-circular cylinders, it is instructive to see the results obtained from a more quantitative perspective. Assuming that the radius of the circular cross section is R > 0, and noticing that |X ⊥ |2 = r 2 = X 12 + X 22 , a simple calculation gives  J0 =

 Σ

X 12 + X 22 d A =



 Σ

2π

r2 dA = 0



R

r 3 dr dθ =

0

1 π R4, 2

and then the direct application of (7.16) produces α=

2M M = . μJ0 μπ R 4

According to this formula, if M is kept constant while the radius R of the cross section is being increased, then the twist per unit length will decrease as R −4 ; on the other hand, if the size of the cross section is fixed, but M is variable, we see that α increases proportionally with this quantity. Another question of interest would be to find the points in a normal cross section that experience maximum tangential stresses. This question is easily answered with the help of (7.15), since t(a) = αμ(a ∧ X ⊥ ) = and hence |t(a)| =

2M (a ∧ X ⊥ ), π R4

2M 2Mr 2M |a||X ⊥ | sin ∠(a, X ⊥ ) = ≤ . π R4 π R4 π R3

The maximum is attained for r = R, i.e. maximum shear traction on a typical normal cross section occurs at the boundary ∂Σ. We are going to see in Sect. 7.10 that this is a generic feature shared by a large class of cross-sectional geometries.

7.5 Non-circular Cylinder Our next immediate objective is to modify the solution of Sect. 7.4 so as to be able to account for the warping of the transverse cross sections that normally accompanies the torsion of non-circular cylinders. Because the problem we are dealing with

328

7 Torsion

involves small rotations and displacements, intuitively, we expect the warping effect to be an additive contribution (a corrective term) to the solution already found in the previous section. Making use again of the semi-inverse method, we shall look for a displacement field u = u(X) in the form u(X) = u⊥ (X) + u|| (X),

(7.17)

u⊥ (X) := α(a · X)(a ∧ X), u|| (X) := αΨ (X ⊥ )a.

(7.18a) (7.18b)

where

Note that u⊥ is just the displacement field that was used for the torsion of the circular cylinder; the new term, u|| , describes the warping of the normal cross sections. According to (7.18b), the displacements associated with this effect are parallel to the axis of the cylinder (whose direction is defined by the unit vector a), and their magnitude and direction are described by the so-called Saint-Venant warping function Ψ .5 This is an unknown of the problem that will have to be determined before we can relate the twist per unit length α to the applied twisting moment M. The constant α in front of this function in Eq. (7.18b) is a normalising factor introduced for the sole purpose of simplifying the algebraic manipulation later in this section. It might seem odd at this stage that Ψ in (7.18b) is taken to depend on X ⊥ rather than X. Geometrically, this means that all normal cross sections will experience the same amount of warping, the nature of this warping effect being determined by Ψ and its magnitude by α. This choice of independent variables for Ψ is not arbitrary, but follows directly from the Navier–Lamé system and is an immediate consequence of (7.18a). Before we go any further, it is instructive to see the arguments that lead to this conclusion; to this end, we shall consider our usual right-handed Cartesian system of coordinates with the positive X 3 -axis pointing in the same direction as a. We start by first observing that, according to (7.18a), u1 = −α X 2 X 3

and

u2 = α X 1 X 3 ,

so that ∇ · u = u3, 3 and ∇ 2 u = (∇ 2 u3 )e3 . Hence the Navier–Lamé system (7.8) simplifies to u3,13 = 0,

(7.19a)

u3,23 = 0,

(7.19b)

u3, 11 + u3, 22 + 5 If

2(1 − ν) u3, 33 = 0. 1 − 2ν

(7.19c)

the displacements in the cylinder are given by (7.17) and (7.18), then the initially plane cross sections perpendicular to a rotate rigidly about this vector and, in addition, warp so that the plane X 3 = X 30 becomes the surface X 3 = X 30 + αΨ (X 1 , X 2 ).

7.5 Non-circular Cylinder

329

Integrating (7.19a) and (7.19b) with respect to X 3 we find that u3, 1 = f (X 1 , X 2 ),

u3, 2 = g(X 1 , X 2 ),

where f and g are arbitrary functions at this stage. But (u3, 3 ), 1 = 0 from (7.19a), and hence u3, 3 is a function of X 2 and X 3 only. Furthermore, from (7.19b) we also have (u3, 3 ), 2 = 0, which tell us that u3,3 does not depend on X 2 . In conclusion, u3, 3 depends only on X 3 , and so does u3, 33 , the last term in (7.19c). On the other hand, the underlined terms in the same equation are functions of X 1 and X 2 only. We are thus led to conclude that −

1 − 2ν (u3,11 + u3,22 ) = u3,33 = 2b, 2(1 − ν)

(7.20)

where b ∈ R is an arbitrary constant. Integrating twice with respect to X 3 the second equation above, yields u3 (X) = bX 32 + cX 3 + w0 (X 1 , X 2 ),

(7.21)

for some c ∈ R and w0 a function that depends on X 1 and X 2 . Since u3 in this notation is the counterpart of αΨ in (7.18b), it must be clear that (7.21) does not match our original assumption unless b = c = 0. To justify this fact we need to recall the boundary condition (7.5c); in our present notation,   T33 d A = T33 d A = 0. {X 3 =0}

{X 3 =L}

From Hooke’s Law (7.4d), T33 = 2μE 33 + λ(∇ · u) = 2μu3, 3 + λu3, 3 or T33 = (λ + 2μ)(2bX 3 + c). On making the appropriate substitution in the boundary condition recorded above results in a linear system of homogeneous equations for b and c, whence b = c = 0. This completes our justification as to why the Saint-Venant warping function does not depend on X 3 , and we are now ready to go back to explaining how this function is determined (in principle, the first equation in (7.20) already shows that ∇ 2 Ψ = 0, but we shall make a brief detour to obtain some additional information as well). Because the infinitesimal deformation tensor E = E(u) is linear in u, and we have already calculated in Sect. 7.4 the corresponding E associated to u⊥ , all that remains to be done is find E associated with u|| . Direct calculations show that   ∇ ⊗ u|| = α∇ ⊗ Ψ (X ⊥ )a = α∇ ⊥ Ψ (X ⊥ ) ⊗ a, which in conjunction with (7.13) leads to the expression of E for the kinematics adopted in (7.17) and (7.18),

330

7 Torsion

E=

1  α (a ⊗ a) ∧ X − X ∧ (a ⊗ a) 2  + a ⊗ (∇ ⊥ Ψ ) + (∇ ⊥ Ψ ) ⊗ a .

(7.22)

Since ∇ ⊥ Ψ and a are orthogonal vectors, the trace of E will again be zero, i.e. |E| = 0. Unsurprisingly, the implication on the constitutive law is that T = 2μE, so the Cauchy stress vector on the transverse cross sections will be   t(a) = a · T = αμ(a ∧ X) + αμ (a · a)∇ ⊥ Ψ + (a · ∇ ⊥ Ψ )a = αμ(a ∧ X + ∇ ⊥ Ψ ).

(7.23)

Similarly, the stress vector on the lateral surface of the cylinder is found to be   t(n) = αμ n · (a ∧ X + ∇ ⊥ Ψ ) a,

(7.24)

where n is the outward unit normal to Σ . In Sect. 7.4 we have seen that the kinematics (θ in Eq. (7.7)) was fixed after demanding that our assumed displacement field must satisfy the Navier–Lamé system. It turns out that this is true in the present case as well; more precisely, by following the same route, we are going to discover that Ψ must be a plane-harmonic function. The corresponding details are presented next. Using the linearity of (7.8), together with (7.17) and the fact that u⊥ already satisfies the Navier–Lamé system, suggests that we need only concentrate on μ∇ 2 u|| + (λ + μ)∇(∇ · u|| ) = 0. The second term in this equation vanishes, because ∇ · u|| = 0, and therefore we are left with ∇ 2 u|| = 0 or ∇⊥2 Ψ (X ⊥ ) = 0,

(7.25)

since ∇ ⊗ u|| = (∇ ⊥ Ψ ) ⊗ a, and then   ∇ 2 u|| = ∇ · (∇ ⊗ u|| ) = ∇ · (∇ ⊥ Ψ ) ⊗ a) = (∇⊥2 Ψ ) ⊗ a. In summary, we have found that the Saint-Venant warping function must satisfy the interior Neumann boundary-value problem ⎧ 2 ⎪ in Σ, ⎨∇⊥ Ψ = 0 (7.26) ∂Ψ ⎪ ⎩ = −n · (a ∧ X ⊥ ) on ∂Σ. ∂n Interestingly, Ψ is determined purely by the geometry of the cylinder cross section, and will be the same for all cylinders that are made of different linearly isotropic materials, but have the same cross section.

7.5 Non-circular Cylinder

331

A general solution of (7.26) that works for all possible cross sections is not practical and, in general, this task is best handled on a case-by-case basis. We are going to see in Sect. 7.12 how to solve this boundary-value problem and its equivalent formulations (described in Sects. 7.7 and 7.8) in some particular cases of interest, but for now the main priority is to deal with the remaining boundary conditions in (7.5). The problem of solving (7.26) for a given planar domain Σ ⊂ E2 is sometimes referred to as the ‘torsion problem’ for the cross section Σ. This terminology is justified in light of the kinematics adopted in (7.18) and the results presented in the remaining of this chapter. In the interest of not overloading the notation in what follows, we write Y := X ⊥ (this means that Y will be a two-dimensional position vector of a point located in the transverse cross section of the cylinder). Let us start by checking that (7.5c) is satisfied, i.e.   t(a) d A ≡ μα (a ∧ Y + ∇ ⊥ Ψ ) d A = 0; (7.27) Σ+

Σ+

only the condition for the upper face of the cylinder (Σ + ) is considered here because the argument for the lower face (Σ − ) is identically similar. We shall assume that ∂Σ + (and implicitly ∂Σ − ) is a smooth curve parametrised by the arclength s; in other words, Y = Y (s) for all Y ∈ ∂Σ + .6 In this case the outward unit normal to the curve bounding the normal cross section is given by n = Y s ∧ a,

with Y s :=

dY . ds

(7.28)

Using this new piece of information we can then rewrite the boundary condition in (7.26) as follows (Y s ∧ a) · (a ∧ Y ) + (Y s ∧ a) · ∇ ⊥ Ψ = 0.

(7.29)

The first term on the left-hand side is just   Y s · a Y s · Y   = −Y s · Y ,  a·a a·Y  since a ⊥ Y and a · a = |a|2 = 1. Substituting back into (7.29) what we have just found, yields Y s · (a ∧ ∇ ⊥ Ψ − Y ) = 0. After multiplying this equation by Y it is easy to see that the resulting expression can be cast as 6 The

same parametrisation of the boundary curve will apply to all transverse cross sections (Σ) of the cylinder.

332

7 Torsion

  Y s · (a ∧ ∇ ⊥ Ψ ) ⊗ Y − Y ⊗ Y = 0. As this is supposed to hold on ∂Σ + , we can integrate along the boundary of Σ + to get    dY · (a ∧ ∇ ⊥ Ψ ) ⊗ Y − Y ⊗ Y = 0. (7.30) ∂Σ +

Next, the path integral in (7.30) will be transformed into an area integral by using Stokes Theorem that is recalled below for our particular setting   dY · B = a · (∇ ⊥ ∧ B) d A, (7.31) ∂Σ +

Σ+

where B := (a ∧ ∇ ⊥ Ψ ) ⊗ Y − Y ⊗ Y . To calculate the ‘curl’ of B, we note that this tensor is the sum of two simpler tensors, so we calculate the ‘curl’ of each of these individual terms separately. The corresponding details follow next. Note that from (A.3c), ∇ ⊥ ∧ (Y ⊗ Y ) = (∇ ⊥ ∧ Y ) ⊗ Y − Y ∧ (∇ ⊥ ⊗ Y ) = −Y ∧ I 2 ,

(7.32)

since ∇ ⊥ ∧ Y = 0 and ∇ ⊥ ⊗ Y = I 2 . As for the other term in the definition of B,     ∇ ⊥ ∧ (a ∧ ∇ ⊥ Ψ ) ⊗ Y = ∇ ⊥ ∧ (a ∧ ∇ ⊥ Ψ ) ⊗ Y − (a ∧ ∇ ⊥ Ψ ) ∧ (∇ ⊥ ⊗ Y )   = a(∇ ⊥ · ∇ ⊥ Ψ ) − a · ∇ ⊥ ⊗ (∇ ⊥ Ψ ) ⊗ Y − (a ∧ ∇ ⊥ Ψ ) ∧ I 2 = −(a ∧ ∇ ⊥ Ψ ) ∧ I 2 ,

(7.33)

where use has been made of ∇ ⊥ · ∇ ⊥ Ψ = ∇⊥2 Ψ = 0 and a · ∇ ⊥ ⊗ (∇ ⊥ Ψ ) = 0. Combining (7.32) and (7.33) gives ∇ ⊥ ∧ B = −(a ∧ ∇ ⊥ Ψ − Y ) ∧ I 2 . Finally, all that remains to be done is take the dot-product of this equation with a in order to evaluate the integrand of the term on the right-hand side of (7.31). Thus,   a · (∇ ⊥ ∧ B) = I 2 ∧ (a ∧ ∇ ⊥ Ψ − Y ) · a = (a ∧ ∇ ⊥ Ψ − Y ) ∧ a = (a ∧ ∇ ⊥ Ψ ) ∧ a − Y ∧ a = −(a · ∇ ⊥ Ψ )a + (a · a)∇ ⊥ Ψ − Y ∧ a = ∇⊥ Ψ + a ∧ Y .

(7.34)

Substituting (7.34) into (7.31), and using (7.30), we discover that (7.27) is satisfied.

7.5 Non-circular Cylinder

333

The boundary condition (7.5d) which specifies the loading of the cylinder will provide the link between α and M, exactly as in the case of the circular cylinder in Sect. 7.4, but we can already anticipate that the Saint-Venant warping function will feature in that relationship. By taking into account the expression of t(a) found in (7.23) we can write    X ⊥ ∧ (a ∧ X ⊥ ) + X ⊥ ∧ (∇ ⊥ Ψ ) d A X ⊥ ∧ t(a) d A = μα Σ+ Σ+    = μα |X ⊥ |2 a + X ⊥ ∧ (∇ ⊥ Ψ ) d A ; 

Ma =

Σ+

after taking the dot-product of this equation with a, we finally obtain M = μα

   |X ⊥ |2 + (X ⊥ ∧ ∇ ⊥ Ψ ) · a d A.

(7.35)

Σ

Note that we have replaced Σ + (the upper face of the cylinder) with Σ (a generic transverse cross section) because these are identical two-dimensional domains. By letting    |X ⊥ |2 + (X ⊥ ∧ ∇ ⊥ Ψ ) · a d A, (7.36) J := Σ

Equation (7.35) becomes M = (μJ )α or α=

M . μJ

(7.37)

The quantity J defined in Eq. (7.36) is known as the torsion constant, while D := μJ represents the torsional rigidity or inertia of the cylinder; this name is justified on account of the relationship M = Dα that follows immediately from (7.37). Assuming the usual system of Cartesian coordinates, X ⊥ = X 1 e1 + X 2 e2 and ∇ ⊥ is given by (7.2). Elementary algebra then shows that the evaluation of (7.36) reduces to finding the integral    J= (7.38) X 1 X 1 + Ψ, 2 + X 2 X 2 − Ψ, 1 d A ; Σ

its calculation is routinely done once the Saint-Venant warping function Ψ ≡ Ψ (X 1 , X 2 ) has been found by solving (7.26). It is of interest to know if circular cylinders are the only cylindrical bodies for which warping of the transverse cross sections can be neglected. This fact can be checked by using the boundary-value problem that defines the Saint-Venant warping function—see Eq. (7.26). Clearly, warping will be absent only if Ψ = constant, for in this situation the displacement in the a-direction is a rigid translation that preserves the planarity of the transverse cross sections. The normal derivative in (7.26) is then zero, so n · (a ∧ Y ) = 0 or, according to (7.28), (Y s ∧ a) · (a ∧ Y ) = 0. The left-

334

7 Torsion

hand side of this last equation can be expanded following the same strategy as in our work on (7.29), leading to the conclusion that Ys · Y = 0

or

d (Y · Y ) = 0 ds

on ∂Σ,

whence |Y |2 ≡ Y12 + Y22 = constant, i.e. ∂Σ is described by the equation of a circle, and our earlier assertion is thus validated. A summary of the component form for the solution of the torsion problem in the case of non-circular cylinder is included below for convenience: u1 = −α X 2 X 3 , u2 = α X 1 X 3 , u3 = αΨ (X 1 , X 2 ),

DISPLACEMENTS:

TORSION CONSTANT:

(7.39a) Ψ,11 + Ψ,22 = 0 in Σ, (7.39b) dX 1  dX 2  Ψ, 1 − X 2 − Ψ, 2 + X 1 = 0 on ∂Σ, ds ds (7.39c)   1 1 E 31 = α Ψ, 1 − X 2 , E 32 = α Ψ, 2 + X 1 , 2 2 (7.39d)   T31 = μα Ψ, 1 − X 2 , T32 = μα Ψ, 2 + X 1 , (7.39e)   2 J= X 1 + X 22 + X 1 Ψ, 2 − X 2 Ψ, 1 d A, (7.39f)

TWIST ANGLE:

M = Dα,

WARPING FUNCTION:

STRAINS:

STRESSES:

Σ

D ≡ μJ is the torsional rigidity.

(7.39g)

7.6 A Closer Look at the Torsional Rigidity Proposition 7.1 The torsion constant defined in (7.36) can be expressed as  J= |∇ ⊥ Ψ + a ∧ X ⊥ |2 d A; (7.40) Σ

in particular, D ≡ μJ > 0, i.e. the torsional rigidity is a positive constant (and hence the applied twisting moment, M, and the twist per unit length, α, have the same sign). We start checking formula (7.40) by first expanding the integrand on the right-handside of that equation, |∇ ⊥ Ψ + a ∧ X ⊥ |2 = (∇ ⊥ Ψ + a ∧ X ⊥ ) · (∇ ⊥ Ψ + a ∧ X ⊥ ) = |∇ ⊥ Ψ |2 + 2(∇ ⊥ Ψ ) · (a ∧ X ⊥ ) + (a ∧ X ⊥ ) · (a ∧ X ⊥ ) = |∇ ⊥ Ψ |2 + 2(∇ ⊥ Ψ ) · (a ∧ X ⊥ ) + |X ⊥ |2 ,

(7.41)

7.6 A Closer Look at the Torsional Rigidity

335

since a ⊥ X ⊥ . Noticing the following property    ∇ ⊥ Ψ · (a ∧ X ⊥ ) = ∇ ⊥ Ψ, a, X ⊥ = a, X ⊥ , ∇ ⊥ Ψ ] = a · (X ⊥ ∧ ∇ ⊥ Ψ ), we come to to the conclusion that |∇ ⊥ Ψ + a ∧ X ⊥ |2 =|X ⊥ |2 + a · (X ⊥ ∧ ∇ ⊥ Ψ ) + |∇ ⊥ Ψ |2 + (∇ ⊥ Ψ ) · (a ∧ X ⊥ ). Thus, in light of this last result, the equivalence of (7.36) and (7.40) is reduced to checking  |∇ ⊥ Ψ |2 + (∇ ⊥ Ψ ) · (a ∧ X ⊥ ) d A = 0. (7.42) Σ

To this end, we shall use one of Green’s identities (see Exercise 7.10) for the planeharmonic function Ψ ,   ∂Ψ ds. |∇ ⊥ Ψ |2 d A = Ψ ∂n Σ ∂Σ The normal derivative in the integrand on the right-hand side of the above identity is available from (7.26), so the (two-dimensional version) of the Divergence Theorem leads to    2 |∇ ⊥ Ψ | d A = − n · (Ψ a ∧ X ⊥ ) ds = − ∇ ⊥ · (Ψ a ∧ X ⊥ ) d A Σ Σ ∂Σ (7.43) = − (∇ ⊥ Ψ ) · (a ∧ X ⊥ ) d A, Σ

since according to (A.2e)   ∇ ⊥ · (Ψ a ∧ X ⊥ ) = ∇ ⊥ ∧ (Ψ a) · X ⊥ − (Ψ a) · (∇ ⊥ ∧ X ⊥ )   = (∇ ⊥ Ψ ) ∧ a · X ⊥ = [X ⊥ , ∇ ⊥ Ψ, a] = [∇ ⊥ Ψ, a, X ⊥ ] = (∇ ⊥ Ψ ) · (a ∧ X ⊥ ). The justification of (7.42) is now completed because that result corresponds to what was found in (7.43), after moving all the terms to the left of the equality sign. Note that we still have to show that J > 0. It is clear from (7.40) that J ≥ 0, but the question is whether the equality with zero can happen in certain special cases. To prove that this is not the case, let us assume by contradiction that J = 0. Since the integrand in (7.40) is positive this can only happen if the integrand is identically zero, which in turn implies that ∇ ⊥ Ψ + a ∧ X ⊥ ≡ 0 or, what is the same, ∇ ⊥ Ψ = X ⊥ ∧ a. Taking the curl of both sides in this last equation we discover that

336

7 Torsion

0 ≡ ∇ ⊥ ∧ (∇ ⊥ Ψ ) = ∇ ⊥ ∧ (X ⊥ ∧ a) = −2a, where use has been made of the identity (A.3d). Clearly, this is impossible because a is a unit vector, so it remains that J > 0. The proof of Proposition 7.1 is now complete. Proposition 7.2 The torsion constant J defined in (7.36) is an intrinsic geometrical characteristic of the cross section, i.e. it is independent of the choice of the system of coordinates. Let Ψi (i = 1, 2) be the solutions of (7.26) for two different choices O 1 and O 2 of the origin, and note (see Fig. 7.3) that −−−→ (2) X (1) ⊥ − X ⊥ = O 1 O2,

(7.44)

−−→ −−→ (2) where X (1) ⊥ ≡ O 1 P and X ⊥ ≡ O 2 P. The warping functions in the two cases will satisfy ⎧ 2 in Σ, ⎨ ∇⊥ Ψ 2 = 0 ⎩ ∂Ψ 2 = −n ·  a ∧ X (2)  on ∂Σ. ⊥ ∂n

⎧ 2 in Σ, ⎨∇⊥ Ψ 1 = 0 ⎩ ∂Ψ 1 = −n ·  a ∧ X (1)  on ∂Σ, ⊥ ∂n

 := Ψ 1 − Ψ 2 , by subtracting these equations term by term we discover that If Ψ  = 0 in Σ ∇⊥2 Ψ

and

  −−−→ ∂Ψ = n · O 1 O 2 ∧ a on ∂Σ. ∂n

(7.45)

The difference between (7.45) and (7.26) lies in the boundary condition: in the  along ∂Σ must be constant, so we suspect that former problem the gradient of Ψ this function has a particularly simple expression. Indeed, introducing further  −−→  := Ψ − − Ψ O 1 O 2 ∧ a · X (2) ⊥ , −−→  = ∇⊥ Ψ −− it follows immediately that ∇ ⊥ Ψ O 1 O 2 ∧ a, and then  = 0 in Σ ∇⊥2 Ψ

and

 ∂Ψ = 0 on ∂Σ. ∂n

(7.46)

 ≡ 0, so The solution of (7.46) is Ψ

and, furthermore,

−−−→ Ψ 2 = Ψ 1 + (a ∧ O 1 O 2 ) · X (2) ⊥ ,

(7.47)

−−−→ ∇⊥ Ψ 2 = ∇⊥ Ψ 1 + a ∧ O 1 O 2 .

(7.48)

7.6 A Closer Look at the Torsional Rigidity

337

X3 = X3 X3 = X3

Σ+

O2

O1

X1

P

X2

X2 X1 X2

O

X2

O X1

X1

Σ−

Fig. 7.3 The effect of changing the origin of the coordinate system. The original Cartesian coordi

nate system O X 1 X 2 X 3 is replaced with O X 1 X 2 X 3 , in which the corresponding axes are parallel. As a result, in a generic cross section Σ, the associated system of coordinates are O1 X 1 X 2 X 3 and, respectively, O2 X 1 X 2 X 3 and the question here is how this particular change affects the solution of the boundary-value problem (7.26) for the Saint-Venant warping function Ψ

We are now ready to verify that the torsional rigidity is invariant when the origin of our system of coordinates is changed. Let us denote by J (1) and J (2) the values of the torsion constant (7.36) corresponding to the warping functions Ψ1 and, respectively, Ψ2 , defined above. Using Proposition 7.1 in conjunction with (7.44) and (7.48), we have successively       −−−→   (2) 2 (2) 2 ∇ ⊥ Ψ 2 + a ∧ X ⊥  d A = ∇ ⊥ Ψ 1 + a ∧ O 1 O 2 + a ∧ X ⊥  d A Σ Σ      −−−→   (2) 2 (1) 2 = ∇ ⊥ Ψ 1 + a ∧ O 1 O 2 + X ⊥  d A = ∇ ⊥ Ψ 1 + a ∧ X ⊥  d A = J (1) .

J (2) =

Σ

Σ

This completes the proof of Proposition 7.2. It is possible to take advantage of the above calculations in order to ascertain the effect that a change of origin will have on the stress and displacement fields in the twisted cylinder. In light of (7.48), (2) ∇ ⊥ Ψ 1 + a ∧ X (1) ⊥ = ∇⊥ Ψ 2 + a ∧ X ⊥ ,

(7.49)

and then the invariance of stresses follows at once by noticing that from (7.22),   T = μα a ⊗ (a ∧ X ⊥ + ∇ ⊥ Ψ ) + (∇ ⊥ Ψ + a ∧ X ⊥ ) ⊗ a .

(7.50)

For displacements the situation is somewhat different, in the sense that the displacement fields u(1) and u(2) associated with two distinct positions of the origin of our

338

7 Torsion

Cartesian system of coordinates will differ from each other through a rigid-body transformation; the justification for this assertion is included below. From (7.17) to (7.18) the displacement field corresponding to Ψ p ( p = 1, 2) will be given by u( p) =

  M  ( p)  ( p) a · X ( p) a ∧ X ⊥ + aΨ p (X ⊥ ) μJ

(no summation).

(7.51)

−−−→ (1) On the other hand, thanks to (7.44), we can write a ∧ X (2) ⊥ = a ∧ X ⊥ − a ∧ O 1 O 2, and it is also evident that a · X (1) = a · X (2) . Using these facts, and remembering (7.47), it transpires that u(2) = u(1) +

  M  −−−→ (2)  −−−→  a, O 1 O 2 , X ⊥ a − a · X (2) a ∧ O1 O2 . ⊥ μJ

(7.52)

We are therefore left with the task of simplifying the second term in the above equation. To this end we use the identity (1.28) twice, as indicated in the following set of calculations, 

   −−−→ −−−→  −−−→ (2)  a ∧ O 1 O 2 = X (2) a, O1 O2 , X (2) ⊥ a − a · X⊥ ⊥ ∧ a ∧ (a ∧ O 1 O 2 )  −−−→ −−−→ = X (2) ⊥ ∧ (a · O 1 O 2 )a − (a · a) O 1 O 2 −−−→ = O 1 O 2 ∧ X (2) (7.53) ⊥ ,

−−−→ where use has been made of a ⊥ O1 O2 and |a| = 1. Finally, combining (7.52) and (7.53) it transpires that u(2) = u(1) +

 M  −−−→ O 1 O 2 ∧ X (2) ⊥ , μJ

(7.54)

thus confirming our earlier assertion that the two displacement fields differ from each other through a rigid-body transformation. It is worth examining (7.54) in a bit more detail. To fix things, let us assume −−−→ that the coordinates of O 2 in Σ are (X 10 , X 20 ), i.e. O 1 O 2 = X 10 e1 + X 20 e2 . Due to the change in origin of our initial system of coordinates, if X (1) ⊥ = X 1 e1 + X 2 e2 (2) 0 0 then X ⊥ = (X 1 − X 1 )e1 + (X 2 − X 2 )e2 . By using these component representa−−−→ 0 0 tions, routine manipulations show that O 1 O 2 ∧ X (2) ⊥ = a(X 1 X 2 − X 2 X 1 ), and thus (7.54) becomes u(2) (P) = u(1) (P) +

M (X 0 X 2 − X 20 X 1 )a. μJ 1

(7.55)

Since in the original system of coordinates u1(1) (P) = −α X 2 X 3 and u2(1) (P) = α X 1 X 3 , it follows from (7.55) that the displacement field u(2) in the new system

7.6 A Closer Look at the Torsional Rigidity

339

of coordinates will have the components u1(2) (P) = −α(X 2 − X 20 )X 3 , u2(2) (P) = α(X 1 − X 10 )X 3 , u3(2) (P) = αΨ (X 1 , X 2 ) + α(X 10 X 2 − X 20 X 1 ). We have already seen in Sect. 7.4 that for a circular cylinder, if the origin of the coordinate system in Σ coincides with the centre of the circle, then Ψ ≡ 0 and there is no displacement along the X 3 -direction. If the origin in Σ is changed, which is equivalent to considering another axis, parallel to the original one, then displacements along the X 3 -axis at all points except the new X 3 -axis are non-zero and equal to u3 = α(X 10 X 2 − X 20 X 1 ). This means that the plane X 3 = k (0 ≤ k ≤ L) becomes the plane X 3 = k + α(X 10 X 2 − X 20 X 1 ) or, (α X 20 )X 1 − (α X 10 )X 2 + X 3 = k, which is no longer orthogonal to the original X 3 -axis.

7.7 Prandtl Stress Function The Saint-Venant semi-inverse approach to the torsion problem, based on the special choice of displacement function (7.17) and (7.18), has revealed a particularly simple expression for the Cauchy stress tensor T —see Eq. (7.50). Motivated by this result, it is possible to implement the semi-inverse approach by starting with a predefined expression for T . The idea is to choose the form of T such that the equilibrium equation (7.4a) is automatically satisfied, but still include enough arbitrariness to ensure that the boundary conditions (7.5) could be enforced in a consistent way. The above approach was originally proposed by L. Prandtl (1903, 1904). The departure point in his analysis was the stress tensor T = (∇ ⊥ φ) ∧ (a ⊗ a) − (a ⊗ a) ∧ (∇ ⊥ φ),

(7.56)

where φ = φ(X ⊥ ) is known as the Prandtl stress function and has to be determined as part of the solution of the torsion problem. It is fairly easy to check that ∇ · T = 0. Indeed, by taking the divergence of both sides in (7.56) gives     ∇ · T = ∇ · (∇ ⊥ φ ∧ a) ⊗ a − ∇ · a ⊗ (a ∧ ∇ ⊥ φ) .

(7.57)

A direct application of formula (A.2c) with u → (∇ ⊥ φ) ∧ a and v → a then  shows that the first term on the left-hand side of (7.57) is just ∇ ⊥ · (∇ ⊥ φ ∧ a) a. A further application of formula (A.2e) with u → ∇ ⊥ φ and v → a indicates that this expression is in fact identically zero. As for the second divergence term in (7.57), by using again (A.2c) with u → a and v → a ∧ ∇ ⊥ φ shows that this is equal to a · ∇ ⊥ ⊗ (a ∧ ∇ ⊥ φ), which is also identically zero.

340

7 Torsion

To find the differential equation that must be satisfied by φ we need to look at the compatibility equation (7.4c) or, what is the same, the Beltrami–Michell equation (6.39). Note that  in  our case body  forces are ignored and, according to (7.56),  |T | = a, ∇ ⊥ φ, a − a, a, ∇ ⊥ φ ≡ 0, the equality to zero being an immediate consequence of standard properties of the scalar triple product. Thus, the Beltrami– Michell equation is simply ∇⊥2 T = O or, with the help of (7.56),     2 ∇⊥ (∇ ⊥ φ ∧ a) ⊗ a − a ⊗ ∇⊥2 (a ∧ ∇ ⊥ φ) = O. Taking the (right) dot-product of this equation with a, leads further to     2 ∇⊥ (∇ ⊥ φ) ∧ a − a a, a, ∇⊥2 (∇ ⊥ φ) = 0,   or ∇⊥2 (∇ ⊥ φ) ∧ a = 0. Since ∇⊥2 and ∇ ⊥ commute with each other, after vectorially multiplying by a, the last equation can be written ∇ ⊥ (∇⊥2 φ) = 0. This implies that the Laplacian of φ must be constant. In conclusion, we have discovered that ∇⊥2 φ = K 0 in Σ,

(7.58)

where K 0 ∈ R is a constant (unknown at present). To fully determine the Prandtl stress function we also need boundary conditions that specify the behaviour of φ along ∂Σ. These will come from the requirement that the lateral surface of the cylinder is traction free—see Eq. (7.5a). From (7.28) to (7.56),   t(n) = n · T = n · (∇ ⊥ φ ∧ a) a − (n · a)(a ∧ ∇ ⊥ φ)   = n · (∇ ⊥ φ ∧ a) a = (Y s ∧ a) · (∇ ⊥ φ ∧ a)a   = Y s · (∇ ⊥ φ) a, that is

  t(n) = Y s · (∇ ⊥ φ) a.

(7.59)

Since φ ≡ φ(Y (s)) along ∂Σ, differentiating this relation yields dφ = dY · (∇ ⊥ φ), and thus from (7.59),   (dφ/ds)a = Y s · (∇ ⊥ φ) a ≡ t(n). In terms of the Prandtl stress function, the boundary condition t(n) = 0 on ∂Σ becomes dφ = 0 on ∂Σ. (7.60) ds In summary, we have found that the Prandtl stress function φ is the solution of the boundary-value problem consisting of (7.58) and (7.60). This represents an interior Dirichlet problem because (7.60) is equivalent to φ(Y ) = K 1 for all Y ∈ ∂Σ, where

7.7 Prandtl Stress Function

341

K 1 ∈ R is another constant. For a simply connected cross section we can always take K 1 = 0, but for multiply connected domains (see Fig. 7.4), when ∂Σ is made up of several closed curves ∂Σi (i = 1, 2, . . . , n), the boundary condition (7.60) becomes φ(Y ) = K i for Y ∈ ∂Σi (i = 1, 2, . . . , n), where K i ∈ R are some constants that must satisfy additional requirements; in general, K i = K j for i = j, so at most one of these constants can be taken to be zero. We shall say more about the case of multiply connected cross sections in Sect. 7.9, but for now let us go back to the boundary-value problem that gives the Prandtl stress function. Before one can solve it, the constant K 0 in (7.58) must be specified. This is our next goal. It must be clear that the Saint-Venant warping function Ψ and the Prandtl stress function φ are not independent of each other. The link between the two can be easily established by comparing the expressions of the Cauchy stress tensors in (7.50) and (7.56). As T must be the same, it transpires that we must have μα(a ∧ X ⊥ + ∇ ⊥ Ψ ) = (∇ ⊥ φ) ∧ a or, after re-arranging,  ∇⊥ Ψ =

 1 ∇ ⊥ φ + X ⊥ ∧ a, μα

(7.61)

which provides the required connection between Ψ and φ. This represents a firstorder system of linear partial differential equations for Ψ . For the purpose of finding K 0 ≡ ∇⊥2 φ, we are going to use (7.61) to calculate the Laplacian of Ψ and φ. Attempting to naively take the divergence of both sides in (7.61) fails, as it leads to a trivial result (0 = 0). Instead, we start by first taking the (left) vector product with a of both sides in (7.61)   1 a ∧ (∇ ⊥ φ) ∧ a + a ∧ (X ⊥ ∧ a) μα   1  (a · a)∇ ⊥ φ − a · ∇ ⊥ φ a + (a · a)X ⊥ − (a · X ⊥ )a = μα 1 = ∇⊥ φ + X ⊥ . μα

a ∧ (∇ ⊥ Ψ ) =

Since ∇⊥2 φ ≡ ∇ ⊥ · (∇ ⊥ φ), the next step is to take the divergence of the last equation above. Note that ∇ ⊥ · (a ∧ ∇ ⊥ Ψ ) = 0 and ∇ ⊥ · X ⊥ = 2, so the outcome is α=−

1 2 ∇ φ, 2μ ⊥

(7.62)

and we conclude that K 0 = −2μα. For the sake of completeness, we shall state below the boundary-value problem for the determination of φ in the case of simply connected cross sections,

342

7 Torsion



∇⊥2 φ = −2μα in Σ, φ=0 on ∂Σ.

(7.63)

In closing this section the last item on the agenda is to provide an expression for the twisting moment M in terms of the Prandtl stress function—the equivalent of (7.35) in Sect. 7.5. From (7.56) it follows immediately that t(a) = (∇ ⊥ φ) ∧ a,

(7.64)

and then (7.5d) can be rewritten as  M =a· 

Σ

  X ⊥ ∧ (∇ ⊥ φ) ∧ a d A



 =a· (X ⊥ · a)∇ ⊥ φ − (X ⊥ · ∇ ⊥ φ)a d A    Σ X ⊥ · (∇ ⊥ φ) d A = − ∇ ⊥ · (φ X ⊥ ) d A + 2 φ dA =− Σ Σ Σ  =− φ X ⊥ · n ds + 2 φ d A, ∂Σ

(7.65)

Σ

where use has been made of the (two-dimensional version of the) Divergence Theorem and the identity (A.2a). In the case of our usual choice of Cartesian coordinates this last result can be written in a form more suitable for explicit computations by remembering (7.28), according to which nds = e1 dX 2 − e2 dX 1 , and hence 

 M =−

∂Σ

φ(X 1 dX 2 − X 2 dX 1 ) + 2

Σ

φ d A.

(7.66)

For a solid cylinder (i.e. when Σ ⊂ E2 is simply connected) φ = 0 on ∂Σ, so Eq. (7.66) becomes  M =2

Σ

φ d A.

(7.67)

Formula (7.66), unlike its earlier counterpart—see Eq. (7.35), does not provide an immediately explicit relationship between M and α because the latter parameter enters implicitly in the expression of φ through the solution of (7.63).

7.8 Modified Stress Function We are going to slightly modify the Prandtl stress function discussed in Sect. 7.7 in order to eliminate the constant (−2μα) that appears in the boundary-value problem stated in (7.63). To this end, let Φ ∗ be the function defined according to

7.8 Modified Stress Function

343

1 μαΦ ∗ := φ + μα|X ⊥ |2 . 2

(7.68)

From (A.1b), ∇ ⊥ (X ⊥ · X ⊥ ) = 2(∇ ⊥ ⊗ X ⊥ ) · X ⊥ = 2(I 2 · X ⊥ ) = 2X ⊥ , and then ∇⊥2 (X ⊥ · X ⊥ ) = 2(∇ ⊥ · X ⊥ ) = 4. Applying ∇⊥2 to both sides of (7.68), it follows immediately that Φ ∗ is a plane-harmonic function. In terms of this modified stress function, the boundary-value problem (7.63) translates into 

∇⊥2 Φ ∗ = 0 in Σ, 1 ∗ 2 Φ = 2 |X ⊥ | on ∂Σ.

(7.69)

Proposition 7.3 The modified stress function Φ ∗ is the harmonic conjugate of the Saint-Venant warping function Ψ , that is, ∇ ⊥ Ψ = −a ∧ (∇ ⊥ Φ ∗ )

in Σ.

(7.70)

In the usual system of Cartesian coordinates, the vectorial relation (7.70) reduces to the familiar Cauchy–Riemann equations, Ψ, 1 = Φ,∗2

and

Ψ, 2 = −Φ,∗1 .

(7.71)

The justification for this result is straightforward. Applying ∇ ⊥ to both sides of (7.68) we discover that 1 ∇⊥ φ + X ⊥ , ∇⊥ Φ ∗ = (7.72) μα and then (7.70) follows by replacing (7.72) into (7.61). Since the modified stress function is the harmonic conjugate of the √warping function, by introducing the complex variable z = X 1 + iX 2 , where i ≡ −1 represents the imaginary unit in the complex plane, we see that the complex-valued function F(z) ≡ F(X 1 + iX 2 ) := Ψ (X 1 , X 2 ) + iΦ ∗ (X 1 , X 2 ) is analytic in z ∈ C. This is a remarkable feature that has far-reaching implications as we shall see in Sect. 7.11, but for now let us explore an entirely elementary aspect of this observation. If we start with a known analytic function F(z), we may then use either its real or imaginary part as a possible stress function. The disadvantage of this route is that we do not know a priori the shape of the cross sections for which these functions represent the solution of the torsion problem. However, in certain cases we can work our way backwards and identify the equation of ∂Σ. The following examples outline how this naive approach works in practice. Consider the analytic function z 2 = (X 1 + iX 2 )2 = X 12 − X 22 + 2iX 1 X 2 , and let us take Φ ∗ to be its real part.7 Note that multiplicative and additive constants do not 7 If

F(z) is analytic then ±iF(z) share the same property.

344

7 Torsion

affect the harmonic character of a function. Therefore, we can use Φ ∗ = m(X 12 − X 22 ) + p, where m, p ∈ R are arbitrary constants that can be ‘fine-tuned’ so that the above equation represents the bounding curve of a known cross section. Since m(X 12 − X 22 ) + p ≡ Φ ∗ =

1 2 (X + X 22 ) 2 1

on ∂Σ,

we deduce that 

   1 1 − m X 12 + + m X 22 = p 2 2

on ∂Σ,

and it must be clear that this curve can be matched to the ellipse X 12 /a 2 + X 22 /b2 = 1 (a > b > 0) by taking m :=

a 2 − b2 , 2(a 2 + b2 )

p :=

a 2 b2 . a 2 + b2

Similarly, consider the analytic function z 3 = (X 1 + iX 2 )3 = (X 13 − 3X 1 X 22 ) + i(3X 12 X 2 − X 23 ), and take

Φ ∗ = m(X 13 − 3X 1 X 23 ) + p,

with m, p playing again the role of ‘fine-tuning’ constants. This function determines the solution of the torsion problem for a cylinder whose cross section has the equation Φ ∗ = m(X 13 − 3X 1 X 23 ) + p =

1 2 (X + X 22 ). 2 1

If we set m := −1/6h and p := 2h 2 /3 (hardly an obvious choice!), then the above equation can be factorised as √ √ (X 1 − h)(X 1 − X 2 3 + 2h)(X 1 + X 2 3 + 2h) = 0. This represents a piecewise-linear curve formed by the segments √ √ belonging to the lines of equations X 1 = h, X 1 − X 2 3 + 2h = 0, and X 1 + X 2 3 + 2h = 0. It can be shown routinely that the region enclosed by this curve is an equilateral triangle of altitude 3h. Both the elliptic and the triangular cross sections will be discussed in more detail in Sect. 7.12.

7.9 Multiply Connected Domains

345

7.9 Multiply Connected Domains It was pointed out in Sect. 7.7—see Eq. (7.60), that the Prandtl stress function must be constant on the boundary ∂Σ of the cross section Σ ⊂ E2 . If this domain is multiply connected, then Σ = Σ0 \

N 

 Σi ,

Σi ⊂ Σ0 (i = 1, 2, . . . , N ),

i=1

i.e. Σ consists of a simply connected domain (Σ0 ) with a number of ‘holes’ in it (Σi , i = 1, 2, . . . , N ). As a consequence, ∂Σ = ∂Σ0 ∪ ∂Σ1 ∪ · · · ∪ ∂Σ N , which means that the boundary of Σ will be comprised of a number of disjoint closed curves—see Fig. 7.4 for an example with N = 3. In the application of Stokes’ Theorem to multiply connected domains it is important to remember the usual convention regarding the orientation of these curves. The outer boundary ∂Σ0 is described in the positive (anticlockwise) direction, while the internal boundaries ∂Σi are described in a clockwise direction. The direction in which the boundary ∂Σ is described is intimately related to the direction of the outward unit normal. The rule here is that the vector product of the outward unit normal with the unit tangent vector to ∂Σ0 in the direction of the arrow must be equal to e3 (i.e. it must be in the positive direction of the axis of the cylinder). A similar convention holds for internal boundaries as well. The developments of Sect. 7.7 are still valid when the cross section of the twisted cylinder is multiply connected, but the boundary-value problem (7.63) now assumes the form ⎧ 2 ⎪ ⎨∇⊥ φ = −2μα in Σ, φ=0 on ∂Σ0 , ⎪ ⎩ on ∂Σi (i = 1, 2, . . . , N ), φ = φi

∂Σ3

∂Σ0

Σ3

Σ1

(7.73)

∂Σ2

∂Σ1 Σ0

Σ2

Fig. 7.4 An example of multiply connected planar domain; the regions shown in white colour are excluded from it. We can regard this planar cross section as consisting of a simply connected domain Σ0 bounded by ∂Σ0 , from which the subdomains Σi (i = 1, 2, 3) have been removed

346

7 Torsion

where φi ∈ R (i = 1, 2, . . . , N ) are constants. Note that we have taken, without loss of generality, φ0 = 0 on the outer boundary. However, the remaining constants are not arbitrary and must be determined such that the out-of-plane (warping) displacement field is single-valued; this condition is equivalent to  ∂Σi

dΨ = 0

for i = 1, 2, . . . , N .

(7.74)

With the help of (7.61) the left-hand side of the above equation becomes      1 ∇ ⊥ φ + X ⊥ ∧ a, dΨ = dX ⊥ · (∇ ⊥ Ψ ) = dX ⊥ · μα ∂Σi ∂Σi ∂Σi and then (7.74) can be re-arranged in the form  ∂Σi

 dX ⊥ · (∇ ⊥ φ ∧ a) = −μα

∂Σi

dX ⊥ · (X ⊥ ∧ a).

(7.75)

The right-hand side is still amenable to further transformations by using Stokes formula,     dX ⊥ · (X ⊥ ∧ a) = ∇ ⊥ ∧ (X ⊥ ∧ a) · a d A ∂Σi Σ  i  = (−2a) · a d A = −2 d A = −2 area (Σi ), (7.76) Σi

Σi

where use has been made by (A.3d). Substituting (7.76) back into (7.75), we discover that  dX ⊥ · (∇ ⊥ φ ∧ a) = 2μα · area (Σi ), i = 1, 2, . . . , N . (7.77) ∂Σi

These represent N additional conditions that must be added to the boundary-value problem (7.73) in order to fix the constants φi (i = 1, 2, . . . , N ). Assuming for now that (7.73) subject to (7.77) has been solved, the next task is to find the relationship between the twisting moment M and the twist per unit length α. It turns out that the work done in (7.65) is still relevant here. Since the sense of integration on the internal boundaries is reversed with respect to the sense of integration on the external boundary, from (7.65) we find that8

8 The

summation convention is tacitly employed here (i ranges from 1 to N ).

7.9 Multiply Connected Domains

347

 M =−

 ∂Σ0

φ X ⊥ · n ds +

Σi

∂Σi



 = φi

 φi X ⊥ · n ds + 2

(∇ ⊥ · X ⊥ ) d A + 2

Σ

Σ

φ dA

φ d A,

and taking into account ∇ ⊥ · X ⊥ = 2, we finally arrive at the required formula  M =2

Σ

φ d A + 2φi area (Σi ).

(7.78)

7.10 The Shear Stress Definition 7.1 A line or trajectory of shear stress in a planar domain Σ ⊂ E2 is a curve C such that the resultant shear stress at any point of the curve is in the direction of the tangent to the curve. This concept is particularly relevant to the torsion problem that we have been discussing so far in this chapter. We have already seen that the only non-zero components of the infinitesimal stress tensor are T31 ≡ T31 (X ⊥ ) and T32 ≡ T32 (X ⊥ ), and hence the traction vector on any normal cross section Σ ⊂ E2 , t(X ⊥ , e3 ) = T31 e1 + T32 e2 ,

(7.79)

lies within its plane. Once the torsion problem for a given cross section shape has been solved, one can draw a line of shear stress starting from any point in that cross section. Since the component shear stress at any point of such curve, in the direction perpendicular to the tangent at that point, must be zero, by using arguments similar to those that led to (7.60) it follows that dφ/ds = 0 on C , where s is the arclength along the shear line. The conclusion that emerges is that the lines of shear stress are actually the level sets of the Prandtl stress function, i.e. φ = K on C (for some K ∈ R). It should be clear from the developments of Sect. 7.7 that the boundary of the cross section, ∂Σ, is also such a line of shear stress. As seen in Fig. 7.5, the shear stress lines form a system of non-intersecting closed curves, starting from the outer boundary, as one of the lines, and ending with one or more infinitesimal closed curves at some point inside Σ, if the cross section is simply connected. For a circular cross section the lines of shear stress are concentric circles; for an elliptic section they are a set of similar concentric ellipses; and for a rectangular cross section they are a set of curves which may be described as rectangles with rounded corners, with the boundary itself at one extreme, and an infinitesimal ellipse at the other extreme. For multiply connected cross sections this picture is less clear-cut, for in this case the inner boundaries act also as lines of shear stress. Nevertheless, for a circular/elliptical hollow cylinder whose cross section is

348

7 Torsion

Fig. 7.5 Examples of ‘lines’ of shear stress for two different cross-sectional geometries. As explained in the text, these are the level curves of the Prandtl stress function φ

bounded by two concentric circles/ellipses, the lines of shear stress are also concentric circles/ellipses which ‘fill up’ the entire annular region between its inner and outer boundaries. A line of shear stress can neither intersect any component of the boundary of the cross section, nor can it end at any point inside it. Each line of stress must therefore be a closed curve. Furthermore, two distinct shear lines (of equations φ = C1 and φ = C2 , with C1 = C2 real constants) cannot intersect, nor can two branches of the same line intersect, for this would give two different directions for the resultant stress at the same point. The only exception to the last statement occurs at points where the shear stress vanishes. Two branches of the same shear line may touch at such a point, or the shear line may reduce to a closed infinitesimal curve. Let us return now to the stress vector in (7.79). From a practical point of view, it is the magnitude of this vector that is most important. Also of equal interest within this context are the locations where the maxima of this magnitude occur. Knowledge of this information can help with the better design of shafts and cylindrical elements that are meant to sustain torsional loads. We focus on these two aspects next, trying to formulate and justify some of the fundamental results that underlie them. To this end, let us introduce τ , the magnitude of the stress vector (7.79),  2 2 1/2 + T32 . τ = T31

(7.80)

It follows directly from (7.56) that T31 = φ, 2 and T32 = −φ, 1 , so τ 2 = |∇ ⊥ φ|2 . Furthermore, from Exercise 7.8 at the end of this chapter it can be shown that ∇⊥2 (τ 2 ) ≥ 0. We shall use this property to establish that the maxima of τ 2 (and hence those of τ ) will always occur for some point(s) on the boundary of the cross section.

7.10 The Shear Stress

349

Proposition 7.4 Let Σ be a two-dimensional bounded domain and consider ϕ : Σ ⊂ E2 → R, that satisfies: (i) ϕ ∈ C 2 (Σ), and (ii) ∇⊥2 ϕ ≥ 0 in Σ. Then the points for which ϕ attains its maximum are situated on ∂Σ. Instead of a formal proof of this classical result, we shall sketch a heuristic justification. Let us assume that there exists a point X 0⊥ = (X 10 , X 20 ) ∈ Σ such that ϕ(X ⊥ ) ≤ ϕ(X 0⊥ ) for all X ⊥ ∈ Σ. Consider now the functions ϕ1 (X 1 ) := ϕ(X 1 , X 20 ) and ϕ2 (X 2 ) := ϕ(X 10 , X 2 ). Then ϕ1 will have a maximum at X 1 = X 10 , while ϕ2 will similarly be maximised by X 2 = X 20 . However, according to elementary calculus facts, this implies that ϕ,11 (X 10 , X 20 ) ≡ ϕ1, 11 (X 10 ) < 0 and ϕ,22 (X 10 , X 20 ) ≡ ϕ2, 22 (X 20 ) < 0. Adding up these two inequalities results in ∇⊥2 φ(X 10 , X 20 ) < 0, which contradicts property (ii) in the hypothesis of Proposition 7.4. It remains that if ϕ is maximised at a point X 0⊥ , then this point has to be on ∂Σ, and we are done.9 Due to the rather simple expression of the stress tensor in the torsion problem for non-circular cross sections, it is possible to easily work out the spectral representation of this tensor in the general case (i.e. irrespective of the shape of the cross section). Let us start by introducing the unit vectors b and s, defined according to b :=

∇⊥ φ |∇ ⊥ φ|

and

s := b ∧ a.

The vector b is normal to the families of curves φ = constant (i.e. to the lines of shear stress), while s is just the tangent to the same curves and is constructed such that the set of vectors {a, s, b} forms a right-handed orthonormal triad. From the definition of the Prandtl stress function—see Eq. (7.56), together with the definitions of the two new vectors introduced above, we can write successively    ∇⊥ φ ∇⊥ φ ∧a ⊗a−a⊗ a∧ |∇ ⊥ φ| |∇ ⊥ φ|   = |∇ ⊥ φ| (b ∧ a) ⊗ a − a ⊗ (a ∧ b) 

T = |∇ ⊥ φ|

= |∇ ⊥ φ|(s ⊗ a + a ⊗ s).

(7.81)

Since (s + a) ⊗ (s + a) − (s − a) ⊗ (s − a) = 2(s ⊗ a + a ⊗ s),

fact that there exists a maximum of ϕ in Σ ∪ ∂Σ is a direct consequence of the boundedness assumption on Σ and the continuity of ϕ. 9 The

350

7 Torsion

and |∇ ⊥ φ| = b · (∇ ⊥ φ), the representation (7.81) can be cast in a form that allows us to identify the principal directions and the principal values of T by inspection, T = t1 ( p1 ⊗ p1 ) + t2 ( p2 ⊗ p2 ) + t3 ( p3 ⊗ p3 ),

(7.82)

where p1 = b, t1 = 0,

1 1 p2 = √ (s + a), p3 = √ (s − a), 2 2 t2 = b · (∇ ⊥ φ), t3 = −b · (∇ ⊥ φ).

Here, pi (i = 1, 2, 3) correspond to the principal axes of the infinitesimal stress tensor and ti (i = 1, 2, 3) represent the corresponding principal values. The interpretation of (7.82) is straightforward. First, note that p2 and p3 are inclined at 45◦ with respect to s and a, respectively. Second, since t1 = 0, there are no normal forces acting on surface elements normal to b. Third, there are normal stress forces equal to ±b · (∇ ⊥ φ) on surface elements perpendicular to p2 and p3 .

7.11 Complex Variables Formulation The relevance of complex variables and analytic function theory to torsion problems was mentioned briefly at the end of Sect. 7.8. A more systematic, but still elementary, treatment of this aspect will be pursued in what follows. Let us recall that starting from the Saint-Venant warping function Ψ and the modified stress function Φ ∗ , we introduced F(z) = Ψ (X 1 , X 2 ) + iΦ ∗ (X 1 , X 2 ),

with z = X 1 + iX 2 ,

(7.83)

and it was argued that, since ∇⊥2 Ψ = ∇⊥2 Φ ∗ = 0, this new function will be analytic; we shall refer to F(z) in (7.83) as the complex potential for the torsion problem. The main idea in adopting the complex variables formalism involves replacing the determination of the plane-harmonic functions Ψ and Φ ∗ , with that of the analytic function F(z). As we already have at our disposal three equivalent formulations of the torsion problem, on the surface, this might seem like a pointless exercise. However, this is far from being so. What makes a difference here is the analytic character of the complex potential and the fact that, under fairly general circumstances, an analytic function defined on a domain Σ is completely determined by the values it takes on the boundary of that domain, ∂Σ. With this in mind, let us start off by trying to find out the boundary condition satisfied by F(z) in (7.83). Since z = X 1 + iX 2 and z = X 1 − iX 2 , by adding and subtracting these two relations we get X1 =

1 (z + z), 2

i X 2 = − (z − z). 2

(7.84)

7.11 Complex Variables Formulation

351

In a similar way,  1 F(z) + F(z) , 2  i  ∗ Φ ≡ Im{F(z)} = − F(z) − F(z) , 2 Ψ ≡ Re{F(z)} =

(7.85a) (7.85b)

where Re{. . . } and Im{. . . } stand for the real and, respectively, the imaginary parts of the expressions between the curly braces. From (7.69), we know that Φ ∗ = 21 (X 12 + X 22 ) on ∂Σ, which can be rewritten in complex form with the help of (7.85b) and by noticing that zz = X 12 + X 22 . Thus, F(z) − F(z) = izz + C

on ∂Σ,

(7.86)

where C ∈ C is an arbitrary constant; for simply connected cross sections this constant can be taken equal to zero. In summary, solving the torsion problem for a cylinder of arbitrary transverse cross section Σ ⊂ E2 is reduced to finding an analytic function F : Σ ⊂ C → C, which satisfies the condition (7.86) along the boundary of this domain. In order to make the most of this simplification we need to show that it is possible to express the unknowns (T31 , T32 , M, etc) in terms of the complex potential defined in (7.83). Let and S := T31 + iT32 , (7.87) U := u1 + iu2 which will be referred to as the complex displacement and the complex stress, respectively. Recalling that Ψ and Φ ∗ are harmonic conjugate to each other—see Eq. (7.71), and using the expression for T31 and T32 recorded in (7.39e), it follows that T31 + iT32 = μα(Φ,∗2 − X 2 ) − μα(iΦ,∗1 − iX 1 )   = μα (iX 1 − X 2 ) + (Φ,∗2 − iΦ,∗1 ) .

(7.88)

Invoking again the Riemann–Cauchy relations (7.71), note also F (z) = Ψ, 1 + iΦ,∗1 = Φ,∗2 + iΦ,∗1 .

(7.89)

From (7.88) to (7.89) we can then conclude that the complex stress S is given by   S = μα iz + F (z) .

(7.90)

Next, we are going to show that the twisting moment M can also be expressed in terms of the complex potential F(z), through either one of the two formulae    

 z F (z) − izz d A , M = Re iμα Σ

(7.91a)

352

7 Torsion

 M = Re

1 μα 4

 ∂Σ

   zz 2F (z) − i z dz .

(7.91b)

It is only (7.91a) that requires justification, as (7.91b) is simply derived from it by an application of the complex version of Stokes’ Theorem (see Exercise 7.11). As the routine calculations that lead to this result are rather tedious, we shall only outline the proof. The starting point is the component form of (7.5d),  M=

Σ

(X 1 T32 − X 2 T31 ) d A.

(7.92)

If the quantities that appear in the integrand are replaced by their corresponding expressions in terms of z and z, one obtains (7.91a). Note that X γ (γ = 1, 2) are immediately available from (7.84), but we still have to work out similar formulae for T31 and T32 . This is accomplished by using (7.90) and its complex conjugate; adding and subtracting these two formulae yields  1  μα i(z − z) + F (z) + F (z) , 2   i T32 = μα − i(z + z) − F (z) + F (z) . 2 T31 =

(7.93a) (7.93b)

These two equations are substituted back into the integrand of (7.91a) to give   i X 1 T32 − X 2 T31 = μα(zz) + μα z F (z) − z F (z) 2   = μα zz − Im{z F (z)}  

= Re iμα z F (z) − i zz , which leads directly to (7.91a). As for the complex displacement, this is independent of the complex potential; it can be checked very easily that U = −iμα X 3 z. Proposition 7.5 If Σ ⊂ E2 is a bounded domain whose boundary ∂Σ can be expressed in the form (7.94) zz = f (z) + f (z), for some analytic function f (z), then the complex torsion potential is given by F(z) = i f (z) + C

(∀) z ∈ Σ,

(7.95)

where C ∈ C is an arbitrary constant (which can be taken to be zero when the cross section is simply connected—see also what was said in relation to Eq. (7.86)). The validity of this result is confirmed by elementary arguments based on the use of (7.86), and the details are omitted. Although the expression of F(z) in (7.95) is rather simple, the formula given there hinges on us being able to write the boundary of

7.11 Complex Variables Formulation

353

the cross section Σ in the way indicated in (7.94). For simple/regular geometries this is possible with relatively little effort, although the calculations tend to be tedious. However, for more complicated geometries this is an impractical task, and more powerful techniques need to be developed. In the meantime, we examine a few particular cases to give the reader a feel for how Proposition 7.5 is applied. These examples will be discussed in more detail in Sect. 7.12, without recourse to complex variables. Ellipse: Σ is an elliptical domain for which ∂Σ is the ellipse b2 X 12 + a 2 X 22 = a 2 b2 , where 0 < a < b are its semi-axes. To translate this equation into complex variables, we use the basic formulae (7.84). After expanding and simplifying the resulting expressions we end up with zz =

2a 2 b2 a 2 − b2 (z 2 + z 2 ), + a 2 + b2 2(a 2 + b2 )

which is clearly of the same form as (7.94) if we set f (z) =

a 2 − b2 2 z , 2(a 2 + b2 )

and then F(z) =

a 2 − b2 i z2. 2(a 2 + b2 )

(7.96)

Equilateral triangle: Σ is a triangular domain bounded by the equilateral triangle shown in √ Fig. 7.6. The triangle is formed by the intersection of three lines of equations √ X 1 = h, 3X 2 = X 1 + 2h, and 3X 2 = −(X 1 + 2h). Hence, the equationof ∂Σ is obtained by  taking the product of these three equations, which is (X 1 − h) (X 1 + 2h)2 − 3X 22 = 0. Using again (7.84), after lengthy, but elementary manipulations, it transpires that 1 3 (z + z 3 ) + 3hz z − 4h 3 = 0. 2 To see that this equation is indeed of the form stated in (7.94) it suffices to take f (z) = −

1 3 z , 6h

and then F(z) = −

i 3 z . 6h

(7.97)

Circle with a groove: Σ in this case corresponds to the geometrical shape included in Fig. 7.9; this configuration is obtained from a disc bounded by the circle of equation (X 1 − a)2 + X 22 = a 2 after removing its overlap with another, smaller disc, whose circumference is described by the equation X 12 + X 22 = b2 (0 < b < a). It must be clear that ∂Σ will be described by the product of the equations for the two circles mentioned above, that is, its equation will be (X 12 + X 22 − 2a X 1 )(X 12 + X 22 − b2 ) = 0. As in the previous examples, our task here is to eliminate the dependence on X 1 and X 2 with the help of (7.84). This is a lengthy exercise in elementary algebra that poses no serious technical challenges, and eventually yields

354

7 Torsion

 z z = b2 + a(z + z) − ab2 Clearly, we should take   b2 f (z) = a z − , z

 1 1 . + z z

  b2 and then F(z) = ia z − . z

(7.98)

7.12 Worked Examples Even though the torsion problem admits a relatively simple mathematical formulation, closed-form solutions are possible only for a limited number of regular geometries. In this section the theory introduced above will be illustrated by several examples (see Fig. 7.6). Example 7.1 (elliptical cross section) Consider an elliptic cylinder C ,

X = (X 1 , X 2 , X 3 ) ∈ E3 | b2 X 12 + a 2 X 22 < a 2 b2 , 0 < X 3 < L where a > b > 0. The cross section of the cylinder is shown in Fig. 7.6 (leftmost sketch) and corresponds to Σ:



X ⊥ = (X 1 , X 2 ) ∈ E2 | b2 X 12 + a 2 X 22 < a 2 b2 ,

X2 X2

(0, b)

A4

A3

X2 B3

(0, b) X1 (−a, 0)

2h

(a, 0) (0, −b)

h

X1 (−a, 0) X1

A1

(a, 0) A2 (0, −b)

B1

B2

Fig. 7.6 Elliptical, triangular, and rectangular cross sections used in the examples discussed below. The triangle is equilateral, has the altitude equal to 3h, and the X 1 -axis bisects ∠B3 B1 B2 . The white markers indicate the locations where the tangential stress τ defined in (7.80) reaches a maximum— more details in the text

7.12 Worked Examples

355

while the curve bounding Σ represents the ellipse ∂Σ :



X ⊥ = (X 1 , X 2 ) ∈ E2 | b2 X 12 + a 2 X 22 = a 2 b2 .

We want to find the warping function for the cross section Σ, and then calculate its torsional rigidity (or the torsion constant). Also, of interest here are the maximum shear stress and the locations within Σ where it occurs. Instead of solving (7.26), we are going to look for a warping function of the form Ψ (X 1 , X 2 ) = β X 1 X 2 , where β ∈ R will be determined such that the boundary condition (7.39c) is satisfied. Note that the chosen Ψ is already a plane-harmonic function. Since Ψ, 1 = β X 2 and Ψ, 2 = β X 1 , Eq. (7.39c) becomes

0 = (β − 1)X 2

 1 d  dX 2 dX 1 − (β + 1)X 1 = (β − 1)X 22 − (β + 1)X 12 , ds ds 2 ds

and hence (β − 1)X 22 − (β + 1)X 12 = γ , where γ ∈ R is completely arbitrary. Comparing this equation with that of the ellipse that defines ∂Σ, we are led to 1+β γ b2 = 2 and = b2 . 1−β a β −1 Simple manipulations yield β = (b2 − a 2 )/(b2 + a 2 ) and γ = −2a 2 b2 /(a 2 + b2 ), which lead to the final expression of the desired warping function,  Ψ (X 1 , X 2 ) =

b2 − a 2 b2 + a 2

 X 1 X 2.

(7.99)

The torsional stiffness D ≡ μJ is calculated from (7.39f) to (7.99). Note that   2 2 J = (X 1 + X 2 + X 1 Ψ, 2 − X 2 Ψ, 1 ) d A = (X 12 + X 22 + β X 12 − β X 22 ) d A Σ Σ   2 2 2 = (1 + β) X 1 d A + (1 − β) X2 d A = 2 (a 2 I1 + b2 I2 ), (7.100) 2 a + b Σ Σ 

where I2 :=

Σ

X 12 d A =

π 3 a b 4

 and

I1 :=

Σ

X 22 d A =

π 3 ab 4

are the moments of inertia of Σ about the X 2 - and X 1 -axes, respectively (see Appendix C). Replacing these values back into (7.100) the torsion constant for the elliptical cross section becomes J=

πa 3 b3 . a 2 + b2

(7.101)

356

7 Torsion

The non-zero components of the Cauchy stress tensor are immediately obtained from (7.39e), 2a 2 μα X 2 , a 2 + b2 2b2 μα X 1 , T32 = μα(Ψ, 2 + X 1 ) = (β + 1)μα X 1 = 2 a + b2 T31 = μα(Ψ, 1 − X 2 ) = (β − 1)μα X 2 = −

and the shear stress turns out to be  2 2 1/2 + T32 = τ ≡ T31

2μα 2 a + b2



b4 X 12 + a 4 X 22 .

(7.102)

To maximise this expression it is sufficient to find the maxima of the radicand. Given the particular form of (7.102) it must be clear that such maxima will be attained at points on ∂Σ that are constrained to lie on the ellipse. By introducing the Lagrange multiplier λ ∈ R, we are thus prompted to look for maxima of the function G(X 1 , X 2 ) = (b4 X 12 + a 4 X 22 ) − λ(b2 X 12 + a 2 X 22 − a 2 b2 ). The critical points of G are obtained by solving the simultaneous system of equations G , 1 = G , 2 = 0 or (b2 − λ)X 1 = 0 and (a 2 − λ)X 2 = 0. The solution choice (X 1 , X 2 ) = (0, 0) can be disregarded because it renders the absolute minimum of (7.102). If λ = b2 then G = a 2 (a 2 − b2 )X 22 + a 2 b4 ; if X 2 = 0 as well, this choice leads to a local minimum again. It remains that λ = a 2 , in which case G = b2 (b2 − a 2 )X 12 + a 4 b2 , and it is clear that X 1 = 0 is indeed a point for which G attains its maximum. With X 1 = 0 in the equation of the ellipse that defines ∂Σ it transpires that X 2 = ±b. In conclusion, we have found that the maximum shear stress occurs at the points (X 1 , X 2 ) = (0, ±b), that is, at the ends of the minor axis of the ellipse ∂Σ; the maximum value will be   2 a b τmax = 2μα . (7.103) a 2 + b2 The fact that the maximum shear stress occurs at the points on the boundary ∂Σ closest to the centroid of the cross section is not accidental. It can be shown that if the cross section of the torsioned cylinder is convex, then the maximum shear stress is always attained at such points. Example 7.2 (triangular cross section) Let us consider now a cylinder for which the transverse cross section Σ is an equilateral triangle whose geometry is as seen in Fig. 7.6 (middle sketch). From the information given therein, it √ follows easily that the equations of B1 B3 and B1 B2 are 3X 2 = X 1 + 2h and

7.12 Worked Examples

357

√

3X 2 = −X 1 − 2h, respectively. The cylinder cross section is the domain Σ:

√ √ 

X ⊥ = (X 1 , X 2 ) ∈ E2  X 1 ≤ h, 3X 2 ≤ X 1 + 2h, 3X 2 ≥ −X 1 − 2h ,

which is bounded by the triangle ∂Σ :



 

 X ⊥ = (X 1 , X 2 ) ∈ E2 (X 1 − h) (X 1 + 2h)2 − 3X 22 = 0 .

The question of interest here is as before: assuming that the triangular cylinder is subjected to pure axial twisting moments on its end faces, we aim to find how this loading is related to the twist per unit length. To answer this question we need the torsion constant. Other things of interest include finding the shear stress and the displacement field in the cylinder. We are going to solve this example by using the formalism of Sect. 7.7. As a candidate for our Prandtl stress function we look for a φ ≡ φ(X 1 , X 2 ) that vanishes identically along ∂Σ, in order to ensure that the boundary condition in (7.63) is automatically satisfied. Thus, we choose   φ(X 1 , X 2 ) = β(X 1 − h) (X 1 + 2h)2 − 3X 22   = β X 13 + 3h(X 12 + X 22 ) − 3X 1 X 22 − 4h 3 , where β ∈ R will be determined by demanding that the first equation in (7.63) is satisfied. Routine manipulations show that ∇⊥2 φ = 12βh, so that β = −μα/6h and φ(X 1 , X 2 ) = −

  μα (X 1 − h) (X 1 + 2h)2 − 3X 22 . 6h

(7.104)

Since Σ is simply connected, the twisting moment is given by (7.67)    27 M =2 φ dA = √ μh 4 α, 5 3 Σ where the quantity between parentheses represents the torsional rigidity of the triangular cross section. The non-zero components of the stress tensor are obtained from (7.64), μα (X 1 − h)X 2 , h μα 2 (X − X 22 + 2h X 1 ), = 2h 1

T31 = φ, 2 = T32 = −φ, 1 and

358

τ≡

7 Torsion



2 T31

+

2 1/2 T32

1/2  μα 1 2 2 2 2 2 (X 1 − h) X 2 + (X 1 − X 2 + 2h X 1 ) = . (7.105) h 4

For finding the maximum shear stress we use the fact that this value is attained at points on ∂Σ—cf. the discussion in Sect. 7.10. Hence, we shall consider maximising (7.105) for each side of the triangle, and then compare the maxima obtained in order to identify the absolute maximum of τ . On the side B2 B3 , X 1 = h and hence T31 ≡ 0. In this case, τ = |T32 | =

μα 3 |3h 2 − X 22 | ≤ μαh, 2h 2

and the maximum is attained for X 2 = 0. The point (X 1 , X 2 ) = (h, 0) is the midpoint of B2 B3 , while the corresponding value of the shear stress is τmax =

3 μαh. 2

(7.106)

√ On the side B1 B3 , 3X 2 = X 1 + 2h. Squaring this out we get 3X 22 = (X 1 + 2h)2 , which can be substituted into (7.105), thus leaving us with a function of just one variable (i.e. X 1 ) that needs to be maximised. Note that the maximum of (7.105) in this case corresponds to the maximum of G(X 1 ) =

4 (X 1 + 2h)2 (X 1 − h)2 . 9

The stationary points of G are given by G , 1 = 0 or (2X 1 + h)(X 1 + 2h)(X 1 − h) = 0. One √ possibility is X 1 = −h/2, and since (X√1 , X 2 ) ∈ B1 B3 it follows that X 2 = 3h/2 3. The point (X 1 , X 2 ) = (−h/2, 3h/2 3) is the midpoint of B1 B3 and the corresponding τmax has the same value as in (7.106). The other two remaining choices (X 1 = h and X 1 = −2h) correspond to the ends of the segment B1 B3 , where τ = 0. A similar argument can be used to deal with points on B1 B2 . It should be noted that the shear stress is zero at all the corners of the triangle, and yet at these points the displacement due to the twist has its greatest value. The displacements in the triangular cross section can be found by using the general theory developed in the previous chapter. Example 7.3 (rectangular cross section) The torsion of a cylindrical rod whose transverse cross section is a rectangle presents several new features, in the sense that the warping and/or stress functions cannot be guessed as in the previous examples. Although the solution is still elementary, a simple closedform formula is not possible, and the final result will be in the form of some infinite series. The geometry of interest is that in Fig. 7.6 (rightmost sketch), with the cylinder cross section being formally defined as

7.12 Worked Examples

Σ:



359



X ⊥ = (X 1 , X 2 ) ∈ E2  − a ≤ X 1 ≤ a, −b ≤ X 2 ≤ b ,

and its boundary, the rectangle ∂Σ :





X ⊥ = (X 1 , X 2 ) ∈ E2  X 1 = ±a, X 2 = ±b ,

where 0 < a < b are given constants. Attempting to use the semi-inverse method for a Prandtl stress function of the form φ(X 1 , X 2 ) = β(X 12 − a 2 )(X 22 − b2 ), with β ∈ R, fails because ∇⊥2 φ is no longer a constant as in the previous examples. We choose to solve this classical problem by using the modified stress function of Sect. 7.8. Recall that according to (7.69) we need to look for Φ ∗ ≡ Φ ∗ (X 1 , X 2 ) such that ⎧ 2 ∗ ⎪ in (−a, a) × (−b, b), ⎨∇⊥ Φ = 0 Φ ∗ = 21 (a 2 + X 22 ) for X 1 = ±a, ⎪ ⎩ ∗ Φ = 21 (X 12 + b2 ) for X 2 = ±b.

(7.107)

To simplify these rather complicated boundary conditions we shall introduce the auxiliary function (7.108) ζ (X 1 , X 2 ) := Φ,∗11 + 1 = −Φ,∗22 + 1, where the second equality in (7.108) is a consequence of the fact that Φ ∗ is harmonic. It is immediately obvious that ζ itself is a harmonic function. Also, the boundary conditions in (7.107) can be expressed in terms of ζ by differentiation. This results in a new boundary-value problem for the function ζ ≡ ζ (X 1 , X 2 ), ⎧ 2 ⎪ ⎨∇⊥ ζ = 0 in (−a, a) × (−b, b), ζ =0 for X 1 = ±a, ⎪ ⎩ ζ =2 for X 2 = ±b,

(7.109)

the solution of which will be sought by using separable variables. To this end, let us assume that ∞

ζ (X 1 , X 2 ) =

gn (X 1 )h n (X 2 ), n=0

where gn (X 1 ) and h n (X 2 ) are functions that need to be found. From the condition that the function ζ (X 1 , X 2 ) must be harmonic it follows that gn

(X 1 ) h

(X 2 ) =− n = −λ2n gn (X 1 ) h n (X 2 ) or,

gn

(X 1 ) + λ2n gn (X 1 ) = 0,

h

n (X 2 ) − λ2n h n (X 2 ) = 0 ;

360

7 Torsion

at this stage λ2n (n ≥ 0) are arbitrary real constants. These equations can be integrated at once, with their general solutions admitting the representations gn (X 1 ) = An cos λn X 1 + Bn sin λn X 1 , h n (X 2 ) = Cn cosh λn X 2 + Dn sinh λn X 2 , where An , Bn , Cn , and Dn ∈ R. Since ζ (±a, X 2 ) = 0 and ζ (X 1 , ±b) = 2, the expressions of gn and h n must contain only even terms, and we conclude that Bn = Dn = 0 for n ≥ 0. Letting βn := An Cn , the expression of ζ can now be written as ∞

ζ (X 1 , X 2 ) =

βn cos λn X 1 cosh λn X 2 .

(7.110)

n=0

The next task is to identify the values of βn and λn consistent with the boundary conditions of our problem. Let us start by observing that the equation ∞

ζ (±a, X 2 ) ≡

βn cos λn a cosh λn X 2 = 0,

(∀) X 2 ∈ (−b, b),

n=1

requires that cos λn a = 0, and thus λn =

(2n + 1)π , 2a

n = 0, 1, . . . .

On the other hand, satisfying the other boundary condition in (7.109) boils down to ∞

2=

βn cos λn X 1 cosh λn b,

(∀) X 1 ∈ (−a, a).

(7.111)

n=1

We are going to take advantage of the fact that the system of functions cos λn X 1 (n ≥ 0) is orthogonal on (−a, a), i.e. 

a

−a

 cos λn X 1 cos λm X 1 dX 1 =

0, if m = n, a, if m = n.

Next, multiply both sides of (7.111) by cos λm X 1 , and then integrate over (−a, a). Noticing further that 

a −a

cos λm X 1 dX 1 =

we conclude that βm =

 4a(−1)m 2 π = , sin mπ + λm 2 (2m + 1)π

8(−1)m 1 · , π(2m + 1) cosh λm b

whence (7.110) can finally be stated as

7.12 Worked Examples

361

ζ (X 1 , X 2 ) =

8 π

∞ n=0

(−1)n cosh λn X 2 · cos λn X 1 . 2n + 1 cosh λn b

(7.112)

This series converges uniformly in X 1 and X 2 for |X 2 | < b, so it can be integrated term by term. To find Φ ∗ explicitly requires a double integration because of the way ζ was introduced in (7.108). Once this is done, the non-zero stress components T31 and T32 are recovered by recalling that Φ ∗ and Ψ are harmonic conjugate; more specifically, from (7.39e) it follows that   T31 = μα Ψ, 1 − X 2 = μα Φ,∗2 − X 2 ,   T32 = μα Ψ, 2 + X 1 = μα −Φ,∗1 + X 1 .

(7.113a) (7.113b)

The integration of (7.112) required in these two formulae introduces two arbitrary functions f (X 1 ) and g(X 2 ). It can be shown that these are in fact identically zero, due to the equilibrium equation T13, 1 + T23, 2 = 0 and the boundary conditions T13 (±a, X 2 ) = T23 (X 1 , ±b) = 0. Without elaborating further, the final expressions obtained from formulae (7.113) are recorded below, T31 = −

8aμα π2 !

∞ n=0

(−1)n sinh λn X 2 · cos λn X 1 , 2 (2n + 1) cosh λn b

" (−1)n cosh λn X 2 sin λn X 1 . · (2n + 1)2 cosh λn b

∞

8a T32 = μα 2X 1 − 2 π

n=0

(7.114a)

(7.114b)

The twisting moment is found by writing (7.5d) in Cartesian coordinates, assuming that the X 3 -axis is directed along the generators of the rectangular cylinder and passes through the centroids of the transverse cross sections; in that case,  M=

b

−b



a −a

(X 1 T32 − X 2 T31 ) dX 1 dX 2 .

Substituting T31 and T32 found in (7.114) into this formula, and then making use of the identity ∞ n=0

1 π4 , = 4 (2n + 1) 96

routine manipulations yield !

"  a  ∞ tanh λ b 1 64 n M = 16μαa 3 b − . 3 π 5 b n=0 (2n + 1)5

(7.115)

362

7 Torsion

The torsional rigidity of the rectangular cross section is the expression M/α obtained from (7.115). Despite the rather unwieldy nature of this result, the convergence of the infinite series in (7.115) is quite fast. In fact, the first term of the series provides a good approximation of its value, as explained next. Start by writing ∞ n=0

∞

tanh λn b tanh λn b πb + = tanh , 5 (2n + 1) 2a (2n + 1)5 n=1

and then observe that ∞ n=1

tanh λn b < (2n + 1)5

∞ n=1

1  0.0046. (2n + 1)5

Furthermore, since tanh(π b/2a) ≥ 0.917, it transpires that a reasonable approximation for (7.115) will be provided by the much simpler formula  M  16μαa 3 b

 64  a  πb 1 − 5 tanh . 3 π b 2a

(7.116)

If the rectangular cross section is also very thin, i.e. a  b, then the last result is susceptible to an additional simplification, M

16 μαa 3 b. 3

(7.117)

Example 7.4 (circular annulus) Let us show that the torsional rigidity for a hollow cylinder of cross section (see Fig. 7.7, left sketch) Σ:





(r, θ )  R1 ≤ r ≤ R2 , 0 ≤ θ < 2π

is given by D=

 1 μπ R24 − R14 . 2

(7.118)

We are going to solve this example by using the Prandtl stress function which, according to (7.73) will have to satisfy ⎧ 2 ⎪ ⎨∇⊥ φ = −2μα in Σ, φ=0 on ∂Σ0 , ⎪ ⎩ on ∂Σ1 . φ = φ1

(7.119)

7.12 Worked Examples

363

X2

X2 b

R2

kb

R1

X1

−a −ka

ka ∂Σ1

∂Σ1

−kb ∂Σ0

a X1 ∂Σ0

−b

Fig. 7.7 Simple examples of doubly connected cross sections—explanations in the text

In terms of the modified stress function (7.68), the first equation in (7.119) is equivalent to ∇⊥2 Φ ∗ = 0. Its general solution in polar coordinates, when there is no dependence on the azimuthal variable θ , has the form Φ ∗ (r ) = A00 + A01 log r , where A00 , A01 ∈ R are arbitrary constants at this stage. From (7.68) the corresponding Prandtl stress function will then be   1 (7.120) φ(r ) = μα A00 + A01 log r − r 2 . 2 Recall from Sect. 7.9 that we need to enforce the condition (7.77) on ∂Σ1 . To this end, the integrand on the left-hand side of that equation must be expressed in polar coordinates. By direct calculations,  ∇ ⊥ φ = μα

   A01 A01 − r er and (∇ ⊥ φ) ∧ e3 = −μα − r eθ . r r

Since X ⊥ = r er , we also have ∂ X⊥ ∂ ∂ X⊥ ∂ dr + dθ = (r er )dr + (r er )dθ ∂r ∂θ ∂r ∂θ ∂ er ∂ er dr + r dθ = er dr + r eθ dθ, = er dr + r ∂r ∂θ

dX ⊥ =

and then dX ⊥ · (∇ ⊥ φ ∧ a) = −μα(A01 − r 2 )dθ. Consequently, (7.77) leads to −2π μα(A01 − R12 ) = 2μα(π R12 ), whence A01 = 0. Enforcing the first boundary condition in (7.119), we also find that A00 = R22 /2, so that we can now state the complete expression of the Prandtl stress function for the circular annulus,

364

7 Torsion

φ(r ) =

1 μα(R22 − r 2 ) ; 2

for further reference let us also note that   φ1 ≡ φ 

∂Σ1

=

1 μα(R22 − R12 ). 2

The twisting moment follows from (7.78), which in our simple case reduces to  M =2

Σ

φ d A + 2φ1 area(Σ1 ).

(7.121)

Therefore, 

R2



2π

1 1 μα(R22 − r 2 ) r dr dθ + μα(R22 − R12 )(2π R12 ) 2 2 0 R1  2  R2  4  R2 r r = 2π μα R22 − 2π μα + π μα R12 (R22 − R12 ) 2 R1 4 R1  1 = π μα R24 − R14 . 2

M =2

Since D = M/α, the earlier formula in (7.118) is now fully justified. Example 7.5 (elliptical annulus) Consider a hollow cylinder whose cross section is bounded by two concentric similar ellipses ∂Σ1 and ∂Σ0 , 

X ⊥ = (X 1 , X 2 ) ∈ E2  b2 X 12 + a 2 X 22 = k 2 a 2 b2 , 

∂Σ0 : X ⊥ = (X 1 , X 2 ) ∈ E2  b2 X 12 + a 2 X 22 = a 2 b2 , ∂Σ1 :



where 0 < b < a and 0 < k < 1; Σ represents the domain enclosed between ∂Σ1 and ∂Σ0 (and hence ∂Σ = ∂Σ1 ∪ ∂Σ0 )—see Fig. 7.7. In this case the torsional rigidity has the expression D=

μπa 3 b3 (1 − k 4 ). a 2 + b2

(7.122)

To justify this formula we use a slightly different approach from that employed in the previous example. Let us start by considering the Prandtl stress function corresponding to the torsion problem for a fully solid elliptical cross section (see Exercise 7.3)   μαa 2 b2 X 12 X 22 + 2 −1 . φ(X 1 , X 2 ) = − 2 a + b2 a 2 b

7.12 Worked Examples

365

By direct calculations it can be confirmed that   φ

=0

∂Σ0

and

  φ1 ≡ φ 

∂Σ1

=−

μαa 2 b2 2 (k − 1). a 2 + b2

(7.123)

Next, we are going to check that (7.77) is satisfied. The parametric equations of ∂Σ1 are X 1 = ka cos θ , X 2 = kb sin θ , with 0 ≤ θ < 2π . Hence, X ⊥ = (ka cos θ )e1 + (kb sin θ )e2 and dX ⊥ = (−ka sin θ dθ )e1 + (kb cos θ dθ )e2 . Noticing further that (∇ ⊥ φ) ∧ e3 = φ, 2 e1 − φ, 1 e2 , simple algebra yields dX ⊥ · (∇ ⊥ φ ∧ e3 ) =

2k 2 abμα 2 2 (a sin θ + b2 cos2 θ ) dθ, a 2 + b2

so that  ∂Σ1

dX ⊥ · (∇ ⊥ φ ∧ e3 ) = 2μα(k 2 πab) = 2μα · area(Σ1 ),

because the area of the ellipse having ka and kb as its semi-axes is equal to π(ka)(kb) = k 2 πab. The twisting moment is, again, given by (7.78)—see also (7.121). For evaluating the integral over the elliptical annulus it is convenient to make the change of variables (X 1 , X 2 ) → (r, θ ), with X 1 = ra cos θ,

X 2 = r b sin θ, 0 ≤ θ < 2π, k ≤ r ≤ 1,

whence d A = (ab)r dθ dr . In conclusion,   2μαa 2 b2 2π 1 2 φ dA = − 2 ab(r 2 − 1)r dr dθ a + b2 0 k Σ π μαa 3 b3 (1 − 2k 2 + k 4 ). = 2 a + b2 

Using this result in conjunction with (7.121) and (7.123), we finally arrive at the formula (7.122).

7.13 Exercises 1. The torsional rigidity of a circle, an ellipse, and an equilateral triangle are denoted by DC , D E , and DT , respectively. If the cross-sectional areas of these three sections are equal, prove that

366

7 Torsion

2ab DE = 2 DC , a + b2

√ 2π 3 DT = DC , 15

where 0 < b < a are the semi-axes of the ellipse. 2. The Prandtl stress function for a linearly elastic solid cylinder subjected to pure axial twisting moments on its free flat ends is given by φ(X 1 , X 2 ) = β(a 2 − X 12 + bX 22 )(a 2 + bX 12 − X 22 ), where a, b ∈ R are constants. Determine the value of β ∈ R in terms of the twist per unit length and the material properties. Assuming that 0 < b < 1, sketch the shape of the cylinder cross section for which the above φ gives the solution of the torsion problem. 3. Consider the torsion of the elliptical cylinder discussed in Example 7.1 of Sect. 7.12. Show that  2  X1 X 22 + 2 −1 , (β ∈ R) φ(X 1 , X 2 ) = β a2 b is a suitable candidate for the corresponding Prandtl stress function. Determine the constant β and find the torsion constant from (7.67). 4. A compound shaft consists of two cylinders of different materials welded together, with lengths and radii L 1 , R1 and L 2 , R2 , respectively, where R2 < R1 (see Fig. 7.8). The shaft is held fixed at either end and a twisting couple M is applied at the weld. a. Assuming that the shear moduli of the two materials are μγ (γ = 1, 2), show that the angle of twist at the point C is given by 2 π



L1 L2 M μ1 L 2 R14 + μ2 L 1 R24

 .

b. Suppose now that the total length of the shaft is known and is equal to L. Determine the lengths L 1 and L 2 if the maximum shear stress in both shaft segments is to be the same. 5. For the torsion of the elliptical cross section discussed in Example 7.1 of Sect. 7.12 show that the out-of-plane displacement field is given by u3 (X 1 , X 2 ) =

(b2 − a 2 )M X 1 X 2. μπa 3 b3

6. Consider two cylinders, one having a circular cross section of radius R, the other an elliptic cross section with semi-axes a and b. a. For equal angles of twist, determine which cylinder experiences the larger shearing stresses?

7.13 Exercises

367

X2

L

X1

X3 (C) L1

(C) L2

Fig. 7.8 Side-view (left) and the cross section at the weld (right) for the composite shaft in Exercise 4; the weld is indicated by (C) in both sketches

b. For equal allowable shearing stresses, which cylinder resists a larger twisting moment? 7. If T31 ≡ T31 (X 1 , X 2 ) and T32 ≡ T32 (X 1 , X 2 ), are the non-zero components of the Cauchy stress tensor T in the torsion problem for a cylindrical body, show that in Σ, ∇⊥2 T31 = ∇⊥2 T32 = 0 where Σ ⊂ E2 is the transverse cross section of the cylinder. Deduce that the maxima of both T31 and T32 can only be attained for points on the boundary ∂Σ. 8. a. Let v : Σ ⊂ E2 → R2 be a two-dimensional vector field. Prove the following identity    ∇⊥2 |v|2 = 2 (∇⊥2 v) · v + (∇ ⊥ ⊗ v) : (∇ ⊥ ⊗ v) ,

(7.124)

where |v| is the magnitude of v. b. If φ is the Prandtl stress function corresponding to the torsion of a cylindrical body whose transverse cross section is Σ, by taking v → ∇ ⊥ φ in (7.124), show that the shear stress τ defined in (7.80) satisfies ∇⊥2 (τ 2 ) = 2(∇ ⊥ ⊗ ∇ ⊥ φ) : (∇ ⊥ ⊗ ∇ ⊥ φ)

(7.125)

or, in Cartesian coordinates, ∇⊥2 (τ 2 ) = 2(φ,211 + 2φ,212 + φ,222 ) ≥ 0.

(7.126)

9. The transverse cross section of a cylinder has the shape shown in Fig. 7.9. Formally, this corresponds to the set difference of Σ1 and Σ2 defined below

X ⊥ = (X 1 , X 2 ) ∈ E2 | (X 1 − a)2 + X 22 ≤ a 2 ,

Σ2 : X ⊥ = (X 1 , X 2 ) ∈ E2 | X 12 + X 22 ≤ b2 , Σ1 :



368

7 Torsion

Fig. 7.9 The circular cross section with a groove in Exercise 9

X2

r a

θ O1

b

X1

O2

where 0 < b < a. The cylinder is loaded at its ends by two torsional couples ±M e3 , where e3 is assumed to pass through (X 1 , X 2 ) = (0, 0). a. Considering the polar coordinates (r, θ ) ∈ (0, ∞) × [0, 2π ) defined by X 1 = r cos θ and X 2 = r sin θ , r = (X 12 + X 22 )1/2 , θ = tan−1 (X 2 / X 1 ), show that   2ab2 cos θ , (7.127) φ(r, θ ) = C r 2 − a 2 − 2ar cos θ + r where C ∈ R, may be used as a Prandtl stress function for the cross section seen in Fig. 7.9. Determine the value of the constant C in terms of the twist per unit length and other mechanical/geometrical parameters. b. Calculate the stresses on the boundary of the cross section and establish that on the groove T13 = μα(2a cos θ − b) sin θ, T23 = −μα(2a cos θ − b) cos θ, (7.128) while on the shaft μα 2 sin 2θ (b − 4a 2 cos2 θ ) 2 , 4a cos θ μα 2 cos 2θ T23 = − (b − 4a 2 cos2 θ ) . 4a cos2 θ

T13 =

(7.129a) (7.129b)

c. Using (7.128) and (7.129) show that the shear stress on the groove and the shaft, τg and τs , respectively, are given by   b2 2 sec θ , τg = μα(2a cos θ − b), τs = μα a − 4a and then deduce that τmax = μα(2a − b).

7.13 Exercises

369

d. By remembering that the torsional rigidity D is equal to the ratio M/α, establish the following formula for the torsional rigidity of the cross section shown in Fig. 7.9, ! D = μb

4

# !    "   "  2 b 7 b 3 1 b 4 1 b + − 1−2 θ0 + sin θ0 , a 2 a 2 a 2 a

where 0 ≤ θ0 ≤ π/2 is such that cos θ0 = b/2a. 10. Let Σ ⊂ E2 be an open region bounded by a smooth closed curve ∂Σ. Given the twice continuously differentiable scalar fields u, v : Σ → R, the following identities are true   ∂u ds, ∇⊥2 u d A = ∂n Σ ∂Σ    ∂v u ds, ∇⊥ u · ∇⊥ v d A = − u ∇⊥2 v d A + Σ Σ ∂Σ ∂n      2 ∂v ∂u u −v ds, u ∇⊥ v − v ∇⊥2 u d A = ∂n ∂n Σ ∂Σ where ∂u/∂n ≡ n · (∇ ⊥ u) is the directional derivative in the outward normal direction to ∂Σ and n represents the unit vector that characterises that direction (these results represent a particular case of the so-called Green’s identities). 11. Consider Stokes’ Formula (1.260) in the case when Σ is a planar domain, and let u(x, y) ≡ F1 (x, y)e1 + F2 (x, y)e2 ; this leads to Green’s Theorem,  

 ∂Σ

F1 (x, y) dx + F2 (x, y) dy =

Σ

∂ F2 ∂ F1 − ∂x ∂y

 d A.

If f (z, z) is a function of z = x + iy and z = x − iy, which is continuous and differentiable in Σ, then     ∂f ∂f d A, d A. f (z, z) dz = 2i f (z, z) dz = −2i ∂z ∂Σ Σ ∂Σ Σ ∂z

Bibliography 1. Baldacci R (1983) Scienza delle Costruzioni, vol 1. Unione Tipografico-Editrice Torinese, Torino (in Italian) 2. Solomon L (1968) Élasticité Lineaire. Mason, Paris (in French) 3. Filonenko-Borodich M (1963) Theory of elasticity. Mir Publishers, Moscow 4. Leipholz H (1974) Theory of elasticity. Noordhoff International Publishing, Leyden 5. Little RW (1973) Elasticity. Prentice-Hall Inc, Englewood Cliffs, New Jersey

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7 Torsion

6. Malvern LE (1969) Introduction to the mechanics of a continuum medium. Prentice-Hall Inc, Englewood Cliffs, New Jersey 7. Nadeau G (1964) Introduction to elasticity. Holt, Rinehart and Winston Inc, New York 8. Saada AS (1993) Elasticity: theory and applications. Krieger Publishing Company, Malabar, Florida 9. Slaughter WS (2002) The linearized theory of elasticity. Birkhäuser, Boston 10. Sokolnikoff IV (1956) Mathematical theory of elasticity. McGraw-Hill Book Company Inc, New York 11. Timoshenko SP, Goodier JN (1970) Theory of elasticity, International edn. McGraw-Hill Book Company, Auckland

Chapter 8

Two-Dimensional Approximations

Abstract Here, we return to the plane approximations briefly introduced at the end of Chap. 5, with the aim of exploring in more detail the nature of those solution strategies. Just as the torsion problem can be solved by using a suitably chosen scalar potential, the plane-stress and plane-strain approximations can also be reduced to finding a scalar potential. However, unlike the torsion case, in which the potential was essentially a harmonic function, this time the potential turns out to be bi-harmonic. Several representative examples are used to illustrate the general theory. A number of general facts about the bi-harmonic equation are relegated to Appendix E, which the reader is encouraged to study concomitantly with the present chapter.

8.1 Introduction Both the plane-strain and plane-stress approximations introduced earlier in Chap. 5 (see Sect. 5.11) are concerned with situations in which the interest is primarily directed to what happens in planes perpendicular to a fixed direction. In what follows we shall assume that this direction is specified by the (constant) unit vector a ∈ V , and the domain of interest in the aforementioned planes will be identified as Σ ⊂ E2 (see Fig. 8.1). As a side note, we remark that both approximations are applicable to an original reference configuration Ω0 ≡ Σ × (−, ),  > 0; broadly speaking, for plane-strain  → ∞, while in plane-stress  → 0. The presence of the vector a ∈ V leads to a natural choice for the systems of coordinates employed in this chapter. We select the vectors b, c ∈ V in the plane of  Σ, such that the set b, c, a forms a right-handed orthonormal basis in V —see Fig. 8.1. The origin of the corresponding system of coordinates associated with these vectors is arbitrary (in the plane of Σ) and, in principle, will depend on the particular geometrical features of the configuration at hand. Two choices of coordinate systems prove to be most useful in applications. The first one is a Cartesian system O X 1 X 2 X 3 , with O X 3 directed along a, while the remaining two axes O X 1 , O X 2 are perpendicular to each other and are situated in the plane of Σ. An equally useful choice is a system of cylindrical polar coordinates (r, θ, z), with the z-axis aligned again with the direction of a. © Springer Nature B.V. 2020 C. D. Coman, Continuum Mechanics and Linear Elasticity, Solid Mechanics and Its Applications 238, https://doi.org/10.1007/978-94-024-1771-5_8

371

372

8 Two-Dimensional Approximations

Fig. 8.1 An illustration of the basic geometry for the two-dimensional approximations discussed in this chapter

The position vector X (say) of an arbitrary point in Ω0 ⊂ E3 can always be resolved into a component situated in Σ and another one along a direction parallel to a. Using the notation introduced in the previous chapter, X = X ⊥ + X || . Furthermore, in light of the information provided in Sect. 5.11 for plane-strain/stress approximations, the elastic states will be functions of X ⊥ only. The validity of the two-dimensional approximations discussed in the present chapter rests on an additional number of assumptions regarding the body force (per unit volume), f ≡ ρb. In particular, we shall require that it is independent of the normal coordinate measured along the direction of a, and its component in that direction vanishes identically; for short, we write f = f (X ⊥ ) and f · a = 0. To keep the presentation simple for the time being, we shall assume that f is constant—the case of a general f is dealt with in some of the problems at the end of this chapter (e.g. see Exercise 8.2). In addition to this, our main focus will be on the static case. With the aforementioned planar approximations in mind, it turns out that, by writing ∇ = ∇ ⊥ + ∇ || and using T = T (X ⊥ ), the following are true ∇ ⊥ · T int + f = 02 ,

(8.1a)

∇⊥2 |T int |

(8.1b)

= 0,

where 02 denotes the usual two-dimensional zero vector, and the label ‘int’ identifies the interior part of the corresponding tensor. If P a ≡ I − a ⊗ a is the projection on to planes orthogonal to a, we recall that T int := P a · T · P a (see Sect. 1.21). For the plane-stress approximation, Eq. (8.1a) is trivially satisfied, while (8.1b) is an easy consequence of the Beltrami–Michell equation for statics, ∇2T +

  1 ∇ ⊗ ∇|T | = O , 1+ν

(8.2)

which is obtained by contraction and taking into account that |T | = |T int |. The situation is, however, not so obvious in the case of plane-strain, but this aspect will be elaborated on shortly.

8.1 Introduction

373

It is important to distinguish between the vector fields associated with threedimensional Linear Elasticity and their two-dimensional counterparts employed in this chapter; we shall use the subscript ‘⊥’ to emphasise this distinction. For example, u⊥ stands for the displacement field in the latter case, while t ⊥ represents the corresponding traction vector. For future reference, we record below the component form of (8.1a) in two special cases. In Cartesian coordinates O X 1 X 2 X 3 (with O X 3 in the direction of a), ∂ T11 ∂ T12 + + f1 = 0 , ∂ X1 ∂ X2

∂ T12 ∂ T22 + + f2 = 0 , ∂ X1 ∂ X2

in Σ ,

(8.3)

while in cylindrical polar coordinates (with Oz in the direction of a), 1 ∂ Tr θ 1 ∂ Trr + + (Trr − Tθθ ) + fr = 0 , ∂r r ∂θ r 1 ∂ Tθθ 2 ∂ Tr θ + + Tr θ + f θ = 0 , ∂r r ∂θ r

(8.4a) in Σ .

(8.4b)

The (static) equilibrium equations recorded above can be easily extended to accommodate inertial effects; this is simply done by inserting the corresponding components of the vector ρ(∂ 2 u⊥ /∂t 2 ) on the right-hand side of those equations. The equations recorded above must be solved subject to appropriate boundary conditions; assuming that ∂Σ = ∂Σ u ∪ ∂Σ t ,

∂Σ u ∩ ∂Σ t = ∅ ,

(8.5)

on ∂Σ t ,

(8.6a)

on ∂Σ ,

(8.6b)

these requirements translate into t ⊥ ≡ n⊥ · T int = ˚t , u⊥ = u˚ ,

u

˚ ⊥ ) are both given two-dimensional vector fields in where ˚t = ˚t (X ⊥ ) and u˚ = u(X the plane of Σ (i.e. ˚t · a = u˚ · a = 0).

8.2 Plane Strain Since E 33 = u3, 3 and u3 ≡ 0, it follows that |E| = E 11 + E 22 and hence |E| = |E int |. In this case, Hooke’s law (5.20) can be restated in the reduced form T int = 2μE int + λ|E int |I 2 ,

Tαβ = 2μE αβ + λ(E γ γ )δαβ ,

(8.7)

374

8 Two-Dimensional Approximations

where I 2 = eα ⊗ eα is the two-dimensional identity tensor1 (recall that, according to the convention introduced in Sect. 5.11, Greek subscripts range from 1 to 2). In writing (8.7) we have left out the non-trivial equation corresponding to T33 , according to which T33 = λ(E γ γ ). Note that, by contraction, Eq. (8.7) gives Tαα = 2(λ + μ)E γ γ . Taken together, these last two results can then be combined into T33 =

λ Tγ γ 2(λ + μ)

=⇒

T33 = ν(T11 + T22 ) ,

(8.8)

where use has been made of the definition of ν—see (5.21). The relation between T33 and Tαα is needed for the consistency of our approximation, but will play only a marginal role in formulating any specific plane-strain problem. More specifically, after the solution of such a problem has been found, the stress component perpendicular to the plane of Σ can be found via Eq. (8.8). The presence of this non-zero stress component is one of the unsatisfactory features of the plane-strain theory. Another use of (8.8) is in the justification of (8.1b). Indeed, since |T | = Tαα + T33

=⇒

|T | = (1 + ν)|T int |

(8.9)

and hence, by contraction, Eq. (8.2) leads directly to (8.1b). The inverted form of Hooke’s law (5.23) will also undergo a slight change as a result of (8.9). Note that in plane strain, E int =

 1+ν T int − ν|T int |I 2 , E

E αβ =

 1+ν Tαβ − ν(Tγ γ )δαβ , E

(8.10)

while  1 ∇ ⊥ ⊗ u⊥ + u⊥ ⊗ ∇ ⊥ , 2 1 = (uα, β + uβ, α ) . 2

E int =

(8.11a)

E αβ

(8.11b)

We now have a complete set of equations describing the plane-strain linear framework. For easy reference these are reiterated below, 1. equilibrium equation: (8.1a)—depending on what type of coordinate system is chosen, in component form we use either (8.3) (Cartesian) or (8.4) (polar); 2. compatibility condition: (8.1b); 3. constitutive law: (8.7) or (8.10)—recall that, for the sake of completeness, there is also the additional Eq. (8.8), although this plays only a passive role; 4. kinematic relations (8.11); 5. boundary conditions (8.6). that, by setting e1 → b and e2 → c, we can think of this reduced identity tensor as I 2 ≡ I − a ⊗ a, where I is the full identity tensor associated with the 3D vector space V .

1 Note

8.2 Plane Strain

375

In the case when polar coordinates are chosen to describe the geometry of Σ, the constitutive laws (8.7) and (8.10) remain unchanged (but α, β ∈ {r, θ }), while (8.8) is replaced by Tzz = ν(Trr + Tθθ ). To find the components of the reduced strain tensor we can still use the general definition (8.11), along with u⊥ = ur (r, θ )er + uθ (r, θ )eθ ,

∇ ⊥ ≡ er

∂ 1 ∂ + eθ . ∂r r ∂θ

The corresponding components of E int assume the expressions  ∂ur 1 ∂uθ , E θθ = + ur , Err = ∂r r ∂θ  ∂uθ uθ 1 1 ∂ur + − . Er θ = 2 r ∂θ ∂r r

(8.12)

(8.13a) (8.13b)

8.3 Plane Stress In this case the inverted Hooke’s law (5.23) leads to E int =

1+ν ν T int − |T int |I 2 , E E

E αβ =

1+ν ν Tαβ − (Tγ γ )δαβ , E E

(8.14)

and E 33 = −(ν/E)(Tγ γ ) .

(8.15)

The last equation suggests that, in general, E 33 ≡ 0 (i.e. there will be undesirable deformations in the a-direction, an artefact of the approximate nature of plane-stress theory). Note that (8.14) is of the same form as in the general three-dimensional case (after making the substitutions E → E int and T → T int ). However, Hooke’s law will be slightly different from the original, as shown next. The contracted form of (8.14) in conjunction with (8.15) give E 33 = −

ν E αα 1−ν

=⇒

E γ γ + E 33 =

2μ Eγ γ . λ + 2μ

(8.16)

To get the reduced form of Hooke’s law, we project the full result (5.20) on to the plane of Σ, remembering that E 33 contributes to |E|, and making use of (8.16). This gives eventually 2λμ |E int |I 2 , (8.17) T int = 2μE int + λ + 2μ which is slightly different from the corresponding result for the plane-strain approximation—see (8.7). In spite of this difference, we remark that both formulae can be written in a unified way,

376

8 Two-Dimensional Approximations

T int = 2μE int + λ∗ |E int |I 2 , where we have introduced the modified Lamé coefficient ⎧ ⎨λ (plane strain) , λ∗ := 2λμ ⎩ (plane stress) . λ + 2μ

(8.18)

(8.19)

In a similar fashion, for the inverted form of Hooke’s law, we have E int = with

 1  T int − δ∗ |T int |I 2 , 2μ

⎧ ⎨ ν (plane strain) , ν δ∗ := (plane stress) . ⎩ ν+1

(8.20)

(8.21)

There are several different ways to write Eq. (8.18), and we refer to Exercises 8.10 and 8.11 for two other versions widely used in the literature. Unlike the approximation outlined in the previous section, the plane-stress theory has some more serious shortcomings. These become apparent when one tries to work out the displacement components. For example, the kinematic relations 0 ≡ 2E 3α = u3, α + uα, 3 (α = 1, 2) suggest that uα are functions of X 3 . Similarly, by integrating E 33 = u3, 3 it is found that u3 is a linear function of X 3 as well (see Exercise 8.4). These inconsistencies arise from the fact that some of the compatibility equations cannot be satisfied by simply making a priori assumptions on the stress tensor (for the obvious reason that compatibility is directly related to deformation, i.e. E). Nonetheless, the plane-stress theory represents a valid and useful approximation, provided that its limitation are understood and one disregards the three-dimensional effects pointed out above. A strong argument in favour of this approximate theory is supported by an indirect strategy in which the fields of interest are averaged over the plate thickness. Although the resulting equations turn out to be identical to the ones already discussed, the upshot is that the unwanted three-dimensional ‘side-effects’ are eliminated from the theory. An outline of this so-called generalised plane-stress approximation is presented next.

8.4 Generalised Plane Stress We consider a thin elastic plate of thickness 2h > 0, as shown in Fig. 8.2. The plate is essentially a right cylindrical body, the generators of which are parallel to our constant vector a, and orthogonal to the two flat end faces. It will also be assumed that the geometry of the plate is symmetric with respect to its median surface, which coincides with the O X 1 X 2 -plane.

8.4 Generalised Plane Stress

377

Fig. 8.2 The deformation of this thin cylindrical body in directions parallel to the O X 1 X 2 -plane is reduced to what happens in the cross section Σ ⊂ E2 (shaded), here assumed to be equidistant from the two flat end faces. This reductive process involves through-thickness averaging for all mechanical fields of interest

Suppose now that the plate is loaded on the lateral surface by tractions and displacements parallel to its end faces and symmetrically distributed with respect to the midplane; the end faces are traction-free and the body force vector f ≡ ρb is even in X 3 , while its component in the vertical direction vanishes identically, i.e. f = f α eα ,

f α = f α (X ⊥ , X 3 ) ,

f α (X ⊥ , −X 3 ) = f α (X ⊥ , X 3 ) ,

f3 ≡ 0 ,

(∀) − h ≤ X 3 ≤ h ,

with X ⊥ = (X 1 , X 2 ) ∈ Σ ⊂ E2 . Noting also that the foregoing hypotheses about the boundary data translate into the following conditions u˚ α (X ⊥ , −X 3 ) = u˚ α (X ⊥ , X 3 ) ,

t˚α (X ⊥ , −X 3 ) = t˚α (X ⊥ , X 3 ) ,

(∀) X 3

and u˚ 3 = t˚3 ≡ 0, it can be checked that uα , E αα (no sum), E 33 , Tαβ , T33 are even (symmetric) in X 3 , whereas u3 , E α3 and Tα3 will be odd (skew-symmetric) functions in the same variable. Since t(±e3 ) = 0 for X 3 = ±h, T13 = T23 = T33 = 0 ,

for X 3 = ±h .

(8.22)

Differentiating (8.22) with respect to X i (i = 1, 2), and using the third equilibrium equation, Ti3,i = 0, written for X 3 = ±h, shows that T33,3 = 0 ,

for X 3 = ±h .

(8.23)

The Taylor expansion of T33 (X ⊥ , X 3 ) around an arbitrary point (X ⊥ , X 3 ) reveals that T33 = O(h 2 ) because of (8.22) and (8.23). Hence, if the plate thickness is sufficiently small, we may assume that T33 (X) ≡ 0 throughout the plate. Next, let us introduce the new mean fields u∗ (X ⊥ ) :=

1 2h

+h −h

u(X ⊥ , X 3 ) dX 3 , T ∗ (X ⊥ ) :=

1 2h

+h

−h

T (X ⊥ , X 3 ) dX 3 .

378

8 Two-Dimensional Approximations

Integrating the first two equilibrium equations, Ti j, i + f j = 0 ( j = 1, 2), with respect to X 3 and then dividing through the resulting expressions by 2h, gives ∗ ∗ ∗ T11, 1 + T21, 2 + f 1 = 0 ,

where f α∗ ≡ f α∗ (X ⊥ ) :=

1 2h

∗ ∗ ∗ T12, 1 + T22, 2 + f 2 = 0 ,

+h −h

(8.24)

f α (X ⊥ , X 3 ) dX 3 .

The terms T3∗j, 3 ( j = 1, 2) have dropped from (8.24) because T3∗j, 3

+h 1 := T3 j, 3 (X ⊥ , X 3 ) dX 3 2h −h  1  T3 j (X 1 , X 2 , h) − T3 j (X 1 , X 2 , −h) = 0 , = 2h

the last equality being a direct consequence of (8.22). Since u3 and Tα3 (α = 1, 2) are skew-symmetric with respect to X 3 = 0, and the averaged fields are defined by ∗ ≡0 integrals over the symmetric interval [−h, h], it transpires that u3∗ ≡ 0 and Tα3 (α = 1, 2). This explains why there is no need to consider the equilibrium equation in the vertical direction (orthogonal to Σ). A close look at (8.24) indicates that we can write ∇ ⊥ · T ∗int + f ∗⊥ = 02 , where ∗ ∗ eα ⊗ eβ represents the interior part of the averaged tensor T ∗ , and f ∗⊥ is T int := Tαβ the two-dimensional vector of components f α∗ . We can also derive an averaged version of Hooke’s law. First, since 0  T33 = λ(E ii ) + 2μE 33 , we can express E 33 in terms of E αα and substitute the result back in Ti j = 2μE i j + λ(E kk )δi j to discover that, in particular, Tαβ = 2μE αβ + λ∗ (E 11 + E 22 )δαβ ,

α, β ∈ {1, 2} .

Finally, we can now take averages in this last equation to find T ∗int = 2μE ∗int + λ∗ |E ∗int |I 2 , where we have set E ∗ (X ⊥ ) :=

1 2h

+h −h

E(X ⊥ , X 3 ) dX 3 ,

∗ with E ∗int := P a · E ∗ · P a ≡ E αβ eα ⊗ eβ , and we note that λ∗ has already been defined in (8.19); the remaining equations in Hooke’s law need not be considered as their averaged form is identically satisfied.

8.4 Generalised Plane Stress

379

The main equations obtained above have exactly the same form as those found in the previous section. We remark in passing that the boundary conditions are also amenable to a straightforward averaging process leading to equations identical in form to (8.6), but for the averaged fields T ∗ and u∗ defined above.

8.5 The Airy Stress Function 8.5.1 Main Definitions The main appeal of the two-dimensional approximations discussed in this chapter lies in the fact that the solution of the equilibrium equations (8.1a) and the compatibility condition (8.1b) can be found in terms of just one scalar field, known as the Airy stress function (or potential). This is introduced in a somewhat indirect manner by assuming that the reduced stress tensor T int has the form   T int = ∇⊥2 Φ I 2 − ∇ ⊥ ⊗ ∇ ⊥ Φ ,

(8.25)

for some scalar field Φ = Φ(X ⊥ ), which represents the actual Airy stress function. Before we carry on with various formal manipulations of this formula, it is useful to see first how Tαβ and Φ are related to each other in two special cases. If Σ is described by using Cartesian coordinates, then Φ = Φ(X 1 , X 2 ) and (8.25) is equivalent to the following three scalar equations T11 =

∂ 2Φ , ∂ X 22

T22 =

∂ 2Φ , ∂ X 12

T12 = −

∂ 2Φ . ∂ X 1∂ X 2

(8.26)

If instead we use polar coordinates, then Φ = Φ(r, θ ) and Trr =

1 ∂ 2Φ 1 ∂Φ + 2 , r ∂r r ∂θ 2

Tθθ =

∂ 2Φ , ∂r 2

Tr θ = −

∂ ∂r



1 ∂Φ r ∂θ

.

(8.27)

8.5.2 The Governing Equation for Φ We are going to see next that (8.25) satisfies the equilibrium equation (8.1a) identically when the body force is absent. Thus, by simply casting the stresses in terms of Φ as done above, we automatically satisfy the equation of equilibrium. It will also be established below that the governing equation for the actual expression of the Airy stress function comes from the compatibility relation (8.1b).

380

8 Two-Dimensional Approximations

Note that     ∇ ⊥ · T int = ∇ ⊥ · (∇⊥2 Φ)I 2 − ∇ ⊥ · ∇ ⊥ ⊗ ∇ ⊥ Φ     = ∇ ⊥ ∇⊥2 Φ − ∇⊥2 ∇ ⊥ Φ = 02 , where we have used the fact that ∇⊥2 and ∇ ⊥ commute with each other, and the property ∇ ⊥ · (ϕ I 2 ) = ∇ ⊥ ϕ, valid for any smooth ϕ = ϕ(X ⊥ ) (see Exercise 8.1). This completes the justification of the first assertion above. Next, we look at the implications of (8.25) on the compatibility condition. The trace of T int in (8.25) is immediately calculated by replacing ‘⊗’ with ‘·’ and taking into account that |I 2 | = 2. Thus, |T int | = ∇⊥2 Φ and the compatibility condition (8.1b) becomes   ∇⊥2 ∇⊥2 Φ = 0

=⇒

∇⊥4 Φ = 0 ,

in Σ .

(8.28)

This represents the main governing equation for the Airy stress function. It is known as the homogeneous bi-harmonic equation because it involves an iterated Laplacian (or harmonic operator). In a Cartesian system of coordinates, O X 1 X 2 , Eq. (8.28) becomes ∂ 4Φ ∂ 4Φ ∂ 4Φ + 2 + ≡ ∂ X 14 ∂ X 12 ∂ X 22 ∂ X 24



∂2 ∂2 + ∂ X 12 ∂ X 22



∂ 2Φ ∂ 2Φ + ∂ X 12 ∂ X 22

= 0,

(8.29)

where Φ = Φ(X 1 , X 2 ), while in polar coordinates (8.29) is replaced by 

∂2 1 ∂2 1 ∂ + + ∂r 2 r ∂r r 2 ∂θ 2



∂ 2Φ 1 ∂ 2Φ 1 ∂Φ + + ∂r 2 r ∂r r 2 ∂θ 2

= 0,

(8.30)

In this latter case, Φ = Φ(r, θ ) and we recall that ∇⊥2 ≡

1 ∂2 ∂2 1 ∂ + + . ∂r 2 r ∂r r 2 ∂θ 2

The solutions of (8.29) and (8.30) are reviewed in Appendix E. Finally, the traction boundary condition t ⊥ ≡ n⊥ · T int = ˚t can be re-arranged so that it features only the Airy stress function,     − E · ∇ ⊥ ⊗ ∇ ⊥ Φ · E · n⊥ = ˚t ,

on ∂Σ t ,

(8.31)

where E := I ∧ a = I ∧ (b ∧ c) = c ⊗ b − b ⊗ c represents a 90◦ -rotation tensor around the axis defined by the unit vector a. In the case of Cartesian coordinates (b → e1 and c → e2 ), Eq. (8.31) boils down to

8.5 The Airy Stress Function

381

∂ 2Φ ∂ 2Φ n − n 2 = t˚1 , 2 1 ∂ X 1∂ X 2 ∂ X2

∂ 2Φ ∂ 2Φ n − n 1 = t˚2 2 2 ∂ X 1∂ X 2 ∂ X1

(8.32)

on ∂Σ t , where n⊥ = n α eα represents the outward unit normal along ∂Σ t . We are going to say more about the boundary conditions in the next section. In the interest of simplicity, at the beginning of this section we have taken the body force to be identically zero. It is easy to restore this effect, but the Airy stress function can only be used if the body force (per unit volume) f ≡ ρb is conservative; more specifically, we require   (8.33) f = − ∇⊥ V , for some given scalar field V = V (X ⊥ ). In Cartesian coordinates, V = V (X 1 , X 2 ) and (8.33) amounts to f α = −V, α (α = 1, 2); for polar coordinates, V = V (r, θ ), with fr = −V, r and f θ = −r −1 V, θ , where f = fr er + f θ eθ . Fortunately, most body forces encountered in practical situations can be expressed in the above form (e.g. gravitational and centrifugal forces). We refer to Exercise 8.8 for the definition of the Airy stress potential and its governing equation when the body force satisfies the condition (8.33).

8.5.3 Physical Interpretation of the Boundary Conditions The expressions in the boundary conditions (8.32) are rather opaque. It is possible to rewrite them in a form that facilitates their interpretation, and this is our next main goal. In what follows it will be assumed that Φ ∈ C 4 (Σ) ∩ C 1 (Σ), where Σ ≡ Σ ∪ ∂Σ. For the sake of generality we are also going to allow Σ to be multiply connected, but we shall take ∂Σ u = ∅ in (8.5), so there are only traction boundary conditions imposed on ∂Σ; in consequence, ∂Σ = C0 ∪ C1 ∪ · · · ∪ C p , where C0 is the external boundary of Σ and C j (for j = 1, 2, . . . , p) represent the internal boundaries, as seen in Fig. 8.3. These are reasonably smooth closed curves described by the parametric equations X 1 = X 1 (s) ,

X 2 = X 2 (s) ,

0 ≤ s ≤ s∗ ,

(8.34)

where the arc length s is measured in the positive sense along each contour, and X α (0) = X α (s∗ ) for some s∗ > 0 (α = 1, 2). With this in mind, the outward unit normal on ∂Σ is given by2 d X⊥ . (8.35) n⊥ = −E · ds 2n

⊥

is obtained by a 90◦ clockwise rotation of the tangent vector τ .

382

8 Two-Dimensional Approximations

Fig. 8.3 An example of multiply connected planar domain Σ that illustrates the nomenclature used in the main text. The curved arrows indicate the positive direction of the contours C j , with j = 0, 1, . . . , p (here p = 2)

Since E · E = −I 2 , replacing (8.35) in (8.31) and then left-multiplying the resulting equation by E, yields  d  ∇ ⊥ Φ = E · ˚t . (8.36) ds We are interested to know whether the values of the Airy stress function Φ and its derivatives on the curves C j ( j = 0, 1, . . . , p) can be determined from the boundary ‘data’, ˚t (s) ≡ ˚t (X ⊥ (s)), given along the same curve. To this end, we shall take two points A, B ∈ C j , with A fixed (but arbitrary) and B ≡ B(s) variable along the curve. Integration of (8.36) from A to B gives   ∇ ⊥ Φ  B = l j + E · R AB ,

(8.37)

where we have introduced the auxiliary notations   l j := ∇ ⊥ Φ  A

and

B

R AB :=

˚t (s) ds ,

A

with the latter representing the resultant external force acting on the arc AB ⊂ C j . This result confirms that in passing from A to B (along a C j ) the increment of the derivatives is determined by the resultant vector of the external forces applied to the contour between the two points. The formula (8.37) allows us to calculate the directional derivative of Φ in any direction in the plane of Σ. For instance, if τ denotes the unit tangent vector to C j ,     d X⊥ dΦ ≡ ∇⊥ Φ · τ = ∇⊥ Φ · dτ ds    d X⊥   dΦ   , = l j + E · R AB  · =⇒ dτ  B  ds  B  where B  ∈ C j is situated between A and B. This last equation can be integrated from A to B to give

8.5 The Airy Stress Function

383

    Φ B = m j + l j · X ⊥B + where

B

  E · R AB  (ζ ) · d X ⊥ (ζ ) ,

(8.38)

A

    m j := Φ  A − l j · X ⊥  A

is a constant (i.e. independent of B). Equation (8.38) shows that given the values of ˚t (s) on any C j , the value of the Airy stress function can be calculated at any point on that contour. The non-integral terms in (8.38) are of the form A0 j + A1 j X 1 + A2 j X 2 ,

Ai j ∈ R , (i = 0, 1, 2) ,

and are thus irrelevant for calculating the stresses (since the Airy stress formulae are defined in terms of second-order derivatives). For a simply connected domain Σ there is only one group of (A0 j , A1 j , A2 j ), and it can be taken to be zero right from the outset. However, in the case of a multiply connected domain this is no longer true; although one can still set to zero one group of constants, the other groups cannot be chosen arbitrarily. Returning back to (8.37), we now want to find the normal derivative of Φ,      d X⊥ dΦ ≡ ∇ ⊥ Φ · n⊥ = ∇ ⊥ Φ · −E · dn ds      dΦ  d X⊥  =⇒ = (∇ ⊥ Φ) A + E · R AB · −E · dn  B ds or

    dΦ  = E · l j − R AB · τ  B .  dn B

(8.39)

Formula (8.39) suggests that the normal derivative of the Airy stress function is fully determined by the traction applied on the boundary. As it was already pointed out above, for a simply connected domain Σ one can take l j = 0, and then (8.39) shows that the normal derivative of Φ at an arbitrary point on ∂Σ is equal to the negative of the projection of R AB on the tangent τ to the curve ∂Σ at the same point. The arguments presented in this section suggest that in the case of a twodimensional traction boundary-value problem the Airy stress function Φ is determined by the bi-harmonic equation (8.28) from prescribed contour values of this function and its normal derivative. While this observation helps to establish the well-posedness of the boundary-value problem for the Airy stress function, simple closed-form solutions are only possible if the geometry of Σ is fairly regular. Based on the foregoing developments we can make a few remarks about the conditions that need to be satisfied in order for Φ and its first-order derivatives to be single-valued. The idea here is to look at the variation of these functions around the closed loops C j . For a function ψ : E2 → R, [ψ] AB := ψ(B) − ψ(A) denotes the variation of this function between the points A, B ∈ E2 . Every closed curve in

384

8 Two-Dimensional Approximations

the plane can be regarded as being traversed starting from a given point on it, P − (say), and ending in P + , where these two points are geometrically coincident. The variation of ψ along C j will then be defined as [ψ]C j := ψ(P + ) − ψ(P − ) ,

(P ∈ C j , arbitrary) .

(8.40)

It follows directly from (8.37) and (8.38) that

where

   ∇ ⊥ Φ C j = E · RC j ,

(8.41a)

      Φ C j = −a · X 0⊥ ∧ RC j + M O C j ,

(8.41b)

 RC j :=

 Cj

˚t (s) ds ,

 MO 

 Cj

:=

Cj

X ⊥ (s) ∧ ˚t (s) ds ,

represent the resultant force and, respectively, the resultant moment of all external forces applied on the closed contour C j ; X 0⊥ denotes the position vector of an arbitrary point on C j , from which a complete circuit is started.  The variation of Φ and its first-order derivatives will be zero if RC j = 02 and  M O  = 0. If some of these conditions are violated, then Φ and/or its derivatives Cj

Φ, α cannot be single-valued in a multiply connected region Σ ⊂ E2 . For example, if R j ≡ 02 for all 0 ≤ j ≤ p then [∇ ⊥ Φ]C j = 02 and [Φ]C j = 0 (in general); hence, Φ, α are single-valued, but Φ is not. Since the stresses are defined in terms of the second-order derivatives of Φ and, on physical grounds, they should be single-valued, it is possible to have multi-valued Airy stress functions in certain applications; see Appendix E.

8.5.4 The Displacement Field The Airy stress function formalism sketched above relies only on the Beltrami– Michell system and circumvents the need of dealing directly with the displacement field. This is one of the strengths of the method that makes it particularly attractive for solving large classes of two-dimensional traction boundary-value problems. In principle, once Φ has been found, we get the stresses from either (8.26) or (8.27), and use of the reduced Hooke’s law (8.20) will then yield the two-dimensional deformation tensor E int . At this stage one can integrate the kinematic relations (8.11) to find the two components of the displacement field (as explained in detail in Chap. 6). In the present section we do not follow this general route, but instead adopt a change of tack that hinges on introducing a new scalar field in the problem (somewhat similar to Φ, but related directly to the displacement field).

8.5 The Airy Stress Function

385

We start by assuming that Φ has been determined in one way or another; according to the foregoing remarks, the two-dimensional strain tensor is known, in particular,   2μE int = (1 − δ∗ ) ∇⊥2 Φ I 2 − ∇ ⊥ ⊗ ∇ ⊥ Φ or, in component form, 2μu1,1 = −δ∗ Φ,11 + (1 − δ∗ )Φ, 22 , 2μu2,2 = (1 − δ∗ )Φ,11 − δ∗ Φ, 22 ,   μ u1,2 + u2,1 = −Φ,12 ,

(8.42a) (8.42b) (8.42c)

where we recall that δ∗ was defined in (8.21). Multiplying (8.42c) by a factor of 2 and re-arranging gives   ∂  ∂  2μu2 + Φ,2 + 2μu1 + Φ,1 = 0 . ∂ X1 ∂ X2 This equation is identically satisfied if we set 2μu1 + Φ,1 = (1 − δ∗ )Ψ,1

2μu2 + Φ, 2 = −(1 − δ∗ )Ψ, 2 ,

and

(8.43)

for some smooth function Ψ = Ψ (X 1 , X 2 ) that we aim to find by using the two unused equations in (8.42); Ψ is called the displacement function (or potential). We differentiate the first equation in (8.43) with respect to X 1 and use the result in (8.42a) to eliminate u1, 1 ; in a similar way, u2, 2 is eliminated from (8.42b) after we differentiate the second equation in (8.43) with respect to X 2 . Eventually, we get the governing equations for the unknown potential Ψ , Ψ, 11 = ∇⊥2 Φ

and

Ψ, 22 = −∇⊥2 Φ ,

which are usually combined and stated in the equivalent form Ψ, 11 = ∇⊥2 Φ

and

∇⊥2 Ψ = 0 .

(8.44)

8.6 Worked Examples In this section we present a number of representative examples for the theory developed earlier in this chapter. As we shall see shortly, most theoretical solutions require further simplifying assumptions or the presence of certain symmetry properties. To avoid overloading the notation we shall drop the subscript ‘⊥’ whenever there is no risk of confusion.

386

8 Two-Dimensional Approximations

2

1

Fig. 8.4 Cantilever bar subject to a vertical force on its free end. The rectangular geometry of the cross section is illustrated on the right; see Example 8.1

Example 8.1 As an illustration of how the Airy stress function works in practice, we first consider a simple example that can be conveniently expressed in Cartesian coordinates. We are interested in the deformation of a rectangular cantilever bar of length L > 0 subjected to a vertical end load P > 0 as seen in Fig. 8.4. It is required to find the elastic state in the cantilever bar, assuming that its weight can be neglected and the system is in a state of plane stress.

We start by formulating the relevant boundary conditions. To this end, note that the boundary of the two-dimensional domain Σ ≡ (0, L) × (−h/2, h/2) contains four parts, i.e. ∂Σ = ∂Σ1+ ∪ ∂Σ1− ∪ ∂Σ2+ ∪ ∂Σ2− , with

    ∂Σ2± ≡ (X 1 , X 2 ) ∈ E2  0 ≤ X 1 ≤ L , X 2 = ±(h/2) ,

and     ∂Σ1− ≡ (X 1 , X 2 ) ∈ E2  X 1 = 0 , −(h/2) ≤ X 2 ≤ h/2 ,     ∂Σ1+ ≡ (X 1 , X 2 ) ∈ E2  X 1 = L , −(h/2) ≤ X 2 ≤ h/2 . The boundary conditions are   t(±e2 )

∂Σ2±

=0

and

+h/2

b −h/2

 t(−e1 )∂Σ − dX 2 = P e2 ;

(8.45)

1

the first two express the traction-free state of ∂Σ2± , while the third requires that the resultant of the traction vector on ∂Σ1− matches the applied vertical force. To make

8.6 Worked Examples

387

use of these constraints we need some assumptions about the stress tensor T int . Let us pick the stress function Φ(X 1 , X 2 ) =

1 C X 1 X 23 + B X 1 X 2 , 6

(B, C ∈ R) ,

(8.46)

which satisfies the bi-harmonic equation (8.29) since each one of the two individual monomials that appear in (8.46) share the same property. Note that we have two degrees of freedom included in the assumed form of Φ, and so we hope that this would allow us to satisfy (8.45). Use of (8.26) immediately gives  T11 = C X 1 X 2 ,

T22 ≡ 0 ,

T12 = −

1 C X 22 + B 2

.

(8.47)

Enforcing one of the first two boundary constraints (8.45), we obtain T12 = 0 at X 2 = h/2, whence C = −(8B/ h 2 ). Similarly, the integral constraint in (8.45) becomes

−P = b

+h/2 −h/2

T12 (0, X 2 ) dX 2

=⇒

B=

3P . 2bh

As we now have both B and C in (8.46), the explicit form of the stress field can be finally stated. With I := bh 3 /12, Eqs. (8.47) become  T11 = −

P I

X1 X2 ,

T22 ≡ 0 ,

P T12 = 2I

 X 22

 2  h − , 2

(8.48)

while the strains are easily calculated from (8.20). To find the components of the displacement field we need to integrate the kinematic relations, ∂u1 =− ∂ X1



 ∂u2 νP X1 X2 , = ∂ x2 EI   2  ∂u1 ∂u2 P h 2 + = . X2 − ∂ X2 ∂ X1 2μI 2

P EI



X1 X2 ,

(8.49a) (8.49b)

Referring to Chap. 6 for the details of similar examples, it can be shown that we eventually get   ν P P 1 X 12 X 2 + − X 23 + α X 2 + β , u1 = − 2E I 6I μ E    2 νP P Ph 2 3 X1 X2 + X1 − + α X1 + γ , u2 = 2E I 6E I 8μI

(8.50a) (8.50b)

388

8 Two-Dimensional Approximations

where α, β, γ ∈ R are constants associated with a rigid-body deformation. To suppress this indeterminacy we can impose the additional constraints  ∂u1  u1 (L , 0) = u2 (L , 0) = = 0, ∂ X 2 (L ,0) but we do not pursue the matter further. A close scrutiny of the solution presented above reveals that nothing was said about the boundary conditions on ∂Σ1+ . Since our cantilever bar is in static equilibrium, we expect

+h/2

b −h/2

 t(e1 )∂Σ + dX 2 = −P e2 ,

+h/2

−h/2

1



   X ⊥ ∂Σ + ∧ t(e1 )∂Σ + dX 2 = 0 . 1

1

These two different conditions are easily checked by using the results found in (8.48). We note in passing that the second integral condition above can be rewritten as

−L P = b

+h/2

−h/2

X 2 T11 (L , X 2 ) dX 2 .

Example 8.2 We shall examine next the case of a thick-walled circular cylinder under internal pressure P > 0, with the external curved lateral surface tractionfree. The inner and outer radii are denoted by R1 and R2 , respectively. It is required to find the elastic state in the cylinder, assuming a condition of plane strain (Fig. 5.8 is still relevant here, with the caveat that the setting is entirely two-dimensional, with p1 ≡ P and p2 ≡ 0). The plane-strain assumption allows us to reduce the problem to the annular domain    Σ ≡ (r, θ ) ∈ E2  R1 < r < R2 , 0 ≤ θ < 2π , whose boundary has two parts, ∂Σ ± , defined by ∂Σ + := {r = R2 }

and

∂Σ − := {r = R1 } .

The boundary constraints are  t(er )∂Σ + = 0

and

 t(−er )∂Σ − = P er .

(8.51)

A key assumption in what follows is the cylindrical symmetry of the elastic state, i.e. all fields of interest will be considered to be independent of ‘θ ’ and ‘z’, T int = T int (r ) ,

E int = E int (r ) ,

u⊥ = u⊥ (r ) .

8.6 Worked Examples

389

We are going to use the Navier–Lamé system in which we set ∇ → ∇ ⊥ , and write ur = f (r )er ,

uθ ≡ 0 ,

(8.52)

for some scalar function f = f (r ) that is yet to be found. From Exercise 8.5 it is known that f (r ) = Ar + Br −1 , for some arbitrary A, B ∈ R, and the corresponding components of E int and T int can be found in terms of these constants—see (8.108) and (8.109). All we need to do next is use the boundary conditions (8.51) to fix the arbitrary constants A and B. Noting that t(±er ) = (±er ) · T int = ±(Trr er + Tr θ eθ ) = ±Trr er , it is a routine matter to verify that Eqs. (8.51) lead to 2(λ + μ)A −

2μ B = −P , R12

whence 1 A= 2(λ + μ)



P R12 R22 − R12

2(λ + μ)A −

,

1 B= 2μ



2μ B = 0, R22

P R12 R22 R22 − R12

.

Substituting these constants in the expression of f (r ) and then using (8.109), we arrive at the final values of the stresses   R2 R2 P R2 P R2 Trr = 2 1 2 1 − 22 , Tθθ = 2 1 2 1 + 22 , Tr θ ≡ 0 . r r R2 − R1 R2 − R1 Recall that the plane-strain theory is just an approximation and we still have a stress component acting along the axis of the cylinder (cf. (8.8)), Tzz = ν(Trr + Tθθ ) =

2ν P R12 . R22 − R12

This represents the magnitude of the normal compressive forces that would have to be applied on the ends of the cylinder in order to maintain a true plane-strain state in its cross sections, away from those locations. Finally, the displacement follows immediately from the radial symmetry assumption (8.52),    R22 P R12 1+ν (1 − 2ν)r + . ur = E r R22 − R12

Example 8.3 Consider an infinite thin elastic plate with a circular hole of radius a > 0, which is subjected to the constant uniaxial traction T0 > 0 at infinity (as seen in Fig. 8.5). Assuming that the boundary of the hole is traction-free, it is required to find the stress distribution in the plate.

390

8 Two-Dimensional Approximations

Fig. 8.5 An infinite elastic plate with a small hole is subjected to uniaxial tension T0 at infinity (as X 1 → ±∞); see Example 8.3

We start by choosing a Cartesian system of coordinates such that the X 1 -axis represents the direction in which the plate is being stretched, while the origin of the coordinates is at the centre of the hole. The boundary of this hole is the circle X 12 + X 22 = a 2 , while the domain of the plate can be formally represented as   Σ ≡ (X 1 , X 2 ) ∈ E2 | X 12 + X 22 > a 2 . This is a doubly connected domain in E2 , whose boundary is made up of two parts, ∂Σ = {r = a} ∪ {+∞}; if we imagine a very large circle concentric with the hole, then the part we identified as {+∞} would correspond to the circumference of this circle. The constraints imposed on this part of the boundary are commonly referred to as far-field conditions because they obviously have a different status from the standard boundary conditions applied on a finite curve. The situation described in this problem requires the following boundary and farfield conditions, t(−er ) = 0 , t(±e1 ) → ±T0 e1 ,

on r = a , as r → +∞ .

(8.53a) (8.53b)

An unsatisfactory aspect here is the presence of two different types of coordinate systems in the formulation of (8.53). This is easily rectified by making use of Exercise 8.15, which shows that the far-field condition (8.53b) translates into the following polar-coordinate version, Trr →

T0 T0 T0 (1 + cos 2θ ) , Tθθ → (1 − cos 2θ ) , Tr θ → − sin 2θ . (8.54) 2 2 2

Next, let us assume that the plate is stressed as in (8.53b), but without a hole. An appropriate choice for the Airy stress function in this trivial case is Φ0 =

1 1 T0 X 22 = T0 r 2 (1 − cos 2θ ) , 2 4

(8.55)

8.6 Worked Examples

391

since X 2 = r sin θ . For this particular choice, T11 = T0 , T22 = T12 ≡ 0. Clearly, in the far-field this stress distribution matches the original loading condition in the problem (8.54); the trouble is that if the hole is present, the stress field will be locally perturbed in the vicinity of the hole. This suggests that a potentially correct choice for the Airy stress function in this latter situation will be a perturbation of (8.55), i.e. Φ = Φ0 (r ) + Φ2 (r ) cos 2θ .

(8.56)

Substituting (8.56) into (8.30), after the separation of variables, we end up with two Euler-type ordinary differential equations, 2 d 3 Φ0 1 d 2 Φ0 1 dΦ0 d 4 Φ0 = 0, + − + 3 dr 4 r dr 3 r 2 dr 2 r dr d 4 Φ2 2 d 3 Φ2 9 d 2 Φ2 9 dΦ2 = 0, + − 2 + 3 4 3 dr r dr r dr 2 r dr whose solutions are readily available (see Appendix E), Φ0 (r ) = A1r 2 log r + A2 r 2 + A3 log r + A4 , A7 Φ2 (r ) = A5r 2 + A6r 4 + 2 + A8 , r

(8.57a) (8.57b)

for some arbitrary constants A j ∈ R ( j = 1, 2, . . . , 8). Substituting (8.57) into (8.56) and then using the formulae (8.27), yields  A3 6A7 4 A8 cos 2θ , (8.58a) − 2 A + + 5 r2 r4 r2  A3 6A7 Tθθ = A1 (3 + 2 log r ) + 2 A2 − 2 + 2 A5 + 12 A6 r 2 + 4 cos 2θ , (8.58b) r r  6A7 2 A8 (8.58c) Tr θ = 2 A5 + 6A6 r 2 − 4 − 2 sin 2θ . r r Trr = A1 (1 + 2 log r ) + 2 A2 +

Since Ti j (i, j ∈ {r, θ }) are bounded as r → +∞, it follows that A1 = A6 = 0, and Trr → 2 A2 − 2 A5 cos 2θ ,

Tθθ → 2 A2 + 2 A5 cos 2θ ,

Tr θ → 2 A5 sin 2θ ,

as r → +∞. Matching these expressions with those in (8.54) it transpires that A2 = −A5 = T0 /4. The remaining constants in the expression of the Airy stress function are obtained by imposing the boundary conditions (8.53) in conjunction with (8.58); this results in three algebraic equations, T0 A3 + 2 = 0, 2 a

−

T0 4 A8 6A7 + 4 + 2 = 0, 2 a a

−

T0 2 A8 6A7 − 4 − 2 = 0, 2 a a

392

8 Two-Dimensional Approximations

Fig. 8.6 An infinite elastic plate with a small circular hole is subjected to pure shear at infinity (see Example 8.4)

whence A3 = −A8 = −a 2 T0 /2 and A7 = −a 4 T0 /4. We can eventually state the stress components in their final form,   T0 a2 3a 4 T0 4a 2 1− 2 + 1 + 4 − 2 cos 2θ , 2 r 2 r r   2 4 T0 a 3a T0 1+ 2 − 1 + 4 cos 2θ , Tθθ = 2 r 2 r  4 2 3a T0 2a 1 − 4 + 2 sin 2θ . Tr θ = − 2 r r Trr =

(8.59a) (8.59b) (8.59c)

The reader has not failed to notice that the azimuthal stress component Tθθ did not participate in the local boundary conditions (8.53a), and hence need not vanish. Evaluating it on the edge of the hole, we get Tθθ (a, θ ) = T0 (1 − 2 cos 2θ ) .

(8.60)

Clearly, this is a function of ‘θ ’ only, and it is easy to see that the minimum of the right-hand side in (8.60) is (−T0 ), a value attained for θ ∈ {0◦ , 180◦ }. The maximum of Tθθ is 3T0 and corresponds to the direction perpendicular to the applied load (θ ∈ {90◦ , 270◦ }). It can be concluded that the presence of the hole (essentially a geometrical ‘discontinuity’) plays the role of amplifying the applied load. This local phenomenon is commonly referred to as stress concentration, and the stress concentration factor (SC F, for short) is defined as the ratio of the highest stress in the perturbed configuration (3T0 ) to a reference stress value (T0 ); that is, the stress that would exists in the unperturbed configuration as a result of identical loading conditions. In the present problem SC F = 3.

8.6 Worked Examples

393

Example 8.4 Consider an infinite thin elastic plate with a circular hole of radius a > 0, and which is subjected to pure shear at infinity, as seen in Fig. 8.6. Assuming that the boundary of the hole is traction-free, it is required to find the stress distribution in the plate.

This is similar to the previous example, although the choice of stress function will be rather different; the domain of the plate and the boundary condition on r = a are exactly the same as in the Example 8.3 above. Equations (8.53) are replaced by t(−er ) = 0 ,

on r = a ,

(8.61a)

T int → S0 (e1 ⊗ e2 + e2 ⊗ e1 ) ,

as r → +∞ ,

(8.61b)

and we note in passing that the far-field condition (8.61b) is equivalent to 1/2  T11 , T22 → 0 , T12 → S0 , as r ≡ X 12 + X 22 → +∞ .

(8.62)

Just as before, let us first consider the plate without a hole, but under the same loading. An appropriate choice for the Airy stress function in this particular case is 1 Φ0 = −S0 X 1 X 2 = − S0 r 2 sin 2θ ; 2 with the help of (8.26) it is immediately checked that the far-field condition (8.62) is satisfied. Making use again of Exercise 8.15, the far-field condition translated into polar coordinates becomes Trr → S0 sin 2θ ,

Tθθ → −S0 sin 2θ ,

Tr θ → S0 cos 2θ ,

(8.63)

as r → +∞. Equations (8.63) suggest that a sensible choice for the Airy stress function in the general case would be Φ = Φ 2 (r ) sin 2θ .

(8.64)

The amplitude function Φ2 satisfies the same differential equation as in the preceding example; in fact, we shall adopt exactly the same notation. For this particular choice of stress potential the general formulae (8.27) show that 6A7 4 A8 Trr = − 2 A5 + 4 + 2 sin 2θ , r r  6A7 2 Tθθ = 2 A5 + 12 A6r + 4 sin 2θ , r  6A7 2 A8 Tr θ = − 2 A5 + 6A6r 2 − 4 − 2 cos 2θ . r r 

(8.65a) (8.65b) (8.65c)

394

8 Two-Dimensional Approximations

Eliminating the unbounded terms (as r → +∞) from these expressions and matching with (8.63) leads to A6 = 0 and A5 = −(S0 /2). The other two constants are found from the local boundary condition (8.61a), A7 = −(a 4 S0 /2) and A8 = a 2 S0 . Replacing the values just found in (8.65) yields the final expressions of the stresses,  4a 2 3a 4 Trr = S0 1 − 2 + 4 sin 2θ , r r  3a 4 Tθθ = −S0 1 + 4 sin 2θ , r  2a 2 3a 4 Tr θ = S0 1 + 2 − 4 cos 2θ . r r

(8.66a) (8.66b) (8.66c)

The evaluation of the function Tθθ on the boundary of the hole gives Tθθ (a, θ ) = −4S0 sin 2θ .

(8.67)

It should be clear that the maximum value of the azimuthal stress is 4S0 , and hence SC F = 4. This maximum is attained at two points on the circle r = a, which are situated on a diameter inclined at 135◦ to the positive side of the X 1 -axis. The intersection between the same circle and the diameter perpendicular to this direction gives the location of two other points that render the right-hand side of (8.67) a minimum. Example 8.5 A flat circular bar of outer radius R2 and inner radius R1 (0 < R1 < R2 ), with a cross section in the form of a narrow rectangle (of thickness b = 1, for convenience), is bent by moments M applied at the flat ends (see Fig. 8.7). It is required to find the elastic state in the bar assuming a condition of plane stress.

The domain of interest will be   Σ ≡ (r, θ ) ∈ E2 | R1 < r < R2 , −β < θ < +β , where β ∈ (0, π/2) is a given angle. Note that the boundary of this region consists of four parts, ∂Σ = ∂Σ1 ∪ ∂Σ2 ∪ ∂Σ + ∪ ∂Σ − , with   ∂Σ j ≡ r = R j , −β ≤ θ ≤ +β , and

( j = 1, 2) ,

  ∂Σ ± ≡ R1 ≤ r ≤ R2 , θ = ±β .

8.6 Worked Examples

395

Fig. 8.7 Curved bar subjected to in-plane pure bending by bending moments applied at its two ends (θ = ±β). The curved boundaries are arcs of circles and the cross–section corresponds to the rectangle [0, b] × [R1 , R2 ]; see Example 8.5

The outward unit normals on ∂Σ1 and ∂Σ2 are (−er ) and er , respectively, while for ∂Σ ± this role is fulfilled by (±eθ ). The boundary conditions for the curved bar are     = t(+er ) = 0, t(−er ) ∂Σ1 ∂Σ2

R2  t(±eθ )∂Σ ± dr = 0 , R1

R2 R1

 X ⊥

∂Σ ±

 ∧ t(±eθ )∂Σ ± dr = ∓M ez

or, in component form, Trr = Tr θ = 0 ,

R2 Tθθ dr = 0 , R1

for r = R1 , R2 , R2

r Tθθ dr = −M .

(8.68a) (8.68b)

R1

We introduce the simplifying assumption that the elastic state depends only on the radial coordinate. We have already found the expression of the bi-harmonic equation in this case—see (8.57a). Leaving out the constant term, Φ = A log r + Br 2 log r + Cr 2 , for some arbitrary constants A, B, C ∈ R. Calculating the stresses associated to this Φ, with the help of (8.27), we find A + 2B log r + (3B + 2C) . r2 (8.69) Enforcing the radial conditions in (8.68a) leads to two algebraic equations, Trr =

A + 2B log r + (B + 2C) , r2

Tθθ = −

396

8 Two-Dimensional Approximations

A + 2B log R1 + (B + 2C) = 0 , R12

A + 2B log R2 + (B + 2C) = 0 . R22 (8.70) If these conditions are fulfilled, then the first requirement in (8.68b) will be automatically satisfied as well. As for the second integral condition, note that

R2

r Tθθ dr =

R1

R2 R1

 r =R r Φ  (r ) dr = r Φ  (r ) r =R21 −

r =R  r =R  = r Φ  (r ) r =R21 − Φ r =R21 =

R2

R1  r =R2 − Φ r =R1

Φ  (r ) dr

,

where we have used integration by parts and the fact that the first term on the second line turns out to be zero, cf. (8.70); the ‘dash’ in the above equations indicates differentiation with respect to the radial coordinate. Thus, the second integral condition in (8.68b) is equivalent to Φ(R2 ) − Φ(R1 ) = M or, in expanded form, A log

    R2 + B R22 log R2 − R12 log R1 + C R22 − R12 = M . R1

(8.71)

The relations (8.70) and (8.71) represent a linear system of three equations in three unknowns. Routine algebraic manipulations yield  4M 2 2 R2 2M  2 R R log R2 − R12 , , B=− N 1 2 R1 N  M 2 C= (R2 − R12 ) + 2(R22 log R2 − R12 log R1 ) , N

A=−

where

  2 R2 2 N := R22 − R12 − 4R12 R22 log . R1

Replacing these constants in (8.69) provides the final form of the stresses,   4M R12 R22 R2 r R1 2 2 , log + R log + R log 2 1 N r2 R1 R2 r    R2 R2 4M R2 r R1  2 − 1 2 2 log + R2 − R12 , =− + R22 log + R12 log N r R1 R2 r

Trr = − Tθθ

and the strain tensor follows directly from (8.20)—recall that Hooke’s law has the same form in polar coordinates. All that is left is to compute the displacement field. As usual, this requires the integration of the kinematic relations (8.13), which here assume the form

8.6 Worked Examples

397

   1 A(1 + ν)  ∂ur = + 2(1 − ν) log r + (1 − 3ν) B + 2(1 − ν)C , ∂r E r2    1 ∂uθ ur 1 A(1 + ν)  + = − + 2(1 − ν) log r + (3 − ν) B + 2(1 − ν)C , r ∂θ r E r2 1 ∂ur ∂uθ uθ + − = 0. r ∂θ ∂r r By integrating the first two equations it is found that    A(1 + ν)  1 d f1 − + 2(1 − ν) log r − (1 + ν) Br + 2(1 − ν)Cr + , E r dθ 4B r θ − f 1 (θ ) + f 2 (r ) , uθ = E

ur =

where the functions f 1 = f 1 (θ ) and f 2 = f 2 (r ) are arbitrary at this stage. They are found by substituting these expressions in the third kinematic equation above, with the outcome f2 d 2 f1 d f2 − = 0, + f 1 (θ ) = 0 , dr r dθ 2 whereby f 1 (θ ) = A2 sin θ − A1 cos θ

and

f 2 (r ) = A3r .

The constants Ai (i = 1, 2, 3) reflect the presence of a rigid-body deformation in the expression of the displacement field. They can be determined, for instance, by assuming that the bar is fixed at the point O1 (coordinates: r = (R1 + R2 )/2, θ = 0); that is, ur = uθ = uθ, r = 0 at that point. Example 8.6 Consider a curved cantilever bar loaded by a distributed shear force at the lower edge, as shown in Fig. 8.8. It is required to find the elastic state in this system, if the curved boundaries are traction-free and assuming that the plane-stress approximation is applicable.

In this case the boundary of the domain occupied by the bar consists of four parts ∂Σ = ∂Σ + ∪ ∂Σ − ∪ ∂Σ0 ∪ ∂Σ90 , with     ∂Σ − = r = R1 , 0 ≤ θ ≤ π/2 , ∂Σ + = r = R2 , 0 ≤ θ ≤ π/2 ,     ∂Σ0 = R1 ≤ r ≤ R2 , θ = 0 , ∂Σ90 = R1 ≤ r ≤ R2 , θ = π/2 . The outward unit normal on ∂Σ90 is (−e1 ), while on ∂Σ0 that role is played by (−e2 ); the curved boundaries ∂Σ ± have outward unit normals corresponding to ±er . We can now list the relevant boundary conditions,

398

8 Two-Dimensional Approximations

Fig. 8.8 Curved cantilever bar subjected to shear forces on one of its straight edges. The curved boundaries are arcs of circles; see Example 8.6

2

1

2 1

1

t(±er ) = 0 , on ∂Σ ± ,

t(−e2 ) dr = −P e1

(8.73b)

Trr = Tr θ = 0 ,

(8.74a)

(8.73a)

∂Σ0

or, in component form,

R2 R1

  Tr θ 

θ=0

dr = P ,

for r = R1 , R2 ,   Tθθ  dr = 0 .

R2 R1

θ=0

(8.74b)

Since the bending moment at any transverse cross section (θ = constant) is proportional to sin θ , and the normal stress acting on these planes (Tθθ ) is also proportional to the bending moment, it seems sensible to look for an Airy stress function that has a similar sort of dependence on ‘θ ’. We tentatively assume Φ = Φ1 (r ) sin θ ,

(8.75)

with Φ1 to be found. Separating the variables in the bi-harmonic equation (8.30) reveals a familiar Euler-type ordinary differential equation, r4

d 4 Φ1 d 3 Φ1 d 2 Φ1 dΦ1 − 3Φ1 = 0 , + 2r 3 − 3r 2 + 3r 4 3 2 dr dr dr dr

which admits the general solution Φ1 (r ) = C1r 3 + C2 r + C3r log r +

C4 , (C j ∈ R, j = 1, 2, 3, 4) . r

8.6 Worked Examples

399

The stresses corresponding to the Airy potential (8.75) are then easily calculated,   2C4 C3 2C4 C3 − 3 sin θ , Tθθ = 6C1r + + 3 sin θ , (8.76a) Trr = 2C1r + r r r r  C3 2C4 (8.76b) Tr θ = − 2C1r + − 3 cos θ . r r Applying the constraints (8.74a) for the stress components in (8.76) results in 2C1 R1 +

C3 2C4 − 3 = 0, R1 R1

2C1 R2 +

C3 2C4 − 3 = 0, R2 R2

which can be solved for C3 and C4 in terms of C1 , i.e. C3 /C1 = −2(R12 + R22 ) and C4 /C1 = −R12 R22 . On substituting these values back into (8.76), Tr θ becomes  R12 + R22 R12 R22 cos θ . + Tr θ = −2C1 r − r r3 It should be now clear that, by using this expression in the first constraint in (8.74b), we can fix C1 ; the second integral condition is automatically satisfied because Tθθ = 0 on ∂Σ0 (since θ = 0). A simple calculation gives C1 =

    R2 P , with N := R12 − R22 + R12 + R22 log , 2N R1

and then the components of the stress tensor (T int ) in (8.76) become  R12 + R22 R12 R22 P sin θ, r− + Trr = N r r3  R 2 + R22 R2 R2 P 3r − 1 − 1 3 2 sin θ, Tθθ = N r r  R12 + R22 R12 R22 P cos θ. r− + Tr θ = − N r r3

(8.77a) (8.77b) (8.77c)

Example 8.7 We are interested in finding the stress distribution in a plate subjected to an in-plane concentrated load P at a point on its boundary, as seen in the left panel of Fig. 8.9. We shall consider that P represents a force per unit width of the plate (here taken to be unity, for simplicity), and assume that the plate is in a state of plane stress. The problem can therefore be regarded as being two dimensional, with Σ ≡ {(X 1 , X 2 ) ∈ E2 | X 2 < 0} ,

400

8 Two-Dimensional Approximations

Fig. 8.9 Concentrated load acting normal to the edge of a semi-infinite plate; see Example 8.7

and we shall find it useful to use a polar system of coordinates set up as indicated in Fig. 8.9. Our approach to solving this problem is somewhat different from the previous examples. It will be based on the a priori assumption that the state of stress generated by this particular type of loading is purely radial on any semi-circle around the point where the load is applied, i.e. the constraints Tθθ = Tr θ ≡ 0

(8.78)

hold true on the dashed semi-circle (whose radius is arbitrary). Although this hypothesis could be justified (more or less) rigorously, we shall not do that here. By ignoring the body force in the equilibrium equations (8.4) and taking into account (8.78), leaves us with Trr ∂ Trr + = 0. (8.79) ∂r r This is easily integrated to give Trr = r −1 H (θ ) ,

(8.80)

for some arbitrary function H that depends on θ only. Next, let us express the conditions (8.78) in terms of the Airy stress function, ∂ 2Φ = 0 in Σ , ∂r 2 1 ∂Φ 1 ∂ 2Φ Tr θ ≡ 0 =⇒ 2 − = 0 in Σ . r ∂θ r ∂r ∂θ

Tθθ ≡ 0 =⇒

(8.81a) (8.81b)

8.6 Worked Examples

401

Our aim is to use (8.81) to simplify the bi-harmonic equation expressed in polar coordinates. Use of the first of these constraints allows us to write  2  ∂ 1 ∂Φ 1 ∂ 1 ∂2 1 ∂ 2Φ =0 (8.82) + + + ∂r 2 r ∂r r 2 ∂θ 2 r ∂r r 2 ∂θ 2 or, after suitably expanding, ∂2 ∂r 2



1 ∂Φ 1 ∂ 2Φ + 2 r ∂r r ∂θ 2



 1 ∂ 1 ∂Φ 1 ∂ 2Φ + 2 r ∂r r ∂r r ∂θ 2  1 ∂ 2Φ 1 ∂ 2 1 ∂Φ = 0. + 2 + 2 2 r ∂θ r ∂r r ∂θ 2

+

(8.83)

Note that,   ∂ 2 1 ∂Φ 2 ∂Φ ∂ 1 ∂Φ 1 ∂ 2Φ = 3 = − 2 + , 2 2 ∂r r ∂r ∂r r ∂r r ∂r r ∂r   ∂2 1 ∂ 2Φ ∂ 6 ∂ 2Φ 2 ∂ 2Φ 1 ∂ 3Φ 4 ∂ 3Φ = = · · · = − + − , ∂r 2 r 2 ∂θ 2 ∂r r 3 ∂θ 2 r 2 ∂r ∂θ 2 r 4 ∂θ 2 r 3 ∂r ∂θ 2 and following the same strategy we also get the additional results 1 ∂ r ∂r



 2 ∂ 2Φ 1 ∂Φ 1 ∂Φ 1 ∂ 1 ∂ 2Φ 1 ∂ 3Φ = − =− 3 , + , r ∂r r ∂r r ∂r r 2 ∂θ 2 r 4 ∂θ 2 r 3 ∂r ∂θ 2   1 ∂ 2 1 ∂Φ 1 ∂ 2Φ 1 ∂ 4Φ 1 ∂ 3Φ 1 ∂2 = = , . r 2 ∂θ 2 r ∂r r 3 ∂r ∂θ 2 r 2 ∂θ 2 r 2 ∂θ 2 r 4 ∂θ 4

Replacing all these expressions back into the original Eq. (8.83) gives r

∂ 2Φ ∂ 4Φ ∂Φ +2 2 + = 0. ∂r ∂θ ∂θ 4

(8.84)

This equation is amenable to a solution with separable variables. To this end, take Φ(r, θ ) = F(r )G(θ ) ,

(8.85)

where F and G will be determined by substituting this assumed form of the solution back into (8.84). The outcome is G  (θ ) + 2G  (θ ) r F  (r ) =− , F(r ) G(θ )

(8.86)

where the dash has been used to denote differentiation with respect to either ‘r ’ or ‘θ ’ (no confusion should arise as the functions on the right-hand side in (8.85) depend

402

8 Two-Dimensional Approximations

on only one of these variables). Equation (8.86) suggests that each one of the two fractions must be equal to a constant, k (say), so that we end up with two simple ordinary differential equations, r F  (r ) − k F(r ) = 0

and

G  (θ ) + 2G  (θ ) + kG(θ ) = 0 .

(8.87)

The first of these is easily integrated by inspection, giving F(r ) = A0 r k , for some arbitrary A0 ∈ R. In conjunction with formula (8.27), this result yields   Trr = A0 r k−2 kG(θ ) + G  (θ ) .

(8.88)

Comparing (8.88) with the earlier result (8.80) suggests that k − 2 = −1, whence k = 1 and   (8.89) H (θ ) = A0 G(θ ) + G  (θ ) . Thus, the second equation in (8.87) becomes G  (θ ) + 2G  (θ ) + G(θ ) = 0 , whose general solution is G(θ ) = (A1 θ + A2 )(A3 sin θ + A4 cos θ ) ,

A j ∈ R ( j = 1, 2, 3, 4) .

(8.90)

This expression can now be used in (8.88) to find Trr =

2 A0 (A1 A3 cos θ − A1 A4 sin θ ) . r

(8.91)

As Trr is expected to be symmetric about the vertical axis (θ = 0), the sine term in the above expression has to go, so we set A4 = 0. Furthermore, we note that A2 does not feature in (8.91); without loss of generality we can therefore take A2 = 0 as well. Finally, remembering (8.85), the final form of the Airy stress function becomes Φ(r, θ ) = Br θ sin θ ,

(8.92)

where B := A0 A1 A3 is a constant that has yet to be found. This is done by requiring that the resultant vertical force on any semi-circle of radius r > 0 must equal the applied force P,

P=−

+π/2 −π/2

(Trr cos θ ) r dθ = −

whence B = −P/π , and then

+π/2 −π/2

2B cos2 θ dθ = −Bπ ,

8.6 Worked Examples

403

Φ(r, θ ) = −

P r θ sin θ . π

(8.93)

For future reference we state again the final form of the stress distribution,  Trr = −

2P π



cos θ , r

Tθθ ≡ 0 ,

Tr θ ≡ 0 .

(8.94)

These formulae can be expressed in Cartesian coordinates (using Exercise 8.16)  T11 = −

2P π



X 12 X 2 , (X 12 + X 22 )2 T12 = −

 T22 = − 

2P π



2P π



X 23 , (X 12 + X 22 )2

X 1 X 22 . + X 22 )2

(X 12

(8.95a) (8.95b)

Equations (8.94), or the above Cartesian representation of the same formulae, are usually referred to as the Flamant solution (for plane isotropic elasticity).3 We close this example with an interesting property of the solution derived above. Consider an imaginary disk of diameter d and centre O  , which is tangent to the straight edge at O, as seen in the right panel of Fig. 8.9. The arbitrary point M(r, θ ) is situated on the circumference of this disk, while n is the outward unit normal at that point. Clearly, the system of vectors {er , eθ } is obtained from {e1 , e2 } by a counterclockwise rotation of angle (π/2 + θ ), and hence e2 = er cos θ − eθ sin θ . −−−→ We also note further that (d/2)n = O  M = −(d/2)e2 + r er , so that by combining these results we find n = er cos θ + eθ sin θ , since cos θ = r/d. This allows us to calculate the traction vector t(n) acting at the point M, t(n) ≡ n · T = (er cos θ + eθ sin θ ) · (Trr er ⊗ er ) 2P cos θ er . = Trr cos θ er = − πd We can now conclude that the point M on the circumference of our imaginary circle experiences a non-uniform compression of magnitude (2P/π d) cos θ in the direction specified by the vector (−er ). On the other hand, on a surface element whose normal is (+er ), the stress force is given by t(er ) = −(2P/π d)er ;

3 The

solution given here is a particular case of a more general result obtained in 1892 by AlfredAimé Flamant, a French engineer and pupil of B. de Saint-Venant.

404

8 Two-Dimensional Approximations

this represents a radial force whose magnitude is inversely proportional to the diameter of the circle (d).

8.7 Volterra Distortions We end the chapter with a few more two-dimensional examples that share a special feature: they are all related to the concept of incompatibility discussed in Chap. 6. Two examples were already mentioned in passing at the end of that chapter; here, we want to illustrate how the corresponding elastic states can be explicitly calculated. The situations considered in this section involve infinitely long hollow cylinders, and fall within the scope of the plane-strain theory. For this reason the attention will be confined to what happens in a generic transverse cross section, as seen in Fig. 8.10. The left sketch corresponds to a cylinder that has been cut along an axial halfplane (X 2 = 0, X 1 > 0); the two faces of the cut are then symmetrically displaced with respect to each other by a small rigid displacement. Next, the gap between the open faces is closed by inserting additional material (of the same type as that of the cylinder), and the external forces that have acted on the cylinder during these operations are released. Then, the strains in the resulting new configuration will be continuous, but the displacement field will suffer a jump discontinuity at points situated on the positive semi-axis. The second scenario, which appears in the centre of Fig. 8.10 is the so-called edge dislocation mentioned earlier in the book, and we refer to Sect. 6.9 for details. The given constant b ∈ R denotes the magnitude of the displacement jump in the horizontal direction. The last sketch in the same figure is concerned with a somewhat similar situation: the two faces of the cut are symmetrically displaced with respect to the horizontal axis, but this time in the vertical direction; additional material is added to close the gap; and then the forces applied to maintain this state of deformation are removed. It is clear that the strains will be continuous, whereas the displacement

Fig. 8.10 Examples of two-dimensional systems that can accommodate incompatible strain fields; see text for further details

8.7 Volterra Distortions

405

will experience a discontinuity as it crosses the positive semi-axis (the vertical jump in displacement is a known quantity, d ∈ R). In the first case mentioned above the deformation can be regarded as having radial symmetry, so we choose an Airy function that has no azimuthal dependence,4 Φ = A1r 2 + A2 log r + A3r 2 log r .

(8.96)

With the help of the information in the Tables included in Appendix E it can be checked that (8.96) generates a stress distribution of the form A2 + A3 (2 log r + 1) , r2 A2 Tθθ = 2 A1 − 2 + A3 (2 log r + 3) , r Tr θ = 0,

Trr = 2 A1 +

(8.97a) (8.97b) (8.97c)

with the corresponding displacements given by 1 + ν A2 E r    1+ν 2(1 − ν) r log r − r A3 , + E E

2(1 − ν) ur = A1 r − E

uθ =



4 A3 rθ . E

The arbitrary constants in the expression of the Airy stress function are determined from the boundary conditions. On the inner and outer surfaces of the cylinder Trr and Tr θ must vanish. There is an additional constraint that has to do with the perturbation created by inserting the wedge,   uθ 

θ=0

  − uθ 

θ=2π

= ωr .

(8.98)

From these three equations the constants are found to be A1 =

(R 2 − 1) + 2R 2 log R Eω , 16π(R 2 − 1)

A2 = −

2R 2 log R Eω , 8π(R 2 − 1)

A3 = −

ωE . 8π

Substituting these values into (8.97) gives the stress distribution in the deformed configuration. The stresses acting in any transverse cross section of the deformed cylinder are statically equivalent to a couple of moment M given by

4 See

(E.55) for the general form of the Airy function in polar coordinates.

406

8 Two-Dimensional Approximations

M =−

4R 2 (log R)2 − (R 2 − 1)2 Eω . 16π(R 2 − 1)

(8.99)

The case of the edge dislocation is more complicated because we can no longer rely on radial symmetry; both the radial and the azimuthal coordinates must now be present in Φ. Based on physical considerations we might expect Tθθ to have a sin θ dependence and Tr θ a cos θ one, which are both accomplished by taking Φ ∝ sin θ . Selecting the appropriate terms from (E.55), a potential candidate for Φ is Φ = A1r 3 sin θ + A2

sin θ + A3r log r sin θ , r

(8.100)

where Ai ∈ R (i = 1, 2, 3) are to be found. The components of the stress tensor and the displacement field associated with this stress potential are recorded below, sin θ sin θ , + A3 r3 r sin θ sin θ Tθθ = 6A1r sin θ + 2 A2 3 + A3 , r r cos θ cos θ Tr θ = −2 A1r cos θ + 2 A2 3 − A3 , r r Trr = 2 A1r sin θ − 2 A2

(8.101a) (8.101b) (8.101c)

and  1 + ν sin θ r 2 sin θ + A2 E r2    2 1+ν 1−ν + A3 log r sin θ − θ cos θ − sin θ , E E 2E   ν+5 2 1 + ν cos θ r cos θ − A2 uθ = −A1 E E r2    1−ν 2 1+ν log r cos θ + θ sin θ + cos θ . + A3 E E 2E 

ur = A1

1 − 3ν E



Again, the two curved boundaries must be traction-free, while (8.98) is replaced by   ur 

θ=0

  − ur 

θ=2π

= b,

which results in A1 = −

Eb , 8π(R 2 + 1)

A2 =

R 2 Eb , 8π(R 2 + 1)

A3 =

Eb . 4π

It can be checked without difficulty that on the faces of the cut (θ = 0) the state of stress (8.101) is statically equivalent to a radial force parallel to the X 1 -axis; its magnitude is

8.7 Volterra Distortions

407

Eb F1 = 4π



R2 − 1 − log R R2 + 1

.

(8.102)

Finally, the last case we consider corresponds to the sketch on the right in Fig. 8.10. This is similar to the situation we have just discussed, except that now Φ ∝ cos θ . Therefore, we replace (8.100) by Φ = A1r 3 cos θ + A2

cos θ + A3r log r cos θ , r

(8.103)

where the arbitrary constants A j ( j = 1, 2, 3) are found as before. To this end, we note that cos θ cos θ , + A3 3 r r cos θ cos θ Tθθ = 6A1r cos θ + 2 A2 3 + A3 , r r sin θ sin θ Tr θ = 2 A1r sin θ − 2 A2 3 + A3 , r r Trr = 2 A1r cos θ − 2 A2

(8.104a) (8.104b) (8.104c)

and 

 1 + ν cos θ r 2 cos θ + A2 E r2    2 1+ν 1−ν log r cos θ + θ sin θ − cos θ A3 , + E E 2E   ν+5 2 1 + ν sin θ uθ = A 1 r sin θ + A2 E E r2     1−ν 1+ν 2 + − log r sin θ + θ cos θ − sin θ A3 . E E 2E

ur = A1

1 − 3ν E



The last function must satisfy   uθ 

θ=0

  − uθ 

θ=2π

=d,

which in conjunction with the other two constraints, Trr (r = 1) = Tr θ (r = R) = 0, is used to identify the values of the foregoing constants, A1 =

Ed , 8π(R 2 + 1)

A2 = −

R 2 Ed , 8π(R 2 + 1)

A3 = −

Eb . 4π

It can be shown that on the faces of the cut (θ = 0) the stress distribution obtained above is statically equivalent to a force parallel to the X 2 -axis that has the magnitude

408

8 Two-Dimensional Approximations

Fig. 8.11 The loading required for creating the incompatible deformations shown earlier in Fig. 8.10

F2 =

Ed 4π



R2 − 1 − log R R2 + 1

.

(8.105)

Figure 8.11 illustrates the action of the statically equivalent forces calculated in all three examples above.

8.8 Exercises 8.1. Consider ∇ the usual gradient operator in E3 and ∇ 2 ≡ ∇ · ∇ the corresponding Laplacian. Show that ∇(∇ 2 ϕ) = ∇ 2 (∇ϕ) , for any smooth scalar field ϕ : E3 → R. Hence, deduce that ∇ ⊥ (∇⊥2 ϕ) = ∇⊥2 (∇ ⊥ ϕ) for ϕ = ϕ(X ⊥ ), with X ⊥ ∈ Σ ⊂ E2 . 8.2. Show that in the case of a general body force (per unit volume), f = f (X ⊥ ) and f · a ≡ 0, the compatibility equation (8.1b) must be replaced by   ∇⊥2 |T int | + β ∇ ⊥ f = 0 ,

(8.106)

where β = (1 + ν)−1 (plane strain) or β = 1 + ν (plane stress). 8.3. Use Eqs. (8.11) and (8.12) to derive (8.13). 8.4. Use Eq. (8.15) together with the Beltrami–Michell system (8.2) to deduce that, under the plane-stress assumption, E 33 = A0 + A1 X 1 + A2 X 2 ,

(Ai ∈ R, i = 0, 1, 2) .

8.5. Consider the Navier–Lamé system (5.51) for elastostatics under the assumption of plane strain in a domain Σ ⊂ E2 , and in the absence of body forces. A

8.8 Exercises

409

cylindrical polar system of coordinates is chosen so that the z-axis is perpendicular to the plane of Σ, while the r - and θ -axes are situated in that plane. a. Assuming that u = f (r )er , show that this system is reduced to just one Euler-type ordinary differential equation, whose solution is f (r ) = Ar +

B , r

(A, B ∈ R) .

(8.107)

b. Check that the components of the plane deformation tensor E int are Err = f  (r ) ,

E θθ =

f (r ) , r

Er θ ≡ 0 ,

(8.108)

where the dash stands for differentiation with respect to the radial coordinate. Use (8.108) to further show that Trr = (λ + 2μ) f  (r ) + Tθθ = (λ + 2μ)

λ f (r ) , r

f (r ) + λ f  (r ) , r

Tr θ ≡ 0 .

(8.109a) (8.109b) (8.109c)

8.6. Use the coordinate-free Eq. (8.25) to derive the formulae (8.27). 8.7. Starting from Eq. (8.31) express the boundary conditions for the Airy stress function in polar coordinates. 8.8. If the body force is present in the equilibrium equation and satisfies the additional requirement (8.33), show that the expression of the Airy stress function must be changed to   T int = ∇⊥2 Φ + V I 2 − ∇ ⊥ ⊗ ∇ ⊥ Φ .

(8.110)

In this case Φ satisfies the inhomogeneous bi-harmonic equation ∇⊥4 Φ = −C(ν)∇⊥2 V , where

⎧ ⎨ 1 − 2ν C(ν) := 1 − ν ⎩1 − ν

(plane strain) , (plane stress) .

(8.111)

(8.112)

8.9. In Eqs. (8.4) assume fr = f θ ≡ 0. From (8.4b) deduce that there exists a function F = F(r, θ ) such that Tr θ = −(F/r 2 ), θ and r Tθθ = F, r . Using this result in (8.4a) show that there exists a function H = H (r, θ ) such that r Trr − G = H, θ and Tr θ = −H, r ; here, G = G(r, θ ) is defined by the equa-

410

8 Two-Dimensional Approximations

tion r G , r = F, r . Hence, deduce the representation (8.27) for a suitably defined scalar field Φ = Φ(r, θ ). 8.10. Show that Hooke’s law for the plane-strain/stress approximations can be expressed in the following unified way, 



T int = 2μ E int + where

⎧ ⎨1 − ν Δ # := 1 ⎩ 1+ν

 1 − Δ# |E int |I 2 , 2Δ # − 1 (plane strain) , (plane stress) .

(8.113)

(8.114)

Find the inverted form of (8.113) in terms of μ and Δ # , and then compare your results with those in Sects. 8.2 and 8.3. 8.11. Show that Hooke’s law for the two-dimensional approximations discussed in this chapter can be cast as μ [(κ + 1)E 11 + (3 − κ)E 22 ] , κ −1 μ T22 = [(κ + 1)E 22 + (3 − κ)E 11 ] , κ −1 T12 = 2μE 12 ,

T11 =

where

⎧ ⎨3 − ν κ := 1 + ν ⎩3 − 4ν

(8.115) (8.116) (8.117)

(plane stress) , (plane strain) .

Convince yourself that a solution to a plane-strain problem with a Poisson’s ratio ν is equivalent to that of a plane- stress problem with a Poisson’s ratio ν/(1 − ν). 8.12. Verify the following Cauchy–Riemann relations for the case of (static) plane strain with constant body force, ∂ ∂ X2 ∂ ∂ X1

 2  ∂  2  ∇⊥ u1 = ∇ u2 , ∂ X1 ⊥  2  ∂  2  ∇⊥ u1 = − ∇ u2 , ∂ X2 ⊥

where u⊥ ≡ u1 (X 1 , X 2 )e1 + u2 (X 1 , X 2 )e2 is the two-dimensional displacement field (assumed to be sufficiently smooth). Hence, deduce that ∇⊥4 u1 = ∇⊥4 u2 = 0. The first two relations above show that ∇⊥2 u1 and ∇⊥2 u2 are, respectively, the real and the imaginary parts of an analytic function. 8.13. a. Consider the two-dimensional scalar fields ϕ j : Σ ⊂ E2 → R, with the property that ∇⊥2 ϕ j = 0 for j = 1, 2 (i.e. ϕ j are harmonic fields). Show

8.8 Exercises

411

that

    ∇⊥2 ϕ1 ϕ2 = 2 ∇ ⊥ ϕ1 · (∇ ⊥ ϕ2 ) .

(8.118)

How does this formula change if ϕ j are not harmonic? b. Show that the following functions are harmonic X 2 log r + X 1 θ , 2X 1 X 2 log r +

X 1 log r − X 2 θ ,

(X 12

− X 22 )θ ,

(X 12 − X 22 ) log r − 2X 1 X 2 θ ,

where (r, θ ) are the polar coordinates associated with (X 1 , X 2 ). 8.14. If A is a tensor field, u, v are two vector fields, and φ represents a scalar field, then check the following identities, ∇ 2 (φu) = (∇ 2 φ)u + 2(∇φ) · (∇ ⊗ u) + φ(∇ 2 u) , ∇ (u · v) = (∇ u) · v + 2(∇ ⊗ u) : (∇ ⊗ v) + (∇ v) · u ,   ∇ 2 ( A · u) = (∇ 2 A) · u + 2(∇ ⊗ u) : ∇ ⊗ ( AT ) + A · (∇ 2 u) . 2

2

2

(8.119a) (8.119b) (8.119c)

Will these results remain true if the fields in question are two-dimensional (i.e. A = Aαβ (X ⊥ )eα ⊗ eβ , u = uα (X ⊥ )eα , v = vα (X ⊥ )eα , and φ = φ(X ⊥ ))? 8.15. Show that the polar and Cartesian representations of the normal and shear components of the infinitesimal stress tensor associated with the two-dimensional approximations discussed in this chapter are related by 1 1 (T11 + T22 ) + (T11 − T22 ) cos 2θ + T12 sin 2θ , 2 2 1 1 Tθθ = (T11 + T22 ) − (T11 − T22 ) cos 2θ − T12 sin 2θ , 2 2 1 Tr θ = (T22 − T11 ) sin 2θ + T12 cos 2θ . 2 Trr =

(8.120a) (8.120b) (8.120c)

8.16. In the Eqs. (8.120) set Tθθ = Tr θ = 0, and then express Tαβ (α, β ∈ {1, 2}) in terms of Trr ; in particular, you should find T11 = Trr cos2 θ ,

T22 = Trr sin2 θ ,

T12 = Trr sin θ cos θ .

Use these results to check the Flamant solution described by (8.95); note that the definitions of ‘θ ’ in the above formulae and in (8.94) are not the same. 8.17. The midplane of a linearly elastic thin plate has the shape of a circular annulus,    Σ ≡ (r, θ ) ∈ E2  R1 ≤ r ≤ R2 , 0 ≤ θ < 2π . The plate is uniformly stretched by applying in-plane radial displacements of magnitude Uα along the boundary r = Rα , α = 1, 2 (see Fig. 8.12).

412

8 Two-Dimensional Approximations

a. By assuming a radially symmetric state of stress and deformation, show that the stress distribution in the plate is given by    E U1 R 2 + U2 R 1 R 1 R 2 U1 R 1 + U2 R 2 Trr = + (1 − ν) (1 + ν) , 1 − ν2 r2 R22 − R12 R22 − R12    E U1 R 2 + U2 R 1 R 1 R 2 U1 R 1 + U2 R 2 − (1 − ν) (1 + ν) . Tθθ = 1 − ν2 r2 R22 − R12 R22 − R12



b. Let R1 η := , R2

U1 , λ := U2

η∗ :=

1−ν . 1+ν

Prove that if η < η∗ , then the above azimuthal stresses Tθθ become compressive in the annular region    Σ0 ≡ (r, θ ) ∈ E2  R1 ≤ r ≤ ξ R2 , 0 ≤ θ < 2π , 

where ξ :=

1−ν 1+ν



η2 + λη . 1 + λη

(8.121)

Find the minimum value of λ that corresponds to the onset of compressive azimuthal stresses in the annulus. Are there any compressive stresses when η > η∗ ? Justify your answer. 8.18. A linearly elastic annular thin plate of inner radius R1 , outer radius R2 and thickness h (h/R2  1) is initially stretched in the radial direction by imposing the uniform displacement field U0 > 0 along the outer edge (r = R2 ), while the inner boundary (r = R1 ) is uniformly rotated through a small angle by

Fig. 8.12 An annular plate subjected to uniform radial stretching by prescribing the constant displacements (U1 and U2 ) along the two circular boundaries. The qualitative behaviour of the stresses in the plate is shown superimposed on the plate geometry

U2 Trr

R2 Tθθ

+

U1

+

R1

−

h

8.8 Exercises

413

(a)

(b)

U0 Trr

R2

U0 T1

Tθθ +

+

M

M

R1

T2

+

+

−

h Fig. 8.13 Radially stretched annular plate subjected to azimuthal shear along the inner rim. The qualitative behaviour of the stresses in the plate is shown superimposed on the plate geometry: a Trr , Tθ θ ; b principal stresses, T1 , T2

applying a torque per unit volume of magnitude equal to M (see Fig. 8.13). This in-plane rotation is achieved by means of a concentric rigid shaft. a. Use a cylindrical system of coordinates (r, θ, z) defined in an obvious manner, and assume radial symmetry, to show that the stress distribution in the plate is given by a stress tensor T int with components  R12 + 2 , r    1+ν U0 R 2 R12 E Tθθ (r ) = , − 1 + ν R22 − R12 1−ν r2  1 M Tr θ (r ) = . 2π h r 2 E Trr (r ) = 1+ν



U0 R 2 R22 − R12



1+ν 1−ν



b. Investigate further whether there are any directions in the plate along which compressive stresses may develop (Hint: you must consider the principal stresses corresponding to the above stress distribution; see Fig. 8.13b). 8.19. Solve the plane-strain equations by assuming that the components of the displacement field u⊥ are of the form u1 = X 1 f (r ) ,

u2 = X 2 f (r ) ,

 1/2 r = X 12 + X 22 .

Show that f  + 3 f  /r = 0 and hence find f (r ) (the ‘dash’ indicates differentiation with respect to radial variable ‘r ’). What boundary-value problems can be solved using this solution? 8.20. Find the plane-strain solutions for which ur ≡ 0 and uθ = g(r ) depends only on the radial coordinate ‘r ’. A hollow circular cylinder is deformed in plane

414

8 Two-Dimensional Approximations

Fig. 8.14 Illustrations for Exercise 8.21 (left) and Exercise 8.22 (right)

strain by means of shear tractions uniformly distributed over its inner and outer surfaces. Find the resulting displacement field. 8.21. A rectangular cantilever bar is subjected to a uniform pressure P0 on its upper face, as shown in the left panel of Fig. 8.14. The right end is attached to a wall, while the left end and the lower face are traction-free. It is further assumed that the thickness of the beam is b = 1 and the system is in a state of plane stress. a. Show that the boundary conditions associated with this situation are T22 (X 1 , h/2) = P0 ,

+h/2 −h/2

T22 (X 1 , −h/2) = 0 , T12 (X 1 , ±h/2) = 0 ,

+h/2 T11 (0, X 2 ) dX 2 = X 2 T11 (0, X 2 ) dX 2 = 0 . −h/2

b. Determine the elastic state in the bar by assuming that T12 ≡ −Φ, 12 = X 1 f (X 2 ), where f is a suitably chosen function. 8.22. Consider again the rectangular cantilever beam from the previous question, but this time subject to a non-uniform lateral P = P0 (X 1 /L); see the right panel of Fig. 8.14. a. Show that the boundary conditions can be expressed in the form T22 (X 1 , h/2) = P0 (X 1 /L) , T22 (X 1 , −h/2) = 0 , T12 (X 1 , ±h/2) = 0 ,

+h/2 T12 (0, X 2 ) dX 2 = 0 . T11 (0, X 2 ) = 0, −h/2

b. Determine the elastic state in the bar by assuming that T11 ≡ Φ, 22 = X 13 f (X 2 ) + X 1 g(X 2 ), where f = f (X 2 ) and g = g(X 2 ) are functions that you must find. 8.23. This problem revisits Example 8.6—the only change is that now the circular cantilever bar is subjected to end loads statically equivalent to a force T0 and an

8.8 Exercises

415

Fig. 8.15 Circular cantilever bar subjected to the simultaneous action of tension and bending at the free end

2

1

1

0

0

Fig. 8.16 Circular cantilever bars subjected to various terminal loads

end moment M0 ; see Fig. 8.15 for details of the geometry. Show that a suitable Airy stress function in this case is  Φ=

B Ar + + Cr + Dr log r cos θ , r 3

(8.122)

where A, B, C, D ∈ R are constants that you must determine by imposing the relevant boundary conditions for the situation at hand. 8.24. Use the principle of superposition in conjunction with the results of Example 8.6 and the above problem, to obtain the solutions for the two separate situations shown in Fig. 8.16. 8.25. A circular cantilever bar is loaded in pure bending by terminal loads statically equivalent to a moment M0 , as seen in Fig. 8.17. At the fixed end of the bar the point O1 is constrained such that ur = uθ = ur,θ = 0. Assuming a state of plane stress, find the displacement of the point O2 whose polar coordinates are (r0 , θ0 ), with r0 = (R1 + R2 )/2 and θ0 = π/4.

416

8 Two-Dimensional Approximations 2 1

2

45° 0

1

Fig. 8.17 Curved cantilever bar under a terminal bending moment M0 . In polar coordinates, the geometry of the bar is described as the set of points (r, θ), with R1 ≤ r ≤ R2 and 0 ≤ θ ≤ π/4

Fig. 8.18 Infinite elastic plate with a small rigid core subjected to uniaxial tension at infinity (as X 1 → ±∞); see Exercise 8.26

8.26. An infinite isotropic elastic plate contains a perfectly bonded, rigid core of radius r = a, and is under uniform tension T0 in the X 1 -direction as shown in Fig. 8.18. Using an Airy stress function of the form Φ = Φ1 , where  T0  2 2 2r sin θ + A1 a 2 log r + A2 a 4 r −2 cos 2θ + A3 a 2 cos 2θ , 4 (8.123) for some suitably chosen A j ( j = 1, 2, 3), find the stress distribution in the plate. 8.27. Consider again the situation described in the previous problem, but now assume that the rigid core is replaced by a soft core made of an isotropic elastic material with different properties from the plate. By considering an Airy stress function of the form  if a < r < ∞ , Φ1 (r, θ ) , (8.124) Φ= if 0 < r < a , Φ2 (r, θ ) , Φ1 :=

with Φ1 of the same form as in (8.123) and

8.8 Exercises

Φ2 :=

417

T0 4

 A4 r 2 + A5r 2 cos 2θ + A6

r4 cos 2θ a2

, (A j ∈ R, j = 4, 5, 6) ,

find the stress distribution in the plate. 8.28. Consider again Example 8.7 and assume that points on the X 2 -axis are constrained so that there is no lateral displacement (i.e. uθ = 0 when θ = 0). Show that the components of the displacement field can be expressed in the form ur = − uθ =

(1 − ν)P 2P cos θ log r − θ sin θ + C cos θ , πE πE

(1 − ν)P (1 + ν)P 2P log r sin θ − θ cos θ + sin θ − C sin θ . πE πE πE

The constant C ∈ R is still arbitrary; to eliminate it one can set ur (a, 0) = 0, i.e. the radial displacement becomes zero on the vertical axis, sufficiently far away from the point O (a  1). Note that the expressions of the displacement field are singular for both r → 0+ and r → +∞, which is an unsatisfactory feature of the solution (however, the stresses do attenuate as we move away from the point where the load is applied). 8.29. A circular hole of radius R in an infinite plate is filled with a disk of slightly greater radius R(1 + ε), where 0 < ε  1 (a given parameter). As a result of this process the common radius of the hole and disk becomes R(1 + δ), for some δ > 0 that must be found. Both the disk and the plate are made of elastic materials with the constitutive constants (E 1 , ν1 ) and (E 2 , ν2 ), respectively. Assuming that this is a plane-stress situation with axial symmetry, find the elastic state for the disk-plate system by using a stress function of the form   r Φ(r ) = A1 + A2 r 2 log + A3r 2 , R

(A j ∈ R , j = 1, 2, 3) .

In particular, show that A2 = 0 and  −1 , δ = εE 1 (1 + ν2 ) E 1 (1 + ν2 ) + E 2 (1 − ν1 )  −1 2 A3 = −εE 1 E 2 (1 + ε)E 1 (1 + ν2 ) + E 2 (1 − ν1 ) , A1 = 2R 2 (1 + δ)2 A3 . 8.30. Consider a two-dimensional stress distribution with T3i ≡ 0 (i = 1, 2, 3), and Tαβ = Tαβ (X) for α, β ∈ {1, 2}. Assume that these non-zero stress components are given by the same definitions as in the Airy formulae (8.26), but now Φ = Φ(X). If the external forces are distributed symmetrically with respect to the midplane X 3 = 0, show that the equilibrium and compatibility relations are satisfied by the stress potential

418

8 Two-Dimensional Approximations

Φ = Φ0 −

 2  2 ν ∇ Φ0 X 3 , 2(1 + ν) ⊥

where Φ0 = Φ0 (X ⊥ ) and ∇ 4⊥ Φ0 = 0.

Bibliography 1. Amenzade YuA (1979) Theory of elasticity. Mir Publishers, Moscow 2. Barber JR (2002) Elasticity. Kluwer Academic Publishers, Dordrecht, The Netherlands 3. Boyce WE, DiPrima RC (1996) Elementary differential equations and boundary value problems. Wiley Sons, New York 4. Chou PC, Pagano NJ (1967) Elasticity: tensor, dyadic, and engineering approaches. D. Van Nostrand Company Inc, Princeton, New Jersey 5. Coddington EA (1989) An introduction to ordinary differential equations. Dover Publications, New York 6. Fraeijs de Veubeke BM (1979) A course in elasticity. Springer, New York 7. Filonenko-Borodich M (1963) Theory of elasticity. Mir Publishers, Moscow 8. Karasudhi P (1991) Foundations of solid mechanics. Kluwer Academic Publishers, Dordrecht 9. Leipholz H (1974) Theory of elasticity. Noordhoff International Publishing, Leyden 10. Little RW (1973) Elasticity. Prentice-Hall Inc, Englewood Cliffs, New Jersey 11. Rekach VG (1979) Manual of the theory of elasticity. Mir Publishers, Moscow 12. Saada AS (1993) Elasticity: theory and applications. Krieger Publishing Company, Malabar, Florida 13. Segel LA (1987) Mathematics applied to continuum mechanics. Dover Publications Inc, New York 14. Slaughter WS (2002) The linearized theory of elasticity. Birkhäuser, Boston 15. Sobrero L (1942) Elasticidade. Livraria Boffoni, Rio de Janeiro (in Portuguese) 16. Sokolnikoff IV (1956) Mathematical theory of elasticity. McGraw-Hill Book Company Inc, New York

Chapter 9

Special Two-Dimensional Problems: Unbounded Domains

Abstract In the previous chapter we saw how the solutions of two-dimensional approximations in isotropic Linear Elasticity can be obtained from the Airy stress potential Φ that satisfies a bi-harmonic equation (either homogeneous or inhomogeneous). When the region occupied by the elastic material is unbounded (see Fig. 9.1), integral transforms provide a robust analytical framework for solving the various boundary-value problems satisfied by Φ. We are going to illustrate this technique by focusing on the particular case of Fourier transforms. By confining our attention to two-dimensional approximations the corresponding calculations are relatively straightforward. It should be kept in mind that these integral transforms are applicable to three-dimensional problems as well (although the calculation do tend to become unwieldy quickly).

9.1 The Bi-harmonic Equation via Fourier Transforms We start by sketching the general approach before examining more closely several representative examples. The main aim below will be to solve the bi-harmonic equation ∇ 4 Φ = 0 in an unbounded domain Σ ⊂ E2 ; for definiteness, we take Σ to be a half-plane (X 1 > 0); see Fig. 9.1c. Assuming that points (X 1 , X 2 ) ∈ Σ satisfy −∞ < X 2 < +∞, the onedimensional Fourier transform with respect to this variable is introduced in the usual way (see Appendix D),  +∞ 1 Φ(X 1 ; α) ≡ F2 [Φ(X 1 , X 2 ); α] := √ Φ(X 1 , X 2 ) eiα X 2 dX 2 ; (9.1) 2π −∞ note that this is still a function of two variables, X 1 and α. However, by Fourier transforming the bi-harmonic equation we end up with, 2 ∂4 Φ 2 ∂ Φ − 2α + α4 Φ = 0 , ∂ X 14 ∂ X 12

© Springer Nature B.V. 2020 C. D. Coman, Continuum Mechanics and Linear Elasticity, Solid Mechanics and Its Applications 238, https://doi.org/10.1007/978-94-024-1771-5_9

(9.2) 419

420

9 Special Two-Dimensional Problems: Unbounded Domains

(a)

(b)

(c)

(d)

Fig. 9.1 Typical examples of unbounded two-dimensional domains for which Fourier transforms are applicable: elastic plane (a); quarter plane (b); half-plane (c); strip (d)

which can now be regarded as an ordinary differential equation with constant coefficients (by momentarily treating α as a constant). The characteristic equation (ζ 2 − α 2 )2 = 0 has two pairs of repeated roots, ζ1 = ζ2 = |α| and ζ3 = ζ4 = −|α|. Consequently, the solution of (9.2) will be of the form Φ(X 1 ; α) = (A + B X 1 )e−|α|X 1 + (C + D X 1 )e|α|X 1 , where A, B, C, D are arbitrary functions of α. To eliminate unbounded solutions, we need to set C = D = 0, which leaves us with the much simpler expression   Φ(X 1 ; α) = A + B X 1 e−|α|X 1 .

(9.3)

The functions A = A(α) and B = B(α) must be found by using the boundary conditions; in particular, this will require using the expression (9.3) to calculate the stresses. By taking the Fourier transforms of the formulae (8.26) from the previous chapter, in conjunction with the properties listed in Appendix D, yields T 11 (X 1 ; α) = − α 2 Φ(X 1 ; α) ,

(9.4a)

2

d Φ(X 1 ; α) , d X 12 d T 12 (X 1 ; α) = iα Φ(X 1 ; α) , d X1 T 22 (X 1 ; α) =

(9.4b) (9.4c)

and note that the inversion of (9.4) leads to the physical stresses, 1 Ti j (X 1 , X 2 ) = √ 2π



+∞ −∞

T i j (X 1 ; α) e−iα X 2 dα ,

(i, j ∈ {1, 2}) .

(9.5)

9.1 The Bi-harmonic Equation via Fourier Transforms

421

Fig. 9.2 An illustration of the basic geometry for the two-dimensional approximation discussed in Example 9.1. The half-plane pictured on the right is subjected to arbitrary normal tractions on its boundary X1 = 0

elastic half-plane

Broadly speaking, the Eqs. (9.4) can now be replaced into (9.5), but we are not going to pursue this here because no further simplifications are possible unless we are a bit more specific about the loading and the corresponding boundary conditions involved; this is what we do next. Example 9.1 As a first example, let us consider the problem of finding the stress distribution in an elastic half-plane loaded by normal forces acting on its boundary; no body forces are considered and the geometry is defined in Fig. 9.2. The relevant boundary conditions are, T12 (0, X 2 ) = 0 , −∞ < X 2 < +∞ , (9.6a) T11 (0, X 2 ) = − p(X 2 ) , Ti j (X) → 0 , Ti j, k (X) → 0 , as |X| → +∞ , (i, j, k = 1, 2) , (9.6b) where1 X ≡ X 1 e1 + X 2 e2 and p(X 2 ) is a function that admits a Fourier transform. We start by taking the Fourier transform of (9.6a) to find T 11 (0; α) = − p(α) ¯ ,

T 12 (0; α) = 0 ,

−∞ < α < +∞ .

These new equations are then used in conjunction with (9.3) and (9.4) to identify the ‘constants’ A and B, which turn out to be A=

1 p(α) ¯ α2

and

B=

1 |α| p(α) ¯ , α2

so the transformed Airy stress function becomes Φ(X 1 ; α) =

  p(α) ¯ 1 + |α|X 1 e−|α|X 1 . 2 α

this chapter we do not use X ⊥ for the position vector of points in E2 as most of the work will be done in components.

1 In

422

9 Special Two-Dimensional Problems: Unbounded Domains

With this information in hand, the inverted form of Eq. (9.5) is given by  +∞   1 p(α) ¯ 1 + |α|X 1 e−|α|X 1 −iα X 2 dα , T11 (X 1 , X 2 ) = − √ 2π −∞  +∞   1 p(α) ¯ 1 − |α|X 1 e−|α|X 1 −iα X 2 dα , T22 (X 1 , X 2 ) = − √ 2π −∞  +∞ iX 1 −|α|X 1 −iα X 2 p(α)αe ¯ dα . T12 (X 1 , X 2 ) = − √ 2π −∞

(9.7a) (9.7b) (9.7c)

Unless we make further particular choices for the distribution of the tractions on the boundary, these integrals cannot be simplified further. However, depending on the parity of p = p(X 2 ), we can routinely reduce the integrals to (0, ∞) and also rearrange the integrands in a more sensible way. The key idea is to split up the integrals in (9.7) into two ‘sub-integrals’ over (−∞, 0) and (0, +∞). Then, by exploiting the symmetry properties of the integrands and making the simple substitution t = −α in the former integral, the two individual integrals can be combined; this eliminates the complex exponentials and leads to either a sine or a cosine function instead. If p(X 2 ) is symmetric (with respect to X 2 = 0), i.e. p(−X 2 ) = p(X 2 ) for all X 2 , then (9.7) assume the form  T11 = − 

2 π



∞

  p(α) ¯ 1 + α X 1 e−α X 1 cos(α X 2 ) dα ,

(9.8a)

0

   2 ∞ p(α) ¯ 1 − α X 1 e−α X 1 cos(α X 2 ) dα , π 0   ∞ 2 −α X 1 α p(α)e ¯ sin(α X 2 ) dα , T12 = −X 1 π 0

T22 = −

where now,

 p(α) ¯ ≡

2 π



∞

p(X 2 ) cos(α X 2 ) dX 2 .

(9.8b) (9.8c)

(9.9)

0

On the other hand, for an antisymmetric p(X 2 ), i.e. p(−X 2 ) = − p(X 2 ) for all X 2 , the above results must be modified according to 

   2 ∞ T11 = − p(α) ¯ 1 + α X 1 e−α X 1 sin(α X 2 ) dα , π 0   ∞   2 p(α) ¯ 1 − α X 1 e−α X 1 sin(α X 2 ) dα , T22 = − π  0 ∞ 2 −α X 1 α p(α)e ¯ cos(α X 2 ) dα , T12 = X 1 π 0

(9.10a) (9.10b) (9.10c)

9.1 The Bi-harmonic Equation via Fourier Transforms

where this time,

 p(α) ¯ ≡

2 π



∞

423

p(X 2 ) sin(α X 2 ) dX 2 .

(9.11)

0

Several particular cases of these formulae are considered below. Example 9.2 The loading is uniform and acts over a finite part of the free surface, symmetrically with respect to the origin of the system of coordinates, so p(X 2 ) = p0 for |X 2 | ≤ a and is zero otherwise (see Fig. 9.3). A straightforward calculation shows that  p(α) ¯ = p0

2 sin(αa) , π α

(9.12)

which is an even function. Next, we apply directly the formulae (9.8) to find    2 p0 ∞ 1 + α X 1 −α X 1 e sin(αa) cos(α X 2 ) dα , π 0 α    2 p0 ∞ 1 − α X 1 −α X 1 e sin(αa) cos(α X 2 ) dα , T22 = − π 0 α  2 p0 X 1 ∞ −α X 1 T12 = − e sin(αa) sin(α X 2 ) dα . π 0 T11 = −

(9.13a) (9.13b) (9.13c)

All the integrals in (9.13) can be evaluated in closed form by using the formulae listed in Appendix D. This is facilitated by first transforming the products of trigonometric functions into sums, with the help of the following elementary identities, 2 sin(αa) cos(α X 2 ) = sin α(a + X 2 ) + sin α(a − X 2 ) , 2 sin(αa) sin(α X 2 ) = cos α(a − X 2 ) − cos α(a + X 2 ) .

(9.14a) (9.14b)

The full details of these routine calculations are left to the reader, and here we only outline how (9.13) can be presented in the alternative way, p0 [2(θ1 − θ2 ) + sin(2θ1 ) − sin(2θ2 )] , 2π p0 T22 = [2(θ1 − θ2 ) − sin(2θ1 ) + sin(2θ2 )] , 2π p0 T12 = [cos(2θ2 ) − cos(2θ1 )] , 2π

T11 =

where θ1 := tan−1



X2 − a X1

 ,

θ2 := tan−1



X2 + a X1

(9.15a) (9.15b) (9.15c)  .

424

9 Special Two-Dimensional Problems: Unbounded Domains

Fig. 9.3 Elastic half-plane loaded by uniform normal tractions acting over a one-dimensional finite region (−a ≤ X 2 ≤ +a, X 1 = 0) symmetrically placed with respect to the origin O of the system of coordinates

To justify (9.15a) we note that, with the help of (9.14a), the integral in the expression of T11 in (9.13) can be split up into the sum of two integrals,  Jk :=

∞

 −1   α + X 1 e−α X 1 sin α(a + (−1)k X 2 ) dα , (k = 1, 2) .

0

Using the results from Appendix D, we have Jk = tan

−1



  X 1 a + (−1)k X 2 a + (−1)k X 2 + 2 , (k = 1, 2) . X1 X 1 + [a + (−1)k X 2 ]2

(9.16)

With reference to Fig. 9.3, the following relations are evident, X 2 + (−1)k a X1 , cos θk = , rk rk  2 rk2 = X 12 + a + (−1)k X 2 , (k = 1, 2) . sin θk =

Plugging these back into (9.16) gives

Jk = (−1)

k

 1 θk + sin 2θk , 2

(k = 1, 2) ,

whence the expression for T11 in (9.15a) follows at once. In exactly the same way one can rewrite the other two stress components.

9.1 The Bi-harmonic Equation via Fourier Transforms

425

An important particular case of the above example is the Flamant solution (8.96) obtained before (see Example 8.7 in the previous chapter). By taking  p0 =

π 2



P 2a

 ,

we immediately find from (9.12),  p(α) ¯ =

P 2



sin(αa) P → , αa 2

as a → 0 .

Replacing p¯ in (9.8) by this value, and following the same procedure as explained above, the corresponding integrals can be reduced to a form that is readily evaluated with the results recorded in Appendix D. The final formulae turn out to be    2P X 13 2P X 1 X 22 , T = − , 22 π r4 π r4     2 2P X 12 X 2 , r = X 12 + X 22 , T12 = − π r4 

T11 = −

and we note in passing that, compared to Fig. 8.9, the current labelling of the axes is the other way around (i.e. X 1 → X 2 and X 2 → X 1 ); in polar coordinates we also recover the equations recorded in (8.95). The above expressions for the Flamant solution can also be recovered by choosing p(X 2 ) = Pδ(X 2 ) right from the outset and relying on the well-known properties of the Dirac delta function δ. Example 9.3 Another interesting illustration of (9.7) is the situation of an elastic half-plane whose edge is acted upon by a constant load that extends indefinitely only on a part of the boundary, as seen in Fig. 9.4. Such a load is conveniently represented with the help of the Heaviside unit function, H (say), in the form p(X 2 ) = p0 H (X 2 ), with

Fig. 9.4 The general two-dimensional setting for Example 9.3

426

9 Special Two-Dimensional Problems: Unbounded Domains

H (X 2 ) =

1 if X 2 > 0 , 0 if X 2 < 0 .

(9.17)

To apply the general result (9.7) we first need to find the Fourier transform of the load. By direct calculation, 

+∞ −∞

δ+ (α) e

−iα X 2

 1 ∞ sin(α X 2 ) 1 dα dα = + 2 π 0 α 1 if X 2 > 0 , = = H (X 2 ) , 0 if X 2 < 0

where δ+ (α) :=

(9.18)

1 1 δ(α) − 2 2π iα

represents the Heisenberg delta function. Fourier inversion of (9.18) yields 

+∞ −∞

H (X 2 ) eiα X 2 dX 2 = 2π δ+ (α) ,

√ and hence p(α) ¯ = p0 2π δ+ (α). Plugging this back into (9.7), by using the usual sifting property of the Dirac delta function in conjunction with the strategy that led to (9.16), we eventually find

   1 X2 X1 X2 1 −1 + + tan , T11 (X 1 , X 2 ) = − p0 2 π X1 π(X 12 + X 22 )

   1 X2 X1 X2 1 T22 (X 1 , X 2 ) = − p0 − + tan−1 , 2 π X1 π(X 12 + X 22 )

p  X 12 0 T12 (X 1 , X 2 ) = . 2 π X 1 + X 22

(9.19a) (9.19b) (9.19c)

In polar coordinates these expressions reduce to2 T11 = −

  p0  p0  π + 2θ + sin 2θ , T12 = 1 + cos 2θ , 2π 2π  p0  T22 = − π + 2θ − sin 2θ . 2π

2 These are still Cartesian-coordinate stresses, but expressed in terms of the polar coordinates defined

in Fig. 9.4, X 1 = r cos θ and X 2 = r sin θ.

9.1 The Bi-harmonic Equation via Fourier Transforms

427

   loaded by two Fig. 9.5 Elastic disk Σ ≡ (X 1 , X 2 ) ∈ E2  X 12 + X 22 < a 2 equal diametrically opposite concentrated forces situated on its circumference,  ∂Σ ≡ (X 1 , X 2 ) ∈ E2  X 12 + X 22 = a 2 . Two local polar systems of coordinates (r j , θ j ), with j = 1, 2, are used in the solution of Example 9.4

Example 9.4 An application of the Flamant solution is provided by the problem of a circular disk of thickness b = 1 subjected to equal opposite forces P ± ≡ ±P e2 (P > 0) acting on the circumference of the disk at two diametrically opposite points, as seen in Fig. 9.5. We choose a global Cartesian system of coordinates attached to the centre of the disk and with the X 2 -axis directed along the line of action of the loading. In addition, two local polar system of coordinates are introduced, as well: (r j , θ j ) is attached to O j ( j = 1, 2). Let us start by considering the elastic state obtained by superimposing the solutions for a concentrated load acting orthogonally to the boundary of a half-plane, 1. first solution: P − = −P e2 applied at (0, +a); 2. second solution: P + = +P e2 applied at (0, −a). Since we already have the half-plane solution expressed in a local system of coordinates, the result of the aforementioned superposition will be a new state of stress described by the stress tensor T  , with 2P T =− π 



cos θ1 r1



er(1)

⊗

er(1)

 +

cos θ2 r2



er(2)

⊗

er(2)

 .

(9.20)

However, (9.20) cannot be the full stress solution of our problem because the circumference ∂Σ will not be traction free; this is confirmed next.

428

9 Special Two-Dimensional Problems: Unbounded Domains

If M ∈ ∂Σ then cos θ1 = r1 /d, cos θ2 = r2 /d, and er(2) = e(1) θ , so that (9.20) becomes   2P 2P  (1)  (1) er ⊗ er(1) + e(1) I2 , =− ⊗ e (9.21) T  = − θ θ ∂Σ πd πd with d := 2a being the diameter of the disk. This corresponds to uniform radial compression of magnitude (2P/π d) applied on ∂Σ. Thus, to arrange for the circumference of the disk to be traction-free we must superimpose an additional radial tension 2P I2 . (9.22) T  = πd The complete stress solution of the original problem will then be given by T := T  + T  .

(9.23)

It is helpful to rewrite this result in the global (Cartesian) system of coordinates indicated in Fig. 9.5. To this end, it is noted that the polar radii r j > 0 ( j = 1, 2) can be calculated from r12 = X 12 + (a − X 2 )2 ,

r22 = X 12 + (a + X 2 )2 ,

while routine algebraic manipulations show that (9.23) can be cast in the form

(a − X 2 )X 12 (a + X 2 )X 12 1 + − 4 d r1 r24

 2P (a − X 2 )3 (a + X 2 )3 1 T22 (X 1 , X 2 ) = − , + − π d r14 r24 

2P (a − X 2 )2 X 1 (a + X 2 )2 X 1 T12 (X 1 , X 2 ) = . − π r14 r24

T11 (X 1 , X 2 ) = −

2P π

 ,

(9.24a) (9.24b) (9.24c)

The stresses at the centre of the disk are T11 (0, 0) =

2P , πd

T22 (0, 0) = −

6P , πd

T12 (0, 0) = 0 .

It is interesting to note that T11 (0, 0) above corresponds to tension; in fact, on the X 2 -axis (−a < X 2 < a), we have T12 (0, X 2 ) ≡ 0 and T11 (0, X 2 ) =

2P , πd

T22 (0, X 2 ) = −

4P π



1 1 1 + − d − 2X 2 d + 2X 2 2d

 .

In light of these expressions it is clear that the points situated on the line O1 O2 (except O1 and O2 themselves) will experience a uniform tensile stress of magnitude (2P/π d) in a direction parallel to the X 1 -axis.

9.2 Remarks on the Displacement Field

429

9.2 Remarks on the Displacement Field Some general observations about the displacement field are in order. For a traction boundary-value problem this quantity is found from the inverted form of Hooke’s law by using the already determined Airy stress function. Recall that, from Exercise 8.11, we have  E 12 = T12 /(2μ) , E 11 = (κ + 1)T11 − (3 − κ)T22 /(8μ) ,  E 22 = (κ + 1)T22 − (3 − κ)T11 /(8μ) .

(9.25a) (9.25b)

Since u2, 2 = E 22 , by taking the one-dimensional Fourier transforms with respect to X 2 of this equation and (9.25b), after inversion we discover i u2 = √ 8μ 2π



+∞

−∞

 (κ + 1)α −1 Φ ,11 (X 1 ; α) + (3 − κ)αΦ(X 1 ; α) e−iα X 2 dα .

To find u1 , we use the second equation in (9.25a) and u1, 2 + u2, 1 = 2E 12 . We can take again the one-dimensional Fourier transform (with respect to X 2 ) of this equation and replace the Fourier transform of u2 found previously. By inverting the relation thus obtained, it eventually transpires that 1 u1 = √ 8μ 2π



+∞ −∞

 (κ + 1)α −2 Φ ,111 (X 1 ; α) − (κ + 5)Φ ,1 (X 1 ; α) e−iα X 2 dα .

Depending on the symmetry of the integrands, the two integrals for ui (i = 1, 2) can be further reduced to Fourier sine or cosine transforms that sometimes can be evaluated in closed form. However, more often than not these integrals cannot be calculated directly because they are divergent. We shall return to this issue at the end of the next section. The above simple-minded strategy is quite natural in problems in which the displacements are not involved in setting up the boundary constraints. The case when they do appear in the boundary conditions can be dealt with more expediently by using the Papkovitch–Neuber representation discussed in Sect. 9.6.

9.3 The Direct Approach In this section we explore the solution of the two-dimensional elasticity approximations by means of Fourier transforms, but without using the Airy stress function. This approach has the advantage of being slightly more direct than the method presented earlier in this chapter, especially in the case of traction boundary-value problems. We outline below the general solution strategy for a half-plane subjected to prescribed tractions on its boundary. In a Cartesian system of coordinates O X 1 X 2 , the domain

430

9 Special Two-Dimensional Problems: Unbounded Domains

of interest will correspond to the upper half-plane (X 2 > 0); its finite boundary is obviously the X 1 -axis (X 2 = 0). The starting point of the solution is the stress formulation of two-dimensional Linear Elasticity. Let us recall that this consists of two equilibrium equations plus a compatibility relation; more specifically, in the absence of body forces, T11, 1 + T12, 2 = 0 ,

T12, 1 + T22, 2 = 0 ,

T11, 22 − 2 T12, 12 + T22, 11 = 0 .

(9.26a) (9.26b)

Altogether, these are three equations for three unknowns (T11 , T22 , T12 ), so a solution is possible (in principle, at least). The Eqs. (9.26) are also subject to the prescribed boundary conditions  T22  X 2 =0 = f 0 (X 1 )

 T12  X 2 =0 = g0 (X 1 ) ,

and

(9.27)

where f 0 and g0 are given ‘good’ functions (see Appendix D). We are going to be looking for solutions that, together with their derivatives, decay to zero as |X| → +∞ (far-field requirement). Let us start by introducing the partial Fourier transforms of the unknowns and the boundary data,  +∞ 1 T i j (α; X 2 ) := √ Ti j (X 1 , X 2 ) eiα X 1 dX 1 , (i, j = 1, 2) , 2π −∞  +∞  +∞ 1 1 f 0 (X 1 ) eiα X 1 dX 1 , g¯ 0 (α) := √ g0 (X 1 ) eiα X 1 dX 1 . f¯0 (α) := √ 2π −∞ 2π −∞ Applying the Fourier transform to (9.26) and (9.27), the corresponding differential equations are transformed into − iαT 11 + T 12, 2 = 0 ,

−iαT 12 + T 22, 2 = 0 ,

T 11, 22 + 2iαT 12, 2 − α T 22 = 0 , 2

(9.28a) (9.28b)

while the boundary conditions (9.27) become T 22 (α; 0) = f¯0 (α) ,

T 12 (α; 0) = g¯ 0 (α) .

(9.29)

Next, we aim to reduce (9.28) to just one equation. To this end, note that the two relations in (9.28a) give T 11 = (iα)−1 T 12, 2 ,

T 12 = (iα)−1 T 22, 2 ,

(9.30)

9.3 The Direct Approach

431

whence T 11 = − α −2 T 22, 22 . Replacing this result in the transformed compatibility Eq. (9.28b) leads to a fourth-order differential equation for T 22 , 2 ∂ 4 T 22 2 ∂ T 22 − 2α + α 4 T 22 = 0 . ∂ X 24 ∂ X 22

Guided by our previous experience with (9.2), the general solution of the above equation can be written down directly,  T 22 (α; X 2 ) = A(α) + X 2 B(α) e−|α|X 2 ,

(9.31)

for some functions A = A(α) and B = B(α) that will be found from enforcing the transformed boundary conditions (9.29); straightforward manipulations give A = f¯0 and B = f¯0 |α| + iα g¯ 0 . Finally, we can state the general form of the transformed stresses. Use of (9.31) in conjunction with (9.30) yields      |α| ¯ − α X 2 ig¯ 0 e−|α|X 2 , T 11 (α; X 2 ) = 1 − |α|X 2 f 0 + 2 α   T 22 (α; X 2 ) = 1 + |α|X 2 f¯0 + iα X 2 g¯ 0 e−|α|X 2 ,    T 12 (α; X 2 ) = iα X 2 f¯0 + 1 − |α|X 2 g¯ 0 e−|α|X 2 .

(9.32a) (9.32b) (9.32c)

The problem is almost solved—all that remains to be done is to invert (9.32) by using the familiar formulae, 1 Ti j (X 1 , X 2 ) = √ 2π



+∞ −∞

T i j (α; X 2 ) e−iα X 1 dα ,

(i, j ∈ {1, 2}) .

(9.33)

The complicated integrals obtained by plugging (9.32) into (9.33) are simplified with the help of the convolution theorem. The detailed calculations are somewhat lengthy and are partly relegated to Appendix D. Below, we only outline the strategy for finding T11 = T11 (X 1 , X 2 ). According to (9.32a) and (9.33), we can write 1 T11 (X 1 , X 2 ) = √ 2π

+√

i 2π



+∞ −∞



+∞

−∞

  f¯0 (α) 1 − |α|X 2 e−|α|X 2 · e−iα X 1 dα

  |α| − α X 2 e−|α|X 2 · e−iα X 1 dα . g¯ 0 (α) 2 α

(9.34)

432

9 Special Two-Dimensional Problems: Unbounded Domains

In order to apply the convolution theorem the underlined terms in the two integrals above must be cast as the Fourier transforms of some concrete functions. It can be shown that in this case

   2X 12 X 2 π 1 − |α|X 2 e−|α|X 2 , ; α = (9.35a) F1 2 2 2 2 (X 1 + X 2 )   

 2X 13 π |α| − α X 2 e−|α|X 2 . ;α =i (9.35b) F1 2 2 α (X 12 + X 22 )2 After replacing (9.35) in (9.34) we get the desired simplification; an entirely similar process applies to the inversion of T22 and T12 . The final inverted results are  2 +∞ Π (X 1 , X 2 ; η)(X 1 − η)2 dη , T11 (X 1 , X 2 ) = π −∞  2 +∞ Π (X 1 , X 2 ; η)(X 1 − η)X 2 dη , T12 (X 1 , X 2 ) = π −∞  2 +∞ Π (X 1 , X 2 ; η)X 22 dη , T22 (X 1 , X 2 ) = π −∞ where Π (X 1 , X 2 ; η) :=

(9.36a) (9.36b) (9.36c)

X 2 f 0 (η) + (X 1 − η)g0 (η) .  2 (X 1 − η)2 + X 22

(9.37)

These formulae represent the solution for the elastic half-plane {X 2 > 0} subjected to the tractions   = −g0 (X 1 )e1 − f 0 (X 1 )e2 t(−e2 ) X 2 =0

along its edge (the negative sign comes from (9.27)). For the case of a concentrated load F = F1 e1 + F2 e2 applied at the origin, f 0 (X 1 ) = −F2 δ(X 1 ) and g0 (X 1 ) = −F1 δ(X 1 ), where δ(X 1 ) is the usual Dirac delta function. Taking advantage of the properties of this function the integrals in (9.36) are immediately evaluated leading to T11 = −

2X 12 π



X 2 F2 + X 1 F1 (X 12 + X 22 )2

 ,

2X 2 T22 = − 2 π

T12 = −

2X 1 X 2 π

X 2 F2 + X 1 F1 (X 12 + X 22 )2



X 2 F2 + X 1 F1 (X 12 + X 22 )2

 , (9.38a)

 .

(9.38b)

In the remaining of this section we return to the determination of the displacement field. The situation of interest is the same as before, except that we take g0 ≡ 0 in the boundary conditions  (9.27) and restrict the discussion to plane strain. If u j (α; X 2 ) = F1 u j (X 1 , X 2 ); α ( j = 1, 2), taking the corresponding Fourier transform of (9.25) the results obtained earlier in this section show that

9.3 The Direct Approach

433

 2μ¯u1 = iα −1 f¯0 (α) (1 − 2ν) − |α|X 2 e−|α|X 2 ,  2μ¯u2 = −|α|−1 f¯0 (α) 2(1 − ν) + |α|X 2 e−|α|X 2 .

(9.39a) (9.39b)

The first equation is inverted by using the convolution theorem and the formula F1−1

 −1 iα (1 − 2ν − |α|X 2 )e−|α|X 2 ; X 1 =





2 X1 X1 X2 , (1 − 2ν) tan−1 − 2 π X2 X 1 + X 22

which can be checked with the help of the integrals listed in Appendix D. The final result is     +∞

X1 − η 1 (X 1 − η)X 2 −1 u1 = (1 − 2ν) tan − f 0 (η) dη . 2π μ −∞ X2 (X 1 − η)2 + X 22 The inversion of (9.39b) is more problematic because the integral defining the inverse transform of the function multiplying f¯0 is divergent (hence the convolution result cannot be applied as before). To circumvent this hurdle we introduce an auxiliary function  X1 f 0 (ζ ) dζ , f 1 (X 1 ) := −∞

and note that its Fourier transform, f¯1 , satisfies f¯0 (α) = −iα f¯1 (α). This observation is then used to cast (9.39b) in the alternative form  2μu2 = i f¯1 (α)sgn(α) 2(1 − ν) + |α|X 2 e−|α|X 2 . Now, the inversion of the new right-hand side poses no difficulties. The convolution theorem together with    F1−1 i sgn(α) 2(1 − ν) + |α|X 2 e−|α|X 2 ; X 1 = 2



2 π



(1 − ν)X 1 X 12 + X 22

+

X 1 X 22 (X 12 + X 22 )2

 ,

show that u2 can be expressed as u2 =

1 πμ



+∞

−∞



(1 − ν)(X 1 − η) (X 1 − η)X 22 + 2 2 (X 1 − η) + X 2 [(X 1 − η)2 + X 22 ]2

 f 1 (η) dη .

The above representations for the components of the displacement field can be simplified if the loading enjoys additional symmetry properties. Without going into details, we record below the corresponding formulae in the case in which f 0 (X 1 ) is an even function,  ∞ 1 α −1 FC (α)(1 − 2ν − α X 2 )e−α X 2 sin(α X 1 ) dα , (9.40a) u1 = √ μ 2π 0

434

9 Special Two-Dimensional Problems: Unbounded Domains

1 u2 = − √ μ 2π



∞

 α −1 FC (α) 2(1 − ν) + α X 2 e−α X 2 cos(α X 2 ) dα ,

(9.40b)

0

 where FC (α) := FC f 0 (X 1 ); α is the Fourier-cosine transform of the load.

9.4 A Modification of the Method Our earlier discussion of the bi-harmonic equation in Sect. 9.1 has revealed the structure of the Fourier transform of the Airy stress function—cf. (9.3). That result can be used to find the solution of any plane problem in domains that include at least one unbounded direction. We illustrate this modified approach by considering a simple example: a concentrated force acting in an elastic plane; this problem is sometimes referred to as the 2D Kelvin problem. Let us start by choosing a Cartesian system of coordinates based at the point where the force is concentrated, and with the positive direction of the X 1 -axis in the direction of the force. The nature of this particular configuration suggests that the state of stress in the loaded plane will be symmetric with respect to the X 1 -axis and antisymmetric with respect to the X 2 -axis. By symmetry considerations we can therefore restrict our attention to a simpler half-plane problem, as indicated in Fig. 9.6; for the sake of the argument we shall deal with the upper half-plane (X 2 > 0). First, let us note that the edge of this half-plane will be considered as being acted upon by a concentrated force (P/2)δ(X 1 )e1 . Since the corresponding traction vector is t(−e2 ) = (−e2 ) · T = −T21 e1 − T22 e2 , a first set of boundary conditions is

/

elasƟc plane

/

Fig. 9.6 Elastic plane with a concentrated load Pδ(X 1 )e1 acting at the origin. By symmetry considerations this situation can be reduced to a (constrained) traction boundary-value problem for a half-plane, as explained in the text

9.4 A Modification of the Method

435

 T12  X 2 =0 = −(P/2)δ(X 1 )

and

 T22  X 2 =0 = 0 .

(9.41)

A second requirement comes from demanding that there is no vertical displacement along the X 1 -axis; more precisely, we ask for  u2  X 2 =0 = 0 .

(9.42)

By taking into account the symmetry conditions mentioned above, we then look for an Airy stress function of the form 

∞

Φ(X 1 , X 2 ) =

α −2 (A + α X 2 B) e−α X 2 sin(α X 1 ) dα ,

(9.43)

0

where A = A(α) and B = B(α) are functions to be determined from the boundary conditions (9.41) and (9.42). For this we are going to need the stresses corresponding to (9.43), 

∞

T11 (X 1 , X 2 ) = 0



A − (2 − α X 2 )B e−α X 2 sin(α X 1 ) dα ,

(9.44a)



∞  T22 (X 1 , X 2 ) = − A + α X 2 B e−α X 2 sin(α X 1 ) dα ,  ∞0  T12 (X 1 , X 2 ) = A − (1 − α X 2 )B e−α X 2 cos(α X 1 ) dα .

(9.44b) (9.44c)

0

Integration of the kinematic relations (obtained from (9.44) with the help of Hooke’s law), as explained in Chap. 6, shows that  u2 (X 1 , X 2 ) =

∞

G(α; X 2 )e−α X 2 sin(α X 1 ) dα + f (X 1 ) ,

(9.45)

0

where    G(α; X 2 ) := (α E)−1 (1 + ν)A(α) + (1 − ν) + (1 + ν)α X 2 B(α) .

(9.46)

Applying the boundary condition (9.42) suggests that f ≡ 0 and 

1−ν A=− 1+ν

 B.

(9.47)

To get a second relation between A and B the first traction boundary condition (9.41) must be expressed in terms of an integral over (0, ∞). This is achieved by taking into account that the Dirac delta function is even and (see Appendix D.3) δ(X 1 ) =

1 2π



+∞ −∞

cos(α X 1 ) dα .

(9.48)

436

9 Special Two-Dimensional Problems: Unbounded Domains

Thus, the constraint (9.41) yields A−B =−

P . 2π

(9.49)

From (9.47) and (9.49) it follows that A=−

(1 − ν)P , 4π

B=

(1 + ν)P , 4π

which are then substituted back into (9.44); this results in  ∞  P (3 + ν) − (1 + ν)α X 2 )] e−α X 2 sin(α X 1 ) dα , (9.50a) 4π 0  ∞  P T22 (X 1 , X 2 ) = (1 − ν) − (1 + ν)α X 2 e−α X 2 sin(α X 1 ) dα , (9.50b) 4π 0  ∞  P T12 (X 1 , X 2 ) = − 2 − (1 + ν)α X 2 ) e−α X 2 cos(α X 1 ) dα . (9.50c) 4π 0

T11 (X 1 , X 2 ) = −

By using the formulae from Appendix D the integrals that feature in (9.50) are readily evaluated in closed form,

 X 12 P X1 (1 − ν) + 2(1 + ν) , 4π |X|2 |X|2

 X 22 P X1 (1 − ν) − 2(1 + ν) , T22 (X 1 , X 2 ) = 4π |X|2 |X|2

 X 12 P X2 (1 − ν) + 2(1 + ν) . T12 (X 1 , X 2 ) = − 4π |X|2 |X|2 T11 (X 1 , X 2 ) = −

(9.51a) (9.51b) (9.51c)

where |X|2 = X 12 + X 22 . These expressions describe the state of stress in an unbounded ‘two-dimensional’ elastic solid subjected to an in-plane concentrated load P = Pδ(X 1 )e1 acting at X ≡ (X 1 , X 2 ) = (0, 0). If this is replaced by a concentrated load Q = Qδ(X 1 )e2 , based again at the origin, then (9.51) will assume the form

 X 12 Q X2 (1 − ν) − 2(1 + ν) , (9.52a) T11 (X 1 , X 2 ) = 4π |X|2 |X|2

 X 22 Q X2 (1 − ν) + 2(1 + ν) , (9.52b) T22 (X 1 , X 2 ) = − 4π |X|2 |X|2

 X 22 Q X1 (1 − ν) + 2(1 + ν) . (9.52c) T12 (X 1 , X 2 ) = 4π |X|2 |X|2

9.4 A Modification of the Method

437

Superimposing the solutions (9.51) and (9.52) we can get the stress state in an unbounded solid subjected to a concentrated load Pδ(X 1 )e1 + Qδ(X 1 )e2 that has an arbitrary orientation with respect to the system of coordinates. Also, if the loading is applied at a point X 0 ≡ (X 10 , X 20 ), then in the above expressions we need to make the substitutions X 1 → X 1 − X 10 and X 2 → X 2 − X 20 ; in the interest of brevity these formulae are not recorded separately.

9.5 The Elastic Quarter-Plane The solutions of elastostatic boundary-value problems for the quarter-plane are more difficult than those for the elastic plane or half-plane. One of the main reasons responsible for these complications can be traced back to the presence of two distinct parts for its finite boundary. As we shall see shortly, in this case the Fourier representation of the Airy stress function can no longer be given in the form stated in (9.43). Quarter-plane problems, generally speaking, lead to Fredholm integral equations of the second kind that admit closed-form solutions only in some special circumstances. One such situation will be described below. We start by first presenting several general considerations that are valid for large classes of traction boundary-value problems involving quarter planes. Let us consider the region defined by X 1 ≥ 0 and X 2 ≥ 0, as seen in Fig. 9.7. The lower boundary (X 2 = 0) is subjected to an arbitrary normal load distribution p(X 1 ), while the vertical boundary (X 1 = 0) is taken to be traction-free; these constraints can be expressed compactly as indicated below, X1 = 0 :

t(−e1 ) = 0 ,

(9.53a)

X2 = 0 :

t(−e2 ) = − p(X 1 )e2 .

(9.53b)

Fig. 9.7 Elastic quarter-plane loaded along one of its edges (X 2 = 0), while the other is taken to be traction-free (X 1 = 0). The dashed straight line (Δ) bisects the right angle of the boundary; n and s are orthogonal unit vectors

438

9 Special Two-Dimensional Problems: Unbounded Domains

Without loss of generality, the load is assumed to admit the representation 

∞

p(X 1 ) =

q(α) cos(α X 1 ) dα ,

(9.54)

0

and the Airy stress function will be sought in the form3 

∞

Φ(X 1 , X 2 ) = 0

 α −2 A(α) + α X 2 B(α) e−α X 2 cos(α X 1 ) dα



∞

+

 β −2 C(β) + β X 1 D(β) e−β X 1 cos(β X 2 ) dβ ,

(9.55)

0

where A, B, C and D are functions to be determined. This assumed form of the Airy stress potential is suggested by the observation that the quarter-plane can be viewed as the intersection of two half-planes, each of which can be obtained from the other by a ±90◦ rotation. The two integrals in (9.55) correspond to the form of the Φ found in our discussion of half-planes, the roles of X 1 and X 2 being interchanged. The stress components associated with (9.55) are 

∞

T11 (X 1 , X 2 ) =



0

A − (2 − α X 2 )B e−α X 2 cos(α X 1 ) dα  ∞   − C + β X 1 D e−β X 1 cos(β X 2 ) dβ ,

(9.56)

0



∞

T22 (X 1 , X 2 ) = − 0



 A + α X 2 B e−α X 2 cos(α X 1 ) dα  ∞  + C − (2 − β X 1 )D e−β X 1 cos(β X 2 ) dβ ,

(9.57)

0

 T12 (X 1 , X 2 ) = − 0

∞



A − (1 − α X 2 )B e−α X 2 sin(α X 1 ) dα  ∞  − C − (1 − β X 1 )D e−β X 1 sin(β X 2 ) dβ .

(9.58)

0

From (9.53) it transpires that T12 = 0 for X 1 = 0 or X 2 = 0. By using the above formulae we then conclude that A(α) ≡ B(α) (for all α ≥ 0) and C(β) ≡ D(β) (for all β ≥ 0). Thus, the Fourier representations of the stresses are reduced to

3 Compare

this with (9.43).

9.5 The Elastic Quarter-Plane



439

∞

T11 (X 1 , X 2 ) = −

A(α)(1 − α X 2 ) e−α X 2 cos(α X 1 ) dα  ∞ − C(β)(1 + β X 1 ) e−β X 1 cos(β X 2 ) dβ ,

0



∞

A(α)(1 + α X 2 ) e−α X 2 cos(α X 1 ) dα  ∞ − C(β)(1 − β X 1 ) e−β X 1 cos(β X 2 ) dβ ,  ∞0 T12 (X 1 , X 2 ) = −X 2 α A(α) e−α X 2 sin(α X 1 ) dα 0  ∞ − X1 βC(β) e−β X 1 sin(β X 2 ) dβ . T22 (X 1 , X 2 ) = −

(9.59a)

0

0

(9.59b)

(9.59c)

0

These expressions can now be used in the remaining two boundary conditions, T22 (X 1 , 0) = p(X 1 ) and T11 (0, X 2 ) = 0, to produce the integral relations  

∞ 0

0

∞

 ∞ A(α)(1 − α X 2 ) e−α X 2 dα + C(β) cos(β X 2 ) dβ = 0 , 0  ∞ A(α) cos(α X 1 ) dα + C(β)(1 − β X 1 ) e−β X 1 dβ 0  ∞ =− q(α) cos(α X 1 ) dα .

(9.60a)

(9.60b)

0

The functions X 2 → (1 − α X 2 )e−α X 2 and X 1 → (1 − β X 1 )e−β X 1 can be expressed as Fourier-cosine integrals. Using these representations in (9.60a) and (9.60b), followed by an interchange of the order of integration, we are led to two coupled integral equations,  4β 2 ∞ α A(α) dα = 0 , π 0 (α 2 + β 2 )2  4α 2 ∞ βC(β) A(α) + dβ = −q(α) . π 0 (α 2 + β 2 )2

C(β) +

(9.61a) (9.61b)

For an arbitrary load p(X 1 ) the above equations are not solvable in closed form and require a numerical solution. In fact, (9.61) can be reduced to a single Fredholm-type integral equation of the second kind for just one of the two unknowns, as it will be shown next; once these functions are found, the stresses become available through the formulae (9.59). Eliminating C(α) between the two equations in (9.61) yields 16α 2 A(α) − 2 π

 0

∞



∞ 0

ξ x 3 A(ξ ) dξ dx = −q(α) (α 2 + x 2 )2 (ξ 2 + x 2 )2

440

9 Special Two-Dimensional Problems: Unbounded Domains



or

∞

A(α) +

K(α, ξ )A(ξ ) dξ = −q(α) ,

0

where  x 3 dx 16α 2 ξ ∞ π2 (α 2 + x 2 )2 (ξ 2 + x 2 )2 0

 ξ  16α 2 ξ   2 2 2 2 (α − ξ ) + (α + ξ ) log =− 2 2   . π (α − ξ 2 )3 α

K(α, ξ ) := −

(9.62) (9.63)

The solution of the elastic quarter-plane by superposition and reduction to integral equations appears in references [1–3], to which we refer the reader for additional pointers to the relevant literature; the presentation in this section is based on [2]. The symmetry of the pair of integral equations (9.61) suggests some simplification in finding the solution of the original problem; this is achieved by using the principle of superposition. Let us consider two sub-problems for the elastic quarter-plane that correspond to the following boundary conditions, (P± ) :

t(−e2 ) = − 21 p(X 1 )e2 , on X 2 = 0 , t(−e1 ) = ∓ 21 p(X 2 )e1 , on X 1 = 0 .

The loading in the problem (P− ) is symmetric with respect to the bisector (Δ) of the quarter plane, while in (P+ ) it is antisymmetric. Note that the superposition of (P+ ) and (P− ) leads to the initial problem—see (9.53). By analogy with the Fourier decomposition (9.54) we can also write a similar formula in which we change X 1 into X 2 . By repeating the calculations that led to (9.61), we can derive a similar pair of integrals for the new problems (P± ),  1 4β 2 ∞ α A(α) dα = ∓ q(β) , 2 2 2 π 0 (α + β ) 2  4α 2 ∞ βC(β) 1 A(α) + dβ = − q(α) , π 0 (α 2 + β 2 )2 2

C(β) +

(9.64a) (9.64b)

where the ‘±’ in (9.64a) have identical counterparts in the notation (P± ). Next, let us calculate the stresses associated with the directions defined by the orthonormal vectors {n, s} in Fig. 9.7; these are obviously obtained by a suitable rotation of the Cartesian-coordinate representation of the stress tensor (e.g. one can use θ = −45◦ in the formulae (8.121)). Thus,4 N45◦ =

4 Recall

1 (T11 + T22 ) − T12 , 2

the notations N and T introduced in Sect. 3.7.

1 T45◦ = − (T22 − T11 ) , 2

(9.65)

9.5 The Elastic Quarter-Plane

441

Fig. 9.8 Elastic quarter-plane loaded by a concentrated force Pδ(X 1 )e2 at its vertex. This loading can be regarded as the superposition of two states of stress that are either symmetric (lower left corner) or antisymmetric (lower right corner) with respect to the bisector of ∠ X 1 O X 2 . In the notation employed earlier in this section, the scenario on the left corresponds to (P− ), while that on the right to (P+ )

which can be written in a more detailed form with the help of (9.59), although we leave out the details in the interest of brevity. In the case of symmetric loading T45◦ = 0 for X 1 = X 2 , and we can use the second formula in (9.65) to find that A(α) = C(α). Similarly, for antisymmetric loading N45◦ = 0 for X 1 = X 2 , wherefrom A(α) + C(α) = 0, because of the first relation in (9.65). It can be concluded that the linear systems (9.64) can be replaced by the individual equations 4α 2 A(α) ± 2 π



∞ 0

ξ A(ξ ) 1 dξ = − q(α) , (α 2 + ξ 2 )2 2

(9.66)

with the ‘plus’ sign for (P− ) and the ‘minus’ sign for (P+ ). In general, we are still left with having to solve numerically these two integral equations, but there are several special cases in which (9.66) are amenable to simple closed-form solutions. One of these will be discussed next. Example 9.5 An elastic quarter-plane X 1 ≥ 0, X 2 ≥ 0 is subjected to a concentrated load Pδ(X 1 )e2 at its vertex, as shown in Fig. 9.8. Our interest is in finding the corresponding stress distribution. Note that the associated sub-problems (P∓ ) appear in the lower part of Fig. 9.8. By Fourier-expanding the load, p(X 1 ) ≡ −Pδ(X 1 ) = − the integral equations (9.66) become

2P π

 0

∞

cos(α X 1 ) dα ,

(9.67)

442

9 Special Two-Dimensional Problems: Unbounded Domains

4α 2 A(α) ± π



∞

0

ξ A(ξ ) P dξ = . (α 2 + ξ 2 )2 π

(9.68)

We look for constant solutions of these equations, A(α) = K , for some K ∈ R is yet to be determined. Taking into account the standard integral 

∞

ξ dξ 1 = , 2 2 +ξ ) 2α 2

(α 2

0

it is found that K = P/(π ± 2); more precisely, for (P− ) we have A(α) = C(α) = P/(π + 2), while for (P+ ) the solution is A(β) = −C(β) = P/(π − 2). We can now calculate the state of stress from (9.59), 2P  π 2 π −4



∞

(1 − α X 2 ) e−α X 2 cos(α X 1 ) dα 0  ∞  −2 (1 + β X 1 ) e−β X 1 cos(β X 2 ) dβ ,  ∞ 0 2P  T22 (X 1 , X 2 ) = − 2 π (1 + α X 2 ) e−α X 2 cos(α X 1 ) dα π −4 0  ∞  −2 (1 − β X 1 ) e−β X 1 cos(β X 2 ) dβ , 0  ∞ 2P  T12 (X 1 , X 2 ) = − 2 π X2 α e−α X 2 sin(α X 1 ) dα π −4 0  ∞  − 2X 1 β e−β X 1 sin(β X 2 ) dβ . T11 (X 1 , X 2 ) = −

(9.69a)

(9.69b)

(9.69c)

0

All the integrals appearing in (9.69) are routinely evaluated using the formulae given in Appendix D; the final expressions are recorded below for completeness,



  (2X 1 − π X 2 )X 12 (2X 1 − π X 2 )X 22 4P 4P , T22 = 2 , T11 = 2 π −4 π −4 (X 12 + X 22 )2 (X 12 + X 22 )2 

(2X 1 − π X 2 )X 1 X 2 4P . T12 = 2 π −4 (X 12 + X 22 )2 The problem of a concentrated tangential force acting at the origin can be handled in a very similar way to the above example.

9.6 Displacement Boundary-Value Problems If the boundary constraints are formulated in terms of displacements, the Airy stress function approach is rather inefficient (see Sect. 9.2). In this case the solution can be

9.6 Displacement Boundary-Value Problems

443

found more efficiently by employing a general representation of the displacement field in terms of several scalar functions.

9.6.1 The Papkovitch–Neuber Representation This result applies to three-dimensional Linear Elasticity—the two-dimensional scenario we are interested in will emerge as a particular case of the general theory. Before we embark on its justification, it is instructive to first look at the special case when no body forces are present. Our starting point will be the Navier–Lamé equation (see Sect. 5.7) in which we set f ≡ 0; dividing that equation by (λ + μ) gives (1 − 2ν)∇ 2 u + ∇(∇ · u) = 0 ,

(9.70)

since μ/(λ + μ) = 1 − 2ν. The Navier–Lamé equation is in fact a system of three coupled partial differential equations. The idea behind the Papkovitch–Neuber approach is to decouple and reduce them to either Laplace or Poisson equations. Let us first look for solutions of (9.70) in the form u = ∇ϕ + . . . , for some scalar field ϕ = ϕ(X); the dots stand for other (solenoidal) terms. By taking the divergence of this equation, ∇ · u = ∇ 2 ϕ, and then (9.70) becomes  ∇ 2 (1 − 2ν)u + ∇ϕ = 0 . Thus, there is a harmonic vector function B = B(X) such that (1 − 2ν)u + ∇ϕ = B ,

(∇ 2 B = 0)

(9.71)

or, after taking the divergence of both sides, 2(1 − ν)∇ 2 ϕ − ∇ · B = 0 .

(9.72)

On the other hand, a direct calculation shows that   ∇(X · B) = B + ∇ ⊗ B · X ,

(9.73)

and by further taking the divergence of this equation, ∇·B=

 1 2 ∇ X·B . 2

Substituting (9.74) into (9.72) leads to

∇

2

 X·B ϕ− = 0, 4(1 − ν)

(9.74)

444

9 Special Two-Dimensional Problems: Unbounded Domains

i.e. the function within the square brackets is harmonic. If we define ψ := ϕ −

X·B 4(1 − ν)

=⇒

∇2ψ = 0 ,

(9.75)

so that (9.71) can be cast in the form

 X·B (1 − 2ν)u = B − ∇ ψ + ; 4(1 − ν)

(9.76)

this is a particular version of the Papkovitch–Neuber representation. To bring formula (9.76) more in line with the version that is usually stated in the literature, we rescale the two potentials by letting B → −4(1 − 2ν)(1 − ν)B and ψ → −(1 − 2ν)ψ, to get   u = ∇ ψ + X · B − 4(1 − ν)B ,

(9.77)

where ψ and B satisfy ∇ 2 ψ = 0 and ∇ 2 B = 0. Once these two harmonic functions are found the displacement is given by (9.77), and we can then easily calculate the infinitesimal deformation tensor E and the corresponding stress tensor T . We shall return to these matters shortly, but in the meantime let us note that the above approach does not work for the general case when the body force f ≡ ρb is present in (9.70). As shown below, the general form of (9.77) remains unchanged; however, the two potentials are found to satisfy Poisson-type partial differential equations whose solutions depend crucially on the body force. First, note that in the general case Eq. (9.70) acquires an extra term on the left-hand side, (9.78) (1 − 2ν)∇ 2 u + ∇(∇ · u) + f /(λ + μ) = 0 . Since the differential operator remains unchanged it seems sensible to look for a solution of (9.78) in the form   u = C1 ∇ ψ + X · B + C2 B ,

(9.79)

where C j ∈ R ( j = 1, 2) and ψ ≡ ψ(X), B ≡ B(X) are to be determined (to keep avoiding the proliferation of unnecessary notation we recycle the labels used earlier in this section). The main idea is to calculate the first two terms in (9.78) by using the above assumed form of the solution, and then see how we could fine-tune the various arbitrary elements in (9.79) so that (9.78) is satisfied. Note that     ∇ · (∇ ⊗ B) · X = ∇ · (∇ ⊗ B) · X + ∇ ⊗ B : I   = ∇2 B · X + ∇ · B .

(9.80)

9.6 Displacement Boundary-Value Problems

445

Also, by taking the divergence of (9.73) and using (9.80) yields     ∇ 2 (X · B) = ∇ 2 B · X + 2 ∇ · B .

(9.81)

Next, we have successively    ∇ · u = C1 ∇ · ∇(ψ + X · B) + C2 ∇ · B    = C1 ∇ 2 ψ + ∇ 2 (X · B) + C2 ∇ · B  = C1 ∇ 2 ψ + (2C1 + C2 )∇ · B + C1 (∇ 2 B) · X ,

(9.82) (9.83)

where use has been made of (9.81). By replacing (9.79)–(9.82) back into (9.78), after some rearrangements, we end up with     2C1 (1 − ν)∇ ∇ 2 ψ + (∇ 2 B) · X + 2(1 − 2ν)C1 + (C2 + 2C1 ) ∇ ∇ · B  + C2 (1 − 2ν)∇ 2 B + f /(λ + μ) = 0 . To satisfy this equation we set   ∇2ψ + ∇2 B · X = 0 ,

2(1 − 2ν)C1 + (C2 + 2C1 ) = 0 ,

C2 (1 − 2ν)∇ B + f /(λ + μ) = 0 . 2

(9.84a) (9.84b)

generality we can take C1 = 1. With this result in (9.84b) we obtain ∇2 B =

f . 4μ(1 − ν)

(9.85)

Finally, substituting (9.85) in the first equation of (9.84a) yields ∇2ψ = −

f ·X . 4μ(1 − ν)

(9.86)

The new representation is identical to (9.77), but this time the two potentials ψ and B are the solutions of the Poisson-type equations (9.85) and (9.86), respectively. There are some caveats attached to the Papkovitch–Neuber representation. While the displacement field has only three scalar components, ψ and B are in fact four scalar fields. Fortunately, in the three-dimensional case, for most domains of interest, one of those functions is arbitrary and can be set to zero; furthermore, in plane elasticity we only need two out of the four original functions. Other disadvantages include the non-uniqueness of the potentials, and the lack of their translation-invariance under a change of origin of the chosen system of coordinates.

446

9 Special Two-Dimensional Problems: Unbounded Domains

9.6.2 The Stress Tensor in Terms of ψ and B Having an expression for the displacement field allows us to directly calculate the deformation (E) and hence the stress distribution (T ). This is what we do next. To simplify the subsequent calculations the Papkovitch–Neuber representation (9.77) is cast in the equivalent form   2μu = ∇ψ + ∇ ⊗ B · X − (3 − 4ν)B ,

(9.87)

where the potentials have been rescaled according to ψ → ψ/2μ, B → B/2μ, and use has been made of (9.73). Starting from (9.87) it is possible to express the stress tensor T in terms of ψ and B. For simplicity, we shall ignore the body force ( f ≡ 0); adding this effect presents no particular difficulties.5 Thus, the two potentials in (9.87) are harmonic functions, i.e. ∇2ψ = 0

and

∇2 B = 0 .

(9.88)

To find the stress tensor we use Hooke’s law T = 2μE + λ|E|I ,

(9.89)

and note that all we have to do is to calculate E (the symmetric part of ∇ ⊗ u) and |E| ≡ tr(E) = ∇ · u by using the expression (9.87). Use of (9.80) and (9.88) leads to       2μ ∇ · u = ∇ 2 ψ + ∇ 2 B · X − 2(1 − 2ν) ∇ · B   = −2(1 − 2ν) ∇ · B . Since

(9.90)

  ∇ ⊗ (∇ ⊗ B) · X = (∇ ⊗ ∇ ⊗ B · X + B ⊗ ∇ ,

we can also write     2μ ∇ ⊗ u = ∇ ⊗ ∇ψ+ ∇ ⊗ ∇ ⊗ B · X

 + B ⊗ ∇ − (3 − 4ν) ∇ ⊗ B) .

By adding up this last result and its transpose it follows that   2μE = ∇ ⊗ ∇ψ+ ∇ ⊗ ∇ ⊗ B · X   − (1 − 2ν) ∇ ⊗ B + B ⊗ ∇ . 5 It

(9.91)

must be kept in mind that if f is brought back onto the stage, then the form (9.87) demands a slight amendment to the governing equations for ψ and B—see (9.85) and (9.86).

9.6 Displacement Boundary-Value Problems

447

The final expression for the stress tensor follows by going back into (9.89) with (9.90) and (9.91), with the outcome,   T = ∇ ⊗ ∇ψ + ∇ ⊗ ∇ ⊗ B · X     − (1 − 2ν) ∇ ⊗ B + B ⊗ ∇ − 2ν ∇ · B I .

(9.92)

It is reiterated here that (9.92) represents a fully three-dimensional result. However, the validity of this last formula remains true for the plane-strain approximation as well, but it must be suitably modified for the plane-stress case. In both situations we must impose the obvious two-dimensional restrictions on T , ∇ and X.

9.6.3 Particular Case: Plane Elasticity The two-dimensional version of the above results can be readily obtained by taking B to be two-dimensional, that is B = B1 e1 + B2 e2 , with both B j ( j = 1, 2) being functions of (X 1 , X 2 ). Since one of the three functions ψ, B1 and B2 is arbitrary, we can set B1 ≡ 0, so the Papkovitch–Neuber representation for plane elasticity can be given in terms of only two functions, ψ = ψ(X 1 , X 2 )

and

B = B2 (X 1 , X 2 )e2 ,

(9.93)

which in the absence of body forces will be harmonic. The simplified form of (9.87), is also recorded below, 2μu1 = ψ ,1 + X 2 B2, 1 ,

2μu2 = ψ ,2 + X 2 B2, 2 − (3 − 4ν)B2 .

(9.94)

Thus, rewriting (9.92) with ψ and B given by (9.93), the components of the twodimensional (plane-strain) stress tensor are T11 = ψ ,11 − 2ν B2, 2 + X 2 B2, 11 , T12 = ψ ,12 − (1 − 2ν)B2, 1 + X 2 B2, 12 , T22 = ψ ,22 − 2(1 − ν)B2, 2 + X 2 B2, 22 .

(9.95a) (9.95b) (9.95c)

We illustrate the application of the method outlined above with a typical problem that involves only displacement boundary conditions. Let us consider an infinite elastic strip of width b > 0,    Σ ≡ (X 1 , X 2 ) ∈ E2  − ∞ < X 1 < +∞, 0 < X 2 < b . On the two edges of the strip, {X 2 = 0} and {X 2 = b}, the displacements are taken to be prescribed, i.e.

448

9 Special Two-Dimensional Problems: Unbounded Domains

u1 (X 1 , 0) = u10 (X 1 ) , u2 (X 1 , 0) = u20 (X 1 ) ,

(9.96a)

u1 (X 1 , b) = u1b (X 1 ) , u2 (X 1 , b) = u2b (X 1 ) ,

(9.96b)

−∞ < X 1 < +∞, where u j0 and u jb are given ‘good’ functions. The main goal is to describe the stress distribution in Σ. We look for solutions in the form (9.94). Since both ψ and B2 are harmonic, they admit the Fourier-integral representations, 1 ψ=√ 2π



+∞

−∞

α −1 Γ0 (α; X 2 ) e−iα X 1 dα ,

 +∞ 1 Γ2 (α; X 2 ) e−iα X 1 dα , B2 = √ 2π −∞ Γ0 (α; X 2 ) := A(α) cosh(α X 2 ) + B(α) sinh(α X 2 ) , Γ2 (α; X 2 ) := C(α) cosh(α X 2 ) + D(α) sinh(α X 2 ) , and the functions A, B, C and D are to be found as explained next. Replacing these expressions in (9.94) yields 2μu1 (X 1 , X 2 ) = − √



i 2π 

1 2μu2 (X 1 , X 2 ) = √ 2π

+∞

−∞ +∞

−∞

U1 (α; X 2 ) e−iα X 1 dα ,

U2 (α; X 2 ) e−iα X 1 dα ,

(9.97a) (9.97b)

where  U1 := A cosh(α X 2 ) + B sinh(α X 2 ) + α X 2 C cosh(α X 2 ) + D sinh(α X 2 ) ,  U2 := A sinh(α X 2 ) + B cosh(α X 2 ) − C κ cosh(α X 2 ) − α X 2 sinh(α X 2 )  − D κ sinh(α X 2 ) − α X 2 cosh(α X 2 ) , and κ := 3 − 4ν. To apply the boundary conditions (9.96) using the Fourier representations (9.97) we need to Fourier transform the boundary data,  u¯ j0 (α) := F1 u j0 (X 1 ); α ,

 u¯ jb (α) := F1 u jb (X 1 ); α ,

( j = 1, 2) .

The result of enforcing the boundary conditions turns out to be a set of four algebraic equations for the functions A, B, C and D. The two equations in (9.96a) give − i A = 2μ¯u10 ,

B = κC + 2μ¯u20 ,

while the other two boundary conditions in (9.96b) lead to

(9.98)

9.6 Displacement Boundary-Value Problems

449

  A cosh ζ + B sinh ζ + ζ C cosh ζ + D sinh ζ = 2iμ¯u1b ,   A sinh ζ + B cosh ζ − C κ cosh ζ − ζ sinh ζ   − D κ sinh ζ − ζ cosh ζ = 2μ¯u2b ,

(9.99a) (9.99b)

where ζ := αb. Clearly, A and B can be eliminated from (9.99a) and (9.99b) by using (9.98), thus leading to a linear system in two unknowns (C and D), A11 C + A12 D = R1 ,

A12 C + A22 D = R2 ,

(9.100)

where A11 := κ sin ζ +ζ cosh ζ ,

A12 := ζ sinh ζ , A22 := −(κ sinh ζ − ζ cosh ζ ) ,   R1 := 2μ i¯u1b − i¯u10 cosh ζ − u¯ 20 sinh ζ ,   R2 := 2μ u¯ 2b − i¯u10 sinh ζ − u¯ 20 cosh ζ .

The solution of the linear system (9.100) is given by   R2 ζ sinh ζ + R1 κ sinh ζ − ζ cosh ζ C= , (κ sinh ζ − ζ )(κ sinh ζ + ζ )   R1 ζ sinh ζ − R2 κ sinh ζ + ζ cosh ζ . D= (κ sinh ζ − ζ )(κ sinh ζ + ζ )

(9.101a) (9.101b)

Together with (9.98) these expressions completely determine the two displacement potentials Ψ and B2 . Once they are known, the displacements and the stresses are given by (9.94) and (9.95), respectively. It is not always possible to evaluate in closed-form the integrals that appear in the solution found with the help of Fourier transforms. There are two ways in which such solutions can be used for practical purposes. One approach relies on numerical methods to approximate the integrals; depending on the complexity of the integrands, this is not a routine task. The alternative is suggested by the presence of the expression κ 2 sinh2 ζ − ζ 2 in the denominators of U j ( j = 1, 2). Note that both integrals in (9.97) will be of the form 

+∞

−∞

ζ X1 H j (ζ ) e−i b dζ , κ 2 sinh2 ζ − ζ 2

( j = 1, 2) ,

(9.102)

for some functions H j that admit a straightforward analytic continuation in the complex plane. The transcendental equations κ sinh z + z = 0 and κ sinh z − z = 0 have an infinite number of solutions z = σ + iτ ∈ C, which suggests that one can apply the Residue Theorem from Appendix D around contours similar to those in Fig. D.2. This process transforms each integral in (9.102) into a sum over all the aforementioned roots, in which the summand is a rather complicated function. A critical discussion of the pros and cons of both approaches is contained in [4] (see Chaps. 9 and 10 in that reference).

450

9 Special Two-Dimensional Problems: Unbounded Domains

2

2

=+

1 2

=−

Fig. 9.9 Elastic strip loaded by a concentrated moment at the origin and having its edges restrained against in-plane displacements, i.e. u(0, ±b) = 0

Example 9.6 An elastic strip −∞ < X 1 < +∞, −b ≤ X 2 ≤ b is clamped along its edges and is loaded by a concentrated moment at the point (X 1 , X 2 ) = (0, 0), as shown in Fig. 9.9. It is required to determine the stress distribution in the strip, in particular, the stresses along the clamped boundaries (the concentrated moment we have in mind here corresponds to the situation described in Exercise 9.11). Let us start by writing down the boundary conditions,  u1  X 2 =±b = 0

and

 u2  X 2 =±b = 0 ,

−∞ < X 1 < +∞ .

(9.103)

The particular type of loading considered, together with the symmetry of the strip geometry, suggests that u2 and T12 are even functions with respect to X 2 , while u1 and T11 , T22 are odd with respect to the same coordinate. Also, u1 is symmetric with respect to the X 1 and u2 is antisymmetric. These observations will greatly simplify the Fourier analysis in the present case. We approach the solution of this problem via the Principle of Superposition (discussed in Sect. 5.8). On the known solution of the concentrated moment in an unbounded plane (see Exercise 9.11) we are going to superimpose a second (corrective) solution that is selected such that the boundary conditions (9.103) are satisfied. If u(1) is the displacement field for the former solution, while u(2) represents its counterpart for the latter, then the overall displacement field will be u := u(1) + u(2) . In light of (9.103) the boundary conditions for the corrective solution must satisfy   u(2)  X 2 =±b = −u(1)  X 2 =±b ,

−∞ < X 1 < +∞ .

(9.104)

Due to symmetry considerations it is sufficient to consider (9.104) only for X 2 = b. Therefore, with the help of (9.112) we get  2X 12 κ =: f 0 (X 1 ) , + (X 12 + b2 )2 X 12 + b2 

X 1 (X 12 − b2 ) =: g0 (X 1 ) . = −N (X 12 + b2 )2

  u1(2)  X 2 =b = −u(1)  X 2 =b = bN   u(2)  X 2 =b = −u2(1)  X 2 =b



(9.105a) (9.105b)

9.6 Displacement Boundary-Value Problems

451

We seek u(2) in the form given by the Papkovitch–Neuber representation (9.94). Because of the aforementioned symmetry properties we take 

 2 ∞ −1 α A(α) sinh(α X 2 ) sin(α X 1 ) dα , π 0   ∞ 2 B(α) cosh(α X 2 ) sin(α X 1 ) dα , B2 (X 1 , X 2 ) = π 0

ψ(X 1 , X 2 ) =

(9.106a) (9.106b)

where A ≡ A(α) and B ≡ B(α) are to be found from the constraints (9.105). Direct calculations show that (9.94) can be expressed as  2μu1(2) (X 1 ,

X 2) = 

2μu2(2) (X 1 , X 2 ) =

2 π 2 π



∞

U1 (α; X 2 ) cos(α X 1 ) dα ,

(9.107a)

U2 (α; X 2 ) sin(α X 1 ) dα ,

(9.107b)

0



∞

0

where U1 := A sinh(α X 2 ) + α X 2 B cosh(α X 2 ) ,  U2 := A cosh(α X 2 ) − B κ cosh(α X 2 ) − α X 2 sinh(α X 2 ) . Enforcing the boundary conditions (9.105) results in a linear system for A and B, A sin ζ + Bζ cosh ζ = 2μ f¯0 ,

A cosh ζ + B(ζ sinh ζ − κ cosh ζ ) = 2μg¯ 0 ,

with the solution 

g¯ 0 ζ cosh ζ + f¯0 (κ cosh ζ − ζ sinh ζ ) , A = 4μ κ sinh 2ζ + 2ζ f¯0 cosh ζ − g¯ 0 sinh ζ B = 4μ ; κ sinh 2ζ + 2ζ

(9.108a) (9.108b)

note that f¯0 and g¯ 0 can be found explicitly, 



∞



π −ζ e (1 + κ − ζ ) , 2 0   ∞  2 π −ζ e (1 − ζ ) . g0 (X 1 ) sin(α X 1 ) dα = −N g¯ 0 (α) := π 0 2 f¯0 (α) :=

2 π

f 0 (X 1 ) cos(α X 1 ) dα = N

With A(α) and B(α) determined, the displacement potentials ψ and B2 in (9.106) are fully known and the stress components can finally be obtained from (9.95). The outcome of these substitutions will be some rather complicated integrals, so in the interest of brevity we give below only the expression of the normal and shear stresses

452

9 Special Two-Dimensional Problems: Unbounded Domains

along the upper boundary (X 2 = b),    ∞ G 22 (ζ ) M ζ X1 −ζ ζ e sin dζ , T22 (X 1 , b) = 2π b2 (1 − ν) 0 κsinh2ζ + 2ζ b    ∞ G 12 (ζ ) M ζ X1 dζ , T12 (X 1 , b) = ζ e−ζ cos 2π b2 (1 − ν) 0 κsinh2ζ + 2ζ b  G 22 (ζ ) := (1 − 2ν) cosh ζ sinh ζ + ζ (1 + κ − ζ ) − 2(1 − ν)(1 − ζ ) sinh2 ζ ,  G 12 (ζ ) := 2(1 − ν)(1 + κ − ζ ) cosh2 ζ − (1 − ζ ) (1 − 2ν) cosh ζ sinh ζ + ζ .

9.7 Exercises 1. Use the convolution theorem for Fourier transforms to show that the formulae (9.7) can also be expressed in the equivalent way, 2X 3 T11 (X 1 , X 2 ) = − 1 π T22 (X 1 , X 2 ) = − T12 (X 1 , X 2 ) = −

2X 1 π



+∞

−∞  +∞

−∞ 2  +∞ 2X 1

π

−∞

p(X 2 − ζ ) dζ , (X 12 + ζ 2 )2

(9.109a)

ζ 2 p(X 2 − ζ ) dζ , (X 12 + ζ 2 )2

(9.109b)

ζ p(X 2 − ζ ) dζ . (X 12 + ζ 2 )2

(9.109c)

2. Consider an elastic half-plane loaded by a non-uniformly distributed shear load q = q(X 2 ). Assuming that the main geometry is the same as that in Fig. 9.2, the boundary conditions are t(−e2 ) = q(X 2 )e2 for −∞ < X 2 < +∞. By following a similar approach to the analysis that led to (9.7), show that the stress distribution is given by the following expressions,  2X 12 +∞ (X 2 − ζ )q(ζ ) dζ , π −∞ (X 12 + (X 2 − ζ )2 )2  2 +∞ (X 2 − ζ )3 q(ζ ) T22 (X 1 , X 2 ) = − dζ , π −∞ (X 12 + (X 2 − ζ )2 )2  2X 1 +∞ (X 2 − ζ )2 q(ζ ) T12 (X 1 , X 2 ) = − dζ . π −∞ (X 12 + (X 2 − ζ )2 )2 T11 (X 1 , X 2 ) = −

(9.110a) (9.110b) (9.110c)

3. Revisit the example of the disk subjected to opposite compressive forces. In the notation of Fig. 9.5a, show that this problem can be solved by choosing the Airy stress function

9.7 Exercises

453

Fig. 9.10 Elastic disk of radius a > 0 loaded by two equal opposite concentrated forces situated on its circumference and acting along a chord parallel to the X 1 -axis

P Φ= π



r12 − r1 θ1 sin θ1 − r2 θ 2 sin θ 2 2a

 .

4. An elastic plate in the shape of a thin disk is subjected to two equal but opposite forces acting along a chord, as shown in Fig. 9.10. Assuming a state of plane stress, show that this problem is solved by an Airy stress function of the form   γ r2 r1 θ1 sin θ1 − r2 θ 2 sin θ 2 + , a

P Φ(r, θ ) = − π

where γ ∈ R is a constant that you must determine from the boundary conditions along the circumference of the disk. Show further that the azimuthal stress along this boundary is given by   Tθθ 

r =a

=

P sin 2ω . πa(sin θ + sin ω)

5. The static transverse deflection u = u(X 1 , X 2 ) of a thin elastic plate in the form of a quarter-plane satisfies the boundary-value problem ∇ 4u = 0 ,

0 < X1 < ∞ , 0 < X2 < ∞ ,

u(0, X 2 ) =

∂ u (0, X 2 ) = 0 , ∂ X 12

u(X 1 , 0) =

AX 1 ∂ 2u and (X 1 , 0) = 0 , 2 1 + X1 ∂ X 22

2

0 < X2 < ∞ , 0 < X1 < ∞ ,

where A ∈ R (constant); in addition, u(X 1 , X 2 ) and its derivatives are assumed to decay to zero as X 1 , X 2 → ∞. Use the Fourier sine transform to show that

454

9 Special Two-Dimensional Problems: Unbounded Domains

u(X 1 , X 2 ) = =

A 2



X 12

∞

(2 + α X 2 ) e−α(1+X 2 ) sin(α X 1 ) dα

0

AX 1 AX 1 X 2 (1 + X 2 ) + 2 . 2 + (1 + X 2 ) [X 1 + (1 + X 2 )2 ]2

6. In the previous question replace the conditions on X 2 = 0 with the requirements u(X 1 , 0) = 0 ,

∂ 2u AX 1 (X 1 , 0) = , 2 ∂ X2 (1 + X 12 )2

0 < X1 < ∞ .

Show that the solution in this case is  AX 1 ∞ −α(1+X 2 ) e sin(α X 1 ) dα 4 

0 AX 1 X 2 1 . =− 4 X 12 + (1 + X 2 )2

u(X 1 , X 2 ) = −

7. Consider the stress distribution for the 2D Kelvin problem—see (9.51). a. Show that    κ + 3 cos θ P κ − 1 cos θ , Tθθ = , κ +1 r 2π κ + 1 r   3−ν P κ − 1 sin θ , κ≡ . Tr θ = 2π κ + 1 r 1+ν

P Trr = − 2π



b. Check that these expressions can be obtained from the following Airy potential,

   P κ −1 Φ(r, θ ) = −r θ sin θ + r log r cos θ . 2π κ +1 c. By integrating the corresponding kinematic relations, show that the components of the displacement field u = ur er + uθ eθ have the form P(−2κ log r + 1) cos θ + C cos θ , 4π μ(κ + 1) P(2κ log r + 1) sin θ uθ = − C sin θ , (C ∈ R) . 4π μ(κ + 1) ur =

(Note: By taking κ = 3 − 4ν the above results become applicable to the corresponding plane-strain problem.) 8. Consider a thin elastic plate that occupies the right half-plane bounded by X 1 = 0. The plate is compressed by in-plane tractions normal to its edge, corresponding to the boundary conditions

9.7 Exercises

455

Fig. 9.11 Elastic half-plane subjected to uniform shear over an unbounded part of its boundary. This constraint corresponds to: t(−e2 ) = τ0 e1 for X 1 ≤ 0, and t(−e2 ) = 0 for X 1 > 0 (τ0 > 0 is a given constant)

T11 (0, X 2 ) = − p(X 2 ) , where p(X 2 ) =

T12 (0, X 2 ) = 0 ,

−∞ < X 2 < +∞ ,

−1/2 p0  2 0 − X 22 H (0 − |X 2 |) , π

0 , p0 are given constants, and H denotes the usual Heaviside function (9.17). Check the following, 

a. T11 + T22 = − p0

2 π





b. T22 − T11 + 2iT12 = p0

∞

J0 (0 α)e−α X 1 cos(α X 2 ) dα ;

0

2 X1 π



∞

α J0 (0 α)e−α(X 1 +iX 2 ) dα ,

0

where J0 is the Bessel function of the first kind and order 0. c. On the X 1 -axis show that the previous two formulae reduce to  −1/2  , T11 + T22 = − p0 2/π X 12 + 20   2  2 2 −3/2 T22 − T11 + 2i T12 = p0 2/π X 1 X 1 + 0 . 9. Consider the elastic half-plane shown in Fig. 9.11. Its edge is subjected to a uniform shear load that extends over the whole region −∞ < X 1 ≤ 0, while the other part (0 < X 1 < +∞) is traction-free. a. Use an appropriate Fourier transform to show that the stress distribution in the half-plane is given by

456

9 Special Two-Dimensional Problems: Unbounded Domains

τ0 T11 = π





τ  X 22 X 22 0 2 2 + log(X 1 + X 2 ) , T22 = − , 2 2 2 2 X1 + X2   π X 1 + X 2

X2 X1 X2 τ0 . T12 = − + tan−1 π X 12 + X 22 X1

b. Show that in polar coordinates these formulae become τ0  (2 log r − 1) + (2 log r + 1) cos 2θ − 2θ sin 2θ , 2π τ0  (2 log r + 1) sin θ + 2θ cos θ sin θ , Tθθ = π τ0  (2 log r + 1) sin 2θ + 2θ cos 2θ . Tr θ = − 2π

Trr =

(9.111a) (9.111b) (9.111c)

10. Consider an elastic quarter-plane loaded by a uniform shear distribution over the edge X 1 = 0, while its other boundary (X 2 = 0) is traction-free (see Fig. 9.12). a. By employing a suitably defined Fourier transform show that the stress distribution in the quarter-plane is given by

  π X2 X1 X2 τ0 X 22 −1 , T − 2 T11 = τ0 − tan = − , 12 2 X1 X 1 + X 22 X 12 + X 22

  X1 X2 X2 . T22 = τ0 − tan−1 2 2 X1 X1 + X2 b. Hence, or otherwise, establish the polar-coordinate form of the above results, τ0  π(1 + cos 2θ ) − 4θ − 2 sin 2θ , 4 τ0  Tθθ = π(1 − cos 2θ ) − 4θ + 2 sin 2θ , 4 τ0  Tr θ = 2(1 − cos 2θ ) − π sin 2θ . 4

Trr =

11. Consider an elastic plane loaded by a concentrated moment at the origin of the system of coordinates, as explained in Fig. 9.13. a. Assuming a state of plane strain, show that the stress distribution is X1 X2  2 4X 1 + (1 − 2ν)|X|2 , (9.112a) 6 |X| X1 X2  2 4X 2 − (3 − 2ν)|X|2 , (9.112b) T22 = 4μN |X|6    2μN (X 12 − X 22 ) 2X 12 + (1 − 2ν)|X|2 − 4X 12 X 22 , (9.112c) T12 = − 6 |X|

T11 = 4μN

9.7 Exercises

457

Fig. 9.12 Elastic quarter-plane subjected to uniform shear over one of its boundaries. The constraint on X 1 = 0 corresponds to: t(−e1 ) = τ0 e2 for X 2 ≥ 0, while on the horizontal boundary, t(−e2 ) = 0 for X 1 > 0 (τ0 > 0 is a given constant)

where N = M/8π μ(1 − ν) and X = X 1 e1 + X 2 e2 is the position vector of an arbitrary point in the plane. b. Check that the displacement field has the expression u1 = −

N  2 N 2X 1 X 2 + (3 − 4ν)X 2 |X|2 , u2 = X 1 (X 12 − X 22 ) . |X|4 |X|4

12. Consider the elastic strip shown in Fig. 9.14, which is compressed by two concentrated forces P applied at the points (0, ±b). Find the normal stress T22 along the symmetry axis of the strip, in particular, show that T22 (X 1 , 0) = −

2P πb

 0

∞

sinh ζ + ζ cosh ζ cos (ζ X 1 ) dζ . sinh 2ζ + 2ζ

Fig. 9.13 Elastic plane subjected to a concentrated (directional) moment of magnitude M and located at the origin O. Two concentrated loads ±P e1 act at the points (0, ∓ ε) in opposite directions; the concentrated moment is obtained by taking the limits ε → 0+ , P → +∞, while 2Pε remains constant and equal to M

458

9 Special Two-Dimensional Problems: Unbounded Domains

Fig. 9.14 Elastic strip loaded by two equal concentrated forces ±P e2 acting along the same line, but in opposite directions

Bibliography 1. Babuska I, Rektorys K, Vycichlo F (1960) Mathematische elastizitätstheorie der ebenen probleme. Akademie-Verlag, Berlin 2. Teodorescu PP (1960) Plane problems in the theory of elasticity. Editura Academiei RPR, Bucharest (in Romanian) 3. Teodorescu PP (2013) Treatise on classical elasticity. Springer, Dordrecht, The Netherlands 4. Sneddon IN (1951) Fourier transforms. McGraw-Hill Book Company Inc, New York 5. Amenzade YuA (1979) Theory of elasticity. Mir Publishers, Moscow 6. Churchill RV (1972) Operational methods, 3rd edn. McGraw-Hill Book Company, New York 7. Debnath L, Bhatta D (2014) Integral transform and their applications, 2nd edn. CRC Press, Boca Raton, Florida 8. Filonenko-Borodich M (1963) Theory of elasticity. Mir Publishers, Moscow 9. Girkmann K (1959) Flächentragwertke, 2nd edn. Springer-Verlag, Wien (in German) 10. Kecs W, Teodorescu PP (1974) Applications of the theory of distributions in mechanics. Abacus Press, Tunbridge Wells, Kent (UK) 11. Leipholz H (1974) Theory of elasticity. Noordhoff International Publishing, Leyden 12. Little RWm (1973) Elasticity. Prentice-Hall Inc, Englewood Cliffs, New Jersey 13. Lurie AI (2005) Theory of elasticity. Springer, Berlin 14. Skalskaya IP, Lebedev NN, Uflyand YaS (1965) Problems of mathematical physics. PrenticeHall Inc, Englewood Cliffs, New Jersey 15. Solomon L (1968) Élasticité Lineaire. Mason, Paris (in French) 16. Timoshenko SP, Goodier JN (1970) Theory of elasticity, International edn. McGraw-Hill Book Company, Auckland 17. Tolstov GP (1976) Fourier Series, Dover Publications. Mineola, New York 18. Uflyand YaS (1968) Survey of articles on the applications of integral transforms in the theory of elasticity. Nauka, Leningrad, URSS (in Russian) 19. Ditkin VA, Prudnikov AP (1962) Operational calculus in two variables and its applications. Pergamon Press, New York 20. Mikhlin SG (1964) Integral equations. Pergamon Press, Oxford

Appendix A

Vector and Tensor Identities

Abstract A number of useful vector and tensor identities are listed below. It is assumed that φ, ϕ are scalar fields, u, v represent vector fields, while T , T 1 , T 2 denote tensor fields. The reader is referred to the books [1–8], where other similar formulae may be found. GRAD formulae: ∇ ⊗ (φu) = (∇φ) ⊗ u + φ(∇ ⊗ u) , ∇(u · v) = (∇ ⊗ u) · v + (∇ ⊗ v) · u ,

(A.1a) (A.1b)

∇ ⊗ (u ∧ v) = (∇ ⊗ u) ∧ v − (∇ ⊗ v) ∧ u ,

(A.1c)

∇ ⊗ (T · u) = (∇ ⊗ T ) · u + (∇ ⊗ u) · T ,

(A.1d)

∇(φϕ) = (∇φ)ϕ + φ(∇ϕ) ,

(A.1e)

• (∇ ⊗ u) × • I = ∇ ∧ u = I × (∇ ⊗ u) ,

(A.1f)

(∇ ⊗ u) × × I = I(∇ · u) − u ⊗ ∇ ,

(A.1g)

∇(T 1 : T 2 ) = T 1 : (∇ ⊗ T 2 ) + T 2 : (∇ ⊗ T 1 ) .

(A.1h)

∇ · (φu) = (∇φ) · u + φ(∇ · u) ,

(A.2a)

∇ · (φT ) = (∇φ) · T + φ(∇ · T ) , ∇ · (u ⊗ v) = (∇ · u)v + u · (∇ ⊗ v) ,

(A.2b) (A.2c)

∇ · (T · u) = (∇ · T ) · u + T : (∇ ⊗ u) , ∇ · (u ∧ v) = (∇ ∧ u) · v − u · (∇ ∧ v) ,

(A.2d) (A.2e)

∇ · (T 1 · T 2 ) = (∇ · T 1 ) · T 2 + T 1 : (∇ ⊗ T 2 ) ,

(A.2f)

• (∇ ⊗ u) , ∇ · (T ∧ u) = (∇ · T ) ∧ u + T ×

(A.2g)

∇ · (∇ ∧ u) = 0 .

(A.2h)

T

DIV formulae

© Springer Nature B.V. 2020 C. D. Coman, Continuum Mechanics and Linear Elasticity, Solid Mechanics and Its Applications 238, https://doi.org/10.1007/978-94-024-1771-5

459

460

Appendix A: Vector and Tensor Identities

CURL formulae: ∇ ∧ (φT ) = (∇φ) ∧ T + φ(∇ ∧ T ) , ∇ ∧ (φu) = (∇φ) ∧ u + φ(∇ ∧ u) ,

(A.3a) (A.3b)

∇ ∧ (u ⊗ v) = (∇ ∧ u) ⊗ v − u ∧ (∇ ⊗ v) , ∇ ∧ (u ∧ v) = u(∇ · v) + v · (∇ ⊗ u) − u · (∇ ⊗ v) − v(∇ · u) ,

(A.3c) (A.3d)

∇ ∧ (u ∧ v) = ∇ · (v ⊗ u − u ⊗ v) ,

(A.3e)

∇ ∧ (∇ ∧ u) = ∇(∇ · u) − ∇ u ,

(A.3f)

∇ ∧ (T · u) = (∇ ∧ T ) · u + (∇ ⊗ u) × •T,

(A.3g)

∇ ∧ (T ∧ u) = (∇ ∧ T ) ∧ u − (∇ ⊗ u) × ×T ,

(A.3h)

∇ ∧ (I ∧ u) = u ⊗ ∇ − (∇ · u)I , I ∧ (∇ ∧ u) = u ⊗ ∇ − ∇ ⊗ u ,

(A.3i) (A.3j)

∇ ∧ (φ I) ∧ ∇ = ∇ ⊗ ∇φ − (∇ 2 φ)I ,

(A.3k)

∇ ∧ (∇φ) = 0 , ∇ ∧ (∇ ⊗ T ) = O ,   I : ∇ ∧ (T 1 · T 2 ) = T 2T : (∇ ∧ T 1 ) − T 1 : ∇ ∧ (T 2T ) .

(A.3l) (A.3m)

2

(A.3n)

Laplacian formulae: ∇ 2 (φu) = (∇ 2 φ)u + 2(∇φ) · (∇ ⊗ u) + φ(∇ 2 u) ,

(A.4a)

∇ 2 (φϕ) = φ(∇ 2 ϕ) + ϕ(∇ 2 φ) + 2(∇φ) · (∇ϕ) ,

(A.4b)

∇ (u · v) = (∇ u) · v + 2(∇ ⊗ u) : (∇ ⊗ v) + (∇ v) · u ,

(A.4c)

∇ (u ⊗ v) = (∇ u) ⊗ v + u ⊗ (∇ v) + 2(∇ ⊗ u) · (∇ ⊗ v) ,

(A.4d)

∇(∇ φ) = ∇ (∇φ) ,

(A.4e)

∇ · (∇ u) = ∇ (∇ · u) ,

(A.4f)

∇ ∧ (∇ u) = ∇ (∇ ∧ u) .

(A.4g)

2

2

2

2

2

2

2

2

2

2

2

2

T

Appendix B

Cylindrical and Spherical Coordinates

Abstract For easy reference, we record in this Appendix the expressions of the usual differential operators G R AD, D I V , and CU R L in cylindrical and spherical polar coordinates. The corresponding forms of the Cauchy’s equation of motion are included as well.

B.1

Cylindrical Coordinates

Some of the most commonly encountered differential operators introduced in Sect. 1.22.2 are expressed in cylindrical polar coordinates (r, θ, z) and listed below. The definition of these coordinates can be found in Sect. 1.17.1, while the method used in deriving them was explained in Sect. 1.22.4. Nomenclature: f ≡ f (r, θ, z) , u ≡ u(r, θ, z) = ur er + uθ eθ + uz ez ,  Ti j ei ⊗ e j . T ≡ T (r, θ, z) = i, j∈{r,θ,z}

General formulae: ∂ 1 ∂ ∂ + eθ + ez , ∂r r ∂θ ∂z ∂f 1 ∂f ∂f er + eθ + ez , ∇f = ∂r r ∂θ ∂z ∂uz 1 ∂uθ 1 ∂ (r ur ) + + . ∇·u= r ∂r r ∂θ ∂z ∇ = er

© Springer Nature B.V. 2020 C. D. Coman, Continuum Mechanics and Linear Elasticity, Solid Mechanics and Its Applications 238, https://doi.org/10.1007/978-94-024-1771-5

(B.1a) (B.1b) (B.1c)

461

462

Laplacian:

Appendix B: Cylindrical and Spherical Coordinates

1 ∂ ∇ f = r ∂r 2

  ∂f 1 ∂2 f ∂2 f r + 2 2 + 2 , ∂r r ∂θ ∂z

∇ 2 u = (∇ 2 u)r er + (∇ 2 u)θ eθ + (∇ 2 u)z ez ,   ∂uθ 1 + ur , (∇ 2 u)r = ∇ 2 ur − 2 2 r ∂θ   1 ∂u r (∇ 2 u)θ = ∇ 2 uθ + 2 2 + uθ , r ∂θ (∇ 2 u)z = ∇ 2 uz .

(B.2)

(B.3a) (B.3b) (B.3c) (B.3d)

Curl of a vector field: ∇ ∧ u = (∇ ∧ u)r er + (∇ ∧ u)θ eθ + (∇ ∧ u)z ez , ∂uθ 1 ∂uz − , (∇ ∧ u)r = r ∂θ ∂z ∂uz ∂ur − , (∇ ∧ u)θ = ∂z ∂r 1 ∂ 1 ∂ur (r uθ ) − . (∇ ∧ u)z = r ∂r r ∂θ

(B.4a) (B.4b) (B.4c) (B.4d)

Divergence of a tensor field: ∇ · T = (∇ · T )r er + (∇ · T )θ eθ + (∇ · T )z ez , ∂ Tzr 1 1 ∂ 1 ∂ Tθr (r Trr ) + + − Tθθ , (∇ · T )r = r ∂r r ∂θ ∂z r ∂ Tzθ 1 ∂ 1 ∂ Tθθ 1 + + (r Tr θ ) + Tθr , (∇ · T )θ = r ∂θ ∂z r ∂r r 1 ∂ ∂ Tzz 1 ∂ Tθ z + (r Tr z ) + . (∇ · T )z = ∂z r ∂r r ∂θ

(B.5a) (B.5b) (B.5c) (B.5d)

Gradient of a vector field: A ≡ u ⊗ ∇ = grad u =



A i j ei ⊗ e j ,

(B.6a)

i, j∈{r,θ,z}

∇ ⊗ u ≡ (u ⊗ ∇)T ,

(B.6b)

Appendix B: Cylindrical and Spherical Coordinates

∂ur , ∂r ∂uθ , Aθr = ∂r ∂uz A zr = , ∂r

Arr =

  1 ∂ur − uθ , r ∂θ   1 ∂uθ + ur , = r ∂θ 1 ∂uz = , r ∂θ

Ar θ = Aθθ A zθ

463

∂ur , ∂z ∂uθ Aθ z = , ∂z ∂uz A zz = . ∂z Ar z =

(B.7a) (B.7b) (B.7c)

Cauchy’s first equation of motion: div σ + ρb = ρa: 1 ∂σθr ∂σzr 1 ∂σrr + + + (σrr − σθθ ) + ρbr = ρar , ∂r r ∂θ ∂z r 1 ∂σθθ ∂σzθ 1 ∂σr θ + + + (σr θ + σθr ) + ρbθ = ρaθ , ∂r r ∂θ ∂z r 1 ∂σθ z ∂σzz 1 ∂σr z + + + σr z + ρbz = ρaz , ∂r r ∂θ ∂z r

(B.8a) (B.8b) (B.8c)

where the acceleration a = ar er + aθ eθ + az ez can be given in terms of the Eulerian description of the velocity field v by using the formula a = ∂v/∂t + (v · ∇)v, vθ ∂vr v2 ∂vr ∂vr ∂vr + vr + + vz − θ , ∂t ∂r r ∂θ ∂z r vθ ∂vθ vr vθ ∂vθ ∂vθ ∂vθ + vr + + vz + , aθ = ∂t ∂r r ∂θ ∂z r vθ ∂vz ∂vz ∂vz ∂vz + vr + + vz . az = ∂t ∂r r ∂θ ∂z ar =

(B.9a) (B.9b) (B.9c)

Infinitesimal strain tensor E: E=

 1 (∇ ⊗ u + u ⊗ ∇) ≡ E i j ei ⊗ e j , 2 i, j∈{r,θ,z}

ur ∂ur 1 ∂uθ ∂uz , E θθ = + , E zz = , ∂r r ∂θ r ∂z     ∂uθ uθ ∂uz 1 1 ∂ur 1 ∂ur + − , Er z = + , = 2 r ∂θ ∂r r 2 ∂z ∂r   1 ∂uz 1 ∂uθ + . = 2 ∂z r ∂θ

(B.10)

Err =

(B.11a)

Er θ

(B.11b)

Eθ z

(B.11c)

464

Appendix B: Cylindrical and Spherical Coordinates

B.2

Spherical Coordinates

The information included below is similar to that of the previous sections, but the same differential operators are now expressed in spherical polar coordinates (see Sect. 1.17.2 for definition). Nomenclature: f ≡ f (r, θ, ϕ) , u ≡ u(r, θ, ϕ) = ur er + uθ eθ + uϕ eϕ ,  Ti j ei ⊗ e j . T ≡ T (r, θ, ϕ) = i, j∈{r,θ,ϕ}

General formulae: ∂ ∂ 1 ∂ 1 + eθ + eϕ , ∇ = er ∂r r ∂θ r sin θ ∂ϕ ∂f 1 ∂f 1 ∂f ∇f = er + eθ + eϕ , ∂r r ∂θ r sin θ ∂ϕ  ∂  1 ∂uϕ 1 1 ∂  2  r ur + uθ sin θ + . ∇·u= 2 r ∂r r sin θ ∂θ r sin θ ∂ϕ

(B.12a) (B.12b) (B.12c)

Laplacian: 1 ∂ ∇ f = 2 r ∂r 2

    1 ∂ ∂f 1 ∂2 f 2∂f r + 2 sin θ + 2 2 . ∂r r sin θ ∂θ ∂θ r sin θ ∂ϕ 2

∇ 2 u = (∇ 2 u)r er + (∇ 2 u)θ eθ + (∇ 2 u)ϕ eϕ ,   1 ∂uϕ 2 ∂uθ + , (∇ 2 u)r = ∇ 2 ur − 2 ur + uθ cot θ + r ∂θ sin θ ∂ϕ   ∂ur uθ cos θ ∂uϕ 1 − , −2 2 (∇ 2 u)θ = ∇ 2 uθ + 2 2 2 r ∂θ sin θ sin θ ∂ϕ   ∂ur ∂uθ uϕ 1 2 2 2 + 2 cot θ − . (∇ u)ϕ = ∇ uϕ + 2 r sin θ ∂ϕ ∂ϕ sin θ

(B.13)

(B.14a) (B.14b) (B.14c) (B.14d)

Curl of a vector field: ∇ ∧ u = (∇ ∧ u)r er + (∇ ∧ u)θ eθ + (∇ ∧ u)ϕ eϕ , ∂ 1 1 ∂uθ (uϕ sin θ ) − , (∇ ∧ u)r = r sin θ ∂θ r sin θ ∂ϕ 1 ∂ur 1 ∂ (∇ ∧ u)θ = − (r uϕ ) , r sin θ ∂ϕ r ∂r

(B.15a) (B.15b) (B.15c)

Appendix B: Cylindrical and Spherical Coordinates

465

1 ∂ 1 ∂ur (r uθ ) − . r ∂r r ∂θ

(B.15d)

(∇ ∧ u)ϕ =

Divergence of a tensor field: ∇ · T = (∇ · T )r er + (∇ · T )θ eθ + (∇ · T )ϕ eϕ , 1 (∇ · T )r = 2 r 1 (∇ · T )θ = 2 r 1 (∇ · T )ϕ = 2 r

(B.16)

∂ 1 ∂ 2 1 1 ∂ Tϕr (r Trr ) + (Tθr sin θ ) + − (Tθθ + Tϕϕ ) , ∂r r sin θ ∂θ r sin θ ∂ϕ r ∂ Tθr cos θ ∂ 2 1 1 ∂ Tϕθ (r Tr θ ) + (Tθθ sin θ ) + + − Tϕϕ , ∂r r sin θ ∂θ r sin θ ∂ϕ r r 1 ∂ 2 1 ∂ 1 ∂ Tϕϕ (r Tr ϕ ) + (Tθϕ sin θ ) + + (Tϕr + Tϕθ ) . ∂r r ∂θ r sin θ ∂ϕ r

Gradient of a vector field: A ≡ u ⊗ ∇ = grad u =



A i j ei ⊗ e j ,

(B.17a)

i, j∈{r,θ,ϕ}

∇ ⊗ u ≡ (u ⊗ ∇)T , ∂ur , ∂r ∂uθ Aθr = , ∂r ∂uϕ Aϕr = , ∂r Arr =

  1 ∂ur − uθ , r ∂θ   1 ∂uθ Aθθ = + ur , r ∂θ 1 ∂uϕ Aϕθ = , r ∂θ Ar θ =

(B.17b) uϕ 1 ∂ur − , r sin θ ∂ϕ r uϕ 1 ∂uθ − cot θ , Aθϕ = r sin θ ∂ϕ r 1 ∂uϕ uθ ur Aϕϕ = + cot θ + . r sin θ ∂ϕ r r Ar ϕ =

Cauchy’s first equation of motion: div σ + ρb = ρa: 1 ∂σθr 1 ∂σϕr ∂σrr + + ∂r r ∂θ r sin θ ∂ϕ  1 + 2σrr − σθθ − σϕϕ + σθr cot θ + ρbr = ρar , r 1 ∂σθθ 1 ∂σϕθ ∂σr θ + + ∂r r ∂θ r sin θ ∂ϕ  1 3σr θ + (σθθ − σϕϕ ) cot θ + ρbθ = ρaθ , + r

∂σr ϕ 1 ∂σθϕ 1 ∂σϕϕ + + ∂r r ∂θ r sin θ ∂ϕ  1 + 3σr ϕ + (2σϕθ ) cot θ + ρbϕ = ρaϕ , r

(B.19)

(B.20)

(B.21)

466

Appendix B: Cylindrical and Spherical Coordinates

where the acceleration a = ar er + aθ eθ + az ez can be given in terms of the Eulerian description of the velocity field v by using the formula a = ∂v/∂t + (v · ∇)v, vθ2 + vϕ2 ∂vr ∂vr vθ ∂vr vϕ ∂vr + vr + + − , ∂t ∂r r ∂θ r sin θ ∂ϕ r vϕ2 vθ ∂vθ vϕ ∂vθ vr vθ ∂vθ ∂vθ + vr + + + − cot θ , aθ = ∂t ∂r r ∂θ r sin ϕ ∂ϕ r r ∂vϕ ∂vϕ vθ ∂vϕ vϕ ∂vϕ vr vϕ vθ vϕ cot θ aϕ = + vr + + + + . ∂t ∂r r ∂θ r sin ϕ ∂ϕ r r ar =

(B.22a) (B.22b) (B.22c)

Infinitesimal strain tensor E: E=

 1 (∇ ⊗ u + u ⊗ ∇) ≡ E i j ei ⊗ e j , 2 i, j∈{r,θ,ϕ}

ur ∂ur 1 ∂uθ , E θθ = + , ∂r r ∂θ r 1 ∂uϕ uθ ur E ϕϕ = + cot θ + , r sin θ ∂ϕ r r   ∂uθ uθ 1 1 ∂ur + − , Er θ = 2 r ∂θ ∂r r   1 ∂ur ∂uϕ uϕ 1 + − , Er ϕ = 2 r sin θ ∂ϕ ∂r r   1 ∂uθ 1 1 ∂uϕ uϕ E θϕ = + − cot ϕ . 2 r sin θ ∂ϕ r ∂θ r Err =

(B.23)

(B.24a) (B.24b) (B.24c) (B.24d) (B.24e)

Appendix C

Geometry of Areas

Abstract In the solution of the Saint-Venant problem (e.g., see Sect. 5.11.2) it is necessary to take into account certain geometrical features of the transverse cross sections for slender cylindrical or prismatic bodies. In particular, such features include the centroid of a plane area and its principal inertia axes. The material we touch upon below provides a brief overview of the main definitions and formulas pertaining to these concepts. Particular attention will be given to making the presentation consistent with the tensorial notation used in the rest of the book. We shall be interested exclusively in two-dimensional bounded domains, Σ ⊂ E2 (say); the position vector of an arbitrary point in Σ will be denoted by X ⊥ , the origin of this vector being tacitly assumed to lie in the same plane. If a Cartesian system of coordinates X 1 O X 2 is chosen in E2 , e1 and e2 will correspond to the standard vectors that define the directions of the X 1 - and X 2 -axes, respectively; in this case X ⊥ = X 1 e1 + X 2 e2 , for some X j ∈ R ( j = 1, 2). Definition C.1 (Tensor of inertia) Let Σ be a bounded planar region and O an arbitrary point in the plane of Σ. The tensor of inertia of Σ relative to O, denoted by I O , corresponds to the expression I O :=

 Σ

 |X ⊥ |2 I 2 − X ⊥ ⊗ X ⊥ d A ,

(C.1)

where I 2 is the usual two-dimensional identity tensor, and X ⊥ denotes the position vector (with respect to O ∈ E2 ) of a generic point in Σ. Equation (C.1) can also be written in the form I O = IO I 2 − E O , in terms of the two quantities © Springer Nature B.V. 2020 C. D. Coman, Continuum Mechanics and Linear Elasticity, Solid Mechanics and Its Applications 238, https://doi.org/10.1007/978-94-024-1771-5

467

468

Appendix C: Geometry of Areas

I O :=

|X ⊥ | d A 2

Σ

and

E O :=

Σ

X⊥ ⊗ X⊥ d A ,

known as the polar and the Euler moments of inertia, respectively. In the literature the prevalent notation for I O is J0 (which is also what we have used in the earlier parts of the book). Definition C.2 If Σ ⊂ E2 is a bounded domain, its centroid C is defined as the geometric point in the plane of Σ whose position vector X C is given by1 1 X C := area(Σ)

Σ

X⊥ d A .

(C.2)

The centroid is an intrinsic property of the domain Σ, in the sense that it is altogether independent of the choice of any system of coordinates. If Y ⊥ is the position vector with respect to C of an arbitrary point in Σ, it follows immediately from formula (C.2) that Σ

Y⊥ dA = 0 .

(C.3)

This observation can be used to understand how the moments of inertia I O and I C are related to each other; the next result clarifies this aspect. Proposition C.1 (Huygens–Steiner) Let Σ ⊂ E2 be a bounded planar domain. If O ∈ E2 is an arbitrary point and C ∈ E2 denotes the centroid of Σ, then   I O = I C + Ad 2 I 2 − d ⊗ d ,

(C.4)

−−→ −−→ where A = area(Σ), d is a unit vector along OC , and d = | OC | . I C is known as the centroidal (or central) tensor of inertia. The justification of −−→ (C.4) requires rewriting (C.1) by replacing X ⊥ = OC + Y ⊥ , expanding the various expressions therein, and then taking (C.3) into account; these routine calculations are omitted here. Definition C.3 (Moment of inertia) (a) Let Δ be a line passing through a point O and contained within the plane of the bounded domain Σ ⊂ E2 . If u is a unit direction vector for this line, then the moment of inertia, IΔ , of Σ about the line Δ, is defined as the real number (C.5) IΔ := u · I O · u > 0 . (b) Let Δ and Δ be two perpendicular lines in the plane of Σ (as above), which intersect in a point O and have unit direction vectors u and v, respectively. The 1 The

integral on the right-hand side of (C.2) is known as the static moment vector relative to Σ.

Appendix C: Geometry of Areas

469

Fig. C.1 Left: illustration of the Parallel-Axis Theorem—see formulae (C.9) and (C.10). Right: a particular case corresponding to (C.11) and (C.12)

product of inertia, IΔΔ , of the area Σ with respect to the foregoing axes, corresponds to the scalar (C.6) IΔΔ := u · I O · v . As a particular case, let us consider e1 and e2 , the two unit vectors associated with the axes of the Cartesian system of coordinates X 1 O X 2 . In this case, IX1 X1 =

Σ

X 22

dA

IX2 X2 =

and

Σ

X 12 d A

(C.7)

represent the moments of inertia of Σ about the X 1 - and X 2 -axes, respectively. The products of inertia relative to the two coordinate axes are IX1 X2 = IX2 X1 = −

Σ

X1 X2 d A .

(C.8)

We record below a number of further properties related to the above concepts. The first such result is known as the Parallel-Axis Theorem; if Δ passes through the centroid of Σ, and Δ1 is another line parallel to Δ, the use of (C.4) shows that IΔ1 = IΔ + Ad12 ,

(C.9)

with d1 being the perpendicular distance between Δ and Δ1 . As a corollary, IΔ2 = IΔ1 + A(d22 − d12 ) ,

(C.10)

where Δ j ( j = 1, 2) are two arbitrary parallel lines in the plane of Σ, and d j represent the distances between Δ and Δ j ( j = 1, 2); see the left sketch included in Fig. C.1. If we consider two parallel systems of coordinates X 1 O X 2 and Y1 C Y2 , a direct application of (C.9) leads to

470

Appendix C: Geometry of Areas

I X 1 X 1 = IY1 Y1 + Aa 2 ,

I X 2 X 2 = IY2 Y2 + Ab2 ,

(C.11)

where (a, b) are the centroidal coordinates of Σ (the lines CY1 and CY2 are called centroidal axes); the right sketch in Fig. C.1 illustrates the geometrical setting for these formulae. The counterpart of (C.11) for the products of inertia (C.8) can also be found, namely, (C.12) I X 1 X 2 = IY1 Y2 + Aab . Two important remarks can be made vis-à-vis (C.11) and (C.12). First, the centroidal moment of inertia is the minimum of all those corresponding to an infinite number of parallel lines. Second, if a = 0 or b = 0 then the product of inertia remains unchanged. Note that a and b can be positive or negative, so the aforementioned minimum property does not hold for products of inertia. The inertia tensor (C.1) is clearly symmetric. By the Spectral Representation Theorem (see Sect. 1.20) there exists a set of two orthonormal vectors {a1 , a2 } such that I O admits a diagonal representation in this basis. The lines passing through O and parallel to a j ( j = 1, 2) are the principal axes of inertia of Σ relative to O. If Ik represents the eigenvalue of I O corresponding to the eigenvector ak (k = 1, 2), then the inertia tensor will assume the form I O = I1 a 1 ⊗ a 1 + I2 a 2 ⊗ a 2 . The numbers I1 , I2 ≥ 0 (I12 + I22 = 0) are known as the principal moments of inertia of Σ relative to O; they are in fact the moments of inertia of Σ about the principal axes, i.e. Iak = Ik for k = 1, 2. The existence of a system of Cartesian axes relative to which the product of inertia vanishes can also be established by employing the general transformation formulae for a tensor (see Sect. 1.9). For example, we can use the results of Problem 8.15 from Chap. 8 to deduce, 1 1 (I X 1 X 1 + I X 2 X 2 ) + (I X 1 X 1 − I X 2 X 2 ) cos 2θ − I X 1 X 2 sin 2θ , 2 2 1 1 = (I X 1 X 1 + I X 2 X 2 ) − (I X 1 X 1 − I X 2 X 2 ) cos 2θ + I X 1 X 2 sin 2θ , 2 2 1 = (I X 1 X 1 − I X 2 X 2 ) sin 2θ + I X 1 X 2 cos 2θ , 2

I X 1 X 1 = I X 2 X 2

I X 1 X 2

where X 1 O X 2 and X 1 O X 2 are two Cartesian systems of coordinates as in Fig. C.2. To choose the latter system so that I X 1 X 2 = 0 would amount to selecting a rotation angle, 0◦ ≤ θ ≤ 90◦ , that solves the equation2

this equation applies only if I X 2 X 2 = I X 1 X 1 . If this requirement is violated, the principal axes will correspond to the bisectors of the first two quadrants; that is, one principal axis makes an angle of 45◦ with the positive direction of the X 1 -axis, while for the other this angle is 135◦ .

2 Clearly,

Appendix C: Geometry of Areas

471

Fig. C.2 Rotation of the Cartesian axes for finding the angle θ which leads to the vanishing of the product of inertia (C.8); see text for further details

tan 2θ =

2I X 1 X 2 . IX2 X2 − IX1 X1

(C.13)

Note that (C.13) defines two values 2θ which are 180◦ apart, and thus two values of θ which are 90◦ apart. These angles specify the directions of the principal axes relative to the initial Cartesian system X 1 O X 2 . What has been said above about the principal axes of inertia is true for any point O located in the plane of the Σ. If this point is chosen to coincide with the centroid of this planar domain, the two principal axes of inertia about its centroid are referred to as the principal centroidal axes of Σ. In relation to (C.3) it is worth emphasising that the static moment vector is zero with respect to any pair of centroidal orthogonal axes. However, if the origin of the system of coordinates does not coincide with the centroid C of Σ, there is no rotation as in Fig. C.2 which could possibly render the static moment vector identically zero. The Huygens–Steiner formula (C.4) can help us draw some general conclusions about the eigenvectors of I O and I C . Let us note that the case when the unit vector d is an eigenvector of I C corresponds to O being situated on one of the principal centroidal axes. Since I O · d = I C · d, it is clear that d will be an eigenvector for I O , too. Since the other eigenvector of I C is orthogonal to d, from (C.4) it also follows that this will be an eigenvector for I O , albeit associated with a different eigenvalue. Therefore, we conclude that I O and I C share the same eigenvectors or, in other words, the principal axes of inertia for I O are parallel to the principal centroidal axes. The locations of the centroid and the principal axes of inertia are particularly easy to work out if the region Σ enjoys certain symmetry properties. For instance, consider the case of axial symmetry. Mathematically, this means that there exists an axis Δ ⊂ E2 (in the plane of Σ) defined by a unit direction vector s, and such that: if P ∈ Σ then P∗ ∈ Σ; here, the point P∗ is defined by the conditions −−→ −−→ O P∗ = SΔ · O P ,

SΔ ≡ 2s ⊗ s − I 2 ,

(C.14)

472

Appendix C: Geometry of Areas

Fig. C.3 The role of axial symmetry. The sketch on the left illustrates the axial symmetry condition (C.14). Shown in the middle is a planar domain with one axis of symmetry (the X 1 -axis), while the example on the right has two axes of symmetry (both coordinate axes)

Fig. C.4 Polar symmetry of a planar domain Σ is characterised by the existence of a point C (known as the centre of symmetry) with the following property. If P ∈ Σ (arbitrary) and P∗ is −−→ −−→ constructed such that PC = C P∗ , then P∗ ∈ Σ. Note that this particular shape does not admit any axial symmetry

with O ∈ Δ an arbitrary point (see Fig. C.3). For such axially symmetric planar regions the centroid lies on the axis of symmetry, which is also one of the principal centroidal axes; the other principal axis is obviously perpendicular to Δ. If there are two or more axes of symmetry, as the centroid must be situated on each of them, it coincides with their intersection; in this case the centroid can be determined by inspection. A region Σ is said to have polar symmetry if it is symmetric about a point (or “pole”), called the centre of symmetry. This has the property that every line through that point intersects Σ in a symmetrical manner. For regions with polar symmetry the centroid coincides with the centre of symmetry (see Fig. C.4).

Appendix C: Geometry of Areas

473

Regarding the comments made above about the interplay between symmetry, centroids, and moments of inertia, a caveat is necessary. A principal axis does not need to be an axis of symmetry; irrespective of whether a planar region possesses any axes of symmetry, it will invariably have two principal axes of inertia about any point O. In keeping with the widely accepted nomenclature in Mechanics we have repeatedly employed the appellative ‘inertia’ in a somewhat improper way. This requires some further clarification. Strictly speaking, the inertia of a body is one of the primary manifestations of its mass (a property that was touched upon in Chap. 3). With this in mind, the expressions (C.8) would be more aptly called second moments of area rather than moments of inertia, since the latter terminology is typically used when the integration is with respect to the element of mass ‘dm’ (instead of ‘d A’, the element of area). However, as dm = ρd A, if the density per unit area (i.e., ρ) is constant (as it is the case for homogeneous bodies), then

Σ

X 2j dm =

Σ

ρ X 2j d A = ρ

Σ

X 2j d A ,

( j = 1, 2) .

Thus, the (true) moment of inertia is proportional to the earlier integral expressions (C.8), and it is for this reason that our terminology is justified. A clearer distinction is sometimes drawn by referring to the former as mechanical moments of inertia, while the latter are labelled geometrical moments of inertia. Our definition of the inertia tensor in (C.1) was two-dimensional because we were exclusively interested in geometrical properties for the transverse cross sections of slender cylindrical or prismatic bodies. By contrast, the general inertia tensor used in Classical Mechanics is three-dimensional and is associated with regions in E3 . Details about the properties of this tensor and its uses are well documented and can be found in any book dealing with the mechanics of rigid bodies (e.g., see [9] for a modern treatment); reference [10] deals with discrete mechanical systems and uses a notation similar to that in this book. The calculation of the moments of inertia for various cross-sectional shapes is usually discussed in first-year Calculus courses as an application to double integrals (e.g., see [11, 12]). In addition, engineering textbooks such as [13, 14] contain numerous detailed examples and tables of the moments of inertia for some of the most commonly encountered shapes.

Appendix D

Fourier Transforms

Abstract This Appendix collects information on Fourier integral transforms which is required for reading Chap. 9. Several bibliographical recommendations are made at various points in the presentation, for the benefit of those who are unfamiliar with this key mathematical technique.

D.1

Definitions and General Properties

Let f : R → R be a continuous and differentiable function except at a finite number of points; for any discontinuity point x0 ∈ R we set f (x0+ ) := lim+ f (x0 + ε)

f (x0− ) := lim+ f (x0 − ε) .

and

ε→0

ε→0

Fourier Integral Theorem: Let f and f be piecewise

+∞ continuous functions on every interval in R. If f is absolutely integrable, i.e., −∞ | f (x)| dx < +∞, then 1 f (x) = 2π



+∞

e −∞

−iαx





+∞

e

iαξ

−∞

f (ξ ) dξ dα

(D.1)

at points where f is continuous; if x is a point of discontinuity for f , the left-hand side of (D.1) must be replaced by (1/2)[ f (x + ) + f (x − )]. Functions that satisfy the conditions of this theorem will be referred to as ‘good functions’. The above formula motivates the introduction of the Fourier transform   1 f¯(α) ≡ F f (x); α := √ 2π



+∞ −∞

f (x) eiαx dx ,

(D.2)

and its corresponding inverse, © Springer Nature B.V. 2020 C. D. Coman, Continuum Mechanics and Linear Elasticity, Solid Mechanics and Its Applications 238, https://doi.org/10.1007/978-94-024-1771-5

475

476

Appendix D: Fourier Transforms

  1 f (x) ≡ F−1 f¯(α); x := √ 2π



+∞

−∞

f¯(α) e−iαx dα .

(D.3)

The Eq. (D.1) can be simplified when f is either even or odd. In the former case, f (x) = f (−x) for all x ∈ R, and (D.1) becomes f (x) =

2 π





∞

∞

cos(αx) 0

f (ξ ) cos(αξ ) dξ dα .

(D.4)

0

We can thus introduce the so-called Fourier cosine transform 



FC (α) ≡ FC f (x); α :=

2 π



∞

f (x) cos(αx) dx ,

α > 0,

(D.5)

0

and its inverse, f (x) =

FC−1





FC (α); x :=

2 π



∞

FC (α) cos(αx) dα ,

x > 0.

(D.6)

0

In a similar way we can define the Fourier sine transform when the function is odd. Since f (−x) = − f (x) for all x ∈ R, (D.1) now becomes f (x) =

2 π





∞

∞

sin(αx) 0

f (ξ ) sin(αξ ) dξ dα .

(D.7)

0

This pair of transforms are defined according to 



FS (α) ≡ F S f (x); α :=

2 π



∞

f (x) sin(αx) dx ,

α > 0,

(D.8)

0

and f (x) =

F−1 S





FS (α); x :=

2 π



∞

FS (α) sin(αx) dα ,

x > 0.

(D.9)

0

In the case when the original function f is defined only on (0, ∞) both the sine and cosine transforms might exist because we can construct the even and odd extensions of this function to the entire real axis. The even extension is defined by setting f even (x) := f (|x|) for all x ∈ R, and it can be shown that     F f even (x); α = FC f (x); α ,

−∞ < α < +∞ .

(D.10)

The odd extension corresponds to f odd (x) := f (|x|)sgn(x) for x ∈ R, where the signum function (‘sgn’) is defined in the usual way: (−1) for x < 0 and (+1) for x > 0. In this case,

Appendix D: Fourier Transforms

477

    F f odd (x); α = iF S f (x); |α| sgn(α) ,

−∞ < α < +∞ .

(D.11)

Shifting property: If f = f (x) is a good function and a ∈ R,     F eiax f (x); α = F f (x); α + a ;     F f (x − a); α = eiaα F f (x); α .

(D.12a) (D.12b)

Differentiation property: Assume that the function f and its first (k − 1) derivatives f ( j) (1 ≤ j ≤ k − 1) are continuous everywhere and absolutely integrable. If f ( j) also satisfy lim|x|→∞ f ( j) (x) = 0 for 0 ≤ j ≤ k − 1, then     F f (k) (x); α = (−iα)k F f (x); α .

(D.13)

Convolution property: If f = f (x) and g = g(x) are two good functions, their convolution is a new function denoted by f  g and defined according to 1 ( f  g)(x) := √ 2π



+∞

−∞

f (ξ )g(x − ξ ) dξ .

If F(α) := F[ f (x); α] and G(α) := F[g(x); α], then   F−1 F(α)G(α); x = ( f  g)(x)

(D.14)

or, in expanded form,

+∞

−∞

D.2

F(α)G(α) e−iαx dα =



+∞ −∞

f (ξ )g(x − ξ ) dξ .

(D.15)

Fourier Transforms for Multivariate Functions

Fourier transforms of functions depending on more than one variable are not uncommon in Linear Elasticity. For instance, such transforms are relevant to the study of concentrated forces or moments acting inside an elastic plane; other examples include three-dimensional situations such as the various half-space problems discussed in [4, 15]. We limit the scope of the following presentation to sketching only a few broad features of the two-dimensional Fourier transform. Assuming that f (x1 , x2 ) is a function defined for −∞ < x1 < +∞, one can introduce the following pair of Fourier transforms,   1 f¯(α1 ; x2 ) ≡ F1 f (x1 , x2 ); α1 := √ 2π



+∞ −∞

f (x1 , x2 ) eiα1 x1 dx1 , (D.16a)

478

Appendix D: Fourier Transforms

  1 f (x1 , x2 ) = F1−1 f¯(α1 ; x2 ); x1 := √ 2π



+∞

−∞

f¯(α1 ; x2 ) e−iα1 x1 dα1 , (D.16b)

provided that the above integrals exist. Similarly, for a function f (x1 , x2 ) defined for −∞ < x2 < +∞ one can consider the Fourier transform pair, +∞   1 f (x1 , x2 ) eiα2 x2 dx2 , (D.17a) f¯(x1 ; α2 ) ≡ F2 f (x1 , x2 ); α2 := √ 2π −∞ +∞   1 f¯(x1 ; α2 ) e−iα2 x2 dα2 . (D.17b) f (x1 , x2 ) = F2−1 f¯(x1 ; α2 ); x2 := √ 2π −∞   We shall refer to F j f (x1 , x2 ); α j as the partial Fourier transforms of f (x1 , x2 ) with respect to x j ( j = 1, 2); clearly, the expressions in (D.16b) and (D.17b) represent the corresponding inverse operations. Everything that was said in the previous section remains valid for partial Fourier transforms as well. The definition of the partial transforms mentioned above requires at least one of the independent variables x j ( j = 1, 2) being allowed to move freely between −∞ and +∞. If both independent variables satisfy this requirement, i.e. −∞ < x j < +∞ for j = 1, 2, then one can apply (D.17a) to the function x2 → f¯(α1 ; x2 ). The result will be a new function,  f =  f (α1 , α2 ), which no longer depends on x1 or x2 ,   1  f (α1 , α2 ) := F2 f¯(α1 , x2 ); α2 = √ 2π or

  1  f (α) ≡ F2D f (x); α := 2π



+∞



+∞ −∞



−∞

+∞

−∞

f¯(α1 , x2 ) eiα2 x2 dx2

(D.18)

f (x) eiα·x dx ,

(D.19)

where we have set α ≡ (α1 , α2 ), x ≡ (x1 , x2 ), and dx ≡ dx1 dx2 . The Eq. (D.19) defines the two-dimensional Fourier transform of the function f (x) ≡ f (x1 , x2 )   (indicated by the notation F2D f (x); α ). The corresponding inversion formula is immediately obtained from the repeated application of (D.17b) and (D.16b), and can be stated in the form +∞ +∞   1 −1   (D.20) f (α) e−iα·x dα . f (x) ≡ F2D f (α); x := 2π −∞ −∞ Convolution property for multivariate functions: If f = f (x) and g = g(x) are two functions that admit two-dimensional Fourier transforms, then their convolution f  g is defined according to ( f  g)(x) :=

1 2π



+∞ −∞



+∞ −∞

f ( y)g(x − y) d y ,

Appendix D: Fourier Transforms

479

where y ≡ (y1 , y2 ) and d y ≡ dy1 dy2 . The earlier result (D.14) also holds true for multivariate functions: if F(α) := F2D [ f (x); α] and G(α) := F2D [g(x); α], then   −1 F(α)G(α); x = ( f  g)(x) F2D

(D.21)

or, in expanded form,

+∞



+∞

F(α)G(α) e −∞

−∞

−iα·x

dα =

+∞ −∞



+∞ −∞

f ( y)g(x − y) d y .

(D.22)

The basic theory of Fourier transforms is included in most textbooks on advanced engineering mathematics or elementary partial differential equations (e.g., see [16– 18]). Sneddon’s text [19] still remains one of the best intermediate references on the subject, due to a balanced mixture of theory and interesting examples (some of which come from Linear Elasticity). Other useful books on the general theory include [20, 21], while references [4, 15, 22–24] deal with various applications to Linear Elasticity.

D.3

The Dirac Delta Function

Concentrated forces in Continuum Mechanics require special treatment because the area over which they act is, strictly speaking, zero. To overcome such shortcomings, historically, concentrated forces have been introduced through an ad-hoc limiting process. By first considering the mechanical response to a uniform load distribution p (say) over a small area A (say), one can investigate the limit of the obtained solution by letting A → 0 and p → ∞, while keeping fixed the resultant P ≡ pA. The response to the concentrated force P is then “defined” as the limit of that solution. These heuristic arguments can be made rigorous by introducing the concept of generalised functions and, in particular, the Dirac delta function. We give a quick review of the latter. An important role in the mathematical representation of piecewise functions is played by the so-called Heaviside function,  H (x) =

1 , if x > 0 , 0 , if x < 0 .

(D.23)

Note that if x0 ∈ R, then H (x − x0 ) = 0 for x < x0 and H (x − x0 ) = 1 for x > x0 . This property can be used to represent any given piecewise expression as a linear combination of suitably chosen Heaviside functions. Let us now return to what was said in the opening paragraph above, with the aim of constructing a unit one-dimensional concentrated force at a point x = a (say). To this end we consider the uniform load distribution (see Fig. D.1)

480

Appendix D: Fourier Transforms

area = 1

Fig. D.1 Rectangular “pulse” function used for defining the Dirac delta function

 Pa (x, ε) =

1/2ε , a − ε < x < a + ε , 0, otherwise ,

(D.24)

which can also be expressed in terms of the Heaviside function (D.23) as Pa (x, ε) =

 1 H (x − a + ε) − H (x − a − ε) . 2ε

Clearly, the resultant of this load will be P :=

+∞ −∞

Pa (x, ε) dx = 1 ,

(∀) ε > 0 .

(D.25)

If we now “shrink” the region over which Pa acts, i.e. let ε → 0+ , the form (D.24) will peak strongly at x = a; in the limit, Pa becomes an infinitely high, thin spike at x = a. With this in mind, we can loosely define an idealised concentrated unit force, δ(x − a), by writing (D.26) δ(x − a) := lim+ Pa (x, ε) , ε→0

the expression δ(x − a) being known as the Dirac delta function (at x = a). From a purely mathematical standpoint the situation is unsatisfactory because the above limit does not exist; that is, δ(x − a) is not a conventional mathematical function. Such difficulties could be avoided by not proceeding to the limit ε = 0 until after the completion of whatever mathematical process is to be applied to δ(x − a). In this sense, the notation on the left-hand side in (D.26) corresponds to a symbolic representation of a deferred limiting process. In light of the above remarks, δ(x − a) can be defined in terms of its behaviour as part of an integrand, in the sense that the meaning of (D.26) is taken to be

+∞

−∞

δ(x − a) f (x) dx := lim+ ε→0

+∞ −∞

Pa (x, ε) f (x) dx ,

(D.27)

Appendix D: Fourier Transforms

481

for all continuous and bounded functions f (x). This is the conventional point of view adopted in Mathematics, according to which δ(x − a) is a generalised function (also known as a distribution). To illustrate the possible simplification of (D.27), let us evaluate the right-hand limit. Thus,

+∞ −∞

δ(x − a) f (x) dx = lim+ ε→0

+∞

−∞

1 = lim+ ε→0 2ε



Pa (x, ε) f (x) dx a+ε

f (x) dx .

(D.28)

a−ε

According to the mean value theorem of integral calculus,

B

f (x) dx = f (ζ )(B − A) , for some ζ ∈ (A, B) .

A

Using this result in (D.28),

+∞

δ(x − a) f (x) dx = lim+ ε→0

−∞

1 (2ε) f (ζ ) = lim+ f (ζ ) , ε→0 2ε

a −ε < ζ < a +ε;

since f is continuous and ζ → a as ε → 0, we conclude that

+∞

−∞

δ(x − a) f (x) dx = f (a) ,

(D.29)

which is known as the sifting property of the Dirac delta function. In the Physics literature it is customary to introduce δ(x − a) by the following two defining properties, δ(x − a) = 0 , x = a , +∞ δ(x − a) dx = 1 ;

(D.30a) (D.30b)

−∞

it is clear that (D.30b) is just a particular case of (D.29). The derivative of δ(x), denoted by δ (x), is not directly determined from the definition of δ(x) itself. Instead, one defines it so that δ (x) = 0 for x = 0, and so that the ordinary rules for integration by parts continue to hold for the product of δ (x) with any differentiable function f (x) that satisfies f (x) → 0 as |x| → ∞. For example,

+∞ −∞

 x=+∞ δ (x) f (x) dx := f (x)δ(x) − x=−∞

+∞

−∞

δ(x) f (x) dx = − f (0) (D.31)

482

Appendix D: Fourier Transforms

or, more generally,



+∞

−∞

δ (x − a) f (x) dx = − f (a) .

Higher order derivatives δ (n) (x) (n = 2, 3, . . . ) can be introduced similarly; they are required to vanish for x = 0, and so that, if the function f (x) and its first n derivatives are continuous and satisfy f ( j) (x) → 0 for |x| → ∞ ( j = 0, 1, . . . , n), then

+∞

−∞

δ (n) (x) f (x) dx := (−1)n



+∞ −∞

δ(x) f (n) (x) dx = (−1)n f (n) (0) .

(D.32)

Other useful properties for δ(x): 1 1. δ(ax) = δ(x) for all a ∈ R, a = 0; |a| 2. if g(x) is a monotonic function for which g(a) = 0, then δ(g(x)) =

1 δ(x − a) , |g (a)|

3. δ(x 2 − a 2 ) =

g ≡ dg/d x ;

 1  δ(x + a) + δ(x − a) ; 2|a|

dH , where H = H (x) is the Heaviside function (D.23); dx 5. δ (−x) = −δ (x); d sgn(x) = 2δ(x); 6. dx 7. if f (x) is continuously differentiable, 4. δ(x) =

 d  δ(x) f (x) = δ(x) f (x) + δ (x) f (x) ; dx 8. the delta function can be represented symbolically in the form δ(x) =

1 2π



+∞

−∞

eiαx dα

or

δ(α) =

1 2π



+∞ −∞

eiαx dx .

We close this section with two further observations. First, it must be emphasised that all of the above properties have meaning only when considered under an integral sign. Second, the delta function was introduced as the limit as ε → 0+ of the particular family of functions Pa (x, ε) defined in (D.24). Many other functions can be used for a limiting representation of δ(x). For example,  2   x −1/2 δ(x) = lim+ (π ε) exp − , ε→0 ε

Appendix D: Fourier Transforms

483

 δ(x) = lim

or

ε→0

ε , π(ε2 + x 2 )

and we mention in passing that

+∞ −∞

 2 √ x dx = π ε , exp − ε



+∞ −∞

ε2

dx π = . 2 +x ε

Note that these last two properties mirror closely (D.25). The application of integral Fourier transforms in Linear Elasticity is deceptively simple. Unfortunately, some of the representations obtained via this approach turn out to involve improper integrals that are not convergent in the usual sense. Such difficulties can be circumvented by working with distributions (rather than standard functions), and re-interpreting the results within the scope of that theory. Basic introductions to this topic are given, for example, in [17, 25], while a much broader modern treatment can be found in Kanwal’s book [26]. The classic references in this area include [27–29], all of which are significantly more abstract than the books already mentioned.

D.4

The Use of Complex Variables

Recall that a complex function f is √ a function that depends on a complex variable z := x + iy, with x, y ∈ R and i ≡ −1 (the imaginary unit); that is, f : C → C, with f = f (z). Such functions are typically expressed in the form f (z) = u(x, y) + iv(x, y) ,

(D.33)

where u = u(x, y) and v = v(x, y) are real functions known as the ‘real’ and ‘imaginary’ parts of f , respectively (for short, u = Re( f ) and v = Im( f )). The function (D.33) is called analytic in a region D ⊂ C, if u(x, y) and v(x, y) are continuous and have continuous first-order partial derivatives throughout D that satisfy the so-called Cauchy–Riemann equations, ∂v ∂u = , ∂x ∂y

∂u ∂v =− , (∀) z = x + iy ∈ D . ∂y ∂x

(D.34)

If these conditions are fulfilled, then the functions f (z) is differentiable (as many times as we please) and its first-order derivative f (z) can be calculated from the formulae ∂v ∂v ∂u ∂u +i = −i . f (z) = ∂x ∂x ∂y ∂y

484

Appendix D: Fourier Transforms

A point z = a ∈ D where f (z) fails to be analytic is called a singular point. Without going into a detailed classification of such points, we mention here only isolated singular points. These are characterised by the property that there exists a neighbourhood of z = a in which the function is analytic, except at the point itself. Associated with an isolated singular point any complex function is characterised by a numerical quantity called its residue at that particular point; more precisely,   1 Res f (z); a := 2π i

 C

f (z) dz ,

(D.35)

where C is any smooth closed simple contour in D that encircles the isolated singularity z = a. If z = a is an isolated singularity for f (z) such that lim z→a | f (z)| = +∞, but the function (z − a)m f (z) is analytic for some positive integer m ≥ 1 (and m is the least value with this property), we say that f (z) has a pole of order m at z = a. A simple pole corresponds to m = 1, a double pole is obtained for m = 2, and so on. The residues of such singularities are easily computed by the general formula   Res f (z); a =

 d m−1  1 lim m−1 (z − a)m f (z) . (m − 1)! z→a dz

(D.36)

As a particular case, if f (z) = P(z)/Q(z) has a simple pole at z = a,   P(a) . Res f (z); a =

Q (a)

(D.37)

The concept of residue reviewed above plays a crucial part in the evaluation of closed contour integrals for which the corresponding integrands have a finite number of poles. In this respect, we have the following main result. Residue Theorem: Let f (z) be a single-valued complex function. If D ⊂ C represents the region enclosed by a closed contour C , and f (z) is analytic in D ∪ C , except for finitely many isolated singular points ak ∈ D (k = 1, 2, . . . , n), then  C

f (z) dz = 2π i

n 

  Res f (z); ak ;

(D.38)

k=1

the direction in which the contour C is traversed follows the same conventions as in Stokes’ Theorem (see Sect. 1.22.5). This formula can be adapted to evaluate the integrals that appear in the expression of Fourier transforms and their inverses. The idea is to use the continuation of f (x) into the complex plane, leading to a complex function that we shall denote by f (z). Before we can state the key results, some assumptions are necessary3 : 3 For a more precise statement of the various technical conditions that must be satisfied by

refer to standard texts on Complex Analysis mentioned at the end of this section.

f (z), we

Appendix D: Fourier Transforms

485

Fig. D.2 Integration contours for evaluating Fourier transforms and their inverses. The one on the left consists of a large semi-circle of radius R, together with several small semi-circles of radius rk and centred at (bk , 0), which are connected by straight line segments along the x-axis. The limit mentioned in the text corresponds to R → ∞ and rk → 0+ . A similar interpretation applies to the contour on the right

1. f (z) has a finite number of poles z = ak (k = 1, 2, . . . , n) in the upper half-plane Im(z) > 0; 2. f (z) is analytic along the real axis except at the points x = bk (k = 1, 2, . . . , m), which are simple poles; 3. z f (z)eiαz → 0 as z → ∞ and Im(z) > 0, where α > 0 is fixed. If there are no poles on the real axis the last condition can be relaxed; it suffices that f (z) → 0 as z → ∞, Im(z) > 0. By integrating the function z → f (z)eiαz , α > 0 around the contour shown on the left in Fig. D.2, making use of the above Residue Theorem, and then passing to the limit, it can be shown that   n m   √     1 , F f (x); α = i 2π Res f (z) eiαz ; ak + Res f (z) eiαz ; bk 2 k=1 k=1 (D.39) which holds only for α > 0. If f (x) is either even or odd, one can construct and even or odd extension (as explained in our discussion of (D.10) and (D.11)), so (D.39) can be made relevant to the case α < 0 as well. However, if f (x) has no symmetry properties, the aforementioned integration contour must be chosen in the lower half-plane, Im(z) < 0, as seen in the other sketch in Fig. D.2. As a first illustration of (D.39), let us consider the integral 





+∞ −∞

x eiax dx , x 2 + b2

a, b > 0 .

(D.40)

486

Appendix D: Fourier Transforms

The function f (z) = z/(z 2 + b2 ) satisfies the condition f (z) → 0 as z → ∞ and has two simple poles z 1 = ib, z 2 = −ib, which are not situated on the real axis (since b > 0); hence, f (z) is analytic in C, except at these two locations. Then, 

Res f (z)e ; z 1 iaz



 z(z − ib)eiaz  1 = = e−ab ,  (z − ib)(z + ib) z=ib 2

so, by formula (D.39), we get that for a > 0

+∞

−∞

x eiax dx = iπ e−ab . x 2 + b2

Splitting up the integral on the left into its real and imaginary parts gives

+∞

−∞

x cos(ax) dx = 0 , x 2 + b2



+∞ −∞

x sin(ax) dx = π e−ab . x 2 + b2

Another Fourier integral that can be easily evaluated in a similar way is

+∞ −∞

eiax dx , + b 2 )2

x(x 2

a, b > 0 .

(D.41)

This time f (z) = 1/z(z 2 + b2 )2 , which satisfies z f (z) → 0 as z → ∞, Im(z) > 0. There is one simple pole on the real axis (z 1 = 0) and two other double poles (z 2 = ib, z 3 = −ib). As b > 0 we need only consider the residues of f (z) at z = z 1 and z = z 2 . According to (D.36), 

Res f (z)e ; z 2 iaz



d = lim z→ib dz



1 (z − ib)2 eiaz = · · · = − 4 (2 + ab) e−ab . z(z − ib)2 (z + ib)2 4b

while, formula (D.37) gives   1 Res f (z)eiaz ; z 1 = 4 . b These last two results are combined with the help of (D.39) to give 1 √ 2π



+∞

−∞

√  eiax i 2π  2 − (2 + ab)e−ab ; dx = x(x 2 + b2 )2 4b4

separating the real and imaginary parts of the integrand on the left-hand side we notice that the latter is an even function, so ∞  sin(ax) π  dx = 4 2 − (2 + ab)e−ab . 2 2 2 x(x + b ) 4b 0

Appendix D: Fourier Transforms

487

Applications of the theory of residues to evaluating Fourier-type integrals can be found in most introductory texts on complex analysis (e.g., [30]). Some of the less recent ones that can be recommended include references [31–34] (the last one has a particularly detailed treatment of the Cauchy principal value for singular integrals).

D.5

Some Calculations for Chap. 9

In applying the convolution theorem to simplify expressions such as those in Eq. (9.35), one has to be able to rearrange the corresponding integrands as a product of some suitable Fourier transforms. This is explained next in the particular context of the results stated in (9.36). From the formulae in Sect. D.6 it follows immediately that  F1

X2 π −|α|X 2 e ;α = . 2 2 2 X1 + X2

(D.42)

Differentiation with respect to X 2 of (D.42) yields

    X2 d X2 π −|α|X 2 d ;α e −|α| = F1 ; α = F1 2 d X2 d X 2 X 12 + X 22 X 12 + X 22  2 X 1 − X 22 ; α . (D.43) = F1 (X 12 + X 22 )2 Because of the linearity of the Fourier transforms, (D.43) can be multiplied by X 2 and the result can then be added to (D.42); the outcome is

 X2 π X 2 (X 12 − X 22 ) + ; α (1 − |α|X 2 ) e−|α|X 2 = F1 2 X 12 + X 22 (X 12 + X 22 )2  2X 12 X 2 = F1 ;α , (X 12 + X 22 )2

(D.44)

which corresponds to (9.36a) in Chap. 9. The justification of the other formula (9.36b) proceeds in a similar way. The starting point is the self-evident result d tan−1 d X1



X2 X1

 =−

X2 X 12 + X 22

so, with the help of Eq. (D.42),

      d X2 X2 π −|α|X 2 ; α = (−iα)F1 tan−1 ;α , e − = F1 tan−1 2 d X1 X1 X1

488

Appendix D: Fourier Transforms

   X2 i π −|α|X 2 ;α =− e F1 tan−1 . X1 α 2

whence

(D.45)

Differentiating (D.45) with respect to X 2 , in the same way as we did in (D.43), gives i

|α| α

      X2 d X2 π −|α|X 2 d ; α = F1 ;α = F1 tan−1 tan−1 e 2 d X2 X1 d X2 X1  X1 = F1 ; α . (D.46) X 12 + X 22

We can also differentiate with respect to X 1 inside the Fourier transform that appears in Eq. (D.42),  F1

d d X1



X2 X 12 + X 22



 ; α = (−iα)F1

X2 π −|α|X 2 αe ; α = −i , 2 X 12 + X 22

whence

    X2 2X 1 X 22 π d −|α|X 2 −i ; α = −F1 α X 2e = F1 X 2 ;α . 2 d X 1 X 12 + X 22 (X 12 + X 22 )2 Combining this last result with (D.46), we finally obtain

i

π 2

   |α| 2X 1 2X 1 X 22 2 − α X 2 e−|α|X 2 = F1 − ; α α X 12 + X 22 (X 12 + X 22 )2  2X 13 ; α , (D.47) = F1 (X 12 + X 22 )2

which is precisely the same as the formula stated in (9.36b).

D.6

Useful Sine and Cosine Fourier Transforms

Several common integrals that keep recurring in the applications featured in Chap. 9 are listed below. 0



0

∞

∞

α sin(α X 1 ) π dα = e−β X 1 , α2 + β 2 2 cos(α X 1 ) π −β X 1 e dα = , α2 + β 2 2β

(D.48a) (D.48b)

Appendix D: Fourier Transforms



∞

0



∞

0 ∞ 0



∞

0



0

∞

489

α 3 sin(α X 1 ) π dα = (2 − β X 1 )e−β X 1 , (α 2 + β 2 )2 4 2 α cos(α X 1 ) π dα = (1 − β X 1 )e−β X 1 , (α 2 + β 2 )2 4β α sin(α X 1 ) π dα = X 1 e−β X 1 , (α 2 + β 2 )2 4β cos(α X 1 ) π dα = (1 + β X 1 )e−β X 1 , 2 2 2 (α + β ) 4β 3  α −1 sin(α X 1 ) π  2 − (2 + β X 1 )e−β X 1 . dα = 2 2 2 4 (α + β ) 4β

(D.48c) (D.48d) (D.48e) (D.48f) (D.48g)

In all of the above formulae β > 0; these results are easily checked by following the same arguments as those employed in the examples (D.40) and (D.41). The additional formulae recorded next are easily obtained by direct integration, and are recorded here for reference purposes.

∞

α 2 e−α X 2 cos(α X 1 ) dα =

0 ∞

2X 2 (X 22 − 3X 12 ) , (X 12 + X 22 )3

X 22 − X 12 , (X 12 + X 22 )2 0 ∞ X2 e−α X 2 cos(α X 1 ) dα = 2 , 2 X 0 1 + X2 ∞ 1 α −1 e−α X 2 cos(α X 1 ) dα = − log(X 12 + X 22 ) , 2 0 ∞ 2X 1 (3X 22 − X 12 ) α 2 e−α X 2 sin(α X 1 ) dα = , (X 12 + X 22 )3 0 ∞ 2X 1 X 2 αe−α X 2 sin(α X 1 ) dα = , 2 2 2 (X 0 1 + X2) ∞ X1 e−α X 2 sin(α X 1 ) dα = 2 , 2 X 0 1 + X2   ∞ X1 . α −1 e−α X 2 sin(α X 1 ) dα = tan−1 X2 0 αe−α X 2 cos(α X 1 ) dα =

(D.49a) (D.49b) (D.49c) (D.49d) (D.49e) (D.49f) (D.49g) (D.49h)

Appendix E

The Bi-harmonic Equation

Abstract Most of the examples and exercises in Chap. 8 are intimately related to the solution of the bi-harmonic equation in simple two-dimensional geometries. We review here several elementary aspects relevant to three different formulations of this equation: Cartesian coordinates, polar coordinates, and complex variables. In particular, our discussion includes the Almansi and Goursat representations for bi-harmonic functions, as well as the Michell solution which can be used to solve a large number of problems formulated in terms of polar coordinates.

E.1

Cartesian Coordinates

It is recalled that in the absence of body forces the Airy stress function Φ introduced in Chap. 8 is bi-harmonic, i.e. it satisfies   ∇2 ∇2Φ = 0

=⇒

∇ 4Φ = 0 ,

in Σ ,

(E.1)

where Σ ⊂ E2 is a two-dimensional domain; to simplify the subsequent calculations the two-dimensional Laplacian will be written as ∇ 2 (instead of ∇⊥2 ). The domain Σ in which Eq. (E.1) is considered is described either by the Cartesian coordinates (x, y) or the polar coordinates (r, θ ), with x = r cos θ and y = r sin θ (x, y ∈ R and r ≥ 0, 0 ≤ θ < 2π ). The semi-inverse solution method for many two-dimensional Elasticity problems relies on being able to construct a suitable representation for the bi-harmonic function Φ. We review below a number of useful results for carrying out this task. The Almansi representation: Any bi-harmonic function Φ = Φ(x, y) can be represented in terms of two harmonic functions Φ j = Φ j (x, y) (for j = 1, 2), ∇2Φ 1 = ∇2Φ 2 = 0 , © Springer Nature B.V. 2020 C. D. Coman, Continuum Mechanics and Linear Elasticity, Solid Mechanics and Its Applications 238, https://doi.org/10.1007/978-94-024-1771-5

491

492

Appendix E: The Bi-harmonic Equation

according to the formula   Φ(x, y) = Φ 1 (x, y) + A + Bx + C y + D(x 2 + y 2 ) Φ 2 (x, y) ;

(E.2)

A, B, C, D ∈ R are arbitrary constants assumed to satisfy B 2 + C 2 + D 2 = 0. The fact that the expression on the right-hand side in Eq. (E.2) is a bi-harmonic function can be checked very easily. For example, if f = f (x, y) is harmonic (∇ 2 f = 0), then ∂f ∂f , ∇ 2 (y f ) = 2 , ∂x ∂y     ∂f ∂f ∇ 2 (x 2 + y 2 ) f = 4 f + x +y . ∂x ∂y ∇ 2 (x f ) = 2

By applying ∇ 2 to these equations we discover that each of the functions x f (x, y) ,

y f (x, y) ,

(x 2 + y 2 ) f (x, y) ,

(E.3)

is bi-harmonic. The above observation can be used to generate some important bi-harmonic functions relevant to practical situations. To this end, we first identify a special class of harmonic functions f (x, y) for which the independent variables can be separated. By writing f (x, y) = X (x) Y (y) and using the equation ∇ 2 f (x, y) = 0, we find −

1 d 2Y 1 d2 X = = ±α 2 , 2 X (x) d x Y (y) dy 2

(E.4)

for some arbitrary α ∈ R. Choosing the ‘plus’ sign in (E.4) yields two ordinary differential equations, d2 X + α 2 X (x) = 0 , dx2

d 2Y − α 2 Y (y) = 0 , dy 2

(E.5)

with the solutions X (x) = C1 cos(αx) + C2 sin(αx) ,

Y (y) = C3 eαy + C4 e−αy ,

(E.6)

where C j ∈ R ( j = 1, 2, 3, 4) are arbitrary constants. Taking the ‘minus’ sign in (E.4) does not reveal any new information as we get the same expressions, but the roles of ‘x’ and ‘y’ are reversed. Therefore, it is concluded that the functions e±αx sin(αy) , e±αx cos(αy) , e±αy sin(αx) , e±αy cos(αx)

(E.7)

Appendix E: The Bi-harmonic Equation

493

are harmonic. Since the exponential function is neither even nor odd, these solutions do not have any symmetry properties. This shortcoming can be easily rectified by replacing the exponentials with hyperbolic trigonometric functions, with the help of the well-known formulae e±αy = cosh(αy) ± sinh(αy). Hence, (E.7) suggests the following harmonic functions cosh(αx) sin(αy) , sinh(αx) cos(αy) , cosh(αx) cos(αy) , sinh(αx) sin(αy) , cosh(αy) sin(αx) , sinh(αy) cos(αx) , cosh(αy) cos(αx) , sinh(αy) sin(αx) , which are either even or odd. Each one of these can be multiplied by ‘x’ or ‘y’ to produce a bi-harmonic function—in the interest of brevity we do not record the results here. It is also possible to construct more general bi-harmonic functions by considering the formal superposition of the various groups of functions found above; for example, Φ(x, y) =

∞    (An + Bn y) cosh(αn y) + (Cn + Dn y) sinh(αn y) n=0

  × E n cos(αn x) + Fn sin(αn x) ,

(E.8)

where αn , An , Bn , Cn , Dn , E n , Fn ∈ R are constants that have to be determined from the boundary conditions. This is achieved by expanding the boundary loads in appropriate Fourier series in ‘x’ or ‘y’. We refer to the standard works [22, 35–38], which contain additional details and several examples vis-à-vis (E.8). Another important class of solutions for Eq. (E.1) consists of bi-harmonic polynomials. It can be checked by inspection that ‘x’, ‘y’ and ‘x y’ are harmonic polynomials. Then, a simple application of Almansi’s formula (E.2) shows that x2 ,

y2 , x 3 ,

y3 , x 2 y , x y2 , x 3 y , x y3

are bi-harmonic, and any linear combination of these functions will be yet another bi-harmonic polynomial. To find more general expressions that share the same property as these simple functions one can use a special form of representation for harmonic functions. To this end, let us note that the equation ∇ 2 (x, y) = 0 can be cast in the form 

∂ ∂ +i ∂x ∂y



∂ ∂ −i ∂x ∂y

 (x, y) = 0 ,

√ where i = −1 is the imaginary unit. The general solution of this equation can be shown to be (x, y) = 1 (x, y) + 2 (x, y), where

494

Appendix E: The Bi-harmonic Equation



∂ ∂ +i ∂x ∂y



 1 = 0,

∂ ∂ −i ∂x ∂y

 2 = 0 .

(E.9)

By making a simple change of independent variables (see the beginning of Sect. E.2) these equations can be integrated and we obtain the following representation for our harmonic function, (x, y) = 1 (x + iy) + 2 (x − iy) ,

(E.10)

where  j ( j = 1, 2) are arbitrary functions. Motivated by equation (E.10) and the Almansi formula, let us consider Pn (x, y) = An (x + iy)n + Bn (x − iy)n   + (x 2 + y 2 ) Cn (x + iy)n−2 + Dn (x − iy)n−2 , (E.11) with An , Bn , Cn , Dn ∈ R constants, n ≥ 2. Writing Pn (x, y) = Rn (x, y) + iIn (x, y), then both Rn and In are bi-harmonic polynomials that have certain symmetry properties depending on the parity of n. For example, if n = 2m (m ≥ 1) one of these polynomials will be even in ‘x’ and ‘y’, while the other will be odd in the same variables, namely,    m − k 2m  an x 2m−2k y 2k + bn x 2k y 2m−2k , m 2k k=1   m−2   2m  k m −k −1 cn x 2m−2k−1 y 2k+1 + dn x 2k+1 y 2m−2k−1 , (−1) 2m(m − 1) 2k + 1 k=0 m−1 

(−1)k

where an , bn , cn , dn ∈ R are arbitrary constants,   p p! := , ( p, q ∈ N) . q q!( p − q)! In a similar way, if n = 2m + 1, then one of the polynomials will be even in ‘x’ and odd in ‘y’, while the other will have the opposite symmetry properties. These are not recorded here because in practice it is easier to work directly with (E.11) by employing symbolic manipulation software to quickly generate bi-harmonic polynomials of any desired order. The arbitrary constants are fixed by the boundary conditions after calculating the stresses from the definitions mentioned in Chap. 8, T11 =

∂ 2Φ , ∂ y2

T22 =

∂ 2Φ , ∂x2

T12 = −

∂ 2Φ . ∂ x∂ y

(E.12)

Appendix E: The Bi-harmonic Equation

495

Historically, bi-harmonic polynomials have been mainly used to solve various plane stress/strain engineering applications involving beams of rectangular cross sections. Unfortunately, their effectiveness is limited by the fact that the boundary conditions must be expressed in a similar form (which is possible only in particular instances).

E.2

Complex Variables

Equation (E.1) is amenable to re-formulation in terms of the complex variable z ≡ x + iy = r eiθ and its conjugate z ≡ x − iy = r e−iθ . To this end, we define the operations     1 ∂ ∂ ∂ 1 ∂ ∂ ∂ := −i , := +i , (E.13) ∂z 2 ∂x ∂y ∂z 2 ∂x ∂y motivated by the chain rule for partial differentiation together with the elementary relations x = (z + z)/2 and y = (z − z)/2i. By adding and subtracting the two equations in (E.13), we also find   ∂ ∂ ∂ =i − . ∂y ∂z ∂z

∂ ∂ ∂ = + , ∂x ∂z ∂z

(E.14)

With this in mind Φ(x, y) becomes a function of z and z (regarded here as independent variables), which we shall continue to write as Φ ≡ Φ(z, z). Thus,  ∇ = 2

∂ ∂ +i ∂x ∂y



∂ ∂ −i ∂x ∂y

 =4

∂2 ∂z ∂z

=⇒

∇ 4 = 16

∂4 . (E.15) ∂z 2 ∂z 2

The upshot of this change of variables is that now the bi-harmonic equation can be formally integrated as indicated below, ∂ 3Φ ∂ 2Φ ∂ 4Φ = g = 0 ⇒ (z) ⇒ = z g1 (z) + g2 (z) 1 ∂z 2 ∂z ∂z 2 ∂z 2 ∂z 2

∂Φ = z g1∗ (z) + g2∗ (z) + g3 (z) ⇒ Φ = z g1∗∗ (z) + g2∗∗ (z) + z g3 (z) + g4 (z) , ∂z

∗ where g ∗j (z) := g j (z) dz, g ∗∗ g j (z) dz, and g j (z) ( j = 1, 2, 3, 4) are arbij (z) := trary analytic functions in Σ ⊂ C. In conclusion, the bi-harmonic function Φ can be represented as (E.16) Φ(z, z) = f 1 (z) + z f 2 (z) + f 3 (z) + z f 4 (z) , ⇒

in terms of four arbitrary analytic functions f j : Σ → C ( j = 1, 2, 3, 4). As the Airy stress function must be real-valued, the expression of Φ in the above equation

496

Appendix E: The Bi-harmonic Equation

must match its complex conjugate, and so4 f 3 (z) = f 1 (z) ,

f 4 (z) = f 2 (z) .

Rewriting (E.16) in light of these new findings we eventually arrive at the Goursat representation for a bi-harmonic function,   Φ(z, z) = 2 Re z f 2 (z) + f 1 (z) ,

(E.17)

where it is recalled that ‘Re(a)’ stands for the real part of a ∈ C. In Eq. (E.17) f 1 (z) and f 2 (z) are arbitrary analytic functions in Σ ⊂ C. The components of the stress tensor are easily found from (E.12) and (E.14),  ∂2 ∂2 ∂2 + 2 Φ(z, z) , +2 T22 = ∂z 2 ∂z∂z ∂z   2 ∂2 ∂ ∂2 + 2 Φ(z, z), −2 T11 = − ∂z 2 ∂z∂z ∂z   2 ∂ ∂2 − 2 Φ(z, z) . T12 = −i ∂z 2 ∂z 

(E.18a) (E.18b) (E.18c)

Furthermore, one can then use (E.17) to express these formulae in terms of various complex derivatives of f 1 (z) and f 2 (z). We postpone this for the moment in order to give an alternative derivation for the Goursat representation above. Returning to (E.1), if we set P 1 (x, y) := ∇ 2 Φ(x, y) then ∇ 2 P 1 (x, y) = 0, and hence P 1 is a harmonic function. Now, take P 2 = P 2 (x, y) to be the harmonic conjugate of P 1 (x, y), that is, the pair (P 1 , P 2 ) satisfies the Cauchy–Riemann relations (D.34), in which we make the obvious changes u → P 1 , v → P 2 . By defining a new function F = F(z), with F(z) := P 1 (x, y) + iP 2 (x, y) ,

z = x + iy ,

it is clear that this will be analytic in Σ ⊂ C; its indefinite integral will of course enjoy the same property. Thus, let us also introduce ϕ(z) ≡ Q 1 + iQ 2 :=

1 4

F(z) dz ,

(E.19)

for which ϕ = ϕ(z) is a complex-valued function, then the notation ϕ(z) means that in the expression of ϕ(z), we simply change z → z. However, ϕ(z) means that we take the complex conjugate of everything that makes up the expression of ϕ (not just the independent variable z).

4 If

Appendix E: The Bi-harmonic Equation

ϕ (z) =

497

 ∂ Q2 ∂ Q2 ∂ Q1 1 ∂ Q1 +i = −i = P1 + i P2 , ∂x ∂x ∂y ∂y 4

and therefore ∂Q1 ∂Q2 1 = = P1 , ∂x ∂y 4

∂Q2 1 ∂Q1 =− = − P2 . ∂y ∂x 4

(E.20)

In view of these last two equations, U1 := Φ − x Q 1 − y Q 2 satisfies ∇ 2 U1 = 0, and we conclude that any bi-harmonic function Φ can be represented in the form Φ(x, y) = U1 (x, y) + x Q 1 (x, y) + y Q 2 (x, y) , involving three harmonic functions: U1 , Q 1 and Q 2 . We digress for a moment to point out that one of these functions is actually redundant. Indeed, from (E.20) it transpires that ∇ 2 (Φ − 2x Q 1 ) = 0, suggesting the representation Φ = H1 + 2x Q 1 , where both H1 and Q 1 are harmonic. Similarly, since ∇ 2 (Φ − 2y Q 2 ) = 0, we also have the alternative form Φ = H2 + 2y Q 2 in which H2 and Q 2 are harmonic. Next, let us record a simple observation,   z ϕ(z) = (x − iy)(Q 1 + iQ 2 ) ⇒ x Q 1 + y Q 2 = Re z ϕ(z) .

(E.21)

We also introduce U2 = U2 (x, y) as the harmonic conjugate of U1 mentioned above; therefore, the new function χ (z) := U1 (x, y) + iU2 (x, y) ,

z = x + iy

(E.22)

  is analytic in Σ and U1 = Re χ (z) . Finally, combining this with (E.21) gives   Φ = Re z ϕ(z) + χ (z) ,

(E.23)

which matches the Goursat representation found earlier in (E.17). The two complex functions that feature in (E.23) are known as the complex potentials for the plane stress/strain approximation. We shall give here a very brief sketch of the complex variable formulation for Plane Elasticity. This provides further insight into the nature of the analytic functions that feature in (E.23), and in particular the Airy stress function Φ. The definitive account on this topic remains the monograph [39] (written by one of the main contributors); a somewhat different comprehensive treatment is contained in [15], while elementary introductions to the subject are included in many other Elasticity books (e.g., see [4, 22, 38, 40–43]). We start by noticing that (E.23) can be rewritten as 2 Φ = z ϕ(z) + z ϕ(z) + χ (z) + χ (z) .

(E.24)

498

Appendix E: The Bi-harmonic Equation

To simplify the writing, in what follows we shall resort to the subscript notation for indicating partial differentiation with respect to ‘x’ and ‘y’; the conventions ‘x → 1’ and ‘y → 2’ are tacitly implied below. From (E.19) it follows that ϕ(z) = Q 1 − i Q 2 , and hence ϕ , 1 = Q 1, 1 − i Q 2, 1 = ϕ (z) , ϕ , 2 = Q 1, 2 − i Q 2, 2 = −i (Q 1, 1 − i Q 2, 1 ) = −i ϕ (z) , together with the more obvious relations ϕ , 1 = ϕ (z) and ϕ , 2 = i ϕ (z); here, the ‘dash’ stands for the usual differentiation operation in C. Taking into account (E.24) we then have 2 Φ , 1 = ϕ(z) + z ϕ (z) + ϕ(z) + z ϕ (z) + χ (z) + χ (z) ,   2 Φ , 2 = i −ϕ(z) + z ϕ (z) + ϕ(z) − z ϕ (z) + χ (z) − χ (z) , whence Φ , 1 + i Φ , 2 = ϕ(z) + z ϕ (z) + ψ(z) ,

(E.25)

with ψ(z) := χ (z) being a shorthand notation (rather than a new complex potential). The components of the stress tensor can be transcribed in terms of the complex potentials since T11 + i T12 = Φ , 22 − i Φ , 12 = −i (Φ , 1 + i Φ , 2 ) , 2 , T22 − i T12 = Φ , 11 + i Φ , 12 = (Φ , 1 + i Φ , 2 ) , 1 , and then use of (E.25) yields T11 + i T12 = ϕ (z) + ϕ (z) − z ϕ

(z) − ψ (z) , T22 − i T12 = ϕ (z) + ϕ (z) + z ϕ

(z) + ψ (z) . Combining together the last two equations we finally obtain     T11 + T22 = 2 ϕ (z) + ϕ (z) ≡ 4 Re ϕ (z) ,   T22 − T11 + 2i T12 = 2 z ϕ

(z) + ψ (z) ;

(E.26a) (E.26b)

these are known as Kolosov’s stress formulae. They show that the stress distribution is completely determined by the complex potentials ϕ(z) and ψ(z). It is noted in passing that the particular expressions ϕ(z) = ci z + α ,

χ (z) = γ z + β , (c ∈ R, α, β, γ ∈ C)

correspond to a rigid-body deformation (T11 = T22 = T12 ≡ 0).

Appendix E: The Bi-harmonic Equation

499

The components of the displacement field u = u1 e1 + u2 e2 are amenable to a similar representation as the stresses. From (8.42), earlier in Chap. 8 (see Sect. 8.5.4), 2μu1,1 = −Φ, 11 + (1 − δ∗ )∇ 2 Φ = −Φ, 11 + (1 − δ∗ )P 1 , 2μu2,2 = −Φ, 22 + (1 − δ∗ )∇ 2 Φ = −Φ, 22 + (1 − δ∗ )P 1 , which can be easily integrated by remembering that P 1 = 4Q 1, 1 = 4Q 2, 2 —viz. (E.20); the outcome is 2μu1 = −Φ , 1 + 4(1 − δ∗ )Q 1 + f (y) , 2μu2 = −Φ , 2 + 4(1 − δ∗ )Q 2 + g(x) , (E.27) for some arbitrary functions f (y) and g(x). These are combined with the help of the third equation in (8.42), μ(u1,2 + u2,1 ) = −Φ , 12 . In conjunction with (E.27), the former equation leads to f (y) + g (x) = 0 since Q 1, 2 + Q 2, 1 = 0. It is then immediately found that f (y) = ay + b and g(x) = −ax + c, where a, b, c ∈ R. Note that, without loss of generality, we can take these constants to be zero because f and g correspond to a rigid-body displacement. Finally, by using (E.19) and (E.25) the displacements in (E.27) can be written more compactly as, 2μ(u1 + iu2 ) = κ ϕ(z) − z ϕ (z) − ψ(z) ,

(E.28)

where κ = 3 − 4δ∗ . For plane stress κ = (3 − ν)/(1 + ν), while for plane strain κ = 3 − 4ν; see the definition of δ∗ in (8.21). Equation (E.28) represents Kolosov’s displacement formula. The Goursat representation of the Airy function (E.23), involving the analytic complex potentials ϕ(z) and χ (z), allows the stresses and displacements to be expressed in terms of these functions and their derivatives—see (E.26) and (E.28). The aforementioned analytic functions are determined from the boundary and/or farfield conditions specific to the each particular situation considered. To complete the formulation of the problems satisfied by these complex functions, some additional results are needed. Suppose that the domain Σ ⊂ C is bounded by a curve ∂Σ, and the arc AB constitutes a part of it. The total force and its corresponding moment (about the origin) of the surface tractions acting on AB can be given in terms of the complex potentials; in the terminology of Sect. 8.5.3 (Chap. 8) these were denoted by R AB and M AB , respectively. However, here we shall work with the complex versions of those quantities, which will be denoted by R AB and M AB , respectively (their definitions will be given shortly). If AB has the parametric arclength representation x = x(s) and y = y(s), then the outward unit normal n to AB is given by n = (y (s), −x (s))—see Eq. (8.35). The traction vector t ≡ t1 e1 + t2 e2 = n · T can be shown to satisfy t1 + i t2 = −i

 d  Φ , 1 + iΦ , 2 , ds

500

Appendix E: The Bi-harmonic Equation

whence5

B

R AB :=

 B (t1 + i t2 ) ds = −i ϕ(z) + z ϕ (z) + ψ(z) , A

A

(E.29)

using (E.25). In particular, if the boundary is traction-free then the complex potentials must satisfy ϕ(z) + z ϕ (z) + ψ(z) = constant on the boundary. The constant on the right-hand side above can be taken to be zero if the domain Σ is simply connected. As for the resultant moment, it is easy to check that

B

M AB :=



 x t2 − y t1 ds = −



A

B

 x

A

  d  d  Φ,1 + y Φ , 2 ds . ds ds

Integrating by parts in the last integral and remembering again (E.25), one eventually finds  B (E.30) M AB = Re χ (z) − z ψ(z) − zz ϕ (z) A . Simply connected domains: The formulae (E.26) confirm that in a simply connected and bounded domain Σ the state of stress remains unaltered if we make the substitutions6 ϕ(z) → ϕ(z) + ci z + α ,

ψ(z) → ψ(z) + β , (α, β ∈ C, c ∈ R) .

By making the same changes in the displacement formula (E.28), its invariance ¯ In other words, the addition of a constant to ϕ(z) requires that c = 0 and κα = β. corresponds only to a rigid-body displacement. Because of the degree of arbitrariness present in the complex potentials some simplifying assumptions can be introduced right from the outset. For instance, in a traction boundary-value problem one can take: ϕ(z 0 ) = 0, Im ϕ (z 0 ) = 0, ψ(z 0 ) = 0, with z 0 ∈ Σ being an arbitrary point. If the boundary conditions contain both tractions and displacements, one can enforce a priori either ϕ(z 0 ) = 0 or ψ(z 0 ) = 0, but not both of them. An important theorem in Complex Analysis (known as the Monodromy Theorem) guarantees that any analytic function in a simply connected and bounded domain is single-valued. Thus, ϕ(z) and ψ(z) are both single-valued in this case, and they can be taken in the form ϕ(z) =

∞  k=0

Ak z k ,

ψ(z) =

∞ 

Bk z k ,

(Ak , Bk ∈ C) .

k=0

≡ [ f ] AB := f (B) − f (A), and we refer to Sect. 8.5.3 where this notation was initially introduced. If C is a closed curve, then [ f ]C has the meaning stated in Eq. (8.40). 6 The converse statement is also true. 5 Here, [ f ] B A

Appendix E: The Bi-harmonic Equation

501

For multiply connected or infinite domains this is no longer true. More about this aspect will be said shortly, but first we return to the governing equations for the complex potentials. If the traction vector ˚t = ˚t (s) ≡ t˚1 (s)e1 + t˚2 (s)e2 is prescribed on ∂Σ, then the analytic functions ϕ(z), ψ(z) are found by solving the functional equation ϕ(ζ ) + ζ ϕ (ζ ) + ψ(ζ ) = f 1 + i f 2 + Γ ,

on ∂Σ ,

(E.31)

which follows from (E.29). Here, ζ = ζ (s) ≡ x(s) + i y(s) represents the complex variable along ∂Σ, while the right-hand side corresponds to

  t˚1 (s) + i t˚2 (s) ds .

s

f 1 + i f 2 := i 0

In principle, the constant Γ ∈ C in (E.31) has to be determined such that the stresses and displacements are single-valued and have appropriate continuity properties in Σ. In this particular case (simply connected and bounded domain) one can set Γ = 0, as well as ϕ(0) = 0, Im ϕ (0) = 0 (assuming that 0 ∈ Σ). A closed-form solution of (E.31) for the unit circle can be found in [32] (pp. 352–355). ˚ In the case when the displacements are known on ∂Σ, i.e. u˚ = u(s) ≡ u˚ 1 (s)e1 + u˚ 2 (s)e2 is a given field, the earlier Eq. (E.31) is replaced by   κ ϕ(ζ ) − ζ ϕ (ζ ) − ψ(ζ ) = 2μ u˚ 1 + i u˚ 2 ,

on ∂Σ ,

(E.32)

There is also the mixed boundary conditions scenario that involves solving both (E.31) and (E.32) on disjoint parts of ∂Σ, but we do not pursue this further. Multiply connected domains: Let us consider first the case of a bounded doubly connected domain (see Fig. E.1a). In this case the complex potentials need not be single-valued and it can be shown (e.g., [39, 40, 44]) that they have the form  ϕ(z) = − ψ(z) = where

R1 + i R2

 C1

2π(1 − κ)   κ R 1 − i R 2 C1 2π(1 + κ)

log(z − z 1 ) + ϕ0 (z) ,

(E.33a)

log(z − z 1 ) + ψ0 (z) ,

(E.33b)

  (R1 + i R2 )C1 ≡ −i ϕ(z) + z ϕ (z) + ψ(z)

C1

(E.34)

represents the total force of the surface tractions on C1 , and ϕ0 (z), ψ0 (z) are singlevalued analytic functions in Σ; in particular they admit the representations ϕ0 (z) =

∞  k=−∞

Ak (z − z 1 ) , k

ψ0 (z) =

∞  k=−∞

Bk (z − z 1 )k ,

502

Appendix E: The Bi-harmonic Equation

(a)

(b) 0

1

1 1

1

Σ

Σ

Fig. E.1 Doubly connected domains: a bounded; b unbounded. In the former case Σ is a noncircular annulus with ∂Σ = C0 ∪ C1 , while in the latter situation Σ corresponds to the exterior of the finite region bounded by the curve C1

with Ak , Bk ∈ C. In the case of multiply connected domains of the type considered in Sect. 8.5.3— see Fig. 8.3,   1 R1 + iR2 Ck log(z − z k ) + ϕ0 (z) , 2π(1 − κ) k=1

(E.35a)

  κ R1 − i R2 Ck log(z − z k ) + ψ0 (z) , 2π(1 + κ) k=1

(E.35b)

p

ϕ(z) = −

p

ψ(z) =

where / Σ are arbitrarily located inside the closed contours Ck for 1 ≤ k ≤ p,  zk ∈ and R1 + i R2 Ck has a similar connotation to (E.34). Also, ϕ0 (z) and ϕ0 (z) are single-valued analytic functions in Σ. For an infinite multiply connected domain, like the one seen in Fig. E.1b, the above formulae must be amended. It can be proved that for |z| → ∞,  ϕ(z) = − ψ(z) =

R1 + i R2

 C1

2π(1 + κ)   κ R 1 − i R 2 C1 2π(1 + κ)

log z + (B + i C)z + ϕ0∗ (z) ,

(E.36a)

log z + (B + i C ) + ψ0∗ (z) ,

(E.36b)

assuming that the origin of the coordinate axes is taken outside Σ. Here, ϕ0∗ (z) and ψ0∗ (z) are single-valued analytic functions in Σ, including the point z = ∞, so they admit the representation ϕ0∗ (z)

∞  ak = , k z k=0

ψ0∗ (z)

∞  bk = , k z k=0

Appendix E: The Bi-harmonic Equation

503

with ak , bk ∈ C. If Σ contains several “holes” bounded by the closed contours Ck p (k = 0, 1, . . . , p), then the term (R1 + i R2 )C1 is replaced by k=1 (R1 + i R2 )Ck . Taking the limit |z| → ∞ in (E.26) it can be further shown that B=

 1 ∞ ∞ T11 + T22 , 4

B = −

 1 ∞ ∞ T11 − T22 , 2

∞ C = T12 ,

with Ti∞ j := lim |z|→∞ Ti j (z) for i, j = 1, 2. The constant C in (E.36a) follows from (E.28) by using a similar limit, (1 + κ)C = μ lim|z|→∞ (u2, 1 − u1, 2 ); it is usually taken to be zero since it is related to a rigid-body rotation at infinity. A closed-form formula for the case when Σ is a half-plane can be found in [32] (pp. 356–357). The polar-coordinate version of (E.26) is easily obtained by using the transformation formulae (8.121); in particular,   Trr + Tθθ = 4 Re ϕ (z) ,   Tθθ − Trr + 2i Tr θ = 2e2iθ z ϕ

(z) + ψ (z) .

(E.37a) (E.37b)

Similarly, the complex displacement field in polar coordinates can be cast as   2μ(ur + iuθ ) = e−iθ κ ϕ(z) − z ϕ (z) − ψ(z) .

(E.38)

Finally, from (E.37) one can also find the traction vector t(±er ) = (±er ) · T acting on a circular boundary whose outward unit normal is ±er , Trr + i Tr θ = −z ϕ

(z) − (z/z) ψ (z) + ϕ (z) + ϕ (z) .

(E.39)

Broadly speaking the analytic functions (ϕ0 , ψ0 ) and (ϕ0∗ , ψ0∗ ) in (E.35) and (E.36), respectively, will satisfy equations similar to either (E.31) or (E.32). However, the right-hand sides of those equations will be more complicated owing to the presence of the additional terms in the expressions of ϕ(z) and ψ(z). We also note that for a multiply connected domain the constant Γ in (E.31) will have to be different for each component Ck of ∂Σ. One of these constants can arbitrarily be set to zero, but the remaining constants must be found by enforcing the single-valuedness requirement for the corresponding stress and displacement fields. We refer the interested reader to Muskhelishvili’s original volume [39] for a detailed discussion of these and related matters. Particularly clear and concise accounts of the functional equations omitted here can also be found in [32, 40]. We list below a number of complex potentials used for solving some of the examples discussed in Chaps. 8 and 9. 1. Annular plate a ≤ |z| ≤ b with the edges |z| = a and |z| = b subjected to uniform pressures P 1 and P 2 , respectively: ϕ(z) = Az, χ (z) = B log z (A, B ∈ R), A :=

b2 P 2 − a 2 P 1 , 2(a 2 − b2 )

B :=

a 2 b2 (P1 − P2 ) . a 2 − b2

504

Appendix E: The Bi-harmonic Equation

2. Same annular domain as above, the edge |z| = b is held fixed, while the edge |z| = a undergoes a rigid-body infinitesimal displacement d > 0 in the positive x-direction: ϕ(z) = A log z + Bz 2 , χ (z) = C z log z + Dz + F/z (A, B, C, D, F ∈ R). 3. Infinite plate with a traction-free hole |z| = a and a uniform pressure P applied at infinity: ϕ(z) = Az, χ (z) = B log z (A, B ∈ R), A := −

P , 2

B := a 2 P .

4. Infinite plate with a traction-free hole |z| = a subjected to a uniaxial tension T11 = T0 , T22 = T12 = 0 applied at infinity: ϕ(z) = Az + B/z, χ (z) = C z 2 + D log z + F/z 2 (A, B, C, D, F ∈ R), A = −C :=

T0 , 4

1 2 a T0 , 2

B = −D :=

1 F := − a 4 T0 . 4

5. Same infinite plate with a circular hole as above, but subjected to a pure shear T11 = T22 = 0, T12 = S0 > 0 at infinity: ϕ(z) = i A/z, χ (z) = i Bz 2 + i C/z 2 (A, B, C ∈ R), A := a 2 S0 ,

B :=

S0 , 2

1 C := − a 4 S0 . 2

6. Elastic half-plane y ≥ 0, subjected to a concentrated force of magnitude Q at the origin and directed along the positive direction of the y-axis, while the boundary y = 0 is traction-free: ϕ(z) = A log z, χ (z) = Bz log z, A = B := −

iQ . 2π

7. Infinite elastic plate subjected to an in-plane concentrated force P + i Q at the origin: ϕ(z) = A log z, χ (z) = Bz log z, A := −

E.3

P +iQ , 2π(1 + κ)

B :=

κ(P − i Q) . 2π(1 + κ)

Polar Coordinates

We explore various particular solutions of the bi-harmonic equation (E.1) when the Airy stress function is expressed in polar coordinates, i.e. Φ = Φ(r, θ ). We recall that, in this case, ∇ ≡ er

∂ 1 ∂ + eθ , ∂r r ∂θ

∇2 ≡ ∇ · ∇ =

1 ∂2 ∂2 1 ∂ + + . ∂r 2 r ∂r r 2 ∂θ 2

(E.40)

Appendix E: The Bi-harmonic Equation

505

As a first observation, let us note that the functions r cos θ F(r, θ ) ,

r 2 F(r, θ ) ,

r sin θ F(r, θ ) ,

(E.41)

are bi-harmonic in Σ ⊂ C provided that ∇ 2 F(r, θ ) = 0 for (r, θ ) ∈ Σ; this follows directly from (E.3). The linear function θ is trivially harmonic, so by taking F → θ in (E.41) we find a new set of bi-harmonic functions r θ cos θ ,

r θ sin θ ,

r 2θ .

(E.42)

Similarly, with F → log r in (E.41), it transpires that r log r cos θ ,

r log r sin θ ,

r 2 log r

(E.43)

are also particular solutions of the bi-harmonic equation (E.1). The Laplacian of the product of two real-valued functions F j ( j = 1, 2) can be readily calculated with the help of the formula     ∇ 2 (F1 F2 ) = F1 ∇ 2 F2 + F2 ∇ 2 F1 + 2 ∇ F1 · ∇ F2 . Letting F1 → θ , F2 → log r in this result confirms that θ log r is a harmonic function, and then F → θ log r in (E.41) generates three further bi-harmonic functions, θr log r cos θ ,

θr log r sin θ ,

θr 2 log r .

(E.44)

The Almansi representation (E.2) can also be adapted to the polar-coordinate setting. In its modified form it asserts that any bi-harmonic function Φ can be expressed in terms of two arbitrary harmonic functions Φ j = Φ j (r, θ ) ( j = 1, 2), according to Φ(r, θ ) = Φ 1 (r, θ ) + r 2 Φ 2 (r, θ ) .

(E.45)

Straightforward direct calculations indicate that ∂ ∂ 2 ∇ (. . . ) = ∇ 2 (. . . ) , ∂θ ∂θ

   ∂ ∂ ∇2 r (. . . ) = 2 + 2 ∇ 2 (. . . ) , ∂r ∂r

where the ‘dots’ stand for any expression depending on the polar coordinates (r, θ ). An immediate consequence is that, if F = F(r, θ ) is a harmonic function, then ∂F ∂θ

and

r

∂F ∂r

represent harmonic functions as well. Furthermore, by using mathematical induction one can check that the same is true for r n (∂ n F/∂r n ) for n ≥ 1. A more systematic way of obtaining harmonic functions is by looking for solutions with separable variables for the Laplace equation ∇ 2 F = 0. By writing

506

Appendix E: The Bi-harmonic Equation

we find

F(r, θ ) = R(r )Θ(θ ) ,

(E.46)

 2  2 d R 1 d Θ 1 dR r2 = ±k 2 , = − + 2 R(r ) dr r dr Θ(θ ) dθ 2

(E.47)

for some arbitrary k ∈ R. There are three cases to be discussed. If the common value of the first two expressions is taken to be +k 2 (k = 0), then 1 dR d2 R k2 + R = 0, − dr 2 r dr r2

d 2Θ + k2Θ = 0 , dθ 2

whence R(r ) = C1 r k + C2 r −k ,

Θ(θ ) = C3 cos(kθ ) + C4 sin(kθ ) ,

(E.48)

with C j ∈ R ( j = 1, 2, 3, 4) arbitrary constants. If k = 0, R(r ) = C1 + C2 log r ,

Θ(θ ) = C2 + C4 θ .

(E.49)

Considering (−k 2 ) for the last term in (E.47), the equations become k2 1 dR d2 R + 2 R = 0, + 2 dr r dr r

d 2Θ − k2Θ = 0 , dθ 2

with solutions R(r ) = C1 cos(k log r ) + C2 sin(k log r ) ,

Θ(θ ) = C3 ekθ + C4 e−kθ . (E.50)

Not all separable solutions found above have practical value in constructing Airy stress functions. For example, the solutions (E.48) with k ≥ 1 a positive integer, and (E.49) are the ones that play an important role in the examples discussed in Chap. 8. It should also be clear that in this case one can formally consider the superposition of the k-dependent solutions to generate yet another solution. Up until now, we have constructed solutions of the bi-harmonic equation by taking advantage of the Almansi representation and a number of simple observations. However, this is just one of several approaches available. Since all the solutions found so far are in fact separable, it makes sense to look for such a solution directly in the bi-harmonic equation ∇ 4 Φ = 0. Let us note that by expanding the product of the Laplacian operators, the equation to be solved becomes 

 ∂ 4Φ 2 ∂ 3Φ 1 ∂ 2Φ 1 ∂Φ + − + ∂r 4 r ∂r 3 r 2 ∂r 2 r 3 ∂r  4    ∂ Φ 1 ∂ 4Φ 2 1 ∂ 3Φ ∂ 2Φ + 4 + 2 − +4 2 = 0. r ∂r 2 ∂θ 2 r ∂r ∂θ 2 r ∂θ 4 ∂θ

(E.51)

Appendix E: The Bi-harmonic Equation

507

Radially symmetric solutions correspond to Φ being independent of the azimuthal coordinate θ , so (E.51) reduces to d 4Φ 2 d 3Φ 1 d 2Φ 1 dΦ + − + 3 = 0, dr 4 r dr 3 r 2 dr 2 r dr

(E.52)

which is easily seen to be an Euler differential equation in which the unknown can be taken to be (dΦ/dr ). Such equations are routinely solved by making the substitution r = e ζ , which transforms (E.52) into a constant-coefficient equation for a new function of ζ (e.g., see [45]). Without going into details, after returning to the original independent variable ζ , the final result turns out to be Φ(r ) = C1 log r + C2 r 2 + C3r 2 log r + C4 ,

(Ci ∈ R, i = 1, 2, 3, 4) . (E.53)

If the situation of interest involves a configuration in which the θ -dependence is periodic, i.e. Φ(r, θ + 2π ) = Φ(r, θ ), it makes sense to look for solutions of the form Φ(r, θ ) = RC (r ) cos(nθ ) or Φ(r, θ ) = R S (r ) sin(nθ ), with RC , R S functions to be determined and n a positive integer (n ≥ 1). Substituting this assumed form of solution into (E.51) yields the same equations for both RC and R S (identified as R below), d4 R 2 d3 R + − dr 4 r dr 3



2n 2 + 1 r2



d2 R + dr 2



2n 2 + 1 r3



n 2 (n 2 − 4) dR + R = 0. dr r4

This is, again, an Euler differential equation, and we write r = e ζ , so that it becomes dR d3 R d2 R d4 R 2 + n 2 (n 2 − 4) R = 0 . − 4 − 2(n − 2) + 4n 2 4 3 2 dζ dζ dζ dζ Next, we look for an exponential solution R(ζ ) = eλζ . Routine calculations show that the characteristic equation is (λ2 − n 2 )(λ2 − 4λ + 4 − n 2 ) = 0 , whence the characteristic roots are λ1,2 = ±n and λ3,4 = 2 ± n . For n ≥ 2 these roots lead to the four distinct solutions r n cos(nθ ) ,

r −n cos(nθ ) ,

r n+2 cos(nθ ) ,

r −n+2 cos(nθ ) ,

(E.54)

and another similar set corresponding to sin(nθ ). For n = 1 the situation is complicated by the fact that two of the characteristic roots coalesce; this leads to the appearance of a logarithmic term (e.g., see [46]), so in this case (E.54) are replaced by r 3 cos(θ ) , r log r cos(θ ) . (E.55) r cos(θ ) , r −1 cos(θ ) ,

508

Appendix E: The Bi-harmonic Equation

The formal superposition of a linear combinations of the functions (E.54) for n = 1, 2, . . . leads to a (particular) solution of the Airy stress function. Of course, the resulting solution can be combined with the radially symmetric expression (E.53), as well as a linear combination of (E.55), to obtain an even more general solution of the bi-harmonic equation. We state below such a representation result, which includes a few additional (multi-valued) terms,   Φ(r, θ ) = A + Br cos θ + Cr sin θ + Dr 2 θ + A0 + B0 r 2 + C0 log r + D0 r 2 log r   + A1c r + B1c r 3 + C1c r −1 + D1c r log r cos θ   + A1s r + B1s r 3 + C1s r −1 + D1s r log r sin θ + +

∞   n=2 ∞ 



 Anc r n + Bnc r n+2 + Cnc r −n + Dnc r −n+2 cos (nθ )  Ans r n + Bns r n+2 + Cns r −n + Dns r −n+2 sin (nθ ) ,

(E.56)

n=2

where A, B, C, D and their indexed counterparts are all arbitrary constants. The above result was originally derived by J. H. Michell in 1899 and has sufficient flexibility to accommodate the solution of a fairly large number of particular problems. As seen in Chap. 8, many interesting problems involving configurations whose boundaries are very simple in cylindrical polar coordinates can be solved by taking only a few of the terms of expression (E.56). The arbitrary constants that appear therein are determined by expanding the given boundary conditions in suitable Fourier θ -series, and then using the definitions of the stresses or displacements based on the Airy stress function given above; recall that the former can be calculated from   1 ∂ 2Φ 1 ∂Φ ∂ 2Φ ∂ 1 ∂Φ + 2 . (E.57) , T = , T = − Trr = θθ r θ r ∂r r ∂θ 2 ∂r 2 ∂r r ∂θ Choosing the right form of Φ for a specific situation is usually a trial-and-error process. This step is facilitated by knowledge of the type of stresses generated by each term in (E.56); for convenience, these expressions are summarised in Table E.1. Also, we include the expressions of the corresponding displacements in Tables E.2 and E.3—this latter piece of information is particularly useful if the boundary conditions or any symmetry considerations are directly linked to kinematics (the information in these tables is adapted from Sobrero’s book [47]). A less obvious feature of the polar-coordinate formulation of Plane Elasticity stems from the fact that such an approach is also relevant for various unbounded + ≡ {(x, y) ∈ E2 | y ≥ 0} can be domains. For example, the upper half-plane Ω∞ + regarded as the “limit” of the semi-disk ΩΔ ≡ {(x, y) ∈ E2 | x 2 + y 2 < Δ2 , y ≥ 0}

Appendix E: The Bi-harmonic Equation

509

Table E.1 Stress components associated with the various individual terms in the expression of the Airy stress function (E.56) Φ(r, θ)

Trr

Tθθ

log r

r −2

−r −2

0

r 2 log r

2 log r + 1

2 log r + 3

0

r log r cos θ

r −1 cos θ

r −1 cos θ

r −1 sin θ

r log r sin θ

r −1 sin θ

r −1 sin θ

−r −1 cos θ

θ

0

0

r −2

θr 2

2θ

2θ

−1

θr cos θ

−2r −1 sin θ

0

0

θr sin θ

2r −1 cos θ

0

0

r −1 cos θ

−2r −3 cos θ

2r −3 cos θ

−2r −3 sin θ

r −1 sin θ

−2r −3 sin θ

2r −3 sin θ

2r −3 cos θ

r 3 cos θ

2r cos θ

6r cos θ

2r sin θ

r 3 sin θ

2r sin θ

6r sin θ

−2r cos θ

r n cos(nθ)

−n(n − 1)r n−2 cos(nθ)

n(n − 1)r n−2 cos(nθ)

n(n − 1)r n−2 sin(nθ)

r n sin(nθ)

−n(n − 1)r n−2 sin(nθ)

n(n − 1)r n−2 sin(nθ)

−n(n − 1)r n−2 cos(nθ)

r n+2 cos(nθ)

−(n + 1)(n − 2)r n cos(nθ)

(n + 2)(n + 1)r n cos(nθ)

n(n + 1)r n sin(nθ)

r n+2 sin(nθ)

−(n + 1)(n − 2)r n sin(nθ)

(n + 2)(n + 1)r n sin(nθ)

−n(n + 1)r n cos(nθ)

r −n cos(nθ)

−n(n + 1)r −n−2 cos(nθ)

n(n + 1)r −n−2 cos(nθ)

−n(n + 1)r −n−2 sin(nθ)

Tr θ

r −n sin(nθ)

−n(n + 1)r −n−2 sin(nθ)

n(n + 1)r −n−2 sin(nθ)

n(n + 1)r −n−2 cos(nθ)

r −n+2 cos(nθ)

−(n + 2)(n − 1)r −n cos(nθ)

(n − 2)(n − 1)r −n cos(nθ)

−n(n − 1)r −n sin(nθ)

r −n+2 sin(nθ)

−(n + 2)(n − 1)r −n sin(nθ)

(n − 2)(n − 1)r −n sin(nθ)

n(n − 1)r −n cos(nθ)

as Δ → ∞; similar statements can be made about an elastic plane or a quarter-plane (with the obvious modifications). An important caveat regarding the applicability of formula (E.56) concerns the class of infinite wedge-shaped two-dimensional domains. For such configurations it is only in a few isolated cases that we can still resort to the Michell solution in its original form (as stated above). Strictly speaking, the separation of variables that led to (E.3) was based on taking Φ(r, θ ) = R(r ) exp(inθ ) with a positive n ∈ Z; by changing n → s ∈ R the corresponding solutions can be cast as,   Φ(r, θ ) = r s A sin(sθ ) + B cos(sθ ) + C sin((s − 2)θ ) + D cos((s − 2)θ ) , where A, B, C, D ∈ R. Several simple examples on how this solution is applied in concrete examples can be found in [22, 38, 48]; a more sophisticated approach based on a special type of (Mellin) integral transforms (e.g., see [25, 29, 49]) is included in [4, 24].

510

Appendix E: The Bi-harmonic Equation

Table E.2 The components of plane stress displacement fields associated with the various individual terms in the expression of the Airy stress function (E.56). The corresponding components for plane strain are obtained by making the substitutions ν → ν/(1 − ν) and E → E/(1 − ν 2 ) in the formulae recorded below (see Chap. 8) Φ(r, θ)

ur

uθ

log r

1 + ν −1 − r E

0

2(1 − ν) 1+ν r log r − r E E 1−ν 2 log r cos θ + θ sin θ E E 1+ν cos θ − 2E 1−ν 2 log r sin θ − θ cos θ E E 1+ν sin θ − 2E

4 θr E 1−ν 2 − log r sin θ+ θ cos θ E E 1+ν sin θ − 2E 1−ν 2 log r cos θ + θ sin θ E E 1+ν cos θ + 2E

0

−

2(1 − ν) θr E 2 1−ν − log r sin θ + θ cos θ E E 1+ν sin θ + 2E 2 1−ν log r cos θ + θ sin θ E E 1+ν − cos θ 2E

4 r log r E 2 1−ν − log r cos θ − θ sin θ E E 1+ν cos θ − 2E 2 1−ν − log r sin θ + θ cos θ E E 1+ν sin θ − 2E

r −1 cos θ

1 + ν −2 r cos θ E

1 + ν −2 r sin θ E

r −1 sin θ

1 + ν −2 r sin θ E

−

r 3 cos θ

1 − 3ν 2 r cos θ E

5+ν 2 r sin θ E

r 3 sin θ

1 − 3ν 2 r sin θ E

−

r 2 log r

r log r cos θ

r log r sin θ

θ θr 2

θr cos θ

θr sin θ

1 + ν −1 r E

−

1 + ν −2 r cos θ E

5+ν 2 r cos θ E

Appendix E: The Bi-harmonic Equation

511

Table E.3 Same as Table E.2 Φ(r, θ)

ur

r n cos(nθ)

−

r n sin(nθ) r n+2 cos(nθ) r n+2 sin(nθ) r −n cos(nθ) r −n sin(nθ) r −n+2 cos(nθ) r −n+2 sin(nθ)

n(1 + ν) n−1 r cos(nθ) E n(1 + ν) n−1 − sin(nθ) r E 2 − n − (2 + n)ν n+1 cos(nθ) r E 2 − n − (2 + n)ν n+1 sin(nθ) r E n(1 + ν) −n−1 cos(nθ) r E n(1 + ν) −n−1 sin(nθ) r E 2 + n + (n − 2)ν −n+1 cos(nθ) r E 2 + n + (n − 2)ν −n+1 sin(nθ) r E

uθ n(1 + ν) n−1 r sin(nθ) E n(1 + ν) n−1 − cos(nθ) r E 4 + n + nν n+1 sin(nθ) r E 4 + n + nν n+1 − cos(nθ) r E n(1 + ν) −n−1 sin(nθ) r E n(1 + ν) −n−1 − cos(nθ) r E 4 − n − νn −n+1 − sin(nθ) r E 4 − n − νn −n+1 cos(nθ) r E

References

1. Chen HC (1985) Theory of electromagnetic waves: a coordinate-free approach. McGraw-Hill Book Company, New York 2. Gibbs JW, Wilson EB (1901) Vector analysis. Yale University Press, New Haven 3. Gurtin ME (1972) Theory linear theory of elasticity. In: Truesdell C (ed) Handbuch der Physik. Springer, Berlin, pp 1–273 4. Lurie AI (2005) Theory of elasticity. Springer, Berlin 5. Pach K, Frey T (1964) Vector and tensor analysis. Terra, Budapest 6. Schumann W, Dubas M (1979) Holographic interferometry. Springer, Berlin 7. Schumann W, Zürcher J-P, Cuche D (1985) Holography and deformation analysis. Springer, Berlin 8. Weatherburn CE (1960) Advanced vector analysis. G. Bell and Sons Ltd, London 9. Howland RA (2006) Intermediate dynamics. Springer, New York 10. Biscari P, Poggi C, Virga EG (1999) Mechanics notebook. Liguori Editore, Napoli 11. Adams RA (1999) Calculus: a complete course. Addison-Wesley, Longman Ltd., Ontario 12. Salas SL, Hille E (1995) Calculus: one and several variables. Wiley Inc, New York 13. Beer FP, Johnston ER, Eisenberg ER (2004) Vector mechanics for engineers (statics), vol 1, 7th edn. McGraw-Hill, New York 14. Gere JM (2001) Mechanics of materials, 5th edn. Brooks/Cole, Pacific Grove 15. Solomon L (1968) Élasticité Lineaire. Mason, Paris (in French) 16. Kreyszig E (2011) Advanced engineering mathematics, 10th edn. Wiley Inc, New York 17. Myint-U T, Debnath L (2007) Linear partial differential equations for scientists and engineers, 4th edn. Birkhäuser, Boston 18. Trim DW (1990) Applied partial differential equations. PWS-Kent Pub. Co., Boston 19. Sneddon IN (1951) Fourier transforms. McGraw-Hill Book Company Inc, New York 20. Skalskaya IP, Lebedev NN, Uflyand YaS (1965) Problems of mathematical physics. PrenticeHall Inc, Englewood Cliffs 21. Tolstov GP (1976) Fourier series. Dover Publications, Mineola 22. Little WR (1973) Elasticity, Prentice-Hall Inc, Englewood Cliffs, New Jersey 23. Teodorescu PP (2013) Treatise on classical elasticity. Springer, Dordrecht 24. Uflyand YaS (1968) Survey of articles on the applications of integral transforms in the theory of elasticity. Nauka, Leningrad, URSS (in Russian) 25. Debnath L, Bhatta D (2014) Integral transforms and their applications, 2nd edn. CRC Press, Boca Raton 26. Kanwal RP (2004) Generalized functions: theory and applications, 3rd edn. Birkhäuser, Boston 27. Gelfand IM, Shilov GE (1964) Generalized functions: properties and operations, vol 1. Academic Press, New York © Springer Nature B.V. 2020 C. D. Coman, Continuum Mechanics and Linear Elasticity, Solid Mechanics and Its Applications 238, https://doi.org/10.1007/978-94-024-1771-5

513

514

References

28. Vladimirov VS (1979) Generalized functions in mathematical physics. Mir Publishers, Moscow 29. Zemanian AH (2003) Distribution theory and transform analysis. Dover Publications, New York 30. Brown JW, Churchill RV (2008) Complex variables and applications. McGraw-Hill, New York 31. Ablowitz MJ, Fokas AS (1997) Complex variables. Cambridge University Press, Cambridge 32. Lavrentiev M, Shabat B (1977) Méthodes de la Théorie des Fonctions d’une Variable Complexe. Mir Publishers, Moscow 33. Sidorov YuV, Fedoryuk MV, Shabunin MI (1985) Lectures on the theory of functions of a complex variable. Mir Publishers, Moscow 34. Saff EB, Snider AD (1976) Fundamentals of complex analysis for mathematics, science and engineering, Prentice Hall Inc, Englewood Cliffs 35. Biezeno CB, Grammel R (1955) Engineering dynamics, vol 1 (Theory of Elasticity). Blackie & Son Ltd., London 36. Girkmann K (1959) Flächentragwertke, 2nd edn. Springer, Wien (in German) 37. Filonenko-Borodich M (1963) Theory of elasticity. Mir Publishers, Moscow 38. Timoshenko SP, Goodier JN (1970) Theory of elasticity, International edn. McGraw-Hill Book Company, Auckland 39. Muskhelishvili NI (1963) Some basic problems of the mathematical theory of elasticity. P.Noordhoof Ltd., Groningen 40. Amenzade YuA (1979) Theory of elasticity. Mir Publishers, Moscow 41. Leipholz H (1974) Theory of elasticity. Noordhoff International Publishing, Leyden 42. Saada AS (1993) Elasticity: theory and applications. Krieger Publishing Company, Malabar 43. Sokolnikoff IV (1956) Mathematical theory of elasticity. McGraw-Hill Book Company Inc, New York 44. England AE (1971) Complex variable methods in elasticity. Wiley-Interscience, London 45. Boyce WE, DiPrima RC (1996) Elementary differential equations and boundary value problems. Wiley, New York 46. Coddington EA (1989) An introduction to ordinary differential equations. Dover Publications, New York 47. Sobrero L (1942) Elasticidade. Livraria Boffoni, Rio de Janeiro (in Portuguese) 48. Barber JR (2002) Elasticity. Kluwer Academic Publishers, Dordrecht 49. Churchill RV (1972) Operational methods, 3rd edn. McGraw-Hill Book Company, New York

Index

A Acceleration field, 128 Airy stress function, 379 Almansi Representation Theorem, 491 Alternator, see Ricci’s permutation symbol Analytic function (in C), 483 Angular momentum (of Pt ), 170 Anisotropic (Cauchy-elastic material), 214 Anticlastic curvature, 279 Antiplane stress/strain, 274 Axial vector, 31 B Bar bending plane, 268 flexural rigidity, 269 neutral axis, 268 neutral plane, 268 Beltrami–Michell equation, 299 Bi-harmonic complex potentials, 497 equation, 380 inhomogeneous, 409 Michell solution, 508 Goursat representation, 496 Blatz-Ko elastic material, 221 Boundary conditions linear elasticity, 253 plane elasticity, 373 Boundary-value problem linear elasticity, 253, 254 plane elasticity, 373 Bulk modulus, 251

C Cauchy-elastic materials, 209 Cauchy–Green left deformation tensor, 141 right deformation tensor, 139 Cauchy–Riemann equations, 483 Cauchy’s Stress Theorem, 172 Cauchy traction vector, 168 Cayley–Hamilton equation, 74 Centroid, 468 axial symmetry, 471 polar symmetry, 472 Cesàro–Volterra formula, 293 Change of observer, 198 Co-axial tensors, 78 Compatible (infinitesimal) strain field, 289 Complex torsion potential, 350 Complex variable methods plane elasticity, 495–504 torsion, 350–354 Compliance tensor, 250 Concentrated force problems, 399, 425, 427, 432, 434–437, 441–442, 453, 457 Configuration current, 122 reference, 121 Conservative fields, 288 Constitutive Principle for Elastic Materials with Constraints, 228 Constitutive relation, 197 Contracted product definition, 43 double-dot, 44 triple-dot, 45 Convected rates of change, 237

© Springer Nature B.V. 2020 C. D. Coman, Continuum Mechanics and Linear Elasticity, Solid Mechanics and Its Applications 238, https://doi.org/10.1007/978-94-024-1771-5

515

516 Conversion between plane stress and plane strain, 375–376, 410 Convolution, see Fourier transforms Coordinate transformation, 35–38, 411 Curl, 93, 95

D Deformation, 122 gradient, 133 rate, 150 Del operator, 91 Dilatation, 126 Dirac delta function, 479 integral representation, 482 sifting property, 481 Displacement field, 136 Divergence operator, 92, 94 Divergence Theorem basic form, 102 generalised, 104 Double-dot (contracted) products, 44 Double-vector product, 49 Dyad, dyadic, 21

E Edge dislocation, 311 Eigenvalues and eigenvectors definition of, 71 for symmetric tensors, 73 Elastic stored energy, 219 Elastostatics, linear, 254 Energy-balance equation, 219 Equation of continuity, 165 Equipollent, see forces, statically equivalent Equivalent motions, 199 Euclidean point space, 18 Euler–Bernoulli bending law, 269 Euler differential equations, 391, 409, 507 Eulerian description, 123 principal axes, 143 Exponential of a second-order tensor, 239 External power, 218 Extra stress, 228 Extremal tangential stresses, 183

F Flamant solution (half-plane), 399–403, 427 Forces body & surface, 166 statically equivalent, 262

Index Fourier transforms, 475 evaluation using integrals in C, 485 examples, 487 for multivariate functions, 477 properties convolution, 477 differentiation, 477 shifting, 477 useful integrals, 488

G Galilean transformation, 199 Generalised plane stress, 376 Gradient, 89 Green’s identities, 369 Green’s Theorem, 369

H Hadamard–Green elastic material, 220 Half-plane, elastic, 421, 423, 425 Heisenberg delta function, 426 Hessian, 90 Hooke’s Law anisotropic solid, 247 inverted form, 250 isotropic solid, 248 plane strain, 373–374 plane stress, 375–376 Huygens–Steiner Theorem, see inertia tensor Hydrostatic stress, 184 Hyperelastic (or Green-elastic) material, 220

I Ideal fluid, 233 Incompatibility operator definition, 285 properties, 287 Incompressibility, 229 Inertia tensor, 467 centroidal, 468 Huygens–Steiner Theorem, 468 moment of inertia Euler, 467 geometrical vs. mechanical, 473 Parallel-Axis Theorem, 469 polar, 467 products of inertia, 469 relative to an axis, 468 principal axes, 470 second moment of area, 473

Index

517

Inextensibility, 230 Infinitesimal rotation tensor, 245 Infinitesimal strain tensor, 244 Internal constraint, 227 Invariants, principal, 68 Isochoric, see motion Isotropic Cauchy-elastic material, 214 fields, 216 functional linear, 216 representation theorems, 216 Isotropic tensor, see tensor (second-order)

J Jacobian matrix, 87 of the motion, 134 derivative of, 149 Jaumann stress rate, 238

K Kelvin problem (2D), 434–437 Kinematics, 121 Kinetic energy, 218 Kirsch problem, 389, 393 Kolosov complex potentials, see harmonic, complex potentials Kolosov’s formulae for displacements, 499 for stresses, 498 Kronecker delta, 4

L Lagrangian description, 123 principal axes, 142 strain tensor, 141 Lamé constants, 248 parameters, 63 Laplacian, 94, 95 Linear momentum (of Pt ), 170 Localization Theorem, 165

M Mass density, 164 Material symmetry, 211 Material time derivative, 128 Matrix

adjugate, 7 cofactor of an element, 6 determinant, 6 identity, 4 minor determinant, 6 (multiplicative) inverse, 7 singular, 7 Michell solution, see bi-harmonic equation Modified stress function (for the torsion problem), 342 Moduli, elastic, 247 Modulus of compression, see bulk modulus Moment of inertia, see inertia tensor Mooney–Rivlin material, 231 Motion, 122 homogeneous, 143 isochoric, 138

N Nanson’s formula, 138 Navier–Lamé equation, 259 Navier–Stokes equation, 236 Neo-Hookean material, 231 Newtonian viscous fluid, 235 Normal stress, 179

bi-

O Objective fields, 200 Objectively equivalent motions, see equivalent motions Observer invariance, see Principle of Material Frame-Indifference Octahedral planes/stresses, 193 Ogden material, 231 Orthogonal tensor, 27

P Papkovitch–Neuber representation, 444, 445 plane elasticity, 447 stress tensor, 447 Parallel-Axis Theorem, see inertia tensor Piola–Kirchhoff stress tensor, 186 Planar reversal, 58 Plane strain, 373 Plane stress, 375 Poisson’s ratio, 248 Polar Decomposition Theorem, 79 Pole (of order m ≥ 1), 484 Prandtl stress function, 339 Principal stress, 178 Principal stretches, 142

518 Principle of Action and Reaction, 171 of Conservation of Angular Momentum, 170 of Conservation of Linear Momentum, 170 of Determinism for the Stress, 206 of Local Action, 207 of Mass Conservation, 162 of Material Frame-Indifference, 202 of Superposition, 261 Projection of a vector along a given direction, 25 of a vector onto a given plane, 25 Pseudo-scalars, vectors, tensors, 41 Pure bending (cuboid), 127

Q Quadratic form, 257 Quarter-plane, elastic, 437

R Rate-of-strain tensor, see stretching tensor Reference triad, 20 Reiner–Rivlin fluid, 235 Representation theorems, see isotropic Residue, 484 Resultant force (on Pt ), 169 Resultant moment (on Pt ), 170 Reynolds’ Transport Theorem, 153 Ricci’s permutation symbol, 15 Rigid-body motion, 123 Rivlin–Ericksen tensors, 238 Rotation finite, 56 representation (involving two reference triads), 33 Rodrigues’ formula, 58 Rotation tensor, 142

S Saint-Venant’s compatibility equations, 290 Principle, 262 warping function, 328 Scalar triple product, 12, 13, 17 Screw dislocation, 311 Second moment of area, see inertia tensor Semi-inverse method of solution, 263 Shear modulus, 249 Shear (or tangential) stress, 179

Index Simple fluid, 232 materials, 207 shear, 125 torsion, 126 Simple extension, 125 Simply and multiply connected domains, 283 Solid (Cauchy-elastic) material, 214 Spectral Representation Theorem, 76 Spin tensor, 149 State of stress, 169 Stiffness tensor, 246 Stokes’ Theorem, 104 complex version, 369 Strain infinitesimal distortional, 245 volumetric/spherical, 245 Stress concentration, 392, 394 infinitesimal distortional, 250 spherical, 250 power, 218 Stretch in a given direction, 139 tensors (left & right), 142 Stretching tensor, 149 Summation convention, 14 Superposition Principle, see Principle of Superposition Sylvester’s criterion, 257 Symmetry group, 212 System of coordinates Cartesian, 20 curvilinear, 61 cylindrical, 63 spherical, 65

T Taylor’s expansion, 90 Tensor cofactor, 27 fourth-order, 50 conjugation (or square tensor) product, 54 generalised tensor product, 53 identity in Lin 4 , 51 identity in SSym 4 , 52 invertible, 52 positive-definite, 276

Index SSym 4 , 52 strongly elliptic, 276 Sym 4 , 51 transpose, 51 identities, see vector & tensor identities second-order, 20 invertible, 27 isotropic, 40 orthogonal, 27 positive (semi-) definite, 27 skew-symmetric, 27 skew-symmetric part, 30 Square-Root Theorem, 78 symmetric, 27 symmetric part, 30 transpose, 27 third-order, 38 Tensor of inertia, see inertia tensor Torsion constant, 333 Torsion of cylinders annular cross section circular, 362–364 elliptical, 364–365 elliptical cross section, 354–356 multiply connected cross section, 345– 347 rectangular cross section, 358–362 triangular cross section, 356–358 Torsional rigidity, 333 Trace, 44, 68 Trajectory of shear stress, 347 Transport formulae, 151 Triclinic material, 213 Twist per unit length, 325

519 U Uniaxial tension test, 249 Uniqueness, 255

V Vector & tensor identities, 96 Vector of a tensor, 32 Vector product between a vector and a tensor, 46 between two tensors, 48 between two vectors, 12, 17 Vector space definition, 8 Euclidean, 11 finite-dimensional, 9 linear transformation, 9 standard basis, 9 subspace, 11 Vector triple product, 17 Velocity field, 128 Velocity gradient, 148 Volumetric dilatation, 245

W Weingarten-Volterra dislocation, 308

Y Young’s modulus, 248