CONTINUUM MECHANICSdl.booktolearn.com/ebooks2/engineering/mechanical/... · 2019-06-24 · Published Titles ADVANCED THERMODYNAMICS ENGINEERING Kalyan Annamalai and Ishwar K. Puri

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ADVANCED THERMODYNAMICS ENGINEERINGKalyan Annamalai and Ishwar K. Puri

APPLIED FUNCTIONAL ANALYSISJ. Tinsley Oden and Leszek F. Demkowicz

COMBUSTION SCIENCE AND ENGINEERINGKalyan Annamalai and Ishwar K. Puri

CONTINUUM MECHANICS FOR ENGINEERS, Third EditionThomas Mase, Ronald E. Smelser, and George E. Mase

EXACT SOLUTIONS FOR BUCKLING OF STRUCTURAL MEMBERSC.M. Wang, C.Y. Wang, and J.N. Reddy

THE FINITE ELEMENT METHOD IN HEAT TRANSFER AND FLUID DYNAMICS,Second Edition

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SOLVING ORDINARY and PARTIAL BOUNDARY VALUE PROBLEMSin SCIENCE and ENGINEERING

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GEORGE E. MASE

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Mase, George Thomas.Continuum mechanics for engineers / G. Thomas Mase, George E. Mase. -- 3rd ed. / Ronald E.

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Includes bibliographical references and index.ISBN 978-1-4200-8538-9 (hardcover : alk. paper)1. Continuum mechanics. I. Mase, George E. II. Smelser, Ronald M., 1942- III. Title. IV. Series.

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Contents

List of Figures

List of Tables

Preface to the Third Edition

Preface to the Second Edition

Preface to the First Edition

Acknowledgments

Authors

Nomenclature

1 Continuum Theory 11.1 Continuum Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Starting Over . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2 Essential Mathematics 52.1 Scalars, Vectors and Cartesian Tensors . . . . . . . . . . . . . . . . . . . . . 52.2 Tensor Algebra in Symbolic Notation - Summation Convention . . . . . . 7

2.2.1 Kronecker Delta . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.2.2 Permutation Symbol . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.2.3 ε - δ Identity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.2.4 Tensor/Vector Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.3 Indicial Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.4 Matrices and Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.5 Transformations of Cartesian Tensors . . . . . . . . . . . . . . . . . . . . . 252.6 Principal Values and Principal Directions . . . . . . . . . . . . . . . . . . . 302.7 Tensor Fields, Tensor Calculus . . . . . . . . . . . . . . . . . . . . . . . . . 372.8 Integral Theorems of Gauss and Stokes . . . . . . . . . . . . . . . . . . . . 40Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

3 Stress Principles 533.1 Body and Surface Forces, Mass Density . . . . . . . . . . . . . . . . . . . . 533.2 Cauchy Stress Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 543.3 The Stress Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 563.4 Force and Moment Equilibrium; Stress Tensor Symmetry . . . . . . . . . . 613.5 Stress Transformation Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . 633.6 Principal Stresses; Principal Stress Directions . . . . . . . . . . . . . . . . . 663.7 Maximum and Minimum Stress Values . . . . . . . . . . . . . . . . . . . . 71

3.8 Mohr’s Circles for Stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 743.9 Plane Stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 803.10 Deviator and Spherical Stress States . . . . . . . . . . . . . . . . . . . . . . 853.11 Octahedral Shear Stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

4 Kinematics of Deformation and Motion 1034.1 Particles, Configurations, Deformations and Motion . . . . . . . . . . . . . 1034.2 Material and Spatial Coordinates . . . . . . . . . . . . . . . . . . . . . . . . 1044.3 Langrangian and Eulerian Descriptions . . . . . . . . . . . . . . . . . . . . 1084.4 The Displacement Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1104.5 The Material Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1114.6 Deformation Gradients, Finite Strain Tensors . . . . . . . . . . . . . . . . . 1164.7 Infinitesimal Deformation Theory . . . . . . . . . . . . . . . . . . . . . . . 1204.8 Compatibility Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1284.9 Stretch Ratios . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1314.10 Rotation Tensor, Stretch Tensors . . . . . . . . . . . . . . . . . . . . . . . . . 1344.11 Velocity Gradient, Rate of Deformation, Vorticity . . . . . . . . . . . . . . . 1374.12 Material Derivative of Line Elements, Areas, Volumes . . . . . . . . . . . . 143Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147

5 Fundamental Laws and Equations 1675.1 Material Derivatives of Line, Surface and Volume Integrals . . . . . . . . . 1675.2 Conservation of Mass, Continuity Equation . . . . . . . . . . . . . . . . . . 1695.3 Linear Momentum Principle, Equations of Motion . . . . . . . . . . . . . . 1715.4 Piola-Kirchhoff Stress Tensors, Lagrangian Equations of Motion . . . . . . 1725.5 Moment of Momentum (Angular Momentum) Principle . . . . . . . . . . 1765.6 Law of Conservation of Energy, The Energy Equation . . . . . . . . . . . . 1775.7 Entropy and the Clausius-Duhem Equation . . . . . . . . . . . . . . . . . . 1795.8 The General Balance Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1825.9 Restrictions on Elastic Materials by the Second Law of Thermodynamics . 1865.10 Invariance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1895.11 Restrictions on Constitutive Equations from Invariance . . . . . . . . . . . 1965.12 Constitutive Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202

6 Linear Elasticity 2116.1 Elasticity, Hooke’s Law, Strain Energy . . . . . . . . . . . . . . . . . . . . . 2116.2 Hooke’s Law for Isotropic Media, Elastic Constants . . . . . . . . . . . . . 2146.3 Elastic Symmetry; Hooke’s Law for Anisotropic Media . . . . . . . . . . . 2196.4 Isotropic Elastostatics and Elastodynamics, Superposition Principle . . . 2236.5 Saint-Venant Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226

6.5.1 Extension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2276.5.2 Torsion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2286.5.3 Pure Bending . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2346.5.4 Flexure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236

6.6 Plane Elasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2386.7 Airy Stress Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2426.8 Linear Thermoelasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2526.9 Three-Dimensional Elasticity . . . . . . . . . . . . . . . . . . . . . . . . . . 253

Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 260

7 Classical Fluids 2717.1 Viscous Stress Tensor, Stokesian, and Newtonian Fluids . . . . . . . . . . . 2717.2 Basic Equations of Viscous Flow, Navier-Stokes Equations . . . . . . . . . 2737.3 Specialized Fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2757.4 Steady Flow, Irrotational Flow, Potential Flow . . . . . . . . . . . . . . . . 2767.5 The Bernoulli Equation, Kelvin’s Theorem . . . . . . . . . . . . . . . . . . 280Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282

8 Nonlinear Elasticity 2858.1 Molecular Approach to Rubber Elasticity . . . . . . . . . . . . . . . . . . . 2878.2 A Strain Energy Theory for Nonlinear Elasticity . . . . . . . . . . . . . . . 2928.3 Specific Forms of the Strain Energy . . . . . . . . . . . . . . . . . . . . . . . 2968.4 Exact Solution for an Incompressible, Neo-Hookean Material . . . . . . . 297Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304

9 Linear Viscoelasticity 3099.1 Viscoelastic Constitutive Equations in Linear Differential Operator Form . 3099.2 One-Dimensional Theory, Mechanical Models . . . . . . . . . . . . . . . . 3119.3 Creep and Relaxation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3159.4 Superposition Principle, Hereditary Integrals . . . . . . . . . . . . . . . . . 3189.5 Harmonic Loadings, Complex Modulus, and Complex Compliance . . . . 3209.6 Three-Dimensional Problems, The Correspondence Principle . . . . . . . 324References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 330Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331

Appendix A: General Tensors 343A.1 Representation of Vectors in General Bases . . . . . . . . . . . . . . . . . . 343A.2 The Dot Product and the Reciprocal Basis . . . . . . . . . . . . . . . . . . . 345A.3 Components of a Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 346A.4 Determination of the Base Vectors . . . . . . . . . . . . . . . . . . . . . . . 348A.5 Derivatives of Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 350

A.5.1 Time Derivative of a Vector . . . . . . . . . . . . . . . . . . . . . . . 350A.5.2 Covariant Derivative of a Vector . . . . . . . . . . . . . . . . . . . . 351

A.6 Christoffel Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353A.6.1 Types of Christoffel Symbols . . . . . . . . . . . . . . . . . . . . . . 353A.6.2 Calculation of the Christoffel Symbols . . . . . . . . . . . . . . . . . 354

A.7 Covariant Derivatives of Tensors . . . . . . . . . . . . . . . . . . . . . . . . 355A.8 General Tensor Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 356A.9 General Tensors and Physical Components . . . . . . . . . . . . . . . . . . 358References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 360

Appendix B: Viscoelastic Creep and Relaxation 361

Index 365

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List of Figures

2.1 Base vectors and components of a Cartesian vector. . . . . . . . . . . . . . . 82.2 Rectangular coordinate system Ox ′1x

′2x′3 relative to Ox1x2x3. Direction

cosines shown for coordinate x ′1 relative to unprimed coordinates. Simi-lar direction cosines are defined for x ′2 and x ′3 coordinates. . . . . . . . . . . 26

2.3 Rotation and reflection of reference axes. . . . . . . . . . . . . . . . . . . . . 282.4 Principal axes Ox∗1x

∗2x∗3 relative to axes Ox1x2x3. . . . . . . . . . . . . . . . . 32

2.5 Volume V with infinitesimal element dSi having a unit normal ni. . . . . . 402.6 Bounding space curve C with tangential vector dxi and surface element dSi

for partial volume. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

3.1 Typical continuum volume V with infinitesimal element ∆V having mass∆m at point P. Point P would be in the center of the infinitesimal volume. 54

3.2 Typical continuum volume with cutting plane. . . . . . . . . . . . . . . . . . 553.3 Traction vector t(n)

i acting at point P of plane element ∆Si whose normalis ni. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

3.4 Traction vectors on the three coordinate planes at point P. . . . . . . . . . . 573.5 Free body diagram of tetrahedron element having its vertex at point P. . . 573.6 Cartesian stress components shown in their positive sense. . . . . . . . . . . 603.7 Material volume showing surface traction vector t(n)

i on an infinitesimalarea element dS at position xi, and body force vector bi acting on an in-finitesimal volume element dV at position yi. Two positions are taken sep-arately for ease of illustration. When applying equilibrium the traction andbody forces are taken at the same point. . . . . . . . . . . . . . . . . . . . . . 62

3.8 Rectangular coordinate axes Px ′1x′2x′3 relative to Px1x2x3 at point P. . . . . 63

3.9 Traction vector and normal for a general continuum and a prismatic beam. 663.10 Principal axes Px∗1x

∗2x∗3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

3.11 Traction vector components normal and in-plane (shear) at point P on theplane whose normal is ni. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

3.12 Normal and shear components at P to plane referred to principal axes. . . 733.13 Typical Mohr’s circle for stress. . . . . . . . . . . . . . . . . . . . . . . . . . . 753.14 Typical Mohr’s circle representation. . . . . . . . . . . . . . . . . . . . . . . . 773.15 Typical 3-D Mohr’s circle and associated geometry. . . . . . . . . . . . . . . 783.16 Mohr’s circle for plane stress. . . . . . . . . . . . . . . . . . . . . . . . . . . . 813.17 Mohr’s circle for plane stress. . . . . . . . . . . . . . . . . . . . . . . . . . . . 833.18 Representative rotation of axes for plane stress. . . . . . . . . . . . . . . . . 843.19 Octahedral plane (ABC) with traction vector t(n)

i , and octahedral normaland shear stresses, σN and σS. . . . . . . . . . . . . . . . . . . . . . . . . . . 87

4.1 Position of typical particle in reference configuration XA and current con-figuration xi. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

4.2 Vector dXA, between points P and Q in reference configuration, becomesdxi, between points p and q, in the current configuration. Displacementvector u is the vector between points p and P. . . . . . . . . . . . . . . . . . 116

4.3 The right angle between line segments AP and BP in the reference configu-ration becomes θ, the angle between segments ap and bp, in the deformedconfiguration. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122

4.4 A rectangular parallelpiped with edge lengths dX(1), dX(2) and dX(3) in thereference configuration becomes a skewed parallelpiped with edge lengthsdx(1), dx(2) and dx(3) in the deformed configuration. . . . . . . . . . . . . . 124

4.5 Typical Mohr’s circle for strain. . . . . . . . . . . . . . . . . . . . . . . . . . . 1254.6 Rotation of axes for plane strain. . . . . . . . . . . . . . . . . . . . . . . . . . 1254.7 Differential velocity field at point p. . . . . . . . . . . . . . . . . . . . . . . . 1384.8 Area dS0 between vectors dX(1) and dX(2) in the reference configuration

becomes dS between dx(1) and dx(2) in the deformed configuration. . . . . 1434.9 Volume of parallelpiped defined by vectors dX(1), dX(2) and dX(3) in the

reference configuration deforms into volume defined by parallelpiped de-fined by vectors dx(1), dx(2) and dx(3) in the deformed configuration. . . . 145

5.1 Material body in motion subjected to body and surface forces. . . . . . . . . 1725.2 Reference frames Ox1x2x3 and O+x+

1 x+2 x

+3 differing by a superposed rigid

body motion. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191

6.1 Uniaxial loading-unloading stress-strain curves for various material behav-iors. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211

6.2 Simple stress states. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2176.3 Axes rotations for plane stress. . . . . . . . . . . . . . . . . . . . . . . . . . . 2206.4 Geometry and transformation tables for reducing the elastic stiffness to the

isotropic case. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2226.5 Beam geometry for the Saint-Venant problem. . . . . . . . . . . . . . . . . . 2266.6 Geometry and kinematic definitions for torsion of a circular shaft. . . . . . 2296.7 The more general torsion case of a prismatic beam loaded by self equili-

brating moments. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2306.8 Representative figures for plane stress and plain strain. . . . . . . . . . . . . 2396.9 Differential stress element in polar coordinates. . . . . . . . . . . . . . . . . 245

8.1 Nominal stress-stretch curves for rubber and steel. Note the same data isplotted in each figure, however, the stress axes have different scale and adifferent strain range is represented. . . . . . . . . . . . . . . . . . . . . . . . 286

8.2 A schematic comparison of molecular conformations as the distance be-tween molecule’s ends varies. Dashed lines indicate other possible confor-mations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 288

8.3 A freely connected chain with end-to-end vector r. . . . . . . . . . . . . . . 2888.4 Rubber specimen having original length L0 and cross-section area A0 stretched

into deformed shape of length L and cross section area A. . . . . . . . . . . 2918.5 Rhomboid rubber specimen compressed by platens. . . . . . . . . . . . . . . 3018.6 Rhomboid rubber specimen compressed by platens. . . . . . . . . . . . . . . 302

9.1 Simple shear element representing a material cube undergoing pure shearloading. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311

9.2 Mechanical analogy for simple shear. . . . . . . . . . . . . . . . . . . . . . . 3129.3 Viscous flow analogy. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313

9.4 Representations of Kelvin and Maxwell models for a viscoelastic solid andfluid, respectively. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314

9.5 Three parameter standard linear solid and fluid models. . . . . . . . . . . . 3149.6 Generalized Kelvin and Maxwell models constructed by combining basic

models. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3159.7 Graphic representation of the unit step function (often called the Heaviside

step function). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3169.8 Different types of applied stress histories. . . . . . . . . . . . . . . . . . . . . 3199.9 Stress history with an initial discontinuity. . . . . . . . . . . . . . . . . . . . 3199.10 Different types of applied stress histories. . . . . . . . . . . . . . . . . . . . . 322A.1 A set of non-orthonormal base vectors. . . . . . . . . . . . . . . . . . . . . . 344A.2 Circular-cylindrical coordinate system for x3 = 0. . . . . . . . . . . . . . . . 349


List of Tables

1.1 Historical notation for stress. . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2.1 Indicial form for a variety of tensor quantities. . . . . . . . . . . . . . . . . . 162.2 Forms for inner and outer products. . . . . . . . . . . . . . . . . . . . . . . . 172.3 Transformation table between Ox1x2x3 and Ox ′1x

′2x′3. . . . . . . . . . . . . . 25

3.1 Table displaying direction cosines of principal axes Px∗1x∗2x∗3 relative to axes

Px1x2x3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 693.2 Transformation table for general plane stress. . . . . . . . . . . . . . . . . . . 82

4.1 Transformation table for general plane strain. . . . . . . . . . . . . . . . . . 126

5.1 Fundamental equations in global and local forms. . . . . . . . . . . . . . . . 1835.2 Identification of quantities in the balance laws. . . . . . . . . . . . . . . . . . 184

6.1 Relations between elastic constants. . . . . . . . . . . . . . . . . . . . . . . . 218

A.1 Converting from Cartesian tensor notation to general tensor notation. Sum-mation over only subscript and superscript pairs. . . . . . . . . . . . . . . . 357

B.1 Creep and relaxation responses for various viscoelastic models. . . . . . . . 362


Preface to the Third Edition

First, a thank you to all the users of the second edition over the past years. We hopethat you will find the updates made in the text make it a more valuable introduction forstudents to continuum mechanics. The changes made in this edition were substantial butwe did not change the basic concept of the book. We seek to provide engineering studentswith a complete, concise introduction to continuum mechanics that is not intimidating.

Just like previous editions, the third edition is an outgrowth of course notes and prob-lems used to teach the topic to senior undergraduate or first year graduate students. Theimpetus to do the third edition was to expand it into a text suitable for a two quartergraduate course sequence at Cal Poly. This course sequence introduces continuum me-chanics and subsequently covers linear elasticity, nonlinear elastcity, and viscoelasticity.At Cal Poly the terminal degree is a masters degree so the combination of these topics isessential.

One of the things that students struggle with in continuum mechanics and subsequenttopics is notation. In the third edition, we have made some changes in notation makingthe book more consistent with modern continuum mechanics literature. Minor additionswere made in many places in the text. The chapter on elasticity was rearranged and ex-panded to give Saint-Venant’s solutions more complete coverage. The extension, torsion,pure bending and flexure subsections give the student a good foundation for posing andsolving basic elasticity problems. We have also added some new applications applyingcontinuum mechanics to biological materials in light of their current importance. Finally,a limited amount of material using Matlab

r has been introduced in this edition. We didnot want to minimize the fundamental principles of continuum mechanics by making thetopic seem like it can be mastered by learning mathematical software. Yet at the sametime, these tools can provide valuable help allowing one to stay focused on fundamen-tals. In addition, most current graduate students are quite proficient at using tools suchas Matlab

r, so we did not feel we had to emphasize that topic.There are many people to acknowledge in the writing of this edition, and we ask the

reader to see the Acknowledgments so these people receive their well deserved recogni-tion.

G. Thomas MaseSan Luis Obispo, California, USA

Ronald E. SmelserCharlotte, North Carolina, USA

George E. MaseEast Lansing, Michigan, USA


Preface to the Second Edition

(Note: Some chapter reference information has changed in the Third Edition.)

It is fitting to start this, the preface to our second editions, by thanking all of those whoused the text over the last six years. Thanks also to those of you who have inquired aboutthis revised and expanded version. We hope that you find this edition as helpful as thefirst to introduce seniors or graduate students to continuum mechanics.

The second edition, like its predecessor, is an outgrowth of teaching continuum me-chanics to first- or second-year graduate students. Since my father is now fully retired,the course is being taught to students whose final degree will most likely be a Masters atKettering University. A substantial percentage of these students are working in industry,or have worked in industry, when they take this class. Because of this, the course has toprovide the students with the fundamentals of continuum mechanics and demonstrate itsapplications.

Very often, students are interested in using sophisticated simulation programs thatuse nonlinear kinematics and a variety of constitutive relationships. Additions to thesecond edition have been made with these needs in mind. A student who masters itscontents should have the mechanics foundation necessary to be a skilled user of today’sadvanced design tools such as nonlinear, explicit finite elements. Of course, studentsneed to augment the mechanics foundation provided herein with rigorous finite elementtraining.

Major highlights of the second edition include two new chapters, as well as significantexpansion of two other chapters. First, Chapter Five, Fundamental Laws and Equations,was expanded to add materials regarding constitutive equation development. This in-cludes material on the second law of thermodynamics and invariance with respect torestrictions on constitutive equations. The first edition applications chapter covering elas-ticity and fluids has been split into two separate chapters. Elasticity coverage has beenexpanded by adding sections on Airy stress functions, torsion of non-circular cross sec-tions, and three dimensional solutions. A chapter on nonlinear elasticity has been addedto give students a molecular and phenomenological introduction to rubber-like materi-als. Finally, a chapter introducing students to linear viscoelasticity is given since manyimportant modern polymer applications involve some sort of rate dependent materialresponse.

It is not easy singling out certain people in order to acknowledge their help whilenot citing others; however, a few individuals should be thanked. Ms. Sheri Burton wasinstrumental in preparation of the second edition manuscript. We wish to acknowledgethe many useful suggestions by users of the previous edition, especially Prof. MortezaM. Mehrabadi, Tulane University, for his detailed comments. Thanks also go to Prof.Charles Davis, Kettering University, for helpful comments on the molecular approach torubber and thermoplastic elastomers. Finally, our families deserve sincerest thanks fortheir encouragement.

It has been a great thrill to be able to work as a father-son team in publishing this text,so again we thank you, the reader, for your interest.

G. Thomas MaseFlint, Michigan, USA



Preface to the First Edition

(Note: Some chapter reference information has changed in the Third Edition.)

Continuum mechanics is the fundamental basis upon which several graduate coursesin engineering science such as elasticity, plasticity, viscoelasticity and fluid mechanicsare founded. With that in mind, this introductory treatment of the principles of contin-uum mechanics is written as a text suitable for a first course that provides the studentwith the necessary background in continuum theory to pursue a formal course in any ofthe aforementioned subjects. We believe that first-year graduate students, or upper-levelundergraduates, in engineering or applied mathematics with a working knowledge ofcalculus and vector analysis, and a reasonable competency in elementary mechanics willbe attracted to such a course.

This text evolved from the course notes of an introductory graduate continuum me-chanics course at Michigan State University, which was being taught on a quarter basis.We feel that this text is well suited for either a quarter or semester course in continuummechanics. Under a semester system, more time can be devoted to later chapters dealingwith elasticity and fluid mechanics. For either a quarter or a semester system, the text isintended to be used in conjunction with a lecture course.

The mathematics employed in developing the continuum concepts in the text is thealgebra and calculus of Cartesian tensors; these are introduced and discussed in somedetail in Chapter Two, along with a review of matrix methods, which are useful forcomputational purposes in problem solving. Because of the introductory nature of thetext, curvilinear coordinates are not introduced and so no effort has been made to involvegeneral tensors in this work. There are several books listed in the Reference Sectionthat a student may refer to for a discussion of continuum mechanics in terms of generaltensors. Both indicial and symbolic notations are used in deriving the various equationsand formula of importance.

Aside from the essential mathematics presented in Chapter Two, the book can be seenas divided into two parts. The first part develops the principles of stress, strain and mo-tion in Chapters Three and Four, followed by the derivation of the fundamental physicallaws relating to continuity, energy and momentum in Chapter Five. The second portion,Chapter Six, presents some elementary applications of continuum mechanics to linearelasticity and classic fluids behavior. Since this text is meant to be a first text in contin-uum mechanics, these topics are presented as constitutive models without any discussionas to the theory of how the specific constitutive equations was derived. Interested read-ers should pursue more advanced texts listed in the Reference Section for constitutiveequation development. At the end of each chapter (with the exception of Chapter One)there appears a collection of problems, with answers to most, by which the student mayreinforce her/his understanding of the material presented in the text. In all, 186 suchpractice problems are provided, along with numerous worked examples in the text itself.

Like most authors, we are indebted to may people who have assisted in the preparationof this book. Although we are unable to cite each of them individually, we are pleasedto acknowledge the contributions of all. In addition, sincere thanks must go to the stu-dents who have given feedback for the classroom notes which served as the forerunner

to the book. Finally, and most sincerely of all, we express thanks to our family for theirencouragement from beginning to end of this work.

G. Thomas MaseFlint, Michigan, USA


Acknowledgments

There are too many people to thank for their help in preparing this third edition. Wecan only mention the key contributors. Ryan Miller was a superb help in moving theearly manuscript into LATEX 2εbefore his masters research redirected his focus. We werefortunate that one of us (GTM) was teaching ME 501 and 503 in the fall and winterquarters at Cal Poly while preparing the manuscript. The class was quite helpful inproofreading the manuscript. Specifically, Nickolai Volkoff-Shoemaker, Peter Brennen,Roger Sharpe, John Wildharbor, Kevin Ng, and Jason Luther found many typographicalerrors and suggested helpful corrections and clarifications. Nickolai Volkoff-Shoemaker,Peter Brennen and Roger Sharpe helped in creating some of the figures.

One author (GTM) is very appreciative of Don Bently’s generous gift to Cal Poly al-lowing for partial release time during the Fall 2009 quarter. In addition, many thanks tothe devoted teachers that shaped him as a student including George E. Mase, George C.Johnson, Paul M. Naghdi, Michael M. Carroll and David B. Bogy. The other (RES) wasprivileged to benefit from interactions with several outstanding colleagues and teachersincluding Ronald Huston, University of Cincinnati, William J. Shack, MIT and ArgonneNational Laboratories, Morton E. Gurtin, Carnegie Mellon University and the late OwenRichmond, US Steel Research Laboratories and Alcoa Technical Center.

Of course, our greatest thanks go to our families who very patiently kept asking if thebook was done. Now it is done; so we can spend more time with the ones we love.


Authors

G. Thomas Mase, Ph.D., is Associate Professor of Mechanical Engineering at CaliforniaPolytechnic State University, San Luis Obispo, California. Dr. Mase received his B.S.degree from Michigan State University in 1980 from the Department of Metallurgy, Me-chanics and Materials Science. He obtained his M.S. and Ph.D. degrees in 1982 and 1985,respectively, from the Department of Mechanical Engineering at the University of Califor-nia, Berkeley. After graduate school, he has worked at several positions in industry andacademia. Industrial companies Dr. Mase has worked full time for include General Mo-tors Research Laboratories, Callaway Golf and Acushnet Golf Company. He has taughtor held research positions at the University of Wyoming, Kettering University, Michi-gan State University and California Polytechnic State University. Dr. Mase is a memberof numerous professional societies including the American Society of Mechanical En-gineers, American Society for Engineering Education, International Sports EngineeringAssociation, Society of Experimental Mechanics, Pi Tau Sigma and Sigma Xi. He receivedan ASEE/NASA Summer Faculty Fellowship in 1990 and 1991 to work at NASA LewisResearch Center (currently NASA Glenn Research Center). While at the University ofCalifornia, he twice received a distinguished teaching assistant award in the Departmentof Mechanical Engineering. His research interests include mechanics, design and appli-cations of explicit finite element simulation. Specific areas include golf equipment designand performance and vehicle crashworthiness.

Ronald E. Smelser, Ph.D., P.E., is Professor and Associate Dean for Academic Affairsin the William States Lee College of Engineering at the University of North Carolina atCharlotte. Dr. Smelser received his B.S.M.E. from the University of Cincinnati in 1971.He was awarded the S.M.M.E. in 1972 from M.I.T. and completed his Ph.D. (1978) inmechanical engineering at Carnegie Mellon University. He gained industrial experienceworking for the United States Steel Research Laboratory, the Alcoa Technical Center, andConcurrent Technologies Corporation. Dr. Smelser served as a fulltime or adjunct facultymember at the University of Pittsburgh, Carnegie Mellon University, and the Universityof Idaho and was a visiting research scientist at Colorado State University. Dr. Smelser isa member of the American Academy of Mechanics, the American Society for EngineeringEducation, Pi Tau Sigma, Sigma Xi, and Tau Beta Pi. He is also a member and Fellowof the American Society of Mechanical Engineers. Dr. Smelser’s research interests are inthe areas of process modeling including rolling, casting, drawing and extrusion of singleand multi-phase materials, the micromechanics of material behavior and the inclusion ofmaterial structure into process models, and the failure of materials.

George E. Mase (1920-2007), Ph.D., was Emeritus Professor, Department of Metallurgy,Mechanics and Materials Science (MMM), College of Engineering, at Michigan State Uni-versity. Dr. Mase received a B.M.E in Mechanical Engineering (1948) from the OhioState University, Columbus. He completed his Ph.D. in Mechanics at Virginia PolytechnicInstitute and State University (VPI), Blacksburg, Virginia (1958). Previous to his initialappointment as Assistant Professor in the Department of Applied Mechanics at MichiganState University in 1955, Dr. Mase taught at Pennsylvania State University (instructor),

1950-1951, and at Washington University, St. Louis, Missouri (assistant professor), 1951-1954. He was appointed associate professor in 1959 and professor in 1965, and servedas acting chairperson of the MMM Department 1965-1966 and again in 1978 to 1979. Hetaught as visiting assistant professor at VPI during the summer terms, 1953 through 1956.Dr. Mase held membership in Tau Beta Pi and Sigma Xi. His research interests andpublications were in the areas of continuum mechanics, viscoelasticity and biomechanics.

Nomenclature

x or xi Spatial or current coordinates

X or XA Material or referential coordinates

u or ui Displacement components or displacementvector

v or vi Velocity or general vector

a or ai Acceleration or general vector

x∗1, x∗2, x

∗3 Principal axes

ei Unit vectors along coordinate axes

IA Unit vectors along coordinate axes in refer-ence configuration

δij Kronecker delta

A or aij Transformation or general matrix

I Identity matrix

εijk Permutation symbol

∂t Partial derivative with respect to time

˙(·) Derivative with respect to time

∇ or ∂x Spatial gradient operator

∇φ = grad φ = φ,j Scalar gradient

∇v = ∂jvi = vi,j Vector gradient

∇ · v or vi,i Divergence of a vector v

∇× v or εijkvi,j Curl of a vector v

d/dt = ∂/∂t+ vk∂/∂xk Material derivative operator

b or bi Body force (force per unit mass)

p or pi Body force (force per unit volume)

f or fi Surface force (force per unit area)

V, V0 Current and referential total volumes

∆V, dV Small and infinitesimal element of volumes

S, S0 Current and referential total surfaces

∆S, dS Small and infinitesimal elements of surface

ρ Density

n or ni Unit normal in current configuration

N or NA Unit normal in reference configuration

t(n) or t(n)i Traction vector

σN, σS Normal and shear components of tractionvector

T or tij Cauchy stress or general tensor

T∗ or t∗ij Cauchy stress referred to principal axes

p0(N) or p0(N)i Piola-Kirchhoff stress vector referred to ref-

erential area

P or PiA First Piola-Kirchhoff stress

s or sAB Second Piola-kirchhoff stress

σ(1), σ(2), σ(3) Principal stress values

IT , IIT , IIIT First, second and third stress invariants

σM = 13tii Mean normal stress

Sij Deviatoric stress components

ηij Deviatoric strain components

J1 = 0, J2, J3 Deviatoric stress invariants

σoct Octahedral shear stress

F or FiA Deformation gradient tensor

C or CAB Right Cauchy-Green deformation tensor

E or EAB Lagrangian finite strain tensor

c or cij Cauchy deformation tensor

e or eij Eulerian finite strain tensor

ε or εij Infinitesimal strain tensor

ε(1), ε(2), ε(3) Principal strain values

Iε, IIε, IIIε First, second and third infinitesimal straininvariants

B or Bij Left Cauchy-Green deformation tensor

IB, IIB, IIIB, orI1, I2, I3

Invariants of right deformation tensor

W Strain energy per unit volume or strain en-ergy density

eN Normal strain in the N direction

γij Engineering shear strain

e = ∆V/V = εii = εI Cubical dilatation

ω or ωij Infinitesimal rotation tensor

ω or ωj Rotation vector

ΛN = dx/dX Stretch ratio or stretch in the direction of N

λn = dX/dx Stretch ratio in the direction of n

R or Rij Rotation tensor

U or UAB Right stretch tensor

V or Vij Left stretch tensor

L or Lij Spatial velocity gradient

D or dij Rate of deformation tensor

W or wij Vorticity, or spin tensor

J = det F Jacobian

P(t) or Pi Linear momentum vector

K(t) Kinetic energy

P(t) Mechanical power or rate of work done byforces

S(t) Stress work

Q Heat input rate

r Heat supply per unit mass

q or qi Heat flux vector

θ Temperature or angle

g = grad θ or gi = θ,i Temperature gradient

u Specific internal energy

η Specific entropy or viscoelastic viscosity

ψ Gibbs’ free energy

ζ Free enthalpy

χ Enthalpy

γ Specific entropy production

E modulus of elasticity or Young’s modulus

G or µ shear modulus

K Bulk modulus

ν Poisson’s ratio

λ, µ Lame constants

Cijkl General elastic constants

τij Viscous stress tensor

βij Deviatoric rate of deformation

λ∗, µ∗ Viscosity coefficients

κ∗ Bulk viscosity coefficient

1Continuum Theory

The atomic/molecular composition of matter is well established. On a small enoughscale, a body of aluminum, is really a collection of discrete aluminum atoms stackedon one another in a particular repetitive lattice. And on an even smaller scale, the atomsconsist of a core of protons and neutrons around which electrons orbit. Thus matter is notcontinuous. At the same time, the physical space in which we live is truly a continuum,for mathematics teaches us that between any two points in space we can always findanother point regardless of how close together we choose the original pair. Clearly then,although we may speak of a material body as “occupying” a region of physical space, itis evident that the body does not totally “fill” the space it occupies. It is this “occupying”of space that will be the basis of our study of continuum mechanics.

1.1 Continuum Mechanics

If we accept the continuum concept of matter, we agree to ignore the discrete composi-tion of material bodies, and to assume that the substance of such bodies is distributeduniformly throughout, and completely fills the space it occupies. In keeping with thiscontinuum model, we assert that matter may be divided indefinitely into smaller andsmaller portions, each of which retains all of the physical properties of the parent body.Accordingly, we are able to ascribe field quantities such as density and velocity to eachand every point of the region of space which the body occupies.

The continuum model for material bodies is important to engineers for two very goodreasons. On the scale by which we consider bodies of steel, aluminum, concrete, etc., thecharacteristic dimensions are extremely large compared to molecular distances so thatthe continuum model provides a very useful and reliable representation. Additionally,our knowledge of the mechanical behavior of materials is based almost entirely uponexperimental data gathered by tests on relatively large specimens.

The analysis of the kinematic and mechanical behavior of materials modeled on thecontinuum assumption is what we know as Continuum Mechanics. There are two mainthemes into which the topics of continuum mechanics are divided. In the first, emphasisis on the derivation of fundamental equations which are valid for all continuous media.These equations are based upon universal laws of physics such as the conservation ofmass, the principles of energy and momentum, etc. In the second, the focus of attentionis on the development of the constitutive equations characterizing the behavior of specificidealized materials; the perfectly elastic solid and the viscous fluid being the best knownexamples. These equations provide the focal points around which studies in elasticity,plasticity, viscoelasticity and fluid mechanics proceed.

Mathematically, the fundamental equations of continuum mechanics mentioned abovemay be developed in two separate but essentially equivalent formulations. One, the

1

2 Continuum Mechanics for Engineers

integral, or global form, derives from a consideration of the basic principles being appliedto a finite volume of the material. The other, a differential, or field approach, leads toequations resulting from the basic principles being applied to a very small (infinitesimal)element of volume. In practice, it is often useful and convenient to deduce the fieldequations from their global counterparts.

As a result of the continuum assumption, field quantities such as density and velocitywhich reflect the mechanical or kinematic properties of continuum bodies are expressedmathematically as continuous functions, or at worst as piecewise continuous functions,of the space and time variables. Moreover, the derivatives of such functions, if they enterinto the theory at all, will be likewise continuous.

Inasmuch as this is an introductory textbook, we shall make two further assumptionson the materials we discuss in addition to the principal one of continuity. First, we requirethe materials to be homogeneous, that is to have identical properties at all locations. Andsecondly, that the materials be isotropic with respect to certain mechanical properties,meaning that those properties are the same in all directions at a given point. Later,we will relax this isotropy restriction to discuss briefly anisotropic materials which haveimportant meaning in the study of composite materials.

1.2 Starting Over

The topic of continuum mechanics typically comes at the end of an undergraduate or atthe beginning of a graduate program. Continuum mechanics has a reputation of beinga theoretical course without many applications (during the course). The first of part ofcontinuum mechanics’ reputation is correct: it is based on fundamental mathematics andmechanics. However, the second part is not founded. There are many, many applicationsfor continuum mechanics, but it is hard to cover the basics and develop the applications ina single quarter or semester. Continuum mechanics takes all the mathematical, physicaland engineering principles and casts them in a single structure from which the student isprepared to pursue advanced engineering topics. After having a course in continuum me-chanics many applications become accessible to the student: elasticity, nonlinear elasticity,plasticity, crashworthiness, biomechanics, polymers and more. Many of the sophisticatedsimulation programs such as LS-DYNAr become a playground for advanced design andanalysis once continuum mechanics has been mastered.

Some students find continuum mechanics a difficult subject. However, outside of anew notation, the topics studied should be very familiar to the student. Vectors have tobe written in component form, and we need to be able to use “dot” and “cross” products.These are skills from the sophomore level statics course. Also needed will be a descriptionfor conservation of linear and angular momentum. Taking the time rate of change ofthese quantities is really no different than what was done in an undergraduate coursein dynamics. Just like dynamics, a description of the energy equation will be examined.Rather than study only rigid bodies as done in undergraduate statics and dynamics,deformation is allowed. This requires defining stress and strain that were first introducedin a mechanics of solids class. Stress and strain are tensors which are an order morecomplex than vectors. When looking at strain and the resulting stress one needs to havea material model. At the undergraduate level students have studied linear elastic, fluidand gas behavior. But the topics generally are taught in separate courses, and often thecommon, underlying theory is not noticed. Also, the methods used to determine therelationship between stress and strain are not considered.

Continuum Theory 3

So continuum mechanics is not a new or challenging topic. Rather it is a chance tostart over and put all that was studied previously under a single umbrella. A course incontinuum mechanics is a chance to synthesize what was learned during an undergradu-ate education into a coherent structure. One of the challenges of doing this is developinga common notation, but no new physics is presented. Continuum mechanics is just theprocess of confirming the foundation for all that was done in undergraduate studies.

1.3 NotationAs one would imagine, building a theoretical foundation for the study of continuum me-chanics creates notational difficulties. This is especially true for the student just learningcontinuum mechanics. In the pages that follow there are many different symbols usedfor all the quantities of interest. There are more symbols than a student experienced asan undergraduate because a general, nonlinear theory is being contructed. For instance,consider stress. There are different measures of stress that are indistinguishable in thelinear theory: Cauchy, first Piola-Kirchhoff, and second Piola-Kirchhoff. In addition, vonMises and octahedral stresses are defined to help analyze yield and failure theory. Finally,it is often advantageous to subdivide stress into deviatoric and spherical parts because ofthe different role the two have in deformed bodies.

With all these quantities it is hard to come up with symbols for each of them. Often, onesymbol is very close to another symbol, and the context has to be used to fully understandthe meaning. This is one of the things that makes continuum mechanics difficult for thebeginner.

Finally, as students continue beyond this course, they find the disheartening realitythat not everybody uses the same notation. Often notational marks will have to be madein margins when reading the literature. The reason there is not a single notation comesfrom the fact that people were not flying in airplanes to technical conferences when thismaterial was being developed. Even when reading a single author’s works spanning adecade the notation can change. For example, Table 1.1 1 shows some historical notationfor stress.

1Adapted from Nonlinear Theory of Continuous Media, A. Cemal Eringen, McGraw-Hill Inc., (1962)


TABLE 1.1Historical notation for stress.

Naghdi, Eringen, Clebsch, Truesdell t11 t22 t33 t12 t23 t31

Cauchy (early work) A B C D E F

Cauchy (later work), St. Venant, Maxwell pxx pyy pzz pxy pyz pzx

F. Neumann, Kirchhoff, Love Xx Yy Zz Xy Yz Zx

Green and Zerna, Russian and German writers τ11 τ22 τ33 τ12 τ23 τ31

Karman, Timoshenko σx σy σz τxy τyz τzx

Some English and American writers σ11 σ22 σ33 σ12 σ23 σ31

σxx σyy σzz σxy σyz σzx

2Essential Mathematics

Learning a discipline’s language is the first step a student takes towards becoming compe-tent in that discipline. The language of continuum mechanics is the algebra and calculusof tensors. Here, tensor is the generic name for those mathematical entities which areused to represent the important physical quantities of continuum mechanics. A tensor, orlinear transformation, assigns any vector v to another vector Tv such that

T (v+w) = Tv+ Tw , (2.1a)

andT (αv) = αTv (2.1b)

for all v and w. Furthermore, the sum of two tensors and the scalar multiple of a tensoris defined by

(T + S) v = Tv+ Sv , (2.1c)

and(αT ) v = α (Tv) . (2.1d)

Because of these properties, tensors constitute a vector space.Tensors have a most useful property in the way that they transform from one basis

(reference frame) to another. Having the tensor defined with respect to one referenceframe, the tensor quantity (components) can be written in any admissible reference frame.An example of this would be stress defined in principal and non-principal components.Both representations are of the same stress tensor even though the individual componentsmay be different. As long as the relationship between the reference frames is known, thecomponents with respect to one frame may be found from the other.

Only that category of tensors known as Cartesian tensors is used in this text, and defini-tions of these will be given in the pages that follow. General tensor notation is presentedin the Appendix for completeness, but it is not necessary for the main text. The ten-sor equations used to develop the fundamental theory of continuum mechanics may bewritten in either of two distinct notations; the symbolic notation, or the indicial notation.We shall make use of both notations, employing whichever is more convenient for thederivation or analysis at hand but taking care to establish the inter-relationships betweenthe two. However, an effort to emphasize indicial notation in most of the text has beenmade. An introductory course must teach indicial notation to the student who may havelittle prior exposure to the topic.

2.1 Scalars, Vectors and Cartesian TensorsA considerable variety of physical and geometrical quantities have important roles incontinuum mechanics, and fortunately, each of these may be represented by some form

5


of tensor. For example, such quantities as density and temperature may be specified com-pletely by giving their magnitude, i.e., by stating a numerical value. These quantitiesare represented mathematically by scalars, which are referred to as zero-order tensors. Itshould be emphasized that scalars are not constants, but may actually be functions ofposition and/or time. Also, the exact numerical value of a scalar will depend upon theunits in which it is expressed. Thus, the temperature may be given by either 68F, or 20Cat a certain location. As a general rule, lower-case Greek letters in italic print such as α,β, λ, etc. will be used as symbols for scalars in both the indicial and symbolic notations.

Several physical quantities of mechanics such as force and velocity require not only anassignment of magnitude, but also a specification of direction for their complete charac-terization. As a trivial example, a 20 N force acting vertically at a point is substantiallydifferent than a 20 N force acting horizontally at the point. Quantities possessing suchdirectional properties are represented by vectors, which are first-order tensors. Geometri-cally, vectors are generally displayed as arrows, having a definite length (the magnitude),a specified orientation (the direction), and also a sense of action as indicated by the headand the tail of the arrow. In this text arrow lengths are not to scale with vector magnitude.Certain quantities in mechanics which are not truly vectors are also portrayed by arrows,for example, finite rotations.

Consequently, in addition to the magnitude and direction characterization, the com-plete definition of a vector requires the further statement: vectors add (and subtract) inaccordance with the triangle rule by which the arrow representing the vector sum of twovectors extends from the tail of the first component arrow to the head of the second whenthe component arrows are arranged ”head-to-tail”.

Although vectors are independent of any particular coordinate system, it is often usefulto define a vector in terms of its coordinate components, and in this respect it is neces-sary to reference the vector to an appropriate set of axes. In view of our restriction toCartesian tensors, we limit ourselves to consideration of Cartesian coordinate systems fordesignating the components of a vector.

A significant number of physical quantities having important status in continuum me-chanics require mathematical entities of higher order than vectors for their representationin the hierarchy of tensors. As we shall see, among the best known of these are the stressand the strain tensors. These particular tensors are second-order tensors, and are said tohave a rank of two. Third-order and fourth-order tensors are not uncommon in contin-uum mechanics but they are not nearly as plentiful as second-order tensors. Accordingly,the unqualified use of the word tensor in this text will be interpreted to mean second-ordertensor. With only a few exceptions, primarily those representing the stress and straintensors, we shall denote second-order tensors by upper-case sans serif Latin letters inbold-faced print, a typical example being the tensor T . The components of the said tensorwill, in general, be denoted by lower-case Latin letters with appropriate indices: tij.

Tensors, like vectors, are independent of any coordinate system, but just as with vectors,when we wish to specify a tensor by its components we are obliged to refer to a suitableset of reference axes. The precise definitions of tensors of various order will be givensubsequently in terms of the transformation properties of their components between tworelated sets of Cartesian coordinate axes.

As a quick notation summary, the International Standards Organization (ISO) conven-tions for typesetting mathematics are summarized below:

1. Scalar variables are written as italic letters. The letters may be either Roman orGreek style fonts depending on the physical quantity they represent. The followingexamples are a partial list of scalar notation:

(a) a – magnitude of acceleration

Essential Mathematics 7

(b) v – magnitude of velocity(c) r – radius(d) θ – temperature or angle depending on context(e) α – coefficient of thermal expansion(f) σ – principal value of stress(g) λ – eigenvalue or stretch

2. Vectors are written as boldface italic. Examples are as follows:

(a) x – position(b) v – velocity(c) a – acceleration(d) e1 – base vector in x1 direction

3. Second- and higher-order tensors are designated by uppercase fonts. Additionally,matrices are shown in the calligraphic form to differentiate them from tensors. Ten-sors can be represented by matrices, but not all matrices are tensors. In the caseof several well known engineering quantities this convention will not be accommo-dated. For example, linear strain has been chosen to be represented by ε. Here aresome samples of tensor and matrix symbols:

(a) Q – orthogonal matrix(b) E – finite strain(c) T – Cauchy stress tensor(d) ε – infinitesimal strain tensor(e) R – rotation matrix

2.2 Tensor Algebra in Symbolic Notation - Summation ConventionThe three-dimensional physical space of everyday life is the space in which many of theevents of continuum mechanics occur. Mathematically, this space is known as a Euclideanthree-space , and its geometry can be referenced to a system of Cartesian coordinate axes.In some instances, higher order dimension spaces play integral roles in continuum topics.Because a scalar has only a single component, it will have the same value in every systemof axes, but the components of vectors and tensors will have different component values,in general, for each set of axes.

In order to represent vectors and tensors in component form we introduce in our physi-cal space a right-handed system of rectangular Cartesian axes Ox1x2x3, and identify withthese axes the triad of unit base vectors, e1, e2, e3, shown in Fig. 2.1(a). All unit vectorsin this text will be written with a caret placed above the bold-faced symbol. Due to themutual perpendicularity of these base vectors they form an orthogonal basis, and further-more, because they are unit vectors, the basis is said to be orthonormal. In terms of thisbasis an arbitrary vector v is given in component form by

v = v1e1 + v2e2 + v3e3 =

3∑i=1

viei . (2.2)


x1

x2

x3

O

e2

e1

e3

(a) Unit vectors in the coordinate directions x1,x2 and x3.

x1

x2

x3

Ov

1

v2

v3

v

(b) Rectangular Cartesian components of thevector v.

FIGURE 2.1Base vectors and components of a Cartesian vector.

This vector and its coordinate components are pictured in Fig. 2.1(b). For the symbolicdescription, vectors will usually be given by lower-case Latin letters in bold-faced print,with the vector magnitude denoted by the same letter. Thus v is the magnitude of v.

At this juncture of our discussion it is helpful to introduce a notational device called thesummation convention that will greatly simplify the writing of the equations of continuummechanics. Stated briefly, we agree that whenever a subscript appears exactly twice in agiven term, that subscript will take on the values 1, 2, 3 successively, and the resultingterms summed. For example, using this scheme, we may now write Eq 2.2 in the simpleform

v = viei , (2.3)

and delete entirely the summation symbol∑

. For Cartesian tensors, only subscripts arerequired on the components; for general tensors both subscripts and superscripts areused. The summed subscripts are called dummy indices since it is immaterial which par-ticular letter is used. Thus viei is completely equivalent to vjej, or to vkek, when thesummation convention is used. A word of caution, however; no subscript may appearmore than twice in a singe term. But as we shall soon see, more than one pair of dummyindices may appear in a given expression that is the summation of terms (see Exam-ple 2.2). Note also that the summation convention may involve subscripts from both theunit vectors and the scalar coefficients.

Example 2.1

Without regard for meaning as far as mechanics is concerned, expand thefollowing expressions according to the summation convention:

(a) uiviwjej (b) tijviej (c) tiivjej


Solution

(a) Summing first on i, and then on j

uiviwjej = (u1v1 + u2v2 + u3v3)(w1e1 +w2e2 +w3e3)

(b) Summing on i, then on j and collecting terms on the unit vectors.

tijvjei = t1jvje1 + t2jvje2 + t3jvje3

= (t11v1 + t12v2 + t13v3) e1 + (t21v1 + t22v2 + t23v3) e2

+(t31v1 + t32v2 + t33v3) e3

(c) Summing on i, then on j,

tiivjej = (t11 + t22 + t33) (v1e1 + v2e2 + v3e3)

Note the similarity between (a) and (c).

With the above background in place it is now possible, using symbolic notation, topresent many useful definitions from vector/tensor algebra. There are two symbolsneeded prior to writing out all of the vector and tensor algebra necessary. These twosymbols are the Kronecker delta and the permutation symbol. Additionally, there areseveral useful relationships between the Kronecker delta and permutation symbol thatare used throughout continuum mechanics. The following three subsections introducethe Kronecker delta, permutation symbol and their relationships. Following that, vec-tor/tensor algebra is presented.

The Kronecker delta is similar to the identity matrix, so the reader should quicklyembrace this new entity. However, the permutation symbol is a little more abstract thanthe Kronecker delta since it cannot be represented by a matrix. In subsequent chaptersthe Kronecker delta and the permutation symbol play integral roles in describing howforces are carried by continuum bodies and how the position of a particle is described.

2.2.1 Kronecker Delta

Since the base vectors ei (i = 1,2,3) are unit vectors and orthogonal

ei · ej =

1 if numerical value of i = numerical value of j0 if numerical value of i 6= numerical value of j

.

Therefore, if we introduce the Kronecker delta defined by

δij =

1 if numerical value of i = numerical value of j0 if numerical value of i 6= numerical value of j

we see that

ei · ej = δij (i, j = 1, 2, 3) . (2.4)


Also, note that by the summation convention

δii = δjj = δ11 + δ22 + δ33 = 1+ 1+ 1 = 3 ,

and furthermore, we call attention to the substitution property of the Kronecker delta byexpanding (summing on j) the expression

δijej = δi1e1 + δi2e2 + δi3e3 .

But for a given value of i in this equation, only one of the Kronecker deltas on the righthand side is non-zero, and it has the value one. Therefore,

δijej = ei ,

and the Kronecker delta in δijej causes the summed subscript j of ej to be replaced by ireducing the expression to simply ei.

2.2.2 Permutation Symbol

By introducing the permutation symbol εijk defined by

εijk =

1 if numerical values of ijk appear as in the sequence 12312

−1 if numerical values of ijk appear as in the sequence 321320 if numerical values of ijk appear in any other sequence

(2.5)

we may express the cross products of the base vectors (i=1,2,3) by the use of Eq 2.5 as

ei × ej = εijkek (i, j, k = 1, 2, 3) . (2.6)

Also, note from its definition that the interchange of any two subscripts in εijk causes asign change so that for example,

εijk = −εkji = εkij = −εikj ,

and, furthermore, that for repeated subscripts is zero as in

ε113 = ε212 = ε133 = ε222 = 0 .

2.2.3 ε - δ Identity

The product of permutation symbols εmiqεjkq may be expressed in terms of Kroneckerdeltas by the ε - δ identity

εmiqεjkq = δmjδik − δmkδij (2.7a)

as may be proven by direct expansion. This is a most important formula used throughoutthis text and is well worth memorizing. Also, by the sign-change property of εijk,

εmiqεjkq = εmiqεqjk = εqmiεqjk = εqmiεjkq .

Additionally, it is easy to show from Eq 2.7a that

εjkqεmkq = 2δjm , (2.7b)

by setting i = k, andεjkqεjkq = 6 . (2.7c)


2.2.4 Tensor/Vector Algebra

To begin with, vector addition is easily written in indicial form

w = u+ v or wiei = (ui + vi) ei (2.8)

where the components simply add together.Simple vector multiplication can take one of several forms. The specific form depends

on the type of entity multiplying the vector. For now, two forms of vector multiplicationcan be defined in symbolic form. Multiplication of a vector by a scalar is written as

λv = λviei , (2.9)

and the dot (scalar) product between of two vectors is

u · v = v · u = uv cos θ (2.10)

where θ is the smaller angle between the two vectors when drawn from a common origin.From the definition of δij and its substitution property the dot product u · v may be

written as

u · v = uiei · vjej = uivjei · ej = uivjδij = uivi . (2.11)

Note that scalar components pass through the dot product since it is a vector operator.The vector cross (vector) product of two vectors is defined by

u× v = −v× u = (uv sin θ) e

where 0 6 θ 6 π, is the angle between the two vectors when drawn from a commonorigin, and where e is a unit vector perpendicular to their plane such that a right-handedrotation about e through the angle θ carries u into v .

The vector cross product may be written in terms of the permutation symbol (Eq 2.5) asfollows:

u× v = uiei × vjej = uivj (ei × ej) = εijkuivjek . (2.12)

Again, notice how the scalar components pass through the vector cross product operator.There are a couple of useful ways three vectors can be multiplied. The triple scalar

product (box product) isu · v×w = u× v ·w = [u, v,w] ,

or

[u, v,w] = uiei · (vjej ×wkek) = uiei · εjkqvjwkeq (2.13)= εjkquivjwkδiq = εijkuivjwk

where in the final step we have used both the substitution property of δiq and the sign-change property of εijk. The triple cross product is similar to the triple scalar product

u× (v×w) = uiei × (vjej ×wkek) = uiei × (εjkqvjwkeq) (2.14)= εiqmεjkquivjwkem = εmiqεjkquivjwkem

which may be written as

u× (v×w) = (δmjδik − δmkδij)uivjwkem (2.15)= (uivmwi − uiviwm) em = uiwivmem − uiviwmem .


Observation of the indices in Eq 2.15 admits

u× (v×w) = (u ·w) v− (u · v)w

a well-known identity from vector algebra.In addition to the common vector products above, two vectors can be multiplied to-

gether to yield a tensor. The tensor product of two vectors creates a dyad

uv = uieivjej = uivjeiej (2.16)

which in expanded form, summing first on i, yields

uivjeiej = u1vje1ej + u2vje2ej + u3vje3ej ,

and then summing on j

uivjeiej = u1v1e1e1 + u1v2e1e2 + u1v3e1e3

+ u2v1e2e1 + u2v2e2e2 + u2v3e2e3 (2.17)+ u3v1e3e1 + u3v2e3e2 + u3v3e3e3 .

This nine-term sum is called the nonion form of the dyad, uv. A sum of dyads such as

u1v1 + u2v2 + · · ·+ uNvN (2.18)

is called a dyadic.A common, alternative notation frequently used for the dyad product is

a⊗ b = aiei ⊗ bjej = aibjei ⊗ ej (2.19)

which is called a tensor product. The tensor product of vectors a and b is defined by howa⊗ b maps all vectors u:

(a⊗ b)u = a (b · u) . (2.20)

If one takes vectors a, b and u to be vectors from a Euclidian 3-space, the expanded formfor the tensor product may be written as

(a⊗ b)u = a (b · u) =

a1a2a3

(b1u1 + b2u2 + b3u3)

=

a1b1u1 + a1b2u2 + a1b3u3

a2b1u1 + a2b2u2 + a2b3u3

a3b1u1 + a3b2u2 + a3b3u3

=

a1b1 a1b2 a1b3

a2b1 a2b2 a2b3

a3b1 a3b2 a3b3

u1u2u3

=

a1a2a3

[b1 b2 b3

]u1u2u3

.

(2.21)


The last line of Eq 2.21 shows why the tensor product is sometimes called the outerproduct. The inner product, of course, is defined by

a · b =[a1 a2 a3

]b1b2b3

= a1b1 + a2b2 + a3b3 .

The second line of Eq 2.21 shows the tensor product to be equivalent to the dyad productof two vectors. Both notations will be used in the book.

A dyad can be multiplied by a vector giving a vector-dyad product:

1. u · (vw) = uiei · (vjejwkek) = uiviwkek (2.22)

2. (uv) ·w = (uieivjej) ·wkek = uivjwjei (2.23)

3. u× (vw) = (uiei × vjej)wkek = εijquivjwkeqek (2.24)

4. (uv)×w = uiei (vjej ×wkek) = εjkquivjwkeieq (2.25)

(Note that in products 3 and 4 the order of the base vectors ei is important.)Dyads can be multiplied by each other to yield another dyad

(uv) · (ws) = uiei (vjej ·wkek) sqeq = uivjwjsqeieq . (2.26)

Vectors can be multiplied by a tensor to give a vector. The reduction in the order of thetensor is why the “dot” is used in direct notation.

1. v · T = viei · tjkejek = vitjkδijek = vitikek (2.27)

2. T · v = tijeiej · vkek = tijeiδjkvk = tijvjei (2.28)

(Note that these products are also written as simply vT and Tv .)Finally, two tensors can be multiplied resulting in a tensor

T · S = tijeiej · spqepeq = tijspqδjpeieq = tijsjpeieq . (2.29)

Example 2.2

Let the vector v be given by v = (a · n) n+ n× (a× n) where a is an arbitraryvector, and n is a unit vector. Express v in terms of the base vectors ei , expand,and simplify. (Note that n · n = niei · njej = ninjδij = nini = 1.)

SolutionIn terms of the base vectors ei, the given vector v is expressed by the equation

v = (aiei · njej)nkek + niei × (ajej × nkek) .

We note here that indices i, j, and k appear four times in this line, however, thesummation convention has not been violated. Terms that are separated by a plusor a minus sign are considered different terms each having summation convention


rules applicable within them. Vectors joined by a dot or cross product are notdistinct terms, and the summation convention must be adhered to in that case.Carrying out the indicated multiplications, we see that

v = (ainjδij)nkek + niei × (εjkqajnkeq)

= aininkek + εiqmεjkqniajnkem

= aininkek + εmiqεjkqniajnkem

= aininkek + (δmjδik − δmkδij)niajnkem

= aininkek + niajniej − niainkek

= niniajej = ajej = a .

Since a must equal v, this example demonstrates that the vector v may beresolved into a component (v · n) n in the direction of n, and a componentn× (v× n) perpendicular to n.

Example 2.3

Using Eq 2.7 show that (a) εmkqεjkq = 2δmj and that (b) εjkqεjkq = 6. (Recallthat δkk = 3 and δmkδkj = δmj.)

Solution(a) Write out Eq 2.7a with index i replaced by k to get

εmkqεjkq = δmjδkk − δmkδkj

= 3δmj − δmj = 2δmj .

(b) Start with the first equation in Part (a) and replace the index m with j,giving

εjkqεjkq = δjjδkk − δjkδjk

= (3) (3) − δjj = 9− 3 = 6 .

Example 2.4

Double dot products of dyads are defined by

(a) (uv) · · (ws) = (v ·w) (u · s)

(b) (uv) : (ws) = (u ·w) (v · s) .

Expand these products and compare the component forms.


Solution

(a) (uv) · · (ws) = (viei ·wjej) (ukek · sqeq) = viwiuksk

(b) (uv) : (ws) = (uiei ·wjej) (vkek · sqeq) = uiwivksk

Summary of Symbolic Notations

1. Addition of vectors, Eq 2.8:

w = u+ v or wiei = (ui + vi) ei

2. Multiplication:

(a) of a vector by a scalar, Eq 2.9:λv = λviei

(b) dot (scalar) product of two vectors, Eq 2.11:

u · v = v · u = uv cos θ = uivi

(c) cross (vector) product of two vectors, Eq 2.12:

u× v = −v× u = (uv sin θ) e = εijkuivjek

(d) triple scalar product (box product), Eq 2.13:

[u, v,w] = uiei · (vjej ×wkek) = uiei · εjkqvjwkeq= εjkquivjwkδiq = εijkuivjwk

(e) triple cross product, Eq 2.14:

u× (v×w) = uiei × (vjej ×wkek) = uiei × (εjkqvjwkeq)

= εiqmεjkquivjwkem = εmiqεjkquivjwkem

(f ) tensor product of two vectors (dyad), Eq 2.16:

uv = uieivjej = uivjeiej = uiei ⊗ vjej = uivjei ⊗ ej

(g) Vector-dyad products, Eqs 2.22–2.25:

1. u · (vw) = uiei · (vjejwkek) = uiviwkek

2. (uv) ·w = (uieivjej) ·wkek = uivjwjei

3. u× (vw) = (uiei × vjej)wkek = εijquivjwkeqek

4. (uv)×w = uiei (vjej ×wkek) = εjkquivjwkeieq


TABLE 2.1Indicial form for a variety of tensor quantities.

λ = scalar (zeroth-order tensor) λ

vi = vector (first-order tensor) v, or equivalently, its 3 components

uivj = dyad (second-order tensor) uv, or its 9 components

tij = dyadic (second-order tensor) T , or its 9 components

Qijk = triadic (third-order tensor) Q, or its 27 components

Cijkm = tetradic (forth-order tensor) C, or its 81 components

(h) dyad-dyad product, Eq 2.26:

(uv) · (ws) = uiei (vjej ·wkek) sqeq = uivjwjsqeieq

(i) vector-tensor products, Eqs 2.27–2.28:

1. v · T = viei · tjkejek = vitjkδijek = vitikek

2. T · v = tijeiej · vkek = tijeiδjkvk = tijvjei

(j) tensor-tensor product, Eq 2.29:

T · S = tijeiej · spqepeq = tijsjpeieq

(k) double dot product:

1. (uv) · · (ws) = (viei ·wjej) (ukek · sqeq) = viwiuksk

2. (uv) : (ws) = (uiei ·wjej) (vkek · sqeq) = uiwivksk

2.3 Indicial NotationBy assigning special meaning to the subscripts, indicial notation permits us to carry outthe tensor operations of addition, multiplication, differentiation, etc. without the use, oreven the appearance of the base vectors ei in the equations. We simply agree that thetensor rank (order) of a term is indicated by the number of “free,” that is, unrepeated,subscripts appearing in that term. Accordingly, a term with no free indices represents ascalar, a term with one free index a vector, a term having two free indices a second ordertensor, and so on. The specific meaning of these symbols are given in Table 2.1.

For tensors defined in a three-dimensional space, the free indices take on the values1,2,3 successively, and we say that these indices have a range of three. If N is the numberof free indices in a tensor, that tensor has 3N components in three space.

We must emphasize that in the indicial notation exactly two types of subscripts appear:


TABLE 2.2Forms for inner and outer products.

Outer Products: Contraction(s): Inner Products:

uivj i = j uivi (vector dot product)

εijkuqvm j = q, k = m εijkujvk (vector cross product)

εijkuqvmwn i = q, j = m,k = n εijkuivjwk (box product)

1. “free” indices, which are represented by letters that occur only once in a given term,

2. “summed,” or “dummy” indices which are represented by letters that appear onlytwice in a given term.

Furthermore, every term in a valid equation must have the same letter subscripts for thefree indices. No letter subscript may appear more than twice in any given term.

Mathematical operations among tensors are readily carried out using the indicial nota-tion. Thus addition (and subtraction) among tensors of equal rank follows according tothe typical equations; ui+vi−wi = si for vectors, and tij−vij+sij = qij for second-ordertensors. Multiplication of two tensors to produce an outer tensor product is accomplishedby simply setting down the tensor symbols side by side with no dummy indices appear-ing in the expression. As a typical example, the outer product of the vector vi and tensortjk is the third-order tensor vitjk. Contraction is the process of identifying (that is, settingequal to one another) any two indices of a tensor term. An inner tensor product is formedfrom an outer tensor product by one or more contractions involving indices from separatetensors in the outer product. We note that the rank of a given tensor is reduced by two foreach contraction. Some outer products, which contract, form well-known inner productslisted in Table 2.2.

A tensor is symmetric in any two indices if interchange of those indices leaves the tensorvalue unchanged. For example, if sij = sji and cijm = cjim, both of these tensors are saidto be symmetric in the indices i and j. A tensor is anti-symmetric (or skew-symmetric) inany two indices if interchange of those indices causes a sign change in the value of thetensor. Thus if aij = −aji , it is anti-symmetric in i and j. Also, recall that by definition,εijk = −εjik = εjki, etc., and hence the permutation symbol is anti-symmetric in allindices.

Example 2.5

Show that the inner product sijaij of a symmetric tensor sij = sji , and ananti-symmetric tensor aij = −aji is zero.

SolutionBy definition of symmetric tensor sij and skew-symmetric tensor aij, we have

sijaij = −sjiaji = −smnamn = −sijaij

where the last two steps are the result of all indices being dummy indices.Therefore, 2sijaij = 0 , or sijaij = 0.


One of the most important advantages of the indicial notation is the compactness itprovides in expressing equations in three dimensions. A brief listing of typical equationsof continuum mechanics is presented below to illustrate this feature.

1. φ = sijtij − siitjj (1 equation, 18 terms on RHS)2. ti = qijnj (3 equations, 3 terms on RHS of each)3. tij = λδijEkk + 2µEij (9 equations, 4 terms on RHS of each)

Example 2.6

By direct expansion of the expression vi = εijkwjk determine the componentsof the vector vi in terms of the components of the tensor wjk.

SolutionBy summing first on j and then on k and then omitting the zero terms, we findthat

vi = εi1kw1k + εi2kw2k + εi3kw3k

= εi12w12 + εi13w13 + εi21w21 + εi23w23 + εi31w31 + εi32w32 .

Therefore,v1 = ε123w23 + ε132w32 = w23 −w32 ,

v2 = ε213w13 + ε231w31 = w31 −w13 ,

v3 = ε312w12 + ε321w21 = w12 −w21 .

Note that if the tensor wjk were symmetric, the vector vi would be a null (zero)vector.

To end this section, consider a skew–symmetric second order tensor W = wijeiej.All skew–symmetric tensors can be represented in terms of an axial vector by using thepermutation symbol. Let the axial vector for wij be ωi defined by

ωi = −1

2εijkwjk . (2.30)

It is an easy exercise in indicial manipulation to find the inverse of Eq 2.30. We can writewjk in terms of ωi as follows:

εimnωi = −12εimnεijkwjk

= −12 (δmjδnk − δmkδnj)wjk

= −12 (wmn −wnm)

= −12 (2wmn)

= −wmn

(2.31)

where Eq 2.7a and the skew–symmetric property of wij were used.


2.4 Matrices and Determinants

For computational purposes it is often expedient to use the matrix representation of vectorsand tensors. Accordingly, we review here several definitions and operations of elementarymatrix theory.

A matrix is an ordered rectangular array of elements enclosed by square brackets andsubjected to certain operational rules. The typical element Aij of the matrix is located inthe ith (horizontal) row, and in the jth (vertical) column of the array. A matrix havingelements Aij, which may be numbers, variables, functions, or any of several mathematicalentities, is designated by [Aij], or symbolically by the kernel letter A. An M by N matrix(written M×N) has M rows and N columns, and may be displayed as

A = [Aij] =

A11 A12 · · · A1N

A21 A22 · · · A2N...

......

AM1 AM2 · · · AMN

. (2.32)

IfM =N, the matrix is a square matrix. A 1×Nmatrix [A1N] is an row matrix, and anM×1matrix [AM1] is a column matrix. Row and column matrices represent vectors, whereas a3 × 3 square matrix represents a second-order tensor. A scalar is represented by a 1 × 1matrix (a single element). The unqualified use of the word matrix in this text is understoodto mean a 3× 3 square matrix, that is, the matrix representation of a second-order tensoror a matrix that is not a tensor.

A zero, or null matrix has all elements equal to zero. A diagonal matrix is a square matrixwhose elements not on the principal diagonal, which extends from A11 to ANN, are allzeros. Thus for a diagonal matrix, Aij = 0 for i 6= j. The unit or identity matrix I, which,incidentally, is the matrix representation of the Kronecker delta, is a diagonal matrixwhose diagonal elements all have the value one.

The N×M matrix formed by interchanging the rows and columns of the M×N matrixA is called the transpose of A, and is written as AT , or [Aij]

T . By definition, the elementsof a matrix A and its transpose are related by the equation ATij = Aji. A square matrixfor which A = AT , or in element form, Aij = Aji, is called a symmetric matrix; one forwhich A = −A , or, Aij = −Aji, is called an anti-symmetric, or skew-symmetric matrix.The elements of the principal diagonal of a skew-symmetric matrix are all zeros. Twomatrices are equal if they are identical element by element. Matrices having the samenumber of rows and columns may be added (or subtracted) element by element. Thus ifA = B + C, the elements of A are given by

Aij = Bij + Cij . (2.33)

Addition of matrices is commutative, A + B = B + A, and associative, A + (B + C) =(A + B) + C .


Example 2.7

Show that the square matrix A can be expressed as the sum of a symmetricand a skew-symmetric matrix by the decomposition

A =A + AT

2+

A − AT

2.

SolutionLet the decomposition be written as A = B + C where B = 1

2

(A + AT

)and

C = 12

(A − AT

). Then writing B and C in element form,

Bij =Aij + ATij

2=

Aij + Aji

2=

ATji + Aji

2= Bji = BTij (symmetric) ,

Cij =Aij − ATij

2=

Aij − Aji

2= −

ATji − Aji

2= −Cji = −CTij (skew-symmetric) .

Therefore B is symmetric, and C skew-symmetric.

Multiplication of the matrix A by the scalar λ results in the matrix λA, or [λAij] =λ [Aij]. The product of two matrices A and B, denoted by AB, is defined only if thematrices are conformable, that is, if the prefactor matrix A has the same number of columnsas the postfactor matrix B has rows. Thus the product of an M ×Q matrix multiplied bya Q×N matrix is an M×N matrix. The product matrix C = AB has elements given by

Cij = AikBkj (2.34)

in which k is, of course, a summed index. Therefore each element Cij of the productmatrix is an inner product of the ith row of the prefactor matrix with the jth column ofthe postfactor matrix. In general, matrix multiplication is not commutative, AB 6= BA,but the associative and distributive laws of multiplication do hold for matrices. Theproduct of a matrix with itself is the square of the matrix, and is written AA = A2.Likewise the cube of the matrix is AAA = A3, and in general, matrix products obey theexponent rule

AmAn = AnAm = Am+n (2.35)

where m and n are positive integers, or zero. Also, we note that

(An)T =(AT)n

, (2.36)

and if BB = A then

B =√

A = A12 (2.37)

but the square root is not unique.


Example 2.8

Use indicial notation to show that for arbitrary matrices A and B:

(a) (A + B)T = AT + BT

(b) (AB)T = BTAT

(c) IB = BI = B where I is the identity matrix.

Solution

(a) Let A + B = C, then in element form Cij = Aij + Bij and therefore CT isgiven by

CTij = Cji = Aji + Bji = ATij + BTji ,

orCT = (A + B)T = AT + BT .

(b) Let AB = C, then in element form

Cij = AikBkj = ATkiBTjk = BTjkA

Tki = CTji .

Hence (AB)T = BTAT . Note that exchanging the order ATkiBTjk = BTjkA

Tki

is not necessary when using indicial notation. It is done for clarity. In director matrix notation the order of the terms is critical.

(c) Let IB = C, then in element form

Cij = δikBkj = Bij = Bikδkj

by the substitution property. Thus IB = BI = B.

The determinant of a square matrix is formed from the square array of elements ofthe matrix and this array evaluated according to established mathematical rules. Thedeterminant of the matrix A is designated by either det A, or by |Aij|, and for a 3 × 3matrix A,

det A = |Aij| =

∣∣∣∣∣∣∣A11 A12 A13

A21 A22 A23

A31 A32 A33

∣∣∣∣∣∣∣ . (2.38)

A minor of det A is another determinant |Mij| formed by deleting the ith row and jth

column of |Aij|. The cofactor of the element Aij (sometimes referred to as the signedminor) is defined by

A(c)ij = (−1)

i+j|Mij| (2.39)

where superscript (c) denotes cofactor of matrix A.Evaluation of a determinant may be carried out by a standard method called expansion

by cofactors. In this method, any row (or column) of the determinant is chosen, and eachelement in that row (or column) is multiplied by its cofactor. The sum of these products


gives the value of the determinant. For example, expansion of the determinant of Eq 2.38by the first row becomes

det A = A11

∣∣∣∣∣ A22 A23

A32 A33

∣∣∣∣∣− A12

∣∣∣∣∣ A21 A23

A31 A33

∣∣∣∣∣+ A13

∣∣∣∣∣ A21 A22

A31 A32

∣∣∣∣∣ (2.40)

which upon complete expansion gives

det A = A11 (A22A33 − A23A32) − A12 (A21A33 − A23A31) (2.41)+A13 (A21A32 − A22A31) .

Several interesting properties of determinants are worth mentioning at this point. Tobegin with, the interchange of any two rows (or columns) of a determinant causes a signchange in its value. Because of this property and because of the sign-change property ofthe permutation symbol, the det A of Eq 2.38 may be expressed in the indicial notationby the alternative forms (see Prob. 2.13)

det A = εijkAi1Aj2Ak3 = εijkA1iA2jA3k . (2.42)

Furthermore, following an arbitrary number of column interchanges with the accompa-nying sign change of the determinant for each, it can be shown from the first form ofEq 2.42 that, (see Prob 2.14)

εqmn det A = εijkAiqAjmAkn . (2.43)

Finally, we note that if the det A = 0, the matrix is said to be singular. It may be eas-ily shown that every 3 × 3 skew-symmetric matrix is singular. Also, the determinantof the diagonal matrix, D, is simply the product of its diagonal elements: det D =D11D22 · · ·DNN.

Example 2.9

Show that for matrices A and B, det(AB) = det(BA) = det(A) det(B).

SolutionLet C = AB, then Cij = AikBkj and from Eq 2.42

det C = εijkCi1Cj2Ck3

= εijkAiqBq1AjmBm2AknBn3

= εijkAiqAjmAknBq1Bm2Bn3 .

But from Eq 2.43

εijkAiqAjmAkn = εqmn det A ,

so now


det C = det AB = εqmnBq1Bm2Bn3 det A = det A det B .

By a direct interchange of A and B, det AB = det BA.

Example 2.10

Use Eq 2.40 and Eq 2.41 to show that det A = det AT .

SolutionSince

∣∣AT∣∣ =

∣∣∣∣∣∣∣A11 A21 A31

A12 A22 A32

A13 A23 A33

∣∣∣∣∣∣∣cofactor expansion by the first column here yields

det A = A11

∣∣∣∣∣ A22 A32

A23 A33

∣∣∣∣∣− A12

∣∣∣∣∣ A21 A31

A23 A33

∣∣∣∣∣+ A13

∣∣∣∣∣ A21 A31

A22 A32

∣∣∣∣∣which is equal to Eq 2.41.

The inverse of the matrix A is written A−1, and is defined by

AA−1 = A−1A = I (2.44)

where I is the identity matrix. Thus if AB = I, then B = A−1, and A = B−1. The adjointmatrix A∗ is defined as the transpose of the cofactor matrix

A∗ =[A(c)

]T. (2.45)

In terms of the adjoint matrix the inverse matrix is expressed by

A−1 =A∗

det A(2.46)

which is actually a working formula by which an inverse matrix may be calculated. Thisformula shows that the inverse matrix exists only if det A 6= 0, i.e., only if the matrix A isnon-singular. In particular, a 3× 3 skew-symmetric matrix has no inverse.


Example 2.11

Show from the definition of the inverse, Eq 2.44 that

(a) (AB)−1= B−1A−1 ,

(b)(AT)−1

=(A−1

)T.

Solution

(a) By pre-multiplying the matrix product AB by B−1A−1, we have (us-ing Eq 2.44),

B−1A−1AB = B−1IB = B−1B = I

and therefore B−1A−1 = (AB)−1.(b) Taking the transpose of both sides of Eq 2.44 and using the result of

Example 2.8 we have(AA−1

)T=(A−1

)TAT = IT = I .

Hence(A−1

)Tmust be the inverse of AT , or

(A−1

)T=(AT)−1

.

An orthogonal matrix, call it Q, is a square matrix for which Q−1 = QT . From thisdefinition we note that a symmetric orthogonal matrix is its own inverse since in this case

Q−1 = QT = Q . (2.47)

Also, if A and B are orthogonal matrices.

(AB)−1= B−1A−1 = BTAT = (AB)T , (2.48)

so that the product matrix is likewise orthogonal. Furthermore, if A is orthogonal it mayshown (see Prob. 2.18) that

det A = ±1 . (2.49)

As mentioned near the beginning of this section, a vector may be represented by a rowor column matrix, a second-order tensor by a square 3 × 3 matrix. For computationalpurposes, it is frequently advantageous to transcribe vector/tensor equations into theirmatrix form. As a very simple example, the vector-tensor product, u = Tv (symbolicnotation) or ui = Tijvj (indicial notation) appears in matrix form as

[ui1] = [Tij] [vj1] or

u1

u2

u3

=

T11 T12 T13

T21 T22 T23

T31 T32 T33

v1

v2

v3

. (2.50)


TABLE 2.3Transformation table between Ox1x2x3 andOx ′1x

′2x′3.

e1, x1 e2, x2 e3, x3

e ′1, x ′1 a11 a12 a13

e ′2, x ′2 a21 a22 a23

e ′3, x ′3 a31 a32 a33

In much the same way the product

w = v · T or wi = vjTji

appears as

[w1i] = [v1j] [Tji] or[w1 w2 w3

]=[v1 v2 v3

] T11 T21 T31

T12 T22 T32

T13 T23 T33

. (2.51)

Note that in Eq 2.51 the order of the subscripts are exchanged since the transpose of thematrix is being written. Equations 2.51 and 2.50 could be written in the direct form thefollowing way:

w = v · T = (T · v)T = vT · TT .

From this, it is clear thatw = uT

since u = T · v.

2.5 Transformations of Cartesian TensorsAlthough, as already mentioned, vectors and tensors have an identity independent of anyparticular reference or coordinate system, the relative values of their respective compo-nents do depend upon the specific axes to which they are referred. The relationshipsamong these various components when given with respect to two separate sets of co-ordinate axes are known as the transformation equations. In developing these transfor-mation equations for Cartesian tensors, we consider two sets of rectangular Cartesianaxes, Ox1x2x3 and Ox ′1x

′2x′3, sharing a common origin, and oriented relative to one an-

other so that the direction cosines between the primed and unprimed axes are given byaij = cos(x ′i, xj) as shown in Fig. 2.2.

The square array of the nine direction cosines displayed in Table 2.3 is useful in relat-ing the unit base vectors ei and e ′i to one another, as well as relating the primed andunprimed coordinates x ′i and xi of a point. Thus the primed base vectors e ′i are givenin terms of the unprimed vectors ei by the equations (as is also easily verified from thegeometry of the vectors in the diagram of Fig. 2.2),


x1

x 1

x2

x 2

x3

x 3

cos−1 a11

cos−1 a12

cos−1 a13

O

e3

e2

e1

e 3

e 2e 1

FIGURE 2.2Rectangular coordinate system Ox ′1x

′2x′3 relative to Ox1x2x3. Direction cosines shown for

coordinate x ′1 relative to unprimed coordinates. Similar direction cosines are defined forx ′2 and x ′3 coordinates.

e ′1 = a11e1 + a12e2 + a13e3 = a1jej , (2.52a)e ′2 = a21e1 + a22e2 + a23e3 = a2jej , (2.52b)e ′3 = a31e1 + a32e2 + a33e3 = a3jej , (2.52c)

or in compact indicial forme ′i = aijej . (2.53)

By defining the matrix A whose elements are the direction cosines aij, Eq 2.53 can bewritten in matrix form as

[e ′i1] = [aij] [ej1] or

e ′1e ′2e ′3

=

a11 a12 a13

a21 a22 a23

a31 a32 a33

e1e2e3

(2.54)

where the elements of the column matrices are unit vectors. The matrix A is called thetransformation matrix because, as we shall see, of its role in transforming the componentsof a vector (or tensor) referred to one set of axes into the components of the same vector(or tensor) in a rotated set.

Because of the perpendicularity of the primed axes, e ′i · e ′j = δij. But also, in view ofEq 2.53,

e ′i · e ′j = aiqeq · ajmem = aiqajmδqm = aiqajq = δij

from which we extract the orthogonality condition on the direction cosines (given here inboth indicial and matrix form),

aiqajq = δij or AAT = I . (2.55)

Note that this is simply the inner product of the ith row with the jth row of the matrixA. By an analogous derivation to that leading to Eq 2.53, but using the columns of A, weobtain


ei = ajie′j , (2.56)

which in matrix form is

[e1i] =[e ′1j][aij] or

[e1 e2 e3

]=[e ′1 e ′2 e ′3

] a11 a12 a13

a21 a22 a23

a31 a32 a33

. (2.57)

Note that using the transpose AT , Eq 2.57 may also be written

[e1i] = [aij]T [e ′1j]

or

e1e2e3

=

a11 a21 a31

a12 a22 a32

a13 a23 a33

e ′1e ′2e ′3

(2.58)

in which column matrices are used for the vectors ei and e ′i. By a consideration of thedot product and Eq 2.56 we obtain a second orthogonality condition

aijaik = δjk or ATA = I (2.59)

which is the inner product of the jth column with the kth column of A.Consider next an arbitrary vector v having components vi in the unprimed system, and

v ′i in the primed system. Then using Eq 2.56,

v = v ′je′j = viei = viajie

′j

from which, by matching coefficients on e ′j, we have (in both the indicial and matrixforms),

v ′j = ajivi or v ′ = Av = vAT (2.60)

which is the transformation law expressing the primed components of an arbitrary vector interms of its unprimed components. Although the elements of the transformation matrixare written as aij we must emphasize that they are not the components of a second-order Cartesian tensor as it might appear. Multiplication of Eq 2.60 by ajk and using theorthogonality condition Eq 2.59 we obtain the inverse law

vk = ajkv′j or v = v ′A = ATv ′ (2.61)

giving the unprimed components in terms of the primed.By a direct application of Eq 2.61 to the dyad uv we have

uivj = aqiu′qamjv

′m = aqiamju

′qv′m . (2.62)

But a dyad is, after all, one form of a second-order tensor, and so by an obvious adaptationof Eq 2.62 we obtain the transformation law for a second-order tensor, T as

tij = aqiamjt′qm or T = ATT ′A (2.63)

which may be readily inverted with the help of the orthogonality conditions to yield

t ′ij = aiqajmtqm or T ′ = ATAT . (2.64)


O

2π

2π

x 1

x3

x1, x 3

x2, x 2

(a) Ox ′1x′2x′3 axes relative to Ox1x2x3 axes

following a 90o counterclockwise rotationabout the x2 axis.

Ox2, x

2

x3, x 3

x 1

x1

(b) Ox ′1x′2x′3 axes relative to Ox1x2x3 axes

following a reflection about the x2x3 plane.

FIGURE 2.3Rotation and reflection of reference axes.

Note carefully the location of the summed indices q and m in Equations 2.63 and 2.64.Finally, by a logical generalization of the pattern of the transformation rules developedthus far we state that for an arbitrary Cartesian tensor of any order

R ′ij...k = aiqajm · · ·aknRqm...n . (2.65)

The primed axes may be related to the unprimed axes through either a rotation about anaxis through the origin, or by a reflection of the axes in one of the coordinate planes. (Or bya combination of such changes). As a simple example, consider a 90 (counterclockwise)rotation about the x2 axis shown in Fig. 2.3(a). The matrix of direction cosines for thisrotation is

[aij] =

0 0 −1

0 1 0

1 0 0

and det A = 1. The transformation of tensor components in this case is called a properorthogonal transformation. For a reflection of axes in the x2x3 plane shown in Fig 2.3(b) thetransformation matrix is

[aij] =

−1 0 0

0 1 0

0 0 1

where det A = −1, and we have an improper orthogonal transformation. It may be shownthat true (polar) vectors transform by the rules v ′i = aijvj and vi = aijv

′j regardless of

whether the axes transformation is proper or improper. However, pseudo (axial) vectorstransform correctly only according to v ′i = (det A)aijvj and vj = (det A)aijv

′i under an

improper transformation of axes.


Example 2.12

Let the primed axes Ox ′1x′2x′3 be given with respect to the unprimed axes by

a 45(counterclockwise) rotation about the x2 axis as shown. Determine theprimed components of the vector given by v = e1 + e2 + e3.

O

β

45ο

45ο

v

β

β

e1

e3

e 1

e 3

e2, e 2

x2, x 2

x1

x 1

x3

x 3

Vector v with respect to Ox ′1x′2x′3 and Ox1x2x3.

SolutionHere the transformation matrix is

[aij] =

1√2

0 − 1√2

0 1 0

1√2

0 1√2

,

and from Eq 2.60 in matrix form

v ′1

v ′2

v ′3

=

1√2

0 − 1√2

0 1 0

1√2

0 1√2

1

1

1

=

0

1√2

.


Example 2.13

Determine the primed components of the tensor

[tij] =

2 6 4

0 8 0

4 2 0

under the rotation of axes described in Example 2.12.

SolutionHere Eq 2.64 may be used. Thus in matrix form

[t ′ij]

=

1√2

0 − 1√2

0 1 01√2

0 1√2

2 6 4

0 8 0

4 2 0

1√2

0 1√2

0 1 0

− 1√2

0 1√2

=

−3 4√2

1

0 8 0

1 8√2

5

.

2.6 Principal Values and Principal Directions of Symmetric Second-Order Tensors

First, let us note that in view of the form of the inner product of a second-order tensorT with the arbitrary vector u (which we write here in both the indicial and symbolicnotation),

tijuj = vi or T · u = v (2.66)

any second-order tensor may be thought of as a linear transformation which transformsthe antecedent vector u into the image vector v in a Euclidean three-space. In particular,for every symmetric tensor T having real components tij, and defined at some point inphysical space, there is associated with each direction at that point (identified by the unitvector ni), an image vector vi given by

tijnj = vi or T · n = v . (2.67)

If the vector vi determined by Eq 2.67 happens to be a scalar multiple of ni, that is, if

tijnj = λni or T · n = λn , (2.68)

the direction defined by ni is called a principal direction, or eigenvector, of T , and the scalarλ is called a principal value, or eigenvalue of T . Using the substitution property of theKronecker delta, Eq 2.68 may be rewritten as


(tij − λδij)nj = 0 or (T − λI) · n = 0 , (2.69)

or in expanded form

(t11 − λ)n1 + t12n2 + t13n3 = 0 , (2.70a)

t21n1 + (t22 − λ)n2 + t23n3 = 0 , (2.70b)

t31n1 + t32n2 + (t33 − λ)n3 = 0 . (2.70c)

This system of homogeneous equations for the unknown direction ni and the unknownλ’s will have non-trivial solutions only if the determinant of coefficients vanishes. Thus,

|tij − λδij| = 0 (2.71)

which upon expansion leads to the cubic in λ (called the characteristic equation)

λ3 − ITλ2 + IITλ− IIIT = 0 (2.72)

where the coefficients here are expressed in terms of the known components of T by

IT = tii = tr T , (2.73a)

IIT =1

2(tiitjj − tijtji) =

1

2

[(tr T )2 − tr

(T2)], (2.73b)

IIIT = εijkt1it2jt3k = det T , (2.73c)

and are known as the first, second, and third invariants, respectively, of the tensor T . 1 Thesum of the elements on the principal diagonal of the matrix form of any tensor is calledthe trace of that tensor, and for the tensor T is written tr T as in Eq 2.73.

The roots λ(1), λ(2), and λ(3) of Eq 2.72 are all real for a symmetric tensor T havingreal components. With each of these roots λ(q) (q = 1, 2, 3) we can determine a principaldirection n(q)

i (q = 1, 2, 3) by solving Eq 2.69 together with the normalizing conditionnini = 1. Thus, Eq 2.69 is satisfied by[

tij − λ(q)δij]n

(q)i = 0 (q = 1, 2, 3) (2.74)

withn

(q)i n

(q)i = 1 (q = 1, 2, 3). (2.75)

If the λ(q)’s are distinct the principal directions are unique and mutually perpendicular.If, however, there is a pair of equal roots, say λ(1) = λ(2), then only the direction associatedwith λ(3) will be unique. In this case any other two directions which are orthogonal ton

(3)i , and to one another so as to form a right-handed system, may be taken as principal

directions. If λ(1) = λ(2) = λ(3), every set of right-handed orthogonal axes qualifies asprincipal axes, and every direction is said to be a principal direction.

In order to reinforce the concept of principal directions, let the components of the tensorT be given initially with respect to arbitrary Cartesian axes Ox1x2x3, and let the principalaxes of T be designated by Ox∗1x

∗2x∗3, as shown in Fig. 2.4. The transformation matrix

A between these two sets of axes is established by taking the direction cosines n(q)i as

calculated from Eq 2.74 and Eq 2.75 as the elements of the qth row of A. Therefore, bydefinition, aij ≡ n(i)

j as detailed in the table below.

1Note that some authors use the negative of the second invariant as defined in Eq 2.73b. This changes the signon the IITλ term in Eq 2.72 and must be kept consistent throughout the use of the invariants.


cos−1 n(1)3

cos−1 n(1)2

cos−1 n(1)1

x3

x2

x1

x2∗

O

x3∗

x1∗

FIGURE 2.4Principal axes Ox∗1x

∗2x∗3 relative to axes Ox1x2x3.

x1 or e1 x2 or e2 x3 or e3

x∗1 or e∗1 a11 = n

(1)1 a12 = n

(1)2 a13 = n

(1)3

x∗2 or e∗2 a21 = n

(2)1 a22 = n

(2)2 a23 = n

(2)3

x∗3 or e∗3 a31 = n

(3)1 a32 = n

(3)2 a33 = n

(3)3

(2.76)

The transformation matrix here is orthogonal and in accordance with the transformationlaw for second-order tensors

t∗ij = aiqajmtqm or T∗ = ATAT (2.77)

where T∗ is a diagonal matrix whose elements are the principal values λ(q).

Example 2.14

Determine the principal values and principal directions of the second-ordertensor T whose matrix representation is

[tij] =

5 2 0

2 2 0

0 0 3

.

SolutionHere Eq 2.71 is given by∣∣∣∣∣∣∣

5− λ 2 0

2 2− λ 0

0 0 3− λ

∣∣∣∣∣∣∣ = 0

which upon expansion by the third row becomes

(3− λ)(10− 7λ+ λ2 − 4

)= 0 ,


or

(3− λ) (6− λ) (1− λ) = 0 .

Hence, λ(1) = 3, λ(2) = 6, λ(3) = 1 are the principal values of T . For λ(1) = 3,Eq 2.70 yields the equations

2n1 + 2n2 = 0 ,

2n1 − n2 = 0 ,

which are satisfied only if n1 = n2 = 0, and so from nini = 1 we have n3 = ±1.For λ(2) = 6, Eq 2.70 yields

−n1 + 2n2 = 0 ,

2n1 − 4n2 = 0 ,

−3n3 = 0 ,

so that n1 = 2n2 and since n3 = 0, we have (2n2)2

+ n22 = 1, or n2 = ±1/√5,

and n1 = ±2/√5. For λ(3) = 1, Eq 2.70 yields

4n1 + 2n2 = 0 ,

2n1 + n2 = 0 ,

together with 2n3 = 0. Again n3 = 0, and here n21 + (−2n1)2

= 1 so thatn1 = ±1/

√5 and n2 = ∓2/

√5. From these results the transformation matrix

A is given by

[aij] =

0 0 ±1

± 2√5± 1√

50

± 1√5∓ 2√

50

which identifies two sets of principal direction axes, one a reflection of the otherwith respect to the origin. Also, it may be easily verified that A is orthogonalby multiplying it with its transpose AT to obtain the identity matrix. Finally,from Eq 2.77 we see that using the upper set of the ± signs,

0 0 1

2√5

1√5

0

1√5

− 2√5

0

5 2 0

2 2 0

0 0 3

0 2√

51√5

0 1√5

− 2√5

1 0 0

=

3 0 0

0 6 0

0 0 1

.


Example 2.15

Show that the principal values for the tensor having the matrix

[tij] =

5 1√2

1 5√2√

2√2 6

have a multiplicity of two, and determine the principal directions.

SolutionHere Eq 2.71 is given by ∣∣∣∣∣∣∣

5− λ 1√2

1 5− λ√2√

2√2 6− λ

∣∣∣∣∣∣∣ = 0

for which the characteristic equation becomes

λ3 − 16λ2 + 80λ− 128 = 0 ,

or(λ− 8) (λ− 4)

2= 0 .

For λ(1) = 8, Eq 2.70 yields

−3n1 + n2 +√2n3 = 0 ,

n1 − 3n2 +√2n3 = 0 ,√

2n1 +√2n2 − 2n3 = 0 .

From the first two of these equations n1 = n2, and from the second and thirdequations n3 =

√2n2. Therefore, using, nini = 1, we have

(n2)2

+ (n2)2

+(√2n2

)2= 1 ,

and so n1 = n2 = ±1/2 and n3 = ±1/√2, from which the unit vector in the

principal direction associated with λ(1) = 8 (the so-called normalized eigenvec-tor) is

n(1) =1

2

(e1 + e2 +

√2e3

)= e∗3 .

For n(2), we choose any unit vector perpendicular to n(1); an obvious choicebeing

n(2) =−e1 + e2√

2= e∗2 .

Then n(3) is constructed from n(3) = n(1) × n(2), so that

n(3) =1

2

(−e1 − e2 +

√2e3

)= e∗3 .


Thus, the transformation matrix A is given by Eq 2.76 as

[aij] =

12

12

1√2

− 1√2

1√2

0

− 1√2

−121√2

.

It is worthwhile to revisit the invariants of Eq 2.73 within the context of the character-istic equation because of their extensive uses. The first invariant is just the trace of thetensor which is easy to calculate. Similarly, the third invariant may be calculated by acofactor expansion to evaluate the determinant. Calculating the second invariant may bedone two ways. First is the straight forward calculation. Half the trace squared minus thetrace of the square may be done using matrix operations. For example, take the genericmatrix

G =

a b c

d e f

g h i

. (2.78)

Clearly, from Eq 2.73a, the first invariant is

IG = a+ e+ i . (2.79)

Finding the second invariant requires two terms: 12 I2G and −12 tr

(G2). The first term is

expanded as1

2(a+ e+ i)

2=1

2

(a2 + e2 + i2

)+ ae+ ai+ ei .

Matrix multiplication of the diagonal terms will evaluate the second term:

1

2tr

a b c

d e f

g h i

a b c

d e f

g h i

=

1

2( a2 + bd+ gc︸︷︷︸

1,1 term

+bd+ e2 + fh︸︷︷︸2,2 term

+gc+ hf+ i2 )︸︷︷︸3,3 term

.

Combining these two terms according to Eq 2.73b results in

IIG = ei− hf+ ai− gc+ ae− db . (2.80)

A second way to find the second invariant is to sum the cofactors along the diagonal ofmatrix G. Using this method on Eq 2.78 results in Eq 2.80. This second way is an easierway to calculate the second invariant.

If a problem is given with specific numbers for the matrix, or tensor, there are compu-tational tools that allow for quick invariant calculations. For instance, using Matlab

r,the matrix of Example 2.14 can be entered and the invariants found with the followingcode:


>> T = [5 2 0; 2 2 0; 0 0 3]

T =5 2 02 2 00 0 3

>> Tsqrd = T*T

Tsqrd =29 14 014 8 00 0 9

>> I_T = trace(T)

I_T = 10

>> II_T = 0.5*(I_T^2 - trace(Tsqrd))

II_T = 27

>> III_T = det(T)

III_T = 18

Furthermore, most of these computational tools can use symbolic notation as well asnumbers. Consider the analysis done on the matrix defined in Eq 2.78. All of thosecalculations may be completed with Matlab

r as follows:

>> syms a b c d e f g h i>> A = [a b c; d e f; g h i]

A =[ a, b, c][ d, e, f][ g, h, i]

>> Asqrd = A*A

Asqrd =[ a^2+b*d+c*g, a*b+b*e+c*h, a*c+b*f+c*i][ d*a+e*d+f*g, b*d+e^2+f*h, d*c+e*f+f*i][ g*a+h*d+i*g, g*b+h*e+i*h, c*g+f*h+i^2]

>> I_A = trace(A)I_A = a+e+i

>> II_A = 0.5*(I_A^2 - trace(Asqrd))

II_A = 1/2*(a+e+i)^2-1/2*a^2-b*d-c*g-1/2*e^2-f*h-1/2*i^2


>> simplify(II_A)

ans = a*e+a*i+e*i-b*d-c*g-f*h

>> III_A = det(A)

III_A = a*e*i-a*f*h-d*b*i+d*c*h+g*b*f-g*c*e

In concluding this section, we mention several interesting properties of symmetricsecond-order tensors:

1. The principal values and principal directions of T and TT are the same.

2. The principal values of T−1 are reciprocals of the principal values of T , and bothhave the same principal directions.

3. The product tensors TQ and QT have the same principal values.

4. A symmetric tensor is said to be positive (negative) definite if all of its principal valuesare positive (negative); and positive (negative) semi-definite if one principal value iszero and the others positive (negative).

2.7 Tensor Fields, Tensor CalculusA tensor field assigns to every location x, at every instant of time t, a tensor tij...k(x, t), forwhich x ranges over a finite region of space, and t varies over some interval of time. Thefield is continuous and hence differentiable if the components tij...k(x, t) are continuousfunctions of x and t. Tensor fields may be of any order. For example, we may denotetypical scalar, vector, and tensor fields by the notations φ(x, t), vi(x, t), and tij(x, t), re-spectively.

Partial differentiation of a tensor field with respect to the variable t is symbolized bythe operator ∂/∂t and follows the usual rules of calculus. On the other hand, partialdifferentiation with respect to the coordinate xq will be indicated by the operator ∂/∂xq,which may be abbreviated as simply ∂q. Likewise, the second partial ∂2/∂xq∂xmmaybe written ∂qm, and so on. As an additional measure in notational compactness it iscustomary in continuum mechanics to introduce the subscript comma to denote partialdifferentiation with respect to the coordinate variables. For example, we write φ,i for∂φ/∂xi; vi,j for ∂vi/∂xj; tij,k for ∂tij/∂xk; and ui,jk for ∂2ui/∂xj∂xk. We note from theseexamples that differentiation with respect to a coordinate produces a tensor of one orderhigher. Also, a useful identity results from the derivative ∂xi/∂xj, viz.,

∂xi

∂xj= δij . (2.81)

In the notation adopted here the operator ∇ (del) of vector calculus, which in symbolicnotation appears as

∇ =∂

∂x1e1 +

∂

∂x2e2 +

∂

∂x3e3 =

∂

∂xiei (2.82)


takes on the simple form ∂i. Therefore, we may write the scalar gradient∇φ = gradφ as

∂iφ = φ,i , (2.83)

the vector gradient ∇v as∂ivj = vj,i , (2.84)

the divergence of v, ∇ · v as∂ivi = vi,i , (2.85)

and the curl of v, ∇× v asεijk∂jvk = εijkvk,j . (2.86)

Note in passing that many of the identities of vector analysis can be verified withrelative ease by manipulations using the indicial notation. For example, to show thatdiv(curl v) = 0 for any vector v we write from Eqs 2.86 and Eq 2.85

∂i (εijkvk,j) = εijkvk,ji = 0 ,

and because the first term of this inner product is skew–symmetric in i and j, whereas thesecond term is symmetric in the same indices, (since vk is assumed to have continuousspatial gradients), their product is zero.

Example 2.16

Use indicial notation to verify the following identities:

curl∇u = 0 (2.87a)

andcurl

(∇uT

)=∇ curlu (2.87b)

where u is any continuously differentiable vector field.

SolutionFor the first, write curl∇u = εipkuj,kpeiej where the order of partial differ-entiation of u does not matter since u is continuously differentiable. Thus,uj,kp = uj,pk. By definition, the permutation symbol is skew–symmetric inkp and the product of a skew–symmetric and symmetric tensor is zero (seeExample 2.5).

The second identity is just as easily verified with indicial notation. Writethe left-hand side of the expression in indicial notation, exchange the order ofpartial differentiation on u. Since the permutation symbol is a constant we maywrite εipkuk,jp = (εipkuk,p),j

There is also a need for using the divergence and curl of second order tensors in contin-uum mechanics. Let sjk be a second order tensor that is differentiable at x. The divergenceof sjk at x is defined as

divS = sjk,kej (2.88)


wheresij,k =

∂sij

∂xk= ei · (∇Sek) ej . (2.89)

Just as the divergence of a vector reduces the tensor order by one to a scalar, the diver-gence of a second order tensor reduces the tensor order by one to a vector. The curl of asecond order tensor sjk is defined as

(curlS)ij = εipksjk,p (2.90)

in a manner that parallels the curl of a vector. The curlu returns a vector and the curlSresults in a second order tensor.

Example 2.17

Consider sij = sji to be a continuously differentiable, second order tensor. Showthat

tr (curlS) = 0 . (2.91)

SolutionWrite the curl of S in indicial form:

(curlS)ij = εipksjk,p .

The trace of this is found by setting i = j:

(tr (curlS))ii = εipksik,p .

Since sik = ski and εipk = −εkpi we see that tr (curlS) = 0.

Example 2.18

LetW = wijeiej = −wjieiej be a skew–symmetric, second order tensor havingcontinuous derivatives. SinceW is skew–symmetric it has an axial vector u suchthat wjk = −εqjkuq. Show that

curlW = Idivu−∇u , (2.92a)

or, equivalentlyεipkwjk,p = δijuq,q − ui,j . (2.92b)

SolutionUsing the definition of an axial vector as well as Eq 2.7a we find that:

εipkwjk,p = εipk (−εqjkuq),p

= −(δiqδpj − δijδpq)uq,p

= δijuq,q − ui,j .


x1

x2

x3

dSi=n

idS

ni

V

O

FIGURE 2.5Volume V with infinitesimal element dSi having a unit normal ni.

2.8 Integral Theorems of Gauss and StokesConsider an arbitrary continuously differentiable tensor field tij...k defined on some fi-nite region of physical space. Let V be a volume in this space with a closed surface S

bounding the volume, and let the outward normal to this bounding surface be ni asshown in Fig. 2.5 so that the element of surface is given by dSi = nidS. The divergencetheorem of Gauss establishes a relationship between the surface integral having tij...k as in-tegrand to the volume integral for which a coordinate derivative of tij...k is the integrand.Specifically, ∫

S

tij...knqdS =

∫V

tij...k,q dV . (2.93)

Several important special cases of this theorem for scalar and vector fields are worthnoting, and are given here in both indicial and symbolic notation:

∫S

λnq dS =

∫V

λ,q dV or∫S

λndS =

∫V

grad λdV , (2.94)

∫S

vqnq dS =

∫V

vq,q dV or∫S

v · ndS =

∫V

div vdV , (2.95)

∫S

εijknjvk dS =

∫V

εijkvk,j dV or∫S

n× vdS =

∫V

curl vdV . (2.96)

Called Gauss’s divergence theorem, Eq 2.96 is the one presented in a traditional vector cal-culus course.

Whereas Gauss’s theorem relates an integral over a closed volume to an integral overits bounding surface, Stokes’ theorem relates an integral over an open surface (a so-calledcap) to a line integral around the bounding curve of the surface. Therefore, let C be the


x1

x2

x3

ni

O

dxi

C

xi

dSi=n

idS

FIGURE 2.6Bounding space curve C with tangential vector dxi and surface element dSi for partialvolume.

bounding space curve to the surface S, and let dxi be the differential tangent vector to C

as shown in Fig. 2.6. (A hemispherical surface having a circular bounding curve C is aclassic example). If ni is the outward normal to the surface S, and vi is any vector fielddefined on S and C, Stokes’ theorem asserts that∫

S

εijknivk,j dS =

∫C

vk dxk or∫S

n · (∇× v) dS =

∫C

v · dx . (2.97)

The integral on the right-hand side of this equation is often referred to as the circulationwhen the vector v is the velocity vector.


Problems

Problem 2.1Let v = a× b, or in indicial notation,

viei = ajej × bkek = εijkajbkei

Using indicial notation, show that,

(a) v · v = a2b2 sin2 θ ,

(b) a× b · a = 0 ,

(c) a× b · b = 0 .

Problem 2.2With respect to the triad of base vectors u1, u2, and u3 (not necessarily unit vectors), the

triad u1,u2, and u3 is said to be a reciprocal basis if ui · uj = δij (i, j = 1, 2, 3). Show thatto satisfy these conditions,

u1 =u2 × u3

[u1,u2,u3]; u2 =

u3 × u1[u1,u2,u3]

; u3 =u1 × u2

[u1,u2,u3]

and determine the reciprocal basis for the specific base vectors

u1 = 2e1 + e2 ,

u2 = 2e2 − e3 ,

u3 = e1 + e2 + e3 .

Answer

u1 = 15

(3e1 − e2 − 2e3)

u2 = 15

(−e1 + 2e2 − e3)

u3 = 15

(−e1 + 2e2 + 4e3)

Problem 2.3Let the position vector of an arbitrary point P (x1x2x3) be x = xiei, and let b = biei be

a constant vector. Show that (x− b) · x = 0 is the vector equation of a spherical surfacehaving its center at x = 1

2b with a radius of 12b.

Problem 2.4Using the notations A(ij) = 1

2 (Aij +Aji) and A[ij] = 12 (Aij −Aji) show that

(a) the tensor A having components Aij can always be decomposed into a sum ofits symmetric A(ij) and skew-symmetric A[ij] parts, respectively, by the decom-position,

Aij = A(ij) +A[ij] ,


(b) the trace of A is expressed in terms of A(ij) by

Aii = A(ii) ,

(c) for arbitrary tensors A and B,

AijBij = A(ij)B(ij) +A[ij]B[ij] .

Problem 2.5Expand the following expressions involving Kronecker deltas, and simplify where possible.

(a) δijδij, (b) δijδjkδki, (c) δijδjk, (d) δijAik

Answer

(a) 3, (b) 3, (c) δik, (d) Ajk

Problem 2.6If ai = εijkbjck and bi = εijkgjhk, substitute bj into the expression for ai to show that

ai = gkckhi − hkckgi ,

or in symbolic notation, a = (c · g)h− (c · h)g.

Problem 2.7By summing on the repeated subscripts determine the simplest form of

(a) ε3jkajak, (b) εijkδkj, (c) ε1jka2Tkj, (d) ε1jkδ3jvk .

Answer

(a) 0, (b) 0, (c) a2(T32 − T23), (d) −v2

Problem 2.8Consider the tensor Bik = εijkvj.

(a) Show that Bik is skew-symmetric.

(b) Let Bij be skew-symmetric, and consider the vector defined by vi = εijkBjk(often called the dual vector of the tensor B). Show that Bmq = 1

2εmqivi.

Problem 2.9Use indicial notation to show that

Amiεmjk +Amjεimk +Amkεijm = Ammεijk

where A is any tensor and εijk is the permutation symbol.


Problem 2.10If Aij = δijBkk + 3Bij, determine Bkk and using that solve for Bij in terms of Aij and its

first invariant, Aii.

Answer

Bkk = 16Akk; Bij = 1

3Aij − 118δijAkk

Problem 2.11Show that the value of the quadratic form Tijxixj is unchanged if Tij is replaced by its

symmetric part, 12 (Tij + Tji).

Problem 2.12With the aid of Eq 2.7, show that any skew–symmetric tensor W may be written in terms

of an axial vector ωi given by

ωi = −1

2εijkwjk

where wjk are the components of W.

Problem 2.13Show by direct expansion (or otherwise) that the box product λ = εijkaibjck is equal to

the determinant ∣∣∣∣∣∣∣a1 a2 a3

b1 b2 b3

c1 c2 c3

∣∣∣∣∣∣∣ .Thus, by substituting A1i for ai, A2j for bj and A3k for ck, derive Eq 2.42 in the formdet A = εijkA1iA2jA3k where Aij are the elements of A.

Problem 2.14Starting with Eq 2.42 of the text in the form

det A = εijkAi1Aj2Ak3

show that by an arbitrary number of interchanges of columns of Aij we obtain

εqmn det A = εijkAiqAjmAkn

which is Eq 2.43. Further, multiply this equation by the appropriate permutation symbolto derive the formula

6det A = εqmnεijkAiqAjmAkn .

Problem 2.15Let the determinant of the matrix Aij be given by

det A =

∣∣∣∣∣∣∣A11 A12 A13

A21 A22 A23

A31 A32 A33

∣∣∣∣∣∣∣ .


Since the interchange of any two rows or any two columns causes a sign change in the valueof the determinant, show that after an arbitrary number of row and column interchanges∣∣∣∣∣∣∣

Amq Amr Ams

Anq Anr Ans

Apq Apr Aps

∣∣∣∣∣∣∣ = εmnpεqrs det A .

Now let Aij = δij in the above determinant which results in det A = 1 and, upon expansion,yields

εmnpεqrs = δmq(δnrδps − δnsδpr) − δmr(δnqδps − δnsδpq) + δms(δnqδpr − δnrδpq) .

Thus, by setting p = q, establish Eq 2.7 in the form

εmnqεqrs = δmrδns − δmsδnr .

Problem 2.16Show that the square matrices

[Bij] =

1 0 0

0 −1 0

0 0 1

and [Cij] =

[5 2

−12 −5

]

are both square roots of the identity matrix.

Problem 2.17Using the square matrices below, demonstrate

(a) that the transpose of the square of a matrix is equal to the square of its transpose(Eq 2.36 with n = 2),

(b) that (AB)T = BTAT as was proven in Example 2.8

[Aij] =

3 0 1

0 2 4

5 1 2

, [Bij] =

1 3 1

2 2 5

4 0 3

.

Problem 2.18Let A be any orthogonal matrix, i.e., AAT = AA−1 = I, where I is the identity matrix.

Thus, by using the results in Examples 2.9 and 2.10, show that det A = ±1.

Problem 2.19A tensor is called isotropic if its components have the same set of values in every Cartesian

coordinate system at a point. Assume that T is an isotropic tensor of rank two withcomponents tij relative to axes Ox1x2x3. Let axes Ox

′

1x′

2x′

3 be obtained with respect toOx1x2x3 by a righthand rotation of 120 about the axis along n = (e+ e+ e) /

√3. Show

by the transformation between these axes that t11 = t22 = t33, as well as other relationships.Further, let axes Ox

′′

1x′′

2x′′

3 be obtained with respect to Ox1x2x3 by a right-hand rotation


of 90 about x3. Thus, show by the additional considerations of this transformation that ifT is any isotropic tensor of second order, it can be written as λI where λ is a scalar and Iis the identity tensor.

Problem 2.20For a proper orthogonal transformation between axes Ox1x2x3 and Ox

′

1x′

2x′

3 show theinvariance of δij and εijk. That is, show that

(a) δ′

ij = δij ,

(b) ε′

ijk = εijk.

Hint: For part (b) let ε′

ijk = aiqajmaknεqmn and make use of Eq 2.43.

Problem 2.21The angles between the respective axes of the Ox

′

1x′

2x′

3 and the Ox1x2x3 Cartesian systemsare given by the table below

x1 x2 x3

x ′1 45 90 45

x ′2 60 45 120

x ′3 120 45 60

Determine

(a) the transformation matrix between the two sets of axes, and show that it is aproper orthogonal transformation,

(b) the equation of the plane x1 + x2 + x3 = 1/√2 in its primed axes form, that is,

in the form b1x′

1 + b2x′

2 + b3x′

3 = b.

Answer

(a) [aij] =

1√2

0 1√2

1√2

1√2

− 1√2

− 1√2

1√2

1√2

(b) 2x

′

1 + x′

2 + x′

3 = 1

Problem 2.22Making use of Eq 2.42 of the text in the form det A = εijkA1iA2jA3k write Eq 2.71 as

|tij − λδij| = εijk (t1i − δ1i) (t2j − δ2j) (t3k − δ3j) = 0

and show by expansion of this equation that

λ3 − tiiλ2 +

[1

2(tiitjj − tijtji)

]λ− εijkt1it2jt3k = 0

to verify Eq 2.72 of the text.


Problem 2.23

For the matrix representation of tensor B shown below,

[bij] =

17 0 0

0 −23 28

0 28 10

determine the principal values (eigenvalues) and the principal directions (eigenvectors) ofthe tensor.

Answer

λ1 = 17, λ2 = 26, λ3 = −39

n(1) = e1, n(2) = (4e2 + 7e3) /√65, n(3) = (−7e2 + 4e3) /

√65

Problem 2.24

Consider the symmetrical matrix

[Bij] =

54 0 3

2

0 4 0

32 0 5

2

.(a) Show that a multiplicity of two occurs among the principal values of this matrix.

(b) Let λ1 be the unique principal value and show that the transformation matrix

[aij] =

1√2

0 − 1√2

0 1 0

1√2

0 1√2

gives B∗ according to B∗ij = aiqajmBqm.

(c) Taking the square root of[B∗ij

]and transforming back to Ox1x2x3 axes show

that

[√Bij

]=

32 0 1

2

0 2 0

12 0 3

2

.(d) Verify that the matrix

[Cij] =

−12 0 −32

0 2 0

−32 0 −12

is also a square root of [Bij].


Problem 2.25Determine the principal values of the matrix

[Kij] =

4 0 0

0 11 −√3

0 −√3 9

,and show that the principal axes Ox∗1x

∗2x∗3 are obtained from Ox1x2x3 by a rotation of 60

about the x1 axis.

Answer

λ1 = 4, λ2 = 8, λ3 = 12.

Problem 2.26Determine the principal values λ(q) (q = 1, 2, 3) and principal directions n(q) (q = 1, 2, 3)

for the symmetric matrix

[Tij] =1

2

3 − 1√

21√2

− 1√2

92

32

1√2

32

92

Answer

λ(1) = 1, λ(2) = 2 λ(3) = 3

n(1) = 12

(√2e1 + e2 − e3

)n(2) = 1

2

(√2e1 − e2 + e3

)n(3) = −(e2 + e3) /

√2

Problem 2.27For the second-order tensor Cij = uivj:

(a) Calculate the principal invariants IC, IIC, and IIIC . Reduce your answer to itssimplest form, but leave it in index notation. State the reasons for simplification.

(b) If u =[1 2 3

]and v =

[5 10 4

]write out the matrix form for Cij.

(c) Use the numbers of (b) to validate the answer of (a).

Problem 2.28Let D be a constant tensor whose components do not depend upon the coordinates. Show

that∇ (x ·D) = D

where x = x iei is the position vector.


Problem 2.29Consider the vector x = x iei having a magnitude squared x2 = x21 + x22 + x33. Determine

(a) grad x ,

(b) grad (x−n) ,

(c) ∇2 (1/x) ,

(d) div (xnx) ,

(e) curl (xnx), where n is a positive integer.

Answer

(a) xi/x, (b) −nxi/x(n+2), (c) 0, (d) xn(n+ 3), (e) 0.

Problem 2.30If λ and ϕ are scalar functions of the coordinates xi, verify the following vector identities.

Transcribe the left-hand side of the equations into indicial notation and, following theindicated operations, show that the result is the right-hand side.

(a) v× (∇× v) = 12∇(v · v) − (v ·∇)v

(b) v · u×w = v× u ·w(c) ∇× (∇× v) =∇(∇ · v) −∇2v(d) ∇ · (λ∇ϕ) = λ∇2ϕ+∇λ ·∇ϕ(e) ∇2(λϕ) = λ∇2ϕ+ 2(∇λ) · (∇ϕ) +ϕ∇2λ(f) ∇ · (u× v) = (∇× u) · v− u · (∇× v)

Problem 2.31Let the vector v = b × x be one for which b does not depend upon the coordinates. Use

indicial notation to show that

(a) curl v = 2b ,

(b) div v = 0 .

Problem 2.32Transcribe the left-hand side of the following equations into indicial notation and verify thatthe indicated operations result in the expressions on the right-hand side of the equationsfor the scalar ϕ, and vectors u and v.

(a) div (ϕv) = ϕdiv v+ v · gradϕ

(b) u× curl v+ v× curl u = −(u · grad)v− (v · grad)u+ grad(u · v)(c) div (u× v) = v · curl u− u · curl v

(d) curl(u× v) = (v · grad)u− (u · grad)v+ udiv v− vdivu

(e) curl(curlu) = grad(divu) −∇2u


Problem 2.33Let the volume V have a bounding surface S with an outward unit normal ni. Let xi be

the position vector to any point in the volume or on its surface. Show that

(a)∫

S

xinjdS = δijV ,

(b)∫

S

∇ (x · x) · ndS = 6V ,

(c)∫

S

λw · ndS =

∫V

w · grad λdV, where w = curl v and λ = λ(x) ,

(d)∫

S

[ei × x, ej, n]dS = 2Vδij where ei and ej are coordinate base vectors.

Hint: Write the box product

[ei × x, ej, n] = (ei × x) · (ej × n)

and transcribe into indicial notation.

Problem 2.34Use Stokes’ theorem to show that upon integrating around the space curve C having a

differential tangential vector dxi that for ϕ(x).∮C

ϕ,idxi = 0

Problem 2.35For the position vector xi having a magnitude x, show that x,j = xj/x and therefore,

(a) x,ij =δij

x−xixj

x3,

(b)(x−1

),ij

=3xixj

x5−δij

x3,

(c) x,ii =2

x.

Problem 2.36Show that for arbitrary tensors A and B, and arbitrary vectors a and b,

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

(b) b× a = 12 (B− BT ) · a, if 2bi = εijkBkj ,

(c) a ·A · b = b ·AT · a .

Problem 2.37Use Eqs 2.42 and 2.43 as necessary to prove the identities

(a) [Aa,Ab,Ac] = (det A)[a,b, c] ,


(b) AT · (Aa×Ab) = (det A)(a× b) ,

for arbitrary vectors a, b, c, and matrix A.

Problem 2.38Let ϕ = ϕ(xi) and ψ = ψ(xi) be scalar functions of the coordinates. Recall that in the

indicial notation ϕ,i represents ∇ϕ and ϕ,ii represents ∇2ϕ. Now apply the divergencetheorem, Eq 2.93, to the field ϕψ,i to obtain∫

S

ϕψ,ini dS =

∫V

(ϕ,iψ,i +ϕψ,ii) dV .

Transcribe this result into symbolic notation as∫S

ϕ∇ψ · ndS =

∫S

ϕ∂ψ

∂ndS =

∫V

(∇ϕ ·∇ψ+ϕ∇2ψ

)dV

which is known as Green’s first identity. Show also by the divergence theorem that∫S

(ϕψ,i −ψϕ,i)ni dS =

∫V

(ϕψ,ii −ψϕ,ii) dV ,

and transcribe into symbolic notation as∫S

(ϕ∂ψ

∂n−ψ

∂ϕ

∂n

)dS =

∫V

(ϕ∇2ψ−ψ∇2ϕ

)dV

which is known as Green’s second identity.


3Stress Principles

Stress is a measure of force intensity, either within, or on the bounding surface of a bodysubjected to loads. It should be noted that in continuum mechanics a body is consideredstress free if the only forces present are those inter-atomic forces required to hold thebody together. And so it follows that the stresses which concern us here are those thatresult from the application of forces by an external agent.

3.1 Body and Surface Forces, Mass DensityTwo basic types of forces are easily distinguished from one another and are defined asfollows. First, those forces acting on all volume elements, and are distributed throughoutthe body, are known as body forces. Gravity and inertia forces are the best known examples.We designate body forces by the vector symbol bi (force per unit mass), or by the symbolpi (force per unit volume). Secondly, those forces which act upon, and are distributedin some fashion over a surface element of the body, regardless of whether that elementis part of the bounding surface, or an arbitrary element of surface within the body, arecalled surface forces. These are denoted by the vector symbol fi, and have dimensionsof force per unit area. Forces which occur on the outer surfaces of two bodies pressedagainst one another (contact forces), or those which result from the transmission of forcesacross an internal surface are examples of this type of force.

Next, let us consider a material body B having a volume V enclosed by a surfaceS, and occupying a regular region R0 of physical space. Let P be an interior point ofthe body located in the small element of volume ∆V whose mass is ∆m as indicated inFig. 3.1. Recall that mass is that property of a material body by virtue of which the bodypossesses inertia, that is, the opposition which the body offers to any change in its motion.We define the average density of this volume element by the ratio

ρave =∆m

∆V, (3.1)

and the density ρ at point P by the limit of this ratio as the volume shrinks to the point P,

ρ = lim∆V→0

∆m

∆V=dm

dV. (3.2)

The units of density are kilograms per cubic meter, (kg/m3). Notice that the two measuresof body forces, bi having units of Newton per kilogram (N/kg), and pi having units ofNewtons per meter cubed (N/m3), are related through the density by the equation

ρbi = pi or ρb = p . (3.3)

53


∆V

V

O

x1

x2

x3

P

FIGURE 3.1Typical continuum volume V with infinitesimal element ∆V having mass ∆m at point P.Point P would be in the center of the infinitesimal volume.

Of course, the density is, in general, a scalar function of position and time as indicatedby

ρ = ρ(xi, t) or ρ = ρ(x, t) (3.4)

and thus may vary from point to point within a given body.

3.2 Cauchy Stress Principle

Consider a homogeneous, isotropic material body B having a bounding surface S, and avolume V, which is subjected to arbitrary surface forces fi and body forces bi. Let P be aninterior point of B and imagine a plane surface S∗ passing through point P (sometimesreferred to as a cutting plane) so as to partition the body into two portions, designatedI and II (Fig. 3.2(a)). Point P is in the small element of area ∆S∗ of the cutting plane,which is defined by the unit normal pointing in the direction from portion I into portionII as shown by the free body diagram of portion I in 3.2(b). The internal forces beingtransmitted across the cutting plane due to the action of portion II upon portion I willgive rise to a force distribution on S∗ equivalent to a resultant force ∆fi and a resultantmoment ∆Mi at P, as is also shown in Fig. 3.2(b). (For simplicity body forces bi andsurface forces fi acting on the body as a whole are not drawn in Figs. 3.2(a) and 3.2(b).)Notice that ∆fi and ∆Mi are not necessarily in the direction of the unit normal vector niat P. The Cauchy stress principle asserts that in the limit as the area S∗ shrinks to zero withP remaining an interior point, we obtain

lim∆S∗→0

∆fi

∆S∗=dfi

dS∗= t

(n)i (3.5)

and

lim∆S∗→0

∆Mi

∆S∗= 0 . (3.6)

Stress Principles 55

x1

x2

x3

S∗

O

III

P

(a) Typical continuum volume showing cuttingplane S∗ passing through point P.

x1

x2

x3

∆S∗

O

In

i

∆Mi

∆fi

(b) Force and moment acting at point P in sur-face element ∆S∗.

FIGURE 3.2Typical continuum volume with cutting plane.

The vector dfi/dS∗ = t(n)i is called the stress vector, or sometimes the traction vector. In

Eq 3.6 we have made the assumption that in the limit at P, the moment vector vanishes,and there is no remaining concentrated moment, or couple stress as it is called.

The appearance of n in the symbol t(n)i for the stress vector serves to remind us that

this is a special vector in that it is meaningful only in conjunction with its associatednormal vector n at P. Thus, for the infinity of cutting planes imaginable through point P,each identified by a specific n, there is also an infinity of associated stress vectors t(n)

i fora given loading of the body. The totality of pairs of the companion vectors t(n)

i and n atP, as illustrated by a typical pair in Fig. 3.3, defines the state of stress at that point.

By applying Newton’s third law of action and reaction across the cutting plane we ob-serve that the force exerted by portion I upon portion II is equal and opposite to the forceof portion II upon portion I. Additionally, from the principle of linear momentum (New-ton’s second law) we know that the time rate of change of the linear momentum of anyportion of a continuum body is equal to the resultant force acting upon that portion. Forportions I and II, this principle may be expressed in integral form by the respective equa-tions (these equations are derived in Section 5.3 from the principle of linear momentum),

∫SI

t(n)i dS +

∫VI

ρbidV =d

dt

∫VI

ρvidV , (3.7a)

∫SII

t(n)i dS +

∫VII

ρbidV =d

dt

∫VII

ρvidV (3.7b)

where SI and SII are the bounding surfaces and VI and VII are the volumes of portionsI and II, respectively. Also, bi are the body forces, ρ is the density and vi is the velocityfield for the two portions. We note that SI and SII each contain S∗ as part of their totalareas.


∆S∗

ni

P

t(n)i

FIGURE 3.3Traction vector t(n)

i acting at point P of plane element ∆Si whose normal is ni.

The linear momentum principle may also be applied to the body B as a whole, so that∫S

t(n)i dS +

∫V

ρbidV =d

dt

∫V

ρvidV . (3.8)

If we add Eq 3.7a and Eq 3.7b and use Eq 3.8, noting that the normal to S∗ for portion Iis n, whereas for portion II it is −n, we arrive at the equation∫

S∗

[t(n)i + t

(−n)i

]dS = 0 (3.9)

since both SI and SII contain a surface integral over S∗. This equation must hold forarbitrary partitioning of the body (that is, for every imaginable cutting plane throughpoint P) which means that the integrand must be identically zero. Hence,

t(n)i = −t

(−n)i (3.10)

indicating that if portion II had been chosen as the free body in Fig. 3.2(b) instead ofportion I, the resulting stress vector would have been −t

(n)i .

3.3 The Stress TensorAs noted in Section 3.2, the Cauchy stress principle associates with each direction n atpoint P a stress vector t(n)

i . In particular, if we introduce a rectangular Cartesian referenceframe at P, there is associated with each of the area elements dSi(i = 1, 2, 3) located in thecoordinate planes and having unit normals ei(i = 1, 2, 3), respectively, a stress vector t(n)

i

as shown in Fig. 3.4. In terms of their coordinate components these three stress vectorsassociated with the coordinate planes are expressed by

t(e1) = t(e1)1 e1 + t

(e1)2 e2 + t

(e1)3 e3 , (3.11a)

t(e2) = t(e2)1 e1 + t

(e2)2 e2 + t

(e2)3 e3 , (3.11b)

t(e3) = t(e3)1 e1 + t

(e3)2 e2 + t

(e3)3 e3 , (3.11c)


x1

x2

x3

P

x1

x2

x3

P

x1

x2

x3

P

e1

e2

e3t(e2)i

t(e3)i

t(e1)i

FIGURE 3.4Traction vectors on the three coordinate planes at point P.

P

A

B

C

ρ∗bi

ni

x1

x2

x3

−∗t(e3)i

−∗t(e2)i

−∗t(e1)i

∗t(n)i

FIGURE 3.5Free body diagram of tetrahedron element having its vertex at point P.

or more compactly, using the summation convention

t(ei) = t(ei)j ej . (3.12)

This equation expresses the stress vector at P for a given coordinate plane in terms ofits rectangular Cartesian components, but what is really needed is an expression for thecoordinate components of the stress vector at P associated with an arbitrarily orientedplane. For this purpose we consider the equilibrium of a small portion of the body inthe shape of a tetrahedron having its vertex at P, and its base ABC perpendicular to anarbitrarily oriented normal n = niei as shown by Fig. 3.5. The coordinate directionsare chosen so that the three faces BPC, CPA and APB of the tetrahedron are situated inthe coordinate planes. If the area of the base is assigned the value dS, the areas of therespective faces will be the projected areas dSi = dS cos(n, ei), (i = 1, 2, 3) or specifically,

for BPC dS1 = n1dS , (3.13a)for CPA dS2 = n2dS , (3.13b)for APB dS3 = n3dS . (3.13c)

The stress vectors shown on the surfaces of the tetrahedron of Fig. 3.5 represent averagevalues over the areas on which they act. This is indicated by a superscript asterisk infront of the stress vector symbols (remember that the stress vector is a point quantity).


Equilibrium requires the vector sum of all forces acting on the tetrahedron to be zero, thatis, for,

∗t(n)i dS − ∗t

(e1)i dS1 − ∗t

(e2)i dS2 − ∗t

(e3)i dS3 + ρ ∗bidV = 0 (3.14)

where ∗bi is an average body force which acts throughout the body. The negative signson the coordinate-face tractions result from the outward unit normals on those facespointing in the negative coordinate axes directions. (Recall that t(n)

i = −t(−n)i .) Taking

into consideration Eq 3.13, we can write Eq 3.14 as

∗t(n)i dS −∗ t

(ej)i njdS + ρ ∗bidV = 0 , (3.15)

if we permit the indices on the unit vectors of the ∗t(ej)i term to participate in the sum-mation process. The volume of the tetrahedron is given by dV = 1

3hdS, where h is theperpendicular distance from point P to the base ABC. Inserting this into Eq 3.15 andcanceling the common factor dS we obtain

∗t(n)i =∗ t

(ej)i nj −

1

3ρ ∗bih . (3.16)

Now, letting the tetrahedron shrink to point P by taking the limit as h → 0 and notingthat in this limiting process the starred (averaged) quantities take on the actual values ofthose same quantities at point P, we have

t(n)i = t

(ej)i nj , (3.17)

or, by defining tji ≡ t(ej)i ,

t(n)i = tjinj or t(n) = n · T (3.18)

which is the Cauchy stress formula. We can obtain this same result for bodies which areaccelerating by using the conservation of linear momentum instead of a balance of forceson the tetrahedron of Fig. 3.5.

Stress as a TensorThe quantities tji ≡ t

(ej)i are the components of a second order tensor T

known as the stress tensor. This can be shown by considering the transforma-tion of the components of the stress vector t(n)

i between coordinate systemsPx1x2x3 and Px ′1x

′2x′3 as given by the transformation matrix having elements

(see Section 2.5)aij = e ′i · ej . (3.19)

Since t(n) can be expressed in terms of its components in either coordinatesystem,

t(n) = t(n)i ei = t ′

(n′)i e ′i (3.20a)

or, from Eq 3.18,t(n) = tjinjei = t ′jin

′je′i . (3.20b)

But from Eq 2.53, e ′i = aijej and from Eq 2.60, n ′j = ajsns, so that nowEq 3.20b becomes, after some manipulations of the summed indices,(

tsr − ajsairt′ji

)nser = 0 . (3.21)


Because the vectors er are linearly independent and since Eq 3.21 must bevalid for all vectors ns, we see that

tsr = ajsairt′ji . (3.22)

But this is the transformation equation for a second-order tensor, and thus byEq 2.63 the tensor character of the stress components is clearly established.

The Cauchy stress formula given by Eq 3.18 expresses the stress vector associated withthe element of area having an outward normal ni at point P in terms of the stress tensorcomponents tji at that point. And although the state of stress at P has been describedas the totality of pairs of the associated normal and traction vectors at that point, wesee from the analysis of the tetrahedron element that if we know the stress vectors onthe three coordinate planes of any Cartesian system at P, or equivalently, the nine stresstensor components tji at that point, we can determine the stress vector for any plane atthat point. For computational purposes it is often convenient to express Eq 3.18 in thematrix form [

t(n)1 t

(n)2 t

(n)3

]= [n1 n2 n3]

t11 t12 t13

t21 t22 t23

t31 t32 t33

. (3.23)

The nine components of tji are often displayed by arrows on the coordinate faces of arectangular parallelpiped, as shown in Fig. 3.6. We emphasize that this parallelpiped isnot a block of material from the continuum body (note that no dimensions are given to theparallelpiped), but is simply a convenient schematic device for displaying the stress tensorcomponents. In an actual physical body B, all nine stress components act at the singlepoint P. The three stress components shown by arrows acting perpendicular (normal) tothe respective coordinate planes and labeled t11, t22, and t33 are called normal stresses. Thesix arrows lying in the coordinate planes and pointing in the directions of the coordinateaxes, namely, t12, t21, t23, t32, t13, and t31 are called shear stresses. Note that, for these,the first subscript designates the coordinate plane on which the shear stress acts, and thesecond subscript identifies the coordinate direction in which it acts. A stress componentis positive when its vector arrow points in the positive direction of one of the coordinateaxes while acting on a plane whose outward normal also points in a positive coordinatedirection. All of the stress components displayed in Fig. 3.6 are positive. In general,positive normal stresses are called tensile stresses, and negative normal stresses are referredto as compressive stresses. The units of stress are Newtons per square meter (N/m2) inthe SI system, and pounds per square inch (psi) in the English system. One Newton persquare meter is called a Pascal, but because this is a rather small stress from an engineeringpoint of view, stresses are usually expressed as mega-pascals (MPa) or in English unitsas kilo-pounds per square inch (ksi).

Example 3.1

Let the components of the stress tensor at P be given in matrix form by

[tij] =

21 −63 42

−63 0 84

42 84 −21

in units of mega-Pascals. Determine


x1

x2

x3

t12

t33

t22

t32t

31

t21

t23

t11

t13

FIGURE 3.6Cartesian stress components shown in their positive sense.

(a) the stress vector on the plane at P having the unit normal

n =1

7(2e1 − 3e2 + 6e3) ,

(b) the stress vector on a plane at P parallel to the plane ABC shown inthe sketch.

A(1,0,0)

B(0,1,0)

C(0,0,2)

x1

x2

x3

Solution

(a) From Eq 3.23 for the given data,

[t(n)1 t

(n)2 t

(n)3

]=

[2

7−3

7

6

7

] 21 −63 42

−63 0 84

42 84 −21

= [69 54 − 42] ,


or, in vector form, t(n) = 69e1+54e2−42e3. This vector represents thecomponents of the force per unit area (traction) on the plane definedby 1

7 [2, −3, 6] .(b) The equation of the plane ABC in the sketch is easily verified to be

2x1 + 2x2 + x3 = 2, and the unit outward normal to this plane isn = 1

3(2e1 + 2e2 + e3) so that, again from Eq 3.23,

[t(n)1 t

(n)2 t

(n)3

]=

[23

23

13

] 21 −63 42

−63 0 84

42 84 −21

= [−14 − 14 77]

or, in vector form, t(n) = −14e1 − 14e2 + 77e3.

In this example, we clearly see the dependency of the cutting plane and thestress vector. Here, we have considered two different cutting planes at the samepoint and found that two distinct traction vectors arose from the given stresstensor components.

3.4 Force and Moment Equilibrium; Stress Tensor SymmetryIn the previous section, we used a balance-of-forces condition for a tetrahedron elementof a body in equilibrium to define the stress tensor and to develop the Cauchy stressformula. Here, we employ a force balance on the body as a whole to derive what areknown as the local equilibrium equations. This set of three differential equations must holdfor every point in a continuum body that is in equilibrium. As is well known, equilibriumalso requires the sum of moments to be zero with respect to any fixed point, and we usethis condition, together with the local equilibrium equations, to deduce the fact that thestress tensor is symmetric in the absence of concentrated body moments.

Consider a material body having a volume V and a bounding surface S. Let the bodybe subjected to surface tractions t(n)

i and body forces bi (force per unit mass), as shownby Fig. 3.7. As before, we exclude concentrated body moments from consideration. Equi-librium requires that the summation of all forces acting on the body be equal to zero.This condition is expressed by the global (integral) equation representing the sum of thetotal surface and body forces acting on the body,∫

S

t(n)i dS +

∫V

ρbidV = 0 (3.24)

where dS is the differential element of the surface S and dV that of volume V.Because t(n)

i = tjinj as a result of Eq 3.18, the divergence theorem Eq 2.93 allows thefirst term of Eq 3.24 to be written as∫

S

tjinjdS =

∫V

tji,jdV , (3.25)


n

x y

x1

x2

x3

ρb

dS

O

dV

t(n)

FIGURE 3.7Material volume showing surface traction vector t(n)

i on an infinitesimal area element dSat position xi, and body force vector bi acting on an infinitesimal volume element dV atposition yi. Two positions are taken separately for ease of illustration. When applyingequilibrium the traction and body forces are taken at the same point.

so that Eq 3.24 becomes ∫V

(tji,j + ρbi)dV = 0 . (3.26)

This equation must be valid for an arbitrary volume V (every portion of the body is inequilibrium), which requires the integrand itself to vanish, and we obtain the so-calledlocal equilibrium equations

tji,j + ρbi = 0 . (3.27)

In addition to the balance of forces expressed by Eq 3.24, equilibrium requires that thesummation of moments with respect to an arbitrary point must also be zero. Recall thatthe moment of a force about a point is defined by the cross product of its position vectorwith the force. Therefore, taking the origin of coordinates as the center for moments,and noting that xi is the position vector for the typical elements of surface and volume(Fig. 3.7), we express the balance of moments for the body as a whole by∫

S

εijkxjt(n)i dS +

∫V

εijk xjρbidV = 0 . (3.28)

As before, using the identity t(n)k = tqknq and Gauss’s divergence theorem, we obtain∫

V

εijk

[(xjtqk),q + xjρbk

]dV = 0 ,

or ∫V

εijk [xj,qtqk + xj (tqk,q + ρbk)]dV = 0 .

But xj,q = δjq and by Eq 3.27, tkq,k + ρbk = 0, so that the latter equation immediatelyabove reduces to ∫

V

εijktjkdV = 0 .


P

x1

x2

x3

cos−1 a11

cos−1 a12

cos−1 a13x 1

x 2

x 3

FIGURE 3.8Rectangular coordinate axes Px ′1x

′2x′3 relative to Px1x2x3 at point P.

Again, since volume V is arbitrary, the integrand here must vanish, or

εijktjk = 0 . (3.29)

By a direct expansion of the left-hand side of this equation, we obtain for the free indexi = 1, (omitting zero terms) ε123t23+ε132t32 = 0, or t23−t32 = 0 implying that t23 = t32.In the same way for i = 2 and i = 3 we find that t13 = t31 and t12 = t21, respectively, sothat in general

tjk = tkj . (3.30)

Thus, we conclude from the balance of moments for a body in which concentrated bodymoments are absent, that the stress tensor is symmetric, and Eq 3.27 may now be writtenin the form

tij,j + ρbi = 0 or ∇ · T + ρb = 0 . (3.31)

Also, because of this symmetry of the stress tensor, Eq 3.18 may now be expressed in theslightly altered form

t(n)i = tijnj or t(n) = T · n . (3.32)

In the matrix form of Eq 3.32 the vectors t(n)i and nj are represented by column matrices.

3.5 Stress Transformation LawsLet the state of stress at point P be given with respect to Cartesian axes Px1x2x3 shown inFig. 3.8 by the stress tensor T having components tij. We introduce a second set of axesPx ′1x

′2x′3, which is obtained from Px1x2x3 by a rotation of axes so that the transformation

matrix [aij] relating the two is a proper orthogonal matrix. Because T is a second-orderCartesian tensor, its components tij in the primed system are expressed in terms of theunprimed components by Eq 2.64 as


t ′ij = aiqtqmajm or T ′ = ATAT . (3.33)

The matrix formulation of Eq 3.33 is very convenient for computing the primed compo-nents of stress as demonstrated by the two following examples.

Example 3.2

Let the stress components (in MPa) at point P with respect to axes Px1x2x3be expressed by the matrix

[tij] =

1 3 2

3 1 0

2 0 −2

,

and let the primed axes Px ′1x′2x′3 be obtained by a 45o counterclockwise rotation

about the x3 axis. Determine the stress components t ′ij.

x3 ,

θ θP

x 3

x 2

x 1

x2x1

SolutionFor a positive rotation θ about x3 as shown by the sketch, the transformationmatrix [aij] has the general form

[aij] =

cos θ sin θ 0

− sin θ cos θ 0

0 0 1

.

Thus, from Eq 3.33 expressed in matrix form, a 45o rotation of axes requires

[tij] =

1/√2 1/

√2 0

−1/√2 1/

√2 0

0 0 1

1 3 2

3 1 0

2 0 −2

1/√2 −1/

√2 0

1/√2 1/

√2 0

0 0 1

=

4 0√2

0 −2 −√2√

2 −√2 −2

.


Example 3.3

Assume the stress tensor T (in ksi) at P with respect to axes Px1x2x3 isrepresented by the matrix

[tij] =

18 0 −12

0 6 0

−12 0 24

.

If the x ′1 axis makes equal angles with the three unprimed axes, and the x ′2axis lies in the plane of x ′1x3, as shown by the sketch, determine the primedcomponents of T assuming Px ′1x

′2x′3 is a right-handed system.

φ

β

β

P

βx 3

x1

x 1

x2

x 2

x3

SolutionWe must first determine the transformation matrix [aij]. Let β be the commonangle which x ′1 makes with the unprimed axes, as shown in the figure. Thena11 = a12 = a13 = cosβ and from the orthogonality condition Eq 2.55 withi = j = 1, cosβ = 1/

√3 . Next, let φ be the angle between x ′2and x3. Then

a23 = cosφ = sinβ = 2/√6. As seen from the obvious symmetry of the axes

arrangement, x ′2 makes equal angles with x1 and x2, which means that a21 =

a22. Thus, again from Eq 2.55, with i = 1, j = 2, we have a21 = a22 = −1/√6

(the minus sign is required because of the positive sign chosen for a23). For theprimed axes to be a right-handed system we require e ′3 = e ′1 × e ′2, with theresult that a31 = 1/

√2, a32 = −1/

√2, and a33 = 0. Finally, from Eq 3.33,

[t ′ij

]=

1√3

1√3

1√3

− 1√6

− 1√6

2√6

1√2

− 1√2

0

18 0 −12

0 6 0

−12 0 24

1√3

− 1√6

1√2

1√3

− 1√6

− 1√2

1√3

2√6

0

=

8 2

√2 0

2√2 28 −6

√3

0 −6√3 12

.


ni

P

P

∆S

ni

t(n)i

t(n)i

(a) Traction vector at point P for an arbitrary planewhose normal is ni.

P

P

∆S

ni∗

t(n)i

t(n)i

ni∗

(b) Traction vector at point P for a principal planewhose normal is n∗i .

FIGURE 3.9Traction vector and normal for a general continuum and a prismatic beam.

3.6 Principal Stresses; Principal Stress Directions

Let us turn our attention once more to the state of stress at point P and assume it is givenwith respect to axes Px1x2x3 by the stress tensor tij. As we saw in Example 3.1, for eachplane element of area ∆S at P having an outward normal ni, a stress vector t(n)

i is definedby Eq 3.32. In addition, as indicated by Fig. 3.9(a), this stress vector is not generally inthe direction of ni. However, for certain special directions at P, the stress vector doesindeed act in the direction of ni and may therefore be expressed as a scalar multiple ofthat normal. Thus, as shown in Fig. 3.9(b), for such directions

t(n)i = σni (3.34)

where σ is the scalar multiple of ni. Directions designated by ni for which Eq 3.34 isvalid are called principal stress directions, and the scalar σ is called a principal stress valueof tij. Also, the plane at P perpendicular to ni is referred to as a principal stress plane.We see from Fig. 3.9(b), that, because of the perpendicularity of t(n) to the principal planesthere are no shear stresses acting in these planes.

The determination of principal stress values and principal stress directions followsprecisely the same procedure developed in Section 2.6 for determining principal valuesand principal directions of any symmetric second-order tensor. In properly formulatingthe eigenvalue problem for the stress tensor we use the identity t(n)

i = tjinj and thesubstitution property of the Kronecker delta to rewrite Eq 3.34 as

tijnj − σδijnj = 0 (3.35)


or, in expanded form, using tij = tji,

(t11 − σ)n1 + t12n2 + t13n3 = 0 , (3.36a)t12n1 + (t22 − σ)n2 + t23n3 = 0 , (3.36b)t13n1 + t23n2 + (t33 − σ)n3 = 0 . (3.36c)

In the three linear homogeneous equations expressed by Eq 3.36, the tensor componentstij are assumed known; the unknowns are the three components of the principal normalni, and the corresponding principal stress σ. To complete the system of equations forthese four unknowns, we use the normalizing condition on the direction cosines,

nini = 1 . (3.37)

For non-trivial solutions of Eq 3.35 (the solution nj = 0 is not compatible with Eq 3.37),the determinant of coefficients on nj must vanish. That is,

|tij − δijσ| = 0 , (3.38)

which upon expansion yields a cubic equation in σ (called the characteristic equation of thestress tensor),

σ3 − ITσ2 + IITσ− IIIT = 0 (3.39)

whose roots σ(1), σ(2), σ(3) are the principal stress values of tij. The coefficients IT , IITand IIIT are known as the first, second, and third invariants, respectively, of tij and may beexpressed in terms of its components by

IT = tii = tr T , (3.40a)

IIT =1

2[tiitjj − tijtij] =

1

2

[(tr T )2 − tr T2

], (3.40b)

IIIT = εijkt1it2jt3k = det T . (3.40c)

Because the stress tensor tij is a symmetric tensor having real components, the threestress invariants are real, and likewise, the principal stresses being roots of Eq 3.39 arealso real. To show this, we recall from the theory of equations that for a cubic with realcoefficients at least one root is real, call it σ(1), and let the associated principal direction bedesignated by n(1)

i . Introduce a second set of Cartesian axes Px ′1x′2x′3 so that x ′1 is in the

direction of n(1)i . In this system the shear stresses, t ′12 = t ′13 = 0, so that the characteristic

equation of t ′ij relative to these axes results from the expansion of the determinant∣∣∣∣∣∣∣σ(1) − σ 0 0

0 t ′22 − σ t ′230 t ′23 t ′33 − σ

∣∣∣∣∣∣∣ = 0 , (3.41)

or [σ(1) − σ

] [σ2 − (t ′22 + t ′33)σ+ t ′22t

′33 − (t ′23)

2]

= 0 . (3.42)

From this equation, the remaining two principal stresses σ(2) and σ(3) are roots of thequadratic in brackets.


But the discriminant of this quadratic is

(t ′22 + t ′33)2

− 4[t ′22t

′33 − (t ′23)

2]

= (t ′22 − t ′33)2

+ 4 (t ′23)2,

which is clearly positive, indicating that both σ(2) and σ(3) are real.If the principal stress values σ(1), σ(2), and σ(3) are distinct, the principal directions

associated with these stresses are mutually orthogonal. To see why this is true, let n(1)i

and n(2)i be the normalized principal direction vectors (eigenvectors) corresponding to

σ(1) and σ(2), respectively. Then, from Eq 3.35, σijn(1)j = σ(1)nj and σijn

(2)j = σ(2)nj,

which, upon forming the inner products, that is, multiplying in turn by n(2)i and n(1)

i ,become

tijn(1)j n

(2)i = σ(1)n

(1)i n

(2)i , (3.43a)

tijn(2)j n

(1)i = σ(2)n

(2)i n

(1)i . (3.43b)

Furthermore, because the stress tensor is symmetric, and since i and j are dummy indices,

tijn(1)j n

(2)i = tjin

(1)i n

(2)j = tijn

(1)i n

(2)j ,

so that by the subtraction of Eq 3.43b from Eq 3.43a, the left hand side of the resultingdifference is zero, or

0 =[σ(1) − σ(2)

]n

(1)i n

(2)i . (3.44)

But since we assumed that the principal stresses were distinct, or σ(1) 6= σ(2) , it followsthat

n(1)i n

(2)i = 0 (3.45)

which expresses orthogonality between n(1)i and n

(2)i . By similar arguments, we may

show that n(3)i is perpendicular to both n(1)

i and n(2)i .

If two principal stress values happen to be equal, say σ(1) = σ(2), the principal directionn

(3)i associated with σ(3) will still be unique, and because of the linearity of Eq 3.35, any

direction in the plane perpendicular to n(3)i may serve as a principal direction. Accord-

ingly, we may determine n(3)i uniquely and then choose n(1)

i and n(2)i so as to establish a

right-handed system of principal axes. If it happens that all three principal stresses areequal, any direction may be taken as a principal direction, and as a result every set ofright-handed Cartesian axes at P constitutes a set of principal axes in this case.

We give the coordinate axes in the principal stress directions special status by labelingthem Px∗1x

∗2x∗3, as shown in Fig. 3.10. Thus, for example, σ(1) acts on the plane perpendic-

ular to x∗1 and is positive (tension) if it acts in the positive direction, negative (compres-sion) if it acts in the negative x∗1 direction. Also, if n(q)

i is the unit normal conjugate to theprincipal stress σ(q) (q = 1, 2, 3), the transformation matrix relating the principal stressaxes to arbitrary axes Px1x2x3 has elements defined by aqj ≡ n(q)

j , as indicated by Table3.1. Accordingly, the transformation equation expressing principal stress components interms of arbitrary stresses at P is given by Eq 2.77 in the form

t∗ij = aiqajmtqm or T∗ = ATAT . (3.46)

In addition, notice that Eq 3.35 is satisfied by n(q)i and σ(q) so that

tijn(q)j = σ(q)n

(q)i (3.47)


cos−1 n(1)3

cos−1 n(1)2

cos−1 n(1)1

x3

x2

x1

x2∗

P

x3∗

x1∗

FIGURE 3.10Principal axes Px∗1x

∗2x∗3.

TABLE 3.1Table displaying direction cosines of principal axesPx∗1x

∗2x∗3 relative to axes Px1x2x3.

x1 x2 x3

x∗1 a11 = n(1)1 a12 = n

(1)2 a13 = n

(1)3

x∗2 a21 = n(2)1 a22 = n

(2)2 a23 = n

(2)3

x∗3 a31 = n(3)1 a32 = n

(3)2 a33 = n

(3)3

for (q = 1, 2, 3), which upon introducing the identity aqi ≡ n(q)i becomes tijaqj =

σ(q)aqi. Now, multiplying each side of this equation by ami and using the symmetryproperty of the stress tensor, we have

tjiaqjami = σ(q)aqiami .

The left-hand side of this expression is simply t∗qm, from Eq 3.46. Since, by orthogonality,aqiami on the right hand side, the final result is

t∗qm = δqmσ(q) , (3.48)

which demonstrates that when referred to principal axes, the stress tensor is a diagonaltensor with principal stress values on the main diagonal. In matrix form, therefore,

[t∗ij]

=

σ(1) 0 0

0 σ(2) 0

0 0 σ(3)

or[t∗ij]

=

σI 0 0

0 σII 0

0 0 σIII

(3.49)

where in the second equation the notation serves to indicate that the principal stresses areordered, σI > σII > σIII, with positive stresses considered greater than negative stressesregardless of numerical values. In terms of the principal stresses, the stress invariants


may be written

IT = σ(1) + σ(2) + σ(3) = σI + σII + σIII , (3.50a)IIT = σ(1)σ(2) + σ(2)σ(3) + σ(3)σ(1) = σIσII + σIIσIII + σIIIσI , (3.50b)IIIT = σ(1)σ(2)σ(3) = σIσIIσIII . (3.50c)

Example 3.4

The components of the stress tensor at P are given in MPa with respect toaxes Px1x2x3 by the matrix

[tij] =

57 0 24

0 50 0

24 0 43

.

Determine the principal stresses and the principal stress directions at P.

SolutionFor the given stress tensor, Eq 3.38 takes the form of the determinant∣∣∣∣∣∣

57− σ 0 24

0 50− σ 0

24 0 43− σ

∣∣∣∣∣∣ = 0

which, upon cofactor expansion about the first row, results in the equation

(57− σ)(50− σ)(43− σ) − (24)2(50− σ) = 0 ,

or in its readily factored form

(50− σ)(σ− 25)(σ− 75) = 0 .

Hence, the principal stress values are σ(1) = 25 MPa, σ(2) = 50 MPa, andσ(3) = 75 MPa. Note that, in keeping with Eqs 3.40a and 3.50a, we confirmthat the first stress invariant,

IT = 57+ 50+ 43 = 25+ 50+ 75 = 150 MPa .

To determine the principal directions we first consider σ(1) = 25 MPa, forwhich Eq 3.36 provides three equations for the direction cosines of the principaldirection of σ(1), namely,

32n(1)1 + 24n

(1)3 = 0 ,

25n(1)2 = 0 ,

24n(1)1 + 18n

(1)3 = 0 .

Obviously, n(1)2 = 0 from the second of these equations, and from the other two,

n(1)3 = −43n

(1)1 so that, from the normalizing condition, nini = 1, we see that(

n(1)1

)2= 925 which gives n(1)

1 = ±35 and n(1)3 = ∓45 . The fact that the first


and third equations result in the same relationship is the reason the normalizingcondition must be used.

Next for σ(2) = 50 MPa, Eq 3.36 gives

7n(2)1 + 24n

(2)3 = 0 ,

24n(2)1 − 7n

(2)3 = 0

which are satisfied only when n(2)1 = n

(2)3 = 0. Then from the normalizing

condition, nini = 1, n(2)2 = ±1.

Finally, for σ(3) = 75 MPa, Eq 3.36 gives

− 18n(3)1 + 24n

(3)3 = 0 ,

− 25n(3)2 = 0

as well as24n

(3)1 − 32n

(3)3 = 0 .

Here, from the second equation n(3)2 = 0, and from either of the other two

equations 4n(3)3 = 3n

(3)1 , so that from nini = 1 we have n(3)

1 = ±45 and n(3)3 =

±35 .From these values of n(q)

i , we now construct the transformation matrix [aij]in accordance with Table 3.1, keeping in mind that to assure a right-handed sys-tem of principal axes we must have n(3) = n(1)× n(2). Thus the transformationmatrix has the general form

[aij] =

±35 0 ∓450 ±1 0

±45 0 ±35

.

Therefore, from Eq 3.48, when the upper signs in the above matrix are used,

[tij] =

35 0 −45

0 1 0

45 0 3

5

57 0 24

0 50 0

24 0 43

35 0 −45

0 1 0

45 0 3

5

=

25 0 0

0 50 0

0 0 75

MPa .

3.7 Maximum and Minimum Stress ValuesThe stress vector t(n)

i on an arbitrary plane at Pmay be resolved into a component normalto the plane having a magnitude σN, along with a shear component which acts in theplane and has a magnitude σS, as shown in Fig. 3.11. (Here, σN and σS are not vectors,but scalar magnitudes of vector components. The subscripts N and S are to be taken aspart of the component symbols). Clearly, from Fig. 3.11, it is seen that σN is given by thedot product, σN = t

(n)i ni, and inasmuch as t(n)

i = tijnj, it follows that


σN

σS

ni

∆S

P

t(n)i

FIGURE 3.11Traction vector components normal and in-plane (shear) at point P on the plane whosenormal is ni.

σN = tijninj or σN = t(n) · n . (3.51)

Also, from the geometry of the decomposition, we see that

σ2S = t(n)i t

(n)i − σ2N or σ2S = t(n) · t(n) − σ2N . (3.52)

In seeking the maximum and minimum (the so-called extremal) values of the abovecomponents, let us consider first σN. As the normal ni assumes all possible orientationsat P, the values of σN will be prescribed by the functional relation in Eq 3.51 subjectto the condition that nini = 1. Accordingly, we may use to advantage the Lagrangianmultiplier method to obtain extremal values of σN. To do so we construct the functionf(ni) = tijninj − σ(nini − 1), where the scalar σ is called the Lagrangian multiplier. Themethod requires the derivative of f(ni) with respect to nk to vanish; and, noting that∂ni/∂nk = δik, we have

∂f

∂nk= tij (δiknj + δjkni) − σ (2niδik) = 0 .

But tij = tji, and δkjnj = nk, so that this equation reduces to

(tij − σδij)nj = 0 (3.53)

which is identical to Eq 3.35, the eigenvalue formulation for principal stresses. Therefore,we conclude that the Lagrangian multiplier σ assumes the role of a principal stress and,furthermore, that the principal stresses include both the maximum and minimum normalstress values.

With regard to the maximum and minimum values of the shear component σS, it isuseful to refer the state of stress to principal axes Px∗1x

∗2x∗3, as shown in Fig. 3.12. Let the

principal stresses be ordered in the sequence σI > σII > σIII so that t(n)i is expressed in

vector form by


P

∆S

σS

σN

x2

∗

x3

∗

x1

∗

t(n)i

FIGURE 3.12Normal and shear components at P to plane referred to principal axes.

t(n) = T · n = σIn1e∗1 + σIIn2e

∗2 + σIIIn3e

∗3 , (3.54)

and similarly, σN = t(n) · n by

σN = σIn21 + σIIn

22 + σIIIn

23 . (3.55)

Then, substituting Eqs 3.54 and 3.55 into Eq 3.52, we have

σ2S = σ2In21 + σ2IIn

22 + σ2IIIn

23 − (σIn

21 + σIIn

22 + σIIIn

23) (3.56)

which expresses σ2S in terms of the direction cosines ni. But nini = 1, so that n23 =1−n21 −n22 and we are able to eliminate n3 from Eq 3.56, which then becomes a functionof n1 and n2 only,

σ2S = (σI − σIII)n21 +

(σ2II − σ2III

)n22 + σ2III

−[(σI − σIII)n

21 + (σII − σIII)n

22 + σIII

]2. (3.57)

In order to obtain the stationary, that is, the extremal values of σ2S, we must equate thederivatives of the right-hand side of this equation with respect to both n1 and n2 to zero,and solve simultaneously. After some algebraic manipulations, we obtain

∂σ2S∂n1

= n1σI − σIIIσI − σIII − 2

[σI − σIIIn

21 + σII − σIIIn

22

]= 0 , (3.58a)

∂σ2S∂n2

= n2σII − σIIIσII − σIII − 2

[σI − σIIIn

21 + σII − σIIIn

22

]= 0 . (3.58b)

An obvious solution to Eq 3.58 is n1 = n2 = 0 for which n3 = ±1, and the corre-sponding value of σ2S is observed from Eq 3.57 to be zero. This is an expected result sincen3 = ±1 designates a principal plane upon which the shear is zero. A similar calcula-tion made with n1 and n3, or with n2 and n3 as the variables, would lead to the othertwo principal planes as minimum (zero) shear stress planes. It is easily verified that asecond solution to Eq 3.58 is obtained by taking n1 = 0 and solving for n2. The result is


n2 = ±1/√2 and, from orthogonality, n3 = ±

√2 also. For this solution, Eq 3.57 yields

the results

σ2S =1

4(σII − σIII)

2 or σS = ±12

(σII − σIII) . (3.59)

As before, if we consider in turn the formulation having n1 and n3, or n2 and n3 asthe variable pairs, and assume n3 = 0, and n2 = 0, respectively, we obtain the completesolution which is presented below,

n1 = 0, n2 = ± 1√2, n3 = ± 1√

2; σS =

1

2|σII − σIII| , (3.60a)

n1 = ± 1√2, n2 = 0, n3 = ± 1√

2; σS =

1

2|σIII − σI| , (3.60b)

n1 = ± 1√2, n2 = ± 1√

2, n3 = 0; σS =

1

2|σI − σII| (3.60c)

where the vertical bars in the formulas for σS indicate absolute values of the enclosedexpressions. Because σI > σII > σIII, it is clear that the largest shear stress value is

σmaxS =1

2|σIII − σI| . (3.61)

It may be shown that, for distinct principal stresses, only the two solutions presented inthis section satisfy Eq 3.58.

3.8 Mohr’s Circles for StressMohr’s circle is a method the student should remember from undergraduate mechanicsof materials. It provides a graphic means for the transformation of a second order tensorlike that discussed in Section 2.5. Mohr’s circle is commonly used to transform stress, areaor mass moment of inertia tensors. While this book gives the student the mathematicaltools to transform tensors, we feel a discussion of Mohr’s circle is worthy of inclusionin this introduction to continuum mechanics. Most of the students prior use of Mohr’scircle has likely been with two dimensional problems: plane stress or transformation ofthe area or mass inertia tensors. Mohr’s circle allows for an easy means to derive andcompute transformation and principal values in two dimensions.

In three dimensions, the convenience and simplicity of Mohr’s circle to transform stresscomponents to an arbitrary reference frame vanishes. When working with three dimen-sional problems the advantage of using the mathematics of continuum mechanics becomeevident. This is the case for many problems in mechanics. In two dimensions, intuitivediagrams and relatively simplified equations greet the engineer. Adding the third dimen-sions leaves most all engineers relying on the proper application of mathematical models.This is the case for mechanics of materials and dynamics; two topics where the studentmay have experienced this.

Consider again the state of stress at P referenced to principal axes (Fig. 3.12) and let theprincipal stresses be ordered according to σI > σII > σIII. As before, we may express σNand σS on any plane at P in terms of the components of the normal to that plane by the


O

σS

σNσIII

σI

σII

C1

C2

C3

FIGURE 3.13Typical Mohr’s circle for stress.

equations

σN = σIn21 + σIIn

22 + σIIIn

23 , (3.62a)

σ2N + σ2S = σ2In21 + σ2IIn

22 + σ2IIIn

23 (3.62b)

which, along with the condition

n21 + n22 + n23 = 1 ,

provide us with three equations for the three direction cosines n1, n2 and n3. Solvingthese equations, we obtain

n21 =(σN − σII) (σN − σIII) + σ2S

(σI − σII) (σI − σIII), (3.63a)

n22 =(σN − σIII) (σN − σI) + σ2S

(σII − σIII) (σII − σI), (3.63b)

n23 =(σN − σI) (σN − σII) + σ2S

(σIII − σI) (σIII − σII). (3.63c)

In these equations, σI, σII and σIII are known; σN and σS are functions of the directioncosines ni. Our intention here is to interpret these equations graphically by representingconjugate pairs of σN, σS values, which satisfy Eq 3.63, as a point in the stress plane havingσN as absicca and σS as ordinate (see Fig. 3.13).

To develop this graphical interpretation of the three-dimensional stress state in termsof σN and σS, we note that the denominator of Eq 3.63a is positive since both σI−σII > 0and σI − σIII > 0, and also that n21 > 0, all of which tells us that

(σN − σII)(σN − σIII) + σ2S > 0 . (3.64)


For the case where the equality sign holds, this equation may be rewritten, after somesimple algebraic manipulations, to read[

σN −1

2(σII + σIII)

]2+ σ2S =

[1

2(σII − σIII)

]2(3.65)

which is the equation of a circle in the σN, σS plane, with its center at the point 12 (σII +

σIII) on the σN axis, and having a radius 12 (σII−σIII). We label this circle C1 and displayit in Fig. 3.13. For the case in which the inequality sign holds for Eq 3.64, we observethat conjugate pairs of values of σN and σS which satisfy this relationship result in stresspoints having coordinates exterior to circle C1. Thus, combinations of σN and σS whichsatisfy Eq 3.63a lie on, or exterior to, circle C1 in Fig. 3.13.

Examining Eq 3.63b, we note that the denominator is negative since σII − σIII > 0, andσII − σI < 0. The direction cosines are real numbers, so that n22 > 0 and we have

(σN − σIII)(σN − σI) + σ2S 6 0 (3.66)

which for the case of the equality sign defines the circle[σN −

1

2(σI + σIII)

]2=

[1

2(σI − σIII)

]2(3.67)

in the σN, σS plane. This circle is labeled C2 in Fig. 3.13, and the stress points whichsatisfy the inequality of Eq 3.66 lie interior to it. Following the same general procedure,we rearrange Eq 3.63c into an expression from which we extract the equation of the thirdcircle, C3 in Fig. 3.13, namely,[

σN −1

2(σI + σII)

]2+ σ2S =

[1

2(σI − σII)

]2. (3.68)

Admissible stress points in the σN, σS plane lie on or exterior to this circle. The three circlesdefined above, and shown in Fig. 3.13, are called Mohr’s circles for stress. All possible pairsof values of σN and σS at P which satisfy Eq 3.63 lie on these circles or between the areasenclosed by them. Actually, in conformance with Fig. 3.11 (which is the physical basis forFig. 3.13), we see that the sign of the shear component is arbitrary so that only the tophalf of the circle diagram need be drawn, a practice we will occasionally follow hereafter.In addition, it is clear from the Mohr’s circles diagram that the maximum shear stressvalue at P is the radius of circle C2, which confirms the result presented in Eq 3.61.

In order to relate a typical stress point having coordinates σN and σS in the stress planeof Fig. 3.13 to the orientation of the area element ∆S (denoted by ni in Fig. 3.12) uponwhich the stress components σN and σS act, we consider a small spherical portion of thecontinuum body centered at P. As the unit normal ni assumes all possible directions atP, the point of intersection of its line of action with the sphere will move over the surfaceof the sphere. However, as seen from Eqs 3.55 and Eq 3.56 the values of σN and σS arefunctions of the squares of the direction cosines, and hence do not change for ni reflectedin the principal planes. Accordingly, we may restrict our attention to the first octant ofthe spherical body, as shown in Fig. 3.14(a). Let Q be the point of intersection of the lineof action of ni with the spherical surface ABC in Fig. 3.14(a) and note that

n = cosφ e∗1 + cosβ e∗2 + cos θ e∗3 . (3.69)

If n = e∗1 so that its intersection point Q coincides with A, σN = σI. Likewise, whenQ coincides with B, σN = σII, and with C, σN = σIII. In all three cases, σS will be zero.


β

θ

φ

A

B

C

ni

x2

∗

x3

∗

x1

∗

Q

Pe∗2

e∗3

e∗1

(a) Octant of small spherical portion of body to-gether with plane at P with normal ni referred toprincipal axes Ox∗1x

∗2x∗3.

σIII

σIσ

II

C1

C2

C3

O

σS

σN

abc

(b) Mohr’s stress semicircle for octant ofFig. 3.14(a).

FIGURE 3.14Typical Mohr’s circle representation.

In the Mohr’s circle diagram (Fig. 3.14(b)), these stress values are located at points a, band c, respectively. If now θ is set equal to π

2 and φ allowed to vary from zero to π2 (β

will concurrently go from π2 to zero), Q will move along the quarter-circle arc AB from A

to B. In the stress space of Fig. 3.14(b), the stress point q (the image point of Q) havingcoordinates σN and σS will simultaneously move along the semicircle of C3 from a to b.(Note that as Q moves 90o along AB in physical space, q moves 180o along the semicircle,joining a to b in stress space.) Similarly, when Q is located on the quarter circle BC, orCA of Fig. 3.14(a), point q will occupy a corresponding position on the semicircles of bcand ca, respectively, in Fig. 3.14(b).

Now let the angle φ be given some fixed value less than π2 , say φ = φ1, and imagine

that β and θ take on all values compatible with the movement of Q along the circlearc KD of Fig. 3.15(a). For this case, Eq 3.63a becomes (σN − σII)(σN − σIII) + σ2S =(σI − σII)(σI − σIII) cos2φ1, which may be cast into the standard form of a circle as

[σN −

1

2(σII + σIII)

]2+ σ2S

= (σI − σII)(σI − σIII) cos2φ1 +

[1

2(σII − σIII)

]2= R21 . (3.70)

This circle is seen to have its center coincident with that of circle C1 in stress space andto have a radius R1 indicated by Eq 3.70. Therefore, as Q moves on circle arc KD inFig. 3.15(a), the stress point q traces the circle arc kd shown in Fig. 3.15(b). (Notice thatif φ1 = π

2 so that cosφ1 = 0, R1 reduces to 12 (σII − σIII), the radius of circle C1). Next,

let β = β1 <π2 and then, as φ and θ range through all admissible values, point Q moves

along the circle arc EG of Fig. 3.15(a). For this case Eq 3.63b may be restructured into the


x2

∗

x3

∗

x1

∗ φ

φβ

βD

E

GK

A

B

C

ni

Q

P

e∗1

e∗2

e∗3

(a) Reference angles φ and β for intersection pointQ on surface of body octant.

k

ge

d

q

c abOσ

IIIσ

IσII

σS

σN

2β 2β

2φ

2φ

(b) Mohr’s stress semicircle for octant ofFig. 3.15(a).

FIGURE 3.15Typical 3-D Mohr’s circle and associated geometry.

form [σN −

1

2(σI + σIII)

]2+ σ2S

= (σII − σIII)(σII − σI) cos2 β1 +

[1

2(σI − σIII)

]2= R22 (3.71)

which defines a circle whose center is coincident with that of circle C2, and having aradius R2. Here, when β1 = π

2 , the radius R2 reduces to 12 (σI−σIII) , which is the radius

of circle C2. As Q moves on the circle arc EG of Fig. 3.15(a), the stress point q traces outthe circle arc eg in Fig. 3.15(b).

In summary, for a specific n at point P in the body, point Q, where the line of action ofn intersects the spherical octant of the body (Fig. 3.15(a)), is located at the common pointof circle arcs KD and EG, and at the same time, the corresponding stress point q (havingcoordinates σN and σS) is located at the intersection of circle arcs kd and eg in the stressplane of Fig. 3.15(b). The following example provides details of the procedure.

Example 3.5

The state of stress at point P is given in MPa with respect to axes Px1x2x3 bythe matrix

[tij] =

25 0 0

0 −30 −600 −60 5

.


(a) Determine the stress vector on the plane whose unit normal isn = 1

3 (2e1 + e2 + 2e3).

(b) Determine the normal stress component σN and shear component σS onthe same plane.

(c)Verify the results of part (b) by the Mohr’s circle construction of Fig. 3.15(b).

Solution

(a) Using Eq 3.32 in matrix form gives the stress vector t(n)i

t(n)1

t(n)2

t(n)3

= [tij][nj] =

25 0 0

0 −30 −60

0 −60 5

231313

=1

3

50

−150

−50

,

or

t(n) =1

3(50e1 − 150e2 − 50e3) .

(b) Making use of Eq 3.51, we can calculate σN conveniently from the matrixproduct [

23

13

23

]25 0 0

0 −30 −60

0 −60 5

231323

= σN

σN =100

9−150

9−100

9

so that σN = −1509 = −16.67 MPa. Note that the same result could havebeen obtained by the dot product

σN = t(n) · n =1

3(50e1 − 150e2 − 50e3) ·

1

3(2e1 + e2 + 2e3) .

The shear component σS is given by Eq 3.52, which for the values of σNand t(n)

i calculated above, results in the equation,

σ2S =2 500+ 22 500+ 2 500

9−22 500

81= 2 777 ,

or, finally,σS = 52.7 MPa .

(c) Using the procedure of Example 3.4 the student should verify that for thestress tensor tij given here the principal stress values are σI = 50 MPa,σII = 25 MPa, and σIII = −75 MPa. Also, the transformation matrixfrom axes Px1x2x3 to Px∗1x

∗2x∗3 is

[aij] =

0 −35

45

1 0 0

0 45

35

,


so that the components of n are given relative to the principal axes byn∗1

n∗2

n∗3

=

0 −35

45

1 0 0

0 45

35

231323

=

132323

.

Therefore, with respect to Fig. 3.14(a), φ = cos−1(1/3) = 70.53o; β = θ =cos−1(2/3) = 48.19o, so that—following the procedure outlined for constructionof Fig. 3.15(b)—we obtain the Mohr’s circle below, from which we may measurethe coordinates of the stress point q and confirm the values σN = −16.7 andσS = 52.7, both in MPa.

O

C3

C1

C2

142°96°96°

142°

k

c ab

d e

q(−16.7, 52.7)

σIII = −75 σΙ = 50σ

II = 25

σS

σN

3.9 Plane Stress

When one, and only one principal stress is zero, we have a state of plane stress for whichthe plane of the two nonzero principal stresses is the designated plane. This is an importantstate of stress because it represents the physical situation occurring at an unloaded pointon the bounding surface of a body under stress. The zero principal stress may be anyone of the three principal stresses as indicated by the corresponding Mohr’s circles ofFig. 3.16.

If the principal stresses are not ordered and the direction of the zero principal stressis arbitrarily chosen as x3, we have plane stress parallel to the x1x2-plane and the stressmatrix takes the form

[tij] =

t11 t12 0

t12 t22 0

0 0 0

, (3.72)


σIII

σI

σII

σS

σN

(a) σI = 0

σIII

σI

σII

σS

σN

(b) σII = 0

σIII

σI

σII

σS

σN

(c) σIII = 0

FIGURE 3.16Mohr’s circle for plane stress.


TABLE 3.2Transformation table for general planestress.

x1 x2 x3

x∗1 cos θ sin θ 0

x∗2 − sin θ cos θ 0

x∗3 0 0 1

or, with respect to principal axes, the form

[t∗ij] =

σ(1) 0 0

0 σ(2) 0

0 0 0

. (3.73)

The pictorial description of this plane stress situation is portrayed by the block elementof a continuum body shown in Fig. 3.17(a), and is sometimes represented by a singleMohr’s circle (Fig. 3.17(b)), the locus of which identifies stress points (having coordinatesσN and σS) for unit normals lying in the x1 − x2 plane only. The equation of the circle inFig. 3.17(b) is [

σN −t11 + t22

2

]2+ σ2S =

[t11 − t22

2

]2+ t212 (3.74)

from which the center of the circle is noted to be at σN = 12 (t11 + t22), σS = 0, and the

maximum shear stress in the x1x2-plane to be the radius of the circle, that is, the squareroot of the right-hand side of Eq 3.74. Points A and B on the circle represent the stressstates for area elements having unit normals e1 and e2, respectively. For an element ofarea having a unit normal in an arbitrary direction at point P, we must include the twodashed circles shown in Fig. 3.17(c) to completely specify the stress state.

With respect to axes Ox ′1x′2x′3 rotated by the angle θ about the x3 axis relative to

Ox1x2x3 as shown in Fig. 3.18, the transformation equations for plane stress in the x1x2-plane are given by the general tensor transformation formula, Eq 2.64. Using the table ofdirection cosines for this situation as listed in Table 3.2, we may express the primed stresscomponents in terms of the rotation angle θ and the unprimed components by

t ′11 =t11 + t22

2+t11 − t22

2cos 2θ+ t12 sin 2θ , (3.75a)

t ′22 =t11 + t22

2−t11 − t22

2cos 2θ− t12 sin 2θ , (3.75b)

t ′12 = −t11 − t22

2sin 2θ+ t12 cos 2θ . (3.75c)

In addition, if the principal axes of stress are chosen for the primed directions, it is easilyshown that the two nonzero principal stress values are given by

σ(1), σ(2) =t11 + t22

2±

√[t11 − t22

2

]2+ t212 (3.76)


x1

x2

x3

t12

t22

t21

t11

(a) Plane stress element having nonzero x1 andx2 components.

σI

σII

σS

σN

A(t11, t

12)

B(t22, t

12)

(b) Mohr’s circle for in-plane stress components.

(σS)max

(σS)max

σN

σS

σII

σI

σIII

(c) General Mohr’s circles for the plane stress element.Dashed lines represent out-of-plane Mohr’s circles. Note themaximum shear can occur out-of-plane.

FIGURE 3.17Mohr’s circle for plane stress.


x3 ,

θ θP

x 3

x 2

x 1

x2x1

FIGURE 3.18Representative rotation of axes for plane stress.

Example 3.6

A specimen is loaded with equal tensile and shear stresses. This case of planestress may be represented by the matrix

[tij] =

σ0 σ0 0

σ0 σ0 0

0 0 0

where σ0 is a constant stress. Determine the principal stress values and plot theMohr’s circles.

SolutionFor this stress state, the determinant Eq 3.38 is given by∣∣∣∣∣∣

σ0 − σ σ0 0

σ0 σ0 − σ 0

0 0 −σ

∣∣∣∣∣∣ = 0

which results in a cubic having roots (principal stress values) σ(1) = 2σ0, σ(2) =σ(3) = 0 (as may be readily verified by Eq 3.76) so that, in principal axes form,the stress matrix is

[t∗ij] =

2σ0 0 0

0 0 0

0 0 0

.

The Mohr’s circle diagram is shown below.Here, because of the double-zero root, one of the three Mohr’s circles degen-

erates into a point (the origin) and the other two circles coincide. Also, we notethat physically this is simply a one-dimensional tension in the x∗1 direction andthat the maximum shear stress values (shown by points A and B) occur on thex1 and x2 coordinate planes which make 45 with the principal x∗1 direction.


.

.

+

A

B

σS

σN

+σ0

−σ0

σII=σ

III=0 σ

I=2σο

Mohr’s circle for principal stresses σ1 = 2σo, σ2 = σ3 = 0.

3.10 Deviator and Spherical Stress StatesThe arithmetic mean of the normal stresses,

σM =1

3(t11 + t22 + t33) =

1

3tii (3.77)

is referred to as the mean normal stress. The state of stress having all three principal stressesequal (and therefore equal to σM) is called a spherical state of stress and is represented bythe diagonal matrix

[tij] =

σM 0 0

0 σM 0

0 0 σM

(3.78)

for which all directions are principal directions as explained in Section 3.6. The classicalphysical example for this is the stress in a fluid at rest which is termed hydrostatic stress,and for which σM = −p0, the static pressure.

Every state of stress tij may be decomposed into a spherical portion and a portion Sijknown as the deviator stress in accordance with the equation

tij = Sij + δijσM = Sij +1

3δijtkk (3.79)

where δij is the Kronecker delta. This equation may be solved for Sij, which then appearsin the symmetric matrix formS11 S12 S13

S12 S22 S23S13 S23 S33

=

t11 − σM t12 t13t12 t22 − σM t23t13 t23 t33 − σM

. (3.80)

Also from Eq 3.79, we notice immediately that the first invariant of the deviator stress is

Sii = tii −1

3δiitkk = 0 (3.81)


(since δii = 3), so that the characteristic equation for the deviator stress (analogous toEq 3.39 for tij) is

S3 + IISS− IIIS = 0 (3.82)

for which the deviator stress invariants are

IIS = −1

2SijSji = SISII + SIISIII + SIIISI , (3.83a)

IIIS = εijkS1iS2jS3k = SISIISIII . (3.83b)

Finally, consider a principal direction n(q)j of tij such that the eigenvalue equation[

tij − σ(q)δij]n

(q)j = 0 is satisfied. Then, from the definition of Sij, we have

[Sij + σMδij − σqδij]n(q)j = 0 ,

or[Sij − (σq − σM) δij]n

(q)j = 0 (3.84)

which demonstrates that n(q)j is also a principal direction of Sij, and furthermore, the

principal values of Sij are given in terms of the principal values of tij by

Sq = σq − σM , (q = 1, 2, 3) . (3.85)

Example 3.7

Decompose the stress tensor T of Example 3.4 into its deviator and sphericalportions and determine the principal stress values of the deviator portion.

[tij] =

57 0 24

0 50 0

24 0 43

[MPa] .

SolutionBy Eq 3.77, σM for the given stress is

σM =1

3(57+ 50+ 43) = 50 [MPa]

Thus, decomposition by Eq 3.79 leads to the matrix sum57 0 24

0 50 0

24 0 43

=

7 0 24

0 0 0

24 0 −7

+

50 0 0

0 50 0

0 0 50

[MPa] .

Principal stress values of the deviator portion result from the determinant∣∣∣∣∣∣7− S 0 24

0 −S 0

24 0 −7− S

∣∣∣∣∣∣ = −S[(7− S)(−7− S) − 242] = 0

which is readily factored to yield SI = 25 MPa, SII = 0 and SIII = −25 MPa.These results are easily verified using the principal values determined in Example3.4 together with Eq 3.85.


t(n)i

P

A

B

Cσ

N

σS

x2

∗

x3

∗

x1

∗

ni =e1 + e2 + e3√

3

FIGURE 3.19Octahedral plane (ABC) with traction vector t(n)

i , and octahedral normal and shearstresses, σN and σS.

3.11 Octahedral Shear StressConsider the plane at P whose unit normal makes equal angles with the principal stressdirections. That plane, called the octahedral plane, may be pictured as the triangular surfaceABC of Fig. 3.19 and imagined to be the face in the first octant of a regular octahedron.The traction vector on this plane is

t(n) = T∗ · n =σ(1)e

∗1 + σ(2)e

∗2 + σ(3)e

∗3√

3, (3.86)

and its component in the direction of n is

σN = t(n) · n =1

3

[σ(1) + σ(2) + σ(3)

]=1

3σii . (3.87)

Thus, from Eq 3.52, the square of the shear stress on the octahedral plane, known as theoctahedral shear stress, is

σ2oct = t(n) · t(n) − σ2N =1

3[σ2(1) + σ2(2) + σ2(3)] −

1

9[σ(1) + σ(2) + σ(3)]

2 (3.88)

which may be reduced to either the form (see Problem 3.27)

σoct =1

3

√(σ(1) − σ(2)

)2+(σ(2) − σ(3)

)2+(σ(3) − σ(1)

)2, (3.89)

or

σoct =

√S21 + S22 + S23√

3=

√−2IIS

3. (3.90)


Example 3.8

Determine directly the normal and shear components, σN and σoct, on theoctahedral plane for the state of stress in Example 3.4, and verify the result forσoct by Eq 3.89.

SolutionFrom Example 3.4, the stress vector on the octahedral plane is given by thematrix product

75 0 0

0 50 0

0 0 25

1√3

1√3

1√3

=

75√3

50√3

25√3

,

or

t(n) =75e∗1 + 50e∗2 + 25e∗3√

3,

so that

σN = t(n) · n =1

3(75+ 50+ 25) = 50 MPa .

Also, from Eq 3.52,

σ2oct = t(n) · t(n) − σ2N =1

3(752 + 502 + 252) − 502 = 417 MPa ,

and so σoct = 20.4 MPa. By Eq 3.89, we verify directly that

σoct =1

3

√(75− 50)2 + (50− 25)2 + (25− 75)2 = 20.41MPa

Example 3.9

Show that the octahedral shear stress may be written as

σoct =1√3

[(σ(1) − σM)2 + (σ(2) − σM)2 + (σ(3) − σM)2

]1/2,

and, from this, show that

σoct =

√2

3

(I2T + 3IIT

)1/2is another form of the octahedral shear stress.

SolutionStart by multiplying the squared terms under the radical

σoct =1√3

[σ2(1) + σ2(2) + σ2(3) + 3σ2M − 2

(σ(1) + σ(2) + σ(3)

)σM

]1/2.


Recall Eq 3.77 in the form

σ(1) + σ(2) + σ(3) = 3σM

and substitute to get

σ2oct =

(σ2(1) + σ2(2) + σ2(3)

3− σ2M

).

Substitution of Eq 3.77 again for σM results in Eq 3.88. Multiply out the meanstress terms and simplify to get

σ2oct =

[3(σ2(1) + σ2(2) + σ2(3))

9

−σ2(1) + σ2(2) + σ2(3) + 2(σ(1)σ(2) + σ(2)σ(3) + σ(1)σ(3))

9

]

= 29

[σ2(1) + σ2(2) + σ2(3) −

(σ(1)σ(2) + σ(2)σ(3) + σ(1)σ(3)

)].

Finally, add and subtract 2(σ(1)σ(2) + σ(2)σ(3) + σ(1)σ(3)

)in the bracket

σ2oct = 29

[σ2(1) + σ2(2) + σ2(3) + 2(σ(1)σ(2) + σ(2)σ(3) + σ(1)σ(3))

−3(σ(1)σ(2) + σ(2)σ(3) + σ(1)σ(3))],

and note that, for principal axes

I2T =(σ(1) + σ(2) + σ(3)

)2IIT = σ(1)σ(2) + σ(2)σ(3) + σ(1)σ(3)

leavingσ2oct = 2

9

[I2T + 3IIT

],

orσoct =

√23

[I2T + 3IIT

]1/2.


Problems

Problem 3.1At a point P, the stress tensor relative to axes Px1x2x3 has components tij. On the area

element dS(1) having the unit normal n1, the stress vector is t(n1), and on area elementdS(2) with normal n2 the stress vector is t(n2). Show that the component of t(n1) in thedirection of n2 is equal to the component of t(n2) in the direction of n1.

Problem 3.2Verify the result established in Problem 3.1 for the area elements having normals

n1 =1

7(2e1 + 3e2 + 6e3)

n2 =1

7(3e1 − 6e2 + 2e3)

if the stress matrix at P is given with respect to axes Px1x2x3 by

[tij] =

35 0 21

0 49 0

21 0 14

.

Problem 3.3The stress tensor at P relative to axes Px1x2x3 has components in MPa given by the matrixrepresentation

[tij] =

t11 2 1

2 0 2

1 2 0

where t11 is unspecified. Determine a direction n at P for which the plane perpendicular ton will be stress-free, that is, for which t(n) = 0 on that plane. What is the required valueof t11 for this condition?

Answer

n =1

3(2e1 − e2 − 2e3), t11 = 2 MPa

Problem 3.4The stress tensor has components at point P in ksi units as specified by the matrix

[tij] =

−9 3 −63 6 9

−6 9 −6

.


Determine:

(a) the stress vector on the plane at P whose normal vector is

n =1

9(e1 + 4e2 + 8e3) ,

(b) the magnitude of this stress vector,

(c) the component of the stress vector in the direction of the normal,

(d) the angle in degrees between the stress vector and the normal.

Answer

(a) t(n) = −5e1 + 11e2 − 2e3

(b) t(n) =√150

(c)23

9(d) 77.96

Problem 3.5Let the stress tensor components at a point be given by tij = ±σ0ninj where σ0 is a

positive constant. Show that this represents a uniaxial state of stress having a magnitude±σ0 and acting in the direction of ni.

Problem 3.6Show that the sum of squares of the magnitudes of the stress vectors on the coordinate

planes is independent of the orientation of the coordinate axes, that is, show that the sum

t(e1)i t

(e1)i + t

(e2)i t

(e2)i + t

(e3)i t

(e3)i

is an invariant.

x1

x2

x3

t(e2)i

t(e1)i

t(e3)i

Problem 3.7With respect to axes Ox1x2x3 the stress state is given in terms of the coordinates by the

matrix

[tij] =

x1x2 x22 0

x22 x2x3 x230 x23 x3x1

.


Determine

(a) the body force components as functions of the coordinates if the equilibriumequations are to be satisfied everywhere, and

(b) the stress vector at point P(1,2,3) on the plane whose outward unit normal makesequal angles with the positive coordinate axes.

Answer

(a) b1 =−3x2

ρ, b2 =

−3x3

ρ, b3 =

−x1

ρ

(b) t(n) =(6e1 + 19e2 + 12e3)√

3

Problem 3.8Relative to the Cartesian axes Ox1x2x3 a stress field is given by the matrix

[tij] =

(1− x21

)x2 +

2

3x32 −

(4− x22

)x1 0

−(4− x22

)x1 −

1

3

(x32 − 12x2

)0

0 0(3− x21

)x2

.(a) Show that the equilibrium equations are satisfied everywhere for zero body forces.(b) Determine the stress vector at the point P(2,-1,6) of the plane whose equation is

3x1 + 6x2 + 2x3 = 12.

Answer

(b) t(n) =1

7(−29e1 − 40e2 + 2e3)

Problem 3.9The stress components in a circular cylinder of length L and radius r are given by

[tij] =

Ax2 + Bx3 Cx3 −Cx2Cx3 0 0

−C2 0 0

.(a) Verify that in the absence of body forces the equilibrium equations are satisfied.(b) Show that the stress vector vanishes at all points on the curved surface of the

cylinder.

r

L

x1

x3

x2


Problem 3.10Rotated axes Px′1x

′2x′3 are obtained from axes Px1x2x3 by a right-handed rotation about the

line PQ that makes equal angles with respect to the Px1x2x3 axes (see sketch). Determinethe primed stress components for the stress tensor in (MPa)

[tij] =

3 0 6

0 0 0

6 0 −3

if the angle of rotation is (a) 120, or (b) 60.

Q

β

β

βP

x1 x

2

x3

Answer

(a)[t′ij

]=

0 0 0

0 −3 6

0 6 3

MPa, (b)[t′ij

]=1

3

−5 10 10

10 −11 −210 −2 16

MPa

Problem 3.11At the point P, rotated axes Px′1x

′2x′3 are related to the axes Px1x2x3 by the transformation

matrix

[aij] =1

3

a 1−√3 1+

√3

1+√3 b 1−

√3

1−√3 1+

√3 c

where a, b, and c are to be determined. Determine

[t′ij

]if the stress matrix relative to

axes Px1x2x3 is given in MPa by

[tij] =

1 0 1

0 1 0

1 0 1

.Answer[

t′ij

]=1

9

11+ 2√3 5+

√3 −1

5+√3 5 5−

√3

−1 5−√3 11− 2

√3

MPa

Problem 3.12The stress matrix referred to axes Px1x2x3 is given in ksi by

[tij] =

14 0 21

0 21 0

21 0 7


Let rotated axes Px′1x′2x′3 be defined with respect to axes Px1x2x3 by the table of base

vectorse1 e2 e3

e′1 2/7 3/7 6/7

e′2 3/7 −6/7 2/7

e′3 6/7 2/7 −3/7

(a) Determine the stress vectors on planes at P perpendicular to the primed axes;determine t(e

′1), t(e

′2), and t(e

′3) in terms of base vectors e1, e2, and e3.

(b) Project each of the stress vectors obtained in (a) onto the primed axes to deter-

mine the nine components of[t′ij

].

(c) Verify the result obtained in (b) by a direct application of Eq 3.33 of the text.

Answer

[t′ij

]=1

7

143 36 114

36 166 3

114 3 −15

ksi

Problem 3.13At point P, the stress matrix is given in MPa with respect to axes Px1x2x3 by

Case 1: [tij] =

6 4 0

4 6 0

0 0 −2

Case 2: [tij] =

2 1 1

1 2 1

1 1 2

Determine for each case

(a) the principal stress values,

(b) the principal stress directions.

Answer

(a) Case 1: σ(1) = 10 MPa, σ(2) = 2 MPa, σ(3) = −2 MPaCase 2: σ(1) = 4 MPa, σ(2) = σ(3) = 1 MPa

(b) Case 1: n(1) = ± e1 + e2√2

, n(2) = ± e1 − e2√2

, n(3) = ∓e3

Case 2: n(1) =e1 + e2 + e3√

3, n(2) =

−e1 + e2√2

, n(3) =−e1 − e2 + 2e3√

6

Problem 3.14When referred to principal axes at P, the stress matrix in ksi units is

[t∗ij]

=

2 0 0

0 7 0

0 0 12


If the transformation matrix between the principal axes and axes Px1x2x3 is

[aij] =1√2

−35 1 −45a21 a22 a23−35 −1 −45

where a21, a22, and a23 are to be determined, calculate [tij].

Answer

[tij] =

7 3 0

3 7 4

0 4 7

ksi

Problem 3.15The stress matrix in MPa when referred to axes Px1x2x3 is

[tij] =

3 −10 0

−10 0 30

0 30 −27

.Determine

(a) the principal stresses, σI, σII, σIII,

(b) the principal stress directions.

Answer

(a) σI = 23 MPa, σII = 0 MPa, σIII = −47 MPa(b) n(1) = −0.394e1 + 0.788e2 + 0.473e3

n(2) = 0.931e1 + 0.274e2 + 0.304e3n(3) = 0.110e1 + 0.551e2 − 0.827e3

Problem 3.16At point P, the stress matrix relative to axes Px1x2x3 is given in MPa by

[tij] =

5 a −aa 0 b

−a b 0

where a and b are unspecified. At the same point relative to axes Px∗1x

∗2x∗3 the matrix is

[t∗ij]

=

σI 0 0

0 2 0

0 0 σIII

.If the magnitude of the maximum shear stress at P is 5.5 MPa, determine σI and σIII.

Answer

σI = 7 MPa, σIII = −4 MPa


Problem 3.17The state of stress at point P is given in ksi with respect to axes Px1x2x3 by the matrix

[tij] =

1 0 2

0 1 0

2 0 −2

.Determine

(a) the principal stress values and principal stress directions at P,

(b) the maximum shear stress value at P,

(c) the normal n = niei to the plane at P on which the maximum shear stress acts.

Answer

(a) σI = 2 ksi, σII= 1 ksi, σIII = –3 ksi

n(1) =2e1 + e3√

5, n(2) = e2, n(3) =

−e1 + 2e3√5

(b) (σS)max = ± 2.5 ksi

(c) n =e1 + 3e3√

10

Problem 3.18The stress tensor at P is given with respect to Ox1x2x3 in matrix form with units of MPa

by

[tij] =

4 b b

b 7 2

b 2 4

where b is unspecified. If σIII = 3 MPa and σI = 2σII, determine

(a) the principal stress values,

(b) the value of b,

(c) the principal stress direction of σII .

Answer

(a) σI = 8 MPa, σII = 4 MPa, σIII = 3 MPa(b) b = 0, (c) n(2) = e1

Problem 3.19The state of stress at P, when referred to axes Px1x2x3 is given in ksi units by the matrix

[tij] =

9 3 0

3 9 0

0 0 18


Determine

(a) the principal stress values at P,

(b) the unit normal n∗ = nie∗i of the plane on which σN = 12 ksi and σS = 3 ksi.

Answer

(a) σI = 18 ksi, σII = 12 ksi, σIII = 6 ksi

(b) n∗ =e∗1 +

√6e∗2 + e∗3

2√2

Problem 3.20

Verify the result listed for Problem 3.19b above by use of Eq 3.63.

Problem 3.21

Sketch the Mohr’s circles for the various stress states shown on the cube which is orientedalong the coordinate axes.

σ0

σ0

σ0

σ0

(a)

σ0

σ0

(b)

σ0

σ0

σ0

σ0

(c)

2σ0

σ0

2σ0

σ0

(d)


Problem 3.22The state of stress referred to axes Px1x2x3 is given in MPa by the matrix

[tij] =

9 12 0

12 −9 0

0 0 5

.Determine

(a) the normal and shear components, σN and σS, respectively, on the plane at Pwhose unit normal is

n =1

5(4e1 + 3e2) ,

(b) verify the result determined in (a) by a Mohr’s circle construction similar to thatshown in the figure from Example 3.5.

Answer

σN = 14.04 MPa, σS = 5.28 MPa

Problem 3.23Sketch the Mohr’s circles for the simple states of stress given by

(a) [tij] =

σ0 0 σ00 σ0 0

σ0 0 σ0

,(b) [tij] =

σ0 0 0

0 2σ0 0

0 0 −σ0

,and determine the maximum shear stress in each case.

Answer

(a) (σS)max = σ0

(b) (σS)max = 32σ0

Problem 3.24Relative to axes Ox1x2x3, the state of stress at O is represented by the matrix

[tij] =

6 −3 0

−3 6 0

0 0 0

[ksi] .

Show that, relative to principal axes Ox∗1x∗2x∗3, the stress matrix is

[tij] =

3 0 0

0 9 0

0 0 0

[ksi] ,


and that these axes result from a rotation of 45 about the x3 axis. Verify these results byEq 3.75.

Problem 3.25The stress matrix representation at P is given by

[tij] =

29 0 0

0 −26 6

0 6 9

[ksi] .

Decompose this matrix into its spherical and deviator parts, and determine the principaldeviator stress values.

Answer

SI = 25 ksi, SII = 6 ksi, SIII = −31 ksi

Problem 3.26Let the second invariant of the stress deviator be expressed in terms of its principal values,

that is, byIIS = SISII + SIISIII + SIIISI .

Show that this sum is the negative of two-thirds the sum of squares of the principal shearstresses, as given by Eq 3.60.

Problem 3.27Verify the results presented in Eqs 3.89 and 3.90 for the octahedral shear stress.

Problem 3.28At point P in a continuum body, the stress tensor components are given in MPa with

respect to axes Px1x2x3 by the matrix

[tij] =

1 −3√2

−3 1 −√2√

2 −√2 4

.Determine

(a) the principal stress values σI, σII, and σIII, together with the corresponding prin-cipal stress directions,

(b) the stress invariants IT , IIT , and IIIT ,

(c) the maximum shear stress value and the normal to the plane on which it acts,

(d) the principal deviator stress values,

(e) the stress vector on the octahedral plane together with its normal and shearcomponents,

(f) the stress matrix for axes rotated 60 counterclockwise with respect to the axisPQ, which makes equal angles relative to the coordinate axes Px1x2x3.


Answer

(a) σI = 6 MPa, σII = 2 MPa, σIII = −2 MPan(1) = 1

2

(e1 − e2 +

√2e3

)n(2) = 1

2

(e1 − e2 −

√2e3

)n(3) =

e1 + e2√2

(b) IT = 6 MPa, IIT = −4 MPa, IIIT = −24 MPa

(c) (σS)max = 4 MPa nmax =

(1+√2)e1 −

(1−√2)e2 +

√2e3

2√2

(d) SI = 4 MPa, SII = 0 MPa, SIII = −4 MPa

(e) t(n) =6e∗1 + 2e∗2 − 2e∗3√

3, σN = 2, σoct =

√323

(f)[t′ij

]=1

9

−12 −12+ 3√2 −12− 3

√2

−12+ 3√2 33− 12

√2 −3

−12− 3√2 −3 33+ 12

√2

MPa

Problem 3.29In a continuum, the stress field relative to axes Ox1x2x3 is given by

[tij] =

x21x2 x1

(1− x22

)0

x1(1− x22

)13

(x22 − 3x2

)0

0 0 2x23

.Determine

(a) the body force distribution if the equilibrium equations are to be satisfied through-out the field,

(b) the principal stresses at P(a, 0, 2√a),

(c) the maximum shear stress at P,(d) the principal deviator stresses at P.

Answer

(a) b1 = b2 = 0, b3 = −4x3

ρ

(b) σI = 8a, σII = a, σIII = −a

(c) (σS)max = ±4.5a

(d) SI =16

3a, SII = −

5

3a, SIII = −

11

3a

Problem 3.30In describing the yield surface in plasticity the second invariant of the deviator stress, oftendenoted by J2, plays an important role. Starting with the second invariant of the deviatorstress

J2 = IIS = −(S(1)S(2) + S(2)S(3) + S(1)S(3)

)


derive the formulaJ2 =

3

2σ2oct .

Problem 3.31Show that

J2 = IIS = −(S(1)S(2) + S(2)S(3) + S(1)S(3)) =1

2

(S2(1) + S2(2) + S2(3)

)where J2 is the second invariant of the deviator stress and S(1), S(2), S(3) are its principalvalues.

Problem 3.32Let the stress tensor components tij be derivable from the symmetric tensor field ϕij by theequation tij = εiqkεjpmϕkm,qp. Show that, in the absence of body forces, the equilibriumequations are satisfied. Recall from Problem 2.15 that

εiqkεjpm =

∣∣∣∣∣∣δji δjq δjkδpi δpq δpkδmi δmq δmk

∣∣∣∣∣∣ .

Problem 3.33Verify that ∂tij/∂tmn = δimδjn and use this result (or otherwise) to show that

∂IIS

∂tij= −Sij .

that is, the derivative of the second invariant of the deviatoric stress with respect to thestress components is equal to the negative of the corresponding component of the deviatoricstress.


4Kinematics of Deformation and Motion

Kinematics is the study of motion. In Continuum Mechanics it has the same meaning as indynamics or fluid mechanics. It is the basis for most all constitutive theories. Recall fromChapter 1 that a constitutive response, or material model, describes how stress developsaccording to strain or strain rate. Strain and strain rate are a measures of kinematics.

4.1 Particles, Configurations, Deformations and Motion

In continuum mechanics we consider material bodies in the form of solids, liquids andgases. Let us begin by describing the model we use to represent such bodies. For thispurpose we define a material body B as the set of elements X, called particles, or materialpoints, which can be put into a one-to-one correspondence with the points of a regularregion of physical space. Note that whereas a particle of classical mechanics has anassigned mass, a continuum particle is essentially a material point for which a density isdefined.

The specification of the position of all of the particles of B with respect to a fixedorigin at some instant of time is said to define the configuration of the body at that instant.Mathematically, this is expressed by the mapping

x = κ(X) (4.1)

in which the vector function κ assigns the position x relative to some origin of eachparticle X of the body. Assume that this mapping is uniquely invertible and differentiableas many times as required; in general two or three times will suffice. The inverse iswritten

X = κ−1(x) (4.2)

and identifies the particle X located at position x.A change in configuration is the result of a displacement of the body. For example, a

rigid-body displacement is one consisting of a simultaneous translation and rotation whichproduces a new configuration but causes no changes in the size or shape of the body; onlychanges in its position and/or orientation. On the other hand, an arbitrary displacementwill usually include both a rigid-body displacement and a deformation which results in achange in size, or shape, or possibly both.

A motion of body B is a continuous time sequence of displacements that carries the setof particles X into various configurations in a stationary space. Such a motion may beexpressed by the equation

x = κ(X, t) (4.3)

103


Reference

Current

x3, X

3

x2, X

2

x1, X

1

O

P

p

u

e1, I1 e2, I2

e3, I3

x = κΦ−1 (X) , t

= χ (X, t)

X = Φ(X)

FIGURE 4.1Position of typical particle in reference configuration XA and current configuration xi.

which gives the position x for each particle X for all times t, where t ranges from −∞ to∞. As with configuration mappings, we assume the motion function in Eq 4.3 is uniquelyinvertible and differentiable, so that we may write the inverse

X = κ−1(x, t) (4.4)

which identifies the particle X located at position x at time t.We give special meaning to certain configurations of the body. In particular, we single

out a reference configuration from which all displacements are reckoned. For the purposeit serves, the reference configuration needs not be one the body ever actually occupies.Often however, the initial configuration, that is, the one which the body occupies at timet = 0, is chosen as the reference configuration, and the ensuing deformations and motionsrelated to it. The current configuration is that one which the body occupies at the currenttime t.

In developing the concepts of strain we confine attention to two specific configurationswithout any regard for the sequence by which the second configuration is reached fromthe first. It is customary to call the first (reference) state the undeformed configuration, andthe second state the deformed configuration. Additionally, time is not a factor in derivingthe various strain tensors so that both configurations are considered independent of time.

In fluid mechanics the idea of specific configurations has very little meaning since fluidsdo not possess a natural geometry, and because of this it is the velocity field of a fluid thatassumes the fundamental kinematic role.

4.2 Material and Spatial Coordinates

Consider now the reference configuration prescribed by some mapping function Φ suchthat the position vector X of particle X relative to the axes OX1X2X3 of Fig. 4.1 is given by

Kinematics of Deformation and Motion 105

X =Φ(X) (4.5)

In this case we may express X in terms of the base vectors I1, I2, I3 shown in Fig. 4.1 bythe equation

X = XAIA , (4.6)

and we call the components XA the material coordinates, or sometimes the referential co-ordinates, of the particle X. Uppercase letters which are used as subscripts on materialcoordinates, or on any quantity expressed in terms of material coordinates, obey all therules of indicial notation. It is customary to designate the material coordinates (that is,the position vector X) of each particle as the name, or label of that particle, so that in allsubsequent configurations every particle can be identified by the position X it occupiedin the reference configuration. As usual, we assume an inverse mapping

X = Φ−1(X) , (4.7)

so that upon substitution of Eq 4.7 into Eq 4.3 we obtain

x = κ[Φ−1(X), t

]= χ (X, t) (4.8)

which defines the motion of the body in physical space relative to the reference configu-ration prescribed by the mapping function Φ.

Notice that Eq 4.8 maps the particle at X in the reference configuration onto the pointx in the current configuration at time t as indicated in Fig. 4.1. With respect to the usualCartesian axes Ox1x2x3 the current position vector is

x = xiei (4.9)

where the components xi are called the spatial coordinates of the particle. Although it isnot necessary to superpose the material and spatial coordinate axes as we have done inFig. 4.1, it is convenient to do so, and there are no serious restrictions from this practicein the derivations which follow. We emphasize, however, that the material coordinatesare used in conjunction with the reference configuration only, and the spatial coordi-nates serve for all other configurations. As already remarked the material coordinates aretherefore time independent.

We may express Eq 4.8 in either a Cartesian component or a coordinate-free notationby the equivalent equations

xi = χi(XA, t) or x = χ(X, t) . (4.10)

It is common practice in continuum mechanics to write these equations in the alternativeforms

xi = xi(XA, t) or x = x(X, t) (4.11)

with the understanding that the symbol xi (or x) on the right hand side of the equationrepresents the function whose arguments are X and t, while the same symbol on the left-hand side represents the value of the function, that is, a point in space. We shall use thisnotation frequently in the text that follows.

Notice that as X ranges over its assigned values corresponding to the reference config-uration, while t simultaneously varies over some designated interval of time, the vectorfunction χ gives the spatial position x occupied at any instant of time for every particleof the body. At a specific time, say at t = t1, the function χ defines the configuration


x1 = χ(X, t1) . (4.12)

In particular, at t = 0, Eq 4.10 defines the initial configuration which is often adoptedas the reference configuration, and this results in the initial spatial coordinates beingidentical in value with the material coordinates, so that in this case

x = χ(X, 0) = X (4.13)

at time t = 0.If we focus attention on a specific particle XP having the material position vector XP ,

Eq 4.10 takes the formxP = χ(XP, t) (4.14)

and describes the path, or trajectory of that particle as a function of time. The velocity vP

of the particle along its path is defined as the time rate of change of position, or

vP =dxP

dt= χP =

(∂χ

∂t

)X=XP

(4.15)

where the notation in the last form indicates that the variable X is held constant in takingthe partial derivative of χ. Also, as is standard practice, the super-positioned dot has beenintroduced to denote differentiation with respect to time. In an obvious generalizationwe may define the velocity field of the total body as the derivative

v = x =dx

dt=∂χ(X, t)

∂t=∂x(X, t)

∂t. (4.16)

Similarly, the acceleration field is given by

a = v = x =d2x

dt2=∂2χ(X, t)

∂t2, (4.17)

and the acceleration of any particular particle determined by substituting its materialcoordinates into Eq 4.17.

Of course, the individual particles of a body cannot execute arbitrary motions inde-pendent of one another. In particular, no two particles can occupy the same location inspace at a given time (the axiom of impenetrability), and furthermore, in the smooth mo-tions we consider here, any two particles arbitrarily close in the reference configurationremain arbitrarily close in all other configurations. For these reasons, the function χ inEq 4.10 must be single-valued and continuous, and must possess continuous derivativeswith respect to space and time to whatever order is required; usually to the second orthird. Moreover, we require the inverse function χ−1 in the equation

X = χ−1(x, t) (4.18)

to be endowed with the same properties as χ. Conceptually, Eq 4.18 allows us to “reverse”the motion and trace backwards to discover where the particle, now at x, was located inthe reference configuration. The mathematical condition that guarantees the existence ofsuch an inverse function is the non-vanishing of the Jacobian determinant J. That is, forthe equation

J =

∣∣∣∣ ∂χi∂XA

∣∣∣∣ 6= 0 (4.19)

to be valid. This determinant may also be written as

J =

∣∣∣∣ ∂xi∂XA

∣∣∣∣ . (4.20)


Example 4.1

Let the motion of a body be given by Eq 4.10 in component form as

x1 = X1 + t2X2 ,

x2 = X2 + t2X1 ,

x3 = X3 .

Determine

(a) the path of the particle originally at X = (1, 2, 1) and

(b) the velocity and acceleration components of the same particle whent = 2 s.

Solution

(a) For the particle X = (1, 2, 1) the motion equations are

x1 = 1+ 2t2; x2 = 2+ t2; x3 = 1

which upon elimination of the variable t gives x1 − 2x2 = −3 as wellas x3 = 1 so that the particle under consideration moves on a straightline path in the plane x3 = 1.

(b) By Eqs 4.16 and 4.17 the velocity and acceleration fields are given incomponent form, respectively, by

v1 = 2tX2 a1 = 2X2v2 = 2tX1 and a2 = 2X1v3 = 0 a3 = 0 ,

so that for the particle X = (1, 2, 1) at t = 2

v1 = 8 a1 = 4

v2 = 4 and a2 = 2

v3 = 0 a3 = 0 .

Example 4.2

Invert the motion equations of Example 4.1 to obtain X = χ−1(x, t) and de-termine the velocity and acceleration components of the particle at x = (1, 0, 1)when t = 2s.

SolutionBy inverting the motion equations directly we obtain

X1 =x1 − t2x2

1− t4; X2 =

x2 − t2x1

1− t4; X3 = x3


which upon substitution into the velocity and acceleration expressions of Exam-ple 4.1 yields

v1 =2t(x2 − t2x1)

1− t4a1 =

2(x2 − t2x1)

1− t4

v2 =2t(x1 − t2x2)

1− t4and a2 =

2(x1 − t2x2)

1− t4

v3 = 0 a3 = 0 .

For the particle at x = (1, 0, 1) when t = 2s

v1 =16

15a1 =

8

15

v2 = −4

15and a2 = −

2

15

v3 = 0 a3 = 0 .

Example 4.3

Determine the Jacobian J for the motion of a continuum be given by

x1 = X1 + f(t) coskX3 ,

x2 = X2 + f(t) sinkX3 ,

x3 = X3 .

SolutionForm and compute the determinant of Eq 4.20

det |xi,A| = det

∣∣∣∣∣∣∣∣∣∣∣∣

∂x1

∂X1

∂x1

∂X2

∂x1

∂X3∂x2

∂X1

∂x2

∂X2

∂x2

∂X3∂x3

∂X1

∂x3

∂X2

∂x3

∂X3

∣∣∣∣∣∣∣∣∣∣∣∣= det

∣∣∣∣∣∣1 0 f(t)k sinkX30 1 −f(t)k coskX30 0 1

∣∣∣∣∣∣ = 1 .

4.3 Langrangian and Eulerian DescriptionsIf a physical property of the body B such as its density ρ, or a kinematic property of itsmotion such as the velocity v, is expressed in terms of the material coordinates X, and


the time t, we say that property is given by the referential or material description. When thereferential configuration is taken as the actual configuration at time t = 0, this descriptionis usually called the Lagrangian description. Thus, the equations

ρ = ρ(XA, t) or ρ = ρ(X, t) , (4.21a)

andvi = vi(XA, t) or v = v(X, t) (4.21b)

chronicle a time history of these properties for each particle of the body. In contrast, ifthe properties ρ and v are given as functions of the spatial coordinates x and time t, wesay that those properties are expressed by a spatial description, or as it is sometimes called,by the Eulerian description. In view of Eq 4.18 it is clear that Eq 4.21 may be converted toexpress the same properties in the spatial description. Accordingly, we write

ρ = ρ (X, t) = ρ[χ−1 (x, t) , t

]= ρ (x, t) (4.22a)

andv = v (X, t) = v

[χ−1 (x, t) , t

]= v (x, t) (4.22b)

where the tilde is added solely for the purpose of emphasizing that different functionalforms result from the switch in variables. We note that in the material description at-tention is focused on what is happening to the individual particles during the motion,whereas in the spatial description the emphasis is directed to the events taking place atspecific points in space.

Example 4.4

Let the motion equations be given in component form by the Lagrangian de-scription

x1 = X1et + X3(e

t − 1) ,

x2 = X2 + X3(et − e−t) ,

x3 = X3 .

Determine the Eulerian description of this motion.

SolutionNotice first that for the given motion x1 = X1 , x2 = X2 and x3 = X3 at t = 0,so that the initial configuration has been taken as the reference configuration.Because of the simplicity of these Lagrangian equations of the motion, we maysubstitute x3 for X3 into the first two equations and solve these directly to obtainthe inverse equations

X1 = x1e−t + x3(e

−t − 1) ,

X2 = x2 + x3(e−t − et) ,

X3 = x3 .


Example 4.5

For the motion of Example 4.4 determine the velocity and acceleration fields,and express these in both Lagrangian and Eulerian forms.

SolutionFrom the given motion equations and the velocity definition Eq 4.16 we obtainthe Lagrangian velocity components,

v1 = X1et + X3e

t ,

v2 = X3(et + e−t) ,

v3 = 0 ,

and from Eq 4.17 the acceleration components

a1 = (X1 + X3)et ,

a2 = X3(et − e−t) ,

a3 = 0 .

Therefore, by introducing the inverse mapping equations determined in Example4.4 we obtain the velocity and acceleration equations in Eulerian form,

v1 = x1 + x3 a1 = x1 + x3v2 = x3(e

t + e−t) and a2 = x3(et − e−t)

v3 = 0 a3 = 0 .

4.4 The Displacement FieldAs may be seen from Fig. 4.1, the typical particle of body B undergoes a displacement

u = x − X (4.23)

in the transition from the reference configuration to the current configuration. Becausethis relationship holds for all particles it is often useful to analyze deformation or motionin terms of the displacement field of the body. We may write the displacement vector u incomponent form by either of the equivalent expressions

u = uiei = uAIA . (4.24)

Additionally with regard to the material and spatial descriptions we may interpret Eq 4.23in either the material form

u(X, t) = x(X, t) − X , (4.25)

or the spatial formu(x, t) = x− X(x, t) . (4.26)


In the first of this pair of equations we are describing the displacement that will occur tothe particle that starts at X, and in the second equation we present the displacement thatthe particle now at x has undergone. Recalling that since the material coordinates relateto positions in the reference configuration only, and hence are independent of time, wemay take the time rate of change of displacement as an alternative definition for velocity.Thus

du

dt=d(x− X)

dt=dx

dt= v . (4.27)

Example 4.6

Obtain the displacement field for the motion of Example 4.4 in both materialand spatial descriptions.

SolutionFrom the motion equations of Example 4.4, namely,

x1 = X1et + X3(e

t − 1) ,

x2 = X2 + X3(et − e−t) ,

x3 = X3 ,

we may compute the displacement field in material form directly as

u1 = x1 − X1 = (X1 + X3)(et − 1) ,

u2 = x2 − X2 = X3(et − e−t) ,

u3 = x3 − X3 = 0 ,

and by using the inverse equations from Example 4.4, namely,

X1 = x1e−t + x3(e

−t − 1) ,

X2 = x2 + x3(e−t − et) ,

X3 = x3 ,

in Eq 4.26 we obtain the spatial description of the displacement field in compo-nent form

u1 = (x1 + x3)(1− e−t) ,

u2 = x3(et − e−t) ,

u3 = 0 .

4.5 The Material DerivativeIn this section let us consider any physical or kinematic property of a continuum body. Itmay be a scalar, vector or tensor property, and so we represent it by the general symbolPij... with the understanding that it may be expressed in either the material description

Pij... = Pij...(X, t) , (4.28a)


or in the spatial descriptionPij... = Pij...(x, t) , (4.28b)

The material derivative of any such property is the time rate of change of that property fora specific collection of particles (one or more) of the continuum body. This derivative canbe thought of as the rate at which Pij... changes when measured by an observer attachedto, and traveling with, the particle, or group of particles. We use the differential operatord/dt, or the superpositioned dot, to denote a material derivative, and note that velocityand acceleration as we have previously defined them are material derivatives.

When Pij... is given in the material description of Eq 4.28a, the material derivative issimply the partial derivative with respect to time,

d

dt[Pij...(X, t)] =

∂

∂t[Pij...(X, t)] , (4.29)

since, as explained earlier, the material coordinates X are labels and do not change withtime. If, however, Pij... is given in the spatial form of Eq 4.28b we recognize that thespecific collection of particles of interest will be changing position in space, and we mustuse the chain rule of differentiation of the calculus to obtain

d

dt[Pij...(x, t)] =

∂

∂t[Pij...(x, t)] +

∂

∂xk[Pij...(x, t)]

dxk

dt. (4.30)

In this equation the first term on the right hand side gives the change occurring in theproperty at position x, known as the local rate of change; the second term results from theparticles changing position in space and is referred to as the convective rate of change. Sinceby Eq 4.16 the velocity is defined as v = dx/dt (or vk = dxk/dt), Eq 4.30 may be writtenas

d

dt[Pij...(x, t)] =

∂

∂t[Pij...(x, t)] +

∂

∂xk[Pij...(x, t)] vk (4.31)

from which we deduce the material derivative operator for properties expressed in the spatialdescription

d

dt=∂

∂t+ vk

∂

∂xkor

d

dt=∂

∂t+ v ·∇ . (4.32)

The first form of Eq 4.32 is for rectangular Cartesian coordinates, while the second form iscoordinate-free. The del operator (∇) will always indicate partial derivatives with respectto the spatial variables unless specifically stated.

Example 4.7

Let a certain motion of a continuum be given by the component equations,

x1 = X1e−t, x2 = X2e

t, x3 = X3 + X2(e−t − 1) ,

and let the temperature field of the body be given by the spatial description,

θ = e−t(x1 − 2x2 + 3x3) ,

Determine the velocity field in spatial form, and using that, compute the mate-rial derivative dθ/dt of the temperature field.


SolutionNote again here that the initial configuration serves as the reference configura-tion so that Eq 4.13 is satisfied. When Eq 4.30 is used, the velocity componentsin material form are readily determined to be

v1 = −X1e−t, v2 = X2e

t, v3 = −X2e−t .

Also, the motion equations can be inverted directly to give,

X1 = x1et, X2 = x2e

−t, X3 = x3 − x2(e−2t − e−t)

which upon substitution into the above velocity expressions yields the spatialcomponents,

v1 = −x1, v2 = x2, v3 = −x2e−2t .

Therefore, we may now calculate dθ/dt in spatial form using Eq 4.31,

dθ

dt= −e−t(x1 − 2x2 + 3x3) − x1e

−t − 2x2e−t − 3x3e

−t

which may be converted to its material form using the original motion equations,resulting in

dθ

dt= −2X1e

−2t − 3X2(2e−2t − e−t) − 3X3e

−t .

An interesting and rather unique situation arises when we wish to determine the ve-locity field in spatial form by a direct application of Eq 4.31 to the displacement field inits spatial form. The following example illustrates the point.

Example 4.8

Verify the spatial velocity components determined in Example 4.7 by applyingEq 4.31 directly to the displacement components in spatial form for the motionin that example.

SolutionWe may determine the displacement components in material form directly fromthe motion equations given in Example 4.7,

u1 = x1 − X1 = X1(e−t − 1) ,

u2 = x2 − X2 = X2(et − 1) ,

u3 = x3 − X3 = X2(e−t − 1) ,

and, using the inverse equations X = χ−1(x, t) computed in Example 4.7, weobtain the spatial displacements

u1 = x1(1− et) ,

u2 = x2(1− e−t) ,

u3 = x2(e−2t − e−t) .


Therefore, substituting ui for Pij... in Eq 4.31 yields

vi =dui

dt=∂ui

∂t+ vk

∂ui

∂xk,

so that by differentiating the above displacement components

v1 = −x1et + v1(1− et) ,

v2 = x2e−t + v2(1− e−t) ,

v3 = −x2(2e−2t − e−t) + v2(e

−2t − e−t) ,

which results in a set of equations having the desired velocity components onboth sides of the equations. In general, this set of equations must be solvedsimultaneously. In this case the solution is quite easily obtained, yielding

v1 = −x1, v2 = x2, v3 = −x2e−2t

to confirm the results of Example 4.7.

Example 4.9

Cilia are motile cells that are responsible for locomotion of single-cell bodies,hearing, and moving fluid in the body. The cilia beat in a manner that propelsthe surrounding fluid. The motion of the cilia is complex. A model for the ciliamotion is to enclose the tips by a flexible envelope that moves in a fashion similarto the tips of the cilia. This is reasonable given that in many biological processesfluid motions are slow. Two possible descriptions of the envelope motion are 1

Case 1:x1 = X01 + εα cos (x1 − ct) ,

x2 = X02 + εβ sin (x1 − ct) ,

Case 2:x1 = X01 + εα cos

(X01 − ct

),

x2 = X02 + εβ sin(X01 − ct

).

Envelope

Cilia

Envelope containing the tips of the beating cilia.

Here x is the position of the envelope and X0 is an initial position of the particles.

(a) Show that the both descriptions indicate that particles move in anelliptical path centered about

(X01, X

02

).

1see T. J. Lardner and W. J. Shack (1972) Cilia transport, Bulletin of Mathematical Biophysics, 34 (3) 325-335.


(b) Fluid mechanics has a no-slip condition that requires the velocity ofthe fluid be the same as the velocity of the envelope. Determine thevelocity of the particles in the two descriptions.

Solution

(a) For description 1, the particle paths are(x1 − X01

)2(εβ)2

+

(x2 − X02

)2(εα)2

= sin2 (x1 − ct) + cos2 (x1 − ct) = 1 .

For description 2, the particle paths are(x1 − X01

)2(εβ)2

+

(x2 − X02

)2(εα)2

= sin2(X01 − ct

)+ cos2

(X01 − ct

)= 1 .

These are both descriptions of ellipses centered at(X01, X

02

).

(b) For description 1, the velocity v1 is

v1 =dx1

dt=

[∂

∂t+ v1

∂

∂x1+ v2

∂

∂x2

] [X01 + εα cos (x1 − ct)

]=∂

∂t

[X01 + εα cos (x1 − ct)

]+ v1

∂

∂x1

[X01 + εα cos (x1 − ct)

]= cεα sin (x1 − ct) − v1εα sin (x1 − ct) ,

and

v1 [1+ εα sin (x1 − ct)] = cεα sin (x1 − ct) ,

v1 =cεα sin (x1 − ct)

1+ εα sin (x1 − ct).

The velocity v2 is

v2 =dx2

dt=

[∂

∂t+ v1

∂

∂x1+ v2

∂

∂x2

] [X02 + εβ sin (x1 − ct)

]=∂

∂t

[X02 + εβ sin (x1 − ct)

]+ v1

∂

∂x1

[X02 + εβ sin (x1 − ct)

]= −cεα cos (x1 − ct) + v1εβ cos (x1 − ct)

= −cεα cos (x1 − ct) + εβ cos (x1 − ct)cεα sin (x1 − ct)

1+ εα sin (x1 − ct),

or

v2 =−cεβ cos (x1 − ct)

1+ εα sin (x1 − ct).

For description 2, the velocity v1 is

v1 =dx1

dt=

[∂

∂t+ v1

∂

∂x1+ v2

∂

∂x2

] [X01 + εα cos

(X01 − ct

)]=∂

∂t

[X01 + εα cos

(X01 − ct

)]= cεα sin

(X01 − ct

).


Reference

Current

e1, I1 e2, I2

e3, I3

x3, X

3

x2, X

2

x1, X

1

O

P

pX

x

u+du

dXA

dxi

Q

qu

FIGURE 4.2Vector dXA, between points P and Q in reference configuration, becomes dxi, betweenpoints p and q, in the current configuration. Displacement vector u is the vector betweenpoints p and P.

The velocity v2 is

v2 =dx2

dt=

[∂

∂t+ v1

∂

∂x1+ v2

∂

∂x2

] [X02 + εβ sin

(X01 − ct

)]=∂

∂t

[X02 + εβ sin

(X01 − ct

)]= −cεβ cos

(X01 − ct

).

4.6 Deformation Gradients, Finite Strain TensorsIn deformation analysis we confine our attention to two stationary configurations anddisregard any consideration for the particular sequence by which the final deformed con-figuration is reached from the initial undeformed configuration. Accordingly, the mappingfunction is not dependent upon time as a variable, so that Eq 4.10 takes the form

xi = χi(X) or x = χ(X) . (4.33)

Consider, therefore, two neighboring particles of the body situated at the points P and Qin the undeformed configuration such that Q is located with respect to P by the relativedifferential position vector

dX = dXAIA (4.34)

as shown in Fig. 4.2. The magnitude squared of dX is

(dX)2 = dX · dX = dXAdXA . (4.35)


Under the displacement field prescribed by the function χ of Eq 4.35 the particles origi-nally at P and Q move to the positions p and q, respectively, in the deformed configura-tion such that their relative position vector is now

dx = dxiei (4.36)

having a magnitude squared

(dx)2 = dx · dx = dxidxi . (4.37)

We assume the mapping function χi of Eq 4.33 is continuous so that

dxi =∂χi

∂XAdXA , (4.38)

or as it is more often written,

dxi =∂χi

∂XAdXA = xi,AdXA (4.39)

wherexi,A ≡ FiA (4.40)

is called the deformation gradient tensor or simply the deformation gradient. The tensor Fcharacterizes the local deformation at X, and may depend explicitly upon X, in whichcase the deformation is termed inhomogeneous. If F is independent of X, the deformationis called homogeneous. In symbolic notation Eq 4.39 appears in either of the forms

dx = F · dX or dx = FdX (4.41)

where, as indicated by the second equation, the dot is often omitted for convenience. Inview of the smoothness conditions we have imposed on the mapping function χ we knowthat F is invertible so that the inverse F−1 exists such that

dXA = XA,idxi or dX = F−1 · dx . (4.42)

In describing motions and deformations, several measures of deformation are com-monly used. Let us consider first one based upon the change during the deformationin the magnitude squared of the distance between the particles originally at P and Q.Namely,

(dx)2 − (dX)2 = dxidxi − dXAdXA

which from Eq 4.39 and the substitution property of the Kronecker delta, δAB may bedeveloped as follows,

(dx)2 − (dX)2 = (xi,AdXA)(xi,BdXB) − δABdXAdXB

= (xi,Axi,B − δAB)dXAdXB

= (CAB − δAB)dXAdXB (4.43)

where the symmetric tensor

CAB = xi,Axi,B or C = FT · F (4.44)

is called the Green’s deformation tensor. From this we immediately define the Lagrangianfinite strain tensor EAB as

2EAB = CAB − δAB or 2E = C− I (4.45)


where the factor of two is introduced for convenience in later calculations. Finally we canwrite,

(dx)2 − (dX)2 = 2EABdXAdXB = dX · 2E · dX . (4.46)

The difference (dx)2 − (dX)2 may also be developed in terms of the spatial variables ina similar way as

(dx)2 − (dX)2 = δijdxidxj − (XA,idxi)(XA,jdxj)

= (δij − XA,iXA,j)dxidxj

= (δij − cij)dxidxj (4.47)

where the symmetric tensor

cij = XA,iXA,j or c = (F−1)T · (F−1) (4.48)

is called the Cauchy deformation tensor. From Eq 4.48 we define the Eulerian finite straintensor e as

2eij = (δij − cij) or 2e = (I− c) , (4.49)

so that now(dx)2 − (dX)2 = 2eijdxidxj = dx · 2e · dx . (4.50)

Both EAB and eij are, of course, symmetric second-order tensors, as can be observed fromtheir definitions.

For any two arbitrary differential vectors dX(1) and dX(2) which deform into dx(1) anddx(2), respectively, we have from Eq 4.41 together with Eqs 4.44 and 4.45,

dx(1) · dx(2) = F · dX(1) · F · dX(2) = dX(1) · FT · F · dX(2)

= dX(1) ·C · dX(2) = dX(1) · (I+ 2E) · dX(2)

= dX(1) · dX(2) + dX(1) · 2E · dX(2). (4.51)

If E is identically zero (no strain), Eq 4.51 asserts that the lengths of all line elements areunchanged [we may choose dX(1) = dX(2) = dX so that (dx)2 = (dX)2], and in view ofthe definition dx(1) · dx(2) = dx(1)dx(2) cos θ, the angle between any two elements willalso be unchanged. Thus in the absence of strain, only a rigid body displacement canoccur.

The Lagrangian and Eulerian finite strain tensors expressed by Eqs 4.45 and 4.49, re-spectively, are given in terms of the appropriate deformation gradients. These same ten-sors may also be developed in terms of displacement gradients. For this purpose we beginby writing Eq 4.25 in its time-independent form consistent with deformation analysis. Incomponent notation, the material description is

ui(XA) = xi(XA) − Xi , (4.52)

and the spatial description is

uA(xi) = xA − XA(xi) . (4.53)

From the first of these, Eq 4.45 becomes

2EAB = xi,Axi,B − δAB = (ui,A + δiA)(ui,B + δiB) − δAB (4.54)

which reduces to2EAB = uA,B + uB,A + ui,Aui,B , (4.55)


and from the second, Eq 4.49 becomes

2eij = δij − XA,iXA,j = δij − (δAi − uA,i)(δAj − uA,j) (4.56)

which reduces to2eij = ui,j + uj,i − uA,iuA,j . (4.57)

Example 4.10

Let the simple shear deformation x1 = X1; x2 = X2+kX3; x3 = X3+kX2, wherek is a constant, be applied to the small cube of edge dimensions dL shown in thefigure below. Draw the deformed shape of face ABGH of the cube and determinethe difference (dx)2 − (dX)2 for the diagonals AG, BH and OG of the cube.

x1, X

1 x2, X

2

x3, X

3

OA

B

G

H

dL dL

dL

Reference position of cubeundergoing simple shear.

x3, X

3

x2, X

2a

b(dL,dL,kdL)

g(dL,(1+k)dL,(1+k)dL)

h(dL,kdL,dL)

Small cube geometry.

SolutionFrom the mapping equations directly, the origin O is seen to remain in place,and the particles originally at points A, B, G and H are displaced to the pointsa(dL,O,O), b(dL, dL, kdL), g(dL, (1+ k)dL, (1+ k)dL) and h(dL, kdL, dL), re-spectively, so that particles in planes parallel to the X2X3 remain in those planes,and the square face ABGH becomes the diamond shaped parallelogram abgh

shown above. Also from the mapping equations and Eq 4.40, we see that thedeformation gradient F has the matrix form

[FiA] =

1 0 0

0 1 k

0 k 1

,

and since C = FT · F

[CAB] =

1 0 0

0 1+ k2 2k

0 2k 1+ k2

from which we determine 2E = C− I,

[2EAB] =

0 0 0

0 k2 2k

0 2k k2

.


In general, (dx)2 − (dX)2 = dX · 2E · dX so that for diagonal AG,

(dx)2 − (dX)2 = [0, dL, dL]

0 0 0

0 k2 2k

0 2k k2

0dLdL

= 2(2k+ k2)(dL)2 .

For diagonal BH,

(dx)2 − (dX)2 = [0,−dL, dL]

0 0 0

0 k2 2k

0 2k k2

0

−dL

dL

= 2(−2k+ k2)(dL)2 ,

and for diagonal OG,

(dx)2 − (dX)2 = [dL, dL, dL]

0 0 0

0 k2 2k

0 2k k2

dLdLdL

= 2(2k+ k2)(dL)2 ,

Note: All of these results may be calculated directly from the geometry of thedeformed cube for this simple deformation.

4.7 Infinitesimal Deformation TheoryIf the numerical values of all the components of the displacement and the displacementgradient tensors are very small we may neglect the squares and products of these quan-tities in comparison to the gradients themselves so that Eqs 4.55 and 4.57 reduce to

2EAB = uA,B + uB,A , (4.58)

and2eij = ui,j + uj,i . (4.59)

These expressions are known as the linearized Lagrangian and Eulerian strain tensors,respectively. Furthermore, to the same order of approximation,

∂ui

∂XA=∂ui

∂xk

∂xk

∂XA=∂ui

∂xk

(∂uk

∂XA+ δkA

)≈ ∂ui

∂xkδkA (4.60)

where we have used the relationship

∂xk

∂XA=∂uk

∂XA+ δkA


obtained by differentiating Eq 4.52. Therefore, to the first order of approximation forthe case of small displacement gradients, it is unimportant whether we differentiate thedisplacement components with respect to the material or spatial coordinates. In view ofthis, we may display the equivalent relative displacement gradients for small deformationtheory as either ui,A or ui,j. Similarly, it can be shown that in the linear theory uA,B anduA,j are equivalent. It follows that to the same order of approximation, from Eqs 4.58 and4.59,

EAB ≈ eijδiAδjB , (4.61)

and it is customary to define a single infinitesimal strain tensor for which we introducethe symbol εij as

2εij =∂ui

∂XAδAj +

∂uj

∂XBδBi =

∂ui

∂xj+∂uj

∂xi= ui,j + uj,i . (4.62)

Because the strain tensors EAB, eij, and εij are all symmetric, second-order tensors, theentire development for principal strains, strain invariants and principal strain directionsmay be carried out exactly as was done for the stress tensor in Chapter 3. Thus, taking,εij as the typical tensor of the group, we summarize these results by displaying its matrixrelative to principal axes in the alternative forms,

[ε∗ij] =

ε1 0 0

0 ε2 0

0 0 ε3

=

εI 0 0

0 εII 0

0 0 εIII

(4.63)

together with the strain invariants

Iε = εii = trε = εI + εII + εIII , (4.64a)

IIε =1

2(εiiεjj − εijεij) = εIεII + εIIεIII + εIIIεI , (4.64b)

IIIε = εijkε1iε2jε3k = εIεIIεIII . (4.64c)

The components of ε have specific physical interpretations which we now consider.Within the context of small deformation theory we express Eq 4.46 in the modified form

(dx)2 − (dX)2 = 2εijdXidXj = dX · 2ε · dX (4.65)

which, upon factoring the left hand side and dividing by (dX)2, becomes

dx− dX

dX· dx+ dX

dX= 2εij

dXi

dX

dXj

dX.

But dXi/dX = Ni , a unit vector in the direction of dX, and for small deformations wemay assume (dx+ dX)/dX ≈ 2, so that

dx− dX

dX= εijNiNj = N · ε · N . (4.66)

The scalar ratio on the left-hand side of this equation is clearly the change in length perunit original length for the element in the direction of N . It is known as the longitudinalstrain, or the normal strain and we denote it by e(N). If, for example, N is taken in the X1direction so that N = I1 , then

e(I1) = I1 · ε · I1 = ε11 .


dX(1)

dX(2)

dx(1)

dx(2)

P A

B

b

p

aθ

π/2

DE

FORM

S

FIGURE 4.3The right angle between line segments AP and BP in the reference configuration becomesθ, the angle between segments ap and bp, in the deformed configuration.

Likewise, for N = I2 , or N = I3 the normal strains are found to be ε22 and ε33, re-spectively. Thus the diagonal elements of the small (infinitesimal) strain tensor representnormal strains in the coordinate directions.

To gain an insight into the physical meaning of the off-diagonal elements of the in-finitesimal strain tensor we consider differential vectors dX(1) and dX(2) at position Pwhich are deformed into vectors dx(1) and dx(2) , respectively. In this case, Eq 4.51 maybe written,

dx(1) · dx(2) = dX(1) · dX(2) + dX(1) · 2ε · dX(2) (4.67)

which, if we choose dX(1) and dX(2) perpendicular to one another, reduces to

dx(1) · dx(2) = dx(1)dx(2) cos θ = dX(1) · 2ε · dX(2) (4.68)

where θ is the angle between the deformed vectors as shown in Fig. 4.3. If we let, θ = π2−γ

, the angle γ measures the small change in the original right angle between dX(1) anddX(2). Also

cos θ = cos(π2

− γ)

= sinγ ≈ γ

since γ is very small for infinitesimal deformations. Therefore, assuming as before thatdx(1) ≈ dX(1) and dx(2) ≈ dX(2) because of small deformations

γ ≈ cos θ =dX(1)

dx(1)· 2ε · dX

(2)

dx(2)≈ N(1) · 2ε · N(2) . (4.69)

Here, if we take N(1) = I1 and N(2) = I2 and designate the angle γ as γ12 , we obtain

γ12 = 2 [1, 0, 0]

ε11 ε12 ε13ε12 ε22 ε23ε13 ε23 ε33

010

= 2ε12 , (4.70)

so that by choosing the undeformed vector pairs in Eq 4.68 in coordinate directions wemay generalize Eq 4.70 to obtain


γij = 2εij (i 6= j) . (4.71)

This establishes the relationship between the off-diagonal components of εij and the so-called engineering shear strain components γij, which represent the changes in the originalright angles between the coordinate axes in the undeformed configuration. Note thatsince ε can be defined with respect to any set of Cartesian axes at P, this result holdsfor any pair of perpendicular vectors at that point. In engineering texts, the infinitesimalstrain tensor is frequently written in matrix form as

[εij] =

ε11

12γ12

12γ13

12γ12 ε22

12γ23

12γ13

12γ23 ε33

. (4.72)

If N(1) and N(2) are chosen in principal strain directions, Eq 4.69 becomes

γ = N(1) · 2ε∗ · N(2) = 0 (4.73)

from which we may generalize to conclude that principal strain directions remain or-thogonal under infinitesimal deformation. Therefore, a small rectangular parallelpipedof undeformed edge dimensions dX(1), dX(2) and dX(3) taken in the principal strain di-rections will be deformed into another rectangular parallelpiped having edge lengths

dx(i) = (1+ ε(i))dX(i), (i = 1, 2, 3) (4.74)

as shown in Fig. 4.4, where ε(i) are the normal strains in principal directions. The changein volume per unit original volume of the parallelpiped is

∆V

V=

[1+ ε(1)]dX(1)[1+ ε(2)]dX

(2)[1+ ε(3)]dX(3) − dX(1)dX(2)dX(3)

dX(1)dX(2)dX(3)

≈ ε(1) + ε(2) + ε(3) (4.75)

where terms involving products of the principal strains have been neglected. The ratio∆V/V, being the first invariant of ε, is called the cubical dilatation. We shall denote it bythe symbol e, and write

e = ∆V/V = εii = Iε . (4.76)

Because ε is a symmetrical second-order tensor the development of Mohr’s circles forsmall strain, as well as the decomposition of ε into its spherical and deviator componenttensors follows in much the same way as the analogous concepts for stress in Chapter 3.One distinct difference is that for the Mohr’s circles, the shear strain axis (ordinate) hasunits of 12γ as shown by the typical diagram of Fig. 4.5. The infinitesimal spherical straintensor is represented by a diagonal matrix having equal elements denoted by εM = 1

3εii =13e, known as the mean normal strain. The infinitesimal deviator strain tensor, η is defined by

ηij = εij −1

3δijεkk = εij − δijεM , (4.77)

and in matrix formη11 η12 η13η21 η22 η23η31 η32 η33

=

ε11 − εM ε12 ε13ε12 ε22 − εM ε23ε13 ε23 ε33 − ε33

. (4.78)


dX(1)dX(2)

dX(3)

X3

X1

X2

x1

∗

x3

∗

x2

∗

dx(1)

dx(3)

dx(2)

DEFO

RMS

FIGURE 4.4A rectangular parallelpiped with edge lengths dX(1), dX(2) and dX(3) in the referenceconfiguration becomes a skewed parallelpiped with edge lengths dx(1), dx(2) and dx(3)

in the deformed configuration.

Note that as with its stress counterpart, the first invariant of the deviator strain is zero, or

ηii = 0 , (4.79)

and the principal deviator strains are given by

η(q) = ε(q) − εM, (q = 1, 2, 3) (4.80)

where ε(q) is a principal value of the infinitesimal strain tensor.A state of plane strain parallel to the X1X2 plane exists at P if

ε33 = γ13 = γ31 = γ23 = γ32 = 0 (4.81)

at that point. Also, plane strain relative to the X1X2 plane in the continuum body as awhole exists if Eq 4.81 is satisfied everywhere in the body, and if in addition, the remain-ing non-zero components are independent of X3. With respect to axes OX ′1X

′2X′3 rotated

about X3 by the angle θ relative to OX1X2X3 as shown by Fig. 4.6, the transformationequations for plane strain (analogous to Eq 3.75 for plane stress) follow the tensor trans-formation formula, Eq 2.64. In conjunction with the table of direction cosines of Table 4.1the results are

ε ′11 =ε11 + ε22

2+ε11 − ε22

2cos 2θ+

γ12

2sin 2θ , (4.82a)

ε ′22 =ε11 + ε22

2−ε11 − ε22

2cos 2θ−

γ12

2sin 2θ , (4.82b)

γ ′12 = −(ε11 − ε22) sin 2θ+ γ12 cos 2θ . (4.82c)


γ

2

III II I

FIGURE 4.5Typical Mohr’s circle for strain.

X3 , X3

X2

X1

X2

X1

θ θO

FIGURE 4.6Rotation of axes for plane strain.

Also, the non-zero principal strain values for plane strain are given by

ε(1), ε(2) =ε11 + ε22

2±

√(ε11 − ε22

2

)2+(γ122

)2. (4.83)

Because shear strains are very difficult to measure experimentally, the state of strain ata point is usually determined by recording three separate longitudinal strains at the point(using a strain gauge rosette) and substituting these values into Eqs 4.82 to calculate γ12.

Example 4.11

A delta rosette has the shape of an equilateral triangle, and records longitudinalstrains in the directions x1, x ′1 and x ′′1 shown in the sketch. If the measured


TABLE 4.1Transformation table for general plane strain.

X1 X2 X3

X ′1 cos θ sin θ 0

X ′2 − sin θ cos θ 0

X ′3 0 0 1

strains in these directions are ε11 = −3 × 10−4, ε ′11 = 4 × 10−4, and ε ′′11 =2×10−4 where the units are m/m (dimensionless), determine ε22, γ12 and ε ′22.Show that ε11 + ε22 = ε ′11 + ε ′22 as the first strain invariant requires.

60º

60º 60º

30º30º

x1

x2′′x1′x1

SolutionWe need only Eq 4.82 here, which we write for x ′1 and x ′′1 in turn (omittingthe common factor 10−4 for convenience). Thus, for θ = 60, and θ = 120,respectively, we have

4 =−3+ ε22

2+

−3− ε22

2

(−1

2

)+γ12

2

√3

2,

2 =−3+ ε22

2+

−3− ε22

2

(−1

2

)−γ12

2

√3

2.

Adding these two equations to eliminate γ12 we determine ε22 = 5; subtractingthe second from the first to eliminate ε22 we determine γ12 = 4/

√3 . Next,

using θ = 150 we determine ε ′22 from Eq 4.82

ε ′22 =−3+ 5

2+

−3− 5

2+

2√3

(−

√3

2

)= −2 ,

and by the first invariant of the small strain tensor we check that

ε11 + ε22 = −3+ 5 = ε ′11 + ε ′22 = 4− 2 = 2 .


Consider once more the two neighboring particles which were at positions P and Qin the undeformed configuration, and are now at positions p and q, respectively, in thedeformed configuration (see Fig. 4.2). In general, an arbitrary displacement will includeboth deformation (strain) and rigid body displacements. To separate these we considerthe differential displacement vector du. Assuming conditions on the displacement fieldthat guarantee the existence of a derivative, the displacement differential dui is written

dui =

(∂ui

∂Xj

)P

dXj (4.84)

where the derivative is evaluated at P as indicated by the notation. From this we maydefine the unit relative displacement of the particle at Q with respect to the one at P by theequation

dui

dX=dui

dXj

dXj

dX=dui

dXjNj (4.85)

where Nj is the unit vector in the direction from P toward Q. By decomposing thedisplacement gradient in Eq 4.84 into its symmetric and skew-symmetric parts we obtain

dui =

[1

2

(∂ui

∂Xj+∂uj

∂Xi

)+1

2

(∂ui

∂Xj−∂uj

∂Xi

)]dXj

= (εij +ωij)dXj (4.86)

in which εij is recognized as the infinitesimal strain tensor, andωij is called the infinitesimalrotation tensor.

If εij happens to be identically zero, there is no strain, and the displacement is a rigidbody displacement. For this case we define the rotation vector

ωi =1

2εijkωkj (4.87)

which may be readily inverted since ωkj = −ωjk to yield

ωij = εkjiωk . (4.88)

Therefore, Eq 4.86 with εij ≡ 0 becomes

dui = εkjlωkdXj = εikjωkdXj or du =ω× dX , (4.89)

so that the relative differential displacement is seen to be the result of a rigid body rotationabout the axis of the rotation vector ω. On the other hand, if ωij ≡ 0, the relativedisplacement will be the result of pure strain.

A comment is appropriate about the notation ωij for the skew–symmetric part of thedisplacement gradient in the linear theory. It might have been convenient to denote thisquantity as W = wijei ⊗ ej; however, this notation has been reserved to represent thespin tensor in Section 4.11. The spin tensor is the skew–symmetric part of the velocitygradient. It is a bit awkward to have the tensor ωij and its axial vector ωj denoted withω, but this is not as bad as having the same symbol representing the infinitesimal rotationand spin tensors.


4.8 Compatibility EquationsThe compatibility equations merit a section because of the important role played in linearelasticity problems. Very often, solving a linear elasticity problem (see Chapter 6) involvessolving the compatibility equations discussed below.

If we consider the six independent strain-displacement relations, Eq 4.62

∂ui

∂xj+∂uj

∂xi= 2εij

as a system of partial differential equations for determining the three displacement com-ponents ui (assuming the εij are known as functions of xi), the system is over-determined,and we cannot in general find three single-valued functions ui = ui(xj) satisfying the sixpartial differential equations. Therefore, some restrictive conditions must be imposedupon the strain components (actually upon derivatives of the strain components) if theequations above are to be satisfied by a single-valued displacement field. Such conditionsare expressed by the strain compatibility equations

εij,km + εkm,ij − εik,jm − εjm,ik = 0 . (4.90)

There are 34 = 81 equations in all (four free indices) in Eq 4.90 but only six of these aredistinct

ε11,23 + ε23,11 − ε12,13 − ε13,12 = 0 ,

ε22,31 + ε31,22 − ε23,21 − ε21,23 = 0 ,

ε33,12 + ε12,33 − ε31,32 − ε32,31 = 0 ,

2ε12,12 − ε11,22 − ε22,11 = 0 ,

2ε23,23 − ε22,33 − ε33,22 = 0 ,

2ε31,31 − ε33,11 − ε11,33 = 0 .

(4.91)

For plane strain in the x1-x2 plane, the six unique equations in Eq 4.90 reduce to asingle equation,

ε11,22 + ε22,11 = 2ε12,12 , (4.92)

which may be easily verified as a necessary condition by a simple differentiation of Eq 4.62for a range of two on the indices i and j.

It may be shown that the compatibility equations, either Eq 4.90 or Eq 4.91, are bothnecessary and sufficient for a single-valued displacement field of a body occupying asimply connected domain. The compatibility equations seem daunting, but it is not toodifficult to demonstrate the necessity and sufficiency.

For the necessity, given the strain εij = 12

(ui,j + uj,i) with continuously differentiabledisplacements we write that

curl curlε = 0 . (4.93)

This equation is a more compact way of writing Eqs 4.90 and 4.91. By the definition oflinear strain and the continuity of the displacements, we see

εijkεlmnεjm,kn = εijkεlmn (uj,mkn + um,jkn) = 0

where the first displacement term, symmetric in mn, cancels with skew–symmetric εlmn,and the second displacement term, symmetric in jk, cancels with skew–symmetric εijk.


As with many proofs, sufficiency is more difficult to demonstrate. Here it must beshown that starting from curl curlε = 0 that the linear strain εij = 1

2(ui,j + uj,i) can be

constructed. To start, letA = curlε (4.94)

from which compatibility in the form of Eq 4.93 gives

curlA = 0 . (4.95)

Since ε is symmetric,trA = 0 (4.96)

based on the identity Eq 2.91 from Example 2.17. From Eqs 4.95 and 4.96 there exists askew–symmetric tensor ω such that 2

A = − curlω . (4.97)

Equating Eqs 4.97 and 4.94 on A gives

curl (ε+ω) = 0 (4.98)

from which 3

ε+ω =∇u . (4.99)

Taking the symmetric part of this equation gives us the strain

ε =1

2

(∇u+∇uT

), (4.100a)

orεij =

1

2(ui,j + uj,i) . (4.100b)

The compatibility equations give a convenient way to compute the displacements fromthe strain components. A line integral form of the compatibility equations gives

ui(ξ) =

∫ξξ0

εij(ξ) +

(ξk − ξk

) [εij,k(ξ) − εjk,i(ξ)

]dξj

+uj(ξ0) +wij(ξ

0)(ξj − ξ0j

).

(4.101)

The last two terms represent an infinitesimal rigid motion and are often omitted.

Example 4.12

For a given state of strain, εij, the compatibility equations are necessary andsufficient conditions for unique displacements.

(a) The compatibility condition may be written as εkmnεpqrεmq,nr = 0

where εpqr is the alternating tensor. Show that the compatibilityequation in this form is symmetric in k and p. This means that thereare only six independent conditions.

2This is not a difficult step, however, it is best to refer to ME Gurtin, ”The Linear Theory of Elasticty”, inMechanics of Solids, Volume II, Editor C Truesdell, Springer-Verlag, c. 1973, pp17-18 and p 40.3ibid, p 17


(b) For displacements to be path independent the integrand of Eq 4.101must be exact. If

Uij = εij(ξ) +(ξk − ξk

) [εij,k(ξ) − εjk,i(ξ)

],

a necessary and sufficient condition for the integral to be exact is

Uij,m +Uim,j = 0 .

Show this condition leads to

εij,km + εkm,ij − εjk,im − εim,jk = 0 .

(c) If the displacement field has three continuous derivatives, show thatεkmnεpqrεmq,nr = 0 .

Solution

(a) This can be shown through index manipulation. Noting strain is symmetricand twice continuously differentiable and m, n, q and r are dummy indices:

εkmnεpqrεmq,nr = εkmnεpqrεqm,rn = εkqrεpmnεmq,nr .

This expression is a symmetric, second-order tensor having six independentterms.

(b) Differentiation and substitution leads to

εij,m − δkm (εij,k − εjk,i) − εim,j + δkj (εim,k − εmk,i)

+(ξk − ξk

)[εij,km − εjk,im − εim,kj + εmk,ij] = 0 .

Since strain is symmetric and ξ and ξk can be chosen arbitrarily

εij,km + εkm,ij − εjk,im − εim,jk = 0 .

(c) Substitute the strain-displacement relations into the compatibility equation

εkmnεpqrεmq,nr = 12εkmnεpqr (um,qnr + uq,mnr) .

The alternating tensors are skew-symmetric in mn and qr. Additionally,the displacement is assumed to be continuously differentiable and so orderof differentiation is immaterial. Thus, the parenthetical term is symmetricin mn and qr. The product of skew-symmetric and symmetric tensors iszero. So,

εkmnεpqrεmq,nr = 12εkmnεpqr (um,qnr + uq,mnr) = 0 .


4.9 Stretch RatiosReferring again to Fig. 4.2, define the ratio of the magnitudes of dx and dX to be thestretch ratio, Λ (or simply the stretch). In particular, for the differential element in thedirection of the unit vector N at P, we write

Λ(N) =dx

dX(4.102)

where dx is the deformed magnitude of dX = dXN. As a matter of convenience we oftenprefer to work with stretch-squared values,

Λ2(N)

=

(dx

dX

)2. (4.103)

Thus, from Eqs 4.41 and 4.44,

(dx)2 = dx · dx = F · dX · F · dX = dX ·C · dX , (4.104)

so that after dividing by (dX)2,

Λ2(N)

=dX

dX·C · dX

dX= N ·C · N (4.105)

for the element originally in the direction of N .In an analogous way, we define the stretch ratio, λ(n) in the direction of n = dx/dx at

p by the equation,

1

λ(n)=dX

dx. (4.106)

Here, recalling from Eq 4.42 that dX = F−1 · dx and by using Eq 4.48, we obtain

(dX)2 = dX · dX = F−1 · dx · F−1 · dx = dx · c · dx (4.107)

which upon dividing by (dx)2 becomes

1

λ2(n)

=dx

dx· c · dx

dx= n · c · n . (4.108)

In general, Λ(N) 6= λ(n). However if n is a unit vector in the direction that N assumes in

the deformed configuration, the two stretches are the same. For N = I1 ,

Λ2(I1)

= I1 ·C · I1 = C11 = 1+ 2E11 , (4.109)

and for n = e1,

1

λ2(e1)

= e1 · c · e1 = c11 = 1− 2e11 (4.110)

with analogous expressions for N and n in the other coordinate directions.


Consider next the unit extension (longitudinal strain) in any direction N at P. This maybe expressed in terms of the stretch as

e(N) =dx− dX

dX= Λ(N) − 1 =

√N ·C · N− 1 . (4.111)

Notice that the unit extension is zero when the stretch is unity, as occurs with a rigidbody displacement. If N = I1 ,

e(I1) =

√I1 ·C · I1 − 1 =

√C11 − 1 =

√1+ 2E11 − 1 , (4.112)

or, solving for E11,

E11 = e(I1) +1

2e2

(I1). (4.113)

For small deformation theory where E11 → ε11, and for which e2(N)

may be neglected incomparison to e(N) , the above equation asserts that E11 = ε11 = e(I1).

The change in angle between any two line elements may also be given in terms ofstretch. Let dX(1) and dX(2) be arbitrary vectors which become dx(1) and dx(2), respec-tively, during a deformation. By the dot product, dx(1) · dx(2) = dx(1)dx(2) cos θ, we maycompute the angle θ between dx(1) and dx(2) from its cosine, which, with the help ofEq 4.104, takes the form,

cos θ =dx(1)

dx(1)· dx

(2)

dx(2)=

F · dX(1)

√dX(1) ·C · dX(1)

· F · dX(2)

√dX(2) ·C · dX(2)

,

or upon dividing the numerator and denominator by the scalar product dX(1)dX(2) andmaking use of Eqs 4.105 and 4.44 we obtain

cos θ =N1 ·C · N2Λ(N1)Λ(N2)

. (4.114)

Thus, for elements originally in the I1 and I2 directions, the angle between them in thedeformed configuration may be determined from

cos θ12 =C12

Λ(I1)Λ(I2)

=C12√C11C22

. (4.115)

In a similar fashion, from dX(1) · dX(2) = dX(1)dX(2) cosΘ, where Θ is the angle betweendX(1) and dX(2), we obtain from Eqs 4.107 and 4.108,

cosΘ =dX(1)

dX(1)· dX

(2)

dX(2)=

n1 · c · n2√n1 · c · n1

√n2 · c · n2

= λ(n1)λ(n2)(n1 · c · n2)

which gives the original angle between elements in the directions n1 and n2 of the currentconfiguration.

Example 4.13

A homogeneous deformation is given by the mapping equations, x1 = X1−X2+X3 , x2 = X2 − X3 + X1 and x3 = X3 − X1 + X2. Determine


(a) the stretch ratio in the direction of N1 = (I1 + I2)/√2, and

(b) the angle θ12 in the deformed configuration between elements thatwere originally in the directions of N1, and N2 = I2.

SolutionFor the given deformation (as the student should verify),

[FiA] =

1 −1 1

1 1 −1−1 1 1

and [CAB] =

3 −1 −1−1 3 −1−1 −1 3

.

(a) Therefore, from Eq 4.105,

Λ2(N1)

=[1√2

1√2

0

]3 −1 −1

−1 3 −1

−1 −1 3

1√2

1√2

0

= 2 ,

andΛ(N1) =

√2 .

(b) For N2 = I2 , Λ2(I2)

= I2 · C · I2 = C22 = 3 so that from Eq 4.114,using the result in part (a),

cos θ12 =(I1 + I2)/

√2 ·C · I2√

2√3

=2/√2√6,

and θ12 = 54.7. Thus, the original 45 angle is enlarged by 9.7.

It is evident from Eq 4.115 that if the coordinate axes are chosen in the principal di-rections of C the deformed angle θ12 is a right angle (C12 = 0 in this case) and therehas been no change in the angle between elements in the X1 and X2 directions. By thesame argument, any three mutually perpendicular principal axes of C at P are deformedinto three mutually perpendicular axes at p. Consider, therefore, the volume element ofa rectangular parallelepiped whose edges are in the principal directions of C (and thusalso of E). Since there is no shear strain between any two of these edges the new volumeis still a rectangular parallelopiped, and in the edge directions Ni (i = 1, 2, 3) the unitstrains are

e(Ni)= Λ(Ni)

− 1 (i = 1, 2, 3) , (4.116)

so thatdx(i) = dX(i) + dX(i)[Λ(Ni)

− 1] = dX(i)Λ(Ni), (i = 1, 2, 3) (4.117)

and the ratio of the deformed volume to the original becomes

dV

dV0=dx(1)dx(2)dx(3)

dX(1)dX(2)dX(3)= Λ(N1)Λ(N2)Λ(N3) (4.118a)

which, when Eq 4.105 is used, becomes,


dV

dV0=√C(1)C(2)C(3) =

√IIIC . (4.118b)

The importance of the second form of Eq 4.118b is that it is an invariant expression andcan be calculated without reference to the principal axes of C.

Example 4.14

Determine the volume ratio dV/dV0 for the deformation of Example 4.13 usingEq 4.118a, and verify using Eq 4.118b.

SolutionAs the student should show, a set of principal axes for the C tensor of Ex-ample 4.13 are N1 = (I1 + I2 + I3)/

√3, N2 = (I1 − I2)/

√2, and N3 =

(I1 + I2 − 2I3)/√6. Thus from Eq 4.105 the principal stretches are Λ(N1) = 1,

Λ(N2) = 2, and Λ(N3) = 2, respectively. Using these results, Eq 4.118a givesdV/dV0 = 4. By Eq 4.118b,

IIIC = detC =

∣∣∣∣∣∣3 −1 −1

−1 3 −1−1 −1 3

∣∣∣∣∣∣ = 16 .

and dV/dV0 =√16 = 4.

4.10 Rotation Tensor, Stretch TensorsIn Chapter 2 we noted that an arbitrary second-order tensor may be resolved by an addi-tive decomposition into its symmetric and skew-symmetric parts. Here, we introduce amultiplicative decomposition known as the polar decomposition by which any non-singulartensor can be decomposed into a product of two component tensors. Recall that the de-formation gradient F is a non-singular (invertible) tensor. Because of this nonsingularity,the deformation gradient can be decomposed into either of the two products

F = R ·U = V · R (4.119)

where R is the orthogonal rotation tensor, and U and V are symmetric, positive-definitetensors called the right and left stretch tensors, respectively. Moreover, U and V have thesame eigenvalues (see Problem 4.32).

The deformation gradient can be thought of as a mapping of the infinitesimal vector dXof the reference configuration into the infinitesimal vector dx of the current configuration.Note that the first decomposition in Eq 4.119 replaces the linear transformation dx = F·dXof Eq 4.41 by two sequential transformations,


dx ′ = U · dX (4.120a)

followed bydx = R · dx ′ . (4.120b)

The tensor U has three positive eigenvalues, U(1), U(2) and U(3) called the principalstretches, and associated with each is a principal stretch direction, N1, N2 and N3, respec-tively. These unit vectors form an orthogonal triad known as the right principal directions ofstretch. By the transformation Eq 4.120a, line elements along the directions Ni, (i = 1, 2, 3)are stretched by an amount U(i), (i = 1, 2, 3), respectively, with no change in direction.This is followed by a rigid body rotation given by Eq 4.120b. The second decompositionof Eq 4.120b reverses the sequence; first a rotation by R, then the stretching by V. In ageneral deformation, a rigid body translation may also be involved as well as the rotationand stretching described here.

As a preliminary to determining the rotation and stretch tensors, we note that an ar-bitrary tensor T is positive definite if v · T · v > 0 for all vectors v 6= 0. A necessary andsufficient condition for T to be positive definite is for all its eigenvalues to be positive. Inthis regard, consider the tensor C = FT · F. Inasmuch as F is non-singular (det F 6= 0) andF · v 6= 0 if v 6= 0, we see that (F · v) · (F · v) is a sum of squares and hence greater thanzero. Thus

(F · v) · (F · v) = v · FT · F · v = v ·C · v > 0 , (4.121)

and C is positive definite. Furthermore,

(FT · F)T = FT · (FT )T = FT · F (4.122)

which proves that C is also symmetric. By the same arguments we may show thatc = (F−1)T · (F−1) is also symmetric and positive definite.

Now let C be given in principal axes form by the matrix

[C∗AB] =

C(1) 0 0

0 C(2) 0

0 0 C(3)

(4.123)

and let [aMN] be the orthogonal transformation that relates the components of C∗ to thecomponents of C in any other set of axes through the equation expressed here in bothindicial and matrix form

C∗AB = aAQaBPCQP or C∗ = ACAT . (4.124)

We defineU as the square root of C, that is,U =√C orU ·U = C, and since the principal

values C(i), (i = 1, 2, 3) are all positive we may write

[√C∗AB

]=

√C(1) 0 0

0√C(2) 0

0 0√C(3)

= [U∗AB] , (4.125)

and, as is obvious, the inverse (U∗)−1 by

[(U∗AB)−1

]=

1/√C(1) 0 0

0 1/√C(2) 0

0 0 1/√C(3)

. (4.126)


Note that both U and U−1 are symmetric positive definite tensors given by

UAB = aQAaPBU∗QP or U = ATU∗A , (4.127)

andU−1AB = aQAaPB(U∗QP)

−1 or U−1 = AT (U∗)−1A (4.128)

respectively.Therefore, from the first decomposition in Eq 4.119,

R = F ·U−1 , (4.129)

so that

RT · R = (F ·U−1)T · (F ·U−1) = (U−1)T · FT · F ·U−1 (4.130)

= U−1 ·C ·U−1 = U−1 ·U ·U ·U−1 = I (4.131)

which shows that R is proper orthogonal.The second decomposition in Eq 4.119 may be confirmed by a similar development

using C−1 = F · FT = V2.

Example 4.15

A homogeneous deformation is given by the equations; x1 = 2X1 − 2X2, x2 =X1 + X2 and x3 = X3. Determine the polar decomposition F = R ·U for thisdeformation.

SolutionThe matrix form of the tensor FiA = xi,A is easily determined to be,

[FiA] =

2 −2 0

1 1 0

0 0 1

from which we calculate C = FT · F,

[CAB] =

5 −3 0

−3 5 0

0 0 1

.

In principal axes form this matrix becomes

[C∗AB] =

8 0 0

0 2 0

0 0 1

with an orthogonal transformation matrix found to be

[aMN] =

1√2

− 1√2

0

1√2

1√2

0

0 0 1

.


This is found by calculating the eigenvectors of C. Therefore, from Eqs 4.125and 4.126

[U∗AB] =

2√2 0 0

0√2 0

0 0 1

and [(U∗AB)−1] =1

2√2

1 0 0

0 2 0

0 0 2√2

,

and by use of the transformation equations Eqs 4.127 and 4.128 (as the studentshould verify), we determine

[UAB] =

3/√2 −√2 0

−√2 3

√2 0

0 0 1

,

and

[U−1AB] =

1

4√2

3 1 0

1 3 0

0 0 4√2

.

Finally, from Eq 4.129,

[RAB] =

2 −2 0

1 1 0

0 0 1

3 1 0

1 3 0

0 0 4√2

1

4√2

=

1/√2 −1/√2 0

1/√2 1/

√2 0

0 0 1

It is readily confirmed using these results that F = RU and that RT · R = I.

4.11 Velocity Gradient, Rate of Deformation, VorticityLet the velocity field of a continuum be given in some region of space by vi = vi(x, t).The spatial velocity gradient is defined by

Lij =∂vi

∂xj. (4.132)

An additive decomposition of this tensor into its symmetric and skew-symmetric parts iswritten as

Lij = dij +wij (4.133)

where the symmetric portion

dij =1

2

(∂vi

∂xj+∂vj

∂xi

)(4.134)


p

q

dx i

vi+dv i

vi

O

x2

x3

x1

FIGURE 4.7Differential velocity field at point p.

is the rate of deformation tensor, and the skew-symmetric portion

wij =1

2

(∂vi

∂xj−∂vj

∂xi

)(4.135)

is the vorticity, or spin tensor. This decomposition makes no assumption on the velocitygradient components being small, and is valid for finite components ∂vi/∂xj.

Consider the velocity components at two neighboring points p and q. Let the particlecurrently at p have a velocity vi, and the particle at q a velocity vi + dvi as shown inFig. 4.7. Thus the particle at q has a velocity relative to the particle at p of

dvi =∂vi

∂xjdxj or dv = L · dx . (4.136)

Note that∂vi

∂xj=∂vi

∂XA

∂XA

∂xj=d

dt

(∂xi

∂XA

)∂XA

∂xj, (4.137)

or in symbolic notationL = F · F−1 (4.138)

where we have used the fact that material time derivatives and material gradients com-mute. Therefore,

F = L · F . (4.139)

Consider next the stretch ratio Λ = dx/dX where Λ is as defined in Eq 4.105, that is, thestretch of the line element dX initially along N and currently along n. By the definitionof the deformation gradient, dxi = xi,AdXA, along with the unit vectors ni = dxi/dx andNA = dXA/dX we may write

dxni = xi,AdXNA


which becomes (after dividing both sides by the scalar dX)

niΛ = xi,ANA or nΛ = F · N . (4.140)

If we take the material derivative of this equation (using the symbolic notation for conve-nience),

˙nΛ+ nΛ = F · N = L · F · N = L · nΛ ,so that

˙n+ nΛ/Λ = L · n . (4.141)

By forming the inner product of this equation with n we obtain

n · ˙n+ n · nΛ/Λ = n · L · n .

But n · n = 1 and so ˙n · n = 0, resulting in

Λ/Λ = n · L · n or Λ/Λ = vi,jninj (4.142)

which represents the rate of stretching per unit stretch of the element that originated in thedirection of N, and is in the direction of n of the current configuration. Note further thatEq 4.142 may be simplified since W is skew-symmetric which means that

Lijninj = (dij +wij)ninj = dijninj ,

and soΛ/Λ = n ·D · n or Λ/Λ = dijninj . (4.143)

For example, for the element in the x1 direction, n = e1 and

Λ/Λ =[1 0 0

] d11 d12 d13d12 d22 d23d13 d23 d33

100

= d11 .

Likewise, for n = e2, Λ/Λ = d22 and for n = e3, Λ/Λ = d33. Thus the diagonal elementsof the rate of deformation tensor represent rates of extension, or rates of stretching in thecoordinate (spatial) directions.

In order to interpret the off-diagonal elements of the rate of deformation tensor we con-sider two arbitrary differential vectors dx(1)

i and dx(2)i at p. The material derivative of the

inner product of these two vectors is (using the superposed dot to indicate differentiationwith respect to time,

˙dx

(1)i dx

(2)i =

˙dx

(1)i dx

(2)i + dx

(1)i

˙dx

(2)i

= dv(1)i dx

(2)i + dx

(1)i dv

(2)i

= vi,jdx(1)j dx

(2)i + dx

(1)i vi,jdx

(2)j

= (vi,j + vj,i)dx(1)i dx

(2)j

= 2dijdx(1)i dx

(2)j . (4.144)

But dx(1)i dx

(2)i = dx(1)dx(2) cos θ, and

˙dx(1)dx(2) cos θ =

˙dx

(1)i dx

(2)i cos θ+ dx(1) ˙

dx(2) cos θ− dx(1)dx(2)θ sin θ

=

( ˙dx(1)

dx(1)+

˙dx(2)

dx(2)

cos θ− θ sin θ

)dx(1)dx(2) . (4.145)


Equating Eqs 4.144 and 4.145 gives

2dijdx(1)i dx

(2)i =

( ˙dx(1)

dx(1)+

˙dx(2)

dx(2)

cos θ− θ sin θ

)dx(1)dx(2) . (4.146)

If dx(1)i = dx

(2)i = dxi, then θ = 0, and cos θ = 1, sin θ = 0 and dx(1) = dx(2) = dx so that

Eq 4.146 reduces to

dijdxi

dx

dxj

dx= dijninj =

dx

dx(4.147)

which is seen to be the rate of extension per unit length of the element currently in thedirection of ni (compare with Eq 4.143). If, however, dx(1)

i is perpendicular to dx(2)i , so

that θ = π2 , cos θ = 0, sin θ = 1, then Eq 4.146 becomes

2dijn(1)i n

(2)i = n1· 2D · n2 = −θ . (4.148)

This rate of decrease in the angle θ is a measure of the shear rate between the elementsin the directions of n1 and n2. In the engineering literature it is customary to define therate of shear as half the change (increase or decrease) between two material line elementsinstantaneously at right angles to one another. Thus for n1 = e1 and n2 = e2,

−1

2θ12 = e1 ·D · e2 = d12 ,

and, in general, the off-diagonal elements of the rate of deformation tensor are seen torepresent shear rates for the three pairs of coordinate axes.

BecauseD is a symmetric, second-order tensor, the derivation of principal values, prin-cipal directions, a Mohr’s circles representation, a rate of deformation deviator tensor,etc., may be carried out as with all such tensors. Also, it is useful to develop the relation-ship between D and the material derivative of the strain tensor E. Recall that

2E = C− I = FT · F− I ,

so that, using Eq 4.139

2E = FT · F+ FT · F = (L · F)T · F+ FT · (L · F)= FT · LT · F+ FT · L · F = FT · (LT + L) · F = FT ·(2D) · F ,

orE = FT ·D · F . (4.149)

Note also that from ui + Xi = xi we have ui,A + δi,A = xi,A and if the displacementgradients ui,A are very small, ui,A 1 and may be neglected, then δi,A ≈ xi,A (I ≈ F),and of course, FT = IT = I. At the same time for ui,A very small in magnitude, byEq 4.61, E ≈ ε and Eq 4.74 reduces to

ε = I ·D · I = D (4.150)

for the infinitesimal theory. Finally, taking the material derivative of the difference (dx)2−

(dX)2 = dX · 2E · dX, and noting that ˙(dx)2 − (dX)2 =

˙(dx)2 since ˙

(dX)2 = 0, we obtain

˙(dx)2 = dX · 2E · dX = dX · FT · 2D · F · dX = dx · 2D · dx (4.151)


which shows that the local motion at some point x is a rigid body motion if and only ifD = 0 at x.

Solving Eq 4.141 for ni and using Eq 4.143 we may write

ni = vi,jnj − niΛ/Λ = (dij +wij)nj − dqknqnkni .

If now, ni is chosen along a principal direction of D so that dijn(p)j = D(p)n

(p)i (p =

1, 2, 3) where D(p) represents a principal value of D, then

ni = D(p)n(p)i +wijn

(p)j −D(p)n

(p)q n(p)

q n(p)i = wijn

(p)j (4.152)

since n(p)q n

(p)q = 1. Because a unit vector can change only in direction, Eq 4.152 indicates

that wij gives the rate of change in direction of the principal axes of D. Hence the names,vorticity, or spin given to W. Additionally, we associate with W the vector

wi =1

2εijkvk,j or w =

1

2curl v (4.153)

called the vorticity vector, by the following calculation,

εpqiwi =1

2εpqiεijkvk,j =

1

2(δpjδqk − δpkδqj)vk,j

=1

2(vq,p − vp,q) = wqp . (4.154)

Thus if D ≡ 0 so that Lij = wij, it follows that dvi = Lijdxj = wijdxj = εjikwkdxj andsince εjik = −εijk = εikj,

dvi = εijkwjdxk or dv = w× dx (4.155)

according to which the relative velocity in the vicinity of p corresponds to a rigid bodyrotation about an axis through p. The vectorw indicates the angular velocity, the directionand the sense of this rotation.

To summarize the physical interpretation of the velocity gradient L, we note that iteffects a separation of the local instantaneous motion into two parts:

1. The so-called logarithmic rates of stretching, D(p), (p = 1, 2, 3), that is the eigenval-ues of D along the mutually orthogonal principal axes of D,

Λ/Λ =d(lnΛ)

dt= dijn

(p)i n

(p)j = n

(p)i D(p)n

(p)i = D(p) ,

and

2. A rigid body rotation of the principal axes of D with angular velocity w.

Example 4.16

Consider a continuum having a shearing motion defined by the mapping

x1 = X1 + κX2 ,x2 = X2 ,

x3 = X3 .


(a) Invert the mapping to find XA in terms of xi(b) Determine xj,B and XA,j(c) Compute XA,jxj,B and xj,BXB,i

(d) Use the result of (c) to determine the indicial expression for ˙XA,i forall general motions (not just this pure shear case).

Solution

(a) The mapping can be inverted, find XA in terms of xi, by inspection toget

X1 = x1 − κx2 ,X2 = x2 ,

X3 = x3 .

(b) Partial differentiation of the two forms of the motion give us F = xj,Band F−1 = XA,j

xj,B =

1 κ 0

0 1 0

0 0 1

and XA,j =

1 −κ 0

0 1 0

0 0 1

.

(c) Multiplying these two expressions will solidify that XA,jxj,B is equiv-alent to F−1F and xj,BXB,i is equivalent to FF−1. In indicial notationthis fact is sometimes not obvious when so many new symbols havebeen introduced.

XA,jxj,B =

1 −κ 0

0 1 0

0 0 1

1 κ 0

0 1 0

0 0 1

=

1 κ− κ 0

0 1 0

0 0 1

= δAB

xj,BXB,i =

1 κ 0

0 1 0

0 0 1

1 −κ 0

0 1 0

0 0 1

=

1 −κ+ κ 0

0 1 0

0 0 1

= δij

(d) Take the material derivative of the first form noting that δAB is aconstant, and use Eq 4.138

˙XA,jxj,B = −XA,j ˙xj,B = −XA,jLjmxm,B .

Multiply both sides by XB,i which is the inverse of xj,B to get ˙XA,j:

˙XA,jxj,BXB,i = ˙XA,jδji = ˙XA,i = −XA,jLjmxm,BXB,i= −XA,jLjmδmi = −XA,jLji .

Thus,˙XA,i = −XA,jLji = −XA,jvj,i .


dX(2)

dX(1)

dx(2)

dx(1)

DEFO

R

MS

dS

dS

dS0

dS0

FIGURE 4.8Area dS0 between vectors dX(1) and dX(2) in the reference configuration becomes dSbetween dx(1) and dx(2) in the deformed configuration.

4.12 Material Derivative of Line Elements, Areas, VolumesConsider first the material derivative of the differential line element dx = F · dX. Clearly,dx = F · dX and by Eq 4.139,

dx = F · dX = L · F · dX = L · dx or ˙dxi = vi,jdxj . (4.156)

Note further, that from Eq 4.156 the material derivative of the dot product dx · dx is

˙dx · dx = 2dx · dx = dx · 2L · dx = dx · 2(D+W) · dx = dx · 2D · dx

in agreement with Eq 4.151.It remains to develop expressions for the material derivatives of area and volume ele-

ments. Consider the plane area defined in the reference configuration by the differentialline elements dX(1)

A and dX(2)A as shown in Fig. 4.8. The parallelogram area dS0 may be

represented by the vectordS0A = εABCdX

(1)B dX

(2)C . (4.157)

As a result of the motion x = x(X, t) this area is carried into the current area dSi shownin Fig. 4.8 and is given by

dSi = εijkdx(1)j dx

(2)k = εijkxj,BdX

(1)B xk,CdX

(2)C (4.158)

which upon multiplication by xi,A results in

xi,AdSi = εijkxi,Axj,Bxk,CdX(1)B dX

(2)C = εijkFiAFjBFkCdX

(1)B dX

(2)C .

Recall that detF = J (the Jacobian) and from Eq 2.43

εijkFiAFjBFkC = εABC det F = εABCJ .


Therefore, by inserting this result into the above equation for xi,AdSi and multiplyingboth sides by XA,q, we obtain,

xi,AXA,qdSi = εABCJdX(1)B dX

(2)C XA,q

But xi,AXA,q = δiq so thatδiqdSi = dSq = XA,qJdS

0A (4.159)

which expresses the current area in terms of the original area.To determine the material derivative of dSi, we need the following identity

˙detA = tr(·A ·A−1

)detA (4.160)

where A is an arbitrary tensor. Substituting F for A we obtain

˙det F = J = (det F) tr(F · F−1) = J tr(L) ,

orJ = Jvi,i = Jdiv v . (4.161)

Noting that Eq 4.159 may be written

dSq = JXA,qdS0A

and using symbolic notation to take advantage of Eq 4.160 we obtain

dS = J(F−1)T · dS0 = JdS0 · F−1 ,

and sodS · F = JdS0

which upon differentiating becomes

dS · F+ dS · F = JdS0 = J(trL)dS0 ,

dS + dS · F · F−1 = J(trL)dS0 · F−1 = (trL)dS ,

and finallydS = (trL)dS − dS · L or dSi = vk,kdSi − dSjvj,i (4.162)

which gives the rate of change of the current element of area in terms of the current area,the trace of the velocity gradient, and of the components of L.

Consider next the volume element defined in the referential configuration by the boxproduct,

dV0 = dX(1) · dX(2) × dX(3) = εABCdX(1)A dX

(2)B dX

(3)C =

[dX(1), dX(2), dX(3)

]and pictured in Fig. 4.9, and let the deformed volume element, also shown in Fig. 4.9 begiven by

dV = dx(1) · dx(2) × dx(3) = εijkdx(1)i dx

(2)j dx

(3)k =

[dx(1), dx(2), dx(3)

].


dX(2)

dX(1)

dx(2)

dx(1)

dX(3)

dx(3)

DEF

ORM

FIGURE 4.9Volume of parallelpiped defined by vectors dX(1), dX(2) and dX(3) in the reference con-figuration deforms into volume defined by parallelpiped defined by vectors dx(1), dx(2)

and dx(3) in the deformed configuration.

For the motion x = x(X, t), dx = F · dX so the current volume is the box product

dV =[F · dX(1), F · dX(2), F · dX(3)

]= εijkxi,Axj,Bxk,CdX

(1)A dX

(2)B dX

(3)C

= det(F)[dX(1), dX(2), dX(3)

]= JdV0 (4.163)

which gives the current volume element in terms of its original size. Since J 6= 0 (F isinvertible), we have either J < 0 or J > 0. Mathematically, J < 0 is possible, but physicallyit corresponds to a negative volume, so we reject it. Henceforth, we assume J > 0. If J = 1,then dV = dV0 and the volume magnitude is preserved. If J is equal to unity for all X wesay the motion is isochoric.

To determine the time rate of change of dV we take the material derivative as follows,

dV = JdV0 = J tr (L)dV0 = Jvi,idV0 = vi,idV . (4.164)

Thus a necessary and sufficient condition for a motion to be isochoric is that

vi,i = div v = 0 . (4.165)

In summary, we observe that the deformation gradient F governs the stretch of a lineelement, the change of an area element, and the change of a volume element. But it is thevelocity gradient L that determines the rate at which these changes occur.


Example 4.17

To derive the material derivative of the Jacobian, J, the identity Eq 4.160 wasused in the text. As an alternative derivation, use

J =1

6εABCεijkFiAFjBFkC

to derive Eq 4.161.

SolutionDifferentiate and use Eq 4.139, Eq 2.43 and Eq 2.7b

J = 16εABCεijk

(FiAFjBFkC + FiAFjBFkC + FiAFjBFkC

)= 1

6εABCεijk (LimFmAFjBFkC + FiALjmFmBFkC + FiAFjBLkmFmC)

= 16εijk (LimεmjkJ+ LjmεimkJ+ LkmεijmJ)

= 16

(2δimLim + 2δmjLjm + 2δkmLkm) J

= JLii = Jvi,i .


Problems

Problem 4.1The motion of a continuous medium is specified by the component equations

x1 = 12 (X1 + X2)e

t + 12 (X1 − X2)e

−t ,

x2 = 12 (X1 + X2)e

t − 12 (X1 − X2)e

−t ,

x3 = X3 .

(a) Show that the Jacobian determinant J does not vanish, and solve for the inverseequations X = X(x, t).

(b) Calculate the velocity and acceleration components in terms of the materialcoordinates.

(c) Using the inverse equations developed in part (a), express the velocity and ac-celeration components in terms of spatial coordinates.

Answer

(a) J = cosh2 t− sinh2 t = 1

X1 = 12 (x1 + x2)e

−t + 12 (x1 − x2)e

t

X2 = 12 (x1 + x2)e

−t − 12 (x1 − x2)e

t

X3 = x3

(b) v1 = 12 (X1 + X2)e

t − 12 (X1 − X2)e

−t

v2 = 12 (X1 + X2)e

t + 12 (X1 − X2)e

−t

v3 = 0

a1 = 12 (X1 + X2)e

t + 12 (X1 − X2)e

−t

a2 = 12 (X1 + X2)e

t − 12 (X1 − X2)e

−t

a3 = 0

(c) v1 = x2, v2 = x1, v3 = 0

a1 = x1, a2 = x2, a3 = 0

Problem 4.2Let the motion of a continuum be given in component form by the equations

x1 = X1 + X2t+ X3t2 ,

x2 = X2 + X3t+ X1t2 ,

x3 = X3 + X1t+ X2t2 .

(a) Show that J 6= 0, and solve for the inverse equations.

(b) Determine the velocity and acceleration

(1) at time t = 1 s for the particle which was at point (2.75, 3.75, 4.00) whent = 0.5 s.

(2) at time t = 2 s for the particle which was at point (1, 2,−1) when t = 0.


Answer

(a) J = (1− t3)2

X1 = (x1 − x2t)/(1− t3)X2 = (x2 − x3t)/(1− t3)X3 = (x3 − x1t)/(1− t3)

(b) (1) v = 8e1 + 5e2 + 5e3, a = 6e1 + 2e2 + 4e3

(2) v = −2e1 + 3e2 + 9e3, a = −2e1 + 2e2 + 4e3

Problem 4.3A continuum body has a motion defined by the equations

x1 = X1 + 2X2t2 ,

x2 = X2 + 2X1t2 ,

x3 = X3 .

(a) Determine the velocity components at t = 1.5 s of the particle which occupiedthe point (2, 3, 4) when t = 1.0 s.

(b) Determine the equation of the path along which the particle designated in part(a) moves.

(c) Calculate the acceleration components of the same particle at time t = 2 s.

Answer

(a) v1 = 2, v2 = 8, v3 = 0

(b) 4x1 − x2 = 5 in the plane x3 = 4

(c) a1 = 4/3, a2 = 16/3, a3 = 0.

Problem 4.4If the motion x = x(X, t) is given in component form by the equations

x1 = X1(1+ t), x2 = X2(1+ t)2, x3 = X3(1+ t2) ,

determine expressions for the velocity and acceleration components in terms of both La-grangian and Eulerian coordinates.

Answer

v1 = X1 = x1/(1+ t)

v2 = 2X2(1+ t) = 2x2/(1+ t)

v3 = 2X3t = 2x3t/(1+ t2)

a1 = 0

a2 = 2X2 = 2x2/(1+ t)2

a3 = 2X3 = 2x3/(1+ t2)


Problem 4.5

The Lagrangian description of a continuum motion is given by

x1 = X1e−t + X3(e

−t − 1) ,

x2 = X2et − X3(1− e−t) ,

x3 = X3et .

Show that these equations are invertible and determine the Eulerian description of themotion.

Answer

X1 = x1et − x3(e

t − 1)

X2 = x2e−t + x3(e

−2t − e−3t)

X3 = x3e−t

Problem 4.6

A velocity field is given in Lagrangian form by

v1 = 2t+ X1, v2 = X2et, v3 = X3 − t .

Integrate these equations to obtain x = x(X, t) with x = X at t = 0, and using that resultcompute the velocity and acceleration components in the Eulerian (spatial) form.

Answer

v1 = (x1 + 2t+ t2)/(1+ t)

v2 = x2

v3 = (2x3 − 2t− t2)/2(1+ t)

a1 = 2, a2 = x2, a3 = −1

Problem 4.7

If the motion of a continuous medium is given by

x1 = X1et − X3(e

t − 1) ,

x2 = X2e−t + X3(1− e−t) ,

x3 = X3 ,

determine the displacement field in both material and spatial descriptions.

Answer

u1 = (X1 − X3)(et − 1) = (x1 − x3)(1− e−t)

u2 = (X2 − X3)(e−t − 1) = (x2 − x3)(1− et)

u3 = 0


Problem 4.8The temperature field in a continuum is given by the expression

θ = e−3t/x2 where x2 = x21 + x22 + x23 .

The velocity field of the medium has components

v1 = x2 + 2x3, v2 = x3 − x1, v3 = x1 + 3x2 .

Determine the material derivative dθ/dt of the temperature field.

Answer

dθ/dt = −e−3t(3x2 + 6x1x3 + 8x2x3)/x4

Problem 4.9In a certain region of a fluid the flow velocity has components

v1 = A(x31 + x1x22)e

−kt, v2 = A(x21x2 + x33)e−kt, v3 = 0

where A and k are constants. Use the (spatial) material derivative operator to determinethe acceleration components at the point (1, 1, 0) when t = 0.

Answer

a1 = −2A(k− 5A), a2 = −A(k− 5A), a3 = 0

Problem 4.10A displacement field is given in terms of the spatial variables and time by the equations

u1 = x2t2, u2 = x3t, u3 = x1t .

Using the (spatial) material derivative operator, determine the velocity components.

Answer

v1 = (2x2t+ x3t2 + x1t

3)/(1− t4)

v2 = (x3 + x1t+ 2x2t3)/(1− t4)

v3 = (x1 + 2x2t2 + x3t

3)/(1− t4)

Problem 4.11For the motion given by the equations

x1 = X1 cosωt+ X2 sinωt ,x2 = −X1 sinωt+ X2 cosωt ,x3 = (1+ kt)X3 ,

where ω and k are constants, determine the displacement field in Eulerian form.


Answer

u1 = x1(1− cosωt) + x2 sinωt

u2 = −x1 sinωt+ x2(1− cosωt)

u3 = x3kt/(1+ kt)

Problem 4.12Show that the displacement field for the motion analyzed in Problem 4.1 has the Eulerian

form

u1 = x1 − (x1 + x2)e−t/2− (x1 − x2)e

t/2 ,

u2 = −x2 − (x1 + x2)e−t/2+ (x1 − x2)e

t/2 ,

and by using the material derivative operator (dui/dt = ∂ui/∂t + vj∂ui/∂xj), verify thevelocity and acceleration components calculated in Problem 4.1.

Problem 4.13The Lagrangian description of a deformation is given by

x1 = X1 + X3(e2 − e−2) ,

x2 = X2 − X3(e2 − 1) ,

x3 = X3e2 .

Determine the components of the deformation matrix FiA and from it show that the Jaco-bian J does not vanish. Invert the mapping equations to obtain the Eulerian description ofthe deformation.

Answer

J = e2

X1 = x1 − x3(1− e−4)

X2 = x2 + x3(1− e−2)

X3 = x3e−2

Problem 4.14A homogeneous deformation has been described as one for which all of the deformation

and strain tensors are independent of the coordinates, and may therefore be expressed ingeneral by the displacement field ui = AijXj where the Aij are constants (or in the caseof a motion, functions of time). Show that for a homogeneous deformation with the Aijconstant:

(a) plane material surfaces remain plane,

(b) straight line particle elements remain straight,

(c) material surfaces which are spherical in the reference configuration become ellip-soidal surfaces in the deformed configuration.


Problem 4.15

An infinitesimal homogeneous deformation ui = AijXj is one for which the constants Aij areso small that their products may be neglected. Show that for two sequential infinitesimaldeformations the total displacement is the sum of the individual displacements regardlessof the order in which the deformations are applied.

Problem 4.16

For the homogeneous deformation defined by

x1 = αX1 + βX2 ,

x2 = −βX1 + αX2 ,

x3 = µX3 ,

where α, β and µ are constants, calculate the Lagrangian finite strain tensor E. Show thatif α = cos θ, β = sin θ and µ = 1 the strain is zero and the mapping corresponds to a rigidbody rotation of magnitude θ about the X3 axis.

Answer

EAB = 12

α2 + β2 − 1 0 0

0 α2 + β2 − 1 0

0 0 µ2 − 1

Problem 4.17

Given the deformation defined by

x1 = X1, x2 = X2 +1

2X23, x3 = X3

(a) Sketch the deformed shape of the unit square OABC in the plane X1 = 0.

(b) Determine the differential vectors dx(2) and dx(3) which are the deformed vectorsresulting from dX(2) = dX(2)I2 and dX(3) = dX(3)I3, respectively, that wereoriginally at corner C.

(c) Calculate the dot product dx(2) · dx(3), and from it determine the change in theoriginal right angle between dX(2) and dX(3) at C due to the deformation.

(d) Compute the stretch Λ at B in the direction of the unit normal

N =(I2 + I3

)/√2 .


X3

X2

X1

OA

BC

Unit square OABC in the reference configuration.

Answer

(b) dx(2) = dX(2)e2, dx(3) = dX(3) (e2 + e3)

(c) ∆θ = −45

(d) Λ(N) =√2.5

Problem 4.18Given the deformation expressed by

x1 = X1 +AX22, x2 = X2, x3 = X3 −AX22

where A is a constant (not necessarily small), determine the finite strain tensors E and e,and show that if the displacements are small so that x ≈ X, and if squares of A may beneglected, both tensors reduce to the infinitesimal strain tensor ε.

Answer

[εij] =

0 Ax2 0

Ax2 0 −Ax20 −Ax2 0

Problem 4.19For the infinitesimal homogeneous deformation xi = Xi + AijXj where the constants Aij

are very small, determine the small strain tensor ε, and from it the longitudinal (normal)strain in the direction of the unit vector N =

(I1 − I3

)/√2.

Answer

2e(N) = A11 −A13 −A31 +A33

Problem 4.20A deformation is defined by

x1 = X1/(X21 + X22

), x2 = X2/

(X21 + X22

), x3 = X3 .


Determine the deformation tensor C together with its principal values.

Answer

C(1) = C(2) =(X21 + X22

)−2, C(3) = 1

Problem 4.21For the deformation field given by

x1 = X1 + αX2, x2 = X2 − αX1, x3 = X3

where α is a constant, determine the matrix form of the tensors E and e, and show thatthe circle of particles X21 + X22 = 1 deforms into the circle x21 + x22 = 1+ α2 .

Answer

[Eij] = 12

α2 0 0

0 α2 0

0 0 0

[eij] =1

2 (1+ α2)2

−α2 0 0

0 −α2 0

0 0 0

Problem 4.22Let the deformation of a continuum be given by the equations

x1 = X1 + kX22, x2 = X2 − kX21, x3 = X3

where k is a constant. Determine the Lagrangian finite strain tensor E, and from it, assumingk is very small, deduce the infinitesimal strain tensor ε. Verify this by calculating thedisplacement field and using the definition 2εij = ui,j + uj,i for the infinitesimal theory.

Problem 4.23Given the displacement field

u1 = AX2X3, u2 = AX23, u3 = AX21

where A is a very small constant, determine

(a) the components of the infinitesimal strain tensor ε, and the infinitesimal rotationtensor ω.

(b) the principal values of ε, at the point (1, 1, 0).

Answer

(a) ε11 = ε22 = ε33 = 0, ε12 = AX3, ε13 = 12A(X2+2X1) and ε23 = AX3

ω11 = ω22 = ω33 = 0, ω12 = −ω21 = 12AX3,

ω13 = −ω31 = 12AX1 −AX1 and ω23 = −ω32 = AX3

(b) ε(I) = 32A, ε(II) = 0, ε(III) = −32A


Problem 4.24A 45 strain rosette measures longitudinal strains along the X1, X2, and X′1 axes shown

below. At point O the strains recorded are

ε11 = 6× 10−4, ε22 = 4× 10−4 and ε′11 = 8× 10−4

Determine the shear strain γ12 at O, together with ε′22, and verify that ε11+ε22 = ε′11+ε′22.

See Eq 4.82.

X1

X2

X2‛

X1‛

45o

45o

O

Problem 4.25By a direct expansion of Eq 4.87, 2ωi = εijkωkj, show that ω1 = ω32 = −ω23, etc. Also,

show that only if A is a very small constant does the mapping

x1 = X1 −AX2 +AX3 ,

x2 = X2 −AX3 +AX1 ,

x3 = X3 −AX1 +AX2 ,

represent a rigid body rotation (E ≡ 0). Additionally, determine the infinitesimal rotationtensor ωij in this case; from it, using the result proven above, deduce the rotation vectorωi.

Answer

ω = A(e1 + e2 + e3)

Problem 4.26For the displacement field

u1 = kX1X2, u2 = kX1X2, u3 = 2k(X1 + X2)X3

where k is a very small constant, determine the rotation tensor ω, and show that it hasonly one real principal value at the point (0, 0, 1).


Answer

ω(1) = 0, ω(2) = −ω(3) = ik√2, where i =

√−1

Problem 4.27Let the deformation

x1 = X1 +AX2X3 ,

x2 = X2 +AX23 ,

x3 = X3 +AX21 ,

where A is a constant be applied to a continuum body. For the unit square of material lineelements OBCD as shown, calculate at point C

(a) the stretch and unit elongation for the element in the direction of diagonal OC,

(b) the change in the right angle at C if A = 1; if A = 0.1.

DX

3

X2

X1

OB

C

1

1

Unit square OBCD in the reference configuration.

Answer

(a) Λ2(OC) = 1+ 2A+ 4A2, e(OC) =√1+ 2A+ 4A2 − 1

(b) ∆θ(A=1) = 60, ∆θ(A=0.1) = 11.77

Problem 4.28For the homogeneous deformation expressed by the equations

x1 =√2X1 +

3√2

4X2 ,

x2 = −X1 +3

4X2 +

√2

4X3 ,

x3 = X1 −3

4X2 +

√2

4X3 ,

determine


(a) the unit normal n for the line element originally in the direction ofN =

(I1 − I2 + I3

)/√3.

(b) the stretch Λ(N) of this element.

(c) the maximum and minimum stretches at the point X1 = 1, X2 = 0, X3 = −2 inthe reference configuration.

Answer

(a) n =

√2e1 +

(√2− 7

)e2 +

(√2+ 7

)e3

√104

(b) Λ(N) = 1.472

(c) Λ(max) = 2; Λ(min) = 0.5

Problem 4.29Let the deformation

x1 = a1(X1 + 2X2), x2 = a2X2, x3 = a3X3

where a1, a2, and a3 are constants be applied to the unit cube of material shown in thesketch. Determine

(a) the deformed length l of diagonal OC,

(b) the angle between edges OA and OG after deformation,

(c) the conditions which the constants must satisfy for the deformation to be possibleif

(1) the material is incompressible,(2) the angle between elements OC and OB is to remain unchanged.

X1 , x

1

X2 , x

2

X3 , x

3

O

A

B

G

C

F

D

E

Unit cube having diagonal OC.


Answer

(a) l2 = 9a21 + a22 + a23

(b) cos θ =2a1√4a21 + a22

(c) (1) a1a2 a3 = 1,(2) 9a21 + a22 = 2a23

Problem 4.30A homogeneous deformation is defined by

x1 = αX1 + βX2, x2 = −αX1 + βX2, x3 = µX3

where α, β and µ are constants. Determine

(a) the magnitudes and directions of the principal stretches,(b) the matrix representation of the rotation tensor R,(c) the direction of the axis of the rotation vector, and the magnitude of the angle

of rotation.

Answer

(a) Λ2(1) = Λ2(e1) = 2a2, Λ2(2) = Λ2(e2) = 2β2, Λ2(3) = Λ2(e3) = µ2

(b) [Rij] =

1√2

1√2

0

− 1√2

1√2

0

0 0 1

(c) n = I3; Φ = 45

Problem 4.31Consider the deformation field

x1 = X1 −AX2 +AX3 ,

x2 = X2 −AX3 +AX1 ,

x3 = X3 −AX1 +AX2 ,

where A is a constant. Show that the principal values of the right stretch tensor have amultiplicity of two, and that the axis of the rotation tensor is along N =

(I1 + I2 + I3

)/√3.

Determine the matrix of the rotation vector together with the angle of rotation φ.

Answer

Λ(1) = 1, Λ(2) = Λ(3) =√1+ 3A2 = β

[Rij] =1

3β

β+ 2 β− 1− 3A β− 1+ 3Aβ− 1+ 3A β+ 2 β− 1− 3Aβ− 1− 3A β− 1+ 3A β+ 2

φ = cos−1(1/β)


Problem 4.32For the deformation field

x1 =√3X1 + X2 ,

x2 = 2X2 ,

x3 = X3 .

determine

(a) the matrix representation of the rotation tensor R,

(b) the right stretch tensor U and the left stretch tensor V, then show that theprincipal values of U and V are equal,

(c) the direction of the axis of rotation and the magnitude of the angle of rotation.

Answer

(a) [Rij] =1

2√2

√3+ 1

√3− 1 0

−√3+ 1

√3+ 1 0

0 0 2√2

(b) Λ(1) =

√6, Λ(2) =

√2, Λ(3) = 1

(c) N = I3; φ = 15

Problem 4.33Let a displacement field be given by

u1 =1

4(X3 − X2), u2 =

1

4(X1 − X3), u3 =

1

4(X2 − X1) .

Determine

(a) the volume ratio dV/dV0,

(b) the change in the right angle between line elements originally along the unitvectors N1 =

(3I1 − 2I2 − I3

)/√14 and N2 =

(I1 + 4I2 − 5I3

)/√42. Explain

your answer.

Answer

(a) dV/dV0 = 1.1875

(b) ∆θ = 0

Problem 4.34Consider again the deformation given in Example 4.15, namely

x1 = 2(X1 − X2), x2 = X1 + X2, x3 = X3 .


Determine

(a) the left stretch tensor V,

(b) the direction normals of the principal stretches of V.

Answer

(a) [VAB] =

2√2 0 0

0√2 0

0 0 1

(b) N1 = I1, N2 = I2, N3 = I3

Problem 4.35A deformation field is expressed by

x1 = µ(X1 cosβX3 + X2 sinβX3) ,x2 = µ(−X1 sinβX3 + X2 cosβX3) ,x3 = νX3 ,

where µ, β, and ν are constants.

(a) Determine the relationship between these constants if the deformation is to be apossible one for an incompressible medium.

(b) If the above deformation is applied to the circular cylinder shown by the sketch,determine

(1) the deformed length l in terms of L, the dimension a, and the constants µ,β, and ν of an element of the lateral surface which has unit length and isparallel to the cylinder axis in the reference configuration, and

(2) the initial length L of a line element on the lateral surface which has unitlength and is parallel to the cylinder axis after deformation.

a

L

X3

X2

X1


Answer

(a) µ2ν = 1

(b) (1) l =√µ2β2a2 + ν2

(2) L =1

ν

√β2a2 + 1

Problem 4.36A velocity field is defined in terms of the spatial coordinates and time by the equations,

v1 = 2tx1 sin x3, v2 = 2tx2 cos x3, v3 = 0 .

At the point (1,−1, 0) at time t = 1, determine

(a) the rate of deformation tensor and the vorticity tensor,

(b) the stretch rate per unit length in the direction of the normaln = (e1 + e2 + e) /

√3,

(c) the maximum stretch rate per unit length and the direction in which it occurs,

(d) the maximum shear strain rate.

Answer

(a) [dij] =

0 0 1

0 2 0

1 0 0

, [wij] =

0 0 1

0 0 0

−1 0 0

(b) Λ/Λ = 4/3

(c)(Λ/Λ

)max = 2, n = e2

(d) γmax = 1.5

Problem 4.37Let NA and ni denote direction cosines of a material line element in the reference and

current configurations, respectively. Beginning with Eq 4.140, niΛ = xi,ANA, and usingthe indicial notation throughout, show that

(a) Λ/Λ = Dijninj,

(b) Λ/Λ = Qijninj+ ninj where Qij = 12 (ai,j+aj,i) with ai being the components

of acceleration.

Problem 4.38In a certain region of flow the velocity components are

v1 =(x31 + x1x

22

)e−kt, v2 =

(x32 − x21x2

)e−kt, v3 = 0

where k is a constant, and t is time in s. Determine at the point (1, 1, 1) when t = 0,

(a) the components of acceleration,

(b) the principal values of the rate of deformation tensor,


(c) the maximum shear rate of deformation.

Answer

(a) a1 = 2(4− k), a2 = −4, a3 = 0

(b) D(1) = 4, D(2) = 2, D(3) = 0

(c) γmax = ±2

Problem 4.39For the motion

x1 = X1 ,

x2 = X2et + X1(e

t − 1) ,

x3 = X1(et − e−t) + X3 ,

determine the velocity field vi = vi(x), and show that for this motion

(a) L = F · F−1,(b) D = ε at t = 0.

Problem 4.40Determine an expression for the material derivative d (lndx) /dt in terms of the rate of

deformation tensor D and the unit normal n = dx/dx.

Answer

d (lndx) /dt = n ·D · n

Problem 4.41A velocity field is given in spatial form by

v1 = x1x3, v2 = x22t, v3 = x2x3t .

(a) Determine the vorticity tensor W and the vorticity vector w.(b) Verify the equation εpqiwi = wqp for the results of part (a).(c) Show that at the point (1, 0, 1) when t = 1, the vorticity tensor has only one real

root.

Answer

(a) w1 = 12x3t, w2 = 1

2x1, w3 = 0

(c) W(1) = 0, W(2) = −W(3) = i/√2, where i =

√−1

Problem 4.42Consider the velocity field

v1 = ex3−ct cosωt, v2 = ex3−kt sinωt, v3 = c

where c and ω are constants.


(a) Show that the speed of every particle is constant.

(b) Determine the acceleration components ai.

(c) Calculate the logarithmic stretching, Λ/Λ = d (lnΛ) /dt, for the element in thecurrent configuration in the direction of n = (e1 + e2) /

√2 at x = 0.

Answer

(b) a1 = −ωv2, a2 = ωv1,a3 = 0

(c) 12e

−ct cosωt

Problem 4.43Show that the velocity field

v1 = 1.5x3 − 3x2, v2 = 3x1 − x3, v3 = x2 − 1.5x1

corresponds to a rigid body rotation, and determine the axis of spin (the vorticity vector).

Answer

w = e1 + 1.5e2 + 3e3

Problem 4.44For the steady velocity field

v1 = x21x2, v2 = 2x22x3, v3 = 3x1x2x3

determine the rate of extension at (2, 0, 1) in the direction of the unit vector (4e1 − 3e3) /5.

Answer

Λ/Λ = −48

25

Problem 4.45Prove that d(ln J) /dt = div v and, in particular, verify that this relationship is satisfied

for the motion

x1 = X1 + ktX3, x2 = X2 + ktX3, x3 = X3 − kt(X1 + X2)

where k is a constant.

Answer

J/J = div v = 4k2t/(1+ 2k2t2)

Problem 4.46Prove the identity (

xi,A

J

),i

= 0 , J = det(xi,A)


Problem 4.47Equation 4.151 gives the material derivative of dx2 in terms of dij . Using that equation

as the starting point, show that d2(dx2)/dt2 is given in terms of dij and its time derivativeby

d2(dx2)/dt2 = 2(dij + vk,idkj + vk,jdik)dxidxj .

Problem 4.48A continuum body in the form of the unit cube shown by the sketch undergoes the homo-

geneous deformationx1 = λ1X1, x2 = λ2X2, x3 = λ3X3

where λ1, λ2, and λ3 are constants.Determine the relationships among λ1, λ2, and λ3 if

(a) the length of diagonal OC remains unchanged,

(b) the rectangular area ABFE remains unchanged,

(c) the triangular area ACE remains unchanged.

X1 , x

1

X2 , x

2

X3 , x

3

O

A

B

G

C

F

D

E

(a)

X1 , x

1

X2 , x

2

X3 , x

3

O

A

B

G

C

F

D

E

(b)

X1 , x

1

X2 , x

2

X3 , x

3

O

A

B

G

C

F

D

E

(c)


Answer

(a) λ21 + λ22 + λ23 = 3

(b) λ22(λ21 + λ23

)= 2

(c) λ21λ22 + λ22λ

23 + λ23λ

21 = 3

Problem 4.49Let the unit cube shown in Problem 4.47 be given the motion

x1 = X1 +1

2t2X2, x2 = X2 +

1

2t2X1, x3 = X3

Determine, at time t,

(a) the rate-of-change of area ABFE,

(b) the volume of the body.

Answer

(a).

dS = t3e1/(1− 14t4) − te2 − t3e3

(b) V = (1− 14t4)

Problem 4.50For the homogeneous deformation

x1 = X1 + αX2 + αβX3 ,

x2 = αβX1 + X2 + β2X3 ,

x3 = X1 + X2 + X3 ,

where α and β are constants, determine the relationship between these constants if thedeformation is isochoric.

Answer

β = (α2 + α)/(α2 + α+ 1)

Problem 4.51Show that for any velocity field v derived from a vector potential ψ by v = curlψ, the

flow is isochoric. Also, for the velocity field

v1 = ax1x3 − 2x3, v2 = −bx2x3, v3 = 2x1x2

determine the relationship between the constants a and b if the flow is isochoric.

Answer

a = b


5Fundamental Laws and Equations

A number of the fundamental laws of continuum mechanics are expressions of the con-servation of some physical quantity. These balance laws, as they are often called, areapplicable to all material continua and result in equations that must always be satisfied.In this introductory text, we consider only the conservation laws dealing with mass, lin-ear and angular momentum, and energy. With respect to energy, we shall first develop apurely mechanical energy balance and follow that by an energy balance that includes bothmechanical and thermal energies, that is, a statement of the first law of thermodynamics.In addition to that, the Clausius-Duhem form of the second law of thermodynamics iscovered.

The balance laws are usually formulated in the context of global (integral) relationshipsderived by a consideration of the conservation of some property of the body as a whole.As explained in Chapter 1, the global equations may then be used to develop associatedfield equations which are valid at all points within the body and on its boundary. Forexample, we shall derive the local equations of motion from a global statement of theconservation of linear momentum.

Constitutive equations, which reflect the internal constitution of a material, define specifictypes of material behavior. They are fundamental in the sense that they serve as thestarting point for studies in the disciplines of elasticity, plasticity and various idealizedfluids. Constitutive equations establish the relationship between kinematics and stressesfor a material. These equations are the topic of the final section of this chapter.

Before we begin a discussion of the global conservation laws, it is useful to developexpressions for the material derivatives of certain integrals.

5.1 Material Derivatives of Line, Surface and Volume IntegralsLet any scalar, vector or tensor property of the collection of particles occupying the currentvolume V be represented by the integral

Pij...(t) =

∫V

P∗ij...(x, t)dV (5.1)

where P∗ij... represents the distribution of the property per unit volume and has continu-ous derivatives as necessary. The ellipsis in the subscript indicate that P∗ij... could be anyorder tensor. The material derivative of this property is given in both spatial and materialform, using Eq 4.163, by

Pij...(t) =d

dt

∫V

P∗ij...(x, t)dV =d

dt

∫V0P∗ij...[x(X, t), t]J dV

0 .

Since V0 is a fixed volume in the referential configuration, the differentiation and inte-gration commute and the differentiation can be performed inside the integral sign. Thus,

167


from Eq 4.161, using the notation.

[·] to indicate differentiation with respect to time,

∫V0

.[P∗ij...(X, t)J

]dV0 =

∫V0

(P∗ij...J+ P∗ij...J

)dV0

=

∫V0

(P∗ij... + vk,kP

∗ij...

)J dV0 ,

and converting back to the spatial formulation

Pij...(t) =

∫V

[P∗ij...(x, t) + vk,kP

∗ij...(x, t)

]dV . (5.2)

With the help of the material derivative operator given in Eq 4.32, this equation may bewritten (we omit listing the independent variables x and t for convenience),

Pij...(t) =

∫V

[∂P∗ij...

∂t+ vk

∂P∗ij...

∂xk+ vk,kP

∗ij...

]dV

=

∫V

[∂P∗ij...

∂t+(vkP

∗ij...k

),k

]dV

which upon application of the divergence theorem becomes

Pij...(t) =

∫V

∂P∗ij...

∂tdV +

∫S

vkP∗ij...nk dS . (5.3)

This equation gives the time rate of change of the property Pij... as the sum of the amountcreated in the volume V, plus the amount entering through the bounding surface S, andis often spoken of as the transport theorem.

Time derivatives of integrals over material surfaces and material curves may also bederived in an analogous fashion. First, we consider a tensorial property Qij... of theparticles which make up the current surface S, as given by

Qij...(t) =

∫S

Q∗ij...(x, t)dSp =

∫S

Q∗ij...(x, t)np dS (5.4)

where Q∗ij...(x, t) is the distribution of the property over the surface. From Eq 4.162, wehave in Eulerian form (again omitting the variables x and t),

Qij...(t) =

∫S

(Q∗ij... + vk,kQ

∗ij...

)dSp −

∫S

Q∗ij...vq,p dSq

=

∫S

[(Q∗ij... + vk,kQ

∗ij...

)δpq −Q∗ij...vq,p

]dSq . (5.5)

Similarly, for properties of particles lying on the spatial curve C and expressed by the lineintegral

Rij...(t) =

∫C

R∗ij...(x, t)dxp (5.6)

Fundamental Laws and Equations 169

we have, using Eq 4.156,

Rij...(t) =

∫C

R∗ij... dxp +

∫C

vp,qR∗ij... dxq

=

∫C

(R∗ij...δpq + vp,qR

∗ij...

)dxq . (5.7)

5.2 Conservation of Mass, Continuity EquationEvery material body, as well as every portion of such a body, is endowed with a non-negative, scalar measure, called the mass of the body, or of the portion under consider-ation. Physically, the mass is associated with the inertia property of the body, that is,its tendency to resist a change in motion. The measure of mass may be a function ofthe space variables and time. If ∆m is the mass of a small volume ∆V in the currentconfiguration, and if we assume that ∆m is absolutely continuous, the limit

ρ = lim∆V→0

∆m

∆V(5.8)

defines the scalar field ρ = ρ(x, t) called the mass density of the body for that configurationat time t. Therefore, the mass m of the entire body is given by

m =

∫V

ρ(x, t)dV . (5.9)

In the same way, we define the mass of the body in the referential (initial) configurationin terms of the density field ρ0 = ρ0(X, t) by the integral

m =

∫V0ρ0(X, t)dV

0 . (5.10)

The law of conservation of mass asserts that the mass of a body, or of any portion of thebody, is invariant under motion, that is, remains constant in every configuration. Thus,the material derivative of Eq 5.9 is zero,

m =d

dt

∫V

ρ(x, t)dV = 0 (5.11)

which upon application of Eq 5.2 with P∗ij... ≡ ρ becomes,

m =

∫V

(ρ+ ρvi,i)dV = 0 , (5.12)

and since V is an arbitrary part of the continuum, the integrand here must vanish, result-ing in

ρ+ ρvi,i = 0 (5.13)

which is known as the continuity equation in Eulerian form. But the material derivative ofρ can be written as

ρ =∂ρ

∂t+ vi

∂ρ

∂xi,


so that Eq 5.13 may be rewritten in the alternative forms,

∂ρ

∂t+ vi

∂ρ

∂xi+ ρvi,i = 0 , (5.14a)

or∂ρ

∂t+ (ρvi),i = 0 . (5.14b)

If the density of the individual particles is constant so that ρ = 0, the material is said tobe incompressible, and thus it follows from Eq 5.13 that

vi,i = 0 or div v = 0 (5.15)

for incompressible media.Since the law of conservation of mass requires the mass to be the same in all configu-

rations, we may derive the continuity equation from a comparison of the expressions form in the referential and current configurations. Therefore, if we equate Eqs 5.9 and 5.10,

m =

∫V

ρ(x, t)dV =

∫V0ρ0(X, t)dV

0 , (5.16)

and, noting that for the motion x = x(X, t), we have∫V

ρ [x (X, t) , t] dV =

∫V0ρ (X, t) J dV0 .

The tilde in Eq 5.16 is there because the functional form may be different when writtenas a function of x and t than the functional form written in terms of X and t. Now if wesubstitute the right-hand side of this equation for the left-hand side of Eq 5.16 and collectterms, ∫

V0[ρ(X, t)J− ρ0(X, t)]dV

0 = 0 .

But V0 is arbitrary, and so in the material description

ρJ = ρ0 , (5.17a)

and, furthermore, ρ0 = 0, from which we conclude that.

(ρJ) = 0 . (5.17b)

Equations 5.17 are called the Lagrangian, or material, form of the continuity equation.

Example 5.1

Show that the spatial form of the continuity equation follows from the materialform.

SolutionCarrying out the indicated differentiation in Eq 5.17b,

.

(ρJ) = ρJ+ ρJ = 0

and by Eq 4.161,J = vi,iJ


so now .

(ρJ) = J (ρ+ ρvi,i) = 0 .

But J = det F 6= 0 (an invertible tensor), which requires ρ+ ρvi,i = 0 the spatialcontinuity equation.

As a consequence of the continuity equation, we are able to derive a useful result forthe material derivative of the integral in Eq 5.1 when P∗ij... is equal to the product ρA∗ij...,where A∗ij... is the distribution of any property per unit mass. Accordingly, let

Pij...(t) =d

dt

∫V

A∗ij...(x, t)ρdV =d

dt

∫V0A∗ij...(X, t)ρJ dV0

=

∫V0

[A∗ij... (ρJ) +A∗ij...

.

(ρJ)

]dV0

which, because of Eq 5.17b, reduces to

P∗ij...(t) =

∫V0A∗ij...ρdV

0 =

∫V

A∗ij...ρdV ,

and sod

dt

∫V

A∗ij...(x, t)ρdV =

∫V

A∗ij...ρdV . (5.18)

We shall have numerous occasions to make use of this very important equation.

5.3 Linear Momentum Principle, Equations of MotionLet a material continuum body having a current volume V and bounding surface S besubjected to surface traction t(n)

i and distributed body forces ρbi as shown in Fig. 5.1.In addition, let the body be in motion under the velocity field vi = vi(x, t). The linearmomentum of the body is defined by the vector

Pi(t) =

∫V

ρvi dV , (5.19)

and the principle of linear momentum states that the time rate of change of the linear mo-mentum is equal to the resultant force acting on the body. Therefore, in global form, withreference to Fig. 5.1,

d

dt

∫V

ρvi dV =

∫S

t(n)i dS+

∫V

ρbi dV , (5.20)

and because t(n)i = tjinj, we can convert the surface integral to a volume integral having

the integrand tji,j. By the use of Eq 5.18 on the left-hand side of Eq 5.20 we have, aftercollecting terms, ∫

V

(ρvi − tji,j − ρbi) dV = 0 (5.21)


O

dV

ni

dS

x3

x2

x1

x y ρbi

t(n)i

FIGURE 5.1Material body in motion subjected to body and surface forces.

where vi is the acceleration field of the body. Again, V is arbitrary and so the integrandmust vanish, and we obtain,

tji,j + ρbi = ρvi (5.22)

which are known as the local equations of motion in Eulerian form.When the velocity field is zero, or constant so that vi = 0, the equations of motion

reduce to the equilibrium equations,

tji,j + ρbi = 0 (5.23)

which are important in solid mechanics, especially elastostatics.

5.4 Piola-Kirchhoff Stress Tensors, Lagrangian Equations of MotionAs mentioned in the previous section, the equations of motion Eq 5.22 are in Eulerianform. These equations may also be cast in the referential form based upon the Piola-Kirchhoff tensor, which we now introduce.

Recall that in Section 3.3 we defined the stress components tij of the Cauchy stress ten-sor T as the ith component of the stress vector t(ej)i acting on the material surface havingthe unit normal n = ej. Notice that this unit normal is defined in the current configura-tion. It is also possible to define a stress vector that is referred to a material surface in thereference configuration and from it construct a stress tensor that is associated with thatconfiguration. In doing this, we parallel the development in Section 3.3 for the Cauchystress tensor associated with the current configuration.

Let the vector p0(N)be defined as the stress vector referred to the area element ∆S0 inthe plane perpendicular to the unit normal N = NAIA. Just as we defined the Cauchystress vector in Eq 3.5, we write

lim∆S0→0

∆f

∆S0=df

dS0= p0(N) (5.24)


where ∆f is the resultant force acting on the material surface which in the referenceconfiguration was ∆S0.

The principle of linear momentum can also be written in terms of quantities which arereferred to in the referential configuration as∫

S0p0(N)(X, t)dS0 +

∫V0ρ0b

0(X, t)dV0 =

∫V0ρ0a

0(X, t)dV0 (5.25)

where S0, V0 and ρ0 are the material surface, volume and density, respectively, referred tothe reference configuration. The superscript zero after the variable is used to emphasizethe fact that the function is written in terms of the reference configuration. For example,

ai(x, t) = ai [χ(X, t), t)] = a0i (X, t) .

Notice that, since all quantities are in terms of material coordinates, we have moved thedifferential operator d/dt of Eq 3.8 inside the integral to give rise to the acceleration a0.In a similar procedure to that carried out in Section 3.2, we apply Eq 5.25 to portions Iand II of the body (as defined in Fig. 3.2(a)) and to the body as a whole to arrive at theequation ∫

S0

[p0(N) + p0(−N)

]dS0 = 0 . (5.26)

This equation must hold for arbitrary portions of the body surface, and so

p0(N) = −p0(−N) (5.27)

which is the analog of Eq 3.10.The stress vector p0(N) can be written out in components associated with the referential

coordinate planes as

p0(IA) = p0i(IA)ei, (A = 1, 2, 3) . (5.28)

This describes the components of the stress vector p0(N) with respect to the referentialcoordinate planes, and to determine its components with respect to an arbitrary planedefined by the unit vector N, we apply a force balance to an infinitesimal tetrahedron ofthe body. As we let the tetrahedron shrink to the point, we have

p0i(N) = p0i

(IA)NA , (5.29)

and definingPAi ≡ p0i (IA) (5.30)

we obtainp0(N)i = PAiNA (5.31)

where PAi are the components of the first Piola-Kirchhoff stress tensor. These represent thexi components of the force per unit area of a surface whose referential normal is N .

Using the first Piola-Kirchhoff stress tensor, we can derive the equations of motion, andhence the equilibrium equations in the referential formulation. Starting with Eq 5.25, weintroduce Eq 5.31 to obtain∫

S0PAiNA dS

0 +

∫V0ρ0b

0i dV

0 =

∫V0ρ0a

0i dV

0 , (5.32)


and using the divergence theorem on the surface integral, we consolidate Eq 5.32 as∫V0

(PAi,A + ρ0b0i − ρ0a

0i )dV

0 = 0 .

This equation must hold for arbitrary portions of the body so that the integrand is equalto zero, or

PAi,A + ρ0b0i = ρ0a

0i (5.33a)

which are the equations of motion in referential form. If the acceleration field is zero, theseequations reduce to the equilibrium equations in referential form

PAi,A + ρ0b0i = 0 . (5.33b)

We note that the partial derivatives of the Piola-Kirchhoff stress components are withrespect to the material coordinates because this stress tensor is referred to a surface in thereference configuration.

Equilibrium also requires a balance of moments about every point. Summing momentsabout the origin (Fig. 5.1 may be useful in visualizing this operation) gives us∫

S0εijkxjp

0k

(N) dS0 +

∫V0εijkxjρ0b

0k dV

0 = 0 (5.34)

which reduces to ∫V0εijk

[(xjPAk),A + xjρ0b

0k

]dV0 = 0

where we have used Eq 5.31 and the divergence theorem. Carrying out the indicatedpartial differentiation we obtain∫

V0εijk

[xj,APAk + xj

(PAk,A + ρ0b

0k

)]dV0 = 0

and by Eq 5.33b this reduces to ∫V0

(εijkxj,APAk)dV0 = 0 (5.35)

since the term in parentheses is zero on account of the balance of momentum. Again, thisequation must hold for all portions V0 of the body, so the integrand must vanish, giving

εijkxj,APAk = 0 (5.36)

Following a similar argument to that presented in Section 3.4, we conclude that Eq 5.36implies

xj,APAk = xk,APAj . (5.37)

If we now introduce the definition for sAB

PAi = xi,BsBA , (5.38)

and substitute into Eq 5.37 we observe that

sAB = sBA (5.39)

which is called the second Piola-Kirchhoff stress tensor, or sometimes the symmetric Piola-Kirchhoff stress tensor.


The Piola-Kirchhoff stresses can be related to the Cauchy stress by considering thedifferential force exerted on an element of deformed surface dS as

dfi = tjinj dS . (5.40)

This force can also be written in terms of the first Piola-Kirchhoff stress tensor as

dfi = PAiNA dS0 . (5.41)

Recall from Eq 4.159 that the surface element in the deformed configuration is related tothe surface element in the reference configuration by

nq dS = XA,qJNA dS0 .

Using this, along with Eqs 5.40 and 5.41, we obtain

dfi = tjinj dS = tjiXA,jJNA dS0 = PAiNA dS

0 (5.42)

which can be rewritten as

(tjiXA,jJ− PAi)NA dS0 = 0 . (5.43)

From this we see that the Cauchy stress and the first Piola-Kirchhoff stress are relatedthrough

Jtji = PAixj,A . (5.44)

Also, from Eq 5.38 we can write

Jtji = xj,Axi,BsAB (5.45)

which relates the Cauchy stress to the second Piola-Kirchhoff stress.In Chapter 4 we showed that the difference between Eulerian and Lagrangian strains

disappears when linear deformations are considered. Here, we will show that in lin-ear theories the distinction between Cauchy and Piola-Kirchhoff stress measures is notnecessary.

To show the equivalence of Cauchy and Piola-Kirchhoff stresses in linear theories, wehave to recall some kinematic results from Section 4.7 and also derive a few more. In-troducing a positive number ε that is a measure of smallness such that the displacementgradients ui,A are of the same order of magnitude as ε, we may write

ui,A = O(ε) as ε→ 0 . (5.46)

As we discovered in Section 4.7, the Eulerian and Lagrangian strains are equivalent asε→ 0, so from Eqs 4.58 and 4.59 we have

EABδiAδjB = eij = O(ε) .

Examination of Eqs 5.44 and 5.45 relates stress measures tji, PAi, and sAB. To discussthis relationship in the linear case, we must find an expression for the Jacobian as we letε→ 0. Starting with the definition of J in the form

J =1

6εijkεABCFiAFjBFkC

we substitute FiA = ui,A + δiA, etc., to get

J =1

6εijkεABC(ui,A + δi,A)(uj,B + δjB)(uk,C + δkC) .


Carrying out the algebra and after some manipulation of the indices

J =1

6εijkεABC

[δiAδjBδkC + 3uk,CδiAδjB +O(ε2)

]where terms on the order of ε2 and higher have not been written out explicitly. Sinceεijkεijk = 6 and εijkεijC = 2δkC,

J = 1+ uk,k +O(ε2) . (5.47)

Now we can evaluate Eqs 5.44 and 5.45 as ε → 0, that is, for the case of a linear theory.Since uk,k is O(ε).

tji +O(ε) = PAiδAj +O(ε) . (5.48)

With a similar argument for Eq 5.45, we find

tji = SABδAiδBj as ε→ 0 . (5.49)

Equations 5.48 and 5.49 demonstrate that, in linear theory, Cauchy, Piola-Kirchhoff andsymmetric Piola-Kirchhoff stress measures are all equivalent.

5.5 Moment of Momentum (Angular Momentum) PrincipleMoment of momentum is the phrase used to designate the moment of the linear momentumwith respect to some point. This vector quantity is also frequently called the angularmomentum of the body. The principle of angular momentum states that the time rate ofchange of the moment of momentum of a body with respect to a given point is equalto the moment of the surface and body forces with respect to that point. For the bodyshown in Fig 5.1, if we take the origin as the point of reference, the angular momentumprinciple has the mathematical form

d

dt

∫V

εijkxjρvk dV =

∫S

εijkxjt(n)k dS +

∫V

εijkxjρbk dV . (5.50)

Making use of Eq 5.18 in taking the derivative on the left-hand side of the equation andapplying the divergence theorem to the surface integral, after introducing the identityt(n)k = tqknq, results in∫

V

εijk[xj(ρvk − tqk,q − ρbk) − tjk]dV = 0

which reduces to ∫V

εijktkj dV = 0 (5.51)

because of Eq 5.22 (the equations of motion) and the sign-change property of the permu-tation symbol. Again, with V arbitrary, the integrand must vanish so that

εijktkj = 0 (5.52)

which by direct expansion demonstrates that tkj = tjk, and the stress tensor is symmetric.Note that in formulating the angular momentum principle by Eq 5.50 we have assumedthat no body or surface couples act on the body. If any such concentrated moments doact, the material is said to be a polar material, and the symmetry property of T no longerholds. But as mentioned in Chapter 3, this is a rather specialized situation and we shallnot consider it here.


5.6 Law of Conservation of Energy, The Energy EquationThe statement we adopt for the law of conservation of energy is the following: the mate-rial time derivative of the kinetic plus internal energies is equal to the sum of the rate ofwork of the surface and body forces, plus all other energies that enter or leave the bodyper unit time. Other energies may include, for example, thermal, electrical, magnetic orchemical energies. In this text we consider only mechanical and thermal energies, andalso, we require the continuum material to be non-polar (free of body or traction couples).

If only mechanical energy is considered, the energy balance can be derived from theequations of motion, (Eq 5.22). Here we take a different approach and proceed as follows.By definition, the kinetic energy of the material occupying an arbitrary volume V of thebody in Fig. 5.1 is

K(t) =1

2

∫V

ρv · vdV =1

2

∫V

ρvivi dV . (5.53)

Also, the mechanical power, or rate of work of the body and surface forces shown in thefigure is defined by the scalar

P(t) =

∫S

t(n)i vi dS +

∫V

ρbivi dV . (5.54)

Consider now, the material derivative of the kinetic energy integral

K =d

dt

∫V

1

2ρvivi dV =

1

2

∫V

ρ.

(vivi)dV

=

∫V

ρ(vivi)dV =

∫V

vi (tji,j + ρbi) dV

where Eq 5.22 has been used to obtain the final form of the integrand. But vitji,j =(vitji),j − vi,jtji and so

K =

∫V

[ρbivi + (vitij),j − vi,jtij

]dV

which, if we convert the middle term by the divergence theorem and make use of thedecomposition vi,j = dij +wij, may be written

K =

∫V

ρbivi dV +

∫S

t(n)i vi dS −

∫V

tijdij dV . (5.55)

By the definition Eq 5.54 this may be expressed as

K+ S = P (5.56)

where the integral

S =

∫V

tijdij dV =

∫V

(T : D)dV (5.57)

is known as the stress work, and its integrand tijdij as the stress power. The balance ofmechanical energy given by Eq 5.56 shows that, of the total work done by the external


forces, a portion goes toward increasing the kinetic energy, and the remainder appears aswork done by the internal stresses.

In general, S cannot be expressed as the material derivative of a volume integral, thatis,

S 6= d

dt

∫V

(· · · ) dV (5.58)

because there is no known function we could insert as the integrand of this equation.However, in the special situation when

S = U =d

dt

∫V

ρudV =

∫V

ρu dV (5.59)

where U is called the internal energy and u the specific internal energy, or energy density (perunit mass), Eq 5.56 becomes

d

dt

∫V

ρ

(1

2vivi + u

)dV =

∫V

ρbivi dV =

∫S

t(n)i vi dS , (5.60a)

or, briefly,K+ U = P . (5.60b)

Note that Eq 5.60 indicates that part of the external work P causes an increase in kineticenergy, and the remainder is stored as internal energy. As we shall see in Chapter 6, idealelastic materials respond to forces in this fashion. 1

For a thermomechanical continuum, we represent the rate at which thermal energy isadded to a body by

Q =

∫V

ρr dV −

∫S

qini dS . (5.61)

The scalar field r specifies the rate at which heat per unit mass is produced by internalsources and is known as the heat supply. The vector qi, called the heat flux vector, is ameasure of the rate at which heat is conducted into the body per unit area per unit timeacross the element of surface dS whose outward normal is ni (hence the minus sign inEq 5.61). The heat flux qi is often assumed to obey Fourier’s law of heat conduction;

qi = −κθ,i or q = −κ∇θ (5.62)

where κ is the thermal conductivity and θ,i is the temperature gradient. But, since not allmaterials obey this conduction “law,” it is not universally valid.

With the addition of the thermal energy consideration, the complete energy balancerequires modification of Eq 5.60 which now takes the form

K+ U = P +Q , (5.63a)

or, when written out in detail,

d

dt

∫V

ρ

(1

2vivi + u

)dV =

∫V

ρ (bivi + r) dV +

∫S

[t(n)i vi − qini

]dS . (5.63b)

1The symbol u is used for specific internal energy because of its widespread acceptance in the literature.There appears to be very little chance that it might be misinterpreted in this context as the magnitude of thedisplacement vector u.


If we convert the surface integral to a volume integral and make use of the equations ofmotion (Eq 5.22), the reduced form of Eq 5.63b is readily seen to be∫

V

(ρu− tijdij − ρr+ qi,i) dV = 0 or∫V

(ρu− T : D− ρr+∇ · q) dV , (5.64a)

or, briefly,S = U−Q (5.64b)

which is sometimes referred to as the thermal energy balance, in analogy with Eq 5.56 thatrelates to the mechanical energy balance. Thus, we observe that the rate of work of theinternal forces equals the rate at which internal energy is increasing minus the rate atwhich heat enters the body. As usual for an arbitrary volume V, by the argument whichis standard now, upon setting the integrand of Eq 5.64a equal to zero, we obtain the fieldequation,

ρu− tijdij − ρr+ qi,i = 0 or ρu− T : D− ρr+ divq = 0 (5.65)

which is called the energy equation.In summary, then, the mechanical energy balance Eq 5.55 is derivable directly from the

equations of motion (linear momentum principle) and is but one part of the complete en-ergy picture. When thermal energy is included, the global balance Eq 5.63 is a statementof the first law of thermodynamics.

5.7 Entropy and the Clausius-Duhem Equation

The conservation of energy as formulated in Section 5.6 is a statement of the intercon-vertibility of heat and work. However, there is not total interconvertibility for irreversibleprocesses. For instance, the case of mechanical work being converted to heat via frictionis understood, but the converse does not hold. That is, heat cannot be utilized to directlygenerate work. This, of course, is the motivation for the second law of thermodynamics.

Continuum mechanics uses the second law in a different way than classical thermody-namics. In that discipline, the second law is used to draw restrictions on the direction ofthe flow of heat and energy. In the Kelvin-Plank statement, a device cannot be constructedto operate in a cycle and produce no other effect besides mechanical work through theexchange of heat with a single reservoir. Alternatively, in the Clausius statement, it isimpossible to construct a device operating in a cycle and producing no effect other thanthe transfer of heat from a cooler body to a hotter body (van Wylen and Sonntag, 1965).In continuum mechanics, a statement of the Second Law is made to place restrictions oncontinua. However, in the case of continuum mechanics the restrictions are placed on thematerial response functions called constitutive responses.

In this section, a thermodynamic parameter called entropy is introduced as a way tolink mechanical and thermal responses. Using this parameter the second law of thermo-dynamics is stated in the form of the Clausius-Duhem equation. This equation is usedin later sections to place functional restrictions on postulated constitutive responses forvarious materials.

At any given state for the continuum there are various quantities that effect the internalenergy. These might be the volume of an ideal gas or the components of the deformationgradient of a solid. In the case of the deformation gradient, the nine components represent


a deformation in the body that is storing energy. The collection of these parameters iscalled the thermodynamic substate and will be denoted by v1, v2, · · · ,vn. 2

While the thermodynamic substate influences the internal energy of the body it doesnot completely define it. Assume that the substate plus an additional independent scalarparameter, η, is sufficient to define the internal energy. This definition may be made inthe form of

u = f(η,v1,v2, · · · ,vn) (5.66)

which is often referred to as the caloric equation of state. Parameter η is called the specificentropy. Since the internal energy is unambiguously defined once entropy is adjoined tothe substate, the combination η plus v1, v2, · · · ,vn constitutes the thermodynamic state.

Temperature is the result of the change in internal energy with respect to entropy

θ =∂u

∂η. (5.67)

Furthermore, partial differentiation of the internal energy with respect to the thermody-namic substate variables results in thermodynamic tensions

τa =∂u

∂va. (5.68)

The preceding equations can be used to write a differential form of the internal energy asfollows:

du = θdη+∑a

τadva . (5.69)

From Eqs 5.66 and 5.67 we see that both temperature and thermodynamic tensions arefunctions of entropy and the substate parameters.

Assuming that all the functions defined in this section are continuously differentiableas many times as necessary, it is possible to solve for entropy in terms of temperature

η = η (θ,va) . (5.70)

This result may be substituted into the caloric equation of state to yield internal energyas a function of temperature and substate parameters

u = u (θ,va) . (5.71)

Using this result in Eq 5.68 allows the definition of the thermal equations of state

τa = τa (θ,va) (5.72)

which inverts to give the substate parameters

va = va (θ, τa) . (5.73)

2Notation warning: Do not confuse these thermodynamic substate parameters with particle velocity


The principles of thermodynamics are often posed in terms of thermodynamic poten-tials which may be defined as follows:

internal energy u , (5.74a)

free energy ψ = u− ηθ , (5.74b)

enthalpy χ = u−∑aτava , (5.74c)

free enthalpy ζ = χ− ηθ = u− ηθ−∑aτava . (5.74d)

These potentials are related through the relationship

u−ψ+ ζ− χ = 0 . (5.75)

All of the energy potentials may be written in terms of any one of the following indepen-dent variable sets

η,va; θ,va; η, τa; θ, τa . (5.76)

In order to describe the motion of a purely mechanical continuum the function xi =xi(XA, t) is needed. Adding the thermodynamic response requires the addition of tem-perature, θ, or, equivalently, entropy, η, both being a function of position and time

θ = θ(XA, t) or η = η(XA, t) (5.77)

When considered for a portion P of the body, the total entropy is given as

H =

∫P

ρηdV , (5.78)

and the entropy production in the portion P is given by

Γ =

∫P

ργdV (5.79)

where the scalar γ is the specific entropy production. The second law can be stated asfollows: the time rate-of-change in the entropy equals the change in entropy due to heatsupply, heat flux entering the portion, plus the internal entropy production. For a portionP of the body, this is written as

d

dt

∫P

ρηdV =

∫P

ρr

θdV −

∫∂P

qini

θdS +

∫P

ργdV . (5.80)

The entropy production is always positive which leads to a statement of the second lawin the form of the Clausius-Duhem inequality

d

dt

∫P

ρηdV >∫P

ρr

θdV −

∫∂P

qini

θdS . (5.81)

This global form can easily be posed locally by the now-familiar procedures. Applyingthe divergence theorem to the heat flux term yields∫

∂P

qini

θdS =

∫P

(qiθ

),idV .


Furthermore, the differentiation of the entropy term is simplified by the fact that it is aspecific quantity (see Section 5.2, Eq 5.18). Thus, we write∫

P

[ρη− ρ

r

θ+(qiθ

),i

]dV > 0 , (5.82)

and since this must hold for all arbitrary portions of the body, and the integrand iscontinuous, then

ρη− ρr

θ+(qiθ

),i

= ρη− ρr

θ+qi,i

θ−1

θ2qiθ,i > 0 .

Thus, the local form of the Clausius-Duhem equation is

ρθη− ρr+ qi,i −1

θqiθ,i > 0 . (5.83a)

Often, the gradient of the temperature is written as gi = θ,i in which case Eq 5.83abecomes


θqigi > 0 . (5.83b)

Combining this result with Eq 5.65 brings the stress power and internal energy into theexpression giving a reduced form of the Clausius-Duhem

ρθη− ρu+ dijtij −1

θqigi > 0 . (5.84)

One final form of the Clausius-Duhem equation is obtained by using Eq 5.74b to obtainthe local dissipation inequality

−ρ(ψ+ ηθ

)+ dijtij −

1

θqigi > 0 . (5.85)

Finally, the fundamental laws are summarized in a Table 5.1. This table shows theglobal and local forms of conservation of mass, linear momentum, angular momentum,energy and the Clausius-Duhem equation. Details on these fundamental equations dis-cussed in preceding sections provide the foundation from which engineering problemsare solved. All that is needed in addition to these fundamental equations is a constitutiveresponse.

5.8 The General Balance LawThe above discussion leads to a general expression for the balance law.3 The balance lawis an integral expression that applies to the body as a whole or some part of the body.The expression for the balance law is

d

dt

∫P

Ψ (x, t) dV =

∫∂P

Γ (x, t) dS +

∫P

Σ (x, t) dV . (5.86)

3In the literature, some authors use master balance law rather than general balance law.

FundamentalLaw

sand

Equations183

TABLE 5.1Fundamental equations in global and local forms.

Global Form Local Form

Conservation of Mass m =

∫V

ρ(x, t)dV = 0 ρ+ vi,iρ = 0

∂ρ

∂t+ (ρvi),i = 0

ρJ = ρ0

Linear Momentumd

dt

∫V

ρvidV =

∫S

t(n)i dS +

∫V

ρbidV ρvi = tji,j + ρbi

Angular Momentumd

dt

∫V

εijkxj(ρvk)dV =

∫S

εijkxjt(n)k dS +

∫V

εijkxj(ρbk)dV εijktij = 0 or tij = tji

Conservation of Energy

d

dt

∫V

(1

2vivi + u

)dV =

∫V

(ρbivi + r)dV +

∫S

(t(n)i vi − qini

)dS

ρu− tijdij − ρr+ qi,i = 0

Clausius-Duhem Equationd

dt

∫V

ρηdV >∫

V

ρr

θdV −

∫S

qi

θnidS ρη− ρ

r

θ+(qiθ

),i

> 0

ρθη− ρu+ tijdij −1

θqiθ,i > 0

ψ+ ηθ− tijdij −1

θqiθ,i > 0


TABLE 5.2Identification of quantities in the balance laws.

Balance Law Property Flux Source

Mass ρ 0 0

Momentum ρv t(n) (x, t, ) ρb

Angular Momentum x× ρv x× t(n) (x, t) x× ρb

Energy 12ρv · v+ u t(n) (x, t) · v− q · n ρb+ r

Entropy ρη −q · nθ

ρr

θ

Physically this expression states that the time rate of change of the field Ψ is equal to theflux of Γ through the boundary ∂P plus the rate of increase or generation, Σ, of Ψ withinthe volume V. Σ is the source of Ψ within the volume. Similarly, by changing the equalityin Eq 5.86, the balance law can be recast as an inequality with a similar interpretation.

d

dt

∫P

Ψ (x, t) dV >∫∂P

Γ (x, t) dS +

∫P

Σ (x, t) dV . (5.87)

These expressions are in the current or deformed configuration.If the flux depends on the normal to the boundary, n, the flux, Γ (x, t, n), will be a tensor

one order higher than the field quantity Ψ or the source term Σ. The surface integral inEq 5.86 can be converted to a volume integral by using the divergence theorem Eq 2.93.This leads through the localization principle to a partial differential equation that appliespointwise to determine the field Ψ (x, t).

dΨ

dt+ Ψdiv v = div Γ + Σ . (5.88)

Table 5.2 shows the relationship of the quantities in the various balance laws to the masterbalance law.

The discussion to this point focused on a continuum body with a fixed set of materialpoints. This gives rise to the balance of mass that has no mass flux or source terms.One very important class of materials that is gaining increasing importance is biologicalmaterials. In biological systems, mass increases by the growth of additional material. Thisrequires a recasting of the continuum balance equations to accommodate the addition ofmass through the growth of new material.

Growth can occur in biological systems through either volumetric growth as in softtissue or by surface growth as in bone. The focus in the following will be on materialsthat exhibit volumetric growth. The mass balance equation requires an added source termto model the volumetric growth

d

dt

∫P

ρdV =

∫P

ρcdV (5.89)

where c is the mass growth per unit density. This results in a local balance of massequation

dρ

dt+ ρvk,k = ρc or

dρ

dt+ ρdiv v = ρc . (5.90)


The balance of linear momentum requires an additional source term to accommodatethe momentum that the added mass brings to the system

d

dt

∫P

ρvi dV =

∫∂P

t(n)i dS +

∫P

ρbi dV +

∫P

ρcvi dV , (5.91)

ord

dt

∫P

ρvdV =

∫∂P

t(n) dS +

∫P

ρbdV +

∫P

ρcvdV .

To determine the local form of the balance of linear momentum, it is necessary to deter-mine the derivative of an integral quantity per unit mass

d

dt

∫P

ρf dV =

∫P

(ρf+ ρcf

)dV . (5.92)

The last result follows from Eq 5.90 and is equivalent to Eq 5.18 when c is zero. Thisresult together with t(n)

i = tjinj yields the local balance of linear momentum

ρdvi

dt= tji,j + ρbi . (5.93)

The local form for the balance of linear momentum is unchanged by the addition of themass. Balance of angular momentum also requires an additional source term for theadded system mass. That is

d

dt

∫P

εijkxjρvk dV =

∫∂P

εijkxjt(n)k dS +

∫P

εijkxjρ (bk + cvk) dV , (5.94)

ord

dt

∫P

(x− 0)× ρvdV =

∫∂P

(x− 0)× t(n) dS +

∫P

(x− 0)× ρ (b+ cv) dV .

This results in a symmetric stress tensor upon using the balance of mass and linear mo-mentum. This is the case since the volumetric growth is not coupled to other mechanismslike species diffusion or micromolecular processes.

The energy and entropy equations must be modified for the body as growing massis added. The added mass brings with it an associated kinetic energy and internal en-ergy. The assumption that the added material is mechanically equivalent to the existingmaterial at the point gives

d

dt

∫V

ρ

(1

2vivi + u

)dV =

∫P

ρ (bivi + r) dV +

∫∂P

(t(n)i vi − qini

)dS

+

∫P

ρc

(1

2vivi + u

)dV +

∫P

ρch dV . (5.95)

The final term represents a rate of growth energy needed to account for any additional en-ergy required to create material with the same internal and kinetic energy as the originalmaterial. The local form of the energy equation is

ρu− tijdij − ρr+ qi,i − ρch = 0 or ρu− T : D− ρr+ divq− ρch = 0 , (5.96)

where balance of mass, linear momentum and symmetry of the stress tensor were usedin arriving at this result.


The entropy inequality is

d

dt

∫P

ρηdV >∫P

ρr

θdV −

∫∂P

qini

θdS +

∫P

ρc

η+

=

h

θ

dV .

The added mass brings additional entropy and an additional entropy production,=

h, likethat in the energy equation. Using the standard procedure, the local form of the entropyinequality is


θqiθ,i − ρc

=

h

θ> 0 . (5.97)

The specific free energy, ψ = u−ηθ, allows the entropy production rate to be eliminatedfrom 5.97 with Eq 5.96 giving the energy rate u. This yields an entropy productioninequality of

−(ρψ− ηθ

)− tijdij −

1

θqiθ,i + ρch > 0 .

The value of the variable h is h = h −=

h. The reduced entropy equation can be used torestrict constitutive equations for this special class of biologically growing material.

5.9 Restrictions on Elastic Materials by the Second Law of Thermody-namics

In general, the thermomechanical continuum body must be specified by response func-tions that involve mechanical and thermodynamic quantities. To completely specify thecontinuum, a thermodynamic process must be defined. For a continuum body B havingmaterial points X a thermodynamic process is described by eight functions of the materialpoint and time. These functions would be as follows:

1. Spatial position xi = χi(X, t),

2. Stress tensor tij = tij(X, t),

3. Body force per unit mass bi = bi(X, t),

4. Specific internal energy u = u(X, t),

5. Heat flux vector qi = qi(X, t),

6. Heat supply per unit mass r = r(X, t),

7. Specific entropy η = η(X, t),

8. Temperature (always positive) θ = θ(X, t).

A set of these eight functions which are compatible with the balance of linear momentumand the conservation of energy makes up a thermodynamic process. These two balancelaws are given in their local form in Eqs 5.22 and 5.65 and are repeated below in a slightlydifferent form:

tji,j − ρvi = −ρbi ,

ρu− tijdij + qi,i = ρr .(5.98)


In writing the balance laws this way, the external influences on the body, heat supply andbody force, have been placed on the right-hand side of the equal signs. From this it isnoted that it is sufficient to specify xi, tij, u, qi, η,and θ and the remaining two processfunctions r and bi are determined from Eqs 5.98.

One of the uses for the Clausius-Duhem form of the second law is to infer restrictions onthe constitutive responses. Taking Eq 5.85 as the form of the Clausius-Duhem equationwe see that functions for stress, free energy, entropy, and heat flux must be specified.The starting point for a constitutive response for a particular material is the principle ofequipresence (Coleman and Mizel, 1963):

An independent variable present in one constitutive equation of a materialshould be so present in all, unless its presence is in direct contradiction withthe assumed symmetry of the material, the principle of material objectivity, orthe laws of thermodynamics.

For an elastic material, it is assumed that the response functions will depend on thedeformation gradient, the temperature, and the temperature gradient. Thus, we assume

tij = tij (FiA, θ, gi) , ψ = ψ (FiA, θ, gi) ,

η = η (FiA, θ, gi) , q = q (FiA, θ, gi) .

(5.99)

These response functions are written to distinguish between the functions and their value.A superposed tilde is used to designate the response function rather than the responsevalue. If an independent variable of one of the response functions is shown to contradictmaterial symmetry, material frame indifference, or the Clausius-Duhem inequality it isremoved from that function’s list.

In using Eq 5.85, the derivative of ψ must be formed in terms of its independent vari-ables

ψ =∂ψ

∂FiAFiA +

∂ψ

∂θθ+

∂ψ

∂gigi . (5.100)

This equation is simplified by using Eq 4.139 to replace the time derivative of the defor-mation gradient in terms of the velocity gradient and deformation gradient

ψ =∂ψ

∂FiALijFjA +

∂ψ

∂θθ+

∂ψ

∂gigi . (5.101)

Substitution of Eq 5.100 into 5.85 and factoring common terms results in

−ρ

(∂ψ

∂θ+ η

)θ+

(tij − ρ

∂ψ

∂FiAFjA

)Lij − ρ

∂ψ

∂gigi −

1

θqigi > 0 . (5.102)

Note that in writing Eq 5.102 the stress power has been written as tijLij rather than tijdij.This can be done because stress is symmetric and adding the skew–symmetric part of Lijis essentially adding zero to the inequality. The velocity gradient is used because thepartial derivative of the free energy with respect to the deformation gradient times thetransposed deformation gradient is not, in general, symmetric.

The second law must hold for every thermodynamic process, which means a specialcase may be chosen which might result in further restrictions placed on the responsefunctions. That this is the case may be demonstrated by constructing displacement andtemperature fields as such a special case. Define the deformation and temperature fieldas follows:

xi = χ (XA, t) = YA +AiA (t) [XA − YA] ,


θ = θ (XA, t) = α (t) + [AAi (t)ai (t)] [XA − YA] . (5.103)

Here, XA and YA are the positions in the reference configuration of material points X andY , and function AiA(t) is an invertible tensor, ai(t) is a time dependent vector, and α(t)is a scalar function of time. At the spatial position YA, the following is readily computed

θ (YA, t) = α (t) , (5.104)

FiA (YA, t) = AiA (t) . (5.105)

Note that Eq 5.103 may be written in terms of the current configuration as

θ (yi, t) = α (t) + ai (t) [xi − yi] . (5.106)

Thus, the gradient of the temperature at material point Y is written as

θ,i = g,i = ai (t) . (5.107)

From Eqs 5.104, 5.105, and 5.106 it is clear that quantities θ, gi, and FiA can be inde-pendently chosen. Furthermore, the time derivatives of these quantities may also bearbitrarily chosen. Because of the assumed continuity on the response functions, it ispossible to arbitrarily specify functions u, gi, and FiA and their time derivatives.

Returning to an elastic material and Eq 5.101, for a given material point in the con-tinuum, consider the case where the velocity gradient, Lij, is identically zero and thetemperature is constant. This means Lij = 0 and θ = 0. Furthermore, assume the temper-ature gradient to be some arbitrary constant gi = g0i . Eq 5.102 becomes

−ρ∂ψ(FiA, θ0, g

0i

)∂gi

gi −1

θqi(FiA, θ0, g

0i

)gi > 0 .

Again, take advantage of the fact that the second law must hold for all processes. Sincegi is arbitrary it may be chosen to violate the inequality. Thus, the temperature gradienttime derivative coefficient must be zero

∂ψ(FiA, θ0, g

0i

)∂gi

= 0 . (5.108)

Since g0i was taken to be an arbitrary temperature gradient, Eq 5.108 implies that the freeenergy is not a function of the temperature gradient. That is,

∂ψ (FiA, θ0, gi)

∂gi= 0

which immediately leads to ψ = ψ (FiA, θ). This fact eliminates the third term of Eq 5.102.Further information about the constitutive assumptions can be deduced by applying

additional special cases to the now reduced Eq 5.102. For the next special process, con-sider an arbitrary material point at an arbitrary time in which Lij = 0 and g = 0, but thetemperature gradient is an arbitrary constant gi = g0i . For this case, the Clausius-Duheminequality is written as

−ρ

(∂ψ (FkB, θ)

∂θ+ η

(FkB, θ, g

0k

))θ−

1

θqi(FkB, θ, g

0k

)g0i > 0 (5.109)

which must hold for all temperature rates, θ. Thus, the entropy response function maybe solved in terms of the free energy

η(FkB, θ, g

0k

)= −

∂ψ (FkB, θ)

∂θ,


and since the free energy is only a function of the deformation gradient and temperaturethe entropy must be a function of only those two as well. That is,

η = η (FkB, θ) = −∂ψ

∂θ. (5.110)

One more application of the Clausius-Duhem inequality for a special process will leadto an expression for the Cauchy stress in terms of the free energy. For this process, selectthe temperature gradient to be an arbitrary constant and the time rate-of-change of thetemperature gradient to be identically zero. Eq 5.102 becomes(

tij(FkB, θ, g

0k

)− ρ

∂ψ

∂FiAFjA

)Lij −

1

θqi(FkB, θ, g

0k

)g0i > 0 (5.111)

which must hold for all velocity gradients Lij. Picking a Lij that would violate the in-equality unless the coefficient of Lij vanishes implies

tij = tij (FkB, θ) = ρ∂ψ

∂FiAFjA . (5.112)

In arriving at Eq 5.111 it was noted that the free energy had already been shown to beindependent of the temperature gradient by virtue of Eq 5.108. Thus, gk was droppedfrom the independent variable list in Eq 5.111.

Finally, substituting the results of Eqs 5.112, 5.110, and 5.108 into 5.102 a last restrictionfor an elastic material is found to be

qigi 6 0 . (5.113)

5.10 InvarianceThe concept of invariance has been discussed in Section 2.5 with respect to tensors. A ten-sor quantity is one which remains invariant under admissible coordinate transformations.For example, all the different stress components represented by Mohr’s circle refer to asingle stress state. This invariance is crucial for consolidating different stress componentsinto a yield criterion such as the maximum shearing stress (or Tresca criterion).

Invariance plays another important role in continuum mechanics. Requiring a contin-uum to be invariant with regards to reference frame, or have unchanged response whena superposed rigid body motion is applied to all material points, has significant resultsassociated with them. The most important of these consequences might be restrictionsplaced on constitutive models as discussed in the next section.

There are two basic methods for examining invariance of constitutive response func-tions: material frame indifference and superposed rigid body motion. In the first, a continuumbody’s response to applied forces or prescribed motion must be the same as observedfrom two different reference frames. The body and the applied forces remain the same;only the observers reference frame changes. In superposing a rigid body motion to thebody the observer maintains the same reference frame. Here, each material point hasa superposed motion added to it. The forces applied to the body are rotated with thesuperposed motion.


Both of these methods produce the same restrictions on constitutive responses withinthe context of this book. In this text, the method of superposed rigid body motion will bepresented as the means for enforcing invariance.

The superposition of a rigid motion along with a time shift can be applied to the basicdefinition of the motion

xi = χi(XA, t) . (5.114)

From this state a superposed rigid body motion is applied maintaining all relative dis-tances between material points. Also, since the motion is a function of time, a time shiftis imposed on the motion. After the application of the superposed motion, the position ofthe material point becomes x+

i at time t+ = t+a where a is a constant. A superscript “+”is used for quantities having the superposed motion. Some literature uses a superscript“∗” to denote this, but since we use this to represent principal stress quantities the “+” isused. The motion in terms of the superposed motion is written as

x+i = χ+

i (XA, t) . (5.115)

Assuming sufficient continuity, the motion can be written in terms of the current config-uration since XA = χ−1(xi, t). That is to say

x+i = χ+

i (XA, t) = χ+i (xi, t) (5.116)

where χ+i is written because the substitution of XA = χ−1(xi, t) results in a different

function than χ+i .

To represent the relative distance between two particles, a second material point isselected. In an analogous manner, it is straight-forward to determine

y+i = χ+

i (YA, t) = χ+i (yi, t) . (5.117)

Relative distance between material particles XA and YA is written as(x+i − y+

i

) (x+i − y+

i

)=[χ+i (xi, t) − χ+

i (yi, t)] [χ+i (xi, t) − χ+

i (yi, t)]

(5.118)

which is the dot product of the vector between positions xi and yi. Since the materialpoints XA and YA were arbitrarily chosen, quantities xi and yi are independent. Subse-quent differentiation of Eq 5.118 with respect to xi then yi results in

∂χ+i (xj, t)

∂xp

∂χ+i (yj, t)

∂yq= δpq . (5.119)

Since this must hold for all pairs of material points, it is possible to set

∂χ+i (xj, t)

∂xp=∂χ+i (yj, t)

∂yp= Qip (t) . (5.120)

Use of Eq 5.120 in Eq 5.119 shows that the matrixQip(t) is orthogonal. Furthermore, sincethe special case of the superposed motion as a null motion, that is χ+

i (xi, t) = xi, thenmatrix Qip(t) must be proper orthogonal having Qip(t)Qiq(t) = δpq and det(Qip) = +1.

To come up with a particular form for the superposed motion, Eq 5.120 may be spatiallyintegrated to obtain

χ+i (xi, t) = a (t) +Qim (t) xm , (5.121a)

orx+i = ai +Qimxm . (5.121b)


x 2

x 1

x1

x2

P

P

P ‛

θθ

c(t)

FIGURE 5.2Reference frames Ox1x2x3 and O+x+

1 x+2 x

+3 differing by a superposed rigid body motion.

Vector ai may be written in the alternative form

ai (t) = c+i

(t+)

−Qim (t) cm (t) (5.122)

yieldingx+i = c+

i +Qim[xm − cm] (5.123)

where

QimQin = QmiQni = δmn and det (Qij) = 1 . (5.124)

A similar development of the superposed motion can be obtained by assuming twoCartesian reference frames Ox1x2x3 and O+x+

1 x+2 x

+3 which are separated by vector ci(t)

and rotated by an admissible coordinate transformation defined by Qim (Malvern, 1969).Rather than integrating differential Eq 5.120, the superposed motion can be written as

p+i = ci (t) +Qim (t)pm (5.125)

where vectors p+i and pm are defined as shown in Fig. 5.2. Here, Qim is simply the matrix

of the direction cosines between Ox1x2x3 and O+x+1 x

+2 x

+3 .

Example 5.2

Show that the superposed rigid body motion defined by Eq 5.123 is distanceand angle preserving.

SolutionConsider the distance squared between material points XA and YA in terms ofthe superposed motion(

x+i − y+

i

) (x+i − y+

i

)= Qim (xm − ym)Qin (xn − yn)

= QimQin (xm − ym) (xn − yn) .


Since Qim is orthogonal(x+i − y+

i

) (x+i − y+

i

)= δnm (xm − ym) (xn − yn)

= (xm − ym) (xm − ym)

where the delta substitution property has been used. Thus, distance is preservedin the superposed motion.

Three material points, XA, YA, and ZA, are used to show that angles arepreserved in the superposed rigid body motion. Let θ+ be the angle includedbetween vectors x+

i − y+i and x+

i − z+i . Use of the definition of the dot productgives a convenient way to represent the angle θ+

cos θ+ =x+i − y+

i

|x+n − y+

n |

x+i − z+i

|x+n − z+n |

.

Note the “n” indices in the denominator do not participate in the summationsince they are inside the vector magnitude operator. Direct substitution forEq 5.121 followed by utilizing the orthogonality of Qim and use of results fromthe first part of this example gives

cos θ+ =Qim (xm − ym)Qin (xn − zn)

|xn − yn||xn − zn|

=QimQin (xm − ym) (xn − zn)


=δmn (xm − ym) (xn − zn)


=(xm − ym) (xm − zm)


= cos θ .

Next, consider how superposed rigid body motion affects the continuum’s velocity.Define the velocity in the superposed configuration as the time derivative with respect tot+

v+i = x+

i =dx+i

dt+. (5.126)

By the result of Eq 5.121 along with the chain rule, the velocity is given by

v+i =

d

dt+[ai (t) +Qim (t) xm]

=d

dt[ai (t) +Qim (t) xm]

dt

dt+.


Recall the definition t+ = t + a from which it is obvious that dt/dt+ = 1. As a result ofthis, an expression for the velocity under a superposed rigid body motion is obtained bytaking the time derivatives of the bracketed term of the preceding equation:

v+i = ai (t) + Qim (t) xm +Qimxm

= ai (t) + Qim (t) xm +Qimvm . (5.127)

DefineΩij (t) = Qim (t)Qjm (t) , (5.128)

or, written an alternative way by post-multiplying by Qjk

Ωij (t)Qjk (t) = Qim (t)Qjm (t)Qjk (t) = Qik (t) . (5.129)

Substituting Eq 5.129 into the second of Eq 5.127 yields an expression for the velocity fieldof a continuum undergoing a superposed rigid body rotation

v+i = ai +ΩijQjkxk +Qimvm . (5.130)

Note that Ωij is skew–symmetric by taking the time derivative of the orthogonalitycondition for Qij. That is,

d

dt[QimQin] =

d

dt[1] = 0 .

When the derivatives are taken

QimQin +QimQin = 0 ,

and the definition of Qim used from Eq 5.129 substituted into this expression yields

Qim[Ωji +Ωij]Qjm = 0 .

This is true only if the bracketed term is zero, thus

Ωij = −Ωji . (5.131)

The fact that Ωij is skew–symmetric means that it has an axial vector defined by

ωk = −1

2εijkΩij (5.132)

which may be inverted to giveΩij = −εijkωk . (5.133)

In the case of rigid body dynamics, the axial vector ωk can be shown to be the angularvelocity of the body (see Problem 5.37).

For later use in constitutive modeling, various kinematic quantities’ properties undersuperposed rigid body motions will be needed. Here, a derivation of the superposed rigidbody motion’s effect on vorticity and rate-of-deformation tensors will be demonstrated.

Start with the velocity given in Eq 5.130 and substitute for xi using Eq 5.121b to write

v+i = ai +Ωij[x

+j − aj] +Qijvj (5.134)

= Qij + Qimxm + ai −Ωijaj (5.135)

= Qijvj + Qimxm + ci (5.136)


where Eq 5.128 has been used in going from Eq 5.134 to 5.135, and ci = a − Ωijaj hasbeen used in going from Eq 5.135 to 5.136. Writing the velocity in this form allows for amore convenient computation of the velocity gradient in the superposed reference frame

∂v+i

∂x+m

= Ωij∂x+j

∂x+m

+Qij∂vj

∂x+m. (5.137)

Note that∂x+j

∂x+m

= δjm ,

and∂vj

∂x+m

=∂vj

∂xn

∂xn

∂x+m

=∂vj

∂xn

∂

∂x+m

[Qkn(x+k − ak

)]

where the last substitution comes from solving for x+k in Eq 5.121b. Use of these in

Eq 5.137 followed by the delta substitution property yields

∂v+i

x+m

= Ωim +QijQmnvj,n = Ωim +QijQmn[djn +wjn] (5.138)

where djn and wjn are the rate-of-deformation and vorticity, respectively. In the su-perposed motion the velocity gradient may be decomposed into symmetric and skew–symmetric parts

∂v+i

∂x+m

= d+im +w+

im

which are the rate-of-deformation and vorticity in the superposed rigid body frame. FromEqs 5.138 and 5.131 it is clear that

d+ij = QimQjndmn , (5.139)

andw+ij = Ωij +QimQjnwmn . (5.140)

Recalling the general transformation equations introduced in Chapter 2 it would be ex-pected that vectors and second-order tensors transform according to the generic formulae

u+i = Qimum, U+

ij = QimQjnUmn .

But preceding results show us that this is not the case. Velocity and vorticity transform inmore complex ways. All that is left to ready ourselves for the study of constitutive equa-tion theory is to determine how stress transforms under superposed rigid body motion.

The transformation of the stress vector and the stress components is not as clear cut asthe kinematic quantities demonstrated above. This is a result of starting with the stressvector which is a force. It is assumed that forces transform as a generic vector in the formof

t+i = Qijtj .

Since stress is a measure of force per unit area, a regression back to kinematic results isnecessary to determine how a differential area transforms under superposed rigid bodymotion. A differential element of area in the current configuration may be written as

dak = J∂XK

∂xkdAK .

Under the superposed rigid body motion the differential area element is written as


da+k = da+n+

k J+ ∂XK

∂x+k

dAK (5.141)

where reference quantities do not change, da+ is the infinitesimal area, and n+k is the unit

normal vector to the area all in the superposed rigid body motion state. This expressionwill be further reduced in several steps.

First, the Jacobian can be shown to transform according to J+ = J (see Problem 5.33).Next, the quantity ∂XK/∂x+

k is the inverse of the deformation gradient as represented inthe superposed rigid body rotated frame. Application of the chain rule yields

∂XK

∂x+k

=∂XK

∂xj

∂xj

∂x+k

. (5.142)

The last term of this equation is evaluated by referring to Eq 5.121b and solving for xk bypre-multiplying by Qik. Differentiation of the result shows that the last term of Eq 5.142is Qik. Thus

∂XK

∂x+k

=∂XK

∂xjQkj ,

or(FkK)−1

= F−1jKQkj . (5.143)

Substitution of the results from the preceding paragraph into Eq 5.141 results in

da+k = da+n+

k = Qkjdaj = Qkj (da)nj . (5.144)

Squaring the second and fourth terms of the above equation and equating them leads to(da+)2 = (da)2, and since area is always a positive number

da+ = da . (5.145)

With the use of Eqs 5.144 and 5.145, it is evident that

n+k = Qkjnj . (5.146)

All that remains is to determine how the stress components transform under super-posed rigid body motion. All the results are now in place to find this transformation. Inthe superposed rigid body motion frame, the stress vector t+i can be written as

t+i = t+ijn+j = t+ijQjknk (5.147)

where Eq 5.146 has been used. The assumed transformation for the stress vector yields

t+i = Qijtj = Qijtjknk (5.148)

where, as in Eq 5.147, Cauchy’s stress formula has been used. Equating the stress vectorin the superposed rigid body reference frame leads to(

t+ijQjk −Qijtjk

)nk = 0

which holds for all nk. Thus, the terms in parenthesis must equal zero. Multiplyingthe remaining terms by Qmk results in the following expression for stress componenttransformation under superposed rigid body motion:

t+ij = QimQjntmn . (5.149)


In plasticity, as well as explicit finite element formulations, the stress constitutive re-sponse is usually formulated in an incremental form. This means that the stress rate isused. The stress rate must be objective meaning

t+ij = QimQjntmn +QimQjntmn +QimQjntmn . (5.150)

It is clear from this equation that the stress rate is not objective even though the stress isobjective. This result is serious since the stress rate as shown in Eq 5.150 could not beused as a response function or in the independent variable list of a response function.Luckily, there are several ways to express a form of the stress rate in an invariant manner.

One way to obtain an objective stress rate is found from using the spin tensor wij.Using Eqs 5.140 and 5.128 to solve for Qij in terms of the spin

Qip = w+ijQjp −Qimwmp (5.151)

which is then substituted into Eq 5.150 to give

t+ij = w+iqQqmtmnQjn +QimtmnQqnw

+jq

+Qim[tmn −wnqtmq −wmqtqn]Qjn .(5.152)

Placing all quantities referred to the superposed rigid body motion to the left-hand sideof the equal sign and using Eq 5.149 results in

t+ij −w+iqt

+qj − t+iqwjq = Qim[tmn −wmqtqn − tmqwnq]Qjn . (5.153)

It is clear that the quantity tmn − wnqtmq − wmqtqn is objective which leads to thedefinition of the so called Jaumann stress rate

∇t ij = tij −Wiqtqj − tiqWjq . (5.154)

There are several other stress rate definitions satisfying objectivity. For example, theGreen-Naghdi stress rate is given as

∆tij = tij +Ωiqtqj + tiqΩjq (5.155)

where Ωiq is defined in Eq 5.128.

5.11 Restrictions on Constitutive Equations from InvarianceIt was seen in Section 5.9 that the second law of thermodynamics places restrictions on theform of constitutive response functions. Material frame indifference, or superposed rigidbody motion, may also place restrictions on the independent variables of the responsefunction as was stated in the principle of equipresence. In Section 5.10, the behaviorof many of the quantities used in continuum mechanics undergoing a superposed rigidbody motion was presented. This section will examine the response functions of bodiesundergoing a superposed rigid body motion. In particular, what restrictions are placedon the constitutive independent variables given in Eq 5.99.

Under a superposed rigid body motion, scalars are unaffected allowing the followingto be written


u+ = u, η+ = η, θ+ = θ . (5.156)

This being the case, it is clear from Eq 5.74a through 5.74d that ψ, χ, and ζ would be un-affected by the superposed rigid body motion. The remaining quantities of the responsefunctions of Eq 5.99 and their independent variables are affected in different ways fromthe superposed rigid body motion. Under a superposed rigid body motion, these func-tions transform as follows (some of these are repeated from Section 5.10 or repeated inslightly different form):

F+iA = QijFjA , (5.157)

F+iA = QijFjA +ΩijQjkFkA , (5.158)

L+im = QijLjnQnm +Ωim , (5.159)

t+ij = QiktklQlj , (5.160)

q+i = Qijqj , (5.161)

g+i = Qijgj , (5.162)

where Ωij is a skew–symmetric tensor defined by Eq 5.128.Constitutive equations are objective if and only if they transform under a superposed

rigid body motion as follows:

u+(η+, F+

iA

)= u (η, FiA) , (5.163a)

θ+(η+, F+

iA

)= θ (η, FiA) , (5.163b)

t+ij(η+, F+

mA

)= Qiktkl (η, FmA)Qlk , (5.163c)

q+i

(η+, F+

iA, g+i , L

+kl

)= Qimqm (η, FiA, gi, Lkl) . (5.163d)

In writing these equations it is noted that the roles of η and θ can be interchanged be-cause of assumed continuity. Also, restricting the independent variable list of Eqs 5.163athrough 5.163c to the entropy (or temperature) and the deformation gradient has beenshown to be a general case (Coleman and Mizel, 1964). Finally, the set of constitutivefunctions may be in terms of different response functions. That is, the elastic materialconsidered in Section 5.9 had response functions ψ, tij, η, and θi postulated. However, itcould have just as easily been postulated as a function of u, θ, tij, and qi.

As an example of how Eqs 5.163 could restrict the independent variables consider afluid whose stress response function is assumed to be a function of density, ρ, velocity, vi,and velocity gradient, Lij. With these assumptions, the restrictions of Eq 5.163c would be

t+ij(ρ+, v+

k , d+mn, w

+mn

)= Qiptpl (ρ, vk, dmn, wmn)Qlj (5.164)

where the velocity gradient has been decomposed into its symmetric and skew–symmetricparts, dmn and wmn, respectively. Using the results of Problem 5.34, Eqs 5.136, 5.139, and


5.140 results in

tij (ρ, vk + ck, QmpdpqQqn, QmpwpqQqn +Ωmn) = Qiptpl (ρ, vk, dmn, wmn)Qlj .(5.165)

Since Eq 5.165 must hold for all motions, a specific rigid body rotation may be chosen toreduce the constitutive assumption of Eq 5.164. For this purpose, suppose that Qij = δijand thus Qij = 0 . Using this motion and Eq 5.165 implies

tij (ρ, vk + ck, dmn, wmn) = tij (ρ, vk, dmn, wmn) (5.166)

where it is noted that Eq 5.128 has been used. In this case, vector ck is simply equalto ak the time derivative of the superposed rigid body motion integration factor (seeEq 5.121). This arbitrary nature would allow for Eq 5.166 to be violated if the stressfunction has velocity as an independent variable. Thus, velocity must be removed fromthe independent variable list for stress leaving

tij = tij (ρ, dmn, wmn) . (5.167)

Again, with the modified response function Eq 5.167, the invariance condition undersuperposed rigid body motion may be written as

tij (ρ,QmpdpqQqn, QmpwpqQqn +Ωmn) = Qiptpl (ρ, dmn, wmn)Qlj (5.168)

which must hold for all motions. Select a motion such that Qij = δij as before, but nowrequire Ωij 6= 0 . Substitution of this into Eq 5.168 leaves

tij (ρ, dmn, wmn +Ωmn) = tij (ρ, dmn, wmn) (5.169)

as the invariance requirement on the stress. For this to be true for all motions, the skew–symmetric part of the velocity gradient must not be an independent variable.

Applying the superposed rigid body motion to the twice reduced response functionwill yield further information. In this case,

tij (ρ,QmpdpqQqn) = Qiptpl (ρ, dmn)Qlj . (5.170)

SinceQij is a proper orthogonal tensor, and this equation would hold ifQij were replacedby its negative, so the stress must be an isotropic function of dmn. The most general formof a second-order, isotropic tensor function of dmn may now represent the stress response

tij = −p (ρ) δij + λ (ρ)dkkδij + 2µ (ρ)dij (5.171)

where p, λ, and µ are functions of density and would represent the viscosity coefficients.

5.12 Constitutive EquationsThe global balance laws and resulting field equations developed earlier in this chapter areapplicable to all continuous media, but say nothing about the response of specific mate-rials to force or temperature loadings. To fill this need, we introduce constitutive equationswhich specify the mechanical and thermal properties of particular materials based upontheir internal constitution. Mathematically, the usefulness of these constitutive equations


is to describe the relationships among the kinematic, mechanical, and thermal field equa-tions and to permit the formulations of well-posed problems in continuum mechanics.Physically, the constitutive equations define various idealized materials which serve asmodels for the behavior of real materials. However, it is not possible to write down oneequation which is capable of representing a given material over its entire range of appli-cation, since many materials behave quite differently under changing levels of loading,such as, elastic-plastic response due to increasing stress. And so in this sense it is perhapsbetter to think of constitutive equations as being representative of a particular behaviorrather than of a particular material.

In previous sections of this Chapter, the foundations of constitutive assumptions havebeen addressed. On that fundamental level, it is not possible to discuss in detail thefundamental derivation of all the constitutive models that an engineer would want to befamiliar with. Instead, the theoretical background and definitions were given along witha few specific cases examined. This allows the inductive student to grasp the conceptspresented.

Since only a few constitutive models were considered from a fundamental basis, thissection is devoted to a brief survey of constitutive equations. This acts as an introductionfor subsequent chapters when various constitutive models are discussed as applicationsof continuum mechanics.

A brief listing of some well-known constitutive equations is as follows:

(a) the stress-strain equations for a linear elastic solid assuming infinitesimal strains,

tij = Cijkmεkm (5.172)

where the Cijkm are the elastic constants representing the properties of thebody. For isotropic behavior, Eq 5.172 takes the special form

tij = λδijεkk + 2µεij (5.173)

in which λ and µ are coefficients that express the elastic properties of the ma-terial.

(b) the linear viscous fluid,

τij = Kijmndmn (5.174)

where τij is the shearing stress in the fluid and the constants Kijmn representsits viscous properties. For a Newtonian fluid,

τij = λ∗δijdkk + 2µ∗dij (5.175)

where λ∗ and µ∗ are viscosity coefficients.

(c) plastic stress-strain equation,

dεPij = Sijdλ (5.176)

where dεPij is the plastic strain increment, Sij the deviator stress, and dλ a pro-portionality constant.

(d) linear viscoelastic differential-operator equations


PSij = 2Qηij (5.177a)

tii = 3Kεii (5.177b)

where P and Q are differential time operators of the form

P =

N∑i=0

pi∂i

∂ti(5.178a)

Q =

M∑i=0

qi∂i

∂ti(5.178b)

and in which the coefficients pi and qi (not necessarily constants) representthe viscoelastic properties. Also, K is the bulk modulus. Note further that thispair of equations specifies separately the deviatoric and volumetric responses.

(e) linear viscoelastic integral equations

ηij =

∫t0

ψs(t− t′)∂Sij

∂t′dt′ (5.179a)

εii =

∫t0

ψv(t− t′)∂tij

∂t′dt′ (5.179b)

where the properties are represented by ψs and ψv, the shear and volumetriccreep functions, respectively.

In formulating a well-posed problem in continuum mechanics, we need the field equa-tions together with whatever equations of state are necessary, plus the appropriate con-stitutive equations and boundary conditions. As a point of reference we list again, as agroup, the important field equations in both indicial and symbolic notation:

(a) the continuity equation (Eq 5.14)

∂ρ

∂t+ (ρvk),k = 0 or

∂ρ

∂t+∇ · (ρv) = 0 (5.180)

(b) the equations of motion (Eq 5.22)

tji,j + ρbi = ρvi or ∇ · T + ρb = ρv (5.181)

(c) the energy equation (Eq 5.65)

ρu− tijdij − ρr+ qi,i = 0 or ρu− T : D− ρr+∇ · q = 0 (5.182)


If we assume the body forces bi and distributed heat sources r are prescribed, the abovecollection consists of five independent equations involving fourteen unknowns, namely,ρ, vi, tij, qi andu. In addition, in a non-isothermal situation, the entropy η and tempera-ture field θ = θ(x, t) have to be taken into consideration. For the isothermal theory, elevenequations are needed in conjunction with the five field equations listed above. Of these,six are constitutive equations, three are temperature-heat conduction equations (Fourier’slaw), and two are equations of state.

If the mechanical and thermal fields are uncoupled and isothermal conditions pre-vail, the continuity equation along with the equations of motion and the six constitutiveequations provide a determinate set of ten kinematic-mechanical equations for the tenunknowns ρ, vi and tij. It is with such rather simple continuum problems that we shallconcern ourselves in subsequent chapters.

References[1] Carlson, D. E. (1984), “Linear thermoelasticity,” in S. Flugge’s Handbuch der Physik,

Vol. VIa/2 (edited by C. Truesdell), Springer-Verlag, pp. 297–345

[2] Coleman, B. D. and Noll, W. (1963), “The thermodynamics of elastic materials withheat conduction and viscosity,” Arch. Rational Mech. Anal., Vol. 13, pp. 167–178 (4)

[3] Coleman, B. D. and Mizel, V. J. (1964), “Existence of caloric equations of state inthermodynamics,” J. Chem. Phys., Vol. 40, pp. 1116–1125 (4)

[4] Dienes, J. K. (1979), ”On the Analysis of Rotation and Stress Rate in DeformingBodies,” Acta Mech., Vol. 32, pp. 217-232

[5] Flanagan, D. P. and Taylor, L. M., (1987), ”An Accurate Numerical Algorithm forStress Integration with Finite Rotations,” Computer Methods in Applied Mechanics andEngineering, Vol. 62, pp. 305-320

[6] Green, A. E. and Naghdi, P. M., (1979), “A note on invariance under superposedrigid body motions,” J. of Elasticity, Vol. 9, pp. 1–8

[7] Johnson, G. C. and Bammann, D. C., (1984), ”A discussion of stress rates in finitedeformation problems,” Int J Solids and Structures, Vol. 20, pp 725-737

[8] Malvern, L. E. (1969), Introduction to the Mechanics of a Continuous Medium, Prentice-Hall, Inc., Englewood Cliffs, NJ

[9] Naghdi, P. M. (1984), “The theory of shells and plates,” in S. Flugge’s Handbuch derPhysik, Vol. II (edited by C. Truesdell), Springer-Verlag, pp. 425–640

[10] Van Wylen, G. J. and Sonntag, R. E. (1965), Fundamentals of Classic Thermodymanics,Wiley, Inc., New York


Problems

Problem 5.1Determine the material derivative of the flux of any vector property Q∗i through the spatialarea S. Specifically, show that

d

dt

∫S

Q∗ini dS =

∫S

(Q∗i +Q∗ivk,k −Q∗kvi,k

)ni dS

in agreement with Eq 5.5.

Problem 5.2Let the property P∗ij... in Eq 5.1 be the scalar 1 so that the integral in that equation

represents the instantaneous volume V. Show that in this case

Pij... =d

dt

∫V

dV =

∫V

vi,i dV .

Problem 5.3Verify the identity

εijkak,j = 2 (wi +wiwj,j −wjvi,j) ,

and by using this identity as well as the result of Problem 5.1, prove that the materialderivative of the vorticity flux equals one half the flux of the curl of the acceleration; thatis, show that

d

dt

∫S

wini dS =1

2

∫S

εijkak,jni dS .

Problem 5.4Making use of the divergence theorem of Gauss together with the identity

∂wi

∂t=1

2εijkak,j − εijkεkmq (wmvq),j

show that∂

∂t

∫V

wi dV =

∫S

(1

2εijkak +wjvi −wivj

)nj dS .

Problem 5.5Show that the material derivative of the vorticity of the material contained in a volume V

is given byd

dt

∫V

wi dV =

∫S

(1

2εijkak +wjvi

)nj dS .

Problem 5.6Given the velocity field

v1 = ax1 − bx2, v2 = bx1 + ax2, v3 = c

√x21 + x22


where a, b, and c are constants, determine

(a) whether or not the continuity equation is satisfied,

(b) whether the motion is isochoric.

Answer

(a) only when ρ = ρ0e−2at (b) only if a = 0

Problem 5.7For a certain continuum at rest, the stress is given by

tij = −p0δij

where p0 is a constant. Use the continuity equation to show that for this case the stresspower may be expressed as

tijdij =p0ρ

ρ.

Problem 5.8Consider the motion xi = (1 + t/k)Xi where k is a constant. From the conservation of

mass and the initial condition ρ = ρ0 at t = 0, determine ρ as a function of ρ0, t, and k.

Answer

ρ =ρ0k

3

(k+ t)3

Problem 5.9By combining Eqs 5.17b and 5.13, verify the result presented in Eq 4.161.

Problem 5.10Using the identity

εijkak,j = 2 (wi +wivj,j −wjvi,j)

as well as the continuity equation, show that

d

dt

(wi

ρ

)=εijkak,j + 2wjvi,j

2ρ.

Problem 5.11State the equations of motion and from them show by the use of the material derivative

vi =∂vi

∂t+ vjvi,j ,

and the continuity equation that

∂ (ρvi)

∂t= (tij − ρvivj),j + ρbi .


Problem 5.12Determine the form which the equations of motion take if the stress components are given

by tij = −pδij where p = p(x, t).

Answer

ρai = −p,i + ρbi

Problem 5.13Let a material continuum have the constitutive equation

tij = αδijdkk + 2βdij

where α and β are constants. Determine the form which the equations of motion take interms of the velocity gradients for this material.

Answer

ρvi = ρbi + (α+ β)vj,ij + βvi,jj

Problem 5.14Assume that distributed body moments mi act throughout a continuum in motion. Show

that the equations of motion are still valid in the form of Eq 5.22, but that the angularmomentum principle now requires

εijktjk +mi = 0

implying that the stress tensor can no longer be taken as symmetric.

Problem 5.15For a rigid body rotation about the origin, the velocity field may be expressed by vi =εijkωjxk where ωj is the angular velocity vector. Show that for this situation the angularmomentum principle is given by

Mi =.

(ωjIij)

where Mi is the total moment about the origin of all surface and body forces, and Iij is themoment of inertia of the body defined by the tensor

Iij =

∫V

ρ (δijxkxk − xixj) dV .

Problem 5.16Determine expressions for the stress power tijdij in terms of

(a) the first Piola-Kirchoff stress tensor,

(b) the second Piola-Kirchoff stress tensor.


Answer

(a) tijdij = ρFiAPiA/ρ0 (b) tijdij = ρsABCAB/2ρ0

Problem 5.17Show that, for a rigid body rotation about the origin, the kinetic energy integral Eq 5.53

reduces to the form given in rigid body dynamics, that is,

K =1

2ωiωjIij

where Iij is the inertia tensor defined in Problem 5.15.

Problem 5.18Show that one way to express the rate of change of kinetic energy of the material currently

occupying the volume V is by the equation

K =

∫V

ρbivi dV −

∫V

tijvi,j dV +

∫S

vit(n)i dS

and give an interpretation of each of the above integrals.

Problem 5.19Consider a continuum for which the stress is tij = −p0δij and which obeys the heat

conduction law qi = −κθ,i. Show that for this medium the energy equation takes the form

ρu = −p0vi,i + ρr+ κθ,ii .

Problem 5.20If mechanical energy only is considered, the energy balance can be derived from the equa-

tions of motion. Thus, by forming the scalar product of each term of Eq 5.22 with thevelocity vi and integrating the resulting equation term-by-term over the volume V, weobtain the energy equation. Verify that one form of the result is

1

2ρ

.

(v · v) + tr (T ·D) − ρb · v− div (T · v) = 0 .

Problem 5.21Show that for a continuum body experiencing volumetric growth

d

dt

∫P

ρf dV =

∫P

(ρf+ ρcf

)dV .

Problem 5.22Using the local form of the balance of mass, Eq 5.90, show that the balance of linear

momentum for a material with volumetric growth is Eq 5.93

ρdvi

dt= tji,j + ρbi .


Problem 5.23Using the local form of the balance of mass, Eq 5.90, and linear momentum, Eq 5.93, show

that the stress in a volumetrically growing body is symmetric.

Problem 5.24If a continuum has the constitutive equation

tij = −pδij + αdij + βdikdkj

where p, α and β are constants, and if the material is incompressible (dii = 0), show that

tii = −3p− 2βIID

where IID is the second invariant of the rate of deformation tensor.

Problem 5.25Starting with Eq 5.173 for isotropic elastic behavior, show that

tii = (3λ+ 2µ)εii ,

and using this result, deduce that

εij =1

2µ

(tij −

λ

3λ+ 2µδijtkk

).

Problem 5.26For a Newtonian fluid, the constitutive equation is given by

tij = −pδij + τij

= −pδij + λ∗δijdkk + 2µ∗dij (see Eq 5.175) .

By substituting this constitutive equation into the equations of motion, derive the equation

ρvi = ρbi − p,i + (λ∗ + µ∗)vj,ji + µ∗vi,jj .

Problem 5.27Combine Eqs 5.177a and Eq 5.177b into a single viscoelastic constitutive equation having

the formtij = δijRεkk + Sεij

where the linear time operators R and S are given, respectively,

R =3K P − 2 Q

3 Pand S =

2 Q

P.

Problem 5.28Assume a viscoelastic medium is governed by the constitutive equations Eq 5.177. Let a

slender bar of such material be subjected to the axial tension stress t11 = σ0f(t) where σ0 isa constant and f(t) some function of time. Assuming that t22 = t33 = t13 = t23 = t12 = 0,determine ε11, ε22, and ε33 as functions of P, Q, and f(t).


Answer

ε11 =3K P + Q

9K Qσ0f (t)

ε22 = ε33 =2 Q − 3K P

18K Qσ0f (t)

Problem 5.29

Use the definition of the free energy along with the reduced form of the Clausius-Duhemequation to derive the local dissipation inequality.

Problem 5.30

The constitutive model for a compressible, viscous, and heat-conducting material is definedby

ψ = ψ(θ, gk, FiA, FiA

),

η = η(θ, gk, FiA, FiA

),

tij = tij(θ, gk, FiA, FiA

),

qi = qi(θ, gk, FiA, FiA

).

Deduce the following restrictions on these constitutive response functions:

(a)∂ψ(θ, gk, FkB, FkB

)∂gi

= 0,

(b)∂ψ(θ, gk, FkB, FkB

)∂FiA

= 0,

(c) η (θ, FiA) = −∂ψ (θ, FkB)

∂θ,

(d) tij = ρFjA∂ψ (θ, FkB)

∂FiA,

(e) −1

θqi (θ, gk, FiA, 0)gi > 0.

Problem 5.31

Assume the constitutive relationships

u = u (CAB, η) ,

θ = θ (CAB, η) ,

tij = tij (CAB, η) ,

qi = qi (CAB, η, gk) ,


for an elastic material. Use the Clausius-Duhem inequality to show

θ =∂u

∂η,

u = u (CAB, η) ,

tij =1

2ρFiA

(∂u

∂CAB+

∂u

∂CBA

)FjB ,

qigi > 0 .

Problem 5.32Use the basic kinematic result of superposed rigid body motion given in Eq 5.121b to show

the following:

(a)∂x+i

∂xk= Qik,

(b) F+iA = QijFjA.

Problem 5.33Show that the Jacobian transforms as follows under a superposed rigid body motion:

J+ = J .

Problem 5.34Utilize the result of Problem 5.33 along with the law of conservation of mass to show thatρ+ = ρ.

Problem 5.35Show that the gradient of the stress components transforms under superposed rigid body

motion as follows:∂t+ij

∂x+j

= Qim∂tmn

∂xn.

Problem 5.36Use the superposed rigid body motion definitions to show the following relationships:

(a) C+AB = CAB,

(b) U+AB = UAB,

(c) R+iA = QijRjA,

(d) B+ij = QimBmkQkj.

Problem 5.37In the context of rigid body dynamics, consider the motion defined by

x (X, t) = X


along withx = QT (t)

[x+ − a (t)

]show that

v+ = a (t) +ω (t)×[x+ − a (t)

]where ω is the angular velocity of the body.


6Linear Elasticity

Elastic behavior is characterized by the following two conditions: (1) where the stress ina material is a unique function of the strain, and (2) where the material has the propertyfor complete recovery to a “natural” shape upon removal of the applied forces. If thebehavior of a material is not elastic, we say that it is inelastic. Also, we acknowledge thatelastic behavior may be linear or non-linear.

6.1 Elasticity, Hooke’s Law, Strain EnergyFigure 6.1 shows geometrically these behavior patterns by simple stress-strain curves,with the relevant loading and unloading paths indicated. For many engineering appli-cations, especially those involving structural materials such as metals and concrete, theconditions for elastic behavior are realized, and for these cases the theory of elasticity of-fers a very useful and reliable model for design. Symbolically, we write the constitutiveequation for elastic behavior in its most general form as

T = G (ε) (6.1)

where G is a symmetric tensor-valued function and ε is any one of the various straintensors we introduced earlier. However, for the response function G in this text we con-sider only that case of Eq 6.1 for which the stress is a linear function of strain. Also, weassume that, in the deformed material, the displacement gradients are everywhere smallcompared with unity. Thus, the distinction between the Lagrangian and Eulerian descrip-tions is negligible, and following the argument of Eq 4.60 we make use of the infinitesimal

Str

ess

Strain

LOAD

UNLO

AD

(a) Linear elastic

Str

ess

Strain

LOAD

UNLO

AD

(b) Nonlinear elastic

Str

ess

Strain

LOAD

UN

LO

AD

(c) Inelastic

FIGURE 6.1Uniaxial loading-unloading stress-strain curves for various material behaviors.

211


strain tensor defined in Eq 4.62, which we repeat here:

εij =1

2

(∂ui

∂xj+∂uj

∂xi

)=1

2(ui,j + uj,i) . (6.2)

Within the context of the above assumptions, we write the constitutive equation for linearelastic behavior as

tij = Cijkmεkm or T = Cε (6.3)

where the tensor of elastic coefficients Cijkm has 34 = 81 components. However, due tothe symmetry of both the stress and strain tensors, it is clear that

Cijkm = Cjikm = Cijmk (6.4)

which reduces the 81 possibilities to 36 distinct coefficients at most.We may demonstrate the tensor character of C by a consideration of the elastic consti-

tutive equation when expressed in a rotated (primed) coordinate system in which it hasthe form

t′ij = C′ijkmε′km . (6.5)

But by the transformation laws for second-order tensors, along with Eq 6.3,

t′ij = aiqajstqs = aiqajsCqskmεkm

= aiqajsCqskmapkanmε′pn

which by a direct comparison with Eq 6.5 provides the result

C′ijkm = aiqajsapkanmCqskm , (6.6)

that is, the transformation rule for a fourth-order Cartesian tensor.In general, the Cijkm coefficients may depend upon temperature, but here we assume

adiabatic (no heat gain or loss) and isothermal (constant temperature) conditions. We alsoignore strain-rate effects and consider the components Cijkm to be at most a function ofposition. If the elastic coefficients are constants, the material is said to be homogeneous.These constants are those describing the elastic properties of the material. The constitutivelaw given by Eq 6.3 is known as the generalized Hooke’s law.

For certain purposes it is convenient to write Hooke’s law using a single subscript onthe stress and strain components and double subscripts on the elastic constants. To thisend, we define

t11 = t1 , t23 = t32 = t4 ,

t22 = t2 , t13 = t31 = t5 , (6.7a)t33 = t3 , t12 = t21 = t6 ,

and

ε11 = ε1 , 2ε23 = 2ε32 = ε4 ,

ε22 = ε2 , 2ε13 = 2ε31 = ε5 , (6.7b)ε33 = ε3 , 2ε12 = 2ε21 = ε6 ,

where the factor of two on the shear strain components is introduced in keeping withEq 4.71. From these definitions, Hooke’s law is now written

tα = Cαβεβ or T = Cε (6.8)

Linear Elasticity 213

with Greek subscripts having a range of six. Note the font change in Cαβ to that ofa matrix. This is done to account for the fact that the reduced representation stiffnessmatrix does not transform as a tensor. In matrix form Eq 6.8 appears as

t1t2t3t4t5t6

=

C11 C12 C13 C14 C15 C16C21 C22 C23 C24 C25 C26C31 C32 C33 C34 C35 C36C41 C42 C43 C44 C45 C46C51 C52 C53 C54 C55 C56C61 C62 C63 C64 C65 C66

ε1ε2ε3ε4ε5ε6

. (6.9)

We repeat that the array of the 36 constants Cαβ does not constitute a tensor.In view of our assumption to neglect thermal effects at this point, the energy balance

Eq 5.65 is reduced to the form

u =1

ρtijdij (6.10a)

which for small-deformation theory, by Eq 4.150, becomes

u =1

ρtijεij (6.10b)

The internal energy u in these equations is purely mechanical and is called the strainenergy (per unit mass). Recall now that, by the continuity equation in Lagrangian form,ρ0 = ρJ and also that to the first order of approximation

J = det F = det(δiA +

∂ui

∂XA

)= 1+

∂ui

∂XA. (6.11)

Therefore, from our assumption of small displacement gradients, namely ∂ui/∂XA 1,we may take J ≈ 1 in the continuity equation to give ρ = ρ0 a constant in Eqs 6.10a and6.10b.

For elastic behavior under the assumptions we have imposed, the strain energy is afunction of the strain components only, and we write

u = u (εij) (6.12)

so that

u =∂u

∂εijεij , (6.13)

and by a direct comparison with Eq 6.10b we obtain

1

ρtij =

∂u

∂εij. (6.14)

The strain energy density, W (strain energy per unit volume) is defined by

W = ρ0u , (6.15)

and since ρ = ρ0, a constant, under the assumptions we have made, it follows fromEq 6.14 that

tij = ρ∂u

∂εij=∂W

∂εij. (6.16)


It is worthwhile noting at this point that elastic behavior is sometimes defined on thebasis of the existence of a strain energy function from which the stresses may be de-termined by the differentiation in Eq 6.16. A material defined in this way is called ahyperelastic material. The stress is still a unique function of strain so that this energy ap-proach is compatible with our earlier definition of elastic behavior. Thus, in keeping withour basic restriction to infinitesimal deformations, we shall develop the linearized formof Eq 6.16. Expanding W about the origin, we have

W (εij) = W (0) +∂W (0)

∂εijεij +

1

2

∂2W (0)

∂εij∂εkmεijεkm + · · · , (6.17)

and, from Eq 6.16,

tij =∂W

∂εij=∂W (0)

∂εij+∂2W (0)

∂εij∂εkmεkm + · · · . (6.18)

It is customary to assume that there are no residual stresses in the unstrained state of thematerial so that tij = 0 when εij = 0. Thus, by retaining only the linear term of the aboveexpansion, we may express the linear elastic constitutive equation as

tij =∂2W (0)

∂εij∂εkmεkm = Cijkmεkm (6.19)

based on the strain energy function. This equation appears to be identical to Eq 6.3,but there is one very important difference between the two – not only do we have thesymmetries expressed by Eq 6.4, but now we also have

Cijkm = Ckmij (6.20)

due to the fact that∂2W (0)

∂εij∂εkm=∂2W (0)

∂εkm∂εij.

Thus, the existence of a strain energy function reduces the number of distinct componentsof Cijkm from 36 to 21. Further reductions for special types of elastic behavior are ob-tained from material symmetry properties in the next section. Note that by substitutingEq 6.19 into Eq 6.17 and assuming a linear stress-strain relation, we may now write

W (εij) =1

2Cijkmεijεkm =

1

2tijεij (6.21a)

which in the notation of Eq 6.8 becomes

W (εα) =1

2Cαβεαεβ =

1

2tαεα , (6.21b)

and by the symmetry condition we have only 21 distinct constants out of the 36 possible.

6.2 Hooke’s Law for Isotropic Media, Elastic ConstantsIf the behavior of a material is elastic under a given set of circumstances, it is custom-arily spoken of as an elastic material when discussing that situation even though under


a different set of circumstances its behavior may not be elastic. Furthermore, if a body’selastic properties as described by the coefficients Cijkm are the same in every set of ref-erence axes at any point for a given situation, we call it an isotropic elastic material. Forsuch materials, the constitutive equation has only two elastic constants. A material thatis not isotropic is called anisotropic; we shall define some of these based upon the degreeof elastic symmetry each possesses.

In general, an isotropic tensor is defined as one whose components are unchangedby any orthogonal transformation from one set of Cartesian axes to another. Zero-ordertensors of any order – and all zeroth-order tensors (scalars) – are isotropic, but there areno first-order isotropic tensors (vectors). The unit tensor I, having Kronecker deltas ascomponents, and any scalar multiple of I are the only second-order isotropic tensors (seeProblem 6.5). The only nontrivial third-order isotropic tensor is the permutation symbol.The most general fourth-order isotropic tensor may be shown to have a form in terms ofKronecker deltas which we now introduce as the prototype for C, namely,

Cijkm = λδijδkm + µ (δikδjm + δimδjk) + β (δikδjm − δimδjk) (6.22)

where λ, µ, and β are scalars. But by Eq 6.4, Cijkm = Cjikm = Cijmk. This implies that βmust be zero for the stated symmetries since by interchanging i and j in the expression

β(δikδjm − δimδjk) = β(δjkδim − δjmδik)

we see that β = −β and, consequently, β = 0. Therefore, inserting the reduced Eq 6.22into Eq 6.3, we have

tij = (λδijδkm + µδikδjm + µδimδjk) εkm .

But by the substitution property of δij, this reduces to

tij = λδijεkk + 2µεij (6.23)

which is Hooke’s law for isotropic elastic behavior. As mentioned earlier, we see that forisotropic elastic behavior the 21 constants of the generalized law have been reduced totwo, λ and µ, known as the Lame constants. Note that for an isotropic elastic materialCijkm = Ckmij; that is, an isotropic elastic material is necessarily hyperelastic.

Example 6.1

Show that for an isotropic linear elastic solid the principal axes of the stressand strain tensors coincide, and develop an expression for the relationship amongtheir principal values.

SolutionLet n(q) (q = 1, 2, 3) be unit normals in the principal directions of εij, andassociated with these normals the corresponding principal values are ε(q) where(q = 1, 2, 3). From Eq 6.23 we form the dot products

tijn(q)j = (λδijεkk + 2µεij)n

(q)j

= λn(q)i εkk + 2µεijn

(q)j .


But n(q)j and ε(q) satisfy the fundamental equation for the eigenvalue problem,

namely, εijn(q)j = ε(q)δijn

(q)j , so that now

tijn(q)j = λεkkn

(q)i + 2µε(q)n

(q)i

=[λεkk + 2µε(q)

]n

(q)i ,

and because εkk = ε(1) +ε(2) +ε(3) is the first invariant of strain, it is constantfor all n(q)

i so that

tijn(q)j =

(λ[ε(1) + ε(2) + ε(3)

]+ 2µε(q)

)n

(q)j .

This indicates that n(q)i (q = 1, 2, 3) are principal directions of stress also, with

principal stress values

t(q) = λ[ε(1) + ε(2) + ε(3)

]+ 2µε(q) (q = 1, 2, 3) .

We may easily invert Eq 6.23 to express the strain components in terms of the stresses.To this end, we first determine εii in terms of tii from Eq 6.23 by setting i = j to yield

tii = 3λεkk + 2µεii = (3λ+ 2µ)εii . (6.24)

Now, by solving Eq 6.23 for εij and substituting from Eq 6.24, we obtain the inverse formof the isotropic constitutive equation,

εij =1

2µ

(tij +

λ

3λ+ 2µδijtkk

). (6.25)

By a formal – although admittedly not obvious – rearrangement of this equation, we maywrite

εij =λ+ µ

µ (3λ+ 2µ)

[1+

λ

2 (λ+ µ)

]tij −

λ

2 (λ+ µ)δijtkk

(6.26)

from which, if we define,

E =µ (3λ+ 2µ)

λ+ µ(6.27a)

andν =

λ

2 (λ+ µ), (6.27b)

we obtain the following form of Hooke’s law for isotropic behavior in terms of the engi-neering constants E and ν,

εij =1

E[(1+ ν) Tij − νδijTkk] . (6.28)

Here E is called Young’s modulus, or simply the modulus of elasticity, and ν is knownas Poisson’s ratio. By suitable combinations of these two constants, we may define twoadditional constants of importance in engineering elasticity. First, the shear modulus, ormodulus of rigidity, is defined as


t11

t11

(a) Uniaxial tension

t12

t12

(b) Simple shear

t11

t11

t22 t

22

t33

t33

(c) Uniform triaxial tension

FIGURE 6.2Simple stress states.

G =E

2 (1+ ν)= µ (6.29a)

which, as noted, is identical to the Lame constant µ. Second, the bulk modulus is definedas

K =E

3 (1− 2ν). (6.29b)

For isotropic elastic materials, any two elastic constants completely define the material’sresponse. In addition to that, any elastic constant can be determined in terms of any twoother constants. A listing of all elastic constants in terms of other pairs of constants isgiven in Table 6.1.

The physical interpretations of the constants E, ν, G, and K introduced above can bedetermined from a consideration of the special states of stress displayed in Fig. 6.2. Inthe case of a uniaxial state of stress (tension or compression), say in the x1 directionwith t11 = ±σ0, and all other stress components zero (Fig. 6.2(a)), Eq 6.28 yields (sincetii = ±σ0),

ε11 =t11

E=±σ0E

for (i = j = 1) , (6.30a)

ε22 = −νε11 =∓νσ0E

for (i = j = 2) , (6.30b)

ε33 = −νε11 =∓νσ0E

for (i = j = 3) , (6.30c)

as well as zero shear strains for i 6= j. Thus, E is the proportionality factor betweenaxial (normal) stresses and strains. Geometrically, it is the slope of the one-dimensional

218

Con

tinuu

mM

echa

nics

for

Engi

neer

s

TABLE 6.1Relations between elastic constants.

λ µ E ν K

λ, µ µ (3λ+ 2µ)

λ+ µ

λ

2 (λ+ µ)

3λ+ 2µ

3

λ, E (E− 2λ) +1

4

√(E− 3λ)

2+ 8λE − (E− 2λ) +

1

4λ

√(E− 3λ)

2+ 8λ2 (3λ+ E) +

1

6

√(3λ+ E)

2− 4λE

λ, νλ (1− 2ν)

2ν

λ (1+ ν) (1− 2ν)

ν

λ (1+ ν)

3ν

λ, K3 (K− λ)

2

9K (K− λ)

3K− λ

λ

3K− λ

µ, Eµ (2µ− E)

E− 3µ

E− 2µ

2µ

µE

3 (3µ− E)

µ, ν2µν

1− 2ν2µ (1+ ν)

2µ (1+ ν)

3 (1− 2ν)

µ,K3K− 2µ

3

9Kµ

3K+ µ

3K− 2µ

2 (3K+ µ)

E, ννE

(1+ ν) (1− 2ν)

E

2 (1+ ν)

E

3 (1− 2ν)

E, K3K (3K− E)

(9K− E)

3KE

9K− E

3K− E

6K

ν, K3Kν

1+ ν3K (1− 2ν)

2 (1+ ν)3K (1− 2ν)


linear stress-strain diagram (Fig. 6.1(a)). Note that E > 0; a specimen will elongate undertension, shorten in compression. From the second and third part of Eq 6.30 above, ν isseen to be the ratio of the unit lateral contraction to unit longitudinal extension for tension,and vice versa for compression. For the simple shear case shown in Fig. 6.2(b) where, say,t12 = τ0, all other stresses zero, we have, from Eq 6.28,

ε12 =1+ ν

Et12 =

τ0

2G, (6.31a)

or for engineering strains, using Eq 4.71,

γ12 =t12

G=τ0

G(6.31b)

which casts G into the same role for simple shear as E assumes for axial tension (orcompression). Hence, the name shear modulus for G. Finally, for the case of uniformtriaxial tension (or hydrostatic compression) of Fig. 6.2(c), we take tij = ±pδij with p > 0.For this, Eq 6.28 indicates that

εii =1− 2ν

Etii =

±3 (1− 2ν)

Ep =

±pK

(6.32)

by which we infer that the bulk modulus K relates the pressure p to the volume changegiven by the cubical dilation εii (see Eq 4.76).

It should also be pointed out that, by use of the constants G and K, Hooke’s law maybe expressed in terms of the spherical and deviator components of the stress and straintensors. Thus, the pair of equations

Sij = 2Gηij , (6.33a)

tii = 3Kεii , (6.33b)

may be shown to be equivalent to Eq 6.28 (see Problem 6.6).

6.3 Elastic Symmetry; Hooke’s Law for Anisotropic MediaHooke’s law for isotropic behavior was established in Section 6.2 on the basis of C beinga fourth-order isotropic tensor. The same result may be achieved from the concepts ofelastic symmetry. To do so, we first define equivalent elastic directions as those specifiedby Cartesian axes Ox1x2x3 and Ox′1x

′2x′3 at a point such that the elastic constants Cαβ

are unchanged by a transformation between the two sets of axes. If the transformationrepresents a rotation about an axis, we say the material has axial elastic symmetry withrespect to that axis. If the transformation is a reflection of the axes with respect to someplane, we say the material has a plane of elastic symmetry. Figure 6.3(a) shows the casefor x3 being the axis of elastic symmetry, whereas Fig. 6.3(b) shows the case for the x1x2plane as the plane of elastic symmetry. The fact that the transformation for the reflectionin Fig. 6.3(b) is an improper one (resulting in the axes being a left-handed coordinatesystem) does not invalidate the symmetry considerations to be used. Also, the x3 axis inFig. 6.3(a) is said to be of order N where N = 2π/θ. It is also noteworthy that a point of


x3,x

3‛

x2

x1

x2‛

x1‛

θ θO

(a) Rotation through angle θ about x3 axis.

O

x1,x

1‛ x

2,x

2‛

x3

x3‛

(b) Reflection in the x1x2 plane.

FIGURE 6.3Axes rotations for plane stress.

elastic symmetry would imply isotropic behavior, since the elastic constants would remainunchanged for any two sets of Cartesian axes at the point.

Let us consider the consequences of the x1-x2 plane being a plane of elastic symmetryas shown in Fig. 6.3(b). The transformation matrix for this is clearly

[aij] =

1 0 0

0 1 0

0 0 −1

, (6.34)

so, in the single subscript notation for stress and strain components, the transformationsin matrix form are t′1 t′6 t′5

t′6 t′2 t′4t′5 t′4 t′3

=

1 0 0

0 1 0

0 0 −1

t1 t6 t5t6 t2 t4t5 t4 t3

1 0 0

0 1 0

0 0 −1

(6.35a)

=

t1 t6 −t5t6 t2 −t4

−t5 −t4 t3

,and ε′1

12ε′6

12ε′5

12ε′6 ε′2

12ε′4

12ε′5

12ε′4 ε′3

=

1 0 0

0 1 0

0 0 −1

ε112ε6

12ε5

12ε6 ε2

12ε4

12ε5

12ε4 ε3

1 0 0

0 1 0

0 0 −1

(6.35b)

=

ε112ε6 −12ε5

12ε6 ε2 −12ε4

−12ε5 −12ε4 ε3

.Therefore, assuming all 36 constants in Eq 6.8 are distinct, we note that for axes Ox1x2x3

t1 = C11ε1 + C12ε2 + C13ε3 + C14ε4 + C15ε5 + C16ε6 , (6.36)

whereas for axes Ox′1x′2x′3, under the condition that x1x2 is a plane of symmetry such

that the Cαβ are unchanged in this system, we have

t′1 = C11ε′1 + C12ε

′2 + C13ε

′3 + C14ε

′4 + C15ε

′5 + C16ε

′6 . (6.37)


But from Eq 6.35a, t′α = tα (a = 1, 2, 3, 6) and t′α = −tα (α = 4, 5). Likewise, ε′α = εα(α = 1, 2, 3, 6) and ε′α = εα (α = 4, 5), so that Eq 6.37 becomes

t′1 = t1 = C11ε1 + C12ε2 + C13ε3 − C14ε4 − C15ε5 + C16ε6 . (6.38)

Comparing Eq 6.36 with Eq 6.38, we see that, for these expressions (each representingt1) to be equal, we must have C14 = C15 = 0. Following the same procedure, we learnthat, if we compare expressions for t′α = tα (α = 2, 3, 6) and for t′α = −tα (α = 4, 5), theadditional elastic constants C24, C25, C34, C35, C41, C42, C43, C46, C51, C52, C53, C56, C64,and C65 must also be zero for the x1x2 plane to be one of elastic symmetry. Accordingly,the elastic constant matrix for this case has the form

[Cαβ] =

C11 C12 C13 0 0 C16C21 C22 C23 0 0 C26C31 C32 C33 0 0 C360 0 0 C44 C45 0

0 0 0 C54 C55 0

C61 C62 C63 0 0 C66

, (6.39)

and the original 36 constants are reduced to 20. Also, if a strain energy functions exists,Cαβ = Cβα and these 20 nonzero constants would be further reduced to 13.

If the x2-x3 plane is also one of elastic symmetry at the same time as the x1-x2 plane ata point and we repeat the procedure outlined above, we find that C16, C26, C36, C45, C54,C61, C62, and C63 must also be zero, and the C matrix is further reduced to

[Cαβ] =

C11 C12 C13 0 0 0

C21 C22 C23 0 0 0

C31 C32 C33 0 0 0

0 0 0 C44 0 0

0 0 0 0 C55 0

0 0 0 0 0 C66

(6.40)

having 12 nonzero coefficients, or 9 if a strain energy function exists. Interestinglyenough, if x1x3 is also a plane of elastic symmetry along with the two considered above,no further reduction in the Cαβ matrix occurs. A material possessing three mutuallyperpendicular planes of elastic symmetry is called an orthotropic material, and its elasticconstants matrix is that given in Eq 6.40.

The reduction of the orthotropic elastic matrix to that of the isotropic matrix may becompleted by successive consideration of the three axes of elastic symmetry shown inFig. 6.4 as well as their respective transformation matrices. By the rotation of 90 aboutthe x1 axis (Fig. 6.4(a)), we find that C12 = C13, C21 = C31, C22 = C33, C23 = C32, andC55 = C66. For the rotation of 90 about the x3 axis (Fig. 6.4(b)), we see that C12 = C21,C11 = C22, C12 = C23, C31 = C32, and C44 = C55. Finally, by a rotation of 45 aboutthe x3 axis (Fig. 6.4(c)), we obtain 2C44 = C11 − C12. Therefore, by setting C44 = µ andC12 = λ, we may identify these remaining three Cαβ’s with the Lame constants and writethe elastic coefficient matrix for isotropic behavior as

[Cαβ] =

λ+ 2µ λ λ 0 0 0

λ λ+ 2µ λ 0 0 0

λ λ λ+ 2µ 0 0 0

0 0 0 µ 0 0

0 0 0 0 µ 0

0 0 0 0 0 µ

. (6.41)


x3 , x2

‛

π2

π2

x3‛

x1 , x1

‛

x1

x1‛

x3‛

x2‛

x3 x

2

x2

01

−10

0 0 1

0

0

(a) 90 rotation about x1 axis

x3 , x3

‛

π2

π2

x2‛

x1

x1

x1‛

x3‛

x2‛

x3 x

2

x2 , x

1‛

0 1

-10 0

0

1

0

0

(b) 90 rotation about x3 axis

x3 , x3

‛

π4

π4

x2‛

x1

x1

x1‛

x3‛

x2‛

x3 x

2

x2

0 0 1

0

0

x1‛

12

12

−

12

12

(c) 45 rotation about x3 axis

FIGURE 6.4Geometry and transformation tables for reducing the elastic stiffness to the isotropic

case.


From the definitions given in Eqs 6.27a and 6.27b, this matrix may be expressed interms of the engineering constants E and ν so that Hooke’s law for an isotropic bodyappears in matrix form ast1t2t3t4t5t6

=E

(1+ ν) (1− 2ν)

1− ν ν ν 0 0 0

ν 1− ν ν 0 0 0

ν ν 1− ν 0 0 0

0 0 0 12

(1− 2ν) 0 0

0 0 0 0 12

(1− 2ν) 0

0 0 0 0 0 12

(1− 2ν)

ε1ε2ε3ε4ε5ε6

.(6.42)

6.4 Isotropic Elastostatics and Elastodynamics, Superposition Principle

The formulation and solution of the basic problems of linear elasticity comprise the sub-jects we call elastostatics and elastodynamics. Elastostatics is restricted to those situations inwhich inertia forces may be neglected. In both elastostatics and elastodynamics, certainfield equations have to be satisfied at all interior points of the elastic body under consid-eration, and at the same time the field variables must satisfy specific conditions on theboundary. In the case of elastodynamics problems, initial conditions on velocities anddisplacements must also be satisfied.

We begin our discussion with elastostatics for which the appropriate field equationsare:

(a) Equilibrium equationstji,j + ρbi = 0 , (6.43)

(b) Strain-displacement relation

2εij = ui,j + uj,i , (6.44)

(c) Hooke’s lawtij = λδijεkk + 2µεij , (6.45a)

orεij =

1

E[(1+ ν) tij − νδijtkk] . (6.45b)

It is usually assumed that the body forces bi are known so that the solution we seek fromthe fifteen equations listed here is for the six stresses tij, the six strains εij, and the threedisplacements ui. The conditions to be satisfied on the boundary surface S will appear inone of the following statements:

1. displacements prescribed everywhere,

ui = u∗i (x) on S (6.46)

where the asterisk denotes a prescribed quantity,


2. tractions prescribed everywhere,

t(n)i = t

∗(n)i on S , (6.47)

3. displacements prescribed on portion S1 of S,

ui = u∗i (x) on S1 , (6.48a)

with tractions prescribed on the remainder S2,

t(n)i = t

∗(n)i on S2 . (6.48b)

A most important feature of the field Eqs 6.43 through Eq 6.45a is that they are linearin the unknowns. Consequently, if t(1)ij , and ε(1)

ij , u(1)i are a solution for body forces 1b∗i

and surface tractions 1t∗(n)i , whereas t(2)ij , ε(2)

ij , and u(2)i are a solution for body forces 2b∗i

and surface tractions 2t∗(n)i , then

tij = t(1)ij + t

(2)ij , εij = ε

(1)ij + ε

(2)ij , and ui = u

(1)i + u

(2)i

offer a solution for the situation where bi = 1b∗i+2b∗i and t(n)

i = 1t∗(n)i + 2t

∗(n)i . This is a

statement of the principle of superposition, which is extremely useful for the developmentof solutions in linear elasticity.

For those problems in which the boundary conditions are given in terms of displace-ments by Eq 6.46, it is convenient for us to eliminate the stress and strain unknowns fromthe field equations so as to state the problem solely in terms of the unknown displace-ment components. Thus, by substituting Eq 6.44 into Hooke’s law (Eq 6.45a) and thatresult into the equilibrium equations (Eq 6.43), we obtain the three second-order partialdifferential equations

µui,jj + (λ+ µ)uj,ji + ρbi = 0 (6.49)

which are known as the Navier equations. If a solution can be determined for these equa-tions that also satisfies the boundary condition Eq 6.46, that result may be substitutedinto Eq 6.44 to generate the strains and those in turn substituted into Eq 6.45a to obtainthe stresses.

When the boundary conditions are given in terms of surface tractions (Eq 6.47), theequations of compatibility for infinitesimal strains (Eq 4.90) may be combined with Hooke’slaw (Eq 6.45b) and the equilibrium equations to arrive, after a certain number of algebraicmanipulations, at the equations

tij,kk +1

1+ νtkk,ij + ρ (bi,j + bj,i) +

ν

1− νδijρbk,k = 0 (6.50)

which are known as the Beltrami-Michell stress equations of compatibility. In combinationwith the equilibrium equations, these equations comprise a system for the solution of thestress components, but it is not an especially easy system to solve. As was the case withthe infinitesimal strain equation of compatibility, the body must be simply connected.

The solution for elastostatics problems is unique. Suppose that there exists two solu-tions to the governing equations Eq 6.43-6.45a with boundary conditions Eq 6.46-6.48b.The solutions will be denoted T (1), ε(1), u(1) and T (2), ε(2), u(2) corresponding to thesame body forces and the appropriate boundary conditions. The difference of these solu-tions, T = T (1) − T (2), ε = ε(1) − ε(2), u = u(1) − u(2), is also a solution to the governing


equations with b = 0 since the system of equations is linear and superposition applies.The solution T , ε, u satisfies the boundary conditions

u = 0 on S1 or t = 0 on S2 .

Integrating over the boundary S of the body gives∫S

u · tdS = 0 .

Substituting t = T · n for a symmetric stress tensor, we have∫S

u · T · ndS = 0

and applying the divergence theorem yields∫V

div (u · T ) dV = 0 . (6.51)

Writing the integrand in Cartesian tensor notation gives

div (u · T ) = (uitij),j = ui,jtij + uitij,j .

The last term is zero from the equilibrium equations tij,j = 0 since b = 0. The first termcan be written as

ui,jtij = tij (εij +ωij) = tijεij (6.52)

since tij is symmetric and ωij is skew-symmetric. From Eq 6.21a and 6.52, Eq 6.51becomes ∫

V

2W (ε) dV = 0 .

This integral is zero when ε = 0 provided that W (ε) is positive definite in Eq 6.21a. Fora linear, isotropic body this requires that E > 0 and −1 < ν < 1

2 . Since ε = 0, we haveε(1) = ε(2) and T (1) = T (2) from Hooke’s law. There cannot be two different solutions tothe linear equations of elastostatics for the same body forces and boundary conditions. Asolution that satisfies the governing equations and boundary conditions is unique.

In elastodynamics, the equilibrium equations must be replaced by the equations ofmotion (Eq 5.27) in the system of basic field equations. Therefore, all field quantitiesare now considered functions of time as well as of the coordinates, so that a solution forthe displacement field, for example, appears in the form ui = ui(x, t). In addition, thesolution must satisfy not only boundary conditions which may be functions of time as in

ui = u∗i (x, t) on S (6.53a)

ort(n)i = t

∗(n)i (x, t) on S , (6.53b)

but also initial conditions, which usually are taken as

ui = u∗i (x, 0) (6.54a)

andui = u∗i (x, 0) . (6.54b)

Analogous to Eq 6.49 for elastostatics, it is easily shown that the governing equations fordisplacements in elastodynamics theory are

µui,jj + (λ+ µ)uj,ji + ρbi = ρui (6.55)

which are also called Navier’s equations.


BL

L

x3

x2

C1

C2

t(n)i

t(n)i

x1

FIGURE 6.5Beam geometry for the Saint-Venant problem.

6.5 Saint-Venant Problem

Solving an elasticity problem by satisfying Eqs 6.43, 6.44 and 6.45a along with the bound-ary conditions is a daunting task. This is even before the boundary conditions are con-sidered. In the 1800s Saint-Venant studied long beams that were loaded in a variety ofways: extension, torsion, pure bending and flexure. Rather than satisfy exact bound-ary conditions on the lateral surface and the ends, Saint-Venant solved the problems byconsidering relaxed boundary conditions on the beam ends. If one considers staticallyequivalent force and moment systems, the solution sufficiently far away from the applica-tion of the load will be the same. This is known as the Saint-Venant principle. The relaxedboundary conditions on the beam ends requires that the resultant force and moment beconsidered rather than the exact traction distribution over the end.

To demonstrate Saint-Venant’s solution consider a beam that is represented by a rightcylinder extending in the x3 direction as shown in Fig. 6.5. The lateral surface of thecylinder, BL, will be stress free. This condition is called the lateral boundary conditionand can be written as

t(n)i = ti1n1 + ti2n2 = 0 on BL . (6.56)

Let the cylinder be generated by a simply connected cross section C which is perpen-dicular to the x3 axis. For the different problems (extension, torsion, pure bending andflexure) the ends, C1 and C2, will have relaxed boundary conditions. The sum of the stressover the end’s area will give a force vector Ti∫

C1

t3i dS = Ti . (6.57)


Bending moments on the ends are expressed in terms of the axial stress t33 as follows:∫C1

t33x2 dS = M1 ,

∫C1

t33x1 dS = −M2 . (6.58)

Finally, the torque on the beam’s end may be written as∫C1

(t32x1 − t31x2) dS = M3 = Mt . (6.59)

Similar equations hold for end C2.Saint-Venant used a semi-inverse method in solving these problems. A reasonable as-

sumption was made regarding the stress components, strain or displacements. Fromthis assumption the equilibrium and compatibility equations were shown to be satisfied.Boundary conditions, exact and relaxed, were checked. Because solutions to elasticityproblems are unique the solution obtained was, of course, the only solution.

In the following subsections Saint-Venant solutions for extension, torsion, pure bendingand flexure are developed. The different problems are solved by setting the componentsof the end force Ti and moment Mi to different values:

(I) T1 = T2 = 0, T3 = T, Mi = 0 Pure Extension(II) Ti = 0, M1 = M2 = 0, M3 = Mt Pure Torsion

(III) Ti = 0, M1 = M3 = 0, M2 = M Pure Bending(IV) T1 = T, T2 = T3 = 0, Mi = 0 Pure Flexure

More complex elasticity problems can be obtained by superposing the basic solutions.

6.5.1 Extension

For the case of pure extension the nonzero stress components are assumed to be

t11 =T

A(6.60)

where the area is given by

A =

∫C

d S ,

and all other components are zero. These stress components clearly satisfy the equilib-rium conditions without body forces, Eq 6.43. Also, the relaxed boundary conditions aresatisfied.

The strains can be easily found from Eq 6.28 to be

ε33 =T

EA, ε11 = ε22 = −

νT

EA(6.61)

with all other strain components being zero. Clearly the compatibility condition, Eq 4.90is satisfied since the strains are constant. Using the strain-displacement relationship,Eq 6.44, the displacements can be found to be

u3 =T

EAx3 , u1 = −

νT

EAx1 , u2 = −

νT

EAx2 (6.62)

by direct integration.


6.5.2 Torsion

We begin with a brief review of the solution to the simplest of torsion problems, thecase of a shaft having a constant circular cross section when subjected to equilibratingend couples, Mt as shown in Fig. 6.6(a). Let the end face at x3 = 0 be fixed while theface at x3 = L is allowed to rotate about the axis of the shaft. It is assumed that planesections perpendicular to the axis remain plane under the twisting, and that each rotatesthrough an angle proportional to its distance from the fixed end. Accordingly, a point inthe cross section at coordinate x3 will rotate an angle of θx3where θ is the angle of twistper unit length. Each point, say point P, in the cross section travels a distance θx3R whichis proportional to the distance R from the x3 axis as shown in Fig. 6.6(b). The distancesquared to point P is the square of the x1 and x2 coordinates (Fig. 6.6(c)). Using thisdistance, it is easy to define the cosine and sine of angle β. It is straightforward to writeout displacements u1 and u2 in terms of θ since cos (90− β) = x2/R and sin (90− β) =x1/R. Thus,

u1 = −θx2x3 , u2 = θx1x3 , u3 = 0 , (6.63)

where θ is the angle of twist per unit length of the shaft.Recall from Eq 4.62 that 2εij = ui,j + uj,i by which we may calculate the strains from

Eq 6.63. The resulting strains are then inserted into Eq 6.23 to obtain the stress compo-nents (since µ ≡ G by Eq 6.29a)

t23 = Gθx1 , t13 = −Gθx2 , t11 = t22 = t33 = t12 = 0 . (6.64)

Because these stress components, as well as the strains from which they were derived,are either linear functions of the coordinates or zero, the compatibility equations Eq 4.90are satisfied. Likewise, for zero body forces, the equilibrium equations Eq 6.43 are clearlysatisfied. The lateral surface of the shaft is stress free. To verify this, consider the stresscomponents in the direction of the normal at a point on the cross-section perimeter des-ignated in Fig. 6.6(b). At a radius R = a,

t13x1

a+ t23

x2

a=Gθ

a(−x2x1 + x1x2) = 0 . (6.65)

At the same time, the total shearing stress at any point of the cross section is the resultant

τ =

√t213 + t223 = Gθ

√x21 + x22 = GθR (6.66)

which indicates that the shear is proportional to the radius at the point, and perpendicularto that radius.

By summing the moments of the shear forces on either end face of the shaft, we findthat

Mt =

∫ ∫(x1t23 + x2t13)dx1dx2 = Gθ

∫ ∫R2dx1dx2 = GθIP (6.67)

where IP is the polar moment of inertia of the cross section.For a prismatic shaft of any cross section other than circular, shown by the schematic

contour of Fig. 6.7, plane sections do not remain plane under twisting, and warping willoccur. For such cases we must modify Eq 6.63 by expressing the displacements in theform

u1 = −θx2x3 , u2 = θx1x3 , u3 = θψ (x1, x2) , (6.68)

where ψ(x1, x2) is called the warping function. Note that the warping is independent of x3and therefore the same for all cross sections. Also, we assume that x3 is a centroidal axisalthough this condition is not absolutely necessary.


Mt

Mt

L

a

x2

x1

x3

(a) Cylinder with self-equilibrating momentsMt.

P∗

P

R

x2

x1

u1

u2

θx3R

(b) Displacement of point P to P∗ incross section.

R

β

θx3R

x1

x2

RP

P∗

90−β

β

(c) Detail of cross section twist β.

FIGURE 6.6Geometry and kinematic definitions for torsion of a circular shaft.


Mt

Mt

x2

x1

x3

L

(a) Prismatic cylinder with self-equilibrating moments Mt.

ns

x2

x1

dx1

dx290−θ θ

(b) Displacement of point P to P∗ in cross sec-tion.

FIGURE 6.7The more general torsion case of a prismatic beam loaded by self equilibrating moments.


As in the analysis of the circular shaft we may again use Eq 4.62 along with Eq 6.23 tocalculate the stress components from Eq 6.68. Thus

t11 = t22 = t33 = t12 = 0 , t13 = Gθ (ψ,1 − x2) , t23 = Gθ (ψ,2 − x1) . (6.69)

It is clear from these stress components that there are no normal stresses between thelongitudinal elements of the shaft. The first two of the equilibrium equations, Eq 6.43,are satisfied identically by Eq 6.69 in the absence of body forces, and substitution into thethird equilibrium equation yields

t13,1 + t23,2 + t33,3 = Gθ (ψ,11 +ψ,22) = 0 (6.70)

which indicates that ψ must be harmonic,

∇2ψ = 0 , (6.71)

on the cross section of the shaft.Boundary conditions on the surfaces of the shaft must also be satisfied. On the lateral

surface, which is stress free, the following conditions based upon Eq 3.32, must prevail

t11n1 + t12n2 = 0 , t21n1 + t22n2 = 0 , t31n1 + t32n2 = 0 , (6.72)

noting that here n3 = 0. The first two of these equations are satisfied identically whilethe third requires

Gθ (ψ,1 − x2)n1 +Gθ (ψ,2−x1)n2 = 0 (6.73a)

which reduces immediately to

ψ,1n1 +ψ,2n2 =dψ

dn= x2n1 + x1n2 . (6.73b)

Therefore, ψ (x1, x2) must be harmonic in the cross section of the shaft shown in Fig. 6.7,and its derivative with respect to the normal of the lateral surface must satisfy Eq 6.73bon the perimeter C of the cross section.

We note further that in order for all cross sections to be force free, that is, in simpleshear over those cross sections∫ ∫

t13dx1dx2 =

∫ ∫t23dx1dx2 =

∫ ∫t33dx1dx2 = 0 . (6.74)

Since t33 = 0, the third integral here is trivial. Considering the first integral we may write

Gθ

∫ ∫(ψ,1 − x2)dx1dx2 = Gθ

∫ ∫ ∂

∂x1[x1 (ψ,1 − x2)] +

∂

∂x2[x1 (ψ,2 − x1)]

dx1dx2

(6.75)where the condition∇2ψ = 0 has been used. Green’s theorem allows us to convert to theline integral taken around the perimeter C

Gθ

∫C

x1 (ψ,1 − x2)n1 + (ψ,2 − x1)n2ds = 0 (6.76)

which by Eq 6.73b is clearly satisfied. By an analogous calculation we find that the secondintegral of Eq 6.74 is also satisfied.

On the end faces of the shaft, x3 = 0 or x3 = L , the following conditions must besatisfied∫ ∫

x2t33dx1dx2 =

∫ ∫x1t33dx1dx2 = 0 ,

∫ ∫(x1t23 − x2t13)dx1dx2 = Mt . (6.77)


Again, since t33 = 0, the first two of these are trivial. The third leads to

Mt = Gθ

∫ ∫ (x21 + x22 + x1ψ,2 − x2ψ,1

)dx1dx2 . (6.78)

Defining the torsional rigidity as

K = G

∫ ∫ (x21 + x22 + x1ψ,2 − x2ψ,1

)dx1dx2 (6.79)

which can be evaluated once ψ (x1, x2) is known, we express the angle of twist as

θ =Mt

K. (6.80)

A second approach to the general torsion problem rests upon the introduction of atorsion stress function, designated here by Φ and defined so that the non-zero stressesare related to it by the definitions

t13 =∂Φ

∂x2, t23 = −

∂Φ

∂x1. (6.81)

Thus,∂Φ

∂x2= Gθ (ψ,1 − x2) ,

∂Φ

∂x1= −Gθ (ψ,2 − x1) . (6.82)

By eliminating ψ from this pair of equations we obtain

∇2Φ = −2Gθ . (6.83)

As already noted, the lateral surface of the shaft parallel to the axis must remain stressfree, that is, the third of Eq 6.72 must be satisfied. However, it is advantageous to writethis condition in terms of the unit vector s along the boundary rather than unit nor-mal n as shown in Fig. 6.6(b). It follows directly from geometry that n1 = dx2/ds andn2 = −dx1/ds. In terms of ds, the differential distance along the perimeter C, the stresscomponents in the normal direction will be given by

t13dx2

ds− t23

dx1

ds= 0 (6.84)

which in terms of Φ becomes

∂Φ

∂x1

dx1

ds+∂Φ

∂x2

dx2

ds=dΦ

ds= 0 . (6.85)

Thus, Φ is a constant along the perimeter of the cross section and will be assigned thevalue of zero here.

Finally, conditions on the end faces of the shaft must be satisfied. Beginning with thefirst of Eq 6.74, we have in terms of Φ∫ ∫

∂Φ

∂x2dx1dx2 =

∫ ∫ (∂Φ

∂x2dx2

)dx1 =

∫Φ|ba dx1 = 0 (6.86)

since Φ is constant on the perimeter. Likewise, by the same reasoning, the second ofEq 6.74 is satisfied, while the third is satisfied since t33 = 0. The first two conditions inEq 6.77 are also satisfied identically and the third becomes∫ ∫ (

−x1∂Φ

∂x1− x2

∂Φ

∂x2

)dx1dx2 = Mt . (6.87)


Integrating by parts and using the fact that Φ is assumed zero on the perimeter C yields

Mt = 2

∫ ∫Φdx1dx2 . (6.88)

Thus, the solution by this approach consists of determining the stress function Φ whichis zero on the cross-section perimeter, and satisfies Eq 6.83. Based upon that result wemay determine θ from Eq 6.88.

Example 6.2

Determine the stresses and the angle of twist for a solid elliptical shaft of thedimensions shown when subjected to end couples Mt.

x2

x1

2a

2b

SolutionThe equation of this ellipse is given by

x21a2

+x22b2

= 1 .

Therefore, take the stress function Φ in the form

Φ = λ

(x21a2

+x22b2

− 1

)(6.89)

where λ is a constant. Thus, Φ is zero on the cross-section perimeter. FromEq 6.83

2λ

(1

a2+1

b2

)= −2Gθ ,

so that

λ = −a2b2Gθ

a2 + b2. (6.90)

Now from Eq 6.88

Mt = −2a2b2Gθ

a2 + b2

∫ ∫ (x21a2

+x22b2

− 1

)dx1dx2 ,

and noting that ∫ ∫x21dx1dx2 = Ix2 =

1

4πba3 ,∫ ∫

x22dx1dx2 = Ix1 =1

4πab3 ,


and ∫ ∫dx1dx2 = πab

(the area of the cross section) we may solve for Mt which is

Mt =πa3b3Gθ

a2 + b2.

From this result,

θ =a2 + b2

πa3b3GMt

which when substituted into Eq 6.90 and that into Eq 6.89 gives

Φ = −Mt

πab

(x21a2

+x22b2

− 1

).

Now, by definition

t13 =∂Φ

∂x2= −

2Mt

πab3x2 , t23 = −

∂Φ

∂x1= −

2Mt

πa3bx1 .

The maximum stress occurs at the ends of the minor axis, and equals

τmax = ± 2Mt

πab2,

and the torsional rigidity is easily calculated to be

K =Mt

θ=πa3b3G

a2 + b2=G (A)4

4π2IP.

Note also that for a = b (circular cross section), the resultant stress at any pointis

τ =

√t213 + t223 =

Mt

IPr

in agreement with elementary theory.

It should be pointed out that for shafts having perimeters that are not expressible bysimple equations, solutions may be obtained by using stress functions in the form ofinfinite series. Such analyses are beyond the scope of this introductory section.

6.5.3 Pure Bending

Consider the beam to be subject to end conditions

Ti = 0, M1 = M3 = 0, M2 = M , (6.91)

and assume the stresses to be

t33 = −Mx1

I(6.92)


with all other components zero. The compatibility equations are satisfied since thenonzero stresses are linear in the coordinates. The equilibrium equations, tji,j = 0,only have one equation which is not identically satisfied, and the remaining equation,t33,3 = 0, is satisfied from the assumed stress Eq 6.92.

The end boundary conditions can be shown as satisfied. First, T1 = T2 = 0 is triviallysatisfied by considering Eq 6.92. Considering the torque on the end, Eq 6.59, it is clearthat M3 = Mt = 0 since t31 = t32 = 0. The axial force on the ends is given by Eq 6.57 as

T3 =

∫C1

−Mx1

Id S = −

M

I

∫C1

x1d S = 0 (6.93)

provided that the axes origin is chosen at the centroid of the cross section C. A similarcalculation for M1 gives

M1 =

∫C1

−Mx1

Ix2d S = −

M

I

∫C1

x1x2d S = 0 (6.94)

as long as the principal axes are used. Finally,

M2 = −

∫C1

Mx1

Ix1d S =

Mx1

I

∫C1

x21d S = M (6.95)

where

I =

∫C1

x21d S .

Strain components are found from Eq 6.28 giving three nonzero components

ε11 = ε22 =νMx1

EI; ε33 = −

Mx1

EI. (6.96)

Displacements can be found using Eq 4.101. The term uj(ξ0) + wij(ξ

0)(ξj − ξ0j

)rep-

resents a rigid displacement and will be set to zero. Taking the component u1 as anexample for finding the displacements yields

u1 =

∫x0

[ε11dξ1 + ε12dξ2 + ε13dξ3

+(xk − ξk) ε11,kdξ1 + (xk − ξk) ε12,kdξ2 + (xk − ξk) ε13,kdξ3

− (xk − ξk) εk1,1dξ1 − (xk − ξk) εk2,1dξ2 − (xk − ξk) εk3,1dξ3]

=

∫x0

[ε11dξ1 + (x1 − ξ1) ε11,1dξ1

− (x1 − ξ1) ε11,1dξ1 − (x2 − ξ2) ε22,1dξ2 − (x3 − ξ3) ε33,1dξ3]

=

∫x0

[νM

EIξ1dξ1 − (x2 − ξ2)

νM

EIdξ2 + (x3 − ξ3)

M

EIdξ3

]=

νM

2EIx21 −

νM

EI

(x22 − 1

2x22

)+M

EI

(x23 − 1

2x23

)=

M

2EI

(νx21 − νx22 + x23

).

(6.97)


Similar calculations for u2 and u3 give the displacements for pure bending as

u1 =M

2EI(νx21 − νx22 + x23) ,

u2 =νM

EIx1x2 ,

u3 = −M

EIx1x3 ,

(6.98)

to within an infinitesimal rigid displacement.In the pure bending problem, x1 = 0 represents the neutral surface, and the centerline

displacements match elementary beam theory.

6.5.4 Flexure

As the final Saint-Venant problem, consider the beam to have end C1 fixed and a flexureload of T acting on C2. The relaxed boundary conditions are

T1 = T, T2 = T3 = 0, Mi = 0 .

Assume the stresses to be

t11 = t22 = t12 = 0 ; t33 = −T

I(L− x3)x1 (6.99)

with no assumption made on the components t31 and t32. Equilibrium equations are

t13,3 = 0, t23,3 = 0, t31,1 + t32,2 = −Tx1

I(6.100)

which impliest13 = t13(x1, x2); t23 = t23(x1, x2) . (6.101)

The strain may be calculated from Eq 6.28 to obtain

ε11 = ε22 =νT

EI(L− x3)x1, ε33 = −

T

EI(L− x3)x1

ε23 = t23/2µ; ε31 = t31/2µ; ε12 = 0

(6.102)

The compatibility equations must be satisfied since the problem is posed in terms ofstresses. Eq 4.91 results in two nontrivial conditions

∂

∂x1(ε23,1 − ε13,2) = 0 , (6.103a)

∂

∂x2(ε31,2 − ε23,1) =

νT

EI, (6.103b)

Integration of Eq 6.103a gives

ε23,1 − ε31,2 = α−νT

EIx2 (6.104)

where α is taken as a constant by virtue of Eq 6.103a. Equation 6.104 may be written as(ε23 −

1

2αx1

),1

=

(ε31 +

1

2αx2 −

1

2

νT

EIx22

),2

(6.105)


from which the shear strains may be found in terms of a potential function f

ε32 −1

2αx1 =

1

2f,2; ε31 +

1

2αx2 −

1

2

νT

EIx22 =

1

2f,1 (6.106)

The shear stresses may be written using Hooke’s Law and Eq 6.106

t31 = 2µε31 = −µαx2 +νµT

EIx22 + µf,1 ,

t32 = 2µε32 = µαx1 + µf,2 .

(6.107)

The remaining equilibrium condition, Eq 6.100 is

∇2f (x1, x2) = −Tx1

µI= −

2(1+ ν)T

EIx1 on C (6.108)

subject to the lateral boundary condition tiαnα = 0 in the form

∂f

∂n= α(x2n2 − x1n2) −

νT

EIx22n1 on ∂C . (6.109)

To solve the boundary value problem given by Eqs 6.108 and 6.109, define a functionF(x1, x2) by

f = αψ−T

EI

F+

νx316

+(1+

ν

2

)x1x

22

(6.110)

where ψ is the warping function from torsion

∇2ψ = 0 on C;∂ψ

∂n= x2n1 − x1n2 on ∂C . (6.111)

Stress components may be written in terms of ψ, α and F as

t31 = µα(ψ,1 − x2) −T

2(1+ ν)I

F,1 +

νx212

+ (1+ν

2)x22

,

t32 = µα(ψ,2 + x1) −T

2(1+ ν)IF,2 + (2+ ν)x1x2 ,

t33 = −T

I(L− x3)x1 ,

(6.112)

subject to the lateral boundary condition

∂F

∂n= −

(νx212

+(1+

ν

2

)x22

)n1 − (2+ ν) x1x2n2 on ∂C . (6.113)

Since ψ and F are harmonic it is possible to show the stresses in Eq 6.112 satisfy theequilibrium equations. Substituting the stresses from 6.112 into Eq 6.57 and using Eq 6.101it can be shown that T3 = T2 = 0 and T1 = T . Bending moments M1 = M2 = 0 sincet33 (x3 = L) = 0, and the torque, M3 = Mt, is set to zero by expressing α in terms of F:

α = −T

2(1+ ν)IK

∫C2

x2F,1 − x1F,2 + (1+

ν

2)x32 − 2(1+

ν

2)x21x2

dA (6.114)

where K is the torsional rigidity. The constant α may be shown to be the average rate ofrotation over a cross section.


In general, there is a twist accompanying the bending when the load T is applied at thecentroid of C2. By superposing an appropriate torsion solution with equal and oppositeα it is possible to determine a line of action parallel to Ti such that there is not twist.

The displacements are determined from the line integral Eq 4.101

u1 = −αx2x3 +T

EI

12 (L− x3)ν(x

21 − x22) + 1

2Lx23 − 1

6x33

,

u2 = αx1x3 +Tν

EI(L− x3)x1x2 ,

u3 = αψ(x1, x2) −T

EI

x1x3(L− 1

2x3) + x1x22 + F(x1, x2)

,

(6.115)

to within an infinitesimal rigid displacement. The centerline deflection is the same as thatobtained in beam theory, but the cross-sections do not remain plane in the Saint-Venantsolution.

6.6 Plane ElasticityIn a number of engineering applications, specific body geometry and loading patternslead to a reduced, essentially two-dimensional form of the equations of elasticity, and thestudy of these situations is referred to as plane elasticity. Although the two basic types ofproblems constituting the core of this plane analysis may be defined formally by statingcertain assumptions on the stresses and displacements, we introduce them here in termsof their typical physical prototypes. In plane stress problems, the geometry of the bodyis that of a thin plate with one dimension very much smaller than the other two. Theloading in this case is in the plane of the plate and is assumed to be uniform acrossthe thickness, as shown in Fig. 6.8(a). In plane strain problems, the geometry is that of aprismatic cylinder having one dimension very much larger than the other two and havingthe loads perpendicular to and distributed uniformly with respect to this large dimension(Fig. 6.8(b)). In this case, because conditions are the same at all cross sections, the analysismay be focused on a thin slice of the cylinder.

For the plane stress situation, Fig. 6.8(a), the stress components t33, t31, and t32 aretaken as zero everywhere and the remaining components considered functions of only x1and x2. Thus,

tij = tij(x1, x2) (i, j = 1, 2) , (6.116)

and as a result, the equilibrium equations, Eq 6.43, reduce to the specific equations

t11,1 + t12,2 + ρb1 = 0 , (6.117a)

t21,1 + t22,2 + ρb2 = 0 . (6.117b)

The strain-displacement relations, Eq 6.44, become

ε11 = u1,1; ε22 = u2,2; 2ε12 = u1,2 + u2,1 , (6.118)

and at the same time the strain compatibility equations, Eqs 4.90, take on the form givenby Eq 4.92 and repeated here for convenience,

ε11,22 + ε22,11 = 2ε12,12 . (6.119)


Ox

3

x1

x2

(a) Plane stress problems generally involve bodies that are thinin dimensions with loads perpendicular to that dimension.

x3

(b) Plane strain problems involve bodies that are long in on di-mension with loads applied along that dimension.

FIGURE 6.8Representative figures for plane stress and plain strain.


Hooke’s law equations, Eq 6.45b, for plane stress are written

ε11 =1

E(t11 − νt22) , (6.120a)

ε22 =1

E(t22 − νt11) , (6.120b)

ε12 =1+ ν

Et12 =

t12

2G=γ12

2, (6.120c)

along with

ε33 = −ν

E(t11 + t22) =

−ν

1− ν(ε11 + ε22) . (6.121)

By inverting Eq 6.120a, we express the stress components in terms of the strains as

t11 =E

1− ν2(ε11 + νε22) , (6.122a)

t22 =E

1− ν2(ε22 + νε11) , (6.122b)

t12 =E

1+ νε12 =

E

2 (1+ ν)γ12 = Gγ12 . (6.122c)

These equations may be conveniently cast into the matrix formulation

t11t22t12

=E

1− ν2

1 ν 0

ν 1 0

0 0 1− ν

ε11ε22ε12

, (6.123)

In terms of the displacement components, ui (i = 1, 2), the plane stress field equationsmay be combined to develop a Navier-type equation for elastostatics, namely,

E

2 (1+ ν)ui,jj +

E

2 (1− ν)uj,ji + ρbi = 0 (i, j = 1, 2) . (6.124)

For the plane strain situation (Fig. 6.8(b)) we assume that u3 = 0 and that the remainingtwo displacement components are functions of only x1 and x2,

ui = ui(x1, x2) (i = 1, 2) . (6.125)

In this case, the equilibrium equations, the strain-displacement relations, and the straincompatibility equations all retain the same form as for plane stress, that is, Eqs 6.117a,6.117b, 6.118, and 6.119, respectively. Here, Hooke’s law (Eq 6.45a) may be written in


terms of engineering constants as

t11 =E

(1+ ν) (1− 2ν)[(1− ν) ε11 + νε22] , (6.126a)

t22 =E

(1+ ν) (1− 2ν)[(1− ν) ε22 + νε11] , (6.126b)

t12 =E

1+ νε12 =

E

1+ ν

γ12

2= Gγ12 , (6.126c)

along with

t33 =Eν

(1+ ν) (1− 2ν)(ε11 + ε22) = ν (t11 + t22) . (6.127)

The first three of these equations may be expressed in matrix form by t11t22t12

=E

(1+ ν) (1− 2ν)

1− ν ν 0

ν 1− ν 0

0 0 1− 2ν

ε11ε22ε12

. (6.128)

Furthermore, by inverting the same three equations, we may express Hooke’s law forplane strain by the equations

ε11 =(1+ ν)

E[(1− ν) t11 − νt22] , (6.129a)

ε22 =(1+ ν)

E[(1− ν) t22 + νt11] , (6.129b)

ε12 =1+ ν

Et12 =

2 (1+ ν)

E

t12

2=t12

2G. (6.129c)

By combining the field equations with Hooke’s law for elastostatic plane strain, we obtainthe appropriate Navier equation as

E

2 (1+ ν)ui,jj +

E

2 (1+ ν) (1− 2ν)uj,ji + ρbi = 0 (i, j = 1, 2) . (6.130)

It is noteworthy that Eqs 6.120a and 6.122a for plane stress become identical with theplane strain equations, Eqs 6.129 and 6.126, respectively, if in the plane stress equationswe replace E with E/(1 − ν2) and ν by ν/(1 − ν). Note that, if the forces applied to theedge of the plate in Fig. 6.8(a) are not uniform across the thickness but are symmetricalwith respect to the middle plane of the plate, the situation is sometimes described as astate of generalized plane stress. In such a case, we consider the stress and strain variables tobe averaged values over the thickness. Also, a case of generalized plane strain is sometimesreferred to in elasticity textbooks if the strain component ε33 in Fig. 6.8(b) is taken assome constant other that zero.


6.7 Airy Stress Function

As stated in Section 6.6, the underlying equations for two-dimensional problems in isotropicelasticity consist of the equilibrium relations, Eq 6.117, the compatibility condition, Eq 6.119and Hooke’s law, either in the form of Eq 6.120a (plane stress), or as Eq 6.129 (planestrain). When body forces in Eq 6.117 are conservative with a potential function V =V(x1, x2) such that bi = −V,i, we may introduce the Airy stress function, φ = φ(x1, x2) interms of which the stresses are given by

t11 = φ,22 + ρV ; t22 = φ,11 + ρV ; t12 = −φ,12 . (6.131)

Note that by using this definition the equilibrium equations are satisfied identically.For the case of plane stress we insert Eq 6.120a into Eq 6.119 to obtain

t11,22 + t22,11 − ν(t11,11 + t22,22) = 2(1+ ν)t12,12 (6.132)

which in terms of φ becomes

φ,1111 + 2φ,1212 + φ,2222 = −(1− ν)ρ(V,11 + V,22) . (6.133)

Similarly, for the case of plane strain, when Eq 6.129 is introduced into Eq 6.119 the resultis

(1− ν)(t11,22 + t22,11) − ν(t11,11 + t22,22) = 2t12,12 , (6.134)

or in terms of φ

φ,1111 + 2φ,1212 + φ,2222 = −(1− 2ν)ρ(V,11 − V,22)/(1− ν) . (6.135)

If the body forces consist of gravitational forces only, or if they are constant forces, theright-hand sides of both Eqs 6.133 and 6.135 reduce to zero and φ must then satisfy thebi-harmonic equation

φ,1111 + 2φ,1212 + φ,2222 =∇4φ = 0 . (6.136)

In each case, of course, boundary conditions on the stresses must be satisfied to completethe solution to a particular problem. For bodies having a rectangular geometry, stressfunctions in the form of polynomials in x1 and x2 are especially useful as shown by theexamples that follow.

Example 6.3

For a thin rectangular plate of the dimensions shown, consider the generalpolynomial of the third degree as the Airy stress function and from it determinethe stresses. Assume body forces are zero.


x1

x2

C

C

L

(a)

x1

x2 6D

3c

6D3c

(b)

B3c

B3c

B3L

x2

x1

(c)

SolutionSelect a polynomial stress function of the form φ3 = A3x

31+B3x

21x2+C3x1x

22+

D3x32. Choosing this particular polynomial form for the stress function is not

arbitrary; the choice is based on many trials of different order polynomials.After a certain amount of experience in observing a polynomial’s effect on thestress components computed using Eq 6.131, an educated guess can be madeas to what terms should be considered for a specific problem. For this reason,problems like this are often called semi-inverse problems.

By direct substitution into Eq 6.136 we confirm that φ3 is bi-harmonic.Further, the stresses are given as

t11 = 2C3x1 + 6D3x2 ,

t22 = 6A3x1 + 2B3x2 ,

t12 = −2B3x1 − 2C3x2 .

By selecting different constants to be zero and nonzero, different physical prob-lems may be solved. Here, two specific cases will be considered.

(a) Assume all coefficients in φ3 are zero except D3. This may be shownto solve the case of pure bending of a beam by equilibrating momentson the ends as shown by Fig. (b). Stress in the fiber direction of thebeam varies linearly with the distance from the x1 axis

t11 = 6D3x2; t22 = t12 = 0

as is the case for simple bending. Similarly, by taking only A3 asnonzero, the solution is for bending moments applied to a beam whoselengthwise direction is taken to be x2 rather than x1 direction.


(b) If only B3 (or C3) is non-zero, both shear and normal stresses arepresent. Fig. (c) shows the stress pattern for B3 6= 0.

Example 6.4

Consider a special stress function having the form

φ∗ = B2x1x2 +D4x1x32 .

Show that this stress function may be adapted to solve for the stresses in anend-loaded cantilever beam shown in the sketch. Assume the body forces arezero for this problem.

x2

x1

P

c

c

1 L

Vector v with respect to Ox ′1x′2x′3 and Ox1x2x3.

SolutionIt is easily verified, by direct substitution, that ∇4φ∗ = 0. The stress compo-nents are directly computed from Eq 6.131

t11 = 6D4x1x2 ,

t22 = 0 ,

t12 = −B2 − 3D4x22 .

These stress components are consistent with an end-loaded cantilever beam,and the constants B2 and D4 can be determined by considering the boundaryconditions. In order for the top and bottom surfaces of the beam to be stress-free, t12 must be zero at x2 = ±c. Using this condition B2 is determined interms of D4 as B2 = −3D4c

2. The shear stress is thus given in terms of singleconstant B2

t12 = −B2 +B2x

22

c2.

The concentrated load is modeled as the totality of the shear stress t12 on thefree end of the beam. Thus, the result of integrating this stress over the freeend of the beam at x1 = 0 yields the applied force P. In equation form

P = −

∫c−c

[−B2 + B2

x22c2

]dx2


θ

dθ

dr

r

x2

x1

t θθθθ dθ

θ∂t+∂

r

r

tt

θθ dθ

θ∂

+∂

trr drr

∂+

∂

trθ

drr

∂+

∂

rt θr rttθθ

trr

trθ

rt θ

FIGURE 6.9Differential stress element in polar coordinates.

where the minus sign is required due to the sign convention on shear stress.Carrying out the integration we have B2 = 3P/4c so that stress componentsmay now be written as

t11 = −3P

2c3x1x2 ,

t22 = 0 ,

t12 = −3P

4c

(1−

x22c2

).

But for this beam the plane moment of inertia of the cross section is I = 2c3/3

so that now

t11 = −P

Ix1x2; t22 = 0; t12 = −

P

2I

(c2 − x22

)in agreement with the results of elementary beam bending theory.

Several important solutions in plane elasticity are obtained by the Airy stress functionapproach when expressed in terms of polar coordinates. To this end we introduce herethe basic material element together with the relevant stress components in terms of thecoordinates r and θ as shown on Fig. 6.9. Using this element and summing forces in theradial direction results in the equilibrium equation

∂trr

∂r+1

r

∂trθ

∂θ+trr − tθθ

r+ R = 0 , (6.137a)


and summing forces tangentially yields

1

r

∂tθθ

∂θ+∂trθ

∂r+ 2

trθ

r+Θ = 0 (6.137b)

in which R and Θ represent body forces. In the absence of such forces Eq 6.137a and6.137b are satisfied by

trr =1

r

∂φ

∂r+1

r2∂2φ

∂θ2, (6.138a)

tθθ =∂2φ

∂r2, (6.138b)

trθ =1

r2∂φ

∂θ−1

r

∂2φ

∂r∂θ= −

∂

∂r

(1

r

∂φ

∂φ

), (6.138c)

in which φ = φ(r, θ). To qualify as an Airy stress function φ must once again satisfy thecondition ∇4φ = 0 which in polar form is obtained from Eq 6.136 as

∇4φ =

(∂2

∂r2+1

r

∂

∂r+1

r2∂2

∂θ2

)(∂2φ

∂r2+1

r

∂φ

∂r+1

r2∂2φ

∂θ2

)= 0 . (6.139)

For stress fields symmetrical to the polar axis Eq 6.139 reduces to

∇4φ =

(∂2

∂r2+1

r

∂

∂r

)(∂2φ

∂r2+1

r

∂φ

∂r

)= 0 , (6.140a)

or

r4∂4φ

∂r4+ 2r3

∂3φ

∂r3− r2

∂2φ

∂r2+ r

∂φ

∂r= 0 . (6.140b)

It may be shown that the general solution to this differential equation is given by

φ = A ln r+ Br2 ln r+ Cr2 +D , (6.141)

so that for the symmetrical case the stress components take the form

trr =1

r

∂φ

∂r=A

r2+ B (1+ 2 ln r) + 2C , (6.142a)

tθθ =∂2φ

∂r2= −

A

r2+ B (3+ 2 ln r) + 2C , (6.142b)

trθ = 0 . (6.142c)

When there is no hole at the origin in the elastic body under consideration, A and B

must be zero since otherwise infinite stresses would result at that point. Thus, for a platewithout a hole only uniform tension or compression can exist as a symmetrical case.

A geometry that does qualify as a case with a hole (absence of material) at the origin isthat of a curved beam subjected to end moments as discussed in the following example.

Example 6.5

Determine the stresses in a curved beam of the dimensions shown in the figurewhen subjected to constant equilibrating moments.


a

b

r

M

M

x1

θ

x2

SolutionFrom symmetry, the stresses are as given by Eq 6.142. Boundary conditionsrequire

(1) trr = 0 at r = a, and at r = b ,

(2)∫ba

tθθdr = 0 on the end faces ,

(3)∫ba

r trrdr = −M on the end faces ,

(4) trθ = 0 everywhere on the boundary .

These conditions result in the following equations which are used to evaluatethe constants A, B, and C. The inner and outer radii are free of normal stresswhich can be written in terms of boundary condition (1) as

A/a2 + B(1+ 2 lna) + 2C = 0 ,

A/b2 + B(1+ 2 lnb) + 2C = 0 .

No transverse loading is present on the ends of the curved beam which may bewritten in terms of boundary condition (2) as∫b

a

tθθdr =

∫ba

∂2φ

∂r2dr =

∂φ

∂r

∣∣∣∣ba

= 0 .

Evaluation of this integral at the limits is automatically satisfied as a conse-quence of boundary condition (1). Finally, the applied moments on the ends


may be written in terms of boundary condition (3)∫ba

r∂2φ

∂r2dr =

[r∂φ

∂r

]ba

−

∫ba

∂φ

∂rdr = −M .

Because of condition (1) the bracketed term here is zero and from the integralterm

φb − φa = M ,

orA lnb/a+ B(b2 lnb− a2 lna) + C(b2 − a2) = M .

This expression, together with the two stress equations arising from condition(1) may be solved for the constants A, B, and C, which are

A = −4M

Na2b2 ln

b

a,

B = −2M

N

(b2 − a2

),

C =M

N

[b2 − a2 + 2

(b2 lnb− a2 lna

)],

where N = (b2 − a2)2 − 4a2b2[ln(b/a)]2. Finally, the stress components maybe written in terms of the radii and applied moment by substitution of theconstants into Eq 6.142

trr = −4M

N

(a2b2

r2lnb

a+ b2 ln

r

b+ a2 ln

a

r

),

tθθ = −4M

N

(−a2b2

r2lnb

a+ b2 ln

r

b+ a2 ln

a

r+ b2 − a2

),

trθ = 0 .

Verification of these results can be made by reference to numerous strength ofmaterials textbooks.

If φ is taken as a function of both r and θ, it is useful to assume

φ(r, θ) = f(r)einθ (6.143)

in order to obtain a function periodic in θ. For n = 0, the general solution is, as expected,the same as given in Eq 6.141. As an example of the case where n = 1 we consider φ(r, θ)in the form

φ = (Ar2 + B/r+ Cr+Dr ln r) sin θ . (6.144)

Example 6.6

Show that the stress function given by Eq 6.144 may be used to solve thequarter-circle beam shown under an end load P.


b

a

P

θ

x1

x2

r

SolutionFrom Eq 6.138 the stress components are

trr =

(2Ar−

2B

r3+D

r

)sin θ ,

tθθ =

(6Ar+

2B

r3+D

r

)sin θ ,

trθ = −

(2Ar−

2B

r3+D

r

)cos θ .

The inner and outer radii of the beam are stress-free surfaces leading to boundaryconditions of trr = trθ = 0 at r = a, and at r = b. Also, the applied force Pmay be taken to be the summation of the shear stress acting over the free endθ = 0: ∫b

a

trθ|θ=0 dr = −P .

These conditions lead to the three equations from which constants A, B, and Dmay be determined

2Aa−2B

a3+D

a= 0 ,

2Ab−2B

b3+D

b= 0 ,

−A(b2 − a2

)+ B

(b2 − a2

)a2b2

−D lnb

a= P .

Solving these three equations in three unknowns, the constants are determinedto be

A =P

2N, B = −

Pa2b2

2N, D = −

P(a2 + b2

)N

,

where N = a2 − b2 + (a2 + b2) ln(b/a). Finally, use of these constants in the


stress component equations gives

trr =P

N

(r+

a2b2

r3−a2 + b2

r

)sin θ ,

tθθ =P

N

(3r−

a2b2

r3+a2 + b2

r

)sin θ ,

trθ = −P

N

(r+

a2b2

r3−a2 + b2

r

)cos θ .

Note that, when θ = 0

trr = tθθ = 0 and trθ = −P

N

(r+

a2b2

r3−a2 + b2

r

).

And when θ = π/2, trθ = 0 while

trr =P

N

(r+

a2b2

r3−a2 + b2

r

),

tθθ =P

N

(3r−

a2b2

r3+a2 + b2

r

).

We close this section with an example of the case when n = 2 in Eq 6.143.

Example 6.7

Use the stress function

φ = (Ar2 + Br4 + C/r2 +D) cos 2θ

to solve the stress problem of a large flat plate under a uniform axial stress T ,and having a small circular hole at the origin as shown in the figure.

E

a

b TT

trθ

trr

θ

x1

x2


SolutionFrom Eq 6.138 the stress components have the form

trr = −

(2A+

6C

r4+4D

r2

)cos θ ,

tθθ =

(2A+ 12Br2 +

6C

r2

)cos θ ,

trθ =

(2A+ 6Br2 −

6C

r4−2D

r2

)sin 2θ .

Assume the width of the plate b is large compared to the radius of the hole.From consideration of the small triangular element at a distance b from theorigin, the following boundary conditions must hold

1. trr = T cos2 θ at r = b,

2. trθ = −T

2sin 2θ at r = b,

and the inner surface of the hole is stress free, which may be written as

3. trr = 0 at r = a,

4. trθ = 0 at r = a.

These conditions, when combined with the stress expressions, yield

2A+6C

b4+4D

b2= −

T

2,

2A+ 6Bb2 −6C

b4−2D

b2= −

T

2,

2A+6C

a4+4D

a2= 0 ,

2A+ 6Ba2 −6C

a4−2D

a2= 0 .

Letting b→∞, the above equations may be solved to determine

A = −T

4, B = 0 , C = −

a4T

4, D =

a2T

2,

so that now the stresses are given by

trr =T

2

(1−

a2

r2

)+T

2

(1+

3a4

r4+4a2

r2

)cos θ ,

tθθ =T

2

(1+

a2

r2

)−T

2

(1+

3a4

r4

)cos θ ,

trθ = −T

2

(1−

3a4

r4+2a2

r2

)sin 2θ .

Note that as r tends towards infinity and at θ = 0 the stresses are given bytrr = T ; tθθ = trθ = 0 (simple tension). At r = a, trr = trθ = 0 andtθθ = T − 2T cos 2θ, which indicates that when θ = π/2, or θ = 3π/2, thestresses become tθ = T − 2T(−1) = 3T , a well-known stress concentration factor


used in design. Also, when θ = 0, or θ = π, tθθ = −T a compression factor atthe centerline of the hole.

It should be pointed out that in this brief section only stresses have been determined.These, together with Hooke’s law, can be used to determine the displacements for theproblems considered.

6.8 Linear ThermoelasticityWhen we give consideration to the effects of temperature as well as to mechanical forceson the behavior of elastic bodies, we become involved with thermoelasticity. Here, we ad-dress only the relatively simple uncoupled theory for which temperature changes broughtabout by elastic straining are neglected. Also, within the context of linearity we assumethat the total strain is the sum

εij = ε(M)ij + ε

(T)ij (6.145)

where ε(M)ij is the contribution from the mechanical forces and ε(T)

ij are the temperature-induced strains. If θ0 is taken as a reference temperature and θ as an arbitrary tem-perature, the thermal strains resulting from a change in temperature of a completelyunconstrained isotropic volume are given by

ε(T)ij = α (θ− θ0) δij (6.146)

where α is the linear coefficient of thermal expansion, having units of meters per meter perdegree Celsius (m/m/C). The presence of the Kronecker delta in Eq 6.146 indicates thatshear strains are not induced by a temperature change in an unconstrained, homogenous,isotropic body.

By inserting Eq 6.146 into Eq 6.145 and using Hooke’s law for the mechanical strains inthat equation, we arrive at the thermoelastic constitutive equation

εij =1+ ν

Etij −

ν

Eδijtkk + α (θ− θ0) δij . (6.147)

This equation may be easily inverted to express the stresses in terms of the strains as

tij =E

(1+ ν) (1− 2ν)[νδijεkk + (1− 2ν) εij − (1+ ν)α (θ− θo) δij] . (6.148)

Also, in terms of the deviatoric and spherical components of stress and strain, the ther-moelastic constitutive relations appear as the pair of equations

Sij =E

(1+ ν)ηij , (6.149a)

tii =E

(1− 2ν)[εii − 3α (θ− θ0)] . (6.149b)


If the heat conduction in an elastic solid is governed by the Fourier law, Eq 5.62, whichwe write here as

qi = −κθ,i (6.150)

where κ is the thermal conductivity of the body (a positive constant), and if we introducethe specific heat constant c through the equation

−qi,i = ρcθ (6.151)

the heat conduction equation for the uncoupled theory becomes

κθ,ii = ρcθ . (6.152)

This equation, along with the thermoelastic stress-strain equations Eq 6.147 or Eq 6.148,the equilibrium equations Eq 6.43, and the strain-displacement relations Eq 6.44, consti-tute the basic set of field equations for uncoupled, quasi-static, thermoelastic problems.Of course, boundary conditions and the strain compatibility equations must also be sat-isfied.

6.9 Three-Dimensional ElasticitySolutions of three-dimensional elasticity problems traditionally focus on two distinct for-mulations. First, the displacement formulation is based upon solutions of the Navier equa-tions which were presented in Section 6.4, and which are developed again in the follow-ing paragraph. The second formulation, called the stress formulation, utilizes solutions ofthe equilibrium equations in association with the Beltrami-Michell stress equations pre-viously derived in Section 6.4 and based upon the compatibility equations in terms ofstrains. These equations are also reviewed in the following paragraph.

Starting with the fundamental equations of elastostatics as listed in Section 6.4 andrepeated here under revised numbering, we have:

Equilibrium equations, Eq 6.43

tij,j + ρbi = 0 , (6.153)

Strain-displacement equations, Eq 6.44

2εij = ui,j + uj,i , (6.154)

Hooke’s law, Eq 6.45a, or Eq 6.45b

tij = λδijεkk + 2µεij , (6.155a)

or

εij =1

E[(1− ν) tij − νδijtkk] . (6.155b)


By substituting Eq 6.154 into Hooke’s law, Eq 6.155a, and that result in turn into Eq 6.153,we obtain

µui,jj + (λ+ µ)uj,ji + ρbi = 0 (6.156)

which comprise three second-order partial differential equations known as the Navierequations. For the stress formulation we convert the strain equations of compatibilityintroduced in Section 4.7 as Eq 4.90 and repeated here as

εij,km + εkm,ij − εik,jm − εjm,ik = 0 (6.157)

into the equivalent expression in terms of stresses using Eq 6.155a, and combine thatresult with Eq 6.153 to obtain

tij,kk +1

1+ νtkk,ij + ρ (bi,j + bj,i) +

ν

1+ νδijρbk,k = 0 (6.158)

which are the Beltrami-Michell equations of compatibility. In seeking solutions by either thedisplacement or stress formulation, we consider only the cases for which body forces arezero. The contribution of such forces, often either gravitational or centrifugal in nature,can be appended to the homogeneous solution, usually in the form of a particular integralbased upon the boundary conditions.

Let us first consider solutions developed through the displacement formulation. Ratherthan attempt to solve the Navier equations directly, we express the displacement field interms of scalar and vector potentials and derive equations whose solutions result in therequired potentials. Thus, by inserting an expression for ui in terms of the proposedpotentials into the Navier equations we obtain the governing equations for the appropri-ate potentials. Often such potentials are harmonic, or bi-harmonic functions. We shallpresent three separate methods for arriving at solutions of the Navier equations.

The method used in the first approach rests upon the well-known theorem of Helmholtzwhich states that any vector function that is continuous and finite, and which vanishes atinfinity, may be resolved into a pair of components: one a rotational vector, the other anirrotational vector. Thus, if the curl of an arbitrary vector a is zero, then a is the gradientof a scalar φ, and a is irrotational, or as it is sometimes called, solenoidal. At the same time,if the divergence of the vector a is zero, then a is the curl of another vector ψ, and is arotational vector. Accordingly, in keeping with the Helmholtz theorem, we assume thatthe displacement field is given by

ui = φ,i + εipqψq,p (6.159)

where φ,i is representative of the irrotational portion, and curlψ the rotational portion.Substituting this displacement vector into Eq 6.156 with bi in that equation taken as zero,namely

µui,jj + (λ+ µ)uj,ji = 0 , (6.160)

we obtainµφ,ijj + µεipqψq,pjj + (λ+ µ)φ.ijj + (λ+ µ) εjpqψq,pji = 0 (6.161)

which reduces to(λ+ 2µ)φ,ijj + µεipqψq,pjj = 0 , (6.162)

since εjpqψq,pji = 0. In coordinate-free notation Eq 6.162 becomes

(λ+ 2µ)∇∇2φ+ µ∇×∇2ψ = 0 . (6.163)


Any set of φ and ψ which satisfies Eq 6.162 provides (when substituted into Eq 6.159)a displacement field satisfying the Navier equation, Eq 6.160. Clearly, one such set isobtained by requiring φ and ψ to be harmonic:

∇2φ = 0 , (6.164a)

∇2ψ = 0 . (6.164b)

It should be pointed out that while Eq 6.164 is a solution of Eq 6.160, it is not the generalsolution of the Navier equations. If we choose ∇2φ = constant, and ψ = 0 in Eq 6.164,the scalar function φ is known as the Lame strain potential. By taking the divergence ofEq 6.162, and remembering that the divergence of a curl vanishes, we see that

∇4φ = 0 (6.165)

is a solution of the resulting equation so that a bi-harmonic function as φ also yields asolution for ui. Similarly, by taking the curl of Eq 6.162 we find that

∇4ψ = 0 (6.166)

also provides for a solution ui.The second approach for solving the Navier equations is based on the premise of ex-

pressing the displacement field in terms of the second derivatives of a vector known asthe Galerkin vector, and designated here by F = Fiei. In this approach we assume thedisplacement ui is given in terms of the Galerkin vector specifically by the equation

ui = 2 (1− ν) Fi,jj − Fj,ji (6.167)

which is substituted directly into Eq 6.160. Carrying out the indicated differentiation andreducing the resulting equations with the help of the identity λ = 2νµ/ (1− 2ν) we findthat the Navier equations are satisfied if

∇4F = 0 . (6.168)

Thus, any bi-harmonic vector is suitable as a Galerkin vector. As should be expected,because they are solutions to the same equation, there is a relationship between φ and ψwith F. It has been shown that

φ = −Fi,i , (6.169a)

andεijkψk,j = 2 (1− ν) Fi,jj . (6.169b)

If Fi is not only bi-harmonic, but harmonic as well, Eq 6.169b reduces to

εijkψk,j = 0 , (6.170a)

and the relationship between φ and Fi becomes

φ,ii = −Fi,jji . (6.170b)

In this case φ is called the Lame strain potential.

Example 6.8

Consider a Galerkin vector of the form F = Fiei where F3 is a function of thecoordinates, that is F3 = F3 (x1, x2, x3). Apply this vector to obtain the solution


to the problem of a concentrated force acting at the origin of coordinates in thedirection of the positive x3 axis of a very large elastic body. This is called theKelvin problem.

SolutionLet ui = 2 (1− ν) Fi,jj − Fj,ji as given by Eq 6.167. Accordingly,

u1 = −F3,31 ,

u2 = −F3,32 ,

u3 = 2 (1− ν) (F3,11 + F3,22 + F3,33) − F3,33 .

Take F3 to be proportional to the distance squared from the origin as defined byF3 = BR where B is a constant and R2 = x21 + x22 + x23. Thus, the displacementsare

u1 =Bx3x1

R3,

u2 =Bx3x2

R3,

u3 = B

[4 (1− ν)

R−x21 + x22R3

].

From these displacement components the stresses may be computed using Hooke’slaw. In particular, it may be shown that

t33 =∂

∂x3

[(2− ν)

(∂2

∂x21+∂2

∂x22+∂2

∂x23

)−∂2

∂x23

]BR

which upon carrying out the indicated differentiation and combining terms be-comes

t33 = −B

[2 (2− ν) x3

R3−3(x21 + x22

)x3

R5

].

This equation may be written in a more suitable form for the integration thatfollows by noting that R2 = r2 + x23 where r2 = x21 + x22. The modified equationis

t33 = −B

[(1− 2ν) x3

R3+3x33R5

].

Summing forces in the x3 direction over the plane x3 = constant allows us todetermine B in terms of the applied force P. The required integral is

P =

∫∞0

(−t33) 2πr dr .

But rdr = RdR and so

P = 2πB

[(1− 2ν) x3

∫∞0

dR

R2+ 3x33

∫∞0

dR

R4

]from which we find

B =P

4π (1− ν).


The third approach for solving the Navier equations is called the Papkovich-Neuber so-lution which results in equations in terms of harmonic functions of a scalar, and a vectorpotential. For this we take the displacement vector to be represented by a scalar potentialB, the vector potential V, and the position vector xi in the form

ui = Vi − B,i −(Vkxk),i4 (1− ν)

(6.171)

which when substituted into the homogeneous Navier equations and simplified using theidentity λ = 2νµ/ (1− 2ν) we obtain

µVi,jj − (λ+ 2µ)B,ijj −1

2(λ+ µ) (Vk,ijj + Vi,jj) = 0 . (6.172)

These equations are clearly satisfied when

∇2V = 0 , (6.173a)

and∇2B = 0 (6.173b)

which indicates that any four harmonic functions, Vi with i = 1, 2, 3 and B, will serve toprovide a displacement vector ui from Eq 6.171 that satisfies the Navier equations. Sincethe displacement vector has only three components, the four scalar functions, Vi, and Bare not completely independent and may be reduced to three. It can be shown that thesepotentials are related to the Galerkin vector through the expressions

V = 2 (1− ν)∇2F , (6.174a)

andB =∇ · F−

V · x4 (1− ν)

. (6.174b)

Whereas it is not usually possible to solve the Navier equations directly for problemsinvolving a body of arbitrary geometry, in certain cases of spherical symmetry an ele-mentary solution is available. Consider the case of a hollow spherical geometry of innerradius r1 and outer radius r2 that is subjected to an internal pressure p1 and an externalpressure p2. Due to the symmetry condition here we assume a displacement field

ui = φ (r) xi (6.175)

where r2 = xixi and φ depends solely upon r. By direct substitution of Eq 6.175 intoEq 6.160 we arrive at the ordinary differential equation

d2φ

dr2+4

r

dφ

dr= 0 (6.176)

for which the general solution may be written as

φ (r) = A1 +A2

r3(6.177)

where A1 and A2 are constants of integration depending on the boundary conditions.Eq 6.155 written in terms of displacement derivatives has the form

tij = λδijuk,k + µ (ui,j + uj,i) . (6.178)


It follows by making use of Eqs 6.175 and 6.177 that

tij = 3λA1δij + 2µ

[(A1 +

A2

r3

)δij −

3A2xixj

r3

]. (6.179)

Recall that the traction vector in the radial direction (see Eq 3.51) is σN = tijninj whichupon substitution of Eq 6.177 becomes

σN = (3λ+ 2µ)A1 +4µA2

r3(6.180)

where the identity xi = rni has been used. Similarly, the tangential traction can becalculated using σS = tijvivj where the unit vectors vi are perpendicular to ni. Theresult is

σS = (3λ+ 2µ)A1 +2µA2

r3(6.181)

since vixi = 0.The constants A1 and A2 are determined from boundary conditions on the tractions.

Clearly,

σN = −p1 at r = r1 ,

σN = −p2 at r = r2 .

Carrying out the indicated algebra, we have the well-known formulas

σN =p1r

31 − p2r

32

r32 − r31−r31r

32

r3p1 − p2

r32 − r31, (6.182a)

σS =p1r

31 − p2r

32

r32 − r31+r31r

32

2r3p1 − p2

r32 − r31. (6.182b)

These equations may be easily modified to cover the case where p1 = 0, or the case wherep2 = 0.

We conclude this section with a brief discussion of three dimensional stress functions.These functions are designed to provide solutions of the equilibrium equations. Addi-tionally, in order for the solution to be complete it must be compatible with the Beltrami-Michell equations. Beginning with the equilibrium equations in the absence of bodyforces

tij,j = 0 (6.183)

we propose the stress fieldtij = εipqεjkmΦqk,pm (6.184)

where Φqk is a symmetric tensor function of the coordinates. By a direct expansion ofthis equation the stress components may be expressed in terms of the potential Φqk. Forexample,

t11 = ε1pqε1kmΦqk,pm (6.185)

which when summed over the repeated indices, keeping in mind the properties of thepermutation symbol, becomes

t11 = 2Φ23,23 −Φ22,33 −Φ33,22 . (6.186a)


Similarly,

t22 = 2Φ13,13 −Φ11,33 −Φ33,11 , (6.186b)t33 = 2Φ12,12 −Φ11,22 −Φ22,11 , (6.186c)t12 = Φ12,33 +Φ33,12 −Φ23,31 −Φ13,32 , (6.186d)t23 = Φ23,11 +Φ11,23 −Φ12,13 −Φ13,12 , (6.186e)t31 = Φ31,22 +Φ22,31 −Φ12,23 −Φ32,21 . (6.186f)

It may be shown by direct substitution that the equilibrium equations are satisfied bythese stress components.

Upon setting the off-diagonal terms of Φqk to zero, that is, if Φ12 = Φ23 = Φ31 = 0,we obtain the solution proposed by Maxwell. By setting the diagonal terms of Φqk tozero, namely, Φ11 = Φ22 = Φ33 = 0 we obtain the solution proposed by Morera whichis known by that name. It is interesting to note that if all the components of Φqk exceptΦ33 are zero, that component is the Airy stress function introduced in Section 6.7 ascan be verified by Eq 6.186. Although the potential Φqk provides us with a solutionof the equilibrium equations, that solution is not compatible with the Beltrami-Michellequations except under certain conditions.


Problems

Problem 6.1In general, the strain energy density W may be expressed in the form

W = C∗αβεαεβ (α,β = 1, . . . , 6)

where C∗αβ is not necessarily symmetric. Show that this equation may be rearranged toappear in the form

W =1

2Cαβεαεβ

where Cαβ is symmetric, so that now

∂W

∂εβ= Cαβεβ = tβ


Problem 6.2Let the stress and strain tensors be decomposed into their respective spherical and deviatorcomponents. Determine an expression for the strain energy density W as the sum of adilatation energy density W(1) and a distortion energy density W(2).

Answer

W = W(1) +W(2) =1

6tiiεjj +

1

2Sijηij

Problem 6.3If the strain energy density W is generalized in the sense that it is assumed to be a functionof the deformation gradient components instead of the small strain components, that is, ifW = W(FiA), make use of the energy equation and the continuity equation to show that inthis case Eq 6.16 is replaced by

Jtij =∂W

∂FiAFjA .

Problem 6.4For an isotropic elastic medium as defined by Eq 6.23, express the strain energy density in

terms of

(a) the components of εij,

(b) the components of tij,

(c) the invariants of εij.


Answer

(a) W =1

2(λεiiεjj + 2µεijεij),

(b) W =(3λ+ 2µ) tijtij − λtiitjj

4µ (3λ+ 2µ),

(c) W =(12λ+ µ

)(Iε)

2− 2µIIε

Problem 6.5Let tij be any second-order isotropic tensor such that

t′ij = aimajntmn = tij

for any proper orthogonal transformation aij. Show that by successive applications of thetransformations

[aij] =

0 0 −1−1 0 0

0 1 0

and [aij] =

0 0 1

−1 0 0

0 −1 0

every second-order isotropic tensor is a scalar multiple of the Kronecker delta, δij.

Problem 6.6Verify that Eqs 6.33a and 6.33b when combined result in Eq 6.28 when Eqs 6.29 are used.

Problem 6.7For an elastic medium, use Eq 6.33 to express the result obtained in Problem 6.2 in terms

of the engineering elastic constants K and G.

Answer

W = 12Kεiiεjj +G

(εijεij − 1

3εiiεjj)

Problem 6.8Show that the distortion energy density W(2) (see Problem 6.2) for a linear elastic medium

may be expressed in terms of (a) the principal stresses, σ(1), σ(2), σ(3) and (b) the principalstrains, ε(1), ε(2), ε(3) in the form

(a) W(2) =

(σ(1) − σ(2)

)2+(σ(2) − σ(3)

)2+(σ(3) − σ(1)

)212G

,

(b) W(2) =1

3

[(ε(1) − ε(2)

)2+(ε(2) − ε(3)

)2+(ε(3) − ε(1)

)2]G.

Problem 6.9Beginning with the definition forW (Eq 6.21a), show for a linear elastic material representedby Eq 6.23 or by Eq 6.28 that ∂W/∂εij = tij and ∂W/∂tij = εij (Note that ∂εij/∂εmn =δimδjn.)


Problem 6.10

For an isotropic, linear elastic solid, the principal axes of stress and strain coincide, as wasshown in Example 6.1. Show that, in terms of engineering constants E and ν this result isgiven by

ε(q) =(1+ ν)σ(q) − ν

[σ(1) + σ(2) + σ(3)

]E

, (q = 1, 2, 3).

Thus, let E = 106 psi and ν = 0.25, and determine the principal strains for a body subjectedto the stress field (in ksi)

[tij] =

12 0 4

0 0 0

4 0 6

.

Answer

ε(1) = −4.5× 10−6, ε(2) = 0.5× 10−6, ε(3) = 13× 10−6

Problem 6.11

Show for an isotropic elastic medium that

(a)1

1+ ν=2 (λ+ µ)

3λ+ 2µ, (b)

ν

1+ ν=

λ

λ+ 2µ, (c)

2µν

1− 2ν=3Kν

1+ ν,

(d) 2µ(1+ ν) = 3K(1− 2ν).

Problem 6.12

Let the x1x3 plane be a plane of elastic symmetry such that the transformation matrixbetween Ox1x2x3 and Ox′1x

′2x′3 is

[aij] =

1 0 0

0 −1 0

0 0 1

.

Show that, as the text asserts, this additional symmetry does not result in a further reduc-tion in the elastic constant matrix, Eq 6.40.

Problem 6.13

Let the x1 axis be an axis of elastic symmetry of order N = 2. Determine the form of theelastic constant matrix Cαβ, assuming Cαβ = Cβα.

Answer

[Cαβ] =

C11 C12 C13 C14 0 0

C12 C22 C23 C34 0 0

C13 C23 C33 C34 0 0

C14 C24 C34 C44 0 0

0 0 0 0 C55 C560 0 0 0 C56 C66


Problem 6.14Assume that, by the arguments of elastic symmetry, the elastic constant matrix for an

isotropic body has been reduced to the form

[Cαβ] =

C11 C12 C12 0 0 0

C12 C11 C12 0 0 0

C12 C12 C11 0 0 0

0 0 0 C44 0 0

0 0 0 0 C44 0

0 0 0 0 0 C44

.

Show that, if the x1 axis is taken as an axis of elastic symmetry of any order (θ is arbitrary),C11 = C12 + 2C44. (Hint: Expand t′23 = a2qa3mtqm and ε′23 = a2qa3mεqm.)

Problem 6.15If the axis which makes equal angles with the coordinate axes is an axis of elastic symmetryof order N = 3, show that there are twelve independent elastic constants and that the elasticmatrix has the form

[Cαβ] =

C11 C12 C13 C14 C15 C16C13 C11 C12 C16 C14 C15C12 C13 C11 C15 C16 C14C41 C42 C43 C44 C45 C46C43 C41 C42 C46 C44 C45C42 C43 C41 C45 C46 C44

.

Problem 6.16For an elastic body whose x3 axis is an axis of elastic symmetry of order N = 6, show

that the nonzero elastic constants are C11 = C22, C33, C55 = C44, C66 = 12

(C11 − C12), andC13 = C23.

Problem 6.17Develop a formula in terms of the strain components for the strain energy density W for

the case of an orthotropic elastic medium.

Answer

W = 12

(C11ε1 + 2C12ε2 + 2C13ε3) ε1 + 12

(C22ε2 + 2C23ε3) ε2

+12(C33ε

23 + C44ε

24 + C55ε

25 + C66ε

26

)Problem 6.18Show that, for an elastic continuum having x1 as an axis of elastic symmetry of orderN = 2, the strain energy density has the same form as for a continuum which has an x2x3plane of elastic symmetry.

Answer

2W = (C11ε1 + 2C12ε2 + 2C13ε3 + 2C14ε4) ε1 + (C22ε2 + 2C23ε3 + 2C24ε4) ε2

+(C33ε3 + 2C34ε4) ε3 + C44ε24 + (C55ε5 + 2C56ε6) ε5 + C66ε

26


Problem 6.19Let the stress field for a continuum be given by

[tij] =

x1 + x2 t12 0

t12 x1 − x2 0

0 0 x2

where t12 is a function of x1 and x2. If the equilibrium equations are satisfied in theabsence of body forces and if the stress vector on the plane x1 = 1 is given by t(e1) =(1+ x2) e1 + (6− x2) e2, determine t12 as a function of x1 and x2.

Answer

t12 = x1 − x2 + 5

Problem 6.20Invert Eq 6.45b to obtain Hooke’s law in the form

tij = 2G

(εij +

ν

1− 2νεkkδij

)which, upon combination with Eqs 6.44 and 6.43, leads to the Navier equation

G

(ui,jj +

1

1− 2νuj,ij

)+ ρbi = 0 .

This equation is clearly indeterminate for ν = 0.5. However, show that in this case Hooke’slaw and the equilibrium equations yield the result

Gui,jj +1

3tjj,i + ρbi = 0 .

Problem 6.21Let the displacement field be given in terms of some vector qi by the equation

ui =2 (1− ν)qi,jj − qj,ji

G.

Show that the Navier equation (Eq 6.49) is satisfied providing bi ≡ 0 and qi is bi-harmonicso that qi,jjkk = 0. If q1 = x2/r and q2 = −x1/r where r2 = xixi, determine the resultingstress field.

Answer

t11 = −t22 = 6QGx1x2/r5 ; t33 = 0

t12 = t21 = 3QG(x22 − x21

)/r5

t13 = t31 = 3QGx2x3/r5; t23 = t32 = −3QGx1x2/r

5;where Q = 4(1− ν)/G


Problem 6.22If body forces are zero, show that the elastodynamic Navier equation (Eq 6.55) will be

satisfied by the displacement field

ui = φ,i + εijkψk,j

provided the potential functions φ and ψk satisfy the three-dimensional wave equation.

Problem 6.23Show that, for plane stress, Hooke’s law Eq 6.120a and Eq 6.121 may be expressed in termsof the Lame constants λ and µ by

εij =1

2µ

(tij −

λ

3λ+ 2µδijtkk

)(i, j, k = 1, 2) ,

εkk = −λ

2µ (3λ+ 2µ)tii (i = 1, 2) .

Problem 6.24For the case of plane stress, let the stress components be defined in terms of the functionφ = φ(x1, x2), known as the Airy stress function, by the relationships,

t11 = φ,22, t22 = φ,11, t12 = −φ,12 .

Show that φ must satisfy the biharmonic equation ∇4φ = 0 and that, in the absence ofbody forces, the equilibrium equations are satisfied identically by these stress components.If φ = Ax31x

22 − Bx51 where A and B are constants, determine the relationship between A

and B for this to be a valid stress function.

Answer

A = 5B

Problem 6.25Develop an expression for the strain energy density, W, for an elastic medium in (a) plane

stress and (b) plane strain.

Answer

(a) W =[t211 + t222 − 2νt11t22 + 2 (1+ ν) t212

]/2E

(b) W =(µ+ 1

2λ) (ε211 + ε222

)+ λε11ε22 + 2µε212

Problem 6.26Show that φ = x41x2 + 4x21x

32 − x52 is a valid Airy stress function, that is, that ∇4φ = 0,

and compute the stress tensor for this case assuming a state of plane strain with ν = 0.25.

Answer

[tij] =

24x21x2 − 20x32 −4x31 − 24x1x22 0

−4x31 − 24x1x22 12x21x2 + 8x32 0

0 0 14

(9x21x2 − 3x32

)


Problem 6.27Verify the inversion of Eq 6.147 into Eq 6.148. Also, show that the two equations of

Eq 6.149 may be combined to produce Eq 6.148.

Problem 6.28Develop appropriate constitutive equations for thermoelasticity in the case of (a) plane

stress and (b) plane strain.

Answer

(a) εij = [(1+ ν)tij − νδijtkk]/E+ δij(θ− θ0)α (i, j, k = 1, 2)ε33 = νtii/E+ α(θ− θ0) (i = 1, 2)

(b) tij = λδijεkk + 2µεij − δij(3λ+ 2µ)α(θ− θ0) (i, j, k = 1, 2)t33 = νtii − αE(θ− θ0) = λεii − (3λ+ 2µ)α(θ− θ0) (i = 1, 2)

Problem 6.29Consider the Airy stress function

φ5 = D5x21x32 + F5x

52 .

(a) Show that for this to be valid stress function, F5 = −D5/5.(b) Construct the composite stress function

φ = φ5 + φ3 + φ2

where

φ = D5

(x21x

32 −

1

5x52

)+1

2B3x

21x2 +

1

2A2x

21 .

For this stress function show that the stress components are

t11 = D5(6x21x2 − 4x32

),

t22 = 2D5x32 + B3x2 +A2 ,

t12 = −6D5x1x22 − B3x1 .

Problem 6.30A rectangular beam of width unity and length 2L carries a uniformly distributed load ofq lb/ft as shown. Shear forces V support the beam at both ends. List the six boundaryconditions for this beam the stresses must satisfy using stresses determined in Problem 6.29.

x2

x1

c

cO

q [lb/ft]

V V

L L


Answer

1. t22 = −q at x2 = +c

2. t22 = 0 at x2 = −c

3. t12 = 0 at x2 = ±c4.∫+c

−c t12dx2 = qL at x1 = ±L

5.∫+c

−c t11dx2 = 0 at x1 = ±L

6.∫+c

−c t11x2dx2 = 0 at x1 = ±L

Problem 6.31Using boundary conditions 1, 2, and 3 listed in Problem 6.30, show that the stresses in

Problem 6.29 require that

A2 = −q

2, B3 = −

3q

4c, D5 =

q

8c3.

Thus, for the beam shown the stresses are

t11 =q

2I

(x21x2 −

2

3x32

),

t22 =q

2I

(1

3x32 − c2x2 −

2

3c3),

t12 = −q

2I

(x1x

22 − c2x1

),

where I = 23c3 is the plane moment of inertia of the beam cross section.

Problem 6.32Show that, using the stresses calculated in Problem 6.31, the boundary conditions 4 and 5

are satisfied, but boundary condition 6 is not satisfied.

Problem 6.33Continuing Problems 6.31 and 6.32, in order for boundary condition 6 to be satisfied an

additional term is added to the stress function, namely

φ3 = D3x32 .

Show that, from boundary condition 6,

D3 =3q

4c

(1

15−L2

6c2

),

so that finally

t11 =q

21

[x21 −

2

3x22 +

6

15c2 − L2

]x2 .

Problem 6.34Show that for the shaft having a cross section in the form of an equilateral triangle the

warping function isψ (x1, x2) = λ

(x32 − 3x21x2

).


Determine

(a) the constant λ in terms of the shaft dimension,(b) the torsional rigidity K,(c) the maximum shearing stress.

Mt

O

x2

x1

2a a

Answer

(a) λ = −1

6a

(b) K =9G√3a4

5

(c) t23|max =3Mta

2Kat x1 = a, x2 = 0

Problem 6.35Consider the Galerkin vector that is the sum of three double forces, that is, let

ψ

x2

x1

x3

θ

r

O

eψ

eθ

er


F = B(x1re1 +

x2

re2 +

x3

re3

)where B is a constant and r2 = xixi. Show that the displacement components are given by

ui = −2B (1− 2ν) xi

r3.

Using the sketch above, (spherical coordinates) observe that the radial displacement ur(subscript r not a summed indice, rather indicating the radial component of displacement)

ur =uixi

r,

and show that ur = −2B (1− 2ν) r−2. Also, show that uψ = uθ = 0. Thus

εr =∂ur

∂r=4B (1− 2ν)

r3and εψ = εθ = −

2B (1− 2ν)

r3,

so that the cubical dilatation is εr + εψ + εθ = 0. From Hooke’s law

tij = λδijuk,k + 2µ (ui,j + uj,i)

which reduces here to tij = 2µεij so that

trr =8BG (1− 2ν)

r3,

and

tψψ = tθθ = −4BG (1− 2ν)

r3.


7Classical Fluids

Often, continuum mechanics is taken to gain a foundation for solid mechanics like elas-ticity discussed in the last chapter. However, the foundation of continuum mechanicsallows for the study of fluids. A fundamental characteristic of any fluid – be it a liquid ora gas – is that the action of shear stresses, no matter how small they may be, will causethe fluid to deform continuously as long as the stresses act.

7.1 Viscous Stress Tensor, Stokesian, and Newtonian FluidsA fluid at rest (or in a state of rigid body motion) is incapable of sustaining any shearstress whatsoever. This implies that the stress vector on an arbitrary element of surface atany point in a fluid at rest is proportional to the normal ni of that element, but indepen-dent of its direction. Thus, we write

t(n)i = tijnj = −p0ni (7.1)

where the (positive) proportionality constant p0 is the thermostatic pressure or, as it isfrequently called, the hydrostatic pressure. We note from Eq 7.1 that

tij = −p0δij (7.2)

which indicates that for a fluid at rest the stress is everywhere compressive, that everydirection is a principal stress direction at any point, and that the hydrostatic pressure isequal to the mean normal stress,

p0 = −1

3tii . (7.3)

This pressure is related to the temperature θ and density ρ by an equation of state havingthe form

F (p0, ρ, θ) = 0 . (7.4)

For a fluid in motion the shear stresses are not usually zero, and in this case we write

tij = −pδij + τij (7.5)

where τij is called the viscous stress tensor, which is a function of the motion and vanisheswhen the fluid is at rest. In this equation, the pressure p is called the thermodynamicpressure and is given by the same functional relationship with respect to θ and ρ as thatfor the static pressure p0 in the equilibrium state, that is, by

F (p, ρ, θ) = 0 . (7.6)

271


Note from Eq 7.5 that, for a fluid in motion, p is not equal to the mean normal stress, butinstead is given by

p = −1

3(tii − τii) , (7.7)

so that, for a fluid at rest (τij = 0), p equates to p0.In developing constitutive equations for viscous fluids, we first remind ourselves that

this viscous stress tensor must vanish for fluids at rest, and following the usual practice,we assume that τij is a function of the rate of deformation tensor dij. Expressing thissymbolically, we write

τij = fij (D) . (7.8)

If the functional relationship in this equation is nonlinear, the fluid is called a Stokesianfluid. When fij defines τij as a linear function of dij, the fluid is known as a Newtonianfluid, and we represent it by the equation

τij = Kijpqdpq (7.9)

in which the coefficients Kijpq reflect the viscous properties of the fluid.As may be verified experimentally, most fluids are isotropic. Therefore, Kijpq in Eq 7.9

is an isotropic tensor; this, along with the symmetry properties of dij and τij, allow usto reduce the 81 coefficients Kijpq to 2. We conclude that, for a homogeneous, isotropicNewtonian fluid, the constitutive equation is

tij = −pδij + λ∗δijdkk + 2µ∗dij (7.10)

where λ∗ and µ∗ are viscosity coefficients which denote the viscous properties of the fluid.From this equation we see that the mean normal stress for a Newtonian fluid is

1

3tii = −p+

1

3(3λ∗ + 2µ∗)dii = −p+ κ∗dii (7.11)

where κ∗ = 13

(3λ∗ + 2µ∗) is known as the coefficient of bulk viscosity. The condition

κ∗ =1

3(3λ∗ + 2µ∗) (7.12a)

or, equivalently,

λ∗ = −2

3µ∗ (7.12b)

is known as Stokes condition, and we see from Eq 7.11 that this condition assures us that,for a Newtonian fluid at rest, the mean normal stress equals the (negative) pressure p.

If we introduce the deviator tensors

Sij = tij −1

3δijtkk (7.13a)

for stress and

βij = dij −1

3δijdkk (7.13b)

for rate of deformation into Eq 7.10, we obtain

Sij +1

3δijtkk = −pδij +

1

3(3λ∗ + 2µ∗) δijdkk + 2µ∗βij (7.14)

Classical Fluids 273

which may be conveniently split into the pair of constitutive equations

Sij = 2µ∗βij , (7.15a)

tii = −3 (p− κ∗dkk) . (7.15b)

The first of this pair relates the shear effect of the motion with the stress deviator, and thesecond associates the mean normal stress with the thermodynamic pressure and the bulkviscosity.

7.2 Basic Equations of Viscous Flow, Navier-Stokes EquationsInasmuch as fluids do not possess a “natural state” to which they return upon removalof applied forces, and because the viscous forces are related directly to the velocity field,it is customary to employ the Eulerian description in writing the governing equationsfor boundary value problems in viscous fluid theory. Thus, for the thermomechanicalbehavior of a Newtonian fluid, the following field equations must be satisfied:

(a) the continuity equation (Eq 5.13)

ρ+ ρvi,i = 0 , (7.16)

(b) the equations of motion (Eq 5.22)

tij,j + ρbi = ρvi , (7.17)

(c) the constitutive equations (Eq 7.10)

tij = −pδij + λ∗δijdkk + 2µ∗dij , (7.18)

(d) the energy equation (Eq 5.65)

ρu = tijdij − qi,i + ρr , (7.19)

(e) the kinetic equation of state (Eq 7.6)

p = p (ρ, θ) , (7.20)

(f) the caloric equation of state (Eq 5.66)

u = u (ρ, θ) , (7.21)

(g) the heat conduction equation (Eq 5.62)

qi = −κθ,i . (7.22)

This system, Eqs 7.16 through 7.22, together with the definition of the rate of deforma-tion tensor,

dij =1

2(vi,j + vj,i) (7.23)


represents 22 equations in the 22 unknowns, tij, ρ, vi, dij, u, qi, p, and θ. If thermaleffects are neglected and a purely mechanical problem is proposed, we need only Eqs 7.16through 7.18 as well as Eq 7.23 and a temperature independent form of Eq 7.20, whichwe state as

p = p (ρ) . (7.24)

This provides a system of 17 equations in the 17 unknowns, tij, ρ, vi, dij, and p.Certain of the above field equations may be combined to offer a more compact formu-

lation of viscous fluid problems. Thus, by substituting Eq 7.18 into Eq 7.17 and makinguse of the definition Eq 7.23, we obtain

ρvi = ρbi − p,i + (λ∗ + µ∗) vj,ji + µ∗vi,jj (7.25)

which are known as the Navier-Stokes equations for fluids. These equations, along withEqs 7.19, 7.20, and 7.21, provide a system of seven equations for the seven unknowns,vi, ρ, p, u, and θ. Notice that even though Eq 7.18 is a linear constitutive equation, theNavier-Stokes equations are nonlinear because in the Eulerian formulation

vi =∂vi

∂t+ vjvi,j .

If Stokes condition(λ∗ = −23µ

∗) is assumed, Eq 7.25 reduces to the form

ρvi = ρbi − p,i +1

3µ∗ (vj,ji + 3vi,jj) . (7.26)

Also, if the kinetic equation of state has the form of Eq 7.24, the Navier-Stokes equationsalong with the continuity equation form a complete set of four equations in the fourunknowns, vi and ρ.

In all of the various formulations for viscous fluid problems stated above, the solutionsmust satisfy the appropriate field equations as well as boundary and initial conditions onboth traction and velocity components. The boundary conditions at a fixed surface requirenot only the normal, but also the tangential component of velocity to vanish because ofthe “boundary layer” effect of viscous fluids. The initial and boundary conditions are

1. The velocity must be specifiedvi = v∗i (x, 0)

throughout the volume V. For problems where the density and temperature arechanging the initial conditions at t = 0 are

ρ = ρ∗ (x, 0) ,

θ = θ∗ (x, 0) .

2. The boundary conditions on the surface of the body V for t > 0 are

vi = v∗i (x, t) on Sv ,

tijnj = T∗i (x, t) on ST ,

on those parts of the boundary Sv and ST where the velocity and traction are speci-fied. In addition, the temperature and heat flux must also be specified for problemsinvolving temperature changes.

θ = θ∗ (x, t) on Sθ .

θ,ini = q∗ (x, t) on Sq .


The boundary conditions on traction and velocity at solid and free surfaces occur fre-quently in problems and deserve special comment. On a solid surface, the velocity as-sumes the velocity of the surface. This is known as the no-slip condition for viscous fluids.The condition changes if the fluid is inviscid. Since the viscosity is zero, the shear stressis everywhere zero, and the component along the boundary will be non-zero. The normalcomponent of the velocity will be that of the body if the surface is impermeable. Also, atfree surfaces the normal traction is equal to the ambient pressure when surface tensioneffects are negligible. Additionally the shear stress is zero at this surface. It should alsobe pointed out that the formulations posed in this section are relevant only for laminarflows. Turbulent flows require additional considerations.

7.3 Specialized FluidsAlthough the study of viscous fluids in the context of the equations presented in Section7.2 occupies a major role in fluid mechanics, there is also a number of specialized situa-tions resulting from simplifying assumptions that provide us with problems of practicalinterest. Here, we list some of the assumptions that are commonly made and considerbriefly their meaning with respect to specific fluids.

(a) Barotropic fluids – If the equation of state happens to be independent of temper-ature as expressed by Eq 7.24, the changes of state are termed barotropic, andfluids which obey these conditions are called barotropic fluids. In particular,we may cite both isothermal changes (in which the temperature is constant) andadiabatic changes (for which no heat enters or leaves the fluid) as barotropicchanges.

(b) Incompressible fluids – If the density of a fluid particle is constant, the equationof state becomes

ρ = constant (7.27)

which describes incompressibility. This implies ρ = 0 and, by the continuityequation, vi,i = 0 for incompressible flows. Physically, incompressibility meansthat the elements of a fluid undergo no change in density (or volume) whensubjected to a change in pressure. For incompressible flows, the Navier-Stokesequations become

ρvi = ρbi − p,i + µ∗vi,jj (7.28)

due to the vi,i = 0 condition. Water and oil, among others, are generallyassumed to be incompressible, whereas most gases are highly compressible.

(c) Inviscid (frictionless) fluids – A fluid that cannot sustain shear stresses even whenin motion is called an inviscid, or sometimes a perfect fluid. Clearly, if thecoefficients λ∗ and µ∗ in Eq 7.10 are equal to zero, that equation describes aperfect fluid and the Navier-Stokes equations reduce to

ρvi = ρbi − p,i (7.29)

which are often referred to as the Euler equation of motion. An ideal gas is aperfect fluid that obeys the gas law

p = ρRθ (7.30)


where R is the gas constant for the particular gas under consideration. It shouldbe pointed out that all real fluids are compressible and viscous to some degree.

7.4 Steady Flow, Irrotational Flow, Potential FlowIf the velocity components of a fluid are independent of time, the motion is called a steadyflow. In such cases, the material derivative of the velocity,

vi =∂vi

∂t+ vjvi,j

reduces to the simpler formv = vjvi,j .

Thus, for a steady flow, the Euler equation is modified to read

ρvjvi,j = ρbi − p,i . (7.31)

Furthermore, if the velocity field is constant and equal to zero everywhere, the fluid isat rest, and the theory for this condition is called hydrostatics. For this, the Navier-Stokesequations are simply

ρbi − p,i = 0 . (7.32)

Assuming a barotropic condition between ρ and p, it is possible to define a pressurefunction in the form

P (p) =

∫pp0

dp

ρ. (7.33)

In addition, if the body forces are conservative, we may express them in terms of a scalarpotential function Ω by the relationship

bi = −Ω,i . (7.34)

From the definition Eq 7.34, it follows that

P,i =1

ρp,i or ∇P =

∇pρ, (7.35)

so that now Eq 7.32 may be written

(Ω+ P),i = 0 (7.36)

as the governing equation for steady flow of a barotropic fluid with conservative bodyforces.

Example 7.1

An incompressible, Newtonian fluid maintains a steady flow under the actionof gravity down an inclined plane of slope β. If the thickness of the fluid per-pendicular to the plane is h and the pressure on the free surface is p = p0 (aconstant), determine the pressure and velocity fields for this flow.


h

x2

x3

0p p=

β

v2 =ρg sinβ2µ∗

h2

SolutionAssume v1 = v3 = 0, v2 = v2 (x2, x3). By the continuity equation for incom-pressible flow, vi,i = 0. Hence, v2,2 = 0 and v2 = v2(x3). Thus, the rate ofdeformation tensor has components d23 = d32 = 1

2(∂v2/∂x3) and all others

equal to zero. The Newtonian constitutive equation is given in this case by

tij = −pδij + 2µ∗dij

from which we calculate

[tij] =

−p 0 0

0 −p µ∗∂v2

∂x3

0 µ∗∂v2

∂x3−p

.

The equations of motion having the steady flow form

tij,j + ρbi = ρvjvi,j

result in component equations

(for i = 1) − p,1 = 0 , (7.37a)

(for i = 2) − p,2 + µ∗∂2v2

∂x23+ ρg sinβ = 0 , (7.37b)

(for i = 3) − p,3 − ρg cosβ = 0 , (7.37c)

since gravity is the only body force,

b = g (sinβe2 − cosβe3)

From Eq 7.37a it is easy to see that the pressure is only a function of x2. Namely,p = −f(x2). Integrating the Eq 7.37c gives

p = −(ρg cosβ) x3 + f (x2)


where f(x2) is an arbitrary function of integration. At the free surface (x3 = h),p = p0, and so

f (x2) = p0 + ρgh cosβ

and thusp = p0 + (ρg cosβ) (h− x3)

which describes the pressure in the fluid.Next, by integrating Eq 7.37b twice with respect to x3, we obtain

v2 =−ρg sinβ2µ∗

x23 + ax3 + b

with a and b constants of integration. But from the boundary conditions,1. v2 = 0 when x3 = 0, therefore b = 0,

2. t23 = 0 when x3 = h, therefore a =ρgh

µ∗sinβ.

Finally, therefore, from the equation for v2 we have by the substitution of a =ρgh

µ∗sinβ,

v2 =ρg sinβ2µ∗

(2h− x3) x3

having the profile shown in the figure.

If the velocity field of a fluid is one for which the tensor W vanishes identically, wesay the flow is irrotational. In this case the vorticity vector w, which is related to W byEq 4.154, is also zero everywhere, so that for irrotational flow

wi =1

2εijkvk,j = 0 or w =

1

2∇× v =

1

2curl v = 0 . (7.38)

Finally, from the identity curl(gradφ) = 0, we conclude that, for a flow satisfying Eq 7.38,the velocity field may be given in terms of a velocity potential, which we write as

vi = φ,i or v =∇φ . (7.39)

Indeed, it may be shown that the condition curl v = 0 is a necessary and sufficient condi-tion for irrotational flow and the consequence expressed in Eq 7.39 accounts for the namepotential flow often associated with this situation.

For a compressible irrotational flow, the Euler equation and the continuity equationmay be linearized and combined to yield the wave equation

φ = c2φ,ii (7.40)

where c is the velocity of sound in the fluid. For a steady irrotational flow of a compress-ible barotropic fluid, the Euler equation and the continuity equation may be combined togive (

c2δij − vivj)vj,i = 0 (7.41)

which is called the gas dynamics equation. For incompressible potential flow the continuityequation reduces to a Laplace equation,

φ,ii = 0 or ∇2φ = 0 (7.42)


solutions of which may then be used to generate the velocity field using Eq 7.39. It isworthwhile to mention here that the Laplace equation is linear, so that superposition ofsolutions is available.

Example 7.2

Peristaltic pumping is a form of fluid transport resulting from a wave pass-ing along a distensible fluid containing tube. This form of transport is com-mon in biological systems. The figure below shows a two-dimensional repre-sentation of the tube. The shape of the wall is described by x2 = h (x1, t) =a+b sin 2π

λ(x1 − ct). In the x1-x2 frame shown, the fluid flow is unsteady. How-

ever, in a frame traveling with the wave, x1-x2, the problem becomes steady.The two frames are related by x1 = x1 + ct, x2 = x2 and v1 = v1 + c. Thex1-momentum equation for flows with negligible fluid inertia is

−dp

dx1+ µ

∂2v1

∂x22= 0

with the boundary conditions

v1 = −c on x2 = h and∂v1

∂x2= 0 on x2 = 0 .

a

b c

h(x,t)

x1

x2

l

Two-dimensional figure modeling peristaltic pumping.

Determine the velocity profile v1 (x, t) in the x1-x2 frame. 1

SolutionIntegrating the governing equation twice gives

v1 =1

2µ

dp

dx1x22 + Cx2 +D .

From the symmetry condition,

∂v1

∂x2=1

µ

dp

dx1(0) + C (xx) = 0 and C (x2) = 0 .

On x2 = h, we have

v1 (h) = −c =1

2µ

dp

dx1h2 +D (x2) and D (x2) = −c−

1

2µ

dp

dx1h2 .

1see A. H. Shapiro, M. Y. Jaffrin, and S. L. Weinberg (1969) Peristaltic pumping with long wavelengths at lowReynolds number, Journal of Fluid Mechanics, 37 (4), 799-825.


The velocity profile is

v1 (x, t) =1

2µ

dp

dx1

(x22 − h2

)− c ,

which is transformed to the stationary frame giving

v1 (x, t) =1

2µ

dp

dx1

(x22 − h2

)since v1 = v1 + c and dx1 = dx1. The velocity profile is a simple parabolicdistribution.

7.5 The Bernoulli Equation, Kelvin’s TheoremIf a fluid is barotropic with conservative body forces, Eq 7.36 may be substituted on theright-hand side of Euler’s equation (Eq 7.29), giving

vi = −(Ω+ P),i . (7.43)

As a step in obtaining a solution to this differential equation, we define a streamline asthat space curve of which the tangent vector, at each point, has the direction of the fluidvelocity (vector). For a steady flow, the fluid particle paths are along streamlines. Byintegrating Eq 7.43 along a streamline (see Problem 7.15), we can show that∫x2

x1

∂vi

∂tdxi +

v2

2+Ω+ P = G (t) (7.44)

where dxi is a differential tangent vector along the streamline. This is the well-knownBernoulli equation. If the motion is steady, the time function G(t) resulting from the in-tegration reduces to a constant G, which may vary from one streamline to another. Fur-thermore, if the flow is also irrotational, a unique constant G0 is valid throughout theflow.

When gravity is the only force acting on the body, we write Ω = gh where g = 9.81

m/s2 is the gravitational constant and h is a measure of the height above a reference levelin the fluid. If hp = P/g is defined as the pressure head and hv = v2/2g as the velocity head,Bernoulli’s equation for incompressible fluids becomes

h+ hP + hv = h+p

ρg+v2

2g= G0 (7.45)

Recall that by Eq 2.97 in Chapter 2 we introduced Stoke’s theorem, which relates theline integral around a closed curve to the surface integral over its cap. By this theoremwe define the velocity circulation ΓC around a closed path in the fluid as

ΓC =

∮vidxi =

∫S

εijkvk,jnidS (7.46)


where ni is the unit normal to the surface S bounded by C and dxi is the differentialtangent element to the curve C. Note that, when the flow is irrotational, curl v = 0 andthe circulation vanishes. If we take the material derivative of the circulation by applyingEq 5.7 to Eq 7.46 we obtain

ΓC =

∮(vidxi + vidvi) . (7.47)

For a barotropic, inviscid fluid with conservative body forces, this integral may be shownto vanish, leading to what is known as Kelvin’s theorem of constant circulation.


Problems

Problem 7.1Introduce the stress deviator Sij and the viscous stress deviator τ∗ij = τij − 1

3δijτkk intoEq 7.5 to prove that Sij = τ∗ij.

Problem 7.2Determine an expression for the stress power (a) tijdij and (b) τijdij for a Newtonian

fluid. First, show that

τij =

(κ∗ −

2

3µ∗)δijdkk + 2µ∗dij .

Answer

(a) tijdij = −pdii + κ∗diidjj + 2µ∗βijβij

(b) τijdij = κ∗ (trD)2 + 2µ∗βijβij

Problem 7.3Determine the constitutive equation for a Newtonian fluid for which Stokes condition holds,that is, for κ∗ = 0.

Answer

tij = −pδij + 2µ∗βij

Problem 7.4Develop an expression of the energy equation for a Newtonian fluid assuming the heat

conduction follows Fourier’s law.

Answer

ρu = −pvi,i + λ∗vi,ivj,j + 12µ∗ (vi,j + vj,i) + κ∗θ,ii + ρr

Problem 7.5The dissipation potential Ψ for a Newtonian fluid is defined as a function of D and β by

Ψ =1

2κ∗djjdii + µ∗βijβij, where κ∗ = λ∗ +

2

3µ∗ .

Show that ∂Ψ/∂dij = τij.

Problem 7.6Verify the derivation of the Navier-Stokes equations for a Newtonian fluid as given by

Eq 7.25.


Problem 7.7Consider a two-dimensional flow parallel to the x2x3 plane so that v1 = 0 throughout the

fluid. Assuming that an incompressible, Newtonian fluid undergoes this flow, develop aNavier-Stokes equation and a continuity equation for the fluid.

Answer

(Navier-Stokes) ρvi = ρbi − p,i + µ∗vi,jj (i, j = 2, 3)

(Continuity) vi,i = 0 (i = 2, 3)

Problem 7.8Consider a barotropic, inviscid fluid under the action of conservative body forces. Show

that the material derivative of the vorticity of the fluid in the current volume V is

d

dt

∫V

widV =

∫S

viwjnjdS .

Problem 7.9Show that for an incompressible, inviscid fluid the stress power vanishes identically as one

would expect.

Problem 7.10Show that the vorticity and velocity of a barotropic fluid of constant density moving under

conservative body forces are related through the equation wi = wjvi,j. Deduce that for asteady flow of this fluid vjwi,j = wjvi,j.

Problem 7.11In terms of the vorticity vector w, the Navier-Stokes equations for an incompressible fluid

may be written asρvi = ρbi − p,i − 2µ∗εijkwk,j .

Show that, for an irrotational motion, this equation reduces to the Euler equation

ρvi = ρbi − p,i .

Problem 7.12Carry out the derivation of Eq 7.41 by combining the Euler equation with the continuity

equation, as suggested in the text.

Problem 7.13Consider the velocity potential φ = x2x3/r

2 where r2 = x21 + x22. Show that this satisfiesthe Laplace equation φ,ii = 0. Derive the velocity field and show that this flow is bothincompressible and irrotational.

Problem 7.14If the equation of state of a barotropic fluid has the form p = λρk where k and λ are

constants, the flow is termed isentropic. Show that the Bernoulli equation for a steady


motion in this case becomes

Ω+kp

(k+ 1) ρ+1

2v2 = constant .

Also, show that for isothermal flow the Bernoulli equation takes the form

Ω+p ln ρρ

+1

2v2 = constant .

Problem 7.15Derive Eq 7.44 by taking the scalar product of dxi (the differential displacement along a

streamline) with Eq 7.43 and integrating along the streamline, that is, by the integration of∫x2x1

(vi +Ω,i + P,i)dxi .

Problem 7.16Verify that Eq 7.47 is the material derivative of Eq 7.46. Also, show that for a barotropic,

inviscid fluid subjected to conservative body forces the rate of change of the circulation iszero (See Eq 7.47).

Problem 7.17Determine the circulation ΓC around the square in the x2x3 plane shown in the figure if

the velocity field is given by

v =(x3 − x22

)e2 + (x3 + x2) e3 .

(1,−1)

(−1,1)

(−1,−1)

(1,1)x

3

x2O

Answer

ΓC = 0

8Nonlinear Elasticity

Many of today’s challenging design problems involve materials such as butadiene rubber(BR), natural rubber (NR), or elastomers. Rubber materials might be most easily character-ized by the stretching and relaxing of a rubber band. The resilience of rubber, the abilityto recover initial dimensions after large strain, was not possible with natural latex untilCharles Goodyear discovered vulcanization in 1939. Vulcanization is a chemical reactionknown as cross-linking which turned liquid latex into a non-meltable solid (thermoset) .Cross-linked rubber would also allow considerable stretching with low damping; strongand stiff at full extension, it would then retract rapidly (rebound). One of the first applica-tions was rubber-impregnated cloth, which was used to make the sailor’s “mackintosh.”Tires continue to be the largest single product of rubber although there are many, manyother applications. These applications exhibit some or all of rubber’s four characteristics,viz. damping in motor mounts, rebound/resilience in golf ball cores, or simple stretchingin a glove or bladder. While thermoset rubber remains dominant in rubber production,processing difficulties have led to the development and application of thermoplastic elas-tomers (TPEs). These materials are easier to process and are directly recyclable. WhileTPEs are not as rubberlike as the thermosets, they have found wide application in auto-motive fascia and as energy-absorbing materials.

There are several reasons why designing with plastic and rubber materials is moredifficult than with metals. For starters, the stress-strain response, that is, the constitutiveresponse, is quite different. Figure 8.1(a) shows the stress-strain curves for a mild steelspecimen along with the response of a natural rubber used in an engine mount. Notethat the rubber specimen strain achieves a much higher stretch value than the steel. Thedashed vertical line in Fig. 8.1(b) represents the strain value of the mild steel at failure.This value is much less than the 200% strain the rubber underwent without failing. Infact, many rubber and elastomer materials can obtain 300 to 500% strain. Highly cross-linked and filled rubbers can result in materials not intended for such large strains. Golfball cores are much stiffer than a rubber band, for instance. Each of these products isdesigned for different strain regimes. A golf ball’s maximum strain would be on theorder of 30 to 40%. Its highly crosslinked constitution is made for resilience, not for largestrain. The rubber stress-strain curve exhibits nonlinear behavior from the very beginningof its deformation, whereas steel has a linear regime below the yield stress.

The reason rubber materials exhibit drastically different behavior than metals resultsfrom their sub-microscopic characteristics. Metals are crystalline lattices of atoms all be-ing, more or less, well ordered: in contrast, rubber material molecules are made up ofcarbon atoms bonded into a long chain resembling a tangled collection of yarn scraps.Since the carbon-carbon (C—C) bond can rotate, it is possible for these entangled longchain polymers to rearrange themselves into an infinite number of different conforma-tions. While the random coil can be treated as a spring, true resilience requires a cross-link to stop viscous flow. In a thermoset rubber, a chemical bond, often with sulfur,affords the tie while physical entanglements effect the same function in a TPE material.The degree of cross-linking is used to control the rubber’s stiffness.

285


Stretch

Stee

l Stre

ss, P

a (d

ashe

d lin

e)

0

1000

2000

1 1.002 1.004 1.006 1.008 1.010

5

10x 105

Rub

ber S

tress

, Pa

(sol

id li

ne)

(a) Steel and rubber on strain scale appropriate for steel. The rubber stress is lowand nearly coinciding with the abscissa.

2000

1000

0

10x 105

2.5 30

5

21.51Stretch

Stee

l Stre

ss, P

a (d

ashe

d lin

e)

Rub

ber S

tress

, Pa

(sol

id li

ne)

(b) Steel and rubber on strain scale appropriate for rubber. The strain values forthe steel are small making the curve to be nearly coincides with the ordinate.

FIGURE 8.1Nominal stress-stretch curves for rubber and steel. Note the same data is plotted ineach figure, however, the stress axes have different scale and a different strain range isrepresented.

Nonlinear Elasticity 287

In the context of this book’s coverage of continuum mechanics, material make-up on themicro-scale is inconsistent with the continuum assumption discussed in Chapter 1. How-ever, a rubber elasticity model can be derived from the molecular level which somewhatrepresents material behavior at the macroscopic level. In this chapter, rubber elasticitywill be developed from a first-principle basis. Following that, the traditional continuumapproach is developed by assuming a form of the strain energy density and using re-strictions on the constitutive response imposed by the second law of thermodynamics toobtain stress-stretch response.

8.1 Molecular Approach to Rubber ElasticityOne of the major differences between a crystalline metal and an amorphous polymer isthat the polymer chains have the freedom to rearrange themselves. The term conforma-tion is used to describe the different spatial orientations of the chain. Physically, the easeat which different conformations are achieved results from the bonding between carbonatoms. As the carbon atoms join to form the polymer chain , the bonding angle is 109.5,but there is also a rotational degree of freedom around the bond axis. For most macro-molecules, the number of carbon atoms can range from 1, 000 to 100, 000. With eachbond having a certain degree of freedom to orient itself, the number of conformationsbecomes quite large. Because of this substantial amount, the use of statistical thermo-dynamics may be used to arrive at rubber elasticity equations from first principles. Inaddition to the large number of conformations for a single chain, there is another reasonthe statistical approach is appropriate: the actual polymer has a large number of differentindividual chains making up the bulk of the material. For instance, a cubic meter of anamorphous polymer having 10,000 carbon atoms per molecule would have on the orderof 1024 molecules (McCrum et al., 1997). Clearly, the sample is large enough to justify astatistical approach.

At this point, consider one particular molecule, or polymer chain, and its conforma-tions. The number of different conformations the chain can obtain depends on the dis-tance separating the chain’s ends. If a molecule is formed of n segments each havinglength l, the total length would be L = nl. Separating the molecule’s ends, the lengthL would mean there is only one possible conformation keeping the chain intact. As themolecule’s ends get closer together, there are more possible conformations that can beobtained. Thus, a Gaussian distribution of conformations as a function of the distancebetween chain ends is appropriate. Figure 8.2 demonstrates how more conformations arepossible as the distance between the molecule’s ends is reduced.

Molecule end-to-end distance, r, is found by adding up all the segment lengths, l, as isshown in Fig. 8.3. Adding the segment lengths algebraically gives distance r from end-to-end, but it does not give an indication of the length of the chain. If the ends are relativelyclose and the molecule is long there will be the possibility of many conformations. Form-ing the magnitude squared of the end-to-end vector, r, in terms of the vector addition ofindividual segments

r2 = r · r = (l1 + l2 + · · ·+ ln) · (l1 + l2 + · · ·+ ln) (8.1)

where li is the vector defining the ith segment of the molecule chain. Multiplying out theright-hand side of Eq 8.1 leads to

r2 = nl2 + [l1 · l2 + l1 · l3 + · · ·+ ln−1 · ln] . (8.2)


L

FIGURE 8.2A schematic comparison of molecular conformations as the distance between molecule’sends varies. Dashed lines indicate other possible conformations.

x3

x2

x1

r

dV = dx1dx

2dx

3

FIGURE 8.3A freely connected chain with end-to-end vector r.


This would be the square of the end-to-end vectors for one molecule of the polymer. Ina representative volume of the material there would be many chains from which we mayform the mean square end-to-end distance

⟨r2⟩

=1

N

N∑1

(nl2 + [l1 · l2 + l1 · l3 + · · ·+ ln−1 · ln]

). (8.3)

The bracketed term in Eq 8.3 is argued to be zero from the following logic. Since a largenumber of molecules is taken in the sample, it is reasonable that, for every individualproduct l1 · l2, there will be another segment pair product which will equal its negative.The canceling segment pair does not necessarily have to come from the same molecule.Thus, the bracketed term in Eq 8.3 is deemed to sum to zero, leaving a simple expressionfor the mean end-to-end distance ⟨

r2⟩

= nl2 (8.4)

The mean end-to-end distance indicates how many segments, or carbon atoms, are ina specific chain. To address the issue of how the end-to-end distances are distributedthroughout the polymer, a Gaussian distribution is assumed. Pick the coordinate’s originto be at one end of a representative chain. Figure 8.3 shows this for a single chain, withthe other end of the chain in an infinitesimal volume dV located by the vector r. Theprobability of the chain’s end lying in the volume dV is given by

P (r)dr =e−( rρ )

2(√πρ)3dr (8.5)

where ρ is a parameter of the distribution. Using this assumed distribution of meanend-to-end distances, it is straightforward to find

⟨r2⟩0

=

∫∞0

r2P (r)dr =3

2ρ2 (8.6)

where the subscript 0 denotes that this is an intrinsic property of the chain since it wasconsidered alone. When the chain is placed back into a crosslinked network of chains, themean end-to-end distance is written as

⟨r2⟩i. This latter designation takes into account

the fact the chain has restrictions placed upon it by being packed into a volume withother chains. Equating Eqs 8.4 and 8.6 the distribution parameter ρ is found to be

ρ =

√2n

3l . (8.7)

Similar to the results of Section 5.9, the force created by stretching a uniaxial specimenis given in terms of the Helmholtz free energy

F =∂ψ

∂L

∣∣∣∣θ,V

(8.8)

where F is the force, L is the length, and subscripts θ and V designate that the changein length occurs at constant temperature and volume. Substitution of Eq 5.74b yields theforce in terms of the internal energy and entropy

F =∂u

∂L− θ

∂η

∂L(8.9)


where the constant temperature and volume subscripts have not been written for conve-nience.

Examination of Eq 8.9 offers an informative comparison between metals and ideal rub-bers. In metals, the crystalline structure remains intact as the material is deformed. Atomsare moved closer, or further, from adjacent atoms creating a restoring force, but the rel-ative order among the atoms remains the same. The last term of Eq 8.9 has no forcecontribution since the relative order of the atoms stays unchanged. For an ideal rub-ber, a change in length has no effect on the internal energy. Thus, the first derivativeterm of Eq 8.9 is zero. However, stretching of the specimen increases the mean end-to-end distance, thus reducing the possible conformations for the chains. This reduction inconformations gives rise to a negative change in entropy as the length is increased.

The entropy for a single chain will be related to the conformation through the meanend-to-end length. Noting the number of configurations is proportional to the probabilityper unit volume, P(r), and using Boltzmann’s equation, the entropy may be written as

η = η0 + k lnP (r) = η0 + k

[3 ln

(√πρ)

+

(r

ρ

)2](8.10)

where k is Boltzmann’s constant. Use of this in Eq 8.9 for an ideal rubber gives a singlechain retractive force given by

F =2kθ

ρ2r2 . (8.11)

Consider a polymer having forces applied resulting in stretch ratios λ1, λ2, and λ3. Thework done on each chain of the material is the sum of the work done in each coordinatedirection xi

W(i) =

∫λ(i)xi

xi

fidxi =2kθ

ρ2

∫λ(i)xi

xi

xidxi =kθ

ρ2

[(λ(i)

)2− 1

]x2i (no sum on i) .

Taking into consideration the work done on all of the chains gives total work in eachcoordinate direction∑

n

W(i) =kθ

ρ2

[(λ(i)

)2− 1

]∑n

x2i (no sum on i) (8.12)

where the last summed term is the number of chains, n, times the initial mean end-to-enddistance in the xi direction. Assuming the rubber is initially isotropic yields∑

n

x2i = n⟨x2i⟩i=n

3

⟨r2⟩i

(no sum on i) .

Substituting this and ρ from Eq 8.6 into Eq 8.12 and adding all three coordinate workterms gives

W =nkθ

2

⟨r2⟩i

〈r2〉0

[λ21 + λ22 + λ23 − 3

]. (8.13)

For convenience, this equation may be written as

W =VG

2

[λ21 + λ22 + λ23 − 3

](8.14)

where V is the volume and G is the shear modulus which are given by

N =n

V,


L0

L

A0

A

FIGURE 8.4Rubber specimen having original length L0 and cross-section area A0 stretched into de-formed shape of length L and cross section area A.

and

G = Nkθ

⟨r2⟩i

〈r2〉0.

Next, considering a uniaxial tension of a specimen (Fig. 8.4) the stretching ratios reduceto

λ1 = λ, λ2 = λ3 =1√λ

(8.15)

where the nearly incompressible nature has been used in the form λ1λ2λ3 = 1. The totalwork done is

W =VG

2

[λ2 +

2

λ− 3

]. (8.16)

This is the work done on the polymer chains, but it is the same work done by externalforces since an ideal rubber is assumed. Thus, the work shown in Eq 8.16 is equal tothe change in Helmholtz free energy. Recalling the force from deformation as defined byEq 8.8, the force resulting from a stretch λ is given by

F =∂ψ

∂L=dW

dL=dW

dλ

dλ

dL.

The deformation is volume preserving, having V = A0L0 = AL where the subscript 0denotes the initial area and length. Since the stretch ratio in this case is λ = L/L0, it isclear that dλ/dL = 1/L0. Using this result and differentiating Eq 8.16 results in

F =VG

L0

[λ−

1

λ2

]= A0G

[λ−

1

λ2

]


which may be written as

f =F

A0= G

[λ−

1

λ2

]. (8.17)

Materials satisfying this equation are called neo-Hookean.

8.2 A Strain Energy Theory for Nonlinear ElasticityThe theory developed in the previous section does not represent most experimental datawell at large strains. A better approach to modeling the response of rubbers comes fromassuming the existence of a strain energy which is a function of the deformation gradientin the form of the left deformation tensor Bij = Fi,AFj,A. This approach, first published byMooney (1940) and furthered by Rivlin (1948), actually predates the molecular approachdiscussed in Section 8.1. The basis of Mooney’s and subsequent theories is the initiallyisotropic material has to obey certain symmetries with regards to the functional form ofthe strain energy function.

Assume the strain energy per unit volume to be an isotropic function of the strain inthe form of the right deformation tensor invariants I1, I2, and I3

W = W (I1, I2, I3) (8.18)

where

I1 = CAA ,

I2 =1

2(CAACBB − CABCAB) , (8.19)

I3 = εMNOC1MC2NC3O = det [CAB] .

Note that if principal axes of CAB are chosen the invariants of Eq 8.19 are written in termsof the stretch ratios λ1, λ2, and λ3 as follows:

I1 = λ21 + λ22 + λ23 ,

I2 = λ21λ22 + λ22λ

33 + λ21λ

23 , (8.20)

I3 = λ21λ22λ23 .

Also, it is easy to show that I1 = CAA = Bii = IB = IC, I2 = IIB = IIC and I3 = IIIB =IIIC by using the definitions of the right and left deformation tensors.

An expression for the Cauchy stress in terms of the strain energy density comes fromthe local energy equation in the form

u =1

ρtijdij . (8.21)

Since ρ0u ≡ W and W is a function of the right deformation tensor the derivative of umay be written as

u =1

ρ0

∂W

CABCAB . (8.22)

Noting thatCAB = dij (FjAFiB + FiAFjB)


allows us to writeu =

2

ρJFiA

∂W

CABFjBdij . (8.23)

Equating Eqs 8.21 and 8.23 for u and factoring out dij leaves

dij

(tij − 2J−1FiA

∂W

∂CABFjB

)= 0 (8.24)

which must hold for all motions. Thus, the stress in a compressible material may bewritten in terms of the strain energy density as

tij = 2J−1FiA∂W

∂CABFjB . (8.25)

Differentiation of the strain energy, along with the chain rule, gives stress components

tij = 2J−1FiA

[∂W

∂I1

∂I1

∂CAB+∂W

∂I2

∂I2

∂CAB+∂W

∂I3

∂I3

∂CAB

]FjB (8.26)

where the invariants I1, I2 and I3 are as given in Eq 8.19. Partial derivatives of theinvariants I1, I2 and I3 may be completed by noting the components are independentquantities. Thus,

∂CAB

∂CMN= δAMδBN , (8.27)

and the partial derivatives of the invariants may be written as

∂I1

∂CAB=

∂CMM

∂CAB= δMAδMB = δAB (8.28a)

∂I2

∂CAB=

1

2δABCNN +

1

2CNNδAB − δMAδNBCMN

= I1δAB − CAB (8.28b)

∂I3

∂CAB=

1

2εAMNεBQRCMQCNR = I3C

−1AB . (8.28c)

These terms are further simplified when put into Eq 8.26 since they are multiplied byFiAFjB:

FiAδABFjB = FiAFjA = Bij

FiACABFjB = FiAFkAFkBFjB = BikBkj .

Noting that J−1 = I12

3 , the stress components may be written as

tij = α0δij + α1Bij + α2BikBkj (8.29)

where

α0 = 2I12

3

∂W

∂I3, (8.30a)

α1 = 2I− 123

(∂W

∂I1+ I1

∂W

∂I2

), (8.30b)

α2 = −2I− 123

∂W

∂I2. (8.30c)


A second form for the Cauchy stress can be found by use of the Cayley-Hamilton theorem:

tij = β0δij + β1Bij + β−1B−1ij (8.31)

where

β0 = 2I− 123

(I2∂W

∂I2+ I3

∂W

∂I3

), (8.32a)

β1 = 2I− 123

∂W

∂I1, (8.32b)

β−1 = −2I− 123

∂W

∂I2. (8.32c)

When a specific material model is chosen defining the specific form of the strain energydensity (Eq 8.18), Eq 8.31 can be interpreted. The material parameters are represented byβ0, β1 and β−1, and the deformation is represented by δij, Bij and B−1

ij .Many rubber or elastomeric materials have a mechanical response that is often nearly

incompressible. Even though a perfectly incompressible material is not possible, a varietyof problems can be solved by assuming incompressibility. The incompressible response ofthe material can be thought of as a constraint on the deformation gradient. That is, theincompressible nature of the material is modeled by an addition functional dependencebetween the deformation gradient components. For an incompressible material the den-sity remains constant. This was expressed in Section 5.2 by vi,i = 0 (Eq 5.15) or ρJ = ρ0(Eq 5.17a). With the density remaining constant and Eq 5.17a as the expression of thecontinuity equation it is clear that

J = det [FiA] = 1 . (8.33)

At this point, we digress from specific incompressibility conditions to a more generalcase of a continuum with an internal constraint. Assume a general constraint of the form

φ (FiA) = 0 , (8.34)

or, since CAB = FiAFiB,φ (CAB) = 0 . (8.35)

This second form of the internal constraint has the advantage that it is invariant undersuperposed rigid body motions. Differentiation of φ results in

∂φ

∂CABCAB =

∂φ

∂CAB

(FiAFiB + FiAFiB

)(8.36)

where it is understood partial differentiation with respect to a symmetric tensor resultsin a symmetric tensor. That is,

∂φ

∂CAB≡ 12

(∂φ

∂CAB+

∂φ

∂CBA

). (8.37)

Note that CAB is essentially the same as EAB which is given in Eq 4.149. Thus, Eq 8.36may be written as

φijdij = 0 (8.38)

where φij is defined by

φij =1

2FiA

(∂φ

∂CAB+

∂φ

∂CBA

)FjB . (8.39)


Assume the stress is formed by adding stress components tij, derivable from a con-stitutive response, to components of an arbitrary stress tij resulting from the internalconstraint:

tij = tij + tij . (8.40)

The added arbitrary stress components tij are assumed to be workless, that is,

tijdij = 0 . (8.41)

Comparing Eqs 8.38 and 8.41 shows that both φij and tij are orthogonal to dij, so tij maybe written as

tij = λφij . (8.42)

For an incompressible material, dii = 0 which may be written as

δijdij = 0 , (8.43)

or equivalently by choosing φij = δij in Eq 8.38. Substituting this into Eq 8.42 yields

tij = −pδij (8.44)

where the scalar has been changed to reflect the pressure term it represents.The stress in an incompressible material may now be written as

tij = −pδij + tij (8.45)

where the second term in Eq 8.45 is determined from the constitutive response for Cauchystress. For an incompressible material J = 1, hence I3 = 1, and Eq 8.18 may be written as

W = W (I1, I2) . (8.46)

With the use of Eqs 8.29, the Cauchy stress components for an isotropic, incompressiblematerial are found to be

tij = −pδij + α1Bij + α2BikBkj (8.47)

where

α1 = 2

(∂W

∂I1+ I1

∂W

∂I2

), (8.48a)

α2 = −2∂W

∂I2. (8.48b)

From Eq 8.31, the stress for an isotropic, incompressible material may be written in thealternative form

tij = −pδij + β1Bij + β−1B−1ij (8.49)

where

β1 = 2∂W

∂I1, (8.50a)

β−1 = −2∂W

∂I2. (8.50b)

At this point, the strain energy has been assumed to be a function of I1 and I2, but theexact functional form has not been specified. Rivlin (1948) postulated the strain energyshould be represented as a general polynomial in I1 and I2

W =∑

Cαβ (I1 − 3)α (I2 − 3)

β. (8.51)


It is noted that the strain energy is written in terms of I1− 3 and I2− 3 rather than I1 andI2 to ensure zero strain corresponds to zero strain energy.

Depending on the type of material and deformation, that is, experimental test data,the number of terms used in Eq 8.51 is chosen. For instance, choosing C10 = G and allother coefficients zero results in a neo-Hookean response where G is the shear modulus.Stresses are evaluated with Eq 8.49 and used in the equations of motion, Eq 5.22. Thisresults in a set of differential equations solved with the use of the problem’s appropriateboundary conditions. The indeterminate pressure is coupled with the equations of motionand is found by satisfying the boundary conditions.

8.3 Specific Forms of the Strain EnergyWhen confronted with the design of a specific part made of a rubber-like material, say thejacket or lumbar of an automotive impact dummy or a golf ball core, testing must be doneto evaluate the various constants of the strain energy function. But before the testing canbe analyzed, the specific form of the strain energy must be chosen. This is tantamount tochoosing how many terms of the Mooney-Rivlin strain energy, Eq 8.51, will be assumednonzero to effectively represent the material response. There are other common forms ofthe strain energy which are all somewhat equivalent to the form put forward by Mooney(see Rivlin, 1976). Since constants Cαβ of Eq 8.51 do not represent physical quantitieslike, for example, modulus of elasticity, the constants are essentially curve fit parameters.

The simplest form of the strain energy for a rubber-like material is a one parametermodel called a neo-Hookean material. The single parameter is taken to be the shear modu-lus, G, and the strain energy depends only on the first invariant of the deformation tensor,Bij

W = G (I1 − 3) . (8.52)

Assuming principal axes for the left deformation tensor, Bij, for a motion having principalstretches λ1, λ2, and λ3 means Eq 8.52 is written as

W = G(λ21 + λ22 + λ23 − 3

). (8.53)

For the case of uniaxial tension the stretches are λ21 = λ2, λ22 = λ23 = λ−1 and furthermore,if the material is incompressible λ21λ

22λ23 = 1. Thus,

W = G

(λ2 +

2

λ− 3

).

Using this expression in Eq 8.47 for the case of uniaxial tension yields the stress per unitundeformed area as

f = P11 =∂W

∂λ= 2G

(λ−

1

λ2

)(8.54)

where P11 is the 11-component of the first Piola-Kirchhoff stress.The neo-Hookean material is the simplest form of the strain energy function and makes

exact solutions much more tractable. A slightly more general model is a simple, or two-term, Mooney-Rivlin model. In this case, the strain energy function is assumed to belinear in the first and second invariants of the left deformation tensor. Again, assumingthe material is isotropic and incompressible, the strain energy may be written as

W = C1(λ21 + λ22 + λ23 − 3

)+ C2

(1

λ21+1

λ22+1

λ23− 3

). (8.55)


For a uniaxial test, the principal stretches are λ21 = λ2, λ22 = λ23 = λ−1. Substitution ofthese stretches into Eq 8.55 and differentiating with respect to λ gives the uniaxial stressper unit undeformed area

f = P11 = 2

(λ−

1

λ2

)(C1 +

C2

λ

). (8.56)

The Cauchy stress is easily formed by noting the area in the deformed configurationwould be found by scaling dimensions in the x2 and x3 direction by λ2 and λ3, respec-tively. For the uniaxial case, this means that multiplying the force per undeformed areaby stretch λ results in the uniaxial Cauchy stress

t11 = 2

(λ2 −

1

λ

)(C1 +

C2

λ

). (8.57)

Making use of Eq 4.109 while noting Λ in that equation is λ in Eq 8.57, a formula forstress in terms of strain is obtained

t11 = 2C1

(1+ 2E11 +

1√1+ 2E11

)+ 2C2

(√1+ 2E11 +

1

1+ 2E11

). (8.58)

Series expansion of the last three terms of Eq 8.58 followed by assuming E11 is smallresults in

t11 = 6 (C1 + C2)E11 . (8.59)

Hence, for small strain the modulus of elasticity may be written as E = 6(C1 + C2).Furthermore, since an incompressible material is assumed, Poisson’s ratio is equal to 0.5.This means, by virtue of Eq 6.29a, that the shear modulus is given by G = 3(C1 + C2) forsmall strain.

The simple, two-term Mooney-Rivlin model represents material response well for smallto moderate stretch values, but for large stretch higher order terms are needed. Some ofthese different forms for the strain energy function are associated with specific names.For instance, an Ogden material (Ogden, 1972) assumes a strain energy in the form

W =∑n

µn

αn

[λαn1 + λαn2 + λαn3 − 3

]. (8.60)

This form reduces to the Mooney-Rivlin material if n takes on values of 1 and 2 withα1 = 2, α2 = −2, µ1 = 2C1, and µ2 = −2C2.

8.4 Exact Solution for an Incompressible, Neo-Hookean MaterialExact solutions for nonlinear elasticity problems come from using the equations of mo-tion, Eq 5.22, or for equilibrium, Eq 5.23, along with appropriate boundary conditions. Aneo-Hookean material is one whose uniaxial stress response is proportional to the combi-nation of stretch λ− 1/λ2 as was stated in Section 8.1. The proportionality constant is theshear modulus, G. The strain energy function in the neo-Hookean case is proportional toI1 − 3 where I1 is the first invariant of the left deformation tensor. A strain energy of thisform leads to stress components given by

tij = −pδij +GBij (8.61)


where p is the indeterminate pressure term, G is the shear modulus, and Bij is the leftdeformation tensor.

The stress components must satisfy the equilibrium equations which, in the absence ofbody forces, are given as

tij,j = 0 . (8.62)

Substitution of Eqs 4.48 and 8.61 into Eq 8.62 gives a differential equation governingequilibrium

∂tij

∂xj= −

∂p

∂xj+G

[(∂2xi

∂XA∂XB

∂XB

∂xj

)∂xj

∂XA+

(∂2xj

∂XA∂XB

∂XB

∂xj

)∂xi

∂XA

]= 0 . (8.63)

This simplifies to∂p

∂xj= G

∂2xi

∂XA∂XA= G∇2xi (8.64)

where∂XB

∂xj

∂xj

∂XA= δAB ,

and∂2xj

∂XA∂XB

∂XB

∂xj= 0

have been used. Eq 8.64 may be made more convenient by writing the pressure term as afunction of the reference configuration and using the chain rule

∂P

∂XA= G

∂xj

∂XA∇2xi (8.65)

where P is a function of XA and t.Consider the case of plane strain defined by

x = x (X, Y) ; y = y (X, Y) ; z = Z (8.66)

from which the deformation gradient follows as

[FiA] =

∂x

∂X

∂x

∂Y0

∂y

∂X

∂y

∂Y0

0 0 1

. (8.67)

Note that we have departed from the strict continuum summation convention for clarity insolving this specific problem. Substitution of the deformation gradient into the definitionfor the left deformation tensor and use of Eq 8.61 yields

txx = −p+G

[(∂x

∂X

)2+

(∂x

∂Y

)2], (8.68a)

tyy = −p+G

[(∂y

∂X

)2+

(∂y

∂Y

)2], (8.68b)

txy = G

[∂x

∂X

∂y

∂X+∂x

∂Y

∂y

∂Y

], (8.68c)


where again it is noted that strict continuum notation is not used for clarity in this par-ticular solution. The incompressibility condition, J = det[FiA] = 1 may be written as

∂x

∂X

∂y

∂Y−∂x

∂Y

∂y

∂X= 1 (8.69)

The inverse of the deformation gradient can be directly calculated from Eq 8.67 to be

[F−1iA

]=

∂X

∂x

∂X

∂y

∂X

∂z

∂Y

∂x

∂Y

∂y

∂Y

∂z

∂Z

∂x

∂Z

∂y

∂Z

∂z

=

∂y

∂Y−∂x

∂Y0

−∂y

∂X

∂x

∂X0

0 0 1

. (8.70)

Thus, in the case of incompressible material undergoing a plane strain motion the follow-ing relationships must hold:

∂x

∂X=∂Y

∂y;

∂x

∂Y= −

∂X

∂y;

∂y

∂X= −

∂y

∂x;

∂y

∂Y=∂X

∂x. (8.71)

An alternate form of the stress components may be written by factoring out the quantityΠ = (txx + tyy) /2 for future convenience

txx = Π+1

2G

[(∂X

∂y

)2+

(∂Y

∂y

)2−

(∂X

∂x

)2−

(∂Y

∂x

)2], (8.72a)

tyy = Π−1

2G

[(∂X

∂y

)2+

(∂Y

∂y

)2−

(∂X

∂x

)2−

(∂Y

∂x

)2], (8.72b)

txy = −G

[∂X

∂x

∂X

∂y+∂Y

∂x

∂Y

∂x

], (8.72c)

where the results of Eq 8.71 have also been utilized. The nontrivial equilibrium equationsare associated with the x and y directions. Using the stress components as defined inEq 8.72, the equilibrium conditions are

∂Π

∂x= G

[∂X

∂x∇2X+

∂Y

∂x∇2X

], (8.73a)

∂Π

∂y= G

[∂X

∂y∇2X+

∂Y

∂y∇2X

](8.73b)

The incompressibility condition may also be written as

∂X

∂x

∂Y

∂y−∂X

∂y

∂Y

∂x= 1 . (8.74)

Eqs 8.73 and 8.74 are three nonlinear partial differential equations that must be satisfiedto ensure equilibrium and the constraint of incompressibility. To actually solve a problem,a solution must be found that satisfies the appropriate boundary conditions. Often theway this is done is by assuming a specific form for X(x, y) and Y(x, y) and demonstratingthat the aforementioned equations are satisfied. Following that, the boundary conditionsare defined and demonstrated, completing the solution. In a sense, this is similar to


the semi-inverse method discussed in Chapter 6. The assumed “guess” of the functionalform of X(x, y) and Y(x, y) requires some experience and or familiarity with the problemat hand. Here, a function form will be assumed in pursuit of the solution of a rectangularrubber specimen being compressed in the x direction. The faces of the specimen areperfectly attached to rigid platens. In this case, assume that X = f(x). After a modestamount of work, this will be shown to lead to a solution of a neo-Hookean materialcompressed between rigid platens.

Substituting the assumed form of X into the incompressibility equation, Eq 8.74, yieldsthe ordinary differential equation

f′∂Y

∂y= 1 (8.75)

where the prime following f represents differentiation with respect to x. Defining q(x) =[f′(x)]−1, Eq 8.75 may be integrated to obtain

Y = q (x)y+ g (x) (8.76)

where function g(x) is an arbitrary function of integration.Use of the given functions X and Y, the incompressibility condition is satisfied, and the

equilibrium equations may be written as follows:

∂Π

∂x= G [f′f′′ + (yq′ + g′) (yq′′ + g′′)] , (8.77a)

∂Π

∂y= G [q (yq′′ + g′′)] , (8.77b)

where, again, primes after a symbol represent differentiation with respect to x. Thesecond equation, 8.77b, is easily integrated to obtain an expression for Π

Π

G=1

2qq′′y2 + qg′′y+M (x) (8.78)

whereM(x) is a function of integration. Differentiation of Eq 8.78 with respect to x resultsin an equation that must be consistent with Eq 8.77a. This comparison gives rise to thefollowing three ordinary differential equations:

1

2

d

dx(qq′′) = q′q′′ , (8.79a)

d

dx(qq′′) = g′q′′ + q′g′′ , (8.79b)

M′ (x) = f′f′′ + g′g′′ . (8.79c)

Eq 8.79a may be expanded and integrated to obtain q′′ = k2q from which the functionq(x) is determined to within the constants A and B:

q (x) = Aekx + Be−kx . (8.80)

Eqs 8.79a and 8.79b may be combined along with the fact that q′′ = k2q to obtain thefunction g(x)

g (x) = Cekx +De−kx (8.81)

where C and D are constants. With functions q(x) and g(x) found it is now possible towrite functions X(x, y) and Y(x, y) as

X =

∫dx

Aekx + Be−kx, (8.82a)


2H

2L

X

Y

(a) Rubber specimen in reference config-uration.

2H

2l

x

y

(b) Rubber specimen in deformed config-uration. The dashed line represents theundeformed shape.

FIGURE 8.5Rhomboid rubber specimen compressed by platens.

Y =(Aekx + Be−kx

)y+ Cekx +De−kx . (8.82b)

In pursuit of the compressed rectangular rubber solution, constants C and D are takento be identically zero and constants A and B are set equal. Thus,

X = (kA)−1 [tan−1(ekx)

+ C′], (8.83a)

Y = 2Ay coshkx , (8.83b)

where k, A, and C′ are constants that must be determined. The deformation consideredhere has the rubber rhomboid deforming symmetrically about the origin of both the ref-erential and deformed coordinate system (Fig. 8.5(a) and 8.5(b)). Thus, set X = 0 whenx = 0 to obtain constant C′, of Eq 8.83a, to have the value 0.25π. Since the platens are fixedto the specimen, Y = y on planes x = ±l the constant A is found to be 1

2[cosh (kl)]

−1. Fi-nally, the constant k may be evaluated in terms of the deformed and undeformed lengthsl and L by

kL = 2 cosh (kl)[tan−1

(ekl)

−π

4

]. (8.84)

To obtain an exact solution it must be shown that the barreled surfaces initially atY = ±H are stress free in the deformed configuration. Unfortunately, this is not possiblefor this problem. Instead, a relaxed boundary condition may be satisfied by enforcing theresultant force on the boundary to be zero rather than have zero traction at every point.

With all the constants of Eq 8.84 determined, the stress components may be written as

txx = −p+1

2G

cosh2 (kx)

cosh2 (kl), (8.85a)

tyy = −p+G

[k2y2

sinh2 (kx)

sinh2 (kl)+ 4

cosh2 (kl)

cosh2 (kx)

], (8.85b)

txy = −1

2Gky

sinh2 (2kx)

cosh2 (kl). (8.85c)


txx

tyy

txy

txy

txx

x

y

FIGURE 8.6Rhomboid rubber specimen compressed by platens.

At this point, the stresses are known to within the additive pressure term p. A solutionis sought by picking p such that the average force over the barreled edge is zero.

Symmetry of the deformation allows for considering the top half, y > 0, in determiningthe condition for zero resultant stress on the barreled top edge (Fig. 8.6). Integrating thestresses yields ∫ l

0

tyy|y=0 dx =

∫H0

txy|x=l dy , (8.86)

or

p =1

4

GkH2

l

sinh (2kl)

cosh2 (kl)−4G

kcosh2 (kl) tanh (kl) . (8.87)

Knowing the pressure term, the compressive force may be determined from

F =

∫H0

txx|x=l dy . (8.88)

Bibliography[1] Carroll, M. M. (1988), “Finite strain solutions in compressible isotropic elasticity,” J.

Elas., Vol. 20, No. 1, pp. 65-92

[2] Kao, B. G. and L. Razgunas (1986), “On the Determination of Strain Energy Func-tions of Rubbers,” Proceedings of the Sixth Intl. Conference on Vehicle Structural Dynam-ics P-178, Society of Automotive Engineering, Warrendale, PA

[3] McCrum, N. G., C. P. Buckley, and C. B. Bucknall (1997), Principles of Polymer Engi-neering, Second Edition, Oxford University Press, Oxford, UK

[4] Mooney, M. (1940), “A Theory of Large Elastic Deformation,” J. Appl. Phys., Vol. 11,pp. 582-592

[5] Rivlin, R.S. (1948), “Large Elastic Deformations of Isotropic Materials: IV. FurtherDevelopments of the General Theory,” Phil. Trans. Roy. Soc., A241, pp. 379-397

[6] Rivlin, R.S. (1949), “Large Elastic Deformations of Isotropic Materials: VI. FurtherResults in the Theory of Torsion, Shear and Flexure,” Phil. Trans. Roy. Soc., A242, pp.173-195


[7] Rivlin, R.S. and K.N. Sawyers (1976), “The Strain-Energy Function for Elastomers,”Trans. of the Society of Rheology, 20:4, pp. 545-557

[8] Sperling, L. H. (1992), Introduction to Physical Polymer Science, Second Edition, Wiley& Sons, Inc., New York

[9] Ward, I. M. and D. W. Hadley (1993), An Introduction to the Mechanical Properties ofSolid Polymers, Wiley & Sons, Inc., Chichester, UK


Problems

Problem 8.1Referred to principal axes, the invariants of the Green deformation tensor CAB are

I1 = λ21 + λ22 + λ23 ,

I2 = λ21λ22 + λ21λ

23 + λ22λ

23 ,

I3 = λ21λ22λ23 .

For an isotropic, incompressible material, show that

I1 = λ21 + λ22 +1

λ21λ22

,

I2 =1

λ21+1

λ22+ λ21λ

22 .

Problem 8.2Derive the following relationships between invariants I1, I2, and I3, and the deformation

gradient, CAB:

(a)∂I1

∂CAB= δAB,

(b)∂I2

∂CAB= CAB − I1δAB,

(c)∂I3

∂CAB= I3C

−1AB.

Problem 8.3Use the definitions of I1 and I2 in terms of the principal stretches λ1, λ2, and λ3 to show

(a)∂W

∂λ1=2

λ1

(λ21 − λ23

)(∂W∂I1

+ λ22∂W

∂I2

),

(b)∂W

∂λ2=2

λ2

(λ22 − λ23

)(∂W∂I1

+ λ21∂W

∂I2

).

Problem 8.4LetW(I1, I2, I3) be the strain energy per unit volume for a homogeneous, isotropic material.Show that the first Piola-Kirchhoff stress components may be written as follows:

PiA ≡∂W

∂FiA= 2

∂W

∂I1FiA + 2

∂W

∂I2(Bij − I1δij) FjA + 2I3

∂W

∂I3F−1iA .

Problem 8.5The Cauchy stress is given by

tij =1

JFjAPiA .


Start with the result of Problem 8.4 to show that

tij =2

J

[∂W

∂I1Bij +

∂W

∂I2(Bik − I1δik)Bjk + I3

∂W

∂I3δij

].

Problem 8.6Assuming a strain energy of the form

W = w (λ1) +w (λ2) +w

(1

λ1λ2

)for an isotropic, incompressible material, show that

λ1∂3W

∂λ21∂λ2= λ2

∂3W

∂λ22∂λ1.

Problem 8.7For biaxial loading of a thin vulcanized rubber sheet the strain energy may be written as

W = C1 (I1 − 3) + C2 (I2 − 3) + C3 (I2 − 3)2

(Rivlin and Saunders, 1951).

(a) Use the definitions of invariants I1, I2, and I3 in terms of stretches λ1, λ2, andλ3 to show

(I2 − 3)2

=1

λ41+1

λ42+1

λ43+ 2I1 − 6I2 + 9 .

(b) Substitute the results from (a) into the strain energy above to obtain

w (λ1) = (C1 + 2C3) λ21 + (C2 − 6C3) λ

−21 + C3λ

−43 − (C1 + C2 − 3C3)

where

W = w (λ1) +w (λ2) +w

(1

λ1λ2

).

Problem 8.8Consider a material having reference configuration coordinates (R,Θ, Z) and current con-

figuration coordinates (r, θ, z). Assume a motion defined by

r =R

g (Θ), θ = f (Θ) , z = Z .

Determine FiA and Bij in terms of g, g′, and f′.

Answer

FiA =

1

g−g′

g20

0f′

g0

0 0 1

, Bij =

1

g2+

(g′

g2

)2−f′g′

g30

−f′g′

g3

(f′

g

)20

0 0 1


Problem 8.9Show

t11 = 2(λ21 − λ−2

1 λ−22

) ∂W∂I1

− 2(λ−21 − λ21λ

22

) ∂W∂I2

,

t22 = 2(λ22 − λ−2

1 λ−22

) ∂W∂I1

− 2(λ−22 − λ21λ

22

) ∂W∂I2

,

are the nonzero Cauchy stress components for biaxial tensile loading of a homogeneous,isotropic, incompressible rubber-like material by solving for the indeterminant pressure pfrom the fact that t33 = 0 condition.

Problem 8.10The following compression force-deflection data was obtained for a highly filled, polybuta-

diene rubber having initial gauge length of 0.490 in and an undeformed cross-section areaof 1 in2

Displ. (in) Force (lbs)

–8.95E–04 –7.33E+00–5.45E–03 –4.15E+01–8.06E–03 –7.57E+01–1.13E–02 –1.29E+02–1.44E–02 –1.83E+02–1.80E–02 –2.52E+02–2.55E–02 –3.93E+02–2.94E–02 –4.62E+02–3.28E–02 –5.35E+02–4.00E–02 –6.72E+02–4.39E–02 –7.40E+02–4.75E–02 –8.13E+02–5.46E–02 –9.45E+02–5.80E–02 –1.01E+03–6.58E–02 –1.15E+03–7.30E–02 –1.28E+03–7.69E–02 –1.35E+03–8.05E–02 –1.42E+03–8.80E–02 –1.56E+03–9.19E–02 –1.63E+03–9.55E–02 –1.70E+03–1.03E–01 –1.86E+03–1.06E–01 –1.93E+03–1.17E–01 –2.20E+03–1.28E–01 –2.50E+03–1.38E–01 –2.85E+03–1.50E–01 –3.26E+03–1.60E–01 –3.73E+03–1.71E–01 –4.22E+03–1.77E–01 –4.42E+03–1.83E–01 –4.48E+03

(a) Generate a stress-stretch plot.(b) Using a spreadsheet, or other appropriate tool, show that simple Mooney-Rivlin

constants C1 = 1, 550 and C2 = −500 represent the material for the range andtype of loading given.


(c) Generate a set of significantly different constants C1 and C2 which might equallywell model the material for this range and type of loading, showing that C1 andC2 are not unique.


9Linear Viscoelasticity

The previous chapters have considered constitutive equations that deal primarily withtwo different types of material behavior: elastic response of solids and viscous flow offluids. Examples of materials that behave elastically under modest loading, and at mod-erate temperatures, are metals such as steel, aluminum, and copper, certain polymers,and even cortical bone. Examples of viscous flow may involve a variety of fluids rangingfrom water to polymers under certain conditions of temperature and loading.

Polymers are especially interesting because they may behave (respond) in either elastic,viscous, or combined manners. At a relative moderate temperature and loading, a poly-mer such as polymethylmethacrylate (PMMA), may be effectively modeled by a linearelastic constitutive equation. However, at a somewhat elevated temperature, the samematerial may have to be modeled as a viscous fluid.

Polymers are by no means the only materials that exhibit different behavior underaltered temperature/frequency conditions. Steel, as well as aluminum, copper, and othermetals, becomes molten at high temperatures and can be poured into molds to formingots. Additionally, at a high enough deformation rate, for example, at the 48 km/hrrate of a vehicle crash, steel will exhibit considerably altered stiffness properties.

Just as continuum mechanics is the basis for constitutive models as distinct as elasticsolids (stress/strain laws) and viscous fluids (stress/strain-rate laws), it also serves as thebasis for constitutive relations that describe material behavior over a range of tempera-ture/frequency and time. One of the simplest models for this combined behavior is thatof linear viscoelasticity.

9.1 Viscoelastic Constitutive Equations in Linear Differential OperatorForm

One of the principal features of elastic behavior is the capacity for materials to storemechanical energy when deformed by loading, and to release this energy totally uponremoval of the loads. Conversely, in viscous flow, mechanical energy is continuously dis-sipated with none stored. A number of important engineering materials simultaneouslystore and dissipate mechanical energy when subjected to applied forces. In fact, all ac-tual materials store and dissipate energy in varying degrees during a loading/unloadingcycle. This behavior is referred to as viscoelastic. In general, viscoelastic behavior may beimagined as a spectrum with elastic deformation as one limiting case and viscous flow theother extreme case, with varying combinations of the two spread over the range between.Thus, valid constitutive equations for viscoelastic behavior embody elastic deformationand viscous flow as special cases, and at the same time provide for response patterns thatcharacterize behavior blends of the two. Intrinsically, such equations will involve not onlystress and strain, but time-rates of both stress and strain as well.

309


In developing the linear differential operator form of constitutive equations for viscoelas-tic behavior as presented in Eq 5.177 we draw upon the pair of constitutive equations forelastic behavior, Eq 6.33, repeated here,

Sij = 2Gηij , (9.1a)

tii = 3Kεii , (9.1b)

together with those for viscous flow, Eq 7.15,

Sij = 2µ∗βij , (9.2a)

tii = −3 (p− κ∗dii) , (9.2b)

each expressed in terms of their deviatoric and dilatational responses. These equationsare valid for isotropic media only. For linear viscoelastic theory we assume that displace-ment gradients, ui,A, are small, and as shown by Eq 4.150, this results in

εij ≈ dij (9.3a)

from which we immediately conclude that

εii = dii (9.3b)

so that now, from Eq 4.77 and Eq 7.13

ηij ≈ βij . (9.4)

If the pressure p in Eq 9.2b is relatively small and may be neglected, or if we considerthe pressure as a uniform dilatational body force that may be added as required to thedilatational effect of the rate of deformation term dii when circumstances require, Eq 9.2may be modified in view of Eq 9.3 and Eq 9.4 to read

Sij = 2µ∗ηij , (9.5a)

tii = 3κ∗εii . (9.5b)

A comparison of Eqs 9.5 and Eq 9.1 indicates that they differ primarily in the physicalconstants listed and in the fact that in Eq 9.5 the stress tensors are expressed in terms ofstrain rates. Therefore, a generalization of both sets of equations is provided by intro-ducing linear differential operators of the form given by Eq 5.178 in place of the physicalconstants G, K, µ∗, and κ∗. In order to make the generalization complete we add similardifferential operators to the left-hand side of the equations to obtain

PSij = 2 Qηij , (9.6a)

M tii = 3 N εii , (9.6b)

where the numerical factors have been retained for convenience in relating to traditionalelasticity and viscous flow equations. As noted, the linear differential time operators, P,Q, M, and N, are of the same form as in Eq 5.178 with the associated coefficients pi,qi, mi, and ni representing the physical properties of the material under consideration.Although these coefficients may in general be functions of temperature or other parame-ters, in the simple linear theory described here they are taken as constants. As stated atthe outset, we verify that for the specific choices of operators P = 1, Q = G, M = 1,

Linear Viscoelasticity 311

t12

γ12

t12

x1

x2

t12

t12

FIGURE 9.1Simple shear element representing a material cube undergoing pure shear loading.

and N = K, Eqs 9.6 define elastic behavior, whereas for P = 1, Q = µ∗∂/∂t, M = 1,and N = κ∗∂/∂t, linear viscous behavior is indicated.

Extensive experimental evidence has shown that practically all engineering materialsbehave elastically in dilatation so without serious loss of generality we may assume thefundamental constitutive equations for linear viscoelastic behavior in differential operatorform to be

PSij = 2 Qηij , (9.7a)

tii = 3Kεii , (9.7b)

for isotropic media. For anisotropic behavior, the operators P and Q must be aug-mented by additional operators up to a total of as many as twelve as indicated by Pi

and Qi with the index i ranging from 1 to 6, and Eq 9.7a expanded to six separateequations.

9.2 One-Dimensional Theory, Mechanical ModelsMany of the basic ideas of viscoelasticity can be introduced within the context of a one-dimensional state of stress. For this reason, and because the viscoelastic response of amaterial is associated directly with the deviatoric response as was pointed out in arrivingat Eq 9.7, we choose the simple shear state of stress as the logical one for explainingfundamental concepts. Thus, taking a material cube subjected to simple shear, as shownby Fig. 9.1, we note that for this case Eq 9.7 reduces to the single equation

P t12 = 2 Qη12 = 2 Q ε12 = Qγ12 (9.8)

where γ12 is the engineering shear strain as shown in Fig. 9.1. If the deformationalresponse of the material cube is linearly elastic, the operators P and Q in Eq 9.8 areconstants (P = 1, Q = G) and that equation becomes the familiar

t12 = Gγ12 (9.9)


t12

G

1

γ12

(a)

t12

t12

(b)

k

1

δ

F

(c)

k

F F

k

δ

(d)

FIGURE 9.2Mechanical analogy for simple shear.

where G is the elastic shear modulus. This equation has the same form as the one whichrelates the elongation (or shortening) δ of a linear mechanical spring to the applied forceF as given by the equation

F = kδ (9.10)

where k is the spring constant. Because Eqs 9.9 and 9.10 are identical in form, the linearspring is adopted as the mechanical analog of simple elastic shearing with k assumingthe role of G. The analogy is depicted graphically by the plots shown in Fig. 9.2.

In similar fashion, if the response of the material cube is viscous flow, Eq 9.8 is written

t12 = µ∗γ12 (9.11a)

where µ∗ is the coefficient of viscosity. In viscoelastic theory it is a longstanding practicethat the coefficient of viscosity be represented by the symbol η, and it is in keepingwith this practice that we hereafter use the scalar η for the coefficient of viscosity. Thus,Eq 9.11a becomes

t12 = ηγ12 (9.11b)

in all subsequent sections of this chapter. The mechanical analog for this situation is thedashpot (a loose fitting piston sliding in a cylinder filled with a viscous fluid) subjected toan axial force F. Here

F = ηδ (9.12)

where δ is the time-rate of extension. This analogy is illustrated in Fig. 9.3.Based upon the two fundamental elements described above, it is easy to construct

viscoelastic models by suitable combinations of this pair of elements. Two especially


η

1

t12

δ

(a)

FF

η

(b)

η

1

t12

γ12

(c)

t12

t12

(d)

FIGURE 9.3Viscous flow analogy.

simple combinations immediately come to mind. The first, that of the spring and dashpotin parallel, Fig. 9.4(a), portrays the Kelvin solid for which Eq 9.8 becomes

T = t12 = G+ η∂tγ12 (9.13)

where the partial derivative with respect to time is denoted by ∂t ≡ ∂/∂t. The second,the spring and the dashpot connected in series, Fig. 9.4(b), represents the Maxwell fluidhaving the constitutive equation

∂t +1

τ

t12 = G∂tγ12 (9.14)

where τ = η/G. In this chapter, the uppercase T is used to designate the stress appliedto the model. Using this notation makes the differential equations that arise from themodels easier to write than if t12 were used. Viscoelastic materials exhibit time dependentbehavior and designating stress with a lowercase t creates confusion. Throughout thischapter T , or T(t), are used interchangeably with t12 to denote the stress.

Models composed of more than two elements are readily constructed. When a Kelvinunit is combined in series with the linear spring element, Fig. 9.5(a), the resulting model issaid to represent the standard linear solid. If the same Kelvin unit is joined in series witha dashpot, Fig. 9.5(b), the model represents a three-parameter fluid. In general, the modelof a fluid has a “free dashpot” as one of its elements. Other three-parameter models areeasily imagined, for example, a Maxwell unit in parallel with a spring, or a Maxwell unitin parallel with a dashpot.

Four-parameter and higher order models may also be constructed. There are two basicpatterns for systematically designing higher order models. One, leading to the generalized


T T

G

η

(a) Kelvin Model

TTG

η

(b) Maxwell Model

FIGURE 9.4Representations of Kelvin and Maxwell models for a viscoelastic solid and fluid, respec-tively.

T T

G1

G2

η1

(a) Standard Linear Solid

T T

G1

η1

η2

(b) Three Parameter Fluid

FIGURE 9.5Three parameter standard linear solid and fluid models.


T T

Gn

ηn

G2

η2

G1

η1

(a) Generalized Kelvin

Gn

G2

ηn

T T

η1G

1

η2

(b) Generalized Maxwell

FIGURE 9.6Generalized Kelvin and Maxwell models constructed by combining basic models.

Kelvin model, has n-Kelvin units in series, Fig. 9.6(a). The second consists of n-Maxwellunits in parallel and is called the generalized Maxwell model, Fig. 9.6(b).

For these models the constitutive equations (Eq 9.8) in operator form are

Kelvin γ12 =t12

G1 + η1∂t+

t12

G2 + η2∂t+ · · ·+ t12

Gn + ηn∂t, (9.15a)

Maxwell t12 =G1γ12

∂t +1

τ1

+G2γ12

∂t +1

τ2

+ · · ·+ Gnγ12

∂t +1

τn

. (9.15b)

In these generalized model equations, one or more of the constants Gi and ηi may beassigned the values 0 or∞ in order to represent behavior for a particular material. Thus,with η2 and all of the constants which follow it in Eq 9.15a set equal to zero, the consti-tutive equation for the standard solid, Fig. 9.5(a), will be given.

9.3 Creep and RelaxationFurther insight into the viscoelastic nature of the material making up the cube shownin Fig. 9.1 is provided by two basic experiments, the creep test and the stress relaxationtest. The creep test consists of instantaneously subjecting the material cube to a simpleshear stress t12 of magnitude T0 and maintaining that stress constant thereafter whilemeasuring the shear strain as a function of time. The resulting strain γ12(t), is called thecreep. In the stress relaxation test, an instantaneous shear strain γ12 of magnitude γ0 isimposed on the cube and maintained at that value while the resulting stress, t12(t), isrecorded as a function of time. The decrease in the stress values over the duration of thetest is referred to as the stress relaxation. Expressing these test loadings mathematicallyis accomplished by use of the unit step function, U(t− t1), defined by the equation

U(t− t1) =

1 t > t10 t 6 t1

(9.16)


U(t-t1)

1

t=t1

t

FIGURE 9.7Graphic representation of the unit step function (often called the Heaviside step function).

which is shown by the diagram in Fig. 9.7. If the creep loading is applied at time t = 0,the stress is written as

t12 = T0U (t) . (9.17)

By inserting this stress into the constitutive equation for a Kelvin material model, Eq 9.13,the resulting differential equation

T0U (t) = Gγ12 + ηγ12 (9.18)

may be integrated to yield the creep response

γ12 (t) = T0

(1− e−t/τ

) U (t)

G(9.19)

where e is the base of the natural logarithm system. It is interesting to note that ast → ∞, the strain approaches a terminal value of T0/G. Also, when t = 0, the strain rateγ12 equals T0/η and if the creep were to continue at this rate it would reach its terminalvalue at time t = τ. For this reason, τ is called the retardation time. Eq 9.19, as well as anycreep response, may always be written in the general form

γ12 (t) = J (t) T0U (t) (9.20)

in which J(t) is called the creep function.1 Thus, for the Kelvin solid, the creep function isseen to be

J (t) =1− e−t/τ

G= J

(1− e−t/τ

)(9.21)

where the constant J, the reciprocal of the shear modulus G, is called the shear compliance.As a general rule, the creep function of any viscoelastic model is the sum of the creepfunctions of its series-connected units. Thus, for the standard linear solid of Fig. 9.5(a),

J (t) = J1

(1− e−t/τ1

)+ J2 , (9.22)

and for the generalized Kelvin model of Fig. 9.6(a)

J (t) =

n∑i=1

Ji

(1− e−t/τi

). (9.23)

1Note: The creep function, J(t), is different from the Jacobian defined earlier.


With respect to creep loading for the Maxwell model, Eq 9.17 is substituted into theconstitutive equation, Eq 9.14, resulting in the differential equation

γ12 = T0δ (t)

G+ T0

U (t)

η(9.24)

where δ(t) is the Dirac delta function, the time derivative of the unit step function. Ingeneral,

δ (t− t1) =dU (t− t1)

dt, (9.25)

and is defined by the equations

δ (t− t1) = 0, t 6= t1 , (9.26a)∫t+1

t−1

δ (t− t1)dt = 1 , (9.26b)

from which it may be shown that∫t−∞ f (t′) δ (t′ − t1)dt

′ = f (t1)U (t− t1) for t > t1 (9.27)

for any continuous function, f(t). Accordingly, Eq 9.24 integrates to yield the Maxwellcreep response as

γ12 = T0J

(1+

t

τ

)U (t) (9.28)

from which the Maxwell creep function is

J (t) = J

(1+

t

τ

). (9.29)

Development of details relative to the stress relaxation test follows closely that of thecreep test. With an imposed strain at time t = 0

γ12 = γ0U (t) (9.30)

the resulting stress associated with Kelvin behavior is given directly by inserting γ12 =γ0δ (t) into Eq 9.13 resulting in

t12 = γ0 [GU (t) + ηδ (t)] . (9.31)

The delta function in this equation indicates that it would require an infinite stress at timet = 0 to produce the instantaneous strain γ0. For Maxwell behavior, when the instanta-neous strain, Eq 9.30, is substituted into Eq 9.14, the stress relaxation is the solution to thedifferential equation

t12 +1

τt12 = Gγ0δ (t) (9.32)

which upon integration using Eq 9.27 yields

t12 (t) = γ0Ge−t/τU (t) . (9.33)

The initial time-rate of decay of this stress is seen to be γ0G/τ, which if it were to continuewould reduce the stress to zero at time t = τ. Thus, τ is called the relaxation time for theMaxwell model.


Analogous to the creep function J(t) associated with the creep test we define the stressrelaxation function, G(t), for any material by expressing t12(t) in its most general form

t12 (t) = G (t)γ0U (t) (9.34)

From Eq 9.33, the stress relaxation function for the Maxwell model is

G (t) = Ge−t/τ (9.35)

and for the generalized Maxwell model it is

G (t) =

n∑i=1

Gie−t/τi . (9.36)

Creep compliance and relaxation modulus for several simple viscoelastic models aregiven in Table B.1. The differential equations governing the various models are also givenin this table.

9.4 Superposition Principle, Hereditary IntegralsFor linear viscoelasticity, the principle of superposition is valid just as in elasticity. In thecontext of stress/strain relationships under discussion here, the principle asserts that thetotal strain (stress) resulting from the application of a sequence of stresses (strains) isequal to the sum of the strains (stresses) caused by the individual stresses (strains). Thus,for the stepped stress history in simple shear displayed in Fig. 9.8(a) when applied to amaterial having a creep function J(t), the resulting strain will be

γ12 (t) = ∆T0J (t) + ∆T1J (t− t1) + ∆T2J (t− t2) =

2∑i=0

∆TiJ (t− ti) , (9.37)

and by an obvious generalization to the arbitrary stress loading considered as an infinityof infinitesimal step loadings, Fig. 9.8(b), the strain is given by

γ12 (t) =

∫t−∞ J (t− t′)

[dt12 (t′)

dt′

]dt′ (9.38)

which is called a hereditary integral since it expresses the strain at time t as a function ofthe entire stress history from time t = −∞. If there is an initial discontinuity in the stressat time t = 0, and if the stress is zero up until that time (Fig. 9.9), the strain becomes

γ12 (t) = T0J (t) +

∫t0

J (t− t′)

[dt12 (t′)

dt′

]dt′ . (9.39)

Upon integrating the integral in this equation by parts and inserting the assigned limitsof integration, the alternative form

γ12 (t) = J0t12 (t) +

∫t0

[t12 (t′)

dJ (t− t′)

d (t− t′)

]dt′ (9.40)

is obtained where J0 = J(0).


t12

∆T0

∆T1

∆T2

tt=t1

t=t2

(a) Stepped Stress

dt

dt12

t12

∆T0

t

(b) Arbitrary Stress

FIGURE 9.8Different types of applied stress histories.

dt

dt12

t12

T0

t

FIGURE 9.9Stress history with an initial discontinuity.


In a completely analogous way, we may develop hereditary integrals expressing stressas the result of arbitrary strains. The basic forms are as follows:

t12 (t) =

∫t−∞G (t− t′)

[dγ12 (t′)

dt′

]dt′ , (9.41a)

t12 (t) = γ0G (t) +

∫t0

G (t− t′)

[dγ12 (t′)

dt′

]dt′ , (9.41b)

t12 (t) = G0γ12 (t) +

∫t0

γ12 (t′)

[dG (t− t′)

d (t− t′)

]dt′ , (9.41c)

where G0 = G(0).The hereditary integral Eq 9.38, derived on the basis of simple shear, is a special case

of the general viscoelastic constitutive equations in hereditary integral form, as given bythe pair below expressed in terms of the distortional and dilatational responses

2ηij (t) =

∫t−∞ JS (t− t′)

[dSij (t

′)

dt′

]dt′ , (9.42a)

3εkk (t) =

∫t−∞ JV (t− t′)

[dtkk (t′)

dt′

]dt′ , (9.42b)

where JS is the shear compliance and JV the volumetric compliance. Likewise, Eq 9.41ais the simple shear form of the general equations

Sij (t) =

∫t−∞ 2GS (t− t′)

[dηij (t

′)

dt′

]dt′ , (9.43a)

tkk (t) =

∫t−∞ 3GV (t− t′)

[dεkk (t′)

dt′

]dt′ , (9.43b)

where GS is the relaxation modulus in shear and GV the relaxation modulus in dilatation.Because we have assumed elastic behavior in dilatation, JV = 1/GV = K and both Eq 9.42band Eq 9.43b reduce to the form tkk = 3Kεkk in keeping with Eq 9.7b.

9.5 Harmonic Loadings, Complex Modulus, and Complex ComplianceThe behavior of viscoelastic bodies when subjected to harmonic stress or strain is anotherimportant part of the theory of viscoelasticity. To investigate this aspect of the theory, weconsider the response of the material cube shown in Fig. 9.1 under an applied harmonicshear strain of frequency ω as expressed by

γ12 (t) = γ0 sinωt , (9.44a)

or byγ12 (t) = γ0 cosωt . (9.44b)

Mathematically, it is advantageous to combine these two by assuming the strain in thecomplex form

γ12 (t) = γ0 (cosωt+ i sinωt) = γ0eiωt (9.45)


where i =√

−1. It is understood that physically the real part of the resulting stresscorresponds to the real part of the applied strain, and likewise, the imaginary parts ofeach are directly related. The stress resulting from the excitation prescribed by Eq 9.45will have the same frequency, ω, as the imposed strain. Therefore, expressing the stressas

t12 (t) = T∗eiωt (9.46)

where T∗ is complex, the response will consist of two parts: a steady-state response whichwill be a function of the frequency ω, and a transient response that decays exponentiallywith time. It is solely with the steady-state response that we concern ourselves in theremainder of this section.

Substituting Eq 9.45 and Eq 9.46 into the fundamental viscoelastic constitutive equationgiven by Eq 9.8, and keeping in mind the form of the operators P and Q as listed byEq 5.178 we obtain

N∑k=0

T∗pk (iω)k eiωt =

M∑k=0

γ0qk (iω)k eiωt . (9.47)

Canceling the common factor eiωt we solve for the ratio

T∗

γ0=

∑Nk=0 qk (iω)k∑Mk=0 pk (iω)k

(9.48)

which we define as the complex modulus, G∗(iω), and write it in the form

T∗

γ0= G∗ (iω) = G′ (iω) + iG′′ (iω) . (9.49)

The real part, G′(iω), of this modulus is associated with the amount of energy stored inthe cube during a complete loading cycle and is called the storage modulus. The imaginarypart, G′′(iω), relates to the energy dissipated per cycle and is called the loss modulus.

In terms of the complex modulus, the stress t12 as assumed in Eq 9.46 may now bewritten

t12 = G∗ (iω)γ0eiωt = [G′ (ω) + iG′′ (ω)]γ0e

iωt , (9.50)

and by defining the absolute modulus, G (ω) as the magnitude of G∗(iω) according to

G (ω) =

√[G′ (ω)]2 + [G′′ (ω)]2 (9.51)

together with the loss angle, δ between G (ω) and G′(ω) as given by its tangent

tan δ =G′′ (ω)

G′ (ω)(9.52)

the stress t12 (Eq 9.50), may now be written

t12 = G (ω) eiδγ0eiωt = G (ω)γ0e

i(ωt+δ) . (9.53)

From this equation we see that the peak value of the stress is

T0 = G (ω)γ0 , (9.54)

and that the strain lags behind the stress by the loss angle δ. Figure 9.10 provides agraphical interpretation of this phenomenon. In Fig. 9.10(a), the two constant magnitude


δ ωtγ

0

ωt-δ

ωO

T0

(a)

2π

ω

π

ω

δ

ω

Ot

γ0 sin (ωt− δ)

T0 sinωt

(b)

FIGURE 9.10Different types of applied stress histories.

stress and strain vectors, separated by the constant angle δ, rotate about a fixed originwith a constant angular velocity ω. The vertical projections of these vectors, representingthe physical values of the stress and strain, are plotted against time in Fig. 9.10(b). FromFig. 9.10(a), the portion of the stress in phase with the strain is T0 cos δ and by Eq 9.54together with Eq 9.52, the storage modulus may be expressed as

G′ =T0 cos δγ0

. (9.55a)

Similarly, the loss modulus is written as

G′′ =T0 sin δγ0

(9.55b)

Consistent with the duality present in all of viscoelastic theory we reverse the roles ofstress and strain in the preceding portion of this section to define the complex compliance,J∗(iω) along with its associated real and imaginary parts. Briefly, we assume an appliedstress

t12 = T0eiωt (9.56)

together with the resulting strainγ12 = γ∗eiωt (9.57)

which when substituted into Eq 9.8 leads to

γ∗

T0= J∗ (iω) = J′ (ω) − iJ′′ (ω) (9.58)

where the minus sign reflects the fact that the strain lags the stress by the loss angle δ,defined in this case by

tan δ =J′′ (ω)

J′ (ω). (9.59)

In analogy with the complex modulus components, J′(ω) is called the storage compliance,J′′(ω) the loss compliance, and

J (ω) =

√[J′ (ω)]2 + [J′′ (ω)]2 =

γ0

T0(9.60)


the absolute compliance, in which γ0 is the peak value of the strain as given by

γ0 = J (ω) T0 (9.61)

Based upon the definitions of G∗ and J∗, it is clear that these complex quantities arereciprocals of one another. Thus

G∗ (iω) J∗ (iω) = 1 . (9.62)

A simple procedure for calculating G∗ (or J∗) of a specific model or material is to replacethe partial differential operator ∂t in the material’s constitutive equation by iω and solvethe resulting algebraic equation for the ratio t12/γ12 = G∗, (or γ12/t12 = J∗ if that isthe quantity required). Accordingly, from Eq 9.13 for the Kelvin solid, the constitutiveequation t12 = G + η∂tγ12 becomes t12 = (G + iηω)γ12 which yields t12/γ12 = G(1 +iτω) = G∗. Likewise, from Eq 9.14 for the Maxwell fluid, ∂t + 1/τt12 = G∂tγ12becomes (iω+1/τ)t12 = (Giω)γ12 from which t12/γ12 = G(τ2ω2+iτω)/(1+τ2ω2) = G∗.

In developing the formulas for the complex modulus and the complex compliancewe have used the differential operator form of the fundamental viscoelastic constitutiveequations. Equivalent expressions for these complex quantities may also be derived usingthe hereditary integral form of constitutive equations. To this end we substitute Eq 9.45into Eq 9.41a. However, before making this substitution it is necessary to decompose thestress relaxation function G(t) into two parts as follows,

G(t) = G0[1− φ(t)] (9.63)

where G0 = G(0), the value of G(t) at time t = 0. Following this decomposition and theindicated substitution, Eq 9.41a becomes

t12 (t) = iωγ0G0

∫t−∞ eiωt

′[1− φ (t− t′)]dt′ . (9.64)

In this equation, let t − t′ = ξ so that dt′ = −dξ and such that when t′ = t, ξ = 0, andwhen t′ = −∞, ξ =∞. Now

t12 (t) = iωγ0G0

eiωt

iω−

∫∞0

eiωξeiωtφ (ξ)dξ

(9.65)

which reduces to

t12 (t) = γ0

G0 −G0

∫∞0

(iω cosωξ+ω sinωξ)φ (ξ)dξ

eiωt . (9.66)

But by Eq 9.50, t12(t) = γ0[G′(ω) + iG′′(ω)]eiωt so that from Eq 9.66

G′ (ω) = G0 −ωG0

∫∞0

sinωξφ (ξ)dξ , (9.67a)

G′′ (ω) = −ωG0

∫∞0

cosωξφ (ξ)dξ . (9.67b)

For a Kelvin material, G(t) = G[1+ τδ(t)] so that G0 = G and φ(t) = −τδ(t). Thus, fora Kelvin solid, Eqs 9.67 yield

G′ (ω) = G−Gω

∫∞0

−τδ(ξ) sinωξdξ = G ,


G′′ (ω) = −Gω

∫∞0

−τδ(ξ) cosωξdξ = Gωτ = ωη .

For a Maxwell material, G(t) = G[1 − (1 − e−t/τ)] so that G0 = G and φ(t) = 1 − e−t/τ.Thus, for a Maxwell fluid, Eqs 9.67 yield

G′ (ω) = G−Gω

∫∞0

(1− e−ξ/τ

)sinωξdξ

= G−Gω

[1

ω−

ωτ2

1+ τ2ω2

]=

Gω2τ2

1+ τ2ω2,

G′′ (ω) = −Gω

∫∞0

(1− e−ξ/τ

)cosωξdξ =

Gωτ

1+ τ2ω2.

These values for G′ and G′′ for the Kelvin and Maxwell models agree with those calcu-lated in the previous paragraph.

In a completely analogous fashion, if we adopt the hereditary form Eq 9.38 as theconstitutive equation of choice, and substitute into that equation t12 = T0e

iωt, togetherwith the decomposition of J(t) in the form J(t) = J0[1+ψ(t)], we obtain

γ12 (t) =

∫t−∞ J0 [1−ψ (t− t′)] iωT0e

iωt′dt′ . (9.68)

Upon making the same change in variable of integration, t−t′ = ξ, this equation becomes

γ12 (t) = T0

[J0 + J0ω

∫∞0

sinωξψ (ξ)dξ+ iJ0ω

∫∞0

cosωξψ (ξ)dξ

]eiωt (9.69)

from which we extract

J′ (ω) = J0 + J0ω

∫∞0

sinωξψ (ξ)dξ , (9.70a)

J′′ (ω) = −J0ω

∫∞0

cosωξψ (ξ)dξ . (9.70b)

These expressions may be specialized to obtain J′(ω) and J′′(ω) for any particular modelfor which the creep function J(t) is known.

9.6 Three-Dimensional Problems, The Correspondence PrincipleThe fundamental viscoelastic constitutive equations in differential operator form as ex-pressed by Eqs 9.6 distinguish between the distortional response (a change in shape atconstant volume due to the deviatoric portion of the applied stress, Eq 9.6a), and thedilatational response (a change in volume without a change in shape due to the sphericalportion of the applied stress, Eq 9.6b). Extensive experimental evidence indicates thatpractically all materials of engineering importance behave elastically in the dilatationalmode, and for this reason the reduced form of Eq 9.6 as given by Eq 9.7 is used in thisbook. In expanded component notation these equations appear as

P(t11 − 1

3tii)

= 2 Q(ε11 − 1

3εii), (9.71a)


P(t22 − 1

3tii)

= 2 Q(ε22 − 1

3εii), (9.71b)

P(t33 − 1

3tii)

= 2 Q(ε33 − 1

3εii), (9.71c)

P t12 = 2 Q ε12 , (9.71d)

P t23 = 2 Q ε23 , (9.71e)

P t31 = 2 Q ε31 , (9.71f)

tii = 3Kεii . (9.71g)

Depending upon the particular state of applied stress, some of Eq 9.71 may be satisfiedidentically. For example, the state of simple shear in the x1-x2 plane, introduced earlierto develop the basic concepts of viscoelastic behavior, results in Eq 9.71 being reducedto a single equation, Eq 9.71d as expressed by Eq 9.8. Similarly, for a hydrostatic stateof stress with t11 = t22 = t33 = p0, and t12 = t23 = t31 = 0 (or for a uniform triaxialtension having t11 = t22 = t33 = T0, with t12 = t23 = t31 = 0), the behavior is simplyelastic. On the other hand, for a simple one-dimensional tension or compression in oneof the coordinate directions, several of Eq 9.71 enter into the analysis as discussed in thefollowing paragraph.

Let an instantaneously applied constant stress T0 be imposed uniformly in the x1 direc-tion on a member having a constant cross section perpendicular to that direction. Thus,let t11 = T0U(t) with all other components zero which results in tii = T0U(t) so that fromEq 9.71g, εii = T0U(t)/3K. For this situation the first three of Eq 9.71 become

P(T0U (t) − 1

3T0U (t))

= 2 Q

(ε11 −

T0U (t)

9K

), (9.72a)

P(−13T0U (t)

)= 2 Q

(ε22 −

T0U (t)

9K

), (9.72b)

P(−13T0U (t)

)= 2 Q

(ε33 −

T0U (t)

9K

), (9.72c)

and the next three of the set indicate zero shear strains. Clearly, from Eq 9.72b andEq 9.72c we see that ε22 = ε33. Furthermore, Eq 9.72a may be solved directly for ε11 toyield

ε11 = T0U (t)3K P + Q

9K Q=T0U (t)

E(9.73)

where E = 9KQ/(3KP + Q) is the operator form of E, Young’s modulus. Similarly,from Eq 9.72 using ε22 = ε33 we find that

ε22

ε11=ε33

ε11= −

3K P − 2 Q

6K P + 2 Q= − ν (9.74)

which designates the operator form of Poisson’s ratio, ν.In order to compute a detailed solution for a particular material we need the specific

form of the operators P and Q for that material as illustrated by the following example.

Example 9.1

Let the stress t11 = T0U(t) be applied uniformly to a bar of constant crosssection made of a Kelvin material and situated along the x1 axis. Determineε11 and ε22 as functions of time.


SolutionFrom the constitutive relation for a Kelvin material, Eq 9.13, we note thatP = 1 and Q = G + η∂t, which when inserted into Eq 9.72a results (aftersome algebraic manipulations) in the differential equation

ε11 +ε11

τ= T0U (t)

[3K+G

9KG

]+T0δ (t)

9K. (9.75)

This differential equation may be solved by standard procedures to yield thesolution

ε11 (t) = T0U (t)

3K+G

9KG

(1− e−t/τ

)+e−t/τ

9K

. (9.76)

When t = 0, ε11 = T0U(t)/9K which is the result of elastic behavior in bulk.As t→∞, ε11 → T0[(3K+G)/9KG] = T0/E, the terminal elastic response.

From Eq 9.72b the governing differential equation for determining ε22 is(when expressed in its standard form)

ε22 +ε22

τ= T0U (t)

[G

9K−1

6

]1

η+T0δ (t)

9K(9.77)

which upon integration and simplification yields

ε22 (t) = −T0U (t)3K− 2G

18KG

(1− e−t/τ

)− T0U(t)

e−t/τ

9K. (9.78)

When t = 0, ε22 = T0/9K which, due to the elastic dilatation effect, is identicalwith the initial value of ε11. As t→∞, ε22 → (2G− 3K)/18KG.

Up until now in this section we have discussed three-dimensional problems from thepoint of view of constitutive equations in differential operator form, but our analysis canbe developed equally well on the basis of the hereditary integral form of constitutiveequations as given by Eq 9.42 or Eq 9.43. With respect to the uniaxial stress loadinganalyzed above, Eq 9.42a (assuming elastic behavior in dilatation with tkk = 3Kεkk andεii = T0U(t)/3K, along with zero stress at time t = 0) results in the equations

2

[ε11 −

T0U (t)

9K

]=

∫t0

(T0 −

T0

3

)δ (t′) JS (t− t′)dt′ , (9.79a)

2

[ε22 −

T0U (t)

9K

]=

∫t0

(−T0

3

)δ (t′) JS (t− t′)dt′ , (9.79b)

2

[ε33 −

T0U (t)

9K

]=

∫t0

(−T0

3

)δ (t′) JS (t− t′)dt′ . (9.79c)

From these equations it is again apparent that ε22 = ε33, and that in order to developthe solution details for a particular material we need the expression for JS, the shearcompliance of that material as shown by the example that follows.


Example 9.2

Develop the solution for the problem of Example 9.7-1 using the hereditaryintegral form of constitutive equations for a bar that is Kelvin in distortion,elastic in dilatation.

SolutionFor a Kelvin material, the shear (creep) compliance, Eq 9.21, is JS = (1 −e−t/τ)/G so that Eq 9.79a becomes for t11 = T0U(t)

2

[ε11 −

T0U (t)

9K

]=

∫t0

2T0δ (t′)

3G

(1− e− t−t′

τ

)dt′ (9.80)

which may be integrated directly using Eq 9.27 to yield

ε11 (t) = T0U (t)

[1− e− t

τ

3G+1

9K

], (9.81)

or by a simple rearrangement

ε11 (t) = T0U (t)

[3K+G

9KG−e− t

τ

3G

]

in agreement with Eq 9.76. Likewise, from Eq 9.79b we obtain for this loading

2

[ε22 −

T0U (t)

9K

]=

∫t0

−T0δ (t′)

3G

(1− e− t−t′

τ

)dt′ (9.82)

which also integrates directly using Eq 9.27 to yield

ε22 (t) = −T0U (t)

[1− e− t

τ

6G+1

9K

](9.83)


The number of problems in viscoelasticity that may be solved by direct integration asin the examples above is certainly quite limited. For situations involving more generalstress fields, or for bodies of a more complicated geometry, the correspondence principlemay be used to advantage. This approach rests upon the analogy between the basicequations of an associated problem in elasticity and those of the Laplace transforms ofthe fundamental equations of the viscoelastic problem under consideration. For the caseof quasi-static viscoelastic problems in which inertia forces due to displacements may beneglected, and for which the elastic and viscoelastic bodies have the same geometry, thecorrespondence method is relatively straightforward as described below. In an elasticbody under constant load, the stresses, strains, and displacements are independent oftime, whereas in the associated viscoelastic problem, even though the loading is constantor a slowly varying function of time, the governing equations are time dependent andmay be subjected to the Laplace transformation. By definition, the Laplace transform of


an arbitrary continuous time-dependent function, say the stress tij(x, t) for example, isgiven by

tij (x, s) =

∫t0

tij (x, t) e−stdt (9.84)

in which s is the transform variable, and barred quantities indicate transform. Standardtextbooks on the Laplace transform list tables giving the transforms of a wide varietyof time-dependent functions. Of primary importance to the following discussion, theLaplace transforms of the derivatives of a given function are essential. Thus, for the firsttwo derivatives of stress∫t

0

dtij (x, t)

dte−stdt = −tij (0) + stij (x, s) , (9.85a)

∫t0

d2tij (x, t)

dt2e−stdt = −tij (0) − stij (0) + s2tij (x, s) , (9.85b)

and so on for higher derivatives. Note that in these equations the values of the stress andits derivatives at time t = 0 become part of the transform so that initial conditions arebuilt into the solution.

We may now examine the details of the correspondence method by considering thecorrespondence between the basic elasticity equations (as listed in Chapter 6 and repeatedhere)

Equilibrium (Eq 6.43) tij,j(x) + ρbi(x) = 0 ,

Strain− displacement (Eq 6.44) 2εij(x) = ui,j(x) + uj,i(x) ,

Constitutive relations (Eq 6.33a) Sij(x) = 2Gηij(x) ,

(Eq 6.33b) tii(x) = 3Kεii(x) ,

and the Laplace transforms of the associated time-dependent viscoelastic equations

Equilibrium tij,j(x, s) + ρbi(x, s) = 0 , (9.86a)

Strain− displacement 2εij(x, s) = ui,j(x, s) + uj,i(x, s) , (9.86b)

Constitutive relations P (s) Sij(x, s) = 2Q (s) ηij(x, s) , (9.86c)

tii(x, s) = 3Kεii(x, s) , (9.86d)

where barred quantities are transforms, and in which P(s) and Q(s) are polynomials inthe transform variable s in accordance with Eq 9.85. A comparison of Eqs 9.86, whichare algebraic, time-independent relations, with the elasticity equations above indicates acomplete analogy between the barred and unbarred entities if we assign the equivalencyof the ratio Q(s)/P(s) to the shear modulus G. This allows us to state the correspondenceprinciple as follows:

If the solution of a problem in elasticity is known, the Laplace transform of thesolution of the associated viscoelastic problem is constructed by substitutingthe quotient Q(s)/P(s) of the transformed operator polynomials in place ofthe shear modulus G, and the actual time-dependent loads by their Laplacetransforms.

Since many elasticity solutions are written in terms of Young’s modulus, E and Poisson’sratio, ν, it is useful to extract from Eqs 9.73 and 9.74 the transform replacements for theseconstants which are

E (s)→ 9KQ

3KP + Q, (9.87)


ν (s)→ 3KP − 2Q

6KP + 2Q. (9.88)

As an illustration of how the correspondence method works, we consider the followingproblem.

Example 9.3

The radial stress trr in an elastic half-space under the action of a concentratedconstant force F0 acting at the origin as shown in the figure below is

trr (r, z) =F0

2π[(1− 2ν)A (r, z) − B (r, z)]

where A(r, z) and B(r, z) are known functions of the coordinates. Determinethe time-dependent viscoelastic stress trr(r, z, t) for a half-space that is Kelvinin shear and elastic in dilatation if the force at the origin is given by the steploading F(t) = F0U(t).

z

xφ

y r

F0U(t)

SolutionFrom Eq 9.88 for ν(s) we may directly calculate the Laplace transform of theexpression for 1 − 2ν appearing in the elastic solution as 3Q/

(3KP + Q

). The

Laplace transform of the load function F0U(t) is given by F0/s. Thus, theLaplace transform of the associated viscoelastic solution is

trr (r, z, s) =F0

2πs

[3Q

3KP + QA (r, z) − B (r, z)

].

From the Kelvin model constitutive equation, Eq 9.13, we have Q/P = G+η∂t for which Q(s)/P(s) = G+ ηs so that

trr (r, z, s) =F0

2πs

[3 (G+ ηs)

3K+G+ ηA (r, z) − B (r, z)

].

This expression may be inverted with the help of a table of transforms from anystandard text on Laplace transforms to give the time-dependent viscoelastic


solution

trr (r, z, t) =F0

2π

3G

3K+G+ 9Ke− (3K+G)t

η

3K+G

A (r, z) − B (r, z)

.

Notice that when t = 0

trr (r, z, 0) =F0

2π[3A (r, z) − B (r, z)] ,

and as t→∞trr (r, z, t→∞) =

F0

2π

[3G

3K+GA (r, z) − B (r, z)

]which is the elastic solution.

References[1] Ferry, J. D. (1961), Viscoelastic Properties of Polymers, Wiley and Sons, New York

[2] Findley, W. N., Lai, J. S., and Onanran, O. (1976), Creep and Relaxation of NonlinearViscoelastic Materials, North-Holland Publishing Company, London

[3] Flugge, W. (1967), Viscoelasticity, Blaisdell Publishing Company, Waltham, MA

[4] Fried, J.R. (1995), Polymer Science and Technology, Prentice Hall PTR, Upper SaddleRiver, NJ

[5] McCrum, N. G., Buckley, C. P., and Bucknall, C. B. (1997), Principles of PolymerEngineering, Second Edition, Oxford University Press, New York

[6] Pipkin, A. C. (1972), Lectures on Viscoelasticity Theory, Springer-Verlag, New York


Problems

Problem 9.1By substituting Sij = tij − 1

3δijtkk and ηij = εij − 13δijεkk into Eq 9.7 and combining

those two equations, determine expressions in operator form for

(a) the Lame constant, λ,(b) Young’s modulus, E,(c) Poisson’s ratio, ν.

Answer

(a) λ = K− 2Q/3P

(b) E = 9KQ/(3KP + Q)

(c) ν = (3KP − 2Q)/(6KP + 2Q)

Problem 9.2Compliances are reciprocals of moduli. Thus, in elasticity theory D = 1/E, J = 1/G, andB = 1/K. Show from the stress-strain equations of a simple one-dimensional tension that

D =1

3J+

1

9B .

Problem 9.3The four-parameter model shown consists of a Kelvin unit in series with a Maxwell unit.

Knowing that γMODEL = γKELVIN +γMAXWELL, together with the operator equations Eqs 9.13and 9.14, determine the constitutive equation for this model.

G2

η2

G1

η1

T T

Answer

G2η1γ+G1G2γ = η1T + (G1 +G2 + η1/τ2) T + (G1/τ2) T

Problem 9.4Develop the constitutive equations for the three-parameter models shown.

G1

η1

η2T T

(a)

G2

G1

η2

T T

(b)


G2

η1

η2

TT

(c)

Answer

(a) γ+ γ/τ1 = [(η1 + η2) /η1η2] T + (1/τ1η2) T

(b) T + T/τ2 = (G2 +G1) γ+ (G1/τ2)γ

(c) T + T/τ2 = η1γ+ (G2 + η1/τ2) γ

Problem 9.5A proposed model consists of a Kelvin unit in parallel with a Maxwell unit. Determine the

constitutive equation for this model.

G2

G1

η1

η2

T T

Answer

T + T/τ2 = η1γ+ (G1 +G2 + η1/τ2) γ+ (G1/τ2)γ

Problem 9.6For the four-parameter model shown, determine

(a) the constitutive equation,

(b) the relaxation function, G(t). [Note that G(t) is the sum of the G(t)’s of theparallel joined units.]

G1

η3

G2

η2T T

Answer

(a) T + T/τ2 = η3γ+ (G1 +G2 + η3/τ2) γ+ (G1/τ2)γ

(b) G(t) = G1 +G2e−t/τ2 + η3δ (t)


Problem 9.7For the model shown the stress history is given by the accompanying diagram. Determine

the strain γ(t) for this loading during the intervals

(a) 0 6 t/τ 6 2,(b) 0 6 t/τ 6 4.

Use superposition to obtain answer (b).

G

η

GT T

10

T0

t

t/τ2 3 4 5

Answer

(a) γ(t) = T0J(2− e−t/τ

)U (t)

(b) γ(t) = T0J(2− e−t/τ

)U (t) − T0J

(2− e−(t−2τ)/τ

)U (t− 2τ)

Problem 9.8For the model shown determine

(a) the constitutive equation,(b) the relaxation function, G(t),(c) the stress, T(t) for 0 6 t 6 t1, when the strain is given by the accompanying

graph.

G

η

G ηT T

γ

t

1

0

λ

Answer

(a) T + T/τ = ηγ+ 3Gγ+ (G/τ)γ

(b) G(t) = ηδ (t) +G(1+ e−t/τ2

)(c) T (t) = λ

(2η− 2e−t/τGt

)U (t)

Problem 9.9For the model shown in Problem 9.8, determine T(t) when γ(t) is given by the diagram

shown here.

γ

γ0

t

1

0

λ


Answer

T(t) = γ0[ηδ (t) +G

(1+ e−t/τ

)]U (t) + λ

[η(2− e−t/τ

)+Gt

]U (t)

Problem 9.10The three-parameter model shown is subjected to the strain history pictured in the graph.

Use superposition to obtain T(t) for the t > t1, from T(t) for t 6 t1. Let γ0/t1 = λ.

G2

G1

η1

T T

γ

γ0

tt10

Answer

For t 6 t1; T(t) = λ[η1(1− e−t/τ1

)+G2t

]U(t)

For t > t1; T(t) = λ[η1(1− e−t/τ1

)+G2t

]U(t)

−λ[η1(1− e−(t−t1)/τ1

)+G2 (t− t1)

]U(t− t1)

Problem 9.11For the model shown determine the stress, T(t) at (a) t = t1; (b) t = 2t1; and (c) t = 3t1,

if the applied strain is given by the diagram. Use superposition for (b) and (c).

G

η

η

T T

γ

γ0

3t1

t1

2t1

0

Answer

(a) T(t1) = (γ0η/t1)[2− e−t1/τ

](b) T(2t1) = (γ0η/t1)

[−2+ 2e−t1/τ − e−2t1/τ

](c) T(3t1) = (γ0η/t1)

[−e−t1/τ + 2e−2t1/τ − 3e−3t1/τ

]Problem 9.12If the model shown in the sketch is subjected to the strain history γ (t) = (σ0/2G)

[2− e−t/2τ

]U (t),

as pictured in the time diagram, determine the stress, T(t).

G

G

η

T T

γ

t0

γ (t) =T0

2G

2− e−

t2τ

T0

2G


Answer

T(t) = T0(e−t/τ +U(t)

)Problem 9.13For the hereditary integral, Eq 9.38

γ (t) =

∫t−∞ J (t− t′) (dT (t′) /dt′)dt′

assume T(t) = est where s is a constant. Let τ = t − t′ be the “elapsed time” of the loadapplication and show that γ(t) = sestJ (s) where J (s) is the Laplace transform of J(t).

Problem 9.14Using T(t) = est as in Problem 9.13, together with the hereditary integral Eq 9.41a

T (t) =

∫t−∞G (t− t′) (dγ (t′) /dt′)dt′ ,

and the result of Problem 9.13 show that G (s) J (s) = 1/s2 where G (s) is the Laplacetransform of G(t). Assume s is real.

Problem 9.15Taking the hereditary integrals for viscoelastic behavior in the form Eq 9.40

γ (t) = J0T (t) +

∫t0

T (t′) [dJ (t− t′) /d (t− t′)]dt′

and Eq 9.41c

T (t) = G0γ (t) +

∫t0

γ (t′) [dG (t− t′) /d (t− t′)]dt′ ,

show that for the stress loading T(t) = est and with τ = t− t′, the expression

G0A (s) + J0B (s) + A (s) B (s) = 0

results, where here

A (s) =

∫∞0

e−sτ (dJ/dτ)dτ

andB (s) =

∫∞0

e−sτ (dG/dτ)dτ .

Problem 9.16Let the stress relaxation function be given as G(t) = a(b/t)m where a, b, and m

are constants and t is time. Show that the creep function for this material is J (t) =1

amπ sinmx(tb

)m with m < 1. Use the identity G (s) J (s) = 1/s2 where barred quantitiesare Laplace transforms.

Problem 9.17A three-parameter solid has the model shown. Derive the constitutive equation for this

model and from it determine (a) the relaxation function, and (b) the creep function for themodel.


G1

G2

η1

T T

Answer

(a) G(t) = G2 +G1e−t/τ1

(b) J (t) = (1/ (G1 +G2)) e−t/τ∗1 + (1/G2)

(1− e−t/τ∗1

)where τ∗1 = (G1 +G2) τ1/G2

Problem 9.18

A material is modeled as shown by the sketch. (a) For this model determine the relaxationfunction, G(t). (b) If a ramp function strain as shown by the diagram is imposed on themodel, determine the stress, using the appropriate hereditary integral involving G(t).

G

G

2G

2η

ηT T

γ(t)

t

1

0

λ

Answer

(a) G(t) = G+ 2Ge−2t/τ +Ge−t/2τ

(b) T(t) = Gλ[t+ 3τ− τe−2t/τ − 2τe−t/2τ

]U(t)

Problem 9.19

Determine the complex modulus, G∗(iω) for the model shown using the substitution iωfor ∂t in the constitutive equation.

G1

G2

η1

T T

Answer

G∗ (iω) =[G2 + (G1 +G2) τ

21ω

2 + iG1τ1ω]/(1+ τ21ω

2)

Problem 9.20

Show that, in general, J′ = 1/G′(1+ tan 2δ) and verify that G′ and J′ for the Kelvin modelsatisfies this identity. (Hint: Begin with G∗J∗ = 1.)


Problem 9.21Let the complex viscosity (denoted here by η∗(iω)) be defined through the equation

T0eiωt = η∗

[iωγ0e

iωt].

Determine η∗ (iω) in terms of G∗(iω) (see Eq 9.49) and calculate η∗ (iω) for the modelshown below.

G

η

η

T T

Answer

η∗ (iω) =[η(2+ω2τ2

)− iτηω

]/(1+ω2τ2

)Problem 9.22From Eq 9.55a in which G′ = T0 cos δ/γ0 and 9.55b in which G′′ = (T0 sin δ)/γ0 show thatJ′ = (γ0 cos δ)/T0 and that J′′ = (γ0 sin δ)/T0. Use G∗J∗ = 1.

Problem 9.23Show that the energy dissipated per cycle is related directly to the loss compliance, J′′

by evaluating the integral∫Tdγ over one complete cycle assuming T(t) = T0 sinωt. (See

Fig. 9.10.)

Answer∫Tdγ = T20πJ

′′

Problem 9.24For the rather complicated model shown here, determine the constitutive equation and

from it G∗(iω). Sketch a few points on the curve G′′ vs. ln(ωτ).

G

G

2G

2η

ηT T

Answer

T + (5/2τ) T +(1/τ2

)T = 4Gγ+ (11G/2τ) γ+

(G/τ2

)γ

G∗(iω) = G′ + iG′′ where

G′ = G[1+ (35/4) τ2ω2 + 4τ4ω4

]/[1+ (17/4) τ2ω2 + τ4ω4

]


G′′ = G[3τω+ (9/2) τ3ω3

]/[1+ (17/4) τ2ω2 + τ4ω4

]For ln(ωτ) = 0, G′′ = 1.2G

For ln(ωτ) = 1, G′′ = 1.13G

For ln(ωτ) = 2, G′′ = 0.572G

For ln(ωτ) =∞, G′′ = 0

Problem 9.25

A block of viscoelastic material in the shape of a cube fits snugly into a rigid container.A uniformly distributed load p = −p0U(t) is applied to the top surface of the cube. If thematerial is Maxwell in shear and elastic in dilatation determine the stress component t11(t)using Eq 9.71. Evaluate t11(0) and t11(∞).

x1

x2

p = −p0U (t)

Answer

t11(t) = −p0[1− (6G/(3K+ 4G))]e−(3K/(3K+4G)τ)t]U(t)

t11(0) = −p0[(3K− 2G)/(3K+ 4G)]

t11(∞) = −p0

Problem 9.26

A slender viscoelastic bar is loaded in simple tension with the stress T(t) = t11(t) = T0U(t).The material may be modeled as a standard linear solid in shear having the model shown,and as elastic in dilatation. Using the hereditary integrals, Eq 9.42, determine the axialstrain ε11(t) and the lateral strain ε22(t).

GG

η

T T t11

t11

Answer

ε11(t) = T0[(6K+G)/3K− e−t/τ]/3GU(t)

ε22(t) = T0[(1/9K) − (2− e−t/τ)/6G)]U(t)


Problem 9.27A cylinder of viscoelastic material fits snugly into a rigid container so that ε11 = ε22 =εrr = 0 (no radial strain). The body is elastic in dilatation and has a creep complianceJS = J0(1+ t) with J0 a constant. Determine t33(t) if ε33 = A (a constant).

x1

x2

r

x3

Answer

t33(t) = A[Kt+ 4(1− e−t)/3J0]U(t)

Problem 9.28A viscoelastic body in the form of a block is elastic in dilatation and obeys the Maxwell

law in distortion. The block is subjected to a pressure impulse t11 = −p0δ(t) distributeduniformly over the x1 face. If the block is constrained so that ε22 = ε33 = 0, determineε11(t) and t22(t).

t11

= −poδ(t)

t22

t33

Answer

t22(t) = p0[(2G− 3K)δ(t)/(3K+ 4G) − (6G/(3K+ 4G))e[−3K/(3K+4G)τ]t]U(t)

ε11(t) = p0[3δ(t)/(3K+ 4G) − [4G/(3K+ 4G)K]e[−3K/(3K+4G)τ]t]U(t)

Problem 9.29A viscoelastic cylinder is inserted into a snug fitting cavity of a rigid container. A flat,

smooth plunger is applied to the surface x1 = 0 of the cylinder and forced downward at


a constant strain rate ε11 = ε0. If the material is modeled as the three-parameter solidshown in shear and as elastic in dilatation, determine t11(t) and t22(t) during the downwardmotion of the plunger.

G

G η

T T

x1

x2

x3

Answer

t11(t) = −ε0[(4Gτ/3)(1− e−t/τ) + (K+ 4G/3)t]U(t)

t22(t) = ε0[(2Gτ/3)(1− e−t/τ) + (−K+ 2G/3)t]U(t)

Problem 9.30For a thick-walled elastic cylinder under internal pressure p0, the stresses are trr = A−B/r2;tθθ = A+B/r2 and the radial displacement is given by ur = 1+ν

E [A(1−2ν)r+B/r] where Aand B are constants involving p0, E is Young’s modulus, and ν is Poisson’s ratio. Determinetrr, tθθ , and ur for a viscoelastic cylinder of the same dimensions that is Kelvin in shearand elastic dilatation if p = p0U(t).

tθθ

tθθ

trr

trr

p0

r

Answer

trr = same as elastic solution but with p0 now p0U(t)

tθθ = same as elastic solution but with p0 now p0U(t)

ur(t) = (3Ar/(6K+ 2G))(1− e−(3K+G)t/Gτ))U(t) + (B/2Gr)(1− e−t/τ)U(t)

Problem 9.31A viscoelastic half-space is modeled as Kelvin in shear, elastic in dilatation. If the point

force P = P0e−t is applied at the origin of a stress free material at time t = 0, determine

trr(t) knowing that for an elastic half-space the radial stress is

trr = P0[(1− 2ν)A− B]/2π

where A and B are functions of the coordinates only.


x3

x2

x1

r

P = P0e−t

θ

Answer

trr = (P0/2π)[(3(G− η)e−t + 9Ke−(3K+G)t/η]A/(3K+G− η) − e−tB

Problem 9.32The deflection at x = L for an end-loaded cantilever elastic beam is w = P0L

3/3EI.Determine the deflection w(L, t) for a viscoelastic beam of the same dimensions if P =P0U(t) assuming (a) one-dimensional analysis based on Kelvin material, and (b) three-dimensional analysis with the beam material Kelvin in shear, elastic in dilatation. Checkw(L,∞) in each case.

P0

L

Answer

(a) w(L, t) = (P0L3/3EI)(1− e−t/τE)U(t), τE = η/E

w(L,∞) = (P0L3/3EI), the elastic deflection.

(b) w(L, t) = (P0L3/3EI)[(1− e−t/τ) + (1/9K)e−t/τ]U(t)

w(L,∞) = (P0L3/3EI), the elastic deflection.

Problem 9.33A simply-supported viscoelastic beam is subjected to the time-dependent loading f(x, t) =q0t where q0 is a constant and t is time. Determine the beam deflection w(x, t) in termsof the elastic beam shape X(x) if the beam material is assumed to be (a) one-dimensionalKelvin, and (b) three-dimensional Kelvin in shear, and elastic in dilatation. Compare theresults.

q0t

L


Answer

(a) w(x, t) = X(x)[t− τE(1− e−t/τE)

]U(t)

(b) w(x, t) = X(x)[t− (3τE/ (3K+G)) (1− e−t/τE)

]U(t)

where τE = η/E

Appendix A: General Tensors

The balance laws and the constitutive relations describing the behavior of continuousmedia were formulated using direct or coordinate free notation. This is a recognition ofthe invariance of the physical principles. These laws were written using indicial notationin a rectangular-Cartesian coordinate frame. The study of tensors in other coordinatesystems may be better suited for solving particular boundary value problems. Thesecoordinate frames might provide ways to describe vectors and tensors that better describethe geometry of the body. The study of general tensors also aids in understanding muchof the literature that uses general tensors.

It would be convenient if there was a way to convert the Cartesian equations to generalcoordinate systems without a great deal of effort. It is the purpose of this appendix togive such a scheme. The following section begins by developing ideas about the generalrepresentation of vectors. Next methods are given for taking derivatives in general coor-dinate systems. Then a set of rules is introduced that will allow one to take an expressionin Cartesian coordinates and convert it to an expression in a general coordinate system.

A.1 Representation of Vectors in General BasesVectors are quantities that do not depend on a coordinate system. The representation ofvectors does depend on the basis for computing the components of a vector. Any set ofthree non-coplanar vectors can be used to represent a vector v. For a Cartesian coordinatesystem e1, e2, e3 forms a basis that allows the vector v to be written as

v = v1e1 + v2e2 + v3e3 = viei .

The basis e1, e2, e3 is the same for every point in the space defined in the Cartesiancoordinate system. The coordinate system is said to be homogenous. It is also possible tochoose a set of base vectors g1,g2,g3 that is not orthonormal, Fig A.1. The vectors g1and g2 lie in a plane, and the vector g3 is out of the plane. These base vectors may alsovary from point to point in the space. Such a system is called nonhomogeneous.

A vector can be represented by

v = v1g1 + v2g2 + v3g3 = vigi .

The summation is over a subscript and superscript. The superscripts are indices not ex-ponents. An exponent would be represented by vi × vi =

(vi)2. The components of the

vector v are found by projecting the vector onto the planes formed by g1-g2, g2-g3, andg1-g3. This gives the unique scalar components v1, v2, and v3 with respect to the basisgi. The calculation of the components vi are not as straight forward as in the Cartesiansystem. The basis vectors are not orthonormal and

gi · gj 6=0 for i 6= j

1 for i = j

as in the Cartesian system.343


g1

g2

g3

FIGURE A.1A set of non-orthonormal base vectors.

Example A.1

Given the set of base vectors gi with Cartesian components

g1 = 3e1 + 2e2 + e3g2 = e1 − 2e3g3 = e1 + e3

and a vectorv = 4e1 + e2 + 8e3 .

Find v1, v2, and v3.

SolutionWe know that

v = v1g1 + v2g2 + v3g3 .

Taking the dot product v · ei, gives

v · e1 4 = 3v1 + v2 + v3

v · e2 1 = 2v1 + v3

v · e3 8 = v1 − 2v3

Solving yieldsv1 = 2, v2 = 1, v3 = −3 .

The vector isv = 2g1 + g2 − 3g3 .

Determining whether a set of vectors forms a basis is not a simple task. A simple testfor linear independence is to form a matrix, G, from the basis vectors expressed in theCartesian components.

G = [g1,g2,g3]

The determinate of G being non-zero assures that the basis is linearly independent. Themathematical condition for this is

det G 6= 0

and is called the Jacobian matrix of the basis gi.

Appendix A: General Tensors 345

Example A.2

From the previous example, show that the basis gi is linearly independent.

SolutionThe matrix G is

G =

3 1 1

2 0 1

1 0 −2

and

det G =

∣∣∣∣∣∣3 1 1

2 0 1

1 0 −2

∣∣∣∣∣∣ = 5 6= 0 .

The basis is linearly independent. Note that the determinate of G being non-zerois just the condition that the equations for the components vi has a solution.

A.2 The Dot Product and the Reciprocal BasisThe dot product between two vectors u and v is

u · v = uigi · vjgj = uivjgi · gj .

If the basis is Cartesian, the dot product gi ·gj is just the Kronecker delta. However, sincethe basis vectors are not orthonormal, the dot product becomes

u · v =u1v1g1 · g1 + u1v2g1 · g2 + u1v3g1 · g3 + · · ·+ u3v1g3 · g1 + u3v2g3 · g2 + u3v3g3 · g3 .

The dot product of the base vectors is a set of scalars and possesses the symmetry

gi · gj = gij = gj · gi = gji .

The dot product would be simplified considerably if the dot product of the basis vectorswere orthogonal. This can be accomplished by computing a reciprocal or dual basis. Thereciprocal vectors can be computed in a number of ways but have the property that

gi · gj = δji =

1 if i = j

0 if i 6= j.

δji is the Kronecker delta. One method for the computation of the reciprocal basis is given

in Problem 2.2. The reciprocal basis allows us to write a vector as

v = vigi = vigi .


Note that the implied summation is always across a subscript and superscript pair. Withthe reciprocal basis, the dot product becomes

u · v = uivjgi · gj = uivjδji = uivi .

It is also possible to form the dot product of the base vectors with themselves. The dotproducts of the basis vectors give

gi · gj = gij

gi · gj = gij

gi · gj = gji = δ

ji

gi · gj = gij = δij

These equations give the metric tensors of the space.The metric tensor is associated with distances or lengths in the space. For a vector

v = vigi, we havev · v = |v|

2= vivjgi · gj = gijv

ivj

orv · v = |v|

2= vivjg

i · gj = gijvivj .

The magnitude of the vector is related to the metric tensors gij and gij.The products

gij = gi · gj

are the dual, reciprocal or inverse metric tensor. This latter designation is seen from thefollowing calculation. The bases are related by

gi = gijgj and gi = gijgj .

Now, this givesgi · gj = δ

ji = gilg

l · gjkgk = gilgjkδlk = gilg

jl

andgilg

jl = gilglj = δ

ji .

Hence the designation inverse metric.The components of a vector are related by

vi = v · gi = vkgk · gi = gkivk = gikv

k

from the symmetry of the metric tensor. This can also be repeated for the gi base vectors.The results are

vi = gijvj and vi = gijvj .

This is termed raising and lowering the indices and will be useful when we describe therules for generating general tensor expressions from Cartesian tensor expression.

A.3 Components of a TensorThe components of a tensor in an orthonormal basis are defined by the operation of thetensor on the base vectors ei. The components are

Tij = ei · T · ej .


This process can be extended to the general basis. However, unlike the orthonormalbasis, the general basis gives four different component representations. Two of thesetensor components are

Tij = gi · T · gjT ij = gi · T · gj .

The components are formed in the standard way. In addition, there are two sets of mixedcomponents. These are2

T ij = gi · T · gjTji = gi · T · gj

Thus, the four representations of a tensor are

T = T ijgigj = Tijgigj = T ijgig

j = Tji g

igj .

The mixed components of a tensor are not related. In general

Tji 6= T ij .

However, for a symmetric tensorT = TT

the result is

Tij = Tji

T ij = T ji

Tji = T ij

T ij = Tji .

Just as with orthonormal bases, transformation rules exist between two sets of generalbasis vectors gi and g′i. The transformation follows the same patterns established pre-viously. The only change being that summation occurs over subscripts and superscripts.The transformation of a vector is

v′i = ajivj

andvi =

(a−1

)jiv′j .

A second order tensor transforms as

T ′ij = api akj Tpk .

For the other components, we need to have the transformation rules for the reciprocalbasis

g′i =(a−1

)ijgj

andgi = aijg

′j .

These can be used to determine the transformation laws for the other tensor components.

2Sometimes dots are used to hold the positions of the indices, T ji = T ·ji . This is most commonly done whenwriting the mixed components. Typesetting offers the opportunity to clearly separate the superscript andsubscript.


A.4 Determination of the Base VectorsWe have seen that vectors can be represented in terms of two sets of base vectors gi orgi

.v = vigi = vig

i .

The base vectors gi are called the covariant basis vectors, andgi

is called a set ofcontravariant basis vectors. The covariant basis vectors are defined by the derivatives of acoordinate transformation

x = x(uj)

= xi(uj)ei .

The vectors ei are the orthonormal base vectors of the rectangular system. The functionsxi(uj)

are the coordinate functions relating the rectangular coordinates, xi, to the newcoordinate variables, uj. The covariant base vectors are

gj =∂x

∂uj=∂xi

∂ujei .

The base vectors are the derivatives along the coordinate curves. They vary along thecoordinate curves. The dual or reciprocal basis is the contravariant basis vectors. Thesevectors have the property that

gi · gj = δji =

1 if i = j

0 if i 6= j

δji is the Kronecker delta.

Example A.3

Determine the base vectors for the circular-cylindrical coordinate system.

SolutionFigure A.3 shows the circular-cylindrical coordinate system for x3 = 0. Theposition of the point p is

x = x1e1 + x2e2 + x3e3

= r cos θe1 + r sin θe2 + ze3 .

This shows that x = x (r, θ, z). Now

gr =∂x

∂r= cos θe1 + sin θe2

gθ =∂x

∂θ= −r sin θe1 + r cos θe2

andgz =

∂x

∂z= e3 .

The basis vectors gi can be shown to be a basis by forming the Jacobian,G (gr,gθ,gz). This gives

G = det

∣∣∣∣∣∣cos θ sin θ 0

−r sin θ r cos θ 0

0 0 1

∣∣∣∣∣∣ = r(cos2 θ+ sin2 θ

)= r .


grgθ

x

r

x2

x1

r−curve

θ

θ−curve

p

FIGURE A.2Circular-cylindrical coordinate system for x3 = 0.

The vectors form a basis except for the point r = 0. The vector

gr =∂x


lies along the r-coordinate. Similarly, gθ is a vector that is perpendicular to gr.The basis vectors form a non-homogeneous basis. The two vectors are tangentto the r and θ curves. This is consistent with the definition of the covariantbasis vectors.

The dual basis is found from the computation in Problem 2.2. These vectorsare

gr = cos θe1 + sin θe2

gθ =1

r(− sin θe1 + cos θe2)

and

gz = e3 .

The properties of the bases can be verified.


A.5 Derivatives of VectorsA.5.1 Time Derivative of a Vector

Suppose that in the circular cylindrical system of the above example that r = r (t), θ =θ (t), and z = z (t). This gives

x = r (t) cos θ (t) e1 + r (t) sin θ (t) e2 + z (t) e3 .

The time derivative of x (t) is the velocity v (t) and is

v (t) =dx (t)

dt=∂x

∂r

dr

dt+∂x

∂θ

dθ

dt+∂x

∂z

dz

dt

using the chain rule. From the previous section, the partial derivatives of x are the basevectors gi and

v (t) =dx (t)

dt=dr

dtgr +

dθ

dtgθ +

dz

dtgz = vrgr + vθgθ + vzgz .

The result isvr =

dr

dtvθ =

dθ

dtand vz =

dz

dt.

The first and third terms have the dimensions of a velocity. However, the second termdoes not. It only has units of s−1. This raises the need to develop the physical componentsof a vector.

The physical components of a vector w in the direction u are

w · u .

Here u is a unit vector in the direction of u.

u =u

|u|

The physical components of v denoted by v<i> are the components of the vector alongthe covariant unit vectors. The unit vectors gi are

gi =gi√gii

no sum on i .

Now for a vector v = vigi,

v = vigi = v1√g11g1 + v2

√g22g2 + v3

√g33g3 .

The physical components of v are

v<i> =√giiv

i no sum on i

andv = v<k>gk .

For the circular cylindrical system

v<r> = v · gr =dr

dt

andv<θ> = v · gθ = v · rgθ = r

dθ

dt.

These are the familiar components of the velocity that appear in elementary dynamics.


A.5.2 Covariant Derivative of a Vector

The spatial derivative of a vector field frequently arises in applications. This is the casefor gradients of displacement fields defining strain fields or the divergence of a velocityfield in a fluid. The derivative of a vector v with respect to a curvilinear coordinate ui is

∂v

∂uj=∂vi

∂ujgi + vi

∂gi

∂uj.

The latter term arises because the base vectors, gi, can vary from point to point in thespace.

It is desirable to express the derivative of the base vector in terms of components alongthe base vector. This allows the writing of an expression analogous to that in a Cartesiancoordinate system.

∂v

∂xj=∂vi

∂xjei = vi,jei .

Here, the comma notation denotes partial differentiation.The derivative of the base vector is

∂gi

∂uj=

∂

∂uj

(∂xk

∂uiek

)=

∂2xk

∂uj∂uiek .

The Cartesian base vectors are expressible in terms of the covariant base vectors.

ei =∂uj

∂xigj .

This is the inverse of the original expression defining gi. Inserting this expression for eigives

∂gi

∂uj=

∂2xk

∂ui∂uj∂ul

∂xkgl .

The coefficients in the above expression are known as the Christoffel symbols of the secondkind. These are given a special notation

∂2xk

∂ui∂uj∂ul

∂xk= Γ lij =

l

i j

Using this notation gives,

∂gi

∂uj= Γ lijgl =

l

i j

gl .

The different notations for the Christoffel symbols appear in the literature and are pre-sented to aid in understanding the literature on general tensors.

This expression gives one way to compute the Christoffel symbols. Taking the dotproduct of both sides with gpgives

Γpij =

∂gi

∂uj· gp

after rearranging the indices. The process for computing the Christoffel symbols is1. From the coordinate transformation x = xi

(uj)ei, find gk = ∂xi

∂ukei and compute

gi,j =∂2xk

∂uj∂uiek .


2. Determine the contravariant base vectors gk.3. Form the dot product to find the Christoffel symbol.

Γ lij =∂gi

∂uj· gl .

Another method is available for finding the Christoffel symbols that is not dependent onthe coordinate transformation. It will be presented subsequently.

Example A.4

Compute the Christoffel symbols for the circular cylindrical coordinate system.

SolutionThe covariant base vectors are

gr =∂x


gθ =∂x

∂θ= −r sin θe1 + r cos θe2

andgz =

∂x

∂z= e3 .

The computation is not as daunting as it might first appear. Many of theChristoffel symbols are zero. Differentiating the base vectors we have

∂gr

∂r=∂gr

∂z= 0,

∂gr

∂θ=∂gθ

∂r= − sin θe1 + cos θe2

∂gθ

∂z= 0,

∂gθ

∂θ= −r cos θe1 − r sin θe2

and∂gz

∂r=∂gz

∂θ=∂gz

∂z= 0 .

The reciprocal base vectors are

gr = cos θe1 + sin θe2

gθ =1

r(− sin θe1 + cos θe2)

andgz = e3 .

The Christoffel symbols found from Γ lij = ∂gi∂uj· gl give

Γrrr = Γθrr = Γzrr = Γrrz = Γθrz = Γzrz = 0

Γrrθ = Γrθr = Γzrθ = Γzθr = 0 Γθrθ = Γθθr =1

r

Γθθθ = 0 Γrθθ = −r .

The remaining components are zero.


The derivative of the vector can now be written as∂v

∂uj=∂vi

∂ujgi + vi

∂gi

∂uj=∂vi

∂ujgi + viΓ lijgl .

This can be rearranged to give

∂v

∂uj=∂vi

∂ujgi + vkΓ ikjgi =

(∂vi

∂uj+ vkΓ ikj

)gi .

The last form is the result of changing the dummy indices. The coefficients of the covari-ant base vectors are

vi|j =∂vi

∂uj+ vkΓ ikj .

This is the covariant derivative of the contravariant components of a vector. It is a naturalgeneralization of the result for partial differentiation in Cartesian systems.

To this point, all calculation was done using the contravariant components of the vector.If v = vig

i represents the vector, then

∂v

∂uj=∂vi

∂ujgi + vi

∂gi

∂uj.

This presents the problem of computing the derivative of the second term. To do this,note that

gi · gj = δji

and∂

∂uk

(gi · gj

)=∂gi

∂uk· gj + gi ·

∂gj

∂uk=∂δji

∂uk= 0 .

This gives∂gj

∂uk· gi = −

∂gi

∂uk· gj = −Γ likgl · gj = −Γ jik .

The right hand side is the covariant component of a vector and

∂gj

∂uk= −Γ jikg

i .

This result gives∂v

∂uj=

(∂vi

∂uj− vkΓ

kij

)gi .

The covariant derivative of the covariant components is

vi|j =∂vi

∂uj− vkΓ

kij .

Compare this to the previous result.

A.6 Christoffel SymbolsA.6.1 Types of Christoffel Symbols

The definition of the Christoffel symbols of the second kind indicates that there is sym-metry in the indices i and j.

∂2xk

∂ui∂uj∂ul

∂xk=

∂2xk

∂uj∂ui∂ul

∂xk


Γ lij = Γ lji =

l

i j

=

l

j i

The Christoffel symbols of the second kind are the partial derivatives of the basis vectorsin a curvilinear coordinate system with respect to the coordinate variables. Also, it isuseful to look at the placement of the indices. The indices are all free indices. The upperindex corresponds to the term in the numerator and the lower indices correspond to theterms in the denominator.

The above discussion indicates that the Christoffel symbols are of two types. TheChristoffel symbols of the first kind are denoted by [ij, l]. These are computed from thefollowing formula

[ij, l] = glmΓmij

Similarly, using the metric glp, we find

Γpij = glp [ij, l] .

The Christoffel symbols of the first kind have a representation similar to those of thesecond kind. To find this representation, note that

glm = gl · gm =∂xq

∂uleq ·

∂xk

∂umek =

∂xq

∂ul∂xk

∂umδqk .

The Christoffel symbols of the first kind are

[ij, l] = glmΓmij =

∂xq

∂ul∂xk

∂umδqk

∂2xp

∂ui∂uj∂um

∂xp

but∂xk

∂um∂um

∂xp=∂xk

∂xp= δkp and δkpδqk = δpq .

This gives

[ij, l] =∂2xp

∂ui∂uj∂xp

∂ul.

Notice that all the indices are in the denominator. Also, the grouping appears in a logicalsequence.

A.6.2 Calculation of the Christoffel Symbols

From the above discussion, it would appear that the Christoffel symbols are connected toa particular rectangular Cartesian coordinate system. However, the Christoffel symbolscan be computed directly from the metric and the inverse metric for the space. This isshown in the following.

The metric of the space isgij = gi · gj .

We differentiate this with respect to the coordinate uk and obtain

∂gij

∂uk=∂gi

∂uk· gj + gi ·

∂gj

∂uk.

This can be written as

∂gij

∂uk= Γ likgl · gj + gi · gpΓpjk = Γ likglj + gipΓ

pjk .


This is∂gij

∂uk= [ik, j] + [jk, i] .

Cyclically permuting the indices generates two more formulae.

∂gki

∂uj= [kj, i] + [ij, k]

∂gjk

∂ui= [ji, k] + [ki, j]

Adding the first 2 expressions and subtracting the last expression gives

[jk, i] =1

2

(∂gij

∂uk+∂gki

∂uj−∂gjk

∂ui

).

The symmetry of the Christoffel symbols was exploited in obtaining this result.The Christoffel symbols of the second kind can be obtained from the above expression

and the relation Γpjk = gip [jk, i]. The result is

Γpjk = gip [jk, i] =

1

2gip

(∂gij

∂uk+∂gki

∂uj−∂gjk

∂ui

).

This demonstrates that the Christoffel symbols are not dependent on the Cartesian basisthat was used to motivate their introduction.

The use of the metric tensor is especially useful in determining the Christoffel symbolswhen the general coordinate bases are orthogonal. In this case, we have gmn = 0 ifm 6= n. This immediately shows that many of the Christoffel symbols will be zero in anorthogonal basis. If i, j, and k are all distinct, then Γkij = 0. The only non-zero values willoccur when two of the indices are equal in a general, orthogonal coordinate system.

A.7 Covariant Derivatives of TensorsIn a manner similar to that used to find the covariant derivative of a vector, the covariantderivative of a tensor can be found. To calculate ∂T

∂uk, begin with the representation of a

tensor in the basis gi , T = T ijgigj. Then

∂T

∂uk=

∂

∂uk

(T ijgigj

)=∂T ij

∂ukgigj + T ij

∂gi

∂ukgj + T ijgi

∂gj

∂uk.

Using the Christoffel symbols, this can be written as

∂T

∂uk=∂T ij

∂ukgigj + T ijΓ likglgj + T ijΓ ljkgigl .

Rearranging the dummy indices gives

∂T

∂uk=

(∂T ij

∂uk+ T ljΓ ilk + T ilΓ jlk

)gigj

= T ij|kgigj .


The term in parentheses is the covariant derivative of the contravariant components ofthe tensor.

T ij|k =∂T ij

∂uk+ T ljΓ ilk + T ilΓ jlk .

Other covariant derivatives can be formed. These are

Tij|k =∂Tij

∂uk− TljΓ

lik − TilΓ

ljk

T ij|k =∂T ij

∂uk+ T ljΓ

ilk − T ilΓ

ljk .

They are computed using the same techniques as were employed previously.

A.8 General Tensor EquationsTensors and tensor equations describing physical phenomena are invariant with respectto coordinate systems. A number of relations in direct and Cartesian tensor notationwere developed. These relations span the spectrum from inner products of vectors tolocal forms of the First Law of Thermodynamics.

The following presents a way to convert these relations from Cartesian tensor notationto general tensor notation. Vectors and tensors are represented in terms of covariant andcontravariant base vectors. The techniques presented allow us to compute the derivativesof vector fields and tensor fields. Three rules allow the conversion of Cartesian tensorexpressions to expressions in general coordinate systems.

Rule 1: Rewrite summations over dummy indices in Cartesian expressions so that the summationis only over raised and lowered indices.

This rule is a direct consequence of the way that summation was introduced for vectorand tensor representation in a general coordinate system. We defined vectors and tensorsin terms of covariant and contravariant bases. This led to the convention that impliedsummation occurs only over a subscript and a superscript pair.

Summation across pairs of indices may be accomplished in a number of ways. We maysimply change an index in the Cartesian expression from a subscript to a superscript. Thisis possible since we have Kronecker deltas δij and δij in the Cartesian system. We mayraise or lower indices by noting that the metric or inverse metric acts as a generalized Kro-necker delta. Table A.1 gives examples of Cartesian expressions and the correspondinggeneral tensor expressions.

Rule 2: Replace partial differentiation in Cartesian tensor expressions with covariant differentia-tion.

This rule is useful in converting derivatives and differential operators. An example isthe divergence of a vector field. In Cartesian tensor notation, we have

div v = vi,i .

This becomesdiv v = vi|i


TABLE A.1Converting from Cartesian tensor notation to general tensornotation. Summation over only subscript and superscript pairs.

Cartesian Tensor Expression General Tensor Expression

dkk d kk = gkmd

km = gkmdkm

aibi aibi = gika

kbi = gikaibk

didi didi = gijd

idj = gijdidj

vi = wikk vi = w kik = gkmw

kmi = gkmwikm

in general tensor notation. Note that the summation is over a raised and lowered index.Another example is the equilibrium equations. In Cartesian notation, they read

Tij,j + bi = ρai .

In general tensor notation, this becomes

T ij|j + bi = ρai .

Are there other forms that could be used to represent these equations?These general tensor forms appear to be quite simple. However, it must be remembered

that the covariant derivative involves Christoffel symbols. For general coordinate systems,these are quite complicated. In orthogonal systems, things improve since the non-zeroChristoffel symbols have a pair of matching indices. The distinction disappears in aCartesian system where the Christoffel symbols are all zero. Things are not always assimple as they first appear!

Rule 3: Replace the Cartesian tensor permutation symbol, εijk, by the general tensor permutationsymbol, eijk or eijk, and rewrite summations over raised and lowered indices.

This rule is useful when considering the cross product of vectors or vector operators.The cross product of two vectors in a general coordinate system is

u× v = uivjgi × gj

butgi × gj = eijkg

k .

This givesu× v = eijku

ivjgk .

Also note thateijk = (gi × gj) · gk = εijk

√g

whereg = det |gij| .

These three rules allow one to conveniently change Cartesian tensor expressions intogeneral tensor expressions. Now the laws and equations that are valid in Cartesian refer-ence frames can be rewritten so that they have forms appropriate for general coordinatesystems. This gives us a very powerful methodology for exploring the mechanics ofcontinuous media.


A.9 General Tensors and Physical ComponentsA tensor has the following representations

T = T iigigj = Tijgigj = T<ij>gigj .

The equilibrium equations are given by the derivatives of the physical components T<ij>.For cylindrical-polar coordinates, these are

g1 = er∂T<rr>

∂r+1

r

∂T<rθ>

∂θ+∂T<rz>

∂z+T<rr> − T<θθ>

r= 0

g2 = eθ1

r2∂(r2T<θr>

)∂r

+1

r

∂T<θθ>

∂θ+∂T<θz>

∂z= 0

g3 = ez1

r

∂ (rT<zr>)

∂r+1

r

∂T<θz>

∂θ+∂T<zz>

∂z= 0 .

The covariant derivative of the tensors for computing the equilibrium equation can bewritten in two forms.

Form 1:

T ji∣∣i=∂T ji

∂ui+ T jlΓ ili + T liΓ jli = 0

This is an acceptable form for the equilibrium equations since the tensor components arethe contravariant components. These are similar to the contravariant physical componentsused above.

g1 (j = 1)∂T1i

∂ui+ T1lΓ ili + T liΓ1li = 0

This equation contains many terms. However, all but two of the Christoffel symbols arezero. The non-zero Christoffel symbols are

Γ212 = Γ221 =1

rΓ122 = −r .

This gives∂T1i

∂ui+ T11Γ212 + T22Γ122 = 0 ;

∂T11

∂u1+∂T12

∂u2+∂T13

∂u3+1

rT11 − rT22 = 0 .

The contravariant components in this expression are not the physical components of thetensor. What is the relationship between the two? The representation of a tensor abovegives

T iigigj = T<ij>gigj = T<ij>gi

gi

gj

gj.

gi and gj are the magnitudes of the vectors. These are

gi =√gi · gi =

√gii no sum on i .


gij is the metric tensor. For cylindrical coordinates, this is

[gij] =

1 0 0

0 r2 0

0 0 1

.The connection between the components are

T<ij> =√gii√gjj T

ij no sum on i or j . T<rr> T<rθ> T<rz>

T<θr> T<θθ> T<θz>

T<zr> T<zθ> T<zz>

=

T11 rT12 T13

rT21 r2T22 rT23

T31 rT32 T33

This gives

∂T<rr>

∂u1+

∂

∂u2

(1

rT<rθ>

)+∂T<rz>

∂u3+1

rT<rr> − r

(1

r2T<θθ>

)= 0 .

This is the er equilibrium equation where(u1, u2, u3

)is (r, θ, z).

g2 (j = 2)∂T2i


or∂T2i

∂ui+ T21Γ212 + T12Γ212 + T21Γ221 =

∂T2i

∂ui+3

rT21 = 0 .

This leads to∂T21

∂u1+∂T22

∂u2+∂T23

∂u3+3

rT21 = 0 .

Writing this in terms of the physical components and the coordinates (r, θ, z) gives

∂

∂r

(1

rT<θr>

)+∂

∂θ

(1

r2T<θθ>

)+∂

∂z

(1

rT<θz>

)+3

r

(1

rT<θr>

)= 0

and1

r2∂

∂r

(r2T<θr>

)+∂

∂θ

(1

rT<θθ>

)+∂

∂z

(T<θz>

)= 0

g3 (j = 3)∂T3i


or∂T3i

∂ui+ T31Γ212 + T li (0) = 0 .

This yields∂T31

∂u1+∂T32

∂u2+∂T33

∂u3+1

rT31 = 0 .

Converting to physical components gives

1

r

∂ (rT<zr>)

∂r+∂

∂θ

(1

rT<zθ>

)+∂T<zz>

∂z= 0


Form 2:

This form uses the covariant tensor components

Tij|i =∂Tij

∂uj− TmjΓ

mij + TimΓ

mjj = 0

This is not an acceptable form for the equilibrium equations for the physical componentssince the tensor components are the covariant components rather than the contravariantcomponents. These are not the same as the contravariant physical components used inthe equilibrium equations. While these expressions will give partial differential equationsfor the covariant components, they will not give an equivalent set of equations to thosealready presented. The contravariant and covariant components and associated equationsbecome equivalent in an orthonormal rectangular coordinate system.

References[1] A. C. Eringen (1962) Nonlinear Theory of Continuous Media, McGraw-Hill Book Co.,

New York.

[2] Y. C. Fung (1965) Foundations of Solid Mechanics, Prentice Hall, Englewood Cliffs.

[3] L. E. Malvern (1969) Introduction to the Mechanics of a Continuous Media, Prentice Hall,Englewood Cliffs.

[4] M. N. L. Narasimhan (1993) Principles of Continuum Mechanics, John Wiley & Sons,Inc., New York.

[5] J. G. Simmonds (1994) A Brief on Tensor Analysis, Springer, New York.

Appendix B: Viscoelastic Creep and Relaxation

361

362C

ontinuumM

echanicsfor

EngineersTABLE B.1: Creep and relaxation responses for various viscoelastic models.

Model/Name Constitutive Equation Creep Compliance, J(t) Relaxation Modulus, G(t)

Elastic SolidG T = Gγ

1

G= J =

U(t)

GG(t) = GU(t)

Viscous Fluidη

T = ηγ

T = ηε

t

η=

t

Gτηδ(t)

Maxwell (Fluid)

G η

T +1

τT = Gγ

∂t +1

t

T = G∂tγ

1

G+t

η=τ+ t

ηGe−t/τ

Kelvin (Solid)

G

η

T = Gγ+ ηγ

T = G+ η∂tγ

1

ηT =

1

τ+ ∂t

γ

1

G

(1− e−t/τ

)G+ ηδ(t)

Three Parameter SolidG

1 G2

η

G1G2γ+G1η2γ =

(G1 +G2)T + η2T

e−t/τ

G1+G1 +G2

G1G2

(1− et/τ2

)or

1

G1+1

G2

(1− e−t/τ2

)G1e

−t/τ′ +G1G2

G1 +G2

(1− e−t/τ′

)or

G1

G1 +G2

(G2 +G1e

−t/τ′)

where τ ′ =η2

G1 +G2

Continued on next page

Appendix

B:Viscoelastic

Creep

andR

elaxation363

TABLE B.1 – continued from previous page


G1

G2

η

T+T

τ2= (G1+G2)γ+

G1

τ2γ

e−t/τ′

G1 +G2+1

G1

(1− e−t/τ′

)or

1

G1−

G2

G1(G1 +G2)e−t/τ′

where τ ′ =G1 +G2

G1τ2

G1 +G2e−t/τ2

G

η1

η2

G2η1γ+ η1η2γ =

G2T + (η1 + η2)T

γ+γ

τ2=η1 + η2

η1η2T +

T

τ2η1

1

G1

(1− e−t/τ′

)+t

η2

η1η2

η1 + η2δ(t) +

G2

η1 + η2

(η1 +

η1η2

η1 + η2

)eτ′

where τ ′ =η1 + η2

G2

G

η2

η1

T +T

τ2=(

G2 +η1

τ2

)γ+ η1γ

t

η1 + η2+1

G2

(η2

η1 + η2

)2 (1− e−λt

)where λ =

G2 (η1 + η2)

η1η2

η1δ(t) +G2e−t/τ2

G2

η2

G1

η1

γ+γ

τ2=T

G1+(

1

η2+

1

G1τ2+1

η1

)T +

1

η1τ2T

1

G1+t

η1+1

G2

(1− e−t/τ2

)

Continued on next page

364C

ontinuumM

echanicsfor

EngineersTABLE B.1 – continued from previous page


G1

η1

G2

η2

(η1 + η2)T + (G1 +G2)T =

G1G2γ+ (η2G1 +

η1G2)γ+ η1η2γ

1

η1 + η2

η1η2

(δ(t) −

e−t/τ′

τ ′

)+

G1G2τ′ (1− e−t/τ′

)

Index

ε - δ identity, 10Matlab

r

invariants, 35symbolic

invariants, 36

acceleration field, 106, 172adiabatic, 212, 275angular momentum, 176axiom of impenetrability, 106

balance laws, see conservation ofbarotropic, 278, 280

fluid, 275base vectors, 7basis

orthogonal, 7Bernoulli equation, 280bi-harmonic equation, 242, 255

polar form, 246box product, see triple scalar product

caloric equation of state, 180Cartesian axes, 7Cauchy deformation tensor, 118, 208

left, 292right, 133, 136, 292

Cayley-Hamilton theorem, 294characteristic equation, 31Christoffel symbols

of the first kind, 354of the second kind, 351

circulation, 41, 281Clausius-Duhem equation, 167, 179, 181,

182local form, 182

coefficient ofbulk viscosity, 272thermal expansion, 252

compatibilityBeltrami-Michell, 224, 253, 254, 258equations, 254strain, 128

compliance

creep, 318shear, 316, 320volumetric, 320

concentrated moment, 55configuration, 103

current, 104, 105, 116initial, 106reference, 104–106

conservation ofangular momentum, 167, 182energy, 167, 177, 179, 186linear momentum, 167, 182, 186mass, 167, 169, 182

constitutive equations, 167, 179, 196, 198,272

continuity equation, 169, 213, 278contraction, 17convective rate of change, 112covariant

base vectors, 353derivative, 353derivative of tensors, 355

creep, 315function, 316response

Kelvin, 316Maxwell, 317

cross product, 11, 15cubical dilatation, 123curl, 38

decompositionadditive, 134multiplicative, 134

deformation, 103, 117gradient, 116, 117, 134homogeneous/inhomogeneous, 117

del operator, 38, 112density, 53, 54, 109determinant

cofactor, 21deviatoric response, 310, 311dilatational response, 310, 320

365


Dirac delta function, 317direction cosines, 25displacement, 103, 110, 120

differential, 127field, 110

displacement gradient, 120distortional response, 320divergence, 38

theorem of Gauss, 40, 62dot product, 11double dot product, 14dual basis, 345dummy index, 8dyad, see tensor productdyad-dyad product, 13

eigen problem, 216characteristic equation, 31eigenvalue, 30eigenvector, 30

normalized, 34elastic materials, 178elasticity, 211

Airy stress function, 242, 246, 259constantsCijkl, 212anisotropic, 215, 219bulk modulus, 217isotropic, 215, 223Lame, 215, 217, 221modulus of elasticity, 216orthotropic, 221Poisson’s ratio, 216shear modulus, 216, 219

elastodynamics, 223, 225elastostatics, 223linear, 211linear solid, 199nonlinear, 211, 285plane, 238semi-inverse, 243symmetry, 219, 220three dimensional, 253

displacement formulation, 253stress formulation, 253

energybalance, 213

mechanical, 179thermal, 179

density, 178equation, 179

internal, 178specific internal, 178

entropy, 179equation of state

fluid, 271equations of motion, 174

Eulerian, 172Lagrangian, 172local, 172

equilibrium, 61force and moment, 61balance of moments, 62equation, 174, 253local equations, 61

Euclidian space, 7Euler equation, 275, 276, 278, 280Eulerian

description, 109, 273strain, 118

field equations, 167first law of thermodynamics, 167, 179fluid, 271

linear viscous, 199Newtonian, 271, 272Stokesian, 271, 272

forcebody, 53, 223contact, 53surface, 53

Fourier’s law, 178, 253

Galerkin vector, 255, 256gas dynamics equation, 278gradient, 38

scalar, 38vector, 38

Green’s deformation tensor, 117Green’s second identity, 51

harmonic, 231heat, 179

conduction, 253flux, 178supply, 178

Helmholtz, theorem of, 254homogeneous, 212

body, 54deformation, 117

Hooke’s law, 211, 214–216, 219, 240, 241,253

Index 367

isotropic body, 223plane strain, 241

hydrostatics, 276hyperelastic material, 214

incompressiblefluid, 275material, 294

indicesraising and lowering, 346

indicial notation, 5, 16free index, 16

inelastic, 211infinitesimal

deformation theory, 120rotation tensor, 127strain tensor, 127

inner product, 17internal

constraint, 294energy, 213

invariance, 189invariants, 31, 35inverse mapping, 105inviscid fluid, 275irrotational, 254, 280

flow, 278isochoric, 145isothermal, 212, 275

Jacobiandeterminant, 106matrix, 344

Kelvin problem, 256Kelvin’s theorem, 280, 281kinematics, 103kinetic energy, 177Kronecker delta, 9, 19

Lagrange multiplierstress, 72

Lagrangiandescription, 109equations of motion, see equations

of motionLame strain potential, 255laminar flows, 275Laplace equation, 279left stretch tensor, 134linear

independence, 344momentum, 171transformation, 5, 30

antecedent, 30image, 30

local rate of change, 112

mapping, 103mass, 169

density, 169material

body, 103coordinates, 105, 109, 121derivative, 111, 112

of a line element, 143of a line segment, 143of a volume, 145of an area, 144of Jacobian, 144operator, 112

description, 109, see referential de-scription, 110, 118

material frame indifference, 189matrix, 19

column matrix, 19addition, 19adjoint, 23decomposition, 20determinant

diagonal matrix, 22expansion by cofactors, 22minor, 21signed minor, 21

determinant of square matrix, 21diagonal, 19exponent rule, 20identity, 19inverse, 23multiplication, 20

associative, 20communitive, 20

null, 19orthogonal, 24prefactor/postfactor, 20row matrix, 19scalar multiplication, 20singular, 22skew–symmetric, 19square, 19square of, 20square root, 20


symmetric, 19trace, 31transpose, 19vector-tensor product, 24zero, 19

mechanical power, 177Mohr’s circle, 189

for stress, 76moment

of momentum, 176vector, 55

Mooney-Rivlin material, 292, 296motion, 103, 117multiplication

cross product, 15double dot product, 16dyad product, 15dyad-dyad product, 16tensor-tensor product, 16triple cross product, 15triple scalar product, 15vector by scalar, 15vector by vector, 15vector-dyad product, 15vector-tensor products, 16

Navier equation, 224, 225, 240, 253, 254,257

plane strain, 241Navier-Stokes equations, 273–275neo-Hookean material, 292, 296Newton’s laws, 55nonion form, 12

Ogden material, 297orthonormal, 7orthotropic, 221outer product, 17

Papkovich-Neuber, 257partial differentiation

subscript comma notation, 37particles, 103path, 106permutation symbol, 9, 10plane strain, 124, 238

generalized, 241plane stress, 80, 238

generalized, 241plastic material, 199polar

coordinates, 245decomposition, 134material, 176

polymer, 309chain, 287conformation, 287

positive/negative definite, 37, 134potential flow, 278pressure

function, 276head, 280hydrostatic, 271thermostatic, 271

principle of linear momentum, 55

rate ofdeformation, 137, 138extension, 139shear, 140stretching, 139

reciprocal vectors, 345referential

coordinates, 105description, see material description

relaxation, 315function, 318

residual stress, 214resilience, 285retardation time, 316right

principal direction of stretch, 135stretch tensor, 134

rigid body, 103displacement, 132translation, 135

rotationtensor, 134vector, 127

rubberbutadiene, 285molecular approach, 287natural, 285

Saint-Venantextension, 227flexure, 236pure bending, 234torsion, 228

scalar, 6field, 37product, 11

Index 369

second law of thermodynamics, 179, 186simple shear deformation, 119solenoidal, 254spatial

coordinates, 105, 121description, 110, 118position, 105velocity gradient, 137

specificentropy, 180entropy production, 181free energy, 186heat, 253internal energy, 178

steady flow, 276Stokes’

condition, 272theorem, 41

strain, 104axial, 217deviator tensor, 123energy, 211, 213energy density, 213engineering shear, 123Eulerian, 118, 175

linearized, 120finite strain tensor, 116gauge rosette, 125infinitesimal, 123invariants, 121Lagrangian, 117, 118, 175

linearized, 120linear, 120longitudinal, 121, 132mean normal, 123normal, 121principal, 121

deviator, 124directions, 121, 123

pure, 127rate of, 140shear, 125

strain rate, see strain, rate of, 310strain-displacement equation, 253streamline, 280stress, 53

as a tensor, 58axial, 217Cauchy

formula, 58, 59, 61principle, 54, 56

characteristic equation, 67components, 59compressive, 59deviatoric, 85, 252

principal values, 86extremal values, 72first Piola-Kirchhoff, 173hydrostatic, 85invariants, 67maximum

and minimum values, 71shear, 76

mean, 85, 271Mohr’s circle, 74normal, 59, 71octahedral, 87

normal, 87shear, 87

orthogonal principal directions, 68Piola-Kirchhoff, 172plane, 80power, 178principal

directions, 66real values, 67stresses, 66values, 66

rate, 196Green-Naghdi, 196Jaumann, 196

second Piola-Kirchhoff, 174shear, 59

magnitude, 72spherical, 85, 252symmetry, 61tensor, 56three dimensional stress function, 258transformation, 63, 69units, 59vector, 55, 57, 66

stretchratio, 131tensor, 134

summation convention, 8superposed rigid body motion, 189superposition principle, 223, 224symbolic notation, 5, 11

temperaturereference, 252

tensor, 5


-tensor products, 13anti-symmetric, 17Cartesian, 5component, 7curl, 38definition of, 5divergence, 38field, 37metric, 346order, 6physical components, 358positive/negative definite, 37product, 12skew-symmetric, 17spin, 138, 141symmetric, 17trace, 31transformation, 25, 27

thermodynamic state, 180thermoelasticity, 252thermomechanical continuum, 178thermoplastic elastomers, 285thermoset, 285traction vector, 55trajectory, 106transformation, 25

fourth-order tensor, 212arbitrary Cartesian tensor, 28improper orthogonal, 28law, 27orthogonality condition, 26tensor, 25, 212vectors, 25

transport theorem, 168triple

cross product, 11, 15scalar product, 11, 15

undeformed configuration, 104unit

base vectors, 7extension, 132

vector, 6-dyad product, 13-tensor product, 24-tensor products, 13addition, 11, 15axial, 18, 44axial or pseudo, 28components, 7

curl, 38divergence, 38field, 37multiplication, 11physical components, 350transformation, 25, 27triangle rule, 6true or polar, 28

velocity, 106field, 55, 104, 106gradient, 137head, 280

viscoelastic, 309absolute compliance, 323complex

compliance, 320modulus, 320

correspondence principle, 324, 327creep function, 318dashpot, 312harmonic loading, 320hereditary integral, 318, 323Kelvin model, 313, 323

generalized, 315linear

integral equations, 200material, 199

linear differential operator, 310loss


Maxwell model, 313, 323, 324generalized, 315

mechanical models, 311relaxation modulus, 320retardation time, 316shear compliance, 320standard linear solid, 313storage


superposition principle, 318three-dimensional problems, 324volumetric compliance, 320

viscosity coefficients, 272viscous stress tensor, 271vorticity, 137, 283

tensor, 138, 141vector, 141

work, 179

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CONTINUUM MECHANICSdl.booktolearn.com/ebooks2/engineering/mechanical/... · 2019-06-24 · Published Titles ADVANCED THERMODYNAMICS ENGINEERING Kalyan Annamalai and Ishwar K. Puri