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Introduction to
Continuum Mechanics
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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
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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
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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)
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