Introduction to Continuum Mechanics This page intentionally left blank Introduction to Continuum Mechanics Third Edition W. MICHAEL LAI Professor of Mechanical Engineering and Orthopaedic Bioengineering Columbia University, New York, NY, USA DAVID RUBIN Principal Weidlinger Associates, New York, NY, USA ERHARD KREMPL Rosalind and John J. Redfern, Jr, Professor of Engineering Rensselaer Polytechnic Institute, Troy, NY, USA U T T E R W O R T H E I N E M A N N First published by Pergamon Press Ltd 1993 Reprinted 1996 © Butterworth-Heinemann Ltd 1993 Reprinted in 1999 by Butterworth-Heinemann Ό^ A member of the Reed Elsevier group All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. © Recognizing the importance of preserving what has been written, Butterworth-Heinemann prints its books on acid-free paper whenever possible. n ' f i R A l ' Butterworth-Heinemann supports the efforts of American E3MJEAT Forests and the Global ReLeaf program in its campaign for ^^2000 the betterment of trees, forests, and our environment. Library of Congress Cataloging-in-Publication Data Lai, W. Michael, 1930Introduction to continuum mechanics/W. Michael Lai, David Rubin, Erhard Krempl - 3rd ed. p. cm. ISBN 0 7506 2894 4 1. Contiuum mechanics I. Rubin, David, 1942II. Krempl, Erhard III. Title QA808.2.L3 1993 531-dc20 93-30117 for Library of Congress CIP British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library. The publisher offers special discounts on bulk orders of this book. For information, please contact: Manager of Special Sales Butterworth-Heinemann 225 Wildwood Avenue Woburn, MA 01801-2041 Tel: 781-960-2500 Fax: 781-960-2620 For information on all Butterworth-Heinemann publications available, contact our World Wide Web home page at: http://www.bh.com 10 9 8 7 6 5 4 3 Printed in the United States of America Contents Preface to the Third Edition xii Preface to the First Edition xiii The Authors xiv Introduction Chapter 1 1.1 Continuum Theory 1.2 Contents of Continuum Mechanics 1 1 1 Chapter 2 3 Part A 2A1 2A2 2A3 2A4 2A5 PartB 2B1 2B2 2B3 2B4 2B5 2B6 2B7 Tensors The Indicial Notation 3 Summation Convention, Dummy Indices Free Indices Kronecker Delta Permutation Symbol Manipulations with the Indicial Notation 3 5 6 7 8 Tensors 11 Tensor: A Linear Transformation Components of a Tensor Components of a Transformed Vector Sum of Tensors Product of Two Tensors Transpose of a Tensor Dyadic Product of Two Vectors 11 13 16 17 18 20 21 V vi Contents 2B8 2B9 2B10 2B11 Trace of a Tensor Identity Tensor and Tensor Inverse Orthogonal Tensor Transformation Matrix Between Two Rectangular Cartesian Coordinate Systems Transformation Laws for Cartesian Components of Vectors Transformation Law for Cartesian Components of a Tensor Defining Tensors by Transformation Laws Symmetric and Antisymmetric Tensors The Dual Vector of an Antisymmetric Tensor Eigenvalues and Eigenvectors of a Tensor Principal Values and Principal Directions of Real Symmetric Tensors Matrix of a Tensor with Respect to Principal Directions Principal Scalar Invariants of a Tensor 22 23 24 Tensor Calculus 47 Tensor-valued functions of a Scalar Scalar Field, Gradient of a Scalar Function Vector Field, Gradient of a Vector Field Divergence of a Vector Field and Divergence of a Tensor Field Curl of a Vector Field 47 49 53 54 55 Curvilinear Coordinates 57 2D1 Polar Coordinates 2D2 Cylindrical Coordinates 2D3 Spherical Coordinates Problems 57 61 63 68 2B12 2B13 2B14 2B15 2B16 2B17 2B18 2B19 2B20 Part C 2C1 2C2 2C3 2C4 2C5 Part D Chapter 3 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 Kinematics of a Continuum Description of Motions of a Continuum Material Description and Spatial Description Material Derivative Acceleration of a Particle in a Continuum Displacement Field Kinematic Equations For Rigid Body Motion Infinitesimal Deformations Geometrical Meaning of the Components of the Infinitesimal Strain Tensor 26 28 30 32 35 36 38 43 44 45 79 79 83 85 87 92 93 94 99 Contents vii 3.9 Principal Strain 3.10 Dilatation 3.11 The Infinitesimal Rotation Tensor 3.12 Time Rate of Change of a Material Element 3.13 The Rate of Deformation Tensor 3.14 The Spin Tensor and the Angular Velocity Vector 3.15 Equation of Conservation Of Mass 3.16 Compatibility Conditions for Infinitesimal Strain Components 3.17 Compatibility Conditions for the Rate of Deformation Components 3.18 Deformation Gradient 3.19 Local Rigid Body Displacements 3.20 Finite Deformation 3.21 Polar Decomposition Theorem 3.22 Calculation of the Stretch Tensor from the Deformation Gradient 3.23 Right Cauchy-Green Deformation Tensor 3.24 Lagrangian Strain Tensor 3.25 Left Cauchy-Green Deformation Tensor 3.26 Eulerian Strain Tensor 3.27 Compatibility Conditions for Components of Finite Deformation Tensor 3.28 Change of Area due to Deformation 3.29 Change of Volume due to Deformation 3.30 Components of Deformation Tensors in other Coordinates 3.31 Current Configuration as the Reference Configuration Problems Chapter 4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10 Stress Stress Vector Stress Tensor Components of Stress Tensor Symmetry of Stress Tensor - Principle of Moment of Momentum Principal Stresses Maximum Shearing Stress Equations of Motion - Principle of Linear Momentum Equations of Motion in Cylindrical and Spherical Coordinates Boundary Condition for the Stress Tensor Piola Kirchhoff Stress Tensors 105 105 106 106 108 111 112 114 119 120 121 121 124 126 128 134 138 141 144 145 146 149 158 160 173 173 174 176 178 182 182 187 190 192 195 viii Contents 4.11 Equations of Motion Written With Respect to the Reference Configuration 4.12 Stress Power 4.13 Rate of Heat Flow Into an Element by Conduction 4.14 Energy Equation 4.15 Entropy Inequality Problems Chapter 5 The Elastic Solid 5.1 Mechanical Properties 5.2 Linear Elastic Solid Part A 5.3 5.4 5.5 5.6 5.7 5.8 5.9 5.10 5.11 5.12 5.13 5.14 5.15 5.16 5.17 5.18 5.19 5.20 5.21 Part B 5.22 5.23 5.24 201 203 207 208 209 210 217 217 220 Linear Isotropie Elastic Solid 225 Linear Isotropie Elastic Solid Young's Modulus, Poisson's Ratio, Shear Modulus, and Bulk Modulus Equations of the Infinitesimal Theory of Elasticity Navier Equation in Cylindrical and Spherical Coordinates Principle of Superposition Plane Irrotational Wave Plane Equivoluminal Wave Reflection of Plane Elastic Waves Vibration of an Infinite Plate Simple Extension Torsion of a Circular Cylinder Torsion of a Noncircular Cylinder Pure Bending of a Beam Plane Strain Plane Strain Problem in Polar Coordinates Thick-walled Circular Cylinder under Internal and External Pressure Pure Bending of a Curved Beam Stress Concentration due to a Small Circular Hole in a Plate under Tension Hollow Sphere Subjected to Internal and External Pressure 225 228 232 236 238 238 242 248 251 254 258 266 269 275 281 284 285 287 291 Linear Anisotropie Elastic Solid 293 Constitutive Equations for Anisotropie Elastic Solid Plane of Material Symmetry Constitutive Equation for a Monoclinic Anisotropie Elastic Solid 293 296 299 Contents ix 5.25 5.26 5.27 5.28 5.29 5.30 5.31 Part C Constitutive Equations for an Orthotropic Elastic Solid Constitutive Equation for a Transversely Isotropie Elastic Material Constitutive Equation for Isotropie Elastic Solid Engineering Constants for Isotropie Elastic Solid. Engineering Constants for Transversely Isotropie Elastic Solid Engineering Constants for Orthotropic Elastic Solid Engineering Constants for a Monoclinic Elastic Solid. Constitutive Equation For Isotropie Elastic Solid Under Large Deformation 314 5.32 Change of Frame 5.33 Constitutive Equation for an Elastic Medium under Large Deformation. 5.34 Constitutive Equation for an Isotropie Elastic Medium 5.35 Simple Extension of an Incompressible Isotropie Elastic Solid 5.36 Simple Shear of an Incompressible Isotropie Elastic Rectangular Block 5.37 Bending of a Incompressible Rectangular Bar. 5.38 Torsion and Tension of an Incompressible Solid Cylinder Problems Chapter 6 6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9 6.10 6.11 6.12 6.13 6.14 6.15 6.16 301 303 306 307 308 311 312 Newtonian Viscous Fluid Fluids Compressible and Incompressible Fluids Equations of Hydrostatics Newtonian Fluid Interpretation of A and μ Incompressible Newtonian Fluid Navier-Stokes Equation for Incompressible Fluids Navier-Stokes Equations for In compressible Fluids in Cylindrical and Spherical Coordinates Boundary Conditions Streamline, Pathline, Streakline, Steady, Unsteady, Laminar and Turbulent Flow Plane Couette Flow Plane Poiseuille Flow Hagen Poiseuille Flow Plane Couette Flow of Two Layers of Incompressible Fluids Couette Flow Flow Near an Oscillating Plate 314 319 322 324 325 327 331 335 348 348 349 350 355 357 359 360 364 365 366 371 372 374 377 380 381 x Contents 6.17 6.18 6.19 6.20 6.21 Dissipation Functions for Newtonian Fluids Energy Equation for a Newtonian Fluid Vorticity Vector Irrotational Flow Irrotational Flow of an Inviscid, Incompressible Fluid of Homogeneous Density 6.22 Irrotational Flows as Solutions of Navier-Stokes Equation 6.23 Vorticity Transport Equation for Incompressible Viscous Fluid with a Constant Density 6.24 Concept of a Boundary Layer 6.25 Compressible Newtonian Fluid 6.26 Energy Equation in Terms of Enthalpy 6.27 Acoustic Wave 6.28 Irrotational, Barotropic Flows of Inviscid Compressible Fluid 6.29 One-Dimensional Flow of a Compressible Fluid Problems 383 384 387 390 391 394 396 399 401 402 404 408 412 419 Chapter 7 Integral Formulation of General Principles 7.1 Green's Theorem 7.2 Divergence Theorem 7.3 Integrals over a Control Volume and Integrals over a Material Volume 7.4 Reynolds Transport Theorem 7.5 Principle of Conservation of Mass 7.6 Principle of Linear Momentum 7.7 Moving Frames 7.8 Control Volume Fixed with Respect to a Moving Frame 7.9 Principle of Moment of Momentum 7.10 Principle of Conservation of Energy Problems 427 427 430 433 435 437 440 447 449 451 454 458 Chapter 8 Non-Newtonian Fluids 462 Linear Viscoelastic Fluid 464 Linear Maxwell Fluid Generalized Linear Maxwell Fluid with Discrete Relaxation Spectra Integral Form of the Linear Maxwell Fluid and of the Generalized Linear Maxwell Fluid with Discrete Relaxation Spectra Generalized Linear Maxwell Fluid with a Continuous Relaxation Spectrum 464 471 Part A 8.1 8.2 8.3 8.4 473 474 Contents xi Part B 8.5 8.6 8.7 8.8 8.9 8.10 8.11 8.12 8.13 8.14 8.15 8.16 8.17 8.18 8.19 Part C Nonlinear Viscoelastic Fluid 476 Current Configuration as Reference Configuration Relative Deformation Gradient Relative Deformation Tensors Calculations of the Relative Deformation Tensor History of Deformation Tensor. Rivlin-Ericksen Tensors Rivlin-Ericksen Tensor in Terms of Velocity Gradients The Recursive Formulas Relation Between Velocity Gradient and Deformation Gradient Transformation Laws for the Relative Deformation Tensors under a Change of Frame Transformation law for the Rivlin-Ericksen Tensors under a Change of Frame Incompressible Simple Fluid Special Single Integral Type Nonlinear Constitutive Equations General Single Integral Type Nonlinear Constitutive Equations Differential Type Constitutive Equations Objective Rate of Stress The Rate Type Constitutive Equations 476 477 478 480 486 Viscometric Flow Of Simple Fluid 516 8.20 Viscometric Flow 8.21 Stresses in Viscometric Flow of an Incompressible Simple Fluid 8.22 Channel Flow 8.23 Couette Flow Problems Appendix: Matrices 491 493 494 496 497 498 503 503 506 511 516 520 523 526 532 537 Answer to Problems 543 References 550 Index 552 Preface to the Third Edition The first edition of this book was published in 1974, nearly twenty years ago. It was written as a text book for an introductory course in continuum mechanics and aimed specifically at the junior and senior level of undergraduate engineering curricula which choose to introduce to the students at the undergraduate level the general approach to the subject matter of continuum mechanics. We are pleased that many instructors of continuum mechanics have found this little book serves that purpose well. However, we have also understood that many instructors have used this book as one of the texts for a beginning graduate course in continuum mechanics. It is this latter knowledge that has motivated us to write this new edition. In this present edition, we have included materials which we feel are suitable for a beginning graduate course in continuum mechanics. The following are examples of the additions: 1. Anisotropie elastic solid which includes the concept of material symmetry and the constitutive equations for monoclinic, orthotropic, transversely isotropic and isotropic materials. 2. Finite deformation theory which includes derivations of the various finite deformation tensors, the Piola-Kirchhoff stress tensors, the constitutive equations for an incompressible nonlinear elastic solid together with some boundary value problems. 3. Some solutions of classical elasticity problems such as thick-walled pressure vessels (cylinders and spheres), stress concentrations and bending of curved bars. 4. Objective tensors and objective time derivatives of tensors including corotational derivative and convected derivatives. 5. Differential type, rate type and integral type linear and nonlinear constitutive equations for viscoelastic fluids and some solutions for the simple fluid in viscometric flows. 6. Equations in cylindrical and spherical coordinates are provided including the use of different coordinates for the deformed and the undeformed states. We wish to state that notwithstanding the additions, the present edition is still intended to be "introductory" in nature, so that the coverage is not extensive. We hope that this new edition can serve a dual purpose: for an introductory course at the undergraduate level by omitting some of the "intermediate level" material in the book and for a beginning graduate course in continuum mechanics at the graduate level. W. Michael Lai David Rubin Erhard Krempl July, 1993 xii Preface to the First Edition This text is prepared for the purpose of introducing the concept of continuum mechanics to beginners in the field. Special attention and care have been given to the presentation of the subject matter so that it is within the grasp of those readers who have had a good background in calculus, some differential equations, and some rigid body mechanics. For pedagogical reasons the coverage of the subject matter is far from being extensive, only enough to provide for a better understanding of later courses in the various branches of continuum mechanics and related fields. The major portion of the material has been successfully class-tested at Rensselaer Polytechnic Institute for undergraduate students. However, the authors believe the text may also be suitable for a beginning graduate course in continuum mechanics. We take the liberty to say a few words about the second chapter. This chapter introduces second-order tensors as linear transformations of vectors in a three dimensional space. From our teaching experience, the concept of linear transformation is the most effective way of introducing the subject. It is a self-contained chapter so that prior knowledge of linear transformations, though helpful, is not required of the students. The third-and higher-order tensors are introduced through the generalization of the transformation laws for the secondorder tensor. Indicial notation is employed whenever it economizes the writing of equations. Matrices are also used in order to facilitate computations. An appendix on matrices is included at the end of the text for those who are not familiar with matrices. Also, let us say a few words about the presentation of the basic principles of continuum physics. Both the differential and integral formulation of the principles are presented, the differential formulations are given in Chapters 3,4, and 6, at places where quantities needed in the formulation are defined while the integral formulations are given later in Chapter 7. This is done for a pedagogical reason: the integral formulations as presented required slightly more mathematical sophistication on the part of a beginner and may be either postponed or omitted without affecting the main part of the text. This text would never have been completed without the constant encouragement and advice from Professor F. F. Ling, Chairman of Mechanics Division at RPI, to whom the authors wish to express their heartfelt thanks. Gratefully acknowledged is the financial support of the Ford Foundation under a grant which is directed by Dr. S. W. Yerazunis, Associate Dean of Engineering. The authors also wish to thank Drs. V. C. Mow and W. B. Browner, Jr. for their many useful suggestions. Special thanks are given to Dr. H. A. Scarton for painstakingly compiling a list of errata and suggestions on the preliminary edition. Finally, they are indebted to Mrs. Geri Frank who typed the entire manuscript. W. Michael Lai David Rubin Erhard Krempl Division of Mechanics, Rensselaer Polytechnic Institute September, 1973 XIII The Authors W. Michael Lai (Ph.D., University of Michigan) is Professor of Mechanical Engineering and Orthopaedic Bioengineering at Columbia University, New York, New York. He is a member of ASME (Fellow), AIMBE (Fellow), ASCE, AAM, ASB,ORS, AAAS, Sigma Xi and Phi Kappa Phi. David Rubin (Ph.D., Brown University) is a principal at Weidlinger Associates, New York, New York. He is a member of ASME, Sigma Xi, Tau Beta Pi and Chi Epsilon. Erhard Krempl (Dr.-Ing., Technische Hochschule München) is Rosalind and John J. Refern Jr. Professor of Engineering at Rensselaer Polytechnic Institute. He is a member of ASME (Fellow), AAM (Fellow), ASTM, ASEE, SEM, SES and Sigma Xi. XIV 1 Introduction 1.1 CONTINUUM THEORY Matter is formed of molecules which in turn consist of atoms and sub-atomic particles. Thus matter is not continuous. However, there are many aspects of everyday experience regarding the behaviors of materials, such as the deflection of a structure under loads, the rate of discharge of water in a pipe under a pressure gradient or the drag force experienced by a body moving in the air etc., which can be described and predicted with theories that pay no attention to the molecular structure of materials. The theory which aims at describing relationships between gross phenomena, neglecting the structure of material on a smaller scale, is known as continuum theory. The continuum theory regards matter as indefinitely divisible. Thus, within this theory, one accepts the idea of an infinitesimal volume of materials referred to as a particle in the continuum, and in every neighborhood of a particle there are always neighbor particles. Whether the continuum theory is justified or not depends on the given situation; for example, while the continuum approach adequately describes the behavior of real materials in many circumstances, it does not yield results that are in accord with experimental observations in the propagation of waves of extremely small wavelength. On the other hand, a rarefied gas may be adequately described by a continuum in certain circumstances. At any case, it is misleading to justify the continuum approach on the basis of the number of molecules in a given volume. After all, an infinitesimal volume in the limit contains no molecules at all. Neither is it necessary to infer that quantities occurring in continuum theory must be interpreted as certain particular statistical averages. In fact, it has been known that the same continuum equation can be arrived at by different hypothesis about the molecular structure and definitions of gross variables. While molecular-statistical theory, whenever available, does enhance the understanding of the continuum theory, the point to be made is simply that whether the continuum theory is justified in a given situation is a matter of experimental test, not of philosophy. Suffice it to say that more than a hundred years of experience have justified such a theory in a wide variety of situations. 1.2 Contents of Continuum Mechanics Continuum mechanics studies the response of materials to different loading conditions. Its subject matter can be divided into two main parts: (1) general principles common to all media, 1 2 Introduction and (2) constitutive equations defining idealized materials. The general principles are axioms considered to be self-evident from our experience with the physical world, such as conservation of mass, balance of linear momentum, of moment of momentum, of energy, and the entropy inequality law. Mathematically, there are two equivalent forms of the general principles: (1) the integral form, formulated for a finite volume of material in the continuum, and (2) the field equations for differential volume of material (particle) at every point of the field of interest. Field equations are often derived from the integral form. They can also be derived directly from the free body of a differential volume. The latter approach seems to suit beginners. In this text both approaches are presented, with the integral form given toward the end of the text. Field equations are important wherever the variations of the variables in the field are either of interest by itself or are needed to get the desired information. On the other hand, the integral forms of conservation laws lend themselves readily to certain approximate solutions. The second major part of the theory of continuum mechanics concerns the "constitutive equations" which are used to define idealized material. Idealized materials represent certain aspects of the mechanical behavior of the natural materials. For example, for many materials under restricted conditions, the deformation caused by the application of loads disappears with the removal of the loads. This aspect of the material behavior is represented by the constitutive equation of an elastic body. Under even more restricted conditions, the state of stress at a point depends linearly on the changes of lengths and mutual angle suffered by elements at the point measured from the state where the external and internal forces vanish. The above expression defines the linearly elastic solid. Another example is supplied by the classical definition of viscosity which is based on the assumption that the state of stress depends linearly on the instantaneous rates of change of length and mutual angle. Such a constitutive equation defines the linearly viscous fluid. The mechanical behavior of real materials varies not only from material to material but also with different loading conditions for a given material. This leads to the formulation of many constitutive equations defining the many different aspects of material behavior. In this text, we shall present four idealized models and study the behavior they represent by means of some solutions of simple boundary-value problems. The idealized materials chosen are (1) the linear isotropic and anisotropic elastic solid (2) the incompressible nonlinear isotropic elastic solid (3) the linearly viscous fluid including the inviscid fluid, and (4) the Non-Newtonian incompressible fluid. One important requirement which must be satisfied by all quantities used in the formulation of a physical law is that they be coordinate-invariant. In the following chapter, we discuss such quantities. 2 Tensors As mentioned in the introduction, all laws of continuum mechanics must be formulated in terms of quantities that are independent of coordinates. It is the purpose of this chapter to introduce such mathematical entities. We shall begin by introducing a short-hand notation - the indicial notation - in Part A of this chapter, which will be followed by the concept of tensors introduced as a linear transformation in Part B. The basic field operations needed for continuum formulations are presented in Part C and their representations in curvilinear coordinates in Part D. Part A The Indicial Notation 2A1 Summation Convention, Dummy Indices Consider the sum s = a\X\ + #2*2 + a3*3 + · · · + cif^n We can write the above equation in a compact form by using the summation sign: n * = Σαίχί (2A1.1) (2Α12) ι= 1 It is obvious that the following equations have exactly the same meaning as Eq. (2A1.2) 3 = Σαΐχΐ ;'-ι n s = Yiamxm m=l etc. 3 ( 2Α1 · 3 > (2A1.4) 4 Indicial Notation The index i in Eq. (2A1.2), or; in Eq. (2A1.3), or m in Eq. (2A1.4) is a dummy index in the sense that the sum is independent of the letter used. We can further simplify the writing of Eq.(2Al.l) if we adopt the following convention: Whenever an index is repeated once, it is a dummy index indicating a summation with the index running through the integers 1,2,..., n. This convention is known as Einstein's summation convention. Using the convention, Eq. (2A1.1) shortens to s = diXi (2A1.5) «/*/ = amXm = "jxj = - (2A1.6) We also note that It is emphasized that expressions such as αφρί are not defined within this convention. That is, an index should never be repeated more than once when the summation convention is used. Therefore, an expression of the form n ^aibiXi 1=1 must retain its summation sign. In the following we shall always take n to be 3 so that, for example, a,·*,· = amxm = αχχχ + a&2 + «3*3 «// = a mm = «11 + «22 + «33 afi[ = a\ ei + #2 e 2 + a3 e 3 The summation convention obviously can be used to express a double sum, a triple sum, etc. For example, we can write α χ χ 2 Σνίί (2A1.7) ι= 1 ;= 1 simply as ayXiXj (2A1.8) Expanding in full, the expression (2A1.8) gives a sum of nine terms, i.e., ayXiXj = 011*1*1 + ^12*1*2 + «13*1*3 + «21*2*1 + «22*2*2 +«23*2*3 + «31*3*1 + «32*3*2 + «33*3*3 (2A1.9) For beginners, it is probably better to perform the above expansion in two steps, first, sum over i and then sum over j (or vice versa), i.e., «/;*/*; = «l/*l*y + «2/*2*y + «3/*3*; Part A Free Indices 5 where a x x lj l j = 011*1*1 + 012*1*2 + 013*1*3 etc. Similarly, the triple sum 3 3 3 Σ Σ Σ */w** ( 2AL1 °) cijfi xi Xj Xfr (2A1.11) 1= 1 ; = 1 A : = l will simply be written as The expression (2A1.11) represents the sum of 27 terms. We emphasize again that expressions such as an JC/ Xj Xj or a^ x\ jt,· Xj x^ are not defined in the summation convention, they do not represent 3 3 a x x x Σ Σ Uijj 3 0Γ 3 Σ Σ Σ /=1 ; = 1 2A2 3 a /=1 / = 1 k=l ijkxixixjxk Free Indices Consider the following system of three equations x[ = α η χχ + a12x2 + 013*3 *2 = «21*1 + ^22*2 + 023*3 *3 = 031*1 + <*32*2 + 033*3 (2A2.1) Using the summation convention, Eqs. (2A2.1) can be written as *1 = a lmxm *2 = aimtm *3 = a?>nSm (2A2.2) which can be shortened into xi = aimxm. i=W (2A2.3) An index which appears only once in each term of an equation such as the index i in Eq. (2A2.3) is called a "free index." A free index takes on the integral number 1, 2, or 3 one at a time. Thus Eq. (2A2.3) is short-hand for three equations each having a sum of three terms on its right-hand side [i.e., Eqs. (2A2.1)]. A further example is given by 6 Indicial Notation «i = ßmÄm i = 1,23 (2A2·4) representing «i = ö n e i + 021*2 + 03i e 3 «2 = Ö l 2 e l + Ö22 e 2 + Ö32 e 3 *3 = Ö l 3 e l + Ö23 e 2 + 033^3 (2A2.5) We note that xj = a]n#,m,]= 1,2,3, is the same as Eq. (2A2.3) and ej = Qmfim, j= 1,2,3 is the same as Eq. (2A2.4). However, a/ = bj is a meaningless equation. The free index appearing in every term of an equation must be the same. Thus the following equations are meaningful a i + fy = ci ai + bjCjdj = 0 If there are two free indices appearing in an equation such as Τη = Aim Ajm i = 1,2,3;; = 1,2,3 (2A2.6) then the equation is a short-hand writing of 9 equations; each has a sum of 3 terms on the right-hand side. In fact, Tn =AlmAlm =Aiy4n T = ^11^21 + ^12^22 + ^13^23 \2 = ^ληΑτιη + A12A12+A13A13 A 7"l3 = ^\m ?>m = ^ 1 1 ^ 3 1 + ^ 1 2 ^ 3 2 + ^ 1 3 ^ 3 3 r 33 =^3mA3m = A3}A31 + A32A32+A3y433 Again, equations such as T ij = T ik have no meaning. 2A3 Kronecker Delta The Kronecker delta, denoted by ö,y, is defined as «»-{iSi? That is, (2A3i) Part A Permutation Symbol 7 ό 11 = ό 22 = ό 33 = 1 = ό 1 3 = <521 = 0 2 3 = ö 31 = ό 32 = 0 όη In other words, the matrix of the Kronecker delta is the identity matrix, i.e., [<*«>·] = (2A3.2) 1 0 θ" = 0 1 0 0 0 lj ό 11 ό 12 ö 13 21 ö 22 ό 23 ό 31 ό 32 ό 33 ö We note the following: (a) öü = όη + ό 22 + ό 33 = 1 + 1 + 1 = 3 (2A3.3) (b) δλιηαηχ = ό η α ι + ό12α2 + ό13α3 = αχ d 2mam = ö2\a\ + ö2lß2 + ö 23 ß 3 = α 2 ö 3mam = ό31αχ + ό32α2 + <533α3 = α3 Or, in general ^imam (c)dlmTmj i = 0llTlj+d12T2j+d13T3j ö 2mTmj = T ö T = T 3m mj (2A3.4) a = T lj 2j 3j or, in general Vim*mj *■ ij (2A3.5) °imPmj "ij (2A3.6) In particular, *inßmrfinj = *ij etc. (d) If e1,e2,e3 are unit vectors perpendicular to each other, then (2A3.7) 2A4 Permutation Symbol The permutation symbol, denoted by eijk is defined by b Sal· ijk — form an even +1 - 1 =according to whether i,j,k form an odd do not form a 0 permutation of 1,2,3 (2A4.1) 8 Indicial Notation i.e., ε 123 = ε 231 = ε 312 = ε 132 ε = ε 321 = ε +1 213= " 1 111 = ε 112 = ' ' ' = 0 We note that E ijk = Ejki = Ekij = ~Ejik = ~Eikj = ~Ekji (2A4.2) If βι,β2,β3 form a right-handed triad, then e i * e 2 = e3, e 2 Xe 3 = eh e 2 Xe! = - e 3 , βχΧβχ = 0,... which can be written for short as e Xe / , = *ijk*k = Ejki*k = 8kijek (2A4.3) Now, if a = aft, and b = bi% then a x b = (afii)x(bfi) = Ojbfaxej) = ajbjEijkek i.e., a x b = aibjEijkzk (2 A4.4) The following useful identity can be proven (see Prob. 2A7) E 2A5 ijmEklm = dikßjrdifijk (2A4.5) Manipulations with the Indicial Notation (a) Substitution If β/ = Uimbm (0 bi = Vimcm (ii) and then, in order to substitute thefcj'sin (ii) into (i) we first change the free index in (ii) from i to m and the dummy index m to some other letter, say n so that bm = Vnufin (iii) */ = UfaVnnCn (iv) Now, (i) and (iii) give Note (iv) represents three equations each having the sum of nine terms on its right-hand side. Part A Manipulations with the Indicial Notation 9 (b) Multiplication If P = ambm (i) q = cmdm (ii) and then, M = ^mbmcndn ("0 It is important to note thatpq * ambmcmdm. In fact, the right hand side of this expression is not even defined in the summation convention and further it is obvious that 3 P9 φ X m= l a mbmcmdm· Since the dot product of vectors is distributive, therefore, if a = α,-e,· and b = bi% then a · b = (afij)· (bfi) = αβ^ ■ e;) (iv) In particular, if el5e2,e3 are unit vectors perpendicular to one another, then e,· · ey = ό^ so that a-b = apj^ij = afii - afy = αφ\+αφ2+αΦ'}> (ν) T^nj-Xni = 0 (i) (c) Factoring If then, using the Kronecker delta, we can write tt; = Öißlj (Ü) Tijnj-Xdijnj = 0 (iii) (Tij-Mfinj = 0 (iv) so that (i) becomes Thus, (d) Contraction The operation of identifying two indices and so summing on them is known as contraction. For example, 7^ is the contraction of T^ Τα = Τη+Τ22+Τ33 (i) 10 Indicial Notation If Τ^λθδ^ΊμΕ^ (ϋ) then Tü = λθδα+ΊμΕα = 3λθ+2μΕα (iii) Part B Tensor - A Linear Transformation 11 Part B Tensors 2B1 Tensor - A Linear Transformation Let T be a transformation, which transforms any vector into another vector. If T transforms a into c and b into d, we write Ta = c and Tb = d. If T has the following linear properties: T(a+b) = Ta+Tb (2Bl.la) T(aa) = «Ta (2Bl.lb) where a and b are two arbitrary vectors and a is an arbitrary scalar then T is called a linear transformation. It is also called a second-order tensor or simply a tensor. An alternative and equivalent definition of a linear transformation is given by the single linear property: T(aa+ßb) = aTa+ßTb (2B1.2) where a and b are two arbitrary vectors and a and/J are arbitrary scalars. If two tensors, T and S, transform any arbitrary vector a in an identical way, then these tensors are equal to each other, i.e., Ta=Sa -» T=S. Example 2B1.1 Let T be a transformation which transforms every vector into a fixed vector n. Is this transformation a tensor? Solution. Let a and b be any two vectors, then by the definition of T, Ta = n, Tb = n and T(a+b) = n Clearly, T(a+b) * Ta+Tb Thus, T is not a linear transformation. In other words, it is not a tensor. t Scalars and vectors are sometimes called the zeroth and first order tensor, respectively. Even though they can also be defined algebraically, in terms of certain operational rules, we choose not to do so. The geometrical concept of scalars and vectors, which we assume that the students are familiar with, is quite sufficient for our purpose. 12 Tensors Example 2B1.2 Let T be a transformation which transforms every vector into a vector that is k times the original vector. Is this transformation a tensor? Solution. Let a and b be arbitrary vectors and a and ß be arbitrary scalars, then by the definition of T, Ta = fca, Tb = kb, and T(aa+j8b) = fc(aa+/3b) Clearly, T(aa+j3b) = a(k*)+ß(kb) = aTa+ßTb Thus, by Eq. (2B1.2), T is a linear transformation. In other words, it is a tensor. In the previous example, if k=0 then the tensor T transforms all vectors into zero. This tensor is the zero tensor and is symbolized by 0. Example 2B 1.3 Consider a transformation T that transforms every vector into its mirror image with respect to a fixed plane. Is T a tensor? Solution. Consider a parallelogram in space with its sides represented by vectors a and b and its diagonal represented the resultant a + b. Since the parallelogram remains a parallelogram after the reflection, the diagonal (the resultant vector) of the reflected parallelogram is clearly both T(a + b ) , the reflected (a + b), and Ta + Tb, the sum of the reflected a and the reflected b . That is, T(a + b) = Ta + Tb. Also, for an arbitrary scalar a , the reflection of aa is obviously the same as a times the reflection of a (i.e., T(aa )= «Ta) because both vectors have the same magnitude given by a times the magnitude of a and the same direction. Thus, by Eqs. (2B1.1), T is a tensor. Example 2B1.4 When a rigid body undergoes a rotation about some axis, vectors drawn in the rigid body in general change their directions. That is, the rotation transforms vectors drawn in the rigid body into other vectors. Denote this transformation by R. Is R a tensor? Solution. Consider a parallelogram embedded in the rigid body with its sides representing vectors a and b and its diagonal representing the resultant a + b. Since the parallelogram remains a parallelogram after a rotation about any axis, the diagonal (the resultant vector) of the rotated parallelogram is clearly both R(a + b) , the rotated (a + b) , and Ra 4- Rb, the sum of the rotated a and the rotated b . That is R(a + b) = Ra + Rb. A similar argument as that used in the previous example leads to R(aa )= «Ra . Thus, R is a tensor. Part B Components of a Tensor 13 Example 2B1.5 Let T be a tensor that transforms the specific vectors a and b according to Ta = a+2b, Tb = a - b Given a vector c = 2a+b, find Tc. Solution. Using the linearity property of tensors Tc = T(2a+b) = 2Ta+Tb = 2(a+2b)+(a-b) = 3a+3b 2B2 Components of a Tensor The components of a vector depend on the base vectors used to describe the components. This will also be true for tensors. Let e1? e2, e3 be unit vectors in the direction of the x\-, X2~, £3-axes respectively, of a rectangular Cartesian coordinate system. Under a transformation T, these vectors, el5 ^ e3 become Tel5 Te2, and Te3. Each of these Te/ (i = 1,2,3), being a vector, can be written as: T*i = Tne1+T21e2+T3ie3 Te 2 = ΤΏ?1 + Τ 22*2+Τ32*3 Te3 = Γ13β1+Γ23β2+Γ33β3 (2B2.1a) Te/ = Tjfij (2B2.1b) or It is clear from Eqs. (2B2.1a) that T11 = e1-Te1, 7712 = e1-Te2, r 2 1 = e2-Tex, ... or in general 7» = e/-Tey (2B2.2) The components Ty in the above equations are defined as the components of the tensor T. These components can be put in a matrix as follows: Mil T\2 7i3 [T] = T2\ ^22 T2$ M31 T32 ^33 This matrix is called the matrix of the tensor T with respect to the set of base vectors {ei> e2, e3} o r {e/} f° r short. We note that, because of the way we have chosen to denote the components of transformation of the base vectors, the elements of the first column are components of the vector Te1? those in the second column are the components of the vector Τβ2, and those in the third column are the components of Te3. 14 Tensors Example 2B2.1 Obtain the matrix for the tensor T which transforms the base vectors as follows: T e i = 4 e i +e 2 Te2 = 2e!+3e3 Te3 = -e 1 +3e 2 +e 3 Solution. By Eq. (2B2.1a) it is clear that: m= 4 2 -ll 1 0 3 0 3 1 Example 2B2.2 Let T transform every vector into its mirror image with respect to a fixed plane. If βχ is normal to the reflection plane (e2 and e3 are parallel to this plane), find a matrix of T. Mirror Fig. 2B.1 Solution. Since the normal to the reflection plane is transformed into its negative and vectors parallel to the plane are not altered: Τβχ = - Te2 = e2 Te3 = e3 Thus, e i Part B Components of a Tensor 15 [T] = -1 0 0 0 1 0 0 0 1 We note that this is only one of the infinitely many matrices of the tensor T, each depends on a particular choice of base vectors. In the above matrix, the choice of e,· is indicated at the bottom right corner of the matrix. If we choose e^ and e2 to be on a plane perpendicular to the mirror with each making 45° with the mirror as shown in Fig. 2B.1 and e3 points straight out from the paper. Then we have Tei = e2 Τβ3 = e 3 Thus, with respect to {e/ }, the matrix of the tensor is 0 1 0 [T]' = 1 0 0 0 0 1 Jej Throughout this book, we shall denote the matrix of a tensor T with respect to the basis e, by either [T] or [Tjj\ and with respect to the basis e,·' by either [T]' or[7)y] The last two matrices should not be confused with [T ' ] , which represents the matrix of the tensor T ' with respect to the basis e,·. Example 2B2.3 Let R correspond to a right-hand rotation of a rigid body about the Jt3-axis by an angle 0. Find a matrix of R. Solution. From Fig. 2B.2 it is clear that Rex = cos0e1+sin0e2 Re2 = -sin0ei+cos0e2 Re3 = e3 Thus, ]cos0 -sin0 0 [R] = I sin0 cos0 0 0 0 1 16 Tensors Fig. 2B.2 2B3 Components of a Transformed Vector Given the vector a and the tensor T, we wish to compute the components of b=Ta from the components of a and the components of T. Let the components of a with respect to {^1,^,^3} be [ah a2, a3], i.e., a = <ziei+a2e2+a3e3 0) b = Ta = Τ(α1β1+α2β24-α3β3) = a1Te1+a2Te2+a3Te3 (ϋ) then Thus, bl = e r b = a i ( e r Te 1 )+a 2 (erTe2)+a3(e r Te 3 ) b2 = e2-b = a1(erTe1)+a2(e2'Te2)+a3(erTe3) (iii) b3 = e 3 .b = a1(e3-Te1)+a2(e3-Te2)+a3(e3-Te3) By Eq. (2B2.2), we have, bi = Τ11α1+Τ12α2+Τ13α3 b2 = Τ21αλ + ^22^2+ ^23^3 h = 731fll + 732«2+ Γ33β3 We can write the above three equations in matrix form as: (2B3.1a) PartB Sum of Tensors 17 τ η ^21 ^22 723 ^31 ^32 ^33 or W Τ\2 Τχ$\ «2 H [b] = [T][a] (2B3.1b) (2B3.1C) We can concisely derive Eq. (2B3.1a) using indicial notation as follows: From a = α,-β/, we get Ta = Tafii = α/Γβ/. Since Te, = 7}/ey, (Eq. (2B2.1b)), therefore, h = h*k = T a e* = aiTjfij'ek = aiTjfijk = a{Tki i.e., bk = Tm (2B3.1d) Eq. (2B3.1d) is nothing but Eq. (2B3.1a) in indicial notation. We see that for the tensorial equation b = Ta, there corresponds a matrix equation of exactly the same form, i.e., [b] = [T][a]. This is the reason we adopted the convention that Tei = 7nei+72162+73103, etc. If we had adopted the convention Tex = 7nei+7i 2 e2+7i3e3, etc., then we would have obtained [b] = [T] [a] for the tensorial equation b = Ta, which would not be as natural. Example 2B3.1 Given that a tensor T which transforms the base vectors as follows: Tex = 2ei-6e 2 +4e 3 Te2 = 3ei+4e 2 -e 3 Τβ3 = -2βι+β2+2β3 How does this tensor transform the vector a = ei+2e 2 +3e 3 ? Solution. Using Eq. (2B3.1b) M= N M or 2B4 2 3 6 4 4 -1 - 2 ] Γι 2 1 2 = 5 8 2J h b = 2e!+5e 2 +8e 3 Sum of Tensors Let T and S be two tensors and a be an arbitrary vector. The sum of T and S, denoted by T + S, is defined by: (T+S)a = Ta+Sa (2B4.1) 18 Tensors It is easily seen that by this definition T + S is indeed a tensor. To find the components of T + S, let W=T+S (2B4.2a) Using Eqs. (2B2.2) and (2B4.1), the components of W are obtained to be Wij = ef-(T+S)ey = e/'Tey+e/'Sey i.e., Wij = Tij+Sq (2B4.2b) [W] = [T] + [S] (2B4.2c) In matrix notation, we have 2B5 Product of Two Tensors Let T and S be two tensors and a be an arbitrary vector, then TS and ST are defined to be the transformations (easily seen to be tensors) (TS)a = T(Sa) (2B5.1) (ST)a = S(Ta) (2B5.2) and Thus the components of TS are (TS),y=e,· · (TS)ey=ez· · T(Sey)=e/ · TSmyem=5mye,· · Tem = 7/m5my i.e., (TS)Ö. = TimSmj (2B5.3) (ST),y = SimTmj (2B5.4) Similarly, In fact, Eq. (2B5.3) is equivalent to the matrix equation: [TS] = [T][S] (2B5.5) whereas, Eq. (2B5.4) is equivalent to the matrix equation: [ST] = [S][T] (2B5.6) The two matrix products are in general different. Thus, it is clear that in general, the tensor product is not commutative (i.e., TS ^ ST). If T,S, and V are three tensors, then (T(SV))a = T((SV)a) = T(S(Va)) PartB Product of Two Tensors 19 and (TS)(Va) = T(S(Va)) i.e., T(SV) = (TS)V (2B5.7) Thus, the tensor product is associative. It is, therefore, natural to define the integral positive powers of a transformation by these simple products, so that T 2 = TT, T 3 = TTT, .... (2B5.8) Example 2B5.1 (a)Let R correspond to a 90° right-hand rigid body rotation about the Jt3-axis. Find the matrix ofR. (b)Let S correspond to a 90° right-hand rigid body rotation about thej^-axis. Find the matrix ofS. (c)Find the matrix of the tensor that corresponds to the rotation (a) then (b). (d)Find the matrix of the tensor that corresponds to the rotation (b) then (a). (e)Consider a point P whose initial coordinates are (1,1,0). Find the new position of this point after the rotations of part (c). Also find the new position of this point after the rotations of part (d). Solution, (a) For this rotation the transformation of the base vectors is given by Rei = e2 Re2 = - e i Re3 = e3 so that, "o [R] = 1 0 - 1 θ" 0 0 0 1 (b)In a similar manner to (a) the transformation of the base vectors is given by Se^ej Se2 = e3 Se3 = - e 2 so that, 1 0 o" [S] = 0 0 - 1 0 1 ol 20 Tensors (c)Since S(Ra) = (SR)a, the resultant rotation is given by the single transformation SR whose components are given by the matrix 0 - 1 0 1 0 0 0-10 [SR] = 0 0 - 1 1 0 0 0 0-1 1 0 0 0 1 0 0 0 1 (d)In a manner similar to (c) the resultant rotation is given by the single transformation RS whose components are given by the matrix "o - 1 0 [RS] = 1 1° 0 0 0 1 1 0 0 o o il 0 0 - 1 = i o ol 0 1 0 0 1 0 (e)Let r be the initial position of the point P. Let r and r after the rotations of part (c) and part (d) respectively. Then 0 -1 0 [r·] = [SR][r] = 0 0 - 1 1 0 0 "l" 1 = 0 be the rotated position of P -l] 0 1 i.e., r = -e!+e3 and ol Ό 0 1 Y [I·"] = [RS][r] = 1 0 0 1 = 1 1 0 1 0 0 i.e., r =e 2 +e 3 This example further illustrates that the order of rotations is important. 2B6 Transpose of a Tensor The transpose of a tensor T, denoted by TrT, ·is defined to be the tensor which satisfies the following identity for all vectors a and b: a Tb = b T a (2B6.1) It can be easily seen that T is a tensor. From the above definition, we have erTe^eyT'e; Thus, T- = T? or (2B6.2) PartB Dyadic Product of Two Vectors 21 [T7] = [Τ] Γ pT i.e., the matrix of T : is the transpose of the matrix of T. We also note that by Eq. (2B6.1), we have a-T r b = b-(T r ) r a Ρ7\Γ Thus, b · Ta = b · (ΊΛ) a for any a and b, so that T = (T 7 ) 7 (2B6.3) It can also be established that (see Prob. 2B13) (TS)T = S r T r (2B6.4) That is, the transpose of a product of the tensors is equal to the product of transposed tensors in reverse order. More generally, (ABC ...D) r = D r ...C 7 B r A r 2B7 (2B6.5) Dyadic Product of Two Vectors The dyadic product of vectors a and b, denoted by ab, is defined to be the transformation which transforms an arbitrary vector c according to the rule: (ab)c = a(b-c) (2B7.1) Now, for any c, d, a and/?, we have, from the above definition: (ab)(ac+ßd) = a(b-(ac+j8d)) = a((ab-c)+08b-d)) = a(ab)c+£(ab)d Thus, ab is a tensor. Letting W = ab, then the components of W are: Wij = e/-Wey = e/-(ab)ef = e;-a(b-e;) = aty i.e., (2B7.2a) In matrix notation, Eq. (2B7.2a) is \ai\ αφι αφ2 αΦ?> [W] = \a2\ [bhb2,b3] = U2&i a2b2 αφΔ \a3\ αφχ αφ2 a3b3 In particular, the components of the dyadic product of the base vectors ej are: [0 1 0' Γι 0 0 [*iei] = 0 0 0 . tele2] = 0 0 0 0 0 ol 0 0 0 Thus, it is clear that any tensor T can be expressed as: (2B7.2b) 22 Tensors T = r 1 1 e 1 e 1 +r 1 2 e i e 2 +... 4- 7 ^ 3 (2B7.3a) T = η,-e/e,· (2B7.3b) i.e., We note that another commonly used notation for the dyadic product of a and b is a (9b. 2B8 Trace of a Tensor The trace of any dyad ab is defined to be a scalar given by a · b. That is, trab = a b (2B8.1) Furthermore, the trace is defined to be a linear operator that satisfies the relation: tr(aab+ßcd) = atr ab+ßtr cd (2B8.2) Using Eq. (2B7.3b), the trace of T can, therefore, be obtained as tr T = tr^ye/e,·) = Ttfx(efij) = 7 » e ^ = Τ^ = TÜ that is, tr T = Tu ~ 7Ίι + 722 + ^33 = sum of diagonal elements (2B8.3) It is obvious that trT T =trT (2B8.4) Example 2B8.1 Show that for any second-order tensor A and B tr(AB)=tr(BA) (2B8.5) Solution. Let C=AB, then Cij=AimBmj. Thus, trAB=tr C=Cii=AimBnti (i) Let D=BA, thenDy—5/^^y, and trE\=tvO=DirBintAmi But BimAmi-Bm\Aim (ii) (change of dummy indices), that is trBA=trAB (iii) Part B Identity Tensor and Tensor Inverse 23 2B9 Identity Tensor and Tensor Inverse The linear transformation which transforms every vector into itself is called an identity tensor. Denoting this special tensor by I, we have, for any vector a, la = a (2B9.1) and in particular, Ιβχ = e x Ie2 = e2 1*3 = e3 Thus, the components of the identity tensor are: I i r e r l e j = erej = oij ( 2 B 9 .2a) i.e., 1 0 [I] = 10 0 1 0 0 0 00| 1 (2B9.2b) It is obvious that the identity matrix is the matrix of I for all rectangular Cartesian coordinates and that TI = IT = T for any tensor T. We also note that if Ta = a for any arbitrary a, then T = I. Example 2B9.1 Write the tensor T, defined by the equation Ta = fca, where k is a constant and a is arbitrary, in terms of the identity tensor and find its components. Solution. Using Eq. (2B9.1) we can write A: a asfclaso that Ta = fca becomes Ta = fcla and since a is arbitrary T= H The components of this tensor are clearly, Tij = kdij Given a tensor T, if a tensor S exists such that ST=I then we call S the inverse of T or S=T _ 1 . (Note: With Τ _ 1 Τ=Τ" 1 + 1 =Τ°=Ι, the zeroth power of a tensor is the identity tensor). To find the components of the inverse of a tensor T is to find the inverse of the matrix of T. From the study of matrices we know that the inverse exists as long as detT^O (that is, T 24 Tensors is non-singular) and in this case, [T] * [T] = [T] [T] l = [I]. Thus, the inverse of a tensor satisfies the following reciprocal relation: (2B9.3) T-lT = TT-l = j We can easily show (see Prob. 2B15) that for the tensor inverse the following relations are satisfied, (T 7 )- 1 = ( I " 1 ) 7 (2B9·4) (CTT^T-V1 ^ 2B9 · 5 ) and We note that if the inverse exists then we have the reciprocal relation that Ta = b and a = T _ 1 b This indicates that when a tensor is invertible there is a one to one mapping of vectors a and b. On the other hand, if a tensor T does not have an inverse, then, for a given b , there are in general more than one a which transforms into b . For example, consider the singular tensor T = cd (the dyadic product of c and d , which does not have an inverse because its determinant is zero), we have Ta = c(da) Ξ b Now, let h be any vector perpendicular to d (i.e., d · h = 0), then T(a+h) = c(d-a) = b That is, all vectors a + h transform under T into the same vector b . 2B10 Orthogonal Tensor An orthogonal tensor is a linear transformation, under which the transformed vectors preserve their lengths and angles. Let Q denote an orthogonal tensor, then by definition, | Qa | = | a | and cos(a,b) = cos(Qa,Qb) for any a and b, Thus, Qa Qb = a b (2B10.1) for any a and b. Using the definitions of the transpose and the product of tensors: (Qa)-(Qb) = b Q7(Qa) = b (QrQ)a (0 b (QrQ)a = a b = b a = b la (ü) Therefore, Since a and b are arbitrary, it follows that QrQ = I (üi) Part B Orthogonal Tensor 25 This means that Q 1=QT and from Eq. (2B9.3), (2B10.2a) QrQ = Q Q 7 = I In matrix notation, Eqs. (2B 10.2a) take the form: (2B10.2b) [Q][Qf=[Q] r [Q] = [i] and in subscript notation, these equations take the form: QimQjm = QmiQmj = (2B10.2c) ^ij Example 2B10.1 The tensor given in Example 2B2.2, being a reflection, is obviously an orthogonal tensor. Verify that [T][T]7= [I] for the [T] in that example. Also, find the determinant of [T]. Solution. Using the matrix of Example 2B7.1: [T][Tf = Γ-i o o 0 1 0 0 0 1 -10 0 0 1 0 0 0 1 = 1 0 0 0 1 0 0 0 1 The determinant of [T] is |T| = -1 0 0 0 1 0 = -1 0 0 1 Example 2B10.2 The tensor given in Example 2B2.3, being a rigid body rotation, is obviously an orthogonal tensor. Verify that [R][R] = [I] for the [R] in that example. Also find the determinant of [R]. Solution. It is clear that [R][R]r = cos0 sinö 0 1 0 0 -sinö cosö 0 0 1 0 0 0 1 0 0 1 cosö -sinö 0 det[R]= |R| = sinö cosö 01 = +1 0 0 1 cosö -sinö 0 sinö cosö 0 0 0 1 The determinant of the matrix of any orthogonal tensor Q is easily shown to be equal to either + 1 or -1. In fact, [Q][Q] T =[i] 26 Tensors therefore, |[Q][Q]7l = IQIIQ r l = | i | Now, |Q| = |Q 7 |,and |I| = 1, therefore, | Q | 2 = l,thus |Q| = ± 1 (2B10.3) From the previous examples we can see that the value of +1 corresponds to rotation and -1 corresponds to reflection. 2B11 Transformation Matrix Between Two Rectangular Cartesian Coordinate Systems. Suppose {e/} and {e,· } are unit vectors corresponding to two rectangular Cartesian coordinate systems (see Fig. 2B.3). It is clear that {e/} can be made to coincide with {e/ } through either a rigid body rotation (if both bases are same handed) or a rotation followed by a reflection (if different handed). That is {e,·} and {ez: } can be related by an orthogonal tensor Q through the equations e,; = Qe/ = ö,n/em (2Bll.la) i.e., «ί = 011β1+021*2+031*3 «2 = Öl2el+Ö22e2+Ö32e3 «3 = 013el+023e2+033e3 (2Bll.lb) where SJimüjm ~~ \2mvJmj ~" ^ij or QQr=Q7Q = I We note that Qn = e^Qe! = e^ei = cosine of the angle between ex and e[, 012 = e i * Qe2 = e i ' e2 = cosine of the angle between ex and e2, etc. In general, ßfy = cosine of the angle between e,· and ej which may be written: Qij = cos(e/,ej) = e/-ej The matrix of these directional cosines, i.e., the matrix Toll 012 013] [Q]= 021 022 023 031 032 033 (2B11.2) Part B Transformation Matrix Between Two Rectangular Cartesian Coordinate Systems. 27 is called the transformation matrix between {e/} and {e/ }. Using this matrix, we shall obtain, in the following sections, the relationship between the two sets of components, with respect to these two sets of base vectors, of either a vector or a tensor. Fig.2B.3 Example 2B 11.1 Let {ez: } be obtained by rotating the basis {e/} about the e$ axis through 30° as shown in Fig. 2B.4. We note that in this figure, e3 and e^ coincide. Solution. We can obtain the transformation matrix in two ways. (i) Using Eq. (2B11.2), we have 1 Fx ßn=cos(ei,ei)=cos30°=—, Q\2~cos(ei,e2)=cos 120° = - - , Ö13=cos(ei,e3)=cos90°=0 1 Fx Ö21=cos(e2,ei)=cos60°=-, (322= cos ( e 2. e 2)=cos30°=—, 023= cos ( e 2. e 3)=cos90°=0 031=cos(e3,ei)=cos9O°=O, 032=cos(e3,e2)=cos90°=0, ß 33 =cos(e3,e3)=cos0 o =1 (ii) It is easier to simply look at Fig. 2B.4 and decompose each of the e,: 's into its components in the {βιφ,ββ} directions, i.e., - ei = e V3 1 ei+ e2 ^" 2 1 ^ 2=-2ei+V2 28 Tensors e3 = e3 Thus, by either method, the transformation matrix is V3 2 1 [Q] = 2 4° fo 0 0 1 Fig.2B.4 2B12 Transformation Laws for Cartesian Components of Vectors Consider any vector a, then the components of a with respect to {e/} are a/ = a · e/ and its components with respect to {e,: }are a] = a · e,· Now, e/ = Qmfim, [Eq. (2Bll.la)], therefore, a i — &'Qmiem ~ Qmi\a'em) i.e., a i In matrix notation, Eq. (2B12.1a) is = SJmflm (2B12.1a) Part B Transformation Laws for Cartesian Components of Vectors 29 011 Ö21 Qhi Ö12 Ö22 Ö32 Ö13 Ö23 Ö33 «1 «2 «3 or (2B12.1b) (2B12.1c) [a]' = [Q]7[a] Equation (2B12.1) is the transformation law relating components of the same vector with respect to different rectangular Cartesian unit bases. It is very important to note that in Eq. (2B 12.1c), [a]' denote the matrix of the vector a with respect to the primed basis ez: and [a] denote that with respect to the unprimed basis ez. Eq. (2B12.1) is not the same as T a'=Q a. The distinction is that [a] and [a]' are matrices of the same vector, where a and a' Tare two different vectors; a' being the transformed vector of a (through the transformation Q ), If we premultiply Eq. (2B12.1c) with [Q], we get (2B12.2a) [a] = [Q][a]' The indicial notation equation for Eq.(2B12.2a) is a (2B12.2b) i ~ S2itnan Example 2B 12.1 Given that the components of a vector a with respect to {ez·} are given by (2,0,0), (i.e., a = 2ei), find its components with respect to {e; }, where the e/ axes are obtained by a 90° counter-clockwise rotation of the e,· axes about the e3-axis. Solution. The answer to the question is obvious from Fig. 2B.5, that is a = 2βχ = - 2 β 2 We can also obtain the answer by using Eq. (2B12.2a). First we find the transformation matrix. With e^ = e2, e2 = - e j and e3 = e3, by Eq. (2B 11.1b), we have 0 -1 0 [Q] = 1 0 0. 0 0 1 Thus, 0 1 0 [a]' = [QHa] = - 1 0 0 0 0 1 i.e., a = -2e? 2 0 0 = -2 0 0 30 Tensors eae 2 ei a Fig. 2B.5 2B13 Transformation Law for Cartesian Components of a Tensor Consider any tensor T, then the components of T with respect to the basis {e(}are: Its components with respect to {e(: }are: With e,: = Qmiem, * ij ~ sJmfim' *SJnj*n ~~ fJmi>Jnj\^m ' *e/iJ i.e., Tij = QmiQnjTmn (2B13.1a) In matrix notation, Eq. (2B13.1a) reads fai Ί'η Γύΐ 721 722 723 731 Γ32 Γ33 = fan Ö2i Ö3i] [7*11 7-12 7 B ] [ ß u Öi2 ö i 3 l ( 2 B 1 3 - l b > Öl2 Qll Ö32 Τ^ι Γ 22 Τ2-Λ 0 2 1 Q22 £?23 013 Ö23 Ö33 7*3! Γ 32 T33 ö 3 1 ζ?32 Ö33 or [T]' = [Q]'[T][Q] (2B13.1c) Part B Transformation Law for Cartesian Components of a Tensor 31 We can also express the unprimed components in terms of the primed components. Indeed, premultiply Eq. (2B13.1c) with [Q] and postmultiply it with [Q] , we obtain, since [Q][Q] r =[Q] r [Q] = [i], (2B13.2a) [T] = [Q][T]'[Q]y Using indicial notation, Eq. (2B 13.2a) reads *ij = (2B13.2b) \2im&jn*mn Equations (2B13.1& 2B13.2) are the transformation laws relating the components of the same tensor with respect to different Cartesian unit bases. It is important to note that in these equations, [T] and [T]'are different matrices of the same tensor T. We note that the equation [T]' = [Q]r[T][Q] differs from the equation T ' = Q7TQ in that the former relates the components of the same tensor T whereas the latter relates the two different tensors T and T '. Example 2B13.1 Given the matrix of a tensor T in respect to the basis {e/}: "o i 0 1 2 0 0 0 1 [T] = Find [T]e;, i.e., find the matrix of T with respect to the {e,· } basis, where {e; } is obtained by rotating {e/} about β3 through 90°. (see Fig. 2B.5). Solution. Since e^ = ^ e^ = -βχ and e'>$ = e$, by Eq. (2Bll.lb), we have Γο -1 0 0 0 0 1 0 1 0 1 2 0 0 0 1 0 -1 0 1 0 0 = 0 0 1 lo Thus, Eq. (2B 13.1c) gives [T]' = [Q] = 1 0 1 0 -10 0 0 0 1 2 - 1 0 - 1 0 0 0 0 1 i.e., T[x = 2, T{2 = - 1 , 7i3 = 0,7^ = - 1 , etc. Example 2B13.2 Given a tensor T and its components Tq and T(j with respect to two sets of bases {e/} and {e/ }. Show that 7); is invariant with respect to this change of bases, i.e., 7},· = 7)/. 32 Tensors Solution. The primed components are related to the unprimed components by Eq. (2B13.1a) * ij s2mvJnj*mn *ii QmiQni*n Thus, But, QmiQni = dmn (Eq. (2B10.2c)), therefore, *ii vmnlmn lrnm i.e., Τ η+Τ72+ΤΉ ~ Tn + T22+T?ß We see from Example 2B13.1, that we can calculate all nine components of a tensor T with respect to e(: from the matrix [T]e., by using Eq. (2B13.1c). However, there are often times when we need only a few components. Then it is more convenient to use the Eq. (2B2.2) (T(j = el - Tej ) which defines each of the specific components. In matrix form this equation is written as: r9 = K f m t e j ] (2B13.4) where [e'] denotes a row matrix whose elements are the components of e,' with respect to the basis {e; }. Example 2B13.3 Obtain T{2 for the tensor T and the bases e,· and e; given in Example 2B13.1 Solution. Since ei = ^ and e2 = -e^, thus 7l2 = *ί·Τβ2 = e 2 -T(-e 1 ) = - e 2 - T e i = - Γ 2 1 = - 1 Alternatively, using Eq. (2B13.4) Γο l ol Γ-i T{2 = [βίΠΤ][β2] = [0,1,0] 1 2 0 o o 11 0 0 = [0, 1, 0] - 1 L° = -1 0 2B14 Defining Tensors by Transformation Laws Equations (2B12.1) or (2B13.1) state that when the components of a vector or a tensor with respect to {e,·} are known, then its components with respect to any {e,: } are uniquely determined from them. In other words, the components a,· or 7^· with respect to one set of {e,·} Part B Defining Tensors by Transformation Laws 33 completely characterizes a vector or a tensor. Thus, it is perfectly meaningful to use a statement such as "consider a tensor 7 y meaning consider the tensor T whose components with respect to some set of {e;} are Τφ In fact, an alternative way of defining a tensor is through the use of transformation laws relating the components of a tensor with respect to different bases. Confining ourselves to only rectangular Cartesian coordinate systems and using unit vectors along positive coordinate directions as base vectors, we now define Cartesian components of tensors of different orders in terms of their transformation laws in the following where the primed quantities are referred to basis {e; } and unprimed quantities to basis {e/}, the e} and e/ are related by e;=Qe;, Q being an orthogonal transformation a' = a zeroth-order tensor(or scalar) a\ = Qmiam first-order tensor (or vector) Tij = QmiQnjTmn second-order tensor(or tensor) Tijk = QmiQnjQrkTmnr third-order tensor etc. Using the above transformation laws, one can easily establish the following three rules (a)the addition rule (b) the multiplication rule and (c) the quotient rule. (a)The addition rule: If Ty and 5,y are components of any two tensors, then 7^+5^ are components of a tensor. Similarly if T^k and Syk are components of any two third order tensors, then 7 ^ + S ^ are components of a third order tensor. To prove this rule, we note that since Tijk=QmiQnjQrkTmnr and Sijk=QmiQnjQrkSmnr have, *ijk~^~^ijk ~ QmiQnjQrk*mnr^Qmi>2nj>2rh*mnr ~~ we Slmv2np2rk\^mnr^~^mnr) Letting W/jk = 7»*+S»* and Wmnr=Tmnr+Smnn we have, **ijk — S2mv2nj\2rk}^mnr i.e, Wjjk are components of a third order tensor. (b)The multiplication rule: Let a\ be components of any vector and T^ be components of any tensor. We can form many kinds of products from these components. Examples are (a)a/af* {^)afljak (c) 7/.·Γ^/, etc. It can be proved that each of these products are components of a tensor, whose order is equal to the number of the free indices. For example, α,-α,- is a scalar (zeroth order tensor), αμρ,^ are components of a third order tensor, T^Ty are components of a fourth order tensor. To prove that 7^7^/ are components of a fourth-order tensor, let Α/^/=7^·Γ^/, then 34 Tensors M/jid - T-jT^-QmiQnjTmnQrkQsiTrs-QmiQnjQrkQslTmnTrs i.e., M/ju = QmiQnjQrkQsMmnrs which is the transformation law for a fourth order tensor. It is quite clear from the proof given above that the order of the tensor whose components are obtained from the multiplication of components of tensors is determined by the number of free indices; no free index corresponds to a scalar, one free index corresponds to a vector, two free indices correspond a second-order tensor, etc. (c) The quotient rule: If a\ are components of an arbitrary vector and Ty are components of an arbitrary tensor and a\ = Τφ^ for all coordinates, then bj are components of a vector. To prove this, we note that since a,· are components of a vector, and Ty are components of a second-order tensor, therefore, <H = Qinflm W and Tq = QimQjnTmn (") Now, substituting Eqs. (i) and (ii) into the equation a\ - T[pj, we have Qinflln = QimQjnTmnbj (üi) But, the equation a\ = Τφ^ is true for all coordinates, thus, we also have *m = Tmnb'n (iv) Thus, Eq. (iii) becomes Qim^mn^n = QimQjnTmnbj W Multiplying the above equation with Qlk and noting that Q[kQim - dkm, we get TM = QjnUnbj (v0 i.e., TknVn-Qjnbj)=0 (vii) Since the above equation is to be true for any tensor T, therefore, the parenthesis must be identically zero. Thus, bn=Qjnbj ( viii ) PartB Symmetric and Antisymmetric Tensors 35 This is the transformation law for the components of a vector. Thus,ft;are components of a vector. Another example which will be important later when we discuss the relationship between stress and strain for an elastic body is the following: If 7)y and Εή are components of arbitrary second order tensors T and E then T C ij = ijklEkl for all coordinates, then Q^/ are components of a fourth order tensor. The proof for this example follows that of the previous example. 2B15 Symmetric and Antisymmetric Tensors A tensor is said to be symmetric if'Τ = ΤΓ. Thus, the components of a symmetric tensor have the property, (2B15.1) i.e., A tensor is said to be antisymmetic if r p tensor have the property rw\l T - -TTI J ij ij Thus, the components of an antisymmetric (2B15.2) -T J ji i.e., ^ll = Γ 22 = Γ 33 = 0 and I V2-"'I2h y 13-~73b y 23-~y32. Any tensor T can always be decomposed into the sum of a symmetric tensor and an antisymmetric tensor. In fact, T = T^+T 4 where T+T7 . r = —-— is symmetric and 7 A T-T A = —-— is antisymmetric It is not difficult to prove that the decomposition is unique (see Prob. 2B27) (2B15.3) 36 Tensors Example 2B 15.1 Show that if T is symmetric and W is antisymmetric, then tr(TW)=0. Solution. We have, [see Example 2B8.4] tr(TW)=tr(TW) 7 =tr(W r T r ) (0 T Ύ Since T is symmetric and W is antisymmetric, therefore, by definition, T=T , W= - W . Thus, (see Example 2B8.1) tr (TW)=-tr(WT)=-tr(TW) (ii) Consequently, 2tr(TW)=0. That is, tr(TW)=0 (iii) 2B16 The Dual Vector of an Antisymmetric Tensor The diagonal elements of an antisymmetric tensor are always zero, and, of the six nondiagonal elements, only three are independent, because Γ 12 = ~Τχ2,Τι3 = ~?3i and Γ 23 = — T32· Thus, an antisymmetric tensor has really only three components, just like a vector. Indeed, it does behavior like a vector. More specifically, for every antisymmetric tensor T, there corresponds a vector 1r, such that for every vector a the transformed vector, Ta, can be obtained from the cross product of 1 with a. That is, Ta = f 4 xa (2B16.1) This vector, V, is called the dual vector (or axial vector ) of the antisymmetric tensor. The form of the dual vector is given below: FromEq.(2B16.1), we have, since a-bxc = b-cxa, T12 = e r T e 2 = e 1 -f 4 xe 2 = ί ^ Χ β χ = ~^'^ 4 4 = ~4 T31 = *3 Tex = β 3 · ^ χ β ι = f -e 1 xe 3 = - f ^ = T23 = e 2 'Te 3 = e 2 -f 4 xe 3 = <*e3Xe2 = ~^·*ι = ~4 -4 Similar derivations will give T21 = 4 > Ttt = nJzi = 4 and Tn = T22 = T33 = 0. Thus, only an antisymmetric tensor has a dual vector defined by Eq.(2B16.1). It is given by: r4 = -(r 2 3 e 1 + r 3 1 e 2 +r 1 2 e 3 ) = (r 3 2 e 1 + r 1 3 e 2 +r 2 1 e 3 ) (2B16.2a) or, in indicial notation 2f = -epTj* (2B16.2b) Part B The Dual Vector of an Antisymmetric Tensor 37 Example 2B16.1 Given Γι 2 3~ [T] = 4 2 1 1 1 (a)Decompose the tensor into a symmetric and an antisymmetric part. li (b)Find the dual vector for the antisymmetric part. (c)Verify T^a = f^xa for a = βχ+β3. Solution, (a) [T] = [T^+fT4 ], where ] _[T]±[IL = S [T 2 1 3 2 3 2 1 1 1 = rrVßtpl- r 0 - 1 l" 1 0 0 -1 0 0 (b)The dual vector of T 4 is ί 4 = - ( ^ 3 β ΐ + 7 ^ ΐ β 2 + ^ 3 ) = - ( 0 e i - e 2 - e 3 ) = e 2 +e 3 . (c) Let b = T V then [b] = i.e., 0 1 1 -l il Γι 0 0 0 = o oj li l" 1 -1 b = e!+e 2 -e 3 On the other hand, i Xa = (e2+e 3 )x(ei+e 3 ) = ~e 3 +ei+e2 = b Example 2B16.2 Given that R is a rotation tensor and that m is a unit vector in the direction of the axis of rotation, prove that the dual vector q of K4 is parallel to m. Solution. Since m is parallel to the axis of rotation, therefore, Rm = m (i) 38 Tensors Thus, (RrR)m = R r m. Since R7R = I, we have Rrm = m (V) (R-R r )m = 0 (m) Thus, (i) and (ii) gives But (R-R )m = 2qxm, where q is the dual vector of I * . Thus, qXm = 0 (iv) i.e., q is parallel to m. We note that it can be shown (see Prob. 2B29 or Prob. 2B36) that if Θ denotes the right-hand rotation angle, then q = (sinÖ)m (2B16.3) 2B17 Eigenvalues and Eigenvectors of a Tensor Consider a tensor T. If a is a vector which transforms under T into a vector parallel to itself, i.e., Ta=Aa (2B17.1) then a is an eigenvector and A is the corresponding eigenvalue. If a is an eigenvector with corresponding eigenvalue A of the linear transformation T, then any vector parallel to a is also an eigenvector with the same eigenvalue A. In fact, for any scalar a, T(aa) = aTa = a(Aa) = A(aa) (i) Thus, an eigenvector, as defined by Eq. (2B17.1), has an arbitrary length. For definiteness, we shall agree that all eigenvectors sought will be of unit length. A tensor may have infinitely many eigenvectors. In fact, since la = a, any vector is an eigenvector for the identity tensor I, with eigenvalues all equal to unity. For the tensor ßl9 the same is true, except that the eigenvalues are all equal to/?. Some tensors have eigenvectors in only one direction. For example, for any rotation tensor, which effects a rigid body rotation about an axis through an angle not equal to integral multiples of π, only those vectors which are parallel to the axis of rotation will remain parallel to themselves. Let n be a unit eigenvector, then Tn = An = AIn (2B17.2) (T-AI)n = 0 (2B17.3a) Thus, Part B Eigenvalues and Eigenvectors of a Tensor 39 Let n = afii, then in component form (Tij-Mi-yzj = o (2B17.3b) In long form, we have (Γ 11 -Α)α 1 +7 12 α 2 +Γ 13 α 3 = 0 T21<*l + (T22-*)a2+T2p3 = 0 Τ3ΐ^ι^Τ32α2+(Τ33-λ)α3 =0 (2B17.3c) Equations (2B17.3c) are a system of linear homogeneous equations in ah a2,anda3. Obviously, regardless of the values of A, a solution for this system isa1=a2=a3=0. This is know as the trivial solution. This solution simply states the obvious fact that a = 0 satisfies the equation Ta = Aa, independent of the value of A. To find the nontrivial eigenvectors for T, we note that a homogeneous system of equations admits nontrivial solution only if the determinant of its coefficients vanishes. That is | T—AI | = 0 (2B17.4a) Tl3 (2B17.4b) i.e., ^11 "Λ- 7*12 T ^21 22"^ T T ?>1 ?>2 T = 0 2$ λ ^33 ~ For a given T, the above equation is a cubic equation in A. It is called the characteristic equation of T. The roots of this characteristic equation are the eigenvalues of T. Equations (2B17.3), together with the equation 1 1 1 a{+a2+a3 = 1 (2B17.5) allow us to obtain eigenvectors of unit length. The following examples illustrate how eigenvectors and eigenvalues of a tensor can be obtained. Example 2B 17.1 If, with respect to some basis {e,·}, the matrix of T is \2 0 0 0 2 0 0 0 2 find the eigenvalues and eigenvectors for this tensor. Solution. We note that this tensor is 21, so that Ta = 21a = 2a, for any vector a. Therefore, by the definition of eigenvector,(see Eq. (2B17.1)), any direction is a direction for an eigenvector. The eigenvalues for all the directions are the same, which is 2. However, we can also [T] = 40 Tensors use Eq. (2B17.3) to find the eigenvalues and Eqs. (2B17.4) to find the eigenvectors. Indeed, Eq. (2B17.3) gives, for this tensor the following characteristic equation: (2-A) 3 = 0. So we have a triple root A=2. Substituting A=2 in Eqs. (2B17.3c), we obtain (2-2)«! = 0 (2-2)a 2 = 0 (2-2)« 3 = 0 Thus, all three equations are automatically satisfied for arbitrary values of ah a2, and a3, so that vectors in all directions are eigenvectors. We can choose any three directions as the three independent eigenvectors. In particular, we can choose the basis {e,·} as a set of linearly independent eigenvectors. Example 2B17.2 = = Show that if 72l ^3l 0> then ±βχ is an eigenvector of T with eigenvalue Tu. Solution. From Τβι=7\ιβι+72ΐβ2+ T^e^, we have T e i = Γ η β ι and T ( - e i ) = Tn(-ei) Thus, by definition, Eq. (2B17.1), ±e x are eigenvectors with T n as its eigenvalue. Similarly, if 7x2=732=0, then ±e 2 are eigenvectors with corresponding eigenvalue T22 and if 7^3=723=0, then ±e3 are eigenvectors with corresponding eigenvalue T33. Example 2B17.3 Given that [T] "2 0 0 0 2 0 0 0 3 Find the eigenvalues and their corresponding eigenvectors. Solution. The characteristic equation is (2-A)2(3-A) = 0 Thus, Αχ=3, A2=A3=2. (note the ordering of the eigenvalues is arbitrary). These results are obvious in view of Example 2B17.2. In fact, that example also tells us that the eigenvector corresponding to Αχ=3 is e$ and eigenvectors corresponding to A2=A3=2 are βχ and e^ How- Part B Eigenvalues and Eigenvectors of a Tensor 41 ever, there are actually infinitely many eigenvectors corresponding to the double root. In fact, since Te 1 =2e 1 and Te2=2e2, therefore, Τ(αβ!+ββ2) = aTe 1 +^Te 2 = 2ae 1 +33e 2 =2(ae 1 +ße 2 ) i.e., ae1+j8e2 is an eigenvector with eigenvalue 2. This fact can also be obtained from Eqs.(2B17.3c). With A=2 these equations give 0«1 = 0 0a 2 =0 «3=0 Thus, a1 and a2 are arbitrary and a 3 =0 so that any vector perpendicular to e3, i.e., n=a 1 e 1 +a 2 e 2 is an eigenvector. Example 2B 17.4 Find the eigenvalues and eigenvectors for the tensor '2 0 0 [T] = 0 3 4 0 4 -3 Solution. The characteristic equation gives Γ2-Α 0 0 [T-AI] = 0 3-A 4 = (2-A)(A -25) = 0 0 4 -3-A Thus, there are three distinct eigenvalues, λχ=2, Α2=5 and A3= - 5 . Corresponding to Aj=2, Eqs. (2B17.3c) give 0^ = 0 α 2 +4α 3 = 0 4α 2 -5α 3 = 0 and Eq. (2B17.5) gives «ΐ+α 2 +α 3 =1 Thus, α 2 =α 3 =0 and αχ=±1, so that the eigenvector corresponding to AX=2 is n 1 =±e 1 . We note that from the Example 2B17.2, this eigenvalue 2 and the corresponding eigenvector ±*i can be written down by inspection without computation. Corresponding to A2=5, we have 3αι = 0 42 Tensors -2α 2 +4α 3 =0 4α 2 -8α 3 =0 Thus (note the second and third equations are the same), «l = 0, a2 = ±2/^5, a3 = ±1/V5 and the eigenvector corresponding toA2=5 is n2 =± 75"( 2e 2+ e 3) Corresponding to λ 3 = - 5 , similar computations give n 3 = ± 75"(- e 2+ 2 e 3) All the examples given above have three eigenvalues that are real. It can be shown that if a tensor is real (i.e., with real components) and symmetric, then all its eigenvalues are real. If a tensor is real but not symmetric, then two of the eigenvalues may be complex conjugates. The following example illustrates this possibility. Example 2B17.5 Find the eigenvalues and eigenvectors for the rotation tensor R corresponding to a 90° rotation about the e3-axis (see Example 2B5.1(a)). Solution. The characteristic equation is 10—A - 1 0 1 0-λ 0 0 1-λ 0 =0 i.e., λ 2 (1-λ)+(1-λ) = (1-λ)(λ 2 +1) = 0 Thus, only one eigenvalue is real, namely A x =l, the other two are imaginary, A 2 3 =±V^1. Correspondingly, there is only one real eigenvector. Only real eigenvectors are of interest to us, we shall therefore compute only the eigenvector corresponding to Ax=1. From (0-1)^-02=0 ax-a2 =0 (1-1)α 3 = 0 and 9 9 ? αΐ+α 2 +α 3 =1 Part B Principal Values and Principal Directions of Real Symmetric tensors 43 We obtainax=0, «2=0, «3= ± 1, i.e., n= ±e 3 , which, of course, is parallel to the axis of rotation. 2B18 Principal Values and Principal Directions of Real Symmetric tensors In the following chapters, we shall encounter several tensors (stress tensor, strain tensor, rate of deformation tensor, etc.) which are symmetric, for which the following theorem, stated without proof, is important: "the eigenvalues of any real symmetric tensor are all real." Thus, for a real symmetric tensor, there always exist at least three real eigenvectors which we shall also call the principal directions. The corresponding eigenvalues are called the principal values. We now prove that there always exist three principal directions which are mutually perpendicular. Let nx and n 2 be two eigenvectors corresponding to the eigenvalues λ1 and A2 respectively of a tensor T. Then Tn^ = A^nj (i) Tn2 = A2n2 (ϋ) and Thus, ^ini'n2 (iii) i*l Tn2 = λ 2 η 2 ·η! (iv) n 2'Tnl = T The definition of the transpose of T gives , ni-Tn2 = η2·Τ ηχ, thus for a symmetric tensor T, T=T T , so that η χ ·Τη 2 = n ^ T ^ . Thus, from Eqs. (iii) and (iv), we have (v) It follows that if λ1 is not equal to A2, then nx · n2 = 0, i.e., nx and n 2 are perpendicular to each other. We have thus proven that if the eigenvalues are all distinct, then the three principal directions are mutually perpendicular. Next, let us suppose that nx and n2 are two eigenvectors corresponding to the same eigenvalue A. Then, by definition, Tnx = A^ and Tn2 = An2 so that for any a, and β, Τ(αηχ +βη2)—αΎιΐι +/?Τη2=λ(αη1 +/ϊη2). That is ατΐι +/ϊη2 is also an eigenvector with the same eigenvalue A . In other words, if there are two distinct eigenvectors with the same eigenvalue, then, there are infinitely many eigenvectors (which forms a plane) with the same eigenvalue. This situation arises when the characteristic equation has a repeated root. Suppose the characteristic equation has roots λ1 and A2=A3=A (Ax distinct from A). Let nx be the eigenvector corresponding to A1? then nx is perpendicular to any eigenvector of A. Now, corresponding to A, the equations (2B18.1a)

1/--pages