Continuum_Mechanics_for_Engineers_Mase_3

CONTINUUM MECHANICS FOR ENGINEERS 7HP SEDit|ON RONAUDIE, SMELSER SEORGE E)MASE | CONTINUUM MECHANICS FOR ENGINEERS THIRD EDITION 85387_FM.indd 1 6/26/09 3:02:53 PM Published Titles ADVANCED THERMODYNAMICS ENGINEERING Kalyan Annamalai and Ishwar K. Puri APPLIED FUNCTIONAL ANALYSIS J. Tinsley Oden and Leszek F. Demkowicz COMBUSTION SCIENCE AND ENGINEERING Kalyan Annamalai and Ishwar K. Puri CONTINUUM MECHANICS FOR ENGINEERS, Third Edition Thomas Mase, Ronald E. Smelser, and George E. Mase EXACT SOLUTIONS FOR BUCKLING OF STRUCTURAL MEMBERS C.M. Wang, C.Y. Wang, and J.N. Reddy THE FINITE ELEMENT METHOD IN HEAT TRANSFER AND FLUID DYNAMICS, Second Edition J.N. Reddy and D.K. Gartling MECHANICS OF LAMINATED COMPOSITE PLATES AND SHELLS: THEORY AND ANALYSIS, Second Edition J.N. Reddy PRACTICAL ANALYSIS OF COMPOSITE LAMINATES J.N. Reddy and Antonio Miravete SOLVING ORDINARY and PARTIAL BOUNDARY VALUE PROBLEMS in SCIENCE and ENGINEERING Karel RektorysCRC Series in COMPUTATIONAL MECHANICS and APPLIED ANALYSIS Series Editor: J.N. Reddy Texas A&M University 85387_FM.indd 2 6/26/09 3:02:53 PM CRC Press is an imprint of the Taylor & Francis Group, an informa businessBoca Raton London New YorkG. THOMAS MASE RONALD E. SMELSER GEORGE E. MASECONTINUUM MECHANICS FOR ENGINEERS THIRD EDITION 85387_FM.indd 3 6/26/09 3:02:53 PM CRC Press Taylor & Francis Group6000 Broken Sound Parkway NW, Suite 300Boca Raton, FL 33487-2742 © 2010 by Taylor and Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government worksPrinted in the United States of America on acid-free paper 10 9 8 7 6 5 4 3 2 1 International Standard Book Number: 978-1-4200-8538-9 (Hardback)This book contains information obtained from authentic and highly regarded sources. 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I. Mase, George E. II. Smelser, Ronald M., 1942- III. Title. IV. Series. QA808.2.M364 2009 531--dc22 2009022575 Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com 85387_FM.indd 4 6/26/09 3:02:53 PM 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 1 1.1 Continuum Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Starting Over . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.3 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 2 Essential Mathematics 5 2.1 Scalars, Vectors and Cartesian Tensors . . . . . . . . . . . . . . . . . . . . . 5 2.2 Tensor Algebra in Symbolic Notation - Summation Convention . . . . . . 7 2.2.1 Kronecker Delta . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 2.2.2 Permutation Symbol . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 2.2.3"-Identity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 2.2.4 Tensor/Vector Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . 11 2.3 Indicial Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 2.4 Matrices and Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 2.5 Transformations of Cartesian Tensors . . . . . . . . . . . . . . . . . . . . . 25 2.6 Principal Values and Principal Directions . . . . . . . . . . . . . . . . . . . 30 2.7 Tensor Fields, Tensor Calculus . . . . . . . . . . . . . . . . . . . . . . . . . 37 2.8 Integral Theorems of Gauss and Stokes . . . . . . . . . . . . . . . . . . . . 40 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 3 Stress Principles 53 3.1 Body and Surface Forces, Mass Density . . . . . . . . . . . . . . . . . . . . 53 3.2 Cauchy Stress Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54 3.3 The Stress Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56 3.4 Force and Moment Equilibrium; Stress Tensor Symmetry . . . . . . . . . . 61 3.5 Stress Transformation Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 3.6 Principal Stresses; Principal Stress Directions . . . . . . . . . . . . . . . . . 66 3.7 Maximum and Minimum Stress Values . . . . . . . . . . . . . . . . . . . . 71 3.8 Mohr’s Circles for Stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 3.9 Plane Stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 3.10 Deviator and Spherical Stress States . . . . . . . . . . . . . . . . . . . . . . 85 3.11 Octahedral Shear Stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90 4 Kinematics of Deformation and Motion 103 4.1 Particles, Configurations, Deformations and Motion . . . . . . . . . . . . . 103 4.2 Material and Spatial Coordinates . . . . . . . . . . . . . . . . . . . . . . . . 104 4.3 Langrangian and Eulerian Descriptions . . . . . . . . . . . . . . . . . . . . 108 4.4 The Displacement Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 4.5 The Material Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111 4.6 Deformation Gradients, Finite Strain Tensors . . . . . . . . . . . . . . . . . 116 4.7 Infinitesimal Deformation Theory . . . . . . . . . . . . . . . . . . . . . . . 120 4.8 Compatibility Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128 4.9 Stretch Ratios . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131 4.10 Rotation Tensor, Stretch Tensors . . . . . . . . . . . . . . . . . . . . . . . . . 134 4.11 Velocity Gradient, Rate of Deformation, Vorticity . . . . . . . . . . . . . . . 137 4.12 Material Derivative of Line Elements, Areas, Volumes . . . . . . . . . . . . 143 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 5 Fundamental Laws and Equations 167 5.1 Material Derivatives of Line, Surface and Volume Integrals . . . . . . . . . 167 5.2 Conservation of Mass, Continuity Equation . . . . . . . . . . . . . . . . . . 169 5.3 Linear Momentum Principle, Equations of Motion . . . . . . . . . . . . . . 171 5.4 Piola-Kirchhoff Stress Tensors, Lagrangian Equations of Motion . . . . . . 172 5.5 Moment of Momentum (Angular Momentum) Principle . . . . . . . . . . 176 5.6 Law of Conservation of Energy, The Energy Equation . . . . . . . . . . . . 177 5.7 Entropy and the Clausius-Duhem Equation . . . . . . . . . . . . . . . . . . 179 5.8 The General Balance Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182 5.9 Restrictions on Elastic Materials by the Second Law of Thermodynamics . 186 5.10 Invariance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189 5.11 Restrictions on Constitutive Equations from Invariance . . . . . . . . . . . 196 5.12 Constitutive Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202 6 Linear Elasticity 211 6.1 Elasticity, Hooke’s Law, Strain Energy . . . . . . . . . . . . . . . . . . . . . 211 6.2 Hooke’s Law for Isotropic Media, Elastic Constants . . . . . . . . . . . . . 214 6.3 Elastic Symmetry; Hooke’s Law for Anisotropic Media . . . . . . . . . . . 219 6.4 Isotropic Elastostatics and Elastodynamics, Superposition Principle . . . 223 6.5 Saint-Venant Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226 6.5.1 Extension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227 6.5.2 Torsion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228 6.5.3 Pure Bending . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234 6.5.4 Flexure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236 6.6 Plane Elasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238 6.7 Airy Stress Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242 6.8 Linear Thermoelasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252 6.9 Three-Dimensional Elasticity . . . . . . . . . . . . . . . . . . . . . . . . . . 253 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 260 7 Classical Fluids 271 7.1 Viscous Stress Tensor, Stokesian, and Newtonian Fluids . . . . . . . . . . . 271 7.2 Basic Equations of Viscous Flow, Navier-Stokes Equations . . . . . . . . . 273 7.3 Specialized Fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275 7.4 Steady Flow, Irrotational Flow, Potential Flow . . . . . . . . . . . . . . . . 276 7.5 The Bernoulli Equation, Kelvin’s Theorem . . . . . . . . . . . . . . . . . . 280 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282 8 Nonlinear Elasticity 285 8.1 Molecular Approach to Rubber Elasticity . . . . . . . . . . . . . . . . . . . 287 8.2 A Strain Energy Theory for Nonlinear Elasticity . . . . . . . . . . . . . . . 292 8.3 Specific Forms of the Strain Energy . . . . . . . . . . . . . . . . . . . . . . . 296 8.4 Exact Solution for an Incompressible, Neo-Hookean Material . . . . . . . 297 Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304 9 Linear Viscoelasticity 309 9.1 Viscoelastic Constitutive Equations in Linear Differential Operator Form . 309 9.2 One-Dimensional Theory, Mechanical Models . . . . . . . . . . . . . . . . 311 9.3 Creep and Relaxation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315 9.4 Superposition Principle, Hereditary Integrals . . . . . . . . . . . . . . . . . 318 9.5 Harmonic Loadings, Complex Modulus, and Complex Compliance . . . . 320 9.6 Three-Dimensional Problems, The Correspondence Principle . . . . . . . 324 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 330 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331 Appendix A: General Tensors 343 A.1 Representation of Vectors in General Bases . . . . . . . . . . . . . . . . . . 343 A.2 The Dot Product and the Reciprocal Basis . . . . . . . . . . . . . . . . . . . 345 A.3 Components of a Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 346 A.4 Determination of the Base Vectors . . . . . . . . . . . . . . . . . . . . . . . 348 A.5 Derivatives of Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 350 A.5.1 Time Derivative of a Vector . . . . . . . . . . . . . . . . . . . . . . . 350 A.5.2 Covariant Derivative of a Vector . . . . . . . . . . . . . . . . . . . . 351 A.6 Christoffel Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353 A.6.1 Types of Christoffel Symbols . . . . . . . . . . . . . . . . . . . . . . 353 A.6.2 Calculation of the Christoffel Symbols . . . . . . . . . . . . . . . . . 354 A.7 Covariant Derivatives of Tensors . . . . . . . . . . . . . . . . . . . . . . . . 355 A.8 General Tensor Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 356 A.9 General Tensors and Physical Components . . . . . . . . . . . . . . . . . . 358 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 360 Appendix B: Viscoelastic Creep and Relaxation 361 Index 365 This page intentionally left blank List of Figures 2.1 Base vectors and components of a Cartesian vector. . . . . . . . . . . . . . . 8 2.2 Rectangular coordinate system Ox0 1x0 2x0 3relative toOx1x2x3. Direction cosines shown for coordinate x0 1relative to unprimed coordinates. Simi- lar direction cosines are defined for x0 2andx0 3coordinates. . . . . . . . . . . 26 2.3 Rotation and reflection of reference axes. . . . . . . . . . . . . . . . . . . . . 28 2.4 Principal axes Ox 1x 2x 3relative to axes Ox1x2x3. . . . . . . . . . . . . . . . . 32 2.5 Volume Vwith infinitesimal element dSihaving a unit normal ni. . . . . . 40 2.6 Bounding space curve Cwith tangential vector dxiand surface element dSi for partial volume. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 3.1 Typical continuum volume Vwith infinitesimal element
Vhaving mass
m at pointP. PointPwould be in the center of the infinitesimal volume. 54 3.2 Typical continuum volume with cutting plane. . . . . . . . . . . . . . . . . . 55 3.3 Traction vector t(^n) iacting at point Pof plane element
Siwhose normal isni. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56 3.4 Traction vectors on the three coordinate planes at point P. . . . . . . . . . . 57 3.5 Free body diagram of tetrahedron element having its vertex at point P. . . 57 3.6 Cartesian stress components shown in their positive sense. . . . . . . . . . . 60 3.7 Material volume showing surface traction vector t(^n) ion an infinitesimal area element dSat positionxi, and body force vector biacting on an in- finitesimal volume element dVat positionyi. Two positions are taken sep- arately for ease of illustration. When applying equilibrium the traction and body forces are taken at the same point. . . . . . . . . . . . . . . . . . . . . . 62 3.8 Rectangular coordinate axes Px0 1x0 2x0 3relative toPx1x2x3at pointP. . . . . 63 3.9 Traction vector and normal for a general continuum and a prismatic beam. 66 3.10 Principal axes Px 1x 2x 3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 3.11 Traction vector components normal and in-plane (shear) at point Pon the plane whose normal is ni. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 3.12 Normal and shear components at Pto plane referred to principal axes. . . 73 3.13 Typical Mohr’s circle for stress. . . . . . . . . . . . . . . . . . . . . . . . . . . 75 3.14 Typical Mohr’s circle representation. . . . . . . . . . . . . . . . . . . . . . . . 77 3.15 Typical 3-D Mohr’s circle and associated geometry. . . . . . . . . . . . . . . 78 3.16 Mohr’s circle for plane stress. . . . . . . . . . . . . . . . . . . . . . . . . . . . 81 3.17 Mohr’s circle for plane stress. . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 3.18 Representative rotation of axes for plane stress. . . . . . . . . . . . . . . . . 84 3.19 Octahedral plane (ABC) with traction vector t(^n) i, and octahedral normal and shear stresses, NandS. . . . . . . . . . . . . . . . . . . . . . . . . . . 87 4.1 Position of typical particle in reference configuration XAand current con- figurationxi. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104 4.2 VectordXA, between points PandQin reference configuration, becomes dxi, between points pandq, in the current configuration. Displacement vectoruis the vector between points pandP. . . . . . . . . . . . . . . . . . 116 4.3 The right angle between line segments APandBPin the reference configu- ration becomes , the angle between segments apandbp, in the deformed configuration. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122 4.4 A rectangular parallelpiped with edge lengths dX(1),dX(2)anddX(3)in the reference configuration becomes a skewed parallelpiped with edge lengths dx(1),dx(2)anddx(3)in the deformed configuration. . . . . . . . . . . . . . 124 4.5 Typical Mohr’s circle for strain. . . . . . . . . . . . . . . . . . . . . . . . . . . 125 4.6 Rotation of axes for plane strain. . . . . . . . . . . . . . . . . . . . . . . . . . 125 4.7 Differential velocity field at point p. . . . . . . . . . . . . . . . . . . . . . . . 138 4.8 AreadS0between vectors dX(1)anddX(2)in the reference configuration becomesdSbetweendx(1)anddx(2)in the deformed configuration. . . . . 143 4.9 Volume of parallelpiped defined by vectors dX(1),dX(2)anddX(3)in the reference configuration deforms into volume defined by parallelpiped de- fined by vectors dx(1),dx(2)anddx(3)in the deformed configuration. . . . 145 5.1 Material body in motion subjected to body and surface forces. . . . . . . . . 172 5.2 Reference frames Ox1x2x3andO+x+ 1x+ 2x+ 3differing 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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217 6.3 Axes rotations for plane stress. . . . . . . . . . . . . . . . . . . . . . . . . . . 220 6.4 Geometry and transformation tables for reducing the elastic stiffness to the isotropic case. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222 6.5 Beam geometry for the Saint-Venant problem. . . . . . . . . . . . . . . . . . 226 6.6 Geometry and kinematic definitions for torsion of a circular shaft. . . . . . 229 6.7 The more general torsion case of a prismatic beam loaded by self equili- brating moments. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230 6.8 Representative figures for plane stress and plain strain. . . . . . . . . . . . . 239 6.9 Differential stress element in polar coordinates. . . . . . . . . . . . . . . . . 245 8.1 Nominal stress-stretch curves for rubber and steel. Note the same data is plotted in each figure, however, the stress axes have different scale and a different 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. . . . . . . . . . . . . . . 288 8.4 Rubber specimen having original length L0and cross-section area A0stretched into deformed shape of length Land cross section area A. . . . . . . . . . . 291 8.5 Rhomboid rubber specimen compressed by platens. . . . . . . . . . . . . . . 301 8.6 Rhomboid rubber specimen compressed by platens. . . . . . . . . . . . . . . 302 9.1 Simple shear element representing a material cube undergoing pure shear loading. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311 9.2 Mechanical analogy for simple shear. . . . . . . . . . . . . . . . . . . . . . . 312 9.3 Viscous flow analogy. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313 9.4 Representations of Kelvin and Maxwell models for a viscoelastic solid and fluid, respectively. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314 9.5 Three parameter standard linear solid and fluid models. . . . . . . . . . . . 314 9.6 Generalized Kelvin and Maxwell models constructed by combining basic models. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315 9.7 Graphic representation of the unit step function (often called the Heaviside step function). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 316 9.8 Different types of applied stress histories. . . . . . . . . . . . . . . . . . . . . 319 9.9 Stress history with an initial discontinuity. . . . . . . . . . . . . . . . . . . . 319 9.10 Different types of applied stress histories. . . . . . . . . . . . . . . . . . . . . 322 A.1 A set of non-orthonormal base vectors. . . . . . . . . . . . . . . . . . . . . . 344 A.2 Circular-cylindrical coordinate system for x3=0. . . . . . . . . . . . . . . . 349 This page intentionally left blank List of Tables 1.1 Historical notation for stress. . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 2.1 Indicial form for a variety of tensor quantities. . . . . . . . . . . . . . . . . . 16 2.2 Forms for inner and outer products. . . . . . . . . . . . . . . . . . . . . . . . 17 2.3 Transformation table between Ox1x2x3andOx0 1x0 2x0 3. . . . . . . . . . . . . . 25 3.1 Table displaying direction cosines of principal axes Px 1x 2x 3relative to axes Px1x2x3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 3.2 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. . . . . . . . . . . . . . . . 183 5.2 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 This page intentionally left blank Preface to the Third Edition First, a thank you to all the users of the second edition over the past years. We hope that you will find the updates made in the text make it a more valuable introduction for students to continuum mechanics. The changes made in this edition were substantial but we did not change the basic concept of the book. We seek to provide engineering students with 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. The impetus to do the third edition was to expand it into a text suitable for a two quarter graduate 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 is essential. One of the things that students struggle with in continuum mechanics and subsequent topics is notation. In the third edition, we have made some changes in notation making the book more consistent with modern continuum mechanics literature. Minor additions were 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 and solving basic elasticity problems. We have also added some new applications applying continuum mechanics to biological materials in light of their current importance. Finally, a limited amount of material using Matlabrhas been introduced in this edition. We did not want to minimize the fundamental principles of continuum mechanics by making the topic seem like it can be mastered by learning mathematical software. Yet at the same time, 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 such as Matlabr, 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 Mase San Luis Obispo, California, USA Ronald E. Smelser Charlotte, North Carolina, USA George E. Mase East Lansing, Michigan, USA This page intentionally left blank 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 who used the text over the last six years. Thanks also to those of you who have inquired about this revised and expanded version. We hope that you find this edition as helpful as the first 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 at Kettering 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 to provide the students with the fundamentals of continuum mechanics and demonstrate its applications. Very often, students are interested in using sophisticated simulation programs that use nonlinear kinematics and a variety of constitutive relationships. Additions to the second edition have been made with these needs in mind. A student who masters its contents should have the mechanics foundation necessary to be a skilled user of today’s advanced design tools such as nonlinear, explicit finite elements. Of course, students need to augment the mechanics foundation provided herein with rigorous finite element training. Major highlights of the second edition include two new chapters, as well as significant expansion 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 to restrictions on constitutive equations. The first edition applications chapter covering elas- ticity and fluids has been split into two separate chapters. Elasticity coverage has been expanded 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 added to give students a molecular and phenomenological introduction to rubber-like materi- als. Finally, a chapter introducing students to linear viscoelasticity is given since many important modern polymer applications involve some sort of rate dependent material response. It is not easy singling out certain people in order to acknowledge their help while not citing others; however, a few individuals should be thanked. Ms. Sheri Burton was instrumental in preparation of the second edition manuscript. We wish to acknowledge the many useful suggestions by users of the previous edition, especially Prof. Morteza M. Mehrabadi, Tulane University, for his detailed comments. Thanks also go to Prof. Charles Davis, Kettering University, for helpful comments on the molecular approach to rubber and thermoplastic elastomers. Finally, our families deserve sincerest thanks for their 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 Mase Flint, Michigan, USAGeorge E. Mase East Lansing, Michigan, USA This page intentionally left blank 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 courses in engineering science such as elasticity, plasticity, viscoelasticity and fluid mechanics are 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 student with the necessary background in continuum theory to pursue a formal course in any of the aforementioned subjects. We believe that first-year graduate students, or upper-level undergraduates, in engineering or applied mathematics with a working knowledge of calculus and vector analysis, and a reasonable competency in elementary mechanics will be 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 continuum mechanics. Under a semester system, more time can be devoted to later chapters dealing with elasticity and fluid mechanics. For either a quarter or a semester system, the text is intended to be used in conjunction with a lecture course. The mathematics employed in developing the continuum concepts in the text is the algebra and calculus of Cartesian tensors; these are introduced and discussed in some detail in Chapter Two, along with a review of matrix methods, which are useful for computational purposes in problem solving. Because of the introductory nature of the text, curvilinear coordinates are not introduced and so no effort has been made to involve general tensors in this work. There are several books listed in the Reference Section that a student may refer to for a discussion of continuum mechanics in terms of general tensors. Both indicial and symbolic notations are used in deriving the various equations and formula of importance. Aside from the essential mathematics presented in Chapter Two, the book can be seen as 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 physical laws relating to continuity, energy and momentum in Chapter Five. The second portion, Chapter Six, presents some elementary applications of continuum mechanics to linear elasticity 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 discussion as 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 constitutive equation 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 may reinforce her/his understanding of the material presented in the text. In all, 186 such practice 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 preparation of this book. Although we are unable to cite each of them individually, we are pleased to 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 their encouragement from beginning to end of this work. G. Thomas Mase Flint, Michigan, USAGeorge E. Mase East Lansing, Michigan, USA Acknowledgments There are too many people to thank for their help in preparing this third edition. We can only mention the key contributors. Ryan Miller was a superb help in moving the early manuscript into L ATEX 2"before his masters research redirected his focus. We were fortunate that one of us (GTM) was teaching ME 501 and 503 in the fall and winter quarters at Cal Poly while preparing the manuscript. The class was quite helpful in proofreading the manuscript. Specifically, Nickolai Volkoff-Shoemaker, Peter Brennen, Roger Sharpe, John Wildharbor, Kevin Ng, and Jason Luther found many typographical errors 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 to the 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) was privileged to benefit from interactions with several outstanding colleagues and teachers including Ronald Huston, University of Cincinnati, William J. Shack, MIT and Argonne National Laboratories, Morton E. Gurtin, Carnegie Mellon University and the late Owen Richmond, US Steel Research Laboratories and Alcoa Technical Center. Of course, our greatest thanks go to our families who very patiently kept asking if the book was done. Now it is done; so we can spend more time with the ones we love. This page intentionally left blank Authors G. Thomas Mase, Ph.D., is Associate Professor of Mechanical Engineering at California Polytechnic 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 and academia. Industrial companies Dr. Mase has worked full time for include General Mo- tors Research Laboratories, Callaway Golf and Acushnet Golf Company. He has taught or held research positions at the University of Wyoming, Kettering University, Michi- gan State University and California Polytechnic State University. Dr. Mase is a member of numerous professional societies including the American Society of Mechanical En- gineers, American Society for Engineering Education, International Sports Engineering Association, Society of Experimental Mechanics, Pi Tau Sigma and Sigma Xi. He received an ASEE/NASA Summer Faculty Fellowship in 1990 and 1991 to work at NASA Lewis Research Center (currently NASA Glenn Research Center). While at the University of California, he twice received a distinguished teaching assistant award in the Department of Mechanical Engineering. His research interests include mechanics, design and appli- cations of explicit finite element simulation. Specific areas include golf equipment design and performance and vehicle crashworthiness. Ronald E. Smelser, Ph.D., P .E., is Professor and Associate Dean for Academic Affairs in the William States Lee College of Engineering at the University of North Carolina at Charlotte. 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) in mechanical engineering at Carnegie Mellon University. He gained industrial experience working for the United States Steel Research Laboratory, the Alcoa Technical Center, and Concurrent Technologies Corporation. Dr. Smelser served as a fulltime or adjunct faculty member at the University of Pittsburgh, Carnegie Mellon University, and the University of Idaho and was a visiting research scientist at Colorado State University. Dr. Smelser is a member of the American Academy of Mechanics, the American Society for Engineering Education, Pi Tau Sigma, Sigma Xi, and Tau Beta Pi. He is also a member and Fellow of the American Society of Mechanical Engineers. Dr. Smelser’s research interests are in the areas of process modeling including rolling, casting, drawing and extrusion of single and multi-phase materials, the micromechanics of material behavior and the inclusion of material 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 Ohio State University, Columbus. He completed his Ph.D. in Mechanics at Virginia Polytechnic Institute and State University (VPI), Blacksburg, Virginia (1958). Previous to his initial appointment as Assistant Professor in the Department of Applied Mechanics at Michigan State 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 served as acting chairperson of the MMM Department 1965-1966 and again in 1978 to 1979. He taught 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 and publications were in the areas of continuum mechanics, viscoelasticity and biomechanics. Nomenclature xorxi Spatial or current coordinates XorXA Material or referential coordinates uorui Displacement components or displacement vector vorvi Velocity or general vector aorai 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 ij Kronecker delta Aoraij Transformation or general matrix I Identity matrix "ijk Permutation symbol @t Partial derivative with respect to time ˙(
) Derivative with respect to time ror@x Spatial gradient operator r=grad=;j Scalar gradient rv=@jvi=vi;j Vector gradient r
vorvi;i Divergence of a vector v rvor"ijkvi;j Curl of a vector v d=dt =@=@t+vk@=@xk Material derivative operator borbi Body force (force per unit mass) porpi Body force (force per unit volume) forfi Surface force (force per unit area) V,V0Current and referential total volumes
V,dV Small and infinitesimal element of volumes S,S0Current and referential total surfaces
S,dS Small and infinitesimal elements of surface  Density ^norni Unit normal in current configuration ^NorNA Unit normal in reference configuration t(^n)ort(^n) i Traction vector N,S Normal and shear components of traction vector Tortij Cauchy stress or general tensor Tort ij Cauchy stress referred to principal axes p0(^N)orp0(^N) i Piola-Kirchhoff stress vector referred to ref- erential area PorPiA First Piola-Kirchhoff stress sorsAB Second Piola-kirchhoff stress (1); (2); (3) Principal stress values IT; IIT; IIIT First, second and third stress invariants M=1 3tii Mean normal stress Sij Deviatoric stress components ij Deviatoric strain components J1=0; J2; J3 Deviatoric stress invariants oct Octahedral shear stress ForFiA Deformation gradient tensor CorCAB Right Cauchy-Green deformation tensor EorEAB Lagrangian finite strain tensor corcij Cauchy deformation tensor eoreij Eulerian finite strain tensor orij Infinitesimal strain tensor (1); (2); (3) Principal strain values I; II; III First, second and third infinitesimal strain invariants BorBij Left Cauchy-Green deformation tensor IB; IIB; IIIB, or I1; I2; I3Invariants of right deformation tensor W Strain energy per unit volume or strain en- ergy density e^NNormal strain in the ^Ndirection ij Engineering shear strain e=
V=V =ii=I Cubical dilatation !or!ij Infinitesimal rotation tensor !or!j Rotation vector ^N=dx=dX Stretch ratio or stretch in the direction of ^N ^n=dX=dx Stretch ratio in the direction of ^n RorRij Rotation tensor UorUAB Right stretch tensor VorVij Left stretch tensor LorLij Spatial velocity gradient Dordij Rate of deformation tensor Worwij Vorticity, or spin tensor J=detF Jacobian P(t) orPi Linear momentum vector K(t) Kinetic energy P(t) Mechanical power or rate of work done by forces S(t) Stress work Q Heat input rate r Heat supply per unit mass qorqi Heat flux vector  Temperature or angle g=gradorgi=;i Temperature gradient u Specific internal energy  Specific entropy or viscoelastic viscosity Gibbs’ free energy  Free enthalpy  Enthalpy Specific entropy production E modulus of elasticity or Young’s modulus Gor shear modulus K Bulk modulus  Poisson’s ratio , Lam ´e constants Cijkl General elastic constants ij Viscous stress tensor ij Deviatoric rate of deformation ;Viscosity coefficients Bulk viscosity coefficient 1 Continuum Theory The atomic/molecular composition of matter is well established. On a small enough scale, a body of aluminum, is really a collection of discrete aluminum atoms stacked on one another in a particular repetitive lattice. And on an even smaller scale, the atoms consist of a core of protons and neutrons around which electrons orbit. Thus matter is not continuous. 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 find another 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, it is 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 distributed uniformly throughout, and completely fills the space it occupies. In keeping with this continuum model, we assert that matter may be divided indefinitely into smaller and smaller 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 each and every point of the region of space which the body occupies. The continuum model for material bodies is important to engineers for two very good reasons. On the scale by which we consider bodies of steel, aluminum, concrete, etc., the characteristic dimensions are extremely large compared to molecular distances so that the continuum model provides a very useful and reliable representation. Additionally, our knowledge of the mechanical behavior of materials is based almost entirely upon experimental data gathered by tests on relatively large specimens. The analysis of the kinematic and mechanical behavior of materials modeled on the continuum assumption is what we know as Continuum Mechanics. There are two main themes into which the topics of continuum mechanics are divided. In the first, emphasis is 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 of mass, the principles of energy and momentum, etc. In the second, the focus of attention is on the development of the constitutive equations characterizing the behavior of specific idealized materials; the perfectly elastic solid and the viscous fluid being the best known examples. 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 above may 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 applied to a finite volume of the material. The other, a differential, or field approach, leads to equations 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 field equations from their global counterparts. As a result of the continuum assumption, field quantities such as density and velocity which reflect the mechanical or kinematic properties of continuum bodies are expressed mathematically as continuous functions, or at worst as piecewise continuous functions, of the space and time variables. Moreover, the derivatives of such functions, if they enter into the theory at all, will be likewise continuous. Inasmuch as this is an introductory textbook, we shall make two further assumptions on the materials we discuss in addition to the principal one of continuity. First, we require the materials to be homogeneous, that is to have identical properties at all locations. And secondly, 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 have important 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 at the beginning of a graduate program. Continuum mechanics has a reputation of being a theoretical course without many applications (during the course). The first of part of continuum mechanics’ reputation is correct: it is based on fundamental mathematics and mechanics. However, the second part is not founded. There are many, many applications for continuum mechanics, but it is hard to cover the basics and develop the applications in a single quarter or semester. Continuum mechanics takes all the mathematical, physical and engineering principles and casts them in a single structure from which the student is prepared 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 sophisticated simulation programs such as LS-DYNArbecome a playground for advanced design and analysis once continuum mechanics has been mastered. Some students find continuum mechanics a difficult subject. However, outside of a new notation, the topics studied should be very familiar to the student. Vectors have to be 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 description for conservation of linear and angular momentum. Taking the time rate of change of these quantities is really no different than what was done in an undergraduate course in 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 introduced in a mechanics of solids class. Stress and strain are tensors which are an order more complex than vectors. When looking at strain and the resulting stress one needs to have a material model. At the undergraduate level students have studied linear elastic, fluid and gas behavior. But the topics generally are taught in separate courses, and often the common, underlying theory is not noticed. Also, the methods used to determine the relationship 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 to start over and put all that was studied previously under a single umbrella. A course in continuum 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 developing a common notation, but no new physics is presented. Continuum mechanics is just the process of confirming the foundation for all that was done in undergraduate studies. 1.3 Notation As 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 learning continuum mechanics. In the pages that follow there are many different symbols used for all the quantities of interest. There are more symbols than a student experienced as an undergraduate because a general, nonlinear theory is being contructed. For instance, consider stress. There are different measures of stress that are indistinguishable in the linear theory: Cauchy, first Piola-Kirchhoff, and second Piola-Kirchhoff. In addition, von Mises 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 of the 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, one symbol is very close to another symbol, and the context has to be used to fully understand the meaning. This is one of the things that makes continuum mechanics difficult for the beginner. Finally, as students continue beyond this course, they find the disheartening reality that not everybody uses the same notation. Often notational marks will have to be made in margins when reading the literature. The reason there is not a single notation comes from the fact that people were not flying in airplanes to technical conferences when this material was being developed. Even when reading a single author’s works spanning a decade the notation can change. For example, Table 1.11shows some historical notation for stress. 1Adapted from Nonlinear Theory of Continuous Media, A. Cemal Eringen, McGraw-Hill Inc., (1962) 4 Continuum Mechanics for Engineers TABLE 1.1 Historical notation for stress. Naghdi, Eringen, Clebsch, Truesdell t11t22t33t12t23t31 Cauchy (early work) ABCDEF Cauchy (later work), St. Venant, Maxwell pxxpyypzzpxypyzpzx F. Neumann, Kirchhoff, Love XxYyZzXyYzZx Green and Zerna, Russian and German writers 112233122331 Karman, Timoshenko xyzxyyzzx Some English and American writers 112233122331 xxyyzzxyyzzx 2 Essential 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 calculus oftensors. Here, tensor is the generic name for those mathematical entities which are used to represent the important physical quantities of continuum mechanics. A tensor, or linear transformation, assigns any vector vto another vector Tvsuch that T(v+w)=Tv+Tw; (2.1a) and T(v)=Tv (2.1b) for allvandw. Furthermore, the sum of two tensors and the scalar multiple of a tensor is defined by (T+S)v=Tv+Sv; (2.1c) and (T)v=(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 reference frame, 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 components may be different. As long as the relationship between the reference frames is known, the components 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 presented in 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 be written 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 the derivation or analysis at hand but taking care to establish the inter-relationships between the two. However, an effort to emphasize indicial notation in most of the text has been made. An introductory course must teach indicial notation to the student who may have little prior exposure to the topic. 2.1 Scalars, Vectors and Cartesian Tensors A considerable variety of physical and geometrical quantities have important roles in continuum mechanics, and fortunately, each of these may be represented by some form 5 6 Continuum Mechanics for Engineers 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 quantities are represented mathematically by scalars, which are referred to as zero-order tensors. It should be emphasized that scalars are not constants, but may actually be functions of position and/or time. Also, the exact numerical value of a scalar will depend upon the units in which it is expressed. Thus, the temperature may be given by either 68F, or 20C at a certain location. As a general rule, lower-case Greek letters in italic print such as , ,, 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 an assignment 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 substantially different than a 20 N force acting horizontally at the point. Quantities possessing such directional 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 head and 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) in accordance with the triangle rule by which the arrow representing the vector sum of two vectors extends from the tail of the first component arrow to the head of the second when the component arrows are arranged ”head-to-tail”. Although vectors are independent of any particular coordinate system, it is often useful to 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 to Cartesian tensors, we limit ourselves to consideration of Cartesian coordinate systems for designating 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 representation in the hierarchy of tensors. As we shall see, among the best known of these are the stress and the strain tensors. These particular tensors are second-order tensors, and are said to have 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-order tensor. With only a few exceptions, primarily those representing the stress and strain tensors, we shall denote second-order tensors by upper-case sans serif Latin letters in bold-faced print, a typical example being the tensor T. The components of the said tensor will, 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 suitable set of reference axes. The precise definitions of tensors of various order will be given subsequently in terms of the transformation properties of their components between two related 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 or Greek style fonts depending on the physical quantity they represent. The following examples are a partial list of scalar notation: (a)a– magnitude of acceleration Essential Mathematics 7 (b)v– magnitude of velocity (c)r– radius (d)– temperature or angle depending on context (e)– coefficient of thermal expansion (f)– principal value of stress (g)– 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 x1direction 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 case of several well known engineering quantities this convention will not be accommo- dated. For example, linear strain has been chosen to be represented by . Here are some samples of tensor and matrix symbols: (a)Q– orthogonal matrix (b)E– finite strain (c)T– Cauchy stress tensor (d)– infinitesimal strain tensor (e)R– rotation matrix 2.2 Tensor Algebra in Symbolic Notation - Summation Convention The three-dimensional physical space of everyday life is the space in which many of the events of continuum mechanics occur. Mathematically, this space is known as a Euclidean three-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 system of 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 with these axes the triad of unit base vectors, ^e1,^e2,^e3, shown in Fig. 2.1(a). All unit vectors in this text will be written with a caret placed above the bold-faced symbol. Due to the mutual 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 this basis an arbitrary vector vis given in component form by v=v1^e1+v2^e2+v3^e3=3X i=1vi^ei: (2.2) 8 Continuum Mechanics for Engineers x1x2x3 O ^e2 ^e1^e3 (a) Unit vectors in the coordinate directions x1, x2andx3. x1x2x3 O v1v2v3 v(b) Rectangular Cartesian components of the vectorv. FIGURE 2.1 Base vectors and components of a Cartesian vector. This vector and its coordinate components are pictured in Fig. 2.1(b). For the symbolic description, vectors will usually be given by lower-case Latin letters in bold-faced print, with the vector magnitude denoted by the same letter. Thus vis the magnitude of v. At this juncture of our discussion it is helpful to introduce a notational device called the summation convention that will greatly simplify the writing of the equations of continuum mechanics. Stated briefly, we agree that whenever a subscript appears exactly twice in a given term, that subscript will take on the values 1, 2, 3 successively, and the resulting terms summed. For example, using this scheme, we may now write Eq 2.2 in the simple form v=vi^ei; (2.3) and delete entirely the summation symbolP. For Cartesian tensors, only subscripts are required on the components; for general tensors both subscripts and superscripts are used. The summed subscripts are called dummy indices since it is immaterial which par- ticular letter is used. Thus vi^eiis completely equivalent to vj^ej, or tovk^ek, when the summation convention is used. A word of caution, however; no subscript may appear more than twice in a singe term. But as we shall soon see, more than one pair of dummy indices 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 the unit vectors and the scalar coefficients. Example 2.1 Without regard for meaning as far as mechanics is concerned, expand the following expressions according to the summation convention: (a)uiviwj^ej(b)tijvi^ej(c)tiivj^ej Essential Mathematics 9 Solution (a) Summing rst on i, and then on j uiviwj^ej= (u1v1+u2v2+u3v3)(w1^e1+w2^e2+w3^e3) (b) Summing on i, then onjand collecting terms on the unit vectors. tijvj^ei=t1jvj^e1+t2jvj^e2+t3jvj^e3 =(t11v1+t12v2+t13v3)^e1+(t21v1+t22v2+t23v3)^e2 +(t31v1+t32v2+t33v3)^e3 (c) Summing on i, then onj, tiivj^ej=(t11+t22+t33) (v1^e1+v2^e2+v3^e3) Note the similarity between (a) and (c). With the above background in place it is now possible, using symbolic notation, to present many useful definitions from vector/tensor algebra. There are two symbols needed prior to writing out all of the vector and tensor algebra necessary. These two symbols are the Kronecker delta and the permutation symbol. Additionally, there are several useful relationships between the Kronecker delta and permutation symbol that are used throughout continuum mechanics. The following three subsections introduce the 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 quickly embrace this new entity. However, the permutation symbol is a little more abstract than the Kronecker delta since it cannot be represented by a matrix. In subsequent chapters the Kronecker delta and the permutation symbol play integral roles in describing how forces 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 j 0 if numerical value of i6=numerical value of j: Therefore, if we introduce the Kronecker delta defined by ij= 1 if numerical value of i= numerical value of j 0 if numerical value of i6=numerical value of j we see that ^ei
^ej=ij (i;j=1;2;3 ): (2.4) 10 Continuum Mechanics for Engineers Also, note that by the summation convention ii=jj=11+22+33=1+1+1=3; and furthermore, we call attention to the substitution property of the Kronecker delta by expanding (summing on j) the expression ij^ej=i1^e1+i2^e2+i3^e3: But for a given value of iin this equation, only one of the Kronecker deltas on the right hand side is non-zero, and it has the value one. Therefore, ij^ej=^ei; and the Kronecker delta in ij^ejcauses the summed subscript jof^ejto be replaced by i reducing the expression to simply ^ei. 2.2.2 Permutation Symbol By introducing the permutation symbol "ijkdefined by "ijk=8 >< >:1 if numerical values of ijkappear as in the sequence 12312 -1 if numerical values of ijkappear as in the sequence 32132 0 if numerical values of ijkappear 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="ijk^ek (i;j;k =1;2;3 ): (2.6) Also, note from its definition that the interchange of any two subscripts in "ijkcauses a sign change so that for example, "ijk= -"kji="kij= -"ikj; and, furthermore, that for repeated subscripts is zero as in "113="212="133="222=0: 2.2.3"-Identity The product of permutation symbols "miq"jkqmay be expressed in terms of Kronecker deltas by the "-identity "miq"jkq=mjik-mkij (2.7a) as may be proven by direct expansion. This is a most important formula used throughout this text and is well worth memorizing. Also, by the sign-change property of "ijk, "miq"jkq="miq"qjk="qmi"qjk="qmi"jkq: Additionally, it is easy to show from Eq 2.7a that "jkq"mkq=2jm; (2.7b) by settingi=k, and "jkq"jkq=6: (2.7c) Essential Mathematics 11 2.2.4 Tensor/Vector Algebra To begin with, vector addition is easily written in indicial form w=u+vor w i^ei=(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 multiplication can be defined in symbolic form. Multiplication of a vector by a scalar is written as v=vi^ei; (2.9) and the dot (scalar) product between of two vectors is u
v=v
u=uvcos (2.10) whereis the smaller angle between the two vectors when drawn from a common origin. From the definition of ijand its substitution property the dot productu
vmay be written as u
v=ui^ei
vj^ej=uivj^ei
^ej=uivjij=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 uv= -vu=(uvsin)^e where066, is the angle between the two vectors when drawn from a common origin, and where ^eis a unit vector perpendicular to their plane such that a right-handed rotation about ^ethrough the angle carriesuintov. The vector cross product may be written in terms of the permutation symbol (Eq 2.5) as follows: uv=ui^eivj^ej=uivj(^ei^ej)="ijkuivj^ek: (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) is u
vw=uv
w=[u;v;w]; or [u;v;w]=ui^ei
(vj^ejwk^ek)=ui^ei
"jkqvjwk^eq (2.13) ="jkquivjwkiq="ijkuivjwk where in the final step we have used both the substitution property of iqand the sign- change property of "ijk. The triple cross product is similar to the triple scalar product u(vw)=ui^ei(vj^ejwk^ek)=ui^ei("jkqvjwk^eq) (2.14) ="iqm"jkquivjwk^em="miq"jkquivjwk^em which may be written as u(vw)=(mjik-mkij)uivjwk^em (2.15) =(uivmwi-uiviwm)^em=uiwivm^em-uiviwm^em: 12 Continuum Mechanics for Engineers Observation of the indices in Eq 2.15 admits u(vw)=(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=ui^eivj^ej=uivj^ei^ej (2.16) which in expanded form, summing first on i, yields uivj^ei^ej=u1vj^e1^ej+u2vj^e2^ej+u3vj^e3^ej; and then summing on j uivj^ei^ej=u1v1^e1^e1+u1v2^e1^e2+u1v3^e1^e3 +u2v1^e2^e1+u2v2^e2^e2+u2v3^e2^e3 (2.17) +u3v1^e3^e1+u3v2^e3^e2+u3v3^e3^e3: 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=ai^ei bj^ej=aibj^ei ^ej (2.19) which is called a tensor product. The tensor product of vectors aandbis defined by how a bmaps all vectors u: (a b)u=a(b
u): (2.20) If one takes vectors a,banduto be vectors from a Euclidian 3-space, the expanded form for the tensor product may be written as (a b)u=a(b
u)=2 64a1 a2 a33 75(b1u1+b2u2+b3u3) =2 64a1b1u1+a1b2u2+a1b3u3 a2b1u1+a2b2u2+a2b3u3 a3b1u1+a3b2u2+a3b3u33 75 =2 64a1b1a1b2a1b3 a2b1a2b2a2b3 a3b1a3b2a3b33 752 64u1 u2 u33 75 =0 B@2 64a1 a2 a33 75h b1b2b3i1 CA2 64u1 u2 u33 75:(2.21) Essential Mathematics 13 The last line of Eq 2.21 shows why the tensor product is sometimes called the outer product. The inner product, of course, is defined by a
b=h a1a2a3i2 64b1 b2 b33 75=a1b1+a2b2+a3b3: The second line of Eq 2.21 shows the tensor product to be equivalent to the dyad product of 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)=ui^ei
(vj^ejwk^ek)=uiviwk^ek (2.22) 2. (uv)
w=(ui^eivj^ej)
wk^ek=uivjwj^ei (2.23) 3.u(vw)=(ui^eivj^ej)wk^ek="ijquivjwk^eq^ek (2.24) 4. (uv)w=ui^ei(vj^ejwk^ek)="jkquivjwk^ei^eq (2.25) (Note that in products 3 and 4 the order of the base vectors ^eiis important.) Dyads can be multiplied by each other to yield another dyad (uv)
(ws)=ui^ei(vj^ej
wk^ek)sq^eq=uivjwjsq^ei^eq: (2.26) Vectors can be multiplied by a tensor to give a vector. The reduction in the order of the tensor is why the “dot” is used in direct notation. 1.v
T=vi^ei
tjk^ej^ek=vitjkij^ek=vitik^ek (2.27) 2.T
v=tij^ei^ej
vk^ek=tij^eijkvk=tijvj^ei (2.28) (Note that these products are also written as simply vTandTv.) Finally, two tensors can be multiplied resulting in a tensor T
S=tij^ei^ej
spq^ep^eq=tijspqjp^ei^eq=tijsjp^ei^eq: (2.29) Example 2.2 Let the vector vbe given byv=(a
^n)^n+^n(a^n)whereais an arbitrary vector, and ^nis a unit vector. Express vin terms of the base vectors ^ei, expand, and simplify. (Note that ^n
^n=ni^ei
nj^ej=ninjij=nini=1.) Solution In terms of the base vectors ^ei, the given vector vis expressed by the equation v=(ai^ei
nj^ej)nk^ek+ni^ei(aj^ejnk^ek): We note here that indices i,j, andkappear four times in this line, however, the summation convention has not been violated. Terms that are separated by a plus or a minus sign are considered dierent terms each having summation convention 14 Continuum Mechanics for Engineers rules applicable within them. Vectors joined by a dot or cross product are not distinct terms, and the summation convention must be adhered to in that case. Carrying out the indicated multiplications, we see that v=(ainjij)nk^ek+ni^ei("jkqajnk^eq) =ainink^ek+"iqm"jkqniajnk^em =ainink^ek+"miq"jkqniajnk^em =ainink^ek+(mjik-mkij)niajnk^em =ainink^ek+niajni^ej-niaink^ek =niniaj^ej=aj^ej=a: Sinceamust equalv, this example demonstrates that the vector vmay be resolved into a component (v
^n)^nin the direction of ^n, and a component ^n(v^n)perpendicular to ^n. Example 2.3 Using Eq 2.7 show that (a) "mkq"jkq=2mjand that (b)"jkq"jkq=6. (Recall thatkk=3andmkkj=mj.) Solution (a) Write out Eq 2.7a with index ireplaced bykto get "mkq"jkq =mjkk-mkkj =3mj-mj=2mj: (b) Start with the rst equation in Part (a) and replace the index mwithj, giving "jkq"jkq =jjkk-jkjk =(3) (3)-jj=9-3=6: Example 2.4 Double dot products of dyads are dened by (a)(uv)

(ws)=(v
w) (u
s) (b)(uv):(ws)=(u
w) (v
s): Expand these products and compare the component forms. Essential Mathematics 15 Solution (a)(uv)

(ws)=(vi^ei
wj^ej) (uk^ek
sq^eq)=viwiuksk (b)(uv):(ws)=(ui^ei
wj^ej) (vk^ek
sq^eq)=uiwivksk Summary of Symbolic Notations 1.Addition of vectors, Eq 2.8: w=u+vor w i^ei=(ui+vi)^ei 2.Multiplication: (a)of a vector by a scalar, Eq 2.9: v=vi^ei (b) dot (scalar) product of two vectors, Eq 2.11: u
v=v
u=uvcos=uivi (c)cross (vector) product of two vectors, Eq 2.12: uv= -vu=(uvsin)^e="ijkuivj^ek (d)triple scalar product (box product), Eq 2.13: [u;v;w]=ui^ei
(vj^ejwk^ek)=ui^ei
"jkqvjwk^eq ="jkquivjwkiq="ijkuivjwk (e)triple cross product, Eq 2.14: u(vw)=ui^ei(vj^ejwk^ek)=ui^ei("jkqvjwk^eq) ="iqm"jkquivjwk^em="miq"jkquivjwk^em (f)tensor product of two vectors (dyad), Eq 2.16: uv=ui^eivj^ej=uivj^ei^ej=ui^ei vj^ej=uivj^ei ^ej (g)Vector-dyad products, Eqs 2.22–2.25: 1.u
(vw)=ui^ei
(vj^ejwk^ek)=uiviwk^ek 2. (uv)
w=(ui^eivj^ej)
wk^ek=uivjwj^ei 3.u(vw)=(ui^eivj^ej)wk^ek="ijquivjwk^eq^ek 4. (uv)w=ui^ei(vj^ejwk^ek)="jkquivjwk^ei^eq 16 Continuum Mechanics for Engineers TABLE 2.1 Indicial form for a variety of tensor quantities. =scalar (zeroth-order tensor)  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)=ui^ei(vj^ej
wk^ek)sq^eq=uivjwjsq^ei^eq (i)vector-tensor products, Eqs 2.27–2.28: 1.v
T=vi^ei
tjk^ej^ek=vitjkij^ek=vitik^ek 2.T
v=tij^ei^ej
vk^ek=tij^eijkvk=tijvj^ei (j)tensor-tensor product, Eq 2.29: T
S=tij^ei^ej
spq^ep^eq=tijsjp^ei^eq (k)double dot product: 1.(uv)

(ws)=(vi^ei
wj^ej) (uk^ek
sq^eq)=viwiuksk 2.(uv):(ws)=(ui^ei
wj^ej) (vk^ek
sq^eq)=uiwivksk 2.3 Indicial Notation By assigning special meaning to the subscripts, indicial notation permits us to carry out the tensor operations of addition, multiplication, differentiation, etc. without the use, or even the appearance of the base vectors ^eiin the equations. We simply agree that the tensor 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 a scalar, a term with one free index a vector, a term having two free indices a second order tensor, 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 values 1,2,3 successively, and we say that these indices have a range of three. IfNis the number of free indices in a tensor, that tensor has 3Ncomponents in three space. We must emphasize that in the indicial notation exactly two types of subscripts appear: Essential Mathematics 17 TABLE 2.2 Forms for inner and outer products. Outer Products: Contraction(s): Inner Products: uivj i=j uivi(vector dot product) "ijkuqvmj=q;k=m " ijkujvk(vector cross product) "ijkuqvmwni=q;j=m;k=n "ijkuivjwk(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 only twice in a given term. Furthermore, every term in a valid equation must have the same letter subscripts for the free 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 to the typical equations; ui+vi-wi=sifor vectors, and tij-vij+sij=qijfor second-order tensors. Multiplication of two tensors to produce an outer tensor product is accomplished by 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 viand tensor tjkis the third-order tensor vitjk.Contraction is the process of identifying (that is, setting equal to one another) any two indices of a tensor term. An inner tensor product is formed from an outer tensor product by one or more contractions involving indices from separate tensors in the outer product. We note that the rank of a given tensor is reduced by twofor each contraction. Some outer products, which contract, form well-known inner products listed in Table 2.2. A tensor is symmetric in any two indices if interchange of those indices leaves the tensor value unchanged. For example, if sij=sjiandcijm=cjim, both of these tensors are said to be symmetric in the indices iandj. A tensor is anti-symmetric (orskew-symmetric) in any two indices if interchange of those indices causes a sign change in the value of the tensor. Thus if aij= -aji, it is anti-symmetric in iandj. Also, recall that by definition, "ijk= -"jik="jki, etc., and hence the permutation symbol is anti-symmetric in all indices. Example 2.5 Show that the inner product sijaijof a symmetric tensor sij=sji, and an anti-symmetric tensor aij= -ajiis zero. Solution By denition of symmetric tensor sijand 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, orsijaij=0. 18 Continuum Mechanics for Engineers One of the most important advantages of the indicial notation is the compactness it provides in expressing equations in three dimensions. A brief listing of typical equations of continuum mechanics is presented below to illustrate this feature. 1:  =sijtij-siitjj (1 equation, 18 terms on RHS) 2: ti=qijnj (3 equations, 3 terms on RHS of each) 3: tij=ijEkk+2Eij (9 equations, 4 terms on RHS of each) Example 2.6 By direct expansion of the expression vi="ijkwjkdetermine the components of the vector viin terms of the components of the tensor wjk. Solution By summing rst on jand then onkand then omitting the zero terms, we nd that vi="i1kw1k+"i2kw2k+"i3kw3k ="i12w12+"i13w13+"i21w21+"i23w23+"i31w31+"i32w32: Therefore, v1="123w23+"132w32=w23-w32; v2="213w13+"231w31=w31-w13; v3="312w12+"321w21=w12-w21: Note that if the tensor wjkwere symmetric, the vector viwould be a null (zero) vector. To end this section, consider a skew–symmetric second order tensor W=wij^ei^ej. All skew–symmetric tensors can be represented in terms of an axial vector by using the permutation symbol. Let the axial vector for wijbe!idefined by !i= -1 2"ijkwjk: (2.30) It is an easy exercise in indicial manipulation to find the inverse of Eq 2.30. We can write wjkin terms of!ias follows: "imn!i= -1 2"imn"ijkwjk = -1 2(mjnk-mknj)wjk = -1 2(wmn-wnm) = -1 2(2wmn) = -wmn(2.31) where Eq 2.7a and the skew–symmetric property of wijwere used. Essential Mathematics 19 2.4 Matrices and Determinants For computational purposes it is often expedient to use the matrix representation of vectors and tensors. Accordingly, we review here several definitions and operations of elementary matrix theory. Amatrix is an ordered rectangular array of elements enclosed by square brackets and subjected to certain operational rules. The typical element Aijof the matrix is located in theith(horizontal) row, and in the jth(vertical) column of the array. A matrix having elements Aij, which may be numbers, variables, functions, or any of several mathematical entities, is designated by [Aij], or symbolically by the kernel letter A. AnMbyNmatrix (writtenMN) hasMrows andNcolumns, and may be displayed as A=[Aij]=2 66664A11A12


A1N A21A22


A2N ......... AM1AM2


AMN3 77775: (2.32) IfM=N, the matrix is a square matrix. A 1Nmatrix [A1N]is an row matrix, and an M1 matrix [AM1]is acolumn matrix. Row and column matrices represent vectors, whereas a 33square matrix represents a second-order tensor. A scalar is represented by a 11 matrix (a single element). The unqualified use of the word matrix in this text is understood to mean a33square matrix, that is, the matrix representation of a second-order tensor or a matrix that is not a tensor. Azero, or null matrix has all elements equal to zero. A diagonal matrix is asquare matrix whose elements not on the principal diagonal, which extends from A11toANN, are all zeros. Thus for a diagonal matrix, Aij=0fori6=j. The unit or identity matrix I, which, incidentally, is the matrix representation of the Kronecker delta, is a diagonal matrix whose diagonal elements all have the value one. TheNMmatrix formed by interchanging the rows and columns of the MNmatrix Ais called the transpose ofA, and is written as AT, or[Aij]T. By definition, the elements of a matrix Aand its transpose are related by the equation AT ij=Aji. A square matrix for which A=AT, or in element form, Aij=Aji, is called a symmetric matrix; one for which 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. Two matrices are equal if they are identical element by element. Matrices having the same number of rows and columns may be added (or subtracted) element by element. Thus if A=B+C, the elements of Aare given by Aij=Bij+Cij: (2.33) Addition of matrices is commutative, A+B=B+A, and associative, A+(B+C)= (A+B)+C. 20 Continuum Mechanics for Engineers Example 2.7 Show that the square matrix Acan be expressed as the sum of a symmetric and a skew-symmetric matrix by the decomposition A=A+AT 2+A-AT 2: Solution Let the decomposition be written as A=B+CwhereB=1 2 A+AT
and C=1 2 A-AT
. Then writing BandCin element form, Bij=Aij+AT ij 2=Aij+Aji 2=AT ji+Aji 2=Bji=BT ij (symmetric); Cij=Aij-AT ij 2=Aij-Aji 2= -AT ji-Aji 2= -Cji= -CT ij(skew-symmetric) . Therefore Bis symmetric, and Cskew-symmetric. Multiplication of the matrix Aby the scalar results in the matrix A, or [Aij]= [Aij]. The product of two matrices AandB, denoted by AB, is defined only if the matrices are conformable, that is, if the prefactor matrix Ahas the same number of columns as the postfactor matrix Bhas rows. Thus the product of an MQmatrix multiplied by aQNmatrix is anMNmatrix. The product matrix C=AB has elements given by Cij=AikBkj (2.34) in whichkis, of course, a summed index. Therefore each element Cijof the product matrix is an inner product of the ithrow of the prefactor matrix with the jthcolumn of the postfactor matrix. In general, matrix multiplication is not commutative, AB6=BA, but the associative and distributive laws of multiplication do hold for matrices. The product 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 the exponent rule AmAn=AnAm=Am+n(2.35) wheremandnare positive integers, or zero. Also, we note that (An)T= AT
n; (2.36) and ifBB=Athen B=p A=A1 2 (2.37) but the square root is not unique. Essential Mathematics 21 Example 2.8 Use indicial notation to show that for arbitrary matrices AandB: (a)(A+B)T=AT+BT (b)(AB)T=BTAT (c)IB=BI=BwhereIis the identity matrix. Solution (a) Let A+B=C, then in element form Cij=Aij+Bijand therefore CTis given by CT ij=Cji=Aji+Bji=AT ij+BT ji; or CT=(A+B)T=AT+BT: (b) Let AB=C, then in element form Cij=AikBkj=AT kiBT jk=BT jkAT ki=CT ji: Hence (AB)T=BTAT. Note that exchanging the order AT kiBT jk=BT jkAT ki is not necessary when using indicial notation. It is done for clarity. In direct or matrix notation the order of the terms is critical. (c) LetIB=C, then in element form Cij=ikBkj=Bij=Bikkj by the substitution property. Thus IB=BI=B. The determinant of a square matrix is formed from the square array of elements of the matrix and this array evaluated according to established mathematical rules. The determinant of the matrix Ais designated by either det A, or by jAijj, and for a33 matrix A, detA=jAijj=A11A12A13 A21A22A23 A31A32A33: (2.38) A minor of det Ais another determinant jMijjformed by deleting the ithrow andjth column of jAijj. The cofactor of the element Aij(sometimes referred to as the signed minor) is defined by A(c) ij=(-1)i+jjMijj (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 each element in that row (or column) is multiplied by its cofactor. The sum of these products 22 Continuum Mechanics for Engineers gives the value of the determinant. For example, expansion of the determinant of Eq 2.38 by the first row becomes detA=A11A22A23 A32A33-A12A21A23 A31A33+A13A21A22 A31A32(2.40) which upon complete expansion gives detA=A11(A22A33-A23A32)-A12(A21A33-A23A31) (2.41) +A13(A21A32-A22A31): Several interesting properties of determinants are worth mentioning at this point. To begin with, the interchange of any two rows (or columns) of a determinant causes a sign change in its value. Because of this property and because of the sign-change property of the permutation symbol, the det Aof Eq 2.38 may be expressed in the indicial notation by the alternative forms (see Prob. 2.13) detA="ijkAi1Aj2Ak3="ijkA1iA2jA3k: (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 of Eq 2.42 that, (see Prob 2.14) "qmn detA="ijkAiqAjmAkn: (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 33skew-symmetric matrix is singular. Also, the determinant of the diagonal matrix, D, is simply the product of its diagonal elements: det D= D11D22


DNN. Example 2.9 Show that for matrices AandB,det(AB) = det(BA) = det(A) det(B). Solution LetC=AB, then Cij=AikBkjand from Eq 2.42 detC="ijkCi1Cj2Ck3 ="ijkAiqBq1AjmBm2AknBn3 ="ijkAiqAjmAknBq1Bm2Bn3: But from Eq 2.43 "ijkAiqAjmAkn="qmn detA; so now Essential Mathematics 23 detC=detAB="qmnBq1Bm2Bn3detA=detAdetB: By a direct interchange of AandB,detAB=detBA. Example 2.10 Use Eq 2.40 and Eq 2.41 to show that detA=detAT. Solution Since AT=A11A21A31 A12A22A32 A13A23A33 cofactor expansion by the rst column here yields detA=A11A22A32 A23A33-A12A21A31 A23A33+A13A21A31 A22A32 which is equal to Eq 2.41. The inverse of the matrix Ais written A-1, and is defined by AA-1=A-1A=I (2.44) whereIis the identity matrix. Thus if AB=I, then B=A-1, andA=B-1. The adjoint matrix Ais defined as the transpose of the cofactor matrix A=h A(c)iT : (2.45) In terms of the adjoint matrix the inverse matrix is expressed by A-1=A detA(2.46) which is actually a working formula by which an inverse matrix may be calculated. This formula shows that the inverse matrix exists only if det A6=0, i.e., only if the matrix Ais non-singular. In particular, a 33skew-symmetric matrix has no inverse. 24 Continuum Mechanics for Engineers Example 2.11 Show from the denition 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 ABbyB-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. Anorthogonal matrix, call it Q, is a square matrix for which Q-1=QT. From this definition we note that a symmetric orthogonal matrix is its own inverse since in this case Q-1=QT=Q: (2.47) Also, if AandBare orthogonal matrices. (AB)-1=B-1A-1=BTAT=(AB)T; (2.48) so that the product matrix is likewise orthogonal. Furthermore, if Ais orthogonal it may shown (see Prob. 2.18) that detA=1: (2.49) As mentioned near the beginning of this section, a vector may be represented by a row or column matrix, a second-order tensor by a square 33matrix. For computational purposes, it is frequently advantageous to transcribe vector/tensor equations into their matrix form. As a very simple example, the vector-tensor product, u=Tv(symbolic notation) orui=Tijvj(indicial notation) appears in matrix form as [ui1]=[Tij] [vj1] or2 64u1 u2 u33 75=2 64T11T12T13 T21T22T23 T31T32T333 752 64v1 v2 v33 75: (2.50) Essential Mathematics 25 TABLE 2.3 Transformation table between Ox1x2x3and Ox0 1x0 2x0 3. ^e1; x1 ^e2; x2 ^e3; x3 ^e0 1; x0 1a11a12a13 ^e0 2; x0 2a21a22a23 ^e0 3; x0 3a31a32a33 In much the same way the product w=v
T orwi=vjTji appears as [w1i]=[v1j] [Tji]orh w1w2w3i =h v1v2v3i2 64T11T21T31 T12T22T32 T13T23T333 75:(2.51) Note that in Eq 2.51 the order of the subscripts are exchanged since the transpose of the matrix is being written. Equations 2.51 and 2.50 could be written in the direct form the following way: w=v
T=(T
v)T=vT
TT: From this, it is clear that w=uT sinceu=T
v. 2.5 Transformations of Cartesian Tensors Although, as already mentioned, vectors and tensors have an identity independent of any particular reference or coordinate system, the relative values of their respective compo- nents do depend upon the specific axes to which they are referred. The relationships among 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 Cartesian axes,Ox1x2x3andOx0 1x0 2x0 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 by aij=cos(x0 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 ^eiand^e0 ito one another, as well as relating the primed and unprimed coordinates x0 iandxiof a point. Thus the primed base vectors ^e0 iare given in terms of the unprimed vectors ^eiby the equations (as is also easily verified from the geometry of the vectors in the diagram of Fig. 2.2), 26 Continuum Mechanics for Engineers x1x 1 x2x 2x3 x 3 cos−1a11cos−1a12cos−1a13 O^e3 ^e2 ^e1^e 3^e 2^e 1 FIGURE 2.2 Rectangular coordinate system Ox0 1x0 2x0 3relative toOx1x2x3. Direction cosines shown for coordinatex0 1relative to unprimed coordinates. Similar direction cosines are defined for x0 2andx0 3coordinates. ^e0 1=a11^e1+a12^e2+a13^e3=a1j^ej; (2.52a) ^e0 2=a21^e1+a22^e2+a23^e3=a2j^ej; (2.52b) ^e0 3=a31^e1+a32^e2+a33^e3=a3j^ej; (2.52c) or in compact indicial form ^e0 i=aij^ej: (2.53) By defining the matrix Awhose elements are the direction cosines aij, Eq 2.53 can be written in matrix form as [^e0 i1]=[aij] [^ej1] or2 64^e0 1 ^e0 2 ^e0 33 75=2 64a11a12a13 a21a22a23 a31a32a333 752 64^e1 ^e2 ^e33 75 (2.54) where the elements of the column matrices are unit vectors. The matrix Ais called the transformation matrix because, as we shall see, of its role in transforming the components of 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, ^e0 i
^e0 j=ij. But also, in view of Eq 2.53, ^e0 i
^e0 j=aiq^eq
ajm^em=aiqajmqm=aiqajq=ij from which we extract the orthogonality condition on the direction cosines (given here in both indicial and matrix form), aiqajq=ij orAAT=I: (2.55) Note that this is simply the inner product of the ithrow with the jthrow of the matrix A. By an analogous derivation to that leading to Eq 2.53, but using the columns of A, we obtain Essential Mathematics 27 ^ei=aji^e0 j; (2.56) which in matrix form is [^e1i]=^e0 1j [aij]orh ^e1^e2^e3i =h ^e0 1^e0 2^e0 3i2 64a11a12a13 a21a22a23 a31a32a333 75:(2.57) Note that using the transpose AT, Eq 2.57 may also be written [^e1i]=[aij]T^e0 1j or2 64^e1 ^e2 ^e33 75=2 64a11a21a31 a12a22a32 a13a23a333 752 64^e0 1 ^e0 2 ^e0 33 75 (2.58) in which column matrices are used for the vectors ^eiand^e0 i. By a consideration of the dot product and Eq 2.56 we obtain a second orthogonality condition aijaik=jk orATA=I (2.59) which is the inner product of the jthcolumn with the kthcolumn of A. Consider next an arbitrary vector vhaving components viin the unprimed system, and v0 iin the primed system. Then using Eq 2.56, v=v0 j^e0 j=vi^ei=viaji^e0 j from which, by matching coefficients on ^e0 j, we have (in both the indicial and matrix forms), v0 j=ajivi orv0=Av=vAT(2.60) which is the transformation law expressing the primed components of an arbitrary vector in terms of its unprimed components. Although the elements of the transformation matrix are written as aijwe must emphasize that they are not the components of a second- order Cartesian tensor as it might appear. Multiplication of Eq 2.60 by ajkand using the orthogonality condition Eq 2.59 we obtain the inverse law vk=ajkv0 j orv=v0A=ATv0(2.61) giving the unprimed components in terms of the primed. By a direct application of Eq 2.61 to the dyad uvwe have uivj=aqiu0 qamjv0 m=aqiamju0 qv0 m: (2.62) But a dyad is, after all, one form of a second-order tensor, and so by an obvious adaptation of Eq 2.62 we obtain the transformation law for a second-order tensor, Tas tij=aqiamjt0 qm orT=ATT0A (2.63) which may be readily inverted with the help of the orthogonality conditions to yield t0 ij=aiqajmtqm orT0=ATAT: (2.64) 28 Continuum Mechanics for Engineers O 2π 2π x 1x3 x1,x 3x2,x 2 (a)Ox0 1x0 2x0 3axes relative to Ox1x2x3axes following a90ocounterclockwise rotation about thex2axis. O x2,x 2x3,x 3 x 1 x1(b)Ox0 1x0 2x0 3axes relative to Ox1x2x3axes following a reflection about the x2x3plane. FIGURE 2.3 Rotation and reflection of reference axes. Note carefully the location of the summed indices qandmin Equations 2.63 and 2.64. Finally, by a logical generalization of the pattern of the transformation rules developed thus far we state that for an arbitrary Cartesian tensor of any order R0 ij:::k=aiqajm


aknRqm:::n: (2.65) The primed axes may be related to the unprimed axes through either a rotation about an axis through the origin, or by a reflection of the axes in one of the coordinate planes. (Or by a combination of such changes). As a simple example, consider a 90(counterclockwise) rotation about the x2axis shown in Fig. 2.3(a). The matrix of direction cosines for this rotation is [aij]=2 640 0 -1 0 1 0 1 0 03 75 and det A=1. The transformation of tensor components in this case is called a proper orthogonal transformation. For a reflection of axes in the x2x3plane shown in Fig 2.3(b) the transformation matrix is [aij]=2 64-1 0 0 0 1 0 0 0 13 75 where det A= -1, and we have an improper orthogonal transformation. It may be shown that true (polar) vectors transform by the rules v0 i=aijvjandvi=aijv0 jregardless of whether the axes transformation is proper or improper. However, pseudo (axial) vectors transform correctly only according to v0 i=(detA)aijvjandvj=(detA)aijv0 iunder an improper transformation of axes. Essential Mathematics 29 Example 2.12 Let the primed axes Ox0 1x0 2x0 3be given with respect to the unprimed axes by a45(counterclockwise) rotation about the x2axis as shown. Determine the primed components of the vector given by v=^e1+^e2+^e3. O β45ο 45οv ββ ^e1^e3 ^e 1^e 3 ^e2,^e 2x2,x 2 x1 x 1x3 x 3 Vectorvwith respect to Ox0 1x0 2x0 3andOx1x2x3. Solution Here the transformation matrix is [aij]=2 66641p 20 -1p 2 0 1 0 1p 201p 23 7775; and from Eq 2.60 in matrix form 2 6664v0 1 v0 2 v0 33 7775=2 66641p 20 -1p 2 0 1 0 1p 201p 23 77752 66641 1 13 7775=2 66640 1 p 23 7775: 30 Continuum Mechanics for Engineers Example 2.13 Determine the primed components of the tensor [tij]=2 642 6 4 0 8 0 4 2 03 75 under the rotation of axes described in Example 2.12. Solution Here Eq 2.64 may be used. Thus in matrix form  t0 ij =2 641p 20 -1p 2 0 1 0 1p 201p 23 752 642 6 4 0 8 0 4 2 03 752 641p 201p 2 0 1 0 -1p 201p 23 75 =2 64-34p 21 0 8 0 18p 253 75: 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 tensor Twith the arbitrary vector u(which we write here in both the indicial and symbolic notation), tijuj=vi orT
u=v (2.66) any second-order tensor may be thought of as a linear transformation which transforms theantecedent vectoruinto the image vectorvin a Euclidean three-space. In particular, for every symmetric tensor Thaving real components tij, and defined at some point in physical space, there is associated with each direction at that point (identified by the unit vectorni), an image vector vigiven by tijnj=vi orT
^n=v: (2.67) If the vectorvidetermined by Eq 2.67 happens to be a scalar multiple of ni, that is, if tijnj=ni orT
^n=^n; (2.68) the direction defined by niis called a principal direction, or eigenvector, of T, and the scalar is called a principal value, or eigenvalue ofT. Using the substitution property of the Kronecker delta, Eq 2.68 may be rewritten as Essential Mathematics 31 (tij-ij)nj=0or(T-I)
^n=0; (2.69) or in expanded form (t11-)n1+t12n2+t13n3=0; (2.70a) t21n1+(t22-)n2+t23n3=0; (2.70b) t31n1+t32n2+(t33-)n3=0: (2.70c) This system of homogeneous equations for the unknown direction niand the unknown ’s will have non-trivial solutions only if the determinant of coefficients vanishes. Thus, jtij-ijj=0 (2.71) which upon expansion leads to the cubic in (called the characteristic equation) 3-IT2+IIT-IIIT=0 (2.72) where the coefficients here are expressed in terms of the known components of Tby IT=tii=trT; (2.73a) IIT=1 2(tiitjj-tijtji)=1 2h (trT)2-tr T2
i ; (2.73b) IIIT="ijkt1it2jt3k=detT; (2.73c) and are known as the first, second, and third invariants, respectively, of the tensor T.1The sum of the elements on the principal diagonal of the matrix form of any tensor is called thetrace of that tensor, and for the tensor Tis written trTas in Eq 2.73. The roots(1),(2), and(3)of Eq 2.72 are all real for a symmetric tensor Thaving real components. With each of these roots (q)(q=1;2;3 )we can determine a principal directionn(q) i(q=1;2;3 )by solving Eq 2.69 together with the normalizing condition nini=1. Thus, Eq 2.69 is satisfied by  tij-(q)ij n(q) i=0 (q=1;2;3 ) (2.74) with n(q) in(q) i=1 (q=1;2;3 ): (2.75) If the(q)’s are distinct the principal directions are unique and mutually perpendicular. If, however, there is a pair of equal roots, say (1)=(2), then only the direction associated with(3)will be unique. In this case any other two directions which are orthogonal to n(3) i, and to one another so as to form a right-handed system, may be taken as principal directions. If (1)=(2)=(3), every set of right-handed orthogonal axes qualifies as principal 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 tensor Tbe given initially with respect to arbitrary Cartesian axes Ox1x2x3, and let the principal axes ofTbe designated by Ox 1x 2x 3, as shown in Fig. 2.4. The transformation matrix Abetween these two sets of axes is established by taking the direction cosines n(q) ias calculated from Eq 2.74 and Eq 2.75 as the elements of the qthrow of A. Therefore, by definition,aijn(i) jas 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 sign on theIITterm in Eq 2.72 and must be kept consistent throughout the use of the invariants. 32 Continuum Mechanics for Engineers cos−1n(1) 3 cos−1n(1) 2 cos−1n(1) 1x3 x2 x1x2∗ Ox3∗ x1∗ FIGURE 2.4 Principal axes Ox 1x 2x 3relative to axes Ox1x2x3. x1or^e1x2or^e2x3or^e3 x 1or^e 1a11=n(1) 1a12=n(1) 2a13=n(1) 3 x 2or^e 2a21=n(2) 1a22=n(2) 2a23=n(2) 3 x 3or^e 3a31=n(3) 1a32=n(3) 2a33=n(3) 3(2.76) The transformation matrix here is orthogonal and in accordance with the transformation law for second-order tensors t ij=aiqajmtqm orT=ATAT(2.77) where Tis a diagonal matrix whose elements are the principal values (q). Example 2.14 Determine the principal values and principal directions of the second-order tensorTwhose matrix representation is [tij]=2 645 2 0 2 2 0 0 0 33 75: Solution Here Eq 2.71 is given by 5- 2 0 2 2 - 0 0 0 3 -=0 which upon expansion by the third row becomes (3-) 10-7+2-4
=0; Essential Mathematics 33 or (3-) (6-) (1-)=0: Hence,(1)=3,(2)=6,(3)=1are the principal values of T. For(1)=3, Eq 2.70 yields the equations 2n1+2n2=0; 2n1-n2=0; which are satised only if n1=n2=0, and so from nini=1we haven3=1. For(2)=6, Eq 2.70 yields -n1+2n2=0; 2n1-4n2=0; -3n3=0; so thatn1=2n2and sincen3=0, we have (2n2)2+n2 2=1, orn2=1=p 5, andn1=2=p 5. For(3)=1, Eq 2.70 yields 4n1+2n2=0; 2n1+n2=0; together with 2n3=0. Againn3=0, and heren2 1+(-2n1)2=1so that n1=1=p 5andn2=2=p 5. From these results the transformation matrix Ais given by [aij]=2 66640 01 2p 51p 50 1p 52p 503 7775 which identies two sets of principal direction axes, one a re ection of the other with respect to the origin. Also, it may be easily veried that Ais orthogonal by multiplying it with its transpose ATto obtain the identity matrix. Finally, from Eq 2.77 we see that using the upper set of the signs, 2 66640 0 1 2p 51p 50 1p 5-2p 503 77752 66645 2 0 2 2 0 0 0 33 77752 666402p 51p 5 01p 5-2p 5 1 0 03 7775=2 66643 0 0 0 6 0 0 0 13 7775: 34 Continuum Mechanics for Engineers Example 2.15 Show that the principal values for the tensor having the matrix [tij]=2 645 1p 2 1 5p 2p 2p 2 63 75 have a multiplicity of two, and determine the principal directions. Solution Here Eq 2.71 is given by 5- 1p 2 1 5 -p 2p 2p 2 6 -=0 for which the characteristic equation becomes 3-162+80-128=0; or (-8) (-4)2=0: For(1)=8, Eq 2.70 yields -3n1+n2+p 2n3=0; n1-3n2+p 2n3=0;p 2n1+p 2n2-2n3=0: From the rst two of these equations n1=n2, and from the second and third equationsn3=p 2n2. Therefore, using, nini=1, we have (n2)2+(n2)2+p 2n22 =1; and son1=n2=1=2 andn3=1=p 2, from which the unit vector in the principal direction associated with (1)=8(the so-called normalized eigenvec- tor) is ^n(1)=1 2 ^e1+^e2+p 2^e3 =^e 3: For^n(2), we choose any unit vector perpendicular to ^n(1); an obvious choice being ^n(2)=-^e1+^e2p 2=^e 2: Then^n(3)is constructed from ^n(3)=^n(1)^n(2), so that ^n(3)=1 2 -^e1-^e2+p 2^e3 =^e 3: Essential Mathematics 35 Thus, the transformation matrix Ais given by Eq 2.76 as [aij]=2 66641 21 21p 2 -1p 21p 20 -1p 2-1 21p 23 7775: 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 the tensor which is easy to calculate. Similarly, the third invariant may be calculated by a cofactor expansion to evaluate the determinant. Calculating the second invariant may be done two ways. First is the straight forward calculation. Half the trace squared minus the trace of the square may be done using matrix operations. For example, take the generic matrix G=2 64a b c d e f g h i3 75: (2.78) Clearly, from Eq 2.73a, the first invariant is IG=a+e+i: (2.79) Finding the second invariant requires two terms:1 2I2 Gand-1 2tr G2
. The first term is expanded as 1 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 2tr0 B@2 64a b c d e f g h i3 752 64a b c d e f g h i3 751 CA=1 2(a2+bd+gc|{z} 1;1 term+bd+e2+fh|{z} 2;2 term+gc+hf+i2)|{z} 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 of matrix G. Using this method on Eq 2.78 results in Eq 2.80. This second way is an easier way 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 Matlabr, the matrix of Example 2.14 can be entered and the invariants found with the following code: 36 Continuum Mechanics for Engineers >> T = [5 2 0; 2 2 0; 0 0 3] T = 5 2 0 2 2 0 0 0 3 >> Tsqrd = T*T Tsqrd = 29 14 0 14 8 0 0 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 as numbers. Consider the analysis done on the matrix defined in Eq 2.78. All of those calculations may be completed with Matlabras 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 Essential Mathematics 37 >> 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 symmetric second-order tensors: 1. The principal values and principal directions of TandTTare the same. 2. The principal values of T-1are reciprocals of the principal values of T, and both have the same principal directions. 3. The product tensors TQandQThave the same principal values. 4. A symmetric tensor is said to be positive (negative) definite if all of its principal values are positive (negative); and positive (negative) semi-definite if one principal value is zero and the others positive (negative). 2.7 Tensor Fields, Tensor Calculus Atensor field assigns to every location x, at every instant of time t, a tensortij:::k(x;t), for whichxranges over a finite region of space, and t varies over some interval of time. The field is continuous and hence differentiable if the components tij:::k(x;t)are continuous functions ofxandt. Tensor fields may be of any order. For example, we may denote typical scalar, vector, and tensor fields by the notations (x;t),vi(x;t), andtij(x;t), re- spectively. Partial differentiation of a tensor field with respect to the variable tis symbolized by the operator @=@t and follows the usual rules of calculus. On the other hand, partial differentiation with respect to the coordinate xqwill be indicated by the operator @=@xq, which may be abbreviated as simply @q. Likewise, the second partial @2=@xq@xmmay be written@qm, and so on. As an additional measure in notational compactness it is customary in continuum mechanics to introduce the subscript comma to denote partial differentiation with respect to the coordinate variables. For example, we write ;ifor @=@xi;vi;jfor@vi=@xj;tij;kfor@tij=@xk; andui;jkfor@2ui=@xj@xk. We note from these examples that differentiation with respect to a coordinate produces a tensor of one order higher. Also, a useful identity results from the derivative @xi=@xj, viz., @xi @xj=ij: (2.81) In the notation adopted here the operator r(del) of vector calculus, which in symbolic notation appears as r=@ @x1^e1+@ @x2^e2+@ @x3^e3=@ @xi^ei (2.82) 38 Continuum Mechanics for Engineers takes on the simple form @i. Therefore, we may write the scalar gradientr=gradas @i=;i; (2.83) the vector gradient rvas @ivj=vj;i; (2.84) the divergence of v,r
vas @ivi=vi;i; (2.85) and the curl of v,rvas "ijk@jvk="ijkvk;j: (2.86) Note in passing that many of the identities of vector analysis can be verified with relative ease by manipulations using the indicial notation. For example, to show that div(curlv) =0for any vector vwe write from Eqs 2.86 and Eq 2.85 @i("ijkvk;j)="ijkvk;ji=0; and because the first term of this inner product is skew–symmetric in iandj, whereas the second term is symmetric in the same indices, (since vkis assumed to have continuous spatial gradients), their product is zero. Example 2.16 Use indicial notation to verify the following identities: curlru=0 (2.87a) and curl ruT
=rcurlu (2.87b) whereuis any continuously dierentiable vector eld. Solution For the rst, write curlru ="ipkuj;kp^ei^ejwhere the order of partial dier- entiation ofudoes not matter since uis continuously dierentiable. Thus, uj;kp=uj;pk. By denition, the permutation symbol is skew{symmetric in kpand the product of a skew{symmetric and symmetric tensor is zero (see Example 2.5). The second identity is just as easily veried with indicial notation. Write the left-hand side of the expression in indicial notation, exchange the order of partial dierentiation on u. Since the permutation symbol is a constant we may write"ipkuk;jp= ("ipkuk;p);j There is also a need for using the divergence and curl of second order tensors in contin- uum mechanics. Let sjkbe a second order tensor that is differentiable at x. The divergence ofsjkatxis defined as divS=sjk;k^ej (2.88) Essential Mathematics 39 where sij;k=@sij @xk=^ei
(rS^ek)^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 a second order tensor sjkis defined as (curlS)ij="ipksjk;p (2.90) in a manner that parallels the curl of a vector. The curl ureturns a vector and the curl S results in a second order tensor. Example 2.17 Considersij=sjito be a continuously dierentiable, second order tensor. Show that tr(curlS)=0: (2.91) Solution Write the curl of Sin indicial form: (curlS)ij="ipksjk;p: The trace of this is found by setting i=j: (tr(curlS))ii="ipksik;p: Sincesik=skiand"ipk= -"kpiwe see that tr(curlS)=0. Example 2.18 LetW=wij^ei^ej= -wji^ei^ejbe a skew{symmetric, second order tensor having continuous derivatives. Since Wis skew{symmetric it has an axial vector usuch thatwjk= -"qjkuq. Show that curlW=Idivu-ru; (2.92a) or, equivalently "ipkwjk;p=ijuq;q-ui;j: (2.92b) Solution Using the denition of an axial vector as well as Eq 2.7a we nd that: "ipkwjk;p ="ipk(-"qjkuq);p = - (iqpj-ijpq)uq;p =ijuq;q-ui;j: 40 Continuum Mechanics for Engineers x1x2x3 dSi=nidSni V O FIGURE 2.5 Volume Vwith infinitesimal element dSihaving a unit normal ni. 2.8 Integral Theorems of Gauss and Stokes Consider an arbitrary continuously differentiable tensor field tij:::k defined on some fi- nite region of physical space. Let Vbe a volume in this space with a closed surface S bounding the volume, and let the outward normal to this bounding surface be nias shown in Fig. 2.5 so that the element of surface is given by dSi=nidS. The divergence theorem 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, Z Stij:::knqdS=Z Vtij:::k;qdV: (2.93) Several important special cases of this theorem for scalar and vector fields are worth noting, and are given here in both indicial and symbolic notation: Z SnqdS=Z V;qdV orZ S^ndS=Z VgraddV; (2.94) Z SvqnqdS=Z Vvq;qdV orZ Sv
^ndS=Z VdivvdV; (2.95) Z S"ijknjvkdS=Z V"ijkvk;jdV orZ S^nvdS=Z VcurlvdV: (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 over its bounding surface, Stokes’ theorem relates an integral over an open surface (a so-called cap) to a line integral around the bounding curve of the surface. Therefore, let Cbe the Essential Mathematics 41 x1x2x3ni OdxiC xidSi=nidS FIGURE 2.6 Bounding space curve Cwith tangential vector dxiand surface element dSifor partial volume. bounding space curve to the surface S, and letdxibe the differential tangent vector toC as shown in Fig. 2.6. (A hemispherical surface having a circular bounding curve Cis a classic example). If niis the outward normal to the surface S, andviis any vector field defined on SandC, Stokes’ theorem asserts that Z S"ijknivk;jdS=Z CvkdxkorZ S^n
(rv)dS=Z Cv
dx: (2.97) The integral on the right-hand side of this equation is often referred to as the circulation when the vector vis the velocity vector. 42 Continuum Mechanics for Engineers Problems Problem 2.1 Letv=ab, or in indicial notation, vi^ei=aj^ejbk^ek="ijkajbk^ei Using indicial notation, show that, (a)v
v=a2b2sin2, (b)ab
a=0, (c)ab
b=0. Problem 2.2 With respect to the triad of base vectors u1,u2, andu3(not necessarily unit vectors), the triadu1,u2, andu3is said to be a reciprocal basis ifui
uj=ij(i;j=1;2;3 ). Show that to satisfy these conditions, u1=u2u3 [u1;u2;u3];u2=u3u1 [u1;u2;u3];u3=u1u2 [u1;u2;u3] and determine the reciprocal basis for the specic base vectors u1=2^e1+^e2; u2=2^e2-^e3; u3=^e1+^e2+^e3: Answer u1=1 5(3^e1-^e2-2^e3) u2=1 5(-^e1+2^e2-^e3) u3=1 5(-^e1+2^e2+4^e3) Problem 2.3 Let the position vector of an arbitrary point P(x1x2x3)bex=xi^ei, and letb=bi^eibe aconstant vector. Show that (x-b)
x=0is the vector equation of a spherical surface having its center at x=1 2bwith a radius of1 2b. Problem 2.4 Using the notations A(ij)=1 2(Aij+Aji)andA[ij]=1 2(Aij-Aji)show that (a) the tensor Ahaving components Aijcan always be decomposed into a sum of its symmetric A(ij)and skew-symmetric A[ij]parts, respectively, by the decom- position, Aij=A(ij)+A[ij]; Essential Mathematics 43 (b) the trace of Ais expressed in terms of A(ij)by Aii=A(ii); (c) for arbitrary tensors AandB, AijBij=A(ij)B(ij)+A[ij]B[ij]: Problem 2.5 Expand the following expressions involving Kronecker deltas, and simplify where possible. (a)ijij;(b)ijjkki;(c)ijjk;(d)ijAik Answer (a) 3, (b) 3, (c) ik, (d)Ajk Problem 2.6 Ifai="ijkbjckandbi="ijkgjhk, substitutebjinto the expression for aito show that ai=gkckhi-hkckgi; or in symbolic notation, a= (c
g)h- (c
h)g. Problem 2.7 By summing on the repeated subscripts determine the simplest form of (a)"3jkajak;(b)"ijkkj;(c)"1jka2Tkj;(d)"1jk3jvk: Answer (a)0, (b)0, (c)a2(T32-T23), (d) -v2 Problem 2.8 Consider the tensor Bik="ijkvj. (a) Show that Bikis skew-symmetric. (b) LetBijbe skew-symmetric, and consider the vector dened by vi="ijkBjk (often called the dual vector of the tensorB). Show that Bmq=1 2"mqivi. Problem 2.9 Use indicial notation to show that Ami"mjk+Amj"imk+Amk"ijm=Amm"ijk whereAis any tensor and "ijkis the permutation symbol. 44 Continuum Mechanics for Engineers Problem 2.10 IfAij=ijBkk+3Bij, determineBkkand using that solve for Bijin terms ofAijand its rst invariant, Aii. Answer Bkk=1 6Akk;Bij=1 3Aij-1 18ijAkk Problem 2.11 Show that the value of the quadratic form Tijxixjis unchanged if Tijis replaced by its symmetric part,1 2(Tij+Tji). Problem 2.12 With the aid of Eq 2.7, show that any skew{symmetric tensor Wmay be written in terms of an axial vector!igiven by !i= -1 2"ijkwjk wherewjkare the components of W. Problem 2.13 Show by direct expansion (or otherwise) that the box product ="ijkaibjckis equal to the determinant a1a2a3 b1b2b3 c1c2c3: Thus, by substituting A1iforai,A2jforbjandA3kforck, derive Eq 2.42 in the form detA="ijkA1iA2jA3kwhere Aijare the elements of A. Problem 2.14 Starting with Eq 2.42 of the text in the form detA="ijkAi1Aj2Ak3 show that by an arbitrary number of interchanges of columns of Aijwe obtain "qmn detA="ijkAiqAjmAkn which is Eq 2.43. Further, multiply this equation by the appropriate permutation symbol to derive the formula 6detA="qmn"ijkAiqAjmAkn: Problem 2.15 Let the determinant of the matrix Aijbe given by detA=A11A12A13 A21A22A23 A31A32A33: Essential Mathematics 45 Since the interchange of any two rows or any two columns causes a sign change in the value of the determinant, show that after an arbitrary number of row and column interchanges AmqAmrAms AnqAnrAns ApqAprAps="mnp"qrsdetA: Now let Aij=ijin the above determinant which results in detA=1and, upon expansion, yields "mnp"qrs=mq(nrps-nspr) -mr(nqps-nspq) +ms(nqpr-nrpq): Thus, by setting p=q, establish Eq 2.7 in the form "mnq"qrs=mrns-msnr: Problem 2.16 Show that the square matrices [Bij]=2 641 0 0 0-1 0 0 0 13 75and [Cij]=" 5 2 -12 -5# are both square roots of the identity matrix. Problem 2.17 Using 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=BTATas was proven in Example 2.8 [Aij]=2 643 0 1 0 2 4 5 1 23 75;[Bij]=2 641 3 1 2 2 5 4 0 33 75: Problem 2.18 LetAbe any orthogonal matrix, i.e., AAT=AA-1=I, whereIis the identity matrix. Thus, by using the results in Examples 2.9 and 2.10, show that detA=1. Problem 2.19 A tensor is called isotropic if its components have the same set of values in every Cartesian coordinate system at a point. Assume that Tis an isotropic tensor of rank two with componentstijrelative to axes Ox1x2x3. Let axesOx0 1x0 2x0 3be obtained with respect to Ox1x2x3by a righthand rotation of 120about the axis along ^n=(^e+^e+^e)=p 3. Show by the transformation between these axes that t11=t22=t33, as well as other relationships. Further, let axes Ox00 1x00 2x00 3be obtained with respect to Ox1x2x3by a right-hand rotation 46 Continuum Mechanics for Engineers of90aboutx3. Thus, show by the additional considerations of this transformation that if Tis any isotropic tensor of second order, it can be written as Iwhereis a scalar and I is the identity tensor. Problem 2.20 For a proper orthogonal transformation between axes Ox1x2x3andOx0 1x0 2x0 3show the invariance of ijand"ijk. That is, show that (a)0 ij=ij, (b)"0 ijk="ijk. Hint: For part (b) let "0 ijk=aiqajmakn"qmn and make use of Eq 2.43. Problem 2.21 The angles between the respective axes of the Ox0 1x0 2x0 3and theOx1x2x3Cartesian systems are given by the table below x1x2x3 x0 1459045 x0 26045120 x0 31204560 Determine (a) the transformation matrix between the two sets of axes, and show that it is a proper orthogonal transformation, (b) the equation of the plane x1+x2+x3=1=p 2in its primed axes form, that is, in the formb1x0 1+b2x0 2+b3x0 3=b. Answer (a) [aij]=2 66641p 201p 2 1p 21p 2-1p 2 -1p 21p 21p 23 7775 (b)2x0 1+x0 2+x0 3=1 Problem 2.22 Making use of Eq 2.42 of the text in the form detA="ijkA1iA2jA3kwrite Eq 2.71 as jtij-ijj="ijk(t1i-1i) (t2j-2j) (t3k-3j)=0 and show by expansion of this equation that 3-tii2+1 2(tiitjj-tijtji) -"ijkt1it2jt3k=0 to verify Eq 2.72 of the text. Essential Mathematics 47 Problem 2.23 For the matrix representation of tensor Bshown below, [bij]=2 6417 0 0 0-23 28 0 28 103 75 determine the principal values (eigenvalues) and the principal directions (eigenvectors) of the tensor. Answer 1=17,2=26,3= -39 ^n(1)=^e1,^n(2)=(4^e2+7^e3)=p 65,^n(3)=(-7^e2+4^e3)=p 65 Problem 2.24 Consider the symmetrical matrix [Bij] =2 66645 403 2 0 4 0 3 205 23 7775: (a) Show that a multiplicity of two occurs among the principal values of this matrix. (b) Let1be the unique principal value and show that the transformation matrix [aij]=2 66641p 20-1p 2 0 1 0 1p 201p 23 7775 givesBaccording to B ij=aiqajmBqm. (c) Taking the square root ofh B iji and transforming back to Ox1x2x3axes show that hp Biji =2 66643 201 2 0 2 0 1 203 23 7775: (d) Verify that the matrix [Cij]=2 6664-1 20-3 2 0 2 0 -3 20-1 23 7775 is also a square root of [Bij]. 48 Continuum Mechanics for Engineers Problem 2.25 Determine the principal values of the matrix [Kij]=2 644 0 0 0 11 -p 3 0-p 3 93 75; and show that the principal axes Ox 1x 2x 3are obtained from Ox1x2x3by a rotation of 60 about thex1axis. Answer 1=4,2=8,3=12. Problem 2.26 Determine the principal values (q)(q=1;2;3 )and principal directions ^n(q)(q=1;2;3 ) for the symmetric matrix [Tij]=1 22 66643 -1p 21p 2 -1p 29 23 2 1p 23 29 23 7775 Answer (1)=1;  (2)=2  (3)=3 ^n(1)=1 2p 2^e1+^e2-^e3 ^n(2)=1 2p 2^e1-^e2+^e3 ^n(3)= - (^e2+^e3)=p 2 Problem 2.27 For the second-order tensor Cij=uivj: (a) Calculate the principal invariants IC,IIC, andIIIC. Reduce your answer to its simplest form, but leave it in index notation. State the reasons for simplication. (b) Ifu=h 1 2 3i andv=h 5 10 4i write out the matrix form for Cij. (c) Use the numbers of (b) to validate the answer of (a). Problem 2.28 LetDbe a constant tensor whose components do not depend upon the coordinates. Show that r(x
D)=D wherex=xi^eiis the position vector. Essential Mathematics 49 Problem 2.29 Consider the vector x=xi^eihaving a magnitude squared x2=x2 1+x2 2+x3 3. Determine (a)gradx, (b)grad (x-n), (c)r2(1=x), (d)div(xnx), (e)curl(xnx), wherenis a positive integer. Answer (a)xi=x, (b) -nxi=x(n+2), (c) 0, (d) xn(n+3), (e)0. Problem 2.30 Ifand'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 the indicated operations, show that the result is the right-hand side. (a)v(rv) =1 2r(v
v) - (v
r)v (b)v
uw=vu
w (c)r(rv) =r(r
v) -r2v (d)r
(r') =r2'+r
r' (e)r2(') =r2'+2(r)
(r') +'r2 (f)r
(uv) = (ru)
v-u
(rv) Problem 2.31 Let the vector v=bxbe one for which bdoes not depend upon the coordinates. Use indicial notation to show that (a)curlv=2b, (b)divv=0. Problem 2.32 Transcribe the left-hand side of the following equations into indicial notation and verify that the indicated operations result in the expressions on the right-hand side of the equations for the scalar ', and vectors uandv. (a)div('v) ='divv+v
grad' (b)ucurlv+vcurlu= -(u
grad)v - (v
grad)u +grad(u
v) (c)div(uv) =v
curlu-u
curlv (d)curl(uv) = (v
grad)u - (u
grad)v +udivv-vdivu (e)curl(curlu) = grad(divu) -r2u 50 Continuum Mechanics for Engineers Problem 2.33 Let the volume Vhave a bounding surface Swith an outward unit normal ni. Letxibe the position vector to any point in the volume or on its surface. Show that (a)Z SxinjdS=ijV, (b)Z Sr(x
x)
^ndS=6V, (c)Z Sw
^ndS=Z Vw
graddV, wherew=curlvand=(x), (d)Z S[^eix;^ej;^n]dS=2Vijwhere ^eiand^ejare coordinate base vectors. Hint: Write the box product [^eix;^ej;^n]=(^eix)
(^ej^n) and transcribe into indicial notation. Problem 2.34 Use Stokes' theorem to show that upon integrating around the space curve Chaving a dierential tangential vector dxithat for'(x). I C';idxi=0 Problem 2.35 For the position vector xihaving a magnitude x, show that x;j=xj=xand therefore, (a)x;ij=ij x-xixj x3, (b) x-1
;ij=3xixj x5-ij x3, (c)x;ii=2 x. Problem 2.36 Show that for arbitrary tensors AandB, and arbitrary vectors aandb, (a)(A
a)
(B
b) =a
(AT
B)
b, (b)ba=1 2(B-BT)
a, if2bi="ijkBkj, (c)a
A
b=b
AT
a. Problem 2.37 Use Eqs 2.42 and 2.43 as necessary to prove the identities (a)[Aa;Ab;Ac] = ( detA)[a;b;c], Essential Mathematics 51 (b)AT
(AaAb) = (det A)(ab), for arbitrary vectors a,b,c, and matrix A. Problem 2.38 Let'='(xi)and = (xi)be scalar functions of the coordinates. Recall that in the indicial notation ';irepresentsr' and';iirepresentsr2'. Now apply the divergence theorem, Eq 2.93, to the eld ' ;ito obtain Z S' ;inidS=Z V(';i ;i+' ;ii)dV: Transcribe this result into symbolic notation as Z S'r
^ndS=Z S'@ @ndS=Z V r'
r +'r2
dV which is known as Green’s first identity. Show also by the divergence theorem that Z S(' ;i- ';i)nidS=Z V(' ;ii- ';ii)dV; and transcribe into symbolic notation as Z S '@ @n- @' @n dS=Z V 'r2 - r2'
dV which is known as Green’s second identity. This page intentionally left blank 3 Stress Principles Stress is a measure of force intensity, either within, or on the bounding surface of a body subjected to loads. It should be noted that in continuum mechanics a body is considered stress free if the only forces present are those inter-atomic forces required to hold the body together. And so it follows that the stresses which concern us here are those that result from the application of forces by an external agent. 3.1 Body and Surface Forces, Mass Density Two basic types of forces are easily distinguished from one another and are defined as follows. First, those forces acting on all volume elements, and are distributed throughout the 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 symbol pi(force per unit volume). Secondly, those forces which act upon, and are distributed in some fashion over a surface element of the body, regardless of whether that element is part of the bounding surface, or an arbitrary element of surface within the body, are called surface forces. These are denoted by the vector symbol fi, and have dimensions of force per unit area. Forces which occur on the outer surfaces of two bodies pressed against one another (contact forces), or those which result from the transmission of forces across an internal surface are examples of this type of force. Next, let us consider a material body Bhaving a volume Venclosed by a surface S, and occupying a regular region R0of physical space. Let Pbe an interior point of the body located in the small element of volume
Vwhose mass is
m as indicated in Fig. 3.1. Recall that mass is that property of a material body by virtue of which the body possesses 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 ave=
m
V; (3.1) and the density at pointPby the limit of this ratio as the volume shrinks to the point P, =lim
V!0
m
V=dm dV: (3.2) The units of density are kilograms per cubic meter, (kg=m3). Notice that the two measures of body forces, bihaving units of Newton per kilogram (N=kg), and pihaving units of Newtons per meter cubed (N=m3), are related through the density by the equation bi=pi orb=p: (3.3) 53 54 Continuum Mechanics for Engineers ∆V V O x1x2x3P FIGURE 3.1 Typical continuum volume Vwith infinitesimal element
Vhaving mass
m at pointP. PointPwould be in the center of the infinitesimal volume. Of course, the density is, in general, a scalar function of position and time as indicated by =(xi;t) or=(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 Bhaving a bounding surface S, and a volume V, which is subjected to arbitrary surface forces fiand body forces bi. LetPbe an interior point of Band imagine a plane surface Spassing through point P(sometimes referred to as a cutting plane ) so as to partition the body into two portions, designated I and II (Fig. 3.2(a)). Point Pis in the small element of area
Sof the cutting plane, which is defined by the unit normal pointing in the direction from portion I into portion II as shown by the free body diagram of portion I in 3.2(b). The internal forces being transmitted across the cutting plane due to the action of portion II upon portion I will give rise to a force distribution on Sequivalent to a resultant force
fiand a resultant moment
MiatP, as is also shown in Fig. 3.2(b). (For simplicity body forces biand surface forces fiacting on the body as a whole are not drawn in Figs. 3.2(a) and 3.2(b).) Notice that
fiand
Miare not necessarily in the direction of the unit normal vector ni atP. The Cauchy stress principle asserts that in the limit as the area Sshrinks to zero with Premaining 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 x1x2x3 S∗ OIIIP (a) Typical continuum volume showing cutting planeSpassing through point P. x1x2x3 ∆S∗ OIni∆Mi ∆fi(b) Force and moment acting at point Pin sur- face element
S. FIGURE 3.2 Typical continuum volume with cutting plane. The vectordfi=dS=t(^n) iis 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 ^nin the symbol t(^n) ifor the stress vector serves to remind us that this is a special vector in that it is meaningful only in conjunction with its associated normal vector ^natP. 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) ifor a given loading of the body. The totality of pairs of the companion vectors t(^n) iand^nat P, 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 force of 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 any portion of a continuum body is equal to the resultant force acting upon that portion. For portions 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), Z SIt(^n) idS+Z VIbidV=d dtZ VIvidV; (3.7a) Z SIIt(^n) idS+Z VIIbidV=d dtZ VIIvidV (3.7b) where SIandSIIare the bounding surfaces and VIandVIIare the volumes of portions I and II, respectively. Also, biare the body forces, is the density and viis the velocity field for the two portions. We note that SIandSIIeach contain Sas part of their total areas. 56 Continuum Mechanics for Engineers ∆S∗niPt(^n) i FIGURE 3.3 Traction vector t(^n) iacting at point Pof plane element
Siwhose normal is ni. The linear momentum principle may also be applied to the body Bas a whole, so that Z St(^n) idS+Z VbidV=d dtZ VvidV: (3.8) If we add Eq 3.7a and Eq 3.7b and use Eq 3.8, noting that the normal to Sfor portion I is^n, whereas for portion II it is -^n, we arrive at the equation Z Sh t(^n) i+t(-^n) ii dS=0 (3.9) since both SIandSIIcontain a surface integral over S. This equation must hold for arbitrary partitioning of the body (that is, for every imaginable cutting plane through pointP) which means that the integrand must be identically zero. Hence, t(^n) i= -t(-^n) i (3.10) indicating that if portion IIhad been chosen as the free body in Fig. 3.2(b) instead of portionI, the resulting stress vector would have been -t(^n) i. 3.3 The Stress Tensor As noted in Section 3.2, the Cauchy stress principle associates with each direction ^nat pointPa stress vector t(^n) i. In particular, if we introduce a rectangular Cartesian reference frame atP, there is associated with each of the area elements dSi(i=1;2;3 )located in the coordinate 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 vectors associated 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) Stress Principles 57 x1x2x3 P x1x2x3 P x1x2x3 P ^e1^e2^e3 t(^e2) it(^e3) i t(^e1) i FIGURE 3.4 Traction vectors on the three coordinate planes at point P. P ABC ρ∗bini x1x2x3 −∗t(^e3) i−∗t(^e2) i−∗t(^e1) i ∗t(^n) i FIGURE 3.5 Free 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 Pfor a given coordinate plane in terms of its rectangular Cartesian components, but what is really needed is an expression for the coordinate components of the stress vector at Passociated with an arbitrarily oriented plane. For this purpose we consider the equilibrium of a small portion of the body in the shape of a tetrahedron having its vertex at P, and its base ABC perpendicular to an arbitrarily oriented normal ^n=ni^eias shown by Fig. 3.5. The coordinate directions are chosen so that the three faces BPC,CPA andAPB of the tetrahedron are situated in the coordinate planes. If the area of the base is assigned the value dS, the areas of the respective faces will be the projected areas dSi=dScos(^n;^ei);(i=1;2;3 )or specifically, forBPC d S1=n1dS; (3.13a) forCPA d S2=n2dS; (3.13b) forAPB d S3=n3dS: (3.13c) The stress vectors shown on the surfaces of the tetrahedron of Fig. 3.5 represent average values over the areas on which they act. This is indicated by a superscript asterisk in front of the stress vector symbols (remember that the stress vector is a point quantity). 58 Continuum Mechanics for Engineers Equilibrium requires the vector sum of all forces acting on the tetrahedron to be zero, that is, for, t(^n) idS-t(^e1) idS1-t(^e2) idS2-t(^e3) idS3+bidV=0 (3.14) wherebiis an average body force which acts throughout the body. The negative signs on the coordinate-face tractions result from the outward unit normals on those faces pointing 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) idS-t(^ej) injdS+bidV=0; (3.15) if we permit the indices on the unit vectors of thet(^ej) i term to participate in the sum- mation process. The volume of the tetrahedron is given by dV=1 3hdS, wherehis the perpendicular distance from point Pto the baseABC . Inserting this into Eq 3.15 and canceling the common factor dSwe obtain t(^n) i=t(^ej) inj-1 3bih: (3.16) Now, letting the tetrahedron shrink to point Pby taking the limit as h!0and noting that in this limiting process the starred (averaged) quantities take on the actual values of those same quantities at point P, we have t(^n) i=t(^ej) inj; (3.17) or, by defining tjit(^ej) i, t(^n) i=tjinj ort(^n)=^n
T (3.18) which is the Cauchy stress formula. We can obtain this same result for bodies which are accelerating by using the conservation of linear momentum instead of a balance of forces on the tetrahedron of Fig. 3.5. Stress as a Tensor The quantities tjit(^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) ibetween coordinate systems Px1x2x3andPx0 1x0 2x0 3as given by the transformation matrix having elements (see Section 2.5) aij=^e0 i
^ej: (3.19) Sincet(^n)can be expressed in terms of its components in either coordinate system, t(n)=t(^n) i^ei=t0(^n0) i^e0 i (3.20a) or, from Eq 3.18, t(^n)=tjinj^ei=t0 jin0 j^e0 i: (3.20b) But from Eq 2.53, ^e0 i=aij^ejand from Eq 2.60, n0 j=ajsns, so that now Eq 3.20b becomes, after some manipulations of the summed indices, tsr-ajsairt0 ji
ns^er=0: (3.21) Stress Principles 59 Because the vectors ^erare linearly independent and since Eq 3.21 must be valid for all vectors ns, we see that tsr=ajsairt0 ji: (3.22) But this is the transformation equation for a second-order tensor, and thus by Eq 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 with the element of area having an outward normal niat pointPin terms of the stress tensor components tjiat that point. And although the state of stress at Phas been described as the totality of pairs of the associated normal and traction vectors at that point, we see from the analysis of the tetrahedron element that if we know the stress vectors on the three coordinate planes of any Cartesian system at P, or equivalently, the nine stress tensor components tjiat that point, we can determine the stress vector for any plane at that point. For computational purposes it is often convenient to express Eq 3.18 in the matrix form h t(^n) 1t(^n) 2t(^n) 3i =[n1n2n3]2 64t11t12t13 t21t22t23 t31t32t333 75: (3.23) The nine components of tjiare often displayed by arrows on the coordinate faces of a rectangular parallelpiped, as shown in Fig. 3.6. We emphasize that this parallelpiped is not a block of material from the continuum body (note that no dimensions are given to the parallelpiped), but is simply a convenient schematic device for displaying the stress tensor components. In an actual physical body B, all nine stress components act at the single pointP. The three stress components shown by arrows acting perpendicular (normal) to the respective coordinate planes and labeled t11,t22, andt33are called normal stresses. The six arrows lying in the coordinate planes and pointing in the directions of the coordinate axes, namely, t12,t21,t23,t32,t13, andt31are called shear stresses. Note that, for these, the first subscript designates the coordinate plane on which the shear stress acts, and the second subscript identifies the coordinate direction in which it acts. A stress component is positive when its vector arrow points in the positive direction of one of the coordinate axes while acting on a plane whose outward normal also points in a positive coordinate direction. 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 referred to as compressive stresses. The units of stress are Newtons per square meter (N=m2)in the SI system, and pounds per square inch (psi)in the English system. One Newton per square meter is called a Pascal, but because this is a rather small stress from an engineering point of view, stresses are usually expressed as mega-pascals (MPa) or in English units as kilo-pounds per square inch (ksi). Example 3.1 Let the components of the stress tensor at Pbe given in matrix form by [tij]=2 6421-63 42 -63 0 84 42 84 -213 75 in units of mega-Pascals. Determine 60 Continuum Mechanics for Engineers x1x2x3 t12t33 t22t32 t31 t21t23 t11t13 FIGURE 3.6 Cartesian stress components shown in their positive sense. (a) the stress vector on the plane at Phaving the unit normal ^n=1 7(2^e1-3^e2+6^e3); (b) the stress vector on a plane at Pparallel to the plane ABC shown in the sketch. A(1,0,0)B(0,1,0)C(0,0,2) x1x2x3 Solution (a) From Eq 3.23 for the given data, h t(^n) 1t(^n) 2t(^n) 3i =2 7-3 76 72 6421-63 42 -63 0 84 42 84 -213 75=[69 54 -42]; Stress Principles 61 or, in vector form, t(^n)=69^e1+54^e2-42^e3. This vector represents the components of the force per unit area (traction) on the plane dened by1 7[2;-3; 6 ]. (b) The equation of the plane ABC in the sketch is easily veried to be 2x1+2x2+x3=2, and the unit outward normal to this plane is ^n=1 3(2^e1+2^e2+^e3)so that, again from Eq 3.23, h t(^n) 1t(^n) 2t(^n) 3i =2 32 31 32 6421-63 42 -63 0 84 42 84 -213 75 =[-14 -14 77 ] or, in vector form, t(^n)= -14 ^e1-14^e2+77^e3. In this example, we clearly see the dependency of the cutting plane and the stress vector. Here, we have considered two dierent cutting planes at the same point and found that two distinct traction vectors arose from the given stress tensor components. 3.4 Force and Moment Equilibrium; Stress Tensor Symmetry In the previous section, we used a balance-of-forces condition for a tetrahedron element of a body in equilibrium to define the stress tensor and to develop the Cauchy stress formula. Here, we employ a force balance on the body as a whole to derive what are known as the local equilibrium equations. This set of three differential equations must hold for every point in a continuum body that is in equilibrium. As is well known, equilibrium also requires the sum of moments to be zero with respect to any fixed point, and we use this condition, together with the local equilibrium equations, to deduce the fact that the stress tensor is symmetric in the absence of concentrated body moments. Consider a material body having a volume Vand a bounding surface S. Let the body be subjected to surface tractions t(^n) iand body forces bi(force per unit mass), as shown by 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 the total surface and body forces acting on the body, Z St(^n) idS+Z VbidV=0 (3.24) wheredSis the differential element of the surface SanddVthat of volume V. Becauset(^n) i=tjinjas a result of Eq 3.18, the divergence theorem Eq 2.93 allows the first term of Eq 3.24 to be written as Z StjinjdS=Z Vtji;jdV; (3.25) 62 Continuum Mechanics for Engineers n� xy x1x2x3 ρbdS OdVt(^n) FIGURE 3.7 Material volume showing surface traction vector t(^n) ion an infinitesimal area element dS at positionxi, and body force vector biacting on an infinitesimal volume element dVat positionyi. Two positions are taken separately for ease of illustration. When applying equilibrium the traction and body forces are taken at the same point. so that Eq 3.24 becomesZ V(tji;j+bi)dV=0: (3.26) This equation must be valid for an arbitrary volume V(every portion of the body is in equilibrium), which requires the integrand itself to vanish, and we obtain the so-called local equilibrium equations tji;j+bi=0: (3.27) In addition to the balance of forces expressed by Eq 3.24, equilibrium requires that the summation of moments with respect to an arbitrary point must also be zero. Recall that the moment of a force about a point is defined by the cross product of its position vector with the force. Therefore, taking the origin of coordinates as the center for moments, and noting that xiis 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 Z S"ijkxjt(^n) idS+Z V"ijkxjbidV=0: (3.28) As before, using the identity t(^n) k=tqknqand Gauss’s divergence theorem, we obtain Z V"ijkh (xjtqk);q+xjbki dV=0; or Z V"ijk[xj;qtqk+xj(tqk;q+bk)]dV=0: Butxj;q=jqand by Eq 3.27, tkq;k+bk=0, so that the latter equation immediately above reduces to Z V"ijktjkdV=0: Stress Principles 63 P x1x2x3 cos−1a11cos−1a12cos−1a13 x 1x 2 x 3 FIGURE 3.8 Rectangular coordinate axes Px0 1x0 2x0 3relative toPx1x2x3at pointP. Again, since volume Vis arbitrary, the integrand here must vanish, or "ijktjk=0: (3.29) By a direct expansion of the left-hand side of this equation, we obtain for the free index i=1, (omitting zero terms) "123t23+"132t32=0, ort23-t32=0implying that t23=t32. In the same way for i=2andi=3we find that t13=t31andt12=t21, respectively, so that in general tjk=tkj: (3.30) Thus, we conclude from the balance of moments for a body in which concentrated body moments are absent, that the stress tensor is symmetric, and Eq 3.27 may now be written in the form tij;j+bi=0 orr
T+b=0: (3.31) Also, because of this symmetry of the stress tensor, Eq 3.18 may now be expressed in the slightly 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) iandnjare represented by column matrices. 3.5 Stress Transformation Laws Let the state of stress at point Pbe given with respect to Cartesian axes Px1x2x3shown in Fig. 3.8 by the stress tensor Thaving components tij. We introduce a second set of axes Px0 1x0 2x0 3, which is obtained from Px1x2x3by a rotation of axes so that the transformation matrix [aij]relating the two is a proper orthogonal matrix. Because Tis a second-order Cartesian tensor, its components tijin the primed system are expressed in terms of the unprimed components by Eq 2.64 as 64 Continuum Mechanics for Engineers t0 ij=aiqtqmajm orT0=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 Pwith respect to axes Px1x2x3 be expressed by the matrix [tij]=2 641 3 2 3 1 0 2 0 -23 75; and let the primed axes Px0 1x0 2x0 3be obtained by a 45ocounterclockwise rotation about thex3axis. Determine the stress components t0 ij. x3 ,� θθPx 3 x 2 x 1x2 x1 Solution For a positive rotation aboutx3as shown by the sketch, the transformation matrix [aij]has the general form [aij] =2 64cossin 0 -sincos 0 0 0 13 75: Thus, from Eq 3.33 expressed in matrix form, a 45orotation of axes requires [tij]=2 641=p 2 1=p 2 0 -1=p 2 1=p 2 0 0 0 13 752 641 3 2 3 1 0 2 0 -23 752 641=p 2-1=p 2 0 1=p 2 1=p 2 0 0 0 13 75 =2 644 0p 2 0-2 -p 2p 2-p 2-23 75: Stress Principles 65 Example 3.3 Assume the stress tensor T(inksi) atPwith respect to axes Px1x2x3is represented by the matrix [tij]=2 6418 0 -12 0 6 0 -12 0 243 75: If thex0 1axis makes equal angles with the three unprimed axes, and the x0 2 axis lies in the plane of x0 1x3, as shown by the sketch, determine the primed components of TassumingPx0 1x0 2x0 3is a right-handed system. φ ββ P βx 3 x1x 1 x2x 2x3 Solution We must rst determine the transformation matrix [aij]. Letbe the common angle whichx0 1makes with the unprimed axes, as shown in the gure. Then a11=a12=a13=cosand from the orthogonality condition Eq 2.55 with i=j=1,cos=1=p 3. Next, letbe the angle between x0 2andx3. Then a23=cos=sin=2=p 6. As seen from the obvious symmetry of the axes arrangement, x0 2makes equal angles with x1andx2, which means that a21= a22. Thus, again from Eq 2.55, with i=1,j=2, we havea21=a22= -1=p 6 (the minus sign is required because of the positive sign chosen for a23). For the primed axes to be a right-handed system we require ^e0 3=^e0 1^e0 2, with the result thata31=1=p 2,a32= -1=p 2, anda33=0. Finally, from Eq 3.33, h t0 iji =2 66641p 31p 31p 3 -1p 6-1p 62p 6 1p 2-1p 203 77752 666418 0 -12 0 6 0 -12 0 243 77752 66641p 3-1p 61p 2 1p 3-1p 6-1p 2 1p 32p 603 7775 =2 66648 2p 2 0 2p 2 28 -6p 3 0 -6p 3 123 7775: 66 Continuum Mechanics for Engineers niP P ∆Snit(^n) i t(^n) i (a) Traction vector at point Pfor an arbitrary plane whose normal is ni. P P ∆Sni∗ t(^n) i t(^n) i ni∗(b) Traction vector at point Pfor a principal plane whose normal is n i. FIGURE 3.9 Traction 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 Pand assume it is given with respect to axes Px1x2x3by the stress tensor tij. As we saw in Example 3.1, for each plane element of area
SatPhaving an outward normal ni, a stress vector t(^n) iis defined by Eq 3.32. In addition, as indicated by Fig. 3.9(a), this stress vector is not generally in the direction of ni. However, for certain special directions at P, the stress vector does indeed act in the direction of niand may therefore be expressed as a scalar multiple of that normal. Thus, as shown in Fig. 3.9(b), for such directions t(^n) i=ni (3.34) whereis the scalar multiple of ni. Directions designated by nifor which Eq 3.34 is valid are called principal stress directions, and the scalar is called a principal stress value oftij. Also, the plane at Pperpendicular to niis 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 planes there are no shear stresses acting in these planes. The determination of principal stress values and principal stress directions follows precisely the same procedure developed in Section 2.6 for determining principal values and principal directions of any symmetric second-order tensor. In properly formulating the eigenvalue problem for the stress tensor we use the identity t(^n) i=tjinjand the substitution property of the Kronecker delta to rewrite Eq 3.34 as tijnj-ijnj=0 (3.35) Stress Principles 67 or, in expanded form, using tij=tji, (t11-)n1+t12n2+t13n3=0; (3.36a) t12n1+ (t22-)n2+t23n3=0; (3.36b) t13n1+t23n2+ (t33-)n3=0: (3.36c) In the three linear homogeneous equations expressed by Eq 3.36, the tensor components tijare assumed known; the unknowns are the three components of the principal normal ni, and the corresponding principal stress . To complete the system of equations for these 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=0is not compatible with Eq 3.37), the determinant of coefficients on njmust vanish. That is, jtij-ijj=0; (3.38) which upon expansion yields a cubic equation in (called the characteristic equation of the stress tensor), 3-IT2+IIT-IIIT=0 (3.39) whose roots (1);(2);(3)are the principal stress values of tij. The coefficients IT,IIT andIIITare known as the first, second, and third invariants, respectively, of tijand may be expressed in terms of its components by IT=tii=trT; (3.40a) IIT=1 2[tiitjj-tijtij]=1 2h (trT)2-trT2i ; (3.40b) IIIT="ijkt1it2jt3k=detT: (3.40c) Because the stress tensor tijis a symmetric tensor having real components, the three stress invariants are real, and likewise, the principal stresses being roots of Eq 3.39 are also real. To show this, we recall from the theory of equations that for a cubic with real coefficients at least one root is real, call it (1), and let the associated principal direction be designated by n(1) i. Introduce a second set of Cartesian axes Px0 1x0 2x0 3so thatx0 1is in the direction ofn(1) i. In this system the shear stresses, t0 12=t0 13=0, so that the characteristic equation oft0 ijrelative to these axes results from the expansion of the determinant (1)- 0 0 0 t0 22- t0 23 0 t0 23t0 33-=0; (3.41) or (1)-h 2-(t0 22+t0 33)+t0 22t0 33-(t0 23)2i =0: (3.42) From this equation, the remaining two principal stresses (2)and(3)are roots of the quadratic in brackets. 68 Continuum Mechanics for Engineers But the discriminant of this quadratic is (t0 22+t0 33)2-4h t0 22t0 33-(t0 23)2i =(t0 22-t0 33)2+4(t0 23)2; which is clearly positive, indicating that both (2)and(3)are real. If the principal stress values (1),(2), and(3)are distinct, the principal directions associated with these stresses are mutually orthogonal. To see why this is true, let n(1) i andn(2) ibe the normalized principal direction vectors (eigenvectors) corresponding to (1)and(2), respectively. Then, from Eq 3.35, ijn(1) j=(1)njandijn(2) j=(2)nj, which, upon forming the inner products, that is, multiplying in turn by n(2) iandn(1) i, become tijn(1) jn(2) i=(1)n(1) in(2) i; (3.43a) tijn(2) jn(1) i=(2)n(2) in(1) i: (3.43b) Furthermore, because the stress tensor is symmetric, and since iandjare dummy indices, tijn(1) jn(2) i=tjin(1) in(2) j=tijn(1) in(2) j; so that by the subtraction of Eq 3.43b from Eq 3.43a, the left hand side of the resulting difference is zero, or 0= (1)-(2) n(1) in(2) i: (3.44) But since we assumed that the principal stresses were distinct, or (1)6=(2), it follows that n(1) in(2) i=0 (3.45) which expresses orthogonality between n(1) iandn(2) i. By similar arguments, we may show thatn(3) iis perpendicular to both n(1) iandn(2) i. If two principal stress values happen to be equal, say (1)=(2), the principal direction n(3) iassociated with (3)will still be unique, and because of the linearity of Eq 3.35, any direction in the plane perpendicular to n(3) imay serve as a principal direction. Accord- ingly, we may determine n(3) iuniquely and then choosen(1) iandn(2) iso as to establish a right-handed system of principal axes. If it happens that all three principal stresses are equal, any direction may be taken as a principal direction, and as a result every set of right-handed Cartesian axes at Pconstitutes a set of principal axes in this case. We give the coordinate axes in the principal stress directions special status by labeling themPx 1x 2x 3, as shown in Fig. 3.10. Thus, for example, (1)acts on the plane perpendic- ular tox 1and is positive (tension) if it acts in the positive direction, negative (compres- sion) if it acts in the negative x 1direction. Also, if n(q) iis the unit normal conjugate to the principal stress (q)(q=1;2;3 ), the transformation matrix relating the principal stress axes to arbitrary axes Px1x2x3has elements defined by aqjn(q) j, as indicated by Table 3.1. Accordingly, the transformation equation expressing principal stress components in terms of arbitrary stresses at Pis given by Eq 2.77 in the form t ij=aiqajmtqm orT=ATAT: (3.46) In addition, notice that Eq 3.35 is satisfied by n(q) iand(q)so that tijn(q) j=(q)n(q) i (3.47) Stress Principles 69 cos−1n(1) 3 cos−1n(1) 2 cos−1n(1) 1x3 x2 x1x2∗ Px3∗ x1∗ FIGURE 3.10 Principal axes Px 1x 2x 3. TABLE 3.1 Table displaying direction cosines of principal axes Px 1x 2x 3relative to axes Px1x2x3. x1 x2 x3 x 1a11=n(1) 1a12=n(1) 2a13=n(1) 3 x 2a21=n(2) 1a22=n(2) 2a23=n(2) 3 x 3a31=n(3) 1a32=n(3) 2a33=n(3) 3 for(q=1;2;3 ), which upon introducing the identity aqin(q) ibecomestijaqj= (q)aqi. Now, multiplying each side of this equation by amiand using the symmetry property of the stress tensor, we have tjiaqjami=(q)aqiami: The left-hand side of this expression is simply t qm, from Eq 3.46. Since, by orthogonality, aqiamion the right hand side, the final result is t qm=qm(q); (3.48) which demonstrates that when referred to principal axes, the stress tensor is a diagonal tensor with principal stress values on the main diagonal. In matrix form, therefore,  t ij =2 4(1)0 0 0  (2)0 0 0  (3)3 5or t ij =2 4I0 0 0 II0 0 0 III3 5 (3.49) where in the second equation the notation serves to indicate that the principal stresses are ordered,I>II>III, with positive stresses considered greater than negative stresses regardless of numerical values. In terms of the principal stresses, the stress invariants 70 Continuum Mechanics for Engineers may be written IT=(1)+(2)+(3)=I+II+III; (3.50a) IIT=(1)(2)+(2)(3)+(3)(1)=III+IIIII+IIII; (3.50b) IIIT=(1)(2)(3)=IIIIII: (3.50c) Example 3.4 The components of the stress tensor at Pare given in MPa with respect to axesPx1x2x3by the matrix [tij] =2 457 0 24 0 50 0 24 0 433 5: Determine the principal stresses and the principal stress directions at P. Solution For the given stress tensor, Eq 3.38 takes the form of the determinant 57- 0 24 0 50 - 0 24 0 43 -=0 which, upon cofactor expansion about the rst row, results in the equation (57-)(50-)(43-) - (24 )2(50-) =0; or in its readily factored form (50-)(-25)(-75) =0: Hence, the principal stress values are (1)=25 MPa ,(2)=50 MPa , and (3)=75 MPa. Note that, in keeping with Eqs 3.40a and 3.50a, we conrm that the rst stress invariant, IT=57+50+43=25+50+75=150MPa: To determine the principal directions we rst consider (1)=25 MPa, for which Eq 3.36 provides three equations for the direction cosines of the principal direction of(1), namely, 32n(1) 1+24n(1) 3=0; 25n(1) 2=0; 24n(1) 1+18n(1) 3=0: Obviously,n(1) 2=0from the second of these equations, and from the other two, n(1) 3= -4 3n(1) 1so that, from the normalizing condition, nini=1, we see that n(1) 12 =9 25which givesn(1) 1=3 5andn(1) 3=4 5. The fact that the rst Stress Principles 71 and third equations result in the same relationship is the reason the normalizing condition must be used. Next for(2)=50MPa, Eq 3.36 gives 7n(2) 1+24n(2) 3=0; 24n(2) 1-7n(2) 3=0 which are satised only when n(2) 1=n(2) 3=0. Then from the normalizing condition,nini=1,n(2) 2=1. Finally, for(3)=75MPa, Eq 3.36 gives -18n(3) 1+24n(3) 3=0; -25n(3) 2=0 as well as 24n(3) 1-32n(3) 3=0: Here, from the second equation n(3) 2=0, and from either of the other two equations4n(3) 3=3n(3) 1, so that from nini=1we haven(3) 1=4 5andn(3) 3= 3 5. 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 transformation matrix has the general form [aij]=2 6643 504 5 01 0 4 503 53 775: Therefore, from Eq 3.48, when the upper signs in the above matrix are used, [tij]=2 6643 50-4 5 0 1 0 4 503 53 7752 66457 0 24 0 50 0 24 0 433 7752 6643 50-4 5 0 1 0 4 503 53 775=2 66425 0 0 0 50 0 0 0 753 775MPa: 3.7 Maximum and Minimum Stress Values The stress vector t(^n) ion an arbitrary plane at Pmay be resolved into a component normal to the plane having a magnitude N, along with a shear component which acts in the plane and has a magnitude S, as shown in Fig. 3.11. (Here, NandSarenotvectors, but scalar magnitudes of vector components. The subscripts NandSare to be taken as part of the component symbols). Clearly, from Fig. 3.11, it is seen that Nis given by the dot product, N=t(^n) ini, and inasmuch as t(^n) i=tijnj, it follows that 72 Continuum Mechanics for Engineers σN σSni� ∆S Pt(^n) i FIGURE 3.11 Traction vector components normal and in-plane (shear) at point Pon the plane whose normal isni. N=tijninj orN=t(^n)
^n: (3.51) Also, from the geometry of the decomposition, we see that 2 S=t(^n) it(^n) i-2 Nor2 S=t(^n)
t(^n)-2 N: (3.52) In seeking the maximum and minimum (the so-called extremal) values of the above components, let us consider first N. As the normal niassumes all possible orientations atP, the values of Nwill be prescribed by the functional relation in Eq 3.51 subject to the condition that nini=1. Accordingly, we may use to advantage the Lagrangian multiplier method to obtain extremal values of N. To do so we construct the function f(ni) =tijninj-(nini-1), where the scalar is called the Lagrangian multiplier. The method requires the derivative of f(ni)with respect to nkto vanish; and, noting that @ni=@nk=ik, we have @f @nk=tij(iknj+jkni)-(2niik)=0: Buttij=tji, andkjnj=nk, so that this equation reduces to (tij-ij)nj=0 (3.53) which is identical to Eq 3.35, the eigenvalue formulation for principal stresses. Therefore, we conclude that the Lagrangian multiplier assumes the role of a principal stress and, furthermore, that the principal stresses include both the maximum and minimum normal stress values. With regard to the maximum and minimum values of the shear component S, it is useful 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 I>II>IIIso thatt(^n) iis expressed in vector form by Stress Principles 73 P∆S σSσN x2∗x3∗ x1∗t(^n) i FIGURE 3.12 Normal and shear components at Pto plane referred to principal axes. t(^n)=T
^n=In1^e 1+IIn2^e 2+IIIn3^e 3; (3.54) and similarly, N=t(^n)
^nby N=In2 1+IIn2 2+IIIn2 3: (3.55) Then, substituting Eqs 3.54 and 3.55 into Eq 3.52, we have 2 S=2 In2 1+2 IIn2 2+2 IIIn2 3- (In2 1+IIn2 2+IIIn2 3) (3.56) which expresses 2 Sin terms of the direction cosines ni. Butnini=1, so thatn2 3= 1-n2 1-n2 2and we are able to eliminate n3from Eq 3.56, which then becomes a function ofn1andn2only, 2 S=(I-III)n2 1+ 2 II-2 III
n2 2+2 III - (I-III)n2 1+(II-III)n2 2+III2: (3.57) In order to obtain the stationary , that is, the extremal values of2 S, we must equate the derivatives of the right-hand side of this equation with respect to both n1andn2to zero, and solve simultaneously. After some algebraic manipulations, we obtain @2 S @n1=n1I-III I-III-2 I-IIIn2 1+II-IIIn2 2 =0; (3.58a) @2 S @n2=n2II-III II-III-2 I-IIIn2 1+II-IIIn2 2 =0: (3.58b) An obvious solution to Eq 3.58 is n1=n2=0for whichn3=1, and the corre- sponding value of 2 Sis observed from Eq 3.57 to be zero. This is an expected result since n3=1designates a principal plane upon which the shear is zero. A similar calcula- tion made with n1andn3, or withn2andn3as the variables, would lead to the other two principal planes as minimum (zero) shear stress planes. It is easily verified that a second solution to Eq 3.58 is obtained by taking n1=0and solving for n2. The result is 74 Continuum Mechanics for Engineers n2=1=p 2and, from orthogonality, n3=p 2also. For this solution, Eq 3.57 yields the results 2 S=1 4(II-III)2orS=1 2(II-III): (3.59) As before, if we consider in turn the formulation having n1andn3, orn2andn3as the variable pairs, and assume n3=0, andn2=0, respectively, we obtain the complete solution which is presented below, n1=0; n2=1p 2; n3=1p 2;S=1 2jII-IIIj; (3.60a) n1=1p 2; n2=0; n3=1p 2;S=1 2jIII-Ij; (3.60b) n1=1p 2; n2=1p 2; n3=0;S=1 2jI-IIj (3.60c) where the vertical bars in the formulas for Sindicate absolute values of the enclosed expressions. Because I>II>III, it is clear that the largest shear stress value is max S=1 2jIII-Ij: (3.61) It may be shown that, for distinct principal stresses, only the two solutions presented in this section satisfy Eq 3.58. 3.8 Mohr’s Circles for Stress Mohr’s circle is a method the student should remember from undergraduate mechanics of materials. It provides a graphic means for the transformation of a second order tensor like that discussed in Section 2.5. Mohr’s circle is commonly used to transform stress, area or mass moment of inertia tensors. While this book gives the student the mathematical tools to transform tensors, we feel a discussion of Mohr’s circle is worthy of inclusion in this introduction to continuum mechanics. Most of the students prior use of Mohr’s circle has likely been with two dimensional problems: plane stress or transformation of the area or mass inertia tensors. Mohr’s circle allows for an easy means to derive and compute transformation and principal values in two dimensions. In three dimensions, the convenience and simplicity of Mohr’s circle to transform stress components to an arbitrary reference frame vanishes. When working with three dimen- sional problems the advantage of using the mathematics of continuum mechanics become evident. This is the case for many problems in mechanics. In two dimensions, intuitive diagrams 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 student may have experienced this. Consider again the state of stress at Preferenced to principal axes (Fig. 3.12) and let the principal stresses be ordered according to I>II>III. As before, we may express N andSon any plane at Pin terms of the components of the normal to that plane by the Stress Principles 75 OσS σNσIIIσI σIIC1C2 C3 FIGURE 3.13 Typical Mohr’s circle for stress. equations N=In2 1+IIn2 2+IIIn2 3; (3.62a) 2 N+2 S=2 In2 1+2 IIn2 2+2 IIIn2 3 (3.62b) which, along with the condition n2 1+n2 2+n2 3=1; provide us with three equations for the three direction cosines n1,n2andn3. Solving these equations, we obtain n2 1=(N-II) (N-III)+2 S (I-II) (I-III); (3.63a) n2 2=(N-III) (N-I)+2 S (II-III) (II-I); (3.63b) n2 3=(N-I) (N-II)+2 S (III-I) (III-II): (3.63c) In these equations, I,IIandIIIare known;NandSare functions of the direction cosinesni. Our intention here is to interpret these equations graphically by representing conjugate pairs of N,Svalues, which satisfy Eq 3.63, as a point in the stress plane having Nas absicca and Sas ordinate (see Fig. 3.13). To develop this graphical interpretation of the three-dimensional stress state in terms ofNandS, we note that the denominator of Eq 3.63a is positive since both I-II>0 andI-III>0, and also that n2 1>0, all of which tells us that (N-II)(N-III) +2 S>0: (3.64) 76 Continuum Mechanics for Engineers For the case where the equality sign holds, this equation may be rewritten, after some simple algebraic manipulations, to read  N-1 2(II+III)2 +2 S=1 2(II-III)2 (3.65) which is the equation of a circle in the N,Splane, with its center at the point1 2(II+ III)on theNaxis, and having a radius1 2(II-III). We label this circle C1and display it in Fig. 3.13. For the case in which the inequality sign holds for Eq 3.64, we observe that conjugate pairs of values of NandSwhich satisfy this relationship result in stress points having coordinates exterior to circle C1. Thus, combinations of NandSwhich satisfy Eq 3.63a lie on, or exterior to, circleC1in Fig. 3.13. Examining Eq 3.63b, we note that the denominator is negative since II-III>0, and II-I<0. The direction cosines are real numbers, so that n2 2>0and we have (N-III)(N-I) +2 S60 (3.66) which for the case of the equality sign defines the circle  N-1 2(I+III)2 =1 2(I-III)2 (3.67) in theN,Splane. This circle is labeled C2in Fig. 3.13, and the stress points which satisfy 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 third circle,C3in Fig. 3.13, namely,  N-1 2(I+II)2 +2 S=1 2(I-II)2 : (3.68) Admissible stress points in the N,Splane lie onorexterior to this circle. The three circles defined above, and shown in Fig. 3.13, are called Mohr’s circles for stress. All possible pairs of values ofNandSatPwhich satisfy Eq 3.63 lie on these circles or between the areas enclosed by them. Actually, in conformance with Fig. 3.11 (which is the physical basis for Fig. 3.13), we see that the sign of the shear component is arbitrary so that only the top half 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 stress value atPis the radius of circle C2, which confirms the result presented in Eq 3.61. In order to relate a typical stress point having coordinates NandSin the stress plane of Fig. 3.13 to the orientation of the area element
S(denoted by niin Fig. 3.12) upon which the stress components NandSact, we consider a small spherical portion of the continuum body centered at P. As the unit normal niassumes all possible directions at P, the point of intersection of its line of action with the sphere will move over the surface of the sphere. However, as seen from Eqs 3.55 and Eq 3.56 the values of NandSare functions of the squares of the direction cosines, and hence do not change for nireflected in the principal planes. Accordingly, we may restrict our attention to the first octant of the spherical body, as shown in Fig. 3.14(a). Let Qbe the point of intersection of the line of action ofniwith the spherical surface ABC in Fig. 3.14(a) and note that ^n=cos^e 1+cos^e 2+cos^e 3: (3.69) If^n=^e 1so that its intersection point Qcoincides with A,N=I. Likewise, when Qcoincides with B,N=II, and withC,N=III. In all three cases, Swill be zero. Stress Principles 77 βθ φ ABC ni x2∗x3∗ x1∗Q P ^e∗ 2^e∗ 3 ^e∗ 1 (a) Octant of small spherical portion of body to- gether with plane at Pwith normalnireferred to principal axesOx 1x 2x 3. σIIIσI σIIC1C2 C3OσS σNa b c(b) Mohr’s stress semicircle for octant of Fig. 3.14(a). FIGURE 3.14 Typical Mohr’s circle representation. In the Mohr’s circle diagram (Fig. 3.14(b)), these stress values are located at points a,b andc, respectively. If now is set equal to 2andallowed to vary from zero to 2( will concurrently go from 2to zero),Qwill move along the quarter-circle arc ABfromA toB. In the stress space of Fig. 3.14(b), the stress point q(the image point of Q) having coordinatesNandSwill simultaneously move along the semicircle of C3fromatob. (Note that as Qmoves90oalongABin physical space, qmoves180oalong the semicircle, joiningatobin stress space.) Similarly, when Qis located on the quarter circle BC, or CAof Fig. 3.14(a), point qwill occupy a corresponding position on the semicircles of bc andca, respectively, in Fig. 3.14(b). Now let the angle be given some fixed value less than 2, say=1, and imagine thatandtake on all values compatible with the movement of Qalong the circle arcKD of Fig. 3.15(a). For this case, Eq 3.63a becomes (N-II)(N-III) +2 S= (I-II)(I-III)cos21, which may be cast into the standard form of a circle as  N-1 2(II+III)2 +2 S = (I-II)(I-III)cos21+1 2(II-III)2 =R2 1: (3.70) This circle is seen to have its center coincident with that of circle C1in stress space and to have a radius R1indicated by Eq 3.70. Therefore, as Qmoves on circle arc KD in Fig. 3.15(a), the stress point qtraces the circle arc kdshown in Fig. 3.15(b). (Notice that if1= 2so that cos1=0,R1reduces to1 2(II-III), the radius of circle C1). Next, let=1< 2and then, as andrange through all admissible values, point Qmoves along the circle arc EGof Fig. 3.15(a). For this case Eq 3.63b may be restructured into the 78 Continuum Mechanics for Engineers x2∗x3∗ x1∗ φφβ βD EG K ABC niQ P ^e∗ 1^e∗ 2^e∗ 3 (a) Reference angles andfor intersection point Qon surface of body octant. k g e dq c a b O σIIIσI σIIσS σN2β2β2φ 2φ(b) Mohr’s stress semicircle for octant of Fig. 3.15(a). FIGURE 3.15 Typical 3-D Mohr’s circle and associated geometry. form  N-1 2(I+III)2 +2 S = (II-III)(II-I)cos21+1 2(I-III)2 =R2 2 (3.71) which defines a circle whose center is coincident with that of circle C2, and having a radiusR2. Here, when 1= 2, the radiusR2reduces to1 2(I-III), which is the radius of circleC2. AsQmoves on the circle arc EGof Fig. 3.15(a), the stress point qtraces out the circle arc egin Fig. 3.15(b). In summary, for a specific ^nat pointPin the body, point Q, where the line of action of ^nintersects the spherical octant of the body (Fig. 3.15(a)), is located at the common point of circle arcs KDandEG, and at the same time, the corresponding stress point q(having coordinatesNandS) is located at the intersection of circle arcs kdandegin the stress plane of Fig. 3.15(b). The following example provides details of the procedure. Example 3.5 The state of stress at point Pis given inMPa with respect to axes Px1x2x3by the matrix [tij] =2 425 0 0 0-30 -60 0-60 53 5: Stress Principles 79 (a) Determine the stress vector on the plane whose unit normal is ^n=1 3(2^e1+^e2+2^e3). (b) Determine the normal stress component Nand shear component Son the 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 2 664t(^n) 1 t(^n) 2 t(^n) 33 775= [tij][nj] =2 66425 0 0 0-30 -60 0-60 53 7752 6642 3 1 3 1 33 775=1 32 66450 -150 -503 775; or t(^n)=1 3(50^e1-150^e2-50^e3): (b) Making use of Eq 3.51, we can calculate Nconveniently from the matrix product h 2 31 32 3i2 66425 0 0 0-30 -60 0-60 53 7752 6642 3 1 3 2 33 775=N N=100 9-150 9-100 9 so thatN= -150 9= -16:67MPa . Note that the same result could have been obtained by the dot product N=t(^n)
^n=1 3(50^e1-150^e2-50^e3)
1 3(2^e1+^e2+2^e3): The shear component Sis given by Eq 3.52, which for the values of N andt(^n) icalculated above, results in the equation, 2 S=2500 +22500 +2500 9-22500 81=2777; or, nally, S=52:7MPa: (c) Using the procedure of Example 3.4 the student should verify that for the stress tensor tijgiven here the principal stress values are I=50 MPa , II=25 MPa , andIII= -75 MPa . Also, the transformation matrix from axesPx1x2x3toPx 1x 2x 3is [aij] =2 6640-3 54 5 1 0 0 04 53 53 775; 80 Continuum Mechanics for Engineers so that the components of ^nare given relative to the principal axes by 2 664n 1 n 2 n 33 775=2 6640-3 54 5 1 0 0 04 53 53 7752 6642 3 1 3 2 33 775=2 6641 3 2 3 2 33 775: Therefore, with respect to Fig. 3.14(a), =cos-1(1=3) =70:53o;== cos-1(2=3) =48:19o, so that|following the procedure outlined for construction of Fig. 3.15(b)|we obtain the Mohr's circle below, from which we may measure the coordinates of the stress point qand conrm the values N= -16:7 and S=52:7, both inMPa . OC3 C1C2 142° 96° 96°142°k c a bdeq(−16.7, 52.7) σIII = −75 σΙ = 50 σII = 25σS σN 3.9 Plane Stress When one, and only one principal stress is zero, we have a state of plane stress for which the plane of the two nonzero principal stresses is the designated plane . This is an important state of stress because it represents the physical situation occurring at an unloaded point on the bounding surface of a body under stress. The zero principal stress may be any one of the three principal stresses as indicated by the corresponding Mohr’s circles of Fig. 3.16. If the principal stresses are not ordered and the direction of the zero principal stress is arbitrarily chosen as x3, we have plane stress parallel to the x1x2-plane and the stress matrix takes the form [tij] =2 4t11t120 t12t220 0 0 03 5; (3.72) Stress Principles 81 σIIIσIσIIσS σN (a)I=0 σIIIσIσIIσS σN (b)II=0 σIIIσIσIIσS σN (c)III=0 FIGURE 3.16 Mohr’s circle for plane stress. 82 Continuum Mechanics for Engineers TABLE 3.2 Transformation table for general plane stress. x1x2x3 x 1 cos sin 0 x 2 -sin cos 0 x 30 0 1 or, with respect to principal axes, the form [t ij] =2 4(1)0 0 0  (2)0 0 0 03 5: (3.73) The pictorial description of this plane stress situation is portrayed by the block element of a continuum body shown in Fig. 3.17(a), and is sometimes represented by a single Mohr’s circle (Fig. 3.17(b)), the locus of which identifies stress points (having coordinates NandS) for unit normals lying in the x1-x2plane only. The equation of the circle in Fig. 3.17(b) is  N-t11+t22 22 +2 S=t11-t22 22 +t2 12 (3.74) from which the center of the circle is noted to be at N=1 2(t11+t22),S=0, and the maximum shear stress in the x1x2-plane to be the radius of the circle, that is, the square root of the right-hand side of Eq 3.74. Points AandBon the circle represent the stress states for area elements having unit normals ^e1and^e2, respectively. For an element of area having a unit normal in an arbitrary direction at point P, we must include the two dashed circles shown in Fig. 3.17(c) to completely specify the stress state. With respect to axes Ox0 1x0 2x0 3rotated by the angle about thex3axis relative to Ox1x2x3as 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 of direction cosines for this situation as listed in Table 3.2, we may express the primed stress components in terms of the rotation angle and the unprimed components by t0 11=t11+t22 2+t11-t22 2cos2+t12sin2; (3.75a) t0 22=t11+t22 2-t11-t22 2cos2-t12sin2; (3.75b) t0 12= -t11-t22 2sin2+t12cos2: (3.75c) In addition, if the principal axes of stress are chosen for the primed directions, it is easily shown that the two nonzero principal stress values are given by (1);(2)=t11+t22 2st11-t22 22 +t2 12 (3.76) Stress Principles 83 x1x2x3 t12t22 t21 t11 (a) Plane stress element having nonzero x1and x2components. σI σIIσS σNA(t11, t12) B(t22, t12)(b) Mohr’s circle for in-plane stress components. (σS)max(σS)max σNσS σIIσI σIII (c) General Mohr’s circles for the plane stress element. Dashed lines represent out-of-plane Mohr’s circles. Note the maximum shear can occur out-of-plane. FIGURE 3.17 Mohr’s circle for plane stress. 84 Continuum Mechanics for Engineers x3 ,� θθPx 3 x 2 x 1x2 x1 FIGURE 3.18 Representative rotation of axes for plane stress. Example 3.6 A specimen is loaded with equal tensile and shear stresses. This case of plane stress may be represented by the matrix [tij] =2 4000 000 0 0 03 5 where0is a constant stress. Determine the principal stress values and plot the Mohr's circles. Solution For this stress state, the determinant Eq 3.38 is given by 0- 00 00- 0 0 0 -=0 which results in a cubic having roots (principal stress values) (1)=20,(2)= (3)=0(as may be readily veried by Eq 3.76) so that, in principal axes form, the stress matrix is [t ij] =2 4200 0 0 0 0 0 0 03 5: 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 note that physically this is simply a one-dimensional tension in the x 1direction and that the maximum shear stress values (shown by points AandB) occur on the x1andx2coordinate planes which make 45with the principal x 1direction. Stress Principles 85 .. +A BσS σN+σ0 −σ0σII=σIII=0 σI=2σο Mohr's circle for principal stresses 1=2o,2=3=0. 3.10 Deviator and Spherical Stress States The arithmetic mean of the normal stresses, M=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 stresses equal (and therefore equal to M) is called a spherical state of stress and is represented by the diagonal matrix [tij] =2 4M0 0 0 M0 0 0  M3 5 (3.78) for which all directions are principal directions as explained in Section 3.6. The classical physical example for this is the stress in a fluid at rest which is termed hydrostatic stress , and for which M= -p0, the static pressure. Every state of stress tijmay be decomposed into a spherical portion and a portion Sij known as the deviator stress in accordance with the equation tij=Sij+ijM=Sij+1 3ijtkk (3.79) whereijis the Kronecker delta. This equation may be solved for Sij, which then appears in the symmetric matrix form 2 4S11S12S13 S12S22S23 S13S23S333 5=2 4t11-Mt12t13 t12t22-Mt23 t13t23t33-M3 5: (3.80) Also from Eq 3.79, we notice immediately that the first invariant of the deviator stress is Sii=tii-1 3iitkk=0 (3.81) 86 Continuum Mechanics for Engineers (sinceii=3), so that the characteristic equation for the deviator stress (analogous to Eq 3.39 fortij) is S3+IISS-IIIS=0 (3.82) for which the deviator stress invariants are IIS= -1 2SijSji=SISII+SIISIII+SIIISI; (3.83a) IIIS="ijkS1iS2jS3k=SISIISIII: (3.83b) Finally, consider a principal direction n(q) joftijsuch that the eigenvalue equation tij-(q)ij n(q) j=0is satisfied. Then, from the definition of Sij, we have [Sij+Mij-qij]n(q) j=0; or [Sij-(q-M)ij]n(q) j=0 (3.84) which demonstrates that n(q) jis also a principal direction of Sij, and furthermore, the principal values of Sijare given in terms of the principal values of tijby Sq=q-M; (q=1;2;3 ): (3.85) Example 3.7 Decompose the stress tensor Tof Example 3.4 into its deviator and spherical portions and determine the principal stress values of the deviator portion. [tij] =2 457 0 24 0 50 0 24 0 433 5[MPa]: Solution By Eq 3.77,Mfor the given stress is M=1 3(57+50+43)=50[MPa] Thus, decomposition by Eq 3.79 leads to the matrix sum 2 457 0 24 0 50 0 24 0 433 5=2 47 0 24 0 0 0 24 0 -73 5+2 450 0 0 0 50 0 0 0 503 5[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=25MPa ,SII=0andSIII= -25MPa . These results are easily veried using the principal values determined in Example 3.4 together with Eq 3.85. Stress Principles 87 t(^n) i P ABCσN σS x2∗x3∗ x1∗ni=^e1+^e2+^e3√ 3 FIGURE 3.19 Octahedral plane (ABC) with traction vector t(^n) i, and octahedral normal and shear stresses,NandS. 3.11 Octahedral Shear Stress Consider the plane at Pwhose unit normal makes equal angles with the principal stress directions. That plane, called the octahedral plane, may be pictured as the triangular surface ABC 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=(1)^e 1+(2)^e 2+(3)^e 3p 3; (3.86) and its component in the direction of ^nis N=t(^n)
^n=1 3 (1)+(2)+(3) =1 3ii: (3.87) Thus, from Eq 3.52, the square of the shear stress on the octahedral plane, known as the octahedral shear stress, is 2 oct=t(^n)
t(^n)-2 N=1 3[2 (1)+2 (2)+2 (3)] -1 9[(1)+(2)+(3)]2(3.88) which may be reduced to either the form (see Problem 3.27) oct=1 3q (1)-(2)
2+ (2)-(3)
2+ (3)-(1)
2; (3.89) or oct=q S2 1+S2 2+S2 3p 3=r -2IIS 3: (3.90) 88 Continuum Mechanics for Engineers Example 3.8 Determine directly the normal and shear components, Nandoct, on the octahedral plane for the state of stress in Example 3.4, and verify the result for octby Eq 3.89. Solution From Example 3.4, the stress vector on the octahedral plane is given by the matrix product2 666475 0 0 0 50 0 0 0 253 77752 66641p 3 1p 3 1p 33 7775=2 666475p 3 50p 3 25p 33 7775; or t(^n)=75^e 1+50^e 2+25^e 3p 3; so that N=t(^n)
^n=1 3(75+50+25) =50MPa: Also, from Eq 3.52, 2 oct=t(^n)
t(^n)-2 N=1 3(752+502+252) -502=417MPa; and sooct=20:4MPa . By Eq 3.89, we verify directly that oct=1 3q (75-50)2+ (50 -25)2+ (25 -75)2=20:41 MPa Example 3.9 Show that the octahedral shear stress may be written as oct=1p 3 ((1)-M)2+ ( (2)-M)2+ ( (3)-M)21=2; and, from this, show that oct=p 2 3 I2 T+3IIT
1=2 is another form of the octahedral shear stress. Solution Start by multiplying the squared terms under the radical oct=1p 3h 2 (1)+2 (2)+2 (3)+32 M-2 (1)+(2)+(3)
Mi1=2 : Stress Principles 89 Recall Eq 3.77 in the form (1)+(2)+(3)=3M and substitute to get 2 oct= 2 (1)+2 (2)+2 (3) 3-2 M! : Substitution of Eq 3.77 again for Mresults in Eq 3.88. Multiply out the mean stress terms and simplify to get 2 oct =" 3(2 (1)+2 (2)+2 (3)) 9 -2 (1)+2 (2)+2 (3)+2((1)(2)+(2)(3)+(1)(3)) 9# =2 9h 2 (1)+2 (2)+2 (3)- (1)(2)+(2)(3)+(1)(3)
i : Finally, add and subtract 2 (1)(2)+(2)(3)+(1)(3)
in the bracket 2 oct =2 9h 2 (1)+2 (2)+2 (3)+2((1)(2)+(2)(3)+(1)(3)) -3( (1)(2)+(2)(3)+(1)(3)) ; and note that, for principal axes I2 T= (1)+(2)+(3)
2 IIT=(1)(2)+(2)(3)+(1)(3) leaving 2 oct=2 9 I2 T+3IIT ; or oct=p 2 3 I2 T+3IIT1=2: 90 Continuum Mechanics for Engineers Problems Problem 3.1 At a pointP, the stress tensor relative to axes Px1x2x3has components tij. On the area elementdS(1)having the unit normal ^n1, the stress vector is t(^n1), and on area element dS(2)with normal ^n2the stress vector is t(^n2). Show that the component of t(^n1)in the direction of ^n2is equal to the component of t(^n2)in the direction of ^n1. Problem 3.2 Verify the result established in Problem 3.1 for the area elements having normals ^n1=1 7(2^e1+3^e2+6^e3) ^n2=1 7(3^e1-6^e2+2^e3) if the stress matrix at Pis given with respect to axes Px1x2x3by [tij]=2 435 0 21 0 49 0 21 0 143 5: Problem 3.3 The stress tensor at Prelative to axes Px1x2x3has components in MPa given by the matrix representation [tij]=2 4t112 1 2 0 2 1 2 03 5 wheret11is unspecied. Determine a direction ^natPfor which the plane perpendicular to ^nwill be stress-free, that is, for which t(^n)=0on that plane. What is the required value oft11for this condition? Answer ^n=1 3(2^e1-^e2-2^e3),t11=2MPa Problem 3.4 The stress tensor has components at point Pin ksi units as specied by the matrix [tij]=2 4-9 3 -6 3 6 9 -6 9 -63 5: Stress Principles 91 Determine: (a) the stress vector on the plane at Pwhose normal vector is ^n=1 9(^e1+4^e2+8^e3); (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)=-5^e1+11^e2-2^e3 (b)t(^n)=p 150 (c)23 9 (d)77:96 Problem 3.5 Let the stress tensor components at a point be given by tij=0ninjwhere0is a positive constant. Show that this represents a uniaxial state of stress having a magnitude 0and acting in the direction of ni. Problem 3.6 Show 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) it(^e1) i+t(^e2) it(^e2) i+t(^e3) it(^e3) i is an invariant. x1x2x3 t(^e2) i t(^e1) it(^e3) i Problem 3.7 With respect to axes Ox1x2x3the stress state is given in terms of the coordinates by the matrix [tij]=2 4x1x2x2 20 x2 2x2x3x2 3 0 x2 3x3x13 5: 92 Continuum Mechanics for Engineers Determine (a) the body force components as functions of the coordinates if the equilibrium equations are to be satised everywhere, and (b) the stress vector at point P(1,2,3) on the plane whose outward unit normal makes equal angles with the positive coordinate axes. Answer (a)b1=-3x2 ,b2=-3x3 ,b3=-x1  (b)t(^n)=(6^e1+19^e2+12^e3)p 3 Problem 3.8 Relative to the Cartesian axes Ox1x2x3a stress eld is given by the matrix [tij]=2 66664 1-x2 1
x2+2 3x3 2 - 4-x2 2
x10 - 4-x2 2
x1 -1 3 x3 2-12x2
0 0 0 3-x2 1
x23 77775: (a) Show that the equilibrium equations are satised 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(-29^e1-40^e2+2^e3) Problem 3.9 The stress components in a circular cylinder of length Land radiusrare given by [tij]=2 4Ax2+Bx3Cx3-Cx2 Cx30 0 -C20 03 5: (a) Verify that in the absence of body forces the equilibrium equations are satised. (b) Show that the stress vector vanishes at all points on the curved surface of the cylinder. rL x1x3 x2 Stress Principles 93 Problem 3.10 Rotated axes Px0 1x0 2x0 3are obtained from axes Px1x2x3by a right-handed rotation about the linePQthat makes equal angles with respect to the Px1x2x3axes (see sketch). Determine the primed stress components for the stress tensor in (MPa) [tij]=2 43 0 6 0 0 0 6 0 -33 5 if the angle of rotation is (a) 120, or (b)60. Q ββ βP x1 x2x3 Answer (a)h t0 iji =2 40 0 0 0-3 6 0 6 33 5MPa, (b)h t0 iji =1 32 4-5 10 10 10-11 -2 10 -2 163 5MPa Problem 3.11 At the pointP, rotated axes Px0 1x0 2x0 3are related to the axes Px1x2x3by the transformation matrix [aij]=1 32 4a 1 -p 3 1 +p 3 1+p 3 b 1 -p 3 1-p 3 1 +p 3 c3 5 wherea,b, andcare to be determined. Determineh t0 iji if the stress matrix relative to axesPx1x2x3is given in MPa by [tij]=2 41 0 1 0 1 0 1 0 13 5: Answer h t0 iji =1 92 411+2p 3 5 +p 3 -1 5+p 3 5 5 -p 3 -1 5 -p 3 11 -2p 33 5MPa Problem 3.12 The stress matrix referred to axes Px1x2x3is given in ksi by [tij]=2 414 0 21 0 21 0 21 0 73 5 94 Continuum Mechanics for Engineers Let rotated axes Px0 1x0 2x0 3be dened with respect to axes Px1x2x3by the table of base vectors ^e1^e2 ^e3 ^e0 12=7 3=7 6=7 ^e0 23=7 -6=7 2=7 ^e0 36=7 2=7 -3=7 (a) Determine the stress vectors on planes at Pperpendicular to the primed axes; determinet(^e0 1),t(^e0 2), andt(^e0 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 ofh t0 iji . (c) Verify the result obtained in (b) by a direct application of Eq 3.33 of the text. Answer h t0 iji =1 72 4143 36 114 36 166 3 114 3 -153 5ksi Problem 3.13 At pointP, the stress matrix is given in MPa with respect to axes Px1x2x3by Case 1: [tij]=2 46 4 0 4 6 0 0 0 -23 5Case 2: [tij]=2 42 1 1 1 2 1 1 1 23 5 Determine for each case (a) the principal stress values, (b) the principal stress directions. Answer (a) Case 1:(1)=10MPa,(2)=2MPa,(3)= -2 MPa Case 2:(1)=4MPa,(2)=(3)=1MPa (b) Case 1: ^n(1)=^e1+^e2p 2,^n(2)=^e1-^e2p 2,^n(3)=^e3 Case 2: ^n(1)=^e1+^e2+^e3p 3,^n(2)=-^e1+^e2p 2,^n(3)=-^e1-^e2+2^e3p 6 Problem 3.14 When referred to principal axes at P, the stress matrix in ksi units is  t ij =2 42 0 0 0 7 0 0 0 123 5 Stress Principles 95 If the transformation matrix between the principal axes and axes Px1x2x3is [aij]=1p 22 4-3 51 -4 5 a21a22a23 -3 5-1 -4 53 5 wherea21,a22, anda23are to be determined, calculate [tij]. Answer [tij]=2 47 3 0 3 7 4 0 4 73 5ksi Problem 3.15 The stress matrix in MPa when referred to axes Px1x2x3is [tij]=2 43-10 0 -10 0 30 0 30 -273 5: Determine (a) the principal stresses, I,II,III, (b) the principal stress directions. Answer (a)I=23MPa,II=0MPa,III= -47 MPa (b)^n(1)= -0:394 ^e1+0:788 ^e2+0:473 ^e3 ^n(2)=0:931 ^e1+0:274 ^e2+0:304 ^e3 ^n(3)=0:110 ^e1+0:551 ^e2-0:827 ^e3 Problem 3.16 At pointP, the stress matrix relative to axes Px1x2x3is given in MPa by [tij]=2 45 a -a a 0 b -a b 03 5 whereaandbare unspecied. At the same point relative to axes Px 1x 2x 3the matrix is  t ij =2 4I0 0 0 2 0 0 0  III3 5: If the magnitude of the maximum shear stress at Pis5:5MPa, determine IandIII. Answer I=7MPa,III= -4 MPa 96 Continuum Mechanics for Engineers Problem 3.17 The state of stress at point Pis given in ksi with respect to axes Px1x2x3by the matrix [tij]=2 41 0 2 0 1 0 2 0 -23 5: Determine (a) the principal stress values and principal stress directions at P, (b) the maximum shear stress value at P, (c) the normal ^n=ni^eito the plane at Pon which the maximum shear stress acts. Answer (a)I=2ksi,II= 1 ksi,III= {3 ksi ^n(1)=2^e1+^e3p 5,^n(2)=^e2,^n(3)=-^e1+2^e3p 5 (b)(S)max=2:5ksi (c)^n=^e1+3^e3p 10 Problem 3.18 The stress tensor at Pis given with respect to Ox1x2x3in matrix form with units of MPa by [tij]=2 44 b b b 7 2 b 2 43 5 wherebis unspecied. If III=3MPa andI=2II, determine (a) the principal stress values, (b) the value of b, (c) the principal stress direction of II. Answer (a)I=8MPa,II=4MPa,III=3MPa (b)b=0, (c)^n(2)=^e1 Problem 3.19 The state of stress at P, when referred to axes Px1x2x3is given in ksi units by the matrix [tij]=2 49 3 0 3 9 0 0 0 183 5 Stress Principles 97 Determine (a) the principal stress values at P, (b) the unit normal ^n=ni^e iof the plane on which N=12ksi andS=3ksi. Answer (a)I=18ksi,II=12ksi,III=6ksi (b)^n=^e 1+p 6^e 2+^e 3 2p 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 oriented along the coordinate axes. σ0σ0 σ0 σ0 (a) σ0 σ0 (b) σ0σ0 σ0 σ0 (c) 2σ0σ0 2σ0 σ0 (d) 98 Continuum Mechanics for Engineers Problem 3.22 The state of stress referred to axes Px1x2x3is given in MPa by the matrix [tij]=2 49 12 0 12-9 0 0 0 53 5: Determine (a) the normal and shear components, NandS, respectively, on the plane at P whose unit normal is ^n=1 5(4^e1+3^e2); (b) verify the result determined in (a) by a Mohr's circle construction similar to that shown in the gure from Example 3.5. Answer N=14:04 MPa,S=5:28 MPa Problem 3.23 Sketch the Mohr's circles for the simple states of stress given by (a)[tij]=2 400 0 0 00 00 03 5; (b)[tij]=2 400 0 0 200 0 0 -03 5; and determine the maximum shear stress in each case. Answer (a)(S)max=0 (b)(S)max=3 20 Problem 3.24 Relative to axes Ox1x2x3, the state of stress at Ois represented by the matrix [tij]=2 46-3 0 -3 6 0 0 0 03 5[ksi]: Show that, relative to principal axes Ox 1x 2x 3, the stress matrix is [tij]=2 43 0 0 0 9 0 0 0 03 5[ksi]; Stress Principles 99 and that these axes result from a rotation of 45about thex3axis. Verify these results by Eq 3.75. Problem 3.25 The stress matrix representation at Pis given by [tij]=2 429 0 0 0-26 6 0 6 93 5[ksi]: Decompose this matrix into its spherical and deviator parts, and determine the principal deviator stress values. Answer SI=25ksi,SII=6ksi,SIII= -31 ksi Problem 3.26 Let the second invariant of the stress deviator be expressed in terms of its principal values, that is, by IIS=SISII+SIISIII+SIIISI: Show that this sum is the negative of two-thirds the sum of squares of the principal shear stresses, as given by Eq 3.60. Problem 3.27 Verify the results presented in Eqs 3.89 and 3.90 for the octahedral shear stress. Problem 3.28 At pointPin a continuum body, the stress tensor components are given in MPa with respect to axes Px1x2x3by the matrix [tij]=2 41 -3p 2 -3 1 -p 2p 2-p 2 43 5: Determine (a) the principal stress values I,II, andIII, together with the corresponding prin- cipal stress directions, (b) the stress invariants IT,IIT, andIIIT, (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 shear components, (f) the stress matrix for axes rotated 60counterclockwise with respect to the axis PQ, which makes equal angles relative to the coordinate axes Px1x2x3. 100 Continuum Mechanics for Engineers Answer (a)I=6MPa,II=2MPa,III= -2 MPa ^n(1)=1 2 ^e1-^e2+p 2^e3 ^n(2)=1 2 ^e1-^e2-p 2^e3 ^n(3)=^e1+^e2p 2 (b)IT=6MPa,IIT= -4 MPa,IIIT= -24 MPa (c)(S)max=4MPa ^nmax= 1+p 2 ^e1- 1-p 2 ^e2+p 2^e3 2p 2 (d)SI=4MPa,SII=0MPa,SIII= -4 MPa (e)^t(^n)=6^e 1+2^e 2-2^e 3p 3,N=2,oct=q 32 3 (f)h t0 iji =1 92 4-12 -12+3p 2-12-3p 2 -12+3p 2 33 -12p 2 -3 -12-3p 2 -3 33 +12p 23 5MPa Problem 3.29 In a continuum, the stress eld relative to axes Ox1x2x3is given by [tij]=2 6664x2 1x2x1 1-x2 2
0 x1 1-x2 2
1 3 x2 2-3x2
0 0 0 2x2 33 7775: Determine (a) the body force distribution if the equilibrium equations are to be satised through- out the eld, (b) the principal stresses at P(a;0;2pa), (c) the maximum shear stress at P, (d) the principal deviator stresses at P. Answer (a)b1=b2=0,b3= -4x3  (b)I=8a,II=a,III= -a (c)(S)max=4:5a (d)SI=16 3a,SII= -5 3a,SIII= -11 3a Problem 3.30 In describing the yield surface in plasticity the second invariant of the deviator stress, often denoted byJ2, plays an important role. Starting with the second invariant of the deviator stress J2=IIS= - S(1)S(2)+S(2)S(3)+S(1)S(3)
Stress Principles 101 derive the formula J2=3 22 oct: Problem 3.31 Show that J2=IIS= -(S (1)S(2)+S(2)S(3)+S(1)S(3)) =1 2 S2 (1)+S2 (2)+S2 (3) whereJ2is the second invariant of the deviator stress and S(1); S(2); S(3)are its principal values. Problem 3.32 Let the stress tensor components tijbe derivable from the symmetric tensor eld 'ijby the equationtij="iqk"jpm'km;qp . Show that, in the absence of body forces, the equilibrium equations are satised. Recall from Problem 2.15 that "iqk"jpm=jijqjk pipqpk mimqmk: Problem 3.33 Verify that@tij=@tmn=imjnand 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 the stress components is equal to the negative of the corresponding component of the deviatoric stress. This page intentionally left blank 4 Kinematics of Deformation and Motion Kinematics is the study of motion. In Continuum Mechanics it has the same meaning as in dynamics or fluid mechanics. It is the basis for most all constitutive theories. Recall from Chapter 1 that a constitutive response, or material model, describes how stress develops according 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 and gases. Let us begin by describing the model we use to represent such bodies. For this purpose we define a material body Bas the set of elements X, called particles, or material points, which can be put into a one-to-one correspondence with the points of a regular region of physical space. Note that whereas a particle of classical mechanics has an assigned mass, a continuum particle is essentially a material point for which a density is defined. The specification of the position of all of the particles of Bwith respect to a fixed origin 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=(X) (4.1) in which the vector function assigns the position xrelative to some origin of each particleXof the body. Assume that this mapping is uniquely invertible and differentiable as many times as required; in general two or three times will suffice. The inverse is written X=-1(x) (4.2) and identifies the particle Xlocated 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 which produces a new configuration but causes no changes in the size or shape of the body; only changes in its position and/or orientation. On the other hand, an arbitrary displacement will usually include both a rigid-body displacement and a deformation which results in a change in size, or shape, or possibly both. Amotion of body Bis a continuous time sequence of displacements that carries the set of particlesXinto various configurations in a stationary space. Such a motion may be expressed by the equation x=(X;t) (4.3) 103 104 Continuum Mechanics for Engineers Reference Currentx3, X3 x2, X2 x1, X1OP pu ^e1,^I1^e2,^I2^e3,^I3 x=κ� Φ−1(X),t =χ(X,t)X=Φ(X) FIGURE 4.1 Position of typical particle in reference configuration XAand current configuration xi. which gives the position xfor each particle Xfor all times t, wheretranges from -1 to 1. As with configuration mappings, we assume the motion function in Eq 4.3 is uniquely invertible and differentiable, so that we may write the inverse X=-1(x;t) (4.4) which identifies the particle Xlocated at position xat timet. 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 purpose it 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 time t=0, is chosen as the reference configuration, and the ensuing deformations and motions related to it. The current configuration is that one which the body occupies at the current timet. In developing the concepts of strain we confine attention to two specific configurations without any regard for the sequence by which the second configuration is reached from the first. It is customary to call the first (reference) state the undeformed configuration, and the second state the deformed configuration. Additionally, time is not a factor in deriving the 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 fluids do not possess a natural geometry, and because of this it is the velocity field of a fluid that assumes the fundamental kinematic role. 4.2 Material and Spatial Coordinates Consider now the reference configuration prescribed by some mapping function such that the position vector Xof particleXrelative to the axes OX1X2X3of Fig. 4.1 is given by Kinematics of Deformation and Motion 105 X=(X) (4.5) In this case we may express Xin terms of the base vectors ^I1;^I2;^I3shown in Fig. 4.1 by the equation X=XA^IA; (4.6) and we call the components XAthematerial coordinates, or sometimes the referential co- ordinates, of the particle X. Uppercase letters which are used as subscripts on material coordinates, or on any quantity expressed in terms of material coordinates, obey all the rules 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 all subsequent configurations every particle can be identified by the position Xit occupied in the reference configuration. As usual, we assume an inverse mapping X=-1(X); (4.7) so that upon substitution of Eq 4.7 into Eq 4.3 we obtain x= -1(X);t =(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 . Notice that Eq 4.8 maps the particle at Xin the reference configuration onto the point xin the current configuration at time tas indicated in Fig. 4.1. With respect to the usual Cartesian axes Ox1x2x3the current position vector is x=xi^ei (4.9) where the components xiare called the spatial coordinates of the particle. Although it is not necessary to superpose the material and spatial coordinate axes as we have done in Fig. 4.1, it is convenient to do so, and there are no serious restrictions from this practice in the derivations which follow. We emphasize, however, that the material coordinates are used in conjunction with the reference configuration only, and the spatial coordi- nates serve for all other configurations. As already remarked the material coordinates are therefore time independent. We may express Eq 4.8 in either a Cartesian component or a coordinate-free notation by the equivalent equations xi=i(XA;t) orx=(X;t): (4.10) It is common practice in continuum mechanics to write these equations in the alternative forms xi=xi(XA;t) orx=x(X;t) (4.11) with the understanding that the symbol xi(orx) on the right hand side of the equation represents the function whose arguments are Xandt, while the same symbol on the left- hand side represents the value of the function, that is, a point in space. We shall use this notation frequently in the text that follows. Notice that as Xranges over its assigned values corresponding to the reference config- uration, while tsimultaneously varies over some designated interval of time, the vector functiongives the spatial position xoccupied at any instant of time for every particle of the body. At a specific time, say at t=t1, the function defines the configuration 106 Continuum Mechanics for Engineers x1=(X;t1): (4.12) In particular, at t=0, Eq 4.10 defines the initial configuration which is often adopted as the reference configuration, and this results in the initial spatial coordinates being identical in value with the material coordinates, so that in this case x=(X;0) =X (4.13) at timet=0. If we focus attention on a specific particle XPhaving the material position vector XP, Eq 4.10 takes the form xP=(XP;t) (4.14) and describes the path, or trajectory of that particle as a function of time. The velocityvP of the particle along its path is defined as the time rate of change of position, or vP=dxP dt=˙P=@ @t X=XP(4.15) where the notation in the last form indicates that the variable Xis held constant in taking the partial derivative of . Also, as is standard practice, the super-positioned dot has been introduced to denote differentiation with respect to time. In an obvious generalization we may define the velocity field of the total body as the derivative v=˙x=dx dt=@(X;t) @t=@x(X;t) @t: (4.16) Similarly, the acceleration field is given by a=˙v=¨x=d2x dt2=@2(X;t) @t2; (4.17) and the acceleration of any particular particle determined by substituting its material coordinates 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 in space 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 configuration remain arbitrarily close in all other configurations. For these reasons, the function in Eq 4.10 must be single-valued and continuous, and must possess continuous derivatives with respect to space and time to whatever order is required; usually to the second or third. Moreover, we require the inverse function -1in the equation X=-1(x;t) (4.18) to be endowed with the same properties as . Conceptually, Eq 4.18 allows us to “reverse” the motion and trace backwards to discover where the particle, now at x, was located in the reference configuration. The mathematical condition that guarantees the existence of such an inverse function is the non-vanishing of the Jacobian determinant J. That is, for the equation J=@i @XA6=0 (4.19) to be valid. This determinant may also be written as J=@xi @XA: (4.20) Kinematics of Deformation and Motion 107 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 when t=2s: 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 tgivesx1-2x2= -3as well asx3=1so that the particle under consideration moves on a straight line path in the plane x3=1. (b) By Eqs 4.16 and 4.17 the velocity and acceleration elds are given in component form, respectively, by v1=2tX2 a1=2X2 v2=2tX1 anda2=2X1 v3=0 a3=0; so that for the particle X= (1;2;1 )att=2 v1=8 a1=4 v2=4 anda2=2 v3=0 a3=0: Example 4.2 Invert the motion equations of Example 4.1 to obtain X=-1(x;t)and de- termine the velocity and acceleration components of the particle at x= (1;0;1 ) whent=2s. Solution By inverting the motion equations directly we obtain X1=x1-t2x2 1-t4;X2=x2-t2x1 1-t4;X3=x3 108 Continuum Mechanics for Engineers 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 )whent=2s v1=16 15a1=8 15 v2= -4 15and a2= -2 15 v3=0 a3=0: Example 4.3 Determine the Jacobian Jfor the motion of a continuum be given by x1=X1+f(t)coskX3; x2=X2+f(t)sinkX3; x3=X3: Solution Form and compute the determinant of Eq 4.20 detjxi;Aj=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 sinkX3 0 1 -f(t)k coskX3 0 0 1=1: 4.3 Langrangian and Eulerian Descriptions If a physical property of the body Bsuch as its density , or a kinematic property of its motion such as the velocity v, is expressed in terms of the material coordinates X, and Kinematics of Deformation and Motion 109 the timet, we say that property is given by the referential ormaterial description. When the referential configuration is taken as the actual configuration at time t=0, this description is usually called the Lagrangian description. Thus, the equations =(XA;t) or=(X;t); (4.21a) and vi=vi(XA;t) orv=v(X;t) (4.21b) chronicle a time history of these properties for each particle of the body. In contrast, if the properties andvare given as functions of the spatial coordinates xand timet, we say 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 to express the same properties in the spatial description. Accordingly, we write =(X;t)= -1(x;t);t =˜(x;t) (4.22a) and v=v(X;t)=v -1(x;t);t =˜v(x;t) (4.22b) where the tilde is added solely for the purpose of emphasizing that different functional forms 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 at specific points in space. Example 4.4 Let the motion equations be given in component form by the Lagrangian de- scription x1=X1et+X3(et-1); x2=X2+X3(et-e-t); x3=X3: Determine the Eulerian description of this motion. Solution Notice rst that for the given motion x1=X1,x2=X2andx3=X3att=0, so that the initial conguration has been taken as the reference conguration. Because of the simplicity of these Lagrangian equations of the motion, we may substitutex3forX3into the rst two equations and solve these directly to obtain the inverse equations X1=x1e-t+x3(e-t-1); X2=x2+x3(e-t-et); X3=x3: 110 Continuum Mechanics for Engineers Example 4.5 For the motion of Example 4.4 determine the velocity and acceleration elds, and express these in both Lagrangian and Eulerian forms. Solution From the given motion equations and the velocity denition Eq 4.16 we obtain the Lagrangian velocity components, v1=X1et+X3et; 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 Example 4.4 we obtain the velocity and acceleration equations in Eulerian form, v1=x1+x3 a1=x1+x3 v2=x3(et+e-t) anda2=x3(et-e-t) v3=0 a3=0: 4.4 The Displacement Field As may be seen from Fig. 4.1, the typical particle of body Bundergoes a displacement u=x-X (4.23) in the transition from the reference configuration to the current configuration. Because this relationship holds for all particles it is often useful to analyze deformation or motion in terms of the displacement field of the body. We may write the displacement vector uin component form by either of the equivalent expressions u=ui^ei=uA^IA: (4.24) Additionally with regard to the material and spatial descriptions we may interpret Eq 4.23 in either the material form u(X;t) =x(X;t) -X; (4.25) or the spatial form u(x;t) =x-X(x;t): (4.26) Kinematics of Deformation and Motion 111 In the first of this pair of equations we are describing the displacement that will occur to the particle that starts at X, and in the second equation we present the displacement that the particle now at xhas undergone. Recalling that since the material coordinates relate to positions in the reference configuration only, and hence are independent of time, we may 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 eld for the motion of Example 4.4 in both material and spatial descriptions. Solution From the motion equations of Example 4.4, namely, x1=X1et+X3(et-1); x2=X2+X3(et-e-t); x3=X3; we may compute the displacement eld 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 eld in compo- nent form u1= (x1+x3)(1-e-t); u2=x3(et-e-t); u3=0: 4.5 The Material Derivative In this section let us consider any physical or kinematic property of a continuum body. It may be a scalar, vector or tensor property, and so we represent it by the general symbol Pij:::with the understanding that it may be expressed in either the material description Pij:::=Pij:::(X;t); (4.28a) 112 Continuum Mechanics for Engineers or in the spatial description Pij:::=Pij:::(x;t); (4.28b) The material derivative of any such property is the time rate of change of that property for a specific collection of particles (one or more) of the continuum body. This derivative can be thought of as the rate at which Pij:::changes when measured by an observer attached to, and traveling with, the particle, or group of particles. We use the differential operator d=dt , or the superpositioned dot, to denote a material derivative, and note that velocity and acceleration as we have previously defined them are material derivatives. WhenPij:::is given in the material description of Eq 4.28a, the material derivative is simply 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 Xare labels and do not change with time. If, however, Pij::: is given in the spatial form of Eq 4.28b we recognize that the specific collection of particles of interest will be changing position in space, and we must use 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 the property at position x, known as the local rate of change; the second term results from the particles changing position in space and is referred to as the convective rate of change. Since by Eq 4.16 the velocity is defined as v=dx=dt(orvk=dxk=dt), Eq 4.30 may be written as 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 spatial description d dt=@ @t+vk@ @xkord dt=@ @t+v
r: (4.32) The first form of Eq 4.32 is for rectangular Cartesian coordinates, while the second form is coordinate-free. The del operator (r) will always indicate partial derivatives with respect to 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=X2et; x3=X3+X2(e-t-1); and let the temperature eld of the body be given by the spatial description, =e-t(x1-2x2+3x3); Determine the velocity eld in spatial form, and using that, compute the mate- rial derivative d=dt of the temperature eld. Kinematics of Deformation and Motion 113 Solution Note again here that the initial conguration serves as the reference congura- tion so that Eq 4.13 is satised. When Eq 4.30 is used, the velocity components in material form are readily determined to be v1= -X1e-t; v2=X2et; 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 spatial components, v1= -x1; v2=x2; v3= -x2e-2t: Therefore, we may now calculate d=dt in spatial form using Eq 4.31, d 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 d 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 in its spatial form. The following example illustrates the point. Example 4.8 Verify the spatial velocity components determined in Example 4.7 by applying Eq 4.31 directly to the displacement components in spatial form for the motion in that example. Solution We may determine the displacement components in material form directly from the 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=-1(x;t)computed in Example 4.7, we obtain the spatial displacements u1=x1(1-et); u2=x2(1-e-t); u3=x2(e-2t-e-t): 114 Continuum Mechanics for Engineers Therefore, substituting uiforPij:::in Eq 4.31 yields vi=dui dt=@ui @t+vk@ui @xk; so that by dierentiating 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 on both sides of the equations. In general, this set of equations must be solved simultaneously. In this case the solution is quite easily obtained, yielding v1= -x1; v2=x2; v3= -x2e-2t to conrm 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 uid in the body. The cilia beat in a manner that propels the surrounding uid. The motion of the cilia is complex. A model for the cilia motion is to enclose the tips by a exible envelope that moves in a fashion similar to the tips of the cilia. This is reasonable given that in many biological processes uid motions are slow. Two possible descriptions of the envelope motion are1 Case 1: x1=X0 1+cos(x1-ct); x2=X0 2+sin(x1-ct); Case 2: x1=X0 1+cos X0 1-ct
; x2=X0 2+sin X0 1-ct
: Envelope Cilia Envelope containing the tips of the beating cilia. Herexis the position of the envelope and X0is an initial position of the particles. (a) Show that the both descriptions indicate that particles move in an elliptical path centered about X0 1;X0 2
. 1see T. J. Lardner and W. J. Shack (1972) Cilia transport, Bulletin of Mathematical Biophysics ,34(3) 325-335. Kinematics of Deformation and Motion 115 (b) Fluid mechanics has a no-slip condition that requires the velocity of the uid be the same as the velocity of the envelope. Determine the velocity of the particles in the two descriptions. Solution (a) For description 1, the particle paths are x1-X0 1
2 ()2+ x2-X0 2
2 ()2=sin2(x1-ct)+cos2(x1-ct)=1: For description 2, the particle paths are x1-X0 1
2 ()2+ x2-X0 2
2 ()2=sin2 X0 1-ct
+cos2 X0 1-ct
=1: These are both descriptions of ellipses centered at X0 1;X0 2
. (b) For description 1, the velocity v1is v1=dx1 dt=@ @t+v1@ @x1+v2@ @x2 X0 1+cos(x1-ct) =@ @t X0 1+cos(x1-ct) +v1@ @x1 X0 1+cos(x1-ct) =csin(x1-ct)-v1sin(x1-ct); and v1[1+sin(x1-ct)]=csin(x1-ct); v1=csin(x1-ct) 1+sin(x1-ct): The velocity v2is v2=dx2 dt=@ @t+v1@ @x1+v2@ @x2 X0 2+sin(x1-ct) =@ @t X0 2+sin(x1-ct) +v1@ @x1 X0 2+sin(x1-ct) = -c cos(x1-ct)+v1cos(x1-ct) = -c cos(x1-ct)+cos(x1-ct)csin(x1-ct) 1+sin(x1-ct); or v2=-c cos(x1-ct) 1+sin(x1-ct): For description 2, the velocity v1is v1=dx1 dt=@ @t+v1@ @x1+v2@ @x2 X0 1+cos X0 1-ct
 =@ @t X0 1+cos X0 1-ct
 =csin X0 1-ct
: 116 Continuum Mechanics for Engineers Reference Current ^e1,^I1 ^e2,^I2^e3,^I3x3, X3 x2, X2 x1, X1OP pX xu+dudXA dxiQ qu FIGURE 4.2 VectordXA, between points PandQin reference configuration, becomes dxi, between pointspandq, in the current configuration. Displacement vector uis the vector between pointspandP. The velocity v2is v2=dx2 dt=@ @t+v1@ @x1+v2@ @x2 X0 2+sin X0 1-ct
 =@ @t X0 2+sin X0 1-ct
 = -ccos X0 1-ct
: 4.6 Deformation Gradients, Finite Strain Tensors In deformation analysis we confine our attention to two stationary configurations and disregard any consideration for the particular sequence by which the final deformed con- figuration is reached from the initial undeformed configuration . Accordingly, the mapping function is not dependent upon time as a variable, so that Eq 4.10 takes the form xi=i(X) orx=(X): (4.33) Consider, therefore, two neighboring particles of the body situated at the points PandQ in the undeformed configuration such that Qis located with respect to Pby the relative differential position vector dX=dXA^IA (4.34) as shown in Fig. 4.2. The magnitude squared of dXis (dX)2=dX
dX=dXAdXA: (4.35) Kinematics of Deformation and Motion 117 Under the displacement field prescribed by the function of Eq 4.35 the particles origi- nally atPandQmove to the positions pandq, respectively, in the deformed configura- tion such that their relative position vector is now dx=dxi^ei (4.36) having a magnitude squared (dx)2=dx
dx=dxidxi: (4.37) We assume the mapping function iof Eq 4.33 is continuous so that dxi=@i @XAdXA; (4.38) or as it is more often written, dxi=@i @XAdXA=xi;AdXA (4.39) where xi;AFiA (4.40) is called the deformation gradient tensor or simply the deformation gradient. The tensor F characterizes the local deformation at X, and may depend explicitly upon X, in which case the deformation is termed inhomogeneous. If Fis independent of X, the deformation is called homogeneous. In symbolic notation Eq 4.39 appears in either of the forms dx=F
dX ordx=FdX (4.41) where, as indicated by the second equation, the dot is often omitted for convenience. In view of the smoothness conditions we have imposed on the mapping function we know thatFis invertible so that the inverse F-1exists such that dXA=XA;idxi ordX=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 deformation in the magnitude squared of the distance between the particles originally at PandQ. Namely, (dx)2- (dX)2=dxidxi-dXAdXA which from Eq 4.39 and the substitution property of the Kronecker delta, ABmay be developed as follows, (dx)2- (dX)2= (xi;AdXA)(xi;BdXB) -ABdXAdXB = (xi;Axi;B-AB)dXAdXB = (CAB-AB)dXAdXB (4.43) where the symmetric tensor CAB=xi;Axi;B orC=FT
F (4.44) is called the Green’s deformation tensor. From this we immediately define the Lagrangian finite strain tensor EABas 2EAB=CAB-AB or2E=C-I (4.45) 118 Continuum Mechanics for Engineers where the factor of two is introduced for convenience in later calculations. Finally we can write, (dx)2- (dX)2=2EABdXAdXB=dX
2E
dX: (4.46) The difference (dx)2- (dX)2may also be developed in terms of the spatial variables in a similar way as (dx)2- (dX)2=ijdxidxj- (XA;idxi)(XA;jdxj) = (ij-XA;iXA;j)dxidxj = (ij-cij)dxidxj (4.47) where the symmetric tensor cij=XA;iXA;j orc= (F-1)T
(F-1) (4.48) is called the Cauchy deformation tensor. From Eq 4.48 we define the Eulerian finite strain tensoreas 2eij= (ij-cij) or2e= (I-c); (4.49) so that now (dx)2- (dX)2=2eijdxidxj=dx
2e
dx: (4.50) BothEABandeijare, of course, symmetric second-order tensors, as can be observed from their definitions. For any two arbitrary differential vectors dX(1)anddX(2)which deform into dx(1)and dx(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) IfEis identically zero (no strain), Eq 4.51 asserts that the lengths of all line elements are unchanged [we may choose dX(1)=dX(2)=dXso that (dx)2= (dX)2], and in view of the definition dx(1)
dx(2)=dx(1)dx(2)cos, the angle between any two elements will also be unchanged. Thus in the absence of strain, only a rigid body displacement can occur. 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 begin by writing Eq 4.25 in its time-independent form consistent with deformation analysis. In component 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-AB= (ui;A+iA)(ui;B+iB) -AB (4.54) which reduces to 2EAB=uA;B+uB;A+ui;Aui;B; (4.55) Kinematics of Deformation and Motion 119 and from the second, Eq 4.49 becomes 2eij=ij-XA;iXA;j=ij- (Ai-uA;i)(Aj-uA;j) (4.56) which reduces to 2eij=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, where kis a constant, be applied to the small cube of edge dimensions dLshown in the gure below. Draw the deformed shape of face ABGH of the cube and determine the dierence (dx)2- (dX)2for the diagonals AG,BHandOGof the cube. x1, X1 x2, X2x3, X3 O A BGH dL dLdL Reference position of cube undergoing simple shear. x3, X3 x2, X2ab(dL,dL,kdL)g(dL,(1+k)dL,(1+k)dL) h(dL,kdL,dL)Small cube geometry. Solution From the mapping equations directly, the origin Ois seen to remain in place, and the particles originally at points A,B,GandHare displaced to the points a(dL;O;O ),b(dL;dL;kdL ),g(dL; (1+k)dL; (1+k)dL) andh(dL;kdL;dL), re- spectively, so that particles in planes parallel to the X2X3remain 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 the deformation gradient Fhas the matrix form [FiA] =2 41 0 0 0 1 k 0 k 13 5; and sinceC=FT
F [CAB] =2 41 0 0 0 1 +k22k 0 2k 1 +k23 5 from which we determine 2E=C-I, [2EAB] =2 40 0 0 0 k22k 0 2k k23 5: 120 Continuum Mechanics for Engineers In general, (dx)2- (dX)2=dX
2E
dXso that for diagonal AG, (dx)2- (dX)2= [0;dL;dL ]2 640 0 0 0 k22k 0 2k k23 752 640 dL dL3 75 =2(2k+k2)(dL)2: For diagonal BH, (dx)2- (dX)2= [0; -dL;dL]2 640 0 0 0 k22k 0 2k k23 752 640 -dL dL3 75 =2(-2k +k2)(dL)2; and for diagonal OG, (dx)2- (dX)2= [dL;dL;dL ]2 640 0 0 0 k22k 0 2k k23 752 64dL dL dL3 75 =2(2k+k2)(dL)2; Note: All of these results may be calculated directly from the geometry of the deformed cube for this simple deformation. 4.7 Infinitesimal Deformation Theory If the numerical values of all the components of the displacement and the displacement gradient 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) and 2eij=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+kA @ui @xkkA (4.60) where we have used the relationship @xk @XA=@uk @XA+kA Kinematics of Deformation and Motion 121 obtained by differentiating Eq 4.52. Therefore, to the first order of approximation for the case of small displacement gradients, it is unimportant whether we differentiate the displacement components with respect to the material or spatial coordinates. In view of this, we may display the equivalent relative displacement gradients for small deformation theory as either ui;Aorui;j. Similarly, it can be shown that in the linear theory uA;Band uA;jare equivalent. It follows that to the same order of approximation, from Eqs 4.58 and 4.59, EABeijiAjB; (4.61) and it is customary to define a single infinitesimal strain tensor for which we introduce the symbolijas 2ij=@ui @XAAj+@uj @XBBi=@ui @xj+@uj @xi=ui;j+uj;i: (4.62) Because the strain tensors EAB,eij, andijare all symmetric, second-order tensors, the entire development for principal strains, strain invariants and principal strain directions may be carried out exactly as was done for the stress tensor in Chapter 3. Thus, taking, ijas the typical tensor of the group, we summarize these results by displaying its matrix relative to principal axes in the alternative forms, [ ij] =2 410 0 0 20 0 0 33 5=2 4I0 0 0 II0 0 0 III3 5 (4.63) together with the strain invariants I=ii=tr=I+II+III; (4.64a) II=1 2(iijj-ijij)=III+IIIII+IIII; (4.64b) III="ijk1i2j3k=IIIIII: (4.64c) The components of 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=2ijdXidXj=dX
2
dX (4.65) which, upon factoring the left hand side and dividing by (dX)2, becomes dx-dX dX
dx+dX dX=2ijdXi dXdXj dX: ButdXi=dX=Ni, a unit vector in the direction of dX, and for small deformations we may assume (dx+dX)=dX2, so that dx-dX dX=ijNiNj=^N

^N: (4.66) The scalar ratio on the left-hand side of this equation is clearly the change in length per unit original length for the element in the direction of ^N. It is known as the longitudinal strain, or the normal strain and we denote it by e(^N). If, for example, ^Nis taken in the X1 direction so that ^N=^I1, then e(^I1)=^I1

^I1=11: 122 Continuum Mechanics for Engineers dX(1)dX(2)dx(1)dx(2) P ABb pa θ π/2DEFORMS FIGURE 4.3 The right angle between line segments APandBPin the reference configuration becomes , the angle between segments apandbp, in the deformed configuration. Likewise, for ^N=^I2, or^N=^I3the normal strains are found to be 22and33, re- spectively. Thus the diagonal elements of the small (infinitesimal) strain tensor represent normal 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)anddX(2)at positionP which are deformed into vectors dx(1)anddx(2), respectively. In this case, Eq 4.51 may be written, dx(1)
dx(2)=dX(1)
dX(2)+dX(1)
2
dX(2)(4.67) which, if we choose dX(1)anddX(2)perpendicular to one another, reduces to dx(1)
dx(2)=dx(1)dx(2)cos=dX(1)
2
dX(2)(4.68) whereis the angle between the deformed vectors as shown in Fig. 4.3. If we let, = 2- , the angle measures the small change in the original right angle between dX(1)and dX(2). Also cos=cos 2-  =sin  since is very small for infinitesimal deformations. Therefore, assuming as before that dx(1)dX(1)anddx(2)dX(2)because of small deformations cos=dX(1) dx(1)
2
dX(2) dx(2)^N(1)
2
^N(2): (4.69) Here, if we take ^N(1)=^I1and^N(2)=^I2and designate the angle as 12, we obtain 12=2[1;0;0 ]2 4111213 122223 1323333 52 40 1 03 5=212; (4.70) so that by choosing the undeformed vector pairs in Eq 4.68 in coordinate directions we may generalize Eq 4.70 to obtain Kinematics of Deformation and Motion 123 ij=2ij (i6=j): (4.71) This establishes the relationship between the off-diagonal components of ijand the so- called engineering shear strain components ij, which represent the changes in the original right angles between the coordinate axes in the undeformed configuration. Note that sincecan be defined with respect to any set of Cartesian axes at P, this result holds for any pair of perpendicular vectors at that point. In engineering texts, the infinitesimal strain tensor is frequently written in matrix form as [ij] =2 664111 2 121 2 13 1 2 12221 2 23 1 2 131 2 23333 775: (4.72) If^N(1)and^N(2)are chosen in principal strain directions, Eq 4.69 becomes =^N(1)
2
^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 parallelpiped of undeformed edge dimensions dX(1),dX(2)anddX(3)taken in the principal strain di- rections will be deformed into another rectangular parallelpiped having edge lengths dx(i)= (1+(i))dX(i);(i=1;2;3 ) (4.74) as shown in Fig. 4.4, where (i)are the normal strains in principal directions. The change in volume per unit original volume of the parallelpiped is
V V=[1+(1)]dX(1)[1+(2)]dX(2)[1+(3)]dX(3)-dX(1)dX(2)dX(3) dX(1)dX(2)dX(3) (1)+(2)+(3) (4.75) where terms involving products of the principal strains have been neglected. The ratio
V=V, being the first invariant of , is called the cubical dilatation. We shall denote it by the symbole, and write e=
V=V =ii=I: (4.76) Becauseis a symmetrical second-order tensor the development of Mohr’s circles for small strain, as well as the decomposition of into its spherical and deviator component tensors 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) has units of1 2 as shown by the typical diagram of Fig. 4.5. The infinitesimal spherical strain tensor is represented by a diagonal matrix having equal elements denoted by M=1 3ii= 1 3e, known as the mean normal strain. The infinitesimal deviator strain tensor, is defined by ij=ij-1 3ijkk=ij-ijM; (4.77) and in matrix form 2 4111213 212223 3132333 5=2 411-M1213 1222-M23 132333-333 5: (4.78) 124 Continuum Mechanics for Engineers dX(1)dX(2) dX(3)X3� X1�X2�x1∗x3∗ x2∗dx(1) dx(3)dx(2) DEFORMS FIGURE 4.4 A rectangular parallelpiped with edge lengths dX(1),dX(2)anddX(3)in the reference configuration becomes a skewed parallelpiped with edge lengths dx(1),dx(2)anddx(3) in the deformed configuration. Note that as with its stress counterpart, the first invariant of the deviator strain is zero, or ii=0; (4.79) and the principal deviator strains are given by (q)=(q)-M; (q=1;2;3 ) (4.80) where(q)is a principal value of the infinitesimal strain tensor. A state of plane strain parallel to the X1X2plane exists at Pif 33= 13= 31= 23= 32=0 (4.81) at that point. Also, plane strain relative to the X1X2plane in the continuum body as a whole 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 OX0 1X0 2X0 3rotated aboutX3by the angle relative toOX1X2X3as shown by Fig. 4.6, the transformation equations 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 4.1 the results are 0 11=11+22 2+11-22 2cos2+ 12 2sin2; (4.82a) 0 22=11+22 2-11-22 2cos2- 12 2sin2; (4.82b) 0 12= -(11-22)sin2+ 12cos2: (4.82c) Kinematics of Deformation and Motion 125 γ 2 III II I FIGURE 4.5 Typical Mohr’s circle for strain. X3 , X3� X2 X1X2� X1�θθO FIGURE 4.6 Rotation of axes for plane strain. Also, the non-zero principal strain values for plane strain are given by (1);(2)=11+22 2s11-22 22 + 12 22 : (4.83) Because shear strains are very difficult to measure experimentally, the state of strain at a 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 12. Example 4.11 A delta rosette has the shape of an equilateral triangle, and records longitudinal strains in the directions x1,x0 1andx00 1shown in the sketch. If the measured 126 Continuum Mechanics for Engineers TABLE 4.1 Transformation table for general plane strain. X1X2X3 X0 1 cos sin 0 X0 2 -sin cos 0 X0 30 0 1 strains in these directions are 11= -310-4,0 11=410-4, and00 11= 210-4where the units are m=m (dimensionless), determine 22, 12and0 22. Show that11+22=0 11+0 22as the rst strain invariant requires. 60º 60º 60º30º 30º x1x2′′x1′x1 Solution We need only Eq 4.82 here, which we write for x0 1andx00 1in turn (omitting the common factor 10-4for convenience). Thus, for =60, and=120, respectively, we have 4=-3+22 2+-3-22 2 -1 2 + 12 2p 3 2; 2=-3+22 2+-3-22 2 -1 2 - 12 2p 3 2: Adding these two equations to eliminate 12we determine 22=5; subtracting the second from the rst to eliminate 22we determine 12=4=p 3. Next, using=150we determine 0 22from Eq 4.82 0 22=-3+5 2+-3-5 2+2p 3 -p 3 2! = -2; and by the rst invariant of the small strain tensor we check that 11+22= -3+5=0 11+0 22=4-2=2: Kinematics of Deformation and Motion 127 Consider once more the two neighboring particles which were at positions PandQ in the undeformed configuration, and are now at positions pandq, respectively, in the deformed configuration (see Fig. 4.2). In general, an arbitrary displacement will include both deformation (strain) and rigid body displacements. To separate these we consider the differential displacement vector du. Assuming conditions on the displacement field that guarantee the existence of a derivative, the displacement differential duiis written dui=@ui @Xj PdXj (4.84) where the derivative is evaluated at Pas indicated by the notation. From this we may define the unit relative displacement of the particle at Qwith respect to the one at Pby the equation dui dX=dui dXjdXj dX=dui dXjNj (4.85) whereNjis the unit vector in the direction from PtowardQ. By decomposing the displacement 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 = (ij+!ij)dXj (4.86) in whichijis recognized as the infinitesimal strain tensor, and !ijis called the infinitesimal rotation tensor. Ifijhappens to be identically zero, there is no strain, and the displacement is a rigid body displacement. For this case we define the rotation vector !i=1 2"ijk!kj (4.87) which may be readily inverted since !kj= -!jkto yield !ij="kji!k: (4.88) Therefore, Eq 4.86 with ij0becomes dui="kjl!kdXj="ikj!kdXj ordu=!dX; (4.89) so that the relative differential displacement is seen to be the result of a rigid body rotation about the axis of the rotation vector !. On the other hand, if !ij0, the relative displacement will be the result of pure strain. A comment is appropriate about the notation !ijfor the skew–symmetric part of the displacement gradient in the linear theory. It might have been convenient to denote this quantity asW=wij^ei ^ej; however, this notation has been reserved to represent the spin tensor in Section 4.11. The spin tensor is the skew–symmetric part of the velocity gradient. It is a bit awkward to have the tensor !ijand its axial vector !jdenoted with !, but this is not as bad as having the same symbol representing the infinitesimal rotation and spin tensors. 128 Continuum Mechanics for Engineers 4.8 Compatibility Equations The compatibility equations merit a section because of the important role played in linear elasticity problems. Very often, solving a linear elasticity problem (see Chapter 6) involves solving the compatibility equations discussed below. If we consider the six independent strain-displacement relations, Eq 4.62 @ui @xj+@uj @xi=2ij as a system of partial differential equations for determining the three displacement com- ponentsui(assuming the ijare 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 six partial differential equations. Therefore, some restrictive conditions must be imposed upon the strain components (actually upon derivatives of the strain components) if the equations above are to be satisfied by a single-valued displacement field. Such conditions are expressed by the strain compatibility equations ij;km +km;ij -ik;jm -jm;ik =0: (4.90) There are34=81equations in all (four free indices) in Eq 4.90 but only six of these are distinct 11;23 +23;11 -12;13 -13;12 =0; 22;31 +31;22 -23;21 -21;23 =0; 33;12 +12;33 -31;32 -32;31 =0; 212;12 -11;22 -22;11 =0; 223;23 -22;33 -33;22 =0; 231;31 -33;11 -11;33 =0:(4.91) For plane strain in the x1-x2plane, the six unique equations in Eq 4.90 reduce to a single equation, 11;22 +22;11 =212;12; (4.92) which may be easily verified as a necessary condition by a simple differentiation of Eq 4.62 for a range of two on the indices iandj. It may be shown that the compatibility equations, either Eq 4.90 or Eq 4.91, are both necessary and sufficient for a single-valued displacement field of a body occupying a simply connected domain. The compatibility equations seem daunting, but it is not too difficult to demonstrate the necessity and sufficiency. For the necessity, given the strain ij=1 2(ui;j+uj;i)with continuously differentiable displacements we write that curl curl=0: (4.93) This equation is a more compact way of writing Eqs 4.90 and 4.91. By the definition of linear strain and the continuity of the displacements, we see "ijk"lmnjm;kn ="ijk"lmn(uj;mkn +um;jkn )=0 where the first displacement term, symmetric in mn, cancels with skew–symmetric "lmn, and the second displacement term, symmetric in jk, cancels with skew–symmetric "ijk. Kinematics of Deformation and Motion 129 As with many proofs, sufficiency is more difficult to demonstrate. Here it must be shown that starting from curl curl =0that the linear strain ij=1 2(ui;j+uj;i)can be constructed. To start, let A=curl (4.94) from which compatibility in the form of Eq 4.93 gives curlA=0: (4.95) Sinceis 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 a skew–symmetric tensor !such that2 A= - curl!: (4.97) Equating Eqs 4.97 and 4.94 on Agives curl(+!)=0 (4.98) from which3 +!=ru: (4.99) Taking the symmetric part of this equation gives us the strain =1 2 ru+ruT
; (4.100a) or ij=1 2(ui;j+uj;i): (4.100b) The compatibility equations give a convenient way to compute the displacements from the strain components. A line integral form of the compatibility equations gives ui() =Z 0 ij(^) + k-^kh ij;k(^) -jk;i(^)i d^j +uj(0) +wij(0) j-0 j :(4.101) The last two terms represent an infinitesimal rigid motion and are often omitted. Example 4.12 For a given state of strain, ij, the compatibility equations are necessary and sucient conditions for unique displacements. (a) The compatibility condition may be written as "kmn"pqrmq;nr =0 where"pqris the alternating tensor. Show that the compatibility equation in this form is symmetric in kandp. This means that there are only six independent conditions. 2This is not a difficult step, however, it is best to refer to ME Gurtin, ”The Linear Theory of Elasticty”, in Mechanics of Solids, Volume II, Editor C Truesdell, Springer-Verlag, c. 1973, pp17-18 and p 40. 3ibid, p 17 130 Continuum Mechanics for Engineers (b) For displacements to be path independent the integrand of Eq 4.101 must be exact. If Uij=ij(^) + k-^kh ij;k(^) -jk;i(^)i ; a necessary and sucient condition for the integral to be exact is Uij;m+Uim;j=0: Show this condition leads to ij;km +km;ij -jk;im -im;jk =0: (c) If the displacement eld has three continuous derivatives, show that "kmn"pqrmq;nr =0. Solution (a) This can be shown through index manipulation. Noting strain is symmetric and twice continuously dierentiable and m,n,qandrare dummy indices: "kmn"pqrmq;nr ="kmn"pqrqm;rn ="kqr"pmnmq;nr: This expression is a symmetric, second-order tensor having six independent terms. (b) Dierentiation and substitution leads to ij;m -km(ij;k-jk;i)-im;j+kj(im;k-mk;i) + k-^k [ij;km -jk;im -im;kj +mk;ij]=0: Since strain is symmetric and and^kcan be chosen arbitrarily ij;km +km;ij -jk;im -im;jk =0: (c) Substitute the strain-displacement relations into the compatibility equation "kmn"pqrmq;nr =1 2"kmn"pqr(um;qnr +uq;mnr ): The alternating tensors are skew-symmetric in mnandqr. Additionally, the displacement is assumed to be continuously dierentiable and so order of dierentiation is immaterial. Thus, the parenthetical term is symmetric inmnandqr. The product of skew-symmetric and symmetric tensors is zero. So, "kmn"pqrmq;nr =1 2"kmn"pqr(um;qnr +uq;mnr )=0: Kinematics of Deformation and Motion 131 4.9 Stretch Ratios Referring again to Fig. 4.2, define the ratio of the magnitudes of dxanddXto be the stretch ratio, (or simply the stretch). In particular, for the differential element in the direction of the unit vector ^NatP, we write (^N)=dx dX(4.102) wheredxis the deformed magnitude of dX=dX^N. As a matter of convenience we often prefer to work with stretch-squared values, 2 (^N)=dx dX2 : (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, 2 (^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, (^n)in the direction of ^n=dx=dx at pby the equation, 1 (^n)=dX dx: (4.106) Here, recalling from Eq 4.42 that dX=F-1
dxand 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)2becomes 1 2 (^n)=dx dx
c
dx dx=^n
c
^n: (4.108) In general,(^N)6=(^n). However if ^nis a unit vector in the direction that ^Nassumes in the deformed configuration, the two stretches are the same. For ^N=^I1, 2 (^I1)=^I1
C
^I1=C11=1+2E11; (4.109) and for ^n=^e1, 1 2 (^e1)=^e1
c
^e1=c11=1-2e11 (4.110) with analogous expressions for ^Nand^nin the other coordinate directions. 132 Continuum Mechanics for Engineers Consider next the unit extension (longitudinal strain) in any direction ^NatP. This may be expressed in terms of the stretch as e(^N)=dx-dX dX=(^N)-1=p ^N
C
^N-1: (4.111) Notice that the unit extension is zero when the stretch is unity, as occurs with a rigid body displacement. If ^N=^I1, e(^I1)=q ^I1
C
^I1-1=p C11-1=p 1+2E11-1; (4.112) or, solving for E11, E11=e(^I1)+1 2e2 (^I1): (4.113) For small deformation theory where E11!11, and for which e2 (^N)may be neglected in comparison to e(^N), the above equation asserts that E11=11=e(^I1). The change in angle between any two line elements may also be given in terms of stretch. LetdX(1)anddX(2)be arbitrary vectors which become dx(1)anddx(2), respec- tively, during a deformation. By the dot product, dx(1)
dx(2)=dx(1)dx(2)cos, we may compute the angle betweendx(1)anddx(2)from its cosine, which, with the help of Eq 4.104, takes the form, cos=dx(1) dx(1)
dx(2) dx(2)=F
dX(1) p dX(1)
C
dX(1)
F
dX(2) p dX(2)
C
dX(2); or upon dividing the numerator and denominator by the scalar product dX(1)dX(2)and making use of Eqs 4.105 and 4.44 we obtain cos=^N1
C
^N2 (^N1)(^N2): (4.114) Thus, for elements originally in the ^I1and^I2directions, the angle between them in the deformed configuration may be determined from cos12=C12 (^I1)(^I2)=C12pC11C22: (4.115) In a similar fashion, from dX(1)
dX(2)=dX(1)dX(2)cos, whereis the angle between dX(1)anddX(2), we obtain from Eqs 4.107 and 4.108, cos=dX(1) dX(1)
dX(2) dX(2)=^n1
c
^n2p^n1
c
^n1p^n2
c
^n2 =(^n1)(^n2)(^n1
c
^n2) which gives the original angle between elements in the directions ^n1and^n2of the current configuration. Example 4.13 A homogeneous deformation is given by the mapping equations, x1=X1-X2+ X3,x2=X2-X3+X1andx3=X3-X1+X2. Determine Kinematics of Deformation and Motion 133 (a) the stretch ratio in the direction of ^N1= (^I1+^I2)=p 2, and (b) the angle 12in the deformed conguration between elements that were originally in the directions of ^N1, and ^N2=^I2. Solution For the given deformation (as the student should verify), [FiA] =2 41-1 1 1 1 -1 -1 1 13 5and [CAB] =2 43-1-1 -1 3 -1 -1-1 33 5: (a) Therefore, from Eq 4.105, 2 (^N1)=h 1p 21p 20i2 66643-1-1 -1 3 -1 -1-1 33 77752 66641p 2 1p 2 03 7775=2; and (^N1)=p 2: (b) For ^N2=^I2,2 (^I2)=^I2
C
^I2=C22=3so that from Eq 4.114, using the result in part (a), cos12=(^I1+^I2)=p 2
C
^I2p 2p 3=2=p 2p 6; and12=54:7. Thus, the original 45angle is enlarged by 9:7. It is evident from Eq 4.115 that if the coordinate axes are chosen in the principal di- rections ofCthe deformed angle 12is a right angle (C12=0in this case) and there has been no change in the angle between elements in the X1andX2directions. By the same argument, any three mutually perpendicular principal axes ofCatPare deformed into three mutually perpendicular axes at p. Consider, therefore, the volume element of a rectangular parallelepiped whose edges are in the principal directions of C(and thus also ofE). Since there is no shear strain between any two of these edges the new volume is still a rectangular parallelopiped, and in the edge directions ^Ni(i=1;2;3 )the unit strains are e(^Ni)=(^Ni)-1 (i=1;2;3 ); (4.116) so that dx(i)=dX(i)+dX(i)[(^Ni)-1] =dX(i)(^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)=(^N1)(^N2)(^N3)(4.118a) which, when Eq 4.105 is used, becomes, 134 Continuum Mechanics for Engineers dV dV0=q C(1)C(2)C(3)=p IIIC: (4.118b) The importance of the second form of Eq 4.118b is that it is an invariant expression and can be calculated without reference to the principal axes of C. Example 4.14 Determine the volume ratio dV=dV0for the deformation of Example 4.13 using Eq 4.118a, and verify using Eq 4.118b. Solution As the student should show, a set of principal axes for the Ctensor of Ex- ample 4.13 are ^N1= (^I1+^I2+^I3)=p 3,^N2= (^I1-^I2)=p 2, and ^N3= (^I1+^I2-2^I3)=p 6. Thus from Eq 4.105 the principal stretches are (^N1)=1, (^N2)=2, and(^N3)=2, respectively. Using these results, Eq 4.118a gives dV=dV0=4. By Eq 4.118b, IIIC=detC=3-1-1 -1 3 -1 -1-1 3=16: anddV=dV0=p 16=4. 4.10 Rotation Tensor, Stretch Tensors In 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 a multiplicative decomposition known as the polar decomposition by which any non-singular tensor can be decomposed into a product of two component tensors. Recall that the de- formation gradient Fis 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) whereRis the orthogonal rotation tensor, and UandVare symmetric, positive-definite tensors called the right and left stretch tensors, respectively. Moreover, UandVhave the same eigenvalues (see Problem 4.32). The deformation gradient can be thought of as a mapping of the infinitesimal vector dX of the reference configuration into the infinitesimal vector dxof the current configuration. Note that the first decomposition in Eq 4.119 replaces the linear transformation dx=F
dX of Eq 4.41 by two sequential transformations, Kinematics of Deformation and Motion 135 dx0=U
dX (4.120a) followed by dx=R
dx0: (4.120b) The tensorUhas three positive eigenvalues, U(1),U(2)andU(3)called the principal stretches, and associated with each is a principal stretch direction, ^N1,^N2and^N3, respec- tively. These unit vectors form an orthogonal triad known as the right principal directions of stretch. 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 decomposition of Eq 4.120b reverses the sequence; first a rotation by R, then the stretching by V. In a general deformation, a rigid body translation may also be involved as well as the rotation and stretching described here. As a preliminary to determining the rotation and stretch tensors, we note that an ar- bitrary tensor Tis positive definite if v
T
v>0 for all vectors v6=0. A necessary and sufficient condition for Tto be positive definite is for all its eigenvalues to be positive. In this regard, consider the tensor C=FT
F. Inasmuch as Fis non-singular (detF6=0)and F
v6=0ifv6=0, we see that (F
v)
(F
v)is a sum of squares and hence greater than zero. Thus (F
v)
(F
v) =v
FT
F
v=v
C
v>0; (4.121) andCis positive definite. Furthermore, (FT
F)T=FT
(FT)T=FT
F (4.122) which proves that Cis also symmetric. By the same arguments we may show that c= (F-1)T
(F-1)is also symmetric and positive definite. Now letCbe given in principal axes form by the matrix [C AB] =2 4C(1)0 0 0 C (2)0 0 0 C (3)3 5 (4.123) and let [aMN]be the orthogonal transformation that relates the components of Cto the components of C in any other set of axes through the equation expressed here in both indicial and matrix form C AB=aAQaBPCQP orC=ACAT: (4.124) We defineUas the square root of C, that is,U=p CorU
U=C, and since the principal valuesC(i);(i=1;2;3 )are all positive we may write hp C ABi =2 4pC(1)0 0 0pC(2)0 0 0pC(3)3 5= [U AB]; (4.125) and, as is obvious, the inverse (U)-1by  (U AB)-1 =2 41=pC(1)0 0 0 1=pC(2)0 0 0 1=pC(3)3 5: (4.126) 136 Continuum Mechanics for Engineers Note that both UandU-1are symmetric positive definite tensors given by UAB=aQAaPBU QP orU=ATUA; (4.127) and U-1 AB=aQAaPB(U QP)-1orU-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 Ris proper orthogonal. The second decomposition in Eq 4.119 may be confirmed by a similar development usingC-1=F
FT=V2. Example 4.15 A homogeneous deformation is given by the equations; x1=2X1-2X2,x2= X1+X2andx3=X3. Determine the polar decomposition F=R
Ufor this deformation. Solution The matrix form of the tensor FiA=xi;Ais easily determined to be, [FiA] =2 42 -2 0 1 1 0 0 0 13 5 from which we calculate C=FT
F, [CAB] =2 45 -3 0 -3 5 0 0 0 13 5: In principal axes form this matrix becomes [C AB] =2 48 0 0 0 2 0 0 0 13 5 with an orthogonal transformation matrix found to be [aMN] =2 66641p 2-1p 20 1p 21p 20 0 0 13 7775: Kinematics of Deformation and Motion 137 This is found by calculating the eigenvectors of C. Therefore, from Eqs 4.125 and 4.126 [U AB] =2 42p 2 0 0 0p 2 0 0 0 13 5and [(U AB)-1] =1 2p 22 41 0 0 0 2 0 0 0 2p 23 5; and by use of the transformation equations Eqs 4.127 and 4.128 (as the student should verify), we determine [UAB] =2 43=p 2-p 2 0 -p 2 3p 2 0 0 0 13 5; and [U-1 AB] =1 4p 22 43 1 0 1 3 0 0 0 4p 23 5: Finally, from Eq 4.129, [RAB] =2 42 -2 0 1 1 0 0 0 13 52 43 1 0 1 3 0 0 0 4p 23 51 4p 2 =2 41=p 2 -1=p 2 0 1=p 2 1=p 2 0 0 0 13 5 It is readily conrmed using these results that F=RUand thatRT
R=I. 4.11 Velocity Gradient, Rate of Deformation, Vorticity Let 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 is written as Lij=dij+wij (4.133) where the symmetric portion dij=1 2@vi @xj+@vj @xi (4.134) 138 Continuum Mechanics for Engineers pq dx ivi+dv i vi O x2x3x1 FIGURE 4.7 Differential 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 velocity gradient components being small, and is valid for finite components @vi=@xj. Consider the velocity components at two neighboring points pandq. Let the particle currently atphave a velocity vi, and the particle at qa velocityvi+dvias shown in Fig. 4.7. Thus the particle at qhas a velocity relative to the particle at pof dvi=@vi @xjdxj ordv=L
dx: (4.136) Note that @vi @xj=@vi @XA@XA @xj=d dt@xi @XA@XA @xj; (4.137) or in symbolic notation L=˙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 =dx=dX whereis as defined in Eq 4.105, that is, the stretch of the line element dXinitially along ^Nand currently along ^n. By the definition of the deformation gradient, dxi=xi;AdXA, along with the unit vectors ni=dxi=dx and NA=dXA=dX we may write dxni=xi;AdXNA Kinematics of Deformation and Motion 139 which becomes (after dividing both sides by the scalar dX) ni=xi;ANA or ^n=F
^N: (4.140) If we take the material derivative of this equation (using the symbolic notation for conve- nience), ˙^n+^n˙=˙F
^N=L
F
^N=L
^n; so that ˙^n+^n˙==L
^n: (4.141) By forming the inner product of this equation with ^nwe obtain ^n
˙^n+^n
^n˙==^n
L
^n: But^n
^n=1and so ˙^n
^n=0, resulting in ˙==^n
L
^n or ˙==vi;jninj (4.142) which represents the rate of stretching per unit stretch of the element that originated in the direction of ^N, and is in the direction of ^nof the current configuration. Note further that Eq 4.142 may be simplified since Wis skew-symmetric which means that Lijninj= (dij+wij)ninj=dijninj; and so ˙==^n
D
^n or ˙==dijninj: (4.143) For example, for the element in the x1direction, ^n=^e1and ˙== 1 0 02 4d11d12d13 d12d22d23 d13d23d333 52 41 0 03 5=d11: Likewise, for ^n=^e2,˙==d22and for ^n=^e3,˙==d33. Thus the diagonal elements of the rate of deformation tensor represent rates of extension, or rates of stretching in the coordinate (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) ianddx(2) iatp. The material derivative of the inner product of these two vectors is (using the superposed dot to indicate differentiation with respect to time, ˙dx(1) idx(2) i=˙dx(1) idx(2) i+dx(1) i˙dx(2) i =dv(1) idx(2) i+dx(1) idv(2) i =vi;jdx(1) jdx(2) i+dx(1) ivi;jdx(2) j =(vi;j+vj;i)dx(1) idx(2) j =2dijdx(1) idx(2) j: (4.144) Butdx(1) idx(2) i=dx(1)dx(2)cos, and ˙dx(1)dx(2)cos=˙dx(1) idx(2) icos+dx(1) ˙dx(2)cos-dx(1)dx(2)˙sin = ˙dx(1) dx(1)+˙dx(2) dx(2) cos-˙sin! dx(1)dx(2): (4.145) 140 Continuum Mechanics for Engineers Equating Eqs 4.144 and 4.145 gives 2dijdx(1) idx(2) i= ˙dx(1) dx(1)+˙dx(2) dx(2) cos-˙sin! dx(1)dx(2): (4.146) Ifdx(1) i=dx(2) i=dxi, then=0, and cos=1, sin=0anddx(1)=dx(2)=dxso that Eq 4.146 reduces to dijdxi dxdxj dx=dijninj=˙dx dx(4.147) which is seen to be the rate of extension per unit length of the element currently in the direction ofni(compare with Eq 4.143). If, however, dx(1) iis perpendicular to dx(2) i, so that= 2, cos=0, sin=1, then Eq 4.146 becomes 2dijn(1) in(2) i=^n1
2D
^n2= - ˙: (4.148) This rate of decrease in the angle is a measure of the shear rate between the elements in the directions of ^n1and^n2. In the engineering literature it is customary to define the rate of shear ashalf the change (increase or decrease) between two material line elements instantaneously at right angles to one another. Thus for ^n1=^e1and^n2=^e2, -1 2˙12=^e1
D
^e2=d12; and, in general, the off-diagonal elements of the rate of deformation tensor are seen to represent shear rates for the three pairs of coordinate axes. BecauseDis 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 Dand the material derivative of the strain tensor E. Recall that 2E=C-I=FT
F-I; so that, using Eq 4.139 2˙E=˙FT
F+FT
˙F= (L
F)T
F+FT
(L
F) =FT
LT
F+FT
L
F=FT
(LT+L)
F=FT
(2D)
F; or ˙E=FT
D
F: (4.149) Note also that from ui+Xi=xiwe haveui;A+i;A=xi;Aand if the displacement gradientsui;Aare very small, ui;A1and may be neglected, then i;Axi;A(IF), and of course, FT=IT=I. At the same time for ui;Avery small in magnitude, by Eq 4.61,Eand Eq 4.74 reduces to ˙=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)2since˙(dX)2=0, we obtain ˙(dx)2=dX
2˙E
dX=dX
FT
2D
F
dX=dx
2D
dx (4.151) Kinematics of Deformation and Motion 141 which shows that the local motion at some point xis a rigid body motion if and only if D=0atx. Solving Eq 4.141 for ˙ niand using Eq 4.143 we may write ˙ni=vi;jnj-ni˙== (dij+wij)nj-dqknqnkni: If now,niis chosen along a principal direction of Dso thatdijn(p) j=D(p)n(p) i(p= 1;2;3 )whereD(p)represents a principal value of D, then ˙ni=D(p)˙n(p) i+wij˙n(p) j-D(p)n(p) qn(p) qn(p) i=wijn(p) j (4.152) sincen(p) qn(p) q=1. Because a unit vector can change only in direction, Eq 4.152 indicates thatwijgives the rate of change in direction of the principal axes of D. Hence the names, vorticity, or spin given toW. Additionally, we associate with Wthe vector wi=1 2"ijkvk;j orw=1 2curlv (4.153) called the vorticity vector, by the following calculation, "pqiwi=1 2"pqi"ijkvk;j=1 2(pjqk-pkqj)vk;j =1 2(vq;p-vp;q) =wqp: (4.154) Thus ifD0so thatLij=wij, it follows that dvi=Lijdxj=wijdxj="jikwkdxjand since"jik= -"ijk="ikj, dvi="ijkwjdxk ordv=wdx (4.155) according to which the relative velocity in the vicinity of pcorresponds to a rigid body rotation about an axis through p. The vectorwindicates the angular velocity, the direction and the sense of this rotation. To summarize the physical interpretation of the velocity gradient L, we note that it effects 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 ofDalong the mutually orthogonal principal axes of D, ˙==d(ln) dt=dijn(p) in(p) j=n(p) iD(p)n(p) i=D(p); and 2. A rigid body rotation of the principal axes of Dwith angular velocity w. Example 4.16 Consider a continuum having a shearing motion dened by the mapping x1=X1+X2; x2=X2; x3=X3: 142 Continuum Mechanics for Engineers (a) Invert the mapping to nd XAin terms ofxi (b) Determine xj;BandXA;j (c) Compute XA;jxj;Bandxj;BXB;i (d) Use the result of (c) to determine the indicial expression for˙XA;ifor all general motions (not just this pure shear case). Solution (a) The mapping can be inverted, nd XAin terms ofxi, by inspection to get X1=x1-x2; X2=x2; X3=x3: (b) Partial dierentiation of the two forms of the motion give us F=xj;B andF-1=XA;j xj;B=2 41  0 0 1 0 0 0 13 5 andXA;j=2 41- 0 0 1 0 0 0 13 5: (c) Multiplying these two expressions will solidify that XA;jxj;Bis equiv- alent toF-1Fandxj;BXB;iis equivalent to FF-1. In indicial notation this fact is sometimes not obvious when so many new symbols have been introduced. XA;jxj;B=2 41- 0 0 1 0 0 0 13 52 41  0 0 1 0 0 0 13 5=2 41  - 0 0 1 0 0 0 13 5=AB xj;BXB;i=2 41  0 0 1 0 0 0 13 52 41- 0 0 1 0 0 0 13 5=2 41-+ 0 0 1 0 0 0 13 5=ij (d) Take the material derivative of the rst form noting that ABis a constant, and use Eq 4.138 ˙XA;jxj;B= -XA;j˙xj;B= -XA;jLjmxm;B: Multiply both sides by XB;iwhich is the inverse of xj;Bto get˙XA;j: ˙XA;jxj;BXB;i=˙XA;jji=˙XA;i = -XA;jLjmxm;BXB;i = -XA;jLjmmi= -XA;jLji: Thus, ˙XA;i= -XA;jLji= -XA;jvj;i: Kinematics of Deformation and Motion 143 dX(2) dX(1)dx(2) dx(1) DEFORMSdS dS dS0 dS0 FIGURE 4.8 AreadS0between vectors dX(1)anddX(2)in the reference configuration becomes dS betweendx(1)anddx(2)in the deformed configuration. 4.12 Material Derivative of Line Elements, Areas, Volumes Consider first the material derivative of the differential line element dx=F
dX. Clearly, ˙dx=˙F
dXand 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
dxis ˙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 differential line elements dX(1) AanddX(2) Aas shown in Fig. 4.8. The parallelogram area dS0may be represented by the vector dS0 A="ABCdX(1) BdX(2) C: (4.157) As a result of the motion x=x(X;t) this area is carried into the current area dSishown in Fig. 4.8 and is given by dSi="ijkdx(1) jdx(2) k="ijkxj;BdX(1) Bxk;CdX(2) C (4.158) which upon multiplication by xi;Aresults in xi;AdSi="ijkxi;Axj;Bxk;CdX(1) BdX(2) C="ijkFiAFjBFkCdX(1) BdX(2) C: Recall that detF =J(the Jacobian) and from Eq 2.43 "ijkFiAFjBFkC="ABC detF="ABCJ: 144 Continuum Mechanics for Engineers Therefore, by inserting this result into the above equation for xi;AdSiand multiplying both sides by XA;q, we obtain, xi;AXA;qdSi="ABCJdX(1) BdX(2) CXA;q Butxi;AXA;q=iqso that iqdSi=dSq=XA;qJdS0 A (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) whereAis an arbitrary tensor. Substituting FforAwe obtain ˙detF=˙J= (detF)tr(˙F
F-1) =Jtr(L); or ˙J=Jvi;i=Jdivv: (4.161) Noting that Eq 4.159 may be written dSq=JXA;qdS0 A and using symbolic notation to take advantage of Eq 4.160 we obtain dS=J(F-1)T
dS0=JdS0
F-1; and so dS
F=JdS0 which upon differentiating becomes d˙S
F+dS
˙F=˙JdS0=J(trL)dS0; d˙S+dS
˙F
F-1=J(trL)dS0
F-1= (trL)dS; and finally d˙S= (trL)dS-dS
Lord˙Si=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 box product, dV0=dX(1)
dX(2)dX(3)="ABCdX(1) AdX(2) BdX(3) C=h dX(1);dX(2);dX(3)i and pictured in Fig. 4.9, and let the deformed volume element, also shown in Fig. 4.9 be given by dV=dx(1)
dx(2)dx(3)="ijkdx(1) idx(2) jdx(3) k=h dx(1);dx(2);dx(3)i : Kinematics of Deformation and Motion 145 dX(2) dX(1)dx(2) dx(1) dX(3)dx(3) DEFORM FIGURE 4.9 Volume of parallelpiped defined by vectors dX(1),dX(2)anddX(3)in the reference con- figuration deforms into volume defined by parallelpiped defined by vectors dx(1),dx(2) anddx(3)in the deformed configuration. For the motion x=x(X;t);dx =F
dXso the current volume is the box product dV=h F
dX(1);F
dX(2);F
dX(3)i ="ijkxi;Axj;Bxk;CdX(1) AdX(2) BdX(3) C =det(F )h dX(1);dX(2);dX(3)i =JdV0(4.163) which gives the current volume element in terms of its original size. Since J6=0(Fis invertible), we have either J<0 orJ>0. Mathematically, J<0 is possible, but physically it corresponds to a negative volume, so we reject it. Henceforth, we assume J>0 . IfJ=1, thendV=dV0and the volume magnitude is preserved. If Jis equal to unity for all Xwe say the motion is isochoric. To determine the time rate of change of dVwe take the material derivative as follows, ˙dV=˙JdV0=Jtr(L)dV0=Jvi;idV0=vi;idV: (4.164) Thus a necessary and sufficient condition for a motion to be isochoric is that vi;i=divv=0: (4.165) In summary, we observe that the deformation gradient Fgoverns the stretch of a line element, the change of an area element, and the change of a volume element. But it is the velocity gradient Lthat determines the rate at which these changes occur. 146 Continuum Mechanics for Engineers Example 4.17 To derive the material derivative of the Jacobian, ˙J, the identity Eq 4.160 was used in the text. As an alternative derivation, use J=1 6"ABC"ijkFiAFjBFkC to derive Eq 4.161. Solution Dierentiate and use Eq 4.139, Eq 2.43 and Eq 2.7b ˙J=1 6"ABC"ijk˙FiAFjBFkC+FiA˙FjBFkC+FiAFjB˙FkC
=1 6"ABC"ijk(LimFmAFjBFkC+FiALjmFmBFkC+FiAFjBLkmFmC) =1 6"ijk(Lim"mjkJ+Ljm"imkJ+Lkm"ijmJ) =1 6(2imLim+2mjLjm+2kmLkm)J =JLii=Jvi;i: Kinematics of Deformation and Motion 147 Problems Problem 4.1 The motion of a continuous medium is specied by the component equations x1=1 2(X1+X2)et+1 2(X1-X2)e-t; x2=1 2(X1+X2)et-1 2(X1-X2)e-t; x3=X3: (a) Show that the Jacobian determinant Jdoes not vanish, and solve for the inverse equationsX=X(x;t). (b) Calculate the velocity and acceleration components in terms of the material coordinates. (c) Using the inverse equations developed in part (a), express the velocity and ac- celeration components in terms of spatial coordinates. Answer (a)J=cosh2t-sinh2t=1 X1=1 2(x1+x2)e-t+1 2(x1-x2)et X2=1 2(x1+x2)e-t-1 2(x1-x2)et X3=x3 (b)v1=1 2(X1+X2)et-1 2(X1-X2)e-t v2=1 2(X1+X2)et+1 2(X1-X2)e-t v3=0 a1=1 2(X1+X2)et+1 2(X1-X2)e-t a2=1 2(X1+X2)et-1 2(X1-X2)e-t a3=0 (c)v1=x2;v2=x1;v3=0 a1=x1;a2=x2;a3=0 Problem 4.2 Let 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 J6=0, and solve for the inverse equations. (b) Determine the velocity and acceleration (1) at timet=1s for the particle which was at point (2:75;3:75;4:00 )when t=0:5s. (2) at timet=2s for the particle which was at point (1;2; -1)whent=0. 148 Continuum Mechanics for Engineers Answer (a)J= (1-t3)2 X1= (x1-x2t)=(1 -t3) X2= (x2-x3t)=(1 -t3) X3= (x3-x1t)=(1 -t3) (b) (1)v=8^e1+5^e2+5^e3,a=6^e1+2^e2+4^e3 (2)v= -2^e1+3^e2+9^e3,a= -2^e1+2^e2+4^e3 Problem 4.3 A continuum body has a motion dened by the equations x1=X1+2X2t2; x2=X2+2X1t2; x3=X3: (a) Determine the velocity components at t=1:5s of the particle which occupied the point (2;3;4 )whent=1:0s. (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=2s. Answer (a)v1=2,v2=8,v3=0 (b)4x1-x2=5in the planex3=4 (c)a1=4=3,a2=16=3,a3=0. Problem 4.4 If the motionx=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) Kinematics of Deformation and Motion 149 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 the motion. Answer X1=x1et-x3(et-1) X2=x2e-t+x3(e-2t-e-3t) X3=x3e-t Problem 4.6 A velocity eld is given in Lagrangian form by v1=2t+X1; v2=X2et; v3=X3-t: Integrate these equations to obtain x=x(X;t) withx=Xatt=0, and using that result compute 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(et-1); x2=X2e-t+X3(1-e-t); x3=X3; determine the displacement eld 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=0 150 Continuum Mechanics for Engineers Problem 4.8 The temperature eld in a continuum is given by the expression =e-3t=x2wherex2=x2 1+x2 2+x2 3: The velocity eld of the medium has components v1=x2+2x3; v2=x3-x1; v3=x1+3x2: Determine the material derivative d=dt of the temperature eld. Answer d=dt = -e-3t(3x2+6x1x3+8x2x3)=x4 Problem 4.9 In a certain region of a uid the ow velocity has components v1=A(x3 1+x1x2 2)e-kt; v2=A(x2 1x2+x3 3)e-kt; v3=0 whereAandkare constants. Use the (spatial) material derivative operator to determine the acceleration components at the point (1;1;0 )whent=0. Answer a1= -2A (k-5A),a2= -A(k -5A),a3=0 Problem 4.10 A displacement eld 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+x1t3)=(1-t4) v2= (x3+x1t+2x2t3)=(1-t4) v3= (x1+2x2t2+x3t3)=(1-t4) Problem 4.11 For the motion given by the equations x1=X1cos!t+X2sin!t; x2= -X1sin!t+X2cos!t; x3= (1+kt)X3; where!andkare constants, determine the displacement eld in Eulerian form. Kinematics of Deformation and Motion 151 Answer u1=x1(1-cos!t) +x2sin!t u2= -x1sin!t+x2(1-cos!t) u3=x3kt=(1+kt) Problem 4.12 Show that the displacement eld for the motion analyzed in Problem 4.1 has the Eulerian form u1=x1- (x1+x2)e-t=2- (x1-x2)et=2; u2= -x2- (x1+x2)e-t=2+ (x1-x2)et=2; and by using the material derivative operator (dui=dt=@ui=@t+vj@ui=@xj), verify the velocity and acceleration components calculated in Problem 4.1. Problem 4.13 The 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 FiAand from it show that the Jaco- bianJdoes not vanish. Invert the mapping equations to obtain the Eulerian description of the deformation. Answer J=e2 X1=x1-x3(1-e-4) X2=x2+x3(1-e-2) X3=x3e-2 Problem 4.14 Ahomogeneous 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 in general by the displacement eld ui=AijXjwhere theAijare constants (or in the case of a motion, functions of time). Show that for a homogeneous deformation with the Aij constant: (a) plane material surfaces remain plane, (b) straight line particle elements remain straight, (c) material surfaces which are spherical in the reference conguration become ellip- soidal surfaces in the deformed conguration. 152 Continuum Mechanics for Engineers Problem 4.15 Aninfinitesimal homogeneous deformation ui=AijXjis one for which the constants Aijare so small that their products may be neglected. Show that for two sequential innitesimal deformations the total displacement is the sum of the individual displacements regardless of the order in which the deformations are applied. Problem 4.16 For the homogeneous deformation dened by x1=X1+X2; x2= -X1+X2; x3=X3; where,andare constants, calculate the Lagrangian nite strain tensor E. Show that if=cos,=sinand=1the strain is zero and the mapping corresponds to a rigid body rotation of magnitude about theX3axis. Answer EAB=1 22 42+2-1 0 0 0 2+2-1 0 0 0 2-13 5 Problem 4.17 Given the deformation dened by x1=X1; x2=X2+1 2X2 3; x3=X3 (a) Sketch the deformed shape of the unit square OABC in the planeX1=0. (b) Determine the dierential vectors dx(2)anddx(3)which are the deformed vectors resulting from dX(2)=dX(2)^I2anddX(3)=dX(3)^I3, respectively, that were originally at corner C. (c) Calculate the dot product dx(2)
dx(3), and from it determine the change in the original right angle between dX(2)anddX(3)atCdue to the deformation. (d) Compute the stretch atBin the direction of the unit normal ^N= ^I2+^I3 =p 2: Kinematics of Deformation and Motion 153 X3 X2 X1O ABC Unit squareOABC in the reference conguration. Answer (b)dx(2)=dX(2)^e2,dx(3)=dX(3)(^e2+^e3) (c)
= -45 (d)(^N)=p 2:5 Problem 4.18 Given the deformation expressed by x1=X1+AX2 2; x2=X2; x3=X3-AX2 2 whereAis a constant (not necessarily small), determine the nite strain tensors Eande, and show that if the displacements are small so that xX, and if squares of Amay be neglected, both tensors reduce to the innitesimal strain tensor . Answer [ij]=2 40 Ax20 Ax20 -Ax2 0-Ax203 5 Problem 4.19 For the innitesimal homogeneous deformation xi=Xi+AijXjwhere the constants Aij are very small, determine the small strain tensor , and from it the longitudinal (normal) strain in the direction of the unit vector ^N= ^I1-^I3 =p 2. Answer 2e(^N)=A11-A13-A31+A33 Problem 4.20 A deformation is dened by x1=X1= X2 1+X2 2
; x2=X2= X2 1+X2 2
; x3=X3: 154 Continuum Mechanics for Engineers Determine the deformation tensor Ctogether with its principal values. Answer C(1)=C(2)= X2 1+X2 2
-2; C (3)=1 Problem 4.21 For the deformation eld given by x1=X1+X2; x2=X2-X1; x3=X3 whereis a constant, determine the matrix form of the tensors Eande, and show that the circle of particles X2 1+X2 2=1deforms into the circle x2 1+x2 2=1+2. Answer [Eij]=1 22 420 0 0 20 0 0 03 5 [eij]=1 2(1+2)22 4-20 0 0-20 0 0 03 5 Problem 4.22 Let the deformation of a continuum be given by the equations x1=X1+kX2 2; x2=X2-kX2 1; x3=X3 wherekis a constant. Determine the Lagrangian nite strain tensor E, and from it, assuming kis very small, deduce the innitesimal strain tensor . Verify this by calculating the displacement eld and using the denition 2ij=ui;j+uj;ifor the innitesimal theory. Problem 4.23 Given the displacement eld u1=AX2X3; u2=AX2 3; u3=AX2 1 whereAis a very small constant, determine (a) the components of the innitesimal strain tensor , and the innitesimal rotation tensor!. (b) the principal values of , at the point (1;1;0 ). Answer (a)11=22=33=0,12=AX3,13=1 2A(X2+2X1)and23=AX3 !11=!22=!33=0,!12= -!21=1 2AX3, !13= -!31=1 2AX1-AX1and!23= -!32=AX3 (b)(I)=3 2A,(II)=0,(III)= -3 2A Kinematics of Deformation and Motion 155 Problem 4.24 A45strain rosette measures longitudinal strains along the X1,X2, andX0 1axes shown below. At point Othe strains recorded are 11=610-4; 22=410-4and0 11=810-4 Determine the shear strain 12atO, together with "0 22, and verify that "11+"22="0 11+"0 22. See Eq 4.82. X1X2 X2‛X1‛ 45o45o O Problem 4.25 By a direct expansion of Eq 4.87, 2!i="ijk!kj, show that!1=!32= -!23, etc. Also, show that only if Ais a very small constant does the mapping x1=X1-AX2+AX3; x2=X2-AX3+AX1; x3=X3-AX1+AX2; represent a rigid body rotation (E0). Additionally, determine the innitesimal rotation tensor!ijin this case; from it, using the result proven above, deduce the rotation vector !i. Answer !=A(^e1+^e2+^e3) Problem 4.26 For the displacement eld u1=kX1X2; u2=kX1X2; u3=2k(X1+X2)X3 wherekis a very small constant, determine the rotation tensor !, and show that it has only one real principal value at the point (0;0;1 ). 156 Continuum Mechanics for Engineers Answer !(1)=0;! (2)= -! (3)=ikp 2, wherei=p -1 Problem 4.27 Let the deformation x1=X1+AX2X3; x2=X2+AX2 3; x3=X3+AX2 1; whereAis a constant be applied to a continuum body. For the unit square of material line elementsOBCD 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 CifA=1; ifA=0:1. DX3 X2 X1O BC 11 Unit squareOBCD in the reference conguration. Answer (a)2 (OC)=1+2A+4A2,e(OC)=p 1+2A+4A2-1 (b)
(A=1) =60,
(A=0:1) =11:77 Problem 4.28 For the homogeneous deformation expressed by the equations x1=p 2X1+3p 2 4X2; x2= -X1+3 4X2+p 2 4X3; x3=X1-3 4X2+p 2 4X3; determine Kinematics of Deformation and Motion 157 (a) the unit normal ^nfor the line element originally in the direction of ^N= ^I1-^I2+^I3 =p 3. (b) the stretch (^N)of this element. (c) the maximum and minimum stretches at the point X1=1,X2=0,X3= -2in the reference conguration. Answer (a) ^n=p 2^e1+p 2-7 ^e2+p 2+7 ^e3 p 104 (b)(^N)=1:472 (c)(max)=2;(min)=0:5 Problem 4.29 Let the deformation x1=a1(X1+2X2); x2=a2X2; x3=a3X3 wherea1,a2, anda3are constants be applied to the unit cube of material shown in the sketch. Determine (a) the deformed length lof diagonalOC, (b) the angle between edges OAandOGafter deformation, (c) the conditions which the constants must satisfy for the deformation to be possible if (1) the material is incompressible, (2) the angle between elements OCandOBis to remain unchanged. X1 , x1X2 , x2X3 , x3 O A BGCF DE Unit cube having diagonal OC. 158 Continuum Mechanics for Engineers Answer (a)l2=9a2 1+a2 2+a2 3 (b)cos=2a1q 4a2 1+a2 2 (c) (1)a1a2a3=1, (2)9a2 1+a2 2=2a2 3 Problem 4.30 A homogeneous deformation is dened by x1=X1+X2; x2= -X1+X2; x3=X3 where,andare 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)2 (1)=2 (^e1)=2a2,2 (2)=2 (^e2)=22,2 (3)=2 (^e3)=2 (b)[Rij]=2 66641p 21p 20 -1p 21p 20 0 0 13 7775 (c)^n=^I3;=45 Problem 4.31 Consider the deformation eld x1=X1-AX2+AX3; x2=X2-AX3+AX1; x3=X3-AX1+AX2; whereAis a constant. Show that the principal values of the right stretch tensor have a multiplicity of two, and that the axis of the rotation tensor is along ^N= ^I1+^I2+^I3 =p 3. Determine the matrix of the rotation vector together with the angle of rotation . Answer (1)=1,(2)=(3)=p 1+3A2= [Rij]=1 32 4+2  -1-3A  -1+3A -1+3A  +2  -1-3A -1-3A  -1+3A  +23 5 =cos-1(1=) Kinematics of Deformation and Motion 159 Problem 4.32 For the deformation eld x1=p 3X1+X2; x2=2X2; x3=X3: determine (a) the matrix representation of the rotation tensor R, (b) the right stretch tensor Uand the left stretch tensor V, then show that the principal values of UandVare equal, (c) the direction of the axis of rotation and the magnitude of the angle of rotation. Answer (a)[Rij]=1 2p 22 64p 3+1p 3-1 0 -p 3+1p 3+1 0 0 0 2p 23 75 (b)(1)=p 6,(2)=p 2,(3)=1 (c)^N=^I3;=15 Problem 4.33 Let a displacement eld 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 unit vectors ^N1= 3^I1-2^I2-^I3 =p 14and^N2= ^I1+4^I2-5^I3 =p 42. Explain your answer. Answer (a)dV=dV0=1:1875 (b)
=0 Problem 4.34 Consider again the deformation given in Example 4.15, namely x1=2(X1-X2); x2=X1+X2; x3=X3: 160 Continuum Mechanics for Engineers Determine (a) the left stretch tensor V, (b) the direction normals of the principal stretches of V. Answer (a)[VAB]=2 42p 2 0 0 0p 2 0 0 0 13 5 (b)^N1=^I1,^N2=^I2,^N3=^I3 Problem 4.35 A deformation eld is expressed by x1=(X1cosX3+X2sinX3); x2=(-X1sinX3+X2cosX3); x3=X3; where,, andare constants. (a) Determine the relationship between these constants if the deformation is to be a possible 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 lin terms ofL, the dimension a, and the constants , , andof an element of the lateral surface which has unit length and is parallel to the cylinder axis in the reference conguration, and (2) the initial length Lof a line element on the lateral surface which has unit length and is parallel to the cylinder axis after deformation. a LX3 X2 X1 Kinematics of Deformation and Motion 161 Answer (a)2=1 (b) (1)l=p 22a2+2 (2)L=1 p 2a2+1 Problem 4.36 A velocity eld is dened in terms of the spatial coordinates and time by the equations, v1=2tx1sinx3; v2=2tx2cosx3; v3=0: At the point (1;-1;0)at timet=1, determine (a) the rate of deformation tensor and the vorticity tensor, (b) the stretch rate per unit length in the direction of the normal ^n=(^e1+^e2+^e)=p 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]=2 40 0 1 0 2 0 1 0 03 5,[wij]=2 40 0 1 0 0 0 -1 0 03 5 (b) ˙==4=3 (c)˙=
max=2,^n=^e2 (d) ˙ max=1:5 Problem 4.37 LetNAandnidenote direction cosines of a material line element in the reference and current congurations, respectively. Beginning with Eq 4.140, ni=xi;ANA, and using the indicial notation throughout, show that (a) ˙==Dijninj, (b) ¨==Qijninj+˙ninjwhereQij=1 2(ai;j+aj;i)withaibeing the components of acceleration. Problem 4.38 In a certain region of ow the velocity components are v1= x3 1+x1x2 2
e-kt; v2= x3 2-x2 1x2
e-kt; v3=0 wherekis a constant, and tis time in s. Determine at the point (1;1;1 )whent=0, (a) the components of acceleration, (b) the principal values of the rate of deformation tensor, 162 Continuum Mechanics for Engineers (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) ˙ max=2 Problem 4.39 For the motion x1=X1; x2=X2et+X1(et-1); x3=X1(et-e-t) +X3; determine the velocity eld vi=vi(x), and show that for this motion (a)L=˙F
F-1, (b)D=˙att=0. Problem 4.40 Determine an expression for the material derivative d(lndx)=dtin terms of the rate of deformation tensor Dand the unit normal ^n=dx=dx. Answer d(lndx)=dt=^n
D
^n Problem 4.41 A velocity eld is given in spatial form by v1=x1x3; v2=x2 2t; v3=x2x3t: (a) Determine the vorticity tensor Wand the vorticity vector w. (b) Verify the equation "pqiwi=wqpfor the results of part (a). (c) Show that at the point (1;0;1 )whent=1, the vorticity tensor has only one real root. Answer (a)w1=1 2x3t,w2=1 2x1,w3=0 (c)W(1)=0,W(2)= -W (3)=i=p 2, wherei=p -1 Problem 4.42 Consider the velocity eld v1=ex3-ctcos!t; v2=ex3-ktsin!t; v3=c wherecand!are constants. Kinematics of Deformation and Motion 163 (a) Show that the speed of every particle is constant. (b) Determine the acceleration components ai. (c) Calculate the logarithmic stretching, ˙= =d(ln)=dt, for the element in the current conguration in the direction of ^n=(^e1+^e2)=p 2atx=0. Answer (b)a1= -!v2,a2=!v1,a3=0 (c)1 2e-ctcos!t Problem 4.43 Show that the velocity eld 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+1:5^e2+3^e3 Problem 4.44 For the steady velocity eld v1=x2 1x2; v2=2x2 2x3; v3=3x1x2x3 determine the rate of extension at (2;0;1 )in the direction of the unit vector (4^e1-3^e3)=5. Answer ˙== -48 25 Problem 4.45 Prove that d (lnJ)=dt=divvand, in particular, verify that this relationship is satised for the motion x1=X1+ktX3; x2=X2+ktX3; x3=X3-kt(X1+X2) wherekis a constant. Answer ˙J=J=divv=4k2t=(1+2k2t2) Problem 4.46 Prove the identityxi;A J ;i=0; J =det(xi;A) 164 Continuum Mechanics for Engineers Problem 4.47 Equation 4.151 gives the material derivative of dx2in terms ofdij. Using that equation as the starting point, show that d2(dx2)=dt2is given in terms of dijand its time derivative by d2(dx2)=dt2=2(˙dij+vk;idkj+vk;jdik)dxidxj: Problem 4.48 A continuum body in the form of the unit cube shown by the sketch undergoes the homo- geneous deformation x1=1X1; x2=2X2; x3=3X3 where1,2, and3are constants. Determine the relationships among 1,2, and3if (a) the length of diagonal OCremains unchanged, (b) the rectangular area ABFE remains unchanged, (c) the triangular area ACE remains unchanged. X1 , x1X2 , x2X3 , x3 O A BGCF DE (a) X1 , x1X2 , x2X3 , x3 O A BGCF DE (b) X1 , x1X2 , x2X3 , x3 O A BGCF DE (c) Kinematics of Deformation and Motion 165 Answer (a)2 1+2 2+2 3=3 (b)2 2 2 1+2 3
=2 (c)2 12 2+2 22 3+2 32 1=3 Problem 4.49 Let 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=t3^e1=(1-1 4t4) -t^e2-t3^e3 (b)V= (1-1 4t4) Problem 4.50 For the homogeneous deformation x1=X1+X2+X3; x2=X1+X2+2X3; x3=X1+X2+X3; whereandare constants, determine the relationship between these constants if the deformation is isochoric. Answer = (2+)=(2++1) Problem 4.51 Show that for any velocity eld vderived from a vector potential byv=curl , the ow is isochoric. Also, for the velocity eld v1=ax1x3-2x3; v2= -bx2x3; v3=2x1x2 determine the relationship between the constants aandbif the ow is isochoric. Answer a=b This page intentionally left blank 5 Fundamental 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, are applicable 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 a purely mechanical energy balance and follow that by an energy balance that includes both mechanical 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 is covered. The balance laws are usually formulated in the context of global (integral) relationships derived 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 associated field equations which are valid at all points within the body and on its boundary. For example, we shall derive the local equations of motion from a global statement of the conservation of linear momentum. Constitutive equations, which reflect the internal constitution of a material, define specific types of material behavior. They are fundamental in the sense that they serve as the starting point for studies in the disciplines of elasticity, plasticity and various idealized fluids. Constitutive equations establish the relationship between kinematics and stresses for 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 develop expressions for the material derivatives of certain integrals. 5.1 Material Derivatives of Line, Surface and Volume Integrals Let any scalar, vector or tensor property of the collection of particles occupying the current volume Vbe represented by the integral Pij:::(t) =Z VP ij:::(x;t)dV (5.1) whereP 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 any order tensor. The material derivative of this property is given in both spatial and material form, using Eq 4.163, by ˙Pij:::(t) =d dtZ VP ij:::(x;t)dV=d dtZ V0P ij:::[x(X;t);t]JdV0: Since V0is a fixed volume in the referential configuration, the differentiation and inte- gration commute and the differentiation can be performed inside the integral sign. Thus, 167 168 Continuum Mechanics for Engineers from Eq 4.161, using the notation: [
]to indicate differentiation with respect to time, Z V0:h P ij:::(X;t)Ji dV0=Z V0˙P ij:::J+P ij:::˙J
dV0 =Z V0˙P ij:::+vk;kP ij:::
JdV0; and converting back to the spatial formulation ˙Pij:::(t) =Z 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 be written (we omit listing the independent variables xandtfor convenience), ˙Pij:::(t) =Z V@P ij::: @t+vk@P ij::: @xk+vk;kP ij::: dV =Z V@P ij::: @t+ vkP ij:::k
;k dV which upon application of the divergence theorem becomes ˙Pij:::(t) =Z V@P ij::: @tdV+Z SvkP ij:::nkdS: (5.3) This equation gives the time rate of change of the property Pij:::as the sum of the amount created in the volume V, plus the amount entering through the bounding surface S, and is often spoken of as the transport theorem. Time derivatives of integrals over material surfaces and material curves may also be derived in an analogous fashion. First, we consider a tensorial property Qij::: of the particles which make up the current surface S, as given by Qij:::(t) =Z SQ ij:::(x;t)dSp=Z SQ ij:::(x;t)npdS (5.4) whereQ ij:::(x;t)is the distribution of the property over the surface. From Eq 4.162, we have in Eulerian form (again omitting the variables xandt), ˙Qij:::(t) =Z S˙Q ij:::+vk;kQ ij:::
dSp-Z SQ ij:::vq;pdSq =Z S˙Q ij:::+vk;kQ ij:::
pq-Q ij:::vq;p dSq: (5.5) Similarly, for properties of particles lying on the spatial curve Cand expressed by the line integral Rij:::(t) =Z CR ij:::(x;t)dxp (5.6) Fundamental Laws and Equations 169 we have, using Eq 4.156, ˙Rij:::(t) =Z C˙R ij:::dxp+Z Cvp;qR ij:::dxq =Z C˙R ij:::pq+vp;qR ij:::
dxq: (5.7) 5.2 Conservation of Mass, Continuity Equation Every 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 of the space variables and time. If
m is the mass of a small volume
Vin the current configuration, and if we assume that
m is absolutely continuous, the limit =lim
V!0
m
V(5.8) defines the scalar field =(x;t)called the mass density of the body for that configuration at timet. Therefore, the mass mof the entire body is given by m=Z V(x;t)dV: (5.9) In the same way, we define the mass of the body in the referential (initial) configuration in terms of the density field 0=0(X;t) by the integral m=Z V00(X;t)dV0: (5.10) The law of conservation of mass asserts that the mass of a body, or of any portion of the body, is invariant under motion, that is, remains constant in every configuration. Thus, the material derivative of Eq 5.9 is zero, ˙m=d dtZ V(x;t)dV=0 (5.11) which upon application of Eq 5.2 with P ij:::becomes, ˙m=Z V(˙+vi;i)dV=0; (5.12) and since Vis an arbitrary part of the continuum, the integrand here must vanish, result- ing in ˙+vi;i=0 (5.13) which is known as the continuity equation in Eulerian form. But the material derivative of can be written as ˙=@ @t+vi@ @xi; 170 Continuum Mechanics for Engineers so that Eq 5.13 may be rewritten in the alternative forms, @ @t+vi@ @xi+vi;i=0; (5.14a) or @ @t+ (vi);i=0: (5.14b) If the density of the individual particles is constant so that ˙ =0, the material is said to be incompressible, and thus it follows from Eq 5.13 that vi;i=0 or divv=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 for min the referential and current configurations. Therefore, if we equate Eqs 5.9 and 5.10, m=Z V˜(x;t)dV=Z V00(X;t)dV0; (5.16) and, noting that for the motion x=x(X;t), we have Z V[x(X;t);t]dV=Z V0(X;t)JdV0: The tilde in Eq 5.16 is there because the functional form may be different when written as a function of xandtthan the functional form written in terms of Xandt. Now if we substitute the right-hand side of this equation for the left-hand side of Eq 5.16 and collect terms, Z V0[(X;t)J -0(X;t)]dV0=0: ButV0is arbitrary, and so in the material description J=0; (5.17a) and, furthermore, ˙ 0=0, from which we conclude that : (J)=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 material form. Solution Carrying out the indicated dierentiation in Eq 5.17b, : (J)=˙J+˙J=0 and by Eq 4.161, ˙J=vi;iJ Fundamental Laws and Equations 171 so now: (J)=J(˙+vi;i)=0: ButJ=detF6=0(an invertible tensor), which requires ˙+vi;i=0the spatial continuity equation. As a consequence of the continuity equation, we are able to derive a useful result for the material derivative of the integral in Eq 5.1 when P ij:::is equal to the product A ij:::, whereA ij:::is the distribution of any property per unit mass. Accordingly, let ˙Pij:::(t) =d dtZ VA ij:::(x;t)dV =d dtZ V0A ij:::(X;t)Jd V0 =Z V0 ˙A ij:::(J)+A ij:::: (J) dV0 which, because of Eq 5.17b, reduces to ˙P ij:::(t) =Z V0˙A ij:::dV0=Z V˙A ij:::dV; and so d dtZ VA ij:::(x;t)dV =Z V˙A ij:::dV: (5.18) We shall have numerous occasions to make use of this very important equation. 5.3 Linear Momentum Principle, Equations of Motion Let a material continuum body having a current volume Vand bounding surface Sbe subjected to surface traction t(^n) iand distributed body forces bias shown in Fig. 5.1. In addition, let the body be in motion under the velocity field vi=vi(x;t). The linear momentum of the body is defined by the vector Pi(t) =Z VvidV; (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, with reference to Fig. 5.1, d dtZ VvidV=Z St(^n) idS+Z VbidV; (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, after collecting terms,Z V(˙vi-tji;j-bi)dV=0 (5.21) 172 Continuum Mechanics for Engineers OdV ni dSx3 x2 x1xyρbit(^n) i FIGURE 5.1 Material body in motion subjected to body and surface forces. where ˙viis the acceleration field of the body. Again, Vis arbitrary and so the integrand must vanish, and we obtain, tji;j+bi=˙vi (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+bi=0 (5.23) which are important in solid mechanics, especially elastostatics. 5.4 Piola-Kirchhoff Stress Tensors, Lagrangian Equations of Motion As mentioned in the previous section, the equations of motion Eq 5.22 are in Eulerian form. 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 tijof the Cauchy stress ten- sorTas theithcomponent of the stress vector t(^ej) iacting on the material surface having the 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 the reference configuration and from it construct a stress tensor that is associated with that configuration. In doing this, we parallel the development in Section 3.3 for the Cauchy stress tensor associated with the current configuration. Let the vector p0(^N)be defined as the stress vector referred to the area element
S0in the plane perpendicular to the unit normal ^N=NA^IA. Just as we defined the Cauchy stress vector in Eq 3.5, we write lim
S0!0
f
S0=df dS0=p0(^N)(5.24) Fundamental Laws and Equations 173 where
fis the resultant force acting on the material surface which in the reference configuration was
S0. The principle of linear momentum can also be written in terms of quantities which are referred to in the referential configuration as Z S0p0(^N)(X;t)dS0+Z V00b0(X;t)dV0=Z V00a0(X;t)dV0(5.25) where S0,V0and0are the material surface, volume and density, respectively, referred to the reference configuration. The superscript zero after the variable is used to emphasize the fact that the function is written in terms of the reference configuration. For example, ai(x;t) =ai[(X;t);t) ]=a0 i(X;t): Notice that, since all quantities are in terms of material coordinates, we have moved the differential 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 I and II of the body (as defined in Fig. 3.2(a)) and to the body as a whole to arrive at the equationZ S0h p0(^N)+p0(-^N)i 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)=p0 i(^IA)^ei; (A=1;2;3 ): (5.28) This describes the components of the stress vector p0(^N)with respect to the referential coordinate planes, and to determine its components with respect to an arbitrary plane defined by the unit vector ^N, we apply a force balance to an infinitesimal tetrahedron of the body. As we let the tetrahedron shrink to the point, we have p0 i(^N)=p0 i(^IA)NA; (5.29) and defining PAip0 i(^IA)(5.30) we obtain p0(^N) i=PAiNA (5.31) wherePAiare the components of the first Piola-Kirchhoff stress tensor. These represent the xicomponents 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, and hence the equilibrium equations in the referential formulation. Starting with Eq 5.25, we introduce Eq 5.31 to obtain Z S0PAiNAdS0+Z V00b0 idV0=Z V00a0 idV0; (5.32) 174 Continuum Mechanics for Engineers and using the divergence theorem on the surface integral, we consolidate Eq 5.32 as Z V0(PAi;A+0b0 i-0a0 i)dV0=0: This equation must hold for arbitrary portions of the body so that the integrand is equal to zero, or PAi;A+0b0 i=0a0 i (5.33a) which are the equations of motion in referential form. If the acceleration field is zero, these equations reduce to the equilibrium equations in referential form PAi;A+0b0 i=0: (5.33b) We note that the partial derivatives of the Piola-Kirchhoff stress components are with respect to the material coordinates because this stress tensor is referred to a surface in the reference configuration. Equilibrium also requires a balance of moments about every point. Summing moments about the origin (Fig. 5.1 may be useful in visualizing this operation) gives us Z S0"ijkxjp0 k(^N)dS0+Z V0"ijkxj0b0 kdV0=0 (5.34) which reduces to Z V0"ijk (xjPAk);A+xj0b0 k dV0=0 where we have used Eq 5.31 and the divergence theorem. Carrying out the indicated partial differentiation we obtain Z V0"ijk xj;APAk+xj PAk;A +0b0 k
 dV0=0 and by Eq 5.33b this reduces to Z V0("ijkxj;APAk)dV0=0 (5.35) since the term in parentheses is zero on account of the balance of momentum. Again, this equation must hold for all portions V0of the body, so the integrand must vanish, giving "ijkxj;APAk=0 (5.36) Following a similar argument to that presented in Section 3.4, we conclude that Eq 5.36 implies 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. Fundamental Laws and Equations 175 The Piola-Kirchhoff stresses can be related to the Cauchy stress by considering the differential force exerted on an element of deformed surface dSas dfi=tjinjdS: (5.40) This force can also be written in terms of the first Piola-Kirchhoff stress tensor as dfi=PAiNAdS0: (5.41) Recall from Eq 4.159 that the surface element in the deformed configuration is related to the surface element in the reference configuration by nqdS=XA;qJNAdS0: Using this, along with Eqs 5.40 and 5.41, we obtain dfi=tjinjdS=tjiXA;jJNAdS0=PAiNAdS0(5.42) which can be rewritten as (tjiXA;jJ-PAi)NAdS0=0: (5.43) From this we see that the Cauchy stress and the first Piola-Kirchhoff stress are related through 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 not necessary. To show the equivalence of Cauchy and Piola-Kirchhoff stresses in linear theories, we have to recall some kinematic results from Section 4.7 and also derive a few more. In- troducing a positive number "that is a measure of smallness such that the displacement gradientsui;Aare of the same order of magnitude as ", we may write ui;A=O(") as"!0: (5.46) As we discovered in Section 4.7, the Eulerian and Lagrangian strains are equivalent as "!0, so from Eqs 4.58 and 4.59 we have EABiAjB=eij=O("): Examination of Eqs 5.44 and 5.45 relates stress measures tji,PAi, andsAB. To discuss this relationship in the linear case, we must find an expression for the Jacobian as we let "!0. Starting with the definition of Jin the form J=1 6"ijk"ABCFiAFjBFkC we substitute FiA=ui;A+iA, etc., to get J=1 6"ijk"ABC(ui;A+i;A)(uj;B+jB)(uk;C+kC): 176 Continuum Mechanics for Engineers Carrying out the algebra and after some manipulation of the indices J=1 6"ijk"ABC iAjBkC+3uk;CiAjB+O("2) where terms on the order of "2and higher have not been written out explicitly. Since "ijk"ijk=6and"ijk"ijC=2kC, J=1+uk;k+O("2): (5.47) Now we can evaluate Eqs 5.44 and 5.45 as "!0, that is, for the case of a linear theory. Sinceuk;kisO("). tji+O() =PAiAj+O("): (5.48) With a similar argument for Eq 5.45, we find tji=SABAiBj as"!0: (5.49) Equations 5.48 and 5.49 demonstrate that, in linear theory, Cauchy, Piola-Kirchhoff and symmetric Piola-Kirchhoff stress measures are all equivalent. 5.5 Moment of Momentum (Angular Momentum) Principle Moment of momentum is the phrase used to designate the moment of the linear momentum with respect to some point. This vector quantity is also frequently called the angular momentum of the body. The principle of angular momentum states that the time rate of change of the moment of momentum of a body with respect to a given point is equal to the moment of the surface and body forces with respect to that point. For the body shown in Fig 5.1, if we take the origin as the point of reference, the angular momentum principle has the mathematical form d dtZ V"ijkxjvkdV=Z S"ijkxjt(^n) kdS+Z V"ijkxjbkdV: (5.50) Making use of Eq 5.18 in taking the derivative on the left-hand side of the equation and applying the divergence theorem to the surface integral, after introducing the identity t(^n) k=tqknq, results in Z V"ijk[xj(˙vk-tqk;q-bk) -tjk]dV=0 which reduces to Z V"ijktkjdV=0 (5.51) because of Eq 5.22 (the equations of motion) and the sign-change property of the permu- tation symbol. Again, with Varbitrary, the integrand must vanish so that "ijktkj=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 assumed that no body or surface couples act on the body. If any such concentrated moments do act, the material is said to be a polar material, and the symmetry property of Tno longer holds. But as mentioned in Chapter 3, this is a rather specialized situation and we shall not consider it here. Fundamental Laws and Equations 177 5.6 Law of Conservation of Energy, The Energy Equation The 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 of work of the surface and body forces, plus all other energies that enter or leave the body per unit time. Other energies may include, for example, thermal, electrical, magnetic or chemical energies. In this text we consider only mechanical and thermal energies, and also, 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 the equations 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 Vof the body in Fig. 5.1 is K(t) =1 2Z Vv
vdV=1 2Z VvividV: (5.53) Also, the mechanical power, or rate of work of the body and surface forces shown in the figure is defined by the scalar P(t) =Z St(^n) ividS+Z VbividV: (5.54) Consider now, the material derivative of the kinetic energy integral ˙K=d dtZ V1 2vividV=1 2Z V: (vivi)dV =Z V(vi˙vi)dV=Z Vvi(tji;j+bi)dV where Eq 5.22 has been used to obtain the final form of the integrand. But vitji;j= (vitji);j-vi;jtjiand so ˙K=Z Vh bivi+(vitij);j-vi;jtiji dV which, if we convert the middle term by the divergence theorem and make use of the decomposition vi;j=dij+wij, may be written ˙K=Z VbividV+Z St(^n) ividS-Z VtijdijdV: (5.55) By the definition Eq 5.54 this may be expressed as ˙K+S=P (5.56) where the integral S=Z VtijdijdV=Z V(T:D)dV (5.57) is known as the stress work, and its integrand tijdijas the stress power. The balance of mechanical energy given by Eq 5.56 shows that, of the total work done by the external 178 Continuum Mechanics for Engineers forces, a portion goes toward increasing the kinetic energy, and the remainder appears as work done by the internal stresses. In general,Scannot be expressed as the material derivative of a volume integral, that is, S6=d dtZ 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 dtZ VudV =Z V˙udV (5.59) whereUis called the internal energy anduthespecific internal energy, or energy density (per unit mass), Eq 5.56 becomes d dtZ V1 2vivi+u dV=Z VbividV=Z St(^n) ividS; (5.60a) or, briefly, ˙K+˙U=P: (5.60b) Note that Eq 5.60 indicates that part of the external work Pcauses an increase in kinetic energy, and the remainder is stored as internal energy. As we shall see in Chapter 6, ideal elastic materials respond to forces in this fashion.1 For a thermomechanical continuum, we represent the rate at which thermal energy is added to a body by Q=Z VrdV -Z SqinidS: (5.61) The scalar field rspecifies the rate at which heat per unit mass is produced by internal sources and is known as the heat supply. The vector qi, called the heat flux vector, is a measure of the rate at which heat is conducted into the body per unit area per unit time across the element of surface dSwhose outward normal is ni(hence the minus sign in Eq 5.61). The heat flux qiis often assumed to obey Fourier’s law of heat conduction; qi= -;i orq= -r (5.62) whereis the thermal conductivity and ;iis the temperature gradient. But, since not all materials obey this conduction “law,” it is not universally valid. With the addition of the thermal energy consideration, the complete energy balance requires modification of Eq 5.60 which now takes the form ˙K+˙U=P+Q; (5.63a) or, when written out in detail, d dtZ V1 2vivi+u dV=Z V(bivi+r)dV+Z Sh t(^n) ivi-qinii dS: (5.63b) 1The symboluis 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 the displacement vector u. Fundamental Laws and Equations 179 If we convert the surface integral to a volume integral and make use of the equations of motion (Eq 5.22), the reduced form of Eq 5.63b is readily seen to be Z V(˙u-tijdij-r+qi;i)dV=0orZ V(˙u-T:D-r+r
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 that relates to the mechanical energy balance. Thus, we observe that the rate of work of the internal forces equals the rate at which internal energy is increasing minus the rate at which heat enters the body. As usual for an arbitrary volume V, by the argument which is standard now, upon setting the integrand of Eq 5.64a equal to zero, we obtain the field equation, ˙u-tijdij-r+qi;i=0or˙u-T:D-r+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 statement of 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 irreversible processes. For instance, the case of mechanical work being converted to heat via friction is understood, but the converse does not hold. That is, heat cannot be utilized to directly generate 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 of the flow of heat and energy. In the Kelvin-Plank statement, a device cannot be constructed to operate in a cycle and produce no other effect besides mechanical work through the exchange of heat with a single reservoir. Alternatively, in the Clausius statement, it is impossible to construct a device operating in a cycle and producing no effect other than the 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 on continua. However, in the case of continuum mechanics the restrictions are placed on the material response functions called constitutive responses. In this section, a thermodynamic parameter called entropy is introduced as a way to link 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 used in later sections to place functional restrictions on postulated constitutive responses for various materials. At any given state for the continuum there are various quantities that effect the internal energy. These might be the volume of an ideal gas or the components of the deformation gradient of a solid. In the case of the deformation gradient, the nine components represent 180 Continuum Mechanics for Engineers a deformation in the body that is storing energy. The collection of these parameters is called the thermodynamic substate and will be denoted by v 1;v2;


;vn.2 While the thermodynamic substate influences the internal energy of the body it does not completely define it. Assume that the substate plus an additional independent scalar parameter,, is sufficient to define the internal energy. This definition may be made in the form of u=f(;v1;v2;


;vn) (5.66) which is often referred to as the caloric equation of state. Parameter is called the specific entropy. Since the internal energy is unambiguously defined once entropy is adjoined to the substate, the combination plus v1;v2;


;vnconstitutes the thermodynamic state. Temperature is the result of the change in internal energy with respect to entropy =@u @: (5.67) Furthermore, partial differentiation of the internal energy with respect to the thermody- namic substate variables results in thermodynamic tensions a=@u @va: (5.68) The preceding equations can be used to write a differential form of the internal energy as follows: du=d+X aadva: (5.69) From Eqs 5.66 and 5.67 we see that both temperature and thermodynamic tensions are functions of entropy and the substate parameters. Assuming that all the functions defined in this section are continuously differentiable as many times as necessary, it is possible to solve for entropy in terms of temperature =(;va): (5.70) This result may be substituted into the caloric equation of state to yield internal energy as a function of temperature and substate parameters u=u(;va): (5.71) Using this result in Eq 5.68 allows the definition of the thermal equations of state a=a(;va) (5.72) which inverts to give the substate parameters va=va(;a): (5.73) 2Notation warning: Do not confuse these thermodynamic substate parameters with particle velocity Fundamental Laws and Equations 181 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 =u-; (5.74b) enthalpy =u-P aava;(5.74c) free enthalpy =-=u--P aava:(5.74d) These potentials are related through the relationship u- +-=0: (5.75) All of the energy potentials may be written in terms of any one of the following indepen- dent variable sets ;va;;va;;a;;a: (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,, or, equivalently, entropy, , both being a function of position and time =(XA;t) or=(XA;t) (5.77) When considered for a portion Pof the body, the total entropy is given as H=Z PdV; (5.78) and the entropy production in the portion Pis given by =Z P dV (5.79) where the scalar is the specific entropy production . The second law can be stated as follows: the time rate-of-change in the entropy equals the change in entropy due to heat supply, heat flux entering the portion, plus the internal entropy production. For a portion Pof the body, this is written as d dtZ PdV =Z Pr dV-Z @Pqini dS+Z P dV: (5.80) The entropy production is always positive which leads to a statement of the second law in the form of the Clausius-Duhem inequality d dtZ PdV >Z Pr dV-Z @Pqini dS: (5.81) This global form can easily be posed locally by the now-familiar procedures. Applying the divergence theorem to the heat flux term yields Z @Pqini dS=Z Pqi  ;idV: 182 Continuum Mechanics for Engineers Furthermore, the differentiation of the entropy term is simplified by the fact that it is a specific quantity (see Section 5.2, Eq 5.18). Thus, we write Z Ph ˙-r +qi  ;ii dV>0; (5.82) and since this must hold for all arbitrary portions of the body, and the integrand is continuous, then ˙-r +qi  ;i=˙-r +qi;i -1 2qi;i>0: Thus, the local form of the Clausius-Duhem equation is ˙-r+qi;i-1 qi;i>0: (5.83a) Often, the gradient of the temperature is written as gi=;iin which case Eq 5.83a becomes ˙-r+qi;i-1 qigi>0: (5.83b) Combining this result with Eq 5.65 brings the stress power and internal energy into the expression giving a reduced form of the Clausius-Duhem ˙-˙u+dijtij-1 qigi>0: (5.84) One final form of the Clausius-Duhem equation is obtained by using Eq 5.74b to obtain the local dissipation inequality -˙ +˙
+dijtij-1 qigi>0: (5.85) Finally, the fundamental laws are summarized in a Table 5.1. This table shows the global 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 problems are solved. All that is needed in addition to these fundamental equations is a constitutive response. 5.8 The General Balance Law The above discussion leads to a general expression for the balance law.3The balance law is 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 dtZ P (x;t)dV=Z @P(x;t)dS+Z P(x;t)dV: (5.86) 3In the literature, some authors use master balance law rather than general balance law. Fundamental Laws and Equations 183TABLE 5.1 Fundamental equations in global and local forms. Global Form Local Form Conser vation of Mass ˙m=Z V(x;t)dV =0 ˙+vi;i=0 @ @t+(vi);i=0 J=0 Linear Momentumd dtZ VvidV=Z St(^n) idS+Z VbidV ˙vi=tji;j+bi Angular Momentumd dtZ V"ijkxj(vk)dV=Z S"ijkxjt(^n) kdS+Z V"ijkxj(bk)dV "ijktij=0ortij=tji Conser vation of Energyd dtZ V1 2vivi+u dV= Z V(bivi+r)dV+Z S t(^n) ivi-qini dS˙u-tijdij-r+qi;i=0 Clausius-Duhem Equationd dtZ VdV >Z Vr dV-Z Sqi nidS ˙-r +qi  ;i>0 ˙-˙u+tijdij-1 qi;i>0 ˙ +˙-tijdij-1 qi;i>0 184 Continuum Mechanics for Engineers TABLE 5.2 Identification of quantities in the balance laws. Balance Law Property Flux Source Mass  0 0 Momentum vt(^n)(x;t;)b Angular Momentum xvxt(^n)(x;t)xb Energy1 2v
v+ut(^n)(x;t)
v-q
nb+r Entropy  -q
n r  Physically this expression states that the time rate of change of the field is equal to the flux ofthrough the boundary @Pplus the rate of increase or generation, , of within the volume V.is the source of within the volume. Similarly, by changing the equality in Eq 5.86, the balance law can be recast as an inequality with a similar interpretation. d dtZ P (x;t)dV>Z @P(x;t)dS+Z P(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,(x;t;^n), will be a tensor one order higher than the field quantity or the source term . The surface integral in Eq 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 applies pointwise to determine the field (x;t). d dt+ divv=div+: (5.88) Table 5.2 shows the relationship of the quantities in the various balance laws to the master balance law. The discussion to this point focused on a continuum body with a fixed set of material points. 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 biological materials. In biological systems, mass increases by the growth of additional material. This requires a recasting of the continuum balance equations to accommodate the addition of mass through the growth of new material. Growth can occur in biological systems through either volumetric growth as in soft tissue or by surface growth as in bone. The focus in the following will be on materials that exhibit volumetric growth. The mass balance equation requires an added source term to model the volumetric growth d dtZ PdV=Z PcdV (5.89) wherecis the mass growth per unit density. This results in a local balance of mass equation d dt+vk;k=c ord dt+divv=c: (5.90) Fundamental Laws and Equations 185 The balance of linear momentum requires an additional source term to accommodate the momentum that the added mass brings to the system d dtZ PvidV=Z @Pt(^n) idS+Z PbidV+Z PcvidV; (5.91) or d dtZ PvdV=Z @Pt(^n)dS+Z PbdV+Z PcvdV: 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 dtZ PfdV =Z P ˙f+cf
dV: (5.92) The last result follows from Eq 5.90 and is equivalent to Eq 5.18 when cis zero. This result together with t(^n) i=tjinjyields the local balance of linear momentum dvi dt=tji;j+bi: (5.93) The local form for the balance of linear momentum is unchanged by the addition of the mass. Balance of angular momentum also requires an additional source term for the added system mass. That is d dtZ P"ijkxjvkdV=Z @P"ijkxjt(^n) kdS+Z P"ijkxj(bk+cvk)dV; (5.94) or d dtZ P(x-0)vdV=Z @P(x-0)t(^n)dS+Z P(x-0)(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 mechanisms like species diffusion or micromolecular processes. The energy and entropy equations must be modified for the body as growing mass is 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 existing material at the point gives d dtZ V1 2vivi+u dV=Z P(bivi+r)dV+Z @P t(^n) ivi-qini dS +Z Pc1 2vivi+u dV+Z Pc¯hdV: (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 original material. The local form of the energy equation is ˙u-tijdij-r+qi;i-c¯h=0or˙u-T:D-r+divq-c¯h=0; (5.96) where balance of mass, linear momentum and symmetry of the stress tensor were used in arriving at this result. 186 Continuum Mechanics for Engineers The entropy inequality is d dtZ PdV >Z Pr dV-Z @Pqini dS+Z Pc0 @+= h 1 AdV: The added mass brings additional entropy and an additional entropy production,= h, like that in the energy equation. Using the standard procedure, the local form of the entropy inequality is ˙-r+qi;i-1 qi;i-c= h >0: (5.97) The specific free energy, =u-, allows the entropy production rate to be eliminated from 5.97 with Eq 5.96 giving the energy rate ˙ u. This yields an entropy production inequality of - ˙ -˙
-tijdij-1 qi;i+ch>0: The value of the variable hish=¯h-= h. The reduced entropy equation can be used to restrict 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 the continuum, a thermodynamic process must be defined. For a continuum body Bhaving material points Xa thermodynamic process is described by eight functions of the material point and time. These functions would be as follows: 1. Spatial position xi=i(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 =(X;t), 8. Temperature (always positive) =(X;t). A set of these eight functions which are compatible with the balance of linear momentum and the conservation of energy makes up a thermodynamic process. These two balance laws are given in their local form in Eqs 5.22 and 5.65 and are repeated below in a slightly different form: tji;j-˙vi= -bi; ˙u-tijdij+qi;i=r:(5.98) Fundamental Laws and Equations 187 In writing the balance laws this way, the external influences on the body, heat supply and body force, have been placed on the right-hand side of the equal signs. From this it is noted that it is sufficient to specify xi;tij;u;qi;;andand the remaining two process functionsrandbiare determined from Eqs 5.98. One of the uses for the Clausius-Duhem form of the second law is to infer restrictions on the constitutive responses. Taking Eq 5.85 as the form of the Clausius-Duhem equation we 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 of equipresence (Coleman and Mizel, 1963): An independent variable present in one constitutive equation of a material should be so present in all, unless its presence is in direct contradiction with the assumed symmetry of the material, the principle of material objectivity, or the laws of thermodynamics. For an elastic material, it is assumed that the response functions will depend on the deformation gradient, the temperature, and the temperature gradient. Thus, we assume tij=˜tij(FiA;;gi); =˜ (FiA;;gi); =˜(FiA;;gi); q =˜q(FiA;;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 response value. If an independent variable of one of the response functions is shown to contradict material symmetry, material frame indifference, or the Clausius-Duhem inequality it is removed from that function’s list. In using Eq 5.85, the derivative of must be formed in terms of its independent vari- ables ˙ =@˜ @FiA˙FiA+@˜ @˙+@˜ @gi˙gi: (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 ˙ =@˜ @FiALijFjA+@˜ @˙+@˜ @gi˙gi: (5.101) Substitution of Eq 5.100 into 5.85 and factoring common terms results in -@˜ @+˜ ˙+ tij-@˜ @FiAFjA Lij-@˜ @gi˙gi-1 ˜qigi>0: (5.102) Note that in writing Eq 5.102 the stress power has been written as tijLijrather thantijdij. This can be done because stress is symmetric and adding the skew–symmetric part of Lij is essentially adding zero to the inequality. The velocity gradient is used because the partial derivative of the free energy with respect to the deformation gradient times the transposed deformation gradient is not, in general, symmetric. The second law must hold for every thermodynamic process, which means a special case may be chosen which might result in further restrictions placed on the response functions. That this is the case may be demonstrated by constructing displacement and temperature fields as such a special case. Define the deformation and temperature field as follows: xi=(XA;t)=YA+AiA(t) [XA-YA]; 188 Continuum Mechanics for Engineers =(XA;t)=(t)+[AAi(t)ai(t)] [XA-YA]: (5.103) Here,XAandYAare the positions in the reference configuration of material points Xand Y, and function AiA(t)is an invertible tensor, ai(t)is a time dependent vector, and (t) is a scalar function of time. At the spatial position YA, the following is readily computed (YA;t)=(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 (yi;t)=(t)+ai(t) [xi-yi]: (5.106) Thus, the gradient of the temperature at material point Yis written as ;i=g;i=ai(t): (5.107) From Eqs 5.104, 5.105, and 5.106 it is clear that quantities ,gi;andFiAcan be inde- pendently chosen. Furthermore, the time derivatives of these quantities may also be arbitrarily chosen. Because of the assumed continuity on the response functions, it is possible to arbitrarily specify functions u,gi;andFiAand 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 the temperature is constant. This means Lij=0and ˙=0. Furthermore, assume the temper- ature gradient to be some arbitrary constant gi=g0 i. Eq 5.102 becomes -@˜ FiA;0;g0 i
@gi˙gi-1 ˜qi FiA;0;g0 i
gi>0: Again, take advantage of the fact that the second law must hold for all processes. Since ˙giis arbitrary it may be chosen to violate the inequality. Thus, the temperature gradient time derivative coefficient must be zero @˜ FiA;0;g0 i
@gi=0: (5.108) Sinceg0 iwas taken to be an arbitrary temperature gradient, Eq 5.108 implies that the free energy is not a function of the temperature gradient. That is, @˜ (FiA;0;gi) @gi=0 which immediately leads to =˜ (FiA;). 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=0and ˙g=0, but the temperature gradient is an arbitrary constant gi=g0 i. For this case, the Clausius-Duhem inequality is written as -@˜ (FkB;) @+˜ FkB;;g0 k
 ˙-1 ˜qi FkB;;g0 k
g0 i>0 (5.109) which must hold for all temperature rates, ˙. Thus, the entropy response function may be solved in terms of the free energy ˜ FkB;;g0 k
= -@˜ (FkB;) @; Fundamental Laws and Equations 189 and since the free energy is only a function of the deformation gradient and temperature the entropy must be a function of only those two as well. That is, =˜(FkB;)= -@˜ @: (5.110) One more application of the Clausius-Duhem inequality for a special process will lead to an expression for the Cauchy stress in terms of the free energy. For this process, select the temperature gradient to be an arbitrary constant and the time rate-of-change of the temperature gradient to be identically zero. Eq 5.102 becomes  ˜tij FkB;;g0 k
-@˜ @FiAFjA Lij-1 ˜qi FkB;;g0 k
g0 i>0 (5.111) which must hold for all velocity gradients Lij. Picking aLijthat would violate the in- equality unless the coefficient of Lijvanishes implies tij=˜tij(FkB;)=@˜ @FiAFjA: (5.112) In arriving at Eq 5.111 it was noted that the free energy had already been shown to be independent of the temperature gradient by virtue of Eq 5.108. Thus, gkwas dropped from 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 restriction for an elastic material is found to be ˜qigi60: (5.113) 5.10 Invariance The 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 a single stress state. This invariance is crucial for consolidating different stress components into 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 when a superposed rigid body motion is applied to all material points, has significant results associated with them. The most important of these consequences might be restrictions placed 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 continuum body’s response to applied forces or prescribed motion must be the same as observed from 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 the body the observer maintains the same reference frame. Here, each material point has a superposed motion added to it. The forces applied to the body are rotated with the superposed motion. 190 Continuum Mechanics for Engineers Both of these methods produce the same restrictions on constitutive responses within the context of this book. In this text, the method of superposed rigid body motion will be presented as the means for enforcing invariance. The superposition of a rigid motion along with a time shift can be applied to the basic definition of the motion xi=i(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 shift is imposed on the motion. After the application of the superposed motion, the position of the material point becomes x+ iat timet+=t+awhereais 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 “+” is used. The motion in terms of the superposed motion is written as x+ i=+ i(XA;t): (5.115) Assuming sufficient continuity, the motion can be written in terms of the current config- uration since XA=-1(xi;t). That is to say x+ i=+ i(XA;t) = ˜+ i(xi;t) (5.116) where ˜+ iis written because the substitution of XA=-1(xi;t)results in a different function than + i. To represent the relative distance between two particles, a second material point is selected. In an analogous manner, it is straight-forward to determine y+ i=+ i(YA;t) = ˜+ i(yi;t): (5.117) Relative distance between material particles XAandYAis written as x+ i-y+ i
x+ i-y+ i
= ˜+ i(xi;t)-˜+ i(yi;t) ˜+ i(xi;t)-˜+ i(yi;t) (5.118) which is the dot product of the vector between positions xiandyi. Since the material pointsXAandYAwere arbitrarily chosen, quantities xiandyiare independent. Subse- quent differentiation of Eq 5.118 with respect to xithenyiresults in @˜+ i(xj;t) @xp@˜+ i(yj;t) @yq=pq: (5.119) Since this must hold for all pairs of material points, it is possible to set @˜+ i(xj;t) @xp=@˜+ i(yj;t) @yp=Qip(t): (5.120) Use of Eq 5.120 in Eq 5.119 shows that the matrix Qip(t)is orthogonal. Furthermore, since the special case of the superposed motion as a null motion, that is ˜ + i(xi;t) =xi, then matrixQip(t)must be proper orthogonal having Qip(t)Qiq(t) =pqand det(Qip) = +1. To come up with a particular form for the superposed motion, Eq 5.120 may be spatially integrated to obtain ˜+ i(xi;t) =a(t)+Qim(t)xm; (5.121a) or x+ i=ai+Qimxm: (5.121b) Fundamental Laws and Equations 191 x2 x1 x1�x2�P P�P ‛ θθ c(t) FIGURE 5.2 Reference frames Ox1x2x3andO+x+ 1x+ 2x+ 3differing by a superposed rigid body motion. Vectoraimay be written in the alternative form ai(t)=c+ i t+
-Qim(t)cm(t) (5.122) yielding x+ i=c+ i+Qim[xm-cm] (5.123) where QimQin=QmiQni=mn and det (Qij)=1: (5.124) A similar development of the superposed motion can be obtained by assuming two Cartesian reference frames Ox1x2x3andO+x+ 1x+ 2x+ 3which 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+ iandpmare defined as shown in Fig. 5.2. Here, Qimis simply the matrix of the direction cosines between Ox1x2x3andO+x+ 1x+ 2x+ 3. Example 5.2 Show that the superposed rigid body motion dened by Eq 5.123 is distance and angle preserving. Solution Consider the distance squared between material points XAandYAin terms of the superposed motion x+ i-y+ i
x+ i-y+ i
=Qim(xm-ym)Qin(xn-yn) =QimQin(xm-ym) (xn-yn): 192 Continuum Mechanics for Engineers SinceQimis orthogonal x+ i-y+ i
x+ i-y+ i
=nm(xm-ym) (xn-yn) =(xm-ym) (xm-ym) where the delta substitution property has been used. Thus, distance is preserved in the superposed motion. Three material points, XA,YA, andZA, are used to show that angles are preserved in the superposed rigid body motion. Let +be the angle included between vectors x+ i-y+ iandx+ i-z+ i. Use of the denition of the dot product gives a convenient way to represent the angle + cos+=x+ i-y+ i jx+n-y+njx+ i-z+ i jx+n-z+nj: Note the \n" indices in the denominator do not participate in the summation since they are inside the vector magnitude operator. Direct substitution for Eq 5.121 followed by utilizing the orthogonality of Qimand use of results from the rst part of this example gives cos+=Qim(xm-ym)Qin(xn-zn) jxn-ynjjxn-znj =QimQin(xm-ym) (xn-zn) jxn-ynjjxn-znj =mn(xm-ym) (xn-zn) jxn-ynjjxn-znj =(xm-ym) (xm-zm) jxn-ynjjxn-znj =cos: 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 to t+ 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+: Fundamental Laws and Equations 193 Recall the definition t+=t+afrom which it is obvious that dt=dt+=1. As a result of this, an expression for the velocity under a superposed rigid body motion is obtained by taking the time derivatives of the bracketed term of the preceding equation: v+ i=˙ai(t)+˙Qim(t)xm+Qim˙xm =˙ai(t)+˙Qim(t)xm+Qimvm: (5.127) Define ij(t)=˙Qim(t)Qjm(t); (5.128) or, written an alternative way by post-multiplying by Qjk ij(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 field of a continuum undergoing a superposed rigid body rotation v+ i=˙ai+ ijQjkxk+Qimvm: (5.130) Note that ijis skew–symmetric by taking the time derivative of the orthogonality condition for Qij. That is, d dt[QimQin] =d dt[1] =0: When the derivatives are taken ˙QimQin+Qim˙Qin=0; and the definition of ˙Qimused from Eq 5.129 substituted into this expression yields Qim[ ji+ ij]Qjm=0: This is true only if the bracketed term is zero, thus ij= - ji: (5.131) The fact that ijis skew–symmetric means that it has an axial vector defined by !k= -1 2"ijk ij (5.132) which may be inverted to give ij= -"ijk!k: (5.133) In the case of rigid body dynamics, the axial vector !kcan be shown to be the angular velocity of the body (see Problem 5.37). For later use in constitutive modeling, various kinematic quantities’ properties under superposed rigid body motions will be needed. Here, a derivation of the superposed rigid body 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 xiusing Eq 5.121b to write v+ i=˙ai+ ij[x+ j-aj] +Qijvj (5.134) =Qij+˙Qimxm+˙ai- ijaj (5.135) =Qijvj+˙Qimxm+ci (5.136) 194 Continuum Mechanics for Engineers where Eq 5.128 has been used in going from Eq 5.134 to 5.135, and ci=˙a- ijajhas been used in going from Eq 5.135 to 5.136. Writing the velocity in this form allows for a more convenient computation of the velocity gradient in the superposed reference frame @v+ i @x+m= ij@x+ j @x+m+Qij@vj @x+m: (5.137) Note that @x+ j @x+m=jm; 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+ kin Eq 5.121b. Use of these in Eq 5.137 followed by the delta substitution property yields @v+ i x+m= im+QijQmnvj;n= im+QijQmn[djn+wjn] (5.138) wheredjnandwjnare 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. From Eqs 5.138 and 5.131 it is clear that d+ ij=QimQjndmn; (5.139) and w+ ij= ij+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 in more 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 as the kinematic quantities demonstrated above. This is a result of starting with the stress vector which is a force. It is assumed that forces transform as a generic vector in the form of t+ i=Qijtj: Since stress is a measure of force per unit area, a regression back to kinematic results is necessary to determine how a differential area transforms under superposed rigid body motion. 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 Fundamental Laws and Equations 195 da+ k=da+n+ kJ+@XK @x+ kdAK (5.141) where reference quantities do not change, da+is the infinitesimal area, and n+ kis the unit normal vector to the area all in the superposed rigid body motion state. This expression will 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+ kis the inverse of the deformation gradient as represented in the 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 xkby pre-multiplying by Qik. Differentiation of the result shows that the last term of Eq 5.142 isQik. Thus @XK @x+ k=@XK @xjQkj; or (FkK)-1=F-1 jKQkj: (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. In the superposed rigid body motion frame, the stress vector t+ ican 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 vector in 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. Multiplying the remaining terms by Qmkresults in the following expression for stress component transformation under superposed rigid body motion: t+ ij=QimQjntmn: (5.149) 196 Continuum Mechanics for Engineers 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 is used. The stress rate must be objective meaning ˙t+ ij=˙QimQjntmn+Qim˙Qjntmn+QimQjn˙tmn: (5.150) It is clear from this equation that the stress rate is not objective even though the stress is objective. This result is serious since the stress rate as shown in Eq 5.150 could not be used 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 ˙Qijin 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 side of 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-wmqtqnis objective which leads to the definition of the so called Jaumann stress rate r tij=˙tij-Wiqtqj-tiqWjq: (5.154) There are several other stress rate definitions satisfying objectivity. For example, the Green-Naghdi stress rate is given as
tij=˙tij+ iqtqj+tiq jq (5.155) where iqis defined in Eq 5.128. 5.11 Restrictions on Constitutive Equations from Invariance It was seen in Section 5.9 that the second law of thermodynamics places restrictions on the form of constitutive response functions. Material frame indifference, or superposed rigid body motion, may also place restrictions on the independent variables of the response function as was stated in the principle of equipresence. In Section 5.10, the behavior of many of the quantities used in continuum mechanics undergoing a superposed rigid body motion was presented. This section will examine the response functions of bodies undergoing a superposed rigid body motion. In particular, what restrictions are placed on the constitutive independent variables given in Eq 5.99. Under a superposed rigid body motion, scalars are unaffected allowing the following to be written Fundamental Laws and Equations 197 u+=u; +=; +=: (5.156) This being the case, it is clear from Eq 5.74a through 5.74d that ,, andwould be un- affected by the superposed rigid body motion. The remaining quantities of the response functions of Eq 5.99 and their independent variables are affected in different ways from the 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 in slightly different form): F+ iA=QijFjA; (5.157) ˙F+ iA=Qij˙FjA+ ijQjkFkA; (5.158) L+ im=QijLjnQnm+ im; (5.159) t+ ij=QiktklQlj; (5.160) q+ i=Qijqj; (5.161) g+ i=Qijgj; (5.162) where ijis 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+ +;F+ iA
=u(;FiA); (5.163a) + +;F+ iA
=(;FiA); (5.163b) t+ ij +;F+ mA
=Qiktkl(;FmA)Qlk; (5.163c) q+ i +;F+ iA;g+ i;L+ kl
=Qimqm(;FiA;gi;Lkl): (5.163d) In writing these equations it is noted that the roles of andcan be interchanged be- cause of assumed continuity. Also, restricting the independent variable list of Eqs 5.163a through 5.163c to the entropy (or temperature) and the deformation gradient has been shown to be a general case (Coleman and Mizel, 1964). Finally, the set of constitutive functions may be in terms of different response functions. That is, the elastic material considered in Section 5.9 had response functions ,tij,, andipostulated. However, it could have just as easily been postulated as a function of u,,tij, andqi. As an example of how Eqs 5.163 could restrict the independent variables consider a fluid whose stress response function is assumed to be a function of density, , velocity,vi, and velocity gradient, Lij. With these assumptions, the restrictions of Eq 5.163c would be t+ ij +;v+ k;d+ mn;w+ mn
=Qiptpl(;vk;dmn;wmn)Qlj (5.164) where the velocity gradient has been decomposed into its symmetric and skew–symmetric parts,dmnandwmn, respectively. Using the results of Problem 5.34, Eqs 5.136, 5.139, and 198 Continuum Mechanics for Engineers 5.140 results in ˜tij(;vk+ck;QmpdpqQqn;QmpwpqQqn+ mn)=Qip˜tpl(;vk;dmn;wmn)Qlj: (5.165) Since Eq 5.165 must hold for all motions, a specific rigid body rotation may be chosen to reduce the constitutive assumption of Eq 5.164. For this purpose, suppose that Qij=ij and thus ˙Qij=0. Using this motion and Eq 5.165 implies ˜tij(;vk+ck;dmn;wmn)=˜tij(;vk;dmn;wmn) (5.166) where it is noted that Eq 5.128 has been used. In this case, vector ckis simply equal to ˙akthe time derivative of the superposed rigid body motion integration factor (see Eq 5.121). This arbitrary nature would allow for Eq 5.166 to be violated if the stress function has velocity as an independent variable. Thus, velocity must be removed from the independent variable list for stress leaving tij=˜tij(;dmn;wmn): (5.167) Again, with the modified response function Eq 5.167, the invariance condition under superposed rigid body motion may be written as ˜tij(;QmpdpqQqn;QmpwpqQqn+ mn)=Qip˜tpl(;dmn;wmn)Qlj (5.168) which must hold for all motions. Select a motion such that Qij=ijas before, but now require ij6=0. Substitution of this into Eq 5.168 leaves ˜tij(;dmn;wmn+ mn)=˜tij(;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 function will yield further information. In this case, ˜tij(;QmpdpqQqn)=Qip˜tpl(;dmn)Qlj: (5.170) SinceQijis a proper orthogonal tensor, and this equation would hold if Qijwere replaced by its negative, so the stress must be an isotropic function of dmn. The most general form of a second-order, isotropic tensor function of dmnmay now represent the stress response tij= -p ()ij+()dkkij+2()dij (5.171) wherep,, andare functions of density and would represent the viscosity coefficients. 5.12 Constitutive Equations The global balance laws and resulting field equations developed earlier in this chapter are applicable 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 equations which specify the mechanical and thermal properties of particular materials based upon their internal constitution. Mathematically, the usefulness of these constitutive equations Fundamental Laws and Equations 199 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 as models for the behavior of real materials. However, it is not possible to write down one equation 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 perhaps better to think of constitutive equations as being representative of a particular behavior rather than of a particular material. In previous sections of this Chapter, the foundations of constitutive assumptions have been addressed. On that fundamental level, it is not possible to discuss in detail the fundamental derivation of all the constitutive models that an engineer would want to be familiar with. Instead, the theoretical background and definitions were given along with a few specific cases examined. This allows the inductive student to grasp the concepts presented. Since only a few constitutive models were considered from a fundamental basis, this section is devoted to a brief survey of constitutive equations. This acts as an introduction for subsequent chapters when various constitutive models are discussed as applications of 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=Cijkmkm (5.172) where theCijkm are the elastic constants representing the properties of the body. For isotropic behavior, Eq 5.172 takes the special form tij=ijkk+2ij (5.173) in whichandare coefficients that express the elastic properties of the ma- terial. (b) the linear viscous fluid, ij=Kijmndmn (5.174) whereijis the shearing stress in the fluid and the constants Kijmn represents its viscous properties. For a Newtonian fluid, ij=ijdkk+2dij (5.175) whereandare viscosity coefficients. (c)plastic stress-strain equation, dP ij=Sijd (5.176) wheredP ijis the plastic strain increment, Sijthe deviator stress, and da pro- portionality constant. (d)linear viscoelastic differential-operator equations 200 Continuum Mechanics for Engineers fPgSij=2fQgij (5.177a) tii=3Kii (5.177b) where fPgandfQgare differential time operators of the form fPg=NX i=0pi@i @ti(5.178a) fQg=MX i=0qi@i @ti(5.178b) and in which the coefficients piandqi(not necessarily constants) represent the viscoelastic properties. Also, Kis the bulk modulus. Note further that this pair of equations specifies separately the deviatoric and volumetric responses. (e)linear viscoelastic integral equations ij=Zt 0 s(t-t0)@Sij @t0dt0(5.179a) ii=Zt 0 v(t-t0)@tij @t0dt0(5.179b) where the properties are represented by sand v, the shear and volumetric creep 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 a group, the important field equations in both indicial and symbolic notation: (a) the continuity equation (Eq 5.14) @ @t+ (vk);k=0or@ @t+r
(v)=0 (5.180) (b) the equations of motion (Eq 5.22) tji;j+bi=˙viorr
T+b=˙v (5.181) (c) the energy equation (Eq 5.65) ˙u-tijdij-r+qi;i=0or˙u-T:D-r+r
q=0 (5.182) Fundamental Laws and Equations 201 If we assume the body forces biand distributed heat sources rare prescribed, the above collection consists of fiveindependent equations involving fourteen unknowns, namely, ;vi;tij;qiandu. In addition, in a non-isothermal situation, the entropy and tempera- ture field=(x;t)have to be taken into consideration. For the isothermal theory, eleven equations are needed in conjunction with the five field equations listed above. Of these, six are constitutive equations, three are temperature-heat conduction equations (Fourier’s law), 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 constitutive equations provide a determinate set of ten kinematic-mechanical equations for the ten unknowns;viandtij. It is with such rather simple continuum problems that we shall concern 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 with heat 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 in thermodynamics,” J. Chem. Phys., Vol. 40, pp. 1116–1125 (4) [4] Dienes, J. K. (1979), ”On the Analysis of Rotation and Stress Rate in Deforming Bodies,” Acta Mech., Vol. 32, pp. 217-232 [5] Flanagan, D. P . and Taylor, L. M., (1987), ”An Accurate Numerical Algorithm for Stress Integration with Finite Rotations,” Computer Methods in Applied Mechanics and Engineering, Vol. 62, pp. 305-320 [6] Green, A. E. and Naghdi, P . M., (1979), “A note on invariance under superposed rigid 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 finite deformation 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 der Physik, 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 202 Continuum Mechanics for Engineers Problems Problem 5.1 Determine the material derivative of the ux of any vector property Q ithrough the spatial areaS. Specically, show that d dtZ SQ inidS=Z S˙Q i+Q ivk;k-Q kvi;k
nidS in agreement with Eq 5.5. Problem 5.2 Let the property P ij:::in Eq 5.1 be the scalar 1so that the integral in that equation represents the instantaneous volume V. Show that in this case ˙Pij:::=d dtZ VdV=Z Vvi;idV: Problem 5.3 Verify the identity "ijkak;j=2(˙wi+wiwj;j-wjvi;j); and by using this identity as well as the result of Problem 5.1, prove that the material derivative of the vorticity ux equals one half the ux of the curl of the acceleration; that is, show that d dtZ SwinidS=1 2Z S"ijkak;jnidS: Problem 5.4 Making use of the divergence theorem of Gauss together with the identity @wi @t=1 2"ijkak;j-"ijk"kmq(wmvq);j show that @ @tZ VwidV=Z S1 2"ijkak+wjvi-wivj njdS: Problem 5.5 Show that the material derivative of the vorticity of the material contained in a volume V is given by d dtZ VwidV=Z S1 2"ijkak+wjvi njdS: Problem 5.6 Given the velocity eld v1=ax1-bx2,v2=bx1+ax2,v3=cq x2 1+x2 2 Fundamental Laws and Equations 203 wherea,b, andcare constants, determine (a) whether or not the continuity equation is satised, (b) whether the motion is isochoric. Answer (a) only when =0e-2at(b) only ifa=0 Problem 5.7 For a certain continuum at rest, the stress is given by tij= -p0ij wherep0is a constant. Use the continuity equation to show that for this case the stress power may be expressed as tijdij=p0˙ : Problem 5.8 Consider the motion xi= (1 +t=k)Xiwherekis a constant. From the conservation of mass and the initial condition =0att=0, determine as a function of 0,t, andk. Answer =0k3 (k+t)3 Problem 5.9 By combining Eqs 5.17b and 5.13, verify the result presented in Eq 4.161. Problem 5.10 Using the identity "ijkak;j=2(˙wi+wivj;j-wjvi;j) as well as the continuity equation, show that d dtwi  ="ijkak;j+2wjvi;j 2: Problem 5.11 State 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 @(vi) @t=(tij-vivj);j+bi: 204 Continuum Mechanics for Engineers Problem 5.12 Determine the form which the equations of motion take if the stress components are given bytij= -pijwherep=p(x;t). Answer ai= -p;i+bi Problem 5.13 Let a material continuum have the constitutive equation tij=ijdkk+2dij whereandare constants. Determine the form which the equations of motion take in terms of the velocity gradients for this material. Answer ˙vi=bi+ (+)vj;ij+vi;jj Problem 5.14 Assume that distributed body moments miact throughout a continuum in motion. Show that the equations of motion are still valid in the form of Eq 5:22, but that the angular momentum principle now requires "ijktjk+mi=0 implying that the stress tensor can no longer be taken as symmetric. Problem 5.15 For a rigid body rotation about the origin, the velocity eld may be expressed by vi= "ijk!jxkwhere!jis the angular velocity vector. Show that for this situation the angular momentum principle is given by Mi=: (!jIij) whereMiis the total moment about the origin of all surface and body forces, and Iijis the moment of inertia of the body dened by the tensor Iij=Z V(ijxkxk-xixj)dV: Problem 5.16 Determine expressions for the stress power tijdijin terms of (a) the rst Piola-Kircho stress tensor, (b) the second Piola-Kircho stress tensor. Fundamental Laws and Equations 205 Answer (a)tijdij=˙FiAPiA=0(b)tijdij=sAB˙CAB=20 Problem 5.17 Show 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 2!i!jIij whereIijis the inertia tensor dened in Problem 5.15. Problem 5.18 Show that one way to express the rate of change of kinetic energy of the material currently occupying the volume Vis by the equation ˙K=Z VbividV-Z Vtijvi;jdV+Z Svit(^n) idS and give an interpretation of each of the above integrals. Problem 5.19 Consider a continuum for which the stress is tij= -p0ijand which obeys the heat conduction law qi= -;i. Show that for this medium the energy equation takes the form ˙u= -p0vi;i+r+;ii: Problem 5.20 If 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 the velocityviand integrating the resulting equation term-by-term over the volume V, we obtain the energy equation. Verify that one form of the result is 1 2: (v
v)+tr(T
D)-b
v-div(T
v)=0: Problem 5.21 Show that for a continuum body experiencing volumetric growth d dtZ PfdV =Z P ˙f+cf
dV: Problem 5.22 Using 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 dvi dt=tji;j+bi: 206 Continuum Mechanics for Engineers Problem 5.23 Using 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 5.24 If a continuum has the constitutive equation tij= -pij+dij+dikdkj wherep,andare constants, and if the material is incompressible (dii=0), show that tii= -3p -2IID whereIIDis the second invariant of the rate of deformation tensor. Problem 5.25 Starting with Eq 5.173 for isotropic elastic behavior, show that tii= (3 +2)ii; and using this result, deduce that ij=1 2 tij- 3+2ijtkk : Problem 5.26 For a Newtonian uid, the constitutive equation is given by tij= -pij+ij = -pij+ijdkk+2dij(see Eq 5.175) : By substituting this constitutive equation into the equations of motion, derive the equation ˙vi=bi-p;i+ (+)vj;ji+vi;jj: Problem 5.27 Combine Eqs 5.177a and Eq 5.177b into a single viscoelastic constitutive equation having the form tij=ijfRgkk+fSgij where the linear time operators fRgandfSgare given, respectively, fRg=3KfPg-2fQg 3fPgand fSg=2fQg fPg: Problem 5.28 Assume 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=0f(t)where0is a constant and f(t)some function of time. Assuming that t22=t33=t13=t23=t12=0, determine11,22, and33as functions of fPg,fQg, andf(t). Fundamental Laws and Equations 207 Answer 11=3KfPg+fQg 9KfQg0f(t) 22=33=2fQg-3KfPg 18KfQg0f(t) Problem 5.29 Use the denition of the free energy along with the reduced form of the Clausius-Duhem equation to derive the local dissipation inequality. Problem 5.30 The constitutive model for a compressible, viscous, and heat-conducting material is dened by =˜ ;gk;FiA;˙FiA
; =˜ ;gk;FiA;˙FiA
; tij=˜tij ;gk;FiA;˙FiA
; qi=˜qi ;gk;FiA;˙FiA
: Deduce the following restrictions on these constitutive response functions: (a)@˜ ;gk;FkB;˙FkB
@gi=0, (b)@˜ ;gk;FkB;˙FkB
@˙FiA=0, (c)˜(;FiA)= -@˜ (;FkB) @, (d) ˜tij=FjA@˜ (;FkB) @FiA, (e)-1 ˜qi(;gk;FiA;0)gi>0. Problem 5.31 Assume the constitutive relationships u=˜u(CAB;); =˜(CAB;); tij=˜tij(CAB;); qi=˜qi(CAB;;gk); 208 Continuum Mechanics for Engineers for an elastic material. Use the Clausius-Duhem inequality to show =@˜u @; u=˜u(CAB;); tij=1 2FiA@˜u @CAB+@˜u @CBA FjB; ˜qigi>0: Problem 5.32 Use 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 5.33 Show that the Jacobian transforms as follows under a superposed rigid body motion: J+=J: Problem 5.34 Utilize the result of Problem 5.33 along with the law of conservation of mass to show that +=. Problem 5.35 Show that the gradient of the stress components transforms under superposed rigid body motion as follows: @t+ ij @x+ j=Qim@tmn @xn: Problem 5.36 Use the superposed rigid body motion denitions to show the following relationships: (a)C+ AB=CAB, (b)U+ AB=UAB, (c)R+ iA=QijRjA, (d)B+ ij=QimBmkQkj. Problem 5.37 In the context of rigid body dynamics, consider the motion dened by x(X;t)=X Fundamental Laws and Equations 209 along with x=QT(t) x+-a(t) show that v+=˙a(t)+!(t) x+-a(t) where!is the angular velocity of the body. This page intentionally left blank 6 Linear Elasticity Elastic behavior is characterized by the following two conditions: (1) where the stress in a material is a unique function of the strain, and (2) where the material has the property for complete recovery to a “natural” shape upon removal of the applied forces. If the behavior of a material is not elastic, we say that it is inelastic. Also, we acknowledge that elastic behavior may be linear ornon-linear. 6.1 Elasticity, Hooke’s Law, Strain Energy Figure 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, the conditions 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 constitutive equation for elastic behavior in its most general form as T=G() (6.1) whereGis a symmetric tensor-valued function and is any one of the various strain tensors we introduced earlier. However, for the response function Gin this text we con- sider only that case of Eq 6.1 for which the stress is a linear function of strain. Also, we assume that, in the deformed material, the displacement gradients are everywhere small compared 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 Stress StrainLOAD UNLOAD (a) Linear elastic Stress StrainLOAD UNLOAD (b) Nonlinear elastic Stress StrainLOAD UNLOAD (c) Inelastic FIGURE 6.1 Uniaxial loading-unloading stress-strain curves for various material behaviors. 211 212 Continuum Mechanics for Engineers strain tensor defined in Eq 4.62, which we repeat here: ij=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 linear elastic behavior as tij=Cijkmkm orT=C (6.3) where the tensor of elastic coefficients Cijkm has34=81components. However, due to the symmetry of both the stress and strain tensors, it is clear that Cijkm=Cjikm=Cijmk (6.4) which reduces the 81possibilities to 36distinct coefficients at most. We may demonstrate the tensor character of Cby a consideration of the elastic consti- tutive equation when expressed in a rotated (primed) coordinate system in which it has the form t0 ij=C0 ijkm0 km: (6.5) But by the transformation laws for second-order tensors, along with Eq 6.3, t0 ij=aiqajstqs=aiqajsCqskmkm =aiqajsCqskmapkanm0 pn which by a direct comparison with Eq 6.5 provides the result C0 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 also ignore strain-rate effects and consider the components Cijkm to be at most a function of position. 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 constitutive law 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 on the stress and strain components and double subscripts on the elastic constants. To this end, we define t11=t1; t23=t32=t4; t22=t2; t13=t31=t5; (6.7a) t33=t3; t12=t21=t6; and 11=1; 223=232=4; 22=2; 213=231=5; (6.7b) 33=3; 212=221=6; where the factor of two on the shear strain components is introduced in keeping with Eq 4.71. From these definitions, Hooke’s law is now written t=CorT=C (6.8) Linear Elasticity 213 with Greek subscripts having a range of six. Note the font change in Cto that of a matrix. This is done to account for the fact that the reduced representation stiffness matrix does not transform as a tensor. In matrix form Eq 6.8 appears as 2 6666664t1 t2 t3 t4 t5 t63 7777775=2 6666664C11C12C13C14C15C16 C21C22C23C24C25C26 C31C32C33C34C35C36 C41C42C43C44C45C46 C51C52C53C54C55C56 C61C62C63C64C65C663 77777752 66666641 2 3 4 5 63 7777775: (6.9) We repeat that the array of the 36 constants C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 tijdij (6.10a) which for small-deformation theory, by Eq 4.150, becomes ˙u=1 tij˙ij (6.10b) The internal energy uin these equations is purely mechanical and is called the strain energy (per unit mass). Recall now that, by the continuity equation in Lagrangian form, 0=Jand also that to the first order of approximation J=detF=det iA+@ui @XA =1+@ui @XA: (6.11) Therefore, from our assumption of small displacement gradients, namely @ui=@XA1, we may take J1in the continuity equation to give =0a constant in Eqs 6.10a and 6.10b. For elastic behavior under the assumptions we have imposed, the strain energy is a function of the strain components only, and we write u=u(ij) (6.12) so that ˙u=@u @ij˙ij; (6.13) and by a direct comparison with Eq 6.10b we obtain 1 tij=@u @ij: (6.14) The strain energy density, W(strain energy per unit volume) is defined by W=0u; (6.15) and since=0, a constant, under the assumptions we have made, it follows from Eq 6.14 that tij=@u @ij=@W @ij: (6.16) 214 Continuum Mechanics for Engineers It is worthwhile noting at this point that elastic behavior is sometimes defined on the basis 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 a hyperelastic 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 with our basic restriction to infinitesimal deformations, we shall develop the linearized form of Eq 6.16. Expanding Wabout the origin, we have W(ij)=W(0)+@W(0) @ijij+1 2@2W(0) @ij@kmijkm+


; (6.17) and, from Eq 6.16, tij=@W @ij=@W(0) @ij+@2W(0) @ij@kmkm+


: (6.18) It is customary to assume that there are no residual stresses in the unstrained state of the material so that tij=0whenij=0. Thus, by retaining only the linear term of the above expansion, we may express the linear elastic constitutive equation as tij=@2W(0) @ij@kmkm=Cijkmkm (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 the symmetries expressed by Eq 6.4, but now we also have Cijkm=Ckmij (6.20) due to the fact that @2W(0) @ij@km=@2W(0) @km@ij: Thus, the existence of a strain energy function reduces the number of distinct components ofCijkm from36to21. Further reductions for special types of elastic behavior are ob- tained from material symmetry properties in the next section. Note that by substituting Eq 6.19 into Eq 6.17 and assuming a linear stress-strain relation, we may now write W(ij)=1 2Cijkmijkm=1 2tijij (6.21a) which in the notation of Eq 6.8 becomes W()=1 2C=1 2t; (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 Constants If 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 Linear Elasticity 215 a different set of circumstances its behavior may not be elastic. Furthermore, if a body’s elastic 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. For such materials, the constitutive equation has only two elastic constants. A material that is not isotropic is called anisotropic; we shall define some of these based upon the degree of elastic symmetry each possesses. In general, an isotropic tensor is defined as one whose components are unchanged by any orthogonal transformation from one set of Cartesian axes to another. Zero-order tensors of any order – and all zeroth-order tensors (scalars) – are isotropic, but there are no first-order isotropic tensors (vectors). The unit tensor I, having Kronecker deltas as components, and any scalar multiple of Iare the only second-order isotropic tensors (see Problem 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 of Kronecker deltas which we now introduce as the prototype for C, namely, Cijkm=ijkm+(ikjm+imjk)+(ikjm-imjk) (6.22) where,, andare scalars. But by Eq 6.4, Cijkm=Cjikm=Cijmk . This implies that  must be zero for the stated symmetries since by interchanging iandjin the expression (ikjm-imjk) =(jkim-jmik) we see that= - and, consequently, =0. Therefore, inserting the reduced Eq 6.22 into Eq 6.3, we have tij=(ijkm+ikjm+imjk)km: But by the substitution property of ij, this reduces to tij=ijkk+2ij (6.23) which is Hooke’s law forisotropic elastic behavior. As mentioned earlier, we see that for isotropic elastic behavior the 21constants of the generalized law have been reduced to two,and, known as the Lam´ e constants. Note that for an isotropic elastic material Cijkm=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 stress and strain tensors coincide, and develop an expression for the relationship among their principal values. Solution Let^n(q)(q=1;2;3 )be unit normals in the principal directions of ij, and associated with these normals the corresponding principal values are (q)where (q=1;2;3 ). From Eq 6.23 we form the dot products tijn(q) j=(ijkk+2ij)n(q) j =n(q) ikk+2ijn(q) j: 216 Continuum Mechanics for Engineers Butn(q) jand(q)satisfy the fundamental equation for the eigenvalue problem, namely,ijn(q) j=(q)ijn(q) j, so that now tijn(q) j=kkn(q) i+2(q)n(q) i = kk+2(q) n(q) i; and becausekk=(1)+(2)+(3)is the rst invariant of strain, it is constant for alln(q) iso that tijn(q) j=  (1)+(2)+(3) +2(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)= (1)+(2)+(3) +2(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 iiin terms oftiifrom Eq 6.23 by setting i=jto yield tii=3kk+2ii= (3 +2)ii: (6.24) Now, by solving Eq 6.23 for ijand substituting from Eq 6.24, we obtain the inverse form of the isotropic constitutive equation, ij=1 2 tij+ 3+2ijtkk : (6.25) By a formal – although admittedly not obvious – rearrangement of this equation, we may write ij=+ (3+2) 1+ 2(+) tij- 2(+)ijtkk (6.26) from which, if we define, E=(3+2) +(6.27a) and = 2(+); (6.27b) we obtain the following form of Hooke’s law for isotropic behavior in terms of the engi- neering constants Eand, ij=1 E[(1+)Tij-ijTkk]: (6.28) HereEis called Young’s modulus, or simply the modulus of elasticity , andis known asPoisson’s ratio. By suitable combinations of these two constants, we may define two additional constants of importance in engineering elasticity. First, the shear modulus, or modulus of rigidity, is defined as Linear Elasticity 217 t11 t11 (a) Uniaxial tension t12t12 (b) Simple shear t11t11 t22 t22 t33t33 (c) Uniform triaxial tension FIGURE 6.2 Simple stress states. G=E 2(1+)= (6.29a) which, as noted, is identical to the Lam ´e constant. Second, the bulk modulus is defined as K=E 3(1-2): (6.29b) For isotropic elastic materials, any two elastic constants completely define the material’s response. In addition to that, any elastic constant can be determined in terms of any two other constants. A listing of all elastic constants in terms of other pairs of constants is given in Table 6.1. The physical interpretations of the constants E,,G, andKintroduced above can be determined from a consideration of the special states of stress displayed in Fig. 6.2. In the case of a uniaxial state of stress (tension or compression), say in the x1direction witht11=0, and all other stress components zero (Fig. 6.2(a)), Eq 6.28 yields (since tii=0), 11=t11 E=0 Efor(i=j=1); (6.30a) 22= -11=0 Efor(i=j=2); (6.30b) 33= -11=0 Efor(i=j=3); (6.30c) as well as zero shear strains for i6=j. Thus,Eis the proportionality factor between axial (normal) stresses and strains. Geometrically, it is the slope of the one-dimensional 218 Continuum Mechanics for EngineersTABLE 6.1 Relations between elastic constants.   E  K ; (3+2) + 2(+)3+2 3 ;E (E-2)+1 4q (E-3)2+8E -(E-2)+1 4q (E-3)2+82(3+E)+1 6q (3+E)2-4E ;(1-2) 2(1+)(1-2) (1+) 3 ;K3(K-) 29K(K-) 3K- 3K- ;E(2-E) E-3E-2 2E 3(3-E) ;2 1-22(1+)2(1+) 3(1-2) ;K3K-2 39K 3K+3K-2 2(3K+) E;E (1+)(1-2)E 2(1+)E 3(1-2) E;K3K(3K-E) (9K-E)3KE 9K-E3K-E 6K ;K3K 1+3K(1-2) 2(1+)3K(1-2) Linear Elasticity 219 linear stress-strain diagram (Fig. 6.1(a)). Note that E>0; a specimen will elongate under tension, shorten in compression. From the second and third part of Eq 6.30 above, is seen 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=0, all other stresses zero, we have, from Eq 6.28, 12=1+ Et12=0 2G; (6.31a) or for engineering strains, using Eq 4.71, 12=t12 G=0 G(6.31b) which casts Ginto the same role for simple shear as Eassumes for axial tension (or compression). Hence, the name shear modulus forG. Finally, for the case of uniform triaxial tension (or hydrostatic compression) of Fig. 6.2(c), we take tij=pijwithp>0 . For this, Eq 6.28 indicates that ii=1-2 Etii=3(1-2) Ep=p K(6.32) by which we infer that the bulk modulus Krelates the pressure pto the volume change given by the cubical dilation ii(see Eq 4.76). It should also be pointed out that, by use of the constants GandK, Hooke’s law may be expressed in terms of the spherical and deviator components of the stress and strain tensors. Thus, the pair of equations Sij=2Gij; (6.33a) tii=3Kii; (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 Media Hooke’s law for isotropic behavior was established in Section 6.2 on the basis of Cbeing a fourth-order isotropic tensor. The same result may be achieved from the concepts of elastic symmetry. To do so, we first define equivalent elastic directions as those specified by Cartesian axes Ox1x2x3andOx0 1x0 2x0 3at a point such that the elastic constants C are unchanged by a transformation between the two sets of axes. If the transformation represents a rotation about an axis, we say the material has axial elastic symmetry with respect to that axis. If the transformation is a reflection of the axes with respect to some plane, we say the material has a plane of elastic symmetry. Figure 6.3(a) shows the case forx3being the axis of elastic symmetry, whereas Fig. 6.3(b) shows the case for the x1x2 plane as the plane of elastic symmetry. The fact that the transformation for the reflection in Fig. 6.3(b) is an improper one (resulting in the axes being a left-handed coordinate system) does not invalidate the symmetry considerations to be used. Also, the x3axis in Fig. 6.3(a) is said to be of order NwhereN=2= . It is also noteworthy that a point of 220 Continuum Mechanics for Engineers x3,x3‛ x2x1x2‛ x1‛θθO (a) Rotation through angle aboutx3axis. O x1,x1‛x2,x2‛x3 x3‛ (b) Reflection in the x1x2plane. FIGURE 6.3 Axes rotations for plane stress. elastic symmetry would imply isotropic behavior, since the elastic constants would remain unchanged for any two sets of Cartesian axes at the point. Let us consider the consequences of the x1-x2plane being a plane of elastic symmetry as shown in Fig. 6.3(b). The transformation matrix for this is clearly [aij]=2 41 0 0 0 1 0 0 0 -13 5; (6.34) so, in the single subscript notation for stress and strain components, the transformations in matrix form are 2 4t0 1t0 6t0 5 t0 6t0 2t0 4 t0 5t0 4t0 33 5=2 41 0 0 0 1 0 0 0 -13 52 4t1t6t5 t6t2t4 t5t4t33 52 41 0 0 0 1 0 0 0 -13 5 (6.35a) =2 4t1t6-t5 t6t2-t4 -t5-t4t33 5; and 2 40 11 20 61 20 5 1 20 60 21 20 4 1 20 51 20 40 33 5=2 41 0 0 0 1 0 0 0 -13 52 411 261 25 1 2621 24 1 251 2433 52 41 0 0 0 1 0 0 0 -13 5 (6.35b) =2 411 26-1 25 1 262-1 24 -1 25-1 2433 5: Therefore, assuming all 36 constants in Eq 6.8 are distinct, we note that for axes Ox1x2x3 t1=C111+C122+C133+C144+C155+C166; (6.36) whereas for axes Ox0 1x0 2x0 3, under the condition that x1x2is a plane of symmetry such that the Care unchanged in this system, we have t0 1=C110 1+C120 2+C130 3+C140 4+C150 5+C160 6: (6.37) Linear Elasticity 221 But from Eq 6.35a, t0 =t(a=1;2;3;6) andt0 = -t(=4;5). Likewise, 0 = (=1;2;3;6) and0 =(=4;5), so that Eq 6.37 becomes t0 1=t1=C111+C122+C133-C144-C155+C166: (6.38) Comparing Eq 6.36 with Eq 6.38, we see that, for these expressions (each representing t1) to be equal, we must have C14=C15=0. Following the same procedure, we learn that, if we compare expressions for t0 =t(=2;3;6 )and fort0 = -t(=4;5), the additional elastic constants C24,C25,C34,C35,C41,C42,C43,C46,C51,C52,C53,C56,C64, andC65must also be zero for the x1x2plane to be one of elastic symmetry. Accordingly, the elastic constant matrix for this case has the form [C]=2 6666664C11C12C130 0 C16 C21C22C230 0 C26 C31C32C330 0 C36 0 0 0 C44C450 0 0 0 C54C550 C61C62C630 0 C663 7777775; (6.39) and the original 36 constants are reduced to 20. Also, if a strain energy functions exists, C=Cand these20nonzero constants would be further reduced to 13. If thex2-x3plane is also one of elastic symmetry at the same time as the x1-x2plane at a point and we repeat the procedure outlined above, we find that C16,C26,C36,C45,C54, C61,C62, and C63must also be zero, and the Cmatrix is further reduced to [C]=2 6666664C11C12C130 0 0 C21C22C230 0 0 C31C32C330 0 0 0 0 0 C440 0 0 0 0 0 C550 0 0 0 0 0 C663 7777775(6.40) having 12 nonzero coefficients, or 9 if a strain energy function exists. Interestingly enough, ifx1x3is also a plane of elastic symmetry along with the two considered above, no further reduction in the Cmatrix occurs. A material possessing three mutually perpendicular planes of elastic symmetry is called an orthotropic material, and its elastic constants matrix is that given in Eq 6.40. The reduction of the orthotropic elastic matrix to that of the isotropic matrix may be completed by successive consideration of the three axes of elastic symmetry shown in Fig. 6.4 as well as their respective transformation matrices. By the rotation of 90about thex1axis (Fig. 6.4(a)), we find that C12=C13,C21=C31,C22=C33,C23=C32, and C55=C66. For the rotation of 90about thex3axis (Fig. 6.4(b)), we see that C12=C21, C11=C22,C12=C23,C31=C32, and C44=C55. Finally, by a rotation of 45about thex3axis (Fig. 6.4(c)), we obtain 2C44=C11-C12. Therefore, by setting C44=and C12=, we may identify these remaining three C’s with the Lam ´e constants and write the elastic coefficient matrix for isotropic behavior as [C]=2 6666664+2   0 0 0   +2  0 0 0    +2 0 0 0 0 0 0  0 0 0 0 0 0  0 0 0 0 0 0 3 7777775: (6.41) 222 Continuum Mechanics for Engineers x3 , x2‛ π2π2 x3‛ x1 , x1‛x1 x1‛ x3‛x2‛x3 x2 x2 01 −10001 00 (a) 90rotation aboutx1axis x3 , x3‛ π2 π2x2‛ x1x1 x1‛ x3‛x2‛x3 x2 x2 , x1‛ 01 -1 000 100 (b) 90rotation aboutx3axis x3 , x3‛ π4 π4x2‛ x1x1 x1‛ x3‛x2‛x3 x2 x200100 x1‛ 1 2 1 2−1 2 1 2 (c) 45rotation aboutx3axis FIGURE 6.4 Geometry and transformation tables for reducing the elastic stiffness to the isotropic case. Linear Elasticity 223 From the definitions given in Eqs 6.27a and 6.27b, this matrix may be expressed in terms of the engineering constants Eandso that Hooke’s law for an isotropic body appears in matrix form as 2 6666664t1 t2 t3 t4 t5 t63 7777775=E (1+) (1-2)2 66666641-   0 0 0  1 -  0 0 0   1 - 0 0 0 0 0 01 2(1-2)0 0 0 0 0 01 2(1-2)0 0 0 0 0 01 2(1-2)3 77777752 66666641 2 3 4 5 63 7777775: (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 in which inertia forces may be neglected. In both elastostatics and elastodynamics, certain field 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 the boundary. In the case of elastodynamics problems, initial conditions on velocities and displacements must also be satisfied. We begin our discussion with elastostatics for which the appropriate field equations are: (a) Equilibrium equations tji;j+bi=0; (6.43) (b) Strain-displacement relation 2ij=ui;j+uj;i; (6.44) (c) Hooke’s law tij=ijkk+2ij; (6.45a) or ij=1 E[(1+)tij-ijtkk]: (6.45b) It is usually assumed that the body forces biare known so that the solution we seek from the fifteen equations listed here is for the six stresses tij, the six strains ij, and the three displacements ui. The conditions to be satisfied on the boundary surface Swill appear in one of the following statements: 1. displacements prescribed everywhere, ui=u i(x)onS (6.46) where the asterisk denotes a prescribed quantity, 224 Continuum Mechanics for Engineers 2. tractions prescribed everywhere, t(^n) i=t(^n) i onS; (6.47) 3. displacements prescribed on portion S1ofS, ui=u i(x)onS1; (6.48a) with tractions prescribed on the remainder S2, t(^n) i=t(^n) i onS2: (6.48b) A most important feature of the field Eqs 6.43 through Eq 6.45a is that they are linear in the unknowns. Consequently, if t(1) ij, and(1) ij,u(1) iare a solution for body forces1b i and surface tractions1t(^n) i, whereast(2) ij,(2) ij, andu(2) iare a solution for body forces2b i and surface tractions2t(^n) i, then tij=t(1) ij+t(2) ij; ij=(1) ij+(2) ij;andui=u(1) i+u(2) i offer a solution for the situation where bi=1b i+2b iandt(^n) i=1t(^n) i+2t(^n) i. This is a statement of the principle of superposition, which is extremely useful for the development of 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 from the 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 that result into the equilibrium equations (Eq 6.43), we obtain the three second-order partial differential equations ui;jj+ (+)uj;ji+bi=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 substituted into Eq 6.44 to generate the strains and those in turn substituted into Eq 6.45a to obtain the stresses. When the boundary conditions are given in terms of surface tractions (Eq 6.47), the equations of compatibility for infinitesimal strains (Eq 4.90) may be combined with Hooke’s law (Eq 6.45b) and the equilibrium equations to arrive, after a certain number of algebraic manipulations, at the equations tij;kk+1 1+tkk;ij+(bi;j+bj;i)+ 1-ijbk;k=0 (6.50) which are known as the Beltrami-Michell stress equations of compatibility. In combination with the equilibrium equations, these equations comprise a system for the solution of the stress components, but it is not an especially easy system to solve. As was the case with the 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),(1),u(1)andT(2),(2),u(2)corresponding to the same body forces and the appropriate boundary conditions. The difference of these solu- tions,T=T(1)-T(2),=(1)-(2),u=u(1)-u(2), is also a solution to the governing Linear Elasticity 225 equations with b=0since the system of equations is linear and superposition applies. The solutionT,,usatisfies the boundary conditions u=0onS1 ort=0onS2: Integrating over the boundary Sof the body gives Z Su
tdS=0: Substitutingt=T
nfor a symmetric stress tensor, we have Z Su
T
ndS=0 and applying the divergence theorem yields Z Vdiv(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=0sinceb=0. The first term can be written as ui;jtij=tij(ij+!ij)=tijij (6.52) sincetijis symmetric and !ijis skew-symmetric. From Eq 6.21a and 6.52, Eq 6.51 becomes Z V2W()dV=0: This integral is zero when =0provided that W()is positive definite in Eq 6.21a. For a linear, isotropic body this requires that E>0 and-1<<1 2. Since=0, we have (1)=(2)andT(1)=T(2)from Hooke’s law. There cannot be two different solutions to the linear equations of elastostatics for the same body forces and boundary conditions. A solution that satisfies the governing equations and boundary conditions is unique. In elastodynamics, the equilibrium equations must be replaced by the equations of motion (Eq 5.27) in the system of basic field equations. Therefore, all field quantities are now considered functions of time as well as of the coordinates, so that a solution for the displacement field, for example, appears in the form ui=ui(x;t). In addition, the solution must satisfy not only boundary conditions which may be functions of time as in ui=u i(x;t)onS (6.53a) or t(^n) i=t(^n) i(x;t)onS; (6.53b) but also initial conditions, which usually are taken as ui=u i(x;0) (6.54a) and ˙ui=˙u i(x;0): (6.54b) Analogous to Eq 6.49 for elastostatics, it is easily shown that the governing equations for displacements in elastodynamics theory are ui;jj+ (+)uj;ji+bi=¨ui (6.55) which are also called Navier’s equations. 226 Continuum Mechanics for Engineers BLL x3x2C1 C2t(^n) it(^n) i x1 FIGURE 6.5 Beam 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 of ways: 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 by considering relaxed boundary conditions on the beam ends. If one considers statically equivalent 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 relaxed boundary conditions on the beam ends requires that the resultant force and moment be considered rather than the exact traction distribution over the end. To demonstrate Saint-Venant’s solution consider a beam that is represented by a right cylinder extending in the x3direction as shown in Fig. 6.5. The lateral surface of the cylinder, BL, will be stress free. This condition is called the lateral boundary condition and can be written as t(^n) i=ti1n1+ti2n2=0 onBL: (6.56) Let the cylinder be generated by a simply connected cross section Cwhich is perpen- dicular to the x3axis. For the different problems (extension, torsion, pure bending and flexure) the ends, C1andC2, will have relaxed boundary conditions. The sum of the stress over the end’s area will give a force vector Ti Z C1t3idS=Ti: (6.57) Linear Elasticity 227 Bending moments on the ends are expressed in terms of the axial stress t33as follows: Z C1t33x2dS=M1;Z C1t33x1dS= -M2: (6.58) Finally, the torque on the beam’s end may be written as Z 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. From this assumption the equilibrium and compatibility equations were shown to be satisfied. Boundary conditions, exact and relaxed, were checked. Because solutions to elasticity problems are unique the solution obtained was, of course, the only solution. In the following subsections Saint-Venant solutions for extension, torsion, pure bending and flexure are developed. The different problems are solved by setting the components of the end force Tiand moment Mito different values: (I)T1=T2=0; T3=T; Mi=0 Pure Extension (II)Ti=0; M1=M2=0; M3=MtPure 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=Z CdS; 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 are satisfied. The strains can be easily found from Eq 6.28 to be 33=T EA; 11=22= -T EA(6.61) with all other strain components being zero. Clearly the compatibility condition, Eq 4.90 is 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= -T EAx1; u2= -T EAx2 (6.62) by direct integration. 228 Continuum Mechanics for Engineers 6.5.2 Torsion We begin with a brief review of the solution to the simplest of torsion problems, the case of a shaft having a constant circular cross section when subjected to equilibrating end couples, Mtas shown in Fig. 6.6(a). Let the end face at x3=0be fixed while the face atx3=Lis allowed to rotate about the axis of the shaft. It is assumed that plane sections perpendicular to the axis remain plane under the twisting, and that each rotates through an angle proportional to its distance from the fixed end. Accordingly, a point in the cross section at coordinate x3will rotate an angle of x3whereis the angle of twist per unit length. Each point, say point P, in the cross section travels a distance x3Rwhich is proportional to the distance Rfrom thex3axis as shown in Fig. 6.6(b). The distance squared to point Pis the square of the x1andx2coordinates (Fig. 6.6(c)). Using this distance, it is easy to define the cosine and sine of angle . It is straightforward to write out displacements u1andu2in terms ofsince cos (90-)=x2=Rand sin (90-)= x1=R. Thus, u1= -x2x3; u2=x1x3; u3=0; (6.63) whereis the angle of twist per unit length of the shaft. Recall from Eq 4.62 that 2ij=ui;j+uj;iby 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 (sinceGby Eq 6.29a) t23=Gx1; t13= -Gx2; 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 4.90 are satisfied. Likewise, for zero body forces, the equilibrium equations Eq 6.43 are clearly satisfied. The lateral surface of the shaft is stress free. To verify this, consider the stress components 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+t23x2 a=G a(-x2x1+x1x2)=0: (6.65) At the same time, the total shearing stress at any point of the cross section is the resultant =q t2 13+t2 23=Gq x2 1+x2 2=GR (6.66) which indicates that the shear is proportional to the radius at the point, and perpendicular to that radius. By summing the moments of the shear forces on either end face of the shaft, we find that Mt=ZZ (x1t23+x2t13)dx1dx2=GZZ R2dx1dx2=GIP (6.67) whereIPis 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 will occur. For such cases we must modify Eq 6.63 by expressing the displacements in the form u1= -x2x3; u2=x1x3; u3= (x1;x2); (6.68) where (x1;x2)is called the warping function. Note that the warping is independent of x3 and therefore the same for all cross sections. Also, we assume that x3is a centroidal axis although this condition is not absolutely necessary. Linear Elasticity 229 Mt MtL ax2 x1 x3 (a) Cylinder with self-equilibrating moments Mt. P∗ P Rx2 x1u1u2θx3R (b) Displacement of point PtoPin cross section. R βθx3R x1x2RPP∗ 90−β β(c) Detail of cross section twist . FIGURE 6.6 Geometry and kinematic definitions for torsion of a circular shaft. 230 Continuum Mechanics for Engineers Mt Mtx2 x1 x3L (a) Prismatic cylinder with self-equilibrating moments Mt. ^n ^s x2 x1dx1dx290−θ θ (b) Displacement of point PtoPin cross sec- tion. FIGURE 6.7 The more general torsion case of a prismatic beam loaded by self equilibrating moments. Linear Elasticity 231 As in the analysis of the circular shaft we may again use Eq 4.62 along with Eq 6.23 to calculate the stress components from Eq 6.68. Thus t11=t22=t33=t12=0; t13=G( ;1-x2); t23=G( ;2-x1): (6.69) It is clear from these stress components that there are no normal stresses between the longitudinal 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 the third equilibrium equation yields t13;1+t23;2+t33;3=G( ;11+ ;22)=0 (6.70) which indicates that must be harmonic, r2 =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 while the third requires G( ;1-x2)n1+G( ;2-x1)n2=0 (6.73a) which reduces immediately to ;1n1+ ;2n2=d dn=x2n1+x1n2: (6.73b) Therefore, (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 6.73b on the perimeter Cof the cross section. We note further that in order for all cross sections to be force free, that is, in simple shear over those cross sections ZZ t13dx1dx2=ZZ t23dx1dx2=ZZ t33dx1dx2=0: (6.74) Sincet33=0, the third integral here is trivial. Considering the first integral we may write GZZ ( ;1-x2)dx1dx2=GZZ@ @x1[x1( ;1-x2)]+@ @x2[x1( ;2-x1)] dx1dx2 (6.75) where the condition r2 =0has been used. Green’s theorem allows us to convert to the line integral taken around the perimeter C GZ Cx1f( ;1-x2)n1+( ;2-x1)n2gds=0 (6.76) which by Eq 6.73b is clearly satisfied. By an analogous calculation we find that the second integral of Eq 6.74 is also satisfied. On the end faces of the shaft, x3=0orx3=L, the following conditions must be satisfied ZZ x2t33dx1dx2=ZZ x1t33dx1dx2=0;ZZ (x1t23-x2t13)dx1dx2=Mt: (6.77) 232 Continuum Mechanics for Engineers Again, since t33=0, the first two of these are trivial. The third leads to Mt=GZZ x2 1+x2 2+x1 ;2-x2 ;1
dx1dx2: (6.78) Defining the torsional rigidity as K=GZZ x2 1+x2 2+x1 ;2-x2 ;1
dx1dx2 (6.79) which can be evaluated once (x1;x2)is known, we express the angle of twist as =Mt K: (6.80) A second approach to the general torsion problem rests upon the introduction of a torsion stress function, designated here by and defined so that the non-zero stresses are related to it by the definitions t13=@ @x2; t23= -@ @x1: (6.81) Thus, @ @x2=G( ;1-x2);@ @x1= -G ( ;2-x1): (6.82) By eliminating from this pair of equations we obtain r2= -2G: (6.83) As already noted, the lateral surface of the shaft parallel to the axis must remain stress free, that is, the third of Eq 6.72 must be satisfied. However, it is advantageous to write this condition in terms of the unit vector ^salong the boundary rather than unit nor- mal^nas shown in Fig. 6.6(b). It follows directly from geometry that n1=dx2=ds and n2= -dx1=ds. In terms of ds, the differential distance along the perimeter C, the stress components in the normal direction will be given by t13dx2 ds-t23dx1 ds=0 (6.84) which in terms of becomes @ @x1dx1 ds+@ @x2dx2 ds=d ds=0: (6.85) Thus,is a constant along the perimeter of the cross section and will be assigned the value of zero here. Finally, conditions on the end faces of the shaft must be satisfied. Beginning with the first of Eq 6.74, we have in terms of  ZZ@ @x2dx1dx2=ZZ@ @x2dx2 dx1=Z jb adx1=0 (6.86) sinceis constant on the perimeter. Likewise, by the same reasoning, the second of Eq 6.74 is satisfied, while the third is satisfied since t33=0. The first two conditions in Eq 6.77 are also satisfied identically and the third becomes ZZ -x1@ @x1-x2@ @x2 dx1dx2=Mt: (6.87) Linear Elasticity 233 Integrating by parts and using the fact that is assumed zero on the perimeter Cyields Mt=2ZZ dx1dx2: (6.88) Thus, the solution by this approach consists of determining the stress function which is zero on the cross-section perimeter, and satisfies Eq 6.83. Based upon that result we may determine from Eq 6.88. Example 6.2 Determine the stresses and the angle of twist for a solid elliptical shaft of the dimensions shown when subjected to end couples Mt. x2 x1 2a2b Solution The equation of this ellipse is given by x2 1 a2+x2 2 b2=1: Therefore, take the stress function in the form =x2 1 a2+x2 2 b2-1 (6.89) whereis a constant. Thus, is zero on the cross-section perimeter. From Eq 6.83 21 a2+1 b2 = -2G; so that = -a2b2G a2+b2: (6.90) Now from Eq 6.88 Mt= -2a2b2G a2+b2ZZx2 1 a2+x2 2 b2-1 dx1dx2; and noting that ZZ x2 1dx1dx2=Ix2=1 4ba3; ZZ x2 2dx1dx2=Ix1=1 4ab3; 234 Continuum Mechanics for Engineers and ZZ dx1dx2=ab (the area of the cross section) we may solve for Mtwhich is Mt=a3b3G a2+b2: From this result, =a2+b2 a3b3GMt which when substituted into Eq 6.90 and that into Eq 6.89 gives = -Mt abx2 1 a2+x2 2 b2-1 : Now, by denition t13=@ @x2= -2Mt ab3x2; t23= -@ @x1= -2Mt a3bx1: The maximum stress occurs at the ends of the minor axis, and equals max=2Mt ab2; and the torsional rigidity is easily calculated to be K=Mt =a3b3G a2+b2=G(A)4 42IP: Note also that for a=b(circular cross section), the resultant stress at any point is =q t2 13+t2 23=Mt IPr in agreement with elementary theory. It should be pointed out that for shafts having perimeters that are not expressible by simple equations, solutions may be obtained by using stress functions in the form of infinite 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) Linear Elasticity 235 with all other components zero. The compatibility equations are satisfied since the nonzero 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=0is trivially satisfied by considering Eq 6.92. Considering the torque on the end, Eq 6.59, it is clear thatM3=Mt=0sincet31=t32=0. The axial force on the ends is given by Eq 6.57 as T3=Z C1-Mx1 IdS= -M IZ C1x1dS=0 (6.93) provided that the axes origin is chosen at the centroid of the cross section C. A similar calculation for M1gives M1=Z C1-Mx1 Ix2dS= -M IZ C1x1x2dS=0 (6.94) as long as the principal axes are used. Finally, M2= -Z C1Mx1 Ix1dS=Mx1 IZ C1x2 1dS=M (6.95) where I=Z C1x2 1dS: Strain components are found from Eq 6.28 giving three nonzero components 11=22=Mx1 EI;33= -Mx1 EI: (6.96) Displacements can be found using Eq 4.101. The term uj(0) +wij(0) j-0 j rep- resents a rigid displacement and will be set to zero. Taking the component u1as an example for finding the displacements yields u1=Zx 0[11d1+12d2+13d3 +(xk-k)11;kd1+(xk-k)12;kd2+(xk-k)13;kd3 -(xk-k)k1;1d1-(xk-k)k2;1d2-(xk-k)k3;1d3] =Zx 0[11d1+(x1-1)11;1d1 -(x1-1)11;1d1-(x2-2)22;1d2-(x3-3)33;1d3] =Zx 0M EI1d1-(x2-2)M EId2+(x3-3)M EId3 =M 2EIx2 1-M EI x2 2-1 2x2 2
+M EI x2 3-1 2x2 3
=M 2EI x2 1-x2 2+x2 3
:(6.97) 236 Continuum Mechanics for Engineers Similar calculations for u2andu3give the displacements for pure bending as u1=M 2EI(x2 1-x2 2+x2 3); u2=M EIx1x2; u3= -M EIx1x3;(6.98) to within an infinitesimal rigid displacement. In the pure bending problem, x1=0represents 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 C1fixed and a flexure load ofTacting 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 t31andt32. Equilibrium equations are t13;3=0; t23;3=0; t31;1+t32;2= -Tx1 I(6.100) which implies t13=t13(x1;x2);t23=t23(x1;x2): (6.101) The strain may be calculated from Eq 6.28 to obtain 11=22=T EI(L-x3)x1; 33= -T EI(L-x3)x1 23=t23=2;31=t31=2;12=0(6.102) The compatibility equations must be satisfied since the problem is posed in terms of stresses. Eq 4.91 results in two nontrivial conditions @ @x1(23;1-13;2)=0; (6.103a) @ @x2(31;2-23;1)=T EI; (6.103b) Integration of Eq 6.103a gives 23;1-31;2=-T EIx2 (6.104) whereis taken as a constant by virtue of Eq 6.103a. Equation 6.104 may be written as  23-1 2x1 ;1= 31+1 2x2-1 2T EIx2 2 ;2(6.105) Linear Elasticity 237 from which the shear strains may be found in terms of a potential function f 32-1 2x1=1 2f;2;31+1 2x2-1 2T EIx2 2=1 2f;1 (6.106) The shear stresses may be written using Hooke’s Law and Eq 6.106 t31=231= -x2+T EIx2 2+f;1; t32=232=x1+f;2:(6.107) The remaining equilibrium condition, Eq 6.100 is r2f(x1;x2)= -Tx1 I= -2(1+)T EIx1onC (6.108) subject to the lateral boundary condition tin=0in the form @f @n=(x2n2-x1n2) -T EIx2 2n1on@C: (6.109) To solve the boundary value problem given by Eqs 6.108 and 6.109, define a function F(x1;x2)by f= -T EI F+x3 1 6+ 1+ 2 x1x2 2 (6.110) where is the warping function from torsion r2 =0onC;@ @n=x2n1-x1n2on@C: (6.111) Stress components may be written in terms of ,andFas t31=( ;1-x2) -T 2(1+)I F;1+x2 1 2+ (1+ 2)x2 2 ; t32=( ;2+x1) -T 2(1+)IfF;2+ (2+)x1x2g; t33= -T I(L-x3)x1;(6.112) subject to the lateral boundary condition @F @n= -x2 1 2+ 1+ 2 x2 2 n1-(2+)x1x2n2on@C: (6.113) Since andFare harmonic it is possible to show the stresses in Eq 6.112 satisfy the equilibrium equations. Substituting the stresses from 6.112 into Eq 6.57 and using Eq 6.101 it can be shown that T3=T2=0andT1=T. Bending moments M1=M2=0since t33(x3=L)=0, and the torque, M3=Mt, is set to zero by expressing in terms ofF: = -T 2(1+)IKZ C2 x2F;1-x1F;2+ (1+ 2)x3 2-2(1+ 2)x2 1x2 dA (6.114) whereKis the torsional rigidity. The constant may be shown to be the average rate of rotation over a cross section. 238 Continuum Mechanics for Engineers In general, there is a twist accompanying the bending when the load Tis applied at the centroid of C2. By superposing an appropriate torsion solution with equal and opposite it is possible to determine a line of action parallel to Tisuch that there is not twist. The displacements are determined from the line integral Eq 4.101 u1= -x2x3+T EI1 2(L-x3)(x2 1-x2 2) +1 2Lx2 3-1 6x3 3 ; u2=x1x3+T EI(L-x3)x1x2; u3= (x1;x2) -T EI x1x3(L-1 2x3) +x1x2 2+F(x1;x2) ;(6.115) to within an infinitesimal rigid displacement. The centerline deflection is the same as that obtained in beam theory, but the cross-sections do not remain plane in the Saint-Venant solution. 6.6 Plane Elasticity In a number of engineering applications, specific body geometry and loading patterns lead to a reduced, essentially two-dimensional form of the equations of elasticity, and the study of these situations is referred to as plane elasticity. Although the two basic types of problems constituting the core of this plane analysis may be defined formally by stating certain assumptions on the stresses and displacements, we introduce them here in terms of their typical physical prototypes. In plane stress problems, the geometry of the body is that of a thin plate with one dimension very much smaller than the other two. The loading in this case is in the plane of the plate and is assumed to be uniform across the thickness, as shown in Fig. 6.8(a). In plane strain problems, the geometry is that of a prismatic cylinder having one dimension very much larger than the other two and having the 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 analysis may be focused on a thin slice of the cylinder. For the plane stress situation, Fig. 6.8(a), the stress components t33,t31, andt32are taken as zero everywhere and the remaining components considered functions of only x1 andx2. 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+b1=0; (6.117a) t21;1+t22;2+b2=0: (6.117b) The strain-displacement relations, Eq 6.44, become 11=u1;1;22=u2;2;212=u1;2+u2;1; (6.118) and at the same time the strain compatibility equations, Eqs 4.90, take on the form given by Eq 4.92 and repeated here for convenience, 11;22 +22;11 =212;12: (6.119) Linear Elasticity 239 O x3x1x2 (a) Plane stress problems generally involve bodies that are thin in 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 6.8 Representative figures for plane stress and plain strain. 240 Continuum Mechanics for Engineers Hooke’s law equations, Eq 6.45b, for plane stress are written 11=1 E(t11-t22); (6.120a) 22=1 E(t22-t11); (6.120b) 12=1+ Et12=t12 2G= 12 2; (6.120c) along with 33= - E(t11+t22)=- 1-(11+22): (6.121) By inverting Eq 6.120a, we express the stress components in terms of the strains as t11=E 1-2(11+22); (6.122a) t22=E 1-2(22+11); (6.122b) t12=E 1+12=E 2(1+) 12=G 12: (6.122c) These equations may be conveniently cast into the matrix formulation 2 4t11 t22 t123 5=E 1-22 41  0  1 0 0 0 1 -3 52 411 22 123 5; (6.123) In terms of the displacement components, ui(i=1;2), the plane stress field equations may be combined to develop a Navier-type equation for elastostatics, namely, E 2(1+)ui;jj+E 2(1-)uj;ji+bi=0(i;j=1;2): (6.124) For the plane strain situation (Fig. 6.8(b)) we assume that u3=0and that the remaining two displacement components are functions of only x1andx2, ui=ui(x1;x2) (i =1;2): (6.125) In this case, the equilibrium equations, the strain-displacement relations, and the strain compatibility 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 Linear Elasticity 241 terms of engineering constants as t11=E (1+) (1-2)[(1-)11+22]; (6.126a) t22=E (1+) (1-2)[(1-)22+11]; (6.126b) t12=E 1+12=E 1+ 12 2=G 12; (6.126c) along with t33=E (1+) (1-2)(11+22)=(t11+t22): (6.127) The first three of these equations may be expressed in matrix form by 2 4t11 t22 t123 5=E (1+) (1-2)2 41-  0  1 - 0 0 0 1 -23 52 411 22 123 5: (6.128) Furthermore, by inverting the same three equations, we may express Hooke’s law for plane strain by the equations 11=(1+) E[(1-)t11-t22]; (6.129a) 22=(1+) E[(1-)t22+t11]; (6.129b) 12=1+ Et12=2(1+) Et12 2=t12 2G: (6.129c) By combining the field equations with Hooke’s law for elastostatic plane strain, we obtain the appropriate Navier equation as E 2(1+)ui;jj+E 2(1+) (1-2)uj;ji+bi=0(i;j=1;2): (6.130) It is noteworthy that Eqs 6.120a and 6.122a for plane stress become identical with the plane strain equations, Eqs 6.129 and 6.126, respectively, if in the plane stress equations we replaceEwithE=(1-2)andby=(1-). Note that, if the forces applied to the edge of the plate in Fig. 6.8(a) are not uniform across the thickness but are symmetrical with respect to the middle plane of the plate, the situation is sometimes described as a state of generalized plane stress . In such a case, we consider the stress and strain variables to be averaged values over the thickness. Also, a case of generalized plane strain is sometimes referred to in elasticity textbooks if the strain component 33in Fig. 6.8(b) is taken as some constant other that zero. 242 Continuum Mechanics for Engineers 6.7 Airy Stress Function As stated in Section 6.6, the underlying equations for two-dimensional problems in isotropic elasticity consist of the equilibrium relations, Eq 6.117, the compatibility condition, Eq 6.119 and Hooke’s law, either in the form of Eq 6.120a (plane stress), or as Eq 6.129 (plane strain). When body forces in Eq 6.117 are conservative with a potential function V= V(x1;x2)such thatbi= -V;i, we may introduce the Airy stress function, =(x1;x2)in terms of which the stresses are given by t11=;22+V;t22=;11+V;t12= -;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 -(t11;11 +t22;22) =2(1+)t12;12 (6.132) which in terms of becomes ;1111 +2;1212 +;2222 = -(1 -)(V;11+V;22): (6.133) Similarly, for the case of plane strain, when Eq 6.129 is introduced into Eq 6.119 the result is (1-)(t11;22 +t22;11) -(t11;11 +t22;22) =2t12;12; (6.134) or in terms of  ;1111 +2;1212 +;2222 = -(1 -2)(V;11-V;22)=(1-): (6.135) If the body forces consist of gravitational forces only, or if they are constant forces, the right-hand sides of both Eqs 6.133 and 6.135 reduce to zero and must then satisfy the bi-harmonic equation ;1111 +2;1212 +;2222 =r4=0: (6.136) In each case, of course, boundary conditions on the stresses must be satisfied to complete the solution to a particular problem. For bodies having a rectangular geometry, stress functions in the form of polynomials in x1andx2are especially useful as shown by the examples that follow. Example 6.3 For a thin rectangular plate of the dimensions shown, consider the general polynomial of the third degree as the Airy stress function and from it determine the stresses. Assume body forces are zero. Linear Elasticity 243 x1x2 CC L (a) x1x26D3c 6D3c(b) �B3c �B3c�B3Lx2 x1 (c) Solution Select a polynomial stress function of the form 3=A3x3 1+B3x2 1x2+C3x1x2 2+ D3x3 2. Choosing this particular polynomial form for the stress function is not arbitrary; the choice is based on many trials of dierent order polynomials. After a certain amount of experience in observing a polynomial's eect on the stress components computed using Eq 6.131, an educated guess can be made as to what terms should be considered for a specic problem. For this reason, problems like this are often called semi-inverse problems. By direct substitution into Eq 6.136 we conrm that 3is bi-harmonic. Further, the stresses are given as t11=2C3x1+6D3x2; t22=6A3x1+2B3x2; t12= -2B3x1-2C3x2: By selecting dierent constants to be zero and nonzero, dierent physical prob- lems may be solved. Here, two specic cases will be considered. (a) Assume all coecients in 3are zero except D3. This may be shown to solve the case of pure bending of a beam by equilibrating moments on the ends as shown by Fig. (b). Stress in the ber direction of the beam varies linearly with the distance from the x1axis t11=6D3x2;t22=t12=0 as is the case for simple bending. Similarly, by taking only A3as nonzero, the solution is for bending moments applied to a beam whose lengthwise direction is taken to be x2rather thanx1direction. 244 Continuum Mechanics for Engineers (b) If onlyB3(orC3) is non-zero, both shear and normal stresses are present. Fig. (c) shows the stress pattern for B36=0. Example 6.4 Consider a special stress function having the form =B2x1x2+D4x1x3 2: Show that this stress function may be adapted to solve for the stresses in an end-loaded cantilever beam shown in the sketch. Assume the body forces are zero for this problem. x2 x1P cc 1 L Vectorvwith respect to Ox0 1x0 2x0 3andOx1x2x3. Solution It is easily veried, by direct substitution, that r4=0. The stress compo- nents are directly computed from Eq 6.131 t11=6D4x1x2; t22=0; t12= -B2-3D4x2 2: These stress components are consistent with an end-loaded cantilever beam, and the constants B2andD4can be determined by considering the boundary conditions. In order for the top and bottom surfaces of the beam to be stress- free,t12must be zero at x2=c. Using this condition B2is determined in terms ofD4asB2= -3D4c2. The shear stress is thus given in terms of single constantB2 t12= -B2+B2x2 2 c2: The concentrated load is modeled as the totality of the shear stress t12on the free end of the beam. Thus, the result of integrating this stress over the free end of the beam at x1=0yields the applied force P. In equation form P= -Zc -c -B2+B2x2 2 c2 dx2 Linear Elasticity 245 θdθdr rx2 x1tθθ θθdθ θ∂t+∂r rttθ θdθ θ∂+∂trrdr r∂+∂ trθdrr∂+∂ rtθ rrt tθθtrr trθ rtθ FIGURE 6.9 Differential 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 components may now be written as t11= -3P 2c3x1x2; t22=0; t12= -3P 4c 1-x2 2 c2 : 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-x2 2
in agreement with the results of elementary beam bending theory. Several important solutions in plane elasticity are obtained by the Airy stress function approach when expressed in terms of polar coordinates. To this end we introduce here the basic material element together with the relevant stress components in terms of the coordinatesrandas shown on Fig. 6.9. Using this element and summing forces in the radial direction results in the equilibrium equation @trr @r+1 r@tr @+trr-t r+R=0; (6.137a) 246 Continuum Mechanics for Engineers and summing forces tangentially yields 1 r@t @+@tr @r+2tr r+=0 (6.137b) in whichRandrepresent body forces. In the absence of such forces Eq 6.137a and 6.137b are satisfied by trr=1 r@ @r+1 r2@2 @2; (6.138a) t=@2 @r2; (6.138b) tr=1 r2@ @-1 r@2 @r@= -@ @r1 r@ @ ; (6.138c) in which=(r;). To qualify as an Airy stress function must once again satisfy the conditionr4=0which in polar form is obtained from Eq 6.136 as r4=@2 @r2+1 r@ @r+1 r2@2 @2@2 @r2+1 r@ @r+1 r2@2 @2 =0: (6.139) For stress fields symmetrical to the polar axis Eq 6.139 reduces to r4=@2 @r2+1 r@ @r@2 @r2+1 r@ @r =0; (6.140a) or r4@4 @r4+2r3@3 @r3-r2@2 @r2+r@ @r=0: (6.140b) It may be shown that the general solution to this differential equation is given by =Alnr+Br2lnr+Cr2+D; (6.141) so that for the symmetrical case the stress components take the form trr=1 r@ @r=A r2+B(1+2lnr)+2C; (6.142a) t=@2 @r2= -A r2+B(3+2lnr)+2C; (6.142b) tr=0: (6.142c) When there is no hole at the origin in the elastic body under consideration, AandB must be zero since otherwise infinite stresses would result at that point. Thus, for a plate without 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 is that 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 gure when subjected to constant equilibrating moments. Linear Elasticity 247 a br MM x1θx2 Solution From symmetry, the stresses are as given by Eq 6.142. Boundary conditions require (1)trr=0atr=a;and atr=b; (2)Zb atdr=0on the end faces ; (3)Zb artrrdr= -M on the end faces ; (4)tr=0everywhere on the boundary : These conditions result in the following equations which are used to evaluate the constants A,B, andC. The inner and outer radii are free of normal stress which can be written in terms of boundary condition (1) as A=a2+B(1+2lna) +2C=0; A=b2+B(1+2lnb) +2C=0: No transverse loading is present on the ends of the curved beam which may be written in terms of boundary condition (2) as Zb atdr=Zb a@2 @r2dr=@ @rb a=0: Evaluation of this integral at the limits is automatically satised as a conse- quence of boundary condition (1). Finally, the applied moments on the ends 248 Continuum Mechanics for Engineers may be written in terms of boundary condition (3) Zb ar@2 @r2dr= r@ @rb a-Zb a@ @rdr= -M: Because of condition (1) the bracketed term here is zero and from the integral term b-a=M; or Alnb=a+B(b2lnb-a2lna) +C(b2-a2) =M: This expression, together with the two stress equations arising from condition (1) may be solved for the constants A,B, andC, which are A= -4M Na2b2lnb a; B= -2M N b2-a2
; C=M N b2-a2+2 b2lnb-a2lna
 ; whereN= (b2-a2)2-4a2b2[ln(b=a )]2. Finally, the stress components may be written in terms of the radii and applied moment by substitution of the constants into Eq 6.142 trr= -4M Na2b2 r2lnb a+b2lnr b+a2lna r ; t= -4M N -a2b2 r2lnb a+b2lnr b+a2lna r+b2-a2 ; tr=0: Verication of these results can be made by reference to numerous strength of materials textbooks. Ifis taken as a function of both rand, it is useful to assume (r;) =f(r)ein(6.143) in order to obtain a function periodic in . Forn=0, the general solution is, as expected, the same as given in Eq 6.141. As an example of the case where n=1we consider(r;) in the form = (Ar2+B=r+Cr+Drlnr)sin: (6.144) Example 6.6 Show that the stress function given by Eq 6.144 may be used to solve the quarter-circle beam shown under an end load P. Linear Elasticity 249 b a Pθ x1x2 r Solution From Eq 6.138 the stress components are trr= 2Ar-2B r3+D r sin; t= 6Ar+2B r3+D r sin; tr= - 2Ar-2B r3+D r cos: The inner and outer radii of the beam are stress-free surfaces leading to boundary conditions of trr=tr=0atr=a, and atr=b. Also, the applied force P may be taken to be the summation of the shear stress acting over the free end =0: Zb atrj=0dr= -P: These conditions lead to the three equations from which constants A,B, andD may be determined 2Aa-2B a3+D a=0; 2Ab-2B b3+D b=0; -A b2-a2
+B b2-a2
a2b2-Dlnb a=P: Solving these three equations in three unknowns, the constants are determined to be A=P 2N; B = -Pa2b2 2N; D = -P a2+b2
N; whereN=a2-b2+ (a2+b2)ln(b=a). Finally, use of these constants in the 250 Continuum Mechanics for Engineers stress component equations gives trr=P N r+a2b2 r3-a2+b2 r sin; t=P N 3r-a2b2 r3+a2+b2 r sin; tr= -P N r+a2b2 r3-a2+b2 r cos: Note that, when =0 trr=t=0andtr= -P N r+a2b2 r3-a2+b2 r : And when==2,tr=0while trr=P N r+a2b2 r3-a2+b2 r ; t=P N 3r-a2b2 r3+a2+b2 r : We close this section with an example of the case when n=2in Eq 6.143. Example 6.7 Use the stress function = (Ar2+Br4+C=r2+D)cos2 to solve the stress problem of a large at plate under a uniform axial stress T, and having a small circular hole at the origin as shown in the gure. E abT Ttrθ trr θ x1x2 Linear Elasticity 251 Solution From Eq 6.138 the stress components have the form trr= - 2A+6C r4+4D r2 cos; t= 2A+12Br2+6C r2 cos; tr= 2A+6Br2-6C r4-2D r2 sin2: Assume the width of the plate bis large compared to the radius of the hole. From consideration of the small triangular element at a distance bfrom the origin, the following boundary conditions must hold 1.trr=Tcos2atr=b, 2.tr= -T 2sin2atr=b, and the inner surface of the hole is stress free, which may be written as 3.trr=0atr=a, 4.tr=0atr=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: Lettingb!1, 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; t=T 2 1+a2 r2 -T 2 1+3a4 r4 cos; tr= -T 2 1-3a4 r4+2a2 r2 sin2: Note that as rtends towards innity and at =0the stresses are given by trr=T;t=tr=0(simple tension). At r=a,trr=tr=0and t=T-2Tcos2, which indicates that when ==2, or=3=2 , the stresses become t=T-2T(-1) =3T, a well-known stress concentration factor 252 Continuum Mechanics for Engineers used in design. Also, when =0, or=,t= -T a compression factor at the 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 the problems considered. 6.8 Linear Thermoelasticity When we give consideration to the effects of temperature as well as to mechanical forces on the behavior of elastic bodies, we become involved with thermoelasticity. Here, we ad- dress only the relatively simple uncoupled theory for which temperature changes brought about by elastic straining are neglected. Also, within the context of linearity we assume that the total strain is the sum ij=(M) ij+(T) ij (6.145) where(M) ij is the contribution from the mechanical forces and (T) ijare the temperature- induced strains. If 0is taken as a reference temperature and as an arbitrary tem- perature, the thermal strains resulting from a change in temperature of a completely unconstrained isotropic volume are given by (T) ij=(-0)ij (6.146) whereis the linear coefficient of thermal expansion, having units of meters per meter per degree Celsius (m/m/C). The presence of the Kronecker delta in Eq 6.146 indicates that shear 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 in that equation, we arrive at the thermoelastic constitutive equation ij=1+ Etij- Eijtkk+(-0)ij: (6.147) This equation may be easily inverted to express the stresses in terms of the strains as tij=E (1+) (1-2)[ijkk+(1-2)ij-(1+)(-o)ij]: (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+)ij; (6.149a) tii=E (1-2)[ii-3(-0)]: (6.149b) Linear Elasticity 253 If the heat conduction in an elastic solid is governed by the Fourier law, Eq 5.62, which we write here as qi= -;i (6.150) whereis the thermal conductivity of the body (a positive constant), and if we introduce thespecific heat constant cthrough the equation -qi;i=c˙ (6.151) the heat conduction equation for the uncoupled theory becomes ;ii=c˙: (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 Elasticity Solutions 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 of the 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 of strains. These equations are also reviewed in the following paragraph. Starting with the fundamental equations of elastostatics as listed in Section 6.4 and repeated here under revised numbering, we have: Equilibrium equations, Eq 6.43 tij;j+bi=0; (6.153) Strain-displacement equations, Eq 6.44 2ij=ui;j+uj;i; (6.154) Hooke’s law, Eq 6.45a, or Eq 6.45b tij=ijkk+2ij; (6.155a) or ij=1 E[(1-)tij-ijtkk]: (6.155b) 254 Continuum Mechanics for Engineers By substituting Eq 6.154 into Hooke’s law, Eq 6.155a, and that result in turn into Eq 6.153, we obtain ui;jj+(+)uj;ji+bi=0 (6.156) which comprise three second-order partial differential equations known as the Navier equations. For the stress formulation we convert the strain equations of compatibility introduced in Section 4.7 as Eq 4.90 and repeated here as ij;km +km;ij -ik;jm -jm;ik =0 (6.157) into the equivalent expression in terms of stresses using Eq 6.155a, and combine that result with Eq 6.153 to obtain tij;kk+1 1+tkk;ij+(bi;j+bj;i)+ 1+ijbk;k=0 (6.158) which are the Beltrami-Michell equations of compatibility. In seeking solutions by either the displacement or stress formulation, we consider only the cases for which body forces are zero. 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 integral based upon the boundary conditions. Let us first consider solutions developed through the displacement formulation. Rather than attempt to solve the Navier equations directly, we express the displacement field in terms of scalar and vector potentials and derive equations whose solutions result in the required potentials. Thus, by inserting an expression for uiin terms of the proposed potentials into the Navier equations we obtain the governing equations for the appropri- ate potentials. Often such potentials are harmonic, or bi-harmonic functions. We shall present 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 Helmholtz which states that any vector function that is continuous and finite, and which vanishes at infinity, may be resolved into a pair of components: one a rotational vector, the other an irrotational vector. Thus, if the curl of an arbitrary vector ais zero, thenais the gradient of a scalar, andaisirrotational, or as it is sometimes called, solenoidal. At the same time, if the divergence of the vector ais zero, thenais the curl of another vector , and is a rotational vector. Accordingly, in keeping with the Helmholtz theorem, we assume that the displacement field is given by ui=;i+ipq q;p (6.159) where;iis representative of the irrotational portion, and curl the rotational portion. Substituting this displacement vector into Eq 6.156 with biin that equation taken as zero, namely ui;jj+(+)uj;ji=0; (6.160) we obtain ;ijj+"ipq q;pjj+(+):ijj+(+)"jpq q;pji=0 (6.161) which reduces to (+2);ijj+"ipq q;pjj=0; (6.162) since"jpq q;pji=0. In coordinate-free notation Eq 6.162 becomes (+2)rr2+rr2 =0: (6.163) Linear Elasticity 255 Any set ofand 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 is obtained by requiring and to be harmonic: r2=0; (6.164a) r2 =0: (6.164b) It should be pointed out that while Eq 6.164 is a solution of Eq 6.160, it is not the general solution of the Navier equations. If we choose r2=constant, and =0in Eq 6.164, the scalar function is known as the Lam´ e strain potential. By taking the divergence of Eq 6.162, and remembering that the divergence of a curl vanishes, we see that r4=0 (6.165) is a solution of the resulting equation so that a bi-harmonic function as also yields a solution forui. Similarly, by taking the curl of Eq 6.162 we find that r4 =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 as theGalerkin vector, and designated here by F=Fi^ei. In this approach we assume the displacement uiis given in terms of the Galerkin vector specifically by the equation ui=2(1-)Fi;jj-Fj;ji (6.167) which is substituted directly into Eq 6.160. Carrying out the indicated differentiation and reducing the resulting equations with the help of the identity =2= (1-2)we find that the Navier equations are satisfied if r4F=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 and withF. It has been shown that = -Fi;i; (6.169a) and "ijk k;j=2(1-)Fi;jj: (6.169b) IfFiis not only bi-harmonic, but harmonic as well, Eq 6.169b reduces to "ijk k;j=0; (6.170a) and the relationship between andFibecomes ;ii= -Fi;jji: (6.170b) In this caseis called the Lam ´e strain potential. Example 6.8 Consider a Galerkin vector of the form F=Fi^eiwhereF3is a function of the coordinates, that is F3=F3(x1;x2;x3). Apply this vector to obtain the solution 256 Continuum Mechanics for Engineers to the problem of a concentrated force acting at the origin of coordinates in the direction of the positive x3axis of a very large elastic body. This is called the Kelvin problem. Solution Letui=2(1-)Fi;jj-Fj;jias given by Eq 6.167. Accordingly, u1= -F3;31; u2= -F3;32; u3=2(1-) (F3;11+F3;22+F3;33)-F3;33: TakeF3to be proportional to the distance squared from the origin as dened by F3=BRwhereBis a constant and R2=x2 1+x2 2+x2 3. Thus, the displacements are u1=Bx3x1 R3; u2=Bx3x2 R3; u3=B4(1-) R-x2 1+x2 2 R3 : From these displacement components the stresses may be computed using Hooke's law. In particular, it may be shown that t33=@ @x3 (2-)@2 @x2 1+@2 @x2 2+@2 @x2 3 -@2 @x2 3 BR which upon carrying out the indicated dierentiation and combining terms be- comes t33= -B" 2(2-)x3 R3-3 x2 1+x2 2
x3 R5# : This equation may be written in a more suitable form for the integration that follows by noting that R2=r2+x2 3wherer2=x2 1+x2 2. The modied equation is t33= -B(1-2)x3 R3+3x3 3 R5 : Summing forces in the x3direction over the plane x3=constant allows us to determineBin terms of the applied force P. The required integral is P=Z1 0(-t33)2rdr: Butrdr=RdR and so P=2B (1-2)x3Z1 0dR R2+3x3 3Z1 0dR R4 from which we nd B=P 4(1-): Linear Elasticity 257 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 vector potential. For this we take the displacement vector to be represented by a scalar potential B, the vector potential V, and the position vector xiin the form ui=Vi-B;i-(Vkxk);i 4(1-)(6.171) which when substituted into the homogeneous Navier equations and simplified using the identity=2= (1-2)we obtain Vi;jj-(+2)B;ijj-1 2(+) (Vk;ijj+Vi;jj)=0: (6.172) These equations are clearly satisfied when r2V=0; (6.173a) and r2B=0 (6.173b) which indicates that any four harmonic functions, Viwithi=1;2;3 andB, will serve to provide a displacement vector uifrom Eq 6.171 that satisfies the Navier equations. Since the displacement vector has only three components, the four scalar functions, Vi, andB are not completely independent and may be reduced to three. It can be shown that these potentials are related to the Galerkin vector through the expressions V=2(1-)r2F; (6.174a) and B=r
F-V
x 4(1-): (6.174b) Whereas it is not usually possible to solve the Navier equations directly for problems involving 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 inner radiusr1and outer radius r2that is subjected to an internal pressure p1and an external pressurep2. Due to the symmetry condition here we assume a displacement field ui=(r)xi (6.175) wherer2=xixianddepends solely upon r. By direct substitution of Eq 6.175 into Eq 6.160 we arrive at the ordinary differential equation d2 dr2+4 rd dr=0 (6.176) for which the general solution may be written as (r)=A1+A2 r3(6.177) whereA1andA2are constants of integration depending on the boundary conditions. Eq 6.155 written in terms of displacement derivatives has the form tij=ijuk;k+(ui;j+uj;i): (6.178) 258 Continuum Mechanics for Engineers It follows by making use of Eqs 6.175 and 6.177 that tij=3A1ij+2 A1+A2 r3 ij-3A2xixj r3 : (6.179) Recall that the traction vector in the radial direction (see Eq 3.51) is N=tijninjwhich upon substitution of Eq 6.177 becomes N=(3+2)A1+4A2 r3(6.180) where the identity xi=rnihas been used. Similarly, the tangential traction can be calculated using S=tijvivjwhere the unit vectors viare perpendicular to ni. The result is S=(3+2)A1+2A2 r3(6.181) sincevixi=0. The constants A1andA2are determined from boundary conditions on the tractions. Clearly, N= -p1atr=r1; N= -p2atr=r2: Carrying out the indicated algebra, we have the well-known formulas N=p1r3 1-p2r3 2 r3 2-r3 1-r3 1r3 2 r3p1-p2 r3 2-r3 1; (6.182a) S=p1r3 1-p2r3 2 r3 2-r3 1+r3 1r3 2 2r3p1-p2 r3 2-r3 1: (6.182b) These equations may be easily modified to cover the case where p1=0, or the case where p2=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 body forces tij;j=0 (6.183) we propose the stress field tij="ipq"jkmqk;pm (6.184) whereqkis a symmetric tensor function of the coordinates. By a direct expansion of this equation the stress components may be expressed in terms of the potential qk. For example, t11="1pq"1kmqk;pm (6.185) which when summed over the repeated indices, keeping in mind the properties of the permutation symbol, becomes t11=223;23 -22;33 -33;22: (6.186a) Linear Elasticity 259 Similarly, t22=213;13 -11;33 -33;11; (6.186b) t33=212;12 -11;22 -22;11; (6.186c) t12=12;33 +33;12 -23;31 -13;32; (6.186d) t23=23;11 +11;23 -12;13 -13;12; (6.186e) t31=31;22 +22;31 -12;23 -32;21: (6.186f) It may be shown by direct substitution that the equilibrium equations are satisfied by these stress components. Upon setting the off-diagonal terms of qkto zero, that is, if 12=23=31=0, we obtain the solution proposed by Maxwell. By setting the diagonal terms of qkto zero, namely, 11=22=33=0we obtain the solution proposed by Morera which is known by that name. It is interesting to note that if all the components of qkexcept 33are zero, that component is the Airy stress function introduced in Section 6.7 as can be verified by Eq 6.186. Although the potential qkprovides us with a solution of the equilibrium equations, that solution is not compatible with the Beltrami-Michell equations except under certain conditions. 260 Continuum Mechanics for Engineers Problems Problem 6.1 In general, the strain energy density Wmay be expressed in the form W=C (; =1;:::;6 ) where C is not necessarily symmetric. Show that this equation may be rearranged to appear in the form W=1 2C where Cis symmetric, so that now @W @=C=t in agreement with Eq 6.8. Problem 6.2 Let the stress and strain tensors be decomposed into their respective spherical and deviator components. Determine an expression for the strain energy density Was the sum of a dilatation energy density W(1)and a distortion energy density W(2). Answer W=W(1)+W(2)=1 6tiijj+1 2Sijij Problem 6.3 If the strain energy density Wis generalized in the sense that it is assumed to be a function of the deformation gradient components instead of the small strain components, that is, if W=W(FiA), make use of the energy equation and the continuity equation to show that in this case Eq 6.16 is replaced by Jtij=@W @FiAFjA: Problem 6.4 For an isotropic elastic medium as dened by Eq 6.23, express the strain energy density in terms of (a) the components of ij, (b) the components of tij, (c) the invariants of ij. Linear Elasticity 261 Answer (a)W=1 2(iijj+2ijij), (b)W=(3+2)tijtij-tiitjj 4(3+2), (c)W=1 2+
(I)2-2II Problem 6.5 Lettijbe any second-order isotropic tensor such that t0 ij=aimajntmn=tij for any proper orthogonal transformation aij. Show that by successive applications of the transformations [aij]=2 40 0 -1 -1 0 0 0 1 03 5and [aij]=2 40 0 1 -1 0 0 0-1 03 5 every second-order isotropic tensor is a scalar multiple of the Kronecker delta, ij. Problem 6.6 Verify that Eqs 6.33a and 6.33b when combined result in Eq 6.28 when Eqs 6.29 are used. Problem 6.7 For an elastic medium, use Eq 6.33 to express the result obtained in Problem 6.2 in terms of the engineering elastic constants KandG. Answer W=1 2Kiijj+G ijij-1 3iijj
Problem 6.8 Show 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, (1),(2),(3)and (b) the principal strains,(1),(2),(3)in the form (a)W(2)= (1)-(2)
2+ (2)-(3)
2+ (3)-(1)
2 12G, (b)W(2)=1 3h (1)-(2)
2+ (2)-(3)
2+ (3)-(1)
2i G. Problem 6.9 Beginning with the denition for W(Eq 6.21a), show for a linear elastic material represented by Eq 6.23 or by Eq 6.28 that @W=@ij=tijand@W=@tij=ij(Note that@ij=@mn= imjn.) 262 Continuum Mechanics for Engineers Problem 6.10 For an isotropic, linear elastic solid, the principal axes of stress and strain coincide, as was shown in Example 6.1. Show that, in terms of engineering constants Eandthis result is given by (q)=(1+)(q)- (1)+(2)+(3) E;(q=1;2;3 ): Thus, letE=106psi and=0:25, and determine the principal strains for a body subjected to the stress eld (in ksi) [tij]=2 412 0 4 0 0 0 4 0 63 5: Answer (1)= -4:510-6,(2)=0:510-6,(3)=1310-6 Problem 6.11 Show for an isotropic elastic medium that (a)1 1+=2(+) 3+2, (b) 1+= +2, (c)2 1-2=3K 1+, (d)2(1+) =3K(1-2): Problem 6.12 Let thex1x3plane be a plane of elastic symmetry such that the transformation matrix betweenOx1x2x3andOx0 1x0 2x0 3is [aij]=2 41 0 0 0-1 0 0 0 13 5: 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 thex1axis be an axis of elastic symmetry of order N=2. Determine the form of the elastic constant matrix C, assuming C=C. Answer [C]=2 6666664C11C12C13C140 0 C12C22C23C340 0 C13C23C33C340 0 C14C24C34C440 0 0 0 0 0 C55C56 0 0 0 0 C56C663 7777775 Linear Elasticity 263 Problem 6.14 Assume that, by the arguments of elastic symmetry, the elastic constant matrix for an isotropic body has been reduced to the form [C]=2 6666664C11C12C120 0 0 C12C11C120 0 0 C12C12C110 0 0 0 0 0 C440 0 0 0 0 0 C440 0 0 0 0 0 C443 7777775: Show that, if the x1axis is taken as an axis of elastic symmetry of any order ( is arbitrary), C11=C12+2C44.(Hint: Expandt0 23=a2qa3mtqmand0 23=a2qa3mqm:) Problem 6.15 If the axis which makes equal angles with the coordinate axes is an axis of elastic symmetry of orderN=3, show that there are twelve independent elastic constants and that the elastic matrix has the form [C]=2 6666664C11C12C13C14C15C16 C13C11C12C16C14C15 C12C13C11C15C16C14 C41C42C43C44C45C46 C43C41C42C46C44C45 C42C43C41C45C46C443 7777775: Problem 6.16 For an elastic body whose x3axis is an axis of elastic symmetry of order N=6, show that the nonzero elastic constants are C11=C22,C33,C55=C44,C66=1 2(C11-C12), and C13=C23. Problem 6.17 Develop a formula in terms of the strain components for the strain energy density Wfor the case of an orthotropic elastic medium. Answer W=1 2(C111+2C122+2C133)1+1 2(C222+2C233)2 +1 2 C332 3+C442 4+C552 5+C662 6
Problem 6.18 Show that, for an elastic continuum having x1as an axis of elastic symmetry of order N=2, the strain energy density has the same form as for a continuum which has an x2x3 plane of elastic symmetry. Answer 2W =(C111+2C122+2C133+2C144)1+(C222+2C233+2C244)2 +(C333+2C344)3+C442 4+(C555+2C566)5+C662 6 264 Continuum Mechanics for Engineers Problem 6.19 Let the stress eld for a continuum be given by [tij]=2 4x1+x2t120 t12x1-x20 0 0 x 23 5 wheret12is a function of x1andx2. If the equilibrium equations are satised in the absence of body forces and if the stress vector on the plane x1=1is given byt(^e1)= (1+x2)^e1+(6-x2)^e2, determinet12as a function of x1andx2. Answer t12=x1-x2+5 Problem 6.20 Invert Eq 6.45b to obtain Hooke's law in the form tij=2G ij+ 1-2kkij which, upon combination with Eqs 6.44 and 6.43, leads to the Navier equation G ui;jj+1 1-2uj;ij +bi=0: This equation is clearly indeterminate for =0:5. However, show that in this case Hooke's law and the equilibrium equations yield the result Gui;jj+1 3tjj;i+bi=0: Problem 6.21 Let the displacement eld be given in terms of some vector qiby the equation ui=2(1-)qi;jj-qj;ji G: Show that the Navier equation (Eq 6.49) is satised providing bi0andqiis bi-harmonic so thatqi;jjkk =0. Ifq1=x2=randq2= -x1=rwherer2=xixi, determine the resulting stress eld. Answer t11= -t22=6QGx1x2=r5;t33=0 t12=t21=3QG x2 2-x2 1
=r5 t13=t31=3QGx2x3=r5;t23=t32= -3QGx1x2=r5; whereQ=4(1-)=G Linear Elasticity 265 Problem 6.22 If body forces are zero, show that the elastodynamic Navier equation (Eq 6.55) will be satised by the displacement eld ui=;i+"ijk k;j provided the potential functions and ksatisfy the three-dimensional wave equation. Problem 6.23 Show that, for plane stress, Hooke's law Eq 6.120a and Eq 6.121 may be expressed in terms of the Lam e constants andby ij=1 2 tij- 3+2ijtkk (i;j;k =1;2); kk= - 2(3+2)tii(i=1;2): Problem 6.24 For the case of plane stress, let the stress components be dened in terms of the function =(x1;x2), known as the Airy stress function, by the relationships, t11=;22; t22=;11; t12= -;12: Show thatmust satisfy the biharmonic equation r4=0and that, in the absence of body forces, the equilibrium equations are satised identically by these stress components. If=Ax3 1x2 2-Bx5 1whereAandBare constants, determine the relationship between A andBfor this to be a valid stress function. Answer A=5B Problem 6.25 Develop an expression for the strain energy density, W, for an elastic medium in (a) plane stress and (b) plane strain. Answer (a)W= t2 11+t2 22-2t11t22+2(1+)t2 12 =2E (b)W= +1 2
2 11+2 22
+1122+22 12 Problem 6.26 Show that=x4 1x2+4x2 1x3 2-x5 2is a valid Airy stress function, that is, that r4=0, and compute the stress tensor for this case assuming a state of plane strain with =0:25. Answer [tij]=2 424x2 1x2-20x3 2-4x3 1-24x1x2 20 -4x3 1-24x1x2 212x2 1x2+8x3 20 0 01 4 9x2 1x2-3x3 2
3 5 266 Continuum Mechanics for Engineers Problem 6.27 Verify 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 6.28 Develop appropriate constitutive equations for thermoelasticity in the case of (a) plane stress and (b) plane strain. Answer (a)ij= [(1+)tij-ijtkk]=E+ij(-0)(i;j;k =1;2) 33=tii=E+(-0) (i =1;2) (b)tij=ijkk+2ij-ij(3+2)(-0) (i;j;k =1;2) t33=tii-E( -0) =ii- (3 +2)(-0) (i =1;2) Problem 6.29 Consider the Airy stress function 5=D5x2 1x3 2+F5x5 2: (a) Show that for this to be valid stress function, F5= -D5=5. (b) Construct the composite stress function =5+3+2 where =D5 x2 1x3 2-1 5x5 2 +1 2B3x2 1x2+1 2A2x2 1: For this stress function show that the stress components are t11=D5 6x2 1x2-4x3 2
; t22=2D5x3 2+B3x2+A2; t12= -6D5x1x2 2-B3x1: Problem 6.30 A rectangular beam of width unity and length 2Lcarries a uniformly distributed load of qlb/ft as shown. Shear forces Vsupport the beam at both ends. List the six boundary conditions for this beam the stresses must satisfy using stresses determined in Problem 6.29. x2 x1c cOq [lb/ft] VV L L Linear Elasticity 267 Answer 1.t22= -q atx2= +c 2.t22=0atx2= -c 3.t12=0atx2=c 4.R+c -ct12dx2=qLatx1=L 5.R+c -ct11dx2=0 atx1=L 6.R+c -ct11x2dx2=0 atx1=L Problem 6.31 Using 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 x2 1x2-2 3x3 2 ; t22=q 2I1 3x3 2-c2x2-2 3c3 ; t12= -q 2I x1x2 2-c2x1
; whereI=2 3c3is the plane moment of inertia of the beam cross section. Problem 6.32 Show that, using the stresses calculated in Problem 6.31, the boundary conditions 4 and 5 are satised, but boundary condition 6 is not satised. Problem 6.33 Continuing Problems 6.31 and 6.32, in order for boundary condition 6 to be satised an additional term is added to the stress function, namely 3=D3x3 2: Show that, from boundary condition 6, D3=3q 4c1 15-L2 6c2 ; so that nally t11=q 21 x2 1-2 3x2 2+6 15c2-L2 x2: Problem 6.34 Show that for the shaft having a cross section in the form of an equilateral triangle the warping function is (x1;x2)= x3 2-3x2 1x2
: 268 Continuum Mechanics for Engineers Determine (a) the constant in terms of the shaft dimension, (b) the torsional rigidity K, (c) the maximum shearing stress. MtOx2 x12aa Answer (a)= -1 6a (b)K=9Gp 3a4 5 (c)t23jmax=3Mta 2Katx1=a,x2=0 Problem 6.35 Consider the Galerkin vector that is the sum of three double forces, that is, let ψ x2 x1x3 θr O^eψ^eθ^er Linear Elasticity 269 F=Bx1 r^e1+x2 r^e2+x3 r^e3 whereBis a constant and r2=xixi. Show that the displacement components are given by ui= -2B(1-2)xi r3: Using the sketch above, (spherical coordinates) observe that the radial displacement ur (subscriptrnot a summed indice, rather indicating the radial component of displacement) ur=uixi r; and show that ur= -2B (1-2)r-2. Also, show that u =u=0. Thus r=@ur @r=4B(1-2) r3and == -2B(1-2) r3; so that the cubical dilatation is r+ +=0. From Hooke's law tij=ijuk;k+2(ui;j+uj;i) which reduces here to tij=2ijso that trr=8BG(1-2) r3; and t =t= -4BG(1-2) r3: This page intentionally left blank 7 Classical 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 mechanics allows for the study of fluids. A fundamental characteristic of any fluid – be it a liquid or a gas – is that the action of shear stresses, no matter how small they may be, will cause the fluid to deform continuously as long as the stresses act. 7.1 Viscous Stress Tensor, Stokesian, and Newtonian Fluids A fluid at rest (or in a state of rigid body motion) is incapable of sustaining any shear stress whatsoever. This implies that the stress vector on an arbitrary element of surface at any point in a fluid at rest is proportional to the normal niof that element, but indepen- dent of its direction. Thus, we write t(^n) i=tijnj= -p0ni (7.1) where the (positive) proportionality constant p0is the thermostatic pressure or, as it is frequently called, the hydrostatic pressure. We note from Eq 7.1 that tij= -p0ij (7.2) which indicates that for a fluid at rest the stress is everywhere compressive, that every direction is a principal stress direction at any point, and that the hydrostatic pressure is equal to the mean normal stress, p0= -1 3tii: (7.3) This pressure is related to the temperature and densityby an equation of state having the form F(p0;;)=0: (7.4) For a fluid in motion the shear stresses are not usually zero, and in this case we write tij= -pij+ij (7.5) whereijis called the viscous stress tensor, which is a function of the motion and vanishes when the fluid is at rest. In this equation, the pressure pis called the thermodynamic pressure and is given by the same functional relationship with respect to andas that for the static pressure p0in the equilibrium state, that is, by F(p;; )=0: (7.6) 271 272 Continuum Mechanics for Engineers Note from Eq 7.5 that, for a fluid in motion, pis not equal to the mean normal stress, but instead is given by p= -1 3(tii-ii); (7.7) so that, for a fluid at rest (ij=0),pequates top0. 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 ijis a function of the rate of deformation tensor dij. Expressing this symbolically, we write ij=fij(D): (7.8) If the functional relationship in this equation is nonlinear, the fluid is called a Stokesian fluid. When fijdefinesijas a linear function of dij, the fluid is known as a Newtonian fluid, and we represent it by the equation ij=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 dijandij, allow us to reduce the 81 coefficients Kijpq to 2. We conclude that, for a homogeneous, isotropic Newtonian fluid, the constitutive equation is tij= -pij+ijdkk+2dij (7.10) whereandareviscosity 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(3+2)dii= -p +dii (7.11) where=1 3(3+2)is known as the coefficient of bulk viscosity. The condition =1 3(3+2) (7.12a) or, equivalently, = -2 3(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 3ijtkk (7.13a) for stress and ij=dij-1 3ijdkk (7.13b) for rate of deformation into Eq 7.10, we obtain Sij+1 3ijtkk= -pij+1 3(3+2)ijdkk+2ij (7.14) Classical Fluids 273 which may be conveniently split into the pair of constitutive equations Sij=2ij; (7.15a) tii= -3 (p-dkk): (7.15b) The first of this pair relates the shear effect of the motion with the stress deviator, and the second associates the mean normal stress with the thermodynamic pressure and the bulk viscosity. 7.2 Basic Equations of Viscous Flow, Navier-Stokes Equations Inasmuch as fluids do not possess a “natural state” to which they return upon removal of 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 equations for boundary value problems in viscous fluid theory. Thus, for the thermomechanical behavior of a Newtonian fluid, the following field equations must be satisfied: (a) the continuity equation (Eq 5.13) ˙+vi;i=0; (7.16) (b) the equations of motion (Eq 5.22) tij;j+bi=˙vi; (7.17) (c) the constitutive equations (Eq 7.10) tij= -pij+ijdkk+2dij; (7.18) (d) the energy equation (Eq 5.65) ˙u=tijdij-qi;i+r; (7.19) (e) the kinetic equation of state (Eq 7.6) p=p(;); (7.20) (f) the caloric equation of state (Eq 5.66) u=u(;); (7.21) (g) the heat conduction equation (Eq 5.62) qi= -;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) 274 Continuum Mechanics for Engineers represents 22 equations in the 22 unknowns, tij,,vi,dij,u,qi,p, and. If thermal effects are neglected and a purely mechanical problem is proposed, we need only Eqs 7.16 through 7.18 as well as Eq 7.23 and a temperature independent form of Eq 7.20, which we state as p=p(): (7.24) This provides a system of 17 equations in the 17 unknowns, tij,,vi,dij, andp. 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 making use of the definition Eq 7.23, we obtain ˙vi=bi-p;i+(+)vj;ji+vi;jj (7.25) which are known as the Navier-Stokes equations for fluids. These equations, along with Eqs 7.19, 7.20, and 7.21, provide a system of seven equations for the seven unknowns, vi,,p,u, and. Notice that even though Eq 7.18 is a linear constitutive equation, the Navier-Stokes equations are nonlinear because in the Eulerian formulation ˙vi=@vi @t+vjvi;j: If Stokes condition = -2 3
is assumed, Eq 7.25 reduces to the form ˙vi=bi-p;i+1 3(vj;ji+3vi;jj): (7.26) Also, if the kinetic equation of state has the form of Eq 7.24, the Navier-Stokes equations along with the continuity equation form a complete set of four equations in the four unknowns,viand. In all of the various formulations for viscous fluid problems stated above, the solutions must satisfy the appropriate field equations as well as boundary and initial conditions on both traction and velocity components. The boundary conditions at a fixed surface require not only the normal, but also the tangential component of velocity to vanish because of the “boundary layer” effect of viscous fluids. The initial and boundary conditions are 1. The velocity must be specified vi=v i(x;0) throughout the volume V. For problems where the density and temperature are changing the initial conditions at t=0are =(x;0); =(x;0): 2. The boundary conditions on the surface of the body Vfort>0 are vi=v i(x;t) onSv; tijnj=T i(x;t) onST; on those parts of the boundary SvandSTwhere the velocity and traction are speci- fied. In addition, the temperature and heat flux must also be specified for problems involving temperature changes. =(x;t) onS: ;ini=q(x;t) onSq: Classical Fluids 275 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 stress is everywhere zero, and the component along the boundary will be non-zero. The normal component of the velocity will be that of the body if the surface is impermeable. Also, at free surfaces the normal traction is equal to the ambient pressure when surface tension effects are negligible. Additionally the shear stress is zero at this surface. It should also be pointed out that the formulations posed in this section are relevant only for laminar flows. Turbulent flows require additional considerations. 7.3 Specialized Fluids Although the study of viscous fluids in the context of the equations presented in Section 7.2 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 practical interest. Here, we list some of the assumptions that are commonly made and consider briefly 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, and fluids which obey these conditions are called barotropic fluids. In particular, we may cite both isothermal changes (in which the temperature is constant) and adiabatic changes (for which no heat enters or leaves the fluid) as barotropic changes. (b)Incompressible fluids – If the density of a fluid particle is constant, the equation of state becomes =constant (7.27) which describes incompressibility. This implies ˙ =0and, by the continuity equation,vi;i=0for incompressible flows. Physically, incompressibility means that the elements of a fluid undergo no change in density (or volume) when subjected to a change in pressure. For incompressible flows, the Navier-Stokes equations become ˙vi=bi-p;i+vi;jj (7.28) due to thevi;i=0condition. Water and oil, among others, are generally assumed to be incompressible, whereas most gases are highly compressible. (c)Inviscid (frictionless) fluids – A fluid that cannot sustain shear stresses even when in motion is called an inviscid, or sometimes a perfect fluid. Clearly, if the coefficientsandin Eq 7.10 are equal to zero, that equation describes a perfect fluid and the Navier-Stokes equations reduce to ˙vi=bi-p;i (7.29) which are often referred to as the Euler equation of motion. An ideal gas is a perfect fluid that obeys the gas law p=R (7.30) 276 Continuum Mechanics for Engineers whereRis the gas constant for the particular gas under consideration. It should be pointed out that all real fluids are compressible and viscous to some degree. 7.4 Steady Flow, Irrotational Flow, Potential Flow If the velocity components of a fluid are independent of time, the motion is called a steady flow. In such cases, the material derivative of the velocity, ˙vi=@vi @t+vjvi;j reduces to the simpler form ˙v=vjvi;j: Thus, for a steady flow, the Euler equation is modified to read vjvi;j=bi-p;i: (7.31) Furthermore, if the velocity field is constant and equal to zero everywhere, the fluid is at rest, and the theory for this condition is called hydrostatics. For this, the Navier-Stokes equations are simply bi-p;i=0: (7.32) Assuming a barotropic condition between andp, it is possible to define a pressure function in the form P(p)=Zp p0dp : (7.33) In addition, if the body forces are conservative, we may express them in terms of a scalar potential function by the relationship bi= - ;i: (7.34) From the definition Eq 7.34, it follows that P;i=1 p;iorrP=rp ; (7.35) so that now Eq 7.32 may be written ( +P);i=0 (7.36) as the governing equation for steady flow of a barotropic fluid with conservative body forces. Example 7.1 An incompressible, Newtonian uid maintains a steady ow under the action of gravity down an inclined plane of slope . If the thickness of the uid per- pendicular to the plane is hand the pressure on the free surface is p=p0(a constant), determine the pressure and velocity elds for this ow. Classical Fluids 277 h x2 x3 0pp= βv2=ρgsinβ 2µ∗h2 Solution Assumev1=v3=0,v2=v2(x2;x3). By the continuity equation for incom- pressible ow, vi;i=0. Hence,v2;2=0andv2=v2(x3). Thus, the rate of deformation 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= -pij+2dij from which we calculate [tij]=2 6664-p 0 0 0 -p @v2 @x3 0 @v2 @x3-p3 7775: The equations of motion having the steady ow form tij;j+bi=vjvi;j result in component equations (fori=1) -p;1=0; (7.37a) (fori=2) -p;2+@2v2 @x2 3+gsin=0; (7.37b) (fori=3) -p;3-gcos=0; (7.37c) since gravity is the only body force, b=g(sin^e2-cos^e3) 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= -(gcos)x3+f(x2) 278 Continuum Mechanics for Engineers wheref(x2)is an arbitrary function of integration. At the free surface (x3=h), p=p0, and so f(x2)=p0+gh cos and thus p=p0+(gcos) (h-x3) which describes the pressure in the uid. Next, by integrating Eq 7.37b twice with respect to x3, we obtain v2=-g sin 2x2 3+ax3+b withaandbconstants of integration. But from the boundary conditions, 1.v2=0whenx3=0, thereforeb=0, 2.t23=0whenx3=h, thereforea=gh sin. Finally, therefore, from the equation for v2we have by the substitution of a= gh sin, v2=gsin 2(2h-x3)x3 having the prole shown in the gure. If the velocity field of a fluid is one for which the tensor Wvanishes identically, we say the flow is irrotational. In this case the vorticity vector w, which is related to Wby Eq4:154 , is also zero everywhere, so that for irrotational flow wi=1 2"ijkvk;j=0orw=1 2rv=1 2curlv=0: (7.38) Finally, from the identity curl(grad ) =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=;iorv=r: (7.39) Indeed, it may be shown that the condition curl v=0is a necessary and sufficient condi- tion for irrotational flow and the consequence expressed in Eq 7.39 accounts for the name potential flow often associated with this situation. For a compressible irrotational flow, the Euler equation and the continuity equation may be linearized and combined to yield the wave equation ¨=c2;ii (7.40) wherecis 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 to give c2ij-vivj
vj;i=0 (7.41) which is called the gas dynamics equation. For incompressible potential flow the continuity equation reduces to a Laplace equation, ;ii=0orr2=0 (7.42) Classical Fluids 279 solutions of which may then be used to generate the velocity field using Eq 7.39. It is worthwhile to mention here that the Laplace equation is linear, so that superposition of solutions is available. Example 7.2 Peristaltic pumping is a form of uid transport resulting from a wave pass- ing along a distensible uid containing tube. This form of transport is com- mon in biological systems. The gure below shows a two-dimensional repre- sentation of the tube. The shape of the wall is described by x2=h(x1;t)= a+bsin2 (x1-ct). In thex1-x2frame shown, the uid ow 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=^x2andv1=^v1+c. The ^x1-momentum equation for ows with negligible uid inertia is -dp d^x1+@2^v1 @^x2 2=0 with the boundary conditions ^v1= -c on^x2=h and@^v1 @^x2=0 on^x2=0: abc h(x,t) x1x2 l Two-dimensional gure modeling peristaltic pumping. Determine the velocity prole v1(x;t)in thex1-x2frame.1 Solution Integrating the governing equation twice gives ^v1=1 2dp d^x1^x2 2+C^x2+D: From the symmetry condition, @^v1 @^x2=1 dp d^x1(0)+C(^xx)=0 andC(^x2)=0: On^x2=h, we have ^v1(h)= -c =1 2dp d^x1h2+D(^x2) andD(^x2)= -c -1 2dp d^x1h2: 1see A. H. Shapiro, M. Y. Jaffrin, and S. L. Weinberg (1969) Peristaltic pumping with long wavelengths at low Reynolds number, Journal of Fluid Mechanics, 37(4), 799-825. 280 Continuum Mechanics for Engineers The velocity prole is ^v1(^x;t)=1 2dp d^x1^x2 2-h2
-c; which is transformed to the stationary frame giving v1(x;t)=1 2dp dx1 x2 2-h2
sincev1=^v1+canddx1=d^x1. The velocity prole is a simple parabolic distribution. 7.5 The Bernoulli Equation, Kelvin’s Theorem If a fluid is barotropic with conservative body forces, Eq 7.36 may be substituted on the right-hand side of Euler’s equation (Eq 7.29), giving ˙vi= -( +P);i: (7.43) As a step in obtaining a solution to this differential equation, we define a streamline as that space curve of which the tangent vector, at each point, has the direction of the fluid velocity (vector). For a steady flow, the fluid particle paths are along streamlines. By integrating Eq 7.43 along a streamline (see Problem 7.15), we can show that Zx2 x1@vi @tdxi+v2 2+ +P=G(t) (7.44) wheredxiis a differential tangent vector along the streamline. This is the well-known Bernoulli 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 G0is valid throughout the flow. When gravity is the only force acting on the body, we write =ghwhereg=9:81 m/s2is the gravitational constant and his a measure of the height above a reference level in the fluid. If hp=P=g is defined as the pressure head andhv=v2=2gas the velocity head, Bernoulli’s equation for incompressible fluids becomes h+hP+hv=h+p g+v2 2g=G0 (7.45) Recall that by Eq 2.97 in Chapter 2 we introduced Stoke’s theorem, which relates the line integral around a closed curve to the surface integral over its cap. By this theorem we define the velocity circulation Caround a closed path in the fluid as C=I vidxi=Z S"ijkvk;jnidS (7.46) Classical Fluids 281 whereniis the unit normal to the surface Sbounded by Canddxiis the differential tangent element to the curve C. Note that, when the flow is irrotational, curl v=0and the circulation vanishes. If we take the material derivative of the circulation by applying Eq 5.7 to Eq 7.46 we obtain ˙C=I (˙vidxi+vidvi): (7.47) For a barotropic, inviscid fluid with conservative body forces, this integral may be shown to vanish, leading to what is known as Kelvin’s theorem of constant circulation. 282 Continuum Mechanics for Engineers Problems Problem 7.1 Introduce the stress deviator Sijand the viscous stress deviator  ij=ij-1 3ijkkinto Eq 7.5 to prove that Sij= ij. Problem 7.2 Determine an expression for the stress power (a) tijdijand (b)ijdijfor a Newtonian uid. First, show that ij= -2 3 ijdkk+2dij: Answer (a)tijdij= -pdii+diidjj+2ijij (b)ijdij=(trD)2+2ijij Problem 7.3 Determine the constitutive equation for a Newtonian uid for which Stokes condition holds, that is, for=0. Answer tij= -pij+2ij Problem 7.4 Develop an expression of the energy equation for a Newtonian uid assuming the heat conduction follows Fourier's law. Answer ˙u= -pvi;i+vi;ivj;j+1 2(vi;j+vj;i)+;ii+r Problem 7.5 The dissipation potential for a Newtonian uid is dened as a function of Dandby =1 2djjdii+ijij, where=+2 3: Show that@ =@dij=ij. Problem 7.6 Verify the derivation of the Navier-Stokes equations for a Newtonian uid as given by Eq 7.25. Classical Fluids 283 Problem 7.7 Consider a two-dimensional ow parallel to the x2x3plane so that v1=0throughout the uid. Assuming that an incompressible, Newtonian uid undergoes this ow, develop a Navier-Stokes equation and a continuity equation for the uid. Answer (Navier-Stokes) ˙vi=bi-p;i+vi;jj(i;j=2;3) (Continuity) vi;i=0(i=2;3) Problem 7.8 Consider a barotropic, inviscid uid under the action of conservative body forces. Show that the material derivative of the vorticity of the uid in the current volume Vis d dtZ VwidV=Z SviwjnjdS: Problem 7.9 Show that for an incompressible, inviscid uid the stress power vanishes identically as one would expect. Problem 7.10 Show that the vorticity and velocity of a barotropic uid of constant density moving under conservative body forces are related through the equation ˙wi=wjvi;j. Deduce that for a steady ow of this uid vjwi;j=wjvi;j. Problem 7.11 In terms of the vorticity vector w, the Navier-Stokes equations for an incompressible uid may be written as ˙vi=bi-p;i-2"ijkwk;j: Show that, for an irrotational motion, this equation reduces to the Euler equation ˙vi=bi-p;i: Problem 7.12 Carry out the derivation of Eq 7.41 by combining the Euler equation with the continuity equation, as suggested in the text. Problem 7.13 Consider the velocity potential =x2x3=r2wherer2=x2 1+x2 2. Show that this satises the Laplace equation ;ii=0. Derive the velocity eld and show that this ow is both incompressible and irrotational. Problem 7.14 If the equation of state of a barotropic uid has the form p=kwherekandare constants, the ow is termed isentropic. Show that the Bernoulli equation for a steady 284 Continuum Mechanics for Engineers motion in this case becomes +kp (k+1)+1 2v2=constant: Also, show that for isothermal ow the Bernoulli equation takes the form +pln +1 2v2=constant: Problem 7.15 Derive Eq 7.44 by taking the scalar product of dxi(the dierential displacement along a streamline) with Eq 7.43 and integrating along the streamline, that is, by the integration of Zx2 x1(˙vi+ ;i+P;i)dxi: Problem 7.16 Verify that Eq 7.47 is the material derivative of Eq 7.46. Also, show that for a barotropic, inviscid uid subjected to conservative body forces the rate of change of the circulation is zero (See Eq 7.47). Problem 7.17 Determine the circulation Caround the square in the x2x3plane shown in the gure if the velocity eld is given by v= x3-x2 2
^e2+(x3+x2)^e3: (1,−1)(−1,1) (−1,−1)(1,1)x3 x2 O Answer C=0 8 Nonlinear 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 ability to recover initial dimensions after large strain, was not possible with natural latex until Charles Goodyear discovered vulcanization in 1939. Vulcanization is a chemical reaction known as cross-linking which turned liquid latex into a non-meltable solid (thermoset) . Cross-linked rubber would also allow considerable stretching with low damping; strong and 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, many other 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 stretching in 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. While TPEs 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 more difficult than with metals. For starters, the stress-strain response, that is, the constitutive response, is quite different. Figure 8.1(a) shows the stress-strain curves for a mild steel specimen along with the response of a natural rubber used in an engine mount. Note that the rubber specimen strain achieves a much higher stretch value than the steel. The dashed 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. In fact, 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. Golf ball cores are much stiffer than a rubber band, for instance. Each of these products is designed for different strain regimes. A golf ball’s maximum strain would be on the order of 30 to 40%. Its highly crosslinked constitution is made for resilience, not for large strain. The rubber stress-strain curve exhibits nonlinear behavior from the very beginning of its deformation, whereas steel has a linear regime below the yield stress. The reason rubber materials exhibit drastically different behavior than metals results from 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 of carbon 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 long chain 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. 285 286 Continuum Mechanics for Engineers StretchSteel Stress, Pa (dashed line) 010002000 1 1.002 1.004 1.006 1.008 1.010510x 105 Rubber Stress, Pa (solid line) (a) Steel and rubber on strain scale appropriate for steel. The rubber stress is low and nearly coinciding with the abscissa. 2000 1000 010x 105 2.5 305 2 1.5 1 StretchSteel Stress, Pa (dashed line) Rubber Stress, Pa (solid line) (b) Steel and rubber on strain scale appropriate for rubber. The strain values for the steel are small making the curve to be nearly coincides with the ordinate. FIGURE 8.1 Nominal stress-stretch curves for rubber and steel. Note the same data is plotted in each figure, however, the stress axes have different scale and a different strain range is represented. Nonlinear Elasticity 287 In the context of this book’s coverage of continuum mechanics, material make-up on the micro-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 somewhat represents material behavior at the macroscopic level. In this chapter, rubber elasticity will be developed from a first-principle basis. Following that, the traditional continuum approach 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 to obtain stress-stretch response. 8.1 Molecular Approach to Rubber Elasticity One of the major differences between a crystalline metal and an amorphous polymer is that 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 ease at which different conformations are achieved results from the bonding between carbon atoms. 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 to100;000 . With each bond having a certain degree of freedom to orient itself, the number of conformations becomes 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. In addition to the large number of conformations for a single chain, there is another reason the statistical approach is appropriate: the actual polymer has a large number of different individual chains making up the bulk of the material. For instance, a cubic meter of an amorphous polymer having 10,000 carbon atoms per molecule would have on the order of1024molecules (McCrum et al., 1997). Clearly, the sample is large enough to justify a statistical 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 nsegments each having lengthl, the total length would be L=nl. Separating the molecule’s ends, the length Lwould mean there is only one possible conformation keeping the chain intact. As the molecule’s ends get closer together, there are more possible conformations that can be obtained. Thus, a Gaussian distribution of conformations as a function of the distance between chain ends is appropriate. Figure 8.2 demonstrates how more conformations are possible 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 is shown in Fig. 8.3. Adding the segment lengths algebraically gives distance rfrom end-to- end, but it does not give an indication of the length of the chain. If the ends are relatively close 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 of individual segments r2=r
r=(l1+l2+


+ln)
(l1+l2+


+ln) (8.1) whereliis the vector defining the ithsegment of the molecule chain. Multiplying out the right-hand side of Eq 8.1 leads to r2=nl2+[l1
l2+l1
l3+


+ln-1
ln]: (8.2) 288 Continuum Mechanics for Engineers L FIGURE 8.2 A schematic comparison of molecular conformations as the distance between molecule’s ends varies. Dashed lines indicate other possible conformations. x3 x2 x1 rdV = dx1dx2dx3 FIGURE 8.3 A freely connected chain with end-to-end vector r. Nonlinear Elasticity 289 This would be the square of the end-to-end vectors for one molecule of the polymer. In a representative volume of the material there would be many chains from which we may form the mean square end-to-end distance r2 =1 NNX 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 large number of molecules is taken in the sample, it is reasonable that, for every individual productl1
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 expression for the mean end-to-end distance r2 =nl2(8.4) The mean end-to-end distance indicates how many segments, or carbon atoms, are in a specific chain. To address the issue of how the end-to-end distances are distributed throughout the polymer, a Gaussian distribution is assumed. Pick the coordinate’s origin to be at one end of a representative chain. Figure 8.3 shows this for a single chain, with the other end of the chain in an infinitesimal volume dVlocated by the vector r. The probability of the chain’s end lying in the volume dVis given by P(r)dr=e-(r )2 p
3dr (8.5) whereis a parameter of the distribution. Using this assumed distribution of mean end-to-end distances, it is straightforward to find r2 0=Z1 0r2P(r)dr=3 22(8.6) where the subscript 0denotes that this is an intrinsic property of the chain since it was considered alone. When the chain is placed back into a crosslinked network of chains, the mean 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 with other chains. Equating Eqs 8.4 and 8.6 the distribution parameter is found to be =r 2n 3l: (8.7) Similar to the results of Section 5.9, the force created by stretching a uniaxial specimen is given in terms of the Helmholtz free energy F=@ @L ;V(8.8) whereFis the force, Lis the length, and subscripts andVdesignate that the change in length occurs at constant temperature and volume. Substitution of Eq 5.74b yields the force in terms of the internal energy and entropy F=@u @L-@ @L(8.9) 290 Continuum Mechanics for Engineers 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. Atoms are 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 force contribution 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 derivative term 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 in conformations 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 mean end-to-end length. Noting the number of configurations is proportional to the probability per unit volume, P(r), and using Boltzmann’s equation, the entropy may be written as =0+klnP(r)=0+k" 3lnp
+r 2# (8.10) wherekis Boltzmann’s constant. Use of this in Eq 8.9 for an ideal rubber gives a single chain retractive force given by F=2k 2r2: (8.11) Consider a polymer having forces applied resulting in stretch ratios 1,2, and3. The work done on each chain of the material is the sum of the work done in each coordinate directionxi W(i)=Z(i)xi xifidxi=2k 2Z(i)xi xixidxi=k 2 (i)2 -1 x2 i(no sum oni): Taking into consideration the work done on all of the chains gives total work in each coordinate direction X nW(i)=k 2 (i)2 -1X nx2 i(no sum oni) (8.12) where the last summed term is the number of chains, n, times the initial mean end-to-end distance in the xidirection. Assuming the rubber is initially isotropic yields X nx2 i=n x2 i i=n 3 r2 i(no sum oni): Substituting this and from Eq 8.6 into Eq 8.12 and adding all three coordinate work terms gives W=nk 2 r2 i hr2i0 2 1+2 2+2 3-3 : (8.13) For convenience, this equation may be written as W=VG 2 2 1+2 2+2 3-3 (8.14) where Vis the volume and Gis the shear modulus which are given by N=n V; Nonlinear Elasticity 291 L0LA0A FIGURE 8.4 Rubber specimen having original length L0and cross-section area A0stretched into de- formed shape of length Land cross section area A. and G=Nk r2 i hr2i0: Next, considering a uniaxial tension of a specimen (Fig. 8.4) the stretching ratios reduce to 1=; 2=3=1p (8.15) where the nearly incompressible nature has been used in the form 123=1. The total work done is W=VG 2 2+2 -3 : (8.16) This is the work done on the polymer chains, but it is the same work done by external forces since an ideal rubber is assumed. Thus, the work shown in Eq 8.16 is equal to the change in Helmholtz free energy. Recalling the force from deformation as defined by Eq 8.8, the force resulting from a stretch is given by F=@ @L=dW dL=dW dd dL: The deformation is volume preserving, having V=A0L0=ALwhere the subscript 0 denotes the initial area and length. Since the stretch ratio in this case is =L=L0, it is clear thatd=dL =1=L0. Using this result and differentiating Eq 8.16 results in F=VG L0 -1 2 =A0G -1 2 292 Continuum Mechanics for Engineers which may be written as f=F A0=G -1 2 : (8.17) Materials satisfying this equation are called neo-Hookean. 8.2 A Strain Energy Theory for Nonlinear Elasticity The theory developed in the previous section does not represent most experimental data well at large strains. A better approach to modeling the response of rubbers comes from assuming the existence of a strain energy which is a function of the deformation gradient in the form of the left deformation tensor Bij=Fi;AFj;A. This approach, first published by Mooney (1940) and furthered by Rivlin (1948), actually predates the molecular approach discussed in Section 8.1. The basis of Mooney’s and subsequent theories is the initially isotropic material has to obey certain symmetries with regards to the functional form of the strain energy function. Assume the strain energy per unit volume to be an isotropic function of the strain in the form of the right deformation tensor invariants I1,I2, andI3 W=W(I1;I2;I3) (8.18) where I1=CAA; I2=1 2(CAACBB-CABCAB); (8.19) I3="MNOC1MC2NC3O=det[CAB]: Note that if principal axes of CABare chosen the invariants of Eq 8.19 are written in terms of the stretch ratios 1,2, and3as follows: I1=2 1+2 2+2 3; I2=2 12 2+2 23 3+2 12 3; (8.20) I3=2 12 22 3: Also, it is easy to show that I1=CAA=Bii=IB=IC,I2=IIB=IICandI3=IIIB= IIICby using the definitions of the right and left deformation tensors. An expression for the Cauchy stress in terms of the strain energy density comes from the local energy equation in the form ˙u=1 tijdij: (8.21) Since0uWandWis a function of the right deformation tensor the derivative of u may be written as ˙u=1 0@W CAB˙CAB: (8.22) Noting that ˙CAB=dij(FjAFiB+FiAFjB) Nonlinear Elasticity 293 allows us to write ˙u=2 JFiA@W CABFjBdij: (8.23) Equating Eqs 8.21 and 8.23 for ˙ uand factoring out dijleaves dij tij-2J-1FiA@W @CABFjB =0 (8.24) which must hold for all motions. Thus, the stress in a compressible material may be written 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,I2andI3are as given in Eq 8.19. Partial derivatives of the invariantsI1,I2andI3may be completed by noting the components are independent quantities. Thus, @CAB @CMN=AMBN; (8.27) and the partial derivatives of the invariants may be written as @I1 @CAB=@CMM @CAB=MAMB=AB (8.28a) @I2 @CAB=1 2ABCNN+1 2CNNAB-MANBCMN =I1AB-CAB (8.28b) @I3 @CAB=1 2"AMN"BQRCMQCNR=I3C-1 AB: (8.28c) These terms are further simplified when put into Eq 8.26 since they are multiplied by FiAFjB: FiAABFjB=FiAFjA=Bij FiACABFjB=FiAFkAFkBFjB=BikBkj: Noting thatJ-1=I1 2 3, the stress components may be written as tij=0ij+1Bij+2BikBkj (8.29) where 0=2I1 2 3@W @I3; (8.30a) 1=2I-1 2 3@W @I1+I1@W @I2 ; (8.30b) 2= -2I-1 2 3@W @I2: (8.30c) 294 Continuum Mechanics for Engineers A second form for the Cauchy stress can be found by use of the Cayley-Hamilton theorem: tij=0ij+1Bij+-1B-1 ij (8.31) where 0=2I-1 2 3 I2@W @I2+I3@W @I3 ; (8.32a) 1=2I-1 2 3@W @I1; (8.32b) -1= -2I-1 2 3@W @I2: (8.32c) When a specific material model is chosen defining the specific form of the strain energy density (Eq 8.18), Eq 8.31 can be interpreted. The material parameters are represented by 0,1and-1, and the deformation is represented by ij,BijandB-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 variety of problems can be solved by assuming incompressibility. The incompressible response of the material can be thought of as a constraint on the deformation gradient. That is, the incompressible nature of the material is modeled by an addition functional dependence between 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) orJ=0 (Eq 5.17a). With the density remaining constant and Eq 5.17a as the expression of the continuity equation it is clear that J=det[FiA]=1: (8.33) At this point, we digress from specific incompressibility conditions to a more general case of a continuum with an internal constraint. Assume a general constraint of the form (FiA)=0; (8.34) or, sinceCAB=FiAFiB, ˜(CAB)=0: (8.35) This second form of the internal constraint has the advantage that it is invariant under superposed rigid body motions. Differentiation of results in @ @CAB˙CAB=@ @CAB˙FiAFiB+FiA˙FiB
(8.36) where it is understood partial differentiation with respect to a symmetric tensor results in a symmetric tensor. That is, @ @CAB1 2@ @CAB+@ @CBA : (8.37) Note that ˙CABis essentially the same as ˙EABwhich is given in Eq 4.149. Thus, Eq 8.36 may be written as ijdij=0 (8.38) whereijis defined by ij=1 2FiA@ @CAB+@ @CBA FjB: (8.39) Nonlinear Elasticity 295 Assume the stress is formed by adding stress components ^tij, derivable from a con- stitutive response, to components of an arbitrary stress ¯tijresulting from the internal constraint: tij=^tij+¯tij: (8.40) The added arbitrary stress components ¯tijare assumed to be workless, that is, ¯tijdij=0: (8.41) Comparing Eqs 8.38 and 8.41 shows that both ijand ¯tijare orthogonal to dij, so¯tijmay be written as ¯tij=ij: (8.42) For an incompressible material, dii=0which may be written as ijdij=0; (8.43) or equivalently by choosing ij=ijin Eq 8.38. Substituting this into Eq 8.42 yields ¯tij= -pij (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= -pij+^tij (8.45) where the second term in Eq 8.45 is determined from the constitutive response for Cauchy stress. For an incompressible material J=1, henceI3=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, incompressible material are found to be tij= -pij+1Bij+2BikBkj (8.47) where 1=2@W @I1+I1@W @I2 ; (8.48a) 2= -2@W @I2: (8.48b) From Eq 8.31, the stress for an isotropic, incompressible material may be written in the alternative form tij= -pij+1Bij+-1B-1 ij (8.49) where 1=2@W @I1; (8.50a) -1= -2@W @I2: (8.50b) At this point, the strain energy has been assumed to be a function of I1andI2, but the exact functional form has not been specified. Rivlin (1948) postulated the strain energy should be represented as a general polynomial in I1andI2 W=X C(I1-3)(I2-3): (8.51) 296 Continuum Mechanics for Engineers It is noted that the strain energy is written in terms of I1-3andI2-3rather thanI1and I2to 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=Gand all other coefficients zero results in a neo-Hookean response where Gis the shear modulus. Stresses are evaluated with Eq 8.49 and used in the equations of motion, Eq 5.22. This results in a set of differential equations solved with the use of the problem’s appropriate boundary conditions. The indeterminate pressure is coupled with the equations of motion and is found by satisfying the boundary conditions. 8.3 Specific Forms of the Strain Energy When confronted with the design of a specific part made of a rubber-like material, say the jacket or lumbar of an automotive impact dummy or a golf ball core, testing must be done to evaluate the various constants of the strain energy function. But before the testing can be analyzed, the specific form of the strain energy must be chosen. This is tantamount to choosing how many terms of the Mooney-Rivlin strain energy, Eq 8.51, will be assumed nonzero to effectively represent the material response. There are other common forms of the strain energy which are all somewhat equivalent to the form put forward by Mooney (see Rivlin, 1976). Since constants Cof Eq 8.51 do not represent physical quantities like, 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 parameter model 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 principal stretches1,2, and3means Eq 8.52 is written as W=G 2 1+2 2+2 3-3
: (8.53) For the case of uniaxial tension the stretches are 2 1=2,2 2=2 3=-1and furthermore, if the material is incompressible 2 12 22 3=1. Thus, W=G 2+2 -3 : Using this expression in Eq 8.47 for the case of uniaxial tension yields the stress per unit undeformed area as f=P11=@W @=2G -1 2 (8.54) whereP11is the11-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 be linear in the first and second invariants of the left deformation tensor. Again, assuming the material is isotropic and incompressible, the strain energy may be written as W=C1 2 1+2 2+2 3-3
+C21 2 1+1 2 2+1 2 3-3 : (8.55) Nonlinear Elasticity 297 For a uniaxial test, the principal stretches are 2 1=2,2 2=2 3=-1. Substitution of these stretches into Eq 8.55 and differentiating with respect to gives the uniaxial stress per unit undeformed area f=P11=2 -1 2 C1+C2  : (8.56) The Cauchy stress is easily formed by noting the area in the deformed configuration would be found by scaling dimensions in the x2andx3direction by 2and3, respec- tively. For the uniaxial case, this means that multiplying the force per undeformed area by stretchresults in the uniaxial Cauchy stress t11=2 2-1  C1+C2  : (8.57) Making use of Eq 4.109 while noting in that equation is in Eq 8.57, a formula for stress in terms of strain is obtained t11=2C1 1+2E11+1p1+2E11 +2C2p 1+2E11+1 1+2E11 : (8.58) Series expansion of the last three terms of Eq 8.58 followed by assuming E11is small results 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)for small strain. The simple, two-term Mooney-Rivlin model represents material response well for small to moderate stretch values, but for large stretch higher order terms are needed. Some of these 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=X nn n n 1+n 2+n 3-3 : (8.60) This form reduces to the Mooney-Rivlin material if ntakes on values of 1and2with 1=2,2= -2,1=2C1, and2= -2C2. 8.4 Exact Solution for an Incompressible, Neo-Hookean Material Exact 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. A neo-Hookean material is one whose uniaxial stress response is proportional to the combi- nation of stretch -1=2as was stated in Section 8.1. The proportionality constant is the shear modulus, G. The strain energy function in the neo-Hookean case is proportional to I1-3whereI1is the first invariant of the left deformation tensor. A strain energy of this form leads to stress components given by tij= -pij+GBij (8.61) 298 Continuum Mechanics for Engineers wherepis the indeterminate pressure term, Gis the shear modulus, and Bijis the left deformation tensor. The stress components must satisfy the equilibrium equations which, in the absence of body 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 governing equilibrium @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=Gr2xi (8.64) where @XB @xj@xj @XA=AB; and @2xj @XA@XB@XB @xj=0 have been used. Eq 8.64 may be made more convenient by writing the pressure term as a function of the reference configuration and using the chain rule @P @XA=G@xj @XAr2xi (8.65) wherePis a function of XAandt. 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]=2 666664@x @X@x @Y0 @y @X@y @Y0 0 0 13 777775: (8.67) Note that we have departed from the strict continuum summation convention for clarity in solving this specific problem. Substitution of the deformation gradient into the definition for the left deformation tensor and use of Eq 8.61 yields txx= -p +G"@x @X2 +@x @Y2# ; (8.68a) tyy= -p +G"@y @X2 +@y @Y2# ; (8.68b) txy=G@x @X@y @X+@x @Y@y @Y ; (8.68c) Nonlinear Elasticity 299 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] =1may 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-1 iA =2 6666664@X @x@X @y@X @z @Y @x@Y @y@Y @z @Z @x@Z @y@Z @z3 7777775=2 666664@y @Y-@x @Y0 -@y @X@x @X0 0 0 13 777775: (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 quantity =(txx+tyy)=2for future convenience txx=+1 2G"@X @y2 +@Y @y2 -@X @x2 -@Y @x2# ; (8.72a) tyy=-1 2G"@X @y2 +@Y @y2 -@X @x2 -@Y @x2# ; (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 equations are associated with the xandydirections. Using the stress components as defined in Eq 8.72, the equilibrium conditions are @ @x=G@X @xr2X+@Y @xr2X ; (8.73a) @ @y=G@X @yr2X+@Y @yr2X (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 satisfied to ensure equilibrium and the constraint of incompressibility. To actually solve a problem, a solution must be found that satisfies the appropriate boundary conditions. Often the way this is done is by assuming a specific form for X(x;y) andY(x;y) and demonstrating that the aforementioned equations are satisfied. Following that, the boundary conditions are defined and demonstrated, completing the solution. In a sense, this is similar to 300 Continuum Mechanics for Engineers the semi-inverse method discussed in Chapter 6. The assumed “guess” of the functional form ofX(x;y) andY(x;y) requires some experience and or familiarity with the problem at hand. Here, a function form will be assumed in pursuit of the solution of a rectangular rubber specimen being compressed in the xdirection. The faces of the specimen are perfectly attached to rigid platens. In this case, assume that X=f(x). After a modest amount of work, this will be shown to lead to a solution of a neo-Hookean material compressed between rigid platens. Substituting the assumed form of Xinto the incompressibility equation, Eq 8.74, yields the ordinary differential equation f0@Y @y=1 (8.75) where the prime following frepresents differentiation with respect to x. Definingq(x) = [f0(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 XandY, the incompressibility condition is satisfied, and the equilibrium equations may be written as follows: @ @x=G[f0f00+(yq0+g0) (yq00+g00)]; (8.77a) @ @y=G[q(yq00+g00)]; (8.77b) where, again, primes after a symbol represent differentiation with respect to x. The second equation, 8.77b, is easily integrated to obtain an expression for   G=1 2qq00y2+qg00y+M(x) (8.78) whereM(x) is a function of integration. Differentiation of Eq 8.78 with respect to xresults in an equation that must be consistent with Eq 8.77a. This comparison gives rise to the following three ordinary differential equations: 1 2d dx(qq00)=q0q00; (8.79a) d dx(qq00)=g0q00+q0g00; (8.79b) M0(x)=f0f00+g0g00: (8.79c) Eq 8.79a may be expanded and integrated to obtain q00=k2qfrom which the function q(x) is determined to within the constants AandB: q(x)=Aekx+Be-kx: (8.80) Eqs 8.79a and 8.79b may be combined along with the fact that q00=k2qto obtain the functiong(x) g(x)=Cekx+De-kx(8.81) whereCandDare constants. With functions q(x) andg(x) found it is now possible to write functions X(x;y) andY(x;y) as X=Zdx Aekx+Be-kx; (8.82a) Nonlinear Elasticity 301 2H 2LX Y (a) Rubber specimen in reference config- uration. 2H 2lx y(b) Rubber specimen in deformed config- uration. The dashed line represents the undeformed shape. FIGURE 8.5 Rhomboid rubber specimen compressed by platens. Y= Aekx+Be-kx
y+Cekx+De-kx: (8.82b) In pursuit of the compressed rectangular rubber solution, constants CandDare taken to be identically zero and constants AandBare set equal. Thus, X=(kA)-1 tan-1 ekx
+C0 ; (8.83a) Y=2Ay coshkx; (8.83b) wherek,A, andC0are constants that must be determined. The deformation considered here 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=0when x=0to obtain constant C0, of Eq 8.83a, to have the value 0:25 . Since the platens are fixed to the specimen, Y=yon planesx=lthe constant Ais found to be1 2[cosh(kl)]-1. Fi- nally, the constant kmay be evaluated in terms of the deformed and undeformed lengths landLby kL=2cosh(kl)h tan-1 ekl
- 4i : (8.84) To obtain an exact solution it must be shown that the barreled surfaces initially at Y=Hare stress free in the deformed configuration. Unfortunately, this is not possible for this problem. Instead, a relaxed boundary condition may be satisfied by enforcing the resultant 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 2Gcosh2(kx) cosh2(kl); (8.85a) tyy= -p +G" k2y2sinh2(kx) sinh2(kl)+4cosh2(kl) cosh2(kx)# ; (8.85b) txy= -1 2Gkysinh2(2kx) cosh2(kl): (8.85c) 302 Continuum Mechanics for Engineers txx tyytxy txy txxx y FIGURE 8.6 Rhomboid rubber specimen compressed by platens. At this point, the stresses are known to within the additive pressure term p. A solution is sought by picking psuch that the average force over the barreled edge is zero. Symmetry of the deformation allows for considering the top half, y>0, in determining the condition for zero resultant stress on the barreled top edge (Fig. 8.6). Integrating the stresses yieldsZl 0tyyjy=0dx=ZH 0txyjx=ldy; (8.86) or p=1 4GkH2 lsinh(2kl) cosh2(kl)-4G kcosh2(kl)tanh(kl): (8.87) Knowing the pressure term, the compressive force may be determined from F=ZH 0txxjx=ldy: (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- icsP-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 . Further Developments of the General Theory,” Phil. Trans. Roy. Soc., A241, pp. 379-397 [6] Rivlin, R.S. (1949), “Large Elastic Deformations of Isotropic Materials: VI. Further Results in the Theory of Torsion, Shear and Flexure,” Phil. Trans. Roy. Soc., A242, pp. 173-195 Nonlinear Elasticity 303 [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 of Solid Polymers, Wiley & Sons, Inc., Chichester, UK 304 Continuum Mechanics for Engineers Problems Problem 8.1 Referred to principal axes, the invariants of the Green deformation tensor CABare I1=2 1+2 2+2 3; I2=2 12 2+2 12 3+2 22 3; I3=2 12 22 3: For an isotropic, incompressible material, show that I1=2 1+2 2+1 2 12 2; I2=1 2 1+1 2 2+2 12 2: Problem 8.2 Derive the following relationships between invariants I1,I2, andI3, and the deformation gradient,CAB: (a)@I1 @CAB=AB, (b)@I2 @CAB=CAB-I1AB, (c)@I3 @CAB=I3C-1 AB. Problem 8.3 Use the denitions of I1andI2in terms of the principal stretches 1,2, and3to show (a)@W @1=2 1 2 1-2 3
@W @I1+2 2@W @I2 , (b)@W @2=2 2 2 2-2 3
@W @I1+2 1@W @I2 . Problem 8.4 LetW(I1;I2;I3)be the strain energy per unit volume for a homogeneous, isotropic material. Show that the rst Piola-Kirchho stress components may be written as follows: PiA@W @FiA=2@W @I1FiA+2@W @I2(Bij-I1ij)FjA+2I3@W @I3F-1 iA: Problem 8.5 The Cauchy stress is given by tij=1 JFjAPiA: Nonlinear Elasticity 305 Start with the result of Problem 8.4 to show that tij=2 J@W @I1Bij+@W @I2(Bik-I1ik)Bjk+I3@W @I3ij : Problem 8.6 Assuming a strain energy of the form W=w(1)+w(2)+w1 12 for an isotropic, incompressible material, show that 1@3W @2 1@2=2@3W @2 2@1: Problem 8.7 For 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 denitions of invariants I1,I2, andI3in terms of stretches 1,2, and 3to show (I2-3)2=1 4 1+1 4 2+1 4 3+2I1-6I2+9: (b) Substitute the results from (a) into the strain energy above to obtain w(1)=(C1+2C3)2 1+(C2-6C3)-2 1+C3-4 3-(C1+C2-3C3) where W=w(1)+w(2)+w1 12 : Problem 8.8 Consider a material having reference conguration coordinates (R;;Z )and current con- guration coordinates (r;;z ). Assume a motion dened by r=R g(),=f(),z=Z: DetermineFiAandBijin terms ofg,g0, andf0. Answer FiA=2 6666641 g-g0 g20 0f0 g0 0 0 13 777775,Bij=2 6666641 g2+g0 g22 -f0g0 g30 -f0g0 g3f0 g2 0 0 0 13 777775 306 Continuum Mechanics for Engineers Problem 8.9 Show t11=2 2 1--2 1-2 2
@W @I1-2 -2 1-2 12 2
@W @I2; t22=2 2 2--2 1-2 2
@W @I1-2 -2 2-2 12 2
@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 p from the fact that t33=0condition. Problem 8.10 The following compression force-de ection data was obtained for a highly lled, polybuta- diene rubber having initial gauge length of 0:490 inand an undeformed cross-section area of1in2 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 constantsC1=1;550 andC2= -500represent the material for the range and type of loading given. Nonlinear Elasticity 307 (c) Generate a set of signicantly dierent constants C1andC2which might equally well model the material for this range and type of loading, showing that C1and C2are not unique. This page intentionally left blank 9 Linear Viscoelasticity The previous chapters have considered constitutive equations that deal primarily with two different types of material behavior: elastic response of solids and viscous flow of fluids. 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 ranging from 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 linear elastic constitutive equation. However, at a somewhat elevated temperature, the same material may have to be modeled as a viscous fluid. Polymers are by no means the only materials that exhibit different behavior under altered temperature/frequency conditions. Steel, as well as aluminum, copper, and other metals, becomes molten at high temperatures and can be poured into molds to form ingots. Additionally, at a high enough deformation rate, for example, at the 48 km/hr rate of a vehicle crash, steel will exhibit considerably altered stiffness properties. Just as continuum mechanics is the basis for constitutive models as distinct as elastic solids (stress/strain laws) and viscous fluids (stress/strain-rate laws), it also serves as the basis 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 that of linear viscoelasticity. 9.1 Viscoelastic Constitutive Equations in Linear Differential Operator Form One of the principal features of elastic behavior is the capacity for materials to store mechanical energy when deformed by loading, and to release this energy totally upon removal of the loads. Conversely, in viscous flow, mechanical energy is continuously dis- sipated with none stored. A number of important engineering materials simultaneously store 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/unloading cycle. This behavior is referred to as viscoelastic. In general, viscoelastic behavior may be imagined as a spectrum with elastic deformation as one limiting case and viscous flow the other extreme case, with varying combinations of the two spread over the range between. Thus, valid constitutive equations for viscoelastic behavior embody elastic deformation and viscous flow as special cases, and at the same time provide for response patterns that characterize behavior blends of the two. Intrinsically, such equations will involve not only stress and strain, but time-rates of both stress and strain as well. 309 310 Continuum Mechanics for Engineers 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 for elastic behavior, Eq 6.33, repeated here, Sij=2Gij; (9.1a) tii=3Kii; (9.1b) together with those for viscous flow, Eq 7.15, Sij=2ij; (9.2a) tii= -3 (p-dii); (9.2b) each expressed in terms of their deviatoric and dilatational responses. These equations are 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 ˙ijdij (9.3a) from which we immediately conclude that ˙ii=dii (9.3b) so that now, from Eq 4.77 and Eq 7.13 ˙ijij: (9.4) If the pressure pin Eq 9.2b is relatively small and may be neglected, or if we consider the pressure as a uniform dilatational body force that may be added as required to the dilatational effect of the rate of deformation term diiwhen circumstances require, Eq 9.2 may be modified in view of Eq 9.3 and Eq 9.4 to read Sij=2˙ij; (9.5a) tii=3˙ii: (9.5b) A comparison of Eqs 9.5 and Eq 9.1 indicates that they differ primarily in the physical constants listed and in the fact that in Eq 9.5 the stress tensors are expressed in terms of strain 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 physical constantsG,K,, and. In order to make the generalization complete we add similar differential operators to the left-hand side of the equations to obtain fPgSij=2fQgij; (9.6a) fMgtii=3fNgii; (9.6b) where the numerical factors have been retained for convenience in relating to traditional elasticity and viscous flow equations. As noted, the linear differential time operators, fPg, fQg,fMg, and fNg, are of the same form as in Eq 5.178 with the associated coefficients pi, qi,mi, andnirepresenting 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 at the outset, we verify that for the specific choices of operators fPg=1,fQg=G,fMg=1, Linear Viscoelasticity 311 t12γ12 t12 x1x2 t12 t12 FIGURE 9.1 Simple shear element representing a material cube undergoing pure shear loading. andfNg=K, Eqs 9.6 define elastic behavior, whereas for fPg=1,fQg=@=@t, fMg=1, andfNg=@=@t, linear viscous behavior is indicated. Extensive experimental evidence has shown that practically all engineering materials behave elastically in dilatation so without serious loss of generality we may assume the fundamental constitutive equations for linear viscoelastic behavior in differential operator form to be fPgSij=2fQgij; (9.7a) tii=3Kii; (9.7b) for isotropic media. For anisotropic behavior, the operators fPgandfQg must be aug- mented by additional operators up to a total of as many as twelve as indicated by fPig andfQigwith the index iranging from 1to6, and Eq 9.7a expanded to six separate equations. 9.2 One-Dimensional Theory, Mechanical Models Many 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 a material is associated directly with the deviatoric response as was pointed out in arriving at Eq 9.7, we choose the simple shear state of stress as the logical one for explaining fundamental concepts. Thus, taking a material cube subjected to simple shear, as shown by Fig. 9.1, we note that for this case Eq 9.7 reduces to the single equation fPgt12=2fQg12=2fQg12=fQg 12 (9.8) where 12is the engineering shear strain as shown in Fig. 9.1. If the deformational response of the material cube is linearly elastic, the operators fPgandfQgin Eq 9.8 are constants (P=1,Q=G)and that equation becomes the familiar t12=G 12 (9.9) 312 Continuum Mechanics for Engineers t12 G 1 γ12 (a) t12t12 (b) k 1 δF (c) k F F kδ (d) FIGURE 9.2 Mechanical analogy for simple shear. whereGis the elastic shear modulus. This equation has the same form as the one which relates the elongation (or shortening) of a linear mechanical spring to the applied force Fas given by the equation F=k (9.10) wherekis the spring constant. Because Eqs 9.9 and 9.10 are identical in form, the linear spring is adopted as the mechanical analog of simple elastic shearing with kassuming the role ofG. 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=˙ 12 (9.11a) whereis the coefficient of viscosity. In viscoelastic theory it is a longstanding practice that the coefficient of viscosity be represented by the symbol , and it is in keeping with this practice that we hereafter use the scalar for the coefficient of viscosity. Thus, Eq 9.11a becomes t12=˙ 12 (9.11b) in all subsequent sections of this chapter. The mechanical analog for this situation is the dashpot (a loose fitting piston sliding in a cylinder filled with a viscous fluid) subjected to an axial force F. Here F=˙ (9.12) where ˙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 Linear Viscoelasticity 313 η 1t12 δ (a) F Fη (b) η 1t12 γ12 (c) t12 t12 (d) FIGURE 9.3 Viscous flow analogy. simple combinations immediately come to mind. The first, that of the spring and dashpot in parallel, Fig. 9.4(a), portrays the Kelvin solid for which Eq 9.8 becomes T=t12=fG+@tg 12 (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 fluid having the constitutive equation  @t+1  t12=fG@tg 12 (9.14) where==G. In this chapter, the uppercase Tis used to designate the stress applied to the model. Using this notation makes the differential equations that arise from the models easier to write than if t12were used. Viscoelastic materials exhibit time dependent behavior and designating stress with a lowercase tcreates confusion. Throughout this chapterT, orT(t), are used interchangeably with t12to denote the stress. Models composed of more than two elements are readily constructed. When a Kelvin unit is combined in series with the linear spring element, Fig. 9.5(a), the resulting model is said to represent the standard linear solid. If the same Kelvin unit is joined in series with a dashpot, Fig. 9.5(b), the model represents a three-parameter fluid. In general, the model of a fluid has a “free dashpot” as one of its elements. Other three-parameter models are easily imagined, for example, a Maxwell unit in parallel with a spring, or a Maxwell unit in parallel with a dashpot. Four-parameter and higher order models may also be constructed. There are two basic patterns for systematically designing higher order models. One, leading to the generalized 314 Continuum Mechanics for Engineers T TG η (a) Kelvin Model T TGη (b) Maxwell Model FIGURE 9.4 Representations of Kelvin and Maxwell models for a viscoelastic solid and fluid, respec- tively. T TG1G2 η1 (a) Standard Linear Solid T TG1 η1η2 (b) Three Parameter Fluid FIGURE 9.5 Three parameter standard linear solid and fluid models. Linear Viscoelasticity 315 T TGn ηnG2 η2G1 η1 (a) Generalized Kelvin GnG2 ηnT Tη1 G1 η2 (b) Generalized Maxwell FIGURE 9.6 Generalized Kelvin and Maxwell models constructed by combining basic models. Kelvin model, hasn-Kelvin units in series, Fig. 9.6(a). The second consists of n-Maxwell units 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 12=t12 G1+1@t+t12 G2+2@t+


+t12 Gn+n@t; (9.15a) Maxwell t12=G1˙ 12 @t+1 1+G2˙ 12 @t+1 2+


+Gn˙ 12 @t+1 n: (9.15b) In these generalized model equations, one or more of the constants Giandimay be assigned the values 0or1in order to represent behavior for a particular material. Thus, with2and 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 Relaxation Further insight into the viscoelastic nature of the material making up the cube shown in Fig. 9.1 is provided by two basic experiments, the creep test and the stress relaxation test. The creep test consists of instantaneously subjecting the material cube to a simple shear stress t12of magnitude T0and maintaining that stress constant thereafter while measuring the shear strain as a function of time. The resulting strain 12(t), is called the creep. In the stress relaxation test, an instantaneous shear strain 12of magnitude 0is imposed on the cube and maintained at that value while the resulting stress, t12(t), is recorded as a function of time. The decrease in the stress values over the duration of the test is referred to as the stress relaxation. Expressing these test loadings mathematically is accomplished by use of the unit step function, U(t-t1), defined by the equation U(t-t1) =1 t>t1 0 t 6t1(9.16) 316 Continuum Mechanics for Engineers U(t-t1)� 1 t=t1t FIGURE 9.7 Graphic 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)=G 12+˙ 12 (9.18) may be integrated to yield the creep response 12(t)=T0 1-e-t=U(t) G(9.19) whereeis the base of the natural logarithm system. It is interesting to note that as t!1 , the strain approaches a terminal value of T0=G. Also, when t=0, the strain rate ˙ 12equalsT0=and if the creep were to continue at this rate it would reach its terminal value at time t=. For this reason, is called the retardation time . Eq 9.19, as well as any creep response, may always be written in the general form 12(t)=J(t)T0U(t) (9.20) in whichJ(t)is called the creep function .1Thus, for the Kelvin solid, the creep function is seen to be J(t)=1-e-t= G=J 1-e-t= (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 creep functions of its series-connected units. Thus, for the standard linear solid of Fig. 9.5(a), J(t)=J1 1-e-t=1 +J2; (9.22) and for the generalized Kelvin model of Fig. 9.6(a) J(t)=nX i=1Ji 1-e-t=i : (9.23) 1Note: The creep function, J(t), is different from the Jacobian defined earlier. Linear Viscoelasticity 317 With respect to creep loading for the Maxwell model, Eq 9.17 is substituted into the constitutive equation, Eq 9.14, resulting in the differential equation ˙ 12=T0(t) G+T0U(t) (9.24) where(t) is the Dirac delta function, the time derivative of the unit step function. In general, (t-t1)=dU(t-t1) dt; (9.25) and is defined by the equations (t-t1)=0; t6=t1; (9.26a) Zt+ 1 t- 1(t-t1)dt=1; (9.26b) from which it may be shown that Zt -1f(t0)(t0-t1)dt0=f(t1)U(t-t1)fort>t1 (9.27) for any continuous function, f(t). Accordingly, Eq 9.24 integrates to yield the Maxwell creep response as 12=T0J 1+t  U(t) (9.28) from which the Maxwell creep function is J(t)=J 1+t  : (9.29) Development of details relative to the stress relaxation test follows closely that of the creep test. With an imposed strain at time t=0 12= 0U(t) (9.30) the resulting stress associated with Kelvin behavior is given directly by inserting ˙ 12= 0(t)into Eq 9.13 resulting in t12= 0[GU(t)+(t)]: (9.31) The delta function in this equation indicates that it would require an infinite stress at time t=0to produce the instantaneous strain 0. For Maxwell behavior, when the instanta- neous strain, Eq 9.30, is substituted into Eq 9.14, the stress relaxation is the solution to the differential equation ˙t12+1 t12=G 0(t) (9.32) which upon integration using Eq 9.27 yields t12(t)= 0Ge-t=U(t): (9.33) The initial time-rate of decay of this stress is seen to be 0G=, which if it were to continue would reduce the stress to zero at time t=. Thus,is called the relaxation time for the Maxwell model. 318 Continuum Mechanics for Engineers Analogous to the creep function J(t) associated with the creep test we define the stress relaxation function, G(t), for any material by expressing t12(t)in its most general form t12(t)=G(t) 0U(t) (9.34) From Eq 9.33, the stress relaxation function for the Maxwell model is G(t)=Ge-t=(9.35) and for the generalized Maxwell model it is G(t)=nX i=1Gie-t=i: (9.36) Creep compliance and relaxation modulus for several simple viscoelastic models are given in Table B.1. The differential equations governing the various models are also given in this table. 9.4 Superposition Principle, Hereditary Integrals For linear viscoelasticity, the principle of superposition is valid just as in elasticity. In the context of stress/strain relationships under discussion here, the principle asserts that the total strain (stress) resulting from the application of a sequence of stresses (strains) is equal 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 a material having a creep function J(t), the resulting strain will be 12(t)=
T0J(t)+
T1J(t-t1)+
T2J(t-t2)=2X i=0
TiJ(t-ti); (9.37) and by an obvious generalization to the arbitrary stress loading considered as an infinity of infinitesimal step loadings, Fig. 9.8(b), the strain is given by 12(t)=Zt -1J(t-t0)dt12(t0) dt0 dt0(9.38) which is called a hereditary integral since it expresses the strain at time tas a function of the entire stress history from time t= -1. If there is an initial discontinuity in the stress at timet=0, and if the stress is zero up until that time (Fig. 9.9), the strain becomes 12(t)=T0J(t)+Zt 0J(t-t0)dt12(t0) dt0 dt0: (9.39) Upon integrating the integral in this equation by parts and inserting the assigned limits of integration, the alternative form 12(t)=J0t12(t)+Zt 0 t12(t0)dJ(t-t0) d(t-t0) dt0(9.40) is obtained where J0=J(0). Linear Viscoelasticity 319 t12 ∆T0∆T1∆T2 t t=t1t=t2 (a) Stepped Stress dtdt12t12 ∆T0 t (b) Arbitrary Stress FIGURE 9.8 Different types of applied stress histories. dtdt12t12 T0 t FIGURE 9.9 Stress history with an initial discontinuity. 320 Continuum Mechanics for Engineers In a completely analogous way, we may develop hereditary integrals expressing stress as the result of arbitrary strains. The basic forms are as follows: t12(t)=Zt -1G(t-t0)d 12(t0) dt0 dt0; (9.41a) t12(t)= 0G(t)+Zt 0G(t-t0)d 12(t0) dt0 dt0; (9.41b) t12(t)=G0 12(t)+Zt 0 12(t0)dG(t-t0) d(t-t0) dt0; (9.41c) whereG0=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 by the pair below expressed in terms of the distortional and dilatational responses 2ij(t)=Zt -1JS(t-t0)dSij(t0) dt0 dt0; (9.42a) 3kk(t)=Zt -1JV(t-t0)dtkk(t0) dt0 dt0; (9.42b) whereJSis the shear compliance and JVthe volumetric compliance. Likewise, Eq 9.41a is the simple shear form of the general equations Sij(t)=Zt -12GS(t-t0)dij(t0) dt0 dt0; (9.43a) tkk(t)=Zt -13GV(t-t0)dkk(t0) dt0 dt0; (9.43b) whereGSis the relaxation modulus in shear and GVthe relaxation modulus in dilatation. Because we have assumed elastic behavior in dilatation, JV=1=GV=Kand both Eq 9.42b and Eq 9.43b reduce to the form tkk=3K"kkin keeping with Eq 9.7b. 9.5 Harmonic Loadings, Complex Modulus, and Complex Compliance The behavior of viscoelastic bodies when subjected to harmonic stress or strain is another important part of the theory of viscoelasticity. To investigate this aspect of the theory, we consider the response of the material cube shown in Fig. 9.1 under an applied harmonic shear strain of frequency !as expressed by 12(t)= 0sin!t; (9.44a) or by 12(t)= 0cos!t: (9.44b) Mathematically, it is advantageous to combine these two by assuming the strain in the complex form 12(t)= 0(cos!t+isin!t)= 0ei!t(9.45) Linear Viscoelasticity 321 wherei=p -1. It is understood that physically the real part of the resulting stress corresponds to the real part of the applied strain, and likewise, the imaginary parts of each are directly related. The stress resulting from the excitation prescribed by Eq 9.45 will have the same frequency, !, as the imposed strain. Therefore, expressing the stress as t12(t)=Tei!t(9.46) whereTis complex, the response will consist of two parts: a steady-state response which will be a function of the frequency !, and a transient response that decays exponentially with time. It is solely with the steady-state response that we concern ourselves in the remainder of this section. Substituting Eq 9.45 and Eq 9.46 into the fundamental viscoelastic constitutive equation given by Eq 9.8, and keeping in mind the form of the operators fPgandfQgas listed by Eq 5.178 we obtain NX k=0Tpk(i!)kei!t=MX k=0 0qk(i!)kei!t: (9.47) Canceling the common factor ei!twe solve for the ratio T 0=PN k=0qk(i!)k PM k=0pk(i!)k(9.48) which we define as the complex modulus, G(i!), and write it in the form T 0=G(i!)=G0(i!)+iG00(i!): (9.49) The real part, G0(i!), of this modulus is associated with the amount of energy stored in the cube during a complete loading cycle and is called the storage modulus. The imaginary part,G00(i!), relates to the energy dissipated per cycle and is called the loss modulus. In terms of the complex modulus, the stress t12as assumed in Eq 9.46 may now be written t12=G(i!) 0ei!t=[G0(!)+iG00(!)] 0ei!t; (9.50) and by defining the absolute modulus, ˜G(!)as the magnitude of G(i!) according to ˜G(!)=q [G0(!)]2+[G00(!)]2(9.51) together with the loss angle, between ˜G(!)andG0(!) as given by its tangent tan=G00(!) G0(!)(9.52) the stresst12(Eq 9.50), may now be written t12=˜G(!)ei 0ei!t=˜G(!) 0ei(!t+ ): (9.53) From this equation we see that the peak value of the stress is T0=˜G(!) 0; (9.54) and that the strain lags behind the stress by the loss angle . Figure 9.10 provides a graphical interpretation of this phenomenon. In Fig. 9.10(a), the two constant magnitude 322 Continuum Mechanics for Engineers δωtγ0 ωt-δ ωOT0 (a) 2π ωπ ωδ ωOtγ0sin(ωt−δ)T0sinωt (b) FIGURE 9.10 Different types of applied stress histories. stress and strain vectors, separated by the constant angle , rotate about a fixed origin with a constant angular velocity !. The vertical projections of these vectors, representing the physical values of the stress and strain, are plotted against time in Fig. 9.10(b). From Fig. 9.10(a), the portion of the stress in phase with the strain is T0cosand by Eq 9.54 together with Eq 9.52, the storage modulus may be expressed as G0=T0cos 0: (9.55a) Similarly, the loss modulus is written as G00=T0sin 0(9.55b) Consistent with the duality present in all of viscoelastic theory we reverse the roles of stress and strain in the preceding portion of this section to define the complex compliance, J(i!)along with its associated real and imaginary parts. Briefly, we assume an applied stress t12=T0ei!t(9.56) together with the resulting strain 12= ei!t(9.57) which when substituted into Eq 9.8 leads to  T0=J(i!)=J0(!)-iJ00(!) (9.58) where the minus sign reflects the fact that the strain lags the stress by the loss angle , defined in this case by tan=J00(!) J0(!): (9.59) In analogy with the complex modulus components, J0(!)is called the storage compliance , J00(!)theloss compliance , and ˜J(!)=q [J0(!)]2+[J00(!)]2= 0 T0(9.60) Linear Viscoelasticity 323 theabsolute compliance, in which 0is the peak value of the strain as given by 0=˜J(!)T0 (9.61) Based upon the definitions of GandJ, it is clear that these complex quantities are reciprocals of one another. Thus G(i!)J(i!)=1: (9.62) A simple procedure for calculating G(orJ) of a specific model or material is to replace the partial differential operator @tin the material’s constitutive equation by i!and solve the resulting algebraic equation for the ratio t12= 12=G, (or 12=t12=Jif that is the quantity required). Accordingly, from Eq 9.13 for the Kelvin solid, the constitutive equationt12=fG+@tg 12becomest12= (G+i!) 12which yields t12= 12=G(1+ i!) =G. Likewise, from Eq 9.14 for the Maxwell fluid, f@t+1=gt12=fG@tg 12 becomes (i!+1=)t12= (Gi!) 12from whicht12= 12=G(2!2+i!)=(1 +2!2) =G. In developing the formulas for the complex modulus and the complex compliance we have used the differential operator form of the fundamental viscoelastic constitutive equations. Equivalent expressions for these complex quantities may also be derived using the hereditary integral form of constitutive equations. To this end we substitute Eq 9.45 into Eq 9.41a. However, before making this substitution it is necessary to decompose the stress relaxation function G(t) into two parts as follows, G(t) =G0[1-(t)] (9.63) whereG0=G(0), the value of G(t) at timet=0. Following this decomposition and the indicated substitution, Eq 9.41a becomes t12(t)=i! 0G0Zt -1ei!t0[1-(t-t0)]dt0: (9.64) In this equation, let t-t0=so thatdt0= -d and such that when t0=t,=0, and whent0= -1,=1. Now t12(t)=i! 0G0ei!t i!-Z1 0ei!ei!t()d (9.65) which reduces to t12(t)= 0 G0-G0Z1 0(i!cos!+!sin!)()d ei!t: (9.66) But by Eq 9.50, t12(t) = 0[G0(!) +iG00(!)]ei!tso that from Eq 9.66 G0(!)=G0-!G0Z1 0sin! ()d; (9.67a) G00(!)= -!G0Z1 0cos! ()d: (9.67b) For a Kelvin material, G(t) =G[1+(t)] so thatG0=Gand(t) = -(t). Thus, for a Kelvin solid, Eqs 9.67 yield G0(!)=G-G!Z1 0-() sin!d =G; 324 Continuum Mechanics for Engineers G00(!)= -G!Z1 0-() cos!d =G! =!: For a Maxwell material, G(t) =G[1- (1-e-t=)]so thatG0=Gand(t) =1-e-t=. Thus, for a Maxwell fluid, Eqs 9.67 yield G0(!)=G-G!Z1 0 1-e-= sin!d =G-G!1 !-!2 1+2!2 =G!22 1+2!2; G00(!)= -G!Z1 0 1-e-= cos!d =G! 1+2!2: These values for G0andG00for 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 the constitutive equation of choice, and substitute into that equation t12=T0ei!t, together with the decomposition of J(t) in the formJ(t) =J0[1+ (t)] , we obtain 12(t)=Zt -1J0[1- (t-t0)]i!T0ei!t0dt0: (9.68) Upon making the same change in variable of integration, t-t0=, this equation becomes 12(t)=T0 J0+J0!Z1 0sin! ()d+iJ0!Z1 0cos! ()d ei!t(9.69) from which we extract J0(!)=J0+J0!Z1 0sin! ()d; (9.70a) J00(!)= -J0!Z1 0cos! ()d: (9.70b) These expressions may be specialized to obtain J0(!) andJ00(!) for any particular model for which the creep function J(t) is known. 9.6 Three-Dimensional Problems, The Correspondence Principle The fundamental viscoelastic constitutive equations in differential operator form as ex- pressed by Eqs 9.6 distinguish between the distortional response (a change in shape at constant volume due to the deviatoric portion of the applied stress, Eq 9.6a), and the dilatational response (a change in volume without a change in shape due to the spherical portion of the applied stress, Eq 9.6b). Extensive experimental evidence indicates that practically all materials of engineering importance behave elastically in the dilatational mode, and for this reason the reduced form of Eq 9.6 as given by Eq 9.7 is used in this book. In expanded component notation these equations appear as fPg t11-1 3tii
=2fQg 11-1 3ii
; (9.71a) Linear Viscoelasticity 325 fPg t22-1 3tii
=2fQg 22-1 3ii
; (9.71b) fPg t33-1 3tii
=2fQg 33-1 3ii
; (9.71c) fPgt12=2fQg12; (9.71d) fPgt23=2fQg23; (9.71e) fPgt31=2fQg31; (9.71f) tii=3Kii: (9.71g) Depending upon the particular state of applied stress, some of Eq 9.71 may be satisfied identically. For example, the state of simple shear in the x1-x2plane, introduced earlier to develop the basic concepts of viscoelastic behavior, results in Eq 9.71 being reduced to a single equation, Eq 9.71d as expressed by Eq 9.8. Similarly, for a hydrostatic state of stress with t11=t22=t33=p0, andt12=t23=t31=0(or for a uniform triaxial tension having t11=t22=t33=T0, witht12=t23=t31=0), the behavior is simply elastic. On the other hand, for a simple one-dimensional tension or compression in one of the coordinate directions, several of Eq 9.71 enter into the analysis as discussed in the following paragraph. Let an instantaneously applied constant stress T0be imposed uniformly in the x1direc- tion on a member having a constant cross section perpendicular to that direction. Thus, lett11=T0U(t) with all other components zero which results in tii=T0U(t) so that from Eq 9.71g,ii=T0U(t)=3K . For this situation the first three of Eq 9.71 become fPg T0U(t)-1 3T0U(t)
=2fQg 11-T0U(t) 9K ; (9.72a) fPg -1 3T0U(t)
=2fQg 22-T0U(t) 9K ; (9.72b) fPg -1 3T0U(t)
=2fQg 33-T0U(t) 9K ; (9.72c) and the next three of the set indicate zero shear strains. Clearly, from Eq 9.72b and Eq 9.72c we see that 22=33. Furthermore, Eq 9.72a may be solved directly for 11to yield 11=T0U(t)3KfPg+fQg 9KfQg=T0U(t) fEg(9.73) where fEg=9KfQg=(3K fPg+fQg) is the operator form of E, Young’s modulus. Similarly, from Eq 9.72 using 22=33we find that 22 11=33 11= -3KfPg-2fQg 6KfPg+2fQg= -fg (9.74) which designates the operator form of Poisson’s ratio, . In order to compute a detailed solution for a particular material we need the specific form of the operators fPgandfQgfor 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 cross section made of a Kelvin material and situated along the x1axis. Determine 11and22as functions of time. 326 Continuum Mechanics for Engineers Solution From the constitutive relation for a Kelvin material, Eq 9.13, we note that fPg=1andfQg=fG+@tg, which when inserted into Eq 9.72a results (after some algebraic manipulations) in the dierential equation ˙11+11 =T0U(t)3K+G 9KG +T0(t) 9K: (9.75) This dierential equation may be solved by standard procedures to yield the solution 11(t)=T0U(t)3K+G 9KG 1-e-t= +e-t= 9K : (9.76) Whent=0,11=T0U(t)=9K which is the result of elastic behavior in bulk. Ast!1,11!T0[(3K+G)=9KG] =T0=E, the terminal elastic response. From Eq 9.72b the governing dierential equation for determining 22is (when expressed in its standard form) ˙22+22 =T0U(t)G 9K-1 61 +T0(t) 9K(9.77) which upon integration and simplication yields 22(t)= -T0U(t)3K-2G 18KG 1-e-t= -T0U(t)e-t= 9K: (9.78) Whent=0,22=T0=9Kwhich, due to the elastic dilatation eect, is identical with the initial value of 11. Ast!1,22!(2G-3K)=18KG . Up until now in this section we have discussed three-dimensional problems from the point of view of constitutive equations in differential operator form, but our analysis can be developed equally well on the basis of the hereditary integral form of constitutive equations as given by Eq 9.42 or Eq 9.43. With respect to the uniaxial stress loading analyzed above, Eq 9.42a (assuming elastic behavior in dilatation with tkk=3Kkkand ii=T0U(t)=3K , along with zero stress at time t=0) results in the equations 2 11-T0U(t) 9K =Zt 0 T0-T0 3 (t0)JS(t-t0)dt0; (9.79a) 2 22-T0U(t) 9K =Zt 0 -T0 3 (t0)JS(t-t0)dt0; (9.79b) 2 33-T0U(t) 9K =Zt 0 -T0 3 (t0)JS(t-t0)dt0: (9.79c) From these equations it is again apparent that 22=33, and that in order to develop the solution details for a particular material we need the expression for JS, the shear compliance of that material as shown by the example that follows. Linear Viscoelasticity 327 Example 9.2 Develop the solution for the problem of Example 9.7-1 using the hereditary integral form of constitutive equations for a bar that is Kelvin in distortion, elastic in dilatation. Solution For a Kelvin material, the shear (creep) compliance, Eq 9.21, is JS= (1- e-t=)=Gso that Eq 9.79a becomes for t11=T0U(t) 2 11-T0U(t) 9K =Zt 02T0(t0) 3G 1-e-t-t0  dt0(9.80) which may be integrated directly using Eq 9.27 to yield 11(t)=T0U(t)" 1-e-t  3G+1 9K# ; (9.81) or by a simple rearrangement 11(t)=T0U(t)" 3K+G 9KG-e-t  3G# in agreement with Eq 9.76. Likewise, from Eq 9.79b we obtain for this loading 2 22-T0U(t) 9K =Zt 0-T0(t0) 3G 1-e-t-t0  dt0(9.82) which also integrates directly using Eq 9.27 to yield 22(t)= -T0U(t)" 1-e-t  6G+1 9K# (9.83) in agreement with Eq 9.78. The number of problems in viscoelasticity that may be solved by direct integration as in the examples above is certainly quite limited. For situations involving more general stress fields, or for bodies of a more complicated geometry, the correspondence principle may be used to advantage. This approach rests upon the analogy between the basic equations of an associated problem in elasticity and those of the Laplace transforms of the fundamental equations of the viscoelastic problem under consideration. For the case of quasi-static viscoelastic problems in which inertia forces due to displacements may be neglected, and for which the elastic and viscoelastic bodies have the same geometry, the correspondence method is relatively straightforward as described below. In an elastic body under constant load, the stresses, strains, and displacements are independent of time, whereas in the associated viscoelastic problem, even though the loading is constant or a slowly varying function of time, the governing equations are time dependent and may be subjected to the Laplace transformation. By definition, the Laplace transform of 328 Continuum Mechanics for Engineers an arbitrary continuous time-dependent function, say the stress tij(x;t)for example, is given by ¯tij(x;s)=Zt 0tij(x;t)e-stdt (9.84) in whichsis the transform variable, and barred quantities indicate transform. Standard textbooks on the Laplace transform list tables giving the transforms of a wide variety of time-dependent functions. Of primary importance to the following discussion, the Laplace transforms of the derivatives of a given function are essential. Thus, for the first two derivatives of stress Zt 0dtij(x;t) dte-stdt= -tij(0)+s¯tij(x;s); (9.85a) Zt 0d2tij(x;t) dt2e-stdt= -tij(0)-stij(0)+s2¯tij(x;s); (9.85b) and so on for higher derivatives. Note that in these equations the values of the stress and its derivatives at time t=0become part of the transform so that initial conditions are built into the solution. We may now examine the details of the correspondence method by considering the correspondence between the basic elasticity equations (as listed in Chapter 6 and repeated here) Equilibrium (Eq6:43) tij;j(x) +bi(x) =0; Strain -displacement (Eq6:44) 2ij(x) =ui;j(x) +uj;i(x); Constitutiverelations (Eq6:33a )Sij(x) =2Gij(x); (Eq6:33b) tii(x) =3Kii(x); and the Laplace transforms of the associated time-dependent viscoelastic equations Equilibrium ¯tij;j(x;s) +¯bi(x;s) =0; (9.86a) Strain -displacement 2 ¯ij(x;s) = ¯ui;j(x;s) + ¯uj;i(x;s); (9.86b) Constitutiverelations ¯P(s)¯Sij(x;s) =2¯Q(s)¯ij(x;s); (9.86c) ¯tii(x;s) =3K¯ii(x;s); (9.86d) where barred quantities are transforms, and in which ¯P(s)and ¯Q(s) are polynomials in the transform variable sin accordance with Eq 9.85. A comparison of Eqs 9.86, which are algebraic, time-independent relations, with the elasticity equations above indicates a complete analogy between the barred and unbarred entities if we assign the equivalency of the ratio ¯Q(s)= ¯P(s)to the shear modulus G. This allows us to state the correspondence principle as follows: If the solution of a problem in elasticity is known, the Laplace transform of the solution of the associated viscoelastic problem is constructed by substituting the quotient ¯Q(s)= ¯P(s)of the transformed operator polynomials in place of the shear modulus G, and the actual time-dependent loads by their Laplace transforms. Since many elasticity solutions are written in terms of Young’s modulus, Eand Poisson’s ratio,, it is useful to extract from Eqs 9.73 and 9.74 the transform replacements for these constants which are ¯E(s)!9K¯Q 3K¯P+¯Q; (9.87) Linear Viscoelasticity 329 ¯(s)!3K¯P-2¯Q 6K¯P+2¯Q: (9.88) As an illustration of how the correspondence method works, we consider the following problem. Example 9.3 The radial stress trrin an elastic half-space under the action of a concentrated constant force F0acting at the origin as shown in the gure below is trr(r;z)=F0 2[(1-2)A(r;z)-B(r;z)] whereA(r;z)andB(r;z)are known functions of the coordinates. Determine the time-dependent viscoelastic stress trr(r;z;t)for a half-space that is Kelvin in shear and elastic in dilatation if the force at the origin is given by the step loadingF(t) =F0U(t). zx φ yrF0U(t) Solution From Eq 9.88 for ¯(s)we may directly calculate the Laplace transform of the expression for 1-2appearing in the elastic solution as 3¯Q= 3K¯P+¯Q
. The Laplace transform of the load function F0U(t)is given byF0=s. Thus, the Laplace transform of the associated viscoelastic solution is ¯trr(r;z;s)=F0 2s3¯Q 3K¯P+¯QA(r;z)-B(r;z) : From the Kelvin model constitutive equation, Eq 9.13, we have fQg=fPg=fG+ @tgfor whichQ(s)=P(s) =G+sso that ¯trr(r;z;s)=F0 2s3(G+s) 3K+G+A(r;z)-B(r;z) : This expression may be inverted with the help of a table of transforms from any standard text on Laplace transforms to give the time-dependent viscoelastic 330 Continuum Mechanics for Engineers solution trr(r;z;t)=F0 28 >< >:2 643G 3K+G+9Ke-(3K+G )t  3K+G3 75A(r;z)-B(r;z)9 >= >;: Notice that when t=0 trr(r;z;0 )=F0 2[3A(r;z)-B(r;z)]; and ast!1 trr(r;z;t!1 )=F0 23G 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 Nonlinear Viscoelastic 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 Saddle River, NJ [5] McCrum, N. G., Buckley, C. P ., and Bucknall, C. B. (1997), Principles of Polymer Engineering, Second Edition, Oxford University Press, New York [6] Pipkin, A. C. (1972), Lectures on Viscoelasticity Theory, Springer-Verlag, New York Linear Viscoelasticity 331 Problems Problem 9.1 By substituting Sij=tij-1 3ijtkkandij=ij-1 3ijkkinto Eq 9.7 and combining those two equations, determine expressions in operator form for (a) the Lam e constant, , (b) Young's modulus, E, (c) Poisson's ratio, . Answer (a)fg=K-2fQg=3fP g (b)fEg=9KfQg=(3K fPg+fQg) (c)fg= (3K fPg-2fQg)=(6K fPg+2fQg) Problem 9.2 Compliances are reciprocals of moduli. Thus, in elasticity theory D=1=E,J=1=G, and B=1=K. Show from the stress-strain equations of a simple one-dimensional tension that D=1 3J+1 9B: Problem 9.3 The four-parameter model shown consists of a Kelvin unit in series with a Maxwell unit. Knowing that MODEL = KELVIN + MAXWELL , together with the operator equations Eqs 9.13 and 9.14, determine the constitutive equation for this model. G2η2G1 η1T T Answer G21¨ +G1G2˙ =1¨T+(G1+G2+1=2)˙T+(G1=2)T Problem 9.4 Develop the constitutive equations for the three-parameter models shown. G1 η1η2 T T (a) G2 G1η2 T T (b) 332 Continuum Mechanics for Engineers G2 η1η2 T T (c) Answer (a) ¨ +˙ =1=[(1+2)=12]˙T+(1=12)T (b) ˙T+T=2=(G2+G1)˙ +(G1=2) (c) ˙T+T=2=1¨ +(G2+1=2)˙ Problem 9.5 A proposed model consists of a Kelvin unit in parallel with a Maxwell unit. Determine the constitutive equation for this model. G2 G1 η1η2 T T Answer ˙T+T=2=1¨ +(G1+G2+1=2)˙ +(G1=2) Problem 9.6 For 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 the parallel joined units.] G1 η3G2η2 T T Answer (a) ˙T+T=2=3¨ +(G1+G2+3=2)˙ +(G1=2) (b)G(t) =G1+G2e-t=2+3(t) Linear Viscoelasticity 333 Problem 9.7 For the model shown the stress history is given by the accompanying diagram. Determine the strain (t)for this loading during the intervals (a)06t=62, (b)06t=64. Use superposition to obtain answer (b). G ηG T T 10T0t t/τ2345 Answer (a) (t) =T0J 2-e-t=
U(t) (b) (t) =T0J 2-e-t=
U(t)-T0J 2-e-(t-2 )=
U(t-2) Problem 9.8 For the model shown determine (a) the constitutive equation, (b) the relaxation function, G(t), (c) the stress, T(t)for06t6t1, when the strain is given by the accompanying graph. G ηG ηT T γ t1 0λ Answer (a) ˙T+T==¨ +3G˙ +(G=) (b)G(t) =(t)+G 1+e-t=2
(c)T(t)= 2-2e-t=Gt
U(t) Problem 9.9 For the model shown in Problem 9.8, determine T(t)when (t)is given by the diagram shown here. γ γ0 t1 0λ 334 Continuum Mechanics for Engineers Answer T(t) = 0 (t)+G 1+e-t=
 U(t)+  2-e-t=
+Gt U(t) Problem 9.10 The three-parameter model shown is subjected to the strain history pictured in the graph. Use superposition to obtain T(t)for thet>t1, fromT(t)fort6t1. Let 0=t1=. G2 G1 η1T T γ γ0 t t1 0 Answer Fort6t1;T(t) = 1 1-e-t=1
+G2t U(t) Fort>t1;T(t) = 1 1-e-t=1
+G2t U(t) - 1 1-e-(t-t1)=1
+G2(t-t1) U(t-t1) Problem 9.11 For 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 ηη T T γ γ0 3t1t12t10 Answer (a)T(t1) =( 0=t1) 2-e-t1= (b)T(2t1) =( 0=t1) -2+2e-t1=-e-2t1= (c)T(3t1) =( 0=t1) -e-t1=+2e-2t1=-3e-3t1= Problem 9.12 If the model shown in the sketch is subjected to the strain history (t)=(0=2G) 2-e-t=2 U(t), as pictured in the time diagram, determine the stress, T(t). G Gη T T γ t 0γ(t)=T0 2G 2−e−t 2τ T0 2G Linear Viscoelasticity 335 Answer T(t) =T0 e-t=+U(t)
Problem 9.13 For the hereditary integral, Eq 9.38 (t)=Zt -1J(t-t0) (dT(t0)=dt0)dt0 assumeT(t) =estwheresis a constant. Let =t-t0be the \elapsed time" of the load application and show that (t) =sest¯J(s)where ¯J(s)is the Laplace transform of J(t). Problem 9.14 UsingT(t) =estas in Problem 9.13, together with the hereditary integral Eq 9.41a T(t)=Zt -1G(t-t0) (d (t0)=dt0)dt0; and the result of Problem 9.13 show that ¯G(s)¯J(s)=1=s2where ¯G(s)is the Laplace transform of G(t). Assume sis real. Problem 9.15 Taking the hereditary integrals for viscoelastic behavior in the form Eq 9.40 (t)=J0T(t)+Zt 0T(t0) [dJ(t-t0)=d(t-t0)]dt0 and Eq 9.41c T(t)=G0 (t)+Zt 0 (t0) [dG(t-t0)=d(t-t0)]dt0; show that for the stress loading T(t) =estand with=t-t0, the expression G0¯A(s)+J0¯B(s)+¯A(s)¯B(s)=0 results, where here ¯A(s)=Z1 0e-s(dJ=d )d and ¯B(s)=Z1 0e-s(dG=d )d: Problem 9.16 Let the stress relaxation function be given as G(t) =a(b=t)mwherea,b, andm are constants and tis time. Show that the creep function for this material is J(t)= 1 amsinmxt b
mwithm<1. Use the identity ¯G(s)¯J(s)=1=s2where barred quantities are Laplace transforms. Problem 9.17 A 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 the model. 336 Continuum Mechanics for Engineers G1 G2η1 T T Answer (a)G(t) =G2+G1e-t=1 (b)J(t)=(1=(G1+G2))e-t= 1+(1=G2) 1-e-t= 1
where 1=(G1+G2)1=G2 Problem 9.18 A material is modeled as shown by the sketch. (a) For this model determine the relaxation function,G(t). (b) If a ramp function strain as shown by the diagram is imposed on the model, determine the stress, using the appropriate hereditary integral involving G(t). G G2G 2ηηT T γ(t) t1 0λ Answer (a)G(t) =G+2Ge-2t=+Ge-t=2 (b)T(t) =G t+3-e-2t=-2e-t=2 U(t) Problem 9.19 Determine the complex modulus, G(i!)for the model shown using the substitution i! for@tin the constitutive equation. G1 G2η1 T T Answer G(i!)= G2+(G1+G2)2 1!2+iG11! = 1+2 1!2
Problem 9.20 Show that, in general, J0=1=G0(1+tan2)and verify that G0andJ0for the Kelvin model satises this identity. (Hint: Begin with GJ=1.) Linear Viscoelasticity 337 Problem 9.21 Let the complex viscosity (denoted here by (i!)) be dened through the equation T0ei!t= i! 0ei!t : Determine(i!)in terms ofG(i!) (see Eq 9.49) and calculate (i!)for the model shown below. G ηη T T Answer (i!)=  2+!22
-i! = 1+!22
Problem 9.22 From Eq 9.55a in which G0=T0cos= 0and 9.55b in which G00= (T0sin)= 0show that J0= ( 0cos)=T0and thatJ00= ( 0sin)=T0. UseGJ=1. Problem 9.23 Show that the energy dissipated per cycle is related directly to the loss compliance, J00 by evaluating the integralR Td over one complete cycle assuming T(t) =T0sin!t. (See Fig. 9.10.) Answer Z Td =T2 0J00 Problem 9.24 For the rather complicated model shown here, determine the constitutive equation and from itG(i!). Sketch a few points on the curve G00vs.ln(!). G G2G 2ηηT T Answer ¨T+(5=2)˙T+ 1=2
T=4G¨ +(11G=2 )˙ + G=2
G(i!) =G0+iG00where G0=G 1+(35=4)2!2+44!4 = 1+(17=4)2!2+4!4 338 Continuum Mechanics for Engineers G00=G 3!+(9=2)3!3 = 1+(17=4)2!2+4!4 Forln(!) =0,G00=1:2G Forln(!) =1,G00=1:13G Forln(!) =2,G00=0:572G Forln(!) =1,G00=0 Problem 9.25 A block of viscoelastic material in the shape of a cube ts snugly into a rigid container. A uniformly distributed load p= -p0U(t) is applied to the top surface of the cube. If the material is Maxwell in shear and elastic in dilatation determine the stress component t11(t) using Eq 9.71. Evaluate t11(0)andt11(1). x1x2p=−p0U(t) Answer t11(t) = -p0[1- (6G=(3K +4G))]e-(3K=(3K +4G))t]U(t) t11(0) = -p0[(3K-2G)=(3K +4G)] t11(1) = -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 axial strain11(t)and the lateral strain 22(t). GG ηT T t11t11 Answer 11(t) =T0f[(6K+G)=3K -e-t=]=3GgU(t) 22(t) =T0[(1=9K) - (2 -e-t=)=6G)]U(t) Linear Viscoelasticity 339 Problem 9.27 A cylinder of viscoelastic material ts snugly into a rigid container so that 11=22= rr=0(no radial strain). The body is elastic in dilatation and has a creep compliance JS=J0(1+t)withJ0a constant. Determine t33(t)if˙33=A(a constant). x1x2 rx3 Answer t33(t) = fA[Kt +4(1-e-t)=3J0]gU(t) Problem 9.28 A 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= -p0(t)distributed uniformly over the x1face. If the block is constrained so that 22=33=0, determine 11(t)andt22(t). t11= −poδ(t)t22t33 Answer t22(t) =p0[(2G-3K)(t)=(3K +4G) - (6G=(3K +4G))e[-3K=(3K +4G)]t]U(t) 11(t) =p0[3(t)=(3K +4G) - [4G=(3K +4G)K]e[-3K=(3K +4G)]t]U(t) Problem 9.29 A viscoelastic cylinder is inserted into a snug tting cavity of a rigid container. A at, smooth plunger is applied to the surface x1=0of the cylinder and forced downward at 340 Continuum Mechanics for Engineers a constant strain rate ˙11=0. If the material is modeled as the three-parameter solid shown in shear and as elastic in dilatation, determine t11(t)andt22(t)during the downward motion of the plunger. G GηT T x1x2 x3 Answer t11(t) = -0[(4G=3)(1-e-t=) + (K+4G=3)t]U(t) t22(t) =0[(2G=3)(1-e-t=) + (-K +2G=3)t]U(t) Problem 9.30 For a thick-walled elastic cylinder under internal pressure p0, the stresses are trr=A-B=r2; t=A+B=r2and the radial displacement is given by ur=1+ E[A(1-2)r+B=r] whereA andBare constants involving p0,Eis Young's modulus, and is Poisson's ratio. Determine trr,t, andurfor a viscoelastic cylinder of the same dimensions that is Kelvin in shear and elastic dilatation if p=p0U(t). tθθtθθ trrtrr p0r Answer trr=same as elastic solution but with p0nowp0U(t) t=same as elastic solution but with p0nowp0U(t) ur(t) = (3Ar=(6K +2G))(1 -e-(3K +G)t=G))U(t) + (B=2Gr )(1-e-t=)U(t) Problem 9.31 A viscoelastic half-space is modeled as Kelvin in shear, elastic in dilatation. If the point forceP=P0e-tis 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[(1-2)A -B]=2 whereAandBare functions of the coordinates only. Linear Viscoelasticity 341 x3x2 x1rP = P0e−t θ Answer trr= (P0=2)[(3(G -)e-t+9Ke-(3K +G)t=]A=(3K+G-) -e-tB Problem 9.32 The de ection at x=Lfor an end-loaded cantilever elastic beam is w=P0L3=3EI. Determine the de ection 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. Check w(L;1)in each case. P0 L Answer (a)w(L;t) = (P 0L3=3EI)(1-e-t=E)U(t);E==E w(L;1) = (P0L3=3EI), the elastic de ection. (b)w(L;t) = (P 0L3=3EI)[(1-e-t=) + (1=9K)e-t=]U(t) w(L;1) = (P0L3=3EI), the elastic de ection. Problem 9.33 A simply-supported viscoelastic beam is subjected to the time-dependent loading f(x;t) = q0twhereq0is a constant and tis time. Determine the beam de ection w(x;t)in terms of the elastic beam shape X(x)if the beam material is assumed to be (a) one-dimensional Kelvin, and (b) three-dimensional Kelvin in shear, and elastic in dilatation. Compare the results. q0t L 342 Continuum Mechanics for Engineers Answer (a)w(x;t) =X(x) t-E(1-e-t=E) U(t) (b)w(x;t) =X(x) t-(3E=(3K+G))(1-e-t=E) U(t) whereE==E Appendix A: General Tensors The balance laws and the constitutive relations describing the behavior of continuous media were formulated using direct or coordinate free notation. This is a recognition of the invariance of the physical principles. These laws were written using indicial notation in a rectangular-Cartesian coordinate frame. The study of tensors in other coordinate systems may be better suited for solving particular boundary value problems. These coordinate frames might provide ways to describe vectors and tensors that better describe the geometry of the body. The study of general tensors also aids in understanding much of the literature that uses general tensors. It would be convenient if there was a way to convert the Cartesian equations to general coordinate systems without a great deal of effort. It is the purpose of this appendix to give such a scheme. The following section begins by developing ideas about the general representation 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 expression in Cartesian coordinates and convert it to an expression in a general coordinate system. A.1 Representation of Vectors in General Bases Vectors are quantities that do not depend on a coordinate system. The representation of vectors does depend on the basis for computing the components of a vector. Any set of three non-coplanar vectors can be used to represent a vector v. For a Cartesian coordinate system f^e1;^e2;^e3gforms a basis that allows the vector vto be written as v=v1^e1+v2^e2+v3^e3=vi^ei: The basis f^e1;^e2;^e3gis the same for every point in the space defined in the Cartesian coordinate system. The coordinate system is said to be homogenous. It is also possible to choose a set of base vectors fg1;g2;g3gthat is not orthonormal, Fig A.1. The vectors g1 andg2lie in a plane, and the vector g3is out of the plane. These base vectors may also vary 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 vivi= vi
2. The components of the vectorvare found by projecting the vector onto the planes formed by g1-g2,g2-g3, and g1-g3. This gives the unique scalar components v1,v2, andv3with respect to the basis fgig. The calculation of the components viare not as straight forward as in the Cartesian system. The basis vectors are not orthonormal and gi
gj6=0fori6=j 1fori=j as in the Cartesian system. 343 344 Continuum Mechanics for Engineers g1g2g3 FIGURE A.1 A set of non-orthonormal base vectors. Example A.1 Given the set of base vectors fgigwith Cartesian components g1=3^e1+2^e2+^e3 g2=^e1-2^e3 g3=^e1+^e3 and a vector v=4^e1+^e2+8^e3: Findv1,v2, andv3. Solution We know that v=v1g1+v2g2+v3g3: Taking the dot product v
^ei, gives v
^e14=3v1+v2+v3 v
^e21=2v1+v3 v
^e38=v1-2v3 Solving yields v1=2; v2=1; v3= -3: The vector is v=2g1+g2-3g3: Determining whether a set of vectors forms a basis is not a simple task. A simple test for linear independence is to form a matrix, G, from the basis vectors expressed in the Cartesian components. G=[g1;g2;g3] The determinate of Gbeing non-zero assures that the basis is linearly independent. The mathematical condition for this is detG6=0 and is called the Jacobian matrix of the basis fgig. Appendix A: General Tensors 345 Example A.2 From the previous example, show that the basis giis linearly independent. Solution The matrix Gis G=2 43 1 1 2 0 1 1 0 -23 5 and detG=3 1 1 2 0 1 1 0 -2=56=0: The basis is linearly independent. Note that the determinate of Gbeing non-zero is just the condition that the equations for the components vihas a solution. A.2 The Dot Product and the Reciprocal Basis The dot product between two vectors uandvis u
v=uigi
vjgj=uivjgi
gj: If the basis is Cartesian, the dot product gi
gjis just the Kronecker delta. However, since the 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 vectors were orthogonal. This can be accomplished by computing a reciprocal or dual basis. The reciprocal vectors can be computed in a number of ways but have the property that gi
gj=j i=1ifi=j 0ifi6=j: j iis 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: 346 Continuum Mechanics for Engineers Note that the implied summation is always across a subscript and superscript pair. With the reciprocal basis, the dot product becomes u
v=uivjgi
gj=uivjj i=uivi: It is also possible to form the dot product of the base vectors with themselves. The dot products of the basis vectors give gi
gj=gij gi
gj=gij gi
gj=gj i=j i gi
gj=gi j=i j 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 have v
v=jvj2=vivjgi
gj=gijvivj or v
v=jvj2=vivjgi
gj=gijvivj: The magnitude of the vector is related to the metric tensors gijandgij. The products gij=gi
gj are the dual, reciprocal or inverse metric tensor. This latter designation is seen from the following calculation. The bases are related by gi=gijgjandgi=gijgj: Now, this gives gi
gj=j i=gilgl
gjkgk=gilgjkl k=gilgjl and gilgjl=gilglj=j i: Hence the designation inverse metric. The components of a vector are related by vi=v
gi=vkgk
gi=gkivk=gikvk from the symmetry of the metric tensor. This can also be repeated for the gibase vectors. The results are vi=gijvjandvi=gijvj: This is termed raising and lowering the indices and will be useful when we describe the rules for generating general tensor expressions from Cartesian tensor expression. A.3 Components of a Tensor The components of a tensor in an orthonormal basis are defined by the operation of the tensor on the base vectors ^ei. The components are Tij=^ei
T
^ej: Appendix A: General Tensors 347 This process can be extended to the general basis. However, unlike the orthonormal basis, the general basis gives four different component representations. Two of these tensor components are Tij=gi
T
gj Tij=gi
T
gj: The components are formed in the standard way. In addition, there are two sets of mixed components. These are2 Ti j=gi
T
gj Tj i=gi
T
gj Thus, the four representations of a tensor are T=Tijgigj=Tijgigj=Ti jgigj=Tj igigj: The mixed components of a tensor are not related. In general Tj i6=Ti j: However, for a symmetric tensor T=TT the result is Tij=Tji Tij=Tji Tj i=Ti j Ti j=Tj i: Just as with orthonormal bases, transformation rules exist between two sets of general basis vectors fgigandfg0 ig. 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 v0 i=aj ivj and vi= a-1
j iv0 j: A second order tensor transforms as T0 ij=ap iak jTpk: For the other components, we need to have the transformation rules for the reciprocal basis g0i= a-1
i jgj and gi=ai jg0j: 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, Tj i=T
j i. This is most commonly done when writing the mixed components. Typesetting offers the opportunity to clearly separate the superscript and subscript. 348 Continuum Mechanics for Engineers A.4 Determination of the Base Vectors We have seen that vectors can be represented in terms of two sets of base vectors fgigor gi . v=vigi=vigi: The base vectors fgigare called the covariant basis vectors, and gi is called a set of contravariant basis vectors. The covariant basis vectors are defined by the derivatives of a coordinate transformation x=˜x uj
=˜xi uj
^ei: The vectors f^eigare the orthonormal base vectors of the rectangular system. The functions ˜xi uj
are the coordinate functions relating the rectangular coordinates, xi, to the new coordinate variables, uj. The covariant base vectors are gj=@˜x @uj=@˜xi @uj^ei: The base vectors are the derivatives along the coordinate curves. They vary along the coordinate curves. The dual or reciprocal basis is the contravariant basis vectors. These vectors have the property that gi
gj=j i=1ifi=j 0ifi6=j j iis the Kronecker delta. Example A.3 Determine the base vectors for the circular-cylindrical coordinate system. Solution Figure A.3 shows the circular-cylindrical coordinate system for x3=0. The position of the point pis x=x1^e1+x2^e2+x3^e3 =rcos^e1+rsin^e2+z^e3: This shows that x=˜x(r;;z ). Now gr=@˜x @r=cos^e1+sin^e2 g=@˜x @= -r sin^e1+rcos^e2 and gz=@˜x @z=^e3: The basis vectors fgigcan be shown to be a basis by forming the Jacobian, G(gr;g;gz). This gives G=detcos sin 0 -rsin r cos 0 0 0 1=r cos2+sin2
=r: Appendix A: General Tensors 349 gr gθ xrx2 x1r−curve θ θ−curvep FIGURE A.2 Circular-cylindrical coordinate system for x3=0. The vectors form a basis except for the point r=0. The vector gr=@˜x @r=cos^e1+sin^e2 lies along the r-coordinate. Similarly, gis a vector that is perpendicular to gr. The basis vectors form a non-homogeneous basis. The two vectors are tangent to therandcurves. This is consistent with the denition of the covariant basis vectors. The dual basis is found from the computation in Problem 2.2. These vectors are gr=cos^e1+sin^e2 g=1 r(-sin^e1+cos^e2) and gz=^e3: The properties of the bases can be veried. 350 Continuum Mechanics for Engineers A.5 Derivatives of Vectors A.5.1 Time Derivative of a Vector Suppose that in the circular cylindrical system of the above example that r=r(t),= (t), andz=z(t). This gives x=r(t)cos(t)^e1+r(t)sin(t)^e2+z(t)^e3: The time derivative of x(t)is the velocity v(t)and is v(t)=dx(t) dt=@x @rdr dt+@x @d dt+@x @zdz dt using the chain rule. From the previous section, the partial derivatives of xare the base vectorsgiand v(t)=dx(t) dt=dr dtgr+d dtg+dz dtgz=vrgr+vg+vzgz: The result is vr=dr dtv=d dtandvz=dz dt: The first and third terms have the dimensions of a velocity. However, the second term does not. It only has units of s-1. This raises the need to develop the physical components of a vector. The physical components of a vector win the direction uare w
^u: Here^uis a unit vector in the direction of u. ^u=u juj The physical components of vdenoted byv<i>are the components of the vector along the covariant unit vectors. The unit vectors ^giare ^gi=gipgiino sum oni: Now for a vector v=vigi, v=vigi=v1pg11^g1+v2pg22^g2+v3pg33^g3: The physical components of vare v<i>=pgiivino sum oni and v=v<k>^gk: For the circular cylindrical system v<r>=v
^gr=dr dt and v<>=v
^g=v
rg=rd dt: These are the familiar components of the velocity that appear in elementary dynamics. Appendix A: General Tensors 351 A.5.2 Covariant Derivative of a Vector The spatial derivative of a vector field frequently arises in applications. This is the case for gradients of displacement fields defining strain fields or the divergence of a velocity field in a fluid. The derivative of a vector vwith respect to a curvilinear coordinate uiis @v @uj=@vi @ujgi+vi@gi @uj: The latter term arises because the base vectors, gi, can vary from point to point in the space. It is desirable to express the derivative of the base vector in terms of components along the base vector. This allows the writing of an expression analogous to that in a Cartesian coordinate system. @v @xj=@vi @xj^ei=vi;j^ei: Here, the comma notation denotes partial differentiation. The derivative of the base vector is @gi @uj=@ @uj@˜xk @ui^ek =@2˜xk @uj@ui^ek: 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 ^ei gives @gi @uj=@2˜xk @ui@uj@ul @˜xkgl: The coefficients in the above expression are known as the Christoffel symbols of the second kind. These are given a special notation @2˜xk @ui@uj@ul @˜xk=l ij= l ij Using this notation gives, @gi @uj=l ijgl=l ij 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 dot product of both sides with gpgives p ij=@gi @uj
gp after rearranging the indices. The process for computing the Christoffel symbols is 1. From the coordinate transformation x=˜xi uj
^ei, findgk=@˜xi @uk^eiand compute gi;j=@2˜xk @uj@ui^ek: 352 Continuum Mechanics for Engineers 2. Determine the contravariant base vectors gk. 3. Form the dot product to find the Christoffel symbol. l ij=@gi @uj
gl: Another method is available for finding the Christoffel symbols that is not dependent on the coordinate transformation. It will be presented subsequently. Example A.4 Compute the Christoel symbols for the circular cylindrical coordinate system. Solution The covariant base vectors are gr=@˜x @r=cos^e1+sin^e2 g=@˜x @= -r sin^e1+rcos^e2 and gz=@˜x @z=^e3: The computation is not as daunting as it might rst appear. Many of the Christoel symbols are zero. Dierentiating the base vectors we have @gr @r=@gr @z=0;@gr @=@g @r= - sin^e1+cos^e2 @g @z=0;@g @= -r cos^e1-rsin^e2 and@gz @r=@gz @=@gz @z=0: The reciprocal base vectors are gr=cos^e1+sin^e2 g=1 r(-sin^e1+cos^e2) and gz=^e3: The Christoel symbols found from l ij=@gi @uj
glgive r rr= rr=z rr=r rz= rz=z rz=0 r r=r r=z r=z r=0  r= r=1 r  =0 r = -r: The remaining components are zero. Appendix A: General Tensors 353 The derivative of the vector can now be written as @v @uj=@vi @ujgi+vi@gi @uj=@vi @ujgi+vil ijgl: This can be rearranged to give @v @uj=@vi @ujgi+vki kjgi=@vi @uj+vki kj gi: The last form is the result of changing the dummy indices. The coefficients of the covari- ant base vectors are vijj=@vi @uj+vki kj: This is the covariant derivative of the contravariant components of a vector. It is a natural generalization of the result for partial differentiation in Cartesian systems. To this point, all calculation was done using the contravariant components of the vector. Ifv=vigirepresents 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=j i and @ @uk gi
gj
=@gi @uk
gj+gi
@gj @uk=@j i @uk=0: This gives @gj @uk
gi= -@gi @uk
gj= -l ikgl
gj= -j ik: The right hand side is the covariant component of a vector and @gj @uk= -j ikgi: This result gives @v @uj=@vi @uj-vkk ij gi: The covariant derivative of the covariant components is vijj=@vi @uj-vkk ij: Compare this to the previous result. A.6 Christoffel Symbols A.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 iandj: @2˜xk @ui@uj@ul @˜xk=@2˜xk @uj@ui@ul @˜xk 354 Continuum Mechanics for Engineers l ij=l ji=l ij =l ji The Christoffel symbols of the second kind are the partial derivatives of the basis vectors in a curvilinear coordinate system with respect to the coordinate variables. Also, it is useful to look at the placement of the indices. The indices are all free indices. The upper index corresponds to the term in the numerator and the lower indices correspond to the terms in the denominator. The above discussion indicates that the Christoffel symbols are of two types. The Christoffel symbols of the first kind are denoted by [ij;l]. These are computed from the following formula [ij;l]=glmm ij Similarly, using the metric glp, we find p ij=glp[ij;l]: The Christoffel symbols of the first kind have a representation similar to those of the second kind. To find this representation, note that glm=gl
gm=@˜xq @ul^eq
@˜xk @um^ek=@˜xq @ul@˜xk @umqk: The Christoffel symbols of the first kind are [ij;l]=glmm ij=@˜xq @ul@˜xk @umqk@2˜xp @ui@uj@um @˜xp but @˜xk @um@um @˜xp=@˜xk @˜xp=kp andkpqk=pq: This gives [ij;l]=@2˜xp @ui@uj@˜xp @ul: Notice that all the indices are in the denominator. Also, the grouping appears in a logical sequence. A.6.2 Calculation of the Christoffel Symbols From the above discussion, it would appear that the Christoffel symbols are connected to a particular rectangular Cartesian coordinate system. However, the Christoffel symbols can be computed directly from the metric and the inverse metric for the space. This is shown in the following. The metric of the space is gij=gi
gj: We differentiate this with respect to the coordinate ukand obtain @gij @uk=@gi @uk
gj+gi
@gj @uk: This can be written as @gij @uk=l ikgl
gj+gi
gpp jk=l ikglj+gipp jk: Appendix A: General Tensors 355 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 p jk=gip[jk;i]. The result is p jk=gip[jk;i]=1 2gip@gij @uk+@gki @uj-@gjk @ui : This demonstrates that the Christoffel symbols are not dependent on the Cartesian basis that was used to motivate their introduction. The use of the metric tensor is especially useful in determining the Christoffel symbols when the general coordinate bases are orthogonal. In this case, we have gmn=0if m6=n. This immediately shows that many of the Christoffel symbols will be zero in an orthogonal basis. If i,j, andkare all distinct, then k ij=0. The only non-zero values will occur when two of the indices are equal in a general, orthogonal coordinate system. A.7 Covariant Derivatives of Tensors In a manner similar to that used to find the covariant derivative of a vector, the covariant derivative of a tensor can be found. To calculate@T @uk, begin with the representation of a tensor in the basis fgig;T=Tijgigj. Then @T @uk=@ @uk Tijgigj
=@Tij @ukgigj+Tij@gi @ukgj+Tijgi@gj @uk: Using the Christoffel symbols, this can be written as @T @uk=@Tij @ukgigj+Tijl ikglgj+Tijl jkgigl: Rearranging the dummy indices gives @T @uk=@Tij @uk+Tlji lk+Tilj lk gigj =Tijjkgigj: 356 Continuum Mechanics for Engineers The term in parentheses is the covariant derivative of the contravariant components of the tensor. Tijjk=@Tij @uk+Tlji lk+Tilj lk: Other covariant derivatives can be formed. These are Tijjk=@Tij @uk-Tljl ik-Till jk Ti jjk=@Ti j @uk+Tl ji lk-Ti ll jk: They are computed using the same techniques as were employed previously. A.8 General Tensor Equations Tensors and tensor equations describing physical phenomena are invariant with respect to coordinate systems. A number of relations in direct and Cartesian tensor notation were developed. These relations span the spectrum from inner products of vectors to local forms of the First Law of Thermodynamics. The following presents a way to convert these relations from Cartesian tensor notation to general tensor notation. Vectors and tensors are represented in terms of covariant and contravariant base vectors. The techniques presented allow us to compute the derivatives of vector fields and tensor fields. Three rules allow the conversion of Cartesian tensor expressions to expressions in general coordinate systems. Rule 1: Rewrite summations over dummy indices in Cartesian expressions so that the summation is only over raised and lowered indices. This rule is a direct consequence of the way that summation was introduced for vector and tensor representation in a general coordinate system. We defined vectors and tensors in terms of covariant and contravariant bases. This led to the convention that implied summation occurs only over a subscript and a superscript pair. Summation across pairs of indices may be accomplished in a number of ways. We may simply change an index in the Cartesian expression from a subscript to a superscript. This is possible since we have Kronecker deltas ijandijin the Cartesian system. We may raise 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 corresponding general 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 is the divergence of a vector field. In Cartesian tensor notation, we have divv=vi;i: This becomes divv=viji Appendix A: General Tensors 357 TABLE A.1 Converting from Cartesian tensor notation to general tensor notation. Summation over only subscript and superscript pairs. Cartesian Tensor Expression General Tensor Expression dkk dk k=gkmdkm=gkmdkm aibi aibi=gikakbi=gikaibk didi didi=gijdidj=gijdidj vi=wikk vi=wk ik=gkmwkm i=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=ai: In general tensor notation, this becomes Tijjj+bi=ai: 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-zero Christoffel symbols have a pair of matching indices. The distinction disappears in a Cartesian system where the Christoffel symbols are all zero. Things are not always as simple as they first appear! Rule 3: Replace the Cartesian tensor permutation symbol, "ijk, by the general tensor permutation symbol,eijkoreijk, 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 uv=uivjgigj but gigj=eijkgk: This gives uv=eijkuivjgk: Also note that eijk=(gigj)
gk="ijkpg where g=detjgijj: These three rules allow one to conveniently change Cartesian tensor expressions into general 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 coordinate systems. This gives us a very powerful methodology for exploring the mechanics of continuous media. 358 Continuum Mechanics for Engineers A.9 General Tensors and Physical Components A tensor has the following representations T=Tiigigj=Tijgigj=T<ij>^gi^gj: 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<r> @+@T<rz> @z+T<rr>-T<> r=0 ^g2=^e 1 r2@ r2T<r>
@r+1 r@T<> @+@T<z> @z=0 ^g3=^ez 1 r@(rT<zr>) @r+1 r@T<z> @+@T<zz> @z=0: The covariant derivative of the tensors for computing the equilibrium equation can be written in two forms. Form 1: Tji i=@Tji @ui+Tjli li+Tlij li=0 This is an acceptable form for the equilibrium equations since the tensor components are the contravariant components. These are similar to the contravariant physical components used above. g1(j=1) @T1i @ui+T1li li+Tli1 li=0 This equation contains many terms. However, all but two of the Christoffel symbols are zero. The non-zero Christoffel symbols are 2 12=2 21=1 r1 22= -r: This gives @T1i @ui+T112 12+T221 22=0; @T11 @u1+@T12 @u2+@T13 @u3+1 rT11-rT22=0: The contravariant components in this expression are not the physical components of the tensor. What is the relationship between the two? The representation of a tensor above gives Tiigigj=T<ij>^gi^gj=T<ij>gi gigj gj: giandgjare the magnitudes of the vectors. These are gi=pgi
gi=pgiino sum oni: Appendix A: General Tensors 359 gijis the metric tensor. For cylindrical coordinates, this is [gij]=2 41 0 0 0 r20 0 0 13 5: The connection between the components are T<ij>=pgiipgjjTijno sum oniorj: 2 4T<rr>T<r>T<rz> T<r>T<>T<z> T<zr>T<z>T<zz>3 5=2 4T11rT12T13 rT21r2T22rT23 T31rT32T333 5 This gives @T<rr> @u1+@ @u21 rT<r> +@T<rz> @u3+1 rT<rr>-r1 r2T<> =0: This is the ^erequilibrium equation where u1;u2;u3
is(r;;z ). g2(j=2) @T2i @ui+T2li li+Tli2 li=0 or @T2i @ui+T212 12+T122 12+T212 21=@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;;z )gives @ @r1 rT<r> +@ @1 r2T<> +@ @z1 rT<z> +3 r1 rT<r> =0 and 1 r2@ @r r2T<r>
+@ @1 rT<> +@ @z T<z>
=0 g3(j=3) @T3i @ui+T3li li+Tli3 li=0 or @T3i @ui+T312 12+Tli(0)=0: This yields @T31 @u1+@T32 @u2+@T33 @u3+1 rT31=0: Converting to physical components gives 1 r@(rT<zr>) @r+@ @1 rT<z> +@T<zz> @z=0 360 Continuum Mechanics for Engineers Form 2: This form uses the covariant tensor components Tijji=@Tij @uj-Tmjm ij+Timm jj=0 This is not an acceptable form for the equilibrium equations for the physical components since the tensor components are the covariant components rather than the contravariant components. These are not the same as the contravariant physical components used in the equilibrium equations. While these expressions will give partial differential equations for the covariant components, they will not give an equivalent set of equations to those already presented. The contravariant and covariant components and associated equations become 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 361 362 Continuum Mechanics for EngineersTABLE B.1: Creep and relaxation responses for various viscoelastic models. Model/Name Constitutive Equation Creep Compliance, J(t) Relaxation Modulus, G(t) Elastic Solid G T=G 1 G=J=U(t) GG(t)=GU(t) Viscous Fluid η T=˙ T=˙t =t G(t) Maxw ell (Fluid) Gη˙T+1 T=G˙  @t+1 t T=fG@tg 1 G+t =+t Ge-t= Kelvin (Solid) G η T=G +˙ T=fG+@tg 1 T=1 +@t 1 G 1-e-t=
G+(t) Three Parameter Solid G1G2 η G1G2 +G12˙ = (G1+G2)T+2˙Te-t= G1+G1+G2 G1G2 1-et=2
or1 G1+1 G2 1-e-t=2
G1e-t=0+G1G2 G1+G2 1-e-t=0
orG1 G1+G2 G2+G1e-t=0
wher e0=2 G1+G2 Continued on next page Appendix B: Viscoelastic Creep and Relaxation 363TABLE B.1 – continued from previous page Model/Name Constitutive Equation Creep Compliance, J(t) Relaxation Modulus, G(t) G1 G2η ˙T+T 2=(G1+G2)˙ +G1 2 e-t=0 G1+G2+1 G1 1-e-t=0
or1 G1-G2 G1(G1+G2)e-t=0 wher e0=G1+G2 G12G1+G2e-t=2 G η1η2 G21˙ +12¨ = G2T+(1+2)˙T ¨ +˙ 2=1+2 12˙T+T 211 G1 1-e-t=0
+t 212 1+2(t)+G2 1+2 1+12 1+2 e0 wher e0=1+2 G2 G η2η1 ˙T+T 2=  G2+1 2 ˙ +1¨ t 1+2+1 G22 1+22 1-e-t
wher e=G2(1+2) 121(t)+G2e-t=2 G2η2G1 η1 ¨ +˙ 2=¨T G1+ 1 2+1 G12+1 1 ˙T+ 1 12T1 G1+t 1+1 G2 1-e-t=2
Continued on next page 364 Continuum Mechanics for EngineersTABLE B.1 – continued from previous page Model/Name Constitutive Equation Creep Compliance, J(t) Relaxation Modulus, G(t) G1 η1G2 η2 (1+2)˙T+(G1+G2)T= G1G2 + (2G1+ 1G2)˙ +12¨ 1 1+2 12 (t)-e-t=0 0! + G1G20 1-e-t=0
Index "-identity, 10 Matlabr invariants, 35 symbolic invariants, 36 acceleration field, 106, 172 adiabatic, 212, 275 angular momentum, 176 axiom of impenetrability, 106 balance laws, seeconservation of barotropic, 278, 280 fluid, 275 base vectors, 7 basis orthogonal, 7 Bernoulli equation, 280 bi-harmonic equation, 242, 255 polar form, 246 box product, seetriple scalar product caloric equation of state, 180 Cartesian axes, 7 Cauchy deformation tensor, 118, 208 left, 292 right, 133, 136, 292 Cayley-Hamilton theorem, 294 characteristic equation, 31 Christoffel symbols of the first kind, 354 of the second kind, 351 circulation, 41, 281 Clausius-Duhem equation, 167, 179, 181, 182 local form, 182 coefficient of bulk viscosity, 272 thermal expansion, 252 compatibility Beltrami-Michell, 224, 253, 254, 258 equations, 254 strain, 128 compliancecreep, 318 shear, 316, 320 volumetric, 320 concentrated moment, 55 configuration, 103 current, 104, 105, 116 initial, 106 reference, 104–106 conservation of angular momentum, 167, 182 energy, 167, 177, 179, 186 linear momentum, 167, 182, 186 mass, 167, 169, 182 constitutive equations, 167, 179, 196, 198, 272 continuity equation, 169, 213, 278 contraction, 17 convective rate of change, 112 covariant base vectors, 353 derivative, 353 derivative of tensors, 355 creep, 315 function, 316 response Kelvin, 316 Maxwell, 317 cross product, 11, 15 cubical dilatation, 123 curl, 38 decomposition additive, 134 multiplicative, 134 deformation, 103, 117 gradient, 116, 117, 134 homogeneous/inhomogeneous, 117 del operator, 38, 112 density, 53, 54, 109 determinant cofactor, 21 deviatoric response, 310, 311 dilatational response, 310, 320 365 366 Continuum Mechanics for Engineers Dirac delta function, 317 direction cosines, 25 displacement, 103, 110, 120 differential, 127 field, 110 displacement gradient, 120 distortional response, 320 divergence, 38 theorem of Gauss, 40, 62 dot product, 11 double dot product, 14 dual basis, 345 dummy index, 8 dyad, seetensor product dyad-dyad product, 13 eigen problem, 216 characteristic equation, 31 eigenvalue, 30 eigenvector, 30 normalized, 34 elastic materials, 178 elasticity, 211 Airy stress function, 242, 246, 259 constants Cijkl, 212 anisotropic, 215, 219 bulk modulus, 217 isotropic, 215, 223 Lam ´e, 215, 217, 221 modulus of elasticity, 216 orthotropic, 221 Poisson’s ratio, 216 shear modulus, 216, 219 elastodynamics, 223, 225 elastostatics, 223 linear, 211 linear solid, 199 nonlinear, 211, 285 plane, 238 semi-inverse, 243 symmetry, 219, 220 three dimensional, 253 displacement formulation, 253 stress formulation, 253 energy balance, 213 mechanical, 179 thermal, 179 density, 178 equation, 179internal, 178 specific internal, 178 entropy, 179 equation of state fluid, 271 equations of motion, 174 Eulerian, 172 Lagrangian, 172 local, 172 equilibrium, 61 force and moment, 61 balance of moments, 62 equation, 174, 253 local equations, 61 Euclidian space, 7 Euler equation, 275, 276, 278, 280 Eulerian description, 109, 273 strain, 118 field equations, 167 first law of thermodynamics, 167, 179 fluid, 271 linear viscous, 199 Newtonian, 271, 272 Stokesian, 271, 272 force body, 53, 223 contact, 53 surface, 53 Fourier’s law, 178, 253 Galerkin vector, 255, 256 gas dynamics equation, 278 gradient, 38 scalar, 38 vector, 38 Green’s deformation tensor, 117 Green’s second identity, 51 harmonic, 231 heat, 179 conduction, 253 flux, 178 supply, 178 Helmholtz, theorem of, 254 homogeneous, 212 body, 54 deformation, 117 Hooke’s law, 211, 214–216, 219, 240, 241, 253 Index 367 isotropic body, 223 plane strain, 241 hydrostatics, 276 hyperelastic material, 214 incompressible fluid, 275 material, 294 indices raising and lowering, 346 indicial notation, 5, 16 free index, 16 inelastic, 211 infinitesimal deformation theory, 120 rotation tensor, 127 strain tensor, 127 inner product, 17 internal constraint, 294 energy, 213 invariance, 189 invariants, 31, 35 inverse mapping, 105 inviscid fluid, 275 irrotational, 254, 280 flow, 278 isochoric, 145 isothermal, 212, 275 Jacobian determinant, 106 matrix, 344 Kelvin problem, 256 Kelvin’s theorem, 280, 281 kinematics, 103 kinetic energy, 177 Kronecker delta, 9, 19 Lagrange multiplier stress, 72 Lagrangian description, 109 equations of motion, seeequations of motion Lam ´e strain potential, 255 laminar flows, 275 Laplace equation, 279 left stretch tensor, 134 linearindependence, 344 momentum, 171 transformation, 5, 30 antecedent, 30 image, 30 local rate of change, 112 mapping, 103 mass, 169 density, 169 material body, 103 coordinates, 105, 109, 121 derivative, 111, 112 of a line element, 143 of a line segment, 143 of a volume, 145 of an area, 144 of Jacobian, 144 operator, 112 description, 109, seereferential de- scription, 110, 118 material frame indifference, 189 matrix, 19 column matrix, 19 addition, 19 adjoint, 23 decomposition, 20 determinant diagonal matrix, 22 expansion by cofactors, 22 minor, 21 signed minor, 21 determinant of square matrix, 21 diagonal, 19 exponent rule, 20 identity, 19 inverse, 23 multiplication, 20 associative, 20 communitive, 20 null, 19 orthogonal, 24 prefactor/postfactor, 20 row matrix, 19 scalar multiplication, 20 singular, 22 skew–symmetric, 19 square, 19 square of, 20 square root, 20 368 Continuum Mechanics for Engineers symmetric, 19 trace, 31 transpose, 19 vector-tensor product, 24 zero, 19 mechanical power, 177 Mohr’s circle, 189 for stress, 76 moment of momentum, 176 vector, 55 Mooney-Rivlin material, 292, 296 motion, 103, 117 multiplication cross product, 15 double dot product, 16 dyad product, 15 dyad-dyad product, 16 tensor-tensor product, 16 triple cross product, 15 triple scalar product, 15 vector by scalar, 15 vector by vector, 15 vector-dyad product, 15 vector-tensor products, 16 Navier equation, 224, 225, 240, 253, 254, 257 plane strain, 241 Navier-Stokes equations, 273–275 neo-Hookean material, 292, 296 Newton’s laws, 55 nonion form, 12 Ogden material, 297 orthonormal, 7 orthotropic, 221 outer product, 17 Papkovich-Neuber, 257 partial differentiation subscript comma notation, 37 particles, 103 path, 106 permutation symbol, 9, 10 plane strain, 124, 238 generalized, 241 plane stress, 80, 238 generalized, 241 plastic material, 199 polarcoordinates, 245 decomposition, 134 material, 176 polymer, 309 chain, 287 conformation, 287 positive/negative definite, 37, 134 potential flow, 278 pressure function, 276 head, 280 hydrostatic, 271 thermostatic, 271 principle of linear momentum, 55 rate of deformation, 137, 138 extension, 139 shear, 140 stretching, 139 reciprocal vectors, 345 referential coordinates, 105 description, seematerial description relaxation, 315 function, 318 residual stress, 214 resilience, 285 retardation time, 316 right principal direction of stretch, 135 stretch tensor, 134 rigid body, 103 displacement, 132 translation, 135 rotation tensor, 134 vector, 127 rubber butadiene, 285 molecular approach, 287 natural, 285 Saint-Venant extension, 227 flexure, 236 pure bending, 234 torsion, 228 scalar, 6 field, 37 product, 11 Index 369 second law of thermodynamics, 179, 186 simple shear deformation, 119 solenoidal, 254 spatial coordinates, 105, 121 description, 110, 118 position, 105 velocity gradient, 137 specific entropy, 180 entropy production, 181 free energy, 186 heat, 253 internal energy, 178 steady flow, 276 Stokes’ condition, 272 theorem, 41 strain, 104 axial, 217 deviator tensor, 123 energy, 211, 213 energy density, 213 engineering shear, 123 Eulerian, 118, 175 linearized, 120 finite strain tensor, 116 gauge rosette, 125 infinitesimal, 123 invariants, 121 Lagrangian, 117, 118, 175 linearized, 120 linear, 120 longitudinal, 121, 132 mean normal, 123 normal, 121 principal, 121 deviator, 124 directions, 121, 123 pure, 127 rate of, 140 shear, 125 strain rate, seestrain, rate of, 310 strain-displacement equation, 253 streamline, 280 stress, 53 as a tensor, 58 axial, 217 Cauchy formula, 58, 59, 61 principle, 54, 56characteristic equation, 67 components, 59 compressive, 59 deviatoric, 85, 252 principal values, 86 extremal values, 72 first Piola-Kirchhoff, 173 hydrostatic, 85 invariants, 67 maximum and minimum values, 71 shear, 76 mean, 85, 271 Mohr’s circle, 74 normal, 59, 71 octahedral, 87 normal, 87 shear, 87 orthogonal principal directions, 68 Piola-Kirchhoff, 172 plane, 80 power, 178 principal directions, 66 real values, 67 stresses, 66 values, 66 rate, 196 Green-Naghdi, 196 Jaumann, 196 second Piola-Kirchhoff, 174 shear, 59 magnitude, 72 spherical, 85, 252 symmetry, 61 tensor, 56 three dimensional stress function, 258 transformation, 63, 69 units, 59 vector, 55, 57, 66 stretch ratio, 131 tensor, 134 summation convention, 8 superposed rigid body motion, 189 superposition principle, 223, 224 symbolic notation, 5, 11 temperature reference, 252 tensor, 5 370 Continuum Mechanics for Engineers -tensor products, 13 anti-symmetric, 17 Cartesian, 5 component, 7 curl, 38 definition of, 5 divergence, 38 field, 37 metric, 346 order, 6 physical components, 358 positive/negative definite, 37 product, 12 skew-symmetric, 17 spin, 138, 141 symmetric, 17 trace, 31 transformation, 25, 27 thermodynamic state, 180 thermoelasticity, 252 thermomechanical continuum, 178 thermoplastic elastomers, 285 thermoset, 285 traction vector, 55 trajectory, 106 transformation, 25 fourth-order tensor, 212 arbitrary Cartesian tensor, 28 improper orthogonal, 28 law, 27 orthogonality condition, 26 tensor, 25, 212 vectors, 25 transport theorem, 168 triple cross product, 11, 15 scalar product, 11, 15 undeformed configuration, 104 unit base vectors, 7 extension, 132 vector, 6 -dyad product, 13 -tensor product, 24 -tensor products, 13 addition, 11, 15 axial, 18, 44 axial or pseudo, 28 components, 7curl, 38 divergence, 38 field, 37 multiplication, 11 physical components, 350 transformation, 25, 27 triangle rule, 6 true or polar, 28 velocity, 106 field, 55, 104, 106 gradient, 137 head, 280 viscoelastic, 309 absolute compliance, 323 complex compliance, 320 modulus, 320 correspondence principle, 324, 327 creep function, 318 dashpot, 312 harmonic loading, 320 hereditary integral, 318, 323 Kelvin model, 313, 323 generalized, 315 linear integral equations, 200 material, 199 linear differential operator, 310 loss compliance, 322 modulus, 321 Maxwell model, 313, 323, 324 generalized, 315 mechanical models, 311 relaxation modulus, 320 retardation time, 316 shear compliance, 320 standard linear solid, 313 storage compliance, 322 modulus, 321 superposition principle, 318 three-dimensional problems, 324 volumetric compliance, 320 viscosity coefficients, 272 viscous stress tensor, 271 vorticity, 137, 283 tensor, 138, 141 vector, 141 work, 179