Applied Solid Mechanics (Cambridge Texts in Applied Mathematics)

APPLIED SOLID MECHANICS Much of the world around us, both natural and man-made, is built from and held together by sol...

Author: Peter Howell | Gregory Kozyreff | John Ockendon

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APPLIED SOLID MECHANICS

Much of the world around us, both natural and man-made, is built from and held together by solid materials. Understanding how they behave is the task of solid mechanics, which can in turn be applied to a wide range of areas from earthquake mechanics and the construction industry to biomechanics. The variety of materials (such as metals, rocks, glasses, sand, flesh and bone) and their properties (such as porosity, viscosity, elasticity, plasticity) are reflected by the concepts and techniques needed to understand them, which are a rich mixture of mathematics, physics, experiment and intuition. These are all brought to bear in this distinctive book, which is based on years of experience in research and teaching. Theory is related to practical applications, where surprising phenomena occur and where innovative mathematical methods are needed to understand features such as fracture. Starting from the very simplest situations, based on elementary observations in engineering and physics, models of increasing sophistication are derived and applied. The emphasis is on problem solving and on building an intuitive understanding, rather than on a technical presentation of theoretical topics. The text is complemented by over 100 carefully chosen exercises, and the minimal prerequisites make it an ideal companion for mathematics students taking advanced courses, for those undertaking research in the area or for those working in other disciplines in which solid mechanics plays a crucial role.

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 Fonds de la Recherche Scientifique—FNRS and Universit´e Libre de Bruxelles

JOHN OCKENDON University of Oxford

CAMBRIDGE UNIVERSITY PRESS

Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, São Paulo Cambridge University Press The Edinburgh Building, Cambridge CB2 8RU, UK Published in the United States of America by Cambridge University Press, New York www.cambridge.org Information on this title: www.cambridge.org/9780521854894 © P. D. Howell, G. Kozyreff and J. R. Ockendon 2009 This publication is in copyright. Subject to statutory exception and to the provision of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published in print format 2008

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978-0-521-85489-4

hardback

ISBN-13

978-0-521-67109-5

paperback

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Contents

List of illustrations Prologue

page viii xiii

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 Exercises

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

Linear elastostatics 2.1 Introduction 2.2 Linear displacements 2.3 Antiplane strain 2.4 Torsion 2.5 Multiply-connected domains 2.6 Plane strain 2.7 Compatibility 2.8 Generalised stress functions 2.9 Singular solutions in elastostatics 2.10 Concluding remark Exercises

28 28 29 37 39 42 47 68 70 82 93 93

v

vi

Contents

Linear elastodynamics 3.1 Introduction 3.2 Normal modes and plane waves 3.3 Dynamic stress functions 3.4 Waves in cylinders and spheres 3.5 Initial-value problems 3.6 Moving singularities 3.7 Concluding remarks Exercises

103 103 104 121 124 132 138 143 143

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 4.5 Linear rod theory 4.6 Linear plate theory 4.7 Von K´ arm´an plate theory 4.8 Weakly curved shell theory 4.9 Nonlinear beam theory 4.10 Nonlinear rod theory 4.11 Geometrically nonlinear wave propagation 4.12 Concluding remarks Exercises

150 150 151 152 153 158 162 172 177 187 195 198 204 205

Nonlinear elasticity 5.1 Introduction 5.2 Stress and strain revisited 5.3 The constitutive relation 5.4 Examples 5.5 Concluding remarks Exercises

215 215 216 221 233 239 239

Asymptotic analysis 6.1 Introduction 6.2 Antiplane strain in a thin plate 6.3 The linear plate equation 6.4 Boundary conditions and Saint-Venant’s principle 6.5 The von K´ arm´an plate equations 6.6 The Euler–Bernoulli plate equations 6.7 The linear rod equations 6.8 Linear shell theory

245 245 246 248 253 261 267 273 278

Contents

vii

6.9 Concluding remarks Exercises

282 283

Fracture and contact 7.1 Introduction 7.2 Static brittle fracture 7.3 Contact 7.4 Concluding remarks Exercises

287 287 288 309 320 321

Plasticity 8.1 Introduction 8.2 Models for granular material 8.3 Dislocation theory 8.4 Perfect plasticity theory for metals 8.5 Kinematics 8.6 Conservation of momentum 8.7 Conservation of energy 8.8 The ﬂow rule 8.9 Simultaneous elasticity and plasticity 8.10 Examples 8.11 Concluding remarks Exercises

328 328 330 337 344 358 360 360 362 364 365 370 372

More general theories 9.1 Introduction 9.2 Viscoelasticity 9.3 Thermoelasticity 9.4 Composite materials and homogenisation 9.5 Poroelasticity 9.6 Anisotropy 9.7 Concluding remarks Exercises Epilogue

378 378 379 388 391 408 413 417 417 426

Appendix References Index

428 440 442

Orthogonal curvilinear coordinates

Illustrations

1.1 1.2 1.3 1.4 1.5 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 3.4 3.5 3.6 3.7

A reference tetrahedron. page 8 The forces acting on a small two-dimensional element. 9 A small pill-box-shaped region at the boundary between two elastic solids. 18 Forces acting on a polar element of solid. 22 A system of masses connected by springs. 25 A unit cube undergoing (a) uniform expansion, (b) one-dimensional shear, (c) uniaxial stretching. 30 A uniform bar being stretched under a tensile force. 32 A paper model with negative Poisson’s ratio. 33 A strained plate. 34 A bar in a state of antiplane strain. 38 A twisted bar. 39 A uniform tubular torsion bar. 43 The cross-section of (a) a circular cylindrical tube; (b) a cut tube. 44 The unit normal and tangent to the boundary of a plane region. 49 A plane annulus being inﬂated by an internal pressure. 53 A plane rectangular region subject to tangential tractions on its faces. 57 The tractions applied to the edge of a semi-inﬁnite strip. 59 The surface displacement of a half-space and corresponding surface pressure. 65 83 A family of functions δε (x) that approach a delta-function as ε → 0. Contours of the maximum shear stress created by a point force acting at the origin. 85 Four point forces. 91 Plots of the ﬁrst three Bessel functions. 108 A P -wave reﬂecting from a rigid boundary. 116 A layered elastic medium. 117 Dispersion relation for symmetric and antisymmetric Love waves. 120 Illustration of ﬂexural waves. 128 The one-dimensional fundamental solution. 134 The two-dimensional fundamental solution. 135

viii

List of illustrations 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 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

ix

The cone x2 + y 2 = c2 t2 tangent to the plane k1 x + k2 y = ωt. 138 The two-sheeted characteristic cone for the Navier equation. 138 The response of a string to a point force moving at speed V . 139 Wave-fronts generated by a moving force on an elastic membrane. 141 P -wave- and S -wave-fronts generated by a point force moving at speed V in plane strain. 142 Group velocity versus wave-number for symmetric and antisymmetric Love waves. 146 The forces acting on a small length of a uniform bar. 151 The forces acting on a small length of an elastic string. 153 The forces and moments acting on a small segment of an elastic beam. 154 The end of a beam under clamped, simply supported and free conditions.155 The ﬁrst three buckling modes of a clamped elastic beam. 157 The internal force components in a thin elastic rod. 159 Cross-section through a rod showing the bending moment components. 159 Examples of cross-sections in the (y, z)-plane and their bending stiﬀnesses. 161 The forces acting on a small section of an elastic plate. 163 The bending moments acting on a section of an elastic plate. 164 The displacement of a simply supported rectangular plate sagging under gravity. 169 (a) A cylinder, (b) a cone, (c) another developable surface, (d) a hyperboloid. 175 Typical surface shapes with (a) zero, (b) negative and (c) positive Gaussian curvature. 179 Deformations of a cylindrical shell. 184 Deformations of an anticlastic shell. 185 Deformations of a synclastic shell. 186 A beam (a) before and (b) after bending; (c) a close-up of the displacement ﬁeld. 187 (a) The forces and moments acting on a small segment of a beam. (b) The sign convention for the forces at the ends of the beam. 188 (a) Final angle of a diving board versus applied force parameter. (b) Deﬂection of a diving board for various values of the force parameter.191 (a) Response diagram of the amplitude of the linearised solution for a buckling beam versus the force parameter. (b) Corresponding response of the weakly nonlinear solution. 193 (a) Pitchfork bifurcation diagram of leading-order amplitude versus forcing parameter. (b) The corresponding diagram when asymmetry is introduced. 195 A system of pendulums attached to a twisting rubber band. 199 A kink propagating along a series of pendulums attached to a rod. 200 Travelling wave solution of the nonlinear beam equations. 201 A beam clamped near the edge of a table. 206 A beam supported at two points. 207 The ﬁrst three buckling modes of a vertically clamped beam. 213

x 5.1 5.2 5.3 5.4 5.5 5.6 6.1 6.2 7.1 7.2 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 8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8 8.9 8.10 8.11 8.12

List of illustrations The deformation of a small scalene cylinder. Typical force–strain graphs for uniaxial tests on various materials. 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. The edge of a plate subject to tractions. The geometry of a deformed two-dimensional plate. Deﬁnition sketch of a thin crack. Deﬁnition sketch for contact between two solids. (a) A Mode III crack. (b) A cross-section in the (x, y)-plane. √ Deﬁnition sketch for the function z 2 − c2 . Displacement ﬁeld for a Mode III crack. (a) A planar Mode II crack. (b) The regularised problem of a thin elliptical crack. Contour plot of the maximum shear stress around a Mode II crack. The displacement of a Mode II crack under increasing shear stress. A Mode I crack. Contour plot of the maximum shear stress around a Mode I crack. The displacement of a Mode I crack under increasing normal stress. Solution for the contact between a string and a level surface. Three candidate solutions for a contact problem. The contact between a beam and a horizontal surface under a uniform pressure. Contact between a rigid body and an elastic half-space. The penetration of a quadratic punch into an elastic half-space. A ﬂexible ruler ﬂattened against a table. A wave travelling along a rope on the ground. A typical stress–strain relationship for a plastic material. The stress–strain relationship for a perfectly plastic material. The forces acting on a particle at the surface of a granular material. The normal force and frictional force acting on a surface element inside a granular material. The Mohr circle. The triaxial stress factor versus angle of friction. An antiplane cut-and-weld operation. The displacement ﬁeld in an edge dislocation. An edge dislocation in a square crystal lattice. A moving edge dislocation. The normalised torque versus twist applied to an elastic-plastic cylindrical bar. The normalised torque versus twist applied to an elastic-plastic cylindrical bar, showing the recovery phase.

218 233 234 235 236 238 254 269 288 288 290 292 293 297 301 303 304 306 307 310 311 314 317 319 326 326 329 330 331 332 333 336 339 340 341 342 347 348

List of illustrations 8.13 8.14

xi

The free-boundary problem for an elastic-perfectly plastic torsion bar. 349 Residual shear stress in a gun barrel versus radial distance for diﬀerent values of the maximum internal pressurisation. 352 8.15 The Tresca yield surface. 355 8.16 The von Mises yield surface. 356 8.17 The Coulomb yield surface. 358 8.18 L¨ uders bands in a thin sheet of metal. 369 8.19 The Mohr surface for three-dimensional granular ﬂow. 373 8.20 The normalised torque versus twist applied to an elastic-plastic cylindrical bar undergoing a loading cycle. 375 9.1 (a) A spring; (b) a dashpot; (c) a spring and dashpot connected in parallel; (d) a spring and dashpot connected in series. 380 9.2 (a) Applied tension as a function of time. (b) Resultant displacement of a linear elastic spring. (c) Resultant displacement of a linear dashpot.381 9.3 Displacement of a Voigt element due to the applied tension shown in Figure 9.2(a). 382 9.4 Displacement of a Maxwell element due to the applied tension shown in Figure 9.2(a). 383 9.5 (a) The variation of Young’s modulus with position in a bar. (b) The corresponding longitudinal displacement. 392 9.6 (a) The variation of Young’s modulus with position in a bar. (b) The corresponding longitudinal displacement. 395 9.7 A periodic microstructured shear modulus. 396 9.8 A symmetric, piecewise constant shear modulus distribution. 400 9.9 Some modulus distributions that are antisymmetric about the diagonals of a square. 402 9.10 Dimensionless wavenumber versus the Young’s modulus non-uniformity parameter. 407 9.11 The one-dimensional squeezing of a sponge. 411 9.12 Dimensionless stress applied to a sponge versus dimensionless time for diﬀerent values of the P´eclet number. 412 9.13 A Jeﬀreys viscoelastic element. 418 9.14 A system of masses connected by springs and dashpots in parallel. 418 9.15 A system of masses connected by springs and dashpots in series. 419 9.16 Dimensionless wavenumber versus Young’s modulus contrast for a piecewise uniform bar. 424 A1.1 A small reference box. 432 A1.2 Cylindrical polar coordinates. 437 A1.3 Spherical polar coordinates. 438

Prologue

Although solid mechanics is a vitally important branch of applied mechanics, it is often less popular, at least among students, than its close relative, ﬂuid mechanics. Several reasons can be advanced for this disparity, such as the prevalence of tensors in models for solids or the especial diﬃculty of handling nonlinearity. Perhaps the most daunting prospect for the student is the multitude of diﬀerent behaviours that can occur and cause elementary theories of elasticity to become irrelevant in practice. Examples include fracture, buckling and plasticity, and these 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 objective in this book is to demonstrate this fact to undergraduate and beginning graduate students. 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 do this by relating the theory to practical applications where surprising phenomena occur and where innovative mathematical methods are needed. Our layout is essentially pragmatic. Although more advanced texts in solid mechanics often begin with quite general theories founded on basic mechanical and thermodynamic principles, we start from the very simplest models, based on elementary observations in engineering and physics, and build our way towards models that are the basis for current applied research in solid mechanics. Hence, we begin by deriving the basic Navier equations of linear elasticity, before illustrating the mathematical techniques that allow these equations to be solved in many diﬀerent practically relevant situations, both static and dynamic. We then proceed to describe some approximate theories for the elastic deformation of thin solids, namely bars, strings, beams, rods, plates and shells. We soon discover that many everyday phenomena, such as the buckling of a beam under a compressive load, cannot be fully described xiii

xiv

Prologue

using linear theories. We therefore give a brief exposition of the general theory of nonlinear elasticity, and then show how formal asymptotic methods allow simpliﬁed linear and weakly nonlinear models to be systematically deduced. Although we regard such asymptotic techniques as invaluable to any applied mathematician, these last two topics may both be omitted on a ﬁrst reading without loss of continuity. We go on to present simple models for fracture and contact, comparing and contrasting these apparently similar phenomena. Next, we show how plasticity theory can be used to describe situations where a solid yields under a suﬃciently high stress. Finally, we show how elasticity theory may be generalised to include further physical effects, such as thermal stresses, viscoelasticity 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 techniques that will emerge in this book, we will usually try to present the theoretical developments ab initio. Nonetheless, the book is very far from being self-contained. Any student who aspires to becoming a solid mechanics specialist will have to delve further into the literature, and we will provide references to help with this. We assume only that the reader has a reasonable familiarity with the calculus of several variables. Fluency with the more advanced techniques required for Chapters 6 and 7, in particular, will readily be acquired by a student who works through the exercises in the early Chapters, especially those cited in the text. Indeed, we ﬁrmly believe that solid mechanics provides a wonderful arena in which to build an understanding of such important mathematical areas as linear algebra, partial diﬀerential equations, complex variable theory, diﬀerential geometry and the calculus of variations. Our hope is that, having read this book, a student should be able to confront 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. During the writing of this book, we received a great deal of help and inspiration as a result of discussions with David Allwright, Jon Chapman, Sam Howison, L. Mahadevan, Roman Novokshanov and Domingo Salazar, as well as many other colleagues and students too numerous to thank individually. We would like to express our particular gratitude to Gareth Jones, Hilary Ockendon and Tom Witelski who gave invaluable advice on draft Chapters. We are also indebted to David Tranah and his colleagues at Cambridge University Press for helping to make this book a reality.

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. Hence, 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. When we refer to a solid “particle”, we will be thinking of a very small region of matter but one whose dimension is nonetheless much greater than an atomic spacing. 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-andfast 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 reference state, from which it is displaced by a distance that depends on the applied forces. This is in contrast with a ﬂuid, which has no special rest state and responds to forces via a velocity distribution. 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 elastic deformation of a continuous solid. Following this simple approach, we will ﬁnd that we can construct solid mechanics theories for phenomena as diverse as earthquakes, ultrasonic testing and the buckling of railway tracks. 1

2

Modelling solids

By basing our theory 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. 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 when combined with heat transfer (leading to thermoelasticity) and with genuine ﬂuid eﬀects, in cases where the material ﬂows even in the absence of large applied forces (leading to viscoelasticity) or when the material is porous (leading to poroelasticity). We will defer consideration of all these combined ﬁelds until the ﬁnal chapter. 1.2 Hooke’s law Robert 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 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),

(1.2.1)

where the constant of proportionality k is called the spring constant.

1.3 Lagrangian and Eulerian coordinates

3

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 (1.2.1) in the form −L , (1.2.2) T = k L where k 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 (1.2.2) is the simplest example of the all-important constitutive law relating the force to displacement. As shown in Exercise 1.3, it is possible to construct a one-dimensional continuum model for an elastic solid from this law, but, to generalise it 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 time 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 the solid 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.

(1.3.1)

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

4

Modelling solids

This will be the case provided the Jacobian of the transformation from X to x is bounded away from zero: ∂xi . (1.3.2) 0 < J < ∞, where J = det ∂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

(1.3.3)

as shorthand. The positivity of J means that we exclude the possibility that the solid turns itself inside-out. We can use (1.3.3) 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 ρ dx = ρJ dX. (1.3.4) M (t) = V (t)

V (0)

Since V (t) designates a ﬁxed set of material points, M (t) must be a constant, namely its initial value M (0): ρJ dX = M (t) = M (0) = ρ0 dX, (1.3.5) V (0)

V (0)

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

(1.3.6)

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 (1.3.6) also applies if ρ0 = ρ0 (X).

1.4 Strain To generalise the concept of strain introduced in Section 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) respectively. Thus we can use Taylor’s theorem to show that the line element

1.4 Strain

5

δX that joins the two particles is transformed to δx = δX + u(X + δX, t) − u(X, t) = δX + (δX · ∇)u(X, t) + · · · , (1.4.1) where (δX · ∇) = δX1

∂ ∂ ∂ + δX2 + δX3 . ∂X1 ∂X2 ∂X3

(1.4.2)

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 .

(1.4.3)

Although we will try in subsequent chapters to minimise the use of suﬃces, it is helpful at this stage to introduce components so that X = (Xi ) = (X1 , X2 , X3 )T and similarly for u. Then (1.4.3) may be written in the form 3 2 − L2 = 2 Eij δXi δXj , (1.4.4) i,j=1

where 1 Eij = 2

∂uk ∂uk ∂uj ∂ui + + ∂Xj ∂Xi ∂Xi ∂Xj 3

.

(1.4.5)

k=1

By way of introduction to some notation that will be useful later, we point out that (1.4.4) may be written in at least two alternative ways. First, we may invoke the summation convention, in which one automatically sums over any repeated suﬃx. This avoids the annoyance of having to write explicit summation, so (1.4.4) is simply ∂uj 1 ∂ui ∂uk ∂uk 2 2 . + + = L + 2Eij δXi δXj , where Eij = 2 ∂Xj ∂Xi ∂Xi ∂Xj (1.4.6) 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.

(1.4.7)

It is clear from (1.4.4) 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 happens when we perform the same calculation in a coordinate system rotated by an

6

Modelling solids

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 and u in the new coordinate system, where X = P X,

u = P u.

(1.4.8)

Since P is orthogonal, (1.4.8) may be inverted to give X = P T X . Alternatively, using suﬃx notation, we have Xβ = pjβ Xj ,

ui = piα uα .

The strain in the new coordinate system is denoted by ∂uj ∂uk ∂uk 1 ∂ui , + + Eij = 2 ∂Xj ∂Xi ∂Xi ∂Xj

(1.4.9)

(1.4.10)

which may be manipulated using the chain rule, as shown in Exercise 1.4, to give Eij = piα pjβ Eαβ .

(1.4.11)

In matrix notation, (1.4.11) takes the form E = P EP T ,

(1.4.12)

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 (1.4.11) 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 + (Q − I)X,

(1.4.13)

where I is the identity matrix while the vector c and orthogonal matrix Q are constant, then E is identically zero. This result follows directly from substituting (1.4.13) into (1.4.6) and using the fact that QQT = 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 (1.4.11) 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 an inﬁnitesimal 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. (By Newton’s third law, the material on the other side will also exert a force equal to −df .) In the expectation that the force should be proportional to the area da, we write df = σ da,

(1.5.1)

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.

(1.5.2)

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), each 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 ,

(1.5.3)

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 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 .

8

Modelling solids

x3

a1 n a2 x2

a3 x1 Fig. 1.1 A reference tetrahedron; ai is the area of the face orthogonal to the xi -axis.

The outward normal to the face with area a1 is in the negative x1 -direction 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 .

(1.5.4)

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

(1.5.5)

Now we shrink the tetrahedron to zero volume. 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 †

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 (1.5.6) can be more rigorously justiﬁed provided the macroscopic dimensions of the solid are large compared to any atomistic length-scale.

1.5 Stress

9

τ22 τ12 τ21 τ11 δx2

G

τ11

x2 δx1

τ21

τ12 x1

τ22

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

expression for σ: σi = τij nj ,

or σ = τ n.

(1.5.6)

This important result enables us to ﬁnd the stress on any surface element in terms of the nine quantities (τij ) = τ . Now let us follow Section 1.4 and examine what happens to τij when we rotate the axes by an orthogonal matrix P . In the new frame, (1.5.6) will become σ = τ n

(1.5.7)

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

n = P n.

(1.5.8)

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

or τij = piα pjβ ταβ .

(1.5.9)

Thus τij , like Eij , 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) 2 (τ21 δx2 )

δx1 δx2 − 2 (τ12 δx1 ) , 2 2

10

Modelling solids

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 (see Exercise 1.5) and it shows that τij ≡ τji

(1.5.10)

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: ∂ui d ρ dx = gi ρ dx + τij nj da. (1.6.1) dt V (t) ∂t V (t) ∂V (t) The terms in (1.6.1) 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 apply the divergence theorem to the ﬁnal term to obtain ∂τij ∂ 2 ui ρ dx = gi ρ dx + dx. (1.6.2) 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 + . 2 ∂t ∂xj

(1.6.3)

This may alternatively be written in vector form by adopting the following notation for the divergence of a tensor: we deﬁne the ith component of ∇ · τ to be ∂τji (∇ · τ )i = . (1.6.4) ∂xj Since τ is symmetric, we may thus write Cauchy’s equation as ∂2u = ρg + ∇ · τ. (1.6.5) ∂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 ρ

1.7 Linear elasticity

11

or some other continuum comes when we impose an empirical constitutive relation between τ and u. For solids, (1.6.5) 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 recall that E was deﬁned in Section 1.4 in terms of the Lagrangian variables X; indeed, the time derivative in (1.6.5) 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 (1.6.5) with respect to the Eulerian variable x. It is not immediately clear, therefore, how the stress and strain, which are deﬁned in diﬀerent frames of reference, may be self-consistently 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 (1.3.1) 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 (1.3.6) 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 (1.4.5) and hence obtain the linearised strain tensor ∂uj 1 ∂ui Eij ≈ eij = + . (1.7.1) 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 (1.4.13), we note that it is possible to approximate E by (1.7.1) 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 (1.4.13); however eij does not vanish for such rigid-body motions,

12

Modelling solids

but rather for displacements of the form u = c + ω × x,

(1.7.2)

where c and ω are constant (see Exercise 1.6). Assuming the validity of (1.7.1), we can now generalise Hooke’s law by postulating a linear relationship between the stress and strain tensors. We 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 8. Even with this assumption, we apparently are led to the problem of deﬁning 81 material parameters Cijk (i, j, k, = 1, 2, 3) such that τij = Cijk ek .

(1.7.3)

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 pii pjj pkk p ≡ Ci j k

(1.7.4)

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 Cijk to just two scalar quantities λ and µ, such that Cijk = λδij δk + 2µδik δj ,

(1.7.5)

where δij is the usual Kronecker delta, which represents the identity matrix; consequently, τij = λ (ekk ) δij + 2µeij .

(1.7.6)

This relation can also be inverted to give the strain corresponding to a given stress, that is 1 λ(τkk ) (1.7.7) eij = τij − δij . 2µ (3λ + 2µ) In Chapter 9 we will consider solids, such as wood or ﬁbre-reinforced materials, that are not isotropic, and for which (1.7.6) must be generalised. The material parameters λ and µ are known as the Lam´e 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 elastic deformation. 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”

1.8 The incompressibility approximation

Cartilage Rubber Polystyrene Granite Glass Copper Steel Diamond

λ (GPa)

µ (GPa)

3 × 10−5 0.04 2.3 10 28 86 100 270

9 × 10−5 0.003 1.2 30 28 37 78 400

13

Table 1.1 Typical values of the Lam´e constants λ and µ for some everyday materials (1 GPa = 109 N m−2 = 104 atmospheres; a typical car tyre pressure is two atmospheres). 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 (1.7.6) into the momentum equation (1.6.3) and replace X with x to obtain the Navier equation, also known as the Lam´e equation, ∂2u = ρg + (λ + µ) grad div u + µ∇2 u. (1.7.8) ∂t2 Recall that ρ does not vary to leading order, so (1.7.8) comprises three equations for the three components of u. It may alternatively be written in component form ρ

ρ

∂ 2 uj ∂ 2 ui ∂ 2 ui = ρg + (λ + µ) + µ , i ∂t2 ∂xi ∂xj ∂x2j

(1.7.9)

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.”

(1.7.10)

(1.7.11)

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 Lam´e constant λ is much larger than the shear modulus µ.

14

Modelling solids

The values given in Table 1.1 show that rubber has this property, as do many biomaterials such as muscle. If a material is almost incompressible, we can set 1 λ = , µ ε

(1.8.1)

where ε is a small parameter. From (1.7.8), we expect that, in the limit ε → 0, div u will be of order ε. Hence, if we deﬁne a scalar function p such that µ (1.8.2) pε = − div u, ε then pε will approach a ﬁnite limit p as ε → 0. When we now substitute (1.8.1) and (1.8.2) into the Navier equation (1.7.8) and let ε → 0, we obtain ∂2u = ρg − ∇p + µ∇2 u, ∂t2 along with the limit of (1.8.2), that is ρ

div u = 0.

(1.8.3a)

(1.8.3b)

The condition (1.8.3b) 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 (1.6.3) by taking the dot product with ∂u/∂t and integrating over an arbitrary volume V : ∂τij ∂ui ∂ui ∂ 2 ui ∂ui dx = dx + dx. (1.9.1) ρ 2 ρgi ∂t ∂t ∂t V V V ∂xj ∂t The ﬁnal term may be rearranged, using the divergence theorem, to ∂τij ∂ui ∂eij ∂ui τij dx = τij nj da − dx. (1.9.2) ∂t V ∂xj ∂t ∂V ∂t V

1.9 Energy

15

Hence (1.9.1) may be written in the form d dt

V

1 ∂u 2 ρ dx + W dx 2 ∂t V ∂ui ∂ui dx + τij nj da, (1.9.3) ρgi = ∂t V ∂V ∂t

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

(1.9.4)

With τij given by (1.7.6), we can integrate (1.9.4) to determine W up to an arbitrary constant as 1 1 W = τij eij = λ (ekk )2 + µ (eij eij ). 2 2

(1.9.5)

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 (1.9.3) 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. Hence, in the absence of other energy sources resulting from, say, chemical or thermal eﬀects, we can interpret equation (1.9.3) 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 (1.9.5) that W is a non-negative function of the strain components, whose unique global minimum is attained when eij = 0. In fact, Exercise 1.7 demonstrates that it is only necessary to have µ, (λ + 2µ/3) > 0. The net conservation of energy implied by (1.9.3) reﬂects the fact that the Navier equation is not dissipative. Furthermore, even without the constitutive relation (1.7.6), the steady Navier equation is a necessary condition for the net gravitational and strain energy in an elastic body D, namely U= W − ρg · u dx, (1.9.6) D

16

Modelling solids

to be minimised, as shown in Exercise 1.8. However, the situation changes when thermal eﬀects are important, as we will see in Chapter 9.

1.10 Boundary conditions and well-posedness Suppose that we wish to solve (1.7.8) 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 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 left-hand side of (1.7.8) 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.

(1.10.1)

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

(1.10.2)

and we will now show that, if a solution u of (1.7.6), (1.10.2) with the boundary condition (1.10.1) 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 homogeneous problem, with ub = g = 0. Now, by multiplying (1.10.2) by ui , integrating over D and using the divergence theorem, we obtain ui τij nj da = eij τij dx = 2 W dx, (1.10.3) ∂D

D

D

where W is given by (1.9.5). The left-hand side of (1.10.3) 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

1.10 Boundary conditions and well-posedness

17

is identically zero in D, and the displacement can therefore only be a rigidbody motion (i.e. a uniform translation and rotation; see Exercise 1.6). Since u is zero on ∂D, we deduce that it must be zero everywhere and, hence, that u(1) ≡ u(2) . Now we attempt the same calculation when the surface traction, rather than the displacement, is speciﬁed: τ n = σ(x)

on ∂D.

(1.10.4)

Like the Neumann problem for a scalar elliptic partial diﬀerential equation (Ockendon et al., 2003, p. 154), the Navier equation only admits solutions satisfying (1.10.4) if so-called solvability conditions are satisﬁed. If we integrate (1.10.2) over D and use the divergence theorem, we ﬁnd that τij nj da + ρgi dx = 0 (1.10.5) ∂D

and hence that

D

σ da +

∂D

ρg dx = 0.

(1.10.6)

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 (1.10.2) before integrating, to give x×σ da + ρx×g dx = 0, (1.10.7) ∂D

D

as shown in Exercise 1.9. As well as representing physical balances on the system, (1.10.6) and (1.10.7) may be interpreted as instances of the Fredholm Alternative (see Ockendon et al., 2003, p. 43). Now suppose the solvability conditions (1.10.6) and (1.10.7) are satisﬁed and that two solutions u(1) and u(2) of (1.10.2) and the boundary condition (1.10.4) 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 rigid-body motion, as shown in Exercise 1.6. Thus the solution of (1.10.2) subject to the applied traction (1.10.4) is determined only up to the addition of an arbitrary translation and rotation. As well as the boundary conditions (1.10.1) and (1.10.4), there are generalisations in which the traction is speciﬁed on some parts of the boundary

18

Modelling solids n

solid 2 solid 1

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

and the displacement on others, for example in contact problems and in fracture, as described in Chapter 7. Another common generalisation of (1.10.1) and (1.10.4) 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,

(1.10.8)

where τ (1) and τ (2) are the values of τ on either side of the boundary. 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, as we will see in Chapter 7. For elastodynamic problems, we may anticipate that (1.7.8) 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

(1.10.9)

which describes small transverse waves on a string with tension T and line density (see Section 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 (1.10.1) or (1.10.4).

1.11 Coordinate systems

19

1.11 Coordinate systems In the next two chapters, we will construct some elementary solutions of the Navier equation (1.7.8). In doing so, it is often useful to employ 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 the Appendix. 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 (1.7.6), which is invariant under any such rotation. Hence the constitutive relation (1.7.6) 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 (1.7.6) gives τxx = (λ + 2µ)exx + λeyy + λezz ,

τxy = 2µexy ,

τyy

= λexx + (λ + 2µ)eyy + λezz ,

τxz = 2µexz ,

τzz

= λexx + λeyy + (λ + 2µ)ezz ,

τyz = 2µeyz ,

(1.11.1)

where ∂u , ∂x ∂v , eyy = ∂y ∂w ezz = , ∂z exx =

∂u ∂v + , ∂y ∂x ∂v ∂w 2eyz = + , ∂z ∂x ∂u ∂w 2exz = + , ∂z ∂x 2exy =

(1.11.2)

20

Modelling solids

and the three components of Cauchy’s momentum equation are ∂τxy ∂τxz ∂2u ∂τxx + + , = ρgx + ∂t2 ∂x ∂y ∂z ∂τyy ∂τyz ∂τxy ∂2v ρ 2 = ρgy + + + , ∂t ∂x ∂y ∂z ∂τyz ∂τzz ∂2w ∂τxz + + , ρ 2 = ρgz + ∂t ∂x ∂y ∂z ρ

(1.11.3)

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 (∇ · u) + µ∇2 u, = ρgx + (λ + µ) 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

(1.11.4)

1.11.2 Cylindrical polar coordinates We deﬁne cylindrical polar coordinates (r, θ, z) in the usual way and denote the displacements in the r-, θ- and z-directions 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 Section 1.5, τij is deﬁned to be the i-component of stress on a surface element whose normal points in the jdirection. As noted above, the constitutive relation (1.7.6) applies directly to this coordinate system, so that τrr = (λ + 2µ)err + λeθθ + λezz ,

τrθ = 2µerθ ,

τθθ = λerr + (λ + 2µ)eθθ + λezz ,

τrz = 2µerz ,

τzz = λerr + λeθθ + (λ + 2µ)ezz ,

τθz = 2µeθz ,

(1.11.5)

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

∂uθ uθ 1 ∂ur + − , r ∂θ ∂r r ∂uz ∂ur + , = ∂z ∂r 1 ∂uz ∂uθ + . (1.11.6) = ∂z r ∂θ

err =

2erθ =

eθθ

2erz

ezz

2eθz

1.11 Coordinate systems

21

The three components of Cauchy’s momentum equation (1.6.3) read ∂ 2 ur ∂τrz τθθ 1 ∂ 1 ∂τrθ (rτrr ) + + − , = ρgr + 2 ∂t r ∂r r ∂θ ∂z r ∂ 2 uθ ∂τθz τrθ 1 ∂ 1 ∂τθθ ρ 2 = ρgθ + (rτrθ ) + + + , ∂t r ∂r r ∂θ ∂z r ∂ 2 uz ∂τzz 1 ∂ 1 ∂τθz ρ 2 = ρgz + (rτrz ) + + , ∂t r ∂r r ∂θ ∂z ρ

(1.11.7)

where the body force is g = gr er + gθ eθ + gz ez . Written out in terms of displacements, these become ∂ 2 ur ∂ ur 2 ∂uθ 2 ρ 2 = ρgr + (λ + µ) (∇ · u) + µ ∇ ur − 2 − 2 , ∂t ∂r r r ∂θ ∂ 2 uθ (λ + µ) ∂ uθ 2 ∂ur 2 (∇ · u) + µ ∇ uθ − 2 + 2 , ρ 2 = ρgθ + ∂t r ∂θ r r ∂θ ∂ 2 uz ∂ ρ 2 = ρgz + (λ + µ) (∇ · u) + µ∇2 uz , (1.11.8) ∂t ∂z where 1 ∂uθ ∂uz 1 ∂ (rur ) + + , r ∂r r ∂θ ∂z 1 ∂ ∂ui 1 ∂ 2 ui ∂ 2 ui ∇2 u i = + r + 2 r ∂r ∂r r ∂θ2 ∂z 2

∇·u =

(1.11.9)

are the divergence of u and the Laplacian of ui respectively, expressed in cylindrical polars. Detailed derivations of (1.11.6) and (1.11.7) are given in the Appendix. Notice the undiﬀerentiated terms proportional to 1/r which are not present in the corresponding Cartesian expressions (1.11.2) and (1.11.3). The origin of these terms may be understood in two dimensions (r, θ) by considering the equilibrium 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 τˆαβ = ταβ (r, θ + δθ) = ταβ

(1.11.10)

when we expand using Taylor’s theorem. 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, τˆθθ δr cos δθ − τθθ δr + τˆrθ δr sin δθ + τ˜rθ (r + δr) δθ − τrθ rδθ = 0.

(1.11.11)

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 (1.11.10), we obtain 1 ∂τrθ τrr − τθθ ∂τrθ 1 ∂τθθ 2τrθ ∂τrr + + = 0, + + = 0, (1.11.12) ∂r r ∂θ r ∂r r ∂θ r which are the components of the two-dimensional steady Navier equation in plane polar coordinates with no body force; cf (1.11.7). 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 Section 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  . (1.11.13) r cos θ Again, we can apply the constitutive relation (1.7.6) literally, to obtain τrr = (λ + 2µ)err + λeθθ + λeφφ ,

τrθ = 2µerθ ,

τθθ = λerr + (λ + 2µ)eθθ + λeφφ ,

τrφ = 2µerφ ,

τφφ = λerr + λeθθ + (λ + 2µ)eφφ ,

τθφ = 2µeθφ .

(1.11.14)

1.11 Coordinate systems

23

The linearised strain components are now given by ∂uθ uθ ∂ur 1 ∂ur , 2erθ = + − , ∂r r ∂θ ∂r r ∂uφ uφ 1 ∂uθ 1 ∂ur 2erφ = + ur , + − , = r ∂θ r sin θ ∂φ ∂r r uφ cot θ ur uθ cot θ 1 ∂uφ 1 ∂uφ 1 ∂uθ + + , 2eθφ = + − . = r sin θ ∂φ r r r sin θ ∂φ r ∂θ r (1.11.15)

err = eθθ eφφ

Cauchy’s equation of motion leads to the three equations 1 ∂(sin θτrθ ) 1 ∂(r2 τrr ) ∂ 2 ur + = ρg + r 2 2 ∂t r ∂r r sin θ ∂θ τθθ + τφφ 1 ∂τrφ − , + r sin θ ∂φ r 1 ∂(sin θτθθ ) ∂ 2 uθ 1 ∂(r2 τrθ ) + ρ 2 = ρgθ + 2 ∂t r ∂r r sin θ ∂θ τrθ − cot θτφφ 1 ∂τθφ + , + r sin θ ∂φ r ∂ 2 uφ 1 ∂(sin θτθφ ) 1 ∂(r2 τrφ ) + ρ 2 = ρgφ + 2 ∂t r ∂r r sin θ ∂θ τrφ + cot θτθφ 1 ∂τφφ + , + r sin θ ∂φ r ρ

(1.11.16)

where the body force is g = gr er +gθ eθ +gφ eφ . Again, (1.11.15) and (1.11.16) may be derived using the general approach given in the Appendix or more directly by analysing a small polar element. In terms of displacements, the Navier equation reads ρ

ρ

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

(λ + µ) ∂ ∂ 2 uθ (∇ · u) = ρgθ + ∂t2 r ∂θ uθ 2 ∂ur 2 cos θ ∂uφ 2 − 2 2 − 2 2 , + µ ∇ uθ + 2 r ∂θ r sin θ r sin θ ∂φ

24

ρ

Modelling solids

∂2u

φ ∂t2

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

(1.11.17)

where 1 ∂uφ 1 ∂ 2 1 ∂ (sin θuθ ) + , r ur + 2 r ∂r r sin θ ∂θ r sin θ ∂φ 1 ∂ ∂ui 1 ∂ ∂ui 1 ∂ 2 ui ∇2 u i = 2 . r2 + 2 sin θ + 2 2 r ∂r ∂r r sin θ ∂θ ∂θ r sin θ ∂φ2 (1.11.18)

∇·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 Deduce that 1 m 2

1.2

1.3

d dt

2

d2 + k( − L) − mg = 0. dt2 1 + k( − L)2 + mg(L − ) = const. 2

and interpret this result in terms of energy. 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 (1.10.9), 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,

b ∂w ∂w b 1 ∂w 2 d 1 b ∂w 2 dx + T dx = T . dt 2 a ∂t ∂x ∂x ∂t a a 2 Interpret this result in terms of conservation of energy. A system of masses m along the x-axis at positions Xn = nL (n = 0, 1, 2, . . .) are linked by springs satisfying Hooke’s law (1.2.1), as shown in Figure 1.5. If each mass is displaced by a distance un (t), show that the tension Tn joining Xn to Xn+1 satisﬁes Tn − Tn−1 = m

d2 un , dt2

where

Tn = k (un+1 − un ).

Exercises Tn−1

Tn−1

m

Tn

25 Tn

m

un−1

un

m un+1

Fig. 1.5 A system of masses connected by springs along the x-axis.

Deduce that m

d2 un = k (un+1 − 2un + un−1 ) dt2

and show that this is a spatial discretisation of the partial diﬀerential equation ∂2u ∂2u = c2 , 2 ∂t ∂X 2

1.4

1.5

1.6

where X = nL, un (t) = u(X, t) and c2 = kL2 /m. [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 the Navier equation as the limit of a lattice of masses joined by springs aligned along three orthogonal axes.] If x, x , u and u are related by (1.4.9), use the chain rule to show that ∂uα ∂ui = piα pjβ . ∂xj ∂xβ Hence establish equation (1.4.11) 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 .] 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 ∂u d 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 (1.6.3), deduce that τij ≡ τji . Show that the linearised strain eij , given by (1.7.1), is identically zero if and only if u = c + ω×x, where c and ω are spatially-uniform vectors. Show that this approximates the rigid-body motion (1.4.13) when P is close to I. If u is of this form, and known to be zero at three non-collinear points, deduce that c = ω = 0.

26

1.7

1.8

Modelling solids

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 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. Suppose a solid body occupies the region D and the displacement u is prescribed on ∂D. Let W − ρg · u dx, U= D

where W is the strain energy density. Show that, if ui is changed by a small virtual displacement ηi , then the corresponding leading-order change in U is ∂ηi τij − ρgi ηi dx. δU = ∂xj D [Hint: use the fact that τij is symmetric.] Use the divergence theorem to show that ∂τij ηi τij nj da − + ρgi ηi dx, δU = ∂xj ∂D D

1.9

1.10

and deduce that the minimisation of U with respect to all displacements satisfying the given boundary condition leads to the steady Navier equation. Deduce also that, if no boundary condition is imposed, the natural boundary condition is the vanishing of the traction τij nj on ∂D. Show also that the minimisation of U subject to the constraint div u ≡ 0 leads to the steady incompressible Navier equation (1.8.3a), where p is a Lagrange multiplier. 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 (1.10.2) before integrating over D, derive (1.10.7) and deduce that the net moment acting on D must be zero. Suppose that u satisﬁes the steady Navier equation (1.10.2) 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.

Exercises

1.11

27

[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 partial diﬀerential equations (Ockendon et al., 2003, p. 154).] (a) Show that, in plane polar coordinates (r, θ), the basis vectors satisfy deθ der = 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 is imposed, with u = ur (r, θ)er (θ) + uθ (r, θ)eθ (θ), show that the line element δX is displaced to ∂ur ∂ur δr + δθ − uθ δθ er δx = δX + ∂r ∂θ ∂uθ ∂uθ δr + δθ + ur δθ eθ . + ∂r ∂θ (d) Hence show that |δx| = |δX| + 2

2

(δr, rδθ)

err erθ

erθ eθθ

δr rδθ

where, to leading order in the displacements, err

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

∂uθ + ur . ∂θ

[These are the elements of the linearised strain tensor in plane polar coordinates, as in (1.11.6).]

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 electrostatic or gravitational potentials, say, or the stream function in hydrodynamics. We 28

2.2 Linear displacements

29

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 tensor describes the response of an 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 (1.7.8), many diﬀerent kinds of singular solutions can be constructed using stress functions and Green’s tensors, 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 reduces to ∇ · τ = (λ + µ) grad div u + µ∇2 u = 0.

(2.2.1)

This vector partial diﬀerential equation for u is the starting point for all we will say in this chapter. As discussed in Section 1.10, it needs to be supplemented with suitable boundary conditions, which will vary depending on the situation being modelled. 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 (2.2.1). Such solutions provide considerable insight into the predictions of (2.2.1) 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 u=

α x, 3

(2.2.2)

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

30

Linear elastostatics

y

1

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 (2.2.2), (b) one-dimensional shear (2.2.6), (c) uniaxial stretching (2.2.8).

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 α 3 − 1 ∼ α. (2.2.3) 1+ 3 The strain and stress tensors corresponding to this displacement ﬁeld are given by α 2 (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

(2.2.5)

measures the resistance to expansion or compression and is called the bulk modulus or modulus of compression; from Exercise 1.7, we know that K is positive.

2.2 Linear displacements

31

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

(2.2.6)

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 E= τ = αµ 1 0 0 . (2.2.7) 1 0 0 , 2 0 0 0 0 0 0 Note that λ does not aﬀect the stress, so the 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  , (2.2.8) −νz 0 0 −ν 

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

(2.2.9)

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 . (2.2.10) ν= 2 (λ + µ)

32

Linear elastostatics

y τn = 0 T

x

A

T

z Fig. 2.2 A uniform bar being stretched under a tensile force T .

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. With ν given by (2.2.10), the stress tensor has just one non-zero element, namely τxx = Eα,

(2.2.11)

where E=

µ(3λ + 2µ) λ+µ

(2.2.12)

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;

(2.2.13)

thus AE is the elastic modulus k referred to in (1.2.2). 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 Lam´e constants may then be evaluated using λ=

νE , (1 + ν)(1 − 2ν)

µ=

E . 2(1 + ν)

(2.2.14)

We note that the constitutive relation (1.7.7) can be written in terms of E and ν as Eeij = (1 + ν)τij − ντkk δij .

(2.2.15)

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

2.2 Linear displacements

33 (b)

(a)

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

required for convexity of the strain energy, as shown in Exercise 1.7. The upper bound ν → 1/2 is approached as λ → ∞, in other words as the material becomes incompressible, as discussed in Section 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 ν < 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 pulled in only one (see Lakes, 1987).

2.2.4 Biaxial strain As a ﬁnal illustrative example, we generalise Section 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     u ax + by u =  v  = cx + dy  , w −γz

(2.2.16)

and, as in Section 2.2.3, the stress and strain tensors are both constant. Here we choose γ to satisfy the condition τzz = 0 on the traction-free upper and

34

Linear elastostatics Tyy Tyx

y

Txy Txx

z Txx

x

h

Txy

Tyx Tyy

Fig. 2.4 A plate being strained under tensions Txx , Ty y and shear forces Txy , Ty x .

lower surfaces of the plate, so that ν λ (a + d) = (a + d), γ= λ + 2µ 1−ν

(2.2.17)

where ν again denotes Poisson’s ratio. With this choice, and with E again denoting the Young’s modulus, the only non-zero stress components are τxx =

E(a + νd) , 1 − ν2

τxy =

E(b + c) , 2(1 + ν)

τyy =

E(νa + d) . 1 − ν2

(2.2.18)

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 (2.2.18) to relate these to the in-plane strain components by Eh (exx + νeyy ), 1 − ν2 Eh exy , = 1+ν Eh = (νexx + eyy ). 1 − ν2

Txx = Txy = Tyx Tyy

(2.2.19a) (2.2.19b) (2.2.19c)

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 (2.2.19) 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

(2.2.20)

2.2 Linear displacements

35

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

2.2.5 General linear displacements The simple linear examples considered above shed 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: T u(x) = u(x0 ) + ∇u (x0 ) (x − x0 ) + · · · . (2.2.21) Here ∇u is the displacement gradient matrix, whose entries are conventionally deﬁned to be ∇u = (∂uj /∂xi ), and may be written as the sum of symmetric and skew-symmetric parts: ∂uj ∂uj 1 ∂uj 1 ∂ui ∂ui + . (2.2.22) = + − ∂xi 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   0 −ω3 ω2 1 ∇u − ∇uT =  ω3 (2.2.23) 0 −ω1  , 2 −ω2 ω1 0 so that, with ω = (ω1 , ω2 , ω3 )T , (2.2.21) becomes u(x) = u0 + ω×(x − x0 ) + E(x − x0 ) + · · · .

(2.2.24)

The terms on the right-hand side represent a small 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 {1 , 2 , 3 }, 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 = x 0 + P T x ,

(2.2.25)

then E, u, ω are transformed to E , u , ω such that u = u0 + ω ×x + E x + · · · ,

(2.2.26)

36

Linear elastostatics

where E = P EP T

  1 0 0 =  0 2 0  . 0 0 3

(2.2.27)

We can hence think of the strain at any point as comprising three orthogonal expansions or contractions, depending on the signs of {1 , 2 , 3 }, followed by a small rigid-body rotation. Generalising Section 2.2.1, we note that the net relative volume change associated with this expansion/contraction is (1 + 1 )(1 + 2 )(1 + 3 ) − 1 = 1 + 2 + 3 + · · · .

(2.2.28)

The sum of the eigenvalues is the invariant trace of the matrix E so that the relative volume change is Tr (E) = ekk = ∇ · u.

(2.2.29)

With respect to the principal axes, the stress tensor is also diagonal, namely    τ1 0 0  τ1 = λ(1 + 2 + 3 ) + 2µ1 , τ =  0 τ2 0  , where τ = λ(1 + 2 + 3 ) + 2µ2 ,  2 (2.2.30) 0 0 τ3 τ3 = λ(1 + 2 + 3 ) + 2µ3 , and the eigenvalues {τ1 , τ2 , τ3 } are called the principal stresses. As an illustration, let us re-examine the simple shear of Section 2.2.2. The eigenvalues and corresponding eigenvectors v i of E are readily found to be α , 2   1 v 1 = 1 , 0 1 =

α 2 = − , 2   1 v 2 = −1 , 0

3 = 0,   0 v 3 = 0 . 1

(2.2.31a) (2.2.31b)

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 (2.2.32) u =  0  = −x + x + y  − y − x . 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.

2.3 Antiplane strain

37

In general, the strain tensor E and, therefore, the principal strains i and principal stresses τi are functions of position x. At any such position, the strain and stress tensors are diagonal only with respect to the principal axes; if these axes are rotated by another orthogonal matrix Q, then the strain and stress tensors become E and τ respectively, where E = QE QT ,

τ = Qτ QT .

(2.2.33)

It is easiest to understand the consequence of such a rotation in two dimensions, when we can take cos θ sin θ (2.2.34) Q= −sin θ cos θ and (2.2.33b) becomes 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

(2.2.35)

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 is shown in 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

(2.2.36)

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 Section 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 8.

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). One way to create this state of antiplane strain is to apply a tangential traction σ(x, y), in the axial direction only,

38

Linear elastostatics

z

surface traction σ D

y

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

to the curved boundary of a cylindrical bar, as illustrated in Figure 2.5. Of course, σ(x) must exert no net force or moment on the bar, as pointed out in Section 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   µ

(2.3.1)

0

At any point of the bar, the traction on an element normal to (cos θ, sin θ, 0)T is µ (cos θ∂w/∂x + sin θ∂w/∂y), whose maximum value as θ varies is µ|∇w|. If we neglect gravity, the Navier equation reduces to the two-dimensional Laplace’s equation ∂2w ∂2w + = ∇2 w = 0, ∂x2 ∂y 2

(2.3.2)

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 lies in the (x, y)-plane, so the applied traction is simply a shear force σ in the z-direction that is related to w by σ(x) = τzn = µ

∂w ∂n

on ∂D.

(2.3.3)

2.4 Torsion

z

39

M

M

y

D x Fig. 2.6 A bar being twisted under a moment M .

The solvability condition for the Neumann problem (2.3.2), (2.3.3) is σ ds = 0, (2.3.4) ∂D

which conﬁrms that no net traction may be applied to any cross-section (cf equation (1.10.6)). If, instead of the traction, we were to specify the displacement w on ∂D, we would obtain the Dirichlet boundary condition w = f (x, y)

on ∂D

(2.3.5)

instead of (2.3.3). 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, (2.3.3) with σ = 0 is the natural boundary condition (as in Exercise 1.8). Despite its simplicity, we will ﬁnd this model a very useful paradigm when we come to consider fracture in Chapter 7. We note that (2.3.2) also describes the small transverse displacement w(x, y) of a nearly planar membrane under tension (Exercise 2.2); (2.3.3) and (2.3.5) 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 ﬁeld of the form     u −Ωyz u =  v  =  Ωxz  , (2.4.1) w Ωψ(x, y)

40

Linear elastostatics

where Ω is a constant representing the twist of the bar about its axis. As in Section 2.3, the stress tensor is of the form   0 0 τxz τ = 0 (2.4.2) 0 τyz  τxz τyz 0 where, now,

τxz = µΩ

∂ψ −y , ∂x

τyz = µΩ

∂ψ +x . ∂y

(2.4.3)

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

in D,

(2.4.4)

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, ds ds and the corresponding boundary condition for ψ is τxz

1 d 2 ∂ψ = X +Y2 ∂n 2 ds

on ∂D.

(2.4.5)

(2.4.6)

Notice that the solvability condition for the Neumann problem (2.4.4), (2.4.6) is satisﬁed identically. Hence, given the shape of the cross-section 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 (xτyz − yτxz ) dxdy = RΩ, (2.4.7) M= D

say, where

∂ψ 2 ∂ψ 2 −y + x +y R=µ x dxdy. ∂y ∂x D

(2.4.8)

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.

2.4 Torsion

41

The boundary-value problem (2.4.4), (2.4.6) for ψ may be usefully reformulated as follows. Whenever the stress tensor is of the form (2.4.2) and there is no body force, the steady Navier equation reduces to ∂τyz ∂τxz + = 0. ∂x ∂y

(2.4.9)

We can guarantee that (2.4.9) is satisﬁed by postulating the existence of a stress function φ (x, y) such that τxz = µΩ

∂φ , ∂y

τyz = −µΩ

∂φ , ∂x

(2.4.10)

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 (2.4.9) 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 never uniquely deﬁned; in our case, φ is unique only up to the addition of an arbitrary constant. By comparing (2.4.10) with (2.4.3), we see that φ and ψ are related by ∂ψ ∂φ = + y, ∂x ∂y

∂ψ ∂φ =− − x. ∂y ∂x

(2.4.11)

Elimination of φ retrieves (2.4.4), while elimination of ψ reveals that φ satisﬁes Poisson’s equation ∇2 φ = −2

in D.

(2.4.12)

Indeed, it is easy to see from (2.4.11) that ψ and φ+(x2 +y 2 )/2 are harmonic conjugates. The zero-stress boundary condition (2.4.5) implies that φ is constant on ∂D and, without loss of generality, we may take φ=0

on ∂D.

(2.4.13)

The advantage of introducing the stress function φ is that the Neumann problem for ψ has been converted to the Dirichlet problem (2.4.12), (2.4.13), which has a unique solution; exactly the same procedure could have been applied to the antiplane strain problem (2.3.2), (2.3.3). The torque M is then given in terms of φ by ∂φ ∂φ −y φ dxdy, (2.4.14) −x dxdy = 2µΩ M = µΩ ∂x ∂y D D

42

Linear elastostatics

the ﬁnal step being a consequence of Green’s theorem and (2.4.13). An alternative formula to (2.4.8) for the torsional rigidity is, therefore, φ dxdy. (2.4.15) R = 2µ 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 (2.4.4), (2.4.6) becomes r < a, (2.4.16a) ∇2 ψ = 0, ∂ψ = 0, r = a. (2.4.16b) ∂r It follows that ψ is a constant and, hence, that the torsional rigidity is 2π a 2 πµa4 2 . (2.4.17) R=µ r3 drdθ = x + y dxdy = µ 2 D 0 0 Note that a circular cylinder is the only case in which the right-hand side of (2.4.6) is zero. For any other cross-sectional shape, twisting a bar inevitably results in a non-zero axial displacement w = Ωψ. Alternatively, we may solve the Dirichlet problem (2.4.12), (2.4.13) for φ. Assuming that φ is a function only of r, we have dφ 1 d 2 r = −2, r < a, (2.4.18a) ∇ φ= r dr dr φ = 0,

r = a,

(2.4.18b)

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

a2 − r2 . 2

The torsional rigidity is then given by (2.4.15): 2π a πµa4 R=µ , (a2 − r2 )r drdθ = 2 0 0

(2.4.19)

(2.4.20)

which reproduces (2.4.17) as expected. This result can easily be generalised to bars of elliptical cross section (see Exercise 2.4).

2.5 Multiply-connected domains In Section 2.4, we have implicitly assumed that D is simply connected. Many torsion bars are tubular in practice, and the resulting change in topology

2.5 Multiply-connected domains

43

∂Do D ∂Di z

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

makes a big diﬀerence to the integration of our mathematical model. We now have to apply the condition (2.4.5) on two stress-free boundaries, namely the inner (∂Di ) and outer (∂Do ) 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 only choose φ to satisfy φ=0

on ∂Do ,

φ=k

on

∂Di ,

(2.5.1)

where k is a 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 ∂φ ∂φ +y dxdy x R = −µ ∂x ∂y D ∂ ∂ = −µ (xφ) + (yφ) − 2φ dxdy ∂x ∂y D = 2µ φ dxdy + 2µkAi , (2.5.2) D

after using Green’s theorem, where Ai is the area of the hole inside ∂Di . To obtain φ and thus R 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 (2.4.12) with

44

Linear elastostatics

y

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.

Dirichlet boundary conditions (2.5.1) 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 (2.4.11). To be physically acceptable, ψ must be a single-valued function of (x, y), which implies that ∂ψ ∂ψ dx + dy ≡ 0 (2.5.3) ∂y C ∂x for all simple closed paths C contained in D. In fact, it is readily shown that (2.5.3) holds for all such paths if it holds when C is the inner boundary ∂Di (see Exercise 2.5). By substituting for ψ in favour of φ, we therefore obtain ∂φ ∂φ ∂φ dy − dx = ds = −2Ai (2.5.4) ∂y ∂D i ∂x ∂D i ∂n 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 R. 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 (2.5.4) 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 dφ 1 d 2 r = −2, a < r < b, (2.5.5a) ∇ φ= r dr dr φ = k,

r = a,

(2.5.5b)

φ = 0,

r = b,

(2.5.5c)

2.5 Multiply-connected domains

45

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

(2.5.6)

We can determine k by substituting this expression for φ into (2.5.4) to give 2π 2π b2 − a2 ∂φ dφ 2 a dθ = −2πa + ds = k − , dr r=a log (a/b) 2 ∂D i ∂n 0 (2.5.7) while Ai = πa2 , so (2.5.4) tells us that k=

b2 − a2 . 2

(2.5.8)

This value of k eliminates the logarithmic term from (2.5.6), and we might have anticipated that this would be necessary by recalling that ψ and φ+r2 /2 are harmonic conjugates. Since b2 − a2 r2 = const. + k − log r (2.5.9) φ+ 2 2 and −log r is the harmonic conjugate of θ, we deduce that b2 − a2 θ, ψ = const. − k − 2

(2.5.10)

which is evidently a single-valued function only if k satisﬁes (2.5.8). We therefore ﬁnd that φ=

b2 − r 2 , 2

(2.5.11)

and it is straightforward to substitute this into (2.5.2) and obtain the torsional rigidity µπ 4 R= (2.5.12) b − a4 . 2 Had we chosen to use ψ instead of φ, we would have discovered that ψ is constant and then quickly reproduced (2.5.12) using (2.4.8). 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.

46

Linear elastostatics

Notice that (2.5.12) reproduces the result (2.4.17) 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 (2.5.13) R = 2µπa4 ε 1 + O (ε) , where ε = b/a − 1 1. We can now compare this with the torsional rigidity of a tube, such as part 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 ∂φ 1 ∂2φ 1 ∂ 2 r + 2 2 = −2, a < r < b, (2.5.14a) ∇ φ= r ∂r ∂r r ∂θ with φ=0

on r = a, b,

(2.5.14b)

on θ = 0, 2π.

(2.5.14c)

and φ=0

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 + εξ) ,

(2.5.15)

where ε is again small. To lowest order in ε, (2.5.14a,b) reduces to ∂2φ = −2ε2 a2 , ∂ξ 2

0 < ξ < 1,

(2.5.16a)

with φ=0

on ξ = 0, 1,

(2.5.16b)

whose solution is φ = ε2 a2 ξ(1 − ξ).

(2.5.17)

Notice that (2.5.17) 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 must be retained in (2.5.14). For this simply-connected cross-section, we may use (2.4.15) to determine the torsional rigidity and thus obtain, to lowest order in ε, 1 2π 4 3 2 µa ε . φ dξ = (2.5.18) R = 4πµa ε 3 0

2.6 Plane strain

47

Comparing (2.5.18) with (2.5.13), we see that the change in topology caused by introducing the cut has a dramatic inﬂuence on the strength of the tube, reducing its torsional rigidity by two orders of magnitude! 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) u = v (x, y) , (2.6.1) 0 we ﬁnd that the stress tensor takes the form   τxx τxy 0 τ = τxy τyy 0  , 0 0 τzz where τxx τyy

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

∂u ∂v + , =µ ∂y ∂x ∂u ∂v + . =λ ∂x ∂y

(2.6.2)

τxy τzz

(2.6.3a) (2.6.3b)

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 (2.6.2), and gravity neglected, the Navier equation reduces to ∂τxy ∂τxx + = 0, ∂x ∂y

∂τxy ∂τyy + = 0, ∂x ∂y

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

(2.6.4)

(2.6.5a) (2.6.5b)

48

Linear elastostatics

By cross-diﬀerentiating, it is straightforward to eliminate u and v in turn from (2.6.5) and thus show that each satisﬁes the biharmonic equation ∇4 u = ∇4 v = 0,

(2.6.6)

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

(2.6.7)

In fact, we can reduce (2.6.5) to a single biharmonic equation by making use of another potential function as follows. By inspection, the integrability condition for the system (2.6.4) is the existence of an Airy stress function A such that τxx =

∂2A , ∂y 2

τxy = −

∂2A , ∂x∂y

τyy =

∂2A . ∂x2

(2.6.8)

In the same way that the stress function φ in (2.4.10) 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 (2.6.3), we obtain the following expressions for the displacement gradients ∂2A ∂2A ∂u = −ν 2 + (1 − ν) 2 , ∂x ∂x ∂y 2 2 ∂ A ∂ A ∂v = (1 − ν) −ν 2, 2µ 2 ∂y ∂x ∂y 2 ∂ A ∂u ∂v + =− . µ ∂y ∂x ∂x∂y 2µ

(2.6.9a) (2.6.9b) (2.6.9c)

Now, by taking ∂ 2 /∂x∂y of (2.6.9c) and using (2.6.9a) and (2.6.9b), we can eliminate u and v and hence ﬁnd that A satisﬁes the biharmonic equation: ∇4 A = 0.

(2.6.10)

Moreover, given A, (2.6.9) forms a compatible system of three equations that determine u and v to within a small 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 in two dimensions may be written in the form φ = Re f (z) , (2.6.11)

2.6 Plane strain

49

y t n D x ∂D (x, y) = X(s), Y (s) Fig. 2.9 The unit normal n and tangent t to the boundary ∂D of a plane region D.

where f is an arbitrary analytic function of z = x + iy. Similarly, the general solution of (2.6.10) has the Goursat representation A = Re z¯f (z) + g (z) ,

(2.6.12)

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 (2.6.12) 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 (2.6.10) in some region D, on whose boundary a prescribed traction σ is imposed, that is τn = σ

on

∂D.

(2.6.13)

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 X , t= Y

n=

Y , −X

(2.6.14)

50

Linear elastostatics

where is shorthand for d/ds. Using (2.6.8) to write the stress components in terms of A, we thus ﬁnd that (2.6.13) can be written in the form d ∂A/∂y = σ. (2.6.15) ds −∂A/∂x If no surface traction is applied, that is σ = 0, then it follows from (2.6.15) 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 ∂A dA = =0 ds ∂n

(2.6.16)

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 arrive at the boundary conditions ∂A =0 ∂n

A = 0,

(2.6.17)

to be imposed on a stress-free boundary. We note that the divergence theorem on any closed region D yields ∂ 2 ∂A 2 A ∇ A − ∇ A ds ∂n ∂n ∂D div A grad(∇2 A) − div ∇2 A grad A dxdy = D 2 A∇4 A − ∇2 A dxdy. (2.6.18) = D

Hence, if A satisﬁes the biharmonic equation in D and the boundary conditions (2.6.17) on ∂D, then ∇2 A = 0 and a second use of (2.6.17) reveals that A ≡ 0. This result conﬁrms that the stresses inside a closed body in plane strain are uniquely determined by the tractions applied to the boundary. We can combine (2.6.8) and (2.6.9) to obtain the strain energy density in the form 1 (τxx exx + 2τxy exy + τyy eyy ) 2

2 2 2 2 ∂ A ∂2A ∂2A 1 . (1 − ν) ∇ A + 2 −2 2 = 4µ ∂x∂y ∂x ∂y 2

W=

(2.6.19)

2.6 Plane strain

51

Then the calculus of variations can be used to show that minimisers of the net strain energy W dxdy U= D

in a closed region D satisfy the biharmonic equation (2.6.10). 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 because the displacements are related to A by (2.6.9), so boundary conditions for u and v cannot usually be expressed simply in terms of A. We will show in Section 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. We begin by deriving the relations between the Airy stress function and the stress components in polar coordinates. Recalling (1.11.7) with gi = 0 and all variables depending only on r and θ, we try writing τθθ =

∂2A . ∂r2

(2.6.20a)

Then (1.11.7b) gives −∂ 3 A/∂r2 ∂θ = 2τrθ + r∂τrθ /∂r, which suggests that we write τrθ = ∂B/∂r. Since ∂ 2 /∂r2 (rB) is now equal to −∂ 3 A/∂r2 ∂θ, we ﬁnd that ∂ 1 ∂A , (2.6.20b) τrθ = − ∂r r ∂θ and then (1.11.7a) gives 1 ∂ 2 A 1 ∂A . (2.6.20c) + r2 ∂θ2 r ∂r Equations (2.6.20) are the analogues of (2.6.8) in plane polar coordinates. The biharmonic equation for A reads 2 2 1 ∂2 ∂ 1 ∂ 4 + + A = 0. (2.6.21) ∇ A= ∂r2 r ∂r r2 ∂θ2 τrr =

We will only consider here cases where a purely normal pressure P is applied, so the boundary conditions on r = a are ∂ A 1 ∂ 2 A 1 ∂A = −P, =0 on r = a; (2.6.22) + r2 ∂θ2 r ∂r ∂r r

52

Linear elastostatics

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 2 1 d d + A = 0, (2.6.23) dr2 r dr subject to the boundary conditions A=r

dA = −P a2 dr

on r = a.

(2.6.24)

It is then straightforward to solve (2.6.23) in the form A = c1 r2 + c2 + c3 r2 log r + c4 log r.

(2.6.25)

For the stresses to exist throughout the circle, we require A to be twice diﬀerentiable as r → 0 and hence, applying the boundary conditions (2.6.24), we obtain P 2 (2.6.26) r + a2 . A=− 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 (2.6.21) in the form A(r, θ) = f (r) sin(nθ), where n is a positive integer, we ﬁnd that 2 2 n2 1 d d − 2 + f = 0. dr2 r dr r

(2.6.27)

(2.6.28)

This Euler diﬀerential equation admits the solution f (r) = rk , where k satisﬁes 2 (2.6.29) k − n2 (k − 2)2 − n2 = 0. We must again ensure that the stress components (2.6.20) 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 ,

(2.6.30)

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

(2.6.31)

2.6 Plane strain

r=a

53

r=b

P

Fig. 2.10 A plane annulus being inﬂated by an internal pressure P .

We can take a linear combination of separable solutions that satisfy (2.6.24) and the radially-symmetric solution (2.6.26) to obtain 2 r + a2 A0 A=− 4 ∞ rn 1 rn+2 + − An cos(nθ) + Bn sin(nθ) , (2.6.32) n−2 n 2 (n − 1)a (n + 1)a n=2

where An and Bn are the Fourier coeﬃcients of P , that is 1 2π 1 2π P (θ) cos(nθ) dθ, Bn = P (θ) sin(nθ) dθ. An = π 0 π 0

(2.6.33)

Notice that the n = 1 term does not appear in the series in (2.6.32), so it is possible to satisfy the boundary condition (2.6.22) only if 2π cos θ P (θ) dθ = 0. (2.6.34) sin θ 0 This is simply a manifestation of the solvability condition (1.10.6) 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.

54

Linear elastostatics

Assuming that A is a function of r alone, it is given by (2.6.25), 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 in (2.6.25) to be determined from the boundary conditions. The condition of zero traction on the outer boundary r = b means, as explained in Section 2.6.3, that we may without loss of generality set dA =0 on r = b. (2.6.35a) dr With A independent of θ, τrθ is identically zero, so the speciﬁed internal pressure gives us just one more boundary condition, namely A=

1 dA = −P r dr

on r = a.

(2.6.35b)

As in Section 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 singlevalued, where, from the constitutive relations (1.11.5), dur ur d2 A ur 1 dA dur = (λ + 2µ) +λ , + (λ + 2µ) . (2.6.36) =λ r dr dr r dr2 dr r By cross-diﬀerentiation, we ﬁnd that that the compatibility condition for ur to exist is d3 A 1 d2 A 1 dA = 0, (2.6.37) + − 2 dr3 r dr2 r dr which is satisﬁed by (2.6.25) only if c3 = 0. The boundary conditions (2.6.35) may then be used to evaluate the remaining three constants and hence show that r 2 P a2 b2 P a2 2 − a log − A= r . (2.6.38) 2 (b2 − a2 ) b2 − a2 b In a plane strain problem with multiple traction-free holes, we would have ∂A (2.6.39) = Ki ∂n 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 Section 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 A = ki ,

2.6 Plane strain

55

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 obtained the considerably more tractable problem d 1 d (rur ) = 0, a < r < b, (2.6.40a) dr r dr ur dur + λ = −P, r = a, (2.6.40b) (λ + 2µ) dr r ur dur + λ = 0, r = b. (2.6.40c) (λ + 2µ) dr r This quickly yields the solution r b2 P a2 + , (2.6.41) ur = 2 (b2 − a2 ) λ + µ µr and we can then recover the stress components using the constitutive relations (1.11.5). 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 b2 b2 P a2 P a2 2νP a2 1 − , τ 1 + , τ = = , τrr = 2 zz θθ b − a2 r2 b2 − a2 r2 b2 − a2 (2.6.42) where ν denotes Poisson’s ratio, as usual. These results are in accordance with (2.6.38). We can now use the above solution to predict the failure of a gun barrel as P increases. As explained in Section 2.2.5, the Tresca criterion predicts that the material will fail when |τi − τj | = τY , (2.6.43) S = max i,j 2 where τY is the yield stress and τi are the principal stresses. Here the stress tensor is diagonal, so τi are just the three stress components given in (2.6.42). If we assume that the material has positive Poisson’s ratio† then, since ν < 1/2 and r < b, we can deduce from (2.6.42) the inequalities τrr < 0 < τzz < τθθ ,

(2.6.44)

and so the maximum shear stress is S= †

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

(2.6.45)

for an auxetic material with ν < 0, the second inequality in (2.6.44) is reversed and the maximum shear stress may be τ θ θ − τ z z instead of (2.6.45)

56

Linear elastostatics

This is maximised at r = a, so we would expect a gun barrel always to yield ﬁrst on its inner surface, along a generator. We can also use (2.6.45) to predict the maximum pressure P ∗ that the barrel can withstand, namely a2 ∗ (2.6.46) P = τY 1 − 2 . 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 8. 2.6.6 Plane strain in a rectangle In Section 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, solving 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,

(2.6.47a)

τyy = 0,

τxy = f± (x)

on y = ±b,

(2.6.47b)

as shown schematically in Figure 2.11. Of course the applied tractions must exert no net force or moment, so that b a f+ (x) − f− (x) dx = g+ (y) − g− (y) dy = 0, (2.6.48a) b

−a a

−a

f+ (x) + f− (x) dx − a

−b b

−b

g+ (y) + g− (y) dy = 0.

(2.6.48b)

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,

(2.6.49a)

on y = ±b,

(2.6.49b)

after integrating the normal stress conditions as in Section 2.6.3.

2.6 Plane strain

57

y f+ g+ b a

x

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

The seemingly simplest solution procedure is to separate the variables, writing A(x, y) = F (x)G(y) where F G G F +2 + = 0. F F G G

(2.6.50)

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 G + 2α2 G + α4 G = 0

(2.6.51)

and hence to solutions of the form A = a1 y cos (αy) + a2 y sin (αy) + a3 cos (αy) + a4 sin (αy) eαx + b1 y cos (αy) + b2 y sin (αy) + b3 cos (αy) + b4 sin (αy) e−αx . (2.6.52) Clearly a second class of separable solutions arises from exchanging the roles of x, y. All these solutions can also be derived from (2.6.12) by putting f = Aeαz + Be−αz , g = Ceαz + De−αz . We now need to select from (2.6.52) 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 solutions of 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 integer n. By interchanging x and y, we can therefore construct a

58

Linear elastostatics

solution of the form nπy nπx A= An Fn (x) cos + Bn Gn (y) cos , 2b 2a

(2.6.53)

n odd

where An and Bn are constants and the functions Fn and Gn are deﬁned by nπa nπx nπa nπx Fn (x) = a sinh cosh − x cosh sinh , (2.6.54a) 2b 2b 2b 2b nπy nπy nπb nπb cosh sinh − y cosh . (2.6.54b) Gn (y) = b sinh 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 nπy nπ πn g+ (y) = aFn (a)An sin + b sin Bn Gn (y) , (2.6.55) 2ab 2b 2 n odd

but {sin(nπy/2b), Gn (y)} do not form an orthogonal set on −b < y < b. Hence, if we were to multiply (2.6.55) by Gm (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 below. 2.6.7 Plane strain in a semi-inﬁnite strip We now present an example which is technically easier than the rectangular geometry discussed above 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 < h/2, with zero stress on y = ±h/2 and with prescribed tractions τxx = σx (y),

τxy = σy (y)

(2.6.56)

on x = 0, as illustrated in Figure 2.12. Then the Airy stress function satisﬁes the biharmonic equation along with the boundary conditions ∂2A (0, y) = σx (y), ∂y 2

∂2A (0, y) = −σy (y), ∂x∂y

(2.6.57)

and, as in Section 2.6.3, we can choose A such that A(x, h/2) =

∂A (x, h/2) = 0. ∂y

(2.6.58)

However, because the end x = 0 is not traction-free, we cannot use the same conditions on the other face y = −h/2. All we can safely deduce from the

2.6 Plane strain

59

y stress free

x

τxx = σx (y)

τxy = σy (y)

stress free

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

zero-stress condition is that ∂2A ∂2A (x, −h/2) = 0, (x, −h/2) = ∂x2 ∂x∂y

(2.6.59a)

which implies that A(x, −h/2) = a + bx,

∂A (x, −h/2) = c, ∂y

(2.6.59b)

where a, b and c are constants. We can calculate these constants by insisting that A and ∇A be continuous at the corners (0, ±h/2); otherwise we would, in general, have point forces or moments acting there, rather than just jump discontinuities in the tractions. By integrating (2.6.57) with respect to y, we ﬁnd that h/2 h/2 h/2 ∂A ∂A A= = =− (η − y)σx (η) dη, σy (η) dη, σx (η) dη ∂x ∂y y y y (2.6.60) on x = 0, and we deduce that h/2 (y + h/2) σx (y) dy, b = a= −h/2

h/2 −h/2

σy (y) dy, c = −

h/2

−h/2

σx (y) dy. (2.6.61)

We would like to try to solve this problem by separating the variables, as in Section 2.6.6, but this is diﬃcult because of the inhomogeneous boundary condition (2.6.59b). However, we note that the function A∞ (x, y) =

(h − 2y)2 4a(h + y) + 4bx(h + y) + ch(h + 2y) 3 8h

(2.6.62)

60

Linear elastostatics

satisﬁes both the biharmonic equation and the boundary conditions (2.6.58), (2.6.59). Hence, if we set ˜ A = A∞ + A,

(2.6.63)

˜ is biharmonic and satisﬁes homogeneous boundary conditions then A ˜ ˜ ∂A ∂A ˜ ˜ (x, −h/2) = A(x, h/2) = (x, h/2) = 0. A(x, −h/2) = ∂y ∂y

(2.6.64)

Using (2.6.52), we can write the solution as a superposition of functions of the form ˜ = a1 y cos (αy) + a2 y sin (αy) + a3 cos (αy) + a4 sin (αy) e−αx , (2.6.65) A with Re(α) > 0. The boundary conditions (2.6.64) lead to four homogeneous linear equations of the form h ± cos 2

αh 2

h a1 + sin 2

αh 2

a2 + cos

αh 2

a3 ± sin

αh 2

a4 = 0,

(2.6.66a) αh αh αh αh αh αh sin cos − a1 ± sin + a2 cos 2 2 2 2 2 2 αh αh a3 + α cos a4 = 0, ∓ α sin 2 2 (2.6.66b) which admit nontrivial solutions for the constants ai only if the relevant determinant is zero. This eventually leads to the transcendental equation† sin (αh) = ±αh,

(2.6.67)

and the “eigenvalues” α satisfying (2.6.67) are complex. This occurs because the ordinary diﬀerential equation (2.6.51), along with the boundary conditions G = G = 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 calculation is much simpler if the problem is posed on the strip 0 < y < h

2.6 Plane strain

61

the “eigenfunctions” are not mutually orthogonal, which seriously complicates the task of ﬁtting the initial conditions, as in (2.6.55). 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 Section 4.6 but, unfortunately, specifying ∂ 2 A/∂n2 on the boundary has no obvious physical signiﬁcance in plane strain. ˜ can in fact be A completeness argument can be given to show that A expressed as a sum of terms of the form (2.6.65), with Re(α) > 0, although, because of the diﬃculties described above, the coeﬃcients can only be computed numerically by inverting an inﬁnite matrix. Thus, whatever tractions ˜ will decay exponentially as x → ∞. The stress far are imposed on x = 0, A from the edge of the strip will then be characterised entirely by the stress function A∞ , which corresponds to a far-ﬁeld stress, as x → ∞, τxx →

12y(a + bx) + h(6y − h)c , h3

τxy →

3b 2 2 − 4y h , 2h3

τyy → 0. (2.6.68)

The net tensile force T , shear force N and bending moment M exerted on any section x = const. by the stress ﬁeld (2.6.68) are given by h/2 h/2 τxx dy = −c, N (x) = τxy dy = b, (2.6.69a) T (x) = −h/2

−h/2

h/2

M (x) = −h/2

yτxx dy = a + bx +

ch . 2

(2.6.69b)

From (2.6.61), we deduce that T , N and M satisfy the diﬀerential equation dM = N, dx and the boundary conditions h/2 T (0) = σx (y) dy,

(2.6.70)

h/2

N (0) =

−h/2

−h/2

σy (y) dy,

(2.6.71a)

h/2

M (0) = −h/2

yσx (y) dy.

(2.6.71b)

Note that (2.6.71) does not correspond to setting x = 0 as in (2.6.57); rather it is the matching condition between (2.6.68) and the solution for A in the form (2.6.63). 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

62

Linear elastostatics

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 exerts the same net force and moment, just as long as we are suﬃciently far from the edge. This phenomenon occurs quite generally in elasticity and is known as Saint-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 Chapters 4 and 6, when considering the boundary conditions to be imposed on thin solids such as plates and rods, and we will also re-encounter the relation (2.6.70) between the shear force and bending moment. Finally, let us examine the far-ﬁeld displacement by using (2.6.62) and (2.6.9). From (2.6.9), we ﬁnd that u and v satisfy −

2µ ∂u 2µ ∂v = = ν ∂y 1 − ν ∂x ∂u ∂v + = µ ∂y ∂x

T 12M (x)y + , h h3 3N 2 2 h . − 4y 2h3

(2.6.72a) (2.6.72b)

We can eliminate u by diﬀerentiating (2.6.72b) with respect to x to obtain 6(1 − ν) ∂2u ∂2v =− M (x). (2.6.73) =− ∂x2 ∂x∂y µh3 This relationship between the bending moment and the curvature of the displaced strip will also 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.

(2.6.74)

Assuming that P and σ decrease suﬃciently rapidly as |x| → ∞, we deﬁne the Fourier transform ∞ ! A (x, y) eikx dx. (2.6.75) A (k; y) = −∞

Assuming further that this integral converges, we also recall the inversion formula ∞ 1 ˆ −ikx dx. Ae (2.6.76) A= 2π −∞

2.6 Plane strain

63

From the biharmonic equation (2.6.10) we ﬁnd that 2 2 d 2 !=0 A y > 0, −k dy 2

(2.6.77a)

and, from (2.6.74), ˆ = P!(k) k2 A ! dA =σ !(k) ik dy !→0 A

y = 0,

(2.6.77b)

y = 0,

(2.6.77c)

y → ∞.

(2.6.77d)

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

(2.6.78)

The presence of inverse powers of k in (2.6.78) indicates that the integral in (2.6.75) 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, when we deduce that τ!xx = −P!(k) (1 − y|k|) e−|k|y , τ!xy = P!(k)iyke−|k|y ,

(2.6.79b)

τ!yy = −P!(k) (1 + y|k|) e

(2.6.79c)

−|k|y

,

(2.6.79a)

which, as shown in Exercise 2.11(a), may be inverted as convolutions of the form 2 ∞ s2 y ds P (x − s) , (2.6.80a) τxx = − π −∞ (s2 + y 2 )2 2 ∞ sy 2 ds P (x − s) , (2.6.80b) τxy = − π −∞ (s2 + y 2 )2 2 ∞ y 3 ds P (x − s) . (2.6.80c) τyy = − π −∞ (s2 + y 2 )2 We can now use (2.6.9) to calculate the displacements. For example, v! satisﬁes 2µ

2! d! v ! − ν d A = ((2ν − 1) − y|k|) Pˆ (k)e−y|k| , = −(1 − ν)k 2 A dy dy 2

(2.6.81)

64

Linear elastostatics

which may be integrated, in the case σ = 0, to obtain 1−ν ! 2µ! v = y+2 P (k) e−y|k| . |k|

(2.6.82)

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 1 ∞ (1 − ν)s2 + (2 − ν)y 2 ∂v =− P (x − s)s ds. (2.6.83) µ ∂x π −∞ (s2 + y 2 )2 Alternatively, we can also obtain (2.6.83) from the identity E

∂τyy ∂τxy ∂τxx ∂2v − +ν , = 2(1 + ν) ∂x2 ∂x ∂y ∂y

(2.6.84)

by using (2.6.80) and then integrating with respect to x. 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 Exercise 2.12. When we do so, the right-hand side of (2.6.83) reduces to the singular integral 1 ∞ P (s) ds dv0 = (1 − ν) − = (1 − ν)H[P ], (2.6.85) µ dx π −∞ s − x " where H[P ] is called the Hilbert transform of P , − being the principal value integral deﬁned in Exercise 2.12. As shown # $in Exercise 2.13, this transform may be inverted using the formula H H[f ] ≡ −f , and hence we can calculate the pressure required to achieve a given surface displacement v0 (x), namely # $ µ P =− H v0 . (2.6.86) 1−ν As an illustrative example, the surface displacement a 1 + x2 /L2

(2.6.87a)

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

(2.6.87b)

v0 (x) = corresponds to a surface pressure P (x) =

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.

2.6 Plane strain

65

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 v0 (x) of a half-space and corresponding surface pressure P (x) deﬁned by (2.6.87).

One can deduce generally from (2.6.85) that it is only possible to achieve a localised displacement, with v0 → 0 as x → ±∞, if ∞ P (x) dx = 0, (2.6.88) −∞

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, and we will see in Section 2.9.3 that it does not occur in radially symmetric problems. The general question of the indentation produced by traction applied to the boundary of a half-plane will be discussed further in connection with “punch” 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 ∂V ∂τxx + =ρ , ∂x ∂y ∂x

∂τxy ∂τyy ∂V + =ρ , ∂x ∂y ∂y

(2.6.89)

and (2.6.8) therefore generalises to τxx = ρV +

˜ ∂2A , ∂y 2

τxy = −

˜ ∂2A , ∂x∂y

τyy = ρV +

˜ ∂2A . ∂x2

(2.6.90)

66

Linear elastostatics

˜ satisﬁes the Now when we eliminate the displacements, we discover that A inhomogeneous biharmonic equation 1 − 2ν 4˜ ∇2 V = 0. (2.6.91) ∇ A+ 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 (2.6.90) is used to transform these con˜ In other words, the substitution (2.6.90) ditions into boundary data for A. eliminates the body force from the problem, but replaces it with boundary tractions. This substitution is often useful in solving elastostatic problems numerically. As an illustration, let us try to ﬁnd the stress in the elastic medium surrounding a tunnel, a problem of importance say 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 (2.6.20) by setting ˜ ˜ 1 ∂2A 1 ∂A , τrr = ρg (r sin θ − H) + 2 2 + r ∂θ r ∂r ˜ ∂ 1 ∂A , τrθ = − ∂r r ∂θ τθθ = ρg (r sin θ − H) +

˜ ∂2A , ∂r2

(2.6.92a) (2.6.92b) (2.6.92c)

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 ˜ ˜ ˜ 1 ∂2A 1 ∂A ∂ A = 0, = ρg (H − r sin θ) (2.6.93) + ∂r r r2 ∂θ2 r ∂r on r = a. ˜ By eliminating ∂ A/∂r from the boundary conditions (2.6.93), we obtain ˜ ∂2A ˜ = ρga2 (H − a sin θ) +A ∂θ2

on r = a.

(2.6.94)

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

(2.6.95a)

2.6 Plane strain

67

where A and B are arbitrary constants, and (2.6.93) then gives us the second condition ˜ ρga2 ∂A = ρgaH + θ cos θ + A cos θ + B sin θ ∂r 2

on r = a.

(2.6.95b)

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

(2.6.96)

The functions fj (r) can, in principle, be found by substituting (2.6.96) into the biharmonic equation and into the boundary conditions (2.6.95). We now make the important simplifying assumption that H 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 2 a2 A cos θ + B sin θ + + 2r log r . (2.6.97) 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 Section 2.6.5, the key to closing the problem in this multiply-connected domain is to ensure that the displacement ﬁeld is single-valued. 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)

(2.6.98)

We thus evaluate the stress components as τrr τrθ τθθ

(1 − 2ν) a2 a2 −H + r sin θ 1 + , = ρg 1 − 2 r 4(1 − ν) r2 a2 ρga2 cos θ (1 − 2ν) 1− 2 , =− r 4(1 − ν) r 2 (1 − 2ν) a2 sin θ a . −H + = ρgr sin θ + ρg 1 + 2 r 4(1 − ν) r

(2.6.99a) (2.6.99b) (2.6.99c)

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Linear elastostatics

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

3 − 4ν ρga sin θ 2(1 − ν)

on r = a.

(2.6.100)

Hence a small tunnel suﬀers twice as much compressive stress as would occur in its absence. Moreover, the minimum compressive stress 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 ﬁnal result remains valid for all values of H and a (see Mindlin, 1939).

2.7 Compatibility In Section 1.6, Newton’s second law led us to write down the equation ∂τij + ρgi = 0 ∂xj

(2.7.1)

for the stress tensor in elastostatics in the presence of a body force gi . Viewed as a system of equations for the stress components, (2.7.1) is underdetermined, comprising just three equations for six unknowns. It was only by using the constitutive relations (1.7.6), ∂uj ∂uk ∂ui δij + µ (2.7.2) + 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 , (2.7.2) is itself an over-determined system for the displacements. If we were in the very fortunate position of knowing all the stress components τij , which of course must satisfy (2.7.1), then it would be straightforward to obtain the corresponding linearised strain tensor from (2.2.15). Thus, given τij , we could view our expression ∂uj 1 ∂ui (2.7.3) + eij = 2 ∂xj ∂xi as a system of equations for the displacement components ui , rather than a deﬁnition of eij . In this light, however, (2.7.3) 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.

2.7 Compatibility

69

T

First consider a plane strain problem, in which u = u(x, y), v(x, y), 0 and ∂u 1 ∂u ∂v ∂v , exy = + , eyy = . (2.7.4) exx = ∂x 2 ∂y ∂x ∂y Since the three strain components are functions of just two displacements, (2.7.4) 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

(2.7.5a)

Equation (2.7.5a) is the compatibility condition which ensures that (2.7.4) 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, (2.7.5b) ∂z 2 ∂y∂z ∂y 2 ∂ 2 ezx ∂ 2 exx ∂ 2 ezz − 2 = 0, (2.7.5c) + ∂x2 ∂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 into account, we obtain the additional three compatibility relations ∂exy ∂eyz ∂ ∂exz ∂ 2 exx = + − , (2.7.5d) ∂y∂z ∂x ∂y ∂z ∂x ∂ 2 eyy ∂eyz ∂ ∂eyx ∂ezx = + − , (2.7.5e) ∂z∂x ∂y ∂z ∂x ∂y ∂exy ∂ 2 ezz ∂ ∂ezy ∂ezx = + − . (2.7.5f) ∂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 (2.7.5) inevitably leads to just one independent equation. However, all six compatibility conditions are necessary for a single-valued displacement to exist. This situation is analogous

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Linear elastostatics

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 shown in Exercise 2.21, the compatibility conditions (2.7.5) are equivalent to the components of the tensor equation   0 −∂/∂z ∂/∂y curl (curl E)T = 0, where curl =  ∂/∂z 0 −∂/∂x . (2.7.6) −∂/∂y ∂/∂x 0 We also remark that any three independent compatibility equations for eij are just as under-determined as the three equilibrium equations (2.7.1) are for τij . Hence, the fact that the compatibility conditions ensure the existence of three physically acceptable displacement components suggests that (2.7.1) will similarly guarantee the existence of three scalar stress functions. We have already encountered an application of this idea to plane strain in Section 2.6, and we will show below in Section 2.8 how it may be generalised to non-planar problems. It is important to note that, if eij does not satisfy (2.7.5), then there is no physically acceptable displacement ﬁeld that gives rise to such a strain and it is, therefore, incompatible with linear elasticity. In Section 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 non-zero boundary tractions, then, 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 (2.7.5), and we will see speciﬁc examples of this in Section 8.4.

2.8 Generalised stress functions 2.8.1 General observations By taking the divergence of the steady Navier equation (2.2.1) in the absence of body forces, we see that div u must be a harmonic function: ∇2 (div u) = 0.

(2.8.1)

If instead we take the curl of (2.2.1), we ﬁnd that curl3 u = 0.

(2.8.2)

2.8 Generalised stress functions

71

and so, by taking the curl once more and using (2.8.1), we obtain ∇4 u = 0.

(2.8.3)

Each Cartesian component of the displacement must therefore satisfy the biharmonic equation, as shown in Section 2.6 for plane strain. We emphasise that (2.8.3) can only be written component-wise as  4  ∇ u ∇4 u =  ∇4 v  = 0 (2.8.4) 4 ∇ w in Cartesian coordinates; in any other coordinate systems we must use the identity (1.7.11). 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.

(2.8.5)

(It may be shown that this decomposition applies to any continuously differentiable vector ﬁeld u.) Substitution of (2.8.5) into (2.2.1) reveals that the potential functions φ and A must also satisfy the biharmonic equation ∇4 φ = 0,

∇4 A = 0.

(2.8.6)

These results all suggest that we may be able to construct large classes of solutions of (2.2.1) by considering suitable harmonic or biharmonic scalar problems. We have already seen some examples of this approach in plane and antiplane strain (Sections 2.3–2.6) and we will now show how analogous ideas may be applied in some other commonly-occurring conﬁgurations. 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 Section 2.7. As we showed in Section 2.6, the two conservation laws ∂τxy ∂τxx + = 0, ∂x ∂y

∂τxy ∂τyy + =0 ∂x ∂y

(2.8.7)

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 (2.7.5a) which, when written in terms of the stress

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Linear elastostatics

components, is ∂ 2 τyy ∂ 2 τxy ∂ 2 τxx + − 2 − ν∇2 (τxx + τyy ) = 0. ∂y 2 ∂x∂y ∂x2

(2.8.8)

Now (2.8.7) and (2.8.8) form a closed system of partial diﬀerential equations for the stress components (τxx , τxy , τyy ). If, as in Section 2.6, we write τxx =

∂2A , ∂y 2

τxy = −

∂2A , ∂x∂y

τyy =

∂2A , ∂x2

(2.8.9)

then the stress balance equations (2.8.7) are satisﬁed identically, and substitution of (2.8.9) into (2.8.8) quickly retrieves the biharmonic equation for A. It is worth noting that an inhomogeneous equation for A of the form ∇4 A = f

(2.8.10)

does not, as might have been suspected from (2.6.91), correspond to the imposition of a body force. The mere existence of A implies that (2.8.7) are satisﬁed, so there is no such force. Instead the right-hand side f of (2.8.10) corresponds to a distribution of incompatibility through the material, and we will discuss in Section 2.9.4 and Chapter 8 what the physical signiﬁcance of such a conﬁguration might be. Now let us suppose that we do impose a body force in, say, the x-direction so that the ﬁrst Navier equation (2.8.7a) is modiﬁed to ∂τxy ∂τxx + = −f. ∂x ∂y

(2.8.11)

Now there is no longer an Airy stress function, but we can instead try to obtain an integrability condition from the two remaining homogeneous equations (2.8.7b) and (2.8.8). From (2.8.7b) we deduce the existence of a function A1 (x, y) such that τxy =

∂A1 , ∂y

τyy = −

∂A1 , ∂x

so the compatibility condition (2.8.8) becomes 2 2 ∂ ∂ 2 2 ∂A1 = 0. − ν∇ τxx − + (1 − ν)∇ ∂y 2 ∂y 2 ∂x

(2.8.12)

(2.8.13)

To write this in conservation form, we ﬁrst let A2 be such that τxx = ∂A2 /∂x, so that, without loss of generality, 2 2 ∂ ∂ 2 2 − ν∇ A2 − + (1 − ν)∇ A1 = 0. (2.8.14) ∂y 2 ∂y 2

2.8 Generalised stress functions

73

We must now spot that, if we set A2 = A1 + ∇2 L, then (2.8.14) is satisﬁed identically by 2 ∂ 2 L. (2.8.15) − ν∇ A1 = ∂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 2 ∂ ∂ ∂A2 2 (2.8.16a) = + (1 − ν)∇ L, τxx = ∂x ∂x ∂y 2 2 ∂ ∂ ∂A1 2 τxy = (2.8.16b) = − ν∇ L, ∂y ∂y ∂y 2 2 ∂ ∂ ∂A1 τyy = − (2.8.16c) =− − ν∇2 L, ∂x ∂x ∂y 2 and, when we ﬁnally substitute these into the x-momentum equation (2.8.11), we ﬁnd that L is such that ∂τxy ∂τxx + = (1 − ν)∇4 L = −f. ∂x ∂y

(2.8.17)

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

(2.8.18a) (2.8.18b) (2.8.18c)

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 ∂2L 1 ∂2L 1 2 . (2.8.19) , v = − (1 − 2ν)∇ L + u= 2µ ∂y 2 2µ ∂x∂y

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Linear elastostatics

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. 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. For a more direct, though perhaps less instructive, derivation of (2.8.19) we note that the second Navier equation (2.6.5b), written out in terms of the displacements, may be rearranged to ∂2 ∂2u 2 (2.8.20) + (1 − 2ν)∇ + 2 v = 0. ∂x∂y ∂y Now we can argue that (2.8.19) is just the integrability condition for (2.8.20). However, this intuitively appealing derivation is deceptively simple; it is clear that (2.8.19) guarantees (2.8.20) 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 non-zero components of the stress tensor are τxx , τxy and τyy . Such a state might exist, for example, in a plate with 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 Section 2.6, the normal stress τzz is in general non-zero. We again introduce an Airy stress function A as in (2.6.8), although now we must allow A to be a function of x, y and z. We can still use the compatibility condition (2.7.5a) to deduce that A satisﬁes the two-dimensional biharmonic equation 2 2 2 ∂ ∂ 4 % A= + 2 A = 0, (2.8.21) ∇ ∂x2 ∂y 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

2.8 Generalised stress functions

75

must ensure that all ﬁve remaining compatibility conditions are also satisﬁed. To this end we ﬁnd that the non-zero 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 % 2 A; Eezz = −ν (τxx + τyy ) = −ν ∇ Eexx = τxx − ντyy =

(2.8.22a) (2.8.22b) (2.8.22c) (2.8.22d)

note in particular that ezz is in general non-zero, in contrast with plane strain. First considering (2.7.5d) and (2.7.5e), we obtain ∂2 % 2 ∂2 % 2 ∇ A = ∇ A = 0, (2.8.23) ∂x∂z ∂y∂z from which we deduce that % 2 A = f (z) + φ(x, y), ∇

(2.8.24)

% 2 φ = 0 and f is arbitrary. Next, by adding (2.7.5c) and (2.7.5b), we where ∇ obtain ∂ 2 % 2 d2 f (2.8.25) ∇ A = 2 = 0, ∂z 2 dz so, without loss of generality, we may set f (z) = βz for some constant β. Finally, taking (2.7.5c), (2.7.5b) and (2.7.5f) individually, we ﬁnd that ∂2A ∂2A ∂2 ∂2 (1 + ν) (1 + ν) + νφ = + νφ ∂x2 ∂z 2 ∂x∂y ∂z 2 2 2 ∂ A ∂ (1 + ν) 2 + νφ = 0. (2.8.26) = ∂y 2 ∂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

(2.8.27)

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)

(2.8.28)

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Linear elastostatics

where % 2 χ0 = φ, ∇

% 2 φ = 0, ∇

% 2 χ1 = β = const.. ∇

(2.8.29)

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 strain, introduced in Section 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

(2.8.30a)

(2.8.30b)

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 cylindrical coordinates, the Navier equation does not 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 Section 2.8.2. We will instead follow the more direct, although slightly cavalier, approach suggested at the end of Section 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

(2.8.31)

and there is a unidirectional body force in the z-direction. The homogeneous er -component of the Navier equation, given by (1.11.8), may be written in the form 2 ∂ 1 1 ∂ ∂2 ∂ 2 uz + 2(1 − ν) − + + (1 − 2ν) 2 ur = 0. (2.8.32) ∂r∂z ∂r2 r ∂r r2 ∂z

2.8 Generalised stress functions

77

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), the relation ∂ 1 ∂f 1 ∂2f 1 ∂ 1 ∂f 1 ∂f = = − (2.8.33) − 2 ∂r r ∂r r ∂r2 r ∂r r ∂r r2 ∂r implies that 2 2 ∂ ∂2 ∂ f 1 1 ∂ 2(1 − ν) − 2 + (1 − 2ν) 2 + 2 ∂r r ∂r r ∂z ∂r∂z 2 2 ∂2 ∂ ∂ 1 ∂ 2(1 − ν) + (1 − 2ν) 2 f. (2.8.34) = + ∂r∂z ∂r2 r ∂r ∂z Since the right-hand side may be written as ∂2 1 ∂ ∂ 2 (1 − 2ν)∇ + r f, ∂r∂z r ∂r ∂r we may infer from (2.8.32) the existence of a function L(r, z) such that ∂L 1 ∂2L 1 1 ∂ 2 ur = − , uz = (1 − 2ν)∇ L + r . (2.8.35) 2µ ∂r∂z 2µ r ∂r ∂r With this displacement, it is straightforward to evaluate the stress components from (1.11.5) as ∂ ∂2L 2 τrr = − − ν∇ L , (2.8.36a) ∂z ∂r2 ∂L ∂ 1 ∂ τrz = r − ν∇2 L , (2.8.36b) ∂r r ∂r ∂r ∂L ∂ 1 ∂ 2 r + (1 − ν)∇ L , (2.8.36c) τzz = ∂z r ∂r ∂r ∂ 1 ∂L (2.8.36d) − + ν∇2 L . τθθ = ∂z r ∂r 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, (2.8.37) r ∂r ∂z 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,

(2.8.38)

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Linear elastostatics

which is identical to the equation (2.8.17) for the planar Love function. We will now show that this is no accident and, in Section 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 ,

(2.8.39)

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,

(2.8.40)

for some vector ﬁeld B(x) and suitably chosen constants a and b.† The advantage of this representation is that, when we substitute (2.8.40) into (2.8.39), we ﬁnd that B satisﬁes (λ + 2µ)(a − b) (grad div)2 B + µa curl4 B = −f ,

(2.8.41)

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

(2.8.42)

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

(2.8.43)

2µu = 2(1 − ν)∇2 B − grad div B.

(2.8.44)

and (2.8.40) 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 (2.8.43) if the body force f shares the corresponding symmetry. †

This is reasonable since we expect the curl of (2.8.40) to determine curl B given u, and, if B = ∇ϕ, (2.8.40) is eﬀectively a Poisson equation for ϕ. Note also that, in view of (1.7.11), (2.8.44) may be viewed as an example of a Helmholtz representation (2.8.5).

2.8 Generalised stress functions

79

However, in cases where there is no body force, (2.8.43) opens up the possibility of ﬁnding a wealth of scalar stress functions describing problems with various symmetries, simply by posing suitable forms for B.

2.8.6 Papkovich–Neuber potentials Next we will show that (2.8.44) 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 equation or Poisson’s equation. We can obtain an immediate simpliﬁcation by deﬁning Ψ = ∇2 B,

(2.8.45)

so that (2.8.43) becomes ∇2 Ψ = −

f . 1−ν

Now, using the vector identity (1.7.11), we easily ﬁnd that ∇2 (div B) ≡ div ∇2 B = div Ψ,

(2.8.46)

(2.8.47)

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 (2.8.47) in the form 1 x·f ∇2 div B − (x · Ψ) = 2 2(1 − ν)

(2.8.48)

(2.8.49)

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

(2.8.50)

so that φ satisﬁes a simple Poisson equation. The deﬁnitions (2.8.45) and (2.8.50) transform (2.8.44) into the Papkovich–Neuber representation 1 (2.8.51) 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 scalar harmonic functions. In plane strain, we can represent the displacement in terms of a single biharmonic Love function, while three

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Linear elastostatics

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 (2.8.51) can prove extremely useful, as we will see in Section 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))T , we simply take T z ∂ψ ∂ψ ∂ψ ∂ψ and φ = Ψ= z ,z , (3 − 4ν)ψ x +y −ψ , ∂x ∂y 2 ∂x ∂y (2.8.52) whence, after a simple calculation, (2.8.51) gives T 2µu = 0, 0, 4(1 − ν)(1 − 2ν)ψ ,

(2.8.53)

with the anticipated result that ∇2 w = 0. Concerning plane strain, we anticipate that T φ = φ(x, y) and Ψ = ψ1 (x, y), ψ2 (x, y), 0 ,

(2.8.54)

so that (2.8.51) gives

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

and

(2.8.55a) (2.8.55b)

∂2φ ∂2 − (xψ1 + yψ2 ). ∂x∂y ∂x∂y (2.8.56) Hence (2.6.9c) suggests that we choose ∂ψ1 /∂y + ∂ψ2 /∂x to be zero and this in turn suggests that we take ψ1 and ψ2 to be harmonic conjugates. Then (2.6.9c) gives the Goursat representation 2µ

∂u ∂v + ∂y ∂x

= 2(1 − ν)

∂ψ1 ∂ψ2 + ∂y ∂x

−2

1 A = φ + (xψ1 + yψ2 ), 2

(2.8.57)

which is equivalent to (2.6.12). 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 y 1 A = φ + yψ2 − (1 − ν) ψ2 (x, y ) dy , (2.8.58) 2

2.8 Generalised stress functions

81

where ψ2 is a harmonic function. This representation will be found to be very useful in Chapter 7. The uniqueness of the Papkovich–Neuber representation is discussed in Barber (1993, Section 15.4).

2.8.7 Maxwell and Morera potentials We recall our observation in Section 2.7 that the equilibrium Navier equation (2.7.1) 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 (2.7.1) with gi = 0. As usual, there is some freedom in selecting a particular set of stress functions. One well-known possibility is the Maxwell stress functions χi , deﬁned by τxx =

∂ 2 χ3 ∂ 2 χ2 + , ∂y 2 ∂z 2

τxy = −

∂ 2 χ3 , ∂x∂y

τyy =

∂ 2 χ1 ∂ 2 χ3 + , ∂z 2 ∂x2

τyz = −

∂ 2 χ1 , ∂y∂z

τzz =

∂ 2 χ2 ∂ 2 χ1 + , ∂x2 ∂y 2 (2.8.59a)

τzx = −

∂ 2 χ2 . ∂z∂x

(2.8.59b)

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 ∂ψ1 ∂ψ2 ∂ 2 ψ1 ∂ ∂ψ3 , τxy = − − , (2.8.60a) τxx = 2 ∂y∂z ∂z ∂z ∂x ∂y ∂ψ2 ∂ψ3 ∂ 2 ψ2 ∂ ∂ψ1 τyy = 2 , τyz = − − , (2.8.60b) ∂z∂x ∂x ∂x ∂y ∂z ∂ψ3 ∂ψ1 ∂ 2 ψ3 ∂ ∂ψ2 τzz = 2 , τzy = − − . (2.8.60c) ∂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 can construct a symmetric stress tensor τ (x) with zero divergence using the formula τij =

k,l,m,n

ikl jmn

∂ 2 Akm , ∂xl ∂xn

(2.8.61)

82

Linear elastostatics

where ijk is the alternating symbol. Using (2.7.6), we can formally write (2.8.61) in the form τ = curl (curl A)T ,

(2.8.62)

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  , (2.8.63) 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 from Section 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 of the equilibrium equation. This is why it is often referred to as a strain function rather than a stress function. Finally, we note the strong analogy between the equations (2.8.59), (2.8.60) satisﬁed by the Maxwell and Morera stress potentials and the compatibility conditions that we found for the strain components in Section 2.7. The formulation (2.8.62) shows that this resemblance is no accident. By using (2.7.6), we can obtain the compatibility condition for τ , namely T = 0, (2.8.64) curl curl (1 + ν)τ − ντkk I and the equations satisﬁed by the Maxwell functions χi , for example, can then be found by substituting for τ from (2.8.62). 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.

2.9 Singular solutions in elastostatics

83

δε (x) 1.2

1

0.8

ε = 0.25

0.6

ε = 0.5

0.4

ε = 1.0

0.2

-4

-3

-2

-1

1

2

3

4

x

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

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 family of functions ε (2.9.1) δε (x) = 2 π (x + ε2 ) have the property that

∞ −∞

δε (x) dx ≡ 1

(2.9.2)

for any ε > 0, although δε (x) → 0 as ε → 0 at any ﬁxed non-zero value of x, as illustrated in Figure 2.14. If we “deﬁne” δ(x) = limε→0 δε (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 ∞ δ(x − ξ)ψ(ξ) dξ = ψ(x) (2.9.3) −∞

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),

(2.9.4)

84

Linear elastostatics

in three dimensions. By construction, this function possesses a property analogous to (2.9.3), that is δ(x − ξ)ψ(ξ) dξ = ψ(x). (2.9.5) 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).

(2.9.6)

Suitable singular solutions of this equation can be constructed using the following key results, derived in Exercise 2.23: |x| one dimension: ∇2 = δ(x), (2.9.7a) 2 2 1 2 2 = δ(x)δ(y), (2.9.7b) log x + y two dimensions: ∇ 2π −1 2 three dimensions: ∇ = δ(x)δ(y)δ(z). (2.9.7c) 4π x2 + y 2 + z 2

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 ∂τxy ∂τxx + = −f δ(x)δ(y), ∂x ∂y

∂τxy ∂τyy + = 0. ∂x ∂y

(2.9.8)

As described in Section 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).

(2.9.9)

Now we can read oﬀ from (2.9.7b) a suitable radially-symmetric solution for ∇2 L, namely 1 d dL f 2 log r, (2.9.10) ∇ L= r =− r dr dr 2π(1 − ν)

2.9 Singular solutions in elastostatics

85

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.

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)

(2.9.11)

The displacement ﬁeld is then found by substitution into (2.8.19) to give   2 x − y2 − (3 − 4ν) log r f ,  2r2 (2.9.12) u= xy 8πµ(1 − ν) r2 and the stress is recovered from (2.8.16) as 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 (2.9.13) We can hence ﬁnd the maximum shear stress S deﬁned in (2.2.36), by calculating the eigenvalues τi of τ : S=

f |τ1 − τ2 | = 4(1 − ν)2 x2 + (1 − 2ν)2 y 2 . 2 2 2π(1 − ν)r

(2.9.14)

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. The response to a point force in three dimensions may be found in an analogous fashion by using the axisymmetric Love function; see Exercise 2.19. 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

86

Linear elastostatics

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 (2.9.13) 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    (2.9.15) τ = τ0 (x, y − h) + τ0 (x, y + h) +  , 2 ˜ ˜  ∂2A ∂ 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, = (2.9.16) ∂y ∂x2 2π(1 − ν) (h2 + x2 )2 on y = 0. This is now precisely the half-space problem we have already analysed in Section 2.6.8, and may be solved by substituting the right-hand side of (2.9.16) into (2.6.80) to give −2hxy f ˜ − (1 − 2ν)x log x2 + (y + h)2 A= 2 2 4π(1 − ν) x + (y + h) −1 y + h . (2.9.17) + 4h(1 − ν) tan 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 non-trivial 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) in y > 0, with ∂ ∂2L − ν∇2 L = 2 ∂y ∂y

∂2L 2 − ν∇ L = 0 ∂y 2

on y = 0.

(2.9.18a)

(2.9.18b)

2.9 Singular solutions in elastostatics

87

This may be solved by subtracting oﬀ the full-space solution (2.9.11), that is, writing L=−

f ˜ R2 (log R − 1) + L, 8π(1 − ν)

R2 = x2 + (y − h)2 ,

(2.9.19)

and then taking a Fourier transform in x. We can also quite quickly ﬁnd the response to a point force at the boundary of a half-space by using the Papkovich–Neuber representation (2.8.51). Let us consider the axisymmetric version of the half-space problem of Section 2.6.8 in cylindrical polar coordinates (r, z). The half-space is z > 0 and we suppose the tractions on z = 0 consist purely of a normal point force P at the origin, so that τrz = 0 and τzz = −P δ(x)δ(y). Because these conditions are imposed on z = 0, in (2.8.51) we take T Ψ = 0, 0, ψ(r, z) ,

φ = φ(r, z),

(2.9.20)

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

∂φ 1 ∂ψ − z , ∂r 2 ∂r

2µuz = 2(1 − ν)ψ −

∂φ 1 ∂(zψ) − . ∂z 2 ∂z

(2.9.21)

From the formulae of Section 1.11.2, the vanishing of the shear stress requires that ∂ur ∂uz + =0 on z = 0, (2.9.22) ∂z ∂r so that 2

∂ψ ∂2φ − (1 − 2ν) = 0. ∂r∂z ∂r

(2.9.23)

Hence, without loss of generality, we can choose 2

∂φ − (1 − 2ν)ψ = 0 ∂z

(2.9.24)

on z = 0 and, since the left-hand side is a harmonic function that also vanishes at inﬁnity, it is identically zero. The normal traction is thus ∂ψ ∂2φ λ ∂ ∂uz (rur ) + (λ + 2µ) = 2(1 − ν) −2 2 r ∂r ∂z ∂z ∂z ∂ψ . = ∂z

τzz =

(2.9.25)

88

Linear elastostatics

Our problem has been reduced to ﬁnding a harmonic function ψ whose z-derivative is equal to −P δ(x)δ(y) on z = 0. It can readily be veriﬁed that the solution is P , (2.9.26) ψ= √ 2π r2 + z 2 and uz , which is proportional to ψ on z = 0, does not display the nonmonotonicity of Figure 2.13. 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)

(2.9.27)

to an arbitrary forcing function f . The standard mathematical approach to linear diﬀerential equations like (2.9.27) is to use a Green’s function. For a scalar equation of the form Lu = −f (x),

(2.9.28)

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),

(2.9.29)

subject to suitable boundary conditions. We can then use the key property (2.9.5) of the delta-function and formal diﬀerentiation under the integral to deduce that u(x) = R3

G(x − ξ)f (ξ) dξ

(2.9.30)

is a solution of the inhomogeneous equation (2.9.28). To adapt the above procedure to the vector equation (2.9.27), one needs to deﬁne a Green’s tensor G(x) such that (λ + µ) grad div G + µ∇2 G = −δ(x)I,

(2.9.31)

where I = (δij ) is the identity matrix. Equation (2.9.31) is shorthand for the scalar equations (λ + µ)

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

(2.9.32)

2.9 Singular solutions in elastostatics

89

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 (2.9.30) to G(x − ξ)f (ξ) dξ. (2.9.33) u(x) = 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, (2.9.33) 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 Section 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 tensor directly from the point force solutions found in Section 2.9.2. In plane strain, the displacement caused by a unit point force in the x-direction is given by (2.9.12) with f = 1: 2 x − y2 1 G11 = − (3 − 4ν) log r , (2.9.34a) 8πµ(1 − ν) 2r2 xy 1 , (2.9.34b) G12 = 8πµ(1 − ν) r2 and, by symmetry, we deduce that the Green’s tensor is given by  2  xy x − y2 1  2r2 − (3 − 4ν) log r  r2 G(x, y) =  . xy y 2 − x2 8πµ(1 − ν) − (3 − 4ν) log r r2 2r2 (2.9.35) The corresponding three-dimensional result is derived in Exercise 2.18. Finding the Green’s tensor 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 (2.9.31) along with boundary conditions for G corresponding to those that are imposed on u (so that (2.9.33) 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 (2.9.31). Hence we can no longer assume that G is a function only of x − ξ, and we must rewrite (2.9.29) as LG(x, ξ) = −δ(x − ξ).

(2.9.36)

90

Linear elastostatics

We note that the full-space Green’s tensor, which is (2.9.35) in plane strain, can be used to reduce the Navier equation on an arbitrary domain D to an integral equation satisﬁed on the boundary of D. This reduction in dimension can facilitate an 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 ∂u ∂v 2 2 −v· u · ∇ v − v · ∇ u dx = u· dS (2.9.37) ∂n ∂n D ∂D and D

{u · (grad div v) − v · (grad div u)} dx {(div v) u · n − (div u) v · n} dS, (2.9.38) = ∂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 = −δ(x − ξ) δij , ∂τjk = 0, (λ + µ) (grad div u)j + µ∇2 uj = ∂xk

(2.9.39) (2.9.40)

where we have neglected the body force for simplicity. Hence, multiplying (2.9.39) by ui , (2.9.40) by Gij , then subtracting and integrating, we ﬁnd that ui (ξ) = G i (x − ξ) · (τ n (x)) − u(x) · τ i n(x − ξ) dS, (2.9.41) ∂D

where G i = (Gi1 , Gi2 , Gi3 )T and τ i is the stress ﬁeld associated with the displacement G i . 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, (2.9.41) 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, 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 care must be taken with the singularities that occur in G i as ξ approaches ∂D.

2.9 Singular solutions in elastostatics

91

y

f f

a

f x

f

Fig. 2.16 Four point forces.

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 Section 2.6.5 and, when we let b → ∞ and a → 0 in (2.6.41), we ﬁnd that the radial displacement is given by ur =

P a2 . 2µr

(2.9.42)

The same displacement ﬁeld may also be produced by superimposing four point forces, as indicated in Figure 2.16, and letting the distance between them tend to zero (Exercise 2.26). However, it is more illuminating to consider the incompatibility of the displacement ﬁeld (2.9.42). Since the vanishing of ∇4 A is the condition for compatibility, we begin by trying to compute this quantity for (2.9.42). It is

92

Exercises

easy to see that A = −a2 P log r and hence, using (2.9.7), that ∇2 A = −2πa2 P δ(x)δ(y).

(2.9.43)

Thus the incompatibility associated with the displacement ﬁeld (2.9.42) is given by (2.9.44) ∇4 A = −2πa2 P δ (x)δ(y) + δ(x)δ (y) , where, by analogy with (2.9.3), the derivative of a delta-function can be deﬁned by ∞ δ (x − ξ)ψ(ξ) dξ = ψ (x) (2.9.45) −∞

for any suitably smooth test function ψ. Hence (2.9.42) 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 the stress ﬁeld associated with (2.9.42), and the material is therefore said to be in a state of self-stress. More generally, in Exercise 2.21 it is shown that the compatibility conditions (2.7.5) are the six components of the symmetric tensor equation ηij = ikl jmn

∂ 2 emk ≡ 0, ∂xn ∂xl

(2.9.46)

where ijk 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 (2.9.46) is ∂ 2 eyy ∂ 2 exy 1−ν 4 ∂ 2 exx + ∇ A = 0, − 2 = 2 2 ∂y ∂x∂y ∂x 2µ

(2.9.47)

which reproduces (2.7.5a) and shows explicitly how ∇4 A is related to compatibility. Thus the right-hand side of (2.9.44) can be thought of as a point singularity in the incompatibility tensor, and this prompts us to wonder what the physical signiﬁcance might be of introducing other point singularities on the right-hand side of (2.9.46). For example, the physical interpretation of solutions of ∂ 2 eyy ∂ 2 exy ∂ 2 exx + − 2 = δ(x)δ (y) (2.9.48) ∂y 2 ∂x∂y ∂x2 has fundamental implications for the theory of metal plasticity, to which we will return in Chapter 8.

2.10 Concluding remark

93

2.10 Concluding remark All the solutions in this chapter were derived for purely static situations but, in fact, they also apply to unsteady problems that evolve on a suﬃciently long time-scale. For example, the analysis of a statically loaded torsion bar carried out in Section 2.4 applies equally well when the applied twist Ω is a slowly varying function of t. In such quasi-static situations, time appears in the solution only as a parameter. This approach fails, however, to describe elastic waves, in which the displacement varies rapidly with time, and the temporal and spatial dependence of the solution are coupled, as we will see in the following chapter.

Exercises 2.1

2.2

2.3

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. 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 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 ∇2 w =

2.4

ρg µ

in the cross-section D of the bar, with w = 0 on ∂D. If D is the ellipse x2 /a2 + y 2 /b2 < 1, show that the solution of (2.4.12), (2.4.13) is x2 y2 a2 b2 1− 2 − 2 . φ= 2 (a + b2 ) a b Hence show that the torsional rigidity of a bar with uniform elliptical cross-section is πµa3 b3 T = 2 . a + b2 Deduce that a circular bar has a higher rigidity than an elliptical bar with the same cross-sectional area.

94

2.5

2.6

Linear elastostatics

Derive the expression (2.5.2) for the torsional rigidity of a tubular bar. By contour deformation show that, if (2.5.3) holds when C is ∂Di , then it holds for all simple closed paths C contained in D. By separating the variables in polar coordinates, show that the problem (2.5.14) for a cut tubular bar has the solution 2 b − a2 log (r/b) b2 − r 2 + φ= 2 2 log (b/a) ∞ # $ # $ + 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 (2.4.15) to determine the torsional rigidity $ πµ b2 − a2 # 2 R= a + b2 log (b/a) − b2 − a2 2 log (b/a) $ $ # ∞ # 16µ log4 (b/a) b2 − (−1)n a2 tanh nπ 2 / log (b/a) − . $ # 2 π 2 + 4 log2 (b/a) 2 π n n=1 Now suppose that the tube is thin, so that b/a = 1 + ε, where ε 1. By expanding the above expression for small ε, show that 1 93ζ(5) 2π 4 3 µa ε 1 − + ε + ··· , R= 3 2 π6

2.7

where ζ is the Riemann zeta function (Abramowitz & Stegun, 1972, Section 23.2). If z = x + iy and z¯ = x − iy are regarded as independent variables, show that ∂2ψ ∇2 ψ = 4 . ∂z∂ z¯ Deduce that the general solution of Laplace’s equation may be written in the form 1 ψ = {f (z) + h(¯ z )}, 2 where f and h are arbitrary analytic functions. If ψ takes only real values, show that h(¯ z ) = f (z) and hence that ψ = Re f (z) .

Exercises

95

Now solve the biharmonic equation by writing ∇2 A = ψ in the form 4

2.8

2.9

∂2A 1 = F (z) + F¯ (¯ z) ∂z∂ z¯ 2

and integrating with respect to z and z¯ to obtain (2.6.12). 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 from the Goursat representation (2.6.12). Now ﬁnd all separable solutions of the biharmonic equation in polar coordinates of the form A = f (r) cos(mθ), with m = 0, 1, 2, 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. 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 sur& face with unit normal n is i τi mi ni . Hence show that the maximum shear stress S satisﬁes the constrained optimisation problem max S =

3

τi mi ni ,

i=1

subject to

3 i=1

m2i = 1,

3

n2i = 1,

i=1

3

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 m3 = n3 = 0, or m1 = n1 = 0, or m2 = n3 = 0, 2.10

1 m21 = m22 = n21 = n22 = , 2 1 m22 = m23 = n22 = n23 = , 2 1 m23 = m21 = n23 = n21 = , 2

τ1 − τ2 , 2 τ2 − τ3 S=± , 2 τ3 − τ1 S=± . 2 S=±

Prove the identity u∇4 v − v∇4 u ≡ div u∇ ∇2 v − v∇ ∇2 u + ∇2 u∇v − ∇2 v∇u . If the functions Ai (x) and constants λi (i = 1, 2, . . .) satisfy ∇4 Ai = λi Ai

96

Linear elastostatics

in some bounded region V , show that ∂Aj ∂Ai ∇2 Aj − ∇2 Ai Ai Aj dx = (λi − λj ) ∂n ∂n V ∂V ∂ 2 ∂ 2 dS. + Aj ∇ Ai − Ai ∇ Aj ∂n ∂n Deduce that the biharmonic operator has real eigenvalues and orthogonal eigenfunctions when the boundary conditions are ∂A = 0, or (ii) A = ∇2 A = 0, ∂n ∂ 2 ∂A ∂ 2 = or (iii) ∇ A , or (iv) ∇2 A = ∇ A = 0. ∂n ∂n ∂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 ∞ 1 y f1 (x, y) = , e−ikx e−|k|y dk = 2 2π −∞ π (x + y 2 ) (i)

2.11

A=

and i f2 (x, y) = 2π

∞

e−ikx sign(k)e−|k|y dk =

−∞

x , + y2 )

π (x2

respectively. (b) Hence show that (2.6.79) may be written as ∂ ! f!1 (k, y), τ!xx = −P (k) 1 + y ∂y ∂f1 (k, y), τ!xy = P!(k)y ∂x ∂ f!1 (k, y). τ!yy = −P!(k) 1 − y ∂y Using the fact that the inverse Fourier transform of a product f!(k)! g (k) is the convolution of f and g, that is ∞ f (x − s)g(s) ds, (f ∗ g)(x) = −∞

deduce (2.6.80). (c) Show from (2.6.82) that ∂f ∂v 1 = P!(k) y (k) − 2(1 − ν)f!2 (k) , 2µ ∂x ∂x and hence deduce (2.6.83).

Exercises

2.12

97

By making a suitable substitution, write the second term in the integral (2.6.83) as 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 (2.6.83) may be written as −ε ε ∞ P (s)(x − s)3 ds 1−ν + + . − π ((x − s)2 + y 2 )2 −∞ −ε ε Show that the middle integral is approximately equal to (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, (2.6.83) tends to (2.6.85), where the Hilbert transform is deﬁned by −ε ∞ 1 P (s) ds 1 ∞ P (s) ds = lim . + H[P ](x) = − π −∞ s − x π ε↓0 s−x −∞ ε

2.13

Show that, if ∇2 φ = 0 in the half-space y > 0 with φ(x, 0) = f (x), ∂φ/∂y(x, 0) = g(x) and φ → 0 at inﬁnity, then ∞ ' ( 1 g(s) log (s − x)2 + y 2 ds. φ= 2π −∞ Deduce that f (x) = ∂φ/∂x(x, 0) = −H[g]. Show further that ∂ 2 φ/∂y 2 (x, 0) = −f (x), so that ∞ ' ( 1 ∂φ =− f (s) log (s − x)2 + y 2 ds. ∂y 2π −∞

2.14

Hence deduce that g(x) = H[f ]. Reconsider the torsion problem of Section 2.4 and the stress function φ deﬁned by (2.4.10). Using two of the compatibility relations (2.7.5), deduce successively that ∇2 φ = f (y), that ∇2 φ = g(x), and hence that ∇2 φ = const.. [This illustrates that two compatibility conditions that are not independent may nevertheless both be necessary to determine the stress.]

98

2.15

Linear elastostatics

(a) Derive the expression (2.6.97) for the modiﬁed Airy stress function outside a circular underground tunnel. (b) Prove the identity ∂eθθ ∂erθ ∂ 2 ur + rerr + 2r + ur = −r2 2 ∂θ ∂r ∂θ

(E2.1)

relating the strain components to the radial displacement ur in plane polar coordinates (r, θ). Deduce that the coeﬃcients of sin θ and cos θ on 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 (2.6.92), show that the right-hand side of (E2.1) is −ρgh

2.16

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

and hence obtain the expressions (2.6.98). Return to the tunnel example of Section 2.6.9, this time assuming a far-ﬁeld stress distribution given by τxx = τxy = 0, τyy = ρg(y − H). [This might correspond to a cylindrical hole in a vertical column.] In this case, show that (2.6.92) becomes ˜ ˜ 1 ∂2A 1 ∂A , + 2 2 r ∂θ r∂r ˜ 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 H a ρga2 ˜ H + aθ cos θ + cos 2θ − sin 3θ A(a, θ) = 2 2 12 + A cos θ + B sin θ, ˜ ∂A ρga a (a, θ) = H + aθ cos θ + H cos 2θ − sin 3θ ∂r 2 4 A B + cos θ + sin θ, a a

Exercises

99

where A and B are arbitrary constants. Use (E2.1) and the fact that ur is single-valued to show that A = 0,

B = −νρga3 /4,

so that r a2 H 1 + log + rθ cos θ + H 1 − 2 cos 2θ a 2r 4 2 1 a2 r a a + − 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 1 − 2ν sin θ + sin 3θ τθθ = −ρgH (1 + 2 cos 2θ) + ρga 2

2.17

on r = a. Note the contrast with the stress distribution (2.6.100) caused by a hydrostatic far-ﬁeld stress. By comparing (2.6.9) and (2.8.22), show that z-averaged plane stress is mathematically equivalent to plane strain with new elastic constants ν and E given by ν =

2.18

ν , 1+ν

E =

1 + 2ν E. (1 + ν)2

We deﬁne the three-dimensional Fourier transform of a function f (x, y, z) by ∞ ∞ ∞ ! f= f (x, y, z)ei(kx+y+mz) dxdydz −∞ −∞ −∞ f (x)ei(k·x) dx, = R3

where k = (k, , m). Show that the Fourier transform of the Green’s tensor in three dimensions is given by G! = G!ij =

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

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

(E2.2)

100

2.19

Linear elastostatics

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

|x| . 8π (1 − ν)

Hence, show that the corresponding displacement is xz yz 1 z2 , (G31 , G32 , G33 ) = , , 3 − 4ν + 16πµ |x| (1 − ν) |x|2 |x|2 |x|2 2.20

and deduce (E2.2) by symmetry. Show that the two-dimensional Papkovich–Neuber potentials φ = φ(x, y), Ψ = (ψ1 (x, y), ψ2 (x, y), 0)T generate the stress ﬁeld ∂ψ2 ∂ 2 φ x ∂ 2 ψ1 ∂ψ1 y ∂ 2 ψ2 +ν − − − , ∂x ∂y ∂x2 2 ∂x2 2 ∂x2 ∂ψ1 ∂ 2 φ x ∂ 2 ψ1 ∂ψ2 y ∂ 2 ψ2 +ν − 2 − = (1 − ν) − , ∂y ∂x ∂y 2 ∂y 2 2 ∂y 2 ∂ψ2 ∂2φ x ∂ 2 ψ1 y ∂ 2 ψ2 ∂ψ1 1 −ν + − − − . = 2 ∂y ∂x ∂x∂y 2 ∂x∂y 2 ∂x∂y

τxx = (1 − ν) τyy τxy 2.21

Prove the identity mjk

∂ 2 ui ≡0 ∂xk ∂xj

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

2.22

∂ 2 eij ≡0 ∂xl ∂xk

for all m and n, where eij is the linearised strain tensor. Hence obtain the compatibility conditions (2.7.5) and show that they are equivalent to (2.7.6). Show that the compatibility conditions (2.7.5) can be written in terms of the stress components as 2 ∂ 2 τyy ∂ 2 τxy ∂ ν ∂2 ∂ 2 τxx + −2 = + 2 Tr (τ ), ∂y 2 ∂x∂y ∂x2 1 + ν ∂x2 ∂y 2 ∂τxy ∂τyz ∂ τxx ∂ ∂τxz ν ∂2 = + − + Tr (τ ), ∂y∂z ∂x ∂y ∂z ∂x 1 + ν ∂y∂z the other relations being obtained by permuting x, y and z.

Exercises

2.23

Verify (2.9.7a) by writing |x| = limε→0 x2 + ε2 and noting that ε2 d2 2 2 = x + ε , dx2 (x2 + ε2 )3/2 and

∞

ε2 dx

∞

+

ε2 )3/2

=

ds

= 2. (1 + s2 )3/2 Repeat this procedure with log x2 + y 2 + ε2 to derive (2.9.7b) and ﬁnd a similar derivation of (2.9.7c). Show that 2 $ ∂ 2 # 2 ∂ x + y 2 log x2 + y 2 = 8 + 4 log x2 + y 2 + 2 2 ∂x ∂y and deduce that, if A = (1/8π) x2 + y 2 log x2 + y 2 , then A satisﬁes −∞ (x2

2.24

101

√

−∞

∇4 A = δ(x)δ(y). 2.25

The stress components due to a point force f in the x-direction satisfy ∂τxy ∂τxx + = −f δ(x)δ(y), ∂x ∂y

∂τxy ∂τyy + =0 ∂x ∂y

plus the compatibility condition ∂ 2 τyy ∂ 2 τxy ∂ 2 τxx + −2 − ν∇2 (τxx + τyy ) = 0. 2 ∂y ∂x∂y ∂x2 Take a two-dimensional Fourier transform (deﬁned as in Exercise 2.18) to show that k! τxx + ! τxy = −if,

k! τxy + ! τyy = 0, 2 2 τxy − k τ!yy + ν k + (! − τ!xx + 2k! τxx + τ!yy ) = 0. 2

2

Hence obtain τ!xx

ikf = 1−ν

τ!xy =

if 1−ν

τ!yy =

−ikf 1−ν

ν k 2 + 22 , − k 2 + 2 (k 2 + 2 )2 ν 2 , − k 2 + 2 (k 2 + 2 )2 2 ν , − k 2 + 2 (k 2 + 2 )2

the inverse Fourier transform of which is (2.9.13).

102

2.26

Linear elastostatics

Write down the displacement ﬁeld for the four point forces depicted 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 (2.9.42) 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 Section 1.10, the unsteady Navier equation (1.7.8) 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 in an inﬁnite medium 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 (1.10.9). 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 Section 3.2.5 by considering the reﬂection 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. 103

104

Linear elastodynamics

In Section 3.5, we consider some initial-value problems for elastic wave propagation 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 sources. Before we begin, let us brieﬂy recall 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 (3.1.1) 2 =T 2, ∂t ∂x known as the one-dimensional wave equation. The derivation and validity of this equation will be discussed in Chapter 4, where we will show that it also describes longitudinal waves in an elastic bar. 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 (3.1.1) is ς

∂2w = T ∇2 w, ∂t2

(3.1.2)

where ∇2 = ∂ 2 /∂x2 + ∂ 2 /∂y 2 is the two-dimensional Laplacian. The threedimensional version of (3.1.2) governs many familiar wave propagation problems including acoustic waves and light waves. There are two ways in which we can approach (3.1.1) and (3.1.2) 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 (3.1.1) and (3.1.2). 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 of an elastic

3.2 Normal modes and plane waves

105

string by trying solutions of (3.1.1) in the form w(x, t) = f1 (x) cos(ωt) + f2 (x) sin(ωt),

(3.2.1a)

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

(3.2.1b)

where f (x) = f1 (x) + if2 (x). Solutions of the form (3.2.1) are called waves in the frequency domain. Substitution of (3.2.1) into (3.1.1) reveals that f (x) must satisfy the ordinary diﬀerential equation ω 2 d2 f f = 0, + dx2 T

(3.2.2)

whose general solution is

)

f (x) = A cos(kx) + B sin(kx),

where

k=ω

. T

(3.2.3)

Thus (3.2.1b), with f (x) given by (3.2.3), 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,

(3.2.4)

and by applying these to (3.2.3) we discover that nonzero solutions for f exist only if ω takes speciﬁc values, namely * nπ T , (3.2.5) ωn = 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 nπx (3.2.6) e−iωn t wn (x, t) = Re Bn sin 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

106

Linear elastodynamics

(3.2.5), namely that the frequency increases as the tension increases and as the density decreases. Now suppose we are additionally given the initial displacement and velocity of the string, say w(x, 0) = w0 (x),

∂w (x, 0) = v(x). ∂t

(3.2.7)

These may be satisﬁed by trying a linear combination of normal modes, that is ∞ nπx {An cos(ωn t) + Bn sin(ωn t)} , sin (3.2.8a) w(x, t) = a n=1

where the coeﬃcients are found by Fourier analysis of the initial data: a nπx nπx 2 a 2 dx, Bn = dx. w0 (x) sin v(x) sin An = a 0 a ωn a 0 a (3.2.8b) This analysis, although very straightforward, illustrates the following important points that will be useful in analysing more complex models later in this chapter. (i) We will always be able to look for time-harmonic solutions of the form (3.2.1b) whenever the problem we are solving is both linear and autonomous in t: in other words, when there is no explicit dependence on t. (ii) When looking for normal modes, we might as well have ignored the “Re” in (3.2.1b), 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! (iii) Although normal modes may at ﬁrst glance appear to be rather special solutions of the wave equation (3.1.2), it can be shown that the general solution, subject to simple homogeneous boundary conditions, may be written uniquely as a superposition of normal modes. (iv) When a → ∞, it may be shown that (3.2.8) generalises to the half-range Fourier transform representation ∞ sin (kx) {A(k) cos(kct) + B(k) sin(kct)} dk, (3.2.9a) w(x, t) = 0

3.2 Normal modes and plane waves

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

B(k) =

2 πkc

107

∞

v(x) sin(kx) dx. 0

(3.2.9b) We can extend these ideas to the two-dimensional wave equation (3.1.2) by seeking solutions of the form w(x, y, t) = A(x, y)e−iωt ,

(3.2.10)

and we ﬁnd that A must satisfy the Helmholtz equation

∇2 A + k 2 A = 0,

(3.2.11)

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 (3.2.11) 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 mπx nπy (3.2.12a) sin e−iωm , n t , wm,n (x, y, t) = sin a b π 2 T m2 n2 2 ωm,n = + , (3.2.12b) ς a2 b2 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 (3.2.11) reads d2 A 1 dA + k 2 A = 0, + dr2 r dr

(3.2.13a)

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.

(3.2.13b)

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 (3.2.13) will be approximately in balance. Moreover, for small ξ, we can spot that the ﬁrst and

108

Linear elastodynamics

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).

second terms in (3.2.13) 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. 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, . . .),

(3.2.14a)

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

3.2 Normal modes and plane waves

by ξ0,i ωi = a

)

T ς

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

109

(3.2.14b)

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 1 ∂2A ∂ 2 A 1 ∂A + + + k 2 A = 0, (3.2.15) ∂r2 r ∂r r2 ∂θ2 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 us to seeking separable solutions of the form (3.2.16) A(r, θ) = f (r) C1 cos(nθ) + C2 sin(nθ) , 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.

(3.2.17a)

Setting ξ = kr as before, we ﬁnd that (3.2.17a) becomes ξ2

df 2 d2 f + ξ − n2 f = 0 +ξ 2 dξ dξ

with f (ka) = 0.

(3.2.17b)

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 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, . . .),

(3.2.18)

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

(3.2.19)

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).

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Linear elastodynamics

3.2.2 Plane waves As suggested by (3.2.9), when looking for solutions of (3.1.1) on an inﬁnite domain, it is appropriate to seek travelling waves of the form w(x, t) = A exp i(kx − ωt) , (3.2.20) rather than seeking a discrete set of normal modes. The real constants |A|, k and ω represent the amplitude, wave-number and frequency, respectively, of the wave (3.2.20). The wave-number is related to the wave-length λ by k = 2π/λ, so large values of k correspond to short waves and vice versa. We can alternatively write (3.2.20) in the form ω (3.2.21) w(x, t) = A exp ik(x − ct) , where c = ; k the quantity c is known as the phase velocity. From (3.2.21) 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 (3.2.20) 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 (3.1.2) by looking for a plane wave solution of the form w(x, y, t) = A exp i(k1 x + k2 y − ωt) = A exp i k · x − ωt , (3.2.22) where we deﬁne the wave-vector k = (k1 , k2 )T . By writing (3.2.22) in the form k·x w = A exp i |k|X − ωt , where X = , (3.2.23) |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 ωk . (3.2.24) c= |k|2 When we substitute (3.2.20) into (3.1.1), we ﬁnd that the amplitude can be nonzero only if ω 2 = k 2 T / .

(3.2.25a)

This dispersion relation tells us how the frequency of any given wave depends on its wave-length. We can also view (3.2.25a) as an equation for the phase velocity, namely (3.2.25b) c = ± T / ,

3.2 Normal modes and plane waves

111

which tells us that all waves move at the same speed, irrespective of their wave-length. Waves with this property are called non-dispersive, to distinguish them from dispersive waves in which c varies with k. Similarly, substitution of (3.2.22) into the two-dimensional wave equation (3.1.2) leads to the dispersion relation ω 2 = |k|2 T /ς. The phase speed |c| = c =

T /ς

(3.2.26)

(3.2.27)

is still independent of the wave-vector k, so these two-dimensional waves are also non-dispersive. As in (3.2.9), 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 ∞ ∞ A(k1 , k2 ) exp i (k1 x + k2 y − ω(k1 , k2 )t) w(x, y, t) = −∞ −∞ + B(k1 , k2 ) exp i (k1 x + k2 y + ω(k1 , k2 )t) dk1 dk2 , (3.2.28a) or

w(x, t) = R2

A(k) exp i (k · x − ω(k)t) + B(k) exp i (k · x + ω(k)t) dk, (3.2.28b)

where ω(k) is given by the dispersion relation (3.2.26) 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 (3.2.28b), we indicate that this and all the other results in this section apply also to the wave equation in three space dimensions.

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, seismology and ultrasonic testing, where we try to infer, non-invasively, the properties of some inhomogeneities in a bulk elastic medium by measuring the scattered wave-ﬁelds that they produce.

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The basic idea is illustrated by the problem of an elastic string, modelled by (3.1.1), stretched along the x-axis with a point mass m attached at the origin. The equation of motion for the mass leads to the boundary conditions w(0−, t) = w(0+, t),

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

(3.2.29)

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) = A(x)e−iωt , where eikx + cR e−ikx x < 0, A(x) = (3.2.30) x > 0. cT eikx 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 (3.2.29), as shown in Exercise 3.1. 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 (3.2.11) to model plane wave incidence at a scatterer on which A = 0, say, is beyond the scope of this book. The two principal diﬃculties are the following. (i) Assuming an incident wave eikx , as in (3.2.30), we have to solve the Helmholtz equation for the scattered wave A˜ = A − eikx , subject to A˜ = eikx on the scatterer. This means that there is no simple solution by separation of variables, even when the scatterer is a circle. (ii) In the far ﬁeld, it is not good enough simply to say that A˜ → 0 as we did in many of the elastostatic problems of Chapter 2. We now need to capture the physical requirement that A˜ be outgoing from the scatterer, again as in (3.2.30). It can be shown (see Exercise 3.3 and Bleistein, 1984, Section 6.4) that, as the radial coordinate r tends to inﬁnity, all possible solutions of (3.2.11) take the form A˜ → A˜± (θ)r−1/2 e±ikr and, since A˜ must be multiplied by e−iωt to ﬁnd the scattered wave, only the positive sign is appropriate to describe outward-travelling waves. This is equivalent to saying that A˜ satisﬁes the Sommerfeld radiation condition ∂ A˜ 1/2 ˜ − ik A → 0 as r → ∞. (3.2.31) r ∂r

3.2 Normal modes and plane waves

113

Then the canonical tomography problem for (3.2.11) 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 (3.2.16), A depends only on r and is a linear combination of J0 (kr) and Y0 (kr). Using the fact that ) 2 ±i(kr−(π/4)) e as r → ∞, (3.2.32) J0 (kr) ± i Y0 (kr) → πkr we see that, to describe outward-propagating waves, A must tend to a mul(1) tiple of the Hankel function H0 (kr) = J0 (kr) + i Y0 (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 (1.7.8). We begin by seeking travelling-wave solutions in the form u = a exp i (k · x − ωt) , (3.2.33) 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 3.4, 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.

(3.2.34)

Using this decomposition, we ﬁnd that

∇2 u = − |k|2 (Ak + B×k) exp i (k · x − ωt) , grad div u = −A |k|2 k exp i (k · x − ωt) , ∂2u 2 = −ω (Ak + B×k) exp i (k · x − ωt) . ∂t2 Hence the Navier equation (1.7.8) reduces to 2 ρω − µ|k|2 (B×k) + ρω 2 − (λ + 2µ)|k|2 Ak = 0,

(3.2.35a) (3.2.35b) (3.2.35c)

(3.2.36)

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

and

ρω 2 = (λ + 2µ) |k|2

(3.2.37a)

A=0

and

ρω 2 = µ |k|2 .

(3.2.37b)

or

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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 Pressure waves, take the form u = Ak exp i (k · x − ωt) , (3.2.38a) where ω 2 = (λ+2µ)|k|2 /ρ. We recall from Section 1.9 that µ and λ+2µ/3 are both positive, so ω is real. The phase speed is thus given by * λ + 2µ (3.2.38b) cp = ρ and, since cp is independent of k, the waves are non-dispersive. The phase velocity cp k (3.2.38c) c= |k| 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 (3.2.38a). (ii) S -waves, also known as Secondary or Shear waves, take the form u = (B×k) exp i (k · x − ωt) , (3.2.39a) where ω 2 = µ|k|2 /ρ. S -waves are also non-dispersive, with constant phase speed ) µ . (3.2.39b) cs = ρ This time, though, the phase velocity is perpendicular to the displacement, so S -waves are said to be transverse. Since (3.2.39a) satisﬁes div u = 0, we deduce that S -waves conserve volume, and they may thus be referred to as equivoluminal. Evidently cp > cs , so that P -waves always propagate faster than S -waves. This fact is familiar to seismologists: following an earthquake, two distinct initial signals can usually be observed, 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) .

(3.2.40)

3.2 Normal modes and plane waves

115

When u0 is not parallel to k, this is called a plane polarised wave, with u0 and k deﬁning the plane of polarisation. As in the expression (3.2.28) 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 u(x, t) = kA1 (k) exp i (k · x − |k|cp t) R3 + kA2 (k) exp i (k · x + |k|cp t) + k×B 1 (k) exp i (k · x − |k|cs t) (3.2.41) + k×B 2 (k) exp i (k · x + |k|cs t) dk, 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 (3.2.41) 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 (3.2.33) 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 appropriately, we may write the incident wave in the form of a plane strain displacement, # $ cos α exp i kp (x cos α + y sin α) − ωt , (3.2.42) u = ui n c = 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 ur e f such that the net displacement u = ui n c +ur e f is zero on the boundary x = 0. We soon realise that

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Linear elastodynamics y P S

α β α

P

0110 Rigid 1010 boundary 1010 x 1010 1010 1010

Fig. 3.2 A P -wave reﬂecting from a rigid boundary.

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 # $ − cos γ exp i kp (−x cos γ + y sin γ) − ωt u r e f = rp sin γ # $ sin β + rs exp i ks (−x cos β + y sin β) − ωt , (3.2.43) cos β where ks = ω/cs . Recall that S -waves are transverse so the amplitude is orthogonal to the wave-vector. From the condition u = 0 on x = 0, we ﬁnd that the P -wave reﬂection is specular, meaning that the angle of reﬂection γ is equal to the angle of incidence α. However, the S -wave reﬂection angle satisﬁes sin β =

cs sin α cp

(3.2.44)

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

cos(α + β) , cos(α − β)

rs = −

sin(2α) . cos(α − β)

(3.2.45)

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 3.6, the same happens in

3.2 Normal modes and plane waves

117

y

“rock”

µ2 ,

ρ2

µ1 ,

ρ1

h “coal”

x

“rock”

−h

µ2 ,

ρ2

Fig. 3.3 A layered elastic medium.

the case of refracted waves, and the phenomenon of total internal reﬂection becomes correspondingly more complicated. We now describe an example that illustrates how P - and S -waves can interact in a practical situation. 3.2.6 Love waves Let us consider the dynamic version of antiplane strain, the static case of which was introduced T in Section 2.3. If the displacement ﬁeld takes the form u = 0, 0, w(x, y, t) , then w satisﬁes 2 ∂2w ∂ w ∂2w 2 = ρ + , (3.2.46) µ∇ w = µ ∂x2 ∂y 2 ∂t2 which is just the familiar two-dimensional scalar wave equation (3.1.2), 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 (3.2.46). 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. This is in contrast with

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the situation depicted in Figure 3.2 where, if we identify the rigid boundary with a horizontal rock stratum, the displacement is in the vertical direction. For this reason, the S -waves considered in Section 3.2.5 are referred to as SV -waves, while the antiplane S -waves described by (3.2.46) are referred to as SH -waves. 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 2 ∂ wi ∂ 2 wi ∂ 2 wi 2 = cs i + (i = 1, 2), (3.2.47) ∂x2 ∂y 2 ∂t2 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 Section 1.10, so that w1 = w 2 ,

µ1

∂w1 ∂w2 = µ2 , ∂y ∂y

on y = ±h.

(3.2.48)

We seek travelling-wave solutions propagating in the x-direction in which the displacements take the form (3.2.49) wi = fi (y) exp i (kx − ωt) . Substituting (3.2.49) into (3.2.47), we ﬁnd that the functions fi (y) satisfy 2 ω 2 − k fi = 0 (3.2.50) fi + c2si 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 ey

in y < −h,

(3.2.51)

where is real and positive. In such modes, the seam acts as a wave-guide, propagating waves in the x-direction without any energy radiating or “leaking out” to y → ±∞ as discussed in Section 3.2.3. Substitution of (3.2.51) into (3.2.50) reveals that k and ω must be such that c2s2 k 2 − ω 2 = c2s2 2 > 0.

(3.2.52)

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

(3.2.53)

which may be sinusoidal if m is real or exponential if m is pure imaginary. Now (3.2.50) leads to ω 2 = c2s1 k 2 + c2s1 m2 ,

(3.2.54)

3.2 Normal modes and plane waves

so the Love waves propagate at a speed cL given by m2 ω2 2 2 cL = 2 = cs1 1 + 2 . k k

119

(3.2.55)

Let us ﬁrst consider symmetric modes in which B1 = 0 and B2 = A2 , so the boundary conditions (3.2.48) reduce to A2 e−h = A1 cos(mh),

µ2 lA2 e−h = µ1 mA1 sin(mh).

(3.2.56)

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 .

(3.2.57)

For antisymmetric waves, with A1 = 0 and B2 = −A2 , the analogous calculation in Exercise 3.7 leads to µ1 m cot mh = −µ2 .

(3.2.58)

It remains to determine m from either of the transcendental equations (3.2.57) or (3.2.58). 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 3.7. 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 (3.2.52) and (3.2.55), we deduce the inequalities ω 2 < c2s2 , (3.2.59) c2s1 < 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 practice. The phase speed of the waves is then bounded between cs1 and cs2 and the resulting wave-ﬁelds in the rock decay exponentially as we move away from the seam. For each ﬁxed allowable value of m, (3.2.55) shows that the wave-speed cL 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 (3.2.37) were obtained only for plane P - or S -waves in an inﬁnite medium, and (3.2.55) illustrates how the presence of boundaries can often give rise to dispersion.

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Linear elastodynamics

ω

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 .

We see that the lowest-frequency mode is a symmetric wave with mh = π/2 and therefore 2

ω =

c2s1

π2 2 +k . 4h2

(3.2.60)

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 (3.2.52), (3.2.54) and either (3.2.57) or (3.2.58), as shown in Exercise 3.7. Crucially, the values of m satisfying (3.2.57) or (3.2.58) with > 0 are all real, so the inequalities (3.2.59) continue to hold and there is still a cut-oﬀ frequency. As shown in Figure 3.4, 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 plane strain in y < 0 below a stress-free

3.3 Dynamic stress functions

121

boundary at y = 0, where the zero-traction condition is ∂v ∂u ∂v ∂u ∂v + = 2µ +λ + =0 at y = 0. µ ∂y ∂x ∂y ∂x ∂y Now we seek frequency-domain solutions of the form u = (up eκp y + us eκs y ) exp i (kx − ωt) ,

(3.2.61)

(3.2.62)

where the dispersion relations (3.2.37) for P - and S -waves tell us that κ2p = k 2 −

ω2 , c2p

κ2s = k 2 −

ω2 . c2s

(3.2.63)

We assume as usual that k and ω are real and positive, and that Re(κp ) > 0 and Re(κs ) > 0, so that the waves propagate without attenuation in the x-direction 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 < cs < cp ,

(3.2.64)

so that Rayleigh waves inevitably travel more slowly than either P -waves or S -waves. After the rather tortuous manipulations of Exercise 3.8, we ﬁnd an equation for c in the form 2 1/2 1/2 c2 c2 c2 1− 2 =4 1− 2 . (3.2.65) 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 (3.2.64). 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 such a wave to propagate indeﬁnitely in the x-direction near y = 0. This hypothesis is born out by seismological observation: the radius of the earth is large enough for (3.2.61) 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

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Linear elastodynamics

(1.7.8), we obtain 2 ∂ 2 2 − cp ∇ (div u) = 0 ∂t2

and

∂2 2 2 − cs ∇ (curl u) = 0 ∂t2

(3.3.1)

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 (3.3.1a) for div u and then the inhomogeneous wave equation 2 ∂ 2 2 − cs ∇ u = c2p − c2s ∇ (div u) (3.3.2) 2 ∂t for the displacement u. Moreover, by cross-diﬀerentiating (3.3.1), we obtain 2 2 ∂ ∂ 2 2 2 2 (3.3.3) − cp ∇ − cs ∇ u = 0, ∂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 scalar 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, in which the second component of the Navier equation (1.7.8), namely 2 ∂ u ∂2v ∂2v + µ∇2 v, (3.3.4) + ρ 2 = (λ + µ) ∂t ∂x∂y ∂y 2 can be rewritten as

∂2u ∂2v 1 ∂2v 2 + (1 − 2ν) ∇ v − 2 2 + 2 = 0. ∂x∂y cs ∂t ∂y

This is satisﬁed identically if we set 1 ∂2L 1 ∂2L 2 u= , (1 − 2ν) ∇ L − 2 2 + 2µ cs ∂t ∂y 2

v=−

1 ∂2L , 2µ ∂x∂y

(3.3.5)

(3.3.6)

for some function L(x, y, t), which is called the dynamic Love function by analogy with the static Love stress function introduced in Section 2.8.2. It

3.3 Dynamic stress functions

123

is then readily veriﬁed that the ﬁrst Navier equation is satisﬁed if and only if L satisﬁes the scalar wave equation 2 2 ∂ ∂ 2 2 2 2 (3.3.7) − cp ∇ − cs ∇ L = 0. ∂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 (3.3.6), the stress components may be evaluated from L using ∂ ∂2L 1 ∂2L 2 + (1 − ν) ∇ L − 2 2 , (3.3.8a) τxx = ∂x ∂y 2 cs ∂t 2 1 ∂2L ∂ ∂ L 1 ∂2L 2 − , (3.3.8b) − ν ∇ L − τxy = ∂y ∂y 2 c2s ∂t2 2c2s ∂t2 ∂ ∂2L 1 ∂2L 2 τyy = − . (3.3.8c) −ν ∇ L− 2 2 ∂x ∂y 2 cs ∂t For radially symmetric problems, in which the displacement takes the form u = ur er + uz ez and is dependent only on the cylindrical polar coordinates (r, z) and t, it is again possible to deﬁne a Love function L(r, z, t) such that 1 ∂ ∂L 1 ∂2L 1 1 ∂2L 2 , uz = (1 − 2ν) ∇ L − 2 2 + r , ur = − 2µ ∂r∂z 2µ cs ∂t r ∂r ∂r (3.3.9) and again L satisﬁes (3.3.7). In this case, the stress components by ∂ ∂2L 1 ∂2L 2 , τrr = − −ν ∇ L− 2 2 ∂z ∂r2 cs ∂t 1 ∂2L ∂ 1 ∂ 1 ∂2L ∂L 2 τrz = r −ν ∇ L− 2 2 − 2 2 , ∂r r ∂r ∂r c ∂t 2cs ∂t s 2 ∂ 1 ∂ 1 ∂ L ∂L τzz = , r + (1 − ν) ∇2 L − 2 2 ∂z r ∂r ∂r cs ∂t 1 ∂L ∂ 1 ∂2L 2 τθθ = . − +ν ∇ L− 2 2 ∂z r ∂r cs ∂t

are given

(3.3.10a) (3.3.10b) (3.3.10c) (3.3.10d)

We will now use the ideas of Sections 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.

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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 (1.11.8) simpliﬁes to ∂ 2 uθ uθ 1 ∂uθ ∂ 2 uθ 1 ∂ 2 uθ − + + = . ∂r2 r ∂r r2 ∂z 2 c2s ∂t2

(3.4.1)

This scalar equation supports S -waves, as expected for a displacement ﬁeld satisfying ∇ · u = 0. We seek waves travelling in the z-direction with wavenumber k and frequency ω by separation of variables: uθ = f (r) ei(kz−ωt) . Substituting (3.4.2) into (3.4.1), we ﬁnd that f (r) satisﬁes 1 1 2 f + f + m − 2 f = 0, r r

(3.4.2)

(3.4.3)

where m2 =

ω2 − k2 , c2s

(3.4.4)

and we will only consider cases in which m is real. The general solution of (3.4.3) is f (r) = A J1 (mr) + B Y1 (mr) ,

(3.4.5)

where J1 and Y1 again denote Bessel functions, while A and B are arbitrary constants. To describe waves in a circular cylinder 0 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 uθ ∂uθ − =0 (3.4.6) τrθ = µ ∂r r on r = a, which implies that m d 1 J1 (mr) = − J2 (ma) = 0, dr r a r=a

(3.4.7)

the latter equality following from known properties of Bessel functions (Gradshteyn & Ryzhik, 1994, Section 8.4). We deduce that ma must equal one of

3.4 Waves in cylinders and spheres

125

the zeros ξ2,n of J2 as deﬁned in Section 3.2.1. There is thus one family of waves corresponding to each integer value of n. The dispersion relation cs + 2 ξ2,n + k 2 a2 (3.4.8) ω= a tells us that the bar acts as a wave-guide, 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. Also, for a cylinder that is ﬁnite in the z-direction, we can use (3.4.8) 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 Section 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) .

(3.4.9)

The partial diﬀerential equation (3.3.7) satisﬁed by L leads to the ordinary diﬀerential equation 2 2 d d 1 d 1 d 2 2 (3.4.10) + mp + ms f (r) = 0, + + dr2 r dr dr2 r dr for f , where now we deﬁne m2p =

ω2 − k2 , c2p

m2s =

ω2 − k2 , c2s

(3.4.11)

as shorthand. The general solution of (3.4.10) is f (r) = A J0 (mp r) + B J0 (ms r) + C Y0 (mp r) + D Y0 (ms r) ,

(3.4.12)

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 f ω2 d 2 (3.4.13a) (1 − ν) f + + νk + (2 − ν) 2 f = 0, dr r 2cs f (3.4.13b) (1 − ν)f − ν − νm2s f = 0 r on r = a. These give us a homogeneous linear system for the two remaining

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constants A and B, namely (1 − ν)m2p − (1 − 2ν)m2s − k 2 mp J1 (mp a) A + m2s − k 2 ms J1 (ms a) B = 0, (3.4.14a) mp J1 (mp a) − νm2s + (1 − ν)m2p a J0 (mp a) A + ms (J1 (ms a) − ms a J0 (ms a)) B = 0. (3.4.14b) 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 nonzero. There is no Love stress function available in this case, but we can try a particularly simple θ-dependence of the displacement ﬁeld, namely u(r, θ, z, t) = Ur (r) cos θer + Uθ (r) sin θeθ + Uz (r) cos θez ei(kz−ωt) . (3.4.15) 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 (3.4.16)

Hence, when we recall from (3.3.1) that 2 ∂ 2 2 − cp ∇ (div u) = 0, ∂t2 we ﬁnd that f (r) again satisﬁes Bessel’s equation 1 1 f + f + m2p − 2 f = 0. r r

(3.4.17)

(3.4.18)

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

(3.4.19)

Having found div u, we can use (3.3.2) to determine the displacement components. To calculate ∇2 u in polar coordinates, we have to use (1.7.11) and

3.4 Waves in cylinders and spheres

127

the polar coordinate formulae for div, grad and curl given in Appendix A7.1 to obtain 2 Ur 2Uθ 2 2 − Ur + + k Ur − 2 cos θer ∇ u= r r2 r Uθ 2 2Ur 2 − + Uθ + + k Uθ − 2 sin θeθ r r2 r Uz 1 2 + Uz + + k Uz cos θez ei(kz−ωt) . (3.4.20) − r r2 Similarly, by taking the grad of (3.4.16) in polar coordinates, we ﬁnd that f (r) sin θeθ + ikf (r) cos θez ei(kx−ωt) . grad div u = f (r) cos θer − r (3.4.21) Just considering the ez -component of (3.3.2), we ﬁnd that Uz must satisfy c2p 1 1 2 (3.4.22) Uz + Uz + ms − 2 Uz = 1 − 2 ikf, r r cs with f given by (3.4.19). The bounded solution of this inhomogeneous Bessel equation is ikA Uz = − 2 J1 (mp r) + B J1 (ms r), (3.4.23) k + m2p where B is an arbitrary constant. Next we can use (3.4.16) and (3.4.23) to determine Uθ as Uθ =

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

Finally, the er -component of (3.3.2) implies that c2p 2 2 2 r2 f . rUθ + Uθ + ikr Uz + 1 − ms r Ur = c2s

(3.4.24)

(3.4.25)

By substituting for Uθ , Uz and f into this expression, we obtain a single diﬀerential equation for Ur , namely 3 J1 (ms r) Ur + Ur + m2s Ur = −2ikB r r A 2 2 2 2 J1 (mp r) . (3.4.26) mp (mp − ms ) J0 (mp r) + (mp + ms ) + 2 k + m2p r

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Linear elastodynamics

z

t=0

2π/5ω

4π/5ω

6π/5ω

8π/5ω

Fig. 3.5 Illustration of ﬂexural waves.

Since the general bounded solution of (3.4.26) is Ur =

ikB J0 (ms r) − J2 (ms r) 2ms mp A J1 (ms r) − J , (3.4.27) (m r) − J (m r) +C 0 p 2 p 2(k 2 + m2p ) r

where C is a further arbitrary constant, we can ﬁnally obtain Uθ from (3.4.24) as 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 (3.4.28) Now we can 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 this condition leads to the dispersion relation for ﬂexural waves. We illustrate a typical displacement in Figure 3.5 which shows typical waves corresponding to bending, or ﬂexing, of the cylinder propagating in the z-direction.

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 .

(3.4.29)

3.4 Waves in cylinders and spheres

129

The Navier equation (1.11.17) in spherical polar coordinates thus reduces to the scalar equation ur 1 ∂ 2 ur ∂ 2 ur 2 ∂ur −2 2. = + 2 2 2 cp ∂t ∂r r ∂r r

(3.4.30)

One can easily check that the displacement ﬁeld (3.4.29) satisﬁes ∇×u = 0, which is why (3.4.30) 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 (3.4.30) takes the form ∂ 1 ∂ 2 φ ∂ 2 φ 2 ∂φ = 0, − 2 − ∂r c2p ∂t2 ∂r r ∂r

(3.4.31)

(3.4.32)

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

(3.4.33)

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 (3.4.33) yields 2 ∂2 2 ∂ (rφ) − c (rφ) = 0. p ∂t2 ∂r2

(3.4.34)

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 (3.4.34) becomes 2 ω d2 (rf ) = 0, (3.4.35) (rf ) + dr2 c2p whose general solution is f=

A sin(ωr/cp ) + B cos(ωr/cp ) , r

where A and B are two arbitrary constants.

(3.4.36)

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Linear elastodynamics

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 Section 1.11.3, we obtain

2 − 2c2 ) 2(c ur ∂ur p s + 2λ = ρe−iωt c2p f + f (3.4.37) τrr = (λ + 2µ) ∂r r r and, with f (r) given by (3.4.36), we ﬁnd that the zero-stress condition on r = a reduces to −1 tan(ωa/cp ) ω 2 a2 = 1− . (3.4.38) 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 real solutions for ω, corresponding to the natural frequencies for radial oscillations 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 (1.11.17), 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 (1.11.17) reduces to ∂uφ uφ 1 ∂ 1 ∂ 2 uφ 1 ∂ 2 ∂uφ r + 2 sin θ − 2 2 , (3.4.39) = 2 c2s ∂t2 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 .

(3.4.40)

By separating the variables in the usual way, we ﬁnd that the functions f and g must satisfy 2 2 ω r 2 r f + 2rf + − κ f = 0, (3.4.41a) c2s (3.4.41b) g + cot θg + κ − cosec2 θ g = 0, 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

3.4 Waves in cylinders and spheres

131

positive integer.† In this case, the relevant solution of (3.4.41b) is d g(θ) = A Pn (cos θ) , (3.4.42) dθ where A is an arbitrary constant and Pn is the Legendre polynomial of degree n (Gradshteyn & Ryzhik, 1994, Section 8.91). These polynomials are deﬁned by 1 dn 2 n (z , (3.4.43) − 1) Pn (z) = n 2 n! dz n and the ﬁrst few cases are P1 (z) = z,

P2 (z) =

3z 2 − 1 , 2

P3 (z) =

5z 3 − 3z . 2

(3.4.44)

With κ = n(n + 1), the general solution of (3.4.41a) that is bounded as r → 0 is B Jn+1/2 (ωr/cs ) √ f (r) = , (3.4.45) r where B is an arbitrary constant. Fortunately, the Bessel function Jn+1/2 takes a relatively simple form when n is an integer, namely ) 2 n+1/2 1 d n sin z n Jn+1/2 (z) = (−1) z , (3.4.46) π z dz z and the ﬁrst few cases are ) 2 sin z − cos z , J3/2 (z) = πz z ) 2 (3 − z 2 ) sin z 3 cos z , − J5/2 (z) = πz z2 z ) 2 3(5 − 2z 2 ) sin z (z 2 − 15) cos z . + J7/2 (z) = πz z3 z2

(3.4.47a) (3.4.47b) (3.4.47c)

Now, for a stress-free spherical boundary at r = a, the appropriate boundary condition is ∂uφ uφ τrφ = µ − = 0, (3.4.48) ∂r r and the corresponding condition for f is f (a) − †

f (a) = 0. a

(3.4.49)

This may be established by writing g(θ) as a power series in cos θ and then requiring the series to converge as cos θ → ±1.

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Linear elastodynamics

By substituting for f from (3.4.45), with C = 0, and simplifying, we arrive at the equation (n − 1) Jn+1/2 (ωa/c) = (ωa/c) Jn+3/2 (ωa/c)

(3.4.50)

for the natural frequencies ω. For each integer value of n, (3.4.50) 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 the time domain, 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. The simplest example is the one-dimensional wave equation (3.1.1), whose general solution is w(x, t) = F (x − ct) + G(x + ct), (3.5.1) where F and G are arbitrary functions, and c = T / is again the constant wave-speed. By this we mean that any solution of (3.1.1) may be expressed in the form (3.5.1), and this is readily veriﬁed for all the simple solutions given in Section 3.2. If, for example, we are given the initial conditions w = f (x),

∂w = g(x) ∂t

at

t = 0,

(3.5.2)

then, as in Exercise 3.12, we can easily evaluate the corresponding F and G to obtain the d’Alembert solution 1 x+ct 1 f (x − ct) + f (x + ct) + g(s) ds. (3.5.3) w(x, t) = 2 2c x−ct

3.5 Initial-value problems

133

In more than one space dimension, there is usually no general solution akin to (3.5.1) 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 Section 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 Section 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 (3.1.1) 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), (3.5.4) ∂t2 ∂x2 where δ is the Dirac delta-function deﬁned in Section 2.9 and w is assumed to be zero for t < 0. The right-hand side of (3.5.4) indicates that there is a jump in ∂w/∂t at t = 0, and we can therefore restate (3.5.4) as the initial-value problem 2 ∂2w 2∂ w − c = 0, in t > 0, (3.5.5a) ∂t2 ∂x2 ∂w = δ(x) at t = 0. (3.5.5b) w = 0, ∂t Substitution of (3.5.5) into the d’Alembert solution (3.5.3) leads to the fundamental solution   1 −ct < x < ct, (3.5.6a) w = R(x, t) = 2c 0 otherwise,

which is also known as the Riemann function for the one-dimensional wave equation. It may conveniently be recast as 1 R(x, t) = H(x + ct) − H(x − ct) , (3.5.6b) 2c

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R(x, t)

t

x

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

which is explicitly in the form of (3.5.1), where 0 s < 0, H(s) = 1 s > 0,

(3.5.7)

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

(3.5.8)

then we easily ﬁnd that w=

1 δ(x + ct) + δ(x − ct) . 2

(3.5.9)

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 (s) = δ(s), it follows that w = ∂R/∂t, and we can thus write the general d’Alembert solution (3.5.3) as a superposition of fundamental solutions, namely ∞ ∞ ∂ f (s)R(x − s, t) ds + g(s)R(x − s, t) ds. (3.5.10) w(x, t) = ∂t ∞ −∞ The Riemann function thus plays an analogous rˆ ole 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 †

for example noting that d(|s|)/ds = H(s) + 1 and using Exercise 2.23

3.5 Initial-value problems

135

R(r, t)

r

t

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

outside, and that ∂w/∂t is zero everywhere except on the the lines x = ±ct. Using (3.5.3), 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 (3.5.1) to guide us. Nevertheless, the wave equation (3.1.2) can still be solved subject to singular initial data analogous to (3.5.5), for example by seeking a solution in plane polar coordinates, as in Exercise 3.13. This reveals that the solution of (3.1.2) with ∂w = δ(x)δ(y) at t = 0, (3.5.11) w = 0, ∂t is H (ct − r) √ , (3.5.12) w = R(x, y, t) = 2πc c2 t2 − r2 where r2 = x2 +y 2 and H again denotes the Heaviside function. The function R(r, t) deﬁned by (3.5.12) is plotted in Figure 3.7. We see that, like the onedimensional 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 retreating edge or “tail” is not, and ∂R/∂t is nonzero everywhere inside the cone. 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) ∂t

at t = 0,

(3.5.13)

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as a linear superposition of fundamental solutions, namely

∞

w(x, y, t) = −∞

∞

g(ξ, η)R(x − ξ, y − η, t) dξdη ∞ ∞ ∂ f (ξ, η)R(x − ξ, y − η, t) dξdη. (3.5.14) + ∂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 (3.4.33), the radially-symmetric wave equation takes the form ∂ 2 w 2 ∂w 1 ∂2w , = ∇2 w = + 2 2 c ∂t ∂r2 r ∂r

(3.5.15)

where now r2 = x2 + y 2 + z 2 , and the transformation in (3.4.34) leads to the general solution w(r, t) =

F (r − ct) + G(r + ct) , r

(3.5.16)

where F and G are two arbitrary scalar functions. This illustrates that, as in one-dimensional waves, a localised radially-symmetric disturbance propagates outwards with a sharp leading edge 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 (3.5.16) to construct the three-dimensional Riemann function R(x, y, z, t) =

δ(r − ct) . 4πcr

(3.5.17)

This enables us to write the general solution to the three-dimensional initialvalue problem in a form analogous to (3.5.14), known as the retarded potential solution (Ockendon et al., 2003, p. 121). Remarkably, a retarded potential solution can also be written down for the full three-dimensional Navier equation, as shown in Love (1944, Section 211). 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 Initial-value problems

137

3.5.3 Characteristics The concept of characteristics for hyperbolic partial diﬀerential equations gives fundamental insight into wave propagation. There are several ways of thinking about characteristics mathematically but, for our purposes, it is most convenient to generalise (3.5.8), (3.5.9) 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 . As in (3.2.28b), 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 .

(3.5.18)

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 3.14 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 intersections of the characteristic cone through (x0 , t0 ) with the planes t = T , T > t0 are called the wave-fronts of the disturbance originating at (x0 , t0 ). For example, for the one-dimensional wave equation (3.1.1), the planes (3.5.18) are simply x ± ct = x0 ± ct0 , and they coincide with the characteristics. For higher-dimensional scalar wave equations for which ω(k) = ±ck,

(3.5.19)

|x − x0 |2 = c2 (t − t0 )2

(3.5.20)

we can notice that the cone

has normal (k, ω) in (x, t)-space and is hence the envelope of (3.5.18) as k varies. This is illustrated in two dimensions in Figure 3.8 and Exercise 3.14 gives a derivation of this result. The wave-fronts are circles or spheres in two or three dimensions respectively. For the unsteady Navier equation, Section 3.2.4 shows that (3.5.19) generalises to ω 2 − c2s |k|2 ω 2 − c2p |k|2 = 0 and hence the characteristic surface through x = y = z = t = 0 is a two-sheeted cone 2 (3.5.21) x + y 2 + z 2 − c2p t2 x2 + y 2 + z 2 − c2s t2 = 0, as illustrated in Figure 3.9 for just two spatial dimensions. This is 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, namely the speed of light.

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Linear elastodynamics 2

2

2 2

x +y =c t

k1 x + k2 y = ωt

t

y

x

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

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.

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

3.6 Moving singularities

w

139

w

(a)

(b)

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.

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

(3.6.1a)

where, as usual, 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.

(3.6.1b)

This problem may be solved, for example, by changing dependent variables to ξ = x − ct and η = x + ct, where the speed of sound is c = T / , to obtain, as in Exercise 3.15, w=

Pc (c + V )|x − ct| + (c − V )|x + ct| − 2c|x − V t| , (3.6.2) 4T (c2 − V 2 )

which is a single-valued function of x when c > V . 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 approaches the speed of sound c. The assumption of small slope that underpins the model (3.6.1a) 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 3.15 and Figure 3.10(d). In addition, we note that, no matter

140

Linear elastodynamics

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 (3.6.1), 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

(3.6.3)

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 (3.6.4) + = − δ(ξ)δ(y), 1 − M2 2 2 ∂ξ ∂y T where c = T /ς and the Mach number is deﬁned by M=

V . c

(3.6.5)

When M < 1, so the force is moving subsonically, (3.6.4) is an elliptic partial diﬀerential equation, and w is thus just a multiple of the √ Green’s function, as in Section 2.9. By changing variables again to X = ξ/ 1 − M 2 and y, so that (3.6.4) is transformed to Poisson’s equation (see Exercise 3.16), the solution is easily found to be P (x − V t)2 2 w=− √ (3.6.6) log + y + const.. 1 − M2 2T 1 − M 2 Notice that the displacement is singular (as expected) as the point force (V t, 0) is approached, and that 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 (3.6.4) is a hyperbolic partial diﬀerential equation. In fact, it is analogous to the one-dimensional wave equation (3.5.4), and we can infer the solution directly from (3.5.6) as  ξ ξ   √P H y−√ −H y+ √ , ξ < 0, M2 − 1 M2 − 1 w = 2T M 2 − 1  0, ξ > 0. (3.6.7)

3.6 Moving singularities

(a)

y

141

increasing t x

(b)

y x

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

Notice that we have made the “causality” assumption that w = 0 ahead of the point force. In terms of physical variables, (3.6.7) may be written as    √ P , w = T M2 − 1  0,

x−Vt Vt−x √
(3.6.8)

and the inﬂuence the force is thus conﬁned to a Mach cone, whose apex of √ −1 2 angle is tan 1/ M − 1 ≡ sin−1 (1/M ). This may again be visualised by plotting wave-fronts caused by the moving force, as in Figure 3.11(b). Since the force moves supersonically, it overtakes the wave-fronts generated by itself. A similar wave pattern occurs behind a supersonic aircraft and is responsible for the notorious “sonic boom”. As was the case with (3.6.2), the subsonic (3.6.6) and supersonic (3.6.8) solutions both grow without bound as the Mach number approaches unity. This clearly violates the assumption of small displacements under which the model (3.6.3) for an elastic membrane is valid. To describe the behaviour as M passes through unity it is therefore necessary to incorporate nonlinear eﬀects into (3.6.3).

142

(a)

Linear elastodynamics

y

(b)

(c)

x

Fig. 3.12 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 .

Now let us apply the same ideas to plane strain driven by a moving point force in the x-direction, which is described by the equations 2 2 ∂ u ∂2u ∂ u ∂2v ∂2u −µ + + 2 = P δ(x − V t)δ(y), ρ 2 − (λ + µ) ∂t ∂x2 ∂x∂y ∂x2 ∂y (3.6.9a) ∂2v ∂2v ∂2u ∂2v ∂2v = 0. (3.6.9b) + 2 −µ + ρ 2 − (λ + µ) ∂t ∂x∂y ∂y ∂x2 ∂y 2 These may be conveniently cast in terms of the Love stress function L deﬁned by (3.3.6), which must now satisfy P 1 ∂2 1 ∂2 2 2 ∇ − 2 2 ∇ − 2 2 L=− δ(x − V t)δ(y). (3.6.10) cp ∂t cs ∂t 1−ν We again seek a travelling wave solution in which L is a function of ξ = x − V t and y, in which case L must satisfy 2 2 P ∂2 ∂2 2 ∂ 2 ∂ δ(ξ)δ(y), (1 − Ms ) 2 + 2 L = − (1 − Mp ) 2 + 2 ∂ξ ∂y ∂ξ ∂y 1−ν (3.6.11) where we now deﬁne two Mach numbers Mp =

V , cp

Ms =

V . cs

(3.6.12)

Although the solution of (3.6.11) is quite complicated mathematically, the qualitative behaviour may be understood by considering the wave-fronts generated by the moving force, by analogy with Figure 3.11.

Exercises

143

If Mp < Ms < 1, then the motion is subsonic with respect to both P and S -waves, and (3.6.11) 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, (3.6.11) 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. 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, 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 Section 1.8), in which cp → ∞ and div u = 0, so that A = 0 in (3.2.34) 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 chapter. These include ﬁnite-amplitude waves, to be discussed in Chapter 5, and wave propagation through inhomogeneous media, to which we will return brieﬂy in Chapter 9. Exercises 3.1 †

Let a mass m be attached to a string with line density and tension T . Using the boundary conditions (3.2.29) at this point, show

The answer is now known to be no.

144

Linear elastodynamics

that the reﬂection and transmission coeﬃcients for an incoming wave ei(kx−ωt) are cR = 3.2

3.3

imk/2 , 1 − imk/2

cT =

1 . 1 − imk/2

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 continuity of the displacement and the force at x = 0, show that jump in density transmits a wave with am this plitude cT = 2/ 1 + − / + and reﬂects a wave with amplitude cR = cT − 1. The Helmholtz equation in plane polar coordinates is 1 ∂2A ∂ 2 A 1 ∂A + 2 2 + k 2 A = 0. + 2 ∂r r ∂r r ∂θ To study the far ﬁeld scattered from a localised obstacle, write r = R/ε, where 0 < ε 1, and use the ansatz A ∼ A0 eiku(R)/ε , where A0 and u are independent of ε. Show that A0 du 2 ∂A0 + = 0, = 1, and 2 dR ∂R R and deduce that outward-propagating waves take the form A ∼ A+ (θ)r−1/2 eikr

3.4

3.5

as

r → ∞.

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.] Suppose that an elastic medium occupies the half-space x < 0 and that the face x = 0 is held ﬁxed. A plane S-wave is incident from x → −∞ with sin β uinc = exp{iks (x cos β + y sin β) − iωt}, − cos β where ks = ω/cs . Show that the reﬂected wave takes the form sin β exp{iks (−x cos β + y sin β) − iωt} uref = rs cos β − cos α exp{ikp (−x cos α + y sin α) − iωt}, +rp sin α

Exercises

145

where kp = ω/cp and the reﬂection angle α of the P -wave satisﬁes sin β sin α = . cp cs Show also that the reﬂection coeﬃcients are given by rs =

3.6

cos(β + α) , cos(β − α)

rp =

sin(2β) . cos(β − α)

What happens when β > sin−1 (cs /cp )? Suppose that the face x = 0 separates elastic media characterised by Lam´e constants λ− , µ− in x < 0 and λ+ , µ+ in x > 0. Derive the jump conditions ∂u ∂v + ∂u ∂u ∂v + µ = 2µ =0 + +λ + ∂y ∂x − ∂x ∂x ∂y − to be satisﬁed across x = 0. If an incident S -wave as in Exercise 3.5, with wave-vector (cos β− , sin β− ), generates reﬂected and transmitted S - and P -waves with wave-vectors (− cos β− , sin β− ), (− cos α− , sin α− ) and (cos β+ , sin β+ ), (cos α+ , sin α+ ) respectively, show that sin α− sin β− sin α+ sin β+ = = = . cp− cs− cp+ cs+

3.7

How does the number of transmitted waves vary as β− increases from zero? (a) Show from (3.2.52), (3.2.55) and (3.2.57) that the dispersion relation for symmetric Love waves is given implicitly by * * * 2 ω2 ω2 ω 2 tan h 2 2− k − k − k . = µ µ1 2 c2s1 c2s1 c2s2 Write s = h ω 2 /c2s1 − k 2 , α = cs1 /cs2 , β = µ1 /µ2 to obtain the parametric representation α s α2 + β 2 tan2 s hω √ =√ s 1 + β 2 tan2 s, kh = , cs2 1 − α2 1 − α2 (E3.1) where nπ s < (n + 1/2)π and n is an integer. (b) Show that antisymmetric Love waves satisfy the condition (3.2.58) and deduce that their dispersion relation is given parametrically by (E3.1) with tan s replaced by cot s and (n − 1/2)π s < nπ.

146

Linear elastodynamics

cs2

cg

cs1 k Fig. 3.13 Group velocity cg = dω/dk versus wave-number k for symmetric (solid) and antisymmetric (dotted) modes of Love waves.

(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 dω . dk Use (E3.1) to ﬁnd cg as a function of s for Love waves, and hence plot cg versus k. Deduce that, as shown in Figure 3.13, the group velocity 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 (see Tayler, 2002, Chapter 2).] Show that up = ap (k, −iκp )T and us = as (κs , −ik)T for plane P waves and S-waves of the form (3.2.62), where the amplitudes ap,s are unknown. Show also that the boundary conditions (3.2.61) lead to 2kκp ap + (k 2 + κ2s )as = 0, 2 2 cp (κp − k 2 ) + 2c2s k 2 ap + 2c2s kκs as = 0, cg (k) =

3.8

and deduce that nonzero amplitudes exist only 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 the dispersion relation (3.2.65) for Rayleigh waves. Show graphically that there is only one real solution to this equation for the propagation speed c.

Exercises

3.9

147

In dynamic plane strain, suppose that the stress components are given by τxx =

1 ∂2φ ∂2U 1 ∂2U ∂2U + − , τ = − , xy ∂y 2 2c2s ∂t2 ∂x∂y 2c2s ∂t2 ∂2U 1 ∂2U τyy = − , ∂x2 2c2s ∂t2

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 2 2 2 ∂ ∂ ∂ 2 2 2 2 2 2 − cs ∇ φ = − cp ∇ − cs ∇ U = 0. ∂t2 ∂t2 ∂t2 3.10

Show that the general solution of the scalar wave equation 2 2 2 2 ∂ ∂ 2 ∂ 2 ∂ U=0 − c − c p s ∂t2 ∂x2 ∂t2 ∂x2 is U = Fp (x − cp t) + Gp (x + cp t) + Fs (x − cs t) + Gs (x + cs t) ,

3.11

where Fp , Gp , Fs , Gs are arbitrary scalar functions. Show that the dispersion relation (3.4.50) for torsional waves in a sphere reduces to tan η 3 = η 3 − η2 tan η 12 − η 2 = η 12 − 5η 2 tan η 75 − 8η 2 = η 75 − 33η 2 + η 4

when n = 1, when n = 2, 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 each 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 → ∞.

148

3.12

3.13

Linear elastodynamics

Show that, after a suitable change of independent variables, the onedimensional wave equation (3.1.1) can be written as ∂2w = 0. ∂η∂ξ Hence obtain the d’Alembert solution (3.5.1). (a) Show that the axially symmetric wave equation 1 ∂2w ∂ 2 w 1 ∂w 2 = ∇ w = + c2 ∂t2 ∂r2 r ∂r admits similarity solutions of the form 1 r w(r, t) = f , t t where f (η) satisﬁes 2 d 2 df 2 η c −η − η f = 0. dη dη

(E3.2)

(b) Assuming that f is well-deﬁned as η → 0, show that A f (η) = , c2 − η 2

0 η < c,

where A is an arbitrary constant. (c) Show that ct r dr √ = ct 2 c t2 − r 2 0 and deduce that  1  √ , 0 r < ct, 2 w(r, t) = 2πc c t2 − r2  0, r > ct, satisﬁes (E3.2) subject to ∂w = δ(x)δ(y) at t = 0. ∂t The characteristic surface through the origin for the two-dimensional wave equation (3.1.2) is deﬁned to be the envelope of the planes (3.5.18) subject to the dispersion relation (3.5.19). Show that it is equivalent to the envelope of the planes w = 0,

3.14

(x − x0 ) cos θ + (y − y0 ) 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.]

Exercises

149

In higher dimensions, show that xi − x0i =

3.15

cki (t − t0 ) |k|

on the envelope of (3.5.18), and deduce (3.5.20). Show that, in terms of the variables ξ = x − ct and η = x + ct, the one-dimensional wave equation (3.6.1a) subject to a moving point source is transformed to Pc c−V ∂2w = δ ξ+ η . (E3.3) ∂ξ∂η 2T (c + V ) c+V Using the fact that, from (2.9.7a), d2 |z| = 2δ(z), dz 2 deduce that the general solution of (E3.3) is Pc (c + V )ξ + (c − V )η + F (ξ) + G(η), w= 2 2 4T (c − V ) where F and G are arbitrary functions. Transform back to (x, t) and apply the initial conditions to obtain the solution (3.6.2) when c>V. When V > c, write x − V t = X and show that (3.6.1a) becomes

3.16

V 2 − c2

∂2w ∂2w ∂2w P c2 + δ(X). − 2V = ∂X 2 ∂X∂t ∂t2 T

Deduce that ∂w/∂X suﬀers a jump of P c2 /T (V 2 − c2 ) across x = V t and hence obtain (3.6.2). Prove the identity 1 δ(x), δ(ax) ≡ |a| where δ√is the Dirac delta-function. Hence show that, in terms of X = ξ/ 1 − M 2 and y, (3.6.4) reads ∂2w ∂2w P + =− √ δ(X)δ(y) 2 2 ∂X ∂y T 1 − M2 and deduce that

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

4 Approximate theories

4.1 Introduction So far in this book we have been considering 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 Section 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, such as the skin of a drum, which supports only in-plane tensions. A bar, on the other hand, is a nearly one-dimensional solid that can be subject to either tension or compression. However, many thin elastic bodies also have an appreciable bending stiﬀness and therefore 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 pane 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 150

4.2 Longitudinal displacement of a bar

151

δx T (x, t)

T (x + δx, t)

x Fig. 4.1 The forces acting on a small length δx of a uniform bar.

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 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 geometric nonlinearity, when the displacements are large but the strain is small. The nonlinear models that result exhibit interesting non-uniqueness 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 propagating 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 Section 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 ∂2u (4.2.1) T (x + δx, t) − T (x, t) = ρAδx 2 (x, t). ∂t In the limit δx → 0, this reduces to the partial diﬀerential equation ∂2u ∂T = ρA 2 . ∂x ∂t

(4.2.2)

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

152

Approximate theories

∂u/∂x, by analogy with Hooke’s law. Motivated by the exact static solution obtained in Section 2.2.3, we postulate the constitutive relation ∂u , (4.2.3) ∂x where E again denotes Young’s modulus. By substituting (4.2.3) into (4.2.2), we ﬁnd that u(x, t) satisﬁes the wave equation T = EA

E

∂2u ∂2u = ρ , ∂x2 ∂t2

(4.2.4) 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 Section 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 in each cross-section of the bar, and that the constitutive relation (4.2.3) holds, although its derivation in Section 2.2.3 was performed only under static conditions. Neither of these assumptions is exactly true in practice, reﬂecting the fact that solutions of (4.2.4) are not exact solutions of the Navier equations. Intuitively, we expect (4.2.4) 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 force experienced by a small segment of undisturbed length δx is x+δx 0 cos θ δx, + T − g sin θ x where denotes the line density.

4.4 Transverse displacements of a beam

153

T (x + δx, t) θ g

T (x, t) w

x δx Fig. 4.2 The forces acting on a small length δx of an elastic string.

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 ∂w , cos θ ∼ 1. (4.3.1) sin θ ∼ ∂x At the same time, we suppose that there is no longitudinal displacement so that the acceleration is just ∂ 2 w/∂t2 in the vertical direction. By applying Newton’s second law to the segment and letting δx → 0, we thus obtain the equations ∂2w ∂2w ∂T = 0, T 2 − g = 2 . (4.3.2) ∂x ∂x ∂t 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 make the same observation on mathematical grounds by noting that (4.3.2b) 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., 2003, p. 41). 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;

154

Approximate theories 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 The forces and moments acting on a small segment of an elastic beam.

for convenience N is deﬁned to act perpendicular to the x-axis. Again considering small, purely transverse displacements, we obtain the equations ∂ 2 w ∂N ∂2w ∂T = 0, T 2 + − g = 2 . (4.4.1) ∂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 Section 2.6.7 that each section of the beam exerts on its neighbours a bending moment M about the y-axis, as well as the tangential and normal force components T and N . 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 ∂2θ + Iˆ 2 , 2 ∂t

(4.4.2)

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 (4.4.2) is thus negligible and we are left with ∂M + N = 0. (4.4.3) ∂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, this suggests a constitutive relation of the form −

M = −B

∂2w , ∂x2

(4.4.4)

where B is a constant known as the bending stiﬀness. We note that (4.4.3) and (4.4.4) are consistent with the semi-inﬁnite strip results (2.6.70) and (2.6.73) respectively.

4.4 Transverse displacements of a beam

(a)

(b)

155

(c)

x

x

x

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

As in (4.2.3), 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 (2.8.30) in which the displacement ﬁeld and the only nonzero stress component are     −2xz u κ , (4.4.5) u = v  =  τxx = −κEz. 2νyz 2 2 2 2 x − νy + νz w 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 zτxx dydz = −Eκ z 2 dydz, (4.4.6) M= A

A

which corresponds to (4.4.4) with B = EI,

where

z 2 dydz

I=

(4.4.7)

A

is the moment of inertia of the cross-section A about the y-axis. Postulating that the constitutive relation (4.4.4) still holds under dynamic conditions, we combine (4.4.1)–(4.4.7), to obtain the beam equation −EI

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

(4.4.8)

Evidently, (4.4.8) reduces to (4.3.2) as the bending stiﬀness EI tends to zero. Unlike the wave equation, though, (4.4.8) is not immediately classiﬁable as elliptic or hyperbolic.† †

It is interesting to note that the real and imaginary parts of the solution ψ of the Schr¨ odinger equation ∂2 ψ ∂ψ =− ∂t 2m ∂x2 satisfy (4.4.8) with T = g = 0, /2m = EI/ . i

156

Approximate theories

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 (4.4.8) 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 w= = 0. (4.4.9) ∂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 ∂2w = 0. (4.4.10) w= ∂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 (4.4.3) and (4.4.4) we deduce the boundary conditions ∂3w ∂2w = = 0, (4.4.11) ∂x2 ∂x3 and in this case we must also have T ≡ 0. If we specify distributed 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 σx (y, z) dydz, N= σz (y, z) dydz, T = A A zσx (y, z) dydz. (4.4.12) M= A

Here we are eﬀectively invoking Saint-Venant’s principle: all details of the 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 Section 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 (4.4.8) compared with (4.3.2) is that we can now consider situations in which T < 0. We illustrate an

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

1

0.8

x

0.2

-0.5

-0.5

-0.5

-1

-1

-1

157

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).

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 (4.4.8) reduces to the ordinary diﬀerential equation EI

d4 w d2 w + P = 0, dx4 dx2

(4.4.13a)

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

at

x = 0, L.

(4.4.13b)

We can consider (4.4.13) 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 n2 π 2 P = , EI L2

(4.4.14)

where n is an integer, then (4.4.13) admits the eigenfunction solution w = A cos

nπx L

,

(4.4.15)

where A is a constant. The ﬁrst three such eigenfunctions are shown in Figure 4.5; these are the shapes that a beam can adopt as it buckles under a suﬃciently large compressive force. The amplitude A appears to be indeterminate, although we might expect it to be a smoothly increasing function of the applied force. This is a deﬁciency of the model that will be addressed in Section 4.9.

158

Approximate theories

4.4.4 Waves on a beam Concerning the dynamic equation (4.4.8) with g = 0 and adopting the philosophy of Section 3.2.2, we can seek travelling-wave solutions of the form u = A exp i(kx − ωt) , (4.4.16) where the real part is assumed as usual, and hence obtain the dispersion relation ω 2 = EIk 4 + T k 2 .

(4.4.17)

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 wave-length 2π/k provided * 1/2 EIk 2 T ω = 1+ . (4.4.18) k T Evidently this reduces 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 wave-length 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, wave-like solutions of (4.4.8) exist only if k 2 > P/EI. For smaller values of k, (4.4.17) leads to complex values of ω, corresponding to solutions that grow exponentially in time. It follows that the beam is unstable to waves with wave-length greater than 2π EI/P , and we will investigate this further in Section 4.9.3.

4.5 Linear rod theory We next consider the motion of a nearly straight uniform elastic rod that can bend in two transverse T directions, so that the displacement is w(x, 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

4.5 Linear rod theory

z

159

Nz

y

Ny T x

Fig. 4.6 The internal force components T , Ny and Nz in a thin elastic rod.

z

y Mz My

Mx

x

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

uniform and obtain the following generalisation of (4.4.8): T

∂2w ∂ 2 w ∂N + , + f = ∂x2 ∂x ∂t2

(4.5.1)

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 Section 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 moments in the y- and z-directions analogous to (4.4.3) then leads to ∂My − Nz = 0, ∂x

∂Mz + Ny = 0. ∂x

(4.5.2)

Finally, we need to generalise the constitutive relation (4.4.4), and we will again use the plane stress solution (4.4.5) as a guide. By combining (4.4.5) with an analogous expression representing bending in the y-direction, we

160

Approximate theories

obtain the following exact solution of the Navier equations:     −2xz −2xy κ κz   + y x2 + νy 2 − νz 2  , u= 2νyz 2 2 2νyz x2 − νy 2 + νz 2 τxx = −κz Ez − κy Ey.

(4.5.3a) (4.5.3b)

The two constants κy and κz represent bending in the (x, y)-plane and the (x, z)-plane respectively. The moments about the y- and z-axes generated over the cross-section A by this stress ﬁeld are given by zτxx dydz, Mz = − yτxx dydz. (4.5.4) My = A

A

By substituting (4.5.3) into (4.5.4) we obtain the relations ∂2w Mz = EI 2 , −My ∂x where the “inertia tensor” Iyy I= Iyz

Iyz Izz

= A

y 2 yz yz z 2

(4.5.5)

dydz

(4.5.6)

characterises the bending stiﬀness in each direction. Hence the generalisation of (4.4.8) is the coupled system ∂4w ∂2w ∂2w − EI + f = . (4.5.7) ∂x2 ∂x4 ∂t2 Note that we can choose the y- and z-axes such that the symmetric positive deﬁnite 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 (4.5.7) 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 (4.4.8), 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. The conditions corresponding to clamped, simply supported and free ends are obvious generalisations of (4.4.9), (4.4.10) and (4.4.11). Examples of values of Iyy and Izz for various cross-section shapes with respect to principal axes are given in Figure 4.8. These are crucial pieces of information to bear in mind when designing beams or rods in practice. By considering (4.5.4), we observe that Iyy measures resistance to bending about the z-axis, while Izz represents resistance to bending about the y-axis. For example, if one considers the bending of a long strip of paper with T

4.5 Linear rod theory

161

(b) (a)

a

Izz = a3 b/12

b

a Iyy = ab3 /12

b (c)

Iyy = Izz = πab(a2 + 4b2 )/4 Izz Iyy < Izz

Fig. 4.8 Examples of cross-sections in the (y, z)-plane and their bending stiﬀnesses.

transverse dimensions a and b in the z- and y-directions respectively, where b a, it is then easy to see as in Figure 4.8(a) that Izz Iyy . 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. Meanwhile, the “I-beam” in Figure 4.8(c) strongly resists bending about the y-axis. Proceeding along the same lines as for longitudinal waves, we now consider torsion, i.e. twisting a rod about its 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 ∂Mx ∂2θ = (Iyy + Izz ) 2 , ∂x ∂t

(4.5.8)

where θ(x, t) denotes the angle of twist about the x-axis. Now we refer to the exact steady solution of Section 2.4 to motivate the constitutive relation ∂θ (4.5.9) Mx = R ∂x

162

Approximate theories

between the torque and the twisting gradient, where R again denotes the † torsional 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 πa4 µ πa4 , R= (4.5.10) 2 2 and hence that the wave speed is equal to cs . This reproduces the wave speed of the ﬁrst family of torsional waves found in Section 3.4.1; that is, (3.4.8) with n = 0 and ξ2,0 = 0: the theory presented here assumes each section deforms uniformly and thus fails to discern the higher modes. Iyy + Izz =

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 Txx Txy , (4.6.1) 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 Section 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 ∂Txy ∂Txx + = 0, ∂x ∂y and ∂ ∂x

∂Txy ∂Tyy + =0 ∂x ∂y

(4.6.2)

∂w ∂w Nx + Txx + Txy ∂x ∂y ∂ ∂2w ∂w ∂w + Ny + Txy + Tyy − ςg = ς 2 , (4.6.3) ∂y ∂x ∂y ∂t

where ς is now the mass of the plate per unit area. †

For consistency with the notation of this chapter, x should be substituted for z in the results of Section 2.4.

4.6 Linear plate theory

163

Ny Tyy Txy Txx δy Tyx

Nx Tyx

Nx

δx

z

Txx y Txy x

Tyy Ny

Fig. 4.9 The forces acting on a small section of an elastic plate.

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 ∂Mxy ∂Mxx + + Ny = 0, ∂x ∂y

∂Myy ∂Myx + − Nx = 0. ∂x ∂y

(4.6.4)

Next we seek constitutive equations for the bending moments which, by analogy with Section 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 suitable quadratic function of x and y in (2.8.28), we can construct a solution with displacement ﬁeld  1 u= 2

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

(4.6.5)

164

Approximate theories

z Mxy Myx

y Myy Mxx x

Fig. 4.10 The bending moments acting on a section of an elastic plate.

and nonzero stress components τxx = −

E(a + νc)z , 1 − ν2

τxy = −

Ebz , 1+ν

τyy = −

E(νa + c)z . 1 − ν2

(4.6.6)

Now by considering the moments exerted on surfaces x = const. and y = const. respectively, we obtain the expressions h/2 h/2 zτxy dz, Myx = zτxx dz, (4.6.7a) Mxx = − −h/2

Mxy = −

−h/2

h/2

−h/2

zτyy dz,

Myy =

h/2

−h/2

zτxy dz,

(4.6.7b)

where h is the plate thickness. Hence, Mij are the components of the tensor h/2 0 −1 τxx τxy z dz. 1 0 −h/2 τxy τyy By substituting (4.6.5) and (4.6.6) into (4.6.7) we obtain the relations ∂2w Eh3 , 12(1 + ν) ∂x∂y 2 ∂ w Eh3 ∂2w =− +ν 2 , 12 (1 − ν 2 ) ∂x2 ∂y ∂2w ∂2w Eh3 . ν = + 12 (1 − ν 2 ) ∂x2 ∂y 2

Mxx = −Myy = Myx Mxy

(4.6.8a) (4.6.8b) (4.6.8c)

4.6 Linear plate theory

165

It is reassuring that, in the notation of this section, the plane strain result (2.6.73) implies that Myx = −

µh3 ∂ 2 w , 6(1 − ν) ∂x2

(4.6.9)

which is consistent with (4.6.8b) when w = w(x). On the other hand, if we were to bend a plate about the y-axis while applying no moment about the x-axis, then we would have Mxy = 0 and therefore ∂2w ∂2w = −ν . ∂y 2 ∂x2

(4.6.10)

Thus, without the imposition of a transverse bending moment, the plate will automatically bend upwards in the y-direction when we attempt to bend it downwards in the x-direction. Unfortunately, it is diﬃcult to observe this eﬀect using a piece of paper because Poisson’s ratio is close to zero. By substituting (4.6.10) into (4.6.8b), we ﬁnd in this case that Myx =

Eh3 ∂ 2 w , 12 ∂x2

(4.6.11)

which is consistent with (4.4.4) and (4.4.7), since h3 /12 is the moment of inertia per unit length in the y-direction. Hence we see that the constant of proportionality between curvature and bending moment in one dimension depends on whether the transverse curvature or the transverse moment are set to zero. This is analogous to the biaxial strain examples considered in Section 2.2.4, where the eﬀective one-dimensional elastic modulus depends on whether zero stress or zero strain is imposed in the transverse direction. When we use the constitutive relations (4.6.8) in (4.6.3), we are left with the plate equation Txx

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

(4.6.12)

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

Eh3 12 (1 − ν 2 )

(4.6.13)

is the bending stiﬀness of the plate. A membrane is a plate with negligible bending stiﬀness, and is described by (4.6.12) with D = 0. Unfortunately, the problem is still under-determined since (4.6.2) 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 in-plane displacements of the plate. We leave this calculation to Section 4.6.5, here

166

Approximate theories

concentrating on the simple case where the plate is subjected to a known isotropic tension T . This corresponds to the trivial exact solution of (4.6.2) in which Txx = Tyy = T and Txy = 0, and reduces (4.6.12) to T ∇2 w − D∇4 w − ςg = ς

∂2w . ∂t2

(4.6.14)

Comparison with (4.4.8) reveals the diﬀerence in bending stiﬀness between a beam undergoing planar displacements and a plate deforming in two dimensions.

4.6.2 Boundary conditions As initial conditions for (4.6.14), we should specify w and ∂w/∂t. For a clamped plate, the boundary conditions are analogous to those for a beam, namely ∂w = 0, (4.6.15) w= ∂n where ∂/∂n is the normal derivative. For a simply supported plate, we require the displacement and the bending moment about an axis tangential to the boundary to be zero. The latter may be written in the form (−ny , nx ) Mxx Mxy nx = 0, (4.6.16) Msn = Myx Myy ny where n = (nx , ny )T is the outward-pointing unit normal to the boundary. By substituting for the moments from (4.6.8) and using the fact that w is zero on the boundary, we can reduce (4.6.16) to w=

∂w ∂2w = 0, + νκ 2 ∂n ∂n

(4.6.17)

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 (4.4.10) 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 (4.6.14)! This conceptual diﬃculty caused great consternation to the pioneers of plate theory (see Love, 1944, Article 297), and

4.6 Linear plate theory

167

a version of Saint-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 narrow region near the free edge where the assumptions that gave rise to the plate equation fail. The careful analysis of this layer carried out in Section 6.4 shows that the correct conditions at a free edge x = 0 are Myx = 0,

∂Myx ∂Mxx −2 =0 ∂x ∂y

(4.6.18)

∂3w ∂3w + (2 − ν) = 0. ∂x3 ∂x∂y 2

(4.6.19)

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

By using (4.6.4) and (4.6.8), we may also rewrite the second of (4.6.18) as Nx =

∂Mxx , ∂y

(4.6.20)

which shows that the shear stress Nx as calculated from plate theory does not in general vanish at a free edge! Note also that a discontinuity in the bending moment at a corner in the boundary results in a delta-function in the shear stress, corresponding to a point force in the transverse direction (see Timoshenko & Woinowsky-Krieger, 1959). This suggests the existence of a three-dimensional “Saint-Venant” region in the vicinity of the corner. It is also possible to obtain (4.6.18) as the natural boundary conditions associated with minimising the strain energy in the plate. Although it is beyond the scope of this chapter to derive this systematically, we note that, in the exact solution (4.6.5) where tension is neglected, the nonzero stress components are given by (4.6.6) and the corresponding strain components by exx = −az,

exy = −bz,

eyy = −cz.

(4.6.21)

Recalling the deﬁnition (1.9.5) of the strain energy density W, we ﬁnd that the stored energy per unit plate area is h/2 h/2 1 τij ij dz W dz = −h/2 −h/2 2 D 2 a + 2νac + c2 + 2(1 − ν)b2 = 2 D (4.6.22) = (a + c)2 + 2(1 − ν) b2 − ac . 2

168

Approximate theories

This suggests that, more generally, the net strain energy due to bending will take the form 2 2

2 2 ∂2w ∂2w ∂ w D dxdy, − U= ∇ w − 2(1 − ν) 2 2 2 ∂x ∂y ∂x∂y Ω (4.6.23) where the integral is taken over the region Ω in the (x, y)-plane occupied by the plate. The integrand in (4.6.23) shows that the total energy depends on a combination of the mean curvature ∇2 w = κ1 + κ2 and the Gaussian curvature ∂2w ∂2w − ∂x2 ∂y 2

∂2w ∂x∂y

(4.6.24)

2 = κ1 κ2 ,

(4.6.25)

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. Exercise 4.10 demonstrates that, if we minimise (4.6.23) using the calculus of variations, then w must satisfy the biharmonic equation, which agrees with the steady version of (4.6.14), when tension and gravity are neglected. In addition, Exercise 4.10 shows that (4.6.19) 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 (4.6.14) with T = 0, that is ςg (4.6.26) ∇4 w = − . D Solutions of (4.6.26) describe the sagging of a plate under gravity. We will consider both clamped and simply supported boundary conditions; it is impossible to solve (4.6.26) 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 (4.6.26) in cylindrical polar coordinates. With clamped boundary conditions w = dw/dr = 0 at r = a, we ﬁnd that 2 ςg 2 w=− (4.6.27) a − r2 . 64D Alternatively, the simply supported boundary conditions (4.6.17) at r = a lead to the solution 5+ν 2 ςg 2 2 2 a −r . (4.6.28) a −r w=− 64D 1+ν

4.6 Linear plate theory

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

169

0.8

10

Fig. 4.11 The displacement (4.6.29) of a simply supported rectangular plate sagging under gravity, with a = 1, b = 1/2, ςg/D = 1.

Note in particular that the reduced rigidity aﬀorded by the simple support compared with the clamped plate means that (4.6.28) gives a larger displacement at the centre of the disc by a factor of (5 + ν)/(1 + ν) compared with (4.6.27). 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 Section 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 supported boundary conditions w = ∂ 2 w/∂n2 = 0 are now physically realistic and it is straightforward to separate the variables and obtain the solution as a sum of mutually orthogonal trigonometric functions, namely −16ςg sin (mπx/a) sin (nπy/b) w= 6 . (4.6.29) π D mn (m2 /a2 + n2 /b2 )2 m,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 enhances the sag of the glass under gravity and, by suitably distributing the heating, any desired sag can be achieved in principle. If D is temperature- and hence position-dependent, and the in-plane tension is

170

Approximate theories

negligible, then the displacement satisﬁes 2 ∂2w ∂ w ∂2 ∂2w ∂2 D D + 2(1 − ν) + ν ∂x2 ∂x2 ∂y 2 ∂x∂y ∂x∂y 2 2 2 ∂ w ∂ w ∂ + ςg = 0. (4.6.30) +ν 2 + 2 D ∂y ∂y 2 ∂x Given D(x, y), (4.6.30) 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, (4.6.30) is a second-order linear partial diﬀerential 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 (4.6.30) are 2 2 ∂ w ∂2w ∂2D ∂2w ∂2D ∂2w ∂2D ∂ w + +ν 2 + 2(1 − ν) +ν 2 , ∂x2 ∂y ∂x2 ∂x∂y ∂x∂y ∂y 2 ∂x ∂y 2 we can read oﬀ the discriminant 2 2 2 2 ∂ w ∂ w ∂ w ∂2w ∂2w 2 − +ν 2 +ν 2 , ∆ = (1 − ν) ∂x∂y ∂x2 ∂y ∂y 2 ∂x

(4.6.31a)

and simple rearrangements of the right-hand side yield M 2 + Mxy Myx (4.6.31b) ∆ = xx D2 2 2 2 2 ∂ w ∂2w ∂2w ∂ w ∂2w − + . (4.6.31c) − ν = −(1 − ν)2 ∂x2 ∂y 2 ∂x∂y ∂x2 ∂y 2 Now, if there is a simply supported edge at x = 0, then we must have Myx = 0 there and we deduce from (4.6.31b) that ∆ > 0 and (4.6.30) is therefore hyperbolic. On the other hand, the right-hand side of (4.6.31c) 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 (4.6.30) 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 (see Salazar & Westbrook, 2004). 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

4.6 Linear plate theory

171

components Txx , Txy and Tyy . We appeal to the exact biaxial strain solution (2.2.19) to pose the constitutive relations Eh (exx + νeyy ) , 1 − ν2 Eh exy , = 1+ν Eh = (νexx + eyy ) , 1 − ν2

Txx = Txy = Tyx Tyy

(4.6.32a) (4.6.32b) (4.6.32c)

where the linearised strain components are related to in-plane displacements by ∂u 1 ∂u ∂v ∂v , exy = + , eyy = . (4.6.33) exx = ∂x 2 ∂y ∂x ∂y We therefore have a plane strain problem to solve for the two-dimensional displacement ﬁeld (u, v)T . From (4.6.2) we infer the existence of an Airy stress function A such that Txx =

∂2A , ∂y 2

Txy = −

∂2A , ∂x∂y

Tyy =

∂2A . ∂x2

(4.6.34)

Then elimination of u and v from (4.6.32) reveals that A satisﬁes the twodimensional biharmonic equation ∇4 A = 0.

(4.6.35)

The transverse displacement of the plate is therefore governed by ∂2A ∂2w ∂2A ∂2w ∂2w ∂2A ∂2w 4 + − 2 − D∇ w − ςg = ς , ∂y 2 ∂x2 ∂x∂y ∂x∂y ∂x2 ∂y 2 ∂t2

(4.6.36)

where A satisﬁes (4.6.35). The uniform tension T assumed in (4.6.14) corre sponds to the particular solution of (4.6.35) in which A = T x2 + y 2 /2. Equally, an anisotropically stretched membrane corresponds to D = 0, A = T1 x2 + T2 y 2 /2, where T1 and T2 are constant. In general, however, in addition to the displacement and/or bending conditions (4.6.15), (4.6.17) 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 u Txx Txy n = σ(x, y), (4.6.37) = ub (x, y) or Txy Tyy v where n is the unit normal to the edge of the plate.

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Approximate theories

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 Section 4.6, although we will have to wait until Chapter 6 for a systematic exposition on the range of applicability of the theory. We follow the derivations used in the previous section to obtain the governing equations ∂Txy ∂Txx + = 0, ∂x ∂y

∂Txy ∂Tyy + =0 ∂x ∂y

(4.7.1)

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

(4.7.2)

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 (4.6.32) still hold. Our focus, then, is on obtaining improved approximations for the strain components eij . We note with concern that in (4.6.32) we are including the in-plane displacement, although we derived (4.7.1) 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 (4.6.12) requires both the in-plane displacements (u, v) and the transverse displacement w to be suﬃciently small. The von K´ arm´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 (4.6.32).

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 )  . (4.7.3) 0 w(X, Y )

4.7 Von K´ arm´ an plate theory

173

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 )     1 + ∂u/∂X ∂u/∂Y (4.7.4) = x(X) + δX  ∂v/∂X  + δY 1 + ∂v/∂Y  + O |δX|2 ∂w/∂X ∂w/∂Y 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 . ∂w 2 ∂u 2 2 2 |δx| − |δX| = δX 2 + ∂X ∂X . ∂v ∂w ∂w ∂w 2 ∂v ∂u 2 + + + δY 2 + . (4.7.5) + 2δXδY ∂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 (4.7.5) 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 take the in-plane strain components to be ∂u 1 ∂w 2 + exx = , (4.7.6a) ∂x 2 ∂x 1 ∂u ∂v ∂w ∂w + + , (4.7.6b) exy = 2 ∂y ∂x ∂x ∂y ∂v 1 ∂w 2 + eyy = . (4.7.6c) ∂y 2 ∂y Then (4.7.5) gives us the lowest-order change in length of a line element in the form |δx|2 − |δX|2 = 2exx δX 2 + 4exy δXδY + 2eyy δY 2 .

(4.7.7)

Notice that (4.7.6) may also be obtained from (1.4.5) by assuming that w is an order of magnitude larger than u and v. By including the nonlinear terms in (4.7.6) 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 relations (4.6.32) between stress and strain. The von K´ arm´an theory is our ﬁrst example of a theory that is

174

Approximate theories

mechanically linear but geometrically nonlinear, a distinction that we will discuss in more detail in Chapter 5. Before proceeding to incorporate (4.7.6) into the governing equations for the plate, let us pause to contrast them with the corresponding plane strain components (4.6.33) where w ≡ 0. First, we recall that in plane strain (4.6.33) are only soluble for u and v if the compatibility condition (2.7.5a) is satisﬁed. Here, (4.7.6) gives us three equations in the three displacements, so there is no compatibility condition, but (4.7.6) does give us the interesting result that 2 2 ∂ 2 exy ∂ 2 eyy ∂2w ∂2w ∂ w ∂ 2 exx + 2 = − . (4.7.8) − − ∂y 2 ∂x∂y ∂x2 ∂x2 ∂y 2 ∂x∂y Second, let us pose the question: “which displacement ﬁelds give rise to zero in-plane strain?” We see from (4.7.7) 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 (4.7.8) reduces the system to a single equation for w, namely 2 2 ∂2w ∂2w ∂ w − = 0. (4.7.9) 2 2 ∂x ∂y ∂x∂y Thus a surface z = w(x, y) can be isometrically deformed from a plane if and only if w satisﬁes (4.7.9), which is a version of the Monge–Amp`ere equation (Ockendon et al., 2003, p. 380). 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 (see, for example, Kreyszig, 1959, Section 58). Equation (4.7.9) also states that the plate has zero Gaussian curvature. We recall from Section 4.6.2 that the Gaussian curvature is deﬁned to be K = κ1 κ2 ,

(4.7.10)

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,

4.7 Von K´ arm´ an plate theory

175

(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.

so at 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 nondevelopable 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.

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 (4.7.1) to introduce

176

Approximate theories

an Airy stress function A, deﬁned in the usual way, so that (4.7.2) becomes ∂2A ∂2w ∂2w ∂2A ∂2w ∂2A ∂2w 4 − 2 − D∇ w − ςg = ς . + ∂y 2 ∂x2 ∂x∂y ∂x∂y ∂x2 ∂y 2 ∂t2

(4.7.11)

Next, we use (4.6.32) and (4.7.8) to eliminate the strain components and hence obtain 2 2

2w ∂2w ∂ ∂ w . (4.7.12) − ∇4 A = −Eh 2 2 ∂x ∂y ∂x∂y The coupled partial diﬀerential equations (4.7.11) and (4.7.12) for w and A comprise the von K´ arm´an model. Evidently (4.7.11) reproduces the linear plate equation (4.6.36), so the improved approximation for the strain has led to the introduction of just one new term: the right-hand side of (4.7.12). 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 contrast with linear plate theory, where in-plane and transverse deformations are decoupled. It is the nonlinear coupling provided by the right-hand side of (4.7.12) that makes the von K´ arm´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 (4.6.22) for linear bending of a plate, we may use (4.6.23), generalised to include linear stretching, to suggest that the energy per unit area in a von K´ arm´an plate is D 2

2 2

∂2w ∂2w ∂ w − ∂x2 ∂y 2 ∂x∂y 1 2 + , (Txx + Tyy )2 − 2(1 + ν) Txx Tyy − Txy 2Eh

2 2 ∇ w − 2(1 − ν)

where the integrated stress components are given by (4.6.32) and the strain components are again deﬁned by (4.7.6). A lengthy exercise in the calculus of variations shows that when we consider virtual displacements (u, v, w) that minimise the net strain energy in the plate, we do retrieve (4.7.11), (4.7.12). Concerning boundary conditions, we have to impose two conditions on w, as described in Section 4.6.2, and two conditions on A, analogous to those described in Section 2.6.3 for plane strain problems.

4.8 Weakly curved shell theory

177

4.8 Weakly curved shell theory 4.8.1 Strain in a weakly curved shell It is relatively straightforward to generalise the von K´ arm´an model to describe the deformation of a pre-cast 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 Section 4.7.2, considering the displacement of a point on the centre-surface from 

 X  X= Y W (X, Y )



to

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

(4.8.1)

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  + O |δX|2 = X + δX  0 ∂W/∂X ∂W/∂Y

(4.8.2)

is deformed to x + δx, where    ∂u/∂Y 1 + ∂u/∂X  + δY  1 + ∂v/∂Y  + O |δX|2 . (4.8.3) δx = δX  ∂v/∂X ∂(W + w)/∂Y ∂(W + w)/∂X 

Now when we calculate the change in the square of the length of the small line element δX, we ﬁnd to leading order that 2

2

|δx| − |δX| = δX 2 + 2δXδY

∂u ∂W ∂w 2 +2 + ∂X ∂X ∂X

∂w ∂X

∂v ∂W ∂w ∂W ∂u + + + ∂Y ∂X ∂X ∂Y ∂y ∂v ∂W +2 + δY 2 2 ∂Y ∂Y

2 . ∂w ∂w ∂w + ∂X ∂X ∂Y . ∂w ∂w 2 , (4.8.4) + ∂Y ∂Y

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Approximate theories

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

(4.8.5a) (4.8.5b) (4.8.5c)

Again following Section 4.7.2, we can eliminate the horizontal displacement (u, v) from (4.8.5) to obtain 2 2 ∂ 2 eyy ∂ 2 exy ∂2w ∂2w ∂ w ∂ 2 exx − +2 = − − ∂y 2 ∂x∂y ∂x2 ∂x2 ∂y 2 ∂x∂y ∂2W ∂2w ∂2W ∂2w ∂2W ∂2w . (4.8.6) + + − 2 ∂x2 ∂y 2 ∂y 2 ∂x2 ∂x∂y ∂x∂y Notice that the right-hand side of (4.8.6) may be written as 2 2 2

2 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 shell by z = f (x, y) = W (x, y) + w(x, y)

(4.8.7)

and the initial Gaussian curvature by K0 (x, y), then, for the strain to be zero, f must satisfy the inhomogeneous Monge–Amp`ere equation 2 2 ∂ f ∂2f ∂2f − = K0 (X, Y ). (4.8.8) 2 2 ∂X ∂Y ∂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 4.12. If K0 = 0 then the initial shell shape is developable, so at each point it resembles a cylinder, as illustrated in Figure 4.13(a), and we recover (4.7.9), derived previously for an initially ﬂat plate. In this case, (4.8.8) is parabolic, with one repeated set of real characteristics corresponding to the generators of the initial developable surface. This reﬂects the fact that the shell can be arbitrarily bent about the generators without causing any in-plane stress.

4.8 Weakly curved shell theory (a)

(b)

179 (c)

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

If K0 < 0 then the initial shell shape is anticlastic, like the saddle point shown in Figure 4.13(b), and (4.8.8) 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 shell will cause non-local displacements in the region of inﬂuence spanned by these characteristics. Such a shell is therefore somewhat stronger than a developable surface. Finally, if K0 > 0 then the initial shell shape is synclastic, as in Figure 4.13(c), and (4.8.8) is elliptic. Such a shell will be stronger still because any local bending causes deformations throughout the entire shell.

4.8.2 Linearised equations for a weakly curved shell Having obtained expressions for the in-plane strains, let us generalise the force balances (4.6.2) and (4.6.3). As shown in Exercise 4.11, we ﬁnd that the transverse displacement satisﬁes ∂2A ∂2f ∂2A ∂2f ∂2A ∂2f ∂2w 4 + − 2 − D∇ w − ςg = ς , ∂y 2 ∂x2 ∂x∂y ∂x∂y ∂x2 ∂y 2 ∂t2

(4.8.9)

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

(4.8.10a) (4.8.10b) (4.8.10c)

180

Approximate theories

with the strain components now given by (4.8.5). By eliminating the in-plane displacement (u, v) between these, we ﬁnd that A now satisﬁes 2 2 2 2

∂2W ∂2W ∂ W ∂2f ∂2f ∂ f 4 . (4.8.11) − − + ∇ A = Eh 2 2 2 2 ∂x ∂y ∂x∂y ∂x ∂y ∂x∂y In (4.8.9) and (4.8.11) we have two equations for A and w, generalising the von K´ arm´an equations, which may be recovered by setting W ≡ 0, f ≡ w. To make the model 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, (4.8.9) and (4.8.11) reduce to ∂2W ∂y 2 ∂2W ∂y 2

∂2W ∂2A ∂2W ∂2A ∂2A + − 2 = D∇4 w, ∂x2 ∂x∂y ∂x∂y ∂x2 ∂y 2 ∂2W ∂2w ∂2W ∂2w ∂2w + − 2 = −(Eh)−1 ∇4 A, ∂x2 ∂x∂y ∂x∂y ∂x2 ∂y 2

(4.8.12) (4.8.13)

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 coupled 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 (4.8.12), (4.8.13) as follows: x = Lx ,

y = Ly ,

W = W0 W ,

w = w0 w ,

A=

Dw0 A, W0

(4.8.14)

where L is a typical lateral dimension, and w0 a typical boundary displacement. We obtain, after suppressing primes, ∂2W ∂y 2 ∂2W ∂y 2

∂2W ∂2A ∂2A ∂2W ∂2A + − 2 = ∇4 w, ∂x2 ∂x∂y ∂x∂y ∂x2 ∂y 2 ∂2W ∂2w ∂2W ∂2w ∂2w + − 2 = −δ∇4 A, ∂x2 ∂x∂y ∂x∂y ∂x2 ∂y 2

(4.8.15) (4.8.16)

where δ=

h2 . 12(1 − ν 2 )W02

(4.8.17)

4.8 Weakly curved shell theory

181

For a suﬃciently thin shell, we may therefore neglect the right-hand side of (4.8.16) so that the equations decouple: given suitable boundary conditions, our strategy would be to solve ∂2W ∂2w ∂2W ∂2w ∂2W ∂2w + − 2 =0 ∂y 2 ∂x2 ∂x∂y ∂x∂y ∂x2 ∂y 2

(4.8.18)

for w and then substitute the result into (4.8.15) 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 (4.8.18). It is easy to verify that the discriminant ∆ which determines the type of (4.8.18) is equal to the initial Gaussian curvature of the shell, that is 2 2 ∂2W ∂2W ∂ W − . (4.8.19) ∆ = K0 (x, y) = 2 2 ∂x ∂y ∂x∂y The problem is therefore parabolic, hyperbolic or elliptic, according to whether the shell is developable, anticlastic or synclastic. This immediately casts doubt on the model (4.8.18) for, suppose we wished to prescribe w around the perimeter of an anticlastic shell. Then (4.8.18) would present us with an ill-posed Dirichlet problem for a hyperbolic partial diﬀerential equation, and there would probably be no solution at all. On the other hand, to apply clamped boundary conditions on a synclastic shell, we would apparently need to impose both w and ∂w/∂n on a second-order elliptic equation, which again leads to an ill-posed problem. Fortunately we can understand the deﬁciencies in (4.8.18) when we realise from (4.8.6) that, when w is small enough for quadratic terms to be neglected, (4.8.18) 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 our use of (4.8.14) to render the problem dimensionless. One possibility is to rescale A such that ˜ = δA = O (1), A

(4.8.20)

which models deformations other than pure bending which increase the inplane stress by an order of magnitude. Substituting (4.8.20) into (4.8.15) and (4.8.16) before neglecting δ, we ﬁnd that the equations again decouple, ˜ now determined by with A ˜ ˜ ˜ ∂2W ∂2A ∂2W ∂2A ∂2W ∂2A + − 2 =0 ∂y 2 ∂x2 ∂x∂y ∂x∂y ∂x2 ∂y 2

(4.8.21)

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Approximate theories

and w then found from ∂2W ∂2w ∂2W ∂2w ∂2W ∂2w ˜ + − 2 = −∇4 A. ∂y 2 ∂x2 ∂x∂y ∂x∂y ∂x2 ∂y 2

(4.8.22)

Hence we again ﬁnd a reduction from fourth to second order, and the types of the equations (4.8.21) and (4.8.22) are again determined by (4.8.19). Therefore, although the rescaling (4.8.20) allows us to describe in-plane stretching, it does not alter the limitations on the boundary conditions that we may be able to impose. However, any incompatibilities that may arise with the imposed boundary conditions will be resolved over narrow boundary layers, in which both stretching and bending are important. To explore the behaviour in such regions, one possibility is to rescale (x, y) as well as A, setting ˆ A = δ −1/2 A,

(x, y) = δ 1/4 (ξ, η).

(4.8.23)

Recalling that W is a prescribed function of x and y, we see that this transforms (4.8.15) and (4.8.16) into ∂2W ∂2A ∂2W ∂2A ∂2W ∂2A ˆ 4 w, + − 2 =∇ (4.8.24) ∂y 2 ∂ξ 2 ∂x∂y ∂ξ∂η ∂x2 ∂η 2 ∂2W ∂2w ∂2W ∂2w ∂2W ∂2w ˆ 4 A, + − 2 = −∇ (4.8.25) ∂y 2 ∂ξ 2 ∂x∂y ∂ξ∂η ∂x2 ∂η 2 ˆ 2 = ∂ 2 /∂ξ 2 + ∂ 2 /∂η 2 . This is eﬀectively the full problem for w, A, where ∇ except that the coeﬃcients are constant to lowest order on the (ξ, η)-scale. We have therefore identiﬁed the following three alternative classes of deformation that a thin shell may suﬀer. (i) If relatively small stresses are applied at the boundaries, then the deformation described by (4.8.18) is an isometry, and the in-plane stress that it produces may be determined a posteriori from (4.8.15). (ii) When large in-plane stresses are applied, then the bending stiﬀness of ˜ satisﬁes the decouthe shell becomes negligible in comparison, so that A pled equation (4.8.21). The in-plane stretching caused by these stresses then drives the equation (4.8.22) for the transverse displacement. (iii) Both eﬀects of bending stiﬀness and in-plane stretching can balance in narrow regions of width O δ 1/4 . These regions may be close to the boundary of the shell, or they may take the form of creases that are remote from the boundary. In cases (i) or (ii), the reduction of order means that it will in general be impossible to ﬁnd solutions that are everywhere compatible with any

4.8 Weakly curved shell theory

183

boundary conditions imposed on w and A. However, the full system (4.8.15), (4.8.16) is well posed given any physically reasonable boundary conditions. The full solution of the problem as δ → 0 will thus comprise regions described by (i) or (ii), patched together across narrow layers described by (iii). These creases in the shell, and the construction described above, may be visualised by simply crumpling a piece of paper: the resultant shape will consist of developable regions, which have undergone pure bending, joined together by thin creases. Some beautiful experiments illustrating the corresponding behaviour of synclastic, cylindrical and anticlastic shells are described in Vaziri & Mahadevan (2008). We will now illustrate these scenarios with some simple examples. To avoid the complication of solving a boundary-value problem each time, we will simply ask ourselves what kinds of displacements w will be possible for any prescribed shell shape W . Let us ﬁrst consider deformations of the initially cylindrical shell αy 2 , 2 for which (4.8.15) and (4.8.16) become

(4.8.26)

W =

∂2w ∂2A = ∇4 w, α 2 = −δ∇4 A. (4.8.27) 2 ∂x ∂x Now, letting δ → 0, we see that w can only be a combination of two forms: α

w(x, y) = a(y)

and w(x, y) = xb(y),

(4.8.28)

where a and b are arbitrary functions. These displacement ﬁelds correspond, respectively, to the bending of the shell about its generators and to the 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 about an axis perpendicular to the generators, as in Figure 4.14(d), is much more diﬃcult. Indeed, a uniform bending βx2 w= , (4.8.29) 2 inevitably causes a large in-plane stress, with αβy 4 + O (1), (4.8.30) 24δ corresponding to the stress ﬁeld, made dimensionless with Dw0 /W0 L2 , A=−

Txx = −

αβy 2 , 2δ

Txy = 0,

Tyy = 0.

(4.8.31)

184

Approximate theories (a)

(b)

z y x (c)

(d)

Fig. 4.14 Deformations of a cylindrical shell; (a) original shape, (b) bending about the generators, (c) rotating the generators, (d) bending along the generators.

We can associate this stress in the x-direction with compression of the generators. Next, let us consider the anticlastic shell W = αxy,

(4.8.32)

depicted in Figure 4.15(a), which contains two families of straight lines x = const. and y = const. This time the governing equations (4.8.15) and (4.8.16) take the form −2α

∂2A = ∇4 w, ∂x∂y

2α

∂2w = δ∇4 A. ∂x∂y

(4.8.33)

When δ is negligible, the displacement ﬁeld must be a linear combination of w(x, y) = a(x)

and w(x, y) = b(y),

(4.8.34)

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 the neck of a plastic bottle. Large stresses result from any

4.8 Weakly curved shell theory (a)

185

(b)

z y x (c)

(d)

Fig. 4.15 Deformations 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.

other transverse displacement. For example, w=

β(x + y)2 , 2

(4.8.35)

which represents 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

(4.8.36)

The corresponding dimensionless stress ﬁeld Txx =

αβy 2 , 2δ

Txy = 0,

Tyy =

αβx2 2δ

(4.8.37)

shows that both the normal stresses increase quadratically. due to stretching of the straight lines x = const. and y = const. respectively. Finally, let us consider the synclastic shell α x2 + y 2 W = . (4.8.38) 2

186

Approximate theories (a)

(b)

z y x (c)

(d)

Fig. 4.16 Deformations of a synclastic shell; (a) original shape, (b) and (c) two possible deformations, (d) one-dimensional bending.

As may be seen in Figure 4.16(a), there are no straight lines embedded in this surface. The governing equations (4.8.15) and (4.8.13) read α∇2 A = ∇4 w,

α∇2 w = −δ∇4 A,

(4.8.39)

so that, when δ is small, only harmonic displacement ﬁelds are possible. Some examples are 2 β x4 − 6x2 y 2 + y 4 2 , (4.8.40) w(x, y) = β x − y and w(x, y) = − 4 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 ping-pong ball. On the other hand, if we perform the rescaling (4.8.20) before letting δ → 0, we ﬁnd that ˜ satisﬁes Laplace’s equation and thus A ˜ = 0. α∇2 w = −∇4 A

(4.8.41)

Hence, any non-harmonic 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. If we tried to impose boundary

4.9 Nonlinear beam theory

187

conditions that force such a one-dimensional bending response, we would inevitably cause a crease to be formed. Although the above theory only applies to very small transverse displacements of shells whose built-in curvature is already small, it illustrates 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 Section 4.4, now generalising the model to make it applicable to commonly observed phenomena such as the deﬂection of a diving board. Diving boards undergo large displacement without any of the damaging eﬀects that are associated with 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 (a) h

(b) (c)

R

Fig. 4.17 A beam (a) before and (b) after bending; (c) a close-up of the displacement ﬁeld.

188

Approximate theories

N0 (a)

(b) N (s + δs) T0

T (s + δs) T0

M (s) θ

M (s + δs) N0

T (s) N (s)

Fig. 4.18 (a) The forces and moments acting on a small segment of a beam. (b) The sign convention for the forces at the ends of the beam.

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 Section 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 K´ arm´an plate theory from Section 4.7. In equilibrium, it is surprisingly easy to generalise our derivation of linear beam theory in Section 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 order, so that arc-length is conserved by the deformation and we can thus 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 dw = cos θ, = sin θ. (4.9.1) ds ds 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

4.9 Nonlinear beam theory

189

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 (4.9.2)

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 T = T0 cos θ + N0 sin θ,

N = N0 cos θ − T0 sin θ.

(4.9.3)

As a constitutive relation we expect, as in Section 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 of proportionality is the same as in the linear case, that is M = −EI

dθ . ds

(4.9.4)

By collecting (4.9.2c), (4.9.3) and (4.9.4), we obtain the Euler–Bernoulli beam equation EI

d2 θ + N0 cos θ − T0 sin θ = 0. ds2

(4.9.5)

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 secondorder diﬀerential equation (4.9.5). 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 can take the end s = 0 to be 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 (4.9.1) the conditions L L cos θ(s) ds, Z= sin θ(s) ds, (4.9.6) X= 0

0

from which we can in principle recover T0 and N0 . It is reassuring to conﬁrm that the theory above reduces to the linear beam theory derived in Section 4.4 for cases where the deﬂection is small.

190

Approximate theories

For small θ, we can simplify the geometric relations (4.9.1) to lowest order to give x = s,

dw = θ, dx

(4.9.7)

and (4.9.5) thus becomes d3 w dw = 0. (4.9.8) + N 0 − T0 3 dx dx One further x-derivative reproduces the linear beam equation (4.4.8) in the absence of inertia and gravity. EI

4.9.2 Example: deﬂection of a diving board Under a shift in the origin of θ, (4.9.5) is equivalent to theequation for a pendulum whose small-amplitude frequency of oscillation is T02 + N02 /EI. However, in (4.9.5), s is far from being a time-like variable, and the typical boundary conditions discussed above are quite diﬀerent from 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 (4.9.5). Let us ﬁrst consider the example, alluded to previously, of the steady deformation of a diving board, with gravity neglected for simplicity. 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 (4.9.5) becomes d2 θ (4.9.9) EI 2 = F cos θ, ds subject to dθ (L) = 0. (4.9.10) ds Multiplying (4.9.9) through by dθ/ds and integrating once with respect to s, we obtain 2 dθ 2F (sin θ − sin α) , (4.9.11) = 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 ) −θ dφ 2F √ =s . (4.9.12) EI sin α − sin φ 0

L 1

2

4.9 Nonlinear beam theory

191 x

2F/EI

3

4

5

0.2

6

0.4

0.6

0.8

1

-0.1

-0.25

1

-0.2

-0.5

-0.3

−α -0.75

w

-0.4

-0.5

-1

-0.6

-1.25

-0.7

-1.5

-0.8

L

3

2

2F/EI = 4

Fig. 4.19 (a) Final angle −α of a diving board versus applied 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 when L 2F/EI = 1 and 2.

It just remains to determine the angle α from the transcendental equation ) α dφ 2F √ =L , (4.9.13) EI sin α − sin φ 0 which can be written in terms of elliptic integrals. Given the applied force F , we have to solve (4.9.13) for α (numerically), and the deﬂection may then be reconstructed using (4.9.12) and (4.9.1); see Exercise 4.15. In Figure 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 (4.9.8) to be applicable. The linear model that describes our diving board is dw d3 w (0) = 0, = F, w(0) = 3 dx dx and the solution is readily found to be EI

d2 w (L) = 0, dx2

(4.9.14)

F 2 x (3L − x). (4.9.15) 6EI In Figure 4.19(b), (4.9.15) is plotted as dashed curves, and we see that there is excellent agreement with the full nonlinear model provided the deﬂections are reasonably small. w=−

4.9.3 Weakly nonlinear theory and buckling Now let us return to the compression of an elastic beam, which we discussed previously in the linear case in Section 4.4. We focus on the model problem of a beam of length L, clamped at both ends, subject to a compressive force

192

Approximate theories

P = −T0 and zero transverse force. We therefore have to solve the Euler strut equation P d2 θ sin θ = 0, + 2 ds EI

(4.9.16a)

subject to the boundary conditions θ (0) = θ (L) = 0.

(4.9.16b)

This is a nonlinear eigenvalue problem: θ ≡ 0 is always a possibility, and we are seeking values of the applied force P for which (4.9.16) admits nonzero solutions. Such solutions, if they exist, correspond to buckling of the beam. If θ is small, it is easy to show that (4.9.16) reduces to the linear eigenvalue problem (4.4.13) studied previously, and we can read oﬀ from (4.4.15) the solution  nπs L2 P  = n2 , , A sin (4.9.17) θ= L π 2 EI  0, otherwise, 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 can only occur when P = n2 π 2

EI . L2

(4.9.18)

We can try to visualise this behaviour by plotting a response diagram of the magnitude of the solution θ = max |θ| = |A| 0<s
(4.9.19)

versus the control parameter λ=

L2 P . π 2 EI

(4.9.20)

The response diagram corresponding to (4.9.17) is shown in Figure 4.20(a); exactly the same diagram would have resulted had we used any other sensible norm θ instead of (4.9.19). There are several problems with the solution (4.9.17). When λ = n2 , where n is an integer, the solution is not unique, since its amplitude is indeterminate. Also, if the applied force slightly exceeds one of the critical values, we would expect on physical grounds that the buckling would continue, while

4.9 Nonlinear beam theory

(a)

|A|

193

(b)

|A| 1.4

1.2

1

0.8

0.6

0.4

0.2

5

10

15

20

2.5

λ

5

7.5

10 12.5 15 17.5 20

λ

Fig. 4.20 (a) Response diagram of the amplitude θ of the linearised solution (4.9.17) for a buckling beam versus the force parameter λ. (b) Corresponding response of the weakly nonlinear solution; the exact nonlinear solution is shown using dashed lines.

Figure 4.20(a) indicates that the solution instantly reverts to θ ≡ 0. These criticisms can be answered by noting that, if we follow one of the branches λ = n2 , then θ will eventually increase to the point where the linearisation of (4.9.16) is no longer valid. We therefore need to reintroduce the nonlinearity to resolve the behaviour when λ is close to n2 . To do this systematically requires the use of asymptotic analysis, and we must ﬁrst non-dimensionalise the model. The angle θ is already dimensionless, but it is convenient to deﬁne θ = δΘ,

(4.9.21)

where Θ is order unity and δ is a small positive parameter that characterises the smallness of θ. We also set s = Lξ and thus obtain the non-dimensional model d2 Θ sin(δΘ) = 0, + π2 λ 2 dξ δ

θ (0) = θ (1) = 0,

(4.9.22)

where λ is again the dimensionless parameter deﬁned by (4.9.20). We focus on a neighbourhood of λ = n2 by writing λ = n2 + ελ1 ,

(4.9.23)

where ε is a ﬁxed small positive number and λ1 is a control parameter which parametrises the diﬀerence between λ and n2 . We thus have two small parameters in the problem: the amplitude δ of the solution and the deviation ε of λ from its critical value. The vital step is to ﬁnd a relation between these two parameters which yields a balance

194

Approximate theories

between the compressive loading, characterised by ε, and the nonlinearity, characterised by δ, and the correct choice leads to a so-called weakly nonlinear theory. If we use the assumed smallness of δ to expand the sine in (4.9.22), we obtain the Duﬃng equation 3 2 d2 Θ 2 2Θ Θ − δ + · · · = 0, (4.9.24) + π + ελ n 1 dξ 2 6 and we clearly see that the required balance is obtained by choosing √ δ = ε. (4.9.25) Before doing any further calculation, we can already infer from this that the amplitude of the solution will vary as the square root of the excess loading. After making the choice (4.9.25), we write Θ as an asymptotic expansion of the form† Θ ∼ Θ0 (ξ) + εΘ1 (ξ) + · · · . (4.9.26) By substituting (4.9.26) into (4.9.24) and equating successive powers of ε, we ﬁnd that d2 Θ0 + n2 π 2 Θ0 = 0, Θ0 (0) = Θ0 (1) = 0, (4.9.27a) dξ 2 3 d2 Θ1 2 2 2 2 2 Θ0 , Θ1 (0) = Θ1 (1) = 0. (4.9.27b) + n π Θ = −π λ Θ + n π 1 1 0 dξ 2 6 The general solution of (4.9.27a) is Θ0 = A0 sin (nπξ) ,

(4.9.28)

and we seem to have made very little progress from (4.9.17). However, when we examine the problem for Θ1 , we ﬁnd that (4.9.27b) has no solutions unless the right-hand side satisﬁes a solvability condition, namely 1 3 2 Θ0 sin (nπξ) dξ = 0. (4.9.29) λ1 Θ0 − n 6 0 By substituting for Θ0 from (4.9.28), we obtain the following equation for the leading-order amplitude: 8λ1 (4.9.30) A0 A20 − 2 = 0. n When λ1 is negative, the only real solution of (4.9.30) is A0 = 0 but, for √ positive values of λ1 , there are three such solutions: A0 = 0, ±(2/n) 2λ1 . †

see Hinch (1991), for example, for the deﬁnition of an asymptotic expansion; for the purposes of this chapter, we can interpret ∼ as meaning approximates increasingly well as ε → 0

4.10 Nonlinear rod theory

(a)

(b)

A0

195

A0

stable stable

unstable

λ1

λ1

stable Fig. 4.21 (a) Pitchfork bifurcation diagram of leading-order amplitude A0 versus forcing parameter λ1 . (b) The corresponding diagram when asymmetry is introduced.

This behaviour of A0 as a function of λ1 , depicted in Figure 4.21(a), is characteristic of a so-called pitchfork bifurcation. Working backwards, we can easily turn (4.9.30) into an equation for the amplitude of θ; that is, λ 2 −1 = 0, (4.9.31) |A| |A| − 8 n2 and this weakly nonlinear response is plotted in Figure 4.20(b). This gives the physically expected result that the amplitude of each buckled solution increases with the excess applied force. We can also see how Figure 4.20(a) approximates this response for very small |A|. For comparison, we also show the response of the full nonlinear solution of (4.9.16); see Exercise 4.18. We can use the argument after (4.4.18) to show that the solution θ = 0 is stable for λ1 < 0 but loses stability to the nonzero solution as λ1 crosses zero. This justiﬁes the labelling in Figure 4.21(a) and can also be explained on the basis of an energy argument, as in Exercise 4.14. Another key ingredient of bifurcation pictures like Figure 4.21(a) is the symmetry that we have assumed in our strut model. Any imperfection or asymmetry in the model would deform the response diagram into something like Figure 4.21(b), in which there is a smooth dependence of the solution on the control parameter. This is exempliﬁed by the re-introduction of gravity in the model as in Exercise 4.16. 4.10 Nonlinear rod theory We are now in a position to generalise and combine the linear rod theory of Section 4.5 with the nonlinear beam theory of Section 4.9 to derive a model for the equilibrium of a rod that can both twist about its tangent vector

196

Approximate theories

and bend in two perpendicular directions. We let F and M denote the force and moment exerted on any cross-section s = const. of the rod, s being arc-length along the centre-line. As in Section 4.9, we will assume that the rod is inextensible so that s is eﬀectively a Lagrangian variable. Neglecting body forces for simplicity, a force balance gives dF = 0, ds

so that

F = F 0 = const..

(4.10.1)

This equation generalises the ﬁrst two equations of (4.9.2). Also, if r(s) parametrises the centre-line of the rod, then a moment balance analogous to the ﬁnal equation of (4.9.2) gives dr dM + ×F = 0, so that M + r×F 0 = M 0 = const.. (4.10.2) ds ds The equations (4.10.1), (4.10.2) for the vectors F , M and r now need to be closed by a constitutive law for M , and this is what poses the most interesting modelling challenge. Instead of adopting the usual diﬀerential geometry deﬁnition of curvature, we consider what happens to a Lagrangian orthonormal system ei (s) (i = 1, 2, 3) as the rod deforms. This system is ﬁxed in the rod and we choose e1 (s) =

dr , ds

(4.10.3)

with e2 and e3 deﬁning the normal plane. At diﬀerent s-stations, this system will have rotated by diﬀerent amounts, and we infer that there exists a vector Ω(s) such that dei = Ω(s)×ei (s). ds

(4.10.4)

Thinking of s as time for the moment, the vector Ω = (Ω1 , Ω2 , Ω3 )T represents the “angular velocity” of the orthonormal system {e1 , e2 , e3 }, with Ω1 corresponding to twisting of the rod about its axis and Ω2 , Ω3 to bending. This leads us to propose a linear constitutive relation between M and Ω, so that M = BΩ

(4.10.5)

for some 3 × 3 matrix B, whose elements we will choose to ensure that (4.10.5) agrees with (4.5.5), (4.5.9) and (4.9.4) in the appropriate limits. T In linear rod theory, we write r = x, v(x), w(x) , where x = s to lowest order. When we neglect quadratic terms, the orthonormal vectors ei may be

4.10 Nonlinear rod theory

written in the form dv dw T , , e1 = 1, dx dx

e2 =

−

dv , 1, θ dx

197

T ,

e3 =

T dw , −θ, 1 − , dx (4.10.6)

where θ measures the twist of the rod about the x-axis. It is easy to check that these are consistent with (4.10.4) and correspond to T dθ d2 w d2 v Ω= ,− 2 , 2 . (4.10.7) dx dx dx Now we see that we can reproduce (4.5.5) and (4.5.9) by choosing   R 0 0 B =  0 EIyy −EIyz  , (4.10.8) 0 −EIyz EIzz in the notation of Section 4.5. T To retrieve nonlinear beam theory, we write r = x(s), 0, w(s) , so that       cos θ 0 − sin θ e1 =  0  , e2 = 1 , e1 =  0  , (4.10.9) sin θ 0 cos θ and we easily ﬁnd that

T dθ . Ω = 0, , 0 ds

(4.10.10)

Provided the beam is symmetric so that Iyz = 0, (4.10.5) with B given by (4.10.8) thus reduces to the constitutive relation (4.9.4) for a nonlinear beam. The relations (4.10.1)–(4.10.5) comprise a formidable system of 15 equations for F , M , r, ei and Ω. Fortunately, there is a wonderful analogy that relieves us of the task of discussing the behaviour of this system. Consider a rigid body rotating about a ﬁxed origin 0 such that one principal axis e1 at 0 passes through the centre of gravity which is a distance h from 0. The equations of motion for such a body are dei dH H = Iω, (4.10.11) = ω×ei , = he1 ×W , dt dt where ω is the angular velocity, e2 , e3 are the other principal axes, H is the angular momentum, W is the weight and I the inertia tensor. When we identify ω with Ω, H with M , hW with F and, crucially, s with t, these equations are simply (4.10.4), (4.10.2) and (4.10.5). Hence our knowledge

198

Approximate theories

about spinning tops, gyroscopes and “inertia coupling” in classical mechanics can be used to explain the often infuriating coupling between torsion and bending that is only too familiar to those who have to wind cables and garden hoses. This Kirchhoﬀ analogy disappears when we consider the dynamics of a rod subject to a body force g, the model for which is precisely as above with d/ds replaced by ∂/∂s and (4.10.1) replaced by ρ

∂F ∂2r . = ρg + 2 ∂t ∂s

(4.10.12)

4.11 Geometrically nonlinear wave propagation When we include nonlinear eﬀects in models for elastic wave propagation, we will rarely be able to ﬁnd explicit solutions. Hence, in this section, we will simply describe a few nonlinear models that are relatively easy to derive and posess interesting exact solutions. 4.11.1 Gravity-torsional waves Our ﬁrst example of nonlinear wave propagation is a generalisation to include gravity in the model (4.5.8) for torsional waves in a rod. In particular, we will try to model the waves that can propagate when a series of rigid pendulums is attached to a rod which is stiﬀ but has negligible mass. In practice, such a conﬁguration can be realised by attaching paper clips to a taut segment of a rubber band, as shown in Figure 4.22, where the paper clips are drawn as pendulums.† To model this situation, we suppose that each pendulum is acted upon by a moment δM at its point of support. A net moment balance, as illustrated in Figure 4.22(b), leads to ∂2θ , (4.11.1) ∂t2 where θ(x, t) is the angle between the pendulum and the downward vertical while m and are the mass and length of the pendulum respectively. Now we identify the moment δM with b∂M/∂x, where b is the spacing between the pendulums and ∂M/∂x is given by the constitutive relation (4.5.9). We therefore ﬁnd that θ satisﬁes ∂2θ ∂2θ (4.11.2) bR 2 = mg sin θ + m2 2 , ∂x ∂t δM = mg sin θ + m2

†

We are grateful to the late Prof. T. Brooke Benjamin for drawing this experiment to our attention.

4.11 Geometrically nonlinear wave propagation

199

(a) b

(b) z δM

z

y y

θ

x mg Fig. 4.22 (a) A system of pendulums attached to a twisting rubber band. (b) Transverse view deﬁning the angle θ(x, t).

which can easily be rescaled to the sine–Gordon equation ∂2θ ∂2θ − + sin θ = 0. (4.11.3) ∂t2 ∂x2 This is one of the most famous completely integrable nonlinear partial diﬀerential equations.This means that it can be solved explicitly for any given initial data on −∞ < x < ∞ by using inverse scattering theory; see for instance Drazin & Johnson (1989). Moreover, localised travelling waves exist that exhibit strong stability properties; even though their speed depends on their amplitude, they can pass through each other virtually unscathed. Such localised waves are called solitons and, for (4.11.3), we can ﬁnd them by setting θ to be a function only of η = x − vt, where the wave speed v is constant. Then (4.11.3) is reduced to the nonlinear ordinary diﬀerential equation d2 θ 2 + sin θ = 0. (4.11.4) v −1 dη 2 If we seek an isolated soliton solution, in which θ → const. as η → ±∞, then the only possible solutions of (4.11.4) are η − η0 −1 √ when |v| > 1 (4.11.5) tanh θ = 4 tan 2 v2 − 1 or η0 − η when |v| < 1. (4.11.6) θ = 4 tan−1 exp √ 1 − v2 Since (4.11.5) approaches the unstable solutions θ = ±π, in which the paper clips are upright, as η → ±∞, it is not physically realisable and only

200

Approximate theories

t=0 z

t=1

t=2

x y

Fig. 4.23 A kink propagating along a series of pendulums attached to a rod.

the subsonic waves (4.11.6) can be observed in practice. In these so-called kink solutions, the rod undergoes one complete twist, with θ = 0 ahead of the wave (η → +∞) and θ = 2π behind (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 paper clips dangling vertically, and then quickly twisting one end through 2π. 4.11.2 Travelling waves on a beam Our second example concerns the unsteady generalisation of the geometrically nonlinear beam theory derived in Section 4.9. The displacement of the beam is now given parametrically by x = x(s, t), w = w(x, t), where, as in (4.9.1), ∂w ∂x = cos θ, = sin θ, (4.11.7) ∂s ∂s and θ(s, t) is the angle made by the beam with the x-axis. When we include acceleration, the force balances (4.9.2) become ∂ ∂2x (T cos θ − N sin θ) = 2 , ∂s ∂t

∂2w ∂ (N cos θ + T sin θ) = 2 , (4.11.8) ∂s ∂t

where N = −EI

∂2θ . ∂s2

(4.11.9)

Although this is a more complicated model than (4.11.3), 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 T → T∞ ,

N → 0,

θ → 0,

x ∼ s,

w→0

as

s → ∞, (4.11.10)

which would apply, for example, to a long taut rope that had been shaken brieﬂy at one end. As in Section 4.11.1, by looking for a travelling wave, we reduce (4.11.8) to nonlinear ordinary diﬀerential equations, which may be

4.11 Geometrically nonlinear wave propagation

w -2

-1

1

2

201

x

-0.5

-1

-1.5

-2

v Fig. 4.24 Travelling wave solution of the nonlinear beam equations (with k = 1).

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. (4.11.11)

Hence the tension and shear force are given by N = v 2 − T∞ sin θ, T = v 2 + T∞ − v 2 cos θ,

(4.11.12)

and (4.11.9) therefore becomes EI d2 θ 2 + M − 1 sin θ = 0, 2 T∞ dη

(4.11.13)

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 the Mach number of the wave. By analogy with (4.11.4), we see that, as long as M < 1, there is a solution connecting θ → 0 as η → +∞ to θ → 2π as η → −∞, namely ) T∞ −1 k(η 0 −η) (1 − M 2 ), θ = 4 tan (4.11.14) e , where k = EI where the constant η0 represents an arbitrary translation. With θ given by (4.11.14), we can integrate (4.11.7) to obtain the parametric description 1 ξ − 2 tanh ξ x = (4.11.15) 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 loop, in which the rope turns through 2π, propagating at speed v which depends, through (4.11.14), on the amplitude. The multivaluedness of w(x) is only physically realistic if the rope is allowed to deform slightly out of the (x, z)-plane.

202

Approximate theories

4.11.3 Weakly nonlinear waves on a beam If we wish to analyse general unsteady deformations of a beam, we appear to be confronted with the full system (4.11.7)–(4.11.9) 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 (4.11.13) is degenerate when EI = 0 and M 2 = 1: when the 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 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 (s, t) to (ζ, t), where ζ = s − ct, so that (4.11.7)–(4.11.9) become ∂2θ , ∂ζ 2 2 ∂ ∂ x ∂2x , − 2c (T − T∞ ) cos θ − N sin θ = ∂ζ ∂t2 ∂ζ∂t 2 ∂ ∂ w ∂2w . − 2c N cos θ + (T − T∞ ) sin θ = ∂ζ ∂t2 ∂ζ∂t ∂x = cos θ, ∂ζ

∂w = sin θ, ∂ζ

N = −EI

(4.11.16) (4.11.17) (4.11.18)

Now we have to non-dimensionalise 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 (4.11.16). The relevant time-scale is then found by balancing the ﬁnal term in (4.11.17) with the normal stress on the left-hand side. We thus deﬁne the dimensionless variables as     x X w = W  , ζ ξ

T = T∞ T ,

N=

EI 2

N,

t=

2c 3 EI

τ,

(4.11.19) so that (4.11.16)–(4.11.18) become ∂X = cos θ, ∂ξ

∂W = sin θ, ∂ξ

(4.11.20)

4.11 Geometrically nonlinear wave propagation

N =−

∂2θ

, ∂ξ 2 ε2 ∂ 2 X ∂ ∂2X , − ε (T − 1) cos θ − εN sin θ = ∂ξ 4 ∂τ 2 ∂ξ∂τ ε2 ∂ 2 W ∂ ∂2W εN cos θ + (T − 1) sin θ = , − ε ∂ξ 4 ∂τ 2 ∂ξ∂τ

203

(4.11.21) (4.11.22) (4.11.23)

where the only remaining dimensionless parameter in the problem is ε=

EI . T∞ 2

(4.11.24)

We suppose, as in Section 4.11.2, 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 → ∞. (4.11.25)

Now we take the limit ε → 0, which corresponds to the bending stiﬀness of the beam being small. When ε = 0, (4.11.22) and (4.11.23) are satisﬁed identically by T = 1, and this motivates the ansatz T ∼ 1 + εT1 + · · · .

(4.11.26)

When we substitute (4.11.26) into (4.11.22) and (4.11.23) and let ε → 0, we ﬁnd that the leading-order equations may be integrated once with respect to ξ, using the boundary conditions (4.11.25), to obtain T1 cos θ − N sin θ +

∂X = 0, ∂τ

T1 sin θ + N cos θ +

∂W = 0. ∂τ

(4.11.27)

We solve these simultaneously for T1 and N and substitute into (4.11.21) to obtain ∂W ∂ 2 θ ∂X sin θ − cos θ = 0, + 2 ∂ξ ∂τ ∂τ

(4.11.28)

which is to be solved along with (4.11.20) 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 (4.11.28) by ∂θ/∂ξ we discover by virtue of (4.11.20) that it may be integrated with respect to ξ, leading, from (4.11.25), to ∂W 1 ∂X cos θ + sin θ = ∂τ ∂τ 2

∂θ ∂ξ

2 .

(4.11.29)

204

Approximate theories

On the other hand, if we instead diﬀerentiate (4.11.28) with respect to ξ, again using (4.11.20), we ﬁnd that ∂W ∂θ ∂ 3 θ ∂θ ∂X (4.11.30) cos θ + sin θ = − 3. ∂ξ ∂τ ∂τ ∂τ ∂ξ Finally, comparing (4.11.29) and (4.11.30), we obtain ∂θ 1 ∂θ 3 ∂ 3 θ − − 3 = 0. (4.11.31) ∂τ 2 ∂ξ ∂ξ If we diﬀerentiate (4.11.31) 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 equation (see Exercise 4.20). The KdV equation was originally used to model large-amplitude water waves, and its study provided a key stimulus for the theory of solitons. By reversing the scalings (4.11.19), we obtain the dimensional version of (4.11.31), namely 1 ∂θ 3 ∂ 3 θ 2T∞ 1 ∂θ ∂θ + − − 3 = 0. (4.11.32) EI c ∂t ∂s 2 ∂s ∂s Now if we seek a travelling wave, in which θ is again a function only of η = s − V t, then we obtain the ordinary diﬀerential equation dθ d3 θ 1 dθ 3 2T∞ (1 − M ) = 0, (4.11.33) + − 3 dη 2 dη EI dη where M = V /c again. As shown in Exercise 4.21, the solution of this equation satisfying θ → 0 as η → +∞ and θ → 2π as η → −∞ is ) 2T∞ (1 − M ) −1 κ(η 0 −η) . (4.11.34) e , where κ = θ = 4 tan EI We thus reproduce the travelling wave solution (4.11.14) found previously, noting that κ is the leading-order approximation to k as M → 1. 4.12 Concluding remarks In this chapter, 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 twodimensional solids. This challenge is exempliﬁed by the fact that, although (4.3.2) is an apparently “physically obvious” paradigm for linear wave propagation, we will see in Chapter 6 that it can only be derived systematically

Exercises

205

from the equations of nonlinear three-dimensional elasticity. Similarly, our model (4.9.2) 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 (4.3.2) implicitly relied on the string being extensible, yet no mention was made of Young’s modulus. Geometric nonlinearity and inextensibility are both concepts that can be properly understood only by a more thorough mathematical approach, which we will present in Chapters 5 and 6. 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, rather than itself being thin. We will discuss such 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 suﬃciently small bending stiﬀness EI are, approximately, * n3 π 3 EI nπ T + 3√ . ωn ∼ 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 x(L − x) L2 + Lx − x2 w=− 24EI

4.3

and deduce that the maximum transverse deﬂection is 5 g/384EI. 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 : P dw d3 w = 0, + dx3 EI dx

4.4

w(0) =

dw (0) = 0, dx

w(L) =

dw (L) = 0. dx

Find the minimum force P required for w to be nonzero. Why is this value diﬀerent from that found in Section 4.10? 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

206

Approximate theories

z

R

111111111 000000000 000000000 111111111 000000000 111111111 000000000 111111111

x

F Fig. 4.25 A beam clamped near the edge of a table.

is negligible compared with F and that the transverse displacement w(x) is small, justify the following model: d4 w R δ(x − ), = dx4 EI d2 w (L) = 0, dx2

dw (0) = 0, dx d3 w F (L) = , dx3 EI

w(0) =

where δ is the Dirac delta-function and the reaction force R is to be determined from the contact condition w() = 0. Show that the reaction force is given by R = (3L − )F/2 and that the displacement is x < , 6(L − )x2 ( − x), F w(x) = # $ 2 24EI ( − x) 3(L − )(3L − )−(2x − 3L + ) , x > .

4.5

Find the maximum magnitude of the slope |dw/dx| and deduce that these results are valid provided F L(L − )/EI 1. 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 the following dimensionless model for the vertical displacement w(x) of the beam: d4 w = Rδ(x − β), dx4 d3 w dw d2 w d3 w (0) = (0) = (1/2) = (1/2) = w(β/2) = 0, dx dx3 dx2 dx3 where β = /L and δ is the Dirac delta-function. Show that the dimensionless reaction force R is equal to 1/2 and

Exercises

g

207

z x L

Fig. 4.26 A beam supported at two points.

4.6

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. Longitudinal waves in a bar of section A, Young’s modulus E and length L are modelled by (4.2.4). The bar is initially unstressed and at rest. The ends x = 0 and x = L are stress-free except for an impulsive compressive force J applied at x = 0 for 0 < t < T , where T < L/c. Show that   1, x − ct < −cT,   u ct − x = , −cT < x − ct < 0,  U cT   0, 0 < x − ct, for 0 < t < L/c, where the maximum displacement at x = 0 is given by U = JcT /EA. Show that an incident wave u = f (x − ct) impinging on the stressfree boundary at x = L gives rise to a reﬂected wave f (2L − x − ct). Deduce that, as T → 0 with U kept ﬁxed, the resulting displacement after the ﬁrst reﬂection from x = L is u = U H(ct − x) + H(x + ct − 2L) ,

4.7

where H is the Heaviside function deﬁned in (3.5.7). Hence show that the compressive incident wave reﬂects as a tensile wave. [This is the basis of the Hopkinson bar test; the fracture strength of a metal bar is determined by the impulse that must be applied for a piece to be ejected from x = L.] An elastic beam of bending stiﬀness EI and line density is subject to a large compressive force P . Using the approach of Section 4.4.4,

208

4.8

Approximate theories

show that the most unstable waves have wave-length 2π 2EI/P √ and grow over a time-scale t ∼ EI/P . Suppose a strand of uncooked spaghetti is held vertically while a 1 kg weight is placed on the upper end. Explain why it will shatter into fragments roughly 1 cm long. [Typical parameter values for spaghetti are ≈ 2 × 10−3 kg m−1 , EI ≈ 1.5 × 10−4 N m2 .] Writing (nx , ny ) = (sin θ, − cos θ), show that Msn , as given by (4.6.16) and (4.6.8), may be written as ∂2w Eh3 2(1 − ν) sin θ cos θ Msn = 12 (1 − ν 2 ) ∂x∂y 2 2 2 ∂ w ∂ w ∂2w ∂ w 2 2 cos θ . + ν 2 sin θ − ν 2 + − ∂x2 ∂y ∂x ∂y 2 Now use the chain rules cos θ ∂ ∂ ∂ = + sin θ , ∂x 1 + κn ∂s ∂n

∂ sin θ ∂ ∂ = − cos θ , ∂y 1 + κn ∂s ∂n

where the curvature of the boundary is κ = dθ/ds, to show that 2 ∂2w ∂ w Eh3 ∂w Msn = − +ν 2 . + νκ 12 (1 − ν 2 ) ∂n2 ∂n ∂s

4.9

Hence obtain the conditions (4.6.17) that apply on a simply supported boundary. Show that the steady sag of a cylindrically symmetric plate with non-uniform bending stiﬀness D(r) is governed by the equation 2 d w d d2 w 1 dw ςgr2 dw dD r 2 +ν + Dr = − . (E4.1) + dr dr dr dr dr2 r dr 2 For the special case in which we wish w to be given by

2 7+ν 2 2 12a4 2 2 r − a , + w =− a −r 1+ν (1 + ν)2 show that the only solution D(r) of (E4.1) that is bounded as r → a is D=

(1 + ν)ςg . 24(7 − ν) (5 + ν)a2 − (1 + ν)r2

Exercises

4.10

209

Show that, when w is changed to w + η, where |η| |w|, the change in the net strain energy given by (4.6.23) is

∇2 w ∇2 η Ω 2 ∂2w ∂2η ∂ w ∂2η ∂2w ∂2η + − 2 dxdy, − (1 − ν) ∂x2 ∂y 2 ∂y 2 ∂x2 ∂x∂y ∂x∂y

δU = D

and show this can be written as δU = D η∇4 w + div ∇2 w∇η − η∇(∇2 w) Ω

∂ 2 w ∂η ∂ 2 w ∂η ∂ 2 w ∂η ∂ 2 w ∂η − (1 − ν) div − , − ∂y 2 ∂x ∂x∂y ∂y ∂x2 ∂y ∂x∂y ∂x

T

dxdy.

Deduce that ∇4 w = 0 when U is minimised with respect to variations in w. Show that, when Ω is the half-space x > 0 and all variables are assumed to decay at inﬁnity, 3 ∞ ∂η ∂ 2 w ∂ w ∂2w ∂3w dy = 0. +ν 2 −η + (2 − ν) ∂x2 ∂y ∂x3 ∂x∂y 2 −∞ ∂x

4.11

Deduce that (4.6.19) are the natural boundary conditions for the minimisation. 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 it is deformed to z = f (x, y) = W (x, y) + w(x, y). Following the approach adopted in Section 4.6.1, show that the transverse displacement and in-plane stress resultants satisfy Txx

4.12

∂2w ∂2f ∂2f ∂2f 4 + T + 2T − D∇ w − ςg = ς , xy yy ∂x2 ∂x∂y ∂y 2 ∂t2 ∂Txy ∂Txy ∂Tyy ∂Txx + = + = 0, ∂x ∂y ∂x ∂y

where the surface density ς is assumed to be constant. By diﬀerentiating (4.8.8) successively with respect to x and y, show that it can be transformed into the quasi-linear system ∂ p −2s 0 ∂ p ∂K0 /∂x q p + = , q p ∂y q ∂K0 /∂y 0 −2s ∂x q

210

Approximate theories

where p = ∂ 2 f /∂x2 , q = ∂ 2 f /∂y 2 and s = ∂ 2 f /∂x∂y. Hence show that the characteristics are given by −s ± s2 − pq dy = , dx q 4.13

and deduce that the system is hyperbolic if and only if K0 is negative. 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, the tension T and the shear force N in the beam: d (T cos θ − N sin θ) = 0, ds EI

d (T sin θ + N cos θ) = g, ds d2 θ = N. ds2

Consider the special case of a string of length L with zero bending stiﬀness, sagging under gravity and pinned at (x, w) = (0, 0) and (x, w) = (d, 0). Show that 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 the string is a catenary, given by L 2x − d d w= tan θ0 − cosh tan θ0 cosh . 2 tan θ0 L L 4.14

Use the calculus of variations to show that minimisation of the functional

L EI dθ 2 − (1 − cos θ)P ds, U [θ] = 2 ds 0 subject to the boundary conditions θ(0) = θ(L) = 0, leads to the Euler strut equation (4.9.16a). Give physical interpretations of the terms in the integrand.

Exercises

211

Show that the value of U [θ] corresponding to the solution (4.9.17) is approximately U [θ] ≈

4.15

π 2 EIA2 2 n −λ , 4L

where λ = L2 P/π 2 EI. Deduce that θ = 0 is a local minimiser of U , and hence stable, only if λ < 1. 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: d2 θ dθ F cos θ, θ(0) = 0, (L) = 0. = 2 ds EI ds Non-dimensionalise 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 2 ξ θ ∼ − 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 −θ √ dφ √ = ξ 2ε, sin α − sin φ 0 where α is to be determined from α √ dφ √ = 2ε. sin α − sin φ 0 Deduce that the dimensionless shape of the board is given parametrically by x = x(θ), y = y(θ), where ) √ 2 √ sin α − sin α + sin θ , x(θ) = ε −θ 1 sin φ dφ √ , y(θ) = − √ sin α − sin φ 2ε 0

4.16

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. Consider a beam of length L and mass per unit length, clamped horizontally at each end and sagging under gravity. If a horizontal

212

Approximate theories

compressive force P and equal vertical forces are applied at either end, derive the boundary conditions gL , θ(0) = θ(L) = 0, T (0) = T (L) = −P, −N (0) = N (L) = 2 and deduce that the Euler strut equation (4.9.16) becomes, d2 θ L EI 2 + P sin θ + g s − cos θ = 0. ds 2

4.17

By following the non-dimensionalisation of Section 4.9.3 and assuming that gL3 β= = O ε3/2 , EI show that the orthogonality condition (4.9.30) becomes nπ 16β 8λ1 2 . A0 A0 − 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 buckle upwards only if λ 4 + 3β 3/2 /2π 2 . 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 dθ d2 θ EI (s − L) cos θ = 0, θ(0) = π/2, (L) = 0. + 2 ds g ds Now suppose the beam is only slightly perturbed from the vertical, so that θ = π/2 + φ, where |φ| 1. Deﬁning g 1/3 g 1/3 , β=L , ξ = (s − L) EI EI show that φ satisﬁes Airy’s equation d2 φ − ξφ = 0. dξ 2 The solutions of Airy’s equation with zero slope at ξ = 0 are √ φ=a 3 Ai(ξ) + Bi(ξ) , where Ai and Bi are Airy functions (Abramowitz & Stegun, 1972, Section 10.4) and a is an arbitrary constant. Deduce that, for nontrivial solutions, β must satisfy √ 3 Ai(−β) + Bi(−β) = 0,

Exercises

2

z 1

213

3

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.

4.18

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 rebellious hair) before it buckles under its own weight. [See Figure 4.27 for a plot of the ﬁrst three buckling modes.] Show that the exact solution of the buckling problem (4.9.16) satisﬁes −1/2 θ ˆ √ sin2 (θ/2) 2πs 1− dθˆ = λ sin( θ /2) 2 L sin ( θ /2) 0 and deduce that the nth branch of the response diagram is given implicitly by λ=

4.19

2 4n2 2 K sin ( θ /2) , 2 π

where K denotes an elliptic integral (Gradshteyn & Ryzhik, 1994, Section 8.1). An inﬁnite beam initially lying at rest along the x-axis is subject to a transverse point force F (t) acting at the origin. Derive the model ∂2w ∂4w + F (t)δ(x) = , ∂x4 ∂t2 ∂w (x, 0) = 0, w(x, 0) = ∂t

−EI

w→0

as

x → ±∞

and, for example by taking a Fourier transform in x, show that the

214

Approximate theories

displacement of the origin is related to the applied force by t √ 1 F (τ ) t − τ dτ. w(0, t) = √ 2π (EI 3 )1/4 0 If the origin is forced to move at speed V , so w(0, t) = V t, show that the required force is ) 2 3 1/4 . F (t) = 2V EI πt

4.20

[This is a simple model for the action of a lawn-mower blade on a grass stalk, or of a razor blade on a bristle.] Start from the equation (4.11.31) 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 ∂ 3 u ∂u + 6u2 + 3 = 0. ∂τ ∂z ∂z Show that the Miura transform ∂u ∂z converts the mKdV equation into the Korteweg–de Vries (KdV) equation ∂v ∂ 3 v ∂v + 6v + = 0. ∂τ ∂z ∂z 3 v(ζ, τ ) = u2 − i

4.21

By multiplying (4.11.33) by d2 θ/dη 2 , obtain the ﬁrst integral 2 2 2 dθ 1 dθ 4 d θ 2 + −κ = 0, 2 dη 4 dη dη where

) κ=

2T∞ (1 − M ) . EI

By using the substitution dθ = ±2κ sech φ(η) , dη or otherwise, obtain the general solution θ = const. ± 4 tan−1 eκ(η0 −η) .

5 Nonlinear elasticity

5.1 Introduction In Chapters 2 and 3, we have analysed solutions of the steady and unsteady Navier equations, which were derived in Chapter 1 under the two assumptions that underpin linear elasticity. First, we assumed that we could discard the nonlinear terms in the relation (1.4.5) 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 (see Treloar, 2005). Models for such materials require mechanically nonlinear elasticity, in which the stress is a nonlinear function of the strain. As indicated in Section 1.6, 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 will 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 (1.7.6), and we will see that great care has to be taken to avoid models that allow unphysical behaviour. Finally we will show how the nonlinear theory, be it mechanical or geometric or both, can be greatly simpliﬁed for hyperelastic materials for which the stress is related to 215

216

Nonlinear elasticity

the strain via the partial derivatives of an appropriate strain energy density, thereby generalising (1.9.4).

5.2 Stress and strain revisited 5.2.1 Deformation and strain We begin by recalling the analysis of strain from Section 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 ,

(5.2.1)

where the deformation gradient tensor is deﬁned by F = (Fij ) = (∂xi /∂Xj ) .

(5.2.2)

The change in the length of the small element dX is thus given by |dx|2 = dX T C dX,

(5.2.3)

C = F T F,

(5.2.4)

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 x = c + P x,

(5.2.5)

where the vector c and orthogonal matrix P are constant. It is a straightforward matter to substitute for x into (5.2.2) and thus obtain the new deformation gradient as Fij =

∂xi = Pik Fkj ∂Xj

(5.2.6)

or, in matrix notation, F = P F. From the orthogonality of P , it therefore follows that T C = F F = F T F = C,

(5.2.7)

(5.2.8)

5.2 Stress and strain revisited

217

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 → x(X ),

where

X = c + QX,

(5.2.9)

where the vector c and orthogonal matrix Q are constant. Now the chain rule gives ∂xi (5.2.10) = F QT F = ∂Xj and the corresponding Green deformation tensor is thus T C = F F = QCQT .

(5.2.11)

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 1 T E = (C − I) = (5.2.12) F F −I . 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 (5.2.3), written in the form (1.4.4), namely |dx|2 − |dX|2 = 2dX T E dX,

(5.2.13)

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.

(5.2.14)

This is called the left Green deformation tensor or, more commonly, the Finger tensor. We will ﬁnd that B is more natural than C as a measure of deformation for an observer using Eulerian coordinates x.

218

Nonlinear elasticity dX

dx

N n

da

dA

Fig. 5.1 The deformation of a small scalene cylinder.

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 da and normal n, is given by df = τ n da,

(5.2.15)

where τ = (τij ) is the Cauchy stress tensor, as deﬁned in Chapter 1. As in Section 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 (5.2.1). The volumes of the two cylinders are given by dV = dX · dA and dv = dx · da respectively and, since the Jacobian J = det (()F ) 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, (5.2.16) in which the vector element dX is arbitrary, so the area elements are related by −1 dA. (5.2.17) da = det (F ) F T When we apply this result to the expression (5.2.15) for the force on a small area element, we obtain −1 dA. (5.2.18) df = det (F ) τ F T

5.2 Stress and strain revisited

219

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 , (5.2.19) T = det (F )τ F T is a second-rank tensor called the ﬁrst Piola–Kirchhoﬀ stress tensor. As shown in Section 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.

(5.2.20)

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 , (5.2.21) so that S is symmetric.

5.2.3 The momentum equation In Section 1.6 we derived Cauchy’s momentum equation ρ

∂τij ∂ 2 xi = ρgi + , 2 ∂t ∂xj

(5.2.22)

which must apply to any continuum for which a Cauchy stress tensor (τij ) can be deﬁned. In (5.2.22), 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 ). (5.2.23) ρ Now let us derive the Lagrangian equivalent of (5.2.22). We start, as in Section 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 ): ∂xi d ρ dx = gi ρ dx + τij nj da. (5.2.24) dt V (t) ∂t V (t) ∂V (t) Now we change the integration variables to Lagrangian coordinates X, using (5.2.17) to convert the ﬁnal surface integral: d ∂xi −1 ρJ dX = Jτik Fjk Nj dA. gi ρJ dX + dt ∂t V (0) V (0) ∂V (0) (5.2.25)

220

Nonlinear elasticity

With respect to Lagrangian variables, the integration domain V (0) is timeindependent, 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–Kirchhoﬀ tensor Tij and apply the divergence theorem to obtain ∂Tij ∂ 2 xi ρ dX = gi ρ0 dX + dX. (5.2.26) 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: ∂Tij ∂ 2 xi . (5.2.27) ρ0 2 = ρ0 gi + ∂t ∂Xj Although it is not immediately obvious, (5.2.27) may also be obtained directly by using (5.2.19) to replace τ with T in (5.2.22) 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 (5.2.13), we obtain the one-dimensional strain 1 ∂u 2 ∂u + (5.2.28) Exx = ∂X 2 ∂X and hence Txx = (F S)xx

∂x = φ ∂X

∂u 1 + ∂X 2

∂u ∂X

2 .

(5.2.29)

In equilibrium under a constant stress Txx , the displacement is such that ∂u 1 ∂u 2 ∂u φ + = Txx . (5.2.30) 1+ ∂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 ,

(5.2.31)

then (5.2.30) is a cubic equation for ∂u/∂X with only one real root.

5.3 The constitutive relation

221

More exciting possibilities occur when we examine longitudinal waves propagating in the X-direction. Again assuming the linear constitutive relation (5.2.31), we ﬁnd that the momentum equation (5.2.27) in one spatial dimension reads ∂2u ∂u 1 ∂u ∂u 2 ∂ 1+ 1+ , (5.2.32) = cp ∂t2 ∂X ∂X 2 ∂X ∂X where c2p = (λ + 2µ)/ρ0 . This is a nonlinear generalisation 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 characteristics X ± cp t = const. in the (X, t)-plane. However, for (5.2.32), the characteristics are given by the equation dX = ± cp dt

3 ∂u + 1+3 ∂X 2

∂u ∂X

2 1/2 ,

(5.2.33)

and they are generally not straight lines. 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 characteristics are usually incompatible, so that u becomes multi-valued. This results in the formation of a shock, which is a curve in the (X, t)-plane across which the ﬁrst derivatives of u are discontinuous (Bland, 1988, Section 10.1).

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.

(5.3.1)

Hence, if the real, positive eigenvalues of U are denoted λ1 , λ2 and λ3 , there exists an orthogonal matrix R such that  λ1 0 0 Λ =  0 λ2 0  . 0 0 λ3 

U = RT ΛR,

where

(5.3.2)

222

Nonlinear elasticity

The rows of R are the eigenvectors of U , that is  T e1 , R = eT where U ei = λi ei , ei · ej = δij . 2 T e3

(5.3.3)

Since F is a function of X, it follows that M , U , R and Λ likewise vary with X. Substitution of (5.3.1) and (5.3.2) into (5.2.1) leads to the relation dx = M T RT ΛR dX

(5.3.4)

between line elements dX and dx in the reference and deformed conﬁgurations respectively. Using (5.3.3), we can decompose dX into its components along each of the eigenvectors ei : λi ei (ei · dX) . (5.3.5) dx = M T 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 stretches and corresponding eigenvectors ei to be the principal directions corresponding to the deformation; both will, in general, vary with position. By substituting (5.3.1) into (5.2.4), we see that U is related to C by C = U 2 = RT Λ2 R.

(5.3.6)

It follows that C shares the same eigenvectors ei as U , while the eigenvalues of C are λ21 , λ22 and λ23 . Moreover, since C is related to the strain tensor E by (5.2.12), the principal strains Ei , introduced in Section 2.2.5, are related to λi by (5.3.7) λi = 1 + 2Ei . 5.3.2 Strain invariants We may expand out the characteristic polynomial of C as det (ζI − C) = ζ − λ21 ζ − λ22 ζ − λ23 = ζ 3 − I1 (C)ζ 2 + I2 (C)ζ − I3 (C), (5.3.8) 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 . (5.3.9)

5.3 The constitutive relation

223

If C is diagonal (with entries λ21 , λ22 , λ23 ), then it is clear that the three invariants may be written as I1 (C) = Tr (C),

I2 (C) =

1 Tr (C)2 − Tr C 2 , 2

I3 (C) = det (C). (5.3.10)

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 (5.3.10) 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 Section 5.2.1, transformation of the axes by an orthogonal matrix Q converts C to C = QCQT , so the invariance required of φ may be stated as φ QCQT ≡ φ (C) (5.3.11) for all orthogonal matrices Q. In particular, (5.3.11) must hold if Q is equal to R, the orthogonal matrix that diagonalises C, so φ (C) ≡ φ Λ2 ,

(5.3.12)

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 ) ,

(5.3.13)

which represents a considerable restriction of all possible functions of the six independent components of the symmetric tensor C. Note that B in (5.2.14) also has eigenvalues λ2i and invariants Ii , although generally not the same eigenvectors as C. We also recall from the Cayley– Hamilton theorem that C satisﬁes its own characteristic polynomial, namely C 3 − I1 C 2 + I2 C − I3 I = 0.

(5.3.14)

This can be used to express any power of C in terms of just I, C and C 2 , and similarly for B.

224

Nonlinear elasticity

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 );

(5.3.15)

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 (5.3.15) 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 Section 5.2.1, rotation of the Eulerian coordinate axes through an orthogonal matrix P transforms F to F = P F , so this frame indiﬀerence leads to the restriction S (F ) ≡ S (P F )

(5.3.16)

for all orthogonal matrices P . Since the Green deformation tensor C is invariant under rigid-body motions, (5.3.16) is certainly satisﬁed if the constitutive relation takes the form S = S(C),

(5.3.17)

and it may be shown, as in Exercise 5.3, that this condition is necessary as well as suﬃcient. Notice that (5.3.17) is more speciﬁc than (5.3.15) 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 = X 0 + Q (X − X 0 ) .

(5.3.18)

Then, if we take the rotated material and apply exactly the same deformation as before, the net deformation is given by (5.3.19) X → x(X ) = x X 0 + Q (X − X 0 ) . As shown in Section 5.2.1, the Green deformation tensor corresponding to (5.3.19) is C = QCQT , and the corresponding stress tensor at the point X 0 is therefore S (C) = S QCQT . (5.3.20)

5.3 The constitutive relation

225

If the material is isotropic, then this should be identical to the stress which we would have obtained without the original rotation, albeit related to the new coordinates X rather than to X. By applying the usual rule for the rotation of a tensor, we deduce that S (C) = QS(C)QT ,

(5.3.21)

which must apply identically for any ﬁxed point X 0 and any rotation matrix Q. The constitutive relation (5.3.17) between S and C must therefore satisfy the symmetry condition S(QCQT ) ≡ QS(C)QT

(5.3.22)

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 (5.3.22) 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 ,

(5.3.23)

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 strain, when Cij → δij + 2eij in the notation of Chapter 1, the only relevant strain invariant is the trace ekk . Hence the right-hand side of (5.3.23) reduces to a linear combination of ekk δij from the ﬁrst term and eij from the second and third terms, thereby retrieving (1.7.6). We will return to the small displacement limit in Section 5.3.6. We may directly infer from (5.3.23) that in an isotropic material, S has the same eigenvectors as C. On the other hand, using (5.2.19), equation (5.3.23) gives, for the Cauchy stress, $ −1/2 # φ0 B + φ1 B 2 + φ2 B 3 , (5.3.24) τ = I3 so that the principal axes of τ coincide with those of B. Furthermore, using (5.3.14), one can eliminate B 3 and obtain $ −1/2 # τ = I3 I3 φ2 I + (φ0 − I2 φ2 ) B + (φ1 + I1 φ2 ) B 2 . (5.3.25) 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.

(5.3.26)

226

Nonlinear elasticity

However, we will see in the next section that the requirement (5.3.26) is far from suﬃcient for the constitutive relation (5.3.23) 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 (5.2.27) by the velocity ∂xi /∂t and integrating over a control material volume V0 : ∂xi ∂ 2 xi ∂xi ∂xi ∂Tij dX + ρ0 dX = ρ0 gi dX. (5.3.27) 2 ∂t ∂t ∂t V0 V0 V 0 ∂t ∂Xj 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

V0

ρ0 2

∂xi ∂t

2

∂Fij dX + Tij dX ∂t V0 ∂xi ∂xi dX + Tij Nj dA. (5.3.28) ρ0 gi = ∂t V0 ∂V 0 ∂t

Recalling that ρ0 dX ≡ ρ dx, we see that the ﬁrst term on the left-hand side of (5.3.28) 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 (5.3.28) 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 constitutive relation S = λ21 λ22 λ23 − 1 I. (5.3.29) subject to a deformation in which   0 0 λ1 (t) F = 0 λ2 (t) 0 , 0 0 1

(5.3.30)

5.3 The constitutive relation

227

where λ1 , λ2 are periodic functions of t. We use (5.2.21) to calculate the ﬁrst Piola–Kirchhoﬀ stress as   0 0 2 2 λ1 (t) T = F S = λ 1 λ2 − 1  0 (5.3.31) λ2 (t) 0 0 0 1 and hence ﬁnd that 2 2 ∂Fij dλ1 dλ2 dX = V0 λ1 λ2 − 1 λ1 + λ2 Tij , ∂t dt dt V0

(5.3.32)

which is a function whose time integral over a period may be positive or negative. When λ1 = 1 + a cos(ωt), λ2 = 1 + bsin(ωt), for example, the integral of (5.3.32) over a period 2π/ω is 3πab b2 − a2 /2. Hence, if an elastic material were to exist that satisﬁed (5.3.29), then it could be used as a limitless energy source! This behaviour is clearly unacceptable and a suﬃcient condition to preclude it is the existence of a strain energy density W (Fij ) such that Tij =

∂W , ∂Fij

(5.3.33a)

or, introducing the convenient notation for diﬀerentiation by a tensor, T = If so, then we can write Tij V0

∂W . ∂F

∂Fij d dX ≡ ∂t dt

(5.3.33b) W dX,

(5.3.34)

V0

and interpret the right-hand side as the strain energy stored in the medium; the change in this energy automatically integrates to zero over a periodic loading cycle. We will now explore this happy circumstance in more detail. 5.3.5 Hyperelasticity A material is called hyperelastic if there is a strain energy density W (Fij ) satisfying (5.3.33). 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 (5.3.33) 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

228

Nonlinear elasticity

that our experience with linear elasticity in Chapter 1 suggests that W (Fij ) should have a minimum (taken to be zero without loss of generality) at Fij = δij . As in Section 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–Kirchhoﬀ stress is then given by S=2

∂W , ∂C

(5.3.35)

as shown in Exercise 5.8. 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 Section 5.3.2 to assert that W = W (I1 , I2 , I3 ) .

(5.3.36)

The second Piola–Kirchhoﬀ stress tensor is therefore given by S=2

∂W ∂Ik . ∂Ik ∂C

(5.3.37)

k

To ﬁnd the circumstances under which such a strain energy function can exist, we note from Exercise 5.9 that ∂I1 = I, ∂C

∂I2 = I1 I − C, ∂C

∂I3 = I3 C −1 , ∂C

(5.3.38)

where I is the identity. Hence, after a bit of algebra (see Exercise 5.11), we ﬁnd that ∂W 2 I− C+ C , S=2 ∂I3 (5.3.39) which gives us explicit forms for the functions φi in (5.3.23). Since the φi satisfy

φ2 = 2

∂W ∂W ∂W + I1 + I2 ∂I1 ∂I2 ∂I3

∂W ∂W + I1 ∂I2 ∂I3

∂W ∂W ∂W , φ1 + I1 φ2 = −2 , φ0 + I1 (φ1 + I1 φ2 ) − I2 φ2 = 2 , ∂I3 ∂I2 ∂I1 (5.3.40)

5.3 The constitutive relation

229

cross-diﬀerentiation reveals that

∂φ0 ∂I3 ∂φ1 ∂φ0 + I1 ∂I2 ∂I2

∂φ1 ∂φ2 ∂φ2 + + I1 = 0, ∂I3 ∂I2 ∂I3 ∂φ2 ∂φ2 ∂φ1 2 + I1 + I1 − I2 − = 0, ∂I3 ∂I3 ∂I1 ∂φ2 ∂φ1 2 ∂φ2 + + I1 − I2 + I1 = 0. ∂I1 ∂I2 ∂I1

(5.3.41a) (5.3.41b) (5.3.41c)

Any choice of the functions φi that does not satisfy (5.3.41) will lead to behaviour such as that exhibited in (5.3.32); indeed it was these integrability conditions that suggested the counter-example (5.3.29). We note from (5.2.21) that the Cauchy stress is given in terms of W as −1/2 ∂W

τ = I3

∂F

FT

(5.3.42)

and, as shown in Exercise 5.7, this leads to the useful expression −1/2

τi = I3

λi

∂W ∂λi

(5.3.43)

(no summation) for the Cauchy principal stresses. As well as clarifying the characterisation of elastic materials, hyperelasticity also often simpliﬁes the mathematical and numerical analysis of nonlinear elastic problems. The net elastic energy stored in a material volume V0 is given by W dX. (5.3.44) U= V0

By virtue of (5.3.33), a calculation analogous to Exercise 1.8 implies that the steady version of the gravity-free momentum equation (5.2.27) is a necessary condition for U to be minimised with respect to the displacement, assuming of course that W is suﬃciently smooth. Crucially, the minimisation of a functional like (5.3.44) is generally easier computationally than solving the nonlinear partial diﬀerential equation (5.2.27). In particular, minimisation problems of this type are naturally suited to the ﬁnite element method (see Fung & Tong, 2001, Chapter 18). However, for a nonlinear material, ﬁnding the uniqueness and smoothness of a minimum of (5.3.44) now poses a typically diﬃcult challenge in the calculus of variations. It transpires that not only do we require W to be strictly positive except at F = I, but also that it should be quasi-convex, 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

230

Nonlinear elasticity

equilibrium, (5.2.27) should comprise, with suitable boundary conditions, a well-posed system for xi .

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 (5.3.7) that the principal strains are related to the principal stretches by (5.3.45) λi = 1 + 2Ei ∼ 1 + Ei when |Ei | 1. 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 }. Hence, as λ1 , λ2 , λ3 → 1, the local behaviour of W near its minimum must take the form W∼µ

3 i=1

(λi − 1)2 +

3 λ (λi − 1) (λj − 1) , 2

(5.3.46)

i,j=1

where the scalar constants λ and µ, which we will shortly identify with the Lam´e constants of Chapter 1, must satisfy µ > 0 and 3λ+2µ > 0 for (5.3.46) to be positive deﬁnite. Since, in principal axes, W = W(λi ) and C = diag λ2i , (5.3.35) implies that 1 ∂W . (5.3.47) Si = λi ∂λi Substitution from (5.3.45) and (5.3.46) thus leads to Si ∼ λ (λ1 + λ2 + λ3 − 3) + 2µ (λi − 1) ∼ λ (E1 + E2 + E3 ) + 2µEi (5.3.48) as the strain tends to zero. The stress and strain tensors are thus related by S ∼ λ Tr (E) I + 2µE,

(5.3.49)

and, since the left- and right-hand sides both transform as tensors, (5.3.49) 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 (5.2.19) and (5.3.48), both the Piola–Kirchhoﬀ stress tensors and the Cauchy stress tensor are all approximately equal. In addition, the quadratic

5.3 The constitutive relation

231

terms in the strain tensor may be neglected to leading order, and (5.3.49) thus reproduces the linear elastic constitutive relation (1.7.6). However, (5.3.49) 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 (5.3.49), while retaining the geometrically nonlinear relation between x and X. Indeed we have already done this several times in Chapter 4.

5.3.7 Incompressibility Many rubber-like materials are observed to be almost incompressible, and it is common to employ constitutive relations that impose this. From (1.3.6), which holds for nonlinear deformations, the condition of incompressibility is J = ρ0 /ρ = 1 or, in terms of the strain invariants, I3 ≡ 1.

(5.3.50)

As we found in Section 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 into nonlinear elasticity in a self-consistent way, and there are two ways to go about this. The approach followed in Section 1.8 was to characterise the compressibility by a small but nonzero parameter ε and then carefully take the limit ε → 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 (5.3.50), as in Exercise 1.8. We can follow the ﬁrst strategy by constructing a strain energy density that heavily penalises departures of I3 from unity, for example W (I1 , I2 , I3 ) =

µ (I3 − 1)2 ˜ (I1 , I2 ) , +W 2ε

(5.3.51)

where ε 1. The actual functional form of W is unimportant provided it has a very strong minimum at I3 = 1, but the particular example (5.3.51) will help to ﬁx ideas. We anticipate that this strain energy function will force I3 to remain close to unity, and therefore set I3 = 1 −

2ε p, µ

(5.3.52)

where the scalar function p is to be determined. Substituting (5.3.51) into

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Nonlinear elasticity

(5.3.39) and using (5.3.14), we ﬁnd that our desired constitutive relation becomes ∂W ∂W ∂W I −2 + I1 C − pC −1 . (5.3.53) S=2 ∂I1 ∂I2 ∂I2 The status of p as a hydrostatic pressure is clearer in the corresponding expression for the Cauchy stress, namely ∂W ∂W 2 ∂W τ = −pI + 2 B−2 + I1 B . (5.3.54) ∂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

(5.3.55)

In Exercise 5.6, we show that this is equivalent to Fji−1 δFij = 0.

(5.3.56)

We also note from (5.3.28) that no additional work is done by the increment provided Tij δFij = 0.

(5.3.57)

Hence, considering only virtual deformations satisfying the constraint (5.3.56), we deduce that T is only determined up to an unknown scalar −1 . An arbitrary scalar multiple of C −1 may be thus added multiple of F T to S, and (5.3.53) follows immediately. 5.3.8 Examples of constitutive relations Some commonly used examples of constitutive relations for the strain energy density are µ (5.3.58a) neo-Hookean: W = (I1 − 3) , 2 (5.3.58b) Mooney–Rivlin: W = c1 (I1 − 3) + c2 (I2 − 3) , W = 2µ (λ1 + λ2 + λ3 − 3) , (5.3.58c) α α α λ + λ2 + λ3 − 3 , (5.3.58d) Ogden: W = 2µ 1 α where µ, c1 , c2 and α are positive constants. In each case, the material is also assumed to be incompressible, so the constitutive relation is supplemented by the constraint I3 ≡ 1. Varga:

5.4 Examples

F

(a)

F

λ1 F

(b)

233

F

λ1

(d)

F

λ1

(c)

λ1

(e)

F

(f)

λ1

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 suﬃcient 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 ,

(5.4.1)

234

Nonlinear elasticity

F h

F Fig. 5.3 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 (5.3.54) to be τi = λi

∂W − p. ∂λi

(5.4.2)

Here, the λi are treated as independent variables while performing the derivatives. We focus on the Mooney–Rivlin constitutive relation (5.3.58b), for which (5.4.2) reads τi = 2c1 λ2i −

2c2 − p, λ2i

(5.4.3)

where we have made use of the incompressibility condition λ21 λ22 λ23 ≡ 1. 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 , in which case the stresses are also uniform and given by (5.4.2). By relating these to the applied force F , we obtain the equations λ1

F ∂W , −p= ∂λ1 λ2 λ3 h

λ2

F ∂W , −p= ∂λ2 λ1 λ3 h

λ3

∂W − p = 0, ∂λ3

(5.4.4)

and elimination of p leads to ∂W λ3 ∂W ∂W λ3 ∂W F = − = − . h ∂λ1 λ1 ∂λ3 ∂λ2 λ2 ∂λ3

(5.4.5)

To proceed further, we now have to choose a constitutive relation. Using the Mooney–Rivlin model (5.4.3), we ﬁnd that the stretch ratios satisfy $ # (5.4.6) (λ1 − λ2 ) c1 1 + λ31 λ32 + c2 λ21 + λ1 λ2 + λ22 − λ41 λ42 = 0.

5.4 Examples

(a)

λ2

(b)

W

5

235

(c)

F

30 15 25

4

12.5 20 3

10 15 7.5

2

λc 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.

This always has the solution λ1 = λ2 but, whenever c2 = 0, (5.4.6) also admits asymmetric solutions in which λ1 = λ2 . As shown in Figure 5.4(a), there is a critical value λc of the stretches at which symmetric response 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 non-uniqueness 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

236

Nonlinear elasticity

P

c = 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 c = c2 /c1 .

to the stretched radius r by λθ = λφ =

r , R

(5.4.7)

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, ∂W ∂W , (5.4.8) − λr 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 P =

2T . r

(5.4.9)

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 (5.4.3), for example, we ﬁnd 4 1 + c λ2θ λ6θ − 1 RP = P (say) = , (5.4.10) hc1 λ7θ where c is the dimensionless ratio c2 /c1 . With c = 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 c , however, the behaviour is more complicated, with

5.4 Examples

237

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. Increasing c corresponds to making the material stiﬀer at large strains, reﬂecting the fact that polymer molecules can only withstand a ﬁnite extension without breaking. If c exceeds a critical value 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 = λ θ =

r , R

(5.4.11)

and the incompressibility condition λR λθ λφ = 1 gives us a diﬀerential equation for r(R), namely dr R2 = 2. dR r

(5.4.12)

Denoting by a the initial radius of the sphere, we write the solution of (5.4.12) as 1/3 , (5.4.13) r = R3 + λ30 a3 − a3 where λ0 = λθ (0) =

r(a) a

(5.4.14)

is the expansion ratio of the cavity. A displacement of the form (5.4.13) is the only incompressible purely radial deformation. We can read oﬀ from

238

Nonlinear elasticity P/c1

0.2

10

0.1

8

6

c = 0

4

2

2

3

4

5

7

6

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 c = c2 /c1 .

(5.4.11) the principal stretches λR =

R2 R3 + λ30 a3 − a3

2/3 ,

R3 + λ30 a3 − a3 λφ = λ θ = R

1/3 .

(5.4.15)

We next turn to the familiar radial momentum equation from (1.11.16): 2τrr − τθθ − τφφ dτrr + = 0. dr r

(5.4.16)

We can use (5.4.13) to transform to Lagrangian coordinates and substitute for the stress components from (5.4.2), thus obtaining R3 + a3 λ20 − 1 dτrr ∂W ∂W = 2λθ − 2λR . (5.4.17) R2 dR ∂λθ ∂λR Now solving the ﬁrst-order diﬀerential equation (5.4.17) subject to the two boundary conditions τrr (a) = −P,

τrr → 0

as

R → ∞,

(5.4.18)

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 4 1 1 (5.4.19) − 4 − 2c2 1 − 2λ0 + 2 , P = c1 5 − λ0 λ0 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

Exercises

239

to withstand unbounded strains, and the situation is remedied whenever c2 is positive.

5.5 Concluding remarks This chapter has provided an introduction to what is probably the most mathematically challenging branch of elasticity theory. In principle, our basic task is to solve for xi from (5.2.27), 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 (5.3.44). However, even here we have to select from a wide range of functions relating the strain energy density to the strain invariants, and, 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

5.2

Show that the displacement corresponding 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 (5.2.12) is identically zero for this displacement ﬁeld, but that the linearised strain (1.7.1) is not. (a) Prove the identity ∂xj 1 ∂J ∂ , (E5.1) ≡ ∂xj ∂Xk J ∂Xk (assuming the summation convention). [Hint: integrate both

240

Nonlinear elasticity

sides over an arbitrary material volume V (t) and use the divergence theorem.] (b) Substitute the Piola–Kirchhoﬀ stress tensor for the Cauchy stress tensor in the momentum equation (5.2.22) and use the chain rule to obtain −1 ∂ 2 xi −1 ∂ J Til Fjl . (E5.2) ρ0 2 = ρ0 gi + JFkj ∂t ∂Xk

5.3

(c) Expand out the right-hand side of (E5.2) and apply the identity (E5.1) to obtain the Lagrangian momentum equation (5.2.27). Suppose the scalar function f of the two-dimensional matrix F is invariant under the transformation F → P F , where P is any orthogonal matrix. Let f (F ) = g(A, B, α, β), where F is written in the form A cos α B cos β F11 F12 = . F = F21 F22 A sin α B sin β By expressing P as the rotation matrix 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)

5.4

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).] (a) Suppose that two 2 × 2 symmetric matrices S and C satisfy a functional relation S = S(C) which enjoys the symmetry (5.3.22) for all orthogonal matrices Q. By considering 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 0 s1 (λ21 , λ22 ) , S= 0 s2 (λ21 , λ22 )

Exercises

241

where s1 and s2 are arbitrary functions of the eigenvalues λ21 and λ22 of C. Show that this expression may be manipulated to S = φ0 (I1 , I2 )I + φ1 (I1 , I2 )C,

5.5

(E5.3)

where φ0 and φ1 are two arbitrary functions of the invariants I1 and I2 of C (note that there are only two such invariants in two dimensions). (c) Show that (E5.3) is invariant with respect to orthogonal transformations and, hence, is true whether or not C is diagonal. (d) Generalise the above argument to three dimensions. 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 dλ1 dλ2 + λ2 S22 dt dt per unit volume. Deduce that the medium conserves energy if and only if the stress components satisfy q = λ1 S11

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 2π/ω q dt = 2πε2 (α − 1)µ. 0

5.6

Show that ∂det (F ) = det (F )Fji−1 . ∂Fij

5.7

[Hint: recall the inversion formula Fij−1 = cofactor (Fji ) / det(F ).] Recall from Section 5.3.1 the existence of two orthogonal matrices R and M such that F = M T RT ΛR, where Λ = diag (λi ). Use the chain rule to show that ∂W ∂W = M T RT R. T = ∂F ∂Λ

242

Nonlinear elasticity

Now use the relationship (5.2.19) between the Cauchy and Piola– Kirchhoﬀ stresses to show that −1/2 ∂W

RM τ M T RT = I3

5.8

∂Λ

Λ−1 .

Since Λ is diagonal, deduce that the principal Cauchy stresses are given by (5.3.43). If the strain energy density W is a function only of the symmetric Green deformation tensor Cij , show that ∂W ∂W = Fik , ∂Fij ∂Ckj and deduce that the second Piola–Kirchhoﬀ stress tensor is given by S=2

5.9

Use Exercise 5.6 to prove the identities (a)

5.10

∂W . ∂C

∂I1 = δij , ∂Cij

(b)

∂I2 = I1 δij − Cij , ∂Cij

(c)

∂I3 −1 = I3 Cij . ∂Cij

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 F λ2 ∂W ∂W − , = A ∂λ1 λ1 ∂λ2 −1/2

5.11

5.12

where λ2 = λ3 = λ1 . Hence justify the response diagrams shown in Figure 5.2. Use the results from Exercise 5.9 to write out equation (5.3.37) as ∂W ∂W ∂W −1 ∂W S=2 + I1 C + I3 C I− . (E5.4) ∂I1 ∂I2 ∂I2 ∂I3 Using (5.3.14), deduce (5.3.39). Assuming the Mooney–Rivlin form for the strain energy in (5.4.4), deduce (5.4.6). Show that the critical stretch λc , at which bifurcation from symmetric to asymmetric stretching occurs, is given implicitly by 1 + λ6 c2 = c = 2 6 c , c1 λc (λc − 3)

Exercises

243

where λc > 31/6 . By introducing Θ = λ1 λ2 , show that the asymmetric solution satisﬁes λ1 = λ± , where λ2± = 5.13

−P (Θ) ±

λ2 =

+ P (Θ) + 4c 2 Θ2 2c

Θ , λ±

, P (Θ) = 1 + Θ3 + c Θ − Θ4 .

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 Section 5.4.3, show that the internal pressure P and the axial tension Tz are related to the tangential stretches by 2 1 + c λ2z λ4θ λ2z − 1 RP = , P = hc1 λ4θ λ3z 2 1 + c λ2θ λ2θ λ4z − 1 Tz = , hc1 λ3θ λ3z where c = c2 /c1 . (b) If the tube is constrained in the z-direction so that λz = 1, show that only a ﬁnite pressure P = 2 (1 + c ) is needed to inﬂate the tube to inﬁnity. Show also that the axial tension 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 → ∞ as λθ → ∞, although the behaviour is non-monotonic if c >

5.14

√ 2/3 34 + 11 11 ≈ 4.663. 7

(a) By changing independent variables to λθ , show that the differential equation (5.4.17) for the stress outside a spherical cavity may be written in the form dτrr ∂W ∂W = 2λθ − 2λR −λθ λ3θ − 1 dλθ ∂λθ ∂λR

244

Nonlinear elasticity

and, for the Mooney–Rivlin model, derive the equation 4 c1 + c2 λ2θ λ3 + 1 dτrr =− . dλθ λ5θ (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 (5.4.19) 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 4 1 4 1 1 1 + 4 − − 4 −2c2 2λ − 2 − 2λ0 + 2 , P = c1 λ λ λ0 λ λ0 λ0 where

λ =

1+

1/3 3 λ0 − 1 a3 (a + h)3

.

Hence recover the balloon result (5.4.10) 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 and boundary conditions. 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 the tools that are widely useful for analysing more general thin structures. 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 non-dimensionalising 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. Kevorkian & Cole (1981), Hinch (1991) and Bender & Orszag (1978) provide very good general expositions of these methods. In fact we have already encountered an example of this approach when discussing the torsion of a thin-walled tube in Section 2.5, and an even simpler paradigm is the dimensionless model for antiplane strain of a thin plate. We will use this example in Section 6.2 to introduce the asymptotic arguments that form the basis of this chapter. Their systematic nature oﬀers 245

246

Asymptotic analysis

many advantages compared to the less rigorous derivations of Chapter 4. For example, in principle the process 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 unable to resolve the details of the tractions applied to the edges. A boundary layer analysis in the neighbourhood of these edges will allow us to explain how such tractions can eﬀectively be replaced by an equivalent point force/moment system, without having to assume Saint-Venant’s principle. Finally, since we start from the general continuum equations of elasticity, our asymptotic approach will lead to explicit formulae for practically useful parameters, such as the bending stiﬀness, in terms of the underlying elastic parameters, such as the Lam´e constants. In Section 6.3 we re-derive the linear plate equation. As it describes inﬁnitesimal transverse displacements of a plate, we can obtain it starting from linear elasticity. In Section 6.4, we analyse a boundary layer near an edge of the plate and re-establish the correct boundary conditions for this model. Next, in Section 6.5, we re-derive the improved von K´ arm´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 Section 6.3. In Section 6.6, we extend the analysis to describe large two-dimensional deﬂections of a plate and thus derive the Euler strut equation. Finally, in Sections 6.7–6.8, we illustrate the geometrical generalisations needed to obtain the equations governing inﬁnitesimal displacements of rods and shells. 6.2 Antiplane strain in a thin plate We ﬁrst introduce our asymptotic methodology by modelling small antiplane displacements in a thin plate such that the region D in Figure 2.5 is the rectangle 0 < y < h, 0 < x < L. The slenderness parameter in this case is h . (6.2.1) L Suppose the lateral boundaries of this plate are traction-free and that w is prescribed on the ends x = 0, L. The ﬁrst step towards an asymptotic analysis is to non-dimensionalise the spatial coordinates and w by writing ε=

x = Lx ,

y = hy ,

w = W w ,

(6.2.2)

6.2 Antiplane strain in a thin plate

247

where W is a typical displacement at the boundary. From (2.3.2) and (2.3.3) we ﬁnd that the dimensionless displacement w satisﬁes ε2

∂ 2 w ∂ 2 w + =0 ∂x 2 ∂y 2

with

∂w = 0 on y = 0, 1, ∂y

(6.2.3)

along with w = f (y ) at x = 0 and w = g(y ) at x = 1, where both f and g are prescribed functions. To avoid clutter in the equations, we drop the primes henceforth. Away from the ends of the plate, we expand w in an asymptotic expansion in which w ∼ w0 + ε2 w2 + · · · as ε → 0. This gives, to lowest order in ε, ∂ 2 w0 =0 ∂y 2

with

∂w0 = 0 on y = 0, 1. ∂y

(6.2.4)

Hence w0 = w0 (x), but this function is not determined at this stage and we must proceed further in the asymptotic expansion. Collecting all O ε2 terms, we ﬁnd that d2 w0 ∂ 2 w2 = − ∂y 2 dx2

with

∂w2 = 0 on y = 0, 1. ∂y

The solution w2 can only satisfy both boundary conditions if 1 2 ∂w2 1 ∂ w2 dy = . ∂y 2 ∂y 0 0

(6.2.5)

(6.2.6)

This solvability condition for w2 gives rise to the “thin sheet” equation d2 w0 =0 dx2

(6.2.7)

for w0 . To complete the model, we need to derive boundary conditions for (6.2.7). We do this by subjecting the ends of the plate to a mathematical magnifying glass. We ﬁrst focus on the end x = 0 by setting x = εˆ x and letting the displacement in this boundary layer near the edge be given by the function w(x, y) = w(ˆ ˆ x, y).

(6.2.8)

This gives the semi-inﬁnite strip model ˆ ∂2w ˆ ∂2w + =0 2 2 ∂x ˆ ∂y

with

∂w ˆ = 0 on y = 0, 1 ∂y

(6.2.9)

and w ˆ = f (y) prescribed on x ˆ = 0. Letting now w ˆ∼w ˆ 0 + εw ˆ1 + · · · , using

248

Asymptotic analysis

separation of variables and making the assumption that w ˆ0 does not grow as x ˆ → ∞, we ﬁnd ∞

a0 + an e−nπ xˆ cos nπy, w ˆ0 = 2

(6.2.10)

n=1

where the ai are the Fourier cosine coeﬃcients of f (y). Finally, we require that the functions w ˆ0 and w0 match, in the sense that, ˆ0 when the latter is evaluated for large x ˆ. as x → 0, w0 joins smoothly with w More speciﬁcally, we require that ˆ0 , lim w0 = lim w x ˆ →∞

x→0

and we thus obtain

(6.2.11)

1

w0 (0) =

f (y) dy,

(6.2.12)

g(y) dy.

(6.2.13)

0

and similarly

w0 (1) =

1

0

These are the crucial boundary conditions for the thin sheet model (6.2.7): the only information that w0 has about f and g is their average values. Now we undertake the more challenging task of extending this asymptotic argument to the bending of elastic plates.

6.3 The linear plate equation 6.3.1 Non-dimensionalisation and scaling We now turn to small displacements in a horizontal elastic plate subject to a transverse gravitational acceleration g. As for the antiplane strain problem just analysed, 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 (1.11.3), and the components of the Cauchy stress tensor are given by (1.11.1). 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.

(6.3.1)

In linear elasticity we do not distinguish between Eulerian and Lagrangian coordinates, so the boundaries of the plate eﬀectively remain ﬁxed at

6.3 The linear plate equation

249

z = ±h/2. The treatment of larger transverse displacements, comparable to h, say, will be deferred to Section 6.5. Denoting by L a typical dimension in the x- and y-directions, the slenderness parameter is again ε = h/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 again non-dimensionalise the equations before taking the asymptotic limit ε → 0, making the important assumption that λ and µ are held ﬁxed. The spatial coordinates are again scaled with their typical values as x = Lx ,

y = Ly ,

z = hz .

(6.3.2)

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 ). Moreover, from (6.3.1), we know that, on the free surface, ∂u/∂z = −∂w/∂x. Now, we have the order-of-magnitude estimates ∂u/∂z ∼ u/h and ∂w/∂x ∼ w/L, and balancing the two yields the non-dimensionalisation u = εW u ,

v = εW v ,

w = W w .

(6.3.3)

The appropriate non-dimensionalisation 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 (1.11.3), and we give ourselves the freedom to do this by writing , , τxy , τyy (τxx , τxy , τyy ) = τ τxx (τxz , τyz ) = ετ τxz , τyz ,

(6.3.4b)

, τzz = ε2 τ τzz

(6.3.4c)

(6.3.4a)

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 (1.11.1), which leads to τ =

ε2 EW , h

(6.3.5)

where E is Young’s modulus. Finally, by looking for a balance in the

250

Asymptotic analysis

transverse Navier equation (1.11.3c), we infer a suitable time-scale, namely ) h ρ t . t= (6.3.6) 2 ε E To summarise, the geometry of the plate has driven our non-dimensionalisation (6.3.2) of the spatial variables. Our scalings (6.3.3) of the displacements amount to assumptions about how big the transverse and in-plane deﬂections are; thereafter, stress components are rescaled so as to balance as many terms as possible in the momentum equations. We will now derive the leading-order model that results from these scalings.

6.3.2 Dimensionless equations Substituting (6.3.2) and (6.3.4) into (1.11.3) and neglecting terms of order ε2 we obtain the dimensionless momentum equations ∂τxy ∂τxx ∂τxz + + , ∂x ∂y ∂z ∂τyy ∂τyz ∂τxy + + , 0= ∂x ∂y ∂z ∂τyz ∂2w ∂τzz ∂τxz + + − G, = ∂t2 ∂x ∂y ∂z 0=

(6.3.7a) (6.3.7b) (6.3.7c)

where G=

ρgh2 ε4 EW

(6.3.8)

is a dimensionless constant measuring the importance of gravity. Here, and henceforth, we drop the primes to avoid cluttering the equations. The stressfree conditions (6.3.1) on the upper and lower surfaces simply become τxz = τyz = τzz = 0

on z = ±1/2.

(6.3.9)

It is helpful to average the momentum equations by integrating (6.3.7) with respect to z and applying (6.3.9). This yields 0=

∂Txy ∂Txx + , ∂x ∂y

0=

∂Txy ∂Tyy + ∂x ∂y

(6.3.10a)

and ∂Ny ∂2w ∂Nx + − G, = 2 ∂t ∂x ∂y

(6.3.10b)

6.3 The linear plate equation

251

where w denotes the average of w across the plate, and the net in-plane and transverse stresses are deﬁned as in Section 4.6 by 1/2 1/2 Tij = τij dz, Nj = τjz dz. (6.3.11) −1/2

−1/2

Similarly, multiplication through by z before integrating leads to the moment balance equations 0=

∂Myy ∂Myx + − Nx , ∂x ∂y

∂Mxy ∂Mxx + + Ny , ∂x ∂y

0=

where the bending moments are deﬁned as in Section 4.6 by 1/2 1/2 Myx = zτxx dz, Mxy = − zτyy dz, −1/2

Myy = −Mxx =

(6.3.12)

(6.3.13a)

−1/2

1/2

−1/2

zτxy dz.

(6.3.13b)

We may now use (6.3.12) to eliminate Nx and Ny from (6.3.10b) to give ∂ 2 Myx ∂ 2 Mxy ∂2w ∂2 (M = + − M ) − − G. yy xx ∂t2 ∂x2 ∂x∂y ∂y 2

(6.3.14)

Hence the deformation of the plate is governed by the averaged equations (6.3.10a) and (6.3.14). We must now relate the averaged stresses and moments to the displacements by using the dimensionless constitutive equations. 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) u y + vx , 2(1 + ν) ε2 νux + ε2 (1 − ν)vy + νwz , (1 + ν)(1 − 2ν) u z + wx , 2(1 + ν) vz + w y , 2(1 + ν) ε2 νux + ε2 νvy + (1 − ν)wz , (1 + ν)(1 − 2ν)

(6.3.15a) (6.3.15b) (6.3.15c) (6.3.15d) (6.3.15e) (6.3.15f)

where ν is Poisson’s ratio and, in this chapter only, we will occasionally use subscripts as shorthand for partial derivatives of the displacements. When we take the limit ε → 0 in (6.3.15), we treat all the other parameters as

252

Asymptotic analysis

being O (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 Section 1.8.

6.3.3 Leading-order equations If we immediately set ε = 0 in (6.3.15), we get the same equation wz = 0 three times. It is therefore expedient ﬁrst to 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

(6.3.16)

This allows us simply to let ε tend to zero in the above equations without having to write the dependent variables as explicit asymptotic expansions. With w independent of z (to leading order), we may integrate (6.3.15d) and (6.3.15e) to obtain u(x, y, z, t) = u(x, y, t) − zwx (x, y, t),

(6.3.17a)

v(x, y, z, t) = v(x, y, t) − zwy (x, y, t)

(6.3.17b)

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. By substituting (6.3.17) back into (6.3.15) 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)

wxx + νwyy , 12(1 − ν 2 ) Myy = −Mxx

Tyy =

νux + v y , 1 − ν2

νwxx + wyy , 12(1 − ν 2 ) wxy . =− 12(1 + ν)

Mxy =

(6.3.18)

(6.3.19a) (6.3.19b)

We are reassured by the fact that the expressions (6.3.18) for the in-plane tensions agree with the exact biaxial straining solutions (2.2.19). Equally, the relations between the bending moments and the second derivatives of w reproduce those found in (4.6.8) by ad hoc comparison with exact plane stress solutions. Recalling that Txx , Txy and Tyy satisfy (6.3.10a), we see that the in-plane displacements satisfy a plane strain problem. As explained in Section 2.6, one approach to this problem is to introduce an Airy stress function A(x, y)

6.4 Boundary conditions and Saint-Venant’s principle

253

such that Txx =

∂2A , ∂y 2

Txy = −

∂2A , ∂x∂y

Tyy =

∂2A , ∂x2

(6.3.20)

and elimination of u and v shows that A satisﬁes the biharmonic equation ∇4 A = 0.

(6.3.21)

Also, by substituting (6.3.19) into (6.3.14), we ﬁnd that the transverse displacement w satisﬁes the linear plate equation wtt = −

∇4 w − G. 12(1 − ν 2 )

(6.3.22)

Notice that the equation (6.3.22) for w can be solved independently of u and v. In other words, transverse bending completely decouples from inplane deformation of the plate. In summary, our model consists of the linear partial diﬀerential equation (6.3.22) for w and a decoupled biharmonic equation for A. We have derived these equations systematically, without invoking any other assumptions than the scalings in Section 6.3.1. Our derivation is constructive and yields explicit formulae for the other quantities of interest, that is the displacements and stress components. Equation (6.3.22) does not contain the “membrane” term T ∇2 w obtained from a momentum balance argument in Section 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 Section 4.6, it is impossible, starting from the linear Navier equations (1.11.3) and constitutive relations (1.11.1), to obtain any such nonlinear terms. We will show below in Section 6.5 how they arise from a geometrically nonlinear model.

6.4 Boundary conditions and Saint-Venant’s principle 6.4.1 Boundary layer scalings We are now poised to resolve systematically the diﬃculties encountered in Section 4.6.2 concerning the boundary conditions for plate models. In the light of Section 6.2, it will come as no surprise that we must proceed by addressing a boundary layer region near the edge of the plate. In this boundary layer we will ﬁnd a model that is fully three-dimensional, but fortunately we will not need to solve it in detail to obtain the sought-after boundary conditions.

254

Asymptotic analysis

z y

x

σx σy σz

Fig. 6.1 The edge of a plate subject to tractions.

For simplicity, we consider only a straight edge x = 0 which is subject to a given traction, so we impose the boundary conditions τxx = σx ,

τxy = σy ,

τxz = σz

at

x = 0,

(6.4.1)

on the dimensionless stress components; see Figure 6.1 for the sign convention. The tractions σx , σy and σz may in general be any functions of y and z. In the plate solutions constructed in Section 6.3, the in-plane stress components are linear functions of z of the form τxx = Txx + 12zMyx ,

τxy = Txy − 12zMxx ,

τyy = Tyy − 12zMxy , (6.4.2a)

and the transverse shear stresses are found from (6.3.7) to be 1 1 2 2 −z , −z , τyz = 6Ny τxz = 6Nx 4 4 τzz = −2

∂2w 1 2 −z . +G z ∂t2 4

(6.4.2b)

(6.4.2c)

Hence, unless the applied tractions have exactly the same dependence on z, it will be impossible to apply the boundary conditions (6.4.1) directly. Moreover, even if the boundary tractions do by chance have a z-dependence consistent with (6.4.2), we can see that there is still a boundary layer to be negotiated before the conditions (6.4.1) can be imposed. To illustrate the

6.4 Boundary conditions and Saint-Venant’s principle

255

diﬃculty, consider a traction-free edge, so that σx = σy = σz = 0. By naively applying (6.4.1) to (6.4.2), we apparently obtain ﬁve conditions, namely Txx = Txy = Myx = Mxx = Nx = 0.

(6.4.3)

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, as in Section 6.2 we now focus our attention on a thin boundary layer in which x = εˆ x, where x ˆ is O (1). To get a nontrivial balance in the Navier equations (6.3.7), we also ﬁnd it necessary to rescale some of the stress components as follows: τxz =

τˆxz , ε

τyz =

τˆyz , ε

τzz =

τˆzz . ε2

(6.4.4)

In view of (6.3.4), 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.

6.4.2 Equations and boundary conditions After we apply the rescaling (6.4.4) and neglect terms of order ε2 , the Navier equations become ∂ τˆxy ∂ τˆxz ∂ τˆxx +ε + , ∂x ˆ ∂y ∂z ∂ τˆyy ∂ τˆyz ∂ τˆxy +ε + , 0= ∂x ˆ ∂y ∂z ∂ τˆyz ∂ τˆxz ∂ τˆzz 0= +ε + . ∂x ˆ ∂y ∂z

0=

(6.4.5a) (6.4.5b) (6.4.5c)

Here we have used hats to distinguish dependent variables (even unscaled ones) evaluated in the boundary layer from those in the interior. As ε → 0, (6.4.5) eﬀectively reduces 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 = σx , τˆxy = σy , τˆxz = εσz

on z = ±1/2,

(6.4.6a)

on x ˆ = 0.

(6.4.6b)

256

Asymptotic analysis

The rescaled constitutive relations (6.3.15) take the form ∂w ˆ ∂u ˆ ∂ˆ v + ε2 ν +ν , ∂x ˆ ∂y ∂z v ∂u ˆ ∂ˆ 2ε(1 + ν)ˆ τxy = ε + , ∂y ∂x ˆ ∂w ˆ ∂u ˆ ∂ˆ v ε2 (1 + ν)(1 − 2ν)ˆ + ε2 (1 − ν) +ν , τyy = εν ∂x ˆ ∂y ∂z ˆ ∂u ˆ ∂w 2ε2 (1 + ν)ˆ + , τxz = ε ∂z ∂x ˆ ˆ ∂ˆ v ∂w 2ε(1 + ν)ˆ τyz = + , ∂z ∂y ∂w ˆ ∂u ˆ ∂ˆ v ε2 (1 + ν)(1 − 2ν)ˆ + ε2 ν + (1 − ν) . τzz = εν ∂x ˆ ∂y ∂z

ε2 (1 + ν)(1 − 2ν)ˆ τxx = ε(1 − ν)

(6.4.7a) (6.4.7b) (6.4.7c) (6.4.7d) (6.4.7e) (6.4.7f)

As in the antiplane strain example of Section 6.2, we must match the two solutions: the asymptotic behaviour of the inner solution as x ˆ → ∞ must be the same as that of the outer solution as x → 0. In other words, we impose that

∂2w ∂u (0, y) − z 2 (0, y) + · · · , (6.4.8a) ∂x ∂x ∂2w ∂v ∂w (0, y) + εˆ x (0, y) − z (0, y) + · · · , (6.4.8b) vˆ ∼ v(0, y) − z ∂y ∂x ∂x∂y

∂w (0, y) + εˆ x u ˆ ∼ u(0, y) − z ∂x

∂w (0, y) + · · · , ∂x ∼ Txx (0, y) + 12zMyx (0, y) + · · · ,

w ˆ ∼ w(0, y) + εˆ x τˆxx

τˆxy ∼ Txy (0, y) − 12zMxx (0, y) + · · · , 1 2 − z + ··· , τˆxz ∼ 6εNx (0, y) 4 1 2 τˆyz ∼ 6εNy (0, y) − z + ··· , 4 τˆzz ∼ O ε2 as x ˆ → ∞.

(6.4.8c) (6.4.8d) (6.4.8e) (6.4.8f) (6.4.8g) (6.4.8h)

6.4 Boundary conditions and Saint-Venant’s principle

257

6.4.3 Asymptotic expansions Unlike the situation encountered in Section 6.3.3, there is now no way to avoid writing the dependent variables in asymptotic expansions in powers of ε, for example (0) (1) (2) + εˆ τxx + ε2 τˆxx + ··· . τˆxx ∼ τˆxx

(6.4.9)

To leading order in ε, (6.4.5) becomes (0)

(0)

(0)

(0)

(0)

(0)

∂ τˆxx ∂ τˆxz ∂ τˆxy ∂ τˆyz ∂ τˆxz ∂ τˆzz + , 0= + , 0= + , ∂x ˆ ∂z ∂x ˆ ∂z ∂x ˆ ∂z and the integrated versions of these equations are, simply, 1/2 1/2 1/2 ∂ ∂ ∂ (0) (0) τˆ d