Applied solid mechanics

Applied Solid Mechanics Cambridge Texts in Applied Mathematics Editorial Board Mark Ablowitz, University of Colorado, ...

Author: Peter Howell | Gregory Kozyreff | John Ockendon

154 downloads 1400 Views 8MB Size Report

This content was uploaded by our users and we assume good faith they have the permission to share this book. If you own the copyright to this book and it is wrongfully on our website, we offer a simple DMCA procedure to remove your content from our site. Start by pressing the button below!
Report copyright / DMCA form

DOWNLOAD PDF

Applied Solid Mechanics

Cambridge Texts in Applied Mathematics Editorial Board Mark Ablowitz, University of Colorado, Boulder S. Davis, Northwestern University E.J. Hinch, University of Cambridge Arieh Iserles, University of Cambridge John Ockendon, University of Oxford Peter Olver, University of Minnesota

Applied Solid Mechanics Peter Howell University of Oxford

Gregory Kozyreff Universit´ e Libre de Bruxelles

John Ockendon University of Oxford

published by the press syndicate of the university of cambridge The Pitt Building, Trumpington Street, Cambridge, United Kingdom cambridge university press The Edinburgh Building, Cambridge CB2 2RU, UK 40 West 20th Street, New York, NY 10011–4211, USA 477 Williamstown Road, Port Melbourne, VIC 3207, Australia Ruiz de Alarc´ on 13, 28014 Madrid, Spain Dock House, The Waterfront, Cape Town 8001, South Africa http://www.cambridge.org c Cambridge University Press 2008 ° This book is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published 2008. Printed in the United Kingdom at the University Press, Cambridge Typeface Monotype Times 10/13pt

System LATEX 2ε

[author]

A catalogue record for this book is available from the British Library Library of Congress Cataloguing in Publication data available ISBN 987–0521– hardback

Contents

List of Illustrations Prologue

page xi xvii

1

Modelling Solids 1.1 Introduction 1.2 Hooke’s law 1.3 Lagrangian and Eulerian coordinates 1.4 Strain 1.5 Stress 1.6 Conservation of momentum 1.7 Linear elasticity 1.8 The incompressibility approximation 1.9 Energy 1.10 Boundary conditions and well-posedness 1.11 Coordinate systems 1.11.1 Cartesian coordinates 1.11.2 Cylindrical polar coordinates 1.11.3 Spherical polar coordinates Exercises

1 1 2 3 4 7 10 11 13 14 16 19 19 20 22 24

2

Linear Elastostatics 2.1 Introduction 2.2 Linear displacements 2.2.1 Isotropic expansion 2.2.2 Simple shear 2.2.3 Uniaxial stretching 2.2.4 Biaxial strain 2.2.5 General linear displacement 2.3 Antiplane strain

27 27 28 28 30 30 33 34 36

v

vi

Contents

2.4 2.5 2.6

Torsion Multiply-connected domains Plane strain 2.6.1 Definition 2.6.2 The Airy stress function 2.6.3 Boundary conditions 2.6.4 Plane strain in a disc 2.6.5 Plane strain in an annulus 2.6.6 Plane strain in a rectangle 2.6.7 Plane strain in a semi-infinite strip 2.6.8 Plane strain in a half-space 2.6.9 Plane strain with a body force 2.7 Compatibility 2.8 Generalised stress functions 2.8.1 General observations 2.8.2 Plane strain revisited 2.8.3 Plane stress 2.8.4 Axisymmetric geometry 2.8.5 The Galerkin representation 2.8.6 Papkovich–Neuber potentials 2.8.7 Maxwell and Morera potentials 2.9 Singular solutions in elastostatics 2.9.1 The delta function 2.9.2 Point and line forces 2.9.3 The Green’s tensor 2.9.4 Point incompatibility Exercises 3

Linear Elastodynamics 3.1 Introduction 3.2 Normal modes and plane waves 3.2.1 Normal modes 3.2.2 Waves in the frequency domain 3.2.3 Scattering 3.2.4 P -waves and S -waves 3.2.5 Mode conversion in plane strain 3.2.6 Love waves 3.2.7 Rayleigh waves 3.3 Dynamic stress functions 3.4 Waves in cylinders and spheres

38 42 46 46 46 48 50 52 55 57 61 64 66 69 69 70 72 74 76 77 79 81 81 83 85 89 91 100 100 101 101 107 109 110 112 114 118 119 121

Contents

4

vii

3.4.1 Waves in a circular cylinder 3.4.2 Waves in a sphere 3.5 Initial-value problems 3.5.1 Solutions in the time domain 3.5.2 Fundamental solutions 3.5.3 Characteristics 3.6 Moving singularities 3.7 Concluding remarks Exercises

121 125 129 129 130 134 135 140 141

Approximate Theories 4.1 Introduction 4.2 Longitudinal displacement of a bar 4.3 Transverse displacements of a string 4.4 Transverse displacements of a beam 4.4.1 Derivation of the beam equation 4.4.2 Boundary conditions 4.4.3 Compression of a beam 4.4.4 Waves on a beam 4.5 Linear rod theory 4.6 Linear plate theory 4.6.1 Derivation of the plate equation 4.6.2 Boundary conditions 4.6.3 Simple solutions of the plate equation 4.6.4 An inverse plate problem 4.6.5 More general in-plane stresses 4.7 Von Kármán plate theory 4.7.1 Assumptions underlying the theory 4.7.2 The strain components 4.7.3 The Von Kárm´ an equations 4.8 Weakly curved shell theory 4.8.1 Strain in a weakly curved shell 4.8.2 Linearised equations for a weakly curved shell 4.8.3 Solutions for a thin shell 4.9 Nonlinear beam theory 4.9.1 Derivation of the model 4.9.2 Example: deflection of a diving board 4.9.3 Weakly nonlinear theory and buckling 4.10 Nonlinear rod theory 4.11 Geometrically nonlinear wave propagation

147 147 148 149 151 151 153 154 155 156 160 160 162 165 166 167 169 169 169 172 174 174 176 177 183 183 186 188 192 194

viii

Contents

4.11.1 Nonlinearity and solitons 4.11.2 Gravity-torsional waves 4.11.3 Travelling waves on a beam 4.11.4 Weakly nonlinear waves on a beam 4.12 Concluding remarks Exercises

194 195 196 198 201 202

5

Nonlinear Elasticity 5.1 Introduction 5.2 Stress and strain revisited 5.2.1 Deformation and strain 5.2.2 The Piola–Kirchhoff stress tensors 5.2.3 The momentum equation 5.2.4 Example: one-dimensional nonlinear elasticity 5.3 The constitutive relation 5.3.1 Polar decomposition 5.3.2 Strain invariants 5.3.3 Frame indifference and isotropy 5.3.4 The energy equation 5.3.5 Hyperelasticity 5.3.6 Linear elasticity 5.3.7 Incompressibility 5.3.8 Examples of constitutive relations 5.4 Examples 5.4.1 Principal stresses and strains 5.4.2 Biaxial loading of a square membrane 5.4.3 Blowing up a balloon 5.4.4 Cavitation 5.5 Conclusion Exercises

213 213 214 214 216 217 218 219 219 220 222 224 226 228 229 230 231 231 232 233 235 237 237

6

Asymptotic Analysis 6.1 Introduction 6.2 The linear plate equation 6.2.1 Nondimensionalisation and scaling 6.2.2 Dimensionless equations 6.2.3 Leading-order equations 6.3 Boundary conditions and St Venant’s principle 6.3.1 Boundary layer scalings 6.3.2 Equations and boundary conditions 6.3.3 Solvability conditions

243 243 246 246 247 249 251 252 253 255

Contents

ix

6.3.4 Asymptotic expansions 256 The von Kármán plate equations and weakly curved shells 260 6.4.1 Background 260 6.4.2 Scalings 261 6.4.3 Leading-order equations 262 6.4.4 Equations for a weakly curved shell 265 6.5 The Euler–Bernoulli plate equations 267 6.5.1 Dimensionless equations 267 6.5.2 Asymptotic structure of the solution 269 6.5.3 Leading-order equations 269 6.5.4 Longitudinal stretching of a plate 271 6.6 The linear rod equations 273 6.7 Linear shell theory 274 6.7.1 Geometry of the shell 274 6.7.2 Dimensionless equations 275 6.7.3 Leading-order equations 276 6.8 Conclusions 279 Exercises 279 6.4

7

Fracture and Contact 7.1 Introduction 7.2 Static Brittle Fracture 7.2.1 Physical background 7.2.2 Mode III cracks 7.2.3 Mathematical methodologies for crack problems 7.2.4 Mode II cracks 7.2.5 Mode I cracks 7.2.6 7.2.7 Dynamic fracture 7.3 Contact 7.3.1 Contact of elastic strings 7.3.2 Other thin solids 7.3.3 Smooth contact in plane strain 7.4 Conclusions Exercises

283 283 284 284 285 290 292 299 301 304 306 306 310 313 316 317

8

Plasticity 8.1 Introduction 8.2 Models for granular material 8.2.1 Static behaviour 8.2.2 Granular flow

323 323 325 325 326

x

8.3 8.4

8.5 8.6

Dislocation theory Perfect plasticity theory for metals 8.4.1 Torsion problems 8.4.2 Plane strain 8.4.3 Three-dimensional yield conditions Plastic flow Simultaneous Elasticity and Plasticity

332 338 338 343 346 351 357

9

More General Theories 9.1 Introduction 9.2 Viscoelasticity 9.2.1 Introduction 9.2.2 Springs and dashpots 9.2.3 Three-dimensional linear viscoelasticity 9.2.4 Large-strain viscoelasticity 9.3 Thermoelasticity 9.4 Composite Materials and Homogenisation 9.4.1 One-dimensional homogenisation 9.4.2 Two-dimensional homogenisation 9.4.3 Three-dimensional homogenisation 9.5 Poroelasticity 9.6 Anisotropy 9.7 Epilogue Epilogue Appendix

367 367 368 368 369 373 375 378 381 381 385 391 393 398 402 410 412

Appendix

412

Appendix 2

Elementary differential geometry

415

Appendix 3 Orthogonal curvilinear coordinates 419 BibliographyOrthogonal curvilinear coordinatesA3.8 Time-dependent coordinatesBIBLIOGRAPHYBIBLIOGRAPHY 433

List of Illustrations

1.1 1.2 1.3 1.4 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 2.12 2.13 2.14 2.15 2.16 3.1 3.2 3.3

A tetrahedron; ai is the area of the face orthogonal to the xi -axis. 8 The forces acting on a small two-dimensional element. 9 Schematic of a small pill-box-shaped region at the boundary between two elastic solids. 18 Forces acting on a polar element of solid. 22 A unit cube undergoing (a) uniform expansion, (2.2), (b) one-dimension shear, (2.6), (c) uniaxial stretching, (2.8). 29 Schematic of a uniform bar being stretched under a tensile force T . 30 A paper model with negative Poisson’s ratio. Each line segment is a strip of paper viewed end-on. 32 Schematic of a plate being strained under tensions Txx , Tyy and shear forces Txy and Tyx . 32 Schematic of a bar in a state of antiplane strain. 37 Schematic of a bar being twisted under a moment M . 38 Schematic of a uniform tubular torsion bar, with inner and outer free surfaces given by ∂Di and ∂Do respectively. 42 The cross-section of (a) a circular cylindrical tube; (b) a cut tube. 43 Schematic of a region D whose boundary ∂D has unit normal n and tangent t. 48 Schematic of a plane annulus being inflated by an internal pressure P . 52 Schematic of a plane rectangular region subject to tangential tractions on its faces. 55 Schematic of the tractions applied to the edge of a semi-infinite strip. 57 The surface displacement and pressure fields defined by (2.161). 63 A family of functions δ² (x) that approach a delta-function as ² → 0. 81 Contours of the maximum shear stress S created by a point force acting at the origin. Here ν = 1/4 and f /S = 0.2, 0.4, . . . , 2.0. 84 Schematic of four point forces. 90 Plots of the first three Bessel functions Jn (ξ) and Yn (ξ) for n = 0 (solid), n = 1 (dashed), n = 2 (dot-dashed). 105 Schematic of a P -wave reflecting from a rigid boundary. 113 Schematic of a layered elastic medium. 115

xi

xii 3.4

3.5 3.6 3.7 3.8 3.9 3.10 3.11 3.12

3.13 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10 4.11 4.12 4.13 4.14

List of Illustrations Dispersion relation between frequency ω and wavenumber k for symmetric (solid) and antisymmetric (dotted) Love waves. The phase speed ω/k is bounded between cs1 and cs2 . Illustration of flexural waves. The one-dimensional fundamental solution R(x, t). The two-dimensional fundamental solution R(x, y, t) versus radial p distance r = x2 + y 2 . Schematic showing the cone x2 + y 2 = c2 t2 tangent to the plane k1 x + k2 y = ωt. The two-sheeted characteristic cone for the Navier equation. The response of a string to a point force moving at speed V : (a) V = 0, (b) 0 < V < c, (c) V = c, (d) V > c. Schematic of wave fronts generated by a moving force on an elastic membrane with (a) M < 1, (b) M > 1. P -wave (dashed) and S -wave (solid) fronts generated by a point force moving at speed V in plane strain with (a) 0 < V < cs , (b) cs < V < cp , (c) V > cp . Group velocity cg = dω/dk versus wavenumber k for symmetric (solid) and antisymmetric (dotted) Love waves. Schematic of a uniform bar showing the forces acting on a small segment of length δx. Schematic of an elastic string showing the forces acting on a small segment of length δx. Schematic showing the forces and moments acting on a small segment of an elastic beam. Schematic of the end of a beam under (a) clamped, (b) simply supported and (c) free conditions. The first three buckling modes of a clamped elastic beam (with a = 1 and L = 1). Schematic of a thin elastic rod showing the internal force components T , Ny and Nz . Cross-section through a rod showing the bending moment components. Some examples of cross-sections and their moments of inertia. Schematic showing the forces acting on a small section of an elastic plate. Schematic showing the bending moments acting on a section of an elastic face. The solution (4.61) for a plate sagging under gravity, with a = 1, b = 1/2, ςg/D = 1. (a) A cylinder. (b) A cone. (c) Another developable surface. (d) A hyperboloid, which is ruled but not developable. Typical surface shapes with (a) positive, (b) negative and (c) zero Gaussian curvature. Deformation of a cylindrical shell; (a) original shape, (b) bending about the generators, (c) rotating the generators, (d) bending along the generators.

117 126 132 133 135 136 137 138

140 143 148 150 151 152 155 156 156 158 159 161 166 172 176

179

List of Illustrations 4.15

4.16 4.17 4.18

4.19

4.20

4.21

4.22 4.23 4.24 4.25 4.26 4.27 4.28

5.1 5.2 5.3 5.4 5.5 5.6 6.1 6.2 7.1 7.2

Deformation of an anticlastic shell; (a) original shape, (b) bending along the lines x = const., (c) bending along the lines y = const., (d) bending along the lines x − y = const.. Deformation of a convex shell; (a) original shape, (b) and (c) two possible deformations, (d) one-dimensional bending. Schematic of a beam (a) before and (b) after bending; (c) a close-up of the displacement field. (a) Schematic of the forces and moments acting on a small segment of a beam. (b) Definition sketch showing the sign convention for the forces at the ends of the beam. p (a) Final angle −α versus force parameter L 2F/EI. (b) Deflection of a diving board for various values of the force parameter; the dashed lines show the linearised solution. (a) Response diagram of the norm kθk of the solution (4.136) versus the control parameter λ. (b) Corresponding response of the weakly nonlinear solution; the exact nonlinear solution is shown using dashed lines. (a) Pitchfork bifurcation diagram of leading-order amplitude A0 versus forcing parameter λ1 . (b) The corresponding diagram when asymmetry is introduced. (a) Schematic of a system of pendulums attached to a twisting rubber band. (b) Transverse view defining the angle θ(x, t). A kink propagating along a series of pendulums attached to a rod. Travelling wave solution (4.173) of the nonlinear beam equations (with k = 1). Definition sketch of a beam clamped near the edge of a table. Schematic of a beam supported at two points. The first three buckling modes of a vertically clamped beam. Left: a piston containing a tuft of fibres and subjected to a pressure p. Right: detail of the tuft. A piece of fibre is supported at two contact points a distance L apart. A fibre on top is transmitting a force F ∼ δp × dL. The deformation of a small scalene cylinder. Typical force–strain graphs for uniaxial tests on various materials. Schematic of a square membrane subject to an isotropic tensile force. Response diagrams for a biaxially-loaded incompressible sheet of Mooney–Rivlin material. Scaled pressure inside a balloon as a function of the stretch for various values of the Mooney–Rivlin parameter. Gas pressure inside a cavity as a function of inflation coefficient for various values of the Mooney–Rivlin parameter. Schematic of the edge of a plate subject to tractions. Definition sketch of the geometry of a deformed two-dimensional plate. Definition sketch of a thin crack. Definition sketch for contact between two solids with nearly parallel boundaries.

xiii

181 182 184

185

188

189

191 195 197 198 203 203 209

211 216 231 232 233 234 236 251 268 283 283

xiv 7.3 7.4 7.5 7.6 7.7 7.8 7.9 7.10 7.11 7.12 7.13 7.14 7.15 7.16 7.17

7.18 7.19 7.20 7.21 8.1 8.2 8.3 8.4 8.5 8.6 8.7

List of Illustrations (a) A Mode III crack. (b) A close-up of one tip. 285 Definition sketch for r1 , r2 ,θ1 and θ2 in (7.12). 287 Displacement field for a Mode III crack. 288 Sketch of a rounded tip, showing the radius of curvature r0 . 289 Schematic showing cohesion between crack faces. 289 (a) Schematic of a planar Mode II crack. (b) The regularised problem of a thin elliptical crack. 292 Contour plot of the maximum shear stress S around a Mode II crack, with crack length c = 1. 296 The displacement of a Mode II crack of half-length c = 1 under increasing shear stress σII . 297 Schematic of a Mode I crack. 298 Contour plot of the maximum shear stress S around a Mode I crack, with crack length c = 1. 300 The displacement of a Mode I crack of half-length c = 1 under increasing normal stress σI . 300 Schematic of an elastic string being forced against a rigid obstacle Γ. 305 Solution for the contact between a string and a level surface, with applied pressure p/T = 0, 1, 2, 3, 4, 5. tidy up figure 307 Three candidate solutions for the contact problem between a stress-free string and the surface z = 2x2 − 3x4 . tidy up figure 308 The contact between a beam and a horizontal surface under a uniform pressure p. (a) The beam sags towards the surface and makes contact: p/EI = 0, 12/5, 24/5. (b) The beam makes contact at just one point: p/EI = 24/5, 24. (c) The contact set grows: p/EI = 24, 2048/27, 384. 310 Definition sketch for contact between a rigid body and an elastic half-space. 314 The penetration of a quadratic punch into an elastic half-space y > 0. 315 A flexible ruler being flattened against a table with applied bending moment M/EI = 0, 3, 6, 9, 12, 15. 320 (a) A wave travelling along a rope on the ground. (b) Some higher travelling modes. 321 Schematic of a typical stress-strain relationship for a plastic material: (a) below the yield stress τY ; (b) when the yield stress is exceeded. 323 Schematic of the stress-strain relationship for a perfectly plastic material: (a) under a single loading; (b) under periodic loading/unloading. 324 Schematic showing the forces acting on a particle at the surface of a granular material. 325 Schematic of the normal force N and frictional force F acting on a surface element inside a granular material. 327 The Mohr circle in the (N, F )-plane, and the lines where |F | = |N | tan φ.328 Schematic of a cut-and-weld operation leading to the displacement field (8.21). 332 The displacement field in an edge dislocation: (a) pristine material; (b) after the cut-and-weld operation. Nodal points correspond and the cut is shown dotted. 334

8.8 8.9

8.10

List of Illustrations

xv

Schematic of an edge dislocation in a square crystal lattice: (a) pristine crystal; (b) insertion of an extra row of atoms.

335

Schematic of a moving edge dislocation: (a) initial configuration; (b) and (c) the dislocation moves to the left as the atoms realign themselves, eventually leading to a displacement of the upper block of atoms relative to the lower.

336

The normalised torque M versus twist Ω applied to an elastic/plastic cylindrical bar.

340

8.11

The normalised torque M versus twist Ω applied to an elastic/plastic cylindrical bar, showing the recovery phase when the torque is released. 342

8.12

Schematic of the free-boundary problem for an elastic/perfectly plastic torsion bar.

343

(a) The Tresca yield surface (8.61) plotted in the space of principal stress deviators (τ10 , τ20 , τ30 ) (normalised with τY ), showing the intersection with the plane τ10 + τ20 + τ30 = 0. (b) The projection in the (τ10 , τ20 )-plane.

348

8.13

8.14

(a) The Von Mises yield surface plotted in the space of principal stress deviators (τ10 , τ20 , τ30 ) (normalised with τY ), showing the intersection with the plane τ10 + τ20 + τ30 = 0. (b) The projection in the (τ10 , τ20 )-plane.349

8.15

The Coulomb yield surface plotted in the space of principal stresses (τ1 , τ2 , τ3 ) (with θ = π/6).

8.16 8.17

8.18

9.1 9.2

9.3

350 359

The Mohr surface in the (N, F )-plane for three-dimensional granular flow, and the line F = |N | tan φ. The region where solutions exist is shaded.

361

The normalised torque M versus twist Ω applied to an elastic/plastic cylindrical bar undergoing one loading cycle. The bar deforms elastically (1), then yields (2), then recovers elastically (3) when the load is released. When a load is now applied in the opposite direction (4), the bar yields at a lower critical torque (5) and then deforms plastically. Finally, the load is released (6) and a permanent negative twist remains.

363

(a) A spring; (b) a dashpot; (c) a spring and dashpot connected in parallel; (d) a spring and dashpot connected in series.

369

(a) Applied tension T as a function of time t; te is the characteristic time-scale of the experiment. (b) Resultant displacement of a linear elastic spring. (c) Resultant displacement of a linear dashpot.

370

Displacement of a Voigt element due to the applied tension shown in Figure 9.2(a). (a) Small Deborah number; the corresponding response of an elastic spring is shown as a dashed curve. (a) Large Deborah number; the corresponding response of a dashpot is shown as a dashed curve.

371

xvi 9.4

9.5

9.6

9.7 9.8

9.9 9.10 9.11

9.12 9.13 9.14 A2.1 A3.1 A3.2 A3.3

List of Illustrations Displacement of a Maxwell element due to the applied tension shown in Figure 9.2(a). (a) Small Deborah number; the corresponding response of a dashpot is shown as a dashed curve. (a) Large Deborah number; the corresponding response of an elastic spring is shown as a dashed curve. 372 (a) The variation of Young’s modulus with position in a bar. (b) The corresponding longitudinal displacement u(x); the approximation (9.53) is shown as a dashed line. 381 (a) The variation of Young’s modulus with position in a bar; the effective Young’s modulus is shown as a dashed curve. (b) The corresponding longitudinal displacement u(x); the leading-order approximation, given by (9.62), is shown as a dashed curve. 384 A typical microstructured shear modulus. 385 Definition sketch for a symmetric, piecewise constant shear modulus distribution; the symmetry axes are shown as dot-dashed lines. (a) The boundary conditions satified by φ(1) and φ(2) . (b) The corresponding boundary conditions for the harmonic conjugate functions ψ (1) and ψ (2) .389 Some modulus distributions in a square that are antisymmetric about the diagonals. 391 Schematic of the one-dimensional squeezing of a sponge. 396 Dimensionless stress applied to a sponge versus dimensionless time for different values of the Péclet number Pe; the large-t limits are shown as dotted lines. 397 A Jeffreys viscoelastic element. 404 405 405 (i) a straight line in the direction (a, b, c)T ; (ii) a circle of radius a; (iii) a helix of radius a and pitch 2πb. 416 A small box B. 425 Cylindrical polar coordinates. 429 Spherical polar coordinates. 430

Prologue

Although solid mechanics is a vitally important branch of applied mechanics, it is often thought to be less intellectually exciting than its close relative, fluid mechanics. Several reasons can be advanced for this disparagement, such as the daunting prevalence of tensors in models for solids, the especial difficulty of handling nonlinearity, or the many ways in which elementary theories soon become irrelevant in practice because of mysterious changes in the way a solid behaves. In these notes, we aim to give the subject as wide an accessibility as possible to mathematically-minded students and to emphasise the interesting mathematical issues that it raises. We will do this by relating the theory to practical applications where surprising phenomena occur and where innovative mathematical methods are needed. Phenomena such as fracture, buckling and plasticity pose intellectual challenges in solid mechanics that are every bit as fascinating as concepts like flight, shock waves and turbulence in fluid dynamics. Our principal aim in these notes is to demonstrate this fact to undergraduate and beginning graduate students who only have some knowledge of basic mechanics and of the calculus of several variables. Our layout is essentially pragmatic. Although more advanced texts in solid mechanics often begin with quite general theories founded on very basic mechnical and thermodynamic principles, we will start from elementary observations in high-school physics and build our way towards models that are the basis for current applied research in solid mechanics. Hence, we will begin with linear elasticity, static and dynamic, including simple models for fracture and contact. We will then proceed to describe some asymptotic theories for thin solids, namely elastic rods, plates and shells. The detailed formalism of geometrical nonlinearity that lies behind these models is developed in a separate chapter, along with a brief mention of the somewhat recherché subject of mechanically nonlinear elasticity. Both of these may

xviii

Prologue

be omitted on a first reading. Finally, we will show how the theory may be generalised to describe inelastic phenomena, especially plasticity, and to include further physical effects, such as thermal stresses and porosity. These “combined fields” of solid mechanics are increasingly finding applications in industrial and medical processes, and pose ever more elaborate modelling questions. Despite the breadth of the models and relevant technologies that will emerge in these notes, we will always try to present the theoretical developments ab initio. Nonetheless, the notes are very far from being selfcontained. Any student who aspires to becoming a solid mechanics specialist will have to delve much further and we will provide numerous references to help with this. Also we have assumed a reasonable knowledge of the calculus of several variables, and some very basic knowledge of partial differential equations and the calculus of variations. Our hope is that, having read these notes, a student should be able to confront, without too much trepidation, any practical problem that may be encountered in everyday solid mechanics with at least some idea of the basic mathematical modelling that will be required.

1 Modelling Solids

1.1 Introduction In everyday life we regularly encounter physical phenomena that apparently vary continuously in space and time. Examples are the bending of a paper clip, the flow of water or the propagation of sound or light waves. Such phenomena can be described mathematically, to lowest order, by a continuum model, and this book will be concerned with that class of continuum models that describes solids. For our purposes, the diagnostic feature of a solid is the way in which it responds to an applied system of forces and moments. There is no hard and fast rule about this but, for most of this book, we will say that a continuum is a solid when the response consists of displacements distributed through the material. In other words, the material starts at some rest state, from which it is displaced by a distance that depends on the applied forces. This is in contrast to a fluid, which has no special rest state and responds to forces via a velocity distribution. Even from this simple definition we will find that we can construct solid mechanics theories for phenomena as diverse as earthquakes, ultrasonic testing and the buckling of railway tracks. Our modelling philosophy is straightforward. We take the most fundamental pieces of experimental evidence, for example Hooke’s law, and use mathematical ideas to combine this evidence with the basic laws of mechanics to construct a model that describes the deformation of a continuous solid. This approach inevitably leads to systems of partial differential equations, and is epitomised by the well-known description of one-dimensional elastic wave propagation in strings and bars. We emphasise that, at least to begin with, we will avoid all consideration of the “atomistic” structure of solids, even though these ideas lead to great practical insight and also to some beautiful mathematics. Thus, when we refer to a solid “particle”, we will 1

2

Modelling Solids

be thinking of a very small region of matter but one whose dimension is nonetheless much greater than an atomic spacing. By basing our model on Hooke’s law, the simplest model of elasticity, for small enough forces and displacements, we will first be led to a system of differential equations that is both linear, and therefore mathematically tractable, and reversible for time-dependent problems. By this we mean that, when forces and moments are applied and then removed, the system eventually returns to its original state without any significant energy being lost, i.e. the system is not dissipative. Reversibility may apply even when the forces and displacements are so large that the problem ceases to be linear; a rubber band, for example, can undergo large displacements and still return to its initial state. However, nonlinear elasticity encompasses some striking new behaviours not predicted by linear theory, including the possibility of multiple steady states and buckling. However, for many materials, experimental evidence reveals that even more dramatic changes can take place as the load increases, the most striking phenomenon being that of fracture under extreme stress. On the other hand, as can be seen by simply bending a metal paper clip, irreversibility can readily occur and this is associated with plastic flow that is significantly dissipative. In this situation, the solid takes on some of the attributes of a fluid, but the model for its flow is quite different from that for, say, water. Practical solid mechanics encompasses not only all the phenomena mentioned above but also the effects of elasticity and plasticity when combined with heat transfer (leading to thermo-elasticity and thermo-plasticity) and with genuine fluid effects, in cases where the material flows even in the absence of large applied forces (leading to visco-elasticity and visco-plasticity) or when the material is porous (leading to poro-elasticity). We will defer consideration of all these “combined fields” until the end of the book.

1.2 Hooke’s law Robert Hooke Hooke (1678) wrote “ . . . it is . . . evident that the rule or law of nature in every springing body is that the force or power thereof to restore itself to its natural position is always proportionate to the distance or space it is removed therefrom, whether it be by rarefaction, or separation of its parts the one from the other, or by condensation, or crowding of those parts nearer together.”

Hooke’s observation is exemplified by a simple high-school physics experiment in which a tensile force T is applied to a spring whose natural length

1.3 Lagrangian and Eulerian coordinates

3

is L. Hooke’s law states that the resulting extension of the spring is proportional to T : if the new length of the spring is `, then T = k(` − L),

(1.1)

where the constant of proportionality k is called the spring constant. Hooke devised his law while designing clock springs, but noted that it appears to apply to all “springy bodies whatsoever, whether metal, wood, stones, baked earths, hair, horns, silk, bones, sinews, glass and the like.” In practice, it is commonly observed that k scales with 1/L; that is, everything else being equal, a sample that is initially twice as long will stretch twice as far under the same force. It is therefore sensible to write (1.1) in the form T = k0

`−L , L

(1.2)

where k 0 is the elastic modulus of the spring, which will be defined more rigorously in Chapter 2. The dimensionless quantity (` − L)/L, measuring the extension relative to the initial length, is called the strain. Equation (1.2) is simplest example of the all-important constitutive law relating the force to displacement. To generalise this law to a three-dimensional continuum, we first need to generalise the concepts of strain and tension. 1.3 Lagrangian and Eulerian coordinates Suppose that a three-dimensional solid starts, at t = 0, in its rest state, or reference state, in which no macroscopic forces exist in the solid or on its boundary. Under the action of any subsequently applied forces and moments, the solid will be deformed such that, at some later time t, a “particle” in that state whose initial position was the point X is displaced to the point x (X, t). This is a Lagrangian description of the continuum: if the independent variable X is held fixed as t increases, then x(X, t) labels a material particle. In the alternative Eulerian approach, we consider the material point which currently occupies position x at time t, and label its initial position by X(x, t). In short, the Eulerian coordinate x is fixed in space, while the Lagrangian coordinate X is fixed in the material. The displacement u(X, t) is defined in the obvious way to be the difference between the current and initial positions of a particle, that is u(X, t) = x(X, t) − X.

(1.3)

Many basic problems in solid mechanics amount to determining the displacement field u corresponding to a given system of applied forces.

4

Modelling Solids

The mathematical consequence of our statement that the solid is a continuum is that there must be a smooth one-to-one relationship between X and x, i.e. between any particle’s initial position and its current position. This will be the case provided the Jacobian of the transformation from X to x is bounded away from zero: µ ¶ ∂xi 0 <J < ∞, where J = det . (1.4) ∂Xj The physical significance of J is that it measures the change in a small volume compared with its initial volume: dx1 dx2 dx3 = J dX1 dX2 dX3 ,

or dx = J dX

(1.5)

as shorthand. We can use (1.5) to derive a kinematic equation representing conservation of mass. Consider a moving volume V (t) that is always bounded by the same solid particles. Its mass at time t is given, in terms of the density ρ(X, t), by ZZZ ZZZ M (t) = ρ dx = ρJ dX. (1.6) V (t)

V (0)

Since V (t) is a fixed material volume, M (t) must be a constant, namely its initial value M (0): ZZZ ZZZ ρJ dX = M (t) = M (0) = ρ0 dX, (1.7) V (0)

V (0)

where ρ0 is the density in the rest state. Since V is arbitrary, we deduce that ρJ = ρ0 .

(1.8)

Hence, we can calculate the density at any time t in terms of ρ0 and the displacement field. The initial density ρ0 is usually taken as constant, but (1.8) also applies if ρ0 = ρ0 (X).

1.4 Strain To generalise the concept of strain introduced in §1.2, we consider the deformation of a small line segment joining two neighbouring particles with initial positions X and X + δX. At some later time, the solid deforms such that the particles are displaced to X + u(X, t) and X + δX + u(X + δX, t)

1.4 Strain

5

respectively. Thus, the line element δX that joins the two particles is transformed to δx = δX + u(X + δX, t) − u(X, t) ∼ δX + (δX · ∇)u(X, t),

(1.9)

where (δX · ∇) = δX1

∂ ∂ ∂ + δX2 + δX3 . ∂X1 ∂X2 ∂X3

(1.10)

Let L = |δX| and ` = |δx| denote the initial and current lengths respectively of the line segment; the difference ` − L is known as the stretch. Then, to lowest order in L, `2 = |δX + (δX · ∇)u(X, t)|2 .

(1.11)

Although we will try in subsequent chapters to minimise the use of suffices, it is unavoidable at this stage to introduce components so that X = (Xi ) = (X1 , X2 , X3 ) and similarly for u. Then (1.11) may be written in the form `2 − L2 = 2

3 X

Eij δXi δXj ,

(1.12)

i,j=1

where 1 Eij = 2

Ã

3

X ∂uk ∂uk ∂uj ∂ui + + ∂Xj ∂Xi ∂Xi ∂Xj

! .

(1.13)

k=1

By way of introduction to some notation that will be useful later, we point out that (1.12) may be written in at least two alternative ways. First, we may invoke the so-called summation convention, in which one automatically sums over any repeated suffix. This avoids the annoyance of having to write explicit summation, so (1.12) is simply µ ¶ ∂uj 1 ∂ui ∂uk ∂uk 2 2 ` = L + 2Eij δXi δXj , where Eij = + + . 2 ∂Xj ∂Xi ∂Xi ∂Xj (1.14) Second, we note that `2 − L2 is a quadratic form on the symmetric matrix E whose components are (Eij ): `2 − L2 = 2 δX T E δX.

(1.15)

It is clear from (1.12) that the stretch is measured by the quantities Eij ; in particular, the stretch is zero for all line elements if and only if Eij ≡ 0. It is thus natural to identify Eij with the strain. Now let us ask: “what

6

Modelling Solids

happens when we perform the same calculation in a set of axes rotated by an orthogonal matrix P = (pij )?” Intuitively, we might expect the strain to be invariant under such a rotation, and we can verify that this is so as follows. The vectors X and u are transformed to X 0 and u0 in the new coordinate system, where X 0 = P X,

u0 = P u.

(1.16)

Since P is orthogonal, (1.16) may be inverted to give X = P T X 0 . Alternatively, using suffix notation, we have Xβ = pjβ Xj0 ,

u0i = piα uα .

The strain in the new coordinate system is denoted by Ã ! 0 0 ∂u0 0 ∂u ∂u 1 ∂u j i k k Eij0 = + + , 2 ∂Xj0 ∂Xi0 ∂Xi0 ∂Xj0

(1.17)

(1.18)

which may be manipulated, using the chain rule, to give (Exercise 1.3) Eij0 = piα pjβ Eαβ .

(1.19)

In matrix notation, (1.19) takes the form E 0 = P EP T ,

(1.20)

so the 3 × 3 symmetric array (Eij ) transforms exactly like a matrix representing a linear transformation of the vector space R3 . Arrays that obey the transformation law (1.19) are called second-rank Cartesian tensors, and E = (Eij ) is therefore called the strain tensor.† Almost as important as the fact that E is a tensor is the fact that it can vanish without u vanishing. More precisely, if we consider a rigid-body translation and rotation u = c + (P − I)X,

(1.21)

where I is the identity matrix while the vector c and orthogonal matrix P are constant, then E is identically zero. This result follows directly from substituting (1.21) into (1.14) and using the fact that P P T = I, and confirms our intuition that a rigid-body motion induces no deformation. † The word “tensor” as used here is effectively synonymous with “matrix”, but it is easy to generalise (1.19) to a tensor with any number of indices. A vector, for example, is a tensor with just one index.

1.5 Stress

7

1.5 Stress In the absence of any volumetric (e.g. gravitational or electromagnetic) effects, a force can only be transmitted to a solid by being applied to its boundary. It is, therefore, natural to consider the force per unit area or stress applied at that boundary. To do so, we now analyse a surface element, whose area and unit normal are da and n respectively. If it is contained within a stressed medium, then the material on (say) the side into which n points will exert a force df on the element. (The material on the other side will also exert a force equal, by Newton’s third law, to −df .) In the expectation that the force should be proportional to the area da, we write df = σ da,

(1.22)

where σ is called the traction or stress acting on the element. Perhaps the most familiar example is that of an inviscid fluid, in which the stress is related to the pressure p by σ = −pn.

(1.23)

This expression implies that (i ) the stress acts only in a direction normal to the surface element, (ii ) the magnitude of the stress (i.e. p) is independent of the direction of n. In an elastic solid, neither of these simplifying assumptions holds; we must allow for stress which acts in both tangential and normal directions and whose magnitude depends on the orientation of the surface element. First consider a surface element whose normal points in the x1 -direction, and denote the stress acting on such an element by τ 1 = (τ11 , τ21 , τ31 )T . By doing the same for elements with normals in the x2 - and x3 -directions, we generate three vectors τ j (j = 1, 2, 3) representing the stress acting on an element normal to the xj -direction. In total, therefore, we obtain nine scalars τij (i, j = 1, 2, 3), where τij is the i-component of τ j , that is τ j = τij ei ,

(1.24)

where ei is the unit vector in the xi -direction. The scalars τij may be used to determine the stress on an arbitrary surface element by considering the tetrahedron shown in figure 1.1. Here ai denotes the area p of the face orthogonal to the xi -axis. The fourth face has area a = a21 + a22 + a23 ; in fact if this face has unit normal n as shown, with components (ni ), then it is an elementary exercise in trigonometry to show that ai = ani . The outward normal to the face with area a1 is in the negative x1 -direction

8

Modelling Solids x3

x1

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

qq qq qqqqq qq qqqqqqqqq qqqqq qqq qqqqq q qqqqq qq a1 qqqqq qqq qqqqq qq qqqqq qq qqqqq ppppp ppppppp qq qqqqq ppppp ppppppppppppppppppp p n q qpqpqpqpqp ppppp qq p q p p q p qqq p ppppppp qqqqqqqq q ppppppppp qqqqq qq ppppp pppp qqqqq qq a2 qqqqq qqq qqqqq qq qqqqq qq qqqqq qq qq qqqqqqqqqqq q q q q q q q q q q qq q qq q q q q q q q q q q q qq q qqqqqqq q qqqqqqqqqqqqq qq qqqqqqqqqqqq qq qqqqqqqqqqqqq q q q q q q q q q q qqq q qqqqq q qqqqqqqqqqqqq qq qqqqqqqqqqqqq qq qqqqqqqqqqqq qpqqqqqqqqqqqqqqqqqqqqqqq p a3 ppppp ppppp ppppp

x2

Fig. 1.1. A tetrahedron; ai is the area of the face orthogonal to the xi -axis.

and the force on this face is thus −a1 τ 1 . Similar expressions hold for the faces with areas a2 and a3 . Hence, if the stress on the fourth face is denoted by σ, then the total force on the tetrahedron is f = aσ − aj τ j .

(1.25)

When we substitute for aj and τ j , we find that the components of f are given by fi = a (σi − τij nj ) .

(1.26)

Now we shrink the tetrahedron to zero. Since the area a scales with `2 , where ` is a typical edge length, while the volume is proportional to `3 , if we apply Newton’s second law and insist that the acceleration be finite, we see that f /a must tend to zero as ` → 0.† Hence we deduce an expression for σ: σi = τij nj ,

or σ = τ n.

(1.27)

This important result enables us to find the stress on any surface element in terms of the 9 quantities (τij ) = τ . Now let us follow §1.4 and examine what happens to τij when we rotate † Readers of a sensitive disposition may be slightly perturbed by our glibly letting the dimensional variable ` tend to zero: if ` is reduced indefinitely then we will eventually reach an atomic scale on which the solid can no longer be treated as a continuum. We reassure such readers that (1.27) can be more rigorously justified provided the macroscopic dimensions of the solid are large compared to any atomistic length-scale.

1.5 Stress

9

ppp pp ppp pp pp τ22

τ12 ppppppppp

p pppppppppppppp

τ11 ppppppppppppppppppppppp

δx2

x2 pppp pp pp pp pp

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

τ21

ppp pp ppp ppp p

pppp pp pp pp pp τ21

G

τ11 ppppppppp

p pppppppppppppp

δx1

pppppppppppppppppppppppp

τ12 pppppppppppppp pppppppp p

x1

ppp pp ppp ppp p τ22

Fig. 1.2. The forces acting on a small two-dimensional element.

the axes by an orthogonal matrix P . In the new frame, (1.27) will become σ 0 = τ 0 n0

(1.28)

where, since σ and n are vectors, they transform according to σ 0 = P σ,

n0 = P n.

(1.29)

It follows that τ 0 n = (P τ P T )n and so, since n is arbitrary, τ 0 = P τ P T,

or τij0 = piα pjβ ταβ .

(1.30)

Thus, like Eij , we have that τij is a second-rank tensor, called the Cauchy stress tensor. We can make one further observation about τij by considering the angular momentum of the small two-dimensional solid element shown in Figure 1.2. The net anticlockwise moment acting about the centre of mass G is (per unit length in the x3 -direction) δx1 δx2 − 2 (τ12 δx1 ) 2 2 where τ21 and τ12 are evaluated at G to lowest order. By letting the rectangle shrink to zero (see again the footnote on page 8), and insisting that the angular acceleration be finite, we deduce that τ12 = τ21 . This argument can be generalised to three dimensions (Exercise 1.4) and it shows that 2 (τ21 δx2 )

τij ≡ τji

(1.31)

10

Modelling Solids

for all i and j, i.e. that τij , like Eij is a symmetric tensor. 1.6 Conservation of momentum Now we derive the basic governing equation of solid mechanics by applying Newton’s second law to a material volume V (t) that moves with the deforming solid: ZZZ ZZZ ZZ d ∂ui ρ dx = gi ρ dx + τij nj da. (1.32) dt V (t) ∂t V (t) ∂V (t) The terms in (1.32) represent successively the rate of change of momentum of the material in V (t), the force due to an external body force g, such as gravity, and the traction exerted on the boundary of V , whose unit normal is n, by the material around it. We differentiate under the integral (using the fact that ρ dx = ρ0 dX is independent of t) and use the divergence theorem on the final term to obtain ZZZ ZZZ ZZZ ∂τij ∂ 2 ui dx. (1.33) ρ dx = gi ρ dx + 2 V (t) ∂t V (t) V (t) ∂xj Assuming each integrand is continuous, and using the fact that V (t) is arbitrary, we arrive at Cauchy’s momentum equation: ρ

∂τij ∂ 2 ui . = ρgi + ∂t2 ∂xj

(1.34)

This may alternatively be written in vector form by adopting the following simple notation for the divergence of a tensor: we define the ith component of ∇ · τ to be ∂τji (∇ · τ )i = . (1.35) ∂xj Using the fact that τ is symmetric, we may thus write Cauchy’s equation as ∂2u = ρg + ∇ · τ. (1.36) ∂t2 This equation applies to any continuous medium for which a displacement u and stress tensor τ can be defined. The distinction between solid, fluid or some other continuum comes when we impose an empirical constitutive relation between τ and u. For solids, (1.34) already confronts us with a distinctive fundamental difficulty. The most obvious generalisation of Hooke’s law is to suppose that a linear relationship exists between the stress τ and the strain E. But we now ρ

1.7 Linear elasticity

11

recall that E was defined in §1.4 in terms of the Lagrangian variables X; indeed, the time derivative in (1.34) is taken in a Lagrangian frame, with X fixed. On the other hand, the stress tensor τ has been defined relative to Eulerian coordinates and is differentiated in (1.34) with respect to the Eulerian variable xj . It is not immediately clear, therefore, how the stress and strain, which are defined in different frames of reference, may be selfconsistently related. We will postpone the full resolution of this difficulty until Chapter 5 and, for the present, restrict our attention to linear elasticity in which, as we shall see, the two frames are essentially identical. 1.7 Linear elasticity The theory of linear elasticity follows from the assumption that the displacement u is small relative to any other length-scale. This assumption allows the theory developed thus far to be simplified in several ways. First, it means that ∂ui /∂Xj is small for all i and j. Second, we note from (1.3) that x and X are equal to lowest order in u. Hence, if we only consider leading-order terms, there is no need to distinguish between the Eulerian and Lagrangian variables: we can simply replace X by x and ∂ui /∂Xj by ∂ui /∂xj throughout. A corollary is that the Jacobian J is approximately equal to one, so (1.8) tells us that the density ρ is fixed, to leading order, at its initial value ρ0 . Finally, we can use the smallness of ∂ui /∂xj to neglect the quadratic term in (1.13) and hence obtain the linearised strain tensor µ ¶ ∂uj 1 ∂ui Eij ≈ eij = + . (1.37) 2 ∂xj ∂xi Much of this book will be concerned with this approximation. Therefore, and with a slight abuse of notation, we will write E = (eij ). Remembering (1.21), we note that it is possible to approximate E by (1.37) even when u is not small compared with X, just as long as u is close to a rigid-body translation and rotation. This situation is called geometric nonlinearity and we will encounter it frequently in Chapters 4 and 6. It occurs because Eij is identically zero for rigid-body motions of the solid given by (1.21); however eij does not vanish for such rigid-body motions, but rather for displacements of the form u = c + ω×x,

(1.38)

where c and ω are constant (see Exercise 1.5). Assuming the validity of (1.37), we can now generalise Hooke’s law by postulating a linear relationship between the stress and strain tensors. We

12

Modelling Solids

Cartilage Rubber Polystyrene Granite Glass Copper Steel Diamond

λ (GPa)

µ (GPa)

−5

9 × 10−5 0.003 1.2 30 28 37 78 400

3 × 10 0.04 2.3 10 28 86 100 270

Table 1.1. Typical values of the Lamé constants λ and µ for some everyday materials (1 GPa = 109 N m−2 = 104 atmospheres; a typical car tyre pressure is two atmospheres). assume that τ is zero when E is; in other words the stress is zero in the reference state. This is not the case for pre-stressed materials, and we will consider some of the implications of so-called residual stress in Chapter 9. Even with this assumption, we apparently are lead to the problem of defining 81 material parameters Cijαβ (i, j = 1, 2, 3) such that τij = Cijk` ek` .

(1.39)

The symmetry of τij and eij only enables us to reduce the number of unknowns to 36. This can be reduced to a more manageable number by assuming that the solid is isotropic, by which we mean that it behaves the same way in all directions. This implies that Cijk` must satisfy Cijk` pii0 pjj 0 pkk0 p``0 ≡ Ci0 j 0 k0 `0

(1.40)

for all orthogonal matrices P = (pij ). It can be shown (see, for example, Ockendon & Ockendon, 1995, pp. 7–9) that this is sufficient to reduce the specification of C to just two scalar quantities λ and µ, such that Cijk` = λδij δk` + 2µδik δj`

(1.41)

τij = λ (ekk ) δij + 2µeij ,

(1.42)

and, consequently,

and hence 1 eij = 2µ

µ τij −

¶ λ (τkk )δij . (3λ + 2µ)

(1.43)

In Chapter 9 we will consider solids, such as wood or fibre-reinforced materials, that are not isotropic.

1.8 The incompressibility approximation

13

The material parameters λ and µ are known as the Lamé constants, and µ is called the shear modulus. As we shall see in Chapter 2, λ and µ measure a material’s ability to resist compression/expansion and shearing respectively. They have the units of pressure; typical values for a few familiar solid materials are given in Table 1.1. It will be observed that these values may be very large for relatively “hard” materials, the significance being that tractions much less than these values will result in small deformations, so that linear elasticity is valid. Now we substitute our linear constitutive relation (1.42) into the momentum equation (1.34) and replace X with x to obtain the Navier equation, also known as the Lamé equation, ρ

∂2u = ρg + (λ + µ) grad div u + µ∇2 u. ∂t2

(1.44)

Recall that ρ does not vary to leading order, so (1.44) comprises three equations for the three components of u. It may alternatively be written in component form ρ

∂ 2 uj ∂ 2 ui ∂ 2 ui +µ 2 , = ρgi + (λ + µ) 2 ∂t ∂xi ∂xj ∂xj

(1.45)

where the final ∂x2j is treated as a repeated suffix, or ∂2u = ρg + (λ + 2µ) grad div u − µ curl curl u, ∂t2 where we have used the well-known vector identity ρ

“del squared equals grad div minus curl curl.”

(1.46)

(1.47)

1.8 The incompressibility approximation There is an interesting and important class of materials that, although elastic, are virtually incompressible, so they may be sheared elastically but are highly resistant to tension or compression. In linear elasticity, this amounts to saying that the Lamé constant λ is much larger than the shear modulus µ. The values given in Table 1.1 show that rubber has this property, as do many bio-materials such as muscle. If a material is almost incompressible, we can set 1 λ = , µ ²

(1.48)

where ² is a small parameter. From (1.44), we expect that, in the limit

14

Modelling Solids

² → 0, div u will be of order ². Hence, if we define a scalar function p such that µ p² = − div u, (1.49) ² then p² will approach a finite limit p as ² → 0. When we now substitute (1.48) and (1.49) into the Navier equation (1.44) and let ² → 0, we obtain ρ

∂2u = ρg − ∇p + µ∇2 u, ∂t2

(1.50a)

along with the limit of (1.49), that is div u = 0.

(1.50b)

The condition (1.50b) means that each material volume is conserved during the deformation, and it imposes an extra constraint on the Navier equation. The extra unknown p, representing the isotropic pressure in the medium, gives us the extra freedom we need to satisfy this constraint. 1.9 Energy We can obtain an energy equation from (1.34) by taking the dot product with ∂u/∂t and integrating over an arbitrary volume V : ZZZ ZZZ ZZZ ∂τij ∂ui ∂ 2 ui ∂ui ∂ui dx − dx = dx. (1.51) ρ 2 ρgi ∂t ∂t ∂t V V V ∂xj ∂t The final term may be rearranged, using the divergence theorem, to ZZ ZZZ ZZZ ∂eij ∂τij ∂ui ∂ui dx = τij nj da − τij dx. (1.52) ∂t V ∂xj ∂t ∂V ∂t V Hence (1.51) may be written in the form (Z Z Z ) ¯ ¯ ZZZ d 1 ¯¯ ∂u ¯¯2 ρ ¯ ¯ dx + W dx = dt ∂t V 2 V ZZ ZZZ ∂ui dx + ρgi ∂t V

∂V

∂ui τij nj da, ∂t

(1.53)

where W is a scalar function of the strain components that is chosen to satisfy ∂W = τij . ∂eij

(1.54)

1.9 Energy

15

With τij given by (1.42), we can integrate (1.54) to determine W up to an arbitrary constant as 1 1 W = τij eij = λ (ekk )2 + µ (eij eij ) . 2 2

(1.55)

Here the summation convention is invoked such that (ekk )2 is the square of the trace of E, while (eij eij ) is the sum of the squares of the components of E. The first term in braces in (1.53) is the net kinetic energy in V , while the terms on the right-hand side represent the rate of working of the external body force g and the tractions on ∂V respectively. In the absence of other energy sources resulting from, say, chemical or thermal effects, we can interpret equation (1.53) as a statement of conservation of energy. The difference between the rate of working and the rate of change of kinetic energy is the rate at which elastic energy is stored in the material as it deforms; W is therefore called the strain energy density. This is analogous to the energy stored in a stretched spring (see Exercise 1.1) and, at a fundamental scale, is a manifestation of the energy stored in the bonds between the atoms. If µ, λ > 0, we can easily see from (1.55) that W is a non-negative function of the strain components, whose unique global minimum is attained when eij = 0. In fact, Exercise 1.6 demonstrates that it is only necessary to have µ, λ + 2µ/3 > 0. Even without the equation of state (1.42), we can see that, under steady ∂τ gravity-free conditions, the Navier equation ∂xijj = 0 is a necessary condition for ZZZ Wdx (1.56) to be minimised, assuming all the functions involved are smooth. Following the usual calculus of variations procedure, all we need do to minimise (1.56) is change the ui by a small amount, u0i . This induces ³ displacement ´ ∂u0

00 a small change 12 ∂xji + ∂u in eij and hence, from (1.54), W changes by ∂xi ³ 0 ´ ³ ´ ∗ ∂uj ∂ui 1 1 0 ∂τij + u0 ∂τij τ + . This differs from a pure divergence by u i ∂xj j ∂xi 2 ij ∂xj ∂xi 2

∂τ

and hence (1.56) can only be minimised when ∂xijj = 0. The net conservation of energy implied by (1.53) reflects the fact that the Navier equation is not dissipative, and hence it can be expressed as a variational problem, as in Exercise 1.7. However, the situation changes when thermal effects are important, as we will see in Chapter 9.

16

Modelling Solids

1.10 Boundary conditions and well-posedness Suppose that we wish to solve (1.44) for u(x, t) when t is positive and x lies in some prescribed domain D. We now ask: “what sort of boundary conditions may be imposed on ∂D in order to obtain a well-posed mathematical problem, in other words, one for which a solution u exists, is unique and depends continuously on the boundary data?” For boundary-value problems in linear elasticity, it is generally far easier to discuss questions of uniqueness than it is to prove existence. Hence in this section we will focus only on establishing uniqueness. In elastostatic problems, in which the the left-hand side of (1.44) is zero, the Navier system is, roughly speaking, a generalisation of a scalar elliptic equation. By analogy, it seems appropriate for either u or three linearly independent scalar combinations of u and ∂u/∂n to be prescribed on ∂D. In many physical problems, we specify either the displacement u or the traction τ n everywhere on the boundary, and we will now examine each of these in turn. First consider a solid body D on whose boundary the displacement is prescribed, that is u = ub (x) on ∂D.

(1.57)

Inside D, u satisfies the steady Navier equation ∂τij + ρgi = 0, ∂xj

(1.58)

and we will now show that, if a solution u of (1.58) with the boundary condition (1.57) exists, then it is unique. Suppose that two solutions u(1) and u(2) exist and let u = u(1) − u(2) . Thus u satisfies the homogenous problem, with ub = g = 0. Now, by multiplying (1.58) by ui , integrating over D and using the divergence theorem, we obtain ZZ ZZZ ZZZ ui τij nj da = eij τij dx = 2 W dx, (1.59) ∂D

D

D

where W is given by (1.55). The left-hand side of (1.59) is zero by the boundary conditions, while the integrand W on the right-hand side is nonnegative and must, therefore, be zero. It follows that the strain tensor eij is identically zero in D, and the displacement can therefore only be a rigidbody motion (i.e. a uniform translation and rotation; see Exercise 1.5). Since u is zero on ∂D, we deduce that it must be zero everywhere and, hence, that u(1) ≡ u(2) .

1.10 Boundary conditions and well-posedness

17

Now we attempt the same calculation when the surface traction, rather than the displacement, is specified: τ n = σ(x) on

∂D.

(1.60)

Like the Neumann problem for a scalar elliptic PDE (Ockendon et al., 1999, pp. 146–147), the Navier equation only admits solutions satisfying (1.60) if a so-called solvability condition is satisfied. If we integrate (1.58) over D and use the divergence theorem, we find that ZZ ZZZ τij nj da + ρgi dx = 0 (1.61) ∂D

and hence that

D

ZZ

ZZZ σ da + ∂D

ρg dx = 0.

(1.62)

D

This represents a net balance between the forces, namely surface traction and gravity, acting on D. An analogous balance between the moments acting on D may also be obtained by taking the cross product of x with (1.58) before integrating ZZ ZZZ x×σ da + ρx×g dx = 0, (1.63) ∂D

D

as in Exercise 1.8). As well as representing physical balances on the system, (1.62) and (1.63) may be interpreted as instances of the Fredholm Alternative (see Keener, 2000, Chapter 3; Ockendon et al., 1999, pp. 43–44). Now suppose the solvability conditions (1.62) and (1.63) are satisfied and that two solutions u(1) and u(2) of (1.58) and the boundary condition (1.60) exist. As before, the difference u = u(1) − u(2) satisfies the homogeneous version of the problem, with g and σ set to zero. By an argument analogous to that presented above, we deduce that the strain tensor eij must be identically zero. However, since u is now not specified on ∂D, we can only infer from this that the displacement is a small rigid-body motion, as shown in Exercise 1.5. Thus the solution of (1.58) subject to the applied traction (1.60) is determined only up to the addition of an arbitrary translation and rotation. This indeterminacy, which might appear disconcerting at first glance, is really an illustration of Newton’s first law: the body may move with a constant linear and angular momentum in the absence of any applied forces or moment. As well as the boundary conditions (1.57) and (1.60), there are generalisations in which the traction is specified on some parts of the boundary and

18

Modelling Solids n

solid 2 solid 1

Fig. 1.3. Schematic of a small pill-box-shaped region at the boundary between two elastic solids.

the displacement on others, for example in contact problems and in fracture, as described in Chapter 7. Another common generalisation of (1.57) and (1.60) occurs when two solids with different elastic moduli are bonded together across a common boundary ∂D, as shown in Figure 1.3. Then the displacement vectors are the same on either side of ∂D and, by balancing the stresses on the small pill-box-shaped region shown in Figure 1.3, we see that τ (1) n = τ (2) n.

(1.64)

Thus there are six continuity conditions across such a boundary. On the other hand, if two unbonded solids are in smooth contact, only the normal displacement is continuous across ∂D. However, this loss of information is compensated by the fact that the four tangential components of τ (1) n and τ (2) n are zero and the normal components of these tractions are continuous. Frictional contact between rough unbonded surfaces poses serious modelling challenges. For elastodynamic problems, we may anticipate that (1.44) admits wavelike solutions. It may, therefore, be viewed as a generalisation of a scalar wave equation, such as the familiar equation %

∂2w ∂2w = T ∂t2 ∂x2

(1.65)

which describes small transverse waves on a string with tension T and line density % (see §4.3). We will examine elastic waves in more detail in Chapter 3 but, in the meantime, we expect to prescribe Cauchy initial conditions for u and ∂u/∂t at t = 0, as well as elliptic boundary conditions such as (1.57) or (1.60).

1.11 Coordinate systems

19

1.11 Coordinate systems In the next two chapters, we will construct some elementary solutions of the Navier equation (1.44). In doing so, it is often useful to employ alternative coordinate systems particularly chosen to fit the geometry of the problem being considered. A detailed derivation of the Navier equation in an arbitrary orthogonal coordinate system may be found in Appendix 3. Here we state the main results that will be useful in subsequent chapters for the three most popular coordinate systems, namely Cartesian, cylindrical polar and spherical polar coordinates. All three of these coordinate systems are orthogonal ; in other words the tangent vectors obtained by varying each coordinate in turn are mutually perpendicular. This means that the coordinate axes at any fixed point are orthogonal and may thus be obtained by a rotation of the usual Cartesian axes. Under the assumptions of isotropic linear elasticity, the Cartesian stress and strain components are related by (1.42), which is invariant under any such rotation. Hence the constitutive relation (1.42) applies literally to any orthogonal coordinate system.

1.11.1 Cartesian coordinates First we write out in full the results derived thus far using the usual Cartesian coordinates (x, y, z). To avoid the use of suffices, we will denote the displacement components by u = (u, v, w)T . It is also conventional to label the stress components by {τxx , τxy , . . . } rather than {τ11 , τ12 , . . . }, and similarly for the strain components. The linear constitutive relation (1.42) gives τxx = (λ + 2µ)exx + λeyy + λezz ,

τxy = 2µexy ,

(1.66a)

τyy = λexx + (λ + 2µ)eyy + λezz ,

τxz = 2µexz ,

(1.66b)

τzz = λexx + λeyy + (λ + 2µ)ezz ,

τyz = 2µeyz ,

(1.66c)

where ∂u , ∂x ∂u ∂v = + , ∂y ∂x

exx = 2exy

∂v , ∂y ∂u ∂w = + , ∂z ∂x

eyy = 2exz

∂w , ∂z ∂v ∂w = + , ∂z ∂x

ezz = 2eyz

(1.67a) (1.67b)

20

Modelling Solids

and the three components of Cauchy’s momentum equation are ∂2u ∂τxx ∂τxy ∂τxz = ρgx + + + , 2 ∂t ∂x ∂y ∂z ∂τxy ∂τyy ∂τyz ∂2v ρ 2 = ρgy + + + , ∂t ∂x ∂y ∂z ∂τyz ∂τxz ∂2w ∂τzz ρ 2 = ρgz + + + , ∂t ∂x ∂y ∂z ρ

(1.68a) (1.68b) (1.68c)

where the body force is g = (gx , gy , gz )T . In terms of the displacements, the Navier equation reads (assuming that λ and µ are constant) ∂2u ∂ = ρgx + (λ + µ) (∇ · u) + µ∇2 u, 2 ∂t ∂x ∂2v ∂ ρ 2 = ρgy + (λ + µ) (∇ · u) + µ∇2 v, ∂t ∂y 2 ∂ w ∂ ρ 2 = ρgz + (λ + µ) (∇ · u) + µ∇2 w. ∂t ∂z ρ

(1.69a) (1.69b) (1.69c)

1.11.2 Cylindrical polar coordinates We define the cylindrical polar coordinates (r, θ, z) in the usual way (see Appendix 3 for details) and denote the displacements in the r-, θ- and zdirections by ur , uθ and uz respectively. The stress components are denoted by τij where now i and j are equal to either r, θ or z and, as in §1.5, τij is defined to be the i-component of stress on a surface element whose normal points in the j-direction. As noted above, the constitutive relation (1.42) applies directly to this coordinate system, so that τrr = (λ + 2µ)err + λeθθ + λezz ,

τrθ = 2µerθ ,

(1.70a)

τθθ = λerr + (λ + 2µ)eθθ + λezz ,

τrz = 2µerz ,

(1.70b)

τzz = λerr + λeθθ + (λ + 2µ)ezz ,

τθz = 2µeθz ,

(1.70c)

where the strain components are now given by ∂ur , ∂r µ ¶ 1 ∂uθ = + ur , r ∂θ ∂uz = , ∂z

1 ∂ur ∂uθ uθ + − , r ∂θ ∂r r ∂ur ∂uz = + , ∂z ∂r ∂uθ 1 ∂uz = + . ∂z r ∂θ

err =

2erθ =

(1.71a)

eθθ

2erz

(1.71b)

ezz

2eθz

(1.71c)


21

The three components of Cauchy’s momentum equation (1.34) read ∂ 2 ur ∂t2 ∂ 2 uθ ρ 2 ∂t ∂ 2 uz ρ 2 ∂t ρ

1 ∂ 1 ∂τrθ ∂τrz τθθ (rτrr ) + + − , r ∂r r ∂θ ∂z r 1 ∂ 1 ∂τθθ ∂τθz τrθ = ρgθ + (rτrθ ) + + + , r ∂r r ∂θ ∂z r 1 ∂ 1 ∂τθz ∂τzz = ρgz + (rτrz ) + + , r ∂r r ∂θ ∂z = ρgr +

(1.72a) (1.72b) (1.72c)

where the body force is g = gr er + gθ eθ + gz ez . Written out in terms of displacements, these become µ ¶ ∂ 2 ur ∂ ur 2 ∂uθ 2 ρ 2 = ρgr + (λ + µ) (∇ · u) + µ ∇ ur − 2 − 2 (1.73a) , ∂t ∂r r r ∂θ µ ¶ ∂ 2 uθ uθ (λ + µ) ∂ 2 ∂ur 2 ρ 2 = ρgθ + (∇ · u) + µ ∇ uθ − 2 + 2 (1.73b) , ∂t r ∂θ r r ∂θ ∂ 2 uz ∂ ρ 2 = ρgz + (λ + µ) (∇ · u) + µ∇2 uz , (1.73c) ∂t ∂z where ∇·u = and ∇2 ui =

1 ∂ 1 ∂uθ ∂uz (rur ) + + r ∂r µ ¶r ∂θ 2 ∂z 2 1 ∂ ∂ui 1 ∂ ui ∂ ui r + 2 + r ∂r ∂r r ∂θ2 ∂z 2

(1.74a) (1.74b)

are the divergence of u and the Laplacian of ui respectively expressed in cylindrical polars. Detailed derivations of (1.71) and (1.72) are given in Appendix 3. Notice the undifferentiated terms proportional to 1/r which are not present in the corresponding Cartesian expressions (1.67) and (1.68). The origin of these terms may be understood in two dimensions (r, θ) by considering the equlilibrium of a small polar element as illustrated in Figure 1.4, in which ∂ταβ , ∂r ∂ταβ + δθ . ∂θ

τ˜αβ = ταβ (r + δr, θ) ∼ ταβ + δr

(1.75a)

τˆαβ = ταβ (r, θ + δθ) ∼ ταβ

(1.75b)

Summing the resultant forces in the r- and θ-directions to zero results in τ˜rr (r + δr) δθ − τrr rδθ − τˆθθ δr sin δθ + τˆrθ δr cos δθ − τrθ δr ∼ 0,

(1.76a)

τˆθθ δr cos δθ − τθθ δr + τˆrθ δr sin δθ + τ˜rθ (r + δr) δθ − τrθ rδθ ∼ 0.

(1.76b)

22

Modelling Solids

τˆrθ

τ˜rr

τ˜rθ

τˆθθ

eθ τrθ

τrr r δθ

er

δr

τrθ

τθθ

Fig. 1.4. Forces acting on a polar element of solid.

Now letting δθ, δr → 0 and using (1.75), we obtain ∂τrr 1 ∂τrθ τrr − τθθ + + = 0, ∂r r ∂θ r

∂τrθ 1 ∂τθθ 2τrθ + + = 0, ∂r r ∂θ r

(1.77)

which are the components of the two-dimensional steady Navier equation in plane polar coordinates with no body force; cf (1.72). The stress component τθθ is the so-called hoop stress in the θ-direction that results from inflating an elastic object radially; we will see an explicit example of hoop stress in §2.6.

1.11.3 Spherical polar coordinates The spherical polar coordinates (r, θ, φ) are defined in the usual way, such that the position vector of any point is given by   r sin θ cos φ r(r, θ, φ) =  r sin θ sin φ  ; r cos θ

(1.78)


23

see Appendix 3 for details. Again, we can apply the constitutive relation (1.42) literally, to obtain τrr = (λ + 2µ)err + λeθθ + λeφφ ,

τrθ = 2µerθ ,

(1.79a)

τθθ = λerr + (λ + 2µ)eθθ + λeφφ ,

τrφ = 2µerφ ,

(1.79b)

τφφ = λerr + λeθθ + (λ + 2µ)eφφ ,

τθφ = 2µeθφ .

(1.79c)

The linearised strain components are now given by ∂ur , ∂r µ ¶ 1 ∂uθ = + ur , r ∂θ

1 ∂ur ∂uθ uθ + − , (1.80a) r ∂θ ∂r r ∂uφ uφ 1 ∂ur = + − , r sin θ ∂φ ∂r r (1.80b) 1 ∂uθ 1 ∂uφ cot θuφ = + − . r sin θ ∂φ r ∂θ r (1.80c)

err =

2erθ =

eθθ

2erφ

eφφ =

cot θuθ 1 ∂uφ ur + + , 2eθφ r sin θ ∂φ r r

Cauchy’s equation of motion leads to the three equations ρ

ρ

ρ

∂ 2 ur 1 ∂(sin θτrθ ) 1 ∂τrφ τθθ + τφφ 1 ∂(r2 τrr ) + + − , = ρg + r 2 2 ∂t r ∂r r sin θ ∂θ r sin θ ∂φ r (1.81a) ∂ 2 uθ 1 ∂(sin θτθθ ) 1 ∂τθφ τrθ − cot θτφφ 1 ∂(r2 τrθ ) + + + , = ρgθ + 2 2 ∂t r ∂r r sin θ ∂θ r sin θ ∂φ r (1.81b) ∂ 2 uφ 1 ∂(r2 τrφ ) 1 ∂(sin θτθφ ) 1 ∂τφφ τrφ + cot θτθφ = ρg + + + + , φ 2 2 ∂t r ∂r r sin θ ∂θ r sin θ ∂φ r (1.81c)

where the body force is g = gr er + gθ eθ + gφ eφ . Again, (1.80) and (1.81) may be derived using the general approach given in Appendix 3 or more directly by analysing a small polar element. In terms of displacements, the Navier equation reads ρ

∂ 2 ur ∂ = ρgr + (λ + µ) (∇ · u) 2 ∂t ∂r ½ ¾ 2u 2 ∂ 2 ∂uφ r 2 + µ ∇ ur − 2 − 2 (uθ sin θ) − 2 , r r sin θ ∂θ r sin θ ∂φ

(1.82a)

24

Modelling Solids

∂ 2 uθ ρ 2 ∂t

ρ

(λ + µ) ∂ (∇ · u) ∂θ ½r ¾ 2 ∂ur uθ 2 cos θ ∂uφ 2 + µ ∇ uθ + 2 − 2 2 − 2 2 , r ∂θ r sin θ r sin θ ∂φ

= ρgθ +

∂ 2 uφ (λ + µ) ∂ = ρgφ + (∇ · u) 2 ∂t r sin θ ∂φ ¾ ½ uφ 2 ∂ur 2 cos θ ∂uθ 2 , + µ ∇ uφ + 2 + 2 2 − 2 2 r sin θ ∂φ r sin θ ∂φ r sin θ

(1.82b)

(1.82c)

assuming λ and µ are constant, where 1 ∂ ¡ 2 ¢ 1 ∂ 1 ∂uφ r ur + (sin θuθ ) + , (1.83a) 2 r ∂r r sin θ ∂θ r sin θ ∂φ µ ¶ µ ¶ 1 ∂ 1 ∂ ∂ui 1 ∂ 2 ui 2 2 ∂ui ∇ ui = 2 r + 2 sin θ + 2 2 . (1.83b) r ∂r ∂r r sin θ ∂θ ∂θ r sin θ ∂φ2

∇·u=

Exercises 1.1 A light spring of natural length L and spring constant k hangs freely with a mass m attached to one end and the other end fixed. Show that the length ` of the spring satisfies the differential equation m

d2 ` + k(` − L) − mg = 0. dt2

Deduce that 1 m 2

µ

d` dt

¶2

1 + k(` − L)2 + mg(L − `) = const 2

and interpret this result in terms of energy. 1.2 A string, stretched to a tension T along the x-axis, undergoes small transverse displacements such that its position at time t is given by the graph z = w(x, t). Given that w satisfies the wave equation (1.65), where % is the mass per unit length of the string, show that, if x = a and x = b are any two points along the string, ( Z µ ) · ¶ ¶ ¸ Z b µ d 1 b ∂w 2 1 ∂w 2 ∂w ∂w b % dx + T dx = T . dt 2 a ∂t ∂x ∂x ∂t a a 2 Interpret this result in terms of conservation of energy.

EXERCISES

25

1.3 If x, x0 , u and u0 are related by (1.17), use the chain rule to show that ∂u0i ∂uα . = piα pjβ 0 ∂xj ∂xβ Hence establish equation (1.19) to show that Eij transforms as a tensor under a rotation of the coordinate axes. [Hint: note that, since P is orthogonal, pkl pkm ≡ δlm .] 1.4 Consider a volume V (t) that is fixed in a deforming solid body. Show that conservation of angular momentum for V (t) leads to the equation ZZ ZZZ ZZZ d ∂u x× ρ dx = x×gρ dx + x×(τ n) da. dt ∂t ∂V (t) V (t) V (t) From this and Cauchy’s momentum equation (1.34), deduce that τij ≡ τji . 1.5 Show that the linearised strain eij , given by (1.37), is identically zero if and only if u = c + ω×x, where c and ω are spatially-uniform vectors. Interpret c and ω in terms of a rigid-body motion of the solid. If u is of this form, and known to be zero at three non-collinear points, show that c = ω = 0. PUT IN P-I... show linear strain is nonzero. 1.6 By writing the linearised strain tensor eij as the sum of a zerotrace contribution eij −(1/3)δij ekk and a purely diagonal contribution (1/3)δij ekk , show that W can be rewritten as ¶µ ¶ ¶ µ µ 1 1 λ µ 2 eij − δij ekk . + (ekk ) + µ eij − δij ekk W= 2 3 3 3 Deduce that, for W to have a single global minimum at eij = 0, it is sufficient for µ and λ + 2µ/3 to be positive. By considering particular values of eij , show that it is also necessary. 1.7 Suppose a solid body occupies the region D and the displacement u is prescribed on ∂D. Let ZZZ S= W − ρg · u dx, D

where W is the strain energy density. Show that the minimisation of S with respect to all displacements satisfying the given boundary condition leads to the steady Navier equation. Show also that the minimisation of S subject to the constraint div u ≡ 0 leads to the steady incompressible Navier equation (1.50a), where p is a Lagrange multiplier.

26

Modelling Solids

1.8 An elastic body D at rest is subject to a traction τ n = σ(x) on its boundary ∂D. By taking the cross product of x with (1.58) before integrating over D, derive (1.63) and deduce that the net moment acting on V must be zero. 1.9 Suppose that u satisfies the steady Navier equation (1.58) in a region D and the mixed boundary condition α(x)u+β(x)τ n = f (x) on ∂D. Show that, if a solution exists, it is unique provided α > 0 and β > 0. [If α and β take different signs, then there is no guarantee of uniqueness. This is analogous to the difficulty associated with the Robin boundary condition for scalar elliptic PDEs (Ockendon et al., 1999, pp. 157–158).] 1.10 (a) Show that, in plane polar coordinates (r, θ), the basis vectors satisfy der deθ = eθ , = −er . dθ dθ (b) Consider a small line segment joining two particles whose polar coordinates are (r, θ) and (r +δr, θ +δθ). Show that the vector joining the two particles is given to leading order by δX ∼ δrer + rδθeθ . (c) If a two-dimensional displacement field u = ur (r, θ)er (θ) + uθ (r, θ)eθ (θ) is imposed, show that the line element δX is displaced to ¶ µ ¶ µ ∂ur ∂uθ ∂uθ ∂ur δr + δθ − uθ δθ er + δr + δθ + ur δθ eθ . δx = δX + ∂r ∂θ ∂r ∂θ (d) Hence show that µ ¶µ ¶ ¡ ¢ err erθ δr |δx| = |δX| + δr rδθ erθ eθθ rδθ 2

2

where, to leading order in the displacements, err

∂ur 1 ∂ur ∂uθ uθ 1 = , 2erθ = + − , eθθ = ∂r r ∂θ ∂r r r

µ

¶ ∂uθ + ur . ∂θ

[These are the elements of the linearised strain tensor in plane polar coordinates: cf (1.71).]

2 Linear Elastostatics

2.1 Introduction This chapter concerns steady state problems in linear elasticity. This topic may appear to be the simplest in the whole of solid mechanics, but we will find that it offers many interesting mathematical challenges. Moreover the material presented in this chapter will provide crucial underpinning to the more general theories of later chapters. We will begin by listing some very simple explicit solutions which give valuable intuition concerning the role of the elastic moduli introduced in Chapter 1. Our first application of practical importance is elastic torsion, which concerns the twisting of an elastic bar. This leads to a class of exact solutions of the Navier equation in terms of solutions of Laplace’s equation in two dimensions. However, as distinct from the use of Laplace’s equation in, say, hydrodynamics or electromagnetism, the dependent variable is the displacement, which has a direct physical interpretation, rather than a potential, which does not. This means we have to be especially careful to ensure that the solution is single-valued in situations involving multiply-connected bars. These remarks remain important when we move on to another class of two-dimensional problems called plane strain problems. These have even more general practical relevance but involve the biharmonic equation. This equation, which will be seen to be ubiquitous in linear elastostatics, poses significant extra difficulties as compared to Laplace’s equation. In particular, we will find that it is much more difficult to construct explicit solutions using, for example, the method of separation of variables. An interesting technique to emerge from both these classes of problems is the use of stress potentials, which are the elastic analogues of, say, the stream function in hydrodynamics or electrostatic or gravitational potentials. We 27

28

Linear Elastostatics

will find that a large class of elastostatic problems with some symmetry, for example two-dimensional or axisymmetric, can be described using a single scalar potential that satisfies the biharmonic equation. Fully three-dimensional problems are mostly too difficult to be suitable for this chapter. Nonetheless, we will be able to provide a conceptual framework within which to represent the solution by generalising the idea of Green’s functions for scalar ordinary and partial differential equations. The necessary Green’s matrices describe the response of the elastic body to a localised point force applied at some arbitrary position in the body. This idea opens up one of the most distinctive and fascinating aspects of linear elasticity: because of the intricacy of (1.44), many different kinds of singular solutions can be constructed using stress functions and Green’s matrices, each being the response to a different kind of localised forcing, and the catalogue of these different responses is a very helpful toolkit for thinking about solid mechanics more generally. 2.2 Linear displacements We will begin by neglecting the body force g so the steady Navier equation, ∇ · τ = 0, reduces to ∇ · τ = (λ + µ) grad div u + µ∇2 u = 0.

(2.1)

This vector partial differential equation for u is the starting point for all we will say in this chapter. As discussed on page XXX, it needs to be supplemented with suitable boundary conditions, which will vary from case to case. If the displacement u is a linear function of position x, then the strain tensor E is spatially uniform. It follows that the stress tensor τ is also uniform and, therefore, trivially satisfies (2.1). Such solutions provide considerable insight into the predictions of (2.1) and also give a feel for the significance of the parameters λ and µ. To avoid suffices, we will write u = (u, v, w)T and x = (x, y, z)T . 2.2.1 Isotropic expansion As a first example, suppose α x, (2.2) 3 where α is a constant scalar, which must be small for linear elasticity to be valid. When α > 0, this corresponds to a uniform isotropic expansion of the u=

2.2 Linear displacements

y 1

29

1

1

x

z (a)

(b)

(c)

1 + α/3 1 + α/3 α 1 + α/3

1+α 1 − αν 1 − αν

Fig. 2.1. A unit cube undergoing (a) uniform expansion, (2.2), (b) one-dimension shear, (2.6), (c) uniaxial stretching, (2.8).

medium so that, as illustrated in Figure 2.1(a), a unit cube is transformed to a cube with sides of length 1 + α/3. (Of course, if α is negative, the displacement is an isotropic contraction.) Since α is small, the relative change in volume is thus ³ α ´3 1+ − 1 ∼ α. (2.3) 3 The strain and stress tensors corresponding to this displacement field are given by µ ¶ α 2 (2.4) eij = δij and τij = λ + µ αδij . 3 3 This is a so-called hydrostatic situation, in which the stress is characterised by a scalar isotropic pressure p, and τij = −pδij . The pressure is related to the relative volume change by p = −Kα, where 2 K =λ+ µ 3

(2.5)

measures the solid’s resistance to expansion/compression and is called the bulk modulus or modulus of compression; from Exercise 1.6, we know that K is positive.

30

Linear Elastostatics y τn = 0 T

x

A

T

z

Fig. 2.2. Schematic of a uniform bar being stretched under a tensile force T .

2.2.2 Simple shear As our next example, suppose     u αy    u = v = 0 , w 0

(2.6)

where α is again a constant scalar. This corresponds to a simple shear of the solid in the x-direction, as illustrated in Figure 2.1(b). The strain and stress tensors are now given by     0 1 0 0 1 0 α τ = αµ 1 0 0 . (2.7) E = 1 0 0 , 2 0 0 0 0 0 0 Note that λ does not affect the stress, so the solid’s response to shear is accounted for entirely by µ which is, therefore, called the shear modulus. 2.2.3 Uniaxial stretching Our next example is uniaxial stretching in which, as shown in Figure 2.1(c), the solid is stretched by a factor α in (say) the x-direction. We suppose, for reasons that will emerge shortly, that the solid simultaneously shrinks by a factor να in the other two directions. The corresponding displacement, strain and stress are     x 1 0 0 u = α −νy  , E = α 0 −ν 0  , (2.8) −νz 0 0 −ν 

 (1 − 2ν)λ + 2µ 0 0 . τ = α 0 (1 − 2ν)λ − 2νµ 0 0 0 (1 − 2ν)λ − 2νµ

(2.9)


31

This simple solution may be used to describe a uniform elastic bar that is stretched in the x-direction under a tensile force T , as shown in Figure 2.2. Notice that, since the bar is assumed not to vary in the x-direction, the outward normal n to the lateral boundary always lies in the (y, z)-plane. If the curved surface of the bar is stress-free, then the resulting boundary condition τ n = 0 may be satisfied identically by ensuring that τyy = τzz = 0, which occurs if λ ν= . (2.10) 2 (λ + µ) Hence the bar, while stretching by a factor α in the x-direction, must shrink by a factor να in the two transverse directions; if ν happened to be negative, this would correspond to an expansion. The ratio ν between lateral contraction and longitudinal extension is called Poisson’s ratio. We recall that (2.10) becomes ν τkk δij 2µEij = τij − 1+ν With ν given by (2.10), the stress tensor has just one nonzero element, namely τxx = Eα,

(2.11)

where E=

µ(3λ + 2µ) λ+µ

(2.12)

is called Young’s modulus. If the cross-section of the bar has area A, then the tensile force T applied to the bar is related to the stress by T = Aτxx = AEα;

(2.13)

thus AE is the elastic modulus k 0 referred to in (1.2). By measuring T , the corresponding extensional strain α and transverse contraction να, one may thus infer the values of E and ν for a particular solid from a bar-stretching experiment like that illustrated in Figure 2.2. The Lamé constants may then be evaluated using λ=

νE , (1 + ν)(1 − 2ν)

µ=

E . 2(1 + ν)

(2.14)

While E is a positive constant with the dimensions of pressure, ν is dimensionless and constrained on physical grounds to lie in the range −1 < ν < 1/2. The lower bound for ν comes from the condition λ + 2µ/3 > 0 required for convexity of the strain energy, as shown in Exercise 1.6. The

32

Linear Elastostatics (b) (a)

Fig. 2.3. A paper model with negative Poisson’s ratio. Each line segment is a strip of paper viewed end-on.

Tyy Tyx

y

Txy Txx

z Txx Txy

x

h Tyx Tyy

Fig. 2.4. Schematic of a plate being strained under tensions Txx , Tyy and shear forces Txy and Tyx .

upper bound ν → 1/2 is approached as λ → ∞, in other words as the material becomes incompressible, and we will discuss this limit further in §1.8. For most solids, ν is positive, but it is possible (see Figure 2.3) to construct simple hinged paper models with negative Poisson’s ratio; try extending a crumpled piece of paper! So-called auxetic materials, in which λ < 0, have been developed whose microscopic structure mimics such paper models so they also display negative values of ν and they expand in all directions when extended in only one.


33

2.2.4 Biaxial strain As a final illustrative example, we generalise §2.2.3 and consider an elastic plate strained in the (x, y)-plane as illustrated in Figure 2.4. Suppose the plate experiences a linear in-plane distortion while shrinking by a factor γ in the z-direction, so the displacement is given by     ax + by u (2.15) u =  v  = cx + dy  , −γz w and, as in §2.2.3, the stress and strain tensors are both constant. Here we choose γ to satisfy the condition τzz = 0 required on the traction-free upper and lower surfaces of the plate, so that µ ¶ µ ¶ λ ν γ= (a + d) = (a + d), (2.16) λ + 2µ 1−ν where ν again denotes Poisson’s ratio. With this choice, and with E again denoting the Young’s modulus, the only nonzero stress components are τxx =

E(a + νd) , 1 − ν2

τxy =

E(b + c) , 2(1 + ν)

τyy =

E(νa + d) . 1 − ν2

(2.17)

We denote the net in-plane tensions and shear stresses applied to the plate by Tij = hτij , as illustrated in Figure 2.4. We can use (2.17) to relate these to the in-plane strain components by Eh (exx + νeyy ) , 1 − ν2 Eh = exy , 1+ν Eh = (νexx + eyy ) . 1 − ν2

(2.18a)

Txx = Txy = Tyx Tyy

(2.18b) (2.18c)

These will provide useful evidence when constructing more general models for the deformation of plates in Chapter 4. If no force is applied in the y-direction, that is Txy = Tyy = 0, then (2.18) reproduces the results of uniaxial stretching, with d = −νa and Txx = Eha. On the other hand, it is possible for the displacement to be purely in the (x, z)-plane, with b = c = d = 0,

τyy =

Eνa , 1 − ν2

τxx =

Ea . 1 − ν2

(2.19)

Thus a transverse stress τyy must be applied to prevent the plate from contracting in the y-direction when we stretch it in the x-direction. Notice

34


also that the effective elastic modulus E/(1 − ν 2 ) is larger than E whenever ν is nonzero, which shows that purely two-dimensional stretching is always more strenuous than uniaxial stretching.

2.2.5 General linear displacement The simple examples considered above shed a useful light on more general solutions of the Navier equation. Indeed, any displacement field, when expanded in a Taylor series about some point x0 , is linear to leading order: u(x) ∼ u(x0 ) + ∇uT (x0 ) · (x − x0 ) + · · · .

(2.20)

Here ∇u is the displacement gradient matrix, with entries (∇u)ij = ∂uj /∂xi ,† and may be written as the sum of symmetric and a skew-symmetric parts: µ ¶ µ ¶ ∂uj 1 ∂ui 1 ∂uj ∂ui (∇u)ij = + + − . (2.21) 2 ∂xj ∂xi 2 ∂xi ∂xj We recognise the symmetric part of (∇u) as the linear strain tensor E, while the skew-symmetric part may be written in the form   ½ µ ¶¾ 0 −ω3 ω2 1 ∂uj ∂ui Ω= − =  ω3 (2.22) 0 −ω1  , 2 ∂xi ∂xj −ω2 ω1 0 say, so that (2.20) becomes u(x) ∼ u0 + ω×(x − x0 ) + E(x − x0 ) + · · · .

(2.23)

The terms on the right-hand side represent a rigid-body translation and rotation, followed by a linear deformation characterised by the matrix E. Since E is real and symmetric, it has real eigenvalues, say {ε1 , ε2 , ε3 }, and orthogonal eigenvectors. These eigenvalues are referred to as the principal strains, and the directions defined by the eigenvectors as the principal directions. If we use an orthogonal matrix P to align the coordinate axes with these principal directions, that is x = x0 + P T x0

(2.24)

then E is transformed to a diagonal matrix: u0 ∼ u00 + ω 0 ×x0 + E 0 x0 , † convention dictates that the suffices are chosen in this order

(2.25)


where E 0 = P EP T

  ε1 0 0 =  0 ε2 0  . 0 0 ε3

35

(2.26)

We can hence think of the strain at any point as comprising three orthogonal expansions or contractions, depending on the signs of {ε1 , ε2 , ε3 }. By analogy with §2.2.1, the net relative volume change associated with this expansion/contraction is (1 + ε1 )(1 + ε2 )(1 + ε3 ) − 1 ∼ ε1 + ε2 + ε3 .

(2.27)

The sum of the eigenvalues is the invariant trace of the matrix E so that the relative volume change is Tr (E) = ekk = ∇ · u. With respect  τ1 τ0 =  0 0

(2.28)

to the principal axes, the stress tensor is also diagonal,   0 0  τ1 = λ(ε1 + ε2 + ε3 ) + 2µε1 , (2.29) τ2 0  , where τ = λ(ε1 + ε2 + ε3 ) + 2µε2 ,  2 0 τ3 τ3 = λ(ε1 + ε2 + ε3 ) + 2µε3 ,

and the eigenvalues {τ1 , τ2 , τ3 } are called the principal stresses. As an illustration, let us re-examine the simple shear of §2.2.2. The eigenvalues and corresponding eigenvectors of E are readily found to be α α ε1 = , ε2 = − , ε3 = 0, (2.30a) 2 2       1 1 0 v 1 = 1 , v 2 = −1 , v 3 = 0 . (2.30b) 0 0 1 Hence the shear is equivalent to an expansion of magnitude α/2 in the direction (1, 1, 0)T plus an equal and opposite contraction in the direction (1, −1, 0)T . In the original non-principal axes, we can decompose the displacement as follows:         y x+y x−y αy α α α u =  0  = −x + x + y  − y − x . (2.31) 2 4 4 0 0 0 0 The terms on the right-hand side represent a rigid-body rotation, an expansion and an orthogonal contraction. In general, the strain tensor E and, therefore, the principal strains εi and principal stresses τi are functions of position x. At any such position, the

36


strain and stress tensors are diagonal only with respect to the principal axes; if these axes are rotated by an arbitrary orthogonal matrix P then the strain and stress tensors become E 00 and τ 00 respectively, where E 00 = P E 0 P T ,

τ 00 = P τ 0 P T .

(2.32)

It is easiest to understand the consequence of such a rotation in two dimensions when we can take µ ¶ cos θ sin θ P = (2.33) − sin θ cos θ and (2.32) becomes µ ¶ τ1 cos2 θ + τ2 sin2 θ (τ1 − τ2 ) sin θ cos θ τ= , (τ1 − τ2 ) sin θ cos θ τ2 cos2 θ + τ1 sin2 θ

(2.34)

with an analogous expression for E. It follows that the diagonal elements of the stress tensor, known as the normal stresses, always lie between min(τi ) and max(τi ). However, the off-diagonal shear stress takes its maximum value |τ1 − τ2 |/2 when θ = ±π/4. It may be shown (Exercise 2.9) that these properties also hold for threedimensional stress fields. In particular, the maximum shear stress always occurs at an angle of π/4 with two of the principal axes and is given by S = max i,j

|τi − τj | . 2

(2.35)

This is often used as a diagnostic to test for failure, since many elastic materials, particularly metals, are susceptible to plastic deformation when subject to excessive shear stress. The so-called Tresca criterion proposes that a material will yield and become plastic when S exceeds some critical yield stress τY . We will show below in §2.6.5 how this criterion can be applied to predict the failure of a gun barrel, and a fuller account of plasticity will be given in Chapter 9.

2.3 Antiplane strain The simplest two-dimensional model for elastostatics occurs when the displacement u is unidirectional, u = (0, 0, w)T say, and w depends only on the transverse coordinates (x, y) = x. One way to create this state of antiplane strain is to apply a tangential traction f (x), in the axial direction only, to the curved boundary of a cylindrical bar, as illustrated in Figure 2.5. Of

2.3 Antiplane strain

37

z

surface traction f D

y

x Fig. 2.5. Schematic of a bar in a state of antiplane strain.

course, f (x) must exert no net force or moment on the bar, as pointed out in §1.10. With the z-axis parallel to the bar, the stress tensor is  0

   τ = 0   ∂w µ ∂x

0 0 µ

∂w ∂y

∂w  ∂x   ∂w  , µ ∂y   0 µ

(2.36)

so, at any point of the bar, the traction on an element is purely a shear stress whose maximum value is µ|∇w|. The Navier equation reduces to the two-dimensional Laplace’s equation ∂2w ∂2w + = ∇2 w = 0, ∂x2 ∂y 2

(2.37)

to be solved inside the cross-section D of the bar, which is a region of the (x, y)-plane. Since the bar is uniform, the unit normal n to the curved boundary is confined to the (x, y)-plane, so the applied traction is simply a shear force f in the z-direction that is related to w by f (x) = τzn = µ

∂w ∂n

on ∂D.

(2.38)

The solvability condition for the so-called Neumann problem (2.37), (2.38)

38


z

M

M

y

D x

Fig. 2.6. Schematic of a bar being twisted under a moment M .

is I f ds = 0,

(2.39)

∂D

which implies that no net traction may be applied to any cross-section (cf equation (1.62)). If, instead of the traction, we were to specify the displacement w on ∂D, we would clearly obtain the Dirichlet boundary condition w = f (x)

on ∂D

(2.40)

instead of (2.38). We note that the strain energy density is W = µ|∇w|2 and that, when the total elastic energy in the bar is minimised using the calculus of variations, (2.38) is the natural boundary condition for this minimisation. Despite its simplicity, we will find this model a very useful paradigm when we come to consider fracture in Chapter 7. Equation (2.37) also describes the small transverse displacement w(x, y) of a nearly planar membrane under tension (Exercise 2.2); (2.38) and (2.40) then correspond to prescribing either force or displacement at its boundary.

2.4 Torsion Now consider a bar which, instead of stretching or contracting along its axis, twists under the action of moments applied at its ends. Such torsion bars are often used in car suspensions, and may be described by a displacement

2.4 Torsion

field of the form

39

    u −Ωyz u =  v  =  Ωxz  , w Ωψ(x, y)

(2.41)

where Ω is a constant representing the twist of the bar about its axis. As in §2.3, the stress tensor is of the form   0 0 τxz (2.42) τ = 0 0 τyz  τxz τyz 0 where, now,

µ

τxz = µΩ

¶ ∂ψ −y , ∂x

µ τyz = µΩ

¶ ∂ψ +x . ∂y

(2.43)

The Navier equation therefore implies that, as in §2.3, ψ satisfies Laplace’s equation ∇2 ψ = 0

in D,

(2.44)

where D is the cross-section of the bar. Recall that, since the bar is uniform, its unit normal n lies purely in the (x, y)-plane; if ∂D is parametrised by x = x(s), y = y(s), where s is arc-length, then n = (dy/ds, −dx/ds, 0)T . Hence, assuming the curved boundary of the bar is stress-free, we must have dy dx − τyz =0 on ∂D, (2.45) ds ds and the corresponding boundary condition for ψ is ¢ 1 d ¡ 2 ∂ψ = x + y2 on ∂D. (2.46) ∂n 2 ds Notice that the solvability condition (2.39) for the Neumann problem (2.44), (2.46) is satisfied identically. Hence, given the shape of the crosssection D, the displacement w = Ωψ is uniquely determined up to an arbitrary constant, corresponding to an arbitrary uniform translation. Once ψ has been found, the moment applied at each end of the bar is given by ZZ M= (xτyz − yτxz ) dxdy = RΩ, (2.47) τxz

D

where

ZZ R=µ

¾ ½ ¢ ∂ψ ¡ 2 ∂ψ 2 −y + x +y dxdy. x ∂y ∂x D

(2.48)

40


The constant of proportionality R between the applied moment M and the resulting twist Ω is called the torsional rigidity of the bar. The quantity R/µ is proportional to the square of the cross-sectional area of the bar and, as we will see shortly, is readily found for simple cross-section shapes. The boundary-value problem (2.44), (2.46) for ψ may be usefully reformulated as follows. Whenever the stress tensor is of the form (2.42) and there is no body force, the steady Navier equation reduces to ∂τyz ∂τxz + = 0. ∂x ∂y

(2.49)

We can guarantee that (2.49) is satisfied by postulating the existence of a stress function φ (x, y) such that τxz = µΩ

∂φ , ∂y

τyz = −µΩ

∂φ , ∂x

(2.50)

where the factors of µΩ are introduced for later convenience. Moreover, it can be shown that the existence of φ is necessary as well as sufficient for solutions of (2.49) to exist. In the same way, the existence of a stream function in fluid dynamics guarantees mass conservation, and the existence of a magnetic vector potential ensures that the magnetic field is divergencefree in magnetostatics. Such potentials are often called gauges and they are never uniquely defined; in our case, φ is unique only up to the addition of an arbitrary constant. By comparing (2.50) with (2.43), we see that φ and ψ are related by ∂ψ ∂φ = + y, ∂x ∂y

∂ψ ∂φ =− − x. ∂y ∂x

(2.51)

Elimination of φ retrieves (2.44), while elimination of ψ reveals that φ satisfies Poisson’s equation ∇2 φ = −2

in D.

(2.52)

Indeed, it is easy to see from (2.51) that ψ and φ + (x2 + y 2 )/2 are harmonic conjugates. The zero-stress boundary condition (2.45) implies that φ is constant on ∂D and, without loss of generality, we may take φ=0

on ∂D.

(2.53)

The advantage of introducing the stress function φ is that the Neumann problem for ψ has been converted to the Dirichlet problem (2.52), (2.53), which has a unique solution; exactly the same procedure could have been

2.4 Torsion

41

applied to the antiplane strain problem (2.37), (2.38). The torque M is then given in terms of φ by ¶ ZZ µ ZZ ∂φ ∂φ M = µΩ −x −y dxdy = 2µΩ φ dxdy, (2.54) ∂x ∂y D D the final step being a consequence of Green’s theorem and (2.53). An alternative formula to (2.48) for the torsional rigidity is, therefore, ZZ R = 2µ φ dxdy. (2.55) D

To illustrate the theory of this section we will now find the torsional rigidity of a circular bar firstly using ψ and secondly using φ. If D is the disc r < a, where (r, θ) are the usual plane polar coordinates, then the Neumann problem (2.44), (2.46) becomes ∇2 ψ = 0, r < a, (2.56a) ∂ψ = 0, r = a. (2.56b) ∂n It follows that ψ is a constant and, hence, that the torsional rigidity is ZZ Z 2π Z a ¡ 2 ¢ πµa4 r3 drdθ = R=µ x + y 2 dxdy = µ . (2.57) 2 0 0 D Note that a circular cylinder is the only case in which the right-hand side of (2.46) is zero. For any other cross-sectional shape, twisting a bar inevitably results in a nonzero axial displacement w = Ωψ. Alternatively, we may solve the Dirichlet problem (2.52), (2.53) for φ. Assuming that φ is a function only of r, we have µ ¶ 1 d dφ ∇2 φ = r = −2, r < a, (2.58a) r dr dr φ = 0,

r = a,

(2.58b)

whose solution, bounded as r → 0, is φ=

a2 − r 2 . 2

The torsional rigidity is then given by (2.55): Z 2π Z a πµa4 R=µ (a2 − r2 )r drdθ = , 2 0 0

(2.59)

(2.60)

which reproduces (2.57) as expected. This result can easily be generalised to bars of elliptical cross section (see Exercise 2.4).

42


∂Do D ∂Di z

y x Fig. 2.7. Schematic of a uniform tubular torsion bar, with inner and outer free surfaces given by ∂Di and ∂Do respectively.

2.5 Multiply-connected domains In §2.4, we have implicitly assumed that D is simply connected. Many torsion bars are tubular in practice, and the resulting change in topology makes a big difference to the integration of our mathematical model. We now have to apply the condition (2.45) on two stress-free boundaries, namely the inner (∂Di ) and outer (∂D0 ) surfaces of the tube, as illustrated in Figure 2.7. As before, we can deduce that φ must be constant on each of these boundaries, but these two constants are not necessarily equal. Hence, we may choose φ to satisfy φ=0

on ∂Do ,

φ=k

on ∂Di ,

(2.61)

where k is an arbitrary constant. Note that k is not simply additive to φ and hence may not be set to zero without loss of generality. The torsional rigidity is now given by ¶ ZZ µ ZZ ∂φ ∂φ R = −µ x +y dxdy = 2µ φ dxdy + 2µkAi , (2.62) ∂x ∂y D D where Ai is the area of the hole inside ∂Di . To obtain φ and thus M uniquely, we still have to evaluate the unknown constant k. By working with the stress function φ, we have reduced the torsional rigidity problem to a seemingly innocuous Poisson equation (2.52) with Dirichlet boundary conditions (2.61) on the two boundaries of the annular tube. However, we have temporarily lost sight of the displacement field w = Ωψ, which has yet to be determined from (2.51). To be physically acceptable, ψ must

2.5 Multiply-connected domains

y

43

y (b)

(a) b

a

b

a

x

x

Fig. 2.8. The cross-section of (a) a circular cylindrical tube; (b) a cut tube.

be a single-valued function of (x, y), which implies that I ∂ψ ∂ψ dx + dy ≡ 0 ∂x ∂y C

(2.63)

for all simple closed paths C contained in D. In fact, it is readily shown that (2.63) holds for all such paths if it holds when C is the inner boundary ∂Di (Exercise 2.5). By substituting for ψ in favour of φ, we therefore obtain I I ∂φ ∂φ ∂φ dy − dx = ds = 2Ai (2.64) ∂x ∂y ∂n ∂Di ∂Di where, as before, Ai is the area of the void in the tube cross section. This constraint gives the extra information needed to determine the constant k and, hence, φ and M . If there are m holes in the cross section, with φ taking the constant value ki (i = 1, . . . , m) on each hole, then there are m relations of the form (2.64) to determine the ki . As an illustration, suppose that D is the circular annulus a < r < b in plane polar coordinates (r, θ), as illustrated in Figure 2.8(a). Assuming that φ is a function of r alone, it must satisfy the boundary-value problem µ ¶ 1 d dφ 2 ∇ φ= r = −2, a < r < b, (2.65a) r dr dr φ = k,

r = a,

(2.65b)

φ = 0,

r = b,

(2.65c)

whose solution is easily found to be µ ¶ b2 − r2 b2 − a2 log (r/b) φ= + k− . 2 2 log (a/b)

(2.66)

44


We can determine k by substituting this expression for φ into (2.64): ¯ ¶ µ I Z 2π ∂φ dφ ¯¯ a2 + b2 ds = , (2.67) a dθ = π 2k − dr ¯r=a 2 ∂Di ∂n 0 while Ai = πa2 , so (2.64) tells us that k=

b2 − a2 . 2

(2.68)

This value of k eliminates the logarithmic term from (2.66), and we might have anticipated that this would be necessary by recalling that ψ and φ+r2 /2 are harmonic conjugates. Since µ ¶ r2 b2 − a2 φ+ = const + k − log r (2.69) 2 2 and − log r is the harmonic conjugate of θ, we deduce that µ ¶ b2 − a2 ψ = const − k − θ, 2

(2.70)

which is evidently a single-valued function only if k satisfies (2.68). We therefore find that φ=

b2 − r2 , 2

(2.71)

and it is straightforward to substitute this into (2.62) and obtain the torsional rigidity ¢ µπ ¡ 4 R= b − a4 . (2.72) 2 Had we chosen to use ψ instead of φ, we would have discovered that ψ ≡ 0 and then quickly reproduced (2.72) using (2.48). Even in this simple radially symmetric geometry, the bother of finding the arbitrary constant k has outweighed the convenience of introducing a stress function. When considering multiply-connected domains it is therefore often a better idea to return to the physical variable ψ. Notice that (2.72) reproduces the result (2.57) for a solid circular bar as a tends to zero. On the other hand, if the tube is thin, so a and b are nearly equal, then R ≈ 2µπa4 ²,

(2.73)

where ² = b/a − 1 ¿ 1. We can now compare this with the torsional rigidity of a tube, such as part

2.5 Multiply-connected domains

45

of an old bicycle frame, with a thin axial cut, as illustrated in Figure 2.8(b). Here the cross-section is simply connected, so that φ must satisfy µ ¶ 1 ∂ ∂φ 1 ∂2φ 2 ∇ φ= r + 2 2 = −2, a < r < b, (2.74a) r ∂r ∂r r ∂θ with φ=0

on r = a, b,

(2.74b)

φ=0

on θ = 0, 2π.

(2.74c)

and

As shown in Exercise 2.6, this problem can be solved exactly by separating the variables. Alternatively, if we assume that the tube is thin, then an approximate solution may be found by performing the rescaling r = a (1 + ²ξ) ,

(2.75)

where ² is again small. To lowest order in ², (2.74a,b) reduces to ∂2φ ≈ −2²2 a2 , ∂ξ 2

0 < ξ < 1,

(2.76a)

with φ=0

on ξ = 0, 1,

(2.76b)

whose solution is φ = ²2 a2 ξ(1 − ξ).

(2.77)

Notice that (2.77) does not satisfy the boundary conditions on θ = 0, 2π. This is because there are so-called boundary layers near the cut where the θ-derivatives in (2.74) must be solved exactly. For a simply-connected cross-section, we use (2.55) to determine the torsional rigidity and, again using the smallness of ², we obtain Z 1 2π 4 3 µa ² . (2.78) R ≈ 4πµa2 ² φ dξ ≈ 3 0 Comparing (2.78) with (2.73), we see that the change in topology caused by the cut has a dramatic influence on the strength of the tube, reducing its torsional rigidity by two orders of magnitude!

46


2.6 Plane strain 2.6.1 Definition A more common configuration than that of antiplane strain is plane strain, in which a solid is displaced in the (x, y)-plane only, with the displacement being independent of z. Writing   u (x, y) (2.79) u = v (x, y) , 0 we find that the stress tensor takes the form   τxx τxy 0 τ = τxy τyy 0  , 0 0 τzz

(2.80)

where µ

τxx τyy

¶ ∂u ∂v ∂u =λ + + 2µ , ∂x ∂y ∂x µ ¶ ∂u ∂v ∂v =λ + + 2µ , ∂x ∂y ∂y

µ

τxy τzz

¶ ∂u ∂v =µ + , ∂y ∂x µ ¶ ∂u ∂v =λ + . ∂x ∂y

(2.81a) (2.81b)

This configuration arises, for example, when a z-independent traction just in the (x, y)-plane is applied to the curved boundary of a cylindrical bar aligned with the z-axis. The easily-forgotten stress component τzz represents the normal traction that would need to be applied to the ends of such a bar to prevent it expanding or contracting in the z-direction.

2.6.2 The Airy stress function With the stress tensor given by (2.80), the Navier equation reduces to ∂τxx ∂τxy + = 0, ∂x ∂y

∂τxy ∂τyy + = 0, ∂x ∂y

or, in terms of displacements, ¶ µ 2 ¶ µ 2 ∂2v ∂ u ∂2u ∂ u + + µ + = 0, (λ + µ) ∂x2 ∂x∂y ∂x2 ∂y 2 µ 2 ¶ µ 2 ¶ ∂ u ∂2v ∂ v ∂2v (λ + µ) + +µ + = 0. ∂x∂y ∂y 2 ∂x2 ∂y 2

(2.82)

(2.83a) (2.83b)

2.6 Plane strain

47

By cross-differentiating, it is straightforward to eliminate u and v in turn from (2.83) and thus show that each satisfies the biharmonic equation ∇4 u = ∇4 v = 0,

(2.84)

where ∇4 refers here to the two-dimensional biharmonic operator: µ 2 ¶µ 2 ¶ ∂ ∂2 ∂ ∂2 ∂4 ∂4 ∂4 4 ∇ = + + = + 2 + . ∂x2 ∂y 2 ∂x2 ∂y 2 ∂x4 ∂x2 ∂y 2 ∂y 4

(2.85)

In fact, we can reduce (2.83) to a single biharmonic equation by making use of another gauge function as follows. By inspection, the integrability condition for the system (2.82) is the existence of an Airy stress function A such that τxx =

∂2A , ∂y 2

τxy = −

∂2A , ∂x∂y

τyy =

∂2A . ∂x2

(2.86)

In the same way that the stress function φ in (2.50) is only defined to within a constant, A is only defined to within a function that is linear in x and y; in other words, any such function may be added to A without contributing to the stress. By using (2.81), we obtain the following expressions for the displacement gradients ∂u ∂2A ∂2A = −ν 2 + (1 − ν) 2 , ∂x ∂x ∂y ∂2A ∂2A ∂v = (1 − ν) −ν 2, 2µ ∂y ∂x2 ∂y µ ¶ 2 ∂u ∂v ∂ A µ + =− . ∂y ∂x ∂x∂y 2µ

(2.87a) (2.87b) (2.87c)

Now by taking ∂ 2 /∂x∂y of (2.87c) and using (2.87a) and (2.87b), we can eliminate u and v and hence find that A satisfies the biharmonic equation: ∇4 A = 0.

(2.88)

Moreover, given A, (2.87) form a compatible system of three equations that determine u and v to within a rigid body displacement in which u = u0 −ωy, v = v0 − ωx, with u0 , v0 and ω constant. We now recall that the general solution of Laplace’s equation (2.37) in two dimensions may be written in the form © ª φ = Re f (z) , (2.89)

48


y t n D x ∂D ¡ ¢ (x, y) = X(s), Y (s) Fig. 2.9. Schematic of a region D whose boundary ∂D has unit normal n and tangent t.

where f is an arbitrary analytic function of z = x+iy. Similarly, the general solution of (2.88) has the Goursat representation © ª A = Re z¯f (z) + g (z) , (2.90) where z¯ = x − iy, and f , g are analytic (see Exercise 2.7). By choosing different functional forms for f and g, we can use (2.90) to construct many exact solutions of the biharmonic equation. However, making this representation satisfy realistic boundary conditions is usually not easy, as we will see in Chapter 7.

2.6.3 Boundary conditions Suppose we wish to solve (2.88) in some region D, on whose boundary a prescribed traction σ is imposed, that is τn = σ

on ∂D.

(2.91) ¡ ¢ As illustrated in Figure 2.9, we parametrise ∂D using (x, y) = X(s), Y (s) , where s is arc-length, so the unit tangent and outward normal vectors are given by µ 0¶ µ 0 ¶ X Y , (2.92) t= , n= 0 Y −X 0

2.6 Plane strain

49

where 0 is shorthand for d/ds. Using (2.86) to write the stress components in terms of A, we thus find that (2.91) can be written in the form µ ¶ d ∂A/∂y = σ. (2.93) ds −∂A/∂x If no surface traction is applied, that is σ = 0, then it follows from (2.93) that ∇A is constant on ∂D. Since, as noted above, an arbitrary linear function of x and y may be added to A without affecting the stresses, we can, in a simply-connected region, take this constant to be zero without loss of generality. Then, by taking the dot product of ∇A with t and n respectively, we deduce that dA ∂A = =0 (2.94) ds ∂n on ∂D. The former of these tells us that A is constant on ∂D and, again, this constant may without loss of generality be set to zero. Finally, we thus arrive at the boundary conditions ∂A =0 (2.95) ∂n to be imposed on a stress-free boundary. We note that the divergence theorem on any closed region D yields ¶ Z µ ∂ ¡ 2 ¢ ∂A 2 A ∇ A − ∇ A ds ∂n ∂n ∂D ZZ © ¡ ¢ ¡ ¢ª = div grad(∇2 A) − div ∇2 A grad A dxdy ZZ D n ¡ ¢2 o = A∇4 A − ∇2 A (2.96) dxdy. A = 0,

D

Hence, if A satisfies the biharmonic equation in D and the boundary conditions (2.95) on ∂D, then ∇2 A = 0 and a second use of (2.95) reveals that A ≡ 0. In view of the comment made after (2.40), we caution that although RR ¡ 2 ¢2 1 the calculus of variations shows that minimisers of 4µ D ∇ A dxdy satisfy (2.88), this integral is not the strain energy  µ ³ ´ ³ 2 ´2 ¶   ∂2A ∂ A 2 2  µ ¶ Z Z  A ∂x2 + ∂y2 2  2ν ∂∂xA2 ∂∂yA2 ∂2A 1 dx. (2.97) − +2 4µ (1 + ν) (1 + ν) ∂x∂y  d     When the displacement rather than the stress is prescribed on a boundary in plane strain, the usefulness of the stress function A is diminished. This is

50


because the displacements are related to A by (2.87), so boundary conditions for u and v cannot usually be expressed simply in terms of A. We will show below in §2.8 that other stress functions can be defined that are better suited to such problems. 2.6.4 Plane strain in a disc As a first illustrative example, let us consider plane strain in a circular region r < a on whose boundary r = a a prescribed traction is applied. Recalling (1.72) with gi − 0 and all variables depending only or r, θ, we try writing τθθ =

∂2A . ∂r2

(2.98a)

Then (1.72b) gives −

∂3A ∂τrθ = 2τrθ + r , ∂r2 ∂θ ∂r

τrθ

∂B ∂r .

2

3

∂ ∂ A Since ∂r 2 rB is now equal to − ∂r 2 ∂θ , µ ¶ ∂ 1 ∂A =− (2.98b) ∂r r ∂θ

which suggests that we write τrθ =

and then (1.72a) gives 1 ∂ 2 A 1 ∂A . + r2 ∂θ2 r ∂r The biharmonic equation for A reads µ 2 ¶2 ∂ 1 ∂2 1 ∂ ∇4 A = + + A = 0. ∂r2 r ∂r r2 ∂θ2 τrr =

(2.98c)

(2.99)

We will only consider here cases where a purely normal pressure P is applied, so the boundary conditions on r = a are µ ¶ 1 ∂ 2 A 1 ∂A ∂ A = −P, =0 on r = a. (2.100) + r2 ∂θ2 r ∂r ∂r r We obtain the latter equation by integrating the condition τrθ = 0 with respect to θ. The simplest case occurs if P is constant, so we expect the displacement to be purely radial and A to be a function of r alone. The problem thus reduces to ¶2 µ 2 1 d d + A = 0, (2.101) dr2 r dr

2.6 Plane strain

51

subject to the boundary conditions dA = −P a2 on r = a. (2.102) dr For the stresses to exist throughout the circle, we require A to be twice differentiable as r → 0. It is then straightforward to solve (2.101) in the form A=r

A = c1 r2 + c2 + c3 r2 log r + c4 log r

(2.103)

and apply the boundary conditions (2.102) to obtain ¢ P ¡ 2 A=− r + a2 . (2.104) 2 If P is not assumed to be constant, then we can solve the problem by separating the variables in polar coordinates, using the fact that A must be a 2π-periodic function of θ. Seeking a solution of (2.99) in the form A(r, θ) = f (r) sin(nθ), where n is a positive integer, we find that µ 2 ¶2 d n2 1 d − 2 + f = 0. dr2 r dr r

(2.105)

(2.106)

This Euler differential equation admits the solution f (r) = rk , where k satisfies ¡ 2 ¢¡ ¢ k − n2 (k − 2)2 − n2 = 0. (2.107) We must again ensure that the stress components (2.98) are well defined as r → 0, and now this restricts us to the solutions k = n, n + 2, that is f (r) = c1 rn+2 + c2 rn ,

(2.108)

where the ci are again arbitrary constants. For the special case n = 1, the only acceptable solution is f (r) = c1 r3 .

(2.109)

We can take a linear combination of separable solutions that satisfy (2.102) and the radially-symmetric solution (2.104) to obtain µ 2 ¶ r + a2 A=− A0 4 ¶ ∞ µ © ª 1X rn rn+2 + − An cos(nθ) + Bn sin(nθ) , (2.110) n−2 n 2 (n − 1)a (n + 1)a n=2

52


r=a

r=b

P

Fig. 2.10. Schematic of a plane annulus being inflated by an internal pressure P .

where An and Bn are the Fourier coefficients of P , that is 1 An = π

Z

2π

P (θ) cos(nθ) dθ, 0

1 Bn = π

Z

2π

P (θ) sin(nθ) dθ.

(2.111)

0

Notice that the n = 1 term does not appear in the series in (2.110), so it is possible to satisfy the boundary condition (2.100) only if Z 0

2π

µ ¶ cos θ P (θ) dθ = 0. sin θ

(2.112)

This is simply a manifestation of the solvability condition (1.62) and states that the net force on the disc must be zero.

2.6.5 Plane strain in an annulus Our next example concerns the inflation of a circular annulus a < r < b under an applied internal pressure P , as illustrated in Figure 2.10. This practical and important instance of plane strain might model, for example, the response of a gun barrel or a diving cylinder, or even the skin of a Belgian sausage. We will only consider here a constant pressure P , the techniques of the previous section again being appropriate when the pressure depends on θ. Assuming that A is a function of r alone, it is given by (2.103), but, since we are now solving in the annular region a < r < b, there is no a priori justification for eliminating the logarithmic terms that cause the stress to be ill-behaved as r → 0. We therefore have four arbitrary constants ci to be determined from the boundary conditions. The condition of zero traction on the outer boundary r = b means, as

2.6 Plane strain

53

explained in §2.6.3, that we may without loss of generality set A=

dA =0 dr

on r = b.

(2.113a)

With A independent of θ, τrθ is identically zero, so the specified internal pressure gives us just one more boundary condition, namely 1 dA = −P r dr

on r = a.

(2.113b)

As in §2.5, the vital fourth relation needed to determine the constants ci comes from the realisation that the annulus is multiply connected We need A(r) to be such that the displacement ur (r) exists and is single-valued, where, from the constitutive relations (1.70), 1 dA dur ur = (λ + 2µ) +λ , r dr dr r

d2 A ur dur + (λ + 2µ) . =λ dr2 dr r

(2.114)

By cross-differentiation, we find that that the compatibility condition for ur to exist is d3 A 1 d2 A 1 dA = 0, + − 2 3 2 dr r dr r dr

(2.115)

which is satisfied by (2.103) only if c3 = 0. The boundary conditions (2.113) may then be used to evaluate the remaining three constants and hence show that A=

³r´ ¡ 2 ¢ P a2 P a2 b2 2 r − a − log . 2 (b2 − a2 ) b2 − a2 b

(2.116)

In a plane strain problem with multiple traction-free holes, we would have A = ki ,

∂A = Ki ∂n

(2.117)

on the ith hole, where the constants ki and Ki must be determined by ensuring that both displacement components (u, v) are single-valued around each hole. As noted in §2.5, the utility of introducing a stress function in such cases is negated by the difficulty of solving for all the arbitrary constants, and it is often preferable to work with physical variables in multiply-connected domains. By way of illustration, had we posed the annular problem above in terms of the radial displacement ur (r) rather than A, we would have

54


obtained the considerably more tractable problem µ ¶ d 1 d (rur ) = 0, a < r < b, dr r dr dur ur (λ + 2µ) + λ = −P, r = a, dr r dur ur (λ + 2µ) + λ = 0, r = b. dr r This quickly yields the solution µ ¶ r P a2 b2 , ur = + 2 (b2 − a2 ) λ + µ µr

(2.118a) (2.118b) (2.118c)

(2.119)

and we can then recover the stress components using the constitutive relations (1.70). In our radially symmetric geometry, τrθ is zero and the only non-zero stresses are the radial stress τrr , the hoop stress τθθ and the axial stress τzz , given respectively by µ ¶ µ ¶ P a2 b2 P a2 b2 2νP a2 τrr = 2 1 − , τ = 1 + , τ = , zz θθ b − a2 r2 b2 − a2 r2 b2 − a2 (2.120) where ν denotes Poisson’s ratio as usual. These results are in accordance with (2.116). We can now use the above solution to predict the failure of a gun barrel as P increases. As explained in §2.2.5, the Tresca criterion predicts that the material will fail if |τi − τj | S = max > τY , (2.121) i,j 2 where τY is the yield stress and τi are the principal stresses. Here the stress tensor is diagonal, so these are just the three stress components given in (2.120). If we assume that the material has positive Poisson’s ratio† then, since ν < 1/2 and r < b, we can deduce from (2.120) the inequalities τrr < 0 < τzz < τθθ ,

(2.122)

and so the maximum shear stress is S=

P a2 b2 τθθ − τrr = 2 . 2 (b − a2 ) r2

(2.123)

This is maximised at r = a, so we would expect a gun barrel always to yield first on its inner surface. † For an auxetic material with ν < 0, the second inequality in (2.122) is reversed and the maximum shear stress may be τθθ − τzz instead of (2.123).

2.6 Plane strain

55

y f+ g+ b a

x

g− f− Fig. 2.11. Schematic of a plane rectangular region subject to tangential tractions on its faces.

We can also use (2.123) to predict the maximum pressure P ∗ that the barrel can withstand, namely µ ¶ a2 P∗ = τ∗ 1 − 2 . (2.124) b Notice in particular how the barrel becomes more susceptible to failure as its thickness b − a is reduced. It turns out that this is a canonical problem for the theory of plasticity, as will be explained in Chapter 9. 2.6.6 Plane strain in a rectangle In §2.6.4, we showed how plane strain in a disc can be solved by separating the variables in polar coordinates. On the face of it, the procedure was only slightly more complicated than that for solving Laplace’s equation, the extra complication arising from the application of two boundary conditions rather than one on r = a. In other geometries, however, solution of the biharmonic equation generally involves serious practical difficulties that are not encountered with Laplace’s equation. To illustrate the problems that may occur, we will now consider plane strain in a cylinder with rectangular cross-section |x| < a, |y| < b. For simplicity, we suppose that the faces are subject to purely tangential tractions, that is τxx = 0,

τxy = g± (y)

on x = ±a,

(2.125a)

τyy = 0,

τxy = f± (x)

on y = ±b,

(2.125b)

56


as shown schematically in Figure 2.11. Of course the applied tractions must exert no net force or moment, so that Z a Z b ¡ ¢ ¡ ¢ f+ (y) − f− (y) dx = g+ (y) − g− (y) dy = 0, (2.126a) 0 0 Z a Z b ¡ ¢ ¡ ¢ b f+ (y) + f− (y) dy − a g+ (y) + g− (y) dy = 0. (2.126b) 0

0

The Airy stress function therefore satisfies the biharmonic equation subject to A = 0, A = 0,

∂2A = −g± (y) ∂x∂y ∂2A = −f± (x) ∂x∂y

on x = ±a,

(2.127a)

on y = ±b,

(2.127b)

after integrating the normal stress conditions as in §2.6.3. The seemingly simplest solution procedure is to separate the variables, writing A(x, y) = F (x)G(y) where F 00 G00 G0000 F 0000 +2 + = 0. F F G G

(2.128)

For any fixed value of y, this is a linear, constant-coefficient, fourth-order ordinary differential equation for F (x), which suggests writing F as a combination of exponentials eαx . This leads to G0000 + 2α2 G00 + α4 G = 0

(2.129)

and hence to solutions of the form © ª A = a1 y cos (αy) + a2 y sin (αy) + a3 cos (αy) + a4 sin (αy) eαx © ª + b1 y cos (αy) + b2 y sin (αy) + b3 cos (αy) + b4 sin (αy) e−αx . (2.130) Clearly a second class of separable solutions arises from exchanging the roles of x, y. All these solutions can also be derived from (2.90) by putting f = Aeαz + Be−αz , g = Ceαz + De−αz . We now need to select from (2.130) functions which satisfy A = 0 on |x| = a and on |y| = b. For simplicity, we suppose that the tangential tractions f± (x) and g± (y) are both odd functions. We can thus assume that A is an even function of x and y, which is analogous to seeking a Fourier cosine series representation for Laplace’s equation. After a little experimentation, we find that α must be of the form (n + 1/2)π/a or (n + 1/2)π/b, for some

2.6 Plane strain

57

y stress free

x

τxx = σx (y)

τxy = σy (y)

stress free

Fig. 2.12. Schematic of the tractions applied to the edge of a semi-infinite strip.

integer n. We can therefore construct a solution of the form ³ nπy ´ ³ nπx ó X n A= An Fn (x) cos + Bn Gn (y) cos , 2b 2a

(2.131)

n odd

where An and Bn are constants and the functions Fn and Gn are defined by ³ nπa ´ ³ nπx ´ ³ nπa ´ ³ nπx ´ Fn (x) = a sinh cosh − x cosh sinh , (2.132a) 2b 2b ¶ 2b µ 2b ¶ µ ³ nπy ´ ³ nπy ´ nπb nπb Gn (y) = b sinh cosh − y cosh sinh . (2.132b) 2a 2a 2a 2a In contrast to more elementary Fourier series examples, we now encounter a serious difficulty when we try to determine An and Bn in terms of the applied tractions. Considering x = a, for example, we find that ³ nπy ´ ³ nπ ´ o π X n 0 g+ (y) = n aFn (a)An sin + b sin Bn G0n (y) , 2ab 2b 2 n odd

(2.133) but {G0n (y)} do not form an orthogonal set on −b < y < b. Hence, if we were to multiply (2.133) by G0m (y) and integrate over |y| < b in the usual Fourier series procedure, we would end up with an infinite system of equations for the constants An and Bn . We will discuss this difficulty further in §2.6.7.

2.6.7 Plane strain in a semi-infinite strip We now present an example which is technically easier than plane strain in a rectangle and which has wide significance in the theory of linear elasticity. We consider plane strain in the semi-infinite strip 0 < x < ∞, −h/2 < y
0. The constants ai are determined, up to an arbitrary scalar multiple, by the homogeneous boundary conditions (2.142). It is easiest to transfer the boundary condition to y = 0 and y = a which immediately gives that a3 = 0, a1 + αa4 = 0, h cos(αh)a1 + h sin(αh)a2 + sin(αh)a4 = 0, and (cos αh − αh sin αh)a1 + (sin αh + αh cos αh)a2 + cos(αh)a4 = 0. The vanishing of the relevant determinant then gives us the transcendental equation sin (αh) = ±αh

(2.144)

and the “eigenvalues” α satisfying (2.144) are complex. This occurs because the ordinary differential equation (2.129), along with the boundary conditions G = G0 = 0 at y = ±h/2 is not a self-adjoint eigenvalue problem, in contrast with the St¨ urm–Liouville problems which would be encountered were we solving Laplace’s equation. This has the additional implication that the “eigenfunctions” are not mutually orthogonal, which seriously complicates the task of fitting the initial conditions, as in (2.133). Had we instead imposed the conditions A = ∂ 2 A/∂y 2 = 0 on y = ±h/2, we would have found that the eigenvalues are real and the eigenfunctions are orthogonal trigonometric functions. We will see a practically relevant example of this in §4.6 but, unfortunately, specifying ∂ 2 A/∂n2 on the boundary has no obvious physical significance in plane strain. ˜ can in fact be exA completeness argument can be given to show that A pressed as a sum of terms of the form (2.143), with Re(α) > 0, although the difficulties described above make it impractical to compute the coefficients. ˜ will decay exponentially Thus, whatever tractions are imposed on x = 0, A as x → ∞. The stress far from the edge of the strip will then be characterised entirely by the stress function A∞ , which corresponds to a far-field

60


stress τxx →

12y(a + bx) + h(6y − h)c , h3

τxy →

¢ 3b ¡ 2 h − 4y 2 , 3 2h

τyy → 0 (2.145)

as x → ∞. The net tensile force T , shear force N and bending moment M exerted on any section x = const. by the stress field (2.145) are given by Z h/2 Z h/2 T (x) = τxx dy = −c, N (x) = τxy dy = b, (2.146a) −h/2

−h/2

Z M (x) =

h/2

c yτxx dy = a + bx + . 2 −h/2

(2.146b)

From (2.139), we deduce that T , N and M satisfy the differential equations dM = N, dx Z

Z

h/2

T (0) = −h/2

(2.147)

σx (y) dy,

h/2

N (0) = −h/2

Z

σy (y) dy,

(2.148a)

h/2

M (0) = −h/2

yσx (y) dy,

(2.148b)

as x → 0, i.e., as we approach the edge of the strip from the far field. Hence the only information about the tractions applied to x = 0 that is preserved as x → ∞ is the value of the scalars T , N and M , which correspond to the net force and moment ¡ exerted on¢ the edge of the strip. It follows that any system of tractions σx (y), σy (y) applied to the edge of an elastic strip is indistinguishable from a different set of tractions which exert the same net force and moment, when we are sufficiently far from the edge. This phenomenon occurs quite generally in elasticity and is known as St Venant’s principle: any localised system of tractions applied to a sufficiently large elastic body may be characterised in the far field purely by its net force and moment. We will return to this important fact in Chapter 4, when considering the boundary conditions to be imposed at the boundaries of thin solids such as plates and rods, and we will also re-encounter the relation (2.147) between the shear force and bending moment. Finally, let us calculate the far-field displacement by using (2.140) and

2.6 Plane strain

61

(2.87). From (2.87a), (2.87b) we find that u and v satisfy ¢ 2µ ∂v 2µ ∂u 1 ¡ − = = 3 6y(2[a + bx] + ch) − ch2 . ν ∂y 1 − ν ∂x h The functions that arise in integrating these expressions are determined, to within rigid body motion, by (2.87c) which gives µ ¶ ¡ ¢ ∂u ∂v 2µ + = 24b h2 − 4y 2 . ∂y ∂x Since b = N , this gives that, as x → ∞, the transverse displacement is v∼

4(1 − ν) N x3 , µν

(2.149)

another fact that will prove very useful in Chapter 4. 2.6.8 Plane strain in a half-space As long as we are deft with Fourier transforms, all the difficulties associated with the separability of solutions melt away when we consider a half-space, say y > 0, with prescribed traction τyy = −P (x),

τxy = σ(x)

on y = 0.

(2.150)

Assuming that P and σ decrease sufficiently rapidly as |x| → ∞, we define the Fourier transform Z ∞ b (y; k) = A A (x, y) eikx dx. (2.151) −∞

Assuming further that this integral converges, we also recall the inversion formula Z ∞ 1 ˆ −ikx dx. A= Ae 2π −∞ From (2.88) we find that µ 2 ¶2 d 2 b=0 −k A dy 2

y > 0,

(2.152a)

y = 0,

(2.152b)

y = 0,

(2.152c)

y → ∞.

(2.152d)

and, from (2.150), ˆ = Pb(k) k2 A b dA =σ b(k) ik dy b→0 A

62


b is readily solved to give This ordinary differential equation for A µ ¶ 1 + y|k| iy b= A Pb(k) − σ b(k) e−y|k| . k2 k

(2.153)

The presence of inverse powers of k in (2.153) indicates that A is such that the integral in (2.151) fails to converge unless k is a complex variable. However, we can avoid the use of complex k by inverting the formulae for the stress components. For simplicity, we just consider the case σ = 0, whence we deduce that τbxx = −Pb(k) (1 − y|k|) e−|k|y , τbxy = Pb(k)iyke−|k|y ,

(2.154b)

τbyy = −Pb(k) (1 + y|k|) e

(2.154c)

−|k|y

,

(2.154a)

which, as shown in Exercise 2.11(a), may be inverted as convolutions of the form Z 2 ∞ s2 y ds , (2.155a) τxx = − P (x − s) π −∞ (s2 + y 2 )2 Z 2 ∞ sy 2 ds τxy = − P (x − s) , (2.155b) π −∞ (s2 + y 2 )2 Z 2 ∞ y 3 ds . (2.155c) τyy = − P (x − s) π −∞ (s2 + y 2 )2 We can now use (2.87) to calculate the displacements. For example, vb satisfies 2b db v b − ν d A = ((2ν − 1) − y|k|) Pˆ (k)e−y|k| , 2µ = −(1 − ν)k 2 A (2.156) dy dy 2 which may be integrated, in the case σ = 0, using (2.153), to obtain ¶ µ 1−ν b P (k) e−y|k| . (2.157) 2µb v = y+2 |k| Again, this is awkward to invert as it stands (because of the |k| in the denominator) and it is easier first to differentiate with respect to x. As shown in Exercise 2.11(b), this leads to Z 1 ∞ (1 − ν)s2 + (2 − ν)y 2 ∂v =− µ P (x − s)s ds. (2.158) ∂x π −∞ (s2 + y 2 )2 The displacement v0 (x) = v(x, 0) of the surface caused by the imposed pressure can now be found by carefully letting y tend to zero, as shown in

2.6 Plane strain

63

1 0.8 0.6

v0 /a

0.4

(1 − ν)LP/µa

0.2 -4

-2

2

4

x/L

Fig. 2.13. The surface displacement and pressure fields defined by (2.161).

Exercise 2.12. When we do so, as in Exercise 2.11(b), the right-hand side of (2.158) reduces to the singular integral Z 1 ∞ P (s) ds dv0 = (1 − ν) − = (1 − ν)H[P ], (2.159) µ dx π −∞ s − x where H[P ] is called the Hilbert transform of P . This transform may be £ ¤ inverted using the formula H H[f ] ≡ −f REF, and hence we can calculate the pressure required to achieve a given surface displacement v0 (x), namely £ ¤ µ P =− H v00 . (2.160) 1−ν As an illustrative example, the surface displacement a v0 (x) = (2.161a) 1 + x2 /L2 corresponds to a surface pressure P (x) =

µa 1 − x2 /L2 , (1 − ν)L (1 + x2 /L2 )2

(2.161b)

and both are plotted in Figure 2.13. Notice that in some parts a negative pressure must be applied to obtain the given bounded surface displacement. One can deduce generally from (2.158) that it is only possible to achieve a localised displacement, with v0 → 0 as x → ±∞, if Z ∞ P (x) dx = 0, (2.162) −∞

and it is therefore necessary for P to change sign. This phenomenon is an artefact of plane strain which results from the artificial constraint of zero transverse strain. It can be shown not to occur in radially symmetric problems. We will see in §2.9.3 that it does not.

64


The general question of the indentation produced by traction applied to the boundary of a half-plane will be discussed further in connection with “push” problems in Chapter 7 2.6.9 Plane strain with a body force The Airy stress function approach can easily be extended to plane strain under the action of a body force that is conservative. If g may be written as the gradient of a potential V (x, y), then the two-dimensional Navier equation takes the form ∂τxy ∂τyy ∂τxx ∂τxy ∂V ∂V + =ρ , + =ρ , (2.163) ∂x ∂y ∂x ∂x ∂y ∂y and (2.86) therefore generalises to τxx = ρV +

˜ ∂2A , ∂y 2

τxy = −

˜ ∂2A , ∂x∂y

τyy = ρV +

˜ ∂2A . ∂x2

(2.164)

˜ satisfies the Now when we eliminate the displacements, we discover that A forced biharmonic equation µ ¶ 1 − 2ν 4˜ ∇ A+ ∇2 V = 0. (2.165) 1−ν This result makes it appear that gravity, which is modelled by setting ˜ since ∇2 V = 0. However, if traction conditions V = gy, has no effect on A, for τ are given, V will enter when (2.164) is used to transform these con˜ In other words, the substitution (2.164) ditions into boundary data for A. eliminates the body force from the problem, but replaces it with boundary tractions. This approach is often useful in solving elastostatic problems numerically. As an illustration, let us try to find the stress in the elastic medium surrounding an tunnel, a problem of importance to civil engineers involved with underground railways. Supposing that the tunnel is circular, with radius a, and at a depth H, we incorporate gravity into the plane polar stress components (2.98) by setting ˜ ˜ 1 ∂2A 1 ∂A τrr = ρg (r sin θ − H) + 2 2 + , r ∂θ r ∂r Ã ! ˜ ∂ 1 ∂A τrθ = − , ∂r r ∂θ τθθ = ρg (r sin θ − H) +

˜ ∂2A , ∂r2

(2.166a) (2.166b) (2.166c)

2.6 Plane strain

65

where the depth is −y = H − r sin θ. As shown above, the modified Airy stress function satisfies the biharmonic equation, and the zero-traction conditions τrr = τrθ = 0 on the tunnel wall lead to Ã ! ˜ ˜ ˜ 1 ∂A ∂ A 1 ∂2A + = ρg (H − r sin θ) (2.167) = 0, 2 2 ∂r r r ∂θ r ∂r on r = a. By eliminating between the boundary conditions (2.167), we obtain ˜ ∂2A ˜ = ρga2 (H − a sin θ) +A ∂θ2

on r = a.

(2.168)

˜ θ), whose general This is effectively an ordinary differential equation for A(a, solution is 3 ˜ = ρga2 H + ρga θ cos θ + Aa cos θ + Ba sin θ A 2

on r = a,

(2.169a)

where A and B are arbitrary constants, and (2.167) then gives us the second condition ˜ ∂A ρga2 = ρgaH + θ cos θ + A cos θ + B sin θ ∂r 2

on r = a.

(2.169b)

These boundary conditions suggest that we try a solution of the form ˜ = f1 (r) + f2 (r)θ cos θ + f3 (r) cos θ + f4 (r) sin θ. A

(2.170)

The functions fj (r) can, in principle, be found by substituting (2.170) into the biharmonic equation and into the boundary conditions (2.169). We now make the important simplifying assumption that H À a so that, rather than imposing traction-free boundary conditions on y = 0, we simply ensure that ˜ decay sufficiently rapidly as r → ∞ for the far-field the derivatives of A ˜ is only defined up to an stress field to be hydrostatic. We also recall that A arbitrary linear function of x and y, so we can ignore terms proportional to r cos θ and r sin θ to obtain 2 ¡ ¢ ˜ = ρga2 H 1 + log(r/a) + ρga rθ cos θ A 2µ ¶ a2 A cos θ + B sin θ + + 2r log r . (2.171) r 1 + 2 log a

We do not yet have a unique solution, since the constants A and B still appear to be arbitrary. As in §2.6.5, the key to closing the problem in this

66


multiply-connected domain is to ensure that the displacement field is singlevalued. As shown in Exercise 2.15, this condition allows us to determine both remaining constants, and we find that A = 0,

B=

ρga2 (1 − 2ν)(1 + 2 log a) . 8(1 − ν)

We thus evaluate the stress components as µ ¶½ µ ¶¾ a2 (1 − 2ν) a2 τrr = ρg 1 − 2 −H + r sin θ 1 + , r 4(1 − ν) r2 µ ¶ a2 ρga2 cos θ (1 − 2ν) τrθ = − 1− 2 , r 4(1 − ν) r µ ¶½ ¾ 2 a (1 − 2ν) a2 sin θ τθθ = ρgr sin θ + ρg 1 + 2 −H + . r 4(1 − ν) r

(2.172)

(2.173a) (2.173b) (2.173c)

Evidently, τrr and τrθ are both zero on r = a, as required, and the nonzero hoop stress on the tunnel is given by τθθ = −2ρgH +

3 − 4ν ρga sin θ 2(1 − ν)

on r = a.

(2.174)

The maximum tensile stress therefore occurs at θ = π/2, which is the roof of the tunnel, while the maximum compressive stress is at the bottom θ = −π/2. In fact, it can be shown that this result remains valid for all values of H and a. See Mindler (1939). NEED PROPER CITE HERE 2.7 Compatibility In §1.6, Newton’s second law led us to write down the simple equation ∂τij + ρgi = 0 ∂xj

(2.175)

for the stress tensor in elastostatics in the presence of a body force gi . Viewed as a system of equations for the stress components, (2.175) is underdetermined, comprising just three equations for six unknowns. It was only by using the constitutive relations (1.42), ¶ µ ¶ µ ∂uj ∂ui ∂uk δij + µ + (2.176) τij = λ (ekk ) δij + 2µeij = λ ∂xk ∂xj ∂xi that we were able to obtain a closed system of the three Navier equations for the three components of displacement. It follows that, given τij , (2.176) is itself an over-determined system for the displacements, and it is not a trivial matter to obtain ui from (2.176), as we will now explain.

2.7 Compatibility

67

If we were in the very fortunate position of knowing all the stress components τij , which of course must satisfy (2.175), then it would be straightforward to obtain the corresponding linearised strain tensor from (2.176): Eeij = (1 + ν)τij − ντkk δij . Thus, given τij , we could view our expression ¶ µ ∂uj 1 ∂ui eij = + 2 ∂xj ∂xi

(2.177)

(2.178)

as a system of equations for the displacement components ui , rather than a definition of eij . In this light, however, (2.178) is a set of six partial differential equations for the three displacements ui , and this imposes on eij severe restrictions, called compatibility conditions. Since these conditions will soon be seen to have considerable theoretical and practical importance, we will now derive them. ¡ ¢T First consider a plane strain problem, in which u = u(x, y), v(x, y), 0 and µ ¶ 1 ∂u ∂v ∂v ∂u , exy = + , eyy = . (2.179) exx = ∂x 2 ∂y ∂x ∂y Since the three strain components are functions of just two displacements, (2.179) is over-determined when viewed as a system of equations for (u, v). However, whenever u is a twice continuously differentiable single-valued function, we must have ∂ 2 u/∂x∂y ≡ ∂ 2 u/∂y∂x and similarly for v. It then follows from cross-differentiation that ∂ 2 eyy ∂ 2 exy ∂ 2 exx + − 2 = 0. ∂y 2 ∂x∂y ∂x2

(2.180a)

Equation (2.180a) is the compatibility condition which ensures that (2.179) can be solved for single-valued functions u and v. In three dimensions, the same argument applies to the in-plane displacements in the (x, z)- and (y, z)-planes, yielding ∂ 2 eyy ∂ 2 eyz ∂ 2 ezz − 2 + = 0, (2.180b) ∂z 2 ∂y∂z ∂y 2 ∂ 2 ezz ∂ 2 ezx ∂ 2 exx −2 + = 0, (2.180c) 2 ∂x ∂z∂x ∂z 2 and it might look as if we have then found all the compatibility conditions needed for the existence of u, v and w. However, we also have to ensure that displacements transverse to each coordinate plane exist and are singlevalued, so that, for example, ∂ 2 w/∂x∂y ≡ ∂ 2 w/∂y∂x. Once we take this

68


into account, we obtain the additional three compatibility relations µ ¶ ∂exy ∂eyz ∂ 2 exx ∂ ∂exz = + − , (2.180d) ∂y∂z ∂x ∂y ∂z ∂x µ ¶ ∂ 2 eyy ∂ ∂eyx ∂eyz ∂ezx = + − , (2.180e) ∂z∂x ∂y ∂z ∂x ∂y µ ¶ ∂ 2 ezz ∂ ∂ezy ∂ezx ∂exy = + − . (2.180f) ∂x∂y ∂z ∂x ∂y ∂z We thus have a total of six compatibility conditions, which now seems excessive for a system of six equations in the three unknowns ui . In fact, only three of the equations are independent, in the sense that cross-differentiation to eliminate all but three unknowns from (2.180) inevitably leads to just one independent equation. However, we will soon see that all six compatibility conditions are needed in practice. This situation is analogous to that which arises when considering a vector field u which is the gradient of a potential. In this case, it is necessary that all three components of curl u vanish, although only two components of curl u are independent in the above sense. Our situation is a generalisation of this statement, since, as we show in Exercise 2.19, the compatibility conditions (2.180) imply that the components of the tensor equation   0 −∂/∂z ∂/∂y curl (curl E)T = 0 where curl =  ∂/∂z 0 −∂/∂x −∂/∂y ∂/∂x 0 (2.181) all vanish. We also remark that any three independent compatibility equations for eij are just as under-determined as the three equilibrium equations (2.175) are for τij . Hence, the fact that the compatibility conditions ensure the existence of three physically acceptable displacement components suggests that (2.175) will similarly guarantee the existence of three scalar stress functions. We have already encountered an application of this idea to plane strain in §2.6, and we will show below in §2.8 how it may be generalised to non-planar problems. It is important to note that, if eij does not satisfy (2.180), then there is no physically acceptable displacement field that gives rise to such a strain and it is, therefore, incompatible with linear elasticity. In §2.9.4 we will discuss one physical interpretation of incompatibility in detail. Meanwhile, we simply conjecture that, if we are presented with an elastic material which has been deformed under the action of, say, some nonzero boundary tractions, then,

2.8 Generalised stress functions

69

when the tractions are removed, the necessary and sufficient condition for the material to return to its pristine unstrained state is that the strain field in it satisfies (2.180). 2.8 Generalised stress functions 2.8.1 General observations By taking the divergence of the steady Navier equation (2.1) in the absence of body forces, we see that div u must be a harmonic function: ∇2 (div u) = 0.

(2.182)

If instead we take the curl of (2.1), we find that curl3 u = 0

(2.183)

and so, by taking the curl once more and using (2.182, we obtain ∇4 u = 0.

(2.184)

Each Cartesian component of the displacement must therefore satisfy the biharmonic equation, as shown in §2.6 for plane strain. We emphasise that (2.184) can only be written component-wise as  4  ∇ u 4  (2.185) ∇ u = ∇4 v  = 0 ∇4 w in Cartesian coordinates; in any other coordinate systems we must use the identity (1.47). Alternatively, we may use the Helmholtz representation to write u as the sum of a gradient and a curl: u = ∇φ + ∇×A,

with ∇ · A = 0.

(2.186)

(It may be shown that this decomposition applies to any continuously differentiable vector field u.) Substitution of (2.186) into (2.1) reveals that the potential functions φ and A must also satisfy the biharmonic equation ∇4 φ = 0,

∇4 A = 0.

(2.187)

These results all suggest that we may be able to construct large classes of solutions of (2.1) by considering suitable harmonic or biharmonic scalar problems. We have already seen some examples of this approach in plane and antiplane strain (§§2.3–2.6) and we will now show how analogous ideas may be applied in some other commonly-occurring configurations.

70


2.8.2 Plane strain revisited We begin by re-examining the theory of plane strain in the light of the concept of compatibility developed in §2.7. As we showed in §2.6, the two conservation laws ∂τxy ∂τyy ∂τxx ∂τxy + = 0, + =0 (2.188) ∂x ∂y ∂x ∂y imply the existence of an Airy stress function A. We then eliminated the displacement components to deduce that A satisfies the biharmonic equation. We thus implicitly assumed compatibility of the stress tensor with a physically acceptable displacement field. We can make this procedure more systematic by making use of the plane compatibility condition (2.180a) which, when written in terms of the stress components, is ∂ 2 τyy ∂ 2 τxy ∂ 2 τxx + − 2 − ν∇2 (τxx + τyy ) = 0. ∂y 2 ∂x∂y ∂x2

(2.189)

Now (2.188) and (2.189) form a closed system of partial differential equations for the stress components (τxx , τxy , τyy ). If, as in §2.6, we write τxx =

∂2A , ∂y 2

τxy = −

∂2A , ∂x∂y

τyy =

∂2A , ∂x2

(2.190)

then the stress balance equations (2.188) are satisfied identically, and substitution of (2.190) into (2.189) quickly retrieves the biharmonic equation for A. It is worth noting that an inhomogeneous equation for A of the form ∇4 A = f

(2.191)

does not, as might have been suspected from (2.165), correspond to the imposition of a body force. The mere existence of A implies that (2.188) are satisfied, so there is no such force. Instead the right-hand side f of (2.191) corresponds to a distribution of incompatibility through the material, and we will discuss in §2.9.4 and Chapter 9 what the physical significance of such a model might be. Now let us suppose that we do impose a body force in, say, the x-direction so that the first Navier equation (2.188a) is modified to ∂τxx ∂τxy + = −f. ∂x ∂y

(2.192)

Now there is no longer an Airy stress function, but we can instead try to obtain an integrability condition from the two remaining homogeneous


71

equations (2.188b) and (2.189). From (2.188b) we deduce the existence of a function A1 (x, y) such that τxy =

∂A1 , ∂y

τyy = −

∂A1 , ∂x

so the compatibility condition (2.189) becomes ¶ µ 2 ¶ µ 2 ∂ ∂ 2 2 ∂A1 − ν∇ τxx − + (1 − ν)∇ = 0. ∂y 2 ∂y 2 ∂x

(2.193)

(2.194)

To write this in conservation form, we first let τxx = ∂A2 /∂x, so that, without loss of generality, µ 2 ¶ µ 2 ¶ ∂ ∂ 2 2 − ν∇ A2 − + (1 − ν)∇ A1 = 0. (2.195) ∂y 2 ∂y 2 We must now spot that, if we set A2 = A1 + ∇2 L, then (2.195) is satisfied identically by µ 2 ¶ ∂ 2 A1 = − ν∇ L. (2.196) ∂y 2 The existence of the Love function L, also known as the Love stress function or the Love strain function, guarantees that the strain field is compatible and the stress balance in the y-direction is satisfied. It is related to the stress components by µ 2 ¶ ∂ ∂ ∂A2 2 = τxx = + (1 − ν)∇ L, (2.197a) ∂x ∂x ∂y 2 µ 2 ¶ ∂ ∂ ∂A1 = τxy = − ν∇2 L, (2.197b) ∂y ∂y ∂y 2 µ 2 ¶ ∂ ∂ ∂A1 2 =− τyy = − − ν∇ L, (2.197c) ∂x ∂x ∂y 2 and, when we finally substitute these into the x-momentum equation (2.192), we find that L is such that ∂τxx ∂τxy + = (1 − ν)∇4 L = −f. (2.198) ∂x ∂y Even though the Love function is ideally suited to problems with a unidirectional body force, the derivation given above remains valid if there is no body force, in which case f ≡ 0 and L satisfies the biharmonic equation. Hence, since L is also easily related to the stress components by (2.197), it provides an alternative to the Airy stress function for plane strain problems; indeed it can be shown that ∂L/∂x differs from A by a harmonic function.

72


More importantly, when we try to find u and v using ∂u = (1 − ν)τxx − ντyy , ∂x ¶ µ ∂u ∂v µ + = τxy , ∂y ∂x ∂v 2µ = (1 − ν)τyy − ντxx , ∂y 2µ

(2.199a) (2.199b) (2.199c)

the resulting system is guaranteed to be integrable, by our construction of L. Indeed, apart from a rigid body displacement, we may obtain the explicit formulae µ ¶ ∂2L 1 ∂2L 1 2 (1 − 2ν)∇ L + , v = − u= . (2.200) 2µ ∂y 2 2µ ∂x∂y We note that, for a given displacement field, L is determined uniquely up to terms linear in x, y; rigid body displacements create quadratic terms in L and cubic terms proportional to νy 3 − 3(1 − ν)x2 y. Hence, in principle, the difficulties associated with posing displacement boundary conditions in the Airy stress function formulation can be avoided: all you need is Love. For a more direct, though perhaps less instructive, derivation of (2.200) we note that the second Navier equation (2.83b), written out in terms of the displacements, may be rearranged to µ ¶ ∂2u ∂2 2 + (1 − 2ν)∇ + 2 v = 0. (2.201) ∂x∂y ∂y Now we can argue that (2.200) is just the integrability condition for (2.201). However, this intuitively appealing derivation is deceptively simple; it is clear that (2.200) guarantees (2.201) but we have not proved the converse. Finally, we note that our derivation of the Love function depended on initially disregarding the x-momentum equation. If we had instead discarded the y-momentum equation, or considered a body force in the y-direction, we would have obtained a different but mathematically equivalent formulation for L. Clearly we can thus define a one-parameter family of Love functions corresponding to an arbitrarily-oriented unidirectional body force. 2.8.3 Plane stress Plane strain is characterised by the fact that the strain tensor E is purely two-dimensional, that is exz = eyz = ezz = 0. Now we instead look for a state of plane stress, in which the only nonzero components of the stress tensor are τxx , τxy and τyy . Such a state might exist, for example, in a plate with


73

no applied traction on its faces and suitably chosen loading conditions on its edges; indeed we will meet exactly this configuration again in Chapters 4 and 6. It contrasts with plane strain, where, as we recall from §2.6, the normal stress τzz is in general nonzero. We again introduce an Airy stress function A as in (2.86), although now we must allow A to be a function of x, y and z. We can still use the compatibility condition (2.180a) to deduce that A satisfies the two-dimensional biharmonic equation µ 2 ¶2 ∂ ∂2 4 e ∇ A= + A = 0, (2.202) ∂x2 ∂y 2 the tilde denoting that ∂ 2 /∂z 2 has been omitted from the Laplacian. However, this is now insufficient to determine the z-dependence of A, and so we must ensure that all five remaining compatibility conditions are also satisfied. To this end we find that the nonzero strain components are given by ∂2A ∂2A − ν , ∂y 2 ∂x2 ∂2A , Eexy = (1 + ν)τxy = −(1 + ν) ∂x∂y ∂2A ∂2A Eeyy = τyy − ντxx = − ν , ∂x2 ∂y 2 e 2 A; Eezz = −ν (τxx + τyy ) = −ν ∇ Eexx = τxx − ντyy =

(2.203a) (2.203b) (2.203c) (2.203d)

note in particular that ezz is in general nonzero, in contrast with plane strain. First considering (2.180d) and (2.180e), we obtain ∂2 ³ e 2 ´ ∂2 ³ e 2 ´ ∇ A = ∇ A = 0, ∂x∂z ∂y∂z

(2.204)

from which we deduce that e 2 A = f (z) + φ(x, y), ∇

(2.205)

e 2 φ = 0 and f is arbitrary. Next, by adding (2.180c) and (2.180b), where ∇ we obtain ∂ 2 ³ e 2 ´ d2 f (2.206) ∇ A = 2 = 0, ∂z 2 dz so, without loss of generality, we may set f (z) = βz for some constant β.

74


Finally, taking (2.180c), (2.180b) and (2.180f) individually, we find that ∂2 ∂x2

µ ¶ µ ¶ ∂2A ∂2 ∂2A (1 + ν) 2 + νφ = (1 + ν) 2 + νφ ∂z ∂x∂y ∂z µ ¶ 2 ∂ ∂2A = 2 (1 + ν) 2 + νφ = 0. (2.207) ∂y ∂z

Since terms linear in x and y do not contribute to the stress, we lose no generality in setting (1 + ν)

∂2A + νφ = 0, ∂z 2

(2.208)

and it follows that A may be written in the form A=−

νφ(x, y)z 2 + zχ1 (x, y) + χ0 (x, y), 2(1 + ν)

(2.209)

where e 2 χ0 = φ, ∇

e 2 φ = 0, ∇

e 2 χ1 = β = const. ∇

(2.210)

This rather unwieldy representation in terms of three stress potentials φ, χ0 and χ1 is one of the few ways in which three-dimensional stress fields can be generated analytically. A simple example is that of biaxial stretching, introduced in §2.2.4, which may be reproduced by setting χ1 to be zero and χ0 to be a suitable quadratic function of x and y. Another case is obtained by choosing φ = 0, χ0 = 0 and χ1 = −Eκy 2 /2, so that β = −Eκ and τxx = −Eκz,

τyy = τxy = 0,

while the corresponding displacement field is   −2xz κ . u=  2νyz 2 2 2 2 x − νy + νz

(2.211a)

(2.211b)

This solution represents bending of a beam about the y-axis, and will provide a useful check on the approximate beam theory to be developed in Chapters 4 and 6.

2.8.4 Axisymmetric geometry The Love function was first introduced for axisymmetric stress fields, (Love, 1944, Article 188). Indeed, in this geometry, the Navier equation does not


75

take the conservation form required for an analogue of the Airy stress function to be defined. Although technically tedious, the derivation of the axisymmetric Love function follows the same conceptual procedure as for the plane strain Love function described in §2.8.2. We will instead follow the more direct, although slightly cavalier, approach suggested at the end of §2.8.2. Let us consider an axisymmetric problem, in which the displacement takes the form u = ur (r, z)er + uz (r, z)ez

(2.212)

and there is a unidirectional body force in the z-direction. The homogeneous er -component of the Navier equation, given by (1.73), may be written in the form ½ µ 2 ¶ ¾ ∂ 2 uz ∂ 1 1 ∂ ∂2 + 2(1 − ν) − + + (1 − 2ν) 2 ur = 0. ∂r∂z ∂r2 r ∂r r2 ∂z

(2.213)

When seeking an integrability condition for this equation, we must keep in mind that the differential operators acting on uz and ur do not commute. Instead, for any suitably differentiable function f (r, z), we can write down an analogy of (2.201) in the form ½ µ 2 ¶ ¾ 2 ∂ 1 1 ∂ ∂2 ∂ f 2(1 − ν) − + + (1 − 2ν) 2 ∂r2 r ∂r r2 ∂z ∂r∂z ½ µ 2 ¶ ¾ ∂2 ∂ ∂2 1 ∂ = 2(1 − ν) + (1 − 2ν) 2 f. (2.214) + ∂r∂z ∂r2 r ∂r ∂z Since the right-hand side may be written as ∂2 ∂r∂z

½ µ ¶¾ 1 ∂ ∂ 2 (1 − 2ν)∇ + r f, r ∂r ∂r

we may infer from (2.213) the existence of a function L(r, z) such that 1 ∂2L ur = − , 2µ ∂r∂z

1 uz = 2µ

½ µ ¶¾ 1 ∂ ∂L 2 (1 − 2ν)∇ L + r . r ∂r ∂r

(2.215)

With this displacement, it is straightforward to evaluate the stress com-

76


ponents from (1.70): τrr τrz τzz τθθ

½ ¾ ∂ ∂2L 2 =− − ν∇ L , ∂z ∂r2 ½ µ ¶ ¾ ∂ 1 ∂ ∂L = r − ν∇2 L , ∂r r ∂r ∂r ½ µ ¶ ¾ ∂L ∂ 1 ∂ 2 r + (1 − ν)∇ L , = ∂z r ∂r ∂r ½ ¾ ∂ 1 ∂L 2 = − + ν∇ L . ∂z r ∂r

(2.216a) (2.216b) (2.216c) (2.216d)

Now to obtain an equation for L, we must substitute these into the ez -component of the Navier equation, which reads 1 ∂ ∂τzz (rτrz ) + = −f r ∂r ∂z

(2.217)

where the body force is f ez per unit volume. After a lengthy algebraic manipulation, we find that L satisfies (1 − ν)∇4 L = −f,

(2.218)

which is identical to the equation (2.198) for the planar Love function. We will now show that this is no accident and, in §2.9, we will further show how to use Love functions to construct solutions of the Navier equation corresponding to point forces.

2.8.5 The Galerkin representation We will now show briefly how the planar and axisymmetric Love functions may be viewed as instances of a more widely-applicable theory. We start from the three-dimensional Navier equation in the form (λ + µ) grad div u + µ∇2 u = −f ,

(2.219)

where the body force f per unit volume is no longer presumed to be unidirectional. To obtain the Galerkin representation, we make the “out-of-the-blue” hypothesis that the displacement can be written in the form u = a∇2 B − b grad div B,

(2.220)

for some vector field B(x) and suitably chosen constants a and b.† † This is remarkable since we expect (2.220) to determine curl B given u, and if B = ∇ϕ, (2.220) is effectively a Possion equation for ϕ.


77

The advantage of this representation is that, when we substitute (2.220) into (2.219), we find that B satisfies (λ + 2µ)(a − b) (grad div)2 B + µa curl4 B = −f ,

(2.221)

and the differential operator on the left-hand side may be made proportional to the biharmonic operator by choosing a λ + 2µ = = 2(1 − ν). b λ+µ

(2.222)

If we also (arbitrarily) set b = 1/2µ, then we find that B satisfies (1 − ν)∇4 B = −f ,

(2.223)

2µu = 2(1 − ν)∇2 B − grad div B.

(2.224)

and (2.220) reads

The planar and cylindrical Love functions may now be obtained by supposing that B = L(x, y)i or B = L(r, z)ez respectively. Evidently each will only satisfy (2.223) if the body force f shares the corresponding symmetry. However, in cases where there is no body force, (2.223) opens up the possibility of finding a wealth of scalar stress functions describing problems with various symmetries, simply by posing suitable forms for B. Note also that in view of (1.47), (2.224) may also be viewed as an example of a Helmholtz representation (2.186).

2.8.6 Papkovich–Neuber potentials Next we will show that (2.224) can be used to obtain an extremely useful representation for u that avoids all mention of the biharmonic operator and only involves solutions of Laplace’s or Poisson’s equation. We can obtain an immediate simplification by defining Ψ = ∇2 B,

(2.225)

so that (2.223) becomes ∇2 Ψ = −

f . 1−ν

Now, using the vector identity (1.47), we easily find that ¡ ¢ ∇2 (div B) ≡ div ∇2 B = div Ψ,

(2.226)

(2.227)

78


so that, disappointingly, div B turns out not to be a harmonic function. However, the well-known (and easily proved) identity 1 1 div Ψ ≡ ∇2 (x · Ψ) − x · ∇2 Ψ 2 2 allows us to write (2.227) in the form µ ¶ 1 2 ∇ div B − (x · Ψ) = 0 2

(2.228)

(2.229)

and prompts us to define 1 φ = div B − x · Ψ, 2

(2.230)

so that φ is a harmonic function. The definitions (2.225) and (2.230) transform (2.224) into the Papkovich–Neuber representation ¶ µ 1 (2.231) 2µu = 2(1 − ν)Ψ − grad φ + x · Ψ . 2 When there is no body force, φ and the Cartesian components of Ψ all satisfy Laplace’s equation. In other words, we have replaced the job of finding three scalar biharmonic functions (the components of B) with that of finding four harmonic scalar functions. In plane strain, we can represent the displacement in terms of a single biharmonic Love function, while three Papkovich–Neuber potentials are required, namely φ and two components of Ψ. Nevertheless, Laplace’s equation is susceptible to so many more solution techniques than the biharmonic equation that (2.231) can prove extremely useful, as we will see in §2.9.3 for an axisymmetric point force problem and in Chapter 7 for crack problems. To illustrate the Papkovich–Neuber potentials in antiplane strain problems with u = (0, 0, w(x, y)) we simply take   µ ¶ z ∂ψ ∂x z ∂ψ ∂ψ   ∂ψ Ψ= x +y −ψ , (2.232) z ∂y  and φ = 2 pax ∂y (3 − 4ν)ψ where, after a simple calculation, (2.231) gives that ¡ ¢ 2µu = 0, 0, 4(1 − ν)(1 − 2ν)ψ , with the anticipated result that ∇2 w = 0. Concerning plane strain, we anticipate that φ = φ(x, y) and Ψ = Ψ(ψ1 (x, y), ψ2 (x, y), 0)

(2.233)


79

so that (2.231) gives

µ ¶ ∂ 1 2µu = 2(1 − ν)ψ1 − φ + (xψ1 + yψ2 ) ∂x 2 µ ¶ ∂ 1 2µν = 2(1 − ν)ψ2 − φ + (xψ1 + yψ2 ) ∂y 2

and 2µ

µ

∂u ∂v + ∂y ∂x

¶

µ = 2(1 − ν)

∂ψ1 ∂ψ2 + ∂y ∂x

¶ −2

(2.234) (2.235)

∂2φ ∂2 − (xψ1 + yψ2 ). ∂x∂y ∂x∂y (2.236)

∂ψ2 1 Hence (2.87c) implies that ∂ψ pay + ∂x = 0 is one possibility and since ψ1 and ψ2 are harmonic functions, the Cauchy–Riemann equations imply that they are harmonic conjugates. Thus (2.87c) also gives that, apart from a gauge transformation, the Goursat representation

1 A = φ + (xψ1 + yψ2 ). (2.237) 2 However the use of φ and harmonic conjugates ψ1 and ψ2 to represent A is far from unique and we could equally well set ψ1 = 0 and Z y 1 ψ2 (x, y 0 ) dy 0 , (2.238) A = φ + ψ2 − (1 − ν) 2 where ψ2 is a harmonic function. This representation will be found to be very useful in Chapter 7. The question of the uniqueness of the Papkovich–Neuber representation is discussed in [Barber]. 2.8.7 Maxwell and Morera potentials We recall our observation in §2.7 that the equilibrium Navier equation (2.175) comprises just three equations in the six stress components. In three dimensions, we would therefore expect to need three scalar potentials to ensure integrability of (2.175). As with all stress functions or gauges of which the magnetic vector potential is an example, there is some freedom in selecting a particular set of stress functions. One well-known possibility is the Maxwell stress functions χi , defined by ∂ 2 χ1 ∂ 2 χ3 ∂ 2 χ2 ∂ 2 χ1 ∂ 2 χ3 ∂ 2 χ2 + , τ = + , τ = + , (2.239a) yy zz ∂y 2 ∂z 2 ∂z 2 ∂x2 ∂x2 ∂y 2 ∂ 2 χ3 ∂ 2 χ1 ∂ 2 χ2 =− , τyz = − , τzx = − . (2.239b) ∂x∂y ∂y∂z ∂z∂x

τxx = τxy

80


It is easy to verify that the steady Navier equation with no body force is identically satisfied for any stress field of this form. Another popular choice is the Morera stress functions ψi , for which µ ¶ ∂ ∂ψ3 ∂ψ1 ∂ψ2 ∂ 2 ψ1 , τxy = − − , (2.240a) τxx = 2 ∂y∂z ∂z ∂z ∂x ∂y µ ¶ ∂ 2 ψ2 ∂ ∂ψ1 ∂ψ2 ∂ψ3 τyy = 2 , τyz = − − , (2.240b) ∂z∂x ∂x ∂x ∂y ∂z µ ¶ ∂ 2 ψ3 ∂ ∂ψ2 ∂ψ3 ∂ψ1 τzz = 2 . (2.240c) , τzy = − − ∂x∂y ∂y ∂y ∂z ∂x At first glance, little is gained by formulating three-dimensional problems in terms of three stress functions rather than just using the three components of the displacement u. However, we will now show that these potentials suggest a useful unification of all the stress functions that we have thus far encountered. In general, given any symmetric matrix field A(x), we construct a stress tensor τ (x) with zero diagonal using the formula τij =

X

²ikl ²jmn

k,l,m,n

∂ 2 Akm , ∂xl ∂xn

(2.241)

where ²ijk is the alternating symbol. Using (2.181), we can formally write (2.241) in the form τ = curl (curl A)T ,

(2.242)

which makes it clear that the tensor τ thus defined is symmetric and identically satisfies div τ ≡ 0. All the stress functions we have used in this chapter can now be viewed simply as particular choices of the matrix A. For example, the Maxwell and Morera functions are obtained by choosing     χ1 0 0 0 ψ3 ψ2 A =  0 χ2 0  , or A = − ψ3 0 ψ1  , (2.243) 0 0 χ3 ψ2 ψ1 0 respectively. The Airy stress function for plane strain (or plane stress) is obtained by setting† χ1 = χ2 = 0, χ3 = A, while the antiplane stress function φ introduced in §2.4 corresponds to χ3 = 0, ∂χ1 /∂x − ∂χ2 /∂y = Ωµφ(x, y). The Love function L does not fall into the same category, since its existence relies on the compatibility conditions as well as one component † This gives an alternative way of deriving (2.98).

2.9 Singular solutions in elastostatics

81

δ² (x) 1.2 1 0.8

² = 0.25

0.6

² = 0.5

0.4

² = 1.0

0.2 -4

-3

-2

-1

1

2

3

4

x

Fig. 2.14. A family of functions δ² (x) that approach a delta-function as ² → 0.

of the equilibrium equation. This is why it is often refered to as a strain function rather than a stress function. Finally, we note the strong analogy between the equations (2.239), (2.240) satisfied by the Maxwell and Morera stress potentials and the compatibility conditions that we found for the strain components in §2.7. The formulation (2.242) shows that this resemblance is no accident. By using (2.181), we can obtain the compatibility condition for τ , namely © ¡ ¢T ª curl curl (1 + ν)τ − ντkk I = 0,

(2.244)

and the equations satisfied by the Maxwell functions χi , for example, can then be found by substituting for τ from (2.242).

2.9 Singular solutions in elastostatics 2.9.1 The delta function In potential theory, it is often helpful to introduce point singularities such as point charges or dipoles in electrostatics or point masses in gravitation. In inviscid fluid dynamics it is also common to discuss singular solutions describing sources and vortices, in the latter of which the singularity is localised on a line rather than a point. As we shall see, such fundamental solutions are very useful building blocks for constructing more general solutions. We begin with an informal discussion of the Dirac delta-function, which will be used henceforth in describing various point and line singularities. In one dimension, we can view the delta-function as the singular limit of a one-parameter family of well-defined positive functions. For example, the

82


family of functions δ² (x) = have the property that

Z

² + ²2 )

π (x2

(2.245)

∞

−∞

δ² (x) dx ≡ 1

(2.246)

for any ² > 0, although δ² (x) → 0 as ² → 0 at any fixed nonzero value of x, as illustrated in Figure 2.14. If we “define” δ(x) = lim²→0 δ² (x), then δ(x) is zero everywhere except at x = 0, but has unit area. (It therefore is not really a function and is called either a generalised function or a distribution.) Although some care is needed to formalise such a definition mathematically, we can use it to deduce the important property that Z ∞ δ(x − ξ)ψ(ξ) dξ = ψ(x) (2.247) −∞

for any suitably smooth test function ψ. We can then define a multidimensional delta-function δ(x) as a product of one-dimensional delta-functions, for example δ(x) = δ(x)δ(y)δ(z),

(2.248)

in three dimensions. By construction, this function possesses a property analogous to (2.247), that is ZZZ δ(x − ξ)ψ(ξ) dξ = ψ(x). (2.249) R3

By way of illustration, a point charge in electrostatics and a point mass in gravitation are both described by Poisson’s equation with a localised righthand side, namely ∇2 φ = −δ(x).

(2.250)

Suitable singular solutions of this equation can be constructed using the following key results, derived in Exercise 2.22: µ ¶ 2 |x| = δ(x), (2.251a) one dimension: ∇ 2 µ ¶ ¡ ¢ 1 two dimensions: ∇2 log x2 + y 2 = δ(x)δ(y), (2.251b) 2π Ã ! −1 2 p three dimensions: ∇ = δ(x)δ(y)δ(z). (2.251c) 4π x2 + y 2 + z 2


83

2.9.2 Point and line forces Let us first consider plane strain subject to a point force (really a line force in (x, y, z)-space) of magnitude f in the x-direction at the origin. This situation is described by applying a point body force to the two-dimensional Navier equation, that is ∂τxx ∂τxy + = −f δ(x)δ(y), ∂x ∂y

∂τxy ∂τyy + = 0. ∂x ∂y

(2.252)

As described in §2.8.2, this problem may conveniently be posed in terms of a Love function L satisfying (1 − ν)∇4 L = −f δ(x)δ(y).

(2.253)

Now we can read off from (2.251b) a suitable radially-symmetric solution for ∇2 L, namely µ ¶ 1 d dL f 2 ∇ L= r =− log r, (2.254) r dr dr 2π(1 − ν) up to an arbitrary constant, where r2 = x2 + y 2 . Integrating with respect to r and neglecting another arbitrary constant, we obtain the Love function L=−

f r2 (log r − 1) . 8π(1 − ν)

(2.255)

The displacement field is then found by substitution into (2.200) to give   2 x − y2 − (3 − 4ν) log r f ,  2r2 u= (2.256) xy 8πµ(1 − ν) r2 and the stress is recovered from (2.197) as µ ¶ f −(3 − 2ν)x3 − (1 − 2ν)xy 2 −(3 − 2ν)x2 y − (1 − 2ν)y 3 τ= . 4π (1 − ν) r4 −(3 − 2ν)x2 y − (1 − 2ν)y 3 (1 − 2ν)x3 − (1 + 2ν)xy 2 (2.257) We can hence find the maximum shear stress S defined in (2.35), by calculating the eigenvalues τi of τ : p f |τ1 − τ2 | = 4(1 − ν)2 x2 + (1 − 2ν)2 y 2 . (2.258) S= 2 2π(1 − ν)r2 Typical contours of this function are plotted in Figure 2.15 (with ν = 1/4). The shear stress becomes unbounded as the point force at the origin is approached, and we would expect the material to yield plastically on that surface at which S reaches a critical value.

84


y

x

Fig. 2.15. Contours of the maximum shear stress S created by a point force acting at the origin. Here ν = 1/4 and f /S = 0.2, 0.4, . . . , 2.0.

The response to a point force in three dimensions may be found in an analogous fashion by using the axisymmetric Love function. Such problems become very much harder if we attempt to impose any boundary conditions, and analytical solutions are usually impossible to find unless the geometry is very simple. One relatively straightforward example concerns plane strain in the half-space y > 0 with a stress-free boundary at y = 0 and a point force in the x-direction applied at the point (0, h). We recall that, for corresponding half-space problems in potential theory, the method of images provides a simple solution procedure, and it seems sensible to try and apply a suitable generalisation here. If the tensor field defined by (2.257) is denoted by τ0 (x, y), then the stress due to the point force at (0, h) is given by τ0 (x, y − h). At first sight, it seems that we can solve the half-space problem by superimposing this and the response to an identical point force at (0, −h). However, when we add these contributions together, we find a stress field in which τxy is zero at y = 0 but τyy is not. We therefore seek a solution of the form  2  ˜ ˜ ∂ A ∂2A −  ∂y 2 ∂x∂y    τ = τ0 (x, y − h) + τ0 (x, y + h) +  (2.259) , 2 2 ˜  ˜  ∂ A ∂ A − ∂x∂y ∂x2 ˜ now satisfies the two-dimensional biharmonic equation in y > 0 where A and the boundary conditions £ ¤ ˜ ˜ f x (1 + 2ν)h2 − (1 − 2ν)x2 ∂2A ∂A = 0, = (2.260) ∂y ∂x2 2π(1 − ν) (h2 + x2 )2 on y = 0. This is now precisely the half-space problem we have already


85

analysed in §2.6.8, and may be solved by substituting the right-hand side of (2.260) into (2.155) to give ½ ¡ 2 ¢ f −2hxy 2 ˜= A − (1 − 2ν)x log x + (y + h) 4π(1 − ν) x2 + (y + h)2 ¶¾ µ −1 y + h +4h(1 − ν) tan . (2.261) x This example shows, once again, how the vectorial nature of the Navier equation complicates well-known techniques like the method of images from ˜ it is potential theory. Since we still had to solve a nontrivial problem for A, debatable whether the problem was significantly simplified by introducing the image. A more direct attack using the Love function leads to (1 − ν)∇4 L = −f δ(x)δ(y − h)

y > 0,

(2.262a)

with ∂2L − ν∇2 L = 0 ∂y 2 and ∂ ∂y

µ

y = 0,

¶ ∂2L 2 − ν∇ L =0 ∂y 2

y = 0.

(2.262b)

(2.262c)

This may be solved by subtracting off the full-space solution (2.255), that is L=−

f ˜ R2 (log R − 1) + L, 8π(1 − ν)

R2 = x2 + (y − h)2 ,

(2.263)

and then taking a Fourier transform in x. 2.9.3 The Green’s tensor The solutions we have presented above concern the response of an elastic body to a body force concentrated at a point. To model a distributed body force, for example gravity, we must consider the response of the Navier equations (λ + µ) grad div u + µ∇2 u = −f (x)

(2.264)

to a more general right-hand side. The standard mathematical approach to linear differential equations like (2.264) is to use a Green’s function (Ockendon et al., 1999§????). For a scalar equation of the form Lu = −f (x),

(2.265)

86


where L is an autonomous linear differential operator, i.e. one in which x does not appear explicitly, we would define the Green’s function G(x) to be the solution of LG = −δ(x),

(2.266)

subject to suitable boundary conditions. We can then use the key property (2.249) of the delta-function to deduce that ZZZ u(x) =

G(x − ξ)f (ξ) dξ.

(2.267)

R3

is a solution of the inhomogeneous equation (2.265). To adapt the above procedure to the vector equation (2.264), one needs to define a Green’s tensor G(x) such that (λ + µ) grad div G + µ∇2 G = −δ(x)I,

(2.268)

where I = (δij ) is the identity matrix. Equation (2.268) is shorthand for the scalar equations (λ + µ)

∂ 2 Gij ∂ 2 Gki +µ = −δij δ(x), ∂xj ∂xk ∂xk ∂xk

(2.269)

in which the summation convention is invoked, and it is easy to see that G should be symmetric, that is Gij ≡ Gji . Assuming that there are no boundaries and that the stress decays sufficiently rapidly at large distances, it is intuitively reasonable to generalise (2.267) to ZZZ u(x) =

G(x − ξ)f (ξ) dξ.

(2.270)

R3

From a practical viewpoint, Gij may be interpreted physically as the displacement in the j-direction at x in response to a unit point force in the i-direction at ξ. In other words, (2.270) shows that the response to an arbitrary body force f (x) may be written as a superposition of point force responses like those constructed above in §2.9.2. In the case where the elastic medium occupies the whole space, with bounded stress at infinity, we can infer the Green’s matrix directly from the point force solutions found in §2.9.2. In plane strain, the displacement


87

caused by a unit point force in the x-direction is given by (2.256) with f = 1: G11 G12

µ 2 ¶ 1 x − y2 = − (3 − 4ν) log r , 8πµ(1 − ν) 2r2 xy 1 , = 8πµ(1 − ν) r2

(2.271a) (2.271b)

and, by symmetry, we deduce that the Green’s matrix is given by 

x2 − y 2 1  2r2 − (3 − 4ν) log r G(x, y) =  xy 8πµ(1 − ν) r2

xy r



 . y 2 − x2 − (3 − 4ν) log r 2r2 (2.272)

The corresponding three-dimensional result is derived in Exercise 2.17. Finding the Green’s matrix explicitly in a body with finite boundaries is usually impossible unless there is sufficient symmetry to allow some trick to be employed. In general, we must now solve (2.268) along with boundary conditions for G corresponding to those that are imposed on u (so that (2.270) satisfies the required boundary conditions). Since the problem is not translation invariant, we are not at liberty to move the delta-function to the origin, as we did in (2.268). Hence we can no longer assume that G is a function only of x − ξ, and we must rewrite (2.266) as LG(x, ξ) = −δ(x − ξ.

(2.273)

There are, however, situations where we can quite quickly find the response to a point force by using the Papkovich–Neuber representation (2.231). Let us consider the axisymmetric version of the half-space problem of §2.6.8 in cylindrical poalr coordinates (r, z); the half-space is z > 0 and we suppose the tractions on z = 0 are such that τrz = 0 everywhere and τzz = 0 except on the axis r = 0, where there is a normal point force P . Thus τzz kz=0 = P δxδy, where r2 = x2 + y 2 . Because these conditions are imposed on z = 0, in (2.231) we take ¡ ¢ Ψ = 0, 0, ψ(r, z) ,

φ = φ(r, z),

(2.274)

where φ, ψ are harmonic functionms. Hence 2µur = −

∂φ 1 ∂ψ − , ∂r 2 ∂r

2µu2 = 2(1 − ν)ψ −

∂φ 1 ∂(zψ) − . ∂z 2 ∂z

(2.275)

88


From the formulae of §1.11.2, the vanishing of the shear stress requires that ∂ur ∂uz + = 0 on z = 0, ∂z ∂r ∂2φ ∂φ 2 − (1 − 2ν) =0 ∂r∂z ∂r

so that

(2.276) (2.277)

there. Hence, since 2 ∂φ ∂r − (1 − 2ν)ψ is a harmonic function that vanishes on z = 0 and at infinity, it is identically zero. The normal traction is thus λ ∂ ∂u2 ∂φ ∂2φ (rur ) + (λ + 2µ) = (λ + 2µ)(1 − 2ν) − 2µ 2 r ∂r ∂z ∂z ∂z ∂ψ = (λ + µ)(1 − 2ν) . (2.278) ∂z Our problem has been reduced to finding a harmonic function ψ whose zderivative is proportional to δ(x)δ(y) on z = 0. In fact τzz =

ψ

1 P ·√ 2 2πµ r + z2

(2.279)

and uz , which is proportional to φ, does not display the non-monotonicity of Figure 2.13. We note that the full-space Green’s matrix (2.272) can be used to reduce the Navier equation on an arbitrary domain D to an integral equation satisfied on the boundary of D, and this reduction in dimension facilitates efficient numerical solution, and forms the basis of the boundary element method. To illustrate the basic ideas behind this method, we begin by recalling the identities µ ¶ ZZZ ZZ ¡ ¢ ∂v ∂u 2 2 u · ∇ v − v · ∇ u dx = u· −v· dS, (2.280) ∂n ∂n D ∂D and

ZZZ {u · (grad div v) − v · (grad div u)} dx ZZ = {(div v) u · n − (div u) v · n} dS, D

∂D

for any pair of vector fields u and v, Now, reluctantly reverting to suffix notation, we know that (λ + µ) grad div Gij + µ∇2 Gij (λ + µ) (grad div u)j + µ∇2 uj

= −δ(x − ξ) δij , ∂τjk = = 0, ∂xk

(2.281) (2.282)


89

where we have neglected the body force for simplicity. Hence, multiplying (2.281) by ui , (2.282) by Gij , then substracting and integrating, we find that ZZ © ¡ ¢ª ui (ξ) = G i (x − ξ) · (τ n (x)) − u(x) · τ i n(x − ξ) dS, (2.283) ∂D

where G i = (Gi1 , Gi2 , Gi3 )T and τ i is the associated stress field. Hence, given the traction and displacement on the boundary ∂D, we have an explicit representation of u everywhere in D. However, we do not expect both u and τ n to be given simultaneously on the boundary, so let us assume for example that the surface traction τ n is prescribed. In this case, (2.283) becomes an integral equation for u(ξ) as ξ tends to the boundary of the elastic solid. To solve this integral equation rather than the Navier equation numerically is the basic idea behind the boundary element method. This approach may prove advantageous compared to the more commonly used finite element method, for example, since only the boundary of the solid needs to be discretised, rather than the whole body. On the other hand, it also brings its own share of difficulties. For example, the matrices resulting from the discretisation are usually full, and some care must be taken to handle the singularities in G.

2.9.4 Point incompatibility We have shown above how a point force may be described mathematically by incorporating a delta-function into the Navier equation. We then found that the singular solutions so produced are very useful, since they may be superimposed to describe the response to an arbitrary body force. Two further avenues for investigation now suggest themselves. We could use physical motivations other than point forces to suggest useful new singular solutions, or we could ask ourselves whether the introduction of delta-functions elsewhere in the formulation of linear elasticity might have interesting physical interpretations. As an example of the former, suppose we consider the stress field in plane strain when a cavity of small radius a is maintained by a pressure P in an otherwise traction-free material. This is a limiting case of the gun barrel problem solved in §2.6.5 and, when we let b → ∞ and a → 0 in (2.119), we recall that the radial displacement is given by ur =

P a2 . 2µr

(2.284)

The same displacement field may also be produced by superimposing four

90


y

f f

a

f

x

f

Fig. 2.16. Schematic of four point forces.

point forces, as indicated in Figure 2.16, and letting the distance between them tend to zero (Exercise 2.25). However, it is more illuminating to consider the incompatibility of the displacement field (2.284). Since the vanishing of ∇4 A is the condition for compatibility, we begin by trying to compute this quantity for (2.284). It is easy to see that A = −a2 P log r and hence, using (2.251), that ∇2 A = −2πa2 P δ(x)δ(y).

(2.285)

Thus the incompatibility associated with the displacement field (2.284) is given by ¡ ¢ ∇4 A = −2πa2 P δ 00 (x)δ(y) + δ(x)δ 00 (y) , (2.286) with a suitable definition of the derivative of a delta-function (Ockendon et al., 1999§????) or keener. Hence (2.284) satisfies the compatibility relations everywhere except at the origin, where the localised source of incompatibility is known as a nucleus of strain (Love, 1944, Article 132). Note that no traction need be applied away from the origin to maintain to maintain the stress field associated with (2.284), and the material is therefore said to be in a state of self-stress. More generally, in Exercise 2.19 it is shown that the compatibility conditions (2.180) are the six components of the tensor equation ηij = ²ikl ²jmn

∂ 2 emk ≡0 ∂xn ∂xl

(2.287)

EXERCISES

91

where ²ijk is the alternating symbol and we sum over k, l, m and n. We therefore define ηij as the incompatibility tensor. In plane strain, for example, the zz-component of (2.287) is ∂ 2 exy ∂ 2 eyy ∂ 2 exx 1−ν 4 − 2 + = ∇ A = 0, ∂y 2 ∂x∂y ∂x2 2µ

(2.288)

which reproduces (2.180a) and shows explicitly how ∇4 A is related to compatibility. Thus the right-hand side of (2.286) can be thought of as a point singularity in the compatibility tensor, and this prompts us to wonder what the physical significance might be of introducing other point singularities on the right-hand side of (2.287). For example, the physical interpretation of solutions of ∂ 2 exy ∂ 2 eyy ∂ 2 exx − 2 + = δ(x)δ 0 (y) ∂y 2 ∂x∂y ∂x2

(2.289)

has fundamental implications for the theory of metal plasticity, to which we will return in Chapter 9. Exercises 2.1 Use strips of stiff paper and sticky tape to construct the model shown in Figure 2.3(a). Now when you pull it in the direction shown, it expands in the transverse direction, as indicated by Figure 2.3(b). It therefore has negative Poisson’s ratio. 2.2 An elastic membrane is stretched to an isotropic tension T such that its height is given by z = w(x, y). Assuming that the membrane is nearly horizontal (so that |∇w| ¿ 1), show that T must be spatially uniform and that w must satisfy Laplace’s equation. [A rigorous derivation of this model as an asymptotic limit of the Navier equation will be given in Chapter 6.] 2.3 A uniform cylindrical bar is held with its axis along the z-axis and its boundary pinned. If the gravitational acceleration g acts in the −z direction, show that the vertical displacement w(x, y) satisfies ρg ∇2 w = µ in the cross-section D of the bar, with w = 0 on ∂D. 2.4 If D is the ellipse x2 /a2 + y 2 /b2 < 1, show that the solution of (2.52), (2.53) is µ ¶ x2 y 2 a2 b2 1− 2 − 2 . φ= 2(a2 + b2 ) a b

92


Hence show that the torsional rigidity of a bar with uniform elliptical cross-section is T =

πµa3 b3 . a2 + b2

2.5 Derive the expression (2.62) for the torsional rigidity of a tubular bar. By contour deformation show that, if (2.63) holds when C is ∂Di , then it holds for all simple closed paths C contained in D. 2.6 By separating the variables in polar coordinates, show that the problem (2.74) for a cut tubular bar has the solution ¡ 2 ¢ b − a2 log (r/b) b2 − r2 φ= + 2 2 log (b/a) ∞ X £ ¤ £ ¤ + Cn cosh kn (θ − π) sin kn log (b/r) , n=1

where nπ , kn = log (b/a)

£ ¤ 4 b2 − (−1)n a2 log2 (b/a) £ ¤. Cn = − nπ cosh(kn π) n2 π 2 + 4 log2 (b/a)

Use (2.55) to determine the torsional rigidity ¡ ¢ ¢ ¡ ¢¤ πµ b2 − a2 £¡ 2 a + b2 log (b/a) − b2 − a2 R= 2 log (b/a) ¤ £ ¤ ∞ £ 16µ log4 (b/a) X b2 − (−1)n a2 tanh nπ 2 / log (b/a) − . £ ¤2 π n2 π 2 + 4 log2 (b/a) n=1 Now suppose that the tube is thin, so that b/a = 1 + ², where ² ¿ 1. By expanding the above expression for small ², show that ½ µ ¶ ¾ 1 93ζ(5) 2π 4 3 µa ² 1 − + R∼ ² + ··· , 3 2 π6 where ζ is the Riemann zeta function (Abramowitz & Stegun, 1972§23.2). 2.7 If z = x + iy and z¯ = x − iy, show that ∇2 ψ = 4

∂2ψ . ∂z∂ z¯

Deduce that the general solution of Laplace’s equation may be written in the form 1 ψ = {f (z) + h(¯ z )} , 2

EXERCISES

93

where f and h are arbitrary analytic functions. If ψ takes only real values, show that h(¯ z ) = f (z) and hence that © ª ψ = Re f (z) . Now solve the biharmonic equation by writing ∇2 A = ψ in the form 4

ª ∂2A 1© = f (z) + f¯(¯ z) ∂z∂ z¯ 2

and integrating with respect to z and z¯ to obtain (2.90). 2.8 Show that, for radially symmetric problems in two dimensions, A is generally a constant plus a combination of log r, r2 , and r2 log r. Show how the same conclusion may be reached by reasoning in terms of analytic functions only. Now find all separable solutions of the biharmonic equation in polar coordinates of the form A = f (r) cos(mθ), with m = 0, . . . , 3, or A = f (r)θ cos θ. In each case, find the restrictions on f (r) for the stress field to vanish at infinity. the stress field in an elastic gun barrel of inner and outer radius a and b, respectively. 2.9 Suppose that in some basis the stress tensor is diagonal: τ = diag (τk ). Show that the stress in the direction of the unit vector m on a surface with unit normal n is τi mi ni . Hence show that the maximum shear stress S satisfies the constrained optimisation problem max

S=

3 X

τi mi ni ,

i=1

subject to

3 X i=1

m2i

= 1,

3 X i=1

n2i

= 1,

3 X

mi ni = 0.

i=1

Assuming that the principal stresses τi are distinct, use the method of Lagrange multipliers to show that the extreme values occur when τ1 − τ2 1 , m3 = n3 = 0, m21 = m22 = n21 = n22 = √ , S = ± 2 2 1 τ2 − τ3 or m1 = n1 = 0, m22 = m23 = n22 = n23 = √ , S = ± , 2 2 1 τ3 − τ1 or m2 = n3 = 0, m23 = m21 = n23 = n21 = √ , S = ± . 2 2

94


2.10 Prove the identity © ¡ ¢ ¡ ¢ ª u∇4 v − v∇4 u ≡ div u∇ ∇2 v − v∇ ∇2 u + ∇2 u∇v − ∇2 v∇u . Hence show that, if the functions Ai (x) and constants λi (i = 1, 2, . . . ) satisfy ∇4 Ai = λi Ai in some bounded region V , then ZZZ ZZ (λi − λj ) Ai Aj dx =

½ ∂Aj ∂Ai ∇2 Aj − ∇2 Ai ∂n ∂n ∂V ¾ ¡ ¢ ¡ ¢ ∂ ∂ ∇2 Ai − Ai ∇2 Aj dS. +Aj ∂n ∂n

V

Deduce that the biharmonic operator has real eigenvalues and orthogonal eigenfunctions when the boundary conditions are ∂A = 0, ∂n ∂ ¡ 2 ¢ ∂A = ∇ A , ∂n ∂n

(i) A = or (iii) 2.11

or (ii) A = ∇2 A = 0, or (iv) ∇2 A =

∂ ¡ 2 ¢ ∇ A = 0. ∂n

(a) Assuming y > 0, show that the inverse Fourier transforms of e−|k|y and i sign(k)e−y|k| are given by Z ∞ y 1 e−ikx e−|k|y dk = , f1 (x) = 2 2π −∞ π (x + y 2 ) and i f2 (x) = 2π

Z

∞

−∞

e−ikx sign(k)e−|k|y dk =

x , + y2)

π (x2

respectively. (b) Hence show that (2.154) may be written as µ ¶ ∂ b τbxx = −P (k) 1 + y fb1 (k), ∂y [ ∂f1 (k), τbxy = Pb(k)y ∂x µ ¶ ∂ b τbyy = −P (k) 1 − y fb1 (k). ∂y Using the fact that the inverse Fourier transform of a product

EXERCISES

95

fb(k)b g (k) is the convolution of f and g, that is Z ∞ (f ∗ g)(x) = f (x − s)g(s) ds, −∞

deduce (2.155). (c) Show from (2.157) that [ ∂v 2µ = Pb(k) ∂x

Ã

! [ ∂f1 y (k) − 2(1 − ν)fb2 (k) , ∂x

and hence deduce (2.158). 2.12 By making a suitable substitution, write the second term in the integral (2.158) as Z Z 2 − ν ∞ P (x − yt)t dt (2 − ν)y 2 ∞ P (x − s)s ds . − 2 =− π π (1 + t2 )2 −∞ (s2 + y 2 ) −∞ Deduce that, provided P (x) is bounded, this term tends to zero as y → 0. Show further that the first term in (2.158) may be written as µZ −ε Z ε Z ∞ ¶ 1−ν P (s)(x − s)3 ds − + + . π ((x − s)2 + y 2 )2 −∞ −ε ε Show that the middle integral is approximately equal to Z (1 − ν)P (x) ε/y t3 dt − 2, π −ε/y (1 + t2 ) and explain why this integral tends to zero as y tends to zero with y ¿ ε ¿ 1. Finally deduce that, as y → 0, (2.158) tends to (2.159), where the Hilbert transform is defined by µZ −ε Z ∞ ¶ Z 1 ∞ P (s) ds 1 P (s) ds H[P ](x) = − = lim + . π −∞ s − x π ε↓0 s−x −∞ ε 2.13 Reconsider the torsion problem of §2.4 and the stress function φ defined by (2.50). Using two of the compatibility relations (2.180), deduce successively that ∇2 φ = f (y) and that ∇2 φ = g(x), hence that ∇2 φ = const. [This illustrates that two compatibility conditions that are not independent may nevertheless both be necessary to determine the stresses.] 2.14 STILL NEEDED Derive expressions (2.14) for the stress components in terms of A in plane polars.

96

2.15


(a) Derive the expression (2.171) for the modified Airy stress function outside a circular underground tunnel. (b) Prove the identity ∂eθθ ∂ 2 ur ∂erθ + ur = −r2 + rerr + 2r ∂θ2 ∂r ∂θ

(2.290)

relating the strain components to the radial displacement ur in plane polar coordinates (r, θ). Deduce that the right-hand side must be zero if ur is to be a 2π-periodic function of θ. (c) With the stress components given by (2.166), show that the right-hand side of (2.290) is µ − ρgh

¶ ¶ µ a2 8(1 − ν)A cos θ + (1 − 2ν)r + r 1 + 2 log a ¶ µ 8(1 − ν)B 2 − ρg(1 − 2ν)a sin θ, + 1 + 2 log a

and hence obtain the expressions (2.172). 2.16 Return to the tunnel example of §2.6.9, this time assuming a far-field stress distribution given by τxx = τxy = 0, τyy = ρg(y − H). [This may correspond to a hole in a vertical column, for example.] In this case, show that (2.166) becomes ˜ ˜ 1 ∂2A 1 ∂A , + 2 r2 ∂θ r!∂r Ã ˜ sin 2θ ∂ 1 ∂A = ρg (r sin θ − H) − , 2 ∂r r ∂θ

τrr = ρg sin2 θ (r sin θ − H) + τrθ

τθθ = ρg cos2 θ (r sin θ − H) +

˜ ∂2A , ∂r2

and that the boundary conditions on r = a are µ ¶ H a ρga2 ˜ H + aθ cos θ + cos 2θ − sin 3θ + A cos θ + B sin θ, A(a, θ) = 2 2 12 ´ A ˜ θ) ∂ A(a, ρga ³ a B = H + aθ cos θ + H cos 2θ − sin 3θ + cos θ + sin θ, ∂r 2 4 a a where A and B are arbitrary constants. Use the fact that ur is singlevalued to show that A = 0,

B = −νρga3 /4

EXERCISES

97

[hint: use (2.290)], so that ½ ³ µ ¶ r´ a2 H 1 + log + rθ cos θ + H 1 − cos 2θ a 2r µ ¶ ¶ µ 4 ¾ 1 r a2 a a2 + − r(1 − 2ν) log − sin θ + sin 3θ , 2 a r 6r3 4r

2 ˜ = ρga A 2

and deduce that the hoop stress is given by ¶ µ 1 − 2ν sin θ + sin 3θ τθθ = −ρgH (1 + 2 cos 2θ) + ρga 2 on r = a. Note the contrast with the stress distribution caused by a hydrostatic far-field stress. 2.17 We define the three-dimensional Fourier transform of a function f (x, y, z) by Z ∞Z ∞Z ∞ b f (x, y, z)ei(kx+`y+mz) dxdydz f= Z−∞ Z Z −∞ −∞ = f (x)ei(k·x) dx, R3

where k = (k, `, m). Show that the Fourier transform of the Green’s matrix in three dimensions is given by ³ ´ Gb = Gbij =

1 λ + µ kkT I − , µ|k|2 µ (λ + 2µ) |k|4

and, by inverting the transform, deduce that µ ¶ xi xj λ+µ 1 λ + 3µ Gij = δij + . 8πµ (λ + 2µ) |x| λ + µ |x|2

(2.291)

2.18 Show that the axisymmetric Love stress function corresponding to a vertical point force at the origin is given by L=

|x| . 8π (1 − ν)

Hence, show that the corresponding displacement is µ ¶ 1 xz yz z2 (G31 , G32 , G33 ) = , , 3 − 4ν + , 16πµ |x| (1 − ν) |x|2 |x|2 |x|2 and deduce (2.291) by symmetry.

98


2.19 Prove the identity ²mjk

∂ 2 ui ≡0 ∂xk ∂xj

for all i and m, where ²mjk is the usual alternating symbol, and show that this is equivalent to ∇×(∇ui ) ≡ 0. Deduce that ²mjk ²nil

∂ 2 eij ≡0 ∂xl ∂xk

for all m and n, where eij is the linearised strain tensor. Hence obtain the compatibility conditions (2.180) and show that they are equivalent to (2.181). 2.20 Show that the compatibility conditions can be written in terms of the stress components as µ 2 ¶ ∂ 2 τyy ∂ 2 τxy ∂ 2 τxx ν ∂ ∂2 + −2 = + Tr (τ ), ∂y 2 ∂x∂y ∂x2 1 + ν ∂x2 ∂y 2 µ ¶ ∂τxy ∂τyz ∂ ∂τxz ν ∂2 ∂ 2 τxx = + − + Tr (τ ), ∂y∂z ∂x ∂y ∂z ∂x 1 + ν ∂y∂z the other relations being obtained by permuting x, y and z. 2.21 Show that for axisymmetric elastic equilibrium, (2.180a), (2.180c) and (2.180b) imply that ∂eθθ , ∂rµ ¶ ∂ 1 ∂ (rerz ) . = 2 ∂z r ∂r

err − eθθ = r ∂2 (err + eθθ − ezz ) + ∇2 ezz ∂z 2

SOMETHING MISSING HERE... √ 2.22 Verify (2.251a) by writing |x| = limε→0 x2 + ε2 and noting that ´ d2 ³p 2 ε2 2 = x + ε , dx2 (x2 + ε2 )3/2 and

Z

∞

ε2 dx

Z

∞

ds

= 2. (1 + s2 )3/2 ¡ ¢ Repeat this procedure with log x2 + y 2 + ε2 to derive (2.251b) and find a similar derivation of (2.251c). 2.23 Show that µ 2 ¶ ¢ ¡ ¢¤ ¡ ¢ ∂ ∂ 2 £¡ 2 + 2 x + y 2 log x2 + y 2 = 8 + 4 log x2 + y 2 2 ∂x ∂y −∞ (x2

+

ε2 )3/2

=

−∞

EXERCISES

99

¢ ¢ ¡ ¡ and deduce that, if A = (1/8π) x2 + y 2 log x2 + y 2 , then A satisfies ∇4 A = δ(x)δ(y). 2.24 The stress components due to a point force f in the x-direction satisfy ∂τxx ∂τxy + = −f δ(x)δ(y), ∂x ∂y

∂τxy ∂τyy + ∂x ∂y

=0

plus the compatibility condition ∂ 2 τxy ∂ 2 τyy ∂ 2 τxx − 2 − ν∇2 (τxx + τyy ) = 0. + ∂y 2 ∂x∂y ∂x2 Take a two-dimensional Fourier transform (defined as in Exercise 2.17) of these relations to show that kb τxx + `b τxy = −if,

kb τxy + `b τyy = 0, ¡ 2 ¢ 2 −` τbxx + 2k`b τxy − k τbyy + ν k + ` (b τxx + τbyy ) = 0. 2

2

Hence obtain τbxx =

ikf 1−ν

τbxy =

i`f 1−ν

τbyy =

−ikf 1−ν

µ

¶ ν k 2 + 2`2 , − k 2 + `2 (k 2 + `2 )2 µ ¶ ν `2 − , k 2 + `2 (k 2 + `2 )2 µ ¶ ν `2 − , k 2 + `2 (k 2 + `2 )2

the inverse Fourier transform of which is (2.257). 2.25 Write down displacement field for four point forces as in Figure 2.16. Let a → 0 and show that the displacement tends to u=

1 − 2ν af er , 1 − ν 2πµ r

which agrees with (2.284) if f = πa(1 − ν)/(1 − 2ν).

3 Linear Elastodynamics

3.1 Introduction This chapter concerns simple unsteady problems in linear elasticity. As noted in §1.10, the unsteady Navier equation (1.44) bears some similarity to the familiar scalar wave equation governing small transverse displacements of an elastic string or membrane. We therefore start by reviewing the main properties of this equation and some useful solution techniques. This allows us to introduce, in a simple context, important concepts such as normal modes, plane waves and characteristics that underpin most problems in linear elastodynamics. In contrast with the classical scalar wave equation, the Navier equation is a vector wave equation, and this introduces many interesting new properties. The first that we will encounter is that the Navier equation supports two distinct kinds of plane waves which propagate at two different speeds. These are known as P -waves and S -waves, and correspond to compressional and shearing oscillations of the medium respectively. Considered individually, both P - and S -waves behave very much like waves as modelled by the scalar wave equation (1.65). In practice, though, they very rarely exist in isolation since any boundaries present inevitably convert P -waves into S -waves and vice versa. We will illustrate this phenomenon of mode conversion in §3.2.5 by considering the reflection and refraction of waves at a plane boundary. In two-dimensional and axisymmetric problems, we found in Chapter 2 that the steady Navier equation may be transformed into a single biharmonic equation by introducing a suitable stress function. In Sections §3.3 and §3.4 we will find that the same approach often works even for unsteady problems, and pays dividends when we come to analyse normal modes in cylinders and spheres. In §3.5, we consider some initial-value problems for elastic wave propaga100

3.2 Normal modes and plane waves

101

tion using the ideas of characteristics and fundamental solutions. Finally, we will discuss the interesting phenomena that can occur when elastic waves are generated by moving acoustic sources. Before we begin, let us briefly recall on a study of the dynamic Navier equations some familiar wave propagation models for elastic strings and membranes. The simplest such model describes small oscillations of an elastic string of line density % under a tension T . It is well known that the transverse displacement w (x, t) satisfies the partial differential equation %

∂2w ∂2w = T , ∂t2 ∂x2

(3.1)

known as the one-dimensional wave equation. The derivation and validity of this equation will be discussed in Chapter 4. If we instead consider an elastic membrane with surface density ς stretched across the (x, y)-plane under a uniform tension T , then the appropriate generalisation of (3.1) is ς

∂2w = T ∇2 w, ∂t2

(3.2)

where ∇2 = ∂ 2 /∂x2 + ∂ 2 /∂y 2 is the two-dimensional Laplacian. The threedimensional version of (3.2) governs many familiar wave propagation problems including acoustic waves and light waves. There are two ways in which we can approach (3.1) and (3.2) mathematically. One possibility is to try to construct a general solution that will apply whatever mechanism is driving the waves. However, this strategy is only useful in practice when the general solution can be found explicitly and also takes a sufficiently simple form for realistic boundary conditions to be imposed. Alternatively, we can attempt to describe the solution as a linear combination of elementary solutions, as in the method of normal modes in classical mechanics. This second approach can always be applied in principle and ties in with the first when we view the general solution as a superposition of elementary solutions. We therefore begin the next section by discussing some elementary solutions of (3.1) and (3.2).

3.2 Normal modes and plane waves 3.2.1 Normal modes A normal mode of a dynamical system is a motion in which the whole system oscillates at a single frequency ω. We can seek normal modes af an elastic

102

Linear Elastodynamics

string by trying solutions of (3.1) in the form w(x, t) = f1 (x) cos(ωt) + f2 (x) sin(ωt),

(3.3a)

which is often written in the compact form £ ¤ w(x, t) = Re f (x)e−iωt ,

(3.3b)

where f (x) = f1 (x) + if2 (x). Substitution of (3.3) into (3.1) reveals that f (x) must satisfy the ordinary differential equation d2 f %ω 2 + f = 0, dx2 T

(3.4)

whose general solution is r f (x) = A cos(kx) + B sin(kx),

where k = ω

% . T

(3.5)

Thus (3.3b), with f (x) given by (3.5), is a solution of the wave equation for any value of the frequency ω. Particular values of ω may be selected by applying suitable boundary conditions. As a simple example, consider a string fixed at its two ends x = 0 and x = a, so we require w to satisfy w(0, t) = w(a, t) = 0. Clearly f must likewise satisfy f (0) = f (a) = 0,

(3.6)

and by applying these to (3.5) we discover that nonzero solutions for f exist only if ω takes specific values, known as eigenvalues, namely s nπ T ωn = , (3.7) a % where n is any integer. These are known as the natural frequencies of the system; they represent the frequencies at which the string will oscillate in the absence of any external forcing. The corresponding displacements ³ ³ nπx ´ ´ wn (x, t) = Re Bn sin e−iωn t (3.8) a are called the normal modes, and we can see that there is a countably infinite number of them. For waves on a string these modes are manifested as the harmonics that are familiar to anyone who plays a stringed musical instrument; many people will also recognise the properties predicted by (3.7), namely that the frequency increases as the tension increases and as the density decreases.


103

Now suppose we are given the initial displacement and velocity of the string, say w(x, 0) = w0 (x),

∂w (x, 0) = v(x). ∂t

(3.9)

These may be satisfied by trying a linear combination of normal modes, that is w(x, t) =

∞ X n=1

sin

³ nπx ´ a

{An cos(ωn t) + Bn sin(ωn t)} ,

(3.10a)

where the coefficients are found by Fourier analysis of the initial data: Z Z a ³ nπx ´ ³ nπx ´ 2 a 2 An = w0 (x) sin dx, Bn = v(x) sin dx. a 0 a ωn a 0 a (3.10b) This analysis, although very straightforward, illustrates the following important points that will be useful in analysing more complex models later in this chapter. • We will always be able to look for time-harmonic solutions of the form (3.3b) whenever the problem we are solving is both linear and autonomous in t, in other words, when there is no explicit dependence on t. • When looking for normal modes, we might as well have ignored the “Re” in (3.3b), performed all the calculations, and then taken the real part right at the end. This approach works because all the usual linear operations, such as differentiation with respect to x or t or multiplication by a real constant, commute with taking the real part. Henceforth, in this chapter, we will therefore for simplicity follow the convention of assuming the real part. We note, though, that this would be a very dangerous procedure were we to be considering a nonlinear model! • Although normal modes may at first glance appear to be rather special solutions of the wave equation, it can be shown that the general solution, subject to simple homogeneous boundary conditions, may be written as a superposition of normal modes. • When a → ∞, it may be shown that (3.10) generalises to the Fourier transform representation Z ∞ w(x, t) = sin (kx) {A(k) cos(kct) + B(k) sin(kct)} dk, (3.11a) 0

104


where c2 = T /% and Z 2 ∞ A(k) = w0 (x) sin(kx) dx, π 0

2 B(k) = πkc

Z

∞

v(x) sin(kx) dx. 0

(3.11b)

We can extend these ideas to the two-dimensional wave equation (3.2) by seeking solutions of the form w(x, y, t) = A(x, y)e−iωt ,

(3.12)

and we find that A must satisfy the Helmholtz equation ∇2 A + k 2 A = 0,

(3.13) p where k = ω ς/T . For example, to find the normal modes and frequencies of a drum whose skin occupies a region D in the (x, y)-plane, we must solve (3.13) in D subject to A = 0 on the boundary ∂D. This is another eigenvalue problem: A ≡ 0 is always a possible solution and special values of k must be sought such that A may be nonzero. For many simple shapes, these can be found by separating the variables. For example, we can see at a glance that the normal modes and natural frequencies of the rectangular membrane {0 < x < a, 0 < y < b} are ³ mπx ´ ³ nπy ´ wm,n (x, y, t) = sin sin e−iωm,n t , (3.14a) a b µ ¶ π 2 T m2 n2 2 ωm,n = + 2 , (3.14b) ς a2 b where m and n are arbitrary integers. Thus there is a doubly infinite set of normal modes in this case. In general, there is one infinite set of frequencies for each spatial dimension in the problem. Next we consider radially-symmetric vibrations of a circular drum, so that A is a function of the plane polar distance r and (3.13) reads d2 A 1 dA + k 2 A = 0, (3.15a) + dr2 r dr to be solved subject to A(a) = 0. If we set ξ = kr, then the problem for A becomes d2 A 1 dA + +A=0 dξ 2 ξ dξ

with A(ka) = 0.

(3.15b)

This is Bessel’s equation of order zero, and it is easy to see that A(ξ) oscillates for large ξ, because the first and last terms in (3.15) will be approximately in balance. Moreover, for small ξ, we can spot that the first


105

Jn (ξ) 1

0.6

n=0 1

0.4

2

0.8

0.2 5

10

15

20

25

30 ξ

15

20

25

30 ξ

-0.2 -0.4

Yn (ξ)

n=0 1 2

0.4 0.2 5

10

-0.2 -0.4 -0.6 -0.8

Fig. 3.1. Plots of the first three Bessel functions Jn (ξ) and Yn (ξ) for n = 0 (solid), n = 1 (dashed), n = 2 (dot-dashed).

and second terms in (3.15) both balance and dominate the third term when A = log ξ. This suggests that the solution is either well-behaved or logarithmically singular as ξ → 0, and this conclusion may easily be confirmed using Frobenius’ method (see Exercise ??). The bounded solution with A(0) = 1 is denoted by J0 (ξ) while the singular solution with A ∼ (2/π) log ξ is called Y0 (ξ), and these important functions are plotted in Figure 3.1. Here, we require the amplitude A to be bounded at the centre of the drum r = 0 and must therefore choose A = const. J0 (ξ). The condition at r = a is then satisfied if ka = ξ0,i

(i = 1, 2, . . . ),

(3.16a)

where ξ0,1 < ξ0,2 < · · · are the zeros of J0 (ξ). As indicated in Figure 3.1, there are an infinite number of these zeros and ξ0,n → ∞ as n → ∞. The natural frequencies of the drum for radially-symmetric modes are thus given

106


by ξ0,i ωi = a

r

T ς

(n = 1, 2, . . . ).

(3.16b)

Unlike the modes on a string, the natural frequencies here are irrational multiples of each other. In musical terms, the harmonics are out of tune with each other, and this gives rise to the characteristic sound of a drum, which is quite different from that of a piano or a violin, for example. If we drop the assumption of radial symmetry, then the Helmholtz equation becomes ∂ 2 A 1 ∂A 1 ∂2A + + k 2 A = 0, + ∂r2 r ∂r r2 ∂θ2

(3.17)

in plane polar coordinates (r, θ). Again we impose A = 0 on r = a, and now we insist also that A must be a 2π-periodic function of θ. This restricts to seeking separable solutions of the form ¡ ¢ A(r, θ) = f (r) C1 cos(nθ) + C2 sin(nθ) , (3.18) where n is an integer, and it follows that f satisfies r2

¢ df ¡ 2 2 d2 f +r + k r − n2 f = 0 2 dr dr

with f (a) = 0.

(3.19a)

Setting ξ = kr as before, we find that (3.19a) becomes ξ2

¢ d2 f df ¡ 2 + ξ − n2 f = 0 +ξ 2 dξ dξ

with f (ka) = 0.

(3.19b)

This is Bessel’s equation of order n, whose two linearly dependent solutions are denoted Jn (ξ) and Yn (ξ); the cases n = 0, 1, 2 are plotted in Figure 3.1. For any integer n, Yn (ξ) is logarithmically singular as ξ → 0, so we must choose f = const. Jn (ξ). The condition at r = a is then satisfied if ka = ξn,i

(i = 1, 2, . . . ),

(3.20)

where ξn,1 < ξn,2 < · · · are the zeros of Jn (ξ). There are an infinite number of these zeros and ξn,i → ∞ as i → ∞. The natural frequencies of the drum are thus given by 2 ωn,i =

2 T ξn,i , a2 ς

(3.21)

where n and i are arbitrary integers, so there is a doubly-infinite family of normal modes that depend on the two spatial coordinates (r, θ).


107

3.2.2 Waves in the frequency domain As suggested by (3.11), when looking for solutions of (3.1) on an infinite domain, it is appropriate to seek travelling waves of the form ¡ ¢ w(x, t) = A exp i(kx − ωt) , (3.22) where A is a constant, rather than seeking a discrete set of normal modes. This only works because (3.1) is autonomous in both x and t; if, say, % depended on x in (3.1), then (3.22) would need to be generalised to w(x, t) = A(x) exp (−iωt) .

(3.23)

Such waves with harmonic time-dependence are usually referred to as existing in the frequency domain. The real constants |A|, k and ω represent the amplitude, wavenumber and frequency, respectively, of the wave (3.22). The wavenumber is related to the wavelength λ by k = 2π/λ, so large values of k correspond to short waves and vice versa. We can alternatively write (3.22) in the form ¡ ¢ ω w(x, t) = A exp ik(x − ct) , where c = (3.24) k is known as the phase velocity. From (3.24) we can easily see that c represents the speed at which wave crests propagate. We can also see that the minus sign included by convention in (3.22) ensures that the wave propagates in the positive x-direction when ω and k are both positive. We can generalise this approach to the two-dimensional wave equation (3.2) by looking for a solution of the form ³¡ ¡ ¢ ¢´ w(x, y, t) = A exp i(k1 x + k2 y − ωt) = A exp i k · x − ωt , (3.25) where we define the wave vector k = (k1 , k2 )T . By writing (3.25) in the form ³¡ ¢´ k·x w = A exp i |k|X − ωt , where X = , (3.26) |k| we observe that it represents a harmonic wave travelling in the direction of k at speed ω/|k|. The phase velocity, at which the wave crests propagate, is thus given by c=

ωk . |k|2

(3.27)

When we substitute (3.22) into (3.1), we find that the amplitude can be

108


non-zero only if ω 2 = k 2 T /%.

(3.28a)

This so-called dispersion relation tells us how the frequency of any given wave depends on its wavelength. We can also view (3.28a) as an equation for the phase velocity, namely p c = ± T /%, (3.28b) which tells us that all waves move at the same speed, irrespective of their wavelength. Waves with this property are called non-dispersive, to distinguish them from dispersive waves in which c varies with k. Similarly, substitution of (3.25) into the two-dimensional wave equation (3.2) leads to the dispersion relation ω 2 = |k|2 T /ς.

(3.29)

The phase speed |c| = c =

p

T /ς

(3.30)

is still independent of the wave-vector k, so these two-dimensional waves are also non-dispersive. As in (3.11), the general solution of the wave equation on an infinite domain may be written as a linear superposition of harmonic waves, travelling in both directions, in the form of a Fourier integral. In other words, we can write Z ∞Z ∞ n o w(x, y, t) = A(k1 , k2 ) exp i (k1 x + k2 y − ω(k1 , k2 )t) −∞ −∞ n o +B(k1 , k2 ) exp i (k1 x + k2 y + ω(k1 , k2 )t) dk1 dk (3.31a) 2, or

ZZ w(x, t) =

n o A(k) exp i (k · x − ω(k)t) R2 n o +B(k) exp i (k · x + ω(k)t) dk,

(3.31b)

where ω(k) is given by the dispersion relation (3.29) and the amplitude functions A(k) and B(k) can be found from the Fourier transform of the initial conditions. By writing the displacement in the form (3.31b), we emphasise that this and all the other results in this Section apply also to the wave equation in three space dimensions.


109

3.2.3 Scattering Scattering refers generally to the problem of irradiating a target with a known incoming wave-field and trying to determine the resultant scattered field that is the result of the presence of the target. This idea is important, for example, in tomography and seismic testing, where we try to infer, non-invasively, the properties of some submerged features by measuring the scattered wave-fields that they produce. The basic idea is illustrated by the simple problem of an elastic string stretched along the x-axis with a point mass m attached at the origin. Exercise ?? shows that this leads to the boundary conditions w(0−, t) = w(0+, t),

∂w ∂w m ∂2w (0+, t) − (0−, t) = ∂x ∂x T ∂t2

(3.32)

at x = 0, where T is the tension in the string. Now suppose we send in a known incident wave of the form w = ei(kx−ωt) from x = −∞, where we assume that ω > 0 and k > 0 so that the wave is travelling in the positive x-direction. We can work in the frequency domain to write the resulting displacement field in the form w(x, t) = f (x)e−iωt , where ½ ikx e + cR e−ikx x < 0, f (x) = (3.33) cT eikx x > 0. Thus, apart from the prescribed incoming wave, the scattered wave-field is outgoing from the target, with reflection and transmission coefficients cR and cT that can be determined from the boundary conditions (3.32), as shown in Exercise ??. As a very simple example of a tomography problem we could, for instance, work out the size of the mass by measuring the amplitude cR of the reflected waves. The extension to higher-dimensional problems present us with a serious mathematical challenge. Indeed, even the solution of Helmholtz’ equation (3.13) to modal plane wave incidence as a scatterer on which, say A or ∂A ∂n vanishes is beyond the scope of this book. The two principal difficulties are: (a) Assuming an incident wave eikx , as in (3.33), we have to solve a Helmholtz equation for the scattered wave A˜ = A − eikx , in which A˜ involves eikx in the boundary condition on the scatterer. This means that there is no simple solution by separation of variables, even when the scatterer is a circle. (b) In the far-field, it is not good enough simply to say that A˜ → 0 as we did in many of the electrostatic problems of Chapter 2. We now need to capture the physical requirement that A˜ be the “outgoing” from

110


the scatterer, again as in (3.33). Now it can be shown† that, as the radial coordinate r tends to infinity, all possible solutions of (3.13) ±ikr take the form A˜ ∼ A˜± (θ) er1/2 and, since A˜ m ust be multiplied by e−iωt to find the scattered wave, only the positive sign is appropriate. This is equivalent to saying that A˜ satisfies the Sommerfeld radiation condition ¡ 1¢ ∂ A˜ ik ˜ as r → ∞. − A = o r− 2 ∂r r

(3.34)

Then the canonical tomographic problem for (3.13) is to reconstruct the shape of the scatterer given the “directivity function” A˜+ (θ). We note that, in the absence of any incident field, a circular boundary can radiate waves in which, from (3.18), A depends only on r and is a linear condition of J0 (kr) and Y0 (kr). Using the fact that J0 (kr) ± iY0 (kr) tends q to

2 ±i(kr−(π/4)) πkr e

as r → ∞, we see that A must tend to a multiple of (1)

the Helmholtz function H0 (kr) = J0 (kr) + iY0 (kr) as r → ∞. 3.2.4 P-waves and S-waves Now we examine how the ideas developed above for scalar wave equations may be applied to the unsteady Navier equation (1.44). We begin by seeking travelling-wave solutions in the form n o u = a exp i (k · x − ωt) , (3.35) where the complex amplitude a, the wave-vector k and frequency ω are all constant. It is very helpful to use a little hindsight, or Exercise 3.5, to notice that, for any vector a and nonzero k, there is a unique vector B and a scalar A satisfying a = Ak + B×k,

k · B = 0.

(3.36)

Using this decomposition, we find that n o ∇2 u = − |k|2 (Ak + B×k) exp i (k · x − ωt) , n o grad div u = −A |k|2 k exp i (k · x − ωt) , n o ∂2u 2 = −ω (Ak + B×k) exp i (k · x − ωt) . ∂t2 † This involves the use of the WKB method ?).

(3.37a) (3.37b) (3.37c)


Hence the Navier equation (1.44) reduces to ¡ 2 ¢ ¡ ¢ ρω − µ|k|2 (B×k) + ρω 2 − (λ + 2µ)|k|2 Ak = 0,

111

(3.38)

which we can only satisfy for nonzero k if either B=0

and

ρω 2 = (λ + 2µ) |k|2

(3.39a)

ρω 2 = µ |k|2 .

(3.39b)

or A=0

and

The vectorial nature of the Navier equation has thus led to the existence of two dispersion relations, corresponding to two distinct types of waves. (i) P-waves, also known as Primary or comPressional waves, take the form n o u = Ak exp i (k · x − ωt) , (3.40a) where ω 2 = (λ + 2µ)|k|2 /ρ. We recall from §1.9 that µ and λ + 2µ/3 are both positive, so ω is real. The phase speed is thus given by s λ + 2µ (3.40b) cp = ρ and, since cp is independent of k, the waves are non-dispersive. The phase velocity c=

cp k |k|

(3.40c)

is parallel to the displacement u, so P -waves are said to be longitudinal. They are also sometimes described as irrotational since they satisfy curl u = 0, as is readily verified by direct differentiation of (3.40a). (ii) S -waves, also known as Secondary or Shear waves, take the form n o u = (B×k) exp i (k · x − ωt) , (3.41a) where ω 2 = µ|k|2 /ρ. S -waves are also non-dispersive, with constant phase speed r µ cs = . (3.41b) ρ This time, though, the phase velocity is perpendicular to the displacement, so S -waves are said to be transverse. Since (3.41a) satisfies div u = 0, we deduce that S -waves conserve volume, and they may thus be referred to as equivoluminal.

112


Evidently cp > cs , so that P -waves always propagate faster than S -waves. This fact is familiar to seismologists: following an underground earthquake they measure two distinct initial signals corresponding to the arrival of the P -waves followed by the S -waves, with propagation speeds in rock given approximately by cp ≈ 5 km s−1 and cs ≈ 3 km s−1 respectively. P - and S -waves are prototypical examples of polarised waves, which are defined to be solutions to general wave equations in which u = u0 f (k · x − ωt) .

(3.42)

This solution is called a plane polarised wave, with u0 and k defining the plane of polarisation, provided they are not parallel. As in the expression (3.31) for the general solution of the scalar wave equation, the general solution of the Navier equation may also be expressed as a linear superposition of harmonic waves. In this case, the displacement field reads ZZZ n o u(x, t) = kA1 (k) exp i (k · x − |k|cp t) R3 n o + kA2 (k) exp i (k · x + |k|cp t) n o + k×B 1 (k) exp i (k · x − |k|cs t) n o + k×B 2 (k) exp i (k · x + |k|cs t) dk, (3.43) which represents an arbitrary combination of P - and S -waves travelling in all possible directions. Again the amplitude functions Ai (k) and B i (k) can in principle be determined from the Fourier transform of the initial data, although carrying this out in practice when boundary conditions are imposed is far from easy. 3.2.5 Mode conversion in plane strain The explicit solution (3.43) suggests that the response of an elastic medium to arbitrary initial conditions will nearly always involve coupled P - and S waves. This coupling lies at the heart of elastic wave propagation in more than one space dimension. An indication of the unexpected phenomena that can result comes when we reconsider the familiar rules of reflection and refraction of plane waves of the form (3.35) at a planar interface. Here we will initially consider the plane strain problem of reflection of a P -wave that is incident from x < 0 on a rigid barrier at x = 0, so that u = 0 there. We recall that P -waves are longitudinal and so, by choosing the coordinate axes


113

y P S

α β α

P

1 0 0 1 Rigid 0 1 0 1 boundary 0 1 0 1 x 0 1 0 1 0 1 0 1 0 1 0 1 0 1

Fig. 3.2. Schematic of a P -wave reflecting from a rigid boundary.

appropriately, we may write the incident wave in the form of a plane strain displacement, µ ¶ n£ ¤o cos α u = uinc = exp i kp (x cos α + y sin α) − ωt , (3.44) sin α where kp = ω/cp and α is the angle between the incoming wave and the x-axis. Our task now is to find a reflected wave field uref such that the net displacement u = uinc + uref is zero on the boundary x = 0. We soon realise that this is impossible unless we allow for two reflected waves: one P -wave and one S -wave. Otherwise, there are not enough degrees of freedom to make both displacement components zero on x = 0. We therefore seek a reflected wave field of the form µ ¶ n£ ¤o − cos γ uref = r1 exp i kp (−x cos γ + y sin γ) − ωt sin γ µ ¶ n£ ¤o sin β + r2 exp i ks (−x cos β + y sin β) − ωt , (3.45) cos β where ks = ω/cs . Recall that S -waves are transverse so the amplitude is orthogonal to the wavevector. From the condition u = 0 on x = 0, we find that the P -wave reflection is specular with the angle of reflection γ being equal to the angle of incidence α. However, the S -wave reflection angle satisfies cs sin α (3.46) sin β = cp

114


and, since cp > cs , it follows that β < α, as illustrated in Figure 3.2. The reflection coefficients are given by r1 =

cos(α + β) , cos(α − β)

r2 = −

sin(2α) . cos(α − β)

(3.47)

This is our first encounter with the phenomenon of mode conversion. A boundary will usually turn a pure P -wave (or a pure S -wave) into a combination of P - and S -waves. As shown in Exercise 3.9, the same happens in the case of refracted waves, and the phenomenon of total internal reflection becomes correspondingly more complicated. Since we are considering only plane strain, the displacement associated with this S -wave is in the plane of reflection shown in Figure 3.2. This is in contrast with the antiplane shear waves in §3.2.6, where the displacement is orthogonal to the plane of reflection. Thus the S -waves in (3.45) are referred to as SV -waves, while those in antiplane strain are called SH -waves. Any general incident plane wave can be decomposed into a P -wave, an SH -wave and an SV -wave. We now describe an example that illustrates how P - and S -waves can interact in a practical situation.

3.2.6 Love waves First, let us consider the dynamic version of antiplane strain, the static case of which was introduced in §2.3. If the displacement field takes the form ¡ ¢T u = 0, 0, w(x, y, t) , then w satisfies µ 2 ¶ ∂ w ∂2w ∂2w µ∇2 w = µ + = ρ , (3.48) ∂x2 ∂y 2 ∂t2 which is just the familiar two-dimensional scalar wave equation (3.2), with wave speed cs . This is to be expected, since antiplane strain is volumepreserving, with the displacement depending only on the transverse variables. The solution strategies presented above for two-dimensional waves on a membrane are all directly applicable to the solution of (3.48). Here we will illustrate them by describing Love waves, which are antiplane strain waves guided through a particular type of layered medium. As illustrated in Figure 3.3, the geometry is that of a uniform layer of one material, with constant thickness 2h, encased inside an infinite expanse of a second material. This set-up might model, for example, a coal seam in a rock stratum, with the displacement in the horizontal z-direction taking the form of an


115

y

“rock”

µ2 ,

ρ2

µ1 ,

ρ1

µ2 ,

ρ2

h “coal”

“rock”

−h

x

Fig. 3.3. Schematic of a layered elastic medium.

SH -wave. The density and shear modulus of the “coal” are denoted by ρ1 and µ1 respectively, while those of the “rock” are labelled ρ2 and µ2 . The transverse displacement wi (x, y, t) in either medium satisfies µ 2 ¶ ∂ wi ∂ 2 wi ∂ 2 wi 2 cs i + = , i = 1, 2, (3.49) ∂x2 ∂y 2 ∂t2 p where csi = µi /ρi (i = 1, 2) are the S -wave speeds. On the boundaries of the seam, the displacements and tractions must be continuous, as shown in §1.10, so that w1 = w2 ,

µ1

∂w1 ∂w2 = µ2 , ∂y ∂y

on y = ±h.

(3.50)

We seek travelling-wave solutions propagating in the x-direction in which the displacements take the form n o wi = fi (y) exp i (kx − ωt) . (3.51) Substituting (3.51) into (3.49), we find that the functions fi (y) satisfy µ 2 ¶ ω 00 2 fi + 2 − k fi = 0 (3.52) csi and, hence, are either exponential or sinusoidal. We suppose that the amplitude of the waves decays at infinity so that f2 = A2 e−`y

in y > h,

f2 = B2 e`y

in y < −h,

(3.53)

where ` is real and positive. In such modes, the seam acts as a waveguide,

116


propagating waves in the x-direction without any energy radiating or “leaking out” to y → ±∞ as discussed in §3.2.3. Substitution of (3.53) into (3.52) reveals that k and ω must be such that c2s2 k 2 − ω 2 = c2s2 `2 > 0.

(3.54)

In the coal seam, we try a solution f1 = A1 cos(my) + B1 sin(my)

(3.55)

which may be sinusoidal if m is real or exponential if m is pure imaginary. Now (3.52) leads to ω 2 = c2s1 k 2 + c2s1 m2 , so the Love waves propagate at a speed cL given by µ ¶ ω2 m2 2 2 cL = 2 = cs1 1 + 2 . k k

(3.56)

(3.57)

Let us first consider symmetric modes in which B1 = 0 and B2 = A2 , so the boundary conditions (3.50) reduce to A2 e−`h = A1 cos(mh),

µ2 lA2 e−`h = µ1 mA1 sin(mh).

(3.58)

We can view this as a system of simultaneous equations for A1 and A2 , whose solution is in general A1 = A2 = 0. A nonzero solution can only exist if the determinant of the system is zero, and this gives us the condition µ1 m tan mh = µ2 `.

(3.59)

For antisymmetric waves, with A1 = 0 and B2 = −A2 , the analogous calculation in Exercise 3.6 leads to µ1 m cot mh = −µ2 `.

(3.60)

It remains to determine m from either of the transcendental equations (3.59) or (3.60). These are easiest to analyse in the extreme case where the rock is rigid so that µ2 /µ1 → ∞, and we will focus on this limit henceforth, leaving the general case to Exercise 3.6. We then see that there are two infinite families of solutions, with mh = (2n + 1)π/2 for symmetric waves or mh = nπ for antisymmetric waves, where n is an integer. From (3.54) and (3.57), we deduce the inequalities ³ ω ´2 2 < c2s2 , (3.61) cs1 < k which show that the waves can only exist if cs1 < cs2 , that is if the wave speed in the coal is slower than that in the rock, which is typically true in


ω

117

ω = cs2 k

k

ω = cs1 k

Fig. 3.4. Dispersion relation between frequency ω and wavenumber k for symmetric (solid) and antisymmetric (dotted) Love waves. The phase speed ω/k is bounded between cs1 and cs2 .

practice. The phase speed of the waves is then bounded between cs1 and cs2 and the resulting wavefields in the rock decay exponentially as we move away from the seam. For each fixed allowable value of m, (3.57) shows that the wave-speed varies with wavenumber, with long waves travelling faster than short ones. This is our first encounter with dispersive waves, and it seems to be at odds with our knowledge that S -waves are non-dispersive. However, the dispersion relations (3.39) were obtained only for plane P - or S -waves in an infinite medium, and (3.57) illustrates how the presence of boundaries can often give rise to dispersion. Ignoring the degenerate root m = 0, which corresponds to the trivial solution w ≡ 0, we see that the lowest-frequency mode is a symmetric wave with mh = π/2 and therefore µ 2

ω =

c2s1

¶ π2 2 +k . 4h2

(3.62)

Hence, real values of k can only exist if ω exceeds a critical cut-off frequency πcs1 /2h. The existence of a cut-off frequency below which waves cannot propagate without attenuation is a characteristic of all waveguides. If we do not take the limit µ2 /µ1 → ∞, then the dispersion relation between ω and k must be obtained by eliminating `, m between (3.54), (3.56) and either (3.59) or (3.60), as shown in Exercise 3.6. Crucially, the values of m satisfying (3.59) or (3.60) with ` > 0 are all real, so the inequalities (3.61) continue to hold and there is still a cut-off frequency. As shown in Figure 3.4,

118


there is an infinite family of waves, both symmetric and antisymmetric, and, in each case, the phase speed ω/k is bounded between cs1 and cs2 . 3.2.7 Rayleigh waves Rayleigh discovered perhaps the most famous of all elastic waves, which can propagate close to a planar free boundary, decaying exponentially away from the boundary but suffering no decay at all in the direction of propagation. We can model these waves using plain strain in y < 0 below a stress-free boundary at y = 0, where the zero-traction condition is µ ¶ µ ¶ ∂u ∂v ∂v ∂u ∂v µ + = 2µ +λ + =0 at y = 0. (3.63) ∂y ∂x ∂y ∂x ∂y Now we seek frequency-domain solutions of the form © ª u = (up eκp y + us eκs y ) exp i (kx − ωt) ,

(3.64)

where the dispersion relations (3.39) for P - and S -waves tell us that κ2p = k 2 −

ω2 , c2p

κ2s = k 2 −

ω2 . c2s

(3.65)

We assume as usual that k and ω are real and positive, and that Re(κp ), Re(κs ) > 0 so that the waves propagate without attenuation in the xdirection while decaying exponentially as y → −∞. Evidently such waves can only exist if the wavenumber k in the x-direction is greater than both ω/cs and ω/cp . The propagation speed c = ω/k must therefore satisfy c < c s < cp ,

(3.66)

so that Rayleigh waves inevitably travel slower than either P -waves or S waves. After the rather tortuous manipulations of Exercise 3.10, we find an equation for c in the form µ µ ¶2 ¶1/2 µ ¶1/2 c2 c2 c2 =4 1− 2 1− 2 . (3.67) 2− 2 cs cp cs It can be shown that this leads to a sextic equation for c but that only one positive real value for c2 exists and satisfies (3.66). Rayleigh waves are therefore non-dispersive, with a constant propagation speed c. These results led Rayleigh to suggest that, if a P - or S -wave originated at some source, for example an earthquake, in y < 0, mode conversion would cause a Rayleigh wave to propagate indefinitely in the x-direction

3.3 Dynamic stress functions

119

near y = 0. This hypothesis is born out by seismological observation: the radius of the earth is large enough for (3.63) to be a good approximation and many earthquake records reveal the arrival of a large Rayleigh wave after first the P - and then the S -waves have been detected.

3.3 Dynamic stress functions We now describe some of the ways in which elastic wave propagation can be reduced to the study of scalar wave equations by the use of stress functions. By taking the divergence and the curl of the dynamic Navier equation (1.44), we obtain µ 2 ¶ µ 2 ¶ ∂ ∂ 2 2 2 2 − cp ∇ (div u) = 0 and − cs ∇ (curl u) = 0 (3.68) ∂t2 ∂t2 respectively. This reminds us that irrotational waves (with curl u = 0) propagate at speed cp while equivoluminal waves (with div u = 0) move at speed cs . It also provides a route for finding solutions of the Navier equation that often proves handy. We can first solve the scalar wave equation (3.68a) for div u and then the inhomogeneous wave equation µ 2 ¶ ¡ ¢ ∂ 2 2 − cs ∇ u = c2p − c2s ∇ (div u) (3.69) 2 ∂t for the displacement u. Moreover, by cross-differentiating (3.68), we obtain µ 2 ¶µ 2 ¶ ∂ ∂ 2 2 2 2 − c ∇ − c ∇ u = 0, (3.70) p s ∂t2 ∂t2 which we can interpret as a factorisation of the the Navier wave operator into two scalar operators corresponding to P -waves and S -waves respectively. This generalises the result that each component of the displacement u satisfies the biharmonic equation under static conditions. We recall that in Chapter 2 we showed that, when there is two-dimensional or cylindrical symmetry, the problem can be reduced to a single biharmonic equation by using a suitable stress function. Now we show briefly how the same idea may be applied to dynamic problems. First we consider plane strain, for which the dynamic Navier equation reads ρ

∂τ11 ∂τ12 ∂2u = + , 2 ∂t ∂x ∂y

ρ

∂2v ∂τ12 ∂τ22 = + , 2 ∂t ∂x ∂y

(3.71)

120


where ∂u ∂v τ11 = (λ + 2µ) +λ , ∂x ∂y µ ¶ ∂u ∂v τ12 = µ + , ∂y ∂x ∂u ∂v τ22 = λ + (λ + 2µ) . ∂x ∂y

(3.72a) (3.72b) (3.72c)

For such problems, we can define a dynamic Love function, by analogy with the static Love stress function introduced in §2.8.2. From, for example, the second Navier equation (3.71) in the form µ ¶ ∂2u 1 ∂2v ∂2v 2 + (1 − 2ν) ∇ v − 2 2 + 2 = 0, (3.73) ∂x∂y cs ∂t ∂y we deduce the integrability condition ½ µ ¶ ¾ 1 1 ∂2L ∂2L 2 u= (1 − 2ν) ∇ L − 2 2 + , 2µ cs ∂t ∂y 2

v=−

1 ∂2L , 2µ ∂x∂y

(3.74)

for some function L(x, y, t). It is then readily verified that the first Navier equation (3.71) is satisfied if and only if L satisfies the scalar wave equation µ 2 ¶µ 2 ¶ ∂ ∂ 2 2 2 2 − cp ∇ − cs ∇ L = 0. (3.75) ∂t2 ∂t2 As in the static case, we could have started from the x- rather than the y-component of the momentum equation, or indeed any linear combination of the two. Thus one can define a one-parameter family of stress functions analogous to L. With u and v given by (3.74), the stress components may be evaluated from L using ½ µ ¶¾ ∂ ∂2L 1 ∂2L 2 τ11 = + (1 − ν) ∇ L − 2 2 , (3.76a) ∂x ∂y 2 cs ∂t ½ µ ¶ ¾ 1 ∂2L 1 ∂2L ∂ ∂2L 2 −ν ∇ L− 2 2 − 2 2 , (3.76b) τ12 = ∂y ∂y 2 cs ∂t 2cs ∂t ½ µ ¶¾ ∂ ∂2L 1 ∂2L 2 τ22 = − −ν ∇ L− 2 2 . (3.76c) ∂x ∂y 2 cs ∂t For radially symmetric problems, in which the displacement u = ur er + uz ez is dependent only on the cylindrical polar coordinates (r, z) and t, it is again

3.4 Waves in cylinders and spheres

121

possible to define a Love function L(r, z, t) such that ¶ ½ µ µ ¶¾ 1 ∂2L 1 1 ∂2L 1 ∂ ∂L 2 ur = − , uz = (1 − 2ν) ∇ L − 2 2 + r , 2µ ∂r∂z 2µ cs ∂t r ∂r ∂r (3.77) and again L satisfies (3.75). In this case, the stress components are given by ½ µ ¶¾ ∂ ∂2L 1 ∂2L 2 τrr = − −ν ∇ L− 2 2 , (3.78a) ∂z ∂r2 cs ∂t ½ µ ¶ µ ¶ ¾ ∂L 1 ∂2L 1 ∂2L ∂ 1 ∂ 2 r −ν ∇ L− 2 2 − 2 2 , (3.78b) τrz = ∂r r ∂r ∂r cs ∂t 2cs ∂t ½ µ ¶ µ ¶¾ ∂ 1 ∂ ∂L 1 ∂2L 2 τzz = r + (1 − ν) ∇ L − 2 2 , (3.78c) ∂z r ∂r ∂r cs ∂t ½ µ ¶¾ ∂ 1 ∂L 1 ∂2L τθθ = − + ν ∇2 L − 2 2 . (3.78d) ∂z r ∂r cs ∂t We will now use the ideas of §§3.2–3.3 to enumerate some of the rare configurations for which explicit solutions of the multidimensional elastic wave equations can be written down.

3.4 Waves in cylinders and spheres 3.4.1 Waves in a circular cylinder The simplest cylindrically symmetric waves are torsional waves in which ur = uz = 0 and uθ is independent of θ. In this case, we do not need any stress function representation because the cylindrical Navier equation (1.73) simplifies to ∂ 2 uθ 1 ∂uθ uθ ∂ 2 uθ 1 ∂ 2 uθ + − + = . ∂r2 r ∂r r2 ∂z 2 c2s ∂t2

(3.79)

This scalar equation supports S -waves, as expected for a displacement field satisfying ∇ · u = 0. We seek waves travelling in the x-direction with wavenumber k and frequency ω by separation of variables: uθ = f (r) ei(kz−ωt) . Substituting (3.80) into (3.79), we find that f (r) satisfies µ ¶ 1 1 0 2 00 f + f + m − 2 f = 0, r r

(3.80)

(3.81)

122


where m2 =

ω2 − k2 , c2s

(3.82)

and we will only consider cases in which m is real. The general solution of (3.81) is f (r) = A J1 (mr) + B Y1 (mr) ,

(3.83)

where J1 and Y1 again denote Bessel functions, while A and B are arbitrary constants. To describe waves in a circular cylinder 0 6 r < a, we must set B = 0 to eliminate the solution Y1 which is unbounded as r → 0. If the curved boundary of the cylinder, r = a, is free of traction, we require µ ¶ ∂uθ uθ τrθ = µ − =0 (3.84) ∂r r on r = a, which implies that µ ¶¯ ¯ d 1 m J1 (mr) ¯¯ = − J2 (ma) = 0, dr r a r=a

(3.85)

the latter equality following from known properties of Bessel functions (Gradshteyn & Ryzhik, 1994§8.4). We deduce that ma must equal one of the zeros ξ2,n of J2 as defined in §3.2.1. There is thus one family of waves corresponding to each integer value of n. The dispersion relation cs q 2 ω= ξ2,n + k 2 a2 (3.86) a tells us that the bar acts as a waveguide, in which the frequency must exceed the cut-off value ξ2,1 cs /a, while the phase velocity ω/k is not constant, but is bounded below by cs . Torsional waves are therefore dispersive, and always travel faster than S -waves. For a cylinder that is finite in the z-direction, we can use (3.86) to find the normal frequencies. For cylindrically symmetric longitudinal waves, in which uθ = 0 and ur and uz are independent of θ, the Love stress function L introduced in §3.3 is especially useful. We look for harmonic waves propagating in the z-direction by setting L(r, z, t) = f (r)ei(kz−ωt) .

(3.87)

The partial differential equation (3.75) satisfied by L leads to the ordinary


differential equation µ 2 ¶µ 2 ¶ d 1 d d 1 d 2 2 + + mp + + ms f (r) = 0, dr2 r dr dr2 r dr

123

(3.88)

for f , where now we define m2p =

ω2 − k2 , c2p

m2s =

ω2 − k2 , c2s

(3.89)

as shorthand. The general solution of (3.88) is f (r) = A J0 (mp r) + B J0 (ms r) + C Y0 (mp r) + D Y0 (ms r) ,

(3.90)

where A, B, C and D are arbitrary constants. To describe waves in the solid cylinder r < a, we must choose C = D = 0 so that f (r) is well-defined as r → 0. If the boundary is stress-free, then τrr and τrz are zero on r = a, and this leads to the two boundary conditions ½ µ ¶ µ ¶ ¾ d f0 ω2 00 2 (1 − ν) f + + νk + (2 − ν) 2 f = 0, (3.91a) dr r 2cs f0 (1 − ν)f 00 − ν − νm2s f = 0 (3.91b) r on r = a. These give us a homogeneous linear system for the two remaining constants A and B, namely o n¡ ¢ (1 − ν)m2p − (1 − 2ν)m2s − k 2 mp J1 (mp a) A o n¡ ¢ + m2s − k 2 ms J1 (ms a) B = 0 (3.92a) n o ¡ ¢ mp J1 (mp a) − νm2s + (1 − ν)m2p a J0 (mp a) A n o + ms (J1 (ms a) − ms a J0 (ms a)) B = 0. (3.92b) These admit nonzero solutions for A and B only if the determinant of the system is zero, and this gives us a complicated dispersion relation between ω and k. Notice that the wave-field inevitably involves a combination of P waves and S -waves, illustrating the intimate coupling between these waves in practical situations. An even more complicated wave motion is that of flexural waves, where all three displacement components are non-zero. There is no Love stress function available in this case, but we try a particularly simple θ-dependence

124


of the displacement field, namely ³ ´ u(r, θ, z, t) = Ur (r) cos θer + Uθ (r) sin θeθ + Uz (r) cos θez ei(kz−ωt) . (3.93) It follows that div u takes the form div u = f (r) cos θei(kz−ωt) ,

where f (r) =

1 d Uθ (rUr ) + ikUz + . r dr r (3.94)

Hence, when we recall from (3.68) that ¶ µ 2 ∂ 2 2 − c ∇ (div u) = 0, p ∂t2 we find that f (r) again satisfies Bessel’s equation µ ¶ 1 0 1 00 2 f + f + mp − 2 f = 0. r r

(3.95)

(3.96)

Assuming that f (r) is bounded as r → 0, we must therefore take f (r) = A J1 (mp r).

(3.97)

Having found div u, we can use (3.69) to determine the displacement components. To calculate ∇2 u in polar coordinates, we have to use the identity “∇2 ≡ grad div − curl curl” and the formulae for div, grad and curl given in Appendix A3.7.1 to obtain µ ½· ¶ ¸ 2 Ur0 2Uθ 2 00 2 − ∇ u= Ur + + k Ur − 2 cos θer r r2 r µ · ¶ ¸ 0 Uθ 2 2Ur 00 2 + Uθ + − + k Uθ − 2 sin θeθ r r2 r · µ ¶ ¸ ¾ 0 Uz 1 00 2 + Uz + − + k Uz cos θez ei(kz−ωt) . (3.98) r r2 Just considering the ez -component of (3.69), we find that Uz must satisfy Ã ! µ ¶ c2p 1 0 1 00 2 Uz + Uz + ms − 2 Uz = 1 − 2 ikf, (3.99) r r cs with f given by (3.97). The bounded solution of this inhomogeneous Bessel equation is Uz = −

k2

ikA J1 (mp r) + B J1 (ms r), + m2p

(3.100)


125

where B is an arbitrary constant. Now we can use (3.94) and (3.100) to determine Uθ as Uθ =

m2p A r J1 (mp r) − ikBr J1 (ms r) − rUr0 − Ur k 2 + m2p

(3.101)

and, hence, use Exercise ?? to turn the er -component of (3.69) into the single differential equation for Ur 3 J1 (ms r) Ur00 + Ur0 + m2s Ur = −2ikB r r µ ¶ A 2 2 2 2 J1 (mp r) + 2 mp (mp − ms ) J0 (mp r) + (mp + ms ) . (3.102) k + m2p r Since the general bounded solution of (3.102) is Ur =

¢ ¢ mp A ¡ ikB ¡ J0 (ms r) − J2 (ms r) − J0 (mp r) − J2 (mp r) + C J1 (ωs r), 2 2 2ms 2(k + mp ) (3.103)

where C is a further arbitrary constant, we can finally obtain Uθ from (3.101) as µ ¶ J1 (mp r) ikB J1 (ms r) J1 (ms r) A − 2 +C − ms J0 (ms r) . Uθ = 2 k + m2p r ms r r (3.104) Now let us look for waves in a traction-free cylinder r < a by setting τrr = τrθ = τrz = 0 on r = a. This leads to a system of three homogeneous linear equations for the three constants A, B and C. For nontrivial solutions to exist, the determinant of the system must be zero as usual and, as shown in Exercise ??, this condition leads to the dispersion relation for flexural waves. We illustrate a typical displacement in Figure 3.5 which shows waves corresponding to bending, or flexing, of the cylinder propagating in the zdirection.

3.4.2 Waves in a sphere The simplest waves in a sphere are spherically symmetric with a purely radial displacement, that is u = ur (r, t)er .

(3.105)

126


z

t=0

2π/5ω

4π/5ω

6π/5ω

8π/5ω

Fig. 3.5. Illustration of flexural waves.

The Navier equation (1.82) in spherical polar coordinates thus reduces to the scalar equation 1 ∂ 2 ur ur ∂ 2 ur 2 ∂ur −2 2. = + 2 2 2 cp ∂t ∂r r ∂r r

(3.106)

One can easily check that the displacement field (3.105) satisfies ∇×u = 0, which is why (3.106) describes only P -waves. This in turn implies that we can write u = ∇φ for some potential function φ(r, t), that is ur =

∂φ . ∂r

If we do so, then (3.106) takes the form ½ ¾ ∂ 1 ∂ 2 φ ∂ 2 φ 2 ∂φ − 2 − = 0, ∂r c2p ∂t2 ∂r r ∂r

(3.107)

(3.108)

and integration with respect to r leads to 1 ∂ 2 φ ∂ 2 φ 2 ∂φ = F (t), − 2 − c2p ∂t2 ∂r r ∂r

(3.109)

for some function F (t). However, since an arbitrary function of t may be added to φ without affecting the displacement field, we deduce that F may be set to zero without any loss of generality. Then a simple rearrangement of (3.109) yields 2 ∂2 2 ∂ (rφ) − c (rφ) = 0. p ∂t2 ∂r2

(3.110)


127

Thus the combination (rφ) satisfies the standard one-dimensional wave equation, with wave-speed cp . If we seek normal modes in which φ(r, t) = f (r)e−iωt , then (3.110) becomes µ 2¶ d2 ω (rf ) = 0, (3.111) (rf ) + 2 dr c2p whose general solution is f=

A sin(ωr/cp ) + B cos(ωr/cp ) , r

(3.112)

where A and B are two arbitrary constants. To describe normal modes in a sphere r < a, the boundedness of φ at r = 0 implies that B = 0. If the surface r = a is traction-free, then τrr must be zero there. Using the formula given in §1.11.3, we obtain ( ) 2 − 2c2 ) 2(c ur ∂ur p s + 2λ = ρe−iωt c2p f 00 + f0 (3.113) τrr = (λ + 2µ) ∂r r r and, with f (r) given by (3.112), we find that the zero-stress condition on r = a reduces to µ ¶−1 tan(ωa/cp ) ω 2 a2 = 1− . (3.114) ωa/cp 4c2s By plotting the left- and right-hand sides versus ω, one can easily deduce that this transcendental equation admits a countably infinite set of solutions for ω, corresponding the the natural frequencies of the sphere. More complicated formulae for solutions that are not radially symmetric may be found in (Love, 1944, Chapter 12). These involve so-called spherical harmonics, which inevitably arise when one attempts to separate the variables in spherical polar coordinates. We will only discuss torsional waves, in which the displacement consists of just a rotation about the z-axis, so that u = uφ eφ . Using (1.82), one can easily show that this is only possible if uφ is independent of φ, and it follows that div u is identically zero and we have pure S -waves. The Navier equation (1.82) reduces to µ ¶ µ ¶ ∂uφ uφ 1 ∂ 2 uφ 1 ∂ 1 ∂ 2 ∂uφ = r + sin θ − 2 2 , (3.115) 2 2 2 2 cs ∂t r ∂r ∂r r sin θ ∂θ ∂θ r sin θ and we seek a separable normal mode in which uφ (r, θ, t) = f (r)g(θ)e−iωt .

(3.116)

128


By separating the variables in the usual way, we find that the functions f and g must satisfy µ

¶ ω2 r2 − κ f = 0, r f + 2rf + c2s ¡ ¢ g 00 + cot θg 0 + κ − cosec2 θ g = 0, 2 00

0

(3.117a) (3.117b)

where κ is an arbitrary separation constant. Crucially, we require uφ to be a 2π-periodic function of θ and to be wellbehaved at θ = 0, π. We can now use a well-known result from quantum mechanics to assert that this is possible only if κ = n(n + 1), where n is a positive integer. In this case, the solution of (3.117b) that is bounded as θ → 0 is g(θ) = A

´ d³ Pn (cos θ) , dθ

(3.118)

where A is an arbitrary constant and Pn is the Legendre polynomial of degree n (Gradshteyn & Ryzhik, 1994§8.91). These polynomials are defined by Pn (z) =

´ 1 dn ³ 2 n (z − 1) , 2n n! dz n

(3.119)

and the first few cases are P1 (z) = z,

P2 (z) =

3z 2 − 1 , 2

P3 (z) =

5z 3 − 3z . 2

(3.120)

With λ = n(n + 1), the general solution of (3.117a) that is bounded as r → 0 is f (r) =

B Jn+1/2 (ωr/cs ) √ , r

(3.121)

where B is an arbitrary constant. Fortunately, the Bessel function Jn+1/2 takes a relatively simple form when n is an integer, namely r Jn+1/2 (z) = (−1)

n

2 n+1/2 z π

µ

1 d z dz

¶n µ

sin z z

¶ ,

(3.122)

3.5 Initial-value problems

and the first few cases are r µ ¶ 2 sin z J3/2 (z) = − cos z , πz z r µ ¶ 2 (3 − z 2 ) sin z 3 cos z J5/2 (z) = − , πz z2 z r ¶ µ 2 3(5 − 2z 2 ) sin z (z 2 − 15) cos z + . J7/2 (z) = πz z3 z2

129

(3.123a) (3.123b) (3.123c)

Now, for a stress-free spherical boundary at r = a the appropriate boundary condition is µ ¶ ∂uφ uφ τrφ = µ − = 0, (3.124) ∂r r and the corresponding condition for f is f 0 (a) −

f (a) = 0. a

(3.125)

By substituting for f from (3.121), with C = 0, and simplifying, we arrive at the equation (n − 1) Jn+1/2 (ωa/c) = (ωa/c) Jn+3/2 (ωa/c)

(3.126)

for the natural frequencies ω. For each integer value of n, (3.126) has a countably infinite family of solutions, say ωn,m (m = 1, 2, . . . ) (see Exercise 3.11). Thus there is a doubly infinite set of natural frequencies corresponding to all possible values of m and n, as we would expect for a problem in two spatial dimensions (r, θ).

3.5 Initial-value problems 3.5.1 Solutions in the time domain Thus far in this chapter, we have focused on normal modes and monochromatic waves, both of which correspond to frequency-domain solutions that are harmonic in time with a single frequency ω. However, if we wish to model situations like the plucking of a string or the initiation of elastic waves by an earthquake, we have to solve an initial-value problem, in which all frequencies will be excited. Typically this will comprise a given initial displacement and velocity and our task will be to determine how the medium subsequently evolves. To keep the discussion as simple as possible, we will ignore the effects of boundaries in the medium.

130


The simplest example is the one-dimensional wave equation (3.1), whose general solution is w(x, t) = F (x − ct) + G(x + ct), (3.127) p where F and G are arbitrary functions, and c = T /% is again the constant wave-speed. By this we mean that any solution of (3.1) may be expressed in the form (3.127), and this is readily verified for all the simple solutions given in §3.2. If, for example, we are given the initial conditions w = f (x),

∂w = g(x) ∂t

at t = 0,

(3.128)

then, as in Exercise 3.3, we can easily evaluate the corrseponding F and G to obtain the D’Alembert solution Z ´ 1³ 1 x+ct f (x − ct) + f (x + ct) + g(s) ds. (3.129) w(x, t) = 2 2c x−ct Since there is no specific frequency associated with (3.129), it is said to be a solution in the time domain. In more than one space dimension, there is usually no general solution akin to (3.127) and initial-value problems are consequently much more difficult to solve. There are two main approaches, the first of which is to try and superimpose normal modes or harmonic waves in such a way as to satisfy the initial conditions. We have already seen examples of this approach in §3.2. However, given the scarcity of elastic wave problems that can be solved explicitly in the frequency domain, even fewer explicit solutions can be found in the time domain using this approach. The second possibility is to use singular solutions of the wave equation, analogous to the Green’s functions introduced in §2.9, which in principle allow us to write down the solutions of initial-value problems. Although there are very few geometries for which these fundamental solutions can be constructed explicitly, we will see that they demonstrate the key role played by dimensionality in wave propagation. 3.5.2 Fundamental solutions Suppose we were to force the one-dimensional wave equation (3.1) with an impulse at time t = 0 that is localised at x = 0. In such a situation, w satisfies 2 ∂2w 2∂ w − c = δ(x)δ(t), ∂t2 ∂x2

(3.130)


131

where δ is the Dirac delta-function defined in §2.9 and w is assumed to be zero for t < 0. The right-hand side of (3.130) indicates that there is a jump in ∂w/∂t at t = 0, and we can therefore restate (3.130) as the initial-value problem 2 ∂2w 2∂ w − c = 0, for t > 0, (3.131a) ∂t2 ∂x2 ∂w w = 0, = δ(x) at t = 0. (3.131b) ∂t Substitution of (3.131) into the D’Alembert solution (3.129) leads to the fundamental solution   1 −ct < x < ct, (3.132a) w = R(x, t) = 2c 0 otherwise,

which is also known as the Riemann function for the one-dimensional wave equation (Ockendon et al., 1999, page ????). It may conveniently be recast as ´ 1³ R(x, t) = H(x + ct) − H(x − ct) , (3.132b) 2c which is explicitly in the form of (3.127), where ( 0 x < 0, H(s) = (3.133) 1 x > 0, is the so-called Heaviside function or step function. On the other hand, if w starts from rest with the initial displacement a localised singularity of the form w(x, 0) = δ(x),

∂w (x, 0) = 0, ∂t

(3.134)

then we easily find that ´ 1³ δ(x + ct) + δ(x − ct) . (3.135) 2 This solution might represent the behaviour of a string that is plucked at a single point, while the fundamental solution R(x, t) models a string being struck at a point. Since formal differentiation gives H 0 (s) = δ(s), it follows that w = ∂R/∂t, and we can thus write the general D’Alembert solution (3.129) as a superposition of fundamental solutions, namely Z ∞ Z ∞ ∂ f (s)R(x − s, t) ds + g(s)R(x − s, t) ds. (3.136) w(x, t) = ∂t ∞ −∞ w=

132


R(x, t)

t

x

Fig. 3.6. The one-dimensional fundamental solution R(x, t).

The Riemann function thus plays an analogous rôle to that performed by the Green’s function in elastostatic problems. We plot the fundamental solution R(x, t) in Figure 3.6. Notice that the initial disturbance is felt everywhere inside the region |x| < ct, but nowhere outside, and that ∂w/∂t is zero everywhere except on the the lines x = ±ct. Using (3.129), it is straightforward to generalise this observation to arbitrary localised initial data: if f (x) and g(x) are zero outside some interval, say a < x < b, then for all time ∂w/∂t is zero except on the two intervals a − ct < x < b − ct and a + ct < x < b + ct. Now let us consider two-dimensional elastic waves in a membrane or in antiplane strain, when there is no general solution like (3.127) to guide us. Nevertheless, the wave equation (3.2) can still be solved subject to singular initial data analogous to (3.131), for example by seeking a solution in plane polar coordinates, as in Exercise 3.4. This reveals that the solution of (3.2) with ∂w w = 0, = δ(x)δ(y) at t = 0, (3.137) ∂t is w = R(x, y, t) =

H (ct − r) √ , 2πc c2 t2 − r2

(3.138)

where r2 = x2 + y 2 and H again denotes the Heaviside function. The function R(r, t) defined by (3.138) is plotted in Figure 3.7. We see that, like the one-dimensional fundamental solution, R(r, t) is zero outside the cone r2 = c2 t2 . In this case, although the the “leading edge” of the disturbance at r = ct is sharp, the tail is not, and ∂R/∂t is nonzero everywhere inside the cone.


133

R(r, t)

t

r

Fig. p 3.7. The two-dimensional fundamental solution R(x, y, t) versus radial distance r = x2 + y 2 .

As in the one-dimensional case, we can write the general solution of the two-dimensional wave equation, subject to the initial conditions, w = f (x, y),

∂w = g(x, y) at t = 0, ∂t

(3.139)

as a linear superposition of fundamental solutions, namely Z ∞Z ∞ w(x, y, t) = g(ξ, η)R(x − ξ, y − η, t) dξdη −∞ −∞ Z ∞Z ∞ ∂ + f (ξ, η)R(x − ξ, y − η, t) dξdη.(3.140) ∂t −∞ −∞ From this one can infer that the properties of R noted above also apply to arbitrary localised disturbances. If the initial data f and g are zero outside some region, say r < a, then for all time w is zero outside the cone r = a+ct. However, the smooth tail of R implies that the disturbance persists inside the cone for all time. In three dimensions, as typified by (3.109), the radially-symmetric wave equation takes the form 1 ∂2w ∂ 2 w 2 ∂w 2 = ∇ w = + , c2 ∂t2 ∂r2 r ∂r

(3.141)

where now r2 = x2 + y 2 + z 2 , and the transformation in (3.110) leads to the general solution w(r, t) =

F (r − ct) + G(r + ct) , r

(3.142)

where F and G are two arbitrary scalar functions. This illustrates that, as

134


in one-dimensional waves, a localised radially-symmetric disturbance propagates outwards with a sharp front and a sharp tail. Huygens’ principle states that this happens for radially-symmetric waves in any odd number of space dimensions, but not the two-dimensional case considered above. We can use the general radially-symmetric solution (3.142) to construct the three-dimensional Riemann function δ(r − ct) R(x, y, z, t) = . (3.143) 4πcr This enables us to write the general solution to the three-dimensional initialvalue problem in a form analogous to (3.140), known as the retarded potential solution (Ockendon et al., 1999, page ????). Remarkably, a retarded potential solution can also be written down for the full three-dimensional Navier equation, as shown in Love (1944), page 303. However, we emphasise that these solutions only apply in an infinite medium, and the presence of any boundaries usually makes it impossible to construct a Riemann function. 3.5.3 Characteristics The concept of characteristics for hyperbolic partial differential equations gives fundamental insight into wave propagation. As discussed in Ockendon et al. (1999), page ????, there are several ways of thinking about characteristics mathematically but, for our purposes, it is most convenient to generalise (3.134), (3.135) and base our definition on the solution of the equations that is zero for t < t0 and driven by a localised forcing at x = x0 , t = t0 . Such a forcing will excite a superposition of plane waves of the form u = aei(k·x−ω(k)t) where k · x − ωt = k · x0 − ωt0

(3.144)

As k varies, this is a family of planes in (x, t) space through x = x0 , t = t0 and geometric intuition suggests that they will envelop a cone whose vertex is at this point (see Exercise 3.2 for an analytical explanation of this fact). For different values of x0 , t0 , these cones are called the characteristic cones of the system of partial differential equations. The intersection of the characteristic cone through (x0 , t0 ) with planes t = T , T > t0 are called the wavefronts of the disturbance originating at (x0 , t0 ). For example, for the one-dimensional wave equation (3.1), the planes (3.144) are simply x ± ct = x0 ± ct0 , and they coincide with the characteristics. For higher-dimensional scalar wave equations for which ω(k) = ±ck,

(3.145)

3.6 Moving singularities 2

2

2 2

x +y =c t

t

135

k1 x + k2 y = ωt

y

x

Fig. 3.8. Schematic showing the cone x2 +y 2 = c2 t2 tangent to the plane k1 x+k2 y = ωt.

we can notice that the cone |x − x0 |2 = c2 (t − t0 )2

(3.146)

has normal (k, ω) in (x, t)-space and is hence the envelope of 3.144) as k varies. This is illustrated in Figure 3.8 and again Exercise 3.2 gives a derivation of this result. The wavefronts are circles or spheres in two and three dimensions respectively. For the unsteady Navier equation, there are two possible wave speeds, so the characteristic surface through x = y = z = t = 0 is a two-sheeted cone ¡ 2 ¢¡ ¢ x + y 2 + z 2 − c2p t2 x2 + y 2 + z 2 − c2s t2 = 0, (3.147) as illustrated in Figure 3.9 for just two spatial dimensions. Every time we hit an elastic solid, we thereby create two distinct wave fronts, in stark contrast with, for example, the three-dimensional scalar wave equation or with electromagnetic waves in isotropic media, where Maxwell’s equations, although also vectorial, admit only one wave-speed: the speed of light.

3.6 Moving singularities We end this chapter by considering the interesting class of problems in which elastic waves are initiated by a moving source. This can happen, for example, in an overhead cable above the pantograph on an electric train or in an ice sheet underneath a moving vehicle. In such situations, we might anticipate

136


t x2 + y 2 = c2s t2

x2 + y 2 = c2p t2 y

x

Fig. 3.9. The two-sheeted characteristic cone for the Navier equation.

a dramatic change in behaviour when the source velocity passes through one of the speeds of sound in the material. As a first illustration, consider the one-dimensional problem of an elastic string on which a point force of magnitude P is imposed at the moving point x = V t. This situation is modelled by the equation %

∂2w ∂2w − T = P δ(x − V t), ∂t2 ∂x2

(3.148a)

where δ is the Dirac delta function. Assuming that the string starts from rest with zero displacement, we impose the initial conditions w=

∂w =0 ∂t

at t = 0.

(3.148b)

This problem may be solved, for example, by changing dependentp variables to ξ = x − ct and η = x + ct, where the speed of sound is c = T /%, to obtain, as in Exercise 3.12, w=

© ª Pc (c + V )|x − ct| + (c − V )|x + ct| − 2c|x − V t| , 2 −V ) (3.149)

4T (c2

when e > V , which is a single-valued function of x. As illustrated in Figure 3.10, the gradient ∂w/∂x ahead of the force increases as the propagation speed V increases, becoming infinite as V ap-

3.6 Moving singularities

w

(a)

(b)

137

w

increasing t x w

(c)

x (d)

w

x

x

Fig. 3.10. The response of a string to a point force moving at speed V : (a) V = 0, (b) 0 < V < c, (c) V = c, (d) V > c.

proaches the speed of sound c. The assumption of small slope that underpins the model (3.148a) clearly breaks down when V is close to c and, for V > c, the only continuous solution w ceases to be single-valued, as shown in Exercise 3.12 and Figure 3.10(d). In addition, we note that, no matter how long we wait, the solution never settles down to a travelling wave in which w is just a function of x − V t. In two or three space dimensions, it is more difficult to obtain solutions to initial-value problems analogous to (3.148), but it is now sometimes possible to find travelling wave solutions. For example, an elastic membrane subject to a moving point force is modelled by the equation ς

∂2w − T ∇2 w = P δ(x − V t)δ(y). ∂t2

(3.150)

If we suppose that w is a function only of ξ = x − V t and y, we find that it must satisfy ¡ ¢ ∂2w ∂2w P 1 − M2 + = − δ(ξ)δ(y), 2 2 ∂ξ ∂y T

(3.151)

where the Mach number is defined by M=

V . c

(3.152)

When M < 1, so the force is moving subsonically, (3.151) is an elliptic partial differential equation, and w is thus just a multiple√of the Green’s function, as in §2.9. By changing variables again to X = ξ/ 1 − M 2 and y, so that (3.151) is transformed to Poisson’s equation (see Exercise 3.13), the

138


(a)

increasing t

y

x

(b)

y x

Fig. 3.11. Schematic of wave fronts generated by a moving force on an elastic membrane with (a) M < 1, (b) M > 1.

solution is easily found to be P w=− √ log 2T 1 − M 2

µ

(x − V t)2 + y2 1 − M2

¶ + const.

(3.153)

Notice that the displacement is singular (as expected) as the point force (V t, 0) is approached, but the entire membrane is disturbed by the force. Indeed the displacement grows in amplitude as (x, y) approaches infinity, a feature typical of two-dimensional elliptic problems. Had we tried to solve an initial-value problem, we would have found that the influence of the point force spreads out at speed c, eventually invading the whole membrane as shown schematically in Figure 3.11(a). If M > 1, so the force is moving supersonically, then (3.151) is a hyperbolic partial differential equation. In fact, it is analogous to the one-dimensional wave equation (3.130), and we can infer the solution directly from (3.132):   w=



P √ 2T M 2 − 1

½ µ ¶ µ ¶¾ ξ ξ √ √ H y− −H y+ , ξ < 0, M2 − 2 M2 − 2 0, ξ > 0. (3.154)

Notice that we have made the “causality” assumption that w = 0 ahead of the point force.

3.6 Moving singularities

139

In terms of physical variables, (3.154) may be written as  x−Vt Vt−x  √ P √ , cp .

Exercise ??), the qualitative behaviour may be understood by considering the wave fronts generated by the moving force, by analogy with Figure 3.11. If Mp < Ms < 1, then the motion is subsonic with respect to both P and S -waves, and (3.158) is an elliptic equation similar to the biharmonic equation. In this case, as shown in Figure 3.12(a), all waves propagate out to infinity. If Mp < 1 < Ms , then the motion is subsonic with respect to P -waves, which therefore propagate to infinity, but supersonic with respect to S waves, which are thus confined to a Mach cone, as illustrated in Figure 3.12(b). In this transonic case, (3.158) is of mixed hyperbolic-elliptic type. If Ms > Mp > 1, then the motion is entirely supersonic, so both P -waves and S -waves are confined to Mach cones. Notice the analogy with the twosheeted characteristic cone shown in Figure 3.9. A possible application of these ideas concerns the violent penetration of an elastic solid by a shaped charge jet. A shaped charge is a device in which explosive surrounds a thin cone of elastic/plastic material. The detonation of the explosive causes such high stresses in the cone that it is ejected as an effectively liquid jet moving at speeds of up to 104 m s−1 . Hence, when the jet hits an elastic target, it generates a localised force that can penetrate at high speed. 3.7 Concluding remarks The ideas expounded in this chapter describe the fundamentals of many models that are regularly used to simulate elastic waves in situations ranging from tomography to oil exploration. In particular, non-destructive testing of solid structures is frequently carried out by examining elastic wave reflection,

EXERCISES

141

refraction and diffraction. In these studies, it is often the parameters λ and µ (which will usually be functions of position) that have to be determined from the response of a material to elastic waves, and hence they are called inverse problems. A classic example is the question, first posed by Kac (1966), can one hear the shape of a drum? ; in other words, if we know all the natural frequencies of a membrane, can we determine its boundary?† The main complication that arises when considering the Navier equation as opposed to the scalar wave equation is the existence of two distinct wavespeeds cp and cs . Many of the problems analysed in this Chapter are simplified significantly if we consider incompressible materials (cf §1.8), in which cp → ∞ and ÷u = 0 so that A = 0 in (3.36) and only S -waves propagate at finite speed. There are many other fascinating and practically important problems in the theory of elastic waves that fall beyond the scope of this book. These include finite-amplitude waves, to be discussed briefly in Chapter 5, and wave propagation through inhomogeneous media, to which we will return in Chapter 9. Exercises 3.1 Consider a wave with unit amplitude travelling in the positive xdirection on a string under tension T , whose density changes from %− to %+ at x = 0. From the continuity of the displacement and the force, show ³ thatpthis jump ´ in density transmits a wave with amplitude AT = 2/ 1 + %− /%+ and reflects a wave with amplitude AR = AT − 1 3.2 The characteristic surface through the origin for the two-dimensional wave equation (3.2) is defined to be the envelope of the planes (3.144) subject to the dispersion relation (3.145). Show that it is equivalent to the envelope of the planes (x − x0 ) cos θ + (y − u0 ) sin θ = c(t − t0 ) over all values of θ, and deduce that this is the characteristic cone (x − x0 )2 + (y − y0 )2 = c2 (t − t0 )2 . [The envelope of a one-parameter family of surfaces F (x, y, t; θ) may be found from the simultaneous equations F = ∂F/∂θ = 0.] In higher dimensions, show that on the envelope of (3.144), xi − i x0i = ck |k| (t − t0 ) and deduce (3.146). † The answer is now known to be no.

142


3.3 Show that, after a suitable change of independent variables, the onedimensional wave equation (3.1 can be written as ∂2w = 0. ∂η∂ξ Hence obtain the d’Alembert solution (3.127). 3.4 (a) Show that the two-dimensional radially-symmetric wave equation ∂ 2 w 1 ∂w 1 ∂2w = ∇2 w = + 2 2 c ∂t ∂r2 r ∂r

(3.160)

admits similarity solutions of the form 1 ³r´ w(r, t) = f , t t where f (η) satisfies µ ¶ ¡ 2 ¢ d 2 df 2 η c −η − η f = 0. dη dη (b) Assuming that f is well-defined as η → 0, show that A

f (η) = p

c2 − η 2

,

0 6 η < c,

where A is an arbitrary constant. (c) Show that Z ct r dr √ = ct 2 c t2 − r 2 0 and deduce that

 

1 √ 2 w(r, t) = 2πc c t2 − r2  0

0 6 r < ct, r > ct,

satisfies (3.160) subject to w = 0,

∂w = δ(x)δ(y) ∂t

at t = 0.

3.5 Show that, for any vector a and nonzero vector k, there exists a unique vector B and scalar A such that a = Ak+B×k and k·B = 0. [Hint: calculate a · k and a×k.]

EXERCISES

143

cs2

cg

cs1 k Fig. 3.13. Group velocity cg = dω/dk versus wavenumber k for symmetric (solid) and antisymmetric (dotted) Love waves.

3.6

(a) Show that the dispersion relation for symmetric Love waves is given parametrically by p p hω α s α2 + β 2 tan2 s 2 2 √ =√ s 1 + β tan s, kh = , cs2 1 − α2 1 − α2 (3.161a) where α =£ cs1 /cs2 , β = µ1¢/µ2 and the parameter s lies in the range s ∈ nπ, (n + 1/2)π for any integer n. (b) Show that antisymmetric Love waves (see §3.2.6) satisfy the condition (3.60). Deduce that their dispersion relation is given parametrically by p p hω α s α2 + β 2 cot2 s √ =√ s 1 + β 2 cot2 s, kh = , cs2 1 − α2 1 − α2 (3.161b) £ ¢ where now s ∈ (n − 1/2)π, nπ for any integer n. (c) Hence reproduce the dispersion graphs shown in Figure 3.4. (d) The group velocity cg of linear waves with dispersion relation ω = ω(k) is defined by cg (k) =

dω . dk

Use (3.6(b)) to find expressions for cg as a function of s for symmetric and antisymmetric waves, and hence plot cg versus k. Show that, as shown in Figure 3.13, the group velocity

144


is bounded above by cs2 but attains a local minimum before tending to cs1 as k → ∞. [This minimum in the group velocity allows an obstruction in a coal seam to be detected by studying the travel times of reflected waves.] 3.7 In dynamic plane strain, suppose that the stress components are given by τ11 =

1 ∂2U ∂2U − , ∂y 2 2c2s ∂t2

τ12 = −

∂2U 1 ∂2φ + 2 2, ∂x∂y 2cs ∂t

τ22 =

∂2U 1 ∂2U − , ∂x2 2c2s ∂t2 (3.162)

for two functions U(x, y, t) and φ(x, y, t). Deduce from the twodimensional Navier equation and constitutive relations that U and φ may be chosen to satisfy µ 2 ¶ µ 2 ¶µ 2 ¶ ∂ ∂ ∂ 2 2 2 2 2 2 − cs ∇ φ = − cp ∇ − cs ∇ U = 0 ∂t2 ∂t2 ∂t2 3.8 Show that the general solution of the scalar wave equation µ 2 ¶µ 2 ¶ 2 2 ∂ ∂ 2 ∂ 2 ∂ − c − c U=0 p s ∂t2 ∂x2 ∂t2 ∂x2 is U = Fp (x − cp t) + Gp (x + cp t) + Fs (x − cs t) + Gs (x + cs t) , where Fp , Gp , Fs , Gs are arbitrary scalar functions. 3.9 (a) Show that on a stress-free boundary located at x = 0, ρc2p ∂u/∂x+ ρ(c2p −2c2s )∂v/∂y = ρc2s (∂u/∂y+∂v/∂x) = 0. Show that the reflection coefficients for P-wave with wave vector (−kp cos α, kp sin α) are rp =

−c2p cos(2β) + 2c2s sin(α) sin(α + 2β) , c2p cos(2β) − 2c2s sin(α) sin(α − 2β)

rs =

2(cs /cp ) sin(2α)(c2p − 2c2s sin2 (α)) , c2p cos(2β) − 2c2s sin(α) sin(α − 2β)

where sin(α)/cp = sin(β)/cs . (b) write down the boundary conditions between two elastic media separated by the plane x = 0 and characterized by the constants ρ± , cs,± and cp,± . Show that sin β− sin α+ sin β+ sin α− = = = cp,− cs,− cp,+ cs,+

EXERCISES

145

Assume the left medium is slower than the right medium and consider a P-wave coming from x < 0. Discuss the number of transmitted waves as a function of α− . When does total internal reflection occur? 3.10 Show that for P-waves and S-waves of the form (3.64), up = ap (k, −iκp ) and us = as (κs , −ik), where the amplitudes ap,s are unknown, and that their speed is c = ω/k. Show that the boundary condition are 2kκp ap + (k 2 + κ2s )as = 0, ¡ 2 2 ¢ cp (κp − k 2 ) + 2c2s k 2 ap + 2c2s kκs as = 0, and, hence, that non-zero amplitudes only exist if the determinant ¡ ¢ k 4 (c2p − 2c2s ) + k 2 2c2s (2κp − κs )κs + c2p (κ2s − κ2p ) − c2p κ2p κ2s vanishes. Using the definitions of κp,s , derive µ ¶2 µ ¶1/2 µ ¶1/2 c2 c2 c2 2− 2 =4 1− 2 1− 2 . cs cp cs Show graphically that there is only one real solution to this equation for c. 3.11 Show that the dispersion relation (3.126) for torsional waves in a sphere reduces to tan η η tan η η tan η η

=

=

3 3−η 2

when n = 1,

=

12−η 2 12−5η 2

when n = 2,

75−8η 2 75−33η 2 +η 4

when n = 3,

where η = ωa/cs . By sketching the left- and right-hand sides as functions of η, show that there is a countably infinite set of solutions ηn,m in either case. By letting η → ∞, deduce that the highest modes are given approximately by 3 + ··· , mπ 5 ∼ (m + 1/2)π − + ··· , mπ 8 ∼ mπ − + ··· , mπ

η1,m ∼ mπ − η2,m η3,m as m → ∞.

146


3.12 Show that, in terms of the variables ξ = x − ct and η = x + ct, the one-dimensional wave equation (3.148a) subject to a moving point source is transformed to µ ¶ ∂2w Pc c−V = δ ξ+ η . (3.163) ∂ξ∂η 2T (c + V ) c+V Using the fact that, from (??), d2 ³ ´ |z| = 2δ(z), dz 2 deduce that the general solution of (3.163) is ¯ ¯ Pc ¯(c + V )ξ + (c − V )η ¯ + F (ξ) + G(η), w= 4T (c2 − V 2 ) where F and G are arbitrary functions. Transform back to (x, t) and apply the initial conditions to obtain the solution (3.149) when c > V When V > c, write x − V t = X and show that (??) becomes (V 2 − c2 )

∂w ∂2w P c2 δ(X). − 2V = ∂X 2 ∂t2 T

∂w Deduce that ∂X suffers a jump of P c2 /T (V 2 −c2 ) at x = V t and hence that w = AH(x + ct) · (x + ct) + BH(x − ct) · (x − ct) where H is the Heaviside function and A = P c2 /2T V (V − c), B = P c2 /2T V (V − c). 3.13 Prove the identity

δ(ax) ≡

1 δ(x), |a|

where δ√is the Dirac delta-function. Hence show that, in terms of X = ξ/ 1 − M 2 and y, (3.151) reads ∂2w ∂2w P + =− √ δ(X)δ(y) ∂X 2 ∂y 2 T 1 − M2 and deduce that

¡ ¢ P log X 2 + y 2 √ w=− + const. 2T 1 − M 2

4 Approximate Theories

4.1 Introduction So far in this book we have considered linear elasticity only for very simple geometries such as cylinders, spheres and half-spaces. In this chapter, we will consider more general solids under the restriction that they are thin and the equations of elasticity can consequently be simplified. A familiar example that we have already encountered is the wave equation governing the transverse displacements of a thin elastic string, and we will revisit this model below in §4.3. A string is characterised by its inability to withstand any appreciable shear stress, so its only internal force is a tension acting in the tangential direction. Similarly, a membrane is a thin, nearly two-dimensional structure, for example the skin of a drum, which supports only in-plane stresses. However, many thin elastic bodies have an appreciable bending stiffness and therefore do admit internal shear stress as well as tension. A familiar example is a flexible ruler, which clearly resists bending while deforming transversely in two dimensions, and is known as a beam. A thin, nearly one-dimensional object which can bend in both transverse directions, such as a curtain rod or a strand of hair, will be referred to as a rod. On the other hand, a nearly planar elastic structure with significant bending stiffness, for example a plane of glass or a stiff piece of paper, is called a plate. Finally, a shell is a thin, nearly two-dimensional elastic body which is not initially planar, for example a ping-pong ball or the curved panel of a car. In this chapter, we will show how models for all these structures can be derived using net force and moment balances combined with plausible constitutive relations. We will soon find, however, that it is difficult to enunciate all the physical assumptions that are needed. To do this requires a more systematic asymptotic analysis, and it will be deferred to Chapter 6. Here 147

148

Approximate Theories

δx T (x, t)

T (x + δx, t)

x Fig. 4.1. Schematic of a uniform bar showing the forces acting on a small segment of length δx.

we will focus initially on small displacements and hence obtain linear governing equations. However, we will find that the physically-based approach also allows us to look at the effect of geometrical nonlinearity, when the displacements are large but the strain is small. The nonlinear models that result exhibit interesting nonuniqeness properties that help us to understand important physical phenomena such as buckling.

4.2 Longitudinal displacement of a bar As a first illustrative example, let us consider longitudinal waves along a thin bar with uniform cross-section area A. We consider a short segment of the bar between some arbitrary point x and x + δx, where x measures longitudinal distance along the bar; we can think of x as an Eulerian or Lagrangian coordinate since, as explained in §1.7, they are indistinguishable in linear elasticity. The mass of this segment is ρAδx, where ρ is the volume density of the bar, and the dominant force it experiences consists of the tensions T exerted on each of its faces by the sections of the bar on either side, as illustrated in Figure 4.1. Denoting the longitudinal displacement by u(x, t), we thus obtain Newton’s second law in the form T (x + δx, t) − T (x, t) = ρAδx

∂2u (x, t). ∂t2

(4.1)

In the limit δx → 0, this reduces to the partial differential equation ∂2u ∂T = ρA 2 . ∂x ∂t

(4.2)

To close the problem, we now need a constitutive relation between T and u. On physical grounds, we might expect T to be proportional to the stretch ∂u/∂x, by analogy with Hooke’s law. Motivated by the exact static solution

4.3 Transverse displacements of a string

149

obtained in §2.2.3, we postulate the relation T = EA

∂u , ∂x

(4.3)

where E=

µ(3λ + 2µ) λ+µ

(4.4)

again denotes Young’s modulus. By substituting (4.3) into (4.2), we find that u(x, t) satisfies the wave equation E

∂2u ∂2u = ρ . ∂x2 ∂t2

(4.5)

p which says that longitudinal waves in the bar travel with speed E/ρ. Since E < λ+2µ, these waves travel slower than the longitudinal P-waves of §3.2.4. Typical boundary conditions are that u or the axial force EA∂u/∂x should be prescribed at each end of the bar and, of course, u and ∂u/∂t need to be given at t = 0. We should point out that various implicit assumptions underly the derivation given above. We have assumed, for example, that the longitudinal displacement u is uniform across the bar, and that the constitutive relation (4.3) holds although its derivation in §2.2.3 was performed only under static conditions. Neither of these assumptions is exactly true in practice, reflecting the fact that solutions of (4.5) are not exact solutions of the Navier equations. Intuitively, we expect (4.5) to be a good approximate model √ when the bar is thin and the displacement is small, specifically if u ¿ A ¿ L, where L is the length of the bar. However, this simple example illustrates the difficulty of assessing the validity of such an ad hoc model. We will show in Chapter 6 how such models can be derived from the underlying continuum equations in a more rigorous manner that allows the accuracy to be carefully estimated.

4.3 Transverse displacements of a string Before deriving a model for the transverse displacements of a beam, let us remind ourselves briefly of the corresponding derivation for an elastic string, which is characterised by the fact that the only internal force that it can support is a tension T acting in the tangential direction. Assuming that gravity g acts transversely to the string, as illustrated in Figure 4.2, the net

150


T (x + δx, t) θ g

T (x, t) w

x δx Fig. 4.2. Schematic of an elastic string showing the forces acting on a small segment of length δx.

force experienced by a small segment of undisturbed length δx is · µ ¶¸x+δx µ ¶ cos θ 0 T + δx, sin θ x −%g where % denotes the line density. Under the assumption that the transverse displacement w(x, t) is small compared to the length of the string, we can approximate the angle θ between the string and the x-axis by θ ∼ ∂w/∂x ¿ 1 and hence use the leading-order approximations sin θ ∼

∂w , ∂x

cos θ ∼ 1.

(4.6)

At the same time, we suppose that there is no longitudinal displacement so that the acceleration is just ∂ 2 w/∂t2 (1, 0)T . By applying Newton’s second law to the segment and letting δx → 0, we thus obtain the equations (a):

∂T = 0, ∂x

(b) :

T

∂2w ∂2w − %g = % . ∂x2 ∂t2

(4.7)

Hence we find again that the displacement satisfies the wave equation and that the tension T must be spatially uniform. Usually T is assumed to be a known constant although, in principle, it may be a function of time, for example while a guitar string is being tuned. Common experience suggests that a string is unable to withstand a compressive longitudinal force and hence that T should be positive. We can reach the same conclusion on mathematical grounds by noting that (4.7b) would change type from hyperbolic to elliptic if T were negative, and hence cease to be well-posed as an initial-value problem (Ockendon et al., 1999, page ????).

4.4 Transverse displacements of a beam

151

N (x + δx, t) T (x + δx, t) M (x, t)

g M (x + δx, t)

T (x, t) N (x, t)

Fig. 4.3. Schematic showing the forces and moments acting on a small segment of an elastic beam.

4.4 Transverse displacements of a beam 4.4.1 Derivation of the beam equation Now let us consider how to extend the model derived above to describe an elastic beam. In contrast with a string, a beam can support an appreciable transverse shear force N as well as the tension T , as illustrated in Figure 4.3; for convenience N is defined to act perpendicular to the x-axis. Again considering small, purely transverse displacements, we obtain the equations ∂T ∂ 2 w ∂N ∂2w = 0, T 2 + − %g = % 2 . (4.8) ∂x ∂x ∂x ∂t Now our task is to relate N to w, and we start by performing a moment balance on the segment shown in Figure 4.3. When doing so, we recall the suggestion from §2.6.7 that each section of the beam exerts, on its neighbours, as well as the tangential and normal force components T and N , a bending moment M about the y-axis. Conservation of angular momentum about the point x leads to the equation −M (x + δx, t) + M (x, t) + δxN (x + δx, t) = %g

δx2 ˆ∂ 2 θ +I 2, 2 ∂t

(4.9)

where θ ≈ ∂w/∂x is once again the small slope of the beam and the moment of inertia of the segment about its end is given by Iˆ = %(δx)3 /3. When we let δx → 0, the right-hand side of (4.9) is thus negligible and we are left with ∂M + N = 0. (4.10) − ∂x Finally, we need a constitutive relation for M . It is a practical experience, for example when bending a flexible ruler, that applying an increasing moment to a beam results in an increasing curvature. For small displacements,

152


(a)

(b)

(c)

x

x

x

Fig. 4.4. Schematic of the end of a beam under (a) clamped, (b) simply supported and (c) free conditions.

this suggests a constitutive relation of the form M = −B

∂2w , ∂x2

(4.11)

where B is a constant known as the bending stiffness. Again we can use a simple exact solution of the Navier equations from Chapter 2 to derive an equation for B. In this case we consider the plane stress solution (2.211) in which the displacement field and the only nonzero stress component are     u −2xz κ , u = v  =  τxx = −κEz. (4.12) 2νyz 2 2 2 2 w x − νy + νz We identify the transverse displacement of the beam with κx2 /2, neglecting the small νy 2 and νz 2 on account of the slenderness of the beam. The net moment generated by this stress field on the cross-section A is given by ZZ ZZ M= zτxx dydz = −Eκ z 2 dydz, (4.13) A

A

which corresponds to (4.11) with ZZ B = EI,

z 2 dydz

where I =

(4.14)

A

is the moment of inertia of the cross section A about the y-axis. We also note that (4.10), (4.11) are consistent with the asymptotic result (2.149) for a semi-infinite strip. Postulating that the constitutive relation (4.11) still holds under dynamic conditions, we combine (4.8)–(4.14), to obtain the beam equation −EI

∂2w ∂2w ∂4w + T 2 − %g = % 2 . 4 ∂x ∂x ∂t

(4.15)

Evidently, (4.15) reduces to (4.7) as the bending stiffness EI tends to zero.

4.4 Transverse displacements of a beam

153

Unlike the wave equation, though, (4.15) is not immediately classifiable as elliptic or hyperbolic.† 4.4.2 Boundary conditions On physical grounds we expect to need to specify w and ∂w/∂t at t = 0, as for a string. However, since (4.15) has four x-derivatives rather than two, we might anticipate that it requires two boundary conditions at each end of the beam, in contrast with a string. If, for example, a beam is clamped as shown in Figure 4.4(a), then both w and ∂w/∂x would be prescribed at the clamp; in the case illustrated ∂w = 0. (4.16) ∂x Alternatively, we could envisage a simply supported beam end, as shown in Figure 4.4(b), where the displacement is fixed without applying any moment, so that w = M = 0 and thus w=

∂2w = 0. (4.17) ∂x2 In the final example shown in Figure 4.4(c), the end of the beam is free. It therefore experiences no shear force or bending moment, and from (4.10) and (4.11) we deduce the boundary conditions w=

∂2w ∂3w = = 0, (4.18) ∂x2 ∂x3 and in this case we must also have T ≡ 0. If we specify tractions on the end of a beam, then the only information about them required for our model is the net force and moment that they exert. Specifically, if the longitudinal and tangential stress components applied to the section x = 0 are σx (y, z) and σz (y, z) respectively, then the boundary conditions to be imposed on the beam equation are ZZ ZZ ZZ T = σx (y, z) dydz, N = σz (y, z) dydz, M = zσx (y, z) dydz. A

A

A

(4.19)

Here we are effectively invoking St Venant’s principle: all details of the † It is interesting to note that the real and imaginary parts of the solution ψ of the Schr¨ odinger equation ~ ∂2ψ ∂ψ =− ∂t 2m ∂x2 p satisfy (4.15) with T = g = 0, ~/2m = EI/%. i

154


edge tractions apart from the three numbers T , N and M are lost outside a neighbourhood of the edge where they are applied. We showed in §2.6.7 that this is true in plane strain, and here we assume that it applies more generally.

4.4.3 Compression of a beam One of the most dramatic new features of (4.18) compared with (4.7) is that we can now consider situations in which T < 0. We illustrate an unexpected new possibility that arises in such cases by considering steady displacements of a beam of length L subject to negligible gravity. We suppose that the ends are clamped and subject to a compressive force P = −T and zero transverse force. In this case (4.15) reduces to the ordinary differential equation EI

d4 w d2 w + P = 0, dx4 dx2

(4.20a)

subject to the boundary conditions d3 w dw = =0 dx dx3

at x = 0, L.

(4.20b)

We can consider (4.20) as an eigenvalue problem. One possibility is always that the beam remains straight with w ≡ const. but, if the applied force takes one of the discrete values P n2 π 2 = , EI L2

(4.21)

where n is an integer, then (4.20) admits the eigenfunction solution w = A cos

³ nπx ´ L

,

(4.22)

where A is a constant. The first three such eigenfunctions are shown in Figure 4.5; these are the shapes that a beam would adopt as it buckles under a sufficiently large compressive force. The amplitude A appears to be indeterminate, although we might expect it to to be a smoothly increasing function of the applied force. The resolution of the indeterminacy requires us to include geometrically nonlinear effects in the model, and we will show how this is done in §4.9.

4.4 Transverse displacements of a beam n=1

w

n=2

w

1

1

1

0.5

0.5

0.2

0.4

0.6

0.8

1

x

0.2

0.4

0.6

n=3

w

0.5

0.8

1

x

0.2

-0.5

-0.5

-0.5

-1

-1

-1

155

0.4

0.6

0.8

1

x

Fig. 4.5. The first three buckling modes of a clamped elastic beam (with a = 1 and L = 1).

4.4.4 Waves on a beam Concerning the dynamic equation (4.15) with g = 0 and adopting the philosophy of §3.2.2, we can seek travelling-wave solutions of the form ¡ ¢ u = A exp i(kx − ωt) , (4.23) where the real part is assumed as usual, and hence obtain the dispersion relation %ω 2 = EIk 4 + T k 2 .

(4.24)

Elastic waves on a beam are therefore dispersive, unlike P -waves, S -waves or waves on a string. This should not come as too much of a surprise, since we have already discovered in Chapter 3 that elastic waves are usually dispersive in bounded domains. For a beam under positive tension T > 0, waves with frequency ω can thus propagate with wavelength 2π/k provided s µ ¶1/2 ω T EIk 2 = 1+ . (4.25) k % T Evidently this reduces p to the dispersion relation for a string if the waves are long, with k ¿ T /EI. However, the bending stiffness becomes increasingly important, and the waves become increasingly dispersive, as the wavelength decreases. This observation can be used to explain, for example, why the harmonics of a piano string with small but nonzero bending stiffness are slightly sharp compared with the fundamental (see Exercise 4.1). For a beam under compression, with P = −T > 0, wavelike solutions of (4.15) exist only if k 2 > P/EI. For smaller values of k, (4.24) leads to complex values of ω, corresponding to solutions that grow exponentially in time. Itp follows that the beam is unstable to waves with wavelength greater than 2π EI/P , and we will investigate this further in §4.9.3.

156


Nz

z y

Ny T x

Fig. 4.6. Schematic of a thin elastic rod showing the internal force components T , Ny and Nz .

z

y Mz My

Mx

x

Fig. 4.7. Cross-section through a rod showing the bending moment components.

4.5 Linear rod theory We next consider the motion of a nearly straight uniform elastic rod that can bend in two transverse directions, so that the displacement is w(x, t) = ¡ ¢T v(x, t), w(x, t) . Now, in addition to the tension T , there are two shear force components Ny and Nz in the y- and z-directions respectively, as illustrated in Figure 4.6. By applying Newton’s second law and assuming small displacements as above, we again find that T is spatially uniform and obtain the following generalisation of (4.15): T

∂ 2 w ∂N ∂2w + + %f = % , ∂x2 ∂x ∂t2

(4.26)

¡ ¢T where N = Ny , Nz and %f is the body force per unit length. In general we must now allow for three components of the bending moment M = (Mx , My , Mz )T . As shown in Figure 4.7, My is the moment about the y-axis previously referred to simply as M in §4.4, while Mz represents bending in the (x, y)-plane. We will begin by neglecting the first component Mx , which corresponds to twisting of the rod about its axis. A balance of

4.5 Linear rod theory

157

moments in the y- and z-directions analogous to (4.10) then leads to ∂My ∂Mz − Nz = 0, + Ny = 0. (4.27) ∂x ∂x Finally, we need to generalise the constitutive relation (4.11), and we will again use the plane stress solution (4.12) as a guide. By combining (4.12) with an analogous expression representing bending in the y-direction, we obtain the following exact solution of the Navier equations:     −2xz −2xy κ1   + κ2 x2 + νy 2 − νz 2  , (4.28a) u= 2νyz 2 2 x2 − νy 2 + νz 2 2νyz τxx = −κ1 Ez − κ2 Ey.

(4.28b)

The moments, about the y- and z-axes respectively, generated over the crosssection A by this stress field are given by ZZ ZZ My = zτxx dydz, Mz = − yτxx dydz. (4.29) A

A

By substituting (4.28) into (4.29) we obtain the relations µ ¶ ∂2w Mz = EI 2 , −My ∂x where

µ ¶ ZZ Iyy Iyz I= = Iyz Izz

µ A

y 2 yz yz z 2

(4.30)

¶ dydz

(4.31)

is the inertia matrix of the cross-section.† Hence the generalisation of (4.15) is the coupled system ∂2w ∂4w ∂2w − EI + %f = % . (4.32) ∂x2 ∂x4 ∂t2 Note that we can always choose the y- and z-axes such that the symmetric positive definite inertia matrix I is diagonal; these correspond to the directions in which the resistance to bending is maximised and minimised. It follows that the y- and z-components of (4.32) are simply two decoupled beam equations for v and w.(Note, though, that this decoupling does not occur if the rod cross-section is non-uniform in x.) By analogy with (4.15), we therefore expect to have to impose initial conditions on w and ∂w/∂t as well as two boundary conditions each on v and w at either end of the rod. T

† Note that if x is interpreted as time, (4.30) is the equation of motion of a rigid body whose angular velocity is nearly aligned with a princpal axis; this is an example of the Kirchhoff Analogy.

158


(b) (a)

a

Iz = ab3 /12

a

b

Iy = a3 b/12

b (c)

Iz

Iy = Iz = πab(a2 + 4b2 )/4

Iy > I z

Fig. 4.8. Some examples of cross-sections and their moments of inertia.

The conditions corresponding to clamped, simply supported and free ends are obvious generalisations of (4.16), (4.17) and (4.18). Examples of values of Iy , Iz for various cross-section shapes are given in Figure 4.8. These are crucial pieces of informations to bear in mind when designing beams in practice. For example, if one considers the bending of a long strip of paper with transverse dimensions a and b in the z- and ydirections respectively, where b À a, it is then easy to see as in Figure 4.8(a) that Iy ¿ Iz . This explains why it is so much easier to bend such a strip about the y-axis than about the z-axis. On the other hand, if we roll the same paper strip into a long cylinder, it will become become much stiffer in all bending directions as illustrated in Figure 4.8(b). This situation can be seen as a limiting case of a cylindrical shell, and it gives us a hint as to why curved shells are generally much stiffer than flat rods or plates, and we will return to this point in §??. Proceeding along the same lines as for longitudinal waves, we now consider torsion, i.e. twisting about the longitudinal axis. As illustrated in Figure 4.7, we denote by Mx the moment about the x-axis exerted on each cross-section through the rod. By balancing the net torque with the rate of change of angular momentum of a small segment of length δx and then letting δx tend to zero, we obtain ∂2θ ∂Mx = ρ (Iyy + Izz ) 2 , ∂x ∂t where θ(x, t) denotes the angle of twist about the x-axis.

(4.33)

4.6 Linear plate theory

159

Ny Tyy Txy Txx Nx

δy Tyx

Nx

Tyx

δx

z

Txx y Txy x

Tyy Ny

Fig. 4.9. Schematic showing the forces acting on a small section of an elastic plate.

Now we refer to the exact steady solution of §2.4 to pose the constitutive relation

Mx = R

∂θ ∂x

(4.34)

between the torque and the twisting gradient, where R again denotes the torsional p rigidity.† We thus find that θ satisfies the wave equation, with wave speed R/ρ (Iyy + Izz ). For the simple case of a circular cross-section, it is easy to show that

Iyy + Izz =

πa4 , 2

R=

πa4 µ 2

(4.35)

and hence that the wave speed is equal to cs . This reproduces the wave speed of the first family of torsional waves found in §3.4.1, that is (3.86) with n = 0 and ξ2,0 = 0; the theory presented here assumes each section deforms uniformly and thus fails to discern the higher modes.

160


4.6 Linear plate theory 4.6.1 Derivation of the plate equation Now we derive the equation governing the small transverse displacement w(x, y, t) of an elastic plate that is bent out of the plane z = 0. We describe the in-plane forces using a two-dimensional stress tensor ¶ µ Txx Txy , (4.36) T = Tyx Tyy where Tij is the i-component of the force per unit length on a section of the plate with outward normal in the j-direction. We also denote the transverse shear force on a section with normal in the j-direction by Nj . The forces acting on a small rectangular section of the plate are illustrated in Figure 4.9. By balancing moments around the z-axis, as in §1.5, we discover that Tyx ≡ Txy , so that T is symmetric. If we assume that there is no acceleration in the x- and y-directions and that the only body force is gravity acting in the negative z-direction, then Newton’s second law gives ∂Txx ∂Txy + = 0, ∂x ∂y ∂ ∂x

∂Txy ∂Tyy + = 0, ∂x ∂y

µ ¶ ∂w ∂w Nx + Txx + Txy ∂x ∂y µ ¶ ∂ ∂w ∂w ∂2w + Ny + Txy + Tyy − ςg = ς 2 , ∂y ∂x ∂y ∂t

(4.37)

(4.38)

where ς is now the mass of the plate per unit area. Considering the bending moments, we use an analogous notation Mij (i, j = x, y) to denote the i-component of the moment per unit length acting on a section with outward normal in the j-direction, as illustrated schematically in Figure 4.10. We note that Mij is not in general symmetric, although it is a two-dimensional tensor, as will shortly become apparent. By balancing moments on the segment shown in Figure 4.10, we obtain the relations ∂Mxx ∂Mxy + + Ny = 0, ∂x ∂y

∂Myx ∂Myy + − Nx = 0. ∂x ∂y

(4.39)

Next we seek constitutive equations for the bending moments which, by analogy with §4.4, we expect to be linear functions of the second derivatives of w. Again we can use exact plane stress solutions to evaluate the appropriate coefficients. In this case, by choosing χ0 = φ = 0 and χ1 a † For consistency with the notation of this Chapter, x should be substituted for z in the results of §2.4.


161

z Mxy Myx

y Myy Mxx x

Fig. 4.10. Schematic showing the bending moments acting on a section of an elastic face.

suitable quadratic function of x and y in (2.209), we construct a solution with displacement field 

 −2(ax + by)z 1  u=  −2(bx + cy)z 2 2 2 2 ax + 2bxy + cy + ν(a + c)z /(1 − ν)

(4.40)

and nonzero stress components τxx = −

E(a + νc)z , 1 − ν2

τxy = −

Ebz , 1+ν

τyy = −

E(νa + c)z . 1 − ν2

(4.41)

Now by considering the moments exerted on surfaces with x = const. and y = const. respectively, we obtain the expressions Z Mxx = − Mxy = −

Z

h/2

−h/2 Z h/2 −h/2

zτxy dz,

Myx =

zτyy dz,

Myy =

h/2

−h/2 Z h/2 −h/2

zτxx dz,

(4.42a)

zτxy dz,

(4.42b)

where h is the plate thickness. Hence, Mij are the components of the tensor µ ¶ Z h/2 µ ¶ 0 −1 τxx τxy z dz. 1 0 −h/2 τxy τyy

162


By substituting (4.40) and (4.41) into (4.42) we obtain the relations Eh3 ∂2w , 12(1 + ν) ∂x∂y ¶ µ 2 ∂ w ∂2w Eh3 +ν 2 , =− 12 (1 − ν 2 ) ∂x2 ∂y µ ¶ 3 2 2 Eh ∂ w ∂ w = ν 2 + . 12 (1 − ν 2 ) ∂x ∂y 2

Mxx = −Myy = Myx Mxy

(4.43a) (4.43b) (4.43c)

When we use the constitutive relations (4.43) in (4.38), we are left with the plate equation Txx

∂2w ∂2w ∂2w ∂2w 4 + 2T + T − D∇ w − ςg = ς , xy yy ∂x2 ∂x∂y ∂y 2 ∂t2

(4.44)

where ∇2 = ∂ 2 /∂x2 + ∂ 2 /∂y 2 and D=

Eh3 12 (1 − ν 2 )

(4.45)

is the bending stiffness of the plate. A membrane is a plate with negligible bending stiffness, and is described by (4.44) with D = 0. Unfortunately, the problem is still underdetermined since (4.37) gives us only two equations in the three stress components. To close the problem, we also need constitutive relations for Txx , Txy and Tyy in terms of the inplane displacements of the plate. We leave this calculation to §4.6.5, here concentrating on the simple case where the plate is subjected to a known isotropic tension T . This corresponds to the trivial exact solution of (4.37) in which Txx = Tyy = T and Txy = 0, and reduces (4.44) to T ∇2 w − D∇4 w − ςg = ς

∂2w . ∂t2

(4.46)

4.6.2 Boundary conditions As initial conditions for (4.46), we should specify w and ∂w/∂t. For a clamped plate, the boundary conditions are analogous to those for a beam, namely w=

∂w = 0, ∂n

(4.47)

where ∂/∂n is the normal derivative. For a simply supported plate, we require the displacement and the bending


163

moment about an axis tangential to the boundary to be zero. The latter may be written in the form µ ¶µ ¶ ¡ ¢ Mxx Mxy nx Msn = −ny , nx = 0, (4.48) Myx Myy ny ¡ ¢T where n = nx , ny is the outward-pointing unit normal to the boundary. By substituting for the moments from (4.43) and using the fact that w is zero on the boundary, we can reduce (4.48) to w=

∂w ∂2w + νκ = 0, 2 ∂n ∂n

(4.49)

where κ is the curvature of the boundary (see Exercise 4.8). Only for a straight boundary, then, can we apply the obvious generalisation of (4.17) in which w and its second normal derivative are zero. Finally we consider the boundary conditions to be imposed at a free edge, where a horizontal force balance evidently requires that T = 0. If, for the sake of argument, the boundary is at x = 0, then, for there to be zero applied stress, we are apparently led to the conditions Nx = Mxx = Myx = 0. However, it is to impossible to impose three boundary conditions on the biharmonic problem (4.46)! This conceptual difficulty caused great consternation to the pioneers of plate theory (see Love, 1944, Article 297), and a version of St Venant’s principle is commonly invoked to reduce the number of boundary conditions to two. However, a systematic resolution of the difficulty requires us to acknowledge that there is a boundary layer near the free edge where the assumptions that gave rise to the plate equation fail. The careful analysis of this layer carried out in §6.3 shows that the correct conditions at a free edge x = 0 are ∂Myx ∂Mxx −2 =0 ∂x ∂y

Myx = 0,

(4.50)

or, in terms of the displacement, ∂2w ∂2w + ν = 0, ∂x2 ∂y 2

∂3w ∂3w + (2 − ν) = 0. ∂x3 ∂x∂y 2

(4.51)

By using (4.39) and (4.43), we may also rewrite the second of (4.50) as Nx =

∂Mxx , ∂y

(4.52)

which shows that the shear stress Nx as calculated from plate theory does not in general vanish at a free edge!

164


The conditions (4.49) and (4.52) suggest that the “boundary layer” analysis, is even more difficult in the vicinity of a sharp corner in the plate. Were we to assume that a condition of the type (4.52) was to hold on the free edges x = 0, y = 0 of a plate lying in the positive quadrant, the transverse shear stress would usually be discontinuous at the origin; moreover, (4.39) suggests that the curvature of a simplysupported plate may become very large there. A further discussion of this situation is given in ?). Although it is beyond the scope of this chapter to derive systematically the strain energy in the plate, we note that, for the exact solution (4.40) in which tension is neglected, the nonzero stress components are given by (4.41) and the corresponding strain components by εxx = −az,

εxy = −bz,

εyy = −cz.

(4.53)

Recalling the definition (1.55) of the strain energy density W, we find that the stored energy per unit plate area is Z h/2 Z h/2 1 τij εij dz W dz = −h/2 −h/2 2 ª D© 2 = a + 2νac + c2 + 2(1 − ν)b2 2 ¡ ¢ª D© (a + c)2 + 2(1 − ν) b2 − ac . (4.54) = 2 This suggests that more generally, the net strain energy might take the form Ã µ 2 ¶2 !) ZZ ( ¡ 2 ¢2 D ∂2w ∂2w ∂ w ∇ w − 2(1 − ν) dxdy, (4.55) − 2 2 2 ∂x ∂y ∂x∂y Ω where the integral is taken over the region Ω in the (x, y)-plane occupied by the plate. The integrand in (4.55), which represents the strain energy due to bending, is particularly interesting, showing that the total energy is a combination of both the mean curvature ∇2 w = κ1 + κ2

(4.56)

and the Gaussian curvature ∂2w ∂2w − ∂x2 ∂y 2

µ

∂2w ∂x∂y

¶2 = κ1 κ2

(4.57)

where κ1 and κ2 are the principal curvatures of the surface z = w(x, y). We will soon discover that these geometric quantities are fundamental to all theories of plates and shells and we can already see a hint of this when


165

we consider minimising (4.55) by changing w by a small “virtual” displacement. Using the calculus of variations, as in Exercise 4.10, we find that the minimisation of the Gaussian curvature term contributes nothing to leading order, while minimising the mean curvature demands that ∇4 w = 0. Hence bending manifests itself geometrically just as the mean curvature of the plate. ?????????? that (4.51) are the natural boundary conditions for the minimisation.

4.6.3 Simple solutions of the plate equation Let us start by considering steady solutions of (4.46) with T = 0, that is ∇4 w = −

ςg . D

(4.58)

Solutions of (4.58) describe the sagging of a plate under gravity. We will consider both clamped and simply supported boundary conditions; it is impossible to solve (4.58) with entirely free boundaries since some boundary tractions are needed to balance gravity. For example, to describe the sagging of an elastic disc r < a we can easily solve (4.58) in cylindrical polar coordinates. With clamped boundary conditions w = dw/dr = 0 at r = a, we find that w=−

¢2 ςg ¡ 2 a − r2 . 64D

(4.59)

Alternatively, the simply supported boundary conditons (4.49) at r = a lead to the solution µ ¶ ¢ 5+ν 2 ςg ¡ 2 2 2 a −r a −r . (4.60) w=− 64D 1+ν Note in particular that the reduced rigidity afforded by the simple support compared with the clamped plate means that (4.60) gives a larger displacement at the centre of the disc by a factor of (5 + ν)/(1 + ν) compared with (4.59). Next let us consider the sag of a rectangle 0 < x < a, 0 < y < b. For clamped boundary conditions, we encounter exactly the same difficulty as in §2.6.6. The non-orthogonality of the functions encountered in separating the variables in the biharmonic equation with boundary conditions on w and ∂w/∂n makes any analytical solution impractical. However, the simply ∂w supported boundary conditions w = ∂n 2 = 0 are now physically realistic and it is straightforward to separate the variables and obtain the solution

166


0 -0.0002 -0.0004 w -0.0006 0.5 0.4 0.3 0.2 0.1 y

0 0.2 0.4

x 0.6 0.8 10

Fig. 4.11. The solution (4.61) for a plate sagging under gravity, with a = 1, b = 1/2, ςg/D = 1.

as a sum of mutually orthogonal trigonometric functions, namely w=

∞ −16ςg X sin (mπx/a) sin (nπy/b) . 2 2 2 2 2 π6 D m,n=1 mn (m /a + n /b )

(4.61)

m odd n odd

A typical solution is shown in Figure 4.11. 4.6.4 An inverse plate problem Here we describe a novel version of the plate equation that is relevant to the manufacture of curved windscreens by heating horizontal glass plates simply supported on a frame. The heating decreases the bending stiffness D and hence the sag of the glass under gravity and, by suitably distributing the heating, any desired sag can be achieved in principle. If D is temperatureand hence position-dependent, and the in-plane tension is negligible, then the displacement satisfies µ µ 2 ¶¶ µ ¶ ∂2 ∂ w ∂2w ∂2 ∂2w D D +ν 2 + 2(1 − ν) ∂x2 ∂x2 ∂y ∂x∂y ∂x∂y µ µ 2 ¶¶ 2 ∂ ∂ w ∂2w + 2 D +ν 2 + ςg = 0. (4.62) ∂y ∂y 2 ∂x Given D(x, y), (4.62) is a simple generalisation of the biharmonic equation which is straightforward in principle to solve for w(x, y), at least numerically. The corresponding inverse problem is, given a desired sag profile w(x, y), to determine the required distribution of the bending stiffness D(x, y). When viewed as a problem for D, (4.62) is a second-order linear partial differential


167

equation, and two immediate questions arise: “what boundary conditions should be imposed?” and “will D be positive?”. Since the terms involving the highest derivatives of D in (4.62) are µ

∂2w ∂2w + ν ∂x2 ∂y 2

¶

∂2D ∂2w ∂2D + 2(1 − ν) + ∂x2 ∂x∂y ∂x∂y

µ

∂2w ∂2w + ν ∂y 2 ∂x2

¶

∂2D , ∂y 2 (4.63)

we can read off the discriminant µ 2

∆ = (1 − ν)

∂2w ∂x∂y

¶2

µ −

∂2w ∂2w + ν ∂x2 ∂y 2

¶µ

∂2w ∂2w + ν ∂y 2 ∂x2

¶ ,

(4.64a)

and simple rearrangements of the right-hand side yield 2 +M M Mxx xy yx 2 D Ã µ 2 µ 2 ¶2 ! ¶2 ∂ w ∂2w ∂2w ∂2w ∂ w 2 − ν = −(1 − ν) − + . ∂x2 ∂y 2 ∂x∂y ∂x2 ∂y 2

∆=

(4.64b) (4.64c)

Now, if there is a simply-supported edge at x = 0, then we must have Myx = 0 there and we deduce from (4.64b) that ∆ > 0 and (4.62) is therefore hyperbolic. On the other hand, the right-hand side of (4.64c) shows that ∆ < 0 wherever the Gaussian curvature is positive, which we expect to be the case nearly everywhere in a typical windscreen. It follows that (4.62) must change type somewhere between the simply supported boundary and the interior, and it is an interesting open question as to whether smooth solutions to this mixed type partial differential equation exist. With reference to our original plate model of §4.6.1, we remark that, in practice, simply supported windscreens manufactured in this way can easily “lift off” their supporting frame in the vicinity of fairly sharp corners. This is further evidence for the possible failure of our plate theory in the presence of corners.

4.6.5 More general in-plane stresses If we do not make the simplifying assumption that the in-plane stress is isotropic and known, then we need to obtain constitutive relations for the components Txx , Txy and Tyy . We appeal to the exact biaxial strain solution

168


(2.18) to pose the constitutive relations Eh (exx + νeyy ) , 1 − ν2 Eh = exy , 1+ν Eh (νexx + eyy ) , = 1 − ν2

Txx = Txy = Tyx Tyy

(4.65a) (4.65b) (4.65c)

where the linearised strain components are related to in-plane displacements by µ ¶ ∂u 1 ∂u ∂v ∂v exx = , exy = + , eyy = . (4.66) ∂x 2 ∂y ∂x ∂y We therefore have a plane strain problem to solve for the two-dimensional displacement field (u, v)T . From (4.37) we infer the existence of an Airy stress function A such that Txx =

∂2A , ∂y 2

Txy = −

∂2A , ∂x∂y

Tyy =

∂2A . ∂x2

(4.67)

Then elimination of u and v from (4.65) reveals that A satisfies the twodimensional biharmonic equation ∇4 A = 0.

(4.68)

The transverse displacement of the plate is therefore governed by ∂2A ∂2w ∂2A ∂2w ∂2A ∂2w ∂2w + −2 − D∇4 w − ςg = ς 2 , 2 2 2 2 ∂y ∂x ∂x∂y ∂x∂y ∂x ∂y ∂t

(4.69)

where A satisfies (4.68). The isotropic tension T assumed in ¡(4.46) corre¢ sponds to the particular solution of (4.68) in which A = T x2 + y 2 /2. Equally, stretched membrane corresponds to D = 0, ¡ an anisotropically ¢ A = T1 x2 + T2 y 2 /2. In general, however, in addition to the displacement and/or bending conditions (4.47), (4.49) needed to determine w given A, we would need to specify two further boundary conditions around the edge of the plate to determine A. These would typically consist of specified in-plane displacements or tractions, that is µ ¶ µ ¶ u Txx Txy = ub (x, y) or n = σ(x, y) (4.70) v Txy Tyy where n is the unit normal to the edge of the plate.

4.7 Von K´ arm´ an plate theory

169

4.7 Von K´ arm´ an plate theory 4.7.1 Assumptions underlying the theory There is an interesting geometrically nonlinear generalisation of plate theory which can be derived plausibly using the ideas of §4.6, although we will have to wait until Chapter 6 for a systematic exposition on the range of applicablity of the theory. We follow the derivations used in the previous section to obtain the governing equations ∂Txx ∂Txy + = 0, ∂x ∂y

∂Txy ∂Tyy + =0 ∂x ∂y

(4.71)

∂2w ∂2w ∂2w ∂2w 4 + 2T + T − D∇ w − ςg = ς xy yy ∂x2 ∂x∂y ∂y 2 ∂t2

(4.72)

for the in-plane tensions and Txx

for the transverse displacement. Any small segment of the plate will be locally approximately planar, so we also assume that the constitutive relations (4.65) still hold. Our focus, then, is on obtaining improved approximations for the strain components eij . We note with concern that in (4.65) we are including the in-plane displacement, although we derived (4.71) by neglecting the in-plane acceleration. A systematic justification of this step requires us to render the equations dimensionless and take a careful asymptotic limit, as we will demonstrate in Chapter 6. Nevertheless, it is apparent that our model (4.44) requires both the in-plane displacements (u, v) and the transverse displacement w to be sufficiently small. The von Kárm´ an theory emerges when we make a specific choice about exactly how small they must be, and it will transpire that the in-plane displacements must be an order of magnitude smaller than the transverse displacement. This choice means that all the displacement components contribute to the leading-order strain of the plate, as we will discover below, yet it also allows us to retain the linear constitutive relations (4.65).

4.7.2 The strain components A point that starts on the centre-surface of the plate with Lagrangian position vector X will be displaced to a new Eulerian position x, where     X X + u(X, Y ) X = Y  , x(X) =  Y + v(X, Y )  . (4.73) 0 w(X, Y )

170


A nearby point with initial position (X + δX, Y + δY, 0)T will arrive at   X + δX + u(X + δX, Y + δY ) x(X + δX) =  Y + δY + v(X + δX, Y + δY )  w(X + δX, Y + δY )     ∂u/∂Y 1 + ∂u/∂X ³ ´ (4.74) ∼ x(X) + δX  ∂v/∂X  + δY 1 + ∂v/∂Y  + O |δX|2 ∂w/∂Y ∂w/∂X after the deformation. Now, when calculating the change in length of the small line element δX, we retain only the leading terms in u, v and w to obtain " µ ¶ # ∂u ∂w 2 2 2 2 |δx| − |δX| ∼ δX 2 + ∂X ∂X " · ¸ µ ¶2 # ∂u ∂v ∂w ∂w ∂v ∂w + 2δXδY + + + δY 2 2 + . (4.75) ∂Y ∂X ∂X ∂Y ∂Y ∂Y Our assumption that w, although small, greatly exceeds u and v, means that the terms on the right-hand side of (4.75) are all of the same order. It also means that the Lagrangian and Eulerian coordinates are identical to leading order, so we can replace (X, Y ) with (x, y) without incurring any additional errors. We therefore define the in-plane strain components to be µ ¶ ∂u 1 ∂w 2 exx = + , (4.76a) ∂x 2 ∂x · ¸ ∂w ∂w 1 ∂u ∂v + + (4.76b) exy = 2 ∂y ∂x ∂x ∂y µ ¶ ∂v 1 ∂w 2 eyy = + , (4.76c) ∂y 2 ∂y Then (4.75) gives us the change in length of a line element in the form |δx|2 − |δX|2 ∼ 2exx δX 2 + 4exy δXδY + 2eyy δY 2 .

(4.77)

Notice that (4.76) may also be obtained from (1.13) by assuming that w is an order of magnitude larger than u and v. By including the nonlinear terms in (4.76) we are abandoning one of the principal assumptions of linear elasticity, namely that the strain may be approximated as a linear function of the displacement gradients. However, we still plan to use the linear constitutive relation (4.65) between stress and


171

strain. The Von Kármán theory is our first example of a theory that is mechanically linear but geometrically nonlinear. We will discuss the distinction between geometrical and mechanical nonlinearity further in Chapter 5. Before proceeding to incorporate (4.76) into the governing equations for the plate, let us pause to contrast them with the corresponding plane strain components (4.66) where w ≡ 0. First, we recall that in plane strain (4.66) are only soluble for u and v if the compatibility condition (2.180a) is satisfied. Here, (4.76) give us three equations in the three displacements, so there is no compatibility condition, but (4.76) do give us the interesting result that µ 2 ¶2 ∂ 2 exy ∂ 2 eyy ∂ 2 exx ∂2w ∂2w ∂ w − +2 − = − . (4.78) 2 2 2 2 ∂y ∂x∂y ∂x ∂x ∂y ∂x∂y Second, let us pose the question: “which displacement fields give rise to zero in-plane strain?” We see from (4.77) that such deformations will preserve the lengths of all line elements in the plate: |δx| and |δX| will be equal for all δX and δY . Such length-preserving transformations are known as isometries, and, in plane strain, the only isometries are rigid-body motions. When we allow the additional freedom of transverse as well as in-plane displacements, there is another class of isometries, corresponding to bending of the plate. It is, for example, easy to bend this page without significantly changing the lengths of any lines printed upon it. When we have zero in-plane strain, the equations exx = exy = eyy = 0 form three coupled partial differential equations for u, v and w. The elimination of u and v that led to (4.78) reduces the system to a single equation for w, namely µ 2 ¶2 ∂2w ∂2w ∂ w = 0. (4.79) − ∂x2 ∂y 2 ∂x∂y Thus a surface z = w(x, y) can be isometrically deformed from a plane if and only if w satisfies (4.79), which is a version of the Monge–Ampère equation (Ockendon et al., 1999, page ????). A surface possessing this property is known as developable. Roughly speaking, a developable surface is any shape into which a piece of paper may be bent without creasing or tearing, and familiar examples include cylinders and cones. Equation (4.79) also states that the plate has zero Gaussian curvature. We recall from §4.6.2 that the Gaussian curvature is defined to be K = κ1 κ2 ,

(4.80)

where κ1 and κ2 are the principal curvatures of the surface z = w(x, y). A developable surface therefore has either κ1 = 0 or κ2 = 0 everywhere, so at

172

Approximate Theories (a)

(b)

(c)

(d)

Fig. 4.12. (a) A cylinder. (b) A cone. (c) Another developable surface. (d) A hyperboloid, which is ruled but not developable.

each point there is one principal direction in which the surface has zero curvature. These directions define a family of straight lines, known as generators, embedded in the surface. Indeed, we can say that the surface is generated by moving a straight line through a given trajectory. Simple examples of developable surfaces include cylinders (not necessarily circular) and cones, as illustrated in Figure 4.12(a) and (b), where the generators are shown as black lines. However, there are many less familiar cases, such as that shown in Figure 4.12(c). We also caution that there are also non-developable ruled surfaces, that is, surfaces containing a straight line through any point. For example, the hyperboloid shown in Figure 4.12(d) contains two families of straight lines, although its Gaussian curvature is negative everywhere. To be clear: every developable surface is ruled but most ruled surfaces are not developable. [APPENDIX]

4.7.3 The Von K´ arm´ an equations Now we just have to assemble the ingredients collected above to derive the governing equations for the plate. It is convenient to use (4.71) to introduce


173

an Airy stress function A, defined in the usual way, so that (4.72) becomes ∂2A ∂2w ∂2A ∂2w ∂2w ∂2A ∂2w 4 − 2 + − D∇ w − ςg = ς . ∂y 2 ∂x2 ∂x∂y ∂x∂y ∂x2 ∂y 2 ∂t2

(4.81)

Next, we use (4.65) and (4.78) to eliminate the strain components and hence obtain ( µ 2 ¶2 ) 2w ∂2w ∂ ∂ w ∇4 A = −Eh − . (4.82) 2 2 ∂x ∂y ∂x∂y The coupled partial differential equations (4.81) and (4.82) for w and A comprise the Von Kármán model. Evidently (4.81) reproduces the linear plate equation (4.69), so the improved approximation for the strain has led to the introduction of just one new term: the right-hand side of (4.82). We recognise the expression in braces as the Gaussian curvature, which is zero only if the deformed plate is developable. Hence, any transverse displacement that does not consist of a pure bending of the plate inevitably causes in-plane stress, in constrast with linear plate theory, where in-plane and transverse deformations are decoupled. It is the nonlinear coupling provided by the right-hand side of (4.82) that makes the Von Kárm´ an equations so much more difficult to solve than the linear plate equations. In the same way that we were led to the strain energy (4.54) for linear bending of a plate, we may use (4.55), generalised to include linear stretching, to suggest that the energy per unit area in a Von Kárm´ an plate is ( !) Ã µ 2 ¶2 D ¡ 2 ¢2 ∂2w ∂2w ∂ w ∇ w − 2(1 − ν) − 2 2 2 ∂x ∂y ∂x∂y n ¡ ¢o 1 2 + (Txx + Tyy )2 − 2(1 + ν) Txx Tyy − Txy , 2Eh where the integrated stress components are given by E(exx + νeyy ) ∂2A = , 2 ∂y 1 − ν2 Eexy ∂2A =− = , ∂x∂y 1+ν E(νexx + eyy ) ∂2A = = , 2 ∂x 1 − ν2

Txx = Txy Tyy

and the strain components are again defined by (4.76). A lengthy exercise in the calculus of variations, as in Exercise 4.10, shows that when we consider virtual displacements (u, v, w) that minimise the net strain energy in the plate, we do retrieve (4.81), (4.82).

174


Concerning boundary conditions, the remarks made after (4.69) will apply.

4.8 Weakly curved shell theory 4.8.1 Strain in a weakly curved shell It is relatively straightforward to generalise the Von Kárm´ an model to describe the deformation of a precast elastic shell whose curvature is small. Suppose that the centre-surface of the shell is given in the initial unstressed state by Z = W (X, Y ). We follow the same procedure as in §4.7.2, considering the displacement of a point on the centre-surface from 

 X  X= Y W (X, Y )



 X + u(X, Y ) . to x(X) =  Y + v(X, Y ) W (X, Y ) + w(X, Y )

(4.83)

Now a nearby point on the centre-surface with initial position 

 X + δX  X + δX =  Y + δY W (X + δX, Y + δY )     1 0  + δY  1  ∼ X + δX  0 ∂W/∂X ∂W/∂Y

(4.84)

is deformed to x + δx, where 

   1 + ∂u/∂X ∂u/∂Y  + δY  1 + ∂v/∂Y  . δx ∼ δX  ∂v/∂X ∂(W + w)/∂X ∂(W + w)/∂Y

(4.85)

Now when we calculate the change in the square of the length of the small line element δX, we find " µ ¶ # ∂u ∂W ∂w ∂w 2 2 2 2 |δx| − |δX| ∼ δX 2 +2 + ∂X ∂X ∂X ∂X ¸ · ∂v ∂W ∂w ∂W ∂w ∂w ∂w ∂u + + + + +2δXδY ∂Y ∂X ∂X ∂Y ∂y ∂X ∂X ∂Y " µ ¶2 # ∂v ∂W ∂w ∂w +δY 2 2 +2 + , (4.86) ∂Y ∂Y ∂Y ∂Y

4.8 Weakly curved shell theory

which motivates us to define the strain components as µ ¶ ∂u ∂W ∂w 1 ∂w 2 exx = + + , ∂x ∂x ∂x 2 ∂x · ¸ 1 ∂u ∂v ∂W ∂w ∂W ∂w ∂w ∂w exy = + + + + 2 ∂y ∂x ∂x ∂y ∂y ∂x ∂x ∂y µ ¶2 ∂v ∂W ∂w 1 ∂w eyy = + + . ∂y ∂y ∂y 2 ∂y

175

(4.87a) (4.87b) (4.87c)

Again following §4.7.2, we can eliminate the horizontal displacement (u, v) from (4.87) to obtain µ 2 ¶2 ∂ 2 exy ∂ 2 eyy ∂ 2 exx ∂2w ∂2w ∂ w ∂2W ∂2w − + 2 = − + − ∂y 2 ∂x∂y ∂x2 ∂x2 ∂y 2 ∂x∂y ∂x2 ∂y 2 ∂2W ∂2w ∂2W ∂2w + 2 . (4.88) − 2 ∂y ∂x2 ∂x∂y ∂x∂y Notice that the right-hand side of (4.88) may be written as ( ¶2 ) ( 2 ¶2 ) µ 2 µ 2 ∂ 2 (W + w) ∂ 2 (W + w) ∂ W ∂2W ∂ (W + w) ∂ W − , − − ∂X 2 ∂Y 2 ∂X∂Y ∂X 2 ∂Y 2 ∂X∂Y which is the change in the Gaussian curvature compared to that of the initial shape of the plate. The isometries of our initially curved shell are therefore those deformations that preserve the Gaussian curvature. This evidently generalises the deformation of a flat plate, which can be bent without stretching only into a surface with zero Gaussian curvature. If we denote the current shape of the plate by z = f (x, y) = W (x, y) + w(x, y) and the initial Gaussian curvature by K0 (x, y), then, for the strain to be zero, f must satisfy the inhomogeneous Monge–Ampère equation µ 2 ¶2 ∂2f ∂2f ∂ f − = K0 (X, Y ). (4.89) ∂X 2 ∂Y 2 ∂X∂Y This famous nonlinear partial differential equation has the unusual property that its type is determined by the right-hand side, as shown in Exercise ??. If K0 = 0 then the initial plate shape is developable, so at each point it resembles a cylinder, as illustrated in Figure 4.13(c), and we recover (4.79), derived previously for an initially flate plate. In this case, (4.89) is parabolic, with one repeated set of real characteristics corresponding to the generators of the initial developable surface. This reflects the fact that the plate can be arbitrarily bent about the generators without causing any in-plane stress.

176


(b)

(c)

Fig. 4.13. Typical surface shapes with (a) positive, (b) negative and (c) zero Gaussian curvature.

If K0 < 0 then the initial plate shape is anticlastic, like the saddle point shown in Figure 4.13(b), and (4.89) is hyperbolic. Hence there are two distinct families of real characteristics and, for there to be no in-plane strain, any localised bending of the plate will cause nonlocal displacements in the region of influence penetrated by these characteristics. Such a plate is therefore somewhat stronger than a developable surface. Finally, if K0 > 0 then the initial plate shape is synclastic or convex, as in Figure 4.13(a), and (4.89) is elliptic. Such a plate will be stronger still because any local bending causes deformations throughout the entire plate.

4.8.2 Linearised equations for a weakly curved shell Having obtained expressions for the in-plane strains, next we have to generalise the force balances (4.37) and (4.44). As shown in Exercise 4.11, we find that the transverse displacement satisfies ∂2A ∂2f ∂2A ∂2f ∂2A ∂2f ∂2w + −2 − D∇4 w − ςg = ς 2 , 2 2 2 2 ∂y ∂x ∂x∂y ∂x∂y ∂x ∂y ∂t

(4.90)

where f is again used as shorthand for W + w. The Airy stress function is defined as usual by ∂2A Eh = Txx = (exx + νeyy ) , 2 ∂y 1 − ν2 ∂2A Eh − = Txy = exy , ∂x∂y 1+ν ∂2A Eh = Tyy = (νexx + eyy ) , 2 ∂x 1 − ν2

(4.91a) (4.91b) (4.91c)

4.8 Weakly curved shell theory

177

with the strain components now given by (4.87). By eliminating the in-plane displacement (u, v) between these, we find that A now satisfies ( µ 2 ¶2 µ 2 ¶2 ) ∂2W ∂2W ∂ W ∂2f ∂2f ∂ f 4 ∇ A = Eh − − + . (4.92) 2 2 2 2 ∂x ∂y ∂x∂y ∂x ∂y ∂x∂y In (4.90) and (4.92) we have two equations for A and w, generalising the Von Kármán equations, which may be recovered by setting W ≡ 0, f ≡ w. To make the problem more tractable, let us now suppose that the transverse displacement is much smaller than the initial deflection of the centre-surface, so that w ¿ W . Then, if we keep only the lowest order terms in w, then (4.90) and (4.92) reduce to ∂2W ∂y 2 ∂2W ∂y 2

∂2A ∂2W ∂2A ∂2W ∂2A − 2 + ∂x2 ∂x∂y ∂x∂y ∂x2 ∂y 2 ∂2W ∂2w ∂2w ∂2W ∂2w + − 2 ∂x2 ∂x∂y ∂x∂y ∂x2 ∂y 2

= D∇4 w,

(4.93)

= −(Eh)−1 ∇4 A,

(4.94)

where, for additional simplicity, we have also neglected the body force and time dependence. Given the initial shell profile W (x, y), we therefore have a system of linear partial differential equations governing the small transverse displacement w and the Airy stress function A.

4.8.3 Solutions for a thin shell Let us consider a shell whose initial shape W (x, y) is of order W0 . We first try making the variables dimensionless in (4.93), (4.94) as follows: x == Lx0 ,

y = Ly 0 ,

W = W0 W 0 ,

w0 w0 ,

A=

Dw0 0 A, W0

(4.95)

where L is a typical lateral dimension, and w0 a typical loading displacement. We obtain, after suppressing primes, ∂2W ∂y 2 ∂2W ∂y 2

∂2A ∂2W ∂2A ∂2W ∂2A − 2 + ∂x2 ∂x∂y ∂x∂y ∂x2 ∂y 2 ∂2w ∂2W ∂2w ∂2W ∂2w −2 + 2 ∂x ∂x∂y ∂x∂y ∂x2 ∂y 2

= ∇4 w,

(4.96)

= −δ∇4 A,

(4.97)

where δ=

h2 . 12(1 − ν 2 )W02

(4.98)

178


For a sufficiently thin shell, we may therefore neglect the right-hand side of (4.97) so that the equations decouple: given suitable boundary conditions our strategy would be to solve ∂2W ∂2w ∂2W ∂2w ∂2W ∂2w −2 + =0 2 2 ∂y ∂x ∂x∂y ∂x∂y ∂x2 ∂y 2

(4.99)

for w and then substitute the result into (4.96) to obtain an equation for A. We therefore have to solve two linear second-order partial differential equations in succession, each involving the differential operator on the lefthand side of (4.99). It is easy to verify that the discriminant ∆ which determines the type of (4.99) is equal to the initial Gaussian curvature of the shell, that is µ 2 ¶2 ∂2W ∂2W ∂ W ∆ = K0 (x, y) = − . (4.100) ∂x2 ∂y 2 ∂x∂y The problem is therefore parabolic, elliptic or hyperbolic according to whether the shell is developable, anticlastic or synclastic. This immediately casts doubt on the model (4.99) for, suppose we wished to prescribe w around the perimeter of a synclastic shell. Then (4.99) would present us with an ill-posed Dirichlet problem for a hyperbolic partial differential equation, and there would probably be no solution at all. Fortunately we can understand the deficiencies in (4.99) when we realise from (4.88) that, when w is small enough for quadratic terms to be neglected, (4.99) is a necessary condition for there to be no “in-plane” stretching, i.e. only bending is allowed. Hence, if the stresses applied to the shell are violent enough to produce in-plane stretching, then we must reconsider the way, (4.95), we rendered our problem dimensionless; one way of escaping from pure bending is to rescale A such that ˜ = δA = O (1). A

(4.101)

This shows that deformations other than pure bending increase the in-plane stress by an order of magnitude. Substituting (4.101) into (4.96) and (4.97) ˜ now before neglecting δ, we find that the equations again decouple, with A determined by ˜ ˜ ˜ ∂2W ∂2A ∂2W ∂2A ∂2W ∂2A − 2 + =0 ∂y 2 ∂x2 ∂x∂y ∂x∂y ∂x2 ∂y 2

(4.102)

and then w found from ∂2W ∂2w ∂2W ∂2w ∂2W ∂2w ˜ − 2 + = −∇4 A. ∂y 2 ∂x2 ∂x∂y ∂x∂y ∂x2 ∂y 2

(4.103)

4.8 Weakly curved shell theory (a)

179

(b)

z y x (c)

(d)

Fig. 4.14. Deformation of a cylindrical shell; (a) original shape, (b) bending about the generators, (c) rotating the generators, (d) bending along the generators.

We can now build the following scenarios for modelling the deformation of weakly-curved shells for which δ is small. It is necessary first to check the compatibility of the boundary conditions, whatever they may be in terms of A, w, and the pure bending response governed by (4.96), (4.99). It maybe be that either: (i) they are compatible everywhere, in which case we just have to solve (4.96), (4.99); or (ii) they are compatible over some regions of the shell that are adjacent to the boundary, as might be the case for a synclastic surface, and the boundary data only determines the solution within some “region of influence” for the hyperbolic equation (4.99); in such cases there may be thin “creased” regions in which the in-plane stresses are not neglibible; or (iii) they are nowhere compatible, in which case large stress fields, with A of O (()1/δ), are set up over the whole plate. We will now illustrate these scenarios with some examples. To avoid the complication of solving (4.99) each time, we will simply ask ourselves the design (???) question of what kinds of displacements w will be possible for any prescribed shell shape W .

180


Let us first consider deformations of the initially cylindrical shell W =

αy 2 , 2

(4.104)

for which (4.96) and (4.97) become α

∂2A = ∇4 w, ∂x2

α

∂2w = −δ∇4 A. ∂x2

(4.105)

Now, letting δ → 0, we see that w can only be a combination of two forms: w(x, y) = a(y)

and w(x, y) = xb(y),

(4.106)

where a and b are arbitrary functions. These displacement fields correspond respectively to bending of the shell about its generators and rotation of the generators, as shown schematically in Figure 4.14(b) and (c). One can easily verify using a section cut from a cardboard tube that such deformations require very small stresses, while bending along the generators, as in Figure 4.14(d) is much more difficult. Indeed, a uniform bending w=

βx2 , 2

(4.107)

inevitably causes a large in-plane stress, with A=−

αβy 4 + O (1), 24δ

(4.108)

corresponding to the stress field Txx = −

αβy 2 , 2δ

Txy = 0,

Tyy = 0.

(4.109)

We can associate this stress in the x-direction with compression of the generators. We point out that if we tried to impose clamped boundary conditions such as w = w0 (y), ∂w ∂x = 0 on x = 0, we could always find bending solutions in x2 0000 0 which w ≡ w0 (y), A = 2α w0 (y). The same would be true if we clamped any line y + mx = 0 unless m = 0. Indeed if we tried to set w = w0 (x) on y = 0, we would only ever get a solution with the scaling (4.101) if w0 was quadratic, as in (4.107). Next, let us consider the anticlastic shell W = αxy,

(4.110)

depicted in Figure 4.15(a), contains two families of straight lines x = const.

4.8 Weakly curved shell theory (a)

181

(b)

z y x (c)

(d)

Fig. 4.15. Deformation of an anticlastic shell; (a) original shape, (b) bending along the lines x = const., (c) bending along the lines y = const., (d) bending along the lines x − y = const..

and y = const.. This time the governing equations (4.96) and (4.97) take the form −2α

∂2A = ∇4 w, ∂x∂y

2α

∂2w = δ∇4 A. ∂x∂y

(4.111)

When δ is negligible, the displacement field must be a linear combination of w(x, y) = a(x)

and w(x, y) = b(y),

(4.112)

where a and b are again arbitrary functions. As illustrated in Figure 4.15, these displacements correspond to bending along either of the characteristic lines x = const. or y = const., and one can try these in practice using a section cut from a plastic bottle. Large stresses result from any other transverse displacement. For example, w=

β(x + y)2 , 2

(4.113)

which respresents bending at an angle of π/4 to the coordinate axes, as shown in Figure 4.15(d), leads to a stress function A=

¢ αβ ¡ 4 x + y 4 + O (1). 24δ

(4.114)

182


(b)

z y x (c)

(d)

Fig. 4.16. Deformation of a convex shell; (a) original shape, (b) and (c) two possible deformations, (d) one-dimensional bending.

The corresponding stress field Txx =

αβEhy 2 , 2

Txy = 0,

Tyy =

αβEhx2 , 2

(4.115)

shows that both the normal stresses increase quadratically. due to stretching of the straight lines x = const. and y = const. respectively. We point out that transverse displacements such as (4.113) are unlikely to be generated by applied boundary displacements and/or tractions. Since the model (4.96), (4.97) leads to hyperbolic equations (4.111) for which A, we expect to be able to apply two quite general boundary conditions, at least locally, unless the boundary is a characteristic x = const. or y = const. . Finally, let us consider the convex shell ¢ ¡ α x2 + y 2 W = . (4.116) 2 As may be seen in Figure 4.16(a), there are no straight lines embedded in this surface. The governing equations (4.96) and (4.94) read α∇2 A = ∇4 w,

α∇2 w = −δ∇4 A,

(4.117)

so that, when δ is small, only harmonic displacement fields are possible.

4.9 Nonlinear beam theory

Some examples are ¡ ¢ w(x, y) = β x2 − y 2 ,

¡ ¢ β x4 − 6x2 y 2 + y 4 and w(x, y) = − , 4

183

(4.118)

which are illustrated in Figure 4.16(b) and (c) respectively. In the first, the shell is bent upwards in the x-direction and downwards in the y-direction; in the second, the corners are bent upwards and the edges downwards. These can easily be realised using a segment from a plastic ball. On the other hand, if we perform the rescaling (4.101) before letting δ → 0, we find that ˜ satisfies Laplace’s equation and thus A ˜ = 0. α∇2 w = −δ∇4 A

(4.119)

Hence, any arbitrary displacement field, for example the one-dimensional bending w = βx2 illustrated in Figure 4.16, does not just give rise to large in-plane stress but is actually impossible. Indeed the ellipticity of (4.117) as δ → 0 means that we could never apply, say, clamped boundary conditions to w. These rudimentary examples have revealed that in many situations, shells cannot bend in the absence of in-plane stresses that are large compared to those that arise in plate bending. This is especially true for synclastic shells, which are known to be much stronger than plates. Further study of (4.96), (4.97) as δ → 0 is beyond the scope of this book, but this model can be shown to exhibit solutions in which the in-plane stresses are confined to thin creases in the shells, of which there may be many; this fact has been used to explain the predominance of creasing when paper is compressed. Although the above theory only applies to very small transverse displacements of shells whose built-in curvature is already small, they illustrate many of the key differences between the elastic responses of shells and plates. Any theory for more general shells demands more differential geometry than we can conveniently introduce in this chapter, and we will return to this topic in Chapter 6. We can however consider nearly one-dimensional elastic bodies with arbitrary curvature, and this we do in the next section. 4.9 Nonlinear beam theory 4.9.1 Derivation of the model We now return to the beam configuration considered in §4.4, now generalising the model to make it applicable to commonly observed phenomena such as the large deflection of a diving board. Diving boards undergo large displacement without any of the damaging effects that are associated with

184

Approximate Theories (a) h

(b)

(c)

R

Fig. 4.17. Schematic of a beam (a) before and (b) after bending; (c) a close-up of the displacement field.

large strains. This is possible because a thin elastic beam can be bent without significantly stretching its centre-line, as illustrated in Figure 4.17(a,b). The internal displacement field in such a beam is shown schematically in Figure 4.17(c): the net effect is to stretch the outer surface of the beam and compress the inner surface, while the length of the centre-line is virtually unchanged. It is an elementary exercise in trigonometry to show that the strain associated with such a displacement is of order h/R, where h is the thickness of the beam and R the radius of curvature through which it is bent. Provided the beam is sufficiently thin, it can therefore suffer large displacements while the internal strain remains small. As noted above in §4.8, a developable shell has the same property when bent about its generators. If the strain is small, then we can still assume that the stress is a linear function of the strain. However, when the displacement is large, linear elasticity is no longer valid: the Eulerian and Lagrangian coordinates are far from interchangeable, and it is not legitimate to linearise the relationship between strain and displacement. We therefore seek a geometrically nonlinear but mechanically linear theory, in a similar vein to the Von Kármán plate theory from §4.7. In equilibrium, it is surprisingly easy to generalise our derivation of linear beam theory in §4.4 to model large two-dimensional transverse displacements. We describe the deformation of the beam using arc-length s along the centre-line and the angle θ(s) between the centre-line and the x-axis. As discussed above, we assume the centre-line to be inextensible to leading


(a)

185

N0

(b)

N (s + δs)

T0

T (s + δs) M (s)

T0 θ

M (s + δs) N0

T (s) N (s)

Fig. 4.18. (a) Schematic of the forces and moments acting on a small segment of a beam. (b) Definition sketch showing the sign convention for the forces at the ends of the beam.

order, so that arc-length is conserved by the deformation and we can then view s as a Lagrangian coordinate that is fixed in the deforming beam. The transverse displacement is given parametrically by x = x(s), w = w(s), where dx = cos θ, ds

dw = sin θ. ds

(4.120)

Now we balance forces and moments on a small segment of the beam as shown in Figure 4.18(a). We neglect inertia and body forces for the moment and note that N is now defined to be the normal, rather than transverse, force; since θ is no longer assumed to be small this is a significant distinction. We thus obtain the equations d (T cos θ − N sin θ) = 0, ds

d (N cos θ + T sin θ) = 0, ds

dM − N = 0, ds (4.121)

the first two of which simply show that the internal forces in the x- and zdirections are conserved. If a force (T0 , N0 ) is applied at the right-hand end, with an equal and opposite force at the left-hand end, as in Figure 4.18(b), it follows that it follows that T = T0 cos θ + N0 sin θ,

N = N0 cos θ − T0 sin θ.

(4.122)

As a constitutive relation we expect, as in §4.4, the bending moment to be proportional to the curvature which, for a nonlinear deflection, is given by dθ/ds. Since we are considering small strains, we assume that the constant

186


of proportionality is the same as in the linear case, that is M = −EI

dθ . ds

(4.123)

This assumption is justified in Chapter 6. By collecting (4.121c), (4.122) and (4.123), we obtain the celebrated Euler strut equation EI

d2 θ + N0 cos θ − T0 sin θ = 0. ds2

(4.124)

If we are given the applied force components T0 and N0 , then we expect to impose two further boundary conditions (one at each end) on the second-order differential equation (4.124). Typical examples are clamped conditions, where we specify the angle θ, or simple support, where there is zero bending moment and hence dθ/ds = 0. Suppose, instead, that we are told the positions of the ends of the beam. Without loss of generality, we suppose the end s = 0 is fixed at the origin, and denote by (X, Z) the position of the other end s = L, where L is the length of the beam. We then deduce from (4.120) the conditions Z L Z L X= cos θ(s) ds, Z= sin θ(s) ds, (4.125) 0

0

from which we can in principle recover T0 and N0 . It is reassuring to confirm that the Euler strut model reduces to the linear beam theory derived in §4.4 for cases where the deflection is small. For small θ, we can simplify the geometric relations (4.120) to x ∼ s,

dw ∼ θ, dx

(4.126)

and (4.124) thus becomes EI

d3 w dw + N0 − T0 = 0. dx3 dx

(4.127)

One further x-derivative reproduces the linear beam equation (4.15) in the absence of inertia and gravity.

4.9.2 Example: deflection of a diving board Under a simple change in the origin of θ, (4.124) is precisely equivalent to the equation for a pendulum whose small-amplitude frequency of oscillation is p P02 + N02 /EI. However, in (4.124), s is far from being a timelike variable, and the typical boundary conditions discussed above are quite different from


187

the sort of initial conditions we would expect to apply to a pendulum. However, we can use techniques that are familiar from the study of pendulums to construct solutions of (4.124). Let us first consider the example alluded to previously of the steady deformation of a diving board. If the board is clamped horizontally at s = 0 and a downward load F is applied to the other end, which is otherwise free, then we have T0 = 0, N0 = −F and (4.124) becomes EI

d2 θ = F cos θ, ds2

(4.128)

subject to dθ (L) = 0. (4.129) ds Multiplying (4.128) through by dθ/ds and integrating once with respect to s, we obtain µ ¶2 dθ 2F (sin θ − sin α) , (4.130) = ds EI θ(0) = 0,

where α is shorthand for −θ(L). When taking the square root, we note that we expect dθ/ds to be negative and thus obtain the solution in parametric form as r Z −θ dφ 2F √ . (4.131) =s EI sin α − sin φ 0 It just remains to determine the angle α from the transcendental equation r Z α dφ 2F √ =L , (4.132) EI sin α − sin φ 0 which can be written in terms of elliptic integrals. Given the applied force F , we have to solve (4.132) for α (numerically), and the deflection may then be reconstructed using (4.131) and (4.120); see Exercise 4.13. In Figure p 4.19(a), we plot the final angle −α as a function of the force parameter L 2F/EI. As the applied force increases, the angle decreases, tending towards −π/2. In Figure 4.19(b), we show the diving board profile corresponding to different applied forces, and we can see clearly how the force causes the deflection to increase. For small applied forces, we would expect the linear equation (4.127) to be applicable. The linear model that describes our diving board is EI

d3 w = F, dx3

w(0) =

dw (0) = 0, dx

d2 w (L) = 0, dx2

(4.133)

188

Approximate Theories L 1

2

p

EI/2F

3

4

5

x 0.2

6

0.4

0.6

0.8

-0.1

-0.25 -0.5

1

1

-0.2 -0.3

−α -0.75

w

-0.4 -0.5

-1

-0.6

-1.25

-0.7 -1.5

p

-0.8

L

EI/2F = 4

3

2

p Fig. 4.19. (a) Final angle −α versus force parameter L 2F/EI. (b) Deflection of a diving board for various values of the force parameter; the dashed lines show the linearised solution.

and the solution is readily found to be w=−

F 2 x (3L − x). 6EI

(4.134)

In Figure 4.19(b), (4.134) is plotted as dashed curves, and we see that there is excellent agreement with the full nonlinear model provided the deflections are reasonably small. 4.9.3 Weakly nonlinear theory and buckling Now let us return to the buckling of an elastic beam, which we discussed previously in the linear case in §4.4. We focus on the model problem of a beam of length L, clamped at both ends, subject to a compressive force P = −T0 and zero transverse force. We therefore have to solve d2 θ P sin θ = 0, + ds2 EI

(4.135a)

subject to the boundary conditions θ (0) = θ (L) = 0.

(4.135b)

This is a nonlinear eigenvalue problem: θ ≡ 0 is always a possibility, and we are seeking values of the applied force P for which (4.135) admits nonzero solutions. Such solutions, if they exist, correspond to buckling of the beam. If θ is small, it is easy to show that (4.135) reduces to the linear eigenvalue problem (4.20) studied previously, and we can read off from (4.22) the solution  ³ nπs ´ L2 P  A sin = n2 , θ= (4.136) L π 2 EI  0 otherwise,


(a)

|A|

189

(b)

|A| 1.4 1.2 1 0.8 0.6 0.4 0.2

5

15

10

20

2.5

5

λ

7.5

10 12.5 15 17.5 20

λ

Fig. 4.20. (a) Response diagram of the norm kθk of the solution (4.136) versus the control parameter λ. (b) Corresponding response of the weakly nonlinear solution; the exact nonlinear solution is shown using dashed lines.

where n is any integer. Thus the amplitude A of the solution is zero, unless P takes one of a discrete set of critical values, in which case it is arbitrary. Hence buckling will occur when P = π2

EI L2

(4.137)

We can try to visualise this behaviour by plotting a response diagram of the magnitude of the solution kθk = max |θ| = |A| 0<s 1 (4.163) θ = 4 tan tanh 2 v2 − 1 or ½ µ ¶¾ η0 − η −1 θ = 4 tan exp √ when |v| < 1. (4.164) 1 − v2 Since (4.163) approaches the unstable solutions θ = ±π and the paperclips are vertical as η → ±∞, it is not physically realisable and only the subsonic waves (4.164) can be observed in practice. In these so-called kink solutions, the rod undergoes one complete twist, with θ = 0 ahead of the wave (as η → +∞) and θ = 2π behind (as η → −∞), as shown in Figure 4.23. They can easily be realised in the system illustrated in Figure 4.22 by holding the rubber band tight and horizontal, with all the paperclips dangling vertically, and then quickly twisting one end through 2π. 4.11.3 Travelling waves on a beam Our second example concerns the unsteady generalisation of the geometrically nonlinear beam theory derived in §4.9. The displacement of the beam

4.11 Geometrically nonlinear wave propagation

t=0 z

t=1

197

t=2

x y

Fig. 4.23. A kink propagating along a series of pendulums attached to a rod.

is now given parametrically by x = x(s, t), w = w(x, t), where, as in (4.120), ∂x ∂w = cos θ, = sin θ, (4.165) ∂s ∂s and θ(s, t) is the angle made by the beam with the x-axis. When we include acceleration, the force balances (4.121) become ∂2x ∂ (T cos θ − N sin θ) = % 2 , ∂s ∂t where

∂2w ∂ (N cos θ + T sin θ) = % 2 , ∂s ∂t

(4.166)

∂2θ . (4.167) ∂s2 Although this is a more complicated model than (4.161), we can still seek a travelling wave solution in which x, w, T , N and θ are functions only of η = s − vt. We also postulate the boundary conditions N = −EI

T → T∞ ,

N → 0,

θ → 0,

x ∼ s,

w→0

as s → ∞,

(4.168)

which would apply, for example, to a long taut rope that had been shaken briefly at one end. As in §4.11.2, by looking for a travelling wave, we reduce (4.166) to nonlinear ordinary differential equations, which may be integrated to obtain T cos θ − N sin θ = %v 2 (cos θ − 1) + T∞ ,

T sin θ + N cos θ = %v 2 sin θ. (4.169)

Hence the tension and shear force are given by ¡ ¢ ¡ ¢ T = %v 2 + T∞ − %v 2 cos θ, N = %v 2 − T∞ sin θ,

(4.170)

and (4.167) therefore becomes ¢ EI d2 θ ¡ 2 + M − 1 sin θ = 0, T∞ dη 2

(4.171)

198


w -2

-1

1

2

x

-0.5 -1 -1.5 -2

v Fig. 4.24. Travelling wave solution (4.173) of the nonlinear beam equations (with k = 1).

p p where M = |v| %/T ∞. We recall from Chapter 3 that T∞ /% is the phase velocity of small-amplitude transverse waves on a string, and M is known as the Mach number of the wave. By analogy with (4.162), we see that, as long as M < 1, there is a solution connecting θ → 0 as η → +∞ to θ → 2π as η → −∞, namely r ³ ´ T∞ −1 k(η0 −η) θ = 4 tan e , where k = (1 − M 2 ). (4.172) EI With θ given by (4.172), we can integrate (4.165) to obtain the parametric description µ ¶ ¶ µ 1 ξ − 2 tanh ξ x = (4.173) w −2 sech ξ k for the shape of the rope, where ξ = k(η − η0 ). As shown in Figure 4.24, this represents a single loops, in which the rope turns through 2π, propagating at speed c which depends, through (4.172), on the amplitude. 4.11.4 Weakly nonlinear waves on a beam Now, if we wish to analyse general unsteady deformations of a beam, we appear to be confronted with the full system (4.165)–(4.167) of nonlinear equations for x, w, T , N and θ. However, we will now show that, when the solution is close to a travelling wave, we can obtain a single partial differential equation incorporating all the ingredients of unsteadiness, dispersion and nonlinearity. Moreover, it will transpire that this equation is closely related to one of the most famous of all soliton equations. We observe that (4.171) is degenerate when EI = 0 and M 2 = 1: when the

4.11 Geometrically nonlinear wave propagation

199

beam has zero bending stiffness, θ can be an arbitrary function of s−ct. This motivates us to seek a “distinguished limit” in which the bending stiffness is p small but nonzero and the wave is approximately moving at speed c = T∞ /%, where T∞ is the tension applied at infinity as before. To analyse this limit, our first step is to express the equations in a moving reference frame, by transforming from (x, t) to (ζ, t), where ζ = x−ct, so that (4.165)– (4.167) become ∂x = cos θ, ∂ζ

∂w = sin θ, ∂ζ

N = −EI

∂2θ , ∂ζ 2

µ 2 ¶ ¢ ∂ ¡ ∂ x ∂2x (T − T∞ ) cos θ − N sin θ = % − 2c , ∂ζ ∂t2 ∂ζ∂t µ 2 ¶ ¢ ∂ w ∂ ¡ ∂2w N cos θ + (T − T∞ ) sin θ = % . − 2c ∂ζ ∂t2 ∂ζ∂t

(4.174)

(4.175) (4.176)

Now we have to nondimensionalise the equations before we can take a systematic asymptotic limit. We denote by ` a typical amplitude of the wave and obtain a suitable scaling for N from (4.174). The relevant timescale is then found by balancing the final term in (4.175) with the normal stress on the left-hand side. We thus define the dimensionless variables as     µ ¶ ¶ µ x X 3 w = ` W  , T = T∞ T , N = EI N , t = 2c%` τ, (4.177) `2 EI ζ ξ so that (4.174)–(4.176) become ∂X = cos θ, ∂ξ

∂W = sin θ, ∂ξ

(4.178)

∂2θ , ∂ξ 2 ¢ ε2 ∂ 2 X ∂2X ∂ ¡ (T − 1) cos θ − εN sin θ = − ε , ∂ξ 4 ∂τ 2 ∂ξ∂τ ¢ ε2 ∂ 2 W ∂ ¡ ∂2W − ε , εN cos θ + (T − 1) sin θ = ∂ξ 4 ∂τ 2 ∂ξ∂τ N =−

(4.179) (4.180) (4.181)

where the only remaining dimensionless parameter in the problem is ε=

EI . T∞ `2

(4.182)

200


We suppose, as in §4.11.3, that the beam is held straight and horizontal, under a uniform tension at infinity, so that T → 1,

N → 0,

θ → 0,

X ∼ ξ,

W →0

as ξ → ∞.

(4.183)

Now we take the limit ε → 0, which corresponds to the bending stiffness of the beam being small. When ε = 0, (4.180) and (4.181) are satisfied identically by T = 1, and this motivates the ansatz T ∼ 1 + εT1 .

(4.184)

Now when we substitute (4.184) into (4.180) and (4.181) and let ε → 0, we find that the leading-order equations may be integrated once with respect to ξ, using the boundary conditions (4.183), to obtain ∂X ∂W = 0, T1 sin θ + N1 cos θ + = 0. (4.185) ∂τ ∂τ We solve these simultaneously for T1 and N1 and substitute into (4.179) to obtain T1 cos θ − N1 sin θ +

∂ 2 θ ∂X ∂W sin θ − cos θ = 0, + 2 ∂ξ ∂τ ∂τ

(4.186)

which is to be solved along with (4.178) for θ(ξ, τ ), X(ξ, τ ) and W (ξ, τ ). A remarkable first integration is now possible which allows us to eliminate X and W to obtain a single equation for θ. When we multiply (4.186) by ∂θ/∂ξ we discover by virtue of (4.178) that it may be integrated with respect to ξ, leading, from (4.183), to µ ¶ ∂W 1 ∂θ 2 ∂X cos θ + sin θ = . (4.187) ∂τ ∂τ 2 ∂ξ On the other hand, if we instead differentiate (4.186) with respect to ξ, again using (4.178), we find that µ ¶ ∂θ ∂X ∂W ∂θ ∂ 3 θ cos θ + sin θ = − 3. (4.188) ∂ξ ∂τ ∂τ ∂τ ∂ξ Finally, comparing (4.187) and (4.188), we obtain µ ¶ ∂θ 1 ∂θ 3 ∂ 3 θ − − 3 = 0. ∂τ 2 ∂ξ ∂ξ

(4.189)

The governing equation (4.189) for θ is closely related to the most famous of all soliton equations. If we differentiate (4.189) with respect to ξ, we find that the equation satisfied by ∂θ/∂ξ is a version of the modified KdV equation, which can itself be transformed into the celebrated KdV

4.12 Concluding remarks

201

equation. The KdV equation was originally used to model large-amplitude water waves, and its study provided a key stimulus for the theory of solitons [ref Drazin/Johnson]. By reversing the scalings (4.177), we obtain the dimensional version of (4.189), namely µ ¶ µ ¶ 1 ∂θ 3 ∂ 3 θ 2T∞ 1 ∂θ ∂θ − − 3 = 0. (4.190) + EI C ∂t ∂x 2 ∂x ∂x Now if we seek a travelling wave, in which θ is again a function only of η = x − V t, then we obtain the ordinary differential equation µ ¶ d3 θ 1 dθ 3 2T∞ dθ + − (1 − M ) = 0. (4.191) 3 dη 2 dη EI dη As shown in Exercise 4.20, the solution of this equation satisfying θ → 0 as η → +∞ and θ → 2π as η → −∞ is r ³ ´ 2T∞ (1 − M ) −1 κ(η0 −η) . (4.192) θ = 4 tan e , where κ = EI We thus reproduce the travelling wave solution (4.172) found previously, noting that κ is the leading-order approximation to k as M → 1. 4.12 Concluding remarks This chapter has revealed some of the multitude of phenomena that can be modelled by the equations of elasticity under the assumptions of a linear stress-strain relationship. We have emphasised phenomena that can occur when there is some strong geometric constraint of “thinness”, and hence the possibility of geometric nonlinearity. However, what is more remarkable than the phenomena themselves is the mathematical challenge posed by passing from a fully three-dimensional model to simpler models for nearly one- or two-dimensional solids. This challenge is exemplified by the fact that, although (4.7) is an apparently “physically obvious” paradigm for linear wave propagation, we will see in Chapter 6 that it can only be derived systematically from the equations of nonlinear three-dimensional elasticity. Similarly, our model (4.121) for beams relied on the “physically obvious” assumption that the centre line was inextensible, which it clearly could not be were Young’s modulus sufficiently small; equally (4.7) implicitly relied on the string being extensible, yet no mention was made of Young’s modulus. Both geometric nonlinearity and inextensibility are concepts that can only be understood by a more thorough mathematical approach, which is what we will present in Chapters 5 and 6.

202


Finally, we note that there is another class of problems where the geometry suggests that a mathematical simplification may be possible. This is when the elastic body contains a thin void space other than itself being thin. We will discuss such “complementary” geometries in Chapter 7 and use them to model cracks and contact problems between elastic bodies that are pressed together. Exercises 4.1 Show that the natural frequencies of a piano string of length L, simply supported at either end, with line density %, tension T and small bending stiffness EI, are approximately s nπ T n3 π 3 EI ωn ≈ + 3√ . L % L %T 4.2 A beam of length L is simply supported at its two ends x = 0 and x = L and subject to zero tensile force. Show that its displacement as it sags under gravity is given by ¡ ¢ %g w=− x(L − x) L2 + Lx − x2 24EI and deduce that the maximum transverse deflection is 5%g/384EI. 4.3 Justify the following model for the lateral displacement w(x) of a strut of length L, clamped at its two ends and compressed by a force P: d3 w dw dw P dw = 0, w(0) = (0) = 0, w(L) = (L) = 0. + 3 dx EI dx dx dx Find the minimum force P required for w to be nonzero. Why is this value different from that found in §4.10? 4.4 Consider a beam of length L, clamped at a distance ` < L from the edge of a horizontal table, with a weight F hung from the other end, as shown in Figure 4.25. Assuming that the weight of the beam itself is negligible compared with F and that the transverse displacement w(x) is small, justify the following model: R dw d4 w = δ(x − `), w(0) = (0) = 0, 4 dx EI dx d2 w d3 w F (L) = 0, (L) = , 2 3 dx dx EI where δ is the Dirac delta-function and the reaction force R is to be

EXERCISES

z

203

R

11111111 00000000 00000000 11111111 00000000 11111111 ` 00000000 11111111 00000000 11111111

x

F

Fig. 4.25. Definition sketch of a beam clamped near the edge of a table.

g

z x ` L

Fig. 4.26. Schematic of a beam supported at two points.

determined from the contact condition w(`) = 0. Show that the reaction force is given by R = (3L − `)F/2` and that the displacement is ( 6(L − `)x2 (` − x) x < `, F w(x) = £ ¤ 2 24EI` `(` − x) 3(L − `)(3L − `) − (2x − 3L + `) x > `. Find the maximum magnitude of the slope |dw/dx| and deduce that these results are valid provided F L(L − `)/EI ¿ 1. 4.5 Consider a beam of length L that is simply supported at two points and allowed to sag under its own weight, as shown in Figure 4.26. The supports are assumed to be a distance ` < L apart and to be symmetric about the middle of the beam. Derive and clearly justify the following dimensionless model for the vertical displacement w(x) of the beam: d4 w = Rδ(x − β), dx4 dw d3 w d2 w (0) = (0) = (1/2) = w(β/2) = 0, dx dx3 dx2

204


where β = `/L and δ is the Dirac delta function. Show that the reaction force R is equal to 1/2 and explain this result physically. Solve for w(x) and plot the solution as β is varied. Show that, if β is small, then the beam sags down at its ends and goes up in the middle while, if β is close to 1, then it sags down in the middle and its ends move upwards. Explore how the behaviour varies between these two limits. 4.6 Longitudinal waves in a bar 0 < x < L are modelled by (4.5). The bar is initially unstressed and at rest. The ends are sress-free except that an impulsive compressive stress J is applied at x = 0 for 0 < t < T . Show that, for 0 < t < Lc , ( u=

¡ ¢ J − EA t − xc , ct > x > x(t − τ ) √

τ EA ,

c(t − τ ) > x,

where H is the Heaviside function defined in (3.133). ¢ ¡ on the stress-free Noting that if a wave u = f t − xc impinges ¡ ¢ boundary at x = `, the reflected wave is f t − c` + x−` c , deduce that, as τ → 0 and J → ∞ with Jτ = J˜ of order 1, µ ³ ¶¶ µ x´ ` x−` ˜ u ∼ −J H t − +H t− + c c c and deduce that the compressive incident wave reflects as a tensile wave. [This is the basis of the Hopkinson Bar test for the strength of metals; the strength is determined by the impulse that must be applied at x = 0 in order for a free piece of metal to be ejected from x = L.] 4.7 An elastic beam of bending stiffness EI and line density % is subject to a large compressive force P . Using the approach p of §4.4.4, show that the most unstable waves have wavelength 2π 2EI/P and grow over √ a time-scale t ∼ %EI/P . Hence predict the size of the fragments into which a strand of spaghetti will fracture when a 1 kg weight is dropped on one end with a speed U = 3 m s−1 . [Typical parameter values for spaghetti are % ≈ 2 × 10−3 kg m−1 , EI ≈ 1.5 × 10−4 N m2 .] 4.8 Writing (nx , ny ) = (− sin θ, cos θ) and K =

dθ ds ,

show that Msn , as

EXERCISES

205

given by (4.48) and (4.43) is 2 sin θ cos θ ∂ 2 w 1+ν ∂x∂y µ 2 ¶ µµ 2 ¶ ¶ 1 ∂ w ∂2w ∂ w ∂2w 2 2 − + ν sin θ + ν + cos θ . 1 − ν2 ∂x2 ∂y 2 ∂x2 ∂y 2 Note that we use the indentities

∂ ∂y

∂ ∂ = sin θ ∂s +cos θ ∂n ,

∂ sin θ ∂n to show that when w = 0, and hence

Msn

1 =− 1−ν

µ

∂w ∂s

∂w ∂2w + νK 2 ∂n ∂n

=

∂ ∂x

∂2w ∂s2

∂ = cos θ ∂s −

= 0, then

¶ .

4.9 Show that the steady sag of a cylindrically symmetric plate with nonuniform bending stiffness D(r) is governed by the equation dD dr

µ 2 ¶ µ ¶ d w d d2 w 1 dw ςgr2 dw r 2 +ν + Dr = − . + dr dr dr dr2 r dr 2

(4.193)

For the special case in which we wish w to be given by ¡ ¢ w = − a2 − r2

(µ

7+ν 2 r − a 1+ν 2

¶2

12a4 + (1 + ν)2

) ,

show that the only solution D(r) of (4.193) that is bounded as r → a is D=

(1 + ν)ςg ¡ ¢. 24(7 − ν) (5 + ν)a2 − (1 + ν)r2

4.10 Show that when w is changed to w + η, where |η| ¿ |w|, the integral in (4.55) changes to ZZ

½ ¡ ¢¡ ¢ 2 ∇2 w ∇2 η Ω ¶¾ µ 2 ∂2w ∂2η ∂ w ∂2η ∂2w ∂2η + −2 dxdy, − 2(1 − ν) ∂x2 ∂y 2 ∂y 2 ∂x2 ∂x∂y ∂x∂y

206


and show this can be written as ZZ ¡ ¢ ¡ ¢ 2∇2 (η∇2 w) − 4 ∇η · ∇ ∇2 w − 2η∇4 w Ω µ µ ¶ ∂ ∂ 2 w ∂η ∂ 2 w ∂η −2(1 − ν) − ∂x ∂y 2 ∂x ∂x∂y ∂y µ ¶¶ ∂ 2 w ∂η ∂ ∂ 2 w ∂η − dxdy − ∂y ∂x2 ∂y ∂x∂y ∂x ½ ZZ ¡ ¢ =2 − ∇ · ∇η∇2 w + 2η∇4 w Ω µ 2 ¶¾ ∂ w ∂η ∂ 2 w ∂η ∂ 2 w ∂η ∂ 2 w ∂η −(1 − ν)∇ · − , − dxdy. ∂y 2 ∂x ∂x∂y ∂y ∂x2 ∂y ∂x∂y ∂x Use Green’s theorem to deduce that ZZ (i) η∇4 w dxdy = 0 Ω

and hence that ∇4 w = 0. Show that when Ω is the half-space x > 0, and all variables are assumed to decay x → +∞, ¾ ½ ¾ Z +∞ ½ ∂3w ∂ ¡ 2 ¢ ∂ 2 w ∂η 2 ∇ w − (1 − ν) dy = 0. (ii) η + ∇ w − (1 − ν) 2 ∂x ∂x∂y 2 ∂y ∂x −∞ Deduce the natural boundary condition (4.5). 4.11 Suppose the centre-surface of a weakly-curved shell, initially given by z = W (x, y), undergoes a transverse displacement w(x, y), so that the centre-surface is deformed to z = f (x, y) = W (x, y) + w(x, y). Following the approach adopted in §4.6.1, show that the transverse displacement and in-plane stress resultants satisfy Txx

∂2f ∂2f ∂2f ∂2w 4 + T + 2T − D∇ w − ςg = ς , yy xy ∂x2 ∂x∂y ∂y 2 ∂t2 ∂Txy ∂Tyy ∂Txx ∂Txy + = + = 0, ∂x ∂y ∂x ∂y

where the surface density ς is assumed to be constant. 4.12 Consider a thin two-dimensional elastic beam, with bending stiffness EI, subject to a body force −%gk per unit length. Suppose the centre-line of the beam makes an angle θ(s) with the x-axis, where s is arc-length. Derive the following equations for θ and the tension T

EXERCISES

207

and shear force N in the beam: d (T cos θ − N sin θ) = 0, ds

d d2 θ (T sin θ + N cos θ) = %g, EI 2 = N, ds ds

explaining all the necessary assumptions. Consider the special case of a string with zero bending stiffness (EI = 0) of length L, sagging under gravity and attached at (x, w) = (0, 0) and (x, w) = (d, 0). Show that µ ¶ 2s tan θ(s) = tan θ0 1 − , L where θ0 is the angle made by the string with the horizontal line at s = 0. Show further that 2x − d tan θ0 = sinh−1 (tan θ) , L and deduce an equation for tan θ0 . Finally, show that ½ µ ¶ µ ¶¾ 2x − d d L cosh tan θ0 − cosh tan θ0 . y= 2 tan θ0 L L 4.13 Derive the following model for the deflection θ(s) of a diving board clamped horizontally at one end while a downward force F is applied on the other: ¶ µ d2 θ dθ F cos θ, θ(0) = 0, (L) = 0. = ds2 EI ds Nondimensionalise the problem and show that it depends on a single dimensionless parameter ε = F L2 /EI. Show that, when ε is small, the solution may be approximated asymptotically by µ 2 ¶ ¡ ¢ θ ξ ∼ − ξ + ε2 Θ1 + O ε4 , ε 2 where ξ = s/L, and Θ1 (ξ) is to be found. Hence find and plot the leading-order shape of the board. If ε is not assumed to be small, show that θ is related to ξ by Z −θ √ dφ √ = ξ 2ε, sin α − sin φ 0 where α is to be determined from Z α √ dφ √ = 2ε. sin α − sin φ 0

208


Deduce that the shape of the board is given parametrically by x = x(θ), y = y(θ), where x and y are nondimensionalised with L, r n o √ 2 √ x(θ) = sin α − sin α + sin θ , ε Z −θ 1 sin φ dφ √ y(θ) = − √ , sin α − sin φ 2ε 0 and θ varies between −α and 0. Hence plot the shape of the board for various values of ε. Compare these exact solutions with the small-ε approximation. 4.14 Consider a beam of length L and mass % per unit length, clamped horizontally at each end and sagging under gravity. If a horizontal compressive force P and equal vertical forces are applied at either end, derive the boundary conditions θ(0) = θ(L) = 0, T (0) = T (L) = −P, −N (0) = N (L) =

%gL , 2

and deduce that the Euler strut equation (4.135) becomes, ¶ µ d2 θ 1 cos θ = 0. EI 2 + P sin θ + %g s − ds 2 By following the¡ nondimensionalisation of §4.9.3 and assuming β = ¢ 3 3/2 %gAL /EI = O ε , show that the orthogonality condition (4.149) becomes µ ¶ ³ nπ ´ 8λ1 16β 2 A A − 2 = − 3/2 2 2 cos2 . n 2 ε n π For n = 2, draw the resulting response diagram and compare it with the case β = 0. Show that the beam can only sag downwards until λ = 4 + 3β 3/2 /2π 2 . 4.15 A beam of length L and line density % is clamped vertically upwards at one end (s = 0) and free at the other (s = L). Show that the shear force in the beam is N = %g(s − L) cos θ and hence obtain the model d2 θ EI + (s − L) cos θ = 0, ds2 %g

θ(0) = π/2,

dθ (L) = 0. ds

Now suppose the beam is only slightly perturbed from the vertical, so that θ = π/2 + φ, where |φ| ¿ 1. Defining ³ %g ´1/3 ³ %g ´1/3 , β=L , ξ = (s − L) EI EI

EXERCISES

2

z 1

3

209

n=1

0.8

0.6

0.4

0.2

0.05 0.1 0.15 0.2 0.25 0.3

x

Fig. 4.27. The first three buckling modes of a vertically clamped beam.

show that φ satisfies Airy’s equation d2 φ − ξφ = 0. dξ 2 The solution of Airy’s equation with zero slope at ξ = 0 is ´ ³√ φ=a 3 Ai(ξ) + Bi(ξ) , where Ai and Bi are Airy functions (Abramowitz & Stegun, 1972§10.4) and a is an arbitrary constant. Deduce that, for nontrivial solutions, β must satisfy √ 3 Ai(−β) + Bi(−β) = 0, and show (numerically) that the minimum value of β satisfying this condition is βc ≈ 1.986. Thus find the maximum height of a column (or a recalcitrant hair) before it buckles under its own weight. [See Figure 4.27 for a plot of the first three buckling modes.] 4.16 Show that the exact solution of the buckling problem (4.135) is Z 0

θ

Ã

ˆ sin2 (θ/2) 1− sin2 (kθk/2)

!−1/2

√ dθˆ = (2πs/L) λ sin(kθk/2)

and deduce that the nth branch of the response diagram is given

210


implicitly by λ=

¢2 4n2 ¡ 2 K sin (kθk/2) , π2

where K denotes an elliptic integral (Gradshteyn & Ryzhik, 1994§8.1). 4.17 Consider a tuft of wool compressed in a piston with pressure p, as shown in Figure 4.28. We wish to derive a relation between this pressure and the volume fraction φ of the wool. To this end, assume that the tuft is made of fibres with diameter d piled up with contact points distant by an average length L, so that φ ∼ d/L. The distance L (φ) thus describes the level of compaction. Rather than trying to resolve the elastic deformation of the whole tuft, consider a single fibre and explain that its infinitesimal deformation dw in response to a small variation of pressure dp scales as dw ∼

L4 dp Ed3

[Hint: consider (4.15) from a dimensional point of view and recall that, for a circular cross section I = d4 /64.] Identifying the deformation of the fibre with the displacement of the piston, justify the relations dw dφ dL =− =− , w0 φ0 L0 and hence show that L4 dL ∼− dp. L0 Ew0 d3 Integrating this differential equation for p as a function of L, deduce that p∼

¢ Ew0 ¡ 3 ¢ Ew0 d3 ¡ −3 L − L−3 ∼ φ − φ30 . 0 L0 L0

[This is known as van Wyk’s formula, and is used widely as a quality control in the wool industry.] 4.18 A semi-infinite beam initially lying at rest along the x-axis is subject to a point force F (t) acting at the origin. Derive the model ∂4w ∂2w + F (t)δ(x) = % , ∂x4 ∂t2 ∂w w(x, 0) = (x, 0) = 0, w → 0 as x → ±∞ ∂t

−EI

EXERCISES

211

Fig. 4.28. Left: a piston containing a tuft of fibres and subjected to a pressure p. Right: detail of the tuft. A piece of fibre is supported at two contact points a distance L apart. A fibre on top is transmitting a force F ∼ δp × dL.

and, for example by taking a Fourier transform in x, show that the displacement of the origin is related to the applied force by Z t √ 1 w(0, t) = √ F (τ ) t − τ dτ. 1/4 0 2π (EI%3 ) If the origin is forced to move at speed V , so w(0, t) = V t, show that the required force is r ¡ ¢ 2 3 1/4 F (t) = 2V EI% . πt [This is a simple model for the action of a lawnmower blade on a grass stalk, or of a razor blade on a bristle.] 4.19 Start from the equation (4.189) for weakly nonlinear waves on a beam. If ∂θ/∂ξ = 2u(z, τ ), where z = −ξ, show that u satisfies the modified KdV (mKdV) equation ∂u ∂u ∂ 3 u + 6u2 + 3 = 0. ∂τ ∂z ∂z Show that the Miura transform ∂u ∂z converts the mKdV equation into the KdV equation v(ζ, τ ) = u2 − i

∂v ∂v ∂ 3 v + 6v + = 0. ∂τ ∂z ∂z 3 4.20 By multiplying (4.191) by d2 θ/dη 2 , obtain the first integral µ 2 ¶2 µ ¶ µ ¶2 d θ 1 dθ 4 2 dθ + −k = 0, dη 2 4 dη dη

212


where

r k=

2T∞ (1 − M ) . EI

By using the substitution ¡ ¢ dθ = ±2k sech φ(η) , dη or otherwise, obtain the general solution ³ ´ θ = const. ± 4 tan−1 ek(η0 −η) .

5 Nonlinear Elasticity

5.1 Introduction In Chapters 2 and 3, we have analysed solutions of the steady and unsteady Navier equation, which was derived in Chapter 1 under the assumption of linear elasticity. Actually, two fundamental approximations underpin linear elasticity. First, we will assume that we can discard the nonlinear terms in the relation (1.13) between strain and displacement. Geometrically nonlinear elasticity concerns large deformations in which these terms are not negligible, so the strain is a nonlinear function of the displacement gradients. This inevitably leads to the further complication that the Lagrangian and Eulerian variables may no longer be approximated as equal. The second assumption behind linear elasticity is that the stress is a linear function of the strain. This is a reasonable approximation for small strains, but it does not work well for materials such as rubber, which can suffer large strain and still remain elastic as defined in §1.1 CITE TRELOAR. Models for such materials require mechanically nonlinear elasticity, in which the stress is a nonlinear function of the strain. The fundamental difficulty to be confronted is that the balance of stresses is performed in the deformed state, while the constitutive relation must be imposed relative to the reference configuration. As a first step in addressing this difficulty, we now revisit the concepts of stress and strain. We will show how they may both be expressed in a Lagrangian frame of reference, allowing a self-consistent constitutive law to be imposed between them. For mechanically nonlinear materials, such laws are far less easy to specify than (1.42), and we will see that great care has to be taken to avoid models that allow unphysical behaviour such as perpetual motion. Finally we will show how the nonlinear theory, be it mechanical or geometric or both, can be 213

214

Nonlinear Elasticity

greatly simplified for hyperelastic materials for which the stress is related to the strain via the partial derivatives of an appropriate strain energy density, thereby generalising (1.54).

5.2 Stress and strain revisited 5.2.1 Deformation and strain We begin by recalling the analysis of strain from §1.4, using the same notation that a particle at position X in the undeformed reference state is displaced to a new position x(X) in the deformed state. Recall that X and x are referred to as Lagrangian and Eulerian variables respectively. An infinitesimal line element dX is transformed to dx under the deformation, and these are related by the chain rule dxi = Fij dXj ,

(5.1)

where the deformation gradient tensor is defined by F = (Fij ) = (∂xi /∂Xj ) .

(5.2)

The change in the length of the small element dX is thus given by |dx|2 = dX T C dX,

(5.3)

C = F T F,

(5.4)

where

and C, which will shortly be shown to be a tensor, is called the Green deformation tensor. Now suppose that a rigid-body motion is superimposed, so that each point x is further displaced to x0 = c + P x,

(5.5)

where the vector c and orthogonal matrix P are constant. It is a straightforward matter to substitute for x0 into (5.2) and thus obtain the new deformation gradient as Fij0 =

∂x0i = Pik Fkj ∂Xj

(5.6)

or, in matrix notation, F 0 = P F.

(5.7)

5.2 Stress and strain revisited

From the orthogonality of P , it therefore follows that ¡ ¢T C 0 = F 0 F 0 = F T F = C,

215

(5.8)

so that C, unlike F , is invariant under any superimposed rigid-body motion of the material. On the other hand, suppose we perform a rigid-body motion on the material before applying the deformation. In other words we consider the transformation X 7→ x(X 0 ),

where X 0 = c + QX,

(5.9)

where the vector c and orthogonal matrix Q are constant. Now the chain rule gives ! Ã ∂x i = F QT (5.10) F0 = ∂Xj0 and the corresponding Green deformation tensor is thus ¡ ¢T C 0 = F 0 F 0 = QCQT .

(5.11)

This demonstrates that C transforms as a tensor under transformations of the reference configuration. To connect with Chapter 1, let us note that if the displacement from X to x is a rigid-body motion, then F is orthogonal and, hence, C is equal to the identity matrix I. We are therefore led to define the strain tensor E = (Eij ), where ¢ 1¡ T 1 (C − I) = F F −I . (5.12) 2 2 By construction, E is a symmetric tensor which is objective, by which we mean that it has the same definition after arbitrary rotation of the axes, and is zero for a rigid-body displacement. Also (5.3), written in the form (1.12), namely E=

|dx|2 − |dX|2 = 2dX T E dX,

(5.13)

shows how the stretch of line elements in the continuum depends on E. We do not want the reader to drown under a deluge of new notations in this introduction. Nonetheless, we note for future reference that another deformation tensor can also be used, namely B = F F T.

(5.14)

This is called the left Green deformation tensor or, more commonly, the

216

Nonlinear Elasticity dX

dx

N

n

da

dA

Fig. 5.1. The deformation of a small scalene cylinder.

Finger tensor. We will find that B is more natural than C as a measure of deformation for an observer using Eulerian coordinates x.

5.2.2 The Piola–Kirchhoff stress tensors The elastic force exerted on a surface element da = n da in the deformed medium, with area a and normal n, is given by df = τ n da,

(5.15)

where τ = (τij ) is the Cauchy stress tensor, as defined in Chapter 1. As in §1.7, we will need to impose a constitutive relation between the stress and the strain tensor E, and this requires us to express τ relative to a Lagrangian frame. As a first step, let us find how the surface element da is related to the area dA that it occupied in the reference state. One way to do this is to consider a small scalene cylinder in the reference state, with base dA and axis dX. As shown in Figure 5.1, this will be deformed to a new cylinder, with base da and axis dx, which is related to dX by (5.1). The volumes of the two cylinders are given by dV = dX · dA and dv = dx · da respectively and, since the Jacobian J measures the local volume change at any point, these must be related by dv = J dV . We therefore obtain ¡ ¢ dx · da = dX · F T da = JdX · dA, (5.16) in which the vector element dX is arbitrary, so the area elements are related by ¡ ¢−1 dA. (5.17) da = det (F ) F T

5.2 Stress and strain revisited

217

When we apply this result to the expression (5.15) for the force on a small area element, we obtain n ¡ T ¢−1 o df = det (F ) τ F dA. (5.18) This gives the stress in the deformed configuration on an area element which, in the reference configuration, was given by dA. The quantity in braces, ¡ ¢−1 T = det (F )τ F T , (5.19) is a second-rank tensor called the first Piola–Kirchhoff stress tensor. As shown in §1.4, the Cauchy stress tensor τ is symmetric, but T in general is not. We can, though, use the symmetry of τ to deduce the following identity satisfied by T : T F T = F T T.

(5.20)

This motivates the introduction of the second Piola–Kirchhoff stress tensor S, defined by ¡ ¢−1 S = F −1 T = det (F ) F −1 τ F T , (5.21) so that S is symmetric.

5.2.3 The momentum equation In §1.6 we derived Cauchy’s momentum equation ρ

∂τij ∂ 2 xi , = ρgi + 2 ∂t ∂xj

(5.22)

which must apply to any continuum for which a Cauchy stress tensor (τij ) can be defined. In (5.22), gi are the components of the applied body force and ρ is the density of the material, related to the initial density ρ0 by ρ0 = J = det (F ). ρ

(5.23)

Now let us derive a Lagrangian equivalent of (5.22). We start, as in §1.6, by applying Newton’s second law to a material volume V (t) whose boundary ∂V (t) has outward unit normal denoted by n = (ni ): ZZZ ZZZ ZZ d ∂xi ρ dx = gi ρ dx + τij nj da. (5.24) dt V (t) ∂t V (t) ∂V (t)

218


Now we change the integration variables to Lagrangian coordinates X, using (5.17) to convert the final surface integral: ZZZ ZZZ ZZ n o d ∂xi −1 ρJ dX = gi ρJ dX + Jτik Fjk Nj dA. dt V (0) ∂t V (0) ∂V (0) (5.25) With respect to Lagrangian variables, the integration domain V (0) is time-independent, so we may differentiate through the first integral, using the fact that ρJ = ρ0 (X) is independent of t. In the final integral, we recognise the term in braces as the first Piola–Kirchoff tensor Tij and apply the divergence theorem to obtain ZZZ ZZZ ZZZ ∂Tij ∂ 2 xi ρ dX = g ρ dX + dX. (5.26) i 0 2 0 ∂t ∂X j V (0) V (0) V (0) Since this must hold for all reference volumes V (0) we deduce, assuming that the integrand is continuous, the following Lagrangian form of Cauchy’s equation: ρ0

∂Tij ∂ 2 xi . = ρ0 gi + 2 ∂t ∂Xj

(5.27)

Although it is not immediately obvious, (5.27) may also be obtained directly by using (5.19) to replace τ with T in (5.22) and applying the chain rule (Exercise 5.2).

5.2.4 Example: one-dimensional nonlinear elasticity The simplest example that illustrates both mechanical and geometric nonlinearity is a unidirectional displacement u(X, t) in the X-direction. In this very special situation, S and E just involve scalars Sxx and Exx and the most general constitutive law is simply a functional relationship between them, that is Sxx = φ (Exx ). From (5.13), we obtain the one-dimensional strain µ ¶ ∂u 1 ∂u 2 Exx = + (5.28) ∂X 2 ∂X and hence Txx = (F S)xx

∂x = φ ∂X

Ã

∂u 1 + ∂X 2

µ

∂u ∂X

¶2 ! .

(5.29)

5.3 The constitutive relation

219

In equilibrium under a constant sress Txx , the displacement is such that µ ¶ Ã µ ¶ ! ∂u ∂u 1 ∂u 2 1+ φ + = Txx . (5.30) ∂X ∂X 2 ∂X If we suppose that the material is mechanically linear, so that φ is the linear function φ (Exx ) = (λ + 2µ) Exx ,

(5.31)

then (5.30) is a cubic equation with only one real root. More exciting possibilities occur when we examine the possibility of longitudinal waves propagating in the X-direction. Again assuming the linear constitutive relation (5.31), we find that the momentum equation (5.27) in one spatial dimension reads ·µ ¶µ ¶ ¸ ∂2u ∂u 1 ∂u ∂u 2 ∂ 1 + 1 + , (5.32) = c p ∂t2 ∂X ∂X 2 ∂X ∂X where c2p = (λ + 2µ)/ρ0 . This is a nonlinear version of the P -wave equation that we encountered in Chapter 3. We recall that one-dimensional linear P -waves propagate information along the two families of characteristic X ± cp t = const in the (X, t)-plane. However, for (5.32), the characteristics are given by the equation Ã µ ¶ !1/2 dX ∂u 3 ∂u 2 = ± cp 1 + 3 + , (5.33) dt ∂X 2 ∂X and they are generally not straight lines any more. Even more importantly, they depend on the solution ∂u/∂X itself, and it can be shown that characteristics of the same family will almost always eventually cross. When this happens, the pieces of information carried by each of these two charateristics are usually incompatible, so that u becomes multivalued. This results in the formation of a shock, which is in the (X, t) plane across which the first derivatives of u are discontinuous (Bland, 1988, page???). 5.3 The constitutive relation 5.3.1 Polar decomposition The polar decomposition theorem (Strang, 1988) tells us that any nonsingular matrix F may be written as the product of an orthogonal matrix M T and a positive definite symmetric matrix U : F = M T U.

(5.34)

220


The real, positive eigenvalues of U are denoted exists an orthogonal matrix R such that  λ1 T  U = R ΛR, where Λ = 0 0

λ1 , λ2 and λ3 , and there  0 0 λ2 0  . 0 λ3

The rows of R are the eigenvectors of U , that is  T e1  , R = eT where U ei = λi ei , 2 T e3

ei · ej = δij .

(5.35)

(5.36)

Since F is a function of X, it follows that M , U , R and Λ likewise vary with X. Substitution of (5.34) and (5.35) into (5.1) leads to the relation dx = M T RT ΛR dX

(5.37)

between line elements dX and dx in the reference and deformed configurations respectively. Using (5.36), we can decompose dX into its components along each of the eigenvectors ei : X dx = M T λi ei (ei · dX) . (5.38) i

The effect of the transformation may thus be understood as a stretch, by a positive factor λi , along each of the eigenvectors ei , followed by a rotation via the orthogonal matrix M . We therefore define the eigenvalues λi to be the principal extension ratios and corresponding eigenvectors ei the principal directions corresponding to the deformation; both will, in general, vary with position. We emphasise that the λ2i are the eigenvalues of C, which is related to the strain tensor E by (5.12). It follows that the principal strains, introduced in §2.2.5, are given in terms of λi by p λi = 1 + 2Ei . (5.39) 5.3.2 Strain invariants By substituting (5.34) into (5.4), we see that U is related to C by C = U 2 = RT Λ2 R.

(5.40)

It follows that C shares the same eigenvectors ei as U , while the eigenvalues of C are λ21 , λ22 and λ23 . We may therefore expand out the characteristic


221

polynomial of C as ¡ ¢¡ ¢¡ ¢ det (ζI − C) = ζ − λ21 ζ − λ22 ζ − λ23 = ζ 3 − I1 (C)ζ 2 + I2 (C)ζ − I3 (C), (5.41) where Ij (C) are the isotropic invariants of C, given by I1 (C) = λ21 + λ22 + λ23 ,

I2 (C) = λ21 λ22 + λ22 λ23 + λ23 λ21 ,

I3 (C) = λ21 λ22 λ23 . (5.42)

If C is diagonal (with entries λ21 , λ22 , λ23 ), then it is clear that the three invariants may be written as ¡ ¢o 1n I1 (C) = Tr (C), I2 (C) = Tr (C)2 − Tr C 2 , I3 (C) = det (C). (5.43) 2 It is also clear that I1 (C), I2 (C) and I3 (C), like the eigenvalues and characteristic polynomial of C, are invariant under orthogonal transformations of the axes. Hence the equations given in (5.43) are identities, whether or not C is diagonal. Henceforth we will suppress the argument of Ik to avoid cluttering the equations. Now suppose a scalar function φ(C) of the Green deformation tensor is known to be invariant under rigid-body motions of the reference state. As shown in §5.2.1,transformation of the axes by an orthogonal matrix Q converts C to C 0 = QCQT , so the invariance required of φ may be stated as ¡ ¢ φ QCQT ≡ φ (C) (5.44) for all orthogonal matrices Q. In particular, (5.44) must hold if Q is equal to R, the orthogonal matrix that diagonalises C, so ¡ ¢ φ (C) ≡ φ Λ2 , (5.45) and φ can therefore depend only on λ21 , λ22 and λ23 . Moreover, additional rotations of the axes can further permute the diagonal elements of Λ2 and, hence, φ must be invariant under permutation of the λi . It follows that φ must be a function of the strain invariants Ij , that is φ = φ (I1 , I2 , I3 ) ,

(5.46)

which represents a considerable restriction of all possible functions of the six independent components of the symmetric tensor C. Note that B in (5.14) also has eigenvalues λ2i and invariants Ii , although generally not the same eigenvectors as C. We can also deduce from the

222


Cayley–Hamilton theorem† that C satisfies the identity C 3 − I1 C 2 + I2 C − I3 I = 0,

(5.47)

which can be used to express any power of C in terms of just I, C, and C 2 , and similarly for B. 5.3.3 Frame indifference and isotropy Now we are at last in a position to relate stress and strain in the Lagrangian reference frame. A material is called elastic if the second Piola–Kirchhoff stress is a function only of the deformation gradient, that is S = S(F );

(5.48)

S could also depend on X if the elastic properties were spatially nonuniform, but we will ignore this possibility here. The key requirement is that the relation (5.48) be invariant under rigidbody motions; in other words, the way a material responds to stress should not depend on the frame in which it is observed. As demonstrated in §5.2.1, rotation of the Eulerian coordinate axes through an orthogonal matrix P transforms F to F 0 = P F , so this frame indifference leads to the restriction S (F ) ≡ S (P F )

(5.49)

for all orthogonal matrices P . Since the Green deformation tensor C is invariant under rigid-body motions, (5.49) is certainly satisfied if the constitutive relation takes the form S = S(C),

(5.50)

and it may be shown, as in Exercise 5.3, that this condition is necessary as well as sufficient. Notice that (5.50) is more specific than (5.48) since C has only six distinct elements, while F has nine. Further simplification may be achieved by assuming that the material is isotropic, meaning it has no preferred directions. Suppose we take the undeformed material and rotate it about an arbitrary origin X 0 via an orthogonal matrix Q such that X 0 = X 0 + Q (X − X 0 ) .

(5.51)

Then, if we take the rotated material and apply exactly the same deformation as before, the net deformation is given by ¡ ¢ X 7→ x(X 0 ) = x X 0 + Q (X − X 0 ) . (5.52) † This states that every matrix satisfies its own charactistic polynomial.


223

As shown in §5.2.1, the Green deformation tensor corresponding to (5.52) is C 0 QCQT , and the corresponding stress tensor at the point X 0 is therefore ¡ ¢ S 0 (C) = S QCQT . (5.53) If the material is isotropic, then this should be identical to the stress which we would have obtained without the orginal rotation, albeit related to the new coordinates X 0 rather than to X. By applying the usual rule for the rotation of a tensor, we deduce that S 0 (C) = QS(C)QT ,

(5.54)

which must apply identically for any fixed point X 0 and any rotation matrix Q. The constitutive relation (5.50) between S and C must therefore satisfy the symmetry condition S(QCQT ) ≡ QS(C)QT

(5.55)

for all orthogonal matrices Q, at any point X at which the material is isotropic.† It may be shown (Exercise 5.4) that S has the symmetry property (5.55) if and only if it can be written in the form S(C) = φ0 (I1 , I2 , I3 ) I + φ1 (I1 , I2 , I3 ) C + φ2 (I1 , I2 , I3 ) C 2 ,

(5.56)

where φ0 , φ1 and φ2 are functions of the three invariants of C. An isotropic constitutive relation thus amounts to a specification of these three scalar functions. We note that, in the limit of small stress, when Cij → 1 + 2eij . Hence (5.56) differs from the identity by a linear combination of ekk δij from the first term and eij from the second and third terms, thereby retrieving (1.42). We will return to the small displacement limit in §5.3.6. We may directly infer from (5.56) that in an isotropic material, S has the same eigenvectors as C. On the other hand, using (5.19), equation (5.56) gives, for the Cauchy stress ¤ −1/2 £ τ = I3 φ0 B + φ1 B2 + φ2 B3 , (5.57) so that the principal axes of τ are colinear with those of B. Furthermore, using (5.47), one can eliminate B3 and obtain ¤ −1/2 £ τ = I3 I3 φ2 I + (φ0 − I2 φ2 ) B + (φ1 + I1 φ2 ) B 2 . (5.58) † There are many generalisations of theoretical and practical interest. For example, a material exhibits transverse isotropy if it is invariant only under rotations about a fixed axis. This might apply, for example, to fibre-reinforced materials, which are stronger in the direction of the fibres than in the transverse plane.

224


This demonstrates how the Cauchy stress is more elegantly expressed in terms of B, while C is more natural for the Piola–Kirchhoff stress. If the material is not “pre-stressed”, then S should be zero in the reference configuration, that is S(I) = 0, which leads to the condition φ0 (3, 3, 1) + φ1 (3, 3, 1) + φ2 (3, 3, 1) = 0.

(5.59)

However, we will see in the next section that the requirement (5.59) is far from sufficient for the constitutive relation (5.56) to lead to a physically acceptable theory of mechanically nonlinear elasticity.

5.3.4 The energy equation In Chapter 1, we found that the work done on any region of a linear elastic solid is balanced by the change in both the kinetic energy and the strain energy stored by the material as it deforms. Now let us try to repeat this analysis for a nonlinear elastic material. We start by multiplying the momentum equation (5.27) by the velocity ∂xi /∂t and integrating over a control material volume V0 : ZZZ ZZZ ZZZ ∂xi ∂ 2 xi ∂xi ∂Tij ∂xi ρ0 dX + dX. dX = ρ0 gi 2 ∂t ∂t ∂t V0 V0 ∂t ∂Xj V0 (5.60) The time derivative may be taken outside the first integral, while the final integral can be manipulated using the divergence theorem to obtain d dt

ZZZ V0

ρ0 2

µ

∂xi ∂t

¶2

ZZZ

∂Fij dX ∂t V0 ZZ ∂xi ∂xi ρ0 gi dX + Tij Nj dA. (5.61) ∂t V0 ∂V0 ∂t

dX + ZZZ =

Tij

Recalling that ρ0 dX ≡ ρ dx, we see that the first term on the left-hand side of (5.61) is the rate of change of kinetic energy, while the first term on the right-hand side is the rate at which work is done by the body force. Finally, the boundary integral is the rate at which work is done by stress on ∂V . We can therefore interpret the second term on the left-hand side of (5.61) as the rate at which elastic energy is stored by the medium. Unfortunately, this implies that apparently legitimate nonlinear constitutive relations may allow a net extraction of energy from an elastic material under periodic loading cycles. Let us consider, for example, a material with the isotropic


225

constitutive relation ¡ ¢ S = λ21 λ22 λ23 − 1 I.

(5.62)

subject to a deformation in which   λ1 (t) 0 0 F = 0 λ2 (t) 0 , 0 0 1

(5.63)

where λ1 , λ2 are periodic functions of t. We use (5.21) Piola–Kirchhoff stress as  0 ¡ 2 2 ¢ λ1 (t) T = F S = λ1 λ2 − 1  0 λ2 (t) 0 0

to calculate the first  0 0 0

and hence find that µ ¶ ZZZ ¡ 2 2 ¢ ∂Fij dλ1 dλ2 Tij dX = V0 λ1 λ2 − 1 λ1 + λ2 , ∂t dt dt V0

(5.64)

(5.65)

which is a function whose time integral over a period may be positive or negative. When λ1 = 1 + a cos(ωt), λ2 = 1¡ + b sin(ωt), for example, the ¢ integral of (5.65) over a period 2π/ω is 3πab b2 − a2 /2. If an elastic material were to exist that satisfied (5.62), then it could be used as a limitless energy source! This behaviour is clearly unacceptable and a suffcient condition to preclude it is the existence of a strain energy density W (Fij ) such that Tij =

∂W , ∂Fij

(5.66a)

or, introducing the convenient notation for differentiation by a tensor, T = If so, then we can write ZZZ

∂W . ∂F

∂Fij d Tij dX ≡ ∂t dt V0

(5.66b)

ZZZ W dX,

(5.67)

V0

and interpret the right-hand side as the strain energy stored in the medium. We will now explore this happy circumstance in more detail.

226


5.3.5 Hyperelasticity A material is called hyperelastic if there is a strain energy density W (Fij ) satisfying (5.66). The theory of hyperelasticity has proved to be by far the most commonly-used and convenient way to construct constitutive relations for materials like rubber. By posing a functional form for W, and inferring from (5.66) the corresponding stress-strain relation, one automatically avoids thermodynamically unfeasible situations like that described above. However, the mere existence of W is not sufficient to guarantee wellposedness of the resulting elastic model. At the moment all we can say is that our experience with linear elasticity in Chapter 1 suggests that W (Fij ) should have a maximum at Fij and be strictly non-negative. First, as in §5.3.3, we must insist that the constitutive relation be invariant under rigid-body motions of the reference state. In the hyperelastic framework, this implies that W can depend only on the six components of C, and the second Piola–Kirchoff stress is then given by W=2

∂W , ∂C

(5.68)

as shown in Exercise 5.9. Henceforth we limit our attention to isotropic materials, for which W is invariant under rigid-body rotation of the reference state, and we can follow §5.3.2 to assert that W = W (I1 , I2 , I3 ) .

(5.69)

The second Piola–Kirchhoff stress tensor is therefore given by X ∂W ∂Ik S=2 . ∂Ik ∂C

(5.70)

k

To find the circumstances under which such a strain energy function can exist, we note from Exercise 5.8 that ∂I1 = I, ∂C

∂I2 = I1 I − C, ∂C

∂I3 = I3 C −1 , ∂C

(5.71)

where I is the identity. Hence, after a bit of algebra (see Exercise 5.11), we find that ·µ ¶ µ ¶ ¸ ∂W ∂W ∂W ∂W ∂W ∂W 2 S=2 + I1 + I2 I− + I1 C+ C , ∂I1 ∂I2 ∂I3 ∂I2 ∂I3 ∂I3 (5.72) which gives us explicit forms for the functions φi in (5.56). Since the φi


227

satisfy φ2 = 2

∂W ∂W ∂W , φ1 + I1 φ2 = −2 , φ0 + I1 (φ1 + I1 φ2 ) − I2 φ2 = 2 , ∂I3 ∂I2 ∂I1 (5.73)

the Pfaffian conditions for the existence of W are that ∂φ2 ∂φ1 ∂φ2 + + I1 = 0, ∂I3 ∂I2 ∂I3 ¢ ∂φ2 ∂φ2 ∂φ1 ¡ 2 ∂φ0 + I1 + I1 − I2 − = 0, ∂I3 ∂I3 ∂I3 ∂I1 ¢ ∂φ2 ∂φ0 ∂φ1 ∂φ1 ¡ 2 ∂φ2 + I1 + + I1 − I2 + I1 = 0. ∂I2 ∂I2 ∂I1 ∂I2 ∂I1

(5.74a) (5.74b) (5.74c)

Any choice of the functions φi that does not satisfy (5.74) will lead to behaviour such as that exhibited in (5.65); indeed it was these integrability conditions that suggested the counter-example (5.62). We note from (5.21) that the Cauchy stress is given in terms of W as −1/2 ∂W

FT ∂F and, as shown in Exercise 5.7, this leads to the useful expression τ = I3

−1/2

τi = I3

λi

∂W ∂λi

(5.75)

(5.76)

(no summation) for the Cauchy principal stresses. As well as clarifying the characterisation of elastic materials, hyperelasticity also often simplifies the solution of nonlinear elastic problems. By virtue of (5.66), the arguments of §1.9 still imply that the steady version of the gravity-free momentum equation (5.27) is a necessary condition for ZZZ S= WdX (5.77) V0

to be minimised, assuming of course that W is sufficiently smooth. However, for a nonlinear material, W will no longer be quadratic in F and finding the uniqueness and smoothness of a minimum of (5.77) now poses a typically difficult challenge in teh calculus of variations; it transpires that not only do we require W to be strictly positive except at F = 0, but also that it should be quasiconvex, which demands that the level surfaces of W are convex in a suitable function space. It is also difficult to compare this result with the condition that, in equilibrium, (5.27) should comprise, with suitable boundary conditions, a well-posed system for xi . The minimisation of a functional like (5.77) is generally easier computationally than solving the

228


nonlinear partial differential equation (5.27). In particular, minimisation problems of this type are naturally suited to the finite element method of numerical analysis [REF!!!]. 5.3.6 Linear elasticity Now we briefly verify that the hyperelastic formulation of elasticity outlined above reduces to the linear elastic constitutive relation introduced in Chapter 1 when the strains are small. We recall from (5.39) that the principal strains are related to the principal stretches by p λi = 1 + 2Ei ∼ 1 + Ei (5.78) when the strains are small. Also we recall that W is a function of the principal stretches {λ1 , λ2 , λ3 } with a unique global minimum at λj ≡ 1. Moreover, for isotropic materials, W is invariant under permutations of {λ1 , λ2 , λ3 }, and hence, as λ1 , λ2 , λ3 → 1, the local behaviour of W near its minimum must take the form W ∼ µ1

3 X i=1

2

(λi − 1) + µ2

3 X

(λi − 1) (λj − 1) ,

as ,

(5.79)

i,j=1

where µ1 and µ2 are scalar constants. Now, the principal values of the second Piola–Kirchhoff stress tensor are given by Si =

1 ∂W , λi ∂λi

(5.80)

and substitution from (5.78) and (5.79) leads to Si ∼ 2µ2 (λ1 + λ2 + λ3 − 3) + 2µ1 (λi − 1) ∼ 2µ2 (E1 + E2 + E3 ) + 2µ1 Ei (5.81) as the strain tends to zero. With respect to principal axes, the stress and strain tensors are thus related by S ∼ 1µ2 Tr (E) I + 2µ1 E,

(5.82)

and, since the left- and right-hand sides both transform as tensors, (5.82) holds with respect to any chosen axes. If the displacement is small, relative to any other length-scale, then the deformation gradient F is approximately equal to the identity and hence, from (5.19) and (5.81), both the Piola–Kirchoff stress tensor and the Cauchy stress tensor are all approximately equal. In addition, the quadratic terms in


229

the strain tensor may be neglected to leading order, and, when we identify µ2 with λ/2 and µ1 with µ, (5.82) thus reproduces the linear elastic constitutive relation (1.42). However, (5.82) holds just as long as the strain is small, so the displacement need not be small provided it is close to a rigid-body motion. In such situations, we can thus justify the use of the mechanically linear constitutive equation (5.82), while retaining the geometrically nonlinear relation between x and X. Indeed we have already done this implicitly in (4.10). 5.3.7 Incompressibility Many rubber-like materials are observed to be almost incompressible, and it is common to employ constitutive relations that imply this. From (1.8), which holds for nonlinear deformations, the condition of incompressibility is J = ρ0 /ρ = 1 or, in terms of the strain invariatiants, I3 ≡ 1.

(5.83)

As we found in §1.8, the imposition of this constraint on the problem can be achieved at the expense of introducing an additional unknown scalar function p which can be interpreted as an isotropic pressure. Now we will consider how to introduce p in a self-consistent way, and there are two ways to go about this. The approach followed in §1.8 was to characterise the compressibility by a small but nonzero parameter ² and then carefully take the limit ² → 0. Alternatively, we could have examined the variational formulation of the problem in terms of the strain energy density W and then identified p as a Lagrange multiplier associated with the constraint (5.83), as in Exercise 1.7. We can follow the first strategy by constructing a strain energy function that heavily penalises departures of I3 from unity, for example µ (I3 − 1)2 ˜ (I1 , I2 ) , +W (5.84) 2² where µ is some elastic modulus and ² ¿ 1. The actual functional form of W is unimportant provided it has a very strong minimum at I3 = 1, but the particular example (5.84) will help to fix ideas. We anticipate that this strain energy function will force I3 to remain close to unity, and therefore set 2² (5.85) I3 = 1 − p, µ W (I1 , I2 , I3 ) =

where the scalar function p is to be determined. Substituting (5.84) into

230


(5.72) and using (5.47), we find that our desired constitutive relation becomes µ ¶ ∂W ∂W ∂W S=2 + I1 I −2 C − pC −1 . (5.86) ∂I1 ∂I2 ∂I2 The status of p as a hydrostatic pressure is clearer in the corresponding expression for the Cauchy stress, namely µ ¶ ∂W ∂W ∂W 2 τ = −pI + 2 + I1 B−2 B . (5.87) ∂I1 ∂I2 ∂I2 To arrive at the same conclusion from a purely variational viewpoint, we note that incompressibility is satisfied in a small virtual displacement by adding to Fij an increment δFij such that ∂ (det F ) δFij = 0. ∂Fij

(5.88)

In Exercise 5.6, we show that this is equivalent to Fji−1 δFij = 0.

(5.89)

We also note from (5.61) that no additional work is done by the increment provided Tij δFij = 0.

(5.90)

Hence, considering only virtual deformations satisfying the constraint (5.89), that T is only determined up to an unknown scalar multiple of ¡we Tdeduce ¢−1 F . An arbitrary scalar multiple of C −1 may be thus added to S, and (5.86) follows immediately. 5.3.8 Examples of constitutive relations Some commonly used examples of constitutive relations for the strain energy density are µ neo-Hookean: W = (I1 − 3) , (5.91a) 2 Mooney–Rivlin: W = c1 (I1 − 3) + c2 (I2 − 3) , (5.91b) Varga:

W = 2µ (λ1 + λ2 + λ3 − 3) , (5.91c) λα1 + λα2 + λα3 − 3 Ogden: W = 2µ . (5.91d) α In each case, the material is also assumed to be incompressible, so the constitutive relation is supplemented by the constraint I3 ≡ 1.

5.4 Examples

F

(a)

F

λ1 F

231

(b)

F

λ1

(d)

F

λ1

(c)

λ1

(e)

F

λ1

(f)

λ1

Fig. 5.2. Typical force–strain graphs for uniaxial tests on various materials: (a) neoHookean, (b) Mooney–Rivlin, (c) Varga, (d) Ogden (α < 1), (e) Ogden (1 < α < 2), (f) Ogden (α > 2).

In Figure 5.2, we show typical force-versus-strain response curves for uniaxial stress tests on materials described by each of the above constitutive relations (see Exercise 5.10). The neo-Hookean and Mooney–Rivlin materials are characterised by an initial linear response, followed by transition to a less stiff linear behaviour. Varga materials fail at a finite value of the applied force. For Ogden materials, the behaviour depends on α: when α < 1, the force reaches a maximum before decaying towards zero at large strain; thus once again only a finite force is needed for the material to fail completely. For α between 1 and 2, the response is qualitatively similar to that for a Mooney–Rivlin material, while for α > 2 the stiffness always increases with increasing strain.

5.4 Examples 5.4.1 Principal stresses and strains We conclude with some simple examples which have sufficent symmetry for analytical progress to be made. In all cases, we will assume incompressibility of the medium, which leads to the relations I3 = λ21 λ22 λ23 = 1,

I1 = λ21 + λ22 + λ23 ,

−2 −2 I2 = λ−2 1 + λ2 + λ3 ,

(5.92)

232


` F h

`

F Fig. 5.3. Schematic of a square membrane subject to an isotropic tensile force F .

for the strain invariants. In all the examples considered below, the stress and strain tensors are diagonal, and the principal stresses are found from the incompressible version of (5.76), that is τi = λi

∂W − p. ∂λi

(5.93)

We focus on the Mooney–Rivlin constitutive relation (5.91b), for which (5.93) reads τi = 2c1 λ2i −

2c2 − p. λ2i

(5.94)

5.4.2 Biaxial loading of a square membrane In this example, we consider a thin square membrane of initial thickness h and side length ` loaded on its edges by a tensile force F , as shown in Figure 5.3. This leads to the uniform displacement field xi = λi Xi , for which case the stresses are also uniform and given by (5.93). By relating these to the applied force F , we obtain the equations λ1

F ∂W −p= , ∂λ1 λ2 λ3 h

λ2

∂W F −p= , ∂λ2 λ1 λ3 h

λ3

∂W − p = 0, ∂λ3

(5.95)

and elimination of p leads to F ∂W λ3 ∂W ∂W λ3 ∂W = − = − . h ∂λ1 λ1 ∂λ3 ∂λ2 λ2 ∂λ3

(5.96)

To proceed further, we now have to choose a constitutive relation. Using the Mooney–Rivlin model (5.94), we find that the stretch ratios satisfy £ ¡ ¢ ¡ ¢¤ (λ1 − λ2 ) c1 1 + λ31 λ32 + c2 λ21 + λ1 λ2 + λ22 − λ41 λ42 = 0. (5.97) This always has the solution λ1 = λ2 but, whenever c2 6= 0, (5.97) also admits asymmetric solutions in which λ1 6= λ2 . As shown in Figure 5.4(a), there is a critical value λc of the stretches at which symmetric response

5.4 Examples

(a)

λ2

(b)

W

5

233

(c)

F

30 15 25

4

12.5 20 3

10 15 7.5

λc

2

1

10

5

5

2.5

1 1

2

3

λ1

4

5

1

2

3

4

2

3

4

5

5

λ1

λ1

Fig. 5.4. Response diagrams for a biaxially-loaded incompressible sheet of Mooney– Rivlin material with c2 /c1 = 0.5. (a) Locus of compatible stretches with a symmetric square loading. (b) Strain energy density as a function of one of the stretches. (c) Applied force as a function of one of the stretches. The thin lines represent a symmetric elastic response, the thick lines an asymmetric one, and the arrows indicate the evolution of the stable solution as F increases.

bifurcates to an asymmetric one (see Exercise 5.12 for the details). We also show in Figure 5.4(b) that, when λ1 > λc , the strain energy is reduced by breaking the symmetry, so the asymmetric solution becomes energetically favourable. Finally, in Figure 5.4(c), we show how the applied force varies with the stretch. We see that as λ crosses λc , the stiffness of the sheet suddenly reduces as it switches to an asymmetric response. This example illustrates one of the most dramatic consequences of mechanical nonlinearity: the strain response to a given load is generally not unique. We have already encountered nonuniqueness as the result of geometrical nonlinearity in Chapter 4 for the Euler strut, and we note that in both cases the symmetry is broken in the bifurcation.

5.4.3 Blowing up a balloon It is a common experience that blowing up a balloon is more difficult in the early stages than it is later on. This phenomenon can be understood by considering a thin, spherical, incompressible rubber membrane of initial radius R and thickness h ¿ R. Assuming the spherical symmetry is preserved as the balloon inflates, the two tangential stretches λθ and λφ (where θ and φ are the usual spherical polar coordinates) will be equal and related to the stretched radius r by r (5.98) λθ = λφ = , R

234


P

0

c0 = 0.3

6

0.2

5 4

0.1

3 2 1

0 2

4

3

5

6

λθ Fig. 5.5. Scaled pressure inside a balloon as a function of the stretch λθ = R/r for various values of the Mooney–Rivlin parameter c0 = c2 /c1 .

and the normal stretch λr is determined by incompressibility. Since the balloon is assumed to be thin, we can use the biaxial result above to calculate the tension, µ ¶ ∂W ∂W − λr , (5.99) T = Tθ = Tφ = hλr τθ = hλr λθ ∂λθ ∂λr noting that the thickness of the balloon is hλr following the deformation. The final observation is that a normal force balance relates the tension to the internal pressure P and the curvature of the membrane via 2T . (5.100) r It is now a simple matter to assemble these ingredients and thus relate the inflation pressure to the radius of the balloon. Again using the Mooney– Rivlin model (5.94), for example, we find ¡ ¢¡ ¢ 4 1 + c0 λ2θ λ6θ − 1 RP 0 = P (say) = , (5.101) hc1 λ7θ P =

where c0 is the dimensionless ratio c2 /c1 . With c0 = 0, we see in Figure 5.5 that the pressure required to inflate the balloon initially increases before decreasing when the radius is sufficiently large, successfully reproducing the familiar behaviour described above. For small positive values of c0 , however, the behaviour is more complicated, with the pressure reaching a maximum, decreasing, then finally increasing again for very large strains. In an experiment where a controlled pressure is imposed (not generally possible using one’s lungs!), the radius would jump as indicated by the arrow when the pressure reaches its maximum value.

5.4 Examples

235

Increasing c0 corresponds to making the material stiffer at large strains, reflecting the fact that polymer molecules can only withstand a finite extension without breaking. If c0 exceeds a critical value Ã √ !2/3 11 11 − 34 ≈ 0.21446, 25 this increased stiffness removes the maximum in the pressure completely and the behaviour becomes monotonic. 5.4.4 Cavitation Suppose a spherical cavity embedded in a rubber continuum expands as a result of an increasing internal pressure P . Such a situation could be achieved by dissolving a gas in a rubber matrix, for example. Now we use spherical polar coordinates (R, θ, φ) in the rest state and again assume the displacement is purely radial, so that θ and φ are preserved while each radial position R expands to a new radius r(R). We can hence write down the principal stretches λR =

dr , dR

λφ = λθ =

r , R

(5.102)

and the incompressibility condition λR λθ λφ = 1 gives us a differential equation for r(R), namely dr R2 = 2. dR r

(5.103)

Denoting by a the initial radius of the sphere, we write the solution of (5.103) as ¡ ¢1/3 r = R3 + λ30 a3 − a3 , (5.104) where λ0 = λθ (0) =

r(a) a

(5.105)

is the expansion ratio of the cavity. A displacement of the form (5.104) is the only incompressible purely radial deformation. We can read off from (5.102) the principal stretches ¡ 3 ¢1/3 R + λ30 a3 − a3 R2 . (5.106) λR = ¡ ¢2/3 , λφ = λθ = R R3 + λ3 a3 − a3 0

236

Nonlinear Elasticity P/c1

0.2

10

0.1

8 6

c0 = 0

4 2 2

3

4

5

6

7

8

9

λ0

Fig. 5.6. Gas pressure inside a cavity as a function of inflation coefficient for various values of the Mooney–Rivlin parameter c0 = c2 /c1 .

We next turn to the familiar radial momentum equation from (1.81): 2τrr − τθθ − τφφ dτrr + = 0. dr r

(5.107)

We can use (5.104) to transform to Lagrangian coordinates and substitute for the stress components from (5.93), thus obtaining ¡ ¢ R3 + a3 λ20 − 1 dτrr ∂W ∂W = 2λθ − 2λR . 2 R dR ∂λθ ∂λR

(5.108)

Now solving the first-order differential equation (5.108) subject to the two boundary conditions τrr (a) = −P,

τrr → 0 as R → ∞,

(5.109)

yields an expression for the internal pressure P in terms of the inflation coefficient λ0 . For Mooney–Rivlin materials, for example, this gives rise to the relation µ ¶ ¶ µ 4 1 1 P = c1 5 − − − 2c2 1 − 2λ0 + 2 , (5.110) λ0 λ40 λ0 which we plot in Figure 5.6 for various values of the ratio c2 /c1 . When c2 = 0, the material is neo-Hookean, and we see that only a finite pressure P = 5c1 is needed to inflate the cavity to infinity. This unlikely behaviour reflects the failure of the neo-Hookean model to capture the inability of materials to withstand unbounded strains, and the situation is remedied whenever c2 is positive.

5.5 Conclusion

237

5.5 Conclusion This is chapter has provided an introduction to what is probably the most mathematically challeneging branch of elasticity theory. In principle, our basic task is to solve for xi from (5.27), bearing in mind that the stress is now a much more complicated function of the strain than it was for mechanically linear elasticity. Concerning the constitutive law, it is convenient to work with the second Piola–Kirchhoff tensor S but there are now many more physically acceptable relations between S and F than there were for linear elasticity. Nonetheless, any selection of this relation must ultimately lead to a well-posed system of partial differential equations for xi before we can have confidence that we have a good model for the nonlinear evolution of an elastic solid. If we confine ourselves to statics, then it is conceptually appealing to work with hyperelastic materials for which we have the apparently much simpler task of minimising the net strain energy (5.77). However, even here we have to select from a wide range of functions relating the strain energy density to the strain invariants, and even when we have done this, we still have to confront the daunting problem of the well-posedness of the relevant minimisation problem in the calculus of variations. Having at last faced up to the theoretical challenges of nonlinear elasticity, we are now in a position to discuss the implications for some of the practically important geometric configurations that we encountered in Chapter 4.

Exercises 5.1 Show that the displacement correspoding to a rigid-body motion is given by X(x, t) = a(t) + P (t)x, where the vector a and orthogonal matrix P are spatially uniform. Show that the strain E defined by (5.12) is identically zero for this displacement field, but that the linearised strain (1.37) is not. 5.2 (a) Prove the identity µ ¶ ∂xj 1 ∂J ∂ ≡ , (5.111) ∂xj ∂Xk J ∂Xk (assuming the summation convention). [Hint: integrate both sides over an arbitrary material volume V (t) and use the divergence theorem.] (b) Substitute the Piola–Kirchhoff for the Cauchy stress tensor in the momentum equation (5.22) and use the chain rule to

238


obtain ρ0

¡ −1 ¢ ∂ 2 xi −1 ∂ = ρ0 gi + JFkj J Til Fjl . 2 ∂t ∂Xk

(5.112)

(c) Expand out the right-hand side of (5.112) and apply the identity (5.111) to obtain the Lagrangian momentum equation (5.27). 5.3 Suppose the scalar function f of the two-dimensional matrix F is invariant under the transformation F 7→ P F , where P is any orthogonal matrix. Let f (F ) = g(A, B, α, β), where F is written in the form µ ¶ µ ¶ A cos α B cos β F11 F12 F = = . F21 F22 A sin α B sin β By expressing P as the rotation matrix µ ¶ cos θ sin θ P = , − sin θ cos θ show that g must satisfy g(A, B, α, β) ≡ g(A, B, α − θ, β − θ) for all θ. Explain why g(A, B, α, β) ≡ h(A, B, α−β) for some function h of three independent variables. Deduce that f = f (C), where C = F TF . [A conceptually equivalent argument works in three dimensions; see, for example, Noll (1958); Spencer (1970)] 5.4 (a) Suppose that two 2 × 2 symmetric matrices S and C satisfy a functional relation S = S(C) which enjoys the symmetry (5.55) for all orthogonal matrices Q. By considering µ ¶ 1 0 Q= , 0 −1 or otherwise, show that S must be diagonal whenever C is. (b) With respect to coordinates in which C (and hence also S) is diagonal, deduce that S must take the form µ ¶ s1 (λ21 , λ22 ) 0 S= , 0 s2 (λ21 , λ22 ) where s1 and s2 are arbitrary functions of the eigenvalues λ21

EXERCISES

239

and λ22 of C. Show that this expression may be manipulated to S = φ0 (I1 , I2 )I + φ1 (I1 , I2 )C,

(5.113)

where φ0 and φ1 are two arbitrary functions of the invariants I1 and I2 of C (note that there are only two in two dimensions). (c) Show that (5.113) is invariant with respect to orthogonal transformations and, hence, true whether or not C is diagonal. (d) Generalise the above argument to three dimensions. 5.5 Considering only two-dimensional spatially linear displacements, with (x1 , x2 ) = (λ1 (t)X1 , λ2 (t)X2 ) so that F and S are both diagonal, show that energy is consumed at a rate q = λ1 S11

dλ1 dλ2 + λ2 S22 dt dt

per unit volume. Deduce that the medium conserves energy if and only if the stress components satisfy λ1

∂S11 ∂S22 ≡ λ2 . ∂λ2 ∂λ1

For the anisotropic constitutive relation ¡ ¢ ¡ ¢¡ ¢ S11 S22 = µ λ21 + λ22 1 α , where α is a constant, show that, over a periodic cycle in which λ1 = 1 + ε cos(ωt),

λ2 = 1 + ε sin(ωt),

the net energy consumption per unit volume is Z 2π/ω q dt = 2πε2 (α − 1)µ. 0

5.6 Show that ∂det (F ) = det (F )Fji−1 . ∂Fij [Hint: recall the inversion formula Fij−1 = cofactor (Fji ) / det(F ).] 5.7 Recall from §5.3.1 the existence of two orthogonal matrices R and M such that F = M T RT ΛR, where Λ is diagonal (with entries λi ). Use the chain rule to show that T =

∂W ∂W = M T RT R. ∂F ∂Λ

240


Now use the relationship (5.19) between the Cauchy and Piola–Kirchhoff stresses to show that −1/2 ∂W

RM τ M T RT = I3

∂Λ

Λ−1 .

Since Λ is diagonal, deduce that the principal Cauchy stresses are given by (5.76). 5.8 Prove the identities (a)

∂I1 = δij , ∂Cij

(b)

∂I2 = I1 δij − Cij , ∂Cij

(c)

∂I3 −1 = I3 Cij . ∂Cij

[See Exercise 5.6.] 5.9 If the strain energy density W is a function only of the symmetric Green deformation tensor Cij , show that X ∂W ∂W Fik =2 , ∂Fij ∂Ckj k

and deduce that the second Piola–Kirchhoff stress tensor is given by S=2

∂W . ∂C

5.10 Consider uniaxial stretching, in which a bar of initial cross-sectional area A is subject to an axial force F . Show that F is related to the axial stretch λ1 by ∂W λ2 ∂W F = − , A ∂λ1 λ1 ∂λ2 −1/2

where λ2 = λ3 = λ1 . Hence justify the response diagrams shown in Figure 5.2. 5.11 Use the results from Exercise 5.8 to write out equation (5.70) as ·µ ¶ ¸ ∂W ∂W ∂W ∂W −1 S=2 + I1 I− C + I3 C . (5.114) ∂I1 ∂I2 ∂I2 ∂I3 Using the Cayley–Hamilton theorem (5.47), deduce (5.72). 5.12 Assuming the Mooney–Rivlin form for the strain energy in (5.95), deduce (5.97). Show that the critical stretch λc , at which bifurcation from symmetric to asymmetric stretching occurs, is given implicitly by 1 + λ6 c2 = c0 = 2 6 c , c1 λc (λc − 3)

EXERCISES

241

where λc > 31/6 . By introducing Θ = λ1 λ2 , show that the asymmetric solution satisfies Θ , λ1 = λ± , λ2 = λ± where λ2±

=

q −P (Θ) ±

P (Θ) + 4c0 2 Θ2

2c0

¡ ¢ , P (Θ) = 1 + Θ3 + c0 Θ − Θ4 .

5.13 Consider a thin cylindrical tube, of initial radius R and thickness h ¿ R, inflated by an internal air pressure P . (a) Assuming the Mooney–Rivlin model and following the same methodology as in §5.4.3, show that the internal pressure and the axial tension are related to the tangential stretches by ¡ ¢¡ ¢ 2 1 + c0 λ2z λ4θ λ2z − 1 RP 0 P = = , hc1 λ4θ λ3z ¢¡ ¢ ¡ 2 1 + c0 λ2θ λ2θ λ4z − 1 Tz = , hc1 λ3θ λ3z where c0 = c2 /c1 . (b) If the tube is constrained in the z-direction so that λz = 1, show that only a finite pressure P 0 = 2 (1 + c0 ) is needed to inflate the tube to infinity. Show also that the axial tension Tz needed to maintain zero displacement in the z-direction tends to infinity as λθ does. (c) Now, instead suppose that the tube is free to contract in the z-direction, so that Tz = 0. Show that in this case P 0 → ∞ as λθ → ∞, although the behaviour is non-monotonic if Ã √ !2/3 11 34 + 11 c0 > ≈ 4.663. 7 5.14

(a) By changing independent variables to λθ , show that the differential equation (5.108) for the stress outside a spherical cavity may be written in the form ¡ ¢ dτrr ∂W ∂W = 2λθ − 2λR −λθ λ3θ − 1 dλθ ∂λθ ∂λR and, for the Mooney–Rivlin model, derive the equation ¡ ¢¡ ¢ 4 c1 + c2 λ2θ λ3 + 1 dτrr =− . dλθ λ5θ

242


(b) For a spherical cavity in an infinite medium, show that the appropriate boundary conditions are τrr = −P

at λθ = λ0 ,

τrr → 0 as λθ → 1,

and hence obtain the relation (5.110) between the internal pressure P and the inflation parameter λ0 . (c) For a sphere with initial internal radius a and finite thickness h, show that µ µ ¶ ¶ 4 1 4 1 1 1 + − − P = c1 − 2c2 2λ? − 2 − 2λ0 + 2 , λ? λ4? λ0 λ40 λ? λ0 where

Ã λ? =

1+

¡ 3 ¢ !1/3 λ0 − 1 a3 (a + h)3

.

Hence recover the balloon result (5.101) in the limit h/a → 0.

6 Asymptotic Analysis

6.1 Introduction In Chapter 4, we derived various approximate models for thin or slender elastic configurations such as rods and plates. These models were obtained using net force and moment balances combined with ad hoc constitutive relations, for example between the bending moments and the curvatures. In this chapter, we show how such models may be derived systematically from the underlying continuum equations. We concentrate on a few canonical models for plates, beams, rods and shells. Each of these models is important in its own right, and their derivation illustrates tools that are widely useful. Also it is surprising that we will only be able to consider the paradigm string model (3.1) when we are halfway through this chapter. The basic idea is to exploit the slenderness of the geometry so as to simplify the equations of elasticity asymptotically. This process is made systematic by first nondimensionalising the equations, so that all the variables are dimensionless and of order one. This highlights the small slenderness parameter ε = h/L, where h is a typical thickness and L a typical length of the elastic body. A simplified system of equations is then obtained by carefully taking the limit ε → 0. Typically, the solution is sought as an asymptotic expansion in powers of the small parameter ε, and the techniques demonstrated here fall within the general theories of asymptotic expansions and perturbation methods; Hinch (1991) and Bender & Orszag (1978) provide very good general exposition of these methods. In fact we have already encountered an example of this approach when discussing the torsion of a thin-walled tube in §2.5 and an even simpler paradigm is the dimensionless model for antiplane strain of a thin sheet lying in 0 < y < ε, 0 < x < 1. Suppose the lateral boundaries of this ‘slender’ solid are traction-free. From (2.37,??) the displacement w thus 243

244

Asymptotic Analysis

satisfies ∇2 w = 0

with

∂w =0 ∂y

on y = 0 and y = ε,

(6.1)

and either w or ∂w/∂x is prescribed on x = 0, 1. Away from the ends of the sheet, we rescale y = εˆ y and expand w in an asymptotic expansion in which w ∼ w0 + o(1) as ε → 0. This gives, to lowest order in ε, ∂ 2 w0 =0 ∂ yˆ2

with

∂w0 =0 ∂ yˆ

on yˆ = 0, 1.

(6.2)

Hence w0 = w0 (x), but this function is not determined at this stage and we must proceed further in the asymptotic expansion by writing w ∼ w0 (x) + εw1 (x, yˆ) + · · · . Collecting all O(ε) terms, this gives ∂ 2 w1 =0 ∂ yˆ2

with

∂w1 =0 ∂ yˆ

on yˆ = 0, 1,

(6.3)

and, at order ε2 , ∂ 2 w2 d2 w0 = − ∂ yˆ2 dx2

with

∂w2 =0 ∂ yˆ

on yˆ = 0, 1.

The solution of this last problem, w2 = −d2 w0 /dx2 two boundary conditions at yˆ = 0, 1 if

yˆ2 2

(6.4)

can only satisfy the

d2 w0 = 0. (6.5) dx2 This is the solvability condition for w2 and, as is typical with asymptotic methods, it is this condition that gives rise to the ‘thin sheet’ model for w0 . However, to complete the model, we need to derive boundary conditions for (6.5) and to do this, we must subject the ends of the sheet to a mathematical magnifying glass by setting x = εˆ x, say. Let the displacement in this boundary layer near the edge be given by the function w = w(ˆ ˆ x, yˆ) This gives the semi-infinite strip model ∂2w ˆ ∂2w ˆ + = 0 with 2 2 ∂x ˆ ∂ yˆ

∂w ˆ =0 ∂ yˆ

on y = 0, 1

and w ˆ or ∂ w/∂ ˆ x ˆ prescribed on x ˆ = 0. By separation of the variables, w ˆ is determined as ∞ X w = a0 + a1 x ˆ+ an e−nπˆx cos nπ yˆ, (6.6) n=2

6.1 Introduction

245

and, when ∂w (0, yˆ) is prescribed, the ai are determined for i ≥ 1, with R 1 ∂w ∂x a1 = 0 ∂x (O, yˆ) dˆ y. Finally, we require that the functions w and w0 match, in the sense that, as x → 0, w0 joins smoothly with w when the latter is evaluated for large x. This means that the appropriate boundary condition for (6.5) as x → 0 0 is dw ˆ) is prescribed, the impossibility of dx = a1 . However, when w(0, y prescribing R 1two boundary conditions on (6.5) at zero requires that a1 = 0 and a0 = 0 w(), yˆ) dˆ y ; this makes w = a0 the boundary conditions for (6.5) as x → 0. Generalizations of this asymptotic argument form the basis of this chapter. Their systematic nature offers many advantages compared to the less rigorous derivations of Chapter 4. For example, the process in principle allows further corrections to be calculated, resulting in increasingly accurate models. Furthermore, it makes explicit the assumptions behind a simplified model, thus allowing us to estimate the error incurred in making an approximation and to determine when that approximation may become invalid. In this regard, we recall that the beam and plate theories developed in Chapter 4 are ineffective for resolving the details of the tractions applied to the edges. A boundary layer analysis in the neighbourhood of these edges, along the lines described above, allows us to explain how such tractions can effectively be replaced by an equivalent point force/moment system, without having to assume St Venant’s principle. Finally, since we start from the general continuum equation of elasticity, our asymptotic approach leads to explicit formulae for parameters such as bending stiffness in terms of the underlying elastic parameters, such as the Lamé constants. For simplicity, dimensionless versions of these underlying parameters will be held fixed as ε → 0. Quite different models would emerge were these parameters allowed to depend on ε because there is no reason why, say, the slender and incompressibility limits should commute. better put this last comment in conclusion. In §6.2 we rederive the linear plate equation. As it describes infinitesimal transverse displacements of a plate, we can obtain it starting from linear elasticity. In §6.3, we analyze a boundary layer near an edge of the plate and reestablish the correct boundary conditions for this model. Next, in §6.4, we rederive the improved von Kárm´ an plate equations, incorporating geometrically nonlinear terms. This requires us to start from the equations of nonlinear elasticity, although the asymptotic approach is otherwise analogous to that used in §6.2. In §6.5, we extend the analysis to describe large two-dimensional deflections of a plate and thus derive the Euler strut

246

Asymptotic Analysis

equation. Finally, in §§6.6–6.7, we illustrate the geometrical generalisations needed to obtain the equations governing infinitesimal displacements of rods and shells. 6.2 The linear plate equation 6.2.1 Nondimensionalisation and scaling We first illustrate our asymptotic methodology by modelling small displacements of a horizontal elastic plate subject to gravitational acceleration g. We assume that the displacements are small enough for linear elasticity to apply, so we can use the momentum equation, given in component form by (1.68), and the components of the Cauchy stress tensor are given by (1.66). If the plate has uniform thickness h and its centre-surface initially occupies the plane z = 0, then its free surfaces are at z = ±h/2, and we impose the stress-free boundary conditions τxz = τyz = τzz = 0

on z = ±h/2.

(6.7)

In linear elasticity we do not distinguish between Eulerian and Lagrangian coordinates, so the boundaries of the plate effectively remain fixed at z = ±h/2. The treatment of larger transverse displacements, comparable to h, say, will be differed to §6.4. Denoting by L a typical dimension in the x- and y-directions, the slenderness parameter is h , (6.8) L and we will now construct an approximate theory for the deformations of the plate based on the assumption that ε is small. To exploit this fact, we first nondimensionalise the equations and then take the asymptotic limit ε → 0. The spatial coordinates are scaled with their typical values as follows ε=

x = Lx0 ,

y = Ly 0 ,

z = hz 0 ,

(6.9)

so that the primed variables are dimensionless and of order 1. As always happens in asymptotic simplifications, choosing the right scalings for the dependent variables is a matter of trial-and-error guided by experience. Here, simple geometry implies that a transverse displacement of magnitude O(W ) is associated with in-plane displacements of O(εW ),† which motivates the † From (6.7), we know for instance that, on the free surface, ∂u/∂z = −∂w/∂x. Now, assuming that u and w undergo O(1) relative variations as x0 or z 0 increase by one, we have the order of magnitude estimates ∂u/∂z ∼ u/h and ∂w/∂x ∼ w/L and balancing the two yields (6.10).

6.2 The linear plate equation

247

nondimensionalisation u = εW u0 ,

v = εW v 0 ,

w = W w0 .

(6.10)

The appropriate nondimensionalisation for the stress components is slightly less clear. Our best strategy is to seek a scaling that balances as many terms as possible in the Navier equations (1.68), and we give ourselves the freedom to do this by writing ¡ 0 ¢ 0 0 (τxx , τxy , τyy ) = τ? τxx , τxy , τyy , (6.11a) ¡ 0 ¢ (τxz , τyz ) = ετ? τxz , τyz , (6.11b) 0 τzz = ε2 τ? τzz ,

(6.11c)

where τ? is to be determined. A suitable choice may then be obtained by balancing the in-plane stresses and strains in the constitutive relation (1.66), which leads to τ? =

ε2 EW , h

(=

hW E) L2

(6.12)

where E is Young’s modulus. Finally, by looking for a balance in the transverse Navier equation (1.68c), we infer a suitable time-scale, namely µ r ¶ µ 2r ¶ h ρ L ρ 0 t= t. (= t0 ) (6.13) ε2 E h E To summarise, the geometry of the plate has driven our nondimensionalisation (6.9) of the spatial variables. Our scalings (6.10) of the displacements amount to assumptions about how big the transverse and in-plane deflections are; thereafter, stress componenets are rescaled so as to balance as many terms as possible in the momentum equations. We will now derive a model that is consistent with these scalings. 6.2.2 Dimensionless equations Substituting (6.9) and (6.11) into (1.68) and neglecting terms of order ε2 we obtain the dimensionless momentum equations ∂τxx ∂τxy ∂τxz + + , ∂x ∂y ∂z ∂τyy ∂τyz ∂τxy + + , 0= ∂x ∂y ∂z ∂τyz ∂2w ∂τxz ∂τzz = + + − G, 2 ∂t ∂x ∂y ∂z 0=

(6.14a) (6.14b) (6.14c)

248

Asymptotic Analysis

where G=

ρgh2 ε4 EW

(6.15)

is a dimensionless constant measuring the importance of gravity. Here, and henceforth, we drop the primes to avoid cluttering the equations. The stressfree conditions (6.7) on the upper and lower surfaces simply become τxz = τyz = τzz = 0

on z = ±1/2.

(6.16)

It is also helpful to average these momentum equations by integrating (6.14) with respect to z and applying (6.16). This yields 0=

∂Txx ∂Txy + , ∂x ∂y

0=

∂Txy ∂Tyy + , ∂x ∂y

(6.17a)

∂2w ∂Nx ∂Ny + − G, = 2 ∂t ∂x ∂y

(6.17b)

where w denotes the average of w across the plate, and the net in-plane and transverse stresses are defined as in §4.6 by Z 1/2 Z 1/2 Tij = τij dz, Nj = τjz dz. (6.18) −1/2

−1/2

Similarly, multiplication through by z before integrating leads to the moment balance equations 0=

∂Myx ∂Myy + − Nx , ∂x ∂y

0=

∂Mxx ∂Mxy + + Ny , ∂x ∂y

(6.19)

where the bending moments are defined as in §4.6 by Z 1/2 Z 1/2 Z Myx = zτxx dz, Myy = −Mxx = zτxy dz, Mxy = − −1/2

−1/2

1/2

−1/2

zτyy dz.

(6.20) We may now use (6.2.2) to eliminate Nx and Ny from (6.17b) to give ∂ 2 Myx ∂ 2 Mxy ∂2w ∂2 = + (M − M ) − − G. yy xx ∂t2 ∂x2 ∂x∂y ∂y 2

(6.21)

Hence the deformation of the plate is governed by the averaged equations (6.17) and (6.21). We must now use the dimensionless constitutive equations

6.2 The linear plate equation

249

to relate the averaged stresses and moments to the displacements. These take the form ε2 τxx = τxy = ε2 τyy = ε2 τxz = ε2 τyz = ε4 τzz =

ε2 (1 − ν)ux + ε2 νvy + νwz , (1 + ν)(1 − 2ν) uy + vx , 2(1 + ν) ε2 νux + ε2 (1 − ν)vy + νwz , (1 + ν)(1 − 2ν) uz + wx , 2(1 + ν) vz + wy , 2(1 + ν) ε2 νux + ε2 νvy + (1 − ν)wz , (1 + ν)(1 − 2ν)

(6.22a) (6.22b) (6.22c) (6.22d) (6.22e) (6.22f)

where ν is Poisson’s ratio and, in this chapter only, we will occasionally use subscripts as shorthand for partial derivatives. When we take the limit ε → 0 in (6.22), we treat all the other parameters as being of order 1. In particular, we assume that ν is not too close to either −1 or 1/2, so that we preclude situations such as that described in §1.8. 6.2.3 Leading-order equations If we immediately set ε = 0 in (6.22), we get the same equation wz = 0 three times. It is therefore expedient to first eliminate wz between the equations to obtain τxx −

ux + νvy ε2 ν τzz = , 1−ν 1 − ν2

τyy −

νux + vy ε2 ν τzz = . 1−ν 1 − ν2

(6.23)

This allows us to simply let ε tend to zero in the above equations without having to write the dependent variables as explicit asymptotic expansions. As we just noted, wz is zero to leading order; hence, the transverse displacement is approximately uniform across the plate. Since w is independent of z, we may integrate (6.22d) and (6.22e) to obtain u(x, y, z, t) = u(x, y, t) − zwx (x, y, t),

(6.24a)

v(x, y, z, t) = v(x, y, t) − zwy (x, y, t),

(6.24b)

as ε → 0. The first terms on the right-hand sides give purely in-plane displacements, while the terms proportional to z correspond to bending of the plate.

250

Asymptotic Analysis

By substituting (6.24) back into (6.22) and integrating, we find that the averaged stresses and moments satisfy Txx =

ux + νv y , 1 − ν2

Myx = −

Txy =

uy + v x , 2(1 + ν)

νux + v y , 1 − ν2 (6.25) νwxx + wyy =− . 12(1 − ν 2 ) (6.26)

Tyy =

wxx + νwyy wxy , Myy = −Mxx = − , −Mxy 2 12(1 − ν ) 12(1 + ν)

We are reassured by the fact that the expressions (6.25) for the in-plane tensions agree with the exact biaxial straining solutions obtained in §2.2.4. Equally, the relations between the bending moments and the second derivatives of w reproduce those found in §4.6 by ad hoc comparison with exact plane stress solutions. Recalling that Txx , Txy and Tyy satisfy (6.17a), we see that the in-plane displacements satisfy a plane strain problem. As explained in §2.6, one approach to this problem is to introduce an Airy stress function A(x, y) such that Txx =

ux + νv y uy + v x νux + v y ∂2A ∂2A ∂2A = , T = = , T = − = , yy xy 2 2 2 ∂y 1−ν ∂x∂y 2(1 + ν) ∂x 1 − ν2 (6.27)

and elimination of u and v shows that A satisfies the biharmonic equation. Also, by substituting (6.26) into (6.21), we find that the transverse displacement w satisfies the linear plate equation wtt = −

∇4 w − G. 12(1 − ν 2 )

(6.28)

Notice that the equation (6.28) for w can be solved independently of u and v. In other words, transverse bending completely decouples from in-plane deformation of the plate. In summary, our model consists of the linear partial differential equation (6.28) for w and a decoupled biharmonic equation for A. We have derived these equations systematically, without invoking any other assumptions than the scalings in §6.2.1. Our derivation is constructive and yields explicit formulae for the other quantities of interest, that is the displacements and stress components. Redimensionalisation of (6.28) leads to ςwtt = −D∇4 w − ςhg,

(6.29)

6.3 Boundary conditions and St Venant’s principle

z

251

y

x fx fy

fz

Fig. 6.1. Schematic of the edge of a plate subject to tractions.

where ς = ρh is the surface density of the plate and D=

Eh3 , 12(1 − ν 2 )

(6.30)

thus explicitly identifying the bending stiffness of the plate in terms of the elastic constants and the geometrical parameter h. Equation (6.29) deos not contain the “membrane” term T ∇2 w obtained from momentum balance argument in §4.6. Since the tension T is proportional to the plane strain components ux , vy , etc. this product of T with ∇2 w is nonlinear in the displacements. As hinted at in §4.6, it is impossible, starting from the linear Navier equations (1.68) and constitutive relations (1.66), to obtain any such nonlinear terms. We will show below in §6.4 how they arise from a geometrically nonlinear model.

6.3 Boundary conditions and St Venant’s principle We are now poised to systematically resolve the difficulties encountered in §4.6.2 concerning the boundary conditions for plate models. It will come as no surprise that we must proceed by constructing a boundary layer region near the edge of the p;ate. In this boundary layer we must address a model that is fully three-dimensional, but fortunately we will not need to solve it explicitly: the solvability conditions for the inner problem will provide us with the sought-after boundary conditions.

252

Asymptotic Analysis

6.3.1 Boundary layer scalings [NOTATION: by consistency with §2.6.7, fx , fy , fz → σx , σy , σz ?] Typical boundary conditions for the plate equation were discussed in §4.6, where we encountered a conceptual difficulty with the number of edge conditions when stresses, rather than displacements, are specified. Here we will show how careful asymptotic analysis of the edge of a plate resolves the problem. For simplicity, we consider only a straight edge x = 0 which is subject to a given traction, so we impose the boundary conditions τxx = fx ,

τxy = fy ,

τxz = fz

at x = 0,

(6.31)

on the dimensionless stress components [by the way, this means that fz is one order of magnitude smaller than the other applied tractions before scaling]; see Figure 6.1 for the sign convention. The components fx , fy and fz may in general be any functions of y and z. In the plate solutions constructed above, the in-plane stress components are linear functions of z of the form τxx = Txx + 12zMyx ,

τxy = Txy − 12zMxx ,

τyy = Tyy − 12zMxy , (6.32a)

and the transverse shear stresses are found from (6.14) to be ¶ µ 1 − z2 , τxz = 6Nx 4 µ ¶ 1 2 τyz = 6Ny −z , 4 µ 2 ¶ ¶ µ ∂ w 1 2 τzz = −2 −z . +G z ∂t2 4

(6.32b) (6.32c) (6.32d)

Hence, unless the applied tractions have exactly the same dependence on z, it will be impossible to apply the boundary conditions (6.31) directly. Instead, there must be a boundary layer near the edge where the stress components adjust from their imposed boundary values to the limiting values of the functions (6.32) in the interior of the plate. Even if the boundary tractions do by chance have a z-dependence consistent with (6.32), there is still a boundary layer to be negotiated before the conditions (6.31) can be imposed. To illustrate the difficulty, consider a traction-free edge, so that fx = fy = fz = 0. By naively applying (6.31) to (6.32), we apparently obtain five conditions, namely Txx = Txy = Myx = Mxx = Nx = 0.

(6.33)


253

However, only four boundary conditions can be imposed on our model for the plate, so at least one of these must be neglected. To address this issue, we now focus our attention on a thin boundary layer near the edge, where the stresses no longer have the simple polynomial dependence on z exhibited by (6.32). To do this we define a new variable x ˆ=

x ε

(6.34)

so that x ˆ is order 1 within a boundary layer whose thickness is of the order of magnitude of the thickness of the plate. To get a nontrivial balance in the Navier equations (6.14), we also find it necessary to rescale some of the stress components as follows: τxz =

τˆxz , ε

τyz =

τˆyz , ε

τzz =

τˆzz . ε2

(6.35)

In view of (6.11), all stress components are now of the same order of magnitude. In particular, the transverse stress is two orders of magnitude larger near the edge of the plate than it is in the interior. This gives us a strong clue that the boundary layer may have an unexpected influence on the rest of the plate.

6.3.2 Equations and boundary conditions After we apply the rescaling (6.35) and neglect terms of order ε2 , the Navier equations become ∂ τˆxy ∂ τˆxx ∂ τˆxz +ε + , ∂x ˆ ∂y ∂z ∂ τˆxy ∂ τˆyy ∂ τˆyz 0= +ε + , ∂x ˆ ∂y ∂z ∂ τˆyz ∂ τˆzz ∂ τˆxz +ε + . 0= ∂x ˆ ∂y ∂z 0=

(6.36a) (6.36b) (6.36c)

Hence, as ε → 0, they effectively reduce to a set of partial differential equations with only x ˆ and z as dependent variables, y being simply a parameter. The boundary conditions on the edge and the upper and lower surfaces read τˆxz = τˆyz = τˆzz = 0 τˆxx = fx , τˆxy = fy , τˆxz = εfz

on z = ±1/2,

(6.37a)

on x = 0;

(6.37b)

here we have used hats to distinguish dependent variables (even unscaled ones) evaluated in the boundary layer from those in the interior. The

254

Asymptotic Analysis

rescaled constitutive relations (6.22) take the form

∂ˆ v ∂w ˆ ∂u ˆ + ε2 ν +ν , ∂x ˆ ∂y ∂z ∂u ˆ ∂ˆ v 2ε(1 + ν)ˆ τxy = ε + , ∂y ∂ x ˆ ∂u ˆ ∂ˆ v ∂w ˆ ε2 (1 + ν)(1 − 2ν)ˆ τyy = εν + ε2 (1 − ν) +ν , ∂x ˆ ∂y ∂z ∂u ˆ ∂w ˆ 2ε2 (1 + ν)ˆ τxz = ε + , ∂z ∂x ˆ ∂ˆ v ∂w ˆ 2(1 + ν)εˆ τyz = + , ∂z ∂y ∂ˆ v ∂w ˆ ∂u ˆ + ε2 ν + (1 − ν) , ε2 (1 + ν)(1 − 2ν)ˆ τzz = εν ∂x ˆ ∂y ∂z

ε2 (1 + ν)(1 − 2ν)ˆ τxx = ε(1 − ν)

(6.38a) (6.38b) (6.38c) (6.38d) (6.38e) (6.38f)

and we also note the following useful consistency condition that results from eliminating u ˆ and w ˆ between them:

¢ ¢ ∂2 ¡ ∂ 2 τˆxz ∂2 ¡ + − 2 (1 − ν)ˆ τ − ν τ ˆ (1 − ν)ˆ τ − ν τ ˆ zz xx xx zz ∂x ˆ2 ∂z 2 ∂x∂z 2 ˆ ν ∂ ∇ vˆ = ; (6.39) 1 + ν ∂y

this condition can either be verified by direct substitution or it can be obtained by generalising the argument leading to (2.178) to account for the fact that vˆ is nonzero. We have to ensure that the inner solution that we are about to compute in the boundary layer agrees with the outer solution in the interior of the plate. As in the anti-plane strain example of §6.1, this is done by asymptotically matching the two solutions. The basic idea is that the limit of the inner solution as x ˆ → ∞ must be the same as that of the outer solution as x → 0.


255

In other words, we impose that µ ¶ ∂w ∂u ∂2w u ˆ ∼ u(0, y) − z (0, y) + εˆ x (0, y) − z 2 (0, y) + · · · , ∂x ∂x ∂x µ ¶ ∂w ∂v ∂2w vˆ ∼ v(0, y) − z (0, y) + εˆ x (0, y) − z (0, y) + · · · , ∂y ∂x ∂x∂y ∂w w ˆ ∼ w(0, y) + εˆ x (0, y) + · · · , ∂x τˆxx ∼ Txx (0, y) + 12zMyx (0, y) µ ¶ ∂Myx ∂Txx + εˆ x (0, y) + 12z (0, y) + · · · , ∂x ∂x τˆxy ∼ Txy (0, y) − 12zMxx (0, y) + · · · , µ ¶ 1 2 τˆxz ∼ 6εNx (0, y) − z + ··· , 4 µ ¶ 1 2 τˆyz ∼ 6εNy (0, y) − z + ··· , 4 ¡ 2¢ τˆzz ∼ O ε

(6.40a) (6.40b) (6.40c)

(6.40d) (6.40e) (6.40f) (6.40g) (6.40h)

as x ˆ → ∞.

6.3.3 Solvability conditions As we pointed out in §1.10, if the stresses are prescribed on the boundaries of the boundary layer, solvability conditions of the form (1.62,1.63+++1.60,1.61+++) have to satisfied in order to ensure well-posedness. The corresponding conditions are found here by integrating (6.36) over the boundary layer and applying the divergence theorem: Z 0= −1/2

Z

(−ˆ τxx (0, y, z) + τˆxx (∞, y, z)) dz + ε

−1/2

(−ˆ τxy (0, y, z) + τˆxy (∞, y, z)) dz + ε

0= −1/2

(−ˆ τxz (0, y, z) + τˆxz (∞, y, z)) dz + ε

Z

1/2

1/2

−1/2

∞

∂ τˆxy dˆ xdz, ∂y (6.41a)

∞

∂ τˆyy dˆ xdz, ∂y (6.41b)

∞

∂ τˆyz dˆ xdz. ∂y (6.41c)

0

Z

−1/2

Z

1/2

1/2

−1/2

Z

1/2

0= Z

Z

1/2

0

Z 0

256

Asymptotic Analysis

Furthermore, making the cross product with (ˆ x, 0, z) first and then integrating, we get ¶ Z 1/2 Z 1/2 Z ∞ µ ∂ τˆxy ∂ τˆyz 0= z (ˆ τxx (0, y, z) − τˆxx (∞, y, z)) dz + ε z −x dˆ xdz. ∂y ∂y −1/2 −1/2 0 (6.42) In view of (6.40) and (6.37), we can neglect the O(ε) contributions in (6.41a), (6.41b), and (6.42), and immediately obtain Z 1/2 Txx (0, y) = fx (y, z) dz, (6.43a) −1/2 1/2

Z Txy (0, y) = Mxy (0, y) =

−1/2 Z 1/2 −1/2

fy (y, z) dz,

(6.43b)

zfx (y, z) dz.

(6.43c)

This shows that the net in-plane traction and their moment along the edges are balanced by the corresponding stresses on the other side of the boundary layer, as we might have physically expected. We thus already have three of the boundary conditions needed for the interior plate model. On the other hand, all we can say about (6.41c) so far is that Z 1/2 Z 1/2 Z ∞ ∂ τˆyz dˆ xdz, Nx (0, y) − fz (y, z) dz = − ∂y −1/2 −1/2 0 and this motivates us to pursue the boundary layer analysis in more details.

6.3.4 Asymptotic expansions Unlike the situation encountered in §6.2.3, there is no way to avoid using asymptotic expansions here. We write all the dependent variables as expansions in powers of ε, for example u ˆ=u ˆ(0) + εˆ u(1) + ε2 u ˆ(2) + · · · .

(6.44)

From the constitutive relations (6.38), the leading-order displacements are all independent of x ˆ, and the matching conditions (6.40) then give u ˆ(0) = u ¯(0, y) − z

∂w ∂w (0, y), vˆ(0) = v¯(0, y) − z (0, y), w ˆ (0) = w(0, y). ∂x ∂y (6.45)


257

We further deduce from (6.40f) and (6.40e) that ∂w ˆ (1) = 0, ∂z from which it follows that

∂u ˆ(0) ∂ w ˆ (1) + = 0, ∂z ∂x ˆ

(6.46)

∂w (0, y). (6.47) ∂x We now proceed sequentially, first considering the implications of (6.36b), and then those of (6.36a) and (6.36c). First, from the second Navier equation (6.36b), we obtain the antiplane strain problem w ˆ (1) = x ˆ

(0)

(0)

∂ τˆxy ∂ τˆyz + = 0, ∂x ∂z

(6.48)

and, from (6.38), (0) 2(1 + ν)ˆ τxy =

v (1) ∂u ˆ(0) ∂ˆ + , ∂y ∂x ˆ

(0) 2(1 + ν)ˆ τyz =

∂ˆ v (1) ∂ w ˆ (1) + . ∂z ∂y

(6.49)

Hence vˆ(1) satisfies the two-dimensional Laplace equation 2 (1) 2 (1) ˆ 2 vˆ(1) = ∂ vˆ + ∂ vˆ = 0, ∇ ∂x ˆ2 ∂z 2 subject to the boundary conditions

∂u ˆ(0) ∂ˆ v (1) = 2(1 + ν)fy − ∂x ∂y (1) (1) ∂w ˆ ∂ˆ v =− ∂z ∂y

(6.50)

at x ˆ = 0,

(6.51a)

1 at z = ± . 2

(6.51b)

The matching conditions (6.40b) leads to ∂ˆ v (1) ∂2w ∼ −ˆ x (0, y), ∂z ∂x∂y

(6.52)

as x ˆ → ∞, while combining (6.40a) and (6.40e) yields ¡ ¢ ∂u ∂2w ∂ˆ v (1) ∼ − (0, y) + z (0, y) + 2(1 + ν) Txy (0, y) − 12zMxx (0, y) ∂x ˆ ∂y ∂x∂y (6.53) Note that, by cross-differentiation, the two conditions are consistent only if ∂2w (0, y) = 12(1 + ν)Mxx (0, y), ∂x∂y

(6.54)

258

Asymptotic Analysis

and this reassuringly reproduces the outer constitutive relation (6.26). After using this relation, we can write vˆ(1) in the form µ ¶ ∂u (1) vˆ = 2(1 + ν)Txy (0, y) − (0, y) x ˆ ∂y − 12(1 + ν)Mxx (0, y)ˆ xz + 2(1 + ν)φ,

(6.55)

where φ satisfies ˆ 2 φ = 0, ∇ ∂φ =0 ∂z ∂φ = 12Mxx (0, y)z + fy (y, z) − Txy (0, y), ∂x ˆ φ → 0,

(6.56a) 1 z=± , 2

(6.56b)

x ˆ = 0,

(6.56c)

x ˆ → ∞.

(6.56d)

The solvability of this Neuman problem for φ is ensured since Z 1/2 ZZ I ∂φ 2 ˆ ds = Txy (0, y) − fy (y, z) dz, 0= ∇ φ dˆ xdz = ∂n −1/2 and we have already established that the right hand side above vanishes. Hence we can solve for φ by separating the variables: φ=

∞ X

¡ ¢ an (y) cos nπ(z + 1/2) e−nπˆx ,

(6.57)

n=1

where 2 an (y) = − nπ

Z

1/2

−1/2

¡ ¢ Fy (y, z) cos nπ(z + 1/2) dz,

Fy (y, z) = 12Mxx (0, y)z + fy (y, z).

(6.58a) (6.58b)

The stress components are then given in terms of φ as (0) τˆxy = Txy (0, y) − 12Mxx (0, y)z + (0)

∂φ , ∂x ˆ

(0) τˆyz =

∂φ , ∂z

(6.59)

giving the transition of τˆxy from the prescribed function fy on x ˆ = 0 to the expected linear z-dependence in the outer region. St Venant’s principle, namely that the details of the traction applied to the edge decay to zero outside the narrow boundary layer, is thus recovered. Only the net force and moment exerted by the edge traction are transmitted to the interior of the plate.


259

We are now in a position to evaluate (6.41c): Z 1/2 Z ∞ ∂ Nx (0, y) − fz dz = − φ(ˆ x, y, 1/2) − φ(ˆ x, y, −1/2) dˆ x. (6.60) ∂y 0 −1/2 Substituting for φ from (6.58), we get Z

∞

φ(ˆ x, y, 1/2) − φ(ˆ x, y, −1/2) dˆ x=− 0

Z

Ã

1/2

=2 −1/2

Fy (y, z)

∞ X 1 − (−1)n

nπ

n=1 ∞ X 1 − (−1)n

n=1

n2 π 2

an (y)

! ¡ ¢ cos nπ(z + 1/2) dz Z

1/2

=− −1/2

Fy (y, z)z dz,

and (6.60) therefore becomes Ã ! Z 1/2 Z 1/2 ∂ Mxx (0, y) + fy z dz . Nx (0, y) − fz dz = ∂y −1/2 −1/2

(6.61)

(6.62)

We easily recognize in the left-hand side of (6.62) the difference between the vertical shear forces on either side of the boundary layer. The righthand side represents the rate of change of the net bending moment about the x-axis that is produced by the outer moment and the edge traction. All that matters for the plate model of §6.2 is that both contributions balance; this is akin to the angular momentum balance (4.10+++4.10+++) that arise in the beam model. Both sides of (6.62) are statically equivalent in the interior of the plate. Neither the transverse stress nor the twisting moment is conserved across the boundary layer, but instead (6.62) represents a playoff between the two effects. To summarise, we have obtained the four boundary conditions (6.43), and (6.62) that apply at a straight edge of a plate when specified tractions are imposed. At a stress-free edge, for example, these take the form Txx = Txy = Myx = Nx −

∂Mxx =0 ∂y

(6.63)

or, in terms of the displacements, ∂u ∂v ∂u ∂v ∂2w ∂2w ∂3w ∂3w +ν = + = + ν = + (2 − ν) = 0. ∂x ∂y ∂y ∂x ∂x2 ∂y 2 ∂x3 ∂x∂y 2 (6.64) This concludes our systematic derivation of linear plate theory. It has revealed how much more complicated is the derivation of the boundary

260

Asymptotic Analysis

conditions compared to the derivation of the field equations. Indeed the boundary layer near the plate edge are crucial in determining the global response of the plate and the asymptotic approach leading to (6.35) reveals that the normal stresses there can exceed the shear stresses by two orders of magnitude

6.4 The von K´ arm´ an plate equations and weakly curved shells 6.4.1 Background In §6.2, we modelled the deformation of a plate using linear (infinitesimal) elasticity. In other words, we assumed that the strains are sufficiently small for all nonlinear terms to be neglected a priori . We then used another assumption, namely that the plate is geometrically thin, to reduce the threedimensional Navier equations to the linear plate equation (6.28) and a decoupled plane strain problem (6.27) for the horizontal displacements. In practice, the tractions and displacements applied at the edge may well cause coupling between in-plane stretching and transverse bending. For example, putting a membrane under tension through in-plane stretching certainly affects its response to transverse oscillations. Equally, we will see that bending a plate in two orthogonal directions induces net averaged in-plane stresses. To generalise the theory of §6.2, we begin by realising that a plate, with thickness h and longitudinal dimensions of order L, undergoing a transverse displacement of order W , is characterised by two small parameters, namely the slenderness parameter ε = h/L and δ = W/L, which is the size of a typical transverse strain. In §6.2, by using linear elasticity we were implicitly neglecting all terms of order δ 2 , although we had to consider corrections of order ε2 during our derivation. This suggests that there is a régime in which the small parameters δ and ε are roughly equal, so that the transverse displacement is roughly equal to the plate thickness, and the strains, although small, are not negligible. In the language of perturbation methods, this is a distinguished limit, in which we let δ and ε tend to zero simultaneously, rather than setting δ to zero first and only subsequently using the fact that ε is small. By assuming that the strains are small, but not infinitesimal, this approach leads to a model that is intermediate between the linear plate equation and the formidably complicated fully three-dimensional nonlinear elasticity. Again using the language of perturbation methods, the choice of an appropriate coupling between two small parameters leads to a weakly nonlinear model. We have already encountered an example of such an approach in the Euler strut

6.4 The von K´ arm´ an plate equations and weakly curved shells

261

analysis of §4.9.3, where the small parameters were the excess applied force above the buckling force and the amplitude of the induced perturbation. By following this approach, we will derive an improved plate theory that describes explicitly the coupling between bending and stretching. The crucial coupling terms are nonlinear and hence, as demonstrated in §6.2, could not possibly have been obtained from linear elasticity. The analysis carried out below therefore serves as an illustration of the need for geometrically nonlinear elasticity even in situations where the strains are apparently small. Although the details are more complicated, the asymptotic techniques involved are analogous to those used in §6.2, and so we omit many of the details.

6.4.2 Scalings We consider a uniform plate of thickness h whose centre-surface lies in the plane Z = 0, where now (X, Y, Z) are Lagrangian coordinates which must be distinguished from Eulerian coordinates (x, y, z). The coordinates are scaled with their typical values as follows: X = LX 0 ,

Y = LY 0 ,

Z = hZ 0 .

(6.65)

As suggested above, we will suppose that the transverse displacement is the same order as the plate thickness h; in other words, we set W = h. The analysis to follow is simplified by anticipating the fact (demonstrated in §6.2) that the leading-order transverse displacement is uniform across the plate. We therefore nondimensionalise the displacements as follows u = εhu0

¡ ¢ w = h w0 (x, y, t) + ε2 w ˜ .

v = εhv 0

(6.66)

We follow (6.11) in nondimensionalising the first Piola–Kirchhoff stress tensor as ¡ 0 ¢ 0 0 0 (T11 , T12 , T21 , T22 ) = ε2 E T11 , T12 , T21 , T22 , ¡ 0 ¢ 3 0 0 0 (T13 , T23 , T31 , T32 ) = ε E T13 , T23 , T31 , T32 , 4

T33 = ε

0 ET33 ,

(6.67a) (6.67b) (6.67c)

and the dimensionless time is again defined by (6.13). Henceforth we drop the primes to avoid clutter.

262

Asymptotic Analysis

6.4.3 Leading-order equations After nondimensionalising and neglecting terms of order ε2 , we reduce the the nonlinear momentum equation (5.27) to ∂T11 ∂T12 ∂T13 + + , ∂X ∂Y ∂Z ∂T21 ∂T22 ∂T23 0= + + , ∂X ∂Y ∂Z ∂T31 ∂T32 ∂T33 ∂2w = + + − G, ∂t2 ∂X ∂Y ∂Z 0=

(6.68a) (6.68b) (6.68c)

where now ρ0 gh . (6.69) ε4 E The dimensionless boundary conditions on the stress-free surfaces of the plate are G=

T13 = T23 = T33 = 0

on Z = ±1/2.

(6.70)

As in §6.2, it is helpful to integrate (6.68) with respect to Z and apply (6.70) to obtain the averaged stress and moment equations 0=

∂T 11 ∂T 12 + , ∂X ∂Y

0=

∂T 21 ∂T 22 + , ∂X ∂Y

∂2w ∂T 31 ∂T 32 + − G, = 2 ∂t ∂X ∂Y ∂M21 ∂M22 ∂M11 ∂M12 + − T 13 , 0= + + T 23 , ∂X ∂Y ∂X ∂Y where the averaged stresses and bending moments are defined by Z 1/2 T ij = Tij dZ, 0=

(6.71)

(6.72) (6.73)

(6.74)

−1/2

¶ Z 1/2 µ ¶ µ M11 M12 −T21 −T22 = Z dZ. M21 M22 T11 T12 −1/2

(6.75)

In terms of dimensionless variables, the deformation gradient tensor is given by   ε2 uX ε2 uY εuZ F =I + (6.76) ε2 vX ε 2 vY εvZ  . 3 3 2 εwX + ε w ˜X εwY + ε w ˜Y ε w ˜Z


263

Recall from §5.2.2 that the first Piola–Kirchhoff stress tensor is not symmetric, but satisfies T F T = F T T.

(6.77)

This provides three independent scalar equations which, using (6.67) and (6.76), may be reduced to ¡ ¢ T21 = T12 + O ε2 , (6.78a) ¡ 2¢ T31 − T13 = wX T11 + wY T12 + O ε , (6.78b) ¡ 2¢ T32 − T23 = wX T21 + wY T22 + O ε . (6.78c) The in-plane averaged stress tensor is therefore symmetric to leading order, and, as in §6.2, we can deduce from (6.71) the existence of an Airy stress function A(X, Y ) such that T 11 =

∂2A , ∂Y 2

T 21 = T 12 = −

∂2A , ∂X∂Y

T 22 =

∂2A ∂X 2

(6.79)

up to order ε2 . We can also use (6.78b), (6.78c) and (6.73) to eliminate T 31 and T 32 from (6.72) and hence obtain the generalisation of (6.21) as ∂2w ∂ 2 M11 ∂ 2 M12 ∂ 2 M21 ∂ 2 M22 − − + G = + ∂t2 ∂X 2 ∂X∂Y ∂X∂Y ∂Y 2 2 2 2 ∂2w ∂2A ∂ w∂ A ∂ w ∂2A + + − 2 . (6.80) ∂Y 2 ∂X 2 ∂X∂Y ∂X∂Y ∂X 2 ∂Y 2 To close the problem, we must now impose a constitutive relation. We note that the strain tensor, expanded in powers of ε, takes the form   0 0 uZ + wX ¢ ε 1¡ T E= F F −I =  0 0 vZ + wY  2 2 uZ + wX vZ + wY 0   2 2uX + wX uY + vX + wX wY 0 2 ¡ ¢ ε  +O ε3 . + uY + vX + wX wY 2vY + wY2 0 2 0 0 2w ˜Z + u2Z + vZ2 (6.81) The smallness of ε justifies assuming that second Piola–Kirchhoff stress tensor S is linear† and, arguing as in §5.3.6, we write, in dimensional variables, S = λTr (E)I + 2µE.

(6.82)

† Had we used a nonlinear relation of the form (5.56), the smallness of ε would have led to a similar theory in which some of the coefficients in the lowest-order model were modified as in the example of §5.2.4; the correction to the stress–strain relation would be at most of order O(ε2 .

264

Asymptotic Analysis

In dimensionless form of (6.82), this is   S11 S12 εS13 1 ν ε2  S12 S22 εS23  = Tr (E)I + E. (1 + ν)(1 − 2ν) 1 + ν εS13 εS23 ε2 S33

(6.83)

On the other hand, by definition, T = F S and considering first just the lowest-order expressions for T13 and T23 , ε2 T13 =

wX + uZ + O(ε2 ), 2(1 + ν)

ε2 T23 =

wY + vZ + O(ε2 ), 2(1 + ν)

(6.84)

we deduce, as in the linear theory, that u = u(X, Y, t) − ZwX ,

v = v(X, Y, t) − ZwY .

(6.85)

Making use of these expressions for u and v, we then find to lowest order for T33 : ε2 T33 =

2 + w 2 + 2ν(u + v ) + 2(1 − ν)w ¡ ¢ wX ˜Z X Y Y + O ε2 2(1 + ν)(1 − 2ν)

(6.86)

and, hence, w ˜Z = −

2 + w 2 + 2ν(u + v ) wX X Y Y . 2(1 − ν)

(6.87)

This then allows us to evaluate the remaining stress components as 2 + ν(2v + w 2 ) 2uX + wX Y Y , 2(1 − ν 2 ) uY + vX + wX wY = , 2(1 + ν) 2 ) + 2v + w 2 ν(2uX + wX Y Y = , 2 2(1 − ν )

T11 = T21 = T12 T22

(6.88a) (6.88b) (6.88c)

up to order ε2 . We now have all the ingredients we need to construct the von Kárm´ an equations. By averaging (6.88), we obtain 2 + ν(2v + w 2 ) 2uX + wX ∂2A Y Y = , 2 2 ∂Y 2(1 − ν ) ∂2A uY + v X + wX wY T 12 = − = , ∂X∂Y 2(1 + ν) 2 ) + 2v + w 2 ν(2uX + wX ∂2A Y Y = , T 22 = ∂X 2 2(1 − ν 2 )

T 11 =

(6.89a) (6.89b) (6.89c)


265

and the simultaneous solution of (6.89a) and (6.89c) reveals that uX +

2 wX ∂2A ∂2A = − ν , 2 ∂Y 2 ∂X 2

vY +

wY2 ∂2A ∂2A = − ν . 2 ∂X 2 ∂Y 2

(6.90)

Thus u and v may be eliminated by differentiating (6.89b) with respect to X and Y and using (6.90), giving µ 2 ¶2 ∂ w ∂2w ∂2w 4 − ∇ A+ = 0. (6.91) 2 2 ∂X ∂Y ∂X∂Y Expressions for the bending moments are found by multiplying (6.88) with respect to Z and then integrating: µ 2 ¶ ∂ w ∂2w 1 M21 = − +ν , (6.92a) 12(1 − ν 2 ) ∂X 2 ∂Y 2 ∂2w 1 , (6.92b) M11 = −M22 = 12(1 + ν) ∂X∂Y µ 2 ¶ 1 ∂ w ∂2w M12 = ν + . (6.92c) 12(1 − ν 2 ) ∂X 2 ∂Y 2 Thus (6.80) reduces to ∂2w ∂2A ∂2w ∂2A ∂2w ∇4 w ∂2w ∂2A = + + G + − 2 . (6.93) ∂t2 12(1 − ν 2 ) ∂Y 2 ∂X 2 ∂X∂Y ∂X∂Y ∂X 2 ∂Y 2 With a suitable change of notation (6.91) and (6.93), are the von Kárm´ an equations (4.81,4.82) for transverse motions of a thin elastic plate. Concerning boundary conditions, it is unfortunately beyond the scope of this book to carry out a boundary layer analysis of the type presented in §?? or §??, even though such an analysis is needed too validate the assertions made in §4.7, or even before (4.70). However, in many practical situations, such an analysis would be the same as in the linear theory. 6.4.4 Equations for a weakly curved shell A shell is a thin structure which has appreciable curvature in at least one direction. Shells are well known to be stronger than plates, in that their displacement under small transverse loads is usually one order of magnitude less than would be the case for a plate or a rod. Everyday examples are corrugated (WHY CORRUGATED?) cylinders, which strongly resist bending of their generators, or ping-pong balls which resist deformation in any direction. Let us start by generalising the von Kárm´ an equations to plates that are precast so that they are initially weakly curved with an out-of-plane

266

Asymptotic Analysis

displacement comparable to their thickness; fully-fledged curvature will be considered in §6.7. Presently, we assume that the initial elevation of the plate is comparable to it thickness. Hence, in dimensionless variables, the upper and lower surfaces of the plate are given by 1 Z = H(X, Y ) ± , (6.94) 2 so that H(X, Y ) represents the initial centre-surface. We could easily let the plate thickness depend on X and Y also, but we discard this possibility here; our main aim is to explore how curved plates differ from flat ones. The analysis of §6.4 can be followed almost exactly; the Navier equations and constitutive relations are unchanged, but the stress-free boundary conditions now read ∂H ∂H T13 = T11 + T12 , (6.95a) ∂X ∂Y ∂H ∂H T21 + T22 , (6.95b) T23 = ∂X ∂Y ∂H ∂H T33 = T31 + T32 , (6.95c) ∂X ∂Y on Z = H ± 1/2. We leave the details of the derivation to Exercise 6.3 and just record here that the governing equations (6.91) and (6.93) are modified to µ 2 ¶2 ∂2w ∂2w ∂2H ∂2w ∂2H ∂2w ∂ w ∂2H ∂2w 4 −∇ A = + + − − 2 ∂X 2 ∂Y 2 ∂X∂Y ∂Y 2 ∂X 2 ∂X∂Y ∂X∂Y ∂X 2 ∂Y 2 (6.96) and ∇4 w ∂2w + G + ∂t2 12(1 − ν 2 ) ∂ 2 (H + w) ∂ 2 A ∂ 2 (H + w) ∂ 2 A ∂ 2 (H + w) ∂ 2 A = + − 2 , (6.97) ∂Y 2 ∂X 2 ∂X∂Y ∂X∂Y ∂X 2 ∂Y 2 in agreement with §4.8. Note that even in the case where quadratic terms in w and A are negligible, (6.96) and (6.97) are still coupled through H. This shows how curvature couples in-plane and transverse deformation. The right-hand side of (6.97) just represents the in-plane stress, resolved vertically relative to the displaced centre-surface. Note also that the righthand side of (6.96) is ( µ 2 ¶2 ) ( 2 µ 2 ¶2 ) ∂ H ∂2H ∂ (H + w) ∂ H ∂ 2 (H + w) ∂ 2 (H + w) − , − − ∂X 2 ∂Y 2 ∂X∂Y ∂X 2 ∂Y 2 ∂X∂Y

6.5 The Euler–Bernoulli plate equations

267

which is the change in the Gaussian curvature compared to the initial shape of the plate. Hence the only deformations that do not cause in-plane stress are those that preserve the Gaussian curvature. These deformations preserve area and are known as isometries and one example is the aforementioned pure bending of a planar plate.

6.5 The Euler–Bernoulli plate equations 6.5.1 Dimensionless equations Our next task is to provide an asymptotic derivation of the Euler strut model of §4.9. We begin by considering a thin elastic plate whose thickness h is much smaller than its length L. In §6.4, we analysed transverse displacements of order h and found that these induce small, although not infinitesimal, strains, and this allowed us to simplify the equations of nonlinear elasticity. Here we make the simplifying assumption that the deformation is purely two-dimensional. This allows the plate to undergo large deflections, of order L, while suffering only small strains. As usual, a two-dimensional plate is described using the Lagrangian coordinates (X, Z), where −h/2 < Z < h/2, 0 < X < L, the corresponding Eulerian coordinates (x, z) point initially at ¡ being such that a material ¢ (X, Z) occupies the position x(X, Z, t), z(X, Z, t) at any subsequent time t. We nondimensionalise X and Z with their typical values, while allowing for large displacements as follows: X = LX 0 ,

Z = hZ 0 ,

x = Lx0 ,

z = Lz 0 .

(6.98)

In contrast to (6.67), the first Piola–Kirchhoff stress components are nondimensionalised using ¡ 0 ¢ ¡ 0 ¢ 0 0 (T11 , T31 ) = εE T11 , T31 , (T13 , T33 ) = ε2 E T13 , T33 , (6.99) and the time-scale is chosen to make t0 of O (1), where r h ρ0 0 t= 2 t. ε E

(6.100)

The two-dimensional momentum equations become ∂2x ∂T11 ∂T13 = + , 2 ∂t ∂X ∂Z ∂2z ∂T31 ∂T33 ε 2 = + − εG, ∂t ∂X ∂Z

ε

(6.101a) (6.101b)

268

Asymptotic Analysis

Z

z

X θ x

Fig. 6.2. Definition sketch of the geometry of a deformed two-dimensional plate.

where the relevant dimensionless gravity parameter is now G=

ρ0 gh = εG. ε3 E

(6.102)

We can relate the stress components in (6.101) using the symmetry property (6.77) which, in two dimensions, reduces to zX T11 + zZ T13 = xX T31 + xZ T33 .

(6.103)

Using the boundary conditions T13 = T33 = 0

on Z = ±1/2,

(6.104)

we may again obtain integrated versions of (6.101), namely ∂2x ∂T 11 , = 2 ∂t ∂X ∂2z ∂T 31 − εG, ε 2 = ∂t ∂X

ε

(6.105a) (6.105b)

where overbars denote transverse averages as in §6.4. A balance of moments is obtained by subtracting x×(6.101b) from z×(6.101a), that is by taking the cross product between x and the momentum equation, before integrating with respect to Z. Using the symmetry condition (6.103) and again applying the boundary conditions (6.104), we find ε

∂ ∂t

Z

1/2

−1/2

(zxt − xzt ) dZ =

∂ ∂X

Z

1/2

−1/2

(zT11 − xT31 ) dZ + εGx,

which represents net conservation of angular momentum.

(6.106)


269

6.5.2 Asymptotic structure of the solution Now, to impose a constitutive relation, we must calculate the deformation and strain tensors. As in §6.4, this is simplified by anticipating the structure of the solution and then justifying our assumptions a posteriori. Here we suppose that the displacement is, to leading order, uniform across the plate, so that ¡ ¢ x(X, Z, t) = x(0) (X, t) + εx(1) (X, Z, t) + +ε2 x(2) (X, Z, t) + O ε3 , (6.107a) ¡ ¢ z(X, Z, t) = z (0) (X, t) + εz (1) (X, Z, t) + +ε2 z (2) (X, Z, t) + O ε3 . (6.107b) In addition, we assume as in Chapter 4 that the plate is inextensible to leading order, so that the distance between any two material points on the centre-line Z = 0 is approximately conserved; however, we will shortly be able to give an a posteriori justification of this assumption. It follows that, to leading order, X measures arc-length along the deformed plate, so we can write ∂x(0) = cos θ, ∂X

∂z (0) = sin θ, ∂X

(6.108)

where θ(X, t) is the angle between the centre-line of the plate and the xaxis. Finally, we suppose that each transverse section through the plate is to leading order simply rotated through the angle θ, as shown in Figure 6.2 so that x(1) = −Z sin θ,

z (1) = Z cos θ.

(6.109)

As shown in Exercise 6.4, these physically plausible assumptions ensure that the O(1) displacement (6.107) can be achieved with only O(εE) stresses. TO COMMENT: 1. ABOUT CENTRE LINE 2. SMALL E LIMIT 6.5.3 Leading-order equations The deformation gradient tensor takes the form ! Ã ¶ µ µ ¶ (2) ¡ ¢ ∂xi cos θ − sin θ −ZθX cos θ xZ ∼ + O ε2 , F = +ε (2) sin θ cos θ ∂Xj −ZθX sin θ zZ (6.110)

270

Asymptotic Analysis

in which the leading-order term is simply a rotation through the angle θ. Hence, the strain tensor is given by ¢ 1¡ T F F −I 2Ã ! (2) (2) ¡ ¢ ε −2ZθX xZ cos θ + zZ sin θ ∼ + O ε2 . (6.111) (2) (2) (2) (2) 2 xZ cos θ + zZ sin θ −2xZ sin θ + 2zZ cos θ

E=

Since the strain is small, we can again limit our attention to the mechanically linear dimensional constitutive relation S = F −1 T = λTr (E)I + 2µE.

(6.112)

We nondimensionalise (6.112) using (6.99) and expand the stress components in powers of ε, writing ¡ ¢ ¡ ¢ (0) (1) (0) (1) T11 = T11 + εT11 + O ε2 , T31 = T31 + εT31 + O ε2 , (6.113) to obtain (0)

(0)

T11 cos θ + T31 sin θ =

³ ´ (2) (2) −(1 − ν)ZθX + ν zZ cos θ − xZ sin θ (1 + ν)(1 − 2ν)

,

(6.114a) (2)

0= (0) T31 cos θ

−

(2)

(6.114b)

(2)

x cos θ + zZ sin θ = Z , (6.114c) 2(1 + ν) ³ ´ (2) (2) −νZθX + (1 − ν) zZ cos θ − xZ sin θ 0= , (1 + ν)(1 − 2ν) (6.114d)

(0) T11 sin θ

from which it follows that ZθX cos θ (0) T11 = − , 1 − ν2 (0)

(2)

xZ cos θ + zZ sin θ , 2(1 + ν)

(0)

T31 = −

ZθX sin θ . 1 − ν2

(6.115)

(0)

Thus T 11 = T 31 = 0, so the lowest-order averaged stresses are (1)

Tx = T 11 ,

(1)

Tz = T 31 ,

(6.116)

in the x- and z-directions respectively. The averaged momentum equations (6.105) thus give us (now dropping the superscripts) ∂2x ∂Tx = , 2 ∂t ∂X

∂2z ∂Tz = − G. 2 ∂t ∂X

(6.117)


271

Next we use (6.107) and (6.113) to expand x, z and the stress components in the angular momentum equation (6.106) and simplify the result to obtain ∂M + Tx sin θ − Tz cos θ = 0, ∂X where the bending moment is given by Z 1/2 ³ ´ θX (0) (0) . M= T11 cos θ + T31 sin θ Z dZ = − 12(1 − ν 2 ) −1/2

(6.118)

(6.119)

If we neglect gravity and time derivatives, then (6.117) and (6.118) reduce to the Euler strut model (4.121), when we identify the tangential and normal internal stresses with T = Tx cos θ + Tz sin θ,

N = Tz cos θ − Tx sin θ.

(6.120)

Our asymptotic approach has thus provided a reasonably systematic derivation of this equation and the constitutive relation (6.119) between bending moment and curvature, as well as generalising it to include time-dependence and a body force. In addition, we can use our analysis to determine the details of the internal stresses and strains which are not predicted by the ad hoc theory given in §4.9. As an illustration, we will now show briefly how the longitudinal stretching of the plate can be found. 6.5.4 Longitudinal stretching of a plate By substituting (6.115) into the momentum equations (6.101), we find that the other two leading-order stress components are given by µ ¶ µ ¶ (θX cos θ)X (θX sin θ)X 1 1 (0) (0) 2 2 T13 = Z − , T33 = Z − . (6.121) 2(1 − ν 2 ) 4 2(1 − ν 2 ) 4 (2)

(2)

We can also determine xZ and zZ from (6.114) and integrate to obtain x(2) = a(x, t) −

νZ 2 θX sin θ , 2(1 − ν)

z (2) = b(x, t) +

νZ 2 θX cos θ , 2(1 − ν)

(6.122)

where a and b are as yet arbitrary. Note that the centre-line of the plate is deformed to µ ¶ µ (0) ¶ x(X, 0, t) x (X, t) + ε2 a(X, t) + · · · xc (X, t) = ∼ , (6.123) z(X, 0, t) z (0) (X, t) + ε2 b(X, t) + · · · so that ∂xc ∼ ∂X

µ ¶ µ ¶ cos θ 2 aX +ε + ··· . sin θ bX

(6.124)

272

Asymptotic Analysis

and arc-length s along the centre-line is thus related to X by ¯ ¯ ¯ ∂xc ¯ ∂s ¯ ∼ 1 + ε2 (aX cos θ + bX sin θ) + · · · . = ¯¯ ∂X ∂X ¯

(6.125)

PDH; PLEASE COOMENT OF ELASTIC MODULI HERE OR TWO PAGES EARLIER Arc-length is therefore conserved up to order ε2 : this is to be expected since we assumed that the plate is approximately inextensible. ¡ ¢ The small longitudinal stretching that does occur is described by the O ε2 correction on the right-hand side of (6.125), which may be determined by analysing the constitutive relation (6.112) in more detail as follows. ¡ ¢ Considering the expressions for S11 and S33 at O ε2 , we find ¡ ¢ 2 3 − 10ν + 9ν 2 − ν 3 Z 2 θX (1) (1) T11 cos θ + T31 sin θ = 2(1 − ν 2 )(1 − ν)(1 − 2ν) (1 − ν) + (aX cos θ + bX sin θ) (1 + ν)(1 − 2ν) ³ ´ ν (3) (3) + zZ cos θ − xZ sin θ , (6.126a) (1 + ν)(1 − 2ν) and ¡ ¢ 2 1 − 2ν + 4Z 2 (1 − 3ν + ν 2 ) θX 0= 8(1 − ν 2 )(1 − 2ν) ν (aX cos θ + bX sin θ) + (1 + ν)(1 − 2ν) ³ ´ (1 − ν) (3) (3) + zZ cos θ − xZ sin θ , (1 + ν)(1 − 2ν)

(6.126b)

respectively. By eliminating the final terms involving the third-order strains, we find ´ ¡ ¢ ³ (1) (1) aX cos θ + bX sin θ = 1 − ν 2 T11 cos θ + T31 sin θ ¡ ¢ 2 ν − 12(1 − ν)Z 2 θX (6.127) + 8(1 − ν) and integration with respect to Z over (−1/2, 1/2) leads to 2 ¡ ¢ (1 − 2ν)θX aX cos θ + bX sin θ = 1 − ν 2 T − . 8(1 − ν)

(6.128)

Equation (6.128) tells us how much the plate stretches along its centreline. Recall that T = (Tx cos θ + Tz sin θ) is the tension in the plate, so we

6.6 The linear rod equations

273

should not be surprised that the stretch increases linearly with T . Perhaps less intuitive is the final term which is quadratic in the curvature and tells us that bending the plate inevitably causes it to shrink. 6.6 The linear rod equations GK proposes to keep some of it; see below. (The rest of the text is commented out but not erased) We now briefly consider the deflection of a straight rod whose centre-line initially lies along the x-axis. This time, we use as slenderness parameter √ A ε= , (6.129) L where L is the rod length and A denotes the area of the cross section. As with the linear plate equation, we suppose that the rod undergoes transverse displacements of order W , small enough for the linear Navier equations to be valid. The scalings that apply to the present situation are analogous to those employed in §6.2 to nondimensionalise the equations, namely µ r ¶ L ρ t= t0 , (6.130a) ε E (x, y, z) = L(x0 , εy 0 , εz 0 ), 0

0

(6.130b)

0

(u, v, w) = W (εu , v , w ), (6.130c) ¢ EW ¡ 0 0 0 0 0 0 ετxx , τxy , τxz , ετyy , ετyz , ετzz , (6.130d) (τxx , τxy , τxz , τyy , τyz , τzz ) = L so that the dimensionless Navier equations (with primes dropped) take the form ∂2u ∂τxz ∂τxx ∂τxy = ε2 + + , 2 ∂t ∂x ∂y ∂z ∂τxy ∂τyy ∂τyz ∂2v ε2 2 = + + , ∂t ∂x ∂y ∂z ∂τyz ∂2w ∂τxz ∂τzz ε2 2 = + + − ε2 G, ∂t ∂x ∂y ∂z ε4

(6.131a) (6.131b) (6.131c)

where G is defined by (6.15) and represents downwards gravity. Assuming that the boundary is traction-free, we have ny τxy + nz τxz = ny τyy + nz τyz = ny τyz + nz τzz = 0

on ∂D.

(6.132)

As usual, the way to a simplified model passes by the integration of (6.131)

274

Asymptotic Analysis

and (0, y, z) × (6.131) over the cross-section of the rod followed by the application of the divergence theorem. This produces a set of averaged equations of the form (4.26) and (4.26), save for the nonlinear membrane term of (4.26). Next, constitutive relations must be established between the average tension, bending moments, etc... and the displacements or twists. However, we will not dwell on the details here, as this does not bring us any fundamentally new lesson with respect to the other examples. Let us simply note that (6.131a) and (6.131b,6.131c) decouple to a steady torsion problem and a plain strain problem with zero-traction boundary condition, respectively. Also, a rapid glance of the dimensionless stress-strain relation (Exercise 6.5) shows that v(x, y, z, t) ∼ v¯(x, t) − zc(x, t), w(x, y, z, t) ∼ w(x, ¯ t) + yc(x, t),

(6.133)

to leading order. Hence, the problem is naturally decomposed in the study of the uniform displacements v¯ and w ¯ and a rotation c of the cross-section. MAYBE JUST STOP HERE? 6.7 Linear shell theory 6.7.1 Geometry of the shell Our final asymptotic model deals with the generalisation of §6.2 to cover infinitesimal deformations of a fully fledged elastic shell as distinct from the weakly curved shell of §6.4.4. The initial centre-surface of the shell is parametrised by r = r c (ξ1 , ξ2 ),

(6.134)

in which ξ1 and ξ2 are two spatial parameters. Since we restrict our attention to linear elasticity, there is no need to distinguish between Lagrangian and Eulerian coordinates. We will use curvilinear coordinates (ξ1 , ξ2 , n) to describe any point in the plate whose position vector is r = r c (ξ1 , ξ2 ) + nn,

(6.135)

where n is the unit normal to the centre-surface oriented as described below. As described in Appendix 3, the task of writing down the Navier equations is relatively straightforward if the coordinate system (ξ1 , ξ2 , n) is orthogonal. This can be achieved by choosing coordinates ξ1 and ξ2 that parametrise lines of curvature,† with respect to which the first and second fundamental † This can always be achieved except at umbilic points of the surface, which are irrelevant for our discussion.

6.7 Linear shell theory

275

forms of the centre-surface are both diagonal, that is ∂ 2 rc ·n=0 ∂ξ1 ∂ξ2

∂r c ∂r c · = 0, ∂ξ1 ∂ξ2

(6.136)

(see for example Kreyszig, 1959). This choice simplifies the derivations but does not involve any loss of generality, since the final equations that we will obtain may readily be expressed in any other convenient coordinate system. CHECK NOTATION We set ¯ ¯ ¯ ¯ ¯ ∂r c ¯ ¯ ∂r c ¯ ¯ ¯ ¯ ¯ a1 (ξ1 , ξ2 ) = ¯ , a2 (ξ1 , ξ2 ) = ¯ (6.137) ∂ξ1 ¯ ∂ξ2 ¯ and hence define an orthonormal basis {e1 , e2 , n}, where e1 =

1 ∂r c , a1 ∂ξ1

e2 =

1 ∂r c , a2 ∂ξ2

n = e1 ×e2 .

(6.138)

The derivatives of these basis vectors are given by CHECK BACK ∂e1 1 ∂a1 =− e2 + κ1 a1 n, ∂ξ1 a2 ∂ξ2 ∂e2 1 ∂a2 =− e1 + κ2 a2 n, ∂ξ2 a1 ∂ξ1 ∂n = −κ1 a1 e1 , ∂ξ1

∂e1 1 ∂a2 = e2 ∂ξ2 a1 ∂ξ1 ∂e2 1 ∂a1 = e1 , ∂ξ1 a2 ∂ξ2 ∂n = −κ2 a2 e2 , ∂ξ2

(6.139a) (6.139b) (6.139c)

where κ1 and κ2 are the principal curvatures of the centre-surface. Applying these identities to the parametrisation (6.135), we find that ∂r = a1 (1 − κ1 n)e1 , ∂ξ1

∂r = a2 (1 − κ2 n)e2 , ∂ξ2

∂r = n, ∂n

(6.140)

so this coordinate system is indeed orthogonal, with scaling factors h1 = a1 (1 − κ1 n),

h2 = a2 (1 − κ2 n),

h3 = 1.

(6.141)

6.7.2 Dimensionless equations We denote the shell thickness by h, so its surfaces (assumed stress-free) are given by n = ±h/2. For simplicity we will assume here that h is constant, although it is straightforward to apply the same methods to shells with

276

Asymptotic Analysis

nonuniform thickness. We also define L to be a typical radius of curvature, so the lengths in the problem are nondimensionalised via n = hn0 ,

(ξ1 , ξ2 ) = L(ξ10 , ξ20 ),

(κ1 , κ2 ) =

1 0 0 (κ , κ ). L 1 2

(6.142a)

We denote the displacement field by u = u1 e1 +u2 e2 +wn. Because the shell is assumed not to be nearly flat, we must expect the in-plane and transverse displacements to be of the same order, say (u1 , u2 , w) = W (u01 , u02 , w0 ),

(6.142b)

where W is small enough for linear elasticity to apply. The Cauchy stress tensor, relative to the curvilinear coordinate system, is now scaled using (τ11 , τ12 , τ22 ) =

ε2 EW ¡ 0 0 0 ¢ EW ¡ 0 0 0 ¢ τ11 , τ12 , τ22 , (τ13 , τ23 , τ33 ) = τ13 , τ23 , τ33 , L L (6.142c)

as distinct from (6.11). The appropriate time-scale is now r ρ 0 t=L t. E

(6.142d)

Gravity is assumed to act in the direction −k and is represented by its dimensionless components along each of the basis vectors, defined as g1 = −

ρGL2 (k · e1 ) , EW

g2 = −

ρGL2 (k · e2 ) , EW

g3 = −

ρGL2 (k · n) . EW (6.143)

To avoid clutter, we henceforth drop the prime for the dimensionless variables.

6.7.3 Leading-order equations The Navier equations and linear constitutive relations in an arbitrary orthogonal coordinate system may be found in Appendix 3. As in the previous

6.7 Linear shell theory

277

sections, we simplify the dimensionless equations by letting ε → 0 to obtain ½ ¾ ∂ 2 u1 1 ∂ ∂ ∂a1 ∂a2 − g1 = (a2 τ11 ) + (a1 τ12 ) + τ12 − τ22 , ∂t2 a1 a2 ∂ξ1 ∂ξ2 ∂ξ2 ∂ξ1 (6.144a) ½ ¾ 2 ∂a2 ∂a1 1 ∂ ∂ ∂ u2 − g2 = (a2 τ12 ) + (a1 τ22 ) + τ12 − τ11 , ∂t2 a1 a2 ∂ξ1 ∂ξ2 ∂ξ1 ∂ξ2 (6.144b) ∂2w − g3 = κ1 τ11 + κ2 τ22 , (6.144c) ∂t2 at leading order. Once we have used the constitutive relations to express τ11 , τ12 and τ22 in terms of the displacements, (6.144) will give us three equations for u1 , u2 and w. The constitutive relations imply that the displacement components are all independent of n to leading order. Furthermore, the dimensionless constitutive relation for τ33 reads νe11 + νe22 + (1 − ν)e33 = ε2 τ33 , (1 + ν)(1 − 2ν)

(6.145)

which we can use to express e33 in terms of e11 and e22 . This gives, for τ11 and τ22 e11 + νe22 νe11 + e22 τ11 = , τ22 = . (6.146) 2 1−ν 1 − ν2 On the other had, the relevant strain components are µ ¶ 1 ∂u1 u2 ∂a1 e11 = + − κ1 w + O (ε), (6.147a) a1 ∂ξ1 a2 ∂ξ2 µ ¶ 1 ∂u2 u1 ∂a2 e22 = + − κ2 w + O (ε), (6.147b) a2 ∂ξ2 a1 ∂ξ1 1 ∂w , (6.147c) e33 = ε ∂n while the leading-order in-plane shear stress reads ½ ¾ 1 ∂u1 ∂a1 ∂u2 ∂a2 τ12 = a1 − u1 + a2 − u2 . (6.148) 2(1 + ν)a1 a2 ∂ξ2 ∂ξ2 ∂ξ1 ∂ξ1 The equations (6.144) can be cast in invariant form as ˆ ∂2u ˆ · τˆ, ˆ=∇ −g 2 ∂t ∂2w − g3 = K : τ, ∂t2

(6.149a) (6.149b)

278

Asymptotic Analysis

ˆ is surface divergence, ˆ = u − wn is the in-plane displacement, ∇· where u τˆ is in-plane stress, K is curvature tensor. This form is independent of coordinate system chosen. Further, the in-plane stress satisfies the constitutive relation ¡ ¢ ν 1 ˆ E+ Tr Eˆ I, (6.150) τˆ = 1+ν 1 − ν2 where the surface strain tensor is given by o 1nˆ ˆ u)T − wK Eˆ = ∇ˆ u + (∇ˆ 2 or, in component form,

(6.151)

1 (ˆ ui,j + u ˆj,i ) − wκij . (6.152) 2 The theory just derived is restricted to in-plane tensions in the shell, the bending stiffness being neglected here. However, the equations above can sometimes allow for zero-stress displacements. For example, if we let the shell tend to a flat sheet so that κ1 = κ2 = 0, then the normal displacement w decouples entirely from the in-plane stresses and is arbitrary in the absence of bending stiffness. Also cylinders bending about their generators offer no resistance to a large class of displacements (see Exercise 6.6). More generally, any developable shells, for which r c is the envelope of a one-parameter family of planes, can sustain zero-stress deformation in our theory, and in Exercise 6.8, the partial differential equation for such zero-stress modes of deformation is derived. As for weakly curved shells (see §4.8), it is elliptic, parabolic, or hyperbolic, depending on the sign of the Gaussian curvature, and we can expect the existence of a zero-stress mode to depend strongly on this type. We can see this by eliminating τ22 from the steady-state form (6.144). This leads to the following principal parts (i.e. the terms with highest derivatives): eîj =

∂τ11 ∂τ12 + a1 , ∂ξ1 ∂ξ2 ∂τ12 κ1 ∂τ11 0 = a2 − a1 . ∂ξ1 κ2 ∂ξ2

0 = a2

The characteristic curves for this set of partial differential equations are given by r dξ2 a1 κ1 =± − . (6.153) dξ1 a2 κ2 Hence, if the Gaussian curvature is negative, the stress field at one point

6.8 Conclusions

279

of the shell only influences a region bounded by the two real characteristic curves above. Conversely, a positive Gaussian makes (6.144) elliptic, so that the state of stress in one point depends on the stress everywhere in the shell. A systematic treatment of zero-stress displacement would require a new geometrically nonlinear asymptotic treatment of the Navier equations with suitably rescaled stresses but this is beyond the scope of this book. However, it has resulted in the award of an Ignoble Prize. 6.8 Conclusions It is comforting to see that the approximate models in Chapter 4, which were derived on the basis of intuition and plausibility, can be recovered by systematic perturbation procedures. The asymptotic approach outlined here, however, provides us with a much more detailed picture of the stress field than the approximate theories alone. For instance, we have been able to resolve its three-dimensional structure in the vicinity of the boundaries and elucidate the St-Venant principle. Additionally, we have seen the fascinating challenges posed by geometric nonlinearity. Perhaps the biggest surprise is that the familiar “membrane” terms in equations like (4.15) actually arise, and can only be derived,- from nonlinear elasticity. A recurrent pattern in the asymptotic methodology has emerged: first, the geometry of the solid suggests particular scalings for the displacement field; then, the stress components are rescaled so as to balance as many terms as possible in the Navier equations; finally, these equations are analysed asymptotically, or, if we are lucky, by suitable global averaging over boundary layers. We have only been concerned with plates, shells, and rods with uniform thickness or cross-section and comparable elastic moduli. If the thickness of the plate were to vary rapidly with a typical length scale ` ¿ L, a quite new theory would be needed. The same would be true if µ/λ ¿ 1 or ν ≈ 1/2, when we would have to state precisely what are the relative order of magnitude of the various small parameters involved, namely ε, `/L, and µ/λ. In particular, there is no reason to expect that the limits ε → 0 and µ/λ → 0 commute. Finally, our approximate models would certainly need reformulation in the vicinity of a buckling threshold. Exercises 6.1 Resume the antiplane strain example of the introduction, this time assuming that the bottom and top of the sheet are subjected to the

280

Asymptotic Analysis

traction τyz = σ− (x) and τyz = σ+ (x) on y = 0 and y = ε, respectively. • Using the same perturbation scheme far away from the edges, show that a solvability condition arising at O(ε) is that σ+ − σ− = O(ε). • Hence, letting σ+ = σ− + εσ, deduce from a second solvability condition at O(ε2 ) that d2 w0 + σ = 0. dx2 • and what about letting σ = σ(x/δ) with δ ¿ 1 above and do some homogenization? 6.2 Repeat the linear beam derivation from §6.6 using the fast time-scale µ r ¶ ρ t = εL t0 . E Show that the decompositions (6.133), (??) still apply, where the functions a, b, c and U satisfy ∂2a ∂2b ∂2c ∂2c ∂2U ∂2U = 0, = 0, (I + I ) = R , = . yy zz ∂t2 ∂t2 ∂t2 ∂x2 ∂t2 ∂x2 Note these are wave equations not beam equations! transverse oscillations much more slow so not visible on this timescale 6.3 Show that, for a weakly curved shell, (6.95) imply that the in-plane average stress can still be described by an Airy stress function. Furthermore, show that the analysis up to (6.88) still applies and hence that (6.89) becomes T 11 = T 12 = T 22 =

2 − 2Hw 2 2uX + wX XX + ν(2v Y + wY − 2HwY Y ) , 2(1 − ν 2 ) uY + v X + wX wY − 2HwXY , 2(1 + ν) 2 − 2Hw 2 ν(2uX + wX XX ) + 2v Y + wY − 2HwY Y . 2(1 − ν 2 )

and deduce (6.96). 6.4 Show that the strain tensor in §6.5 is given by µ ¶ 2 −1 1 x2X + zX ε−1 (x X xZ + z¢X zZ ) ¡ E= . 2 ε−1 (xX xZ + zX zZ ) ε−2 x2Z + zZ2 − 1 Hence, show that T is O(ε−2 E) unless (6.107) is assumed. Further,

EXERCISES

281

show that the components of S must be at most O(εE) for the scaling (6.99) to hold and therefore that Ã

∂x(0) ∂X

Ã

!2 +

∂z (0) ∂X

Ã

!2 = 1,

∂x(1) ∂Z

Ã

!2 +

∂z (1) ∂Z

!2 = 1,

∂x(0) ∂x(1) ∂z (0) ∂z (1) + = 0. ∂X ∂Z ∂X ∂Z From this, justify (6.108) and (6.109). 6.5 Show that the stress-strain relations for the scalings appropriate to the linear rod theory are ε2 (1 − ν)ux + νvy + νwz , (1 + ν)(1 − 2ν) ε2 νux + (1 − ν)vy + νwz = , (1 + ν)(1 − 2ν) ε2 νux + νvy + (1 − ν)wz = , (1 + ν)(1 − 2ν)

u y + vx , 2(1 + ν) uz + wx = , 2(1 + ν) vz + wy = , 2(1 + ν)

ε2 τxx =

τxy =

ε2 τyy

τxz

ε2 τzz

ε2 τyz

and deduce (6.133). Furthermore, eliminate v and w to show that τxx − ν (τyy + τzz ) = ux . 6.6 Derive the equation for a circular cylindrical shell of radius r: ∂ 2 ur − gr = ∂t2 ∂ 2 uθ − gθ = ∂t2 ∂ 2 ur − gr = ∂t2

τθθ , r ∂τθz 1 ∂τθθ + , r ∂θ ∂z ∂τzz 1 ∂τθz + . r ∂θ ∂z

Furthermore, show that displacements of the form uθ = f (θ), ur = f 0 (θ), uz = 0 generate no strain and no stress in the leading order shell theory. Show that such a displacement exists for cylindrical shell with any cross section. [This shows that a cylinder can undergo deformations that are one order of magnitude larger if bent along a generator than otherwise]

282

Asymptotic Analysis

6.7 Derive the equation for a spherical shell of radius r: ∂ 2 ur − gr = ∂t2 ∂ 2 uθ − gθ = ∂t2 ∂ 2 ur − gr = ∂t2

τθθ + τφφ , r · ¸ ∂τθφ 1 ∂ (sin θτθθ ) + − τφφ cos θ , r sin θ ∂θ ∂φ · ¸ ∂τφφ 1 ∂ (sin θτθφ ) + + τθφ cos θ . r sin θ ∂θ ∂φ

6.8 Show that, within the linear shell theory of §6.7, in-plane displacements satisfying µ ¶ ∂a2 ∂a1 κ2 ∂u1 κ1 ∂u2 1 − = κ1 u1 − κ2 u2 , a1 ∂ξ1 a2 ∂ξ2 a1 a2 ∂ξ1 ∂ξ2 µ ¶ ∂a1 ∂a2 1 ∂u2 1 ∂u1 1 + = u1 + u2 . a1 ∂ξ1 a2 ∂ξ2 a1 a2 ∂ξ2 ∂ξ1 induce no stress. Derive the equation for the characteristic curves (6.153). 6.9 I’m worried about this example... Show that elimination of A from (??) leads to w00000 + 12(1 − ν 2 )κ2 w0 = 0, when w and A are functions of x alone. Solve this equation subject to the boundary conditions (??) ETC... 6.10 Suppose that the strains (??) in a spherical shell are identically zero. Eliminate w and uθ to show that uφ satisfies the equation 2 ∂ 2 uφ ∂uφ 2 ∂ uφ + uφ = 0. + sin θ − sin θ cos θ 2 2 ∂φ ∂θ ∂θ

By seeking separable solutions that are periodic in φ, show that possible solutions are ¡ ¢ uφ ∝ sin θ, sin θ log tan(θ/2) , sin(kφ + δ) sin θ tank (θ/2). Deduce that the only such solution which has bounded displacement as θ → 0, π is uφ ∝ sin θ, and find the corresponding expressions for uθ and w. 6.11 Thin punch (cf elastic lubrication) 6.12 compel fibre? 6.13 crease + folds chap 4

7 Fracture and Contact

7.1 Introduction In this chapter, we discuss two phenomena that may appear at first glance to have little in common. Fracture describes the behaviour of thin cracks, like the one illustrated in Figure 7.1, in an otherwise elastic material. The crack itself is a thin void, whose faces are usually assumed to be stress-free. When a solid containing a crack is stressed, we will find that the stress is localised near the crack tip, becoming singular if the tip is sharp. We will see that the strength of the singularity can be characterised by a stress intensity factor, which depends on the size of the crack and of the applied stress. This factor determines the likelihood that a crack will grow and, therefore, that the solid will fail.† On the other hand, contact refers to the class of problems in which two elastic solid bodies are brought into contact with each other, as illustrated in Figure 7.2. When the material properties are the same in the two bodies, the geometrical configuration near the edge of the contact region is apparently similar to that of fracture, with voids now outside the contact set, that is the set of points at which the two solids are in contact. The mathematical setup of such problems does therefore have some similarity with fracture, but in the steady state, there is one crucial difference: in contact problems, the contact set itself is often not known in advance. They are thus known as free† The reader may be reassured to know that although there are about a dozen cracks emanating from every rivet in an airliner, they are carefully monitored so that none of them is dangerous.

Fig. 7.1. Definition sketch of a thin crack. Fig. 7.2. Definition sketch for contact between two solids with nearly parallel boundaries.

283

284

Fracture and Contact

boundary problems, that is problems whose geometry must be determined along with the solution. An example of such a problem is to determine how the contact area between a car tyre and the road depends on the inflation pressure.

7.2 Static Brittle Fracture 7.2.1 Physical background In §2.2.5, we proposed the idea that the theory of linear elasticity may fail when some stress components in a solid reach a critical value at some point. While this statement is generally true, the way in which failure occurs can be very different for different solids, or even for the same solid under different loading conditions. This section will focus on the simplest mathematical model for one type of failure, namely brittle fracture. The adjective “fractured” is usually used to describe configurations where a continuum solid has developed cracks, which are very thin stress-free voids. These voids are characterised by being thin in only one direction so that they are nearly slits in a two-dimensional solid (see Figure 7.1). Thus a crack has two faces with a common perimeter which is called the crack tip, and the tip is just two points in the two-dimensional case. Cracks are often initiated at the surface of the material, which is usually the only place where they can be discerned by the naked eye. Brittle fracture covers that class of fractures, such as windscreen shattering, where the initiation and growth of the crack do not affect the material modelling, which is still governed by the Navier equations. Such a situation might not be true for a metal, where quite a large region near the crack tip may have to be modelled using plasticity rather than elasticity equations; then the fracture is called ductile, and we will discuss this further in Chapter 9.† We can say little more here about the physical background of this vast and important subject. However, we urge the reader to think intuitively that fracturing, i.e. crack growth,is basically a local stress-relieving mechanism. As a crack grows, the strain energy in the evolving fractured configuration is less than the value it would have taken had the crack not evolved. This argument suggests that the criterion for a crack to evolve under a small increase of loading is that the release, −δW, of strain energy resulting from † check: Many metals exhibit a “brittle-ductile transition” at a critical temperature. Above this temperature it is believed that microfractures . . . inhomogeneities generate large number of dislocations, but not below it.

7.2 Static Brittle Fracture

285

Fig. 7.3. (a) A Mode III crack. (b) A close-up of one tip.

a small growth δA in the crack face area is such that ZZZ −δ W dx > δS,

(7.1)

V

where V denotes the region excluding the crack and δS is the surface energy of the area δA. In principle, δS could be estimated from molecular considerations but, in the theory of Griffith cracks, δS is thought of as resulting from an effective surface tension γ; in this case, δS = γδA. It is interesting to compare fracture with the more familiar failure of a liquid, which cavitates when the pressure is too low, so that the molecular bonds weaken and allow the molecules to form a vapour. This results in the appearance of a spherical bubble, whereas a crack morphology is usually the one preferred by solids. The difference in behaviour arises because it is easier to create new interface between a liquid and a vapour than it is to create new interface between a solid and a void. Hence a crack can grow most easily by creating new surface energy just in the vicinity of the crack tip. It is also instructive to contrast fracture, which occurs as a result of some remote loading process, with the drilling of a thin hole through the solid (for example using a shaped charge; see §3.6). It is an everyday observation that the energy required to drill a hole per unit surface area generated is greater than that required to cleave the material with an axe. 7.2.2 Mode III cracks Let us now investigate the predictions of the Navier model for linear elasticity in the presence of a crack. The very simplest configuration is that of a crack in antiplane shear, which is called a Mode III crack, as illustrated in Figure 7.3. Although this configuration is not so important in practice, it is much simpler mathematically than Mode I and II cracks, which will be introduced later. We will consider a planar crack whose faces lie close to the (x, z)-plane between the crack tips x = ±c, y = 0, as shown in Figure 7.3. The physical set-up is that of a large cracked slab being sheared at infinity in the (y, z)plane with a shear stress σIII . Assuming that the faces of the crack are stress free, we find that the mathematical model for the displacement w(x, y) in the z-direction is ∇2 w = 0

(7.2a)

286


everywhere except on y = 0, |x| < c, with ∂w → σIII as x2 + y 2 → ∞, ∂y ∂w µ =0 on y = 0, |x| < c. ∂y µ

(7.2b) (7.2c)

There is a fundamental difference between (7.2) and most of the boundary value problems we have thus far considered for elastostatics, namely that the boundary on which the Neumann data (7.2c) are prescribed is not smooth at the crack tip. To see the kind of difficulty that this can cause, suppose we were to concentrate on the region near the crack tip illustrated in Figure 7.3(b) and seek a displacement field in which ∇2 w = 0 everywhere except on y = 0, x > 0, with µ

∂w =0 ∂y

on y = 0, x > 0.

(7.3)

By separating the variables in polar coordinates (r, θ), we immediately see that w can be any function of the form ³ ´ w = Arn/2 cos(nθ/2) = A Re (x + iy)n/2 , (7.4) where z = reiθ , A is a constant and n is an integer. The corresponding stress components are given by µnA n/2−1 µnA n/2−1 r cos(nθ/2), τθz = − r sin(nθ/2). (7.5) 2 2 This plethora of solutions gives us the strong hint that we will not be able to solve (7.2) uniquely unless we supply some extra information about the behaviour of w near (±c, 0). We also note that, whenever n is not an even positive integer, the stress is non-analytic at the crack tip r = 0, and, if n < 2, the stress is not even bounded. From a mathematical point of view it is natural to ask whether the solution of (7.2) must be singular at (±c, 0) and, if so, what is the mildest singularity which we have to endure. To answer these questions, one possibility is to round off the crack, that is, to replace it with a thin but smooth boundary. A particularly convenient shape is the ellipse τrz =

y2 x2 + = 1, (7.6) c2 cosh2 ² c2 sinh2 ² which is smooth for all positive ² but approaches the slit geometry of Figure 7.3(a) in the limit ² → 0. Now we can easily solve Laplace’s equation


287

Im z

z

θ2

r1

r

r2 θ

θ1 c

−c

Re z

Fig. 7.4. Definition sketch for r1 , r2 ,θ1 and θ2 in (7.12).

by introducing elliptical coordinates defined by z = x + iy = c cosh ζ,

where ζ = ξ + iη,

(7.7a)

x = c cosh ξ cos η,

y = c sinh ξ sin η.

(7.7b)

so that

Since the map (7.7) from z to ζ is conformal except on y = 0, |x| < c, Laplace’s equation is preserved and the model (7.2) becomes ∂2w ∂2w + = 0, ∂ξ 2 ∂η 2

in ξ > ²

(7.8a)

with ∂w =0 ∂ξ

on ξ = ²,

(7.8b)

and, since y ∼ (c/2)eξ sin η as ξ → ∞, w∼

cσIII ξ e sin η 2µ

as ξ → ∞.

By separating the variables, we easily find the solution ³ ´ cσIII w= sin η eξ + e2²−ξ . 2µ

(7.8c)

(7.9)

288


w

y x Fig. 7.5. Displacement field for a Mode III crack.

We thus have a unique displacement field for any positive value of ² and, when we let ² → 0, we obtain the solution of the crack problem as cσIII w= Im (sinh ζ) . (7.10) µ Using (7.7), this can be written as w=

³p ´ σIII Im z 2 − c2 , µ

(7.11)

where the square root is defined to be p √ z 2 − c2 ≡ r1 r2 ei(θ1 +θ2 )/2 ,

(7.12)

with r1 , r2 , θ1 and θ2 defined as shown in Figure 7.4. The angles are taken to lie in the ranges −π < θj < π, so that a branch cut lies along the crack between z = −c and z = c. As shown in Figure 7.5, the displacement w is discontinuous across this branch cut. Despite its simplicity, the formula (7.11) lies at the very heart of the theory of brittle fracture. It reveals the famous square root singularity in which the elastic displacement varies as the square root of the distance from the crack tip.† By rounding off the crack before taking the limit ² → 0, we have managed to select one of the many possible singular solutions suggested by (7.4). This dependence immediately implies that the stress tensor, whose non-zero components are τxz = µ

∂w ∂x

and

τyz = µ

∂w , ∂y

(7.13)

† A similarly fundamental square root singularity arises in the mathematically analogous theory of aerofoils.


289

Fig. 7.6. Sketch of a rounded tip, showing the radius of curvature r0 . Fig. 7.7. Schematic showing cohesion between crack faces.

diverges as the inverse square root of the distance from the tip. This physically unacceptable consequence renders the Navier model invalid sufficiently close to the tip, but there are several ways of recovering from this setback, including the following: (i) Rounding the tip This is the approach adopted above when considering the elliptical crack (7.6), and it is easily shown, using this example, that the stress at the crack tip is given approximately by r c τyz ∼ 2σIII (7.14) r0 when r0 , the tip radius of curvature, is small (Exercise 7.1). (ii) Plastic zone Even if the tip is rounded, (7.14) shows that the stress may still become large enough for the material to yield plastically if r0 is sufficiently small. This suggests the introduction of a plastic zone near the tip, and this also has the effect of preventing the stress from growing without bound. We will discuss this situation further in Chapter 8 (iii) Tip cohesion In (7.2), we assumed that the two faces of the crack exert no traction on each other. Instead we might suppose that there is an effective cohesion between the faces, as shown schematically in Figure 7.7, so that (7.2c) is replaced by µ

∂w = f (x) ∂y

on y = 0, |x| < c,

(7.15)

where f (x) models the cohesive effect of intermolecular bonding, which will be strongest near the crack tip. As shown in Exercise 7.2, a suitable choice of f can be to ensure that the stress remains finite everywhere. Despite the existence of these possibilities for “regularising” our model near the crack tip, the existence of the square root in (7.11) does not preclude this formula from having great practical value, as we will now explain. The result (7.11) predicts that any critical stress, no matter how large, will always be attained sufficiently close to the crack tip. Hence, let us return to

290


the discussion of (7.1) and consider whether or not the crack is dangerous, that is, whether or not it will propagate. This poses the question: “is the local stress large enough for it to do sufficient work to create new surface energy at the crack tip when σIII is increased slightly?” The following answer to this question has formed the basis of the theory of fracture for many decades. We note that, in the material ahead of the crack tip (c, 0), the only nonzero stress component at the crack plane y = 0 is τyz and that √ √ lim τxz (x, 0) x − c = σIII 2c. (7.16) x↓c

Hence, no matter which of the mechanisms such as (i)–(iii) above prevents the stress from becoming infinite in practice, it will have to be triggered by a local stress intensification that is proportional to the inverse square root of distance from the tip, with the all-important constant of proportionality being √ KIII lim τxz (x, 0) x − c = √ . x↓c 2π

(7.17)

The stress intensity factor KIII is the parameter that characterises the propensity of the crack to propagate. We postulate that the crack will propagate if KIII exceeds a critical value K ∗ , which must be determined either from (i)–(iii) above or from experiment, and which we expect to be proportional to the surface energy γ. Under this postulate, (7.16) immediately reveals that a Mode III crack is more likely to grow if it is longer or subject to greater stress. We emphasise that a consequence of this postulate is that KIII is the only piece of information concerning the global stress field that is relevant for deciding whether or not the crack propagates. We will shortly find that the concept of a stress intensity factor applies to many other crack configurations, but first let us review the mathematical challenges posed by problems like (7.2).

7.2.3 Mathematical methodologies for crack problems Now let us quickly review some different approaches that we might have adopted to obtain (7.11) had we not been able to utilise elliptic coordinates and take the limit as ² → 0. Three possibilities are the following. (i) We could conformally map the region outside the cut y = 0, |x| < c onto the exterior of a circle and solve Laplace’s equation in polar coordinates. This is effectively what we have done to obtain (7.11), but in principle the


291

method applies to any antiplane strain crack problem where the relevant conformal map can be found by inspection. However, it crucially relies on the fact that Laplace’s equation is invariant under conformal mapping, and so is only obviously applicable to antiplane strain. (ii) We could represent w(x, y) as a suitable distribution of functions that are singular on the slit. This approach allows most scope for physical intuition. Noting that tan−1 (y/x) is a solution of Laplace’s equation except at the origin and on a branch cut from the origin, we can directly compute that ¡ ¢ Z c tan−1 (x − ξ)/y p µw = σIII dξ + σIII y (7.18) c2 − ξ 2 −c is equivalent to (7.11), as shown in Exercise 7.3. Thus w can be thought of as resulting from the imposed stress together with a distribution of virtual dislocations, and we will return to this idea in Chapter 9. This observation suggests that we could have proceeded ab initio by trying Z c ¡ ¢ w= f (ξ) tan−1 y/(x − ξ) dξ + σIII y, (7.19) −c

and then the boundary condition (7.2c) would have led to a singular integral equation for the function f (ξ). (iii) Since w is an odd function of y, the problem may be formulated in the half-plane y > 0 with the boundary conditions w(x, 0) = 0,

|x| > c,

∂w (x, 0) = 0, ∂y

|x| < c.

(7.20)

Such problems in which the boundary data switches from Dirichlet to Neumann are called mixed boundary value problems and they can easily pose serious mathematical challenges. Nonetheless, the theory of such mixed problems provides the basis for the most powerful methodology for solving crack problems. The basic idea is to take a Fourier transform in x, which is a simple matter for many planar crack problems until it comes to applying the boundary conditions. However, to take the transform of (7.20), we would have to introduce the transforms of both w(x, 0) and ∂w/∂y(x, 0), neither of which we know explicitly. Nonetheless, if we were bold enough to do this, we would find that it is possible to use analytic continuation in the complex Fourier plane to find w. This is the basis of the Wiener–Hopf method, but we have no space to describe it further here; we simply remark that it is equivalent to the singular integral equation theory that emerges

292


(a)

y

(b) σII

y

Ω

σII

x σII

x σII

∂Ω

Fig. 7.8. (a) Schematic of a planar Mode II crack. (b) The regularised problem of a thin elliptical crack.

from (ii) above (Carrier, Krook & Pearson, 1983, Chapter 8). Fortunately, we can sometimes bypass this complexity by generalising our derivation of (7.11) iv We can exploit the fact that the Papkovich–Neuber representation (2.231) can take a simple form when boundary conditions are applied on y = 0. This can give us a very useful short cut as we will explain below.

7.2.4 Mode II cracks We now move on to study cracks in plane strain. This is inevitably more complicated than antiplane strain since we have to solve the biharmonic equation rather than Laplace’s equation. As pointed out above, there are several approaches to constructing solutions, and we will consider two of them in detail. We will start with the conformal mapping technique, which is analogous to that employed in §7.2.2. As before, we will regularise the problem by considering an elliptical crack with a small but finite diameter of order ² and then letting ² → 0. Then, we will see how Papkovich-Neuber potentials provides another short route to the solution. In a planar Mode II crack, a far-field shear stress σII parallel to the plane of the crack is applied, as shown in Figure 7.8. We denote the two-dimensional region outside the crack by Ω and the elliptical boundary of the crack by ∂Ω. This is evidently a plane strain problem, so we employ an Airy stress function which, as usual, satisfies the biharmonic equation in Ω. If the crack is stressfree then, as shown in §2.6.3, we can without loss of generality impose the boundary conditions ∇A = 0

on partialΩ.

(7.21)

The imposed uniform shear stress far from the crack implies that A has the


293

asymptotic behaviour A ∼ −σII xy

as x2 + y 2 → ∞.

(7.22)

We recall from §2.6 that the general solution of the two-dimensional biharmonic equation may be written in the form © ª A = Re z¯f (z) + g(z) , (7.23) where f and g are any two analytic functions of z = x + iy. To transform the boundary conditions for A into conditions on f and g, it is helpful to write (7.23) in the form A=

ª 1© z¯f (z) + g(z) + z f¯(¯ z ) + g¯(¯ z) , 2

(7.24)

where f¯ is the conjugate function to f , that is f¯(¯ z ) ≡ f (z). ¿From (7.21), we deduce that ∂A/∂z = 0 on ∂Ω and, hence, by using (7.24) we obtain the condition z¯f 0 (z) + f¯(¯ z ) + g 0 (z) = 0

on ∂Ω.

(7.25)

In addition, the far-field behaviour (7.22) implies that f and g satisfy f (z) → 0,

g(z) ∼

iσII 2 z 2

as z → ∞.

(7.26)

Amazingly, the two functions f and g can be determined using only the fact that they are both analytic in Ω and the conditions (7.25) and (7.26). However, it is not immediately obvious how to do this, and the task is greatly simplified by conformally mapping Ω onto a region that is easier to manipulate. Here we note that the region outside our elliptical crack is the image of the unit disc under the conformal mapping b z = ω(ζ) = aζ + , ζ

(7.27)

where a and b are real constants with b > a > 0. The principal radii of the ellipse are (a + b) and (a − b), so that a=

ce² , 2

b=

ce−² 2

in the terminology of §7.2.2. We also introduce the notation ¡ ¢ ¡ ¢ F (ζ) = f ω(ζ) , G(ζ) = g ω(ζ) ,

(7.28)

(7.29)

294


for the transformed versions of f and g. Thus F is analytic in the unit disc |ζ| 6 1, and G is analytic apart from a double pole at the origin, where G(ζ) ∼

iσII b2 2ζ 2

as ζ → 0.

(7.30)

The boundary condition (7.25) is transformed to −F¯ (1/ζ) =

ω ¯ (1/ζ)F 0 (ζ) G0 (ζ) + 0 ω 0 (ζ) ω (ζ)

at |ζ| = 1,

(7.31)

aζ 2 − b , ζ2

(7.32)

where ω ¯ (1/ζ) =

a + bζ 2 , ζ

ω 0 (ζ) =

and we use the fact that ζ¯ = 1/ζ on the unit circle. Provided b > a > 0, the map z = ω(ζ) is conformal and hence ω 0 (ζ) is nonzero in the unit disc. The only singularity in the right-hand side of (7.31) therefore arises from the pole in G(ζ) at the origin, where we find that G0 (ζ) iσII b ∼ 0 ω (ζ) ζ

as ζ → 0.

(7.33)

Let us subtract this from both sides of (7.31) to remove the singularity from the right-hand side and thus obtain ¡ ¢ ζ 2 G0 (ζ) + ζ a + bζ 2 F 0 (ζ) iσII b iσII b ¯ −F (1/ζ) − = − at |ζ| = 1. ζ aζ 2 − b ζ (7.34) Now we arrive at the crux of the argument. The left-hand side of (7.34) is analytic in |ζ| > 1, while the right-hand side is analytic in |ζ| 6 1. We can therefore use (7.34) to analytically continue the left-hand side and hence obtain a function that is analytic on the entire complex plane and, because of (7.26), tends to zero as ζ → ∞. The only function that has these properties is the zero function, and we deduce that the both the leftand right-hand sides of (7.34) must be identically zero. We can therefore evaluate the function F (ζ) as F (ζ) = iσII bζ

(7.35)

and, by using this expression in the right-hand side of (7.34), we obtain ¶ µ 1 0 2 (7.36) G (ζ) = −iσII b ζ + 3 . ζ


295

By integrating with respect to ζ and ignoring an irrelevant constant of integration, we thus obtain µ ¶ iσII b2 1 2 G(ζ) = −ζ . (7.37) 2 ζ2 Finally, by using the inverse mapping √ z − z 2 − 4ab ζ= , 2a

(7.38)

where the square root is again defined by (7.12), we evaluate the functions f and g in terms of the original variable z as ´ p iσII b ³ f (z) = z − z 2 − 4ab , (7.39a) 2a n o √ iσII g(z) = (a2 − b2 )(z 2 − 2ab) + (a2 + b2 )z z − 4ab . (7.39b) 2 4a When substituting these into the formula (7.23) for A, we simplify the resulting expression by letting the crack thickness tend to zero, so that a and b both tend to c/2, and thus obtain n o p σII A= Im (¯ z − z) z 2 − c2 . (7.40) 2 By using the chain rules µ ¶ ∂ ∂ ∂ ∂ ∂ ∂ = + , =i − , (7.41) ∂x ∂z ∂ z¯ ∂y ∂z ∂ z¯ we can recover the stress components as τxx =

τxy = −

© ª ∂2A = Re −¯ z f 00 (z) − g 00 (z) + 2f 0 (z) 2 ∂y ( ) 4z 3 + c2 (¯ z − 5z) σII Im , = 2 (z 2 − c2 )3/2

© ª ∂2A = Im z¯f 00 (z) + g 00 (z) ∂x∂y ( ) σII 2z 3 + c2 (¯ z − 3z) = Re , 2 (z 2 − c2 )3/2

τyy =

© 00 ª ∂2A 00 0 = Re z ¯ f (z) + g (z) + 2f (z) ∂x2 ( ) σII c2 z − z¯ = Im , 2 (z 2 − c2 )3/2

(7.42a)

(7.42b)

(7.42c)

296

Fracture and Contact 2

S/σII 1 1 0.8

y

0.6

0

0.4 0.2

-1

0 -2 -2

-1

0

1

2

x Fig. 7.9. Contour plot of the maximum shear stress S around a Mode II crack, with crack length c = 1.

and, by approximating these in the neighbourhood of the crack tip z = c, we find that ¶ µ ¶¾ √ ½ µ 5θ1 θ1 σII c sin + 7 sin , (7.43a) τxx ∼ − √ 2 2 4 2r1 µ ¶ µ ¶¾ √ ½ σII c 5θ1 θ1 τxy ∼ √ cos + 3 cos , (7.43b) 2 2 4 2r1 ¶ µ √ σII c 3θ1 , (7.43c) τyy ∼ √ sin θ1 cos 2 2 2r1 as r1 → 0, where r1 and θ1 are defined as in Figure 7.4. Hence, as in Mode III cracks, the stress diverges like the inverse square root of the distance from the crack tip. Ahead of the crack, the only nonzero stress component on the plane y = 0 is found, by setting θ1 = 0, to be √ σII c τxy (x, 0) ∼ p as x ↓ c. (7.44) 2(x − c) We therefore define the stress intensity factor for Mode II cracks by p √ KII = lim 2π (x − c)τxy (x, 0) = σII πc. (7.45) x↓c

As in Mode III cracks, we postulate that the crack tip will propagate if KII exceeds some critical value. We visualise the stress field in Figure 7.9 by plotting the contours of the


297

y 0.2

¡ ¢ 4 1 − ν 2 σII /E = 0 -1

0.1

-0.5

0.25

0.5

1

x

-0.1 -0.2

0.5 Fig. 7.10. The displacement of a Mode II crack of half-length c = 1 under increasing shear stress σII .

maximum shear stress sµ ¶ µ ¶ ¯ |τ1 − τ2 | ∂ 2 A 2 1 ∂ 2 A ∂ 2 A 2 ¯¯ 00 S= = + − = z¯f (z) + g 00 (z)¯ . 2 2 2 ∂x∂y 4 ∂x ∂y (7.46) As expected, the stress is concentrated near the crack tips and minimised along the crack faces. Finally, we can recover the displacement of the crack using ¡ ¢ 4 1 − ν2 f (z) on ∂Ω, (7.47) u + iv = E as shown in Exercise 7.4, and we use this to plot the deformation of the crack in Figure 7.10. The crack is deformed into an ellipse that grows and rotates as the applied shear stress is increased. An alternative and much quicker way to proceed is to pose the mode II crack problem as a half space problem in y > 0 as in 7.2.3(iii). The boundary conditions are clearly τyy = τxy = 0

on y = 0, x < |c|

(7.48)

and τyy = 0

on y = 0, x > |c|

(7.49)

but the second boundary condition on y = 0, x > |c| is less obvious. We can either argue on physical grounds that τxx = 0 or use symmetry arguments to say that ∂2A =0 ∂y 2

on y = 0, x > |c|.

(7.50)

298

Fracture and Contact Fig. 7.11. Schematic of a Mode I crack. T

We now follow (7.50) from p... to write† ψ = (0, ψ(x, y), 0) so that the displacements are given by µ ¶ ∂ 1 2µu = − φ + yψ , ∂x 2 µ ¶ ∂ 1 2µv = 2 (1 − ν) ψ − φ + yψ , ∂y 2 while, from (2.87), A satisfies ∂φ 1 ∂ ∂A = − (1 − ν) ψ + + (yψ) . ∂y ∂y 2 ∂y

(7.51)

Hence on y = 0, using the fact that φ is a harmonic function, µ ¶ ∂v ∂u ∂v ∂ψ ∂ 2 φ 0 = τyy = 2µ +λ + = (1 − ν) − 2. ∂y ∂y ∂x ∂y ∂y Since this holds for all x and since (1 − ν) ∂ψ/∂y − ∂ 2 φ/∂y 2 is a harmonic function, we conclude that ∂φ = (1 − ν) ψ ∂y in y > 0. This implies from (7.51) that A = 21 yψ. Finally, the remaining boundary conditions gives that ∂ψ (x, 0) = 0, ∂x

|x| < c

and ψ(x, 0) = 0,

|x| > c

with ψ ∼ −2σII x at infinity. So our problem effectively reduce to the harmonic conjugate of the mode III crack problem, giving np o ψ = −2σII Re z 2 − c2 and this in turn implies (7.40).


299

7.2.5 Mode I cracks We now move onto the configuration of a crack in plane strain under a tension σI , as shown in Figure 7.11. This is arguably a more realistic configuration than that of either Mode II or Mode III but, unfortunately, the solution turns out to be slightly more difficult to derive in this case. We follow the same approach as in §7.2.4, using an Airy stress function A that satisfies the biharmonic equation in the region Ω outside an elliptical stressfree crack, although the condition at infinity is now σI A ∼ x2 as x2 + y 2 → ∞. (7.52) 2 The calculation follows the steps used above in §7.2.4, and the details may be found in Exercise 7.7. This time the functions f and g take the form ´ p σI ³ f (z) = −z + 2 z 2 − c2 , (7.53a) 4 ³ ´ σ p c2 σI I g(z) = log z − z 2 − c2 + z 2 , (7.53b) 2 4 [ I think there are a couple of typos in g, see en of section] and the stress components are therefore ( ) 2z 3 + c2 (¯ z − 3z) τxx = −σI + σI Re 2 (z 2 − c2 )3/2 √ ¢ σI c ¡ √ ∼ (7.54a) 3 cos(θ1 /2) + cos(5θ1 /2) , 4 2r1 ( ) σI c2 z − z¯ τxy = Im 4 (z 2 − c2 )3/2 √ σI c √ cos(3θ1 /2) sin θ1 , ∼ (7.54b) 2 2r1 ( ) σI 2z 3 − c2 (z + z¯) τyy = Re 2 (z 2 − c2 )3/2 √ σI c ∼ √ cos3 (θ1 /2) (3 − 2 cos θ1 ) , (7.54c) 2r1 as r1 → 0. Ahead of the crack on θ1 = 0, the nonzero stress components are therefore √ σI c τxx , τyy ∼ √ as r1 → 0, (7.55) 2r1 † how on earth do you guess that one??

300

Fracture and Contact 2

S/σI 1 1 0.8

y

0.6

0

0.4 0.2

-1

0 -2 -2

-1

1

0

2

x Fig. 7.12. Contour plot of the maximum shear stress S around a Mode I crack, with crack length c = 1.

y 0.2

¡ ¢ 4 1 − ν 2 σI /E = 0 0.25

0.1

-1

-0.5

0.5

1

x

-0.1

0.5

-0.2

Fig. 7.13. The displacement of a Mode I crack of half-length c = 1 under increasing normal stress σI .

so we define the Mode I stress intensity factor as KI = lim x↓c

p

√ 2π (x − c)τyy (x, 0) = σI πc.

(7.56)

We can again use (7.46) and (7.47) to calculate the maximum shear stress and the displacement of the crack. As shown in Figure 7.12, the stress is again concentrated near the crack and minimised along the crack faces. Figure 7.13 shows that the crack again remains elliptical, growing wider and shorter as the applied normal stress increases.


301

7.2.6 We can solve our models for Mode II and Mode I crack more simply by posing them as half-space problems, as in §7.2.3 (iii) above. For the Mode II crack model, (7.21), (7.22), the boundary conditions are clearly τyy = τxy = 0

on y = 0, |x| < c

(7.57a)

and τyy = 0

on y = 0, |x| > c,

(7.57b)

but the second boundary condition on y = 0, |x| > c is less obvious. We can either argue on physical grounds that τxx = 0 or use symmetry to say that ∂2A =0 ∂y 2

on y = 0, |x| > c.

(7.58)

We now follow (2.238) to write Ψ = (0, ψ(x, y), 0), so that the displacements are given by µ ¶ 1 ∂ φ + yψ 2µu = − ∂x 2 µ ¶ ∂ 1 2µv = 2(1 − ν)ψ − φ + yψ , ∂y 2 and A satisfies ∂A ∂φ 1 = −(1 − ν)ψ + + (yψ), ∂y ∂y 2

(7.59)

Hence, on y = 0, τyy

µ ¶ ∂v ∂u ∂v +λ + = 2µ ∂y ∂x ∂y · ¸ λ + 2µ (1 − 2ν paψ ∂ 2 φ λ ∂2φ = − 2 − 2µ 2 ∂y ∂y 2µ ∂x2 · 2 ¸ ∂ φ ∂ψ = − 2 + (1 − 2ν) , ∂y ∂y

since φ is a harmonic function. Since this holds for all x and since (1 −

2ν) ∂ψ ∂y

∂2φ ∂y 2

−

is harmonic, we conclude that ∂φ ≡ (1 − 2ν)ψ ∂y

in y > 0.

(7.60)

302


This implies that 1 A = yψ. 2 Finally, the remaining boundary conditions give that ψ(x, 0) ∂ψ (x, 0) =), ∂y

=

0,

(7.61)

|x| < c

|x| > c.

So, since ψ ∼ 2σx, our problem has effectively been reduced to (7.20) and this in turn implies (7.40). When we adopt the Papkovich–Neuber approach to Mode I cracks, when 2 the boundary condition A ∼ σ12x at ∞, the second boundary condition 7.58 becomes τxy = v = 0 on y = 0, |x| > c. (7.62) ¡ ¢ ¡ ¢ ∂φ 3 1 This implies that ∂φ ∂y = 2 − 2ν and ∂y = 2 − ν ψ on y = 0, |x| > c, so that ψ =

∂φ ∂y

= 0 there. Meanwhile on y = 0, |x| < c, τyy = τxy = 0 so that ∂2φ ∂ψ = (1 − ν) 2 ∂y ∂y

and ∂φ = ∂y

µ

¶ 1 −ν ψ 2

there. Arguing as above, the harmonic function and (7.63) therefore implies that ∂ψ =0 ∂y

∂φ ∂y

−

¡1 2

¢ − ν ψ vanishes in y > 0

on y = 0, |x| < c.

(7.63)

Moreover (7.59) now gives that ∂A ∂y

µ ¶ 1 1 ∂ = −(1 − ν) − ψ+ (yψ) 2 2 ∂y 1 ∂ψ = y . 2 ∂y

With the condition that ψ be proportional to x at ∞, our problem for ψ is again essentially (7.20) so that p ∂A 1 = yRe z 2 − c2 (7.64) ∂y 2


303

and on integration with respect to y yields (7.53). When we adopt the Papkovich–Neuber description, we must assume this time that ¶T µ 1−ν −ν tx, ty + ψ(x, y), 0 (7.65) ψ= 1 − 2ν 1 − 2ν in order to properly model the displacement field. Otherwise, we would eventually obtain a hydrostatic stress distribution at infinity. With the inclusion of t in (7.65), the stress is given by ∂2A ∂ψ ∂ 2 φ y ∂ 2 ψ = t + ν + 2 + ∂y 2 ∂y ∂y 2 ∂y 2 2 2 ∂ A ∂ψ ∂ φ y ∂ 2 ψ = = (1 − ν) − 2 − ∂x2 ∂y ∂y 2 ∂y 2 µ ¶ ∂2A 1 ∂ψ ∂2φ y ∂2ψ = − = −ν − − . ∂x∂y 2 ∂x ∂x∂y 2 ∂x∂y

τxx =

(7.66)

τyy

(7.67)

τxy

(7.68)

Restricting our attention to the half-plane y > 0, we still have the stress-free boundary conditions (7.48) on the crack faces, while symmetry implies τxy = v = 0 on y = 0, |x| > c. (7.69) ¡1 ¢ Hence, from (7.68), 2 − ν ψ − ∂φ/∂y = 0 on y = 0 and, following the same argument as before, we deduce that µ ¶ 1 ∂φ = −ν ψ (7.70) ∂y 2 everywhere on y > 0. The remaining boundary conditions thus become τyy = 0, i.e. ∂ψ = 0, ∂y

on y = 0, |x| < c,

and v = 0, i.e. ψ = 0,

on y = 0, |x| > c.

Hence we have ψ = 2s Im

(7.71)

np o z 2 − c2 .

for some s. On the other hand, φ = Re{h(z)} for some analytic function h and, from (7.70), ¶ µ np o © 0 ª 1 − ν Im z 2 − c2 . − Im h (z) = 2s 2

304


Hence, µ h(z) = s

¶h ³ ´ i p p 1 − ν c2 log z + z 2 − c2 − z z 2 − c2 . 2

Meanwhile, we can integrate (7.66) twice with respect to y and get A = =

ty 2 y 1 + ψ− φ, 2 2 1 − 2ν np o s n p ³ ó p ty 2 + ys Im z 2 − c2 + Re z z 2 − c2 − c2 log z + z 2 − c2 . 2 2 2

At ∞, we have A ∼ σI2x , which yields s = −t = σI . Finally, we obtain ´ h io n ³ p p σI A= Re z¯ 2 z 2 − c2 − z + z 2 − 2c2 log z + z 2 − c2 . 4 We conclude this discussion with two remarks. First, we note that a general planar crack in two space dimensions wiil be a linear superposition of cracks of Modes I, II and III, and hence the stress near its tip will be characterised by three intensity factors; unfortunately there are no general rules about the factors to which it is most vulnerable. Second, the most famous three-dimensionalcrack configuration is the penny-shaped crack which is the radially symmetric version of our earlier Mode I crack and it can be solved explicitly, with a great deal of effort, by transforming to spheroidal coordinates and taking a limit as in §7.2.2. Both these configurations, along with many others, are lucidly described in [ref sneddon!!!].

7.2.7 Dynamic fracture All the discussion hitherto refers to models of cracks that are static or just about to grow. If, however, we wish to predict structural failure or the response of tectonic plates to earthquakes, for example, we need to be able to predict the speed and depth of penetration of a growing crack. Unfortunately this involves synthesising the theories of elastic wave propagation from Chapter 3 with those of mixed boundary value problems from this chapter, and this task is beyond the scope of this book (see Freund, 1989, for a helpful account). We restrict our attention to the simplest possible model for a semi-infinite Mode III crack propagating at constant speed −V along the x-axis, so its tip is at x = −V t. Although the global stress field away from the crack will be a complicated history-dependent function of the crack geometry, it is reasonable to assume that the local stress field may be approximated

7.3 Contact

305

Fig. 7.14. Schematic of an elastic string being forced against a rigid obstacle Γ.

by a travelling wave moving with speed −V , at least on sufficiently short timescales. Hence, the local displacement will satisfy ¶ µ V 2 ∂2w ∂2w + = 0, 1− 2 cs ∂ξ 2 ∂y 2

(7.72a)

everywhere except on y = 0, ξ > 0, with µ

∂w =0 ∂y

on y = 0, ξ > 0,

(7.72b)

where ξ = x + V t and cs is the shear wave speed. Notice that this problem is analogous to the static semi-infinite mode III crack problem considered in (7.3). [SAY MORE] To close the model (7.72), we should impose some conditions at infinity, but this would require us to solve for the stress away from the crack. Even without precise knowledge of this far field, we expect the physically relevant solution to have a square-root singularity analogous to (7.4) with n = 1. The corresponding solution of (7.72) is Ãr w = A Re

! ξ + iy , B

where B 2 = 1 −

V2 , c2s

(7.73)

and it is only through the constant A that the global stress field is felt. Now the stress intensity factor is KIII

µ ¶ µ ¶1/4 p ∂w √ V2 = µ lim ξ (ξ, 0) = A B = A 1 − 2 . ξ↑0 ∂y cs

(7.74)

Hence, provided A is bounded, as in the example ofExercise 7.8, we see that the stress intensity factor tends to zero as the tip speed tends to the shear wave speed. This is comforting because, unless cracks are boosted by internal pressures or other local driving mechanisms, they are always observed to propagate subsonically. Were crack tips to be forced to propagate supersonically, we would expect a shock wave of some sort to be generated at the tip, with quite a different stress concentration nearby as compared to the subsonic case. [MENTION SPLITTING]

306


7.3 Contact 7.3.1 Contact of elastic strings We start by considering the simplest elastic contact problem, namely an elastic string making steady contact under a prescribed body force p(x) against a smooth, nearly flat, rigid obstacle Γ, as shown in Figure 7.14. For future reference, we introduce the terminology contact set to denote the set of points at which the string makes contact with the obstacle, and noncontact set for those points where it does not. If the transverse displacement w(x) is assumed to be small, then a force balance on a small element of the membrane in the non-contact set yields the familiar equation d2 w = p(x), (7.75) dx2 where the membrane tension T is spatially uniform. Provided friction is negligible, T is also constant throughout the contact set, where w is simply equal to the prescribed obstacle height f (x). Now we must consider how to join the solutions in the contact and noncontact sets, which leads us to the question of the smoothness of the solution at points where contact is lost, a question we also had to address in our discussion of fracture. Here this problem is resolved immediately by a local force balance, as shown in Exercise 7.9, which reveals that T and T dw/dx must both be continuous. Thus, our task is to solve (7.75) in the non-contact set, subject to specified boundary conditions at the ends of the string and T

dw df = (7.76) dx dx at the points where the membrane meets the obstacle. Let us illustrate the procedure for a string whose ends x = ±1 are fixed a unit distance above a flat surface z = f (x) = 0. If the applied pressure p is spatially uniform, then we can solve (7.75), subject to the boundary conditions w(−1) = w(1) = 1, to obtain ¢ p ¡ w =1− 1 − x2 . (7.77) 2T As the applied pressure p is increased, the downward displacement at the centre of the string increases, until it first makes contact with z = 0 when p = 2T . If the pressure is increased further, then a contact set forms near the middle of the string. Let us denote the boundaries of this region by x = ±s, where s is to be determined as part of the solution. In this simple problem, we only need to solve for positive x; the solution in x < 0 may then be w = f,

7.3 Contact

307

1 0.8 0.6 0.4 0.2 -1

-0.5

0.5

1

Fig. 7.15. Solution for the contact between a string and a level surface, with applied pressure p/T = 0, 1, 2, 3, 4, 5. tidy up figure

inferred by symmetry. We therefore solve (7.75) in x > s, subject to w(1) = 1 and the continuity conditions dw (s) = 0 dx

(7.78)

p (1 − x)(1 + x − 2s) 2T

(7.79)

w(s) = to obtain w =1− in x > s, where

s s=1−

2T . p

(7.80)

In Figure 7.15 we see how, as increasing pressure is applied to the string, it sags downward, makes contact with the surface beneath, and the contact set then gradually spreads outwards. This simple example shows how the solution of the contact problem hinges on locating the so-called free boundary between the contact and non-contact sets. In other words, the domain in which the field equation (7.75) is to be solved is unknown in advance and must be found as part of the solution. Such problems, known as free boundary problems, are inevitably nonlinear even when, as in (7.75), the governing equation is itself linear. This means that the solutions of contact problems can exhibit phenomena such as nonuniqueness that are impossible for linear problems. To illustrate some of the possible pitfalls, consider now a string, subject to zero body force, stretched over the curved surface z = f (x) = 2x2 − 3x4

(7.81)

308

Fracture and Contact (a)

-1

(b)

-0.5

(c)

0.3

0.3

0.3

0.25

0.25

0.25

0.2

0.2

0.2

0.15

0.15

0.15

0.1

0.1

0.1

0.05

0.05

0.05

0.5

1

-1

-0.5

0.5

1

-1

-0.5

0.5

1

Fig. 7.16. Three candidate solutions for the contact problem between a stress-free string and the surface z = 2x2 − 3x4 . tidy up figure

with its ends fixed at w(−1) = w(1) = 0. With p = 0, (7.75) implies that the string must be straight wherever it is out of contact. The most obvious linear function satisfying both boundary conditions is simply w ≡ 0 but, as can be seen in Figure 7.16(a), this would imply that the string crosses the obstacle, which is impossible. We must therefore allow the string to make contact with the obstacle, as illustrated in Figure 7.16(b). Recalling that the string must meet the obstacle tangentially, the solution in this case is easily constructed as  8 2   (1 + x) −1 6 x 6 − ,   9 3    2 2 w = 2x2 − 3x4 − 6 x 6 , (7.82)  3 3     2   8 (1 − x) 6 x 6 1. 9 3 We can, however, construct a third solution, shown in Figure 7.16(c), in which the string loses contact with the middle of the obstacle. Again noting that the string is linear when out of contact and must always make contact tangentially, we find that this solution takes the form  8 1  (1 + x) −1 6 x 6 − √ ,    9 3     2 1   2x2 − 3x4 − 6 x 6 − √ ,   3  3   1 1 1 w= (7.83) −√ 6 x 6 √ ,  3 3 3      2x2 − 3x4 √1 6 x 6 2 ,   3  3    8 2   (1 − x) 6 x 6 1. 9 3 We would like to have a mathematical formulation that chooses between possible candidate solutions like those shown in Figure 7.16. Figure 7.16(a)

7.3 Contact

309

is clearly unacceptable because it involves penetration of the obstacle by the string, and this can be avoided by incorporating the requirement w > f into our model. Although it is less obvious, Figure 7.16(b) is also physically unrealistic, since there is a tensile normal stress between the string and the obstacle. In general, if R(x) denotes the normal reaction force exerted on the string by the obstacle, then (7.75) is modified to T

d2 w = p − R. dx2

(7.84)

The reaction force R must be zero outside the contact set. The solution shown in Figure 7.16(b) has negative values of R and is therefore unphysical; the correct solution is therefore (7.83). To summarise, we can select a physically relevant solution of the contact problem by insisting that w = f, w > f,

d2 w −p f, T 2 − p 6 0, (7.86) dx dx and this so-called linear complementarity problem can be written in another way which is very convenient computationally. If we define the inner product and if w satisfies µ ¶ Z 1 Z 1 dw dv dw T − dx 6 p(v − w) dx for all v > f, (7.87) dx dx −1 dx −1 then it may be shown that w satisfies (7.75) and (7.76). In addition, (7.87) legislates against situations such as Figure 7.16(b), as shown in Exercise 7.10. Equation (7.87) is called a variational inequality, and it can be shown that w satisfies (7.87) if and only if it minimises the functional Z 1 µ ¶2 T dv min + pv dx. (7.88) v>f −1 2 dx This has the obvious physical interpretation of minimising the elastic energy minus the work done by pressure over the virtual displacements v which prevent interpenetration of the string and the obstacle.

310

Fracture and Contact (a)

-1

(b)

w

(c)

w

w

1

1

1

0.8

0.8

0.8

0.6

0.6

0.6

0.4

0.4

0.4

0.2

0.2

-0.5

0.5

1

-1

x

-0.5

0.2 0.5

1

-1

-0.5

x

0.5

1

x

Fig. 7.17. The contact between a beam and a horizontal surface under a uniform pressure p. (a) The beam sags towards the surface and makes contact: p/EI = 0, 12/5, 24/5. (b) The beam makes contact at just one point: p/EI = 24/5, 24. (c) The contact set grows: p/EI = 24, 2048/27, 384.

7.3.2 Other thin solids It is straightforward to extend the model derived above to describe smooth contact between an elastic beam and a rigid substrate. In the non-contact set, the displacement w of the beam is governed by the equation d2 w d4 w − EI −p=0 (7.89) dx2 dx4 where, as in Chapter 4, EI denotes the bending stiffness. Evidently (7.89) reduces to (7.75) as the bending stiffness tends to zero. In the contact set, (7.89) is simply replaced by the condition w = f . Since (7.89) is fourth-order, we expect to need additional boundary conditions to specify a unique solution. Now, by balancing forces and moments at the boundary between the contact and non-contact sets, we deduce that w, dw/dx and d2 w/dx2 must all be continuous there. Let us illustrate the solution procedure in this case by considering a beam that is simply supported at its two ends x = ±1 at a unit height above the horizontal surface z = 0. This is the equivalent for a beam of the string sagging problem analysed in §7.3.1, and we will see that the inclusion of bending stiffness leads to interesting new behaviour. Assuming no tension is applied, the beam equation (7.89) reduces to T

d4 w p =− , dx4 EI subject to the boundary conditions w(−1) = 1, w(1) = 1,

(7.90)

d2 w (−1) = 0, dx2 d2 w (1) = 0, dx2

(7.91a) (7.91b)

7.3 Contact

311

and this problem is easily solved to give the displacement ¢¡ ¢ p ¡ w =1− 1 − x2 5 − x2 . (7.92) 24EI As shown in Figure 7.17(a), the beam sags under increasing applied pressure until it first makes contact with the surface z = 0 when p/EI = 24/5. The solution (7.92) cannot remain valid for p/EI > 24/5, because it predicts negative values of w. So, let us instead seek a solution in which the beam makes contact with z = 0 over a region −s 6 x 6 s. We need only solve the problem in x > 0, since we can then infer the solution for negative values of x by symmetry. We therefore replace (7.91a) with the free boundary conditions dw d2 w (s) = 0, (s) = 0. (7.91c) dx dx2 Notice again that, compared with (7.91a), we have introduced one more boundary condition to compensate for not knowing the location of the free boundary a priori. Again, the problem is easy to solve and we find that w(s) = 0,

w=

(x − s)3 (2 − s − x) (1 − s)4

(7.93a)

in x > s, with w ≡ 0 in 0 6 x 6 s, where s is related to the applied pressure by ¶ µ p 24 24EI 1/4 = or s = 1 − . (7.93b) EI (1 − s)4 p In Figure 7.17(c) we show how the contact set grows as p increases, with s = 0, 1/4, 1/2 when p/EI = 24, 2048/27, 384 respectively. However, since s must lie between 0 and 1, the solution (7.93) is only valid when p/EI > 24. This begs the question: ”what happens for values of p/EI between 24/5 and 24?” The answer is that the beam makes contact at just one point, namely the origin. If we replace (7.91a) with w(0) = 0,

dw (0) = 0, dx

(7.91d)

then we find x2 (3 − x) p − x2 (1 − x)(3 − 2x), (7.94) 2 48EI in x > 0, with the even extension of (7.94) in x 6 0. As shown in Figure 7.17(b), the beam initially meets z = 0 at one point, then flattens as p increases, until its curvature matches the cuvature of the obstacle. Only w=

312


then does the contact set start to grow. It is easy to verify that (7.94) matches (7.92) and (7.93) when p/EI = 24/5 and 24 respectively. The final surprise comes when we consider the reaction force R exerted by the surface z = 0 on the beam, given by R = p + EI

d4 w . dx4

(7.95)

Consider first the solution (7.94) in which the contact set is the point x = 0. By differentiating (7.94) and using the symmetry about x = 0, we find that d3 w d3 w 5p (0+) = − (0−) = − 3. 3 3 dx dx 8EI

(7.96)

Hence the third derivative of w is discontinuous across x = 0 and (7.95) then implies that there is a delta function in the reaction force: µ ¶ 5p R= − 6EI δ(x). (7.97) 4 Similarly, differentiation of (7.93) reveals that µ ¶1/4 ³ d3 w p ´3/4 12 3 (s+) = = , dx3 (1 − s)3 2 EI

(7.98)

so, again, there is a delta function in the reaction force at x = s with, by symmetry, an equal delta function at x = −s: ¶1/4 µ ¡ ¢ 3EIp3 δ(x − s) + δ(x + s) (x 6 s). (7.99) R=p+ 2 We can readily verify that (7.99) agrees with (7.97) when p/EI = 24. These point forces acting at the boundary of the contact set do not occur in the contact of strings. They are a consequence of the idealisations inherent in beam theory. If we were to focus on a sufficiently small region near the boundary of the contact set, then beam theory would no longer be valid and we would need to consider contact between a linear elastic solid and a rigid boundary. As shown below in §7.3.3, this gives rise to a reaction force that, although concentrated near the edge of the contact set, is finite. We can obtain a variational formulation of the contact problem for a beam by incorporating the elastic bending energy into the integrand of the minimisation problem (7.88), that is ) ¶ µ ¶2 Z 1( µ T dw 2 EI d2 w min + + pw dx. (7.100) w>f −1 2 dx 2 dx2

7.3 Contact

313

It is easy to show that (7.89) is the Euler–Lagrange equation associated with (7.100) in the absence of any contact. To generalise the above theories into three dimensions, we now consider an elastic membrane with transverse displacement w(x, y) making contact with a smooth obstacle z = f (x, y) under an applied body force p(x, y). Clearly the two-dimensional analogue of (7.75) is Poisson’s equation T ∇2 w = p(x, y),

(7.101)

where T is the uniform tension in the membrane. A force balance at the boundary of the contact set now reveals that w and its normal derivative must be continuous there, that is w = f,

∂w ∂f = ∂n ∂n

on the boundary of the contact set.

(7.102)

In other words, w and ∇w must be continuous everywhere. The additional dimension makes analytical progress difficult except in simple cases such as axisymmetric problems (see Exercise 7.13. However, the numerical solution is straightforward in principle using a variational formulation analogous to (7.88), that is ¶ ZZ µ T 2 |∇w| + pw dxdy, (7.103) min w>f 2 D and we can additionally incorporate bending stiffness as follows: ¶ ZZ µ T EI ¡ 2 ¢2 2 min |∇w| + ∇ w + pw dxdy. w>f 2 2 D

(7.104)

7.3.3 Smooth contact in plane strain As an intermediate case between the thin contact problems described above and general three-dimensional elastic contact, we will now make some elementary observations about contact in plane strain. For simplicity, we start by considering the problem of a rigid body, known as a punch, pushed a distance δ into the elastic half-space y > 0, as shown in Figure 7.18. Let us suppose the geometry is symmetric about x = 0 so that the contact set is in the range −c < x < c for some positive c. Inside this region, the normal displacement is given in terms of the shape f (x) of the punch and, assuming smooth contact, the tangential stress is zero. In addition, there must be a positive reaction force, so that v = δ − f (x),

τxy = 0

τyy < 0

on y = 0, |x| < c.

(7.105a)

314

Fracture and Contact y

f (x)

x

punch

c

δ

Fig. 7.18. Definition sketch for contact between a rigid body and an elastic halfspace.

Outside the contact set, the surface traction is zero and there must be no interpenetration, that is τxy = τyy = 0

v < δ − f (x)

on y = 0, |x| > c.

(7.105b)

Our task, then, is to solve plane strain subject to the mixed boundary conditions (7.105) on y = 0 and given behaviour, typically zero stress, at infinity. Such problems are almost always so complicated that they must be solved computationally but, enlightened by (7.88), we can see that this is not as fearsome a task as might be supposed. One must devise a code that minimises the integral ZZ W dxdy, y>0

where W is the strain energy density defined in §1.9, over all displacement fields satisfying the constraint v 6 f (x) − δ. The proof of the equivalence of this proposed algorithm to finding a displacement field satisfying the Navier equation and the boundary conditions (7.105) is beyond the scope of this book. However, as noted at the end of §7.3.1, it relies crucially on the fact that the boundary conditions together with the inequalities in (7.105) determine a unique solution with bounded stress everywhere. One cunning way to construct analytic solutions of the contact problem described above is to suppose that we know in advance both the size c of the contact set and the contact pressure, say p(x), exerted by the punch.

7.3 Contact

315

v0 1 -10

-5

5

10

x

-1 -2 -3 -4

Fig. 7.19. The penetration of a quadratic punch into an elastic half-space y > 0.

We thus replace the boundary conditions (7.105) with τxy = 0,

τyy = −P (x)

where

( P (x) =

on y = 0,

p(x)

|x| < c,

0

|x| > c.

(7.106)

(7.107)

We can now use the solution obtained in §2.6.8 to calculate the displacement v0 (x) of the boundary y = 0, given by Z 1−ν c v0 (x) = − p(s) log |s − x| ds. (7.108) πµ −c In |x| < c, this allows us to recover the shape of the punch that corresponds to the assumed pressure profile p(x). It only remains to check for consistency that there is no interpenetration in |x| > c. As an example (guided by hindsight), let us consider the pressure profile p (7.109) p(x) = c2 − x2 . By calculating the integral in (7.108) for this particular choice, we find that the displacement of the boundary is given by  2h ³ c í x2 c |x| < c, µv0 (x)  4 1 − 2 log 2 − 2 √ = h ³ í −1 2 2 2 2 2  1−ν  c 1 − 2 log c − x − |x| x − c + c cosh (x/c) |x| > c. 4 2 2 (7.110) The suggested pressure profile (7.109) therefore corresponds to a quadratic

316


punch, as illustrated in Figure 7.19. We note that the displacement is not bounded at infinity, instead growing like v0 (x) ∼ −

(1 − ν)c2 log |x| 2µ

as |x| → ∞.

(7.111)

This reflects the weakness in the theory of plane strain noted in §2.6.8 and the same difficulty does not occur for more realistic three-dimensional configurations (Exercise 7.16). We can observe in Figure 7.19 that both v0 and its first derivative are continuous across x = ±c. Indeed, it is readily verified from (7.110) that the thickness of the void between the punch and the elastic body varies like √ ³ c í x2 µv0 (x) c2 h 2 2c − 1 − 2 log + ∼ (x − c)3/2 (7.112) 1−ν 4 2 2 3 as x ↓ c. In contrast, we recall that, in the theory of brittle fracture, the √ crack thickness typically varies as the square root x − c of the distance from the crack tip. Recall also that the stress grows like (x − c)−1/2 as a crack tip is approached, while the above example, by construction, has √ √ p ∼ 2c c − x as x ↑ c. These observations prompt the question: why should the stress singularity in the smooth contact problem be weaker than that in the fracture problem? The answer lies in the fact that the crack geometry is prescribed a priori ; it is perfectly sensible to seek the stress distributions around a crack with arbitrary faces and an arbitrary crack tip surrounding these faces. However the boundary of a smooth contact set is free to decide its location, and it does so in such a way as to prevent the stress intensification that is characteristic of fracture.

7.4 Conclusions The role of friction is of great importance in most real contact problems, but thus far we have not mentioned it. The is because of the serious mathematical complications to which it leads. Suppose for example we introduce frictional effects into the smooth contact problems of §7.3. If we assume Coulomb friction in which the shear stress τnn multiplied by the coefficient of friction, µ, then we immediately introduce a second component into the free boundary; one component separates the non-contact set from the contact set, and the other divides the contact set into regions where τns | < µτnn and |τns | = µτnn respectively. In the former there is no slip, so the displacement is continuous; but in the latter, slip is possible, and only the normal

EXERCISES

317

displacement is continuous. This considerable extra complexity destroys any hope of applying the standard theory of variational inequalities. Unfortunately, theoretical challenges of this sort have to be configured of we are even addressing the apparently simple problem of finding the forces exerted on the ends of a ladder resting against a wall. This is a nontrivial problem unless limiting friction is assumed at both ends of the ladder. As is usual in friction contact, without this commonly-made assumption, the reaction forces are indeterminate, and the only way forward is to consider the geometry of the ends of the ladder in detail, and solve a frictional contact problem along the lines just described. The ladder will be ready to slip when the slipping subset of the contact region is found to occupy almost the whole of the contact region. Strictly speaking, the ladder problem should be solved as an evolution contact problem starting from the time at which the ladder was erected. However, such dynamic contact problems are of greatest applicability in the theory of impact, where predictions are to be made of, say, coefficients of restitution between various different solids. To carry this out, even in the nonfrictional case, requires a combination of the modelling in this chapter and of the elastic waves in Chapter 3. One frictional contact problem of great practical importance which can be solved exactly is the capstan problem, in which a rope is wound around a rough circular cylinder. In polar coordinates, we suppose that the tension is T (θ) and that the cylinder is r = a. Making the assumption of limiting Coulomb friction everywhere, (7.75) shows that the normal reaction of the capstan is T (θ)/a, using which, a force balance equation along the rope gives dT dθ = µT . The resulting exponential dependence of T on θ explains the effectiveness of capstans. Moreover, this kind of configuration can be used as a building block in a theory of knots (real, not mathematical ones!).

Exercises 7.1 For the elliptical Mode III crack solution (7.9), show that the stress components on y = 0, x > c cosh ² are given by µ ¶ x cosh ² ² τxz = 0, τyz = 2τ0 e √ − sinh ² . x2 − c2 Deduce that the stress at the crack tip is τyz = 2τ0 e² cosech ² ∼

2τ0 ²

as ² → 0.

318


Show also that the radius of curvature r0 of the tip is given by 1 cosh ² 1 = ∼ 2 2 r0 ² c c sinh ² and hence that the shear stress at the tip is approximately r c τyz ∼ 2τ0 r0 when r0 is small. 7.2 Calculation of stress for a Mode III crack with cohesion, with ( 0 on y = 0, |x| < c − δ, ∂w = µ ∂y C on y = 0, c − δ < |x| < c. 7.3 Derive the expression (7.18) for the displacement field as a distribution of dislocations along a Mode III crack. Derive singular integral equation for dislocation distribution f (ξ). Show that solution is consistent with (7.18). 7.4 Show that the displacement components and the Airy stress function in plane strain are related by µ ¶ ∂u ∂v 2(λ + µ) + = ∇2 A, ∂x ∂y and deduce that there exists a function ψ(x, y) such that u=

∂A ∂ψ 1 + 2(λ + µ) ∂x ∂y

v=

1 ∂A ∂ψ − . 2(λ + µ) ∂y ∂x

By eliminating u and v from the constitutive relations, show that A and ψ satisfy ∂2A ∂2A 4µ(λ + µ) ∂ 2 ψ , − = − ∂x2 ∂y 2 λ + 2µ ∂x∂y µ ¶ ∂2A 2µ(λ + µ) ∂ 2 ψ ∂ 2 ψ 2 = − . ∂x∂y λ + 2µ ∂x2 ∂y 2 Deduce that ∂2 ∂ z¯2

µ ¶ 2µ(λ + µ) A− iψ = 0, λ + 2µ

where z = x + iy and z¯ = x − iy, and hence that A−

2µ(λ + µ) iψ = z¯f (z) + g(z), λ + 2µ

where f and g are arbitrary functions.

EXERCISES

319

Hence show that the displacement components are given in terms of f and g by u + iv =

¢ 1 ¡ ¯0 λ + 3µ f (z) − z f (¯ z ) + g¯0 (¯ z) . 2µ(λ + µ) 2µ

Recalling the boundary condition (7.25) satisfied by f and g on a stress-free boundary, show that the displacement of such a boundary is given by ¡ ¢ 4 1 − ν2 u + iv = f (z), E where E and ν are the Young’s modulus and Poisson ratio. 7.5 Recalling Cauchy’s relation for any analytic function f (z) = u(x, y)+ iv(x, y): ∂u ∂v = ∂x ∂y

and

∂u ∂v =− , ∂y ∂x

Show that ∂2A ∂x2 ∂2A ∂y 2 ∂2A ∂x∂y

© ª = Re z¯f 00 (z) + 2f 0 (z) + g 00 (z) , © ª = Re −¯ z f 00 (z) + 2f 0 (z) − g 00 (z) , © ª = Im z¯f 00 (z) + g 00 (z) .

7.6 Show that the 2D Papkovitch–Neuber potential φ(x, y), ψ = (ψ1 (x, y), ψ2 (x, y), 0) generates the stress field ∂ψ1 ∂ψ2 ∂ 2 φ x ∂ 2 ψ1 y ∂ 2 ψ2 +ν − − − , ∂x ∂y ∂x2 2 ∂x2 2 ∂x2 ∂ψ1 ∂ 2 φ x ∂ 2 ψ1 y ∂ 2 ψ2 ∂ψ2 +ν − 2 − = (1 − ν) − , ∂y ∂x ∂y 2 ∂y 2 2 ∂y 2 µ ¶µ ¶ 1 ∂ψ1 ∂ψ2 ∂2φ x ∂ 2 ψ1 y ∂ 2 ψ2 = −ν + − − − . 2 ∂y ∂x ∂x∂y 2 ∂x∂y 2 ∂x∂y

τxx = (1 − ν) τyy τxy

7.7 Apply the method of conformal mapping to mode I crack. Show that the far-field traction corresponds to f ∼ σI z/4 and g ∼ −σI z 2 /4 and that (7.34) becomes ¡ ¢ 2 G0 (ζ) + ζ a + bζ 2 F 0 (ζ) ζ σ b σI b σI b σ b I I + = + + . −F¯ (1/ζ) + 2 4ζ 2ζ aζ − b 4ζ 2ζ

T

320


w 1 0.8 0.6

Increasing M/EI

0.4 0.2 0.2

0.4

0.6

0.8

1

x

Fig. 7.20. A flexible ruler being flattened against a table with applied bending moment M/EI = 0, 3, 6, 9, 12, 15.

7.8 7.9 7.10 7.11 7.12

[hum, it doesn’t quite yield the expected answer if I set both sides to zero...] Example of dynamic fracture (Tectonophysics 1997). Local force balance near contact point for (i) string, (ii) beam. Show that variational formulation (7.87) works. Minimisation problem (7.88) Consider the problem of a flexible ruler of unit length being pressed against a horizontal table, as shown in Figure 7.20. Suppose the end x = 0 is simply supported by the table with w = 0, while a bending moment M is applied at the other end, x = 1, which is held at a height w = 1 above the table. Show that the vertical displacement of the ruler is given by w =x−

¢ M ¡ x 1 − x2 , 6EI

for M/EI 6 6. Show also that, when M/EI > 6, the pruler makes contact with the table over a region of length s = 1 − 6EI/M . 7.13 Consider an elastic membrane stretched to a tension T over the horizontal table z = 0. The membrane is fixed with w = 1 at the circular boundary r = 1, where r2 = x2 +y 2 , and sags under a uniform applied pressure p. Show that, as p is increased from zero, the membrane first makes contact with the table when p = 4T . Show also that, when p > 4T , the contact set is given by r < s, [contact set open or closed??] where 4T = 1 − s2 + 2s2 log s. p

EXERCISES

321

w

w 0.01

0.035

V

0.03

0.008

0.025 0.006 0.02 0.004

0.015 0.01

0.002 0.005 -1.5

-1

-0.5

0.5

1

η 1.5

-1.5

-1

-0.5

0.5

1

η

1.5

2`

Fig. 7.21. (a) A wave travelling along a rope on the ground. (b) Some higher travelling modes.

7.14 A rope with line density % and bending stiffness EI lies on the ground. By flicking one end, it is possible to send a wave of length 2` travelling along the rope at speed V , as illustrated in Figure 7.21(a). Defining the travelling wave coordinate η = x − V t, show that the small transverse displacement w(η) of the rope satisfies EI

2 d4 w 2d w + %V + %g = 0, dη 4 dη 2

subject to the contact conditions w=

d2 w dw = = 0, dη dη 2

at η = ±`.

Hence compute the shape of the hump in the rope and show that the speed is given by s k EI V = , b % where k satisfies the transcendental equation tan k = k. [This equation has a countably infinite set of solutions; the displacement corresponding to the first is shown in Figure 7.21(a) and the next three are shown in Figure 7.21(b).] Deduce that flicking one end of the rope up and down over a time τ will produce a wave travelling at speed ¶ µ EI 1/4 . V ≈3 %τ 2 7.15 von Karman plate minimisation problem.

322


7.16 Axisymmetric punch.

8 Plasticity

8.1 Introduction It is unfortunate that what we called plastics are polymeric solids that usually behave elastically at room temperature and viscoelastically when hot; hence the word plasticity should not be thought of as applying just to plastics. Indeed, some of the most important examples of plastic behaviour occur in metals. A convenient way to gain some intuition about this phenomenon is to apply an increasing bending moment to a paper clip. There is always a critical bending moment above which the clip fails to revert to its initial state, which means that it has become inelastic. This observation is encapsulated in Figure 8.1, which shows the qualitative stress/strain curve for a metal that yields at a stress τY . If a stress lower than τY is applied, the material responds elastically (although possibly nonlinearly), returning to its original state when the loading is removed. However, when either a

(a)

(b)

stress

stress

τY

τY

loading

unloading

strain

strain

Fig. 8.1. Schematic of a typical stress-strain relationship for a plastic material: (a) below the yield stress τY ; (b) when the yield stress is exceeded.

323

324

Plasticity

(a)

un

strain

lo loa adin din g g

τY lo loa adin din g g

τY

(b)

stress

un

stress

strain

Fig. 8.2. Schematic of the stress-strain relationship for a perfectly plastic material: (a) under a single loading; (b) under periodic loading/unloading.

tensile or compressive stress greater than τY is applied and then removed, a nonzero permanent strain remains. Macroscopic plasticity modelling thus poses two key challenges. The first is to explain the existence of a critical yield stress, above which a solid ceases to behave elastically, and the second is to predict the irreversible behaviour that occurs at stresses equal to or greater than this critical value. The part of the loading curve in Figure 8.1(b) above the dotted line is where the material is said to be undergoing work hardening. A limiting case of the behaviour depicted in Figure 8.1, which has proved extremely useful in practical models of plastic behaviour, is known as perfect plasticity. In perfectly plastic theories, the stress is never allowed to exceed the yield stress, and the material can thus exist in one of two distinct states. Below the yield stress, it behaves as an elastic solid and is described by the models described previously in this book; at and only at the yield stress, the material becomes plastic and starts to flow irreversibly. Hence the stressstrain relationship of a perfectly plastic material is sketched in Figure 8.2(a): we can view this is as an idealised version of Figure 8.1(a). We will see below that such models arise naturally in describing granular materials and metal plasticity. It is natural to use the word “flow” to describe the permanent deformation that occurs at the yield stress, although, in Chapter 1, we specifically stated that a solid does not flow in response to a stress. A yielding plastic material therefore ceases to be a solid according to the definitions given in Chapter 1. The situation is further complicated by the existence of viscoelastic materials which behave like elastic solids in some situations and like viscous liquids in others, as we will describe in §9.2. In fact, there exist materials with ever more complicated mechanical properties that may be characterised as “elastic-plastic”, “visco-plastic”, “elasto-visco-plastic” and so forth. The

8.2 Models for granular material

325

y R F

11111111111111 00000000000000 00000000000000 11111111111111 00000000000000 11111111111111 00000000000000 11111111111111 W 00000000000000 11111111111111 θ 00000000000000 11111111111111

x

Fig. 8.3. Schematic showing the forces acting on a particle at the surface of a granular material.

mathematical models that emerge in each case may be quite different from each other, and they all differ markedly from those used in classical fluid mechanics in one crucial respect. While the behaviour of a classical inviscid or Newtonian fluid is determined completely by its instantaneous velocity field, plastic and viscoelastic materials all exhibit history-dependence, in which the behaviour at any time depends on the entire strain history that it has so far experienced. This can be anticipated by returning to your paper clip and cycling the stress so that it loads and unloads above and below τY until it eventually fractures; a typical stress-strain trajectory is shown schematically in Figure 8.2(b). We will spend most of this Chapter describing plastic behaviour in metals. However, we will see that the small scale behaviour underlying Figure 8.2 can only be discerned using very high magnification microscopes. We therefore start with a brief discussion plastic flow in granular solids, whose small scale behaviour is more familiar from everyday experience.

8.2 Models for granular material 8.2.1 Static behaviour The mechanics of granular solids is of wide importance, for example in the transport of powders, foodstuffs and coal and in understanding the behaviour of sand dunes. Let us first consider a single particle at the surface of a granular solid, for example a pile of sugar on a spoon. In addition to its own weight W , the particle experiences a normal reaction force R and tangential frictional force F exerted by the pile underneath, as illustrated in Figure 8.3. Coulomb’s law of friction is an empirical law which states

326

Plasticity

that |F | 6 R tan φ, where the material parameter φ is called the angle of friction and we recall from Chapter 7 that R must be non-negative. Indeed, it is an everyday observation that dry granular material cannot withstand any tensile stress (in the absence of any additional physical effects, such as moisture-induced cohesion or electrostatic charges). In addition, Coulomb’s law implies that the particle is held in place by friction if |F | < R tan φ and can slide only when |F | = R tan φ. A force balance on the particle illustrated in Figure 8.3 gives R = W sin θ and F = W sin θ, where θ is the angle that the surface makes with the horizontal. From Coulomb’s law, we therefore deduce that the surface slope is bounded by tan φ and that slippage will occur when θ = φ. This indeed agrees with experimental observations that granular materials possess an “angle of repose”, that is a maximum surface slope which can be sustained without the particles giving way and flowing via a series of avalanches.† It is also characteristic of the perfectly plastic behaviour described in §8.1. When the material is piled maximally, it follows that the surface slope must equal tan φ everywhere, and the pile height z = h(x, y) therefore satisfies the equation µ ¶2 µ ¶2 ∂h ∂h 2 |∇h| = + = tan2 φ. (8.1) ∂x ∂y This is called the eikonal equation for h and it arises in many other contexts, notably in geometrical optics (see ock...). Single-valued solutions of (8.1) predict shapes that are highly reminiscent of our everyday experience of piles of granular materials (for example sugar, sand, etc.) and can be verified to a high degree of accuracy in the laboratory; see Exercise 8.1.

8.2.2 Granular flow This gives us confidence that Coulomb’s law provides a good model for a maximally piled granular material, under static conditions at least. To model a flowing granular material, we must apply Coulomb’s law to particles in the interior, and it is much easier to implement this when the geometry is two-dimensional. Let us then consider the forces acting on a two-dimensional surface element inside a granular medium whose unit normal is n = (cos θ, sin θ)T , as depicted in Figure 8.4. Assuming that the internal stress can be described using a stress tensor τ , the normal traction † In practice the angle of repose may depend on other factors such as the shape of the particles and thus differ slightly from φ.????


y F

327

N

1111111111 0000000000 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 θ 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111

x

Fig. 8.4. Schematic of the normal force N and frictional force F acting on a surface element inside a granular material.

is given by µ ¶µ ¶ τxx τxy cos θ N = (cos θ, sin θ) τxy τyy sin θ 1 1 = (τxx + τyy ) + (τxx − τyy ) cos(2θ) + τxy sin(2θ). (8.2) 2 2 As pointed out in §8.2.1, we expect the particles to exert a compressive force on each other, but never a tensile one, and N must therefore be nonpositive, for all choices of the angle θ. It follows that none of the principal stress components (i.e. the eigenvalues of τ ) can be positive. Similarly, the tangential (frictional) stress on our surface element is given by µ ¶µ ¶ τxx τxy cos θ F = (− sin θ, cos θ) τxy τyy sin θ 1 = (τyy − τxx ) sin(2θ) + τxy cos(2θ). (8.3) 2 Now Coulomb’s law implies that |F | must be bounded by N tan φ, for all θ. In addition, flow can occur only if there is some value of θ for which |F | is equal to N tan φ. If so, this direction defines a slip surface along which we expect flow to occur. From (8.2) and (8.3), we see that, as θ varies, the tractions lie on the so-called Mohr Circle in the (N, F ) plane, given by ¶2 µ (τxx − τyy )2 1 2 2 + τxy , (8.4) F + N − (τxx + τyy ) = 2 4

328

Plasticity

F q

1 4 (τxx

2 − τyy )2 + τxy

φ

−

N

τxx + τyy 2

Fig. 8.5. The Mohr circle in the (N, F )-plane, and the lines where |F | = |N | tan φ.

where different points on the circle correspond to different choices of the angle θ. Since we require N to be non-positive for all θ, the Mohr circle must lie in the half-plane N 6 0, as shown in Figure 8.5. The Coulomb criterion then tells us that |F | 6 |N | tan φ for all θ, and the Mohr circle must therefore lie between the lines N tan φ 6 F 6 −N tan φ. Finally, if the material is flowing, then there must be one value of θ such that |F | = |N | tan φ, and the Mohr circle must therefore be tangent to the lines |F | = |N | tan φ, as shown in Figure 8.5. Elementary trigonometry now tells us that the stress components in a granular material must satisfy ¡ ¢ 2 1/2 2 τxx τyy − τxy 6 −(τxx + τyy ) cos φ,

(8.5)

with equality when the material is flowing. Notice that (8.5) is a relation between the two stress invariants Tr (τ ) and det (τ ). This is reassuring, since it implies that the condition for the material to yield is independent of our choice of coordinate system. We can only make further analytical progress at all easily if we assume that the flow is slow enough for the inertia of the particles to be negligible in comparison with the frictional forces between them and, perhaps, gravity. This assumption is often valid in practice, even in coal hoppers, and allows us to neglect the acceleration term in Cauchy’s momentum equation, which


329

thus reduces to ∂τij + ρgi = 0. ∂xj

(8.6)

Again restricting our attention to two dimensions, we see that (8.6) and the yield criterion (8.5) give us three equations in the three stress components τxx , τxy and τyy , although as on page ?? we must not forget the existence of τzz . It is therefore possible (in principle) to solve for the stress tensor in a flowing granular material without specifying any particular constitutive relation. This is in stark contrast with all theories of elasticity that we have encountered thus far. The easiest case to analyse occurs when gravity is negligible so we can use (8.6) to introduce an Airy stress function A. Then, when the material is flowing, (8.5) becomes a nonlinear partial differential equation for A, namely µ 2 ¶2 ∂2A ∂2A cos2 φ ¡ 2 ¢2 ∂ A = ∇ A . − (8.7) ∂x2 ∂y 2 ∂x∂y 4 When φ = π/2, this reduces to the Monge–Ampère equation encountered previously in Chapter 4. When φ ∈ (0, π/2), (8.7) is unexpectedly a hyperbolic partial differential equation. Even with the body force included, the system (8.5), (8.6) is likewise hyperbolic, as shown in Exercise 8.3. This means that the stress field in the flowing material is confined to certain regions of space called regions of influence [ref: ock]. Outside these regions, the inequality in (8.5) is strict, so the granular material does not flow but behaves like an elastic solid and hence satisfies elliptic equations. By combining these two regimes, we obtain a perfectly plastic theory, as described in §8.1, in which the yield stress is never exceeded. The key to solving such models is to locate the free boundary that separates the flowing and non-flowing regions. The switch in behaviour from hyperbolic to elliptic makes these problems very difficult in general, but there are a handful of symmetric situations where we can find an explicit solution. One such is the tunnel problem of §?? where we now assume that the tunnel is bored through rock that is granular enough to yield when equality if (8.5 is attaained and is deep enough for the rock to simply exert an isotropic pressure p∞ at large distances; hence the far-field boundary conditions are τrθ → 0,

, τrr → −p∞ ,

τθθ → −p∞ .

(8.8)

In radially symmetric plane strain, (8.5) simplifies because, by rotating the (x, y) axes it becomes 2 (τrr τθθ )1/2 6 − cos φ (τrr + τθθ ) .

(8.9)

330

Plasticity

Hence yield occurs when kτrr = τθθ ,

(8.10)

where the so-called triaxial stress factor k satisfies 2k 1/2 = (1 + k) cos φ.

(8.11)

We now need to recall the elastic solution when no yield has occurred. We can do this most easily from (??) with b → ∞, and with (??)c replaced by (λ + 2µ)

dur λur + = −p∞ dr r

as r → ∞;

(8.12)

we intuitively expect that it is going to be very important to retain some positive pressure P at r = a to simulate the tunnel bracing. Then (??) becomes ur = −

p∞ r (P − p∞ )a2 + 2λ + µ 2µr

(8.13)

and (??) simply changes to τrr = −p∞ + (p∞ − P )

a2 , r2

τθθ = −p∞ − (p∞ − P ) ·

a2 r2

(8.14)

and we must not forget that τzz = −p∞ . Hence, at r = a, 2p∞ τθθ = − 1, τrr P

(8.15)

and, as the internal pressurisation is lowered below p∞ , yield will therefore first occur when P drops to 2p∞ ; 1+k

(8.16)

the closer φ is to π/2, the stronger will be the tunnel. Also we stress that, again at s = a, we have τθθ − τrr = 2(P − p∞ ) so that we expect |τrr | = −τrr to be less than |τθθ | = −τθθ . This means that, in (8.11), k should exceed unity, and indeed k=

1 + sin φ . 1 − sin φ

(8.17)

Once P drops below the value (8.16), we can seek a composite elastic/plastic solution in which there is a free boundary r = r∗ separating an elastic region in r > r∗ from a perfectly plastic region in a < r < r∗ ,


331

with kτ rr = τθθ on r = r∗ , the tractions being continuous there. Hence, in r > r∗ , we simply replace P by 2p∞ /(1 + k) and a by r∗ in (8.14) to give µ ∗ ¶2 µ ∗ ¶2 k−1 r k−1 r τrr = −p∞ + p∞ , τθθ = −p∞ − p∞ . (8.18) k+1 r k+1 r Meanwhile the quasistatic balance in r < r∗ is still r

∂τrr + τrr − τθθ = 0, ∂r

which, using (8.10) and the fact that τrr = P on r = a gives ³ r ´k−1 τrr = −P · . a

(8.19)

Finally, matching (8.18) and (8.19) at r = r∗ gives the radius of the plasticised region as ¾ 1 ½ k−1 2p∞ ∗ · a. (8.20) r = (k + 1)P The above calculation leaves open the role played by the out-of-plane stress tensor τzz , and we weill return to this issue shortly. Moreover, we have given no way of predicting the flow velocity itself in the plastic region. We have suggested that the flow might occur along slip surfaces, but, even with that assumption, some additional information is needed to determine the magnitude of the velocity. There are many empirical theories for this, and we shall not attempt to review them here. Of course, matters would be even worse of the velocity was large enough to invalidate our neglecting the acceleration term in (8.6), in which case the stress and velocity components satisfy a fundamentally coupled problem. It is relatively simple to generalise our elastic/plastic theory to granular materials in which cohesive forces act, perhaps as a result of moisture. Then a tensile stress can be sustained, and all we have to do it to replace N by (N + c) in (8.4), where c is a positive cohesive force. However, it is far more difficult to extend the analysis to three dimensions. The first task is to generalise the yield criterion (8.5), as shown in Exercise 8.2. However, this condition, along with the momentum equation (8.6), gives us only four scalar equations in the six stress components, so the three-dimensional problem is inevitably underdetermined without the imposition of a constitutive relation, and we will discuss this situation further in §??. We now turn our attention to metal plasticity, for which the microscopic mathematical theory is somewhat better developed.

332

Plasticity z

x

y

Fig. 8.6. Schematic of a cut-and-weld operation leading to the displacement field (8.21).

8.3 Dislocation theory Metals can exhibit microstructure at several scales, depending how they have been solidified and whether there are alloying components. From the point of view of plasticity, the basic microstructure is that of a periodic lattice of atoms, and the crucial scale is one small enough for the atom spacing to be noticeable but large enough for the atoms to be thought of as points so that quantum effects are assumed to be largely negligible. Many of the macroscopic properties of the material, such as the elastic constants λ and µ, can be predicted from geometric symmetries of the lattice and knowledge of the interatomic forces [REF!]. However, the same calculations predict that the stress which must be overcome to make a row of atoms push one by one past a neighbouring row should be the same order as the shear modulus µ, which is vastly greate than experimentally measured values of the yield stress, by a factor of up to 105 . This discrepancy implies that the mechanism for yield in metals is quite different from that in granular flow, and acted as a key stimulus for the development of the theory of metal plasticity. Although the behaviour of atomistic lattices during plastic deformation has only been observable since the introduction of the electron microscope, these observations were anticipated several decades earlier using, surprisingly, the continuum theory of linear elastostatics. The simplest configuration in which we can understand the¡ basic ideas ¢is antiplane strain, where T the displacement takes the form u = 0, 0, w(x, y) and ∇2 w = 0. We have already seen several solutions of this model, including some with singularities in Chapter 7.

8.3 Dislocation theory

333

Now we ask ourselves what the physical interpretation might be of the displacement field ³y´ b w= tan−1 . (8.21) 2π x This is a function whose Laplacian is zero except at the origin and on some branch cut emanating from the origin, across which there is a jump of magnitude b in the value of w. If we took tan−1 (y/x) = θ, where (r, θ) are the usual plane polar coordinates with the restriction 0 6 θ < 2π, the branch cut would be along the positive x-axis. However, applying the results of §1.11.2 to (8.21), we find that the stress components in cylindrical polar coordinates are all zero except for τθz =

bµ , 2πr

(8.22)

which is defined everywhere except at the origin, whatever branch cut is chosen. The displacement field (8.21) could in principle be realized by a so-called cut-and-weld operation as shown in Figure 8.6. We simply take a circular cylindrical bar, cut it along a diametral plane from the exterior to the axis, displace one side of the cut by a distance b relative to the other side, and finally weld the two cut faces together again. Clearly there will be a region of very large strain close to the axis, in accordance with (8.22). We will thus have created a bar that is in a state of self-stress, so that it is in equilibrium under the action of no external forces, yet there is a nonzero stress distribution in the interior. Such self-stress is of course often induced during metal fabrication. It is also instructive to interpret (8.21) in terms of incompatibility. We recall from §2.7 that the assumption of a single-valued displacement field gives rise to compatibility relations between the strain components. Since (8.21) is not single-valued, we would expect these relations to be violated. Indeed, in Exercise 8.4 we show that w satisfies ∂2w ∂2w − = bδ(x)δ(y), ∂y∂x ∂x∂y

(8.23)

where δ denotes the Dirac delta function defined in §2.9.1. We can interpret (8.23) as a “failure of single-valuedness” or a line of incompatibility along the z-axis. It is also instructive to introduce an antiplane stress function φ, as in §2.4, defined to be the harmonic conjugate of w so that w and φ satisfy the

334

Plasticity

(a)

(b)

y

y

x

x

Fig. 8.7. The displacement field in an edge dislocation: (a) pristine material; (b) after the cut-and-weld operation. Nodal points correspond and the cut is shown dotted.

Cauchy–Riemann equations ∂w ∂φ = , ∂x ∂y

∂w ∂φ =− . ∂y ∂x

(8.24)

Now the harmonic conjugate of (8.21) is φ = −(b/2π) log r, which, as we know from §2.9.1, satisfies ∇2 φ = −bδ(x)δ(y),

(8.25)

and we can easily verify that (8.25) is consistent with (8.23). The singularity in (8.21) at x = y = 0 is thus analogous to the singularity at the core of a line vortex in inviscid fluid dynamics, where a finite circulation is concentrated on a line in an otherwise irrotational flow. Similar behaviour occurs if we displace one cut face radially relative to the other before welding them back together, as shown in Figure 8.7. This configuration can ¢be described using a plain strain displacement field ¡ T u = u(x, y), v(x, y), 0 , in which u is discontinuous across the positive x-axis. We can find such a displacement by using the Papkovich–Neuber representation from §2.8.6, namely   µ ¶ Ψ1 xΨ1 + yΨ2 2µu = 2(1 − ν) Ψ2  − grad φ + , (8.26) 2 0 where Ψ1 , Ψ2 and φ are all harmonic functions of x and y. Now the required discontinuity in u may be obtained by choosing φ = 0, Ψ1 proportional to θ and Ψ2 equal to a constant. We note that this is effectively a very simple version of the Papkovich–Neuber function used on page ??.

8.3 Dislocation theory (a)

335

(b)

Fig. 8.8. Schematic of an edge dislocation in a square crystal lattice: (a) pristine crystal; (b) insertion of an extra row of atoms.

As shown in Exercise 8.5, this approach leads us to the displacement field µ ¶ µ ¶ µ ¶ b θ b sin θ u cos θ = + (8.27) v 2π 0 2π(3 − 4ν) sin θ which is depicted in Figure 8.7. The associated stress field is given in cylindrical polar coordinates by   τrr τrθ τrz τrθ τθθ τθz  τrz τθz τzz   −2ν(sin θ)/r 2(1 − ν)(cos θ)/r 0 µb 2(1 − ν) cos θ/r −2(1 − ν)(sin θ)/r , = 0 π(3 − 4ν) 0 0 −2ν(sin θ)/r (8.28) so we again have a state of self-stress which is infinite on the line r = 0. In Exercise 8.6, we show that (8.28) corresponds to an Airy stress function A=

−µb {(1 − 2ν)rθ cos θ + 2(1 − ν)r log r sin θ} , π(3 − 4ν)

which satisfies the equation ∇4 A =

µ

4µb (3 − 4ν)

(8.29)

¶ δ(x)δ 0 (y),

(8.30)

which is effectively (??) Now we recall from §2.8.2 that the existence of A ensures that the plane strain Navier equations are satisfied, while the biharmonic equation for A was obtained by assuming compatibility between the strain components. Hence, as in (8.23), we can interpret (8.30) as describing a line of incompatibility along the z-axis. A key insight into the physical mechanism for plastic flow in metals comes when we imagine executing the cut-and-weld operation shown in Figure 8.7

336

Plasticity (a)

(b)

(c)

Fig. 8.9. Schematic of a moving edge dislocation: (a) initial configuration; (b) and (c) the dislocation moves to the left as the atoms realign themselves, eventually leading to a displacement of the upper block of atoms relative to the lower.

at an atomic scale. For example, on the square lattice shown in Figure 8.8(a), we can achieve this by inserting an extra column of atoms, as depicted in Figure 8.8(b). Far from the crystal misfit in the dotted circle, it simply seems that the atoms below the positive x-axis have been displaced one atom spacing to the right, as in the displacement field (8.27). We are thus led to conjecture that, with b equal to a single atomic spacing, an atomistic calculation of the displacement based on Figure 8.8(b) would tend to (8.27) on a scale large compared to b. The region along the z-axis near the dotted circle is called the core of an edge dislocation, the corresponding region in the antiplane configuration analogous to Figure 8.6 being the core of a screw dislocation. The dislocation itself, in either case, is the z-axis and the displacement jump as we move around any curve enclosing the dislocation is called the Burgers vector. In our examples, the Burgers vector is (0, 0, b)T for the screw dislocation and (b, 0, 0)T for the edge dislocation. However, it is quite easy to take linear combinations of these to generate a mixed dislocation along the z-axis which is part edge and part screw. In addition, we can envisage dislocations on a macroscopic scale which lie on curved rather than straight lines, by identifying the z-axis with the tangent to the dislocation at any point. In practice, the symmetries of the crystal lattice usually force any dislocations to lie in certain planes, known as slip planes, parallel to the Burgers vector [REF!]. Anything other than an absolutely perfect crystal must contain many dislocations like that shown in Figure 8.8(b). Now the key observation is that just a small realignment of the atoms near the core is needed for such a dislocation to move irreversibly through the lattice. This is illustrated schematically in Figure 8.9, where a dislocation moves to the left, thereby moving the lower block of atoms bodily with respect to the upper block. This can be achieved without forcing large numbers of atoms to slide over each other, and this explains the huge discrepancy noted above between pre-

8.3 Dislocation theory

337

dictions of the yield stress based on inter-atomic forces and experimentally measured values. We also note for future reference that the configurational changes illustrated in Figure 8.9 take place with no appreciable change in volume of the crystal. All these observations led theoreticians in the 1930s and earlier to suggest that macroscopic metal plasticity could be explained by the motion of dislocations,† and it was a great day when, two decades later‡, that electron micrographs of metal crystals that had undergone plastic deformation were able to confirm the theory. They revealed countless (up to 1012 cm−2 ) black curves in regions which would have been invisible had the crystal been perfect. Each of these curves represents a dislocation bounding a plane region of slip in the crystal and, the more the metal had been deformed plastically, the more black lines were observed. Since any realistic sample contains so many dislocations, it is impractical to track each one individually. Instead, any macroscopic theory must use a density, which is in fact a tensor,§ that characterises the direction and Burgers vector averaged over a large number of dislocations. Moreover, it would need a rule for how dislocations move, but this subject is beyond the scope of this book. The theory would be analogous to that used to obtain a model for collective motion of vortices in an inviscid fluid or of superconducting vortices in a type-II superconductor. Unfortunately, it has proved much harder to carry out in practice for dislocations, and a definitive macroscopic model does not yet exist. Rather than delving further into this fascinating open problem, we will simply use microscopic thought-experiments like Figure 8.9 to suggest the following guidelines that will be employed below to obtain a closed macroscopic model for metal plasticity. (a) Dislocations may move under the action of an applied shear stress, but are generally impervious to any isotropic pressure. (b) Collective motion of many dislocations can only occur when a sufficiently high shear stress exists at every point. The critical yield stress is a constant for any given metal. (c) In contrast with granular flow, where there is just one direction of slip at any point, a dislocation core is constrained to move in a slip plane, of which there may be many depending on the symmetries of the crystal structure. These slip planes are set in the metal, whereas the slip planes in a granular material depend on the stress state. † Egon Orowan, Michael Polanyi and G. I. Taylor (1934) ‡ Hirsch et al. (1956) § Furthermore, such a denisyt tensor would have to vanish in regions in which the tensor (??) vanishes.

338

Plasticity

(d) Plastic flow, which results from collective dislocation motion, is essentially incompressible. These conjectures can be tested experimentally or by performing “atomistic” computations on a discrete lattice. We conclude by mentioning two further benefits that accrue from dislocation theory. One concerns what happens when a metal is repeatedly deformed so strongly that many new families of dislocations are created on different slip planes. These impede further movement, since dislocations cannot easily pass through each other. This gives a theoretical explanation of work hardening, whereby a metal becomes more resistant to yield if it has already gone through several stress cycles. There is one other interesting macroscopic interpretation of dislocations which is purely of theoretical interest. This concerns the modelling of brittle fracture when no plastic effects are taken into consideration. It is a simple matter (see Exercise 7.3) to see that the displacement field (7.11) near a mode III crack can be represented as the sum of a distribution of virtual screw dislocations along the crack. Similarly, the displacement fields near a mode II or III crack can be thought of as being caused by a distribution of edge dislocations, all these displacements having closely related Papkovich– Neuber representations. This situation is reminiscent of the theory of flight, in which a thin aerofoil can be regarded as a distribution of vortices in an otherwise irrotational fluid. Of course, the stress intensification near the tips of a crack in the theory of brittle fracture implies the existence of plastic regions near these tips, and these will contain real, not virtual, dislocations.

8.4 Perfect plasticity theory for metals 8.4.1 Torsion problems The observed, and simulated, behaviour of dislocations suggests that metals can be well described using a perfectly plastic theory: either some measure of the stress is below a critical value, in which case the metal is elastic, or the stress is sufficient to cause bulk dislocation motion and flow. The first question we must address is thus the yield criterion that distinguishes elastic from plastic behaviour. We begin as we did for granular plasticity by considering the tractions on all small surface elements through a point P in the metal. However, as explained at the end of §8.3, instead of involving a limiting friction concept, it is more natural to associate dislocation motion with the existence of a critical shear stress independent of the normal stress. We first consider the simplest case of antiplane shear, in which the only

8.4 Perfect plasticity theory for metals

339

stresses are shear stresses which give rise to a traction      0 0 τxz cos θ 0 τn =  0 0 τyz   sin θ  = (τxz cos θ + τyz sin θ) 0 τxz τyz 0 0 1

(8.31)

on a surface element normal to n = (cos θ, sin θ, 0)T . We require the amplitude of the shear stress to be bounded by a critical yield stress τY for all such surface elements, and this leads to the criterion ¯q ¯ ¯ 2 + τ 2 ¯¯ 6 τ , (8.32) ¯ τxz Y yz with equality when the material is flowing. To illustrate the mathematical structure in this relatively simple case, let us return to the torsion bar problem from §2.4. We recall that the displacement field is given by u = Ω (−yz, xz, ψ(x, y))T ,

(8.33)

where Ω represents the twist of the bar about its axis. We also recall the elastic stress function φ defined such that τxz = µΩ

∂φ , ∂y

τyz = −µΩ

∂φ ∂x

(8.34)

where φ satisfies Poisson’s equation ∇2 φ = −2,

(8.35)

subject to φ = 0 on the boundary of the bar. For example, if the crosssection is circular, with radius a, (??) gives a2 − r 2 , (8.36) 2 but for this solution to be valid, we must check that the shear stress does not exceed the critical value τY . The yield condition (8.32) reads φ=

µΩ |∇φ| 6 τY ,

(8.37)

µΩr 6 τY .

(8.38)

which, with (8.36), requires

The left-hand side is maximised when r = a, and we deduce that the bar will first yield at its surface when the twist Ω reaches a critical value τY Ωc = . (8.39) µa

340

Plasticity

2M/πa3 τY 1.4 1.2 1.0 0.8 0.6 0.4 0.2

0

1

2

3

4

Ω/Ωc

Fig. 8.10. The normalised torque M versus twist Ω applied to an elastic/plastic cylindrical bar.

When Ω > Ωc , the condition (8.38) is violated, so our solution (8.36) is no longer valid. Instead, as in the tunnel problem of §8.2.2, there will be a plastic region near the boundary of the bar where the metal has yielded, although we expect the material at the centre still to be elastic. We therefore again have to introduce a free boundary, say r = s, that separates the yielded and unyielded material, where s is to be determined as part of the solution and repeat the key assumption that, even when the material has yielded, it flows slowly enough for the inertia terms as in (8.6). We can thus employ a stress function φ throughout the bar, satisfying (8.35) in 0 6 r < s and the yield condition µΩ |∇φ| = τY

(8.40)

in s < r < a, again subject to φ = 0 on r = a. Continuity of traction requires φ and its normal derivative to be continuous across r = s. We soon find that  s2 + r2  as − 0 6 r < s, (8.41) φ= 2  s(a − r) s < r < a, where s=

τY . Ωµ

(8.42)

We thus see how the plastic region grows as Ω increases past its critical value Ωc , and the nonzero stress components are easily found to be ( ( −µΩy 0 6 r < s, µΩx 0 6 r < s, τxz = τyz = (8.43) −µΩys/r s < r < a, µΩxs/r s < r < a.


341

We note that as τY tends to zero, a stress singularity develops at r=o, which reflects the fact that (8.40) is a hyperbolic equation for φ; the introduction of even a small amount of elasticity removes the singularity. As in §2.4, we can use these to calculate the torque M applied to the bar as a function of the twist ZZ M= (xτyz − yτxz ) dxdy cross-section µ 3 ¶ Ω πa τY   Ω 6 Ωc ,  2 Ωc ¶ = µ 3 ¶µ (8.44)  πa τY Ω3c   4− 3 Ω > Ωc . 6 Ω Figure 8.10 shows how the torque increases linearly with the twist until Ω reaches its critical value Ωc . Thereafter, it tails off rapidly, and we observe that only a finite torque 2πa3 τY /3 is required for the bar to fail completely. Now suppose that, once a maximim twist Ωmax has been applied, so the bar has yielded down to some radius r = s, the applied torque is then removed sufficiently slowly that we can still use a quantitative model. We would expect the bar to recover and twist back towards its starting config˜ ; in other words, u ˜ uration. We denote this subsequent displacement by u is the displacement relative to the state just before we released the torque. Consistent with our assumption of perfect plasticity is the further assumption that, as soon as the torque is decreased, all the once-yielded material instantly returns to being elastic, but we will return to this point in §?? However, it will now start with the nonzero stress calculated in (8.43) when ˜ is zero. u ˜ has the same structure (2.41) as the original disLet us suppose that u placement: that is ˜ (−yz, xz, 0)T , ˜ =Ω u

(8.45)

since, we recall from §8.2, ψ ≡ 0 for a circular bar. We are trying to evaluate ˜ the net twist of the bar is then given by Ωmax + Ω. ˜ the recovery twist Ω; The total stress consists of the elastic stress corresponding to (8.45) added to the initial stress (8.43) reached at the end of the plastic phase, that is ³ ´ ˜ + Ωmax µ ¶ µ ¶  0 6 r < s,  Ω τxz −y =µ × ³ (8.46) ´ τyz x  ˜ + Ωmax s/r  Ω s < r < a. Now the applied torque is supposed to be zero, and a calculation analogous

342

Plasticity

2M/πa3 τY

plastic

Ωmax /Ωc

1.2 1.0 0.8 0.6 0.4 0.2

0.0

elastic 0.5

1.0

³

recovery 1.5

2.0

2.5

3.0

Ω/Ωc

´ ˜ /2 Ωmax + Ω

Fig. 8.11. The normalised torque M versus twist Ω applied to an elastic/plastic cylindrical bar, showing the recovery phase when the torque is released.

˜: to (8.44) then leads to the following formula for the “recovery” torque M µ ¶ ˜ 2M Ω3c Ωc ˜ 4− 3 =Ω=− . (8.47) πµa4 3 Ωmax ˜ = Ωc , When Ωmax = Ωc (so the bar has not yet yielded), we see that Ω but, as the original twist Ωmax is increased, a decreasing fraction of it is recovered when the torque is released, as illustrated in Figure 8.11. The bar ˜ returns behaves elastically until Ω = Ωc and then starts to yield. When M to zero, a nonzero twist remains and this qualitative behaviour is typical of elastic-plastic systems. The final crucial observation is that, even when the bar has recovered and there is no net torque on it, the internal stress components (8.46) are nonzero, and the bar is said to contain residual stress. This is bound to happen because the recovery phase starts with an initial stress field (8.43) that does not satisfy the compatibility conditions. This means that there is no elastic deformation that the material can adopt that will completely relieve the stress. Without the assumption of radial symmetry, the problem must in general be solved numerically, but it still possesses a very helpful mathematical structure. The general scenario is sketched in Figure 8.12, in which there is a plastic region near the boundary in which we have to solve the Eikonal equation (8.40). This equation is hyperbolic and, given φ = 0 on the surface of the bar, can in principle be solved for φ throughout the plastic region. In the elastic region, φ still satisfies Poisson’s equation, subject to two boundary conditions, namely that both φ and its normal derivative are continuous


µ

∂φ ∂x

¶2

343

Plastic µ ¶2 ∂φ τ2 + = 2Y 2 ∂y µΩ

Elastic ∂ 2φ ∂ 2φ + = −2 ∂x2 ∂y 2 · ¸ £ ¤ ∂φ φ = =0 ∂n

y

x

φ=0

Fig. 8.12. Schematic of the free-boundary problem for an elastic/perfectly plastic torsion bar.

across the free boundary. In principle, this problem determines both φ and the position of this free boundary. However, this switch in behaviour across the free boundary is strongly reminiscent of the complementarity conditions encountered when modelling contact problems in Chapter 7. Indeed, the free-boundary problem depicted in Figure 8.12 may be cast as a variational inequality and hence proved to be well posed, despite the nonlinearity and the change of type from hyperbolic to elliptic. Moreover, in most practical ¡ it2 can be ¢ proved to be equivalent to the problem of minimising RR cases |∇φ| − 2φ dA over all sufficiently smooth functions that vanish on D the outer boundary and are such that |∇φ| < τY /µΩ. We also note that, as in our models for granular materials, although the yield condition enables us to determine the stress, it tells us nothing about the displacement in the bar once it has yielded. We will return to this point in §8.5.

8.4.2 Plane strain The next simplest situation is that of plane strain, where we can follow the approach of §8.2.2 to calculate the shear stress on a surface element with unit normal n = (cos θ, sin θ, 0)T . Indeed, we can read off the maximum tangential traction F by inspecting the Mohr circle in Figure 8.5, and hence

344

Plasticity

deduce the yield criterion ¯ ¯q ¯ 1 ¯ 2 2 ¯ ¯ ¯ 4 (τxx − τyy ) + τxy ¯ 6 τY ,

(8.48)

with equality when the material has yielded. Hence, in the absence of any body force, we can therefore conveniently use an Airy stress function A throughout the material. Where the material has not yielded (so the inequality in (8.48) is strict), we obtain ∇4 A = 0, but when yield occurs, and we have equality in (8.48), A satisfies (µ ) ¶2 ¡ 2 ¢2 ∂2A ∂2A ∂2A − ∇ A +4 = 4τY2 . ∂x∂y ∂x2 ∂y 2

(8.49)

(8.50)

As always the key aspect of the problem is to locate the boundary where the switch from (8.49) to (8.50) occurs. Here, a traction balance shows that A and its first and second partial derivatives must all be continuous across such a boundary. The nonlinear partial differential equation (8.50) satisfied by A when the material has yielded is another generalisation of the Monge–Ampère equation, and closely resembles the equation (8.7) for granular flow in plane strain. It is not immediately classifiable using the standard theory of secondorder partial differential equations [cite ock], but differentiation with respect to x (for example) leads to the quasilinear equation µ 2 µ 2 ¶ ¶ ∂ A ∂2A ∂2P ∂ A ∂2A ∂2P ∂2A ∂2P + − +4 − = 0 (8.51) ∂x2 ∂y 2 ∂x2 ∂x∂y ∂x∂y ∂y 2 ∂x2 ∂y 2 for P = ∂A/∂x. Now it is easy to see that (8.51) is hyperbolic, with two families of real characteristics satisfying µ 2 ¶ Áµ 2 ¶ dy ∂ A ∂ A ∂2A =2 ± τY − . (8.52) dx ∂x∂y ∂x2 ∂y 2 These correspond to the directions in which the shear stress is maximal (see Exercise 8.9), and hence the characteristics of (8.50) are the slip surfaces, along which we might expect the material to flow. As in §8.4.1, analytic solutions of this nonlinear free-boundary problem are unlikely to be available unless the geometry is very simple. A very famous example is the gun barrel problem of §?? which we will only sketch briefly because of its similarity to the tunnel problem of §8.2.2. Also, for


345

simplicity, we will only treat thick gun barrels in which b → ∞, so that (??) gives τrr = −

a2 P , r2

τθθ =

a2 P r2

(8.53)

in the fully elastic region. As in the case of granular flow, (8.50) simplifies considerably when there is radial symmetry and it reduces to (τrr + τθθ )2 − 4τrr τθθ = (τrr − τθθ )2 = τY2 . The chocie of square root is dictated by (8.53) and hence yield first occurs when τθθ − τrr = τY , i.e. when r = a and P = τY /2. This yield condition is very convenient because the radial Navier equation ∂τrr τrr − τθθ + =0 ∂r r holds everywhere even when P > τY /2. hence, in the plastic region, r > r∗ > a, where r∗ is to be determined, ³ r r´ τrr = −P + τY log , τθθ = −P + τY 1 + log . a a Meanwhile, in the elastic region a < r < r∗ , (8.53) generalises to µ ∗ ¶2 µ ∗ ¶2 r τY r τY · , τθθ = · . τrr = − 2 r 2 r Hence, when we balance the traction τrr at r = r∗ , we find that ¶ µ 1 P ∗ − . r = a exp τY 2 We can now repeat the unloading analysis of §8.4.1. If P increases to a value PM greater than τY /2, and then reduces through values P˜ to zero, our assumption that the gun barrel becomes instantaneously elastic when P < PM gives that, for r > r∗ , µ ¶ µ ¶ τY r∗ 2 a2 (PM − P˜ ) τY r∗ 2 a2 (PM − P˜ ) + , τθθ = − , τrr = − 2 r r2 2 r r2 while for r < r∗ , µ ¶ 2 r∗ 1 a (PM − P˜ ) τrr = τY log − + , r 2 r2

τθθ

µ ¶ 2 r 1 a (PM − P˜ ) = τY log ∗ + − . r 2 r2

These formulae describe the residual stress field when P˜ = 0, and this residual stress increases with PM .

346

Plasticity

We caution that our earlier caveats about displacements and the role of τzz still apply and we will address the latter next. 8.4.3 Three-dimensional yield conditions As we will soon discover, generalising the yield condition to three dimensions introduces new complications not encountered in the antiplane and plane strain examples considered above. In general, and not just for metals, we expect to impose a criterion of the form f (τij ) 6 τY ,

(8.54)

where the yield function f is some appropriate measure of the stress. Now we ask ourselves what restrictions need to be imposed on the function f to ensure that (8.54) is physically realistic. If we assume that our metal is isotropic, then f must be invariant under rotation of the axes. It follows that f can be written as a function of the three stress invariants (cf §5.3.2) ¡ ¢o 1n I1 (τ ) = Tr (τ ), I2 (τ ) = Tr (τ )2 − Tr τ 2 , I3 (τ ) = det (τ ), (8.55) 2 and we have therefore reduced f from six degrees of freedom (the elements of the symmetric tensor τij ) to three. Recall, for example, that the Coulomb yield criterion (8.5) for granular material in two dimensions depends only on I1 (τ ) and I3 (τ ). For metals, further simplification can be achieved by recalling the experimental evidence that they do not deform plastically under an isotopic stress, for which τij = −pδij , say; indeed it is intuitively reasonable that it could not cause a dislocation like that sketched in Figure 8.8 to move.† Hence we expect that our yield function f should not vary if τij is replaced by τij + cδij , where c is any scalar. We can understand the implications of this constraint on f if we regard it not as a function of the six independent variables τij but as a function of the varibales 1 τij0 = τij − (Trτ )δij (8.56) 3 0

0

and the variable p = − 13 (Trτ ); since τkk = 0, the τ ij comprise only five independent variables. Then we require that 0

f (τij ) = f (τij0 , p), † Other materials, for example glass, can however flow under sufficient hydrostatic pressure.


347

say, where 0

0

f (τij + cδij ) = f (τij0 , p − c) ≡ f (τij0 , p). Since this holds for all c, we see that f is just a function of τij0 , whose norm is a crude measure of the deviation of τij from isotropy. Hence τij0 is called the deviatoric part of τij , or the stress deviator. Also, henceforth we will 0 use f (τij0 ) to mean f (τij0 ), hopefully without causing too much confusion. Provided f is written as a function of τij0 , it is automatically invariant under the addition of any isotropic stress. Note that our definition (??) imposes the constraint 0 τkk ≡0

(8.57)

(here and henceforth invoking the summation convention), so that, while τij has six independent elements, the deviator has only five. Now we can impose invariance under both rigid-body rotation and isotropic pressure by writing ¢ 0 ¡ f = f τ10 , τ20 , τ30 , (8.58) where τk0 are the eigenvalues of τij0 . It is easy to see that the principal axes of τij0 and τij coincide, and that τk0 are related to the principal stresses τk (k = 1, 2, 3) by τ1 + τ2 + τ3 τk0 = τk + p, where p = − . (8.59) 3 Since τ10 + τ20 + τ30 = 0, there are only two independent degrees of freedom 0 remaining in the function f . 0 One physically plausible possibility is again to define f as the maximum shear stress acting on all possible surface elements. As shown in Exercise 2.9 (see also Exercise 8.2), this leads to ¢ 1 © ª 0 ¡ f τ10 , τ20 , τ30 = max |τ10 − τ20 |, |τ20 − τ30 |, |τ30 − τ10 | , (8.60) 2 and the corresponding yield condition © ª max |τ10 − τ20 |, |τ20 − τ30 |, |τ30 − τ10 | = 2τY . (8.61) is called the Tresca condition. Notice the function (8.60) is symmetric with respect to permutation of its arguments, so that there are no preferred directions. In antiplane strain, with the stress tensor given by (8.31), the principal deviatoric stresses are easily found to be q q 0 2 + τ2 , 2 + τ2 , τ = τxz τ30 = 0, (8.62) τ10 = − τxz yz 2 yz

348

Plasticity

(a)

(b)

τ30

τ20

τ10

τ20 τ10 Fig. 8.13. (a) The Tresca yield surface (8.61) plotted in the space of principal stress deviators (τ10 , τ20 , τ30 ) (normalised with τY ), showing the intersection with the plane τ10 + τ20 + τ30 = 0. (b) The projection in the (τ10 , τ20 )-plane.

so that the Tresca condition reproduces (8.32). In plane strain, we find that q ¡ ¢ τxx + τyy − 2τzz 2 −τ τ τ10 = (8.63a) − 14 (τxx + τyy )2 + τxy xx yy , 6 q ¢ ¡ τxx + τyy − 2τzz 2 −τ τ τ20 = + 14 (τxx + τyy )2 + τxy (8.63b) xx yy , 6 2τzz − τxx − τyy τ30 = , (8.63c) 3 so the Tresca condition is consistent with (8.48) provided τ30 lies between τ10 and τ20 . This explains the role played by τzz in the examples on pages ?? and ??, and Exercise 8.10 gives the basis for how τzz should be incorporated into these examples. In three-dimensional problems, we can represent (8.61) as a surface, known as the yield surface, in (τ10 , τ20 , τ30 )-space. As shown in Figure 8.13(a), the Tresca yield surface is a cylinder whose axis is in the direction (1, 1, 1)T and whose cross section viewed along the axis is a hexagon. By also using the constraint τ10 + τ20 + τ30 = 0, we can project this surface onto (for example) the (τ10 , τ20 )-plane as the polygon © ª max |τ10 − τ20 |, |2τ20 + τ10 |, |2τ10 + τ20 | = 2τY , (8.64) depicted in Figure 8.13(b). In antiplane strain, we have τ30 = 0 and hence τ10 +τ20 = 0. This straight line intersects the polygon shown in Figure 8.13(b) at the two points (τ10 , τ20 ) = (±τY , ∓τY ), so the yield surface degenerates to just two points in antiplane


(a)

(b)

349

τ20

τ30

τ10

τ20 τ10 Fig. 8.14. (a) The Von Mises yield surface plotted in the space of principal stress deviators (τ10 , τ20 , τ30 ) (normalised with τY ), showing the intersection with the plane τ10 + τ20 + τ30 = 0. (b) The projection in the (τ10 , τ20 )-plane.

strain. However, even in plane strain, the whole hexagon comes into play in general (see Exercise 8.10), and the situation is even worse in three dimensions. The presence of the corners in the yield surface is worrying, since it implies non-smooth dependence of the solution on the stress components. It also makes the Tresca condition awkward to implement in general, and this has led to the adoption of the Von Mises yield function Ã !1/2 ¢ (τ10 − τ20 )2 + (τ20 − τ30 )2 + (τ30 − τ10 )2 0 ¡ 0 0 0 f τ1 , τ2 , τ3 = . (8.65) 6 By construction, this is a valid mesure of the stress with all the required properties of invariance although, unlike the Tresca yield function, it has no immediate interpretation in terms of some maximal shear stress. The yield surface corresponding to (8.65), shown in Figure 8.14(a), is still a cylinder whose axis points in the direction (1, 1, 1)T . However, elementary geometry implies that its cross section is circular so, in contrast with Figure 8.13, the Von Mises yield surface is smooth. Again we can use the constraint τ10 + τ20 + τ30 to project the surface onto the (τ10 , τ20 )-plane, where it takes the form of an ellipse, namely 2

2

τ10 + τ10 τ20 + τ20 =

(τ10 − τ20 )2 3 (τ10 + τ20 )2 + = 1, 4 4

(8.66)

as shown in Figure 8.14(b). In antiplane strain, the two yield criteria coincide, but in higher dimensions (8.65) is far easier to implement.

350

Plasticity

τ3 τ2 τ1 Fig. 8.15. The Coulomb yield surface plotted in the space of principal stresses (τ1 , τ2 , τ3 ) (with θ = π/6).

The assumed indifference of f to isotropic stress means that the yield surface corresponding to any valid choice of yield function must always be a cylinder pointing in the direction (1, 1, 1)T . The corresponding picture in granular flow is quite different. First, we recall that none of the principal stresses in a granular material may be positive. Second, the yield criterion for a granular material depends explicitly on the normal as well as the shear stress, so no advantage is gained by introducing the stress deviator. Exercise 8.2 shows that, provided the principal stresses are numbered in order, the Coulomb yield condition is equivalent to τ1 − τ3 = (τ1 + τ3 ) sin φ

when τ1 6 τ2 6 τ3 6 0.

(8.67)

This represents a segment of a plane through the origin in (τ1 , τ2 , τ3 )-space, and six other segments are obtained by switching the orders of the principal stress components. Hence, the analogue of the hexagonal cylinder of Figure 8.13 is a six-sided pyramid, the faces of the pyramid corresponding to the six permutations of the indices in (8.67). Indeed, invariance with respect to permutation of the principal stresses implies that the yield surface in any isotropic medium must share the six-fold symmetry evident in Figures 8.13 and 8.15. We have seen above that, in both antiplane and plane strain, the yield criterion and the steady Navier equations form a closed system of equations for the stress components. However, the same is not true of genuinely threedimensional problems, since we now have just one yield criterion and three

8.5 Plastic flow

351

Navier equations for the six stress components. Moreover if the plastic flow is sufficiently strong for the material inertia to be comparable to the yield stress, the resulting system is under-determined even in two dimensions. As in granular flow, the problem must be closed by incorporating a flow rule that determines how the material responds to stress once it has yielded. 8.5 Plastic flow This is the first time in this book that we have tried to write down a mathematical model for irreversible flow, as opposed to elastic strain. Such models are inevitably more readily stated in terms of the velocity of the medium, rather than the displacement. We therefore introduce the velocity vector v(x, t) defined as ¯ ¯ ∂x ¯¯ ∂u ¯¯ v= = , (8.68) ∂t ¯X ∂t ¯X where the time derivative is taken with the Lagrangian coordinate X held fixed. It is convenient to introduce a shorthand for this convective or material derivative, namely ¯ ¯ ∂ ¯¯ D ∂ ¯¯ = ≡ + (v · ∇) , (8.69) Dt ∂t ¯ ∂t ¯ X

x

the latter identity following from the chain rule. By applying the convective derivative to u, we obtain a kinematic relationship, ∂u + (v · ∇) u, (8.70) ∂t between the displacement and velocity. We can rearrange this to an explicit equation for v, namely ¡ ¢−1 ∂u v = I − ∇uT , (8.71) ∂t recalling that ∇u denotes the displacement gradient tensor with entries ∂uj /∂xi . Hence we can linearise, approximating v as simply the Eulerian time derivative of u, provided the displacement gradients are small. This assumption is usually fine for small displacements of elastic materials, and leads to the theory of linear elasticity as we have seen. However, a characteristic of plastic behaviour is the ability to suffer large permanent displacement without breaking. To describe such situations as the bending of a paperclip, we would need to retain the nonlinear terms in (8.70), as in Chapter 5 and distinguish carefully between Lagrangian and Eulerian time derivatives. v=

352

Plasticity

Next, to describe the deformation of a flowing material, we introduce the concept of rate-of-strain. We recall that the strain tensor describes the change in length of a small line element in the material, compared with its initial length. Once a material is flowing, it starts to “forget” its initial rest state, so it no longer makes sense to use the initial length as a benchmark. Instead, it is more natural to examine the instantaneous variation in a line element over a small time increment. This is not a book about fluid mechanics, so we will not spell out all the details here (but see Exercise 8.12). Suffice it to say that the rate of change of the length of a line element δx = (δx1 , δx2 , δx3 )T takes the form ´ d ³ |δx|2 = 2δxT Dδx, (8.72) dt where the rate-of-strain tensor is defined by† µ ¶ ¢ ∂vj 1 ∂vi 1¡ Dij = + or D = ∇v + ∇v T . (8.73) 2 ∂xj ∂xi 2 If the strains are small enough for us to neglect the nonlinear terms in (8.70), then Dij is simply the time derivative of the linear strain tensor eij . Otherwise, the relationship between strain and rate-of-strain is complicated, and it would have been of fundamental importance in describing dynamic nonlinear elasticity, had we done this in Chapter 5. By applying the convective derivative to the velocity vector, we find that the acceleration of the medium is given by Dv ∂v = + (v · ∇) v, (8.74) Dt ∂t and we can therefore write Cauchy’s momentum equation in the form ½ ¾ ∂v ρ + (v · ∇) v = ∇ · τ + ρg. (8.75) ∂t Along with the yield condition f (τ ) = τY ,

(8.76)

which, even with our incompressibility assumption (d) from Page ??, gives us only five scalar equations so far for the nine unkowns τij and vi . Now our task is to obtain a constitutive relation for the stress field associated with a given flow velocity v. This is a difficult challenge, perhaps the greatett in the whole book, and we will now decide just one way of tackling it. We start by recalling our † NB consistency with elastic definition of strain means factor of 2 out from usual fluid definition

8.5 Plastic flow

353

earlier discussions of energy balances. In Chapter 1, (??) was deduced as a consequence of the momentum equaiton (??), and the same procedure was followed in Lagrangian coordinates in (??). In neither case did we need to make any assumptions about the relationship between stress and strain. When we repeat the procedure in Eulerian coordinates starting from (8.75), with gravity neglected for simplicity, we obtain ZZZ ZZ ZZZ d 1 2 ρ|v| dx − vi τij nj dS = − Φ dx, (8.77) dt V (t) 2 ∂V (t) V (t) where our use of Eulerian coordinates means that V must be a material volume composed of particles moving with velocity v(x, t) and Φ = Dij τij

(8.78)

is called the dissipation function. We emphasise that (??), (??) and (8.77) are purely a consequence of conservation of momentum. When we restrict ourselves to linear (or hyper) elasticity, the right-hand side of (8.77) is the time derivative of the integral of the strain energy density ∂W ∂W W (eij ) (W (Fij )), which is such that ∂e = τij ( ∂F ) is the constitutive ij ij relation. Hence the right-hand side of (8.77) is ZZZ d − W dx, dt V (t) which simply tells us that the rate of change of kinetic energy differs from the work done by the tractions on the boundary by the rate of change of strain energy. However, for a plastic material, the observation that heat is generated when we repeatedly bend a paper clip† leads us to consider the total physical energy balance in accordance with the first law of thermodynamics (“work is heat”). From this viewpoint, the energy density per unit volume is, excluding elastic energy, ρ(CT + 12 (V 2 ) where T is the temperature and c the specific heat. Hence, taking into account the transport of heat by conduction, with constant thermal conductivity k, the “thermodynamic” energy balance is µ ¶ ZZZ ZZ ZZ d 1 ∂T ρ cT + |V |2 dx − vi τij nj dS = k ni dS dt 2 ∂x i V (t) ∂V (t) ∂V (t) (8.79) The thermal balance results from subtracting (8.77) from (8.79) and, after † Bend a reasonably large paper clip a few times and put the bent section against your cheek.

354

Plasticity

using Green’s theorem and the arbitrariness of V (t), it gives DT − k∇2 T = φ; (8.80) Dt note we do not need to assume ρ is constant, only that it satisfies the conservation of mass equation ρx

Dρ ∂vi = 0. +ρ Dt ∂xi This equation shows that in an insulated elastic material in which DW Dt = Φ, and in which k = 0, the temperature of any particle varies at a rate proportional to the local rate of change of elastic energy. It simply heats and cools reversibly as the material is loaded and unloaded† In a plastic material, there is no strain energy W and all we can say about Φ is that it is a source in the thermal energy balance. We are a last in a position to propose a simple constitutive law for perfectly plastic flow in which there is no elasticity and f (τij ) = τY . We require the relationship between τij and Dij to be such that the conversion of mechanical to thermal energy be maximised; indeed, we also expect that Φ > 0 so that this conversion is irreversible. This presents us with a relatively straightforward problem in the calculus of variations; given Dij , find τij that maximises ZZZ Dij τij dx subject to f (τij ) = τY . V (t)

For this to happen, small variations ηij in τij must be such that vanishes nfor all ηij such that

RRR

V (t) Dij ηij dx

∂f ηij = 0. ∂τij Hence there is a scalar Lagrange multiplier Λ such that Dij = Λ

∂f , ∂τij

(8.81)

which has the geometric interpretation that the rate of strain is perpendicular to the yield surface when viewed in stress space as in Figure 8.14. Equation (??) is the associated flow rule to the yield function f . If we adopt this rule to relate the stress to the rate-of-strain, then we gain six scalar equations and one further unknown, namely Λ. In total, therefore, the momentum equation (8.75), the yield condition (8.76) and the flow rule † We will discuss this further in Chapter 9.

8.5 Plastic flow

355

(??) comprise ten scalar equations, so we finally have a closed system for τij , vi and Λ, even without assuming incompressibility. However, our insistence that the yield criterion be insensitive to isotropic stress means that f must satisfy ∂f =0 ∂τkk

(8.82)

(summing over k), and then (??) implies that Dkk = div v = 0,

(8.83)

and hence that the flow is indeed incompressible (as shown in Exercise 8.11). This is in encouraging agreement with the experimental evidence that plastic deformation does not appreciably change the density of a metal. We should point out that it is by no means necessary for the stress to be such as to maximise the dissipation. Indeed, there are several alternative non-associative flow rules which are not derived from the yield function. However, (??) is generally found to agree well with experiment and guarantees that the flow shares all the same symmetry properties that we have already imposed on the yield criterion. For example, an isotropic yield function will lead to an isotropic flow rule. As we noted in §8.3, atomic simulations suggest both that dislocations do not respond to isotropic stress and that, when dislocations do flow, they do so with no appreciable volume change. The flow rule (??) suggests that that these two properties of insensitivity to isotropic stress and incompressibility are actually linked at a macroscopic level. We could therefore think of (??) as a hypothesis that the yielding of a plastic material and the way it subsequently flows are essentially controlled by same underlying mechanism. When applying the flow rule (??) in practice, it is usually inconvenient to use the Tresca yield function because of the presence of the corners and we therefore limit our attention henceforth to the Von Mises yield function (8.65). In this case, (??) simply tells us that the rate-of-strain tensor is proportional to the stress deviator, that is µ ¶ Λ Dij = τij0 , (8.84) 2τY as shown in Exercise 8.14. This is called the Levy–von Mises relation, and it states that the principal axes of stress and strain rate coincide. In summary, our mathematical model comprises the three first-order differential equations (8.75), relating τ and v and the seven algebraic relations (8.76, 8.80) which also involve the unknown scalar Ω. Intuitively, we expect

356

Plasticity

three scalar boundary conditions involving vev and τ and an initial condition for v. Analytical progress is almost impossible if the inertial terms are important in (8.75) but frequently they are negligible. For example, a typical left-hand side is of O (()ρU/T ), where U and T are typically velocity and time scales from the boundary conditions, which the right-hand side is of O (()τY /T U ) and in the paper clip example these are about 10−2 and 108 kg/m2 s2 respectively. Hence, in the following example, (8.75) is replaced by ∇ · τ = 0.

Torsion Revisited In the model of §8.4.1, the deviatoric stress coincides with the physical stress and hence (8.75) has been solved in (8.43b). All that remains is to find v = (vx , vy , vz ) and Λ where, from (8.80), ∂vy ∂vx ∂vz = = =0 ∂x ∂y ∂z

(8.85)

∂vx ∂vy ∂vx ∂vz µΩsy + = 0, + = −Λ , ∂y ∂x ∂z ∂x 2τY r ∂v µΩsx z displaystyle ∂zy + ∂v ∂y = Λ 2τY r ,

(8.86)

and

where Ω and s are functions of time, with Ω beign the data and s satisfying (8.42). From (8.85), vz is zero assuming there is no imposed axial velocity at the ends of the bar, and vx = vx (y, z, t), vy = vy (x, z, t). Then (8.86a) gives vx = yV (z, t),

vy = −xV (z, t)

(8.87)

and the problem is reduced to finding V and Λ, which are related by ∂V ∂z = −µΩs Λ/2τY ; Λ is reassuringly positive. To find V we need to find the displacements resulting from this flow by integrating to obtain the particle paths. Writing vx = dx/dt, vy = dy/dt, we see that at once there are the circles x2 + y 2 = r2 = r02

(8.88)

which is only possible if Λ is such that ΛΩs is constant and equal to its value at the time t = t0 that s decreased through r0 , i.e. the time at which the circle (8.88) became plasticised. Further matching of the displacement with the elastic solution (8.85) at time t0 to give that V is a linear function of z, but we will not pursue the details further here. We merely point out that

8.6 Simultaneous Elasticity and Plasticity

357

considerably greater effort would be needed to find the flow in an unloading cycle of the type carried out in §8.4.1.

Tunnels Revisited The determination of the flow of rock in the tunnel problem of §?? is even simpler. We must be careful to remember that, since we are neglecting τ22 , in cylindrical polar coordinates, we must define µ ¶ µ ¶ 1 τrr 0 1 0 0 τ = − (τrr + τθθ ) . 0 τθθ 0 1 2 hence our flow rule gives that the radial velocity v satsifies

where τrr

∂v Λ v Λ = (τrr − τθθ ), = (τθθ − τrr ) ∂r 2τY r 2τY ³ ´k−1 r = −P (t) a(t) and τθθ = kτrr .

We see at once that v = u0r(t) for some function u0 and hence that the displacement in the plastic region is such that the particle which was plasticised at t = t0 , r = a(t0 ) is now at ½ ¾1/2 √ Z t 1 2 r= 2 u0 (τ ) dτ + (a(t0 )) 2 t0 where u0 , Λ are determined by matching to the elastic displacement at r = a.

8.6 Simultaneous Elasticity and Plasticity By even mentioning the idea of a theory of solids that are at once both plastic and elastic, we give ourselves a greater challenge than that of perfectly plastic flow. The best we can do in this text is to briefly cite two proposals that have been used successfully in practice but which still await a mathematical basis for their validity. Both the proposed theories assume that there is no work hardening, so that f (stress) is never allowed to exceed τY , and that f = τY everywhere where there is plastic flow. We recall that in a purely elastic displacement, the strain-rate gradient is (e)

Deij

(e)

(e)

Dτ 1 Dτij λ = − δij kk , Dt 2µ Dt 3λ + 2µ Dt

(e)

(8.89)

where τij is the stress. Concerning the plastic strain rate, we adapt the

358

Plasticity

Von Mises flow rule for simplicity: (p)

Deij Dt

=

Λ (p)0 τ . 2σY ij

(8.90)

We now assume that the molecular structure responds to a single macro(p) (e) scopic stress field τij = τij = τij but that, at least for small strains, the macroscopic strain, eij , is the sum (e)

(p)

eij = eij + eij . hence (8.80) is replced by (e)

Dτ Deij Λ 0 1 Dτij λ = Dij = τ + − δij kk , Dt 2σY ij 2µ Dt (3λ + 2µ) Dt

(8.91)

and (8.76) still holds†. When we neglect inertia and gravity so that ∂τij = 0, ∂xj

(8.92)

we are left with ten equations for vi , τij , the density ρ and Λ, but we can no longer deduce, as in (8.79), that the material is incompressible, and we must append the mass conservation equation (??). Because of the difficulties concerning the definitions of convected derivatives in (8.89) and (8.90), we often write (8.91) and (8.92) as equations for stress and strain increments, rather than as a system of partial differential equations. the increments of de and dτ are related by the Prandtl–Reuss model in which ¶ µ τij0 1 λ ˜+ deij = dΛ dτij − δij dτkk , (8.93) 2σY 2µ (3λ + 2µ) ˜

where DDtΛ = Λ. The iteration involves choosing physically acceptable paths for the stress deviator and the elastic and plastic strains, as described in [Hill, Ch. II.5]. Time does not explicitly in this iteration, which can only predict the sequence of configurations and hence any ultimate steady state. The model (8.93) can be used to describe one of the most famous manifestations of metal plasticity, namely the formations of Luders Bands in bars or sheets under large tensile stress. As illustrated in Figure 8.16, when a polished bar, √ whos free surface is z = 0, is placed udner a tension T (which will equal 3σY under the Von Mises criterion), plastic flow can cause a † Recall an earlier stricture that Dij 6= Deij /Dt for strains of O (()1).

8.6 Simultaneous Elasticity and Plasticity

359

Fig. 8.16.

“necking” region in which surface roughness associated with the large strain ◦ is observed in a band that is inclined at around 45  to the sides  of the bar. T 0 0 The elastic displacement results from a stress  0 0 0  and this is 0 0 0   τ11 τ12 0 T then u(e) = E (x, −νy, −νz). The plastic stress is  τ12 τ22 0  with an 0 0 0 (p) associated strain that we take to be linear in position, u = (ax + by, cx + dy, ez); here we are basiclaly thinking of expanding in a Taylor series near the origin in Figure 8.16??), the top of the band being y = mx. Continuity of displacement with traction at y = mx implies that a + bm =

T , E

c + dm =

−T νm , E

e=

−T ν E

and τ12 = m(τ11 − T ),

τ22 = mτ12 .

Hence (8.93) predicts that, in the ultimate steady state, τ11 + τ22 = 0. We soon find that τ11 =

m2 T , (1 + m2 )

τ12 =

−mT , (1 + m2 )

τ22 =

−m2 T ,. (1 + m2 )

(8.94)

Our final piece of information comes from the Von Mises condition in the plastic region y < mx, which demands that 2 2 2 τ11 + τ22 − τ11 τ22 + 3τ12 = 3σY2 = T 2 .

360

Plasticity

√ Substituting from (8.94) we find that m = 1/ 2 so that, in this mould, the bands are included at about 35◦ . Concluding Remarks Thus far, we have only found it possible to make any analytical progress at all in cases where the flow is sufficiently slow for the left-hand side of Cauchy’s momentum equation (8.75) to be neglected. We note, however, that there is an opposite regime in which the plastic flow is so fast that the inertia term in (8.75) is dominant, and the plastic stress is negligible in comparison. This limit applies, for example, to the violent metal plasticity which occurs when a device called a shaped charge explodes against a target, producing a liquid metal jet moving at enormous speed (up to 14 kilometres per second). This exerts a pressure on the target that can be orders of magnitude larger than its yield stress, and the target therefore gives way plastically as the jet forces its way through. Shaped charges are chillingly effective, able to penetrate several metres of steel. To understand such behaviour, we neglect the body force and write the momentum equation (8.75) in terms of the pressure and deviatoric stress as ½ ¾ ∂v ρ + (v · ∇) v = −∇p + ∇ · τ 0 . (8.95) ∂t As a consequence of the yield condition, we know that the components of τ 0 can be at most of order τY . On the other hand, if the material is flowing at a typical speed V over a region of typical length-scale L, we can see that the left-hand side of (8.95) is of order ρV 2 /L. The deviatoric stress is therefore p negligible if the flow speed is sufficiently large, specifically if V À τY /ρ. If so, then (8.95) and (8.83) reduce to the Euler equations of incompressible inviscid fluid dynamics: the target flows like an incompressible inviscid liquid to lowest order. Thus shaped charge penetration is relatively simple to model compared with plastic penetration at lower stresses. Exercises 8.1

(i) Show that the height of sand piled maximally of a horizontal circular table bounded by x2 + y 2 = a2 is z = (a − r) tan φ where r2 = x2 + y 2 . [This means a-r =0!] (ii) Show that if the table is the rectangle |x| 6 a, |y| 6 b, then the sand pile takes the shape of a roof with ridge line y = 0, |x| < a − b, when a < b.

Exercises

361

F

φ τ1

τ2

τ3

N

Fig. 8.17. The Mohr surface in the (N, F )-plane for three-dimensional granular flow, and the line F = |N | tan φ. The region where solutions exist is shaded.

(iii) Show that if the table is the ellipse x = a cos s, y = b sin s, the 2 2 ridge line is y = 0, |x| < a −b a . 8.2 Consider a three-dimensional stress field expressed with respect to principal axes so that τ = diag (τ1 , τ2 , τ3 ), where τk are the principal stresses. Recall that the normal and tangential tractions on a surface element with unit normal n are given by N = n · (τ n) ,

F = τ n − N n.

(a) Deduce that N and F = |F | satisfy the relations µ ¶2 1 (τ1 − τ2 )2 F 2 + N − (τ1 + τ2 ) = + (τ1 − τ3 ) (τ2 − τ3 ) n23 , 2 4 µ ¶2 1 (τ2 − τ3 )2 F 2 + N − (τ2 + τ3 ) = + (τ2 − τ1 ) (τ3 − τ1 ) n21 , 2 4 µ ¶2 1 (τ3 − τ1 )2 F 2 + N − (τ3 + τ1 ) = + (τ3 − τ1 ) (τ1 − τ2 ) n22 , 2 4 where nk are the components of n. Hence show that N and F lie on points in the (N, F )-plane bounded by three semicircular arcs, as illustrated in Figure 8.17. (b) Assuming (without loss of generality) that the stresses are ordered such that τ1 6 τ2 6 τ3 6 0, deduce that the Coulomb yield criterion is equivalent to τ3 − τ1 = − (τ3 + τ1 ) sin φ, where φ is the angle of friction.

362

Plasticity

(c) Show also that the maximum value of F is given by (τ3 −τ1 )/2. 8.3 Show that the equation (8.7) governing two-dimensional granular flow with no body force is hyperbolic. [One approach is to differentiate the equation with respect to x to obtain a second-order quasi-linear equation for ∂A/∂x.] Reach the same conclusion for the system (8.5), (8.6) including a body force. [Hint: again try differentiating with respect to x.] 8.4 (a) For the displacement field (8.21) corresponding to a screw dislocation, calculate the jump in w as a simple closed contour C in the (x, y)-plane is traversed, and hence show that ¶ ( I µ £ ¤ b if C encloses the origin, ∂w ∂w w C= dx + dy = ∂x ∂y 0 otherwise. C Now assuming that Green’s theorem holds (this can be justified using the theory of distributions; see, for example, Keener, 2000, Chapter 4), show that ( ¶ ZZ µ 2 b if S contains the origin, ∂2w ∂ w − dxdy = ∂x∂y ∂y∂x 0 otherwise, S where S is the interior of C. Deduce that w satisfies (8.23). (b) Alternatively, observe in Figure 8.6 that, although w has a jump of magnitude b across the positive x-axis, ∂w/∂x is continuous. Hence explain why ∂w −by = , ∂x 2π (x2 + y 2 )

∂w bx = − bH(x)δ(y), ∂y 2π (x2 + y 2 )

where H denotes the Heaviside function (3.133), and, by crossdifferentiating, obtain (8.23). 8.5 Calculations needed for edge dislocation calculation. To get (8.27), take Ψ1 =

4µθ , 3 − 4ν

Ψ2 =

4µ . (3 − 4ν)2

8.6 Using the relations (2.98) for the polar stress components in terms of the Airy stress function A, show that (8.28) is consistent with A = c1 rθ cos θ + c2 r log r sin θ, and determine the appropriate values of the constants c1 and c2 .

Exercises

363

2M/πa3 τY 2

1.0

0.5

-3

-2

1 1

-1

-1.0

2

3

Ω/Ωc

4

-0.5

6

3

5

Fig. 8.18. The normalised torque M versus twist Ω applied to an elastic/plastic cylindrical bar undergoing one loading cycle. The bar deforms elastically (1), then yields (2), then recovers elastically (3) when the load is released. When a load is now applied in the opposite direction (4), the bar yields at a lower critical torque (5) and then deforms plastically. Finally, the load is released (6) and a permanent negative twist remains.

Hence show that A satisfies µ ¶ µ ¶ Eb sin θ Eb ∂θ 2 ∇ A= =− , π(1 + ν)(3 − 4ν) r π(1 + ν)(3 − 4ν) ∂x where θ is the usual polar angle. Assuming that 0 6 θ < 2π, and using the results of Exercise 8.4, deduce that A satisfies (8.30). 8.7 Virtual dislocations calculation for mode III crack. 8.8 Return to the calculation of elastic/plastic behaviour in a circular torsion bar from §8.4.1. Show that the net torque M exerted by the stress field (8.46) is given by µ ¶ ˜ 2M 1 Ω3c Ω = 4 − + , πa3 τY 3 Ω3max Ωc and hence confirm that the recovery twist is given by (8.47) when M = 0. Suppose we now twist the bar in the opposite direction, so that M is negative. Use (8.46) to show that the bar yields again at r = a when ˜ = −2Ωc , Ω

2M Ω3c 2 − = − > −1. πa3 τY 3 3Ω3max

[Thus, after the bar has yielded in one direction, a smaller torque is required to make it subsequently yield in the opposite direction. This is known as the Bauschinger effect.]

364

Plasticity

Show that the stress in the bar is then given  ˜ µ ¶ µ ¶  Ω + Ωmax τxz −y ˜ + Ωmax s/r =µ × Ω τyz x   −τY /µr

by 0 6 r < s, s < r < η, η < r < a,

where the position of the new yield boundary is r=η=−

Ωc a . ˜ Ω

Hence show that the associated torque is given by 2M 4 Ω3c 16Ω3c = − − − ˜3 πa3 τY 3 3Ω3max 3Ω ˜ < −2Ωc . Figure 8.18 shows how the twist varies as for −2Ωmax < Ω the torque cycles through positive and negative values. ˜ < −2Ωmax ? What happens when Ω 8.9 Deduce from (8.3) that, in plane strain, the shear stress on a surface element with unit tangent (cos θ, sin θ, 0)T is 1 F = (τyy − τxx ) sin(2θ) − τxy cos(2θ). 2 Show that F takes its maximum and minimum values where τxx − τyy tan(2θ) = . 2τxy If the maximum value of F is equal to τY , deduce that tan θ =

2 (τxy ± τY ) . τxx − τyy

8.10 From the linear elastic constitutive relations in plane strain, show that the normal stress is given by τzz = ν (τxx + τyy ) , where ν is Poisson’s ratio. Deduce that the normal deviatoric stress τ30 is related to the principal in-plane deviatoric stresses and the pressure p by ¡ ¢ τ30 = (1 − 2ν)p + ν τ10 + τ20 and hence obtain the relation τ10

+

τ20

=−

µ

1 − 2ν 1+ν

¶ p.

Exercises

365

For each fixed value of p, this gives us a straight line in the (τ10 , τ20 )plane. By considering the intersection of this line with the polygon in Figure 8.13(b), show that the plane strain Tresca yield condition is |τ10 − τ20 | = 2τY provided |p| 6

2(1 + ν)τY . 3(1 − 2ν)

[If this condition is violated, then the normal stress τzz that must be applied to prevent out-of-plane displacement will cause the material to yield in the z-direction.] 8.11 (a) Recall that ¶ µ ∂xi J = det ∂Xj is the Jacobian of the transformation from Lagrangian to Eulerian variables. Prove Euler’s Identity DJ = J div v, Dt where D/Dt is the convective derivative. Deduce that the density ρ satisfies the continuity equation ∂ρ + div(ρv) = 0. ∂t (b) Hence prove Reynolds’ transport theorem: for any volume V (t) that is convected with velocity v(x, t) and any differentiable function F (x, t), ZZZ ZZZ d DF F ρ dx = ρ dx. dt V (t) V (t) Dt 8.12 Consider a material flowing with velocity field v(x, t). A material point occupying position x at time t is therefore convected to position x + v(x, t)δt after a small time δt. Show that a neighbouring point x + δx is convected to a position with coordinates ¶ µ ∂vi (x, t)δxj δt xi + δxi + vi (x, t) + ∂xj (using the summation convention). Deduce that the rate of change of the line element δx satisfies ∂vi d (δxi ) = δxj , dt ∂xj

366

Plasticity

and, by appropriate choice of the dummy summation indices, obtain the equations δxi

∂vj d ∂vi (δxi ) = δxi δxj = δxi δxj . dt ∂xj ∂xi

Hence show that the length of the line element satisfies (8.72). 8.13 A mechanical energy balance on a material volume V (t) flowing with velocity v(x, t) yields the equation ZZZ ZZZ d 1 ρ |v|2 dx − v · (ρg) dx dt V (t) 2 V (t) ZZ ZZZ vi τij nj dS = − Φ dx. − ∂V (t)

V (t)

The terms on the left-hand side represent the rate of change of kinetic energy, the work done by gravity and the work done by stress on the boundary ∂V , with unit normal n = (n1 , n2 , n3 )T . The right-hand side is the rate at which mechanical energy is dissipated (as heat), and Φ is known as the dissipation function. By using the transport theorem, the divergence theorem and the momentum equation, deduce that Φ = Dij τij , where Dij is the rate-of-strain tensor. 8.14 (a) Show that the flow rule (8.14) is invariant under rotations of the coordinate axes. (b) Show that the Von Mises flow rule leads to (8.84) when we choose principal axes in which the stress tensor is diagonal. (c) Deduce that (8.84) is valid with respect to any chosen axes.

9 More General Theories

9.1 Introduction We willnow mention a wide range of important solid mechanics phenomena that warrant discussion even in a mathematically-oriented book but have been ignored or scarcely mentioned so far. A phenomenon of interest in industries ranging from food to glass is that of viscoelasticity. Here the molecular structure of the material is such that it flows under any applied stress, no matter how small. As we will see in §9.2, viscoelastic materials are quite unlike plastic materials. Their behaviour depends critically on the timescale of the observer; for example when a ball of “silly putty” or suitably dilute custard impacts a wall, it rebounds almost elastically but, if left on a table, it will slowly spread horizontally under the action of gravity. Yet another attribute of solids is well-known to be of great practical importance in the kitchen, when glass can be observed to break in hot water, and in the railway industry, where track can distort in high summer. This is thermoelasticity, which describes the response of elastic solids to temperature variations. To model this response ab initio, even at a macroscopic scale, requires more thermodynamics than is appropriate for this text, and in §9.3 we will only present the simplest ad hoc model that can give useful realistic results. Other fundamental phenomena that could be discussed if space permitted are electromagnetic effects such as piezoelasticity and ferromagnetism (see [ ]), and phase changes in solids, which can be crucial in determining the material properties of metal alloys [ ]. Reluctantly, we will have to omit most of these fascinating theories here. Even the above lengthy list of diverse phenomena only relates to solids that are fairly homogeneous on a macroscopic scale. Hence this book would 367

368

More General Theories

certainly not be a fair introduction to the mathematics of solid mechanics without some discussion of the increasingly important properties of composites. These are solids in which the elastic properties vary on a length-scale much smaller than any macroscopic length-scale of practical relevance but not so small as to render a continuum model inappropriate. They range from ceramics to metallic foams, in which voids may be distributed throughout the material, to fibre-reinforced solids, wood, fabrics and many other biomaterials. In §9.4 we will briefly introduce the mathematical technique of homogenisation, which allows the macroscopic properties of a composite to be determined from knowledge of the geometry of the microstructure. There are also many important materials consisting of a porous elastic matrix where the pores are filled with a viscous fluid. This description applies, for example, to saturated rock or soil, as well as many biological tissues. In these materials, the elastic deformation and fluid flow are intimately coupled: compression of the matrix forces the fluid to flow (as in squeezing a wet sponge), and the flowing fluid in turn exerts a drag on the matrix. In §9.5, we will derive a simple model for these poroelastic materials. Finally, in §9.6, we will discuss briefly what can be said about anisotropy, which is often a consequence of homogenisation. 9.2 Viscoelasticity 9.2.1 Introduction Viscoelasticity concerns materials which respond to an applied stress by exhibiting a combination of an elastic displacement and a viscous flow. This is another vast area of the theoretical mechanics to which we can only provide an elementary introduction. Such materials exist all around us. Consider rock, for example, which is elastic on human time-scales but flows over geological time-scales, for example when mountains are formed by collision between tectonic plates. Indeed, we will discover shortly that the key dimensionless parameter that characterises viscoelastic behaviour is the Deborah number, which was named after the Old Testament prophet Deborah, quoted as saying “The mountains flowed before the Lord” (Judges 5:5). We note two important distinctions between viscoelastic and plastic behaviour. First, a viscoelastic material always exhibits a combination of elasticity and flow: no yield stress needs be exceeded for flow to occur. Second, as the examples above demonstrate, the response of a viscoelastic material to a given stress is crucially dependent on the time-scale over which that stress is applied. In contrast, plastic flow depends only on the magnitude of the stress.

9.2 Viscoelasticity

(a)

(b)

u

T

369

u T

T

T

(d)

(c) Te

Te

ue T

Tv

T uv

Tv u

Fig. 9.1. (a) A spring; (b) a dashpot; (c) a spring and dashpot connected in parallel; (d) a spring and dashpot connected in series.

In Chapter 1, we based our elastic constitutive law between stress and strain on Hooke’s law. We will therefore begin by considering how Hooke’s law can be generalised to describe viscoelastic behaviour. We will then show how such a law may be used to construct a continuum model for a viscoelastic material. 9.2.2 Springs and dashpots Let us first consider a spring subject to a tension T , as illustrated in Figure 9.1(a). According to Hooke’s law, the extension u of the spring is related to T by (??), i.e. T = ku,

(9.1)

where k is the spring constant. In constructing the constitutive equation (1.42) for a linear elastic solid, we appealed to Hooke’s law, arguing that each line element in the solid should behave in a manner analogous to a linear spring. For a viscous fluid, the corresponding fundamental element is the dashpot, illustrated in Figure 9.1(b). Dashpots encompass shock absorbers in motor car suspensions, as well the mechanisms that prevent doors from slamming. They are mechanical devices designed to offer a resistance proportional to the velocity at which it is exended or compressed. The law corresponding to (9.1) for a linear dashpot is T =Y

du , dt

(9.2)

370


T

(a)

te

u

t

(b)

u

t

(c)

t

Fig. 9.2. (a) Applied tension T as a function of time t; te is the characteristic time-scale of the experiment. (b) Resultant displacement of a linear elastic spring. (c) Resultant displacement of a linear dashpot.

where Y is called the impedance of the dashpot. The two simple systems (9.1) and (9.2) encapsulate the contrast between solid and fluid behaviour, as we now illustrate with a thought experiment. Suppose we apply a time-dependent tension T (t) described by a ramp function, as shown in Figure 9.2. For a spring, the displacement is simply proportional to T , so u will behave in exactly the same way as T . In particular, a constant tension causes a constant displacement, and u returns to zero when T does so. The dashpot, however, continues to extend as long as a positive tension is applied to it, and a nonzero residual displacement remains after the tension has been released. Heuristically, we might say that the dashpot has no memory of its initial state: the stress it exerts depends only on the instantaneous velocity applied to it, which is a defining characteristic of a viscous fluid. In contrast, the spring, like an elastic solid, has a fixed rest state at u = 0 which it “remembers”, and the tension is nonzero whenever the spring departs from this state. We can now construct model viscoelastic elements by sticking together springs and dashpots, and two obvious possibilities are to connect them in parallel or in series, as shown in Figure 9.1(c) and (d) respectively. In Figure 9.1(c), which illustrates what is known as a Voigt element, we see that the displacements in the spring and the dashpot are equal, by construction, and the tension T is the sum of the tensions in the spring and in the dashpot. We therefore obtain du T = Te + Tv , where Te = ku, Tv = Y , (9.3) dt where the subscripts e and v stand for the elastic and viscous contributions respectively. These are easily combined to give µ ¶ du Y du + ku = k tr +u , where tr = , (9.4) T =Y dt dt k

9.2 Viscoelasticity

u

(a)

u

371

(b)

t

t

Fig. 9.3. Displacement of a Voigt element due to the applied tension shown in Figure 9.2(a). (a) Small Deborah number; the corresponding response of an elastic spring is shown as a dashed curve. (a) Large Deborah number; the corresponding response of a dashpot is shown as a dashed curve.

and tr is known as the relaxation time of the Voigt element. The qualitative behaviour depends crucially on the size of the relaxation time compared with the characteristic time-scale te of any given experiment, and we therefore define a dimensionless parameter, called the Deborah number, by De =

tr . te

(9.5)

When De is small, the time derivative in (9.4) has a small influence, and the element therefore behaves somewhat like an elastic spring. In Figure 9.3(a) we show the response of such an element to the same applied tension as in Figure 9.2(a). Here we see that the displacement lags slightly behind the corresponding elastic displacement (shown as a dashed curve), where the time lag is of order R. However, when De is large, the time derivative is the dominant term on the right-hand side of (9.4), so the Voigt element responds like a dashpot, as shown in Figure 9.3(b). However, if we wait long enough, the displacement will always decay to zero, over a time-scale tr , when the tension is released. The Voigt constitutive relation (9.4) therefore describes the basic building block for a damped solid. On the other hand, in Figure 9.1(d), illustrates what is known as a Maxwell element, in which the spring and dashpot experience the same tension T and the net displacement is the sum of the elastic and viscous contributions; hence u = ue + uv ,

where

T = kue ,

T =Y

duv . dt

(9.6)

372


u

(a)

u

(b)

t

t

Fig. 9.4. Displacement of a Maxwell element due to the applied tension shown in Figure 9.2(a). (a) Small Deborah number; the corresponding response of a dashpot is shown as a dashed curve. (a) Large Deborah number; the corresponding response of an elastic spring is shown as a dashed curve.

Now differentiation with respect to t leads to µ ¶ d T dT du =Y + T = tr + T, Y dt dt k dt

(9.7)

where tr = Y/k is again the relaxation time. Now, when De ¿ 1, the time derivative term is small, so the Maxwell constitutive equation (9.7) is approximately that of a dashpot, as shown in Figure 9.4(a). The response when the Deborah number is large, shown in Figure 9.4(b), is close to that of an elastic spring except in two crucial respects. First, the element continues to extend slightly, or creep, when the tension is held constant. Second, a nonzero permanent displacement remains when the tension is released. We can therefore think of (9.7) as describing an elastic fluid or a fluid with memory. By inspecting Figures 9.3 and Figures 9.4, we can contrast the Voigt and Maxwell constitutive relations as follows. The Voigt element exhibits viscous behaviour over short time-scales and elastic behaviour over long time-scales. An example of such a response is a soft rubber toy which can initially be deformed like a fluid but then slowly recovers its initial shape. The Maxwell element has the opposite response, behaving elastically over short time-scales but flowing over long time-scales. This might describe rock or silly putty, for example. By combining springs and dashpots in other configurations, one can construct systems with ever more complicated responses. However, all such linear systems may be viewed as special cases of the general linear consitutive law Z t ¡ ¢ dT ¡ 0 ¢ 0 u(t) = K t − t0 t dt , (9.8) dt −∞

9.2 Viscoelasticity

373

where K(t), known as the creep function, characterises the viscoelastic response of the system. For example, the Voigt and Maxwell constitutive laws may be cast in the form (9.8), with the creep function given by K(t) =

1 − e−t/tr k

and K(t) =

tr + t , Y

(9.9)

respectively (see Exercise 9.1). We point out that one-dimensional linear viscoelasticity can be approached by generalising Exercise ?? and taking the continuum limit of a series of spring and dashpot elements†. Some of the different possibilities are described in Exercise ??.

9.2.3 Three-dimensional linear viscoelasticity When constructing the constitutive relationship for a linear elastic medium in Chapter 1, we supposed that each infinitesimal line element in the material obeys a version of Hooke’s law. Now we can incorporate viscoelastic effects by replacing Hooke’s law with a rate-dependent constitutive law such as (9.4) or (9.7). There is no hard-and-fast rule for doing this: for any given material the choice of constitutive law must be based on its qualitative properties, for example whether it behaves like a fluid or a solid over short and long time-scales, and then verified experimentally. The Voigt element from Figure 9.1(a), for example, could be generalised to three dimensions as follows. We imagine that the stress tensor τij at each point is provided by an elastic contribution τije and a viscous contribution τijv . We then assume that these satisfy the standard linear elastic and Newtonian constitutive relations, so that τij = τije + τijv , τije = λ (ekk ) δij + 2µeij , τijv = ξ (Dkk ) δij + 2ηDij , (9.10) where λ and µ are the usual Lamé constants, while ξ and η are the called the dilatational viscosity and shear viscosity respectively. As explained in §8.5, the rate-of-strain tensor Dij may be approximated as simply the time derivative of the strain tensor eij provided the displacement gradients are small enough for linear elasticity to be valid. If we make the simplifying assumption that the elastic and viscous components have the same Poisson’s ratio, then we can define a single relaxation † It is not so easy to use discrete element models in more than one dimension; for example, it can be shown that it is impossible to retrieve our linear elasticity model (??) as the limit of a lattice of masses joined by springs aligned along three orthogonal axes.

374


time tr =

η ξ = . λ µ

(9.11)

Then, by collecting the terms in (9.10), we obtain the constitutive relation µ ¶ ∂ τij = 1 + tr (λ (ekk ) δij + 2µeij ) , (9.12) ∂t for small displacements of a Voigt solid. Alternatively, we can base a Maxwell model on the idea that the stress at any point gives rise to both an elastic strain eeij and a viscous strain evij . Again we assume that these are both small enough to be described by linear theory and that the viscous and elastic components have the same Poisson’s ratio to obtain ¢ ∂ ¡ eij = eeij + evij , τij = λ (eekk ) δij + 2µeeij = tr λ (evkk ) δij + 2µevij . ∂t (9.13) Now these are easily combined to give the linear Maxwell model ∂τij ∂ + τij = tr (λ (ekk ) δij + 2µeij ) . (9.14) ∂t ∂t Let us illustrate the equations that result from a viscoelastic constitutive relation by returning to the model for compressional waves in a bar. We recall from §4.2. that Newton’s second law leads to the equation tr

∂2u ∂T , (9.15) = 2 ∂t ∂x where ρ and A are the density and cross-sectional area of the bar. To close the model we need to specify a constitutive relation between the tension T and the displacement u. The Voigt model (9.12) leads to µ ¶ ∂u ∂2u T = EA + tr , (9.16) ∂x ∂x∂t ρA

where E is Young’s modulus, so that (9.15) reads µ 2 ¶ ∂2u ∂ u ∂3u ρ 2 =E + tr 2 . ∂t ∂x2 ∂x ∂t

(9.17)

The standard wave equation, which is retrieved when tr = 0, conserves energy, so that any initial displacement will give rise to waves that propagate indefinitely. The Voigt model (9.17) incorporates the physically important effect of damping, so that all disturbances eventually decay to zero.

9.2 Viscoelasticity

375

On the other hand, the tension in a Maxwell material satisfies tr

∂T ∂2u + T = EA , ∂t ∂x∂t

(9.18)

so that (9.15) leads to µ 3 ¶ ∂ u ∂2u ∂3u ρ tr 3 + 2 = E 2 . ∂t ∂t ∂x ∂t

(9.19)

This equation admits steady solutions in which u is an arbitrary function of x, so that any displacement of such a bar can give rise to permanent deformation. More generally, we can think of (9.12) and (9.14) as special cases of the general linear three-dimensional constitutive relation Z t ¡ ¢ ∂τk` ¡ 0 ¢ 0 x, t dt , (9.20) eij = Kijk` t − t0 ∂t −∞ where the creep function is now in general a fourth-rank tensor. If the material is isotropic, then (9.20) may be characterised using just two scalar creep functions L(t) and M (t) by writing ¾ Z t ½ ¡ ¢ ¡ 0¢ ¡ ¢ ¡ 0¢ 0 ∂τkk 0 ∂τij eij = L t−t x, t δij + M t − t x, t dt0 . (9.21) ∂t ∂t −∞ In either case, the integral should really be performed with the Lagrangian variable X held fixed, and we must emphasise once more that all these linear theories are only valid for small strains. We can expect them to work well for small displacements of a damped solid or for small creep of an elastic fluid, but for a fluid that undergoes a substantial deformation, a more sophisticated approach is required, as we will now briefly demonstrate. 9.2.4 Large-strain viscoelasticity In viscoelastic fluids, the strains are frequently so large that, as in Chapter 5, we must resort to Lagrangian variables to deal with the geometric nonlinearity. When such materials may undergo large deformations, their compressibility is nearly always small in comparison, and it is therefore usual to model them as incompressible. We therefore introduce a pressure p and a deviatoric stress τ 0 , as in §8.4.3, so that Cauchy’s momentum equation takes the form ¾ ½ ∂v + (v · ∇) v = −∇p + ∇ · τ 0 + ρg, (9.22) ρ ∂t

376


where v is the velocity vector. Since we would not expect a Voigt solid to undergo particularly large displacements, we concentrate on the example of a Maxwell fluid, in which the deformation is decomposed into an elastic component and a viscous component. As described in any book on viscouse flow (for example, Ockendon & Ockendon, 1995) for a Newtonian viscous fluid, the constitutive relation is simply τij0 = 2ηDij ,

(9.23)

where Dij is the rate-of-strain tensor and η is again the shear viscosity. On the other hand, a neo-Hookean solid, with strain energy density given by (5.91a), obeys the constitutive relation τij0 = µ (Bij − δij ) ,

(9.24)

where the Finger strain tensor Bij is defined in (5.14). By differentiating the Finger tensor with respect to t, as shown in Exercise 9.6, we find that DBij ∂vj ∂vi = Bkj + Bik , Dt ∂xk ∂xk

(9.25)

∂vj DB ∂vi = ∇v T B + B∇v. = Bkj + Bik . Dt ∂xk ∂xk

(9.26)

i.e.

The neo-Hookean stress tensor (9.24) therefore satisfies the equation 5

τ 0 = 2µD,

(9.27)

5

where denotes the upper convected derivative, defined for an arbitrary tensor function A by 5

A=

DA − ∇v T A − A∇v. Dt

(9.28)

By combining (9.23) and (9.27), we can describe a viscoelastic fluid whose elastic component is neo-Hookean could be described by 5

5

tr τ 0 ij + τ 0 ij = 2ηDij

(9.29)

which is known as the upper convected Maxwell model. As another example of an elastic constitutive relation, let us consider the

9.2 Viscoelasticity

377

Mooney–Rivlin strain energy density (5.91b), with c1 = 0, c2 = µ/2, which leads to ³ ´ −1 τij0 = µ δij − Bij . (9.30) Again using (9.25), we discover that this stress tensor satisfies 4

τ 0 = 2µD,

(9.31)

where the lower convected derivative of a tensor A is defined by 4

A=

DA + ∇vA + A∇v T . Dt

We obtain the lower convected Maxwell model by using (9.28). 4

(9.32) 4

instead of

5

in

5

Both and are special cases of a one-parameter family of convective derivatives, defined by DA − ΩA + AΩ − a (DA + AD) , (9.33) Dt where D is again the rate-of-strain, while Ω is the rotation tensor, that is µ ¶ µ ¶ ∂vj ∂vj 1 ∂vi 1 ∂vi Dij = + , Ωij = − . (9.34) 2 ∂xj ∂xi 2 ∂xj ∂xi ?

A=

The parameter a must lie in the range [−1, 1]. The upper and lower convected derivatives are obtained when a = 1 and a = −1 respectively, and ◦

another common choice is the Jaumann or corotational derivative A, corresponding to a = 0. In any viscoelastic constitutive relation, a is a material parameter that must be determined from experiments. For any value of a, the derivative defined by (9.33) is objective, that is, it takes the same value for all observers, in contrast with the “obvious” convective derivative DA/Dt. Hence, when we construct a constitutive relation ? using , we can be sure that the resulting mechanical behaviour will be independent of the frame in which it is measured. Nonetheless, whichever member of (9.33) we choose, the resulting large strain viscoelastic model will couple (9.22) with an equally complicated nonlinear evolution equaiton for τ 0 . Hence analytical progress can usually only be made in the case of ? small strains, in which case is simply ∂/∂t. Finally. we must mention that the identification and classification of convective derivatives is one that arises in many branches of mathematics, and it is interesting that some of these derivatives have a geometrical interpretation, as show in Exercise ??.

378


9.3 Thermoelasticity Thermal effects in solids can be dramatic, even if we exclude the possibility of the solid melting into a liquid. For example, the easiest way in which to fracture an ice cube is to put it into even lukewarm water. We note that density variations are already allowed for in the mass- and momentumconservation equations (1.8) and (1.36). Hence it only remains for us to incorporate thermal effects in the constitutive equation (1.42) and then to include heat in the energy calculation from §1.9. We will assume throughout that the displacement gradients are small enough for linear elasticity to be valid. We begin to construct a linear thermoelastic constitutive relation by considering an unconstrained elastic solid whose absolute temperature is raised from T0 in the reference state to a new temperature T . There is ample experimental evidence that many common materials undergoe a uniform isotropic volume change: an expansion if T > T0 or a contraction if T < T0 . For moderate temperature variations, the thermal expansion is found to vary linearly with temperature, so that eij =

α (T − T0 )δij , 3

(9.35)

where α is called the coefficient of thermal expansion. By referring to §2.2.1, we can interpret α as the relative volume expansion caused by a unit increase in temperature. We are therefore prompted to propose the constitutive relation µ ¶ 2 τij = 2µeij + λekk δij − λ + µ α(T − T0 )δij . (9.36) 3 This is the only linear relation between τij , eij and T which reduces to (1.42) when T = T0 and to (9.35) when τij = 0. In a linear theory, the Lamé constants are assumed not to vary significantly over the temperature ranges of interest. When we use (9.36) in the momentum equation (1.36), we find that the Navier equation generalises to µ ¶ 2 ∂2u 2 (9.37) ρ 2 = ρg + (λ + µ) grad div u + µ∇ u − λ + µ α∇T, ∂t 3 so that the temperature gradient generates an effective body force in the Navier equation. At first glance, it might appear from (9.37) that a uniform temperature change, with ∇T = 0, could not drive a displacement. However, whenever boundary conditions are specified on either the displacement or the stress,

9.3 Thermoelasticity

379

the constituive relation (9.36) will lead to nontrivial dislacements. We illustrate this by returning to the steady uniaxial loading of a bar, considered previously in §2.2.3. Here we examine the effect of heating the bar uniformly to some temperature T . If we seek a displacement of the form u = (βx, −γy, −γz)T , then the stress tensor is given by (9.36) as   · µ ¶ ¸ β 0 0 2µ τ = 2µ  0 −γ 0  + λ(β − 2γ) − λ + α(T − T0 ) I. 3 0 0 −γ

(9.38)

(9.39)

By ensuring that the lateral stress components τyy and τzz are zero, we find that the transverse strain is given by γ=

(3λ + 2µ)α(T − T0 ) 1+ν λβ − = νβ − α(T − T0 ), 2(λ + µ) 6(λ + µ) 3

(9.40)

where ν is Poisson’s ratio. The only remaining nonzero stress component is µ ¶ 1 τxx = β − α(T − T0 ) E, (9.41) 3 where E is Young’s modulus. We can use this result to calculate the thermal stress in a length of railway track†. If the two ends of the track are fixed, then we must have β = 0 and hence the net compressive force is given by P =

EA α(T − T0 ), 3

(9.42)

where A is the cross-sectional area. Note that heating the track will cause a compressive force and cooling a tensile force, as expected. From §4.4.3, we know that the track will buckle if the compressive stress is too large, and (9.42) tells us that the maximum temperature that can be sustained before buckling occurs is T − T0 =

3π 2 I , AL2 α

(9.43)

where I is the moment of inertia of the cross-section. In general, we cannot assume that temperature is known, but must determine it as part of the solution. We now obtain an equation for T by † In the days before “continuously-welded” rails, gaps had to be left between sections of track to allow for thermal expansion to take place without stress generation. These gaps cause track and wheels to wear and led to many accidents.

380


performing an energy balance that generalises the on in §8.6. The net energy U in a control volume V is given by ) Z Z Z ( ¯ ¯2 1 ¯¯ ∂u ¯¯ U= ρ + W + ρcT dx, (9.44) 2 ¯ ∂t ¯ V where, referring to §1.9, we recognise the first two terms in the integrand as the kinetic energy and the strain energy. The final term is the thermal energy, with c denoting the specific heat capacity†, a material parameter which we will assume to be constant. As in §8.6, the first law of thermodynamics tells us that all three forms of energy are equivalent and mutually convertible. The energy inside V will change due to (a) the work done by gravity, (b) the work done by stress on the boundary ∂V , and (c) the flux of heat through ∂V . Collecting these effects together, we arrive at the energy conservation equation ZZZ ZZ ZZ dE ∂u ∂u = dx + da − ρg · (τ n) · q · n da, (9.45) dt ∂t ∂t V ∂V ∂V where n is the unit normal to ∂V and q is the heat flux. Now we have to substitute (9.44) into (9.45) and simplify the resulting expression. As shown in Exercise 9.9, for sufficiently small temperature variations, this leads to ρc

∂T + div q = 0, ∂t

(9.46)

which is the standard equation for heat flow in a rigid sample. In other words, there is no leading-order coupling between thermal expansion and heat flow: we can solve (9.46) for T and then substitute the result into (9.37) to determine the resulting displacement. For most materials, the heat flux is given by Fourier’s law, namely q = −k∇T where k is the thermal conductivity. Provided k is constant, (9.46) just becomes the linear heat equation ∂T = κ∇2 T, ∂t

(9.47)

where κ = k/ρc is called the thermal diffusivity. We note in passing that the situation is much more complicated when thermal radiation is significant, as often happens in semi-transparent materials such as glass. † Strictly speaking, this should be defined “at constant volume”.

9.4 Composite Materials and Homogenisation

E E1

(a)

u

E2

381

(b)

x

x

Fig. 9.5. (a) The variation of Young’s modulus with position in a bar. (b) The corresponding longitudinal displacement u(x); the approximation (9.53) is shown as a dashed line.

9.4 Composite Materials and Homogenisation 9.4.1 One-dimensional homogenisation Many practically important materials have properties which vary over a very fine scale, for example composites consisting of alternating layers with two different properties. To describe deformations of such an object, it is impractical to analyse each individual layer. In this Section, we briefly outline a homogenisation technique that allows us to average over the fine scale and hence determine effective mechanical properties for large-scale deformations of the material as a whole. The word homogenisation is often used synonymously with “multiscaling” or “coarse-graining”. We will only consider configurations where the short scale greatly exceeds molecular length scales. Our first example concerns longitudinal displacement of an elastic bar, with uniform cross-sectional area A and length L, whose Young’s modulus E is a rapidly varying function of position x. We recall from §4.2 that the displacement u(x) in such a bar satisfies AE(x)

du = T, dx

(9.48)

where the tension T is constant under static conditions. To fix ideas, we focus for the moment on a composite bar consisting of 2N alternating sections with moduli E1 and E2 , so that ( E(x) =

E1

nL/N < x < (n + φ)L/N,

E2

(n + φ)L/N < x < (n + 1)L/N,

(9.49)

where n = 0, 1, . . . , N − 1 and φ is the volume fraction of phase 1.. In this

382


case, it is easy to solve (9.48) and find  µ ¶ x (1 − φ)nL 1 1   + −  N E2 E1 T  E1 u(x) = µ ¶ A x φ(n + 1)L 1 1   + −  E2 N E1 E2

nL (n + φ)L <x< , N N (n + φ)L (n + 1)L <x< . N N (9.50)

In Figure 9.5, we plot a typical piecewise constant Young’s modulus profile and the corresponding displacement u(x). We can see that u(x) fluctuates rapidly in response to the variations in E. However, on the scale of the whole bar, these fluctuations are barely noticeable, and the displacement appears to be approximately linear in x. Indeed, from the exact solution (9.50), we can see that u(x) satisfies the condition µ ¶ TL φ (1 − φ) u(x + L/n) − u(x) = + N A E1 E2 TL = (1/E), (9.51) NA where Z N L/N dx (1/E) = (9.52) L 0 E(x) is the average of 1/E over a periodic cell. From (9.51) we deduce that u(x) − (T x/A)(1/E) is periodic over the fine scale and hence, on a scale large enough for these periodic fluctuations to be negligible, Tx (1/E) + constant. (9.53) A In Figure 9.5 it is clear that (9.53) approximates the overall behaviour of the full solution (9.51) very well. We can rewrite (9.53) as u(x) ∼

du ∼ T, (9.54) dx where the effective Young’s modulus of the composite bar is the harmonic average of E, that is AEeff

Eeff =

1 (1/E)

.

(9.55)

This illustrates what the point of homogenisation theory is. When we consider the bar as a whole, the complicated fine-scale variations in E become irrelevant, and the bar behaves like a uniform medium whose modulus


383

is given by (9.55). Now we show how a formal asymptotic method, known as the method of multiple scales may be used to reach the same conclusions in more general configurations. These techniques come into their own in more complicated problems where explicit solutions like (9.50) do not exist. As always, before attempting any asymptotic analysis, it is necessary to nondimensionalise the problem. Here, we scale x with the bar length L, E with a typical value E1 and u with T L/E1 A to obtain E

du = 1. dx

(9.56)

Now x lies between 0 and 1, but E varies over a much smaller length-scale, of order ε say. We therefore write E in the form E = E (x, ξ) ,

(9.57)

where ξ = x/ε and E is assumed to periodic, with period one, with respect to its second argument. In our example (9.49), ε = 1/N and E depends only on the fine scale ξ. Here we allow E also to vary on the scale of the whole bar, for example if we had allowed E1 and E2 in (9.49) to vary slowly with x. With E given by (9.57), we must now take u to be a function of the two length-scales x and ξ; this means that we have moved from a model that is one-dimensional in the variable x to one that is two-dimensional in the plane (x, ξ). However, physical space is just the line x = ²ξ in the (x, ξ) plane and, on this line ∂u 1 ∂u du = + . dx ∂x ² ∂ξ

(9.58)

Hence (9.56) becomes µ E(x, ξ)

∂u ∂u +ε ∂ξ ∂x

¶ = ε.

(9.59)

To obtain a unique representation of u in terms of x and ξ, we insist that u, like E, must be periodic with respect to ξ†. Now we take the limit ε → 0 and seek the solution as an asymptotic expansion of the form u ∼ u0 + εu1 + · · · . When we equate terms of O(1), we discover that ∂u0 = 0, ∂ξ

(9.60)

† This is a very convenient assumption, but as discussed in [ ], it is not necessary for more general problems in which the microgeometry is not known in advance.

384


E E1

(a)

(b)

u

E2

x

x

Fig. 9.6. (a) The variation of Young’s modulus with position in a bar; the effective Young’s modulus is shown as a dashed curve. (b) The corresponding longitudinal displacement u(x); the leading-order approximation, given by (9.62), is shown as a dashed curve.

so that u0 is a function of x alone, and then, at order ε, µ ¶ ∂u1 du0 E(x, ξ) + = 1. ∂ξ dx

(9.61)

We can soon see from the explicit solution of these ordinary differential equations for u1 as a function of ξ that u1 will, in general, tend to infinity as ξ → ∞, thereby rendering asymptotic representation ³ ´ invalid. The only du0 way to avoid this is for the mean value of E dx − 1 over 0 < ξ < 1 to vanish, and we note that this implies the u1 has period 1. Hence Eeff (x)

du0 = 1, dx

(9.62)

where Á½Z Eeff (x) = 1

0

1

dξ E(x, ξ)

¾ =

1 (1/E)

.

(9.63)

This is equivalent to (9.55) when E depends on ξ alone, but (9.63) allows for E also to vary slowly with x. To illustrate this result, let us return to the piecewise constant example (9.49), but now allow E1 and E2 to be functions of x. In terms of dimensionless variables, we can therefore write E as a periodic function of ξ given by ( E1 (x) 0 < ξ < φ, E(x, ξ) = (9.64) E2 (x) φ < ξ < 1.


b

385

µ1

η

µ2 ξ

1

Fig. 9.7. A typical microstructured shear modulus.

We show a typical example in Figure 9.6(a), along with the effective modulus ½Z 1 ¾−1 dξ E1 (x)E2 (x) Eeff (x) = = , (9.65) E(x, ξ) (1 − φ)E 1 (x) + φE2 (x) 0 which averages out the fine scale variations. We see in Figure 9.6(b) that the displacement calculated using Eeff (x) (dashed curve) provides an excellent approximation to the exact solution (solid curve), even though ε = 1/5 is not particularly small. 9.4.2 Two-dimensional homogenisation Let us now consider a less trivial application of this method, namely antiplane strain in a bar whose shear modulus µ varies over the cross-section. It is straightforward, following the ideas of §2.3, to see that, in suitable dimensionless variables, the axial displacement w(x, y) satisfies ∇ · (µ∇w) = 0.

(9.66)

We suppose that µ varies both rapidly over a short length-scale ε and gradually across the whole cross-section, so that we can write µ = µ(x, y, ξ, η),

(9.67)

where (ξ, η) = (x, y)/ε. We also assume that µ is periodic in both directions over the fine scale, so that µ(x, y, ξ + 1, η) ≡ µ(x, y, ξ, η),

µ(x, y, ξ, η + b) ≡ µ(x, y, ξ, η).

(9.68)

386


We choose ε such that the period in ξ is equal to 1, and then define b to be the period in the η-direction. A typical setup is illustrated in Figure 9.7, where the modulus µ is piecewise constant on the fine scale, although µ1 (x, y) and µ2 (x, y) might vary over the long scale. Now, as in §9.4.1, we allow the displacement field w(x, y, ξ, η) to vary over both length-scales, provided it shares the same periodicity (9.68) as µ. The chain rule thus transforms (9.66) into ∇ξ · (µ∇ξ w) + ε {∇ξ · (µ∇x w) + ∇x · (µ∇ξ w)} + ε2 ∇x · (µ∇x w) , (9.69) where ∇x = (∂/∂x, ∂/∂y)T , ∇ξ = (∂/∂ξ, ∂/∂η)T . The assumed periodicity means that we need only solve (9.69) on a single rectangular cell [0, 1] × [0, b] and then apply the periodic boundary conditions w(x, y, 1, η) ≡ w(x, y, 0, η), ∂w ∂w (x, y, 1, η) ≡ (x, y, 0, η), ∂x ∂x

w(x, y, ξ, b) ≡ w(x, y, ξ, 0), ∂w ∂w (x, y, ξ, b) ≡ (x, y, ξ, 0). ∂y ∂y

(9.70a) (9.70b)

When we again seek the solution as an asymptotic expansion of the form w ∼ w0 + εw1 + ε2 w2 + · · · ,

(9.71)

we immediately discover that w0 satisfies ∇ξ · (µ∇ξ w0 ) = 0.

(9.72)

This equation and the boundary conditions (9.70) are identically satisfied if w0 is independent of ξ and η, so that w0 = w0 (x, y), and it may be shown (see Exercise 9.10) that this is the only solution. As we might have hoped, the displacement therefore does not vary on the fine length-scale to leading order. Comparing the terms at O (ε), we obtain ∂ ∂ξ

µ ¶ µ ¶ ∂w1 ∂ ∂w1 ∂µ ∂w0 ∂µ ∂w0 µ + µ =− − = R1 (x, y, ξ, η), say, ∂ξ ∂η ∂η ∂ξ ∂x ∂η ∂y (9.73)

which is a linear elliptic partial differential equation for w1 driven by ∇x w0 . As shown in Exercise 9.10, equation (9.73) admits periodic solutions for w1


387

only if the right-hand side has zero mean†, so that Z

1Z b

0

0

R1 (x, y, ξ, η) dξdη.

(9.74)

This requirement is satisfied identically here, and we can write the solution in the form w1 = W (1) (x, y, ξ, η)

∂w0 ∂w0 + W (2) (x, y, ξ, η) , ∂x ∂y

where W (1) and W (2) satisfy the equations Ã ! Ã ! ∂ ∂W (1) ∂ ∂W (1) ∂µ µ + µ =− ∂ξ ∂ξ ∂η ∂η ∂ξ

(9.75)

(9.76)

and ∂ ∂ξ

Ã

∂W (2) µ ∂ξ

!

∂ + ∂η

Ã

∂W (2) µ ∂η

! =−

∂µ , ∂η

(9.77)

with the periodic conditions (9.70). It can be shown that (9.76) and (9.77) have unique periodic solutions for W (1) and W (2) , up to arbitrary additive functions of (x, y). Although it is usually not possible to obtain these solutions in closed form, it is straightforward in principle to compute them numerically for any given modulus function µ(x, y, ξ, η). ¡ ¢ Finally, proceeding to the terms of O ε2 in (9.69), we find that w2 satisfies an equation analogous to (9.73), but with the right-hand side replaced by ½

µ ¶ µ ¶¾ ∂ ∂w1 ∂ ∂w1 R2 (x, y, ξ, η) = − µ + µ ∂ξ ∂x ∂η ∂y ½ µ ¶ µ ¶¾ ½ µ ¶ µ ¶¾ ∂ ∂w1 ∂ ∂w1 ∂ ∂w0 ∂ ∂w0 − µ + µ − µ + µ , ∂x ∂ξ ∂y ∂η ∂x ∂x ∂y ∂y (9.78) and again we need this to have zero mean over our periodic cell in (ξ, η). The first term satisfies this requirement identically, and, when we substitute † This can eb shown to be equivalent to the statement that w1 does not grow as ξ or η tends to infinity.

388


for w1 from (9.75), we find that Z 1Z b 0= R2 (x, y, ξ, η) dξdη 0 0 " Ã ! # Z 1Z b( ∂ ∂W (1) ∂w0 ∂W (2) ∂w0 =− µ +1 +µ ∂x ∂ξ ∂x ∂ξ ∂y 0 0 " Ã ! #) ∂ ∂W (1) ∂w0 ∂W (1) ∂w0 + µ +µ +1 dξdη. (9.79) ∂y ∂η ∂x ∂η ∂y Thus our homogenised displacement w0 (x, y) satisfies the anistropic model µ ¶ µ ¶ µ ¶ µ ¶ ∂ ∂ ∂ ∂ ∂w0 ∂w0 ∂w0 ∂w0 + + + = 0, µ11 µ12 µ21 µ22 ∂x ∂x ∂x ∂y ∂y ∂x ∂y ∂y (9.80) where µ11 µ22

Ã

! Z bZ 1 ∂W (1) ∂W (1) µ 1+ µ = dξdη, µ21 = dξdη, (9.81a) ∂ξ ∂η 0 0 0 0 ! Z bZ 1 Ã Z bZ 1 ∂W (2) ∂W (2) µ 1+ µ = dξdη, µ12 = dξdη, (9.81b) ∂η ∂ξ 0 0 0 0 Z bZ

1

Thankfully, (9.80) can be shown to be elliptic (see Ex ...). Equally reassuringly, when µ is independent of ξ, we can take W (1) = 0 and ∂W (2) /∂ξ = 0, so that ∂W (2) b = Z b − 1. ∂η dη µ 0 µ Hence the effective moduli are given by Z b µ11 = µdη, µ21 = µ12 = 0, 0

(9.82)

ÁZ µ22 = b

2 0

b

dη , µ

(9.83)

which is a generalisation of (9.63). For truly two-dimensional geometries, it is almost never possible to evaluate the effective moduli explicitly. One famous example where significant progress can be made occurs when the shear modulus µ is piecewise constant, taking one value µ2 on some region D2 inside our rectangular cell, and another value µ1 in the complementary region D1 = [0, 1] × [0, b] \ D1 . When we make the substitutions φ(1) = W (1) + ξ,

φ(2) = W (2) + η,

(9.84)

9.4 Composite Materials and Homogenisation η b (1)

φ

=0

∂φ(1) =0 ∂η

(a)

η φ(2) =b

D1

φ(1) =1

389

(b) ∂ψ (2) =0 ψ (1) =β ∂η

ψ (2) =0

ψ (2) =α

D2 ∂φ(2) =0 ∂ξ

µ2 µ1 ∂φ(1) =0 ∂η

∂D2 φ(2) =0 1

∂φ(2) =0 ∂ξ ξ

∂ψ (1) =0 ∂ξ ∂ψ (2) =0 ∂η

ψ (1) =0 1

∂ψ (1) =0 ∂ξ ξ

Fig. 9.8. Definition sketch for a symmetric, piecewise constant shear modulus distribution; the symmetry axes are shown as dot-dashed lines. (a) The boundary conditions satified by φ(1) and φ(2) . (b) The corresponding boundary conditions for the harmonic conjugate functions ψ (1) and ψ (2) .

the equations (9.76) and (9.77) for W (1) and W (2 are transformed to ³ ´ ³ ´ ∇ξ · µ∇ξ φ(1) = ∇ξ · µ∇ξ φ(2) = 0, (9.85) where

( µ=

µ1

(ξ, η) ∈ D1 ,

µ2

(ξ, η) ∈ D2 .

(9.86)

Hence φ(1) and φ(2) both satisfy Laplace’s equation in each of D1 and D2 , along with the continuity conditions # " h i (j) ∂φ =0 (9.87) φ(j) = µ ∂n on the boundary ∂D2 between them. Crucially, we assume that the geometry possesses reflectional symmetry in both the ξ- and the η-direction. Using this symmetry and the periodicity of W (j) , we deduce that φ(j) must satisfy the boundary conditions φ(1) (0, η) = 0,

φ(1) (1, η) = 1,

φ(2) (ξ, 0) = 0,

φ(2) (ξ, b) = b,

∂φ(1) (ξ, 0) = ∂η ∂φ(2) (0, η) = ∂ξ

∂φ(1) (ξ, b) = 0, ∂η ∂φ(2) (1, η) = 0, ∂ξ

(9.88a) (9.88b)

as illustrated in Figure 9.8(a). Now, since φ(j) satisfy Laplace’s equation in each separate region, we can

390


introduce the harmonic conjugates ψ (j) which satisfy the Cauchy–Riemann equations ∂φ(j) ∂ψ (j) = , ∂ξ ∂η

∂φ(j) ∂ψ (j) =− . ∂η ∂ξ

(9.89)

The continuity conditions (9.87) for φ(j) are transformed to " # h i (j) ∂ψ µψ (j) = = 0, (9.90) ∂n £ ¤ where again denotes the jump across ∂D2 . The boundary conditions (9.88) become ψ (1) (ξ, 0) = 0,

ψ (1) (ξ, b) = β,

ψ (2) (0, η) = 0,

ψ (2) (1, η) = α,

∂ψ (1) ∂ψ (1) (0, η) = (1, η) = 0, ∂ξ ∂ξ ∂ψ (2) ∂ψ (2) (ξ, 0) = (ξ, b) = 0, ∂η ∂η

(9.91a) (9.91b)

as shown in Figure 9.8(b), where the constants α and β are unknown in advance. It is straightforward to calculate the effective elastic constants using ψ (j) , because the continuity conditions (9.90) allow us to integrate directly across the cell. For example, Z bZ 1 Z bZ 1 Z 1h iη=b ∂φ(1) ∂ψ (1) µ µ µ11 = dξdη = dξdη = µψ (1) dξ = µ1 β, ∂ξ ∂η η=0 0 0 0 0 0 (9.92) and analogous calculations yield µ22 = −µ1 bα,

µ21 = µ12 = 0,

(9.93)

so the assumed symmetry of the geometry eliminates the off-diagonal terms. Now the key observation is that the problem satisfied by ψ (2) closely resembles that for φ(1) (and vice versa). Indeed, if we define µψ (2) , φe(1) = αµ1

(9.94)

then the boundary conditions for φe(1) are identical to (9.88a), although the continuity conditions on ∂D2 read ¯ ¯ ¯ ¯ e(1) ¯¯ e(1) ¯¯ ∂ φ ∂ φ ¯ ¯ µ2 (9.95) φe(1) ¯ = φe(1) ¯ , ¯ = µ1 ¯ . ∂n ¯ ∂n ¯ D2 D1 D1

D2


391

Fig. 9.9. Some modulus distributions in a square that are antisymmetric about the diagonals.

Hence φe(1) is what φ(1) would be if we switched the values of µ1 and µ2 . We can use the same technique as above to calculate the corresponding elastic constants, noting that the harmonic conjugate of φe(1) is (2)

so that

µφ ψe(1) = − , αµ1

(9.96)

h iη=b µ2 b µ e11 = µ2 ψe(1) . =− α η=0

(9.97)

Eliminating α between (9.93) and (9.97), we obtain the intriguing relation µ e11 µ22 = b2 µ1 µ2

(9.98)

between the stiffness µ22 in the η-direction and the corresponding stiffness µ e11 in the ξ-direction with µ1 and µ2 switched. This has particularly startling implications if b = 1 and the geometry is anti-symmetric about a diagonal in the resulting square cell, as in the examples shown in Figure 9.9. In such configurations, swapping µ1 and µ2 is equivalent to swapping ξ and η, so that µ e11 ≡ µ22 . Hence a homogenised bar consisting of such cells is isotropic with √ µ11 = µ22 = µ1 µ2 . (9.99) Such results provide an invaluable check on the numerical methods to which one usually has to resort to calculate the effective elastic constants. 9.4.3 Three-dimensional homogenisation Now that we have the ideas of §9.4.2 at our disposal, we can derive the homogenised equations for a three-dimensional elastic matrix quite quickly starting from the equilibrium macroscopic model ∇ · τ = 0,

τij = Cijk` ek` = Cijk`

∂uk , ∂x`

(9.100)

392


where C(x) is a symmetric tensor incorporating the elastic constants. As in §1.7, for an isotropic material C is given by Cijk` (x) = λ(x)δij δk` + 2µ(x)δik δj` ,

(9.101)

but it is more convenient here to persist for the moment with the more general case (9.100), which we can write symbolically as a tensor product: τ = C∇u.

(9.102)

We note that it is straightforward to incorporate a body force in (9.100), but unsteady effects can cause significant complications, depending crucially on the ratio of the elastic wavelength to the length-scale of the microstructure. As in §9.4.2, we suppose that the elastic constants in C vary over two length-scales: a macroscopic scale and a micro-scale which is smaller by a factor of ε. In suitable dimensionless variables, we can therefore write C = C(x, ξ), where ξ = x/ε, and similarly for u and τ . We again assume that C, u and τ are all periodic on some representative cell D, for example a cuboid, in ξ-space. The advantage of working with the first-order system (9.100), rather than the Navier equations, is that when we expand u ∼ u0 (x, ξ) + εu1 (x, ξ) + · · · , τ ∼ τ0 (x, ξ) + ετ1 (x, ξ) + · · · ,

(9.103)

we only need to consider¡ terms up to O (ε), in contrast with §9.4.2 where ¢ 2 we had to proceed to O ε . At leading order, we soon find that ∇ξ · τ0 = 0,

(9.104)

C∇ξ u0 = 0,

(9.105)

the second of which implies that u0 is a rigid body motion as a function of ξ, and our assumed periodicity implies that u0 = u0 (x),

(9.106)

as in (9.72). At order ε, we obtain the equations ∇ξ · τ1 + ∇x · τ0 = 0, τ0 = C∇ξ u1 + C∇x u0

(9.107) (9.108)

If we operate with ∇ξ on (9.108) then, by virtue of (9.104), we obtain a linear problem for u1 , namely ∇ξ · (C∇ξ u1 ) = − (∇ξ · C) (∇x u0 ) ,

(9.109)

9.5 Poroelasticity

393

which is driven by a term proportional to ∇x u0 . Hence, as in (9.75), we can can formally write u1 in the form u1 = A(x, ξ)∇x u0 ,

(9.110)

where A = (Aijk ) is rank-three tensor which in principle is determined by solving (9.109); usually this will have to be done numerically. Backsubstitution into (9.108), as demonstrated in Exercise 9.12, gives that τ0 (x, ξ) is related to u0 by τ0 = (C∇ξ A) ∇x u0 + C∇x u0 .

(9.111)

As always, the final step is to demand that τ be periodic on our unit cell D. This means that ZZZ ∇ξ · τ1 dξ ≡ 0, (9.112) D

and we therefore deduce from (9.107) that ∇x · τ 0 = 0,

(9.113)

where τ 0 again denotes the average of τ0 over D. Hence, on the macroscopic scale, the leading-order displacement u0 satisfies the homogenised equation ∇x · (Ceff ∇x u0 ) = 0,

where Ceff (x) = (C∇ξ A + C).

(9.114)

We remark again that the effective elastic constants can rarely be evaluated explicitly, although, for any given microstructure, they can in principle be calculated numerically over a representative unit cell. Another crucial observation is that, even if the matrix is constructed from isotropic material, so that C is given by (9.101), the averaging will in general lead to an anisotropic macroscopic constitutive law. Of course, most natural heterogeneous media do not have a regular periodic microstructure. For such materials, we must view D as a representative cell that is characteristic of the medium as a whole. Unless the microstructure has a particular directionality, we might expect a random geometry to lead to isotropic macroscopic behaviour. 9.5 Poroelasticity The theory of §9.4.3 may be applied, for example, to a solid that takes the form of a homogeneous elastic matrix in which are embedded many smallscale voids or pores. For such a material, the elastic constants Cijk` would be constant in the elastic matrix and zero in the voids. The sitation is much more interesting, though, when the pores are sufficiently connected for a

394


viscous fluid to pass through them, as in a liquid-filled sponge, for example. Other important deformable porous media include oil reservoirs and biological tissue. In principle, one follow a homogenisation procedure to derive an averaged macroscopic model governing the deformation of such a medium. However, the need to couple the Navier equations for the solid matrix to the Navier–Stokes equations for the fluid makes this task considerably more complicated than any of the examples encountered in §9.4, so we will bypass most of the difficulties associated with the fluid dynamics. We know from (9.114) that we can express the elastic contribution to the stress in the form τe = Ceff ∇u,

(9.115)

where u is the displacement of the medium and Ceff is the matrix of effective elastic constants. If the medium is isotropic, then we can write ¡ ¢ (9.116) τe = λeff (∇ · u) I + µeff ∇u + ∇uT , although the effective elastic constants λeff and µeff will be quite different from the Lamé constants of the solid material from which the matrix is constructed. A detailed analysis of the fluid dynamics would require us to consider the shear and normal stresses set up in the fluid as it moves through the complicated pore structure. In order to avoid this complication, we will assume the well-substantiated Terzaghi Principle that, on a scale of many pore sizes, the stress in the fluid averages to an isotropic pore pressure P . Tthe net stress in the medium is thus given by τ = −P I + τe

(9.117)

and the equilibrium Navier equation reads (λeff + µeff ) grad div u + µeff ∇2 u = ∇P.

(9.118)

We therefore have a situation similar to thermoelasticity, with the pore pressure playing the rôle of the temperature. The pressure gradient term in (9.118) represents the drag exerted by the fluid on the solid matrix. The average velocity v at which the fluid flows through the pores is governed by Darcy’s law. This is the physically plausible statement that the difference between the fluid and solid velocities should be proportional to the pressure gradient. In the context of linear elasticity, we can approximate the velocity of the solid matrix as the Eulerian time derivative of the

9.5 Poroelasticity

395

displacement, so that Darcy’s law reads v=

∂u k − ∇P, ∂t η

(9.119)

where η is the viscosity of the fluid and k is known as the permeability of the medium. This law is confirmed by experiments, and the permeability is empirically well characterised for many important porous media. It has the dimensions of (length)2 and is of order d2 , where d is a typical pore radius. It is also possible to derive (9.119) from the underlying Navier– Stokes equations via a homogenisation procedure analogous to that used in §9.4.3. For an anisotropic medium, k must be replaced by a positive definite permeability tensor. To close the problem, we must consider conservation of mass. Each representative cell of the fluid will contain a volume fraction α, say, of fluid, and the corresponding solid fraction is therefore 1 − α. We can thus use the approach of §2.2.1 and the result of Exercise 9.13 to calculate the relative volume change due to a displacement u and hence obtain α = α0 + (1 − α0 )∇ · u.

(9.120)

On the other hand, conservation of fluid leads to the equation ∂α + α0 ∇ · v = 0. ∂t

(9.121)

By combining these two equations, we find that, as also shown in Exercise 9.13 ∇·v =−

1 − α0 ∂ (∇ · u) , α0 ∂t

(9.122)

which shows how compression of the matrix forces the fluid to flow. For simplicity, we have neglected the compressibility of the solid and fluid phases. We emphasise, though, that the matrix as a whole certainly will be compressible; for example, it is quite possible to compress a sponge made out of incompressible rubber. We have also assumed that α is always close to its initial value α0 , which is consistent with our assumption that linear elasticity is valid, and that the deformation is sufficiently slow for the inertia terms in the Navier equations to be ignored. Equations (9.116)–(9.121) thus comprise the very simplest model that couples elastic deformation to fluid flow. If we assume that the parameters λeff , µeff , k and η are all constant, then a straightforward cross-differentiation reveals that div u satisfies the linear

396


V

P =0

11111111 00000000 00000000 11111111 00000000 11111111 00000000L 11111111

F x

Fig. 9.10. Schematic of the one-dimensional squeezing of a sponge.

diffusion equation: ∂ (div u) = ∂t

µ

(λeff + 2µeff ) α0 k η

¶ ∇2 (div u) .

(9.123)

Hence, the effects of elasticity, viscosity and permeability are all collected in the effective diffusion coefficient D=

(λeff + 2µeff ) α0 k . η

(9.124)

If D is large (in a suitable dimensionless sense), then the left-hand side of (9.123) is negligible and the medium behaves like an elastic solid. Increasing the stiffness of the matrix thus has a similar overall effect to decreasing the fluid viscosity or increasing the permeability. As a simple illustration, let ¢us consider one-dimensional ¡ ¡ ¢T squeezing of a T sponge. With u = u(x, t), 0, 0 and v = v(x, t), 0, 0 , we can combine the equilibrium equation (9.118) and Darcy’s law (9.119) to obtain µ ¶ ∂P η ∂u ∂2u = −v . (9.125) (λeff + 2µeff ) 2 = ∂x ∂x k ∂t In one dimension, the mass-conservation equation (9.121) may be integrated directly with respect to x, leading to (1 − α0 )

∂u + α0 v = q(t), ∂t

(9.126)

where q(t) represents the net volume flux of both solid and fluid phases. We can use (9.126) to eliminate v from (9.125) and thus find that u satisfies a forced heat equation ∂2u ∂u − D 2 = q(t), ∂t ∂x

(9.127)

where D is again defined by (9.124); this makes an interesting comparison with (??). To fix ideas, let us consider the situation shown in Figure 9.10, where

9.5 Poroelasticity

397

F/(λeff + 2µeff )

3 2 1 Pe = 0

2.0

1.5

1.0

0.5

0.2

0.4

0.6

0.8

1.0

V t/L Fig. 9.11. Dimensionless stress applied to a sponge versus dimensionless time for different values of the Péclet number Pe; the large-t limits are shown as dotted lines.

one end x = 0 of the sponge is held fixed, although fluid is free to flow into a reservoir at zero pressure, while the other end x = L is squeezed at a constant velocity −V by an impermeable plunger. This implies the boundary conditions u=P =0

at x = 0,

∂u = v = −V ∂t

at x = L,

(9.128)

noting that, in linear elasticity, the motion of the boundary x = L is neglected. It follows that q = −V , and (9.127) may then readily be solved by separation of variables: if u = 0 initially, then µ ¶ ∞ Vx (L − x)(2L − x) 2V L2 X sin (nπx/L) −n2 π2 Dt/L2 u(x, t) = − t+ + 3 e . L 6D π D n3 n=1

(9.129) We can use this to calculate, for example, the net compressive stress F that we would need to apply to the plunger to force it to move at constant speed. From (9.116)–(9.118), we find that ¯ · ¸ ∂u ∂u ¯¯ F = P − (λeff + 2µeff ) = − (λeff + 2µeff ) , (9.130) ∂x x=L ∂x ¯x=0 and the result of substituting (9.129) into this is plotted in Figure 9.11. The behaviour depends on the key dimensionless parameter known as the Péclet number Pe = LV /D. When Pe = 0, the effect of the fluid is negligible, so the stress just increases linearly as the elastic matrix contracts. However, for nonzero Pe, the initial response is dominated by the need to force the fluid through the matrix. After some time, the behaviour becomes linear

398


again, with F V t Pe ∼ − λeff + 2µeff L 3

as

t → ∞,

(9.131)

as indicated by the dotted lines in Figure 9.11. As the Péclet number increases, the fluid has an increasing influence, and a longer time is taken for the solution to settle down to (9.131). 9.6 Anisotropy We conclude the chapter with a brief account of the effects of anisotropy, which, as we have just seen, is an endemic property of macroscopic models of heterogeneous elastic media. Wood and fibre-reinforced materials, for example, have strongly anistropic response to stress. In §1.7, we saw that the most general linear relation between elastic stress τij and linear strain eij is τij = Cijk` ek` ,

(9.132)

which defines Cijk` as a tensor with 81 components. Since there are only six independent components in the symmetric tensors τij and eij , we could also write the relation above as     τxx exx τ  e   yy   yy       τzz   ezz  (9.133)   = A ,  τyz  2eyz      τxz  2exz  τxy 2exy where A = (Aij ) is a 6 × 6 matrix with entries A11 = C1111 , A12 = C1122 , A13 = C1133 , . . . and the factor of 2 multiplying the off-diagonal strain components accounts for the fact that they appear twice in (9.132). The existence of a strain energy density W such that the equality of τij and ∂W/∂eij implies, by cross-differentiation, that the matrix (Aij ) is symmetric. We are thus left with at most 21 independent elastic moduli. In terms of C, this can be summarised by the relations Cijk` = Cjik` = Cij`k = Ckìj .

(9.134)

The only other restriction we can impose in general is that A must be positive definite, to ensure that W is a positive definite function of the strain components.

9.6 Anisotropy

399

Let us now consider a rotation of the coordinate axes by an orthogonal matrix P . In the rotated frame, (9.132) becomes 0 τij0 = Cijk` e0k` ,

0 where Cijk` = Pim Pjn Pkr P`s Cmnrs .

(9.135)

For the isotropic materials considered in §1.7, we asserted that the relationship between τ 0 and e0 is the same as that between τ and e for any orthogonal matrix P . Now that we can only say that this is true for each orthogonal matrix P representing a material symmetry. For every such P , we therefore have 21 independent equations of the form Cijk` = Pim Pjn Pkr P`s Cmnrs .

(9.136)

Usually, most of these equations are satisfied trivially, but some of them may imply nontrivial relationships between the elastic constants Cijk` . The simplest symmetry that we can encounter is reflectional symmetry with respect to the (x, y)-plane. Such materials are called homoclinic. To fix ideas, we will focus on the example of a material containing fibres oriented in the z-direction. In this case, (9.136) should be satisfied with   1 0 0 P = 0 1 0  , (9.137) 0 0 −1 or, in a more compact notation, Pij = (−1)δi3 δij . We must therefore have Cijk` = (−1)δi3 +δj3 +δk3 +δ`3 Cijk` , and Cijk` must therefore vanish if the index 3 appears an odd number of times. This leaves 13 nonzero entries in Aij , namely   A11 A12 A13 0 0 A16 A 0 A26   12 A22 A23 0    0 A36  A13 A23 A33 0 A= (9.138) . 0 0 A44 A45 0   0    0 0 0 A45 A55 0  A16 A26 A36 0 0 A66 Continuing with our example of z-oriented fibres, let us further assume that they are positioned in the (x, y)-plane regularly enough to impose in addition reflectional symmetry with respect to the (x, z)- and (y, z)-planes. Such a material with three orthogonal planes of symmetry is called orthotropic. In this case (9.136) must hold simultaneously with the three transformations Pijz = (−1)δi3 δij , Pijx = (−1)δi1 δij and Pijy = (−1)δi2 δij . Hence,

400


following the same reasoning as in the monoclinic case, Cijk` vanishes whenever any of the indices 1, 2 or 3 appears an odd number of times. In particular, C1112 = C2212 = C3312 = C2313 = 0, so that (9.138) reduces to   A11 A12 A13 0 0 0 A 0 0    12 A22 A23 0   0 0  A13 A23 A33 0 (9.139) A= . 0 0 A44 0 0   0    0 0 0 0 A55 0  0 0 0 0 0 A66 If, on the other hand, the fibres are randomly positioned in the xy-plane, then, on a macroscopic scale, the material is transversely isotropic. The resulting matrix (Aij ) will clearly be a special case of (9.139) and we first note that, in the absence of preferred direction in the (x, y)-plane, A11 = A22 , A23 = A13 and A44 = A55 . Furthermore if we restrict the displacements to be in the (x, y)-plane only, then we have an isotropic problem in plane strain. Thus, as in (1.42), there must exist two constants c1 and c2 such that τxx = c1 (exx + eyy ) + c2 exx , τxy = c2 exy ,τyy = c1 (exx + eyy ) + c2 eyy ,

(9.140) while (9.139) gives us τxx = A11 exx + A12 eyy , τxy = 2A66 exy , τyy = A12 exx + A11 eyy . (9.141) This imposes the condition 2A66 = A11 − A12 , so that only five independent elastic constants remain:   A11 A12 A13 0 0 0 A  0 0  12 A11 A13 0    0 0 A13 A13 A33 0  A= (9.142) . 0 0 A44 0 0  0     0  0 0 0 A44 0 0 0 0 0 0 (A11 − A12 / 2 Note that the same conclusion could be reached by a blind use of (9.136) with the family of orthogonal transformations   cos θ sin θ 0 B = − sin θ cos θ 0 , (9.143) 0 0 1

9.6 Anisotropy

401

but with considerably more effort. As a simple example, let us generalise the analysis of §2.4 to model the torsion of an orthotropic bar. For a displacement field of the form (2.41), the only nonzero strain components are exz and eyz , and according to (9.139), the nonzero stress components are µ ¶ µ ¶ ∂ψ ∂ψ τxz = A55 Ω −y , τyz = A44 Ω +x . (9.144) ∂x ∂y Hence, instead of Laplace’s equation, ψ now satisfies A55

∂2ψ ∂2ψ + A = 0, 44 ∂x2 ∂y 2

(9.145)

where A44 and A55 are both positive, and the no-stress boundary condition is µ ¶ µ ¶ ∂ψ dy ∂ψ dx A55 −y − A44 +x = 0, (9.146) ∂x ds ∂y ds on the boundary of the bar. We recall from §2.4 that, once we have solved for ψ(x, y), we can evaluate the torsional rigidity using ZZ 1 R= (xτyz − yτxz ) dxdy Ω D µ ¶ µ ¶¾ ZZ ½ ∂ψ ∂ψ = A44 x + x − A55 y −y dxdy. (9.147) ∂y ∂x D The most striking implication of this model is that the elastic constants A44 and A55 can be eliminated from the problem by rescaling x and y appropriately. If we define µ ¶ µ ¶ A55 1/4 A44 1/4 x= X, y= Y, (9.148) A44 A55 then, with respect to the new variables, ψ satisfies ∂2ψ ∂2ψ + = 0, ∂X 2 ∂Y 2

(9.149)

¢ dX ∂ψ 1 d ¡ 2 dY ∂ψ − = X +Y2 ds ∂X ds ∂Y 2 ds

(9.150)

subject to

on the boundary of the bar, while (9.147) is transformed to ¾ ZZ ½ p ¢ ∂ψ ¡ 2 ∂ψ 2 R = A44 A55 −Y + X +Y dxdy. X ∂Y ∂X D

(9.151)

402


Equations (9.149)–(9.151) are identical to those found for an isotropic bar in §2.4. Hence an orthotropic bar behaves under torsion just like an isotropic bar whose transverse dimensions are stretched according to the rule (9.148). For example, a circular orthotropic bar is mechanically equivalent to an elliptical isotropic bar. We conclude by mentioning an important consequence of anisotropy for the propagation of elastic waves. For an anisotropic material, the timedependent momentum equation is ρ

∂τij ∂ 2 ui ∂ 2 um = = C . ijmn ∂t2 ∂xj ∂xj ∂xn

(9.152)

As in §3.2.4, we look for a solution in the form of a plane wave u = aei(k·x−ωt) . A simple substitution into (9.152) yields the system Cijmn kj kn am − ρω 2 ai = 0,

(9.153)

and, if nontrivial waves are to exist, they must therefore satisfy the dispersion relation ¢ ¡ (9.154) det Cijmn kj kn − ρω 2 δim = 0. This generalises the dispersion relations (3.39) governing P -waves and S waves in an isotropic solid. One of the main consequences of (9.154) is that the waves are dispersive, so that the phase speed ω/ |k| of plane waves generally depends on the wave vector k. This is in contrast with waves in an infinite isotropic solid, which propagate with phase speed equal to either cp or cs . Recall, though, that the presence of boundaries can give rise to dispersion even in an isotropic medium; for example, Love waves and waves in beams are both dispersive. 9.7 Epilogue The diversity and open-endedness of the topics described in this chapter reflect the fact that we have not done justice to many scientific ideas in theoretical solid mechanics, both here and in the preceding chapters. We are not too concerned about our coverage of fracture and plasticity. These are subjects of such enormous practical importance that many texts exist that describe both the practical and mathematical aspects in much greater detail than the contents of Chapters ?? and 8. However no such voluminous literature concerns, say, the important phase changes that can occur in metals, and their alloys, as a result of heat treatment and/or mechanical processing. For example, steel can have several different crystalline

9.7 Epilogue

403

microstructures which can depend crucially on its chemical composition and these different phases have very different mechanical properties. In certain circumstances the geometry of the phase boundaries can be elegantly described in terms of hyperelasticity with a strain energy density which is corresponding to the different phases. This naturally opens up exciting new problems in the calculation of variations, as does the related, but especially inaccessible subject of “shape memory” alloys, where the strain energy depends on temperature. The resulting relaxation phenomena associated with different phases depends on thermal histories in the same way that the viscoelastic relaxation depends on the stress or strain histories. It is also difficult for mathematicians to find self-contained accounts of theories of solids such as piezoelectric or ferromagnetic materials whose mechanical properties depend on electromagnetic fields. Indeed, a background in electromagnetics is usually a prerequisite for any applied mathematician wishing to understand the modelling of such materials. Finally from the modelling point of view, we must point out that we have avoided any serious discussion of ceramics which could in fact be defined as any solids that are not metals. Although the mechanics of materials such as rock can be perfectly well described by linear elasticity at sufficiently small stress, little of the theories expounded in Chapters 7 and 8 is helpful in understanding how a glass or piece of crockery fails under impact. For example, mathematics has, as yet, done little to improv although many “ad hoc” macroscale models have been proposed. Nonetheless, certain specialised geological fracture processes such as hydo-fracture, which is an important mechanism in oil extraction, have been successfully analysed mathematically. On the theoretical side, easily our most glaring omission is our failure to discuss the many ingenious codes that have been specifically developed to obtain numerical solutions of the kinds of models we have been enumerating in this book. This is a most important aspect of applied solid mechanics because the subject has provided one of the principle stimuli for the widely-used finite element method, which is so appropriate for finding energy minima in complicated real-world configurations of elastic solids. Moreover, computers have now reached the stage where discrete “particlebased” algorithms can be tailored to simulate mechanics of a solid, be it elastic, plastic or any of the combinations mentioned in this chapter. At an even more fundamental level, “atomistic” computations are now routinely available to simulate many, many atoms as points of the solid interacting via suitable force potentials. Such simulations can give new insights into, say, dislocation intersections and many other nanoscale phenomena.

404


Fig. 9.12. A Jeffreys viscoelastic element.

Despite these qualifications, we hope that this book has shown that a little systematic mathematical modelling and analysis can go a long way towards providing a firm theoretical formulation for the science of solid mechanics. Even more, we hope that readers will be attracted by the many topics that await mathematical attention in an area so important to modern natural science.

Exercises 9.1 Show that the constitutive law for the linear Jeffreys viscoelastic element shown in Figure 9.12 is µ ¶ du dT trT + T = k tru +u . (9.155) dt dt Find expressions for the modulus k and the relaxation times trT and tr u , in terms of the spring constants and impedance of the individual components, and hence show that tr u > tr T . Show that the creep function for this element is K(t) =

1 − (1 − tr T /tr u ) e−t/tr u . k

Obtain the Voigt and Maxwell creep functions (9.9) by taking the limits trt → 0 and tru → ∞ respectively. 9.2 beginenumerate 9.3 Suppose parallel spring/dashpot elements separate a linear array of masses m placed at points Xn as in Figure 9.13. Show that µ ¶ dXn dXn−1 e v e v Tn−1 = Tn−1 +Tn−1 , Tn−1 = k(Xn −Xn−1 ), Tn−1 = Y − dt dt

Exercises

405

Fig. 9.13.

Fig. 9.14.

in the notation of §9.2.2. Use the equation of motion for the mass at Xn−1 to show that d2 Xn−1 d = Y (Xn+1 − 2Xn + Xn−1 ) + K (Xn − 2Xn−1 + Xn−2 ) 2 dt dt and show that this is a discretisation of the Voigt model n

Y

2 ∂2u ∂2u 2∂ u + K = mk ∂x2 ∂t ∂x2 ∂t2

as h → 0. 9.4 Repeat the above modelling for the series element in Figure 9.14. Show that ¶ µ d2 Xn dYn−1 dXn−1 e e v Tn −Tn−1 = m , Tn−1 = k(Xn −Yn−1 ), Tn−1 = Y − dt2 dt dt and hence show that ¶ µ 2 1 d Xn 1 d3 Xn d M + = (Xn+1 − 2Xn + Xn−1 ) . 2 3 k dt Y dt dt Show finally that this is a discretisation of the Maxwell model 1 ∂2u 1 ∂3u k2 ∂ 3 u + = . k ∂t2 Y ∂t3 m ∂x2 ∂t 9.5 A standard rheological test¡ is to ¢impose an oscillatory displacement on a sample, say u = Re ae−iωt , and measure the resulting force T (t). For a linear material, T (t) will also be proportional to e−iωt ,

406


and we define the complex modulus G = T /u. The complex modulus is then decomposed into its real and imaginary parts: G = G0 + iG00 , where G0 is called the storage modulus and G00 the loss modulus. Show that the average work done in each period is Z 2π/ω ω du 1 T dt = − |a|2 G00 ω. 2π 0 dt 2 For the Jeffreys model (9.155), show that G0 =

k (tr u − tr T ) ktr u ¡ ¢, − tr T tr T 1 + tr 2T ω 2

G00 = −

k (tr u − tr T ) ω ¡ ¢. tr T 1 + tr 2T ω 2

Deduce that the rate at which the Jeffreys element dissipates energy is a non-negative increasing function of ω. 9.6 Recall from Chapter 5 the definitions of deformation gradient and of the Green and Finger deformation tensors, namely Fij =

∂xi , ∂Xj

C = F T F,

Prove the identities ¡ ¢ DC = F T ∇v + ∇v T F, Dt

B = F F T.

DB = ∇v T B + B∇v, Dt

where ∇v is the velocity gradient tensor with components (∂vj /∂xi ). Show also that the dissipation may be written in terms of the first Piola–Kirchhoff stress tensor T as Φ = Dij τij = (det F )−1

DFij Tij . Dt

Deduce that the energy equation from Exercise 8.13 is consistent with (5.61). 9.7 Recall that a scalar function a(x, t) is conserved by a flow v(x, t) if Da ∂a = + v · ∇a = 0. Dt ∂t Now consider a vector field a(x, t) that convects and rotates with the flow. Following the procedure in Exercise ??, show that a must satisfy Da = (a · ∇) v. Dt Finally, suppose that the tensor field A(x, t) is such that, if Aa = b

Exercises

407

at time t, then Aa = b for all t, whenever a(x, t) and b(x, t) are both convected by the flow in the above sense. Deduce that A must satisfy DA + A∇v T − ∇v T A = 0. Dt ◦

◦

If A is symmetric, deduce that A = 0, where A is the Jaumann derivative (9.33) with a = 0. 9.8 In a linear thermoelastic solid, the strain energy density is zero for a uniform isotropic expansion in which ekk = α(T − T0 ). By referring to Exercise 1.6, obtain the formula ³ ´³ ´ ¢2 λ¡ α α W= ekk − α(T − T0 ) + µ eij − (T − T0 )δij eij − (T − T0 )δij . 2 3 3 Show that, with this definition of W, the linear thermoelastic constitutive relation (9.36) is consistent with τij = ∂W/∂eij (where the derivative is taken with T held fixed). 9.9 By using the divergence theorem and the momentum equation, deduce from (9.45) and (9.44) that T satisfies ¶ µ ∂W ∂T + div q = 0. ρc + ∂T ∂t For the strain energy density calculated in Exercise 9.8, show that ∂W α = − τkk . ∂T 3 We can use (9.36) to estimate the stress induced by a temperature variation ∆T as τkk = O (Eα∆T ), where E is Young’s modulus. Deduce that ∂W/∂T is negligible in comparison with ρc for temperature 2 variations ∆T such that Eαρc∆T ¿ 1. [Steel, for example, has ρ ≈ 7800 kg m−3 , c ≈ 500 J kg−1 K−1 , E ≈ 200 GPa and α ≈ 13 × 10−6 K−1 , so that ∆T must be much less than 105 K for the estimate to be valid.] 9.10 (a) Prove the identity ZZ ZZ I ∂w w∇ξ · (µ∇ξ w) dξdη + µ|∇ξ w|2 dξdη = µw ds, ∂n D D ∂D where w is twice continuously differentiable on the closed bounded region D of the (ξ, η)-plane. Deduce that, if w satisfies (9.72) and periodic boundary conditions on a rectangular cell [0, 1] × [0, b], then w must be independent of ξ and η.

408


(b) Next suppose that w satisfies ∇ξ · (µ∇ξ w) = R(x, y, ξ, η) and, again, periodic conditions on the boundary of the rectangle [0, 1] × [0, b]. Show that R must have zero mean, that is Z 1Z b R dξdη = 0. 0

0

9.11 Consider the solutions of (9.76) and (9.77) with periodic boundary conditions on a rectangular cell in which the modulus µ is piecewise constant, as shown in Figure 9.7. If no symmetry is assumed, show that the function φ(1) defined by (9.84) satisfies the boundary conditions φ(1) (1, η) − φ(1) (0, η) = 1, φ(1) (ξ, b) − φ(1) (ξ, 0) = 0,

∂φ(1) ∂φ(1) (1, η) − (0, η) = 0, ∂ξ ∂ξ ∂φ(1) ∂φ(1) (ξ, b) − (ξ, 0) = 0, ∂η ∂η

and find the corresponding boundary conditions for φ(2) and the harmonic conjugate functions ψ (1) and ψ (2) . Following the procedure used at the end of §9.4.2, derive the following relations between the effective elastic constants µij and the corresponding constants µ eij when µ1 and µ2 are swapped: µ e11 µ e21 µ e12 µ e22 b2 µ1 µ2 = = = = . µ11 µ12 µ21 µ22 µ11 µ22 − µ21 µ12 9.12 Using the symmetry properties of C, show that equation (9.109) may be written in component form as µ ¶ ∂Cijk` ∂u0k ∂ ∂u1k Cijk` =− . ∂ξi ∂ξ` ∂ξi ∂ξ` Deduce that the solution may be written as u1i = Aijk

∂u0j , ∂xk

where Aijk ≡ Aikj and µ ¶ ∂Cijmn ∂A`mn ∂ Cijk` =− . ∂ξj ∂ξk ∂ξi

Exercises

409

Verify that this gives us 18 independent equations for the 18 independent functions Aijk (x, ξ). By substituting for u1 in (9.108), show that τ0 is given by µ ¶ ∂u0m ∂Akmn . τ0ij = Cijk` + Cijmn ∂ξ` ∂xn 9.13

(a) Consider a control volume V in a porous elastic solid, with initial void fraction equal to α0 . Show that, after a small displacement u, the same solid material occupies a volume V (1 + ∇ · u), and deduce that the solid fraction becomes 1−α=

1 − α0 ∼ (1 − α0 ) (1 − ∇ · u) . 1+∇·u

(b) Now consider a fixed volume V , whose boundary ∂V has unit normal n, and suppose that fluid with density ρf occupies a volume fraction α Use the principle of conservation of mass to obtain the equation ZZZ ZZ d ρf α dx = − ρf αv · n da, dt V ∂V where v is the average fluid velocity, and deduce that ∂ (ρf α) + ∇ · (ρf αv) = 0. ∂t Show that this reduces to (9.121) when the fluid density is constant and α ≈ α0 .

Epilogue

blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah

blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah blah

Epilogue

blah blah blah blah blah blah blah blah blah blah









411







412

APPENDIX

Appendix

A1.1 The Fredholm Alternative

413

A1.1 The Fredholm Alternative Consider the linear equation Ax = b,

(1.156)

where A is an n×n real matrix, b is an n-dimensional real vector and the idea is to solve for the n-dimensional real vector x. There is a unique solution, namely x = A−1 b, provided A is invertible, i.e. provided det (A) 6= 0, but what happens if A is not invertible? This occurs if the rank of A is less than n, so that the rows of A are linearly dependent. By the rank-nullity theorem, A must therefore have a nontrivial kernel of dimension k = n − rank A. In other words, there is a k-dimensional subspace of nonzero vectors v such that Av = 0.

(1.157)

Since the ranks of A and AT are equal, there must likewise be a kdimensional space of nonzero vectors y satisfying AT y = 0

⇔

y T A = 0T .

(1.158)

By multiplying (1.156) on the left by y T , we deduce that y T b = y · b = 0 ∀y ∈ ker AT .

(1.159)

Hence (1.156) has no solution when A is singular unless the solvability condition (1.159) is satisfied. This condition states that the right-hand side b of our original equation must be orthogonal to the kernel of AT . when (1.159) is satisfied, then the solution of (1.156) is nonunique since, if x0 satisfies (1.156), then so does x = x0 +

k X

αi v i ,

i=1

where αi are arbitrary constants and v i are any linearly independent solutions of (1.157). There are, therefore, three alternative outcomes of attempting to solve (1.156). 1. The homogeneous problem (1.157) has only the solution v = 0. In this case, (1.156) has a unique solution. 2.(a) If y · b 6= 0 for some nonzero vector y in the kernel of AT , then (1.156) has no solution. 2.(b) If y · b = 0 for all y in the kernel of AT , then (1.156) has a nonunique solution.

414

APPENDIX

This Fredholm Alternative actually applies to any bounded linear operator T on a complete inner-product space (or Hilbert space) V . We simply have to identify the Cartesian inner product u · v with an arbitrary inner product hu, vi and the transpose of the matrix A with the adjoint adj T of the linear operator T (see Keener, 2000, Chapter 2). As an example, let us re-examine a problem from Chapter 1, namely the steady Navier equation (1.58) subject to the traction boundary condition (1.60). The homogeneous version of this self-adjoint problem admits the two families of eigensolutions u1 = a and u2 = ω×x, where a and ω are constant vectors corresponding to an arbitrary translation and rotation repsectively. The two solvability conditions (1.62) and (1.63) may be interpreted as ensuring that the inhomogeneous terms {g, f } are orthogonal to u1 and u2 .

Appendix 2 Elementary differential geometry

415

416


(i)

(ii)

(iii) 2πb

z

(a, b, c)T x

a

a

y Fig. A2.1. (i) a straight line in the direction (a, b, c)T ; (ii) a circle of radius a; (iii) a helix of radius a and pitch 2πb.

A2.1 Geometry of space curves A smooth curve in R3 may be described parametrically by   x(ξ) r = r(ξ) = y(ξ) , z(ξ)

(2.1)

where ξ is a spatial parameter which typically lies in some interval, say ξ ∈ I. In cases where the curve is semi-infinite, I may likewise be an infinite or semi-infinite interval. Some simple familiar examples are       aξ a cos ξ a cos ξ (i) r(ξ) =  bξ  , (ii) r(ξ) =  a sin ξ  , (iii) r(ξ) =  a sin ξ  , (2.2) cξ 0 bξ which describe respectively (i) a straight line in the direction (a, b, c)T , (ii) a circle of radius a, (iii) a helix of radius a and pitch 2πb, as illustrated in Figure A2.1. In case (ii), we can choose ξ ∈ [0, 2π) without loss of generality, while, in the other two examples, ξ may lie in any given interval, depending on the length of the curve. Provided the curve is smooth, we can always choose a parametrisation such that dr(ξ)/dξ is everywhere finite and satisfies dr 6= 0. dξ

(2.3)

This ensures that there is a one-to-one mapping between values of ξ and points on the curve, and avoids introducing unnecessary singularities and degeneracies into the function r(ξ). For example, as an alternative to (2.2(i)), ¡ ¢T we might have tried to use the parametrisation r(ξ) = aξ 2 , bξ 2 , cξ 2 . This, however, fails to satisfy (2.3) at ξ = 0, and consequently suffers from the

A2.1 Geometry of space curves

417

degeneracy that two values of ξ are mapped to each point on the line (except the origin). Now, assuming that (2.3) is satisfied, we can use Taylor’s Theorem to approximate the behaviour of the curve near some arbitrary point r 0 = r(ξ0 ): r(ξ) ∼ r 0 + (ξ − ξ0 )

¡ ¢ dr (ξ0 ) + O (ξ − ξ0 )2 . dξ

(2.4)

This tells us that the curve locally resembles a straight line in the direction dr/dξ(ξ0 ). At any point r(ξ) on the curve, dr/dξ(ξ) thus gives us a vector that is tangent to the curve. It is convenient to turn this into a unit vector and hence define the unit tangent to be Á¯ ¯ dr ¯¯ dr ¯¯ t= . (2.5) dξ ¯ dξ ¯ We can also use (2.4) to estimate the length δs of a small section of the curve between (say) ξ = ξ0 and ξ = ξ0 + δξ: ¯ ¯ ¯ ¯ dr ¡ ¢ ¯ δs = |r(ξ0 + δξ) − r(ξ0 )| ∼ δξ ¯ (ξ0 )¯¯ + O δξ 2 . (2.6) dξ Hence, to calculate the length ` of the curve between two points parametrised by (say) ξ = a and ξ = b, we have to perform the integral ¯ Z b¯ ¯ dr ¯ ¯ (ξ)¯ dξ. `= (2.7) ¯ ¯ a dξ Alternatively, we can define the arc-length s associated with any value of ξ by ¯ ¯ ds ¯¯ dr ¯¯ = (ξ) . (2.8) dξ ¯ dξ ¯ Our condition (2.3) ensures that ds/dξ > 0 and hence that there is a oneto-one identification between s and ξ. Then we can compute the length of the same segment of the curve as above simply using ` = s(b) − s(a). It is often very convenient to use arc-length to parametrise the curve in the first place. Given any smooth parametrisation that satisfies (2.3), we can always in principle solve (2.8) for s(ξ), invert this bijection to obtain ξ in terms of s, and hence write r as a function of s rather than ξ. This has the advantage that ¯ ¯ ¯ dr ¯ ¯ ¯≡1 (2.9) ¯ ds ¯

418


by definition, and the unit tangent vector is thus simply given by dr . (2.10) ds It is a simple exercise to express the three examples (2.2) in terms of arclength as ³ √ ´      2 + b2 a cos s/ a as a cos(ξ/a) ´ ³  1  bs  , (ii) r =  a sin(ξ/a)  , (iii) r =  a sin s/√a2 + b2  , (i) r = √   a2 + b2 + c2 cs √ 0 2 2 bs/ a + b (2.11) t=

so the unit tangents are given by

´ ³ √      −a sin s/ a2 + b2 a − sin(ξ/a)  ³ √ ´ 1   b  , (ii) t =  cos(ξ/a)  , (iii) r = √ 1 2 + b2  . (i) t = √ a cos s/ a  a2 + b2 + c2 c a2 + b2  0 b (2.12) Since t is a unit vector, it satisfies t · t ≡ 1, and when we differentiate this identity with respect to s we find that dt ≡ 0. (2.13) ds Therefore dt/ds is perpendicular to the tangent vector and thus also to the curve. Again it is convenient to turn this into a unit vector by writing t·

dt = κ(s)n(s), ds where n is the unit normal to the curve and ¯ ¯ ¯ 2 ¯ ¯ dt ¯ ¯ d r ¯ κ(s) = ¯¯ ¯¯ = ¯¯ 2 ¯¯ ds ds

(2.14)

(2.15)

is called the curvature. To try and interpret what κ means, we continue the Taylor expansion (2.4) for one further term to obtain r(s) ∼ r(s0 ) + (s − s0 ) t(s0 ) +

¡ ¢ (s − s0 )2 κ(s0 )n(s0 ) + O (s − s0)3 . 2 (2.16)

Appendix 3 Orthogonal curvilinear coordinates

419

420

Orthogonal curvilinear coordinates

A3.1 Setup Consider a general parametrisation of three-dimensional space, in which the position vector r of any point is given in terms of three spatial coordinates ξ1 , ξ2 and ξ3 : r = r(ξ1 , ξ2 , ξ3 ).

(3.1)

For this to form a sensible coordinate system, there should be a one-to-one correspondence between the position vector r of any point and its coordinates (ξ1 , ξ2 , ξ3 ) = ξ. This is the case provided the Jacobian of the transformation from r to ξ is bounded away from zero, that is µ ¶ · ¸ ∂r ∂r ∂r ∂r 0 < J < ∞, J = det = , , , (3.2) ∂ξ ∂ξ1 ∂ξ2 ∂ξ3 where [·, ·, ·] denotes the triple scalar product between three vectors. By ordering the coordinates appropriately, we may assume that the tangent vectors form a right-handed system and, hence, that J is positive. A coordinate system is orthogonal if the three tangent vectors created by varying each of the coordinates in turn are mutually orthogonal, that is ∂r ∂r · =0 ∂xi ∂xj

whenever i 6= j.

In this case we define the scale factors hi by ¯ ¯ ¯ ∂r ¯ ¯ ¯, hi = ¯ ∂ξi ¯

(3.3)

(3.4)

and the Jacobian is simply given by J = h1 h2 h3 .

(3.5)

We can then define a right-handed, mutually orthogonal set of unit vectors {e1 , e2 , e3 } by ei =

1 ∂r hi ∂ξi

(3.6)

(here the summation convention is not invoked). The most straightforward and familiar coordinate system is Cartesian coordinates x = (x1 , x2 , x3 ), with respect to which the position vector is given by   x1 (3.7) r =  x2  . x3

A3.2 Vectors and tensors

421

This system is clearly orthogonal, with basis vectors       1 0 0      e1 = e2 = e3 = 0 , 1 , 0 , 0 0 1

(3.8)

and scale factors all equal to 1. As another example, consider the well-known cylindrical polar coordinates (r, θ, z), with respect to which the position vector is given by   r cos θ r =  r sin θ  . (3.9) z It is straightforward to show that this coordinate system is orthogonal, with scale factors hr = 1,

hθ = r,

hz = 1.

(3.10)

It follows that J = r, so the Jacobian vanishes on the symmetry axis r = 0. This reflects the fact that θ is indeterminate when r = 0 and leads to the possibility of singular behaviour as r → 0 when solutions of (say) Laplace’s equation are sought in cylindrical polar coordinates.

A3.2 Vectors and tensors Consider a vector v with components vi in Cartesian coordinates, that is ¡ ¢T v = v1 v2 v3 . It is natural to define its components ui in curvilinear coordinates ui = v · ei ,

(3.11)

where ei are as before the basis vectors defined by (3.6). If we denote the ¢T ¡ vector formed by these components u = u1 u2 u3 , then we have     u1 e1 · v u =  u 2  =  e2 · v  , (3.12) u3 e3 · v or

 u = Pv

 where P = 

eT 1 eT 2 eT 3

  .

(3.13)

422


At each point in space, therefore, our curvilinear coordinate basis is found by rotating the usual Cartesian basis, the rotation being characterised by the orthogonal matrix P whose rows are the basis vectors. Not too happy with notation here! Now consider a second-rank tensor with elements denoted by B = (bij ) in Cartesian coordinates and A = (aij ) in curvilinear coordinates. If B is indeed a tensor then, by analogy with Chapter 1, we must have A = P BP T ,

(3.14)

so the curvilinear components of B are given by aij = eT i Bej .

(3.15)

It is thus possible to write any tensor B in terms of its curvilinear components and the so-called outer products of the basis vectors: X B= aij ei eT (3.16) j. ij

A3.3 Derivatives of basis vectors In most orthogonal coordinate systems, including cylindrical polars and other familiar examples, the basis vectors ej vary with position, and this is the property that really distinguishes them from Cartesian coordinates. It will be useful for future reference to calculate the derivatives of the basis vectors as follows. In general, we can write ∂ei X = cijk ek , ∂ξj

(3.17)

k

so there are 27 constants cijk to be determined.† Since ei · ej = δij , we must have ∂ej ∂ei · ej + ei · ≡0 ∂ξk ∂ξk

(3.18)

for all i, j and k. This provides 18 independent equations (since it is invariant if i and j are swapped). The further 9 equations may be obtained by calculating ∂r/∂ξi ∂ξj in two different ways as follows: ´ ´ ∂ ³ ∂ ³ hj ej ≡ hi ei (3.19) ∂ξi ∂ξj † These constants are related to the so-called Christoffel symbols; see Aris (1962).

A3.4 Scalar and vector fields; grad, div and curl

423

for all i and j. It is thus a straightforward but tedious exercise in algebraic manipulation to deduce the following: X ek ∂hi ej ∂hj ∂ei = − δij . (3.20) ∂ξj hi ∂ξi hk ∂ξk k

A3.4 Scalar and vector fields; grad, div and curl We can think of any scalar field f that varies with position r as being a function of our coordinate variables: f = f (ξ). Similarly, any vector field may be written as u = u(ξ), of u with respect to this ¡ and the components ¢ system are now written as u1 u2 u3 , where ui = u · ei .

(3.21)

The gradient of a scalar field is defined in the usual way in Cartesian variables, that is   ∂f /∂x1 grad f = ∇f =  ∂f /∂x2  . (3.22) ∂f /∂x3 The chain rule leads to ∂r ∂f = · ∇f = hi ei · ∇f ∂ξi ∂ξi and, since the ei are orthonormal, it follows that X 1 ∂f ∇f = ek . hk ∂ξk

(3.23)

(3.24)

k

A useful example occurs if f is one of the variables ξi , when (3.24) implies that ei ∇ξi = . (3.25) hi Now the divergence of a vector field u may be written in terms of Cartesian coordinates as X ∂u div u = ∇ · u = · ∇xk . (3.26) ∂xk k

If we rewrite the gradient using (3.24) and use the chain rule, we find ¶ X ek ∂u X µ 1 ∂uk u ∂ek ∇·u= · = − · . (3.27) hk ∂ξk hk ∂ξk hk ∂ξk k

k

424


Now we apply (3.20) to the final term and rearrange to obtain ½ ´ ´ ´¾ 1 ∂ ³ ∂ ³ ∂ ³ ∇·u= h2 h3 u1 + h1 h3 u2 + h1 h2 u3 . (3.28) h1 h2 h3 ∂ξ1 ∂ξ2 ∂ξ3 Having derived the gradient and divergence, we may now express the Laplacian operator ∇2 f = ∇ · (∇f ) in terms of arbitrary orthogonal coordinates: ½ µ ¶ µ ¶ µ ¶¾ 1 ∂ h2 h3 ∂f ∂ h1 h3 ∂f ∂ h1 h2 ∂f 2 ∇ f= + + . h1 h2 h3 ∂ξ1 h1 ∂ξ1 ∂ξ2 h2 ∂ξ2 ∂ξ3 h3 ∂ξ3 (3.29) Next, we define the curl of a vector field by curl u = ∇×u =

X k

∇xk ×

X ek ∂u ∂u = × . ∂xk hk ∂ξk

(3.30)

k

By rearranging the right-hand side and using (3.20), this may be transformed to ´ ´ X 1 ∂ ³ X ei ×ej ∂ ³ u ∂ek ∇×u = ek ×u + × = hj uj , (3.31) hk ∂ξk hk ∂ξk hi hj ∂ξi k

i,j

which may be written conveniently as ¯ ¯ h1 e1 h2 e2 h3 e3 ¯ ¯ ∂ ∂ ∂ 1 ¯ ∇×u= ¯ ∂ξ2 ∂ξ3 h1 h2 h3 ¯ ∂ξ1 ¯ ¯ h1 u1 h2 u2 h3 u3

¯ ¯ ¯ ¯ ¯ ¯. ¯ ¯ ¯

(3.32)

An instructive alternative derivation of (3.28) may be obtained by applying the divergence theorem to the small box B illustrated in Figure A3.1: ZZZ ZZ ∇ · u dx = u · n dS (3.33) B Z Z ∂B ZZ = [u1 h2 h3 ]ξξ11 +δξ1 dξ2 dξ3 + [u2 h1 h3 ]ξξ22 +δξ2 dξ1 dξ3 ZZ + [u3 h1 h2 ]ξξ33 +δξ3 dξ1 dξ2 , (3.34) where the integrals on the right-hand side are over the faces ξ1 = const, ξ2 = const and ξ3 = const respectively. In the limit δξ → 0, we therefore

A3.4 Scalar and vector fields; grad, div and curl

e3

425

h2δξ2

h1δξ1 e2

h3δξ3 e1 Fig. A3.1. A small box B.

obtain ZZZ (∇ · u)h1 h2 h3 dξ1 dξ2 dξ3 ∼ ZBZ Z ∂ ∂ ∂ (h2 h3 u1 ) + (h1 h3 u2 ) + (h1 h2 u3 ) dξ1 dξ1 dξ3 (3.35) ∂ξ2 ∂ξ3 B ∂ξ1 and, since this must hold for all such boxes B, (3.28) follows. This approach is useful in finding the divergence of a tensor A in curvilinear coordinates which, by analogy with (3.26), is defined in Cartesian coordinates to be div A = ∇ · A =

X ∂AT k

∂xj

∇xk .

(3.36)

By applying the identity ZZZ

¡ ¢ ∇ · AT dx = B

ZZ An dS

(3.37)

∂B

to the box B and letting δξ → 0 as above, we find that ¡ ¢ ∇ · AT =

1 h1 h2 h3

½

¾ ∂ ∂ ∂ (h2 h3 Ae1 ) + (h1 h3 Ae2 ) + (h1 h2 Ae3 ) . ∂ξ1 ∂ξ2 ∂ξ3 (3.38)

426


Now we expand the derivatives and use (3.20) to obtain ( Ã !) µ ¶ X ek ∂hi ¡ T ¢ X ei ∂ aij J aij ej ∂hj ∇· A = + − δij , J ∂ξj hi hi hi ∂ξi hk ∂ξk i,j

k

(3.39) where aij are the curvilinear components of A. Hence we find the ei component of ∇ · A in the form ½ ¾ ∂ ∂ ∂ 1 ei · (∇ · A) = (h2 h3 a1i ) + (h1 h3 a2i ) + (h1 h2 a3i ) h1 h2 h3 ∂ξ1 ∂ξ2 ∂ξ3 µ ¶ X 1 ∂hi ∂hk + aik − akk . (3.40) hi hk ∂ξk ∂ξi k

A3.5 Strain in curvilinear coordinates Now we repeat the calculation from §1.4 to derive an expression for the strain tensor in an arbitary orthogonal coordinate system. We define the Lagrangian coordinates X = (X1 , X2 , X3 ) of a material point to be the parametrisation corresponding to its initial position X, that is X = r(X ).

(3.41)

At any subsequent time t, the same material point moves to a new position x parameterised by its Eulerian coordinates ξ = (ξ1 , ξ2 , ξ3 ), so that ¡ ¢ x = r ξ(X , t) . (3.42) Consider a small line element joining two points with Lagrangian coordinates X and X + δX . The initial Cartesian position vectors of the end points are thus X and X + δX, where X hk (X )ek (X )δXk , (3.43) δX = k

and the initial length L of the line element is X L2 = |δX|2 = hk (X )2 δXk2 .

(3.44)

k

At a later time t, the line element is transformed to δx, where, by repeated use of the chain rule X ∂xi ∂ξj X ∂ξj δxi = δXk or δx = hj (ξ)ej (ξ) δXk . (3.45) ∂ξj ∂Xk ∂Xk j,k

j,k

Notice that, since ej and hj are not in general constant, we must distinguish

A3.5 Strain in curvilinear coordinates

427

between their values at a particle’s initial position X and at its current position ξ. By comparing (3.43) and (3.45), we see that δx = F δX,

(3.46)

where the deformation gradient tensor F is given by X hi (ξ) ∂ξi ei (ξ)eT F = j (X ). hj (X ) ∂Xj

(3.47)

i,j

The length ` of the deformed line element is thus given by `2 = δxT δx = δX T C δX, where the Green deformation tensor is defined by X C = F TF = Cij ei (X )eT j (X ),

(3.48)

(3.49)

i,j

and the curvilinear components of C are Cij =

X k

hk (ξ)2 ∂ξk ∂ξk . hi (X )hj (X ) ∂Xi ∂Xj

(3.50)

The nonlinear strain tensor is then found using 1 E = (C − I). (3.51) 2 Further simplification may be achieved if the strains are small enough for linear elasticity to be valid. If each material point is only slightly disturbed from its initial position, then the Eulerian and Lagrangian coordinates are approximately equal. We may therefore write ξi = Xi + so that x∼X+

X

ui (X , t) , hi (X )

(3.52)

¡ ¢ ui ei + O |u|2

(3.53)

i

when the displacements ui are small. By substituting (3.52) into (3.50) and linearising, we find µ ¶ µ ¶ X uk ∂hj ¡ ¢ hj ∂ uj ui hi ∂ + + 2δij + O |u|2 , Cij ∼ δij + hj ∂ξj hi hi ∂ξi hj hj hk ∂ξk k

(3.54)

428


where, within the linear theory, it is no longer necessary to distinguish between ξ and X . The linearised strain components in curvilinear coordinates therefore read ½ ¾ X uk ∂hj uj ∂hj ui ∂hi 1 ∂uj 1 1 ∂ui − + − + δij eij = . 2 hj ∂ξj hi hj ∂ξj hi ∂ξi hi hj ∂ξi hj hk ∂ξk k

(3.55) A3.6 Stress in curvilinear coordinates As in §1.5, we defined the Cauchy stress τij to be the component in the ei -direction of the stress acting on a surface element whose normal points along ej . Although the basis vectors ei may now vary with position, at any fixed location, the curvilinear components τij of the stress tensor are found by simply rotating the Cartesian components through the orthogonal matrix P introduced in §A3.2. It is easily shown that the symmetry of τij in Cartesian coordinates is maintained in an arbitrary orthogonal coordinate system. In §1.7, we introduced the constitutive relation τij = λ (ekk ) δij + 2µeij ,

(3.56)

which links the stress and linearised strain tensors in a linear elastic material. Note that, since τij and eij are tensors, they both obey the transformation rule (3.14) if the coordinate axes are rotated by an orthogonal matrix P , which may be a function of position. Hence, although the relation (3.56) was originally posed using fixed Cartesian coordinates, it actually holds in any orthogonal coordinate system. state full linear navier eqns and const relations in arbitrary system Now say something about nonlinear elasticity and piola–Kirchhoff

A3.7 Examples of orthogonal coordinate systems A3.7.1 Cylindrical polar coordinates The Cartesian position vector is given in terms of the coordinate variables (r, θ, z) by   r cos θ r =  r sin θ  , (3.57) z

A3.7 Examples of orthogonal coordinate systems

429

r

z

θ

Fig. A3.2. Cylindrical polar coordinates.

as illustrated in Figure A3.2. It follows that the basis vectors are given by 

 cos θ er =  sin θ  , 0



 − sin θ eθ =  cos θ  , 0



 0 ez =  0  , 1

(3.58)

hz = 1.

(3.59)

and the scale factors are hr = 1,

hθ = r,

The formulae for the gradient, divergence, Laplacian and curl in this coordinate system are, therefore, ∂f 1 ∂f ∂f er + eθ + ez , (3.60a) ∂r r ∂θ ∂z 1 ∂ 1 ∂uθ ∂uz ∇·u= (rur ) + + , (3.60b) r ∂r µ ¶ r ∂θ 2 ∂z 2 1 ∂ ∂f 1 ∂ f ∂ f ∇2 f = r + 2 2 + 2, (3.60c) r ∂r ∂r r ∂θ ∂z µ ¶ µ ¶ µ ¶ 1 ∂uz ∂uθ ∂ur ∂uz 1 ∂ ∂ur ∇×u = − er + − eθ + (ruθ ) − ez , r ∂θ ∂z ∂z ∂r r ∂r ∂θ (3.60d) ∇f =

where f is any scalar field and u = ur er + uθ eθ + uz ez is any vector field (both suitably differentiable). The components of the linearised strain tensor

430


θ

r

φ

Fig. A3.3. Spherical polar coordinates.

corresponding to a displacement field u are given by µ ¶ 1 ∂uθ ∂uz ∂ur , eθθ = + ur , ezz = , (3.61a) err = ∂r r ∂θ ∂z 1 ∂ur ∂uθ uθ ∂ur ∂uz ∂uθ 1 ∂uz 2erθ = + − , 2erz = + , 2eθz = + . r ∂θ ∂r r ∂z ∂r ∂z r ∂θ (3.61b) The divergence of a tensor A with components aij (i, j = r, θ, z) is given by ¾ ½ ∂azr aθθ 1 ∂(rarr ) 1 ∂aθr + + − er ∇·A= r ∂r r ∂θ ∂z r ½ ¾ 1 ∂(rarθ ) 1 ∂aθθ ∂azθ aθr + + + + eθ r ∂r r ∂θ ∂z r ½ ¾ 1 ∂(rarz ) 1 ∂aθz ∂azz + + + ez . (3.62) r ∂r r ∂θ ∂z A3.7.2 Spherical polar coordinates The position vector r of a point is given in terms of the spherical polar coordinates r, θ and φ by   r sin θ cos φ r =  r sin θ sin φ  , (3.63) r cos θ

A3.7 Examples of orthogonal coordinate systems

431

as illustrated in Figure A3.3. The basis vectors and scale factors are       sin θ cos φ cos θ cos φ − sin φ er =  sin θ sin φ  , eθ =  cos θ sin φ  , eφ =  cos φ  , cos θ − sin θ 0 (3.64) hr = 1,

hθ = r,

hφ = r sin θ,

(3.65)

and it follows that ∂f 1 ∂f 1 ∂f er + eθ + eφ , (3.66a) ∂r r ∂θ r sin θ ∂φ 1 ∂ ¡ 2 ¢ 1 ∂ 1 ∂uφ ∇·u= 2 r ur + (sin θuθ ) + , (3.66b) r ∂r r sin θ ∂θ r sin θ ∂φ µ ¶ µ ¶ ∂2f 1 ∂ 1 ∂ ∂f 1 2 2 ∂f ∇ f= 2 r + 2 sin θ + 2 2 , (3.66c) r ∂r ∂r r sin θ ∂θ ∂θ r sin θ ∂φ2 µ ¶ µ ¶ ∂uφ ∂ ∂uθ 1 ∂ur 1 (sin θuφ ) − er + − eθ ∇×u = r sin θ ∂θ ∂φ r sin θ ∂φ ∂r µ ¶ 1 ∂ ∂ur + (ruθ ) − eφ . (3.66d) r ∂r ∂θ ∇f =

The components of the linear strain tensor in spherical polars read ∂ur , ∂r µ ¶ 1 ∂uθ = + ur , r ∂θ

1 ∂ur ∂uθ uθ + − , (3.67a) r ∂θ ∂r r ∂uφ uφ 1 ∂ur = + − , r sin θ ∂φ ∂r r (3.67b) 1 ∂uθ 1 ∂uφ cot θuφ = + − . r sin θ ∂φ r ∂θ r (3.67c)

err =

2erθ =

eθθ

2erφ

eφφ =

cot θuθ 1 ∂uφ ur + + , 2eθφ r sin θ ∂φ r r

The divergence of a tensor A in spherical polars is given by ½ ¾ aθθ + aφφ 1 ∂(r2 arr ) 1 ∂(sin θaθr ) 1 ∂aφr ∇·A= + + − er r2 ∂r r sin θ ∂θ r sin θ ∂φ r ½ ¾ aθr − cot θaφφ 1 ∂(r2 arθ ) 1 ∂(sin θaθθ ) 1 ∂aφθ + + + + eθ r2 ∂r r sin θ ∂θ r sin θ ∂φ r ½ ¾ 1 ∂(r2 arφ ) 1 ∂(sin θaθφ ) 1 ∂aφφ aφr + cot θaφθ + + + + eφ . r2 ∂r r sin θ ∂θ r sin θ ∂φ r (3.68)

432


A3.8 Time-dependent coordinates When describing large elastic deformations of a plate or a shell, it is often convenient to use coordinates that move with object rather than being fixed in space. This introduces a slight complication in that our coordinate system now varies in time as well as space. Let us then suppose that the position vector r of any point in space is parametrised by r = r(ξ), t),

(3.69)

where ξ = (ξ1 , ξ2 , ξ3 ) are spatial coordinates and t is time. For simplicity, we assume again that the system is orthogonal for all t. Hence we can define the orthogonal unit vectors ei and scale factors just as before by hi by (3.4) and (3.6); the only difference is that they now depend on t as well as ξ. For each fixed value of t, the analysis from §§A3.1–A3.4 may then be followed without modification, so that the curvilinear components of vectors and tensors and the definitions of div, grad and curl are unchanged. We have to think a bit harder, though about how to define displacements and strains in such a coordinate system. Suppose a material point has in initial Lagrangian parameterisation X = (X1 , X2 , X3 ), so its initial position is X = r(X , 0).

(3.70)

At a subsequent time t, the same material point has the Eulerian parameterisation ξ(X , t); in other words its position at time t is ¡ ¢ x = r ξ(X , t), t . (3.71)

Bibliography

Abramowitz, M. & Stegun, I. A. 1972 Handbook of Mathematical Functions (tenth printing). Dover. New York. Aris, R. 1962 Vectors, Tensors, and the Basic Equations of Fluid Mechanics. Dover. New York. Bender, C. M. & Orszag, S. A. 1978 Advanced Mathematical Methods for Scientists and Engineers. McGraw–Hill. Singapore. Bland, D. R. 1988 Wave Theory and Applications. Clarendon Press. Oxford. Boyce, W. E. & DiPrima, R. C. 2004 Elementary Differential Equations and Boundary Value Problems. Eighth Edition. Wiley. Hoboken, New Jersey. Carrier, G. F., Krook, M. & Pearson, C. E. 1983 Functions of a Complex Variable: Theory and Technique. Hod Books. Ithaca, New York. England, A. H. 2003 Complex Variable Methods in Elasticity. Dover. New York. Freund, L. B. 1989 Dynamic Fracture Mechanics. Cambridge University Press. Cambridge. Gradshteyn, I. S. & Ryzhik, I. M. 1994 Table of Integrals, Series, and Products. Fifth Edition. Academic Press. San Diego. Hill, R. 1998 The Mathematical Theory of Plasticity. Oxford University Press. Oxford. Hinch, E. J. 1991 Perturbation Methods. Cambridge University Press. Cambridge. Hooke, R. 1678 De Potentia Restitutiva or of Spring Explaining the Power of Springing Bodies. John Martyn at The Bell in St Paul’s Churchyard. London. Howison, S. D. 2005 Practical Applied Mathematics: Modelling, Analysis, Approximation. Cambridge University Press. Cambridge.

434

Bibliography

Jordan, D. W. & Smith, P. 1999 Nonlinear Ordinary Differential Equations: an Introduction to Dynamical Systems. Oxford University Press. Oxford. Kac, M. 1966 Can one hear the shape of a drum? Am. Math. Monthly 73, 1–23. Keener, J. P. 2000 Principles of Applied Mathematics, Transformation and Approximation. Revised Edition. Westview Press. Boulder CO. Kreyszig, E. 1959 Differential Geometry. Dover. New York. Lakes, R. 1987 Foam structures with a negative Poisson’s ratio. Science 235, 1038–1040. Love, A. E. H. 1944 A Treatise on the Mathematical Theory of Elasticity. Dover. New York. Noll, W. 1958 A mathematical theory of the mechanical behaviour of continuous media. Arch. Rational Mech. Anal. 2, 197–226. Ockendon, J. R., Howison, S. D., Lacey, A. A. & Movchan, A. B. 1999 Applied Partial Differential Equations. Oxford University Press. Oxford. Ockendon, H. & Ockendon, J. R. 1995 Viscous Flow. Cambridge University Press. Cambridge. Priestley, H. A. 2003 Introduction to Complex Analysis. Oxford University Press. Oxford. Rae, A. I, M. 1986 Quantum Mechanics. Second Edition. Adam Hilger. Bristol and Philadelphia. Salazar, D. & Westbrook, R. 2004 Inverse problems of Mixed Type in Linear Plate Theory. Euro. J. Appl. Math. 15, 129–146. Spencer, A. J. M. 1970 The static theory of finite elasticity. J. Inst. Maths Applics 6, 164–200. Strang, G. 1988 Linear Algebra and its Applications. Third Edition. Harcourt Brace Jovanovich College Publishers. Orlando. Timoshenko, S. P. & Goodier, J. N. 1970 Theory of Elasticity. McGraw–Hill. Singapore. Timoshenko, S. P. & Woinowsky-Krieger, S. 1959 Theory of Plates and Shells. McGraw–Hill. Singapore.

Applied solid mechanics

Solid Mechanics

Solid Mechanics

Solid Mechanics, Volume 36 (Advances in Applied Mechanics)