Mechanics of Elastic Composites (Modern Mechanics and Mathematics)

MECHANICS OF ELASTIC COMPOSITES Nicolaie Dan Cristescu Eduard-Marius Craciun Eugen Soós

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CRC S ERIES: M ODERN M ECHANICS

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PUBLISHED TITLES

MECHANICS OF ELASTIC COMPOSITES by Nicolaie Dan Cristescu, Eduard-Marius Craciun, and Eugen Soós

F O RT H C O M I N G T I T L E S

BEYOND PERTUBATION: INTRODUCTION

TO THE HOMOTOPY ANALYSIS

by Shijun Liao

CONTINUUM MECHANICS

AND PLASTICITY

by Han-Chin Wu

HYBRID INCOMPATIBLE FINITE ELEMENT METHODS by Theodore H.H. Pian, Chang-Chun Wu

MICROSTRUCTURAL RANDOMNESS by Martin Ostroja Starzewski

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IN

MECHANICS

OF

MATERIALS

METHOD

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Library of Congress Cataloging-in-Publication Data Cristescu, N. Mechanics of elastic composites / Nicolaie Dan Cristescu, Eduard-Marius Craciun, Eugen Soós. p. cm. — (Modern mechanics and mathematics ; 1) Includes index. ISBN 1-58488-442-8 (alk. paper) 1. Composite materials—Mechanical properties. I. Craciun, Eduard-Marius. II. Soós, E. (Eugen) III. Title. IV. Series. TA418.9.C6C73 2003 620.1¢1832—dc22

2003047291

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Visit the CRC Press Web site at www.crcpress.com © 2004 by Chapman & Hall/CRC No claim to original U.S. Government works International Standard Book Number 1-58488-442-8 Library of Congress Card Number 2003047291 Printed in the United States of America 1 2 3 4 5 6 7 8 9 0 Printed on acid-free paper

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MECHANICS OF ELASTIC COMPOSITES Nicolaie Dan Cristescu Eduard-Marius Craciun

Copyright © 2004 by Chapman & Hall/CRC

Eugen So´os

Preface The history of this book is quite complex, as it has been written over an extended period of time, beginning as a book called Mechanics of Elastic Composites, published in Romania in 1983. It was originally written for a classical course on elastic composites taken by students of mechanical engineering and materials science. Over the past few years, the book has been further developed by Professor So´os and the other authors who have continued to add and delete material as needed for a modern book on this topic. One of the most outstanding features of this book is the inclusion of over 400 problems, half of which are solved completely at the end of the book in an answer section. The remaining solutions are provided in an instructor’s manual available to adopters of the text. All of the problems provided in the book are of varying degrees of difficulty and play a useful role in the exposition of each chapter. The current book contains eight chapters; the first two are introductory chapters, with tensor analysis and tensor algebra addressed in chapter one. In this chapter, all the major formulas are provided for review followed by fifty practice problems. The second chapter is devoted to the elements of linear elastostatics. Included in this chapter are the classical problems of symmetry and the main results of elasticity as work and energy, minimum principles of elastostatics, the concentrated forces and Green’s tensor and Eshelby’s inclusion theorem and problem. This chapter concludes with fifty-nine problems aimed at improving the student’s understanding of these concepts. Chapter three deals with composite laminates. The classical aspects such as lamina, micromechanical and macromechanical constitutive equations and boundary conditions, variational and extreme principles, and rectangular laminates are presented here, followed by fifty-three problems. Anyone wishing to learn the classical aspects of elastic composites should begin with this chapter. Fundamental problems concerning the macroscopically homogeneous biphasic linearly elastic composites are covered in chapter four. This chapter contains the basic and classical part of the general theory of micromechanics of composite materials and concludes with fifty-two problems. Chapter five presents the three-dimensional linearized theory of elastic body stability, as well as covering such topics as small deformation superposed on large static deformation, stable and unstable equilibrium configuration, variational and

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extreme principles, bifurcation analysis, dynamic criteria of stability, and homogeneous initial deformations. This chapter concludes with fifty-four problems. The last three chapters of the book concern the buckling of fiber-reinforced composite strips, stability analysis of composite laminates and brittle fracture mechanics. The results and methods presented in these last chapters are from research papers published in Eastern Europe and prior to this time unavailable in the U.S., Canada and other western countries. These final chapters are described below. Chapter six addresses the buckling of fiber-reinforced composite strips and bars, starting with internal and superficial stability of composite materials and continuing with the buckling of fiber-reinforced composite strips, showing the limits of classical plate theory, and continuing again with the buckling of fiber-reinforced composite bars, showing the limits of Euler’s theory. The chapter ends with fortyeight problems aimed at improving this theory. Chapter seven concerns the stability of composite laminates and covers the influence of the initial applied bending moment on the behavior of composite laminates. The chapter concludes with forty-seven problems. The eighth and final chapter concerns the fracture mechanics of fiber-reinforced composites as well as the brittle fracture. Necessary elements of the complex function are provided as well as some incremental fields, the incremental state of pre-stressed composites containing a crack, and the asymptotic behavior of the incremental fields. Forty-nine problems are included at the end of this chapter. Regrettably, Eugen So´os has passed on, but the remaining authors have endeavored to see this book published in the manner in which Dr. E. So´os had originally envisioned it. With a book of this size, some mistakes are inevitable and to be expected. If any errors are found please feel free to contact the authors or the Publisher. We would like to acknowledge all the individuals who have assisted us with this book and thus, want to thank David Steigmann, University of California, and Eveline Baesu, University of Nebraska-Lincoln, who reviewed portions of the draft manuscript and contributed many helpful suggestions. We are also grateful to Bob Stern at CRC Press for his continued support. For periodic advice, patience and skill in the preparation of this manuscript, we would also like to thank Dr. Adrian Rabaea, “Ovidius” University of Constanta.

The Authors (E-mails: [email protected] [email protected] )

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Contents 1 ELEMENTS OF TENSOR CALCULUS 1.1 Tensor Algebra 1.2 Tensor Analysis 1.3 Problems 2 ELEMENTS OF LINEAR ELASTOSTATICS 2.1 Displacement. Deformation. Stress 2.2 Symmetry transformations and groups 2.3 The fundamental system of field equations 2.4 Minimum principles of elastostatics 2.5 Green’s tensor for an infinite medium 2.6 Piece-wise homogeneous bodies 2.7 Eshelby’s inclusion problem 2.8 Problems 3 COMPOSITE LAMINATES 3.1 Macromechanical behavior of a lamina 3.2 Strength of materials approach 3.3 Global constitutive equations 3.4 Special classes of laminates 3.5 Equilibrium equations and boundary conditions 3.6 Variational and extreme principles 3.7 Rectangular laminates 3.8 Problems 4 MACROSCOPICALLY ELASTIC COMPOSITES 4.1 Macroscopically linearly elastic composites 4.2 Hill’s weak and strong assumptions 4.3 Macroscopically isotropic biphasic mixture 4.4 Hashin-Shtrikman principle 4.5 Budiansky’s and Hill’s self-consistent method 4.6 Overall mechanical properties

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4.7 4.8

Hashin’s bounds for the overall moduli Problems

5 THE THREE-DIMENSIONAL LINEARIZED THEORY 5.1 Elements of nonlinear elasticity 5.2 Lagrangean approach 5.3 Updated Lagrangean approach 5.4 Global and local uniqueness. Stability 5.5 Dynamic criteria of stability 5.6 Homogeneous initial deformations 5.7 Problems 6 BUCKLING OF COMPOSITE STRIPS AND BARS 6.1 Internal and superficial stability 6.2 Buckling of fiber-reinforced composite strips 6.3 Buckling of fiber-reinforced composite bars 6.4 Problems 7 STABILITY OF COMPOSITE LAMINATES 7.1 Constitutive equations for incremental fields 7.2 Equilibrium equations. Boundary conditions 7.3 Buckling of rectangular composite laminates 7.4 Buckling of antisymmetric cross-ply laminates 7.5 Problems 8 FRACTURE MECHANICS 8.1 Elements of complex function theory 8.2 Representation of the incremental fields 8.3 The opening, sliding and tearing modes 8.4 Asymptotic behavior of the incremental fields 8.5 Griffith’s criterion and crack propagation 8.6 Problems SOLUTIONS TO SOME PROBLEMS

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List of Figures 1.1 1.2

Change of basis by a rotation about e3 . Concerning the definition of the directional derivative.

2.1 2.2

The singular surface Γ. Biphasic piece-wise homogeneous mixture or composite formed by a matrix and N inclusions. An arbitrary regular subdomain Bx of B.

2.3 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 3.10 3.11 3.12 3.13 3.14 3.15 3.16 3.17 3.18 3.19

Lamina with unidirectional fibers. Exploded view of laminate structure. Positive rotation of principal material axes 1, 2 from arbitrary axes x, y. Normalized moduli for glass/epoxy. Representative volume element loaded in the 1-direction. Representative volume element loaded in 2-direction. Representative volume element loaded in 1-direction. Representative volume element loaded in shear. Shear deformation of a representative volume element. Coordinate system for laminated plate. Geometry of an N -layered laminate. In-plane forces on a flat laminate. Moments on a flat laminate. Exploded (unbounded) view of a three-layered regular symmetric cross-ply laminate. Exploded (unbounded) view of a three-layered regular symmetric angle-ply laminate. Exploded (unbounded) view of a two-layered regular antisymmetric cross-ply laminate. Exploded (unbounded) view of a two-layered regular antisymmetric angle-ply laminate. Regular cross-ply laminate having N (even) layers. The plane domain occupied by the middle surface of the laminate.

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3.20 Simply supported laminated rectangular plate under distributed normal load. 3.21 Maximum deflection of a rectangular regular antisymmetric cross ply laminate under sinusoidal normal load. 3.22 Maximum deflection of a square regulate antisymmetric cross-ply laminated plate under sinusoidal transverse load. 3.23 Exploded view of a [+45/−45/−45/+45] regular angle-ply laminate. 3.24 Exploded view of a [−45/+45/−45/+45] regular angle-ply laminate. 4.1 4.2 4.3 4.4

Scales of analysis: d << l << L. Bounds for bulk modulus of WC-Co alloy. Bounds for shear modulus of WC-Co alloy. Bounds for Young modulus of WC-Co alloy.

5.1 5.2

The configuration B, Bτ and Bt . Example of traction boundary condition, showing possible deformed configurations of a circular cylinder of uniform cross-section under uniaxial nominal compression of magnitude σ: (a) uniform contraction; (b) buckled configuration. Stable solution path (continuous case) bifurcates at the e-configuration ◦ χc into two branches (broken curves) with tangents W and W + U ◦ ◦ at χc , where U is an e-mode at χc . Continuous line: possible path in the complex frequency plane; interrupted lines: impossible paths in the complex frequency plane. Cylindrical coordinates r, θ, z.

5.3

5.4 5.5 6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8

Superficial instability. Incremental plane states. Composite strip acted upon by compressive forces. Strip instability by necking (antisymmetric mode). Strip instability by bending (symmetric mode). Dependence of the correction factor p∗ on thickness ratio b. Fiber-reinforced composite cylindrical bar. Buckled cylindrical bar. Variation of correction factor p* vs. geometrical parameter α. Transversally isotropic circular bar; E1 /E3 = 0.08. 6.9 Variation of correction factor p* vs. geometrical parameter α. Transversally isotropic circular bar; E1 /E3 = 0.20. 6.10 Variation of correction factor p* vs. geometrical parameter α. Transversally isotropic circular bar; E1 /E3 = 0.80. 6.11 Variation of correction factor p* vs. geometrical parameter α. Transversally isotropic circular bar. Exact solution. 6.12 Variation of correction factor p* vs. geometrical parameter α. Transversally isotropic circular bar. Exact and variational solution.

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6.13 Variation of correction factor p* vs. geometrical parameter α. Transversally orthotropic rectangular bar. Variational solution. 6.14 Correction factor p* vs. geometrical parameter α. Transversally isotropic rectangular bar. Variational solution. 6.15 Variation of correction p* vs. geometrical parameter α. Long plate behavior of rectangular bar. 7.1

Rectangular composite laminate under biaxial compression.

7.2 7.3 7.4

Variation of the critical load N c 11 with aspect ratio K = ab . Rectangular composite laminate under uniaxial compression. Compressive buckling coefficient. Simply supposed plies.

8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8 8.9 8.10 8.11 8.12 8.13 8.14

The definition of the logarithmic function. Example of complex integration. √ The definition of the function f (z) = z 2 − a2 Limit values. Cauchy’s principal value. Plemelj-Sohockii theorem. Plemelj-Sohockii theorem, in the lower half plane (y < 0). The cut L and the closed curve Γ. Crack acted by symmetrical normal load. Crack acted by antisymmetrical tangential load. Asymptotic behavior. Basic crack models and extensions. Segment of crack opening. Integral theorems.

o

S.1 S.2 S.3 S.4 S.5 S.6 S.7 S.8 S.9 S.10 S.11

The spherical coordinate system. Normalized moduli for a boron/epoxy composite. Variation of Q11 with θ. Variation of Q22 with θ. Variation of Q12 with θ. Variation of Q66 with θ. Variation of Q16 with θ. Variation of Q26 with θ. Symmetric laminate; N = even number. Symmetric laminate; N = odd number Normal and tangential tractions on the deformed faces of a sheared block. S.12 Cauchy’s fundamental formula. S.13 The function U + = U + (x1 ) ; 0 < m < n. S.14 The function U − = U − (x1 ) ; 0 < m < n

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Chapter 1

ELEMENTS OF TENSOR CALCULUS 1.1

Tensor Algebra

The vector and tensor calculus represents the principal mathematical tool used in mechanics of deformable bodies to express the mean ideas in this field, to formulate and to solve the specific problems in this domain. The basic concepts, results and techniques of vector and tensor calculus are presented for instance in the monograph by Malvern [1.1] and by Gurtin [1.2], [1.3]. The elements of vector and tensor calculus are presented here following So´os and Teodosiu [1.4]. We denote the three-dimensional Euclidean space by E and the three-dimensional vector space of free vectors by V associated to E. We assume that the properties of these two spaces are known. The concept of general vector space was obtained generalizing the properties of the free vectors known in the elementary geometry. We say that a set of V is a (real) vector space, and its elements u, v, w, ... are vectors if the following conditions (axioms) are satisfied: (v1) To any pair u, v ∈ V corresponds a vector u + v ∈ V, named the sum of the vectors u and v, such that: (v2) u + v = v + u (commutativity); (v3) u + (v + w) = (u + v) + w (associativity); (v4) an element 0 ∈ V, named null vector, exists, such that u + 0 = u for any u ∈ V; (v5) to any vector u ∈ V corresponds a vector −u ∈ V, named the vector opposed to u, such that u + (−u) = 0; (v6) to any vector u ∈ V, and to any real number α ∈ R (the set of all real numbers) corresponds a vector αu = uα ∈V, named the product of the real number α and the vector u, such that:

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CHAPTER 1. ELEMENTS OF TENSOR CALCULUS

(v7) 1u = u; (v8) α (βu) = (αβ) u, α, β ∈ R (associativity); (v9) (α + β) u = αu+βu (distributivity relative to number addition); (v10) α (u + v) = αu+αv (distributivity relative to vector addition); It is easy to see that the usual operation with free vectors of the elementary geometry satisfies the above 10 axioms. We say that n vectors u1 , u2 , ..., un ∈ V are linearly independent if the equation α1 u1 + ... + αn un = 0, with α1 , ..., αn ∈ R, can take place only if α1 = α2 = ... = αn = 0. Otherwise, we say that the set of n vectors is linearly dependent. A vector space V is named n-dimensional and is denoted by Vn , if in Vn there exists at least one set containing n linearly independent vectors u1 , ..., un , and if any set u1 , ..., un , v formed with n + 1 vectors is linearly dependent. In a n-dimensional vector space Vn , any set of n linearly independent vectors is named a basis in Vn . Let g1 , ..., gn be a basis in Vn and let u be a vector of Vn . According to the above definition, the vectors g1 , ..., gn , u are linearly dependent. Hence, a set of real numbers α1 , ..., αn , α, not all of them vanishing, exists such that αu + α1 g1 + ... + αn gn = 0. Obviously α 6= 0, since g1 , ..., gn are linearly independent. Thus, we can conclude that u = u1 g1 + ... + un gn , with uk = −αk /α, k = 1, .., n. We say that u is a linear combination of the vectors g1 , ..., gn , forming a basis in Vn . The above result shows that in an n-dimensional vector space Vn any vector can be expressed as a linear combination of the base vectors g1 , ..., gn . In what follows, we shall use Einstein’s summation convention, and we shall write the above sum writing u in a more concentrated form as u = u k gk . Assuming that two linear combinations exist u = u k gk = u k 0 gk , expressing the same vector u as linear combination of g1 , ..., gk we obtain (uk − u0k ) gk = 0. Since g1 , ..., gn are linearly independent, from the above follows that uk = uk 0 for k = 1, ..., n. Thus, if a basis g1 , ..., gn is given, any vector u can be expressed in an unique way as a linear combination of the base vectors g1 , ..., gn . The real numbers u1 , ..., un are named the components of the vector u in the basis g1 , ..., gn . Let V be a (real) vector space. An application associating with any pair of vectors u, v ∈ V a real number denoted by u · v is named (Euclidean) scalar product if it satisfies the following conditions (axioms): (S1) u · v = v · u (commutativity);

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3

1.1. TENSOR ALGEBRA

(S2) (αu) · v = α (u · v) (associativity with respect to multiplication with real numbers); (S3) u · (v + w) = u · v + u · w (distributivity with respect to vector addition); (S4) u · u ≥ 0; (S5) u · u = 0 if and only if u = 0. The real number u · v is named the (Euclidean) scalar product of the vectors u and v. It is easy to see that the usual scalar product of free vectors introduced in elementary geometry satisfies the above five axioms. Actually, the abstract definition of the (Euclidean) scalar product was obtained starting with the wellknown properties of the usual scalar product existing in V , the three-dimensional vector space of free vectors. The magnitude (intensity, norm) kuk of a vector u from V is defined by the equation √ (1.1.1) kuk = u · u.

A vector space V in which is introduced a scalar product is named an Euclidean vector space. A basis e1 , ..., en in an n-dimensional Euclidean vector space Vn is named an orthonormal basis if the elements of the basis are unit vectors and they are reciprocally orthogonal; that is, if  1 if k = m ek · em = δkm = (1.1.2) 0 if k 6= m δkm being the Kronecker’s symbol. Using the orthogonalization procedure of Gram and Schmidt, it can be shown that in any n-dimensional Euclidean vector space there exist at least one orthonormal basis. In what follows, we shall use only orthonormal bases. If e1 , ..., en is an orthonormal basis in Vn , according to (1.1.2) we have: u = uk ek = (u · ek )ek

and

uk = u · ek .

Also if u = u k ek

and

v = v k ek ,

than

u · v = uk vk = u1 v1 + ... + un vn .

Let us assume now that e1 , ..., en = {ek } and e01 , ..., e0n = {e0r } , k, r = 1, ..., n, are two orthonormal basis in Vn . Expressing all vectors of one of the basis as linear combination of the vectors of the other basis, we get: e0r = qkr ek

and

0 ek = qrk e0r .

(1.1.3)

0 Obviously, the matrix Q0 = [qrk ] is the inverse Q−1 of the matrix Q = [qkr ]. Moreover, since both basis are orthonormal, according to (1.1.2) we have

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4

CHAPTER 1. ELEMENTS OF TENSOR CALCULUS

ek · em = δkm and e0r · e0s = δrs , k, m, r, s = 1, ..., n. Hence, we successively obtain δrs = e0r · e0s = (qkr ek ) · (qls el ) = qkr qls ek el = qkr qls δkl = qkr qks . Thus qkr qks = q1r q1s + ... + qnr qns = δrs . The last relation shows that the transformation matrix Q and its transpose QT satisfy the equation: QT Q = U, where



  U = 

1 0 .. .

0 1 .. .

0

0

··· ···

0 0 .. .

··· ···

1

    

is the n-order square unit matrix. The obtained result shows that the matrix Q, describing the change of an orthonormal basis with a new orthonormal basis, is always an orthogonal matrix . At the same time, we can conclude that QT Q = U,

det Q = ±1

and

Q−1 = QT .

Consequently, the equations (1.1.3) describing the change of the basis, become: e0r = qkr ek ,

ek = qkr e0r

with

qkr = ek · e0r = cos (ek , e0r ) .

(1.1.4)

Using the considered orthonormal basis {ek } and {e0k }, we can express the same vector u in two ways: u = u k ek

and

u = u0r e0r .

Taking into account the equation (1.1.4), we get the relations expressing the transformation of components of a vector, when the basis is changed. We obtain u0r = qkr uk

and

uk = qkr u0r .

(1.1.5) T

T

Introducing the matrixes u = [u1 , ..., un ] and u0 = [u01 , ..., u0n ] , the above equations can be expressed in matrix form, which is very useful in many cases. We have u0 = QT u and u = Qu0 . (1.1.6) Particularly, let {e1 , e2 , e3 } be a (right) orthonormal basis in the usual threedimensional Euclidean vector space V , and let us introduce in V a new orthonormal basis {e01 , e02 , e03 }, defined by the equations e01 = e1 cos θ + e2 sin θ, with θ ∈ (0, 2π).

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e02 = −e1 sin θ + e2 cos θ,

e03 = e3

(1.1.7)

5

1.1. TENSOR ALGEBRA

From the equations (1.1.4), it follows that the transformation matrix Q = [qkr ], its transpose QT and its inverse Q−1 , have the following form:     cos θ sin θ 0 cos θ − sin θ 0 cos θ 0  , QT = Q−1 =  − sin θ cos θ 0  . (1.1.8) Q =  sin θ 0 0 1 0 0 1

Accordingly, the transformation of the components of a vector u from V is given by the following equations: u01 = u1 cos θ + u2 sin θ, u02 = −u1 sin θ + u2 cos θ, u03 = u3 , u1 = u01 cos θ − u02 sin θ, u2 = u01 sin θ + u02 cos θ, u3 = u03 .

(1.1.9)

From a geometrical point of view, the new basis is obtained from the initial one, rotating the vectors e1 and e2 with an angle θ around the vector e3 , as in Figure 1.1.

'3

3

'2

O 2

1

e'1

Figure 1.1: Change of basis by a rotation about e3 . As we shall see later on, the above elementary change of basis plays an important role in the theory of material symmetry and in the theory of composite laminates. In what follows, we shall introduce the concept of a second order tensor in an intrinsic manner. The second order tensors play a fundamental role in the mechanics of deformable bodies, since the deformation and the internal forces characterizing the behavior of a deformable material, are described mathematically by second order tensor fields such as Green’s deformation tensor and Cauchy’s stress tensor. Let V be a (real) vector space. A linear function T defined on V and having values in V is named a second order tensor. If the value of T corresponding to a vector v is the vector u, we shall write u = T (v) = Tv.

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CHAPTER 1. ELEMENTS OF TENSOR CALCULUS

The linearity of the function T, involved in the definition of a second order tensor, is expressed by the following equation T(αu + βv) = αTu + βTv,

∀u, v ∈ V,

∀α, β ∈ R.

(1.1.10)

We denote by L the set of all second order tensors. In the following, speaking about a second order tensor, we shall name it simply tensor. We say that two tensors T and U are equal if Tv = Uv,

∀v ∈ V.

The null tensor, denoted by 0, is defined by the equation 0v = 0,

∀v ∈ V.

The sum T + U of two tensors T and U is defined by the equation (T + U)v = Tv + Uv,

∀v ∈ V.

The product αT = Tα of a real number α and a tensor T is defined by the equation (αT) v = α (Tv) , ∀v ∈V. It it easy to see that 0, T + U and αT are linear applications defined on V and having values in V. That is, 0, T + U and αT are (second order) tensors. Also, it can be seen that the above defined operations satisfy the axioms (V1)-(V10), if we take the null tensor 0 as neutral element (null vector) relative to tensor addition and as opposite of the tensor T the tensor −T = −1T. Consequently L, endowed with the above introduced operation, is a (real) vector space. In the following, we shall denote by L just this vector space. We assume now that V is an Euclidean vector space. In order to construct bases in the vector space L, we first introduce the concept of tensor product of two vectors. We name tensor product or diadic product of two vectors u, v ∈ V the function u ⊗ v = uv,defined on V and having values in V, given by the following rule: (u ⊗ v) (w) = uv (w) = u (v · w) ,

∀w ∈V.

(1.1.11)

Using this definition and the properties (S1)-(S5) of a scalar product, we successively get uv(α1 w1 + α2 w2 ) = u(α1 v · w1 + α2 v · w2 ) = α1 u(v · w1 ) + α2 u(v · w2 ) = α1 (uv)w1 + α2 (uv)w2 .

That is, the function uv satisfies the linearity property (1.1.10). Hence, the tensor product uv of the vectors u and v is a second order tensor.

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7

1.1. TENSOR ALGEBRA

In the following, we assume that the underlying vector space V is an ndimensional Euclidean vector space, and as before, by L we denote the vector space of all second tensors, defined on Vn . Let us assume now that {e1 , ..., en } = {ek } is an orthonormal basis in Vn and let us consider the set {ek em } of all possible tensor products for k, m = 1, ..., n. Assuming that there exist n2 real numbers λkm such that λkm ek em = 0, we get 0 = 0el = (λkm ek em ) el = (λkm ek ) (em · el ) = λkm ek δml = λkl ek for l = 1, .., n. Hence, λkl = 0, for k, l = 1, ..., n, since {ek } is a basis in Vn . Consequently, {ek em } , k, m = 1, ..., n, is a linearly independent set of the vector space L. Let us consider now an arbitrary tensor T ∈ L. Since Tem is a vector of Vn it can be expressed as a linear combination of the vectors e1 , ..., en Tem = Tkm ek . Using the orthonormality of the basis {ek }, the properties of the tensor product and the above relation, for an arbitrary vector v = vs es , we successively get (T − Tkm ek em ) v

= (T − Tkm ek em ) (vs es ) = vs Tes − vs Tkm ek (em es ) = vs (Tks ek − Tkm ek δms ) = vs (Tks ek − Tks ek ) =

0,

∀v ∈ Vn .

Hence, T − Tkm ek em = 0 (the second order null tensor). Accordingly, T = Tkl ek el ,

∀T ∈ L.

(1.1.12)

Thus, any tensor T ∈ L can be expressed as a linear combination of the n2 linearly independent second order tensors ek em , k, m = 1, ..., n. Consequently, if {ek } is an orthonormal basis in Vn , then {ek , em } , k, m = 1, ..., n is a basis in L. Since this set contains n2 elements, the dimension of the vector space L is n2 . The quantities Tkm in the equation (1.1.12) are named the components of the tensor T in the basis {ek em }. From (1.1.12), it results that Tkm = ek · Tem ,

(1.1.13)

Recalling that we use Einstein’s summation convention, the equation (1.1.12) actually means: T = Tkm ek em =

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T11 e1 e1 + T12 e1 e2 + ... + T1n e1 en + T21 e2 e1 + T22 e2 e2 + ... + T2n e2 en + ........................................................ Tn1 en e1 + Tn2 en e2 + ... + Tnn en en .

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CHAPTER 1. ELEMENTS OF TENSOR CALCULUS

In many computations, it is useful to present the components Tkm of a tensor T as an n-order square matrix T :   T11 T12 ... T1n  T21 T22 ... T2n  . T = [Tkm ] =   ... ... ... ...  Tn1 Tn2 ... Tnn

In what follows regarding the components Tkm of a tensor T, we mean its components in a basis {ek em }, where {ek } is an orthonormal basis in Vn . If u = Tv and u = uk ek , T = Tkl ek el , v = vm em , it is easy to see that uk = Tkm vm

or

u = Tv.

Now let us assume that {ek } and {e0k } are two orthonormal basis in Vn , connected by the equations (1.1.4). Let us consider the corresponding basis {e k em }, 0 and {e0r e0s } in L. Denoting by Tkm and Trs the components of a tensor T in these bases, we have 0 0 0 T = Tkm ek em = Trs er es and, according to (1.1.4) and (1.1.13), we successively get: 0 Trs = e0r · Te0s = (qkr ek ) · T (qms em ) = qkr qms (ek · Tem ) ,
Tkm = ek · Tem = (qkr e0r ) · T (qms e0s ) = qkr qms e0r · Te0s .

Using again the relation (1.1.13), we can conclude that the relation characterizing the transformation of the components of a tensor, corresponding to a changing of the basis in L, have the following form 0 = qkr qms Tkm Trs

and

0 . Tkm = qkr qms Trs

(1.1.14)

The above relation can be expressed also in matrix form: T 0 = QT T Q

and

T = QT 0 QT .

(1.1.15)

In particular, if the orthogonal matrix Q is given by the equation (1.1.8), we get: 0 T11 0 T22 0 T12 0 T21 0 T13 0 T31

= T11 cos2 θ + (T12 + T21 ) sin θ cos θ + T22 sin2 θ, = T11 sin2 θ − (T12 + T21 ) sin θ cos θ + T22 cos2 θ, = (T22 − T11 ) sin θ cos θ + T12 cos2 θ − T21 sin2 θ, = (T22 − T11 ) sin θ cos θ − T12 sin2 θ + T21 cos2 θ, 0 = T13 cos θ + T23 sin θ, T23 = −T13 sin θ + T23 cos θ, 0 = −T31 sin θ + T32 cos θ, = T31 cos θ + T32 sin θ, T32

(1.1.16) 0 = T33 . T33

In the following, we shall introduce the composition or the product of two tensors and we shall speak also about some special tensors, playing important roles in the mechanics of deformable bodies.

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The composition or product TU of two tensors T and U is defined using the composition rule of two functions; i.e. (TU) v = T (Uv)

for any

v ∈ V.

It is easy to see that TU is a linear function defined on V and having values in V; that is, TU is a second order tensor. Let us denote by Tkm , Ukm and Vkm the components of the tensors T, U and V = TU in the basis {ek , em }, and let T , U , V be the matrixes of these components. It is easy to see that Vkm = Tkl Ulm

and

V = T U.

That is, the matrix of the components of the product V = TU is the product of the matrixes of the components of T and U. Generally, the product of two tensors, as the product of two matrixes, is not commutative; i.e. TU 6= UT. The unit tensor , denoted by 1, is defined by the relation 1v = v

for any

v ∈ V.

Obviously, 1 is a linear function defined in V and having values in the same space. Hence, 1 is a second order tensor. Denoting by 1km , its components in a basis {ek en }, we get 1km = δkm . Also, we have 1T = T1 = T

for any

T ∈L.

Using the Euclidean scalar product introduced in the underlying Euclidean vector space V, we can introduce the transpose tensor of a tensor from L. The function TT defined in V and having values in V is named the transpose tensor of the tensor T if it satisfies the equation
TT u · v = u · Tv for any u, v ∈V.

The above rule defines in an unique way the function TT . Indeed, assuming
T T u · v = u · Tv for any u, v ∈V, exists, such that T that a  second function T 1  1 T T T we get T u − T1 u · v = 0 for any v ∈V. Hence, T u − TT1 u = 0 for any u ∈V.

Consequently, TT u = TT1 u for any u ∈V. Hence, TT = TT1 . At the same time, using the above definition, the linearity of T and the properties of the scalar product in V, we can show that TT is a linear function, hence, a second order tensor.

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CHAPTER 1. ELEMENTS OF TENSOR CALCULUS

T the components of T and TT , in a basis {ek em } , Denoting by Tkm and Tkm according to the definition of the components of a tensor, we get

Tkm = ek · Tem

and

T = e k · TT em . Tkm

Since ek · TT em = em · Tek , we obtain T = Tmk . Tkm

In other words, the matrix of the components of the transpose tensor TT is the transposed matrix of the components of the tensor T. Also, the following relation takes place TT and:


T

= T,

T

(TU) = UT TT

for any

T

u, v ∈V.

(uv) = vu

for any

T, U ∈ L.

We say that a tensor T is symmetric if TT = T, and is antisymmetric if TT = −T. If the tensor T is symmetric, the matrix of its components is also symmetric; if the tensor T is antisymmetric, the matrix of its components is also antisymmetric. Consequently, a symmetric tensor has n(n + 1)/2 independent components, and an antisymmetric tensor has n (n − 1) /2 independent components. In particular, all diagonal components of an antisymmetric tensor are zero. If T is an arbitrary tensor from L, and if we introduce the tensors TS =

we get:

 1 T + TT , 2

TA =

 1 T − TT , 2

T = TS + T A . Obviously, TS is a symmetric tensor and TA is an antisymmetric tensor. The last equation shows that any tensor T can be expressed in a unique form as the sum of a symmetric and of an antisymmetric tensor. That is the reason why T S and TA are named the symmetric and antisymmetric parts of T. The symmetric second order tensors play an essential role in the mechanics of deformable bodies. Particularly, the Green’s deformation tensor and the Cauchy’s stress tensor are symmetric tensors.

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1.1. TENSOR ALGEBRA

Let us observe also, that if T is a symmetric tensor, the transformation formulas (1.1.16) take a more simple form 0 T11 0 T22 0 T12 0 T13

= T11 cos2 θ + T12 sin 2θ + T22 sin θ, = T11 sin2 θ − T12 sin 2θ + T22 cos θ, = 12 (T22 − T11 ) sin 2θ + T12 cos 2θ, 0 0 = T13 cos θ + T23 sin θ, T23 = −T13 sin θ + T23 cos θ, T33 = T33 .

(1.1.17)

We shall introduce now two invariants of a second order tensor T, playing important roles in the mechanics of deformable bodies. The trace trT of the tensor T is a linear function defined on L, having values in R, and satisfying the following equation tr (uv) = u · v

u, v ∈ V.

for any

Consequently, if {ek em } is a basis in L, we have tr (ek em ) = ek · em = δkm . Hence, if T =Tkm ek em , since tr is a linear function, we successively get trT = tr (Tkl ek em ) = Tkm tr (ek em ) = Tkm δkm . The last equation shows that trT = Tmm = T11 + T22 + ... + Tnn . In other words, the value of trT is the sum of the diagonal components of T. Let T be a tensor. The tensor TS defined by the equation TS =

1 (trT) 1 3

is named the spherical part of T, and the tensor TD defined by the equation TD = T − T S = T −

1 (trT) 1 3

is named the deviatoric part of T. Obviously, we have T = T S + TD =

1 (trT) 1 + TD 3

and

trTD = 0.

The above results show that any second order tensor can be decomposed in a unique way into a spherical part and a diagonal part. Only the diagonal components of the spherical part are (generally) non-vanishing and the trace of the deviatoric part is always zero. As we shall see later, the above decomposition plays an essential role in the mechanics of composite materials, concerning the determination of the overall properties of a composite.

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CHAPTER 1. ELEMENTS OF TENSOR CALCULUS

Returning to the properties concerning the trace of a second order tensor, it is easy to prove that tr (TU) = Tkm Ukm ,

trTT = trT,

trTS = trT trTA = 0,

tr1 =n.

The determinant det T of a tensor T is the determinant det T of the matrix T of the components of T in a basis {ek em }, where {ek } is an orthonormal basis in Vn . More exactly, if T = Tkm ek em , then det T = det T = det [Tkm ] . Let us consider now a new basis {e0r e0s } in L, where {e0r } is a new orthogonal basis in Vn , connected to the initial basis {ek } by an orthogonal matrix Q. We have 0 0 0 T =Trs er es and T 0 = QT T Q. Thus, we get det T 0 = det T since QT = Q−1 . It follows that the value of det T does not depend on the basis used in its definition. T rT and det T are intrinsic characteristics of the tensor T and depend on this tensor only. This is the reason why trT and det T are also named invariants of the tensor T. We also observe that from the definition of det T, it follows that det TT = det T,

det (TU) = (det T) (det U) .

We say that a tensor T is singular if det T = 0 and it is non-singular if det T 6= 0. Any non-singular tensor T has an unique inverse tensor denoted by T −1 which satisfies the equations TT−1 = T−1 T = 1. The above definition results that the matrix of the components of the inverse tensor T−1 is the inverse matrix T −1 of the matrix T of the components of the non-singular tensor T−1 . Using the well known rules of the matrix calculus, it is easy to prove that if T and U are non-singular tensors, TT , T−1 and TU are also non-singular tensors and
−1
T −1 det T−1 = 1/ (det T) , (TU) = U−1 T−1 , TT = T−1 = T−T .

The non-singular tensors are important in the mechanics of deformable bodies since the gradient of any deformation is always a non-singular tensor. A tensor Q ∈L is named orthogonal if it satisfies the equation QT Q = 1.

It is obvious that a tensor Q is orthogonal if and only if the matrix Q of its components is an orthogonal matrix. Also, using the above definition, it is easy

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1.1. TENSOR ALGEBRA

to see that any orthogonal tensor Q is non-singular, and satisfies the following equations det Q = ±1, QQT = 1, QT = Q−1 . Moreover, we have Qu · Qv = u · QT Qv = u · 1v = u · v

∀u, v ∈ Vn .

The above property shows that any orthogonal tensor conserves the scalar product of vectors from Vn . Hence, it conserves the magnitude of vectors and the angle between two vectors. In particular, if {ek } is an orthonormal basis in Vn , the set {Qek } also forms an orthonormal basis in Vn , if Q is arbitrary orthogonal tensor from L. The orthogonal tensors are important in the mechanics of deformable bodies, since the rigid motions of the bodies are described by orthogonal tensors. Moreover, as we shall see, the symmetry properties of a solid deformable body are characterized by various orthogonal tensors or transformations. Using the trace of a tensor, the transpose of a tensor and the product or composition of two tensors, we can introduce an Euclidean scalar product in the vectorial space L of the second order tensors. If T and U are two tensors from L, their Euclidean scalar product, denoted by T · U is a real number given by the following rule:   (1.1.18) T · U = tr TUT . It is easy to see that this operation defined on the Cartesian product L × L and having values in R, satisfies the axioms (S1)-(S5) of an Euclidean scalar product. Moreover, if Tkm and Ukm are the components of T and U in a basis {ek em }, their scalar product T · U can be expressed in term of these components, by the following equation T · U = Tkm Ukm . The magnitude (intensity, norm) kTk of a tensor T is defined using the above introduced scalar product; i.e. p 1/2 (1.1.19) kTk = (T · T) = Tkm Tkm .

get

From the definition of the scalar product, for any vectors u, v, a, b ∈ Vn , we (ab) · (uv) = (a · u) (b · v) .

Particularly, if {ek } is an orthonormal basis in Vn , we have  1, if i = k and j = m (ei ej ) · (ek em ) = (ei · ek ) (ej · em ) = δik δjm = 0, otherwise. Hence, {ek em } is an orthonormal basis in L.

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CHAPTER 1. ELEMENTS OF TENSOR CALCULUS

In the mechanics of deformable bodies, an important role is played by the proper values and proper vectors of various second order tensors, such as Green’s deformation tensor and Cauchy’s stress tensor. This is the reason why we shall briefly present some aspects concerning proper values and proper vectors, stressing some important properties concerning the special, but important case of symmetric tensors. A number λ is an eigenvalue of the tensor T ∈L if a non-vanishing vector u ∈L exists, named eigenvector , such that Tu =λu.

(1.1.20)

The set of all vectors u ∈Vn , satisfying the equation (1.1.20) forms a vectorial subspace of Vn , and is named characteristic space of the tensor T, corresponding to the eigenvalue λ. A unit eigenvector of T is named a proper or principal direction of the tensor T. The vector equation (1.1.20) can be written in the equivalent form (T−λ1) u = 0.

(1.1.21)

Let us introduce now in Vn and L the orthonormal bases {ek } and {ek em } , respectively, and let {um } and {Tkm } be the components of u and T in these bases. It is clear that the above vector equation is equivalent to the following linear and homogeneous algebraic system, formed by n equations: (Tkm − λδkm ) um = 0,

k = 1, ..., n

(1.1.22)

the unknown being u1 , ..., un and λ. The above system has non-vanishing solutions if and only if λ satisfies the following algebraic equation of degree n: T11 − λ T12 .. T1n T21 T − λ ... T 22 2n =0 det [Tkm − λδkm ] = (1.1.23) ... ... ... ... Tn1 Tn2 ... Tnn − λ

This relation is the characteristic equation. In mechanical applications, it is very important the special case in which the tensor T is symmetric. In this case, we can prove the following fundamental theorems.

Theorem 1 If T ∈ L, is symmetric, the characteristic equation (1.1.23) has n real roots (distinct or not). √ To prove this property, we assume that a complex root λ = α + iβ (i = −1) of the characteristic equation (1.1.23) exists, and let um = vm + iwm be a nonvanishing solution of the system (1.1.22), corresponding to λ = α + iβ. Since the

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15

1.1. TENSOR ALGEBRA real and imaginary parts of the system (1.1.22) must vanish, we get Tkm vm − αvk + βwk = 0,

Tkm wm − αwk − βvk = 0.

Multiplying the first equation by wk , the second by vk, summing with respect to k, and making the difference of the obtained results, we get β (vk vk + wk wk ) = 0, since, according to the assumption made Tkm = Tmk . But vk vk + wk wk 6= 0, since {u1 , u2 , ..., un } is a non-vanishing solution of the system (1.1.22). Therefore, β = 0, and thus λ ∈ R. On the other side, the algebraic equation (1.1.23) is of degree n, hence it has n roots (distinct or not). Theorem 2 The proper vectors of a symmetric tensor T ∈ L, corresponding to two distinct eigenvalues, are reciprocally orthogonal. Indeed, let λ1 and λ2 be two distinct eigenvalues of T, and let u1 and u2 be two eigenvectors, corresponding to these eigenvalues we have Tu1 = λ1 u1 , Tu2 = λ2 u2 . From the above relation, it results that (Tu1 ) · u2 − (Tu2 ) · u1 = (λ1 − λ2 ) u1 · u2 . The tensor T is symmetric and we get
(Tu1 ) · u2 = TT u2 · u1 = (Tu2 ) · u1 ,

since TT = T. Consequently, we obtain

(λ1 − λ2 ) u1 · u2 = 0, and thus we can conclude that u1 · u2 = 0, since λ1 6= λ2 . Theorem 3 (Theorem of spectral representation) If T is a symmetric tensor from L, an orthonormal basis (n1 , n2 , n3 ) in V and three proper values (distinct or not) λ1 , λ2 , λ3 of T exists, such that Tnk = λk nk , k = 1, 2, 3(no summation !)

(1.1.24)

T = λ 1 n1 n1 + λ 2 n2 n2 + λ 3 n3 n3 .

(1.1.25)

and If λ1 , λ2 and λ3 are distinct, the characteristic spaces of T are one-dimensional vector subspaces of V, generated by the principal directions n1 , n2 and n3 of T. If λ1 6= λ2 = λ3 , T has only two distinct characteristic spaces, namely, the line generated by n1 and the plane perpendicular to n1 . At the same time, the relation (1.1.25) becomes T =λ1 n1 n1 + λ2 (1 − n1 n1 ) .

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(1.1.26)

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CHAPTER 1. ELEMENTS OF TENSOR CALCULUS

If λ1 = λ2 = λ3 , then T has a single characteristic space, represented by the whole vectorial space V . At the same time, the equation (1.1.25) takes the form T =λ1 1.

(1.1.27)

The relations (1.1.25)-(1.1.26) give the spectral decomposition of the tensor T. To prove the above theorem, we recall that according to Theorem 1, the tensor T has three real eigenvalues (distinct or not), λ1 , λ2 , λ3 , which are the roots of the characteristic equation T11 − λ T12 T13 T12 T22 − λ T23 = 0. (1.1.28) T13 T23 T33 − λ We suppose now that the eigenvalues λ1 , λ2 , λ3 are distinct. Let us denote by n1 , n2 , n3 the corresponding principal or proper directions. We have Tn1 = λ1 n1 ,

Tn2 = λ2 n2 ,

Tn3 = λ3 n3 .

(1.1.29)

Also, since n1 , n2 , n3 are unit eigenvectors, the equations kn1 k = kn2 k = kn3 k = 1,

(1.1.30)

are satisfied. For instance, the components α1 , β1 , γ1 of n1 can be determined using the relations (1.1.29)1 and (1.1.30)1 , which have the following component form: (T11 − λ1 ) α1 + T12 β1 + T13 γ1 = 0, T12 α1 + (T22 − λ1 ) β1 + T23 γ1 = 0, T13 α1 + T23 β1 + (T33 − λ1 ) γ1 = 0, α12 + β12 + γ12 = 1. According to Theorem 2, the unit vectors n1 , n2 , n3 are reciprocally orthogonal, hence {n1 , n2 , n3 } is an orthonormal basis in V . On the other hand, multiplying (1.1.24) by nm and using (1.1.13), we can conclude that the components Tkm of the tensor T in the basis {nk nm } are given by the relation Tkm = λk δkm (no summation!). If we introduce the above values in the general formula (1.1.12), we get the relation (1.1.25). Also, we can see that the matrix of the components of T in the basis {n k nm } has a diagonal form; i.e.   λ1 0 0 T =  0 λ2 0  . 0 0 λ3

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We observe also that any vector v ∈V , having the form v = αn1 , α ∈ R, satisfies the equation Tv =αTn1 = λ1 v; hence, v is an element of the characteristic space of T, corresponding to the eigenvalue λ1 . Reciprocally, let us assume that v =v1 n1 + v2 n2 + v3 n3 is a vector from V , satisfying the equation Tv =λ1 v. Introducing in this relation the expression of v and using (1.1.29), we get v2 (λ2 − λ1 )n2 + v3 (λ3 − λ1 )n3 = 0. From the last equation, we get v2 = v3 = 0, v = v1 n1 , since according to the assumption made λ1 , λ2 , λ3 are distinct. In this way, we can conclude that the characteristic space corresponding to λ1 , is, indeed the line generated by the principal direction n1 . The proof for the other two cases can be obtained in a similar manner. The theorem of spectral decomposition shows that a symmetric tensor is completely determined by its eigenvalues and by the corresponding characteristic spaces. We recall now that the eigenvalues of a tensor were defined in an intrinsic manner, using the vector equation (1.1.20). Hence, the eigenvalues of a tensor do not depend on a basis {ek } used to write the characteristic equation (1.1.28). Consequently, the coefficients of the algebraic equation (1.1.28) are also invariants relative to a change of basis in V . It is easy to see that the characteristic equation (1.1.28) can be written in the following equivalent form: λ3 − IT λ2 − IIT λ − IIIT = 0 where: IT IIT

IIIT

=

T11 + T22 + T33 = trT, T22 T23 T11 T13 T11 T12 − − = − T23 T33 T13 T33 T12 T22  1 1 2 T · T− (trT) , (Tkl Tkl − Tkk Tll ) = = 2 2 T11 T12 T13 = T12 T22 T23 = det T. T13 T23 T33



The quantities IT , IIT , IIIT are called the first, second and third principal invariants of the tensor T. According to Vi`ete’s relations, these invariants can be expressed in terms of the eigenvalues λ1 , λ2 , λ3 by the following equations: IT IIT

= =

IIIT

=

trT = λ1 + λ2 + λ3 −(λ1 λ2 + λ2 λ3 + λ3 λ1 ), det T = λ1 λ2 λ3 .

Since Green’s deformation tensor and Cauchy’s stress tensor are symmetric tensors, the above results play an important role in the mechanics of deformable bodies, and can be used to characterize the deformation and the stress state of a continuous, deformable material.

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CHAPTER 1. ELEMENTS OF TENSOR CALCULUS

In order to introduce tensors of superior order, in all what follows, we assume that the underlying vector space is the usual three-dimensional vector space V of the free vectors. The corresponding nine-dimensional vector space of the second order tensors will be denoted by L or by L2 . We shall denote by LS the subset of all symmetric tensors, and by LA the subset of all antisymmetric tensors. It is easy to see that LS is a six-dimensional vector subspace of L, and LA is a tree-dimensional vector subspace of L. A linear function A defined on V and having values in L2 = L is named a third-order tensor . If the value of A, corresponding to a vector v is the second order tensor T; we shall write T = A (v) = Av. We denote by L3 the set of all third order tensors. We say that two third order tensors A and B are equal if ∀v ∈ V.

Av = Bv,

The third order null tensor, designed by 03 , is defined by the equation 03 v =02 ,

∀v ∈V,

where by 02 = 0, we have designed the second order null tensor. The sum A + B of two third order tensor A and B, and the product αA = Aα of a third order tensor and the real number α ∈ R are defined by the equations (A + B) v = Av + Bv,

(αA) v = α (Av) ,

∀v ∈ V, ∀α ∈ R.

As for L = L2 , it is easy to see that L3 , in which the above concepts and operation were introduced, is a vector space, denoted also by L3 . The neutral element is the null tensor 03 , and the opposite to A is the tensor −A = −1A. To construct bases in L3 , we shall introduce the tensor product of three vectors from V . We name tensor product of three vectors u, v, w ∈ V , the function u ⊗ v ⊗ w = uvw, defined on V and having values in L = L2 , by the following rule: (uvw) (a) = uv (w · a)

for any

a ∈V.

Using this definition and the properties of the tensor product uv ∈ L = L2 , it is easy to prove that uvw is a linear function, hence, a third order tensor. Also, it can be proved that if {ek } , k = 1, 2, 3, is an orthonormal basis in V , the set of 27 tensor products {ek el em } , k, l, m = 1, 2, 3, forms a basis in the vector space L3 . Hence, the dimension of the vector space L3 is 27.

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1.1. TENSOR ALGEBRA

Any third order tensor A can be expressed in a unique way as a linear combination of the 27 third order tensors ek el em , k, l, m = 1, 2, 3; i.e. A = Aklm ek el em . The numbers Aklm , k, l, m = 1, 2, 3, are the components of the A in the considered basis. If T = Av and T = Tkl ek el , v = vs es , we find T = Tkl ek el = (Aklm ek el em )(vs es ) = Aklm vs ek el (em · es ) = Aklm vs ek el δms = Aklm vm ek el .

Hence, the components Tkm of the tensor T are expressed in terms of the components Aklm and vs of A and v by the equation Tkl = Aklm vm . If {ek } and {e0r } are two orthonormal basis in V , connected by the relations (1.1.4), the systems {ek el em } and {e0r e0s e0t } are basis in L3 . Hence, we have also: A = A0rst e0r e0s e0t . Using the properties of the tensor product of three vectors and the relations (1.1.4), connecting the bases {ek } and {e0r }, we can conclude that the components Aklm and A0rst of the same third order tensor A are connected by the equations A0rst = qkr qls qmt Aklm ,

Aklm = qkr qls qmt A0rst ,

(1.1.31)

with k, l, m, r, s, t = 1, 2, 3. The third order tensors play an important role in thermomechanics and electromechanics, since the material properties of deformable bodies which are also heat conducting or electrically polarizable are expressed using tensors of second, third and fourth order. In pure mechanical theories, the role of the third order tensors is relatively reduced. However, in order to introduce the gradient of a second order tensor field, and to obtain in this way the divergence of a second order tensor field, we must use third order tensor fields. In turn, the divergence of a second tensor field is indispensable in the mechanics of deformable bodies, since the equilibrium conditions of a deformable body are expressed just using the divergence of Cauchy’s stress tensor. Also, we use third order tensors to introduce fourth order tensors. As we shall see, the material properties of elastic solids can be expressed, in an adequate way, using fourth order tensors. This is the reason we must introduce these entities. A linear function Φ, defined in V and having values in L3 , is named a fourth order tensor . If the value of Φ, corresponding to a vector v from V , is the third order A from L3 , we shall write A = Φ (v) = Φv.

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CHAPTER 1. ELEMENTS OF TENSOR CALCULUS We denote by L4 the set of all fourth order tensors. We say that two fourth order tensors Φ and Ψ are equal if Φv = Ψv,

∀v ∈V.

The fourth order null tensor, designed by 04 , is defined by the equation 04 v = 0 3 ,

∀v ∈ V.

The sum Φ + Ψ of two fourth order tensors Φ and Ψ, and the product αΦ = Φα of a fourth order tensor Φ with a real number α, are defined by the equations (Φ + Ψ)(v) = Φv + Ψv,

(αΦ)(v) = α(Φv),

∀v ∈ V, α ∈ R.

As for L2 and L3 , it is easy to see that L4 , endowed with the above concept and operations is a vector space. Obviously, the neutral element is the null tensor 04 and the opposite of Φ is −Φ = −1Φ. To find bases in L4 , we introduce the tensors product of four vectors. We name tensor product of four vectors a,b,u,v ∈ V , the function a ⊗ b ⊗ u ⊗ v = abuv, defined on V and having values in L3 , given by the following rule: (abuv)(w) = abu(v · w),

∀w ∈ V.

It can be proved that if {ek }, k = 1, 2, 3, is an orthonormal basis in V , the set of 81 tensor products {ek el em en }, k, l, m, n = 1, 2, 3, form a basis in the vector space L4 . Hence, the dimension of L4 is 81, and any fourth order tensor Φ can be expressed in an unique way as a linear combination of the fourth order tensors ek el em en , , k, l, m, n = 1, 2, 3; i.e. Φ = Φklmn ek el em en .

(1.1.32)

The quantities Φklmn , k, l, m, n = 1, 2, 3 are the components of Φ in the considered basis. If A = Φv and A = Aklm ek el em and v = vs es , we successively get A = Aklm ek el em

= (Φklmn ek el em en )(vs es ) = Φklmn vs ek el em (en · es ) = Φklmn vs ek el em δns = Φklmn vn ek el em .

Hence, the components Aklm of the third order tensor A are expressed in terms of the components Φklmn and vs of Φ and v, by the equations Aklm = Φklmn vn .

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1.1. TENSOR ALGEBRA

If {ek } and {e0r } are two orthonormal basis in V , connected by the equations (1.1.4), the systems {ek el em en } and {e0r e0s e0t e0u } , k, l, m, n, r, s, t, u = 1, 2, 3, are basis in L4 . Also, we have Φ = Φ0rstu e0r e0s e0t e0u . (1.1.33) It is easy to show that the equations connecting the components Φklmn and Φ0rstu of Φ, in the two basis have the following form: Φ0rstu = qkr qls qmt qnu Φklmn ,

Φklmn = qkr qls qmt qnu Φ0rstu

(1.1.34)

with k, l, m, n, r, s, t, u = 1, 2, 3. In concluding this part concerning tensor algebra, we shall introduce some contracted products, which will be used in later considerations. The definitions will be given using some bases, but it will be shown that the defined entities are independent on the bases used. For a given vector v =vk ek , a second order tensor T =Tkm ek em and a third order tensor A =Aklm ek el em , we define the left and right dot product of v and T and of v and A, respectively, by the following equations: v · T = vk Tkl el , v · A = vk Aklm el em ,

T · v = Tkl vl ek , A · v = Aklm vm ek el .

(1.1.35) (1.1.36)

Obviously, v · T = TT v,

T · v = Tv,

A · v = Av.

Hence, v · T, T · v and A · v are independent on the bases used. Also, taking into account the equation (1.1.4) converting two orthonormal bases {ek } and {e0r }, denoting by vr0 and A0rst the components of v and A in the bases {e0r } and {e0r e0s e0t }, and using the transformation rules (1.1.5), (1.1.31), together with the orthogonality conditions satisfied by {qkr }, we can see that v · A = vk Aklm el em = vr0 Arst e0s e0t . Thus the definition used to introduce v · A is actually independent on the bases, and v · A is a second order tensor which depend only on the vector v and on the third order tensor A. In a similar way, we shall introduce the contracted product of a fourth order tensor Φ = Φklmn ek el em en and a second order tensor T = Tmn em en , by using the following rule: ΦT = Φklmn Tmn ek el . (1.1.37) It is easy to see that ΦT does not depend on the bases used. 0 0 0 That is, if Φ = Φ0rstu e0r e0s e0t e0u and T = Ttu et eu , we have 0 0 0 er es . ΦT = Φklmn Tmn ek el = Φ0rstu Ttu

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22

CHAPTER 1. ELEMENTS OF TENSOR CALCULUS

Consequently, U = ΦT = Ukl ek el is a second order tensor and its components Ukl in the basis {ek el } are given by the equation Ukl = Φklmn Tmn . The contracted product of a fourth order tensor and a second order tensor play a fundamental role in the theory of anisotropic linearly elastic materials, since in this theory the Cauchy’s stress tensor is the contracted product of the fourth order elasticity tensor (or Hooke’s tensor) and the infinitesimal second order strain tensor. We also observe that if U = ΦT, according to the definition of the scalar product of two second order tensors U and V = Vkl ek el , we have V · U = Vkl Φklmn Umn

(1.1.38)

U · U = U · ΦU = Ukl Φklmn Umn .

(1.1.39)

and The above relations are essential in the theory of linearly elastic solids. The double of the specific elastic energy is expressed as the scalar product of the infinitesimal strain tensor and stress tensor, and the latter is the contracted product of the elasticity tensor and the infinitesimal strain tensor. Examining the equation (1.1.37), defining the contracted product of a fourth order tensor and a second order tensor, we can conclude that a fourth order tensor can be considered as being a linear function defined on the vector space L = L 2 of the second order tensors and having values in the same space. Taking into account this new point of view, we can introduce the product or composition of two fourth order tensors, using the usual rule of composition of two functions. If Φ and Ψ are two fourth order tensors from L4 , their product (or composition) ΦΨ is defined by the equation: (ΦΨ) (T) = Φ (ΨT)

for any T from L2 = L.

It is easy to see that ΦΨ is a linear function on L2 , having values in the same space. In other words, ΦΨ is a fourth order tensor. Also it is easy to see that if Φ = Φklmn ek el em en , Ψ = Ψklmn ek el em en and Λ = ΦΨ = Λklmn ek el em en , then Λklmn = Φklrs Ψrsmn .

(1.1.40)

The tensor I ∈ L4 , defined by the equation I = δkm δln ek el em en = ek el ek el ,

(1.1.41)

is named the fourth order unit tensor. It is easy to see that the components of this tensor are δkm δln in any basis and I has the following properties: IT = T,

Copyright © 2004 by Chapman & Hall/CRC

∀T ∈ L2 ,

I Φ = ΦI = Φ,

∀Φ ∈L4 .

23

1.1. TENSOR ALGEBRA

The fourth order tensor Φ−1 , if it exists, is named the inverse of the fourth order tensor Φ, if ΦΦ−1 = Φ−1 Φ = T. The function ΦT defined on L2 and having values in L2 , is named the transpose of the fourth order tensor Φ, if it satisfies the condition T · ΦT U = U · ΦT,

∀T, U ∈ L2 .

(1.1.42)

It can be shown that the above condition uniquely defines the transpose Φ T and the components ΦTklmn of ΦT are expressed in terms of the components Φklmn of Φ by the relation ΦTklmn = Φmnkl . (1.1.43) A fourth order tensor Φ is symmetric if ΦT = Φ.

(1.1.44)

From (1.1.43), it follows that if Φ is symmetric, its components satisfy the relations Φklmn = Φmnkl , k, l, m, n = 1, 2, 3. (1.1.45) Consequently, a symmetric fourth order tensor has only 45 independent components. As we shall see, tensors of this type play a fundamental role in the linearized, three dimensional theory of elastic bodies, since the incremental constitutive equations, containing the instantaneous elasticities, are expressed just by fourth order tensors having the symmetry property described above. We shall denote by LS4 the set of all symmetric fourth order tensors. As it is easy to see, LS4 is a vector subspace of L4 , and its dimension is 45. In the theory of linearity elastic solids, an important role is played by those symmetric fourth order tensors from LS4 , which also satisfy the following additional condition:
(1.1.46) Φ TT = ΦT, ∀T ∈ L2 .

Denoting by Φklmn the components of Φ ∈ LS4 , the above condition shows that these components must also satisfy the following additional restriction: Φklmn = Φklnm , k, l, m, n = 1, 2, 3. Since Φ is symmetric, we can conclude that its components satisfy the following symmetry relations: Φklmn = Φlkmn = Φklnm = Φmnkl . Hence, Φ has only 21 independent components. Also, from (1.1.46) it results that T

(ΦT) = ΦT,

Copyright © 2004 by Chapman & Hall/CRC

ΦT = ΦTS ,

∀T ∈ L2

(1.1.47)

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CHAPTER 1. ELEMENTS OF TENSOR CALCULUS

and ΦΩ = 0,

∀Ω ∈ LA 2.

The last equation shows that a symmetric fourth order tensor Φ, having additional property (1.1.46), cannot be an injective application from L2 in L2 . However, considering any symmetric Φ with the property (1.1.46) as a function defined on the set LS2 of the symmetric second order tensors, and having values in the same vector space, we can ask ourselves if this new function has an inverse. b 4 the vector space of all symmetric fourth order tensors Let us denote by L having the property (1.1.46), and which are considered as functions on the vector space LS2 and having values in the same space. We now observe that the fourth order tensor bI with components bIklmn = 1 (δkm δln + δkn δlm ) 2

(1.1.48)

b4 . is symmetric and satisfies the property (1.1.46). Hence, bI is an element of L Moreover bIΦ = ΦbI = Φ, ∀Φ ∈ L b4 .

b4 . Taking into account the last equation, we call bI the unit tensor in L −1 b A fourth order tensor Φ ∈ L4 , if it exists, will be named the inverse of the b 4 , if fourth order tensor Φ ∈ L Φ−1 Φ = ΦΦ−1 = bI.

b 4 is non-singular, if its Also, we shall say that the fourth order tensor Φ ∈ L −1 inverse Φ exists. b 4 will be named positive definite if: A fourth order tensor Φ ∈ L T · ΦT ≥ 0,

∀T ∈ LS2

and T · ΦT = 0

if and only if

T = 0.

b 4 is positive definite, it is also It is easy to see that if the tensor Φ ∈ L non-singular, its inverse Φ−1 exists and is also positive definite. b 4 plays an essential role the linear elasticity, since the The vector space L elasticity tensor of any hyperelastic material is an element of this space. Moreover, all elasticity tensors are positive definite, and consequently their inverse also exists and are positive definite.

Copyright © 2004 by Chapman & Hall/CRC

1.2. TENSOR ANALYSIS

1.2

25

Tensor Analysis

Let us denote by D a simply connected bounded open set of the threedimensional Euclidean space E. We assume that the boundary ∂D of D is a closed regular surface. We denote by (O, e1 , e2 , e3 ) a rectangular Cartesian coordinate system in E, defined by a point O from E and by three orthonormal vectors −−→ e1 , e2 , e3 from V . The position vector OP of a point P ∈ E will be denoted by x =xk ek , x1 , x2 , x3 being the Cartesian coordinates of P in the selected coordinate system (O, ek ) . Various scalar, vector and tensor valued functions defined on D, are named fields. More precisely, any function ϕ : D → R is named a scalar field , any function u : D → V represents a vector field, any function T : D → L = L2 is a second order tensor field , any function A : D → L3 is named a third order tensor field and any function Φ : D→L4 represents a fourth order tensor field . We assume that the definition concerning the continuity and differentiability of various scalar, vector and tensor fields are known. We recall that a vector or a tensor field is continuous in D if and only if their components are continuous real valued functions in D. Similarly, a vector or a tensor field is differentiable in D, if and only if their components are differentiable real valued functions in D. We say that a vector or a tensor field is of class C 0 in D if it is continuous in D, and it is of class C 1 in D or smooth in D, if their components are functions of class C 1 in D; i.e. are differentiable in D and their partial derivatives of first order are continuous functions in D. In a similar manner, is defined a vector or a tensor field of class C k on D, for k = 2, 3, .... Let as observe that the above regularity properties do not depend on their basis used to introduce vector or tensor components. In the following, we shall introduce the most important differential operators, playing essential roles in the mechanics of deformable bodies. For this purpose, we shall use various coordinate systems in order to introduce new vector and tensor fields. In such situations, we shall prove that the new fields introduced in this way, are actually independent on the coordinate systems used and these have an intrinsic character. To accomplish this task, we shall use the concept of the directional derivative, having obviously an intrinsic nature. Let ϕ : D → R be a scalar field and let P be a fixed point in D. Let us consider a variable point P 0 ∈ D situated on the right line passing through P and having the direction fixed by a given unit vector s = sk ek as it is shown in Figure 1.2. Let us denote by x = xk ek and x0 = x0k ek = x+αs, the position vectors of P and P 0 , respectively. If the following limit

ϕ(x + αs) − ϕ(x) ϕ (P 0 ) − ϕ (P ) ∂ϕ = lim (P ) = lim 0 −−→0 α→0 P →P α ∂s k PP k

(1.2.1)

exists and is bounded, we say that the scalar field ϕ is differentiable in the direction

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26

CHAPTER 1. ELEMENTS OF TENSOR CALCULUS

P' P s

x

x' = x + as, aÎR, a>0

O Figure 1.2: Concerning the definition of the directional derivative.

s at the point P . The real number ∂ϕ ∂s (P ) is named the derivative of ϕ in the direction s at the point P . It can be shown that if ϕ is of class C 1 in D, ∂ϕ ∂s (P ) exists in any direction s, in any point P ∈ D, and can be calculated by using the relations

∂ϕ ∂ϕ (P ) = sk ϕ,k (P ). (P ) = sk ∂xk ∂s

(1.2.2)

In order to evaluate the right part of the above equation, we assume that the field ϕ is given as a function of the coordinates x1 , x2 , x3 , i.e. ϕ = ϕ(P ) = ϕ(x1 , x2 , x3 ). As used in the equation (1.2.1), the partial derivative of a real valued function ϕ with respect to the coordinate xk will be denoted in the following by ∂ϕ = ϕ,k . ϕ,k ; i.e. ∂x k Let ϕ : D → R be a scalar field of class C 1 in D. The vector field grad ϕ, defined on D by the equation

grad ϕ(P ) = ek

∂ϕ (P ) = ek ϕ,k (P ), ∂xk

∀P ∈ D,

(1.2.3)

is named the gradient of the scalar field ϕ in D. The above equation and the relation (1.2.2) show that the directional derivative ∂ϕ ∂s (P ) can be expressed as the scalar product of the two vectors, in the following form ∂ϕ (P ) = s · grad ϕ(P ). (1.2.4) ∂s The right hand side of the last equation is independent on any coordinate system; the same property has the unit vector s. Consequently, grad ϕ (P ) has also an intrinsic nature. It is independent on any coordinate system in E and on any basis in V , used initially in the equation (1.2.3) by which grad ϕ (P ) was

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27

1.2. TENSOR ANALYSIS

introduced. Hence, grad ϕ is really a vector field, depending only on the scalar field ϕ. In other words, if (O, e0r ) is a new coordinate system in E, {e0r } being an orthonormal basis in V , we have ϕ = ϕ (P ) = ϕ0 (x01 , x02 , x03 ) , and grad ϕ (P ) = e0k

∂ϕ0 (P ) = e0k ϕ0,k (P ) ∂x0k

∀P ∈ D.

Similar considerations and results will be obtained when we introduce the gradient of a vector or of a tensor field, using the directional derivatives of the involved fields and some arbitrarily selected coordinate systems in E. The newly introduced entities will always have an intrinsic character, and depending on the involved vector or tensor fields, they are independent on the coordinate systems and bases used to define the corresponding gradients. Returning to the equation (1.2.1), we note that formally this equation can be written in the following equivalent form: grad ϕ = (ek

∂ )ϕ. ∂xk

(1.2.5)

The linear differential operator ∇ = ek

∂ ∂ ∂ ∂ +e3 + e2 = e1 ∂x ∂x2 ∂x1 ∂xk 3

(1.2.6)

can be considered as a (formal) vector, which multiplied by ϕ gives grad ϕ. This (formal) vector is named Hamilton’s operator or the nabla operator. Using this differential operator, we can express the equations (1.2.3) and (1.2.4) in a more concentrated form as: grad ϕ = ∇ϕ,

∂ϕ = (s · ∇)ϕ, ∂s

s · ∇ = sk

∂ . ∂xk

(1.2.7)

Hamilton’s ∇ (nabla) will be used to introduce various important differential operators. Let v : D → V be a vector field and let P be a fixed point in D. If the following limit

v(x + αs) − v(x) v(P 0 ) − v(P ) ∂v = lim (P ) = lim −−→0 α→0 P 0 →P α ∂s kP P k

(1.2.8)

exists and is bounded, we say that the vector field v is differentiable in the direction s at the point P . The vector ∂v ∂s (P ) is named the derivative of v in the direction s at the point P . It can be shown that v is of class C 1 in D, ∂v ∂s (P ) exists in any direction s, in any point P ∈ D and can be calculated using the relations

∂vm ∂v (P )em = sk vm,k (P )em (P ) = sk ∂xk ∂s

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(1.2.9)

28

CHAPTER 1. ELEMENTS OF TENSOR CALCULUS

where v(P ) = vm (P )em . Let v : D → V be a vector field of class C 1 in D. The second order tensor field grad v defined on D by the equation grad v(P ) = ∇v(P ) = ek

∂v (P ) ∂xk

(1.2.10)

is named the gradient of the vector field v in D. From the above equation, we get grad v(P ) =

∂vm (P )ek em = vm,k (P )ek em . ∂xk

(1.2.11)

On the other hand, using the left dot product of a vector and a second order tensor, from (1.2.9) and (1.2.11), we obtain ∂v (P ) = s · grad v(P ) = (grad v(P ))T s. ∂s

(1.2.12)

We stress here that by introducing the gradient of a vector field or of a tensor field we use the left formalism, as it can be seen from equations (1.2.10) and (1.2.11). This formalism is very useful if we wish to define and express various differential operators in curvilinear coordinate systems, in which case the base vectors are variable entities depending on the place in which they are determined. It is easy to see that grad is a linear differential operator ; i.e. grad(αu+βv) = αgradu + βgradv, for any u, v ∈ V and α, β ∈ R. By the divergence of a vector field v : D → V , of class C 1 in D, we mean the scalar field div v : D → R, defined by the following equivalent relations div v = ∇ · v = ek ·

∂vk ∂v = vk,k = v1,1 + v2,2 + v3,3 . = ∂xk ∂xk

(1.2.13)

Using the trace of a second order tensor, we can express also div v by the equivalent equations div v = tr(∇v) = tr(gradv). (1.2.14) Since tr and grad are linear operators, div is also linear ; i.e. div(αu + βv) = α div u + β div v,

∀u, v ∈ V, α, β ∈ R.

Finally, let T : D → L = L2 be a second order tensor field in D. If the following limit

T(x + αs) − T(x) T(P 0 ) − T(P ) ∂T = lim = (P ) = lim 0 − − → α→0 P →P α ∂s k PP0 k

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(1.2.15)

29

1.2. TENSOR ANALYSIS

exists and is bounded, we say that the tensor field T is differentiable in the direction s at the point P . The second order tensor ∂T ∂s (P ) is named the derivative of T, in the direction s at the point P . It can be shown that if T is of class C 1 in D, ∂T ∂s (P ) exists in any direction s, at any point P ∈ D, and can be calculated using the relations

∂Tlm ∂T (P )el em = sk Tlm,k (P )el em , (P ) = sk ∂xk ∂s

(1.2.16)

where T(P ) = Tlm (P )el em . Let T : D → L2 = L be a second order tensor field of class C 1 on D. The third order tensor field grad T defined by the equations grad T(P ) = ∇T(P ) = ek

∂T (P ), ∂xk

(1.2.17)

is named the gradient of the second order tensor field T on D. From the equation (1.2.16) and (1.2.17), we get grad T(P ) =

∂Tlm (P )ek el em = Tlm,k (P )ek el em . ∂xk

(1.2.18)

At the same time, using the relations (1.2.16) and (1.2.18) and the left dot product of a vector with a third order tensor, we obtain ∂T (P ) = s · grad T. ∂s

(1.2.19)

Again, it is easy to see that grad is a linear differential operator; i.e. grad(αT + βU) = α grad T + β grad U, for any T, U ∈ L2 = L and any α, β ∈ R. By the divergence of a second order tensor field T : D → L2 = L, we mean the vector field div T : D → V , defined by the following relations: div T = ∇ · T = (ek

∂Tkm ∂ em = Tkm,k em , ) · (Tlm el em ) = ∂xk ∂xk

(1.2.20)

where we have used the left dot product of Hamilton’s vector ∇ with the second order tensor T. It is easy to see that div is a linear differential operator; i.e. div(αT+βU) = α div T + β div U, for any T,U ∈ L2 = L and α, β ∈ R. We stress again the fact that in introducing grad T and div T, we have used the left formalism.

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30

CHAPTER 1. ELEMENTS OF TENSOR CALCULUS According to the above definition, the components (div T)m of div T are (div T)1 =

∂T31 ∂T21 ∂T11 ∂Tk1 + + = Tk1,k = ∂x3 ∂x2 ∂x1 ∂xk

(div T)2 =

∂T32 ∂T22 ∂T12 ∂Tk2 + + = Tk2,k = ∂x3 ∂x2 ∂x1 ∂xk

(div T)3 =

∂T33 ∂T23 ∂T13 ∂Tk3 . + + = Tk3,k = ∂x3 ∂x2 ∂x1 ∂xk

(1.2.21)

If T is a symmetric second order tensor field, the left and the right formalism lead to the same divergence of T and we have div T =

∂Tkm ek = Tkm,m ek ∂xm

if TT = T.

(1.2.22)

In a developed form, we get (div T)1 =

∂T13 ∂T12 ∂T11 , + + ∂x3 ∂x2 ∂x1

(div T)2 =

∂T23 ∂T22 ∂T21 , + + ∂x3 ∂x2 ∂x1

(div T)3 =

∂T33 ∂T32 ∂T31 , + + ∂x3 ∂x2 ∂x1

(1.2.23)

if Tkm = Tmk . In the last part of this Section, we shall present those integral theorems which play an important role in the mechanics of deformable bodies, particularly in the theory of elastic composites. We assume that the scalar vector and tensor fields ϕ, u and T are of class C 2 ¯ = D ∪ ∂D of D. We shall denote by n the on D and of class C 1 on the closure D unit outward normal to the closed regular boundary ∂D of D. With the assumed regularity condition, the following equations hold: Z Z Z grad ϕ dv = ∇ϕ dv = ϕ n da, (1.2.24) D

Z

Z

Z

D

grad u dv = D

div u dv = D

div T dv = D

Z

Z

divT dv =

D

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Z

D

∇u dv = D

Z

∇ · u dv =

D

∇ · T dv =

D

∇ · T dv =

Z

∂D

n · u da,

∂D

n · T da =

Z

D

Z

∂D

n ⊗ u da =

Z

Z

Tn da, ∂D

Z

nu da,

(1.2.25)

∂D

(1.2.26) Z

if

TT n da,

(1.2.27)

∂D

TT = T.

(1.2.28)

31

1.2. TENSOR ANALYSIS Z

Z

D

D

u · div T dv = u · div T dv =

Z

Z

T

∂D

∂D

u · T n da − u · Tn da −

Z

Z

D

D

∇u · T da,

T · E(u) dv,

(1.2.29) if

TT = T,

(1.2.30)

b is the symmetric gradient of the vector field u defined by the where E(u) = ∇u equation b = 1 (∇u + ∇uT ). (1.2.31) E(u) =∇u 2 The formulas (1.2.24)-(1.2.26) concerning the scalar and vector fields ϕ and u are classical, and can be proved using the Gauss-Ostrogradsky’s theorem. To establish the equation (1.2.27), we use the component from (1.2.22) of divT, the Gauss-Ostrogradsky theorem, and take into account that ek , k = 1, 2, 3, are constant vectors. In this way, we successively get Z Z Z Z n · T da. nk Tkm em da = Tkm,k ek dv = div T dv = D

D

∂D

D

Equation (1.2.29), in the same way can be proved. We get Z Z Z u · div T dv = um (divT)m dv = um Tkm,k dv D ZD Z D = (um Tkm ),k dv − um,k Tkm dv ZD ZD = nk um Tkm da − (∇u)km Tkm dv D Z∂D Z = um (n · T)m da − ∇u · Tdv Z∂D Z D T = u · (T n)da − ∇u · Tdv. ∂D

D

The equation (1.2.28) plays a fundamental role in the mechanics of deformable bodies in formulating the balance law of momentum, the involved symmetric second order tensor being the Cauchy’s stress tensor. As we shall see, equation (1.2.27) plays an essential role in stability theorems, the involved second order tensor being the non-symmetric incremental stress tensor. Equations (1.2.31) and (1.2.30) are used in order to express the fundamental energetical relations of the classical linear elasticity and of the linearized three-dimensional theory of elastic bodies, respectively. As we shall see, in the mechanics of composite materials, we encounter piecewise homogeneous structures; that is, bodies which are formed by two or more different homogeneous elastic materials. In such cases, various vectors and tensor fields, describing the behavior of the multi-phasic mixture can have finite jumps across the common boundaries of the component phases. The validity of the above

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CHAPTER 1. ELEMENTS OF TENSOR CALCULUS

given integral formulas must be analyzed accordingly. The results of this analysis will be proved in the Section devoted to the piece-wise homogeneous linearly elastic solids.

1.3

Problems

P1.1 Show that the null vector 0 is unique. Show that for any vector, u corresponds exactly one vector −u, such that u+ (−u) = 0. P1.2 Show that 0u = 0, α0 = 0 for any u ∈V and for any α ∈ R. P1.3 Show that any system of p vectors, containing the null vector, is linearity dependent. P1.4 Show that in the usual three-dimensional vector space V of the free vectors, any three vectors, which are not situated in the same plane, form a basis. P1.5 Let Pn+1 be the set of all polynomials having at most the degree n. Show that Pn+1 is a vector space and the system 1, x1 , x2 , ..., xn is a basis in Pn+1 . Find the dimension of Pn+1 . P1.6 Show that the magnitude kuk of a vector u satisfies the following properties: (N1) kuk ≥ 0 ∀u ∈V and u = 0 if and only if u = 0; (N2) kαuk = |α| kuk ∀u ∈V, α ∈ R; (N3) ku + vk ≤ kuk + kvk ∀u, v ∈V. Rb P1.7 Show that the operation hP, Qi = P (x) Q (x) dx, a, b ∈ R, defined for a

any P, Q ∈ Pn+1 , is a scalar product in Pn+1 . P1.8 Show that 0 ·u = 0 for any u ∈ V and |u ·v| ≤ kuk kvk for any u,v ∈ V . The last inequality determined by Schwarz, Cauchy and Buniakowsky, shows that −1 ≤

u·v ≤ 1. kuk kvk

In this way, we can introduce the angle θ ∈ [0, 2π) between the vectors u and v, using the equation: u·v . θ = arccos kuk kvk

P1.9 Let V be an Euclidean vector space and let u and w be two vectors from V. Show that if u · v = w · v for any v ∈ V, then u = w. P1.10 Show that the second order null tensor 0 is a linear function. P1.11 Show that the tensor product of two vectors has the following properties: (αu)v = u(αv) = α(uv), ∀u, v ∈ V, α ∈ R; (u + v)w = uw + vw,

u(v + w) = uv + uw,

∀u, v, w ∈ V.

We have used frequently the above properties in our considerations.

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33

1.3. PROBLEMS

P1.12 Show that all components of the null tensor 0 are vanishing. P1.13 Show that the components of T + U and αT are Tkm +Ukm and αTkm , Tkm and Ukm being the components of T and U, respectively. P1.14 Let u and v be two vectors from Ln , and let {uk } and {vk } be their components in an orthonormal basis {ek } of Vn . Show that in the basis {ek em } of L, the components (uv)km of the tensor product uv, are given by the relations (uv)km = uk vm

k, m = 1, 2, ..., n.

P1.15 Show, using an example, that generally the tensor product uv of two vectors is not commutative; i.e. uv 6= vu. P1.16 Let T and U be two tensors from L, and let TU be their product. Show that TU is a linear function, that is, it is a second order tensor. P1.17 Show that the product of two tensors has the following properties: (TU)V = T(UV), (associativity) T(U + V) = TU + TV, (T + U)V = TV + UV, (distributivity with respect to tensor addition), 1T = T1 = T where T, U, V are arbitrary second order tensors, and 1 is the (second order) unit tensor. P1.18 Using an example, show that generally, the product of two tensors is not commutative; i.e. TU 6= UT. P1.19 Show that TT is a linear function, that it is a second order tensor. P1.20 Show that: (uv)T = vu, ∀u, v ∈ V.

P1.21 Show that:

(Tu)(Uv) = T(uv)UT ,

∀u, v ∈ Vn , T, U ∈ L.

P1.22 Show that: (ab)(uv) = (b · u)av,

∀a, b, u, v ∈ V.

P1.23 Show that the scalar product T · U of two second order tensors T and U satisfies the axioms (S1)-(S5) of an Euclidean scalar product. P1.24 Show that: (ab) · (uv) = (a · u) (b · v)

∀a, b, u, v ∈ Vn .

P1.25 Let {ek } , k = 1, 2, 3 be an orthonormal basis in V , and let T ∈L be a second order tensor defined by the equations Te1 = e1 + e2 ,

Copyright © 2004 by Chapman & Hall/CRC

Te2 = e2 + e3 ,

Te3 = e3 + e1 .

34

CHAPTER 1. ELEMENTS OF TENSOR CALCULUS

(a) Express the tensor T in the basis {ek em } , k, m = 1, 2, 3 of L; (b) Find the vector u = Tv if v = e1 + e2 + e3 ; T (c) Show that T2 = TT = 2TS and find TA ; (d) Find IT = trT, IIT = det T and kTk ; (e) Show that T is non-singular and find T−1 . P1.26 Show that the tensor T ∈L is non-singular if and only if from Tv = 0 if follows v = 0. P1.27 Show that if S is a symmetric tensor and A is an antisymmetric tensor, then S · A = 0. P1.28 Let {ek } be an orthonormal basis in V . Show that the six second order tensors √ √ √ e1 e1 , e2 e2 , e3 e3 , (e1 e2 + e2 e1 ) / 2, (e2 e3 +e3 e2 ) / 2, (e3 e1 +e1 e3 ) / 2

form an orthonormal basis in LS , the vector space of the symmetric second order tensors. Show that the three second order tensors √ √ √ (e1 e2 − e2 e1 ) / 2, (e2 e3 − e3 e2 ) / 2, (e3 e1 − e1 e3 ) / 2

form an orthonormal basis in LA , the vector space of the antisymmetric second order tensors. P1.29 Prove the following relation: (Tu) v = T (uv) (Tu) (Sv) = T (uv) ST , where u, v are arbitrary vectors from Vn and T, S are arbitrary tensors from L. P1.30 Show that the set of all vectors u ∈Vn which satisfies the equation Tu =λu for T ∈L and λ ∈ R fixed, form a vectorial subspace of Vn . P1.31 Let Ω ∈LA be an antisymmetric tensor, the matrix of its components being given by the equation   0 −ω3 ω2 0 −ω1  , Ω = [Ωkm ] =  ω3 −ω2 ω1 0

ω1 , ω2 , ω3 being arbitrary real numbers. Find the eigenvalues of Ω. P1.32 Let us consider the tensor T defined in P1.24 and let TS be the symmetric part of T. (a) Find the eigenvalues and the eigenvectors of TS ; (b) Give the spectral representation of TS . P1.33 Prove the theorem of spectral representation if λ1 6= λ2 = λ3 and if λ1 = λ 2 = λ 3 . P1.34 Show that one of the eigenvalues of any orthogonal tensor Q is +1 or −1. Show that generally the other two eigenvalues of Q are complex, conjugate number.

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35

1.3. PROBLEMS

P1.35 Let {ek } , k = 1, 2, 3 be an orthonormal basis in V . Show that the 27 tensor products {ek el em } , k, l, m = 1, 2, 3 form a basis in L3 , the vector space of third order tensors. P1.36 Show that the tensor product of three vectors satisfies the following properties, frequently used in our considerations (α1 u1 + α2 u2 ) vw = α1 u1 vw + α2 u2 vw, u(α1 v1 + α2 v2 )w = α1 uv1 w + α2 uv2 w, uv(α1 w1 + α2 w2 ) = α1 uvw1 + α2 uvw2, for any u, u1 , u2 , v, v1 , v2 , w, w1 , w2 from V, for any α1 , α2 from R. P1.37 Find and prove the linearity properties of the tensor product of four vectors, using the results given in P1.33. P1.38 Show that the components Iklmn of the fourth order unit tensor I in any basis {ek el em en } are given by the equations Iklmn = δkm δln . P1.39 Show that if Λ = ΦΨ with Φ, Ψ ∈ L4 then Λklmn = Φklrs Ψrsmn. . P1.40 Show that the transpose ΦT of a fourth order tensor Φ is unique. b 4 is positive definite, it is non-singular P1.41 Show that if the tensor Φ ∈L −1 and its inverse Φ is also positive definitive. P1.42 Let c be a fourth order tensor having the following components: cklmn = λδkl δmn + µ (δkm δln + δkn δlm ) , λ, µ ∈ R, µ (3λ + 2µ) 6= 0. b4 . (a) Show that c ∈ L (b) Show that the fourth order tensor k, having the components kklmn = −

1 λ (δkm δln + δkn δlm ) δkl δmn + 4µ 2µ (3λ + 2µ)

b 4 ; i.e. k = c−1 and kc = ck = bI. is the inverse of c in L (c) In which conditions c is positive definite? P1.43 Let ϕ and ψ be two scalar fields. Prove that: grad (ϕψ) = ϕgrad ψ + ψgrad ϕ. P1.44 Let ϕ and v be a scalar and a vector field, respectively. Prove that: div (ϕv) = v·grad ϕ + ϕ div v.

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36

CHAPTER 1. ELEMENTS OF TENSOR CALCULUS P1.45 Let ϕ and T be a scalar and a tensor field, respectively. Prove that: grad (ϕT) = (grad ϕ) T + ϕ grad T, div (ϕT) = (grad ϕ) · T + ϕ div T. P1.46 Let v and T be a vector and a tensor field, respectively. Prove that: div (Tv) = v · div T + T · grad v. P1.47 Let u and v be two vector fields. Prove that: div (uv) = v div u + u · grad v. P1.48 Let us consider the vector field v = x. Find grad x, div x and ∂x ∂s . P1.49 Let us consider the tensor field T = xx. Find grad xx, div xx and

∂xx ∂s .

P1.50 Let P be a fixed point in the domain D. Let Σ be a closed regular surface, bounding a sub-domain B of D, such that P ∈ B. Let us denote by v the volume of B. Let v and T be a vector and a tensor field, of class C 1 on D. Show that: Z 1 v · n da, div v(P ) = lim v→0,P ∈B v Σ

div T(P ) =

lim

v→0,P ∈B

1 v

Z

v · T da,

Σ

where n is the unit outward to Σ. The above results show again that div v (P ) and div T (P ) have an intrinsic character, depending only on the involved fields, and being independent on any coordinate system.

Bibliography [1.1] Malvern, L.E., Introduction to the mechanics of continuous medium, PrenticeHall, Inc., London, 1969. [1.2] Gurtin, M.E., The linear theory of elasticity, Handbuch der Physics, VIa/2, Ed. C.Truesdell, Springer, Berlin, 1972. [1.3] Gurtin, M.E., An introduction to continuous mechanics, Academic Press, San Diego, 1981. [1.4] So´os, E., Teodosiu, C., Tensor calculus with applications in solid mechanics, Ed. Stiintifica si Enciclopedica, Bucuresti, 1983 (in Romanian).

Copyright © 2004 by Chapman & Hall/CRC

Chapter 2

ELEMENTS OF LINEAR ELASTOSTATICS 2.1

Displacement. Deformation. Stress

We consider a body identified by the domain B which occupies a fixed configuration. A deformation of B is a smooth homeomorphism χ of B into a domain χ (B) in the three-dimensional Euclidean space E with det∇χ > 0. The point χ (x) is the place occupied by the material point x in the deformation χ, while u (x) = χ (x) − x,

(2.1.1)

is the displacement of x. The tensor fields F = ∇χ and H = ∇u,

(2.1.2)

are called, the deformation gradient and the displacement gradient, respectively. From (2.1.1) and (2.1.2), it results: H = ∇u = F − 1,

(2.1.3)

1 being the unit tensor. Many different measures of strain exist. The most useful is the finite strain tensor or Green’s tensor
1 T F F−1 . (2.1.4) G= 2

In the linear theory of importance is the infinitesimal strain tensor ε=

Copyright © 2004 by Chapman & Hall/CRC

  1 1 ∇u + ∇uT . H + HT = 2 2

(2.1.5)

38

CHAPTER 2. ELEMENTS OF LINEAR ELASTOSTATICS From (2.1.4) and (2.1.5), it results 1 G = ε+ ∇uT ∇u. 2

(2.1.6)

The linear or infinitesimal theory models are those cases when the displacement gradient H = ∇u is small. An infinitesimal rigid displacement is defined by the equation: u (x) = u0 +ω 0 ×x, (2.1.7) where u0 and ω 0 are constant vectors. It is easy to see that the infinitesimal strain field corresponding to an infinitesimal rigid displacement field is zero on B. Also we have: Kirchhoff ’s theorem. If two displacement fields, u and u0 correspond to the same infinitesimal strain field, then u = u0 + ω,

(2.1.8)

where ω is an infinitesimal rigid displacement field. A homogeneous displacement field is a displacement field of the form u (x) = u0 +Ax,

(2.1.9)

where u0 is a constant vector and A is a non-singular constant tensor. The infinitesimal strain field corresponding to a homogeneous displacement field is a constant symmetric tensor  1 A + AT . (2.1.10) ε (x) = 2

If u0 = 0 and A = E is symmetric; i.e. if

u (x) = Ex and ET = E,

(2.1.11)

then u is called a pure strain, and obviously ε (x) = E,

(2.1.12)

is a constant tensor field. Given a continuous strain field ε on B, we call the symmetric tensor Z 1 εdv, (2.1.13) E= V B

the mean strain. In the above relation, V is the volume of the domain B occupied by the body in its reference configuration. The following theorem holds: Mean strain theorem. Let u be a displacement field, and let ε = ε (u) be the corresponding strain field, and let us assume that u and ε are continuous on

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39

2.1. DISPLACEMENT. DEFORMATION. STRESS

B. Then the mean strain E depends on the boundary values of u only, and is given by: Z 1 (un + nu) da, (2.1.14) E= 2V ∂B

where ∂B is the boundary of B. Indeed, by the divergence theorem (1.2.25) Z  Z Z  εdv. ∇u + ∇uT dv = 2 (un + nu) da = B

∂B

B

Let σ = σ (x) be the Cauchy’s stress tensor, and let b = b (x) be the body force field. These fields must satisfy in B the Cauchy’s equilibrium equation; i.e. div σ + b = 0.

(2.1.15)

Also, σ is a symmetric tensor field; i.e. σ T = σ.

(2.1.16)

For the latter use, we introduce the following two definitions: A vector field u is an admissible displacement field if u is of class C 2 in B, and u and ∇u are continuous on B. A symmetric tensor field σ is an admissible stress field if σ is of class C 1 on B and σ and div σ are continuous on B. The associated surface force field is the vector field sn defined in each regular point x ∈ ∂B by sn (x) = σ (x) n (x) , (2.1.17)

where n (x) is the outward unit normal to ∂B at x. The following lemma is useful in many problems of linear elastostatics: Lemma. Let σ be an admissible stress field, let u be an admissible displacement field, and let ε (u) be the infinitesimal strain field corresponding to u. Then Z Z Z sn · uda = u · divσdv + σ · ε (u) dv, (2.1.18) ∂B

B

B

sn being the surface force field associated to σ. In order to prove the relation (2.1.18), we must use the integral theorem (1.2.30) and the equations (2.1.5) and (2.1.17). If σ satisfies the equilibrium equation (2.1.15), we have div σ = −b, and the given lemma has the following important consequence: Theorem of work expended. Let σ be an admissible stress field, u an admissible displacement field, ε (u) the corresponding strain field, and let us assume that σ satisfies the equilibrium equation. Then Z Z Z sn · uda + b · udv = σ · ε (u) dv. (2.1.19) ∂B

Copyright © 2004 by Chapman & Hall/CRC

B

B

40

CHAPTER 2. ELEMENTS OF LINEAR ELASTOSTATICS

According to this theorem, the work done by the surface force sn = σn associated to σ and by the body force b over the displacement field u is equal to the work done by the stress field σ over the strain field ε (u) corresponding to u. Given an admissible stress field σ, we call the symmetric tensor Z 1 σdv, (2.1.20) Σ= V B

the mean stress. We have the following: Mean stress theorem. The mean stress corresponding to an admissible stress field that satisfies the equation of equilibrium depends on the associated surface force (or traction) and body force fields only, and is given by  Z Z 1 (xb + bx) dv . (2.1.21) (xsn + sn x) da + Σ= 2V B ∂B

In order to prove the above equation, we use the component form of this relation. Taking into account the relation (2.1.17) and the Gauss-Ostrogradski divergence theorem, we successively get Z Z (xk σlm nm + xl σkm nm ) da (xk snl + snk xl ) da = ∂B ∂B Z n o (xk σlm ),m + (xl σkm ),m dv. = B

Since xk,m = δkm , xl,m = δlm and σlm,m + bl = 0, σkm,m + bk = 0, it results Z Z Z (xk snl + snk xl ) da = (δkm σlm − xk bl ) dv + (δlm σkm − xl bk ) dv. ∂B

B

Thus, we get Z Z 2 σkl dv = B

(xk snl + snk xl ) da + ∂B

B

Z

(xk bl + xl bk ) dv. B

Taking into account now the definition (2.1.20) of the mean stress, we obtain the equation (2.1.21) in its component form and the proof is complete. As we shall see later, the two main value theorems play a fundamental role in the theory of macroscopically homogeneous composite materials. In the mechanics of deformable bodies, a specific material is defined by a constitutive relation or by a material law. Here we will be concerned exclusively with linearly elastic materials. For such materials, the stress at any time and point is a linear function of the (infinitesimal) deformation at the same time and any point. This function may depend on the point under consideration, but it is independent on time.

Copyright © 2004 by Chapman & Hall/CRC

2.1. DISPLACEMENT. DEFORMATION. STRESS

41

Therefore, we say that the body is linearly elastic if for each x ∈ B a linear transformation or a fourth order tensor c (x) from the space of all symmetric tensors to the same space exists such that σ (x) = c (x) ε (x) .

(2.1.22)

We call c (x) the elasticity tensor or Hooke’s tensor for x, and the function c defined on B, with values c (x) , is called the elasticity field. If the mass density of the body and its elasticity field c are independent of x, we say that the body is homogeneous. If the mass density and the elasticity field depend on x, the body will be called non-homogeneous or heterogeneous. Since the space of all symmetric tensors has the dimension 6, the matrix of the components cijkl (x) of c (x), relative to any bases, is 6 × 6. If σij , εkl and cijkl are the components of σ, ε and c in the corresponding bases of the involved tensor spaces, the relation (2.1.22) takes the form σij (x) = cijkl (x) εkl (x) , i, j, k, l = 1, 2, 3,

(2.1.23)

and the following symmetric relations take place: cijkl (x) = cjikl (x) = cijlk (x) .

(2.1.24)

We call the 36 independent quantities cijkl (x) elasticities at x. If the elasticity tensor c (x) is invertible, then its inverse k (x) = c−1 (x) ,

(2.1.25)

is called the compliance tensor and we have ε(x) = k(x)σ(x).

(2.1.26)

According to the general definition, we say that the elasticity field c is symmetric if A · c (x) B = B · c (x) A, (2.1.27) for every point x ∈ B and for every pair of symmetric tensors A and B. As we already know, c is symmetric if, and only if, its components cijkl (x) satisfies the restrictions cijkl (x) = cklij (x) . (2.1.28) We say that a linearly elastic material is hyperelastic if its elasticity tensor c is symmetric. The elasticity tensor c of a hyperelastic material is an element of b 4 and has only 21 independent components. the tensor space L b 4 of According to the general definition, we say that the elasticity field c ∈L a hyperelastic materials is positive definite if A · c (x) A > 0,

Copyright © 2004 by Chapman & Hall/CRC

(2.1.29)

42

CHAPTER 2. ELEMENTS OF LINEAR ELASTOSTATICS

for every point x ∈ B and for every non-zero symmetric tensor A. If c is positive defined, it is non-singular, its inverse k = c−1 exists and is also positive defined. In what follows, we shall consider only hyperelastic materials characterized by positive definite elasticity fields c. According to (2.1.23) and (2.1.28), the components of c, in any bases and in any point x, satisfy the following symmetry relations: cijkl (x) = cjikl (x) = cijlk (x) = cklij (x) .

2.2

(2.1.30)

Symmetry transformations and groups

In order to specify the symmetry properties of the material, we introduce the concept of the symmetry transformation. Let us consider a fixed point x of the linearly elastic body B and let c = c (x) be the elasticity tensor for x. An orthogonal tensor Q is called symmetry transformation at x, if for every symmetric tensor ε, the following relation:
(2.2.1) Q (cε) QT = c QεQT ,

is satisfied. Obviously, the above equation represents a restriction on the elasticity field c. The mechanical meaning of the given definition can be rigourously clarified and justified only in the most general frame-work of the non-linear mechanics. We have also the following: Group property of the symmetry transformations. If Q is a symmetry transformation Q−1 = QT is also a symmetry transformation, and if Q and P are two symmetry transformations, Q P is also a symmetry transformation. In order to prove these properties, we observe that the equation (2.2.1) can be written in the following equivalent form:   cε = QT c QεQT Q. (2.2.2) If Q is a symmetry transformation, since QQT = 1, the relation (2.2.2) applied to the symmetric tensor QT εQ yields 

c QT εQ = QT c Q QT εQ QT Q = QT c (ε) Q.

Consequently, we can conclude from (2.2.2) that QT = Q−1 is a symmetry transformation. Next, if Q and P are two symmetry transformations, then applying (2.2.2) twice yields h i
T c (ε) = PT c PεPT P = PT QT c (QP) ε (QP) QP;

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43

2.2. SYMMETRY TRANSFORMATIONS AND GROUPS

thus QP is a symmetry transformation. Since the unit transformation 1 is obviously a symmetry transformation, the proved property shows that the set of all symmetry transformation form a group. We shall denote by Sx the set of all orthogonal transformations Q that satisfy (2.2.1) and shall call Sx the symmetry group of the material at x. We say that the material at x is isotropic if the symmetry group Sx is the orthogonal group, and it is anisotropic if Sx is a proper subgroup of the orthogonal group. Obviously, Sx always contains the two-element group {1, −1} as a subgroup. In fact, it can be proved that Sx is the direct product of this two element group and a group Sx+ which consists only of proper orthogonal transformations for which det Q = 1. Hence, the symmetry group of the material is completely characterized by the group Sx+ . In particular, the material at x is isotropic or anisotropic according to either Sx+ equals, or is a proper subgroup of the proper orthogonal group. Even if an infinite number of subgroups of the proper orthogonal group exists, twelve of them seems to exhaust the kind of symmetries occurring in the theories proposed up to now and which are appropriate to describe the behavior of real anisotropic elastic materials. The first eleven of these subgroups of the proper orthogonal group correspond to the thirty-two crystal classes. The last type of anisotropy, called transverse isotropy (with respect to a unit direction e) is characterized by the symmetry group consisting of 1 and all the rotations by the angle θ ∈ (0, 2π) about an axis in the direction e. We denote by R(e, θ) a right-handed rotation by the angle θ, 0 < θ < π, about an axis oriented in the direction of the unit vector e. According to its definition, the matrix of the components of R(e, θ), in the tensorial basis e i ej , generated by the orthonormal basis (e1 , e2 , e3 = e) , has the following form:   cos θ − sin θ 0 (2.2.3) [Rij (e,θ)] =  sin θ cos θ 0  . 0 0 1 The transformation R (e) = −R (e,π) is called reflection in the plane P (e) having as normal the unit vector e. If e = e3 , that is if P (e3 ) is spanned by the unit vectors e1 and e2 , the matrix of the components of the reflection R (e3 ) has the following form:   1 0 0 (2.2.4) [Rij (e3 )] =  0 1 0  . 0 0 −1 A unit vector e is called an axis of symmetry (for the material at x) if Qe = e, for some Q ∈ Sx , with Q 6= 1.

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(2.2.5)

44

CHAPTER 2. ELEMENTS OF LINEAR ELASTOSTATICS

It is easy to see that e is an axis of symmetry if and only if one of the symmetry transformations is a rotation about the axis spanned by e; i.e. if and only if R (e,θ) ∈ Sx , for some θ ∈ (0, π) . A plane P (e1 , e2 ), spanned by two mutually orthogonal unit vectors e1 , e2 , is called a plane of symmetry (for the material at x) if Qe1 = e1 and Qe2 = e2 ,

(2.2.6)

for some Q ∈ Sx ,with Q 6= 1. It is easy to see that P (e1 ,e2 )is a plane of symmetry if and only if one of the symmetry transformation is a reflection on the plane P (e1 ,e2 ); i.e. if and only if R (e3 ) ∈ Sx , e1 , e2 , e3 are three mutually orthogonal unit vectors. A material is called triclinic if its symmetry group Sx for any x ∈ B is minimal; i.e. Sx is the two element group (1, −1) . As we already know, a triclinic linearly hyperelastic material has 21 independent elasticities. A material is called monoclinic if for any x ∈ B its symmetry group Sx for any x ∈ B is formed by the transformations +1, ±R (e, π) . Consequently, if the material is monoclinic, the plane P (e) , perpendicular to the unit vector e, is a plane of symmetry for the material at any x ∈ B. Or, equivalently, a monoclinic material has a plane of symmetry, the same for any points of the body. A material is called orthotropic if its symmetry group Sx for any x ∈ B consists of the transformations ±1, ±R (e1 , π) , ±R (e2 , π) , ±R (e3 , π) ,

(2.2.7)

where e1 , e2 , e3 are three mutually orthogonal unit vectors. Consequently, if the material is orthotropic its symmetry group contains reflections with respect to three mutually perpendicular planes P (e1 ) , P (e2 ) , P (e3 ) which are the planes of symmetry of the material at any x ∈ B. Or, equivalently, an orthotropic material has three mutually orthogonal symmetry planes, the same for any point of the body. A material is called transversally isotropic with respect to the direction e, if its symmetry group Sx , for any x ∈ B, consists of the transformations ±1 and ±R (e, θ) with θ ∈ (0, π) . That is, Sx contains all the rotations through the angle θ ∈ (0, 2π) about the axis generated by e and the reflection R (e) with respect to the plane P (e) . Obviously, e is an axis of symmetry and P (e) is a plane of symmetry of the material.

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45

2.2. SYMMETRY TRANSFORMATIONS AND GROUPS

In order to express the restrictions due to various material symmetries, we shall use the Voigt’s convention, by which the tensorial indices are replaced by matrix indices, according to the following rule: tensorial indices matrix indices

11 1

22 2

33 3

23 and 32 4

31 and 13 5

12 and 21 6 .

Accordingly, the tensor components σij and εij of the stress σ and strain ε, are denoted in the following way: σ11 = σ1 , σ22 = σ2 , σ33 = σ3 , σ23 = σ4 , σ31 = σ5 , σ12 = σ6 ,

(2.2.8)

ε11 = ε1 , ε22 = ε2 , ε33 = ε3 , 2ε23 = ε4 , 2ε31 = ε5 , 2ε12 = ε6 .

(2.2.9)

The tensorial components cijkl , i, j, k, l = 1, 2, 3 of the elasticity tensor c will be replaced by matrix components Cij , i, j = 1, 2, ..., 6, according to the Voigt’s convention. For instance, we shall have: c1111 c1122 c1123 c1223

= C11 , = c2211 = C12 = C21 , = c1132 = c2311 = c3211 = C14 = C41 , = c1232 = c2123 = c2132 = c2312 = c3212 = c2321 = c3221 = C64 = C46 . (2.2.10) We recall that the tensorial indices take the values 1, 2, 3, whereas the matrix indices take the values 1, 2, ..., 6. Since the material is hyperelastic, we have Cij = Cji , i, j = 1, 2, ..., 6.

(2.2.11)

Now, it is easy to see that the stress-strain relation σ = cε or σij = cijkl εkl , i, j, k, l = 1, 2, 3, can be written in the following equivalent forms: σi = Cij εj , i, j = 1, 2, ..., 6,        

σ1 σ2 σ3 σ4 σ5 σ6





      =      

C11 C21 C31 C41 C51 C61

C12 C22 C32 C42 C52 C62

C13 C23 C33 C43 C53 C63

C14 C24 C34 C44 C54 C64

C15 C25 C35 C45 C55 C65

[σ] = [C] [ε] .

C16 C26 C36 C46 C56 C66

(2.2.12)        

ε1 ε2 ε3 ε4 ε5 ε6



   ,   

(2.2.13)

(2.2.14)

The symmetric 6 × 6 matrix [C] is called the stiffness matrix. Let us establish now the restrictions imposed on the elasticities by the symmetry properties of the body, in the case of monoclinic, orthotropic, transversally isotropic and isotropic materials.

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CHAPTER 2. ELEMENTS OF LINEAR ELASTOSTATICS

Monoclinic materials. Let us assume that P (e3 ) is a plane of symmetry of the material. In this case R (e3 ) = −R (e3 , π) ∈ Sx for any x ∈ B. Consequently, the relation (2.2.2) must be satisfied for Q = R (e3 ). That is, in an orthonormal basis (e1 , e2 , e3 ) the restrictions: cijkl = Qim Qjn Qkp Qlq cmnpq ,

(2.2.15)

must be satisfied, where, according to (2.2.4), Q11 = Q22 = 1, Q33 = −1 and Qij = 0 if, i 6= j.

(2.2.16)

For instance, we must have c1123 = C14 = Q1m Q1n Q2p Q3q cmnpq . Using (2.2.16), it is easy to see that the above condition reduces to c1123 = C14 = −c1123 = −C14 . Accordingly, we must have C14 = 0. In a similar elementary way, it results that the restriction (2.2.15) can be satisfied if and only if C14 = C24 = C34 = C64 = C15 = C25 = C35 = C65 = 0.

(2.2.17)

Consequently, the constitutive equation of a monoclinic material can be written in the following matrix form:      σ1 C11 C12 C13 0 0 C16 ε1  σ2   C12 C22 C23   0 0 C26       ε2   σ3   C13 C23 C33    0 0 C ε 36   3     (2.2.18)  σ4  =  0   . 0 0 C C 0 ε 44 45     4   σ5   0 0 0 C45 C55 0   ε5  σ6 C16 C26 C36 0 0 C66 ε6

According to the above result, a monoclinic material has 13 independent elasticities. Orthotropic materials. Let us assume that P (e1 ) , P (e2 ) , P (e3 ) are three mutually perpendicular symmetry planes of the material. Since P (e3 ) is a plane of symmetry, the restrictions (2.2.17) must be satisfied. Since P (e2 ) is also a plane of symmetry, it results that the following supplementary restrictions must be satisfied: C14 = C24 = C34 = C54 = C16 = C26 = C36 = C56 = 0.

Copyright © 2004 by Chapman & Hall/CRC

(2.2.19)

2.2. SYMMETRY TRANSFORMATIONS AND GROUPS

47

Finally, we take into account that P (e1 ) is also a symmetry plane. Direct examination shows that this fact does not impose supplementary restrictions on the elasticities. In other words, if the material has two mutually perpendicular symmetry planes, the plane perpendicular to them is also a plane of symmetry. From (2.2.17) and (2.2.19), we can conclude that the constitutive equation of an orthotropic material can be written in the following matrix form:      ε1 σ1 C11 C12 C13 0 0 0   σ2   C12 C22 C23  0 0 0     ε2     σ3   C13 C23 C33   ε3  0 0 0   .     σ4  =  0  0 0 C44 0 0     ε4     σ5   0 0 0 0 C55 0   ε5  0 0 0 0 0 C66 ε6 σ6

Accordingly, an orthotropic material has 9 independent elasticities. We stress the fact that the constitutive equation of a monoclinic or orthotropic material can be expressed in the simple forms given by the above and (2.2.18) matrix relations, only if the elasticities cijkl correspond to the tensor basis ei ej ek el generated by that orthonormal basis e1 , e2 , e3 which correspond to the symmetry planes P (e1 ) , P (e2 ) , P (e3 ) . Generally, the elasticities corresponding to a tensor basis e0i e0j e0k e0l generated by an arbitrary orthonormal basis e01 , e02 , e03 , are all non-vanishing. Transversally isotropic materials. Let us assume now that the symmetry group of the material consists by ±1 and ±R (e3 , θ) for any θ ∈ (0, 2π) . In the tensor basis ei ej generated by the orthonormal basis (e1 , e2 , e3 ) , the components of R (e3 , θ) are given by the relation (2.2.3). Long, but elementary algebra shows that in the case of transversally isotropic materials the restrictions (2.2.2) take the following form: C11 = m4 C11 + 2m2 n2 C12 + n4 C22 − 4m3 nC16 − 4mn3 C26 + 4m2 n2 C66 , (2.2.20) C22 = n4 C11 + 2n2 m2 C12 + m4 C22 + 4n3 mC16 + 4nm3 C26 + 4n2 m2 C66 , (2.2.21) C12 = m2 n2 C11 + (m4 + n4 )C12 + m2 n2 C22 + 2mn(m2 − n2 )(C16 − C26 ) −4m2 n2 C66 ,

(2.2.22)


C66 = m2 n2 C11 − 2m2 n2 C12 + m2 n2 C22 + 2mn m2 − n2 (C16 − C26 )
2 (2.2.23) + m2 − n2 C66 ,

C16 = m3 nC11 − mn m2 − n2 C12 − mn3 C22 + m2 m2 − n2 C16

(2.2.24) +n2 3m2 − n2 C26 − 2mn m2 − n2 C66 ,

Copyright © 2004 by Chapman & Hall/CRC

48

CHAPTER 2. ELEMENTS OF LINEAR ELASTOSTATICS
C26 = n3 mC11 + nm m2 − n2 C12 − m3 nC22

+n2 3m2 − n2 C16 + m2 m2 − n2 C26 ,

(2.2.25)

C13 = m2 C13 + n2 C23 − 3mnC36 ,

(2.2.26)

C23 = n C13 + m C23 + 3mnC36 ,
C36 = mn (C13 − C23 ) + m2 − n2 C36 ,

(2.2.27)

2

2

(2.2.28)

C44 = m2 C44 + n2 C55 + 2mnC45 ,

(2.2.29)

C55 = n2 C44 + m2 C55 − 2nmC45 ,
C45 = mn (C55 − C44 ) + m2 − n2 C45 ,

(2.2.30) (2.2.31)

C34 = mC34 + nC35 ,

(2.2.32)

C35 = −nC34 + mC35 ,

(2.2.33)

C14 = m3 C14 + m2 nC15 + mn2 C24 + n3 C25 − 2m2 nC46 − 2mn2 C56 ,

(2.2.34)

C24 = mn2 C14 + n3 C15 + m3 C24 + m2 nC25 + 2m2 nC46 + 2mn2 C56 ,

(2.2.36)

C15 = −m2 nC14 + m3 C15 − n3 C24 + mn2 C25 + 2mn2 C46 − 2m2 nC56 , (2.2.35) C25 = −m3 C14 + mn2 C15 − m2 nC24 + m3 C25 − 2mn2 C46 + 2m2 nC56 , (2.2.37)
C46 = m2 nC14 + mn2 C15 − m2 nC24 − mn2 C25 + n2 − m2 (mC46 − nC56 ) , (2.2.38)
C56 = −mn2 C14 + m2 nC15 + mn2 C24 − m2 nC25 + n2 − m2 (nC46 − mC56 ) . (2.2.39) In these relations, we use the notation m = cos θ, n = sin θ.

(2.2.40)

Adding (2.2.24) and (2.2.25), and subtracting (2.2.20) from (2.2.31), we get
C16 + C26 = (C11 − C22 ) sin θ cos θ + (C16 + C26 ) cos4 θ − sin4 θ ,
C11 − C22 = (C11 − C22 ) cos4 θ − sin4 θ − 4 (C16 + C26 ) sin θ cos θ.

From these equations, it results

(C11 − C22 ) sin2 θ = 0, (C16 + C26 ) sin2 θ = 0. Since for θ ∈ (0, 2π) , and generally sin θ 6= 0, the above restrictions can be satisfied for any θ ∈ (0, 2π) if and only if C11 = C22 , C16 = −C26 .

Copyright © 2004 by Chapman & Hall/CRC

(2.2.41)

49

2.2. SYMMETRY TRANSFORMATIONS AND GROUPS Introducing (2.2.4) in any of the relations (2.2.20)-(2.2.22), we obtain
(C11 − C12 − 2C66 ) sin2 θ cos2 θ + 2C16 sin θ cos θ cos2 θ − sin2 θ = 0. Introducing (2.2.4) in any of the relations (2.2.24), (2.2.25), we get


−8C16 sin2 θ cos2 θ + (C11 − C12 − 2C66 ) sin θ cos θ cos2 θ − sin2 θ = 0.

It is easy to see that the last two restrictions can be satisfied for any θ ∈ (0, 2π) if and only if 1 (2.2.42) C66 = (C11 − C22 ) , C16 = 0. 2 A simple examination shows that the restrictions (2.2.26)-(2.2.28) can be satisfied for any θ ∈ (0, 2π) if and only if

C13 = C23 , C36 = 0.

(2.2.43)

Analogously, the restrictions (2.2.29)-(2.2.31) can be satisfied for any θ ∈ (0, 2π) if and only if C44 = C55 , C45 = 0. (2.2.44) The restrictions (2.2.32) and (2.2.33) will be satisfied for any θ ∈ (0, 2π) if and only if C34 = C35 = 0. (2.2.45) Adding (2.2.34), (2.2.36) and (2.2.35), (2.2.37), respectively, we obtain C14 + C24 = (C14 + C24 ) cos θ + (C15 + C25 ) sin θ, C15 + C25 = − (C14 + C24 ) sin θ + (C15 + C25 ) cos θ. These restrictions will be satisfied for any θ ∈ (0, 2π) if and only if C14 + C24 = 0, C15 + C25 = 0.

(2.2.46)

Using these restrictions, from (2.2.24), (2.2.39) and (2.2.35), (2.2.37), we get C14 − C56 = (C14 − C56 ) cos θ − (C15 + C46 ) sin θ, C15 + C46 = (C14 − C56 ) sin θ + (C15 + C46 ) cos θ. These restrictions can be satisfied for any θ ∈ (0, 2π) if and only if C14 = C56 , C15 = −C46 .

(2.2.47)

Taking into account the restrictions (2.2.46) and (2.2.47), from (2.2.38) and (2.2.39), we obtain the conditions

C46 1 − cos3 θ + 3 sin2 θ cos θ − C56 3 cos2 θ sin θ − sin3 θ = 0,

Copyright © 2004 by Chapman & Hall/CRC

50

CHAPTER 2. ELEMENTS OF LINEAR ELASTOSTATICS

C46 3 cos2 θ sin θ − sin3 θ + C56 1 − cos3 θ + 3 sin2 θ cos θ = 0.

The above relations can be satisfied for any θ ∈ (0, 2π) if and only if C46 = C56 = 0.

(2.2.48)

Examining the conditions (2.2.41)-(2.2.48), we can conclude that if the material is transversally isotropic, the elasticities must satisfy the following restrictions: C16 = C26 = C36 = C46 = C56 = C45 = C14 = C24 = C15 = C25 = C34 = C35 = 0, 1 (C11 − C12 ) . (2.2.49) 2 Consequently, by direct examination it can be seen that if the restrictions (2.2.49) are satisfied, the relations (2.2.20)-(2.2.39) hold for any θ ∈ (0, 2π) . According to (2.2.49), the constitutive equation of a transversally isotropic material can be written in the following matrix form:      C11 C12 C13 0 0 0 ε1 σ1   ε2   σ2   C12 C11 C13 0 0 0        ε3   σ3   C13 C13 C33 0 0 0    = (2.2.50)   ε4  .  σ4   0 0 0 C44 0 0        ε5   σ5   0 0 0 0 C44 0 (C11 −C12 ) ε6 σ6 0 0 0 0 0 2 C11 = C22 , C13 = C23 , C44 = C55 , C66 =

Thus, a transversally isotropic material has 5 independent elasticities. Isotropic materials. In this case, the symmetry group Sx for any x ∈ B is the full orthogonal group. Particularly, for any θ ∈ (0, 2π) , R (e3 , θ) and R (e2 , θ) are elements of the symmetry group. The restrictions imposed by R (e3 , θ) were just established. Taking into account that R (e2 , θ) ∈ Sx , in a similar way we get the following additional restrictions: C11 = C33 , C12 = C13 , C44 =

1 (C11 − C12 ) . 2

(2.2.51)

Consequently, the constitutive equation of an isotropic material can be written in the following matrix form:      C11 C12 C12 0 0 0 ε1 σ1  C C11 C12 0 0 0  σ2     ε2      12  C C C 0 0 0  σ3   12 12 11   ε3  .  = (C −C )  11 12  σ4   0 0 0 0 0   ε4  2      (C11 −C12 )  σ5   0 0 0 0   ε5   0 2 (C11 −C12 ) ε6 σ6 0 0 0 0 0 2 (2.2.52) Accordingly, an isotropic material has 2 independent elasticities.

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2.2. SYMMETRY TRANSFORMATIONS AND GROUPS

51

In the theory of linearly elastic isotropic materials, the following notations are used: C11 = λ + 2µ, C12 = λ, (2.2.53) λ and µ being the Lam´e’s coefficients; µ is called also shear modulus and frequently is denoted by G. It is easy to see that the stress-strain relation for isotropic materials can be written in the following tensorial form: σ = cε = λ (trε) 1+2µε.

(2.2.54)

In the case of a homogeneous isotropic material, λ and µ are constant quantities. If the material is heterogeneous, λ and µ depend on x ∈ B. We denote by

1 1 e = ε− (trε) 1 and s = σ− (trσ) 1, 3 3

(2.2.55)

the deviatoric part of the strain ε and of the stress σ, respectively. From (2.2.54), we get trσ = 3ktrε, k = λ + 2µ/3, s = 2µe. (2.2.56) Since trε characterizes the volumetric changes of an elastic solid, the coefficient k is called bulk modulus. If µ 6= 0 and 3λ + 2µ 6= 0, the stress-strain relation (2.2.54) can be inverted, and   λ 1 (trσ) 1 . (2.2.57) σ− ε = kσ = 3λ + 2µ 2µ

The Young’s modulus E and the Poisson’s ratio ν are defined by E=

λ µ (3λ + 2µ) , , ν= 2 (λ + µ) λ+µ

(2.2.58)

assuming (λ + µ) 6= 0. Thus, the strain-stress relation (2.2.58) becomes ε=

ν 1+ν σ− (trσ) 1. E E

(2.2.59)

From (2.2.59), we can see that E is determined in a tensile test, by dividing the axial stress by the corresponding axial strain, and ν is the ratio of the lateral contraction towards the axial strain of the bar subjected to traction. Since an elastic bar should increase its length when pulled, and should contract in the direction perpendicular to its axis, and since an isotropic elastic body should decreases its volume when acted upon by a hydrostatic pressure, we have E > 0, ν > 0, k > 0.

Copyright © 2004 by Chapman & Hall/CRC

(2.2.60)

52

CHAPTER 2. ELEMENTS OF LINEAR ELASTOSTATICS If 1 − 2ν 6= 0, the relation (2.2.58) can be inverted, and we get λ=

E Eν . , µ=G= 2(1 + ν) (1 + ν) (1 − 2ν)

(2.2.61)

In this way from (2.2.56), we obtain the bulk modulus k expressed in terms of the Young’s modulus E and the Poisson’s ratio ν k=

E . 3(1 − 2ν)

(2.2.62)

Now, from(2.2.60)−(2.2.62), we can conclude that λ > 0, µ > 0, ν <

1 . 2

(2.2.63)

Also, taking into account the isotropic constitutive equation (2.2.54) and the decompositions (2.2.56), (2.2.57), we obtain ε · cε = k(trε)2 + 2µe · e.

(2.2.64)

Since according to the experimental evidence, the bulk modulus k and the shear modulus µ of an isotropic material are always positive, from the equation (2.2.64) we can conclude, that the elasticity field c of an isotropic material is positive definite. Generalizing this fact, we have assumed that the elasticity field c of any elastic material is positive definite. Starting with this assumption, from (2.2.64) we can conclude that the elasticity field of an isotropic material is positive definite if and only if µ > 0 and k > 0.

(2.2.65)

Since c is positive definite, the stress-strain relation can be invented and k = c−1 is also positive definite and we have ε = kσ.

(2.2.66)

ε · cε = [ε][C][ε],

(2.2.67)

Obviously, since the stiffness matrix [C] is also a positive definite, invertible, and symmetric 6 × 6 matrix. Consequently, the inverse matrix [S] = [C]

−1

,

(2.2.68)

named the compliance matrix exists, is positive definite and symmetric. Thus, we can inverse the matrix form (2.2.13) of the constitutive equation getting [ε] = [S][σ].

Copyright © 2004 by Chapman & Hall/CRC

(2.2.69)

2.2. SYMMETRY TRANSFORMATIONS AND GROUPS

tain

form

Particulary, taking into account    S11 S12 S13 ε1  ε2   S12 S22 S23     ε3   S13 S23 S33     ε4  =  0 0 0     ε5   0 0 0 0 0 0 ε6

53

(2.2.19), for an orthotropic material, we ob  0 0 0 σ1   0 0 0    σ2   σ3  0 0 0  .  (2.2.70)   S44 0 0    σ4  0 S55 0   σ5  σ6 0 0 S66

Usually, the compliance matrix [S] is expressed in the following, more useful   1 0 0 − νE212 − νE313 0 E1 1   − ν12 0 0 − νE323 0   E1 E2   − ν13 − ν23 1 0 0 0   E1 E3 E2 (2.2.71) [S] =  . 1 0 0 0 0   0 G23   1 0 0 0 0   0 G31 1 0 0 0 0 0 G12

Here E1 , E2 and E3 are Young’s moduli in 1, 2, 3 directions, respectively; νij are Poisson’s ratio for the transverse strain in the j-direction when stressed in the i-direction, that is νij = −εj /εi for σi = σ and all other stresses are zero.

(2.2.72)

Finally, G23 , G31 and G12 are shear moduli in the 2-3, 3-1 and 1-2 planes, respectively. Since the compliance matrix is symmetric, i.e Sij = Sji , the engineering or technical constants must satisfy the supplementary conditions:

νji νij , i, j = 1, 2, 3. = Ej Ei

(2.2.73)

Since the compliance matrix is positive definite, according to the Sylvester’s criterion, its components must satisfy the following restrictions: E1 , E2 , E3 , G23 , G31 , G12 > 0,

(2.2.74)

ν12 ν21 < 1, ν23 ν32 < 1, ν13 ν31 < 1,

(2.2.75)

νe = 1 − ν12 ν21 − ν23 ν32 − ν31 ν13 − ν12 ν23 ν31 − ν21 ν13 ν32 > 0.

(2.2.76)

Using the symmetry relation (2.2.73), we can express the restrictions (2.2.75) in the following equivalent forms: p p p (2.2.77) |ν12 | < E1 /E2 , |ν23 | < E2 /E3 , |ν31 | < E3 /E1 ,

or

|ν21 | <

p

Copyright © 2004 by Chapman & Hall/CRC

E2 /E1 , |ν32 | <

p

E3 /E2 , |ν13 | <

p

E1 /E3 .

(2.2.78)

54

CHAPTER 2. ELEMENTS OF LINEAR ELASTOSTATICS Similarly, (2.2.76) can be expressed as   1 2 E3 2 E2 2 E1 , − ν13 − ν32 1 − ν21 ν21 ν13 ν32 < E1 E3 E2 2

(2.2.79)

From the above restriction, we can conclude that ν21 ν13 ν32 < 1/2.

(2.2.80)

The Sylvester’s criterion tells us that if the restrictions (2.2.74)−(2.2.76) are satisfied, k and [S], and consequently c and [C] are positive definite, and the converse is also true. Other, independent restriction of the engineering constants cannot be obtained using the assumed positive definiteness of the elasticity field c of an orthotropic material. The obtained restrictions are used to examine experimental data, to see if they are physically consistent within the framework of the usual linearly elastic model based on positive definite elasticity. For example, in testing boron/epoxy composite materials, Dickerson and Di Martino reported Poisson’s ratio as high as 1.97 in the 2 direction due to loading in the 1 -direction (ν12 ). The reported values for thep Young’s moduli for the two direction are Ep 1 = 1.72kP a, E2 = 0.19kP a. Thus, E1 /E2 = 2.99 and the restriction |ν12 | < E1 /E2 is satisfied. Accordingly, ν12 = 1.97 is a reasonable value, even if our intuition based on isotropic materials (0 < ν < 1/2) rejects such large number. Also, the “converse” Poisson’s ratio ν21 was reported as 0.22. This value satisfies the symmetry condition or reciprocal relation (2.2.73). Should the measured material properties satisfy the constraints, we can proceed with confidence to design structures with that material. Otherwise, we have a reason to doubt either the material model or the experimental data or both. As we have seen, even if we assume that the elasticity c of an orthotropic material is positive definite, we cannot deduce that Poisson’s ratios ν ij are positive. However, by assuming that the traction tensile stress acting in the direction of the axis of symmetry of an orthotropic material, produces elongation in that direction but contraction in the directions of corresponding to the other two axis of symmetry, we can conclude that

νij > 0, i, j = 1, 2, 3, i 6= j.

(2.2.81)

If, in addition, we assume that in the above mentioned experiment the volume of the body increases, that is ε1 + ε 2 + ε 3 =

1 (1 − ν12 − ν13 ) σ1 > 0 if only σ1 > 0 is non-vanishing, E1

1 (1 − ν21 − ν23 ) σ2 > 0 if only σ2 > 0 is non-vanishing, (2.2.82) E2 1 (1 − ν32 − ν31 ) σ3 > 0 if only σ3 > 0 is non-vanishing, ε1 + ε 2 + ε 3 = E3 ε1 + ε 2 + ε 3 =

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2.2. SYMMETRY TRANSFORMATIONS AND GROUPS

55

we can conclude that Poisson’s ratios satisfy the following additional restrictions: ν12 + ν13 < 1, ν21 + ν23 < 1, ν32 + ν31 < 1.

(2.2.83)

In this case, from (2.2.81) and (2.2.83), it results 0 < νij < 1, i, j = 1, 2, 3, i 6= j.

(2.2.84)

As the data reported by Dickerson and Di Martino show, the last restrictions are not generally fulfilled for an orthotropic composite material. Consequently, (2.2.82) generally do not take place and the volume of an orthotropic material may decrease during the experiment above considered! Obviously such situation can occur if the differences between Young’s moduli, corresponding to the three symmetry directions, are relatively large. In the case of fiber-reinforced composite materials, this is just the case! By elementary computations, from (2.2.71) we obtain the expressions of the non-vanishing independents components of the stiffness matrix [C] of an orthotropic material, as function of its engineering constants C11 =

ν12+ ν32 ν13 ν21+ ν31 ν23 1 − ν23 ν32 , = , C12 = E1 E3 ∆ E2 E3 ∆ E2 E3 ∆

C13 =

ν13+ ν12 ν23 ν31+ ν21 ν32 , = E1 E2 ∆ E2 E3 ∆

C22 =

ν + ν21 ν13 ν32+ ν12 ν31 1 − ν13 ν31 , = 23 , C23 = E1 E2 ∆ E1 E3 ∆ E1 E3 ∆

C33 =

1 − ν12 ν21 , E1 E2 ∆

(2.2.85)

C44 = G23 , C55 = G13 , C66 = G12 , where ∆=

νe 1 − ν12 ν21 − ν23 ν32 − ν31 ν13 − ν21 ν32 ν13 − ν12 ν23 ν31 . (2.2.86) = E1 E2 E3 E1 E2 E3

If the material is transversally isotropic, the stiffness matrix has the form given by the relation (2.2.50). Consequently, the compliance matrix of the material has the same structure, and using the engineering or technical coefficients, we get   1 0 0 − νE121 − νE131 0 E1 1   − ν12 0 0 − νE131 0   E1 E1 1   − ν13 − ν13 0 0 0   E1 E3 E1 (2.2.87) [S] =  . 1 0 0 0 0   0 G13   1 0 0 0 0   0 G13 2(1+ν12 ) 0 0 0 0 0 E1

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56

CHAPTER 2. ELEMENTS OF LINEAR ELASTOSTATICS The independent and non-vanishing components of the stiffness matrix be-

came C11 =

1 − ν2 ν13 (1 + ν12 ) ν12 + ν13 ν31 1 − ν13 ν31 , C33 = − 2 12 , , C13 = , C12 = 2 2 E1 ∆ E1 ∆ E1 ∆ E1 E3 ∆

C44 = G13 , C55 = G13 , C66 =

where ∆=

1 − ν12 − 2ν13 ν31 , 2E1 E3 ∆

(1 + ν12 )(1 − ν12 − 2ν13 ν31 ) . E3 E12

(2.2.88)

(2.2.89)

Finally, if the material is isotropic, we have S11 = S22 = S33 =

ν 1 , S12 = S23 = S31 = − , E E

S44 = S55 = S66 =

2(1 + ν) 1 1 , = = E G µ

(2.2.90)

and

νE (1 − ν)E , , C12 = C23 = C31 = (1 + ν)(1 − 2ν) (1 + ν)(1 − 2ν)

C11 = C22 = C33 =

C44 = C55 = C66 = µ = G =

E . 2(1 + ν)

(2.2.91)

In concluding this Section, we observe that if the material is isotropic its Young’s modules E and its shear modulus G has the same order of magnitude since 0 < ν < 1/2. On the contrary, the order of magnitude of Young’s moduli and of shear module of an orthotropic material can be quite different. In the case of fiber-reinforced materials this is just the case, and as we shall see, this fact will have important consequences concerning the behavior of such kind of structured composite bodies.

2.3

The fundamental system of field equations

The fundamental system of the field equations for the time independent behavior of a linear elastic body consists of the strain-displacement relation ε=

1 1 (∇u + ∇uT ) or εij = (ui,j + uj,i ), 2 2

(2.3.1)

the stress-strain relation σ = cε or σij = cijkl εkl ,

Copyright © 2004 by Chapman & Hall/CRC

(2.3.2)

2.3. THE FUNDAMENTAL SYSTEM OF FIELD EQUATIONS

57

the equation of equilibrium div σ+b = 0 or σij,j + bi = 0.

(2.3.3)

Here u, ε, σ and b are the displacement, strain, stress and body force, respectively, while c is the symmetric positive definite elasticity field, assumed to be continuous on B. Since σ = cε = c∇u, when the displacement field is sufficiently smooth, the above relation imply the displacement equation of equilibrium div c∇u + b = 0 or (cijkl uk,l ),j +bi = 0.

(2.3.4)

Conversely, if u satisfies the displacement equation of equilibrium, and if ε and σ are defined by the strain-displacement and stress strain relations, respectively, then the stress equation of equilibrium is satisfied. We define the strain-energy U = U (ε) corresponding to a continuous strainfield ε or B Z Z 1 1 εij cijkl εkl dv. (2.3.5) ε · cεdv = U (ε) = 2 B 2 B The quantity 1 1 (2.3.6) u = u(ε) = ε · cε = εij cijkl εkl 2 2 will be named specific strain energy. Since the elasticity field c is positive definite, the specific strain energy u = u(ε) is a positive definite quadratic form. Also, we have the following: Lemma. Let ε and e ε be continuous symmetric tensor fields on B. Then Z U (ε+e ε) = U (ε) + U (e ε) + ε · ce εdv. (2.3.7) B

Indeed, since c is symmetric ε · ce ε=e ε · cε. Thus

(ε+e ε) · c (ε+e ε) = ε · cε + e ε · ce ε + 2ε · ce ε,

which implies the relation (2.3.7). In what follows by an admissible state, we mean an ordered array s = [u, ε, σ] having the following properties: u is an admissible displacement field, ε is a continuous symmetric tensor field on B and σ is an admissible stress field. We observe that the fields u, ε, and σ need not to be related. We say that s = [u, ε, σ] is an elastic state on B, corresponding to the body force field b, is an admissible state, and

ε=


1 ∇u+∇uT , σ = cε, div σ + b = 0. 2

(2.3.8)

The corresponding surface traction sn is defined in any regular point x of ∂B by sn (x) = σ (x) n (x) , (2.3.9)

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CHAPTER 2. ELEMENTS OF LINEAR ELASTOSTATICS

n (x) being the outward unit normal to the boundary ∂B at x. The pair [b, sn ] will be called external force system for s. The linearity of the fundamental field equations implies the validity of the : e Principle of superpositions for elastic states. If [u, ε, σ] and h[e u, e ε, σ i] e are elastic states corresponding to the external force systems [b, s n ] and b, e sn , ei ] respectively, and if α and α e are arbitrary real numbers, then α [u, ε, σ]+ α e [e ε, σ hu, e e e is an elastic state corresponding to the external force system α [b, s n ] + α e b, sn . Here i h e = [αu+e e, e e , α ε+ α e] , ε, b αu ee ε, α σ+ α eσ α [u, ε, σ] + α e u h i h i e e e αsn + α α [b, sn ] + α e b, sn = αb+e αb, ee sn .

The direct consequence of the theorem of work expended,that is of the equation (2.1.19), is the Theorem of work and energy. Let [u, ε, σ]be an elastic state corresponding to the external force system [b, sn ] . Then Z Z sn · udv + b · udv = 2U (ε). (2.3.10) ∂B

B

The quantity on the left-hand side of this equation is the work done by the external forces; our theorem states that this work is equal to twice the strain energy. The positive definiteness of the elasticity field c implies the validity of the: Theorem of positive work. For any elastic state s = [u,ε, σ], corresponding to the external force system [b, sn ], the work done by [b, sn ] is non-negative and vanishes only when the displacement field u is rigid. Indeed, since c is positive definite, U (ε) ≥ 0, and according to (2.3.10), the work done by the external forces is non-negative. If U (ε) = 0, since ε · cε ≥ 0, ε · cε must vanish on B, since it is a continuous function; i.e. ε · cε = 0 on B. Using again the positive definiteness of c, we can conclude that ε must vanish on B; i.e. ε = 0 on B. Consequently, according to Kirchhoff’s theorem u must be an infinitesimal rigid displacement. Conversely, if u is a rigid displacement, the strain corresponding to u is vanishing on B. Consequently the corresponding strain energy is also vanishing. Hence, according to equation (2.3.10), the work done by the external force system is zero. A direct consequence of the above theorem is the following: Lemma. Let [u, ε, σ] be an elastic state corresponding to vanishing body force. We also suppose that the surface traction is also vanishing. Then ε = σ = 0 and u is a rigid displacement field. The following theorem expresses the fact that the fundamental system of field equations of the linear elastostatics is self-adjoint.

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2.3. THE FUNDAMENTAL SYSTEM OF FIELD EQUATIONS

59

e ] be elastic states Betti’s reciprocal theorem. Let [u, ε, σ] and h[e u, e ε,iσ e corresponding to the external force systems [b, sn ] and b, e sn , respectively. Then Z

∂B

e da + sn · u

Z

b·e udv

=

B

=

Z

2

∂B

Z

B

e sn · uda +

Z

σ·e εdv = 2

B

Z

e · udv b

B

e · εdv. σ

(2.3.11)

Indeed, since c is symmetric, from the stress-strain relation, we can conclude that e · ε. σ·e ε=e ε · cε = ε · ce ε=σ

From the theorem of work expected it follows that Z Z Z e dv = e da + σ·e εdv, b· u sn · u Z

∂B

B

B

∂B

and

(2.3.12)

e sn · uda +

Z

B

e · udv = b

Z

B

e · εdv, σ

and the proof is complete. Betti’s theorem asserts that given two elastic states, the work done by the external forces of the first state over the displacements of the second, equals the work done by the external forces of the second over the displacement of the first. We suppose now that it is given an elastic field c on B, a body force field b b on S1 and surface forces b on B, surface displacements u s on S2 , where S1 and S2 are complementary regular subsurface of ∂B. Given the above data, the mixed problem of the elasto-statics is to find an elastic state [u, ε, σ] that corresponds to b and satisfies the displacement condition

and the traction condition

b on S1 , u=u

(2.3.13)

sn = σn = b s on S2 .

(2.3.14)

b on ∂B, u=u

(2.3.15)

sn = σn = b s on ∂B,

(2.3.16)

We call such an elastic state a solution of the mixed problem. When S2 is empty, the above boundary condition reduces to

and the associated problem is called the displacement problem or the first boundary value problem. If S1 is empty, the boundary conditions becomes

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CHAPTER 2. ELEMENTS OF LINEAR ELASTOSTATICS

and the corresponding problem is called the traction problem or the second boundary value problem. For the moment, we assume that c is of class C 1 on B, b is continuous on b is continuous on S1 and b B, u s is piece-wise regular on S2 . By a displacement field corresponding to a solution of the mixed problem, we mean a vector field u with the property that there exist fields ε, σ such that [u, ε, σ] is a solution of the mixed problems. Within the assumed regularity conditions, we have the following: Characterization of the mixed problem in terms of displacements. Let u be an admissible displacement field. Then u corresponds to a solution of the mixed problems if and only if

b on S1 , (c∇u)n = b div c∇u + b = 0 on B, u = u s on S2 .

(2.3.17)

The above relations are necessary and it follows from the equilibrium equation (2.3.3) satisfied by the stress σ, from the stress-strain relation σ = cε and from T . To prove sufficiency, we assume the strain-displacement relation ε = ∇u+∇u 2 that u satisfies the relation (2.3.17). We define ε through the strain-displacement relation and σ through the stress-strain relation. Then, since u is admissible, ε is continuous on B, and σ is continuous on B and smooth, or of class C 1 on B. In addition, div σ + b = 0 on B and sn = σn = b s on S2 . From the equilibrium equation, it results that div σ is continuous on B. That is, σ is an admissible stress field. Therefore, [u,ε, σ] meets all requirements of a solution of mixed problems. In order to discuss the uniqueness questions appropriate to the fundamental boundary value problems of elasto-statics, we shall say that two solutions [u,ε, σ] e ] of the mixed problem are equal modulo a rigid displacement if and [e u,e ε, σ

e ], [u,ε, σ] = [e u + w,e ε, σ

where w is a rigid displacement field. We have: Kirchhoff ’s uniqueness theorem for the mixed problem. Any two solution of the mixed problem are equal modulo a rigid displacement. If S 1 is nonempty, the mixed problem has at the most one solution. e ] be two solution of the mixed problem, Indeed, let s = [u,ε, σ] and se = [e u, e ε, σ and let as take e ]. [u0 , ε0 , σ 0 ] = [u,ε, σ] − [e u, e ε, σ

Then, by the principle of superposition s0 = [u0 , ε0 , σ 0 ] is an elastic state corresponding to vanishing body force, and let u0 = 0 on S1 , sn 0 = σ 0n = 0 on S2 . Thus, since S1 and S2 are complementary subsets of ∂B, we have sn 0 · u0 = 0 on ∂B. Now from the theorem of work and energy and from the theorem of positive work, we can conclude that u0 is a rigid displacement. Consequently ε0 = σ 0 = 0, and the two solutions s and se are equal modulo a rigid displacement. S1 being a regular surface, if it is non-empty, it must contain at least three non-collinear points. Consequently, since the rigid displacement u0 = 0 on S1 , u0 vanishes identically. That is, s = se and the proof is complete.

Copyright © 2004 by Chapman & Hall/CRC

2.4. MINIMUM PRINCIPLES OF ELASTOSTATICS

2.4

61

Minimum principles of elastostatics

The principle of minimum potential energy and the principle of minimum complementary energy completely characterize the solution of the mixed problem presented previously. b and b We suppose that c, b, u s have the regularity properties assumed in the Section 2.3. Let ε and σ be continuous symmetric tensor fields on B. For convenience, we denote by Uc (ε), rather than U (ε) for the strain energy; i.e. Z 1 ε · cεdv. (2.4.1) Uc (ε) = 2

We define the strain-energy Uk (σ) by Z 1 σ · kσdv, Uk (σ) = 2

where k = c−1 is the compliance tensor. If σ = cε, we have Uk (σ) = Uc (ε).

(2.4.2)

(2.4.3)

By a kinematically admissible state, we mean an admissible state that satisfies T , the stress-strain relation σ = cε the strain-displacement relation ε = ∇u+∇u 2 b on S1 . and the displacement boundary condition u = u We have the following: Principle of minimum potential energy. Let A be the set of all kinematically admissible states, and let Φ be the functional on A defined by Z Z b s · uda, (2.4.4) b · udv − Φ(s) = Uc (ε) − S2

B

for every s = [u, ε, σ] ∈ A. Further, let s be a solution of the mixed problem. Then Φ(s) ≤ Φ(e s),

(2.4.5)

for every se ∈ A, and the equality holds only if se = s modulo a rigid displacement. In order to prove this theorem, we consider s, se ∈ A and define s0 = se − s. Then s0 = [u0 , ε0 , σ 0 ] is an admissible state and ε0 =

1 (∇u0 + ∇u0T ), σ 0 = cε0 and u0 = 0 on S1 . 2

Moreover, σ = cε since s ∈ A. Hence, the equation (2.3.7) and the above relation imply Z 0 Uc (e ε) − Uc (ε) = Uc (ε ) + σ · ε0 dv. B

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62

CHAPTER 2. ELEMENTS OF LINEAR ELASTOSTATICS Since u0 = 0 on S1 , using the integral theorem (1.2.30), we get Z Z Z 0 0 σ · ε dv = sn · u da − u0 · div σ dv, sn = σn. B

S2

B

Thus, from (2.4.4), it results Z Z (sn − b s) · u0 da. Φ(e s) − Φ(s) = Uc (ε0 ) − (div σ + b) · u0 dv + B

S2

Since s = [u, ε, σ] is a solution, the last relation implies Φ(e s) − Φ(s) = Uc (ε0 ).

Thus, since c is positive definite, Uc (ε0 ) ≥ 0, and we get Φ(s) ≤ Φ(e s) on A. Also, for the same reason Φ(s) = Φ(e s) if and only if ε0 = e ε − ε = 0. Hence, according to Kirchhoff’s theorem s = se modulo a rigid displacement. In words, the principle of minimum potential energy asserts that the difference between the strain energy and the work done by the prescribed body forces and surface forces, take a smaller value for the solution of the mixed problem, than for any other kinematically admissible state. By a kinematically admissible displacement field, we mean an admissible displacement field that satisfies the displacement boundary condition, and for which T div c∇u is continuous on B. The corresponding strain field ε = ε(u) = ∇u+∇u 2 is then called a kinematically admissible strain field. Obviously, if [u, ε, σ] is a kinematically admissible, then u is a kinematically admissible displacement field. Conversely, the latter assumption implies the former, provided that ε and σ are defined through the strain-displacement and stress-strain relation. In view of these relations, Φ(s) can be considered to be a functional Φ(u) of u Z Z Z 1 b s · uda. (2.4.6) b · udv − ∇u · c∇udv − Φ (u) = 2 B S2 B

Therefore, we have the following new form for the principle of minimum potential energy: Principle of minimum potential energy (displacement formulation). Let u corresponds to a solution of the mixed problem. Then Φ (u) ≤ Φ (e u) ,

(2.4.7)

e. for any kinematically admissible displacement field u The functional Φ, defined by the equation (2.4.4) or by the relation (2.4.6) represents the potential energy of the elastic body, corresponding to the given external forces b and b s.

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2.4. MINIMUM PRINCIPLES OF ELASTOSTATICS

63

By a statically admissible stress field, we mean an admissible stress field that satisfies the equation of equilibrium and the traction boundary condition. The following holds: Principle of minimum complementary energy. Let B denote the set of all statically admissible stress fields, and let Ψ be the functional on B defined by Z b da, sn = σn, sn · u (2.4.8) Ψ (σ) = Uk (σ) − S1

for every σ ∈ B. Let σ be the stress field corresponding to the solution of the mixed problem. Then Ψ (σ) ≤ Ψ (e σ) , (2.4.9) e ∈ B, and the equality holds only if σ = σ e. for every σ To prove the theorem, let us denote by [u, ε, σ] a solution of the mixed e ∈ B be a statically admissible stress field, and let us define problem. Let σ e − σ. σ0 = σ

Then σ 0 satisfies

div σ 0 = 0 on B and s0n = σ 0 n = 0 on S2 . Since ε = kσ, an obvious analog of (2.3.7) implies Z Uk (e σ ) − Uk (σ) = Uk (σ 0 ) + σ 0 εdv. B

Taking into account the homogeneous equilibrium equation satisfied by σ 0 and the homogeneous boundary condition satisfied by the same field on S 2 , and using again the integral theorem (1.2.30), we get Z Z s0n uda. σ 0 εdv = B

S1

0

Then, since u = u on S1 , finally we obtain Ψ (e σ ) − Ψ (σ) = Uk (σ 0 ) . e ) and Ψ (e Therefore, since k is positive definite, Ψ (σ) ≤ Ψ (σ σ ) = Ψ (σ) if and e − σ = 0. only if σ 0 = σ The principle of minimum complementary energy asserts that from all statically admissible stress fields, the one corresponding to a solution of the mixed problem makes a minimum difference between the stress energy and the work done against the prescribed displacement. The functional Ψ, defined by the equation (2.4.8) represents the complementary energy of the elastic body, corresponding to the given surface displacement b. u

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CHAPTER 2. ELEMENTS OF LINEAR ELASTOSTATICS

By using the following minimum principles, we can establish the: Upper and lower bounds for the strain energy. Let U be the strain energy association with a solution of the displacement problem. Assume that the body force vanishes. Then Z b da − Uk (σ ∗ ) ≤ U ≤ Uc (e s∗n · u ε), (2.4.10) ∂B

where e ε is a kinematically admissible strain field, σ ∗ is a statically admissible stress field and s∗n = σ ∗ n is the corresponding surface traction. On the other hand, let U be the strain energy corresponding to a solution of the traction problem. Then Z Z e dv + b e da − Uc (e b·u s·u ε) ≤ U ≤ Uk (σ ∗ ), (2.4.11) B

B

e is a kinematically admissible displacement field, e where u ε is the corresponding strain field and σ ∗ is a statically admissible stress field. To prove the inequalities (2.4.10), let as denote by s = [u, ε, σ] a solution of the displacement problem with b = 0. Let U be the associated strain energy U = Uc (ε) = Uk (σ). Then, since S2 = ∅, we conclude from the principle of minimum potential energy that U = Uc (ε) ≤ Uc (e ε).

On the other hand, since S1 = ∂B, it follows from the principle of work and energy that the last term in the expression of Ψ(σ) in (2.4.8) is equal to −2U . Thus Ψ(σ) = −U, and (2.4.9) implies U≥

Z

∂B

b da − Uk (σ ∗ ). s∗n · u

Next, since S2 = ∂B, we conclude from the principle of work and energy that the last terms is the expression (2.4.4) are equals to −2U . Thus Φ(s) = −U, and (2.4.5) implies U≥ and the proof is complete.

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Z

b·e udv + B

Z

∂B

b e da − Uc (e s·u ε),

65

2.4. MINIMUM PRINCIPLES OF ELASTOSTATICS

By a statically admissible state, we mean an admissible state [u, ε, σ] with σ a statically admissible field. For convenience, we now define Ψ on the set of all statically admissible states s = [u, ε, σ] by writing Ψ(s) = Ψ(σ). Clearly, an admissible state s is a solution of the mixed problem if and only if s is both kinematically and statically admissible. We use this fact in establishing the following results. Theorem 4 : Let s be a solution of the mixed problem. Then Φ(s) + Ψ(s) = 0.

(2.4.12)

Indeed, in view of the definitions (2.4.4) and (2.4.8) of Φ and Ψ Z Z Z b b da − s · uda Φ(s) + Ψ(s) = Uc (ε) + Uk (σ) − sn · u b · udv − B S2 S1 Z Z sn · uda, b · udv − 2Uc (ε) − B

∂B

and the given results follows from the principle of work and energy. Assuming that the mixed problem has a solution, the second theorem is a consequence of the preceding results. Theorem 5 : Let se and s∗ be admissible states with se kinematically admissible and s∗ statically admissible. Then Φ(s) + Φ(s∗ ) ≥ 0.

(2.4.13)

Indeed, let s be a solution of the mixed problem. Then the principles of minimum potential energy and complementary energy imply that Φ(e s) ≥ Φ(s), Ψ(s∗ ) ≥ Ψ(s). Thus, according to (2.4.12) Φ(e s) + Φ(s∗ ) ≥ Φ(s) + Ψ(s) = 0. We shall now use the established results in order to prove the: Converse to the principle of minimum potential energy. Let s be a kinematically admissible state, and let us suppose that Φ(s) ≤ Φ(e s)

(2.4.14)

for every kinematically admissible state se. Then s is a solution of the mixed problem. In order to prove this important result, we consider an arbitrary vector field u0 , of class C ∞ on B, and let us suppose that u0 vanishes near S1 , that part of the

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CHAPTER 2. ELEMENTS OF LINEAR ELASTOSTATICS

boundary ∂B, on which the displacement is given. Let us consider s0 = [u0 , ε0 , σ 0 ], where  1 ∇u0 +∇u0T , σ 0 = cε0 . ε0 = 2 Obviously, se = s + s0 is a kinematically admissible state, and it is not difficult to verify that the equation Z Z 0 0 (sn − b s) · u0 da (2.4.15) Φ (e s) − Φ (s) = Uc (ε ) − (div σ + b) · u da + B

S2

holds in the present circumstances. Thus, since Φ (s) ≤ Φ (e s), Z Z (sn − b s) · u0 da. 0 ≤ Uc (ε0 ) − (div σ + b) · u0 dv + B

S2

It is obvious that this relation must hold with u0 replaced by αu0 and ε0 by 0 αε ; hence Z Z 0 2 0 (sn − b s) · u0 da 0 < α Uc (ε ) − α (div σ + b) · u dv + α B

for every real number α which implies Z Z 0 − (div σ + b) · u dv + α B

S2

S2

(sn − b s) · u0 da = 0,

(2.4.16)

for every C ∞ vector field u0 that vanishes near S1 . If, in addition, u0 vanishes near ∂B, from (2.4.16) we get Z (div σ + b) · u0 dv = 0, B

∞

0

for every C vector field u that vanishes near ∂B. In these circumstances, using the fundamental lemma of variation calculus, we can conclude that div σ + b = 0 on B. Accordingly, from (2.4.16) it results Z (sn − b s) · u0 da = 0, S2

∞

0

for every C field u vanishing near S1 . Hence, again using the fundamental lemma, we get sn = σ n = b s on S2 .

Thus, s is a kinematically admissible state, that satisfies the equilibrium equation and the boundary condition in traction. Hence, s is a solution of the mixed problem and the proof of the converse theorem is ended. The principle of minimum potential energy and its converse characterize completely the solution of the mixed problem. Concerning the principle of minimum complementary energy and its converse, a similar property is true.

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67

2.5. GREEN’S TENSOR FOR AN INFINITE MEDIUM

2.5

Green’s tensor for an infinite medium

We consider an idealized situation assuming a homogeneous elastic body which occupies the entire three-dimensional Euclidean space E. We say that B is an infinite body or an infinite elastic medium. As usual by (O, xk ), k = 1, 2, 3, we denote a rectangular Cartesian coordinate system in E. We call fundamental (singular) solution or Green’s tensor function of the infinite elastic medium a second order tensor field G(x) having the following properties: (i) For any point of E with the position vector x 6= 0 and for each p = 1, 2, 3, (p) the displacement field uk (x) = Gkp (x) defines a regular elastic state corresponding to zero body force. In particular, are satisfied the equilibrium equations expressed in terms of displacements cijkl Gkp,lj (x) = 0 for i, p = 1, 2, 3 and x 6= 0.

(2.5.1)

(ii) G(x) is a homogeneous function of degree −1 in xk . In particular, we have G(x) = O(r −1 ), σ (p) (x) = O(r −2 ) as r → 0 and

(2.5.2)

(p)

also as r → ∞, where r = kxk = (x21 + x22 + x23 )1/2 and σij = cijkl Gkp,l are the components of the stress tensor σ (p) corresponding to the displacement u(p) with (p) components uk = Gkp . (iii) For all η > 0 and p = 1, 2, 3 Z Z (p) (2.5.3) σ (p) nda = ep or σij nj da = δip , i, p = 1, 2, 3, Ση

Ση

where Ση is the sphere with radius η centered at the origin O, n is the inward unit normal to Ση and (e1 , e2 , e3 ) is the orthonormal basis, corresponding to the coordinate system (O, x1 , x2 , x3 ). It can be shown that the properties (i)-(iii) uniquely characterize G(x). Equation (2.5.3)1 shows that the resultant of the stress vector or traction (p) sn = σ (p) n corresponding to the displacement u(p) and acting on any sphere (p) centered at the origin equals the unit vector ep . This is the reason why uk (x) = Gkp (x) is called the component in the direction of the xk -axis of the displacement produced by a unit concentrated force acting at origin O and directed along the xp -axis. Since the elastic medium is homogeneous and occupies the entire space, a unit concentrated force acting at an arbitrary point with position vector x0 and directed (p) along the unit (base) vector ep , produces a displacement field uk = Gkp (x − x0 ). In this way, it results that an arbitrary concentrated force P acting at x 0 produces the displacement field 0

u(x) = G(x − x )P or uk (x) = Gkp (x − x0 )Pp .

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(2.5.4)

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CHAPTER 2. ELEMENTS OF LINEAR ELASTOSTATICS

Let us assume now that the elastic medium is subjected to the action of a body force b(x) of class C 2 in E which satisfies the condition b(x) = O(r −3 ) as r → ∞.

(2.5.5)

The displacement field u(x) produced by this body force field is given by the relation Z 0 0 0 u(x) = G(x − x )b(x )dv(x ), E

or uk (x) =

Z

Gkp (x − x0 )bp (x0 )dv(x0 ).

(2.5.6)

E

The convergence of the above improper integrals is guaranteed by the conditions (2.5.2)1 and (2.5.5). Also, it is easy to see that lim ku(x)k = 0.

r→∞

(2.5.7)

It can be shown that u(x) given by (2.5.6) is the unique solution of the equilibrium equation expressed in displacement div c∇u + b = 0 in E,

(2.5.8)

which satisfies the boundary condition (2.5.7) at large distances. Let us consider now a finite, regular, closed surface Γ given in E. Let us assume that on the infinite elastic body acts an external force distributed over this surface and characterized by the surface force density f (x), assumed to be a continuous function on Γ. Then the displacement field u(x) produced by this system of external forces is given by the equation Z 0 0 u(x) = G(x − x )f (x )da(x0 ). (2.5.9) Γ

It can be shown that: (i) (ii)

u is continuous across the surface Γ; div c∇u = 0 in E − Γ;

u(x) = O(r −1 ) and ∇u(x) = O(r −2 ) as r → ∞; [σn] + f = 0 on Γ,

(iii) (iv) T

(2.5.10)

) , σ = cε and [σn] represents the jump of σn across Γ. where ε = (∇u+∇u 2 The equation (2.5.9) and the properties (i)-(iv) play the fundamental role in Eshelby’s inclusion problem, as we shall see a little later on.

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2.5. GREEN’S TENSOR FOR AN INFINITE MEDIUM

69

Up to this moment, we have considered concentrated forces, acting at a point or on a surface, in an infinite elastic medium, and we have analyzed the singular elastic states produced by such kind of external actions. We consider now the concepts concerning concentrated forces and Green’s tensor function in the case of a finite elastic body B, bounded by the surface ∂B. Let P be a vector valued function whose domain of definition D is a finite set of points of B. Interpreting P as a system of concentrated loads, acting in B, we say that [u, ε, σ] is a singular elastic state corresponding to the external force system [b, sn , P] if: (i) [u, ε, σ] is a regular elastic state in B − D corresponding to the external force system [b, sn ]; (ii) for each x0 ∈ D, we have

u(x) = O(r −1 ), σ(x) = O(r −2 ) as r = kx − x0 k → 0; (iii) for each x0 ∈ D

Z

lim

η→0 B∩Ση (x0 )

0

σnda = P(x ),

(2.5.11)

(2.5.12)

where Ση (x0 ) is the sphere of radius η > 0 and centered at x0 , and n is the inward unit normal to Ση (x0 ). From the above general definition, it follows that 0

u(x) = G(x − x )P, given by equation (2.5.4), represents a singular elastic state in an infinite elastic medium, that corresponds to a concentrated force P applied at x0 and to zero body forces. In the case of singular states, we have the following: Balance of forces and moments for singular states. Let P be a system of concentrated loads, and let s = [u, ε, σ] be a singular elastic state corresponding to the external force system [b, sn , P] with b continuous on B. Then Z Z X sn da + bdv + P(x) = 0, (2.5.13) B

∂B

Z

∂B

x × sn da +

Z

x∈D

x × bdv +

B

X

x∈D

x × P(x) = 0,

In order to prove this theorem, we introduce the set [ Bη = B − Ση (x), x∈D

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(2.5.14)

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CHAPTER 2. ELEMENTS OF LINEAR ELASTOSTATICS

and take η sufficiently small, such that the balls Ση (x), x ∈D and the boundary ∂B be mutually disjoint. This is possible, since the places x ∈ D are points in B, which is an open set. According to the general definition, s = [u, σ, ε] is a regular elastic states on B, hence (see problem P2.21) Z Z sn da + bdv = 0. Bη

∂Bη

Since b is continuous on B lim

Z

η→0 Bη

bdv =

Z

bdv.

B

Further, we have Z

sn da = ∂Bη

Z

XZ

sn da + ∂B

x∈D

sn da. ∂Ση (x)

Combining the last three relations and letting η → 0, with the aid of property (iii), we conclude that the relation (2.5.13) of the theorem holds. The second relation (2.5.14) is derived exactly in the same manner starting with the equation Z Z x × sn da + x × bdv = 0, ∂Bη

Bη

which is true, since s = [u, ε, σ] is a regular elastic state on Bη . For singular elastic states, the following generalization of Betti’s theorem holds: ˜ be systems Reciprocal theorem for singular elastic states. Let P and P ˜ of concentrated loads with disjoint domains D and D. If [u, ε, σ] and [˜ u, ε˜, σ ˜] are singular elastic states corresponding to the external force systems [b, s , P] and n h i ˜ ˜ ˜ respectively, then b, sn , P Z

Z

B

σ·˜ εdv

B

σ ˜ · εdv

= =

Z

Z

∂B

sn · u ˜ da +

∂B

˜ sn ·uda +

Z

Z

B

B

b·u ˜ dv +

˜ · udv + b

X

x0 ∈D

X

˜ x 0 ∈D

0

0

P(x ) · u ˜ (x ),

˜ 0 ) · u(x0 ). P(x

The proof of this theorem is based on Betti’s reciprocal theorem for regular elastic states combined with the definition of the singular elastic states and with the procedure used in order to prove the generalized balance theorem. In order to introduce the Green’s tensor function concerning boundary value problems for finite elastic bodies, we need the following:

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71

2.5. GREEN’S TENSOR FOR AN INFINITE MEDIUM

Lemma. Let f and m be two vectors. Then there exists a unique rigid displacement field w that satisfies the following system of equations: Z Z wda = f , (x − ξ) × wda = m, ∂B

∂B

where ξ is the position vector of the centroid of ∂B. This solution is given by w(x) = u0 +ω 0 ×(x − ξ), u0 =

1 f , ω 0 = J−1 ·m, a

(2.5.15)

where a is the area of ∂B and J is the centroidal inertia tensor, whose components are the moments of inertia of ∂B with respect to the principal axes of ∂B passing through its centroid. For the proof of this lemma, see P2.43. We say that an integrable vector field u on ∂B is normalized if Z Z uda = 0, (x − ξ) × uda = 0. (2.5.16) ∂B

∂B

Let us observe that given a solution u of the traction problem (S1 = ∅, S2 = ∂B), the field u + w, with w a rigid displacement field, is also a solution. According to the lemma, there exists a unique rigid displacement field w such that u + w is normalized. Thus, we may always assume, without loss of generality, that the solutions of the traction problem are normalized. 0 ˜ kp (x, x0 ) is called ˜ A second order tensor field G(x, x ), with components G Green’s tensor function of the body B, provided that: (i) The elastic displacement u(p) (x, x0 ) with components (p) ˜ kp (x, x0 ) k, p = 1, 2, 3, u ˜k (x, x0 ) = G

(2.5.17)

and the corresponding stress tensor σ (p) (x, x0 ) with components ˜ (p) ˜ kp,l (x, x0 ) = cijkl ∂ Gkp (x, x0 ), σ ˜ij (x, x0 ) = cijkl G ∂xl

(2.5.18)

represents a singular elastic state corresponding to vanishing body forces and to a unit concentrated load ep acting at x0 . (ii) If S1 is not empty (mixed or displacement boundary value problem), u ˜ (p) = 0 on S1 and σ ˜ (p) n = 0 on S2 ;

(2.5.19)

if S1 is empty (traction boundary value problem), σ ˜ (p) n = w on S2 = ∂B,

(2.5.20)

where w is given by (2.5.15) with 0

f = −ep and m = −(x −ξ) × ep .

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(2.5.21)

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CHAPTER 2. ELEMENTS OF LINEAR ELASTOSTATICS

The boundary condition (2.5.20), (2.5.21) ensure that the balance of forces and moments are satisfied in the traction problem (S1 = ∅, S2 = ∂B). Indeed, R substituting (2.5.20) into ∂B σ ˜ (p) nda and taking into account the given lemma, according to (2.5.15) and (2.5.21), we get Z σ ˜ (p) nda + ep = 0. (2.5.22) ∂B

Analogous, it results Z (p) (x − ξ) × σ ˜ nda + (x0 −ξ) × ep = 0.

(2.5.23)

∂B

Note that Green’s tensor function defined above depends not only on the material, as in the case of an infinite elastic medium, but also on the domain B occupied by the elastic body, as well as on the boundary conditions. 0 ˜ The Green’s tensor function G(x, x ) admits the decomposition (p)

˜ kp (x, x0 ) = Gkp (x − x0 )+u∗ (x, x0 ), G k

(2.5.24)

0

where G(x − x) is Green’s tensor function of the infinite elastic medium, and ∗ (p)

u (x, x0 ) is the regular displacement field corresponding to vanishing body 0 ˜ forces and concentrated forces and such that G(x, x ) satisfies the boundary conditions (2.5.19) and (2.5.20). The following fundamental theorem gives an integral representation to the boundary value problems of linear elastostatics in terms of Green’s tensor function 0 ˜ G(x, x ): Integral representation theorem. Let u(x) be the solution of one of the boundary value problems corresponding to the external force system [b, s n , P] and let us assume that u is normalized in the traction problem. Then for any x 0 ∈ B−D Z Z (p) 0 b da + u ˜ (p) · b sda σ ˜ n·u up (x ) = − S2 S1 Z X u ˜ (p) · bdv + u ˜ (p) (x, x0 ) · P(x), (2.5.25) + B

x∈D

where u ˜ (p) and σ ˜ (p) are given by (2.5.17) and (2.5.18), respectively. In order to prove the equation (2.5.25), we first assume that S1 is not empty (mixed or displacement problem). Since P = ep , by using the reciprocal theorem (2.5.18) and taking into account (2.5.20), (2.5.21), we get Z Z Z X (p) 0 b ˜ (p) n · uda + up (x0 ) = P(x) · u ˜ (x, x ), b·u ˜ (p) dv + s·u ˜ (p) da + σ S1

and the theorem is proved.

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S2

B

x∈D

73

2.5. GREEN’S TENSOR FOR AN INFINITE MEDIUM

On the other hand, in the traction problem, from (2.5.20), (2.5.21) and (2.5.15) and the normalization condition, we find that Z Z Z (p) (x − ξ) × uda = 0, uda + ω 0 · σ ˜ n · uda = u0 · ∂B

∂B

∂B

and the reciprocal theorem leads again to (2.5.25). ˜ The advantage of using Green’s tensor function G(x, x0 ) is that, after solving ∗ (p)

the particular boundary value problem whose solution is u (x, x0 ), the general solution corresponding to any other boundary value problem may be obtained by quadratures, provided that the subboundaries S1 and S2 remain unchanged. Let us assume now that [u, ε, σ] is the regular solution of the displacement problem, in the absence of body forces and concentrated loads. According to (2.5.25) the displacement field u(x) is given by the equation Z 0 0 b (x0 ) · σ u ˜ (p) (x , x)n(x )da(x0 ). (2.5.26) up (x) = − ∂B

b (x0 ) is a continuous function on We suppose that the given displacement u ∂B. In this case, one can deduce an integral representation for the displacement gradient ∇u(x), and hence for the strain and stress fields ε(x) and σ(x), by differentiating the relation (2.5.26) under the integral sign. The fact that this operation is possible, can be proved in a manner analogous to that used in the classical theory of Newtonian potentials. In particular, for the strain fields ε(x), we get ) ( (p) Z (q) ˜km ∂σ ˜km (x0 , x) ∂ σ 1 0 nm (x0 )da(x0 ). (2.5.27) + u bk (x ) εpq (x) = − ∂xp ∂xq 2 ∂B

Now, let us assume that we have homogeneous boundary condition; i.e. b (x) = Ex, u

E = ET = const. on ∂B. In this particular situation, from (2.5.27), we obtain  Z Γpqkl (x0 , x)da(x0 ) Ekl , εpq (x) = ∂B

with 1 Γpqkl (x0 , x) = − 4

+

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! (q) (p) ˜km (x0 , x) ∂σ ˜km (x0 , x) ∂ σ x0l + ∂xp ∂xq ! ) (q) (p) ˜lm (x0 , x) ∂σ ˜lm (x0 , x) ∂ σ x0k nm (x0 ), + ∂xp ∂xq

(

(2.5.28)

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CHAPTER 2. ELEMENTS OF LINEAR ELASTOSTATICS

with x ∈ B and x0 ∈ ∂B. Finally, denoting Apqkl (x) =

Z

Γpqkl (x0 , x)da(x0 ), ∂B

we get εpq (x) = Apqkl (x)Ekl or ε(x) = A(x)E,

(2.5.29)

Apqkl (x) = Aqpkl (x) = Apqlk (x) .

(2.5.30)

with The fourth order tensor field A = A(x) is called the influence function corresponding to the homogeneous displacement problem. It can be proved (see P2.44) that there also exists a fourth order tensor field B = B(x), named the influence function corresponding to the homogeneous traction problem. This influence function has the following characteristic property: If we assume homogeneous traction boundary condition; i.e. if b s = Σn, Σ = ΣT = const. on ∂B

(2.5.31)

Bpqkl (x) = Bqpkl (x) = Bpqlk (x) .

(2.5.33)

and if [u, ε, σ] is the regular solution of homogeneous traction boundary value problem, then σpq(x) = Bpqkl (x) Σkl or σ(x) = B(x)Σ, (2.5.32) with As we shall see later, the influence tensor fields A(x) and B(x) play an important role in the theory of macroscopically homogeneous composite materials. The above results show that these influence functions there exists, and just their existence is essential in studying the properties of the overall elastic moduli of the macroscopically homogeneous composites. Let us assume now an infinite, homogeneous, linearly elastic and isotropic medium. It can be shown that in this case Green’s tensor function G(x) has the following components: Gjp (x) =

∂r ∂r 1 }, − {2(1 − ν)δjp ∂xj ∂xp ∂xm ∂xm 16πµ (1 − ν)

(2.5.34)

with r = (x21 + x22 + x23 )1/2 . From (2.5.34), we can deduce the following equivalent form of the components: xj xp 1 1 (2.5.35) {(3 − 4ν) δjp + 3 }. Gjp (x) = r r 16πµ (1 − ν)

The tensor form of the above relation is: G(x) =

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xx 1 1 {(3 − 4ν) 1+ 3 }. r r 16πµ (1 − ν)

(2.5.36)

2.6. PIECE-WISE HOMOGENEOUS BODIES

75

In consequence, according to the general relation (2.5.4), the displacement field u(x) produced in an infinite homogeneous and isotropic elastic medium, by a concentrated force P applied in the origin of the coordinate system, has the expression 1 1 1 (2.5.37) {(3 − 4ν) P+ 3 (P · x)x}. u(x) = r r 16πµ (1 − ν)

Accordingly, the corresponding stress field σ(x) is σ(x) = −

1 3 1 − 2ν 1 (P · x) xx}. (2.5.38) · {− (P · x) 1 + Px + xP + 1 − 2ν r2 8π(1 − ν) r3

The associated traction sn = σn is given by the relation

1 3 1 − 2ν 1 (P · x) (x · n)x}. {− (P · x) n + P(x · n) + x(P · n)+ 1 − 2ν r2 8π(1 − ν) r3 (2.5.39) In particular, if P = ep , we have

sn = −

xp 3 1 1 − 2ν · 2 xx}, · 3 {−xp 1 + ep x + xep + 1 − 2ν r 8π(1 − ν) r

(2.5.40)

1 3 1 − 2ν 1 xp xi xj }. {−δij xp +δ pi xj +δ pj xi + 1 − 2ν r2 8π(1 − ν) r3

(2.5.41)

σ (p) (x) = −

or (p)

σij (x) = −

(p)

The traction sn = σ (p) n, associated to σ (p) is s(p) = −

2.6

1 3 1 − 2ν 1 xp (x · n)x}. (2.5.42) {−xp n + ep (x · n) + xnp + 1 − 2ν r2 8π(1 − ν) r3

Piece-wise homogeneous bodies

In order to describe the laws governing the behavior of the piece-wise homogeneous elastic bodies, we begin by introducing some necessary mathematical concepts. Let B be an open set in the three-dimensional Euclidean space E, its boundary ∂B being the union of a finite number of non-intersecting closed regular surfaces. Note that ∂B may have corners and edges. A partition for B is a finite collection B1 , ...., BN mutually disjoint regular subregions of B such that N [ Bα. (2.6.1) B= α=1

A scalar, vector or tensor field f is piece-wise constant on B if there exists a partition B1 , ..., BN such that for each Bα the restriction of f to Bα has constant

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CHAPTER 2. ELEMENTS OF LINEAR ELASTOSTATICS

value; f is piece-wise continuous on B if the restriction of f to Bα is bounded and continuous on Bα ; f is piece-wise smooth on B if ∇f exists almost everywhere on B and is piece-wise continuous on B; and f is piece-wise of class C 2 on B if ∇(∇f ) exists almost everywhere on B and is piece-wise continuous on B. We define now a singular surface. Roughly speaking, this is a surface across which some kinematic or dynamic quantity suffers a jump discontinuity. Let us consider a partition B1 , B2 of B and let us denote by Γ the common boundary as shown in Figure 2.1.

Figure 2.1: The singular surface Γ. Let f be a scalar, vector or tensor field given on B1 ∪ B2 . We call Γ a singular surface of order zero with respect to f , if f (x) is continuous for x ∈B − Γ and f (x) approaches definite and distinct values f (1) (x0 ) and f (2) (x0 ) as x approaches a point x0 on Γ while remaining in B1 , or in B2 , respectively; i.e. f (1) (x0 ) = lim f (x) with x ∈B1 , x→x0

f

(2)

(x0 ) = lim f (x) with x ∈B2 . x→x0

(2.6.2)

The jump of f across Γ is then a function [f ] on Γ defined by [f ] (x0 ) = f (1) (x0 ) − f (2) (x0 ) with x0 ∈ Γ.

(2.6.3)

Let n ≥ 1 be a fixed integer.We call Γ a singular surface of order n with respect to if f is of class C n−1 on B, of class C n on B − Γ, and the derivatives of order n of f suffer jump discontinuities across Γ. Let u be a vector field given on B − Γ, of class C 1 in B1 and B2 , and let Γ be a singular surface of order 0 with respect to u.

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2.6. PIECE-WISE HOMOGENEOUS BODIES

77

Using Figure 2.1 and taking into account the divergence theorem for B1 and B2 , respectively, we get Z Z Z div u dv = u · nda + u(1) nda, ∂B Γ Z Z B1 (2) div u dv = − u ·nda. Γ

B2

Adding these relations leads to Z Z Z div u dv = u · nda + [u · n] da, B = B1 ∪ B2 . B

∂B

(2.6.4)

Γ

In particular, if [u · n] = 0 on Γ, then

Z

div u dv = B

Z

∂B

u · nda.

(2.6.5) (2.6.6)

Hence, if the jump [u · n] of the normal component of the vector field u is vanishing across the singular surface Γ, the divergence theorem rests true in its usual form, even if u has a non-vanishing jump across Γ. Let σ be a symmetric second order tensor field given on B − Γ, of class C 1 in B1 and in B2 , and let Γ be a singular surface of order 0 with respect to σ. Following the above procedure, it is easy to obtain the relation Z Z Z σnda + [σn] da, B = B1 ∪ B2 . div σ dv = ∂B

B

Γ

In particular, if [σn] = 0 on Γ, then

Z

div σ dv = B

Z

σn da.

(2.6.7) (2.6.8)

∂B

Hence, if the jump [σn] is vanishing across the singular surface Γ, the divergence theorem rests true in its usual form, even if σ has a non-vanishing jump across Γ. (∇u+∇uT ) . Let us denote now by ε the symmetric gradient of u; i.e. ε = 2 Using the integral theorem (1.2.30), it is easy to see that, if u and σ have the properties assumed above, then Z Z Z Z σ · εdv = u · σn da− u·div σ dv+ [u · σn] da, B = B1 ∪B2 . (2.6.9) B

∂B

B

Γ

In particular, if [u] = 0 and [σn] = 0 on Γ,

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(2.6.10)

78 then

CHAPTER 2. ELEMENTS OF LINEAR ELASTOSTATICS Z

σ · εdv = B

Z

∂B

u · σnda −

Z

B

u · div σ dv.

(2.6.11)

Hence, if the jumps [u] and [σn] are vanishing across the singular surface Γ, the classical integral theorem (1.2.30) rests true in its usual form, even if σ has a non-vanishing jump but u is continuous across Γ. The relations (2.6.7), (2.6.8) and (2.6.10), (2.6.11) play essential roles in the theory of piece-wise homogeneous bodies, and particularly, in the theory of composite materials formed by a homogeneous matrix in which a great number of homogeneous inclusions are firmly imbedded. The generalization of the above given results to a case in which a finite number of mutually disjoint singular surface are present, is obvious. It is essential to note that if the null-jump conditions, as (2.6.7) and (2.6.10) are satisfied on these singular surfaces, various integral theorems, such as (2.6.8) and (2.6.11), as well as various consequences of such theorems, rest true in their usual forms. Let us consider now a body B formed by two different homogeneous linearly elastic materials, firmly bounded together. We assume that one of the material occupies the domain B1 and the other the domain B2 , B1 and B2 corresponding to a partition of B as shown in Figure 2.1. We say that B is a piece-wise homogeneous body or a mixture or a composite. The two materials forming the body are named phases of the biphasic mixture. The elasticity c = c(x) of this mixture (or composite) is a piece-wise constant function on B (see Figure 2.1)  c1 = const. if x ∈B1 c(x) = and c1 6= c2 . (2.6.12) c2 = const. if x ∈B2 The stress-strain relation, for this biphasic mixture or composite is  c1 ε(x) if x ∈B1 σ(x) = . (2.6.13) c2 ε(x) if x ∈B2 We assume that c1 and c2 are symmetric, positive definite tensors. We say that a vector field u is an admissible displacement field for the mixture B if u is continuous on B, piecewise smooth on B, relative to the partition B 1 , B2 , and the restrictions of u to B1 and B2 are of class C 2 on B1 and B2 . Accordingly, u satisfies the following null-jump condition

[u] = 0 on Γ,

(2.6.14)

expressing the fact, that the two phases of the composite are firmly bound together. T ) However, ∇u can have a finite jump across Γ. We call ε = ε(u) = (∇u+∇u 2 defined on B − Γ, an admissible strain field for the mixture B, corresponding to the admissible displacement field u. Generally, ε(u) has non-vanishing jump across Γ, the common boundary of the phases.

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2.6. PIECE-WISE HOMOGENEOUS BODIES

79

We say that a symmetric tensor field σ is an admissible stress field for the mixture B, if σ and div σ are piece-wise smooth on B, relative to the partition B1 , B2 , the restrictions of σ to B1 and B2 are of class C 1 on B1 and B2 , and the jump of [σn] across Γ is vanishing; i.e. σ satisfies the following null-jump condition: [σn] = 0 on Γ, (2.6.15) expressing the principle of action and reaction in equilibrium problems of continuous deformable bodies. Taking into account the null-jump condition (2.6.14) and (2.6.15), we can conclude that the integral theorems (2.6.6), (2.6.8) and (2.6.12) are true if u is an admissible displacement field, σ is an admissible stress field and ε = ε(u) = (∇u+∇uT ) is an admissible strain field corresponding to the admissible displace2 ment field u. Consequently, the theorem of work expended (2.1.19) is true, if u is an admissible displacement field, ε = ε(u) is the corresponding strain field, and σ is an admissible stress field, which satisfies the equilibrium equation on B 1 and B2 ; that is if div σ+b = 0 on Bα , α = 1, 2. (2.6.16)

In the same way, we can see that the mean strain theorem (2.1.14) is true for a mixture with two homogeneous phases if u is an admissible displacement field and ε = ε(u) is the corresponding strain. Also, the mean stress theorem, (2.1.21) for a mixture holds, if σ is an admissible stress field, satisfying the equilibrium equation on B1 and B2 . In what follows by an admissible state for a two phasic mixture, we mean an ordered array s = [u, ε, σ], having the following properties: u is an admissible displacement field for the mixture, ε is a piece-wise continuous symmetric tensor field, corresponding to the partition B1 , B2 of B and σ is an admissible stress field for the mixture. Analogously, we say that s = [u, ε, σ] is an elastic state of the two phasic mixture, corresponding to the body force field b, if s is an admissible state for the mixture and 1 (2.6.17) ε = ε(u) = (∇u + ∇uT ), on Bα , 2

σ = cα ε on Bα ,

(2.6.18)

div σ + b = 0 on Bα , for α = 1, 2, 3.

(2.6.19)

The corresponding surface traction sn is defined in any regular point of ∂B as previously; i.e. sn = σn on ∂B. As before, the pair [b, sn ] represents the external force system, corresponding to s. It is easy to see that the principle of superposition rests valid for the mixture B.

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CHAPTER 2. ELEMENTS OF LINEAR ELASTOSTATICS

Taking into account the null-jump conditions (2.6.14) and (2.6.15) satisfied by the displacement u and by the stress σ, one can prove without difficulty that the theorem of work and energy (2.3.10) is true for a piece-wise continuous body, if s = [u, ε, σ] is an elastic state of the mixture, corresponding to the external force system [b, sn ], and the strain energy U (ε) is defined in the usual way Z Z Z 1 1 1 ε · c2 εdv. (2.6.20) ε · c1 εdv + ε · cεdv = U (ε) = 2 B2 2 B1 2 B

Also, the theorem of positive work is true, since the constant elasticities c 1 and c2 are positive definite. We stress the fact that the above presented properties are satisfied for the considered heterogeneous, piece-wise homogeneous body, since the null-jump conditions (2.6.14) and (2.6.15) are satisfied by any admissible displacement and stress field. We note that generally a theorem which is true for a homogeneous body does not rest true for a piece-wise homogeneous body. For instance, the second mean stress theorem (see P2.17) and the second mean strain theorem (see P2.18) are not true if the body is heterogeneous. In exchange, Betti’s reciprocal theorem is true for a mixture, if s = [u, ε, σ] and s˜ = [˜ u, ε˜, σ ˜ ] are two helastici states of the mixture, corresponding to external ˜ ˜ force systems [b, sn ] and b, sn , respectively.

Again the null-jump condition play a central role, in proving the validity of Betti’s theorem (2.3.11). We can now formulate the boundary value problem of elastostatics for a biphasic piece-wise continuous mixture B. We assume that a constant elasticity c1 on B1 is given, and also is given a constant elasticity c2 on B2 . We suppose also that is given a body force field b on S1 , and a b on B. Also, we assume that are given a surface displacement u surface force b s on S2 , S1 and S2 being complementary regular subsurfaces of ∂B. b is continuous on S1 and b We suppose that b is continuous on B, u s is continuous on S2 . Given the above data, the mixed problem of elastostatics, for the biphasic piece-wise homogeneous mixture B is to find an elastic state [u, ε, σ] of the mixture that corresponds to b and satisfies the displacement condition on S1 and the traction condition on S2 ; i.e. b on S1 and sn = σn = b u=u s on S2 .

(2.6.21)

We call such an elastic state a solution of the mixed problem for the piece-wise homogenous biphasic mixture or composite B. When S2 is empty so that S1 = ∂B, the above boundary condition reduces to b on ∂B u=u (2.6.22)

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81

2.6. PIECE-WISE HOMOGENEOUS BODIES and the associated problem is called the displacement problem. If S2 = ∂B, the boundary condition becomes sn = σn = b s on ∂B,

(2.6.23)

and we refer to the resulting problem as the traction problem. Summing up, we recall the conditions which must be satisfied by an admissible state s = [u, σ, ε] to be a solution of the mixed problem for a piece-wise homogeneous biphasic elastic mixture or composite ε=

1 (∇u + ∇uT ), σ = cα ε, div σ+b = 0 on Bα , α = 1, 2, 2

(2.6.24)

and [u] = 0, [σn] = 0 on Γ = ∂B1 ∩ ∂B2 .

(2.6.25)

In the theory of composite materials, in many cases, the subject under consideration is a mixture formed by two linearly elastic homogeneous phases firmly bound together. The mixture is of a kind such that one phase can be regarded as representing inclusions in the other one, the matrix. The number of inclusions is very large, and generally no restrictions are placed on the shapes of the inclusions, which may be for example spherical, plate-like or fibrous. To be more exact, we assume that B0 , B1 , ..., BN is a partition of B. The subregion B0 is occupied by the matrix having constant elasticity c1 . The subregions B1 , ..., BN are occupied by the inclusions having constant elasticity c2 , as in Figure 2.2.

Figure 2.2: Biphasic piece-wise homogeneous mixture or composite formed by a matrix and N inclusions. The elasticity c(x) of the mixture or composite considered here, is a piecewise constant function having only two distinct values. For brevity, we write B1 = B 0 , B 2 =

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N [

α=1

Bα , Γ =

N [

α=1

∂Bα ,

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CHAPTER 2. ELEMENTS OF LINEAR ELASTOSTATICS

and for simplicity, we suppose that Γ ∩ ∂B = ∅. Obviously, Γ is the common boundary of the matrix and of the inclusions, and we have  c1 for x ∈B1 c(x) = and c1 6= c2 . c2 for x ∈B2 We suppose that c1 and c2 are symmetric and positive definite fourth order constant tensors. The appropriate definitions for an admissible displacement, stress and strain, as well for an admissible state and elastic state, corresponding to the considered biphasic composite or mixture, can be obtained, by generalizing in an obvious way the definitions given for the piece-wise homogeneous body analyzed in the first part of this Section. We stress only the fact that if u and σ represent an admissible, displacement and stress field, respectively, the following null-jump conditions must be satisfied on the common boundary of the matrix and inclusion [u] = 0 and [σn] = 0 on Γ =

N [

∂Bα .

α=1

Since these conditions are fulfilled, the general theorems, analyzed in the first part of this Section, rest true for our composite composed by a matrix and a large number of inclusions. We now give the conditions which must be satisfied by an ordered array [u, ε, σ], in order to be a solution of the mixed boundary problem for the considered composite material 1 (∇u + ∇uT ), σ = cα ε, div σ+b = 0 on Bα , α = 1, 2, 2 N [ [u] = 0, [σn] = 0 on Γ = ∂Bα , (2.6.26)

ε=

α=1

b on S1 , sn = σn = b u=u s on S2 .

Moreover, u, ε and σ must have the regularity properties, imposed by the definitions of the admissible displacement, strain and stress fields. When S2 is empty, the above boundary condition on ∂B reduces to b on ∂B, u=u

(2.6.27)

sn = σn = b s on ∂B,

(2.6.28)

and, as usual, the associated problem is called displacement problem. When S1 is empty, the boundary condition on ∂B reduces to

and the resulting problem is called the traction problem.

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Further by a displacement field corresponding to a solution of the mixed problem, we mean a vector field u having the property that there exist the fields ε, σ such that [u, ε, σ] is a solution of the mixed problem for the composite. As in the classical case, we have the following characterization of the mixed problem for a composite in terms of displacement. Let u be an admissible displacement field for a composite. Then u corresponds to a solution of the mixed problem if and only if div c1 ∇u + b = 0 on B1 , div c2 ∇u + b = 0 on B2 , [u] = 0, [(c∇u)n] = 0 on Γ,

(2.6.29)

b on S2 , (c1 ∇u)n = 0 on S2 . u=u

That the above relations are necessary, follow from the strain-displacement relation, from the assumed stress-strain relation, from the equilibrium equation, from the assumed null-jump condition and from the boundary condition given on ∂B. To establish sufficiencies, let us assume that the admissible displacement field u satisfies the equations (2.6.29). We define ε on B1 and B2 by the straindisplacement relation and σ by the assumed piece-wise homogeneous stress-strain relation. Then since u is admissible, ε will be piece-wise continuous on B, σ piecewise continuous and piece-wise smooth on B. In addition div σ + b = 0 on B 1 ∪B2 , [σn] = 0 on Γ and σn = b s on S2 . Thus, σ is an admissible stress field. Therefore, [u, ε, σ] meets all requirements of a solution of the mixed boundary value problem for a piece-wise homogeneous biphasic composite. Also, taking into account the assumed regularity properties, the supposed null-jump condition, the symmetry and positive definiteness of the elasticities c 1 and c2 , it is easy to see that Kirchhoff ’s uniqueness theorem for the mixed problem concerning the considered composite rests true. Also, it can be shown that a necessary condition for the existence of a solution for the traction problem concerning a composite, is that the external forces be in equilibrium, as in the classical case (see P2.21). In order to analyze the status of the minimum principles for a composite, we observe that the strain energy Uc (ε) is defined in the usual way Z Z Z 1 1 1 ε · c2 εdv, (2.6.30) ε · c1 εdv + ε · cεdv = Uc (ε) = 2 B2 2 B1 2 B

where B1 is the domain occupied by the matrix, and B2 is the reunions of the domains occupied by the inclusions. Denoting by −1 k1 = c−1 1 and k2 = c2

the constant compliance tensors, corresponding to c1 and c2 , respectively, we define the stress-energy, Uk (σ) as usually by Z Z Z 1 1 1 σ · k2 σdv. (2.6.31) σ · k1 σdv + σ · kσdv = Uk (σ) = 2 B2 2 B1 2 B

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CHAPTER 2. ELEMENTS OF LINEAR ELASTOSTATICS Obviously, Uk (σ) = Uc (ε),

provided that σ = c1 ε on B1 and σ = c2 ε on B2 . By a kinematically admissible state for a composite material B, we mean an admissible state for the composite, that satisfies the strain-displacement relation on B1 and B2 , the stress-strain relation on B1 and B2 , and the displacement boundary b on S1 . condition u = u By a kinematically admissible displacement field for a composite material B we mean an admissible displacement field u for the composite, that satisfies the displacement boundary condition and for which div c∇u is piece-wise continuous on B, relative to the partition B0, B1 , ..., BN . The corresponding strain field ε = T

) , defined on B0, B1,..., BN is called a kinematically admissible ε(u) = (∇u+∇u 2 strain field for the composite. Obviously, if [u, ε, σ] is a kinematically admissible state for a composite, then u is a kinematically admissible displacement field for the composite. Conversely, the later assumption implies the former provided that ε and σ are defined by the strain-displacement relation and by the supposed stress-strain relation, on the domains B0, B1 , ..., BN . By a statically admissible stress field for a composite material B, we mean an admissible stress field for the composite, that satisfies the equilibrium equation div σ+b = 0, on B0 , B1,..., BN . Once the above notions are adequately extended to a composite material, it is easy to prove the validity of the minimum principles and we get the following results. Principle of minimum potential energy for a composite. Let A be the set of all kinematically admissible states for a composite, and let Φ be a functional on A defined by Z Z b s · uda, (2.6.32) b · udv − Φ(s) = Uc (ε) −

B

S2

for every s = [u, ε, σ] ∈ A. Further, let s be a solution of the mixed problem for the composite.Then Φ(s) ≤ Φ(˜ s) (2.6.33)

for every s˜ ∈ A, and equality holds only if s˜ = s modulo a rigid displacement. For a kinematically admissible displacement field u, Φ(s) can be considered as a functional Φ(u) of u Z Z Z 1 b s · uda b · udv − ∇u · c∇udv − Φ(u) = 2 B S2 B Z Z Z Z 1 1 s · uda, (2.6.34) ∇u · c2 ∇udv − b · udv − b ∇u · c1 ∇udv + = 2 B2 2 B1 B S2

and we have the following:

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Property : Let u corresponds to a solution of the mixed problem for a composite. Then Φ(u) ≤Φ (˜ u) (2.6.35)

for every kinematically admissible displacement field u ˜. Also, the following is true. Principle of minimum complementary energy for a composite. Let B denote the set of all statistically admissible stress field for a composite, and let Ψ be a functional on B defined by Z b da, sn · u (2.6.36) Ψ(σ) = Uk (σ) − S1

for every σ ∈ B. Let σ be the stress field corresponding to the solution of the mixed problem. Then Ψ(σ) ≤ Ψ(˜ σ ), (2.6.37)

˜ for every σ ˜ ∈ B, and equality holds only if σ = σ. Let us observe also that in the case of the composite material the properties giving the upper and lower bounds for the strain energy, proved in the classical case for an elastic body formed by a single phase, are also true. By a statistically admissible state for a composite material, we mean an admissible state for the composite s = [u, ε, σ] with σ a statistically admissible stress field for the composite. As before, we define Ψ on the set of all statistically admissible states s by writing Ψ(s) = Ψ(σ). An admissible state s for a composite is a solution of the mixed problem for the composite, if and only if s is both kinematically and statically admissible for the composite. Using this property, it is easy to see that for a biphasic piece-wise homogeneous material the following classical results are true. If s is a solution of the mixed problem, then Φ(s) + Ψ(s) = 0, and if s˜ and s∗ are admissible states for the composite, s˜ being kinematically admissible and s∗ being statically admissible, then Φ(˜ s) + Ψ(s∗ ) ≥ 0. We end this Section, by analyzing some particular cases, which play an important role in the theory of macroscopically homogeneous composite materials. We assume that the body force is vanishing; i.e. b = 0.

An admissible stress field σ which satisfies the equilibrium equation div σ = 0 on Bα , α = 0, 1, ..., N, corresponding to zero body force, is called a self-equilibrated stress field.

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Let us first consider the displacement problem for a composite. In this case, S2 = ∅ and S1 = ∂B and the principle of minimum potential energy takes the following simplified form: Principle of minimum potential energy for the displacement problem for a composite with null body forces. Let A denote the set of all kinematically admissible states for the composite, and let Φ be a functional on A defined by Z Z 1 1 ε · c2 εdv (2.6.38) ε · c1 εdv + Φ(s) = Uc (ε) = 2 B2 2 B1

for every s = [u, ε, σ] ∈ A. Further, let s be a solution of the displacement problem for the composite. Then Φ(s) ≤ Φ(˜ s) or equivalently Uc (ε) ≤ Uc (˜ ε)

(2.6.39)

for every s˜ = [˜ u, ε˜, σ ˜ ] ∈ A and the equality holds only if s˜ = s. Analogously we have the Principle of minimum complementary energy for the displacement problem for a composite with null body forces. Let B denote the set of all statistically admissible stress fields for the composite and let Ψ be the functional on B defined by Z Ψ(σ) = Uk (σ) −

∂B

b da s·u

(2.6.40)

for every s ∈ B. Let σ be the stress field corresponding to a solution of the displacement problem for the composite. Then Ψ(σ) ≤ Ψ(˜ σ)

(2.6.41)

˜ for every σ ˜ ∈ B, the equality holds if σ = σ. Note that in this case, σ ˜ must not satisfy any restriction on the boundary ∂B of the body, in order to be an admissible stress state for the composite. Next, we consider the traction problem for a composite. In this case, S1 = ∅ and S2 = ∂B and the principle of minimum potential energy takes the following simplified form. Principle of minimum potential energy for the traction problem for a composite with null body forces. Let A denote the set of all kinematically admissible states for the composite, and let Φ be the functional on A defined by Z ˜ s · uda, (2.6.42) Φ(s) = Uc (ε) − ∂B

for every s = [u, ε, σ] ∈ A. Further, let s be a solution of the traction problem for the composite. Then Φ(s) ≤ Φ(˜ s) (2.6.43)

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for every s˜ ∈ A and the equality holds only if s = s˜,modulo a rigid displacement. Note that in this case, u ˜ must not satisfy any restriction on the boundary ∂B of the body, in order to be an admissible displacement field for the composite! Analogously, we have the Principle of minimum complementary energy for the traction problem for a composite with null body forces. Let B denote the set of all statistically admissible stress fields for the composite and let Ψ be the functional on B defined by Z Z 1 1 σ · k2 σdv, (2.6.44) σ · k1 σdv + Ψ(σ) = Uk (σ) = 2 B2 2 B1

for every σ ∈ B. Let σ be the stress field corresponding to a solution of the traction problem for the composite. Then Ψ(σ) ≤ Ψ(˜ σ ), or equivalently Uk (σ) ≤ Uk (˜ σ ),

(2.6.45)

for every σ ˜ ∈ B, and the equality holds only if σ = σ ˜. The notions and results which will be presented in this last part of the Section are essential in the mechanics of macroscopically homogeneous composites. Let E be a symmetric constant tensor. We say that ε is a kinematically admissible strain field corresponding to E if T

ε=

(∇u + ∇u ) on Bα , α = 0, 1, ..., N, 2

and if u is a kinematically admissible displacement field which satisfies the homogeneous boundary condition u = Ex on ∂B. (2.6.46) Let Σ be a symmetric constant tensor. We say that σ is a statically admissible stress field corresponding to Σ if σ is a statistically admissible stress field which satisfies the homogeneous boundary condition sn = σn = Σn on ∂B.

(2.6.47)

Let us consider now an integrable scalar, vector or tensor field f given on B. The mean value of f , denoted by f¯, is defined by the equation Z 1 ¯ f dv, (2.6.48) f= v B

where v is the volume of the domain B.

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Taking into account the mean strain theorem for composites and the particular form of the homogeneous boundary condition (2.6.46) for the displacement, it is easy to see that the following is true: Mean value theorem for admissible strain. Let ε be a kinematically admissible strain field corresponding to E. Then, the mean value ε¯ of ε is just E Z 1 εdv = E. (2.6.49) ε¯ = v B

Similarly, taking into account the mean stress theorem for composites and the particular form of the homogeneous boundary condition (2.6.47) for the traction, it is easy to prove the following: Mean value theorem for admissible stress. Let σ be a statistically admissible stress field corresponding to Σ. Then the mean value σ ¯ of σ is just Σ; i.e. Z 1 σdv = Σ. (2.6.50) σ ¯= v B

Note that the above given properties are valid due to the particular homogeneous form of the corresponding boundary conditions and are true for any materials. Moreover, in the same conditions, the following two Hill and Mandel lemmas can be proved. The first Hill-Mandel lemma. If ε is a kinematically admissible strain field corresponding to E and σ is an admissible self-equilibrated stress field; i.e. div σ = 0 on Bα , α = 0, 1, ..., N , then the mean value σ · ε of the product σ · ε is ¯ i.e. equal to the products σ ¯ · ε¯ of the mean values σ ˜ and ε¯ = E;

σ·ε=σ ¯ · ε¯ = σ ¯ · E.

(2.6.51) T

) on Bα , α = The second Hill-Mandel lemma. If ε = (∇u+∇u 2 0, 1, ..., N is an admissible strain field corresponding to the admissible displacement field u and σ is a statically admissible self-equilibrated stress field corresponding ¯ · ε¯ to Σ , then the mean value σ · ε of the product σ · ε is equal to the product σ of the mean values σ ¯ = Σ and ε¯; i.e.

σ·ε=σ ¯ · ε¯ = Σ · ε¯. The reader is left to prove the above lemmas! Generally since the mean value of a product is not equal to the product of mean values, the two lemmas are not true, if the boundary conditions are not homogeneous. In exchange, if this requirement of homogeneity is fulfilled, the lemmas are valid for any material. In this sense, the above given properties do not reflect any material property, since they are true for both homogeneous, as well as for heterogeneous or piecewise homogeneous bodies.

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Let us now consider a point x of the composite material B and let us denote by Bx an arbitrary regular subdomain of B, containing x in its interior. The volume of Bx will be denoted by vx . If f is an integrable field given on B, its mean value on Bx , denoted by hf ; Bx i, is defined as follows (see Figure 2.3):

Figure 2.3: An arbitrary regular subdomain Bx of B.

hf ; Bx i =

1 vx

Z

f (x + ζ)dv(ζ),

(2.6.52)

Bx

Generally, the mean value of f corresponding to x and Bx , depends on x as well as on Bx . Let us suppose that the body B is homogeneous, characterized by its constant elasticity tensor c. In this case, the unique solution corresponding to the homogeneous displacement boundary condition (2.6.46) is u(x) = Ex, ε(x) = E, σ(x) = cε(x) = cE on B,

(2.6.53)

That is, the strain and the stress are constant fields on B. Consequently, for any point x and for any subdomain Bx , containing x, we have hε;Bx i = E = const., hσ;Bx i = cE = const., hσ · ε;Bx i = hσ;Bx i · hε;Bx i = E · cE = const.

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(2.6.54)

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Similarly, the unique solution corresponding to the homogeneous traction boundary condition (2.6.47) is u(x) = kΣx + w(x), ε(x) = kΣ, σ(x) = Σ on B,

(2.6.55)

where w(x) is an arbitrary rigid displacement. Again, the stress and strain are constant fields on B, and for any point x and for any subdomain Bx , we get hε; Bx i = kΣ = const., hσ; Bx i = Σ = const., hσ · ε;Bx i = hσ;Bx i · hε;Bx i = Σ · kΣ = const..

(2.6.56)

There are two reasons for which the mean value properties (2.6.54) and (2.6.56) are true (i) the corresponding boundary conditions are homogeneous, (ii) the body itself is homogeneous. In this sense, the obtained mean value properties reflects the material properties of the body, assumed to be linearly elastic and homogeneous. As we shall see later on, a macroscopically homogeneous elastic composite is a heterogeneous elastic body, characterized by a representative volume element, for which the mean value properties (2.6.53) and (2.6.56), are true, at least approximately, for any x ∈B, if the assumed boundary conditions are homogeneous. In Chapter 4, we shall analyze in detail the way in which a macroscopically homogeneous composite material can be replaced by an equivalent homogeneous body. At the end part of this Section, we present a theorem by Eshelby, which is very useful in the mechanics of composite materials. Eshelby’s energetical theorem. Let us assume a homogeneous body occupying the domain B and having the elasticity c◦ . Let us also assume that the subdomain D of B is replaced by an inclusion having constant elasticity c 1 . The inclusion and the body are firmly connected together along their common boundary Γ. Let us denote by [u◦ , ε◦ , σ ◦ ] and by [u, ε, σ] the elastic states in the homogeneous and heterogeneous body, respectively. First, we assume that on the boundary ∂B of the two bodies the same displacement condition is given; i.e. b and u = u b on ∂B. u◦ = u

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(2.6.57)

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Let [u◦ , ε◦ , σ ◦ ] and [u, ε, σ] be the solutions of these boundary value problems and let us denote by U ◦ = U ◦ (ε◦ ) and U = U (ε), the strain energies in the homogeneous and in the piece-wise homogeneous body, respectively.Then Z 1 ◦ ◦ (σ · ε◦ − σ ◦ · ε)dv U (ε) = U (ε )+ 2 D Z 1 (σn · u◦ − σ ◦ n · u)dv. (2.6.58) = U ◦ (ε◦ )+ 2 Γ

Next, we assume that on the boundary ∂B of the two bodies, the same traction condition is given; i.e.

◦

◦

◦

s and sn = σn = b s on ∂B. s◦n = σ ◦ n = b

(2.6.59)

Let [u , ε , σ ] and [u, ε, σ] be the solutions of the boundary value problem, and let us denote by U ◦ = U ◦ (ε) and U = U (ε) the strain energies in the homogeneous and in the piece-wise homogeneous body, respectively. Then Z 1 ◦ ◦ (σ ◦ · ε − σ · ε◦ )dv U (ε) = U (ε ) + 2 D Z 1 ◦ ◦ (σ ◦ n · u − σn · u◦ )da. (2.6.60) = U (ε )+ 2 Γ

First, we analyze the displacement problem. Since σ ◦ and σ are self-equilibrated; i.e. div σ ◦ = div σ = 0 in B − Γ, satisfy null-jump condition; i.e. [σ ◦ n] = [σn] = 0 on Γ, and since u◦ and u satisfy also null-jump conditions; i.e. [u◦ ] = [u] = 0 on Γ, we can apply the integral theorem (2.6.11). In this way, taking into account the boundary conditions (2.6.57), we get Z Z ◦ σ · (ε − ε )dv = (u − u◦ ) · σnda = 0, B ∂B Z Z ◦ ◦ ◦ ◦ σ · (ε − ε )dv = (u − u ) · σ nda = 0. B

∂B

Hence, Z Z Z Z σ ◦ · ε◦ dv. σ ◦ · εdv = σ · ε◦ dv and σ · εdv = B

B

We recall now that ◦

◦ ◦

σ = c ε in B and σ = cε =



c◦ ε c1 ε

in in

B−D D.

Consequently, from (2.6.61), it results Z Z 1 1 σ ◦ · ε◦ dv σ · εdv − U (ε) − U ◦ (ε◦ ) = 2 B 2 B Z Z 1 1 ◦ σ ◦ · εdv. σ · ε dv − = 2 B 2 B

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(2.6.61)

B

B

(2.6.62)

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CHAPTER 2. ELEMENTS OF LINEAR ELASTOSTATICS Now, according to (2.6.62), we get Z Z Z σ · ε◦ dv = σ · ε◦ dv + σ · ε◦ dv B B−D D Z Z = ε◦ · c◦ εdv + σ · ε◦ dv, B−D

D

and Z

◦

B

σ · εdv

= =

Z

Z

◦

B−D

σ · εdv +

Z

◦ ◦

B−D

ε · c ε dv +

σ ◦ · εdv

D

Z

D

σ ◦ · εdv.

Using the above relations and taking into account the symmetry of c◦ , we successively get Z Z 1 1 (σn · u◦ − σ ◦ n · u)da. (σ · ε◦ − σ ◦ · ε)dv = U (ε)−U ◦ (ε◦ ) = 2 Γ 2 D

In order to obtain the last equality, we have used for the domain D the integral theorem (2.6.11) and the fact that the stress fields σ ◦ and σ are self-equilibrated. The obtained results show that the equation (2.6.58) is true. Now, we analyze the traction problem. In the assumed conditions, we can again use the integral theorem (2.6.11). In this way, taking into account the boundary condition (2.6.59), we obtain Z Z ◦ ◦ (σ − σ ) · εdv = (σn − σ n) · uda = 0 B

and

Z Hence, Z

B

B

∂B

(σ − σ ◦ ) · ε◦ dv =

σ · εdv =

Z

B

Z

σ ◦ · εdv and

◦

∂B

Z

B

◦

(σn − σ n) · u da = 0.

σ · εdv =

Z

B

σ ◦ · ε◦ dv.

Thus, from (2.6.63), we obtain U (ε)−U ◦ (ε◦ )

=

=

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Z Z 1 1 σ ◦ · ε◦ dv σ · εdv − 2 B 2 B Z Z 1 1 ◦ σ · ε◦ dv. σ · εdv − 2 B 2 B

(2.6.63)

93

2.7. ESHELBY’S INCLUSION PROBLEM Also, from (2.6.63), it results Z Z Z ◦ ◦ σ · εdv = σ · εdv + σ ◦ · εdv B B−D D Z Z ◦ = ε · c εdv + σ ◦ · εdv, B−D

D

and Z

B

◦

σ · ε dv

= =

Z

Z

◦

B−D

σ · ε dv + ◦

B−D

◦

Z

ε · c εdv +

D

Z

σ · ε◦ dv D

σ · ε◦ dv.

Now the symmetry of c◦ implies U (ε)−U ◦ (ε◦ )

=

=

Z 1 (σ ◦ · ε − σ · ε◦ )dv 2 D Z 1 (u · σ ◦ n − u◦ · σn)da, 2 Γ

and the equation is proved. Eshelby’s energetic theorem shows that if the solution of one of the boundary value problems for the homogeneous body is known, in order to obtain the strain energy stored in the heterogeneous, piece-wise homogeneous body, it is sufficient to know the strain and stress only in the inclusion. Moreover, it is sufficient to know only the displacement u and traction sn = σn on the boundary Γ of the inclusion. This fact is very useful in the theory of macroscopically homogeneous composite materials and is used to evaluate the overall elastic module of the material, taking into account various simplified models of the composite.

2.7

Eshelby’s inclusion problem

Eshelby’s inclusion problem or transformation problem is one of the most wonderful problems encountered in linear elastostatics. The results obtained by Eshelby are beautiful and have unexpected applications in the theory of macroscopically homogeneous composites, as well as in the theory of dislocations. Further, we present the inclusion or transformation problem. A regular region, the inclusion, in an infinite homogeneous elastic medium, the matrix, undergoes a change of shape and size which in the absence of its surrounding, the matrix, would be an arbitrary homogeneous strain. What is the elastic state of the inclusion and of the matrix? Eshelby has obtained the answer with the help of a set of imaginary cutting, straining and welding operations.

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We denote by D the domain occupied by the inclusion and let Γ = ∂D be the boundary of D. First, we cut out the region D which is to be transformed and remove it from the matrix. Next, we allow the unconstrained homogeneous transformation or deformation of D to take place. Afterward, we apply on the boundary Γ of D the surface tractions chosen so as to restore the domain occupied by the inclusion to its original form. Then, we put the matrix back into the hole and rejoin the material across the cut. At this moment, the stress is vanishing in the matrix and has a constant value in the inclusion. The applied surface tractions were introduced as a layer of body force, applied over the interface Γ between matrix and inclusion. Finally, to obtain the elastic state, that is the solution, this layer will be removed by applying on Γ an equal and opposite layer of body force. The additional elastic field obtained in this way will be found by using the displacement field corresponding to the applied layer of body force distributed on Γ and described by a surface force density. In the following, we shall name stress-free strain the uniform (constant) transformation or deformation εt to which the inclusion would undergo in the absence of the matrix. The main problem is to find the constrained strain εc in the inclusion and in the matrix, when the inclusion transforms while it is embedded in the matrix. Following Eshelby, we carry out the steps outlined above: (i) We remove the inclusion and allow it to undergo the stress-free strain ε t . Let
(2.7.1) σ t = λ trεt 1+2µεt

be the (constant stress) corresponding to εt , according to the elastic constitutive equation. Not that at this stage the stress in the matrix is vanishing. (ii) We apply now the surface traction stn = −σ t n on the boundary of the transformed inclusion. This brings it back to the shape and size it had before the transformation. (iii) We put back the stressed inclusion in the matrix and we reweld across Γ. At this moment, the applied surface traction has become a layer of body forces spread over Γ and completely characterized by the surface density f = σ t n on Γ, n representing the unit outward normal to the boundary Γ of the inclusion. (iv) Finally, we relax the above body forces, or, what represents the same thing, we apply an additional surface distribution f = σ t n on the boundary Γ. At the moment, this whole body is free of external forces, but in a state of self-stress, produced by the transformation εt of the inclusion. According to the relation (2.5.9), the displacement field uc (x) imposed on the material in the final stage (iv) is Z c G (x − x0 ) f (x0 ) da (x0 ) with f (x0 ) = σ t n (x0 ) on Γ. (2.7.2) u (x) = Γ

According to (2.5.34) and (2.5.35), the components of Green’s tensor function

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2.7. ESHELBY’S INCLUSION PROBLEM G (x − x0 ) are

∂ 2 kx − x0 k 1 ∂ 2 kx − x0 k 1 (2.7.3) − δij 16π (1 − ν) ∂xi ∂xj ∂xm ∂xm 8πµ (
) (xi − x0i ) xj − x0j 1 1 0 . (3 − 4v) δij + Gij (x − x ) = 2 16πµ (1 − ν) kx − x0 k kx − x0 k (2.7.4) Hence, from (2.7.2) and (2.7.3) it results Gij (x − x0 ) =

uci (x) =

1 4πµ

Z 

δij

Γ

∂ 2 kx − x0 k 1 ∂ 2 kx − x0 k − 4 (1 − ν) ∂xi ∂xj ∂xm ∂xm



t n0k da0 . σjk

(2.7.5)

It is convenable to take the configuration of the material at the end of the stage (ii) as reference configuration; i.e. as the state of zero displacement. Indeed, at this point, the stress and strain in the matrix are zero, and the inclusion, though not stress free, has just the geometrical form which it had before the transformation occurred. If this convention was made, uc (x) given by (2.7.5) is the actual displacement in the matrix and in the inclusion. Consequently, the strain εc in the matrix and in the inclusion is given by  1 T (2.7.6) ∇uc + (∇uc ) . εc = 2

According to the elastic law, the stress σ c in the matrix will be σ c = λ (trεc ) 1+2µεc .

(2.7.7)

On the other hand, the inclusion had a stress state σ t even before the stage (iv) . Consequently, the stress σ i in the inclusion will ultimately be

(2.7.8) σ i = σ c −σ t = λtr εc −εt 1 + 2µ εc −εt ,

where σ c is the stress derived according to the elastic law from the strain εc in the inclusion. Now, by using Gauss-Ostrogradski theorem and the equivalence of ∂/∂x i and −∂/∂x0i when acting on kx − x0 k, after some simple computations from (2.7.5), we get 1 t 1 t σ Φ,k (2.7.9) σij Ψ,ijk − uci (x) = 4πµ ik 16πµ (1 − ν)

where

Φ (x) =

Z

D

dv 0 and Ψ (x) = kx − x0 k

Z

D

kx − x0 k dv 0

(2.7.10)

are the well known Newtonian potential and the biharmonic potential of attractive matter of unit mass density filling the domain D bounded by Γ.

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96

CHAPTER 2. ELEMENTS OF LINEAR ELASTOSTATICS It is easy to see that ∆Ψ = 2Φ

and ∆Φ =



−4π inside D 0 outside D



, ∆∆Ψ =



(2.7.11) −8π inside D 0 outside D



.

(2.7.12)

From (2.7.6) and (2.7.9) for the strain εc , we get εcil =


1 1 t t t Φ,ki . Φ,kl + σlk σik Ψ,ijkl − σjk 8πµ 16πµ (1 − ν)

(2.7.13)

Taking into account the stress-strain relation (2.7.1), we can express ε c through εt as εcil = −

1 1 t ν εt Ψ,ijkl . (2.7.14) (ε Φ,kl + εtlk Φ,ki ) + εt Φ,il − 8π (1 − ν) jk 4π ik 4π (1 − ν) mm

As an interesting result, we show now that we can find the strain and hence the stress, just outside the inclusion, from their values at an adjacent point just inside, without having to solve the exterior problem at all. To prove this result, we recall a property due to H. Poincar´e (1899) from the theory of Newtonian potential. The second derivatives of a potential function U satisfying ∆U = −4πρ,

(2.7.15)

[U,ij ] = −4π [ρ] ni uj

(2.7.16)

undergo a jump on crossing a singular surface (as Γ) across which the density ρ has a jump [ρ] . In (2.7.16), n is the outward unit normal to Γ (in our inclusion problem). From this general result and from the equation (2.7.12) satisfied by Φ, we find for the jump of Φ,ij across Γ [Φ,ij ] = 4πni uj ,

(2.7.17)

since [ρ] = −1 on Γ, as this can be seen, examining the equation (2.7.12). We observe now, according to (2.7.11)   1 ∆Ψ,ij = 2Φ,ij = −4π − Φ,ij . 2π
1 Φ,ij . Thus, Accordingly, Ψ,ij is the Newtonian potential for the density − 2π the Poincar´e formula (2.7.16) states that   1 [Ψ,ijkl ] = −4π − Φ,ij nk nl on Γ. 2π

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97

2.7. ESHELBY’S INCLUSION PROBLEM

Hence, from (2.7.17), we obtain the following equation giving the jumps of the fourth order derivatives of the biharmonic potential Ψ [Ψ,ijkl ] = 8πni nj nk nl on Γ.

(2.7.18)

Now, from (2.7.14), (2.7.17) and (2.7.18), we can deduce the jump of the strain εc across the inclusion boundary Γ [εcil ] = −

1 t ν ε ni nj nk nl . εtmm ni nl − εtik nk nl − εtlk nk ni + 1 − ν jk (1 − ν)

(2.7.19)

This equation proves the result mentioned above. In order to get the jump of the stress σ c across Γ, we use the relation (2.7.14) and the constitutive equation (2.7.7) to get σilc = −


1 ν t t t σik Φ,kl + σlk Φ,ki σjk Φ,jk δil − 4π 4π (1 − ν) +

1 σ t Ψ,ijkl . 8π (1 − ν) jk

(2.7.20)

Now, with (2.7.17) and (2.7.18), it results [σilc ] = −


1 ν t t t σ t ni nj nk nl on Γ. σjk nj nk δil − σik nk nl+ σlk nk ni + 1 − ν jk (1 − ν)

Using the surface force density f = σ t n, the above relation becomes [σilc ] = −

1 ν fj nj ni nl on Γ. fk nk δil + (fi nl + fl ni ) + 1−ν 1−ν

Now, after simple computations, we find that across the singular surface Γ the traction scn = σ c n satisfies the following jump condition: [σ c n] + f = 0 or [σ c n] + σ t n = 0 on Γ.

(2.7.21)

In this way, we have proven for our particular problem, the main property (iv) of the displacement field defined by the equation (2.5.9) from the Section 2.5. Using the second expression (2.7.4) of Green’s tensor, Eshelby was able to obtain a new, useful expression for the displacement field uc . Inserting (2.7.4) in (2.7.2) and using the Gauss-Ostrogradski theorem to convert to a volume integral, we find Z 1 1 t c fijk (l) dv σ ui (x) = 16πµ (1 − ν) jk D r2 Z 1 1 gijk (l) dv, (2.7.22) εt = 8π (1 − ν) jk D r2

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CHAPTER 2. ELEMENTS OF LINEAR ELASTOSTATICS

where r = kx − x0 k and l = (l1 , l2 , l3 ) are the length and the direction of the line drawn from the volume element dv at x0 towards the point of observation x, and fijk = (1 − 2ν) (δij lk + δik lj ) − δjk li + 3li lj lk ,

(2.7.23)

gijk = (1 − 2ν) (δij lk + δik lj − δki li ) + 3li lj lk .

(2.7.24)

Following Eshelby, we shall analyze in detail the case of an ellipsoidal inclusion. To discuss the elastic field inside the inclusion, it is convenient to redefine l i in (2.7.23) and (2.7.24) to be the direction cosines of a line drawn from the point of observation x towards the volume element dv. Obviously this involves changing the sign of the integrals in (2.7.22). To evaluate the integrals in (2.7.22), we use spherical coordinates centered at the point of observation x. In this way, from (2.7.22), we get Z 8πµ(1 − ν)uci (x) = −εtjk r(l)dw(l)gijk (l), (2.7.25) Σ

where Σ is the unit sphere centered at x, r (l) is the distance from x to the boundary Γ of the inclusion, in the direction l, and dω (l) is the surface element of Σ in the same direction. For instance, we have l1 = sin θ cos ϕ, l2 = sin θ sin ϕ, l3 = cos θ, dω (l) = sin θdθdϕ with θ ∈ [0, π] and ϕ ∈ [0, 2π] . For an ellipsoidal inclusion, having semiaxis a1 , a2 , a3 , r (l) is the positive root of the equation: 2

2

2 (x3 + rl3 ) (x2 + rl2 ) (x1 + rl1 ) = 1, + + a23 a22 a21

where {x1 , x2 , x3 } are the Cartesian coordinates of the point of observation x, in a system of coordinates determined by the center and by the symmetry axes of the ellipsoid. From the above equation, we get: s e f2 f (2.7.26) + , r (l) = − + g g2 g

where

Copyright © 2004 by Chapman & Hall/CRC

g

=

f

=

e

=

l2 l2 l12 + 22 + 32 , 2 a3 a2 a1 l3 x 3 l2 x 2 l1 x 1 + 2 + 2 , a3 a2 a21 2 2 x2 x x 1 − 21 − 22 − 23 . a3 a2 a1

(2.7.27)

99

2.7. ESHELBY’S INCLUSION PROBLEM

The sign of the square root is correct, sinceqe > 0 if x is within the ellipsoid. 2 We observe now that we can omit the term fg2 + ge when (2.7.26) is inserted in (2.7.25), since it is even in l, while gijk (l) is odd. Thus (2.7.25) becomes Z λm gijk dw (l) , (2.7.28) 8πµ (1 − ν) uci (x) = xm etjk g Σ

where λ1 =

l3 l2 l1 , λ2 = 2 , λ3 = 2 . 2 a3 a2 a1

(2.7.29)

Since the integral in (2.7.28) does not depend on x, the displacement field uc (x) is a linear function on x. Consequently, the corresponding strain ε c (x) is constant in the inclusion and depends only on the shape of the ellipsoid, that is on its semi-axes a1 , a2 , and a3 . This homogeneity or uniformity of the strain field in an ellipsoidal inclusion is one of the most important result discovered by Eshelby in the transformed inclusion problem. Later, we shall see some implications of this result in the mechanics of macroscopically homogeneous body. Returning to (2.7.28), we obtain the components of the strain εc (x) in the ellipsoid Z λi gljk + λl gijk 1 dω (l) . (2.7.30) εtjk εcil (x) = g 16π (1 − ν) Σ

For later use, it is convenient to write the relations (2.7.30) between the constrained and stress-free strains in the inclusion in the form εcil = Silmn εtmn .

(2.7.31)

From symmetry considerations, it results that the coefficients coupling an extension and shear (S1112 , S1123 , S2311 , ...) or one shear to another one (S1223 , ...) are zero. By simple examination, from (2.7.30) and (2.7.31), we find S1111 S1122

= =

S1212

=

where Q=

and I a1 =

Z

Σ

Qa21 Ia1 a1 + RIa1 , Qa22 Ia1 a2 − RIa1 ,
1 1 Q a21 + a22 Ia1 a2 + R (Ia1 + Ia2 ) , 2 2

1 1 1 − 2ν 3 , , Q+R= , R= 4π 8π (1 − ν) 3 8π (1 − ν)

l12 dω (l) , I a1 a1 = a21 g

Copyright © 2004 by Chapman & Hall/CRC

Z

Σ

l14 dω (l) , I a1 a2 = a41 g

Z

Σ

l12 l22 dω (l) . a21 a22 g

(2.7.32)

(2.7.33)

(2.7.34)

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CHAPTER 2. ELEMENTS OF LINEAR ELASTOSTATICS

The remaining coefficients are found by simultaneous interchange of (1, 2, 3), (a1 , a2 , a3 ) and (l1 , l2 , l3 ) . Generally, the integrals Ia1 , Ia1 a1 , Ia1 a2 , ... cannot be calculated by elementary methods. Their values can be expressed only using elliptic integrals of the first and second kinds. However, when we use the expression (2.7.27) of g and the relation l12 +l22 +l32 = 1, it is easy to see that the following equations are true: Ia1 + Ia2 + Ia3 = 4π, I a1 a1 + I a1 a2 + I a1 a3 =

4π , 3a21

(2.7.35)

a21 Ia1 a1 + a22 Ia1 a2 + a23 Ia1 a3 = Ia . There exist three particular situations in which all integrals Ia1 , Ia1 a1 , ... can be expressed in an elementary way: this can be done for an oblate spheroid (a1 = a2 > a3 ), a prolate spheroid (a2 = a3 < a1 ) and a sphere (a1 = a2 = a3 ) . The results obtained in these cases can be used to evaluate the overall elastic moduli of various reinforced macroscopically homogeneous composite materials. Many essential results of this type are presented and analyzed in Christensen’s fundamental monograph. In Section 4.5, we shall study only the case of spherical inclusions. If the inclusion in Eshelby’s problem is a sphere of radius a, according to the obvious symmetry, from (2.7.35), we get I a1 = I a2 = I a3 =

4π . 3

(2.7.36)

Also, since (2.7.37)

I a1 a2 = I a1 a3 the relation (2.7.35)2 gives Ia1 a1 + 2Ia1 a2 =

4π . 3a2

(2.7.38)

Obviously, according to the equation (2.7.27) now g=

1 . a2

Consequently, from (2.7.34), we get Z 1 l2 dω (l) . I a1 a1 = 2 a Σ 1

Copyright © 2004 by Chapman & Hall/CRC

(2.7.39)

(2.7.40)

101

2.7. ESHELBY’S INCLUSION PROBLEM This integral can be evaluated in an elementary way and we obtain I a1 a1 = I a2 a2 = I a3 a3 =

4π . 5a2

(2.7.41)

4π . 15a2

(2.7.42)

Now, from (2.7.38), we get I a1 a2 = I a2 a3 = I a3 a1 =

Now, returning to the relations (2.7.32) and (2.7.33), we can express Eshelby’s coefficients Sijkl for a spherical inclusion; we get 7 − 5ν , 15(1 − ν)

S1111

=

S2222 = S3333 =

S1122

=

S1133 = S2211 = S3311 = S2233 = S3322 =

S1212

=

S2112 = S2121 = S2112 = ... =

4 − 5ν . 15(1 − ν)

−1 + 5ν , (2.7.43) 15(1 − ν)

In this way, from (2.7.31) and (2.7.43), for the constrained strain εc inside the transformed sphere, we get the following expressions in terms of the stress-free strain εt : θc = αθ t and ec = βet (2.7.44) where α=

and

2 4 − 5ν 11+ν , , β= 15 1 − ν 31−ν

(2.7.45)

θt = trεt , θc = trεc , et and ec being the deviatoric parts of εt and εc , respectively; i.e. 1 et = εt − θt 1, 3

1 ec = εc − θc 1. 3 We recall Eshelby’s mean result: the strain and the stress in the transformed ellipsoidal inclusion are uniform. Using this result, Eshelby has succeeded in solving the following inhomogeneity problem playing an important role in the mechanics of macroscopically homogeneous composite materials. An ellipsoidal region in an infinite homogeneous elastic medium has elastic constants differing from those of the remainder. How is an applied stress, which is uniform at large distances, disturbed by this inhomogeneity?

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CHAPTER 2. ELEMENTS OF LINEAR ELASTOSTATICS

This inhomogeneity problem was solved by Eshelby in the way outlined at the beginning of this Section. On the elastic field εc due to an ellipsoidal inclusion with arbitrary εt , we superpose a uniform strain εa . Now the deformation of the boundary of the inclusion is characterized by the strain εc + εa . Since the part εt of this strain is not associated with any stress (see Equation (2.7.8)), the uniform strain ε in the inclusion is given by the stress-strain relation not for εc +εa but rather for εc +εa − εt . Hence, the strain ε in the inclusion is ε = εc +εa ,

(2.7.46)

but the stress in it is

or equivalently



σ = λtr εc +εa − εt 1 + 2µ εc +εa − εt ,

where



σ = k θc + θa − θt , s = 2µ ec + ea − et σ=

(2.7.47)

(2.7.48)

1 trσ, θ a = trεa , 3

and

1 s = σ−σ1, ea = εa − θa 1, 3 are the deviatoric parts of the stress σ and of the strain εa , respectively, and k = λ + 2µ 3 is the bulk modulus. Now we take an ellipsoid of the same shape and size as the untransformed ellipsoid, but made of an isotropic material with elastic constants λ1 , µ1 , k1 = λ1 + 2µ1 3 different from those of the matrix and inclusion. We subject this ellipsoid to the uniform strain (2.7.41). If this treatment develops the uniform stress (2.7.46), it may be used to replace the initial (transformed) inclusion with continuity of displacement and surface traction across the interface Γ. Since the stress and strain are constant in the initial (transformed) ellipsoidal inclusion, we can always ensure that the needed stress is developed by choosing suitable λ1 and µ1 . For this purpose, it is only necessary that they should satisfy the conditions
k1 (θc + θa ) = k θc + θa − θt ,
(2.7.49) µ1 (ec + ea ) = µ1 ec + ea − et .

Actually, it is the uniform applied strain εa and the elastic constants λ1 , µ1 , k1 of the ellipsoidal inhomogeneity which are prescribed. Hence, the relations (2.7.49) represent the equations which have to be solved for εt in terms of εa , λ1 , µ1 , λ, µ after eliminating εc with the help of Eshelby’s relations (2.7.31). We observe that the system (2.7.49) is not as simple as it appears to be, since each of the θ c , ec depend simultaneously on θ t and et .

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2.7. ESHELBY’S INCLUSION PROBLEM

103

However, for the shear components the solution is immediate, since as we already know, the coefficients Sijkl do not couple different shears. We get et12 =

µ − µ1 ea , ... . 2 (µ1 − µ) S1212 + µ 12

(2.7.50)

In order to obtain the components εt11 , εt22 and εt33 , we first observe that (2.7.49) can be written in the following equivalent form:

λ1 1tr (εc + εa ) + 2µ1 (εc + εa ) = λ1tr εc +εa − εt + 2µ εc +εa − εt , or

(λ1 − λ) 1trεc + 2 (µ1 − µ) εc + λ1trεt + 2µεt = (λ1 − λ) 1trεa + 2 (µ1 − µ) εa . Now, taking into account Eshelby’s relations (2.7.31) and the symmetry properties of the coefficients Sijkl , we obtain the following three simultaneous equations which must be solved in order to get εt11 , εt22 and εt33 (λ1 − λ) Smmpq εtpq + 2 (µ1 − µ) Sijpq εtpq + λεtmm + 2µεtij

(2.7.51)

= (λ1 − λ) εamm + 2 (µ1 − µ) εtij , for i, j, = 11, 22, 33. Only εt11 , εt22 and εt33 appear in the pq summation since there does not exist a coupling between an extension and a shear. In this way, Eshelby’s ellipsoidal inhomogeneity problem is solved. From the derivation, it results that εt found in the above way is a stressfree strain of certain transformed inclusion which, with the given applied strain εa , could replace the inhomogeneity without altering the stress or displacement anywhere. Eshelby calls this imaginary transformed inclusion the “equivalent inclusion”. Outside the inclusion, the elastic state of the matrix, that is [u, ε, σ] is the sum of the applied field [ua , εa , σ a ] and the field [uc , εc , σ c ] of the equivalent inclusion. Actually, this field is due to the assumed ellipsoidal inhomogeneity, and the field [uc , εc , σ c ] measures the perturbation of the applied field [ua , εa , σ a ] by the inhomogeneity and may be found from εt , using the method presented in the first part of this Section. In the simplest case of the spherical inhomogeneity with elastic constants k1 , µ1 , in an applied field εa , according to (2.7.44) and (2.7.47), the equivalent strain εt is given by the equations θt = Aθ a , e = Bea with A=

Copyright © 2004 by Chapman & Hall/CRC

(µ1 − µ) k1 − k , ,B = (µ − µ1 ) β − µ (k − k1 ) α − k

(2.7.52)

(2.7.53)

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CHAPTER 2. ELEMENTS OF LINEAR ELASTOSTATICS

the coefficients α and β having the expressions (2.7.45). Finally, using (2.7.44) and (2.7.52), for θ = θ c + θa and e = ec +ea , that is for the strain ε = εc + εa in the spherical inclusion, we get the following expressions, in terms of the applied strain εa = θa 1 + ea θ=

2.8

µ k ea . θa , e = (µ1 − µ) β + µ (k1 − k) α + k

(2.7.54)

Problems

P2.1 Show that the infinitesimal strain field ε (x) corresponding to a rigid displacement field u (x) = u0 + ω 0 ×x, u0 = const., ω 0 = const., vanishes on B. P2.2 Show that if the infinitesimal strain field ε (x) , corresponding to a displacement field u (x) is identically zero on B, then u (x) is an infinitesimal rigid displacement field. P2.3 Using P2.2, prove Kirchhoff’s theorem given at the beginning of the Section 2.1. P2.4 Prove Piola’s theorem: Let b be an integrable vector field on B and let b s be an integrable vector field on ∂B. Then Z Z b s · wda + b · wdv = 0, ∂B

B

for any rigid displacement field w on B, if and only if Z Z Z Z b sda + bdv = 0 and x×b sda + x × bdv = 0. ∂B

B

∂B

B

P2.5 Let w (x) = u0 + ω 0 × x be a rigid displacement on B and let x1 , x2 , x3 three non-collinear points of B. Assuming that w (x1 ) = w (x2 ) = w (x3 ) = 0, show that w (x) vanishes identically on B. P2.6 Prove that the elasticity tensor or Hooke’s tensor c is symmetric if and only if its components satisfy the relation

cijkl = cklij , i, j, k, l = 1, 2, 3. b 4 is positive definite, it is nonP2.7 Prove that if the elasticity tensor c ∈L singular. b 4 is positive definite, its inverse P2.8 Prove that if the elasticity tensor c ∈L −1 k = c there exists, is symmetric and positive definite. P2.9 Using the relation (2.2.1) defining a symmetry transformation Q, show that if Q is a symmetry transformation, it satisfies the equation (2.2.3). Conversely, show that if this equation is satisfied for every symmetric tensor, then Q is a symmetry transformation.

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105

2.8. PROBLEMS

P2.10 Let c be the elasticity tensor of an elastic body and let k = c−1 be the corresponding compliance tensor. Show that   Q (kσ) QT = k QσQT ,

for every Q ∈Sx and for every symmetric tensor σ. P2.11 Prove that e is an axis of symmetry at x ∈ B if and only if R (e, θ) ∈ Sx for some θ ∈ (0, 2π) ; i.e. one of the symmetry transformation of the material is a rotation about the axis spanned by e. P2.12 Prove that P (e1 , e2 ) is a plane of symmetry if and only if R (e3 , π) ∈ Sx . P2.13 Prove that P (e1 , e2 ) is a plane of symmetry if and only if one of the symmetry transformation of the material is a reflection on the plane P (e1 , e2 ) . P2.14 Prove that the set of transformations ±1, ± R (e1 , π) , ±R (e2 , π) , ±R (e3 , π) forms a proper subgroup of the orthogonal group. P2.15 Show that if ±1, ± R (e2 , π) , ±R (e3 , π) ∈ Sx than ±R (e1 , π) ∈ Sx . P2.16 Show that if an elastic material at any x ∈ B has two mutually orthogonal symmetry planes P (e2 , e3 ) and P (e3 , e1 ), it is orthotropic. P2.17 Prove the following second mean stress theorem: If the body B is homogeneous, the mean stress Σ corresponding to the elastic state [u, ε, σ] depends only on the boundary values of u and is given by  Z 1 unda . Σ= c v ∂B

P2.18 Prove the following second main strain theorem: If the body B is homogeneous, and [u, ε, σ] is an elastic state corresponding to the external force system [b, sn ] , then the corresponding mean strain E depends only on the external force system and is given by Z  Z 1 xsn da + xbdv . E= k v ∂B B

P2.19 Prove the principle of superposition for elastic states. P2.20 Prove the following generalized reciprocal theorem: Let [u, ε, σ] e ] be elastic states corresponding to external force systems [b, sn ] and ui, e ε, σ hand [e e e ] to u, e ε, σ b, e sn respectively, and let [u, ε, σ] correspond to the elasticity field c, [e the elasticity field e c = cT . Then Z Z Z Z Z Z e · udv = e · εdv, e e dv = e da + σ σ·e εdv = sn · uda + b b·u sn · u ∂B

B

∂B

B

where

e =e σ = cε and σ ce ε.

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B

B

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CHAPTER 2. ELEMENTS OF LINEAR ELASTOSTATICS

P2.21 Show that for the traction problem a necessary condition for the existence of a solution is that the given external forces [b,b s] be in equilibrium; i.e. that Z Z Z Z b sda + bdv = 0 and x×b sda + x × bdv = 0. ∂B

B

∂B

B

P2.22 Let us consider an isotropic material characterized by the constitutive equation (2.2.54). (a) Show that the strain ε and the stress σ have the same eigen directions. (b) Let us denote by ε1 , ε2 , ε3 the eigenvalues of the strain ε, corresponding to its eigen directions n1 , n2 , n3 . And let σ1 , σ2 , σ3 be the eigen values of the stress σ, corresponding to n1 , n2 , n3 , respectively. Express σ1 , σ2 , σ3 , called the principal stresses in terms of Lame’s coefficients λ, µ and of ε1 , ε2 , ε3 called the principal strains. P2.23 The Young’s modulus E and Poisson’s ratio ν of a stainless steel are given by the values E = 206GP a, ν = 0, 3. Find the values of λ, µ and k. P2.24 Prove that if the elasticity c of a material is a positive definite fourth order tensor, then the rigidity matrix [C] of the material is a positive definite 6 × 6 matrix. Is the converse true? P2.25 Prove that if [C] is the rigidity matrix of an orthotropic material, the −1 structure of the compliance matrix [S] = [C] has the form given in the relation (2.2.70). P2.26 Find the compliance matrix [S] for a material having a plane of symmetry. P2.27 Give the mechanical meaning of the inequalities (2.2.74) obtained by us assuming that the elasticity c of an orthotropic material is positive definite. P2.28 Show that the technical or engineering constants of an orthotropic relation satisfy the relations ν12 ν23 ν31 = ν21 ν32 ν13 ,   12 #   12  "   E2 E1 2 E3 2 E2 > 0, + ν32 ν13 − ν21 1 − ν13 1 − ν32 E1 E E E1 2 1 (  12   12 )  12   E2 E2 2 E3 2 E2 < ν21 < 1 − ν13 + 1 − ν32 ν32 ν13 E1 E1 E3 E1 (  12   12 )  12   E2 E E E2 3 2 2 2 . 1 − ν13 − 1 − ν32 < − ν32 ν13 E1 E1 E3 E1

P2.29 Show that the determinant inequality (2.2.76) for orthotropic materials reduces to ν < 1/2 for isotropic materials if 1 + ν 6= 0.

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107

2.8. PROBLEMS

P2.30 Show that the inequalities given in P2.28 reduce to the known bounds 0 < ν < 1/2 for an isotropic material. P2.31 Verify the equations (2.2.85). P2.32 Assuming c positive definite, find the restrictions which must be satisfied by the engineering coefficients of a transversally isotropic material. What kind of assumptions lead to the positiveness of Poisson’s ratios of a transversally isotropic material? P2.33 Let us consider an isotropic material. Show that the components c ijkl of the elasticity field c and the components kijkl of the compliance field k = c−1 , in any basis ei ej ek el are given by the following relations: cijkl = λδij δkl + µ (δik δjl + δil δjk ) ,

1+ν ν (δik δjl + δil δjk ) . δij δkl + 2E E P2.34 Show that the specific strain energy u (ε) of an isotropic material has the expression kijkl = −

u (ε) =

k 2 θ + µe · e, 2

where

1 θ = trε and e = ε − θ1. 3

Using the above results, show that u (ε) > 0 for any ε 6= 0,

if and only if

k > 0 and µ > 0.

P2.35 Formulate and prove the principle of minimum potential energy corresponding to the displacement problem and the traction problem, respectively. P2.36 Formulate and prove the principle of minimum complementary energy corresponding to the traction problem and the displacement problem, respectively. P2.37 Using the theorem of work expended, show that the inequalities (2.4.16) and (2.4.11) from the upper and lower bounds theorems can be written in the following equivalent forms:   Z 1 ε) , ε − kσ ∗ dv ≤ U < Uc (e σ∗ · e 2 B   Z 1 ∗ e ε dv ≤ U < Uk (σ ∗ ) . ε · σ − ce 2 B

P2.38 Using the principle of minimum potential energy, prove Kirchhoff’s uniqueness theorem for the mixed problem. P2.39 Let us consider the displacement field u (x) defined by the equation (2.5.4) in terms of Green’s tensor function G (x − x0 ) and of the constant vector P. Let σ (x) = c∇u (x) be the stress field corresponding to u (x) . (a) Show that div σ (x) = 0 for any x 6= 0 from the three-dimensional Euclidean space E.

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CHAPTER 2. ELEMENTS OF LINEAR ELASTOSTATICS



(b) Show that u (x) = O r−1 and σ (x) = O r−2 for r = kxk → 0 and for r → ∞. R (c) Show that Ση σnda = P for any spherical surface Ση centrad in x = 0 and having the radius η > 0. What is the meaning of this result? P2.40 The displacement field u (x) produced in an infinite, homogeneous, isotropic elastic medium, by a concentrated force P applied at the point x = 0 is given by the equation (2.5.37). (a) Show that the corresponding strain field ε (x) has the expression   3 1 (P · x) 1− (1 − 2ν) (xP + Px) − 2 (P · x) xx . ε (x) = r 16πµ (1 − ν) r 3

(b) Show that the corresponding stress field σ (x) has the expression given by the equation (2.5.38). (c) Prove that this stress field satisfies Cauchy’s homogeneous equilibrium equation div σ (x) = σ for any x 6= 0. (d) Prove that this stress field satisfies the relation Z σnda = P, Ση

where Ση is an arbitrary spherical surface centrad at x = 0 and having a radius η > 0, and n is the inward unit normal to Ση . P2.41 Prove the reciprocal theorem for singular states. That is, prove the validity of the equations (2.5.13). P2.42 Show that the reciprocal theorem for singular elastic states is still valid in the case of an infinite media with finite boundary, provided that

e (x) = O r−2 and e (x) = O r−1 , σ (x) , σ u (x) , u e (x) = O r−3 b (x) , b




as r = kxk → ∞.

P2.43 Prove the lemma concerning the existence of a unique rigid displacement field w (x), given in Section 2.5. e (x, x0 ) for a body, and the integral repP2.44 Using Green’s tensor function G resentation theorem corresponding to the traction boundary value problem prove the existence of the influence function B = B (x) having the properties given in the equations (2.5.32), (2.5.33) and corresponding to homogeneous traction problems. P2.45 Starting from the equation (2.7.13), prove the relations (2.7.14). P2.46 Using the relations (2.7.13) and the constitutive equation for an isotropic body, show that the relations (2.7.20) are true. P2.47 By direct calculations, show the validity of the equations (2.7.22).

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2.8. PROBLEMS

P2.48 Starting with the relations (2.7.22), show that for points x which are far from the inclusion, the displacements uci (x) in the matrix can be expressed approximately by the relations uci (x) ≈

1 t ν σ fijk (l) , 16πµ (1 − ν) r2 jk

where r and l are the distance and the direction from the inclusion of x, and v is the volume of the domain D occupied by the inclusion. P2.49 Show that Eshelby’s coefficients S1112 and S1123 , introduced by the equation (2.7.31), which are coupling an extension (ε11 ) and a shearing (ε12 or ε21 ), are zero. Show also that Eshelby’s coefficient S1223 coupling two shearings (ε12 or ε23 ) is vanishing. P2.50 Prove Eshelby’s relations (2.7.32) by direct computations. P2.51 Show that the equations (2.7.36) and (2.7.42) giving Ia1 , ..., Ia1 a1 , ..., in the case of a spherical inclusion are true. P2.52 Show that the components Sijkl of Eshelby’s tensor S can be expressed in the form 1 1 Sijkl = (α − β) δij δkl + β (δik δjl + δil δjk ) , 2 3

if the transformed inclusion is spherical and the infinite elastic medium is isotropic. In the above relations, α and β are the coefficients given by the equations (2.7.45). P2.53 Show that Eshelby’s coefficients α and β can be expressed in term of the shear modulus µ and of the bulk modulus k in the form α = 3 − 5β =

k
. k + 43 µ

P2.54 Show that Eshelby’s solution for the transformed inclusion problem satisfies the necessary null-jump conditions across the interface Γ separating the transformed inclusion and the matrix. P2.55 Show that Eshelby’s solution for the inhomogeneity problem satisfies the necessary null-jump conditions across the interface Γ separating the inclusion and the matrix. P2.56 Let u be an admissible displacement field for a biphasic piece-wise continuous composite material B. Let ε = ε (u) be the admissible strain field corresponding to u. Let σ be an admissible stress field for the composite. Let ε and σ be the mean values of ε and σ on B, respectively. Prove that

σ · ε − σ·ε = (σ−σ) · (ε−ε),

where the superposed bar denotes the corresponding mean values on B. P2.57 Let u be an admissible displacement field for the above composite, and let ε = ε (u) be the corresponding strain field. Let σ be an admissible stress

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field which is self-equilibrated; i.e. satisfies the equilibrium equations div σ = 0 on Bα , α = 0, 1, ..., N, corresponding to zero body forces. Show that Z 1 (u−εx) · (σn−σn) da σ · ε − σ·ε = v ∂B Z Z 1 1 u · (σn−σn) da, (u−εx) · σnda = = v ∂B v ∂B

where v is the volume of the domain B. P2.58 Using the above result, show that σ · ε = σ·ε if: (i) ε is a kinematically admissible strain field corresponding to E and σ is a self-equilibrated stress field; (ii) σ is a statically admissible self-equilibrated stress field, corresponding to Σ and ε is an admissible strain field. P2.59 Show that if (i) of P2.58 takes place, then ε = E and σ · ε = σ·E, and if (ii) of P2.58 is true, then σ = Σ and σ · ε = Σ·ε.

Bibliography ´ [2.1] Solomon, L., Elasticit´ e lin´eaire, Mason, Paris, 1968. [2.2] Malvern, L.E., Introduction to the mechanics of continuous medium, PrenticeHall, Inc., London, 1969. [2.3] Gurtin, M.E., The linear theory of elasticity, Handbuck der Physik VI a/2, Ed. C. Truesdell, Springer, Berlin, 1972. [2.4] Gurtin, M.E., An introduction to continuum mechanics, Academic Press, San Diego, 1981. [2.5] Teodosiu, C., Elastic models of crystal defects, Ed. Academiei, Bucuresti, Springer, Berlin, 1982. [2.6] Hill, R., Elastic properties of reinforced solids-some theoretical principles, J. Mech. Phys. Solids, 11, 357-372, 1963. [2.7] Mandell, J., Plasticit´e clasique et viscoplasticit´e, Int. Centre Math. Sci., Courses and Lectures No 97, Udine 1971, Springer, Wien, New York, 1972. [2.8] Eshelby, J.D., The determination of the elastic field of an ellipsoidal inclusion, and selected problems, Proc. Roy. Soc. A 241, 376-396, 1957. [2.9] Eshelby, J.D., The elastic field outside an ellipsoidal inclusion, Proc. Roy. Soc. A 252, 561-569, 1959. [2.10] Jones, R.M., Mechanics of composite materials, Hemisphere Publishing Co., New York, 1975.

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Chapter 3

COMPOSITE LAMINATES 3.1

Macromechanical behavior of lamina

In this chapter, we discuss some fundamental problems concerning fiberreinforced composite laminates; i.e. the classical part of the general theory of composite materials. The basic results existing in this field can be found, for instance, in the monographies due to Ashton and Whitney [3.1], Jones [3.2], Christensen [3.3], Tsai and Hahn [3.4], Cristescu [3.5], Whitney [3.6] and Gibson [3.7]. The fiber-reinforced composite laminates are made of fiber-reinforced laminae. The fibers considered here are long and continuous. A lamina is a plane arrangement of unidirectional fibers strongly bounded in a matrix. In Figure 3.1 is shown a typical lamina together with its material symmetry axis, named also principal material axes or directions.

Figure 3.1: Lamina with unidirectional fibers.

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Axis 1 is parallel to the fibers, axis 2 is perpendicular to the fibers in the plane of the lamina and axis 3 is perpendicular to the plane of lamina. The fibers or filaments are the main reinforcing or load-carrying elements. They are generally strong and stiff. The matrix can be organic, ceramic, or metallic. The function of the matrix is to support and protect fibers and to provide a means of distributing and transmitting load among fibers. The fibers generally exhibit linear elastic behavior. Fiber-reinforced composites, such as boron-epoxy and graphite-epoxy are usually considered to be linear elastic materials since the fibers provide most of the stiffness. A laminate is a stack of laminae with various orientations of the principal material directions with respect to the laminae as shown in Figure 3.2.

Figure 3.2: Exploded view of laminate structure. Generally the fiber orientation of the layers cannot be symmetric about the middle surface of the laminate. The layers of a laminate are usually firmly bounded together by the same matrix material that is used in laminae. Laminates can be composed of plates of different materials, or layers of fiber-reinforced laminae, as shown in Figure 3.2. Also, various laminae can have various thicknesses. A major purpose of lamination is to determine the directional dependence of stiffness of a material in accordance with the given loading environment of the structural element. Laminates are suited to this objective since the principal material directions of each layer can be oriented according to the need. For example, six layers of a ten-layer laminate could be oriented in one direction and the other four at 90◦ with respect to that direction. The resulting laminate has an extensional stiffness roughly 50 percent higher in one direction than in the other one. The fiber-reinforced lamina is the basic building block in a laminated fiberreinforced composite or laminate. Thus, the knowledge of the mechanical behavior of a laminae is essential to the understanding of laminated fiber-reinforced structures. Analyzing the macro and micromechanical behavior of a laminae, we assume that the matrix, the fibers and the lamina itself have linear elastic behavior. Also

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3.1. MACROMECHANICAL BEHAVIOR OF A LAMINA

113

we suppose that the fibers and the matrix are firmly bounded together. The same assumption will be made concerning the laminae forming a laminate. At the macro-mechanical level, the fiber-reinforced lamina will be assumed to be an orthotropic linearly elastic material. The symmetry axis are parallel and perpendicular to the fibers direction as shown in Figure 3.2. The most advantageous description of the stress-strain relation involves the (macro-mechanical or effective or equivalent or overall ) technical or engineering constants of the lamina, considered as a homogeneous body. These constants are particulary helpful in describing material behavior since they are determined by obvious and relatively simple mechanical tests. In the following, our attention will be focused on stress-strain relation for orthotropic materials in a plane stress state, the most common condition satisfied by a loaded composite lamina. The constitutive relations, initially formulated using the material symmetry axes, will be expressed later by using coordinate systems that are not aligned along the principal material directions. Such a change is necessary in order to describe the global behavior of various laminates, composed of laminae with various orientations of the reinforcing fibers. Let us consider now a lamina in the 1-2 plane as shown in Figure 3.1. Here the axes 1, 2, 3 are the principal material directions of the laminae, assumed to be (macroscopically) orthotropic. As usual, we say that the lamina is in a plane stress state relative to its symmetry plane 1-2 if the components of the stress tensor σ satisfy the following relations: σ31 = σ32 = σ33 = 0. (3.1.1) Since the material is orthotropic, according to the constitutive equation (2.2.70), from the above relation, it follows that the components of the strain tensor ε satisfy the equations ε31 = ε32 = 0 , ε33 = S13 σ11 + S23 σ22 , and, thus, the stress-strain relation (2.2.70) reduces to      σ1 ε1 S11 S12 0  ε2  =  S12 S22 0   σ2  . 0 0 S66 σ6 ε6

(3.1.2)

We recall that in the above matrix form of the remaining constitutive equation, we have used the Voigt’s convention; i.e. ε1 = ε11 , ε2 = ε22 , ε6 = 2ε12 , σ1 = σ11 , σ2 = σ22 , σ6 = σ12 . Also, we note again that the axis 1, 2, 3 are the principal material directions of the lamina, axis 1 being parallel to the fibers, axis 2 being perpendicular to the fibers and situated in the plane of the lamina and axis 3 being perpendicular to this plane.

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The general relation (2.2.71) shows that the involved components S11 , S12 , S22 and S66 of the compliance matrix [S] can be expressed in terms of the technical constants of the orthotropic lamina by the following equations: S11 =

ν12 ν21 1 , =− , S12 = − E2 E1 E1 1 1 . , S66 = S22 = G12 E2

(3.1.3)

Since the matrix [S] is positive definite, the relation (3.1.2) can be inverted to obtain the inverse stress-strain relations        σ1 Q11 Q12 0 ε1 ε1  σ2  =  Q12 Q22 0   ε2  = [Q]  ε2  . (3.1.4) σ6 0 0 Q66 ε6 ε6

The quantities Q11 , Q12 , Q22 and Q66 are named reduced stiffnesses. They have the following expressions: Q11 =

S11 S12 S22 2 , , Q66 = G12 , S = S11 S22 − S12 , Q22 = , Q12 = − S S S

(3.1.5)

or, in terms of the engineering constants

E2 ν21 E1 ν12 E2 E1 , Q66 = G12 . , Q22 = = , Q12 = 1 − ν12 ν21 1 − ν12 ν21 1 − ν12 ν21 1 − ν12 ν21 (3.1.6) The reduced constitutive equations (3.1.4) represent the basis for the analysis of the behavior of an individual lamina subjected to forces acting in its own plane. For such special loading, the orthotropic lamina is indeed in a plane stress state. We stress again that E1 is Young’s modulus in the fibers direction, E2 is Young’s modulus in the direction perpendicular to the fibers and situated in the lamina plane, ν12 and ν21 are Poisson’s ratios in the same plane, and G12 is the shear modulus in the lamina plane. We now present some numerical values of the involved material parameters for laminae frequently used in applications. The values are taken from the monograph [3.4] by Tsai and Hahn (see pp. 19 and 20). The material constants having physical dimensions (such as E1 , E2 , G12 , S11 , ..., S66 , Q11 , ..., Q66 ) are expressed in GP a = 109 N m−2 . Obviously, if E1 , E2 , ν12 and G12 are known from experimental data S11 , ..., S66 and Q11 , ..., Q66 can be calculated using the relation (3.1.3) and (3.1.6). The data given in Tables 3.1, 3.2 and 3.3 show that for fiber-reinforced laminae, generally E2 << E1 and G12 << E1

Q11 =

and Q22 << Q11

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and

Q66 << Q11 .

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3.1. MACROMECHANICAL BEHAVIOR OF A LAMINA

Type T300/5208 B(4)/5505 AS/3501

Material Graphite/Epoxy Boron/Epoxy Graphite/Epoxy

E1 181 204 138

E2 10.3 18.5 8.96

ν12 0.28 0.23 0.30

G12 7.17 5.59 7.1

Table 3.1: Engineering constants of typical fiber-reinforced laminae.

Type T300/5208 B(4)/5505 AS/3501

S11 5.525 4.902 7.246

S22 97.09 54.05 111.6

S12 -1.547 -1.128 -2.174

S66 139.5 172.7 140.8

Table 3.2: Compliance components of typical fiber-reinforced laminae.

Type T300/5208 B(4)/5505 AS/3501

Q11 181.8 205.0 138.8

Q22 10.34 18.58 9.013

Q12 2.897 4.275 2.704

Q66 7.17 5.75 7.1

Table 3.3: Reduced stiffnesses of typical fiber-reinforced laminae.

We shall see in the Section 5, that the above large differences between the magnitudes of the different rigidity moduli of a fiber-reinforced composite material have essential implications on the stability behavior of these bodies, having obviously an internal structure. We recall that the reduced constitutive relations (3.1.4) are expressed using the stress and strain components corresponding to the material symmetry direction of the lamina. These special directions often do not coincide with the coordinate direction which are geometrically related to a given problem. Hence, we must be able to express the reduced stress-strain relations using arbitrary systems of coordinates x1 = x, x2 = y, x3 = z. For our needs, we assume that the principal material direction 3 and the direction of the axis x3 = z coincide. Also, we suppose that the planes x, y and 1, 2 coincide, and the principal directions 1, 2 are obtained by rotating the axes x, y with an angle θ about the axis z, as shown in Figure 3.3. In the above mentioned case, the orthogonal matrix [qkr ], present in the general lows (1.1.14) characterizing the connections between the components of a tensor in the old and new axes have, according to the relations (1.1.8), the following

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Figure 3.3: Positive rotation of principal material axes 1, 2 from arbitrary axes x, y.

form:



cos θ [qkr ] =  sin θ 0

− sin θ cos θ 0

 0 0 · 1

(3.1.7)

For simplicity, we shall denote by σx , σy , σxy the components σ11 , σ22 , σ12 of the stress tensor σ in the coordinate system (x, y, z), and by εx , εy , εxy = γxy /2 the components ε11 , ε22 , ε12 of the strain ε in the same coordinate system (x, y, z). Taking into account (3.1.7) and the general transformation law (1.1.14) or its special form (1.1.16), we get         σ1 σx ε1 εx  σ2  = [T (θ)]  σy  ,  ε2  = [T (θ)]  εy  , (3.1.8) σ6 σxy ε6 /2 εxy where the 3 × 3 square matrix [T (θ)]  cos2 θ  [T (θ)] = sin2 θ − sin θ cos θ −1

is given by the equation sin2 θ cos2 θ sin θ cos θ

 2 sin θ cos θ −2 sin θ cos θ  . cos2 θ − sin2 θ

Denoting by [T (θ)] the inverse matrix of [T (θ)] from        σx σ1 εx  σy  = [T (θ)]−1  σ2  ,  εy  = [T (θ)]−1  σxy σ6 εxy

(3.1.9)

(3.1.8), we get  ε1 · ε2 (3.1.10) ε6 /2

Taking into account the geometrical significance of the transformation matrix [T (θ)] , or by direct computations, it is easy to see that   cos2 θ sin2 θ −2 sin θ cos θ −1  , (3.1.11) [T (θ)] = [T (−θ)] =  sin2 θ cos2 θ 2 sin θ cos θ 2 2 sin θ cos θ − sin θ cos θ cos θ − sin θ

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Consequently, (3.1.10) can be expressed in the following equivalent form: 

   σx σ1  σy  = [T (−θ)]  σ2  , σxy σ6 Introducing the Reuter’s  1 0 [R] =  0 1 0 0 we have     ε1 ε1  ε2  = [R]  ε2 , ε6 ε6 /2



   εx ε1  εy  = [T (−θ)]  ε2 . εxy ε6 /2

matrices  0 0 , 2

[R]

−1



1 = 0 0

0 1 0

 0  0 1/2



   εx εx  εy  = [R]−1  εy  , since εxy γxy

(3.1.12)

(3.1.13)

εxy = γxy /2.

Now, returning to the primary stress-strain relation (3.1.4) and using the above equations, we successively get         σx σ1 ε1 ε1  σy  = [T (−θ)]  σ2  = [T (−θ)] [Q]  ε2  = [T (−θ)] [Q] [R]  ε2  σxy σ6 ε6 ε6 /2     εx εx −1 = [T (−θ)] [Q] [R] [T (θ)]  εy  = [T (−θ)] [Q] [R] [T (θ)] [R]  εy  . εxy γxy

Using (3.1.9), (3.1.11) and (3.1.13), it is easy to see that

 sin θ cos θ  = [T (−θ)]T . [R] [T (θ)] [R] − sin θ cos θ cos2 θ − sin2 θ (3.1.14) Consequently, the needed stress-strain relation becomes −1



cos2 θ  = sin2 θ −2 sin θ cos θ

sin2 θ cos2 θ 2 sin θ cos θ

    Q11 σx  εx   σy  = Q(θ)  εy  =  Q12 γxy σxy Q16 

with



Q12 Q22 Q26

  Q16 εx Q26   εy  , γxy Q66

 T Q(θ) = [T (−θ)] [Q] [T (−θ)] ·

(3.1.15)

(3.1.16)

Finally, using the relations (3.1.11), (3.1.14) and the last equation, after long,  but elementary computations, we get for the components of the matrix Q(θ) the

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following expressions:

Q11 (θ) = Q11 cos4 θ + 2(Q12 + 2Q66 ) sin2 θ cos2 θ + Q22 sin4 θ,

Q12 (θ) = (Q11 + Q22 − 4Q66 ) sin2 θ cos2 θ + Q12 (sin4 θ + cos4 θ),

Q22 (θ) = Q11 sin4 θ + 2(Q12 + 2Q66 ) sin2 θ cos2 θ + Q22 cos4 θ,

Q16 (θ) = (Q11 − Q12 − 2Q66 ) sin θ cos3 θ + (Q12 − Q22 + 2Q66 ) sin3 θ cos θ,

Q26 (θ) = (Q11 − Q12 − 2Q66 ) sin3 θ cos θ + (Q12 − Q22 + 2Q66 ) sin θ cos3 θ,

Q66 (θ) = (Q11 + Q22 − 2Q12 − 2Q66 ) sin2 θ cos2 θ + Q66 (sin4 θ + cos4 θ). (3.1.17)   The matrix Q(θ) is named the transformed reduced stiffness matrix, and its components Q11 (θ), ..., Q66 (θ) are the transformed reduced stiffness of the fiberreinforced lamina. Note that the transformed reduced stiffness matrix has non-vanishing coefficients in all nine positions in contrast to the zeros existing in the primary reduced stiffness matrix [Q]. However, there are still only four independent material constants since the lamina is orthotropic and it is in a plane stress state. The stress-strain relation (3.1.15) shows that in general, with arbitrary x, y axis, there is coupling between normal stresses and shear strains and between shear stresses and normal strains. Thus, in the coordinates x, y, named in the following body coordinates, even an orthotropic lamina behaves as would a general anisotropic. That is the reason why such a lamina is called general orthotropic lamina, even if it is actually orthotropic. We observe now that, as an alternative to the foregoing procedure, we can express in the body coordinates the strains in terms of stresses, by inverting the relation (3.1.15) and by using the property

[Q]

−1



S11 ≡ [S] =  S12 0

S12 S22 0

 0 0 . S66

Thus, by using also (3.1.16) and the equation [T (−θ)] = [T (θ)]

    S 11 εx  σx   εy  = S(θ)  σy  =  S 12 σxy γxy S 16 

with

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S 12 S 22 S 26

  S 16 σx S 26   σy  , σxy S 66

 T S(θ) = [T (θ)] [S] [T (θ)] .

(3.1.18)

−1

, we obtain

(3.1.19)

(3.1.20)

3.1. MACROMECHANICAL BEHAVIOR OF A LAMINA

119

Now, using (3.1.19), (3.1.18) and the last equation, we obtain

S 11 (θ) = S11 cos4 θ + (2S12 + S66 ) sin2 θ cos2 θ + S22 sin4 θ,

S 12 (θ) = S11 (sin4 θ + cos4 θ) + (S11 + S22 − S66 ) sin2 θ cos2 θ,

S 22 (θ) = S11 sin4 θ + (2S12 + S66 ) sin2 θ cos2 θ + S22 cos4 θ,

S 16 (θ) = (2S11 − 2S12 − S66 ) sin θ cos3 θ − (2S22 − 2S12 − S66 ) sin3 θ cos θ,

S 26 (θ) = (2S11 − 2S12 − S66 ) sin3 θ cos θ − (2S22 − 2S12 − S66 ) sin θ cos3 θ,

S 66 (θ) = 2(2S11 + 2S22 − 4S12 − S66 ) sin2 θ cos2 θ + S66 (sin4 θ + cos4 θ). (3.1.21) Note that because of the presence of Q16 , Q26 in (3.1.15), and of S 16 , S 26 in (3.1.19), there is no difference between the behavior of the general orthotropic lamina and the actually anisotropic lamina in plane stress-state. As for anisotropic lamina, the coefficients S 11 , .., S 66 of the generally orthotropic lamina can be expressed in terms of the apparent technical or engineering coefficients, introduced in the following way (see for instance Jones [3.2] Chapter 2 or Lekhnitski [3.8] Chapter 2):

1 1 νyx νxy 1 , , S 66 = , S 22 = =− , S 12 = − Gxy Ey Ey Ex Ex ηxy,y ηx,xy ηy,xy ηxy,x , S 26 = = = = . (3.1.22) Gxy Ex Ey Gxy

S 11 =

S 16

The mechanical significance of the apparent Young moduli Ex , Ey , the Poisson ratios νxy , νyx and the shear modulus Gxy is the same as in the case of an orthotropic material. Obviously, their usual significance must be related to the coordinate axes x and y. As can be seen, we have also introduced new engineering coefficients η xy,x , ηx,xy , ηxy,y and ηy,xy . These material constants are named by Lekhnitski coefficients of mutual influence and are defined as: ηi,ij = coefficient of mutual influence of the first kind which characterizes the stretching in the i−direction caused by shear in the ij− plane, that is η i,ij = εii /2εij , for σij = τ , all other stresses being zero, and i 6= j; ηij,i = coefficient of mutual influence of the second kind which characterizes the shearing in the ij− plane caused by a normal stress in the i−direction, that is ηij,i = γij /εi , for σii = σ, all other stresses being zero, and i 6= j. Obviously, the apparent technical moduli depend on the angle θ by which the principal mutual directions were rotated. Using the relation (3.1.22) and the equations (3.1.21), the apparent moduli can be expressed in terms of the primary engineering moduli of the lamina and

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the angle θ. Elementary computations give   2ν12 1 1 1 4 sin2 θ cos2 θ + − cos θ + = E1 G12 E1 Ex

νxy = Ex 1 Ey

=

1 Gxy

1 E1

=2

n


 sin4 θ + cos4 θ − E11 +

ν12 E1

sin4 θ +



2 E1

ηxy,x = Ex

ηxy,y = Ey

+

n

n



2 E2

1 G12

+

2ν12 E1

−

4ν12 E1

−



+

2ν12 E1

−

1 G12

2 E1

+

2ν12 E1

−

1 G12

−

sin2 θ cos2 θ +

1 G12

2 E1

1 E2







1 E2

1 G12

1 E2

sin2 θ cos2 θ +

sin θ cos3 θ − sin3 θ cos θ −

sin4 θ,







o sin2 θ cos2 θ ,

cos4 θ, 1 G12


sin4 θ + cos4 θ ,

2 E2

+

2ν12 E1

−

1 G12

2 E2

+

2ν12 E1

−

1 G12





o sin3 θ cos θ ,

o sin θ cos3 θ .

(3.1.23) An important consequence of the presence of the coefficients ηxy,x and ηxy,y is that traction tests in non principal material directions result, not only in axial extensions and lateral contractions, but also in shear deformations. Following Jones (see [3.2], Chapter 2), values typical for a glass/epoxy composite (E1 = 3E2 , E2 = 8.27GP a, G12 = 0.5E2 , ν12 = 0.25) are plotted in Figure 3.4. In Figure 3.4, Ex is divided by E2 and Gxy by G12 . This normalization permits an easier analysis of the behavior of the apparent technical moduli as a function of θ. 3,0

3,0

2,5

E1 =3 E2 G12 =0.5 E

EX E2

EX E2

2

2,0

xy

=0.25

1,5

2,0

12

xy

Gxy G12

xy,x

2,5

1,5

G xy G12 1,0

1,0

0,5

0,5

xy,x 0

0

15

30

45

60

75

0 90

Figure 3.4: Normalized moduli for glass/epoxy. The figure shows that ηxy,x is vanishing at θ = 00 and θ = 900 , as is to

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121

be expected, since the laminae actually is orthotropic. Also it can be seen that at intermediate angles, this coefficient of mutual influence achieves large values as compared to the apparent Poisson ratio νxy . Also, as the first two equations (3.1.23) show, the transverse axial modulus Ey behaves essentially like the longitudinal one Ex , with the exception that Ey is small for θ near 00 and large when θ is near 900 . Similar comments can be made for νyx and ηxy,y . We observe that the behavior presented in the Figure 3.4 is not always typical for all composites, fiber-reinforced laminae. For the considered glass/epoxy composite, the maximal value of Ex is just E1 . There exist cases where Ex can actually exceed both E1 and E2 , or can be smaller than both E1 and E2 , for some orthotropic laminae and some intermediate values of the angle θ (see P.3.8). The reduced stiffnesses given in relation (3.1.17) are relatively complicated functions of the four primary material characteristics E1 , E2 , ν12 , G12 , as well as of the angle of rotation θ. There exists an ingenious recasting of the stiffness transformations equations that enables a more clear understanding of the consequences of rotating a lamina in a laminate (see Jones [3.2], Chapter 2). By using elementary trigonometric identities, the transformed reduced stiffnesses can be expressed in the following way:

Q11 = U1 + U2 cos 2θ + U3 cos 4θ, Q12 = U4 − U3 cos 4θ, Q22 = U1 − U2 cos 2θ + U3 cos 4θ, 1 Q16 = − U2 sin 2θ − U3 sin 4θ, 2 1 Q26 = − U2 sin 2θ + U3 sin 4θ, 2 Q66 = U5 − U3 cos 4θ,

(3.1.24)

where U1 =

U2 =

U3 =

U4 =

U5 =

1 (3Q11 + 3Q22 + 2Q12 + 4Q66 ) , 8 1 (Q11 − Q22 ) , 2 1 (Q11 + Q22 − 2Q12 − 4Q66 ) , 8 1 (Q11 + Q22 + 6Q12 − 4Q66 ) , 8 1 (Q11 + Q22 − 2Q12 + 4Q66 ) . 8

(3.1.25)

The advantage of writing the expressions of the reduced stiffnesses in the above form is that these relations show just those parts of Q11 , .., Q66 which rest invariant under rotation of the lamina. This concept of invariance is useful when examining the prospect of orienting a lamina at various angles to achieve a certain

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stiffness property. For example, the first equation (3.1.24) shows that the value of Q11 is determined by a fixed constant, U1 , plus a quantity of low frequency variation with θ, U2 cos 2θ, plus a third quantity, U3 cos 4θ, of higher frequency variation with θ. Hence, U1 is an effective measure of lamina stiffness in a design application, and it is not being affected by the orientation of the lamina.

3.2

Strength of materials approach

In the Section 3.1, our approach was macromechanical or macroscopic considering the overall properties of a lamina. That is, a large enough piece of the lamina has been considered as being (macroscopically) homogeneous. The fact that the lamina is piece-wise homogeneous, being made of two constituent materials (the matrix and the fibers) was neglected. In this sense, we were able to say that a boron/epoxy composite lamina with unidirectional boron fibers has certain elasticities and stiffnesses which were experimentally determined. In this “homogenized” situation, the following question cannot be asked and cannot be answered: how can the (effective, equivalent, overall) stiffness of the composite be varied by changing the amount of boron fibers in the lamina? Because there must be some rationales (reasons) for selecting a particular stiffness for a particular design application, there must also exist a rationale for determining how to find the best procedure to achieve that stiffness for a fiber-reinforced lamina. That is, how can the percentage or the concentration or the volume fraction of the constituent materials be varied so as to arrive at the desired (overall, macroscopic, equivalent) stiffness? There are two methods to answer the above questions which can be characterized as being either micromechanical or macromechanical. In micromechanics, the composite material behavior is studied taking into account the interaction of the constituent materials, that is the composite is analyzed as being a (piecewise) heterogeneous body. In macromechanics, the composite material behavior is analyzed assuming the body as being homogeneous, and the effects of the actual non-homogeneities are taken into account only as averaged apparent, overall, equivalent properties of the composite. When using micromechanical methods, the properties of a lamina can be mathematically derived on the basis of the properties of the constituent materials. When using macromechanical methods, the properties of a lamina can be experimentally determined is the “as mate” state. That is, we can predict the lamina properties by the procedures of micromechanics and we can measure the lamina properties by mechanical experiments and use the properties obtained by one of the above methods in a macroscopic analysis of the structure. Knowledge of how to predict properties is essential in order to construct composites that must have certain apparent, overall, equivalent or macroscopical properties. Consequently, micromechanics is a natural approach beside macromechanics when viewed from a design rather than an analysis point of view. Obvi-

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ously, the real design efficiency is evidenced when the micromechanical predictions of the properties of the composite agree with the measured properties. Unfortunately, the micromechanical approach has inherent limitation. For example, a perfect bound between fibers and matrix is a usual analysis restriction that might not be satisfied by some composites. Thus, the micromechanical predictions must be validated by careful experimental work. Nowadays there exist two basic approaches in the micromechanics of composite materials: (i) mechanics (strength) of materials; (ii) elasticity. The mechanics of materials approach contains simplifying assumptions concerning the hypothesized behavior of the mechanical system. The elasticity approach is actually: (i) bounding principles; (ii) exact solutions; (iii) approximate solutions. Some of these approaches will be discussed in detail, for some important cases, in Section 4 devoted to macroscopically homogeneous composites. We shall present bounds for the overall moduli, obtained by Hill, Hashin and Shtrikman for macroscopically isotropic and transversally isotropic composites. Exact solutions will also be presented due to Hill and one, derived by Budiansky and Hill. Also we shall discuss briefly some results obtained by taking into account various geometrical models of different composite materials. The final objective of all micromechanical approaches is to determine the overall (equivalent, macroscopic, effective) elastic moduli or stiffness of a composite material in terms of the elastic moduli and concentrations of the constituent bij of materials or phases. For example, the overall elastic moduli, designed by C a fiber-reinforced composite lamina must be expressed in terms of the fibers and matrix moduli and their concentrations bij = C bij (Em , νm , Ef , νf , cm , cf ) , C

where Em , νm and Ef , νf are Young’s moduli and Poisson’s ratios of the matrix and of the fibers, respectively, and cm = vm /v , cf = vf /v represent the concentration or volume fractions of the matrix and of the fibers, respectively, v, vm , vf being the volumes occupied by the lamina, the matrix and the fibers, respectively. As we shall see, the above problem generally cannot be solved without introducing unrealistic assumptions, used in the strengths of materials. The overall properties obtained in this way, generally do not agree with the measured ones. This is the main reason why the much powerful approach formulated on the base of elasticity and on the theory of macroscopically homogeneous composite materials must be involved. In this way, generally, we can derive lower and upper bounds for the overall moduli, and if these bounds are close, the obtained results can be used in the design. According to the micromechanical approach used, we must impose some basic restrictions on the composite material that can be treated, using the methods

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of strength of materials or those of elasticity theory. For instance, in the case of a fiber-reinforced lamina we assume that: (i) the matrix is linearly elastic, homogeneous and isotropic or transversally isotropic; (ii) the fibers are linearly elastic, homogeneous and isotropic or transversally isotropic, and perfectly aligned; (iii) the lamina is macroscopically linearly elastic, homogeneous and transversally isotropic or orthotropic. We suppose also that no voids can exist in the fibers or in the matrix or between them, and the fibers and matrix are firmly bounded together. Basic in the discussion of micro and macromechanics of a macroscopically homogeneous composite is its representative volume element (RVE). Roughly speaking the RVE is the smallest region or piece of composite material over which the stresses and strains are macroscopically uniform. However, it is obvious, that microscopically the stresses and strains are nonuniform in the RVE, due to the heterogeneity of the composite material. Thus, the scale of the RVE is very important. Other concepts concerning the characteristics of the RVE, if they exist, will be presented and discussed in Chapter 4 concerning the elasticity approach for macroscopically homogeneous composites. Here we shall present and discuss briefly only the mechanics of material approach to the micromechanics for the overall material stiffnesses. In this way, we shall obtain very simple, but generally unrealistic approximations, to the effective engineering constants of the fiber-reinforced lamina, assumed to be macroscopically orthotropic. For simplicity, the matrix and the fibers are supposed to be homogeneous and isotropic. In this Section, the mechanical and geometrical characteristics of the matrix will be designed by m, and those of the fibers by f . As we already know, the key feature of the mechanics of material approach is that certain simplifying assumption are made regarding the mechanical behavior of a composite material. Using this procedure we can derive the mechanics of material expression for the overall orthotropic moduli of the unidirectionally reinforced fibrous composite material. It is assumed that the RVE contains only one fiber. b1 Determination of E The first overall modulus to be determined is that of the composite in the fiber direction. We suppose that the axial strain ε1 in the fiber direction is the same in the matrix and in the fiber. Such a hypothesis was first made by Voigt in 1910. From the Figure 3.5, we get ε1 =

4L , L

where ε1 is the axial strains for both the fibers and the matrix, according to the basic Voigt type assumption. Then, the axial stresses σm and σf in the matrix and in the fiber are σ m = E m ε1 , σ f = E f ε1 .

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Figure 3.5: Representative volume element loaded in the 1-direction.

The average axial stress σ1 acts on the cross sectional area S, σm acts on the cross sectional area Sm of the matrix, and σf acts on the cross sectional area Sf of the fiber. Thus, the resultant axial force F on the RVE is F = σ 1 S = σ m Sm + σ f Sf . Using the obtained results and taking into account that according to the b1 , we have definition of the overall axial moduli E we get

b1 ε1 , σ1 = E

b1 = Sm Em + Sf Ef . E S S But the concentrations or volume fractions cm = vm /v and cf = vf /v of the matrix and of the fibers can be expressed as

cm =

Sf Sm , , cf = S S

where v, vm and vf are the volumes occupied by the RVE, by the matrix and by the fiber, respectively. In this way, finally we get b1 = cm Em + cf Ef . E

(3.2.1)

This expression for the overall (apparent, equivalent, macroscopical) Young modulus in direction of the fibers is known as the rule of mixture or as the Voigt type estimate. This rule leads to a simple linear variation of the overall Young modulus b1 from Em to Ef as the fibers concentration cf varies from 0 to 1. E We stress the fact that, according to its definition, the overall axial modulus b1 connects the mean stress and the mean strain, evaluated on the RVE of the E

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composite material. In the elasticity approach of the problem, the overall moduli will be introduced in the same way! b2 Determination of E b2 , in the direction transverse We now consider the overall Young modulus E to the fibers. In the mechanics of the material approach, the same transverse stress σ2 is assumed to be applied to both the matrix and the fiber, as shown in Figure 3.6. Such kind of hypotheses was first made by Reuss in 1929.

Figure 3.6: Representative volume element loaded in 2-direction. The transverse strains εm and εf , in the matrix and in the fiber, respectively, are therefore σ2 σ2 . , εf = εm = Ef Em

The transverse direction over which on the average εm acts is approximately cm W , whereas εf acts on cf W . Thus, the total transverse deformation is ε2 W = c m W εm + c f W εf ; Hence, the mean transverse deformation ε2 becomes ε2 = c m εm + c f εf . Introducing here the stress-strains relations, we get ε2 = c m

σ2 σ2 . + cf Ef Em

b2 must satisfy Recognizing that according to its definition the overall moduli E the material law 1 σ , ε2 = b2 2 E Copyright © 2004 by Chapman & Hall/CRC

3.2. STRENGTH OF MATERIALS APPROACH

127

finally we get

1 1 1 . + cf = cm b E E f m E2

(3.2.2)

b2 in the transverse direction This expression for the overall Young modulus E of fibers is known as the Reuss type estimate. Obviously, we have cm + cf = 1. Hence if cm = 1, that is cf = 0, according to the above rule, the overall modulus predicted is that of the matrix; if cf = 1, hence cm = 0, the modulus predicted is that of the fibers. However, now the rule does not represent a linear b2 as cf goes from 0 to 1. Let us observe variation of the overall Young modulus E also that according to the Reuss type estimate, more than 50 percent by volume of b2 to twice the matrix modulus, fibers is required to raise the transverse modulus E even if Ef = 10Em . That is, if σ2 is a tensile test, the fibers cannot contribute much to the overall transverse modulus unless their percentage is very high, and, obviously the bound between the fibers and the matrix is perfect. In exchange, no such bound is needed if σ2 is a compression test! Obviously, the assumptions involved in the foregoing derivation are not entirely consistent, since the transverse stresses in the matrix and in the fibers are not the same. Indeed, if the Poisson ratios of the matrix and the fibers are not the same, then longitudinal stresses are introduced in the matrix and fiber, with accompanying shear stresses at the matrix-fiber boundary. Such shearing stresses lead to a stress state much more complicated than that assumed in our derivation. The consequence of such inconsistent assumptions can be measured only by comparison with experimental results. Determination of νb12 The overall Poisson ratio νb12 can be determined using the assumption made b1 ; that is, supposing that the axial strains in the matrix and the fiber to obtain E are the same, ε1 . Denoting by ε2 the (mean) transverse strain of the RVE, νb12 is defined by ε2 νb12 = − , ε1

for the stress state σ1 6= 0 and all over stresses are zero. According to the Figure 3.7, the transverse deformation ε2 is ε2 =

∆W = −b ν12 ε1 . W

We also have ∆W = ∆Wm + ∆Wf ,

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Figure 3.7: Representative volume element loaded in 1-direction.

∆Wm and ∆Wf being the transverse displacements of the matrix and of the fiber, respectively. Consequently

∆Wf ∆Wm = − νb12 ε1 . + W W

Following the same procedure as in analysis for the overall transverse Young b2 , we assume that the transverse displacements ∆Wm and ∆Wf are modulus E approximately ∆Wm = −W cm νm ε1 , ∆Wf = −W cf νf ε1 ,

−νm ε1 and −νf ε1 being the average transverse deformations of the matrix and of the fiber, respectively. Combining the last equations, we get νb12 = cm νm + cf νf .

(3.2.3)

The strength of materials rule leads to the mixture rule or to the Voigt type estimate of the overall Poisson ratio νb12 .

b 12 Determination of G b 12 of a lamina is estimated in the meThe overall in-plane shear modulus G chanics of materials approach by assuming a Reuss type hypothesis. It is supposed that the same shear stress τ acts in the matrix and in the fiber. Denoting by γ m and γf , the shear strains in the matrix and fiber, respectively, we get γm =

1 1 τ. τ , γf = µf µm

The loading is shown in Figure 3.8 and the deformations on microscopic scale in Figure 3.9. The total (mean) shear deformation γ is γ=

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∆ , W

129

3.2. STRENGTH OF MATERIALS APPROACH

Figure 3.8: Representative volume element loaded in shear. FIBER MATRIX

m /2

f MATRIX

Figure 3.9: Shear deformation of a representative volume element. and we have ∆ = ∆ m + ∆f , ∆m and ∆f being the horizontal displacements of the matrix and of the fiber, respectively. Denoting by γm and γf the shear strains in the matrix and fiber, respectively, we approximately get ∆ m = c m W γm ∆ f = c f W γf . Hence, γ = c m γm + c f γf . b 12 , connecting the mean strain γ and The overall in-plane shear modulus G the mean stress τ , is defined by γ=

1 τ. b 12 G

Thus, using the above formulas, we obtain

1 1 1 + cf . = cm b 12 µf µm G Copyright © 2004 by Chapman & Hall/CRC

(3.2.4)

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CHAPTER 3. COMPOSITE LAMINATES

Material Ef νf Em νm cf Methods Experimental Mixture rules

Material Ef νf Em νm cf Methods Experimental Mixture rules

b1 E

49.40 47.16

Glass/Epoxy 70 0.17 2.85 0.33 0.66 b2 b 12 νb12 E G 18.00 7.77

7.80 2.95

Boron/Epoxy 413 0.2 4.10 0.35 0.70 b 12 b2 b1 G E E 237.8 290

13.3 26.7

5.5 12.2

0.22 0.224

b1 E

Carbon/Epoxy 234 0.2 3.8 0.33 0.6 b2 G b 12 νb12 E

151 141.8

9.3 9.2

6.2 3.5

0.32 0.25

νb12

0.245

Table 3.4: Experimental and calculated values of the overall elastic coefficients.

The strength of materials approach leads to a Reuss type estimation for the b 12 . overall in-plane shear modulus G b2 , only for a fiber volume greater than 50 percent of the As in the case of E b 12 rise to above twice µm even if µf = 10µm . total volume does G Using the data given by Barran and Loroze [3.9], we present in Table 3.4 the mechanical characteristics of three fiber-reinforced composite materials, giving also the fiber concentrations. We also give the overall elastic coefficients experimentally determined and the values of the overall moduli calculated using the mixture rules b1 and obtained by the strength of materials approach. The axial Young modulus E the transverse Poisson ratio νb12 are evaluated taking into account the Voigt type b2 and the inmixture rules (3.2.1), (3.2.2), and the transverse Young modulus E b plane transverse shear modulus G12 are obtained using the Reuss type mixture rules (2.3.3), (3.2.4). The axial and transverse Young moduli, as well as the shear modulus are expressed in GP a = 109 P a. Examining the above data, we can see that the calculated values of the overall b1 and those of the overall transverse Poisson ratio νb12 are axial Young modulus E acceptable as first approximations. However, the calculated values of the overall b2 and those of the overall transverse shear modulus transverse Young modulus E b G12 are not acceptable, and cannot be used as a first approximation. Generally, we can say that much more powerful methods are necessary to evaluate and/or to bound the overall moduli as those obtained with the strength of materials approach.

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The above estimations are only examples of the type of mechanics of materials approaches that can be used to obtain approximate expression for the overall moduli. Other assumptions of mechanical behavior lead to different estimations for the overall elastic moduli of the lamina. The true significance of the Voigt and Reuss type estimates can be clarified only by using the elasticity approach to get the overall stiffnesses. As we shall see in Section 4.1, the Voigt and Reuss type estimations give universal-bounds for the overall moduli. Generally, these estimates are the worst bounds that can be derived by the elastic approach. We end this Section with some words concerning the approach named netting analysis (see Jones [3.2], Chapter 3, Section 3.3.1). The basic assumption in netting analysis is that the fibers provide all the longitudinal stiffness and the matrix provide all the transverse and shear stiffness as well as the Poisson effect. Even on the base of the above results furnished by the mechanics of material, we can see that the assumptions made by the netting analysis must generally be rejected. In turn, the results due to the strength of materials approach must be carefully analyzed in light of the elasticity approach. Some important results of this analysis will be presented in Chapter 4.

3.3

Global constitutive equations

As we have seen, a laminate is composed of two or more laminae bounded together to act as a structural element. The constituent laminae are oriented to produce a structural element capable of resisting load in several directions. The stiffness of such a composite body results from the properties of the constituent laminae, as well as from their relative orientations. In the following, we present the basic formulation of the classical lamination theory. The major difference between this theory and the classical theory of homogeneous isotropic plates is in the form of the stress strain relationships of the lamina. Other elements of the theory such as the deformation hypothesis, the equilibrium equation and the strain displacement relationships are the same as those used in the classical plate theory. Although the laminate is made up of multiple laminae, it is assumed, that the individual laminae are perfectly bounded together so as to behave as a unitary, nonhomogeneous, anisotropic plate. Interfacial slip is not allowed and the interfacial bounds are not allowed to deform in shear, which mean that the displacement across laminae interfaces are assumed to be continuous. The assumptions imply that deformation hypothesis from the classical homogeneous plate theory can be used for the laminated composite plate. Figure 3.10 shows the coordinate system to be used in developing the laminated plate analysis. The x1 , x2 , x3 coordinate system is assumed to have its origin on the middle surface of the plate, so that the x1 x2 planes lie in the middle plane. The components of the displacement u are u1 , u2 , u3 and they depend on x1 , x2 , x3 . Frequently, x1 , x2 , x3 are denoted by x, y, z and u1 , u2 , u3 by u, v, w.

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Figure 3.10: Coordinate system for laminated plate.

The basic assumptions made in the frame-work of the classical composite laminate theory are the following: (1) The plate consists of orthotropic laminae bounded together, with the principal material axes of the orthotropic laminae oriented at arbitrary direction with respect to the x1 , x2 axes. (2) The thickness h of the plate is much smaller than the length along the plate edges a and b. (3) The displacements u1 , u2 , u3 are small compared with the plate thickness h. (4) The in-plane strains ε11 , ε22 , ε12 are small compared with unity. (5) Transverse shear strains ε13 and ε23 are negligible. (6) The transverse normal strain ε33 is negligible. (7) The normal stress σ33 is small in comparison with the other stress components. (8) The transverse shear stresses σ13 and σ23 vanish on the plate surfaces x3 = ± h2 . (9) Each lamina obeys the reduced stress-strain relation corresponding to plane stress state. The assumption (2) stresses the fact that we develop here the classical thin lamination theory. The assumptions (3) and (4) show that the theory refers to small deformations, that is it is geometrically linear. The assumptions (5) and (6) express the classical Love-Kirchhoff hypothesis, known also as the hypothesis of plane sections: any normal to the middle surface remains straight and normal to the deformed middle surface, and at the same time, its magnitude rests constant during the deformation. The assumptions (7), (8) and (9) express the fact the stresses σ13 , σ23 and σ33 are assumed to be small in comparison with the stresses σ11 , σ22 and σ12 . That is, as is stated in the assumption (9), the stresses σ11 , σ22 and σ12 and the strains ε11 , ε22 and ε12 can be related using the reduced stress-strain relatives corresponding to the plane stress state of the laminae. Since the assumed hypotheses are similar to those used in the classical Love-Kirckhhoff theory of homogeneous isotropic thin plate, the classical lamination theory of composite laminates inherits all internal contradictions and inconsistencies of the Love-Kirckhhoff

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133

theory. This observation concerns all of the internal contradiction existing between the assumptions (7), (8) and (9): though σ13 , σ23 and σ33 are not vanishing, we use the reduced stress-strain relations corresponding to vanishing σ13 , σ23 and σ33 . Obviously, the seriousness and consequences of these inconveniences can be established only by studying the implication of the theory based on the assumptions (1)-(9). For this purpose, we must first develop the classical lamination theory, using the supposed hypothesis. From assumption (6), we obtain ε33 (x1 , x2 , x3 ) =

∂u3 (x1 , x2 , x3 ) = 0. ∂x3

Consequently, the normal displacement u3 (x1 , x2 , x3 ) depends only on x1 and x2 ; i.e. u3 = U3 (x1 , x2 ). (3.3.1) From the assumption (5), we get 2ε13 (x1 , x2 , x3 ) =

∂u3 ∂u1 (x1 , x2 , x3 ) = 0, (x1 , x2 , x3 ) + ∂x1 ∂x3

2ε23 (x1 , x2 , x3 ) =

∂u3 ∂u2 (x1 , x2 , x3 ) = 0. (x1 , x2 , x3 ) + ∂x2 ∂x3

From here, according to (3.3.1), the displacements u1 (x1 , x2 , x3 ) and u2 (x1 , x2 , x3 ) depend linearly on x3 ; i.e. u1 = U1 (x1 , x2 ) − x3

∂U3 (x1 , x2 ) ∂U3 (x1 , x2 ) . , u2 = U2 (x1 , x2 ) − x3 ∂x2 ∂x1

(3.3.2)

In the above relations, U3 (x1 , x2 ) is the normal displacement of the middle surface, and U1 (x1 , x2 ), U2 (x1 , x2 ) characterize the tangential displacement of the same surface. From (3.3.2), we obtain the following expressions for the non-vanishing strain components ε11 , ε22 , ε33 : εαβ = eαβ + x3 kαβ , α, β = 1, 2, where eαβ = eαβ (x1 , x2 ) =

1 ∂Uβ 1 ∂Uα ) = (Uα,β + Uβ,α ) + ( 2 ∂xα 2 ∂xβ

(3.3.3)

(3.3.4)

describe the deformation of the middle surface x3 = 0, and kαβ (x1 , x2 ) = kβα (x1 , x2 ) = −

are the curvatures of the same surface.

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∂U3 = −U3,αβ ∂xα ∂xβ

(3.3.5)

134

CHAPTER 3. COMPOSITE LAMINATES In the following, the Greek indices take the values 1 and 2. For later use, we introduce the following matrix notations:         σx σ11 εx ε11 [σ] =  σy  =  σ22  , [ε] =  εy  =  ε22  , σxy σ12 γxy 2ε12 

 e11 [e] =  e22  , 2e12

Thus from (3.3.3) we get



 k11 [k] =  k22  2k12

[ε] = [e] + x3 [k] .

(3.3.6)

(3.3.7)

In Figure 3.11, we present the geometry of an N-layered laminate, clarifying in this way the relations which will be used in what follows. Occasionally, to simplify some formulas, we shall use the notation x3 = z.

Figure 3.11: Geometry of an N -layered laminate. The k-th lamina occupies the domain defined by zk−1 < z < zk , k = 1, ..., N with x3 = z. Obviously, z0 = − h2 and zN = h2 . We return now to assumption (9). According to this hypothesis, in each lamina the reduced and transformed stress-strain relation (3.1.15) is valid. Hence, we have       Q11 Q12 Q16 εx σx  σy  =  Q12 Q22 Q26   εy  for k = 1, .., N. (3.3.8) γ σxy k Q16 Q26 Q66 k xy k

Using the simplified matrix notation (3.3.6), we get   [σ]k = Q k [εk ] for zk−1 < x3 = z < zk and k = 1, .., N.

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(3.3.9)

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3.3. GLOBAL CONSTITUTIVE EQUATIONS

Since according to (3.3.7), eαβ and kαβ depend on x1, x2 only, the last equation becomes     [σ]k = Q k [e] + x3 Q k [k] for zk−1 < x3 = z < zk and k = 1, .., N. (3.3.10)

The last equation expresses the plane stress σ11 , σ22 , σ12 in the k-th layer, in terms of the laminate middle surface strains and curvatures. Expanded, the equation (3.3.10) becomes           k11 Q11 Q12 Q16 e11 Q11 Q12 Q16 σ11  σ22  =  Q12 Q22 Q26   e22 +x3  Q12 Q22 Q26   k22  σ12 k Q16 Q26 Q66 k 2k12 Q16 Q26 Q66 k 2e12 (3.3.11) for zk−1 < x3 = z < zk and k = 1, .., N. In the above equations, k denotes the k-th lamina, (σαβ )k , α, β = 1, 2 are the stress in the k-th lamina, (Qij )k , i, j = 1, 2, 6 are the transformed reduced stiffness of the k-th lamina, zk−1 and zk are the distances from the middle surface to the inner and to the outer surfaces of the k-th lamina, respectively, and N is the total number of the laminae. We recall that the reduced stiffness Qij , i, j = 1, 2, 6 depend on θ, the angle made by the fibers with axis Ox1 , and we have

(Qij )k = Qij (θk ) for zk−1 < x3 = z < zk and k = 1, ...N,

(3.3.12)

θk representing the angle made by the fibers in the k-th lamina and the body axis x1 . Since (Qij )k can be different for each lamina of the laminate, the stress variation through the thickness is not necessarily linear, even though the strain variation is linear, as can be seen by examining equation (3.3.3). In the laminated plate analysis, it is convenient to use the forces Nαβ and the moments Mαβ per unit length, defined by the following relations:

Nαβ = Nβα =

Z

h 2

−h 2

σαβ dx3 , Mαβ = Mβα =

Z

h 2

−h 2

x3 σαβ dx3 , α, β = 1, 2. (3.3.13)

Let us observe that usually N11 , N12 = N21 , N22 are denoted by Nxx , Nxy = Nyx and Nyy , respectively, and also M11 , M12 = M21 , M22 are denoted by Mxx , Mxy = Myx , Myy , respectively. According the relation (3.3.13)1 N11 , N12 , N22 are forces per unit length of the cross-section. The mechanical meaning of these force resultants are shown in the Figure 3.12. Similarly, equation (3.3.13)2 shows that M11 , M12 , M22 are moments per unit length of the cross-section. In Figure 3.13 is shown the mechanical meaning of these moment resultants.

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Figure 3.12: In-plane forces on a flat laminate.

Figure 3.13: Moments on a flat laminate.

The relations (3.3.13) show that these force and moments resultants do not depend on x3 , but are functions of x1 and x2 , the in the plane coordinates of the laminate middle surface. In more detail, the defining equations (3.3.13) can be written as 

   N Z zk N11 σ11 X  σ22  dz; [N ] =  N22  = z k−1 N12 σ12 k k=1     Z N M11 σ11 zk X [M ] =  M22  = z  σ22  dz. z k−1 M12 σ12 k k=1

(3.3.14)

The integrations indicated in these equations can be rearranged to take advantage of the fact that the stiffness matrix for a lamina is constant within each lamina. Thus, substituting the stress-strain relations (3.3.11) and taking into ac-

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137

3.3. GLOBAL CONSTITUTIVE EQUATIONS count the fact that eαβ and kαβ do not depend on x3 = z, we get ) ) (N (N Z zk Z zk X X zdz [k], dz [e] + [Q]k [N ] = [Q]k

[M ] =

(

zk−1

k=1

N X

[Q]k

k=1

Z

zk

zdz zk−1

)

[e] +

(

zk−1

k=1

N X

k=1

[Q]k

Z

zk

z dz zk−1

Finally, these equations can be written as       N11 A11 A12 A16 e11 B11 [N ] =  N22 = A21 A22 A26  e22 + B21 N12 A61 A62 A66 2e12 B61 

  M11 B11 [M ] =  M22 = B21 M12 B61

B12 B22 B62

2

)

B12 B22 B62

   B16 e11 D11 B26  e22 + D21 B66 2e12 D61

D12 D22 D62

[k].

  B16 k11 B26  k22  , B66 2k12

  D16 k11 D26  k22  , D66 2k12 (3.3.15)

where the coefficients Aij , Bij , Dij , i, j = 1, 2, 6 are defined by N P

(Qij )k (zk − zk−1 ), k=1 N P 2 ), (Qij )k (zk2 − zk−1 Bij = Bji = 12 k=1 N P 3 ). (Qij )k (zk3 − zk−1 Dij = Dji = 13 k=1

Aij = Aji =

Introducing the symmetric 3 × 3 matrices      A11 A12 A16 B11 B12 B16 D11 [A] = A12 A22 A26  , [B] = B12 B22 B26  , [D] = D12 A61 A26 A66 B61 B26 B66 D61

(3.3.16)

D12 D22 D26

 D16 D26  D66 (3.3.17)

the equations (3.1.15) can be expressed in a concentrated matrix form [N ] = [A][e] + [B][k], [M ] = [B][e] + [D][k]

(3.3.18)

the 3 × 1 matrixes [e] and [k] being defined by equation (3.3.6)3,4 . Also, the system (3.3.18) can be replaced by the following matrix equation:      N11 A11 A12 A16 B11 B12 B16 e11  N22   A21 A22 A26 B21 B22 B26   e22        N12   A61 A62 A66 B61 B62 B66   2e12   =    M11   B11 B12 B16 D11 D12 D16   k11  . (3.3.19)       M22   B21 B22 B26 D21 D22 D26   k22  M12 B61 B62 B66 D61 D62 D66 2k12

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This equation describing the global behavior of the laminate, can be expressed in the concentrated form        e e A B N . (3.3.20) = [E] = k k B D M The 6 × 6 symmetric matrix [E] is the global laminate stiffness matrix. The coefficients Aij are called extensional stiffnesses, the coefficients Bij describe the coupling stiffness, and the coefficients Dij are called bending stiffnesses. The presence of the coefficients Bij implies coupling between bending and extension of a laminate. That is, it is impossible to pull on a laminate that has non-vanishing Bij terms, without at the same time bending and/or twisting the laminate. Thus, an extensional force results not only in extensional deformation of the middle surface, but also in twisting and/or bending of the laminate. Also, such a laminate cannot be subjected to a moment without at the same time being subjected to an extension of the middle surface. The experiments made with laminates confirm these theoretical predictions. In spite of this fact, in the stability analysis of laminates, this coupling is generally neglected and we shall discuss this question in Chapter 7. It is easy to see that the matrix equation (3.3.19) can be written in the following tensorial or component form, very useful in many problems Nαβ = Aαβγϕ eγϕ + Bαβγϕ kγϕ , Mαβ = Bαβγϕ eγϕ + Dαβγϕ kγϕ , α, β, γ, ϕ = 1, 2.

(3.3.21)

The coefficients of these equations can be expressed simply and obviously by using the quantities A11 , ..., D66 . For instance, we have A1111 = A11 , A1122 = A2211 = A12 , A1112 = A1121 = A1211 = A2111 = A16 , A1212 = A1221 = A2112 = A2121 = A66 , ..., B1111 = B11 , B1122 = B2211 = B12 , B1112 = B1121 = B1211 = B2111 = B16 , B1212 = B1221 = B2112 = B2121 = B66 , ..., D1111 = D11 , D1122 = D2211 = D12 , D1112 = D1121 = D1211 = D2111 = D16 , D1212 = D1221 = D2112 = D2121 = D66 , ....

(3.3.22)

It is also clear that the following symmetry relations take place Aαβγϕ = Aβαγϕ = Aαβϕγ = Aγϕαβ , Bαβγϕ = Bβαγϕ = Bαβϕγ = Bγϕαβ , Dαβγϕ = Dβαγϕ = Dαβϕγ = Dγϕαβ .

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(3.3.23)

3.3. GLOBAL CONSTITUTIVE EQUATIONS

139

We stress the fact that the constitutive coefficients Aαβγϕ , Bαβγϕ , Dαβγϕ have the same symmetries as the elasticities of a linearly elastic material. As we shall see later, this property will have important consequences. We assume now that the global constitutive equation (3.3.20) is invertible. In Chapter 7, we shall see that this property is always true if the initial configuration of the laminate is (locally) stable. In order to express [e] and [k] in terms of [N ] and [M ], we rewrite (3.3.20) as [N ] = [A][e] + [B][k], [M ] = [B][e] + [D][k].

(3.3.24)

From the first equation we get [e] = [A]−1 [N ] − [A]−1 [B][k].

(3.3.25)

Substitution of (3.3.25) in (3.3.24)2 gives [M ] = [B][A]−1 [N ] − [B][A]−1 [B][k] + [D][k].

(3.3.26)

Equation (3.3.25) and (3.3.26) give a partially inverted form of the equation (3.3.20)    ∗   e A B∗ N = , (3.3.27) M C ∗ D∗ k

with

[A∗ ] = [A]

−1

, [B ∗ ] = − [A]

−1

[B] , [C ∗ ] = [B][A]−1 , [D∗ ] = [D] − [B][A]−1 [B]. (3.3.28) Now, using (3.3.27) for [k], we get [k] = [D ∗ ]−1 [M ] − [D ∗ ]−1 [C ∗ ][N ]. Introducing (3.3.29) in (3.3.25), we obtain  [e] = [A∗ ] − [B ∗ ][D∗ ]−1 [C ∗ ] [N ] + [B ∗ ][D∗ ]−1 [M ].

(3.3.29)

(3.3.30)

Finally, (3.3.29) and (3.3.30) lead to the following inverted global constitutive equation.    0     e A B0 N N −1 = = [E] , (3.3.31) k C 0 D0 M M

where

[A0 ] = [A∗ ] − [B ∗ ][D∗ ]−1 [C ∗ ], [B 0 ] = [B ∗ ][D∗ ]−1 , [C 0 ] = −[D ∗ ]−1 [C ∗ ] = [B 0 ]| = [B 0 ], [D0 ] = [D∗ ]−1 .

(3.3.32)

The last results show that the 6 × 6 global compliance matrix [E]−1 is symmetric. This is an obvious result, since if the global stiffness matrix [E] being symmetric, its inverse, if it exists, must also be symmetric.

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3.4

Special classes of laminates

This Section considers special classes of laminates for which the stiffnesses can easily be calculated. The special classes will be presented in increasing order of complexity. 1. Single - layered configurations For a single isotropic layer with material properties E, ν and thickness h, equations (3.3.16) give A11 = A22 =

1−ν Eh A , Bij = 0 , ≡ A, A12 = νA , A16 = A26 = 0 , A66 = 2 2 1−ν

1−ν Eh3 D. ≡ D , D12 = νD , D16 = D26 = 0 , D66 = 2 2 12(1 − ν ) (3.4.1) In order to obtain the above relations, we must use the equations (3.1.6), supposing an isotropic material. From (3.4.1), we can conclude that the resultant forces depend only on the in-plane strains of the laminate middle surface, and the resultant moments depend only on the curvatures of the middle surface. There is no coupling. The constitutive equations become      A νA 0 e11 N11   e22  ,  N22  =  νA A 0 2e12 N12 0 0 (1 − ν) A2 (3.4.2)      D νD 0 k11 M11   k22  .  M22  =  νD D 0 D 2k12 M12 0 0 (1 − ν) 2 D11 = D22 =

In particular, we have

h2 A. (3.4.3) 12 For a simple specially orthotropic layer of thickness h the lamina stiffnesses are given by equation (3.1.6). Hence, according to (3.3.10), the laminate stiffnesses are D=

A11 = hQ11 , A12 = hQ12 , A22 = hQ22 , A16 = A26 = 0 , A66 = hQ66 , Bij = 0, D11 =

h3 h3 h3 h3 Q66 . Q22 , D16 = D26 = 0 , D66 = Q12 , D22 = Q11 , D12 = 12 12 12 12 (3.4.4)

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3.4. SPECIAL CLASSES OF LAMINATES

141

Again, the resultant forces depend only on the in-plane strains, and the resultant moments depend only on the curvatures. There is no coupling. The constitutive equation becomes      e11 N11 A11 A12 0  N22  =  A12 A22 0   e22  , 0 0 A66 2e12 N12 (3.4.5)      M11 D11 D12 0 k11  M22  =  D12 D22 0   k22  . 0 0 D66 M12 2k12 2. Symmetric laminate For laminates that are symmetric in both geometry and material properties about the middle surface, the general stiffness equations (3.3.16) simplify considerably. Because of the symmetry of the transformed stiffnesses (Qij )k and of the thicknesses hk , it can be shown that all coupling stiffness Bij of the laminate are zero. There is no coupling. Obviously such laminates are much easier to analyze than laminates with coupling. Consequently, symmetric laminates are commonly used unless special circumstances require an unsymmetrical laminate possessing the coupling property. The constitutive equations for a symmetric laminate are      N11 A11 A12 A16 e11  N22  =  A12 A22 A26   e22  , N12 A16 A26 A66 2e12 (3.4.6)      M11 D11 D12 D16 k11  M22  =  D12 D22 D26   k22  . M12 D16 D26 D66 2k12

In the following, we shall present some special cases of symmetric laminates, determining the stiffness Aij and Dij in each case. For symmetric laminates with multiple isotropic layers, multiple isotropic laminae of various thicknesses are arranged symmetrically about the middle surface from both a geometric and a material property standpoint. The resulting laminate does not exhibit coupling between extension and bending. The extensional and bending stiffnesses are calculated from the equations (3.3.16), where, according to (3.1.6), and assuming an isotropic material, for the k-th layer, we get

Ek , (Q16 )k = (Q26 )k = 0 , 1 − νk2 Ek ν k Ek . , (Q66 )k = (Q12 )k = 2(1 + νk2 ) 1 − νk2

(Q11 )k = (Q22 )k =

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(3.4.7)

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CHAPTER 3. COMPOSITE LAMINATES

In these equations, Ek and νk are the Young’s modulus and the Poisson’s ratio for the k-th lamina. It is easy to see that A11 = A22 , A16 = A26 = 0 , D11 = D22 , D16 = D26 = 0. Hence, the constitutive equations become      N11 A11 A12 0 e11  N22  =  A12 A11 0   e22  , 0 0 A66 N12 2e12   D11 M11  M22  =  D12 0 M12 

D12 D11 0

  k11 0 0   k22  . 2k12 D66

(3.4.8)

(3.4.9)

A symmetric laminate with multiple specially orthotropic layers is made of orthotropic layers that have their principal material directions aligned with the laminate axes, and the layers (laminae) having various thicknesses, are arranged symmetrically about the middle surface both from a geometric and a material property standpoint. The stiffnesses of the laminate are calculated from the general equations (3.3.16), whereas, according to (3.1.6) for the k-th lamina (Q11 )k =

k E2k ν21 E1k E1k , , (Q22 )k = , (Q12 )k = k νk k k k k 1 − ν12 1 − ν12 ν21 1 − ν12 ν21 21

(Q66 )k = Gk12 , (Q16 )k = (Q26 )k = 0 ,

(3.4.10)

k k E1k , E2k , ν12 , ν21 and Gk12 being the engineering material constants of the k-th specially orthotropic lamina. Because (Q16 )k and (Q26 )k are zero, it is easy to see that A16 , A26 , D16 and D26 vanish; i.e. A16 = A26 = 0 , D16 = D26 = 0. (3.4.11)

Also, because of symmetry, the coupling stiffnesses Bij are all zero; i.e. Bij = 0.

(3.4.12)

Hence, the constitutive equation for the laminate takes the form      N11 A11 A12 0 e11  N22  =  A12 A22 0   e22  , N12 0 0 A66 2e12 

  M11 D11  M22  =  D12 0 M12

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D12 D22 0

  k11 0 0   k22  . D66 2k12

(3.4.13)

3.4. SPECIAL CLASSES OF LAMINATES

143

Taking into account the above equations, this type of laminate could be called specially orthotropic laminate in analogy to a special orthotopic lamina. A regular symmetric cross-ply laminate represents a very common special case of symmetric laminates with multiple specially orthotropic laminae (layers). The regular symmetric cross-ply laminate occurs when the laminae are all of the same thickness and material properties, and their major principal material direction (that is, the fiber directions) alternate at 00 or 900 with respect to the laminate (body) axes, for examples (00 /900 /00 ) as in Figure 3.14.

Figure 3.14: Exploded (unbounded) view of a three-layered regular symmetric cross-ply laminate. The regular symmetric cross-ply laminate must have an odd number of layers if we wish to satisfy the symmetric requirement by which coupling between bending and extension is eliminated. Cross-ply laminates with an even number of layers are not symmetric and will be discussed a little later on. Before analyzing other special classes of laminates, let us say a few words about the logic to establish various stiffnesses. For simplicity, we denote x3 by z; i.e. x3 = z. Let us consider the extensional stiffnesess Aij =

N X

k=1

Qij




k

(zk − zk−1 ) ,

given by equation (3.3.16)1 . Since zk − zk−1 > 0 for k = 1, ..., N , the above equation shows that the only
way to have an Aij zero is either for all Qij k to be zero, or for some of the
Qij k to be a negative and some positive, so that the sum of their products with their respective thicknesses be zero. From equation (3.1.17) giving the transformed reduced stiffnesses Qij , follows that Q11 , Q12 , Q22 and Q66 are positive, since all

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CHAPTER 3. COMPOSITE LAMINATES

trigonometrical functions are involved with even powers and Q11 , Q12 , Q22 , Q66 are positive. Thus, A11 , A12 , A22 and A66 are all positive since the thicknesses of the laminae are obviously positive. However, (Q16 )k and (Q26 )k are zero for lamina orientation of 00 or 900 to the laminate axes. Thus, A16 and A26 are zero for laminates made of orthotropic laminae oriented at either 00 or 900 to the laminate axes. Next, we consider the coupling stiffnesses N

Bij =


1X 2 , (Qij )k zk2 − zk−1 2 k=1

given by equation (3.3.16)2 . It is easy to see that if the cross-ply laminate is symmetric about the middle surface, then all the Bij vanish. Finally, we consider the bending stiffnesses N

Dij =


1X 3 , (Qij )k zk3 − zk−1 3 k=1

given by equation (3.3.16)3 .



3 Since zk3 − zk−1 > 0 and Q11 k , Q12 k , Q22 k , Q66 k > 0, it results that

D11 , D12 , D22 and D66 are positive. Also Q16 k and Q26 k are zero for laminae having principal material property orientation of 00 or 900 with respect to the laminate coordinates. Thus, D16 and D26 also vanish. Summing up, we can say that the status of the extensional and bending stiffnesses is the same. As we have seen, a laminate of multiple generally orthotropic layers that are symmetrically disposed about the middle surface exhibits no coupling between bending and extension; that is the Bij are zero. Therefore, the force and moments resultants are given by equation (3.4.6) There, all the Aij and Dij are required because of forces and shearing strains, shearing force and normal strains, normal moments and twist, and twisting moment and normal curvatures coupling between normal forces N11 , N12 and shearing strains e12 , shearing force N12 and normal strains e11 , e22 , normal moments M11 , M22 and twist k12 and twisting moment M12 and normal curvatures k11 , k22 . Such coupling is evidenced by the A16 , A26 , D16 and D26 stiffnesses. A special subclass of this class of symmetric laminates is the regular symmetric angle-ply laminate. Such laminates have orthotropic laminae of equal thicknesses. The adjacent laminae have opposite signs of angle of orientation of the principal material properties with respect to the laminates axes, for example +α/ − α/ + α as in Figure 3.15. For symmetry, there must be an odd number of layers. The aforementioned coupling that involves A16 , A26 , D16 and D26 takes a special form for symmetric angle-ply laminates. Those stiffnesses can be shown to

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3.4. SPECIAL CLASSES OF LAMINATES

Figure 3.15: Exploded (unbounded) view of a three-layered regular symmetric angle-ply laminate.

be largest when N = 3 (the lowest N for which this class of laminates exists) and decrease in proportion to 1/N as N increases A16 =

N X

k=1

N

(Q16 )k (zk − zk−1 ) and D16 =


1X 3 , (Q16 )k zk3 − zk−1 3 k=1

obviously, A16 and D16 are sums of terms of alternating signs since (Q16 )+α = −(Q16 )−α .

(3.4.14)

Consequently, for many layered symmetric angle-ply laminates, the values of A 16 , A26 , D16 and D26 can be quite small when compared to the other Aij and Dij , respectively. 3. Antisymmetric laminates As we have seen, symmetry of a laminate about a middle surface is generally desired to avoid coupling between bending and extension. However, many engineering applications of laminated composite require antisymmetric laminates to achieve design requirements. The general class of antisymmetric laminates has an even numbers of layers if adjacent laminae have alternating signs of the principal material directions with respect to the laminate axes. In addition, each pair of laminae must have the same thickness. It can be shown that in the case of an antisymmetric laminate A16 = A26 = D16 = D26

(3.4.15)

(Q16 )α = −(Q16 )−α , (Q26 )α = −(Q26 )−α ,

(3.4.16)

since

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where the above symbols represent the corresponding transformed reduced stiffnesses in the laminae with the orientation α and −α of the fibers. For general antisymmetric laminates, all coupling stiffnesses are non-vanishing. Hence, the constitutive equation (3.3.15) become         N11 A11 A12 0 e11 B11 B12 B16 k11  N22  =  A12 A22 0   e22  +  B12 B22 B26   k22  , N12 0 0 A66 2e12 B16 B26 B66 2k12 

  M11 B11  M22  =  B12 M12 B16

  k11 0 0   k22  . D66 2k12 (3.4.17) We discuss now two important subclasses of antisymmetric laminates. An antisymmetric cross-ply laminate consists of an even number of orthotropic laminae laid down on each other with principal material directions alternating at 00 and 900 to the laminate axes, as in the example given in Figure 3.16. B12 B22 B26

   B16 e11 D11 B26   e22  +  D12 0 B66 2e12

D12 D22 0

Figure 3.16: Exploded (unbounded) view of a two-layered regular antisymmetric cross-ply laminate. The antisymmetric cross-ply laminates do not have non-zero A16 , A26 , D16 , D26 , but do have coupling between extension and bending. More precisely A16 = A26 = 0 , D16 = D26 = 0 , B11 = −B22 , B12 = B16 = B26 = B66 = 0. (3.4.18) Hence, the constitutive equations of the antisymmetric cross-ply laminates become         k11 e11 B11 0 0 N11 A11 A12 0  N22  =  A12 A22 −B11 0   k22  , 0   e22  +  0 0 0 0 0 0 A66 2k12 2e12 N12 

  M11 B11  M22  =  0 M12 0

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0 −B11 0

   0 e11 D11 0   e22  +  D12 0 2e12 0

D12 D22 0

  0 k11 0   k22  . D66 2k12 (3.4.19)

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3.4. SPECIAL CLASSES OF LAMINATES

A regular antisymmetric cross-ply laminate is a special case, which has laminae of equal thickness. This type of laminate is common because of simplicity of fabrication. It can be shown that the coupling stiffness B11 of an antisymmetric cross-ply laminate approaches zero as the number of layers increases, for a fixed laminate thickness. An antisymmetric angle-ply laminate has laminae oriented at +α degrees to the laminate coordinate axes on one side of the middle surface and the corresponding equal thickness laminae on the other side is oriented at −α degrees. A simple example is given in Figure 3.17.

y = x2

x = x1

Figure 3.17: Exploded (unbounded) view of a two-layered regular antisymmetric angle-ply laminate. A regular antisymmetric angle-ply laminate has all laminae of the same thicknesses. It can be shown that for an antisymmetric angle-ply laminate, the following stiffnesses are vanishing: A16 = A26 = 0 , B11 = B12 = B22 = 0 , D16 = D26 = 0.

(3.4.20)

Consequently, the global constitutive equations become         N11 A11 A12 0 e11 0 0 B16 k11  N22  =  A12 A22 0   e22  +  0 0 B26   k22  , 0 0 A66 N12 2e12 B16 B26 0 2k12 

  M11 0  M22  =  0 M12 B16

   B16 e11 D11 B26   e22  +  D12 0 2e12 0

  0 k11 0   k22  . D66 2k12 (3.4.21) It can be shown for a fixed laminate thickness that the coupling stiffnesses B16 and B26 tend towards zero as the number of layers in the laminate increases. Summing up the presented results, concerning some special classes of fiberreinforced composite laminates, we can say the following:

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0 0 B26

D12 D22 0

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(1) Simple layer laminates with a reference surface, at the middle surface do not exhibit coupling between extension and bending. (2) Multilayered laminates, in general, develop coupling between extension and bending. (3) The coupling is influenced by the geometrical as well as by the material properties of the laminae. (4) There exist combinations of the material properties and geometrical characteristics for which there is no coupling between extension and bending. (5) The elastic symmetry of the laminae (transversally isotropy, orthotropy) is not necessarily conserved in the laminate. (6) Moreover, the symmetries can be increased, decreased or remain the same. (7) The symmetry properties of the three stiffness matrices [A], [B] and [D] need not be the same, and are generally different. We stress the fact that the fundamental concept of coupling between extension and bending must be well understood, because there exists many applications of the composite laminates where the neglect of the coupling can be catastrophic. Considering coupling is the key to the correct analysis of eccentrically stiffened plates. The procedure to describe a laminate by use of individual layer thicknesses, principal material property orientations, and overall sequence can be quite involved. However, all pertinent parameters can be represented in a simple way if one uses the following stacking sequence terminology. For regular (equal thickness layers) laminates, a listing of layers and their orientation suffices, for example, [00 /900 /450 ]. Let us observe that only the principal material direction orientations need to be given. For irregular (when the layers do not have the same thickness) laminates, a notation of layers’ thickness must be added to the previous notation, for example, [00 /h1 /900 /h2 /450 /h3 ]. For symmetric laminates, the simplest representation of the laminate [0 0 /900 / 0 45 /450 /900 /00 ] is [00 /900 /450 ] sym. We shall discuss now in greater details the relation by which the cross-ply laminate stiffnesses can be expressed. We recall that a cross-ply laminate has N unidirectionally reinforced orthotropic layers with the principal material directions alternatingly oriented at 00 and 900 with respect to the laminate coordinate axes. The fiber direction of the odd -numbered layers is the x1 direction of the laminate. The fiber direction of the even-numbered layers is the x2 direction of the laminate. We assume that all odd-numbered layers have the same thickness, all even-numbered layers have also equal thicknesses but the odd and even numbered layers do not necessarily have the same thickness. For this special but important case for applications, two geometrical parameters are important: (1) N, the total number of layers, and

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3.4. SPECIAL CLASSES OF LAMINATES

(2) m, the ratio of the total thicknesses of odd numbered layers to the total thickness of the even-numbered layers, called cross-ply ratio. Hence,

m=

P

hk

k=odd

P

hk

.

(3.4.22)

k=even

For instance, for a five-layered cross-ply laminate, which has a lamination or stacking sequence [00 /h1 /900 /2h1 /00 /2h1 /900 /2h1 /00 /h1 ], we get m=

h1 + h 1 + h 1 = 3/4. 2h1 + 2h1

Let us observe that the cross-ply ratio m has a specific meaning only when the laminae (layers) have alternating 00 and 900 orientation! Tsai (see Jones [3.2], Chapter 4) has shown that the laminate stiffnesses Aij , Bij and Dij for cross-ply laminates with odd or even numbers of layers, can be expressed in terms of m and N . In addition, Tsai uses the stiffness ratio f = Q22 /Q11 = E1 /E2 6 1

(3.4.23)

of principal lamina stiffnesses. Tsai has obtained the following expressions for the stiffnesses: Symmetric cross-ply laminates (N odd ).

1 + mf 1 + mf m+f A11 , hQ11 = hQ11 , A12 = hQ12 , A22 = m+f 1+m 1+m A16 = A26 = 0 , A66 = hQ66 , Bij = 0, h3 h2 1+m (f − 1)p + 1 3 Q12 , {(f − 1)p + 1} A11 , D12 = h Q11 = D11 = 12 12 m+f 12 h2 1+m (1 − f )p + f 3 {(1 − f )p + f } A11 , h Q11 = D22 = 12 m+f 12 h3 Q66 , (3.4.24) D16 = D26 = 0 , D66 = 12

A11 =

where p=

m(N − 3){m(N − 1) + 2(N + 1)] 1 . + (N 2 − 1)(1 + m)3 (1 + m)3

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150

CHAPTER 3. COMPOSITE LAMINATES Antisymmetric cross-ply laminates (N even)

1 + mf 1 + mf m+f A11 , hQ11 = hQ11 , A12 = hQ12 , A22 = m+f 1+m 1+m A16 = A26 = 0, A66 = hQ66 , m(f − 1) m(f − 1) 2 hA11 , B22 = −B11 , h Q11 = B11 = N (1 + m)(f + m) N (1 + m)2 B12 = B16 = B26 = B66 = 0, h3 h2 1+m (f − 1)r + 1 3 Q12 , {(f − 1)r + 1} A11 , D12 = h Q11 = D11 = 12 12 m+f 12 h2 1+m (1 − f )r + f 3 {(1 − f )r + f } A11 , h Q11 = D22 = 12 m+f 12 3 h Q66 , (3.4.25) D16 = D26 = 0 , D66 = 12

A11 =

where r=

8m(m − 1) 1 . + 1 + m N 2 (1 + m)3

Concerning the above formulas, we can make the following observations: (1) For both symmetric and antisymmetric cross-ply laminates, the extensional stiffnesses Aij are independent of N , the number of layers. (2) However, the stiffness ratios A11 and A22 depend on the cross-ply ratio m, as well as on the stiffness ratio f . (3) The stiffnesses A12 and A66 are independent on m and f . (4) The remaining stiffnesses A16 and A26 are zero for all cross-ply laminates. (5) All coupling stiffnesses Bij are zero for the symmetric cross-ply laminates. (6) For the antisymmetric cross-ply laminates only, the bending stiffnesses B11 and B22 are not vanishing. (7) The values of B11 and B22 decrease as N , number of laminae, increases, the thickness h of the laminate being fixed. Since N must be even to get any coupling, N = 2 corresponds to the largest coupling between extension and bending. (8) The bending stiffnesses D11 and D22 are involved functions on the numbers of layers, N , the cross-ply ratio, m, and the stiffness ratio, f . (9) The value of D11 approaches A11 h2 /12 and D22 approaches A22 h2 /12 as N gets large or as m gets large, or as f approaches 1. Regular antisymmetric cross-ply laminates (N even) In this case, all layers have the same thickness h/N , hence, the cross-ply ratio m is one; i.e. m = 1.

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3.4. SPECIAL CLASSES OF LAMINATES

Consequently, Tsai’s formulas (3.4.25) are considerably simplified and we get: 1+f hQ11 , A12 = hQ12 , A16 = A26 = 0 , A66 = hQ66 , 2 1 f −1 f −1 2 hA11 , B12 = B16 = B26 = B66 = 0, h Q11 = B11 = −B22 = 2N f + 1 4N h3 h2 f +1 3 Q12 , A11 , D12 = h Q11 = D11 = D22 = 12 12 24 h3 Q66 . (3.4.26) D16 = D26 = 0 , D66 = 12

A11 = A22 =

Since E2 6 E1 , we have f 6 1; thus, B11 < 0

and

B22 > 0.

(3.4.27)

In the following, we shall indicate the way in which Tsai’s formulas (3.4.26) for a regular antisymmetric cross-ply laminate can be obtained. Such a laminate is shown in the Figure 3.18.

Figure 3.18: Regular cross-ply laminate having N (even) layers. Since all laminae have the same thickness h/N , we have

h h for k = 0, 1, ..., N. (3.4.28) +k N 2 According to the definition of a regular antisymmetric cross-ply laminate, we have also θk = 00 if k is odd and θk = 900 if k is even . (3.4.29) zk = −

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CHAPTER 3. COMPOSITE LAMINATES

Conceptually, from the general equations (3.1.17), giving the transformed reduced stiffnesses Qij , we get (Q11 )k = Q11 if k is odd and (Q11 )k = Q22 if k is even, (Q12 )k = Q12 for any k, (Q22 )k = Q22 if k is odd and (Q22 )k = Q11 if k is even, (Q66 )k = Q66 for any k, (Q16 )k = (Q26 )k = 0 for any k. In order to prove Tsai’s relations (3.4.26), the above properties must be used. We begin with A11 , which, according to (3.3.16)1 , is given by the equation (x3 = z!) N X A11 = (Q11 )k (zk − zk−1 ). k=1

Since all layers have the same thickness h/N , we obviously have zk − zk−1 = h/N for any k.

(3.4.30)

Hence, N h X (Q11 )k . N

A11 =

k=1

According to the above results, the reduced transformed thickness (Q11 )k takes N/2-times the value Q11 and N/2-times the value Q22 . Hence,

A11 =

h (Q11 + Q22 ). 2

Using the stiffness ratio f = Q22 /Q11 , we get: A11 =

1+f hQ11 . 2

Hence, we get the first Tsai formula (3.4.26), giving the extensional stiffness A 11 . In the same way all Tsai formulas can be deduced concerning the extensional stiffnesses Aij . Let us analyze now the coupling stiffnesses. According to (3.3.16)2 , we have (x3 = z) N 1X 2 ). (Qij )k (zk2 − zk−1 Bij = 2 k=1

At the same time from (3.4.28) and (3.4.30), we get 2 zk2 − zk−1 =

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h2 {−(1 + N ) + 2k}. N2

153

3.4. SPECIAL CLASSES OF LAMINATES Hence, the formula giving B11 becomes B11 =

N

N

k=1

k=1

X X h2 k(Q11 )k }. (Q11 )k + 2 {−(1 + N ) 2N

We recall again that (Q11 )k = Q11 if k is odd, and (Q11 )k = Q22 if k is even. Thus we obtain

B11 =

X X N (1 + N ) h2 (Q11 + Q22 ) + 2(Q11 k + Q22 k)}. {− 2 2N k=odd

k=even

Now let us use the well known relation n X

l=

l=0

n(n + 1) . 2

(3.4.31)

Thus, after some elementary computations, we get X

k=

N (N + 2) N2 X . , k= 4 4

(3.4.32)

k=even

k=odd

Introducing these values in the last expression of B11 , after some algebra, it results B11 =

h2 (−Q11 + Q22 ). 4N

But Q22 = f Q11 , and the above equation becomes B11 =

f −1 2 h Q11 . 4N

Therefore, we have obtained the first Tsai formula (3.4.26), giving the coupling stiffness B11 . In the same way, all Tsai formulas can be deduced (3.4.26) giving the coupling stiffness Bij. According to (3.3.16)3 , for the bending stiffness D11 , we have (x3 = z) N

D11 =

1X 3 ). (Q11 )k (zk3 − zk−1 3 k=1

Furthermore, from (3.4.28) and (3.4.30), it results 3 = zk3 − zk−1

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h3 N3



 3N 2 + 6N + 4 − 3(N + 1)k + 3k 2 . 4

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CHAPTER 3. COMPOSITE LAMINATES

Also, we know that (Q11 )k takes N/2 times the value Q11 , if k is odd, and N/2-times the value Q22 , if k is even. Hence, the equation giving D11 becomes

D11 =

X X h3 3N 2 + 6N + 4 N (Q + Q ) − 3(N + 1)(Q k + Q k) { 11 22 11 22 2 N 3N 3 k=odd

X

+3(Q11

k 2 + Q22

k=odd

X

k=even

k 2 )}.

k=even

We use again the relation (3.4.31) and the well known formula n X

l2 =

l=o

n(n + 1)(2n + 1) . 6

(3.4.33)

Thus, after some algebra, from (3.4.31) and (3.4.33), we get X

k=odd

k2 =

N (N + 1)(N + 2) N (N 2 − 1) X 2 . , k = 6 6

(3.4.34)

k=even

Introducing (3.4.32) and (3.4.34) in the last expression of D11 , elementary computations leads to D11 =

1+f 3 h2 h Q11 , (Q11 + Q22 ) = 24 24

since Q22 = f Q11 , according to the definition of the rigidity ratio f. In this way we have derived the Tsai formula (3.4.26) giving the bending stiffness D11 . In the same way can be deduced all Tsai formulas concerning the bending stiffnesses Dij .

3.5

Equilibrium equations and boundary conditions

In the theory of thin plate, the Cauchy’s equilibrium equations are replaced by global equilibrium conditions satisfied by the force resultants N αβ , moment resultants Mαβ and by the resultant shear forces Qα defined by the relations Qα =

Z

h/2

σα3 dx3 , α = 1, 2.

(3.5.1)

−h/2

To get these equations, we start with the Cauchy’s (local) equilibrium conditions, assuming absence of body forces σαβ,β + σα3,3 = 0 , σ3α,α + σ33,3 = 0, α, β = 1, 2.

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(3.5.2)

3.5. EQUILIBRIUM EQUATIONS AND BOUNDARY CONDITIONS

155

We integrate these equations with respect to x3 from −h/2 to h/2. At the same time, we take into account that integration with respect to x3 , and differentiation with respect to x1 and x2 are invertible. Thus, from (3.5.2), and using equations (3.3.13)1 and (3.5.1), which are defining Nαβ and Qα , respectively, we get Nαβ + σα3 (h/2) − σα3 (−h/2) = 0 , Qα,α + σ33 (h/2) − σ33 (−h/2) = 0 , α, β = 1, 2. We recall now the assumption (8) made at the beginning of Section 3.3. According to the made assumption, σα3 are vanishing on the faces x3 = ±h/2 of the laminate. Hence, σα3 (±h/2) = 0. Also, we interpret the function q(x1 , x2 ) = σ33 (x1 , x2 , h/2) − σ33 (x1 , x2 , −h/2) as representing a given external surface force distribution, acting normal to the face x3 = −h/2 of the laminate. Taking into account the above observations, we obtain the first three (global) equilibrium equation that must be satisfied by the resultants in-plane forces N αβ and by the resultant shear forces Qα , in the plane domain D occupied by the middle surface of the laminate Nαβ,β = 0 , Qα,α + q = 0 in D, α, β = 1, 2.

(3.5.3)

In order to obtain the (global) equilibrium equations satisfied by the moment resultants Mαβ , we multiply the first two Cauchy equations (3.5.2) by x3 and integrate the obtained results with respect to x3 from −h/2 to h/2. In this way, using equation (3.3.13)2 defining Mαβ , and the invertibility of differentiation with respect to x1 ,x2 and of integration with respect to x3 , we get Mαβ,β +

Z

h/2

x3 σα3,3 dx3 = 0. −h/2

We have x3 σα3,3 = (x3 σα3,3 ) − σα3 . Consequently, Z

h/2

x3 σα3,3 dx3 = −h/2

h h σα3 (h/2) − σα3 (−h/2) − 2 2

Z

h/2

σα3 dx3 . −h/2

We recall that σα3 (±h/2) = 0 and we will use the definition (3.5.1) of the resultant shear forces Qα .

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CHAPTER 3. COMPOSITE LAMINATES

Thus, we finally get the equilibrium condition which must be satisfied by the shear forces resultants Qα and moment resultants Mαβ , in the plane domain D occupied by the middle surface of the laminate Mαβ,β − Qα = 0 in D, α, β = 1, 2.

(3.5.4)

From (3.5.3)2 and (3.5.4), we can now see that the resultant moments Mαβ must satisfy the following second order (global) equilibrium equation Mαβ,αβ + q = 0 in D α, β = 1, 2.

(3.5.5)

It remains to formulate the boundary conditions which can be prescribed and must be satisfied by the displacement Uα , U3 , by the force resultants Nαβ and by the moments resultants Mαβ on the boundary ∂D of the domain D occupied by the middle surface of the laminate in the x1 , x2 plane. In order to obtain the possible boundary condition (assuming uniqueness of the solution), we recall the geometrical equations (3.3.4) and (3.3.5) expressing the plane strains eαβ and the curvatures kαβ in terms of the displacements Uα , U3 , the constitutive relations in tensorial form (3.3.21) expressing the force and moments resultants Nαβ , Mαβ in terms of eαβ , kαβ , and the just established equilibrium conditions (3.5.3), (3.5.4) and (3.5.5). We have Geometrical equations: eαβ =

1 (Uα,β + Uβ,α ), kαβ = −U3,αβ ; 2

(3.5.6)

Constitutive equations: Nαβ = Aαβγϕ eγϕ + Bαβγϕ kγϕ , Mαβ = Bαβγϕ eγϕ + Dαβγϕ kγϕ ;

(3.5.7)

Equilibrium equations: Nαβ,β = 0 , Qα,α + q = 0 , Mαβ,β − Qα = 0 , Mαβ,αβ + q = 0.

(3.5.8)

We recall also that according to (3.3.23), the constitutive coefficients have the following symmetry properties : Aαβγϕ = Aβαγϕ = Aαβϕγ = Aγϕαβ , Bαβγϕ = Bβαγϕ = Bαβϕγ = Bγϕαβ ,

(3.5.9)

Dαβγϕ = Dβαγϕ = Dαβϕγ = Dγϕαβ . Obviously the Greek indices take the values 1, 2 and the Einstein’s summation convention applies.

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3.5. EQUILIBRIUM EQUATIONS AND BOUNDARY CONDITIONS

157

In the following we shall prove a theorem of work and energy, appropriate to the Love-Kirchhoff type of plate theory developed above. As before, we denote by D the plane domain occupied by the middle surface of the plate, ∂D will be the boundary curve of D. Let n be the unit outward normal to ∂D, and let τ be the unit tangent vector to ∂D, as shown in Figure 3.19.

Figure 3.19: The plane domain occupied by the middle surface of the laminate. Denoting by nα and τα the (plane) components of n and τ , we get n1 = τ2 , n2 = −τ1 .

(3.5.10)

We introduce now the two-dimensional vectors N n and M n defined on ∂D, by the following equations, giving their components Nnα and Mnβ : Nnα = Nαβ nβ , Mnα = Mαβ nβ .

(3.5.11)

Obviously, Nn and Mn are corresponding to the Cauchy’s stress vector, and represent the resultant force vector and the resultant moment vector, respectively, acting on the boundary ∂D of the laminate middle surface. The normal and tangential components Nnn , Nnτ and Mnn , Mnτ of these vectors have the following expression: Nnn = nα Nnα , Nnτ = τα Nnα , Mnn = nα Mnα , Mnτ = τα Mnα .

(3.5.12)

In the same way, the normal and tangential components Un and Uτ of the in-plane displacement on the boundary ∂D can be obtained using the equations Un = n α Uα , U τ = τα Uα .

(3.5.13)

Also, for a later use, we introduce on the boundary ∂D the normal and tangential derivatives U3,n and U3,τ of the normal displacement U3 . According to

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CHAPTER 3. COMPOSITE LAMINATES

the general definition of the directional derivative of a scalar field, given in the Section 1.2, we have U3,n = nα U3,α , U3,τ = τα U3,α . (3.5.14) Using these equations, we can express the partial derivatives U3,α through the normal and tangential derivatives U3,n , U3,τ . From (3.5.10) and (3.5.14), we get U3,1 = τ2 U3,n + τ1 U3,τ , U3,2 = −τ1 U3,n + τ2 U3,τ . (3.5.15)

These equations will be used a little later on. Multiplying the equilibrium equation (3.5.8)1 by Uα , integrating the obtained result on D, using Green’s theorem, the symmetry of Nαβ , the geometrical equations (3.5.6)1 and the Cauchy’s type relations (3.5.11)1 , we obtain Z Z Nαβ eαβ da = Uα Nnα ds. (3.5.16) D

∂D

The two-dimensional scalar product Uα Nnα of the two-dimensional vectors (U1 , U2 ) and (Nn1 , Nn2 ) can be expressed in an equivalent form using the normal and tangential components of the involved vectors; i.e. Uα Nnα = Un Nnn + Uτ Nnτ . Thus, equation (3.5.16) becomes Z Z Nαβ eαβ da = D

(Un Nnn + Uτ Nnτ )ds.

(3.5.17)

(3.5.18)

∂D

We multiply now the second equilibrium equation (3.5.8)2 by U3 and integrate the obtained result on D. In this way, Green’s theorem leads to the following relation: Z Z Z Qα U3,α da. (3.5.19) U3 Qα nα ds = qU3 da + D

∂D

D

In order to transform the right hand side term of this equation, we now use the equilibrium condition (3.5.8)3 . We get Z Z Qα U3,α da = U3,α Mαβ,β da. (3.5.20) D

D

Using again the Green’s theorem, the geometrical equations (3.5.6)2 and the Cauchy’s type relations (3.5.11), from (3.5.20) we obtain Z Z Z Qα U3,α da = U3,α Mnα ds + Mαβ kαβ da. (3.5.21) D

∂D

D

Consequently, from (3.5.19) and (3.5.21), it results Z Z Z Z qU3 da + U3 Qn ds = U3,α Mnα ds + Mαβ kαβ da, D

∂D

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∂D

D

(3.5.22)

3.5. EQUILIBRIUM EQUATIONS AND BOUNDARY CONDITIONS

159

where Qn = Q α n α

(3.5.23)

is the normal component of the resultant shear forces (Q1 , Q2 ). In order to obtain the final results, we still have to express in an adequate form the first term from the right hand side of equation (3.5.22). To do this, we use equation (3.5.15). In this way we successively get U3,α Mnα

= = =

U3,1 Mn1 + U3,2 Mn2 (τ2 U3,n + τ1 U3,τ )Mn1 + (−τ1 U3,n + τ2 U3,τ )Mn2 (τ1 Mn1 + τ2 Mn2 )U3,τ + (τ2 Mn1 + τ1 Mn2 )U3,n .

We use now equations (3.5.10) and (3.5.12)3,4 to get U3,α Mnα

=

(τ1 Mn1 + τ2 Mn2 )U3,τ + (n1 Mn1 + n2 Mn2 )U3,n

=

Mnτ U3,τ + Mnn U3,n .

(3.5.24)

Transforming adequately the first term in the right hand side of (3.5.24), we obtain U3,α Mnα = (U3 Mnτ ),τ − U3 Mnτ,τ + U3,n Mnn . The tangential derivatives involved in this equation have the following expressions: (U3 Mnτ ),τ = (U3 Mnτ ),1 τ1 + (U3 Mnτ ),2 τ2 , Mnτ,τ = Mnτ,1 τ1 + Mnτ,2 τ2 . We assume that U3 and Mnτ are uniform function on ∂D. Hence, Z (U3 Mnτ ),τ ds = 0.

(3.5.25)

∂D

since ∂D is a closed curve. Thus from (3.5.24) and (3.5.25) we finally get Z Z U3,α Mnα ds = (−U3 Mnτ,τ + U3,n Mnn )ds. ∂D

Z

∂D

Using this equation in (3.5.22), we obtain Z Z Mαβ kαβ da. {U3 (Qn + Mnτ,τ ) − U3,n Mnn }ds = qU3 da +

D

(3.5.26)

D

∂D

Now, equations (3.5.18) and (3.5.26) lead to the following work relation: Z Z {Nnn Un + Nnτ Uτ + (Qn + Mnτ,τ )U3 − U3,n Mnn }ds + qU3 da ∂D D Z = 2 wda, (3.5.27) D

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where the quadratic form w is defined by the following equation: 2w = Nαβ eαβ + Mαβ kαβ .

(3.5.28)

The work relation (3.5.27) is a direct consequence of the geometrical equation (3.5.6) and equilibrium condition (3.5.8). Using the constitutive relations (3.5.7) and the symmetry properties (3.5.9), w can be expressed in the following form: w=

1 1 eαβ Aαβγϕ eγϕ + eαβ Bαβγϕ kγϕ + kαβ Dαβγϕ kγϕ . 2 2

(3.5.29)

As in the usual elasticity theory, the left-hand side of the work relation (3.5.27) represents the total work of the external force acting on the laminate. Consequently, w is the specific elastic energy stored in the deformed laminate. Hence, the total elastic energy W stored in the laminate is Z W = wda. (3.5.30) D

The above interpretation and equation (3.5.27) represents the content of the announced and proved theorem of work and energy. This theorem ”tells” us what kind of boundary conditions can be given on the boundary ∂D, in order to assume the uniqueness of the solution of various boundary value problems. The structure of the left-hand side of the work relation (3.5.27) shows that the following fields can be prescribed on the boundary ∂D: Un or Nnn , Uτ or Nnτ , U3 or Qn + Mnτ,τ and U3,n or Mnn .

(3.5.31)

The same work relations show that the following theorem takes place. Uniqueness theorem. If the specific elastic energy w is a positive definite quadratic form, the various boundary value problems can have no more than one regular solution, modulo a rigid displacement.

3.6

Variational and extreme principles

We observe that due to the symmetry relations (3.5.9), the differential system (3.5.6)−(3.5.8) is self-adjoint. Due to this fact we can establish various variational and extreme principles, corresponding to various boundary value problem. In order to do this, we introduce first the three-dimensional vectors U and δU having the components (Uα , U3 ) and (δUα , δU3 ), respectively. Also we consider the energy functional Z W (U) = w(U)da (3.6.1) D

where w is the quadratic form given by the equation (3.5.28) and (3.5.29).

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3.6. VARIATIONAL AND EXTREME PRINCIPLES

For brevity we denote simply by δW the first variation of W in U, in the direction δU. According to the usual definition, we have δW =

d W (U + λδU)|λ=0 , λ being a real variable. dλ

(3.6.2)

In order to evaluate the above variation, we take into account the geometric relation (3.5.6), the constitutive equations (3.5.7), the symmetry properties (3.5.8) and the expression (3.5.29) of the specific elastic (strain) energy w. Thus, after some elementary computations, we get Z δW = (Nαβ δUα,β − Mαβ δU3,αβ )da. (3.6.3) D

Using Green’s formula, we obtain Z δW = − (Nαβ,β δUα + Mαβ,αβ δU3 )da D

+

Z

∂D

(δUα Nαβ nβ + δU3 Mαβ,β nα − δU3,α Mαβ nβ )ds.

(3.6.4)

To obtain this equation, we have used the following relations: Mαβ δU3,αβ = (Mαβ δU3,α ),β − Mαβ,β δU3,α = (Mαβ δU3,α ),β − (Mαβ,β δU3 ),α + Mαβ,αβ δU3 . Using the Cauchy’s type relation (3.5.11)1 , we get δUα Nαβ nβ = δUα Nnα = δUn Nnn + δUτ Nnτ ,

(3.6.5)

δUn and δUτ being the normal and tangential components of the in-plane displacement field (δU1 , δU2 ). Similarly, using the equilibrium condition (3.5.8)4 and the equation (3.5.23), we obtain δU3 Mαβ,β nα = δU3 Qα nα = δU3 Qn , (3.6.6) Qn = Qα nα being the normal component of the resultant shear force (Q1 , Q2 ). Using a relation analogous to (3.5.27), we get δU3,α Mαβ nβ = δU3,α Mnα = (δU3 Mnτ ),τ − δU3 Mnτ,τ + δU3,n Mnn .

(3.6.7)

Introducing (3.6.5), (3.6.6), (3.6.7) in the line integral of equation (3.6.4), and observing that Z (δU3 Mnτ ),τ ds = 0, ∂D

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we obtain the expression of the variation δW Z δW = − (Nαβ,β δUα + Mαβ,αβ δU3 )da +

Z

D

∂D

{δUn Nnn + δUτ Nnτ + δU3 (Mnτ,τ + Qn ) − δU3,n Mnn } ds.

(3.6.8)

The last equation can be used to obtain variational principles, corresponding to different boundary value problems.We shall illustrate the procedure by analyzing two possibilities. In order to do this we shall interpret the third equilibrium condition (3.5.8) 3 , that is the relations Qα = Mαβ,β (3.6.9) as representing supplementary constitutive equations, expressing the resultant shear forces Qα in terms of the in-plane deformations eαβ and of the curvatures kαβ .The status of the relations (3.6.5) is similar to that of the constitutive equations (3.5.7) expressing the resultant in-plane forces Nαβ and the resultant bending moments Mαβ in terms of the above mentioned kinematical fields. Also, we suppose that the involved displacement fields U and their variations δU are of class C 2 on D and of class C 1 on D = D ∪∂D, since we take into account only regular solutions. First let us assume that on the boundary ∂D are given Nnn , Nnτ , Mnτ,τ +Qn and Mnn ; i.e.

Nnn = Φ, Nnτ = Ψ, Mnτ,τ + Qn = Γ, Mnn = ∆ on ∂D,

(3.6.10)

where Φ, Ψ, Γ, ∆ are given continuous functions on ∂D. We suppose also that q is a given continuous function on D. In this case, we introduce the functional I(U) defined by the equation Z Z qda. (3.6.11) (ΦUn + ΨUτ + ΓU3 − ∆U3,n )ds − I(U) = W (U) − ∂D

D

Taking into account equation (3.6.8), for the variation δI of I in U, in the direction δU, we get Z δI = − {Nαβ,β δUα + (Mαβ,αβ + q)δU3 } da D Z + {(Nnn − Φ)δUn + (Nnτ − Ψ)δUτ ∂D

+(Mnτ,τ + Qn − Γ)δU3 − (Mnn − ∆)δU3,n }da.

(3.6.12)

Using the above equation, we can prove the following. First variational principle. If U is a regular solution of the boundary value problem (3.6.10), the first variation δI of I is vanishing in U for any direction

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δU. Conversely, if δI is vanishing in U for any direction δU, then U is a regular solution of the boundary value problem (3.6.10). In order to prove this theorem, let us assume that U is a regular solution of the considered boundary value problem. Hence, the equilibrium equations (3.5.8) 1 , (3.5.8)4 and the boundary conditions (3.6.10) are satisfied. Hence, according to (3.6.12), δI = 0 in U for any δU. In order to prove the inverse implication, we assume that δI = 0 in U for any δU. In this case the relation (3.6.12) shows that the equilibrium equations (3.5.8)2 and (3.5.8)4 , as well as the boundary conditions (3.6.10) are satisfied. From the assumed supplementary constitutive relations (3.6.5), it follows that the remaining equilibrium conditions (3.5.8)2 and (3.5.8)3 are also verified and the proof is complete. Let us observe that in the above variational principle, the variations δU are not submitted to any restrictions on the boundary ∂D. In this sense we consider the boundary conditions (3.6.10) as being natural boundary conditions. The situation is entirely analogous to that encountered in the usual elasticity theory if the traction is prescribed on the boundary of the body. Let us suppose now that, on the boundary ∂D, are given Un , Uτ , U3 and U3,n ; Un = ϕ , Uτ = ψ , U3 = γ and U3,n = δ on ∂D, (3.6.13) where ϕ, ψ, γ, δ are continuous functions given on ∂D. We suppose also that q is a continuous function given on D. In this case we shall introduce the functional J(U) defined by the equation Z J(U) = W (U) − qU3 da. (3.6.14) D

We shall calculate the variation δJ of J in U, in a direction δU that satisfies homogeneous boundary conditions; i.e. δUn = δUτ = δU3 = δU3,n = 0 on ∂D. Using again (3.6.8) and taking into account (3.6.15), we get Z {Nαβ,β δUα + (Mαβ,αβ + q)δU3 } da. δJ = −

(3.6.15)

(3.6.16)

D

Based on the last equation, we can prove the following. Second variational principle. If U is a regular solution of the boundary value problem (3.6.13), then the variation δJ of J in U is vanishing for any direction (variation) δU satisfying the homogeneous boundary conditions (3.6.16). Conversely, if U is a solution of the given boundary conditions (3.6.13) and if the variation δJ of J is vanishing in U for any variation (direction) δU, which satisfies the homogeneous boundary conditions (3.6.15), then U is a regular solution of the boundary value problem (3.6.13).

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The proof of the second variational principle is similar to that given for the first one, and is entirely based on equation (3.6.16). Let us observe that in the second variational principle, U and δU are not arbitrary on the boundary ∂D; U must satisfy the given boundary conditions and δU must satisfy homogeneous boundary conditions. In this sense, we consider the boundary conditions (3.6.13) as being essential boundary conditions. The situation is entirely analogous to that existing in the usual elasticity theory where the displacement is prescribed on the boundary of the body. The above boundary conditions are not typical either in the classical plate theory or in the composite laminate theory. In these domains, the most frequently encountered boundary value problems have a mixed character. That is, on boundary ∂D are given some data concerning the displacements and complementary information concerning the force and moment resultants. Jones shows ([3.2], Chapter 5) that the most frequently used boundary conditions in laminate theory are formulated as a choice between simply supported, clamped or free edges. The situation for laminate plates (laminates) is complex because there are actually four types of boundary conditions that can be called simply supported edges and four types of boundary conditions that can be called clamped edges. According to Jones, the eight types of simply supported (prefix S) and clamped (prefix C) boundary conditions are commonly classified as S1 : U3 = 0 , Mnn = 0 , Un = ϕ , Uτ = ψ, S2 : U3 = 0 , Mnn = 0 , Nnn = Φ , Uτ = ψ, S3 : U3 = 0 , Mnn = 0 , Un = ϕ , Nnτ = Ψ, S4 : U3 = 0 , Mnn = 0 , Nnn = Φ , Nnτ = Ψ,

(3.6.17)

and C1 : U3 = 0 , U3,n = 0 , Un = ϕ , Uτ = ψ, C2 : U3 = 0 , U3,n = 0 , Nnn = Φ , Uτ = ψ, C3 : U3 = 0 , U3,n = 0 , Un = ϕ , Nnτ = Ψ, C4 : U3 = 0 , U3,n = 0 , Nnn = Φ , Nnτ = Ψ, on ∂D.

(3.6.18)

In these relations, ϕ, ψ, Φ and Ψ are given functions on the boundary curve ∂D. The functionals and variational principles corresponding to these eight boundary conditions can be obtained as before, examining the structure of equation (3.6.8) together with the structure of the boundary conditions taken into account. Let us observe that the variational principle corresponding to the boundary condition C1 represents a particular case of the second variational principle discussed above. In the following we shall prove a relation which can be used to obtain minimum principles. Let us consider two displacement fields U and U0 , and as denoted 0 by eαβ , kαβ and e0αβ , kαβ , let the in-plane strains and the curvatures correspond

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to U and U0 , respectively. Let us evaluate the functional W corresponding to U + U0 . According to equations (3.5.29) and (3.5.30), we have Z 1 0 { (eαβ + e0αβ )Aαβγϕ (eγϕ + e0γϕ )+ W (U + U ) = D 2

1 0 0 0 )}da. )Dαβγϕ (kγϕ + kγϕ ) + (kαβ + kαβ +(eαβ + e0αβ )Bαβγϕ (kγϕ + kγϕ 2 Taking into account the symmetry relations (3.5.9) and the constitutive equations (3.5.7), we get Z 0 Mαβ )da, W (U + U0 ) = W (U) + W (U0 ) + (e0αβ Nαβ + kαβ D

where Nαβ and Mαβ correspond to U. Now we recall the geometric relations (3.5.6) and use Green’s theorem to transform the surface integral into a line integral. Taking into account supplementary constitutive equations (3.6.5) and using the relations of type (3.5.24) on (3.5.25), we obtain Z W (U + U0 ) = W (U) + W (U0 ) − (Uα0 Nαβ,β + U30 Mαβ,αβ )da D Z 0 Mnn }ds. + {Un0 Nnn + Uτ0 Nnτ + U30 (Qn + Mnτ,τ ) − U3,n ∂D

(3.6.19)

In order to obtain minimum principles, this equation plays the same role as that played by equation (3.6.8) used to obtain variational principles. Let us consider first the boundary value problem (3.6.10) and the functional I(U) defined by equation (3.6.11). As in the usual elasticity theory, we call I(U) the potential energy of the laminate, corresponding to the “traction” boundary value problem (3.6.10). We shall denote by A the set of all regular displacement fields U. Using equation (3.6.19), we can formulate and prove a principle of minimum potential energy appropriate to the composite laminate theory and to the boundary value problem (3.6.10). In order to prove the minimum principles, we assume that the specific elastic (strain) energy w of the laminate, defined by equation (3.5.30), is a positive definite quadratic form. We have: The first principle of minimum potential energy. Let A denote the set of all regular displacement field U and let I(U) be the functional defined on A by equation (3.6.11). Let U be the solution of the “traction” boundary value problem (3.6.10). Then e I(U) ≤ I(U) (3.6.20) e ∈ A, and equality holds only if U e = U modulo a rigid displacement. for every U

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e U ∈ A and define U0 =U e − U. Using the definition (3.6.11) of the Let U, functional I(U), the equation (3.6.19), the equilibrium equation (3.5.8) 1,4 and the fact that U is a solution of the “traction” boundary value problem (3.6.10), after some simple computations, we get e = I(U) + W (U0 ). I(U)

(3.6.21)

e J(U) ≤ J(U)

(3.6.22)

Since w is positive definite, W (U0 ) ≥ 0, hence (3.6.20) takes place. If I(U) = e I(U), then W (U0 ) = 0. Again, since w is positive definite, we get W (U0 ) = 0. Thus, using again the positive definiteness of w, according to equation (3.5.29), 0 we must have e0αβ = 0 and kαβ = 0 in D. Accordingly, U0 is a rigid displacement field of the laminate and the demonstration is complete. We observe again that the admissible displacement fields for which the functional I(U) was defined do satisfy any restriction on the boundary line ∂D. This is so, since the “traction” boundary conditions (3.6.10) are natural boundary conditions. Next, we consider the “displacement” boundary value problem (3.6.13) and the functional J(U) defined by equation (3.6.14). We call J(U) the potential energy of the laminate corresponding to the “displacement” boundary value problem (3.6.13). We shall denote by B the set of all regular displacement fields U satisfying the given displacement boundary conditions (3.6.13). We have: The second principle of minimum potential energy. Let B denote the set of all regular displacement field U that satisfy the boundary conditions (3.6.13), and let J(U) be the functional defined on B by equation (3.6.14). Let U be the solution of the “displacement” boundary value problem (3.6.13). Then

e ∈ B,and the equality holds only if U e = U on D. for every U

e U ∈ B and let us define U0 =U e − U. Then U0 satisfies homogeneous Let U, 0 boundary conditions on ∂D; i.e. Un = U0τ = U03 = U03,n = 0 on ∂D. Using the definition (3.6.14) of the functional J, equation (3.6.19), the facts that U is a solution of the boundary value problem (3.6.13) and U0 satisfies homogeneous boundary conditions, we obtain e = J(U) + W (U0 ). J(U)

(3.6.23)

Since w is positive definite, from (3.6.23) we can conclude that (3.6.22) is true. 0 e If J(U) = J(U), we obtain e0αβ = kαβ = 0 on D, hence U0 is a rigid displacement 0 field. Moreover, U must satisfy the homogeneous boundary conditions on ∂D. Hence U0 = 0 on D and the proof is complete. We note again that now the admissible displacement fields for which the functional J is defined must satisfy the given “displacement” boundary conditions

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on the boundary line ∂D. This is due to the fact that the “displacement” boundary conditions (3.6.13) are essential boundary conditions. We recall that the most frequent boundary conditions in the laminate theory correspond to simple supported or clamped edges and are of mixed type. Using equation (3.6.19), appropriate potential energies can be found, and can be formulated and proved appropriate minimum principles, corresponding to various boundary conditions listed in the relations (3.6.17) and (3.6.18). As in the usual elasticity theory, we can prove the conversers of the given minimum principles. Converse of the first principle of minimum potential energy. Let U ∈ A and suppose that e I(U) ≤ I(U) (3.6.24)

e ∈ A. Then U is a regular solution of the “traction” boundary value for every U problem (3.6.10). e = U0 +U ∈ A. Let U0 be an arbitrary vector field of class C ∞ on D. Then U Using the equation (3.6.19) and the assumption (3.6.29), it is easy to see that Z {Uα0 Nαβ,β + U30 (Mαβ,αβ + q)}da W (U0 ) − D Z 0 {Un (Nnn − Φ) + Uτ0 (Nnτ − Ψ) + U30 (Qn + ∂D

0 +Mnτ,τ − Γ) − U3,n (Mnn − ∆)}ds ≥ 0.

Obviously, this inequality must hold when U0 is replaced by αU0 , α being an arbitrary real number. Hence we must have Z − {Uα0 Nαβ,β + U30 (Mαβ,αβ + q)}da ZD + {Un0 (Nnn − Φ) + Uτ0 (Nnτ − Ψ) + U30 (Qn ∂D

0 (Mnn − ∆)}ds = 0, +Mnτ,τ − Γ) − U3,n

since W (U0 ) = 0. Since U0 is an arbitrary field, from the above equation and from the supplementary constitutive relation (3.6.5), it follows that the equilibrium equations (3.5.8) and the boundary conditions (3.6.10) are satisfied. Hence, U is a regular solution of the “traction” boundary value problem (3.6.10). Analogously, we have the following. Converse of the second principle of minimum potential energy. Let U ∈B. Suppose that e J(U) ≤ J(U) (3.6.25) e ∈ B. Then U is a solution of the ”displacement” boundary value for every U problem (3.6.13).

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Let U0 be an arbitrary vector field of class C ∞ on D, and suppose that U0 e = U0 +U ∈ B. Using equation (3.6.19), the assumption vanishes on ∂D. Then U 0 (3.6.25) and the fact that Un0 = Uτ0 = U30 = U3,n = 0 on ∂D, we get Z {Uα0 Nαβ,β + U30 (Mαβ,αβ + q)}da ≥ 0. W (U0 ) − D

Since this inequality must remain true if U0 is replaced by αU0 , we get Z {Uα0 Nαβ,β + U30 (Mαβ,αβ + q)}da = 0. D

0

Since U is arbitrary in D, from the last equation and from the supplementary constitutive relations (3.6.5), it follows that the equilibrium equations (3.5.8) are verified by U. Moreover, since U ∈ B, U satisfies also the boundary conditions (3.6.13). Hence, U is a regular solution of the “displacement” boundary value problem (3.6.13) and the demonstration is complete. In a similar way, the converses of principles of minimum potential energies corresponding to various, possible boundary value problems can be proved. We note also that, as in the usual elasticity theory, the principles of minimum potential energy can be used to prove uniqueness theorems for the involved boundary value problems. As we have seen, in order to prove uniqueness theorems and minimum principles, we have assumed that the specific deformation (strain) energy w is a positive definite quadratic form. We recall also that from the beginning we have supposed that the elasticity tensor c of any linearly elastic body is positive definite. Using this hypotheses, in the Section 2.2 we have derived various restrictions which must be satisfied by the elasticities in order to ensure the positiveness of c. In a natural way, the following question is raised. If the elasticities ck , k = 1, ..., N of the laminate are positive definite, does the specific strain energy w of the laminate have the same property? Since the reduced transformed stiffnesses (Qij )k of the laminae and the laminate stiffnesses Aij , Bij , Dij are complicated functions of the laminae mechanical and geometrical characteristics, it is difficult to find an answer for the above question for a laminate having arbitrary structure. This is the reason why we shall analyze the problem only for the particular, but important case of the regular antisymmetric cross-ply laminates, for which the extensional, coupling and bending stiffnesses of the composite laminate can be expressed by relatively simple relations in terms of the primary mechanical characteristics of the laminae, using Tsai’s formulas (3.4.26). We recall that from the positive definiteness of the elasticity c of an orthotropic elastic material, it follows that the technical constants of the body must satisfy the restrictions (2.2.74)−(2.2.79). In particular, we have

E1 , E2 , G12 > 0, ν12 , ν21 < 1,

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(3.6.26)

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3.6. VARIATIONAL AND EXTREME PRINCIPLES |ν12 | <

r

E1 , |ν21 | < E2

r

E2 . E1

(3.6.27)

If we assume in addition that a traction tensor stress acting in a principal material direction produces extension in that direction and contractions in the other two principal material directions, we can conclude that ν12 , ν21 > 0.

(3.6.28)

Hence in this case, the Poisson’s ratios ν12 and ν21 satisfy the restrictions 0 < ν12 <

r

E1 , 0 < ν21 < E2

r

E2 . E1

(3.6.29)

We recall that E1 is the Young’s modulus in the fiber’s direction and E2 is the Young’s modulus in the perpendicular, in-plane direction. Consequently, we assume that E1 < 1. (3.6.30) 0
Using the Tsai’s stiffness ratio f , we can express the inequalities (3.6.29) satisfied by Poisson’s ratios ν12 and ν21 in the equivalent form 0 < ν12 <

r

p 1 , 0 < ν21 < f . f

(3.6.31)

We return to equations (3.1.6) giving the (primary) reduced stiffness Q11 , Q12 , Q22 and Q66 in terms of the engineering constants of the lamina; we have

ν21 E1 E1 , , Q12 = 1 − ν12 ν21 1 − ν12 ν21 E2 and Q66 = G12 . = 1 − ν12 ν21

Q11 =

Q22

(3.6.32)

Thus, from (3.6.26) and (3.6.28) we can conclude that all these stiffnesses are positive; i.e. Q11 , Q12 , Q22 , Q66 > 0.

(3.6.33)

The positive definiteness of the elasticity c of the lamina assures only the positivity of the reduced stiffnesses Q11 , Q22 and Q66 . For the positivity of the stiffness Q12 some supplementary assumption, leading to the inequalities (3.6.28), must be also used. We recall now Tsai’s formulas (3.4.26) giving the extensional, coupling and bending stiffnesses of a regular antisymmetric cross-ply laminate. Using also the

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relation (3.6.32), we get

2ν21 E1 1+f A11 , A66 = hG12 , , A12 = h 1+f 1 − ν12 ν21 2 1 f −1 hA11 , B11 = 2N f + 1 h2 h2 2ν12 h2 A66 . (3.6.34) A11 , D66 = A11 , D12 = D11 = 12 12 1 + f 12

A11 =

From (3.6.26), (3.6.28), (3.6.30) and (3.6.34), we can conclude that A11 , A12 , A66 > 0, B11 < 0, D11 , D12 , D66 > 0.

(3.6.35)

We observe now that the positive definiteness of the specific strain energy w of a laminate, defined by equation (3.5.29), is equivalent with the positive definiteness of the global stiffness matrix [E] of the laminate, introduced by equation (3.3.20). According to Tsai’s relations (3.4.26), the stiffness matrix [E] of a regular antisymmetric cross-ply laminate has the following expression:   A11 A12 0 B11 0 0   A12 A11 0 0 −B11 0     0 0 A66 0 0 0 . (3.6.36) [E] =    B11 0 0 D11 D12 0     0 −B11 0 D12 D11 0 0 0 0 0 0 D66

According to the Sylvester’s criterion, [E] is positive definite following six determinants are positive: A11 A12 0 A11 A12 , ∆3 = A12 A11 0 ∆1 = |A11 | , ∆2 = A12 A11 0 0 A66 ∆4 =

A11 A12 0 B11

A12 A11 0 0

0 0 A66 0

∆6 = det [E] .

B11 0 0 D11

A11 A12 , ∆5 = 0 B11 0

A12 A11 0 0 −B11

0 0 A66 0 0

if and only if the ,

B11 0 0 D11 D12

0 −B11 0 D12 D11

,

(3.6.37) Long, but elementary computations lead to the following expression of the above determinants:

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  2 4ν21 A211 , ∆1 = A11 , ∆2 = 1 − (f + 1)2 (   2 4ν21 1 h2 3 − 1− A66 A11 ∆4 = (f + 1)2 3 2 (   2 4ν21 1 h4 4 − 1− A11 A66 ∆5 = (f + 1)2 3 16

∆3 = ∆2 A66 , 2 )  f −1 1 , N2 f + 1 2 )  f −1 1 , N2 f + 1

∆6 = ∆5 D66 .

(3.6.38)

From the (3.6.35) and the first relation (3.6.38), we get ∆1 > 0.

(3.6.39)

We use now the restriction (3.6.31)2 satisfied by ν21 . Thus we obtain 2  2 f −1 4f 4ν21 > 0. = >1− 1− f +1 (f + 1)2 (f + 1)2

Consequently, from the second relation (3.6.38), we get ∆2 > 0.

(3.6.40)

The third relation (3.6.38) and the above equality show that ∆3 > 0,

(3.6.41)

since, according to (3.6.35), A66 > 0. Using again the restriction (3.6.31)2 , we successively get 2  2  2    2 f −1 1 1 f −1 f −1 1 4ν21 1 − 2 > − 2 1− f +1 N 3 f +1 f +1 N (f + 1)2 3 2   f −1 1 1 . − = f +1 3 N

We recall that the analyzed structure is a regular antisymmetric cross-ply laminate. Hence N can take only the even values 2, 4, 6, .... Consequently, N ≥ 2. Hence 31 − N12 > 31 − 41 > 0. Thus, from the above relations, we can conclude that 1 3



2 4ν21 1− (f + 1)2



1 − 2 N



f −1 f +1

2

> 0.

Now the fourth equation (3.6.38) leads to ∆4 > 0,

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(3.6.42)

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Hence, according to the last two equations (3.6.38), we have also ∆5 > 0 , ∆ 6 > 0

(3.6.43)

since the inequalities (3.6.35) take place. Examining the inequalities (3.6.39)−(3.6.43), we can see that all Sylvester’s determinants (3.6.37) are positive. Hence, we have the following. Theorem of positive definiteness. The specific strain (deformation) energy w of a regular antisymmetric cross-ply laminate is positive definite, if the specific strain (deformation) energies of the composing laminae are positive definite and if tensile tests applied in the fibers direction produce, in each laminae, extensions in that direction and contractions in the perpendicular in-plane direction. As the above example shows, the analysis of the positive definiteness of the specific strain energy of a composite laminate can generally be a difficult task. However, such analysis must be done since questions concerning uniqueness and extreme properties are strongly connected with positive definiteness. Moreover, as we shall see in the Chapter 7, the stability of a fiber-reinforced composite laminate can be lost if its specific strain energy ceases to be positive definite.

3.7

Rectangular laminates

Introducing the global constitutive equations (3.5.7) and the geometrical equations (3.5.6) into the equilibrium equations (3.5.8)1 and (3.5.8)4 , we shall obtain the differential system which must be satisfied by the components U 1 , U2 , U3 , of the displacement U in the plane domain D occupied by the middle surface of the laminate. Elementary, but long computations lead to the following displacement equilibrium equations: A11 U1,11 + 2A16 U1,12 + A66 U1,22 + A16 U2,11 + (A12 + A66 )U2,12 + A26 U2,22 −B11 U3,111 − 3B16 U3,221 − (B12 + 2B66 )U3,122 − B26 U3,222 = 0, A16 U1,11 + (A12 + A66 )U1,12 + A26 U1,22 + A66 U2,11 + 2A26 U2,12 + A22 U2,22 −B16 U3,111 − (B12 + 2B66 )U3,112 − 3B26 U3,122 − B22 U3,111 = 0, D11 U3,1111 + 4D16 U3,1112 + 2(D12 + 2D66 )U3,1122 + 4D26 U3,1222 + D22 U3,2222 −B11 U1,111 − 3B16 U1,112 − (B12 + 2B66 )U1,122 − B26 U1,222 −B16 U2,111 − (B12 + 2B66 )U2,112 − 3B26 U2,122 − B22 U2,222 = q. (3.7.1)

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Generally, the relation (3.7.1) is a complicated and coupled differential system. In-plane deformation and distortions of the laminate appear together, in an inseparable way. But, obvious and sometimes important simplifications result when the laminate is symmetric about the middle surface (Bij = 0), specially orthotropic (all the terms with 16 and 26 indices vanish in addition to the B ij ), 2 homogeneous (Bij = 0 and Dij = Aij h12 ), or isotropic. In all these cases, equations (3.7.1)1,2 are uncoupled from the equations (3.7.1)3 . In such situations, the first two equations contain only the in-plane displacements U1 , U2 , and the third one contains only the normal displacement U3 . Accordingly, equation (3.7.1)3 must be solved to obtain the deflections of a plate and the system (3.7.1)1,2 must be integrated to obtain the in-plane deformations of the laminate. Unfortunately, the more general case of nonsymmetric laminates (there exist nonvanishing coupling stiffnesses Bij ) requires the simultaneous integration of the coupled system (3.7.1). Currently, many methods exist to solve the equilibrium equations (3.7.1). The methods range from exact solutions to approximate numerical integration, using finite element or finite difference approaches and various approximate energy methods of Rayliegh-Ritz or Galerkin type, based on the given minimum or variational principles. In what follows, we shall consider only the case of a rectangular laminate, taking into account a small number of special examples. Let us consider first the general class of laminated rectangular plates, as shown in Figure 3.20, that are simply supported along edges x1 = 0, x1 = a1 , x2 = 0, x2 = a2 and subjected to a given normal load q = q(x1 , x2 ). We suppose that the given normal load q = q(x1 , x2 ) can be expanded in a double Fourier series; i.e.

q(x1 , x2 ) =

∞ X ∞ X

m=1 n=1

qmn sin

nπx2 mπx1 . sin a2 a1

(3.7.2)

Figure 3.20: Simply supported laminated rectangular plate under distributed normal load. In what follows, the various types of possible laminates, such as specially orthotropic, symmetric angle-ply, antisymmetric cross-ply and antisymmetric angleply will be analyzed for the loading q = q(x1 , x2 ) given by equation (3.7.2).

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As we know, a specially orthotropic laminate has either a single layer of specially orthotropic material or multiple specially orthotropic layers that are symmetrically placed about the laminate middle surface. For this special case considered, the nonvanishing laminate stiffnesses are A11 , A12 , A22 , A66 , D11 , D12 , D22 , D66 . In other words, neither shear or twist coupling, nor bending-extension coupling exists. Also, the in-plane deformations and bending and twisting are decoupled.Thus, for the laminate problem, the vertical deflection are described only by one differential equation of equilibrium, resulting from the third equation (3.7.1), D11 U3,1111 + 2(D12 + 2D66 )U3,1122 + D22 U3,2222 = q.

(3.7.3)

The boundary conditions corresponding to simply supported edges become U3 = 0 , M11 = −D11 U3,11 − D12 U3,22 = 0 for x1 = 0 and x1 = a1 ,

(3.7.4)

U3 = 0 , M22 = −D12 U3,11 − D22 U3,22 = 0 for x2 = 0 and x2 = a2 . The solution of the boundary value problem can be determined using the separation of variables technique, as in the case of isotropic rectangular plates. In this way, it is easy to check that the solution satisfying the boundary conditions (3.7.4) must have the following relatively simple form: U3 (x1 , x2 ) =

∞ X ∞ X

m=1 n=1

Amn sin

nπx2 mπx1 . sin a2 a1

(3.7.5)

This normal displacement field satisfies the equilibrium equation (3.7.3) only if the Fourier coefficients amn are given by the relation Amn =

qmn 1 . m 4 4 π D11 ( a1 ) + 2(D12 + 2D66 )( am1 )2 ( an2 )2 + D22 ( an2 )4

(3.7.6)

Once the normal displacement or deflection U3 is known, all force and moment resultants can be obtained using the corresponding constitutive equations. The case of a symmetric angle-ply laminate is much more complicated, even if, for these structures, the coupling stiffnesses Bij are also vanishing. But now the shear coupling stiffnesses A16 , A26 and the twist coupling stiffnesses D16 , D26 are not zero. The equilibrium equation (3.7.1)3 , describing the deflection of the plate, is decoupled and takes the form D11 U3,1111 + 4D16 U3,1112 + 2(D12 + 2D66 )U3,1122 + 4D26 U3,1222 + D22 U3,2222 = q. (3.7.7) The simply supported edge condition becomes U3 = 0, M11 = −D11 U3,11 − D12 U3,22 − 2D16 U3,12 = 0 for x1 = 0 and x1 = a1 , U3 = 0, M22 = −D12 U3,11 − D22 U3,22 − 2D26 U3,12 = 0 for x2 = 0 and x2 = a2 . (3.7.8)

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3.7. RECTANGULAR LAMINATES

Now the solution of the equilibrium equation (3.7.7) is not as simple as before, because of the presence of D16 and D26 . Due to the terms involving these nonvanishing coefficients, the method using the separation of variables cannot be applied and a Fourier type expression (3.7.5) does not satisfy the governing equation (3.7.7). Moreover, the expansion (3.7.5) also does not satisfy the boundary conditions (3.7.8), again since the terms involving D16 and D26 are present. Actually, the variables x1 and x2 cannot be separated in the expression of the normal displacement U3 . This is the reason why Ashton (see Jones [3.2], Chapter 5) has solved the problem using the second principle of the minimum potential energy. The problem being decoupled, the involved functional J(U), given in the equation (3.6.14), takes the following simplified form: 1 2

Z

2 2 + 2D12 U3,11 U3,22 + D22 U3,22 (D11 U3,11 Z 2 qU3 da. +4D66 U3,12 + 4D16 U3,11 U3,12 + 4D26 U3,22 U3,12 )da −

J = J(U3 ) =

D

(3.7.9)

D

Ashton has approximated the deflection U3 by a finite number of terms of the Fourier expansion (3.7.5). Such an expression satisfies the displacement boundary conditions (3.7.8), hence is an admissible displacement field. The boundary conditions (3.7.8) concerning the bending moments must not be a` priori satisfied, since they are natural boundary conditions. Ashton’s approach is obviously a RayleighRitz method, applied to our particular case. Minimizing the functional J(U 3 ), taking into account a finite number of terms in the expansion (3.7.5), we are lead to a finite set of simultaneous linear algebraic equations for the chosen unknowns amn . Such a system can be solved using a computer. Let us observe again that only the displacement boundary conditions are exactly satisfied, but the natural boundary conditions will be only approximately satisfied. The convergence of the method may be slow, just because the natural boundary condition are not satisfied exactly. For instance, Ashton has used 49 terms (up to m = 7 and n = 7) to obtain approximatively the deflection U3 . Thus, for a uniformly loaded (q(x1 , x2 ) = 1) (D12 +2D66 ) D26 16 22 = 1, D square laminate (a1 = a2 = a) with D D11 = D11 = −0.5, D11 D11 = 1, the maximum deflection at the center of the plate, found by Ashton, is

U3 max = U3 (0, 0) =

0.00425a4 q . D11

If D16 and D26 are ignored, that is the symmetric angle-ply laminate is approximated by a special orthotropic laminate having

(D12 + 2D66 ) D11 = 1 , D16 = D26 = 0, = 1, D11 D22

the maximum deflection found by Ashton is U3 max = U3 (0, 0) =

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0.0032a4 q . D11

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CHAPTER 3. COMPOSITE LAMINATES

Hence, the error in neglecting the twist coupling terms D16 and D26 , is about 24 percent, which represents a no negligible error. Thus, generally, a specially orthotropic laminate is an unacceptable approximation for a symmetric angle-ply laminate. Let us consider now a regular antisymmetric cross-ply laminate. Such a laminate has nonvanishing extensional stiffnesses A11 , A12 , A11 = A22 and A66 , bending-extensional coupling stiffnesses B11 and B22 = −B11 , and bending stiffnesses D11 , D12 , D22 = D11 and D16 . Since B11 and B22 are not vanishing, the displacement equilibrium equations are coupled. The general system (3.7.1) takes the following simplified form: A11 U1,11 + A66 U1,22 + (A12 + A66 )U2,12 − B11 U3,111 = 0, (A12 + A66 )U1,12 + A66 U2,11 + A11 U2,22 + B11 U3,222 = 0,

(3.7.10)

D11 (U3,1111 + U3,2222 ) + 2(D12 + 2D66 )U3,1122 − B11 (U1,111 − U2,222 ) = 0. Whitney and Leissa (see Jones [3.2], Chapter 5) have solved the problem for simply supported edges taking into account boundary conditions of type S 2 (see Equations (3.6.17)) U3 = 0, M11 = B11 U1,1 − D11 U3,11 − D12 U3,22 = 0, for x1 = 0 and x1 = a1, U2 = 0, N11 = A11 U1,1 + A12 U2,2 − B11 U3,11 = 0, U3 = 0, M22 = B11 U2,2 − D12 U3,22 − D11 U3,22 = 0, for x2 = 0 and x2 = a2 , U1 = 0, N22 = A12 U1,1 + A11 U2,2 + B11 U3,22 = q. (3.7.11) These authors have observed that in this case the variables x1 and x2 can be separated and the displacements U1 , U2 , U3 can be obtained using Fourier’s method; i.e. U1 =

U2 =

U3 =

∞ X ∞ X

m=1 n=1 ∞ X ∞ X

m=1 n=1 ∞ X ∞ X

m=1 n=1

αmn cos

nπx2 mπx1 , sin a2 a1

βmn sin

nπx2 mπx1 , cos a2 a1

γmn sin

nπx2 mπx1 . sin a2 a1

(3.7.12)

It is easy to see that the boundary conditions (3.7.11) are satisfied. Obviously, the Fourier coefficients αmn , βmn and γmn can be obtained taking into account the equilibrium equation (3.7.10). The results obtained by Whitney and Leissa are replotted in Figure 3.21 for the special situation in which the normal load is one term of the Fourier series;

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177

3.7. RECTANGULAR LAMINATES i.e. q = q(x1 , x2 ) = q0 sin

πx2 πx1 , q0 = const. sin a2 a1

Figure 3.21 gives the normalized maximum deflexion for a rectangular (a = b) regular antisymmetric cross-ply graphite/epoxy laminated plate, for 2,4,6 and an infinite number of layers. 20 2

15

U3,MAX E 2 h3 3 10 q 0 a24

E1 G =40 E12=0.5 E2 2

=0.25

12

NUMBER OF LAYERS

10 4

8

6

5

x q(x1,x2)=q0sin ax1 sin a22 1

0

1

3

2 a ASPECT RATIO, a 1 2

4

5

Figure 3.21: Maximum deflection of a rectangular regular antisymmetric cross ply laminate under sinusoidal normal load. Let us observe that the infinite number of layers corresponds to the specially orthotropic laminate, for which the coupling between bending and extension does not exist. As the results show, for a two-layered laminate, the neglect of the coupling results in an under-prediction of the deflection by 64 percent. That is the actual prediction is approximately three times bigger than the prediction corresponding to the orthotropic approximation. However, the effect of coupling on the deflection dies out rapidly as the number of layers increases, independent of the plate aspect ratio ab . We can say that when more than six layers exist, the coupling can be neglected without important error and the orthotropic approximation becomes acceptable. As Whitney and Leissa have shown, for composite laminates, the effect of coupling between bending and extension on the plate deflection depends essentially G12 1 on the stiffness ratio E E2 , whereas the influence of E2 and ν12 is relatively small.

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In Figure 3.22 are given the maximum deflection of a square regulate antisymmetric cross-ply laminate, under sinusoidal transverse load, in terms of the E1 1 orthotropic modulus ratio E E2 . At E2 = 1, the effect of coupling is nonexistent. As E1 E2 increases, the effect of coupling between bending and extension increases. 30

25 3

U 3,MAX E 2 h 2 10 q0a 4

G12 =0.5 E2

20

=0.25

12

15

2 10 NUMBER OF LAYERS

8

4 5

x q(x1,x2)=q sin ax1 sin a 2 0

10

20

30

MODULUS RATIO,

E1 E2

40

50

Figure 3.22: Maximum deflection of a square regulate antisymmetric cross-ply laminated plate under sinusoidal transverse load. Summing up the conclusions resulting from the above presented examples, we can say the following: (1) In few special situations, the variables can be separated and the exact solution can be obtained, but only as a Fourier series expansion. (2) Frequently the Fourier series are lent convergent and many terms must be taken into account to obtain a satisfactory solution. (3) Generally the variables can’t be separated. In these situations RayleighRitz and /or Galerkin type numerical methods can, and must be used, taking into account the corresponding extreme and variational principles, and using computers to solve the resulting linear algebraic systems. The displacement fields taken into account must satisfy exactly all given displacement boundary conditions. (4) The presence in a composite laminate of coupling between bending and extension generally increases deflections. Hence, coupling decreases the effective stiffness of a composite laminate. (5) For laminates exhibiting twist-curvature coupling, the deflections are increased.

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3.8. PROBLEMS

179

(6) In the case of bending-extension coupling and twist-curvature coupling, the effect on deflection decreases rapidly as the number of layers increases. (7) The approximation of a general laminate by a specially orthotropic one can lead to bigger errors by a factor of 3. Thus, the use of such an approximation must be carefully proven to be justified for the case under consideration. (8) The first general rule that can be established analyzing the presented results tells us that coupling should be included in every analysis of composite laminate behavior, unless coupling is proven to be insignificant. (9) The first general rule tells us that for general laminates, specific investigations are necessary to obtain accurate information concerning their behavior.

3.8

Problems

P3.1 Show that for an isotropic lamina and in plane stress state, the strainstress relation are      σ1 ε1 S11 S12 0   σ2   ε2  =  S12 S11 0 0 0 2(S11 − S12 ) σ6 ε6

with S11 = E1 , S12 = − Eν , E and ν being Young’s modulus and the Poisson’s ratio of the laminae, respectively. P3.2 Show that for an isotropic lamina and in plane stress state, the stressstrain relations are      σ1 Q11 Q12 0 ε1  σ2  =  Q12 Q11 0   ε2  0 0 Q66 σ6 ε6

with

Q11 =

E νE E = µ = G. , Q66 = , Q12 = 2 2 2(1 + ν) (1 − ν ) (1 − ν )

P3.3 The engineering constants of a scotchply1002 glass/epoxy fiber-reinforced lamina have the following values: E1 = 38.6GP a , E2 = 8.27GP a , ν12 = 0.26 , G12 = 4.14GP a. Find the compliance components and the reduced stiffnesses of this lamina. P3.4 The engineering constants of a Kevlar 49/E epoxy type aramid/epoxy lamina are E1 = 76GP a , E2 = 5.5GP a , ν12 = 0.34 , G12 = 2.3GP a. Find the compliance components and the reduced stiffnesses of this lamina.

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P3.5 Prove by direct calculations that equations (3.1.14) are true. P3.6 Obtain by direct calculations the relations (3.1.19)−(3.1.21). P3.7 Prove that the equations (3.1.23), giving the apparent technical moduli Ex , νxy , Ey , Gxy , ηxy,x and ηxy,y , as functions of the angle θ ∈ [00 , 900 ], are true. 0 0 x Gxy P3.8 Plot E E2 , G12 , νxy and −ηxy,x as functions on θ ∈ [0 , 90 ] for a boron/epoxy composite, an orthotropic material with

E1 = 10E2 , G12 =

1 E2 , ν12 = 0.3 , E2 = 18.5GP a. 3

Analyze the obtained results. P3.9 Show that the apparent axial or longitudinal modulus Ex of a fiberreinforced orthotropic lamina can be written as function of θ ∈ [00 , 900 ] in the following form: E1 = (1 + a − 4b) cos4 θ + 2 (2b − a) cos2 θ + a Ex

with E1 and b = a= Ex



E1 G12

− 2ν12

4



.

P3.10 Use the above expression of Ex to find its maxima and minima. Show that if E1 , G12 > 2 (1 + ν12 )

Ex is greater than both E1 and E2 for some value of θ, and if G12 <

2



E1 E1 E2

+ ν12

,

Ex is smaller than both E1 and E2 for some value of θ. P3.11 Prove the validity of the equations (3.1.24) and (3.1.25). P3.12 Prove that the combinations Q11 + Q22 + 2Q12 and Q66 − Q12 are invariant under rotations about the x3 = z axis and these invariants have the constant values given by the equations

Q11 + Q22 + 2Q12 = Q11 + Q22 + 2Q12 = 2 (U1 + U4 ) , Q66 − Q12 = Q66 − Q12 = U5 − U6 ,

where U1 , U4 , U5 and U6 are given by equations (3.1.25). P3.13 Show that the reduced stiffnesses as a function of the angle θ can be expressed in the following matrix form:

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181

3.8. PROBLEMS

     Q11 U1 U2 0 U3 0 1   Q22   U1 −U2 0 U3 0   cos 2θ        Q12   U4 0 0 −U 0 3   sin 2θ , =       U5 0  Q 0 −4U3 0   cos 4θ   66     2Q   0 0 U2 0 2U3 16 sin 4θ 0 0 U2 0 −2U3 2Q26 where U1 , U2 , U3 , U4 , U5 are given by the equations (3.1.25). P3.14 The engineering constants of a T300/5208 graphyte/epoxy fiber-reinforced lamina are given in Table 3.1. Find the coefficients U1 , U2 , U3 , U4 and U5 for this material. P3.15 A boron/epoxy fiber-reinforced composite lamina has the following engineering constants: E1 = 206.85GP a , E2 = 20.68GP a , ν12 = 0.3 , G12 = 6.86GP a. (a) Plot the reduced transformed stiffnesses Q11 (θ), Q22 (θ), Q12 (θ) and Q66 (θ)  for this lamina for θ ∈ 00 , 900 . (b) Plot the reduced transformed stiffnesses Q16 (θ) and Q26 (θ) for the same   lamina and for θ ∈ 00 , 900 . (c) Analyze the results obtained in (a) and (b). P3.16 Show that the functions S 11 (θ), ..., S 66 (θ), given by equations (3.1.21) can be expressed in the following matrix form: 

       

S 11 S 22 S 12 S 66 S 16 S 26





      =      

V1 V1 V4 V5 0 0

V2 −V2 0 0 0 0

0 0 0 0 V2 V2

V3 V3 −V3 −4V3 0 0

0 0 0 0 2V3 −2V3



       

1 cos 2θ sin 2θ cos 4θ sin 4θ



  ,  

where V1 =

V2 =

V3 =

V4 =

V5 =

1 (3S11 + 3S22 + 2S12 + S66 ) , 8 1 (S11 − S22 ) , 2 1 (S11 + S22 − 2S12 − S66 ) , 8 1 (S11 + S22 + 6S12 − S66 ) , 8 1 (S11 + S22 − 2S12 + S66 ) . 2

b1 and the variation P3.17 Plot the variation of the overall elastic modulus E of the overall Poisson’s ratio νb12 with the fiber volume fraction cf ∈ [0, 1], using Voigt’s mixture rule.

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CHAPTER 3. COMPOSITE LAMINATES b

P3.18 Plot the variations of the ratios EEm2 and cf ∈ [0, 1], using Reuss mixture rule, assuming that

(a)

Ef = 10; Em

(b)

b12 G Gm

with the volume fraction

Ef = 100. Em

P3.19 What conclusion do you get if you assume for the determination of b2 equal strains in both the fiber and the matrix, the overall transverse modulus E instead of equal stresses. P3.20 Show that if the curvatures kαβ (x1 , x2 ), α, β = 1, 2 are vanishing, the middle surface of a composite laminate rests plane after the deformation of the laminate. In other words, show that the curvatures characterizes the bending and the twisting of the laminate. P3.21 Let us assume that the in-plane deformations eαβ (x1 , x2 ) and the curvatures kαβ (x1 , x2 ), α, β = 1, 2 of a laminate are vanishing. Find in that case the components uk (x1 , x2 , x3 ), k = 1, 2, 3 of the displacement field and give the geometrical meaning of the obtained result. P3.22 Assuming that a laminate is submitted to a rigid displacement, find the components of the in-plane deformation and the components of the curvature. P3.23 Show that the extensional stiffnesses Aij , i, j = 1, 2, 6 of a composite laminate can be expressed by the following relations: A11 = U1 V0A + U2 V1A + U3 V3A , A22 = U1 V0A − U2 V1A + U3 V3A , A12 = U4 V0A − U3 V3A , A66 = U5 V0A − U3 V3A , 1 A16 = − U2 V2A − U3 V4A , 2 1 A26 = − U2 V2A + U3 V4A , 2

where V0A = h, N X V1A = (zk − zk−1 ) cos 2θk , k=1

V2A =

N X

k=1

V3A =

N X

k=1

V4A =

N X

k=1

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(zk − zk−1 ) sin 2θk , (zk − zk−1 ) cos 4θk , (zk − zk−1 ) sin 4θk .

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3.8. PROBLEMS

P3.24 Show that the coupling stiffnesses of a composite laminate can be expressed by the following relations: B11 = U2 V1B + U3 V3B , B22 = −U2 V1B + U3 V3B ,

B12 = −U3 V3B , B66 = −U3 V3B , 1 B16 = − U2 V2B − U3 V4B , 2 1 B26 = − U2 V2B + U3 V4B , 2

where V0B = 0, N

V1B

1X 2 2 (zk − zk−1 ) cos 2θk , = 2 k=1

N

V2B =

1X 2 2 (zk − zk−1 ) sin 2θk , 2 k=1

N

V3B =

1X 2 2 ) cos 4θk , (zk − zk−1 2 k=1

N

V4B =

1X 2 2 ) sin 4θk . (zk − zk−1 2 k=1

P3.25: Show that the bending stiffnesses of a composite laminate can be expressed by the following relations: D11 = U1 V0D + U2 V1D + U3 V3D , D22 = U1 V0D − U2 V1D + U3 V3D , D12 = U4 V0D − U3 V3D ,

D66 = U5 V0D − U3 V3D , 1 D16 = − U2 V2D − U3 V4D , 2 1 D26 = − U2 V2D + U3 V4D , 2

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CHAPTER 3. COMPOSITE LAMINATES

where V0D =

V1D =

h3 , 12 N 1X

3

k=1

3 ) cos 2θk , (zk3 − zk−1

N

V2D =

1X 3 3 ) sin 2θk , (zk − zk−1 3 k=1

N

V3D =

1X 3 3 ) cos 4θk , (zk − zk−1 3 k=1

N

V4D

1X 3 3 ) sin 4θk . (zk − zk−1 = 3 k=1

P3.26 Prove that the relations given in P3.24-P3.26 can be expressed following concentrated matrix form:      U1 U2 0 U3 0 [A11 , B11 , D11 ] V0[A,B,D]   [A22 , B22 , D22 ]   U1 −U2 0 U 0 3   V1[A,B,D]      [A12 , B12 , D12 ]   U4 0 0 −U3 0   V2[A,B,D]      [A66 , B66 , D66 ]  =  U5 0 0 −U 0 3   V3[A,B,D]     2 [A16 , B16 , D16 ]   0 0 −U2 0 −2U3  V4[A,B,D] 0 0 −U2 0 2U3 2 [A26 , B26 , D26 ]

in the 

  .  

P3.27 Let us consider a composite laminate submitted to the constant force ◦

◦

◦

resultants N 11 , N 22 and N 12 . Find the appropriate constant moment resultants ◦ ◦ ◦ M 11 , M 22 and M 12 which must be applied to the laminate to obtain zero curvatures. (a) Solve first the problem using the global constitutive equations expressing [N ] and [M ] in terms of [e] and [k]. (b) Next, solve the problem using the inversed global constitutive equations expressing [e] and [k] in terms of [N ] and [M ]. (c) Compare the results obtained in (a) and (b). (d) Analyze the case in which the coupling stiffnesses are vanishing. P3.28 Find for a single layer isotropic laminate with Young modulus E, Poisson ratio ν and thickness h, the extensional, coupling and bending stiffnesses. P3.29 Find for a single layer specially orthotropic laminate with technical constants E1 , E2 , ν12 , G12 and thickness h, the components of the global stiffness matrix [E]. P3.30 Consider two orthotropic laminae with principal material directions at +α and −α with respect to the laminate reference axis. Prove that



Q16 +α = − Q16 −α and Q26 +α = − Q26 −α .

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3.8. PROBLEMS

185

P3.31 Show that if a cross-ply laminate is symmetric about its middle surface, all of its coupling stiffnesses are vanishing. P3.32 Starting with Tsai’s formulas (3.4.24) and using the definition of a regular symmetric cross-ply laminate, express the stiffnesses of the laminate in 2 terms of the reduced stiffness Q11 , the thickness h and the stiffness ratio f = E E1 . P3.33 Give a direct proof to the results obtained in P3.32. P3.34 Prove that the global stiffness matrix [E] of a regular symmetric crossply laminate is positive definite if the elasticity tensor of the composing laminae are positive definite, and the Poisson’s ratio satisfy the inequalities ν ij > 0 for i, j = 1, 2, 3. P3.35 An expanded view of a [+45/−45/−45/+45] regular angle-ply laminate consisting of 0.25-mm thick unidirectional AS/3501 graphite/epoxy laminae is shown in Figure 3.23. Determine the stiffness matrix [E] of this laminate.The engineering constants of the laminae are

E1 = 138GP a , E2 = 9GP a , ν12 = 0.3 , G12 = 6.9GP a , ν21 = ν12

E2 = 0.0196. E1

Figure 3.23: Exploded view of a [+45/−45/−45/+45] regular angle-ply laminate. P3.36 An exploded view of a [−45/+45/−45/+45] regular angle-ply laminate consisting of the same 0.25-mm thick unidimensional A.S/3501 graphite/epoxy laminae as used in P3.35 is given in Figure 3.24. Determine the global stiffness matrix [E] of the laminate. P3.37 The angle-ply laminate described in P3.35 is subjected to a single uniaxial force per unit length N11 = 50M P a mm−1 . (a) Determine the resulting in-plane deformations and curvatures associated with the x1 and x2 axes.

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Figure 3.24: Exploded view of a [−45/+45/−45/+45] regular angle-ply laminate.

(b) Determine the resulting deformations and displacements associated with the x1 , x2 and x3 axes in each laminae. (c) Determine the resulting stresses associated with the x1 and x2 axes in each laminae. P3.38 The angle-ply laminate described in P3.36 is subjected to a simple uniaxial force per unit length N11 = 50M P a mm−1 . (a) Determine the resulting in-plane deformations and curvatures associated with the x1 and x2 axes. (b) Determine the resulting deformations and displacements associated with the x1 , x2 and x3 axes in each laminae. (c) Determine the resulting stresses associated with the x1 and x2 axes in each laminae. P3.39 A regular angle-ply laminate has N unidirectionally reinforced orthotropic layers having the same thickness and with principal material directions alternatingly oriented at +α and −α to the laminate coordinate axes. The oddnumbered plies are at −α and the even-numbered plies at +α. Show that the transformed reduced stiffnesses of the laminate satisfy the following equations:





Q11 +α = Q11 −α , Q12 +α = Q12 −α ,



Q22 +α = Q22 −α , Q66 +α = Q66 −α ,



Q16 +α = − Q16 −α , Q26 +α = − Q26 −α .

P3.40 Show that for a symmetric (N odd!) regular angle-ply laminate, the

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187

3.8. PROBLEMS laminate stiffnesses are given by the following equations:


(A11 , A12 , A22 , A66 ) = h Q11 , Q12 , Q22 , Q66 ,
h Q16 , Q26 , Bij = 0, i, j = 1, 2, 6, (A16 , A26 ) = N
h3 Q11 , Q12 , Q22 , Q66 , (D11 , D12 , D22 , D66 ) = − 12
h3 2N 2 − 2 Q16 , Q26 . (D16 , D26 ) = 3 12 N

In these equations in Qij , i, j = 1, 2, 6, and α is accounted for. P3.41 Show that for an antisymmetric (N even!) regular angle-ply laminate, the laminate stiffnesses are given by the following equations:
(A11 , A12 , A22 , A66 ) = h Q11 , Q12 , Q22 , Q66 , (A16 , A26 ) = 0,
h2 Q16 , Q26 , (B11 , B12 , B22 , B66 ) = 0, (B16 , B26 ) = − 2N
h3 (D11 , D12 , D22 , D66 ) = Q11 , Q12 , Q22 , Q66 , (D16 , D26 ) = 0. 12

P3.42 Show that A16 , A26 and D16 , D26 for a symmetric regular angle-ply laminate approach zero as the number of layers increases, while the total thickness is held constant. What happens if equal thickness layers are added so that the total laminate thickness increases, too? P3.43 Show that B16 and B26 for an antisymmetric regular angle-ply laminate approach zero as the total number of layers increases, while the total thickness is held constant. What happens if equal thickness layers are added so that the total laminate thickness increases, too? P3.44 In what conditions are the stiffnesses A16 , A26 , D16 , D26 of a symmetric regular angle-ply laminate vanishing? In what conditions are the stiffnesses B16 , B26 of an antisymmetric regular angle-ply laminate vanishing? P3.45 Using the notations introduced in the Section 3.5 shows that Uα Nnα = U1 Nn1 + U2 Nn2 = Un Nnn + Uτ Nnτ . P3.46 Show that the specific strain energy w of a composite laminate can be expressed in the following matrix form: w=

1 T [d] [E] [d] , 2

where

T

T

[d] = [e11 , e22 , 2e12 , k11 , k21 , 2k12 ] .

P3.47 Using the work theorem for composite laminates and assuming the positive definiteness of the specific elastic energy w, formulate and prove the uniqueness theorem corresponding to the classical plate theory of composite laminates. More exactly, prove that if Un or Nnn , Uτ or Mnτ , U3 or Qn + Mnτ,τ and U3,n or Mnn

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are vanishing on the boundary ∂D of the plane domain D occupied by the middle surface of the laminate, then eαβ and kαβ are vanishing on D, assuming usual regularity conditions. Hence, the solution is vanishing, modulo a rigid displacement of the plate. Give conditions in which this displacement is also vanishing! P3.48 Formulate and prove a variational principle corresponding to the simply supported edge boundary conditions S1. P3.49 Formulate and prove a variational principle corresponding to the clamped edge boundary conditions C4. P3.50 Find the appropriate potential energy and prove its minimum property for a simply supported laminate, submitted to the boundary conditions S2. P3.51 Find the appropriate potential energy and prove its minimum property for a clamped laminate, submitted to the boundary conditions C2. P3.52 Using the corresponding principles of minimum potential energy prove the uniqueness theorems corresponding to the boundary value problems S2 and C2, respectively. P3.53 Formulate and prove the converses of the principles of minimum potential energy corresponding to the boundary value problems S2 and C2, respectively.

Bibliography [3.1] Ashton J.E., Whitney J.M., Theory of laminated plates, Progress in Material Science Series, Vol. IV, Technomic Publishing Co., Stanford, 1970. [3.2] Jones, R.M., Mechanics of composite materials, Hemisphere Publishing Co., New York, 1975. [3.3] Christensen, R.M., Mechanics of composite materials, John Wiley and Sons, 1979. [3.4] Tsai, W., Hahn, M.T., Introduction to composite materials, Technomic Publishing Co., Westport, Conneticut, 1980. [3.5] Cristescu, N., Mechanics of composite materials, University of Bucharest, Bucharest, 1983 (in Romanian). [3.6] Whitney, J.M., Analysis of laminated anisotropic plates, Technomic, Lancaster, PA, 1987. [3.7] Gibson, R.F., Principles of composite material mechanics, McGraw-Hill Inc., New York, 1994. [3.8] Lekhnitski, S.G., Theory of elasticity of an anisotropic elastic body, HoldenDay, San Francisco, 1963. ´ [3.9] Barran, J.J., Laroze, S., Calcul des structures en mat´eriaux composite, Ecole National Sup´erieur de l’A´eronautique et de l’´espace, Dept. structures, mat´eriaux, technologie, 1987.

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Chapter 4

MACROSCOPICALLY ELASTIC COMPOSITES 4.1

Macroscopically linearly elastic composites

In this chapter, we discuss some fundamental problems concerning macroscopically homogeneous biphasic linearly elastic composites, currently representing the basic and classical part of the general theory of micromechanics of composite materials. The problems that will be analyzed are founded on the original results obtained by Hill [4.1]−[4.6], Budiansky [4.7], Hashin [4.8]−[4.10], Hashin and Shtrikmann [4.11], [4.12]. Important information concerning the numerical characteristics of macroscopically homogeneous bodies are given in their Lecture Notes by Suquet [4.13] and Zaoui [4.14]. An important result obtained by G˘ar˘ajeu [4.15] in 1995 shows that even in the classical part of the micromechanics of composite materials, there exists important open problems which must be clarified in the future. The subject of the theory of macroscopically or statistically homogeneous linearly elastic composite materials, in the simplest case, is the characterization of the macroscopic or overall or equivalent elastic behavior of a mixture formed by two solid phases firmly bounded together. One of the phases is the matrix, the other represents the inclusions. No restriction is placed on the shape of the inclusions, which may be, for example, fibrous, spherical or plate-like. It is assumed that the mixture is homogeneous on a macroscale, but not necessarily isotropic. The phases themselves are to be homogeneous. Their elastic moduli differ, and so the stress and strain fields through the mixture are microscopically inhomogeneous, whatever the external displacement (constraint) or traction (load). The intensity and the directionality of local heterogeneities should be distributed at the micro-level without overall bias, a condition which does not exclude macroscopic anisotropy. In other words, at a certain level, a macroscopically homogeneous composite is

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heterogeneous and its response is considered to be prescribed pointwise in the spirit of the mechanics of continua. But the microhomogeneities are distributed in such a way that the material volumes of the composite beyond some representative minimum, have comparable macroscopic or overall properties. This property, if it exists, identifies the macroscopic level at which the material can be treated as if it were homogeneous. A real composite material contains a very large number of inclusions in a matrix. Consequently, for arbitrary geometry and concentration of the phases, the task of solving the usual boundary value problem, and of determining the elastic state of the material, is hopelessly complex. A restricted version of the problem is far more tractable but still central. One is asking not for the stress-strain state in detail, but only for the macroscopic or overall or equivalent elastic properties of the macroscopically or statistically homogeneous composite. This involves finding the overall moduli (elasticities and compliances) as functions on the relative concentrations of the phases, the inclusion geometry, the arrangement and degree of order, the distribution of sizes of inclusions of given shapes, etc. Actually, even this restricted theoretical objective is a formidable task and can be more or less successfully approached only if the composite material is macroscopically or statistically homogeneous. Roughly speaking, this means that there exist representative volume elements (RVEs) of the body under consideration. By RVE we mean a relatively small sample of the composite material that is structurally entirely typical of the whole mixture on average and contains a sufficient number of inclusions for the overall moduli to be effectively independent on the surface values of constrains and loading, so long as these are macroscopically homogeneous or uniform. That is, they fluctuate about a mean value with a (spatial) wave-length small compared with the dimension l of the sample, and the effects of such fluctuations become insignificant within a few wave-lengths from the boundary. Accordingly, the contribution of this surface layer to any (volume) average becomes negligible if the sample becomes large enough, resting however much smaller than the characteristic dimension L of the body itself. In theories of macroscopically homogeneous composites, the transition from micro- to macrolevel depends on the finding of connections between suitable defined macro variables and the volume averages of micro fields over a representative sample or RVE. These theories are based on the assumed existence of the overall or macroscopic or equivalent response of some RVE. As we have seen, the microconstituents of a RVE are treated as homogeneous continua whose properties (elastic moduli, concentrations, inclusion shape and arrangement, etc.) are abstracted from experiment. In such theories, the transition from micro- to macrolevels is critical. Two main problems are involved:(i) to define macro-variables and boundary data for the RVE in a physically meaningful way, and (ii) to determine whether and how the macro-variables, alone or in combinations, are related to the volume averages of their micro-counterparts. In the case of linearly elastic macroscopically homogeneous composite mate-

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191

rials, the primary micro-fields are Cauchy’s stress σ and the infinitesimal strain ε. The corresponding macro-variables, intended for overall (equivalent) constitutive equations, are the volume averages σ and ε of these micro-variables on suitably chosen RVEs. The macrofields so defined are obviously easier to handle analytically in the transition between levels. Moreover, as we already know, these macrofields depend only on the corresponding surface data. To quantify (approximate) the characterization of a macroscopically homogeneous composite and of a RVE, let us denote by d a linear order of magnitude measure of the spatial heterogeneity; for example, the inclusions mean dimensions or the spatial distances between inclusions. We take a representative sample of the material (assuming its existence) in the form of a cube with side l, assuming to be much larger than d, but much smaller than L, the characteristic dimension of the whole body (see Figure 4.1).

Figure 4.1: Scales of analysis: d << l << L. We suppose that the distribution of the heterogeneities in the RVE (sample) is such that the following property is obtained for all sufficiently small d/l: over each face, the surface displacement under any homogeneous surface traction, or the surface traction under any homogeneous surface displacement, can be broadly described as a wavelike function about a mean, with a period of order d and an amplitude independent of l. This property, if it takes place, will be considered as a minimal prescription for RVE’s suitability for observing or calculating an equivalent, or macroscopic, or overall constitutive law. A composite material having such RVEs will be called macroscopically or statistically homogeneous. Any surface traction or loading and any surface displacement or constraint, having the above mentioned wavelike character, will be called macroscopically homogeneous

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loading or constraint, respectively. More exact mathematical characterization of a macroscopically homogeneous composite material, having RVEs with nearly identical micro- and macro- properties, is difficult to give without using the apparatus of mathematical statistics. Note that the property of a composite material to be macroscopically homogeneous is a constitutive characteristic of the body. Observational, experimental and theoretical facts can lead us to assume that a given structure is macroscopically homogeneous. Once this hypothesis is accepted, its theoretical and experimental implications can tell us if the supposed property actually takes place. According to the main characteristic of a macroscopically homogeneous composite, these implications can be obtained analyzing the macro- and the equivalent macro- characteristics of a RVE and establishing the connections existing between the suitable chosen micro- and macrofields. As we shall see, the constitutive linkage between the two levels will be first established through the tensor influence functions introduced in Section 2.5. These tensor fields are associated with the microscopic kinematical and dynamical fields existing in a RVE and have a fundamental role for obtaining the macroscopic elastic response, since they act as weighting factors in average, giving the overall elastic moduli. Let us assume that a macroscopically homogeneous composite body B is given. Let us denote by D the regular subregion occupied by a RVE of B, and let v be the volume of D. If f is a microscopic kinematical or dynamical field on D, we denote by Z 1 f dv (4.1.1) f= v D

the mean value of f on D. In the following, we assume zero body forces. We recall some important results, given in Section 2.6 (the first and second Hill-Mandel lemmas), true for any body, in particular for a RVE of B, we have P1 : If u is a kinematically admissible displacement corresponding to the symmetric constant tensor E, that is if u(x)= Ex on ∂D, if ε is the admissible strain field corresponding to u, and if σ is an admissible, self-equilibrated stress field, then (4.1.2) ε = E and σ · ε = σ · ε = σ · E.

P2 : If σ is a statically admissible stress field corresponding to the symmetric constant tensor Σ, that is if σn = Σn on ∂D, and if u is an admissible displacement field, ε being the strain corresponding to u, then

σ = Σ and σ · ε = σ · ε = Σ · ε.

(4.1.3)

P3 : Particularly the relation (4.1.2) and (4.1.3) are true if [u, ε, σ] is the solution of the homogeneous displacement problem, or of the homogeneous traction problem, formulated for the RVE.

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193

The above product theorems or properties play a fundamental role in establishing the overall (equivalent, macroscopic) constitutive properties of a macroscopically homogeneous composite material. Their implication expresses material or constitutive characteristics, since the existence of the RVEs, for which these theorems are applied, is a material or constitutive property of a composite. One can ask if the above means value theorems are true if the constraint or the loading on the boundary of a RVE is only macroscopically homogeneous. The final answer to this question is not yet given. According to Hill’s [5.6] and Mandel’s [5.16] conjecture, the product theorems within a tolerance of order d/l are valid when one of this field is subjected to surface data and this data is merely macroscopically homogeneous. More exactly, for such kind of boundary conditions, the relations (4.1.2) and (4.1.3) become

σ · ε = σ · ε + O(d/l).

(4.1.4)

Hill’s and Mandel’s conjecture is yet an unproved statement. The demonstration of this conjecture represents an important and difficult challenge for further research working to more rigorous formulations of the theory of macroscopically homogeneous materials. Our object is a biphasic macroscopically homogeneous linearly elastic mixture. As before, D is the domain occupied by a RVE, ∂D is the boundary and v the volume of D. We denote by D1 and D2 the complementary subdomain occupied by the matrix and by the inclusions, respectively, v1 and v2 are the volumes of D1 and D2 , Γ = ∂D1 ∩ ∂D2 being the common boundary of two phases. In what follows, the indices 1 and 2 refer to the characteristics of the matrix and of the inclusions, respectively. The phases are assumed to be homogeneous, but may be anisotropic. The elasticity c of the composite is piece-wise constant; i.e.   c1 = const. in D1 , and c1 6= c2 (4.1.5) c(x) =  c2 = const. in D2 , and

k(x) =



k1 = const. in D1 , k1 = c−1 1 , k2 = const. in D2 , k2 = c−1 2 .

(4.1.6)

As we already know (see Section 3.2), the positive numbers c1 = v1 /v and c2 = v2 /v

(4.1.7)

are the concentrations or the volume fractions of the two phases. Since v1 + v2 = v, we obviously have c1 + c2 = 1.

(4.1.8)

We assume that c1 , c2 , c1 and c2 are given, c1 and c2 being symmetric and positive definite fourth order tensors. Using these data, we shall try to characterize

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the overall ( macroscopic, equivalent) properties of the composite. The size and form of the inclusions in the RVE, as well as their spatial distribution, rest arbitrary at this level of generality. If f is a field given on D by Z Z Z 1 1 1 f dv (4.1.9) f dv, f 2 = f dv, f1 = f= v2 D 2 v1 D 1 v D

we denote the mean values of f , over the domains D, D1 , D2 , respectively. According to (4.1.7) and (4.1.9), we have the following fundamental relation between the mean values and concentrations

f = c 1 f 1 + c2 f 2 .

(4.1.10)

In order to define the overall elastic moduli, we use homogeneous displacement or traction boundary conditions on the boundary of the RVE. First let us assume that a ∂D homogeneous displacement condition is prescribed; i.e. u (x) = Ex on ∂D, ET = E = const. (4.1.11) Let s = [u, ε, σ] be the solution of this homogeneous displacement problem corresponding to E. As we know (see Section 2.5), a unique dependence of the strain ε for the given surface displacement condition is completely characterized by E. That is ε (x) = A (x) E, (4.1.12) where A (x) is the influence tensor function corresponding to the homogeneous displacement boundary condition. This fourth order tensor field can be expressed in terms of the corresponding Green’s tensor function. The expression of the Green tensor, and hence the expression of the influence tensor are generally unknown; but their existence is certain. Since as we know (see P1 ) ε = E, (4.1.13)

the fundamental relation (4.1.12) can be written as ε (x) = A (x) ε.

(4.1.14)

Consequently, the mean value A of A (x) satisfies the equation

A=J

(4.1.15)

where J is the fourth order unit tensor with components Jklmn = δkm δln , k,l,m,n = 1,2,3.

(4.1.16)

Now let us recall the assumed stress-strain relation σ(x) = c(x)ε(x). Using (4.1.14) for the mean value σ of the stress we get

σ = cε = cAε = cAε,

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(4.1.17)

4.1. MACROSCOPICALLY LINEARLY ELASTIC COMPOSITES

195

since ε is constant on D. The constant tensor

b c = cA

(4.1.18)

is by definition the overall elasticity of the macroscopically homogeneous composite. Accordingly, equation (4.1.19) σ = cbε

represents the overall stress-strain relation of the composite. As equation (4.1.18) shows, the overall elasticity b c is not the mean value of the elasticity c of the composite. In order to obtain b c, the weightened mean value (4.1.18) must be calculated, and the influence tensor function A (x) plays the role of the weighting factor. Once b c is determined, in further calculi, the initial complex structure is replaced by the equivalent homogeneous linearly elastic body, completely characterized by the overall stress-strain relation (4.1.19). Hence, the main theoretical problem reduces to the determination of the influence tensor function A (x) and to the evaluation of the mean value (4.1.18). This problem represents a formidable task and can be solved only in special cases and/or by using mechanically meaningful supplementary assumptions. However, for a biphasic mixture some simplifications occur even at this level of generality. Indeed, from (4.1.14), using the definitions (4.1.9) we get

ε1 = A1 ε and ε2 = A2 ε,

(4.1.20)

where the constant influence tensors A1 and A2 are the mean values of the influence function A (x) on D1 and D2 , respectively. Also, from the fundamental relation (4.1.10), we obtain

σ = c 1 σ 1 + c2 σ 2 .

(4.1.21)

Further, since c (x) has constant values on D1 and on D2 , using also the equations (4.1.20), we obtain

σ 1 = (c1 ε)1 = c1 ε1 = c1 A1 ε

and

σ 2 = (c2 ε)2 = c2 ε2 = c2 A2 ε.

Consequently, using (4.1.21), we get σ = c1 c1 A1 ε + c2 c2 A2 ε. Hence the overall stress-strain relation (4.1.19) becomes σ=b cε with b c = c 1 c 1 A1 + c 2 c 2 A2 .

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(4.1.22)

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Thus the overall elasticity b c is known, if the influence tensors A1 and A2 are known. However, finding of these constant tensors presents a very difficult problem, even if A1 and A2 are not reciprocally independent. Indeed, from (4.1.10) and (4.1.15), it results that A1 and A2 must satisfy the following equation:

c1 A1 + c2 A2 = J.

(4.1.23)

Hence, in order to obtain the overall stress-strain relation, only one of the constant influence tensors must be determined. From the symmetry of the strain ε it follows that the components Aklmn of the influence tensor function A have the symmetries

Aklmn = Alkmn = Aklnm .

(4.1.24)

The components of the constant influence tensors A1 and A2 have the same symmetries. The extension of the definition of the overall elasticity for a macroscopically homogeneous composite having an arbitrary number of phases is obvious. From (4.1.18), (4.1.24) and from the symmetry properties of the microscopic elasticity c, the following symmetry properties result for the components b c klmn of the overall elasticity b c b cklmn = b clkmn = b cklnm . (4.1.25)

There exists a dual modality to introduce the equivalent homogeneous body, which can replace the composite. To show that, we now assume that on the boundary ∂D of the RVE, homogeneous traction condition is prescribed; i.e. σ (x) n (x) = Σn(x) on ∂D, ΣT = Σ = const.

(4.1.26)

As before, we denote by s = [u, ε, σ] the solution of this special traction problem. There exists (see Section 2.5 and P.2.43) a unique dependence of the stress σ on the given traction condition completely characterized by Σ; i.e. σ (x) = B (x) Σ

(4.1.27)

where B = B (x) is the influence tensor function corresponding to the homogeneous traction boundary condition. Since, we know (see P2 )

σ=Σ

(4.1.28)

σ(x) = B (x) σ.

(4.1.29)

B = J.

(4.1.30)

the relation (4.1.17) becomes

Thus we can conclude that

Now we take into account the assumed micro-strain-stress relation ε (x) = k (x) σ (x), k (x) = c−1 (x). Since σ is a constant tensor, from (4.1.29) we get

ε = kσ = kBε = kBε.

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(4.1.31)

4.1. MACROSCOPICALLY LINEARLY ELASTIC COMPOSITES The constant tensor

b = kB k

197

(4.1.32)

is by definition the overall compliance of the macroscopically homogeneous composite. Accordingly, the equation bσ (4.1.33) ε=k

represents the overall strain-stress relation of the composite. As before, in the case of a biphasic mixture, we get

σ 1 = B1 σ and σ 2 = B2 σ,

(4.1.34)

the constant influence tensors B1 and B2 being the mean values of the influence tensor function B (x) on D1 and D2 , the domains occupied by the matrix and by the inclusions, respectively. Using again the fundamental relation (4.1.10), we can conclude that the overall strain-stress relation for a biphasic mixture has the form

b = c 1 k1 B 1 + c 2 k2 B 2 . bσ with k ε=k

(4.1.35)

c1 B1 + c2 B2 = J.

(4.1.36)

Obviously, equation (4.1.23) is replaced by its dual

Since σ is a symmetric tensor, the components Bklmn of B have the same symmetries as the components of A; i.e. Bklmn = Blkmn = Bklnm .

(4.1.37)

Obviously, the components of the constant influence tensors B1 and B2 have the same symmetry properties. b has the same symmetries as the overall elasAlso, the overall compliance k ticity b c; i.e.

b kklmn = b klkmn = b kklnm .

(4.1.38)

We recall that the microscopic elastic moduli c and k are symmetric tensors and b have the same property. To get the answer we ask if the overall moduli b c and k shall use the product theorems by Hill and Mandel, that is we shall take into account the properties P1 , P2 and P3 , presented at the beginning of this Section. First we consider again the homogeneous displacement problem and use equation (4.1.14).Thus we get

σ · ε = ε · cε = Aε · cAε.

Let us denote by AT the transposed tensor of the fourth order tensor A, AT defined by the equation (1.1.42). Using this relation, from the above equation, we have σ · ε = ε · AT cAε.

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Since the mean value ε is a constant tensor, the above relation becomes

σ · ε = ε · (AT cA)ε.

(4.1.39)

At the same time according to the properties P1 and P3 , we have

σ · ε = σ · ε.

Using the overall stress-strain relation (4.1.19), we obtain

cε. σ ·ε = ε·b

(4.1.40)

b c = AT cA.

(4.1.41)

Comparing the relations (4.1.39) and (4.1.40) valid for any ε, we can conclude that the overall elasticity b c can be expressed by the following equation:

The symmetry of c and the above relation implies the symmetry of b c, as it is easy to see. Hence, the overall elasticity ˆ c has the same symmetries as the micro-elasticity c; i.e. b cT = b c or b cTklmn = b cmnkl . (4.1.42) Recalling again the micro-stress strain relation σ (x) = c(x)ε(x), and taking into account equation (4.1.10), we get

cε. ε · cε = ε · b

(4.1.43)

This equation can be used as a new energetic definition of the overall elasticity ˆ c.

Indeed, if s = [u, ε, σ] is the solution of the homogeneous displacement boundary value problem corresponding to ε, due to the product theorem, the initial definition of ˆ c, using (4.1.19), and the new definition, using (4.1.43), lead to the same value of ˆ c. This fact can be obtained by using the fundamental relation (4.1.14). The new energetic approach, based on the second definition, is more versatile if one wishes to obtain, or to estimate, the overall elasticity b k. Now let us consider the homogeneous traction problem and let us use the relation (4.1.24). Since ε(x)= k (x) σ (x) , we obtain

σ · ε = σ · kσ = Bσ · kBσ.

Let us denote by BT the transposed tensor of B. Using BT , we get

σ · ε = σ · BT kBσ.

Since the mean value σ is a constant tensor, the above relation becomes σ · ε = σ · (BT kB)σ.

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(4.1.44)

4.1. MACROSCOPICALLY LINEARLY ELASTIC COMPOSITES

199

At the same time, the properties P2 and P3 (see Equation (4.1.23)) lead to equation σ · ε = σ · ε.

Now, using the overall strain-stress relation (4.1.33), we get

bσ. σ·ε=σ·k

(4.1.45)

b = BT kB. k

(4.1.46)

bT = k or b k kklmn = b kmnkl .

(4.1.47)

Comparing the relations (4.1.44) and (4.1.45), valid for any σ, we can obtain b can be expressed in the following way: that the overall compliance k

The above relation and the symmetry of k show that the overall compliance b has the same symmetries as the microcompliance k; i.e. k

Recalling again the micro strain-stress relation ε (x) = k (x) σ (x), and using equation (4.1.45), we get bσ. (4.1.48) σ · kσ = σ · k

As before, this equation can be used as a new energetic definition of the overall b compliance k. Indeed, if s = [u, ε, σ] is the solution of the homogeneous traction problem b using corresponding to σ, due to the product theorem, the initial definition of k b (4.1.33) and that using (4.1.45), lead to the same value of k. This can be obtained by using the fundamental relation (4.1.29). This new energetic approach is very useful in order to determine or to estimate b the overall compliance k. We recall now that the microscopic elastic moduli c and k are positive definite. Consequently, ε · cε ≥ 0 and from the equality (4.1.41) and the definition of the mean value, it follows that:

ε·b cε ≥ 0 for any ε = εT .

cε = 0, from (4.1.41) we get ε · σε = 0. Since c is positive definite, Assuming ε · b from the above equation and the definition of the mean value, it results that ε ≡ 0 on D. Since ε is the mean value of ε, we can conclude that ε = 0. Thus, we can conclude that the overall elasticity ˆ c is positive definite; i.e.

and

ε·b cε ≥ 0 for any ε = εT

ε·b cε = 0 if and only if ε = 0.

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(4.1.49)

(4.1.50)

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b is By a similar reasoning, it can be proved that the overall compliance k positive definite; i.e. bσ ≥ 0 for any σ = σ T (4.1.51) σ·k

and

bσ = 0 if and only if σ = 0. σ·k

(4.1.52)

We recall now the principles of minimum potential energy and minimum complementary energy given in Section 2.4 for an usual elastic body and in Section 2.6 for piece-wise homogeneous elastic bodies (see mainly the inequalities (2.6.39) and (2.6.45)!). Using these theorems, we can obtain the third definition of the overall elastic moduli. Let us denote by K (ε) the set of all kinematically admissible displacement fields v corresponding to ε = εT = const. More exactly

K (ε) = {v admissible displacement field and v (x) = εx on ∂D}.

(4.1.53)

Let ε (v) be the strain field corresponding to v ∈K (ε). According to the definition of the mean value and according to the principle of minimum potential energy, if s = [u, ε (u) , σ] is the solution of the homogeneous displacement boundary value problem corresponding to ε; i.e. if u (x) = εx on ∂D, then we have

ε (u) · cε (u) ≤ ε (v) · cε (v) for any v ∈K (ε) .

(4.1.54)

Now, taking into account that ε (u) = ε, and using the second definition of ˆ c, based on the relation (4.1.43), we can see that the overall elasticity ˆ c satisfies the relation  Z  1 ε (v) · cε (v) dv . (4.1.55) ε·b cε = inf ε (v) · cε (v) = inf v∈K(ε) v D v∈K(ε)

Now let us denote by S (σ) the set of all statically admissible stress field τ , corresponding to σ = σ T = const. More exactly, S (σ) = {τ statically admissible stress field and τ n = σn on ∂D}.

(4.1.56)

According to the definition of the mean value and according to the principle of minimum complementary energy, if s = [u, ε, σ] is the solution of the homogeneous traction boundary value problem corresponding to σ; i.e. if σn = σn on ∂D, then we have σ · kσ ≤ τ · kτ for any τ ∈ S (σ) . (4.1.57) Now, taking into account the fact that the mean value of σ is just σ, and using the b based on the relation (4.1.48), we can see that the overall second definition of k, b compliance k satisfies the relation  Z  1 bσ = inf τ · kτ = inf σ·k τ · kτ dv . (4.1.58) τ ∈S(σ) τ ∈S(σ) v D

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201

Note that the energetic relations (4.1.55) and (4.1.58) can be used to define b respectively. It is easy to the overall elasticity ˆ c and the overall compliance k, see that these new definitions and those previously used are equivalent. The new energetic definitions based on equations (4.1.55) and (4.1.58) are useful in order to obtain approximate values for the overall elastic moduli, or to get estimates, concerning these macro-characteristics of a composite material.

4.2

Hill’s weak and strong assumptions

b are positive definite, hence, inverWe recall that the overall moduli ˆ c and k b−1 exist. tible. Thus, b c−1 and k First, we analyze the homogeneous displacement boundary value problem corresponding to ε. Since ˆ c is invertible, using (4.1.19) we can express ε as a function of σ, c−1 σ. Now, the c−1 σ. Thus, from (4.1.14), we obtain ε (x) = A (x) b getting ε = b micro stress-strain relation σ (x) = c (x) ε (x) can be used to express the microstress σ (x) in terms of its mean value σ; we obtain

c−1 . σ (x) = B(x)σ with B (x) = c (x) A(x)b

(4.2.1)

The mean value of the influence tensor function B (x) satisfies the relation

B=J .

(4.2.2)

Now, we return the homogeneous traction problem corresponding to σ. b is invertible, using (4.1.33) we can express σ as a function of ε, Since k b−1 ε. b−1 ε. Consequently, from (4.1.29), we get: σ (x) = B (x) k obtaining σ = k Now, the micro strain-stress relation ε (x) = k (x) σ (x) gives the micro-strain ε (x) as a function of its mean value ε:

b−1 . ε (x) = A (x) ε with A (x) = k (x) B (x) k

(4.2.3)

A=J.

(4.2.4)

The influence tensor function A (x) satisfies the relation

Until now we have obtained two overall constitutive equations (4.1.19) and (4.1.33). If we start with the homogeneous displacement boundary value problem, cε. If we start for the mean values we arrive to the overall material law σ = b with the homogeneous traction boundary value problem, for the mean values we bσ. Our aim is to replace the macroscopically get the overall material law ε = k homogeneous body by an equivalent homogeneous material. This replacement will be meaningful only if the two overall material laws, obtained using two differ−1 b=b ent approaches, are equivalent; that is, only if k c . Consequently, we must

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look for supplementary conditions, which must be fulfilled by a macroscopically homogeneous composite in order to assume that its replacement by an equivalent homogeneous body can be realized in a consistent way. In order to solve this important problem, let us denote by [u, ε, σ] the solution of the homogeneous displacement boundary value problem corresponding to ε, and let [u1 , ε1 , σ 1 ] be the solution of the homogeneous traction problem corresponding to σ 1 . Now let us assume that ε1 = ε and let us check if there exists any relation between σ and σ 1 . In order to answer to this question, we shall suppose that any macroscopically homogeneous composite, which can be replaced by an equivalent homogeneous body, must satisfy the following supplementary restriction, tacitly assumed in the theory of macroscopically homogeneous elastic mixtures. Hill’s weak assumption. If [u, ε, σ] is the solution of the homogeneous displacement boundary value problem corresponding to a given ε, if [u1 , ε1 , σ 1 ] is the solution of the homogeneous traction boundary value problem corresponding to a given σ 1 , and if the mean value of ε1 is just ε; i.e. if ε1 = ε, then the mean value of σ is just σ 1 ; i.e. σ = σ 1 . Hill’s weak assumption, if it is satisfied by a macro-homogeneous mixture, expresses a constitutive property of the composite. Only those macro-homogeneous bodies can be meaningfully replaced by equivalent homogeneous materials, which satisfy this supplementary constitutive restriction. If Hill’s weak assumption is satisfied, and if ε1 = ε, according to the hypothesis made, we also have σ = σ 1 . Hence, the overall constitutive equations (4.1.19) and (4.1.33), obtained by the dual procedures used, become

bσ for any ε . σ=b cε and ε = k

bcε for any ε. Hence, we get Consequently, we shall have ε = kb −1 b=b k c .

(4.2.5)

b is the Hence, if Hill’s weak assumption is fulfilled, the overall compliance k inverse of the overall elasticity b c. Thus, Hill’s weak assumption means the equivalence of the dual overall material laws and the replacement of the composite by a homogeneous body can be realized in a consistent, meaningful way. Also, equations (4.2.1)2 and (4.2.3)2 , connecting the influence tensor function A (x) , B (x) and B (x) , A (x), appearing in the two dual boundary value problems, take the following simplified form: b and A (x) = k (x) B (x) b B (x) = c (x) A (x) k c.

(4.2.6)

However, even if Hill’s weak assumption is fulfilled, until now we are not able to establish any connection between the influence tensor function A (x) and B (x), appearing in the homogeneous displacement problem, and B (x) and A (x) present in the homogeneous traction problem.

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Hill’s weak assumption is a statement concerning the mean values of the solutions of the two dual boundary value problems. We cannot say, at a first look, that the equality ε1 = ε of the mean values implies the equality (modulo a rigid displacement) of the solutions of the dual boundary value problems. The assertion concerning the validity of such an implication represents: Hill’s strong assumption. Let s = [u, ε, σ] be the solution of the homogeneous displacement problem corresponding to a given ε and let s1 = [u1 , ε1 , σ 1 ] be the solution of the homogeneous traction problem corresponding to a given σ 1 . If the mean value of ε1 is just ε; i.e. if ε1 = ε, then s = s1 , modulo a rigid displacement. If Hill’s strong assumption is fulfilled, obviously Hill’s weak assumption is also satisfied. Moreover, if Hill’s strong assumption takes place, the relations (4.1.14), (4.1.25) and (4.2.1)1 , (4.2.3)1 are simultaneously fulfilled with the same ε and σ; i.e. we have ε (x) = A (x) ε = A (x) ε for any ε

and σ (x) = B (x) σ = B (x) σ for any σ .

Consequently, we get A (x) = A (x) and B (x) = B (x) .

(4.2.7)

The above equations show that if Hill’s strong assumption is satisfied, the influence tensor functions A (x) and B (x) are not independent. Indeed, according to the relations (4.2.6), the influence tensor function must satisfy Hill’s compatibility conditions b and A (x) = k (x) B (x) b B (x) = c (x) A (x) k c. (4.2.8)

−1 b=b Since k c , these equations can be written in the following equivalent form: b c (x) A (x) = B (x) b c and k (x) B (x) = A (x) k. (4.2.9)

We can ask now if Hill’s strong assumption is implied by Hill’s weak assumption. As we shall prove, using G˘ar˘ajeu’s reasoning, the answer is yes. To present G˘ar˘ajeu’s theorem, we first introduce some necessary entities and shall prove a convexity and uniqueness theorem. Let us denote by C the set of all admissible displacement fields v. If u ∈C, we design by ε (v) the corresponding admissible strain field. Let ε be a given, constant symmetric tensor. We denote by K (ε) the set of all admissible displacement fields v from C which satisfy the following supplementary restriction: the mean value of the corresponding strain ε (v) is just ε; i.e. o n (4.2.10) K (ε) = v ∈C and ε (v) = ε .

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We have the following: Convexity and uniqueness theorem. (i) The set K (ε) is convex ; i.e. if u, u1 ∈ K (ε) , then αu+(1 − α) u1 ∈ K (ε) for any α ∈ [0, 1]. (ii) The elastic energy U (ε) is strictly convex on K (ε). That is, if u, u 1 ∈ K (ε) and ε = ε (u), ε1 = ε (u1 ), then

U {αε + (1 − α) ε1 } ≤ αU (ε) + (1 − α) U (ε1 ) for any α ∈ [0, 1], and the equality takes place if and only if ε = ε1 . (iii) If u and u1 from K (ε) realize the infimum of U (ε) on K (ε), then u and u1 are equal on D, modulo a rigid displacement. In order to prove (i) we observe that if u, u1 ∈ K (ε), then ε (u) = ε (u1 ) = ε, and we have

ε (αu+ (1 − α) u1 )

= αε (u) + (1 − α) ε (u1 ) = αε (u) + (1 − α) ε (u1 ) = αε + (1 − α) ε = ε for any α ∈ [0, 1] .

Hence, αu+ (1 − α) u1 ∈ K (ε) and (i) is proved. In order to prove (ii), we introduce the symmetric bilinear form V = V (ε, ε1 ) defined by the equation Z Z ε1 · cεdv. ε · cε1 dv = V (ε, ε1 ) = V (ε1 , ε) = D

D

Elementary computations show that 2

U {αε+ (1 − α) ε1 } = α2 U (ε) + α (1 − α) V (ε, ε1 ) + (1 − α) U (ε1 ) . Since, as we know U (ε − ε1 ) ≥ 0, we obtain V (ε, ε1 ) ≤ U (ε) + U (ε1 ) .

Thus, since α ∈ [0, 1], it results

2

U {αε+ (1 − α) ε1 } ≤ α2 U (ε) + α (1 − α) {U (ε) + U (ε1 )} + (1 − α) U (ε) = αU (ε) + (1 − α) U (ε1 ) .

Thus, the inequality of (ii) takes place. If in (ii) there is equality, taking into account the above formulas, we get 2

U {αε+ (1 − α) ε1 } = α2 U (ε) + α (1 − α) V (ε, ε1 ) + (1 − α) U (ε1 ) = αU (ε) + (1 − α) U (ε1 ) .

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205

From this relation it results α (1 − α) {U (ε) − V (ε, ε1 ) + U (ε1 )} = 0 for any α ∈ [0, 1] . Hence, we obtain U (ε) − V (ε, ε1 ) + U (ε1 ) = U (ε − ε1 ) = 0. Accordingly, ε = ε1 , since the elasticity c of the material is positive definite. Now it is clear that u =u1 , modulo a rigid displacement, and thus (ii) is completely proved. In order to prove (iii), let us assume that u and u1 from K (ε) realize the infimum of U = U (ε) on K (ε). Hence, we have

U (ε) = U (ε1 ) =

inf

v∈K(ε)

U {ε (v)} .

Now from (i) and (ii) it results U (ε) ≤ U {αε+ (1 − α) ε1 } ≤ αU (ε) + (1 − α) U (ε1 ) = U (ε) for α ∈ [0, 1] , since U (ε1 ) = U (ε). Consequently, U (αε + (1 − α)ε1 ) = U (ε) = U (ε1 ) for any α ∈ [0, 1] . Hence, U (αε + (1 − α)ε1 ) = αU (ε) + (1 − α)U (ε1 ) for any α ∈ [0, 1] . Thus, according to the second part of the property, (ii) ε = ε1 and (iii) is proved. Now we are ready to prove an important result due to G˘ar˘ajeu [4.15]. Theorem. Hill’s weak assumption implies Hill’s strong assumption. In order to prove this theorem we denote again by C the set of all admissible displacement fields u and let ε (u) be the strain corresponding to u. Let s = [u, ε = ε(u), σ = cε] be the solution of the homogeneous traction problem corresponding to σ. Thus sn (x) = σ (x) n (x) = σn (x) on ∂D. Let s1 = [u1 , ε1 = ε(u1 ), σ 1 = cε1 ] be the solution of the homogeneous displacement problem corresponding to ε1 . Thus, u1 (x) = ε1 x on ∂D. According to the first energetic definitions (4.1.43) and (4.1.48) of the overall b we have elastic moduli ˆ c and k,

and

bσ ε (u) · cε (u) = σ · kσ = σ · k

c ε1 . ε (u1 ) · cε (u1 ) = ε1 · b

(4.2.11)

(4.2.12)

Now let us assume that the basic hypothesis of Hill’s weak assumption is fulfilled; i.e. (4.2.13) ε (u1 ) = ε (u).

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In this case, according to Hill’s weak assumption, we have also

σ 1 (x) = σ (x).

(4.2.14)

At the same time, as we already know, the following relation takes place: −1 b=b k c ,

(4.2.15)

and the two boundary value problems lead to the same overall-stress-strain relation

c ε1 . cε = b σ = σ1 = b

(4.2.16)

cε = ε (u1 ) · cε (u1 ). ε (u) · cε (u) = ε · b

(4.2.17)

From the above consequence of Hill’s weak assumption, we can see that the following relations are true:

We recall that in the first boundary value problem sn = σn = σn on ∂D. In this case, the definition of the mean value and the principle of the minimum potential energy (see the relations (2.6.41) and (2.6.42)) show that the solutions of the homogeneous traction problem corresponding to σ satisfies the relation Z 1 1 u·σnda ε (u) · cε (u) − v ∂D 2   Z 1 1 (4.2.18) v·σnda . ε (v) · cε (v) − = inf v∈C v ∂D 2

As before, we denote by K (ε) the set of all admissible displacement fields v from C, which satisfy the restriction ε (v) = ε. Obviously, K (ε) is a subset of C. Consequently, from (4.2.18), we get   Z Z 1 1 1 1 ε (u) · cε (u) − u·σnda ≤ inf ε (v) · cε (v) − v·σnda . 2 v ∂D v∈K(ε) 2 v ∂D (4.2.19) As we know, u is the displacement field corresponding to the solution s of the traction boundary value problem. Hence ε (u) = ε, since the mean value ε is defined just by this equation. Thus, we see that u ∈K (ε). Consequently, from (4.2.19) we can conclude that   Z Z 1 1 1 1 inf ε (v) · cε (v) − v·σnda ≤ ε (u) · cε (u) − u·σnda. v∈K(ε) 2 v ∂D 2 v ∂D (4.2.20) Comparing (4.2.19) and (4.2.20) we arrive to the equation   Z Z 1 1 1 1 ε (u) · cε (u) − u·σnda = inf ε (v) · cε (v) − v·σnda . 2 v ∂D v∈K(ε) 2 v ∂D (4.2.21)

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207

Since σ is a constant symmetric tensor, the last equation can be written in the following equivalent form: Z 1 1 (un + nu) da σ· ε (u) · cε (u) − 2v 2 ∂D   Z 1 1 (vn + nv) da . (4.2.22) ε (v) · cε (v) − σ · = inf 2v v∈K(ε) 2 ∂D

that

We recall now the mean strain theorem (see Equation (2.1.14)) and conclude Z 1 (4.2.23) (vn + nv) da = ε (v) = ε for any v ∈K (ε) . 2v ∂D

Since u ∈K (ε) from (4.2.22) and (4.2.23), it becomes

ε (u) · cε (u) =

inf

v∈K(ε)

ε (v) · cε (v).

(4.2.24)

Let us return now to u1 , the displacement field corresponding to the solution s1 of the traction problem. According to the assumption (4.2.13), we have

ε (u1 ) = ε1 = ε.

(4.2.25)

Hence u1 ∈K (ε) and from (4.2.25), we get

ε (u) · cε (u) =

inf

v∈K(ε)

ε (v) · cε (v) ≤ ε (u1 ) · cε (u1 ).

(4.2.26)

But, if Hill’s weak assumption is assumed, according to the relation (4.2.17) ε (u) · cε (u) and ε (u1 ) · cε (u1 ) are equal and their common value is ε·b cε. Hence, from (4.2.26) we can conclude that ε·b cε

= ε (u) · cε (u) = ε (u1 ) · cε (u1 )

=

inf

v∈K(ε)

ε (v) · cε (v).

(4.2.27)

The above consequence of Hill’s weak assumption shows that the infimum of the convex function v →ε (v) · cε (v) on the convex set K (ε) is realized by two functions, u and u1 , contained in K (ε). Hence, according to the property (iii) of the convexity and uniqueness theorem, u = u1 modulo a rigid displacement. Hence, s = s1 modulo a rigid displacement and G˘ar˘ajeu’s theorem is proved. At the same time, G˘ar˘ajeu’s result leads to the following important Theorem. Hill’s weak assumption implies Hill’s compatibility relations. Note also that the equations (4.2.27) show that the overall elasticity ˆ c can be defined in an equivalent manner by the following energetical relation: Z 1 ε·b cε = inf ε (v) · cε (v) dv. (4.2.28) v∈K(ε) v D

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This energetic definition, the third one, can be used to evaluate or to estimate the overall elasticity ˆ c of a macro-homogeneous composite, satisfying Hill’s weak assumption. b following Hill As a first step to estimate the overall elastic moduli ˆ c and k, [4.1] we can obtain some universal bounds for the mechanical characteristics of the equivalent homogeneous body. In order to see how this can be done, let us consider first the homogeneous displacement boundary value problem corresponding to ε. According to the relation (4.1.55), we have   Z 1 ε(v) · cε(v)dv . (4.2.29) ε·b cε = inf v∈K(ε) v D

Trying to evaluate b c, Voigt (1912) has supposed a constant strain in the composite, if homogeneous displacement is applied on its boundary. Taking into account this assumption, let us consider the following displacement field: vV (x) = εx in D.

(4.2.30)

Since vV (x) satisfies the given homogeneous displacement condition and the nulljump condition on the common boundary Γ of the matrix and the inclusions, vV (x) is an element of the set K (ε). Hence, from (4.2.29), we obtain

ε·b cε ≤ ε · c (x)ε,

since ε (vV (x)) = ε = const. Let us introduce the constant tensor

cV ≡ c (x) =

1 v

Z

cdv

(4.2.31)

(4.2.32)

D

representing the mean value of the elasticity c of the composite. We get Hill’s first universal estimate

ε·b cε ≤ ε · c V ε

for any ε.

(4.2.33)

Generally, b c 6= cV , and Voigt’s mixture rule is not true. Let us consider now the homogeneous traction boundary value problem corresponding to σ. According to the relation (4.1.58), we have   Z 1 bσ = inf τ · kτ dv . (4.2.34) σ·k τ ∈S(σ) v D

b Reuss (1925) has supposed constant stress in the comIn trying to evaluate k, posite, if homogeneous boundary conditions are applied on its boundary. Taking into account this assumption, let us consider the following stress field: τ R (x) = σ on D.

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(4.2.35)

4.2. HILL’S WEAK AND STRONG ASSUMPTIONS

209

Since τ R (x) satisfies the given homogeneous traction condition and the nulljump condition on Γ, τ R (x) is an element of the set S (σ). Hence, from (4.2.34), we get bσ ≤ σ · k (x)σ (4.2.36) σ·k

since τ R (x) = σ = const. Let us introduce the constant tensor

kR ≡ k (x) =

1 v

Z

kdv,

(4.2.37)

D

representing the mean value of the compliance k of the composite. We get Hill’s second universal estimate

b σ ≤ σ · kR σ σ·k

for any σ.

(4.2.38)

b 6= k , and the Reuss’ mixture rule is not true. Generally k R At the first look these universal estimates are easy to obtain. However, they were obtained at this level of generality by Hill only in 1963. That is, it was necessary, for about a half century, to understand the real significance of the estimates due to Voigt and to Reuss. Following Hill [4.1] again, we shall show that the overall elasticities are increased when either or both of the elasticities are increased in one or both phases. “Increased” means that in each charged element of the material, the specific elastic energy obtained afterward should exceed the one existing before, if the local strains were the same. More exact, let us assume that two composite materials B and B1 have the same geometry for the matrix and for the inclusions. Let c = c (x) and c1 = c1 (x) be the micro-elasticities of B and B1 , respectively. B1 is said to be strengthened in comparison with B if ε · c (x) ε ≤ ε · c1 (x) ε for any ε and any x.

(4.2.39)

In this condition, we have Hill’s comparison theorem. Let B and B1 be two macrohomogeneous composites, having the same geometry. Let B1 be strengthened in comparison with B. Let b c and b c1 be the overall elasticities of B and B1 , respectively. Then,

or, briefly

c1 ε ε·b cε ≤ ε · b

b c≤b c1 .

for any ε,

(4.2.40) (4.2.41)

In order to prove the theorem, we analyze the behavior of the corresponding RVEs. Let us consider for both RVEs the homogeneous displacement boundary value problem corresponding to the same ε. Let us denote by [u, ε = ε (u) , σ = cε] and

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[u1 , ε1 = ε (u1 ) , σ 1 = c1 ε1 ] the corresponding solutions. The principle of minimum potential energy, applied first to the original body and next to the strengthened body, gives (4.2.42) ε·b cε = ε (u) · cε (u) = inf ε (v) · cε (v) v∈K(ε)

and

ε·b c1 ε = ε (u1 ) · cε (u1 ) =

inf ε (v) · c1 ε (v) .

v∈K(ε)

(4.2.43)

Since the two RVEs have the same geometry u1 ∈ K (ε); hence,

ε (u) · cε (u) ≤ ε (u1 ) · cε (u1 ).

(4.2.44)

Also, according to the relation (4.2.39), we have

ε (u1 ) · cε (u1 ) ≤ ε (u1 ) · c1 ε (u1 ).

(4.2.45)

Hence, according to (4.2.44) and (4.2.45), we obtain

ε (u) · cε (u) ≤ ε (u1 ) · c1 ε (u1 ).

(4.2.46)

The last relation (4.2.46) and the formulas (4.2.42) and (4.2.43) show that b c and b c1 satisfy the inequality (4.2.40) and the comparison theorem is proved.

4.3

Macroscopically isotropic biphasic mixture

Let us assume isotropic matrix and inclusions. Using the notations

1 1 θ = trε, e = ε − θ1, σ = trσ, s = σ − σ1 , 3 3

the micro-stress-strain relations are  k1 θ in D1 , σ= k2 θ in D2 ,

and s =



µ1 e in D1 . µ2 e in D2

(4.3.1)

(4.3.2)

Here, k1 , k2 are the bulk moduli, µ1 , µ2 are the shear moduli of the matrix and of the inclusions, respectively. We assume the composite to be macroscopically isotropic; i.e. the overall behavior of the mixture is completely described by its overall bulk modulus b k and shear modulus µ b. In order to estimate these overall moduli, it is no longer necessary to impose general homogeneous boundary conditions on the boundary of the RVE. It is enough to consider just two independent strain states, a pure dilatation and a pure shear characterized by a given constant number θ and by a given constant deviator e, respectively. We take first a pure dilatation θ. Since the composite is macroscopically isotropic, the corresponding overall mean stress will be an all round tension σ.

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211

Accordingly, the overall (equivalent) stress-strain relation, connecting these mean values is 1 (4.3.3) σ=b k θ or θ = σ, b k b k being the unknown overall bulk modulus. Using the fundamental formula (4.1.10), we obtain σ = c1 σ 1 + c2 σ 2 and θ = c1 θ1 + c2 θ2 ,

where c1 , c2 are the concentrations of the matrix and of the inclusions, respectively, σ 1 , σ 2 are the mean values of σ on D1 and D2 , the domains occupied by the matrix and by the inclusions, and θ 1 , θ2 are the mean values of θ on D1 and D2 , respectively. Using the constitutive equations (4.3.2), we obtain

σ 1 = k1 θ 1 , σ 2 = k2 θ 2 , θ 1 =

1 1 σ1 , θ2 = σ2 . k2 k1

Thus we get σ = c1 k1 θ1 + c2 k2 θ2 and θ =

c1 c2 σ1 + σ2 . k1 k2

(4.3.4)

Since θ 1 , θ2 and σ 1 , σ 2 are uniquely determined by θ and σ, respectively, the general tensor equations (4.1.20) and (4.1.34) are now replaced by more simple scalar relations, θ1 = a1 θ, θ2 = a2 θ and σ 1 = b1 σ, σ 2 = b2 σ.

(4.3.5)

Also, the general tensor relations (4.1.23) and (4.1.36) simply become c1 a1 + c2 a2 = 1 and c1 b1 + c2 b2 = 1.

(4.3.6)

The numbers a1 , a2 and b1 , b2 are the influence coefficients corresponding to the homogeneous displacement and traction problems, completely characterized by the pure dilatation θ and by the hydrostatic pressure σ, respectively. As we know (see Equations (4.2.9)), the micro-and macro-moduli are not independent. It is easy to see that Hill’s tensorial compatibility conditions (4.2.9) now take the following more simple scalar form: a 1 k1 = b 1 b k and a2 k2 = b2 b k.

(4.3.7)

Using the influence coefficients, from the equations (4.3.3), (4.3.4) and (4.3.5), we obtain the following expressions for the overall bulk modulus b k: 1 c 1 b1 c 2 b2 b k = c1 a1 k1 + c2 a2 k2 and = + . b k1 k2 k

(4.3.8)

These more simple scalar relations take the place of the more general tensor equations (4.1.22) and (4.1.35), valuable for an arbitrary biphasic composite.

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We observe also that due to Hill’s compatibility relations (4.3.7), the two expressions (4.3.8) for b k are equivalent, as it is easy to see. This equivalence is necessary to have a consistent theory, allowing the replacement of the heterogeneous composite by a single equivalent homogeneous body. This fact reveals the great importance of Hill’s compatibility conditions. As we already know, their validity is implied by Hill’s strong assumption, and this, in turn, follows from Hill’s weak assumption. We can thus conclude that the replacement can be meaningfully realized only if a macrohomogeneous composite also has the supplementary property stipulated in Hill’s weak assumption. Since k1 , k2 , c1 , c2 are supposed to be known, as the relations (4.3.6) and (4.3.8) show, it is necessary to determine one of the influence coefficients a 1 , a2 , b1 or b2 , in order to obtain the overall bulk modulus b k. If we consider a pure shear e and denote by s the corresponding mean deviatoric stress, we get the overall stress-strain relations

s=µ be and e =

1 s. µ b

The overall shear modulus µ b is given by the equivalent relations µ b = c1 α1 µ1 + c2 α2 µ2 and

c 2 β2 c 1 β1 1 . + = µ2 µ1 µ b

(4.3.9)

(4.3.10)

Here α1 , α2 , β1 , β2 are the concentration coefficients corresponding to the involved displacement or traction problem. As we know (see Sections 3.2 and 4.2), a crude approximate treatment of the problem assumes that the strain throughout the mixture is uniform if the displacement boundary condition is homogeneous (Voigt’s assumption). This supposition leads to Voigt’s evaluation or mixture rule kV = c 1 k1 + c 2 k2 , µ V = c 1 µ 1 + c 2 µ 2

(4.3.11)

for the overall moduli. The dual assumption is that the stress is uniform if the traction boundary condition is homogeneous (Reuss’ assumption). This supposition leads to Reuss’ evaluation of mixture rule

c2 c1 c2 1 c1 1 + = + , = µ2 µ1 k2 µ R k1 kR

(4.3.12)

for the overall moduli. The above relations represent special forms of the general equations (4.2.32) and (4.2.37), if the mixture is macroscopically isotropic. In this special case, Hill’s universal estimates (4.2.33) and (4.2.38) become more simple: kR ≤ b k ≤ k V , µR ≤ µ b ≤ µV . (4.3.13)

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213

Elementary computations, using (4.3.11), (4.3.12) and the relation c1 +c2 = 1, show that the differences between Voigt’s and Reuss’ estimates can be expressed as 2

kV − k R =

2 (µ1 − µ2 ) (k1 − k2 ) , µ V − µ R = µ1 µ2 . k2 k1 + c1 + c2 c2 c1

(4.3.14)

Thus the Voigt’s values exceed indeed the Reuss’ ones as required by the Hill’s bound estimates (4.3.13). These differences are only second order quantities with respect to the small differences between the corresponding elastic constants of the phases. Hill’s universal estimates lose their practical importance if one of the phases (the inclusion) is comparatively rigid (k2 , µ2 → ∞), the other having finite moduli, or if one of the phases (the matrix) is comparatively weak (k1 , µ1 → 0). In the first case kV , µV → ∞, while kR , µR rest finite, while in the second case kR , µR → 0 while kV , µV rest finite. In general, Hill’s universal estimates (4.3.13) are rather poor when the phase moduli differ by more than a factor of two, or so. It is known that Young’s modulus E can be expressed in terms of k and µ as

1 1 3 . = + µ 3k E

(4.3.15)

Let ER and EV denote Young’s moduli calculated from (4.3.15) with Reuss’ and Voigt’s estimates, respectively, of the bulk and shear moduli, given by (4.3.11) and (4.3.12); i.e. 1 1 3 1 1 3 . (4.3.16) + = , + = 3kV µV 3kR EV µR ER

From (4.3.13), it follows that

or

1 1 1 , ≤ ≤ b E EV R E

b ≤ EV , ER ≤ E

b is the overall Young modulus; i.e. where E

1 1 3 . = + b µ b 3b E k

(4.3.17)

(4.3.18)

Due to the linearity of the relation (4.3.15) between the reciprocal of Young’s modulus and the reciprocal of the bulk and shear moduli, from (4.3.15) it results in c2 c1 1 . (4.3.19) + = E2 E1 ER

Thus Reuss’ estimate (4.3.16) of Young’s overall modulus is what would be obtained by calculating the average (mean) longitudinal strain in a traction test on the assumption that the microscopic stress in the mixture is uniform.

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CHAPTER 4. MACROSCOPICALLY ELASTIC COMPOSITES By contrast, in general, EV 6= c1 E1 + c2 E2

(4.3.20)

since Young’s modulus is not a linear combination of the bulk and shear moduli (see Equation (4.3.15)). By direct calculation, it can be shown that c1 E1 + c2 E2 ≤ EV and c1 E1 + c2 E2 = EV if and only if ν1 = ν2 ,

(4.3.21)

ν1 and ν2 being Poisson’s ratios of the matrix and inclusions, respectively. Finally we have the following chain of comparisons: −1    b c2 c1 E = ER ≤ ≤ EV . (4.3.22) + c 1 E1 + c 2 E2 E2 E1

b and c1 E1 + c2 E2 . When the Poisson There is no universal ordering of E b ≤ c1 E1 + c2 E2 ; when the rations of the phases are equal, it can be shown that E b shear moduli of the phases are equal, E ≥ c1 E1 + c2 E2 . Let us assume now that the phases have equal shear moduli µ; i.e. µ1 = µ2 = µ,

(4.3.23)

and only their bulk moduli k1 and k2 are different. In this situation, Hill [4.1] was able to determine the overall bulk modulus b k. We now present Hill’s solution and some interesting and important conclusions derivable from Hill’s wonderful result. Let us introduce the Newtonian potential ϕ = ϕ (x) satisfying the following Poisson equation: −c2 in D1 , ∆ϕ = { (4.3.24) c1 in D2 . As it is known from the theory of the Newtonian potential, ϕ satisfying (4.3.24) is a continuous function on D and its gradient grad ϕ has the same property. However, the second derivatives ϕ,ij of ϕ are discontinuous in D. According to equations (2.7.16), (2.7.17) and (4.3.24), these derivatives satisfy the following jump conditions across Γ, the common boundary of the matrix and inclusions [ϕ,ij ] = ni nj on Γ .

(4.3.25)

Following Hill, let us introduce the following displacement field u (x) on D   4 1 µ (4.3.26) c2 k1 + c1 k2 + µ x in D, u = (k1 − k2 ) grad ϕ + 3 3 η

where η is a scalar parameter, to be determined later. The first term in the right hand side of (4.3.26) is an irrotational displacement field generated by ϕ and the

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second one represents an uniform expansion. Since u and grad u are continuous on D, u is an admissible displacement field for the piece-wise homogeneous composite. The corresponding strain ε is given on D by   4 1 µ (4.3.27) c2 k1 + c1 k2 + µ δij . εij = (k1 − k2 ) ϕ,ij + 3 3 η

Due to (4.3.25), ε has nonvanishing jump across Γ. Particularly, its dilatational part θ = trε is piece-wise constant in D. Indeed, from (4.3.25) and (4.3.27), we get    η 4   in D1 , k2 + µ    µ 3 (4.3.28) θ=     η 4   in D2 .  k1 + µ µ 3

Hence, the mean values θ 1 and θ2 of θ, on the subdomains D1 and D2 occupied by the matrix and by the inclusions in RVE are given by     η 4 η 4 . (4.3.29) , θ 2 = k1 + µ θ 1 = k2 + µ µ 3 µ 3

Hence, according to the fundamental relation (4.1.10) the mean (average) dilatation θ = trε in the mixture is given by the equation   η 4 , (4.3.30) θ = c 1 θ 1 + c 2 θ 2 = c 1 k2 + c 2 k1 + µ µ 3

since c1 + c2 = 1. In order to obtain the (microscopic) stress σ, we use the stress-strain relation in the form     2µ 2µ θ1+2µε. (4.3.31) (trε) 1 + 2µε = k − σ= k− 3 3

Thus, using (4.3.27) and (4.3.28), we get     2µ 4µ   k − k + δij + 2µ (k2 − k1 ) ϕ,ij 1 2   3 3    2 4   + (c2 k1 + c1 k2 + µ)µδij in D1 ,   3 3 σij =       2µ 4µ   k2 − k1 + δij + 2µ (k2 − k1 ) ϕ,ij    3 3    2 4  + (c2 k1 + c1 k2 + µ)µδij in D2 . 3 3

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(4.3.32)

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Using these expressions and taking into account equation (4.3.24) satisfied by ϕ, it is easy to see that div σ =0 in D1 and D2 .

(4.3.33)

Hence, σ is a self-equilibrated stress field. Taking into account the jump relations (4.3.25) satisfied by ϕ, it is easy to prove that σ also satisfies the null-jump condition [σn] = 0 on Γ.

(4.3.34)

The obtained results show that we have obtained the real displacement field in the RVE under surface displacement values of type (4.3.26) and with the normal component un = u · n of u given by the equation un = (k1 − k2 )

η ∂ϕ 1 + θx · n on ∂D1 , µ ∂n 3

(4.3.35)

resulting from the relations (4.3.26) and (4.3.30). This normal component is as yet arbitrary. It remains to be established that the obtained family of solutions contains a solution corresponding to a macroscop1 ◦ ically uniform (see Section 4.1) pure dilatation (u (x) = θx) in the considered 3 RVE. To this end, it is enough to prescribe a normal component of the surface dis1 placement equalling that in an uniform expansion (i.e. θx · n on ∂D ); or equiv3 ∂ϕ =0 alently, to prescribe vanishing normal derivative of the potential ϕ (i.e. ∂n on ∂D ), as can be seen from (4.3.35). We assume that the mixture is such that the solution of this Neumann problem for ϕ is a function that fluctuates with a wave-length of the order of the mean dimensions of the inclusions of the mean distance between them. Consequently, the displacement field given by (4.3.26) really corresponds to macroscopically homogeneous displacement boundary conditions. Hence, assuming that Hill’s and Mandel’s conjecture (4.1.4) is right, the product theorem can be applied neglecting terms of order O (d/l). Thus the above selected solution can be used to determine the equivalent bulk modulus. Obviously, if θ is prescribed, η can be determined using (4.3.30). Now we can return to the equations (4.3.28) and (4.3.30). Using them, we get k1 + 4µ k2 + 4µ θ2 θ1 3 3 . (4.3.36) = , = c1 k2 + c2 k1 + 43 µ c1 k2 + c2 k1 + 43 µ θ θ

Comparing (4.3.5)1,2 and (4.3.36), we obtain the influence coefficients a1 and a2 a1 =

k1 + 4µ k2 + 4µ 3 3 , a = . 2 c1 k2 + c2 k1 + 34 µ c1 k2 + c2 k1 + 43 µ

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(4.3.37)

4.3. MACROSCOPICALLY ISOTROPIC BIPHASIC MIXTURE

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Thus, if µ1 = µ2 = µ, our problem is completely solved. Indeed, from (4.3.8) and the above equations, we get the value of the overall bulk modulus b k in terms of the known data (k1 , k2 , µ1 = µ2 = µ, c1 , c2 ): 4µ (c1 k1 + c2 k2 ) + 3k1 k2 b . k= 3 (c1 k2 + c2 k1 ) + 4µ

(4.3.38)

Let us observe that the above relation can be expressed in the following equivalent useful form:

c2 c1 1 = 4µ . 4µ + b k + k + k + 4µ 2 1 3 3 3

(4.3.39)

Hill’s remarkable result shows that the overall bulk modulus b k depends solely on the separated moduli and on the concentrations, and is unaffected by the shapes and distributions of the inclusions. Using Voigt’s and Reuss’ estimates (4.3.11)1 , (4.3.12), the overall bulk modulus b k can be expressed in various equivalent ways: b 1+ k = kR 1+

or

4µkV 3k1 k2 4µkR 3k1 k2

1 kV − k 1 k − kR . = , = 4µkR 3k k 1 2 kV − k R 1 + 4µkR kV − kR 1 + 3k 1 k2

(4.3.40)

(4.3.41)

Of course, the overall shear modulus µ b is just µ itself. This can be seen from (4.3.10) and equation c1 α1 + c2 α2 = 1 satisfied by the concentration coefficients α1 and α2 . We recall that Poisson’s ratio ν can be expressed in terms of k and µ by ν=

µ − 3k µ . 1 + 3k 1 2

(4.3.42)

Using this equation and the relation (4.3.38), it is easy to see that the overall Poisson ratio νb is given by

We recall also the equation

νb =

c 1 ν1 + c 2 ν2 − ν 1 ν2 . 1 − c 1 ν2 − c 2 ν1

(4.3.43)

E = 2µ (1 + ν) .

(4.3.44)

b we get the Using (4.3.43) and the last relation, for the overall Young modulus E, expression b = 4µ (c1 E1 + c2 E2 ) − E1 E2 . (4.3.45) E 4µ − (c1 E2 + c2 E1 )

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CHAPTER 4. MACROSCOPICALLY ELASTIC COMPOSITES This formula shows “the deviation” from Voigt’s mixture rule. Also, from (4.3.45), we get b if µ1 = µ2 = µ. c 1 E1 + c 2 E2 ≤ E

(4.3.46)

µ1 (matrix) < µ2 (inclusions),

(4.3.47)

Using Hill’s results for µ1 = µ2 = µ, it is possible to establish universal bounds on the overall bulk modulus b k for an arbitrary mixture (µ1 6= µ2 ). In order to do this, we recall Hill’s comparison theorem according to which b k would be increased (decreased) if both phases have a common rigidity modulus equal to the larger (smaller) of their actual values. Consequently, the actual overall bulk modulus b k must lie between those of similarly proportional mixtures, in which one of the phases have the same shear rigidity µ2 and, in the other one, the same shear rigidity µ1 (the separate bulk moduli k1 and k2 being fixed). Hence, assuming that

from the relations (4.3.11)2 , (4.3.12)1 and (4.3.38), we get

b 4µ2 kV + 3k1 k2 k 4µ1 kV + 3k1 k2 . ≤ ≤ 4µ2 kR + 3k1 k2 kR 4µ1 kR + 3k1 k2

(4.3.48)

We stress the following fundamental result: Hill’s completeness theorem [4.1]. The bounds (4.3.48) are the best possible, in terms only of the moduli and concentrations, when no regard is paid to geometry. Indeed, each bound in (4.3.48) is exact, for arbitrary geometries when the shear moduli are both equal to µ1 or to µ2 , as shown in Hill’s evaluation (4.3.48). Hence, to obtain best bounds for b k, the geometry of the inclusions must be taken into account. Let us observe that using equation c1 + c2 = 1, the overall bulk modulus b k, given by the relation (4.3.38), can also be expressed in the following equivalent forms: c1 c2 b . (4.3.49) = k2 + k = k1 + 3c2 1 3c1 1 + + 3k2 + 4µ k1 − k 2 3k1 + 4µ k2 − k 1

Thus, Hill’s universal bounds (4.3.49) can be expressed by the inequalities

where

k− ≤ b k ≤ k+ ,

k − = k1 +

k + = k2 +

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(4.3.50)

c2 , 3c1 1 + 3k1 + 4µ1 k2 − k 1

c1

3c2 1 + 3k2 + 4µ2 k1 − k 2

(4.3.51) .

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4.4. HASHIN-SHTRIKMAN PRINCIPLE

4.4

The Hashin-Shtrikman principle

The difficult problem of bounding the overall shear modulus remains. The idea behind the method leading to (4.3.48) or (4.3.49) is valid here, too, but cannot be used since the formula analogous to (4.3.48) is not known there, where only shear moduli are distinct. Bounds for the overall shear modulus have been proposed by Hashin and Shtrikman [4.11], [4.12] via their new variational and extreme principle. We present this important result for the displacement boundary value problem. Consider a homogeneous body which occupies the domain B limited by the b isi prescribed on the boundary ∂B. Let us assume that surface displacement h◦ ◦ u ◦ boundary ∂B and no body forces are present. Let u , ε , σ be the solution of this displacement boundary value problem. According to the assumption made, the stress-strain relation for the body B is ◦

◦ ◦

◦

σ (x) = c ε (x) , c = const.

(4.4.1)

Next we consider a geometrically identic composite body B, with the same b , but which is piecewise homogeneous and anisoprescribed surface displacement u tropic, containing two phases, the matrix and the inclusions. Let [u, ε, σ] be the solution of the considered displacement boundary value problem. According to the assumption made, the stress-strain relation for the body B is  c1 = const. in B1 σ (x) = c (x) ε (x) with c (x) = (4.4.2) c2 = const. in B2 , B1 and B2 being the complementary subdomains occupied by the matrix, and by the inclusions, respectively. Following Hashin and Shtrikman, we introduce the symmetric stress polarization tensor p defined by ◦ σ =c ε + p. (4.4.3) We define also the vector field

◦

u0 = u− u . Hence,

(4.4.4) ◦

ε0 = ε (u0 ) = ε− ε .

(4.4.5) 0

In their analysis Hashin and Shtrikman have chosen p and ε as unknown fields and have formulated their variational principle in terms of these fields. Following them, we introduce the tensor ◦

r = c− c

(4.4.6)

and assume that it is invertible. The inverse of r will be denoted by h; i.e. h = r−1 .

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(4.4.7)

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Let us introduce the functional U, defined on p and ε0 by the following equation: Z   ◦ ◦ 1 ◦ p · hp − p · ε0 − 2p· ε dv (4.4.8) U (p, ε0 ) = U ε − 2 B

where

Z Z ◦ ◦ 1 1 ◦ ◦◦ ◦ ◦ ε ·cεdv. σ·εdv = U ε = 2 B 2 B

(4.4.9)

div (c ε0 + p) = 0 in B1 and in B2 , u0 = 0 on ∂B,

(4.4.11)

The first variation δU of U in (p, ε0 ), in the direction δp, δε0 is defined by relation
d U p+αδp, ε0 + αδε0 |α=0 , (4.4.10) δU = dα α being a real parameter. We give and prove now The Hashin-Shtrikman variational and extreme principle. (i)The functional U, subjected to the subsidiary conditions ◦

◦

[u0 ] = 0 and [(c ε0 + p)n] = 0 on Γ, ◦

div (c δε0 + δp) = 0 in B1 and in B2 , δu0 = 0 on ∂B,

(4.4.12) (4.4.13)

◦

(4.4.14)

p = rε.

(4.4.15)

[δu0 ] = 0 and [(cδε0 + δp)n] = 0 on Γ, is stationary; i.e. δU = 0, when

Here Γ in the common boundary of the matrix and the inclusions. (ii) The stationary value U s = U(p, ε0 )of U is an absolute maximum when r is positive definite and is an absolute minimum when r is negative definite. (iii)The stationary value U s is equal to the strain energy U(ε) stored in the body. (i) First of all we observe that according to (4.4.3) and (4.4.6), the stationary condition (4.4.15) is equivalent to the stress-strain relation (4.4.2) for the heterogeneous body. Next we stress the fact that the divergence theorem can be applied in its usual form since the null-jump conditions (4.4.12) and (4.4.14) are assumed to be satisfied. In order to prove (i), we must show that δU = 0,

(4.4.16)

for any δp, δε0 = ε(δu0 ) satisfying (4.4.13), (4.4.14), if p and ε0 = ε(u0 ) satisfy (4.4.11), (4.4.12).

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221

We observe that h has the same symmetry as c according to (4.4.6) and (4.4.7). We also take into account the equation (4.4.5) defining ε0 . In this way from (4.4.8)−(4.4.10) we get Z 1 ◦ (2δp · hp−δp · ε0 − p·δε0 − 2δp· ε)dv δU = − 2 B Z 1 (2 (hp − ε) · δp + ε0 · δp − p·δε0 )dv. =− 2 B

Since h = r−1 , when the extreme condition (4.4.15) holds, the first term of the integral vanishes. Hence, Z 1 (ε0 · δp − p·δε0 ) dv. (4.4.17) δU = − 2 B

The supplementary conditions (4.4.11)1 , (4.4.12)2 and (4.4.13)1 , (4.4.14)2 can be expressed in the following form: ◦

c ε0 + p = t with div t = 0 in B1 ∪ B2 , [tn] = 0 on Γ,

(4.4.18)

c δε0 + δp = δt with div δt = 0 in B1 ∪ B2 , [δtn] = 0 on Γ.

(4.4.19)

◦

Introducing p and δp from (4.4.18) and (4.4.19) into (4.4.17) and using the ◦ symmetry of c, we get Z 1 (ε0 · δt − t·δε0 ) dv, δU = − 2 B

or, taking into account the divergence theorem and the relations (4.4.18), (4.4.19) Z 1 (u0 · δtn−δu0 · tn) da. δU = − 2 B

Now the restrictions (4.4.12)1 and (4.4.14)1 show that the relation (4.4.16) is true and (i) is proved. In order to prove
(ii) we must evaluate the second variation δ 2 U in (p, ε0 ) in the direction δp, δε0 . We have δ2 U =

From (4.4.8), we obtain 2

δ U =−

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d2 U p+αδp, ε0 + αδε0 |α=0 . 2 dα

Z  B

 ◦ δp · hδp+δε0 · c δε0 − δt·δε0 dv.

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The integral of the last term vanishes since δt and δu satisfy (4.4.19) and (4.4.14), respectively. Hence, Z   ◦ (4.4.20) δ2 U = − δp · hδp+δε0 · c δε0 dv. B

If r is positive definite, so is its inverse h. From (4.4.20) we get, in this case, δ 2 U < 0,

hence the stationary value U s = U (p, ε0 ) of U is an absolute maximum. In order to prove the minimum condition, we consider the integral Z ◦ δI = δp· k δpdv, (4.4.21) B

◦

◦

k being the inverse of c. Using again (4.4.19), we get Z   ◦ ◦ δt· k δt + δε0 · c δε0 − 2δt·δε0 dv. δI = B

The integral of the last term vanishes again; hence,  Z  ◦ 0 ◦ 0 δt ·k δt + δε · c δε dv. δI =

(4.4.22)

B

◦

As we know, k is positive definite. Hence, comparing the expressions (4.4.21) and (4.4.22) of δI, we can conclude that Z Z ◦ ◦ δε0 · c δε0 dv. δp ·k δpdv ≥ (4.4.23) B

B

Now (4.4.20) and (4.4.23) results in Z Z Z ◦ ◦ 2 δ U ≥ − δp · hδpdv − δp· k δpdv = − δp · (h+ k)δpdv. B

B

B

From this inequation, we can see that a sufficient condition for

◦

δ2U > 0

is that h+ k be negative-definite. Because of the relations (4.4.6) and (4.4.7) and ◦ of the positive definiteness of c and c, this condition is equivalent to the negative definiteness of r. Hence, if r is negative definite, the stationary value U s = U (p, ε0 ) of U is an absolute minimum and (ii) is completely proved. In order to prove (iii) we must show that Z 1 s 0 σ · εdv. U = U (p, ε ) = U (ε) = 2 B

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4.4. HASHIN-SHTRIKMAN PRINCIPLE

Introducing the stationary condition (4.4.15) in the first term of the integral (4.4.8), and using (4.4.7), we get successively Z   ◦ ◦ 1 ◦ rε · hp − p · ε0 − 2p· ε dv = U s = U (p, ε0 ) = U ε − 2 B Z   ◦ ◦ 1 ◦ ε · p − p · ε0 − 2p· ε dv. =U ε − 2 B

Taking into account (4.4.5), we get Z   ◦ ◦ 1 ◦ ◦ ε ·p − 2p· ε dv Us = U ε − 2 B Z ◦ ◦ 1 ◦ p· ε dv. =U ε + 2 B

Now, taking into account (4.4.1), (4.4.3) and (4.4.9), we obtain Z   1 ◦ ◦ ◦ ◦ s σ · ε +σ· ε − ε ·cε dv U = 2 B Z   1 ◦ ◦ ◦ ◦ σ · ε +σ· ε − σ ·ε dv. = 2 B ◦

From (4.4.5) ε= ε − ε0 and using this equation, U s becomes Z  Z  1 1 ◦ σ+ σ · ε0 dv. σ · εdv − Us = 2 B 2 B

But

◦

Z

B

σ · ε0 dv =

Z

◦

B

σ ·ε0 dv = 0

since σ and σ are the stress corresponding to the solutions of the two displacement boundary value problems, and u0 vanishes on ∂B, satisfying null-jump conditions on Γ. Consequently, from the last expression of U s , we get Z 1 σ · εdv = U (ε) , (4.4.24) U s = U (p, ε0 ) = 2 B

and (iii) is proved. As was shown by Hashin and Shtrikman, the above principle can be used to obtain bounds for the overall elastic moduli. In order to see how this can be done, we assume in this Section that both components (phases) of the composite are isotropic. Moreover, as in Section 4.3, we suppose that the biphasic mixture is also macroscopically isotropic. All quantities referring to the homogeneous body, which are present in the given theorem, will be marked by a superposed zero. The fields referring to the matrix will be indexed by 1, those concerning the inclusions by 2.

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CHAPTER 4. MACROSCOPICALLY ELASTIC COMPOSITES

We shall use the following notations denoting the decomposition of various fields in their spherical and deviatoric parts ◦

◦

◦

◦

◦

◦

◦

ε = ε 1+ e , σ = σ 1+ s , ε =

1 ◦ 1 ◦ ◦ tr ε , σ = tr σ , 3 3

(4.4.25) 1 1 ε = ε1 + e, σ = σ1 + s, ε = trε, σ = trσ. 3 3 The stress-strain relation for the homogeneous body can be written in two equivalent forms: ◦  ◦ ◦ ◦ ◦ σ = λ trε 1+2 µ ε or

is

◦◦ ◦ ◦ ◦◦ 2 ◦ ◦ ◦ σ = 3 k ε and s = 2 µe with k = λ + µ . (4.4.26) 3   ◦ ◦ ◦ Consequently, the specific strain energy u = u ε of the homogeneous body

  ◦ ◦2 ◦◦ ◦ 1 1◦ ◦ µ ε e e 9 +2 · . (4.4.27) σ·ε= k 2 2 Similarly, we can express the stress-strain relation for the composite in two equivalent forms σ = λ (trε) 1+2µε ◦

u=

or

2 (4.4.28) σ = 3kε and s = 2µe with k = λ + µ. 3 In these equations, λ, µ and k take the constant values λ1 , µ1 , k1 in the domain occupied by the matrix, and the constant values λ2 , µ2 , k2 in the domain occupied by the inclusions.   ◦ −1 We must express now the tensor k = r−1 = c− c . If σ and ε are two symmetric tensor connected by the equation   ◦ σ = rε = c− c ε, ε = hσ,

from (4.4.26) and (4.4.28), we get   ◦ ◦ σ = 3 k− k ε, s = 2 µ− µ e. Hence,

1 1 ε = ε1 + e =  ◦s ◦  σ1+  2 µ− µ 3 k− k       1 1 1   − =   ◦  σ. ◦  σ1+  ◦   3 k− k 2 µ− µ 2 µ− µ 

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225

4.4. HASHIN-SHTRIKMAN PRINCIPLE

Now it is easy to see that the components hijkl of h are       1 1 1 1    −   hijkl = ◦  (δik δjl + δil δjk ) , ◦  δij δkl + ◦ 3  3 k− k 4 µ− µ 2 µ− µ 

or, equivalently,

hijkl =

1 δij δkl  ◦ ◦ +  4 µ− µ 9 k− k



 2 δik δjl + δil δjk − δij δkl . (4.4.29) 3

Let us decompose now the polarization tensor p in its spherical and deviatoric parts: 1 (4.4.30) p = p1 + q, p = trp, trq = 0. 3 Thus from (4.4.29) and (4.4.30), we get

p · hp =

1

◦p

2

+

k− k



1

◦q

2 µ− µ

· q.

(4.4.31)

Let D be the domain occupied by the RVE of the composite, and let ∂D and v the boundary and the volume of D, respectively. We consider also the homogeneous material occupying the same domain D. We suppose that on the boundary ∂D of both bodies, the same homogeneous displacement is prescribed; i.e. ◦

u (x) = εx, u (x) = εx on ∂D

where ε is given, symmetric, constant tensor. The solution of this displacement problem for the homogeneous body is ◦

◦

◦

◦

u (x) = εx, ε (x) = ε, σ (x) =c ε, in D.

(4.4.32) ◦ ◦ Hence, according to (4.4.27), the mean value of the strain energy U ε of the homogeneous body will be  ◦ 1 ◦ 2 1 ◦ ◦ (4.4.33) 9k ε +2µ e·e , U ε = 2 v

where

1 trε, tre = 0. (4.4.34) 3 Now we return to the RVE of the composite. As before, let [u, ε, σ] be the solution corresponding to the homogeneous displacement condition imposed on the boundary of the RVE. In order to introduce the overall elastic moduli b k and ε = ε1+e, ε =

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CHAPTER 4. MACROSCOPICALLY ELASTIC COMPOSITES

µ b of the composite their energetical definition will be used. Thus, if U (ε) is the strain energy of the RVE, we get

 1 b 2 1 µ e·e . 9k ε + 2b U (ε) = 2 v

(4.4.35)

In order to apply the Hashin-Shtrikman principle, the mean value of U (p, ε 0 ) must also be determined. According to (4.4.8), we get Z Z Z 1 1 1 1 ◦ ◦ 1 ◦ 0 0 p· ε dv. p · ε dv + p · hpdv + U (p, ε ) = U ε − 2v D 2v D 2v D v v

(4.4.36) In order to evaluate this mean value first, we use (4.4.31), (4.4.32)2 and (4.4.33). In this way, (4.4.36) becomes   Z   ◦ ◦ 1 1 2 1 2  q · q dv + p +  U = 9 k ε2 + 2 µ e · e − ◦ ◦  v D  k− v 2 µ− µ k Z 1 (4.4.37) p · ε0 dv + 2p · ε. + v D

Here p is the mean value of the polarization tensor p. Following Hashin and Shtrikman, we assume now that the polarization stress field p is piece-wise constant; i.e.  p1 = const. in D1 , p (x) = (4.4.38) p2 = const. in D2 ,

and

1 trpα , trqα = 0, α = 1, 2. 3 Consequently, for the mean value of p, we get p α = p α 1 + q α , pα =

p=

2 X

cα pα = p1+q with p =

α=1

2 X

c α pα , q =

α=1

2 X

c α qα ,

(4.4.39)

(4.4.40)

α=1

c1 and c2 representing the concentration of the matrix and of the inclusions, respectively. From (4.4.40), we get pα · ε = 3ε

Copyright © 2004 by Chapman & Hall/CRC

2 X

α=1

c α pα + e ·

2 X

α=1

c α qα .

227

4.4. HASHIN-SHTRIKMAN PRINCIPLE Now (4.4.37) becomes

2 X ◦ ◦ cα 2 2 (4.4.41) U (p, ε0 ) = 9 k ε2 + 2 µ e · e − ◦ pα v α=1 kα − k 2 2 2 X X X cα e · cα qα + 2U 0 (p, ε0 ) . ε c p + 2 q · q + 6 − α α α α ◦ µ α=1 α=1 α=1 2(µα − )

Here

1 0 W (p, ε0 ) v Z 1 0 0 p · ε0 dv. W (p, ε ) = 2 D U 0 (p, ε0 ) =

and

(4.4.42)

(4.4.43)

It remains the most difficult problem: the evaluation of the mean value U 0 (p, ε0 ) in terms of the polarization field (4.4.38). The problem was solved by Hashin and Shtrikman using Fourier transforms. First of all we introduce the fluctuation p0 of the polarization stress p; i.e. p0 = p−p.

(4.4.44)

Taking into account that the mean value ε0 of the fluctuation ε0 of the strain ε is vanishing, from (4.4.43) we obtain Z 1 0 0 0 0 0 p0 · ε0 dv. (4.4.45) W (p, ε ) = W (p , ε ) = 2 D

Also, since p is constant, the field equations (4.4.11), (4.4.12), which must be satisfied by the fluctuations u0 , ε0 and p0 , take the following form: ◦  div c ε0 + p0 = 0 in B1 ∪ B2 , u0 = 0 on ∂B, h ◦ 0  i (4.4.46) c ε + p0 n = 0 on Γ. [u0 ] = 0,

Introducing now the stress-strain relation (4.4.1), characterizing the homogeneous and isotropic material, we get the equilibrium equations which must be satisfied by the field u0 : ◦ ◦ 0 ◦ 0 0 (4.4.47) λ + µ uj,ij + µ ui,jj + pij,j = 0 in D1 ∪ D2 .

In order to evaluate the mean value (4.4.42), we assume now that the composite has a infinite extent. In this case, according to (4.4.43)-(4.4.47), our problem can be formulated in the following way: Given ◦ ◦ 0 ◦ 0 0 0 (4.4.48) λ + µ uj,ij + µ ui,jj + pij,j = 0 in E − Γ, lim u (x) = 0 |x|→∞

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CHAPTER 4. MACROSCOPICALLY ELASTIC COMPOSITES

find u0 (x), and than evaluate U 0 (p, ε0 ), using the equation Z 1 1 0 0 0 0 0 p0 · ε0 dv. U (p, ε ) = lim W (p , ε ) = lim v →∞ 2v D v →∞ v

(4.4.49)

Recall that E is the whole three-dimensional Euclidean space. Fortunately the complete solution of the system (4.4.48) is not required to obtain the mean value (4.4.49). To see this we recall that if f (x) is a field given on E, its Fourier transform F (ξ) is defined by Z √ 1 (4.4.50) f (x) eiξ·x dx, i = −1, F (ξ) = (2π)3/2

where the integration is over the entire physical space E. We recall also that the Fourier transform of the partial derivative f,i (x) is −iξj F (ξ), where ξj is the j-th component of the wave number vector ξ. Also, if g (x) is a second field given on E and G (ξ) is its Fourier transform, the Parceval equality takes place: Z Z ∗ (4.4.51) F (ξ) G (ξ) d ξ = f (x) g (x) dx.

Here the superposed star denotes complex conjugation and in the left-hand side the integration is over the entire wave number space, while in the right hand side, it is over the entire physical space. Denoting by U(ξ) and P (ξ) the Fourier transform of u0 (x) and p0 (x) and using the above recalled properties of the Fourier transforms, from (4.4.48) we get the equations which must be satisfied by the components of U (ξ) and P (ξ) in the wave-number space ◦ ◦ ◦ 2 2 2 2 2 λ + µ ξi ξj Uj + µ ξ Ui = −iξj Pij , ξ = ξ1 + ξ2 + ξ3 . Solving this system for the unknowns Ui we get ◦ ◦ ◦ ◦ − λ +2 µ iξ 2 Pim ξn + λ + µ iξi Pkl ξk ξl . Ui = ◦ ◦ ◦ µ λ +2 µ ξ 4
Now, for the Fourier transform F u0i,j of u0i,j , we get ◦ ◦ ◦ ◦ − λ +2 µ ξ 2 Pim ξj ξn + λ + µ ξi ξj Pkl ξk ξl
. F u0i,j = −ikj Ui = ◦ ◦ ◦ µ λ +2 µ ξ 4

Returning to the expression (4.4.45) of W 0 (p0 , ε0 ) and using the Parceval equation (4.4.51), we get   ◦  ◦ Z    1 ∗ ∗ λ+µ P l l l l P dξ, (4.4.52) 2W 0 (p0 , ε0 ) = − ◦ Pmi Pmj li lj + ◦  ◦  ij kl i j k l ◦     µ µ λ +2 µ

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229

4.4. HASHIN-SHTRIKMAN PRINCIPLE where li = ξi /ξ.

(4.4.53)

The difficult task to evaluate these integrals still remains. To solve this problem we recall that the composite is macroscopically isotropic. Hence the polarization stress p (x) depends on x only, through the magnitude r = |x| of x, since cannot exist a privileged direction in the mixture. To establish an important consequence of this fact we consider an arbitrary but fixed wave number vector ξ. We introduce in the physical space a Cartesian coordinate system such that the axis x3 has the direction of ξ. Simultaneously, let (r, θ, ϕ) be a spherical system of coordinates such that x1 = r sin θ cos ϕ, x2 = r sin θ sin ϕ, x3 = r cos θ, 0 < r < ∞, 0 ≤ θ < π, 0 ≤ ϕ < 2π. The volume element dx and the scalar product ξ · x become dx = r 2 sin θdrdθdϕ and ξ · x = ξr cos θ. Thus, for the Fourier transform P (ξ) of p0 (x) = p0 (r), we get P (ξ) =

4π

(2π)

3/2

Z

∞

p0 (r)

0

r sin ξ rdr = P (ξ) . ξ

Hence, the Fourier transform P (ξ) depends on ξ only through the magnitude ξ of this wave vector, if the composite is macro-isotropic. Thus, according to (4.4.52), we have to calculate integrals of the following type: I=

Z

J=

∗

Pmi (ξ) Pmj (ξ) li lj dξ,

Z

Pij (ξ) Pkl (ξ) ξi ξj ξk ξl dξ.

In order to do this we introduce the spherical coordinates (ξ, θ, ϕ), 0 ≤ ξ < ∞, 0 ≤ θ < π, 0 ≤ ϕ < 2π in the wave number space. We get l1 = sin θ cos ϕ, l2 = sin θ sin ϕ, l3 = cos θ, dξ = ξ 2 sin θdξdθdϕ. Consequently, I and J become I= J=

Z

∞

2

ξ Pmi (ξ)

∗ Pmj

(ξ) dξ

0

Z

∞ 0

ξ 2 Pij (ξ) Pkl (ξ) dξ

Z

Z

li lj dω,

ω

li lj lk ll dω,

ω

where ω is the unit sphere and dω = sin θdθdϕ is the surface element on ω. Due

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CHAPTER 4. MACROSCOPICALLY ELASTIC COMPOSITES

to the spherical symmetry, we have Z Z Z Z 4π 4π 2 2 2 if i 6= j, and li lj dω = l1 dω = l2 dω = l3 dω = 15 3 ω ω ω ω Z Z Z 4π , l14 dω = l24 dω = l34 dω = 5 ω ω ω Z Z Z 4π 2 2 2 2 , l1 l2 dω = l2 l3 dω = l32 l12 dω = 15 ω ω ω Z li lj lk ll dω = 0, ω

if l1 , l2 , or l3 appears with an odd power. Taking into account these results, and the equation dξ = 4πξ 2 dξ, valid in our case, we get Z ∗ 1 P· P dξ, I= 3  Z  ∗ ∗ ∗ 1 P11 P11 +P22 P22 +P33 P33 dξ J= 5   Z   ∗ ∗ ∗ ∗ ∗ 1 dξ (trP) tr P + 2P· P −3 P11 P11 +P22 P22 +P33 P33 + 15 or Z n  ∗ ∗o 1 (trP) tr P + 2P· P dξ. J= 15 Using again the Parceval equation (4.4.51), we obtain Z n Z o 1 1 2 (trp0 ) + 2p0 · p0 dx. p0 · p0 dx, J = I= 15 3

Taking into account again the relation (4.4.52), W 0 (p0 , ε0 ) becomes 2W 0 = −

1 ◦

3µ

Z

p0 · p0 dx+

◦

◦

λ+µ ◦ ◦ ◦ 15 µ λ +2 µ

Z n

o 2 (trp0 ) + 2p0 · p0 dx.

Now we return to equation (4.4.49) getting 0

0

2U (p, ε ) = −

1

or 2U 0 (p, ε0 ) = −

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◦

3µ

p0

· ◦

p0

◦ ◦ o n λ+µ 0 ) 2 + 2 p 0 · p0 , (trp +   ◦ ◦ ◦ 15 µ λ +2 µ

◦

◦

◦

3 λ +8 µ λ+µ 2 (trp0 ) . p0 · p 0 + ◦ ◦ ◦ ◦ ◦ ◦ 15 µ λ +2 µ 15 µ λ +2 µ

231

4.4. HASHIN-SHTRIKMAN PRINCIPLE Recalling that p0 = p−p, we obtain ◦

◦

3 λ +8 µ {p · p − p · p} 2U (p, ε ) = − ◦  ◦ ◦ 15 µ λ +2 µ ◦ ◦ n
2 o λ+µ 2 trp . (trp) −  + ◦ ◦ ◦ 15 µ λ +2 µ 0

0

Now taking into account the decomposition (4.4.40) of the mean value p into ◦ ◦ ◦ its spherical and deviatoric part, and using the relation λ=k −2 µ /3, we obtain finally the result derived by Hashin and Shtrikman:   (4.4.54) 2U 0 (p, ε0 ) = α0 p2 − p2 + β0 (q · q − q · q),

where the Hashin-Shtrikman coefficients α0 and β0 have the following expressions: ◦ ◦ µ +2 3 k 3 . (4.4.55) , β0 = ◦  ◦ α0 = − ◦ ◦ ◦ 5 µ 3 k +4 µ 3 k +4 µ

Recalling the definition (4.4.38) of the polarization stress p, we get

p2

2

−p =

2 P

cα p2α

−



2 P

2

c α pα ,    2  2 2 P P P c α qα . c α qα · c α qα · q α − q·q−q·q= α=1

α=1

α=1

α=1

α=1

Introducing these results in (4.4.54) and using the relation (4.4.42), we obtain 2 X ◦ ◦ cα 2 2 0 2 µ U (p, ε ) = 9 k ε + 2 e · e − ◦ pα v k − α=1 α k 2 2 2 X X X cα − q · q + 6 ε c p + 2 e · c α qα α α α α ◦ α=1 2(µα − µ) α=1 α=1  !2  2 2 X  X +α0 c α pα cα p2α − (4.4.56)   α=1 α=1 ( 2 ! !) 2 2 X X X +β0 c α qα · q α − c α qα · . c α qα α=1

α=1

α=1

Now, we return to the Hashin-Shtrikman principle to get bounds for the overall moduli b k and µ b. For simplicity, we take v =1.

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CHAPTER 4. MACROSCOPICALLY ELASTIC COMPOSITES From the principle, it results that, if ◦

◦

k ≤ kα and µ≤ µα for α = 1, 2,

(4.4.57)

◦

that is, if the tensor r = c− c is positive definite, then U ≤ U s = U.

(4.4.58)

Analogously, if ◦

◦

k ≥ kα and µ≥ µα r for α = 1, 2,

(4.4.59)

◦

that if the tensor r = c− c is negative definite, then U ≥ U s = U.

(4.4.60)

In this way, taking U in the form (4.4.35) and using (4.4.56), the inequalities (4.4.58) and (4.4.60) lead to bounds for the overall moduli. In order to find the best bounds for a polarization field of type (4.4.38), we have maximized (4.4.56) for conditions (4.4.57) and have minimized the same quantity for the conditions (4.4.59). Differentiating (4.4.56) with respect to pα and qα , α = 1, 2, the extreme conditions are found to be −

−

1

+ α0 pα − α0 p + 3ε = 0,

(4.4.61)

qα + β0 qα − β0 q + e = 0.

(4.4.62)

◦ pα

kα − k

1 ◦

2(µα − µ)

Let us denote by p˜α and q ˜α the (unique) solution of this system. It is easy to see that this solution realizes the maximum of U when (4.4.57) holds, and the minimum of U when (4.4.59) takes place. Also, direct computations show that the value Ue of U for the solution is
◦ 1 ◦ 1 ˜ ·˜ ε. (4.4.63) ˜ · e =U + p Ue =U + 3p˜ε + q 2 2

˜ = c1 q ˜ 1 + c2 q ˜2 can be obtained by The mean values p˜ = c1 p˜1 + c2 p˜2 and q solving (4.4.61) and (4.4.62) for p˜α and q ˜α . The results are

p˜ =

with A=

2 X

α=1

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B 3A e, ˜= ε, q 1 + β0 B 1 + α0 A

cα

1 ◦

kα − k

, − α0

B=

2 X

α=1

(4.4.64)

cα

1 ◦

2(µα − µ)

. − β0

(4.4.65)

233

4.4. HASHIN-SHTRIKMAN PRINCIPLE

In order to found bounds for the bulk modulus b k, we assume that the given strain ε is spherical; i.e. ε = ε1. It follows from (4.4.33), (4.4.35), (4.4.63) and (4.4.64)1 that >◦ A b , (4.4.66) k
◦

where the upper inequality sign applies if kα >k and the lower if kα ◦

µ b <µ +

B 1 , 2 1 + β0 B

(4.4.67) ◦

◦

where the upper inequality sign applies if µα > µ and the lower if µα < µ for α = 1.2. ◦ ◦ It remains to specify the values of k and µ which, when introduced into (4.4.66) and (4.4.67), will yield the highest lower bounds and the lowest upper bounds. It can be proved by differentiation that the expressions in the right sides ◦ ◦ of (4.4.66) and (4.4.67) are monotonically increasing functions of k and µ. Ac◦

cordingly, the highest lower bounds are obtained by taking the largest values of k ◦ and µ which comply with (4.4.57). The lowest upper bounds are obtained for the ◦ ◦ smallest values of k and µ which comply with (4.4.59). Let us assume that k1 ≤ k2 and µ1 ≤ µ2 . (4.4.68)

Then, for the highest lower bounds, denoted by k − and µ− , we must take ◦

◦

k = k1 and µ = µ1 ,

(4.4.69)

and for the lower upper bounds, denoted by k + and µ+ , we must choose ◦

◦

k = k2 and µ = µ2 .

(4.4.70)

Introducing (4.4.69)1 and (4.4.70)1 into the right side of the relations (4.4.66) it is found that c2 , k − = k1 + 3c1 1 + 3k1 + 4µ1 k2 − k 1 (4.4.71) c1 + , k = k2 + 3c2 1 + 3k2 + 4µ2 k1 − k 2

and

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k− ≤ b k ≤ k+ .

(4.4.72)

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CHAPTER 4. MACROSCOPICALLY ELASTIC COMPOSITES

Analogously, introducing (4.4.69)2 and (4.4.70)2 into the right hand side of the relation (4.4.67), we get µ− = µ 1 +

c2 6c1 (k1 + 2µ1 ) 1 + 5µ1 (3k1 + 4µ1 ) µ2 − µ 1

c1 , µ = µ2 + 6c2 (k2 + 2µ2 ) 1 + 5µ2 (3k2 + 4µ2 ) µ1 − µ 2

(4.4.73)

+

and µ− ≤ µ b ≤ µ+ .

(4.4.74)

The relations (4.4.71)−(4.4.74) are true if (4.4.68) takes place. If there exists a different ordering between k1 and k2, µ1 and µ2 , the above formulas must be changed accordingly. Comparing equations(4.3.48) obtained by Hill and the relations (4.4.71) deduced by Hashin and Shtrikman, we can see that k − in the overall bulk modulus of a composite with bulk moduli k1 , k2 and common shear modulus µ1 ; similarly, k + is the overall bulk modulus of a composite with bulk moduli k1, k2 and common shear modulus µ2 . The limits k − and k + obtained by Hashin and Shtrikman, using their variational and extremal principle, are the same as those obtained by Hill, using its comparison theorem and its exact solution for the bulk modulus of a composite for which the two phases have the same shear modulus. Consequently, the bounds k − and k + are the best possible, if only the phases moduli and their concentration, the inclusions geometry and relative distribution resting arbitrary are known. Since it was not possible yet to identify the bounds µ− or µ+ with exact solutions, the question whether or not µ− and µ+ are the most restrictive bounds for µ b remains at present unanswered. Summing up the theoretical results presented up to now, we can say that in order to obtain the exact values of the overall moduli, we must know the strainenergy. In order to evaluate this energy, it is necessary to find the strain and stress fields in the composite, which appears to be an impossible task. Consequently, the attempts made to find the exact values of the overall moduli are found on simplifying assumptions concerning the geometry and the distribution of the inclusions. A more attractive and more general approach consists in the use of the extremal principle, in order to find bounds for the strain energy, and thus for the overall moduli, without making assumptions about phase geometry. The nature of the bounds to be found is such that the distance between them increases with increasing relative stiffnesses of one phase ratio. As can be seen from (4.4.71) and (4.4.73) in the extreme case of a rigid phase (k2 , µ2 converge simultaneously to infinity), the upper bounds k + , µ+ will increase to infinity, where as in the other extreme situation of a very soft matrix (k1 , µ1 converge simultaneously

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235

4.4. HASHIN-SHTRIKMAN PRINCIPLE

to zero), the lower bounds k − , µ− will tend to zero. However, when the ratios between the different phase moduli are not too large, the derived bounds are close enough to provide a good estimate for the overall moduli. In order to see that, we present the results obtained by Hashin and Shtrikman [4.12] for a tungsten carbide-cobalt (W C − Co) alloy for which experimental data are available. According to the adopted notation, moduli of the cobalt have been assigned the subscript 1 and for those of the tungsten carbide the subscript 2. The bounds k − , k + and µ− , µ+ have been numerically evaluated using formulas (4.4.71) and (4.4.74), respectively. The results are given in Figures 4.2 and 4.3.

k 6 10 MPa

6

k 2 = 418.5 10 MPa 400

k+ k 200

-

6

k 1= 172.4 10 MPa

c2 0

0.2

0.6

0.4

0.8

1.0

Figure 4.2: Bounds for bulk modulus of WC-Co alloy.

6

10 MPa

2=

6

288.2 10 MPa

k+

200

k100

1=

6

79.3 10 MPa c2

0

0.2

0.4

0.6

0.8

1.0

Figure 4.3: Bounds for shear modulus of WC-Co alloy.

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236

CHAPTER 4. MACROSCOPICALLY ELASTIC COMPOSITES

The bounds E − and E + for Young’s modulus have been obtained by using the usual relation E = 9kµ/ (3k + µ). These are shown in Figure 4.4 together with experimental results and Hill’s universal bounds ER and EV . The comparison shows that the Hill-Hashin-Shtrikman theory is in very good agreement with the experimental values of the overall Young modulus and that the present bounds E − and E + are a marked improvement with respect to the Voigt’s and Reuss’ ˜V = c1 E1 + c2 E2 and ER = (c1 /E1 + c2 /E2 )−1 . type bounds E PREVIOUS BOUNDS EXPERIMENTAL POINTS

E 6 10 MPa 700

E 2 = 703.3

6

10 MPa ~ EV

600 E+

500

E-

400 300

E1= 206.9 106 MPa

200

ER

100

c2 0

0.2

0.4

0.6

0.8

1.0

Figure 4.4: Bounds for Young modulus of WC-Co alloy. It should be noted that the moduli of tungsten carbide phase were quite large as compared to those of the cobalt phase, the ratio being k2 /k1 = 2.4, µ2 /µ1 = 3.6, E2 /E1 = 3.4. The bounds are nevertheless quite close. This indicates that the bounds obtained by Hill, Hashin, and Shtrikman should give a good estimate of the overall moduli of two-phase components for a great variety of practical cases.

4.5

Budiansky’s and Hill’s self-consistent method

The general predictions obtained in Sections 4.3 and 4.4 are mostly restricted to formulate universal bounds. Such bounds depend only on relative volumes (concentrations) and do not reflect any particular geometry. When one phase is a dispersion of spherical inclusions, a much more direct approach is available. The method draws on Eshelby’s solution (see Section 2.7) concerning a uniform loaded infinite medium containing an ellipsoidal and, in

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4.5. BUDIANSKY’S AND HILL’S SELF-CONSISTENT METHOD

237

particular, a spherical inclusion. Eshelby’s results were used first by Budiansky [4.7] and Hill [4.4] to obtain information about the overall properties of a composite. As before, we assume a biphasic mixture with isotropic components. We suppose also that the mixture is macro homogeneous and macro isotropic, and the inclusions are spherical (or nearly spherical). As before, the bulk and shear moduli will be denoted by k1, µ1 and k2, µ2 for the matrix, and for the inclusions, 1 respectively. If θ = trε, σ = trσ, e and s are the deviatoric parts of the strain ε 3 and of the stress σ, respectively, the constitutive equations of the composite can be written as   1 1     s in D1 , σ in D1 , 2µ k1 1 (4.5.1) and e = θ= 1 1     s in D2 . σ in D2 , 2µ2 k2

Here D1 and D2 are the domains occupied by the matrix and by the inclusions in the considered RVE, occupying the domain D. Consequently, the overall stress-strain relations are

θ=

1 1 s. σ and e = b 2b µ k

(4.5.2)

1 Here θ = trε, σ = trσ, e and s are the deviatoric parts of the mean strain ε and 3 of the mean stress σ, respectively. In order to determine the overall moduli b k and µ b in the first part of this Section, we follow Budiansky. A little more general procedure, applicable also to anisotropic composites was developed by Hill [4.3], [4.4]. Let us consider the homogeneous traction problem corresponding to σ = σ1+s = const. Let us denote by s = [u, ε, σ] the solution of this boundary value problem. Using the energetic definition of the overall compliance, we get Z 1 1 1 σ · kσdv = σ 2 + 2Uk (σ) = s · s. (4.5.3) b v D 2b µ k

Taking into account the special form of the boundary condition (σn =σn on ∂D and σ = const.), using the divergence theorem and the microscopic constitutive equation (4.5.1), we successively obtain Z Z Z Z 1 1 1 1 2Uk (σ) = σ · εdv = u · σnda = u·σnda = (un + nu)da. σ· v D v D v D 2v D From the last result, through the divergence theorem, we get Z 1 2Uk (ε) = σ · εdv. v D

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238

CHAPTER 4. MACROSCOPICALLY ELASTIC COMPOSITES Now we use again the microscopic strain-stress relation (4.5.1) getting

2v Uk (σ) = σ

Z

D1

1 σdv + k1

Z

D2

1 σdv k2



1 + s· 2

Z

D1

1 sdv + µ1

Z

D2

 1 sdv . µ2

Adding and substracting we obtain   Z Z  1 1 1 σdv − σdv + 2v Uk (σ) = σ k1 k2 D2 D k1   Z Z  1 1 1 1 sdv . − sdv + + s· µ1 µ2 2 D2 D µ1

Taking into account again the micro-constitutive equation (4.5.1) and the definition of the mean values, we obtain   Z  Z  1 1 1 1 1 1 1 1 s· s+ σ − 2Uk (σ) = σ 2 + k2 θdv + s· − µ2 edv. v D2 k 2 2µ1 k1 k1 v µ2 µ1 D2 Thus, it results in 2Uk (σ) =

1 2 1 σ + s · s + c 2 k2 k1 2µ1



1 1 − k2 k1



σ θ 2 + c 2 µ2



1 1 − µ2 µ1



s · e2 , (4.5.4)

where θ 2 and e2 are the mean values of θ and e on the domain D2 occupied by the inclusions in the RVE. Comparing (4.5.3) and (4.5.4), we get 1 2 1 1 1 σ + s · s = σ2 + s · s + c 2 k2 b 2b µ k 2µ 1 1 k



1 1 − k2 k1



σ θ 2 + c 2 µ2



1 1 − µ2 µ1



s · e2 .

(4.5.5) b Now it is clear that to obtain k and µ b, the mean values θ 2 and e2 must be expressed in terms of σ and s. In order to solve this problem, we apply the self-consistence method. We approximate θ 2 and e2 by the constant dilatation and the shear that would occur in a single isolated spherical inclusion embedded in an infinite isotropic elastic matrix subjected to the stress σ = σ1+s at infinity, and having the as-yet unknown overall moduli b k and µ b of the composite. The exact solution of Eshelby for this problem was given in the Section 2.7. Since the strain in the inclusion is uniform, according to the equations (2.7.52) and (4.5.2), for the mean values θ 2 and e2 , we get θ2 = 

1 1  o s, σ, e2 = n k2 − b k α b+b k 2 (µ2 − µ b) βb + µ b

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(4.5.6)

4.5. BUDIANSKY’S AND HILL’S SELF-CONSISTENT METHOD

239

with the following expressions for the coefficients α b and βb (see Equations (2.7.45)) α b=

b k + 2b µ 6 b ν k 1 + νb b βb = 2 4 − 5b , = = 3 − 5β, = 4 b 15 1 − ν b 15 3 (1 − νb) 3 k + 4b µ b b k+ µ 3

(4.5.7)

νb being the overall Poisson ratio of the composite. Introducing (4.5.6) in (4.5.5), we obtain two equations satisfied by the overall elastic moduli     c2 µ2 1 1 c2 k2 1 1   . + 1− = , + 1− = b µ1 βb (µ2 − µ µ1 b k1 α k1 b) + µ b k b k2 − b k +b k µ (4.5.8) The above relations were obtained by Budiansky. They, together with the standard relation 3b k − 2b µ (4.5.9) νb = b 6k + 2b µ

provide, in implicit form, the desired dependence of the overall moduli b k and µ b on the elastic constants and concentrations of the matrix and of the spherical inclusions. Using the identity c1 + c2 = 1, after elementary computations, equations (4.5.8) can be expressed in an equivalent, but more symmetric form c1

b k − k2

+

c2

b k − k1

=

βb c1 c2 α b = . + , b µ b µ b − µ1 b − µ2 k µ

(4.5.10)

These relations were first obtained by Hill. Using the second expression (4.5.7) of α b and taking into account that c 1 +c2 = 1, from the equations (4.5.10)1 , we get     4 4 4 c k + µ b + c k + µ b 2 1 1 2 µ b + c 2 k1 + c 1 k2 4 3 3 b  =    b=  3 k+ µ . 4 4 4 4 3 k1 + µ k2 + µ k1 + µ b k2 + µ b 3 3 3 3

Thus, we can express b k parametrically in terms of µ b: 1

4 b k+ µ b 3

=

c1

4 k1 + µ b 3

+

c2

4 k2 + µ b 3

.

(4.5.11)

It is noteworthy to mention that this is identical with Hill’s exact solution (4.3.39) for composites with arbitrary geometry, when the phases have equal shear moduli. However, now we can determine both overall moduli, even if the phases

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CHAPTER 4. MACROSCOPICALLY ELASTIC COMPOSITES

have different shear moduli. But now the geometry (form) of the inclusions is specified. Following Hill [4.4] we shall now analyze some properties of the obtained results. Since the obtained formulas are meaningful only when b k, µ b > 0,

from (4.5.7), we can deduce the following restrictions for α b and βb 0<α b < 1, 2/5 < βb < 3/5.

In order to discuss the behavior of µ b, in general terms, one may retain rameter in view of its restricted range given above. Clearing fractions in (4.5.10), we get   n o b 1 µ2 = 0. f (b µ) ≡ 1 − βb µ b2 + βb (µ1 + µ2 ) − (c1 µ1 + c2 µ2 ) µ b − βµ

βb as paequation (4.5.12)

We find

f (µR ) < 0 and f (µV ) > 0,

µR and µV being the Reuss and Voigt estimates, respectively. Consequently, the required positive root lies between µR and µV , respecting Hill’s universal restriction (4.3.13). Also, it follows that b k is certainly in the interval obtained by substituting µR and µV in the monotonic relation (4.5.11). Hence, b k lies a fortiori between the best-possible bounds for an arbitrary geometry, which, as we know, correspond to µ1 and µ2 in (4.5.11), according to the results obtained in Section 4.3 and founded on Hill’s comparison theorem. The above properties of µ b and b k obtained using the self-consistent method are obviously satisfactory features of this procedure to evaluate overall elastic moduli. Hill has derived also an explicit equation for µ b, expressing both sides of the relation α b = 3 − 5βb using (4.5.10) and (4.5.11). Elementary computations lead to the equation     c2 µ1 c1 µ2 c 2 k2   c 1 k1 + 2 = 0. (4.5.13) + +5 + g (b µ) ≡  4  4 µ b − µ1 µ b − µ2 b b k2 + µ k1 + µ 3 3

This form of the equation satisfied by µ b is useful for iterative or graphical solutions, as well as for qualitative analysis. The equation (4.5.13) could be put in a quadratic form but is far better as it stands, and a solution between µ1 and µ2 can be obtained, by tabulating c1 or c2 as a function of µ b. It is easy to see that as µ increases from 0 to ∞, the first bracketed function decreases monotonically to zero from 1 if k1 k2 6= 0, and from c1 if k2 = 0, from c2 if k1 = 0, and finally vanishes if both k1 and k2 are zero. If µ1 µ2 6= 0 and

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4.5. BUDIANSKY’S AND HILL’S SELF-CONSISTENT METHOD

241

µ2 < µ1 , the second bracketed function decreases monotonically from −1 to −∞ in the range (0, µ1 ); from +∞ to −∞ in (µ1 , µ2 ), with values 0 and −1 at µR and at µV ; and from +∞ to 0 in (µ2 , ∞). It is therefore confirmed again, provided that neither phase rigidity vanishes, that there exists precisely one positive root and that it lies between the Reuss and Voigt estimates. This root can be stated explicitly when the dispersion is diluted ; thus, if c2 << 1. In this case, in which the self-consistent method can give good results, we find µ b ≈ µ1 (1 + kc2 )

where





k1  . 4  k1 + µ 1 3 As can be seen from equation (4.5.13), the obtained approximate value of the overall shear moduli µ b is correct up to zero order in c2 . Finally we analyze some extremal situations. To do this we shall use Budiansky’s equation (4.5.8). From the relations (4.5.7), we obtain 1 µ2 1 − 2 + = 5 µ2 − µ 1 k

1−α b=

4b µ

3b k + 4b µ

, 1 − βb =

9b k + 8b µ  . b 5 3k + 4b µ

(4.5.14)

In what follows, we assume an incompressible matrix ; i.e. k1 = ∞. In this case, from (4.5.8)1 , we get

c 1 1 . = b b k k+α b k2 − b k

(4.5.15)

First we suppose that the inclusions are holes; i.e. k2 = µ2 = 0. In this case, from (4.5.8)2 and (4.5.15), it results

We first assume

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c2 1 1 c2 1  . + , = = b µ1 b (1 − α b) b k µ k 1 − βb µ b

b k 6= 0.

(4.5.16)

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CHAPTER 4. MACROSCOPICALLY ELASTIC COMPOSITES

Consequently, (4.5.16)1 gives Thus, from (4.5.14)1 it results

Hence, (4.5.14)2 gives

c2 = 1 − α b. 4 1 − c2 b µ b. k= 3 c2

4 1 − c2 µ b. 1 − βb = 3 c2

(4.5.17)

(4.5.18)

(4.5.19)

Thus, from (4.5.16)2 and (4.5.18), we get µ b=3

(1 − c2 ) (1 − 2c2 ) 1 − 2c2 µ1 . µ1 , b k=4 c2 (3 − c2 ) 3 − c2

(4.5.20)

Since µ b and b k cannot be negative, the obtained results can be accepted only if c2 ≤ 1/2. If c2 > 1/2, the only acceptable solution of the equation (4.5.16)1 is b k = 0. In this case the equation (4.5.16)2 can be satisfied only by µ b = 0. Summing up we can say that according to the self-consistent method if the matrix is incompressible and the inclusions are holes, the overall moduli are

1 − 2c2 4 (1 − c2 ) (1 − 2c2 ) b µ1 , if 0 ≤ c2 < 1/2, ,µ b=3 k= 3 − c2 c2 (3 − c2 )

(4.5.21)

b k=µ b = 0 if c2 > 1/2.

Next, we suppose that the inclusions are rigid; i.e. k2 = µ2 = ∞. In this case (4.5.15) gives b k = ∞.

(4.5.22)

βb = 2/5.

(4.5.23)

Consequently, from (4.5.14)2 it results

Since µ2 → ∞, from (4.5.8)2 we get

µ b=

µ2 . 1 − 5c2 /2

(4.5.24)

This result can be accepted only for c2 ≤ 2/5, since µ b cannot be negative. If c2 > 1/2, the equation (4.5.8)2 can be satisfied only by µ b = ∞.

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243

4.6. OVERALL MECHANICAL PROPERTIES

Summing up we see that if the matrix is incompressible and the inclusions are rigid, according to the self-consistent method, the overall moduli becomes b k = ∞ and µ b=

(

µ2 1 − 5c2 /2 ∞

if

0 ≤ c2 ≤ 2/5

if

c2 > 2/5

.

(4.5.25)

Analyzing (4.5.21) and (4.5.25) we can see that if the matrix is incompressible and if the inclusions are holes, zero stiffness of the composite is reached with a 50 percent voids ratio, but only a 40 percent concentration of rigid inclusions produces infinite stiffness, that is a rigid composite. These facts are interesting, but they show that the overall moduli obtained using the self-consistent method should be used with caution when extreme ratios of the stiffnesses of the phases are involved, or when the concentration of the inclusions is relatively large. Under extreme conditions, the results are reliable only when the dispersed phase is sufficiently diluted. The last conclusion is in perfect agreement with the solving procedure involved in the self-consistent method, since this procedure uses Eshelby’s results concerning a single inclusion in an infinite matrix.

4.6

Overall mechanical properties

Following Hill [4.2], we consider an anisotropic (aelotropic) composite having a particular geometry and being of a great technological interest. In this structure, the embedded phase consists of continuous and perfectly aligned cylindrical fibers. The matrix and the fibers are assumed to be homogeneous and transversally isotropic about the fiber direction (the direction of the x3 axis). The crosssections and spatial arrangement of fibers are subject merely to the requirement that assures macroscopical homogeneity of the mixture together of its macroscopic transverse isotropy. In these conditions, Hill was able to show that some of the overall elastic moduli are connected by simple universal relations which are independent on the cross-section geometry at given concentrations. Moreover, Hill has obtained the values of some of the overall elastic moduli when the phases have equal transverse shear rigidity. As we know from the results given in Section 2.2, the stress-strain relations of a transversally isotropic elastic material are completely described by five independent elastic moduli and has the following form (see Equations (2.2.87)−(2.2.89)): σ11 = C11 ε11 + C12 ε22 + C13 ε33 , σ23 = 2C44 ε23 , σ22 = C12 ε11 + C11 ε22 + C13 ε33 , σ31 = 2C44 ε31 , σ33 = C13 ε11 + C23 ε22 + C33 ε33 , σ12 = (C11 − C12 ) ε12 .

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(4.6.1)

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CHAPTER 4. MACROSCOPICALLY ELASTIC COMPOSITES

These equations can be inverted. If we use the technical constants, we get ε11 =

ν13 ν12 1 σ33 , σ22 − σ11 − E1 E1 E1

ε22 = −

ν13 1 ν12 σ33 , σ22 − σ11 + E1 E1 E1

ε33 = −

1 ν31 ν31 σ33 , σ22 + σ11 − E3 E3 E3

ν31 ν13 , = E3 E1

ε23 =

1 σ23 , 2G23

ε31 =

1 σ31 , 2G23

ε12 =

1 σ12 . 2G12

(4.6.2)

From (4.6.1) and (4.6.2), we get 1 (σ11 + σ22 ) = K12 (ε11 + ε22 ) + C13 , 2 σ11 − σ22 = 2G12 (ε11 − ε22 ) ,

where K12 =

σ33 = C13 (ε11 + ε22 ) + C33 ε33 , σ12 = 2G12 ε12 , (4.6.3)

E1 C11 − C12 C11 + C12 . = , G12 = 2 (1 + ν12 ) 2 2

(4.6.4)

In these equations, K12 and G12 are the bulk and shear modulus, respectively, governing plane−strain deformation in the isotropy plane x1 , x2 ; C33 is the modulus for longitudinal uniaxial straining in the x3 (fiber) direction perpendicular to the isotropy plane; and C13 is the associated cross-modulus. In Hill’s analysis, concerning only these four elastic moduli, analytically it is convenient to regard K12 , G12 , C33 and C13 as a basic independent set. In terms of them, the axial Young modulus E3 and the associated Poisson ratio ν31 are given by the equations C13 C2 . (4.6.5) E3 = C33 − 13 , ν31 = 2K12 K12

The first two equations (4.6.3) can now be recast as ε11 + ε22 =

1 (σ11 + σ22 ) − 2ν31 ε33 , σ33 = ν31 (σ11 + σ22 ) + E3 ε33 . (4.6.6) 2K12

This unusual rearrangement will be useful a little later on. For simplicity, in this Section, we shall use the following simplified notations: K12 = k, C13 = l, C33 = n, G12 = m, E3 = E, ν31 = ν, e = ε11 + ε22 , s =

1 (σ11 + σ22 ) , ε = ε33 , σ = σ33 . 2

(4.6.7)

Thus the equations (4.6.3), (4.6.5) and (4.6.6) become s = ke + le, σ11 − σ22 = 2m (ε11 − ε22 ) ,

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σ = le + nε, σ12 = 2mε12 ,

(4.6.8)

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4.6. OVERALL MECHANICAL PROPERTIES

1 s − 2νε, σ = 2νs + Eε, (4.6.9) 2k l l2 . (4.6.10) E =n− , ν = 2k k If the elasticity tensor c of a transversally isotropic material is positive definite, it can be shown that k, E and m are all positive. As before, all quantities referring to the first phase (the matrix) will be indexed by 1, and those referring to the second phase (the fibers) will be indexed by 2. We now consider RVE a prismatic specimen containing so many fibers as to be macro-homogeneous. There is no loss in generality supposing that the ends of this specimen are maintained plane and parallel by frictionless constraints which can be moved longitudinally. Then, by symmetry, all transverse sections remain plane and the axial strain component ε33 = ε is independent of position and equal to a constant, ε say, while the shears ε13 and ε23 vanish everywhere, and the remaining components depend only on x1 , x2 . In the following we present Hill’s results obtained from the above assumptions. As usual, the mean values are denoted by a superposed bar and the overall moduli by a superposed. Thus the overall (equivalent) constitutive equations, corresponding to (4.6.7) 1,2 are e=

bε. le + n lε, σ = b s=b ke + b

(4.6.11)

b (c1 ε1 + c2 ε2 ) . l (c1 e1 + c2 e2 ) + n c1 σ 1 + c 2 σ 2 = b

(4.6.12)

Using the fundamental relation (4.1.10) concerning mean values, we get from (4.6.11) l (c1 ε1 + c2 ε2 ) , k (c1 e1 + c2 e2 ) + b c1 s1 + c2 s2 = b

Since ε = ε is constant, obviously ε1 = ε2 = ε. Thus, (4.6.12) become

c1 s1 + c 2 s2 = b k (c1 e1 + c2 e2 ) + b lε,

(4.6.13)

c1 σ 1 + c 2 σ 2 = b l (c1 e1 + c2 e2 ) + n bε .

As we know the elastic moduli have constant values in each phase. Hence, s1 = (k1 e + l1 ε)1 = k1 e1 + l1 ε,

and

σ 1 = (l1 e + n1 ε)1 = l1 e1 + n1 ε,

s2 = k2 e + l2 ε = k2 e2 + l2 ε,

σ 2 = l2 e + n 2 ε = l2 e2 + n 2 ε . (4.6.14) Introducing the above relation in (4.6.13), we obtain c1 (k1 e1 + l1 ε) + c2 (k2 e2 + l2 ε) = b k (c1 e1 + c2 e2 ) + b lε,

c1 (l1 e1 + n1 ε) + c2 (l2 e2 + n2 ε) = b l (c1 e1 + c2 e2 ) + n bε,

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(4.6.15)

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CHAPTER 4. MACROSCOPICALLY ELASTIC COMPOSITES

or, equivalently       l − c1 l1 − c2 l2 ε = 0, k − k 2 e2 + b c1 b k − k 1 e1 + c 2 b     n − c1 n1 − c2 n2 ) ε = 0. l − l2 e2 + (b c1 b l − l 1 e1 + c 2 b

(4.6.16)

We have used here the basic identity c1 + c2 = 1. If the two homogeneous equations (4.6.16), the strain variables e1 , e2 , ε, were distinct, they would define the same ratios e1 : e2 : ε no matter what the triaxial loading of the RVE is. Whereas, on the contrary, the ratio e : ε = (c1 e1 + c2 e2 ) : ε can be varied at will or, for instance, any one of e, e1 , e2 and ε can be made to vanish, preventing lateral or axial deformations by suitable applied rigid enclosures. Thus the two equations (4.6.15) must be similar and hence their corresponding coefficients are proportional

b b b k1 − k 2 l − c 1 l1 − c 2 l2 k − k2 k − k1 , = = = b b l1 − l 2 n b − c n − c n 1 1 2 2 l − l2 l − l1

(4.6.17)

the final ratios following in a trivial way from the first two. As equivalent, more symmetrical version of the relations (4.6.17), is b l − c 1 l1 − c 2 l2

=

 l1 − l2 b k − c 1 k1 − c 2 k2 k1 − k 2

(4.6.18)

k1 − k 2 = (b n − c 1 n1 − c 2 n2 ) . l1 − l 2

These relations characterize the deviations of the overall moduli from the values corresponding to Voigt’s estimates or to the “mixture rule”. The above procedure can be applied to the equation (4.6.9): let every averaged quantity in (4.6.9) be decomposed using the basic formula: f = c1 f1 + c2 f2 , and afterward eliminate the phase averages of strains in terms of the stresses, or the phase averages of stresses of the strain, by means of the local (micro-) constitutive relations. In this, way we get     1 1 1 1 c1 − s1 + c 2 − s2 − 2 (b ν − c1 ν1 − c2 ν2 ) ε = 0, b b k k1 k k2 (4.6.19)   b − c1 E1 − c2 E2 ε = 0. 2c1 (b ν − ν1 ) s1 + 2c2 (b ν − ν 2 ) s2 + E Since these equations must also be similar, we get 1 1 1 1 − − b k k1 = b k k2 = νb − ν1 νb − ν2

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νb − c1 ν1 − c2 ν2 1 b − E − c 1 E1 − c 2 E2 4

1 1 − k k2 . = 1 ν1 − ν 2

(4.6.20)

4.6. OVERALL MECHANICAL PROPERTIES

247

Equivalently,

  c2 c1 ν1 − ν 2 1 − − νb − c1 ν1 − c2 ν2 = 1 b 1 k2 k k1 − k2 k1 1 1 −   1 k1 k2 b E − c 1 E1 − c 2 E2 . =− 4 ν1 − ν 2

(4.6.21)

Hill’s derivation is preferable to the tedious elimination of l and n between the macroscopic analogues of the relations (4.6.10) and (4.6.18). Applying the principle of minimum potential energy to the plane transverse dilatation of the composite, one readily finds Voigt’s and Reuss’ type bounds for the plane-strain overall bulk modulus b k in analogy to those obtained in the Section 4.3 for the three-dimensional overall bulk modulus. We get

c2 c1 1 1 b + . ≡ ≤ k ≤ kV ≡ c1 k1 + c2 k2 and b k k k 2 1 R k

(4.6.22)

It is precisely the deviations of the exact value and its reciprocal from these bounds which appear in connections (4.6.18) and (4.6.21) between the overall moduli. As in the Section 4.3, the intervals c1 c2 2 (k1 − k2 ) , c1 k2 + c2 k 1 2 1 1 c1 c2 1 1 , = c1 − c2 k − k kV kR 2 1 + k1 k2

kV − k R =

(4.6.23)

become small quantities of second order when the phase moduli are only slightly different. Taking into account (4.6.22)2 and returning to (4.6.21), we can derive the important inequality b c1 E1 + c2 E2 ≤ E, (4.6.24)

the equality holding only for ν1 = ν2 . As we have seen in Section 4.3 (see relations (4.3.22)), there is no universal ordering of the overall Young modulus and of the Voigt’s type estimate in the case of macro isotropic composite with inclusion having arbitrary geometry and distribution. Contrary to this fact, for a fiber-reinforced composite, the inequality (4.6.24) is true, but now the geometry and distribution of the inclusions (parallel cylindrical fibers) are known. From (4.6.22)2 and (4.6.21), we can conclude also that νb ≷ c1 ν1 + c2 ν2 according to (ν1 − ν2 ) (k1 − k2 ) ≷ 0.

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(4.6.25)

248

CHAPTER 4. MACROSCOPICALLY ELASTIC COMPOSITES Similarly, from (4.6.22)2 and (4.6.18), we get n b ≤ c 1 n1 + c 2 n2

(4.6.26)

b l ≷ c1 l1 + c2 l2 according to (k1 − k2 ) (l1 − l2 ) ≶ 0.

(4.6.27)

Let us observe that the above analysis concerns universal connections and estimates only for 4 overall moduli: b12 = b b13 = b b33 = n b3 = E b and νb13 = νb. C k, C l, C b, E

Following Hill’s method, further results and exact evaluations for the above four overall moduli can be obtained assuming that the matrix and the fibers have the same transverse shear moduli; i.e. G12 (matrix) = G12 (f ibers) or

m1 = m2 = m.

(4.6.28)

In the last part of this Section, we present Hill’s results for this special case. Let us introduce an irrotational plane displacement field using the potential function f = f (x1 , x2 ) uα = f,α , α = 1, 2. (4.6.29) From (4.6.28), (4.6.29) and the constitutive equations (4.6.8), we obtain   1 l1 ∆f + n1 ε in D1 , k1 ∆f + l1 ε in D1, σ33 = (σ11 + σ22 ) = l2 ∆f + n2 ε in D2 . ε in D , k ∆f + l 2 2 2 2

σ11 − σ22 = 2m (f,22 −f,11 ) , σ12 = 2mf,12 in D1 and D2 .

(4.6.30)

Here D, D1 , D2 are the domains occupied by the RVE by the matrix and by the fibers, respectively. Also, it is easy to see that the above stress field is self-equilibrated if the potential f satisfies Poisson’s equation  e1 in D1, (4.6.31) ∆f = e2 in D2 .

As we know, the potential f , satisfying the equation (4.6.31) is continuous in D and its first order derivatives are also continuous in D. In exchange, its second order derivatives have nonvanishing jumps across Γ, the common boundary of the matrix and of the fibers, we have [f,αβ ] = (e2 − e1 ) nα nβ on Γ, α, β = 1, 2.

(4.6.32)

Since the potential is of class C 1 on D, the displacement corresponding to f has null-jump across the boundary Γ. As we already know, the stress vector must have also null-jump across Γ; i.e. we must have [σn] = 0 or [σαβ nβ ] = 0 on Γ, α, β = 1, 2.

Copyright © 2004 by Chapman & Hall/CRC

(4.6.33)

249

4.6. OVERALL MECHANICAL PROPERTIES

Elementary computation, using the relations (4.6.29)−(4.6.32) shows that the condition (4.6.33) will be satisfied if and only if e1 and e2 satisfy the following restriction: (4.6.34) (k1 + m) e1 + l1 ε = (k2 + m) e2 + l2 ε.

If this condition is fulfilled, [u, ε, σ] represents a possible elastic state of the fiber-reinforced piece-wise homogeneous composite. Obviously the second equation satisfied by the mean values e1 and e2 is the fundamental relation (4.6.35) c1 e1 + c2 e2 = e.

Solving the system (4.6.34), (4.6.35), we get

1 {(k2 + m) e + c2 (l2 − l1 ) ε} , c 1 k2 + c 2 k1 + m 1 {(k1 + m) e − c1 (l2 − l1 ) ε} . e2 = c 1 k2 + c 2 k1 + m

e1 =

(4.6.36)

Similarly, from (4.6.14) and from s = c1 s1 + c2 s2 , it results

s = c1 k1 e1 + c2 k2 e2 + (c1 l1 + c2 l2 ) ε.

(4.6.37)

Now we assume that ε = 0 and use the equations (4.6.36). Thus, we get s=

m (c1 k1 + c2 k2 ) + k1 k2 e. c 1 k2 + c 2 k1 + m

(4.6.38)

Comparing (4.6.11)1 and (4.6.38), and recalling that ε = 0, we can conclude that the overall plane-strain bulk modulus b k=b k12 is given by the following equivalent relations: m (c1 k1 + c2 k2 ) + k1 k2 c1 k1 (k2 + m) + c2 k2 (k1 + m) b k= = . c 1 k2 + c 2 k1 + m c1 (k2 + m) + c2 (k1 + m)

(4.6.39)

Now, the remaining overall moduli follow straightforward from the equations (4.6.18) and (4.6.21). In order to obtain the best bounds for the overall moduli ( without restriction m1 = m2 !), we again use Hill’s comparison theorem. First, we observe that by virtue of the monotonic connections (4.6.18) and (4.6.21) any upper (lower) bound on b k for a particular fiber composite automatically yield upper (lower) bounds b and νb. More specifically, let on b l and n b together with lower (upper) bounds on E b us suppose that such a bound on k is derived as the actual overall plane bulk modulus of another composite that differs only in its plane shear rigidity m 1 and m2 . Then the bounds also coincide with the actual moduli of this comparison composite, since the connections are unaffected. In particular, a change in either of the plane shear rigidities alters b k in the same sense. It follows that b k is bracketed by the plane overall bulk-moduli of two comparison composites, identical with the

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CHAPTER 4. MACROSCOPICALLY ELASTIC COMPOSITES

one under consideration except that the phases have a common transverse shear rigidity either m1 or m2 . Therefore, from the exact solution (4.6.39) for just such a structure, and assuming m1 (matrix) ≤ m2 (f ibers)

(4.6.40)

we get c1 k1 (k2 + m1 ) + c2 k2 (k1 + m1 ) b c1 k1 (k2 + m2 ) + c2 k2 (k1 + m2 ) . ≤k≤ c1 (k2 + m2 ) + c2 (k1 + m2 ) c1 (k2 + m1 ) + c2 (k1 + m1 ) (4.6.41) Also, from the equations (4.6.21), we get, among others,

b − c 1 E1 − c 2 E2 c1 c2 E c1 c2 , ≤ ≤ 2 1 c2 c 1 c2 c1 1 4 (ν − ν ) 1 2 + + + + m2 k1 k2 m1 k1 k2

(4.6.42)

and

c1 c2 νb − c1 ν1 − c2 ν2 c1 c2 ≤  . ≤ 1 c2 c 1 c2 c1 1 1 1 + + + + − (ν1 − ν2 ) m2 k1 k2 m1 k1 k2 k1 k2

(4.6.43)

The above bounds are the best-possible when no regard is paid to the transverse geometry at any given concentration.

4.7

Hashin’s bounds for the overall moduli

Using Hill’s method, bounds were obtained for the overall plane bulk modulus, for the overall axial Young modulus and for the overall axial Poisson ratio. Now we present Hashin’s results [4.9] concerning the bounds for the overall transverse and axial shear moduli. In order to obtain these bounds, Hashin uses the variational and extreme principle presented in Section 4.4. We recall again that a fiber-reinforced material may be considered as a matrix material which contains long cylindrical fibers of another material. The case to be considered again is that of parallel fibers which are so long that the end effects can be neglected. The material is a cylindrical specimen whose cross-section is very large in comparison to fiber cross-sections. The generators of the specimen and the fibers are parallel and because of the neglecting of the end effects, the fibers may be assumed to traverse the specimen continuously from base to base, as was the case in the preceding Section. Also, we assume that the matrix and the fibers are transversally isotropic, the cross-section being the plane of isotropy. Moreover we suppose, as before, that the composite is macroscopically transversally isotropic, with the cross-section as the plane of isotropy. The cross-section of various fibers are arbitrary. Since the phase regions are cylindrical, the geometry is completely defined by that of any

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4.7. HASHIN’S BOUNDS FOR THE OVERALL MODULI

251

transverse plane cutting through the specimen. As before, the objective is to bound the overall elastic moduli in terms of phase moduli and phase concentrations only. In the following, the specimen is referred, as in the Section 4.6, to a Cartesian system of axes x1 , x2 , x3 such that x1 , x2 are in the plane of isotropy and x3 has the fibers’ direction. We recall that a transversally isotropic material has five independent elastic moduli. As in Section 4.6, for our analysis, a convenient choice of the moduli is the plane-strain bulk modulus K12 , the axial Young modulus E3 , the axial Poisson’s ratio ν31 , the transverse shear modulus G12 , and the axial shear modulus G13 . For brevity in this Section, we use the following simplified notations: k = K12 , E = E3 , ν = ν31 , m = G12 , µ = G13 .

(4.7.1)

Since the bounds for E3 , K12 and ν31 were obtained by Hill, here our attention will be concentrated on K12 , G12 and G13 . As we shall see, the bounds obtained b 12 by using Hill’s and Hashin’s method, for the overall plane bulk modulus K respectively, are the same. Obviously, this is a very satisfactory feature of the two theories. Following Hashin, first we shall try to get bounds for k = K12 and m = G12 . As usual, we consider a RVE which occupies in the isotropy plane x1 , x2 the domain D, bounded by the curve ∂D. The characteristics of the matrix will be indexed by 1 and those concerning the fibers by 2. The complementary plane subdomain occupied by the matrix and by the fibers will be denoted by D 1 and D2 , respectively. The common boundary of the matrix and of the fibers will be designed by Γ. We assume that the RVE is in plane strain state; that is, the axial displacement u3 is vanishing and the transverse displacements u1 and u2 depend only on x1 and x2 ; i.e. uα = uα (x1 , x2 ) , α = 1, 2. (4.7.2) In order to apply the Hashin-Shtrikman principle (see Section 4.5), we consider a geometrically identical cylindrical homogeneous and transversally isotropic body. The material characteristics and fields referring to this body will be denoted by a superposed zero. We assume that the homogeneous body is also in plane-strain state; i.e. ◦ ◦ ◦ uα = uα (x1 , x2 ) , α = 1, 2, u3 ≡ 0. (4.7.3) The deformation εαβ and stress σαβ , α, β = 1, 2, in the RVE are decomposed in their isotropic and deviatoric parts as follows: 1 εαα , 2 1 + sαβ , σ = σαα . 2

εαβ = εδαβ + eαβ , ε =

σαβ = σδαβ

As we already know, the concerned stress-strain relations are   2k1 ε in D1 , 2m1 eαβ in D1 , σ= , sαβ = 2k2 ε in D2 2m2 eαβ in D2 .

Copyright © 2004 by Chapman & Hall/CRC

(4.7.4)

(4.7.5)

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CHAPTER 4. MACROSCOPICALLY ELASTIC COMPOSITES

Here k1 and m1 are the plane bulk modulus and transverse shear modulus of the matrix, whereas k2 and m2 are the plane bulk modulus and transverse shear modulus of the fibers. For the homogeneous cylinder, analogous decompositions are used: 1 ◦ εαα , 2 1 ◦ ◦ ◦ ◦ ◦ σ αβ =σ δαβ + sαβ , σ= σ αα . 2 ◦

◦

◦

◦

εαβ =ε δαβ + eαβ , ε=

(4.7.6)

The corresponding stress-strain relations are ◦◦

◦

◦

◦◦

σ = 2 k ε, sαβ = 2 meαβ in D, ◦

(4.7.7)

◦

k and m being the plane bulk modulus and transverse shear modulus of the homogeneous body.   ◦ −1 We shall now express the tensor: h = r−1 = c− c corresponding to our plane-strain problem. To do this we observe that if    ◦ ◦ σ = 2 k− k ε and sαβ = 2 m− m εαβ , then, it is easy to see that εαβ =



1

◦  σδαβ

hαβγϕ =



1

◦  δαβ δγϕ

4 k− k

with

+



1 ◦

2 m− m

2 k− k

Hence,



+

1 ◦

4 m− m

 (σαβ − σδαβ ) .

 (δαγ δβϕ + δαϕ δβγ − δαβ δγϕ ) , (4.7.8)

α, β, γ, ϕ = 1, 2. Let us introduce now the polarization tensor p. As is natural, we assume that its components pαβ are functions only on x1 , x2 ; i.e. pαβ = pαβ (x1 , x2 ) . As before, we use the following decomposition: pαβ = pδαβ + qαβ , p =

1 pαα . 2

(4.7.9)

Taking into account (4.7.8) and (4.7.9), we get pαβ hαβγϕ pγϕ =

p2 ◦

k− k

Copyright © 2004 by Chapman & Hall/CRC

+

1 ◦

m− m

qαβ qαβ .

(4.7.10)

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4.7. HASHIN’S BOUNDS FOR THE OVERALL MODULI

We assume now that the two cylinders are submitted to the same homogeneous displacement conditions corresponding to εαβ = εβα = const.; i.e.

uα (xγ ) = εαβ xβ , and uα (xγ ) = εαβ xβ on ∂D.

In this case the solution in the homogeneous body has the following form: ◦

◦

◦

◦

◦

uα (xβ ) = εαβ xβ , εαβ (xγ ) = εαβ , σ αβ (xγ ) =σ δαβ + sαβ = const. in D . (4.7.11)   ◦

◦

Hence, the mean value of the strain energy U ε of the homogeneous body will be 1 ◦ ◦ ◦ 2 ◦ (4.7.12) U ε =k ε + m eαβ eαβ . S Here S is the area of the plane domain D, and εαβ is decomposed in its spherical and deviatoric parts as

εαβ = εδαβ + eαβ , ε =

1 εαα . 2

Now we return to the RVE of the mixture. If U (ε) is the strain energy of this body, according to the energetical definition of the overall moduli b k and m, b we have 1 (4.7.13) b eαβ eαβ . U (ε) = b k ε2 + m S To apply the Hashin-Shtrikman principle, the mean value U (p, ε0 ) /S also must be determined. We recall that the fluctuations ε0αβ are defined by the relation ◦

ε0αβ = εαβ − εαβ = εαβ − εαβ .

Using the general formula (4.6.8) for our plane-strain state, we get Z 1 1 ◦ ◦ 1 pαβ hαβγϕ pγϕ da U (p, ε0 ) = U ε − 2S D S S Z Z 1 1 ◦ + pαβ εαβ da + pαβ ε0αβ da. S D 2S D ◦

Now we take into account that εαβ (xγ ) = εαβ in D, and use the evaluations (4.7.10) and (4.7.12). Thus the above equation becomes ◦ 1 ◦ U (p, ε0 ) = k ε2 + m eαβ eαβ S   Z 1 1 1   qαβ qαβ  da − p2 +  ◦ 2S D k− ◦ m 2 m− k ◦

+pαβ εαβ +U 0 (p, ε0 ) .

Copyright © 2004 by Chapman & Hall/CRC

(4.7.14)

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CHAPTER 4. MACROSCOPICALLY ELASTIC COMPOSITES

In this equation, we have used the following notations: Z 1 1 0 0 0 0 0 0 0 0 pαβ ε0αβ da. U (p, ε ) = W (p , ε ) and W (p , ε ) = 2 D S

(4.7.15)

Now, following Hashin, we assume a piece-wise constant polarization field  1 pαβ in D1 (4.7.16) pαβ (x) = p2αβ in D2 . Also, we decompose pαβ , p1αβ and p2αβ in their isotropic and deviatoric parts 1 pαα , 2 1 1 + qαβ , p1 = p1αα , 2 1 2 + qαβ , p2 = p2αα . 2

pαβ = pδαβ + qαβ , p =

p1αβ = p1 δαβ

p2αβ = p2 δαβ

(4.7.17)

Thus, since the material is piece-wise homogeneous, the relation (4.7.15) takes the following simplified form: ◦ 1 c1 1 ◦ p2 U (p, ε0 ) = k ε2 + m eαβ eαβ − ( 2 k −◦ 1 S k 1 c2 c1 c2 2 2 1 1 2  qαβ  qαβ qαβ  )+ qαβ +  + ◦ p2 + ◦ ◦ 2 m− m 2 m− m k2 − k

+2pε + q αβ q αβ + U 0 (p, ε0 ) .

(4.7.18)

Obviously, the mean values p and q αβ are given by 2 1 , + c2 qαβ p = c1 p1 + c2 p2 , q αβ = c1 qαβ

(4.7.19)

c1 and c2 representing the concentrations for the matrix and for the fibers, respectively (in the cross-section of the cylinder!). The most difficult problem remains: to evaluate U 0 (p, ε0 ), depending on the unknown field ε0 . In order to solve this problem, we first introduce the fluctuation p0 of the polarization field: p0αβ = pαβ − pαβ .

Since the mean value of the fluctuation ε0 is vanishing, from (4.7.15)2 we can see that W (p, ε0 ) can be expressed as Z 1 0 0 (4.7.20) p0 ε0 da. W (p, ε ) = 2 D αβ αβ

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4.7. HASHIN’S BOUNDS FOR THE OVERALL MODULI

255

Now we recall that the fluctuation ε0 must satisfy the differential system (4.4.11)1 and the fluctuation u0 of the displacement field must satisfy the homogeneous boundary condition (4.4.11)2 . Taking into account the assumptions made, it is easy to see that for our plane-strain state and for our transversally isotropic phases, the general subsidiary restrictions relation (4.4.11)1,2 take the following special forms: ◦

◦

0 0 0 0 k uβ,αβ + m uα,ββ + pαβ,β = 0 in D1 ∪ D2 , uα = 0 on ∂D.

(4.7.21)

As before, in order to evaluate the mean value U 0 (p, ε0 ), we assume that the composite cylinder has an infinite extend in its transverse plane. Then our problem can be formulated as follows: Given ◦ ◦ k u0β,αβ + m u0α,ββ + p0αβ,β = 0 in E2 − Γ, (4.7.22)
1/2 , lim u0α (x) = 0, |x| = x21 + x21 |x|→∞

find u0α (x), and evaluate U 0 (p, ε0 ) by using the equation

1 0 0 0 W (p , ε ) . (4.7.23) S→∞ S Here E2 is the two-dimensional Euclidean space. To solve the problem, we now use two-dimensional Fourier transforms: if f = f (x) is a field depending on x1 , x2 , its two-dimensional Fourier transform F (ξ), depend on ξ1 , ξ2 , the Cartesian components of ξ, and is given by the equation Z 1 f (x) eiξ·x dx, ξ · x = ξ1 x1 + ξ2 x2 . F (ξ) = 2π U 0 (p, ε0 ) = lim

Here the integration is over the entire two-dimensional physical space E 2 . Denoting, by Uα (ξ) and Pαβ (ξ), the two-dimensional Fourier transform of u0α (x) and p0αβ (x) , and using the properties of the Fourier transforms, we can replace the field equation (4.7.22) by the relation ◦

◦

2 2 2 2 k ξα Uβ ξβ + m ξ Uα = −iξβ Pαβ , ξ = ξ1 + ξ2 .

Solving this system for Uα , we get

 ◦ ◦ ◦ k Pγϕ ξγ ξϕ ξα − k + m ξ 2 Pαγ ξγ . Uα = i ◦  ◦ ◦ m k + m ξ4   Accordingly, the Fourier transform F u0α,β of u0α,β is ◦  ◦ ◦ k Pγϕ ξα ξβ ξγ ξϕ − k + m ξ 2 ξβ Pαγ ξγ
. F u0α,β =  ◦ ◦ ◦ m k + m ξ4

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CHAPTER 4. MACROSCOPICALLY ELASTIC COMPOSITES

Thus, from (4.7.15)2 and the Parceval equation, we get

  ∗ ∗ ◦ Z k◦ Pαβ Pγϕ ξα ξβ ξγ ξϕ − k◦ + m ξ 2 Pαβ Pαγ ξβ ξγ 1 0 0 dξ. ◦  W (p, ε ) = ◦ ◦ 2 m k + m ξ4

Here the integration is taken over the entire two-dimensional space of the wave vectors ξ and the superposed star denotes complex conjugation. Introducing ξα lα = ξ

W 0 (p, ε0 ) becomes 0

0

W (p, ε ) = −

where I=

Z

◦

1 ◦

2m

∗

Pαβ Pαγ lβ lγ dξ, J =

I+

Z

k

 J, ◦ ◦ ◦ 2m k+m

(4.7.24)

∗

Pαβ Pγϕ lα lβ lγ lϕ dξ.

(4.7.25)

In order to evaluate these integrals, we recall that the composite is macroisotropic in its transverse plane. Consequently the polarization stress p defined in
1/2 (4.7.16) can depend on x only through its magnitude ||x|| = x21 + x22 . Hence, as in Section 4.4, we can conclude that the Fourier transform Pαβ of pαβ can
1/2 depend on ξ only through its magnitude ξ = ξ12 + ξ22 . Thus, the integrals (4.7.25) become I=

Z∞

∗

2

ξ Pαβ (ξ) P αγ (ξ) dξ

Z

lβ lγ dω, J =

ω

0

Z∞

∗

2

ξ Pαβ (ξ) P γϕ (ξ)

Z

lα lβ lγ lϕ dω.

ω

0

(4.7.26) Here ω is the unit circle and dω is the line element on ω. Due to the existing symmetry, we get Z Z Z 2 2 l1 dω = l2 dω = π, l1 l2 dω = 0, ω

Z

ω

l14 dω =

ω

Z ω

ω

3π , l24 dω = 4

Z ω

if lα , lβ , lγ , lϕ appears at an odd power.

Copyright © 2004 by Chapman & Hall/CRC

l12 l22 dω =

π , 4

Z ω

lα lβ lγ lϕ dω = 0,

257

4.7. HASHIN’S BOUNDS FOR THE OVERALL MODULI In this way, using also the equation d ξ = 2πξdξ, from (4.7.25), we get

Z ∗ 1 Pαβ Pαβ dξ, 2  Z Z  ∗ ∗ ∗ ∗ 1 3 {Pαα P ββ +2Pαβ P αβ P11 P 11 +P22 P22 dξ + J= 8 8   Z ∗ ∗ ∗ ∗ 1 (Pαα P ββ +2Pαβ P αβ )dξ. −3 P11 P 11 +P22 P22 }d ξ = 8

I=

Introducing these values in (4.7.24), after some computations, it results ◦

k W =  ◦ ◦ ◦ 16 m k + m 0

Z

∗

Pαα P ββ

◦

◦

k +2 m dξ −  ◦ ◦ ◦ 8m k+m

Z

∗

Pαβ P αβ dξ.

Using again the Parceval equation, we find ◦

k W = ◦  ◦ ◦ 16 m k + m 0

Z

p0αα p0ββ dx−

◦

◦

k +2 m ◦  ◦ ◦ 8m k+m

Z

p0αβ p0αβ dx.

Now, returning to the equation (4.7.23), we obtain ◦

◦

◦

k +2 m k  p0αβ p0αβ . U = ◦  p0αα p0ββ − ◦  ◦ ◦ ◦ ◦ 8m k+m 16 m k + m 0

Recalling that p0αβ = pαβ − pαβ and pαβ = pδαβ + qαβ , we get

with

 
2U 0 = α0 p2 − p2 + β0 qαβ qαβ − q αβ q αβ ,

1

◦

◦

k +2 m

, , β0 = − ◦ ◦ α0 = − ◦ ◦ ◦ 4 m (k + m) k+m

(4.7.27)

(4.7.28)

and 2

p2 = c1 p21 + c2 p22 , p2 = (c1 p1 + c2 p2 ) , 2 2 1 1 , qαβ + c2 qαβ qαβ qαβ qαβ = c1 qαβ

1 2 1 2 . q αβ q αβ = c1 qαβ + c2 qαβ c1 qαβ + c2 qαβ

Now, we can return to equation (4.7) and can give the final form of the mean

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CHAPTER 4. MACROSCOPICALLY ELASTIC COMPOSITES

value U (p, ε0 ) /S. ◦ 1 ◦ U (p, ε0 ) = k ε2 + m eαβ eαβ S    c c c 1  c1 2 1 2 2 2 1 1 2 2     q q + q q + p + p − αβ αβ αβ αβ ◦ ◦  2  k − ◦ 1 k − ◦ 2 2 m1 − m 2 m2 − m k k 2 1

2 1 2 1 + c2 qαβ c1 qαβ + c2 qαβ +2 (c1 p1 + c2 p2 ) ε + c1 qαβ o n α0 2 c1 p21 + c2 p22 − (c1 p1 + c2 p2 ) + 2

β0  1 1 2 1 2 1 2 2 . (4.7.29) + c2 qαβ c1 qαβ + c2 qαβ − c1 qαβ qαβ c1 qαβ qαβ + c2 qαβ + 2

Now, we use the Hashin-Shtrikman principle to get bounds for the overall moduli b k and m, b taking for simplicity s = 1. From this principle, it results that if ◦

◦

k ≤ kα and m ≤ mα for α = 1, 2,

(4.7.30)

◦

that is, if the tensor r = c− c is positive definite, we have U ≤ U s = U. Also, if

◦

◦

k ≥ kα and m ≥ mα for α = 1, 2,

(4.7.31)

(4.7.32)

◦

hence, if the tensor r = c− c is negative definite, we have U ≥ U s = U.

(4.7.33)

Thus, taking U in the form (4.7.13) and using (4.7), the inequalities (4.7.31) and (4.7.33) lead to bounds for the overall transverse bulk modulus b k and for the overall transverse shear modulus m. b As in Section 4.4, in order to obtain the best bounds, (4.7) must be maximized if (4.7.30) takes place, and must be minimized 1 2 if (4.7.32) is true. Differentiating (4.7) with respect to pα and qαβ , qαβ , we obtain the following extreme conditions: −

1

◦ pα

kα − k

+ α0 pα − α0 p + 2ε = 0, α = 1, 2,

1 γ γ  qαβ + β0 qαβ − β0 q αβ + eαβ = 0, α, β, γ = 1, 2. −  ◦ 2 mα − m

(4.7.34)

(4.7.35)

γ Let us denote by p˜α and q˜αβ the (unique) solution of this system. It is easy to see that the solution leads to a maximum of U if (4.7.30) is true, and to a

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259

4.7. HASHIN’S BOUNDS FOR THE OVERALL MODULI

minimum if (4.7.32) is true. Also, direct computation starting from the system (4.7.34)−(4.7.35) shows that the corresponding value Ue of U has the following simple expression:
◦ 1 (4.7.36) Ue = U + 2p˜ε + q˜αβ eαβ . 2 1 2 + c2 qαβ can be obtained The mean values p˜ = c1 p˜1 + c2 p˜2 and q˜αβ = c1 qαβ γ solving (4.7.34) and (4.7.35) for p˜α and q˜αβ , respectively. The results are

p˜ =

with A=

2 X

α=1

A ◦

1+ α0 A

, q˜αβ =

cα

1 ◦

kα − k

,B= − α0

B eαβ , 1 + β0 B

2 X

α=1

(4.7.37)

cα

,

1



(4.7.38)

◦  − β0 2 µα − µ

α0 and β0 being given by the equations (4.7.28). In order to find bounds for the overall transverse bulk modulus b k, we assume that εαβ = εδαβ . In this case, from (4.7.13), (4.7.36) and (4.7.37), it follows that ◦

b k≷k+

A , 1 + α0 A

(4.7.39)

◦

◦

where the upper sign applies for kα >k and the lower one for kα
m b ≷m+

B 1 , 2 1 + β0 B

(4.7.40)

◦

◦

where the upper sign applies for µα >µ and the lower for µα <µ. ◦

◦

As in the Section 4.4, we must specify the values of k and m, introduced in (4.7.39) and (4.7.40), will give the highest lower bounds and the lowest upper bounds. It can be proved (see P.4.32) that the functions from the right-hand side ◦

of the relations (4.7.39) and (4.7.40) are monotonically increasing functions of k ◦ and µ. Hence, the highest lower bounds are obtained by considering the largest ◦ ◦ values of k and m compatible with (4.7.30). The lowest upper bounds will be ◦

◦

obtained taking into account the smallest values of k and m which rest compatible with (4.7.30). Let us assume that k1 ≤ k2 and m1 ≤ m2 . (4.7.41)

Then for the lowest lower bounds, denoted by k − and m− , we must take ◦ k = k1 , m = m1 , and for the highest upper bounds, denoted by k + and m+ , we ◦

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CHAPTER 4. MACROSCOPICALLY ELASTIC COMPOSITES ◦

◦

must take k = k2 , m = m2 . Introducing this value in (4.7.38)1 and (4.7.39), and taking into account the expression (4.7.28)1 of α0 , we get k − = k1 +

c2 , c1 1 + k1 + m 1 k2 − k 1

and

k + = k2 +

c1 , c2 1 + k2 + m 2 k1 − k 2

k− ≤ b k ≤ k+ .

(4.7.42)

(4.7.43)

Analogously, introducing the adopted values in (4.7.38)2 and (4.7.40), and taking into account the expression (4.7.28)2 of β0 , we get c2 , c1 (k1 + 2m1 ) 1 + 2m1 (k1 + m1 ) m2 − m 1 c 1 m+ = m 2 + c2 (k2 + 2m2 ) 1 + 2m2 (k2 + m2 ) m1 − m 2

m− = m 1 +

(4.7.44)

and m− ≤ m b ≤ m+ .

(4.7.45)

In order to analyze the obtained results, we first observe that taking into account the equation c1 + c2 = 1, after simple computations, we can express the bounds k − and k + in the equivalent form:

c1 k1 (k2 + m2 ) + c2 k2 (k1 + m2 ) c1 k1 (k2 + m1 ) + c2 k2 (k1 + m1 ) . , k+ = c1 (k2 + m2 ) + c2 (k1 + m2 ) c1 (k2 + m1 ) + c2 (k1 + m1 ) (4.7.46) Comparing (4.7.46) with Hill’s inequations (4.6.41), we can see that Hill’s bounds and Hashin’s bounds for the overall plane bulk modulus b k are the same. This coincidence is remarkable, since the approaches used by Hill and by Hashin to obtain these bounds are entirely different. Hill derived an exact solution for a twophase fiber-reinforced material of arbitrary phase geometry under uniform average axial strain and isotropic transverse strain, for the case of equal phase transverse shear moduli, and obtained the bounds using his energetical comparison theorem; Hashin has obtained the bounds using his variational and extreme principle and used piece-wise constant polarization fields. However, there exists a difference between the results obtained by the two approaches: Hashin was able to obtain the bounds only assuming that (k2 − k1 ) (m2 − m1 ) > 0, while Hill could deduce the bounds even if (k2 − k1 ) (µ2 − µ1 ) < 0. In exchange, Hill’s method cannot be used to obtain the bounds for the overall transverse shear modulus m. b We already know that k − and k + are the best bounds that can be obtained if only the phase moduli and their concentrations are taken into account, the geometry of the fibers and their relative distributions in the matrix remaining arbitrary. k− =

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4.7. HASHIN’S BOUNDS FOR THE OVERALL MODULI

Unfortunately, it cannot be concluded that the bounds m− and m+ are the best possible for the entire range of concentrations. In this regard, the situation is similar to that encountered in the case of a macro-isotropic composite. b 13 ). It remains to obtain bounds for the overall axial shear modulus µ b = (G This problem was solved by Hashin, using again the variation and extreme principle. As before, we consider a RVE, occupying the cylindrical domain D, bounded by ∂D. The matrix characteristics will be indexed by 1, those of the fibers by 2. We assume that on the boundary ∂D of the RVE are applied the following homogeneous displacement conditions corresponding to γ =const.

u1 = 0, u2 = γx3 , u3 = 0 on ∂D.

(4.7.47)

In this case in the RVE arises an antiplane strain state described by the following displacement field: u1 = 0, u2 = γx3 , u3 = u3 (x1 , x2 ) in D.

(4.7.48)

Consequently, only ε13 and ε23 are nonvanishing and ε13 =

1 1 u3,1 , ε23 = (γ + u3,2 ) . 2 2

(4.7.49)

Hence, the only nonvanishing components of the stress are   2µ1 ε13 in D1 2µ1 ε23 in D1 σ13 = , σ23 = 2µ2 ε13 in D2 2µ2 ε23 in D2 .

(4.7.50)

Here µ1 and µ2 are the axial shear moduli of the matrix and of the fibers, respectively. Let us consider now a homogeneous material occupying the same domain D. We denote by µ its axial shear moduli. Let us assume that on the boundary ∂D of the homogeneous material, the same displacement conditions are prescribed; i.e. ◦

◦

◦

u1 = 0, u2 = γx3 , u3 = 0 on ∂D.

Since this body is homogeneous, the displacement field in the whole domain D will have the same form: ◦

◦

◦

u1 = 0, u2 = γx3 , u3 = 0 in D.

(4.7.51)

Hence, the only nonvanishing components of the strain and stress are ◦

◦

◦ ◦

◦

ε23 = γ/2, σ23 = 2 µε23 = µ γ in D.

(4.7.52) ◦

From these equations for the mean value of the strain energy U in the homogeneous cylinder, having the volume v, we get the value 1 ◦ 1 ◦ 2 U= µ γ . 2 v

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(4.7.53)

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CHAPTER 4. MACROSCOPICALLY ELASTIC COMPOSITES

We take now into account the equations (4.7.3), (4.7.50) and (4.7.52). Thus, we can see that the polarization stress p has only two nonvanishing components which depend on x1 and x2 p13 = p13 (x1 , x2 ) , p23 = p23 (x1 , x2 ) . (4.7.54)   ◦ −1 In order to find the form of the tensor h = c− c adequate to our analysis,

we use the stress-strain relations (4.7.50) and (4.7.52). Thus we find that if   ◦ ◦ σ13 = 2 µ− µ ε13 , σ23 = 2 µ− µ ε23 ,

then

ε13 =



1

◦  σ13 ,

2 µ− µ

ε23 =



1

Hence, in the considered antiplane strain-state, we get p · hp =

1 ◦

µ− µ

◦  σ23 .

2 µ− µ


p213 + p223 .

(4.7.55)

Thus, the general relation (4.4.8) for the mean value of U (p, ε0 ) becomes Z Z
2 1 1 1 ◦ 1 ◦ 2 2 p23 ε23 dv + U 0 . (4.7.56) dv + + p p U = µ γ2 − 23 13 ◦ v D 2v D µ− µ 2 v

In order to obtain (4.7.56), we have taken into account (4.7.52)−(4.7.54) and have used the following notations: Z 1 (4.7.57) (p13 ε013 + p23 ε023 ) dv, U 0 = W 0 with W 0 = v D

ε013 and ε023 being the involved fluctuations. ◦ Also, since according to (4.7.52)1 ε23 is constant in D, the equation (4.7.56) becomes Z
1 1 1 ◦ 1 0 2 2 (4.7.58) U = µ γ2 − ◦ p13 + p23 dv + p13 γ + U . 2v D µ− µ 2 v

Denoting by U the strain energy of the RVE and designing by µ b the overall axial shear modulus of the composite cylinder, according to the energetical definition of the overall moduli, we get

1 2 1 bγ . U= µ 2 v

(4.7.59)

Now we return to the general relations (4.4.4) and (4.4.5), introducing the fluctuations u0 and ε0 . According to the equations (4.7.48) and (4.7.51), only the

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4.7. HASHIN’S BOUNDS FOR THE OVERALL MODULI

axial component u03 of the fluctuation u0 is nonvanishing and it depends only on x1 and x2 (4.7.60) u01 = 0, u02 = 0, u03 = u03 (x1 , x2 ) . Hence, only the components ε013 and ε023 of the fluctuation ε0 are not zero and ε013 =

1 1 0 u , ε0 = u03,2 . 2 2 3,1 23

(4.7.61)

Let us introduce the fluctuation p013 and p023 p013 = p13 − p13 , p023 = p23 − p23 .

(4.7.62)

Since the mean values of the fluctuations ε013 and ε023 are vanishing, we can express W 0 introduced by the equation (4.7.57)2 in the following equivalent forms: Z Z
1 0 0 0 0 0 (4.7.63) p013 u03,1 + p023 u03,2 dv. W = (p13 ε13 + p23 ε23 ) dv = 2 D D

Now we recall that p and ε0 must satisfy the subsidiary condition (4.4.11). Taking into account (4.7.61) and (4.7.62), it is easy to see that in our antiplane strain state, the first two differential equations and boundary conditions (4.6.11) are identically satisfied and the third one takes the following form:
◦ µ u03,11 + u03,22 + p13,1 + p23,2 = 0 in D − Γ, u03 = 0, on ∂D.

In the differential equation, p13 and p23 can be replaced by the fluctuations p013 and p023 , since the mean values p13 and p23 are constant in D. Thus we get the following field equation and boundary condition, which must be verified by u 03


◦ µ u03,11 + u03,22 + p013,1 + p023,1 = 0 in D − Γ, u03 = 0 on ∂D.

(4.7.64)

Now, following Hashin, we assume piece-wise constant polarization fields   τ11 in D1 τ21 in D1 p13 (x) = , p23 (x) = (4.7.65) τ12 in D2 τ22 in D2 . Hence, the fluctuations p013 , p023 and the mean values p13 and p23 are   τ21 − p23 in D1 τ11 − p13 in D1 , (4.7.66) , p023 = p013 (x) = τ22 − p23 in D2 τ12 − p13 in D2

p13 = c1 τ11 + c2 τ12 , p23 = c1 τ21 + c2 τ22 .

(4.7.67)

At the same time, according to (4.7.58), (4.7.66) and (4.7.67), we get ) (

c2 c1 1 ◦ 2 1 1 2 2 2 2 U = µγ − ◦ τ12 + τ22 ◦ τ11 + τ21 + 2 µ −µ 2 v µ −µ 2

1

0

+ (c1 τ21 + c2 τ22 ) γ + U .

Copyright © 2004 by Chapman & Hall/CRC

(4.7.68)

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CHAPTER 4. MACROSCOPICALLY ELASTIC COMPOSITES

In order to evaluate the mean value U 0 , we suppose that the fiber-reinforced cylinder has an infinite extent in its transverse direction. Now our problem becomes: Given
◦ µ u03,11 + u03,22 + p031,1 + p032,2 = 0 in E2 − Γ, lim u03 (x) = 0, (4.7.69) kxk→∞

find u03 (x1 , x2 ), and obtain U 0 using the equations Z
1 1 p013 u03,1 + p023 u03,2 dv. U 0 = lim W 0 with W 0 = v→∞ v 2

(4.7.70)

E2

In order to solve this problem, we again use the two-dimensional Fourier transform. Denoting by U3 , P13 and P23 the Fourier transform of u03 , p013 , p023 and using the properties of the Fourier transform, we replace the field equation (4.7.69) 1 given in the physical space, by its correspondent in the wave-number space ◦

µ ξ 2 U3 = −i (ξ1 P13 + ξ2 P23 ) . Hence, the Fourier transforms of the derivatives u03,1 , u03,2 are



ξ1 ξ2 P13 + ξ22 P23 ξ 2 P13 + ξ1 ξ2 P23 0 . = − , F u F u03,1 = − 1 3,2 ◦ ◦ µ ξ2 µ ξ2

Now, from (4.7.70)2 and using the Parceval theorem, we get  Z  ∗ ∗ ∗ 1 2 2 0 ξ1 P13 P13 +2ξ1 ξ2 P13 P23 +ξ2 P23 P23 dξ. W =− ◦ 2µ

In order to evaluate this integral, we return to (4.7.65). Since the composite is macro-isotropic in its transverse plane, the polarization fields p13 (x) and p23 (x)
1/2 depend on (x1 , x2 ) only through the magnitude kxk = x21 + x22 . Consequently, as we already know, P13 (ξ) and P23 (ξ) can depend on (ξ1 , ξ2 ) only through the magnitude ξ 2 = ξ12 + ξ22 . Taking into account this fact and the existing cylindrical symmetry, we can conclude that W 0 is given by the following equation:  Z  ∗ ∗ 1 P13 P13 +P23 P23 dξ. W0 = − ◦ 4µ

Thus, a new appeal to the Parceval equality gives Z n o 1 2 2 (p013 ) + (p023 ) dx. W0 = − ◦ 4µ

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4.7. HASHIN’S BOUNDS FOR THE OVERALL MODULI

Now, returning to the equation (4.7.70)1 and using the relations (4.7.66), we find U0 = −

1 n ◦

4µ

2

2

2 c1 (τ11 − p13 ) + c1 (τ21 − p23 ) + c2 (τ12 − p13 ) + c2 (τ22 − p23 )

2

o

.

Using the equation c1 + c2 = 1 and the relations (4.7.67), finally we get o c1 c2 n 2 2 U 0 = − ◦ (τ11 − τ12 ) + (τ22 − τ21 ) . 4µ

The last equation and the relation (4.7.68) give ) (
2
2 c2 c1 1 ◦ 2 1 1 2 2 2 2 + U = µγ − ◦ τ12 + τ22 ◦ τ11 + τ21 2 µ −µ 2 v µ2 − µ 1 o c1 c2 n 2 2 + (c1 τ21 + c2 τ22 ) γ − ◦ (τ11 − τ12 ) + (τ22 − τ21 ) .(4.7.71) 4µ

Now, we return to the Hashin-Shtrikman principle to get bounds for µ b. For simplicity, we take v = 1. According to this principle, if ◦

µ ≤ µ 1 , µ2

(4.7.72)

U ≤ U s = U,

(4.7.73)

we have and if

◦

µ ≥ µ 1 , µ2

(4.7.74)

U ≥ U s = U.

(4.7.75)

we get Thus taking U in the form (4.7.59) and U in the form (4.7.71), the inequalities (4.7.73) and (4.7.75) lead to bounds for the overall axial shear modulus µ b. In order to get the best bounds, (4.7.71) must be maximized if (4.7.72) takes place, and must be minimized if (4.7.74) is true. Differentiating (4.7.71) with respect to τ11 , τ12 , τ21 , τ22 , we get the following extreme conditions: τ11 ◦

µ1 − µ

τ21 ◦

µ1 − µ

+

−

c2 ◦

2µ

c2 ◦

2µ

(τ11 − τ12 ) = 0,

(τ22 − τ21 ) = γ,

τ12 ◦

µ2 − µ

−

τ22 ◦

µ2 − µ

c1 ◦

2µ

+

(τ11 − τ12 ) = 0,

c1 ◦

2µ

(4.7.76) (τ22 − τ21 ) = γ.

Let us denote by τ˜11 , τ˜12 , τ˜21 , τ˜22 the (unique) solution of this system. It can be seen that the solution leads to a maximum of U for (4.7.72) and to its minimum

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for (4.7.74). Also, by some algebra, it can be deduced that the corresponding value Ue of U is 1 1 ◦ (4.7.77) Ue = µ γ 2 + (c1 τ˜21 + c2 τ˜22 ) . 2 2

The mean value p˜32 = c1 τ˜21 + c2 τ˜22 can be obtained solving (4.7.76)3,4 for τ21 and τ22 . We get (4.7.78) c1 τ˜21 + c2 τ˜22 = C γ

with C=

 ◦  ◦ ◦ ◦ c1 µ +µ2 µ1 − µ + c2 µ +µ1 µ2 − µ ◦

µ +c1 µ2 + c2 µ1

.

(4.7.79)

Now it follows from (4.7.59), (4.7.77) and (4.7.78) that ◦

µ b ≷ µ +C,

(4.7.80)

with the upper sign if (4.7.72) is true, and with the lower sign if (4.7.74) takes place. ◦ As before, we must assign those values to µ which, when introduced in (4.7.80), will give the highest lower bound and the lowest upper bound for the overall axial shear modulus µ b. It can be seen that the function from the right◦ hand side of the inequalities (4.7.80) is monotonically increasing with µ. Hence, ◦ the highest lower bound is obtained by taking the largest value of µ compatible with (4.7.72). Similarly, the lowest upper bound will be obtained by taking into ◦ account the smallest value of µ compatible with (4.7.74). Let us assume that µ1 ≤ µ 2 . (4.7.81) Then, for the highest lower bound, denoted by µ− , ◦

µ = µ1 , must be considered, and for the lowest upper bound, denoted by µ+ , ◦

µ = µ2 must be taken. Introducing the above values in (4.7.79), and taking into account (4.7.80), we get, after some elementary computations, µ− = µ 1 +

c2 , c1 1 + 2µ1 µ2 − µ 1

µ+ = µ 2 +

c1 c2 1 + 2µ2 µ1 − µ 2

(4.7.82)

and µ− ≤ µ b ≤ µ+ .

Copyright © 2004 by Chapman & Hall/CRC

(4.7.83)

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4.8. PROBLEMS

Hashin was able to show that µ− and µ+ are the best possible bounds for the overall axial shear modulus µ b in terms of the phase moduli and the concentration only. More exactly, Hashin has shown that µ− and µ+ are the exact solutions for some fiber-reinforced composites having particular structures. Hence, the above bounds can be improved only if supplementary informations concerning the geometry of phases and their distributions are taken into account. Summing up the presented results from the Sections 4.6 and 4.7, due to Hill and Hashin, we can say the following concerning the fiber-reinforced composites which are macro-homogeneous and macroscopically transversally isotropic: (i) Were obtained bounds for the five independent overall moduli for an arbitrary transverse phase geometry. (ii) All of the bounds, except perhaps those for the overall transverse shear modulus, are the best possible in terms of phase moduli and concentrations only. (iii) The obtained bounds are valuable for any concentrations of fibers in the matrix. (iv) If the transverse shear moduli of the matrix and of the fibers are equal, the overall axial Young modulus, the overall plane bulk modulus, the overall axial Poisson ratio and, obviously, the overall transverse shear modulus can be exactly evaluated in terms of the corresponding phase moduli and concentrations, for arbitrary transverse phase geometry and for any value of the fiber concentration in the matrix.

4.8

Problems

P4.1 Give the kinematical meaning of the influence tensor function A = A (x). P4.2 Give the definition of the overall elasticity of a macro-homogeneous composite having three different phases and find the analogous of the equations (4.1.20), (4.1.22) and (4.1.23). P4.3 Give the dynamical meaning of the influence tensor function B = B (x). P4.4 Using the equations (4.1.41) and (4.1.44), show that the overall elastic b are symmetric tensors. moduli b c and k P4.5 Show that the overall compliance b k is positive definite. P4.6 Prove the equivalence of the second and third (energetic) definitions of the overall elasticity b c. P4.7 Give a dual, equivalent formulation of Hill’s weak assumption. P4.8 Show that if Hill’s strong assumption is fulfilled, the influence tensor functions A (x) and B (x) satisfy the equation AT (x) B (x) = J. P4.9 Assuming Hill’s strong assumption, show that the constant influence tensors A1 and B1 satisfy the following Hill compatibility conditions:

b = k1 B1 and B1b c = c 1 A1 . A1 k

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Which compatibility conditions are satisfied by the constant influence tensors A 2 and B2 ? Are the above compatibility conditions satisfied if only Hill’s weak assumption is satisfied? P4.10 Prove Hill’s compatibility conditions (4.3.7) for a macro-homogeneous and macro-isotropic biphasic composite. P4.11 Formulate and prove the dual Hill type comparison theorem for the overall compliances. P4.12 Why must the two composites have the same geometry in Hill’s comparison theorem? P4.13 Show that Voigt’s estimate can be obtained assuming A (x) = J and Reuss’ estimate follows supposing B (x) = J. P4.14 Prove that for a biphasic mixture, Hill’s universal estimates become

b ≤ k V ≡ c 1 k1 + c 2 k2 . b c ≤ c V ≡ c 1 c1 + c 2 c2 , k

P4.15 Prove Hill’s universal estimates (4.3.13) for a biphasic macro-homogeneous and macro-isotropic composite. P4.16 (a) Using the equations (4.3.8)1 , (4.3.11)1 , (4.3.12)1 and the inequalities (4.3.13)1 , show that the influence coefficients a1 and b1 satisfy the relations a1 < 1, b1 > 1 if k2 < k1 ; a1 > 1, b1 < 1 if k1 < k2 . (b) Using the equations (4.3.5), give the mechanical meaning of the results obtained in (a). P4.17 Prove the relations (4.3.15) and (4.3.42). P4.18 For a biphasic macro-homogeneous and macro-isotropic composite, E V is defined by the equation (4.3.16). Prove that c1 E1 + c2 E2 ≤ EV and c1 E1 + c2 E2 = EV if and only if ν1 = ν2 . P4.19 P4.20 P4.21 P4.22

Using (4.3.38) and the relation c1 + c2 = 1, prove (4.3.39). Prove the relation (4.3.43). Prove the relation (4.3.45). Show that b if µ1 = µ2 = µ. c 1 E1 + c 2 E2 ≤ E

P4.23 Show that if µ1 = µ2 < µ, the overall Poisson ratio νb satisfies the relation c2 c1 1 . + = 1 − ν2 1 − ν1 1 − νb

P4.24 Show that if µ1 = µ2 = µ, the following inequations are satisfied: −1  c2 c1 ≤ νR ≤ c1 ν1 + c2 ν2 ≤ νb ≤ νV + ν2 ν1

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269

4.8. PROBLEMS where

µ 1 µ 1 − − 2 3kV 2 3kR νR = µ . µ , νV = 1+ 1+ 3kV 3kR

P4.25 Show that b k given by (4.3.38) satisfies the equation b k − k2 = k1 − k 2

tions

c1 . c2 (k1 − k2 ) 1+ 4 k2 + µ 3

P4.26 Using P4.25, show that the overall bulk modulus b k satisfies the inequa-

b c1 k − k2 c1 , ≤ ≤ c1 (k1 − k2 ) c2 (k1 − k2 ) k1 − k 2 1+ 1+ 4 4 k2 + µ 1 k2 + µ 2 3 3 when sgn (k2 − k1 ) = sgn (µ2 − µ1 ), and with the inequalities reversed when sgn (k2 − k1 ) = −sgn (µ2 − µ1 ) . P4.27 Prove the validity of the relation (4.3.48) giving b k if µ1 = µ2 = µ. P4.28 Hill’s bounds k − and k + are given by the equations (4.3.50). Prove that kR < k − and k + < kV .

P4.29 A biphasic macro-homogeneous and macro-isotropic glass-epoxy composite has the following mechanical and geometrical characteristics: E1 = 3GP a, E2 = 70GP a, ν1 = 0.3, ν = 0.2, c1 = 0.3, c2 = 0.7. (a) Find k1 , µ1 and k2 , µ2 . (b) Find kV , µV and kR , µR . c2 c1 + . (c) Find EV , νV , ER , νR , c1 E1 + c2 E2 , c1 ν1 + c2 ν2 , ν2 ν1 (d) Compare EV and c1 E1 + c2 E2 . (e) Find k − , k + and compare kR , k − and k + , kV . P4.30 A biphasic macro-homogeneous and macro-isotropic mixture has the following mechanical characteristics:

µ1 = µ2 = µ = 30GP a, k1 = 25GP a, k2 = 40GP a. b and νb if c1 = 9 , c2 = 1 . (a) Find the overall elastic moduli b k, µ b, E 10 10 9 1 . , c2 = (b) Find the same quantities if c1 = 10 10 (c) Compare and analyze the results obtained in (a) and (b).

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P4.31 Find the solution of the system (4.4.61), (4.4.62) and prove the relations (4.4.63)−(4.4.65). P4.32 Prove that the functions ◦ ◦ ◦ ◦ ◦ ◦ B A , g = g k , µ =µ + f = f k , µ =k + 2 (1 + β0 B) 1 + α0 A ◦

◦

◦

◦

cannot decrease if k and µ increase. In these relations, α, β are given by the equations (4.4.55) and A, B are given by the equations (4.4.65). P4.33 Show that the Hashin-Shtrikman bounds can be expressed by the equations    +    − µ µ+ − µ 1 µ µ− − µ 2 −1 c2 , = 1 + β2 −1 c1 , = 1 + β1 µ2 µ2 − µ 1 µ1 µ1 − µ 2

in which the concentrations c1 and c2 are isolated , and β1 =

6 (k2 + 2µ2 ) 6 (k1 + 2µ1 ) . , β2 = 5 (3k2 + 4µ2 ) 5 (3k1 + 4µ1 )

P4.34 Prove that the difference µ+ − µ− is given by the following relation

1 − β2 1 − β1 µ2 µ1 − β2 β1 . = c c β β 1 2 1 2 2 {µ1 + c1 β1 (µ2 − µ1 )} {µ2 + c2 β2 (µ1 − µ2 )} (µ2 − µ1 ) µ+ − µ −

P4.35 Find the difference k + − k − of the Hill-Hashin-Shtrikman bounds for the overall bulk modulus b k. P4.36 Give the graphical representation of kR , kV , k − and k + as functions of the inclusion concentration c2 , assuming k1 < k2 and µ1 < µ2 . P4.37 The mechanical and geometrical characteristics of a macro-homogeneous and macro-isotropic carbon/epoxy composite are E1 = 4GP a, E2 = 200GP a, ν1 = 0.3, ν2 = 0.2, c1 = 0.4, c2 = 0.6. (a) Find kR , kV , µR , µV . (b) Find k − , k + , µ− , µ+ . P4.38 (a) Show that if k2 and µ2 converge to infinity, the upper bounds k + , µ+ also converge to infinity. (b) Show that if k1 and k2 converge to zero, the lower bounds k − , µ− also converge to zero. P4.39 Prove that the relations (4.5.7) are true. P4.40 Starting with Budiansky’s relations (4.5.8), show that Hill’s relations (4.5.10) are true. P4.41 Using the equation (4.5.11), express b k as function of µ b and show that b if µ b is increasing, than k is also increasing.

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4.8. PROBLEMS

P4.42 Show that the function f = f (b µ) defined by the equation (4.5.12) satisfies the inequalities f (µR ) < 0 and f (µV ) > 0. What important conclusion can be obtained from the above result? P4.43 Show that the function f = f (b µ) satisfies also the inequalities f (µ1 ) < 0 and f (µ2 ) > 0, if µ1 < µ2 . What important conclusions can be obtained from the above result? P4.44 Show that the overall shear modulus µ b determined by the self-consistent method satisfies the equation   c2 µ1 c1 µ2 c 2 k2 c 1 k1 +2=0 . + +5 + 4 4 µ b − µ1 µ b − µ2 b b k2 + µ k1 + µ 3 3

P4.45 Find the overall moduli b k and µ b using the self-consistent method and Eshelby’s energetical theorem given in the final part of the Section 2.6. P4.46 In the final part of Section 4.5 for a dilute dispersion (c2 << 1) we have obtained the following approximate value for the overall shear modulus:   µ b = µ1 (1 + kc2 ) with

1 µ2 1 − 2 + = 5 µ2 − µ 1 k

k1  . 4  k1 + µ 1 3

Assuming an incompressible matrix (k1 = ∞) and supposing rigid inclusions (k2 = µ2 = ∞), find the approximative value of the overall shear modulus µ b. Compare the obtained result with the relation (4.5.24) giving µ b in the conditions assumed in this problem. P4.47 Show that if the elasticity tensor c of a transversally isotropic material is positive definite, the material parameters E = E3 , k = K12 and m = G12 , introduced by the equations (4.6.4) and (4.6.7) are all positive. P4.48 Prove the inequality b ≤ c 1 E1 + c 2 E2 E

b of a fiber-reinforced composis satisfied by the overall axial Young modulus E ite. Here E1 and E2 are Young’s axial moduli for the matrix and for the fibers, respectively. P4.49 Show that the values obtained for the lower and upper bounds of the overall plane bulk modulus b k, using Hill’s and Hashin’s method are the same. P4.50 Express the difference of the bounds k + and k − of the overall plane bulk modulus b k, in terms of the phase moduli and concentrations.

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CHAPTER 4. MACROSCOPICALLY ELASTIC COMPOSITES

P4.51 Using Hashin’s method, find the overall plane bulk modulus b k for a composite in which the matrix and the fibers have the same transverse shear modulus; i.e. m1 = m2 = m. P4.52 Let us consider a transversally isotropic material. Express the transverse Young modulus E1 , the transverse Poisson ratio ν12 and the axial Poisson ratio ν13 , in terms of the axial Young modulus E3 , of the plane bulk modulus K12 , of the transverse shear modulus G12 and of the axial Poisson ratio ν31 of the material.

Bibliography [4.1] Hill, R., Elastic properties of reinforced solids; some theoretical principles, J. Mech. Phys. Solids, 11, 357-372, 1963. [4.2] Hill, R., Theory of mechanical properties of fiber-strengthened materials: I. Elastic behavior, J. Mech. Phys Solids, 12, 199-221, 1964. [4.3] Hill, R., Theory of mechanical properties of fiber-strengthened materials: III. Self consistent model, J. Mech. Phys. Solids, 13, 189-198, 1965. [4.4] Hill, R., A self-consistent mechanics of composite-material, J. Mech. Phys. Solids, 13, 213-222, 1965. [4.5] Hill, R., The essential structure of constitutive laws for metal composites and policrystals, J. Mech. Phys. Solids, 15, 79-95, 1967. [4.6] Hill, R., On macroscopic effects of heterogeneity in elastoplastic media at finite strain, Math. Proc. Cambr. Phil. Soc., 95, 481-494, 1984. [4.7] Budiansky, B., On the elastic moduli of some heterogeneous materials, J. Mech. Phys. Solids, 13, 223-227, 1965. [4.8] Hashin, Z., Theory of mechanical behavior of heterogeneous media, Appl. Mech. Rev., 17, 1-9, 1964. [4.9] Hashin, Z., On the elastic behavior of fiber reinforced materials of arbitrary transverse phase geometry, J. Mech. Phys. Solids, 13, 119-134, 1965. [4.10] Hashin, Z., Analyse of composite materials, A survey, J. Appl. Mech., 50, 481-504, 1983. [4.11] Hashin, Z., Shtrikman, S., On some variational principles in anisotropic and non-homogenous elasticity, J. Mech. Phys. Solids, 10, 335-342, 1962. [4.12] Hashin, Z., Shtrikman, S., A variational approach to the theory of the elastic behavior of multiphase materials, J. Mech. Phys. Solids, 11, 127-140, 1963. [4.13] Suquet, P., Mat´eriaux composites-Mat´eriaux nouveaux, Lecture Notes, Univ. Marseille, 1989. [4.14] Zaoui, A., Materiaux h´et`erog`enes, Lecture Notes,Univ. Marseille, 1989. [4.15] G˘ar˘ajeu, M., Contribution a` l’´etude du comportement non lineaire de milieux poreaux avec ou sans renfort, Th`ese de Docteur, Univ. Marseille, 1995. [4.16] Mandel, J., Plasticit´e classique et viscoplasticit´e, Int. Centre Mech. Sci., Courses and lectures, No. 97, Udine 1971, Springer, Wien, New York, 1972.

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Chapter 5

THE THREE-DIMENSIONAL LINEARIZED THEORY 5.1

Elements of nonlinear elasticity

The elements of nonlinear elasticity succinctly presented in the following can be found, for instance, in Malvern’s fundamental monograph [5.1]. 7Let us denote by B the reference configuration of a deformable body. Let {ek }, k = 1, 2, 3 be an orthonormal basis in V and let {O, ek } be an orthogonal Cartesian coordinate system in E. We design, by X = Xk ek , the position vector of a particle P of the body; Xk represent the material, referential, or Lagrangean coordinates of the particle P . Let x = xk ek be the position vector of the same particle at the current time t, in a motion x = χ(X, t), xk = χk (Xl , t), k, l = 1, 2, 3

(5.1.1)

of the body. The numbers xk represent the current, spatial or Eulerian coordinates of the particle P . The gradient of the deformation F = F(X,t) corresponding to the motion (5.1.1) is defined by the equations F = F(X, t) = ∇X χ(X, t) = [GradX χ(X, t)]T .

(5.1.2)

We assume that F is a nonsingular tensor and its determinant J is always positive; i.e. J(X,t) = det F(X,t) > 0. (5.1.3) If Fkm are the components of F in the basis {ek em }, k, m = 1, 2, 3; i.e. if F = Fkm ek em ,

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CHAPTER 5. THE THREE-DIMENSIONAL LINEARIZED THEORY

according to the definition (5.1.2), we have Fkm =

∂χk = χk,m . ∂Xm

Let us denote by U = U(X, t) = Uk (Xl , t)ek the displacement vector corresponding to the motion (5.1.1), and by H = H(X, t) = ∇X U(X, t) = [GradX U(X, t)]T = Hkm (Xl , t)ek em

(5.1.4)

the gradient of the displacement. We have the equations x = X + U(X, t), xk = Xk + Uk (Xl , t), Hkm =

∂Uk = Uk,m ∂Xm

(5.1.5)

(5.1.6)

and F = 1 + H, Fkm = δkm + Hkm = δkm + Uk,m .

(5.1.7)

We denote by G = G(X, t) the Green’s deformation or strain tensor defined by the equation 1 (5.1.8) G = (FT F − 1). 2 Using (5.1.7), we get

G=

1 1 (H + HT + HT H), Gkm = (Uk,m + Um,k + Ul,k Ul,m ). 2 2

(5.1.9)

We denote by ρ0 = ρ0 (X) and by ρ = ρ(x, t) the mass densities in the reference configuration B and in the current configuration Bt , respectively. These mass densities are related by the equation ρ0 (X) = J(X, t)ρ(x, t).

(5.1.10)

Let us denote by T = T(x, t) the Cauchy’s symmetric stress tensor , by b = b(x, t) the unit body force in Bt , and by a = a(x, t) the acceleration corresponding to the given motion. These three fields and the mass distribution in B t satisfy Cauchy’s equation of motion divx T(x, t) + ρ(x, t)b(x, t) = ρ(x, t)a(x, t),

(5.1.11)

or in component form, Tkm,m + ρbk = ρak , k, m = 1, 2, 3.

(5.1.12)

Now let us consider a material surface element da in the current configuration of the body and let us denote by n = n(x, t) the unit normal to da, and

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5.1. ELEMENTS OF NONLINEAR ELASTICITY

by df the (surface) force acting on da. Let tn = tn (x, t) be Cauchy’s stress vector corresponding to da and df ; we have df = tn da.

(5.1.13)

Accordingly, the Cauchy’s stress vector tn is the current (surface) force per unit area of the current material surface. As is well known, tn can be expressed in terms of T and n by the Cauchy’s equation tn (x, t) = T(x, t)n(x, t). (5.1.14) Let dA be the considered material surface element in the reference configuration B of the body, and let N = N(X) be the unit normal to dA. Nanson’s formula establishes the connection between the pairs N, dA and n, da; we have JNdA = FT nda.

and

(5.1.15)

From this equation, we get q da = (det C)N · C−1 N dA with C = FT F,

1

n= √

F−T N with F−T ≡ (F

−1 T

T −1

. N·C N Let us denote by Θ = Θ(X, t) = Θkm (Xl , t)ek em the nominal stress, that is the transposed of the first, nonsymmetric Piola-Kirchhoff stress tensor. The nominal stress tensor Θ is defined in terms of Cauchy’s stress tensor T and of the gradient of the deformation F by the equation −1

) = (F )

Θ = JF−1 T

(5.1.16)

and satisfies the equation of motion of Piola-Kirchhoff type ¨ DivX Θ(X, t) + ρ0 (X)b(X, t) = ρ0 (X)U(X, t),

(5.1.17)

or in component form, ¨m , k, m = 1, 2, 3. Θkm,k + ρ0 bm = ρ0 U

(5.1.18)

In these equations, the superposed dot designs the derivative with respect to ¨ = U(X, ¨ the time t, i.e. the material time derivative. Hence U t) is the acceleration field, corresponding to the Lagrangean description of the motion. The current (surface) force df acting on the current material surface element da can be equivalently expressed using the Piola-Kirchhoff stress vector sN = sN (X, t), by the following equation: df = sN (X, t)dA.

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(5.1.19)

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CHAPTER 5. THE THREE-DIMENSIONAL LINEARIZED THEORY

Accordingly, Piola’s and Kirchhoff’s stress vector sN is the current (surface) force reported to the unit area of the material surface element taken in the reference configuration of the body. As is well-known, sN can be expressed in terms of Θ and N by the following relation: sN (X, t) = ΘT (X, t)N(X), sNm = Θkm Nk . (5.1.20) Returning to the equations (5.1.13) and (5.1.19), and using the relation (5.1.14), we can obtain the formula connecting the Cauchy’s stress vector t n and the Piola’s and Kirchhoff’s stress vector sN : tn = q

1

(det C)N · C

−1

sN . N

Let us denote by Π = Π(X, t) = Πkm (Xl , t)ek em the second symmetric Piola-Kirchhoff stress tensor, defined by the following equation: Π = JF−1 TF−T or T = J −1 FΠFT .

(5.1.21)

Examining the relations (5.1.16) and (5.1.21), we can conclude that the nominal stress tensor Θ and the symmetric Piola-Kirchhoff stress tensor Π are connected by the equation Θ = ΠFT or Θkm = Πkl Fml .

(5.1.22)

In the following, we consider only homogeneous, hyperelastic solid bodies. We denote by u = u(G), the specific strain energy or the specific elastic energy of this material, reported to the unit material volume in the reference configuration. We assume that the function u = u(G) is given, and is the constitutive equation of the considered body. Since the material is hyperelastic, the specific strain energy is an elastic potential and Π can be expressed through G by the following constitutive equation or stress-strain relation: Π=

∂U (G). ∂G

(5.1.23)

The above equation satisfies automatically the principle of objectivity or of the material frame indifference. Since both Π and G are symmetric tensors, if the basic constitutive relation u = u(G) is expressed as a function depending on the components Gpq of G, i.e. if u = u(Gpq ), in order to obtain the component form of the stress-strain relation (5.1.23), we must use the following relation:   ∂ ∂ 1 u(Gpq ). (5.1.24) + Πkm = ∂Gmk 2 ∂Gkm

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5.1. ELEMENTS OF NONLINEAR ELASTICITY

Taking into account the equation (5.1.22) and (5.1.23), we can obtain the nominal stress Θ in terms of F and G Θ = H(F) =

∂U ∂U T (G)F . (F) = ∂G ∂F

(5.1.25)

This relation shows the way in which the constitutive function H = H(F) can depend on F, in order to respect the principle of objectivity. The presented relations show that in the spatial or Eulerian description of the motion, the Cauchy’s stress tensor and the actual configuration are involved; in the referential or Lagrangean description of the motion, the nominal stress tensor with respect to the reference configuration and the reference configuration are involved. Sometimes it is useful to take as a reference configuration of the body, its current configuration Bτ corresponding to the time τ. We denote by x = χ(X, τ )

(5.1.26)

the position vector of a particle P at this time, and by y = χ(X, t)

(5.1.27)

we design the position vector of the same particle at an arbitrary moment t in the current configuration Bt , as shown in Figure 5.1. B B

F( )

P

P F (t) F(t) X

x

O

P

Bt

y

Figure 5.1: The configuration B, Bτ and Bt . As we know, the equation (5.1.26) can be inverted to obtain X = χ−1 (x, τ ). Thus, y can be expressed in terms of x and t in the following way: y = χ(X, t) = χ(X

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−1

(x, τ ), t) ≡ χτ (x, t) = χτ (χ(X, τ ), t).

(5.1.28)

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CHAPTER 5. THE THREE-DIMENSIONAL LINEARIZED THEORY

We denote by F(X, t) and F(X, τ ) the involved gradients of deformation, with respect to the reference configuration in B: F(X, t) ≡ F(t) = [GradX χ(X,t)]T ,

(5.1.29)

F(X, τ ) ≡ F(τ ) = [GradX χ(X, τ )]T . Also, we denote by Fτ (x, t) the gradient of deformation reported to the configuration Bτ , taken as reference configuration: Fτ (x, t) ≡ Fτ (t) = [gradx χτ (x, t)]T .

(5.1.30)

Since χτ (x, τ ) = x, we have Fτ (τ ) = 1.

(5.1.31)

Using (5.1.28) and the chain rule, we get the following fundamental relation connecting the deformation gradients F(τ ), F(t) and Fτ (t): F(τ ) = Fτ (t)F(t).

(5.1.32)

Similarly, if J(τ ) = det F(τ ), J (t) = det F(t) and Jτ (t) = det Fτ (t),

(5.1.33)

we get J(t) = Jτ (t)J(t)

(5.1.34)

Jτ (τ ) = 1.

(5.1.35)

and We denote by uτ = uτ (x, t) the displacement vector from Bτ to Bt ; i.e. y = x + uτ (x, t).

(5.1.36)

Hτ = Hτ (x, t) ≡ Hτ (t) = [gradx uτ (x,t)]T

(5.1.37)

Fτ (t) = 1 + Hτ (t).

(5.1.38)

Its transposed gradient is

and we have We return now to equation (5.1.9) giving Green’s deformation tensor G; accordingly, we get G(τ ) =

1 1 T (F (τ )F(τ ) − 1) and G(t) = (FT (t)F(t) − 1). 2 2

(5.1.39)

Taking into account the relations (5.1.32) and (5.1.38), we obtain an equation expressing the current Green tensor G(t) in terms of G(τ ), F(τ ) and Hτ (t) 1 T T G(t) = G(τ ) + FT (τ ){Hτ (t) + Hτ (t) + Hτ (t)Hτ (t)}F(τ ). 2

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(5.1.40)

279

5.1. ELEMENTS OF NONLINEAR ELASTICITY

Now we introduce the nominal stress tensor Θτ = Θτ (x, t) ≡ Θτ (t) reported to Bτ , taken as reference configuration. According to the general rule (5.1.16), we get Θτ (t) = Jτ (t)F−1 (t)T(t).

(5.1.41)

Since the relations (5.1.31) and (5.1.35) take place, we have Θτ (τ ) = T(τ ).

(5.1.42)

Also, using the relations (5.1.22), (5.1.32) and (5.1.34), we obtain the following equation connecting the current values of the nominal stress Θτ (t) corresponding to Bτ and the current values of the symmetric Piola-Kirchhoff stress Π(t) corresponding to B: Θτ (t) = J −1 (τ )F(τ )Π(t)FT (τ )FTτ (t).

(5.1.43)

Also, according to the general rule (5.1.17), the nominal stress tensor Θ τ (x, t) corresponding to Bτ taken as reference configuration, satisfies the following equation of motion: ··

divx Θτ (x, t) + ρ(x, τ )b(x, t) = ρ(x, τ ) uτ (x, t),

(5.1.44)

where ρ(x, τ ) is the mass density in Bτ taken as the reference configuration. The component form of the above equation is ··

Θτ km,k + ρbm = ρ uτ m .

(5.1.45)

The above equation corresponds to the updated Lagrangean approach. In this procedure the involved fields are expressed as functions of xk , the coordinates of the particle P in the configuration Bτ , taken as reference configuration at time τ. Let us denote by sn = sn (x, t) the Piola-Kirchhoff stress vector, reported to Bτ , the chosen reference configuration. According to the general rule (5.1.20), we have sn (x, t) = ΘTτ (x, t)n(x, τ ) or snm (t) = Θτ km (t)nk (τ ),

(5.1.46)

where n = n(x, τ ) is the unit normal to the material surface element in Bτ . We recall that sn is a surface force reported to the unit material surface in the deformed configuration Bτ selected as the reference configuration. According to (5.1.12) and (5.1.43), the following equation is true: sn (x,τ ) = tn (x, τ ) = T(x, τ )n(x, τ ).

(5.1.47)

Now we are ready to derive the laws governing the incremental behavior of a body, where one of its initial deformed equilibrium configuration is changed under the action of small, time-dependent external perturbations.

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CHAPTER 5. THE THREE-DIMENSIONAL LINEARIZED THEORY

5.2

Lagrangean approach

The fundamental aspects concerning the incremental behavior of deformable bodies are presented, for instance, in Guz’s monographs [5.2]−[5.8] for the cases of elastic, viscoelastic, elastic-plastic and elastoviscoplastic materials; in Ogden’s monograph [5.9], analyzed in great detail, is the case of elastic materials; Eringen’s and Maugin’s monographs [5.10] concerns the case of elastic materials interacting with electromagnetic and thermal fields. In presenting the basic results in the case of hyperelastic materials, we follow Guz’s approach exposed in its fundamental monograph [5.8]. For certain problems, the current state of an elastic body can be reached in two steps: initial static deformation followed by superposed, time-dependent, infinitesimal deformations from this static equilibrium state. In this case, it is possible to find the final state of the body, by solving a nonlinear elastic-static problem and a linear dynamic (or static) problem. The theory, corresponding to the second step, will be named linearized mechanics of deformable bodies. The theory of local stability of elastic solids is based on the linearized mechanics. The laws describing the behavior of the incremental fields involved in the linearized mechanics can be obtained using the Lagrangean, or the updated Lagrangean approach. The first procedure is followed, for instance, by Guz (see [5.2]−[5.8]), the second one by Eringen and Maugin [5.10]. Ogden [5.9] presents both alternatives. Depending on the analyzed problem, both variants can be useful. If the initial imposed static deformation is small, both methods lead to the same field equation describing the behavior of the incremental fields, produced by the small supplementary perturbations of the initial deformed equilibrium state. In the Lagrangean approach the external surface tractions applied on the boundary of the body are reported to the unit area of the corresponding material surface elements taken in the reference configuration B. In the updated Lagrangean approach, the same surface forces are reported to the unit area of the corresponding material surface element taken in the initial deformed equilibrium configuration. If the initial applied static deformation is small, the above distinctions become negligible. In the Lagrangean approach, all fields are considered as functions of the Lagrangean coordinates X k of the particles. In the updated Lagrangean approach, the fields concerning the initial deformed equilibrium state are considered as functions of the Lagrangean coordinates Xk of the particles, but the incremental fields, produced by the applied small time dependent perturbations, are considered as functions of the Eulerian coordinates xk of the particles in the initial deformed equilibrium configuration of the body. If the initial applied static deformations are small, the above distinctions become negligible. The component form of the involved field equations will be presented only at the end of each analysis, since most parts of the computations can be realized more easily using direct vector and tensor forms. This intrinsec way is followed by Owen [5.9]; Guz [5.2]−[5.8] and Eringen and Maugin [5.10] utilize the component form in their analysis.

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281

5.2. LAGRANGEAN APPROACH

In this Section we discuss the Lagrangean approach. We assume that in its initial reference configuration B, the body is stressfree. We suppose that at the time τ = 0 the body is statically deformed. We shall ◦

denote the resulting equilibrium configuration B0 by B ; i.e. ◦

B ≡ B0 . The above notation is in accordance with that used in the preceding Section, since ◦

B was identified with B0 , that is, with the configuration Bτ corresponding to the moment τ = 0 at which the initial static deformation was applied. This static deformation will be described by the equation ◦

x = χ(X, 0) ≡ χ (X).

(5.2.1) ◦

Accordingly, the gradient of this deformation will be denoted by F; i.e. ◦

◦

◦

T F = F (X) = [GradX χ (X)]

and

◦

◦

◦

J = J (X) = det F (X). ◦

(5.2.2)

The involved displacement field and its transposed gradient will be denoted ◦

◦

◦

by U = U (X) and H = H (X), respectively; i.e. ◦

◦

x =χ (X) = X+ U (X) and

◦

◦

◦

T H = H (X) = [GradX U (X)] ,

and

(5.2.3)

◦

◦

F = 1+ H .

(5.2.4) ◦

◦

The nominal stress tensor Θ(X, 0) will be simply denoted by Θ = Θ (X) and ◦ ◦ the involved body force density b(X, 0) will be denoted by b = b (X). According ◦

◦

to the general Piola-Kirchhoff equation (5.1.17) we can conclude that ρ◦ , Θ and b must satisfy the following equilibrium equation: ◦

◦

DivX Θ (X) + ρ◦ (X) b (X) = 0.

(5.2.5)

Similarly, the Piola-Kirchhoff stress vector sN (X, 0) will be denoted by ◦ ◦ sN = sN (X), and, according to the general rule (5.1.20), we have ◦

◦

sN (X) = Θ (X)N(X).

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(5.2.6)

282

CHAPTER 5. THE THREE-DIMENSIONAL LINEARIZED THEORY ◦

◦

The involved Green deformation tensor G(X,0), denoted by G=G (X), according to the general rule (5.1.8), is given by the equation ◦

G (X) = ◦

◦ 1 ◦T {F (X) F (X) − 1}. 2

(5.2.7)

◦

Denoting Π(X, 0) by Π=Π (X), and using the hyperelastic constitutive equation (5.1.23), we obtain ◦ ◦ ∂u ∂u ◦ , (5.2.8) ( G) ≡ Π= ∂G ∂G where the superposed “◦” tells us that the involved derivative must be calculated ◦

for G = G. ◦ ◦ Finally, denoting Θ(X, 0) simply by Θ = Θ (X), from the general constitutive equation (5.1.25), we obtain ◦

◦ ◦T

◦

Θ = H(F) = ΠF =

◦

∂ u ◦T F . ∂G

(5.2.9) ◦

Now let us assume that over the initial deformed equilibrium configuration B , small time-dependent external perturbations are applied, resulting in the current configuration Bt . Denoting by y the position vector of a particle P at the time t, we shall have ◦ y = χ(X, t) = χ (X) + U(X, t), (5.2.10) where the incremental field U = U(X, t) represents the incremental displacement ◦

from the initial deformed equilibrium configuration B to the neighboring current configuration Bt . We denote by (5.2.11) H = H(X, t) = [GradX U(X, t)]T

the small transposed gradient of the incremental displacement field. Denoting by F(X,t) the gradient of the deformation (5.2.10), we shall have ◦

F(X, t) = F (X) + H(X, t).

(5.2.12)

By the fact that the applied external time-dependent perturbations are small, we mean that the corresponding transposed gradient H(X, t) of the incremental displacement field is small, that is, it satisfies the following restriction

| H(X, t) |<< 1 for X ∈ B and t ≥ 0.

(5.2.13)

Moreover, we suppose that all products of all perturbations of all fields, due to H(X, t), are negligible with respect to the perturbation itself. The theory obtained using this approximation is just the linearized three-dimensional theory of hyperelastic solids.

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283

5.2. LAGRANGEAN APPROACH

In the following, the small perturbation will be denoted by a superposed bar and, for simplicity, the spatial variable will not be mentioned. In this notation, for Green’s deformation tensor G, we shall have ◦

G(t) = G +G(t).

(5.2.14)

According to the general rule (5.1.8) and the relation (5.2.12), taking into account the assumed approximation, we get

G(t) =

Hence,

◦ 1 ◦T T {F H(t) + H (t) F}. 2

(5.2.15)

◦ ◦ 1 ◦T T G(t) = G + {F H(t) + H (t) F}. 2

Analogously, we take

(5.2.16)

◦

Π(t) = Π +Π(t),

(5.2.17) ◦

Θ(t) = Θ +Θ(t),

and in this way from the general rule (5.1.22) and from the relations (5.2.9) 3 , (5.2.12), we obtain ◦T

◦

T

Θ(t) = Π(t) F + Π H (t).

(5.2.18)

It remains to express, in the adopted approximation, the perturbation Π(t) ◦

of the symmetric Piola-Kirchhoff stress tensor in terms of E and H(t). To this end we use the general hyperelastic constitutive equation (5.1.23) and the expression (5.2.16) of G(t). We have

∂u ◦ 1 ◦ T ∂u T ◦ {G + [F H(t) + H F]}. {G(t)} = 2 ∂G ∂G

◦

Π(t) = Π +Π(t) =

Consequently, it results

Π(t) =

◦

◦ ∂2 u 1 ◦ T T { [F H(t) + H (t) F]}. ∂G∂G 2

(5.2.19)

For simplicity we introduce the tensor ◦

∂2 u ∂2u ◦ = (G). K≡ ∂G∂G ∂G∂G ◦

◦

(5.2.20)

Since G is a second order symmetric tensor, K will be a fourth order symmetric tensor. If u = u(G) is considered as a function u = u(Gpq ) of the components Gpq of G, the components Kklmn of the tensor K ≡

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∂2u (G) ∂G∂G

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must be calculated using the relation Kklmn =

∂ ∂ ∂ 1 ∂ )u(Gpq ). + )( + ( ∂Gnm ∂Glk ∂Gmn 4 ∂Gkl

(5.2.21)

The above equation shows that the tensor K has the usual symmetry properties of the elasticity tensor c of the linear elasticity theory; i.e. Kklmn = Klkmn = Kklnm = Kmnkl .

(5.2.22)

In all that follows, the symmetric fourth order tensor K is considered as a linear function defined on the vector space of all symmetric second order tensor b4 . and having values in the same space. Briefly, K ∈L ◦

As seen, from the symmetry properties of the tensor K, introduced by equation (5.2.20), it results that the tensor equation (5.2.19) can be expressed in the following simplified, but equivalent form ◦

◦T

Π(t) = K {F H(t)}.

(5.2.23)

The component form of this equation is ◦

◦

◦

◦

Πkl = K klmn F pm H pn = K klmn F pm Up,n .

(5.2.24)

Now we can return to the relation (5.2.18) to get the perturbation Θ(t) of ◦

◦

the nominal stress in terms of F, Π and H(t). According to (5.2.23) we obtain   T ◦ ◦T ◦ ◦ T (5.2.25) Θ(t) = {K F H(t) } F + Π H (t).

As it is easy to see, the component form of this equation is ◦

◦

◦

◦

Θkq = F ql F pm K klmn H pn + Πkn H qn .

(5.2.26)

◦

Introducing the fourth order tensor C with components ◦

◦

◦

◦

C kqpn = F ql F pm K klmn ,

(5.2.27)

the above equation becomes ◦

◦

Θkq = C kqpn H pn + Πkn H qn .

(5.2.28)

At the same time, the symmetry relation (5.2.22), defined by the fourth order ◦

◦

tensor K show that C has the following symmetry property: ◦

◦

C kqpn = C npqk .

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(5.2.29)

285

5.2. LAGRANGEAN APPROACH ◦

Moreover, if we introduce the fourth order tensor Ω with components ◦

◦

◦

Ωkqpn = C kqpn + Πkn δqp ,

(5.2.30)

the incremental constitutive equation (5.2.28) takes the following form: ◦

◦

Θkq = Ωkpqn H pn = Ωkqpn Up,n ,

or, in tensor form,

(5.2.31)

◦

Θ = Ω ∇UT .

(5.2.32)

◦

Obviously, generally Ωkqpn is not symmetric neither in (k, q), nor in (p, n); i.e. ◦

◦

◦

◦

Ωkqpn 6= Ωqkpn , Ωkqpn 6= Ωkqnp .

(5.2.33)

This result is not surprising, since neither the incremental nominal stress, ◦

nor the incremental displacement gradient ∇U are symmetric. However, since C ◦ has the symmetry property (5.2.29) and since Π is symmetric, as it is easy to see ◦

examining the relation (5.2.30), we can conclude that the fourth order tensor Ω has the following symmetry property: ◦

◦

Ωkqpn = Ωnpqk .

(5.2.34)

Let us denote by b = b(X,t) the small time-dependent perturbation of the body force density reported to the unit material volume in the stress-free reference configuration B of the body; i.e. ◦

b(X,t) = b (X)+b(X,t).

(5.2.35)

As we know, the nominal stress tensor Θ = Θ(X,t) must satisfy the Piola◦

◦

Kirchhoff equation of motion (5.1.17). At the same time, Θ = Θ (X) satisfies the Piola-Kirchhoff equation of equilibrium (5.2.5). Hence, taking into account the decomposition (5.2.17)2 , we can conclude that the incremental fields Θ(X,t), b(X,t) and U(X,t) must satisfy the following incremental equation of motion: ··

DivX Θ(X, t) + ρ◦ (X)b(X) = ρ◦ (X) U (X, t),

or, in component form,

¨m . Θkm,k + ρ◦ bm = ρ◦ U

(5.2.36)

(5.2.37)

Finally, we shall design by sN = sN (X, t) the incremental Piola-Kirchhoff stress vector; i.e. ◦ (5.2.38) sN (X, t) = sN (X)+sN (X,t).

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According to the general rule, sN (X, t) can be expressed through Θ(X, t) and N(X) by the relation (5.1.20). Hence, the decompositions (5.2.17)2 and (5.2.38) show us that the incremental Piola-Kirchhoff stress vector sN (X, t), in terms of the incremental Piola-Kirchhoff stress tensor Θ(X, t) and of the unit normal N(X), is given by the following Piola-Kirchhoff type equation: T

sN (X, t) = Θ (X, t)N(X), sNm = Θkm Nk .

(5.2.39)

Considerable simplifications of the incremental constitutive equation can be obtained if we assume that the following two assumptions are fulfilled: (i) The initial applied static deformation is infinitesimal (geometrical linearity); i.e. ◦

|H (X) |<< 1 for X ∈ B.

(5.2.40)

(ii) The specific elastic energy u is a quadratic function of the infinitesimal strain tensor ε; i.e. the material is linearly elastic (physical linearity); in this case we have 1 1 (5.2.41) u = u(ε) = ε · cε = εkl cklmn εmn , 2 2 where c is the elasticity tensor of the linear elastic theory. To distinguish this important situation from the general one, we shall denote the Cauchy’s stress tensor T by σ; i.e. ◦

◦

σ = T and σ = T .

(5.2.42) ◦

◦

◦

According to the assumption (5.2.40) in the relations connecting σ = T, Π ◦ and Θ, we can take ◦

◦

◦

F= 1, F km w δkm and J w 1.

(5.2.43)

Consequently, we shall have ◦

◦

◦

◦

T=Π=Θ=σ .

(5.2.44)

Now, we can specify the equilibrium equation, constitutive relation and ge◦

ometrical equation concerning the initial deformed equilibrium configuration B . We shall have   ◦T ◦ 1 ◦ ◦ ◦ ◦ . (5.2.45) ∇ U +∇ U Div σ = 0, σ = c ε, ε = 2 ◦

◦

Also, the Piola-Kirchhoff stress vector sN will be expressed in terms of σ and we obtain ◦ ◦ sN = σ N. (5.2.46)

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5.3. UPDATED LAGRANGEAN APPROACH

The equation of motion (5.2.36) involving the incremental fields Θ, b and U, as well as the relation (5.2.39) connecting the incremental fields sN and Θ are unchanged. In exchange, the incremental constitutive equation (5.2.32) concerning Θ and ∇U takes a simplified form. Indeed, since now the inequality (5.2.40) is true, in ◦

◦

the relation (5.2.27) giving the coefficients C kqpn , we can assume F ql = δql and, in this way, also taking into account the relations (5.2.20) and (5.2.41), we get ◦

C kqpn = ckqpn .

(5.2.47) ◦

Consequently, with (5.2.30) and (5.2.44), the constitutive coefficients Ωkqpn , denoted simply by ωkqpn become ◦

◦

ωkqpn ≡ Ωkqpn = ckqpn + σ kn δqp .

(5.2.48)

Hence, the incremental constitutive equation (5.2.32) becomes

Θ = ω∇UT , Θkq = ωkqpn Up,n .

(5.2.49)

As we know, the elasticities Ckqpn are symmetric in (k, q) and (p, n). However, as equation (5.2.48) shows, even in the geometrically and physically linear case, the incremental constitutive coefficients ωkqpn are not symmetric in (k, q) and (p, n); i.e. generally ωkqpn 6= ωqkpn and ωkqpn 6= ωkqnp . (5.2.50) However, as before, the following symmetry property is true: ωkqpn = ωnpqk .

(5.2.51)

Before analyzing the consequences implied by this symmetry property, as well as those implied by the lack of symmetries (5.2.50), we shall present the updated Lagrangean approach to obtain the incremental field equations.

5.3

Updated Lagrangean approach In the updated Lagrangean approach, the initial deformed equilibrium con◦

figuration B0 = B is taken as reference configuration in order to obtain the linear equations governing the behavior of the incremental fields considered as functions ◦

◦

of x = χ (X), the position vector of a particle P in B . Let us denote the mass density ρ = ρ(x, 0) existing in the initial deformed ◦ ◦ ◦ equilibrium configuration B by ρ = ρ (x). According to the general rule (5.1.11), we have ◦ ◦ ρ (x) J (X) = ρ◦ (X). (5.3.1)

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288

CHAPTER 5. THE THREE-DIMENSIONAL LINEARIZED THEORY The Cauchy’s stress tensor T = T(x, 0), in a similar manner, will be denoted ◦

◦

by T = T (x). Also, for simplicity, the involved body force density b = b(x, 0) ◦ ◦ will be denoted by b = b (x). This notation can be confusing, since in presenting ◦ the Lagrangean approach, the body force density in B was designed by the same ◦

symbol. Then b was considered as depending on the Lagrangean variable X; now ◦ it is taken as a function on the Eulerian variable x = χ (X). Accordingly, we have ◦

◦

◦

◦ −1

b = b (X) = b (χ (x)), and the last composed function was designed again as ◦ ◦ b = b (x)! Taking into account the general Cauchy equation (5.1.10), we can conclude ◦ ◦ ◦ ρ that , T and b satisfy the following equilibrium equation: ◦

◦

◦

divx T (x)+ ρ (x) b (x) = 0.

(5.3.2) ◦

◦

Denoting the involved Cauchy’s stress vector tn = tn (x, 0) by tn = tn (x), according to the general relation (5.1.14), we shall have ◦

◦

tn (x) = T (x)n(x),

(5.3.3)

where n = n(x) is the unit normal to the considered material surface element in ◦

the initial deformed equilibrium configuration B . ◦ ◦ Green’s strain tensor G = G (X) is given by equation (5.2.7), and the sym◦

◦

metric Piola-Kirchhoff stress tensor Π = Π (X) is expressed by the constitutive ◦ relation (5.2.8) as a function of G (X). Hence, using the general rule (5.1.21), for ◦

T, we obtain the following stress-strain relation: ◦

∂ u ◦T (5.3.4) F . ∂G Thus, all field equations involved in the updated Lagrangean approach and concerning the initial deformed equilibrium configuration, were obtained. Let us assume again that to the initial deformed body, at the initial (τ = 0) ◦ −1 ◦ ◦ ◦ T

◦

T=J

◦ −1 ◦

FΠ F = J

F

◦

equilibrium configuration B small time dependent external perturbations are applied, resulting in the current neighboring configuration Bt. The corresponding incremental displacement field u0 = u0 (x, t) will now be denoted for simplicity by u = u(x, t). Thus, using the general notation (5.1.36), and remembering that τ = 0, we shall have J = χ0 (x, t) = x + u(x, t). (5.3.5) Accordingly, we denote by F0 = F0 (x, t) and by H0 = H0 (x, t) the transposed (spatial or updated) gradients corresponding to the incremental motion χ 0 and to the incremental displacement field, respectively; i.e. F0 = F0 (x, t) = [gradx χ0 (x, t)]T , H0 = H0 (x, t) = [gradx u(x, t)]T .

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(5.3.6)

289

5.3. UPDATED LAGRANGEAN APPROACH By J0 = J0 (x, t) we denote the determinant of F0 (x, t); i.e. J0 = J0 (x, t) = det F0 (x, t).

(5.3.7)

F0 (x, t) = 1 + H0 (x, t).

(5.3.8)

Obviously, we have Since χ0 (x, 0)= x, the following equation are true: F0 (x, 0) = 1 and J0 (x, 0) = 1.

(5.3.9)

Taking into account the general equation (5.1.32) and the relations (5.3.7), (5.3.8), we get ◦

◦

F(X, t) = F0 (x, t) F (X) = {1 + H0 (x, t)} F (X) and

(5.3.10)

◦

J(X, t) = det F(X, t) = J0 (x, t) J (X).

(5.3.11)

Here F(X, t) is the gradient of the deformation from the initial reference configuration B to the current one, Bt . In order to get the incremental field equations, applying the updated Lagrangean method, we take as reference configuration the initial deformed equi◦

librium configuration B = B0 . In accordance with the system of notation used, ◦ we designed, by Θ0 = Θ0 (x, t), the nominal stress tensor reported to B taken as reference configuration. Some algebra, the general relations (5.1.21), (5.1.40) and the formula (5.3.10), (5.3.11) show that Θ0 can be expressed by the following equivalent equations: ◦ −1

Θ0 = Θ0 (x, t) = J0 (x, t)F−1 0 (x, t)T(x, t) = J

◦T

◦

T

(X) F (X)Π(X, t) F (X)F0 (x, t). (5.3.12)

In these relations, T = T(x, t) and Π = Π(X, t) are the current Cauchy stress tensor and the current symmetric Piola-Kirchhoff stress tensor, respectively. According to (5.3.9) and (5.3.12)1 , Θ0 (x, t) satisfies the equation ◦

Θ0 (x, 0) = T (x).

(5.3.13)

At the same time the general relation (5.1.44) shows that the updated nominal stress tensor Θ0 (x, t) satisfies the following equation of motion: ◦

◦

··

◦

◦ ··

divx Θ0 (x, t)+ ρ (x)b(x, t) = ρ (x) u (x, t); Θokm,k + ρ bm = ρum ,

(5.3.14)

where b = b(x, t) is the body force density acting in Bt , but reported to the ◦

unit material volume in B taken as reference configuration. Let us denote by

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CHAPTER 5. THE THREE-DIMENSIONAL LINEARIZED THEORY ◦

sno = sno (x, t) the Piola-Kirchhoff stress vector reported to B chosen as reference configuration; sno (x, t) represents a surface force acting in the current configura◦

tion Bt , but reported to the unit material surface in B . According to the general rule (5.1.46), we have sno (x, t) = Θ0 (x,t)n(x)

(5.3.15)

where n = n(x) is the unit normal to the considered material surface element in ◦

B. Let us observe that according to (5.3.3) and (5.3.13), we have ◦

sno (x, o) = tn (x).

(5.3.16)

Now we recall that the applied time-dependent external perturbations, lead◦

ing from B to Bt , are assumed to be small. Consequently, we suppose that H0 = H0 (x, t) satisfies the following restriction: ◦

| H0 (x, t) |<< 1 for x ∈B and t ≥ 0.

(5.3.17)

In this situation, as before, we assume that all products of all perturbations due to H0 (x, t) are negligible with respect to the perturbations. In this approximation, leading to the updated Lagrangean version of the linearized theory, according to the equation (5.3.8), we shall have F0 (t) = 1 + H0 (t), F−1 0 (t) = 1 − H0 (t), J0 (t) = 1 + trH0 (t).

(5.3.18)

In these equations, as well as in the following relations, for simplicity the spatial argument is not mentioned. Also, the small perturbations or increments of various fields will be denoted by a superposed bar. For instance, taking into account the general relation (5.1.8) and the equations (5.2.7), (5.3.10) together with the restriction (5.3.17), in the frame-work of the linearized theory, for the Green’s strain tensor G(t), we get ◦

G(t) = G + G(t),

(5.3.19)

the small perturbation G(t) being given by the equations:

G(t) =

◦ ◦ 1 ◦T 1 ◦T T T F {∇u(t) + ∇u(t) } F . F {H0 (t) + H0 (t)} F = 2 2

(5.3.20)

Analogously, taking into account also (5.3.13), we get ◦

◦

Π(t) = Π + Π(t) and Θ0 (t) = T + Θ0 (t),

(5.3.21)

where Π(t) is the perturbation of the symmetric Piola-Kirchhoff stress-tensor, whereas Θ0 (t) represents the perturbation of the updated nominal stress tensor.

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5.3. UPDATED LAGRANGEAN APPROACH

Now, according to the second relation (5.3.12) and the first formula (5.3.18), from (5.3.20), after some algebra, we get ◦T

◦ −1 ◦

◦

T

F Π(t) F + T H0 (t).

Θ0 (t) = J

(5.3.22)

Consequently, to get the increment Θ0 (t) in terms of the increment H0 (t), we must find the increment Π(t) as a function of H0 (t). To solve the problem, we must use the general constitutive equation (5.1.23) and the relations (5.3.19)−(5.3.21). Thus, following the same procedure as that utilized in the preceding Section, we get   ◦ 1 ◦ ◦T  T (5.3.23) H0 (t) + H0 (t) F , Π(t) = K F 2 ◦

b 4 is the fourth order tensor defined by equation (5.2.20). Since this where K ∈ L tensor has the symmetries (2.2.22), the above equation becomes ◦T

◦

◦

Π(t) = K [F H0 (t) F].

(5.3.24)

Using now the relations (5.3.22) and (5.3.24), we obtain the incremental constitutive equation corresponding to the updated Lagrangean approach ◦ −1 ◦

◦T

◦

◦T

◦

◦

T

F {K [F H0 (t) F]} F + T H0 (t).

Θ0 (t) = J

(5.3.25)

To obtain the component form of this tensor equation, we start with the relation ◦ −1 ◦

Θokq = J

or

◦ −1 ◦

◦

◦T

◦T

◦

◦

T

F kl {K [F H0 F]}lm F mq + T km H omq , ◦

◦

◦T

◦

◦

T

F kl F qm {K [F H0 F]}lm + T km H omq .

Θokq = J

(5.3.26)

Similarly, we have ◦

◦T

◦

◦T

◦

◦

◦

◦T

◦

{K [F H0 F]}lm = K lmnr [F H0 F]nr = K lmnr F np H ops F sr ,

or

◦T

◦

◦

◦

◦

◦

{K [F H0 F]}lm = K lmnr F pn F sr H ops . Introducing the last result in (5.3.26), we get ◦ −1 ◦

Θokq = J

◦

◦

◦

◦

◦

F kl F qm F pn F sr K lmnr H ops + T ks H oqs . ◦

(5.3.27)

Let us introduce now the fourth order tensor c having the following components: ◦ −1 ◦ ◦ ◦ ◦ ◦ ◦ ckqps = J F kl F qm F pn F sr K lmnr . (5.3.28)

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CHAPTER 5. THE THREE-DIMENSIONAL LINEARIZED THEORY

Now the incremental constitutive equation (5.3.27) takes the following form: ◦

◦

Θokq (t) = ckqps H ops (t) + T ks H oqs (t).

(5.3.29)

◦

Also the symmetry properties (5.2.22) of K and the equation (5.3.28) show ◦ that the fourth order tensor c has all symmetries of the usual elasticity tensor c; i.e. ◦ ◦ ◦ ◦ ckqps = cqkps = ckqsp = cpskq . (5.3.30) ◦

Moreover, if we introduce the fourth order tensor ω having the components ◦

◦

◦

ω kqps = ckqps + T ks δqp ,

(5.3.31)

we can express the incremental constitutive equation (5.3.39) in the following more condensed form: ◦ ◦ (5.3.32) Θokq = ω kqps H ops = ω kqps up,s .

Hence, the tensor form of the incremental constitutive equation, appropriate to the updated Lagrangean approach is ◦

Θo = ω ∇uT .

(5.3.33)

◦

In spite of the fact that c has the symmetry properties (5.3.30), generally ω kqps is not symmetric in (k, q) and (p, s); i.e. ◦

◦

◦

◦

◦

ω kqps 6= ω qkps and ω kqps 6= ω kqsp . ◦

However, since c has the symmetry properties (5.3.30) and Cauchy’s stress tensor ◦

◦

T is symmetric, the tensor ω has the following symmetry property: ◦

◦

ω kqps = ω spqk .

(5.3.34)

In order to establish the incremental equation of motion, we assume that ◦

b(x, t) = b (x) + b(x, t),

(5.3.35)

b = b(x, t) representing the small perturbation of the body force density. Taking into account the general equation of motion (5.3.14), the decomposition (5.3.21) 2 and the equilibrium equation (5.3.2), we can conclude that the incremental fields Θ0 , b and u must satisfy the following incremental equation of motion appropriate to the updated Lagrangean approach: ◦

◦

··

◦

◦ ··

divx Θo (x, t) + ρ (x)b(x) = ρ (x) u (x, t), Θokm,k + ρ bm = ρum .

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(5.3.36)

293

5.3. UPDATED LAGRANGEAN APPROACH

Finally, we denote by sn = sn (x, t) the updated incremental Piola-Kirchhoff stress vector; i.e. ◦

◦

sn (x, t) = sn (x) + sn (x, t) = tn (x) + sn (x, t),

(5.3.37)

the last equation following from (5.3.16). Using again the decomposition (5.3.21) and the relation (5.3.3), we get the incremental relation expressing the perturbation sn (x, t) through the perturbation Θo (x, t) and the unit normal n(x)

sn (x, t) = Θ0 (x, t)n(x).

(5.3.38)

As before, further simplification of the incremental field equations can be obtained if we suppose that the initial deformed equilibrium state corresponds to initial applied infinitesimal deformation and the material is linearly elastic; i.e. the restriction (5.2.51) is satisfied. In this case, the field equations concerning the ◦

initial deformed equilibrium configuration B take their form established in the final part of the preceding Section, since the involved incremental fields, corresponding to the Lagrangean or to the updated Lagrangean approach can be considered as ◦ functions of X or as functions of x = χ (X). The equation of motion (5.3.36) involving the incremental fields Θo , b and u, as well as the relation (5.3.38) connecting the incremental fields sn and Θo rests unchanged. Moreover, the unit normal n(x) can be replaced by the unit normal N(X). However, the incremental constitutive equation (5.3.33), concerning Θ o and ∇u takes, as before, a simplified form. Indeed, since now the inequality (5.2.40) ◦ ◦ is true, the components ckqps of the tensor c can be calculated assuming that ◦

F kl = δkl in the relation (5.3.28). Thus, from (5.2.20) and (5.2.41), we get ◦

ckqpm = ckqpn .

(5.3.39)

Consequently, from (5.3.31), and (5.2.44), with (5.2.48), we obtain ◦

◦

◦

ω kqps = ckqps + σ ks δqq = ωkqpn = Ωkqpn .

(5.3.40)

Thus the incremental constitutive equation (5.3.33) becomes

Θokq = ωkqps H ops = ωkqps up,s ,

(5.3.41)

Θo = ω∇uT .

(5.3.42)

or, in tensor form,

Comparing (5.2.48) and (5.3.42), we can conclude that

Θ = Θo .

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(5.3.43)

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CHAPTER 5. THE THREE-DIMENSIONAL LINEARIZED THEORY

To distinguish this important case from the general one, we shall denote Θ = Θo by θ. Hence, if the initial applied deformation is small (infinitesimal), the incremental constitutive equations becomes

θ = ω∇uT or θkq = ωkqps up,s ,

(5.3.44)

the components of ω being given by the relation ◦

ωkqps = ckqps + σ ks δqp ,

(5.3.45)

ckqps being the components of the elasticity tensor c of the considered linearly ◦ ◦ elastic material and σ ks representing the components of Cauchy’s stress tensor σ ◦

describing the stress state in the initial deformed equilibrium configuration B . Summing up the obtained results, we can see that if the initial applied deformation is small and the material is linearly elastic, the Lagrangean approach and the updated Lagrangean approach lead to the same incremental field equation. This conclusion is obviously a natural one and shows the internal consistency of the two distinct ways in which the two variants of the three-dimensional linearized theory were obtained. According to their significance, the two variants are equivalent. As the two equivalent variants of the linearized theory show the incremental ◦

behavior of the body is governed by the constitutive tensor Ω, or equivalently, by ◦ the constitutive tensor ω . At the same time, we observe that the laws governing the incremental behavior of the body are linear. However, generally the incremental boundary value problem can be very difficult, even if the material is homogeneous in its refer◦

◦

ence configuration B, since the constitutive tensors Ω and ω can be complicated ◦ functions of X or of x, respectively, if the initial applied deformation x = χ (X) ◦

◦

is arbitrary, that is if the corresponding gradient of deformation F = F (X) is a complicated function of X. In such a situation, the corresponding incremental equation of motion or of equilibrium lead to linear second order differential systems with variable coefficients. The incremental boundary value problems will be drastically simplified if the initial applied deformation is homogeneous, that is, if ◦

◦

◦

◦

F (X) ≡ F is a constant tensor. In this case the constitutive tensor Ω and ω will also be constant tensors, and the incremental equation of motion or of equilibrium lead to linear second order differential systems with constant coefficients. Further simplifications appear if the initial applied homogeneous deformation is also infinitesimal and the material is linearly elastic, since in this case the involved constitutive tensor ω can be expressed directly in terms of the elasticity tensor c of ◦ the material and of the involved initial applied constant stress tensor σ. However, the boundary value problems involving the incremental fields even in this particular situation are much more complicated than those encountered in the theory of linearly elastic anisotropic solids. Indeed, the stress-strain relation of a linearly

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5.4. GLOBAL AND LOCAL UNIQUENESS. STABILITY

295

elastic generally anisotropic material is completely characterized by, at most, 21 independent elasticities, whereas, in the incremental stress-strain relation, appear generally 45 independent constitutive coefficients ωkqps , since the components of the constitutive tensor ω has only the symmetry properties (5.2.51). However, as we shall see later on, this property plays an essential role in the linearized theory, since it assumes the self-adjointness of the involved incremental field equations, which, at its turn, leads to very useful variational principles, permitting solving approximately various incremental boundary value problems using Ritz type of numerical algorithms. We note that the important symmetry properties (5.2.34), ◦

◦

(5.3.34) and (5.2.51) of the constitutive tensor Ω, ω and ω, respectively, are direct consequences of the fact that the considered material is hyperelastic. Generally, the above mentioned symmetry properties and self-adjointness does not take place in various linearized theories, as was shown for instance by Guz [5.8] in the case of linear viscoelastic materials and by So´os [5.11] in the case of linear piezoelectric crystals. Finally, we observe that the classical elasticity tensor c involved in the linear theory of hyperelastic anisotropic solids is always positive definite. However, the ◦

◦

constitutive tensors Ω, ω and ω, named also instantaneous elasticities, involved in the three dimensional linearized theory of the incremental fields can cease to be positive definite, for some critical values of the initial applied static deformations or stress. In such cases, the instability of the initial deformed and prestressed equilibrium configuration can occur. The three-dimensional linearized theory presented above can be used to detect such dangerous cases. To develop this idea we must clarify what we mean by a stable or unstable equilibrium configuration. In doing this, we follow Pearson [5.12], Hill [5.13], Guz [5.8] and Ogden [5.10].

5.4

Global and local uniqueness. Stability

Following Ogden [5.9] and as a prelude to the discussion of stability, we begin by analyzing the mixed boundary value problem in the framework of nonlinear elastostatics and of linearized (incremental) elastostatics, respectively. Since we consider equilibrium problems, all fields are time independent. We shall use the Lagrangean approach. Consequently, all fields depend on the Lagrangean variable X. We start with nonlinear elastostatics. The gradient F of a static deformation x = χ(X) and Green’s strain tensor G satisfy the following geometric equations: F = ∇χ = [Grad χ]T , G =


1 T F F−1 . 2

(5.4.1)

The nominal stress tensor Θ must satisfy the equilibrium equation corresponding

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CHAPTER 5. THE THREE-DIMENSIONAL LINEARIZED THEORY

to given mass density ρ0 and given body force density b Div Θ + ρ0 b = 0 in B.

(5.4.2)

The nominal stress tensor Θ must satisfy also the hyperelastic constitutive equation corresponding to given specific elastic energy u = u(G) Θ = H(F) =

∂u ∂u (G)FT . (F) = ∂G ∂F

(5.4.3)

The last reduced form of the constitutive equation satisfies the principle of material frame indifference. Consequently, the rotational balance equation is also satisfied. We suppose that S1 and S2 are two complementary subsurface of the boundary ∂B of the body in its stress-free reference configuration B. In mixed boundary value problem on S1 , the boundary condition in displacement must be satisfied, x = χ = V on S1 , (5.4.4) while S2 must be satisfied on the boundary condition in loading or traction sN = ΘT N = L on S2 ,

(5.4.5)

V = V(X) and L = L(X) being a given deformation and a given traction on S 1 and S2 , respectively. Such a loading, which is prescribed and depends only on X, is referred to as dead loading. In any dead loading process, which takes the body from its reference configuration B to its deformed configuration, an increase in the magnitude of the load changes the deformation but, on the other hand, changes in the surface geometry resulting from the deformation do not have the effect of modifying that load. More generally than for dead-loading boundary condition (5.4.5), the traction at a material point P may depend on the deformation χ and possibly on deformation gradient F. Such loading is named configuration dependent. A traction of this type is sensitive to the deformation it has produced, and is modified continually as the deformation process proceeds. This is the reason why Ogden [5.9] names it deformation percutive loading, and Guz [5.5]-[5.8] names it following loading. However, a dead-loading is also named controlled traction. A configuration dependent loading adapts itself to the changing shape of the boundary of the deformed configuration. The loading by a hydrostatic fluid pressure, directed always along the variable unit normal to the boundary of the deforming configuration, represents a typical and important example of deformation sensitive or following loading. The dead loadings are important in applications. As an illustration we consider a circular cylinder of uniform cross-section and with plane ends normal to its axis, consisting of isotropic elastic material. We assume that the circular cylinder is maintained during deformation under the action of nominal compression −σ of magnitude σ > 0 at the ends of the cylinder and parallel to the cylinder axis. The lateral surface is taken free of traction. Figure 5.2.(a) illustrates a possible deformed configuration corresponding to contraction of the cylinder. Another type

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5.4. GLOBAL AND LOCAL UNIQUENESS. STABILITY

297

of possible deformed configuration arises if, at some critical value of σ, the cylinder buckles under the applied compressive stress −σ and the cylinder bows (for example), as illustrated in Figure 5.2(b).

Figure 5.2: Example of traction boundary condition, showing possible deformed configurations of a circular cylinder of uniform cross-section under uniaxial nominal compression of magnitude σ: (a) uniform contraction; (b) buckled configuration. The analyzed case provides an example of nonuniqueness of the solution of the traction-boundary value problem. In the given example, the deformation path (in the space of deformation gradients) is said to bifurcate at the bifurcation point, corresponding to the critical value of the load σ. Beyond the bifurcation, two paths of deformations are possible, as shown by (a) and (b) in Figure 5.2. We discuss now some questions concerning the uniqueness of the mixed boundary value problem of nonlinear elastostatics. ∼

Let χ, χ be two possible solutions of (5.4.1)−(5.4.5), with corresponding de∼ ∼ ˜ body forces b, b on B, surface formation gradients F, F, nominal stresses Θ, Θ, ∼

tractions L, L on S2 and with the same displacement condition on S1 . Then by using the divergence theorem together with (5.4.2), (5.4.4) and (5.4.5), we obtain    T Z v v T dV Θ −Θ · F −F

Z

∼  ∼  ρ0 b −b · χ −χ dV B Z   ∼  ∼ + L −L · χ −χ dA.

=

B

(5.4.6)

S2

If ∼

∼

b = b in B and L = L on S2

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(5.4.7)

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CHAPTER 5. THE THREE-DIMENSIONAL LINEARIZED THEORY

it follows from (5.4.6) that Z  B

   T ∼ ∼ Θ −Θ · F −FT dV = 0.

(5.4.8)

∼

Then, two distinct solutions χ and χ necessarily satisfy (5.4.6) or (5.4.8), as appropriate. We usually say that χ is a kinematically admissible deformation field if χ is a regular vector field on B = B ∪ ∂B, and if χ satisfies the given displacement boundary condition. Let us suppose now that there exists a solution χ to the mixed boundary value problem. From (5.4.8) it follows that this solution is unique if there is no ∼ distinct kinematically admissible deformation field χ for which (5.4.8) holds; i.e.    T Z ∼ ∼ (5.4.9) Θ −Θ · F −FT dV 6= 0 B

∼

for all χ 6≡ χ satisfying (5.4.4). A sufficient condition for uniqueness is that   T  Z ∼ ∼ Θ −Θ · F −FT dV > 0

(5.4.10)

B

∼

for all kinematically admissible deformation fields χ, χ 6≡ χ. An equally valid condition sufficient for uniqueness is provided by (5.4.10) with the inequality reversed. However, in the theory of linearly elastic solids, the inequality (5.4.10) takes place, since the elasticity tensor c is positive definite. Hence, the second alternative must be rejected. A stronger inequality, which implies (5.4.10), is    T  ∼ ∼ Θ −Θ · F −FT > 0 in B (5.4.11) ∼

for all pairs F, F6≡ F (not necessarily gradients of deformation fields). If the elastic material is of Cauchy’s type; i.e. Θ = H(F)

(5.4.12)

where H is a given constitutive function, which is not derived from a specific elastic energy as in (5.4.3), the restriction (5.4.11) states that the constitutive function H, considered as an application depending on F, is a strictly convex function. The buckling of elastic bars shows that uniqueness is not generally expected. Hence, the inequalities (5.4.10) and (5.4.11) are too restrictive for the Cauchy’s type elastic constitutive function H.

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5.4. GLOBAL AND LOCAL UNIQUENESS. STABILITY

For a hyperelastic material the inequality (5.4.11), assuming uniqueness, is replaced by the following restriction concerning the specific elastic energy u = u(F), considered as function of F: ∼T

∼

u(F) − u(F) − Θ · (F −FT ) > 0, with Θ =

∂u (F), ∂F

(5.4.13)

∼

for all F, F 6= F. This condition states that the constitutive function u(F), considered as function of F, is a strictly convex scalar valued function. ∼ By reversing the roles of F and F in (5.4.13)1 and adding the result to (5.4.13)1 , it is easy to see that the restriction (5.4.13) implies the restriction (5.4.11). The converse is not true in general. However, as the above observation shows, the restriction (5.4.13) assures uniqueness for Green’s type hyperelastic materials. By integration of (5.4.13) over B and application of the divergence theorem, with the equation of equilibrium (5.4.7) and boundary conditions (5.4.4), (5.4.5), we have Z Z Z Z Z Z ∼ ∼ ∼ u(F)dV − ρ0 b · χdV − L · χdA. u(F)dV − ρ0 b· χ dV − L· χ dA > B

B

B

S2

B

S2

(5.4.14) Now let us introduce the potential energy functional Φ defined on the set A of all kinematically admissible deformation fields χ by the equation Z Z Z Φ(χ) = u(F)dV − ρ0 bdχdV − L · χdA. (5.4.15) B

B

S2

This inequality (5.4.14) tells us that ∼

Φ(χ) > Φ(χ).

(5.4.16)

Also we can state that if χ is a solution and if the inequality (5.4.16) holds ∼ for all kinematically admissible deformation fields χ 6≡ χ, then uniqueness is guaranteed (see Pearson [5.12] and Hill [5.13]). Moreover let δχ be a (regular) variation of χ, vanishing on S1 . We denote the variation of Φ in χ and in the direction δχ by δΦ. The use of the divergence theorem and of the constitutive equation (5.4.1) gives Z Z δΦ = − (Div Θ + ρ0 b) · δχdV + (ΘT N − L) · δχdA. B

S2

This equation proves the validity of the following variational principle: The kinematically admissible deformation χ is a solution of the mixed boundary value problem (5.4.2)−(5.4.5) if and only if the variation δΦ of Φ vanishes in

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CHAPTER 5. THE THREE-DIMENSIONAL LINEARIZED THEORY

χ for all admissible directions δχ, that is, for all regular variations δχ vanishing on S1 . As in the linear case, this is appropriate to Φ, as the potential energy functional corresponding to the external dead loading system (b, L), and to the above variational principle as the principle of stationary potential energy. Following Pearson [5.12], Hill [5.13], Guz [5.5] , [5.8] and Ogden [5.9], we turn now to the stability interpretation of the restriction (5.4.14), which is written as Z Z Z Z ∼ ∼ ∼ ρ0 b · (χ −χ)dV + L · ( χ −χ)dA. (5.4.17) u(G)dV − u(G)dV > B

B

B

S2

∼

Let χ correspond to an actual solution and let χ be an arbitrary kinematically admissible deformation. The above inequality tells us that the variation of the total elastic energy, in moving from the equilibrium configuration χ to the configuration ∼ χ, exceeds the work done by the prescribed dead body forces and tractions. We say that the equilibrium configuration χ is stable if the above condition is fulfilled ∼ for all configurations χ 6≡ χ. Expressed in the form (5.4.16), the relation (5.4.17) shows that for a stable equilibrium configuration χ, the potential energy functional Φ is minimized by χ in the set A of all kinematically admissible deformations or configurations. In fact, following Owen [5.8], we shall say that an equilibrium configuration χ is stable if the weaker inequality ∼

Φ(χ) ≥ Φ(χ)

(5.4.18) ∼

holds for all kinematically admissible deformations χ, with the possibly of equality ∼ holding for some χ 6≡ χ. It follows that stability does not generally imply uniqueness. Conversely, from (5.4.9) it results that a unique equilibrium configuration need not be stable. However, it is not of practical interest to study uniqueness of unstable equilibrium configurations. The inequality (5.4.18) is a global sufficient condition for stability of the equilibrium configuration determined by χ. It is too restrictive to be regarded as a stability criterion; i.e. as a necessary and sufficient condition for stability of χ. If, ∼ ∼ however, χ is combined to some small neighborhood of χ; i.e. if U = χ − χ is an incremental displacement field, then (5.4.18) is a suitable local stability criterion, often referred to as infinitesimal stability criterion. Just this criterion shall be discussed in the context of linearized theory. Also, we shall examine the question of uniqueness in the framework of the linearized theory. To do this we assume that ◦

the initial deformed equilibrium configuration B corresponds to the deformation ◦

x = χ (X),

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(5.4.19)

5.4. GLOBAL AND LOCAL UNIQUENESS. STABILITY

301

solution of the following mixed boundary value problem: ◦

◦

◦

◦

T F = ∇ χ = [Grad χ] , G =

1 ◦T ◦ (F F −1), 2 ◦

∂ u ◦T Div Θ + ρ0 b = 0, Θ = F , ∂G ◦

◦

◦

◦

◦

◦T

◦

(5.4.20)

◦

x = χ = V on S1 , sN = Θ N = L on S2 . We assume now that in B an increment b(X) of the body force is given, on S1 an increment V(X) of the displacement is prescribed, and on S2 an increment L(X) of the dead loading is imposed. In the following, for simplicity, in writing the above increments, we shall neglect the superposed bars, accordingly, the increment Θ of the nominal stress tensor will be simply denoted by Θ. The resulting neighboring equilibrium configuration B 0 corresponds to the incremental displacement field

U = U(X), solution of the following mixed incremental boundary value problem ◦

Div Θ + ρ0 b = 0, Θ = Ω ∇UT in B, T

U = 0 on S1 , sN = Θ N = L on S2 .

(5.4.21) (5.4.22)

We examine first the question of uniqueness of the mixed incremental bound∼ ary value problem. Let us suppose that U and U are two possible solutions, with ˜ respectively. Let their difcorresponding incremental nominal stresses Θ and Θ, ferences be denoted by ∼ ∼ b = Θ − Θ. b = U − U, Θ (5.4.23) U b and Θ b correspond to the following It follows from (5.4.21) and (5.4.22) that U homogeneous and linear mixed incremental boundary value problem: ◦

b T in B, b = 0, Θ b = Ω ∇U Div Θ b = 0 on S1 , ΘN b = 0 on S2 . U

(5.4.24) (5.4.25)

By use of the divergence theorem, the homogeneous equilibrium equation and boundary conditions lead to Z b · ∇UdV b Θ = 0. (5.4.26) B

With the inverted constitutive equation (5.4.24)2 , the last equation can be replaced by the relation Z ◦ b T dV = 0. b Ω ∇U (5.4.27) ∇U· B

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Hence, by analogy with (5.4.10), the inequality Z

B

b · ∇UdV b Θ > 0 or

Z

◦

B

b Ω ∇U b T dV > 0 ∇U·

(5.4.28)

b satisfying (5.4.25)1 , with ∇U b 6≡ 0, suffices for true for all admissible fields U, uniqueness of the solution of the mixed incremental boundary value problem. Following the pattern of the first part, we consider next the inequality: ◦

b Ω ∇b ∇U· uT > 0 in B

(5.4.29) ◦

b 6≡ 0. If the instantaneous elasticity Ω satisfies in B, the above for all fields ∇U inequality, the uniqueness, of the mixed incremental boundary value problem is assured. According to the general definition, if the inequality (5.4.29) takes place, ◦

the instantaneous elasticity Ω is positive definite in B. Using the incremental constitutive equation (5.4.24)2 , we can express the restriction (5.4.29) in the form b · ∇U b >0 in B Θ

(5.4.30)

b 6≡ 0. for all fields ∇U b = 0, then Θ b = 0 and It follows from (5.4.24)2 and (5.4.30) that if ∇U ◦

conversely. Thus, if the instantaneous elasticity Ω is positive definite in B, the b and the corresponding relation between the incremental displacement gradient ∇U b is one-to-one. incremental nominal stress Θ As the buckling of elastic bars proves uniqueness is not to be expected in general for incremental boundary value problems of the three-dimensional linearized ◦

theory, and the positive definiteness of the instantaneous elasticity Ω may fail for ◦ certain initial applied deformation F. In the first part of this Section, we discussed the stability interpretation of the sufficient conditions for uniqueness in the context of nonlinear elastostatics. We now examine the significance of the global sufficient condition for stability ◦

(5.4.8) for incremental displacements from the initial deformed configuration B . ∼ ◦ ∼ We set, as before, U = χ − χ. If χ is a kinematically admissible deformation, the incremental displacement field U must vanish on S2 ; i.e. U = 0 on S2 .

(5.4.31)

A long, but elementary computation, using Taylor expansion, shows that ◦ ◦ ◦ ∼ 1 3 u(F) = u(F) + Θ ·∇UT + ∇U· Ω ∇UT + O(| ∇U | ). 2

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(5.4.32)

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5.4. GLOBAL AND LOCAL UNIQUENESS. STABILITY

By use of the divergence theorem, the equilibrium equation (5.4.20)3 and the boundary condition (5.4.20)5,6 , (5.4.31) shows that the second order term in ∇U is Z ◦ 1 ∼ ◦ (5.4.33) ∇U· Ω ∇UT dV. Φ(χ) − Φ(χ) = 2 B Thus, according to (5.4.18), the local criterion of stability of the equilibrium ◦

configuration B may be expressed as Z ◦ ∇U· Ω ∇UT dV ≥ 0

(5.4.34)

B

for all incremental displacement fields U which vanishes on S1 . Using the incremental constitutive equation (5.4.21)2 , the restriction (5.4.34) can be equivalently expressed as Z Θ · ∇UdV ≥ 0. (5.4.35) B

It follows also from (5.4.31) or (5.4.35) that the potential energy functional Φ ◦ is rendered a local minimum by a locally stable equilibrium configuration χ within ∼ the considered class of kinematically admissible deformations χ. If equality holds in (5.4.34) for some ∇U 6≡ 0 with U = 0 on S1 , then the local stability is said to be neutral. If the integral (5.4.31) is negative for some ∇U ◦

with U = 0 on S1 , we say that the equilibrium configuration B is unstable. b in (5.4.28) and U in (5.4.34) satisfy the Since the admissible vector fields U same homogeneous boundary condition on S1 , it follows that the restriction placed ◦

on Ω by (5.4.28) and the strict form of (5.4.34); i.e. Z ◦ ∇U· Ω ∇UT dV > 0

(5.4.36)

B

are identical. However, because equality is permitted in (5.4.31), local stability doesn’t generally imply incremental uniqueness. Although inequality (5.4.28) 2 implies local stability, it is not necessary for incremental uniqueness. Therefore, incremental uniqueness, in general, does not entail local stability. The situation is similar to that encountered in discussing global uniqueness and stability. ◦ We note also that if an equilibrium configuration χ is unique in the incremental sense of (5.4.34), it fails to be unique in the global sense. Thus, (5.4.10) ∼ ◦ may hold for χ in some neighborhood of χ, there may exist some kinematically ∼ admissible deformations χ outside that neighborhood for which (5.4.8) holds. This follows, since (5.4.36) cannot hold for every configuration. We suppose now that the arbitrary incremental displacement fields U and ∼ U are associated through the incremental constitutive law (5.4.21)2 with the in˜ respectively, at the same initial deformed cremental nominal stresses Θ and Θ, ◦

equilibrium configuration B .

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CHAPTER 5. THE THREE-DIMENSIONAL LINEARIZED THEORY Thus, ◦

∼

◦

∼T

Θ = Ω ∇UT and Θ = Ω ∇ U .

(5.4.37) ◦

From the symmetry property (5.2.24) of the instantaneous elasticity Ω, we get the local Betti type symmetry relation ∼

∼

Θ · ∇ U = Θ ·∇U.

(5.4.38)

∼

∼

Let b and b be the incremental body forces and let L, L be the incremental ∼ ∼ surface loadings on S2 , associated with U and U, respectively. We take V and V to be the prescribed incremental displacements on S1 . By integration of (5.4.38) over B, use of the divergence theorem, the incremental equilibrium equations and boundary conditions, results in Z Z ∼ ∼ Θ ·∇UdV = Θ · ∇ U dV B B Z Z Z ∼T ∼ ∼ = ρ0 b ·UdV + V· Θ NdA + L ·UdA (5.4.39) B

=

Z

S1

∼

ρ0 b· U dV +

B

Z

S

∼

T V ·Θ NdA +

Z2

∼

L · U dA,

S2

S1

∼

for all solutions U and U of the corresponding mixed incremental boundary value problems. Equations (5.4.39) are referred to as Betti’s reciprocal theorems in the framework of the three-dimensional linearized theory. Its validity relies on the selfadjointness of the governing incremental field equations embodied in the symmetry ◦

of the instantaneous elasticity Ω. The classical Betti theorem of linear elastostatics, ◦

given in Section 2.3, is included in (5.4.39) as a special case, for which Ω is replaced by c, since we have assumed that the reference configuration B is stress free; i.e. ◦ σ = 0 in B. We stress the fact that there is no global nonlinear counterpart of the Betti’s ∼

reciprocal theorem, since the nominal stresses Θ and Θ of the nonlinear elastostatics do not satisfy a symmetry relation as (5.4.38); i.e. generally ∼T

∼

Θ · F 6= Θ · FT even if the material is hyperelastic! Returning to the linearized theory, by following the procedure that led to ∼ (5.4.39) with U = U, we obtain the following work relation: Z Z Z Z Θ · ∇UdV = ρ0 b · UdV + L · UdA + V · ΘT NdA. (5.4.40) B

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B

S2

S1

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5.4. GLOBAL AND LOCAL UNIQUENESS. STABILITY

The above equality is the incremental analogue of the classical work relation given in Section 2.3 and encountered in the linear elastic theory. To obtain a variational principle, appropriate to the three-dimensional lin∼

∼

∼

∼

earized theory, we take U to be a variation δU of U with Θ = δΘ, L = δL, b = δb ∼ and V= 0. The substitution in (5.4.39) leads to the equation Z Z δΘ · ∇UdV = Θ · ∇δUdV B ZB Z Z = ρ0 δb · UdV + U·δLdA + V · δΘT NdA B

=

Z

S

ρ0 b·δUdV + B

Z2

S1

L · δUdA.

(5.4.41)

S2

By making use of the symmetry property (5.2.24) of the instantaneous elastic◦

ity Ω, hence of the self-adjointness of the incremental field equations, from (5.4.41) we obtain  Z  Z Z 1 Θ · ∇UdV = ρ0 b · δUdV + L · δUdA, δ 2 B B S2

for all variations δU vanishing on S1 . Moreover, since b and L are dead-loadings, the above equation can be written in the equivalent form   Z Z  1Z (5.4.42) Θ · ∇UdV − ρ0 b · UdV − L · UdA = 0 δ  2 B B S2

for all variations δU vanishing on S1 . Let us denote by A the set of all kinematically admissible incremental displacement fields, that is, the set of all regular incremental displacement fields that satisfies the given displacement boundary condition on S1 . Let the energy functional Φ be defined on A by the following relation: Z Z Z ◦ 1 (5.4.43) ∇U· Ω ∇UdV − ρ0 b · UdV − L · UdA. Φ(U) = 2 B B S2

As the relation (5.4.42) shows, we have the following incremental variational principle. A solution (unique or not) U of the mixed incremental boundary value problem makes the energy functional Φ stationary on A. The converse is also true, and can be easily established by rearranging (5.4.42) in the form Z Z − δU · (Div Θ + ρb) dV + δU · (ΘT N − L)dA = 0. (5.4.44) B

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S2

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Now it is easy to see that if the variation δΦ of Φ is vanishing in a kinematically admissible incremental displacement field U, for all variation δU of U, vanishing on S1 , then U is a regular solution of the mixed incremental boundary value problem. The above variational principle, appropriate to the linearized theory, is useful to obtain approximate solutions for various incremental boundary value problems. We call the functional Φ incremental potential energy corresponding to the external incremental dead loading system {b, L}. It plays a role similar to that played by the usual potential energy of the classical linear elasticity, and is analyzed in Section 2.4. Now let us assume that the sufficient condition (5.4.28) for incremental uniqueness is satisfied. Let us suppose that U is the unique solution of the mixed incremental boundary value problem (5.4.21), (5.4.22). We consider also an arbib vanishing on S1 ; i.e. trary incremental displacement field U Then

b = 0 on S1 . U

(5.4.45)

e =U+U b U

is a kinematically admissible incremental displacement field. Using the definition (5.4.43) of Φ, the divergence theorem, the incremental equilibrium equation and boundary conditions (5.4.21), (5.4.22), the restriction (5.4.45) and the symmetry ◦

property (5.2.34) of the instantaneous elasticity Ω, after elementary computations, we get Z ◦ b Ω ∇UT dV , e = Φ(U) + 1 ∇U· (5.4.46) Φ(U) 2 B

e ∈ A. for any U Consequently, according to the sufficient incremental uniqueness condition (5.4.28), we can conclude that e > Φ(U) Φ(U)

(5.4.47)

e 6≡ U. Hence, for any kinematically admissible incremental displacement field U the incremental potential energy Φ is minimized by the unique solution U on the set A of all kinematically admissible incremental displacements. Summing up, we can conclude that the incremental variational principle can be strengthened by a minimum principle of the incremental potential energy at initial deformed equilibrium configurations, for which the incremental uniqueness holds. Such equilibrium configurations are locally stable, since (5.4.29) implies (5.4.31), and hence (5.4.34). In the above we have discussed the questions concerning global and local uniqueness and stability using the Lagrangean approach. The same questions can be analyzed using the updated Lagrangean approach. The problems P5.19−P5.24 show how this can be done.

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5.5. DYNAMIC CRITERIA OF STABILITY

5.5

Dynamic criteria of stability

Suppose that the dead-load mixed boundary value problem (5.4.20) with data ◦

◦

◦

◦

b, V, L has a solution χ determining the initial deformed equilibrium configuration ◦

B . We examine now in detail the inhomogeneous mixed incremental boundary value problem (5.4.21), (5.4.22) corresponding to the increments (perturbations) b, V, L. Let us denote by A the set of all regular incremental displacement fields which vanish at S1 , the part of the boundary ∂B on which the incremental displacement is prescribed. Also, we define on A the functional E by the following rule: Z ◦ E(U) = ∇U· Ω ∇UT dV for any U ∈ A. (5.5.1) B

The incremental displacement fields from A will be named admissible and, for reasons which will be clear a little later on, E will be named the exclusion functional . From the results of the preceding Section, the solution of the mixed incremental boundary value problem is unique if E(U) > 0 for any U from A and U 6≡ 0 on B.

(5.5.2)

Also, if the above condition is fulfilled, the initial deformed equilibrium con◦

◦

figuration B , or equivalently χ, is locally stable. Let us consider now the homogeneous mixed incremental boundary value problem (5.4.21), (5.4.22); i.e. the boundary value problem corresponding to null external perturbations (b = V = L = 0). We have ◦

Div Θ = 0, Θ = Ω ∇UT on B,

(5.5.3)

U = 0 on S1 , ΘT N = 0 on S2 . Obviously, the vanishing incremental displacement field U ≡ 0 on B

(5.5.4)

is always a solution of the homogeneous incremental boundary value problem (5.5.3). If the exclusion functional E has the property (5.5.2), U ≡ 0 in B is the only solution. Hence, the underlying dead-loading boundary problem (5.4.20) has ◦ ◦ a unique solution, χ itself, in some neighborhood of χ. ◦

For certain initial deformed equilibrium configurations B , however, there may exist an U 6≡ 0 satisfying the homogeneous incremental boundary value problem

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CHAPTER 5. THE THREE-DIMENSIONAL LINEARIZED THEORY ◦

◦

(5.5.1). In other words, there exists an Ω for some B such that Z ◦ E(U) = ∇U· Ω ∇UT dV = 0,

(5.5.5)

B

U 6≡ 0 being a solution of the homogeneous incremental boundary value problem (5.5.3). If this happens, the solution of the mixed incremental boundary value problem (5.4.21), (5.4.22) is definitely not unique. ◦

An equilibrium configuration B in which the incremental uniqueness fails, is called by Ogden [5.8] an eigen configuration, or an e-configuration for short. Hill [5.13] uses the term eigenstate for an e-configuration. A solution U 6≡ 0 of the homogeneous incremental boundary value problem in an e-configuration is called an eigenmode or e-mode. The inequality (5.5.2) ensures that the considered equilibrium configuration ◦

B is not an e-configuration. Because (5.5.2) excludes e-modes, E is referred to as the exclusion functional for the considered dead-load problem. The divergence theorem and equations (5.5.3) show that if there exists an e-mode U, then Z Z Z ◦ T T E(U) = ∇U· Ω ∇U dV = U · Θ NdA − U · Div ΘdV = 0, (5.5.6) B

B

S2

confirming (5.5.5). The variational counterpart of the relation (5.5.6) is Z Z Z ◦ 1 1 δU · Div ΘdV = 0 (5.5.7) δE = δ ∇U· Ω ∇UT dV = δU · ΘT NdA − 2 B 2 B S2

for all admissible variations δU of U; i.e. for all regular variations which vanish on S1 . Consequently, an e-mode makes the exclusion functional E stationary within the class of all admissible variations. The equations (5.5.7) show that the converse is also true: any admissible incremental displacement field U 6≡ 0 that makes the exclusion functional stationary within the class of admissible variations, is an e-mode. These results are special cases of the incremental variation principle of the preceding subsection. ◦ An e-mode U defines an adjacent equilibrium configuration χ0 = χ + U (i.e. ◦

◦

◦

χ0 is near χ) under the same given loads {b, L}. Since the body, when it is in an eigenstate, would remain in equilibrium if it is slightly displaced in the manner of the eigenmode, we can say that the body is in a neutral equilibrium state and its stability is neutral with respect to such displacements.

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5.5. DYNAMIC CRITERIA OF STABILITY

309

As in any eigenvalue problem, because of the linearity and homogeneity of the incremental boundary value problem (5.5.3), an e-mode U has arbitrary magnitude. It should be remembered that the equation of the linearized theory are valid only if U is incremental in the sense that products of small perturbations can be neglected. The magnitude of an e-mode U must therefore be restricted accordingly. Summing up, we can say that when a body there is in an e-configuration, it remains in equilibrium if it is subjected to an eigenmodal displacement. Moreover, the potential energy functional Φ, defined by the equation (5.4.43), ◦ is stationary not only for the underlying configuration χ but also for each equi◦ ◦ e = χ + U adjacent to χ and defined by an eigenmode U. librium configuration χ This can be seen by noting (5.5.5)−(5.5.7) and recalling from (5.4.33) that Z ◦ 1 1 ◦ ◦ (5.5.8) ∇U· Ω ∇UT dV = E(U). Φ(χ +U) − Φ(χ) = 2 2 B

Now we consider a deformation path generated by some monotonically increasing loading parameter (for example, contained in L). Let us suppose that along the path, the exclusion functional E(U) is positive definite; i.e. (5.5.2) holds for all admissible U, up to some critical value of the parameter in question, beyond which the exclusion functional is indefinite. By continuity, it follows that the set of configurations, where the exclusion functional is positive definite, is bounded by configurations where it is semidefinite; i.e. where Z ◦ E(U) = ∇U· Ω ∇UT dV ≥ 0 (5.5.9) B

for all admissible U, with equality holding for some U 6≡ 0. According to (5.5.9), the exclusion functional E has a local minimum in which U 6≡ 0, hence E is stationary in U for all admissible variations δU. Consequently, U 6≡ 0 is necessarily an e-mode and thus the equilibrium configurations where the exclusion functional is semidefinite are necessarily e-configurations. This is in contrast with the situation for e-configurations where the exclusion functional is indefinite because an admissible incremental displacement U may make the exclusion functional vanish but not be an e-mode; i.e. may not make the exclusion functional stationary. Such e-configurations are locally unstable and therefore of no practical interest. Eigen configurations with the semidefiniteness property (5.5.9) are called primary e-configurations since they are the first e-configurations to be reached on a stable path of deformation of the type considered above. As Hill says, the involved primary e-configurations are at the stability limit. Observe also that according to the accepted classification, in the linearized theory all primary e-configurations are neutrally stable. The above results show that the three dimensional linearized theory can and must be used just to detect the primary eigen configurations and to determine the critical values of the loading parameters for which such configurations occur.

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Since the relations (5.4.28) and (5.4.36) are equivalent, the criticality of the exclusion functional is independent of the nonhomogeneous incremental data b, L, V. It follows that the solutions of the inhomogeneous or homogeneous incremental boundary value problems (5.4.21), (5.4.22) or (5.5.3), respectively, are unique at any stage along the considered stable deformation path, up to a primary e-configuration, at which uniqueness fails and the solution path bifurcates (as for the buckling of a compressed elastic bar). ◦

◦

The critical configuration χC or equivalently B C is named a bifurcation point on the deformation path. In order to emphasize the close connection between the homogeneous and ◦ nonhomogeneous incremental problems we suppose that U is an e-mode at χC ◦ ◦ so that both χC and χC + U are solutions of the underlying dead-load problems ◦

◦

◦

corresponding to critical data bC , LC , VC . The nonhomogeneous incremental ◦ problem at χC , with L 6= 0, b = 0, V = 0, for example, corresponds to L be◦

◦

ing taken beyond its critical value LC up to LC + L. Let W be a solution of this problem. Then W + U is also a solution of the same nonhomogeneous incremental problem, U containing an arbitrary scalar factor. Note however that L is constrained by Z L · UdA = 0 (5.5.10) S2

◦

at χC (when b = V = 0 as assumed). This follows from Betti’s reciprocal relae = W and U the e-mode. Thus, the incremental dead loading tion (5.4.41) with U L in a e-configuration is orthogonal, in the sense of (5.5.10), to each e-mode at that configuration. ◦ By contrast with (5.5.10), since W is not an e-mode at χC , we have Z

S2

L · Wda =

Z

◦

∇W· Ω ∇WT dV > 0. B

If (5.5.10) holds for some L 6≡ 0 on S2 , then W exists and bifurcation occurs. To the first order, L has no component in the direction U (in the sense of (5.5.10)) and therefore the considered loading parameter has a turning point with respect to ◦ the e-mode direction at χc . Any incremental load L that violates (5.5.10) cannot be supported in equilibrium by the material. The branching behavior of a deformation path at the critical configuration ◦ χc is shown in Figure 5.3, which shows the stable path of deformation up to χ c , ◦ and the two solution branches through χc of the local problems, the ”tangents” at ◦ these path at χc being W and W + U. The behavior of the body after branching cannot be predicted by the linearized theory.

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5.5. DYNAMIC CRITERIA OF STABILITY

Figure 5.3: Stable solution path (continuous case) bifurcates at the e-configuration ◦ ◦ χc into two branches (broken curves) with tangents W and W + U at χc , where ◦ U is an e-mode at χc . ◦

◦

Now let us assume that the initial deformation x = χc (X) is homogeneous; ◦

◦

i.e. the corresponding gradient of deformation F = F (X) is a constant tensor ◦

◦

F = F (X) = const. in B. ◦

(5.5.11)

Since the material is homogeneous, it results that the instantaneous elasticity

Ω is a constant tensor ; i.e. ◦

◦

Ω = Ω (X) = const. in B.

(5.5.12)

For all-round dead load, i.e. for the incremental traction boundary value problem (S1 = ∅, S2 = ∂B), we can take as admissible incremental displacement field U, one which has constant gradient on B. Thus, we can see that for all-round dead load, the restriction (5.5.2) is equivalent to ◦

∇U· Ω ∇UT > 0

(5.5.13)

for all gradients ∇U 6≡ 0. In a primary e-configuration, we must have ◦

∇U· Ω ∇UT ≥ 0

(5.5.14)

e 6≡ 0 necessarily holding at each for all (constant) ∇U with equality for some ∇U point of the body. Moreover, the incremental constitutive equation (5.4.21) 2 shows e is subjected to the stationarity property that the corresponding e-mode U ◦

e = Ω ∇U e T = 0 in B Θ

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(5.5.15)

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CHAPTER 5. THE THREE-DIMENSIONAL LINEARIZED THEORY

in a primary e-configuration and for all-round dead loading, since the left-hand e side of (5.5.14) has a local minimum in the corresponding e-mode U. e with a uniform (conIt is also clear that any incremental displacement U stant) nonvanishing gradient satisfying (5.5.15) is an e-mode, but not necessarily a primary one, for all-round dead loading. ◦

Hence, if the initial deformation F is homogeneous (constant), the local sta◦ ◦ bility of the resulting equilibrium configuration B or equivalently χ, relative to any boundary condition implies the positive semidefiniteness of the corresponding ◦

instantaneous elasticity Ω. To obtain this property, it is sufficient to assume local stability for all-round dead loading; i.e. for incremental traction boundary value problems. Up to now we have used a static, energetic criterion of stability. To justify its necessity and sufficiency, dynamical concepts and stability criteria must be taken into account. To do this, we take as fundamental the definition and dynamic test of stability due to Dirichlet and Kelvin (see Hill [5.13]): an equilibrium configuration or state is said to be stable if, in motion, following an arbitrary perturbation, the amplitudes of the additional displacements and velocities are always vanishingly small when the disturbance itself is. If, on the other hand, the amplitudes and velocities are finite for some type of disturbance, however small this may be, the configuration or state is said to be unstable. Hence, we must prove that for a hyperelastic material, a necessary and sufficient condition of stability in the dynamic sense is that in any geometrically possible displacement from the position of equilibrium, the stored elastic energy exceeds the work done on the system by the external dead loads. In thermodynamic context, this question was analyzed by Ericksen [5.14] and Gurtin [5.15]−[5.18]. We shall use Gurtin’s way of reasoning, remaining however in the hyperelastic framework. Let us assume that the time independent fields ρ0 = ρ0 (X), b = b(X), V = V(X) and L = L(X) represent a given mass distribution in B, a given dead body force density in B, a given deformation on S1 , and a given dead traction on S2 , respectively. An elastic process corresponding to ρ0 , b, V, L is an ordered array p = [χ, Θ, u] with χ a regular motion, Θ a nominal stress tensor and u a specific elastic energy defined on B × [0, ∞), such that: (i) the geometric equations F = ∇χT , G =

1 T (F F − 1); 2

(5.5.16)

(ii) the equation of motion ¨ DivΘ + ρ0 b = ρ0 χ;

(5.5.17)

(iii) the hyperelastic constitutive equation Θ=

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∂u ∂u (G)FT (G) = ∂G ∂F

(5.5.18)

313

5.5. DYNAMIC CRITERIA OF STABILITY u = u(G) being a known regular constitutive function; (iv) the mixed boundary conditions χ = V on S1 and sN = ΘT N = L on S2 .

(5.5.19)

Using the geometrical equation (5.5.16)1 and the constitutive equation (5.5.18) for the material time derivative of u, we obtain u˙ = Θ · ∇χ˙ T . Denoting by U = U (χ) =

Z

(5.5.20)

u(G)dV

(5.5.21)

B

the total elastic energy corresponding to χ, and integrating (5.5.20) over B, resulting in Z U˙ (χ) = Θ(t) · ∇χ˙ T (t)dV. (5.5.22) B

Using the divergence theorem, the equation of motion (5.5.17) and the boundary conditions (5.5.19), we can express the material time derivative U˙ of the total elastic energy in the following form, valuable for the considered elastic process Z Z ◦ ρbo · χdV ˙ + L · χdA ˙ − K (χ), ˙ (5.5.23) U˙ (χ) = B

S2

where K(χ) ˙ =

1 2

Z

B

ρ0 χ˙ · χdV ˙

(5.5.24)

is the total kinetic energy of the body corresponding to χ. Integrating the equation (5.5.23) from 0 to t, we obtain Z Z ρ0 b · (χ(t) − χ(0))dV − L · (χ(t) − χ(0))dA U (χ(t)) + K(χ(t)) ˙ − B

S2

= U (χ(0)) + K(χ(0)). ˙ ◦

(5.5.25)

◦

As before, we denote by B or equivalently χ, an initial deformed equilibrium ◦ configuration corresponding to the deformation x =χ (X). ◦

◦

By a kinematically admissible deformation, corresponding to B or χ, we mean now a regular deformation χ with the following properties: (i) ◦ χ =χ on S1 ; ◦

◦

(ii) if χ 6≡χ, then χ and χ are not related by a rigid deformation; i.e. ◦

χ(X) 6= Q χ (X) + a,

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CHAPTER 5. THE THREE-DIMENSIONAL LINEARIZED THEORY

where Q is an orthogonal tensor and a is a vector. On the set of all kinematically admissible deformations χ, we introduce the energy functional Z Z ◦ ◦ Ψ(χ) = U (χ) − U (χ) − ρ0 b · (χ − χ)dV ˙ − L · (χ− χ)dA. (5.5.26) B

S2

◦

We call Ψ(χ) the potential energy of χ, relative to χ, and corresponding to the given dead-load system {b, L}. Note that ◦ Ψ(χ) = 0. (5.5.27) ◦

Let us suppose for the moment, that χ and χ are related by a rigid deformation; i.e. ◦ χ(X) = Q χ (X) + a. ◦

Then G = G and, since u depends on χ only through G, from (5.5.21) it ◦ follows that U (χ) = U (χ). Hence, if b ≡ 0 and if L ≡ 0 on S2 or if S2 = ∅, equation (5.5.26) reduces to Ψ(χ) = 0. (5.5.28) The next theorem is based on the assumption that Ψ(χ) has a strict local ◦ minimum at χ. Using equation (5.5.27), for this assumption to have a meaning we ◦ must rule out situations in which equation (5.5.28) holds for χ 6≡ χ. The requirement (ii) in the definition of a kinematically admissible deformation is introduced with this purpose in mind. Note that (ii) follows from (i) when S2 6= ∂B! We denote by K the set of all kinematically admissible deformations and we assume that a metric d : K × K → R+ is introduced on K. For instance, we can take Z  ◦

1/2

◦

d(χ, χ) =

B

| χ− χ| dV

,

but it is not necessary to specify the form of d, which in applications may vary from problem to problem. We denote by K(a) the following set of kinematically admissible deformations ◦

K(a) = {χ ∈ K : d(χ, χ) = a}, where a is a positive number. We define also the number ψ(a) by the equation ψ(a) =

inf Ψ(χ).

χ∈K(a)

Following Gurtin, we say that the potential energy Ψ has a strict local min◦ ◦ imum at χ, over K and, with respect to d, if there exists a number a > 0 such that ◦ (5.5.29) ψ(a) > 0 for 0 < a
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5.5. DYNAMIC CRITERIA OF STABILITY ◦

If Ψ has strict local minimum at χ, then ◦

Ψ(χ) ≥ ψ{d(χ, χ)} > 0

(5.5.30)

◦

provided that χ ∈ K and d(χ, χ) < a0 . Examining the relations (5.4.17) and (5.5.26), we can see that for the above choice of χ, the classical static, energetic condition of stability holds, if Ψ has a ◦ strict local minimum at χ. By a kinematically admissible motion, we mean a motion x = χ(X, t) with the following properties: (i) for each fixed t ≥ 0, the deformation x = χ(X, t), denoted simply by χ(t), is kinematically admissible; ◦ (ii) the application t → d(χ(t), χ) is continuous. Now we can prove a theorem due to Gurtin stating the sufficiency of the classical static, energetical criterion of stability for dynamic stability. We have the following: ◦ Sufficiency theorem. Assume that Ψ has a strict local minimum at χ. Then given ε > 0 there exists a δ = δ(ε) > 0 such that, if p is an elastic process with the following properties, (i) the motion x = χ(X, t) in p is kinematically admissible; i.e. ◦

χ(X, t) =χ (X) on S1 ; (ii)p satisfies the initial constrains ◦

d(χ(0), χ) < δ, Ψ(χ(0)) < δ, K(χ(0)) ˙ < δ,

(5.5.31)

then p satisfies ◦

d(χ(t), χ) < ε, K(χ(t)) ˙ < ε for all t ≥ 0.

(5.5.32) ◦

That is, if the initial perturbation of the equilibrium state B is small, in the sense that (5.5.31) takes place, the perturbation rests small for all times, ◦

in the sense that (5.5.32) is satisfied. Hence, the equilibrium configuration B is stable according to the Dirichlet-Kelvin dynamic criterion of stability, interpreted quantitatively only in the sense specified by the inequalities (5.5.31) and (5.5.32). For the proof of the above theorem, we chose ε > 0 and, without loss in generality, let ◦ (5.5.33) ε < a, and take δ = min

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ε ψ(ε) , 2 2



.

(5.5.34)

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CHAPTER 5. THE THREE-DIMENSIONAL LINEARIZED THEORY

By (5.5.29), δ > 0. Let p be an elastic process satisfying (i) and (ii). Thus equation (5.5.25) holds for t ≥ 0. In this way, considering the equation (5.5.26) for t = 0 and for t > 0, after elementary computations, we get Ψ(χ(t)) + K(χ(t)) ˙ = Ψ(χ(0)) + K(χ(0)). ˙

(5.5.35)

Since K(χ) ˙ ≥ 0, from (5.3.31), (5.5.34) and the above equality, we get Ψ(χ(t)) < ψ(ε).

(5.5.36)

In view of the relations (5.5.31)1 , (5.5.32) and (5.5.33), ◦

◦

d(χ(t), χ) < ε < a . Let us assume that the time ◦

τ = sup{t : d(χ(t), χ) < ε} ◦

is finite. Since the applications t → d(χ(t), χ) is continuous, we get ◦

◦

d(χ(τ ), χ) = ε < a, and we can conclude from (5.5.30) that ◦

ψ(ε) = ψ(d(χ(τ ), χ)) ≤ Ψ(χ(τ )), which contradicts (5.5.36). Therefore, τ = ∞ and (5.5.32)1 holds. Moreover, (5.5.31)1 and (5.5.33) imply that Ψ(χ(t)) > 0 for all t ≥ 0 and, in this way, the relations (5.5.34), (5.5.35) yield (5.5.32)2 . Hence indeed, the stability in static, energetic sense implies stability in dynamic sense. We now analyze if reciprocal of this property takes place. In order to do this, we consider an elastic process p with the following properties: ◦ (i) the motion x = χ(t) in p is kinematically admissible relative to χ; (ii) p starts from rest; i.e. χ(0) ˙ = 0 in B. Hence, the initial total kinetic energy of the body in this process is vanishing; i.e. K(χ(0)) ˙ = 0. According to (5.5.35) for elastic processes of this type, the current value of ◦ the potential energy Ψ, relative to χ, cannot surpass its initial value; i.e. Ψ(χ(t)) ≤ Ψ(χ(0)).

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(5.5.37)

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5.5. DYNAMIC CRITERIA OF STABILITY

Following Gurtin, by the domain of attraction of the equilibrium configura◦ tion χ we mean the set of all configurations K with the following property. There exists an elastic process p such that ◦ (i) the motion x = χ(t) in p is kinematically admissible relative to χ; (ii) p starts from rest; (iii) p starts from the configuration K; i.e. χ(0) = K in B; ◦

(iv) p end is the equilibrium configuration χ; i.e. ◦

◦

χ(t) →χ , ∇χ(t) → ∇ χ as t → ∞, the convergence being uniform in B. Note that from (i) and (iii), it follows that ◦

K =χ on S1 , ◦

so that K is kinematically admissible relative to χ. Now we are ready to prove a theorem due to Gurtin which shows that in a certain precise sense, the static, energetic criterion of stability supplies a necessary condition for the dynamic elastic stability. We have the following: ◦ Necessity theorem. Let K belong to the domain of attraction of χ. Then, ◦

Ψ(χ) ≤ Ψ(K).

(5.5.38)

To prove this theorem, we use the fact that if K is in the domain of attraction ◦ of χ, then there exists an elastic process p with properties (i)-(iv). According to (i), (ii) and (5.5.38), for the motion x = χ(t) in this process, we have Ψ(χ(t)) ≤ Ψ(K) for t ≥ 0. The last inequality in conjunction with (iv) implies (5.5.37) and the necessity theorem is proved. We recall that according to the static, energetic criterion of local stability, an ◦

equilibrium configuration B is locally stable if other equilibrium configurations do not exist in its neighborhood, and is locally unstable if there exist other equilibrium configurations in its neighborhood. Then, if we use the static, energetic method, the analysis of the stability problem is reduced to searching for the bifurcation points of a deformation path. This is the reason why the static stability analysis is founded on the homogeneous, linear and mixed incremental boundary value problem (5.5.3) of the three-dimensional linearized theory. According to the dynamic ◦

criterion of local stability, the equilibrium configuration B is locally stable if the incremental time dependent fields produced by small time depended perturbations

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◦

of B remain small for any time. Obviously, if we wish to apply the more general dynamic criterion, the homogeneous system (5.5.3) governing the static behavior of the incremental fields must be replaced by the corresponding incremental field equations which take into account the time dependence of the involved fields; i.e. assuming null body forces, the stability analysis must be based on the use of the following dynamic incremental linear and homogeneous system: ◦

¨ Θ = Ω ∇UT , Div Θ = ρ0 U, U = 0 on S1 , sN = ΘT N = 0 on S2 .

(5.5.39)

In dynamic analysis, it is frequently assumed that the small perturbation of ◦

the underlying equilibrium configuration B corresponds to small oscillations in the neighborhood of the analyzed equilibrium state (see for instance Guz [5.5], [5.8], Ogden [5.9] and Truesdell and Noll [5.19]). In this case, according to the ◦

Dirichlet-Kelvin dynamic criterion of stability, B is locally stable if the amplitude of the involved oscillation converges to zero when the time goes to infinity, or when it is a function which varies periodically in time. In the stability analysis of this type, most frequently it is assumed that the time dependence of the involved iΩt oscillations is described by the √ exponential function e , where Ω represents the complex frequency and i = −1. For brevity, we shall call the method based on this assumption the dynamic frequency method. In the framework of this method, the governing system (5.5.39) takes the following simplified homogeneous form: ◦

Div (Ω ∇UT ) + ρ0 Ω2 U = 0 in B, ◦

U = 0 in S1 , N· Ω ∇UT = 0 on S2 ,

(5.5.40)

where, for simplicity, the amplitude U = U(X) of the incremental displacement was denoted by the same symbol U as for the time dependent displacement U = U(X, t). Thus, using the dynamic frequency method, to study local stability we are led to the following eigenvalue problem: those values of the complex frequency Ω must be determined for which there exist nonvanishing solutions of the homoge◦

neous, linear system (5.5.40). We recall that the instantaneous elasticity Ω depends on the material properties of the body, as well as on the initial deformation ◦ χ leading to the reference configuration B to the initial deformed equilibrium con◦

figuration B . In its turn, this deformation is determined by the loading parameters. Consequently the possible eigenvalues ρ0 Ω2 depend on the mass distribution of the body, on the material properties of the body, on the loading parameters, on the geometric characteristics of the body, and through the complementary surfaces S 1 and S2 , on the imposed displacement and dead traction conditions. According to Dirichlet-Kelvin dynamic stability criterion, in the framework of ◦

the dynamic frequency method, the initial deformed equilibrium configuration B

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5.5. DYNAMIC CRITERIA OF STABILITY

is locally stable if the imaginary part Im Ω of the complex frequency Ω is positive; i.e. if Im Ω > 0; (5.5.41) ◦

the equilibrium configuration B is unstable if Im Ω < 0;

(5.5.42)

and the boundary of the stability domain is determined by the condition Im Ω = 0.

(5.5.43)

Since Ω depends on the loading parameters, the last condition determines ◦

the critical values for which local instability of the equilibrium configuration B can occur. In a natural way, we must now raise the following question: what is the connection between the critical values of the local parameters determined using the static (energetic) criterion of local stability, and those obtained using the above presented dynamic frequency method. To find the answer, we first prove the following. Theorem. The eigenvalues ρ0 Ω2 of the eigenvalue problem (5.5.40), corresponding to the dynamic frequency method, are always real numbers. For the proof of this property, we assume that ρ0 Ω2 is a complex number and denote, by a superposed bar, the complex conjugation. ◦

Let ρ0 Ω2 and U be a solution of the eigenvalue problem (5.5.40). Since Ω is 2 a real valued tensor, we can conclude that the complex conjugate quantities ρ 0 Ω and U satisfy the homogeneous linear system ◦

T

2

Div (Ω ∇U ) + ρ0 Ω U = 0 in B, ◦

T

U = 0 on S1 , N· Ω ∇U = 0 on S2 .

(5.5.44)

Multiplying equation (5.5.40)1 by U and by U, respectively, and subtracting the obtained results, we get ◦

◦

T

2

U · Div Ω ∇UT − U · Div Ω ∇U + ρ0 (Ω2 − Ω )U · U = 0.

We integrate this equation on B, use the divergence theorem and the homogeneous boundary conditions (5.5.40)2,3 and (5.5.44)2,3 . Thus we find Z Z ◦ ◦ T 2 (∇U· Ω ∇UT − ∇U· Ω ∇U )dV = ρ0 (Ω2 − Ω ) U · UdV. B

B

◦

Since the instantaneous elasticity Ω has the symmetry property (5.2.34), the left-hand side of the above equality is vanishing. Hence we get Z 2 2 ρ0 (Ω − Ω ) U · UdV = 0. B

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CHAPTER 5. THE THREE-DIMENSIONAL LINEARIZED THEORY

We recall that U is an eigenvector corresponding to the eigenvalue ρ0 Ω2 ; hence, U 6≡ 0 on B. Consequently, the integral in the last equation is nonvanishing. Thus we must have 2 (5.5.45) ρ0 Ω2 = ρ 0 Ω .

Taking into account the above result, we can conclude that all eigenvalues ρ0 Ω2 are real numbers and the theorem is proved. Let us observe that the above property is true since the instantaneous elasticity has the symmetry property (5.2.34); i.e. the involved differential operator is self-adjoint. For instance, for viscoelastic materials or piezoelectric crystals, the above property generally does not take place. According to equation (5.5.45), Ω can be either only a real number, or only an imaginary number; it cannot be a complex number, only if Ω itself is vanishing. As the dynamic frequencies criterion shows, the loss of stability takes place for those critical values of the loading parameters for which Im Ω just pass from positive to negative values. Since Ω 6= 0 cannot be a complex number, assuming continuous dependence of Ω on the loading parameters, we can conclude that in the moment of loss of stability, Ω must be vanishing, as shown in Figure 5.4.

Figure 5.4: Continuous line: possible path in the complex frequency plane; interrupted lines: impossible paths in the complex frequency plane. Hence, the boundary of the stability domain is determined for those critical values of the loading parameters for which Ω = 0.

(5.5.46)

In other words, according to the frequency method, the equilibrium config◦

uration B becomes locally unstable when the eigenvalue problem (5.5.40) has a solution for zero eingevalue. But in this case, the eingenvalues problem (5.5.40) corresponding to the frequency method and the eigenvalue problem (5.5.3) corresponding to the static (energetic) method become identical. This fact shows that the following theorem due to Guz [5.5], [5.8] is true.

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5.5. DYNAMIC CRITERIA OF STABILITY

Equivalent theorem. The static (energetic) method, founded on the static (energetic) criterion of local stability, and the dynamic frequency method, founded on the general dynamic criterion of the local stability, lead to the same critical values of the loading parameters. In the above precise sense, we can say that the static (energetic) criterion of local stability is necessary and sufficient for the dynamic stability, if the material is hyperelastic. An important time-dependent perturbation, frequently used in stability analysis, is a disturbance named progressive or harmonic plane wave, defined by the equation U = Aei(K·X−Ωt) . (5.5.47) Here A, K = KN are constant vectors, N · N = 1 and Ω is a constant number. A is the amplitude of the wave, K represents the wave number, the real unit vector N characterizes the direction of propagation of the wave, and Ω is the frequency. A, K and Ω can be complex quantities; if this is the solution, the incremental displacement U is the real part of the expression from the right-hand side of equation (5.5.47). The quantity U 2 = Ω2 /K 2

(5.5.48)

is the square of the velocity if propagation of the wave. We recall that U 2 is always ◦

a real number, since the instantaneous elasticity Ω is symmetric; i.e. satisfies the property (5.2.34). In general, a progressive plane wave cannot exist in a hyperelastic material, except when the equations describing the incremental motion have constant coefficients. This condition is fulfilled if the initial applied deformation is homogeneous and we assume that this is the case. Also, the displacement (5.5.47) must be compatible with the boundary condition, as, for example, for an infinite medium and we suppose that this is the case. Substitution of (5.5.47) into (5.5.39)1,2 yields ◦

2 Ωklmn Nk Nn Am = ρ0 U Al . ◦

(5.5.49)

◦

The tensor Q = Q (N), having the components ◦

◦

Qlm (N) = Ωklmn Nk Nn

(5.5.50)

represents the acoustic tensor, corresponding to the direction of propagation N. Now the system (5.5.49) becomes ◦

Q (N)A = ρ0 U 2 A. The last equation represents the propagation condition.

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(5.5.51)

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CHAPTER 5. THE THREE-DIMENSIONAL LINEARIZED THEORY ◦

From the symmetry of Ω, it follows that the acoustic tensor is symmetric for any direction of propagation. Equation (5.5.51) determines the possible wave amplitudes and wave speeds for any given direction of propagation in the prestressed material. Since the acoustic tensor is symmetric for a hyperelastic material, the eigenvalues ρ0 U 2 of equation (5.5.51) are real and there exists a triad of mutually orthogonal real eigenvectors A1 , A2 , A3 for each N. However, the wave speed is real for all N if and only if ρ0 U 2 > 0 (5.5.52) for any direction of propagation. From (5.5.51), it results ◦

A· Q (N)A , A 6= 0. ρ0 U = A·A 2

(5.5.53)

Hence, the condition (5.5.52) is fulfilled for any direction of propagation if and ◦

only if the acoustic tensor Q (N) is positive definite for any N; i.e. if and only if ◦

A· Q (N)A > 0 for any A 6= 0 and any N.

(5.5.54)

Taking into account (5.5.50), we can express the last condition in the following equivalent form: ◦

(AN)T · Ω (AN ) > 0 for any AN 6= 0,

(5.5.55)

where AN is the tensor product of the vectors A and N.

◦

If this restriction is fulfilled, we say that the instantaneous elasticity Ω is strongly elliptic. Since not every second order tensor has the form AN, strong ellipticity of the instantaneous elasticity generally does not imply its positive definiteness. Con◦

versely, if Ω is positive definite, it is also strongly elliptic. When the instantaneous elasticity is strongly elliptic, the corresponding incremental equilibrium equation is also named strongly elliptic or briefly elliptic, in accordance with the usual terminology of the theory of partial differential equations. Positive definiteness of the instantaneous elasticity implies the ellipticity of the incremental equation of equilibrium. Conversely, if the instantaneous elasticity ceases to be positive definite, the incremental equation of equilibrium can lose its ellipticity. Since the possible wave speeds are eigenvalues of the acoustic tensor, they are determined from the characteristic equation ◦

det (Q (N) − ρ0 U 2 1) = 0. ◦

(5.5.56) ◦

Since Q (N) is symmetric, any eigenvalue ρ0 U 2 is real. If Ω is strongly elliptic, any eingenvalues is positive, hence all possible velocities of propagation are real.

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5.6. HOMOGENEOUS INITIAL DEFORMATIONS ◦

Particularly, if Ω is positive definite, the initial deformed equilibrium configuration ◦

B is locally stable, all possible velocities of propagation, for any direction, are real. In this case, the amplitudes of all possible incremental plane waves rest small at any time, if they were small at the beginning. In this way we can see that the use of the plane waves offers one advantage in that the nature of roots of the characteristic equation is reflected obviously in a dynamic criterion of stability. First, a negative root leads to a purely imaginary value of U , say U = ±iα, where α > 0. Since the amplitude A is real, a suitable contribution of two solutions of the type (5.5.47) leads to a particular real solution of the form U(X, t) = AeKαt cos K · X, K, α > 0, (5.5.57) the value of which increases without limit as t → ∞. Let us observe also that if 0 ◦ is an eigenvalues of Q (N), the corresponding solution is time independent. Summing up, we can say that, unless all the eingenvalues of the acoustic ◦

tensor Q (N) are nonnegative, there exists in an infinite prestressed hyperelastic medium certain infinitesimal perturbations which amplify with time, according to the three-dimensional linearized theory. Conversely, if the instantaneous elasticity ◦

Ω is positive definite, the prestressed, initial deformed equilibrium configuration ◦ B is locally stable, every plane wave has a real frequency, and its amplitude rests small at all times. We can see again the intimate connection existing between stability in static and dynamic sense, for a hyperelastic material. This does not mean that we should necessarily impose, as a condition to the constitutive function and to the initial deformation, that is, to the corresponding instantaneous elasticity, that the acoustic tensor be such as to have only positive eigenvalues. Rather, the above results indicate that for a given material, any initial ◦

deformed homogeneous configuration giving a negative eigenvalue to Q (N) is not likely to be encountered in practice, since, in such a state, there are certain kinds of perturbation which, though small in magnitude initially, begin to grow and then destroy the given equilibrium configuration. Perhaps in this way, physical instability, such as buckling, ultimately may come to be explained.

5.6

Homogeneous initial deformations We assume the initial applied deformations to be homogeneous. In this case, ◦

◦

the instantaneous elasticities Ω, ω and ω are constant tensors. We assume also vanishing body force. Since the instantaneous elasticities are constant tensors, the incremental equations of equilibrium are a linear system of second order partial differential equations having constant coefficients. In this case, using the associated matrix method, the incremental displacement fields can be expressed by displacement potentials, all of which satisfy the same linear partial differential

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equation with constant coefficients. We shall illustrate the procedure assuming linearly hyperelastic materials and small initial deformations. With appropriate modifications, many of the results which will be presented in this particular, but important case, rest true even if the material is hyperelastic but nonlinear, and the initial applied deformation is large. The case of a nonlinear, isotropic hyperelastic material is analyzed by Guz in his monographies [5.2]−[5.8] in the case in which the initial applied homogeneous deformation is large, but the mixed components of the corresponding Green strain tensor are vanishing. Since the initial applied deformation is small, the involved fields can be considered as functions of X or of x. We shall use the second variant. In the considered case, the involved instantaneous elasticity is ω and its components are given by the equations (5.2.48). For simplicity the initial applied displacement field will be ◦ ◦ denoted by u = u (x) and the incremental displacement field by u = u (x). We recall that if the initial applied deformation is small, the Cauchy’s stress tensor, ◦ ◦ corresponding to the initial applied deformation, is designed by σ = σ(x) and ◦ ◦ the corresponding infinitesimal strain tensor by ε = ε (x). To stress the fact that the initial applied deformation is small, we shall denote the incremental nominal stress tensor by θ = θ (x), the incremental displacement imposed on the part S 1 of the boundary ∂B by v = v (x) and the incremental dead traction imposed in the complementary part S2 of the boundary by l = l (x). Taking into account the introduced simplified notations, and recalling that the material is linearly hyperelastic (physical linearity) and the initial applied deformation is small (geometrical linearity), on the basis of the results presented in the last part of the Section 5.2, we can conclude that the incremental static behavior of the body, if we consider the mixed incremental boundary value problem, is governed by the following incremental field equations: the incremental equilibrium equation div θ = 0 or θkl,k = 0, k,l = 1, 2, 3 in B;

(5.6.1)

the incremental constitutive equation θ = ω∇uT or θkl = ωklmn um,n , k,l,m,n = 1, 2, 3;

(5.6.2)

the expressions of the components of the instantaneous elasticity ◦

ωklmn = cklmn + σ kn δml ,

(5.6.3)

where cklmn are the components of the elasticity tensor c of the considered linear, ◦ ◦ hyperelastic material and σ kn are the components of the Cauchy’s stress tensor σ corresponding to the initial deformation; the mixed incremental boundary conditions u = v or uk = vk on S1 and sn = θT n = l or snl = θkl nk = ll on S2 ,

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(5.6.4)

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5.6. HOMOGENEOUS INITIAL DEFORMATIONS

where v = v(x) is a given incremental displacement on S1 , l = l(x) is a given incremental dead traction on S2 and n = n(x) is the outward unit normal to S2 , sn being the incremental Piola-Kirchhoff stress vector. We recall that the instantaneous elasticity ω is symmetric; i.e. its components satisfy the following relation: ωklmn = ωnmlk .

(5.6.5) ◦

◦

We recall also that the infinitesimal strain tensor ε = ε (x), corresponding to ◦ ◦ the initial applied infinitesimal displacement u = u (x), is given by the classical geometrical relation    1 ◦ 1 ◦T ◦ ◦ ◦ ◦ um,l + ul,m . (5.6.6) or εml = ∇ u +∇ u ε= 2 2 ◦

The initial stress σ, existing in the initial deformed equilibrium configuration, is given by the classical stress-strain relation ◦

◦

◦

◦

σ = c ε or σ kn = cknml εml .

(5.6.7)

Let us observe that all functions depend on the coordinates x1 , x2 , x3 of the particles in the reference configuration of the body and all partial derivatives are considered relative to these coordinates. We recall also that the above used approximation is permitted since the initial applied deformation is small; i.e. ◦ ◦ (5.6.8) ∇ u << 1 or um,n << 1 for m, n = 1, 2, 3. ◦

Ending this introductory part, we recall that Cauchy’s stress vector tn acting ◦ in the initial deformed configuration B on a material surface element with unit normal n is given by the classical relation ◦

◦

◦

◦

tn = σ n or tnk = σ kl nl .

(5.6.9) ◦

Actually, the last equations can be used to determine the traction tn which must be applied on the boundary of the body in order to produce the initial ◦ ◦ displacement u = u (x); i.e. the initial applied deformation of the material. We recall also that Piola’s and Kirchhoff’s incremental stress vector sn , acting on a material surface element with unit normal n and produced by the superposed incremental deformation of the body, can be calculated using the relation sn = θ T n or snl = θkl nk .

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(5.6.10)

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Now we are ready to discuss the displacement potential problem. Since ωklmn are constant quantities, the component form (5.6.1) of the incremental equation of equilibrium can be written in the following equivalent form: Plm um = 0,

(5.6.11)

where Plm are differential operators defined by the equation Plm = ωklmn

∂2 . ∂xk ∂xn

(5.6.12)

Let us introduce the 3 × 3 matrix P = [Plm ]

(5.6.13)

and let us denote by det P its determinant. Also, we design by P −1 the inverse of the matrix P . We recall the following relation, well known in matrix calculus:


∂ det P = P −1 mj det P, ∂Pjm

(5.6.14)


where P −1 mj are the components of the matrix P −1 . Let us consider now three regulated scalar valued functions ϕ(j) = ϕ(j) (x1, x2, x3 ) , j = 1, 2, 3, (j)

and, using the associated matrix method, let us assume that the components u m of the incremental displacement u(j) corresponding to ϕ(j) are expressed by the following equations: ∂ det P (j) ϕ . (5.6.15) u(j) m = ∂Pjm

According to (5.6.12), we get −1 u(j) m ={ P




mj

det P }ϕ(j) .

Introducing the above expression in the operational form (5.6.10) of the incremental equilibrium equation, we successively obtain
(j) Plm um = {Plm P −1 mj det P }ϕ(j) = (δlj det P ) ϕ(j) = ( det P ) ϕ(l) .

Hence, the incremental equilibrium equation (5.6.3) will be satisfied by the incremental displacement fields u(j) given in (5.6.15), if and only if the three functions ϕ(j) satisfy the same differential equation (det P ) ϕ(j) = 0 for j = 1, 2, 3.

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(5.6.16)

5.6. HOMOGENEOUS INITIAL DEFORMATIONS

327

The functions ϕ(j) , given u(j) through the relation (5.6.15) and satisfying the same differential equation (5.6.16), will be called incremental displacement potentials or briefly displacement potentials. The usefulness of the displacement potentials results from the fact that all of them satisfy the same differential equation. In general, in the three-dimensional case (5.6.16), it is a complicated sixth order partial differential equation with constant coefficients. Moreover, even if the displacement potentials satisfy the same differential equation, in the boundary condition (5.6.2) all three potentials appear simultaneously and the problem actually rests coupled in this sense. We can say that even if we use the displacement potentials to solve incremental boundary value problems, generally the mathematical problem to be solved remains extremely complex. Considerable simplification can be achieved only assuming supplementary symmetry properties of the material, particular forms of the initial imposed deformation, supplementary symmetry properties of the geometry of the body and particular forms of the imposed surface displacements and tractions. Concerning the symmetry properties of the material, we shall assume that the material is orthotropic (see Section 2.2), the symmetry planes being the coordinate planes used up to now. The assumption made is sufficiently general to study the stability behavior of a large class of composite materials submitted to various loading conditions. Since the material is linearly elastic and orthotropic, its elasticity tensor c is characterized by the relation (2.2.19) of the Section 2.2. Concerning the particular form of the initial imposed deformation, we shall ◦ assume that the initial imposed displacement u has the following components: ◦

◦

◦

u1 = (λ1 − 1) x1 , u2 = (λ2 − 1) x2 , u3 = (λ3 − 1) x3 ,

(5.6.17)

where λ1 , λ2 , λ3 are given constant quantities. Since the initial deformation is small, λ1 , λ2 , λ3 must satisfy the condition |λk − 1| << 1 for k = 1, 2, 3.

(5.6.18)

The assumption made is sufficiently general to study buckling of a large class of composite materials, as anisotropic bars and plates, or composite laminates. According to (5.6.5) and (5.6.18), only the diagonal components of the initial ◦ infinitesimal strain ε are nonvanishing, and we have ◦

◦

◦

◦

ε11 = λ1 − 1, ε22 = λ2 − 1, ε33 = λ3 − 1, εkl = 0 for k 6= l. Since (5.6.18) takes place, we have ◦ ◦ ◦ ε11 , ε22 , ε33 << 1.

(5.6.19)

(5.6.20)

Using the relation (2.2.19) and the above results, we can conclude that only ◦ the diagonal components of the initial stress σ are nonvanishing and we have the

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CHAPTER 5. THE THREE-DIMENSIONAL LINEARIZED THEORY

relations

◦

◦

◦

◦

◦

◦

◦

◦

σ 11 = C11 ε11 + C12 ε22 + C13 ε33 , σ 22 = C21 ε11 + C22 ε22 + C23 ε33 , ◦ ◦ ◦ ◦ ◦ σ 33 = C31 ε11 + C32 ε22 + C33 ε33 , σ kl = 0 for k 6= l. (5.6.21) Analyzing attentively the structure (2.2.19) of the elasticity tensor c and using the relations (5.6.3) and (5.6.21) we can conclude that the nonvanishing components of the instantaneous elasticity ω are given by the following equations:

◦

◦

◦

ω1111 = C11 + σ 11 , ω2222 = C22 + σ 22 , ω3333 = C33 + σ 33 , ω1122 = ω2211 = C12 = C21 , ω2233 = ω3322 = C23 = C32 , ω3311 = ω1133 = C13 = C31 , ω1212 = ω2121 = C66 , ω2323 = ω3232 = C44 , ω3131 = ω1313 = C55 , ◦

◦

◦

◦

◦

◦

(5.6.22)

ω2112 = C66 + σ 22 , ω1221 = C66 + σ 11 , ω3223 = C44 + σ 33 , ω2332 = C44 + σ 22 , ω1331 = C55 + σ 11 , ω3113 = C55 + σ 33 .

The determination of instantaneous elasticities must be made carefully, since a single, little mistake can lead to completely erroneous results. The behavior of an orthotropic linearly elastic material is governed by 9 independent elasticities. However, even in the special case analyzed by us, the incremental behavior of the same material is governed by 15 independent instantaneous elasticities. Moreover, the elasticity tensor c is always positive definite, but the instantaneous elasticity tensor ω can lose its positiveness if instabilities occur. It is clear that the incremental boundary value problems are much more complicated as those encountered in the usual linear theory at anisotropic elastic bodies. Using the incremental constitutive equation (5.6.2) and the relations (5.6.22), we can express the incremental nominal stress θ through the gradient ∇u of the incremental displacement field. Simple computations lead to the following relations

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5.6. HOMOGENEOUS INITIAL DEFORMATIONS

329

expressing the components of θ in terms of the components of ∇u:   ◦ θ11 = ω1111 u1,1 + ω1122 u2,2 + ω1133 u3,3 = C11 + σ 11 u1,1 + C12 u2,2 + C13 u3,3 ,   ◦ θ12 = ω1212 u1,2 + ω1221 u2,1 = C66 u1,2 + C66 + σ 11 u2,1 ,   ◦ θ13 = ω1313 u1,3 + ω1331 u3,1 = C55 u1,3 + C55 + σ 11 u3,1 ,   ◦ θ21 = ω2112 u1,2 + ω2121 u2,1 = C66 + σ 22 u1,2 + C66 u2,1 ,   ◦ θ22 = ω2211 u1,1 + ω2222 u2,2 + ω2233 u3,3 = C21 u1,1 + C22 + σ 22 u2,2 + C23 u3,3 ,   ◦ θ23 = ω2323 u2,3 + ω2332 u3,2 = C44 u2,3 + C44 + σ 22 u3,2 ,   ◦ θ31 = ω3113 u1,3 + ω3131 u3,1 = C55 + σ 33 u1,3 + C55 u3,1 ,   ◦ θ32 = ω3223 u2,3 + ω3232 u3,2 = C44 + σ 33 u2,3 + C44 u3,2 ,   ◦ θ33 = ω3311 u1,1 + ω3322 u2,2 + ω3333 u3,3 = C31 u1,1 + C32 u2,2 + C33 + σ 33 u3,3 . (5.6.23) It is easy to see that θ12 6= θ21 , θ23 6= θ32 , θ31 6= θ13 . As we have said, further simplifications can be achieved introducing supplementary assumptions concerning the geometry of the body. We assume that the body is a long cylinder, having generators parallel to the Ox3 axis. In this case, the components of the outward unit normal n to the lateral surface Sl of the cylinder are characterized by the following equations: n1 = n1 (x1 , x2 ) , n2 = n2 (x1 , x2 ) , n3 ≡ 0 on Sl .

(5.6.24)

We suppose also that the imposed incremental displacement v and the imposed incremental dead traction l on the lateral surface of the cylinder satisfy the following restrictions: v1 = v2 ≡ 0, v3 = v3 (x1 , x2 ) on Sl1 , l1 = l2 ≡ 0, l3 = l3 (x1 , x2 ) on Sl2 ,

(5.6.25)

where Sl1 and Sl2 are complementary lateral subsurfaces on Sl . In such conditions in an orthotropic cylinder, without initial deformations, if it is sufficiently long relative to its transverse dimensions and the end effects can be neglected, an antiplane state can arise relative to the plane Ox1 x2 , characterized by the following relations: u1 = u2 ≡ 0, u3 = u3 (x1 , x2 ) in B.

(5.6.26)

As it is well known, even if the above conditions concerning the geometry of the body and the applied surface displacements and tractions are fulfilled, in

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an elastic cylinder having general anisotropy cannot appear an antiplane state, that is, the relation (5.6.26) are not fulfilled. We have seen that the incremental stress-strain relations are much more complicated as those encountered in the usual theory of orthotropic materials. Hence, we must verify if the assumption (5.6.26), expressing the existence of an antiplane state, is compatible with the incremental equilibrium equations (5.6.1) and with the imposed external constraints (5.6.25), on the lateral surface Sl of the cylinder. To analyze this compatibility problem, first of all we observe that if the conditions (5.6.26) are fulfilled, according to (5.6.23) the components of the incremental nominal stress depend only on x1 , x2 , and have the following expressions: θ11 ≡ 0, θ21 ≡ 0, θ31 = ω3131 u3,1 , θ12 ≡ 0, θ22 ≡ 0, θ32 = ω3232 u3,2 , θ13 = ω1331 u3,1 , θ23 = ω2332 u3,2 , θ33 ≡ 0.

(5.6.27)

Now it is the easy to see that the first two equilibrium equations (5.6.1) 2 are identically satisfied, and the third equilibrium equation becomes θ13,1 + θ23,2 = 0 or ω1331 u3,11 + ω2332 u3,22 = 0 in S,

(5.6.28)

where S denotes the plane domain bounded by the closed curve Γ, representing the generatrix of the cylinder. Taking into account the relations (5.6.10), (5.6.24) and (5.6.27) on the lateral surface Sl of the cylinder for the components snl of the Piola-Kirchhoff stress vector sn , we obtain the following expressions: sn1 = sn2 ≡ 0, sn3 = θ13 n1 + θ23 n2 on Sl .

(5.6.29)

Examining equations (5.6.4), (5.6.25) and (5.6.29), we can see that on the lateral surface of the cylinder, the first two displacement and traction boundary conditions are identically satisfied, and the third displacement and traction boundary condition will be satisfied if u3 = v3 on Γ1 and sn3 = θ13 n1 + θ23 n2 = l3 on Γ2 ,

(5.6.30)

where Γ1 and Γ2 are complementary subcurves on the curve Γ, corresponding to the lateral complementary subsurfaces Sl1 and Sl2 . Summing up the results of this analysis, we can state that if the assumed conditions are fulfilled, in our cylindrical body can exist an antiplane (incremental) state. The only nonvanishing components of the incremental displacement field are u3 = u3 (x1 , x2 ) .

(5.6.31)

The only nonvanishing components of the incremental nominal stress are θ13 = ω1331 u3,1 , θ23 = ω2332 u3,2 , θ31 = ω3131 u3,1 , θ32 = ω3232 u3,2 .

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(5.6.32)

5.6. HOMOGENEOUS INITIAL DEFORMATIONS

331

The only incremental equation that must be satisfied is θ13,1 + θ23,2 = 0 or ω3131 u3,11 + ω2332 u3,22 = 0 in S.

(5.6.33)

The only boundary conditions that must be satisfied are u3 = v3 on Γ1 , sn3 = θ13 n1 + θ23 n2 = ω1331 u3,1 n1 + ω2332 u3,2 n2 = l3 on Γ2 .

(5.6.34)

The only involved instantaneous elasticities are ◦

◦

ω1331 = C55 + σ 11 , ω2332 = C44 + σ 22 , ω3131 = C55 , ω3232 = C44 .

(5.6.35)

Since the only unknown function is u3 = u3 (x1 , x2 ), the displacement potential problem is automatically solved if we have an antiplane state. Let us assume now that the imposed incremental displacement v and the imposed lead traction l satisfy the following restrictions on the lateral surface S l of the long cylindrical body: v1 = v1 (x1 , x2 ) , v2 = v2 (x1 , x2 ) , v3 ≡ 0 on Sl1 , l1 = l1 (x1 , x2 ) , l2 = l2 (x1 , x2 ) , l3 ≡ 0 on Sl2 .

(5.6.36)

Assuming again a sufficiently long cylinder and neglecting the end effects, we shall analyze if a plane state can exist in the body relative to the plane Ox 1 x2 , and characterized by the relations u1 = u1 (x1 , x2 ) , u2 = u2 (x1 , x2 ) , u3 ≡ 0 in B.

(5.6.37)

To analyze the compatibility problem, we observe that according to (5.6.23) and (5.6.37), the components of the incremental nominal stress depend only in x1 , x2 and are given by the equations θ11 = ω1111 u1,1 + ω1122 u2,2 , θ21 = ω2112 u1,2 + ω2121 u2,1 , θ31 ≡ 0, θ12 = ω1212 u1,2 + ω1221 u2,1 , θ22 = ω2211 u1,1 + ω2222 u2,2 , θ32 ≡ 0, θ13 ≡ 0, θ23 ≡ 0, θ33 = ω3311 u1,1 + +ω3322 u2,2 . (5.6.38) Now it is easy to see that the third incremental equilibrium equation (5.6.1) 2 is identically satisfied, and the first two become θ11,1 + θ21,2 = 0, θ12,1 + θ22,2 = 0 in S.

(5.6.39)

Taking into account (5.6.10), (5.6.24) and (5.6.38), for the components s nl , on the lateral surface Sl , we get sn1 = θ11 n1 + θ21 n2 , sn2 = θ12 n1 + θ22 n2 , sn3 ≡ 0 on Sl .

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(5.6.40)

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CHAPTER 5. THE THREE-DIMENSIONAL LINEARIZED THEORY

Examining equations (5.6.4), (5.6.36) and (5.6.40), we can see that the third displacement and traction boundary conditions are identically satisfied on the lateral surface of the cylinder and the first two will be fulfilled if u1 = v1 , u2 = v2 on Γ1 , (5.6.41) sn1 = θ11 n1 + θ21 n2 = l1 , sn2 = θ12 n1 + θ22 n2 = l2 on Γ2 . Hence, if the assumed conditions are fulfilled, in the considered cylindrical body a plane state can exist . The two nonvanishing components of the incremental displacement field are u1 = u1 (x1 , x2 ) , u2 = u2 (x1 , x2 ) .

(5.6.42)

The only nonvanishing components of the incremental nominal stress are θ11 = w1111 u1,1 + w1122 u2,2 , θ12 = ω1212 u1,2 + ω1221 u2,1 , θ33 = ω3311 u1,1 + ω3322 u2,2 .

θ21 = ω2112 u1,2 + ω2121 u2,1 θ22 = ω2211 u1,1 + ω2222 u2,2

(5.6.43)

The only incremental equilibrium equations that must be satisfied are θ11,1 + θ21,2 = 0, θ12,1 + θ22,2 = 0 in S,

(5.6.44)

or, equivalently, w1111 u1,11 + w1122 u2,21 + ω2112 u1,22 + ω2121 u2,12 = 0, (5.6.45) w1212 u1,21 + w1221 u2,11 + ω2211 u1,12 + ω2222 u2,22 = 0. The only involved boundary conditions that must be satisfied are u1 = v1 , u2 = v2 on Γ1 , (5.6.46) sn1 = θ11 n1 + θ21 n2 = l1 , sn2 = θ12 n1 + θ22 n2 = l2 on Γ2 . The only involved instantaneous elasticities are ◦

ω1111 = C11 + σ 11 , ω1212 = ω2121 + C66 , ω3311 = C13 ,

◦

ω2222 = C22 + σ 22 , ◦ ω1221 = C66 + σ 11 , ω3322 = C23 .

ω1122 = ω2211 + C12 , ◦ ω2112 = C66 + σ 22 ,

(5.6.47)

To analyze the displacement problem, we observe that the equilibrium equations (5.6.45) can be expressed in the following operatorial form: P11 u1 + P12 u2 = 0,

Copyright © 2004 by Chapman & Hall/CRC

P12 u1 + P22 u2 = 0

(5.6.48)

333

5.6. HOMOGENEOUS INITIAL DEFORMATIONS with P11 = ω1111

P21

∂2 ∂2 + ω2112 2 , 2 ∂x2 ∂x1

P12 = (ω1122 + ω2121 )

∂2 , ∂x1 ∂x2

(5.6.49)

∂2 , = (ω1212 + ω2211 ) ∂x1 ∂x2

P22

∂2 ∂2 = ω1221 2 + ω2222 2 . ∂x2 ∂x1

Since ω1122 = ω2211 and ω2121 = ω1212 , we have P12 = P21 .

(5.6.50)

Using the general relations (5.6.15) and (5.6.16) for the two dimensional case, or directly, it can be seen that the components u1 , u2 of the incremental displacement field u can be expressed by two displacement potentials ϕ(1) = ϕ(1) (x1 , x2 ) and ϕ(2) = ϕ(2) (x1 , x2 ) in the following way: u1 = −P12 ϕ(1) + P22 ϕ(2) ,

u2 = P11 ϕ(1) − P12 ϕ(2)

(5.6.51)

and the incremental equilibrium equations (5.6.48) will be identically satisfied if ϕ(1) and ϕ(2) satisfy the differential equation
(α) 2 P11 P22 − P12 ϕ = 0,

α = 1, 2.

(5.6.52)

The explicit form of the relations (5.6.51) and (5.6.52) can be obtained if we use the equations (5.6.49). Thus, we get Guz’s representation [5.8] u1 = − (ω1122 + ω1212 ) ϕ(1) ,12 + ω1221 ϕ(2) ,11 + ω2222 ϕ(2) ,22 ,

(5.6.53)

u2 = ω1111 ϕ(1) ,11 + ω2112 ϕ(1) ,22 − (ω1122 + ω1212 ) ϕ(2) ,12 , and the differential equation which must be satisfied by the displacement potentials ϕ(α) {ω1111 ω1221

∂4 + [ω1111 ω2222 + ω1221 ω2112 − ∂x41 2

(ω1122 + ω1212 ) ]

∂4 ∂4 }ϕ(α) = 0, α = 1, 2. + ω ω 2222 2112 ∂x2 4 ∂x21 ∂x22

Using (5.6.53) and (5.6.43), we obtain the incremental nominal stress com-

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CHAPTER 5. THE THREE-DIMENSIONAL LINEARIZED THEORY

ponents in terms of the potentials ϕ(α) :

∂ 2 ∂ϕ(2) ∂2 θ11 = {ω1111 ω1212 2 + [ω1111 ω2222 − ω1122 (ω2211 + ω1212 )] 2 } ∂x2 ∂x1   ∂x1 ∂ 2 ∂ϕ(1) ∂2 , + −ω1111 ω1212 2 + ω1122 ω2112 2 ∂x1 ∂x2  ∂x1  ∂ 2 ∂ϕ(2) ∂2 θ12 = − ω1221 ω1122 2 + ω1212 ω2222 2 ∂x1 ∂x2 ∂x1 ∂ 2 ∂ϕ(1) ∂2 , +{ω1221 ω1111 2 + [ω1221 ω2112 − ω1212 (ω1122 + ω1212 )] 2 } ∂x2 ∂x1 ∂x1 ∂ 2 ∂ϕ(2) ∂2 } + ω ω θ21 = {[ω2112 ω1221 − ω1212 (ω1122+ ω1221 )] 2112 2222 ∂x22 ∂x2 ∂x21   (1) 2 2 ∂ϕ ∂ ∂ , + ω1111 ω1212 2 − ω2112 ω1122 2 ∂x2 ∂x1 ∂x1  ∂ 2 ∂ϕ(2) ∂2 θ22 = ω1122 ω1221 2 − ω2222 ω1212 2 ∂x1 ∂x2 ∂x1 ∂ 2 ∂ϕ(1) ∂2 . +{[ω1111 ω2222 − ω1122 (ω1122 + ω1212 )] 2 + ω2222 ω2112 2 } ∂x2 ∂x2 ∂x1

(5.6.54) As in the classical theory of anisotropic bodies, there exist incremental boundary value problems which can be solved using a single displacement potential, that is assuming ϕ(1) ≡ 0 or ϕ(2) ≡ 0. However, to solve some important problems, concerning, for instance, cracks in prestressed materials, two independent potentials must be used. We observe that if the instantaneous elasticities satisfy the restriction ω2222 6= 0 and ω2112 6= 0, the equation satisfied by ϕ(α) can be factorized, becoming  2  2  2 2 ∂ ∂ 2 ∂ 2 ∂ + η1 + η2 ϕ(α) = 0, ∂x22 ∂x21 ∂x22 ∂x21

(5.6.55)

(5.6.56)

where the parameters η12 and η22 satisfy equations f (η) ≡ η 4 − 2Aη 2 + B = 0,

(5.6.57)

with 2

A=

ω1111 ω2222 + ω1221 ω2112 − (ω1122 + ω1212 ) ω1111 ω1221 , B= . 2 ω2222 ω2112 ω2222 ω2112

(5.6.58)

If the restriction (5.6.55) does not take place, the factorization procedure can and must be changed accordingly.

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5.6. HOMOGENEOUS INITIAL DEFORMATIONS

335

To study various stability and boundary value problems, the nature of the parameters η12 and η22 must be known. As is clear from the relation (5.6.57), η12 and η22 are given by the relations p 2 (5.6.59) η1,2 = A ± A2 − B.

To study the nature of these roots, we shall analyze the incremental harmonic wave problem, corresponding to a time-dependent plane incremental state. More exactly, we look for solutions of the form uα = aα ei(knβ xβ −ωt) , α, β = 1, 2

(5.6.60)

of the incremental equations of motion, corresponding to the plane state. Taking into account the relations (5.6.45), we can conclude that the incremental equations of motion have the following form: ω1111 u1,11 + ω2112 u1,22 + (ω1122 + ω1212 ) u2,12 = ρ0 u ¨1 , (5.6.61) (ω1122 + ω1212 ) u1,12 + ω1221 u2,11 + ω2222 u2,22 = ρ0 u ¨2 , the instantaneous elasticities being given by equations (5.6.47). In (5.6.60), aα is the constant amplitude of the wave, the constant k > 0 is the wave number, the constant unit vector uα gives the direction of propagation and the constant ω > 0 is the wave frequency. Introducing (5.6.60) in (5.6.61), we obtain the propagation condition
ω1111 n21 + ω2112 n22 a1 + (ω1122 + ω1212 )n1 n2 a2 = ρ0 u2 a1 , (5.6.62)
(ω1122 + ω1212 )n1 n2 a1 + ω1221 n21 + ω2222 n22 a2 = ρ0 v 2 a2 , where:

u2 =

ω2 k2

(5.6.63)

is the square of the wave velocity. The characteristic equation giving the possible wave velocities takes the following form:
(5.6.64) ρ20 u4 − a (n1 , n2 ) ρ0 u2 + ω2222 ω2112 b n1, n2 = 0 with

2a (n1 , n2 ) = (ω1111 + ω1221 ) n21 + (ω2112 + ω2222 ) n22 , b (n1, n2 ) = n42 + 2An22 n21 + Bn21 ,

(5.6.65) (5.6.66)

A and B being given in (5.6.58). To get equation (5.6.64), we have assumed that the restriction (5.6.55) is fulfilled.

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CHAPTER 5. THE THREE-DIMENSIONAL LINEARIZED THEORY

Let us assume now that the initial deformed homogeneous equilibrium configuration is locally stable; i.e. the corresponding instantaneous elasticity ω is positive definite. As Sylvester’s criterion shows, in this case, the involved instantaneous elasticities must satisfy the following restrictions: ω1111 , ω2222 , ω1221 , ω2112 > 0,

2 > 0, ω1111 ω2222 − ω1212

(5.6.67)

2 ω1221 ω2112 − ω1212 > 0.

Let us observe, that in this case, the restrictions (5.6.55) are fulfilled. Also, the positive definiteness of ω implies positive definiteness of the corresponding acoustic tensor involved in the propagation condition (5.6.62) and in the characteristic equation (5.6.64). Consequently, the roots ρ0 u21 and ρ0 u22 of this equation are real and positive for all directions of propagation. Particularly, as seen from (5.6.64) results, we must have

ρ0 u21 + ρ0 u22 = a n1, n2 > 0, ρ0 u21 ρ0 u22 = ω2222 ω2112 b n1, n2 > 0 (5.6.68) for any n1 and n2 such that n21 + n22 = 1. Since w2222 and w2112 are positive, we can conclude from (5.6.68)2 that b(n1 , n2 ) > 0

(5.6.69)

for all n1 , n2 such that n21 + n22 6= 0. In this way, taking n2 = µn1 , where µ is an arbitrary real number, from (5.6.69) and (5.6.66) we can conclude that if the initial deformed homogeneous equilibrium configuration is locally stable, we have g (µ) ≡ µ4 + 2Aµ2 + B > 0 for any real number µ.

(5.6.70)

Let us assume now that µ2 = −η 2 ,

2

(5.6.71)

η being an arbitrary negative number. Taking into account the equation (5.6.57) defining the function f (η), we can conclude that if the initial deformed equilibrium configuration in locally stable, we have f (η) = η 4 − 2Aη 2 + B > 0 for any η 2 < 0.

(5.6.72)

The obtained results can be summed up in the following. Theorem. If the initial deformed homogeneous equilibrium configuration is locally stable, then: (i) The incremental equilibrium equation is elliptic. (ii)The second order algebraic equation g (µ) ≡ µ4 + 2Aµ2 + B = 0 cannot have real roots.

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337

5.6. HOMOGENEOUS INITIAL DEFORMATIONS (iii)The second order algebraic equation f (η) = η 4 − 2Aη 2 + B = 0

cannot have imagining roots. (iv)The parameters η12 and η22 in the factorized equation (5.6.56) are real and positive or complex and conjugate numbers. The above theorem will be used in Section 8 to express the incremental fields by complex potential. Now we return to the three dimensional case and assume that the material in transversally isotropic, Ox1 x2 is the plain of isotropy. We suppose also that the initial applied displacement field satisfies the symmetry condition λ1 = λ 2 .

(5.6.73)

Now the structure of the elasticity tensor c is given by the relation (2.2.50). Consequently, we have C11 = C22 , C13 = C23 , C44 = C55 , 2C66 = C11 − C12 .

(5.6.74)

From (5.6.19) and (5.6.73), we get ◦

◦

◦

ε11 = ε22 = λ1 − 1, ε33 = λ3 − 1.

(5.6.75)

The relation (5.6.22), (5.6.74) and (5.6.75) jointly give ◦

◦

◦

◦

◦

◦

◦

σ 11 = σ 22 = (C11 + C12 ) ε11 + C13 ε33 , σ 33 = 2C13 ε11 + C33 ε33 .

(5.6.76)

Taking into account (5.6.22), (5.6.74), (5.6.75), we conclude that in the considered cylindrical symmetry case, the instantaneous elasticities are given by the following supplementary relations: ◦

◦

ω1111 = ω2222 = C11 + σ 11 , ω3333 = C33 + σ 33 , ω1122 = ω2211 = C12 , ω1133 = ω3311 = ω2233 = ω3322 = C13 , ω1221 = ω2112 =

ω1212 = ω2121 =

1 ◦ (C11 − C12 )+ σ 11 , 2

1 (C11 − C12 ), ω1313 = ω3131 = ω2323 = ω3232 = C44 , 2 ◦

◦

ω1331 = ω2332 = C44 + σ 11 , ω3113 = ω3223 = C44 + σ 33 .

(5.6.77)

Moreover, if follows also that the equation ω1111 − ω1221 = ω1122 + ω1212 is satisfied.

Copyright © 2004 by Chapman & Hall/CRC

(5.6.78)

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CHAPTER 5. THE THREE-DIMENSIONAL LINEARIZED THEORY

In the considered cylindrical symmetric case, these are 8 independent instantaneous elasticities. For simplicity we shall use the notations Yk =

∂2 ∂2 ∂ = Y12 + Y22 = ∆1 . + , k = 1, 2, 3, ∂x22 ∂x21 ∂xk

(5.6.79)

Using (5.6.78) and (5.6.79) for the differential operators Plm introduced by the equation (5.6.12), we find the following simplified expression: P11 = ω1111 Y12 + ω1221 Y22 + ω3113 Y32 , P22 = ω1221 Y12 + ω1111 Y22 + ω3113 Y32 , P33 = ω1331 ∆1 + ω3333 Y32 , P12 = P21 = (ω1111 − ω1221 )Y1 Y2 , P13 = P31 = (ω1133 + ω1313 )Y1 Y3 , P23 = P32 = (ω1133 + ω1313 )Y2 Y3 . Using the above expression, we find
2 2 2 − P22 P13 P33 + 2P12 P13 P23 − P11 P23 det P = P11 P22 − P22 2 = (ω1111 ∆1 + ω3113 Y32 )(ω1221 ∆1 + ω3113 Y32 ), P11 P22 − P12

(5.6.80)

(5.6.81) (5.6.82)

2 2 = −(ω1133 + ω1313 )2 (ω1221 ∆21 + ω3113 ∆Y32 )Y32 . − P22 P13 2P12 P13 P23 − P11 P23 (5.6.83)

Now we assume that ω1111 6= 0, ω1221 6= 0, ω1331 6= 0.

(5.6.84)

If one of these conditions are not fulfilled, the procedure must be changed accordingly. If our assumptions are fulfilled, we get det P = 

ω1111 ω1221 ω1331 { ∆21 +

2 ω3113 ω3113 (ω1111 + ω1221 ) Y4 ∆1 Y32 + ω1111 ω1221 3 ω1111 ω1221

  ω3113 2 (ω1133 + ω1313 )2 ω3333 2 Y ]∆1 Y32 }. [∆1 + Y3 − · ∆1 + ω1221 3 ω1111 ω1331 ω1331

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339

5.6. HOMOGENEOUS INITIAL DEFORMATIONS We also have ∆21 +

2 ω3113 2 ω3113 2 ω3113 ω3113 (ω1111 + ω1221 ) Y ). Y )(∆1 + Y 4 = (∆1 + ∆1 Y32 + ω1221 3 ω1111 3 ω1111 ω1221 3 ω1111 ω1221

In this way, we obtain (ω1111 ω1221 ω1331 )−1 det P

ω3333 2 ω3113 2 Y ) Y3 )(∆1 + ω1331 3 ω1111 2 ω3113 2 (ω1133 + ω1313 ) Y ). ∆1 Y32 ](∆1 + ω1221 3 ω1111 ω1331

=

[(∆1 +

−

Factorizing the first brackets and using the notation (5.6.79), finally we get (ω1111 ω1221 ω1331 )−1 det P = (∆1 + ξ12

2 2 ∂2 2 ∂ 2 ∂ ), (5.6.85) )(∆ + ξ )(∆ + ξ 1 1 3 2 ∂x23 ∂x23 ∂x23

where the parameters ξ12 , ξ22 and ξ32 are given by the following relations: ξ12 = 2 ξ2,3

with c=

=c±

r

ω3113 ω1221

c2 −

ω3113 ω3333 ω1331 ω1111

ω1111 ω3333 + ω1331 ω3113 − (ω1133 + ω1313 )2 . 2ω1111 ω1331

(5.6.86)

(5.6.87)

(5.6.88)

Returning to the relation (5.6.16), according to (5.6.85), we can conclude that the displacement potential ϕ(j) must satisfy the following differential equation established by Guz [5.6], [5.8]: (∆1 + ξ12

∂2 ∂2 ∂2 )(∆1 + ξ22 2 )(∆1 + ξ32 2 )ϕ(j) = 0, j = 1, 2, 3, 2 ∂x3 ∂x3 ∂x3

(5.6.89)

the parameters ξ12 , ξ22 and ξ32 being expressed in terms of the instantaneous elasticities through the relations (5.6.86),(5.6.87). Following Guz, we introduce now two function ψ1 and ψ2 satisfying the following equation: ∂2 (5.6.90) (∆1 + ξ12 2 )ψ1 = 0, ∂x3

(∆1 + ξ22

2 ∂2 2 ∂ )ψ2 = 0. )(∆ + ξ 1 3 ∂x23 ∂x23

(5.6.91)

Let us observe that if ψ1 and ψ2 satisfy the above equations, ψ1 and ψ2 satisfy also Guz’s equation (5.6.89).

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CHAPTER 5. THE THREE-DIMENSIONAL LINEARIZED THEORY

We recall now the representation (5.6.15) of the incremental displacements through the displacement potentials and introduce the following combination of (j) the fields um : (3) (2) um = u(1) m − um + um =

∂ det P (1) ∂ det P (2) ∂ det P (3) ϕ . ϕ + ϕ − ∂P3m ∂P2m ∂P1m

Also we take ϕ(1) = ϕ(2) = ψ1 , ϕ(3) = ψ2 . In this way, we get um = (

∂ det P ∂ det P ∂ det P ψ2 . )ψ1 + − ∂P3m ∂P2m ∂P1m

(5.6.92)

We shall analyze the above relation in detail for m = 3. The computations for the component u1 and u2 corresponding to m = 1 and m = 2 must be made in the same way. For m = 3, equation (5.6.92) becomes u3 = (

∂ det P ∂ det P ∂ det P ψ2 . )ψ1 + − ∂P33 ∂P23 ∂P13

(5.6.93)

In order to evaluate the partial derivatives of det P in relation to Pjm , considered as independent variables, the expression (5.6.85) of det P must be symmetrized in the following manner: det P = (P11 P22 − P12 P21 )P33 + P12 P13 P23 + P21 P31 P32 − P11 P23 P32 − P22 P13 P31 . (5.6.94) After the derivatives were determined, to obtain the final results we can use the symmetry relations Pjm = Pmj . Using the above procedure, we get

∂ det P ∂ det P ∂ det P 2 . = P11 P22 − P12 = P12 P13 − P11 P23 , = P12 P23 − P22 P13 , ∂P33 ∂P23 ∂P13 (5.6.95) Using the expression (5.6.80) of the differential operators Pjm , we get

∂2 ∂ ∂ ∂ ∂ det P ∂ det P (∆1 + ξ12 2 ), (5.6.96) ) − = (ω1133 + ω1313 )ω1221 ( − ∂x3 ∂x2 ∂x1 ∂x3 ∂P23 ∂P13

where ξ12 is given by the equation (5.6.86). According to the assumption made, the potential ψ1 satisfies the differential equation (5.6.90). Hence, the contribution of ψ1 to u3 is vanishing. Consequently, equation (5.6.92), from m = 3, becomes u3 =

∂ det P 2 )ψ2 . ψ2 = (P11 P22 − P12 ∂P33

2 and using the relaReturning now to the expression (5.6.82) of P11 P22 − P12 tion (5.6.80), it is easy to see that this operator has the following expression: 2 = (ω1111 ∆1 + ω3113 Y32 )(ω1221 ∆1 + ω3113 Y32 ). P11 P22 − P12

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341

5.6. HOMOGENEOUS INITIAL DEFORMATIONS The last two equations jointly give u3 = (ω1111 ∆1 + ω3113

∂2 ∂2 )(ω1221 ∆1 + ω3113 2 )ψ2 , 2 ∂x3 ∂x3

∆1 =

∂2 ∂2 . + 2 ∂x22 ∂x1

(5.6.97)

(5.6.98)

After similar calculus for u1 and u2 , results are u1 =

∂2 ∂ ∂ ∂ )[(ω1122 + ω1212 )(ω1331 ∆1 + ω3333 2 ) + ( ∂x3 ∂x2 ∂x2 ∂x1

−(ω1133 + ω1331 )2

u2 = −

∂2 ∂2 ∂2 (ω1221 ∆1 + ω3113 2 )ψ2 . ]ψ1 − (ω1133 + ω1313 ) 2 ∂x3 ∂x1 ∂x3 ∂x3

∂2 ∂ ∂ ∂ )[(ω1122 + ω1212 )(ω1331 ∆1 + ω3333 2 ) + ( ∂x3 ∂x2 ∂x1 ∂x1

−(ω1133 + ω1331 )2

∂2 ∂2 ∂2 (ω1221 ∆1 + ω1313 2 )ψ2 . ]ψ1 − (ω1133 + ω1313 ) 2 ∂x3 ∂x2 ∂x3 ∂x3 (5.6.99)

We suppose now that ω1122 + ω1213 6= 0 and ω1133 + ω1331 6= 0.

(5.6.100)

In this case the relation (5.6.97) and (5.6.99) can be simplified introducing the new potentials ψ≡(

∂2 ∂2 ∂ ∂ )[(ω1122 + ω1212 )(ω1331 ∆1 + ω3333 2 ) − (ω1133 + ω1331 )2 2 ]ψ1 , + ∂x3 ∂x3 ∂x1 ∂x2

χ ≡ (ω1133 + ω1313 )(ω1221 ∆1 + ω3113

∂2 )ψ2 . ∂x23

(5.6.101)

Introducing these relations in equations (5.6.97) and (5.6.99), we finally obtain Guz’s representation of the incremental displacement by two displacement potentials ψ and χ, appropriate to the axially symmetric case u1 =

∂ψ ∂2χ ∂ψ ∂2χ , u2 = − − − ∂x1 ∂x3 ∂x2 ∂x1 ∂x2 ∂x3

u3 =

1 ∂2 (ω1111 ∆1 + ω1313 2 )χ. ω1133 + ω1313 ∂x3

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(5.6.102)

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CHAPTER 5. THE THREE-DIMENSIONAL LINEARIZED THEORY

Since the function ψ1 and ψ2 satisfy the differential equations (5.6.90) and (5.6.91), respectively, Guz’s displacement potentials ψ and χ must satisfy the following differential equations: (∆1 +ξ12

∂2 ∂2 ∂2 ∂2 ∂2 + 2 , (∆1 +ξ22 2 )(∆1 +ξ32 2 )χ = 0. (5.6.103) )ψ = 0, ∆1 = 2 2 ∂x3 ∂x3 ∂x1 ∂x3 ∂x3

If ξ12 6= ξ22 ,

according to Baggio’s theorem, the potential χ can be expressed as χ = χ2 + χ3 ,

(5.6.104)

and the new potential χ2 and χ3 must satisfy the differential equations (∆1 + ξ22

∂2 ∂2 )χ2 = 0, (∆1 + ξ32 2 )χ3 = 0. 2 ∂x3 ∂x3

(5.6.105)

Examining the above results, we can see that Guz’s theorem gives a representation of incremental displacement field through three displacement potentials ψ, χ1 and χ2 , each of them satisfying a second order partial differential equation, if the parameters ξ1 , ξ2 , ξ3 are distinct. Guz’s representation is useful to study a three-dimensional incremental boundary value problem. Particularly, as we shall see in the Section 6, it can be used to study stability problem concerning anisotropic cylindrical bars. As we shall see, if the bar has a circular transverse section, we can find the primary eigenmodes describing the Eulerian buckling of the bar. Since, in the axially symmetric case, all boundary value problems can be more easily analyzed by using cylindrical coordinates, in what follows we shall present Guz’s representation with such coordinates. Let us denote by {er , eθ , ez } the physical basis of a cylindrical system of coordinates and let r, θ, z be the cylindrical coordinates of a point P as in Figure 5.5. We have x1 = r cos θ, x2 = r sin θ, x3 = z, x2 (5.6.106) r = (x21 + x22 )1/2 , θ = arctan , z = x3 , x1 and er = e1 cos θ + e2 sin θ, eθ = −e1 sin θ + e2 cos θ, ez = e3 ,

e1 = er cos θ − eθ sin θ, e2 = er sin θ + eθ cos θ, e3 = ez .

(5.6.107)

u = u 1 e1 + u 2 e2 + u 3 e3 = u r er + u θ eθ + u z ez ,

(5.6.108)

The orthonormal basis {er , eθ , ez } changes when the point P changes, but any vector field depending on P can be expressed as a linear combination of the vector {er , eθ , ez } corresponding to P . Particularly, for the incremental displacement vector u, we have

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343

5.6. HOMOGENEOUS INITIAL DEFORMATIONS

Figure 5.5: Cylindrical coordinates r, θ, z. ur , uθ and uz representing the radial, tangential and axial components of the displacement u. Directly from (5.6.107) and (5.6.108), we obtain ur = u1 cos θ + u2 sin θ, uθ = −u1 sin θ + u2 cos θ, uz = u3 u1 = ur cos θ − uθ sin θ, u2 = ur sin θ + uθ cos θ, u3 = uz .

(5.6.109)

cos θ sin θ ∂θ ∂θ ∂r ∂r . = , =− = sin θ, = cos θ, r ∂x2 r ∂x1 ∂x2 ∂x1

(5.6.110)

Also, from (5.6.106), it results

The above equations show that if the displacement potentials ψ and χ, considered until now as depending on the Cartesian coordinates x1 , x2 , x3 , will be assumed as functions of the cylindrical coordinates r, θ, z, their derivatives relative to x1 , x2 , x3 can be expressed in terms of their derivatives relative to r, θ, z using the relations ∂ ∂ 1 ∂ ∂ ∂ 1 ∂ ∂ ∂ . (5.6.111) = , + cos θ = sin θ , − sin θ = cos θ ∂z r ∂θ ∂x3 ∂r r ∂θ ∂x2 ∂r ∂x1

Considering ψ and χ as depending on r, θ, z and using the relation (5.6.111), from Guz’s representation (5.6.102), we get u1 = (

∂2χ 1 ∂ψ 1 ∂2χ ∂ψ ) cos θ, − ) sin θ + ( + ∂r∂z r ∂θ r ∂θ∂z ∂r

u2 = −(

1 ∂2χ ∂ψ 1 ∂ψ ∂2χ + ) cos θ + ( − ) sin θ, r ∂θ∂z ∂r r ∂θ ∂r∂z

u3 = (ω1133 + ω1313 )−1 (ω1111 ∆1 + ω3113

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∂2 )χ, ∂z 2

(5.6.112)

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CHAPTER 5. THE THREE-DIMENSIONAL LINEARIZED THEORY

where the two-dimensional Laplacian ∆1 is now expressed, using the cylindrical coordinates r and θ, by the equation ∆1 =

1 ∂2 1 ∂ ∂2 . + + r ∂r r2 ∂θ2 ∂r2

(5.6.113)

The above relation can be established using equations (5.6.98), (5.6.110) and (5.6.111). Now, introducing (5.6.112) in (5.6.109)1,2,3 , we obtain Guz’s representation in cylindrical coordinates, appropriate to the cylindrical symmetric case ur =

1 ∂2χ ∂ψ ∂2χ 1 ∂ψ , − , uθ = − − r ∂θ∂z ∂r ∂r∂z r ∂θ

(5.6.114) u3 = (ω1133 + ω1313 )−1 [ω1111 (

2

2

2

∂ 1 ∂ 1 ∂ ∂ ) + ω3113 2 ]χ. + + ∂z r ∂r r2 ∂θ2 ∂r2

In any point P , the corresponding physical basis {er , eθ , ez }, depending on P , can be used to construct in P a tensorial basis, following the usual procedure presented in Section 1.1. Particularly, the incremental nominal stress tensor θ, evaluated in P , can be expressed as a linear combination of the elements forming the constructed terminal basis; i.e. we have θ = θrr er er + θrθ er eθ + θrz er ez + θθr eθ er + θθθ eθ eθ + (5.6.115) θθz eθ ez + θzr ez er + θzθ ez eθ + θzz ez ez = θkl ek el , θkl being the components of θ in the tensorial bases {ek , el }, k, l = 1, 2, 3, and θrr , ..., θzz representing the physical components of θ, corresponding to the introduced cylindrical coordinate system. Now, directly from (5.6.107)4,5,6 and (5.6.115), we obtain θrr = θ11 cos2 θ + (θ12 + θ21 ) sin θ cos θ + θ22 sin2 θ, θθθ = θ11 sin2 θ − (θ12 + θ21 ) sin θ cos θ + θ22 cos2 θ, θrθ = (θ22 − θ11 ) sin θ cos θ + θ12 cos2 θ − θ21 sin2 θ, θθr = (θ22 − θ11 ) sin θ cos θ − θ12 sin2 θ + θ21 cos2 θ, θrz = θ13 cos θ + θ23 sin θ, θθz = −θ13 sin θ + θ23 cos θ, θzr = θ31 cos θ + θ32 sin θ, θzθ = −θ31 sin θ + θ32 cos θ, θzz = θ33 .

(5.6.116)

The above relation will be used to express the physical components θ rr , ..., θzz in terms of ur , uθ , uz considered as functions of r, θ, z. We shall analyze in detail the problem for θrr . For the other components, the procedure is similar and we present only the final results.

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5.6. HOMOGENEOUS INITIAL DEFORMATIONS

345

To solve the problems, we first use the relation (5.6.23) and the restrictions (5.6.77), (5.6.78) satisfied by the instantaneous elasticities in the cylindrical symmetric case. In this way we obtain, in Cartesian coordinates, the following simplified incremental constitutive equations: θ11 = ω1111 u1,1 + ω1122 u2,2 + ω1133 u3,3 , θ12 = ω1212 u1,2 + ω1221 u2,1 , θ13 = ω1313 u1,3 + ω1331 u3,1 , θ21 = ω1221 u1,2 + ω1212 u2,1 , θ22 = ω1122 u1,1 + ω1111 u2,2 + ω1133 u3,3 ,

(5.6.117)

θ23 = ω1313 u2,3 + ω1313 u3,2 , θ31 = ω3113 u1,3 + ω1313 u3,2 , θ32 = ω3113 u1,3 + ω1313 u3,2 , θ33 = ω1133 u1,1 + ω1133 u2,2 + ω3333 u3,3 . Now using the first equation (5.6.116), we get for θrr the following expression: θrr = (ω1111 u1,1 + ω1122 u2,2 + ω1133 u3,3 ) cos2 θ +(ω1122 u1,1 + ω1111 u2,2 + ω1133 u3,3 ) sin2 θ +(ω1111 − ω1122 )(u1,2 + u2,1 ) sin θ cos θ.

(5.6.118)

To get the above relation, we have used also the symmetry restriction (5.6.68) satisfied in the axially symmetric case. Now we return to equations (5.6.109)3,4 and (5.6.110). After long, but elementary computations, we get the following relations: u1,1 =

uθ 1 ∂ur ∂uθ ur 1 ∂uθ ∂ur − ) sin θ cos θ, + + ) sin2 θ − ( cos2 θ + ( r r ∂θ ∂r r r ∂θ ∂r

u1,2 = (

u2,1 =

ur 1 ∂uθ ∂ur ∂uθ uθ 1 ∂ur − ) sin θ cos θ, − sin2 θ + ( − ) cos2 θ − r r ∂θ ∂r ∂r r r ∂θ

ur 1 ∂uθ ∂ur uθ 1 ∂ur ∂uθ − ) sin θ cos θ, − − ) sin2 θ + ( cos2 θ − ( r r ∂θ ∂r r r ∂θ ∂r

u2,2 = (

1 ∂uθ ur ∂ur ∂uθ 1 ∂ur uθ + ) cos2 θ + sin2 θ + ( + − ) sin θ cos θ. r ∂θ r ∂r ∂r r ∂θ r (5.6.119)

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Introducing (5.6.119) in (5.6.118), we get the expression of θrr in terms of the radial, tangential and axial components of the incremental displacement field. Finally, after long but elementary computations, founded in the above procedure, we obtain ∂uz ur 1 ∂uθ ∂ur , + ) + ω1133 + ω1122 ( θrr = ω1111 ∂z r r ∂θ ∂r

θrθ = ω1221

uθ 1 ∂ur ∂uθ − ), + ω1212 ( r r ∂θ ∂r

θrz = ω1313

∂uz ∂ur , + ω1331 ∂r ∂z

θθr = ω1221 (

∂uθ uθ 1 ∂ur , − ) + ω1212 ∂r r r ∂θ

θθθ = ω1111 (

∂uz ∂ur ur 1 ∂uθ , + ω1133 + ) + ω1122 ∂z ∂r r r ∂θ

θθz = ω3113

∂uθ 1 ∂uz , + ω3131 ∂z r ∂θ

θzr = ω3113

∂uz ∂ur , + ω3131 ∂r ∂z

θzθ = ω3113

∂uθ 1 ∂uz + ω3131 , ∂z r ∂θ

∂ur 1 ∂uθ ur ∂uz + + ) + ω3333 . (5.6.120) ∂r r ∂θ r ∂z Finally, let us consider the Piola-Kirchhoff stress vector sn acting on a material surface element with unit normal n. These two vectors can be expressed as a linear combination of er , eθ , ez θzz = ω1133 (

sn = s nr e r + s nθ e θ + s nz e z , n = n r er + n θ eθ + n z ez . snz get

(5.6.121)

In their turn, using the general formula sn = θ T n, the components snr , snθ , can be expressed in terms of the components θrr ,..., θzz , nr , nθ and nz . We snr = θrr nr + θθr nθ + θθz nz , snθ = θrθ nr + θθθ nθ + θzθ nz ,

(5.6.122)

snz = θrz nr + θθz nθ + θzz nz . If one uses cylindrical coordinates to formulate and to solve incremental boundary value problems, the above relations must be taken into account to express boundary conditions in traction.

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347

5.7. PROBLEMS

5.7

Problems ∼

P5.1 Let us consider a body, and let B and B be two reference configurations ∼ of the body, Bt representing its current configuration. Let us denote by X, X and ∼ x the position vectors of a particle P of the body in its configurations B, B and ∼ Bt , respectively. Let us assume that X and X are connected by the equation ∼

X = QX, where Q is a given, constant orthogonal tensor. Let us denote by ∼

∼

x = χ(X) and x = χ (X) ∼

the deformation of the body from B to Bt and from B to Bt , respectively. ∼ (a) Show that F and F, the gradients of deformation from B to Bt and from ∼

B to Bt are connected by the equation ∼

F = F Q. ∼

∼

(b) Let G and G the Green’s strain tensor corresponding to F and F, respectively. Show that ∼ G = QT G Q. P5.2 Let us assume that the body considered in P5.1 is hyperelastic. Let us ∼ assume that uB (G) and u ∼ (G) are the constitutive functions giving the specific B

∼

elastic energy of the body, if B and B are used as reference configurations to describe the deformation of the body; i.e. ∼

u = uB (G) = u ∼ (G). B

Show that the two constitutive functions are connected by the equation ∼

∼

u ∼ (G) = uB (QT G Q) B ∼

for any symmetric tensor G. ∼ P5.3 We say that two reference configurations B and B connected as in P5.1 ∼ are equivalent for a hyperelastic material if the same deformation of B and of B , respectively, produce the same specific elastic energy in the material; i.e. if we have uB (G) = u ∼ (G) B

for any symmetric tensor G. In this case, the orthogonal tensor Q is named a ∼ symmetry transformation from B to B .

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CHAPTER 5. THE THREE-DIMENSIONAL LINEARIZED THEORY ∼

Show that if reference configurations, B and B , connected by the equation ∼ X = QX are equivalent, the constitutive function uB (G) must satisfy the equation uB (QGQT ) = uB (G) for any symmetric tensor G. P5.4 Let us denote by GB the set of all symmetry transformations of the material, corresponding to its reference configuration B. (a) Show that the transformations 1 and −1 are elements of GB for any B. (b) Show that if the orthogonal transformation Q ∈ GB , then its inverse Q−1 = QT is also an element of GB . (c) Show that if the orthogonal transformation Q1 , Q2 ∈ GB , then their composition Q1 Q2 is also an element of GB . Since GB has the properties (a)−(c), its named the symmetry group of the material, corresponding to its reference configuration B. b be two reference configurations of the material and let X P5.5 Let B and B b b respectively. and X be the position vectors of a particle P of the body in B and B, b Let us assume that B and B are connected by the following relation: b = ΛX, X

where Λ is an orthogonal tensor. Let us denote by GB and G B b the symmetry b respectively. Show that the sets groups of the material, corresponding to B and B, GB and G B b are connected by the relation GB b = ΛGB ΛT .

P5.6 Let us assume that the constitutive function uB (G) of a hyperelastic material is a quadratic form of Green’s strain tensor G; i.e. uB (G) =

1 G · CG, 2

where C, the elasticity tensor of the material, is a given constant fourth other tensor, having all symmetries of the elasticity tensor c of a linearly hyperelastic material. (a) Show that if the orthogonal tensor Q is a symmetry transformation of the material, corresponding to the reference configuration of B; i.e. if Q ∈ GB , then the elasticity tensor C of the material must satisfy the restriction (QGQT ) · C(QGQT ) = G · CG for any symmetric tensor G. (b) Show that if C satisfies the above restriction for any symmetric tensor G and for an orthogonal tensor Q, then Q ∈ GB .

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5.7. PROBLEMS

(c) Show that C satisfies the restriction given in (a) if and only if its components Cklmn satisfy the restrictions Cklmn = Qkp Qlq Qmr Qns Cpqrs . (d) Using the results obtained in (a)−(c), justify the definition of a symmetry transformation given in Section 2.2 for a linearly hyperelastic material. P5.7 Let B and Bt be the reference and the current configuration of a body. Let dA be a material surface element of the body in its reference configuration B and let da be the same material surface element in the current configuration B t of the body. Let us denote by N the unit normal real dA and let n be the unit normal to da. Prove Nanson’s formula n da = JF−T NdA with J = det F. P5.8 (a) Let C = FT F be the Cauchy-Green strain tensor. Show that C and its inverse C−1 are symmetric positive definite tensors. b) Using Nanson’s formula, prove the relation q da = ( det C)N · C−1 NdA

and

1

F−T N. N · C−1 N P5.9 Let us denote by ρ0 = ρ0 (X) the mass density of a body in its reference configuration B and by ρ = ρ(x, t) its mass density in its current configuration. Prove the relation ρ0 (X) = J(X, t)ρ(x, t). n= √

P5.10 Starting with the global principle of momentum balance, appropriate to the Eulerian formulation and using the results given in P5.7 and P5.9, give the global principle of momentum balance, appropriate to the Lagrangean formulation. P5.11 Using the results obtained in P5.10, obtain the equation of motion appropriate to the Lagrangean formulation. P5.12 Let {ek } be an orthonormal basis in V and let (O, xk ) be the corresponding Cartesian coordinates system in E. Let us consider a body which, in its reference configuration, is a parallelepiped defined by the inequalities −a k ≤ Xk ≤ ak , k = 1, 2, 3, ak being given positive numbers, and Xk representing the Lagrangean coordinates of a particle P of the body. Let us assume the body submitted to the following static deformation: x 1 = λ 1 X1 , x 2 = λ 2 X2 , x 3 = λ 3 X3 , where xk are the Euclidean coordinates of the particle P , and λ1 , λ2 , λ3 are given real numbers, satisfying the restriction λ1 λ2 λ3 > 0.

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CHAPTER 5. THE THREE-DIMENSIONAL LINEARIZED THEORY ◦

(a) Find the deformed equilibrium configuration B of the body. ◦

◦

◦

◦ ◦

◦

◦ ◦ T

◦

(b) Find F, J = det F, C = FT F and G = 12 (F F −1). (c) Give the interpretation of the restriction imposed by λ1 , λ2 , λ3 .

P5.13 Let us assume that the body given in P5.12 is a nonlinear hyperelastic body, its specific strain energy u = u(G) being given by the following constitutive relation λ u = u(G) = (trG)2 + µG · G, 2 where λ and µ are two material constants and G in the Green’s strain tensor. (a) Show that our material is isotropic. (b) Find the restrictions which must be satisfied by the material constants λ and µ, if the specific strain energy of the material is a positive definite quadratic form.

P5.14 Using the data given in P5.12 and P5.13 find: ◦

◦

(a) The symmetric Piola-Kirchhoff stress Π, the nominal stress Θ and the ◦

◦

Cauchy’s stress T, corresponding to the deformed equilibrium configuration B. ◦ (b)The Cauchy’s stress vector tn acting on the boundary of the deformed ◦ body and the Piola’s and Kirchhoff’s stress vector sN acting on the initial boundary of the body. (c) The body force density b assuming equilibrium of the body in the de◦

formed configuration B. (d) Analyze and interpret the results obtained in (a) and (b), taking into account the mechanical significance of the Cauchy’s stress tensor and vector, and the Piola’s and Kirchhoff’s stresses, tensors and vectors, respectively. P5.15 The parallelepiped considered in P5.12 is submitted to the following deformation named pure shear in the Ox1 x2 plane: x1 = X1 + γX2 , x2 = X 2 , x3 = X 3 , where γ > 0 is a given number.

◦

(a) Find the deformed equilibrium configuration B of the body. ◦

◦

◦

◦ ◦

◦

(b) Find F, J , C = FT F and G =

◦ ◦ 1 T 2 (F F

−1).

◦

(c) Find the eigenvalues and eigendirections of the Green’s strain tensor G. P5.16 Assume now that the parallelepiped deformed as in P5.12 is made up by the hyperelastic material given in P5.13. Find: ◦

◦

◦

◦

(a) The stress tensors Π, Θ and T corresponding to B .

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5.7. PROBLEMS ◦

(b) Cauchy’s stress vector tn acting on the boundary of the deformed body. (c) The force density b assuming the equilibrium of the body in the deformed ◦

configuration B . (d) Analyze and interpret the results obtained in (a). ◦

(e) Find the eingenvalues and eigendirections of the stress tensors Π . (f) What happens if the normal surface forces acting on the boundaries of the body are vanishing? P5.17 Find the condition which must be satisfied by λ1 , λ2 , λ3 from P5.12 ◦

and by γ from P5.15 to have infinitesimal initial deformations from B to B . Solve the problems (a)−(d) of P5.14 and the problems (a)−(e) of P5.16 in this case and compare the obtained results with those obtained for large initial deformation. P5.18 Let us denote by U(X) the incremental displacement field in the Lagrangean approach and by u(x) the (same!) incremental field in the updated Lagrangean approaches. Show that ◦

∇X U = FT ∇x u. ◦

◦

P5.19 Show that the incremental elasticities Ω and ω, corresponding to the Lagrangean and updated Lagrangean approaches, respectively, are related by the equation ◦ −1 ◦

◦

ω klmn = J

◦

◦

F kp F nq Ωplmq .

P5.20 Prove that ◦ −1

◦

∇u· ω ∇uT = J and

Z

◦

B

T

∇U· Ω ·∇U dV =

◦

∇U· Ω ·∇UT Z

◦

∇u· ω ∇uT dv.

◦

B ◦

P5.21 Show that the instantaneous elasticity Ω is positive definite in the ◦ reference configuration B if and only if the instantaneous elasticity ω is positive ◦

definite in the initial deformed equilibrium configuration B . P5.22 Give the formulation of the mixed incremental boundary value problem (5.4.21)−(5.4.22) appropriate to the updated Lagrangean approach and establish the connection existing between the external force system, appropriate to the two equivalent approaches of the incremental problems. P5.23 Give the expression of the incremental potential energy functional Φ = Φ (U), appropriate to the updated Lagragean approach.

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P5.24 Discuss incremental uniqueness, local stability, incremental variational and extremal principles, using the updated Lagrangean formulation of the threedimensional linearized theory. ◦

◦

P5.25 Find the instantaneous elasticities Ω and ω corresponding to the initial ◦

deformed equilibrium configuration B given in P5.12 and P5.13. P5.26 Let us assume that the simple shear given in P5.15 is infinitesimal ; i.e. |γ| << 1. Let us assume also that the considered material is linearly elastic and isotropic. Find the instantaneous elasticity ω corresponding to the initial deformed ◦

equilibrium configuration B . P5.27 Show that the stress free reference configuration B of a linearly hyperelastic anisotropic material is locally stable if its elasticity tensor c is positive definite. P5.28 By direct verification show that the elasticities cklmn of a linearly elastic orthotropic material can be expressed by the relations cklmn = δkl δmn Ckn + (1 − δkl )(δkm δln + δkn δlm )Ckl (!), with G23 = G32 = G44 , G31 = G13 = G55 , G12 = G21 = G66 . In the above relation, there is no summation relative to the indices appearing twice or three times. The sign (!) will be used in what follows to indicate this fact. P5.29 We assume that a linearly elastic orthotropic material is submitted to the initial infinitesimal deformation (5.6.17)−(5.6.18). Verify by direct computations that ω1112 = ω2111 = 0 and ω1223 = ω3221 = 0. What is the general rule giving the vanishing instantaneous elasticities? P5.30 Show that in the case given in P5.29, the instantaneous elasticities are given by the following relation: ◦

ωklmn = δkl δmn Ckn + (1 − δkl )(δkm δln + δkn δlm )Gkl + σ kn δlm (!). P5.31 Let us consider a linearly hyperelastic anisotropic material submitted to an initial applied infinitesimal homogeneous deformation. Find the conditions which must be fulfilled by the elasticity tensor c of the material and by the initial applied infinitesimal homogeneous deformation for which, in the body, can exist an incremental antiplane state relative to the plane Ox1 x2 . P5.32 In the situation given in P5.31, find the conditions which must be fulfilled by the elasticity tensor c and the initial applied deformation for which, in the body, can exist an incremental plane state relative to the plane Ox 1 x2 . ◦

P5.33 Let us assume that the initial deformed equilibrium configuration B of a linearly elastic orthotropic body corresponds to the displacement field given by the relations (5.6.17)−(5.6.18). Let us suppose that the elasticity c of the involved material is positive definite.

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5.7. PROBLEMS ◦

Find sufficient condition assuming local stability of B relative to (a) antiplane incremental states, with respect to the Ox1 x2 plane; (b) plane incremental states, with respect to the Ox1 x2 plane. P5.34 Let us consider a nonlinear hyperelastic and isotropic material. Since the material is isotropic, the specific strain energy u = u(G) can depend on Green’s strain tensor G only through the principal invariants I1, I2 and I3 of G; i.e. we have u = u(G) = u(I1 , I2 , I3 ) with I1 = trG, I2 = trG2 , I3 = trG3 . (a) Show that ∂I1 = 1, ∂G ∂I2 = 2G, ∂G ∂I3 = 3G2 . ∂G

(b) Show that Π=

∂u 2 ∂u ∂u ∂u G . G+3 1+2 (G) = ∂I3 ∂I2 ∂I1 ∂G

(c) Show that if the reference configuration B of the body in stress-free, the constitutive function u = u(I1 , I2 , I3 ) must satisfy the restriction ∂u (0, 0, 0) = 0. ∂I1

P5.35 Let us assume that the nonlinear hyperelastic isotropic material considered in P5.34 is submitted to the initial homogeneous deformation given in P5.12. ◦ ◦ ◦ ◦ (a) Find the invariants I 1 , I 2 , I 3 of G in terms of λ1 , λ2 , λ3 . ◦

◦

(b) Find Π and T in terms of the constitutive function u = u(I1 , I2 , I3 ) and λ1 , λ 2 , λ 3 . ◦

(c) Show that the components of the tensor K, defined by equation (5.2.22) given by the relations ◦

◦

◦

K klmn = δkl δmn akn + (1 − δkl )(δkm δln + δkn δlm ) µkl (!) where

◦

◦

akn = (Σk Σl + 2δkn Bkk ) u (!), ◦

◦

µkl = Bkl u

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CHAPTER 5. THE THREE-DIMENSIONAL LINEARIZED THEORY

and Σk =

◦2 ◦ ∂ ∂ ∂ (!), + 3 Gkk + 2 Gkk ∂I3 ∂I2 ∂I1

Bkl =

◦ ∂ 3 ◦ ∂ (!). + (Gkk + Gll ) ∂I3 2 ∂I2

◦

◦

P5.36 Find the instantaneous elasticity Ωklmn and ω klmn in the case considered in P5.35. ◦

◦

◦

◦

◦

P5.37 Find the nonvanishing components of Ω and ω in terms of akn , µkl , ◦

Πkn , T kn and λ1 , λ2 , λ3 in the case considered in P5.35. P5.38 Using the results obtained in P5.37, find the conditions which must ◦

be fulfilled by λ1 , λ2 , λ3 and Π if we suppose that the instantaneous elasticities ◦

Ωklmn satisfy the following supplementary symmetry relations: ◦

◦

◦

◦

Ωklmn = Ωlkmn and Ωklmn = Ωklnm . P5.39 Analyze problem P5.38 assuming uniform initial expansion; i.e. supposing that λ1 = λ 2 = λ 3 . P5.40 Analyze problem P5.38 assuming linearly hyperelastic material and infinitesimal initial deformation; i.e. |λ1 − 1| , |λ2 − 1| , |λ3 − 1| << 1. P5.41 Show that if, in P5.35, the initial applied deformation satisfies the restriction λ1 = λ 2 , ◦

◦

the instantaneous elasticities Ωklmn and ω klmn satisfy the supplementary symmetry relations (5.6.77), (5.6.78) characterizing the evidence of a cylindrical symmetry ◦

in the initial deformed equilibrium state B . P5.42 Let us consider a body and let Bt be its current configuration. Let us assume that the boundary ∂Bt of the body is submitted to a hydrostatic pressure tn (x, t) = −p(x, t)n(x, t), where p = p(x, t) > 0 is a scalar field and n = n(x, t) is the unit outward normal to ∂Bt . Using the updated Lagrangean method, find the incremental traction boundary condition in this case. Do we have here a dead load incremental traction boundary problem? P5.43 Analyze and solve the above problem using the Lagrangean approach to incremental boundary value problems. P5.44 Prove the validity of equations (5.6.99).

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355

5.7. PROBLEMS P5.45 Prove that ∆1 =

1 ∂2 1 ∂ ∂2 ∂2 ∂2 + 2 2. = 2+ + 2 2 r ∂r r ∂θ ∂r ∂x2 ∂x1

P5.46 Verify equation (5.6.120)4,5,6 expressing θθr , θθθ and θθz in terms of ur , u θ , u z . P5.47 Express θθr , θθθ and θθz in tensor of Guz’s displacement potentials ψ and χ. P5.48 Find the mechanical significance of the Cartesian components θ 11 , θ12 ,..., θ33 of the incremental nominal stress tensor θ. P5.49 Find the mechanical significance of the cylindrical components θ rr , θrθ ,..., θzz of the incremental nominal stress tensor θ. P5.50 Let us consider an infinite body deformed as in P5.12 and P5.13. Find ◦

in this case the acoustic tensor Q (N) and the equation giving the velocity of the incremental progressive or harmonic plane waves. P5.51 Let us consider an infinite body deformed as in P5.15 and P5.13. Find ◦

in this case the acoustic tensor Q (N) and the equation giving the velocity of the incremental progressive or harmonic plane waves. P5.52 Analyze problem P5.51 assuming infinitesimal initial applied shear; i.e. supposing that |γ| << 1. ◦

P5.53 Find the acoustic tensor q (n) appropriate to the updated Lagrangean approach to the incremental problem. P5.54 Using the updated Lagrangean approach, find sufficient condition assuming the existence of the incremental harmonic plane waves for any direction of propagation.

Bibliography [5.1] Malvern, L. F., Introduction to the mechanics of continuous medium, PrentinceHall, Inc. London, 1969. [5.2] Guz, A.N., Stability of three-dimensional deformable bodies, Naukova Dumka, Kiev, 1971 (in Russian). [5.3] Guz, A.N., Stability of elastic bodies at large deformations, Naukova Dumka, Kiev, 1973 (in Russian). [5.4] Guz, A.N., The fundamentals of stability of mine working, Naukova Dumka, Kiev, 1977 (in Russian). [5.5] Guz, A.N., Stability of elastic bodies submitted to hydrostatic pressure, Naukova Dumka, Kiev, 1979 (in Russian).

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[5.6] Guz, A.N., Three dimensional theory of stability of bars, plates and shells, Visha Schola, Kiev, 1980 (in Russian). [5.7] Guz, A.N., Mechanics of brittle fracture of materials, with initial stresses, Naukova Dumka, Kiev, 1983 (in Russian) [5.8] Guz, A.N., Fundamentals of three-dimensional theory of stability of de-formable bodies, Visha Schola, Kiev, 1986 (in Russian). [5.9] Ogden, R.W., Non-linear elastic deformations, John Wiley & Sons, New York, 1984. [5.10] Eringen, A.C., Maugin, G.A., Electrodynamics of continua, Vol. I, Foundations and Solid Media, Springer, New York, 1990. [5.11] So´os, E., Stability, resonance and stress concentration in prestressed piezo electric crystals containing a crack, Int. J. Engn. Sciences, 35, 1997. [5.12] Pearson, C.E., General theory of elastic stability, Q. Appl. Math., 14, 133144, 1956. [5.13] Hill, R., On uniqueness and stability in the theory of finite elastic strain, J. Mech. Phys. Solids, 5, 229-241, 1957. [5.14] Ericksen, J.L., A thermo-kinetic view of elastic stability, Int. J. Solids Structures, 2, 573-580, 1966. [5.15] Gurtin, M.E., Modern continuum thermodynamics, 168-213 in Mechanics Today, Vol.1, Ed. S. Nemat Nasser, Pergamon Press, 1973. [5.16] Gurtin, M.E., Thermodynamics and the potential energy of an elastic body, J. Elasticity, 3, 1-4, 1973. [5.17] Gurtin, M.E., Thermodynamics and the energy criterion for stability, Arch. Rat. Mech. Analysis, 52, 93-102, 1973. [5.18] Gurtin, M.E., Thermodynamics and stability, Arch. Rat. Mech. Analysis, 59, 63-96, 1975. [5.19] Truesdell, C., Noll, W., The non-linear field theories of mechanics, Handbuch der Physik, Band III/3, Ed. S. Fl¨ ugge, Springer, Berlin, Heidelberg, New York, 1965.

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Chapter 6

BUCKLING OF COMPOSITE STRIPS AND BARS 6.1

Internal and superficial stability

We analyze some kinds of instability, characteristic of the bodies having internal structure, such as, for instance, the fiber-reinforced composite materials. We assume a linear orthotropic elastic body, submitted to small, infinitesimal initial deformations. With appropriate modification, the results can be extended to nonlinear hyperelastic solids submitted to arbitrary initial deformations. The phenomenon of internal instability was analyzed for the first time by Biot [6.1]. As Guz [6.2] shows, this phenomenon has a well-defined physical meaning and can be elucidated taking into occurrent the fact that all materials have an internal structure. The phenomenon, named internal instability, concerns the loss of stability of this structure, and depends on the geometrical and mechanical characteristics of the internal structure, but it is independent of geometrical characteristics of the body, as a whole. A rigorous study of this phenomenon must take explicitly into account the parameters describing the internal structure of the body. In continuum approach, it is just these parameters that are neglected. The material is considered as a structureless continuum and its behavior is described by constitutive equations containing material constants, determined using experimental and theoretical procedures. At this phenomenological level, the structural properties of the body are implicitly reflected by the values of the elasticities occurring in the stress-strain relations of the material. Since the internal instability is not related to the geometrical characteristics of the body as a whole, nor to the influence of the boundary conditions on the

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behavior of the body, the problems concerning internal instability lead to the study of the behavior of an infinite body, submitted to well-defined given loads, acting at large distances. This is the reason why we assume that the initial deformed and prestressed infinite body is in a homogeneous equilibrium state. Consequently, its incremental behavior is governed by the differential system (see Equations (5.6.11) and (5.6.12)): ∂2 , (6.1.1) Lml ul = 0, Lml = ωklmn ∂xk ∂xn where the instantaneous elasticities ωklmn depend on the elasticities cklmn of the ◦ material, as well as on the initial applied stress σ . We assume that the stress free reference configuration of the body is locally ◦ stable. Hence, the differential system (6.1.1) corresponding to σ = 0 is elliptic. If, on a given loading path, the instantaneous elasticity ω is positive definite; i.e. the system (6.1.1) conserves its ellipticity, the solutions of various incremental boundary value problems are unique, have a local character, and internal instability does not occur. If for some critical values of the loading parameters, the system (6.1.1) ceases to be elliptic, and becomes hyperbolic, the behavior of the perturbation, described by (6.1.1), changes radically and their local character is lost. A perturbation, appearing in a small domain, can propagate along the characteristics, producing considerable damages in the material. By internal instability we mean just the occurrence of such essential change in the behavior of the perturbation. Hence, internal instability occurs when on a given loading path the differential system (6.1.1) ceases to be elliptic. The corresponding critical values of the loading parameters are determined using the above criterion. According to Guz’s representation theorem (see Equation (5.6.16)), we can replace the system (6.1.1) by a simple equation,

(detP ) ϕ(j) = 0, P = [Plm ] , j = 1, 2, 3,

(6.1.2)

satisfied by the displacement potentials ϕ(j) . Consequently, the critical values of the loading parameters, producing internal instability, are those values for which on a given loading path the ellipticity of the equation (6.1.2) is for the first time lost. As we already know, there exists an internal connection between the loss of internal stability, the loss of incremental uniqueness and the loss of existence of real velocities of propagation of incremental plane harmonic waves. According to the above observation, we can say that the occurrence of the internal instability is guaranteed if the instantaneous elasticities satisfy the condition ζlk ωklmn ζmn = 0 for any ζmn such that ζkn ζmn 6= 0. (6.1.3)

In the following we shall illustrate these general ideas by some special cases. To do this, we observe that the implications of the general criterion (6.1.3) can be more easily determined using the factorized forms of the equations satisfied by the displacement potentials.

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6.1. INTERNAL AND SUPERFICIAL STABILITY

We start the analysis considering incremental plane states. Examining the equation (5.6.56) and the restrictions (5.6.55), it can be seen that internal instability occurs when η1 = 0 or η2 = 0 for ω2222 6= 0 and ω2112 6= 0.

(6.1.4)

According to equations (5.6.57) and (5.6.58), the above condition will be satisfied if B = 0, or, more exactly, if ω1221 = 0 or ω1111 = 0 for ω2222 6= 0 and ω2112 6= 0.

(6.1.5)

Taking into account the values (5.6.47) of the involved instantaneous elasticities, the above conditions become ◦

◦

◦

◦

C66 + σ 11 = 0 or C11 + σ 11 = 0 for C22 + σ 22 6= 0 and C66 + σ 22 6= 0.

(6.1.6)

The reference configuration of the body being assumed locally stable, the elasticities of the material satisfy the inequalities 2 > 0. C11 , C22 , C66 , C11 C22 − C12

(6.1.7) ◦

Thus, the relations (6.1.6) show that internal instability can occur only if σ 11 is a compressive stress; i.e. ◦ σ 11 < 0. We assume now that, on the considered loading path, ◦

σ 22 = 0.

(6.1.8)

In this case, the restrictions (6.1.6)3,4 are satisfied. Internal instability occurs only ◦ if the applied compressive stress σ 11 satisfies the condition ◦

◦

σ 11 = −C66 or σ 11 = −C11 .

(6.1.9)

We suppose that the material is a fiber-reinforced composite, and the fibers have the direction of the applied compressive force. As we know from the results presented in Chapter 3, for such composite materials, the transverse shear rigidity C66 is much smaller as the longitudinal axial rigidity; i.e. C66  C11 .

(6.1.10)

Hence, internal instability occurs if the compressive stress, applied in the fibers direction, reaches the critical value ◦ ci

σ 11 = −C66 .

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(6.1.11)

360

CHAPTER 6. BUCKLING OF COMPOSITE STRIPS AND BARS ◦ ci

The internal deformation produced by σ 11 are infinitesimal, since the compressive stress acts in the fibers direction and (6.1.10) is true. Hence, the conditions in which the used incremental theory is applicable are fulfilled. Consequently, the loss of internal stability actually can occur in a fiber-reinforced composite, if the applied compressive force acts in the fibers direction. If the material is isotropic, C66 and C11 have the same order of magnitude and internal instability cannot occur for compressive force, for which the linear theory of elasticity is applicable. The loss of internal stability for fiber-reinforced composite materials, for relatively small compressive stresses, is a direct consequence of their structured character, reflected at phenomenological level by the relation (5.6.10), expressing the strong anisotropy of the composite. This is the reason why the internal instability is also named structural instability. We analyze now the transversally isotropic, cylindrically symmetric state. According to the restriction (5.6.84) and to the equation (5.6.85), internal instability occurs if ξ1 = 0 or ξ2 = 0 or ξ3 = 0 for ω1111 6= 0, ω1221 6= 0 and ω1331 6= 0.

(6.1.12)

Using the relation (5.6.76), we can conclude that the above condition will be satisfied if ω3113 = 0 or ω3333 = 0 for ω1111 6= 0, ω1221 6= 0 and ω1331 6= 0.

(6.1.13)

Hence, taking into account the expressions of the involved instantaneous elasticities (5.6.77), we can see that the critical values of the loading parameters must satisfy the conditions ◦

◦

C44 + σ 33 = 0 or C33 + σ 33 = 0, ◦ ◦ for 12 (C11 − C12 ) + σ 11 6= 0 and C44 + σ 11 6= 0.

(6.1.14)

Since the reference configuration is assumed to be locally stable, the involved elasticities satisfy the restrictions C11 , C33 , C44 , 2C66 = C11 − C12 > 0.

(6.1.15)

We can see again that internal instability can occur only in the presence of a compressive stress; i.e. only if ◦ σ 33 < 0. We assume that on the loading path ◦

σ 11 = 0.

(6.1.16)

In this case, the restrictions (6.1.14)3,4,5 are satisfied, and internal instability occurs only if ◦ ◦ σ 33 = −C44 or σ 33 = −C33. (6.1.17)

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6.1. INTERNAL AND SUPERFICIAL STABILITY

We suppose that the material is a fiber-reinforced composite, and the compressive force is applied in the direction of fibers. As we know from the results presented in Chapter 3, for such a composite the longitudinal, axial rigidity C 33 is much greater than the axial shear rigidity C44 ; i.e. C44  C33.

(6.1.18)

Hence, internal instability occurs when the applied compressive stress reaches its critical value ◦ ci σ 33 = −C44 . (6.1.19) As before, the loss of the internal stability actually can take place, for relatively small compressive stresses, and it is a direct consequence of the structured character of a fiber-reinforced composite. Summing up the obtained results, we recall that for fiber-reinforced composite materials, their shear moduli have relatively small values. The applied compressive stresses can reach values having the order of magnitude of these moduli, producing however infinitesimal strains. Hence, by compression in the direction of the fibers, the composite can lose its internal stability, for relatively small compressive forces. To avoid such dangerous situations, leading to the destruction of the material, the involved compressive forces must be drastically limited in their magnitude. The second dangerous case, which can occur when we are dealing with composite materials, concerns their superficial instability. This phenomenon is characterized by the appearance of possible eigenmodes having significant amplitudes near the free-boundary of the body, which decay rapidly if the distance from the boundary increases. Rayleigh surface waves, known in classical linear elastodynamics, have similar behavior. Their statical analogue does not exist in classical linear elastostatics for bodies in stress-free configurations, but may exist in a prestressed material. The phenomenon of superficial instability was first analyzed by Biot [6.1] for an incompressible prestressed elastic half space. The case of orthotropic materials, in the framework of the linearized theory, was for the first time analyzed by Guz [6.2]. Let as assume first that the domain occupied by the orthotropic material is the half space −∞ < x1 < ∞, x2 < 0, −∞ < x3 < ∞ as shown in Figure 6.1. We analyze the possibility of superficial instability assuming incremental plane states, relative to the plane x1 x2 . We assume that the material is a fiber-reinforced composite, the fibers having the direction of the axis Ox1 . We assume also that the initial applied stress acts in the direction of fibers; i.e. ◦

◦

σ 11 6= 0 and σ 22 = 0.

(6.1.20)

According to equations (5.6.47), the involved instantaneous elasticities are ◦

ω1111 = C11 + σ 11 , ω1122 = ω2211 = C12 , ω2222 = C22 , ◦ ω1221 = C66 + σ 11 , ω1212 = ω2121 = ω2112 = C66 .

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(6.1.21)

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CHAPTER 6. BUCKLING OF COMPOSITE STRIPS AND BARS

Figure 6.1: Superficial instability. Incremental plane states.

Since the elasticities satisfy the inequalities (6.1.7), the restrictions (5.6.55) are fulfilled. Hence, we can use Guz’s representation (5.6.53). We recall that the displacement potentials ϕ(1) and ϕ(2) must satisfy the differential equation (5.6.56). We suppose that (6.1.22) η12 6= η22 , a restriction satisfied for nearly all orthotropic materials. In this case, according to Baggio’s theorem, the displacement potentials ϕ(α) have the form ϕ(α) = ϕ(α1) + ϕ(α2) , α = 1, 2, where ϕ(αβ) satisfy the equations   2 2 ∂ 2 ∂ + ηβ 2 ϕ(αβ) = 0, α, β = 1, 2. ∂x2 ∂x21

(6.1.23)

(6.1.24)

Since the boundary x2 = 0 of the half space is stress-free, the following boundary condition must be satisfied: θ21 = θ22 = 0 for x2 = 0.

(6.1.25)

Also, for large distances from the boundary of the half-space, we must have lim

x2 →−∞

{u1 (x1 , x2 ) , u2 (x1 , x2 )} = 0.

(6.1.26)

As we know (see Section 5.6), η12 and η22 cannot be negative or vanishing numbers if the considered prestressed equilibrium configuration is locally stable. In order to simplify the analysis, we suppose that η12 and η22 are positive real numbers. If η12 and η22 are complex numbers, the procedure must be modified accordingly. In order to find the appropriate solutions of the equations (6.1.24), we shall consider the equation

∂2ϕ ∂2ϕ + η 2 2 = 0 with η > 0. 2 ∂x2 ∂x1

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(6.1.27)

6.1. INTERNAL AND SUPERFICIAL STABILITY

363

We look for solutions of the form ϕ (x1 , x2 ) = Z1 (x1 ) Z2 (x2 ) . In this case, from (6.1.27), we get

Z 00 Z100 = −η 2 2 = ±a2 , Z2 Z1

where a > 0 is an arbitrary real number. From the above equations, it results that ϕ can have one of the following forms ϕ (x1 , x2 ) = {A exp(−aηx2 ) + B exp(aηx2 )} sin ax1 , ϕ (x1 , x2 ) = {A exp(−aηx2 ) + B exp(aηx2 )} cos ax1 , a ϕ (x1 , x2 ) = {A exp(−ax1 ) + B exp(ax1 )} sin x2 , η a ϕ (x1 , x2 ) = {A exp(−ax1 ) + B exp(ax1 )} cos x2 , η

where A and B are arbitrary constants and a, η > 0. Since the incremental displacement field must converge to zero, when x 2 → −∞, we can use, as possible displacement potentials in our stability analysis only, the functions

a a ϕ (x1 , x2 ) = B1 exp( x2 ) sin ax1, ϕ (x1 , x2 ) = B2 exp( x2 ) cos ax1 , η η

B1 and B2 being arbitrary constants. We take a = π/l > 0,

(6.1.28)

(6.1.29)

where the arbitrary positive number l > 0 is the half wave-length of the possible eigenmode by which the superficial instability manifests. According to equations (6.1.23) and (6.1.28), the displacement potentials can have the following forms: ϕ(1) = (A1 eaη1 x2 + A2 eaη2 x2 ) sin ax1 , ϕ(2) = (B1 eaη1 x2 + B2 eaη2 x2 ) cos ax1 ,

(6.1.30)

where A1 , A2 , B1 , B2 are arbitrary constants, a is given by (6.1.29), and η1 , η2 > 0.

(6.1.31)

First we suppose that ϕ(2) ≡ 0 and look for a possible eigenmode described by ϕ . (1)

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The incremental displacements u1 , u2 are expressed in terms of the potentials by equation (5.6.53). Introducing ϕ(1) in these relations, we get u1 = a2 (ω1122 + ω1212 ) (η1 A1 eaη1 x2 + η2 A2 eaη2 x2 ) cos ax1 ,

−ω1111 + η12 ω2112 A1 eaη1 x2 + −ω1111 + η22 ω2112 A2 eaη2 x2 sin ax1. (6.1.32) The incremental nominal stress components θ21 and θ22 can be evaluated using the incremental constitutive equations (5.6.54)3,4 . We get u2 = a 2



θ21 = −a3 θ22 = a3

2 X

α=1

2 X

α=1


ω1212 ω1111 + ω2112 ω1122 ηα2 Aα eaηα x2 cos ax1 ,

 ηα ω1122 (ω1122 + ω1212 ) − ω2222 ω1111 + ω2222 ω2112 ηα2

·Aα eaηα x2 sin ax1 .

(6.1.33)

From (6.1.33), it follows that the homogeneous boundary conditions (6.1.25) are satisfied if and only if there exists nonvanishing constants A1 and A2 satisfying the following homogeneous system:

ω1212 ω1111 + ω2112 ω1122 η12 A1 + ω1212 ω1111 + ω2112 ω1122 η22 A2 = 0,  η1 ω1122 (ω1122 + ω1212 ) − ω1111 ω2222 + ω2222 ω2112 η12 A1

 + η2 ω1122 (ω1122 + ω1212 ) − ω1111 ω2222 + ω2222 ω2112 η22 A2 = 0.

(6.1.34) Hence, there exists eigenmode, that is, superficial instability occurs, if and only if the determinant ∆ of the system is vanishing. Elementary computations show that ∆ has the expression ∆ = (η2 − η1 ) Γ

(6.1.35)

where Γ = ω1212 ω1111 {ω1122 (ω1122 + ω1212 ) − ω1111 ω2222 } + {−ω2112 ω1122 [ω1122 (ω1122 + ω1212 ) − ω1111 ω2222 ] + ω1212 ω1111 ω2222 ω2112 }
·η1 η2 + ω1212 ω1111 ω2222 ω2112 η12 + η22 + ω2112 ω1122 ω2222 ω2112 η12 η22 .

(6.1.36)

Since η1 6= η2 , from (6.1.35), it follows that ∆ = 0 if and only if Γ = 0.

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(6.1.37)

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6.1. INTERNAL AND SUPERFICIAL STABILITY

The expression (6.1.36) of Γ can be considerably simplified if we take into account the relations (5.6.57) and (5.6.58). According to these equations, 2

η12 + η22 = 2A =

ω1111 ω2222 + ω1221 ω2112 − (ω1122 + ω1212 ) , ω2222 ω2112

η12 η22 = B =

ω1111 ω1221 . ω2222 ω2112

Introducing these expressions in (6.1.37), we obtain r ω1111 Π, Γ = (ω1212 + ω1122 ) ω2222

(6.1.38)

with



√ 2 2 . + ω2112 ω1221 ω1111 ω2222 − ω1122 ω1111 ω2222 ω1221 ω2112 − ω1212 (6.1.39) We suppose that the above expressions have sense; i.e. Π=

√

ω1111 , ω2222 , ω2112 , ω1221 > 0.

(6.1.40)

According to (6.1.20)2 and (5.4.47), the above conditions are satisfied if ◦

◦

C11 + σ 11 > 0, C66 + σ 11 > 0, C22 > 0, C66 > 0.

(6.1.41)

The last two conditions are fulfilled since we have assumed that the stress-free reference configuration is locally stable. According to equations (6.1.9), the first two conditions mean that we assume an internally stable prestressed equilibrium configuration. Now, from (6.1.38) we can conclude that the characteristic equation (6.1.37) is satisfied if ω1122 + ω1212 = 0 or Π = 0. (6.1.42) Equations (6.1.20)2 and (5.6.47) show that the first equation requires C12 + C66 = 0. According to (2.2.74), (2.2.76), (2.2.81) and (2.2.85), this relation cannot he satisfied for composite materials since C66 > 0 and C12 > 0. Accordingly, superficial instability can occur if and only if the second equation (6.1.42) is fulfilled. From (5.6.47), (6.1.20)2 and (6.1.39), we can conclude that this ◦ equation, containing the unknown σ 11 , has the form r r     ◦ ◦ ◦ ◦ 2 + C22 σ 11 = 0. C66 + σ 11 C66 C11 C22 − C12 C11 + σ 11 C22 C66 σ 11 +

(6.1.43)

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If the stress-free reference configuration is locally stable, the inequalities (6.1.7) are satisfied. Then we can conclude that the left-hand side of the above ◦ equation is positive if σ 11 = 0. Hence, a static analogue of Rayleigh’s surface waves cannot exist, if the stress-free reference configuration of the body is locally stable. ◦ The critical value of σ 11 , for which superficial instability can occur in a prestressed equilibrium configuration, must satisfy equation (6.1.43). We shall analyze now if such critical value may exist. After some obvious transformations, (6.1.43) becomes 

2    ◦ ◦ ◦2 ◦ 2 + C22 σ 11 = 0. (6.1.44) C11 + σ 11 C22 C66 σ 11 − C66 + σ 11 C11 C22 − C12 We introduce the dimensionless ratios ◦

x = σ 11 /C11 , ε = C66 /C11 > 0.

(6.1.45)

In this way, (6.1.44) takes the form

+



C22 C11

C22 C11

      C2 C22 C22 C22 C22 x2 − 12 −1 +2 ε − ε x3 + 2 C11 C11 C11 C11 C11 2    2 2 2 C12 C22 C22 C12 C22 C12 = 0. (6.1.46) − 2 x+ε − 2 + 2ε − 2 C11 C11 C11 C11 C11 C11 

We recall that for a fiber-reinforced composite, C66  C11 ; hence, ε  1.

(6.1.47)

Consequently, using an iterative method, we look for a root having the following form: x = x0 + εx1 + ε2 x2 + ε3 x3 . (6.1.48) Introducing (6.1.48) in (6.1.46), and neglecting terms of order ε4 and higher, we determine successively the unknowns x0 , x1 , x2 and x3 . An elementary but long computation gives x0 = 0, x1 = −1, x2 = 0, x3 =

2 2 C22 C11 C11 . 2 )2 C22 (C11 C22 − C12

(6.1.49)

Introducing (6.1.49) in (6.1.48), we get 2 2 C11 C22 C11 x = −ε 1 − ε2 2 )2 C22 (C11 C22 − C12

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!

.

(6.1.50)

6.1. INTERNAL AND SUPERFICIAL STABILITY

367

◦ cs

Using the notation (6.1.45) for the critical value σ 11 for which superficial instability occurs, we obtain the following expression: ! 2 2 2 C22 C11 C66 ◦ cs σ 11 = −C66 1 − . (6.1.51) C11 C22 (C11 C22 − C12 )2 ◦ ci

We recall that the critical value σ 11 , for which internal (structural) instability occurs, is given by the equation (6.1.11). Hence, it results ! 2 2 2 C22 C11 C66 ◦ cs ◦ ci σ 11 = σ 11 1 − . (6.1.52) 2 )2 C11 C22 (C11 C22 − C12

Using the relations (2.2.85), we can express the critical values in terms of the engineering constants of the material. We get   G2 ◦ cs ◦ ci σ 11 = −G12 , σ 11 = −G12 1 − 12 (1 − ν13 ν31 ) (1 − ν23 ν32 ) . (6.1.53) E1 E2

In a fiber-reinforced composite G212  E1 E2 and 0 <

G212 (1 − ν13 ν31 ) (1 − ν23 ν32 ) < 1. E1 E2

Consequently, according to (6.1.53), ◦ si

◦ ci

σ 11 <σ 11 < 0.

(6.1.54)

Hence, the critical load-producing superficial instability is a compressive one, as well as the critical load-producing structural (internal) instability. Moreover, the superficial instability appears before the structural one. Thus, we can see that our assumptions are consistent. Since the critical load leading to superficial instability has the order at magnitude of the transverse shear modulus, the losing of superficial stability, for relatively small compressive stresses, can really occur in a fiber-reinforced composite loaded by compressive forces acting in the fibers direction. To avoid a dangerous situation, due to the structured character of the composites, the admissible compressive forces must be drastically limited. It can be shown that assuming ϕ(1) ≡ 0 and taking ϕ(2) as in equation (6.1.30)2 , we obtain the same results as before. Let us assume now that the domain occupied by the material is the half space −∞ < x1 , x2 < ∞, x3 < 0. We suppose that the material is fiber-reinforced in the planes x3 = const., and in these planes there is no preferred direction. In this case, we can assume that the material is transversally isotropic, x 3 = const. being the isotropy planes. Also, we suppose that ◦

◦

◦

σ 11 = σ 22 , σ 33 = 0.

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(6.1.55)

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Hence, the prestressed equilibrium configuration of the composite is in a transversally isotopic, cylindrically symmetric state. We shall analyze the conditions in which superficial instability can occur through cylindrically symmetric incremental deformations. In order to study the problem, we can use Guz’s representation (5.6.102). Also, we assume that the stress-free reference configuration of the composite is locally stable and its prestressed equilibrium state is structurally (internally) stable. Taking into account the assumptions made, from (5.6.77) we obtain the following values of the nonvanishing instantaneous elasticities: ω1111 ω1133 ω1313 ω1212

= ω2222 = ω3311 = ω3131 = ω2121

◦

= C11 + σ 11 , ω3333 = C33 , ω1122 = ω2211 = C12 , = ω2233 = ω3322 = C13 , = ω2323 = ω3232 = ω3113 = ω3223 = C44 , ◦ ◦ = C66 , ω1331 = ω2332 = C44 + σ 11 , ω1221 = C66 + σ 11 .

(6.1.56)

Since the prestressed equilibrium state is internally stable, the parameters ξ1 , ξ2 , ξ3 given by equation (5.6.86)−(5.6.87) cannot vanish. We shall assume that ξ1 , ξ2 , ξ3 are real numbers, a condition which is satisfied for nearly all composite materials, for relatively small initial applied stresses. Since the equations (5.6.86)−(5.6.87) give only the squares of these parameters, we shall assume that ξ1 , ξ2 , ξ3 > 0.

(6.1.57)

Also, we suppose that ξ1 , ξ2 , ξ3 are distinct; i.e. ξ1 6= ξ2 6= ξ3 6= ξ1 .

(6.1.58)

We know that Guz’s displacement potentials ψ and χ must satisfy the differential equations (5.6.103). Since ξ2 6= ξ3 , χ = χ2 + χ3 , where χ2 and χ3 satisfy the second order differential equations (5.6.105). Before looking for appropriate solutions of equations (5.6.103)1 and (5.6.105), we recall that the boundary x3 = 0 must be stress-free and the amplitudes of the possible eigenmodes converge to zero, when the distance from the boundary increases. Hence, the incremental nominal stresses and displacements must satisfy the following conditions: θ31 = θ32 = θ33 = 0 for x3 = 0,

(6.1.59)

and lim

x3 →−∞

ul (x1 , x2 , x3 ) = 0, l = 1, 2, 3.

(6.1.60)

Examining the structure of equations (5.6.103)1 and (5.6.105), we are ready to study the possible solutions of the following equation:   2 2 ∂2 ∂ 2 ∂ ϕ = 0, ξ > 0. (6.1.61) + ξ + ∂x23 ∂x22 ∂x21

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369

6.1. INTERNAL AND SUPERFICIAL STABILITY We look for a solution having the form ϕ (x1 , x2 , x3 ) = Y (x1 , x2 ) Z (x3 ) .

(6.1.62)

Introducing (6.1.62) in (6.1.61), we get

Z 00 ∆1 Y = −a2 , = −ξ 2 Z Y

(6.1.63)

where a is a positive real number. We consider only this alternative, taking into account the condition (6.1.60) imposed at large distances from the boundary. The equation for Y becomes ∆1 Y + a2 Y = 0.

(6.1.64)

Y (x1 , x2 ) = Y1 (x1 ) Y2 (x2 ) .

(6.1.65)

We assume that Introducing (6.1.65) into (6.1.64), it results in   00 Y2 Y100 2 + a = −b2 , =− Y2 Y1

(6.1.66)

where b is a positive real number, such that c2 = a 2 − b2 ,

(6.1.67)

and c is also a positive real number. Now, for Y1 and Y2 , we find the following possible expressions: Y1 (x1 ) = sin bx1 or Y1 (−x1 ) = cos bx1 , Y2 (x2 ) = sin cx1 or Y2 (x2 ) = sin cx2 .

(6.1.68)

We now return to equation (6.1.63)2 satisfied by Z; with (6.1.66), we get Z 00 −

b2 + c 2 Z = 0. ξ2

(6.1.69)

This equation has the following solutions: a

a

Z1 (x3 ) = e ξ x3 or Z2 (x3 ) = e− ξ x3 .

(6.1.70)

Since ξ > 0, the second alternative must be rejected, as we can see examining the condition (6.1.60) imposed at large distances from the boundary. Consequently, according to (6.1.62), (6.1.68) and (6.1.70), ϕ can have one of the following expressions: a

a

ϕ = e ξ x3 sin bx1 sin cx2 , ϕ = e ξ x3 sin bx1 cos cx2 , a a ϕ = e ξ x3 cos bx1 sin cx2 , ϕ = e ξ x3 cos bx1 cos cx2 .

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(6.1.71)

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CHAPTER 6. BUCKLING OF COMPOSITE STRIPS AND BARS

All these expressions can be used to satisfy the boundary condition (6.1.60). Taking into account also the condition (6.1.59) imposed on the boundary x 3 = 0, and following Guz [6.2], we assume that the displacement potentials ψ and χ have the following form:   a a a x x x ψ = A1 e ξ1 3 sin bx1 sin cx2 , χ = A2 e ξ2 2 + A3 e ξ3 3 cos bx1 cos cx2 , (6.1.72)

where A1 , A2 , A3 are arbitrary constants, and a, b, c are given by the equations q (6.1.73) b = π/l1 , c = π/l2 , a = π 1/l12 + 1/l22 .

The arbitrary positive constants l1 and l2 represent the half wave-lengths of the possible eigenmode, in directions x1 and x2 , respectively. We must determine now the incremental displacements u1 , u2 , u3 , corresponding to the potentials ψ and χ, using to this purpose the Guz’s representation formulas (5.6.102). Elementary computations given o  n a a a x x x sin bx1 cos cx2 , u1 = cA1 e ξ1 3 + ab ξ2−1 A2 e ξ2 3 + ξ3−1 A3 e ξ3 3 o  n a a a x x x cos bx1 sin cx2 , u2 = −bA1 e ξ1 3 + ac ξ2−1 A2 e ξ2 3 + ξ3−1 A3 e ξ3 3
a −1 x u3 = a2 (ω1133 + ω1313 ) { −ω1111 + ξ2−2 ω3113 A2 e ξ2 3
a x (6.1.74) + −ω1111 + ξ3−2 ω3113 A3 e ξ3 3 } cos bx1 cos cx2 .

The involved incremental nominal stress components can now be determined, using equations (5.6.117). We find
a a x −2 ξ2 x 3 θ31 = a{cξ1−1 ω3113 A1 e ξ1 3 + ω1133ab +ω1313 [ ω1313 ω1111 + ξ2 ω3113 ω1133 A2 e


a + ω1313 ω1111 + ξ3−2 ω3113 ω1133 A3 e ξ3 x3 ]} sin bx1 cos cx2 , a

θ32 = a{−bξ1−1 ω3113 A1 e ξ1

θ33 =

x3

+

ac ω1133 +ω1313 [


a x ω1313 ω1111 + ξ2−2 ω3113 ω1133 A2 e ξ2 3


a + ω1313 ω1111 + ξ3−2 ω3113 ω1133 A3 e ξ3 x3 ]} cos bx1 sin cx2 , −1 a3 ω1133 +ω1313 {ξ2


a x 2 + ω1133 ω1313 − ω1111 ω3333 + ξ2−2 ω3333 ω3113 A2 e ξ2 3 ω1133


a x 2 +ξ3−1 ω1133 + ω1133 ω1313 − ω1111 ω3333 + ξ3−2 ω3333 ω3113 A3 e ξ3 3 } cos bx1

· cos cx2 .

(6.1.75) Imposing the boundary conditions (6.1.59), we arrive to the following homogeneous algebraic system which must be satisfied by the unknown constants A1 , A 2 , A 3 :
cξ1−1 ω3113 (ω1133 + ω1313 ) A1 + ab ω1313 ω1111 + ξ2−2 ω3113 ω1133 A2

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371

6.1. INTERNAL AND SUPERFICIAL STABILITY
+ab ω1313 ω1111 + ξ3−2 ω3113 ω1133 A3 = 0,


−bξ1−1 ω3113 (ω1133 + ω1313 ) A1 + ac ω1313 ω1111 + ξ2−2 ω3113 ω1133 A2
+ac ω1313 ω1111 + ξ3−2 ω3113 ω1133 A3 = 0,
2 −ξ2−1 ω1133 + ω1133 ω1313 − ω1111 ω3333 + ξ2−2 ω3333 ω3113 A2
2 +ξ3−1 ω1133 + ω1133 ω1313 − ω1111 ω3333 + ξ3−2 ω3333 ω3113 A3 = 0.

Eigenmodes can exist if and only if the determinant ∆ of this system is vanishing. Long and careful computations lead to the following value of ∆: ∆ = a3 (ξ2 − ξ3 ) (ξ1 ξ2 ξ3 )

−1

ω3113 (ω1133 + ω1313 ) Γ,

(6.1.76)

with
2 + ω1133 ω1313 − ω1111 ω3333 Γ = ω1313 ω1111 ω1133
+ ω1313 ω1111 ω3333 ω3113 ξ22 + ξ32 ξ2−2 ξ3−2
−1 −1 2 ξ2 ξ3 + ω1313 (ω1133 + ω1313 ) ω1111 ω3333 − ω1133 + ω3113 ω1133 ω3333 ω3113 ξ2−2 ξ3−2 .

(6.1.77)

Recalling equations (5.6.86) and (5.6.87), we find ξ12 + ξ22 =

2 ω3333 ω3113 ω1111 ω3333 + ω1331 ω3113 − (ω1133 + ω1313 ) 2 2 . , ξ 1 ξ2 = ω1111 ω1331 ω1111 ω1331

Introducing these values in (6.1.76), we finally obtain 3

∆ = a (ξ2 − ξ3 ) (ξ1 ξ2 ξ3 )

−1

ω3113 (ω1133 + ω1313 )

2

r

ω1111 Π, ω3333

(6.1.78)

with



√ 2 2 . + ω1331 ω3113 ω1111 ω3333 − ω1133 ω1111 ω3333 ω1331 ω3113 − ω1313 (6.1.79) Since the stress-free reference configuration is locally stable and the prestressed equilibrium state is structurally (internally) stable, from (6.1.78) it follows that superficial instability can occur if and only if Π=

√

Π = 0.

(6.1.80)

Introducing the values (6.1.56) of the instantaneous elasticities in this equation, we get the relation which must be satisfied by the critical value of the applied ◦ ◦ stresses σ 11 = σ 22 producing superficial instability of the prestressed half space r r     ◦ ◦ ◦ ◦ 2 + C33 σ 11 = 0. C44 + σ 11 C44 C11 C33 − C13 C11 + σ 11 C33 C44 σ 11 +

(6.1.81)

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372

CHAPTER 6. BUCKLING OF COMPOSITE STRIPS AND BARS As before, we introduce the dimensionless ratios ◦

x = σ 11 /C11 , 0 < ε = C44 /C11  1.

(6.1.82)

The last inequality is true since, for the fiber-reinforced composite, the longitudinal, axial rigidity C11 is greater than the axial shear rigidity C44 . According to (6.1.81), the equation which must be satisfied by x is C33 C11

+





      2 C33 C13 C33 C33 C33 3 − 1 x2 − 2 +ε 2 −ε x + C11 C11 C11 C11 C11

C2 C33 − 13 2 C11 C11



C33 C2 C33 + 2ε − 13 2 C11 C12 C11



x+ε



C2 C33 − 13 2 C11 C11



=0

(6.1.83)

The last inequality is true, since for the fiber-reinforced composites the longitudinal, axial rigidity C11 is much bigger than the axial shear rigidity C44 . The parameters ξ12 , ξ22 , ξ32 are given by the equations (5.6.86) and (5.6.87). Introducing in these relations the values (6.1.16) of the instantaneous elasticities we find v u C44 C33 C44 u 2  2 2 ,  , ξ = c ± ξ1 = tc − 2,3 ◦ ◦ ◦ C66 + σ 11 C44 + σ 11 C11 + σ 11

2c =

◦

2 C11 C33 − C13 − 2C13 C44 + (C33 + C44 ) σ 11    . ◦ ◦ C11 + σ 11 C44 + σ 11

◦

These relations show that for σ 11 = 0, the parameters ξ12 , ξ22 , ξ32 are positive, distinct, real numbers, since the material is transversally isotropic and its stressfree reference configuration is locally stable; i.e. its elasticity tensor c is positive definite. Also, from the equations giving ξ22 and ξ32 , it confirms that ξ22 = ξ32 if and ◦ only if σ 11 satisfies the equation n

◦

2 − 2C13 C44 + (C33 + C44 ) σ 11 C11 C33 − C13    ◦ ◦ = 4C33 C44 C11 + σ 11 C44 + σ 11 .

o2

With the notation (6.1.82), the above equation takes the form 

C33 −ε C11

2

x2 + 2



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C33 +ε C11



  C33 C13 C2 C33 ε (1 + ε) x ε − 2 − 2 − 13 2 C11 C11 C11 C11

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6.1. INTERNAL AND SUPERFICIAL STABILITY

+



C13 C2 C33 − 13 2 − 2C ε C11 C11 11

2

−4

C33 2 ε = 0. C11

(6.1.84)

We must analyze the relationship existing between the roots of equations (6.1.83) and (6.1.84). We recall now the equations (2.2.88) expressing the elasticities of the material in terms of the engineering constants, for a transversally isotropic material. Using these relations, we find 2 ν13 (1 + ν12 ) E3 E3 C13 1 − ν12 C33 = = 2 (E /E ) E , 2 (E /E ) E , C 1 − ν13 1 − ν13 C11 1 3 1 11 1 3 1
2 (E3 /E1 ) G13 (1 + ν12 ) 1 − ν12 − 2ν13 C44 . = ε= 2 E1 1 − ν13 (E3 /E1 ) C11

(6.1.85)

Recalling the assumed structure of the composite, reinforced with fibers in the x3 = const. planes, we can conclude that G13  E1 and E3  E1 .

(6.1.86)

Also, we assume that ν12 and ν13 satisfy the restrictions (2.2.84); i.e. 0 < ν12 < 1 and 0 < ν13 < 1. Thus, we can conclude that ε is a small positive quantity satisfying the assumed property (6.1.82). Consequently, to obtain the approximate value of the root of the characteristic equation (6.1.83), we can apply the iterative procedure used before. Neglecting terms of order ε4 and higher, and observing that the equations (6.1.46) and (6.1.83) have the same structure, we find ! 2 2 C11 C33 2 C11 . x = −ε 1 − ε 2 )2 C33 (C11 C33 − C13

Introducing here the values (6.1.84), we get
2 (E3 /E1 ) G13 (1 + ν12 ) 1 − ν12 − 2ν13 x=− 2 (E /E ) E1 1 − ν13 3 1  2   
G13 2 E3 2 . 1 − ν13 · 1 − 1 − ν12 E1 E1 E3

◦ cs

(6.1.87)

Now, using the expression (2.2.88)1 of C11 , for the critical value σ 11 for which the superficial instability occurs, we find  2   
G13 ◦ cs 2 E3 2 σ 11 = −G13 1 − 1 − ν12 . (6.1.88) 1 − ν13 E1 E1 E3

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CHAPTER 6. BUCKLING OF COMPOSITE STRIPS AND BARS

First of all, let us observe that according to (6.1.86), the quantity in the parenthesis in the right-hand side of the equation (6.1.88) is positive. Hence, superficial instability can occur only if the applied force is compressive. ◦ cs At the same time, we can compare the critical compressive stress σ 11 acting ◦ ◦ in fibers direction and producing superficial instability on a loading path σ 11 = σ 22 ◦ ci

◦

6= 0, σ 33 = 0, and the critical compressive stress σ 33 , acting on the fibers direction ◦ ◦ ◦ and producing internal instability on a loading path σ 11 = σ 22 = 0, σ 33 6= 0. From equations (6.1.19) and (2.2.88), we get ◦ ci

σ 33 = −G13 .

(6.1.89)

Comparing (6.1.88) and (6.1.89), we can conclude that ◦ ci

◦ cs

σ 33 < σ 11 < 0.

(6.1.90)

Hence, in our fiber-reinforced composite, loaded by compressive forces in the fibers direction, surface instability occurs before internal (structural) instability. We have obtained a similar result, analyzing the stability problem concerning the plane states. It remains to compare the absolute value of the negative root of equation ◦ cs (6.1.84) with the absolute value of σ 11 . Solving equation (6.1.84) by our iterative procedure and neglecting terms of the order ε2 and higher, we find for the negative root x1 of the equation (6.1.84) the following value: x1 = −

2 2 (E3 /E1 ) 1 − ν12 − 2ν13 1 C11 C33 − C13 . =− 2 1 − ν13 (E3 /E1 ) 1 − ν12 C11 C33

(6.1.91)

Comparing (6.1.87) and (6.1.91), we get |x| < |x1 | . ◦ cs

(6.1.92)

Hence, for the critical value σ 11 , we still have ξ1 6= ξ2 . Consequently, our assumption made at the beginning concerning the critical value is fulfilled and our procedure is consistent. Summing up, we can conclude that in a fiber-reinforced transversally isotropic composite, in a transversally isotropic prestressed state, for compressive forces acting in planes containing the fibers, superficial instability actually can occur, if the ◦ cs compressive pressures reach their critical value σ 11 . The magnitudes of these critical pressures are of the order of the relatively small axial shear modulus G 13 . As before, the occurrence of the superficial instability, for relatively small compressive pressures, producing infinitesimal deformations, is due to the structured character of the fiber-reinforced composites. Finally, we observe that for plane states, as well as for spatial states, the eigenmodes, that is, the possible ways in which the stability is lost, cannot be

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6.2. BUCKLING OF FIBER-REINFORCED COMPOSITE STRIPS

375

completely determined, since the wave-lengths l, l1 and l2 do not enter in the characteristic equations. As we shall, see such indeterminacy does not occur in the stability problems concerning infinite strips or bars made up by composite materials.

6.2

Buckling of fiber-reinforced composite strips

Following again Guz [6.2], let us assume an infinite strip; i.e. a very long plate, occupying the domain 0 ≤ x1 ≤ l, −h ≤ x2 ≤ h, −∞ < x3 < ∞, as shown in Figure 6.2. 2

h

1

O l

Figure 6.2: Composite strip acted upon by compressive forces. We consider a fiber-reinforced composite,the fibers being directed along the Ox1 axis. We assume that the initial applied loads act along the fibers direction; hence, ◦ ◦ σ 11 6= 0 and σ 22 = 0. (6.2.1) We shall analyze the stability of the strip relative to the plane incremental states, using Guz’s representation theorem. We suppose that the lateral faces x2 = ± h2 are stress-free; i.e.

h θ21 = θ22 = 0 for x2 = ± . 2

(6.2.2)

Also, we assume that on the ends x1 = 0 and x1 = l of the strip, the following mixed boundary conditions hold: u2 = 0 and θ11 = 0 for x1 = 0 and x1 = l.

(6.2.3)

Our strip corresponds to a very long plate in a cylindrical state. Recalling the boundary conditions encountered in the classical plate theory, we can see that the assumptions (6.2.3), taken into account by us in the framework of the three dimensional linearized theory, correspond to those describing a simply supported plate in the framework of the classical plate theory. Obviously, our conditions are more

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CHAPTER 6. BUCKLING OF COMPOSITE STRIPS AND BARS

exact and more restrictive, and have a local character. If they are fulfilled, those of the classical theory having global character, are also fulfilled. The converse is not generally true. Since the boundary conditions of our three-dimensional linearized theory are more exact than those of the classical theory, the results obtained by our more exact modelling can be used to get information concerning the domain of applicability of a less exact modelling; i.e. of the classical plate theory. As usual, we assume that the stress-free reference configuration of the strip is locally stable and its prestressed equilibrium configuration is internally stable. Moreover, we suppose that the parameters η1 , η2 are distinct, real, positive numbers; i.e. η1 , η2 > 0 and η1 6= η2 . (6.2.4)

In order to analyze the stability problem, we assume that one of the displacement potentials ϕ(1) is vanishing, and the other has one of the following forms:  π π  π ϕ(2) ≡ ϕ1 = A1 cosh m η1 x2 + A2 cosh m x2 cos m x1 , ϕ(1) ≡ 0, (6.2.5) l l l or  π π  π (1) ϕ(2) ≡ ϕ2 = A1 sinh m η1 x2 + A2 sinh m x2 cos m x2 , ϕ1 ≡ 0. (6.2.6) l l l

Here, A1 , A2 are arbitrary constants and m = 1, 2, 3, ... is an arbitrary, positive integer number. It is easy to see by direct computation, or by using the results obtained in the superficial stability analysis, that ϕ1 and ϕ2 satisfy equation (5.6.56). Hence, these functions can be used as displacement potential. Using Guz’s representation formulas (5.6.53) and equations (6.7.5), (6.7.6), we get the corresponding incremental displacement fields. For the first mode, corresponding to ϕ(1) ≡ 0 and ϕ(2) = ϕ1 , we obtain
 u1 = a2 −ω1221 + η12 ω2222 A1 cosh aη1 x2
+ −ω1221 + η22 ω2222 A2 cosh aη2 x2 cos ax1 ,

u2 = a2 (ω1122 + ω1212 ) (A1 η1 sinh aη1 x2 + A2 η2 sinh aη2 x2 ) sin ax1 , (6.2.7) with

mπ . (6.2.8) l For the second mode, corresponding to ϕ(1) ≡ 0 and ϕ(2) = ϕ2 , we obtain 
u1 = a2 −ω1221 + η12 ω2222 A1 sinh aη1 x2 a=


+ −ω1221 + η22 ω2222 A2 sinh aη2 x2 cos ax1 ,

u2 = a2 (ω1122 + ω1212 ) (A1 η1 cosh aη1 x2 + A2 η2 cosh aη2 x2 ) sin ax1 . (6.2.9)

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6.2. BUCKLING OF FIBER-REINFORCED COMPOSITE STRIPS

377

The first mode (if it exists) has the following property: u2 (x1 , −x2 ) = −u2 (x1 , x2 ).

(6.2.10)

In this mode, the occurrence of instability leads to necking of the surface of the strip, as shown in Figure 6.3. We call this mode antisymmetric.

O

Figure 6.3: Strip instability by necking (antisymmetric mode).

The second mode (if it exists) has the following property:

u2 (x1 , −x2 ) = u2 (x1 , x2 ).

(6.2.11)

In this mode, the occurrence of instability leads to bending or buckling of the strip as shown in Figure 6.4. We call this mode symmetric.

O

Figure 6.4: Strip instability by bending (symmetric mode).

Now, using the incremental constitutive equation (5.6.43), we determine the involved incremental nominal stress components.

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378

CHAPTER 6. BUCKLING OF COMPOSITE STRIPS AND BARS For the symmetric mode, we find ( 2 X  3 ω1111 ω1221 + (−ω1111 ω2222 + ω1122 (ω1122 + ω1212 )) ηα2 θ11 = a α=1

θ21

· Aα cosh aηα x2 } sin ax1 , ( 2 X  3 =a ω2222 ω2112 ηα2 + (−ω1221 ω2112 + ω1212 (ω1122 + ω1212 )) α=1

θ22

· ηα Aα sinh aηα x2 } cos ax1 , ) ( 2 X  2 3 ω1212 ω2222 ηα + ω1122 ω1221 Aα cosh aηα x2 sin ax1 . =a α=1

(6.2.12)

For the antisymmetric mode, it results ( 2 X  θ11 = a3 (−ω1111 ω2222 + ω1122 (ω1122 + ω1212 )) ηα2 + ω1111 ω1221 α=1

θ21

· Aα sinh aηα x2 } sin ax1 , ( 2 X  3 ω2222 ω2112 ηα2 − ω1221 ω2112 + ω1212 (ω1122 + ω1212 ) =a α=1

θ22

· ηα Aα cosh aηα x2 } cos ax1 , ) ( 2 X  3 2 =a ω1212 ω2222 ηα + ω1122 ω1221 Aα sinh aηα x2 sin ax1 . α=1

(6.2.13)

From (6.2.7)2 , (6.2.9)2 and (6.2.12)1 , (6.2.13)1 , it results that the boundary condition (6.2.3) at the ends x1 = 0 and x1 = l of the strip are satisfied. Imposing the boundary condition (6.2.2) for the symmetric mode, we arrive at the following homogeneous system for the unknowns A1 , A2 and m: α (η1 , m, b) A1 + α (η2 , m, b) A2 = 0, β (η1 , m, b) A1 + β (η2 , m, b) A2 = 0.

(6.2.14)

where

and

  α (η, m, b) = ω2222 ω2112 η 2 − ω1221 ω2112 + ω1221 (ω1212 + ω1122 ) η sinh mbη,
β (η, m, b) = ω1212 ω2222 η 2 + ω1122 ω1221 cosh mbη, (6.2.15) h b=π . l

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(6.2.16)

6.2. BUCKLING OF FIBER-REINFORCED COMPOSITE STRIPS

379

Using the same procedure for the antisymmetric mode, we get the following homogeneous system for the unknowns A1 , A2 and m: γ (η1 , m, b) A1 + γ (η2 , m, b) A2 = 0, δ (η1 , m, b) A1 + δ (η2 , m, b) A2 = 0,

(6.2.17)

where   γ (η, m, b) = ω2222 ω2112 η 2 − ω1221 ω2112 + ω1212 (ω1212 + ω1122 ) η cosh mbη,
δ (η, m, b) = ω1212 ω2222 η 2 + ω1122 ω1221 sinh mbη. (6.2.18)

Using now the expression (5.6.22) of the instantaneous elasticities from (6.2.1), (6.2.15) and (6.2.18), we obtain   ◦ α (η, m, b) = C66 C22 η 2 + C12 − σ 11 η sinh mbη, i  h (6.2.19) ◦ β (η, m, b) = C66 C22 η 2 + C12 C66 + σ 11 cosh mbη, and

  ◦ γ (η, m, b) = C66 C22 η 2 + C12 − σ 11 η cosh mbη, i  h ◦ δ (η, m, b) = C66 C22 η 2 + C12 C66 + σ 11 sinh mbη.

(6.2.20)

From (6.2.14), it results that loss of stability by symmetric mode can occur ◦ if and only if the applied stress σ 11 and the integer positive number m satisfy the following characteristic equation:  α (η , m, b) α (η , m, b) ◦ 1 2 = 0. (6.2.21) ∆s = ∆s σ 11 , m, b = β (η1 , m, b) β (η2 , m, b) Similarly, from (6.2.17), we can conclude that loss of stability by antisym◦ metric mode can occur if and only if σ 11 and m satisfy the characteristic equation  γ (η , m, b) γ (η , m, b) ◦ 1 2 = 0. (6.2.22) ∆a = ∆a σ 11 , m, b = δ (η1 , m, b) δ (η2 , m, b)

Assuming the height h and the breadth l of the strip fixed, to find the critical value of the applied (pressure) force, we must find, as a function of m = 1, 2, 3, ..., ◦ the smallest in absolute value root σ 11 of the characteristic equations (6.2.21) and of (6.2.22). If these two minimization problems were solved, we can say by which mode the stability of the strip is lost. For the strip (having two finite dimensions), beside the critical value of the applied force, we can determine also the wavelength of the mode describing the occurrence of instability. The problem is now much more complex, as in the case of the half space. The minimization problem can be solved only numerically, using a computer. Numerical results and their analysis are reported by Guz and Babitch [6.3]. The obtained results and their

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CHAPTER 6. BUCKLING OF COMPOSITE STRIPS AND BARS

examination show that generally stability loss occurs by antisymmetric (bending, buckling) mode and for m = 1, if b = πh/l is relatively small. Loss of stability by symmetric (necking) mode can occur only if b = πh/l is relatively large. The most frequently encountered situation corresponds to the first possibility. We shall analyze in great detail only the case in which m = 1, using the results due to Guz and Babitch [6.3]. Before doing that, we shall discuss an extremal case, which can be approximatively solved using elementary iterative procedures. In our asymptotic analysis, following Guz and Babitch [6.3], we assume a very thin plate, for which h << l. (6.2.23) In this case, the dimensionless parameter b, introduced by equation (2.6.16), satisfies the restriction b << 1. (6.2.24) We suppose m = 1. That is, we assume that the wave-length of the eigenmode describing the loss of stability, is 2l. Also, taking into account the above observations, we suppose that this eigenmode is an antisymmetric one. That is, we assume that the loss of stability occurs by buckling (bending) of the strip. Using the expressions (6.2.20) of the coefficients, for m = 1 we arrive at the following form of the involved characteristic equation (6.2.22):   C12 (C − C x) (C + C x) η12 η22 + 12 11 66 11 2 C66 C22 · (η1 cosh bη1 sinh bη2 − η2 cosh bη2 sinh bη1 ) C12 − C11 x η1 η2 (η2 cosh bη1 sinh bη2 − η1 cosh bη2 sinh bη1 ) + C22
C12 (C66 + C11 x) η13 cosh bη1 sinh bη2 − η23 cosh bη2 sinh bη1 = 0, + C22 C66

(6.2.25)

with

◦

σ 11 h (6.2.26) and b = π << 1. l C11 To obtain the critical value, we solve approximately the above equation. Since b = π hl << 1, we shall approximate sinh bηα and cosh bηα , α = 1, 2, by their 6 Taylor polynomials, neglecting terms of order (h/l) and higher. Thus we get x=

1 1 5 3 (bηα ) , (bηα ) + 120 6 1 1 4 2 (bηα ) , α = 1, 2. cosh bηα = 1 + (bηα ) + 24 2

sinh bηα = bηα +

(6.2.27)

We introduce these expressions in the characteristic equation (6.2.25). Long, elementary and careful computations, using the same order of approximation, leads

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6.2. BUCKLING OF FIBER-REINFORCED COMPOSITE STRIPS

to the following approximate form of the characteristic equation:    C12 1 2 2 C11 (C66 + C12 ) 2 (C12 − C11 x) (C66 + C11 x) η η + 2 x+b C66 3 1 2 C22 C22 C66  
1 C12 − C11 x C12 1 2 2 (C66 + C11 x) − η + η2 + C22 3 C22 C66 2 1   
C 1 12 2 2 2 2 4 (C12 − C11 x) (C66 + C11 x) η + η 2 η1 η2 + 2 +b C22 C66 30 1  
1 C12 − C11 x C12 1 2 2 (C66 + C11 x) − η + η2 + C22 5 C22 C66 24 1  1 2 2 C11 (C12 + C66 ) x = 0. (6.2.28) η η + 30 1 2 C22 C66

In order to solve this equation, we must use also the relations (5.6.57) and (5.6.58) giving η12 + η22 and η12 η22 . Taking into account the values (6.1.21) of the instantaneous elasticities, we get 2

(1 + x) C11 C22 + (C66 + xC11 ) C66 − (C12 + C66 ) , C22 C66 (1 + x) C11 (C66 + C11 x) . (6.2.29) η12 η22 = C22 C66

η12 + η22 =

In our asymptotic analysis, we solve equation (6.2.28) by an iterative procedure, looking for a root of the form x = x 0 + b2 x1 + b4 x2 .

(6.2.30)

and neglecting terms of order b6 and higher. Introducing (6.2.30) and (6.2.29) into (6.2.28) after elementary, long and careful computations, we obtain 2 C11 C22 − C12 < 0, (6.2.31) 3C11 C22 2 
C11 C22 − C12 2 − 2C12 C66 + 5C66 C22 . 6 C11 C22 − C12 x2 = 2 45C22 C11 C66

x0 = 0, x1 = −

Hence, from (6.2.30) and (6.2.31), we get #) " (
2 2 − 2C12 C66 1 1 6 C11 C22 − C12 C11 C22 − C12 2 2 . + b 1−b x=− 3 C22 C66 15 3C11 C22

(6.2.32) Let us introduce now the quantity p¯E , defined by the equation

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CHAPTER 6. BUCKLING OF COMPOSITE STRIPS AND BARS

p¯E =

2 C11 C22 − C12 1 2 . b C11 C11 C22 3

(6.2.33)

◦c

Thus, using also (6.2.26), for the critical value σ 11 , producing instability by buckling (bending), we obtain the following expression: #) " (
2 1 ◦c 2 6 C11 C22 − C12 − 2C12 C66 σ 11 = −pE 1 + b . (6.2.34) + 3 15C22 C66 ◦c

Before discussing the obtained results, we express p¯E and σ 11 in terms of the engineering constants. Using equation (2.2.85) and (6.2.26)2 , we get 1 p¯E = 3



h π l

2

E1 1 − ν31 ν13

(6.2.35)

and ◦c σ 11

= −pE

(



h 1− π l

2 

1 15



ν12 + ν32 υ13 E1 6 −2 1 − ν13 υ31 1 − ν31 ν13 G12



1 + 3

)

.

(6.2.36)

We observe that the developments (6.2.27), leading to the above approximate ◦c value of the critical pressure σ 11 , are acceptable if, besides the condition (6.2.26)2 satisfied by b, the following restrictions are also fulfilled:   h ηα << 1 for α = 1, 2. (6.2.37) bηα = π l

In order to establish the consequences of this requirement, we return to equation (6.2.29) giving the parameters η12 and η22 . Taking into account the relation (6.2.26)1 and the equation (2.2.85), and using the engineering constants of the material, we get η12 + η22 =

η12 η22 =

◦

ν13 + ν32 ν23 1 − ν13 ν31 + ν˜ (G12 /E2 ) σ 11 E1 1 , −2 + 1 − ν13 ν31 (1 − ν13 ν31 ) (G12 /E2 ) E2 1 − ν13 ν31 G12 ◦

(1 − ν23 ν32 ) (E1 /E2 ) + ν˜ (G12 /E2 ) σ 11 1 − ν23 ν32 E1 + G12 1 − ν13 ν31 1 − ν13 ν31 E2 ◦2

σ 11 ν˜ , − 1 − ν13 ν31 E2 G12

with ν˜ = 1 − ν12 ν21 − ν23 ν32 − ν31 ν13 − 2ν21 ν32 ν13 .

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(6.2.38)

6.2. BUCKLING OF FIBER-REINFORCED COMPOSITE STRIPS

383

The restriction (6.2.38) will be fulfilled if 

π

h l

2


η12 + η22 << 1 and



π

h l

4

η12 η22 << 1.

(6.2.39)

According to the structure of our fiber-reinforced composite, its engineering constants satisfy the following restrictions: G12 << E1 and E2 < E1 .

(6.2.40) ◦c

From(6.2.36) and (6.2.37), we can conclude that near the critical value σ 11 , ◦
2 ◦ the ratios σ 11 /E1 and σ 11 /G12 are of order π hl << 1. Hence, the leading term in the relation (6.2.39) is the first one. Consequently, both restrictions (6.2.40) and, hence, the requirements (6.2.39), will be fulfilled if the thickness ratio b = 2π hl satisfies the restriction r E1 1 h << 1. (6.2.41) π l 1 − ν13 ν31 G12

The classical Love-Kirchhoff plate theory is applicable for composite strips and plates only if this condition is satisfied. For composite plates, the transverse strips modulus G12 is much smaller than the axial Young modulus E1 . Hence, (6.2.40) is a stronger restriction on the thickness ratio b = 2π hl . For an isotropic strip or plate, G12 and E1 have the same order of magnitude. Consequently, the domain of applicability of the classical Love-Kirchhoff plate theory is much larger in this case. The obtained results show how the three-dimensional linearized theory can be used to obtain information concerning the domain of applicability of the classical, more approximative, Love-Kirchhoff type plate theory. Assuming (6.2.41) satisfied, we return now to equation (6.2.34) giving the ◦c critical stress σ 11 . First of all, we observe that the parenthesis in (6.2.41) is positive. Hence, buckling (bending) of the strip can occur only if the applied force is compressive. As we shall see later, in the Section 7.4, p¯E given by equation (6.2.35) is the buckling pressure according to the classical Love-Kirchhoff plate theory. The relation (6.2.36), obtained using the three-dimensional linearized theory, shows that even for very thin composite strips and plates, the classical result is only asymptotically true. If the thickness ratio b = π hl increases, the classical value must be corrected, taking into account terms of higher order in b = π hl . This correction can be important for fiber-reinforced composite plates for which E1 /G12 >> 1, and consequently ◦c (6.2.42) σ 11 < p¯E .

Neglecting the above criteria can lead to a dangerous situation since the actual value of the critical pressure can be much smaller than that predicted by the classical plate theory. This condition is enforced by the detailed analysis due to

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CHAPTER 6. BUCKLING OF COMPOSITE STRIPS AND BARS

Guz and Babitch [6.3] which used numerical methods to get the dependence of the critical buckling pressure on various mechanical and geometrical characteristics of fiber-reinforced composite strips. The two authors have assumed a transversally isotropic strip, Ox2 x3 being the plane of isotropy. In this case, ν13 = ν12 , ν31 = ν21 , E2 = E3 , ν23 = ν32 , and the classical buckling pressure p¯E given by (6.2.33) becomes  2 E1 h 1 (6.2.43) π p¯E = 2 (E /E ) . 1 − ν21 l 3 2 1

In order to present the obtained results, we introduce the dimensionless quantity p∗ =

pc ◦c with pc ≡ − σ 11 p¯E

(6.2.44) ◦c

giving the ratio of the critical buckling pressure pc ≡ − σ 11 obtained using the three-dimensional linearized theory, and the critical buckling pressure p¯E , obtained using the classical plate theory. In Figure 6.5, the dependence of the correction factor p∗ = p∗ (b) on the thickness ratio b = πh/l, for 0 6 b 6 0.32 is given. This dependence was obtained solving the characteristic equation (6.2.25). The curves 1, 2, ..., 8 correspond to the following values of the ratio E1 /G12 : E1 = 6, 8, 10, 20, 30, 50, 70, 100. G12

Figure 6.5: Dependence of the correction factor p∗ on thickness ratio b. The presented results show that the critical values obtained by the classical plate theory are larger than the critical values given by the three-dimensional linearized theory. The errors can be 50% or even bigger. For composite strips and plates with glass fibers, for 0.13 6 πh/l 6 0.23 and E1 /G12 > 30, the errors due

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6.3. BUCKLING OF FIBER-REINFORCED COMPOSITE BARS

385

to the classical theory can surpass 40%. But for this plate (πh/l 6 0.13), and for E1 /G12 6 10, the errors are less than 10%. Guz’s and Babitch’s analysis shows that if we are dealing with fiber-reinforced composite plates, the use of the three-dimensional linearized theory can avoid the occurrence of dangerous situations, which may appear if the design is made using the classical plate theory. However, in the same time, we can see that the classical formula is asymptotically true and represents a very good approximation, even for fiber-reinforced composites, if the strip or plate is relatively thin.

6.3

Buckling of fiber-reinforced composite bars

Let us consider a long cylindrical linearly elastic homogeneous bar. Let us denote by l its length, by S its cross-section and by A the area of this crosssection. We take the origin O of the coordinate system in the centroid of one of the end cross-sections and we assume that the axes Ox1 and Ox2 are the principal centroidal axes of the section. We denote by I1 and I2 the corresponding inertia moments. The axis Ox3 is parallel to the generators of the lateral surface of the bar, as shown in Figure 6.6. X1

O

X3

X2

Figure 6.6: Fiber-reinforced composite cylindrical bar. According to the chosen geometry made, we have Z S

x1 da =

Z

x2 da = 0, I1 =

S

Z

x22 da, I2 =

S

Z S

x21 da,

Z

x1 x2 da = 0.

(6.3.1)

S

We assume that the bar is a fiber-reinforced composite cylinder, the fibers being parallel to the generators of the lateral surface. We assume also that the bar is comprised by compressive forces acting in the direction of the reinforcing fibers. Hence, the stress state in the internal deformed and nonbuckled composite bar is characterized by the following relations: ◦

◦

◦

σ 11 = σ 22 = 0, σ 33 = −p, p = const. > 0.

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(6.3.2)

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CHAPTER 6. BUCKLING OF COMPOSITE STRIPS AND BARS

As usual, we suppose that the stress-free reference configuration of the bar is locally stable. As known for small values of the applied pressure p, the generatrices of the bar remain parallel to the Ox3 axis and the successively deformed and stressed equilibrium configuration of the bar are locally stable. However, for some critical value of the applied pressure, the stability of the bar is lost and a bifurcation of the equilibrium state occurs on the given (lead) loading path. That is, a buckled state of the bar can occur, and the generatrices of the lateral surface become curved lines, ceasing to be parallel to the axis Ox3 . Euler has succeeded in determining the critical value, using an approximate, genial and wonderful procedure. The threedimensional linearized theory can be used to elucidate the domain of applicability of Euler’s result. To see how this can be done, first of all we present briefly Euler’s procedure. We assume I2 ≡ I m < I 1 ≡ I M ,

(6.3.3)

Im and IM being the minimal and maximal centroidal momenta of inertia of the cross-section. If (6.3.3) is true, the buckling, if it appears, takes place in the Ox 1 x3 plane. In Figure 6.7, we show the initial and the assumed buckled state of the bar, P being the applied compressive force. M N U1

P

X1

U1 P

P

X2

O

Figure 6.7: Buckled cylindrical bar. The buckled bar can be in equilibrium only if the following global equilibrium equations are satisfied: Z Z N ≡ σ33 da = −P, M ≡ x1 σ33 da = u1 P. (6.3.4) S

S

In the Euler’s theory, it is assumed that all fields are independent on x2 . It is also assumed that all material fibers orthogonal to the symmetry axis of the bar conserve their length if the bar is buckled. Moreover, the theory is founded on the hypothesis of plane section according to which the transverse section of the bar

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6.3. BUCKLING OF FIBER-REINFORCED COMPOSITE BARS

387

remain plane and orthogonal to the deformed symmetry axis in the buckled state of the bar. The above assumption can be expressed by the equations ε11 = 0 and ε13 = 0.

(6.3.5)

The first relation shows that the transverse displacement u1 depends only on x3 and the second relation leads to the following expression of the axial displacement u3 : ◦ (6.3.6) u3 = u3 (x1 , x3 ) = u3 (x3 ) − x2 u01 (x3 ) , the “prim” denoting derivation with respect to x3 . Let us observe that the second term in the right-hand side of the above equation is an incremental displacement, due to the buckling, the first one corresponds to the underlaying compressed, nonbuckled state of the bar. For the axial deformation ε33 , we get the expression ◦

00

◦0

◦

ε33 = ε33 (x3 ) − x1 u1 (x3 ) , ε33 (x3 ) = u3 (x1 ) .

(6.3.7)

The axial stress σ33 in the buckled state of the bar is ◦

00

σ33 = E3 ε33 = E3 ε33 (x3 ) − E3 x1 u1 (x3 ) ,

(6.3.8)

where E3 is the axial Young modulus of the bar. Introducing the expression (6.3.8) in the first global equilibrium condition (6.3.4) and using the relation (6.3.1)1 we get ◦

AE3 ε33 (x3 ) = −P.

(6.3.9)

◦

This result shows that the axial deformation ε33 , existing also in the buckled ◦ equilibrium state, is a constant quantity. Hence, the axial stress σ 33 in this state is also constant and ◦ ◦ σ 33 = E3 ε33 = −P/A. (6.3.10) Comparing (6.3.2) and (6.3.10), we get the following obvious result connecting the compressive force P and the corresponding compressive pressure p: p = P/A.

(6.3.11)

We return now to the second global equilibrium condition (6.3.4). Introducing (6.3.8) and using again (6.3.1)1 and (6.3.1)4 , we obtain the equation which must be satisfied by the incremental displacement u1 = u1 (x3 ), if buckling occurs u001 +

P u1 = 0 for 0 < x3 < l, Im = I2 . I m E3

The general solution of the above equation is r r P P x3 . x3 + B cos u1 = A sin I m E3 I m E3

Copyright © 2004 by Chapman & Hall/CRC

(6.3.12)

(6.3.13)

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CHAPTER 6. BUCKLING OF COMPOSITE STRIPS AND BARS

We assume a simply supported bar. In the framework of Euler’s theory, this can be expressed by imposing the following boundary conditions: u3 (0) = u3 (l) = 0.

(6.3.14)

Thus, from (6.3.13), we get B = 0, and u3 can be nonvanishing; i.e. buckling and bifurcation of the equilibrium state can occur if and only if r P = mπ, m = 1, 2, ... . (6.3.15) l I m E3

On the given loading path, bifurcation and buckling occur if the compressive force reaches its smallest possible critical value PE =

 π 2

l

I m E3 .

(6.3.16)

From (6.3.11), we get the critical value pE of the applied pressure pE =

 π 2 I E m 3 . A l

(6.3.17)

The last result can be expressed in the following equivalent and useful form : p Im /A α2 , (6.3.18) with α = 2π pE = E 3 l 4

the quantity 2π/α representing the slenderness ratio. The famous Euler’s formula (6.3.17) gives the value of the compressive pressure for which bifurcation of the prestressed equilibrium configuration occurs. In this case, for the same value PE of the applied compressive force, there exist two distinct and neighboring equilibrium states of the bars, characterized by the following relations: u1 ≡ 0 in the underlying equilibrium state,

(6.3.19)

and u1 = A sin πx3 /l 6= 0 in the buckled equilibrium state.

(6.3.20)

The buckled state corresponds to the primary eigenmode u1 6≡ 0 given by the above approximation and corresponding to the smallest eigenvalue P E /Im E3 . Obviously, the used theory, as well as the three-dimensional linearized theory, can tell us nothing about the amplitude of this primary eigenmode. According to the Euler’s theory, the critical buckling force depends only on the axial Young modulus E3 . The other mechanical characteristics of the material have no influence on the critical value. Particularly, two bars having the same geometrical characteristics and the same axial modulus, are buckled by the

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6.3. BUCKLING OF FIBER-REINFORCED COMPOSITE BARS

same compressive force PE , even if one bar is isotropic and the other one is fiberreinforced, hence, orthotropic. The fact that the axial shear modulus G13 and the axial Young modulus E3 have the same order of magnitude in the first case, but G13 is much smaller than E3 in the second case, has no influence on the buckling force according to Euler’s theory. Since the Eulerian approach is based on the hypothesis of plane cross-section (ε13 ≡ 0), the value of the axial shear modulus G13 cannot influence in any way the behavior of the compressed bar. In what follows we shall use the three-dimensional linearized theory to see more exactly in what conditions the results of Euler’s theory can be accepted and in what circumstances they must be improved. For this purpose we suppose that we have a fiber-reinforced composite bar, the fibers having the direction of the symmetry axis of the bar. In this case, the bar is orthotropic and its symmetry planes are the coordinates planes. Consequently, according to (5.6.22) and (6.3.2), the nonvanishing instantaneous elasticities have the following values: ω1111 = C11 , ω2222 = C22 , ω3333 = C33 − p, ω1122 = ω2211 = C12 , ω2233 = ω3322 = C23 , ω2332 = C44 , ω3311 = ω1133 = C31 , ω1212 = ω2121 = C66 , ω2323 = ω3232 = C44 , ω3131 = ω1313 = C55 , ω2112 = ω1221 = C66 , ω3223 = C44 − p, ω1331 = C55 , ω3113 = C55 − p.

(6.3.21)

Using the above results, we can compute the exclusive functional E (u) defined by equation (5.5.1). Since now the material is linearly elastic and the initial applied ◦

deformation is infinitesimal, the instantaneous elasticity Ω must be replaced in (5.5.1) by the instantaneous elasticity ω. Taking into account this fact and using (6.3.22), we get Z E (u) = {C11 u21,1 + C22 u22,2 + (C33 − p) u23,3 B

+2 (C12 u1,1 u2,2 + C23 u2,2 u3,3 + C13 u3,1 u1,1 ) + C66 (u1,2 + u2,1 ) + (C44 − p) u22,3 + 2C44 u2,3 u3,2 + C44 u23,2 + (C55 − p) u21,3 +2C55 u1,3 u3,1 + +C55 u23,1 }dV.

2

(6.3.22)

From the general theory of local stability (see Section 5.5), we know that if the exclusion functional E (u) is positive for any kinematically admissible incremental displacement field u, with ∇u 6≡ 0, the corresponding equilibrium configuration is locally stable. If a primary eigenstate appears on the given loading path, it makes the exclusion functional stationary relative to all admissible variations of the corresponding incremental displacement field. Moveover, in the primary eigenstate, the exclusion functional has a local minimum (zero) relative to the set of all kinematically admissible incremental displacement fields. To find out the critical value

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CHAPTER 6. BUCKLING OF COMPOSITE STRIPS AND BARS

of the loading parameter (p in our case), we must study the above variational and extreme problem. The solution of this problem can be exact or approximate and, in the following, we shall illustrate both cases. First, following Pearson [6.4], we try to solve the buckling problem using simultaneously the three-dimensional linearized theory and the Euler’s assumptions concerning the way on which the buckling takes place. As we know from the relations (6.3.5) and (6.3.6), Euler’s hypothesis lead to the following structure of the admissible incremental displacement fields u1 = u1 (x3 ) , u2 ≡ 0, u3 = −x1 u01 (x3 )

(6.3.23)

assuming that the buckling takes place in the Ox1 x3 plane. The unknown function u1 = u1 (x3 ) must satisfy the homogeneous boundary condition (6.3.14), expressing the fact that the bar is simply supported. Since we now use the three-dimensional linearized theory, we can, and we must, impose vanishing incremental nominal stress θ33 at the ends of the bar, exactly as we have done, by analyzing the stability of a simply supported strip or plate. Hence, we must have θ11 = 0 for x3 = 0 and x3 = l. (6.3.24) Using (6.3.21), (6.3.23), the incremental constitutive equation (5.6.23) 9 , and assuming p < C33 , (6.3.25) it is easy to see that the boundary condition (6.3.24) will be satisfied if and only if u1 (x3 ) fulfills the following restrictions: u001 (0) = u001 (l) = 0.

(6.3.26)

We calculate now in u1 , the first variation δE of the exclusion functional (6.3.22) corresponding to the admissible variation δu1 of u1 . According to (6.3.14) and (6.3.26), these variations must satisfy the restriction δu1 (0) = δu1 (l) = δu001 (0) = δu001 (l) = 0. Since

(6.3.27)

d E (u1 + λδu1 ) , δE = dλ λ=0

simple calculus shows that δE =

Zl 0

00 {(C33 − p) Im u0000 1 + pAu1 } δu1 dx3 , Im = I2 .

(6.3.28)

Accordingly, if u1 is primary eigenmode, it must satisfy the differential equation 00 (C33 − p) Im u0000 1 + pAu1 = 0.

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(6.3.29)

391

6.3. BUCKLING OF FIBER-REINFORCED COMPOSITE BARS We denote by Ω2 the positive quantity Ω2 =

pA . (C33 − p) Im

(6.3.30)

Hence, the general solution of equation (6.3.29) is u1 (x3 ) = C1 sin Ωx3 + C2 cos Ωx3 + C3 x3 + C4 ,

(6.3.31)

where C1, C2 , C3 , C4 are arbitrary real constants. It is easy to see that u1 (x3 ) satisfy the boundary conditions (6.3.14) and (6.3.26) if and only if C1 sin Ωl = 0, C2 = C3 = C4 = 0.

(6.3.32)

The primary eigenmode can exist if and only if C1 6= 0. Consequently, the bifurcation condition must be satisfied Ωl = nπ, n = 1, 2, 3, ... .

(6.3.33)

From (6.3.30) and (6.3.33), we can now deduce the possible critical values of the applied axial pressure p p = C33 Im

n πl


2

A + n πl


2 , n = 1, 2, 3, ... .

(6.3.34)

The right-hand side of the above equation increases when n increases. Hence, the primary eigenmode correspond to the critical value pc =

 π 3 C I 33 m
2 . l A+ π

(6.3.35)

l

∼

Since C33 = (1 − ν12 ν21 ) E3 / ν, the above results can be expressed in terms of the engineering constants pc =

 π 2 E I 1 − ν ν 1 12 21 3 m . ∼ 2 I ) / (Al 2 ) 1 + (π A l m ν

(6.3.36)

Also, using the Eulerian value pE given by the equation (6.3.17), we get pc = p E

1 − ν12 ν21

∼.

{1 + π 2 (Im /l2 A)} ν

(6.3.37)

The above result based on Euler’s geometrical assumptions and on the threedimensional linearized theory shows that Euler’s theory gives good results if Im  1, l2 A

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that is, if the bar is relatively long. However, it can lead to erroneous results if the condition is not fulfilled. Again, the main role in the critical value of the applied pressure is played by the Young’s axial modulus E3 . The Eulerian incremental displacement field (6.3.23) is too simple, and cannot describe the complex real state existing in a buckled bar. To eliminate this fact, and to see more exactly the way in which the internal structure of a fiber-reinforced composite bar influences its performances, we shall analyze now the problem using the three-dimensional linearized theory and trying to obtain the exact solution of the problem in a particular case in which this is possible. As we shall see later on, the results obtained in this approach can be used also to obtain approximate solutions which are much more realistic than those based on Euler’s hypothesis. We assume that the fiber-reinforced composite bar has a circular cross-section with the radius a. We assume also, that it is transversally isotropic, the isotropy plane being perpendicular to the symmetry axis of the bar. The buckling problem in this case was solved and analyzed in great detail by Guz and Babitch [6.3]. First, we observe that for a circular cylindrical bar, the Eulerian buckling presume pE has the following form: pE = E 3

a α2 with α = π . l 4

(6.3.38)

Next, we note that according to (2.2.49) and (6.3.2), the nonvanishing instantaneous elasticities have the following expressions: ω1111 = ω2222 = C11 , ω3333 = C33 − p,

ω1122 = ω2211 = C12 , ω1133 = ω3311 = ω2233 = ω3322 = C13 , 1 ω1221 = ω2112 = ω1212 = ω2121 = (C11 − C12 ) , 2 ω1313 = ω3131 = ω2323 = ω3232 = ω1331 = ω2332 = C44 ,

ω3113 = ω3223 = C44 − p.

(6.3.39)

Assume the cylindrical symmetry of the initial deformed equilibriums state are satisfied. Hence, we can use Guz’s representation (5.6.102). According to the relations (5.6.103)1 , (5.6.105) and (5.6.113), the displacement potentials ψ (r, θ, z), χ1 (r, θ, z) and χ2 (r, θ, z) must satisfy the differential equations   1 ∂2 1 ∂ ∂2 ∂2 + 2 2, ∆1 + ξ12 2 Ψ = 0, ∆1 = 2 + r ∂r r ∂θ ∂r ∂z     2 2 ∂ ∂ (6.3.40) ∆1 + ξ22 2 χ1 = 0, ∆1 + ξ32 2 χ2 = 0. ∂z ∂z

According to (5.6.86)−(5.6.88) and (6.3.39), the parameters ξ12 , ξ22 and ξ32 are given by the equation

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6.3. BUCKLING OF FIBER-REINFORCED COMPOSITE BARS

s

(C44 − p) (C33 − p) , C11 C44

ξ12

=

C44 − p , ξ2 = c ± (C11 − C12 ) /2 2,3

2c

=

C11 (C33 − p) + C44 (C44 − p) − (C13 + C44 ) . C11 C44

c2 −

393

2

(6.3.41)

As before, we assume a simply supported bar. Hence, the boundary conditions at the two ends of the bar are uθ = uz = 0, θzz = 0 for z = 0 and z = l.

(6.3.42)

The lateral surface of the bar is stress-free. Thus the following boundary conditions must be satisfied: θrr = θrθ = θrz = 0 for r = a.

(6.3.43)

Let us observe that these conditions are not fulfilled in the Euler’s theory and the situation is the same in all theories based on Euler’s geometrical assumption. If we use the incremental constitutive equations (5.6.120)1,2,3 and the values (6.3.39) of the instantaneous elasticities, the boundary conditions (6.3.43) become   ∂uz ur 1 ∂uθ ∂ur = 0, + C13 + + C12 C11 ∂z r r ∂θ ∂r uθ 1 ∂ur ∂uθ = 0, for r = a, − + r r ∂θ ∂r ∂uz ∂ur = 0. (6.3.44) + ∂r ∂z

With the instantaneous elasticities given by the equation (6.3.39), Guz’s representation formulas (5.6.114) take the form

1 ∂2χ ∂ψ ∂2χ 1 ∂ψ , − , uθ = − − r ∂θ∂z ∂r∂z r ∂θ   2 ∂r   1 ∂2 1 ∂ ∂ 1 ∂2 + + C11 uz = + (C44 − p) 2 χ, r ∂r r2 ∂θ2 ∂r2 C13 + C44 ∂z χ = χ1 + χ2 . (6.3.45) ur =

As usual, we assume that the stress-free reference configurative of the bar is locally stable. Also, we suppose that the parameters ξ12 , ξ22 and ξ32 are distinct, real, positive numbers. To find the critical buckling pressure, we must first select the solutions of equations (6.3.40) which satisfy the homogeneous boundary conditions (6.3.42) at the ends of the bar and which describe a buckled state of the bar. Hence, we must analyze the nature of the possible solution of the equation  2  2 ∂ 1 ∂ 1 ∂2 2 ∂ + + + ξ f (r, θ, z) = 0, ξ > 0. (6.3.46) ∂r2 r ∂r r2 ∂θ2 ∂z 2

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Frequently in a similar situation, we look for solutions having the following structure: f (r, θ, z) = F (r, θ) Z (z) . (6.3.47) Introducing (6.3.47) in (6.3.46), we get −

Z 00 ∆1 F = ±d2 , d = const. > 0. = Z ξ2F

Since the boundary condition (6.3.42) must be satisfied, the solution corresponding to +d2 are rejected. Thus we get ∆1 F − ξ 2 d2 F = 0, Z 00 + dZ = 0.

(6.3.48)

Examining again the boundary condition (6.3.42), we take π d = m , m = 1, 2, 3, ... l

and it results Z = sin mπ

z z or Z = cos mπ , m = 1, 2, 3, ... . l l

Now the equation (6.3.48)1 becomes 2  ξ F = 0. ∆1 F − mπ l

(6.3.49)

(6.3.50)

We try F = F (r, θ) with the following structure: F (r, θ) = R (r) Θ (θ) .

(6.3.51)

Introducing (6.3.51) into (6.3.50) and separating the variables, we get 2  r2 R00 + rR0 − mπ ξl r2 R Θ00 = ±e2 , e = const. > 0. (6.3.52) = − Θ R

Since the incremental fields must be a uniform function of the angle θ, the solution corresponding to +e2 must be rejected and we must have e2 = n2 , n = 1, 2, 3, ... . Hence, the possible expression for Θ are Θ = sin nθ or Θ = cos nθ, n = 1, 2, 3, ... . The equation which must be satisfied by R becomes ( 2 )  ξ r2 R = 0. r2 R00 + rR0 − n2 + mπ l

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(6.3.53)

(6.3.54)

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This is a Bessel type equation and its bounded solution in the domain 0 ≤ r ≤ a is given by the modified Bessel function of the first kind (for details see, for instance, the monograph [6.5]). Hence, we have   ξ (6.3.55) R = In mπ r , n, m = 1, 2, 3, ... . l

In the following, we take m = n = 1, and as Guz and Babitch have done [6.3], we assume that the displacement potentials Ψ, χ1 , and χ2 are given by the equation π l χ1 = A2 I1 (γξ2 r) cos θ cos γz, χ2 = A3 I1 (γξ3 r) cos θ cos γz, (6.3.56)

Ψ = A1 I1 (γξ1 r) sin θ sin γz, γ =

A1, A2 , A3 being arbitrary constants. To show the mechanical meaning of the above choice, we must determine the corresponding incremental displacement field. Using Guz’s representation formulas (5.6.102) and the relation (6.3.50), we get

 ur = A1 1r I1 (γξ1 r) + γ 2 [A2 ξ2 I10 (γξ2 r) + A3 ξ3 I10 (γξ3 r)] cos θ sin γz,

 uθ = − A1 γξ1 I10 (γξ1 r) + γ 1r [A2 I1 (γξ2 r) + A3 I1 (γξ3 r)] sin θ sin γz, uz =

C11 2 C13 +C44 γ

n

ξ22 −

C44 −p C11



 A2 I1 (γξ2 r) + ξ32 −

C44 −p C11



A3 I1 (γξ3 r)

o

· cos θ cos γz.

(6.3.57) In order to obtain the above expression for uz , we have used the second order differential equation satisfied by I1 (γξ2 r) and I1 (γξ3 r) (see[6.5]). Equation (6.3.57) shows that ur (r, θ + π, z) = −ur (r, θ, z) . Hence, all particles in a cross-section having a symmetric position relative to the centroid of the cross-section, have the same radial displacement. Moreover, this displacement is vanishing at the ends of the bar, and its maximal value is obtained for the particle situated in the middle cross-section of the bar. Thus, Guz’s and Babitch’s exact solution really describes a buckled state of the bar. This state is much more complex than that assumed by Euler, but the essential characteristic of the two states are similar. Using the last incremental constitutive equation (5.6.120) can determine the axial component θzz of the incremental nominal stress, corresponding to the incremental displacement field (6.3.17). The obtained result, together with the relations

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(6.3.57)1,2 , show that the homogeneous boundary conditions (6.3.42) at the ends of the bar are satisfied. Finally, we must impose the homogeneous boundary conditions (6.3.43) at the stress-free lateral surface r = a of the bar. Long, but elementary computations shows that this condition will be satisfied if and only if A1, A2, A3 and p satisfy the following homogeneous algebraic system: mkl Al = 0, mkl = mlk , k, l = 1, 2, 3,

(6.3.58)

where the coefficients mkl , depending on p, have the expressions α (C11 − C12 ) ξ1 I2 (αξ1 ) , a2    C44 − p C13 C11 ξl α2 I1 (αξl ) , ξl2 − m1l = 3 C11 ξl2 I100 (αξl ) + C12 I2 (αξl ) − C11 C13 + C44 α a l = 2, 3, α2 α m21 = 2 2 ξ1 I2 (αξ1 ) − 2 ξl2 I1 (αξl ) , a a α2 m2l = −2 3 I2 (αξl ) , l = 2, 3, a α m31 = 2 I1 (αξ1 ) , a α2 C13 + p + ξl2 C11 , l = 2, 3 m3l = 3 ξl C13 + C44 a

m11 =

(6.3.59) with

a (6.3.60) α=π . l The parameter α characterizes the relation existing between the transversal and longitudinal dimensions of the bar. In order to obtain the coefficients in the form given above, we have used the well-known equation (see [6.5])   1 d 1 I1 (r) = I2 (r) . r dr r

Buckling of the axially compressed bar occurs if and only if the homogenous algebraic system (6.3.58), for the unknowns A1, A2, A3 , has nonvanishing solutions. Imposing this condition, we get to the characteristic equation det M (p) = det [mkl (p)] = 0,

(6.3.61)

M = [mkl ] being a 3 × 3 symmetric matrix. The above relation represents a very complicated equation for the unknown critical presume pc , which is the smallest positive root of the characteristic equation

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397

(6.3.61). To obtain preliminary information about pc , Guz and Babitch [6.3] have used the equations, (see [6.5]) In (r) =

∞ X j=0

 r n+2j 1 j! (n + j)! 2

(6.3.62)

defining the function In (r). Neglecting terms of order (a/l)4 and higher, the two authors were able to show that
(6.3.63) pc = p E + O α 4 , where pE is the Eulerian buckling pressure given by equation (6.3.38). We can see again that the Eulerian formula is asymptotically exact. In order to obtain information about the limits of Euler’s results, Guz and Babitch have analyzed the dependence pc = pc (α) of the buckling pressure on the geometrical parameter α = πh/l. To find this dependence, the two authors have introduced the dimensionless ratio or correction factor pc(α) p∗ = p∗ (α) = pE

and have studied its dependence on α for various values of the mechanical characteristics of the fiber-reinforced composite bar. For various important combinations of these characteristics, the characteristic equation (6.3.61) was numerically solved by Guz and Babitch. In all analyzed situations taken into account, the following resulted: ν12 = 0.3, ν31 = ν32 = 0.2. Shown in Figures 6.8, 6.9 and 6.10 are the results obtained by Guz and Babitch, that in the dependence of the correction factor p∗ (α) on the geometrical parameter α ∈ [0, 0.1], for the following values of the ratio E1 /E3 of the transverse and axial Young’s moduli: E1 /E3 = 0.08, 0.20, 0.80. The curves 1, 2, ..., 7 correspond to the following ratios of the axial Young modulus E3 and of the axial shear modulus G13 : E3 /G13 = 6, 10, 20, 30, 40, 60, 100. These ratios are characteristic for many fiber-reinforced composite bars. The obtained results show that the critical buckling pressure pc depends weakly on E1 /E3 , ν12 and ν31 , but depends strongly on E3 , E3 /G13 and α = πh/l.

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Figure 6.8: Variation of correction factor p* vs. geometrical parameter α. Transversally isotropic circular bar; E1 /E3 = 0.08.

Figure 6.9: Variation of correction factor p* vs. geometrical parameter α. Transversally isotropic circular bar; E1 /E3 = 0.20.

For relatively short bars (α > 0.13) for which the axial shear modulus is relatively small (E3 /G13 > 20), the errors due to the Eulerian formula cannot be neglected. However, for relatively long bars (α ≤ 0.13) for which E3 /G13 ≤ 30, the errors due to the Eulerian formula are less then 10 to 20%. The above results based on the three-dimensional linearized theory show that for fiber-reinforced composite bars, there are situations in which the critical buckling pressure pc is strongly influenced by the shear strains and shear stresses existing in the longitudinal plane perpendicular to the buckling direction. This fact is reflected by the dependence of pc on the ratio E3 /G13 . Just the above strains and stresses are neglected in Euler’s theory based on the hypothesis of plane cross-section. Let us observe also that the numerical results confirm the evaluation (6.3.63) due to Guz and Babitch, since

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Figure 6.10: Variation of correction factor p* vs. geometrical parameter α. Transversally isotropic circular bar; E1 /E3 = 0.80. p∗ (α) converges toward unity when α converges toward zero. We can conclude again that Euler’s formula is asymptotically exact, for any possible combination of the material constants. Let us stress the fact that analytical, exact solution can be obtained for transversally isotropic, circular bars only. This fact emphasizes the importance of variational and extreme principles. These principles can be used in more general circumstances to get approximate solutions for the buckling problem. In order to use the variational and extremal principles, we must select appropriate test functions, taking into account the considered mechanical problem. The assumed set of test functions in a finite dimensional space, and the stationarity or extremal condition is imposed relative to this set. Thus, we are led to finding nonvanishing solutions of homogenous linear algebraic systems. The possible roots of the corresponding characteristic equation give finally the approximate values of the wanted critical loads. These general ideas were applied in the buckling problem by Capanu and So´os [6.6]. In what follows, we present the results obtained by these two authors. Following Guz’s and Babitch’s idea [6.3], we observe that the exact solution (6.3.17) can be used to get “good” test functions. In order to do it, we must utilize equation (6.3.62) defining Bessel’s function In (r). Starting with this relation, and 3 neglecting terms of order (γr) , from (6.3.57) we get    r 2   r 2  sin θ sin γz, cos θ sin γz, uθ = −A1 + A3 ur = A1 + A 2 a a r cos θ cos γz, (6.3.64) uz = A 4 a

where

γ=

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π , γr < γa = πa/l  1, l

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and A1 , A2 , A3 , A4 are arbitrary constants. It is easy to see that the displacements (6.3.64) describe a buckling state of the bar and the displacement ur , uθ satisfy the boundary conditions (6.3.42)1,2 imposed on the two ends of the simply supported bar. Consequently, the equations (6.3.64) furnish “good” test functions in the buckling problem of circular cylindrical bars. Denoting by u1 , u2 , u3 the Cartesian components of the displacement field (6.3.64) and using equations (5.6.97), we get

x1 x2 x2 x1 2 ) − A3 ( )2 } sin γx3 , u2 = (A2 + A3 )( )( ) sin γx3, a a a a x1 u3 = A4 ( ) cos γx3. a

u1 = {A1 + A2 (

Starting with this intermediate result, we introduce the following incremental displacement field:

x1 x2 x2 x1 2 ) + A3 ( )2 } sin γx3 , u2 = A4 ( )( ) sin γx3 , b a b a x1 (6.3.65) u3 = A5 ( ) cos γx3, γ = π/l. a

u1 = {A1 + A2 (

Here A1 , A2 , A3 , A4 , A5 , a and b are arbitrary real constants. Now let us assume that we are dealing with a rectangular cylindrical bar, 2a and 2b are the dimensions of its cross-section. We denote a and b so that a ≤ b.

(6.3.66)

It is easy to see that the displacement field (6.3.65) describes a buckling state of the bar in the Ox1 x3 plane and satisfies the following boundary conditions at the ends of the bar: u1 = u2 = 0 for x3 = 0 and x3 = l.

(6.3.67)

Consequently, equations (6.3.65) furnish “good” test functions for the buckling problem of rectangular cylindrical bars. We begin the analysis, based on the incremental variational principle, assuming a circular bar, having cylindrical orthotropy (see, for instance, Lekhnitski [6.7]). In this case we shall express the exclusion functional E = E(u) given by ◦

equation (5.5.1), using cylindrical coordinates and replacing Ω by ω. In order to do this, we express first E(u) in the following equivalent form: Z E (u) = (∇k ue) ωklmn (∇n um )dV. (6.3.68) B

The components of the differential operator ∇ are in this relation ∇k . In Cartesian coordinates, we have ∇k = ∂x∂ k .

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Now we must express the components of the second order tensor field ∇u in cylindrical coordinates, that is, in the tensor basis formed using the orthonormal set (er , eθ , ez ). As can be shown (see, for instance, Malvern [6.8]) in the indicated tensor basis, ∇u has the following components:

∂uθ ∂ur , ∇ r uz = , ∇ r uθ = ∂r ∂r 1 ∂uθ uθ 1 ∂ur − , ∇ θ uθ = ∇θ ur = r ∂θ r r ∂θ ∂uθ ∂ur , ∇ z uz = , ∇ z uθ = ∇z ur = ∂z ∂z

∇r ur =

∂uz , ∂r 1 ∂uz ur , + , ∇ θ uz = r ∂θ r ∂uz . ∂z

(6.3.69)

In order to express E(u) using cylindrical coordinate, we make the usual identification 1 ↔ r, 2 ↔ θ, 3 ↔ z. Since the material has cylindrical orthotropy, only the instantaneous elasticities given by equations (6.3.21) are nonvanishing. Consequently, from (6.3.68), we get Z E(u) = {C11 (∇r ur )2 + C22 (∇θ uθ )2 + (C33 − p)(∇z uz )2 B

+ 2(C12 ∇r ur ∇θ uθ + C23 ∇θ uθ ∇z uz + C31 ∇z uz ∇θ uθ )

+ C66 (∇r uθ + ∇θ ur )2 + C44 (∇θ uz + ∇z uθ )2 − p(∇z uθ )2 + C55 (∇z ur + ∇r uz )2 − p(∇z ur )2 }dV. (6.3.70)

If the material is transversally isotropic, the elasticities satisfy the supplementary relations C11 = C22 , C23 = C31 , C44 = C55 , 2C66 = C11 − C12.

(6.3.71)

As we know, in any situation Euler’s buckling pressure pE can be expressed by equation (6.3.18)1 , the parameters α being given by equation (6.3.18)2 . In their analysis, realized for various cases, Capanu and So´os have used the correction factor p∗ = p∗ (α) defined by the equation pc = p∗ (E3 α2 /4) = p∗ pE .

(6.3.72)

The connection factor p∗ = p∗ (α) must be determined using exact or approximate methods. The possible values of α depend on the geometrical characteristics of the bar, as can be seen examining the relation (6.3.18)2 . In all numerical tests, Capanu and So´os have assumed the following values of the Poisson’s ratio: ν12 = 0.3, ν31 = ν32 = 0.2.

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First, we consider a circular fiber-reinforced composite bar, as Guz and Babitch have done for the transversally isotropic case. Applying the incremental variational principle, Capanu and So´os were able to treat also the case in which the bar is cylindrically orthotropic. In this case, the nonvanishing instantaneous elasticities are given by equations (6.3.21) as for a rectangular, orthotropic bar. The appropriate test functions are given by equations (6.3.64). Introducing these relations in expression (6.3.70) of the exclusion functional E = E(u), we get a quadratic function E = E (A1 , A2 , A3 , A4 ) depending on four real variables A1 , A2 , A3, A4 . The exclusion functional becomes stationary on the four dimensional space of the assumed test functions, for those values of the variables A1 , A2 , A3 , A4 , for which all first order partial derivatives of the function E = E (A1 , A2 , A3 , A4 ) are vanishing. Imposing this stationarity condition, we are led to the following linear, homogeneous algebraic system which must be satisfied by the unknowns A 1 , A2 , A3 , A4 nij Aj = 0, nij = nji , i, j = 1, 2, 3, 4.

(6.3.73)

Long, but elementary, computations based on the relations (6.3.21), (6.3.69) and (6.3.70) show that the coefficients of the above system are given by the following equations: n11 = (C44 + C55 − p)α2 , n12 = 12 (C55 − p)α2 ,

n13 = − 12 (C44 − p)α2 , n14 = (C44 + C55 )α,

n22 = 2(C11 + C12 ) + 12 (C22 + C66 ) + 13 (C55 − p)α2 ,

n23 = C12 + 12 (C22 − C66 ), n24 = −(C13 + 12 C23 − 12 C55 )α,

n33 =

1 2

(C22 + C66 ) + 13 (C44 − p)α2 , n34 = − 12 (C23 + C44 )α,

n44 = C44 + C55 + 12 (C33 − p)α2 ,

and α = πa/l. In order to find the critical axial pressure pc , we must find the smallest positive root of the characteristic equation det N (p) = det [nij (p)] = 0,

(6.3.74)

where N = [nij ] is a 4 × 4 symmetric matrix. If the above equation is satisfied, the homogeneous system (6.3.73) has a nonvanishing solution, hence, the first eigenmode that exists. Obviously, using the variational principle, we get an approximate value of the critical pressure, since the stationarity condition of the exclusion functional was imposed on a finite dimensional space. To verify the strengths of the used variational principle and of the used test functions, Capanu and So´os have

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403

first determined the dependence of the correction factor p∗ on α, using the exact characteristic equation (6.3.61), due to Guz and Babitch. Afterward, they obtained the dependence of p∗ on α using the approximate characteristic equation (6.3.74). Finally, the two authors have compared the results obtained using the above mentioned two approaches. The result of this test will be presented in what follows. Let us consider now transversally isotropic circular bar. The results concerning the exact characteristic equation (6.3.61) are given in Figure 6.11. The curves given in this Figure 6.11 correspond to the following ratios: E1 /E3 = E2 /E3 = 0.2, and few values of the ration E3 /G13 are taken into account E3 /G13 = 10, 20, 30, ..., 100. The above rations are characteristic for fiber-reinforced composite bars. It was assumed that the parameter α = πh/l varies between 0 and 0.2. The last value α = 0.2 corresponds to l/2a = 7.8. In order to obtain numerical results, only dimensionless quantities were used, the elastic moduli being normed by E 3 . The dependence p∗ = p∗ (α) was obtained using the bisection method, starting with p∗ = 1, corresponding to the Eulerian value of the critical buckling pressure. 1

G13=G23=G 1

0.9

2

0.8

p*

3 4

0.7 0.6 0.5 0.4 0

1 : E3/G=10 2 : E3/G=20 3 : E3/G=30 4 : E3/G=40 5 : E3/G=50 6 : E3/G=60 7 : E3/G=70 8 : E3/G=80 9 : E3/G=90 10 : E3/G=100

0.05

5 6 7 8 9 10

0.1

0.15

Figure 6.11: Variation of correction factor p* vs. geometrical parameter α. Transversally isotropic circular bar. Exact solution. As it is easy to see, there is a very good agreement between the results obtained by Guz and Babitch and by Capanu and So´os. The conclusions obtained by the first two authors are confirmed by the second pair of authors. As we can see again, the errors due to the Eulerian formula can reach 50% for relatively short bars, corresponding to α = 0.2.

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The dependencies obtained using the approximate characteristic equation (6.3.74) are represented in Figure 6.12 by the curves 4, 5 and 6. The curves 1, 2 and 3 correspond to the dependencies obtained using the exact characteristic equation (6.3.61). The curves 1 and 4, 2 and 5, 3 and 6, respectively, correspond to the same ratio E3 /G13 = 10, 50, 100.

1

G13=G23=G 4

0.9

1

0.8

p*

5

0.7

2

0.6 0.5 0.4 0

E3/G= 10: 1=exact sol.;4=variational sol. E3/G= 50: 2=exact sol.;5=variational sol. E3/G=100: 3=exact sol.;6=variational sol.

0.05

0.1

6 3

0.15

Figure 6.12: Variation of correction factor p* vs. geometrical parameter α. Transversally isotropic circular bar. Exact and variational solution. Examining the curves given in the Figure 6.12, it can be seen that the variational method gives accurate results, even if the used test function (6.3.64) has a relatively simple structure. It can be observed also that the critical values furnished by the approximate method are slightly greater than the corresponding values obtained with the exact method. However, the relative errors are less than 0.5% for E3 /G13 = 10 and less than 7% for E3 /G13 = 100. We can conclude that the variational method is able to give accurate results that can be obtained by an appreciable amount of reduced calculus. Using appropriate test functions, this method became very useful in computing the critical buckling values. Consequently, we can expect that the selected test functions (6.3.64) and (6.3.65) can lead to good results also for other materials which have no transverse isotropy, or for those bars which have no circular cross-section. The following results, due to Capanu and So´os, will illustrate this optimistic conviction. We assume now a cylindrically orthotropic circular bar. The dependence of p∗ on α was determined using the approximate characteristic equation (6.3.74). The results for various combinations of the elastic parameter ratios are shown in Figure 6.13.

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1 0.9

1 2 3 4

0.8

5

0.7 0.6 0.5 0.4 0

1:E3/G23=10, E3/G23=10 2:E3/G23=10, E3/G23=10 3:E3/G23=10, E3/G23=70 4:E3/G23=10, E3/G23=100 5:E3/G23=40, E3/G23=40 6:E3/G23=40, E3/G23=70 7:E3/G23=40, E3/G23=100 8:E3/G23=70, E3/G23=70 9:E3/G23=70, E3/G23=100 10:E3/G23=100,E3/G23=100

0.05

6 7 8 9 10

0.1

0.15

0.2

Figure 6.13: Variation of correction factor p* vs. geometrical parameter α. Transversally orthotropic rectangular bar. Variational solution.

Examining the results, we can see that the stability of the bar is equally influenced by the values of the moduli G13 and G23 . The buckling behavior of the bar is influenced in the same manner by the shear strains and stresses existing in the longitudinal cylindrical surfaces and in the radial planes. This fact is not surprising, since the orthotropy is a cylindrical one. The Eulerian theory cannot take into account just the influence of the above mentioned strains and stresses on the critical value of the axial pressure. Let us assume now a transverse isotropic bar, having a rectangular crosssection. In this case, an exact solution does not exist and only the variational method can be used to get the dependence of the involved correction factor p ∗ (α) on the corresponding geometrical parameter α. The exclusion functional E(u) is given by equation (6.3.22) appropriate to Cartesian coordinates and to rectilinear orthotropy. The “good” test functions are given by the relation (6.3.65). Introducing the selected test functions in the expression of the exclusion functional, we obtain a quadratic function E = E(A1 , ..., A5 ) depending on the variables A1 , ..., A5 . Imposing the stationarity condition on the five-dimensional space of the supposed test functions, we are led to the following linear, homogeneous, algebraic system which must be satisfied by the unknowns A1 , ..., A5 :

pij Aj = 0, pij = pji , i, j = 1, 2, ..., 5.

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(6.3.75)

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In these equations, the coefficients pij are given by the following relations: p11 = 6(C55 − p)Kα2 , p12 = 2(C55 − p)Kα2 , √ p13 = 2(C55 − p)Kα2 , p14 = 0, p15 = 4 3C55 Kα, 2 6 32 C11 K + (C55 − p)Kα2 , p23 = (C55 − p)Kα2 , p22 = 3 5 3 4 16 C12 , p25 = − √ (2C13 − C55 )Kα, p24 = 3 3 4 16 6 32 C66 , p35 = √ C55 Kα, C66 K + (C55 − p)α2 , p34 = p33 = 3 5 3 3 4 1 C22 8 + (C44 − p)Kα2 }, p45 = − √ C23 α, p44 = − {C66 K + 4 K 3 3 p55 = 8C55 K + 2(C33 − p)Kα2 . (6.3.76)

In these relations, the geometrical parameter α is given by the equation r π 2γa 2πa 2π Im (6.3.77) =√ = √ , γ= α= l A l 3 3l

and K=

b ≥1 a

is the aspect ratio of the bar. The characteristic equation corresponding to the assumed approximation is det P (p) = det [pij (p)] = 0, where P = [pij ] is a 5 × 5 symmetric matrix. The critical axial pressure pc is the smallest positive root of the above equation. For transversally isotropic rectangular bars, the dependencies of the correction factor p∗ on the parameter α, for various values of the ratio E3 /G13 and for three values K = 1, 2, 3 of the aspect ratio are given in Figure 6.14. The correction factor p∗ is obviously introduced by the general relation (6.3.72), the involved geometrical parameter α being given by equation (6.3.77). For K = 1, that is, for a transverse isotropic bar, having square cross-section, the results for the correction factor p∗ are identical with the results obtained for a transversally isotropic, circular bar, for the same value of the ratio E 3 /G13 and for the same value of the geometrical parameter a. This result suggests the idea that the buckling of a transverse isotropic fiber-reinforced composite bar does not depend on the shape of the cross-section. That is, the curves in Figure 6.14 are

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1.1 1 1 3

0.9

2

0.8

p*

0.7 0.6 0.5 0.4 0

4 6

1 : x=3,E3/G=10 2 : x=2,E3/G=10 3 : x=1,E3/G=10 4 : x=3,E3/G=70 5 : x=2,E3/G=70 6 : x=1,E3/G=70 7 : x=3,E3/G=100 8 : x=2,E3/G=100 9 : x=1,E3/G=100

0.05

7 9

5

8

0.1

0.15

Figure 6.14: Correction factor p* vs. geometrical parameter α. Transversally isotropic rectangular bar. Variational solution.

universal curves, valid for all transversally isotropic bars, if we assume the following equations: r r IM 2π Im , (6.3.78) ,K = α= Im A l

where Im and IM are the principal centroidal moments of inertia of the crosssection and Im ≤ IM . Obviously, this conjecture must be tested by further studies, analyzing, for instance, the buckling behavior of a transversally isotropic bar having an elliptical cross-section. In this case, the test functions (6.3.65) can be used to analyze the dependence of the correction factor p∗ on the geometrical parameters α and K defined by the relations (6.3.78). The curves presented in Figure 6.13 show that if the aspect ratio K = b/a increases, the behavior of the bar is improved. Moreover, let us observe that when α converges to zero (relatively long bars), the correction factor p∗ (α) converges asymptotically toward values greater than 1. The asymptotic value is 1 only for a rectangular bar for which K = 1. In Figure 6.14, the transition from the bar to the plate buckling behavior when the aspect ratio K = b/a becomes steadily greater ( K = 10, 20 and 100 in Figure 6.14) is shown. To analyze the results given in Figure 6.14, we recall that in cylindrical buckling of a very long transversally isotropic plate (see Section 7.3 and the equation (6.2.35) of this Section), the critical pressure pE , given by the classical plate theory has the following value: E3 1 a . (6.3.79) pE = (π )2 3 l 1 − ν13 ν31

As is easy to see, the above relation can be expressed in the following equivalent

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way, having the form of the general equation (6.3.72):

pE = p ∗

α2 E3 = p ∗ p E , 4

(6.3.80)

where the correction factor p∗ is given by the relation p∗ =

1 1 = 2 (E /E ) , 1 − ν31 1 − ν13 ν31 1 3

(6.3.81)

and the geometrical parameter α is given by the equation (6.3.77), 2πa α= √ . 3l

In the above equation, 2a is the thickness (height) of the plate, l is its breadth and the compression force acts in the fibers directions. For the values E1 /E3 = 0.2, ν31 = 0.2, the classical critical pressure given by equations (6.3.80) and (6.3.81) become

pE 1.00806

α2 E3 = p ∗ p E , 4

(6.3.82)

Hence, for a very long bar (α → 0), having very great aspect ratio K = b/a, the classical plate theory gives the value p∗ = 1.00806 > 1 for the correction factor. The numerical results given in Figure 6.15 give, for the correction factor, the value p∗ = 1.008. Thus, we can see that there exists a very good agreement between the theoretical and numerical results even for very great aspect ratios, if we take into account that, in such cases, the bar is in fact a plate, and consequently behaves as a plate. The three-dimensional stability theory, founded on the three-dimensional linearized theory, automatically takes into account the change in the behavior of the body, due to the changes in its geometrical characteristics. Capanu and So´os [6.6] have analyzed also the dependence of the correction factor p∗ on α for an orthotropic rectangular bar using the variational method. The obtained numerical results show that the stability of the bar depends on the axial shear modulus G13 , but it is independent on the axial shear modulus G23 if a ≤ b. In this case, the buckling takes place in the Ox1 x3 plane, and the mentioned behavior is due to the fact that the buckling is essentially influenced only by the shear stresses and strains in those longitudinal planes which are perpendicular to

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6.3. BUCKLING OF FIBER-REINFORCED COMPOSITE BARS

409

1.1 G13=G23=G 1.00

x=10,50,100;E3/G=10 x=1;E3/G=10

0.9 0.8

x=10,50,100;E3/G=100

p*

0.7 x=1;E3/G=100

0.6 0.5 0.4

0

0.05

0.1

0.15

Figure 6.15: Variation of correction p* vs. geometrical parameter α. Long plate behavior of rectangular bar.

the buckling direction. We recall that for cylindrically orthotropic circular bars, the influence of the two axial shear moduli is almost the same. This can be expected since the material is, in this case, cylindrically orthotropic. On the contrary, in the case of rectangular bars we have a plane orthotropy and this fact explain why the modulus G23 has no influence. Summing up the results presented in this Section, we can say the following. The classical Eulerian formula is based on the assumption that the crosssections of the buckled bars remain plane and orthogonal to the curved symmetry axis of the bar. According to this assumption, the critical axial load depends only on the axial (longitudinal) Young modulus of the bar and the other mechanical characteristics have no influence on the buckling of the bar. In contrast, the three-dimensional stability theory can take into account that the buckling produces also shear strains and stresses in those longitudinal planes, which are perpendicular to the direction of buckling. The produced shear strains are greater when the shear moduli in the above planes are smaller. Consequently, their influence on the buckling can be important when the ratio of the longitudinal rigidity modulus and of the axial shear modulus has great values. This is the case if we have a fiber-reinforced composite bar. The presented results show that the above influences decrease when the length of the bar increases. Hence, Euler’s formula gives very good results for relatively long bars, but it must be improved for relatively short fiber-reinforced bars, exhibiting strong anisotropy. The three-dimensional theory of stability can furnish the improved values, using “good” test functions and the stationarity property of the exclusion functional. Thus, the potentially dangerous situations can be avoided. In order to obtain the critical values for strongly anisotropic, relatively short,

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fibers-reinforced composite bars, the Eulerian value must first be evaluated and then it must be multiplied by the corresponding correction factor, established by using the diagrams obtained from the three-dimensional theory and the adequate numerical methods.

6.4

Problems

P6.1 A graphite/epoxy fiber-reinforced orthotropic composite material has the following technical constants: E1 = 190GP a, E2 = E3 = 10GP a, G12 = 7GP a, G13 = G23 = 6GP a, ν12 = 0.3, ν13 = ν23 = 0.2. ◦ ci

Find the critical compressive stress σ 11 acting in the fibers direction and producing the internal (structural) instability of the material. ◦ ci

P6.2 Find the deformations produced by the critical compressive stress σ 11 determined in P6.1 and analyze the obtained result. P6.3 An isotropic carbon/steel has the following engineering constants: E = 206GP, ν = 0.3. ◦ ci

Find the critical compressive stress σ 11 acting in the direction of the x1 axis and producing the internal instability of the material. ◦ ci

P6.4 Find the deformations produced by the critical compressive stress σ 11 determined in P6.1 and analyze the obtained result. P6.5 A boron/epoxy fiber-reinforced transversally isotropic composite material has the following technical constants: E3 = 200GP a, E1 = E2 = 20GP a, G13 = G23 = 6GP a, ν31 = ν32 = 0.2, ν12 = 0.3. ◦ ci

Find the critical compressive stress σ 33 acting in the fibers direction and producing the internal (structural) instability of the material. ◦ ci

P6.6 Find the deformation produced by the critical compressive stress σ 33 determined in P6.5 and analyze the obtained result. P6.7 Let us consider again the fiber-reinforced orthotropic graphite/epoxy ◦ cs composite given in P6.1. Find the critical compressive stress σ 11 acting in the fibers direction and producing superficial instability of the composite through plane incremental states. Compare the results obtained in P6.1 and P6.7. P6.8 Let us consider the isotropic carbonic steel given in P6.3. Find the criti◦ cs cal compressive stress σ 11 acting in the x1 axis direction and producing instability of the material through plane incremental states.

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6.4. PROBLEMS

◦ cs

P6.9 Find the deformation produced by the critical compressive stress σ 11 determined in P6.8 and analyze the obtained result. P6.10 Let us consider a boron/epoxy fiber-reinforced transversally isotropic composite material having the following technical constants: E1 = E2 = 200GP a, E3 = 20GP a, G13 = G23 = 6GP a, ν13 = ν23 = 0.2, ν12 = 0.3. ◦ cs

◦ cs

Find the critical compressive stresses σ 11 = σ 22 acting in the fibers direction and producing superficial instability of the material through cylindrically symmetric incremental states. Compare the obtained result with the one determined in P6.5 and find the meaning of the observed facts. P6.11 Let us consider a transversally isotropic fiber-reinforced infinite strip having the thickness h and the breadth l. Let us assume that the strip is acted by ◦ compressive stresses σ 11 in the fiber direction. We suppose that the engineering constants of the strip are E1 = 200GP a, E2 = 40GP a, G12 = 2GP a, ν23 = 0.3, ν12 = 0.2. For which values of the thickness ratio b = πh/l can we use the classical LoveKirchhoff plate theory to study the buckling behavior of the strip, according to the three-dimensional linearized theory? P6.12 Let us assume that the strip given in P6.11 has the following geometrical characteristics: h = 0.01m, l = 10m. Find the buckling (bending) pressure pE furnished by the classical plate theory. P6.13 For the strip given in P6.11 and P6.12, find the correction factor p∗ (b) furnished by the three-dimensional linearized theory. P6.14 For the strip given in P6.11, find the correction factor p∗ (b) using Table 6.1 and Figure 6.1, for the following values of the thickness ratio b = πh/l: b = πh/l = 0.08, 0.1, 0.12. P6.15 Let us consider the half space −∞ < x1 < ∞, 0 ≤ x2 , −∞ < x3 < ∞, filled by a homogeneous isotropic linearly elastic material. Let us assume that the boundary x2 = 0 of the half-space is stress-free and the body force acting on the material is zero. Show that Lame’s equations of motion has a nonvanishing solution of the form u1 = Ae−αx2 sin k(x1 − vt), u2 = Be−αx2 cos k(x1 − vt), u3 ≡ 0, where A and B are arbitrary constants, and α > 0, k > 0 and v > 0 are arbitrary positive constants. Show that the homogeneous traction boundary conditions can be satisfied for any k > 0, by adequately determined values of the constants A, B, α and v. Give the mechanical significance of the obtained results.

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P6.16 Analyzing the superficial instability of a prestressed half space, we have used the potentials ϕ(1) and ϕ(2) given by the relations (6.1.30), assuming ϕ(1) ≡ 0. Let us change now the role of the functions ϕ(1) and ϕ(2) , assuming that ϕ(1) = (B1 eaη1 x2 + B2 eaη2 x2 ) cos ax1 , ϕ(2) = (A1 eaη1 x2 + A2 eaη2 x2 ) sin ax1 . Supposing ϕ(2) ≡ 0 and, using Guz’s representation formulas (5.6.53), find the incremental displacement u1 and u2 corresponding to ϕ(1) . P6.17 Find the incremental normal stresses θ21 and θ22 corresponding to the incremental displacements determined in P6.16. Find the condition which must be satisfied by the constants B1 and B2 , if the boundary x2 = 0 of the halfspace must be stress-free. Find the condition which must be satisfied by the initial ◦ applied stress σ 11 in order to have a nonzero solution of the above homogeneous traction boundary value problem. Compare the obtained results with those found in Section 6.1 in the problem surface of instability of a prestressed half space. Analyze the consequence resulting from this comparison. P6.18 Analyzing the stability problem of a prestressed strip, we have used the potential ϕ(2) given by equation (6.2.5) or (6.2.6), assuming ϕ(1) = 0. Let us assume now that ϕ(2) ≡ 0 and ϕ(1) is given by one of the following expressions: Φ ≡ ϕ(1) = (A1 cosh aη1 x2 + A2 cosh aη2 x2 ) sin ax1 , or Ψ ≡ ϕ(2) = (A1 sinh aη1 x2 + A2 cosh aη2 x2) sin ax1.

Here A1 , A2 are arbitrary constants, a = mπ/l , m = 1, 2, 3, ... . Show that Φ and Ψ satisfies Guz’s equation (5.6.56). P6.19 Using Guz’s representative formulas (5.6.53), find the incremental displacements u1 and u2 corresponding to ϕ(1) = Φ, ϕ(2) ≡ 0 and to ϕ(1) = Ψ, ϕ(2) ≡ 0, respectively. Give the interpretation of the obtained results. P6.20 Assuming I2 = Im < I1 = IM , suppose that the buckling of a bar takes place in the Ox2 x3 plane (see Figure 6.6). Using Euler’s theory, find the corresponding buckling pressure. Why, however, can we conclude that the buckling takes place in the Ox1 x3 plane? P6.21 Let us consider an isotropic bar having a rectangular cross-section −a ≤ x1 ≤ a, −b ≤ x2 ≤ b and let l be the length of the bar. Let us assume that if the bar is in a buckled state, its incremental displacement is described by the Eulerian relations (6.3.23). Find the corresponding incremental nominal stresses, assuming that the nonzero instantaneous elasticities are given by equation (6.3.21). Express the results using Young’s modulus E and Poisson’s ratio ν. P6.22 Find the incremental traction acting on the lateral surface of the bar and corresponding to the incremental nominal stresses determined in P6.21. P6.23 Analyze the questions proposed in P6.21 and P6.22 assuming u1 (x3 ) = A sin πx3 /l. Why does Euler’s incremental displacement field (6.3.23), not represent even for the above choice of u1 (x3 ), the exact situation of the buckling problem for the rectangular bar considered in P6.21?

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6.4. PROBLEMS

P6.24 Find θzz = θ33 for the incremental displacement field (6.3.57) involved in the buckling of a transversally isotropic circular bar and prove that θzz = 0 for z = 0 and z = l. P6.25 Using the equation

1 d 1 ( I1 (r)) = I2 (r) r dr r show that the coefficient m11 actually has the value given by equation (6.3.59)1 . P6.26 Using the relation (6.3.62) defining the function In (r), give the details leading to the expression (6.3.64) of the test function. P6.27 Show that the incremental displacement field (6.3.64) describes a buckled state of a circular bar. P6.28 Give the details leading from the test function (6.3.64) to the test function (6.3.65). P6.29 Show that the incremental displacement field (6.3.65) describes a buckled state of a rectangular bar in the Ox1 x3 plane. P6.30 Assuming an isotropic rectangular bar and supposing that the nonzero instantaneous elasticities are given by equation (6.3.21), find the incremental nominal stresses corresponding to the incremental displacement field (6.3.65). Express the result using E and ν. P6.31 Find the incremental tractions acting on the lateral surface of the bar and corresponding to the incremental nominal stresses determined in P6.30. P6.32 If the traction determined in P6.31 are not zero on the lateral surface of the bar, can the incremental displacements (6.3.65) be used in the incremental variational principle corresponding to the buckling problem? Justify your answer! P6.33 Find Im , IM and the Eulerian buckling pressure pE for the rectangular bar considered in P6.21. P6.34 Does any connection exist between the Eulerian buckling field (6.3.23) and the approximate buckling field (6.3.65)? Why is the second field more appropriate to study the buckling problem? P6.35 Prove that the last relation (6.3.76) giving the coefficient p55 is true. P6.36 Let us consider a bar having length l and an elliptical cross-section. Let us denote by a ≤ b the semiaxes of the corresponding ellipse. Find Im , IM and the Eulerian buckling pressure pE . P6.37 Find the parameters α and K, defined by equation (6.3.78), for the bar considered in P6.36. P6.38 The test function (6.3.65) can be used to study the buckling problem for the bar considered in P6.36. Justify the given answer! P6.39 Let us consider a transversally isotropic fiber-reinforced circular cylindrical bar. Let us assume that the bar has the following engineering constants:

E3 = 200GP a, E1 = E2 = 0.2E3 , ν12 = 0.3, ν31 = ν32 = 0.2, G13 = G23 = 0.1E3 . We suppose that the geometrical parameters of the bar are l = 10m, a = 0.01m.

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CHAPTER 6. BUCKLING OF COMPOSITE STRIPS AND BARS

(a) Find the Eulerian buckling force pE given by equation (6.3.17). (b) Find the parameter α = πa/l introduced by equation (6.3.60). (c) Using Figure 6.11, find the correction factor p∗ (α) and analyze the results obtained in (a) and (c). P6.40 For the bar considered in P6.39, find the critical pressure pc given by equation (6.3.37) and compare the obtained result with the one obtained in (a) P6.39. P6.41 Let us consider a fiber-reinforced transversally isotropic circular cylindrical bar and let us assume that the bar has the engineering constants E 3 , E1 = E2 , ν12 , ν31 = ν32 given in P6.39. We suppose that the geometrical parameters of the bar are l = 10m, a = 0.5m. (a) Find the Eulerian buckling pressure. (b) Find the parameter α = πa/l. (c) Using Figure 6.11, find the correction factor p∗ (α) assuming that the transverse shear moduli G13 = G23 of the bar have the following values: G31 = G32 = 0.1E3 , 0.02E3 , 0.01E3 . (d) Analyze the results obtained in (a) and (c). P6.42 Let us consider again a transversally isotropic fiber-reinforced circular cylindrical bar and let us assume that its engineering constants E3 , E1 = E2 , ν12 , ν31 = ν32 are those given in P6.39. Let us suppose that the transverse shear moduli G31 = G32 of the bar have the value G31 = G32 = 0.02E3 . We assume that l = 10m and the radius a of the bar can have the following values: a = 0.01m, 0.1m, 0.5m. (a) Find the Eulerian buckling forces pE corresponding to the given data. (b) Find the parameters α = πa/l for the given data. (c) Using Figure 6.6, find the correction factor p∗ (α) corresponding to the given data. (d) Compare and analyze the results obtained in (a) and (c). P6.43 For the data given in P6.42, find the critical buckling pressure pc given by the equation (6.3.37) and compare the obtained results with those found in (a) and (c) of P6.42. P6.44 Let us consider a cylindrical transversally isotropic fiber-reinforced bar having square cross-section (a = b). We assume that the engineering constants of the bar are those given in P6.39. Also we suppose that the geometrical parameters of the bar are l = 10m, a = b = 0.01m.

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415

6.4. PROBLEMS

(a) Find the Eulerian buckling force pE . (b) Find the parameter α defined in equation (6.3.77) and find the aspect ratio K = b/a of the bar. (c) Using Figure 6.9, find the correction factor p∗ (α) and analyze the results obtained in (a) and (c). P6.45 Let us consider a transversally isotropic fiber-reinforced composite bar, having rectangular cross-section. Let us assume that its engineering constants E3 , E1 = E2 , ν12 , ν31 = ν32 are those given in P6.39. We suppose that the geometrical characteristics of the bar are l = 10m, a = 0.5m, b = 1.5m. (a) Find the Eulerian buckling pressure pE . (b) Find the parameter α and the aspect ratio K. (c) Using Figure 6.9, find the correction factors p∗ (α) assuming that the axial shear moduli G13 = G23 of the bar have the following values: G31 = G32 = 0.1E3 , 0.01E3 . (d) Analyze the results obtained in (a) and (c). P6.46 Let us consider again a transversally isotropic fiber-reinforced composite bar having rectangular cross-section, and let us assume that its engineering constants are those given in P6.42. We suppose also that l = 10m, a = 0.5m, and b can have the following values: b = 1m, 1.5m. (a) Find the Eulerian buckling forces pE corresponding to the given data. (b) Find the parameters α and K, for the given data. (c) Using Figure 6.14, find the correction factor p∗ (α) corresponding to the given data. (d) Compare and analyze the results obtained in (a) and (c). P6.47 Let us consider a fiber-reinforced cylindrical composite bar having elliptical cross-section. Let us denote, by a ≤ b, the semiaxes of the ellipse and let l be the length of the bar. To solve approximately the buckling problem, we assume that the test functions are given by equation (6.3.65). Using equation (6.3.22) giving the exclusion functional E(u), find the expression of the function E(A1 , A2 , A3 , A4 , A5 ) corresponding to the assumed test functions. P6.48 Using the incremental variational principle given in Section 5.4, show that the unknown constants A1 , ..., A5 must satisfy the following homogeneous linear algebraic system: qij Aj = 0, qij = qji , i, j = 1, 2, 3, 4, 5

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where the coefficients qij are given by the following relations (see Capanu [6.9]): q11 = 2(C55 − p)α2 K, q12 = 12 (C55 − p)α2 K, q13 = 12 (C55 − p)α2 K,

q14 = 0, q15 = 2C55 αK, q22 = 2C11 K + 14 (C55 − p)α2 K,

q23 =

1 12 (C55

− p)α2 K, q24 = C12 , q25 = −C13 αK + C55 αK,

q33 = 2C66 K −1 + 14 (C55 − p)α2 K, q34 = C66 , q35 = 12 C55 αK,

q44 =

1 2


C66 K + C22 K −1 +

1 12 (C44

− p)α2 K, q45 = − 12 C23 α,

q55 = 2C55 K + 12 (C33 − p)α2 K,

with α=π

a a and K = ≥ 1. b l

Bibliography [6.1] Biot, M. A., Mechanics of incremental deformation, Wiley, New York, 1965. [6.2] Guz, A.N., Fundamentals of the three-dimensional theory of stability of deformable bodies, Wisha Schola, Kiev. 1986 (in Russian). [6.3] Guz, A.N., Babitch, I. In., The three-dimensional theory of stability of bars, plates and shells, Wisha Schola, Kiev, 1980 (in Russian). [6.4] Pearson, C.E., General theory of elastic stability, Q. Appl. Math., 14, 133144, 1956. [6.5] Gray, A., Mathews G.R., A treatise in Bessel functions and their applications to physics, MacMillan Co. Ltd., London, 1952. [6.6] Capanu, M., So´os, E., The three-dimensional theory of buckling of composite elastic bars. The influence of a lateral hydrostatic pressure, Rev. Roum. Sci. Techn.-M`ec. Appl., 41, 193-209, 1996. [6.7] Lekhnitski, S.G., Theory of elasticity of anisotropic elastic body, Holden Day, San Francisco, 1963. [6.8] Malvern, L.E., Introduction to the mechanics of continuous medium, PrenticeHall Inc., London, 1969. [6.9] Capanu, M., Buckling of composite elastic tubes and elliptic bars. The plate effect, Rev. Roum. Sci. Techn.-M`ec Appl., 42, 151-173, 1997.

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Chapter 7

STABILITY OF COMPOSITE LAMINATES 7.1

Constitutive equations for incremental fields

Some elements on stability theory of composite laminates are presented, for instance, in the monographies due to Ashton and Whitney [7.1], Jones [7.2], Whitney [7.3] and Gibson [7.4]. The stability theory of composite laminates, presented in this chapter, is founded on the classical Love-Kirchhoff plate theory and on the three-dimensional linearized theory. The theory obtained in this way also takes into account the influence of the initial applied bending moment on the behavior of composite laminates and is due to Capanu and So´os [7.5] and Capanu [7.6]. In most studies concerning stability of composite laminates, the influence of the initial applied bending moment is neglected and this can lead to dangerous circumstances. The equilibrium theory of laminated composite, in the framework of LoveKirchhoff plate theory, was presented in Chapter 3. For convenience, and using the notations of Chapter 3, we recall here the simplifying assumptions, on which the Love-Kirchhoff theory of equilibrium of composite plates is based (see Section 3.3 and Figure 3.10): (1) The plate consists of orthotropic laminae bounded together, with the principal material axes of the orthotropic laminae oriented at arbitrary directions with respect to the x1 ,x2 axes. (2) The thickness h of the plate is much smaller then the length of the plate edges a and b. (3) The displacements u1 , u2 , u3 are small compared with the plate thickness h. (4) The in-plane strains ε11 , ε22 , ε12 are small compared with unity. (5) Transverse shear strains ε13 and ε23 are negligible.

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(6) The transverse normal strain ε33 is negligible. (7) The normal stress σ33 is small in comparison with the other stress components. (8) The transverse shear stresses σ13 and σ23 vanish on the plate surfaces x3 = ± h2 . (9) Each laminae obeys the reduced stress−strain relation corresponding to plane−stress state. In order to study the local stability of the composite laminates, we assume that the laminate is initially deformed. We suppose that the obtained equilibrium o o configuration B is known. We assume that the configuration B is perturbed by small external forces and we denote by B 0 the neighboring new equilibrium configo uration of the laminate. All quantities related to B will be denoted by a superposed “◦” and all fields related to B 0 will be designed by a superposed 0 . We denote by o o o 0 the nominal stresses in B 0 . θkl =σ kl the Cauchy’s stresses in B , and by θkl According to the adopted notation, we have o

o

o

o

0 0 = σ kl + θkl . = σ kl + σkl ; θkl u0k = uk + uk ; ε0kl = εkl + εkl , σkl

(7.1.1)

In these relations, uk , εkl , σkl and θkl are the incremental displacement, the deformation, the Cauchy’s stress and the nominal stress, respectively. Other incremental fields (perturbations) are introduced in a similar manner. Since the initial applied deformations and rotations are assumed to be infinitesimal, the incremental fields θkl , σkl and uk,l are related by the following relations: o (7.1.2) θkl = ωklmn um,n with ωklmn = cklmn + σ kn δlm . The above incremental constitutive equation can be written in the following equivalent form: o

θkl = σkl + σ km ul,m with σkl = cklmn εmn .

(7.1.3)

Also, as we already know, for vanishing body forces, the incremental nominal stress θkl satisfies the following incremental equilibrium equation: θkl,k = 0.

(7.1.4)

We assume that the hypothesis (1)−(9) are satisfied in the initial deformed o

configuration B , as well as the neighborhood perturbed equilibrium configuration B0. As we already know, from the assumptions (5) and (6) we get the followo ing structure of the initial applied displacement field uk and of the incremental displacement field uk : o

o ∂ U 3 (x1 , x2 ) o uk = U α (x1 , x2 ) − x3 , α = 1, 2, u3 = U 3 (x1 , x2 ) , ∂xα o

o

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(7.1.5)

7.1. CONSTITUTIVE EQUATIONS FOR INCREMENTAL FIELDS

419

and uα = Uα (x1 , x2 ) − x3

o ∂U3 (x1 , x2 ) , α = 1, 2, u3 = U 3 (x1 , x2 ) . ∂xα

(7.1.6)

Thus, from the strain−displacement relations, we get o

o

o

εαβ = eαβ +x3 k αβ ; α, β = 1, 2, where o

eαβ =

(7.1.7)

 o o o o 1o U α,β + U β,α , k αβ = k βα = − U 3,αβ . 2

(7.1.8)

Similarly, for the incremental fields, we obtain εαβ = eαβ + x3 kαβ with eαβ =

(7.1.9)

1 (Uα,β + Uβ, α ) , kαβ = kβα = −U3,αβ . 2

(7.1.10)

In these equations and in all that follows, Greek indices take the values 1, 2. Also, from the hypotheses (7) and (8), it results   h o o σ 3α ± (7.1.11) = 0 and σ 33 ≈ 0. 2

Thus, taking into account equations (7.1.1)2,3 , (7.1.3) and (7.1.11), for the incremental stress fields, we obtain   h = 0, σ33 ≈ 0, θ33 ≈ 0. (7.1.12) θ3α ± 2

From (7.1.11), we can conclude that the resultant forces and moments o

o

N αβ = N βα =

Z

h 2

−h 2

o

o

o

σ αβ dx3 , M αβ = M βα =

Z

h 2

−h 2

o

x3 σ αβ dx3

(7.1.13)

o

corresponding to the initial deformed equilibrium configuration B , are expressed o o in terms of eαβ and k αβ by relations of type (3.3.21). Thus we have o

o

o

o

o

o

N αβ = Aαβγϕ eγϕ +Bαβγϕ k γϕ , M αβ = Bαβγϕ eγϕ +Dαβγϕ k γϕ .

(7.1.14)

The way in which the coefficients Aαβγϕ , Bαβγϕ and Dαβγϕ are expressed in terms of the material and geometrical characteristics of the laminate was presented in Chapter 3.

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CHAPTER 7. STABILITY OF COMPOSITE LAMINATES o

Also, as before, we can conclude that the resultant forces N αβ and the resulo tant moments M αβ must satisfy the global equilibrium conditions o

o

o

o

o

N αβ,α = 0, Qα,α + q = 0, M αβ,α + Qβ = 0,

(7.1.15)

o

where the resultant shear forces Qα , corresponding to the initial deformed equio

librium configuration B , are defined by o

Qα =

Z

h 2

−h 2

o

σ α3 dx3 .

(7.1.16)

Now we take into account equations (3.1.15). These equations and the relations (7.1.1)2,3 tell us that in the k − th lamina, the incremental stress and strain fields σαβ and εαβ are related by the formula       Q11 Q12 Q16 ε11 σ11  σ22  =  Q21 Q22 Q26   ε22  , zk−1 < x3 < zk ; k = 1, 2, ..., N. σ12 k Q61 Q62 Q66 k 2ε12 k (7.1.17)

From (7.1.17) we can conclude that the incremental resultant forces and moments Nαβ = Nβα =

Z

h 2

−h 2

σαβ dx3 , Mαβ = Mβα =

Z

h 2

−h 2

x3 σαβ dx3

(7.1.18)

are expressed in terms of the incremental fields eαβ and kαβ , by equations of type (7.1.14). Thus, we have Nαβ = Aαβγϕ eγϕ + Bαβγϕ kγϕ , Mαβ = Bαβγϕ eγϕ + Dαβγϕ kγϕ .

(7.1.19)

Now we are ready to calculate the perturbations Nαβ and Mαβ of the resultant forces and moments, defined by equations Nαβ =

Z

h 2

−h 2

θαβ dx3 , Mαβ =

Z

h 2

−h 2

x3 θαβ dx3

(7.1.20)

According to (7.1.3)1 , we have o

o

θαβ = σαβ + σ αγ uβ,γ + σ α3 uβ,3 .

(7.1.21)

uβ,γ = Uβ,γ + x3 kβγ , uβ,3 = −U3,β .

(7.1.22)

From (7.1.6), we get

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7.1. CONSTITUTIVE EQUATIONS FOR INCREMENTAL FIELDS

421

Now, from (7.1.13), (7.1.16), (7.1.18)1 and (7.1.20)1 , it results o

o

o

Nαβ = Nαβ + N αγ Uβ,γ + M αγ kβγ − Qα U3,β .

(7.1.23)

Using again equations (7.1.21) and (7.1.22), we get o

o

o

x3 θαβ = x3 σαβ + x3 σ αγ Uβ,γ + x23 σ αγ kβγ − x3 σ α3 U3,β .

(7.1.24)

Taking into account the hypothesis (2), to calculate Mαβ we shall neglect in o (7.1.24) the term containing x23 . According to the hypotheses (5) and (8), σ α3 are o much smaller than σ αβ . Consequently, to determine Mαβ we shall neglect also the last term in (7.1.24). With this approximation, from equations (7.1.18) 2 , (7.1.20)2 and (7.1.24), we get o

Mαβ = Mαβ + M αγ Uβ,γ .

(7.1.25)

For a later use, we shall introduce the incremental shear force resultants Rα and Pα defined by Rα =

Z

h 2

−h 2

θα3 dx3 and Pα =

Z

h 2

−h 2

θ3α dx3 .

(7.1.26)

In order to evaluate Rα , we return to (7.1.3)1 . We have o

o

θα3 = σα3 + σ αβ u3,β + σ 33 uα,3 .

(7.1.27)

Taking into account the hypothesis (7), we replace this equation by the approximate relation o (7.1.28) θα3 = σαβ + σ αβ u3,β . Thus, from (7.1.13)1 , (7.1.18)1 and (7.1.26)1 , we obtain o

Rα = Qα + N αβ U3,β ,

(7.1.29)

the incremental field Qα being defined by Qα =

Z

h 2

−h 2

σα3 dx3 .

(7.1.30)

In order to evaluate Pα , we use again (7.1.3)1 . We have o

o

θ3α = σ3α + σ 3β uα,β + σ 33 uα,3 .

(7.1.31)

Using again the hypothesis (7), we replace the last equation with the appropriate one o (7.1.32) θ3α = σ3α + σ 3β uα,β .

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422

CHAPTER 7. STABILITY OF COMPOSITE LAMINATES

Now, using (7.1.22)1 , we get o

o

θ3α = σ3α + σ 3β Uα,β + x3 σ 3β kαβ .

(7.1.33)

In evaluating Pα , we neglect the last term, as we have done to obtain equation (7.1.25). In this way, using (7.1.16), (7.1.26)2 and (7.1.30), finally we get o

Pα = Qα + Qβ Uα,β .

(7.1.34)

The incremental fields Nαβ and Mαβ generally are not symmetric, and Rα and Pα are generally different. To get the final form of the global incremental constitutive equations, we observe that the equations (7.1.23) and (7.1.25) can be written in the following equivalent form: o

Nαβ = Ωαβγϕ Uγϕ + Γαβγϕ kγϕ − Qα U3,β

(7.1.35)

Mαβ = Γαβγϕ Uγ,ϕ + Dαβγϕ kγϕ , where the global instantaneous elasticities are given by the equations o

o

Ωαβγϕ = Aαβγϕ + N αϕ δβγ , Γαβγϕ = Bαβγϕ + M αϕ δβγ .

(7.1.36)

These quantities are not symmetric in (α, β) and (γ, ϕ). However, as it is easy to see, the above coefficients have the following essential symmetry property: Ωαβγϕ = Ωϕγβα , Γαβγϕ = Γϕγβα . (7.1.37) Also, the resultant shear forces Rα and Pα according to equations (7.1.24) and (7.1.34), are related by the relation o

o

Rα = Pα + N αβ U3,β − Qβ Uα,β .

7.2

(7.1.38)

Equilibrium equations. Boundary conditions o

We recall that in the initial deformed configuration B , the corresponding o stresses σ kl satisfy the Cauchy’s equilibrium equations o

σ kl,l = 0.

(7.2.1)

Consequently, as in Section 3.5, we obtain the global equilibrium equations satisfied o

o

o

by N αβ , M αβ and Qα o

o

o

o

o

N αβ,α = 0, Qα,α + q = 0, M αβ,α − Qβ = 0.

Copyright © 2004 by Chapman & Hall/CRC

(7.2.2)

7.2. EQUILIBRIUM EQUATIONS. BOUNDARY CONDITIONS

423

In order to obtain the equilibrium equations which must be satisfied by the global incremental fields Nαβ , Mαβ , Rα and Pα , we recall that the incremental nominal stress θkl satisfy in each lamina the equilibrium equations (7.1.4). Hence, we have θαβ,α + θ3β,3 = 0, θα3,α + θ33,3 = 0.

(7.2.3)

Using the same procedures as before, taking into account the boundary conditions (7.1.12)1 , and the definitions (7.1.20) and (7.1.26), we get the following global incremental equations: Nαβ,α = 0, Rα,α + q = 0, Mαβ,α − Pβ = 0.

(7.2.4)

In equations (7.2.4)2 , q = q (x1 , x2 ) is the given perturbation of the external normal surface force acting on the face x3 = − h2 of the laminate. From the equations (7.2.2)2 and (7.2.4)3 , we get the relations which must be satisfied by the second derivatives of the resultant incremental moments M αβ

Mαβ,αβ + (Rα − Pα ) ,α +q = 0.

(7.2.5)

In order to obtain an incremental work theorem, we use the global equilibrium conditions (7.2.4). As in Section 3.5, we denote by D the plane domain occupied by the middle surface of the laminate in its reference configuration. Let ∂D be the boundary of D. We denote by nα the components of the external unit normal to ∂D and by τα the components of the unit tangent vector to ∂D, as in Figure 3.19. We have n1 = τ2 and n2 = −τ1 . (7.2.6)

Also, we introduce the two-dimensional vectors N n and Mn , and the scalar field Rn defined on ∂D by equations Nnβ = nα Nαβ , Mnβ = nα Mαβ , Rn = nα Rα .

(7.2.7)

The normal and tangential components of the incremental fields are given by equations Nnn = nβ Nnβ , Nnτ = τβ Nnβ , Mnn = nβ Mnβ , Mnτ = τβ Mnβ . In an obvious way from the equilibrium equations (7.2.4), we get Z Z Nnβ Uβ ds = Uβ,α Nαβ da. ∂D

D

where Un = nα Uα and Uτ = τα Uα

Copyright © 2004 by Chapman & Hall/CRC

(7.2.9)

D

This relation can be written in the equivalent form Z Z (Nnn Un + Nnτ Uτ ) ds = Uβ,α Nαβ da, ∂D

(7.2.8)

(7.2.10)

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CHAPTER 7. STABILITY OF COMPOSITE LAMINATES

are the normal and tangential components of the in-plane incremental displacement field. In the same way, from the equilibrium equations (7.2.4)2 , we obtain Z Z Z qU3 da + Rn U3 ds = Rα U3,α da. (7.2.11) D

∂D

D

In order to transform the right hand side term, we use the obvious identity Z Z Z Pα U3,α da. (7.2.12) (Rα − Pα ) U3,α da + Rα U3,α da = D

D

D

Using the equation (7.2.4)3 , we can express the last term in (7.2.12), in the following form: Z Z Pα U3,α da = Mαβ,α U3,β da. (7.2.13) D

D

Now, using Green’s formula and the equations (7.2.7)2 , we obtain Z Z Z Pα U3,α da = Mnβ U3,β da + Mαβ kαβ da. D

∂D

(7.2.14)

D

To obtain the last relation, we have used kαβ = −U3,αβ . Now, from (7.2.11), (7.2.12) and (7.2.14), it results Z Z Z Z qU3 da+ Rn U3 ds = Mnβ U3,β ds+ {Mαβ kαβ + (Rα − Pα ) U3,α } da. D

∂D

∂D

D

(7.2.15) Let us transform now the first term in the right-hand side of the last equation. For this purpose, we denote by U3,n = nα U3,α and U3,σ = σα U3,α

(7.2.16)

the normal and tangential derivatives of the normal incremental displacement field U3 along the curve ∂D. From (7.2.6) and (7.2.16)2 , we get U3,1 = τ2 U3,n + τ1 U3,τ , U3,2 = −τ1 U3,n + τ2 U3,τ .

(7.2.17)

Using the above equations, we get Mnβ U3,β = (τ1 Mn1 + τ2 Mn2 ) U3,τ + (τ2 Mn1 − τ1 Mn2 ) U3,n .

(7.2.18)

Taking into account (7.2.6) and (7.2.8)3,4 , we can conclude that Mnβ U3,β = (Mnτ U3 ),τ − Mnτ,τ U3 + Mnn U3,n , where Mnτ,τ is the tangential derivative of Mnτ along the curve ∂D.

Copyright © 2004 by Chapman & Hall/CRC

(7.2.19)

7.2. EQUILIBRIUM EQUATIONS. BOUNDARY CONDITIONS Since

Z

finally, we get Z

∂D

∂D

(Mnτ U3 )1τ ds = 0,

Mnβ U3,β ds =

Z

∂D

(Mnn U3,n − Mnτ,τ U3 ) ds.

425

(7.2.20)

(7.2.21)

Adding (7.2.11), (7.2.15) and using the equality (7.2.21), we obtain the following incremental work relation: Z {Uβ,α Nαβ + kαβ Mαβ + (Rα − Pα ) U3,α } da = D

Z

qU3 da + D

Z

∂D

{Nnn Un + Nnσ Uσ + (Rn + Mnσ,σ ) U3 − Mnn U3,n } ds. (7.2.22)

We denote by 2w quadratic form from the left-hand side of the above equation; i.e. 2w (U)

≡ Uβ,α Nαβ + kαβ Mαβ + (Rα − Pα ) U3,α = Uβ,α Ωαβγϕ Uγ,ϕ + 2Uβ,α Γαβγϕ kγϕ + kαβ Dαβγϕ kγϕ o

o

+ U3,α N αβ U3,β − 2Uα,β Qβ U3,α .

(7.2.23)

In this relation, U is three-dimensional with the components (Uα , U3 ). The work relation (7.2.22) shows that the perturbation W (U) of the elastic energy stored in the laminate is given by the equation Z w (U) da. (7.2.24) W (U) = D

At the same time, the relation (7.2.22) tells us that on the boundary ∂D of D, the following incremental fields can be given: Un or Nn , Uτ or Nnτ , U3 or Rn + Mnτ,τ and U3,n orMnn .

(7.2.25)

Accordingly, at least 16 different incremental boundary value problems can exist. The same equation (7.2.22) shows that the various incremental boundary value problems can have no more than one regular solution (modulo a rigid displacement), if the quadratic form w = w (U) is positive definite. Let us denote by δW the first variations of the functional W = W (U) in U and in the direction δU = (δUα , δU3 ). We have δW =

λ being a real variable.

Copyright © 2004 by Chapman & Hall/CRC

d W (U + λδU) |λ=0 , dλ

(7.2.26)

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CHAPTER 7. STABILITY OF COMPOSITE LAMINATES

In order to evaluate this variation, we take into account the constitutive equations (7.1.35), (7.1.38) and the symmetry relations (7.1.37), together with equation (7.2.4), considered as a constitutive relation, expressing Pβ in terms of Mαβ , hence, as functions of Uα,β and U3,α . In this way, we get δW =

Z

D

{Nαβ δUβ,α − Mαβ δU3,αβ + (Rα − Pα ) δU3,α } da.

(7.2.27)

Using Green’s formula, we obtain Z n o i h Nαβ,α δUβ + Mαβ,αβ + (Rα − Pα ),α δU3,α da δW = − D

+

Z

∂D

{Uα Nαβ δUβ + nα (Rα − Pα + Mβα,β ) U3 − nα Mαβ δU3,β } ds.

(7.2.28)

We have also nα Nαβ δUβ = Nnn δUn + Nnτ δUτ , δUn = nα δUα , δUτ = τα δUα ,

(7.2.29)

nα Mαβ δU3,β = (Mnτ δU3 ),τ − Mnτ,τ δU3 + Mnn δU3,n .

(7.2.30)

and Hence, taking into account also the relation (7.2.4), from (7.2.28)−(7.2.30), we get Z n h o i Nαβ,α δUβ + Mαβ,αβ + (Rα − Pα ),α δU3 da δW = − D

+

Z

∂D

{Nnn δUn + Nnτ δUτ + (Rn + Mnτ,τ ) δU3 − Mnn δU3,n } ds.

(7.2.31)

This relation can be used to obtain various incremental variational principles, corresponding to different boundary value problems. We shall illustrate the procedure by analyzing only two possibilities. First, let us assume that on the boundary curve ∂D are given Nnn , Nnτ , Rn + Mnτ,τ and Mnn Nnn = φ, Nnτ = Ψ, Rn + Mnτ,τ = Γ, Mnn = ∆ on ∂D,

(7.2.32)

Φ,Ψ,Γ and ∆ being known external perturbations on ∂D. In this case, we shall introduce the functional (incremental potential energy) I (U) defined by equation Z Z I (U) = W (U) − (φUn + ΨUσ + ΓU3 − ∆U3,n ) ds − qU3 da. (7.2.33) ∂D

Copyright © 2004 by Chapman & Hall/CRC

D

427

7.2. EQUILIBRIUM EQUATIONS. BOUNDARY CONDITIONS Taking into account (7.2.31), we get Z n h o i δI = − Nαβ δUβ + Mαβ,αβ + (Rα − Pα ),α δU3 da ZD {(Nnn − φ) δUn + (Nnτ − Ψ) δUτ + (Rn + Mnτ,τ − Γ) δU3 + ∂D

− (Mnn − ∆) δU3,n } ds.

(7.2.34)

Now assume that U is a regular solution of the considered incremental boundary value problem. Hence, the global equilibrium equations (7.2.4)2 , (7.2.5) and boundary conditions (7.2.32) are satisfied. From (7.2.34), we can see that, in this case, the first variation δI of I (U) in U is vanishing, for any variation δU. Conversely, assuming that δI = 0 (7.2.35) in U for any variation δU, from (7.2.33) we can conclude that U is a regular solution of the considered incremental boundary value problem. Next, let us suppose that on the boundary line ∂D are given Un , Uτ , U3 and U3,n ; i.e. Un = ϕ, Uϕ = Ψ, U3 = γ, U3,n = δ on ∂D, (7.2.36) σ, Ψ, γ and δ being given functions on ∂D. In this case, we shall introduce the functional (incremental potential energy) Z J (U) = W (U) − qU3 da. (7.2.37) D

We shall calculate the first variation δJ of this functional in a direction δU satisfying homogeneous boundary conditions; i.e. δUn = δUτ = δU3 = δU3,n = 0 on ∂D. From (7.2.34), it results Z n o h i δJ = − Nαβ,α δUβ + Mαβ,αβ + (Rα − Pα ),α + q δU3 da.

(7.2.38)

(7.2.39)

D

The last equation shows that if U is a regular solution of the considered incremental boundary value problem, the first variation δJ of J in U is vanishing for any variation δU satisfying the homogeneous boundary conditions (7.2.38). Conversely, assuming that δJ = 0 (7.2.40) in U, for any variation δU satisfying (7.2.38), we can conclude that U, satisfying the given boundary conditions (7.2.36), is a regular solution of the considered incremental boundary value problem.

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428

CHAPTER 7. STABILITY OF COMPOSITE LAMINATES

Using the potential energy I = I (U) and J = J (U), introduced above, and assuming that the specific internal energy w = w (U) is a positive definite quadratic form, appropriate principles of minimum potential energies can be formulated and proved. This statement is true also for other possible incremental boundary value problems and appropriate potential energies. The procedure to formulate and to prove the corresponding extremal principles is similar to the one used in the classical three-dimensional case. Also, to discuss stability problems, concerning composite laminates, we must introduce the corresponding exclusion functional E = E (U) defined analogously to the procedure used in the general three-dimensional case (see Section 5.5 Equation (5.5.1)). Thus, we have Z E (U) = 2

w (U) da,

(7.2.41)

D

where w = w (U) in the quadratic form introduced by equation (7.2.23). All problems concerning local stability, eigenstates and eigenmodes, primary eigenstates and primary eigenmodes, appropriate to the composite laminates, can be discussed and analyzed as in the general three-dimensional case. Obviously, in all reasonings, the exclusion functional E = E (U) introduced by the equation (7.2.41) appears. We shall not repeat here the analysis made in Section 5.5. As before, we shall consider only stability problems concerning dead loads. Hence, we suppose always that the perturbation of the external normal surface force is vanishing; i.e. q = 0. (7.2.42) We end this subsection presenting the most frequently encountered boundary conditions which must be satisfied by the incremental fields in problems concerning (local) stability of composite laminates: -free boundary Nnn = Nnτ = Rn + Mnτ,τ = Mnn = 0 on ∂D;

(7.2.43)

-simply supported boundary (prefix S) (S1) U3 = 0, Mnn = 0, Un = Uτ = 0 on ∂D, (S2) U3 = 0, Mnn = 0, Nnn = 0, Uτ = 0 on ∂D,

(S3) U3 = 0, Mnn = 0, Un = 0, Nnτ = 0 on ∂D, (S4) U3 = 0, Mnn = 0, Nnn = Nnτ = 0 on ∂D;

(7.2.44)

-clamped boundary (prefix C) (C1) U3 = 0, U3,n = 0, Un = Uτ = 0 on ∂D, (C2) U3 = 0, U3,n = 0, Nnn = 0, Uτ = 0 on ∂D,

(C3) U3 = 0, U3,n = 0, Un = 0, Nnτ = 0 on ∂D, (C4) U3 = 0, U3,n = 0, Nnn = Nnτ = 0 on ∂D.

Copyright © 2004 by Chapman & Hall/CRC

(7.2.45)

429

7.2. EQUILIBRIUM EQUATIONS. BOUNDARY CONDITIONS

Let us observe that the boundary conditions could be different for each edge of the laminate, so the number of combinations of possible boundary conditions is huge. However, all possible variants have a common characteristic: all incremental boundary conditions and equilibrium equations are homogeneous. Hence, as in the general three-dimensional case, analyzing various stability problems concerning composite laminates, we are led to study and to solve various eigenvalue problems, to obtain the critical loads for which the stability of the stressed laminate is lost. Solving buckling problems for composite laminates is a very difficult task. To o realize it, let as assume nonvanishing normal surface forces q 6= 0, acting in the o initial deformed configuration B . Examining the equilibrium equations (7.1.15)2 , o

we can see that Qα cannot be constant in that case. Consequently, as the relao

tion (7.1.15)3 shows, M αβ cannot be constant quantities. Hence, according to the o global constitutive equations (7.1.19), N αβ will also be varying fields. Accordingly, from the relations (7.1.36), we can conclude that the instantaneous elasticities will also be point dependent quantities. Hence, the incremental equilibrium equations, describing the buckling of the laminate will be partial differential equations with variable coefficients. To solve the eigenvalue and eigenvector problems, characterized by such field equations, numerical methods and powerful computers must be used. To avoid this difficulty, in the following, we suppose that the exterior normal o surface force q is vanishing; i.e. o

q = 0 in D.

(7.2.46) o

Moreover, we suppose that the initial deformed equilibrium state B is homogenous. That can be expressed by assuming that o

o

N αβ = const. and M αβ = const.,

(7.2.47)

or supposing that o

o

eαβ = const. and k αβ = const., o

o

o

(7.2.48)

o

where N αβ , M αβ or eαβ , k αβ are given constant quantities. The constitutive equations (7.1.14) show that (7.2.48) takes place, then conversely (7.2.47) is also true. Also, from the above assumptions we can see that the equilibrium conditions (7.1.15) are satisfied and, moreover, o

Qα = 0.

(7.2.49)

Consequently, the assumed homogenous state is possible in the composite laminate, if condition (7.2.46) is fulfilled.

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CHAPTER 7. STABILITY OF COMPOSITE LAMINATES

As usual, we assume that the stress-free reference configuration BR of the laminate is locally stable. Hence, the corresponding quadratic form o

o

o

o

o

o

o

2 w = eαβ Aαβγϕ eγϕ +2 eαβ Bαβγϕ k γϕ + k αβ Dαβγϕ k γϕ is positive definite. Therefore the global elasticity matrix     A B E = B D

(7.2.50)

(7.2.51)

is positive definite. For a later use, we shall repeat here the incremental field equations involved in the stability problems of composite laminates, in the cases in which the initial o deformed equilibrium configuration B of the laminate is homogenous; i.e. the restrictions (7.2.46) and (7.2.49) are satisfied together with the hypotheses (7.2.47) or equivalently (7.2.48). In this case, the incremental behavior of the laminate is governed by the following equations: - the geometrical relations uα = Uα (x1 , x2 ) − x3 U3,α , α = 1, 2, u3 = U3 (x1 , x2 ) ;

(7.2.52)

- the equilibrium conditions Nαβ,α = 0, Rα,α = 0, Mαβ,α − Pβ = 0;

(7.2.53)

- the constitutive equations Nαβ = Ωαβγϕ Uγ,ϕ + Γαβγϕ kγϕ , Mαβ = Γαβγϕ Uγ,ϕ + Dαβγϕ kγϕ ,

(7.2.54)

o

Rα = Pα + N αβ U3,β ; - the instantaneous elasticities o

o

Ωαβγϕ = Aαβγϕ + N αϕ δβγ , Γαβγδ = Bαβγϕ + M αϕ δβγ .

(7.2.55)

As follows from (7.2.53)2,3 and (7.2.54)3 , the following equation must also be satisfied: o Mαβ,αβ + N αβ U3,αβ = 0. (7.2.56) Now we use the geometrical relations (7.2.52) and the constitutive equations (7.2.54). Thus, from (7.2.53)1 and (7.2.56), we get the differential equations which must be satisfied by the incremental displacement fields: Ωαβγϕ Uγ,ϕα − Γαβγϕ U3,γϕα = 0,

o

Γαβγϕ Uγ,ϕαβ − Dαβγϕ U3,γϕαβ + N αβ U3,αβ = 0.

Copyright © 2004 by Chapman & Hall/CRC

(7.2.57)

7.2. EQUILIBRIUM EQUATIONS. BOUNDARY CONDITIONS

431

Finally, taking into account the expressions (7.2.55) of the instantaneous elasticities, the above equations become o

o

Aαβγϕ Uγ,αϕ − Bαβγϕ U3,αγϕ + N αϕ Uβ,αϕ − M αϕ U3,αβϕ = 0, o

o

Bαβγϕ Uγ,αβϕ − Dαβγϕ U3,αβγϕ + M αϕ U3,αβϕ + N αβ U3,αβ = 0. o

(7.2.58)

o

In these equations, N αβ and M αβ are assumed to be constant, fixed quantities. The most frequent encountered boundary conditions are given by equations (7.2.43)−(7.2.45). In order to study stability problems concerning composite laminates, we must o o find the critical values of the initial applied forces N αβ and moments M αβ for which the homogenous boundary value problem described by the field equations (7.2.58) and by one of the possible boundary conditions (7.2.43)−(7.2.45), has for the first time, nonvanishing solutions on a given loading path. As it is easy to see, the system (7.2.58) is generally coupled. Without further assumptions concerning the nature of the analyzed composite and the initial applied in-plane deformations and curvatures on the initial applied forces and moments, the above system cannot be simplified. Generally, it is assumed that the initial applied resultant moments are vanishing; i.e. o (7.2.59) M αβ = 0. In this case, the system (7.2.58) takes the following simplified form: o

Aαβγϕ Uγ,αϕ − Bαβγδ U3,αγϕ + N αϕ Uβ,αϕ = 0,

(7.2.60)

o

Bαβγϕ Uγ,αβϕ − Dαβγϕ U3,αβγϕ + N αβ U3,αβ = 0.

o

First, we observe that even if the initial applied moments M αβ are vanishing, the differential system describing the behavior of the incremental displacement o o o field rests coupled. Also, if M αβ = 0, but N αβ 6= 0, generally k αβ , that is, the curvature of the initial stressed equilibrium configuration will be nonvanishing. On the o contrary, if we assume that this curvature is vanishing; i.e. k αβ = 0, then the inio

tial applied moments M αβ cannot be vanishing, if for the analyzed laminate there exists bending−extensional coupling; i.e. if not all coupling coefficients B αβγϕ are vanishing. This fact is usually ignored in studies concerning the buckling of general composite laminates. This problem will be later analyzed, assuming for simplicity a very long laminate in the cylindrical state. We shall see that the error made, o neglecting the influence of M αβ in this case, is negligible. However, as we shall see, an initial applied moment can influence the essential value of the compressive force, by which cylindrical buckling occurs in composite laminates. Considerable simplifications occur if we consider composite laminate without bending−extensional coupling; i.e. if we suppose that Bαβγϕ = 0,

Copyright © 2004 by Chapman & Hall/CRC

(7.2.61)

432

CHAPTER 7. STABILITY OF COMPOSITE LAMINATES

and, at the same time, we assume that the initial applied moments are vanishing; i.e. the condition (7.2.59) is satisfied. In this case, the system (7.2.60) becomes o

o

Aαβγϕ Uγ,αϕ + N αϕ Uβ,αϕ = 0, Dαβγϕ U3,αβγϕ − N αβ U3,αβ = 0.

(7.2.62)

Now we can see that if the conditions (7.2.59) and (7.2.61) are satisfied, the system describing the incremental behavior of the composite becomes decoupled. In this case, in order to study the buckling problem of the laminate, we can assume that Uα = 0 for α = 1, 2, (7.2.63) since, as it is easy to see, that with this choice, equations (7.2.62)1 are satisfied o

for any value of the initial applied resultant forces N αβ . o In order to obtain the critical values of the applied forces N αβ for which the buckling of the composite laminate occurs for the first time, we can concentrate our attention on equation (7.2.62), looking for its nonvanishing solutions, corresponding to various homogenous boundary conditions imposed on the boundary of the laminate. In the following Section, we shall analyze this problem for rectangular laminates.

7.3

Buckling of rectangular composite laminates

We assume that the initial applied resultant moments are vanishing; i.e. equation (7.2.59) is satisfied. We also suppose that the bending−extensional coupling does not exist; i.e. the condition (7.2.61) is fulfilled. Moreover, we assume that the incremental in-plane displacements are vanishing; i.e. the restrictions (7.2.63) is o

satisfied. In this case equations (7.2.62)1 are satisfied for any N αβ . Consequently, the buckling behavior of the laminate is described by equation (7.2.62) 2 . If we use the relations connecting Dαβγϕ to the components D11 , ..., D66 of the matrix [D], we can see that this equation has the following form: D11

∂ 4 U3 ∂ 4 U3 ∂ 4 U3 ∂ 4 U3 ∂ 4 U3 + D + 4D + 2 (D + 2D ) + 4D 22 26 12 66 16 ∂x42 ∂x1 ∂x32 ∂x21 ∂x22 ∂x31 ∂x2 ∂x41 o

− N 11

o o ∂ 2 U3 ∂ 2 U3 ∂ 2 U3 = 0. − − 2 N N 22 12 ∂x22 ∂x1 ∂x2 ∂x21

(7.3.1)

In order to obtain this equation, we have used the following relations: D1111 = D11 , D1122 = D2211 = D12 = D21 , D2222 = D22 , D1112 = D1121 = D1211 = D2111 = D16 = D61 , D2212 = D2221 = D1222 = D2122 = D26 = D62 , D1212 = D2112 = D1221 = D2122 = D66 .

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(7.3.2)

433

7.3. BUCKLING OF RECTANGULAR COMPOSITE LAMINATES

Taking into account these relations, for the components of the incremental resultant moments, we obtain the following expressions:

∂ 2 U3 ∂ 2 U3 ∂ 2 U3 , − 2D − D 16 12 ∂x1 ∂x2 ∂x22 ∂x21 ∂ 2 U3 ∂ 2 U3 ∂ 2 U3 , − 2D − D = M21 = −D16 66 26 ∂x1 ∂x2 ∂x22 ∂x21 ∂ 2 U3 ∂ 2 U3 ∂ 2 U3 . − 2D − D = −D12 26 22 ∂x1 ∂x2 ∂x22 ∂x21

M11 = −D11

M12

M22

(7.3.3)

Now let us consider a composite laminate with the sides a and b simply supported o o along the edges and compressed by uniform resultant forces N 11 and N 12 as shown in Figure 7.1. X2

O X3

-N22

-N11 X1

Figure 7.1: Rectangular composite laminate under biaxial compression. o

o

We would like to find the critical values of the loads N 11 and N 22 such that the initially flat form of the laminate ceases to be a unique equilibrium solution, that is, the values of the loads for which the laminate is buckled. The incremental equation of equilibrium, in this case, is given by the relation (7.3.1). Also, the deflection U3 must satisfy the following boundary conditions corresponding to simply-supported composite plate: U3 = 0, − M11 = D11

U3 = 0, − M22 = D12

∂ 2 U3 ∂ 2 U3 ∂ 2 U3 = 0 for x1 = 0, a, + 2D + D 16 12 ∂x1 ∂x2 ∂x22 ∂x21

∂ 2 U3 ∂ 2 U3 ∂ 2 U3 = 0 for x2 = 0, b. (7.3.4) + D22 + 2D26 2 ∂x22 ∂x1 ∂x2 ∂x1

We seek a nonzero solution of the system (7.3.1) and (7.3.4). Even in this case, after making all simplifying assumptions, the problem remains difficult. That

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CHAPTER 7. STABILITY OF COMPOSITE LAMINATES

is the reason for which we shall first introduce a new hypothesis supposing that stress-twist curvatures coupling does not exist; i.e. we assume that D16 = D26 = 0. In this case, the incremental equilibrium equation (7.3.1) becomes o o ∂ 2 U3 ∂ 2 U3 ∂ 4 U3 ∂ 4 U3 ∂ 4 U3 = 0, − N 22 − N 11 + 2(D12 + 2D66 ) 2 2 + D22 2 2 4 ∂x22 ∂x1 ∂x2 ∂x1 ∂x2 ∂x1 (7.3.5) since we have assumed that o o (7.3.6) N 12 = N 21 = 0.

D11

Also, the boundary conditions (7.3.4) take the following simplified form: U3 = 0, D11

∂ 2 U3 ∂ 2 U3 = 0 for x1 = 0, a, + D 12 ∂x22 ∂x21

U3 = 0, D12

∂ 2 U3 ∂ 2 U3 = 0 for x2 = 0, b. + D 22 ∂x22 ∂x21

(7.3.7)

Looking for various buckled forms of the biaxially compressed composite laminate, we observe that the function U = Amn sin

mπx2 mπx1 , sin b a

(7.3.8)

where Amn are arbitrary constants, and m = 0,1,2,..., n = 0,1,2,... are positive integers, satisfies the homogeneous boundary conditions (7.3.7). Substituting (7.3.8) in (7.3.5), we obtain    n 4   mn 2  m 4 2 + D22 + 2 (D12 + 2D66 ) π D11 b ab a o

+ N 11

 m 2

a

o

+ N 22

 n 2 

b

Amn = 0

(7.3.9)

Consequently, the characteristic equation corresponding to our buckling problem has the following form: π 2 [D11 (

o o n m n mn 2 m 4 ) + D22 ( )4 ]+ N 11 ( )2 + N 22 ( )2 = 0, ) + 2(D12 + 2D66 )( b a b ab a (7.3.10)

o

o

To find the values N 11 and N 22 satisfying this equation, we introduce the ratio o

α=

N 22 o

N 11

Copyright © 2004 by Chapman & Hall/CRC

.

(7.3.11)

435

7.3. BUCKLING OF RECTANGULAR COMPOSITE LAMINATES Thus, from equation (7.3.10), we get o

2 N 11 = −π

n 4 a 2 n 2 2 ) (m) D11 ( m a ) + 2(D12 + 2D66 )( b ) + D22 ( b , m, n = 1, 2, .... a n 2 1 + α( b m ) (7.3.12)

o

o

This equation gives the critical values of the resultant forces N 11 and N 22 for a given ratio α, for all possible combinations of m and n. We recall that equation (7.3.12) was obtained from the relation (7.3.9) such that more than one (the trivial) solution exists. Therefore, each combination of m and n is a buckling load and buckling mode such that more than one solution for U 3 exists. As in any eigen problem, infinite solutions for U 3 exist, for each combination of m and n, since the constants Amn remain undetermined. From all the possible values of m and n, the critical buckling load; i.e. the primary eigenstate, can be determined by selecting that particular combination o of m and n that gives the smallest absolute value of N 11 . Let us analyze several o cases. As a first case, let us assume that N 22 = 0; i.e. α = 0.

(7.3.13)

On this loading path, the plate is compressed only in the x1 direction and equation (7.3.12) becomes o

2 N 11 = −π [D11 (

a n n m 2 ) + 2(D12 + 2D66 )( )2 + D22 ( )4 ( )2 ]. m b b a

(7.3.14) o

Since the value in the above bracket increases as n increases, the smaller N 11 occurs always when n = 1. Hence, the buckling load is given by the smallest value (in module) of the following expression when m is allowed to vary o

2 N 11 = −π [D11 (

1 a 1 m 2 ) + 2(D12 + 2D66 ) 2 + D22 4 ( )2 ]. b m b a

(7.3.15)

For given values of D11 , D12 , D22 , D66 , a and b, we must determine the value o of m which yields the smallest absolute value of N 11 . This value is the critical buckling load. For example (see Ashton and Whitney[7.1]), we consider a fiber-reinforced laminate in the x1 direction with D11 /D22 = 10,

(D12 + 2D66 )/D22 = 1.

The compressive force acts in the fibers direction. For the above values, equation (7.3.15) becomes o

2 N 11 = −π D22 [10(

Copyright © 2004 by Chapman & Hall/CRC

a 1 2 m 2 ) + 2 + ( )2 4 ]. m b b a

(7.3.16)

436

CHAPTER 7. STABILITY OF COMPOSITE LAMINATES o

The absolute value of N 11 for the aspect ratio a K = ≤ 1.78 b

(7.3.17)

has a minimum when m = 1. In particular, for a square plate (K = 1), the critical buckling load in o

N c 11 = −13π 2

D22 . a2

Other plate dimensions can also be considered when analyzing equation (7.3.17). o

c are plotted vs. the aspect ratio K = In Figure 7.2 the values of N11 Ashton and Whitney [7.1]).

a b

(see

24 22

-

O N 11 b2 2 D22

20 18 16

a=1

14 12 10

a=2 8 6 4 2

0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

a/b

o

Figure 7.2: Variation of the critical load N c 11 with aspect ratio K = ab .

Note that at certain values of K = ab , two possible buckled shapes are possible. That is when r p D11 a p (7.3.18) = 1.78 m(m + 1), K = = m(m + 1) 4 D22 b

the two buckled shapes

U3 = Am1 sin

and U3 = Am+1,1 sin o

give identical values for N c 11 .

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πx2 mπx1 sin b a

πx2 (m + 1)πx1 sin b a

(7.3.19)

(7.3.20)

437

7.3. BUCKLING OF RECTANGULAR COMPOSITE LAMINATES

Also, the absolute minimum buckling load occurs for m = m such that r D11 a 4 (7.3.21) = 1.78m, =m D22 b

and, in this case, is given by o

N c 11 = −8.32π 2 D22 /b2 .

(7.3.22)

These results reveal the complexity of stability analysis, when more parameters are involved, and only one of them can have positive integer values. Let us analyze now the case in which α = 1 and a = b.

(7.3.23)

This is the case of a square laminate under biaxial compression. Equation (7.3.22) now becomes 4

o

o

N 11 = N 22 = −

n π 2 D11 m2 + 2(D12 + 2D66 )n2 + D22 m 2 . n 2 ) 1 + (m a2

(7.3.24)

If this equation is examined for the particular values of m and n, we can o o conclude that the smallest absolute value of N 11 = N 22 will occur with m = 1 as long as D11 > D22 . In this case, o

o

N 11 = N 22 = −

D11 /D22 + [2(D12 + 2D66 )/D22 ]n2 + n4 π2 . D 22 1 + n2 a2

(7.3.25)

For the numerical values of the parameters D11 , D12 , D22 and D66 , considered o o in the first case, the critical value of N 11 = N 22 is obtained for n = 1: o

o

N c 11 = N c 22 = −6.5

π2 D22 , a2

(7.3.26)

but if D11 /D22 = 15,

(D12 + 2D66 )/D22 = 1,

the critical value corresponds to n = 2 and we have o

o

N c 11 = N c 22 = −7.8

π2 D22 . a2

(7.3.27)

Finally, we assume that o

o

α = N 22 / N 11 < 0. o

o

(7.3.28) o

In this case, N 11 or N 22 is a tensile load. We suppose that N 22 is tensile; i.e. o

N 22 > 0.

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(7.3.29)

438

CHAPTER 7. STABILITY OF COMPOSITE LAMINATES

Since α is negative, from (7.3.12) we can conclude that the critical compressive o

load will be greater than that for the case N 22 = 0. For the particular numerical example, considered in the first case and with a = b and α = −0.5 (tension equal one-half the compression), we find o

N c 11 = −26

π2 D22 . a2

(7.3.30) o

That is, the critical compressive load is twice that obtained for N 22 = 0. Following also Ashton and Whitney [71], let us analyze now the buckling of a rectangular laminate simply supported along the edges x1 = 0, a and uniformly compressed in the x1 direction as shown in Figure 7.3. X1

O O

N11

O X3

-N11

X2

Figure 7.3: Rectangular composite laminate under uniaxial compression. Until now, the edges x2 = 0, b have not fixed boundary conditions yet. Assuming o (7.3.31) N 22 = 0, we seek the solution of equation (7.3.5) in the form U3 = Y (x2 ) sin

mπx1 , m = 1, 2, ..., a

(7.3.32)

which satisfies the imposed boundary condition on the edges x1 = 0, a; i.e. we have U3 = 0, M11 = −D11

∂ 2 U3 ∂ 2 U3 = 0 for x1 = 0, a. − D12 2 ∂x22 ∂x1

(7.3.33)

Introducing (7.3.32) in (7.3.5), we get the equation which must be satisfied by the function Y (x2 ): D22 Y

0000

− 2(

o 00 mπ 2 mπ 4 mπ 2 ) Y = 0, (7.3.34) ) Y + N 11 ( ) (D12 + 2D66 )Y + D11 ( a a a

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7.3. BUCKLING OF RECTANGULAR COMPOSITE LAMINATES

439

where the prime denotes differentiation with respect to x2 . The general solution of this equation is Y2 (x2 ) = A1 e−sx2 + A2 esx2 + A3 cos rx2 + A4 sin rx2,

(7.3.35)

where A1 , A2 , A3, A4 are arbitrary constants, and

and

vs u o u D12 + 2D66 2 D11 mπ t N 11 a 2 D12 + 2D66 , ) + ( − ) − ( s= D22 D22 mπ D22 D22 a

(7.3.36)

vs u o u D12 + 2D66 2 D11 mπ t N 11 a 2 D12 + 2D66 . ) − ( − ) − ( r= D22 D22 mπ D22 D22 a

(7.3.37)

We suppose that s and r are positive numbers. The constants A1 , A2 , A3 and A4 must be determined from the four conditions on the edges x2 = 0, b. Thus, we are led to four homogeneous linear algebraic equations for the unknown constants. The critical buckling load is determined by setting the determinant of this system at zero and determining the lowest possible o absolute value of N 11 for the given boundary conditions. Ashton and Whitney [7.1] illustrate the method by assuming that the edges x2 = 0, b are clamped. Then the boundary conditions are U3 = 0 and

∂U3 = 0 for x2 = 0, b. ∂x2

(7.3.38)

Using (7.3.35), we get the conditions which must be satisfied by A1 , A2 , A3 and A4 : A1 + A2 + A3 = 0, −sA1 + sA2 + rA4 = 0, e−sb A1 + esb A2 + cos rbA3 + sin rbA4 = 0, −se−sb A1 + sesb A2 − r sin rbA3 + r cos rbA4 = 0. In order to obtain buckling, i.e. U3 6= 0, the determinant of this system must vanish. Imposing this condition, we get the following characteristic equation r s 2(1 − cos rbcoshsb) = ( − ) sin rbsinhsb. s r o

(7.3.39)

The critical value of N 11 can be determined from this equation since s and r o involve N 11 .

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440

CHAPTER 7. STABILITY OF COMPOSITE LAMINATES Consider a square laminate; i.e. a = b, with D11 /D22 = 10 and (D12 + 2D66 )/D22 = 1.67. Then mπ s= a

sr

mπ P −7.22 + 2 + 1.67, r = a m

where

sr

−7.22 +

P − 1.67, m2

(7.3.40)

o

P =−

N 11 a2 . πD22

(7.3.41)

For this case, equation (7.3.39) is sr sr P P −7.22 + 2 + 1.67) −7.22 + 2 − 1.67)cosh(mπ 2 − 2 cos(mπ m m  v q v q u   u P u −7.22 + P2 − 1.67 u   −7.22 + m2 + 1.67  u u m − u rq = tq t     −7.22 + mP2 + 1.67  −7.22 + P2 − 1.67  m

· sin(mπ

sr

−7.22 +

P − 1.67)sinh(mπ m2

sr

−7.22 +

P + 1.67). m2

The minimum value of P occurs when m = 1 and may be found by interactively solving the above equation. The critical buckling load is found to be o

N c 11 = −19.1π 2

D22 . a2

(7.3.42)

Other boundary condition on the edges x2 = 0, b can be handled in similar manner. The previous examples concern the buckling behavior of specially orthotopic laminates for which D16 = D26 = 0. Such laminates are special cases of more general composite laminates, and occurs for plates laminated of laminae that are orthotopic or isotropic, and which are laminated in such a way that the principal material axes are parallel to the laminate axes. We shall assume now that D16 6= 0 are D26 6= 0.

(7.3.43)

These and the terms which couple twisting curvatures to the normal moment resultants will be included in the analysis. However, we assume that the membranebending coupling terms are absent; i.e. B11 = ..... = B66 = 0. Laminate plates for which the membrane-bending coupling does not exist, but which exhibit nonzero

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7.3. BUCKLING OF RECTANGULAR COMPOSITE LAMINATES

441

D16 or D26 terms are an important class of plates. As we know, such plates occur when the laminate possesses midplane symmetry; i.e. when there exists a lamina above the geometrical midplane at the same distance from the midplane and having identical orientation and properties with a lamina below the midplane. For such a symmetrical stacking sequence, the membrane-bending coupling terms are zero. The occurrence of the normal stress-twist coupling terms D16 , D26 in the governing equation increases the complexity of the analysis by a significant manner. One approach, which is sometimes taken in the analyzes of such laminates, is simply to neglect these coupling terms. Following Ashton and Whitney (7.1), we present some solutions to indicate the nature of the error that is introduced when this is done. The governing equation is now the relation (7.3.1). As before, we assume that the rectangular laminate is simply supported. Hence, on the edges of the laminate, the boundary condition (7.3.4) must be satisfied. To obtain the answer concerning the buckling problem for a specially orthotopic laminate, for which D16 = D26 = 0, we have assumed that the incremental normal displacement U3 is given by the equation (7.3.8). Now this form for U3 does not satisfy the boundary conditions (7.3.4)2 and (7.3.4)4 , due to the presence of the coupling terms. Moreover, for the same reason, the field equation (7.3.1) cannot be satisfied if U3 has the simple form given by the relation (7.3.8). The mentioned difficulties can be avoided if we look for an approximate solution of the buckling problem, using, to this end, the corresponding incremental variational or extremal principle and taking into account appropriate test functions, satisfying the imposed boundary conditions for the incremental normal displacement field U3 . We recall, and this is essential if we use variational or extremal principles, that the boundary conditions concerning tractions must not be exactly satisfied! Taking into account the above observations, we choose for U3 the following expression: n m X X jπx2 iπx1 (7.3.44) sin U3 = Aij sin b a i=1 j=1

which satisfies the imposed displacement boundary conditions (7.3.4) 1 and (7.3.4)2 for the normal displacement U3 , and hence, represents an admissible displacement field in our problem. We return now to the exclusion functional E = E(U) defined by equations (7.2.23) and (7.2.41). Since (7.2.61) and (7.2.63) take place, from (7.2.23) and (7.2.41), we can conclude that the appropriate exclusion functional has the following form: E = E(U3 )

=

Z

a 0

Z

b 0

2 2 2 + 4D66 U3,12 + 2D12 U3,11 U3,22 + D22 U3,22 (D11 U3,11 o

+ 4D16 U3,11 U3,12 + 4D26 U3,22 U3,12 + N 11 U3,1 o

o

2 + 2 N 12 U3,1 U3,2 )dx1 dx2 . + N 22 U3,2

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(7.3.45)

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CHAPTER 7. STABILITY OF COMPOSITE LAMINATES

Introducing (7.3.44) in (7.3.45), the functional E = E(U3 ) becomes a function E = E(Aij ) defined on the m × n dimension space of the selected test functions. To obtain approximate solution for one buckling problem, we must impose the stability condition for this function; i.e., we must impose the conditions ∂E = 0 for i = 1, ...., m, j = 1, ...., n. ∂Aij

(7.3.46)

Introducing (7.3.44) in (7.3.45) and imposing the above conditions, we are lead to the following linear and homogeneous algebraic system that must be satisfied by the unknown constants Aij : aklij Aij = 0; k, i = 1, ..., m; l, j = 1, ..., n,

(7.3.47)

aklij = bijkl + cijkl + dijkl

(7.3.48)

where with Z

a

Z

b

ijπ 2 2 iπ 4 ) ) + 2(D12 + 2D66 )( ab a 0 0 lπx2 kπx1 jπx2 iπx1 jπ dx1 dx2 , sin sin sin +D22 ( )4 ]sin b a b a b Z aZ b iπ jπ cijkl = − [4D16 ( )3 b a 0 0 lπx2 kπx1 jπx2 iπx1 iπ jπ 3 dx1 dx2 sin sin cos +4D26 ( ) ]cos b a b a b a Z a kπx1 iπx1 iπ jπ lπ dx1 sin cos − 4D26 a a b b a 0 Z b lπx2 jπx2 iπ jπ kπ sin cos dx2 , − 4D16 b b a b a 0 Z aZ b o iπ iπx1 jπx2 kπx1 lπx2 dijkl =N 11 ( )2 sin sin sin sin dx1 dx2. a a b a b 0 0

bijkl =

[D11 (

The relations (7.3.47) provide m×n homogeneous simultaneous equations for the unknown coefficients Aij . To obtain buckling, the determinant of the coefficient matrix must vanish. The results are presented in Figure 7.4 (see Ashton and Whitney [7.1]) for laminates with properties typical for boron-epoxy composite materials: E22 /E11 = 0.1,

E12 /E11 = 0.03,

ν12 = 0.3.

The aspect ratio K = a/b considered is K = 1.13.

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443

7.3. BUCKLING OF RECTANGULAR COMPOSITE LAMINATES

The results are presented in terms of the buckling coefficient P , defined as o

N 11 a2 . P =− E11 h3

2.4

(7.3.49)

Orthotropic solution

2.2

20 alternate plies at + -

2.0 1.8 1.6 1.4

P

1.2

all +

1.0 0.8

Experimental points 20 Plies + 20 Plies +

0.6 0.4 0.2

0 0

10

20

30

40

50

60

70

80

90

Orientation,

Figure 7.4: Compressive buckling coefficient. Simply supposed plies. The bottom curve in Figure 7.4. represents the results calculated with m = n = 7 in equation (7.3.47) for the case of unidirectional laminate with the material axes oriented at θ degrees to the plate edges. The curves correspond to the orthotopic solution obtained by neglecting the D16 and D26 terms, respectively, the orthotopic solution obtained for laminates made of 10 plies at alternately plus and minus θ degrees followed by 10 plies laminated at minus plus θ degrees. The D16 and D26 terms are smaller for the case of alternating plies and, as evidenced in Figure 7.4, such “interspaced” 20-layer plates behave essentially in a specially orthotropic manner. Also shown in Figure 7.4 are experimentally determined buckling loads obtained on boron-epoxy plates. These experimental results follow the analytical predictions quite acceptably. This fact indicates that the variational approach allows accurate prediction of the buckling load of composite laminates. Furthermore, examining Figure 7.4 we can see that the use of specially orthotropic analyzes, assuming D16 = D26 = 0, can lead to appreciable inaccuracies. Various examples, stressing the last two observations, are presented in the monograph [7.1] by Ashton and Whitney.

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444

CHAPTER 7. STABILITY OF COMPOSITE LAMINATES

7.4

Buckling of antisymmetric cross-ply laminates

In what follows, we assume a very long rectangular laminate (a composite strip), occupying the spatial domain in its stress-free reference configuration: −a ≤ x1 ≤ a,

−b ≤ x2 ≤ b,

−h/2 ≤ x3 ≤ h/2.

(7.4.1)

Here a is the breadth of the laminate, b is its length and h is the height of the strip, and we suppose that h << a and a << b.

(7.4.2)

Also, in what follows, we assume that the global constitutive coefficients A16 , B16 , and D16 are vanishing; i.e. A16 = B16 = D16 = 0.

(7.4.3)

This restriction is fulfilled by many types of laminates. For instance, (7.4.3) is satisfied by a single specially orthotopic or by isotropic layers, by symmetric laminates with isotropic or specially orthotopic layers, by regular symmetric cross-ply laminates, by nonsymmetric laminates with multiple isotropic or specially orthotopic layers and by antisymmetric cross-ply laminates. o Also, we suppose that the initial applied transverse load q is vanishing o

q = 0.

(7.4.4) o

We assume that the initial deformed equilibrium configuration B of the laminate is a homogeneous cylindrical state relative to the x2 direction; more exactly, we suppose that o o o o o (7.4.5) U 1 = U 1 (x1 ), U 2 = 0, U 3 = U 3 (x1 ), and o

o

o

o

o

o

e11 = const. , k 11 = const. , e22 = e12 = 0, k 22 = k 12 = 0.

(7.4.6)

Now taking into account the global constitutive equations (7.3.14), we get o

o

o

N 11 = A11 e11 +B11 k 11 , o

o

o

M 11 = B11 e11 +D11 k 11 ,

o

o

o

o

o

o

N 22 = A12 e11 +B12 k 11 , M 22 = B12 e11 +D12 k 11 , o

o

N 12 = 0, o

M 12 = 0.

(7.4.7) (7.4.8)

o

According to the assumption (7.4.6), N 11 ...., M 12 are constant quantities. Consequently, from (7.1.15)2 , we get o

o

Q1 =Q2 = 0.

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(7.4.9)

7.4. BUCKLING OF ANTISYMMETRIC CROSS-PLY LAMINATES

445

Thus, taking into account the hypotheses (7.4.4), we can see that the equilibrium conditions (7.1.15) are satisfied in the initial deformed homogeneous cono

figuration B . Hence, assuming cylindrical state (7.4.5), (7.4.6) is possible in the laminate. For simplicity, we shall use the following notations: A = A11 , B = B11 , D = D11 , o

o

e = e11 , o

o

N = N 11 ,

o

o

k = k 11 , o

(7.4.10) (7.4.11)

o

M = M 11 .

(7.4.12)

Now the global constitutive equations(7.4.7)1 and (7.4.8)1 become o

o

o

o

o

o

N = A e +B k , M = B e +D k .

(7.4.13) o

From equations (7.2.23) and (7.4.13) it follows that the quadratic form w, corresponding to the homogeneous and cylindrical initial deformation of the laminate, has the following simple form: o

w=

D o2 A o2 oo e +B ek + k . 2 2

(7.4.14)

We assume that the stress-free reference configuration of the laminate is o locally stable. Hence, w is a positive definite quadratic form. Accordingly, the constitutive constants A, B and D satisfy the following restrictions: A > 0, D > 0, AD − B 2 > 0.

(7.4.15)

We suppose that the incremental normal load q is vanishing q = 0.

(7.4.16)

Concerning the incremental state of the laminate, we assume that it is also a cylindrical one, relative to the x2 direction. That is, the incremental displacement field, which satisfies the following restrictions: U1 = U1 (x1 ), U2 = 0, U3 = U3 (x1 ).

(7.4.17)

Consequently, we have e11 = e11 (x1 ), k11 = k11 (x1 ), e22 = e12 = 0, k22 = k12 = 0.

(7.4.18)

Now, taking into account the notations (7.4.3), (7.4.7), (7.4.9) and the relations (7.4.10), we can see that the incremental global constitutive equations (7.1.35) become o

o

o

N 11 = (A+ N )e11 + (B+ M )k11 , N22 = A12 e11 + B12 k11 , N12 = N21 = 0, (7.4.19)

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CHAPTER 7. STABILITY OF COMPOSITE LAMINATES

and o

M11 = (B+ M )e11 + Dk11 , M22 = B12 e11 + D12 k11 , M12 = M21 = 0. (7.4.20) get

Obviously, N11 , .....M21 depend only on x1 . Consequently, from (7.2.4), we P1 = M11,1 and P2 = 0.

(7.4.21)

Now, taking into account (7.4.7), (7.4.9), (7.4.17) and the notations (7.4.12), from the incremental relations (7.1.38), we obtain o

R1 = M11,1 + N U3,1 , R2 = 0.

(7.4.22)

We also observe that the second incremental equilibrium equation (7.2.4) 1 is identically satisfied, and the first one becomes N11,1 = 0.

(7.4.23)

Analogously, according to (7.4.16) and (7.4.22), the incremental equilibrium condition (7.2.4)2 takes the form R1,1 = 0.

(7.4.24)

For simplicity we shall use the notations e = e11 = U1,1 , k = k11 = −U3,11 ,

(7.4.25)

N = N11 , M = M11 , R = R1 .

(7.4.26)

Thus, equations (7.4.19)1 , (7.4.20)1 and (7.4.22)1 take the form o

o

o

o

N = (A+ N )e + (B+ M )k, M = (B+ M )e + Dk, R = M1 + N U3,1 . (7.4.27) The equilibrium equations (7.4.23) and (7.4.24) become N,1 = 0, R,1 = 0.

(7.4.28)

We note also that in the assumed incremental cylindrical state, the quadratic form (7.2.23) takes the following simplified form: o

w=

o

o D A+ N 2 N 2 . e + (B+ M )ek + k 2 + U3.1 2 2 2

(7.4.29)

As we already know, if w is positive definite, the initial deformed homogeo neous cylindrical equilibrium state B is locally stable.

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7.4. BUCKLING OF ANTISYMMETRIC CROSS-PLY LAMINATES

447

According to (7.4.29), w has the above property if and only if the constitutive o

o

coefficients A, B, D and the initial applied homogeneous loads N , M satisfy the following inequalities: o

o

o

o

o

A+ N > 0, D > 0, (A+ N )D − (B + M )2 > 0, N > 0. o

(7.4.30)

Instability, that is, bifurcation (buckling) of the initial deformed configuration

B can occur only when one of these inequalities is violated. o In what follows, we shall assume that the initial applied external load N o can violate the last restriction (7.4.30). In other words, we suppose that N is a compressive force; i.e. o (7.4.31) − N = P > 0. At the same time, we assume that the restrictions (7.4.30)1 and (7.4.30)3 are o

satisfied by the initial applied compressive force P and bending moment M ; i.e. o

A − P > 0, (A − P )D − (B + M )2 > 0 with M = M .

(7.4.32)

Obviously, the second condition (7.4.32) can be satisfied only if the initial o

applied bending moment M = M satisfies the restriction AD − (B + M )2 > 0.

(7.4.33)

o

In what follows, we suppose that M = M satisfies this inequality. Now we return to the incremental equilibrium conditions (7.4.28). According to these relations, we must have N = C1 , R = C2, for − a ≤ x1 ≤ a,

(7.4.34)

C1 and C2 being arbitrary real constants. Taking into account equations (7.4.27)1 , (7.4.34)1 and the simplified notations, we get (A − P )e + (B + M )k = C1 . (7.4.35) Hence, we have

C1 B+M . k+ A−P A−P Introducing this relation into (7.4.27)2 , we get e=−

M=

B+M D(A − P ) − (B + M )2 C1 . k+ A−P A−P

(7.4.36)

(7.4.37)

Also, equations (7.4.27)3 , (7.4.31) and (7.4.32)2 give M,1 − P U3,1 = C2 .

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(7.4.38)

448

CHAPTER 7. STABILITY OF COMPOSITE LAMINATES Hence, we have M − P U3 = C2 x + C3 with x = x1 ,

(7.4.39)

C3 being an arbitrary real constant. Now, taking into account the relations (7.4.25)3,4 , (7.4.37) and (7.4.39), we get the differential equation, which must be satisfied by the incremental normal displacement field U3 −

B+M D(A − P ) − (B + M )2 C1 . U3,11 − P U3 = C2 x − C3 − A−P A−P

(7.4.40)

It is easy to see that this equation can be written in the following equivalent form: U3,11 + Ω2 U3 = −

Ω2 B + M Ω2 C1 − C3 ). ( C3 x + P A−P P

(7.4.41)

In this equation, Ω is defined by the following relation: Ω2 =

(A − P )P . (A − P )D − (B + M )2

(7.4.42)

Since we have assumed that P and M satisfy the inequalities (7.4.32) and (7.4.33), Ω is a real number. For simplicity, we assume that Ω > 0.

(7.4.43)

The general solution of the equation (7.4.41) is U3 = K sin Ωx + L cos Ωx −

1 B+M C2 C1 − C3 ), x+ ( P A−P P

(7.4.44)

where K and L are arbitrary real constants. According to (7.4.44), we have k = −U3,11 = Ω2 (K sin Ωx + L cos Ωx).

(7.4.45)

Now, by using (7.4.36), we can find e and the incremental in-plane displacement U1 . We get U1 =

C1 B+M x + C4 , Ω(K cos Ωx − L sin Ωx) + A−P A−P

(7.4.46)

where C4 is an arbitrary real constant. Thus, from (7.4.37) and (7.4.42), we obtain M = P (K sin Ωx + L cos Ωx) +

Copyright © 2004 by Chapman & Hall/CRC

B+M C1 . A−P

(7.4.47)

7.4. BUCKLING OF ANTISYMMETRIC CROSS-PLY LAMINATES

449

We recall that according to equations (7.4.34), the incremental fields N and R have constant values. The obtained results can be used to get the critical values of the compressive force P for which bifurcation (buckling) of cylindrical type of the initial deformed o

homogeneous equilibrium state B can occur. At the same time, we can study the influence of the initial applied bending moment M on the critical values of the compressive force P . This influence is usually ignored in analyzes concerning stability of composite laminates. However, as we shall see, there are critical values of the bending moment for which dangerous cases can occur. Also, there exist situations in which an initial applied bending moment can improve the performances of a compressed composite laminate. As a first example, we shall analyze the case of free boundary, assuming that the incremental external loads, acting on the boundaries x = ±a of the laminate, are vanishing. According to the general relations (7.2.43), we shall have the following boundary conditions in the considered case: N = R = M = 0 for x = ±a.

(7.4.48)

In this case, from (7.4.33), we get C1 = C2 = 0.

(7.4.49)

Consequently, from (7.4.41) and (7.4.46), we obtain U3 = K sin Ωx + L cos Ωx −

1 C3 , P

B+M Ω (K cos Ωx − L sin Ωx) + C4 . (7.4.50) A−P Obviously, C3 and C4 describe a rigid translation of the laminate, which will be neglected; i.e. we take C3 = C4 = 0. U1 =

Hence, for the incremental displacements, we get U3 = K sin Ωx + L cos Ωx, U1 =

Also (7.4.47) gives

B+M Ω(K cos Ωx + L sin Ωx). A−P

M = P(KsinΩx + L cos Ωx).

(7.4.51) (7.4.52)

Now the third boundary condition tells us that the following restriction must be satisfied: ±K sin Ωa + L cos Ωa = 0. (7.4.53)

Hence, we can have nonvanishing incremental solutions, that is, eigenstates and o bifurcation (bending) of the prestressed equilibrium configuration B , if and only if one of the following conditions is satisfied: cos Ωa = 0 and K = 0

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(7.4.54)

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CHAPTER 7. STABILITY OF COMPOSITE LAMINATES

or sin Ωa = 0 and L = 0.

(7.4.55)

The first condition will be satisfied if 1 Ωa = ( + k)π, k = 0, 1, 2, ..., 2

(7.4.56)

and the second condition will be fulfilled if Ωa = kπ, k = 1, 2, ...

(7.4.57)

In order to detect the occurrence of the primary eigenstate, we must determine the smallest possible positive critical value of the initial applied compressive force P . To do this we first observe that according to equation (7.4.42), the possible critical values of P must satisfy the following second order algebraic equation: h i 2 P 2 − (A + Ω2 D)P + Ω2 AD − (B + M ) = 0. (7.4.58) o

Taking into account the restriction (7.4.33) satisfied by M =M , it is easy to see that this equation has two real positive roots. The root which satisfies for any real number Ω the restriction (7.4.32)1 is given by the equation 2P = F (Ω, M ) ≡ A + Ω2 D −

q

2

(A − Ω2 D) + 4Ω2 (B + M )2 .

(7.4.59)

Taking into account again the restriction (7.4.33), after some elementary computations, we can see that P given by equation (7.4.59) satisfies also the restriction (7.4.32)2 . Also, taking into account the condition (7.4.33), it is easy to see that the function F = F (Ω, M ) increases when Ω increases. Consequently, the smallest possible critical value Pc of P , for which local instability or bifurcation (buckling) can occur, corresponds to the smallest possible critical value Ωc of Ω. Inspecting the conditions (7.4.54) and (7.4.55), we can see that this critical value is given by Ωc =

π . 2a

(7.4.60)

Hence, according to equation (7.4.59), the critical value Pc of the applied compressive force is given by equation. r π2 D 2 π2 π2 D ) + 2 (B + M )2 . (7.4.61) − (A − 2Pc = F (Ωc , M ) ≡ G(M ) = A + 2 a 4a2 4a

We determine now the corresponding initial and incremental displacement fields neglecting infinitesimal rigid translations and rotations.

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451

After elementary computations, taking into account (7.4.13) and the notations (7.4.31), (7.4.36), we get o

o

U 1 = −∆(Pc D + M B)x, U 3 = −∆(Pc B + M A)

with

x2 , 2

(7.4.62)

1 . AD − B 2 Similarly, from (7.4.51), (7.4.54) and (7.4.60), we obtain ∆=

U1 = −

πx πx π B+M . ; U3 = L cos sin L a 2a 2a A − Pc

(7.4.63)

The last two equations describe the way in which the buckling takes place when the applied compressive force reaches its critical value Pc in the presence of the bending-extensional coupling (B 6= 0) and of an initial applied bending moment (M 6= 0). Now we shall study the influence of the initial applied bending moment M on the critical value Pc . Using elementary computations from (7.4.61), it follows that the function G = G(M ), that is, the critical value Pc reaches its maximal value P c when the applied bending moment M has the value M given by equation

M = −B.

(7.4.64)

According to (7.4.61) and (7.4.64), the corresponding maximum value is given by the equation: π2 D ). (7.4.65) P c = min(A, 4a2 From (7.4.63), we get the corresponding incremental displacement fields

U 1 = 0 and U 3 = L cos

πx . 2a

(7.4.66)

The obtained result shows that in the presence of the bending-extensional coupling, an initial applied bending moment M , having values in a neighborhood of the optimal value M , can improve the buckling behavior of the laminate. Also, according to (7.4.61), the function g = g(M ), hence, the critical value Pc , converge to zero, if the initial applied bending moment M converges to some well defined critical values Mc1 or Mc2 . It is easy to see that we have √ √ (7.4.67) Pc = 0 if M = Mc1 = −B − AD or if M = Mc2 = −B + AD.

As it follows from (7.4.63), the corresponding incremental displacement fields are given by equations r πx πx π D if M = Mc1 , (7.4.68) ; U3 = L cos sin U1 = L 2a 2a 2a A

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CHAPTER 7. STABILITY OF COMPOSITE LAMINATES

and π U2 = −L 2a

r

πx πx D if M = Mc2 . ; U3 = L cos sin 2a 2a A

(7.4.69)

The above results show that an initial applied bending moment M , having values in a neighborhood of the critical values Mc1 or Mc2 , can lead to very small values of the critical compressive force and dangerous situations can occur. A little later on, we shall analyze in detail this problem for a particular laminate, characterized by bending-extensional coupling. We remark that in stability analysis of composite laminates, usually it is o assumed that the prestressed equilibrium configuration B has a plane shape; that is, if it is supposed that o

k = 0.

(7.4.70)

For the constitutive equation (7.4.13), it results that, in order to satisfy the restriction (7.4.70), it is necessary to apply a bending moment o

o

M =M = B e .

(7.4.71)

Consequently, the relation (7.4.59), giving the possible critical values of the compressive force P , takes the form q o 2 (7.4.72) 2P = A + Ω D − (A − Ω2 D)2 + 4Ω2 B 2 (1+ e)2 .

But according to the assumption made concerning the initial deformation of the laminate, we have o (7.4.73) e << 1.

Hence, if the initial applied bending moment is used only to satisfy the usual assumption (7.4.70), the exact relation (7.4.59) can be replaced by the approximate equation p (7.4.74) 2P = A + Ω2 D − (A − Ω2 D)2 + 4Ω2 B 2 .

Our result shows that in the above mentioned case, the stability analysis based on the use of the approximate relation (7.4.72) is justified. Also this relation shows that the presence of the bending-extensional coupling (B 6= 0) diminishes the value of the critical buckling force. We stress the fact that at the beginning, we have assumed that the prestressed o equilibrium configuration B is characterized by the homogeneous initial deformations (7.4.6). Obviously, the stability analysis concerning the critical values of the initial applied compressive force and bending moment remains unchanged if we assume from the beginning that o

N 11 = −P < 0,

o

o

N 22 = N 12 = 0,

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o

M 11 = M,

o

o

M 22 = M 12 = 0,

(7.4.75)

7.4. BUCKLING OF ANTISYMMETRIC CROSS-PLY LAMINATES

453

where P and M are given constant quantities. Obviously, in this new situation, o o o o o o nonvanishing initial deformations e11 , e22 , e12 , k 11 , k 22 , k 12 having constant values also will exist. These values can be obtained using the global constitutive equations and the relations (7.4.75). Also, the initial displacement field will be changed and can be determined using the geometrical relations connecting displacements, inplane deformations and curvatures. In practice, the above mentioned alternative is most frequently encountered. In the following, we assume a regular antisymmetric cross-play laminate and we shall investigate in great detail the influence of various mechanical and geometrical behaviors of the laminate, following the analysis made by Capanu and So´os [7.5]. As it is known, the use of a symmetric laminate about the middle surface is often desirable in order to avoid coupling between bending and extension. However, many physical applications of the laminate composites require nonsymmetric ones to achieve design requirements. We recall that an antisymmetric cross-play laminate consists of an even number of orthotopic laminae laid superposed on each other, having their principal material directions alternating at 0o and 90o with respect to the laminate axes. According to equations (3.4.26), in this case, only the following constitutive constants are nonvanishing: A = A11 , A22 , A12 , B = B11 = −B22 , D = D11 , D22 , D12 .

(7.4.76)

Hence, any antisymmetric cross-play laminate satisfies the restrictions (7.4.3). We recall that a regular antisymmetric cross-play laminae is defined to have identical laminae and it is a common procedure because of its simplicity in manufacturing. According to the relation (3.4.26), the stiffnesses A, B and D are given by the following equations:

h2 1 f −1 1 1 A. hA, D = (f − 1)h2 Q = (1 + f )hQ, Q = Q11 , B = 12 2N f + 1 4n 2 (7.4.77) In these equations E2 Q22 (7.4.78) = f= E1 Q11

A=

and n is the total even number of layers, having values 2, 4, 6, ... . According to equations (7.4.77)2,3 , the coupling stiffness B converges to zero; if the number of layers increases, the thickness h of the laminate remains fixed. As before, in the relations (7.4.77) and (7.4.78), we use a right coordinate system Ox1 x2 x3 for which the fiber direction of odd-numbered layers corresponds to the Ox1 direction, and the fiber direction of the even-numbered layers corresponds to the Ox2 direction. Consequently, we have 0 < f ≤ 1 and B < 0.

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(7.4.79)

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Taking into account equations (7.4.59) and the relations (7.4.77), (7.4.78), we can conclude that the critical compression force Pc depends on the parameters Q, f, a, h, n, M characterizing the material and geometrical properties of the laminate and the applied bending moment; i.e. p 2Pc = A + Ω2 D − (A − Ω2 D)2 + 4Ω2 (B + M )2 = φ(Q, f, a, h, n, M ). (7.4.80)

In this equation, Ω is given by the relation (7.4.56) or (7.4.57), and we always

have

1 . (7.4.81) a In what follows, we shall analyze the influence of the parameters Q, f, a, h, N and M on the critical force Pc , assuming that only one of these parameters changes, the other having fixed values. The influence of the bending moment M on Pc was established before. Now we study the influence of the ratio f on the critical bending moments Mc1 and Mc2 for which Pc vanishes. For these values, buckling can take place without any compressive load. Taking into account equations (7.4.67) and (7.4.77), (7.4.78), we get      1 1 1 1 h2 Q , −√ f+ +√ − Mc1 = n n 4 3 3      1 1 1 1 h2 Q . (7.4.82) +√ f+ −√ − Mc2 = n n 4 3 3 Ω∼

From these equations, it results that Mc1 and Mc2 , respectively, are vanishing if f takes the following values:         1 1 1 1 1 1 1 1 / +√ −√ , f2 = +√ / −√ . (7.4.83) f1 = n n n n 3 3 3 3

Since n = 2, 4, 6, ..., f1 and f2 are always negative. Consequently, according to (7.4.79)1 , the critical values Mc1 and Mc2 cannot vanish. Hence, without compressive force and bending moment, buckling cannot take place. This is a natural result, indicating the consistency of the used stability theory. Let us analyze now the influence of n on the critical value Pc. According to (7.4.77)2 where the number n of the laminate increases, the coupling stiffness B decreases (in module), and converges to zero. We recall now that the plate reaches the best stability state when the applied bending moment M reaches its optimal value M , given by equation (7.4.64). According to this relation, M converges toward zero, when n increases. Consequently, when the applied bending moment M is negative or zero, the stability of the composite plate is improved when more laminae are involved, that is, when n is increasing, h being fixed. If the applied bending moment M is positive and n increases, the plate stability is improved until Pc reaches the maximum value P c given by (7.4.65). After this limit, the critical value Pc decreases asymptotically toward the critical

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455

value, which corresponds to a symmetric plate (B = 0). Obviously, our conclusions are based on the equation (7.4.79) and on the inequality (7.4.78)2 , according to which B < 0. In what follows, we analyze the influence of Q, that is, if E1 (Young’s modulus of fibers!) on Pc . To do this, we observe that according to (7.4.77) and (7.4.80), we have q 2 2 2Pc = a0 Q + Ω2 d0 Q − (a0− Ω2 d0 ) Q2 + 4Ω2 (b0 Q + M ) = ϕ1 (Q), (7.4.84)

with

h2 f −1 2 1+f a0 , h , d0 = h, b0 = 12 4n 2

a0 =

(7.4.85)

and A = a0 Q, B = b0 Q, D = d0 Q.

(7.4.86)

According to (7.4.15), a0 , b0 and d0 satisfy the inequalities a0 > 0, d0 > 0, a0 d0 − b20 > 0.

(7.4.87)

Now, by elementary computations, we can conclude that dϕ1 (Q) > 0, dQ

(7.4.88)

for any Q > 0. Consequently, when Q(E1 ) increases, Pc also increases and the plate stability is improved. Next, we study the influence of the plate breadth a when other parameters are fixed. To do this, according to (7.4.80) we must analyze the behavior of the critical value Pc as a function of Ω. Again, by elementary computations using the equation (7.4.59), we get ∂F (Ω, M ) >0 (7.4.89) ∂Ω

for any Ω, since A,B,D and M satisfy the restrictions (7.4.30). Hence, taking into account (7.4.81), we can conclude that when the plate breadth a decreases, that is when Ω increases, and the plate stability is improved. 2 In order to study the influence of the ratio f = E E1 on Pc , we introduce the following notations

α=

π2 δ2 h hα hQ . , δ= , q= , γ= 48 a 12 2

(7.4.90)

Now from (7.4.77) and (7.4.78), we get A = α(1 + f ), B =

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h2 A h(f − 1)α , , D= 12 2n

(7.4.91)

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and taking into account the critical value (7.4.60) of Ω, from the equation (7.4.80), we get r π π π 2 αδ + M )2 ≡ ϕ2 (f ). 2Pc = α(1+f )(1+q)− α2 (1 + f ) (1 − q)2 + ( αδf − a 2n 2n (7.4.92) Elementary computations show that

dϕ2 (f ) > 0, df

(7.4.93)

if and only if

π2 π π αδ(f − 1) + M }2 {(1 + q)2 − 2 δ 2 } 4n a 2n π π π (7.4.94) −2α δ(1 − q)2 (1 + f ){ αδ(f − 1) + M } > 0. a 2n 2n In order to simplify this relation, we recall that according to the assumptions (7.4.2)1 δ << 1 and q << 1. (7.4.95) 4q(1 − q)2 (1 + f )2 − {

Hence, we can take

π2 2 δ ≈ 1. 4n2 In this way, the condition (7.4.94) takes the following simplified form: (1 − q)2 ≈ 1, 1 + q 2 −

(f + 1)2 α2 (4q 2 −

παδ 2 π π2 2 ) > 0. δ )+( M − n α 4n2

(7.4.96)

(7.4.97)

According to (7.4.90)3,4 , we have 4q 2 −

1 π2 2 1 π2 2 δ ( − ) > 0, for n = 2, 4, 6, ... . δ = 2 3 n 4 4n

(7.4.98)

Consequently, the restriction (7.4.97) is fulfilled. Hence, the inequality (7.4.93) takes place for any f ∈ (0, 1]. That is, the critical value Pc is an increasing function 2 of the ratio f = E E1 . The other parameters being fixed, Pc reaches its maximal value for f = 1, that is, if E1 = E2 . Finally, we analyze the influence of the hight h on Pc . To do this, we introduce the following quantities:

A0 =

1 1−f 1+f A0 . Q, D0 = Q, B0 = 12 4n 2

(7.4.99)

According to (7.4.77) and (7.4.78), we get A = hA0 , B = h2 B0 , D = h3 D0 ,

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(7.4.100)

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7.4. BUCKLING OF ANTISYMMETRIC CROSS-PLY LAMINATES

and the critical value Pc , given by equation (7.4.80) becomes q 2 2Pc = A0 h + Ω2 D0 h2 − (A0 h − Ω2 D0 h3 ) + 4Ω2 (B0 h2 + M )2 ≡ ϕ3 (h). (7.4.101) 2

2

We recall again that ha2 << 1. Taking into account this fact and neglecting ha2 with respect to 1 in the expression of (dϕ3 /h)/dh, long but elementary computations shows that dϕ3 (h) >0 (7.4.102) dh if and only if B2 A4 (7.4.103) h4 0 (1 − 12 02 ) − 2h2 A20 B0 M + A20 M 2 > 0. A0 4

According to (7.4.79) and (7.4.10), we have

1 1 1 (f − 1)2 B02 for n = 2, 4, 6, ... and 0 < f ≤ 1. < 2 ≤ = 2 2 2 16 4n 4n (f + 1) A0

(7.4.104)

Taking into account the last conclusion, it is easy too see that (7.4.103), and, hence, (7.4.101) take place. We can conclude that the stability of the plate is improved if its thickness h is increased, for any fixed value of the initial applied bending moment M. Summing up, we can say that for a fixed value of the bending moment M, the stability of a regular antisymmetric cross-ply laminate can be improved by 1 2 increasing one of the parameters Q(E1 ), f ( E E1 ), a , h or n, the other ones being fixed. Taking into account the mechanical and geometrical significance of these parameters, we can see that the above conclusion is plausible, proving again the consistency of the applied stability theory. As the second example, we consider now a simply supported laminate of type (S4), According to equations (4.2.44), we have the following boundary conditions:

U3 = 0, M = M11 = 0, N = N11 = 0 for x = ±a.

(7.4.105)

Taking into account the third condition and the relation (7.4.34), we can conclude that C1 = 0. (7.4.106) Using (7.4.106) and neglecting in (7.4.46) the rigid translation; i.e. taking C4 = 0, we obtain U1 =

B+M Ω(K cos Ωx − L sin Ωx). A−P

(7.4.107)

Using again (7.4.106) from (7.4.44), we get U3 = K sin Ωx + L cos Ωx −

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C3 C2 . x− P P

(7.4.108)

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CHAPTER 7. STABILITY OF COMPOSITE LAMINATES

Now we recall that M is given by the equation (7.4.52). Using this equation, we can conclude that the second boundary condition (7.4.105) will be satisfied if and only if cos Ωa = 0 and K = 0 (7.4.109) or sin Ωa = 0 and L = 0. The first situation occurs when   1 + k , π, k = 0, 1, 2, ... Ωa = 2

(7.4.110)

(7.4.111)

and the second occurs if Ωa = kπ, k = 1, 2, ... .

(7.4.112)

As we already know, the possible critical values of the compressive force P are given by equation (7.4.59). Also we know that f = f (Ω, M ) is an increasing function of Ω > 0. Consequently, the critical value Ωc for which buckling occurs for the first time is given again by equation (7.4.60). Hence, K = 0 and the incremental displacements U1 and U3 , given by (7.4.107) and (7.4.108), become U1 = −

πx π B+M , L sin 2a 2a A − Pc

(7.4.113)

πx C2 x + C3 , − Pc 2a

(7.4.114)

and U3 = L cos

where Pc is given by equation (7.4.61). Imposing now the first boundary condition (7.4.105), we get C2 = C3 = 0. Accordingly U3 becomes U3 = L cos

πx . 2a

(7.4.115)

Taking into account the obtained results, we can see that the cylindrical buckling behavior of the composite laminate (strip) in the case of free boundary (see Equations (7.4.49)) and in the case of simply supporting one (see Equations (7.4.105)) is the same. We shall end this Section analyzing the simplest possible case. We assume that the initial applied bending moment is vanishing and the extension-bending coupling is missing; i.e. M = 0 and B = 0. (7.4.116) As we know for the considered simply supported laminate, the critical value Pc of the applied compressive force is given by equation (7.4.61).

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7.4. BUCKLING OF ANTISYMMETRIC CROSS-PLY LAMINATES

459

Accordingly, from (7.4.112), we get Pc = min(A,

π2 D ). 4a2

(7.4.117)

Also, from (7.4.113) and (7.4.115), we obtain the corresponding incremental displacement fields U1 = 0, U3 = L cos

πx . 2a

(7.4.118)

We assume now that our laminate actually is a single specially orthotropic layer. As results from the equations (3.4.4) in this case, we have A = hQ11 , D =

E1 h3 . Q11 , and Q11 = 1 − ν12 ν21 12

(7.4.119)

Consequently, from (7.4.117), we can conclude that Pc = min(Q11 h,

π2 h 2 ( ) Q11 h). 48 a

(7.4.120)

Since (h/a) << 1, we finally obtain Pc =

π2 48

 2 h Q11 h. a

(7.4.121)

Denoting by

pE = Pc /h

(7.4.122)

the critical pressure producing the buckling of the simply specially orthotropic layer; i.e. of an usual orthotropic (long) plate, we obtain

pE =

E1 1 h 2 . (π ) 48 a 1 − ν12 ν21

(7.4.123)

This is just the classical formula, well known in the classical Love-Kirchhoff plate theory, founded on the Eulerian assumption of plane sections. We already have used two times the above equation, in analyzing the stability of a fiber-reinforced composite strip and in studying the buckling of a fiber-reinforced composite bar. It is easy to see that equations (6.2.43), (6.3.79), and (7.4.123) express the same thing, if the involved coefficients are accordingly changed, taking into account their mechanical and geometrical meanings.

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460

7.5

CHAPTER 7. STABILITY OF COMPOSITE LAMINATES

Problems

P7.1 Using the global incremental constitutive equations (7.1.35) and (7.1.38) together with the expression (7.2.23)2 of the global specific internal strain energy w = w(U), prove the validity of equations: Nαβ =

∂w ∂w ∂w . , Rα − P α = , Mαβ = ∂U3,α ∂kαβ ∂Uβ,α

P7.2 Assuming that the incremental in-plane deformations and the incremental curvatures are vanishing; i.e. supposing that eαβ = 0 and kαβ = 0, find the corresponding incremental displacements Uα , U3 , uα , u3 and give the interpretation of the obtained result. P7.3 Assuming w = w(U) a positive definite quadratic form prove the uniqueness theorem for the incremental traction boundary value problem (7.2.32). P7.4 In the same conditions as in P7.2, prove the uniqueness theorem for the incremental displacement boundary value problem (7.2.36). P7.5 Let us consider the following mixed incremental boundary value problem for a prestressed composite laminate: Un = V, Nnτ = S, U3 = R, Mnn = T on ∂D, where V ,S,W and T are given functions on ∂D. Assuming w = w(U) positive definite show that there exist at most one regular solution modulo a rigid displacement and find this displacement. P7.6 Formulate and prove the principle of minimum potential energy, appropriate to the incremental traction boundary value problem (7.2.32). P7.7 Formulate and prove the principle of minimum potential energy, appropriate to the incremental displacement boundary value problem (7.2.36). P7.8 Formulate and prove the variational principle appropriate to the incremental mixed boundary value problem given in P7.5. P7.9 Formulate and prove the principle of minimum potential energy appropriate to the incremental mixed boundary value problem given in P7.5. o P7.10 Let us consider the possible critical compressive force N 11 defined by o equation (7.3.15). Show that N 11 as a function of b and the aspect ratio K = a/b, is given by the following relation: o

N 11 = −

where Ψ(K, m) =

π 2 D22 Ψ(K, m) b2

2(D12 + 2D66 ) K 2 D11 m2 + 2 , m = 1, 2, 3, ... . + m D22 D22 K 2

P7.11 Analyze the behavior of the function Ψ = Ψ(K, m), assuming m fixed and K > 0 variable.

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461

7.5. PROBLEMS

P7.12 Analyze the behavior of the function Ψ = Ψ(K, m), assuming K fixed and m variable. P7.13 Show that there exist values of the aspect ratio K = a/b for which there are possible two buckled shapes corresponding to m and m+1 and to the o same value of the possible buckling force N 11 , considered in P7.10. o

P7.14 For what values K and m, the minimum buckling load N 11 , considered in P7.10, is reached? Find this minimum value. P7.15 Find the values of the coefficients aklij defined by equations (7.3.47) and (7.3.48). P7.16 Let us consider a rectangular composite laminate and let a, b, h be its breadth, width and thickness, respectively. Let us assume that the plate is biaxially, uniformly compressed. More exactly, let us assume that the initial applied load is characterized by the following relations: o

o

o

o

o

o

o

o

o

N 11 = −P, N 22 = −αP, N 12 = N 21 = 0, M 11 = M 22 = M 12 = M 21 = 0, q = 0. Here P > 0 is given and α > 0 is also a known quantity. o

(a) Show that the obtained initial deformed configuration B is a possible equilibrium state of the plate. o

o

(b) Find the initial in-plane deformations eαβ and initial curvatures k αβ . o

o

(c) Find the initial in-plane displacements U 1 , U 2 and the initial nominal o displacement U 3 . o o o (d) Find the initial displacement fields u1 , u2 , u3 of the prestressed laminate. P7.17 Find the global incremental constitutive equations of the prestressed laminate considered in P7.16. P7.18 Find the global incremental equilibrium equations satisfied by the incremental displacement field in the case of the prestressed laminate considered in P7.16 assuming vanishing incremental normal load. P7.19 Assuming that the laminate considered in P7.16. is simply supported of type (S2) (see Equations (7.2.44)), write the appropriate boundary conditions. P7.20 Let us assume now that the plate considered in P7.16. is an antisymmetric cross-ply laminate. Find the answers to (a) to (d) of P7.16. in this special case. P7.21 For the laminate considered in (P7.16) and (P7.20), find the answer to the question formulated in P7.17. P7.22 Taking into account again the laminate of (P7.16) and (P7.20) and using the results of P7.18, find the corresponding global incremental equilibrium equations. P7.23 Solve the problem (P7.19) for the laminate considered in (P7.16) and (P7.20).

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462

CHAPTER 7. STABILITY OF COMPOSITE LAMINATES P7.24 Let us consider the following incremental displacement fields:

nπx2 mπx1 , sin b a nπx2 mπx1 , cos U2 = a2 sin b a mπx2 nπx1 , sin U3 = a3 sin b a

U1 = a1 cos

where a1 , a2 , a3 are arbitrary real constants and m, n = 1, 2, 3, ... are arbitrary positive integers. Show that the above incremental displacement field satisfies the boundary conditions obtained in P7.23. P7.25 Assuming the incremental displacement field considered in P7.24, find the relations which must be satisfied by the constants a1 , a2 , a3 to satisfy the incremental equilibrium conditions obtained in P7.22. P7.26 Using the result obtained in P7.25, find the characteristic equation corresponding to the eigenvalue problem involved in the buckling problem of the laminate considered in P7.16 and P7.20. P7.27. Find the exclusion functional E = E(U) corresponding to the laminate considered in P7.16. P7.28 Using the result obtained in P7.26, find the exclusion functional E = E(U) corresponding to the laminate considered in P7.20. P7.29 Taking into account the result obtained in P7.27, find the expression of the exclusion functional E = E(U) assuming the admissible incremental displacement field considered in P7.24. P7.30 (a) Find the stationarity conditions of the exclusion functional E = E(U) obtained in P7.28. (b) Compare the result obtained in (P7.23) and in (a), and comment on the obtained result. P7.31 Let us assume that in the initial deformed equilibrium configuration o B of a composite laminate, the applied load is characterized by the following relations: o

o

o

o

o

o

N 11 = const., N 21 = N 12 = 0, M 11 = const., M 22 = M 12 = 0. o

o

o

Find the resulting shear forces Q1 , Q2 and the normal load q . P7.32 For the laminate considered in P7.30, find the corresponding in-plane o o deformations eαβ and curvatures k αβ , assuming that the global constitutive coefficients A16 , B16 and D16 of the laminate are vanishing; i.e. the restrictions (7.4.3) are satisfied. o o o o o o P7.33 Find the displacement fields U 1 , U 2 , U 3 and u1 , u2 , u3 corresponding to the deformation state obtained in P7.31. o o P7.34 Assuming in P7.32. that u3 = U 3 = 0, find the relation which must o o exist between N 11 and M 11 to have vanishing normal displacement.

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463

7.5. PROBLEMS

P7.35 Let us consider the composite strip analyzed in Section 7.4. Show that the function F = F (Ω, M ) defined by the relation (7.4.58) increases, if Ω > 0 increases. P7.36 For the same composite strip, show that the function ϕ1 = ϕ1 (Q) defined by the relation (7.4.83) increases if Q > 0 increases. P7.37 We consider again the composite strip studied in Section 7.4. But now we suppose that the strip is clamped at its edges x = ±a. More exactly (see Equations (7.2.45)) we assume that U3 = 0, U3,1 = 0 and N11 = 0 for x = ±a. (a) Find the critical value Pbc of the composite force producing cylindrical buckling of the strip. (b) Compare the obtained result with that found for the simply supported strip. P7.38 Find the incremental displacement fields U1 and U3 corresponding to the problem P7.37 and compare the obtained results with that obtained for the simply supported strip. P7.39 Let us assume now that the clamped strip of P7.36. is a regular antisymmetric cross-play laminate. Study the dependence of the critical buckling force Pbc on Young’s modulus E1 of the fibers and on the breadth a of the strip. P7.40 Assume now that the clamped strip is a simple specially orthotropic layer. Assuming M = 0, find the critical compressive force Pbc and the critical compressive pressure pbE = Pbc /h for this case. Compare the obtained results with that corresponding to the simply supported case. P7.41 We consider again the composite strip analyzed in Section 7.4, assuming now that the strip is simply supported on the edge x = −a and clamped on the edge x = a. Hence, we have the following boundary conditions: U3 = 0, N11 = 0, M11 = 0, for x = −a and U3 = 0, N11 = 0, U3,1 = 0, for x = a. ∨

(a) Find the critical compressive force Pc producing cylindrical buckling. (b) Compare the obtained result with that corresponding to the simply supported and clamped strip, respectively. P7.42 Find the incremental displacement fields U1 and U3 corresponding to P7.41, and compare the obtained results with that obtained for the simply supported and clamped strip, respectively. P7.43 Assuming the strip of P7.41 is an antisymmetric cross-ply laminate, ∨

determine the dependence if Pc on E1 and a. P7.44 Suppose now that the strip considered in P7.40 is a single specially orthotropic layer. Assuming M = 0, find the critical compressive force Pbc and the ∨

critical compressive pressure pbc =Pc /h in this case. Compare the obtained results with that corresponding to the simply supported and clamped cases, respectively. P7.45 (a) Find the exclusion functional E = E(U1 , U3 ) corresponding to the cylindrical incremental state of the strip considered in the Section 7.4. (b) Find the Euler-Lagrange equation corresponding to the exclusion functional E = E(U1 , U3 ).

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CHAPTER 7. STABILITY OF COMPOSITE LAMINATES

(c) Find the differential equation satisfied by the incremental displacement field U3 and compare the obtained result with that obtained in Section 7.4 (see Equation (7.4.41)). P7.46 Let us consider a composite strip consisting of a Boron/Epoxy specially orthotropic layer. The material properties of this composite are given in Table 3.1. If the strip is simply supported, the critical value Pc of the buckling compressive force is given by the relation (7.4.121). According to the relations (3.1.6) and (3.4.4), the global constitutive coefficient A66 of the strip is given by the equation A66 = hG2 . Find the ratio Pc /A66 and comment on the obtained result taking into account the differential system obtained in P7.22, describing the incremental equilibrium conditions of an antisymmetric cross-ply laminate. P7.47 The critical buckling pressure pE of fiber-reinforced composite plate, acted upon by compressive forces along the fibers directions, is given by the equation (6.2.36), (6.3.79) and (7.4.123), respectively. Even if these three relations give the same critical pressure, they are apparently different. Examining these equations, and considering also the cases to which they correspond, prove that all equations are true and (6.2.36) and (6.3.79) can be obtained from (7.4.123) by appropriate changes.

Bibliography [7.1] Ashton, J.E., Whitney, J.M., Theory of laminate plates, Technomic Publishing Co. Inc., 1970. [7.2] Jones, R.M., Mechanics of composite materials, Scripta Book Company, 1975. [7.3] Whitney, J.M., Structural analysis of laminated anisotropic plates, Technomic Publishing Co. Inc., 1987. [7.4] Gibson, R.F., Principles of composite materials mechanics, McGraw-Hill, Inc. New York, 1994. [7.5] Capanu, M., So´os, E., Stability of composite laminates. The influence of bending moment. Cylindrical buckling, Rev. Roum. Sci. Techn. M`ec. Appl. 42, 1998. [7.6] Capanu, M., Stability of rectangular composite antisymmetric laminates, Rev. Roum. Sci. Techn. M`ec. Appl. 42, 1998.

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Chapter 8

FRACTURE MECHANICS 8.1

Elements of complex function theory

The basic elements of complex function theory, used in fracture mechanics, are presented in Muskhelishvili’s fundamental monograph [8.1]. We denote by z = x + iy and z = x − iy, i =

√

−1,

(8.1.1)

a complex variable and its conjugate in the complex plan C, x and y are real variables, representing the real and imaginary parts of z. Let us consider an analytic function depending on z and having complex values f = f (z) = u(x, y) + iv(x, y), (8.1.2) where u = u(x, y) = Ref (z) and v = v(x, y) = Imf (z)

(8.1.3)

are real valued functions, representing the real and imaginary parts of f. As in (8.1.1), the operation of complex conjugation will be denoted by a superposed bar; i.e. f (z) = u(x, y) − iv(x, y). (8.1.4)

An analytic function can have regular and singular points and can be univalued or multivalued. An univalued branch, having only regular points, of an analytic function, is a holomorphic function. If f = f (z) is a holomorphic function, it has derivatives of any order in its domain of definition. The real and imaginary parts of f (z) = u(x, y) + iv(x, y) satisfy the Cauchy-Riemann conditions

∂v ∂u , = ∂y ∂x

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∂v ∂u =− . ∂x ∂y

(8.1.5)

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CHAPTER 8. FRACTURE MECHANICS

The derivative f 0 = f 0 (z) of a holomorphic function can be calculated using the equations f 0 (z) =

∂u ∂v ∂v ∂u (x, y) − i (x, y). (x, y) + i (x, y) = ∂x ∂y ∂x ∂x

(8.1.6)

The relations (8.1.5) show that the real and imaginary parts of a holomorphic function are harmonic functions; i.e. ∆u = ∆v = 0,

∆=

∂2 ∂2 + 2. 2 ∂y ∂x

The converses of the above properties are also true. If the real and imaginary parts of a function f = f (z) satisfy the Cauchy-Riemann conditions, that function is holomorphic. Also, any harmonic real valued function u = u(x, y) can be the real or the imaginary part of a holomorphic function f = f (z). Any holomorphic function f = f (z) is integrable on any regular curve L, and is path independent. If Z z

F = F (z) =

f (t)dt,

(8.1.7)

z0

then F = F (z) is a holomorphic function and F 0 (z) = f (z).

(8.1.8)

Also, if f = f (z) is a holomorphic function and L is a closed regular curve in its domain of definition, according to Cauchy’s well-known theorem Z f (z)dz = 0. (8.1.9) L

Let us consider now a complex valued function f = f (z, z) = Ref (z, z) + iImf (z, z),

(8.1.10)

depending on z and z, considered as independent variables. Using (8.1.1), we can express the real and imaginary parts of f = f (z, z) as functions of x and y; i.e.

f = f (z, z) = u(x, y) + iv(x, y).

(8.1.11)

Assuming that f = f (z, z) has continuous derivatives with respect to z and z, from (8.1.1), (8.1.11) and the chain rule, we get     ∂u i ∂v 1 ∂u ∂v ∂f , − + + = 2 ∂x ∂y 2 ∂x ∂y ∂z     1 ∂u ∂v i ∂v ∂f ∂u − = + + . (8.1.12) 2 ∂x ∂y 2 ∂x ∂y ∂z

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8.1. ELEMENTS OF COMPLEX FUNCTION THEORY

467

If f = f (z, z) is a holomorphic function, its real and imaginary parts satisfy the Cauchy-Riemann condition (8.1.5). Hence,

∂f (z, z) = 0. ∂z

(8.1.13)

The last relation shows that f = f (z, z), being holomorphic, actually does not depend on z. Conversely, let us assume that the function f = f (z, z) does not depend z. In this case, (8.1.13) takes place. Hence, according to (8.1.12)2 , the real on and imaginary parts of f = f (z) satisfy the Cauchy-Riemann condition (8.1.5). Accordingly, f = f (z) is a holomorphic function. Let us consider now the real valued function

F = F (x, y)

(8.1.14)

depending on the real variables x and y. Using (8.1.1), we get x=

1 (z+ z), 2

y=

1 (z− z). 2i

(8.1.15)

Taking into account the above equations, we can express F = F (x, y) as a function of z and z; i.e.

F = F (x, y) = F (

z+ z z− z ) ≡ f (z, z). , 2i 2

(8.1.16)

Now, assuming that F = F (x, y) has continuous second order partial derivatives, from (8.1.16) we get: ∆F =

1 ∂2f ∂2F ∂2F (z, z). = + 2 2 4 ∂z∂ z ∂y ∂x

(8.1.17)

Let us assume now that F = F (x, y) is a harmonic function; i.e. ∆F (x, y) = 0.

(8.1.18)

According to (8.1.17), in this case f = f (z, z) must satisfy the equation

∂2f (z, z) = 0. ∂z∂ z

(8.1.19)

F (x, y) = f (z, z) = ϕ(z) + ψ(z),

(8.1.20)

From the above relation, we get

where ϕ = ϕ(z) and ψ = ψ(z) are complex valued functions, depending only on z and z, respectively. Moreover, since F = F (x, y) is a real valued function, from (8.1.20) we can conclude that ψ(z) = ϕ(z).

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468

CHAPTER 8. FRACTURE MECHANICS Hence, if F = F (x, y) is a harmonic function, then F (x, y) = ϕ(z) + ϕ(z) = 2Reϕ(z),

(8.1.21)

where ϕ = ϕ(z), depending only on z, is a holomorphic function. The above result shows again that any harmonic function is the real part of a holomorphic function. An important analytic function is f (z) = ln(z − c) defined by the rule f (z) = ln(z − c) = ln |z − c| + i arg(z − c)

(8.1.22)

where c is a fixed complex number (see Figure 8.1). y

z r

r = z-c =arg z-c

c x

O

Figure 8.1: The definition of the logarithmic function. As is known, z = c is a singular point of this analytical, multivalued function, having an unbounded number of branches. Let us assume now that a and b are two real numbers and z is a complex number which is not situated in the real interval (a, b). Then we have (see Figure 8.2) Z

b a

dt t−z

=

|b − z| b−z = ln |a − z| a−z + i {arg(b − z) − arg(a − z)} .

ln(b − z) − ln(a − z) = ln

(8.1.23) √

As a second example, let us consider the analytic function f (z) = z 2 − a2 , where a > 0 is a fixed real number. This function is defined by the relation (see Figure 8.3): p √ i (8.1.24) f (z) = z 2 − a2 = r1 r2 e 2 (θ1 +θ2 )

where

r1 = |z1 − a| , r2 = |z2 + a| ,

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8.1. ELEMENTS OF COMPLEX FUNCTION THEORY

y

z

x O

a

b

Figure 8.2: Example of complex integration.

y

z

r1

+

r2

y=0 2

-a

x

1

O y=0

a

Figure 8.3: The definition of the function f (z) =

θ1 = arg(z − a), θ2 = arg(z + a).

√

z 2 − a2 . (8.1.25)

As is known, z = a and z = −a are singular points of this analytical function which is multivalued and has two branches. To select a branch, which is holomorphic, we must introduce a cut in the complex plane C. This cut is represented by the segment [−a, a] having two faces, as in Figure 8.3. We have y = 0+ on the upper face of the cut, and y = 0− at the corresponding point of the lower face of the cut. In what follows, we denote by L this cut. √ We choose that branch of the function f (z) = z 2 − a2 which satisfies the relation p x2 − a2 = +∞. (8.1.26) lim x→∞

√

Using this choice and the relation (8.1.24) defining the function f (z) = z 2 − a2 , we can see that the selected branch has the following properties: p p if a < x, y = 0, (8.1.27) f (z) = z 2 − a2 = x2 − a2 p p f (z) = z 2 − a2 = i a2 − x2 if − a < x < a, y = 0+ ,

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CHAPTER 8. FRACTURE MECHANICS

(i.e. on the upper face of the cut) p p f (z) = z 2 − a2 = − x2 − a2 if x < −a, y = 0, p p f (z) = z 2 − a2 = −i a2 − x2 if − a < x < a, y = 0−

(i.e. on the lower face of the cut). Let us denote by C 0 = C − L the complex plane without the cut L. Let us assume that F = F (z) is an arbitrary complex valued function defined in C 0 . Let t ∈ L, t 6= −a and t 6= a, be an arbitrary point on the cut L (see Figure 8.4).

y

z x -a

a

O

t

z Figure 8.4: Limit values.

We denote by F + (t) and by F − (t) the following limit values of F = F (t): F + (t) = F − (t) =

lim

lim

z→t,y>0

z→t,y<0

F (z),

F (z), z = x + iy.

(8.1.28)

If the above limits exist and are path independent, we name F + (t) and F − (t) the upper and the lower limits of F (z) in the point t of the cut. We stress the fact that the above limits are not defined at the end point t = −a and t = a of the cut. According to (8.1.27), the Plemelj type function p (8.1.29) Y = Y (z) = z 2 − a2 ,

satisfying the property (8.1.26), fulfills the relations

p t2 − a 2 )+ = i a 2 − t 2 , p p Y − (t) = ( t2 − a2 )− = −i a2 − t2 ,

Y + (t) = (

p

(8.1.30)

on the cut L. Hence, we have

X + (t) + X − (t) = 0

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on L (t 6= −a, t 6= a).

(8.1.31)

471

8.1. ELEMENTS OF COMPLEX FUNCTION THEORY Let us consider now a complex valued function f = f (t) = f1 (t) + if2 (t)

(8.1.32)

defined on the cut L; i.e. on the real segment (−a, a). For the moment, we assume that f = f (t) is absolutely integrable on L in the usual Riemannian sense. Using this function, we define in C 0 = C − L the complex valued function Z a f (t) 1 dt, z∈ / L. (8.1.33) F = F (z) = 2πi −a t − z

The above function represents Cauchy’s integral corresponding to the function f = f (t) and to the cut L. In the domain C 0 representing the complex plane C without the cut L, Cauchy’s integral F = F (z) is a holomorphic function. Moreover, as it is easy to see from (8.1.32), lim F (z) = 0. (8.1.34) z→∞

Until now, we have assumed that z ∈ / L. Let us assume now that z = t0 ∈ L, t0 6= ±a, as in Figure 8.5. y

l -a

O

t0

t0-

x a

t0+

Figure 8.5: Cauchy’s principal value. If f (t0 ) 6= 0, lim f (t)/(t − t0 ) is unbounded and t→t0

1 2πi

Z

a −a

f (t)dt t − t0

is an improper integral, which is generally not convergent. Let us assume that the function f = f (t) satisfies H o¨lder’s condition or the H condition; i.e. µ

|f (t2 ) − f (t1 )| ≤ A |t2 − t1 | , ∀t1 , t2 ∈ L,

(8.1.35)

where A and µ are positive constants, and 0 < µ ≤ 1.

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(8.1.36)

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CHAPTER 8. FRACTURE MECHANICS

In this case, we can give a well-defined sense to the above presented improper integral. To do this, we consider a small positive number ε > 0 (see Figure 8.5) and consider the integral 1 2πi

Z

t0 −ε

−a

f (t)dt + t − t0

Z

a t0 +ε

f (t)dt t − t0



=

1 2πi

Z

L−l

f (t)dt , t − t0

(8.1.37)

defined on the set (−a, t0 − ε) ∪ (t0 + ε, a) ≡ L − l. The integral (8.1.37) is well defined, since, for t ∈ L − l, we have |t − t0 | ≥ δ > 0, δ being a positive number. We suppose now that ε converges to 0. If in this case, the integral (8.1.37) converges to a well-defined finite value, we call this limit Cauchy’s principal value of the improper integral presented above. If this improper integral is convergent in the usual sense, its principal value also exists and is the integral itself. But the converse is not true. The principal value of the considered improper integral, if it exists, will be denoted by Z a f (t)dt . (8.1.38) (P V ) −a t − t0

We shall prove now that the principal value exists, if f = f (t) satisfies the H condition (8.1.35). We have 1 2πi

Z

L−l

1 f (t)dt = 2πi t − t0

and

Z

L−l

f (t0 ) f (t) − f (t0 ) dt + 2πi t − t0

Z

L−l

dt t − t0

f (t) − f (t0 ) ≤ A |t − t0 |µ−1 with µ > 0. t − t0

(8.1.39)

(8.1.40)

Hence, according to the well-known convergence criterion, the improper integral Z a f (t) − f (t0 ) dt (8.1.41) t − t0 −a

is convergent, and

 f (t) − f (t0 ) dt . t − t0 t0 +ε −a −a (8.1.42) According to (8.1.23) for the second integral in the right-hand side of equation (8.1.39), we successively get Z

a

f (t) − f (t0 ) dt = lim ε→0 t − t0

Z

L−l

dt t − t0

=

=

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Z

t0 −ε

f (t) − f (t0 ) dt + t − t0

Z

a

Z a dt dt + t − t0 t − t 0 t0 +ε −a −ε(a − t0 ) a − t0 −ε . = ln + ln ln ε(−a − t0 ) ε −a − t0

Z

t0 −ε

(8.1.43)

8.1. ELEMENTS OF COMPLEX FUNCTION THEORY Hence,

Z

L−l

a − t0 dt . = ln a + t0 t − t0

473

(8.1.44)

The obtained result shows that the above integral does not depend on ε! Returning now to the definition of the principal value, from (8.1.39), (8.1.42) and (8.1.44), we get 1 (P V ) 2πi

Z

a −a

1 f (t) dt = 2πi t − t0

Z

a −a

f (t0 ) a − t0 f (t) − f (t0 ) , ln dt + a + t0 2πi t − t0

t0 ∈ (−a, a).

(8.1.45)

The above relation proves the existence of Cauchy’s principal value and, at the same time, gives its value. In equation (8.1.45), the first integral in the righthand side is taken in the usual Riemannian sense. It is easy to see that even if f = f (t) satisfies the H condition, our improper integral generally is not convergent. Indeed, lim

ε1 →0,ε2 →0

Z

t0 −ε1 −a

dt + t − t0

Z

a t0 +ε2

dt t − t0



= ln

a − t0 −ε1 + ln −a − t0 ε2

(8.1.46)

generally does not exist! We return now to Cauchy’s integral F = F (z) defined by equation (8.1.33). First of all, we observe that for z ∈ / L, this integral can be expressed as Z a Z a f (t) − f (−a) 1 dt f (−a) dt + F (z) = t−z 2πi −a 2πi −a t − z Z a f (t) − f (−a) 1 a−z f (−a) dt, + ln = t−z 2πi −a −a − z 2πi

or F (z)

=

=

Z a Z f (t) − f (a) 1 f (a) a dt dt + t−z 2πi −a 2πi −a t − z Z a a−z f (a) 1 f (t) − f (a) ln + dt. 2πi −a − z 2πi −a t−z

The above equations can be written also in the form F (z) = −

f (−a) ln(−a − z) + F1 (z) 2πi

or F (z) =

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f (a) ln(a − z) + F2 (z), 2πi

(8.1.47)

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CHAPTER 8. FRACTURE MECHANICS

with

Z a f (t) − f (−a) 1 f (a) dt, ln(a − z) + t−z 2πi −a 2πi Z a f (t) − f (a) 1 f (a) dt. ln(−a − z) + F2 (z) = − t−z 2πi −a 2πi

F1 (z) =

(8.1.48)

Since f = f (t) satisfies the H condition, F1 = F1 (z) converges to a welldefined finite limit when z converges to −a; similarly, F2 = F2 (z) converges to a well-defined limit when z converges to a. Hence, the singular behavior of F = F (z) in the neighborhood of z = −a is characterized by the behavior of the function ϕ1 (z) = ln(−a − z); similarly, the singular behavior of F = F (z) in the neighborhood of z = a is characterized by the behavior of the function ϕ2 (z) = ln(a − z). Moreover we have lim (−a − z)µ ln(−a − z) = lim (a − z)µ ln(a − z) = 0, for any µ > 0.

z→−a

z→a

Consequently, in the neighborhood of the singular points z = −a and z = a, the Cauchy’s integral F = F (z) satisfies the following inequalities: |F (z)| <

A A µ, µ and |F (z)| < |a − z| |−a − z|

(8.1.49)

where A and µ are positive constants. The above important inequalities characterize the singular behavior of Cauchy’s integral in the neighborhood of the end points ±a of the cut L. Now we present and prove for our particular case the following basic theorem due to Sohockii and Plemelj. If the function f = f (t) given on the cut L satisfies the H condition, the limit values F + (t0 ) and F − (t0 ), t0 ∈ L, of Cauchy’s integral (8.1.33) exist and are given by the following equations: Z a f (t)dt 1 1 + , (P V ) F (t0 ) = f (t0 ) + 2πi 2 −a t − t0 Z a f (t)dt 1 1 , t0 ∈ L, t0 6= ±a. (8.1.50) (P V ) F − (t0 ) = − f (t0 ) + 2πi 2 −a t − t0

To prove (8.1.50), we first suppose that z = x + iy is in the upper half plane (y > 0) and z → t0 ∈ L. We have Z a Z Z a f (t) − f (t0 ) 1 f (t0 ) a dt f (t)dt 1 dt. (8.1.51) + = F (t) = t−z 2πi −a 2πi −a t − z 2πi −a t − z

We have (see Figure 8.6) Z a a−z a−z dt + i {arg(a − z) − arg(−a − z)} = ln = ln −a − z −a − z −a t − z a − z + i(θ1 − θ2 ). (8.1.52) = ln a + z

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8.1. ELEMENTS OF COMPLEX FUNCTION THEORY

y z

2

1

-a

O

a

t0

x

Figure 8.6: Plemelj-Sohockii theorem.

It is easy to see that when z → t0 , θ1 → 2π, θ2 → π. Hence, according to (8.1.52), Z a a − t0 dt + iπ. (8.1.53) = ln lim z→t0 −a t − z a + t0

Let us consider now the integral I(z) =

Z

a −a

f (t) − f (t0 ) dt − t−z

Z

a −a

f (t) − f (t0 ) dt = t − t0

Z

a −a

f (t) − f (t0 ) z − t0 dt. t−z t − t0 (8.1.54)

We have I(z) =

Z

t0 −ε −a

f (t) − f (t0 ) z − t0 dt + t−z t − t0

+

Z

a t0 +ε

Z

t0 +ε t0 −ε

f (t) − f (t0 ) z − t0 dt+ t−z t − t0

f (t) − f (t0 ) z − t0 dt. t−z t − t0

(8.1.55)

We observe that in the first and the third integral |t − t0 | ≥ δ > 0, δ being a positive number. Hence, for any η > 0 there exists a positive number K1 (η) > 0 such that Z t0 −ε Z a f (t) − f (t0 ) z − t0 η f (t) − f (t0 ) z − t0 < , if |z − t0 | < K1 (η). dt dt + t−z 2 t − t0 t−z t − t0 t0 +ε −a (8.1.56) To evaluate the second integrals in (8.1.55), we assume that z converges to t 0 on the normal to the cut L in t0 (see Figure 8.6). In this case, |z − t0 | ≤ |t − z| for any t ∈ (−a, a). Hence, since f = f (t) satisfies the H condition, we get

f (t) − f (t0 ) z − t0 f (t) − f (t0 ) ≤ A |t − t0 |µ−1 , 0 < µ ≤ 1. ≤ t − t0 t − z t − t0

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CHAPTER 8. FRACTURE MECHANICS

Accordingly, the proper integral analyzed here is convergent. Hence, for η > 0 there exists K2 (η) > 0 such that Z t0 +ε f (t) − f (t0 ) z − t0 η (8.1.57) dt ≤ if ε ≤ K2 (η). t−z 2 t − t0 t0 −ε

From (8.1.56) and (8.1.57) it follows that for any positive number η > 0 there exists K(η) > 0 such that |I(z)| < η if |z − t0 | < K(η) with K(η) = min {K1 (η), K2 (η)} .

(8.1.58)

The above result shows that lim I(z) = 0. z→t0

Hence, according to (8.1.54), Z a Z a f (t) − f (t0 ) f (t) − f (t0 ) dt. dt = lim z→t0 −a t − t0 t−z −a

(8.1.59)

We recall that we have proved (8.1.59) assuming that z converges to t0 on a particular path. It can be shown that (8.1.59) is true for any path in the upper half plane. Using (8.1.51), (8.1.53) and (8.1.59), we get Z a f (t) − f (t0 ) 1 a − t0 1 1 dt. (8.1.60) + ln F + (t0 ) = f (t0 ) + t − t0 2πi −a 2πi a + t0 2

Comparing (8.1.45) and (8.1.60), we get the first Plemelj-Sohockii formula (8.1.50). To prove the second relation (8.1.50), we suppose that z = x + iy is in the lower half plane (y < 0) and z → t0 ∈ L. This case is shown in Figure 8.7.

y

2

1

-a

a

O

x

z Figure 8.7: Plemelj-Sohockii theorem, in the lower half plane (y < 0). We have again Z

Copyright © 2004 by Chapman & Hall/CRC

a −a

a − z dt + i(θ1 − θ2 ), = ln a + z t−z

(8.1.61)

8.1. ELEMENTS OF COMPLEX FUNCTION THEORY

477

but now θ1 and θ2 are the angles given in Figure 8.7. As is easy to see, when z → t0 , θ1 → 0 and θ2 → π. Hence, according to (8.1.61), Z a a − t0 dt − iπ. (8.1.62) = ln lim z→t0 −a t − z a + t0

Using the above equation and repeating the reasoning made in the first case, we get the second Plemelj-Sohockii formula (8.1.50). Let us observe that these formulas can be written in the following equivalent and useful form: F + (t0 ) − F − (t0 ) = f (t0 ) Z a f (t)dt 1 + − . F (t0 ) − F (t0 ) = (P V ) πi −a t − t0

(8.1.63)

Let us assume now that F = F (z) is a function given in C 0 = C − L and has the following properties: (1) F = F (z) is holomorphic in C 0 = C − L. (2) For any t ∈ (−a, a) there exist the limit values F + (t) and F − (t) and they are path independent. (3) In the neighborhood of the end points ±a of the cut L, the function F = F (z) satisfies the inequality |F (z)| <

A µ , A > 0, 0 ≤ µ < 1, |z − c|

(8.1.64)

where c can be −a or a, and A, µ are given positive constants. We can formulate the Riemann-Hilbert or the conjugation problem for the cut L in the following way: Find the function F = F (z) having the properties (1)−(3) and which satisfies, on the cut L, the following boundary condition: F + (t) + F − (t) = f (t) on L, t 6= ±a,

(8.1.65)

where f = f (t) is a function given on L and which satisfies the H condition. First, we try to solve a more simple problem. Find the function F = F (z) having the properties (1)−(3) and which satisfies, on the cut L, the following boundary condition: F + (t) − F − (t) = f (t) on L.

(8.1.66)

This problem can be solved using Cauchy’s integral. Let us consider the function Z a f (t)dt 1 . (8.1.67) F0 = F0 (z) = 2πi −a t − z

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CHAPTER 8. FRACTURE MECHANICS

As we already know, F0 = F0 (z) is a holomorphic function in C 0 and, according to (8.1.49), satisfies the restriction (8.1.64). Also, according to the PlemeljSohockii theorem and to the first equation (8.1.63), F0+ (t), F0− (t) exists and satisfies the condition (8.1.66). Consequently, the Cauchy integral (8.1.67) represents a solution of our boundary value problem (8.1.66). Let us consider the function F1 (z) = F (z) − F0 (z),

(8.1.68)

where F = F (z) is the searched general solution of the boundary value problem (8.1.66). According to (8.1.66) and (8.1.68), we must have F1+ (t) − F1− (t) = 0 on L.

(8.1.69)

Hence, the upper and lower boundary values of the function F = F1 (z) are equal on the cut. Consequently, if we assume that the function F1 = F1 (z) is defined on the cut by its boundary values, the function obtained in this way will be holomorphic in the whole complex plane C, except in the end points ±a of the cut. We recall now the condition (8.1.64). Since F0 = F0 (z) satisfies this restriction, F1 = F1 (z) must have the same property; i.e. we must have |F (z)| <

A µ , 0 ≤ µ < 1. |z − c|

(8.1.70)

According to this inequality, (z − c)F1 (z) is bounded in the neighborhood of the end points. Hence, in these points, the product (z − c)F1 (z) has only an apparent singularity which can be eliminated. Consequently, we can assume that the function F2 (z) = (z − c)F1 (z) is holomorphic in the whole complex plane C. Accordingly, F1 = F1 (z) can have, at most, a first order pole in C. But, according to (8.1.70), this pole actually is missing since (z − c)F1 (z) → 0 if z → c. Consequently, we can conclude that F1 = F1 (z) is holomorphic in the whole complex plane. But in this case, according to Liouville’s theorem, F1 = F1 (z) must be a complex constant K. Accordingly, the general solution of the boundary value problem (8.1.66) is Z a f (t)dt 1 + K, (8.1.71) F (z) = 2πi −a t − z

where K is an arbitrary complex number. If we assume that lim F (z) = 0, z→∞

(8.1.72)

we must take K = 0. Hence, the solution of the boundary value problem (8.1.66) and (8.1.72), having the properties (1)−(3) is unique and is given by Cauchy’s integral Z a f (t)dt 1 . (8.1.73) F (z) = 2πi −a t − z

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8.1. ELEMENTS OF COMPLEX FUNCTION THEORY

479

We return now to the Riemann-Hilbert problem (8.1.65). We try to find a particular solution of this problem using Plemelj function X(z) =

1 1 . =√ 2 Y (z) z − a2

(8.1.74)

As we already have done, we select that branch of the square root that satisfies the property (8.1.26). In this case, according to (8.1.29)−(8.1.31), we get 1 , X + (t) = √ 2 i a − t2

X − (t) =

and

1 √ on L −i a2 − t2

X + (t) + X − (t) = 0 on L.

(8.1.75)

(8.1.76)

At the same time, we can see that X = X(z) satisfies the property |X(z)| <

A µ , A > 0, 0 ≤ µ < 1 |z − c|

(8.1.77)

in the neighborhood of the point c, which can be −a or a, the end points of the cut L. Using the property (8.1.76), we can replace the boundary condition (8.1.65) by the equivalent restriction

f (t) F − (t) F + (t) on L. = + − − + X (t) X (t) X (t)

(8.1.78)

If we introduce the function Φ(z) =

F (z) , X(z)

(8.1.79)

we can see that Φ = Φ(z) is holomorphic in C 0 = C − L and satisfies the jump condition Φ+ (t) − Φ− (t) =

f (t) on L. X + (t)

Comparing this relation with equations (8.1.66) and using (8.1.71), we can conclude that the general solution of the boundary value problem (8.1.78) is Φ(z) =

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1 2πi

Z

a −a

f (t) dt + K. X + (t)(t − z)

(8.1.80)

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CHAPTER 8. FRACTURE MECHANICS

Returning now to the relation (8.1.79), we get the general solution of the Riemann-Hilbert problem (8.1.65): Z f (t) X(z) a dt + KX(z), (8.1.81) F (z) = + 2πi −a X (t)(t − z)

where K is an arbitrary complex constant. From the expression (8.1.74) of Plemelj function, X = X(z), it follows that F = F (z) satisfies the relation lim F (z) = 0.

z→∞

Without additional conditions, the constant K cannot be determined. Let us observe that the function Z p G(z) = X(z)dz = ln(z + z 2 − a2 )

(8.1.82)

(8.1.83)

converges to ∞ when z converges to ∞. Let us denote by ϕ = ϕ(z) the indefinite integral of the function F = F (z); i.e. Z ϕ(z) =

F (z)dz

(8.1.84)

and let us assume that

lim ϕ(z) = 0.

z→∞

(8.1.85)

If we require this supplementary restriction, the above results show that it can be satisfied if and only if K = 0. Thus, we can conclude that the unique solution of the Riemann-Hilbert problem, satisfying the supplementary condition (8.1.84), (8.1.85), is given by the relation Z f (t) X(z) a dt. (8.1.86) F (z) = 2πi −a X + (t)(t − z)

In problems concerning the behavior of an infinite elastic solid containing a right crack, we shall use just the above obtained solution. We shall show now that if f = f (t) is a polynomial of degree m, the integral in (8.1.86) can be evaluated. To see how this can be done, we consider a regular closed curve Γ containing the cut L, such that point z is in the domain B − , the unbounded region limited by Γ, as in Figure 8.8. The bounded region limited by Γ will be designed by B + . The sense of integration on Γ is also indicated in Figure 8.8. Let us consider the integral Z f (t)dt 1 , (8.1.87) =(z) = 2πi Γ X(t)(t − z)

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481

8.1. ELEMENTS OF COMPLEX FUNCTION THEORY

y

L

-a

O

x a

Figure 8.8: The cut L and the closed curve Γ. √ where X(z) = 1/ z 2 − a2 is the Plemelj function; i.e. that holomorphic branch which satisfies the restriction (8.1.26) and f = f (z) is a polynomial of degree m given in C. In the unbounded domain B − , the function f (z)/X(z) can be developed in the following form:

b2 b1 f (z) + 2 + ..., = am+1 z m+1 + am z m + ... + a0 + z z X(z)

(8.1.88)

where am+1 , am , ..., a0 , b1 , b2 , ... are complex numbers. Let us consider the function h = h(z) =

b2 b1 + 2 + ... . z z

(8.1.89)

This function is holomorphic in the unbounded domain B − and h(∞) = lim h(z) = 0. z→∞

(8.1.90)

Also, from (8.1.87), we get f (z) = am+1 z m+1 + am z m + ... + a0 + h(z). X(z)

Thus, from (8.1.87), it results Z Z h(t)dt 1 am+1 z m+1 + am z m + ... + a0 1 . dt + =(z) = 2πi Γ t − z t−z 2πi Γ

(8.1.91)

(8.1.92)

Obviously, p = p(z) = am+1 z m+1 + am z m + ... + a0 is a holomorphic function in the bounded domain B + . Hence, according to Cauchy’s fundamental theorem (see Equation (8.1.9)), we have Z am+1 z m+1 + am z m + ... + a0 1 dt = 0 for z ∈ B − . (8.1.93) t−z 2πi Γ

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Also, we know that h = h(z) is holomorphic in the unbounded domain B − and is zero at infinity. Hence, according to Cauchy’s fundamental formula, Z h(t)dt 1 = h(z) for z ∈ B − . (8.1.94) 2πi Γ t − z

Now, from (8.1.91)−(8.1.94) we can conclude that Z f (z) f (t)dt 1 − am+1 z m+1 − am z m − ... − a0 , for z ∈ B − . = =(z) = X(z) 2πi Γ X + (t)(t − z) (8.1.95) Let us assume now that the closed curve Γ converges to the cut L (see Figure 8.8). In this way, we obtain  Z a Z −a f (t)dt f (t)dt 1 . + =(z) = + X − (t)(t − z) 2πi a −a X (t)(t − z)

We use now the relation (8.1.76) and the above equation becomes Z a f (t)dt 1 . =(z) = + πi −a X (t)(t − z)

Comparing (8.1.95) and the last equation, finally, we get   Z a 1 f (z) f (t)dt 1 m+1 m − am+1 z − am z − ... − a0 . = 2 X(z) 2πi −a X + (t)(t − z)

(8.1.96)

Let us suppose now that f (t) = c = const. for − a < t < a.

(8.1.97)

As is easy to see in this special case, m = 0, a1 = c, a0 = 0 and from (8.1.96), we get   Z a 1 c dt c −z . (8.1.98) = 2 X(z) 2πi −a X + (t)(t − z)

Hence, if (8.1.97) takes place, the solution F = F (z) of the Riemann-Hilbert problem, which also satisfies the supplementary restriction (8.1.84), (8.1.85) is given by the following equation:   z c , (8.1.99) 1− √ F (z) = 2 z 2 − a2

since X(z) = √z21−a2 . Ending this Section, let us consider a complex valued function f = f (z) given − in C = C − L and let us define the function f = f (z) by the following relation:

f (z) = f (z).

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(8.1.100)

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8.2. REPRESENTATION OF THE INCREMENTAL FIELDS

We suppose that, on the cut L, there exist the limits f + (t) and f − (t). In − + this case, we can conclude that there exist also the limits f (t) and f (t) on L. Moreover, as can be seen using (8.1.100), we have +

f (t) = f − (t)

and

−

f (t) = f + (t) on L.

(8.1.101)

Finally, let us assume that the real valued function F = F (x, y) is defined in C 0 = C − L by the equation F (x, y) = 2Ref (z) = f (z) + f (z).

(8.1.102)

According to (8.1.101) for the limit values F + (t) and F − (t) on L, we get −

F + (t) = f + (t) + f (t)

and

+

F − (t) = f − (t) + f (t) on L.

(8.1.103)

The above relations are useful to solve boundary value problems concerning a crack and if we look for the solution by representation of the elastic fields by complex potentials.

8.2

Representation of the incremental fields

The representation of elastic fields by complex potentials in the classical case of anisotropic elastic bodies was given by Leknitskii [8.2]. This representation was used, for instance, by Sih and Leibowitz [8.3] to analyze problems concerning the existence of a crack in an anisotropic elastic solid. The results obtained by Leknitskii were generalized for the case of a prestressed material by Guz [8.4],[8.5], who also has analyzed the influence of the initial stresses on the behavior of a solid body containing cracks. In what follows, we present Guz’s representation of the incremental fields by complex potentials. We assume that the orthotropic, initial deformed composite material is in plane state relative to the x1 x2 plane. As we already know in this case, the incremental displacement field can be expressed by two real potentials ϕ(1) , ϕ(2) , using Guz’s formulas (5.6.53). These potentials must satisfy equations (5.6.56). The parameters η12 , η22 , appearing in equations (5.6.56), are the roots of the algebraic equation (5.6.57) and are given by the relation (5.6.59). We assume η 12 6= η22 . The quantities A and B in (5.6.59) depend on the instantaneous elasticities and are given by the relation (5.6.58). The instantaneous elasticities at their term, are expressed by the elasticities and initial stresses by equations (5.6.47).

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In what follows, we assume ϕ(2) (x1 , x2 ) ≡ 0 and, in order to simplify the writing, we use the notation ϕ(1) (x1 , x2 ) = ϕ(x1 , x2 ). According to (5.6.56), ϕ = ϕ(x1 , x2 ) must satisfy equation   2  2 2 2 ∂ ∂ 2 ∂ 2 ∂ (8.2.1) + η2 2 ϕ = 0, η12 6= η22 . + η1 2 ∂x1 ∂x22 ∂x1 ∂x22

According to (5.6.53), the incremental displacement fields u1 and u2 are expressed in terms of ϕ = ϕ(1) by the relations u1 = − (ω1122 + ω1212 ) ϕ,12 u2 = ω1111 ϕ,11+ ω2112 ϕ,22 .

(8.2.2)

Let us introduce now the quantities ν1 and ν2 defined by ν1 = −η12 ,

ν2 = −η22 .

(8.2.3)

Now, the differential equation (8.2.1) becomes   2  2 ∂2 ∂ ∂2 ∂ ϕ = 0. − ν − ν1 2 2 ∂x21 ∂x22 ∂x1 ∂x22

(8.2.4)

Also, from (5.6.57) and (8.2.3), it follows that ν1 and ν2 satisfy the following algebraic equation: ν 2 + 2Aν + B = 0. (8.2.5) Hence, ν1 = −A −

p

A2 − B, ν2 = −A +

p

A2 − B.

(8.2.6)

Also, it can be seen that the differential equation (8.2.4) can be written in the following equivalent form:      √ √ √ ∂ √ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ϕ = 0. + ν2 − ν2 + ν1 − ν1 ∂x1 ∂x2 ∂x1 ∂x2 ∂x1 ∂x1 ∂x2 ∂x2 (8.2.7) Let us introduce now the parameters µ21 and µ22 defined by equations

µ21 = ν1 , µ22 = ν2 .

(8.2.8)

From (8.2.5), we can conclude that µ1 and µ2 satisfy the algebraic equation µ4 + 2Aµ2 + B = 0.

(8.2.9)

We assume that the initial deformed equilibrium configuration of the body is locally stable. Also, we compare equation (8.2.9) with the relation (5.6.70). Thus, we can conclude that equation (8.2.9) cannot have real roots (see Theorem given in Section 5.6). Consequently, from (8.2.8) we can conclude that the roots ν 1 and ν2 must satisfy one of the following two conditions: (1) Im νj 6= 0 or (2) Im νj = 0 and Re νj < 0, j = 1, 2.

Copyright © 2004 by Chapman & Hall/CRC

(8.2.10)

485

8.2. REPRESENTATION OF THE INCREMENTAL FIELDS

We denote by µ1 , µ2 , µ3 , µ4 , the complex roots of the equation (8.2.9). These roots are determined by the following rule: If Im νj 6= 0, we have ν1 = ν 2 and we take √ √ √ √ (8.2.11) µ 1 = ν 1 , µ 2 = − ν 2 , µ 3 = ν2 = µ 1 , µ 4 = − ν1 = µ 2 .

If Im νj = 0 and Re νj < 0, we take √ √ √ √ µ 1 = ν1 , µ 2 = ν2 , µ 3 = − ν 1 = µ 1 , µ 4 = − ν 2 = µ 2 .

(8.2.12)

Using (8.2.11) and (8.2.12), we can see that in both situations, equation (8.2.7) can be expressed in the following equivalent form: 

∂ ∂ − µ1 ∂x1 ∂x2



∂ ∂ − µ1 ∂x1 ∂x2



∂ ∂ − µ2 ∂x1 ∂x2



∂ ∂ − µ2 ∂x1 ∂x2

µ1 6= µ2 .



ϕ = 0, (8.2.13)

We introduce now the independent complex variables z1 = x 1 + µ 1 x 2 , z 2 = x 1 + µ 2 x 2 .

(8.2.14)

From these relations, we get

z 1 = x 1 + µ 1 x2 , z 2 = x 1 + µ 2 x2 .

(8.2.15)

Since µ1 6= µ2 , we can see now that the differential equation (8.2.13) can be expressed in the following equivalent form: ∂4ϕ = 0. ∂z1 ∂ z 1 ∂z2 ∂ z 2

(8.2.16)

The general solution of this equation is ϕ = ϕ(x1 , x2 ) = f1 (z1 ) + g1 (z 1 ) + f2 (z2 ) + g2 (z 2 ),

(8.2.17)

where fj = fj (zj ) and gj = gj (z j ), j = 1, 2, are arbitrary analytic functions of the complex variables zj and z j , respectively. We recall now that ϕ = ϕ(x1 , x2 ) is a real valued function. Hence, according to (8.1.17), we must have gj (z j ) = fj (zj ), j = 1, 2. Thus, we can conclude that the real displacement potential ϕ = ϕ(x1 , x2 ), satisfying the differential equation (8.2.1) can be expressed in terms of two arbitrary analytic functions f1 = f1 (z1 ) and f2 = f2 (z2 ), by the following relation due to Guz [8.4]: ϕ = ϕ(x1 , x2 )

Copyright © 2004 by Chapman & Hall/CRC

=

f1 (z1 ) + f1 (z1 ) + f2 (z2 ) + f2 (z2 )

=

2Re {f1 (z1 ) + f2 (z2 )} .

(8.2.18)

486

CHAPTER 8. FRACTURE MECHANICS Following Guz [8.3], let us introduce now the analytical functions Fj (zj ) = (ω1122 + ω1212 ) fj (zj ), j = 1, 2.

(8.2.19)

Thus, from (8.2.18), we get ϕ = ϕ(x1 , x2 ) = 2 (ω1122 + ω1212 )

−1

Re {F1 (z1 ) + F2 (z2 )} .

(8.2.20)

We denote, by a prime, the derivative of a function with respect to the variable on which that function depends; i.e. Fj0 (zj ) =

dFj (zj ), j = 1, 2. dzj

(8.2.21)

We return now to the relations (8.2.2) expressing u1, u2 through ϕ, given by equation(8.2.20). Using the chain rule from (8.2.2) and (8.2.20), we get u1 = −2Re {µ1 F100 (z1 ) + µ2 F200 (z2 )} ,



 2 Re ω1111 + µ21 ω2112 F100 (z1 ) + ω1111 + µ22 ω2112 F200 (z2 ) . (ω1122 + ω1212 ) (8.2.22) Now, we introduce the functions Φj = Φj (zj ), j = 1, 2, by the following rule:

u2 =

Φj (zj ) = uj Bj (ω1122 + ω1212 )

−1

Fj00 (zj ),

(8.2.23)

where Bj

= =

ω2222 ω2112 µ2j + ω1111 ω2222 − ω1122 (ω1122 + ω1212 )

−ω1111 ω1221 µ−2 j − ω2112 ω1221 + ω1212 (ω1122 + ω1212 ). (8.2.24)

In order to obtain the second expression of Bj , we have used the fact that µj satisfy the algebraic equation (8.2.9), A and B being given by the relation (5.6.58). Now, taking into account the relation (8.2.22) and the incremental constitutive equations (5.6.43), after long but elementary computations, we get the representation of the incremental fields by two arbitrary analytic complex potentials Φj = Φj (zj ), j = 1, 2 : θ22 = 2Re {Φ01 (z1 ) + Φ02 (z2 )} ,

(8.2.25)

θ21 = −2Re {a1 µ1 Φ01 (z1 ) + a2 µ2 Φ02 (z2 )} ,

(8.2.26)

aj =

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ω2112 ω1122 µ2j − ω1111 ω1212 , Bj µ2j

(8.2.27)

8.2. REPRESENTATION OF THE INCREMENTAL FIELDS

θ12 = −2Re {µ1 Φ01 (z1 ) + µ2 Φ02 (z2 )} ,  θ11 = 2Re a1 µ21 Φ01 (z1 ) + a2 µ22 Φ02 (z2 ) , u1 = 2Re {b1 Φ1 (z1 ) + b2 Φ2 (z2 )} , bj = −

ω1122 + ω1212 , Bj

u2 = 2Re {c1 Φ1 (z1 ) + c2 Φ2 (z2 )} , cj =

ω2112 µ2j + ω1111 . B j µj

487

(8.2.28) (8.2.29) (8.2.30)

(8.2.31) (8.2.32)

(8.2.33)

These relations were first given by Guz [8.4]. Let us assume now that orthotropic initial deformed composite material is in an antiplane state relative to the x1 x2 plane. In this case, the incremental displacement field u3 = u3 (x1 x2 ) must satisfy the differential equation (5.6.28)2 . We assume that the initial deformed equilibrium configuration of the composite is locally stable. Hence, the involved instantaneous elasticities ω1331 and ω2332 must satisfy the restrictions ω1331 , ω2332 > 0. Consequently, the equilibrium equation (5.6.28)2 can be expressed in the equivalent form u3,22 +

ω1331 u3,11 = 0. ω2332

Let us introduce the parameter µ3 defined by the equation r ω1331 . µ3 = i ω2332

(8.2.34)

(8.2.35)

According to the assumed properties of the involved instantaneous elasticities, µ3 is an imaginary number and r ω1331 . (8.2.36) µ3 = −i ω2332

Accordingly, the differential equation (8.2.34) can be written in the following equivalent form:

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488

CHAPTER 8. FRACTURE MECHANICS 

∂ ∂ − µ3 ∂x1 ∂x2



∂ ∂ − µ3 ∂x1 ∂x2



u3 = 0.

(8.2.37)

Let us introduce the complex variable z3 defined by the equation z3 = x 1 + µ 3 x 2 .

(8.2.38)

Using this variable, we can write equation (8.2.37) in the following equivalent form: ∂ 2 u3 = 0. (8.2.39) ∂z3 ∂ z 3

Accordingly, u3 = u3 (x1 , x2 ) = f3 (z3 ) + f3 (z3 ) = 2Ref3 (z3 ),

(8.2.40)

where f3 = f3 (z3 ) is an arbitrary analytic function of the complex variable z3 . To express the involved incremental nominal stresses θ23 and θ13 , we use the incremental constitutive equations (5.6.27). In this way, from (8.2.38) and (8.2.40), we get θ23 = 2ω2332 Reµ3 f30 (z3 ), θ13 = 2ω1331 Ref30 (z3 ).

(8.2.41)

We introduce now the function Φ3 (z3 ) = −ω2332 µ3 f3 (z3 ).

(8.2.42)

Thus, from (8.2.40) and (8.2.41), we get θ23 = −2ReΦ03 (z3 ), θ13 = 2Reµ3 Φ03 (z3 ), −1 u3 = −2ω2332 Reµ−1 3 Φ3 (z3 ).

(8.2.43)

(8.2.44)

The above relations, due to Guz [8.4], express the incremental fields corresponding to the antiplane state, by a single complex potential Φ3 = Φ3 (z3 ) depending on the complex variable z3 = x1 + µ3 x2 . The above representation reduces to that given by Lekhnitski [8.2] if the initial applied deformations and stresses are vanishing. In what follows, we shall use Guz’s results to study the incremental elastic state in an infinite, prestressed composite containing a crack.

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489

8.3. THE OPENING, SLIDING AND TEARING MODES

8.3

The opening, sliding and tearing modes

We assume that the whole space is occupied by an initial deformed orthotropic material. We suppose that the composite contains a crack of length 2a > 0 situated on the x1 axis and has infinite extent in the direction of the x3 axis. As usual in continuum mechanics, we suppose that the crack is represented as a cut having two faces. We assume that the initial deformed equilibrium configuration of the body is homogeneous and locally stable. As was stressed by Guz [8.4] to ensure homogeneity of the initial deformed, prestressed equilibrium configuration, we must suppose that the initial applied ◦ stress σ satisfies the following restriction (see Figure 8.9): ◦

◦

◦

σ 21 = σ 22 = σ 31 = 0.

(8.3.1)

Indeed, as is well known from the classical fracture mechanics (see for instance [8.2]), if the above restriction is not fulfilled, that is, if nonvanishing surface forces act on the faces of the crack, a nonhomogeneous elastic state is produced in the material! With the above given conditions, we shall analyze successively three possible cases, corresponding to the three modes, known in classical fracture mechanics. These problems were analyzed first by Guz [8.4]. The method to obtain the corresponding incremental elastic states is that due to So´os [8.6]. First we formulate and solve the crack problem corresponding to the first, opening mode. We assume that on the two faces of the crack, the incremental tangential forces are vanishing. We assume also that on the upper face of the crack |x 1 | < a, x2 = 0+ is given an incremental normal force, and on the lower face |x1 | < a, x2 = 0− , a symmetrically applied incremental normal force acts as shown in Figure 8.9.

x2

x1 -a

O

a

x3 Figure 8.9: Crack acted by symmetrical normal load.

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490

CHAPTER 8. FRACTURE MECHANICS

According to the assumptions made, the incremental state of the body is a plane state relative to the plane x1 x2 . The involved nominal stresses θ21 and θ22 must satisfy the following boundary conditions on the two faces of the crack (represented as a cut): θ21 (x1, 0+ ) = θ21 (x1, 0− ) = 0 for |x1 | < a, θ22 (x1, 0+ ) = θ22 (x1, 0− ) = −g(x1 ) for |x1 | < a.

(8.3.2) (8.3.3)

In (8.3.3), g = g(x1 ) is the given value of the incremental normal force. We assume that this function satisfies the H condition. The incremental displacement and nominal stresses are vanishing at large distances from the crack. Hence, the following conditions must be satisfied: lim {uα (x1 , x2 ), θαβ (x1 , x2 )} = 0

r→∞

for r =

q

x21 + x22 and α, β = 1, 2.

In order to find the incremental state, we use Guz’s representation (8.2.25)− (8.2.33). First, we analyze the consequences of the above restrictions which must be satisfied at large distances from the crack. According to the relations (8.2.30) and (8.2.32), the complex potentials Φ j = Φj (zj ) must satisfy the condition lim Φj (zj ) = 0, j = 1, 2.

(8.3.4)

|zj |→∞

For further use, we introduce the potentials Ψj (zj ) = Φ0j (zj ), j = 1, 2.

(8.3.5)

Using again the condition imposed at large distances from the crack and taking into account the relations (8.2.25), (8.2.26), (8.2.28), (8.2.29), we can conclude that the function Ψj = Φj (zj ) must satisfy the restriction lim Ψj (zj ) = 0, j = 1, 2.

(8.3.6)

|zj |→∞

Now we return to the condition (8.3.2) and (8.3.3) imposed on the faces of the crack. We use the complex representations (8.2.25) and (8.2.26) of the incremental nominal stresses θ22 and θ21 . Also, we take into account the general relations (8.1.100)−(8.1.103) concerning the properties of the limit values. In this way, we can see that, on the two faces of the crack, the following conditions must be satisfied: −

−

+ a1 µ1 Ψ + 1 (x1 ) + a2 µ2 Ψ2 (x1 ) + a1 µ1 Ψ1 (x1 ) + a2 µ2 Ψ2 (x1 ) = 0,

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8.3. THE OPENING, SLIDING AND TEARING MODES +

+

− a1 µ1 Ψ − 1 (x1 ) + a2 µ2 Ψ2 (x1 ) + a1 µ1 Ψ1 (x1 ) + a2 µ2 Ψ2 (x1 ) = 0, |x1 | < a,

and

491

(8.3.7)

−

−

+ Ψ+ 1 (x1 ) + Ψ2 (x1 ) + Ψ1 (x1 ) + Ψ2 (x1 ) = −g(x1 ),

+

+

− Ψ− 1 (x1 ) + Ψ2 (x1 ) + Ψ1 (x1 ) + Ψ2 (x1 ) = −g(x1 ), |x1 | < a.

(8.3.8)

First, we shall establish the consequences of the boundary condition (8.3.7). Adding and subtracting, we get a1 µ1 Ψ 1 + a 2 µ2 Ψ 2 + a 1 µ1 Ψ 1 + a 2 µ2 Ψ 2


+


− + a1 µ1 Ψ 1 + a 2 µ2 Ψ 2 + a 1 µ1 Ψ 1 + a 2 µ2 Ψ 2 = 0
+ a1 µ1 Ψ 1 + a 2 µ2 Ψ 2 − a 1 µ1 Ψ 1 − a 2 µ2 Ψ 2
− − a1 µ1 Ψ1 + a2 µ2 Ψ2 − a1 µ1 Ψ1 − a2 µ2 Ψ2 = 0.

(8.3.9)

The above relations must be satisfied for |x1 | < a, but this argument was omitted for simplicity. The second equation shows that the analytic function a1 µ1 Ψ1 (z) + a2 µ2 Ψ2 (z) − a1 µ1 Ψ1 (z) − a2 µ2 Ψ2 (z), depending on the complex variable z = x + iy, has a null-jump across the cut L representing the crack. Hence, for this function, we have a homogeneous boundary value problem of type (8.1.66), corresponding to f (t) ≡ 0 on L. The general solution of this problem, which also satisfy the condition (8.3.6), is given by the relation (8.1.73). Consequently, we shall have a1 µ1 Ψ1 (z) + a2 µ2 Ψ2 (z) − a1 µ1 Ψ1 (z) − a2 µ2 Ψ2 (z) = 0 for any z = x + iy. (8.3.10) The first boundary condition (8.3.9) represents a homogeneous RiemannHilbert problem for the analytic function a1 µ1 Ψ1 (z) + a2 µ2 Ψ2 (z) + a1 µ1 Ψ1 (z) + a2 µ2 Ψ2 (z), depending on the complex variable z = x + iy. To see this, we must compare (8.3.9) and equation (8.1.65). In this way we can see that, indeed, f (t) ≡ 0 on L in our Riemann-Hilbert problem. The general solution of our problem, which satisfies also the condition (8.3.4), is given by the equation (8.1.86). Hence, we must have: a1 µ1 Ψ1 (z) + a2 µ2 Ψ2 (z) + a1 µ1 Ψ1 (z) + a2 µ2 Ψ2 (z) = 0 for any z = x1 + ix2 . (8.3.11) From the equations (8.3.10) and (8.3.11), we get a1 µ1 Ψ1 (z) + a2 µ2 Ψ2 (z) = 0, a1 µ1 Ψ1 (z) + a2 µ2 Ψ2 (z) = 0 for any z = x1 + ix2 . (8.3.12)

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492

CHAPTER 8. FRACTURE MECHANICS Particularly, we shall have a1 µ1 Ψ1 (x1 , 0) + a2 µ2 Ψ2 (x1 , 0) = 0, a1 µ1 Ψ1 (x1 , 0) + a2 µ2 Ψ2 (x1 , 0) = 0

for any − ∞ < x1 < ∞.

(8.3.13)

Hence, in our case, the involved upper and lower limits have the same values on the cut L (representing the cut). The above equations can be used to eliminate Ψ2 = Ψ2 (z2 ) from the boundary conditions (8.3.8). In this way, we get −

αΨ+ 1 (x1 ) + αΨ1 (x1 ) = −g(x1 ), +

αΨ− 1 (x1 ) + αΨ1 (x1 ) = −g(x1 ), |x1 | < a,

(8.3.14)

a2 µ2 − a 1 µ1 . a2 µ2

(8.3.15)

with α=

Adding and subtracting in (8.3.14) lead to
−
+ αΨ1 (x1 ) + αΨ1 (x1 ) + αΨ1 (x1 ) + αΨ1 (x1 ) = −2g(x1 ),
−
+ αΨ1 (x1 ) − αΨ1 (x1 ) − αΨ1 (x1 ) − αΨ1 (x1 ) = 0, |x1 | < a.

(8.3.16)

The first relation represents a nonhomogeneous Hilbert-Riemann problem. The general solution, satisfying the conditions imposed at large distances from the crack, is given by the relation (8.1.86). Thus, we get Z g(t)dt X(z1 ) a . (8.3.17) αΨ1 (z1 ) + αΨ1 (z1 ) = − + (t)(t − z) X πi −a

The second relation represents a homogeneous jump problem. The general solution, satisfying the condition imposed at large distances from the crack is given by the relation (8.1.73). Thus, we get αΨ1 (z1 ) − αΨ1 (z1 ) = 0.

(8.3.18)

Now, from the system (8.3.17) and (8.3.18), we can obtain the complex potential Ψ1 = Ψ1 (z1 ). Then, using equation (8.3.12)1 , we get the second complex potential Ψ2 = Ψ2 (z2 ). Thus, after elementary computations, we can conclude that the complex potentials are given by the following relations: Z a2 µ2 X(z1 ) a g(t)dt , + (t)(t − z ) 2πi∆ X 1 −a Z a1 µ1 X(z2 ) a g(t)dt Ψ2 (z2 ) = Φ02 (z2 ) = , + (t)(t − z ) 2πi∆ X 2 −a

Ψ1 (z1 ) = Φ01 (z1 ) = −

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(8.3.19)

493

8.3. THE OPENING, SLIDING AND TEARING MODES where ∆ is given by the equation ∆ = a 2 µ2 − a 1 µ1 .

(8.3.20)

If we use the expression (8.1.74) of the Plemelj function X = X(z) and the relation (8.1.75) giving its limit value X + (t), we can express the relations (8.3.19) in the following equivalent form: √ Z a g(t) a2 − t2 a2 µ2 0 p dt, Ψ1 (z1 ) = Φ1 (z1 ) = − t − z1 2π∆ z12 − a2 −a √ Z a g(t) a2 − t2 a1 µ1 0 p dt. (8.3.21) Ψ2 (z2 ) = Φ2 (z2 ) = t − z2 2π∆ z22 − a2 −a

In the next Section, we shall analyze the most important properties of the obtained solution. For the moment, we shall formulate and solve the crack problem corresponding to the second, sliding mode. More exactly, we assume that the incremental normal forces acting on the two faces of the crack are vanishing. Also, we suppose that on the two faces of the crack, antisymmetrically applied incremental tangential forces act in the direction of the x1 axis, as shown in Figure 8.10.

y

x -a

O

a

Figure 8.10: Crack acted by antisymmetrical tangential load. According to the assumptions made, the incremental state of the body is again a plane state relative to the plane x1 x2 . Now the involved incremental nominal stresses θ21 and θ22 must satisfy the following boundary conditions on the two faces of the crack: θ21 (x1 , 0+ ) = θ21 (x1 , 0− ) = −h(x1 ) for |x1 | < a, θ22 (x1 , 0+ ) = θ22 (x1 , 0− ) = 0 for |x1 | < a.

(8.3.22) (8.3.23)

Here, h = h(x1 ) is the given incremental tangential force. We assume that this function satisfies the H conditions.

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494

CHAPTER 8. FRACTURE MECHANICS

We assume again that the incremental displacement field and the incremental nominal stress is vanishing at large distances from the crack. Hence, the involved complex potentials must satisfy the restrictions (8.3.4) and (8.3.6). In order to solve the problem, we use again Guz’s representation (8.2.25)− (8.2.33). According to (8.2.25), (8.2.26) and (8.3.22), (8.3.23), we can conclude that the complex potentials Ψj = Ψj (zj ), j = 1, 2, 3 must satisfy the following boundary conditions: −

−

+

+

+ Ψ+ 1 (x1 ) + Ψ2 (x1 ) + Ψ1 (x1 ) + Ψ2 (x1 ) = 0,

− Ψ− 1 (x1 ) + Ψ2 (x1 ) + Ψ1 (x1 ) + Ψ2 (x1 ) = 0, |x1 | < a,

(8.3.24)

and −

−

+

+

+ a1 µ1 Ψ + 1 (x1 ) + a2 µ2 Ψ2 (x1 ) + a1 µ1 Ψ1 (x1 ) + a2 µ2 Ψ2 (x1 ) = h(x1 ),

− a1 µ1 Ψ − 1 (x1 ) + a2 µ2 Ψ2 (x1 ) + a1 µ1 Ψ1 (x1 ) + a2 µ2 Ψ2 (x1 ) = h(x1 ), |x1 | < a. (8.3.25)

As before, from the conditions (8.3.24), we can conclude that Ψ1 (z) + Ψ2 (z) = 0, Ψ1 (z) + Ψ2 (z) = 0, for any z = x + iy.

(8.3.26)

Particularly, we shall have Ψ1 (x1 , 0) + Ψ2 (x1 , 0) = 0, Ψ1 (x1, 0) + Ψ2 (x1 , 0) = 0, for − ∞ < x1 < ∞.

(8.3.27)

The above equations can be used to eliminate Ψ2 = Ψ2 (z2 ) from the boundary conditions (8.3.26). Thus, we get −

∆Ψ+ 1 (x1 ) + ∆Ψ1 (x1 ) = −h(x1 ), +

∆Ψ− 1 (x1 ) + ∆Ψ1 (x1 ) = −h(x1 ), |x1 | < a,

∆ being given by equation(8.3.20). Using the same approach as before, finally, we get Z h(t)dt X(z1 ) a Ψ1 (z1 ) = − , 2πi∆ −a X + (t)(t − z1 ) Z X(z2 ) a h(t)dt Ψ2 (z2 ) = . 2πi∆ −a X + (t)(t − z2 )

(8.3.28)

(8.3.29)

Using again (8.1.74) and (8.1.75), the obtained solution can be expressed in the following equivalent form: √ Z a 1 h(t) a2 − t2 0 p Ψ1 (z1 ) = Φ1 (z1 ) = − dt, t − z1 2π∆ z22 − a2 −a √ Z a 1 h(t) a2 − t2 0 p Ψ2 (z2 ) = Φ2 (z2 ) = dt. (8.3.30) t − z2 2π∆ z12 − a2 −a

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8.3. THE OPENING, SLIDING AND TEARING MODES

495

In the next Section, we shall analyze the most important properties of the obtained solution. Now we shall formulate and solve the crack problem corresponding to the third tearing mode. We assume that on the two faces of the crack only tangential forces act, antisymmetrically distributed relative to the plane x2 = 0 and having the direction of the x3 axis. According to the assumptions made, the incremental state of the composite will be an antiplane state relative to the x1 x2 plane. The involved nominal stress θ23 must satisfy the following boundary condition on the two faces of the crack: θ23 (x1 , 0+ ) = θ23 (x1 , 0− ) = −k(x1 ) for |x1 | < a.

(8.3.31)

In (8.3.31), k = k(x1 ) is the given value of the applied incremental tangential surface force. We suppose that this function satisfies the H condition. At large distances from the crack, we must have lim {u3 (x1 , x2 ), θ13 (x1 , x2 ), θ23 (x1 , x2 )} = 0,

r→∞

r=

q

x21 + x22 .

(8.3.32)

In order to obtain the solution, we use Guz’s representation formulas (8.2.43) and (8.2.44) corresponding to the antiplane state. From these relations and from (8.3.31), we can conclude that on the two faces of the crack the complex potential Ψ3 (z3 ) = Φ03 (z3 )

(8.3.33)

must satisfy the following boundary conditions: −

Ψ+ 3 (x1 ) + Ψ3 (x1 ) = k(x1 ), +

Ψ− 3 (x1 ) + Ψ3 (x1 ) = k(x1 ), for |x1 | < a.

(8.3.34)

Also, from (8.3.32), we can see that at large distances from the crack the following condition must be satisfied: lim {Φ3 (z3 ), Ψ3 (z3 )} = 0.

|z3 |→∞

(8.3.35)

Adding and subtracting in (8.3.34), we get the equivalent conditions
−
+ Ψ3 + Ψ3 (x1 ) + Ψ3 + Ψ3 (x1 ) = 2k (x1 ) ,
−
+ (8.3.36) Ψ3 − Ψ3 (x1 ) − Ψ3 − Ψ3 (x1 ) = 0, |x1 | < a.

From the second condition, we get

Ψ3 (z3 ) = Ψ3 (z3 ) for any z3 = x1 + µ3 x2 .

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(8.3.37)

496

CHAPTER 8. FRACTURE MECHANICS Thus, the first condition (8.3.36) gives Z k(t) X(z3 ) a 0 dt. Ψ3 (z3 ) = Φ3 (z3 ) = 2πi −a X + (t)(t − z3 )

(8.3.38)

Obviously, the obtained result can be expressed in the following equivalent form: Z a √ 2 a − t2 k(t) 1 0 p dt. (8.3.39) Ψ3 (z3 ) = Φ3 (z3 ) = t − z3 2π z32 − a2 −a

An analysis of the properties of the obtained solution will be made in the next Section.

8.4

Asymptotic behavior of the incremental fields

We begin our analysis with the case of symmetrically, distributed normal forces, on the face of the crack; i.e. with the first mode. In this case, the involved complex potentials are given by the relation (8.3.19) or (8.3.21). From equations (8.2.26) and (8.3.19)−(8.3.21), we can conclude that θ21 (x1 , 0) = 0 for − ∞ < x1 < ∞.

(8.4.1)

That is, the incremental tangential nominal stress θ21 vanishes on the whole x1 axis containing the crack. A similar result is valuable for an orthotropic material without initial stresses. In order to determine the incremental normal displacement u2 (x1 , x2 ) on the x1 axis for |x1 | > a, we shall use Guz’s representation (8.2.30) and the solution (8.3.21). Thus, we get ∂u2 (x1 , x2 ) = 2Re{c1 Φ01 (z1 ) + c2 Φ02 (z2 )}; ∂x1

hence,

1 c 2 a1 µ1 − c 1 a2 µ2 ∂u2 ) p 2 (x1 , 0) = Re( ∆ ∂x1 π x1 − a 2

Za

−a

√ g(t) a2 − t2 dt. t − x1

(8.4.2)

2 According to (8.4.2), the value of ∂u ∂x1 for |x1 | > a and x2 = 0 depends on −c1 a2 µ2 . the nature of the number c2 a1 µ1 ∆ Consequently, we must evaluate this number. To this end, we use the relations (8.2.24), (8.2.27), (2.2.33) and (2.3.20). After elementary but long manipulations, we get

∆=

Copyright © 2004 by Chapman & Hall/CRC

(µ2 − µ1 )f µ2 − µ 1 . f= 2 2 µ1 µ2 B 1 B 2 µ21 µ22 B1 B2

(8.4.3)

8.4. ASYMPTOTIC BEHAVIOR OF THE INCREMENTAL FIELDS

497

with f

=

ω1122 ω2112 [ω1111 ω2222 − ω1122 (ω1122 + ω1212 ) − ω2222 ω2112 µ1 µ2 ]µ21 µ22 + ω1111 ω1212 {ω1111 ω1221 + [ω1212 (ω1122 + ω1212 ) − ω2112 ω1221 ]µ1 µ2 }.

(8.4.4)

By using the relations (8.2.27) and (8.2.33), we obtain c 1 a2 µ2 − c 2 a1 µ1 = −

(µ2 − µ1 )l (µ1 − µ2 )l = µ1 µ2 B 1 B 2 µ1 µ2 B 1 B 2

(8.4.5)

where l = ω1111 ω2112 (ω1122 + ω1212 )(µ1 + µ2 ).

(8.4.6)

Now, from (8.4.3) and (8.4.5), it results

l c 1 a2 µ2 − c 2 a1 µ1 = µ 1 µ2 . f ∆

(8.4.7)

To determine the value of this number, we must analyze the two possibilities (8.2.11) and (8.2.12). −

If Imνj = 0, we have ν2 = ν 1 and according to (8.2.11), we get µ1 =

√

− √ ν 1 , µ 2 = − ν2 = − µ 1 .

(8.4.8)

Consequently, Re(µ1 + µ2 ) = 0 and Imµ1 µ2 = 0.

(8.4.9)

Thus from (8.4.4) and (8.4.6), we get Imf = 0 and Re l = 0.

(8.4.10)

Now, equation (8.4.7) gives Re

c 1 a2 µ2 − c 2 a1 µ1 = 0. ∆

(8.4.11)

If Imνj = 0 and Reνj < 0, according to (8.2.12) Reµj = 0.

(8.4.12)

Consequently, the relations (8.4.9) are again satisfied. Thus the relation −c2 a1 µ1 is always an imaginary number. (8.4.11) is true, and we can see that c1 a2 µ2 ∆ Hence, according to (8.4.2),

∂u2 (x1 , 0) = 0 for |x1 | > a. ∂x1

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(8.4.13)

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CHAPTER 8. FRACTURE MECHANICS

Accordingly, u2 (x1 , 0) = const. for |x1 | > a. Since u2 (x1 , x2 ) must vanish at large distances from the crack, we can conclude that u2 (x1 , 0) = 0 for |x1 | > a.

(8.4.14)

Thus, we can see that the incremental normal displacement is zero on the line containing the crack, ahead and behind the crack, that is, for |x1 | > a and x2 = 0. Clearly this result is due to the high symmetry of the material, as well as to the high symmetry of its initial stressing and its incremental loading. A similar result holds for an orthotropic material without initial stresses. We continue our analysis with the case of antisymmetrically, distributed tangential forces on the faces of the crack; i.e. with the second mode, when these forces have the direction of the x1 axis. In this case, the involved complex potentials are given by the relations (8.3.24) or (8.3.30). From the equations (8.2.25) and (8.3.29), we can conclude that θ22 (x1 , 0) = 0 for − ∞ < x1 < ∞.

(8.4.15)

Hence, the incremental normal nominal stress θ22 vanishes on the whole x1 axis containing the crack. A similar result is valuable for an orthotropic material without initial stresses. In order to determine the tangential displacement u1 (x1 , x2 ) on the axis x1 for |x1 | > a, we shall use the representation (8.2.30) and the solution (8.3.30). In this way, we get ∂u1 (x1 , x2 ) = 2Re{b1 Φ01 (z1 ) + b2 Φ02 (z2 )}; ∂x1

hence,

1 b2 − b 1 ∂u1 ) p 2 (x1 , 0) = Re( ∆ ∂x1 π x1 − a 2

Za

−a

√ h(t) a2 − t2 dt. t − x1

(8.4.16)

According to (8.2.24) and (8.2.31), we have b2 − b 1 =

(µ2 − µ1 )m , B1 B2

(8.4.17)

where m = (ω1122 + ω1212 )ω2222 ω2112 (µ1 + µ2 ).

(8.4.18)

Thus, the relation (8.4.9)1 and (8.4.18) show that Re m = 0.

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(8.4.19)

8.4. ASYMPTOTIC BEHAVIOR OF THE INCREMENTAL FIELDS

499

Using (8.4.5) and (8.4.17), we get

m b2 − b 1 = µ21 µ22 . f ∆

(8.4.20)

Hence, using (8.4.9)2 , (8.4.10)1 and (8.4.19), we can conclude that Re

Consequently, (8.4.16),

b2 −b1 ∆

b2 − b 1 = 0. ∆

(8.4.21)

is always an imaginary number. That is, according to ∂u1 (x1 , 0) = 0 for |x1 | > a. ∂x1

(8.4.22)

Accordingly, u1 (x1 , 0) = const. for |x1 | > a. Since u1 (x1 , 0) must vanish for a large distance from the crack, we must have u1 (x1 , 0) = 0 for |x1 | > a .

(8.4.23)

Thus, we can see that the tangential displacement is vanishing on the whole line containing the crack, ahead and behind the crack, that is, for |x1 | > a and x2 = 0. As before, the above result is due to the high symmetry of the material, of its initial deformation and of its incremental loading. A similar result is true for an orthotropic material without initial stresses.We finish our analysis considering the case of antisymmetrically distributed tangential forces on the face of the crack, when these forces have the direction of the x3 axis; i.e. with the third mode. The involved complex potential is given in this case by the relations (8.3.38) or (8.3.39). From (8.3.44), we can conclude that

1 1 ∂u3 Re p 2 (x1 , 0) = − ω2332 ∂x1 µ3 x 1 − a 2

Za

−a

√ k(t) a2 − t2 dt. t − x1

(8.4.24)

At the same time, equation (8.2.35) shows that µ3 is an imaginary number. Accordingly, from (8.4.24), we conclude that ∂u3 (x1 , 0) = 0 for |x1 | > a. ∂x1

(8.4.25)

Hence, u3 (x1 , 0) = const. for |x1 | > a. Since u3 must vanish at large distances from the crack, it results that u3 (x1 , 0) = 0 for |x1 | > a.

(8.4.26)

Thus, the tangential displacement is vanishing on the whole line containing the crack, ahead and behind the crack, that is, for |x1 | > a and x2 = 0. A similar result holds for an orthotropic material without initial stresses. In both cases, the

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CHAPTER 8. FRACTURE MECHANICS

result is due to the high symmetry of the material, of its initial deformation and of its initial loading. In the following, we shall analyze the asymptotical behavior of the incremental field in the neighborhood of the crack tips. This analysis is important since, in this way, the relationship between the stresses and the input energy rates in crack extension may be established; hence, the crack stability can be examined (see Sih and Leibowitz[8.3], and Guz[8.4]). The incremental fields distribution around the (right) tip can be obtained by letting x1 = a + r cos ϕ, x2 = r sin ϕ

(8.4.27)

and by assuming that r is small in comparison with the half crack length a. The polar coordinate r and ϕ designate, respectively, the radial distance from the considered crack tip and angle between the radial line and the line extending the crack, as shown in Figure 8.11.

x2

(x1,x2) r x1

-a

O

a

Figure 8.11: Asymptotic behavior. As the first case, we consider the first mode. The solution is given by equations (8.3.21) and the Guz’s representation formulas (8.2.25)−(8.2.32) hold. In a small neighborhood of the crack tip, x1 ≈ a, x2 ≈ 0, we have z1 = z 2 ≈ a

(8.4.28)

and Plemelj functions may be approximated by q

zj2 − a2 =

where χj (ϕ) =

Copyright © 2004 by Chapman & Hall/CRC

p

√

2arχj (ϕ), j = 1, 2,

cos ϕ + µj sin ϕ, j = 1, 2.

(8.4.29)

(8.4.30)

8.4. ASYMPTOTIC BEHAVIOR OF THE INCREMENTAL FIELDS

501

Taking into account these relations for the asymptotic values of the complex potentials (8.3.21), we get the following expressions: KI a 2 µ 2 1 , Ψ1 (z1 ) = √ 2 2πr ∆ χ1 (ϕ)

KI a 1 µ 1 1 . Ψ2 (z2 ) = − √ 2 2πr ∆ χ2 (ϕ) In the above relations,

1 KI = √ πa

Za

−a

g(t)

r

a+t dt a−t

(8.4.31)

(8.4.32)

is the stress intensity factor, corresponding to the first mode. This quantity has the same expression as in the classical theory of brittle fracture of elastic materials without initial stresses. Now, by using the relations (8.4.31) and Guz’s representation formulas (8.2.25) −(8.2.32), we get the asymptotic expressions of the incremental fields corresponding to the first mode,

a1 µ1 1 a2 µ2 KI }, − Re { θ22 = √ ∆ χ1 (ϕ) χ2 (ϕ) 2πr 1 1 a1 a2 µ1 µ2 KI }, − { Re θ21 = − √ χ χ (ϕ) ∆ 2πr 2 (ϕ) 1 a2 µ1 µ2 a1 KI { Re − θ12 = − √ }, ∆ χ1 (ϕ) χ2 (ϕ) 2πr KI a1 a2 µ1 µ2 µ1 µ2 θ11 = √ Re { − }, ∆ χ1 (ϕ) χ2 (ϕ) 2πr r r 1 u1 = 2 KI Re {b1 a2 µ2 χ1 (ϕ) − b2 a1 µ1 χ2 (ϕ)}, 2π ∆ r r 1 u2 = 2 KI Re {c1 a2 µ2 χ1 (ϕ) − c2 a1 µ1 χ2 (ϕ)}. 2π ∆ To get the last two relations, we have used the equations r r a2 µ2 Φ1 (z1 ) = KI χ1 (ϕ), 2π ∆ r r a1 µ1 Φ2 (z2 ) = −KI χ2 (ϕ), 2π ∆

(8.4.33)

(8.4.34) (8.4.35) (8.4.36) (8.4.37) (8.4.38)

(8.4.39)

that follow from (8.4.27) and (8.4.31), since Φ0j (zj ) = Ψj (zj ) and dzj = drχ2j (ϕ), j = 1, 2.

Copyright © 2004 by Chapman & Hall/CRC

(8.4.40)

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CHAPTER 8. FRACTURE MECHANICS

The asymptotic expressions (8.4.33)−(8.4.38) show that the incremental fields in a prestressed body have the same asymptotic behavior as the corresponding fields in an elastic body without initial stresses. This important result was obtained using different approaches by Guz [8.5] and by So´os [8.6]. Particularly, since the incremental nominal stress behaves like √1r in the neighborhood of the crack tip, the total incremental elastic energy is bounded in any finite domain containing the crack. This fact is essential in crack stability analysis based on Griffith’s type criterion of crack propagation. As a second case, we consider the second mode. In this case, the solution given by equations (8.3.30) and the representation formulas (8.2.31) and (8.2.32) hold. Using again the relations (8.4.28)−(8.4.30) for the asymptotic values of the complex potentials (8.3.30), we get the following expressions:

KII 1 1 , (8.4.41) Ψ1 (z1 ) = √ 2 2πr ∆ χ1 (ϕ) KII 1 1 , Ψ2 (z2 ) = − √ 2 2πr ∆ χ2 (ϕ) where r Z a a+t 1 dt (8.4.42) h(t) KII = √ a−t πa −a is the stress intensity factor, corresponding to the second mode. Again, this quantity has the same expression as that encountered in the theory of brittle fracture of elastic materials without initial stresses. Now, using the representation formulas (8.2.25)−(8.2.32), we obtain the asymptotic expression of the incremental fields corresponding to the second mode:

1 1 1 KII }, − Re { θ22 = √ ∆ χ1 (ϕ) χ2 (ϕ) 2πr a2 µ2 1 a1 µ1 KII }, − Re { θ21 = − √ ∆ χ1 (ϕ) χ2 (ϕ) 2πr KII 1 µ1 µ2 θ12 = − √ Re { − }, ∆ χ (ϕ) χ 2πr 1 2 (ϕ) KII 1 a1 µ21 a2 µ22 θ11 = √ Re { − }, 2πr ∆ χ1 (ϕ) χ2 (ϕ) r r 1 u1 = 2 KII Re {b1 χ1 (ϕ) − b2 χ2 (ϕ)}, 2π ∆ r r 1 u2 = 2 KII Re {c1 χ1 (ϕ) − c2 χ2 (ϕ)}. 2π ∆ To get the last two relations, we used equations r r 1 Φ1 (z1 ) = KII χ1 (ϕ), 2π ∆

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(8.4.43)

(8.4.44) (8.4.45) (8.4.46) (8.4.47) (8.4.48)

(8.4.49)

8.4. ASYMPTOTIC BEHAVIOR OF THE INCREMENTAL FIELDS Φ2 (z2 ) = −KII

r

503

r 1 χ2 (ϕ), 2π ∆

which follows from (8.4.27) and (8.4.41). The asymptotic expressions (8.4.43)−(8.4.48) show again that the incremental fields in the prestressed body have the same asymptotic behavior as the corresponding fields, in an elastic body without initial stresses. As the third and last case, we consider the situation corresponding to tangential loading perpendicular to the crack direction, named also the third mode. In this case, the solution is given by equations (8.3.39) and the representation formulas (8.2.43), (8.2.44) hold. From (8.3.39) and (8.4.28)−(8.4.30), we get the following asymptotic expression of the involved complex potential:

1 KIII , Ψ3 (z3 ) = − √ χ 2 2πr 3 (ϕ)

where χ3 (ϕ) =

and KIII

p

1 =√ πa

cos ϕ + µ3 sin ϕ

Za

−a

k(t)

r

a+t dt a−t

(8.4.50)

(8.4.51)

(8.4.52)

is the stress intensity factor corresponding to the third mode. This quantity has the same expression as in the classical theory of brittle fracture of elastic materials without initial stresses. Now, from (8.2.43) and (8.2.44), we obtain the asymptotic expressions of the incremental fields corresponding to the third mode:

1 KIII , Re θ23 = √ χ3 (ϕ) 2πr µ3 KIII , Re θ13 = − √ χ3 (ϕ) 2πr r χ (ϕ) r 2 . KIII Re 3 u3 = µ3 ω2332 2π

To get the last equation, we have used the relation r r χ3 (ϕ), Φ3 (z3 ) = −KIII 2π

(8.4.53)

(8.4.54)

(8.4.55)

(8.4.56)

which follows from (8.4.28), (8.4.29) and (8.4.51). Again, from (8.4.53)−(8.4.55), we can see that the asymptotic behavior of the incremental fields in the prestressed material is the same as in a body without initial stresses.

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Until now, we tacitly have assumed that the quantity ∆, defined by the relation (8.3.20), is not vanishing. Certainly, this is the case if the body is not initially deformed since we assume the stress-free reference configuration to be locally stable. However, if the material is initially deformed, some critical values ◦ of the initial stress σ 11 for which ∆ can be zero may exist. If the initial applied ◦ stress σ 11 converges to this critical value, all incremental fields increase unbounded. This phenomenon was investigated by Guz [8.4]. Taking into account known results from the Eulerian theory of the elastic string excited by periodical force, we say that when ∆→0 (8.4.57) the resonance phenomenon occurs. Following Guz [8.4], we shall analyze now if the occurrence of the resonance is possible for a fiber-reinforced composite material. We assume that the crack has the direction of the reinforcing fibers which are supposed to be parallel to the x 1 axis. Also, in accordance with the basic hypothesis (8.3.1), we suppose that on the considered loading path ◦

σ 22 = 0.

(8.4.58)

Hence, the initial applied loading forces are in the fibers direction, being parallel to the crack. We consider only the first and the second mode, since, for the third mode, the occurrence of the resonance is not possible, as it can be seen examining equation (8.3.39). We recall that in the considered case, the involved instantaneous elasticities are given by equations (5.6.47). Thus, taking into account (8.4.58), we get ◦

ω1111

=

C11 + σ 11 , ω2222 = C22 ,

ω1122 ω1212

= =

ω2211 = C12 , ω2121 = ω2112 = C66 ,

ω1221

=

C66 + σ 11 .

◦

(8.4.59)

Hence, according to (5.6.58), the coefficients A and B of equation (8.2.9) are ◦

◦

(C11 + σ 11 )C22 + (C66 + σ 11 )C66 − (C12 + C66 )2 , 2A = C22 C66 ◦

(8.4.60)

◦

(C11 + σ 11 )(C66 + σ 11 ) . B= C22 C66

We recall that for the considered fiber-reinforced composite, we have 0<ε≡

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C66 << 1. C11

(8.4.61)

8.4. ASYMPTOTIC BEHAVIOR OF THE INCREMENTAL FIELDS

505

In what follows, we assume that the prestressed composite is internally stable. Hence, we have (see Section 6.1) ◦

◦

C11 + σ 11 , C66 + σ 11 > 0.

(8.4.62)

Consequently, according to (8.4.60)2 , B > 0.

(8.4.63)

We suppose also that the quantity A is positive; i.e. A > 0.

(8.4.64)

A2 > B.

(8.4.65)

and the restriction is satisfied:

For an isotropic material, the inequalities (8.4.62)−(8.4.65) are certainly ful◦ filled if σ 11 = 0 and the stress-free reference configuration is locally stable! Hence, ◦ we can assume that they are true also in the neighborhood of σ 11 and, for the most part, orthotropic composites. Examining the relations (8.2.6), we can see that if the restrictions (8.4.63)− (8.4.65) are satisfied, the roots ν1 and ν2 of equation (8.2.5) are real, distinct, and negative numbers. Consequently, the distinct roots µ1 and µ2 of equation (8.2.9) must be selected in accordance with the rule (8.2.12). Thus, we shall have µ1

=

µ2

=

q

p A + A2 − B, q p i A − A2 − B.

i

(8.4.66)

From (8.4.60)2 and (8.4.66), we get √

µ1 µ2 = − B = −

s

◦

◦

(C11 + σ 11 )(C66 + σ 11 ) . C22 C66

(8.4.67)

◦ cr

We try to find the critical value σ 11 of the applied load for which resonance can occur. To do this, we recall equation (8.4.3) giving the quantity ∆. Since µ1 6= µ2 , ∆ can vanish if and only if f = 0,

(8.4.68)

the expression of f being given by equation (8.4.4). Using the relations (8.4.59) ◦ and (8.4.67), we can see that the relation (8.4.68) can be true if and only if σ 11

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satisfies the following equation: 0

where

=

2 C12 C12 C12 + ε) ( − C11 C22 C11 C11 √ C12 p (1 + x)(ε + x)}(1 + x)(ε + x) + ε√ C11 C22 +ε(1 + x){(1 + x)(ε + x) r √ C12 p C11 ) (1 + x)(1 + εx)}, (x − + ε C11 C22

{(1 + x)

◦

σ 11 . x= C11

(8.4.69)

(8.4.70)

Since ε has the property (8.4.61), we solve the above equation using our iterative method and looking for a root x having the following form: x = x0 + εx1 + ε2 x2 + ε3 x3 .

(8.4.71)

Introducing (8.4.71) in (8.4.64) and neglecting terms of order ε4 and higher, we get x0 = 0, x1 = −1, x2 = 0, x3 =

2 2 C11 C22 C11 2 )2 . C22 (C11 C22 − C12

(8.4.72) ◦ cr

Introducing (8.4.72) in (8.4.70) and using (8.4.61), for the critical value σ 11 leading to resonance, we get the following value: ◦ cr

σ 11 = −C66 {1 −

2 2 2 C11 C22 C66 2 )2 }. C11 C22 (C11 C22 − C12

(8.4.73)

We compare now this equation with the relation (6.1.51) giving the critical ◦ cs stress σ 11 for which surface instability of the fiber-reinforced composite occurs. Thus, we have ◦ cs ◦ cr σ 11 = σ 11 . (8.4.74) Thus, we have established the following important result due to Guz [8.4]: in a fiber-reinforced composite, surface instability and resonance occur for the same ◦ critical value of the compressive force σ 11 acting in the direction of the reinforcing fibers. As for an elastic spring, the occurrence of the resonance phenomenon leads to dangerous situations which must be avoided limiting drastically the magnitude of the applied compressive force. We recall that in a fiber-reinforced composite, ◦ cs ◦ cr material stress, having magnitude corresponding to σ 11 = σ 11 , produce infinitesimal deformations; hence, the occurrence of resonance is really possible! Concerning

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8.5. GRIFFITH’S CRITERION AND CRACK PROPAGATION

the case of surface instability, the actual possibility of the resonance for a fiberreinforced composite is a direct consequence of the existing internal structure of such kind of materials! In a linearly elastic isotropic material, the occurrence of resonance, for infinitesimal deformations, is not possible. In the next Section, we shall analyze the strong connection existing between resonance and crack propagation discovered by Guz [8.4].

8.5

Griffith’s criterion and crack propagation

In the preceding Section, we have introduced the stress intensity factors KI , KII and KIII corresponding to the three modes also named opening, sliding and tearing modes, respectively, and shown in Figure 8.12.

Sliding

Opening

Tearing

Figure 8.12: Basic crack models and extensions. The mechanical and physical interpretations of the concept of the stress intensity factor become more clear when the mathematical relationships between the rate of input work (or strain energy release rate) into the fracture process and the singular (asymptotic) elasticity solution in the neighborhood of the crack tip are established. To this end, following Sih and Leibowitz [8.3], let us suppose that our crack of length 2a in a plate of unit thickness is extended by the increment δa at both ends of the crack. This extension creates new surfaces of the crack with 4δa as a gain in area and, hence, the surface energy is increased by 4γδa, γ representing the specific surface energy of the body. Let us denote by U (a) the elastic energy of the body when the length of the crack is 2a, and let U (a + δa) be its elastic energy when the length of the crack is 2(a + δa). According to Griffith’s criterion, a necessary condition for the crack to propagate is that the change in strain energy must satisfy the inequality U (a) − U (a + δa) ≥ 4γδa.

(8.5.1)

The function U (a + δa) may be expanded in Taylor’s series. Neglecting terms of order higher than δa, we get

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∂U (a)δa. ∂a The strain energy release rate G(a) is defined by the equation δU = U (a) − U (a + δa) = −

(8.5.2)

1 ∂U (a). 2 ∂a

(8.5.3)

G(a) = −

Consequently, we get δU = U (a) − U (a + δa) = −2G(a)δa,

(8.5.4)

and Griffith’s crack propagation or crack instability condition (8.5.1) takes the form G(a) ≥ 2γ.

(8.5.5)

The above inequality represents Griffith’s energy criterion for brittle fracture. The energy release rate G(a) may by regarded as the force tending to open the crack. As we shall see, its evaluation requires only the knowledge of the incremental nominal stress and displacements near the crack tips. In the sequel, the ∂U and G(a) values for the three basic modes of crack extension will be distinguished by the subscripts I, II and III as we had in the case of the stress intensity factors. We begin our analysis with the first opening mode (see Sih and Leibowitz [8.3]). In this case, the problem is symmetric about the crack along the x 1 axis. As we already know, only normal nominal stresses shall be present in elements on the x1 axis as shown in Figure 8.13.

x2 CRACK OPENING 22

u2

u2

r

x1

o 22

t a

t da

Figure 8.13: Segment of crack opening. Let the crack be extended by an amount δa. Conceptually, it may be visualized that the elastic body has been cut along the segment from x1 = a up

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8.5. GRIFFITH’S CRITERION AND CRACK PROPAGATION

509

to x1 = a + δa without releasing the stresses along the cut and then reducing their intensity by infinitesimally small increments until the segment of length δa is stress-free and is represented by the dotted shape in Figure 8.13. The strain energy released is the work done in this process by the incremental nominal stress θ22 (δa−t0 , 0) acting by incremental displacement u2 (t, 0+ ), provided that δa is very small such that in the limit as δa → 0, the conditions u2 (t, 0+ ) → u2 (δ − t0 , 0+ ) and t → t0 are fulfilled. The variation δUI of the elastic energy is due to the work done at both ends of the crack. Hence, we get

δUI = −2

Zδa 0

θ22 (δa − t, 0)u2 (t, 0+ )dt.

(8.5.6)

In the above relation, the factor −2 appears since the crack has two faces and the minus sign is present because the segment δa is relaxed! Comparing the general relation (8.5.4) and the last equation, we obtain the following relation which must be satisfied by the strain energy release rate G I (a) corresponding to the first mode:

GI (a)δa =

Zδa 0

θ22 (δa − t, 0)u2 (t, 0+ )dt.

(8.5.7)

In order to evaluate this integral, we use the asymptotic expressions of the incremental fields obtained in the preceding Section. The incremental nominal stress θ22 (δa − t, 0) can be obtained using equations (8.4.30) and (8.4.33) and taking r = δa − t > 0, ϕ = 0. We obtain θ22 (δa − t, 0) = p

KI . 2π(δa − t)

(8.5.8)

The incremental normal displacement u2 (t, 0+ ) of the upper face of the crack can be obtained using equations (8.4.30) and (8.4.38), and taking r = t > 0, ϕ = π (see Figure 8.13). Thus, we obtain r c 1 a2 µ2 − c 2 a1 µ1 t + ). (8.5.9) Re(i u2 (t, 0 ) = 2KI ∆ 2π

We use now the relation (8.4.7) and get r l t + Re(iµ1 µ2 ). u2 (t, 0 ) = ±2KI f 2π

(8.5.10)

According to equations (8.4.9)2 and (8.4.10), µ1 µ2 and f are real numbers and l is an imaginary number. Consequently, (8.5.10) becomes

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CHAPTER 8. FRACTURE MECHANICS

µ1 µ2 u2 (t, 0 ) = ±2KI f +

r

t Re(il). 2π

(8.5.11)

Let us introduce now the real number b l = ±il.

Thus, for u2 (t, 0+ ), we get

(8.5.12)

b l u2 (t, 0 ) = 2KI µ1 µ2 f +

r

t . 2π

(8.5.13)

Consequently, with (8.5.8) and (8.5.13), equation (8.5.7) becomes

Z l KI2 µ1 µ2b GI (a)δa = f π

δa

0

It is easy to see that

Zδa r 0

r

t dt. δa − t

πδa t . dt = 2 δa − t

(8.5.14)

(8.5.15)

Thus, from (8.5.14) and (8.5.15), finally, we get GI (a) =

l KI2 µ1 µ2b . f 2

(8.5.16)

Using a similar approach, we can conclude that the strain energy release rate GII (a) corresponding to the second mode must satisfy the following equation: GII (a)δa =

Zδa 0

θ21 (δa − t, 0)u1 (t, 0+ )dt.

(8.5.17)

The incremental nominal tangential stress θ21 (δa−t, 0) can be obtained using equations (8.4.30) and (8.4.44), and taking r = δa − t > 0, ϕ = 0. We get θ21 (δa − t, 0) = p

KII . 2π(δa − t)

(8.5.18)

The incremental tangential displacement u1 (t, 0+ ) of the upper face of the crack can be obtained using equations (8.4.30), (8.4.40) and taking r = t > 0, ϕ = π. Thus, we obtain r b1 − b 2 t + ). (8.5.19) Re(i u1 (t, 0 ) = 2KII ∆ 2π

Copyright © 2004 by Chapman & Hall/CRC

8.5. GRIFFITH’S CRITERION AND CRACK PROPAGATION We use now the relation (8.4.20) and get r m t + Re(iµ21 µ22 ). u1 (t, 0 ) = −2KII f 2π

511

(8.5.20)

According to equations (8.4.9)2 , (8.4.10) and (8.4.19), µ21 µ22 and f are real numbers and m is an imaginary number. Consequently, (8.5.20) becomes u1 (t, 0+ ) = −2KII

µ21 µ22 Re(im). f

(8.5.21)

Let us introduce the real number m b = −im.

Thus, for u1 (t, 0+ ), we get +

u1 (t, 0 ) =

(8.5.22)

m b 2KII µ21 µ22

f

r

t . 2π

(8.5.23)

Consequently, with (8.5.18) and (8.5.23), (8.5.17) becomes

b K 2 µ2 µ2 m GII (a)δa = II 1 2 f π

Zδa r 0

t dt. δa − t

(8.5.24)

Using again (8.5.7) and (8.5.15), from (8.5.24), we obtain GII (a) =

2 µ21 µ22 m b KII . f π

(8.5.25)

Finally, let us analyze the third mode. Similar reasoning leads to the conclusion that the involved strain energy release rate GIII (a) must satisfy the equation GIII (a)δa =

Zδa 0

θ23 (δa − t, 0)u3 (t, 0+ )dt.

(8.5.26)

The incremental nominal tangential stress θ23 (δa−t, 0) can be obtained using (8.4.51) and (8.4.53) and taking r = δa − t > 0, ϕ = 0. We get θ23 (δa − t, 0) = p

KIII . 2π(δa − t)

(8.5.27)

The incremental tangential displacement u3 (t, 0+ ) of the upper face of the cut can be obtained using (8.4.51) and (8.4.55) and taking r = t > 0, ϕ = π. It results r 1 t KIII + (8.5.28) Re( ). u3 (t, 0 ) = 2 µ3 ω2332 2π

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The root µ3 is given by equation (8.2.36). Using this relation from (8.5.28), we get r t 2KIII + . (8.5.29) u3 (t, 0 ) = √ ω2332 ω1331 2π

Now, (8.5.26), (8.5.27) and (8.5.29) jointly give K2 GIII (a)δa = √ III π ω2332 ω1331

Zδa r 0

t dt. δa − t

(8.5.30)

Taking into account again equation (8.5.15), finally, we obtain K2 . GIII (a) = √ III 2 ω2332 ω1331

(8.5.31)

The total strain energy release rate G(a) is the sum of the release rates corresponding to the three basic modes. Hence, according to (8.5.16), (8.5.25) and (8.5.31), we get G(a) =

1 1 2 1 2 2 2 1 2 (ω2332 ω1331 )− 2 . µ1 µ2 mf b −1 + KIII K µ1 µ2 b lf −1 + KII 2 2 2 I

(8.5.32)

The above result was first obtained by Guz [8.4]. According to Griffith’s criterion (8.5.5), crack instability occurs and crack propagation starts if the applied incremental forces satisfy the propagation condition: 2 2 2 2 (ω2332 ω1331 )− 2 = 4γ. b −1 + KIII µ1 µ2 mf KI2 µ1 µ2b lf −1 + KII 1

(8.5.33)

We recall that the stress concentration factors KI , KII , KIII depend on the crack length and the normal and tangential incremental forces applied on the two faces of the crack. The roots µ1 , µ2 and the coefficients f, b l, m, b ω2332 , ω1331 depend on the elastic properties of the material and on the initial applied stresses. In a prestressed material, the specific surface energy of the material γ > 0 can depend also on the initial applied stresses. This dependence is not known at present. In all that follows, we assume that the dependence of the specific surface energy γ > 0 on the initial applied stresses is negligible, and γ depends only on the considered material. Clearly, this is a relatively strong assumption, and its validity must be checked by further research. To simplify the analysis of the consequences of the Griffith’s propagation criterion (8.5.33), we suppose in the following that the applied external incremental forces have constant values; i.e.: g(x1 )

=

p = const. > 0,

h(x1 ) k(x1 )

= =

τ = const. > 0, κ = const. > 0 for |x1 | < a.

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(8.5.34)

8.5. GRIFFITH’S CRITERION AND CRACK PROPAGATION

513

The assumed case is frequently encountered in a practical application of fracture mechanics. In the considered case, the stress concentration factors can be easily evaluated using the relation Za r a+t dt = aπ. (8.5.35) a−t −a

Using equations (8.4.32), (8.4.42) and (8.4.52), we get √ πap, KI = √ πaτ, KII = √ πaκ. KIII =

(8.5.36)

Introducing these values in equation (8.5.33), Griffith’s propagation criterion takes the following form: 1 p2 µ 1 µ 2 b b −1 + κ2 (ω2332 ω1331 )− 2 = 4γ(πa)−1 . lf −1 + τ 2 µ21 µ22 mf

(8.5.37)

◦

Also, we suppose that on the given loading path only σ 11 is nonvanishing; i.e. we assume that ◦ ◦ σ 22 = σ 33 = 0. (8.5.38) We recall that in this case the instantaneous elasticities involved in the crack propagation problem and concerning the first and the second modes are given by equation (8.4.59). Also, according to the equation (5.6.22) and (8.5.38), the instantaneous elasticities involved in the crack propagation problem, concerning the third mode, are given by the relations ω1331 ω2332

= =

◦

C55 + σ 11 , C44 .

(8.5.39)

As usual, we suppose that the stress-free reference configuration of the material is locally stable and its initial deformed equilibrium configuration is internally (structural) stable. Let us assume that the applied incremental tangential forces are vanishing; i.e. τ = κ = 0. (8.5.40) In this case, Griffith’s propagation criterion (8.5.37) becomes p2 =

4γf

. πaµ1 µ2b l

(8.5.41)

We suppose that the conditions (8.4.64)−(8.4.65) are fulfilled. Hence, the roots µ1 and µ2 are given by the relation (8.4.66). Consequently, µ1 µ2 satisfies equation (8.4.67) and q q p p 2 (8.5.42) µ1 + µ2 = i{ A + A − B + A − A2 − B},

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CHAPTER 8. FRACTURE MECHANICS

or, equivalently,

√ µ1 + µ 2 = i 2

Now, from (8.4.6) and (8.5.12), it follows

q

A+

√

B.

√ b l = − 2ω1111 ω2112 (ω1122 + ω1212 )

Thus, using (8.4.59), we get

q

√ ◦ b l = − 2(C11 + σ 11 )C66 (C12 + C66 )

(8.5.43)

A+

q

√

A+

B.

√

(8.5.44)

B.

(8.5.45)

Thus, from (8.4.66), (8.5.41) and (8.5.45) for the critical value pc for which the crack propagation starts in the first mode, we get the following expression: ◦

p2c = p2c (σ 11 ) = √

4γf

√ . √ p 2πa(C11 + σ 11 )C66 (C12 + C66 ) B A + B ◦

(8.5.46)

◦

We recall that A and B depend on σ 11 and their expressions are given by the ◦ relations (5.6.59). Also, f depends on σ 11 , its expression being given by the relation ◦ (8.4.4). Thus, the relation (4.5.46) shows that the critical value p c = pc (σ 11 ) of the incremental normal critical stress for which the crack propagation starts in the first mode, depends on the material parameters and on the initial applied stress ◦ σ11 . Also, as it results from (8.5.46) ◦

◦

pc (σ 11 ) → 0 when f (σ 11 ) → 0.

(8.5.47)

◦ cp

◦

Let us denote by σ 11 the critical value of the initial applied stress σ 11 for ◦ which f = f (σ 11 ) vanishes; i.e. for which we have ◦ cp

f (σ 11 ) = 0.

(8.5.48) ◦ cp

According to (4.5.46), for the critical value σ 11 , we have also ◦ cp

pc (σ 11 ) = 0.

(8.5.49) ◦ cp

◦

Hence, if the initial applied stress σ 11 reaches its critical value σ 11 , the crack becomes completely unstable and its propagation can start without any incremental normal force applied at the two faces of the crack! Comparing equations (8.4.68) and (8.5.48), we can conclude that the critical ◦ cp value σ 11 really exists. Moreover, as equation (8.4.74) shows, we have ◦ cp

◦ cr

◦ cs

σ 11 = σ 11 = σ 11 .

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(8.5.50)

8.5. GRIFFITH’S CRITERION AND CRACK PROPAGATION

515

Hence, complete instability of the crack, resonance phenomenon and surface instability of a fiber-reinforced composite material occur for the same critical value of the initial applied compressive force, acting in the direction of the reinforcing fibers. This fundamental result is due to Guz [8.4]. To avoid the dangerous situations leading to a completely unstable crack, the initial applied compressive force acting in the direction of the reinforcing fibers must be drastically limited. Again, as in the case of surface instability or in the case of the resonance, the occurrence of a completely unstable crack in a prestressed fiber-reinforced composite is due to the existing internal structure of such a material. As is easy to see, complete instability of a crack in a prestressed isotropic material cannot appear in stress domains leading to infinitesimal deformations. We have encountered similar situations analyzing surface instability or resonance for prestressed isotropic material. We observe that in presenting the above important result, we have taken into account that the quantity standing in the denominator of the expression (8.5.46) ◦ ◦ ◦ cp of pc (σ 11 ) is now vanishing ◦ cp for σ 11 = σ 11 . This can be seen using the relation (8.4.73) telling us that σ 11 < C66 << C11 . Let us assume now that the applied incremental normal stress and the applied incremental tangential stress, corresponding to the second mode, are vanishing; i.e. p=κ=0.

(8.5.51)

In this case, the propagation condition (8.5.37) becomes τ2 =

4γf . πaµ21 µ22 m b

(8.5.52)

Again µ21 µ22 is given by equation (8.4.67) and, from (8.4.18), (8.5.22) and (8.5.43), it results q √ √ (8.5.53) m b = 2ω2222 ω2112 (ω1122 + ω1212 ) A + B.

or using (8.4.59),

m b =

√

2C22 C66 (C12 + C66 )

q

A+

√

B.

(8.5.54)

Thus, from (8.4.67) and (8.5.52) for the critical value τc for which the crack propagation starts in the second mode, we get the following expression: 4γf

◦

τc2 = τc2 (σ 11 ) √

We can see again that

2πaC22 C66 (C12 + C66 )B ◦

◦

τc (σ 11 ) → 0 if f (σ 11 ) → 0.

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p

A+

√ . B

(8.5.55)

(8.5.56)

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CHAPTER 8. FRACTURE MECHANICS ◦

Hence, when the initial applied compressive stress σ 11 reaches its critical value

◦ cp

◦ cr

◦ cs

σ 11 = σ 11 = σ 11 ,

(8.5.57)

the crack becomes completely unstable and its propagation can start without any incremental tangential force applied at the two faces of the crack and in the direction of the crack. As equations (8.5.50) and (8.5.57) show, the critical value of the compressive force leading to a complete instability of the crack is the same for the first and for the second mode. This important result was established by Guz [8.4]. Finally, let us assume that only the applied incremental tangential stresses corresponding to the third mode are nonvanishing; i.e. p = τ = 0.

(8.5.58)

In this case, the propagation condition (8.5.37) becomes √ 4γ ω1331 ω2332 . κ2 = πa

(8.5.59) ◦

Taking into account the relation (8.5.59) and denoting by κc = κc (σ 11 ) the critical value of the applied tangential stress leading to crack propagation, from (8.5.59), we get q 4γ ◦ ◦ (8.5.60) C44 (C55 + σ 11 ). κ2c = κ2c (σ 11 ) = πa ◦

Let us denote by κ ˆ c the critical value corresponding to σ 11 = 0; i.e. κ ˆ 2c = κ2c (0) =

Obviously, we have ◦

ˆ 2c κ2c (σ 11 ) = κ

Hence,

4γ p C44 C55 . πa

s

(8.5.61)

◦

C55 + σ 11 . C55

◦

◦

◦

◦

κc (σ 11 ) > κ bc if σ 11 > 0

κc (σ 11 ) < κ bc if σ 11 < 0.

(8.5.62)

(8.5.63) ◦

Consequently, an initial applied extensional force σ 11 > 0 acting in the direction of the reinforcing fibers improves the crack stability and an initial applied ◦ compressive force σ 11 < 0 acting in the fibers direction diminishes the stability of the crack.

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8.5. GRIFFITH’S CRITERION AND CRACK PROPAGATION

517

We believe that these predictions are in accordance with our ideas concerning the influence of initial applied stress on the behavior of a prestressed fiberreinforced composite. These results due to Guz [8.4] show again the internal consistency and the great power of the linearized three−dimensional theory. Also, equation (8.5.62) shows that ◦

◦

◦ cp

κc (σ 11 ) → 0 if σ 11 → σ 11 = −C55 .

(8.5.64) ◦

Hence, when the initial applied compressive stress σ 11 reaches the critical value, ◦ cp

σ 11 = −C55 ,

(8.5.65)

the crack becomes completely unstable and its propagation by the third mode can start without any incremental tangential force applied at the two faces of the crack. We recall that, for a fiber-reinforced composite, we have C55 = G13 and G13 << E1 .

(8.5.66)

Hence, using (8.5.65) and (8.5.66), we have ◦ cp ◦ cp σ 11 = −G13 and σ 11 << E1 .

(8.5.67)

Complete instability of the crack can occur in a fiber-reinforced composite material if the condition (8.5.67) is fulfilled. The corresponding deformations rest infinitesimal, as can be seen, taking into account the inequality (8.5.67) 2 . At the same time, from (8.2.36), (8.5.39)1 and (8.5.64), it can be seen that ◦ ◦ ◦ cp µ3 = 0 if σ 11 = σ 11 . Hence, if the compressive force σ 11 reaches its critical value, the differential equation (8.2.34) losses its ellipticity. In other words, internal instability of the prestressed fiber-reinforced composite occurs. The simultaneous appearance of internal instability and complete instability of the crack are direct consequences of the internal structure of a fiber-reinforced composite. Neither complete crack instability nor internal instability can occur in the case of an isotropic material in the frame-work of validity of linear elasticity. The analysis made in this Section reveals again the way in which a phenomenological continuum theory can take into account and predict macroscopic effects due to the internal structure of a composite material. At the same time, we can see that due to its internal structure in a fiberreinforced composite material, a dangerous situation can occur if the initial applied forces are not adequately limited. The three-dimensional linearized theory is able to reveal the involved critical situations and can be successfully used to avoid the occurrence of dangerous situations.

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8.6

CHAPTER 8. FRACTURE MECHANICS

Problems P8.1 Let us consider the function u = u(x1 , x2 ) = x21 − x22 . (a) Prove that ∆u = 0. (b) Find the holomorphic function f = f (z) and g = g(z) such that x21 − x22 = Ref (z)

and x21 − x22 = Img(z). P8.2 Let f = f (z) be a holomorphic function and let L be a regular closed curve in the domain of definition B of f = f (z). Prove that I Z f (t)dt 1 f (z)dz = 0 and f (z) = t−z 2πi L

L

where, in the second relation (Cauchy’s formula), z is a point in the finite domain bounded by the curve L. P8.3 Let L be a regular closed curve as in Figure 8.14. Let B + be the finite domain bounded by L and let B − be the infinite domain bounded by L as in Figure 8.14.

BB+ t2 t1

t0

l= t 1t2

Figure 8.14: Integral theorems. + Let us assume that S f = f (z) is a holomorphic function in domain B and it + L. is continuous on B Prove that Z f (t)dt 1 f (z) if z ∈ B − ={ . 0 if z ∈ B + t−z 2πi

L

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519

8.6. PROBLEMS

P8.4 Let us assume that f = f (z) isSa holomorphic function in the unbounded domain B − and it is continuous on B − L. Prove that Z f (t)dt 1 −f (z) + f (∞) if z ∈ B − . ={ f (∞) if z ∈ B + t−z 2πi L

The first equality is named Cauchy’s formula for an unbounded domain. P8.5 Let L be a regular closed curve as shown in Figure 8.14. Let f = f (t) be a complex valued function defined and absolute integrable on L. The function Z f (t)dt 1 ,z ∈ /L F = F (z) = t−z 2πi L

is named Cauchy’s integral corresponding to f and L. Let us denote as before F + (t0 ) =

lim

z∈B + ,z→t0

F (z)

and F − (t0 ) =

lim

z∈B − ,z→t0

F (z), t0 ∈ L

the interior and the exterior limits of F = F (z) is a point t0 ∈ L. As is known, F = F (z) is a holomorphic function in the open finite domain B + bounded by L. Also, f = f (z) is a different holomorphic function in the infinite domain B − bounded by L. Moreover, lim F (z) = 0. z→∞

For a point t0 situated on the curve L (see Figure 8.14), Z f (t)dt 1 , t0 ∈ L t − t0 2πi L

is generally a nonconvergent improper integral and the limit values F + (t0 ) and F − (t0 ), even if they exist, are distinct. Let us select on L (see Figure 8.14) two points t1 and t2 and let us consider the little arc l = td 1 t2 containing t0 . We assume that it satisfies the following restriction: |t1 − t0 | = |t2 − t0 | . We consider also the integral 1 2πi

Z

L−l

f (t)dt , t0 ∈ L t − t0

defined on the complementary part L−l of L. This is usually a well-defined integral, since |t − t0 | ≥ δ > 0 for any t ∈ L − l, δ being a positive number.

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CHAPTER 8. FRACTURE MECHANICS

Let us assume now that t1 , t2 ∈ L converge toward t0 ∈ L such that |t1 − t0 | = |t2 − t0 |. If, in this case, the above integral converges to a well-defined finite value, the corresponding limit is named Cauchy’s principal value of our improper integral. As in Section 8.1, we denote this value by Z f (t)dt . (P V ) t − t0 L

Let us suppose now that the function f = f (t) satisfies H˝older’s condition or the H condition; i.e. µ

|f (t1 ) − f (t2 )| ≤ A |t1 − t2 | , A > 0, 0 < µ ≤ 1, ∀t1 , t2 ∈ L, A and µ being positive numbers. Prove that, in this case, the Cauchy’s principal value exists and Z Z f (t) − f (t0 ) 1 1 f (t)dt 1 dt. = f (t0 ) + (P V ) t − t0 2πi 2 t − t0 2πi L

L

Moreover, in the assumed conditions, it can be shown that the limit values F + (t0 ) and F − (t0 ) exist and are satisfied by the Plemelj-Sohockii formulas: Z f (t)dt 1 1 , (P V ) F + (t0 ) = f (t0 ) + t − t0 2πi 2 L

1 1 (P V ) F (t0 ) = − f (t0 ) + 2πi 2 −

Z L

f (t)dt . t − t0

P8.6 Let us consider the function Za √ 2 t a − t2 dt. I(z) = t−z −a

Using the relation (8.1.96), find the value of the above integral. P8.7 Find the limit values I + (t) and I − (t) for t ∈ (−a, a).

P8.8 Let f = f (z) be a holomorphic function of the complex variable z = x1 + µx2 , where µ is a complex number. Let us assume that u = u(x1 , x2 ) = 2Ref (z) = f (z) + f (z).

Prove that

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∂u = f 0 (z) + f 0 (z) ∂x1

521

8.6. PROBLEMS and

∂u = µf 0 (z) + µf 0 (z). ∂x2

P8.9 Show that equation(8.2.13) can be written in the equivalent form (8.2.16). P8.10 Assume that µ1 = µ2 = µ. Show that, in this case, equation (8.2.13) becomes ∂4ϕ =0 ∂z 2 ∂ z 2 and its general solution is

ϕ = ϕ(x1 , x2 ) = 2Re{zf1 (z) + f2 (z)},

where z = x1 + µx2 and f1 = f1 (z), f2 = f2 (z) are two arbitrary analytic functions. P8.11 Assuming antiplane incremental state and using the representation (8.2.44) of the incremental displacement u3 = u3 (x1 , x2 ) in terms of the complex potential φ3 = φ3 (z3 ), express the incremental stresses θ31 and θ32 by φ3 (z3 ). P8.12 Let us consider a holomorphic function F = F (z) defined in the upper half plane x2 > 0, denoted by B + . Let us define in the lower half plane x1 > 0, denoted by B − , the function F = F (z) using the following rule:

F = F (z), z = x1 + ix2 ∈ B − .

Let U (x1 , x2 ) and V (x1 , x2 ) be the real and the imaginary part of F = F (z); i.e. F (z) = U (x1 , x2 ) + iV (x1 , x2 ), z = x1 + ix2 ∈ B + . Let U1 (x1 , x2 ) and V1 (x1 , x2 ) be the real and the imaginary part of F = F (z); i.e.

F (z) = U1 (x1 , x2 ) + iV1 (x1 , x2 ), z = x1 + ix2 ∈ B − . Show that U1 (x1 , x2 ) V1 (x1 , x2 ) for −∞ < x1 < ∞ and x2 < 0.

= U (x1 , −x2 ), = −V (x1 , −x2 )

P8.13 Show that F = F (z) is a holomorphic function in the lower half plane

B−.

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522

CHAPTER 8. FRACTURE MECHANICS P8.14 Let t be an arbitrary point on the real axis. (a) Prove that if the upper limit exists F + (t) =

lim

z∈B + ,z→t

F (z)

of the function F = F (z), then there exists also the lower limit −

F (t) =

of the function

lim

z∈B − ,z→t

F (z),

−

F = F (z) and F (t) = F + (t).

(b) Changing the role of B + and B − , prove that if the lower limit exists, F − (t) =

lim

z∈B − ,z→t

F (z),

of the function F = F (z), then the upper limit also exists +

F (t) =

of the function

lim

z∈B + ,z→t

F (z)

+

F = F (z) and F (t) = F − (t).

P8.15 Assume that F = F (z) = a0 z n + a1 z n−1 + ... + an z + an+1 , where n is a positive integer and a0 , ..., an+1 are complex constants. Find the function F (z) = F (z).

P8.16 Let us consider the crack problem for the first mode and let us assume that the symmetrically applied normal forces have a constant value; i.e. g(x1 ) = g = const. for −a < x1 < a. Find the complex potentials Ψj (zj ) and Φj (zj ), j = 1, 2 in this case. P8.17 Let us consider the crack problem for the second mode and let us assume that the antisymmetrically applied tangential forces have a constant value; i.e. h(x1 ) = h = const. for −a < x1 < a. Find the complex potentials Ψj (zj ) and Φj (zj ), j = 1, 2 in this case. P8.18 Let us consider the crack problem for the third mode. Using the results of P8.11, find the asymptotic values of the incremental nominal stresses θ 31 and θ32 near the crack tip x1 = a.

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P8.19 Let us consider again the crack problem for the third mode and let us assume that the antisymmetrically applied tangential forces have a constant value; i.e. k(x1 ) = k = const. for −a < x1 < a. Find the complex potentials Ψ3 (z3 ) and Φ3 (z3 ) in this case. P8.20 In the condition of the problem P8.16, find the normal displacements u2 of the crack face and determine the jump of these displacements across the crack. P8.21 In the condition of the problem P8.17, find the tangential displacements u1 of the crack faces and determine the jump of these displacements across the crack. P8.22 In the condition of the problem P8.19, find the tangential displacements u3 of the crack faces and determine the jump of these displacements across the crack. P8.23 The instantaneous elasticities ω1331 and ω2332 are given by equations (8.5.39). Find the dependence of the tangential displacement u3 = u3 (x1 , x2 ) on ◦ the initial applied stress σ 11 if the conditions of the internal stability are fulfilled. P8.24 Let us assume that the considered material is isotropic. Let us assume also that, on the considered loading path, ◦

◦

σ 22 = σ 33 = 0. Using the general relations (5.6.22), express the instantaneous elasticities in ◦ terms of Lam´e’s constants λ, µ and of the initial applied stress σ 11 . P8.25 In the conditions of P8.24, find the roots ν1 and ν2 of the algebraic equation (8.2.6). P8.26 Find the roots µ1 and µ2 of the algebraic equation(8.2.9) corresponding to the problem P8.25. P8.27 Let us assume that, in the conditions of P8.23, the initial applied stress ◦ σ 11 is vanishing. Find the stress energy release rates GI (a), GII (a) and GIII (a) in this case. P8.28 Write Griffith’s crack propagation criterion corresponding to the case considered in P8.27. P8.29 Let us assume that the conditions of problem P8.27 are fulfilled. Let us suppose also that the normal and tangential forces applied on the two faces of the crack have constant values. (a) Give the form of Griffith’s propagation criterion in this case. (b) Find the critical values of the applied forces for which crack propagation takes place by the first, second and third mode, respectively. P8.30 Let us consider a linearly elastic monoclinic material, the x1 , x2 plane being its plane of symmetry. Let us assume that the initial stresses in the material are zero. We suppose also zero body forces.

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(a) Show that the considered material can be in a plane strain equilibrium state, relative to the x1 x2 plane. (b) Give the equilibrium equations satisfied by the plane displacements. (c) Give the equations of motion satisfied by the plane displacements. P8.31 Using the results obtained in P8.30 (b), give the Guz’s type representation of the plane displacements u1 , u2 in terms of two real displacement potentials ϕ(1) , ϕ(2) and find the differential equations satisfied by these potentials. 2 P8.32 (a) Assuming C22 C66 −C26 6= 0, give the explicit form of the differential equation satisfied by the displacement potential introduced in P8.31. (b) Show that the obtained equation can be factorized in the form

(

∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ )ϕ(α) = 0, − µ4 )( − µ3 )( − µ2 )( − µ1 ∂x1 ∂x1 ∂x2 ∂x1 ∂x2 ∂x1 ∂x2 ∂x2

where µ1 , µ2 , µ3 , µ4 are the roots of the algebraic equation 0

=

2 2 l(µ) = (C22 C66 − C26 )µ4 + 2(C22 C16 − C12 C26 )µ3 2 − 2C12 C66 + 2C16 C26 )µ2 +(C11 C22 − C12 2 +2(C11 C26 − C12 C16 )µ + (C11 C66 − C16 ).

P8.33 (a) Study the propagation of harmonic plane waves in the conditions assumed in P8.30. (b) Assuming that the stress-free reference configuration of the considered monoclinic material is locally stable; i.e. its specific strain energy is positive definite, give the interpretation if the result obtained in (a). (c) Using the conclusions of (a), find the restrictions which must be satisfied by the elasticities of the material, if the assumption made in (b) is true. P8.34 Show that if the stress-free reference configuration of the considered monoclinic material is locally stable, the algebraic equation found in P8.32 cannot have real roots. P8.35 In what follows, we suppose µ1 6= µ2 . Assume ϕ(2) = 0 and use the notation ϕ(1) = ϕ. (a) Show that the differential equation given in P8.31 (b) and satisfied by the displacement potential ϕ = ϕ0 can be written in the following equivalent form: ∂4ϕ =0 ∂z1 ∂ z 1 ∂z2 ∂ z 2

where the complex variables z1 and z2 are defined by z1 = x1 + µ1 x2 and z2 = x1 + µ2 x2 ,

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µ1 and µ2 being nonreal roots of the algebraic equation l(µ) = 0 given in P8.32. (b) Find the general solution of the above equation. P8.36 Using the results obtained in P8.30, P8.31 and P8.35, and the reasoning of Section 8.2, give the Leknitskii-Guz type representation of the elastic state in terms of two complex potentials supposing ϕ(1) = ϕ and ϕ(2) = 0. P8.37 Let us assume that the whole space is occupied by a monoclinic material without initial stresses. We suppose (see Section 8.3) that the material contains a crack of length 2a > 0 situated in the x1 axis and having an infinite extent in the direction of the x3 axis. We recall that the symmetry plane of the material is the x1 x2 plane. We suppose that on the two faces of the crack, the tangential stresses are zero. We assume also that on the upper face of the crack a distributed normal stress is given, and a symmetrically distributed normal stress acts on the lower face (see Figure 8.9). Find the complex potentials Φα (zα ) and Ψα (zα ) describing the elastic state of the body according to the representation obtained in P8.36. P8.38 Assuming that the symmetrically applied normal stress has constant value; i.e. g(x1 ) = p = const. for |x1 | < a, find the expressions of the complex potentials Φα (zα ) and Ψα (zα ). P8.39 Assuming the conditions of problem P8.38, determine the asymptotic behavior of the stress σ22 (x1 , x2 ) in a small neighborhood of the right crack tip x1 = a, x2 = 0. P8.40 We assume again the conditions of problem P8.38. (a) Find the normal displacement u2 of the line containing the crack, behind and ahead of the crack. (b) Find the normal displacements of the two faces of the crack. (c) Give the graphical representation of the obtained results and compare the behavior of the normal displacement with that obtained in Section 8.4 concerning a prestressed orthotropic material. P8.41 Let us consider again a linearly elastic monoclinic material, the x1 x2 plane being its symmetry plane. Let us assume as before, that the initial stresses in the material are zero. We suppose also zero body forces. (a) Show that the considered material can be in antiplane strain equilibrium state relative to the x1 x2 plane. (b) Give the equilibrium equations satisfied by the antiplane displacement field. P8.42 Give the expression of the specific strain energy of a monoclinic material in antiplane strain state and, assuming that the reference configuration of the material is locally stable, find the restrictions imposed on the elasticities C 44 , C45 and C55 by this property. P8.43 In the conditions of P8.42, give the Leknitskii-Guz type representation of the elastic state by a complex potential.

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P8.44 We suppose that the whole space is occupied by a monoclinic material with locally stable reference configuration. We assume that the material contains a crack as in P8.37. We suppose that on the two faces of the crack tangential stresses antisymmetrically distributed acts only relative to the plane x 2 = 0 and having the direction of the x3 axis. Find the complex potential Φ3 (z3 ) describing the corresponding antiplane elastic state of the material. P8.45 Assuming that the antisymmetrically applied tangential stress has a constant value; i.e. k(x1 ) = k = const. for |x1 | < a, find the expressions of the complex potentials Ψ3 (z3 ) and Φ3 (z3 ). P8.46 In the condition of P8.45, find the antiplane displacement u3 and the tangential stresses σ13 and σ23 . P8.47 Analyze the asymptotic behavior of the tangential stress σ23 and of the displacement u3 found in P8.46, near the right crack tip. P8.48 Analyze the displacement u3 of the lines x2 = 0 behind and ahead of the crack, find the displacements u3 of the two faces of the crack and compare the obtained results with that corresponding to an orthotropic material. P8.49 Using Griffith’s energetic criterion and, in conditions of P8.45, find the critical tangential stress κc for which crack propagation starts by the third tearing mode.

Bibliography [8.1] Muskhelishvili, N.I., Some basic problems of mathematical theory of elasticity, Nordhoff, Groningen, Holland, 1953. [8.2] Lekhnitski, S.G., Theory of elasticity of aniosotropic elastic body. Holden Day, San Francisco, 1963. [8.3] Sih, G.C., Leibowitz, H., Mathematical theories of brittle fracture, in Fracture - An advanced treatise, Vol.II, Mathematical fundamentals, Editor H. Lebowitz, pp 68-191, Academic Press, New York, 1968. [8.4] Guz, A.N., Mechanics of brittle fracture of prestressed materials, Visha Shcola, Kiev, 1983 (in Russian). [8.5] Guz, A.N., Brittle fracture of materials with initial stress, Vol. 2 of Nonclassical problems of fracture mechanics, Ed. A.N. Guz, Naukova Dumka, Kiev, 1991 (in Russian). [8.6] So´os, E., Resonance and stress concentration in a prestressed elastic solid containing a crack. An apparent paradox. Int. J. Engng. Sci., 34, pp 363374, 1996.

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SOLUTIONS TO SOME PROBLEMS Chapter 1 P1.1 Let us assume that two vectors 0 and 01 exist, which have the following property: u + 0 = u and u + 01 = u for any u in V. Taking u = 01 in the first equation, and u = 0 in the second one, results in 01 + 0 = 01 and 0 + 01 = 0. From these two relations, we can conclude that 01 = 0 and, thus, the uniqueness of the null vector is proved. P1.2 Using the property (V9), we get (0 + 0)u = 0u + 0u. On the other hand, we have also (0 + 0)u = 0u. From the equation obtained, we get 0u = 0, taking into account P1.1. In order to prove the second property, we observe that α(0 + u) = α0 + αu. But α(0 + u) = αu. Hence, again P1.1 leads to α0 = 0 for any real number α. P1.3 Let us consider the system u1 , u2 , u3 , ..., up−1 , 0; we consider also the set α1 = ... = αp−1 = 0 and αp 6= 0. We get α1 u1 + α2 u2 + ... + αp−1 up−1 + αp 0 = 0u1 + 0u2 + ... + 0up−1 + αp 0 = 0 + ... + 0 + αp 0 = αp 0 = 0. Hence, our system u1 , u2 , u3 , ..., up−1 , 0 is linearly dependent, since αp 6= 0. P1.6 In order to prove the properties (N1)−(N3), we start with equation (1.1.1) defining the magnitude kuk of a vector u ∈ V. Also, we use the properties (S1)−(S5) of an Euclidean scalar product. From (S4) and (S5), it follows that (N1) is true. From (1.1.1) and (S2), it results that (N2) takes place. To prove (N3) now, let us consider two vectors u, v from V and an arbitrary real number, α. According to (S4), we have: (αu + v)·(αu + v) ≥ 0. Using (S1),(S2) and (S3), it results: α2 kuk2 +2αu · v +kvk2 ≥ 0 for any real number α. Hence, we must have (u · v)2 ≤ kuk2 · kvk2 or |u · v| ≤ kuk · kvk for any vectors u and v from V. At the same time, we have ku + vk2 = (u + v)·(u + v) = u · u + 2u · v + v · v = kuk2 + 2u · v + kvk2 . Taking into account the above inequality, we get:ku + vk2 ≤ kuk2 + 2 kuk · kvk + kvk2 = (kuk + kvk)2 . The last result shows that the property (N3) is also true. P1.7 We must show that the operation hP, Qi =

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Rb a

P (x)Q(x)dx introduced in Pn+1

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SOLUTIONS TO SOME PROBLEMS

has the properties (S1)−(S5). In order to do this, we must take into account some well-known properties of the usual integrals. Since P (x)Q(x) = Q(x)P (x), the property Rb Rb (S1) is obviously true. Also, since αP (x)Q(x)dx = α P (x)Q(x)dx, the property (S2) a

takes place. We have also

Rb a

a

P (x) {Q(x) + R(x)} dx =

Rb

P (x)Q(x)dx +

a

Rb

P (x)R(x)dx for

a

P, Q, R ∈ Pn+1 . Consequently, the validity of the property (S3) was proved. In the same Rb way we get that P (x)P (x)dx ≥ 0, and this inequality shows that the property (S4) a

is true. Let us assume now that hP, P i =

Rb

P 2 (x)dx = 0. Obviously, if P (x) = 0, then

a

hP, P i = 0. Let us assume now that hP, P i = 0; hence,

Rb a

P 2 (x)dx = 0. Since P 2 (x) ≥ 0

and taking into account the fact that any polynomial P (x) is a continuous function, a well-known theorem of the integral calculus tells us that P (x) = 0. Consequently, (S5) is proved. P1.11 Let us consider an arbitrary vector w ∈ V. Taking into account the equation (1.1.11), defining the tensor product of two vectors and the properties of a scalar product, we get {(αu) v} (w) = αu (v · w) ,

{u (αv)} (w) = u (αv · w) = uα (v · w) = αu (v · w) , {α (uv)} (w) = αu (v · w) .

The above equations show that the first properties given in P1.11 are true, since w is an arbitrary vector from V. Analogously, let us assume that a is an arbitrary vector from V. We get: {(u + v) w} (a) = (u + v) (w · a) = u (w · a) +v (w · a) = (u · w) (a) + (v · w) (a)

and {u (v + w)} (a)

= =

u {(v + w) · a} = u {v · a + w · a}

u (v · a) +u (w · a) = (u · v) (a) + (u · w) (a) .

Since these equations hold for any vector a from V, we can conclude that the last properties given in P1.11 are also true. P1.14 To prove the given property, we use the relation defining the tensor product of two vectors and the equation giving the components of a tensor. Thus, we obtain successively (uv)km = ek · {uv} em = ek · {u (v · em )} = (u · ek ) (v · em ) = uk vm , since u · ek =uk and v · em =vm .

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529

P1.15 To show that the tensor product is not commutative, it is sufficient to give a particular case, proving this property. Let us take u = ek and v = em , ek and em being two different elements of an orthonormal basis in Vn . For an arbitrary vector w, we get (ek em ) (w) = ek (em · w) = ek wm = wm ek , (em ek ) (w) = em (ek · w) = em wk = wk em , where wk and wm are the corresponding components of w in the considered orthonormal basis. Since ek and em are linearly independent, obviously wm ek 6= wk em , if wm or wk are not vanishing. Hence, ek ·em 6= em ·ek , m 6= k. P1.17 Using again the definition of the product of two tensors and considering an arbitrary vector v, we obtain {(TU) V} (v) = {TU} (Vv) = T {U (Vv)} , {T (UV)} (v) = T {(UV) (v)} = T {U (Vv)} , hence, (TU) V = T (UV) since the above equation are true for any vector v. The distributivity relative to tensor addition can be proved in a similar way. If 1 is the unit tensor, we have 1v = v for any vector v. Hence, we get (1T) (v) = 1 (Tv) = Tv and (T1) (v) = T (1v) = Tv, for any vector v. Hence, we have 1T = T1 = T for any second order tensor T. P1.19 In order to prove that the transposed tensor TT is a linear function, we must use the relation defining TT and the properties of a scalar product. Thus, we obtain the following chain of relations, in which α, β are arbitrary real numbers and u, v, w are arbitrary vectors: n o TT (αu + βv) · w = (αu + βv) · Tw = αu · Tw + βv · Tw   = αTT u · w + βTT v · w = αTT u + βTT v · w. Since the above equations are true for any vector w, from the property proved in P1.9, we can conclude that TT (αu+βv) = αTT u + βTT v; hence, the transposed tensor TT is a linear function.

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P1.21 Let w be an arbitrary vector from Vn . Using the corresponding definition, we get successively the following equalities: {(Tu) (Uv)} (w) = Tu (Uv · w) and

n

T (uv) UT

o

  n  o n  o (w) = T (uv) UT w = T (uv) UT w = T u v · UT w = T {u (Uv · w)} = Tu (Uv · w) .

Since the above relations are true for any vector w, we can conclude that (Tu)(Uv) = T(uv)UT . P1.23 To prove that the scalar product of two tensors, defined by the equation (1.1.18), satisfies the properties (S1)-(S5) of an Euclidean scalar product, we must use the linearity of the trace of a second order tensor and the properties of the product of two second order tensors together with the properties of the transposed tensor. First of all, we have n oT = trUTT = U · T. T · U = trTUT = tr TUT Hence, (S1) takes place. We have also

    (αT) · U = tr (αT) UT = trα TUT = αtr TUT = α (T · U) .

Consequently, (S2) is true. Analogously, we get

  n o T · (U + V) = trT (U + V)T = trT UT +VT = tr TUT + TVT = trTUT + trTVT = TU + TV.

Thus, (S3) takes place. In a similar way, we get: T · T = trTTT = Tkm Tkm , where Tkm are the components of T in a basis {ek , em } . The last result shows that (S4) and (S5) are satisfied. P1.25 The tensor T is defined by the equation Te1 = e1 +e2 ; Te2 = e2 +e3 ; Te3 = e3 +e1 . (a) According to the equation (1.1.13), its components Tkm are given by the relation Tkm = ek ·Tem , k, m = 1, 2, 3. In this way, we get T11 = e1 ·Te1 = e1 · (e1 + e2 ) = e1 ·e1 + e1 ·e2 = 1 + 0 = 1

T12 = e1 ·Te2 = e1 · (e2 + e3 ) = e1 ·e2 + e1 ·e3 = 0 + 0 = 0

T13 = e1 ·Te3 = e1 · (e3 + e1 ) = e1 ·e3 + e1 ·e1 = 0 + 1 = 1.

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SOLUTIONS TO SOME PROBLEMS

531

Using the same procedure, we can see that the matrix of the components of T is given by the following relation:   1 0 1 [Tkm ] =  1 1 0  . 0 1 1 Hence, the tensor T in the basis {ek, em } , k, m = 1, 2, 3 has the following expression: T = e1 e1 +e1 e3 +e2 e1 +e2 e2 +e3 e2 +e3 e3 . (b) We have Tv = T(e1 + e2 + e3 ) = Te1 + Te2 + Te3 = 2(e1 + e2 + e3 ) = 2v.  T  of the components of the transposed tensor TT is the transposed (c) The matrix Tkm matrix of the matrix [Tkm ] of the components of T; hence,   1 1 0 h i T Tkm =  0 1 1  . 1 0 1 
 The matrix TTT km of the components of the product TTT is the  T  ; hence, product of the matrixes [Tkm ] and Tkm      1 0 1 1 1 0 2 1 1 h  i T TT =  1 1 0  0 1 1  =  1 2 1 . km 0 1 1 1 0 1 1 1 2 The tensor TS is defined by the relation TS =

 1 T + TT . 2

In this way, we can see that the matrix of the components of 2TS and that of the components of TTT are equal; hence, 2TS = TTT . The tensor TA is defined by the equation  1 T − TT ; TA = 2  A  hence, the matrix Tkm of the components of TA is given by the following equation:   0 −1/2 1/2 h i A 0 −1/2  . Tkm =  1/2 −1/2 1/2 0

Consequently, the tensor TA has the following expression in the basis {ek em } , k, m = 1, 2, 3: TA =

1 (−e1 e2 + e2 e1 + e1 e3 − e3 e1 − e2 e3 + e3 e2 ) . 2

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SOLUTIONS TO SOME PROBLEMS

(d) From the expression of the matrix of the components of the tensor T, we get trT = T11 + T22 + T33 = 3,   1 0 1 det T =  1 1 0  = 2. 0 1 1

Using the same matrix, we obtain

p

√ Tkm Tkm = 6.  −1  (e) Since det T = 2, T is a nonsingular tensor. The matrix Tkm of the components of the tensor T−1 is the inverse matrix of the matrix [Tkm ] . In this way, we get   1 1 −1  −1  1 −1 1 1 , Tkm = 2 1 −1 1 kTk = (T · T)1/2 =

trTTT =

√

and the expression of T−1 in the given basis is: T−1 =

1 (e1 e1 + e1 e2 − e1 e3 − e2 e1 + e2 e2 + e2 e3 + e3 e1 − e3 e2 + e3 e3 ) . 2

P1.28 Let us denote by Sk , k = 1, ..., 6, the following second order tensors: S 1 = e 1 e1 , S 2 = e 2 e2 , S 3 = e 3 e3 ,

1 1 1 S4 = √ (e1 e2 + e2 e1 ) , S5 = √ (e2 e3 + e3 e2 ) , S6 = √ (e3 e1 + e1 e3 ) . 2 2 2 Using the result proved in P1.20, it is easy to see that these tensors are symmetric; i.e.

STk = Sk , k = 1, ..., 6. Hence, S1 , ..., S6 are the elements of LS , the vector space of the symmetric second order tensors. Also, using the result proved in P1.24, it is easy to see that these tensors form an orthonormal system in LS ; i.e. Sk ·Sl = δkl for k, l = 1, ..., 6. Since the set S1 , ..., S6 is an orthonormal system, it is obviously also a linearly independent system. Let us consider now an arbitrary symmetric tensor T from LS . Since {ek em } , k, m = 1, 2, 3 form a basis in L, we have T = Tkm ek em , and, since T is symmetric, its components satisfy the relations Tkm = Tmk .

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SOLUTIONS TO SOME PROBLEMS

Using this restrictions, it is easy to see that T can be expressed in the following equivalent form: T = T11 S1 + T22 S2 + T33 S3 +

√

2T12 S4 +

√

2T23 S5 +

√

2T31 S1 .

Consequently, any symmetric tensor T can be expressed as a linear combination of the symmetric tensor S1 , ..., S6 and these tensors form a linearly independent orthonormal system. Hence, according to the general definition of an orthonormal basis in a vector space, these six tensors S1 , ..., S6 form an orthonormal basis in LS . At the same time, we can conclude that dimLS = 6. Let us denote now by Ak , k = 1, 2, 3, the following second order tensors:

1 1 1 A1 = √ (e1 e2 − e2 e1 ) , A2 = √ (e2 e3 − e3 e1 ) , A3 = √ (e3 e1 − e1 e3 ) . 2 2 2

Using again P1.20, we can see that these tensors are antisymmetric; i.e. ATk = −Ak , k = 1, 2, 3. Hence A1 , A2 , A3 are elements of LA , the vector space of the antisymmetric second order tensors. Taking into account the property proved in P1.24, we can see without difficulty that the above tensors verify the relations Ak ·Al = δkl , k, l = 1, 2, 3. Hence, the tensors A1 , A2 , A3 form an orthonormal system. This property shows also that our antisymmetric tensors A1 , A2 , A3 form a linearly independent system. Let us consider now an arbitrary antisymmetric tensor T from LA . We have T = Tkm ek em , and the components of T satisfy the restrictions Tkm = −Tmk . Using these equations, we can easily conclude that T can be written in the following equivalent form: √ √ √ T = 2T12 A1 + 2T23 A2 + 2T31 A3 .

Consequently, any antisymmetric tensor can be expressed as a linear combination of the antisymmetric tensors A1 , A2 , A3 and these three tensors form a linearly independent orthonormal system. Hence, A1 , A2 , A3 represent an orthonormal basis in LA . At the same time, we get dimLA = 3. P1.30 Let u1 and u2 be two vectors from Vn , which satisfy the equations Tu1 = λu1 and Tu2 = λu2 . Let α1 and α2 be two real numbers and let us consider the linear combination u = α1 u1 +α2 u2 . We have, taking into account the linearity of T and the above equations, Tu = T (α1 u1 + α2 u2 ) = α1 Tu1 + α2 Tu2 = α1 λu1 + α2 λu2 = λ (α1 u1 + α2 u2 ) .

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SOLUTIONS TO SOME PROBLEMS

Hence, u satisfies the equation Tu = λu. We conclude that the set of all vectors u ∈ Vn , which satisfies the above relation, is a subspace vectorial of Vn . P1.32 As we have seen, the matrix of the components of the symmetric part TS , if the tensor T given in P1.25, has the following form:   1 1/2 1/2 h i S 1/2  . Tkm =  1/2 1 1/2 1/2 1 The principal invariants of this tensor can be obtained using the general formulas (see Section 1.1). Thus, we get I TS = 3

IITS = −

9 4

IIITS =

1 . 2

Hence, according to the relation (1.1.28), the characteristic equation corresponding to TS is 1 9 λ3 − 3λ2 + λ − = 0. 2 4 Consequently, the eigenvalues of the tensor TS are

λ1 = 2,

λ 2 = λ3 =

1 . 2

To get the eigendirection n1 corresponding to the eigenvalue λ1 = 2, we must solve the system determined by equations (1.1.29)1 and (1.1.30)1 , where α1 , β1 , γ1 are the components of n and satisfy the restriction α12 + β12 + γ12 = 1. An elementary computation shows that 1 α1 = β 1 = γ 1 = √ . 3 Consequently, the eigendirection n1 , corresponding to the eigenvalue λ1 = 2 is √ n1 = (e1 + e2 + e3 ) / 3.

Let us denote by α, β, γ the components of an eigenvector v, corresponding to the double eigenvalue λ2 = λ3 = 12 . According to the general relations (1.1.29)1 and (1.1.30)1 , α, β, γ must satisfy the following system of algebraic equations:   1 1 1 α + β + γ = 0, 1− 2 2 2   1 1 1 β + γ = 0, α+ 1− 2 2 2   1 1 1 γ = 0. α+ β+ 1− 2 2 2

In the above algebraic system, only one is independent, and has the following form: α + β + γ = 0.

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535

SOLUTIONS TO SOME PROBLEMS Hence, we get γ = − (α + β) , and v must have the following form: v = αe1 + βe2 − (α + β) e3 ,

where α and β are arbitrary real numbers. Consequently, the eigendirections n corresponding to the double eigenvalue λ2 = λ3 = 1/2 have the following expression: n = (αe1 + βe2 − (α + β) e3 ) /

p

2 (α2 + β 2 + αβ).

It is easy to see that n1 and n are reciprocally orthogonal and, concluding, we can say that any unit vector orthogonal to n1, the eigenvector corresponding to the eigenvalue λ1 = 2, is an eigendirection corresponding to the double eigenvalue λ2 = λ3 = 1/2. Let us denote now by n2 and n3 , two eigendirections corresponding to the double eigenvalue λ2 = λ3 = 1/2, which are reciprocally orthogonal. Consequently, according to the general spectral representation theorem (1.1.25), the tensor TS can be expressed in the following form: 1 1 TS = 2n1 n1 + n2 n2 + n3 n3 . 2 2 We observe now that the system n1 , n2 , n3 forms a basis in V. Let us consider a vector v = vk nk . Then we have vk = (v · nk ) . Consequently, the following relations are true:

(n1 n1 +n2 n2 +n3 n3 ) v = n1 (n1 ·v) +n2 (n2 ·v) +n3 (n3 ·v) = v1 n1 +v2 n2 +v3 n3 = v.

Hence, we have the following important result: if n1 , n2 , n3 is an orthonormal basis, then n1 n1 +n2 n2 +n3 n3 = 1, where 1 is the unit tensor. Using this equation, we get the spectral representation of the tensor TS in the following form: 1 TS = 2n1 n1 + (1 − n1 n1 ) . 2

P1.33 In order to prove the theorem of spectral representation if λ1 6= λ2 = λ3 , let us denote by n1 the eigendirection corresponding to the eigenvalue λ1 , and by n2 an eigendirection corresponding to the eigenvalue λ2 = λ3 6= λ1 . Hence, we have Tn1 = λ1 n1 , Tn2 = λ2 n2 = λ3 n2 . Let us consider also an unit vector n3 such that the system n1 , n2 , n3 be an orthonormal basis in V . Let us denote by Tkm , the components of the tensor T in the tensor basis (nk nm ) , k, m = 1, 2, 3. According to the general relation (1.1.13), we have Tkm = nk ·Tnm .

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Hence, T11 = n1 ·Tn1 = n1 ·λ1 n1 = λ1 n1 ·n1 = λ1 ,

T12 = n1 ·Tn2 = n1 ·λ2 n2 = λ2 n1 ·n2 = 0,

T22 = n2 ·Tn2 = n2 ·λ2 n2 = λ2 n2 ·n2 = λ2 ,

T31 = n3 ·Tn1 = n3 ·λ1 n1 = λ1 n3 ·n1 = 0,

T32 = n3 ·Tn2 = n3 ·λ2 n2 = λ2 n3 ·n2 = 0. Since T is a symmetric tensor, we have also T21 = T13 = T32 = 0.

Thus, the matrix of the components of T in the considered basis has the following form:   λ1 0 0 . λ2 0 [Tkm ] =  0 0 0 T33

Consequently, the characteristic equation becomes

(λ1 − λ) (λ2 − λ) (T33 − λ) = 0. Since, according to the made assumption λ2 = λ3 , is a double eigenvalue, we must have T33 = λ2 = λ3 . Consequently, we must also have Tn3 = λ2 n3 = λ3 n3 . Hence, n3 is an eigendirection corresponding to the double eigenvalue λ2 = λ3 . Moreover, it is easy to see that if n is a unit vector situated in the plane determined by n2 and n3 , it is also an eigendirection corresponding to the double eigenvalue λ2 = λ3 . Indeed, if n = αn2 +βn3 , α, β ∈ R then

Tn = T (αn2 + βn3 ) = αTn2 +βTn3 = αλ2 n2 +βλ2 n3 = λ2 (αn2 + βn3 ) = λ2 n. In other words, if λ1 6= λ2 = λ3 , any unit vector orthogonal to n1 is an eigendirection corresponding to the double eigenvalue. Summing up the obtained results, we obtain, for the matrix of the components of T in our basis, the following simple diagonal form:   λ1 0 0 λ2 0  . [Tkm ] =  0 0 0 λ2 Consequently, the tensor T in the basis {nk nm } has the following expression: T = λ1 n1 n1 +λ2 n2 n2 + λ2 n3 n3 .

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537

SOLUTIONS TO SOME PROBLEMS Using again the relation n1 n1 +n2 n2 +n3 n3 = 1, we get the spectral representation of T in the following form: T = λ1 n1 n1 +λ2 (1 − n1 n1 ) .

In order to prove the spectral representation if λ1 = λ2 = λ3 , we denote by n1 the eigendirection corresponding to λ1 ; i.e. Tn1 = λ1 n1 . Let us consider two unit vectors n2 , n3 such that the system n1 , n2 , n3 is an orthonormal basis in V ; i.e. nk · nm = δkm , k, m = 1, 2, 3. Using again the general formula (1.1.13) and the symmetry of T, we can see that the matrix of components of T in the basis {nk nm } becomes 

λ1 [Tkm ] =  0 0

0 T22 T23

 0 T23  . T33

Consequently, the characteristic equation (1.1.28) becomes 2 (λ1 − λ) {λ2 − (T22 + T33 ) λ + T22 T33 − T23 } = 0.

Since we must have λ1 = λ2 = λ3 , the components T22 , T33 and T23 must satisfy the relations T22 = T33 = λ1 and T23 = 0. Thus, the matrix of the components takes the form   λ1 0 0 λ1 0  , [Tkm ] =  0 0 0 λ1

and, hence, T can be expressed by the relation

T = λ1 n1 n1 +λ1 n2 n2 +λ1 n3 n3 or T = λ1 1. It is easy to see now that any unit vector n is an eigendirection of the tensor T, corresponding to the triple eigenvalue λ1 = λ2 = λ3 . P1.34 Since the characteristic equation (1.1.28) has real coefficients and is of degree three, any tensor has a real eigenvalue. Let us denote by λ this real eigenvalue of the orthogonal tensor Q and let n1 be the corresponding eigendirection; we have Qn1 = λn1 .

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SOLUTIONS TO SOME PROBLEMS

From this equation, we get Qn1 · Qn1 = λ2 n1 · n1 = λ2 , since n1 is an unit vector. Using the definition of the transposed tensor, we obtain Qn1 · Qn1 = n1 QT Qn1 . Since Q is an orthogonal tensor QT Q = 1, hence, Qn1 · Qn1 = n1 · 1n1 = n1 · n1 = 1. Consequently, the real eigenvalue λ of any orthogonal tensor Q must satisfy the equation λ2 = 1. Hence, we must have λ = +1

or

λ = −1.

Let us consider now two unit vectors n2 , n3 , such that the system n1 , n2 , n3 is an orthonormal basis in V. Thus, we have nk ·nm = δkm ,

k, m = 1, 2, 3.

Moreover, since n1 is an eigendiretion of Q, corresponding to the eigenvalue λ1 = 1 or λ1 = −1, we must have Qn1 = ±n1 . Let us denote by Qkm the components of Q in the tensor basis {nk nm } . We get Q11 = n1 · Qn1 = n1 · (±n1 ) = ±1,

Q21 = n2 · Qn1 = n2 · (±n1 ) = 0,

Q31 = n3 · Qn1 = n3 · (±n1 ) = 0.

We observe now, that from the equation satisfied by n1 , we obtain QT Qn1 = ±QT n1 . But QT Q = 1, since Q is an orthogonal tensor. Hence, n1 satisfies also the equation QT n1 = ±n1 . Using this result and the definition of the transposed tensor, we get Q12 = n1 · Qn2 = n2 · QT n1 = n2 · (±n1 ) = 0,

Q13 = n1 · Qn3 = n3 · QT n1 = n3 · (±n1 ) = 0.

The obtained results lead to the following form of the matrix of the components Q km of the orthogonal tensor Q:   ±1 0 0 [Qkm ] =  0 Q22 Q23  . 0 Q32 Q33

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We recall again that Q is an orthogonal tensor, hence, satisfies the equation QQ T = 1. Consequently, its components Qkm in any basis must satisfy the following six equations: Qkl Qml = δkm , k, m = 1, 2, 3. In our special basis {nk nm }, the above system takes the following simplified form: Q222 + Q223 = 1, Q22 Q32 + Q23 Q33 = 0, Q232 + Q233 = 1. The general solution of the first equation can be written in the form Q22 = cos θ, Q23 = − sin θ

θ ∈ [0, 2π),

with

and the general solution of the third equation can be expressed in the form Q32 = sin ϕ, Q33 = cos ϕ

with

ϕ ∈ [0, 2π).

It is easy to see that the second equation can be satisfied taking ϕ = θ. Hence, the matrix of the components of an orthogonal tensor in the chosen special basis takes the special form   ±1 0 0 [Qkm ] =  0 cos θ − sin θ  with θ ∈ [0, 2π). 0 sin θ cos θ Consequently, the characteristic equation becomes cos θ − λ − sin θ (±1 − λ) sin θ cos θ − λ

or, equivalently,

= 0,


(±1 − λ) λ2 − 2λ cos θ + 1 = 0.

Hence, the eigenvalues of an orthogonal tensor Q are given by the following relations: λ1 = ±1, λ2 = cos θ + i sin θ,

λ3 = cos θ − i sin θ,

i=

√

−1.

Thus, generally, an orthogonal tensor has two complex-conjugate eigenvalues if and only if θ = 0 or θ = π; all eigenvalues of Q are real numbers. Obviously, in these specials cases, Q is also a symmetric tensor as it is easy to see evaluating its components for θ = 0 or θ = π. For all other values of θ from the domain [0, 2π), Q is not symmetric. P1.35 First we must prove that the system {ek el em } , k, l, m = 1, 2, 3, of 27 tensor product is linearly independent. In order to do this, let us assume that the real numbers λklm , k, l, m = 1, 2, 3 exist such that λklm ek el em = 03 ,

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k, l, m = 1, 2, 3,

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SOLUTIONS TO SOME PROBLEMS

where 03 is the third order null-tensor. Using the definition of this tensor, the definition of the tensor product of three vectors and the above relation, we successively get 02 = 03 es = (λklm ek el em ) (es ) = λklm ek el (em · es ) = λklm ek el δms = λkls ek el ,

k, l, s = 1, 2, 3.

Hence, if our assumption is made, we can conclude that the following system: λkls ek el = 02 , for s = 1, 2, 3, must be verified, where 02 = 0 is the second order null-tensor. But the system {ek el } , k, l = 1, 2, 3, is a basis in the vector space of the second order tensors, consequently it is a linearly independent system. Consequently, the above equations can be satisfied if and only if λkls = 0 for k, l, s = 1, 2, 3. Our result proves the linear independence of the system {ek el em } . Let us consider now a third order tensor Φ. In this case, according to the definition of a third order tensor, Φem is a second order tensor. Consequently, it can be expressed in the following form: Φem = Φklm ek el . Let us consider now an arbitrary vector v = vs es . Using again the definition of the tensor product of three vectors, as well as the linearity of Φ, we successively obtain (Φ−Φklm ek el em ) v = (Φ−Φklm ek el em ) (vs es ) = vs {Φes −Φklm ek el (em · es )}

= vs {Φes −Φklm ek el δms } = vs {Φes −Φkls ek el } . Taking into account the equation defining the quantities Φklm and the above result, we can conclude that (Φ−Φklm ek el em ) v = 02 for any vector v. Consequently, Φ−Φklm ek el em = 03 , and the last equation shows that Φ = Φklm ek el em . Hence, any third order tensor is a linear combination of the 27 tensor products {ek el em } , k, l, m = 1, 2, 3, which form a linearly independent system. Taking into account the general definition of a basis in a vector space, we can conclude that the system {ek el em } is a basis in the vector space L3 of the third order tensors. At the same time, it follows that dimL3 = 27. The quantities Φklm , k, l, m = 1, 2, 3, are the components of the tensor Φ in the basis {ek el em } . i.e.

b 4 is positive definite; P1.41 Let us assume that the fourth order tensor Φ from L T · ΦT ≥0

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SOLUTIONS TO SOME PROBLEMS for any symmetric second order tensor from LS 2 and T · ΦT = 0

if and only if

T 6= 0.

Let us assume also that a second order tensor T 6= 0 exists, such that ΦT = 0 and T 6= 0. In this case, we will have also T · ΦT = 0 and T 6= 0. However, such a situation cannot occur, since we have assumed that the tensor Φ is positive-definite. Consequently, the equation ΦT = 0 can have only the solution T = 0. In other words, if U is a symmetric second order tensor, the equation ΦT = U or, equivalently, the linear algebraic system Φklmn Tmn = Ukl ,

k, l = 1, 2, 3,

has a unique solution and this solution depends linearly on U. Let us denote this solution T in the following manner T = Φ−1 U. b 4 and we have Obviously, Φ−1 ∈ L 
Φ Φ−1 U = ΦΦ−1 (U) = U

for any symmetric second order tensor U. Consequently, ΦΦ−1 = b I,

b 4 . Also, from the above equation, it follows where b I is the fourth order unit tensor in L that Φ−1 Φ = b I.

Hence, the fourth order tensor Φ−1 is the inverse of Φ, consequently, Φ is nonsingular. Let us consider now an arbitrary symmetric tensor U and let T be given by the relation T = Φ−1 U. Since Φ is positive definite, we have

0 ≤ T · ΦT = Φ−1 U · Φ Φ−1 U = Φ−1 U· ΦΦ−1 (U) Consequently,

= Φ−1 U · b IU = Φ

−1

U · Φ−1 U ≥0

U · U = U · Φ−1 U.

for any symmetric second order tensor U. Let us assume now that U · Φ−1 U =0.

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SOLUTIONS TO SOME PROBLEMS

Accordingly, we will have also T · ΦT = 0,

and, hence, T = 0, since Φ is positive definite. But U = ΦT, and thus U = 0. These results show that the inverse tensor Φ−1 is positive definite if Φ is positive definite. P1.42 (a) It is easy to see that the components cklmn of the tensor c satisfy the following symmetry properties: cklmn = clkmn = cklnm = cmnkl , b4 . hence, c is an element of the vector space L (b) Let us denote by ϕ the fourth order tensor defined by the equation ϕ = ck.

According to the property given in P1.39, we have ϕklmn = cklrs krsmn , or taking into account the expression of cklrs and krsmn ϕklmn = {λδkl δrs + µ(δkr δls + δks δlr )} ·   1 λ (δrm δsn + δrn δsm ) δrs δmn + − 4µ 2µ(3λ + 2µ) 1 = (δkm δln + δkn δlm ). 2

Consequently, according to the equation (1.1.48), ϕklmn = Ibklmn,

b4 . where Ibklmn are the component of the unit tensor Ib in the vector space L Hence ck = b I4 , b 4 is invertible, and its inverse and we can conclude that the fourth order tensor c from L b 4 ; i.e. tensor is the fourth order tensor k from L c−1 = k.

(c) To establish the condition in which the tensor c is positive definite, let us consider an arbitrary symmetric second order tensor ε and let us denote by σ the symmetric second order tensor defined by the equation σ = cε. If σnl and εmn are the components of σ and ε, respectively, it is easy to see that σnl are expressed in terms of εmn by the following equation: σkl = λεmm δkl + 2µεkl .

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543

SOLUTIONS TO SOME PROBLEMS Since we have ε · cε = ε · σ = εkl σkl ,

from the above results, we obtain

ε · cε = λεkk εmm + 2µεkl εkl . Let us denote by θ the trace of ε and let e be the deviatoric part of ε; i.e. θ = trε, e = ε −

1 1 θ1 and ε = θ1 + e with tre = 0. 3 3

From the last two equations, we get εkl =

1 θδkl + ekl and ekk = 0, 3

where ekl are the components of e. Introducing the above results in the equation giving ε · cε, we obtain 2µ 2 )θ + 2µe · e. ε · cε = (λ + 3 If we introduce the number k defined by the relation

k =λ+

2µ , 3

the last result takes the following form: ε · cε = kθ 2 + 2µe · e. Since θ and e are reciprocally independent quantities, the last equation shows that the tensor c is positive definite if and only if the numbers λ and µ satisfy the inequalities k =λ+

2µ and µ > 0. 3

Chapter 2 P2.2 Let us assume now that the infinitesimal strain field   ∇u (x) + ∇u (x)T ε (x) = 2

are identically zero on B. Hence, we have ε11 = u1,1 = 0, ε22 = u2,2 = 0, ε33 = u3,3 = 0 and 2ε12 = u1,2 + u2,1 = 0, 2ε23 = u2,3 + u3,2 = 0, 2ε31 = u3,1 + u1,3 = 0. From the first three equations, it follows that u1 (x1 , x2 , x3 ) does not depend on x1 , u2 (x1 , x2 , x3 ) does not depend on x2 and u3 (x1 , x2 , x3 ) does not depend on x3 . Hence, we have u1 = f1 (x2 , x3 ), u2 = f2 (x3 , x1 ) and u3 = f3 (x1 , x2 ),

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SOLUTIONS TO SOME PROBLEMS

where f1 (x2 , x3 ), f2 (x3 , x1 ) and f3 (x1 , x2 ) are arbitrary functions. Using the fact that ε12 , ε23 and ε31 are also vanishing, we conclude that these functions must satisfy the following differential equations:

∂f2 (x3 , x1 ) ∂f1 (x2 , x3 ) + ∂x1 ∂x2 ∂f3 (x1 , x2 ) ∂f2 (x1 , x3 ) + ∂x2 ∂x3 ∂f1 (x2 , x3 ) ∂f3 (x1 , x2 ) + ∂x3 ∂x1

=

0,

=

0,

=

0.

Differentiating the first equation with respect to x3 and the third equation with respect to x3 , we get ∂ 2 f1 (x2 , x3 ) ∂ 2 f1 (x2 , x3 ) = 0. = 0 and 2 ∂x23 ∂x2 From the first equation, we obtain

∂f1 (x2 , x3 ) = g1 (x3 ) , ∂x2

hence, f1 (x2 , x3 ) = g1 (x3 ) x2 + h1 (x3 ) , where g1 (x3 ) and h1 (x3 ) are arbitrary functions, depending only on x3 . Consequently,

dh1 (x3 ) dg1 (x3 ) ∂f1 (x2 , x3 ) x2 + = dx3 dx3 ∂x3

and

d2 h1 (x3 ) d2 g1 (x3 ) ∂ 2 f1 (x2 , x3 ) . x2 + = 2 2 dx23 dx3 ∂x3

Introducing the last result in the second equation satisfied by f1 (x2 , x3 ), we get

d2 h1 (x3 ) d2 g1 (x3 ) = 0. x2 + 2 dx23 dx3

Since the functions g1 (x3 ) and h1 (x3 ) depend only on x3 , we must have

d2 h1 (x3 ) d2 g1 (x3 ) = 0. = 0 and 2 dx23 dx3

Hence, g1 (x3 ) and h1 (x3 ) are linear functions of x3 and we have g1 (x3 ) = α1 x3 + a1 and h1 (x3 ) = β1 x3 + b1 , where, α1 , a1 , β1 and b1 are arbitrary constant numbers. From the above relations, it follows that the function f1 (x2 , x3 ) has the following expression: f1 (x2 , x3 ) = α1 x2 x3 + a1 x2 + β1 x3 + b1 . Consequently, u1 (x1 , x2 , x3 ) becomes u1 (x1 , x2 , x3 ) = α1 x2 x3 + a1 x2 + β1 x3 + b1 .

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545

SOLUTIONS TO SOME PROBLEMS Thus, by circular permutations, we obtain u1 = α 1 x2 x3 + a 1 x2 + β 1 x3 + b 1 , u2 = α 2 x3 x1 + a 2 x3 + β 2 x1 + b 2 , u3 = α 3 x1 x2 + a 3 x1 + β 3 x2 + b 3 ,

where α1 , ..., b3 are arbitrary constants. We recall now that the obtained functions must satisfy the relations ε12 = ε23 = ε31 = 0. From this fact, after elementary computations, we can conclude that the following condition for any x1 , x2 and x3 , must be satisfied: α1 x3 + a1 + α2 x3 + β2 = 0, α2 x1 + a2 + α3 x1 + β3 = 0, α3 x2 + a3 + α1 x2 + β1 = 0. These equations can be satisfied for any x1 , x2 and x3 if and only if α1 , ..., b3 satisfy the following restrictions: α1 + α2 = 0,

α2 + α3 = 0,

α3 + α1 = 0,

a1 + β2 = 0,

a2 + β3 = 0,

a3 + β1 = 0.

and From the above equations, it follows that we must have α1 = α2 = α3 = 0. At the same time, if we take β1 = ω2, β2 = ω3 , β3 = ω1 , we get a1 = −ω3 , a2 = −ω1 , a3 = −ω2 , ω1 , ω2 and ω3 being arbitrary constants. Consequently, we get for the components u1 , u2 , u3 of the displacement field, the following expressions: u1 = ω 2 x3 − ω 3 x2 + b 1 , u2 = ω 3 x1 − ω 1 x3 + b 2 , u3 = ω 1 x2 − ω 2 x1 + b 3 . Introducing the constant vectors ω and b, with components ω1 , ω2 , ω3 and b1 , b2 , b3 , respectively, we can express now the displacement field u (x) in the following equivalent vectorial form: u (x) = ω × x + b. The last result shows that if the infinitesimal strain tensor is vanishing, the corresponding displacement field corresponds to an infinitesimal rigid displacement.

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SOLUTIONS TO SOME PROBLEMS

P2.9 Let us assume that the orthogonal tensor Q is a symmetry transformation. Hence, according to the relation (2.2.1), Q satisfies for every symmetric tensor ε the following equation:   c QεQT = Q (cε) QT .

Consequently, the following relation also is true:   QT c Qε QT Q = QT Q (cε) QT Q,

for every symmetric tensor ε. We recall now that Q is an orthogonal tensor; hence, it satisfies the restrictions QT Q = QQT = 1. Consequently, the above relation becomes   QT c Qε QT Q = cε,

for every symmetric tensor ε. Hence, we can conclude that if the orthogonal tensor Q is a symmetry transformation, and if it satisfies the equation (2.2.1), then it satisfies also the restriction (2.2.2). The converse of this affirmation can be proved in an analogous manner, using again the orthogonality of Q. P2.12 Let us assume that P (e1 , e2 ) is a plane of symmetry of the material at x. According to equations (2.2.7), in this case, there exists a symmetry transformation Q ∈ Sx such that, Qe1 = e1 and Qe2 = e2 . Let us introduce now the orthonormal basis {e1 , e2 , e3 }, where e3 is a unit vector orthogonal to e1 and to e2 . Since, according to the definition of a symmetry plane, the unit vectors e1 and e2 are mutually orthogonal, we shall have ek ·el = δkl for k, l = 1, 2, 3. Let us denote by Qkm the components of Q in the tensor basis {ek , em }, k, m = 1, 2, 3. As we know, these components are given by the equations Qkm = ek · Qem . Using these relations, we get Q11 = e1 · Qe1 = e1 · e1 = 1, Q12 = e1 · Qe2 = e1 · e2 = 0,

Q21 = e2 · Qe1 = e2 · e1 = 0, Q22 = e2 · Qe2 = e2 · e2 = 1,

Q31 = e3 · Qe1 = e3 · e1 = 0, Q32 = e3 · Qe2 = e3 · e2 = 0.

We recall now, that Q is an orthogonal tensor, that is, QQT = QT Q = 1. Consequently, Q, e1 and e2 satisfy also the following equations: QT e1 = e1 and QT e2 = e2 .

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SOLUTIONS TO SOME PROBLEMS This way, using also the definition of the transposed tensor, we get Q13 = e1 · Qe3 = e3 · QT e1 = e3 · e1 = 0,

Q23 = e2 · Qe3 = e3 · QT e2 = e3 · e2 = 0.

Thus, we can conclude that the matrix has the following simple form:  1 [Qkm ] =  0 0

of the components of Q in the chosen basis 0 1 0

 0 0 . Q33

Moreover, since Q is an orthogonal tensor, as we know, its components must satisfy the restrictions Qkl Qml = δkl . From here, we can conclude that Q33 = ±1. Consequently, the matrix of the components becomes   1 0 0 0 . [Qkm ] =  0 1 0 0 ±1 In particular, the transformation Q− having the components,   1 0 0  −  0  Qkm =  0 1 0 0 −1

is an element of the symmetry group Sx . But any symmetry group contains the transformation −1 and since Q− ∈ Sx , the composition (−1) Q− is also a symmetry trans˜ this transformation, it is easy to see that its formation of the material. Denoting by Q components are given by the following equation:   −1 0 0 h i ˜ km =  0 −1 0  . Q 0 0 1 Also, taking into account equation (2.2.3), we can conclude that the matrix of the components of the transformation R (e3 , π) has the following form in our tensor basis:   −1 0 0 −1 0  . [Rkm (e3 , π)] =  0 0 0 1

˜ = R (e3 , π). Hence, if P (e1 , e2 ) is a symmetry transformation of Accordingly, Q the material at x, then R (e3 , π) is a symmetry transformation of the material at the point x. Let us assume now that R (e3 , π) ∈ Sx . Then, taking into account the properties of the symmetry group Q− = (−1) R (e3 , π) ∈ Sx , it is easy to see that Q− , e1 and e2 satisfy the equations Q − e1 = e 1 and Q − e2 = e 2 . Since, as we have seen, Q− is a symmetry transformation of the material at the location x, from the above equations, we can conclude that the plane P (e1 , e2 ) spanned

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SOLUTIONS TO SOME PROBLEMS

by the two mutual orthogonal unit vectors e1 and e2 is a plane of symmetry of the material at the same location, and our proof is complete. P2.14 Let us denote by G the set formed by the orthogonal transformations ±1, ± R (e1 , π) , ± R (e2 , π) and ±R (e3 , π). The matrix of the components of these transformations in the tensor basis {ek , el }, k, l = 1, 2, 3, are given by the following equations:     1 0 0 −1 0 0 −1 0 , [1km ] =  0 1 0  , [−1km ] =  0 0 0 1 0 0 −1     1 0 0 −1 0 0 0  , [−Rkm (e1 , π)] =  0 1 0 , [Rkm (e1 , π)] =  0 −1 0 0 −1 0 0 1     −1 0 0 1 0 0 1 0  , [−Rkm (e2 , π)] =  0 −1 0  , [Rkm (e2 , π)] =  0 0 0 −1 0 0 1     −1 0 0 1 0 0 −1 0  , [−Rkm (e3 , π)] =  0 1 0 . [Rkm (e3 , π)] =  0 0 0 1 0 0 −1 We must show that the product of any two of these 8 transformations is also an element of G. Using the above matrix representations, it is easy to verify that R (e1 , π) R (e2 , π) = R (e2 , π) R (e1 , π) = R (e3 , π) , R (e2 , π) R (e3 , π) = R (e3 , π) R (e2 , π) = R (e1 , π) , R (e3 , π) R (e1 , π) = R (e1 , π) R (e3 , π) = R (e2 , π) . Now, the desired result follows immediately since, according to the assumption made R (ek , π) as well as −R (ek , π), k = 1, 2, 3, are elements of the set G. P2.17 Let [u, ε, σ] be the elastic state of a homogenous body. According to the general definition (2.1.20), the mean stress Σ corresponding to this state is given by the equation Z 1 σdv, Σ= v B

where v is the volume of the domain B occupied by the body in its reference configuration. Since [u, ε, σ] is an elastic state, the stress σ and the strain ε are connected by the constitutive equation σ = cε, c representing the elasticity of the material. Consequently, the above equation becomes Z 1 cε dv. Σ= v B

According to the assumption made, the body is homogenous; hence, its elasticity c is a constant tensor and the relation giving the mean stress Σ takes the form, Z 1 Σ = c εdv. v B

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SOLUTIONS TO SOME PROBLEMS

We recall now that according to the general definition (2.1.13), the mean strain E corresponding to the elastic state [u, ε, σ] is given by the equation Z 1 εdv. E= v B

Taking into account this fact for the mean stress Σ, we get the expression Σ = cE. We recall also the mean strain theorem, according to which the mean strain E depends only on the boundary values of the displacement u and is given by the following equation: E=

1 2v

Z

(un + nu) dv.

∂B

From the last two equations, we can conclude that Σ=

1 c 2v

Z

(un + nu) dv.

∂B

b 4 . Consequently, its As we know, the elasticity c is an element of the vector space L components cklmn satisfy the relations cklmn = cklnm . Using this property, finally we obtain Z 1 Σ = c unda, v ∂B

and the deduction is complete. P2.20 To prove the generalized reciprocal theorem, we observe that according to the definition of an elastic state, we have σ = cε and σ ˜ =˜ cε˜. Consequently, σ · ε˜ = ε˜ · cε and σ ˜ ·ε = ε·˜ cε˜. Recalling now the definition (1.1.42) of the transposed tensor of a tensor from the b 4 , we obtain vector space L ˜ = ε · cT ε˜. σ·ε

According to the assumption made, ˜ c = cT , and we obtain σ · ε˜ = ε · ˜ cε˜. Taking into account the obtained results, we can conclude that σ · ε˜ = σ ˜ · ε. Hence, we have

Z

B

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σ · ε˜dv =

Z

B

σ ˜ · ε dv.

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SOLUTIONS TO SOME PROBLEMS

Also, we know that ε˜ =

and

  1 1 ∇u + ∇uT ∇˜ u + ∇˜ uT , ε = 2 2 ˜ = 0, div σ + b = 0. div σ ˜ +b

Using these equations and the divergence theorem (1.2.30), we can conclude that Z Z Z sn · u ˜ da + b · u ˜ dv σ · ε˜dv = B

and

Z

B

∂B

σ ˜ · εdv =

B

Z

˜ sn · uda +

Z

˜ · udv, b

B

∂B

with sn = σn and ˜ sn = σ ˜ n on ∂B. Comparing the last three integral relations, we obtain the desired result; i.e. Z Z Z Z ˜ · udv. ˜ sn · uda + b sn · u ˜ da + b · u ˜ dv = ∂B

B

∂B

B

P2.21 Let us assume that the traction problem (2.3.16) has a solution. Hence, there exists an elastic state [u, ε, σ] corresponding to [b, b s] which satisfies the equilibrium equation div σ + b = 0 in B and the boundary condition, sn = σn = b s on ∂B,

the body force field b and the traction b s being given in B and on ∂B, respectively. Integrating the first equation on B, we obtain Z Z div σdv + bdv = 0. B

B

Using the Gauss-Ostrogradsky theorem, we get Z Z σnda + bdv = 0. B

B

Taking into account the boundary condition, we can conclude that if the traction problem has a solution, the given external force b and b s must satisfy the restriction Z Z b sda + bdv = 0. ∂B

B

To prove the second, necessary condition, we use the component force of the field equation and use the Ricci’s symbols eklm defined by the following equations:  for klm = 123, 231, 312,  1 −1 for klm = 132, 321, 213 eklm =  0 if at least two indices have the same value.

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551

SOLUTIONS TO SOME PROBLEMS Let us assume that u and v are two vectors from V and w = u × v.

It is easy to see that the components wk of the vector w can be expressed in terms of the components ul and vm of u and v, respectively, by the relations, wk = eklm ul vm , k, l, m = 1, 2, 3. As we know, according to the assumption made, the equilibrium equation and the traction boundary condition are satisfied. The component form of these relations are σmp,p + bm = 0 in B and σmp xp = sbm on ∂B.

From the equilibrium equation, it results that the above equations are also satisfied on B, eklm xl σmp,p + eklm xl bm = 0. Hence, we have also eklm (xl σmp ),p − eklm xl,p σmp + eklm xl bm = 0 on B. As we know xl,p = δlp , we get eklm xl,m σmp = eklm δlp σmp = eklm σml . From the definition of the Ricci’s symbols, it follows that eklm = −ekml . We recall now that the Cauchy’s stress tensor σ is symmetric; i.e. σ T = σ and σml = σlm . The last two properties show that eklm σml = 0. Consequently, we have eklm (xl σmp ),p + eklm xl bm = 0 on B. Integrating these equations on B and using the Gauss-Ostrogradsky theorem, we obtain Z Z eklm xl σmp np da + eklm xl bm dv = 0, k = 1, 2, 3. B

∂B

Using the traction boundary condition satisfied on ∂B these relations become Z Z eklm xl sbm da + eklm xl bm dv = 0, k = 1, 2, 3. B

∂B

These three scalar equations can be written in the following equivalent vector form: Z Z x×b sda + x × bdv = 0. ∂B

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B

552

SOLUTIONS TO SOME PROBLEMS

Now the proof is complete. P2.28 According to equations (2.2.73), the Young’s moduli E1 , E2 , E3 > 0 and the Poisson’s ratios ν12 , ν21 , ν32 , ν23 , ν13 and ν31 must satisfy the following restrictions:

ν31 ν23 ν13 ν21 ν32 ν12 . = , = , = E3 E2 E1 E2 E3 E1

Since E1 , E2 , E3 > 0, from these equations it follows without any difficulty that ν12 ν23 ν31 = ν21 ν32 ν13 . We observe that the inequality (2.2.79) can be expressed in the following equivalent form:

E1 2 E3 2 E2 − 2ν21 ν13 ν32 > 0. − ν13 − ν32 E1 E3 E2 In turn, this relation can be written as    2 E1 2 2 E2 2 E3 2 E2 − 2ν21 ν13 ν32 > 0, − ν21 − ν32 ν13 1 − ν13 1 − ν32 E2 E1 E1 E3 2 1 − ν21

or, equivalently, 

2 1 − ν32

E2 E1



2 1 − ν13

E3 E1



−

(

ν21



E1 E2

1

2

+ ν32 ν13



E2 E1

 1 )2 2

> 0.

From this inequality, we can conclude that

1 1   2 2 2 E3 2 E2 1 − ν13 − 1 − ν32 E1 E1

<

ν21

<





E1 E2

1

2 1 − ν32

2

E2 E1

+ ν32 ν13

1  2



E2 E1

2 1 − ν13

1

E3 E1

2

<

1 2

.

Finally, after elementary computations, we get (  1  1  1 ) 2 2 E2 2 E2 2 E2 2 E3 + 1 − ν32 1 − ν13 − ν32 ν13 < E1 E1 E1 E1

< ν21 <  1  1  1 ) 2 2 E2 2 E2 2 E2 2 E3 − 1 − ν32 1 − ν13 . − ν32 ν13 E1 E1 E1 E1 (

The last inequalities give a constraint on one Poisson’s ratio ν21 , in terms of two others, ν32 and ν13 . P2.32 As we already know, the elasticity tensor c of a material is positive definite if and only if its rigidity matrix [C] is positive definite. In turn, this matrix is positive definite if and only if its inverse, that is, the compliance matrix [S] of the material is positive definite. To find the restrictions that must be fulfilled by the engineering constants of the material, to ensure the positive definiteness of [S], we must use the

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553

SOLUTIONS TO SOME PROBLEMS

Sylvester’s criterion and the equation (2.2.87) expressing [S] in terms of the technical constants. Thus, we are led to the following restrictions: E1 , E3 , G13 > 0, r

|ν12 | < 1, |ν13 | <

and

E1 , E3

2 1 2 2
(1 + ν12 ) ν13 > 0, − 1 − ν12 E1 E3

or

2 2 1 ν13 > 0. (1 − ν12 ) − E1 E3 Taking into account the reciprocity relation

ν31 ν13 , = E3 E1

we can rewrite the last inequality in the following equivalent form: 1 − ν12 − 2ν13 ν31 > 0. Moreover, we can see that ν31 must satisfy the restriction r E3 . |ν31 | < E1

If we assume that a traction tensile stress acting in the isotropy plane of the material produces elongation in the direction of its action, and contraction in the perpendicular direction situated in the isotropy plane, and suppose also that a traction tensile stress acting perpendicular to the isotropy plane produces elongation in the direction of its action and contraction in any perpendicular direction, we can conclude that ν 12 , ν13 and ν31 are positive; i.e. ν12 , ν13 , ν31 > 0. Hence, if our supplementary hypothesis are fulfilled, the Poisson’s ratios ν12 , ν13 and ν31 must satisfy the following inequalities: r r E3 E1 . , 0 < ν31 < 0 < ν12 < 1, 0 < ν13 < E1 E3

P2.33 Let us assume that the material is isotropic. In this case, according to equation (2.2.54), the stress-strain relation in component form can be expressed by the following relations: σij = λεkk δij + µ (εij + εji ) . Obviously, we have εkk = δkl εkl , εij = δik δjl εkl , εji = δil δjk εkl , and the stress-strain relation becomes σij = {λδij δkl + µ (δik δjl + δil δjk )} εkl .

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554

SOLUTIONS TO SOME PROBLEMS We recall that the component form of the general stress-strain relation is σij = cijkl εkl ,

and the elasticities cijkl satisfy the symmetry relations cijkl = cjikl = cijlk = cklij . Comparing the general stress-strain relations and those corresponding to an isotropic material, and taking into account the above symmetry relations, we can see that for an isotropic material, the elasticities cijkl are given by the following equations: cijkl = λδij δkl + µ (δil δjk + δik δjl ) . Similarly, according to equation (2.2.59), the strain−stress relation for an isotropic material can be expressed in component form by the following equations: εij = −

1+ν ν (σij + σji ) . σkk δij + 2E E

Obviously, we have σkk = δkl σkl , σij = δik δjl σkl , σji = δil δjk σkl and the strain-stress relation becomes   1+ν ν (δik δjl + δil δjk ) σkl . εij = − δij δkl + E E

The component form of the general strain−stress relation is εij = kijkl σkl , and the compliances kijkl satisfy the symmetry relations kijkl = kjikl = kijlk = kklij . Comparing the general strain−stress relations and those corresponding to an isotropic material and taking into account the above symmetry relations, we can see that for an isotropic material, the compliance kijkl are given by the following equations: kijkl = −

1+ν ν (δik δjl + δil δjk ) . δij δkl + 2E E

P2.40 (a) In component form, the equation (2.5.37), giving the displacement produced by a concentrated force, becomes

1 1 16πµ (1 − ν) uk = (3 − 4ν) Pk + 3 (Pm xm ) xk , r r
1 where r = x21 + x22 + x23 2 . To obtain the components uk,l of the gradient ∇u (x), we use the well known relations,

xk,l = δkl and r,l =

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xl . r

555

SOLUTIONS TO SOME PROBLEMS In this way, after elementary computations, we get   3 1 16πµ (1 − ν) uk,l = 3 (Pm xm ) δkl − (3 − 4ν)Pk xl − 2 (Pm xm ) xk xl . r r

Since 2εkl = uk,l + ul,k , the above equation gives   3 1 8πµ (1 − ν) εkl = 3 (Pm xm ) δkl − (1 − 2ν) (xk Pl + Pk xl ) − 2 (Pm xm ) xk xl . r r

Now it is easy to see that the above relation is just the component form of the tensor equation given in (a). (b) As we know, the constitutive equation of an isotropic material has the form σ = λ (trε) 1 + 2µε. We know also that the Lam´e’s coefficients λ and µ can be expressed in terms of Young’s modulus E and Poisson’s ratio ν, by the relations λ=

E Eν . , µ= 2 (1 + ν) (1 + ν) (1 − 2ν)

Hence, we also have

2µν , (1 − 2ν) and the stress-strain relation can be expressed in the following equivalent form,   ν (trε) 1 + ε . σ = 2µ 1 − 2ν λ=

After elementary computations, from the expression of ε, and taking into account the equations tr (x · P) = tr (Px) = P · x, we obtain trε = −2 (1 − 2ν)

1 P · x. r3

Using again the expression of ε obtained in (a) and taking into account the above form of the stress-strain relation, we obtain the expression of the stress field in the following form:   1 3 1 − 2ν 1 (P · x) xx . − (P · x) 1 + Px + xP + σ (x) = − 1 − 2ν r2 8π (1 − ν) r3

Hence, the validity of the relation (2.5.38) is proved. (c) The above tensor relation has the following component form: −

1 3 1 1 1 8π (1 − ν) (Pm xm ) xk xl . σkl (x) = − 3 (Pm xm ) δkl + 3 Pk xl + 3 xk Pl + 1 − 2ν r5 r r r 1 − 2ν

To evaluate σkl,l (x), we use the relations xk,l = δkl , r,l = a long but elementary computation, we obtain

σkl,l (x) = 0 for any x 6= 0.

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xl r

and xl xl = r2 . Thus, after

556

SOLUTIONS TO SOME PROBLEMS

Hence, the Cauchy’s equilibrium equations are satisfied in any point x 6= 0 and the statement made in (c) is proved. (d) Taking into account the expression of σ (x) obtained in (b) for the traction sn = σn, we obtain sn = σn =   1 3 1 − 2ν 1 (P · x) x(x · n) . − (P · x) n + P (x · n) + x(P · n) + − 1 − 2ν r2 8π (1 − ν) r3

We recall now that n is the inward unit normal to the sphere Ση and observe that on this surface the following relations are fulfilled: r = η, n = −

x and x · x = η 2 . η

Taking into account these equations, we get sn = σ · n =

1 − 2ν 1 8π (1 − ν) η 2



P+

1 3 (P · x) x 1 − 2ν η 2



on Ση . Let us denote by R the following force resultant: Z R= σnda. Ση

According to the above expression of sn = σn on Ση , we get  Z  1 3 1 − 2ν 1 (P · x) x da. P + R= 1 − 2ν η 2 8π (1 − ν) η 2 Ση

To evaluate this integral, we shall use a spherical coordinate system indicated in Figure S.1. X3

X

P

O j

q

h X2

X1

Figure S.1: The spherical coordinate system.

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557

SOLUTIONS TO SOME PROBLEMS Accordingly, we have for x on Ση x1 = η sin θ cos ϕ, x2 = η sin θ sin ϕ, x3 = η cos θ, θ ∈ [0, π), ϕ ∈ [0, 2π), da = η 2 sin θdθdϕ, P1 = 0, P2 = 0, P3 = P , P · x = P η cos θ. Hence, the components of R become R1

=

R2

=

R3

=

3P 8π (1 − ν)

Zπ

3P 8π (1 − ν)

Zπ

(1 − 2ν) P 8π (1 − ν)

Zπ 

2

sin θ cos θdθ

0

Z2π

cos ϕdϕ,

Z2π

sin ϕdϕ,

0

2

sin θ cos θdθ

0

0

0

1+

 Z2π 3 cos2 θ sin θdθ dϕ. (1 − 2ν) 0

We have Z2π 0

dϕ = 2π,

Z2π

cos ϕdϕ =

0

Z2π

sin ϕdϕ = 0,

0

Zπ

sin θdθ = 2,

0

Zπ

cos2 θ sin θdθ =

2 3

0

and, finally, we get R1 = R2 = 0, R3 = P. Consequently, the force resultant R is just the applied concentrated force P and we have Z Z sn da = σnda = P. Ση

i.e

Ση

In the same way, you can try to prove that the moment resultant M is vanishing; Z Z M= x × sn da = x × σnda = 0. Ση

Ση

P2.41 To prove the reciprocal theorem for singular elastic states, let us assume that ˜ Let us ˜ are two systems of concentrated loads with disjoint domains D and D. P and P denote by s = [u, ε, σ] and s˜ = [˜ u, ε˜, hσ ˜ ] two singular elastic states, corresponding to the i ˜ ˜ external force systems [b, sn , P] and b, ˜ sn , P , respectively. Let us introduce the set Bη = B −

[

x0 ∈D

[

Ση x0 , Σ η x0 − ˜ x0 ∈D

and take η sufficiently small, such that the balls,
 ˜ Ση x0 = x : x − x0 ≤ η , x0 ∈ D or x0 ∈ D

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SOLUTIONS TO SOME PROBLEMS

˜ and the boundary ∂B of the body, be mutually disjoint. This is possible since D and D 0 0 ˜ are disjoint sets, and x ∈ D, x ∈ D are points in B, which is an open set. According to the definition of singular states, s and ˜ s are regular elastic states on B η and, according to the Betti’s reciprocal theorem for regular elastic states, we have Z Z Z Z ˜ · udv = sn ·˜ uda + b · u ˜ sn · uda + b ˜ dv Bη

∂Bη

Z

Bη

∂Bη

σ ˜ · εdv

Z

=

Bη

σ · ε˜dv.

Bη

˜ the Since s is a regular elastic state on B− D and s˜ is a regular elastic state on B− D, property (ii) and (iii) and the mean value theorem of the integral calculus imply that Z Z

˜ sn · uda = 0 if x0 ∈ D, ˜ x0 , lim lim sn · u ˜ da = P x0 · u η→0 ∂Ση (x0 )

η→0 ∂Ση (x0 )

and lim

Z

η→0 ∂Ση (x0 )

Z

˜ sn · u ˜ da = 0, lim

η→0 ∂Ση (x0 )

Also we have obviously, Z

... =

∂Bη

Z

X

... +



˜ ˜ x0 · u x0 if x0 ∈ D. ˜ sn · uda = P Z

... +

x0 ∈D∂Σ (x0 ) η

∂B

X

Z

...,

˜ x0 ∈D ∂Ση (x0 )

and using the above evaluations, we get Z Z X

˜ x0 · u x0 ˜ sn · uda = ˜ sn · uda + P lim η→0 ∂Bη

and lim

Z

η→0 ∂Bη

˜ x0 ∈D

∂B

sn · u ˜ da =

Z

sn · u ˜ da +

∂B

X

x0 ∈D



˜ x0 . P x0 · u

˜ (x) are continuous functions on B, and u and u We recall that b (x) and b ˜ have the
˜ · u are O r−1 near their singularities. Consequently, ˜ and b property (ii). Hence, b · u Z Z ˜ · udv = b ˜ · udv lim b η→0 Bη

and lim

Z

η→0 Bη

B

b·u ˜ dv =

Z

b·u ˜ dv.

B

Also, since c is invertible, we can conclude from the property (ii) that ε and ε˜ are
O r−2 as the distance r from any of their singularities tends to zero. Thus, since D and

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SOLUTIONS TO SOME PROBLEMS


˜ are disjoint, the property (ii) implies that σ · ε˜ and σ D ˜ · ε are also O r−2 near their singularities. Hence, Z Z ˜ · εdv = σ ˜ · εdv, lim σ η→0 Bη

and lim

Z

η→0 Bη

B

σ · ε˜dv =

Z

σ · ε˜dv.

B

Taking into account the obtained evaluation, we can conclude that if η → 0, Betti’s reciprocal relations, written for the domain Bη , have the following form: Z Z Z X

σ · ε˜dv = sn · u ˜ da + b · u ˜ dv + P x0 · u ˜ x0 B

Z

σ ˜ · εdv

=

B

Z

x0 ∈D

B

∂B

˜ sn · uda +

Z

˜ · udv + b

B

∂B

X

˜ x0 ∈D

 0
˜ x0 · u x , P

and the proof is complete. P2.42 To prove that the reciprocal theorem for singular elastic states is still valid for an infinite media with finite boundary, provided that
u (x) , u ˜ (x) = O r−1 ,
σ (x) , σ ˜ (x) = O r−2 ,
˜ (x) = O r−3 , as r = kxk → ∞, b (x) , b

we consider the ball SR = {x : kxk ≤ R} centered in the point x = 0, having the radius ˜ are R > 0, and such that the finite boundary ∂B of the body, and the sets D and D contained in SR . In these conditions, for the finite domain bounded by the boundary ∂B of the body and by the boundary ∂SR of the ball SR , we can use the reciprocity theorem proved in P2.41. In this way, we obtain the equations Z Z Z Z X

P x0 · u ˜ x0 σ · ε˜dv = sn · u ˜ da + sn · u ˜ da + b·u ˜ dv + B∩DR

Z

B∩DR

∂SR

σ ˜ · εdv =

Z

˜ sn · uda +

∂SR

∂B

x0 ∈D

B∩DR

∂B

Z

˜ sn · uda +

Z

˜ · udv + b

B∩DR

X

˜ x0 ∈D



˜ x0 · u x0 . P

Let us denote, as usual, by sn = σn and ˜ sn = σ ˜ n the traction on the spherical surface ∂SR . From the assumptions made, we can conclude that

sn (x) = O r−2 , ˜ sn (x) = O r−2 as r = kxk → ∞.

Accordingly, we get

lim

Z

R→∞ ∂SR

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sn · u ˜ da = lim

Z

R→∞ ∂SR

˜ sn · uda = 0,

560

SOLUTIONS TO SOME PROBLEMS


since u (x) , u ˜ (x) = O r−1 as r = kxk → ∞. Also, we obtain Z Z Z Z σ · ε˜dv = σ ˜ · εdv = σ lim σ · ε˜dv, lim ˜ · εdv, R→∞ B∩DR

lim

Z

R→∞ B∩DR

R→∞ B∩DR

B

b·u ˜ dv

=

Z

b·u ˜ dv, lim

Z

R→∞ B∩DR

B

B

˜ · udv = b

Z

˜ · udv, b

B

where B is the unbounded domain occupied by the body, having finite boundary ∂B. The assumptions made on the behavior of the involved fields at large distances assure the convergence of the above improper integral. Hence, assuming that R → ∞, the reciprocity relations written for the finite domain B ∩ DR , take the following final form: Z Z Z X

σ · ε˜dv = sn · u ˜ da + b · u P x0 · u ˜ dv + ˜ x0 B

Z

σ ˜ · εdv

B

=

Z

x0 ∈D

B

∂B

˜ sn · uda +

Z

˜ · udv + b

B

∂B

X

˜ x0 ∈D



˜ x0 · u x0 . P

The obtained result can be used to obtain representation theorems for infinite elastic media, containing holes having finite boundaries. P2.43 Suppose that the field w (x) exists and let w (x) = u0 + ω 0 × (x − ξ) where ξ=

1 a

Z

xda

∂B

is the position vector of the centroid of the boundary ∂B, a representing the area ∂B. Obviously, we have Z (x − ξ) da = 0, ∂B

and for the rigid displacement field w (x), we get Z w (x) da = au0 . ∂B

Also, it is easy to see that Z Z (x − ξ) × w (x) da = (x − ξ) × {ω 0 × (x − ξ)} da. ∂B

∂B

Using the well known properties of the vector product, we can express the last result in the following equivalent form: Z Z {(x − ξ) · (x − ξ) ω 0 − (x − ξ) [(x − ξ) · ω 0 ]} da. (x − ξ) × w (x) da = ∂B

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∂B

SOLUTIONS TO SOME PROBLEMS

561

We introduce now the centroidal inertia tensor J of the boundary ∂B, defined by the following equation: Z J= {(x − ξ) · (x − ξ) 1 − (x − ξ) · (x − ξ)} da. ∂B

The components of J are given by the equation Z {(xl − ξl ) (xl − ξl ) δkm − (xk − ξk ) (xm − ξm )} da. Jkm = ∂B

The centroidal inertial tensor J is symmetric and positive definite. Hence, its inverse J−1 exists and is unique. Using the tensor J, we obtain Z (x − ξ) × w (x) da = Jω 0 . ∂B

As we have assumed, w (x) satisfies the equations Z Z wda = f and (x − ξ) × wda = m, ∂B

∂B

f , m being given vectors. The obtained results show that if w (x) satisfies the above equations, u0 and ω 0 must satisfy the relations au0 = f and J ω 0 = m. Hence, the unknown constant vectors u0 and ω 0 must be given by the following equations: u0 =

1 f and ω 0 = J−1 m. a

Our results show that the rigid displacement field w (x), having the desired properties, exists and is uniquely determined by f and m. P2.44 Let us assume that s = [u, ε, σ] is the regular solution of the traction problem, in the absence of body forces and concentrated loads. According to the relation (2.5.25), the components of the displacement field u (x) are given by the equations Z

b up (x) = s · u(p) x0 ; x da x0 . ∂B

We suppose that the given traction b s (x0 ) is a continuous function on ∂B. In this case, we can obtain an integral representation theorem for the displacement gradient ∇u (x), and, hence, for the strain and stress fields ε (x) and σ (x) by differentiating the above relations under the integral sign. The possibility of this operation can be proved in a manner analogous to that used in the classical theory of Newtonian potentials.

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562

SOLUTIONS TO SOME PROBLEMS Particularly, for the strain field ε (x), we get ) ( Z (q) (p)
∂u ˜k (x0 ; x) ∂u ˜k (x0 ; x) 1 0
da x0 . + sbk x εpq (x) = ∂xp ∂xq 2 ∂B

Let us assume now that homogenous traction boundary conditions are given; i.e. b s (x) = σ (x) n (x) = Σn, Σ = ΣT = const. on ∂B,

where Σ is a given symmetric constant tensor. Consequently, we have

sbk (x) = Σkl nl (x) on ∂B,

and the expression for εpq (x) becomes    Z

˜ pqkl x0 ; x da x0 Γ Σkl εpq (x) =   ∂B

where

˜ pqkl x0 ; x Γ




=

1 4

! (q) (p)
∂u ˜k (x0 ; x) ∂u ˜k (x0 ; x) n l x0 + ∂xp ∂xq ∂B ) ! (q) (p)
∂u ˜l (x0 ; x) ∂u ˜l (x0 ; x) 0
da x0 . nk x + ∂xp ∂xq Z (

Obviously, we have the following symmetry properties


˜ pqkl x0 ; x = Γ ˜ qpkl x0 ; x = Γ ˜ pqlk x0 ; x , Γ

and to obtain the above results, we have taken into account the symmetry of the given tensor Σ. Since σrs (x) = crspq εpq (x), for the components of the corresponding stress field σ (x), we obtain the following expressions: σrs (x) = Brskl (x) Σkl , where Brskl (x) = crspq

Z

∂B



˜ pqkl x0 ; x da x0 . Γ

˜ pqkl (x0 ; x) imply the The symmetries of the elasticities crskl and of the functions Γ following symmetries of the function Brskl (x): Brskl (x) = Bsrkl (x) = Brslk (x) . In tensor form, the obtained result can be expressed in the following concentrated form: σ (x) = B (x) Σ.

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563

SOLUTIONS TO SOME PROBLEMS

The fourth order tensor field B (x) is named the influence function corresponding to the homogenous traction problem, and we have just proved its existence together with the symmetry properties given above. P2.47 According to equation (2.7.2), we have Z Z
t




t uci (x) = Gij x − x0 σjk nk x0 da x0 = σjk Gij x − x0 nk x0 da x0 . Γ

Γ

Using the Gauss-Ostrogradsky theorem, we obtain Z
∂Gij (x − x0 ) t dv x0 . uci (x) = σjk ∂x0k D

We recall now equation (2.7.4) to obtain



(xi − x0i ) xj − x0j (3 − 4ν) δij

. + 16πµ (1 − ν) Gij x − x0 =

x − x0 kx − x0 k3

Since

xk − x0k 1 ∂ , = 0 0 ∂xk kx − x k kx − x0 k3

it results 16πµ (1 − ν)

∂Gij (x − x0 ) ∂x0k

xj − x0j xk − x0k 3 − δik 0 kx − x0 k3 kx − x k
(xi − x0i ) xj − x0j (xk − x0k ) xi − x0i . + 3 δjk kx − x0 k5 kx − x0 k3

(3 − 4ν) δij

=

−

Denoting by r = kx − x0 k and by l = (l1 , l2 , l3 ) the length and the direction of the line drawn from the volume element dv (x0 ) at x0 ∈ D toward the point of observation x, we will have xi − x0i xi − x0i = li , = 0 r kx − x k

and, in this way, we get 16πµ (1 − ν)

1 ∂Gij (x − x0 ) = 2 {(3 − 4ν) δij lk − δik lj − δjk li + 3li lj lk } . 0 r ∂xk

Hence, we obtain t 16πµ (1 − ν) uci (x) = σjk

Z

1 {(3 − 4ν) δij lk − δik lj − δjk li + 3li lj lk } dv. r2

D

We observe now that t σjk {(3 − 4ν) δij lk − δik lj }

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= =

t t t (3 − 4ν) σik lk − σij lj = 2 (1 − 2ν) σik lk t (1 − 2ν) σjk (δij lk + δik lj ) .

564

SOLUTIONS TO SOME PROBLEMS

Consequently, uci (x) can be expressed in the following equivalent form: Z 1 t fijk (l) dv, 16πµ (1 − ν) uci (x) = σjk r2 D

where fijk (l) = (1 − 2ν) (δij lk + δik lj ) − δjk li + 3li lj lk . Hence, the Eshelby’s formula (2.7.22)1 is proved. To prove the relation (2.7.22)2 , we take into account that   ν t εtmm δjk + εtjk . σjk = 2µ 1 − 2ν

Taking into account this form of the stress−strain relation, elementary computations show that uci (x) can be expressed also by the equation (2.7.22), the coefficients gijk (l) being given by the relation (2.7.24). P2.57 Let u be an admissible displacement field for the above composite and let (∇u+∇uT ) be the corresponding strain field. Let σ be an admissible selfε = ε (u) = 2 equilibrated stress field corresponding to the same composite. Since u and σ are admissible fields, the necessary null-jump conditions are satisfied, hence, various divergence theorems are true in their usual forms. Moreover, since σ is self-equilibrated, it satisfies the Cauchy’s homogeneous equilibrium equation; i.e. we have div σ = 0, in Bα , α = 0, 1, ..., N . Let us evaluate now the mean value of the scalar product σ · ε. Taking into account the general definition of the mean value, and using the component representation of the field u, ε and σ, we successively get Z 1 (σ − σ ¯ ) (ε − ε¯) dv ¯ ) (ε − ε¯) = σ·ε−σ ¯ · ε¯ = (σ − σ v B

1 v

1 v

Z

Z

(σkl − σ ¯kl ) (εkl − ε¯kl ) dv =

1 v

Z

(σkl − σ ¯kl ) (uk − ε¯km xm ),l dv

B

B

{(σkl − σ ¯kl ) (uk − ε¯km xm )},l dv −

1 v

Z

(σkl − σ ¯kl ),l (uk − ε¯km xm ) dv.

B

B

Since σ is self-equilibrated, we have σkl,l = 0; since σ ¯kl are constant quantities, we have also σ ¯kl,l = 0. Hence, the above formula becomes Z 1 {(σkl − σ ¯kl ) (uk − ε¯km xm )},l dv. σ·ε−σ ¯ · ε¯ = v B

Since u and σ are admissible fields for the composite, the Gauss-Ostrogradsky theorem can be applied in its usual form and we get Z 1 (σkl − σ ¯kl ) (uk − ε¯km xm ) nl da. σ·ε−σ ¯ · ε¯ = v ∂B

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565

SOLUTIONS TO SOME PROBLEMS Obviously, this equation can be expressed in the following equivalent tensor form: Z 1 (u − ε¯x) · (σn − σ ¯ n) da. σ·ε−σ ¯ · ε¯ = v ∂B

We have also

Z

(u − ¯ εx) · σ ¯ nda

Z

=

∂B

(uk − ε¯km xm ) σ ¯kl nl da

∂B

=

σ ¯kl

Z

(uk − ε¯km xm ) nl da.

∂B

Using again the Gauss-Ostrogradsky theorem, we get Z Z (u − ¯ εx) · σ ¯ nda = σ ¯kl (uk − ε¯km xm ),l dv B

∂B

=

σ ¯kl

Z

(uk,l − ε¯km δml ) dv.

B

Since σ ¯kl = σ ¯lk , the above result can be expressed in the following equivalent form: Z Z Z (u − ε¯x) · σ ¯ nda = σ ¯kl (εkl − ε¯kl ) dv = σ ¯ · (ε − ε¯) dv. B

∂B

As we know, 1 v

Z

B

εdv = ε¯,

B

and in this way we can conclude that Z (u − ε¯x) · σ ¯ nda = 0. ∂B

Hence, we have:

σ·ε−σ ¯ · ε¯ =

1 v

Z

(u − ε¯x) · σnda.

B

In the same way, it can be shown that Z ε¯x · (σn − σ ¯ n)da = 0, ∂B

and this result leads us to the following equation: Z 1 u · (σn − σ ¯ n)da. σ·ε−σ ¯ · ε¯ = v ∂B

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566

SOLUTIONS TO SOME PROBLEMS

Chapter 3 P3.6 Since γxy = 2εxy , we have     εx εx  εy  = [R]  εy  . γxy εxy

Using now the equation (3.1.12), we obtain       εx ε1 ε1 −1  εy  = [R][T (−θ)]  ε2  = [R][T (−θ)][R]  ε2  . γxy ε6 /2 ε6

In this way, taking into account the equation    εx  εy  = [R][T (−θ)][R]−1  γxy

(3.1.12), we get S11 S12 0

Now, the relation (3.1.18)1 leads us to the following    S11 S12 εx  εy  = [R][T (−θ)][R]−1  S12 S22 γxy 0 0

S12 S22 0

  0 σ1   σ2  . 0 σ6 S66

equation:    0 σx  [T (θ)]  σy  . 0 σxy S66

We recall now the last result, proved in P3.5. According to the equation obtained, we have [T (θ)]T = [R][T (−θ)][R]−1 .

Hence, finally, we get

   σx εx  εy  = [S(θ)]  σy  σxy γxy 

where the matrix [S(θ)] is given by the following equation:

[S(θ)] = [T (θ)]T [S][T (θ)].

Using the above equation, by direct and elementary computations, it can be shown that equation (3.1.21), expressing the components of the matrix [S(θ)] in terms of the compliances S11 , S12 , S22 , S66 and of the angle θ, holds.

P3.8 In order to plot Ex /E2 as functions of θ ∈ [00 , 900 ], we use the result obtained in P3.7. In this way, we get −1    E2 E2 E2 Ex cos4 θ ≡ f (θ). sin2 θ cos2 θ + − 2ν12 = sin4 θ + E1 E1 G12 E2

For the considered boron-epoxy composite, we have E1 = 10E2 , G12 =

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1 E2 , ν12 = 0.3 , E2 = 18.5GP a. 3

SOLUTIONS TO SOME PROBLEMS

567

Consequently, the function f (θ) giving Ex /E2 as a function of θ ∈ [00 , 900 ] becomes −1  1 . cos4 θ f (θ) = sin4 θ + 2.94 sin2 θ cos2 θ + 10

Taking the derivative of this function, we obtain
f 0 (θ) = − sin θ cos θ −1.88 sin2 θ + 5.48 cos2 θ f 2 (θ).

From this result, we can see that for θ ∈ [00 , 900 ], f 0 (θ) is vanishing if θ = 00 , or θ = 600 , or θ = 900 . Moreover, we have f 0 (θ) < 0 if 00 < θ < 600 , f 0 (θ) > 0 if 600 < θ < 900 . Consequently, f (θ) is a decreasing function in the domain [00 ,600 ], and it is an increasing function in the domain [600 ,90◦ ]. Thus we can see that for θ ∈ [00 , 900 ], f (θ) has a minimum for θ=600 . Elementary computations show that
f 00 = 10, f (600 ) = 0.89, f (900 ) = 1.

Obviously, the first and the last results are not surprising since for θ = 900 , Ex = E1 and for θ = 900 , Ex = E2 . Reasoning in similar manner and using equation (3.1.13), we can plot Gxy /G12 , νxy and −ηxy,x as functions on θ ∈ [00 , 900 ] for the given boron-epoxy composite. The results that can be obtained are presented in the Figure S.2.

Figure S.2: Normalized moduli for a boron/epoxy composite. It is interesting to observe that Ex is slightly smaller then E2 in the neighborhood of θ = 600 . This example shows that the extreme (largest and smallest) material properties do not necessarily occur in principal material directions. Therefore, nothing should be taken for granted with a new composite material and its moduli as a function on θ should be carefully examined to truly understand its character.

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568

SOLUTIONS TO SOME PROBLEMS

P3.10 To analyze the behavior of E1 /Ex we must study the behavior of the function ϕ (θ) obtained in P3.9, for θ ∈ [00 , 900 ]. To do this, first of all, we must obtain the derivative ϕ0 (θ) of this function. Elementary computations show that  ϕ0 (θ) = −4 sin θ cos θ (1 + a − 4b) cos2 θ + (2b − a) .

It is clear that ϕ0 (θ) = 0 if θ = 00 or θ = 900 . But we can have ϕ0 (θ) = 0 also if cos2 θ =

a − 2b . 1 + a − 4b

Obviously, this equation can have real roots in the open interval (0,900 ) if and only if a and b satisfy the following inequalities: a − 2b < 1. 1 + a − 4b

0<

Since we have an orthotropic (fiber reinforced) composite, we assume that a=

It is easy to see that a − 2b

and 1 + a − 4b

Since a > 1, we have obviously



E1 > 1. E2

>0 <0



>0 <0

if b < if b >

if b < if b >

a , 2 a , 2

1+a , 4 1+a . 4

a 1+a < . 2 4

Consequently,

a 1+a a − 2b or if b > . > 0 if b < 2 4 1 + a − 4b Let us assume first that 1+a . b< 4 As we know, in this case, 1 + a − 4b > 0

hence,

1 a − 2b < 1 if b < . 2 1 + a − 4b

Thus, we have 0<

But a > 1, hence

1+a 1 a − 2b . and b < < 1 if b < 4 2 1 + a − 4b

1+a 1 , ≤ 4 2

thus, 0<

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1 a − 2b < 1 if b < . 2 1 + a − 4b

569

SOLUTIONS TO SOME PROBLEMS Let us assume now that b>

a . 2

As we know, in this case, 1 + a − 4b < 0,

hence,

1 a − 2b < 1 if b > . 2 1 + a − 4b

Thus, we have

a 1 a − 2b and b > . < 1 if b > 2 2 1 + a − 4b But a > 1, hence a/2 > 1/2; hence, 0<

0<

a a − 2b < 1 if b > . 2 1 + a − 4b

Summing up the obtained results, we can say that 0<

1 a − 2b < 1 if and only if b < 2 1 + a − 4b

or b >

a . 2

Hence, there exists θ ∈ (0, 900 ) such that

ϕ0 (θ) = 0, θ ∈ (0, 900 ) if and only if b <


Now it is easy to see that for θ ∈ 0, 900 ,  >0 (1 + a − 4b) cos2 θ + (2b − a) <0

and 2

(1 + a − 4b) cos θ + (2b − a)



<0 >0

a 1 or b > . 2 2

for for

θ<θ θ<θ

if b <

for for

θ<θ θ<θ

if b >

1 , 2

a . 2

Returning now to the expression of the derivative ϕ0 (θ) and taking into account
0 0 the above results, we can say that for θ ∈ 0 , 90 ,  1 < 0 for θ < θ if b < , ϕ0 (θ) > 0 for θ < θ 2

and ϕ0 (θ)

Hence, if b <

1 2

if b >

a 2

and

We recall now that

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>0 <0

for for

θ<θ θ<θ

if b >

a . 2


ϕ (θ) has a minimum in θ ∈ 00 , 900 ,


ϕ (θ) has a maximum in θ ∈ 00 , 900 . E1 = ϕ (θ) . Ex

570

SOLUTIONS TO SOME PROBLEMS


and, consequently, we can see that, for θ ∈ 00 , 900 , if b <

and

1 2

Ex (θ) has a maximum in θ ∈ 00 , 900





a Ex (θ) has a minimum in θ ∈ 00 , 900 . 2 Moreover, it is easy to see that if b >


ϕ θ =

hence,

a − 4b2 , 1 + a − 4b


Ex θ 1 + a − 4b . = a − 4b2 E1 Since 0 < a < 1, elementary computations show that

1 1 + a − 4b > 1 if b < , 2 a − 4b2

and

a 1 1 + a − 4b if b > . < 2 a a − 4b2

Thus, we have


1 Ex θ > E1 if b < , 2

and

Ex (θ) <

a E1 if b > . 2 a

As we have seen in P3.9, a=

1 E1 and b = 4 E2



 E1 − 2ν12 . G12

Introducing these values in the last inequalities, we get

and


Ex θ > E 1


Ex θ < E 2

if G12 >

if G12 <

2


E1 , θ ∈ 00 , 900 2 (1 + ν12 )



E2 E1 E2

+ ν12


 , θ ∈ 00 , 900 .

Summing up the obtained results and taking into account that E2 < E1 , we can say that if the engineering constants of the considered fiber reinforced orthotropic composite
satisfy the first restriction, given above, then their exists an angle θ ∈ 00 , 900 such
that Ex θ is greater than both E1 and E2 ; on the contrary, if the engineering constants of the composite satisfy the
second restriction, given above, then there exists an angle
θ ∈ 00 , 900 such that Ex θ is less than both E1 and E2 . We observe that in P3.8, the second of these conditions is fulfilled.

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SOLUTIONS TO SOME PROBLEMS

571

P3.11 We shall prove the validity of the first equation (3.1.24). The validity of the other five equations (3.1.24) can be proved in a similar manner. We start with the first equation (3.1.17), according to which

Q11 (θ) = Q11 cos4 θ + 2 (Q12 + 2Q66 ) sin2 θ cos2 θ + Q22 sin4 θ. We use now the following trigonometrical identities:

1 1 1 (1 − cos 2θ) , cos2 θ = (1 + cos 2θ) , sin2 2θ = (1 − cos 4θ) , 2 2 2 1 1 4 4 sin θ = (3 − 4 cos 2θ + cos 4θ) , cos θ = (3 + 4 cos 2θ + cos 4θ) . 8 8

sin2 θ =

Introducing these relations in the above expression of Q11 (θ), after elementary computations, we get

Q11 (θ) =

1 1 (3Q11 + 3Q22 + 2Q12 + 4Q66 ) + (Q11 − Q22 ) cos 2θ 2 8 1 + (Q11 + Q22 − 2Q12 − 4Q66 ) cos 4θ. 8

Hence, if 1 (3Q11 + 3Q22 + 2Q12 + 4Q66 ) , 8 1 U2 = (Q11 − Q22 ) , 2 1 U3 = (Q11 + Q22 − 2Q12 − 4Q66 ) , 8

U1 =

then Q11 (θ) takes the form

Q11 (θ) = U1 + U2 cos 2θ + U3 cos 4θ, and the first equation (3.1.24) is proved, the constant quantities U1 , U2 , U3 being given by the first three equations (3.1.25). To obtain the other five equations (3.1.24) also, the following trigonometrical identities must be used: 1 cos3 θ sin θ = (2 sin 2θ + sin 4θ) , 8 1 cos θ sin3 θ = (2 sin 2θ − sin 4θ) . 8

P3.15 The considered boron/epoxy reinforced composite lamina has the following engineering constants: E1 = 206.85GP a , E2 = 20.68GP a , ν12 = 0.3 , G12 = 6.86P a. From the reciprocity relation recalled in P3.14, we get ν21 = 0.03.

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572

SOLUTIONS TO SOME PROBLEMS

Also, the formulas used in P3.14 lead to the following values of the reduced stiffnesses: Q11 = 204.99GP a , Q22 = 20.87GP a , Q12 = 6.26GP a , Q66 = 6.89GP a. Taking into account again the relations (3.1.25) for the constants U 1 , ..., U5 , we obtain U1 = 97.05GP a , U2 = 92.06GP a , U3 = 23.22GP a , U4 = 29.48GP a , U5 = 30.11GP a.

 (a) To plot the reduced transformed stiffnesses Q11 (θ) for this lamina and for θ ∈ 00 , 1800 , we use equation (3.1.24)1 ; we have

  Q11 (θ) = U1 + U2 cos 2θ + U3 cos 4θ ≡ f (θ) , θ ∈ 00 , 1800 .

The first derivative of this function has the following expression: f 0 (θ) = −2 (U2 + 4U3 cos 2θ) sin 2θ.

  For θ ∈ 00 , 1800 , this derivative is vanishing for θ = 00 , 900 , 1800 and if cos 2θ = −

U2 . 4U3

Obviously, this equation has real solution if and only if U2 4U3 ≤ 1.

For the given boron/epoxy composite, we have:

Q11 (θ) = f (θ) = 97.05 + 92.06 cos 2θ + 23.22 cos 4θ, f 0 (θ) = −2 (92.06 + 92.88 cos 2θ) sin 2θ, and cos 2θ = −0.991.   0 0 For θ ∈ 0 , 180 , this equation has two solutions close to 900 :

θ1 = 86.20 and θ2 = 93.80 .   From the obtained results, it is easy to see that for θ ∈ 00 , 1800 ,  < 0 if 00 < θ < 86.20 ,    > 0 if 86.20 < θ < 900 , f 0 (θ) < 0 if 900 < θ < 93.80 ,    > 0 if 93.80 < θ < 1800 .   Hence, for θ ∈ 00 , 1800 , f (θ) = Q11 (θ) is a decreasing

function in the interval 00 , 86.20 , an increasing function on the interval 86.20 , 900 , a decreasing function on 0 0 the interval (900, 93.80 ) and  an increasing function on the interval (93, 8 ,180 ). Consequently, for θ ∈ 00 , 1800 , Q11 (θ) has a relative maximum
Q11 00 = U1 + U2 + U3 = 212.33 for θ = 00 ,

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573

SOLUTIONS TO SOME PROBLEMS a relative minimum

Q11 (θ1 ) = U1 − 0.991U2 + 0.965U3 = 27.63 for θ = θ1 = 86.20 , a relative maximum

a relative minimum


Q11 900 = U1 − U2 + U3 = 28.21 for θ = 900 ,

Q11 (θ1 ) = U1 − 0.991U2 + 0.965U3 = 27.63 for θ = θ2 = 93.80 , a relative maximum


Q11 1800 = U1 + U2 + U3 = 212.33 for θ = 1800 .

The obtained results show that the minimum values of Q11 (θ), obtained for θ = θ1 = 86.20 and θ = θ2 = 93.80 , are very close to its relative maximum value, obtained for θ = 900 . In the Figure S.3,  the function Q11 (θ) for the considered boron/epoxy laminate  and for θ ∈ 00 , 1800 is plotted. The little flashes directed down show the minimum values of Q11 (θ) and the little flashes directed up show its maximum values. Q11

200

U3

U2

100

U1

86.2 93.8

0

180

90 o

Figure S.3: Variation of Q11 with θ. Using equation (3.1.24)3 and the obtained values for U1 , ..., U5 , we get   Q22 (θ) = 99.05 − 92.06 cos 2θ + 23.22 cos 4θ, θ ∈ 00 , 1800 .

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574

SOLUTIONS TO SOME PROBLEMS

Using the same method as before, we can see that Q22 (θ) has relative maximum for θ = 00 , θ = 900 , θ = 1800 , and minimum for θ1 = 3.80 and θ2 = 176.20 . The minimum values of Q22 (θ) are very close to its relative maximum value corresponding to θ = 00 and to θ = 1800 .   Function Q22 (θ) for θ ∈ 00 , 1800 is plotted in the Figure S.4. Q

22

3.8

0

176.2

90

180 o

Figure S.4: Variation of Q22 with θ. Now using equation (3.1.24)2 , we obtain

  Q12 (θ) = U4 − U3 cos 4θ = 29.48 − 23.22 cos 4θ, θ ∈ 00 , 1800 .

This function has the same minimum value for θ = 00 , 900 , 1800 , and the same maximum value for θ = 450 , 1350 . We have


Q12 00 = Q12 900 = Q12 1800 = U4 − U3 = 6.26,

Q12 450 = Q12 1350 = U4 + U3 = 52.70.   In Figure S.5, the function Q12 (θ) for θ ∈ 00 , 1800 is shown. Taking into account now equation (3.1.14)6 , we obtain   Q66 (θ) = U5 − U3 cos 4θ = 30.11 − 23.22 cos 4θ, θ ∈ 00 , 1800 .

This function has the same minimum value for θ = 00 , 900 , 1800 and the same maximum value for θ = 450 , 1350 , and we have


Q66 00 = Q66 900 = Q66 1800 = U5 − U3 = 6.89,

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575

SOLUTIONS TO SOME PROBLEMS

Q66 450 = Q66 1350 = U5 + U3 = 53.33.

Q12

U3

40

U4 20

0 90

o

180

Figure S.5: Variation of Q12 with θ.   In Figure S.6, the function Q66 (θ) for θ ∈ 00 , 1800 is plotted.

Q66

40

U5 20

0

GXy 90

180 o

Figure S.6: Variation of Q66 with θ. (b) To plot the reduced transformed stiffness Q16 (θ), we start with equation (3.1.24)4 ; we have   1 Q12 (θ) = − U2 sin 2θ − U3 sin 4θ ≡ g (θ) , θ ∈ 00 , 1800 . 2 The first derivative of this function is

g 0 (θ) = −8U3 cos2 2θ − U2 cos 2θ + 4U3 .

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576

SOLUTIONS TO SOME PROBLEMS

  This derivative is vanishing for θ ∈ 00 , 1800 if the equation p −U2 ± U22 + 128U32 cos 2θ = 16U3   0 0 has a real solution in the interval 0 , 180 . For the considered composite, we get

Q12 (θ) = −10.435 sin 2θ − 23.22 sin 4θ,

and the above equation  takes the following form: cos2θ = 0.22 or cos2θ = −0.99. For θ ∈ 00 , 1800 , the first equation has the following solutions θ 1 = 29.950 and θ 2 = 150.05, and in the same interval, the solutions of the second equation are

θ3 = 87.80 and θ 4 = 92.20 .

It is easy to see that for θ 1 , Q16 (θ) has its maximal value, for θ 4 it has a local maximum, and for θ 2 , Q16 (θ) has its minimal value, while for θ 2 it has a local minimum. However, the extremal values corresponding to θ 3 and θ 4 are very close to 0, the value of Q16 (θ) for θ = 900 . Also, it is easy to see that inθ = 900 ,Q16 has an inflexion point. In Figure S.7, the function for Q16 (θ) for θ ∈ 00 , 1800 is plotted.

Q16 60

29.95

87.8

150.05

0 180

o

92.2

-60

Figure S.7: Variation of Q16 with θ. Finally, using equation (3.1.24)5 , for the given composite we get   1 Q26 (θ) = − U2 sin 2θ + U3 sin 4θ = −10.435 sin 2θ + 23.22 sin 4θ, θ ∈ 00 , 1800 2

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SOLUTIONS TO SOME PROBLEMS

In order to analyze the behavior of this function, we must use the same method as before. The results of this analysis are presented in Figure S.8.

Q26 60

0

90

o 180

-60

Figure S.8: Variation of Q26 with θ. (c) Summing up the obtained results, we can observe that (1) The reduced transformed stiffnesses Q11 (θ) , Q22 (θ) , Q12 (θ) and Q66  (θ) for the considered boron/epoxy composite lamina have only positive values for θ ∈ 00 , 1800 . (2) The minimal values of Q11 (θ) do not correspond generally to θ = 900 , and the minimal values of Q22 (θ) do not correspond generally to θ = 00 and to θ = 1800 . (3) The reduced transformed stiffnesses Q16 (θ) and Q26 (θ) can have positive, as well as negative values. These stiffnesses are vanishing for θ = 00 , 900 , 1800 .

P3.16 We start with the first equation (3.1.21)

S 11 (θ) = S11 cos4 θ + (2S12 + S66 ) sin2 θ cos2 θ + S22 sin 4θ. Using the trigonometrical identities given in P3.11, we obtain

S 11 (θ) =

1 1 (3S11 + 3S22 + 2S12 + S66 ) + (S11 − S22 ) cos 2θ + 2 8 1 + (S11 + S22 − 2S12 − S66 ) cos 4θ. 8

Denoting V1 =

1 1 1 (3S11 + 3S22 + 2S12 + S66 ) , V2 = (S11 − S22 ) , V3 = (S11 + S22 − 2S12 − S66 ) , 8 2 8

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SOLUTIONS TO SOME PROBLEMS

we can express S 11 (θ) in the following form:

S 11 (θ) = V1 + V2 cos 2θ + V3 cos 4θ. Using the other equations (3.1.21) and the trigonometrical identities given in P3.11, in a similar manner it can be proved that

S 22 (θ) = V1 − V2 cos 2θ + V3 cos 4θ,

S 12 (θ) = V4 − V3 cos 4θ,

S 66 (θ) = V5 − 4V3 cos 4θ,

S 16 (θ) = V2 sin 2θ + 2V3 sin 4θ,

S 26 (θ) = V2 sin 2θ − 2V3 sin 4θ, where

1 1 (S11 + S22 + 6S12 − S66 ) , V5 = (S11 + S22 − 2S12 + S66 ) . 2 8 Now, the validity of the matrix equation given in this problem can be proved directly using the above obtained results and the well-known rules of the matrix calculus. V4 =

P3.21 According to equations (3.3.4) and (3.3.5), the in-plane deformations e αβ and the curvatures kαβ of a laminate, are given by the following geometrical relations:   ∂ 2 U3 ∂Uβ 1 ∂Uα , α, β = 1, 2, , kαβ = − + eαβ = ∂xα ∂xβ ∂xα 2 ∂xβ

where Uα = Uα (x1 , x2 ) are the components of the in-plane displacement of the middle surface and U3 = U3 (x1 , x2 ) is the normal displacement of the same surface. If the in-plane deformations and curvatures are vanishing, the displacement components U1 , U2 , U3 , depending only on x1 , x2 must satisfy the following equations:

∂U2 ∂U1 ∂U2 ∂U1 = 0, + = = ∂x1 ∂x2 ∂x2 ∂x1

∂ 2 U3 ∂ 2 U3 ∂ 2 U3 = 0. = = ∂x1 ∂x2 ∂x22 ∂x21

Elementary computations show that, in this case, the components U1 , U2 and U3 are given by the following relations: ◦

◦

◦

◦

◦

◦

U1 (x1 , x2 ) = − ω 3 x2 + u1 , U2 (x1 , x2 ) =ω 3 x1 + u2 , ◦

◦

◦

◦

U3 (x1 , x2 ) = − ω 2 x1 + ω 1 x2 + u3 , ◦

◦

◦

where u1 , u2 , u3 and ω 1 , ω 2 , ω 3 are arbitrary constants. The displacements U1 , U2 , U3 of an arbitrary point of the laminate, in the LoveKirchhoff plate theory are given by the equations (3.3.1) and (3.3.2); i.e. u1 = U 1 − x 3

∂U3 ∂U3 , u3 = U 3 . , u2 = U 2 − x 3 ∂x2 ∂x1

Thus, we get ◦

◦

◦

u1 = ω 2 x3 − ω 3 x2 + u1 ,

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◦

◦

◦

u 2 = ω 3 x1 − ω 1 x3 + u 2 ,

◦

◦

◦

u 3 = ω 1 x2 − ω 2 x1 + u 3 ,

579

SOLUTIONS TO SOME PROBLEMS or, in vector form, ◦

◦

u (x1 , x2 , x3 ) = u + ω ×x. The last result shows that if the in-plane deformations and the curvatures are vanishing, the laminate can have only infinitesimal rigid displacements. P3.23 According to equation (3.3.16)1 , the extensional stiffness A11 is given by the following equation: N X
Q11 k (zk − zk−1 ) . A11 = k=1

According to the first equation (3.1.24), the reduced transformed stiffness Q11 by the following relation:
Q11 k = U1 + U2 cos 2θk + U3 cos 4θk .




k

is given

Hence, we obtain A11 = U1

N X

k=1

(zk − zk−1 ) + U2

N X

k=1

We also observe that zo = − h2 , zN = N X

k=1

(zk − zk−1 ) cos 2θk + U3 h 2

N X

k=1

(zk − zk−1 ) cos 4θk .

(see Figure 3.11). Hence, we have

(zk − zk−1 ) = h,

h representing the thickness of the laminate. Let us introduce now the quantities V0A =

N P

k=1

(zk − zk−1 ) = h , V1A = V3A =

N P

k=1

Thus we will have

N P

k=1

(zk − zk−1 ) cos 2θk ,

(zk − zk−1 ) cos 4θk .

A11 = U1 V0A + U2 V1A + U3 V3A . All formulas given in P3.23 can be obtained in a similar manner. P3.24 According to equation (3.3.16)2 , the coupling stiffness B11 is given by the following equation: N

1X 2 Q11 k zk2 − zk−1 . B11 = 2 k=1
Introducing here the expression of Q11 k from P3.23, we obtain

B11 =

N N N


1 X 1 X 2 1 X 2 2 2 2 cos 4θk . zk − zk−1 cos 2θk + U3 zk2 − zk−1 U1 + U2 zk − zk−1 2 2 2 k=1 k=1 k=1

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SOLUTIONS TO SOME PROBLEMS

We also observe that

N X

k=1


2 zk2 − zk−1 = 0,

since z0 = − h2 and zN = h2 . If we introduce now the quantities

V0B = 0 , V1B =

N N
1X 2 1X 2 2 2 (zk − zk−1 ) cos 4θk , cos 2θk , V3B = zk − zk−1 2 2 k=1 k=1

For the coupling stiffness B11 , we obtain the following expression: B11 = U2 V1B + U3 V3B . All formulas given in P2.24 can be derived using the same procedure. P3.25 According to equation (3.3.16)2 , the bending stiffness D11 is given by the following equation: N

1X 3 . Q11 k zk3 − zk−1 D11 = 3 k=1
Using the expression of Q11 k , given in P3.23, we get

D11 =

N N N


1 X 1 X 3 1 X 3 3 3 3 cos 4θk . zk − zk−1 cos 2θk + U3 zk3 − zk−1 + U2 zk − zk−1 U1 3 3 3 k=1 k=1 k=1

We recall that z0 = − h2 and zN =

h . 2

N X

k=1

Hence, we have
h3 3 . zk3 − zk−1 = 4

Let us introduce now the quantities V0D =

N N

1X 3 1 X 3 h3 3 3 zk − zk−1 cos 4θk . zk − zk−1 cos 2θk , V3D = , V1D = 3 3 12 k=1 k=1

In this way, we get D11 = U1 V0D + U2 V1D + U3 V3D . All formulas given in P3.25 can be obtained applying the procedure given above. P3.27 (a) According to the matrix equations (3.3.18), we have h ◦i h ◦i N = [A] [e] + [B] [k] , M = [B] [e] + [D] [k] .

Let us assume now that the curvature is vanishing; i.e. kαβ = 0. In this case, the above equations become h ◦i h ◦i N = [A] [e] , M = [B] [e] .

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581

SOLUTIONS TO SOME PROBLEMS From the first equation, we get

h ◦i [e] = [A]−1 N . ◦

◦

◦

Hence, the constant moments resultants M 11 , M 22 and M 12 , which must be applied to obtain the curvatures of the laminate submitted to the constant forces resultants ◦ ◦ ◦ N 11 , N22 and N12 , are given by the following matrix equation: h ◦i h ◦i −1 M = [B] [A] N . (b) According to the matrix equation (3.3.31), we have  h ◦i  h ◦ i  h ◦i  h ◦ i [e] = A0 N + B 0 M , [k] = C 0 N + D0 M .

Let us assume now that the curvatures are vanishing. In this case, we must have  0 h ◦ i  0 h ◦ i C N + D M = 0.

Hence,

h ◦i  0 −1  0  h ◦ i C N . M =− D

According to the last equation (3.3.32), we have  0 −1 D = [D∗ ] and, in this way, we get

h ◦i h i ∗  0 ◦ M = − [D ] C N .

The third equation (3.3.32) shows that

Thus, we obtain



 −1 C 0 = − [D ∗ ] [C ∗ ] . h ◦i h ◦i ∗ M = [C ] N .

According to the equation (3.3.28)3 , we have

[C ∗ ] = [B] [A]−1 . Consequently, h

get

h ◦i ◦ i −1 M = [B] [A] N .

◦

◦

◦

(c) The results obtained in (a) and (b) lead to the same values for M 11 , M 22 , M 12. (d) If the coupling stiffnesses are vanishing, [B] is the null 3 × 3 matrix , and we ◦

◦

◦

M 11 = M 22 = M 12 = 0. P3.28 For a single layer laminate,

h h N = 1, z0 = − , z1 = . 2 2

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SOLUTIONS TO SOME PROBLEMS

Consequently, equations (3.3.16), reported to the principal material directions of the laminate, take the following simplified forms: Aij = hQij , Bij = 0, Dij =

h3 Qij , i, j = 1, 2, 6. 12

Hence, the coupling stifnesses are vanishing. Using the equations (3.1.6), for an isotropic lamina, we obtain the following expressions for the reduced stiffnesses Qij :

E E νE E , , Q66 = , Q22 = , Q12 = 2(1 + ν) 1 − ν2 1 − ν2 1 − ν2 = Q26 = 0.

Q11 =

Q16

Hence, the in-plane stiffnesses Aij and the bending stiffnesses become

1−ν Eh A, ≡ A , A12 = νA , A66 = 2 1 − ν2 A16 = A26 = 0, 1−ν Eh3 D, ≡ D , D12 = νD , D66 = D11 = D22 = 2 12(1 − ν 2 ) D16 = D26 = 0.

A11 = A22 =

P3.29 For a single layer specially orthotropic laminate, we can use the formulas given at the beginning of P3.28. Since for the specially orthotropic laminate Q16 = Q26 = 0, we get A11 = hQ11 , A12 = hQ12 , A22 = hQ22 , A66 = hQ66 , A16 = A26 = 0, Bij = 0 for i, j = 1, 2, 6, D11 =

h3 h3 h3 h3 , D16 = D26 = 0. Q22 , D66 = Q12 , D22 = Q11 , D12 = 12 12 12 12

Using equations (3.1.6), for a specially orthotropic laminate, we get Q11 =

E2 ν12 E2 E1 , Q66 = G12 . , Q22 = , Q12 = 1 − ν12 ν21 1 − ν12 ν21 1 − ν12 ν21

Thus, for the nonvanishing in-plane stiffness, we obtain the following expressions: A11 =

E2 h ν12 E2 h E1 h , Q66 = hG12 . , A22 = , A12 = 1 − ν12 ν21 1 − ν12 ν21 1 − ν12 ν21

The nonvanishing bending stifnesses are given by the relations D11 =

D22 =

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ν12 E2 h3 E 1 h3 , , D12 = 12(1 − ν12 ν21 ) 12(1 − ν12 ν21 )

G12 h3 E 2 h3 . , D66 = 12 12(1 − ν12 ν21 )

583

SOLUTIONS TO SOME PROBLEMS Since we have the reciprocity relation (see Equation (2.2.73)) ν21 = ν12

E2 , E1

we can eliminate ν21 from the above equation. In this way, we obtain all components of the global stiffness matrix [E] expressed by the technical constants E1 , E2 , ν12 , G12 and by the thickness h of the single-layer, specially orthotropic laminate. P3.31 We recall that laminates that are symmetric in both geometry and material properties are called symmetric laminates. We recall also equation (3.3.16) 3 giving the coupling stiffnesses N 1X 2 ). (Qij )k (zk2 − zk−1 Bij = 2 k=1

Let us assume first that the number N of the laminate is even (for N = 6 see Figure

S.9).

Figure S.9: Symmetric laminate; N = even number. Since the laminate is symmetric for each k = 0, 1, ..., N/2, there exists k = N/2 + 1, ..., N such that −zk−1 = zk .

Also, if lk is the thickness of the k-th lamina, and lk is the thickness of the corresponding k-th lamina, we have lk = l k .

Obviously, zk = zk−1 + lk and zk−1 = zk − lk = −zk−1 − lk .

In this way, we obtain for k and for the corresponding k 2 zk2 − zk−1 = 2zk−1 lk + lk2 ,

2 = − 2zk−1 lk − lk2 ; zk2 − zk−1

hence, for k and for the corresponding k, we get 2 2 ). zk2 − zk−1 = −(zk2 − zk−1

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SOLUTIONS TO SOME PROBLEMS

At the same time, we have Bij =

N/2 1X 2 )+ (Qij )k (zk2 − zk−1 2 k=1

N X

k=N/2+1

2 ). (Qij )k (zk2 − zk−1

We recall again that the laminate is symmetric, thus, for k and the corresponding k, the relation takes place (Qij )k = (Qij )k .

Consequently, N X

N/2

k=N/2+1

2 )=− (Qij )k (zk2 − zk−1

X

k=1

2 ). (Qij )k (zk2 − zk−1

Thus, we obtain Bij = 0, i, j = 1, 2, 6. Let us assume now that the number N of the laminate is odd (for N = 5, see Figure S.10).

Figure S.10: Symmetric laminate; N = odd number. For any k = 1, ..., (N − 1)/2, there exists k = (N + 1)/2, ..., N such that −zk−1 = zk .

Also, denoting by lk and lk the thicknesses of the corresponding laminae, we shall have

lk = lk for k = 1, ..., (N − 1)/2 and k = (N + 3)/2, ..., N.

Obviously, zk = zk−1 + lk for k = 1, ..., (N − 1)/2

and

zk−1 = zk − lk = −zk−1 − lk for k = (N + 3)/2, ..., N, k = 1, ..., (N − 1)/2.

In this way, we get 2 zk2 − zk−1 = 2zk−1 lk + lk2 for k = 1, ..., (N − 1)/2

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585

SOLUTIONS TO SOME PROBLEMS and 2 = −2zk−1 lk − lk2 for k = (N + 3)/2, ..., N, k = 1, ..., (N + 1)/2. zk2 − zk−1

Hence, 2 2 ) for k = 1, ..., (N − 1)/2, k = (N + 3)/2, ..., N. zk2 − zk−1 = −(zk2 − zk−1

At the same time, we have (N −1)/2

Bij =

X

k=1

2 2 2 ) + (Qij )(N +1)/2 (z(N (Qij )k (zk2 − zk−1 +1)/2 − z(N −1)/2 ) N X

+

k=(N +3)/2

2 ). (Qij )k (zk2 − zk−1

Since, for k and the corresponding k,

(Qij )k = (Qij )k ,

we obtain

(N −1)/2

N X

k=(N +3)/2

Hence, we get

But

2 )=− (Qij )k (zk2 − zk−1

X

k=1

2 ). (Qij )k (zk2 − zk−1

2 2 Bij = (Qij )(N +1)/2 (z(N +1)/2 − z(N −1)/2 ).

z(N +1)/2 = −z(N −1)/2 ,

since the laminate is symmetric and N is odd. Consequently, finally, we obtain Bij = 0. Obviously, the obtained results for a symmetric laminate are true also for a symmetric cross-ply laminate. P3.33 To obtain Tsai’s formulas (3.2.24) for a laminate considered in P3.32, we observe first that N can be expressed by m and we have N=

m+1 . m−1

We recall also the relations zk = −

h h for k = 0, 1, ..., N ; +k N 2

hence, h for k = 1, ..., N. N According to the definition of a regular symmetric cross-ply laminate, we have also zk − zk−1 =

θk = 00 if k is odd and θk = 900 if k is even .

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SOLUTIONS TO SOME PROBLEMS

Consequently, from the general equations (3.1.17), giving the transformed reduced stiffnesses Qij , we get

Q11

Q12

Q22

Q66







Q11




k

= Q11 if k is odd and

k

= Q12 for any k ;

k

= Q22 = f Q11 if k is odd and

k

= Q66 for any k ;







k

= Q22 = f Q11 if k is even ;

Q22




k

= Q11 if k is even ;

(Q16 )k = (Q26 )k = 0 for any k .

In order to obtain Tsai’s relations given in P3.32, the above properties must be used. We begin with A11 , which, according to the general formula (3.3.1)2 , is given by the equation N X A11 = (Q11 )k (zk − zk−1 ). k=1

Since zk − zk−1 = h/N , the above relation becomes ) ( X X h (Q11 )k . (Q11 )k + A11 = N k=even k=odd

We recall now that k is odd (N + 1)/2 times and k is even (N − 1)/2 times. Hence, for A11 , we obtain the following expression:   N −1 h N +1 Q22 . Q11 + A11 = 2 2 N

Since Q22 = f Q11 , we get A11 =

h {(N + 1) + f (N − 1)} Q11 . 2N

As we have seen in P3.32, N can be expressed by the cross-ply ratio m and it is easy to see that N=

2 2m m+1 . , N −1= , N +1= m−1 m−1 m−1

If we use these relations, finally we get the following expression of the rigidity A 11 : A11 =

m+f hQ11 . 1+m

Thus, the first Tsai formula (3.4.24) was proved. All Tsai formulas (3.4.24) concerning the extensional stiffness can be deduced in the same manner. We recall now the result proved in P3.31 where in the coupling stiffnesses Bij are vanishing for any symmetric laminate. Hence, we have Bij = 0, i, j = 1, 2, 6

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SOLUTIONS TO SOME PROBLEMS

and Tsai’s relations (3.4.24) concerning the coupling stiffnesses are proved. We analyze now the bending stiffness D11 . According to the general relations (3.3.16)3 , we have N
1X 3 . (Q11 )k zk3 − zk−1 D11 = 3 k=1

Since

zk = −

h h +k , N 2

it is easy to see that 3 zk3 − zk−1 =

h3 N3



 3N 2 + 6N + 4 − 3(N + 1)k + 3k 2 . 4

Hence, for D11 , we get ) ( N N N X X 3N 2 + 6N + 4 X h3 2 k (Q11 )k . k(Q11 )k + 3 (Q11 )k − 3(N + 1) D11 = 4 3N 3 k=1 k=1 k=1

We recall now that (Q11 )k = Q11 if k is odd, and (Q11 )k = Q22 = f Q 11 if k is even. We know also that (N + 1)/2 times k is odd and (N − 1)/2 times k is even. Taking into account these properties, from the above equation, we get    N −1 h3 Q11 3N 2 + 6N + 4 N + 1 f + D11 = 2 2 4 3N 3 ) X X X 2 X 2 −3(N + 1)( k+f k) + 3( k +f k ) . k=odd

k=even

k=odd

k=even

Now we use the relations (N is an odd number!) X

k=odd

X

k2 =

k=odd

k=

(N + 1)2 , 4

X

k=

k=even

N (N + 1)(N + 2) , 6

X

k=even

N2 − 1 , 4

k2 =

N (N 2 + 1) . 6

In this way, for D11 , we obtain the following expression:  h3 Q11 3N 2 + 6N + 4 [(N + 1) + (N − 1)f ] D11 = 8 3N 3

 3(N + 1)  (N + 1)2 + (N + 1)(N − 1)f 4  1 + [N (N + 1)(N + 2) + N (N + 1)(N − 1)f ] . 2

−

After elementary computations, we obtain   h3 Q11 N 2 − 2N − 2 [N + 1 + (N − 1)f ] + 2N (N + 1) , D11 = 12N 3 2

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SOLUTIONS TO SOME PROBLEMS

or D11

h3 Q11 = 12



 N 3 − 3N 2 + 2 (f − 1) + 1 . 2N 3

According to Tsai’s formulas (3.4.24), the parameter p, entering in the expressions of the bending stiffness Dij , has the following value:   m(N − 3)[m(N − 1) + 2(N + 1)] 1 . 1 + p= N2 − 1 (1 + m)3

Since m = (N + 1)/(N − 1),

we get

N 3 − 3N 2 + 2 . 2N 3 can be expressed by the relation p=

Accordingly, D11

D11 =

(f − 1)p + 1 3 h Q11 . 12

In this way, we have proved that the first Tsai formula (3.4.24) giving D11 , is true. In similar manner can be shown that all Tsai’s relations concerning the bending stiffnesses are true for a regular symmetric cross-ply laminate. P3.34 Since the elasticity tensor of the composing laminae is positive-definite, we shall have E1 , E2 , G12 > 0 , ν12 ν21 < 1. Moreover, we assume that 0 < ν12 , ν21 < 1. According to the relation (3.1.6),

Q11 =

E2 ν21 E1 E1 , Q66 = G12 . , Q22 = , Q12 = 1 − ν12 ν21 1 − ν12 ν21 1 − ν12 ν21

Hence, we have Q11 , Q12 , Q22 , Q66 > 0. Since we consider a regular symmetric cross-ply laminate, we know that its coupling stiffnesses are vanishing Bij = 0, i, j = 1, 2, 6. Moreover, we have also A16 = A26 = D16 = D26 = 0. Thus, we can conclude that the global stiffness matrix of the laminate [E] will be positivedefinite if and only if the matrices     A11 A12 0 D11 D12 0  A12 A22 0  and  D12 D22 0  0 0 A66 0 0 D66

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SOLUTIONS TO SOME PROBLEMS

will be positive-definite. According to Sylvester’s criterion, this property will take place if and only if A11 , A22 , A66 > 0,

A11 A12 − A212 > 0,

D11 , D22 , D66 > 0,

2 D11 D22 − D12 > 0.

We recall now Tsai’s relation (3.4.24) giving the extensional stiffnesses

A11 =

m+f hQ11 , 1+m

A12 = hQ12 ,

A22 =

1 + mf hQ11 , 1+m

A66 = hQ66 .

In these equations, m is the cross-ply ratio given by the relation (3.4.22) and obviously m > 0. In the same equation, f is the stiffness ratio given by the relation (3.4.23) and we have: 0 < f < 1. Thus, we can conclude that A11 , A22 , A66 > 0. Elementary computations give   E12 h2 (m + f )(1 + mf ) 2 . − ν A11 A22 − A212 = 21 (1 + m)2 (1 − ν12 ν21 )2

We recall now (see P3.32) that the cross-ply ratio m, for a regular symmetric cross-ply laminate, is given by the equation m=

N +1 , N −1

where N > 1 is the number of layers and is an odd number. Hence, we have m > 1. Now we can conclude that A11 A22 − A212 > 0

if and only if the following supplementary restriction is fulfilled: 2 ν21 <

(m + f )(1 + mf ) . (1 + m)2

We use now Tsai’s relations (3.4.24) giving the bending stiffness D11 =

h3 (f − 1)p + 1 3 Q12 , h Q11 , D12 = 12 12

D12 =

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h3 (1 − f )p + f 3 Q66 . h Q11 , D66 = 12 12

590

SOLUTIONS TO SOME PROBLEMS

We recall (see P3.34) that N 3 − 3N 2 + 2 2N 3 Since for a regular symmetric cross-ply laminate N is an odd number, we have (assuming N >1) N ≥ 3. p=

In this case, it is easy to see that

p > 0 for any N ≥ 3. It is easy to see now that if the supplementary relation p<

1 1−f

is fulfilled, we shall have D11 , D22 , D66 > 0 since 0 < f < 1. Elementary computations give  2 2 D11 D22 − D12 = [(f − 1)p + 1] [(1 − f )p + f ] − ν21

Now we can conclude that

E12 h6 . 144 (1 − ν12 ν21 )2

2 D11 D22 − D12 >0

if the following supplementary condition:

2 ν21 < [(f − 1)p + 1] [(1 − f )p + f ]

is also fulfilled. Summing up the obtained results, we can say that the global stiffness matrix [E] of a regular symmetric cross-ply laminate is positive-definite if the elasticity tensor of the composing laminae is positive-definite, if the ratios ν12 , ν21 , are positive, and if the following supplementary conditions are also fulfilled: 2 ν21 <

(m + f )(1 + mf ) , (1 + m)2

p<

where

1 , 1−f

2 ν21 < [(f − 1)p + 1] [(1 − f )p + f ] ,

N +1 , for N ≥ 3, N −1 E2 < 1, 0
1<m=

p=

Copyright © 2004 by Chapman & Hall/CRC

N 3 − 3N 2 + 2 > 0 , for N ≥ 3, 2N 3

591

SOLUTIONS TO SOME PROBLEMS

N , an odd number, representing the number of layers of the laminate. Hence, the positive-definiteness of the elasticity tensor of the composing laminae and the assumption 0 < ν12 , ν21 cannot assure the positive-definiteness of the global stiffness matrix [E] of a regular symmetric cross-ply laminate! P3.35 The engineering constants of the lamina are E1 = 138GP a , E2 = 9P a , ν12 = 0.3 , ν21 = 0.0196 , G12 = 6.9GP a. Substitution of these values in equation (3.1.6) fields the components of the lamina stiffness matrix associated with the principal material axes.   138.8 2.72 0 0  GP a. [Q] =  2.72 9.05 0 0 6.9

The transformed lamina stiffness matrices for θ = 450 and θ = −450 plies are then found by substituting the above stiffnesses in equations (3.1.17) or in equations (3.1.24), (3.1.25). For the +450 plies,   45.22 31.42 32.44 [Q]+450 =  31.42 45.22 32.44  GP a, 32.44 32.44 35.6

and for the −450 plies,

 45.22   Q −450 =  31.42 −32.44

31.42 45.22 −32.44

 −32.44 −32.44  GP a. 35.6

Note that the only difference between the stiffness matrices for the two plies is that the shear coupling terms (the terms with subscripts 16 and 26) for the −45 0 ply have the opposite sign from the corresponding terms for the +450 ply. Before finding the laminae stiffnesses, we must determine the distance from the middle surface of the various ply interfaces, according to the Figure 3.23. Since the thickness of the laminae is 0.25, we get z0 = −0.5mm,

z1 = −0.25mm,

z2 = 0,

z3 = 0, 25mm,

z4 = 0.5mm.

The laminate extensional coupling and bending stiffnesses are found by substituting these distances, along with the lamina stiffnesses given above, in equations (3.3.16). It results   45.22 31.42 0 0  GP a − mm, [A] =  31.42 45.22 0 0 35.6   0 0 0 [B] =  0 0 0  GP a − mm2 , 0 0 0   3.77 2.62 2.03 [D] =  2.62 3.77 2.03  GP a − mm3 . 2.03 2.03 2.97

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592

SOLUTIONS TO SOME PROBLEMS

Hence, all coupling stiffnesses are vanishing. P3.37 (a) We have (see P3.35) [B] = 0 and from the relation (3.3.28)1 and (3.3.32)1 , we get [A0 ] = [A∗ ] = [A]−1 . Since [M ] = 0, from the equation (3.3.31), we obtain [e] = [A0 ][N ] = [A]−1 [N ]. The inverse of the matrix [A] obtained in P3.35 has the following expression:   0.04276 −0.0297 0  (GP amm)−1 . 0 [A]−1 =  −0.0297 0.04276 0 0 0.02809

We know that

N11 = 50 × 10−3 M P amm−1 , N22 = N12 = 0.

Hence, for the in-plane deformations, we get       0.04276 −0.0297 0 e11 50 0.002138  e22  =  −0.0297 0.04276   0  × 10−3 =  −0.001485  . 0 2e12 0 0 0.02809 0 0 

Since [B] = 0, from the equation (3.3.28), we obtain [B ∗ ] = 0, and the relation (3.3.32) gives [B 0 ] = 0. Analogously, the relation (3.3.28)3 gives [C ∗ ] = 0, and hence, according to the equation (3.3.32)3 , we shall have [C 0 ] = 0. Consequently, the relation (3.3.31) giving [k], becomes [k] = [D 0 ][M ]. But, [M ] = 0, hence, [k] = 0. That is, all curvatures are vanishing. (b) In order to obtain the resulting deformations, we must use the general relations (3.3.3) ε11 = εαβ = eαβ + x3 kαβ . Since kαβ = 0, we get

Copyright © 2004 by Chapman & Hall/CRC

   ε11 0.002138  ε22  =  −0.001485  . 2ε12 0 

593

SOLUTIONS TO SOME PROBLEMS

Hence, the deformation of the laminate is homogeneous. Neglecting an infinitesimal rigid displacement and using the strain-displacement relations for the components of the displacement vector, we obtain the following expressions: u1 = 0.002138x1 , u2 = −0.001485x2 , u3 = 0. (c) Substituting equations (3.3.10), we    45.22 σ11  σ22  =  31.42 σ12 32.44

the above strains and the lamina stiffnesses from P3.35 in the find that the stresses in the +450 plies are     31.42 32.44 0.02138 50 3 45.22 32.44   −0.001485  × 10 M P a =  0  M P a. 32.44 35.6 0 21.2

Similarly, the stresses in    σ11  σ22  =  σ12 

the −450 plies are

45.22 31.42 31.42 45.22 −32.44 −32.44  50  M P a. 0 = −21.2

  −32.44 0.002138 −32.44   −0.001485  × 103 M P a 35.6 0

Note that since all curvatures vanish for this problem, the stresses do not depend on the distance x3 = z. Let us observe also that the normal stress σ22 is vanishing, but the shear stress σ12 is nonzero! P3.38 (a) Since, for the considered laminate, there exist nonvanishing coupling stiffnesses, we must invert the full stiffness matrix   A B [E] = . B D

Composing the full stiffness matrix from the [A], [B], [D] matrices of P3.36, and inverting, we find the resulting midplane strains and curvatures to be     0.043866 −0.02861 0 0 0 −0.02083 e11  e22   −0.02861 0.04386 0 0 0 −0.02083        2e12   0 0 0.03284 −0.02083 −0.02083 0 =     k11   0 0 −0.02083 0.52625 −0.34331 0       k22   0 0 −0.02083 −0.34331 0.52625 0 2k12 −0.02083 −0.02083 0 0 0 0.39356     50 0.02193  0   −0.001430       0    0 −3    . ×  × 10 =   0 0      0    0 0 −0.001042

(b) In order to obtain the resulting deformations, we must use equation (3.3.3); i.e. εαβ = eαβ + x3 kαβ , α, β = 1, 2.

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594

SOLUTIONS TO SOME PROBLEMS

Thus, we get ε11 = e11 = 0.002193, ε22 = e22 = −0.00143, 2ε12 = 2x3 k12 = −0.001042x3 . The curvatures are expressed in terms of the normal displacement U3 (x1 , x2 ) by equations (3.3.5). Thus, we obtain

1 ∂ 2 U3 ∂ 2 U3 ∂ 2 U3 = − 0.001042. 2 = 0, 2 = 2 ∂x1 ∂x2 ∂x2 ∂x1

Hence, negating rigid displacements, we obtain u3 = U3 (x1 , x2 ) −

1 0.001042x1 x2 . 2

Now, using equation (3.3.2), we obtain u1 = U1 (x1 x2 ) −

1 0.001042x2 x3 , 2

1 0.0001042x1 x3 . 2 Using the strain-displacement relations, we get u2 = U2 (x1 x3 ) −

∂U2 (x1 x2 ) ∂U1 (x1 x2 ) = ε22 = −0.00143. = ε11 = 0.002193 , ∂x2 ∂x1

Hence, U1 (x1 x2 ) = 0.002193x1 + V1 (x2 ), U2 (x1 , x2 ) = −0.00143x2 + V2 (x1 ). At the same time, we must satisfy the equation

∂u2 ∂u1 = 2ε12 = −0.001042x3 . + ∂x1 ∂x2

Hence, we must have

1 dV2 (x1 ) 1 dV1 (x2 ) − 0.001042x3 = −0.001042x3 . − 0.001042x3 + 2 dx1 2 dx2

Thus, we get

dV2 (x1 ) dV1 (x2 ) = 0. + dx1 dx2 This relation can be satisfied only if

dV2 (x1 ) dV1 (x2 ) = const. =− dx1 dx2

Hence, V1 (x2 ) = (const.)x2 + const., V2 (x1 ) = −(const.)x1 + const.

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595

SOLUTIONS TO SOME PROBLEMS Obviously, we have here a rigid displacement which will be neglected. Therefore, for the displacement u1 , u2 and u3 , we get u1 = 0.002193x1 − 0.0000521x2 x3 ,

u2 = −0.00143x2 − 0.0000521x1 x3 ,

u3 = −0.000521x1 x2 .

(c) In order to obtain the stresses, we must use equation (3.3.10). Due to nonvanishing curvature, the strains and stress now depend on the distance z = x3 . For example, at the top surface of the 1 ply (−45) z = x3 = −0.5 mm (see Figure 3.24), and according to the results obtained at (b), the resulting total strains are ε11 = 0.002193, ε22 = −0.00143, 2ε12 = 0.000521 for z = x3 = −0.5mm. Similarly, at the bottom surface of the 1 ply (−450 ), or at the top surface of the 2 ply (+450 ), z = x3 = −0, 25mm and the strains are ε11 = 0, 002193,

ε22 = −0, 00143,

2ε22 = 0, 000269.

In order to obtain the stresses at the top surface of the 1 play(-450 ), we can use equation (3.3.10) or the relation (3.1.15). The transformed reduced stiffnesses were de0 termined in P3.35. To obtain the stresses at the top surface of the 1 ply(−45   ), we must use the strains corresponding to z = x3 = −0, 5 and the rigidity matrix Q −450. Thus, equation (3.1.15) gives        45.22 31.42 −32.44 0.002193 σx σ11  σ22  =  σy  =  31.42 45.22 −32.44   −0.001430  × 103 M P a σxy −32.44 −32.44 35.6 0.000521 σ12   37.3 =  −12.7  M P a. −6.2

Similar calculations, for the other plies, yield the values shown in the following Table (see Figure 3.24): Location 1T op 1Bottom 2T op 2Bottom 3T op 3Bottom 4T op 4Bottom

x3 = z(mm) −0.5 −0.25 −0.25 0.0 0.0 0.25 0.25 0.5

σ11 (M P a) 37.3 45.8 62.7 54.2 54.2 62.7 45.8 37.3

σ22 (M P a) −12.7 −4.2 12.7 4.2 4.2 12.7 −4.2 −12.7

σ12 (M P a) −6.2 −15.5 34.0 24.7 −24.7 −34.0 15.5 6.2

These results show that the stress distribution across the thickness of the considered laminate is quite complex, even for simple uniaxial loading. This is typical for all laminates exhibiting coupling.

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596

SOLUTIONS TO SOME PROBLEMS

P3.40 A regular symmetric angle ply laminate has orthotropic laminae of equal thickness. The adjacent laminae have opposite signs of the angle orientation of the principal material directions with respect to the laminate axes. For symmetry, such a laminate must have an odd number of layers (see Figure 3.15). According to the general equation (3.1.16)1 , we have A11 =

N X

k=1

(Q11 )k (zk − zk−1 ).

Also, the results proved in P3.39 show that (Q11 )k = Q11 ≡ (Q11 )−α .

Hence, we get A11 = Q11

N X

k=1

Obviously, N X

k=1

and thus, we obtain

(zk − zk−1 ).

zk − zk−1 = h,

A11 = hQ11 .

With the same procedure, we get (A12 , A22 , A66 ) = h(Q12 , Q22 , Q66 ),

with

Q12 ≡ (Q12 )−α , Q22 ≡ (Q22 )−α , Q66 ≡ (Q66 )−α .

Let us evaluate now the stiffness A16 . We start with the equation A16 =

N X

k=1

(Q16 )k (zk − zk−1 ).

According to the definition of the considered laminate, we have (see P3.38). (Q16 )k = (Q16 )−α = Q16 for k = odd number ,

(Q16 )k = (Q16 )+α = −Q16 for k = even number.

Hence, we get A16 = Q16

(

X

k=odd

(zk − zk−1 ) −

X

k=even

(zk − zk−1 )

Since the laminate is regular zk − zk−1 =

Copyright © 2004 by Chapman & Hall/CRC

k for any k . N

)

.

597

SOLUTIONS TO SOME PROBLEMS

Since the laminate is symmetric, N is an odd number. Hence, k is (N + 1)/2 times odd number and k is (N − 1)/2 times an even number. Hence,   N −1 h N +1 h . − A16 = Q16 2 N 2 N Finally, it results h A16 = Q16 . N In the same way, it can be seen that

A26 =

h Q with Q26 ≡ (Q26 )−α . N 26

We analyze now the behavior of the coupling stiffnesses Bij . According to equation (3.3.16)2 , we have N 1X 2 ). (Qii )k (zk2 − zk−1 B11 = 2 k=1

We know that (see P3.38)

(Q11 )k = (Q11 )−α ≡ Q11 for any k .

Hence, we get B11 =

Since

N X 1 2 (zk2 − zk−1 ). Q11 2 k=1

zk−1 = zk −

we obtain 2 zk2 − zk−1 =

h , N

h2 {2k − (N + 1)} . N2

Hence, B11 becomes B11

N h2 1 h2 X 1 {2k − (N + 1)} = Q11 2 = Q11 2 N 2 N 2 k=1

But

N X

k=

k=1

hence,

(

2

N X

k=1

k − N (N + 1)

N (N + 1) , 2

B11 = 0. In similar manner, we can show that B12 = B22 = B66 = 0. Similarly, we have B16 =

N N 1 h2 X 1X 2 (Q16 )k (zk2 − zk−1 )= (Q16 )k {2k − (N + 1)} . 2 2 N2 k=1

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k=1

)

.

598

SOLUTIONS TO SOME PROBLEMS

According to the results proved in P3.39, (Q16 )k = (Q16 )−α ≡ Q16 if k is an odd number,

and

Thus, we get B16

(Q16 )k = − Q16 if k is an even number.

1 h2 Q = 2 N 2 16

(

X

k=odd

[2k − (N + 1)] −

X

k=even

[2k − (N + 1)]

)

.

We recall now that N is an odd number and k is (N + 1)/2 times an odd number and (N − 1)/2 times an even number. We use also the relations (see P3.33) X

k=

k=odd

(N + 1)2 4

,

X

k=even

k=

N2 − 1 . 4

Consequently, we obtain B16 =

Hence,

 h2 (N + 1)2 − (N + 1)2 − (N 2 − 1) + (N + 1)(N − 1) . Q 4N 2 16 B16 = 0.

Similarly, it can be shown that B26 = 0. Hence, all coupling stiffnesses of a regular symmetric angle ply laminate are vanishing. The same conclusion can be obtained using the general results given in P3.31! We analyze now the bending stiffnessess. According to equation (3.3.16) 3 , we have D11 =

N 1X 3 ). (Q11 )k (zk3 − zk−1 3 k=1

Since

(Q11 )k = (Q11 )−α ≡ Q11 for any k, we obtain

D11 =

It is easy to see that 3 zk3 − zk−1 =

h3 N3



N X 1 3 (zk3 − zk−1 ). Q11 3 k=1

 2N 2 + 6N + 4 − 3(N + 1)k + 3k 2 . 4

Hence, D11 becomes D11

h3 Q = 3N 3 11

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(

N

N

k=1

k=1

X X 2 3N 2 + 6N + 4 N − 3(N + 1) k+3 k 4

)

.

599

SOLUTIONS TO SOME PROBLEMS We use now the well-known relations N X

N X N (N + 1)(2N + 1) N (N + 1) and k2 = 6 2

k=

k=1

k=1

After elementary computations, we obtain D11 = −

h3 Q . 12 11

Similarly, it can be shown that (D12 , D22, D66 ) = −

h3 (Q , Q , Q ). 12 12 22 66

For the evaluation of D16 , we start with the equation D16 =

N 1X 3 ). (Q16 )k (zk3 − zk−1 3 k=1

In this way, we obtain D16 =

  N 2N 2 + 6N + 4 h3 X 2 − 3(N + 1)k + 3k . Q ) ( k 16 4 3N 3 k=1

Since (Q16 )k = Q16 if k is odd and (Q16 )k = −Q16 if k is even,

we get D16

h3 Q16 = 3N 3

−

(

X 2 X 2N 2 + 6N + 4 N + 1 k k+3 − 3(N + 1) 2 4 k=odd k=odd

X 2 X 2N 2 + 6N + 4 N − 1 k k−3 + 3(N + 1) 2 4 k=odd k=even

)

.

Since N is odd number, we have X

k=odd

X

k2 =

k=odd

k=

(N + 1)2 , 4

N (N + 1)(N + 2) , 6

X

k=

k=even

X

k=even

N2 − 1 , 4

k2 =

N (N 2 − 1) . 6

Introducing these values in the above expression of D16 , after long but elementary computations, we get h3 2N 2 − 2 Q16 . D16 = 12 N 3 Similarly, we get h3 2N 2 − 2 Q26 . D26 = 12 N 3

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600

SOLUTIONS TO SOME PROBLEMS

P3.41 In the case of regular antisymmetric angle-ply laminate, N is an even number and all layers have the same thickness h/N . According to the results proved in P3.39, it is easy to see that (A11 , A12 , A22 , A66 ) = h(Q11 , Q12 , Q22 , Q66 ).

For A16 , we get A16 =

N X

k=1

(Q16 )k (zk − zk−1 ).

We know that k is N/2 times an odd number, and N/2 times is an even number. Also, if k is odd then (Q16 )k = (Q16 )−α = Q16 ,

and if k is even then (Q16 )k = (Q16 )+α = −Q16 .

Since

zk − zk−1 = h/N,

we obtain

A16 =

h Q N 16

Hence,



N N − 2 2



.

A16 = 0. In the same way, we obtain A26 = 0. According to a result obtained in P3.40, we have ( N ) X h2 1 k − N (N + 1) . B11 = Q11 2 2 N 2 k=1

Since

N X

k=1

we obtain

k=

N (N + 1) , 2

B11 = 0. Similarly, it can be seen that B12 = B22 = B66 = 0. Another result obtained in P3.40 shows that ( ) X X 1 h2 Q [2k − (N + 1)] − [2k − (N + 1)] . B16 = 2 N 2 16 k=odd

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k=even

601

SOLUTIONS TO SOME PROBLEMS

We recall that N is an even number and, in this case, we have (see Equation (3.4.22)) X

k=

k=odd

N2 4

X

,

k=

k=even

N (N + 2) . 4

We know also that k is N/2 times odd, and N/2 times even. Consequently, we obtain   2 N N (N + 2) N N 1 h2 . + (N + 1) − − (N + 1) Q B16 = 16 2 2 2 2 2 N2

Hence, B16 = −

h2 Q . 2N 16

B26 = −

h2 Q . 2N 16

Similarly, we get

Without any difficulty, it can be shown that (D11 , D12 , D22 , D66 ) =


h3 Q11 , Q12 , Q22 , Q66 . 12

A result obtained in P3.40 shows that   N 2N 2 + 6N + 4 h3 X 2 . − 3(N + 1)k + 3k Q ) ( D16 = k 16 4 3N 3 k=1

Hence, D16

=

( ) X  2N 2 + 6N + 4 h3 2 − 3(N + 1)k + 3k Q 4 3N 3 16 k=odd ( ) X  2N 2 + 6N + 4 2 − 3(N + 1)k + 3k . − 4 k=even

Since N is an even number, we have ( see Equation (3.4.24)) X

k2 =

k=odd

N (N 2 − 1) 6

X

,

k=even

k2 =

N (N + 1)(N + 2) . 6

Hence, we obtain D16 =

    2  N (N + 2) 2N 2 + 6N + 4 N N N h3 Q − − 3(N + 1) − 3N 3 16 4 2 2 4 4   N (N + 1)(N + 2) N (N 2 − 1) − . +3 6 6

Now we can conclude that D16 = 0. Similar reasoning leads to the relation D26 = 0.

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602

SOLUTIONS TO SOME PROBLEMS

P3.48 We recall the simply supported edge boundary conditions S1, given in equations (3.6.17) S1 : U3 = 0, Mnn = 0, Un = ϕ, Uτ = ψ. Hence, ϕ and ψ are given functions on the boundary curve ∂D. We introduce now the functional J1 (U ) defined by the equation Z J1 (U) = W (U) − qU3 da. D

Hence, q is a given continuous function on D and W (U)n the energy functional defined by equation (3.6.1). As we know (see Equation (3.6.8)), the first variation δW of W in U, in the direction δU, is given by the following relation: Z

δW = −

(Nαβ,β δUα + Mαβ,αβ δU3 )da +

D

+

Z

{δUn Nnn + δUτ Nnτ + δU3 (Mnτ,τ + Qn ) − δU3,n Mnn } ds.

∂D

We shall calculate now the variation δJ1 of J1 in U, in a direction that satisfies homogeneous boundary condition; i.e. δUn = δUτ = δU3 = 0 on ∂D. Since for the boundary value problem S1 , Mnn = 0 on ∂D, we shall obtain Z δJ1 = − {Nαβ,β δUα + (Mαβ,αβ + q) δU3 } da. D

Based on the last relation and recalling the equilibrium equations (3.5.3), (3.5.5) and the supplementary constitutive relation (3.5.4), we obtain the following variational principle adequate to the simply supported edge boundary conditions S1. If U is a regular solution of the boundary value problem S1, the variation δJ1 of J1 in U is vanishing for any variation ∂U satisfying the homogeneous boundary conditions δUn = δUτ = δU3 = 0 on ∂D. Conversely, if U satisfies the displacement boundary condition given in S1 and if the variation δJ1 of J1 is vanishing in U for any variation ∂U satisfying the above homogeneous boundary conditions, then U is a regular solution of the boundary value problem S1. P3.53 We try to prove the following converse of the principle of minimum potential energy appropriate to the boundary value problem S2. Let U ∈ B2 and suppose that e J2 (U) ≤ J2 (U),

e ∈ B2 . Then U is a solution of the boundary value problem S2. for every U

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603

SOLUTIONS TO SOME PROBLEMS

Let us consider an arbitrary vector field U0 of class C ∞ on D, and suppose that U satisfies the homogeneous conditions 0

U30 = 0 and Uτ0 = 0 on ∂D. e = U + U0 ∈ B2 and the results obtained in P3.50 show that Then U Z Z 0 W (U0 ) − {Uα0 Nαβ,β + U30 (Mαβ,αβ + q)}da + {Un0 (Nnn − φ) − U3,n Mnn }ds ≥ 0. D

∂D

0

This inequality must hold when U is replaced by αU0 , α being an arbitrary real number. Since W (U0 ) ≥ 0, we must have Z Z  0  0 0 − Uα Nαβ,β + U30 (Mαβ,αβ + q) da + Un (Nnn − φ) − U3,n Mnn ds = 0. D

∂D

0

Since U is an arbitrary vector field, from the above equation and from the supplementary constitutive relation (3.6.5), it follows that the equilibrium equations (3.5.8) and the boundary conditions corresponding to S2 are satisfied. Hence, U is a regular solution of the boundary value problem S2. With similar procedures, it can be proved that the converse of the principle of minimum potential energy appropriate for a clamped laminate submitted to the boundary condition C2 is also true. Chapter 4 P4.1 To obtain the kinematical meaning of the influence tensor function A = A(x), we introduce the elementary second order tensor Iij , i, j = 1, 2, 3 having the following components: 1 ij i, j, k, l = 1, 2, 3. = (δik δjl + δil δjk ) , Ikl 2 It is easy to see that the matrices of the components of these 9 tensors have the following structure:     1 0 0 0 1 0  11   12  = 0 0 0 , I =  0 0 0  , ... I . 0 0 0 0 0 0

It follows that any constant symmetric second order tensor ¯ ε with components ε kl can be expressed in the following form: ¯ ε = ε¯kl Ikl .

According to equation (4.1.14), the microscopic strain ε(x) which exists in the body, if on ¯T = const. its boundary an homogeneous displacement condition corresponding to ¯ ε=ε is imposed, is given by the relation ε(x) = A(x)¯ ε. Thus, we obtain

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ε(x) = ε¯kl A(x)Ikl .

604

SOLUTIONS TO SOME PROBLEMS This equation shows that εkl (x) = A(x)Ikl

is the microscopic strain in the body, if on its boundary homogeneous displacement conditions corresponding to the constant symmetric tensor Ikl are imposed. Also, taking into account the definition of this special tensor, for the components εkl ij (x) of the corresponding micro-strain, we get εkl ij (x) = Aijkl (x) . The last relation clarifies the significance of the components of the influence tensor function A = A(x): Aijkl (x) is the (ij)-component of the micro strain εkl (x) which exists in the body, when on its boundary, the homogeneous displacement corresponding to I kl is imposed. The obtained results have an important consequence: in order to find the influence tensor function A = A(x), we must solve 6 special homogeneous displacement problems corresponding to the elementary strains Ikl , k, l = 1, 2, 3. P4.2 Let us assume a macro-homogeneous composite having three different phases: a matrix and two kinds of inclusions. Let us consider a RVE of the mixture occupying the domain D. Let D1 , D2 and D3 be the mutually disjoint subdomains of D occupied by the matrix, the first and the second kinds of inclusions, respectively. The micro-stress-strain relation of the mixture has the form   c1 ε(x) , in D1 , c1 = const. , c2 ε(x) , in D2 , c2 = const. , σ(x) = c(x)ε(x) =  c3 ε(x) , in D3 , c3 = const. . Let us assume now that on the boundary ∂D of the RVE an homogeneous displacement corresponding to ¯ ε =¯ εT = const. is imposed. Let us denote by A = A(x) the influence tensor function corresponding to this boundary value problem; i.e. ε(x) = A(x)¯ ε.

Let us denote by ¯ ε1 , ¯ ε2 and ¯ ε3 the mean values of ε(x) in D1 , D2 and D3 , respec¯ 1, A ¯ 2 and A ¯ 3 be the mean values of the influence tively, and let the constant tensors A tensor function A(x) in D1 , D2 and D3 , respectively. We get ¯ 1¯ ¯ 2¯ ¯ 3¯ ¯ ε1 = A ε,¯ ε2 = A ε,¯ ε3 = A ε. In order to evaluate the mean value σ ¯ of the corresponding micro-stress σ (x), we use the relation σ ¯ = c1 σ ¯ 1 + c2 σ ¯ 2 + c3 σ ¯3 . ¯ 3 are the mean values of σ(x) in D1 , D2 and D3 , In this equation, σ ¯ 1, σ ¯ 2 and σ v3 v2 v1 are the concentration of the matrix and the , c3 = , c2 = respectively, and c1 = v v v two different phases, v, v1 , v2 , v3 being the volumes of the domain D, D1 , D2 , D3 . Since the composite is piece-wise homogeneous, from the assumed micro-stress stain relation, we obtain ¯ 1 = c1 ε ¯1 , σ ¯ 2 = c2 ε ¯2 , σ ¯ 3 = c3 ε ¯3 , σ

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SOLUTIONS TO SOME PROBLEMS

605

¯ 1, A ¯ 2, A ¯ 3 , we since c1 , c2 and c3 are constant tensors. Using the influence tensors A can express the above mean value as ¯ 3¯ ¯ 2¯ ¯ 1¯ ε. ε,σ ¯ = c3 A ε,σ ¯ 2 = c2 A σ ¯ 1 = c1 A Consequently, the mean stress σ ¯ can be expressed in terms of the mean strain ¯ ε using the following equation: ¯ 3 )¯ ¯ 2 + c3 ¯ ¯ 1 + c2 ¯ ε. c3 A c2 A σ ¯ = (c1¯ c1 A Hence, denoting by b c the overall elasticity of the three-phasic mixture we get ¯ 2 + c 3 c3 A ¯3 ¯ 1 + c 2 c2 A b c = c 1 c1 A

and the overall stress-strain relation is

σ ¯ =b c¯ ε.

In order to determine the overall elasticity, we must know three constant influence ¯ 1, A ¯ 2 and A ¯ 3 . It is easy to see that these tensors are not independent and must tensors A satisfy the restriction ¯ 1 + c2 A ¯ 2 + c3 A ¯3 = I . c1 A Hence, in order to obtain b c, we must determine only two influence tensors.

P4.9 If Hill’s strong assumption is satisfied, according to the relations (4.1.22), (4.1.33), (4.1.20), (4.1.34), (4.2.5), we have, simultaneously, the following relations: −1 bσ , k b=b ¯ 1¯ ¯ 2σ ¯ 1σ ¯ 2σ ¯2 = A ¯1 = B ¯2 = B ε1 = A ε,ε ¯ ,σ ¯ ,σ ¯ . σ ¯ =b c¯ ε,¯ ε = k¯ c ,¯

Obviously, we have also the micro-constitutive equations

σ(x) = c(x)ε(x) , ε(x) = k(x)σ(x) , k−1 (x) = c(x) . Since the material is piece-wise homogeneous (see Equations (4.1.5), (4.1.6)), we have also σ ¯ 1 = c1 ¯ ε1 , σ ¯ 2 = c2 ¯ ε2 , ¯ ε1 = k 1 σ ¯1 , ¯ ε2 = k 2 σ ¯2 and −1 k1 = c−1 . 1 , k2 = c 2

According to these equations, ¯ ε1 can be expressed in two ways: ¯ ε1 = A1 ε and ¯ ε1 = k1 σ ¯ 1. bσ ; hence, According to the overall strain-stress relation, ¯ ε = k¯ bσ . ¯ ε1 = A1 k¯

¯ 1σ Let us observe also that σ ¯1 = B ¯ ; hence,

¯ 1σ ¯. ¯ ε1 = k 1 B

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SOLUTIONS TO SOME PROBLEMS

Comparing the last two expression of ¯ ε1 , and taking into account that σ ¯ is arbitrary, we can conclude that b = k1 B ¯1 . A1 k Starting with the equations

σ ¯ 1 = B1 σ ¯ and σ ¯ 1 = c1 ¯ ε1 , by similar reasoning, we can conclude that ¯ 1b ¯1 . B c = c1 A

Taking into account the mean values corresponding to the inclusions, it can be shown that the influence tensors A2 and B2 are connected by the equations b = k2 B ¯ 2 and B ¯ 2b ¯ 2k ¯ 2. c = c2 A A

We recall that the above compatibility conditions are consequences of Hill’s strong assumption. According to G˘ ar˘ ajeu’s theorem, Hill’s strong assumption is implied by Hill’s weak assumption. Hence, the compatibility conditions are satisfied only if the Hill’s weak assumption is supposed to hold! P4.13 Let us consider the homogeneous displacement boundary value problem corresponding to ¯ ε. Let [u, ε = ε(u), σ] be the solution of this problem. As we know,

ε) . ¯ ε·b cε = ε(u) · cε(v) ≤ ε(v) · cε(v) for any v ∈ K(¯

We also know that

ε(x) = A(x)¯ ε.

Let us assume now that A(x) = J. Let us denote the corresponding micro-strain field by εR (x); i.e. εR (x) ≡ ε¯ = const. We introduce also the homogeneous displacement field vR (x) = ε¯x in D. Obviously, vR (x) ∈K(¯ ε); that is, it represents a kinematically admissible field corresponding to ¯ ε. Hence, we get

ε. ε · c(x)¯ ¯ ε·b c¯ ε ≤ εR (vR ) · c(x)εR (vR ) = ¯

We recall now that Voigt’s estimate cV is just the mean value of the micro-elasticity c(x); i.e. cV = c(x). Thus, the above inequality becomes

ε¯ · b cε¯ ≤ ε¯ · cV ε¯ for any ε¯.

Thus, we can see that Hill’s universal estimate can be obtained assuming A(x) ≡ J. Obviously, this hypothesis corresponds to Voigt’s assumption: the strain field is constant in the RVE, submitted to homogeneous displacement boundary conditions.

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SOLUTIONS TO SOME PROBLEMS

607

The problem concerning the overall compliance can be analyzed in a similar manner, assuming B(x) = J. P4.24 We use the relations k1 =

2µ 1 + ν2 2µ 1 + ν1 , kV = c 1 k1 + c 2 k2 . , k2 = 3 1 − 2ν2 3 1 − 2ν1

In this way, we get kV =

(1 − 2ν1 )(1 − 2ν2 ) 2µ . 3 1 + (3c1 − 2)ν1 + (3c2 − 2)ν2 − 2ν1 ν2

Since νV

µ 1 − 3kV 2 = µ 1+ 3kV

we find

c1 ν1 + c2 ν2 − 2ν1 ν2 . 1 − 2(c2 ν1 + c1 ν2 ) Using the result given in P4.23, we obtain νV =

νV − νb =

(c1 ν1 + c2 ν2 )(c2 ν1 + c1 ν2 ) − ν1 ν2 . {1 − 2(c2 ν1 + c1 ν2 )}{1 − (c2 ν1 + c1 ν2 )}

We take now into account the identity c21 + 2c1 c2 + c22 = 1 and, thus, we obtain for the above difference νV − νb = c1 c2

(ν1 − ν2 )2 . {1 − 2(c2 ν1 + c1 ν2 )}{1 − (c2 ν1 + c1 ν2 )}

As we already know, c2 ν1 + c1 ν2 <

1 1 since 0 < ν1 , ν2 < and c1 + c2 = 1. Hence, 2 2

νb < νV .

The other inequalities given in P4.24 can be verified in a similar manner. P4.29 A biphasic macro-homogeneous and macro-isotropic glass-epoxy composite has the mechanical and geometrical characteristics given by the following relations: E1 = 3GP a , E2 = 70GP a , ν1 = 0.3 , ν2 = 0.2 , c1 = 0.3 , c2 = 0.7 . (a) To find k1 , µ1 , k2 , µ2 we use the well-known relations k=

E E . , µ= 2(1 + ν) 3(1 − 2ν)

Thus, we get k1 = 2.5GP a , k2 = 38.89GP a , µ1 = 1.15GP a , µ2 = 29.17GP a .

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608

SOLUTIONS TO SOME PROBLEMS (b) To find kV , µV , kR , µR , we use the equations (4.3.11) and (4.3.12), and find kV = 27.97GP a, kR = 7.69GP a, µV = 20.76GP a, µR = 3.57GP a .

(c) EV , νV , ER and νR are given by equations (4.3.16) and by the relations given in P4.25. We get

νV

EV = 60GP a, ER = 9.36GP a, c1 E1+ c2 E2 = 49.9GP a , −1  c2 c1 = 0.22. + = 0.20, νR = 0.30, c1 ν1+ c2 ν2 = 0.23, ν2 ν1

(d) The results show that for the considered composite c1 E1 + c 2 E2 < E V .

(e) Hill, Haskin an Shtrikman bounds k − and k + are given by the relations (4.3.50). We find, for the best possible estimates, the following values k − = 10.27GP a, k + = 23.89GP a. Accordingly, the overall bulk modulus b k of our composite satisfies the following restrictions: 10.27GP a ≤ b k[GP a] ≤ 23.89GP a.

The above restrictions represent all that can be obtained, neglecting the geometry of the inclusions and their distribution in the matrix. To obtain the best bounds or to evaluate the overall modulus b k, new information concerning the above geometrical characteristics is needed. According to Hill’s universal bounds, we have also 7.69GP a ≤ b k[GP a] ≤ 27.97GP a.

Hence, the possibility to obtain the best bounds k − and k + represents an improvement, with respect to the estimates founded in Voigt’s and Reuss’ classical results. However, in the analyzed case, the interval (k − , k + ) in which lies the overall bulk modulus b k rests relatively large. P4.31 We start recalling the system (4.4.61), (4.4.62) −

1

◦ pα

kα − k

−

1

◦

µα − µ We recall also that

+ α0 pα − α0 p¯ + 3¯ ε = 0, α = 1, 2,

¯ = 0, α = 1, 2. qα + β 0 qα − β 0 q ¯+e

p¯ = c1 p1 + c2 p2 , q¯ = c1 q1 + c2 q2 . Solving the system corresponding to the unknowns p1 , p2 and denoting the solution by p˜1 , p˜2 , we get, taking into account also the identity c1 + c2 = 1, ! !) ( 1 c2 c1 1 , p˜1 = 3¯ ε ◦ − α0 ◦ ◦ + ◦ ◦ − α0 k2 − k k2 − k k1 − k (k1 − k )(k2 − k )

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609

SOLUTIONS TO SOME PROBLEMS (

1 ◦

◦

(k1 − k )(k2 − k )

c1

− α0

◦

h

c2

+

◦

k2 − k

k1 − k

!)

1

p˜2 = 3¯ ε

◦

k1 − k

− α0

!

Hence, for the mean value p = c1 p1 + c2 p2 , we get !) ( h c2 c1 1 p ◦ ◦ + ◦ ◦ − α0 k2 − k k1 − k (k1 − k )(k2 − k ) !) ! ( 1 1 . + c2 = 3¯ ε c1 ◦ − α0 ◦ − α0 k1 − k k2 − k Dividing the above equation with ! ! ◦ ◦ [1 − α0 (k1 − k )][1 − α0 (k2 − k )] 1 1 − α = − α 0 0 ◦ ◦ ◦ ◦ (k1 − k )(k2 − k ) k2 − k k1 − k

and using, in the right-hand side term, the identity c1 + c2 = 1, we get ! ( ◦ ◦ 1 c1 (k1 − k )(k2 − k ) ◦ − α0 ◦ ◦ ◦ [1 − α0 (k1 − k )][(1 − α0 (k2 − k )] k1 − k k2 − k   !   h c2 c1 1 c2   p = 3¯ + ε + . ◦ − α0 ◦ 1 1   k2 − k k1 − k − α − α 0 0 ◦ ◦ k2 − k k1 − k Introducing the parameter

A=

c1

1 ◦

we get

k1 − k

− α0

◦

c2

+

1

k2 − k

◦

c2 (k2 − k )

+

(

◦

c1

1 ih

◦

c2

+

1 − α0 (k1 − k ) Simple computations show that

◦

1 − α0 (k2 − k ) c1

1 + α0 A =

◦

1 − α0 (k1 − k )

since c1 + c2 = 1! Now we can see that h

p=

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)

h

p = 3A¯ ε.

c2

+

◦

1 − α0 (k2 − k )

3A ε¯ 1 + α0 A

◦

1 − α0 (k2 − k )

◦ i· 1 − α0 (k1 − k ) 1 − α0 (k2 − k ) n   ◦  ◦ o h p = 3¯ · c1 1 − α0 (k2 − k ) + c2 1 − α0 (k1 − k ) εA,

h

or

◦

1 − α0 (k1 − k )

− α0

◦

c1 (k1 − k )

=

,

.

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SOLUTIONS TO SOME PROBLEMS

and equation (4.4.64)1 was obtained. Similar reasoning leads to the relation (4.4.64)2 . In order to obtain the value U˜ corresponding to the solution p˜α , q ˜α , we use the fact that this solution satisfies the system given in the first part of this problem. Using this fact, we get 2 2 X X h2 h cα 2 εp p ˜ = −α cα p˜2α + α0 p −3¯ 0 α ◦ α=1 α=1 kα − k and 2 2 X X h h h cα e· q . q ˜ · q ˜ = −β cα q ˜α · q ˜α + β0 q · q −¯ α α 0 ◦ α=1 α=1 2(µα − µ)

Introducing these relations in equation (4.4.56), we get

◦ ◦ h 1 h 1 h ε, e) = U + p ·¯ U˜ = U + (3 p ε¯+ q ·¯ 2 2

h

since p =

2 P

h

cα p˜α and q=

α=1

2 P

cα q ¯α . Hence, the relation (4.4.63) is proved.

α=1 ◦ ◦

P4.32 According to the given data, the function f = f (k ,µ) is given by the equation ◦ ◦

◦

f = f ( k , µ) = k +

A , 1 + α0 A

where, as we can see from the relations (4.4.65) and (4.4.55), A=

2 X

α=1

◦

cα (kα − k )

◦

1 − α0 (kα − k )

and α0 = −

3 ◦

◦

3 k +4 µ

.

From the last equation, we get ∂α0 ◦

∂µ

=

12 ◦

=

◦

(3 k +4 µ)2

4 2 α0 . 3

Using this relation, we obtain ∂A ◦

∂µ

=

◦ 2 cα (kα − k )2 4 2X . α0 3 α=1 [1 − α (k − ◦)]2 k 0 α

In the same way, we get

∂A ∂f ◦

∂µ

Hence, the sign of

∂f ◦

∂µ

◦

−

µ = ∂ (1 + α0 A)2

is given by the sign of

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4 2 2 α0 A 3

∂A ◦

∂µ

−

.

4 2 2 α0 A . 3

611

SOLUTIONS TO SOME PROBLEMS We obtain

4 4 2 2 α0 A = α02 ◦ − 3 3 ∂µ ∂A

(

c2 (k2 − k )2

+

◦

◦

[1 − α0 (k2 − k )]2

[1 − α0 (k1 − k )]2 ◦

◦

−

◦

◦

c1 (k1 − k )2

c21 (k1 − k )2

◦

[1 − α0 (k1 − k )]2

◦

(k1 − k )(k2 − k )

−2c1 c2

◦

◦

[1 − α0 (k1 − k )][1 − α0 (k2 − k )] ) ◦ (k2 − k )2 2 . −c2 ◦ [1 − α0 (k2 − k )]2 Now, using the fundamental identity by c1 + c2 = 1 finally results in

4 4 2 2 α0 A = α02 c1 c2 ◦ − 3 3 µ ∂ ∂A

(

◦

◦

k1 − k

−

◦

1 − α0 (k1 − k )

k2 − k

◦

1 − α0 (k2 − k )

)2

Hence, ∂f ◦

∂µ In the same manner, it can be proved that ∂f ◦

∂k

≥ 0 and

≥0.

∂g

◦,

∂g ◦

∂k ∂µ

≥0.

P4.33 According to the equation (4.4.73), we have µ ¯ = µ1 +

c2 µ1 (µ2 − µ1 ) c2 = µ1 + c 1 β1 1 µ1 + c1 β1 (µ2 − µ1 ) + µ1 µ2 − µ 1

where β1 =

6(k1 + 2µ1 ) . 5(3k1 + 4µ1 )

Thus, we get

µ1+ β1 (µ2 − µ1 ) c 2 µ1 µ− − µ 2 , = c1 =1− µ1 + c1 β1 (µ2 − µ1 ) µ1 + c1 β1 (µ2 − µ1 ) µ1 − µ 2

since c1 + c2 = 1. Using the same identity, we obtain

µ1+ c1 β1 (µ2 − µ1 ) + c2 β1 (µ2 − µ1 ) µ− − µ 2 = c1 µ1 − µ 2 µ1 + c1 β1 (µ2 − µ1 )   c2 (µ2 − µ1 ) . = c1 1 + β1 µ1 + c1 β1 (µ2 − µ1 )

Using again the expression giving µ− , we obtain

c2 (µ2 − µ1 ) µ− − µ 1 = . µ1 µ1 + c1 β1 (µ2 − µ1 )

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.

612

SOLUTIONS TO SOME PROBLEMS In this way, finally, we get

  −  µ µ− − µ 2 −1 . = c1 1 + β1 µ1 µ1 − µ 2 The relation giving µ+ can be obtained in a similar manner. P4.35 We start with equations (4.4.71) giving k − and k + ; we have k − = k1 +

c1 γ2 (k1 − k2 ) c2 γ1 (k2 − k1 ) , , k + = k2 + γ2 + 3c2 (k1 − k2 ) γ1 + 3c1 (k2 − k1 )

with γ1 = 3k1 + 4µ1 , γ2 = 3k2 + 4µ2 . Using the identity c1 + c2 = 1, we get   1 1 k+ − k− . − = 3c c 1 2 γ2 + 3c2 (k1 − k2 ) γ1 + 3c1 (k2 − k1 ) (k2 − k1 )2

Since γ2 − γ1 = 3(k2 − k1 ) + 4(µ2 − µ1 ), finally, it results

µ2 − µ 1 k+ − k− . = 12c1 c2 [γ1 + 3c1 (k2 − k1 )][γ2 + 3c2 (k1 − k2 )] (k2 − k1 )2

The above result shows that if both phases have the same shear moduli; i.e. µ1 = µ2 = µ, then k + = k − . Obviously, the common value of the bounds is just the overall bulk modulus b k of the composite, found by Hill. Now we can see that Hill’s wonderful result can be obtained also, using Hashin’s and Shtrikman’s variational and extreme principle. P4.45 In order to solve the problem, we use Eshelby’s second energetical theorem given in the final part of Section 2.6. Also, we assume that on the boundary of the RVE, a traction boundary condition corresponding to σ ¯=σ ¯ T = const. is given. The homogeneous body of the theorem is, in our case, the matrix without inclusions and the inhomogeneous material is the matrix containing the special inclusions occupying the domain V2 . Taking into account these observations, the constitutive relations (4.5.1) and the energetical definition of the overall moduli, from Eshelby’s equation (2.6.60), we get Z 1 1 1 2 1 1 2 (¯ σθ + ¯ s · e−σ θ¯ − s · ¯ e)dv . ¯ s ·¯ s+ ¯ + ¯ s ·¯ s= σ σ ¯ + b v 2µ1 k1 2b µ k V2

According to the constitutive relation (4.5.1), we have

1 1 ¯ s , σ = k2 θ and s =2µ2 e. σ ¯ and ¯ e= θ¯ = 2µ1 k1

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613

SOLUTIONS TO SOME PROBLEMS Hence, we get

    Z Z 1 µ2 1 k2 1 1 2 1 1 2 edv . ¯ s· θdv + 1 − σ ¯ ¯ s ·¯ s+ 1 − ¯ + ¯ s ·¯ s= σ σ ¯ + b v µ1 v k1 2µ1 k1 2b µ k V2

V2

Obviously, the last relation can be expressed in the following equivalent form:     1 1 1 1 1 1 2 1 1 2 ¯ ¯ s·¯ e2 , − σ ¯ θ + c µ − ¯ s · ¯ s +c k ¯ + ¯ s ·¯ s= σ σ ¯ + 2 2 2 2 2 b µ1 µ2 k1 k2 2µ1 k1 2b µ k

where θ¯2 and ¯ e2 are the mean values of θ and e, respectively, on the domain V2 occupied by the inclusions in the considered RVE. Now it is easy to see that Budiansky’s relation (4.5.5) and the above equation obtained using Eshelby’s second energetical theorem are identical and, obviously, our problem is solved.

Chapter 5 P5.1 (a) The components Fkm and the components Fekm of the gradients of defore respectively, are given by the equations mation F and F, Fkm =

∂χk , ∂Xm

∂χ ekm . Fekm = em ∂X

e are connected by the relations ek of X and X The components Xk and X ek = Qkm Xm , X

where Qkm are the components of the orthogonal tensor Q. Hence, using the chain rule, we obtain el ∂χ ek ∂ X ∂χk = Fekl Qlm . · = Fkm = el ∂Xm ∂Xm ∂X

The obtained result can be expressed in tensorial form as e F = FQ.

e corresponding to F and F, e (b) As we know, Green’s strain tensors G and G, respectively, are given by the relations    1 T eT F e −1 . e = 1 F F F−1 , G G= 2 2

Using the relation obtained in (a), we have

hence,

eT ; FT = Q T F

e T FQ. e FT F = Q T F

Since, the tensor Q is orthogonal, the equation QT Q = 1

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SOLUTIONS TO SOME PROBLEMS

is satisfied. In this way, finally we obtain e G = QT GQ.

P5.4 Let us denote by GB , the set of all symmetry transformation of a hyperelastic material, corresponding to its reference configuration B. Hence, if the orthogonal transformation Q is an element of the set GB , the relation   uB QGQT = uB (G)

must be satisfied for every symmetric tensor G. (a) Obviously, the above restriction is satisfied for Q = 1 and Q = −1. Accordingly, 1 and −1 are elements of the set GB for any reference configuration B of the material. (b) Let as assume now that the orthogonal transformation Q is an element of the set GB . Consequently, the above restriction is satisfied for Q and for every symmetric e e is an arbitrary symmetric tensor. where G tensor G. Let as assume that G = QT GQ, We have     T e e uB QQT GQQ = uB QT GQ

e Since Q is an orthogonal tensor, QQT = QT Q = 1 and for every symmetric tensor G. the above equation becomes     e = uB QT GQ e uB G

e Accordingly, the orthogonal tensor QT = Q−1 is a symfor every symmetric tensor G. metry transformation; i.e. QT = Q−1 ∈ GB , if Q ∈ GB . (c) Let us assume that the orthogonal transformation Q1 and Q2 are symmetry transformations. In this case, we successively get n o   o n n o uB (Q1 Q2 ) G (Q1 Q2 )T = uB Q1 Q2 GQT2 QT1 = uB Q2 GQT2 = uB (G) ,

for any symmetric tensor G. The last result shows that Q1 , Q2 ∈ GB . The properties (a)−(c) show that the set GB is a group relative to the composition of two transformations. This is the reason why GB is named symmetry group of the material, corresponding to the reference configuration B. If the material has no symmetry, properties GB = {1, −1}. If the material is isotopic GB = O, O being the full orthogonal group; i.e. the set of all orthogonal tensors.

b are connected by the P5.5 Let us assume that the reference configurations B and B deformation b = ΛX, X where Λ is orthogonal tensor. Using the chain rule, it is easy to see that

hence, Also, we have

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b F = FΛ,

b G = ΛT GΛ.

  b , u = uB (G) = uBb G

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SOLUTIONS TO SOME PROBLEMS

uB and uBb being the constitutive functions corresponding to the reference configurations b respectively. The above relations show that these two constitutive functions B and B, are related by the equation     b b = uB ΛT GΛ uBb G

b for any symmetric tensor G. Let us assume now that the orthogonal transformation Q is a symmetry transformation of the material relative to the reference configuration B; i.e. Q ∈GB . According to the above equation, we get      T     T  b ΛQΛT b ΛQΛT Λ ΛQΛT G = uB ΛT ΛQΛT G uBb n o T T b = uB ΛT ΛQΛT GΛQ Λ Λ .

Since Λ is an orthogonal tensor, ΛT Λ = 1; hence, we obtain    n   o T  b ΛQΛT b ΛQΛT G = uB Q ΛT GΛ QT . uBb Recalling that Q ∈ GB , we obtain  n    o b b . QT = uB ΛT GΛ uB Q ΛT GΛ

Using again the relation connecting the constitutive functions uB and uBb , we conclude that     b b . uB ΛT GΛ = ub G B

In this way, from the last two equations, we obtain      T  T T b b = uBb G ΛQΛ G ΛQΛ uBb

b The last equation shows that ΛQΛT is a symmetry transfor every symmetric tensor G. b if Q is a symmetry formation of the material relative to its reference configuration B, transformation relative to B; i.e. if Q ∈ GB , then ΛQΛT ∈ GBb . Hence, we can conclude that the set ΛGB ΛT is contained in the set GBb ; i.e. ΛGB ΛT ⊆ GBb .

Similar reasoning shows that the inverse inclusion

is also true. Hence,

GBb ⊆ ΛGB ΛT GBb = ΛGB ΛT .

P5.6 Let us assume that the constitutive function uB (G) of a hyperelastic material is a quadratic form of Green’s strain tensor G; i.e. uB (G) =

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1 G · CG, 2

616

SOLUTIONS TO SOME PROBLEMS

where C is a given, constant fourth order tensor. (a) Let us assume that Q ∈ GB . In this case, we must have   uB QGQT = uB (G)

for any symmetric tensor G. Since uB (G) has the form given above, we can conclude that if Q ∈ GB , the elasticity tensor C of the material must satisfy the restriction     QGQT · C QGQT = GCG

for any symmetric tensor G. (b) It is obvious that if C satisfies the above restriction for an orthogonal tensor 1 Q and for any symmetric tensor G, then uB (G) = GCGT satisfies the restriction 2
uB QGQT = uB (G) for any symmetric tensor G. Hence, Q ∈ GB . (c) The component form of the restriction supposed on C in (b) is   n  o QGQT C QGQT = Gkl Cklmn Gmn . rs

rs

We have also

QGQT 




rs

C QGQT

= Qrk Gkl QTls = Gkl Qrk Qsl


= Crspq QGQT




pq

= Crspq Qpm Gmn QTnq = Crspq Qpm Qqn Gmn .

Hence, the restriction imposed on the elasticity tensor C takes the form Gkl Qrk Qsl Crspq Qpm Qqn Gmn = Gnl Cklmn Gmn .

Since this equation must be satisfied for every symmetric tensor G, we can conclude that if Q ∈ GB , C must satisfy the restrictions Cklmn = Qrk Qsl Qpm Qqn Crspq . As we already know, if Q ∈GB , then QT ∈GB . Consequently, C must verify also the equivalent restrictions Cklmn = Qkr Qls Qmp Qnq Crspq .

(d) Let us assume now that we have a linearly hyperelastic material, submitted to infinitesimal deformations. As we know, its constitutive functions can be obtained from those given in this problem, replacing Green’s strain tensor G with the infinitesimal strain tensor ε, and the elasticity tensor C with the elasticity tensor c of the linear theory. Hence, we have 1 u = uB (ε) = ε · cε 2 and using (a) and (c) simultaneously, we can conclude that if Q is a symmetry transformation of the material; i.e. if Q ∈ GB , then the elasticity tensor c and its components cklmn must satisfy the restrictions     QεQT · c QεQT = ε · cε

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SOLUTIONS TO SOME PROBLEMS

617

for any symmetric tensor, and cklmn = Qkr Qls Qmp Qnq crspq . Thus, using results of the nonlinear mechanics, we were able to justify in a rigorous manner the definition of a symmetry transformation of a linearly hyperelastic material, given in Section 2.2. As our reasoning shows, to obtain this result, we must surpass the framework of the linear theory, since a symmetry transformation is not an infinitesimal transformation! P5.7 Let us assume that the deformation of the body from its reference configuration B to its current configuration Bt is given by the equation x = χ (X,t) or in component forms xk = χk (Xl , t) . Since the time t is fixed, this variable will not be mentioned in what follows. Let us consider now a material surface in the reference configuration B of the body, given by its parametrical representation X = χ (u, v) , ∂X ∂X dv du and u and v being real parameters. The infinitesimal (material) vectors ∂v ∂u are situated in the tangent plane of the considered material surface and the area dA of the little parallelogram formed by these two vectors is given by the equation

∂X ∂X

dudv.

× dA = ∂v ∂u

∂X ∂X is orthogonal to the tangent plane. Conse× As is well known, the vector ∂v ∂u quently, if N is the unit normal to the considered material surface, we should have

NdA =

∂X ∂X dudv, × ∂v ∂u

or, in component form, Nk dA = εklm

∂Xm ∂Xl dudv, × ∂v ∂u

εklm being Ricci’s symbols. Analogously, for the same material surface of the body, but in its current configuration, we shall have ∂x ∂x dudv, × nda = ∂v ∂u or, in component forms, ∂xs ∂xt dudv. nr da = εrst ∂u ∂v Since x and X are connected by the equation x = χ (X), the parametrical representation of the considered material surface in the current configuration of the body is given by the equation x = χ (X (u, v)) ,

or, in component form, xs = χs (Xk (u, v)) .

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618

SOLUTIONS TO SOME PROBLEMS

Using the above relations and the chain rule, we get nr da = εrst

But

∂xs ∂Xl ∂xt ∂Xm dudv. ∂Xl ∂u ∂Xm ∂v

∂xt ∂xs = Ftm ; = Fsl and ∂Xm ∂Xl

hence,

∂Xl ∂Xm dudv. ∂u ∂v On the other hand, we have J = det F = det FT and, as is well known from the matrix calculus, J can be expressed by the following equation: nr da = εrst Fsl Ftm

εklm J = εrst Frk Fsl Ftm . Consequently, multiplying the relation giving nr da by Frk and summing with respect to r, we get ∂Xl ∂Xm dudv, Frk nr da = Jεklm ∂u ∂v or, equivalently, Frk nr da = JNk dA.

Obviously, the tensor form of this equation FT nda = JNdA. Hence, we finally have proved the Nanson’s relation nda = JF−T NdA with J = det F. P5.12 Let us assume the given body submitted to the homogeneous deformation x1 = λ 1 X1 ,

x 2 = λ 2 X2 ,

x 3 = λ 3 X3 ,

λ1 , λ2 , λ3 being arbitrary real numbers satisfying the restriction λ1 , λ2 , λ3 > 0. (a) Since the reference configuration B of the body is the parallelepiped defined by ◦

the inequalities −ak ≤ Xk ≤ ak , k = 1, 2, 3, it is clear that the deformed configuration B of the body is also a parallelepiped defined by the inequalities −λk ak ≤ xk ≤ λk ak , (!) k = 1, 2, 3. ◦

◦

(b) The components of F kl the deformation gradient F are given by the equation ◦ ∂xk . F kl = ∂Xl Consequently, for the matrix of the components, we obtain   λ1 0 0 h◦ i F kl =  0 λ2 0  . 0 0 λ3 ◦

◦T

◦

◦

◦T ◦

We can conclude that F is a symmetric tensor; i.e. F = F. Since C = F F, the ◦ matrix of the components of the Cauchy-Green strain tensor C is  2  λ1 0 0 h◦ i 2 C kl =  0 λ2 0  . 0 0 λ23

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619

SOLUTIONS TO SOME PROBLEMS ◦

Green’s strain tensor G is given by the equation    ◦ 1 ◦ 1 ◦T ◦ C−1 . F F−1 = G= 2 2 ◦

Hence, the matrix of components of G is  1

◦


λ21 − 1 0 0 h◦ i  2
1 2  λ2 − 1 0 Gkl =  0  2
1 2 λ3 − 1 0 0 2

◦



  , 

Since J = det F , we obtain

◦

J = λ 1 λ2 λ3 . (c) As we know in any possible deformation of a body, the restriction J = det F > 0 must be satisfied. Just this restriction is expressed by the inequality ◦

J = λ1 λ2 λ3 > 0, which must be satisfied by the real numbers λ1 , λ2 , λ3 describing the deformation of the body. P5.14 (a) Since the considered body is made up by the given hyperelastic material, ◦

the symmetric Piola-Kirchhoff strain tensor Π is given by the equation ◦ ∂u  ◦  G . Π= ∂G

Taking into account the given special form of the constitutive equation, we find  ◦ ◦ ◦ Π = λ tr G 1 + 2µ G . ◦

The obtained result shows that the nondiagonal components of Π are vanishing. Also, according to the results obtained in P5.12 (b), we have ◦

tr G = ◦


1 2 λ1 + λ22 + λ23 − 3 . 2

Hence, the components of Π are given by the equations



λ 2 λ1 + λ22 + λ23 − 3 + 2µ λ21 − 1 , 2

◦ λ 2 λ1 + λ22 + λ23 − 3 + 2µ λ22 − 1 , Π22 = 2

◦ λ 2 λ1 + λ22 + λ23 − 3 + 2µ λ23 − 1 , Π33 = 2 ◦ Πkl = 0 if k 6= l. ◦

Π11 =

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620

SOLUTIONS TO SOME PROBLEMS ◦

As we know, the nominal strain Θ existing in the deformed configuration B is given by the equation ◦ ◦T

◦

Θ = ΠF . ◦

This result shows that the nondiagonal components of Θ are vanishing, and its components are given by the equations ◦

◦

◦

◦

◦

◦

Θ22 = λ2 Π22 ,

Θ11 = λ1 Π11 ,

Θ33 = λ3 Π33

◦

Πkl = 0 if k 6= l. ◦

As we have seen in P5.10, the Cauchy’s stress T existing in the deformed configu◦ ration B is given by the equation ◦ −1 ◦ ◦

◦

FΘ .

T=J

◦

This results shows that the nondiagonal components of T are vanishing, and its components are given by the equation: ◦

T 11 =

λ1 ◦ Π11 , λ2 λ3

◦

T kl = 0,

◦

λ3 ◦ Π33 , λ2 λ1

◦

1 ◦ Θ33 , λ2 λ1

◦

λ2 ◦ Π22 , λ3 λ1

T 33 =

◦

1 ◦ Θ22 , λ3 λ1

T 33 =

T 22 =

if k 6= l,

or, equivalently, ◦

T 11 =

1 ◦ Θ11 , λ2 λ3

◦

T kl = 0,

T 22 =

if k 6= l. ◦

(c) As we know, the Cauchy’s stress vector tn acting on a material surface element with unit normal n in the deformed configuration is given by the equation ◦

◦

tn = T n. In what follows, we assume that λ1 , λ2 and λ3 are positive numbers; i.e. λ1 , λ2 , λ3 > 0. The outward unit normal n to the deformed faces x1 = ±λ1 a1 has the components ◦

n1 = ±1, n2 = n3 = 0. Hence, the Cauchy’s stress vector tn , acting on these faces in the ◦ deformed configuration B , has the components ◦

◦

tn1 = ± T11 = ±

1 ◦ Θ11 , λ2 λ3

◦

◦

tn2 = tn3 = 0.

The outward unit normal n to the deformed faces x2 = ±λ2 a2 has the components ◦

n1 = 0, n2 = ±1, n3 = 0. Hence, the Cauchy’s stress vector tn , acting on these faces in ◦ the deformed configuration B , has the components ◦

tn1 = 0,

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◦

◦

tn2 = ± T22 = ±

1 ◦ Θ22 , λ3 λ1

◦

tn3 = 0.

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SOLUTIONS TO SOME PROBLEMS

The orthogonal unit normal n to the deformed faces x3 = ±λ3 a3 has the compo◦

nents n1 = n2 = 0, n3 = ±1. Hence, the Cauchy’s stress vector tn , acting on these faces ◦ in the deformed configuration B , has the components ◦

◦

◦

◦

tn3 = ± T33 = ±

tn1 = tn2 = 0,

1 ◦ Θ33 . λ1 λ2

The Piola’s and Kirchhoff’s stress vector sN , acting on a material surface element with unit normal N in the reference configuration B is given by the equation ◦

◦T

sN = Θ N. The outward unit normal N to the undeformed faces X1 = ±a1 has the components N1 = ±1, N2 = N3 = 0. Hence, ◦

◦

sN1 = ± Θ11 ,

◦

◦

sN2 = sN3 = 0.

The outward unit normal N to the undeformed faces X2 = ±a2 has the components N1 = 0, N2 = ±1, N3 = 0. Hence, ◦

◦

◦

sN1 = 0,

sN2 = ± Θ22 ,

◦

sN3 = 0.

The outward unit normal N to the undeformed faces X3 = ±a3 has the components N1 = N2 = 0, N3 = ±1. Hence, ◦

◦

sN1 = sN2 = 0,

◦

◦

sN3 = ± Θ33 . ◦

(c) As we have seen, the Cauchy’s stress tensor T is a constant tensor. Hence, according to Cauchy’s equilibrium equation, the body force density b, assuming the equi◦

librium of the body in the deformed configuration B , must be vanishing. Also, in order to maintain the body in its deformed equilibrium configuration on the faces xk = ±λk ak (!) of the body must be applied constant normal surface tractions, having magnitudes determined in (b). For instance, on the face x1 = λ1 a1 must be applied on unit surface area of this plane, the normal surface traction   ◦

◦ λ 2 λ1 λ1 ◦ 1 ◦ λ1 + λ22 + λ23 − 3 + 2µ λ21 − 1 . tn1 = T11 = Π11 = Θ11 = λ2 λ3 2 λ2 λ3 λ2 λ3

(d) To analyze the results obtained in (a) and (b), we first observe that according to the formulas obtained in (b), the stress vector tn and sN , corresponding to various faces of the parallelepiped are connected by the equations for the faces X1 = ±a1 and x1 = ±λ1 a1 , we have ◦ 1 ◦ sN1 ; tn1 = ± λ2 λ3 for the faces X2 = ±a2 and x2 = ±λ2 a2 , we get ◦

tn2 = ±

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1 ◦ sN2 ; λ3 λ1

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SOLUTIONS TO SOME PROBLEMS

for the faces X3 = ±a3 and x3 = ±λ3 a3 , we obtain ◦

tn3 = ±

1 ◦ sN3 . λ1 λ2

Let us denote by S1 and s1 , the areas of the faces X1 = ±a1 and x1 = ±λ1 a1 , respectively. Obviously, we have S1 = a2 a3 and s1 = λ2 a2 λ3 a3 ; hence, s 1 = λ 2 λ3 S 1 . ◦

Let us denote by F 1 the total normal force activity on the face x1 = λ1 a1 in the ◦ deformed configuration B of the body. This force is given by the equation ◦

◦

F 1 = tn1 s1 . ◦

Since s1 = λ2 λ3 S1 , the same force F 1 , can be expressed as ◦

◦

F 1 = tn1 λ2 λ3 S1 . ◦

But tn1 =

1 ◦ sN 1 , and, consequently, we get λ2 λ3 ◦

◦

F 1 = sN1 S1 . Hence, we have

◦

◦

◦

F 1 = tn1 s1 = sN1 S1 , or, equivalently,

◦

◦

◦ F1 F . ` 1 and sN1 = tn1 = S1 s1 Summing up the above results, obtained in the special problem here analyzed, we have the following results. ◦

◦

◦

(i) The Cauchy’s stress tn1 is the normal force F 1 acting in the deformed state of the body, reported to the unit area of the deformed boundary x1 = λ1 a1 of the body. ◦ ◦ (ii) The Piola’s and Kirchhoff’s stress sN1 is the same normal force F 1 acting on the same deformed boundary x1 = λ1 a1 of the body, and reported to the unit area of the undeformed boundary X1 = a1 . Obviously, all results obtained in (a) and (b) can be analyzed in the same manner, taking into account the mechanical significance of the stress vectors tn and sN . Actually, as we have seen, in our special problem, we are led to the corresponding mechanical meaning and, in this way, to a better understanding of these essential concepts of the general nonlinear mechanics of the deformable bodies. ◦

P5.16 (a) The symmetric Piola-Kirchhoff stress tensor Π is given again by the constitutive equation  ◦ ◦ ◦ Π = λ tr G 1 + 2µ G,

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623

SOLUTIONS TO SOME PROBLEMS ◦

G being Green’s strain tensor determined in P5.15 (b). As it is easy to see in this case, ◦

tr G = γ 2 /2 ◦

and, for the components of Π, we obtain the following values: ◦ ◦ 1 2 λγ , Π12 = µγ, Π13 = 0, 2 ◦ ◦ ◦ 1 2 Π21 = µγ, Π22 = (λ + 2µ) γ , Π23 = 0, 2 ◦ ◦ ◦ 1 2 Π31 = 0, Π32 = 0, Π33 = λγ . 2 ◦

Π11 =

◦

In order to find the components of Θ, we use the equation ◦

◦

◦

T Θ = ΠF .

Thus, we get ◦

Θ11 ◦

Θ21 ◦

Θ31

◦ ◦ 1 (λ + 2µ) γ 2 , Θ12 = µγ, Θ13 = 0, 2 ◦ 1 1 = µγ + (λ + 2µ) γ 3 , Θ22 = (λ + 2µ) γ 2 , 2 2 ◦ ◦ 1 = 0, Θ32 = 0, Θ33 = λγ 2 . 2

=

◦

Θ23 = 0,

◦

In order to find the components of T, we use the equation ◦ −1 ◦ ◦

◦

T=J

FΘ .

◦

Since J = 1, we obtain ◦

T 11 ◦

T 21 ◦

T 31

◦ ◦ 1 1 1 (λ + 4µ) γ 2 + (λ + 2µ) γ 4 , T 12 = µγ + (λ + 2µ) γ 3 , T 13 = 0, 2 2 2 ◦ ◦ 1 1 = µγ + (λ + 2µ) γ 3 , T 22 = (λ + 2µ) γ 2 , T 23 = 0, 2 2 ◦ ◦ 1 = 0, T 32 = 0, T 33 = λγ 2 . 2

=

◦

As is easy to see, the diagonal components of T are even functions of the shear γ ◦ and the nonvanishing mixed component T 12 is an odd function of the shear γ. ◦

◦

◦

◦

If the shear is infinitesimal, all stresses Π, Θ and Π reduce to the Cauchy’s stress σ of the classical linear elasticity. To obtain the components of this tensor, we must neglect all terms containing γ 2 , γ 3 and γ 4 . In this way, we obtain the well-known result   0 µγ 0 h ◦ i σkl =  µγ 0 0  . 0 0 0 ◦

(b) To find Cauchy’s stress vector tn acting on the boundary of the deformed body, we must determine the unit outward normal n of the deformed boundary of the material.

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SOLUTIONS TO SOME PROBLEMS

This can be made in two ways: (i) we can use the result given in P5.8; (ii) we can solve the problem directly, using the Figure 5.6. This second approach leads to the following obvious results. The outward unit normal n at the deformed boundary −a2 ≤ X2 ≤ a2 ,

x1 = a1 + γX2 ,

−a3 ≤ X3 ≤ a3

has the following components: n1 = p

1 , 1 + γ2

n2 = − p

γ , 1 + γ2

n3 = 0.

The outward unit normal n at the boundary x1 = −a1 + γX2 ,

−a2 ≤ X2 ≤ a2 ,

−a3 ≤ X3 ≤ a3

has the following components: n1 = − p

1 , 1 + γ2

n2 = p

γ , 1 + γ2

n3 = 0.

It is easy to see that the outward unit normals to the boundaries X2 = ±a2 and X3 = ±a3 rest unchanged during the shear. Consequently, we have the following results. The outward unit normal n to the deformed boundaries x2 = ±a2 has the following components: n1 = 0, n2 = ±1, n3 = 0.

The outward unit normal n to the deformed boundaries x3 = ±a3 has the following components: n1 = n2 = 0, n3 = ±1. ◦

As before, for all deformed boundaries, the Cauchy’s stress vector tn can be calculated using the Cauchy’s fundamental equation ◦

◦

tn = T n and taking into account the involved unit normal. For the deformed boundary x1 = a1 + γX1 , −a2 ≤ X2 ≤ a2 , −a3 ≤ X3 ≤ a3 , we find     ◦ ◦ ◦ ◦ ◦ ◦ ◦ γ 1 T21 − γ T 22 , tn3 = 0. T11 − γ T 12 , tn2 = p t n1 = p 2 2 1+γ 1+γ

For the deformed boundary x1 = −a1 + γX1 , −a2 ≤ X2 ≤ a2 , −a3 ≤ X3 ≤ a3 , we

get ◦

t n1

= −p

1 1 + γ2



 ◦ ◦ T11 − γ T 12 ,

◦

t n2

= −p

γ 1 + γ2



 ◦ ◦ T21 − γ T 22 ,

For the deformed boundaries x2 = ±a2 , we find ◦

◦

tn1 = ± T 12 ,

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◦

◦

tn2 = ± T 22 ,

◦

tn3 = 0.

◦

tn3 = 0.

625

SOLUTIONS TO SOME PROBLEMS For the deformed boundaries x3 = ±a3 , the result is ◦

tn1 = 0,

◦

tn2 = 0,

◦

◦

tn3 = ± T 33 . ◦

(c) As in P5.14, the Cauchy’s stress tensor T is again a constant tensor. Consequently, according to Cauchy’s equilibrium equation, the body force density b, assuring ◦

the equilibrium of the body in its deformed configuration B , must be vanishing. To maintain the body in its deformed equilibrium configuration on the deformed faces of the material, constant surface tractions must be applied. The surface force acting on a selected ◦

deformed boundary has a normal component denoted in the following by N k , k = 1, 2, 3 ◦ and a tangential component denoted by T k , k = 1, 2, 3. These components are given in the Figure S.11, for a deformed transverse section X3 = const.

Figure S.11: Normal and tangential tractions on the deformed faces of a sheared block. In the following, we shall evaluate the normal and tangential tractions on various deformed faces of the parallelepiped. To do this, we denote by τ the unit tangent vector on the selected face (see Figure S.11). The unit tangent vector τ on the deformed faces x1 = ±a1 + γX2 , −a2 ≤ X2 ≤ a2 , −a3 ≤ X3 ≤ a3 has the following components: τ1 = ± p

γ , 1 + γ2

τ2 = ± p

1 , 1 + γ2

On the deformed faces x2 = ±a2 , we get τ1 = 0,

τ2 = ±1,

τ3 = 0

and, on the deformed faces x3 = ±a3 , we have τ1 = τ2 = 0,

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τ3 = ±1.

τ3 = 0.

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SOLUTIONS TO SOME PROBLEMS ◦

On the selected face, the corresponding normal traction N and the corresponding ◦ tangential traction T are evaluated using the obvious equations ◦

◦

◦

◦

N = n· tn and T = τ · tn . Using the results obtained in (b), we successively find: −on the deformed faces x1 = ±a1 + γX2 , −a2 ≤ X2 ≤ a2 , −a3 ≤ X3 ≤ a3 , to ◦

maintain the body in its deformed equilibrium configuration B , the following normal and tangential tractions must be applied:   ◦ ◦ ◦ ◦ 1 2 T − 2γ + γ = ± T 12 T 22 , N1 11 1 + γ2     ◦ ◦ ◦
◦ 1 2 ; T + 1 − γ γ T − T T1 = ± 12 11 22 1 + γ2

−on the deformed faces, x2 = ±a2 , the following normal and tangential tractions must be applied: ◦

◦

N 2 = ± T 22 ,

◦

◦

T2 = ± T 12 ;

−on the deformed faces, x3 = ±a3 , the tangential traction, which must be applied ◦

to maintain the body in its deformed equilibrium configuration B , is vanishing and the necessary normal force is given by the equation ◦

◦

N 3 = ± T 33 . (d) Analyzing the results obtained in (a)−(c), we can say the following: ◦

(i) The shear stress T 12 is an odd function of the amount of shear γ. The departure from the classical linear proportionality of shear stress to shear strain is an effect of third order in the amount of shear. (ii) In the classical linear theory, shear stress suffices to produce shear. Our results show that this simple property can never hold exactly even for an isotropic material. In fact normal tractions, negligible only in small shears, are required if a simple shear is to be produced. That is, normal tractions (forces) must act upon all faces of the block in order to maintain it in a state of simple shear. Unless these normal forces are supplied, it is natural to assume that a cube of the material, if subjected to shear stress alone upon its faces, will tend to contract or expand. This phenomenon was first remarked upon by Kelvin, and is named Kelvin effect. (iii) Equally present in our isotropic material is the inequality of the normal tractions ◦

◦

◦

N 1 , N 2 and N 3 . The necessity of these unequal normal tractions in order to produce a simple shear suggests that unless they are supplied, an initially cubical specimen, when subjected to shear traction, tends only to dilate or contract unequally. This phenomenon was first noticed by Poyting, and is called Poyting effect. (e) As the constitutive equation  ◦ ◦ ◦ Π = λ tr G 1 + 2µ G

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SOLUTIONS TO SOME PROBLEMS ◦

shows the eigendirections of Green’s strain tensor G and of the symmetric Piola-Kirchhoff ◦

stress tensor Π are the same: M1 , M2 , M3 found in P5.15(c). ◦

◦

◦

◦

If we denote by π 1 , π 2 , π 3 the corresponding eigenvalues of Π, we obtain ◦

π 1 = λ (g1 + g2 ) + 2µg1 ,

◦

◦

π 2 = λ (g1 + g2 ) + 2µg2 ,

π 3 = λ (g1 + g2 ), ◦

where g1 , g2 and g3 = 1 are the corresponding eigenvalues of G. (f) What happens if the normal surface forces acting on the boundaries of the body are vanishing was clarified in (d). P5.25 According to the assumption made in P5.13, the specific strain energy of the considered nonlinear hyperelastic material is given by the following constitutive equation: u = u(G) =

1 λ( trG)2 1 + 2µG · G. 2

◦

In order to obtain the tensor K, we must use the equation (5.2.20). It is easy to see that the components of this tensor are given by the following relations: ◦

K klmn = λδkl δmn + µ(δkm δln + δkn δlm ). ◦

Let us observe that the components of the tensor K do not depend on Green’s strain ◦

tensor G. That is so, since u(G) is a quadratic form of G. To evaluate the components ◦ of the fourth order tensor C, we must use the relation (5.2.27) ◦

◦

◦

◦

Ckqpn = F ql F pm K klmn . According to the results obtained in P5.12, the components of the deformation ◦

gradient F are given by the relation ◦

F ql = λq δql (!). In the above relations, there is no summation relative to the indices appearing twice. The sign (!) is used to indicate this fact. Using this convention and the above formulas, we get ◦

◦

C kqpn = λq λp K kqpn (!). Hence,

◦

C kqpn = λq λp {λδkq δpn + µ(δkp δqn + δkn δqp )}(!).

◦

For the finding of the components of the instantaneous elasticity Ω, we must use the relation (5.2.30) ◦

◦

◦

Ωkqpn = C kqpn + Πkn δqp . ◦

As the results obtained in P5.13 show, only the diagonal components of the tensor

Π are nonvanishing. Correspondingly, we can write ◦

◦

Πkn = Πkk δkn (!).

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628

SOLUTIONS TO SOME PROBLEMS

We recall again that in this equation, there is no summation with respect to the index k, appearing three times! The above formulas lead us to the following expressions of the components of the ◦

instantaneous elasticity Ω : ◦

◦

Ωkqpn = λq λp {λδkq δpn + µ(δkp δqn + δkn δqp )} + Πkk δkn δqp (!). ◦

The components of the corresponding instantaneous elasticity ω can be evaluated using the result proved in P5.19 ◦ −1 ◦

◦

ω rqps = J

◦

◦

F rk F sn Ωkqpn .

According to the results obtained in P5.12, ◦

J = λ 1 λ2 λ3 and, consequently, finally we obtain the following expressions for the components of ◦ the instantaneous elasticity ω, appropriate to the updated Lagrangean approach of the incremental problems: ◦

ω rqps =

λr λs ◦ λr λq λp λs {λδrq δsn + µ(δrp δqs + δrs δqp )} + Πrr δrs δqp (!). λ1 λ2 λ3 λ1 λ2 λ3

It is left to the reader the pleasure of finding the number of independent and non◦

◦

vanishing components of the instantaneous elasticities Ω and ω. If the initial applied deformation is infinitesimal i.e. if | λk − 1 |<< 1 for k = 1, 2, 3 we can take λk ' 1, k = 1, 2, 3, in the above formulas and in this way for the components of the involved instantaneous elasticity ω, we can obtain the following expressions: ◦

ωrqps = λδrq δsn + µ(δrp δqs + δrs δqp ) + σ rr δrs δqp (!). ◦

P5.26 Obviously, the components of the tensor K are those obtained in P5.25. Using the results obtained in P5.15, it is easy to see that the components of the deformation ◦

gradient F in pure shear are given by the relations ◦

F kl = δkl + γδk1 δl2 . Hence, using the relation (5.2.27), we obtain ◦

◦

◦

◦

◦

2 C kqpn = K kqpn + γ(δq1 K k2pn + δp1 K kq2n ) + γ δq1 δp1 K k22n .

Using now the relations (5.2.30), we obtain ◦

◦

◦

◦

◦

◦

2 Ωkqpn = K kqpn + γ(δq1 K k2pn + δp1 K kq2n ) + γ δq1 δp1 K k22n + Πqn δqp ,

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SOLUTIONS TO SOME PROBLEMS ◦

where the components Πkn of the symmetric Piola-Kirchhoff stress tensor are those determined in P5.16 (a) and ◦

K kqpn = λδkq δpn + µ(δkp δqn + δkn δqp ). ◦

In order to find the components of the corresponding instantaneous elasticity ω, ◦

we use now equation (5.3.28). Since J = 1 in pure shear, we obtain for the components ◦ of the tensor c the following expressions: ◦

◦

ckqps = (δkl + γδl2 δk1 )(δqm + γδq1 δm2 )(δpn + γδp1 δn2 )(δsr + γδs1 δr2 ) K lmnr . Consequently, taking into account the relations (5.3.31), we are led to the following ◦ expressions of the components of the instantaneous elasticity ω: ◦

◦

◦

ω kqps = ckpqs + T ks δqp , ◦

where the components T ks of the Cauchy’s stress tensor are those determined P5.16(a). As our results show, in pure shear, the components of the instantaneous elasticities can have very complicated expressions, even if the hyperelastic constitutive equation has the simplest form, but the initial applied shear is large. As we shall see by solving the following problem, considerable simplifications can be achieved if the applied shear is infinitesimal and the considered material is linearly elastic and isotropic. If the involved material is linearly elastic and isotropic, the components of its elasticity tensor c have the following expressions: ckqpn = λδkq δpn + µ(δkp δqn + δkn δqp ), λ and µ being Lam´e’s constants. To obtain the components of the involved instantaneous ◦ elasticity ω, we must use the relations (5.2.48) ◦

◦

ω kqpn = ckqpn + σ kn δqp . As we know from P5.17, if the initial applied shear γ is infinitesimal, only the ◦ ◦ ◦ ◦ ◦ components σ 12 = σ 21 of the stress σ are nonvanishing and σ 12 = σ 21 = µγ. Consequently, we have ◦

σ kn = µγ(δk1 δn2 + δk2 δn1 ) ◦

and for the components of the instantaneous elasticity ω, we obtain the following expressions: ωkqpn = λδkq δpn + µ(δkp δqn + δkn δqp ) + µγ(δk1 δn2 + δk2 δn1 )δqp . Accordingly, the incremental constitutive equation (5.6.2), appropriate to an initial applied infinitesimal shear γ, becomes θkq = λδkq up,p + µ(uk,q + uq,k ) + µγ(δk1 uq,2 + δk2 uq,1 ), or, equivalently, θkq

=

λδkq εpp + 2µεkq + µγ(δk1 uq,2 + δk2 uq,1 )

=

λδkq εpp + 2µεkq + σ 12 (δk1 uq,2 + δk2 uq,1 ),

Copyright © 2004 by Chapman & Hall/CRC

◦

630

SOLUTIONS TO SOME PROBLEMS

where

1 (uk,q + uq,k ) 2 are the components of the infinitesimal strain tensor ε. The above given results show clearly that the incremental behavior of the initial deformed body depends on its material properties, through λ and µ, its Lam´e constants, as well as on the initial applied deformation, through γ, the initial applied infinitesimal shear. εkq =

P5.33 According to the relations (5.5.1) and (5.5.2), the local stability of an ini◦

tial deformed equilibrium configuration B is assured by the positive definiteness of the corresponding instantaneous elasticity. Since the initial homogeneous deformation (5.6.17)−(5.6.18) is infinitesimal, the involved instantaneous elasticity is ω and it is positive definite if ∇u · ω∇uT = ul,k ωklmn um,n > 0 for any ∇u 6= 0. If the material is orthotropic and the initial deformation is that given in (5.6.17)-(5.6.18), the nonvanishing instantaneous elasticities are those given by the relations (5.6.22). (a) Now, let us assume that we have only antiplane incremental states relative to the plane x1 x2 ; i.e. u1 = u2 ≡ 0 and u3 = u3 (x1 , x2 ), ◦

and let us establish the sufficient condition assuring local stability of B relative to antiplane incremental states. It is easy to see that, in this case, the above inequality takes the following simplified form: u3,α ωα33β u3,β > 0,

α, β = 1, 2

for any u3,α such that u3,α u3,α > 0. According to (5.6.22), the above condition becomes u3,1 ω1331 u3,1 + u3,2 ω2332 u3,2 > 0, for any u3,1 and u3,2 such that u23,1 + u23,2 > 0. Since u3,1 and u3,2 are reciprocally independent, the inequality can be satisfied if and only if the involved instantaneous elasticities satisfy the restrictions ω1331 > 0, ω2332 > 0. Taking into account again the relations (5.6.22), we can conclude that the local ◦

stability of the initial deformed equilibrium configuration B , relative to antiplane incre◦ ◦ mental states, is assured if the initial applied normal stresses σ 11 and σ 22 satisfy the restrictions ◦ ◦ C55 + σ 11 > 0 and C44 + σ 22 > 0. Since the elasticity tensor c of the material is positive definite; i.e. the reference configuration B of the body is locally stable, the elasticities C44 , C55 are positive; i.e. C44 > 0, C55 > 0.

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SOLUTIONS TO SOME PROBLEMS ◦

◦

In this way, we can conclude that local instability can occur only if σ 11 or/and σ 22 are negative; i.e. are compressive stresses. (b) Let us assume now that we have only plane incremental states relative to the plain x1 x2 ; i.e. u1 = u1 (x1 , x2 ), u2 = u2 (x1 , x2 ) and u3 ≡ 0. It is easy to see that, in this case, the considered condition, assuming local stability of ◦

B , takes the following simplified form: uβ,α ωαβγϕ uγ,ϕ > 0, α, β, γ, ϕ = 1, 2, for any uβ,α such that uβ,α uβ,α > 0. According to (5.6.22), after a little algebra, the above restriction takes the form u1,1 ω1111 u1,1 + u1,1 ω1122 u2,2 + u2,2 ω2211 u1,1 + u2,2 ω2222 u2,2 + u1,2 ω2112 u1,2 + u1,2 ω2121 u2,1 + u2,1 ω1212 u1,2 + u2,1 ω1221 u2,1 > 0 for any uα,β such that uα,β uα,β > 0. Since u1,1 , u2,2 , u1,2 and u2,1 are reciprocally independent, the above restriction will be fulfilled if and only if u1,1 ω1111 u1,1 + u1,1 ω1122 u2,2 + u2,2 ω2211 u1,1 + u2,2 ω2222 u2,2 > 0 for any u1,1 , u2,2 such that u21,1 + u22,2 > 0, and u1,2 ω2112 u1,2 + +u1,2 ω2121 u2,1 + u2,1 ω1212 u1,2 + u2,1 ω1221 u2,1 > 0 for any u1,2 , u2,1 such that u21,2 + u22,1 > 0. According to Sylvester’s criterion, since ω1122 = ω2211 , ω2121 = ω1212 , the above conditions are fulfilled if and only if the involved instantaneous elasticities satisfy the restrictions 2 ω1111 > 0, ω2222 > 0, ω1111 ω2222 − ω1122 > 0, and 2 ω2112 > 0, ω1221 > 0, ω2112 ω1221 − ω1212 > 0.

Using again the relations (5.6.22), we can conclude that the local stability of the ini◦

tial deformed equilibrium configuration B relative to plane incremental states is assured ◦ ◦ if the initial applied stresses σ 11 , σ 22 satisfy the following conditions: ◦

◦

◦

◦

◦

◦

◦

◦

2 C11 + σ 11 > 0, C22 + σ 22 > 0, (C11 + σ 11 )(C22 + σ 22 ) − C12 > 0, 2 C66 + σ 11 > 0, C66 + σ 22 > 0, (C66 + σ 11 )(C66 + σ 22 ) − C66 > 0.

Since c is positive definite, the elasticities of the material satisfy the relations 2 C11 > 0, C22 > 0, C66 > 0, C11 C22 − C12 > 0. ◦

◦

In this way, we can conclude that local instability can occur only if σ 11 and/or σ 22 are compressive stresses.

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SOLUTIONS TO SOME PROBLEMS

P5.35 Let us assume now that one nonlinear hyperelastic isotropic material is submitted to the initial homogeneous deformation given in P5.12; i.e. x1 = λ1 X1 , x2 = λ2 X2 , x3 = λ3 X3 , λ1 λ2 λ3 > 0. ◦

(a) As we have seen in P5.12, the corresponding Green strain tensor G has the following components: ◦

G11 =

◦ ◦ ◦ 1 2 (λ1 − 1), G22 (λ22 − 1), G33 (λ23 − 1), Gkl = 0 2

for

k 6= l .

Hence, ◦

◦

◦

◦

I 1 = Gmm =

3 1 X 2 (λm − 1), 2 m=1

◦

I 2 = Gkm Gkm =

3 X

◦2

Gmm =

m=1

◦

◦

◦

◦

I 3 = Gkm Gml Glk =

3 X

◦3

3 1 X 2 (λm − 1)2 , 4 m=1

Gmm =

m=1

3 1 X 2 (λm − 1)3 . 8 m=1 ◦

From P5.34 (b), we can conclude that only the diagonal components of Π are non-vanishing and we have ◦

◦

◦

◦

◦

◦

Πmm =

◦ ∂ u ◦2 ∂u ◦ ∂u +2 Gmm (!), Πkm = 0 Gmm + 3 ∂I3 ∂I2 ∂I1

for

k 6= m.

The superposed “◦” shows that the involved derivatives must be evaluated for ◦

I 1 , I 2 , I 3 and the sign (!) shows that we have no summation with respect to the index ◦ m! Introducing the expression of Gmm in the above relation, we get the components of ◦

Π as a function of λ1 , λ2 and λ3 : ◦

Πmm =

◦ ∂u 3 ∂u ∂u (!), Πkm = 0 + (λ2m − 1)2 + (λ2m − 1) ∂I3 4 ∂I2 ∂I1 ◦

for

k 6= m.

◦

As we know, Π and T are connected by the relation ◦

T=J

◦ ◦ ◦T −1 ◦

◦

◦

FΠF , J = det F

or, in component form, ◦

T kl = J

◦ −1 ◦

◦

◦

F km F ln Πmn . ◦

As we have seen in P5.12, only the diagonal components of F are nonvanishing and we have ◦ ◦ ◦ ◦ 6 l. F 11 = λ1 , F 22 = λ2 , F 33 = λ3 , F kl = 0 for k = Hence,

◦

J = λ 1 λ2 λ3

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SOLUTIONS TO SOME PROBLEMS ◦

In this way, we can conclude that only the diagonal components of T are nonvanishing and we obtain ◦

T 11 =

◦ ◦ ◦ λ3 ◦ λ2 ◦ λ1 ◦ Π33 , T km = 0 Π22 , T 33 = Π11 , T 22 = λ1 λ2 λ3 λ1 λ2 λ3

for

k 6= m.

◦

(d) According to equation (5.2.20), the tensor K is defined by the relation K=

∂2u ◦ (G). ∂G∂G

K=

∂2u (G), ∂G∂G

◦

Since

we shall have Kklmn = (

∂u ∂ 1 ∂ ∂2u )mn . )( + )klmn = ( ∂Glk ∂G 2 ∂Gkl ∂G∂G

According to the result obtained in (b), we get

∂u ∂u ∂u ∂ ∂u ∂ Gmp Gpn ). Gmn + 3 δmn + 2 ( )mn = ( ∂I3 ∂I2 ∂Gkl ∂I1 ∂Gkl ∂G

We know that u(G) depends on G only through the invariants I1 , I2 , I3 . Taking into account this fact and using the chain rule after elementary, but long, computations, we obtain for Kklmn the following expression: Kklmn

=

◦ ∂2u ◦ ◦ ∂2u ◦ ∂2u (Gkl δmn + δkl Gmn ) + 4 2 Gkl Gmn 2 δkl δmn + 2 ∂I2 ∂I2 ∂I1 ∂I1

+

3

◦ ◦ ◦ ◦ ∂2u (δkl Gmp Gpn + Gkp Gpl δmn ) ∂I3 ∂I1

+

6

◦ ◦ ◦ ∂2u ◦ ◦ ◦ (Gkp Gpl Gmn + Gkl Gmp Gpn ) ∂I3 ∂I2

+

9

◦ ∂u ∂2u ◦ ◦ ◦ (δkm δln + δkn δlm ) + 2 Gkp Gpl Gmq Gqn ∂I2 ∂I3

+

◦ ◦ ◦ ◦ 3 ∂u (δkm Gln +δnl Gmk +δlm Gkn +δnk Gml ). 2 ∂I3 ◦

We recall now that only the diagonal components of G are nonvanishing; hence, we have

◦

◦

◦

◦

◦

Gkl = Gkk δkl (!), Gkp Gpl = Gkk δkl (!). Using these relations, we obtain ◦

K klmn

◦ ◦ ◦ ◦ ∂2 ∂2 ∂2 + 2(Gkk + Gnn ) + 4 Gkk Gnn 2 ∂I1 ∂I1 ∂I2 ∂I22

=

δkl δmn {

+

3(Gkk + Gnn )

+

9 Gkk Gnn

◦2

◦

◦2 ◦ ◦ ◦2 ∂2 ∂2 + 6(Gkk Gnn + Gkk Gnn ) ∂I1 ∂I3 ∂I2 ∂I3 ◦ ◦ ∂ 3 ◦ ∂ ∂2 ◦ } u + (δkm δln + δkn δlm ){ + (Gkk + Gll ) } u (!). ∂I32 ∂I2 2 ∂I3

◦2

◦

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SOLUTIONS TO SOME PROBLEMS ◦

◦

Let us introduce now the coefficients akn , µkl defined by the following relations: ◦

◦

◦

◦

◦

akn = (Σk Σn + 2δkn Bkk ) u (!), µkl = µlk = Bkl u, ◦

the differential operations Σk and B kl being defined by the relations ◦2 ◦ ∂ ∂ ∂ (!), + 3 Gkk + 2 Gkk ∂I3 ∂I2 ∂I1 ◦ ∂ 3 ◦ ∂ (!). + (Gkk + Gll ) = Blk = ∂I3 2 ∂I2

Σk =

Bkl

◦

Now it is easy to see that the components of the tensor K can be expressed in the following form: ◦

◦

◦

K klmn = δkl δmn Σk Σn u + (δkm δln + δkn δlm )Bkl u (!). By direct verification, it can be seen that the following relation is true: 2δkl δmn δkn = δkl (δkm δln + δkn δlm )(!). ◦

◦

Now, using this equality and the coefficients akn , µkl introduced above, we can ◦

express the components of K in the following equivalent and final form: ◦

◦

◦

K klmn = δkl δmn akn +(1 − δkl )(δkm δln + δkn δlm ) µkl (!). P5.41 Let us assume now that the initial applied deformation in P5.35 satisfies the restriction λ1 = λ2 . Using the results obtained in P5.35 (c) and the fact that now ◦

◦

G11 = G22 , we can conclude that Σ1 = Σ2 , B11 = B22 , B13 = B23 , hence, ◦

◦

◦

◦

◦

◦

◦

◦

a11 = a22 , a13 = a23 , µ11 = µ22 , µ13 = µ23 . Also, from P5.35 (a), it follows that ◦

◦

Π11 = Π22 . Consequently, according to the relations found in P5.37, we obtain the following expressions for the instantaneous elasticities: ◦

◦

◦

◦

◦

◦

◦

◦

◦

◦

2 2 Ω1111 = Ω2222 = λ1 a11 + Π11 , Ω3333 = λ3 a33 + Π33 , 2 Ω1122 = Ω2211 = λ1 a12 , ◦

Ω1133 = Ω3311 = Ω2233 = Ω3322 = λ1 λ3 a13 , ◦

◦

Ω1221 = Ω2112 = λ21 µ12 + Π11 , ◦

◦

◦

◦

◦

◦

◦

◦

◦

◦

◦

◦

◦

2 Ω1212 = Ω2121 = λ1 µ12 , ◦

◦

◦

Ω1313 = Ω3131 = Ω2323 = Ω3232 = λ1 λ3 µ13 , 2 Ω1331 = Ω2332 = λ3 µ13 + Π11 , 2 Ω3113 = Ω3223 = λ1 µ13 + Π33 .

Copyright © 2004 by Chapman & Hall/CRC

635

SOLUTIONS TO SOME PROBLEMS The above relations show that restrictions (5.6.77) are satisfied. ◦

◦

Moreover, since G11 =G22 , using the results given in P5.37, we can see that ◦

◦

◦

◦

◦

◦

◦

◦

a11 = (Σ1 Σ1 + 2 B 11 ) u, a12 = Σ1 Σ1 u, µ12 = B 12 u, hence, ◦

◦

◦

◦

a12 +2 µ12 = (Σ1 Σ1 + 2 B 12 ) u . We have also B11 = B12 =

◦ ∂ ∂ . + 3 G11 ∂I3 ∂I2

Consequently, ◦

◦

◦

a11 = a12 + 2 µ12 . Using the last result, we find that the instantaneous elasticities satisfy also the equation ◦

◦

◦

◦

Ω1111 − Ω1221 = Ω1122 + Ω1212 . The above relation shows that the restriction (5.6.78) is also satisfied. Concluding, we can say that in the special case considered here, all conditions are fulfilled assuming the existence of a cylindrical symmetry in the initial deformed equilibrium state. Particularly, Guz’s representation (5.6.114) by displacement potentials can be used to analyze ◦

incremental boundary value problems, replacing, obviously, ω by Ω. P5.42 Let us consider a body and let Bt be its current configuration. Let us assume that the boundary ∂Bt of the body is submitted to a hydrostatic pressure tn (x, t) = −p(x, t)n(x, t), where p = p(x, t) > 0 is a scalar field and n = n(x, t) is the unit outward normal to ◦

◦

∂ B t . Let us assume that B is the initial (τ = 0) deformed equilibrium configuration of ◦ ◦ the body and let u = u(x, t) be the incremental displacement field from B to B t . As in Section 5.3, we denote by H0 (x, t) = ∇x u(x, t) the gradient of the involved incremental displacement field, and by F0 (x, t) = ∇x χ0 (x, t) ◦

the gradient of the involved incremental deformation from B to Bt . We have F0 (t) = 1 + H0 (t), and J0 (t) = det F0 (t) = 1 + trH0 (t). Also, the following relation is true: F−1 0 (t) = 1 − H0 (t).

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636

SOLUTIONS TO SOME PROBLEMS ◦

Using the nominal stress tensor Θ0 (x, t) corresponding to B taken as reference ◦ ◦ configuration, we shall have on the boundary ∂ B of B , sn0 (x, t) = ΘT0 (x, t)n(x), n(x) = n(x, 0), ◦

where sn0 (x, t) is the Piola-Kirchhoff stress vector, corresponding to B taken as reference ◦ configuration. Denoting by df (t) the surface force acting the boundary ∂ B t of the current configuration Bt of the body, according to the assumption made, we get df (t) = −p(t)n(t)da(t) = sn0 (t)da(0), where da(t), da(0) are the elementary areas of the considered material surface element in ◦

Bt , respectively, in B0 =B . Using Nanson’s formula, we obtain n(t)da(t) = J0 (t)F−T 0 (t)nda(0). Consequently, we get sn0 (t) = −p(t)J0 (t)F−T 0 (t)n. We denote by p(t) the incremental perturbation of the hydrostatic presume and, by sn0 (t), the incremental perturbation of the Piola-Kirchhoff stress vector. We have ◦

◦

p(t) = p + p(t) and sn0 (t) = sn0 + sn0 (t), ◦

p and sn0 being the values of p(t) and sn0 (t) in the initial deformed equilibrium config◦

uration B . As we know, we have ◦

◦

◦

sn0 = tn = − p n. Using the obtained formulas and taking into account the approximations corresponding to the incremental problem, after elementary computations, we get the following expression of the incremental Piola-Kirchhoff stress vector: ◦

◦

◦

sn0 (t) = −p(t)n − p {trH0 (t)}n + p HT (t)n on ∂ B .

Let Θ0 (t) be the perturbation of the nominal stress tensor Θ0 (t); hence, ◦

Θ0 (t) = Θ0 (0) + Θ(t) = T + Θ0 (t), ◦

◦

where T is Cauchy’s stress tensor corresponding to B . As we know, in the initial deformed equilibrium configuration, we have ◦

◦

◦

◦

◦

sn0 = tn = ΘT0 (0)n = T n = − p n on ∂ B . Thus, we can conclude that the perturbations sn0 (t) and Θ0 (t) are connected by the equation T

◦

sn0 (t) = Θ0 (t)n on ∂ B .

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SOLUTIONS TO SOME PROBLEMS

Hence, the incremental nominal stress tensor Θ0 (t) must satisfy, on the boundary ◦

∂ B of the initial deformed equilibrium configuration, the following incremental traction boundary condition: ◦

T

◦

◦

Θ0 (t)n = −p(t)n − p {trH0 (t)}n + p HT0 (t)n on ∂ B , ◦

n representing the unit outward normal to ∂ B . As we can see, the expression on the right-hand side of the above boundary condition depends on the incremental displacement field u = u(x, t), since H0 = ∇u. It is obvious that the given external load, in the case of the hydrostatic presume, is not a dead load ! On the contrary, we have to deal here with a “following” load, since the hydrostatic pressure always has the direction of the external outward unit normal to the boundary of the body, and this direction changes in time if the body is deformed, even if the involved deformations are incremental! The obtained incremental traction boundary condition just takes into account this characteristic behavior of the external load analyzed in this problem. At the same time, our results show that the incremental boundary value problem involving following loads are much more complicated than those corresponding to dead loads, which can be prescribed since they are independent of the deformations of the body. P5.43 We shall analyze now the case given in P5.42, but using the Lagrangean approach and the notations introduced in Section 5.3. We denote by U = U(x, t) the ◦

◦

incremental displacement field from B to B t and by

H(x, t) = ∇x U(x, t), its gradient. By F(x, t) we denote the gradient of the deformation from B to B t , B being the reference configuration. We have ◦

F(t) = F + H(t), ◦

◦

where F is the gradient of the deformation from B to B . Since we use the approximation corresponding to incremental fields, we get ◦ −1

F−1 (t) = F

◦ −1

−F

◦ −1

H(t) F

,

hence, ◦ −T

F−T (t) = F

◦ −T

−F

T

◦ −T

H (t) F

.

Let us denote by J(t) the incremental variation of J(t) = det F(t); hence, ◦

◦

◦

J(t) = J + J(t), J = det F,

and

J(t) =

∂ det F(t) |t=0 · H(t). ∂F(t)

As we know, ∂ det F = (det F)F−T , ∂F(t)

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638

SOLUTIONS TO SOME PROBLEMS

hence, ◦ ◦ −T

J(t) = J F

H(t).

Using the nominal stress tensor Θ(x, t) corresponding to B taken as reference configuration, we shall have, on the boundary ∂B of the reference configuration B, sN (X, t) = ΘT (X, t)N, where sN (X, t) is the Piola-Kirchhoff stress vector, corresponding to B taken as reference configuration and N is the unit outward normal to ∂B. According to the assumption made, we must have df (t) = −p(t)n(t)da(t) = sN (t)dA, where da(t), and dA are the elementary areas of the considered material surface element in Bt and, respectively, in B. Using Nanson’s formula, we obtain n(t)da(t) = J(t)F−T (t)NdA. Consequently, we get sN (t) = −p(t)J(t)F−T (t)N. We denote again, by p(t), the incremental perturbation of the hydrostatic pres◦

◦

sure, by sN , the involved Piola-Kirchhoff stress vector existing in B and by sN (t), the incremental perturbation of the Piola-Kirchhoff stress vector sN (t). Hence, we have ◦

◦

p(t) = p + p(t), sN (t) = sN + sN (t).

We have

◦ ◦ ◦ −T

◦

sN = − pJ F

N.

In this way, using the obtained formulas and taking into account the approximations corresponding to the incremental problem, after elementary computations, we get the following expression of the incremental Piola-Kirchhoff stress vector: ◦ ◦ −T

sN (t) = −p(t) J F

◦◦

◦ −T

N− pJ (F

◦ −T

· H(t)) F

◦ ◦ ◦ −T

N + pJ F

T

◦T

H (t) F N.

Let Θ(t) be the perturbation of the nominal stress tensor Θ(t); hence, ◦

Θ(t) = Θ + Θ(t), ◦

◦

where Θ is the nominal stress tensor corresponding to B . As we know, in the initial deformed equilibrium configuration, we have ◦

◦T

◦ ◦ ◦ −T

sN = Θ N = − pJ F

N on ∂B.

Thus, we can conclude that the perturbation sN (t) and Θ(t) are connected by the equation T sN (t) = Θ (t)N on ∂B.

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SOLUTIONS TO SOME PROBLEMS

Hence, the incremental nominal stress tensor Θ(t) must satisfy, on the boundary ∂B of the reference configuration of the body, the following incremental traction boundary condition: T

◦ ◦ −T

Θ (t)N = −p(t) J F

◦◦

◦ −T

N − p J (F

◦ ◦ ◦ −T

· H(t))N + pJ F

T

◦T

H (t) F N on ∂B.

Comparing the last results obtained in P5.42 and P5.43, we can see that it is more advantageous to use the updated Lagrangean approach if we wish to analyze incremental boundary problems in which following loads are involved. Thus, the traction boundary value problem can be formulated in a more simplified form. Obviously, in the updated Lagrangean approach, the incremental behavior of the material is governed by the in◦ stantaneous elasticity ω, whereas, if the Lagrangean approach is taken into account, the ◦

instantaneous elasticity Ω must be used. Let us observe that if the initial applied deformation is infinitesimal, there is no difference between the two formulations of the incremental boundary value problem, and it is easy to see that the incremental traction boundary condition which must be satisfied by the incremental nominal stress tensor θ(t) becomes ◦

◦

θT (t)n = −p(t)n − p {tr∇u(t)}n + p ∇uT (t)n on ∂B,

where n is the outward unit normal to ∂B and u = u(x, t) is the incremental displacement field. The above result can be obtained from the given boundary condition taking ◦

F = 1, an admissible approximation since the initial applied deformation is infinitesimal. Obviously, in this case, the incremental behavior of the material is governed by the instantaneous elasticity ω, introduced in the last part of Section 5.3. P5.46 To find θθr , we start with the relation θθr = (θ22 − θ11 ) sin θ cos θ − θ12 sin2 θ + θ21 cos2 θ. Taking into account equations (5.6.117), we obtain θθr = (ω1122 − ω1111 )(u1,1 − u2,2 ) sin θ cos θ + (ω1221 u1,2 + ω1212 u2,1 )(cos2 θ − sin2 θ). According to the restriction (5.6.78), we have ω1122 − ω1111 = −(ω1221 + ω1212 ). Hence, θθr = (ω1221 + ω1212 )(u2,2 − u1,1 ) sin θ cos θ + (ω1221 u1,2 + ω1212 u2,1 )(cos2 θ − sin2 θ), or θθr = (ω1212 + ω1221 )(u2,2 − u1,1 ) sin θ cos θ + ω1221 (u1,2 cos2 θ − u2,1 sin2 θ) +ω1212 (u2,1 cos2 θ − u1,2 sin2 θ).

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640

SOLUTIONS TO SOME PROBLEMS We now use equations (5.6.119) and, in this way, obtain u2,2 − u1,1 = (

uθ 1 ∂ur ∂uθ ∂ur ur 1 ∂uθ ) sin 2θ, − + ) cos 2θ + ( − + r r ∂r ∂r ∂r r r ∂θ

u1,2 cos2 θ − u2,1 sin2 θ = (

−

uθ 1 ∂ur )(sin4 θ + cos4 θ) − r r ∂θ

ur 1 ∂uθ ∂ur ∂uθ ) cos 2θ sin θ cos θ, − − sin 2θ sin θ cos θ + ( r r ∂θ ∂r ∂r

u2,1 cos2 θ − u1,2 sin2 θ =

−(

∂uθ (sin4 θ + cos4 θ) ∂r

ur 1 ∂uθ ∂ur uθ 1 ∂ur ) cos 2θ sin θ cos θ. − − ) sin 2θ sin θ cos θ + ( − r r ∂θ ∂r r r ∂θ

Using now the identity sin4 θ + cos4 θ = 1 − sin 2θ sin θ cos θ, after long, but elementary computations, we find θθr = ω1221 (

uθ ∂uθ 1 ∂ur − ) + ω1212 . r ∂θ r ∂r

In a similar way, θθθ and θθz in terms ur , uθ and uz can be obtained. Chapter 6 P6.15 As we know, if the body force b is zero, the Lam´e’s equation of motion for a linearly, elastic, homogeneous, isotropic material has the following form: ..

µ∆u + (λ + µ)grad divu = ρ0 u, where λ and µ are Lame’s constants and ρ0 is the constant mass density of the material. It is easy to see that the above equation can be written in the following equivalent form: ..

2 vT2 ∆u + (vL − vT2 )grad divu = u,

where

vL =

s

λ + 2µ and vT = ρ0

r

µ ρ0

are the velocities of propagation of the longitudinal and transverse plane elastic waves in an unbounded linear elastic homogenous and isotropic material. In component form, the above equation becomes ..

2 vT2 ∆u1 +(vL − vT2 )θ,1 = u1 , ..

2 vT2 ∆u2 +(vL − vT2 )θ,2 = u2 , θ = u1,1 + u2,2 + u3,3 .

2 2 vL ∆u3 +(vL

Copyright © 2004 by Chapman & Hall/CRC

−

vT2 )θ,3

..

= u3,

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SOLUTIONS TO SOME PROBLEMS

It is easy to see that the given displacement field satisfies identically the third equation of motion. Simple reasoning shows that the first and the second equations become ..

2 vT2 (u1,11 + u1,22 )+(vL − vT2 )θ,1 = u1 , θ = u1,1 + u2,2 , ..

2 vT2 (u2,11 + u2,22 )+(vL − vT2 )θ,2 = u2 .

For the given displacement field, we find u1,1 = Ake−αx2 cos k(x1 − vt), u2,1 = −Bke−αx2 sin k(x1 − vt),

u1,11 = −Ak 2 e−αx2 sin k(x1 − vt), u2,11 = −Bk 2 e−αx2 cos k(x1 − vt), u1,2 = −Aαe−αx2 sin k(x1 − vt), u2,2 = −Bαe−αx2 cos k(x1 − vt),

u1,22 = Aα2 e−αx2 sin k(x1 − vt), u 2,22 = Bα2 e−αx2 cos k(x1 − vt).

Hence, ∆u1 = u1,11 + u1,22 = A(α2 − h2 )e−αx2 sin k(x1 − vt),

∆u2 = u2,11 + u2,22 = B(α2 − h2 )e−αx2 cos k(x1 − vt), θ = u1,1 + u2,2 = (Ak − Bα)e−αx2 cos k(x1 − vt), θ,1 = −(Ak − Bα)ke−αx2 sin k(x1 − vt),

θ,2 = −α(Ak − Bα)e−αx2 cos k(x1 − vt).

In a similar manner, we find ..

u1 = −(vk)2 Ae−αx2 sin k(x1 − vt), ..

u2 = −(vk)2 Be−αx2 cos k(x1 − vt).

Using the above results, we can conclude that Lame’s homogenous equations of motion are satisfied by the given displacement field if the following algebraic equations are fulfilled: 2 2 {vT2 α2 + (v 2 − vL )k 2 }A + (vL − vT2 )kαB = 0,

2 2 2 −(vL − vT2 )kαA + {(v 2 − vT2 )k 2 + vL α }B = 0.

These equation have nonzero solutions; i.e. A 6= 0 and B 6= 0, if and only if the following equation is satisfied: 2 2 2 vT α + (v 2 − vL )k 2 2 −(vL − vT2 )kα

2 (vL − vT2 )kα 2 2 2 (v 2 − vL )k 2 + vL α

= 0.

It is easy to see that this characteristic equation can be expressed in the following equivalent form:   v2 v2 v2 v2 4 α − 2 − 2 − 2 k 2 α2 + (1 − 2 )(1 − 2 )k 4 = 0. vT vL vT vL

This equation has the following positive roots: s s v2 v2 α1 = 1 − 2 k > 0, α2 = 1 − 2 k > 0. vT vL

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SOLUTIONS TO SOME PROBLEMS

Obviously, the above roots are real and positive numbers if and only if the unknown constant quantity v satisfies the restriction 2 v 2 < vT2 and v 2 < vL .

We assume that the above condition is satisfied and ultimately we shall prove that this supposition is true. Denoting by A1 and B1 the constants corresponding for α1 , we get s α1 v2 A1. B1 = 1 − 2 A1 = k vL

Denoting by A2 and B2 the constants corresponding for α2 , we get s α2 v2 B2. A2 = 1 − 2 B2 = k vT

Consequently, we have two linearly independent solutions and we can conclude that the homogenous Lam´e’s equations of motion are satisfied by the following nonzero displacement fields: u1 = (A1 e−α1 x2 +

α2 B2 e−α2 x2 ) sin k(x1 − vt), k

α1 A1 e−α1 x2 + B2 e−α2 x3 ) cos k(x1 − vt), k where A1 and B2 are arbitrary constants and v, h are also constant real numbers. Since the boundary x2 = 0 of the half space is stress-free, the components σ11 , σ22 and σ23 of the Cauchy’s stress tensor σ must satisfy the following homogenous traction boundary conditions: u2 = (

σ21 = σ22 = σ33 = 0 for x2 = 0. According to the constitutive equation of the material, we have σ21 = 2µε21 = µ(u2,1 + u1,2 ), σ22 = λθ + 2µε22 = λ(u1,1 + u2,2 ) + 2µu2,2 σ23 = 2µε23 = µ(u2,3 + u3,1 ). Obviously, ε23 = 0; hence, σ23 = 0, and the third boundary condition is identically satisfied. Using the expressions of the displacement fields u1 and u2 , we get   v2 σ21 = −µ 2α1 A1 e−α1 x2 + (2 − 2 )kB2 e−α2 x2 sin k(x1 − vt), vT   2 λk − (λ + 2µ)α12 A1 e−α1 x2 − 2µα2 B2 e−α2 x2 cos k(x1 − vt). σ22 = k

Now it’s easy to see that the homogeneous traction boundary conditions for σ 21 and σ12 take the following form:

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643

SOLUTIONS TO SOME PROBLEMS

2α1 A1 + (2 −

v2 )kB2 = 0, vT2

λk 2 − (λ + 2µ)α12 A1 + 2µα2 B2 = 0. k

Since

2 λ = (vL − 2vT2 )ρ and µ = k 2 ρ,

the above homogenous algebraic system for the unknowns A1 and A2 can be expressed in the following equivalent form: 2α1 A1 + (2 −

v2 )kB2 = 0, vT2

v2 )kA1 + 2α2 B2 = 0. vT2 This system has nonzero solutions if and only if the following equation for the unknown v is satisfied 2 (2 − vv2 )k 2α1 T = 0. (2 − v2 )k 2α 2 v2 (2 −

T

This equation can be expressed as

v2 2 2 ) k − 4α1 α2 = 0. vT2 If we use the expressions of α1 and α2 , we find that v must satisfy the equation s s 2  v2 v2 v2 =4 1− 2 1− 2 . 2− 2 vT vL vT (2 −

If we denote

vL > 1, vT we can conclude after elementary computations that v must satisfy the equation β≡

f(

8(3β 2 − 2) v 2 16(β 2 − 1) v v v = 0. ( ) − ) ≡ ( )6 − 8( )4 + β2 vT β2 vT vT vT

It is easy to see that f (0) = −16

(β 2 − 1) < 0 and f (1) = 1 > 0. β2

These inequalities show that the positive real root v of the above equation is in the interval (0, vT ); i.e. 0 < v < vT < vL. The other roots of the above equation are negative or purely imaginary numbers. Thus, we can conclude that the homogenous tractions boundary conditions can be satisfied for any k and for v satisfying the above algebraic equation where α 1 and α2 are positive real numbers.

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644

SOLUTIONS TO SOME PROBLEMS For instance, for ν = 1/4, we find β 2 = 3 and the above equation becomes 3ξ 3 − 24ξ 2 + 56ξ − 32 = 0 with ξ =

v2 . vT2

Thus, we get v = 0.9194vT and α1 = 0.8475k, α2 = 0.3933k. Hence, the displacement fields u1 and u2 are u1 = A(e−α1 x2 − 0.5773e−α2 x2 ) sin k(x1 − vt), u2 = A(0.8475e−α1 x2 − 1.4679e−α2 x2 ) cos k(x1 − vt),

where A is an arbitrary constant and k is an arbitrary positive number. We have considered a dynamical problem characterized by a homogenous equation of motion and by homogenous traction boundary conditions. We have shown that this homogenous dynamical problem has nonzero solutions even if the reference configuration of the material is locally stable; i.e. vT and vL are positive real numbers. As we know, if the reference configuration of the body is locally stable, the homogenous equilibrium equation with homogenous boundary conditions can be satisfied only by the zero solution. Such kind of problems of the elastostatics can have nonzero solutions only if the material is initially deformed and the deformed equilibrium configuration of the body becomes locally unstable for some critical values of the loading parameters. As is easy to see, the above presented time dependent displacement field describes a plane wave propagating in a direction parallel to the stress-free boundary of the material, v representing the velocity of propagation of the wave. Since α1 and α2 are positive numbers, the amplitude of this wave quickly decreases as the distance from the stress-free boundary increases. Hence, our displacement field describes a surface wave; its existence was proved for the first time by Rayleigh and it is usually named Rayleigh wave.

P6.18 Let us assume now that in the strip stability problem, we take ϕ(2) = 0 and we suppose that ϕ(1) ≡ φ = (A1 cosh aη1 x1 + A2 cosh aη2 x2 ) sin ax1 , or ϕ(1) ≡ ψ = (A1 sinh aη1 x1 + A2 cosh aη2 x2 ) sin ax1 . Let us introduce the function φ(1) and φ(2) defined by the relations φ(1) = A1 cosh aη1 x2 sin ax1 , φ(2) = A2 cosh aη2 x2 sin ax1. Thus, we have

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645

SOLUTIONS TO SOME PROBLEMS

φ = φ(1) + φ(2) and (1)

φ,11 = −a2 A1 cosh aη1 x2 sin ax1 , (1)

φ,22 = a2 η12 A1 cosh aη1 x2 sin ax1 , (2)

φ,11 = −a2 A2 cosh aη2 x2 sin ax1 , (2)

φ,22 = a2 η22 A2 cosh aη2 x2 sin ax1 . Now, it is easy to see that (1)

(2)

(1)

(2)

φ,22 + η12 φ,11 = 0 and φ,22 + η22 φ,11 = 0. Consequently,

  2 2 2 ∂ ∂2 2 ∂ 2 ∂ φ=0 + η + η 2 1 ∂x21 ∂x22 ∂x21 ∂x22 and Guz’s equation is satisfied by the displacement potential φ. Analogously it can be shown that the displacement potential Ψ satisfy the same equation. 

P6.25 Since m11 is the coefficient of A1 in the system (6.3.58), and since the first equation of this system is the direct consequence of the first homogeneous traction boundary condition (6.3.43), we must evaluate θrr , assuming for a moment that 1 ur = A1 I1 (γξ1 r) cos θ sin γz, r

uθ = −A1 γξ1 I10 (γξ1 r) sin θ sin γz, uz = 0,

as is easy to see, examining the relations (6.3.57). According to the first incremental constitutive equation obtained introducing (5.6.119) in (5.6.118), we have   ∂uz ur 1 ∂uθ ∂ur . + ω1133 + + ω1122 θrr = ω1111 ∂z r r ∂θ ∂r

The nonzero instantaneous elasticities are given by equations (6.3.21). Using this relation, and the above assumed expression of the incremental displacements, we get   ur 1 ∂uθ ∂ur . + + C12 θrr = C11 r r ∂θ ∂r

Now, we find      1 γξ1 0 d 1 I1 (γξ1 r) + 2 I1 (γξ1 r) cos θ sin γz. I1 (γξ1 r) + C12 − θrr = A1 C11 r r dr r

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646

SOLUTIONS TO SOME PROBLEMS Using the identity d dr

we obtain

and

d dr





1 I1 (r) r

1 I1 (γξ1 r) r





=

=

1 I2 (r), r

γξ1 I2 (γξ1 r) r

γξ1 1 γξ1 0 I2 (γξ1 r). I1 (γξ1 r) + 2 I1 (γξ1 r) = − r r r Using the above results, we get −

θrr = A1

1 (C11 − C12 ) γξ1 I2 (γξ1 r) cos θ sin γz. r

Evaluating θrr for r = a, we obtain θrr = A1

γ (C11 − C12 ) ξ1 I2 (γξ1 r) cos θ sin γz for r = a. a

According to the relations (6.3.56)2 and (6.3.60), γ = π/l and α = πa/l. Hence, θrr = A1

α (C11 − C12 ) ξ1 I2 (γξ1 r) cos θ sin γz for r = a. a2

Now, taking into account the expression (6.3.59)1 of the coefficient m11 , we get θrr = m11 A1 cos θ sin γz for r = a. The above result proves that the given expression of m11 is correct. The other coefficients mkl of the homogeneous algebraic system can be obtained in a similar manner.

P6.26 According to the relation (6.3.62) defining the function In (r), we have I1 (r) =

∞ X j=0

 r 1+2j 1 . j!(1 + j)! 2

Hence, for r  1, we obtain the following asymptotic expression of the function I 1 (r): I1 (r) ≈

r3 r . + 16 2

Consequently, we get

3r2 1 . + 16 2 We recall now the expressions (6.3.57) of the incremental displacement fields u r , uθ and uz . Using the above estimates, we obtain I10 (r) ≈

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SOLUTIONS TO SOME PROBLEMS

I1 (γξs r) =

647

(γξs r)3 γξs r , s = 1, 2, 3 + 16 2

and

3(γξs r)2 1 , s = 1, 2, 3. + 16 2 Introducing these asymptotic estimates in (6.3.57) and neglecting terms of order r3 , after elementary computations, we obtain I10 (γξs r) =

 r 2  cos θ sin γz, A1 + A 2 a    r 2 sin θ sin γz, uθ = −A1 + A3 a r cos θ cos γz, uz = A 4 a

ur =



where A1 , A2 , A3 , and A4 define dimensionless constant quantities. For instance, by A1 the following expression was designed: γ (A1 ξ1 + γA2 ξ2 + γA3 ξ3 ) , 2 A1 , A2 and A4 being the constants appearing in the expressions (6.3.57) of the incremental displacement fields ur , uθ and uz .

P6.30 To express the incremental nominal stresses corresponding to the incremental displacement field (6.3.65), we must use the nonzero instantaneous elasticities given by the relations (6.3.21) and the incremental constitutive equations, θkl = ωklmn um,n . Elementary but long computations lead to the following result:   x1 A4 A1 sin γx3 , − ω1133 γA5 + ω1122 θ11 = 2ω1111 a b a   A 4 x2 A3 sin γx3 , + ω1221 θ12 = 2ω1212 b a b     x 2   x 2 A5 2 1 cos γx3 , + ω1331 + A2 θ13 = ω1313 γ A1 + A2 a b a   A 4 x2 A3 sin γx3 , + ω2121 θ21 = 2ω2112 b a b   x1 A4 A2 sin γx3 , − ω2233 γA5 + ω2222 θ22 = 2ω2211 a b a x x  2 1 cos γx3 , θ23 = ω2323 γA4 b a

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648

SOLUTIONS TO SOME PROBLEMS

θ31 =





ω3113 γ A1 + A2

 x 2 1

a

+ A3

 x 2  2

b

+ ω3131

A5 a



cos γx3 ,

x x  2 1 cos γx3 , b a   x1 A4 A2 sin γx3 . − ω3333 γA5 + ω3322 = 2ω3311 a b a θ32 = ω3223 γA4

θ33

Using the expression (6.3.21) of the nonzero instantaneous elasticities, and taking into account the fact that the bar is isotropic, that is, its elasticities can be expressed in terms of E and ν by the relations (2.2.91), finally, we obtain   x  E A4 A1 1 sin γx3 , − νγA5 +ν θ11 = 2(1 − ν) (1 + ν)(1 − 2ν) a b a   x  E A4 A3 2 sin γx3 , + θ12 = 2 2(1 + ν) b a b    x 2  A   x 2 E 5 2 1 cos γx3 , + + A3 θ13 = γ A1 + A2 2(1 + ν) a b a   x  E A4 A3 2 sin γx3 , + θ21 = 2 2(1 + ν) b a b   x  A4 E A2 1 + (1 − ν) − νγA5 sin γx3 , θ22 = 2ν a b (1 + ν)(1 − 2ν) a x x  E 1 2 cos γx3 , θ23 = γA4 2(1 + ν) a b

θ31 =



    x 2  x 2  A E E 1 2 5 − p γ A1 + A 2 + A3 + cos γx3 , 2(1 + ν) a b a 2(1 + ν) θ32 = γA4

θ33

=



   x1 x2 E −p cos γx3 , 2(1 + ν) a b

 νE A4 νE A2 + 2 a (1 + ν)(1 − 2ν) b (1 + ν)(1 − 2ν)     (1 − ν)E x1 −p sin γx3 . −γA5 (1 + ν)(1 − 2ν) a

Comparing the results in P6.21, P6.22 and in P6.30, it is easy to see that the actual stress state of the buckled bar is much more complex as that corresponding to the Eulerian incremental displacement field (6.3.23) founded in Euler’s plane section hypothesis.

P6.41 (a) Since E3 = 200GP a, l = 10m, a = 0.5m for the Eulerian buckling force, we obtain

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SOLUTIONS TO SOME PROBLEMS

649

pE = 1232.5M P a = 1.2325GP a. (b) The parameter α = πa/l has the value α = 0.157. (c) The correction factors p∗ (0.157) for G31 = G32 = 0.1E3 , 0.02E3 and 0.01E3 are given by the curves number 1, 5 and 10, respectively, in the Figure 6.11. Taking into account this fact we obtain p∗ (0.157) = 0.9 if G31 = G32 = 0.1E3 , p∗ (0.157) = 0.7 if G31 = G32 = 0.02E3 , p∗ (0.157) = 0.5 if G31 = G32 = 0.01E3 . (d) Since l = 10m and a = 0.5m, we have a relatively short bar. The obtained results show that, in this case, the correction furnished by the three-dimensional linearized theory can be important. Indeed, as the ratio G31 /E3 decreases, the correction factor decreases also. For our relatively short fiber reinforced bar, if it is strongly anisotropic; i.e. if the ratio G31 /E3 is in the interval (0.01, 0.02), the correction factor is in the interval (0.5, 0.7) and its influence on the critical buckling pressure cannot be neglected to avoid the occurrence of dangerous cases actually leading to buckling of the bar if the critical buckling pressure is determined using Euler’s formula.

P6.45 We have a rectangular cylindrical bar, its geometrical characteristics being l = 10m, a = 0.5m, b = 1.5m. (a) Since a < b, the Eulerian buckling pressure pE is given again by the equation used in P6.44 (a). Hence, we have pE = 1643.25M P a = 1.64325GP a. (b) The parameter α and the aspect ratio K are α = 0.1785, K = 3. (c) Since the aspect ratio K = 3 and the axial shear moduli of the bar G13 = G23 can have the values G13 = G23 = 0.1E3 , 0.01E3 , the corresponding correction factor p∗ (α) are given by the curves number 1 and number 7 of Figure 6.14. Using this fact, we find p∗ (0.1785) = 0.94 if G13 = G23 = 0.1E3 , p∗ (0.1785) = 0.58 if G13 = G23 = 0.01E3 .

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SOLUTIONS TO SOME PROBLEMS

(d) We have a relatively short bar. If G13 = G23 = 0.1E3 , the bar is weakly anisotropic, and the Eulerian formula gives an excellent result. If G13 = G23 = 0.01E3 , the bar is strongly anisotropic and the correction furnished by the three-dimensional theory cannot be neglected and the correct value pc , given by the equation (6.3.72) becomes pc = 0.943GP a, being nearly two times smaller than the value pE = 1.64325GP a given by the Euler’s theory. Chapter 7 P7.1 According to equation (7.2.23)2 , the global specific incremental stress energy w = w (U) has the following expression: w = w (U) =

1 Uσ, ρ Ωρσγϕ Uγ,ϕ + Uσ,ρ Γρσγϕ kγϕ + kρσ Dρσγϕ kγφ + 2 o o 1 + U3,ρ Nρσ U3,σ − Uρ,σ Qσ U3,ρ , 2

where σ, ρ, γ, ϕ = 1, 2. Consequently, we obtain o 1 1 ∂w = Ωαβγϕ Uγ,ϕ + Uσ,ρ Ωρσβα + Γαβγϕ kγϕ − Qα U3,β . 2 2 ∂Uβ,α

Using the symmetry property (7.1.37)1 of the global instantaneous elasticities Ωαβγϕ , we can express the above relation in the following equivalent form: o 1 1 ∂w = Ωαβγϕ Uγ,ϕ + Uσ,ρ Ωαβσρ + Γαβγϕ kγϕ − Qα U3,β , 2 2 ∂Uβ,α

or, changing σ with γ and ρ with ϕ in the second term of the right-hand side, o ∂w = Ωαβγϕ Uγ,ϕ + Γαβγϕ kγϕ − Qα U3,β . ∂Uβ,α

Now, the global incremental constitutive equation (7.1.35)1 shows that ∂w = Nαβ . ∂Uβα

In a similar manner, it can be shown that

∂w ∂w . = Mαβ and Rα −P α = ∂U3,α ∂kαβ

P7.3 To prove the uniqueness theorem for the incremental traction boundary value problem (7.2.32), we shall use the incremental work relation (7.2.22). Let us assume that our boundary value problem has two regular solutions U(1) and U(2) and let us denote by U = U(1) − U(2) the difference of these two solutions. We shall design by eαβ , kαβ , Nαβ , Mαβ , Rα and Pα the incremental fields corresponding to U. Since the two solutions U(1) and U(2) satisfy the same boundary conditions (7.2.32) and correspond

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SOLUTIONS TO SOME PROBLEMS

651

to the same normal force q, from the work identity (7.2.22), we can conclude that the following equation is satisfied: Z {Uβ,α Nαβ + kαβ Mαβ + (Rα − Pα ) U3,α } da = 0. D

Using now the expression (7.2.23)2 of the global incremental strain energy w = w (U) corresponding to the difference U = U(1) − U(2) , from the above relation, we can conclude that Z w (U) da = 0. D

Since, according to the assumption made, w (U) is a positive definite quadratic form and U(1) , U(2) are regular solutions, the last equation shows that w (U) = 0 in D. Using again the positive definiteness of the w (U) and its second expression (7.2.23) 2 , we can conclude that Uβ,α = 0 and U3,α = 0 in D. Accordingly, U1 = C1 , U2 = C2 , U3 = C3 in D, C1 , C2 , C3 being arbitrary real constants. Hence, according to equations (3.1.6), u 1 = C1 , u2 = C2 , u3 = C3 in D; i.e. the incremental displacement field u is a rigid translation of the plate. Thus, we can say that if w = w (U) is a positive definite quadratic form, the incremental traction boundary value problem (7.2.32) can have, at most, one regular solution, modulo an infinitesimal rigid translation of the plate. P7.6 Let U = (Uα , U3 ) be the regular solution of the incremental traction boundary value problem and let V = (Vα , V3 ) a regular vector field defined on D. Then, U + V is a regular admissible displacement field. We use now the expression (7.2.24) of the incremental total strain energy, where the incremental specific strain energy w (U) is defined by the equation (7.2.23)2 . Taking into account the global incremental constitutive equations (7.1.19) and the relation (7.1.38), together with the symmetry relations (7.1.37), after a long, but elementary computation, we find W (U + V) = W (U) + W (V) +

Z

{Vβ,α Nαβ + kαβ (V) Mαβ + V3,α (Rα − Pα )} da.

D

In this equation, Nαβ , Mαβ and Rα −P α are the incremental fields corresponding to the incremental displacement U, and kαβ (V) = −V3,αβ are the curvatures corresponding to the vector field V.

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652

SOLUTIONS TO SOME PROBLEMS Using Green’s formula and the relations (7.2.7) after simple calculus, we get Z Z Z W (U + V) = W (U) + W (V) = Vβ Nαβ ds − Vβ Nαβ,α da + V3 Rn ds −

Z

∂D

V3 Rα,α da −

D

Z

D

V3,β Mnβ ds +

∂D

Z

∂D

V3,β (Mαβ,α − Pβ ) da.

D

We use now the relation (7.2.21), which gives Z Z V3,β Mnβ ds = (Mnn V3,n − Mnτ,τ V3 ) ds. ∂D

∂D

Thus, since Vβ Nnβ = Vn Nnn + Vτ Nnτ on ∂D, we get W (U + V) = W (U) + W (V) +

Z

{Vn Nnn + Vτ Nnτ + V3 (Rn + Mnτ,τ ) − V3,n Mnn } ds

∂D

−

Z

{Vβ Nαβ,α + V3 Rα,α − V3,β (Mαβ,α − Pβ )} da.

D

We recall now that U is a solution of the incremental traction boundary value problem (7.2.32). Hence, Nαβ , Mαβ , Rα and Pα satisfy the global incremental equilibrium conditions (7.2.4) and the involved boundary conditions (7.2.32). Accordingly, from the above equation, we get W (U + V) = W (U) + W (V) +

Z

(Vn Φ + Vτ Ψ + V3 Γ − V3,n ∆) ds +

∂D

Z

qV3 da.

D

We return now to the potential energy functional I (U) defined by the relation (7.2.33). It is easy to see that Z

I (U + V) = W (U + V) − {Φ (Un + Vn ) + Ψ (Uτ + Vτ ) + Γ (U3 + V3 ) − ∆ (U3,n − V3,n )} ds−

∂D

Z

q (U3 + V3 ) da.

D

Hence, using the above established relation, we obtain Z I (U + V) = W (U) − (ΦUn + ΨUτ + ΓU3 − ∆U3,n ) ds + W (V), ∂D

or I (U + V) = I (U) + W (V) . Assuming now that the quadratic form w = w (U) is positive definite, from the above relation, we can conclude that

Copyright © 2004 by Chapman & Hall/CRC

653

SOLUTIONS TO SOME PROBLEMS I (U + V) > I (U) ,

for any V 6= const. Summing up the obtained results, we can say that if U is a regular solution of the incremental traction boundary problem and if the global specific strain energy w = w(U) is positive definite, then the potential energy I (U) has an absolute minimum in U on the set of all admissible displacement fields. We stress the fact that in the analyzed situation, the admissible displacement fields are not restricted by any supplementary conditions. We observe also, that using the above extreme principle, we can prove the uniqueness of the solution of the incremental traction boundary value problem, modulo an infinitesimal rigid displacement of the laminate. P7.11 To analyze the behavior of the function Ψ = Ψ (K, m) for m fixed and K variable, we shall calculate the first derivative of Ψ (K, m) with respect to K > 0. We obtain   D11 4 2 ∂Ψ = m = 2 3 K4 − D22 m K ∂K ! ! ! r r r D11 2 2 2 4 D11 4 D11 m . m K− m K+ = 2 3 K + D22 D22 D22 m K

We recall that D11 , D22 > 0 since the stress-free reference configuration of the laminate is assumed to be locally stable. Hence, for m fixed, r r D11 ∂Ψ D11 ∂Ψ m. < 0 if K < 4 m and > 0 if K > 4 D22 ∂K D22 ∂K

The minimum of the function Ψ = Ψ (K, m) , for m fixed, occurs for r 4

K=

D11 m D22

and for any m = 1, 2, 3, ... the minimum value Ψmin of Ψ (K, m) is given by the equation ! r 2D12 + 2D66 D11 . + Ψmin = 2 D22 D22

P7.15 In order to evaluate the coefficients aklij defined by the relations (7.3.47) and (7.3.48), we use the following relations: Za

sin

Zb

sin

kπx1 iπx1 dx1 = sin a a

Za

cos

a kπx1 iπx1 dx1 = δik , cos a 2 a

Zb

cos

jπx2 lπx2 b cos dx2 = δjl , b b 2

0

0

lπx2 jπx2 sin dx2 = b b

0

Copyright © 2004 by Chapman & Hall/CRC

0

654

SOLUTIONS TO SOME PROBLEMS kπx1 iπx1 dx1 = cos sin a a

cos

0

Za

o an kπx1 iπx1 1 − (−1)i+k , dx1 = sin 2 a a

Zb

lπx2 jπx2 dx2 = cos b b

Zb

cos

o bn lπx2 jπx2 1 − (−1)i+k . dx2 = sin 2 b b

Za

0

sin

0

0

We shall illustrate the procedure evaluating the coefficients bijkl given by the relations (  4 ) 2   4 jπ ijπ 2 iπ + D22 + 2 (D12 + 2D66 ) bijkl = D11 b ab a

·

Za

kπx1 iπx1 dx1 sin sin a a

Zb

sin

lπx2 jπx2 dx2 . sin b b

0

0

Using the above given results, we obtain (  4 ) 2   4 jπ ijπ 2 iπ ab . + D22 + 2 (D12 + 2D66 ) δik δjl Dn bijkl = b ab a 4

P7.16 For the considered composite laminate, the initial applied load is a biaxial uniform compression, characterized by the following relations: o

o

o

o

N 11 = −P, N 22 = −αP, N 12 = N 21 = 0, o

o

o

o

o

M 11 = M 22 = M 12 = M 21 = 0, q = 0. Hence, P > 0 is given and α > 0 is also a known quantity. (a) Since the initial applied resultant forces and moments are constant quantities and since the initial applied normal load is vanishing, the global equilibrium conditions o

o

(7.1.15) are satisfied and the shear forces Q1 , Q2 are vanishing; i.e. o

Qα = 0 , α = 1, 2. o

Consequently, the initial deformed configuration B of the laminate is a possible equilibrium state of the considered plate. o (b) To obtain the initial in-plane deformations eαβ and initial curvatures, we must use the global constitutive equations (7.1.14). According to the given data, we have o

o

A11γϕ eγϕ +B11γϕ k γϕ = −P, o

o

A22γϕ eγϕ +B22γϕ k γϕ = −αP, o

o

o

o

o

o

A12γϕ eγϕ +B12γϕ k γϕ = 0, and

B11γϕ eγϕ +D11γϕ k γϕ = 0, B22γϕ eγϕ +D22γϕ k γϕ = 0,

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655

SOLUTIONS TO SOME PROBLEMS o

o

B12γϕ eγϕ +D12γϕ k γϕ = 0. o

Using the last three equations, we can express the curvatures k γϕ as functions of the o in-plane deformations eγϕ . Introducing the obtained results in the first three equations, o we can find the in-plane deformations eγϕ as functions of P , α and the global constitutive o

o

coefficients of the laminate. For a general laminate, all components eγϕ and k γϕ will be nonvanishing, constant quantities. o o o (c) To find the initial displacement U 1 , U 2 and U 3 , we must use the geometrical reo

o

lations (7.1.8) assuming eαβ and k αβ known constant quantities, found in (b) as functions of P, α and the global mechanical parameters of the laminate. We have o

o

o

o o ∂ U2 ∂ U1 ∂ U1 = 2 e12 , + = e11 , ∂x1 ∂x2 ∂x1

o

o

o

o o o ∂2 U 3 ∂2 U 3 ∂ U2 = − k 12 , = − k 11 , = e22 , 2 2 ∂x1 ∂x2 ∂x1 ∂x2 o

o ∂2 U 3 = − k 22 . 2 ∂x2 Hence, neglecting a rigid displacement of the plate, we find o

o

o

o

o

o

U 1 = e11 x1 + e12 x2 , U 2 = e12 x1 + e22 o

o

o

o k k x2 , U 3 = − 11 x21 − k 12 x1 x2 − 22 x22 . 2 2 o

o

o

(d) Now, the initial displacements u1 , u2 , u3 of the prestressed laminate can be obtained taking into account the general relations (7.1.5) o

o

o

o

o

o

o

o

o

o

u1 = e11 x1 + e12 x2 + k 11 x1 x3 + k 12 x2 x3 , u2 = e12 x1 + e22 x2 + k 12 x1 x3 + k 22 x2 x3 , o

o

u3 = U 3 . P7.20 For an antisymmetric cross-ply laminate, we have (see Equations (3.4.18)) A16 = A26 = 0, D16 = D26 = 0 and B11 = −B22 , B12 = B16 = B26 = B66 = 0.

(a) The answer is obvious; (b) Using Voigt’s type convention, the global constitutive equations of P7.16(b) will be simplified, and become o

o

o

A11 e11 +A12 e22 +B11 k 11 = −P, o o o A21 e11 +A22 e22 −B11 k 22 = −αP, o

o

o

o

A60 e12 = 0, B11 e11 +D11 k 11 +D12 k 22 = 0, o o o −B11 e22 +D21 k 11 +D22 k 12 = 0, o

D66 k 12 = 0.

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656

SOLUTIONS TO SOME PROBLEMS

To solve this system, we assume that the initial deformed equilibrium configurao tion B of the composite is locally stable. Hence, particularly, A66 > 0 and D66 > 0. Consequently, we get o o e12 = 0 and k 12 = 0. o

o

2 We have also D11 D22 − D12 > 0, and we can express the curvatures k 11 and k 22 as o o functions of e11 and e22 . We find   o o o B11 D22 e11 +D12 e22 , k 11 = − 2 D11 D22 − D12

o

k 22 =

  o o B11 D12 e11 +D11 e22 . 2 D11 D22 − D12

Introducing these relations in the first two equations, we obtain the following relations:

2
o 2
o e11 + ∆A12 − D12 B11 e22 = −∆P, ∆A11 − D22 B11 2
o 2
o ∆A12 − D12 B11 e11 + ∆A22 − D11 B11 e22 = −α∆P,

with

2 ∆ = D11 D21 − D12 > 0.

o

o

Solving this system, we get the in-plane deformation e11 and e21 as functions of the o

compressive force P > 0 and of the ratio α > 0. Consequently, the curvatures k 11 and o k 22 can be expressed in terms of the same given quantities P and α. o o (c) Since now e12 = k 12 = 0, from P7.16 (c), we obtain o

o

o

o

o

o

o

U 1 = e11 x1 , U 2 = e22 x2 , U 3 = −

k 11 2 k 22 2 x2 . x1 − 2 2

(d) Hence, according to P7.16 (d) o

o

o

o

o

o

o

o

u1 = e11 x1 + k 11 x1 x3 , u2 = e22 x2 + k 22 x2 x3 , u3 = U 3 . P7.37 Let us assume the composite strip clamped at its edges x = ±a; i.e. we suppose the following boundary condition: U3 = 0, U3,1 = 0, N11 = 0 for x = ±a. (a) According to the equations (7.4.34)1 and (7.4.44), we have N = N 11 = C1

and

U3 = K sin Ωx + L cos Ωx −

From the last relation, we obtain

1 C2 x+ P P



 B+M C1 − C 3 . A−P

U3,1 = ΩK cos Ωx − ΩL sin Ωx −

Copyright © 2004 by Chapman & Hall/CRC

C2 . P

657

SOLUTIONS TO SOME PROBLEMS According to the third boundary condition, we obtain C1 = 0. Consequently, for the normal displacement U3 , we get U3 = K sin Ωx + L cos Ωx −

C3 C2 . x− P P

Imposing the first two boundary conditions, we obtain for the unknown constants K,L,C2 and C3 the following linear, homogeneous and algebraic system:

C3 C2 = 0, a− P P C3 C2 = 0, a− −K sin Ωa + L cos Ωa + P P C2 = 0, ΩK cos Ωa − ΩL sin Ωa − P C2 = 0. ΩK cos Ωa + ΩL sin Ωa − P

K sin Ωa + L cos Ωa −

Let us denote by ∆ the determinant of this system. Elementary computation gives ∆=

4Ω sin Ωa (Ωa cos Ωa − sin Ωa) . P2

Hence, the characteristic equation of our buckling problem is sin Ωa (Ωa cos Ωa − sin Ωa) = 0. Consequently, to obtain bifurcation (buckling), we must have sin Ωa = 0 or tgΩa = Ωa. The first condition gives Ωa = kπ, k = 1, 2, 3, ... . Let us consider the following equation: tgζ = ζ. As is well known, this equation has infinite positive roots ζk , k = 1, 2, 3, ... and the smallest positive root ζ1 satisfies the inequality π < ζ1 . Also, from P7.35 we already know that the function F = F (Ω, M ), giving through the relation (7.4.59) the possible buckling forces, in an increasing function of Ω. Thus, inspecting the possible values of Ω, we can see that the critical value of Ω, for which the primary eigenstate occurs, is given by the relation bc = π . Ω a

Accordingly, the critical value Pbc of the applied compressive force is given by the equation: r 2 2 2 b c , M ) = A + π D − (A − π D )2 + 4π (B + M )2 . 2Pbc = F (Ω 2 2 2 a a a

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658

SOLUTIONS TO SOME PROBLEMS

(b) The critical value of Ω for which the primary eigenstate of a simply supported strip occurs is given by the relation (7.4.60), and we have bc = π . Ω 2a

b c and the function F = F (Ω, M ) is an increasing function of Ω, we can Since Ωc < Ω conclude that Pc < Pbc ,

where Pc , given by the relation (7.4.61), is the critical value of the applied compressive force for which the buckling of the simply supported strip occurs. Accordingly, we can conclude that the compressive force producing buckling of a simply supported strip is smaller than the compressive force producing buckling of the same step, if it is clamped. P7.41 Let us consider again the composite strip analyzed in Section 7.4 and in P7.37−P7.40. But now, we assume that the strip is simply supported in the edge x = −a and clamped on the edge x = a. Thus, the following boundary condition must be satisfied:

U3 = 0, N11 = 0, M11 = 0 for x = −a, U3 = 0, N11 = 0, U3,1 = 0 for x = a.

(a) According to the general relations (7.4.34), we have N = N11 = C1 . Hence, C1 = 0. Consequently, the general equations (7.4.44) and (7.4.47) give

C3 C2 , x− P P = M = P (K sin Ωx + L cos Ωx).

U3 = K sin Ωx + L cos Ωx −

M11 Hence, we have also

C2 . P Imposing the remaining boundary conditions, we are lead to the following linear and homogeneous algebraic system for the unknown constants K, L, C2 and C3 U3,1 = ΩK cos Ωx − ΩL sin Ωx −

C3 C2 a = 0, − P P −P K sin Ωa + P L cos Ωa = 0, C3 C2 a = 0, − K sin Ωa + L cos Ωa − P P C2 = 0. ΩK cos Ωa − ΩL sin Ωa − P

−K sin Ωa + L cos Ωa +

From the first two equations, we get C3 = aC2 . Using this result, we can simplify the problem and we can conclude that the following system must be satisfied:

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659

SOLUTIONS TO SOME PROBLEMS

−K sin Ωa + L cos Ωa = 0, 2a C2 = 0, K sin Ωa + L cos Ωa − P 1 ΩK cos Ωa − ΩL sin Ωa − C2 = 0. P

Let us denote by ∆ the determinant of this system. We find 1 (sin 2Ωa − 2Ωa cos 2Ωa). P Hence, the characteristic equation ∆ = 0 of our buckling problem is ∆=

tg2Ωa = 2Ωa. Let ξ1 be the smallest positive root of the equation tgξ = ξ. As we know, this root satisfies the restriction π < ξ1 <

3π . 2

∨

Consequently, the critical value Ωc for which the primary eigenstate occurs is given by the relation ∨

Ωc =

ξ1 . 2a ∨

Accordingly, from the general relation (7.4.59), we get the critical value P c of the applied compressive force r ∨ ∨ ξ12 D 2 ξ12 ξ12 D ) + 2 (B + M )2 . (A − − 2 P c = F ( Ωc , M ) = A + a 4a2 4a2 ∧

(b) Taking the relation (7.4.60) giving Ωc , the result obtained in P7.38 giving Ωc and ∨ b c . Hence, the critical buckling knowing that π < ξ1 < 3π , we can see that Ωc <Ωc < Ω 2

∨

forces Pc , Pbc and P c corresponding to our three stability problems and ordered as follows ∨ Pc

Za

w (U) dx.

0

According to the relation (7.4.29), the incremental specific strain energy w = w (U) has the following expression: w = w (U1 , U3 , U1,1 , U3,1 , U1,11 , U3,11 ) = =

P 2 D 2 A−P 2 . − U3,11 U3,1 − (B + M ) U1,1 U3,11 + U3,11 2 2 2

Copyright © 2004 by Chapman & Hall/CRC

660

SOLUTIONS TO SOME PROBLEMS To obtain this relation, we have used the geometrical equations e = U1,1 , k = −U3,11 . o

o

Also, we have used the notation N = −P and M = M . (b) As is well known, the Euler-Lagrange equations corresponding to our exclusion functional E = E (U) have the following form: d2 dx21

d2 dx21





∂w ∂U1,11

∂w ∂U3,11





−

d dx1

−

d dx1





∂w ∂U1,1

∂w ∂U3,1





+

∂w = 0, ∂U1

+

∂w = 0. ∂U3

Now, using the expression of w = w (U1 , U3 , U1,1 , U3,1 , U1,11 , U3,11 ), we get the following differential system: − (A − P ) U1,11 + (B + M ) U3,11 = 0,

− (B + M ) U1,111 + DU3,1111 + P U3,11 = 0. (c) From the first equation, we obtain − (A − P ) U1,11 + (B + M ) U3,11 = −C1 , where C1 is an arbitrary constant. Similarly, from the second equation, we get − (B + M ) U1,1 + DU3,11 + P U3 = C2 x + C3 , where C2 and C3 are arbitrary constants. From the above two equations, we can eliminate U1,1 and, in this way, after elementary computations, we obtain the differential equation satisfied by U3 :

B+M (B + M )2 − D (A − P ) C1 . U3,11 − P U3 = −C2 x + C3 − A−P A−P Now, we can see that the last relation coincides with the differential equation (7.4.40) obtained using the global incremental equilibrium conditions of the considered strip, assuming that the strip is in cylindrical incremental state. Obviously, this relation is not surprising, taking into account the existence of the incremental variational principle and the fact that the Euler-Lagrange equations express the stationarity condition of the exclusion functional E = E (U) .

Chapter 8 P8.1 (a) It is easy to see that if u = u (x1 , x2 ) = x21 − x22 , then ∆u =

∂2u ∂2u ≡ 0. + 2 ∂x22 ∂x1

Hence, the function u = u (x1 , x2 ) can be the real or the imaginary part of a holomorphic function.

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661

SOLUTIONS TO SOME PROBLEMS

(b) Let f = f (z) = u (x1 , x2 ) + iv (x1 , x2 ), a holomorphic function such that Ref (z) = u (x1 , x2 ) = x21 − x22 . We know that u = u (x1 , x2 ) and v = v (x1 , x2 ) = Imf (z) satisfy Cauchy’s and Riemann’s conditions:

∂v ∂u ∂v ∂u . =− and = ∂x1 ∂x2 ∂x2 ∂x1

Hence, the function v = v (x1 , x2 ) must satisfy the equations

∂v ∂v = 2x2 . = 2x1 , ∂x1 ∂x2

Thus, we find v = 2x1 x2 + K, where K is an arbitrary complex constant. The obtained results lead to the following function:
f = f (z) = x21 − x22 + 2ix1 x2 + iK = z 2 + iK, z = x1 + ix2 .

Let g = g (z) = U (x1 , x2 ) + iV (x1 , x2 ), a holomorphic function such that Img (z) = V (x1 , x2 ) = u (x1 , x2 ) = x21 − x22 . We know that U = U (x1 , x2 ) = Reg (z) and V = V (x1 , x2 ) satisfy Cauchy’s and Riemann’s conditions

∂V ∂U ∂V ∂U . =− and = ∂x1 ∂x2 ∂x2 ∂x1

Hence, the function U = U (x1 , x2 ) must satisfy the equation

∂U ∂U = −2x1 . = −2x2 , ∂x2 ∂x1

Thus, we find U = −2x1 x2 + L, where L is an arbitrary complex constant. The obtained results give
g = g (z) = −2x1 x2 + i x21 − x22 + L = iz 2 + L. P8.2 Let f = f (z) be a holomorphic function defined in the domain B and let L be a regular closed curve in B. Let us assume that: f (z) = u (x1 , x2 ) + iv (x1 , x2 ) , u and v being the real and imaginary parts of f. Since z = x + iy, we have Z Z Z f (z) dz = udx1 − vdx2 + i vdx1 + udx2 . L

L

L

Using Green’s theorem, we get   Z Z  Z Z Z  ∂v ∂u ∂u ∂v dx1 dx2 , − dx1 dx2 + i − f (z) dz = − ∂x2 ∂x1 ∂x2 ∂x1 D L D

where D is the finite domain bounded by L. Since f = f (z) is a holomorphic function, its real and imaginary parts satisfy Cauchy’s and Riemann’s conditions recalled in P8.1. Thus, from the above equation, we obtain Cauchy’s fundamental theorem: Z f (z) dz = 0. L

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662

SOLUTIONS TO SOME PROBLEMS

Q Q P P

z g

Figure S.12: Cauchy’s fundamental formula.

To prove Cauchy’s fundamental formula, let us consider a circle γ centered in z and having radius ρ as shown in Figure S.12. Let us consider the cut having two faces, P Q and P 0 Q0 as shown in Figure S.12. The senses of integration on various curves is also given in Figure S.12. S S S According to Cauchy’s fundamental theorem on the closed curve L Q0 P 0 γ P Q, we have Z I Z I f (t) dt f (t) dt f (t) dt f (t) dt = 0, + + + t − z t − z t − z 0 0 PQ t − z γ Q P L (t) is a holomorphic function on the domain bounded by the above considered since ft−z closed curve, z being in the exterior of this domain (see Figure S.12) . Obviously, on the considered cut, we have Z Z f (t) dt f (t) dt = 0. + PQ t − z Q0 P 0 t − z

Consequently, we obtain

It is easy to see that I

γ

I

γ

f (t) dt = t−z

f (t) dt = f (z) t−z

I

γ

I

L

f (t) dt . t−z

dt + t−z

I

γ

f (t) − f (z) dt. t−z

Using the definition of the logarithmic function, we obtain I dt = 2πi t − z γ

and, thus, the above relation becomes I I f (t) − f (z) f (t) dt dt. = 2πif (z) + t−z t − z γ γ

Copyright © 2004 by Chapman & Hall/CRC

663

SOLUTIONS TO SOME PROBLEMS Thus, we obtain

I

L

f (t) dt − 2πif (z) = t−z

I

γ

f (t) − f (z) dt. t−z

The left-hand side of this equation does not depend on ρ, the radius of the circle γ. Hence, the right-hand side of the above relation is also independent on ρ. We shall show that this quantity is zero. Indeed, we have Z Z f (t) − f (z) |f (t) − f (z)| ≤ |dt| . dt |t − z| t−z γ γ

But we have (see Figure S.12) |t − z| = ρ, |dz| = ρdϕ since t − z = ρeiϕ , ϕ ∈ (0, 2π) on γ. Also, |f (t) − f (z)| < ε for any ε > 0 if |t − z| = ρ < η (ε), since f = f (t) being holomorphic is also a continuous function. Thus, we get Z Z 2π f (t) − f (z) <ε dt dϕ = 2πε, t−z 0 γ

for any ε > 0, if ρ < η (ε) . Consequently, Z f (t) dt < 2πε, − 2πif (z) L t−z

for any ε > 0. But, as we already know, the quantity on the left-hand side of the above inequality does not depend on ε. Hence, this quantity must be zero; i.e. Z f (t) dt = 2πif (z) . L t−z

P8.6 Let us consider the polynomial function f (z) = z and let X (z) = z 2 − a2


− 12

be the Plemelj function. Using Newton’s generalized binomial formula, we obtain the following development

Hence,

  1 
1 1 a4 1 a2 a2 2 f (z) + ... . − = z2 1 − = z z 2 − a2 2 = z 2 1 − 2 8 z4 2 z2 z X (z)

a4 a2 f (z) − 4 + ... = z2 − 8z 2 X (z)

Comparing the above relation and the general equation (8.1.88), we get m = 1, a2 = 1, a1 = 0, a0 = −

a2 . 2

Thus, taking into account the general relation (8.1.96), we find  p    Z a 1 a2 a2 z 1 tdt 1 2 2 2 2 = z z −a −z + . −z + = 2 2 2 2 X (z) 2πi −a X + (t) (t − z)

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664

SOLUTIONS TO SOME PROBLEMS We use now the relation (8.1.75)2 and obtain 1 . X + (t) = √ 2 i a − t2

Thus, the above equation becomes   p Z a √ 2 a2 1 t a − t2 dt 1 . z z 2 − a2 − z 2 + = 2 2 t−z 2π −a

Hence,

  p a2 , I (z) = π z z 2 − a2 − z 2 + 2

or, equivalently,

I (z) = π



a2 z − z2 + 2 X (z)

P8.9 Let us consider the complex variables



.

zj = x1 + µj x2 , j = 1, 2; where µ1 6= µ2 are two complex numbers. Using the results obtained in P8.8, we have

∂ ∂ ∂ ∂ ∂ ∂ , j = 1, 2. + µj = µj and + = ∂ zj ∂zj ∂x2 ∂ zj ∂zj ∂x1

Taking into account these relations, we obtain


∂ ∂ ∂ , j = 1, 2; = µj − µj − µj ∂ zj ∂x1 ∂x2

and


∂ ∂ ∂ − µj = µj − µ j , j = 1, 2. ∂x2 ∂x1 ∂ zj

Consequently, the equation (8.2.13) can be expressed as (µ1 − µ1 )2 (µ2 − µ2 )2

∂4ϕ = 0. ∂z1 ∂ z 1 ∂z2 ∂ z 2

Since µ1 and µ2 are complex numbers µ1 6= µ1 and µ2 6= µ2 .

Thus, the above equation can be satisfied if and only if ∂4ϕ = 0. ∂z1 ∂ z 1 ∂z2 ∂ z 2 Hence, the equations (8.2.13) and (8.2.16) are equivalent. P8.10 Since µ1 = µ2 = µ, obviously, z1 = z2 = z = x + µy and the equation (8.2.16) becomes ∂4ϕ = 0. ∂z 2 ∂ z 2

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665

Integrating this equation with respect to z, we get ∂3ϕ = H1 (z) , ∂z∂ z 2

H1 = H1 (z) being an arbitrary function of z. A new integration with respect to z gives

∂2ϕ = zH1 (z) + G1 (z) , ∂ z2 G1 = G1 (z) being an arbitrary function of z. Integrating the above equation with respect to z, we obtain

∂ϕ = zH1 (z) + G1 (z) + f1 (z) , ∂z where f1 = f1 (z) is an arbitrary function on z and Z Z H1 (z) = H1 (z) dz, G1 (z) = G1 (z) dz.

A new integration with respect to z gives

ϕ = zh1 (z) + g1 (z) + zf1 (z) + f2 (z) ,

where f2 = f2 (z) is an arbitrary function of z and Z Z h1 (z) = H1 (z) dz, g1 (z) = G1 (z) dz.

We recall now that ϕ = ϕ (x1 , x2 ) is a real valued function. Hence, we must have h1 (z) = f1 (z) and g1 (z) = f2 (z). Thus, finally, we get ϕ = ϕ (x1 , x2 ) = zf1 (z) + z f1 (z) + f2 (z) + f2 (z)

or, equivalently, ϕ = ϕ (x1 , x2 ) = 2Re {zf1 (z) + f2 (z)} . Since f1 = f1 (z) and f2 = f2 (z) depend only on z = x1 + µx2 , they are analytic functions. Thus, we can conclude that the general solution of our differential equation can be expressed by two arbitrary analytic functions using the above equation. In the theory of isotropic elastic material, only a representation of this type is used to solve boundary value problems involving plane states. P8.20 We use the representation formulas (8.2.32) and the results obtained in P8.16. Thus, we obtain q   q  −1 2 2 2 2 z 1 − a − z 1 − c 2 a1 µ 1 . z2 − a − z2 u2 (x1 , x2 ) = gRe∆ c 1 a2 µ 2

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SOLUTIONS TO SOME PROBLEMS

Hence, the normal displacement of the upper face of the crack is !) ( + q c 1 a2 µ 2 − c 2 a1 µ 1 +
2 x1 − a2 , −a < x1 < a. − x1 u2 x1 , 0 = gRe ∆

Using the equations (8.1.29) and (8.1.30)1 , we get  q 
c 1 a2 µ 2 − c 2 a1 µ 1 , −a < x1 < a. i a2 − x21 − x1 u2 x1 , 0+ = gRe ∆

According to equations (8.4.11)

Re

c 1 a2 µ 2 − c 2 a1 µ 1 = 0. ∆

Hence,

c 1 a2 µ 1 − c 2 a1 µ 1 = iG, ∆ where G is a real number. Taking into account this result for the normal displacement of the upper face of the crack, we obtain the following expression: q
u2 x1 , 0+ = −gG a2 − x21 for − a < x1 < a.

Similar reasoning leads to the following expression for the normal displacement of the lower face of the crack: q
u2 x1 , 0− = gG a2 − x21 for − a < x1 < a.

Hence, for the jump



[u2 ] (x1 ) = u2 x1 , 0+ − u2 x1 , 0−

of the normal displacement, we obtain the following expression: q [u2 ] (x1 ) = −2gG a2 − x21 for − a < x1 < a.

P8.23 According to equations (8.5.39), the involved instantaneous elasticities have the following expressions: ◦

ω1331 = C55 + σ 11 , ω2331 = C44 . Hence, µ3 = i

We suppose that

s

◦

C55 + σ 11 . C44 ◦

C55 + σ 11 > 0; hence, we assume that the composite is internally or structurally stable. In this case, from the results given in P8.21, we get    q k 2 2 , u3 (x1 , x2 ) = r   Re i z3 − z3 − a ◦ C44 C55 + σ 11

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667

SOLUTIONS TO SOME PROBLEMS where

s

◦

C55 + σ 11 x2 . C44 The above expression of the tangential displacement can be expressed in the following equivalent form:  q k 2 2 −z z . − a Im u3 (x1 , x2 ) = r 3 3   ◦ C44 C55 + σ 11 z 3 = x 1 + µ 3 x2 = x 1 + i

Particularly, as we already have seen in P8.21, on the upper face of the crack, we get q
k 2 2 u3 x1 , 0 + = r   a − x1 for − a < x1 < a. ◦ C44 C55 + σ 11


The last relation shows that u3 x1 , 0+ > 0 if k > 0. This is a normal result since the tangential force applied on the upper face of the cut is in the direction of the positive x3 axis, if k > 0.
Let us denote by u b x1 , 0+ the tangential displacement of the upper face of the ◦

crack if the initial applied stress is zero; i.e. if σ 11 = 0. We have q
k a2 − x21 . u b3 x1 , 0+ = √ C44 C55 Comparing the obtained results, we can see that

◦ b3 x1 , 0+ if σ 11 < 0 u3 x1 , 0 + > u

and



◦ u3 x1 , 0 + < u b3 x1 , 0+ if σ 11 < 0. ◦

That is, if the initial applied force σ 11 , acting in the direction of the reinforcing fibers of the composite is a tensile force, the rigidity of the material is increasing, and if the initial applied force is a compressive one, the rigidity of the composite is decreasing. These results were first obtained by Guz [8.4]. We believe that the obtained properties are completely meaningful and prove the consistency of the three-dimensional linearized theory used to obtain the incremental behavior of a fiber reinforced and prestressed composite containing a crack. Similar results are valuable concerning the first and the second mode. P8.27 Since according to the assumption made, ◦

σ 11 = 0 for the instantaneous elasticities given in P8.23, we get the following values: ω1111 = ω2222 = ω3333 = λ + 2µ, ω1122 = ω2211 = ω2233 = ω3322 = ω3311 = ω1133 = λ, ω1212 = ω2121 = ω2323 = ω3232 = ω3131 = ω1313 = µ, ω2112 = ω1221 = ω3223 = ω2332 = ω1331 = ω3113 = µ.

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SOLUTIONS TO SOME PROBLEMS

According to the results obtained in P8.26, the roots µ1 and µ2 of the algebraic equation (8.2.9) are equal and µ1 = µ2 = i. Hence, the quantity f given by equation (8.4.4) has the following value: f = 4µ2 (λ + µ)2 . Analogously, for the quantity l given by equation (8.4.6), we find l = 2iµ (λ + µ) (λ + 2µ) . Hence, for b l defined by the relation (8.5.12), we get

b l = ±2µ (λ + µ) (λ + 2µ) .

The strain energy release rate GI (a) for the first mode is given by the relation (8.5.16). Using the obtained results, we find GI (a) = KI2

λ + 2µ . 4µ (λ + µ)

The Lam´e’s coefficients λ and µ can be expressed in terms of Young’s modulus E and Poisson’s ratio ν by the relations λ=

E Eν . , µ= 2 (1 + ν) (1 + ν) (1 − 2ν)

Using these equations, finally, we get 1−ν 2 KI . 2µ This formula is well known in the classical fracture mechanics, concerning isotropic materials without initial applied stresses. For the quantity m given by equation (8.4.18), we get GI (a) =

m = 2iµ (λ + µ) (λ + 2µ) . Hence, for m b defined by equation (8.5.22), we obtain

m b = 2µ (λ + µ) (λ + 2µ) .

The strain energy release rate GII (a) for the second mode is given by the relation (8.5.25). Using the obtained results, we find 1−ν 2 KII . 2µ As above, this formula is well known in the classical fracture mechanics. The strain energy release rate GIII (a) for the third mode is given by equation (8.5.31). Using the values of the involved instantaneous elasticities, we obtain for G III (a) the following well-known expression: GII (a) =

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669

1 2 KIII . 2µ It is interesting to observe that we have obtained the above correct result corresponding to an isotropic material (for which µ1 = µ2 ), even if we have used the formula corresponding to an anisotropic material (for which µ1 6= µ2 )! GIII (a) =

P8.30 For a monoclinic material, the constitutive equations are given in the relation (2.2.18). We have σ11 = C11 ε11 + C12 ε22 + C13 ε33 + 2C16 ε12 , σ22 = C12 ε11 + C22 ε22 + C23 ε33 + 2C26 ε12 , σ33 = C13 ε11 + C23 ε22 + C33 ε33 + 2C36 ε12 , σ23 = 2C44 ε23 + 2C45 ε31 , σ31 = 2C45 ε23 + 2C55 ε31 , σ12 = C16 ε11 + C26 ε22 + C36 ε33 + 2C66 ε12 . (a) Let us assume now that the material is in plane strain equilibrium state, relative to the x1 x2 plane; i.e. we have u1 = u1 (x1 , x2 ) , u2 = u2 (x1 , x2 ) , u3 = 0. In this case, we get ε13 = ε23 = ε33 = 0. Also, we can conclude that ε11 , ε22 , ε12 depend only on x1 , x2 . Also, we have σ23 = σ31 = 0, and σ11 , σ22 , σ33 , σ12 depend only on x1 , x2 . Consequently, the Cauchy’s third equilibrium equation is identically satisfied and the first two equations become σ11,1 + σ12,2 = 0, σ21,1 + σ22,2 = 0. (b) According to the obtained results, the constitutive equations take the form σ11 = C11 ε11 + C12 ε22 + 2C16 ε12 , σ22 = C12 ε11 + C22 ε22 + 2C26 ε12 , σ12 = C16 ε11 + C26 ε22 + 2C66 ε12 , σ33 = C13 ε11 + C23 ε23 + 2C36 ε12 . Using the first three equations, taking into account the strain-displacement relations and considering the equilibrium equations, we can conclude that the plane displacements u1 = u1 (x1 , x2 ) , u2 = u2 (x1 , x2 ) must satisfy the following differential equations: P11 u1 + P12 u2 = 0, P21 u1 + P22 u2 = 0, where the involved differential operators are P11 = C11

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∂2 ∂2 ∂2 , + C66 2 + 2C16 2 ∂x1 ∂x2 ∂x2 ∂x1

670

SOLUTIONS TO SOME PROBLEMS P12 = P21 = (C12 + C66 )

∂2 ∂2 ∂2 + C16 2 + C26 2 , ∂x2 ∂x1 ∂x1 ∂x2

∂2 ∂2 ∂2 . 2 + 2C26 2 + C22 ∂x1 ∂x2 ∂x2 ∂x1 The obtained result shows that in the assumed conditions, a plane strain equilibrium state is possible. (c) Let us suppose now a plane strain state motion; i.e. we assume that P22 = C66

u1 = u1 (x1 , x2 , t) , u2 = u2 (x1 , x2 , t) , u3 = 0. From (b), it is obvious that such a state can exist. Moreover, we can conclude that the plane displacement components must satisfy the following equations of motion: P11 u1 + P12 u2 = ρ¨ u1 , P21 u1 + P22 u2 = ρ¨ u2 , ρ representing the mass density of the material in its stress-free reference configuration. P8.35 (a) According to the assumption made, µ1 6= µ2 . We know also that µ1 and µ2 are complex or imaginary numbers. Hence, z1 = x1 + µ1 x2 and z2 = x1 + µ2 x2 are complex variables. We have also

z 1 = x1 + µ1 x2 and z 2 = x1 + µ2 x2 .

At the same time, we can see that the differential equation satisfied by ϕ = ϕ and given in P8.32 (b) can be written in the following equivalent form:      ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ϕ = 0. − µ2 − µ1 − µ2 − µ1 ∂x1 ∂x2 ∂x1 ∂x2 ∂x1 ∂x2 ∂x1 ∂x2

Now, using the relation

∂ ∂ ∂ ∂ ∂ ∂ and = µα + µα , α = 1, 2; + = ∂x2 ∂zα ∂ zα ∂ zα ∂zα ∂x1

it is easy to see that the above differential equations take the equivalent form (µ1 − µ1 )2 (µ2 − µ2 )2

∂4ϕ = 0. ∂z1 ∂ z 1 ∂z2 ∂ z 2

Since µ1 and µ2 are complex, numbers µ1 6= µ1 and µ2 6= µ2 . Hence, the above equation becomes ∂4ϕ = 0. ∂z1 ∂ z 1 ∂z2 ∂ z 2 (b) Now, it is easy to see that the general solution of this equation is

ϕ = ϕ (x1 , x2 ) = f1 (z1 ) + f2 (z2 ) + f3 (z 1 ) + f4 (z 2 ) ,

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SOLUTIONS TO SOME PROBLEMS

where f1 , f2 , f3 and f4 are arbitrary functions depending on z1 , z2 , z 1 and z 2 , respectively. We recall now that ϕ = ϕ (x1 , x2 ) is a real valued function. Hence, we must have f3 (z 1 ) = f1 (z1 ) and f4 (z 2 ) = f2 (z2 ). Consequently, we get

ϕ = ϕ (x1 , x2 ) = f1 (z1 ) + f1 (z1 ) + f2 (z2 ) + f2 (z2 ) = 2Re {f1 (z1 ) + f2 (z2 )} ,

where f1 = f1 (z1 ) and f2 = f2 (z2 ) depending only on the complex variables z1 and z2 , respectively, are arbitrary analytical functions. The obtained results represent the basis to obtain the representation of Leknitskii and Guz type of the elastic state through two complex potentials. P8.36 According to the results obtained in P8.31, we have   ∂2 ∂2 ∂2 + C26 2 ϕ, u1 = −P12 ϕ = − C16 2 + (C12 + C66 ) ∂x2 ∂x1 ∂x2 ∂x1

u2 = P11 ϕ =



C11

∂2 ∂2 ∂2 + C66 2 + 2C16 2 ∂x2 ∂x1 ∂x2 ∂x1



ϕ.

According to P8.35, we have also ϕ = 2Re {f1 (z1 ) + f2 (z2 )} with z1 = x1 + µ1 x2 , z2 = x1 + µ2 x2 . Hence, using the chain rule, we obtain

u1 = −2Re P1 f100 + P2 f200 , u2 = 2Re Q1 f100 + Q2 f200 ,

where

Pα = C16 + (C12 + C66 ) µα + C26 µ2α , Qα = C11 + 2C16 µα + C66 µ2α , α = 1, 2. Now we shall introduce the following analytical functions: Fα (zα ) = Pα fα00 (zα ) , α = 1, 2. Thus, we obtain u1 = −2Re {F1 (z1 ) + F2 (z2 )} , u2 = 2Re



 Q2 Q1 F2 (z2 ) . F1 (z1 ) + P2 P1

According to P8.30 (b), we have σ22 = C12 u1,1 + C22 u2,2 + C26 (u1,2 + u2,1 ) . Using the expressions of u1 and u2 after elementary computations, we get   Γ2 0 Γ1 0 F2 (z2 ) , F1 (z1 ) + σ22 = 2Re p2 p1

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SOLUTIONS TO SOME PROBLEMS

where Γα

=

2
3 C22 C66 − C26 µα + 2 (C22 C16 − C12 C26 ) µ2α
2 + C11 C22 − C12 − C12 C66 + C16 C26 µα

+ (C11 C26 − C12 C16 ) , α = 1, 2.

Since µ1 and µ2 satisfy the equation l (µ) = 0, given in P8.32, the above expression can be simplified and we get
 2 Γα = (C66 C12 − C16 C26 ) µ2α + (C12 C16 − C11 C26 ) µα + C16 − C11 C66 µ−1 α , α = 1, 2.

Let us introduce now the analytic functions Φα (zα ) =

Γα Fα (zα ) , α = 1, 2. Pα

Using the above results, we obtain u1 = −2Re {p1 Φ1 (z1 ) + p2 Φ2 (z2 )} with

Pα , α = 1, 2; Γα u2 = 2Re {q1 Φ1 (z1 ) + q2 Φ2 (z2 )} , pα =

with

qα =

Qα , α = 1, 2, Γα

and

0

σ22 = 2Re {Ψ1 (z1 ) + Ψ2 (z2 )} with Ψα (zα ) = Φα (zα ).

To obtain the components σ12 = σ21 and σ11 , we use the constitutive equations given in P8.30(b). We have σ12 = C16 u1,1 + C26 u2,2 + C66 (u1,2 + u2,1 ), σ11 = C11 u1,1 + C12 u2,2 + C16 (u1,2 + u2,1 ). Using the above complex representations of u1 and u2 , after long, but elementary computations, we get σ12 = 2Re {r1 Ψ1 (z1 ) + r2 Ψ2 (z2 )} ,

where

ra =

and

Rα , α = 1, 2, Γα

2 Rα = (C11 c66 − C16 ) + (C11 c26 − C12 c16 )uα + (C16 C26 − C66 C12 )u2α , α = 1, 2

where

σ11 = 2Re {s1 Ψ1 (z1 ) + s2 Ψ2 (z2 )} sα =

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Sα , α = 1, 2, Γα

673

SOLUTIONS TO SOME PROBLEMS  2 Sα = C11 (C22 − C12 − C66 ) + C16 + (C16 C22 − C11 C26 ) µα + 2 C16 (C66 − C26 ) µα µα , α = 1, 2.

The above obtained relations give the Lekhniskii-Guz type representation of the elastic state by two complex potentials defined in two different complex planes. A different, but equivalent representation is given by Lekhniskii [8.2] and by Sih and Leibowitz [8.4]. These authors use a different approach using a stress potential (Airy’s function) and taking into account the compatibility condition satisfied by the nonvanishing components of the strain tensor. P8.37 According to the assumptions made, the following boundary conditions must be satisfied on the two faces of the crack.

σ21 x1 , 0+ = σ21 x1 , 0− = 0 for |x1 | < a and



σ22 x1 , 0+ = σ22 x1 , 0− = −g (x1 ) for |x1 | < a.

Here, g = g (x1 ) is a given function and we suppose that this function satisfies the H condition. The displacements and stresses are vanishing at large distances from the crack; i.e. we must have q lim {uα (x1 , x2 ) , σαβ (x1 , x2 )} = 0 for r = x21 + x22 , α, β = 1, 2. r→∞

Taking into account the representation formula obtained in P8.36, we can conclude that the complex potentials must satisfy the following conditions at large distances from the crack: lim {Φα (zα ) , Ψα (zα )} = 0 for α = 1, 2. |zα |→∞

Using the representation formula for σ21 and the boundary conditions which must be satisfied by σ21 at the two faces of the crack, we can conclude that the complex potentials Ψ1 (z1 ) and Ψ2 (z2 ) must satisfy the following boundary conditions: −

−

+

+

+ r1 Ψ + 1 (x1 ) + r2 Ψ2 (x1 ) + r 1 Ψ1 (x1 ) + r 2 Ψ2 (x1 ) = 0,

− r1 Ψ − 1 (x1 ) + r2 Ψ2 (x1 ) + r 1 Ψ1 (x1 ) + r 2 Ψ2 (x1 ) = 0,

for |x1 | < a. Reasoning, as in Section 8.3, we can conclude that our complex potentials must satisfy the following relation: r1 Ψ1 (z) + r2 Ψ2 (z) = 0 for any z = x1 + ix2 . According to the second boundary condition satisfied by σ22 on the two faces of the crack, we get − − + Ψ+ 1 (x1 ) + Ψ2 (x1 ) + Ψ1 (x1 ) + Ψ2 (x1 ) = −g (x1 ) , +

for |x1 | < a.

+

− Ψ− 1 (x1 ) + Ψ2 (x1 ) + Ψ1 (x1 ) + Ψ2 (x1 ) = −g (x1 ) ,

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674

SOLUTIONS TO SOME PROBLEMS Using the above established relation connecting the two potentials, we obtain r1 Ψ2 (z) = − Ψ1 (z) for any z = x1 + ix2 . r2

Taking into account this equation, we can eliminate the potential Ψ2 (z2 ) and for the potential Ψ1 (z1 ), we get the following boundary conditions:

r2 − r1 − r2 − r 1 + Ψ1 (x1 ) = −g (x1 ) , Ψ1 (x1 ) + r2 r2

r2 − r1 + r2 − r 1 − Ψ1 (x1 ) = −g (x1 ) . Ψ1 (x1 ) + r2 r2 Adding and subtracting lead to − +   r2 − r1 r2 − r 1 r2 − r1 r2 − r 1 = −2g, Ψ1 Ψ1 + + Ψ1 Ψ1 + r2 r2 r2 r2 +   − r2 − r1 r2 − r 1 r2 − r 1 r2 − r1 − Ψ1 Ψ1 − Ψ1 − = 0. Ψ1 r2 r2 r2 r2

Reasoning, as in Section 8.3, from the second condition, we can conclude that the following condition must be satisfied: r2 − r 1 r2 − r1 Ψ1 (z1 ) − Ψ1 (z1 ) = 0. r2 r2 The first condition represents a nonhomogeneous Hilbert-Riemann problem. Hence, as in Section 8.3, we get Z r2 − r 1 r2 − r1 X (z1 ) a g (t) dt Ψ1 (z1 ) + Ψ1 (z1 ) = − , + r2 r2 πi −a X (t) (t − z) where X (z1 ) is the Plemelj’s function. The last two relations and the formulas connecting the two potentials give the following result: Z a r2 X (z1 ) g (t) dt Ψ1 (z1 ) = Φ01 (z1 ) = − , 2πi (r2 − r1 ) −a X + (t) (t − z) Z a r1 X (z2 ) g (t) dt Ψ2 (z2 ) = Φ02 (z2 ) = . 2πi (r2 − r1 ) −a X + (t) (t − z) If we use the properties of Plemelj’s function, we can express the complex potentials in the following equivalent form: √ Z a r2 g (t) a2 − t2 dt p Ψ1 (z1 ) = Φ01 (z1 ) = − , t−z 2π (r2 − r1 ) z12 − a2 −a √ Z a r1 g (t) a2 − t2 dt p Ψ2 (z2 ) = Φ02 (z2 ) = . t−z 2π (r2 − r1 ) z22 − a2 −a

P8.40 (a) To solve the problem, we use the complex representation of the normal displacement u2 (x1 , x2 ) given in P8.36 jointly with the expressions of the complex potentials obtained in P8.38. We get  q  q  q1 r2 q2 r1 u2 (x1 , x2 ) = pRe z12 − a2 − z1 − . z22 − a2 − z1 r2 − r 1 r2 − r 1

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SOLUTIONS TO SOME PROBLEMS

675

To find the normal displacement of the line −∞ < x1 < −a, x2 = 0 behind the crack, we use the relations (8.1.27) defining the Plemelj’s function. In this way, after elementary computations, we get  q q1 r2 − q 2 r1 2 2 , for − ∞ < x1 < −a. x1 − a + x1 Re u2 (x1 , 0) = −p r2 − r 1

In the case of a monoclinic material, (q1 r2 − q2 r1 )/(r2 − r1 ) is a complex number. Let us denote by m and n, its real and imaginary parts, respectively; i.e. q1 r2 − q 2 r1 = m − in, r2 − r 1

and we have m 6= 0 and n > 0.

Using the above decomposition, we obtain  q 2 2 for − ∞ < x1 < −a. x1 − a + x1 u2 (x1 , 0) = −pm

To find the normal displacement of the line a < x1 < ∞, x2 = 0 ahead of the crack, we use the same procedure and obtain  q 2 2 x1 − a − x1 for a < x1 < ∞. u2 (x2 , 0) = pm


(b) Using again the relations (8.1.27) for the normal displacement u2 x1 , 0+ of the upper face of the crack, we obtain the following expression:   q
u2 x1 , 0+ = p n a2 − x21 − mx1 for − a < x1 < a.


Analogously, for the normal displacement u2 x1 , 0− of the lower face of the crack, we obtain the following equation:   q
u2 x1 , 0− = p −n a2 − x21 − mx1 for − a < x1 < a.

(c) To give the graphical representations of the obtained results, we assume p > 0, m > 0

and we introduce the functions U + (x1 ) and U − (x1 ) defined in the following way:  −1  p u2 (x1 , 0) for
− ∞ < x1 < −a p−1 u2 x1 , 0+ for − a < x1 < a , U + (x1 ) =  p−1 u2 (x1 , 0) for a < x1 < ∞

and

 −1  p u2 (x1 , 0) for
− ∞ < x1 < −a p−1 u2 x1 , 0− for − a < x1 < a U (x1 ) =  p−1 u2 (x1 , 0) for a < x1 < ∞. −

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676 We get

SOLUTIONS TO SOME PROBLEMS  p  2 2    −m p x1 − a + x1 for − ∞ < x1 < −a U + (x1 ) = n  a2 − x21 − mx1 for − a < x1 < a  p   x2 − a2 − x1 for a < x1 < ∞ m 1

and

 p  2 2 +x  − a for − ∞ < x1 < −a x −m 1  1  p U − (x1 ) = −n a2 − x21 − mx 1 for − a < x1 < a  p  2  2 x1 − a − x1 for a < x1 < ∞. m

Obviously, we have U + (x1 ) = U − (x1 ) for −∞ < x1 < −a and a < x1 < ∞ since in the above intervals U + (x1 ) and U − (x1 ) characterize the normal displacement of the lines −∞ < x1 < −a, x2 = 0 and a < x1 < ∞, x2 = 0 behind and ahead of the crack. Also, it is clear that U + (x1 ) 6= U − (x1 ) for − a < x1 < a since in the above interval U + (x1 ) and U − (x1 ) characterize the normal displacements of the two faces of the crack and these displacements are not equal for the same value of the coordinate x1 ! The Figures S.13 and S.14 give the graphical representations of the functions U + = + U (x1 ) and U − = U − (x1 ) .

U+ m2+n2 a na

ma

x

1

-a

-ma m2 +n2

0

na m2 +n2

a

-ma

Figure S.13: The function U + = U + (x1 ) ; 0 < m < n. (c) We can make the following observations: (i) Since for a monoclinic material m 6= 0, the normal displacement of the lines −∞ < x1 < −a, x2 = 0 and a < x1 < ∞, x2 = 0 behind and ahead of the crack are not zero and have opposite orientations on the two lines.

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677

SOLUTIONS TO SOME PROBLEMS

Uma

x

1

-na m2+n2

0

-a

ma m2 +n2

a

-ma

-na -

m 2+n 2a

Figure S.14: The function U − = U − (x1 ) ; 0 < m < n

(ii) As we have seen in Section 8.3 for an orthotropic material, even if it is prestressed in the considered manner, the normal displacement of the above lines is zero. (iii) Even if the symmetrically applied normal stresses have constant value on the two faces of the crack, for a monoclinic material, the normal displacements are not symmetric relative to the line x2 = 0 and the crack tip have nonzero normal displacements, since m 6= 0. (iv) As we have seen in the Section 8.3, for an orthotropic material, even if it is prestressed in the given manner, the normal displacements of the two faces of the crack are symmetric relative to the x2 = 0 line and the crack tips have zero normal displacements, if the normal stresses are symmetrically applied on the crack faces. (v) To explain the above results, we recall that the normal loads are symmetrically applied on the two faces of the crack relative to the plane x1 x3 . This is a symmetry plane for an orthotropic material even if it is prestressed in the considered manner, but it is not a symmetry plane for our monoclinic material! Obviously, this latter fact produces the observed behavior signaled in (i) and (iv) . P8.43 Since C44 > 0, the equilibrium equation satisfied by u3 can be written in the following equivalent form: u3,22 + 2

C55 C45 u3,11 = 0. u3,12 + C44 C44

Let us try to factorize this equation in the following form:    ∂ ∂ ∂ ∂ u3 = 0, − µ4 − µ3 ∂x1 ∂x2 ∂x1 ∂x2

where µ3 and µ4 are constant quantities. It is easy to see that the above two equations

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678

SOLUTIONS TO SOME PROBLEMS

are equivalent if and only if µ3 and µ4 satisfy the following relations: µ3 + µ4 = −2

C55 C45 . , µ3 µ4 = C44 C44

Hence, µ3 and µ4 are the roots of the algebraic equation µ2 + 2

C55 C45 = 0. µ+ C44 C44

Accordingly, we get     q q 1 1 2 2 . −C45 − i C44 C55 − C45 , µ4 = −C45 + i C44 C55 − C45 µ3 = C44 C44 2 > 0, µ3 and µ4 are conjugate complex numbers; i.e. Since C44 C55 − C45

µ4 = µ 3 .

Hence, the differential equation satisfied by u3 becomes    ∂ ∂ ∂ ∂ u3 = 0. − µ3 − µ3 ∂x1 ∂x2 ∂x1 ∂x2

We introduce the complex variable, z 3 = x 1 + µ 3 x2 . Now, we can see that the above differential equation takes the following equivalent form: ∂ 2 u3 = 0. ∂z3 ∂ z 3

Thus, we can conclude that the antiplane elastic state can be represented in the following way: u3 = 2Ref3 (z3 ) , σ13 = 2Re(ρ1 f30 (z3 )) with ρ1 = C55 + C45 µ3 , σ23 = 2Re(ρ2 f30 (z3 )) with ρ2 = C45 + C44 µ3 , f3 = f3 (z3 ) being an arbitrary analytic function of the complex variable z3 . Introducing the analytic function Φ3 (z3 ) = ρ2 f3 (z3 ) we get the following equivalent representation: u3 = 2Reρ−1 2 Φ3 (z3 ) ; σ13 = 2ReqΨ3 (z3 ) , q =

ρ1 , ρ2

σ23 = 2ReΨ3 (z3 ) , Ψ3 (z3 ) = Φ0 (z3 ) , which is more useful to study problems concerning a crack.

Copyright © 2004 by Chapman & Hall/CRC