Finite elements for large deformation solid mechanics ...

Finite elements for large deformation solid mechanics problems Vinh Phu Nguyen Delft unitversity of Technology Falculty of Civil Engineering and Geosciences Computational Mechanics Group

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Chap. 0

Preface Mastering the nonlinear finite element method is never an easy task. This is particularly the case for students being in short of good documents or tutors. The reason of this difficulty stems from the multidisciplinary charasteristic of the finite element method. Basically, it involves continuum mechanics (providing the mathematical model), approximation methods and linear algebra (providing the discrete model) and programming experiences (to write a piece of code that works), just to name a few important. Usually, undergraduate students are given course to introductory finite elements Vinh Phu Nguyen Delft, July 2008

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Chap. 0

Contents 1

Introduction 1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Continuum mechanics 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 The motion and deformation . . . . . . . . . . . . . . . . . 2.2.1 Definitions . . . . . . . . . . . . . . . . . . . . . . 2.2.2 Motion . . . . . . . . . . . . . . . . . . . . . . . . 2.2.3 Lagrangian and Eulerian descriptions . . . . . . . . 2.2.4 Displacement, velocity and acceleration . . . . . . . 2.2.5 Deformation gradient . . . . . . . . . . . . . . . . . 2.2.6 Conditions on motion . . . . . . . . . . . . . . . . . 2.2.7 Volume change . . . . . . . . . . . . . . . . . . . . 2.2.8 Area change . . . . . . . . . . . . . . . . . . . . . 2.2.9 Rigid body rotation . . . . . . . . . . . . . . . . . . 2.3 Strain measures . . . . . . . . . . . . . . . . . . . . . . . . 2.3.1 Green strain tensor . . . . . . . . . . . . . . . . . . 2.3.2 Almansi strain tensor . . . . . . . . . . . . . . . . . 2.3.3 Bulk-deviatoric decomposition of kinematics tensors 2.3.4 Strain rates . . . . . . . . . . . . . . . . . . . . . . 2.3.5 Polar decomposition . . . . . . . . . . . . . . . . . 2.4 Stress measures . . . . . . . . . . . . . . . . . . . . . . . . 2.4.1 The Cauchy stress . . . . . . . . . . . . . . . . . . 2.4.2 The Kirchhoff stress . . . . . . . . . . . . . . . . . 2.4.3 The nominal stress . . . . . . . . . . . . . . . . . . 2.4.4 The second Piola-Kirchhoff stress . . . . . . . . . . 2.4.5 Deviatoric and pressure components of stress tensors 2.5 Material objectivity . . . . . . . . . . . . . . . . . . . . . . 2.5.1 Objective tensors . . . . . . . . . . . . . . . . . . . 2.5.2 Objective stress rates . . . . . . . . . . . . . . . . . 2.6 Conservation equations . . . . . . . . . . . . . . . . . . . . 2.6.1 Reynold’s transport theorem . . . . . . . . . . . . .

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3 3 3 5 5 5 6 6 6 6 8 9 10 12 12 14 14 15 18 18 22 26 26 27 28 29 30 31 31 32 34 35

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CONTENTS

2.7

3

4

2.6.2 Conservation of mass . . . . . . . . 2.6.3 Conservation of linear momentum . 2.6.4 Conservation of angular momentum 2.6.5 Conservation of energy . . . . . . . Total Lagrangian conservation equations . 2.7.1 Conservation of linear momentum . 2.7.2 Conservation of angular momentum 2.7.3 Conservation of energy . . . . . . .

Chap. 0 . . . . . . . .

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Constitutive models 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Hyperelasticity . . . . . . . . . . . . . . . . . . . . . . . . 3.2.1 Linear hyperelasticity . . . . . . . . . . . . . . . . . 3.2.2 Isotropic hyperelasticity . . . . . . . . . . . . . . . 3.2.3 Isotropic linear hyperelasticity . . . . . . . . . . . . 3.2.4 Matrix form using the Voigt notation . . . . . . . . . 3.2.5 Incompressible isotropic linear elasticity . . . . . . . 3.3 Finite hyperelasticity . . . . . . . . . . . . . . . . . . . . . 3.3.1 Elasticity tensors . . . . . . . . . . . . . . . . . . . 3.3.2 Isotropic hyperelastic materials . . . . . . . . . . . 3.3.3 Isotropic hyperelasticity in principle directions . . . 3.4 Incompressible hyperelasticity . . . . . . . . . . . . . . . . 3.5 Nearly incompressible hyperelasticity . . . . . . . . . . . . 3.5.1 Elasticity tensors . . . . . . . . . . . . . . . . . . . 3.6 Hyperelastic plastic materials . . . . . . . . . . . . . . . . . 3.6.1 Hyperelastic potential . . . . . . . . . . . . . . . . 3.6.2 Decomposition of rate of deformation . . . . . . . . 3.7 Stress update algorithms . . . . . . . . . . . . . . . . . . . 3.7.1 Fully implicit backward Euler scheme . . . . . . . . 3.7.2 Von Mises plasticity with isotropic hardening . . . . 3.7.3 Algorithm tangent moduli . . . . . . . . . . . . . . 3.7.4 Algorithmic tangent modulii of von Mises plasticity 3.7.5 Plane stress von Mises material . . . . . . . . . . . Lagrangian finite elements 4.1 Introduction . . . . . . . . . . . . . . 4.2 Governing equations . . . . . . . . . 4.3 Weak formulation . . . . . . . . . . . 4.3.1 Strong form to weak form . . 4.3.2 Weak form to strong form . . 4.4 Principle of virtual power . . . . . . . 4.5 Finite element discretization . . . . . 4.5.1 Finite element approximation

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36 37 39 40 42 42 43 44

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47 47 47 48 49 49 51 53 53 54 56 59 60 62 62 63 63 63 64 64 66 69 71 73

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75 75 76 78 78 79 79 83 83

Sec. 0.0

4.6

4.7

4.8 5

CONTENTS

7

4.5.2 Internal nodal forces . . . . . . . . . . . . . . . . . . 4.5.3 External nodal forces . . . . . . . . . . . . . . . . . . 4.5.4 Mass matrix and inertial force . . . . . . . . . . . . . 4.5.5 Discrete equations . . . . . . . . . . . . . . . . . . . 4.5.6 Natural coordinates . . . . . . . . . . . . . . . . . . . 4.5.7 Derivatives of functions . . . . . . . . . . . . . . . . Numerical integration . . . . . . . . . . . . . . . . . . . . . . 4.6.1 Newton-Cotes quadrature . . . . . . . . . . . . . . . 4.6.2 Gauss-Legendre quadrature . . . . . . . . . . . . . . Implementation . . . . . . . . . . . . . . . . . . . . . . . . . 4.7.1 Three dimensional problems . . . . . . . . . . . . . . 4.7.2 Two dimensional problems . . . . . . . . . . . . . . . 4.7.3 Computation of external force vector . . . . . . . . . 4.7.4 Examples on some 2D and 3D isoparametric elements Computation of the internal nodal force . . . . . . . . . . . .

Solution procedures 5.1 Introduction . . . . . . . . . . . . . . . . . . . 5.2 Explicit dynamics . . . . . . . . . . . . . . . . 5.2.1 Central difference method . . . . . . . 5.2.2 Implementation . . . . . . . . . . . . . 5.2.3 Computation of diagonal mass matrix . 5.3 Implicit dynamics . . . . . . . . . . . . . . . . 5.3.1 The Newmark time integration method 5.3.2 The Newton-Raphson method . . . . . 5.3.3 Convergence criteria . . . . . . . . . . 5.3.4 Linearization . . . . . . . . . . . . . . 5.3.5 Material tangent stiffness . . . . . . . . 5.3.6 Geometric stiffness . . . . . . . . . . . 5.3.7 Load stiffness matrix . . . . . . . . . . 5.4 Stability . . . . . . . . . . . . . . . . . . . . . 5.4.1 Time step estimatation . . . . . . . . . 5.5 Solvers for linear system of equations . . . . . 5.5.1 Direct solvers . . . . . . . . . . . . . . 5.5.2 Iterative solvers . . . . . . . . . . . . . 5.6 Load control . . . . . . . . . . . . . . . . . . . 5.7 Displacement control . . . . . . . . . . . . . . 5.8 Arc-length control . . . . . . . . . . . . . . . . 5.8.1 General formulation . . . . . . . . . . 5.8.2 Some specific arc-length functions . . . 5.8.3 Energy release control . . . . . . . . . 5.8.4 Automatic load incrementation . . . . . 5.9 Mesh generation programs . . . . . . . . . . .

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117 117 117 117 118 120 120 120 121 123 123 123 124 126 126 126 126 127 127 128 128 128 128 129 130 131 132

8

CONTENTS

Chap. 0

5.9.1 Gmsh . . . . . . . . . . . . . . . . . . . . . . . . . 5.9.2 Salome . . . . . . . . . . . . . . . . . . . . . . . . 5.10 Validation examples . . . . . . . . . . . . . . . . . . . . . . 5.10.1 Compressible finite strain hyperelasticity problems . 5.10.2 Incompressible finite strain hyperelasticity problems 6

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134 134 135 136 136

Robust finite elements for constrained media 6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Element characteristics . . . . . . . . . . . . . . . . . . . . . . . 6.2.1 Q4 element . . . . . . . . . . . . . . . . . . . . . . . . . 6.2.2 Rank deficiency . . . . . . . . . . . . . . . . . . . . . . . 6.3 Lagrange multiplier and penalty methods . . . . . . . . . . . . . 6.3.1 Lagrange multiplier method . . . . . . . . . . . . . . . . 6.3.2 Penalty method . . . . . . . . . . . . . . . . . . . . . . . 6.3.3 Augmented Lagrange multiplier method . . . . . . . . . . 6.3.4 Pertubed Lagrange multiplier method . . . . . . . . . . . 6.4 Multi-field weak forms in small strain context . . . . . . . . . . . 6.4.1 The Reisner-Hellinger variational principle . . . . . . . . 6.4.2 The Hu-Washizu variational principle . . . . . . . . . . . 6.4.3 Hu-Washizu weak form for incompressible linear elasticity 6.4.4 Locking phenomenon . . . . . . . . . . . . . . . . . . . . 6.5 Multi-field weak forms in finite strain context . . . . . . . . . . . 6.5.1 Three-field Hu-Washizu weak form . . . . . . . . . . . . 6.5.2 Three-field weak form for nearly incompressibility . . . . 6.5.3 Simo-Hughes form of the Hu-Washizu weak form . . . . 6.6 Locking-free Q4 elements . . . . . . . . . . . . . . . . . . . . . 6.6.1 Two-field weak form for incompressibility . . . . . . . . 6.7 Generalized displacement formulation-small strain . . . . . . . . 6.7.1 Energy Functional . . . . . . . . . . . . . . . . . . . . . 6.7.2 Finite element approximations . . . . . . . . . . . . . . . 6.7.3 Implementation aspects . . . . . . . . . . . . . . . . . . . 6.7.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . 6.8 Generalized displacement formulation-finite strain . . . . . . . . . 6.8.1 Three-field weak form . . . . . . . . . . . . . . . . . . . 6.9 Nodal integration . . . . . . . . . . . . . . . . . . . . . . . . . .

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137 137 137 139 143 145 145 147 148 148 148 149 150 151 153 154 154 155 155 156 158 158 159 159 163 167 167 167 176

Coupled thermo-mechanical problems 7.1 Introduction . . . . . . . . . . . . 7.2 Governing equations . . . . . . . 7.2.1 Mechanical equations . . . 7.2.2 Thermal equations . . . . 7.3 Weak forms . . . . . . . . . . . . 7.4 Finite element discretization . . .

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179 179 179 179 179 179 180

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Sec. 0.0

7.5 7.6 8

CONTENTS

9

7.4.1 Finite element approximations . . . . . . . . . . . . . . . . . . . . . . 180 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181 Solution of the heat equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 181

Continuum-based structural finite elements 8.1 Introduction . . . . . . . . . . . . . . . 8.2 Beam theories . . . . . . . . . . . . . . 8.2.1 Timoshenko beam theory . . . . 8.2.2 Euler-Bernoulli beam theory . . 8.3 Continuum-based beam elements . . . . 8.3.1 Definitions . . . . . . . . . . . 8.3.2 Assumptions . . . . . . . . . . 8.3.3 Motions . . . . . . . . . . . . . 8.3.4 Nodal forces . . . . . . . . . . 8.3.5 Constitutive law . . . . . . . . 8.3.6 Mass matrix . . . . . . . . . . . 8.4 Shell theories . . . . . . . . . . . . . . 8.5 Inhomogeneous materials . . . . . . . .

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A Math A.1 Directional derivative . . . . . . . . . . . . . . . . . A.1.1 Definition . . . . . . . . . . . . . . . . . . . A.1.2 Properties of the directional derivatives . . . A.2 Derivative of the determinant of a second order tensor A.3 Derivatives of second order tensor’s invariants . . . . A.4 Derivative of the inverse of a tensor . . . . . . . . . A.5 Second-order tensors . . . . . . . . . . . . . . . . . A.5.1 Symmetric and skew symmetric tensors . . . A.5.2 Isotropic and deviatoric tensors . . . . . . . A.5.3 Invariants of second order tensor . . . . . . . A.5.4 Invariants of symmetric second order tensor . A.6 Sheriman-Morrison formula . . . . . . . . . . . . .

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185 185 185 185 187 187 187 187 187 190 191 192 192 192

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195 195 195 196 196 197 198 199 199 199 200 200 201

10

CONTENTS

Chap. 0

List of Figures 2.1 2.2 2.3 2.4 2.5

Initial (undeformed) and current (deformed) configurations of body Initial (undeformed) and current (deformed) configurations of body Internal forces in a body . . . . . . . . . . . . . . . . . . . . . . . Initial (undeformed) and current (deformed) configurations of body Stress vectors acting upon on a tetrahedron . . . . . . . . . . . . .

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5 22 26 26 27

4.1

A typical two dimensional finite element discretization, showing the nodal support and a triangle element . . . . . . . . . . . . . . . . . . . . . . . . . . . . Initial and current configurations and their relation to the parent element domain Non-zero strain components in solid of revolution . . . . . . . . . . . . . . . . Non-zero strain components in solid of revolution . . . . . . . . . . . . . . . . Uniform traction applied on a bilinear quadrilateral element. . . . . . . . . . . Boundary elements for evaluation of external force vector. . . . . . . . . . . .

83 87 106 106 108 109

4.2 4.3 4.4 4.5 4.6 5.1

5.2 5.3 6.1 6.2 6.3 6.4 6.5 8.1 8.2 8.3

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Different techniques to complete trace the equilibrium path: from left to right, load control (fail to pass snap through point), displacement control (can handle snap through behavior) and arclength control (able to trace the complete path). . 128 Three dimensional structured finite element mesh created by Salome. . . . . . . 134 A patch-test for compressible Neo-Hookean material . . . . . . . . . . . . . . 136 Commonly used 2D and 3D finite elements in nonlinear analysis . . . . . . . . Two bending hourglass modes, ui = xy . . . . . . . . . . . . . . . . . . . . . ff . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Constrained extrema of ordinary functions. . . . . . . . . . . . . . . . . . . . Plane strain Cook’s membrane: geometry, boundary conditions, material parameters and quantity of interest. . . . . . . . . . . . . . . . . . . . . . . . . .

137 143 144 145 167

Continuum-based three noded beam element . . . . . . . . . . . . . . . . . . . 186 Continuum-based three noded beam element and the underlying six noded continuum element. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188 Continuum-based three noded beam element and the underlying six noded continuum element. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189

2

LIST OF FIGURES

Chap. 0

Chapter 1 Introduction 1.1

Introduction

1.2

References

The aim of this manuscript is to introduce major issues involved in the nonlinear stress analysis of solid using the finite element method (FEM). Nonlinear behavior of solid takes two forms, namely, material nonlinearity and geometry nonlinearity. Nonlinear materials are widely used in practical applications. The nonlinear elasticity including the hypoelasticity, hyperelasticity and viscoelasticity are presented in details. When the deformation of solids reaches a state for which undeformed and deformed shapes are considerably different, finite deformation occurs. In this case, geometry nonlinearity must be taken into account. The underlying block of the finite element method is the theory of continuum mechanics. It is thefore to The outline of the writting is as follows. We first start with some mathematical preliminaries on vector, tensor and matrix analysis. The next section presents the theory of continuum mechanics. Various material behaviors are introduced in the next section. The finite element method is first introduced in the context of one dimensional linear problem. This writting reflects what the author have self-studied from the following interesting books: 1. Nonlinear finite elements for continua and structures. Ted Belytschko, Wing Kam Liu and Brian Moran. 2. Nonlinear continuum mechanics for finite element analysis. Javier Bonet and Richard D. Wood 3. The finite element methods. O.C Zienkiewic and R. Taylor 4. The finite element method. Linear static and dynamic finite element analysis. Thomas J.R. Hughes

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Introduction 5. Functional and structured tensor analysis for Engineers. R.M. Brannon.

Chap. 1

Chapter 2 Continuum mechanics 2.1

Introduction

Continuum mechanics studies the macroscopic behaviour of where inhomogeneities such as are not taken into account. The deformation gradient tensor is the most important concept in expressing finite deformation. It is used to define many strain measures such as the left Cauchy-Green deformation tensor, the Green strain tensor and the Almansi strain tensor. Rotation plays an important role in nonlinear mechanics and the polar decomposition theorem With large deformation, the initial configuration and the deformed configuration are quite different, which leads to different stress measures. The familiar Cauchy stress is the true stress since it is defined with respect to the deformed configuration. Other stress measures such as the Piola-Kirchhoff stress tensors are defined with respect to the initial configuration. Material objectivity

2.2

The motion and deformation Figure 2.1: Initial (undeformed) and current (deformed) configurations of body

2.2.1

Definitions

The position (vector) of a material point (also called particle) in the initial, undeformed configuration of a body is denoted X relative to some coordinate basis. The position of the same material point in the deformed configuration is denoted x. Since X used to label material points it is named the material or Lagrangian coordinate. On the other hand, x defines the position of a point in space, it is therefore the so-called spatial or Eulerian coordinate.

6

2.2.2

Continuum mechanics

Chap. 2

Motion

The motion (deformation) of a solid is described by a function φ(X, t). A relation between spatial coordinates and material coordinates can be established as follows x = φ(X, t)

2.2.3

(2.2.1)

Lagrangian and Eulerian descriptions

There are two approaches in describing the motion of continuum bodies: Lagrangian and Eulerian descriptions. In the Lagrangian description, the independent variables are the material coordinates and time. In the Eulerian description, the independent variables are the spatial coordinates and time. In the Lagrangian approach, since one follows material points, the treatment of history dependent continuums is trivial. The Lagrangian description is therefore extensively used in solid mechanics. On the other hand, the stress of fluids is independent of its history, the Eulerian description is then widely adopted in describing the motion of fluids.

2.2.4

Displacement, velocity and acceleration

The velocity and acceleration fields of a body are the primary kinematical fields in describing the motion of the body. Since the body is described by two different configurations (material or spatial), the velocity and acceleration fields admit two descriptions. Definition 2.2.1 The displacement of a material point X, denoted by u(X, t), is the difference between its current position φ(X, t) and its initial position φ(X, 0). So, u(X, t) := φ(X, t) − φ(X, 0) = x − X

(2.2.2)

Definition 2.2.2 The velocity of a material point X, denoted by v(X, t), is defined as the rate of change of position of this material point, that is v(X, t) :=

∂φ(X, t) ∂t

This is the Lagrangian velocity field. The corresponding Eulerian velocity field is ∂φ(X, t) v(x, t) := ≡ v(φ−1 (x, t), t) ∂t X=φ−1 (x,t)

(2.2.3)

(2.2.4)

Definition 2.2.3 The acceleration of a material point X is the rate of change of its velocity, i.e., the material time derivative of the velocity, ∂v(X, t) ∂ 2 φ(X, t) a(X, t) := = ∂t ∂t2

(2.2.5)

Sec. 2.2

The motion and deformation

Note that this is a Lagrangian field. Its Eulerian counterpart is ∂ 2 φ(X, t) a(x, t) := ≡ a(φ−1 (x, t), t) ∂t2 X=φ−1 (x,t)

7

(2.2.6)

Next, the concept of material time derivative is introduced. To understand this important concept, considering the situation: assuming a certain field ϕ (scalar, vectorial or tensorial) is defined over the body, and we wish to know its rate of change for a given material point X during the motion. This is known as the material time derivative of ϕ. There are two situations to consider (corresponding to material and spatial descriptions): 1. Lagrangian description. In the Lagrangian description, the independent variables are the material coordinates X and time t, so all we have to do is taking the partial derivative of the given field ϕ with respect to time. For a material field ϕ(X, t), its material time derivative is ∂ϕ(X, t) Dϕ(X, t) ≡ ϕ˙ = (2.2.7) Dt ∂t where the first two equations indicate standard notation for material time derivative. The superposed dot is also used for ordinary time derivative if the variable is only function of time. 2. Eulerian description. The considered field is ϕ(x, t). This case is much more complicated since not only time changes but also does changes the spatial position x of the considered particle . We must calculate the partial derivative with respect to time, of the material description of ϕ(x, t), keeping X fixed. ϕ(φ(X, t + ∆t), t + ∆t) − ϕ(φ(X, t), t) Dϕ(x, t) ≡ ϕ˙ := lim ∆t→0 Dt ∆t

(2.2.8)

Using the chain rule, we obtain Dϕ(x, t) ∂ϕ(x, t) ∂ϕ(x, t) ∂φi (X, t) = + Dt ∂t ∂xi ∂t ∂ϕ(x, t) ∂ϕ(x, t) = + vi (x, t) ∂t ∂xi

(2.2.9)

Writing the above in compact form, we obtain the important formula of the material time derivative for an Eulerian scalar field: Dϕ(x, t) ∂ϕ(x, t) = + ∇ϕ(x, t) · v(x, t) Dt ∂t

(2.2.10)

The term ∂ϕ/∂t is called the spatial time derivative, and the term ϕ,j vj is the convective term, which is also called the transport term.

8

Continuum mechanics

Chap. 2

For an Eulerian vector field, by considering each component vi (x, t) a Eulerian scalar field, we have Dvi (x, t) ∂vi (x, t) ∂vi (x, t) = + vj Dt ∂t ∂xj

(2.2.11)

In tensor notation, the above is written as Dv(x, t) ∂v(x, t) = + v · ∇v (2.2.12) Dt ∂t where ∇v is the left gradient of a vector field 1 which is given by (from the definition of dotting a second-order tensor from the left by a vector), (∇v)ji =

∂vi ∂xj

(2.2.13)

For memory, the component matrix of the left gradient of a vector, in two dimensions, is given vx,x vy,x ∇v = (2.2.14) vx,y vy,y In the same manner, consider an Eulerian second-order tensor field σ(x, t), its material time derivative is given by ∂σij (x, t) ∂σij (x, t) ∂σ(x, t) Dσij (x, t) = + vk = + v · ∇σ Dt ∂t ∂xk ∂t

2.2.5

(2.2.15)

Deformation gradient

Considering two nearby material points P and Q in the reference configuration with material coordinates XP and XP + dX, respectively. After deformation, P and Q occupied the positions p and q whose coordinates are given by xp = φ(XP , t) xq = φ(XP + dX, t)

(2.2.16)

dx = φ(XP + dX, t) − φ(XP , t)

(2.2.17)

Hence

Using the Taylor’s series expansion for φ(XP + dX, t), and truncating the high order terms, we can write dx = 1

∂φ · dX, ∂X

dxi =

∂φi dXj ∂Xj

Obviously there exists the so-called right gradient operator, see Appendix ?? for details

(2.2.18)

Sec. 2.2


9

Defining the deformation gradient tensor F as F :=

∂x ∂φ = ∂X ∂X

or

Fij =

∂xi ∂Xj

(2.2.19)

The Eq. (2.2.18) therefore becomes dx = F · dX,

dxi = Fij dXj

(2.2.20)

Noting that F transform vectors in the initial configuration to vectors in the current configuration and is therefore said to be a two-point tensor. The component matrix of the deformation gradient tensor F is 

∂x1  ∂X1   ∂x  2 F=  ∂X1   ∂x3 ∂X1

∂x1 ∂X2 ∂x2 ∂X2 ∂x3 ∂X2

 ∂x1 ∂X3   ∂x2    ∂X3   ∂x3  ∂X3

(2.2.21)

By expressing the spatial vector in terms of the displacement , Eq.(2.2.19) becomes F=

∂u ∂(u + X) =I+ ∂X ∂X

(2.2.22)

Remark. The deformation gradient discussed so far is in the context of inhomogeneous deformations, i.e., the deformation gradient varies with respect to positions in the solid. A special kind of deformation, homogeneous deformation, is one where the deformation gradient has the same value everywhere in the continuum. From Eq.(2.2.19), we can integrate to get x=F·X+C

2.2.6

(2.2.23)

Conditions on motion

The mapping φ(X, t) which describes the motion of a body is assumed to satisfy the following conditions 1. φ(X, t) is continuous differentiable; 2. φ(X, t) is one-to-one; 3. The Jacobian determinant J is positive.

10

Continuum mechanics

2.2.7

Chap. 2

Volume change

Consider an infinitesimal volume element in the reference configuration made by three vectors dX1 , dX2 and dX3 . The volume dV of this element is given by dV = dX1 · (dX2 × dX3 )

(2.2.24)

After deformation, dV deformed to dv which is defined by dv = dx1 · (dx2 × dx3 )

(2.2.25)

with dxi = F · dXi ,

i = 1, 2, 3

(2.2.26)

Equation (2.2.25) can be written in indicial notation as dv = εijk dx1i dx2j dx3k = εijk FiA dX1A FjB dX2B FkC dX3C

(2.2.27)

where dx1i is the ith component of vector dx1 and in the second equality, Eq. (2.2.26) was used. Using the well known formula for determinant, εijk FiA FjB FkC = εABC det(F)

(2.2.28)

dv = εABC dX1A dX2B dX3C det(F) = dV det(F)

(2.2.29)

Hence

Denoting the Jacobian J as the determinant of F, we obtain the equation for the change of volume dv = JdV

(2.2.30)

This shows that J must be positive, the deformation gradient tensor F is therefore invertible. This allows us to write, from Eq.(2.2.20) dX = F−1 · dx

(2.2.31)

The Jacobian is usually used to relate integrals in the reference and current configurations. That is Z Z f (x, t)dΩ = f (φ(X, t), t)JdΩ0 (2.2.32) Ω

Ω0

Another useful formula is the material time derivative of the Jacobian which is given by J˙ = Jvi,i

(2.2.33)

Sec. 2.2


11

In order to prove the above, we must use the following result of matrix algebraic, d dAij C det A = A (2.2.34) dt dt ij −T where AC is the cofactor matrix of A given by AC ij = (det A)Aij . Apply the above equation for J = det F, we have DJ d ∂xi −T = JFij dt dt ∂Xj ∂vi ∂xk −1 ∂vi −T Fij = J F =J ∂Xj ∂xk ∂Xj ji ∂vi ∂vi =J Fkj Fji−1 = J δki ∂xk ∂xk ∂vi =J = Jvi,i = Jdiv(v) ∂xi Definition 2.2.4 If a motion φ is such that there is no volume change, then the motion is said to be isochoric. In that case J(X, t) ≡ det F(X, t) = 1

div(v) = 0

Example 2.2.1 (An isochoric motion- simple shear) x1 = X1 + γX2 x2 = X2

1 γ F= 0 1

Hence, J = 1.

Push forward and pull back operations The transformation of the Lagrangian vector dX (initial configuration) to the Eulerian vector dx (current configuration) can be considered as a push forward operation dx = φ∗ dX = F · dX

(2.2.35)

Similarly, the inverse transformation of dx to dX is considered as a pull back operation dX = φ∗ dx = F−1 · dx

(2.2.36)

12

2.2.8

Continuum mechanics

Chap. 2

Area change

Consider three vectors in the reference configuration dX1 , dX2 and dX3 . Their counterparts in the current configuration are dx1 , dx2 and dx3 , respectively. The volume generated by dx1 , dx2 and dx3 is dv and given by dv = dx1 · (dx2 × dx3 ) = dx1 · nda

(2.2.37)

where n is the unit normal vector and da is the area of the parallel generated by dx1 and dx2 . Similarly, the volume in the reference configuration is dV = dX1 · n0 dA

(2.2.38)

From the relation between the two volumes given in Eq. (2.2.30), we have dx1 · nda = JdX1 · n0 dA = J(F−1 · dx1 ) · n0 dA = dx1 · JF−T · n0 dA

(2.2.39)

= dx1 · (JF−T · n0 dA) where in the second equality we have used the relation u · A = AT · u. Since dx1 is arbitrary, we come up with the Nanson’s formula nda = JF−T · n0 dA

(2.2.40)

which relates the area in the initial configuration with the area in the current configuration. This equation will be used to derive the Piola-Kirchhoff stress tensor from the Cauchy stress tensor.

2.2.9

Rigid body rotation

Any rigid body motion can be generally written as x(X, t) = R(t) · X + xT (t)

(2.2.41)

which means that the motion consists of a rotation and a translation. Since dxT = 0, from Eq.(2.2.41), we have dx = R · dX

(2.2.42)

dx · dx = dX · (RT · R) · dX

(2.2.43)

Hence,

Sec. 2.3


13

Since length is preserved in rigid body rotation, i.e., dx · dx = dX · dX, it yields RT · R = I

(2.2.44)

The above shows that the inverse of R is equal to its transpose, RT = R−1

(2.2.45)

The rotation tensor R is therefore said to be an orthogonal tensor. The velocity of a rigid body motion can be obtained by taking the time derivative of Eq. (2.2.41). So, ˙ ˙ x(X, t) = R(t) · X + x˙ T (t)

(2.2.46)

The above can be converted to the Eulerian description by expressing X in terms of the spatial coordinates x via Eq. (2.2.41), gives ˙ · RT · (x − xT ) + x˙ T ˙ x(X, t) = R = Ω · (x − xT ) + x˙ T

(2.2.47)

˙ · RT Ω := R

(2.2.48)

where

The tensor Ω is called the angular velocity tensor. It is a skew-symmetric tensor. To demonstrate the skew-symmetry of the angular velocity tensor, taking the material time derivative of Eq. (2.2.44), DI D ˙ · RT + R · R ˙ T = 0 =⇒ Ω = −ΩT (R · RT ) = = 0 =⇒ R (2.2.49) Dt Dt Since any skew-symmetric second-order tensor can be expressed in term of the components of the axial vector: Ωik = εijk ωk

(2.2.50)

vi = Ωik (xk − xT k ) + vT i = εijk ωk (xk − xT k ) + vT i

(2.2.51)

Eq.(2.2.47) can be written as

In tensor notation, the above writes v = ω × (x − xT ) + vT

(2.2.52)

This is the equation for rigid body motion. The first term is due to the rotation about the point xT and the second term denotes the translational velocity.

14

Continuum mechanics

2.3

Chap. 2

Strain measures

Since pure rotation should not induce any stress (or strain) in the continuum, it is therefore convenient to define rotation-independent strain measures. In what follows, we present commonly used strain measures in large deformation, all strain measures, after being derived, are check for their correctness by showing that they are vanished during a rigid motion.

2.3.1

Green strain tensor

Considering the change of square of distances, we can write dx · dx = (F · dX) · (F · dX) = dX · FT · F · dX

(2.3.1)

T

= dX · (F · F) · dX ≡ dX · C · dX where C is the right Cauchy-Green deformation tensor 2 which is defined by C := FT · F;

Cij := Fki Fkj

(2.3.2)

Note that the tensor C operates on material vectors dX and consequently C is called a material tensor or Lagrangian tensor quantity. In addition, C is a symmetric tensor, i.e., CT = C. Consider the change in scalar product of two material vectors dX1 and dX2 as they deform to dx1 and dx2 . This change involves the stretching (change in length) and shearing (change in angles between two vectors). We have dx1 · dx2 = dX1 · C · dX2

(2.3.3)

Therefore, the change in scalar product of these two vectors is dx1 · dx2 − dX1 · dX2 = dX1 · C · dX2 − dX1 · I · dX2 = dX1 · (C − I) · dX2

(2.3.4)

1 (dx1 · dx2 − dX1 · dX2 ) = dX1 · E · dX2 2

(2.3.5)

or,

where E is the Green strain tensor, defined by 1 E := (C − I) 2 2

named after Augustin Louis Cauchy and George Green

(2.3.6)

Sec. 2.3

Strain measures

15

Since C and I are symmetric tensors, so E is. Using Eq. (2.3.2), we can write the Green strain tensor in terms of the deformation gradient tensor F as 1 E = (FT · F − I), 2

1 Eij = (Fki Fkj − δij ) 2

(2.3.7)

We also can express the Green strain tensor E in terms of displacement as follows ∂xk ∂xk ∂Xi ∂Xj ∂uk ∂Xk ∂uk ∂Xk = + + ∂Xi ∂Xi ∂Xj ∂Xj ∂uk ∂uk + δki + δkj = ∂Xi ∂Xj ∂uj ∂ui ∂uk ∂uk = + + + δij ∂Xi ∂Xj ∂Xi ∂Xj

Fki Fkj =

(2.3.8)

where the definition of the displacement given in Eq.(2.2.2) was used and δij is the Kronecker delta. Hence, we obtain 1 ∂ui ∂uj ∂uk ∂uk Eij = ( + + ) 2 ∂Xj ∂Xi ∂Xi ∂Xj

(2.3.9)

In tensor notation, the above is written as 1 T T E= (∇0 u) + ∇0 u + ∇0 u(∇0 u) 2

(2.3.10)

where ∇0 u denotes the left gradient operator in the reference configuration. If strain is small, the high order terms can be neglected, we return to the familiar small engineering strain: 1 T E= ∇0 u + (∇0 u) (2.3.11) 2 In order to check if the Green strain tensor is a valid strain measure or not, we must show that it vanishes in rigid body rotation. In fact, x = R · X + xT , the deformation gradient is then given by F = R. And E = 1/2(RT R − I) = 0.

2.3.2

Almansi strain tensor

In the previous section, the scalar product of two spatial vectors was written in terms of the material vectors through the concept of the right Cauchy-Green tensor C. Inversely, one can

16

Continuum mechanics

Chap. 2

express the scalar product of material vectors in terms of spatial vectors as shown in the following dX1 · dX2 = (F−1 · dx1 ) · (F−1 · dx2 ) = (dx1 · F−T ) · (F−1 · dx2 ) = dx1 · (F−T · F−1 ) · dx2 = dx1 · b−1 · dx2

(2.3.12)

where b is the left Cauchy-Green tensor 3 given by b := F · FT

(2.3.13)

Since b operates on the spatial (Eulerian) vectors, it is an Eulerian tensor. In addition, this is a symmetric tensor. The inverse of the left Cauchy-Green tensor is often called the Finger tensor 4 . The change in the scalar product is then given by 1 1 (dx1 · dx2 − dX1 · dX2 ) = dx1 · (I − b−1 ) · dx2 2 2 where e is the Almansi strain tensor, given by 1 e := (I − b−1 ) 2

(2.3.14)

(2.3.15)

Physical interpretation of E and b The squared lengths of the material dX and spatial vectors dx are dS 2 = dX · dX,

ds2 = dx · dx

(2.3.16)

The change in squared lengths occurring as the body deforms from the initial configuration to the current configuration can then be written in terms of the material vector dX as (see Eq.(2.3.5)) 1 2 (ds − dS 2 ) = dX · E · dX 2

(2.3.17)

ds2 − dS 2 dX dX = · E · 2dS 2 dS dS

(2.3.18)

Dividing the above by dS 2 gives

3 4

in C = FT F, F is on the right, where in b = FFT , F is on the left. named after Josef Finger

Sec. 2.3

Strain measures

17

Denoting N as the unit material vector in the direction of dX, i.e., dX = dSN the above becomes ds2 − dS 2 =N·E·N (2.3.19) 2dS 2 The meaning of diagonal elements of E can be shown if we choose N to be the basis vectors. For example, if N = E1 = {1, 0, 0}, then the above equation gives ds2 − dS 2 2dS 2 For small strain, we have the following approximation E11 =

(2.3.20)

(ds + dS)(ds − dS) ∼ (2dS)(ds − dS) ds − dS (2.3.21) = = 2 2 2dS 2dS dS which shows that, in small strain regime, E11 is exactly the normal strain along X direction. In the same manner, we have the following for the left Cauchy-Green tensor: E11 =

ds2 − dS 2 =n·e·n 2dS 2

(2.3.22)

Push forward and pull back operations In the term of the push forward and pull back operations, the Almansi strain tensor is the pushed forward of the Green strain tensor. From the change in scalar product of two material vectors, we have dx1 · e · dx2 = dX1 · E · dX2

(2.3.23)

Transform the material vectors to spatial vectors, the RHS of the above is written as dx1 · e · dx2 = dx1 · F−T · E · F−1 · dx2

(2.3.24)

Using the fact that the above holds for any vectors dx1 and dx2 , we come up with e = F−T · E · F−1 = φ∗ [E]

(2.3.25)

Therefore, the Almansi strain tensor e is the pushed forward of the Green strain tensor E with the push forward operator given by e := φ∗ [E] = F−T · E · F−1

(2.3.26)

Similarly, the Green strain tensor E is the pulled back of the Almansi strain tensor e with the pull back operator given by E := φ∗ [e] = FT · e · F

(2.3.27)

The two equations Eqs. (2.3.26, 2.3.27) are typical examples of the push forward and pull back operations for kinematic second order tensors.

18

2.3.3

Continuum mechanics

Chap. 2

Bulk-deviatoric decomposition of kinematics tensors

Definition 2.3.1 A multiplicative decomposition of the deformation gradient F is given by F = J 1/3 I J −1/3 F

(2.3.28)

where J = det(F). This is the Flory decomposition. The second term of the above decomposition is called isochoric deformation gradient tensor (distortional or volume-preserving part of the deformation gradient are also names of this part) ¯ since its determinant is unity and is denoted by F ¯ := J −1/3 F F

(2.3.29)

The first term in Eq.(2.3.28) is the so-called dilatational deformation gradient or volumechanging part of the deformation gradient. The right and left Cauchy-Green strain tensors corresponding to the isochoric deformation gradient are called isochoric right and left Cauchy-Green tensors and given by ¯ =F ¯T · F ¯ = J −2/3 C C ¯ = J −2/3 b b

2.3.4

(2.3.30a) (2.3.30b)

Strain rates

Definition 2.3.2 The spatial gradient of velocity or velocity gradient tensor L is defined as the spatial gradient of the velocity, that is ∂v ∂vi , or Lij = (2.3.31) ∂x ∂xj The velocity gradient L allows the material time derivative of the deformation gradient F to be written as L(x, t) :=

∂ ∂φ(X, t) ˙ = F ∂t ∂X ∂v (2.3.32) = ∂X ∂v ∂x = · =L·F ∂x ∂X where in the second equality, we have used the fact that material time derivative of Lagrangian fields commute with material gradient. Noting that, this fact does not hold generally for Eulerian fields. From the above, we obtain ˙ · F−1 L=F

(2.3.33)

Sec. 2.3

Strain measures

19

As previously described, the strain is defined as the change in the scalar product of two vectors. Therefore, the strain rate is defined as the rate of change of the scalar product of two vectors. Recall that dx1 · dx2 = dX1 · C · dX2

(2.3.34)

The rate of change of the scalar product of two vectors is then defined by taking the material time derivative of the above and using the fact that C = 2E − I, that is D ˙ · dX2 = 2dX1 · E ˙ · dX2 (dx1 · dx2 ) = dX1 · C Dt

(2.3.35)

˙ the time derivative of the Green strain tensor, is called the material strain rate tensor where E, and given by ˙ = 1C ˙ = 1 (FT F ˙ +F ˙ T F) E 2 2

(2.3.36)

By introducing the following dX1 = F−1 · dx1 ,

dX2 = F−1 · dx2

(2.3.37)

into Eq. (2.3.35), we obtain the rate of change of the scalar product in terms of spatial vectors as D −T ˙ (dx1 · dx2 ) = 2dx1 · F · {z E · F−1} ·dx2 | Dt

(2.3.38)

D

˙ and is known as the rate The tensor D is simply the pushed forward Eulerian counterpart of E, of deformation tensor which is given by ˙ · F−1 D := F−T · E

(2.3.39)

˙ = FT · D · F E

(2.3.40)

Inversely,

It is interesting that the rate of deformation tensor D is exactly the symmetric part of the velocity gradient tensor L. That is 1 D = (L + LT ) 2

(2.3.41)

This can be verified by substituting Eq. (2.3.32) into Eq. (2.3.36), ˙ = FT 1 (L + LT ) F E 2 and comparing the above to Eq. (2.3.40).

(2.3.42)

20

Continuum mechanics

Chap. 2

Rate of J in term of C With the rate of deformation tensor, we can write the material time derivative of the Jacobian J as J˙ = Jvi,i = Jtrace(D)

(2.3.43)

Substituting Eq. (2.3.39) into the above gives ˙ −1 ) J˙ = Jtrace(F−T EF ˙ = Jtrace(C−1 E) ˙ = JC−1 : E 1 ˙ = JC−1 : C 2

(2.3.44)

The above gives 1 ∂J = JC−1 (2.3.45) ∂C 2 These two equations will be useful in the treatment of incompressible hyperelastic materials presented in Chapter 3, Section 3.4. The velocity gradient tensor can be decomposed into symmetric and skew-symmetric tensors by 1 1 L = (L + LT ) + (L − LT ) (2.3.46) 2 2 where the symmetric tensor has been defined as the rate of deformation D and the spin tensor W is the skew-symmetric part of L. With these definitions, we can write L=D+W

(2.3.47)

1 W := (L − LT ) 2

(2.3.48)

with

Physical interpretation of the rate of deformation tensor A simple physical interpretation of the tensor D can be obtained by taking dx1 = dx2 = dx in Eq.(2.3.38) to give D (dx · dx) = 2dx · D · dx Dt Defining n, a unit vector, such that dx = dsn, the above becomes 1 D (ds2 ) = n · D · n 2ds2 Dt

(2.3.49)

(2.3.50)

Sec. 2.3

Strain measures

21

or, 1 D D(ln(ds)) (ds) = (2.3.51) ds Dt dt which shows that with a basis ei , the diagonal D11 , D22 and D33 are the logarithmic strain rates along e1 , e2 and e3 , respectively. n·D·n=

D (dx1 · dx2 ) = 2dx1 · D · dx2 Dt = 2D · dxT 1 · dx2 = 2Dds1 ds2 cos θ = 2ds1 ds2 n1 · D · n2

(2.3.52)

On the other hand, we have D D (dx1 · dx2 ) = (ds1 ds2 cos θ) Dt Dt D D = cos θ (ds1 ds2 ) + ds1 ds2 (cos θ) Dt Dt D D = cos θ (ds1 ds2 ) − ds1 ds2 sin θ (θ) Dt Dt

(2.3.53)

Comparing Eq.(2.3.52) with Eq.(2.3.53) gives D D (ds1 ds2 ) − ds1 ds2 sin θ (θ) Dt Dt perpendicular (sin θ = 1, cos θ = 0), then 2ds1 ds2 n1 · D · n2 = cos θ

n1 · D · n2 = −

1D (θ) 2 Dt

(2.3.54)

(2.3.55)

Lie derivative Note that the rate of deformation D is the push forward of the material time derivative of the Green strain tensor E (see Eq.2.3.39) which is the pull back of the Almansi strain tensor e. We then could write D ∗ D = φ∗ φ [e] (2.3.56) Dt This operation is known as the Lie derivative of a tensor over the mapping φ and is generally defined by Lφ [g] := φ∗

D ∗ φ [g] Dt

(2.3.57)

22

Continuum mechanics

Chap. 2

Figure 2.2: Initial (undeformed) and current (deformed) configurations of body

2.3.5

Polar decomposition

Deformation consists of stretch and rotation. Rotation is a special kind of deformation in which the material vectors change directions but they do not change lengths. It can be shown that the deformation gradient of a rotation is an orthogonal tensor. A stretch, on the other hand, is a kind of deformation in which there exists three material vectors that will change in length but not in directions. In finite deformation theory, rotation plays an important role. The polar decomposition theorem is a mathematical statement that the deformation may be regarded as a combination of a pure stretch and a pure rotation. Mathematically, the deformation gradient tensor can be multiplicatively decomposed into a rotation tensor and a right stretch tensor, F=R·U

(2.3.58)

where R is the rotation tensor (hence orthogonal), and U is the symmetric right stretch tensor. By substitution Eq. (2.3.58) into the definition equation of the right Cauhcy-Green tensor C, Eq. 2.3.2, we have C = (RU)T RU = UT RT RU

(2.3.59)

= UT U = U2 where we used the orthogonality of tensor R and the symmetry of tensor U. In order to compute

Sec. 2.3

Strain measures

23

U from C, we need a way to determine the square root of a tensor. One way is to use the spectral decomposition to write the tensor C as following C=

3 X

λ2i Ni ⊗ Ni

(2.3.60)

i=1

where λ2i and Ni are the eigenvalues and normalized eigenvectors of C. The right stretch tensor U is now computed by U=

3 X

λi Ni ⊗ N i

or

U = QHQT

(2.3.61)

i=1

where the diagonal matrix H contains the eigenvalues λi and the matrix Q whose columns are the eigenvectors of the tensor C. Noting that in the second equation in the above, matrix notation has been used. Having defined U, the rotation tensor R is simply determined by R = F · U−1

(2.3.62)

In terms of the rotation and stretch tensors, the deformation of a material vector dX is written by dx = F · dX = R · (U · dX)

(2.3.63)

The above shows that the vector dX is first stretched (by U) then rotated to the current configuration by R. The tensor U is therefore a material or Lagrangian tensor whereas R is a two point tensor.

Example 2.3.1 (Polar decomposition) Consider a motion given by 1 x1 = 4X1 + (9 − 3X1 − 5X2 − X1 X2 )t 4 1 x2 = 4X2 + (16 + 8X1 )t 4 The deformation gradient tensor F is 1 4 − (3 + X2 )t −(5 + X1 )t F= 8t 4 4 Consider the material point X(0, 0) at time t = 0, we have 1 1 −5 F= 4 8 4

24

Continuum mechanics

Chap. 2

The right Cauchy-Green tensor C is 1 65 27 C = FF = 16 27 41 T

The next step is to compute the eigenvalues and eigenvectors of C. Eigenvalues are solutions of the equation det(C − λI) = 0 After some manipulations, we obtain the characteristic equation as 256λ4 − 1696λ2 + 1936 = 0 This equation gives two real positive solutions λ21 = 1.4658;

λ22 = 5.1592

λ1 = 1.2107;

λ2 = 2.2714

The eigenvectors are computed by solving the following equation (C − λ2i I)Ni = 0 Writing the above out

65 27 − λi Ni1 + Ni2 = 0 16 16 27 1 41 N + − λi Ni2 = 0 16 i 16

To solve this, we need one more equation ||Ni || = 1 Finally, the normalised eigenvectors are −0.5449 N1 = 0.8385

N2 =

0.8385 0.5449

The stretch tensor U is then computed by U = QHQT where −0.5449 0.8385 Q= , 0.8385 0.5449

1.2107 0 H= 0 2.2714

Sec. 2.3

Strain measures

25

Finally, the stretch tensor is given by

1.9564 0.4846 U= 0.4846 1.5256 and the rotation tensor −1

R=F·U

0.3590 −0.9333 = 0.9333 0.3590

The deformation gradient can be also decomposed in terms of a left stretch tensor and a rotation according to F=V·R

(2.3.64)

dx = F · dX = V · (R · dX)

(2.3.65)

The above shows that the vector dX is first rotated by R then stretched by V. The tensor V is therefore a spatial or Eulerian tensor. The Lagrangian stretch tensor U and Eulerian stretch tensor V are related together by V = R · U · RT

(2.3.66)

In the same manner with the right Cauchy-Green tensor, the left Cauchy-Green tensor can be written in terms of V and R as follows b = F · FT = V · R · RT · VT = V2

(2.3.67)

As a consequence, if the principle directions of b are denoted by ni together with associated ¯ 2 , then V is given by eigenvalues λ i V=

3 X

¯ i ni ⊗ ni λ

(2.3.68)

i=1

From equations (2.3.61) and (2.3.66), we have another way to compute V 3 3 X X T V =R·( λi Ni ⊗ Ni ) · R = λi (RNi ) ⊗ (RNi ) i=1

(2.3.69)

i=1

Comparing the two equations (2.3.68) and (2.3.69), it follows that ¯i, λi = λ

ni = R · Ni

(2.3.70)

The above shows that the left and right Cauchy-Green tensors have the same eigenvalues.

26

Continuum mechanics

Chap. 2

Principal stretches It is also useful in demonstrating the physical meaning of these eigenvalues as the stretches in the principle directions. Indeed, taking a material vector dX1 with length dS1 in the direction of N1 , this vector deformed to the spatial vector dx1 given by dx1 = F · dX1 . Using the polar decomposition of F, we can write dx1 as dx1 = R · U · dS1 N1 = dS1 R · λ1 N1 = λ1 dS1 n1

(2.3.71)

where in the second equality, the relation U·N1 = λ1 N1 has been used and we used Eq.(2.3.70) in the last derivation. The length of the spatial vector dx1 is therefore related to the length of the vector dX1 by ds1 (2.3.72) dS1 Then, the eigenvalues λi are called the principle stretches which are exclusively used in the constitutive equations of hyperelasticity. ds1 = λ1 dS1 =⇒ λ1 =

Deformation gradient in terms of principal stretches ! 3 3 X X F=R·U=R· λi Ni ⊗ Ni = λi ni ⊗ Ni i=1

(2.3.73)

i=1

where Eq.(2.3.70) has been used. The above clearly illustrates the two-point nature of the deformation gradient.

2.4

Stress measures

Roughly speaking, stress is a force per unit area. Precisely, stress is a measure of the internal distribution of force per unit area that balances and reacts to the external loads or boundary conditions applied to a body. When the area is the current (deformed) area, the corresponding stress is the true stress or the Cauchy stress. This is the stress used in small deformation analysis. In large deformation, the initial and current configuration must be differenciated. The stress defined with respect to the initial configuration are also usefull although they are not true stress. These stresses are commonly known as the Piola-Kirchhoff stress tensors. The most widely used stress tensors are 1. The Cauchy stress tensor σ 2. The nominal stress P (closely related to the first Piola-Kirchhoff stress tensor) 3. The second Piola-Kirchhoff stress tensor S 4. The Kirchhoff stress tensor τ

Sec. 2.4

2.4.1

Stress measures

27

The Cauchy stress

Stress vector

Figure 2.3: Internal forces in a body dF ∆F = ∆A→0 ∆A dA

t := lim

(2.4.1)

Stress state t(e1 ) = σ11 e1 + σ12 e2 + σ13 e3 t(e2 ) = σ21 e1 + σ22 e2 + σ23 e3 t(e3 ) = σ31 e1 + σ32 e2 + σ33 e3

(2.4.2)

Traction-stress relation t = n · σ = σ T · n,

ti = σji nj = σij nj

(2.4.3)

where t is the traction vector, n denotes a unit normal vector to the surface dΓ, df = tdΓ = n · σdΓ

(2.4.4)

As will be shown later, from the conservation of energy, we have Dwint =D:σ (2.4.5) Dt which leads to the following definition of pair of work conjugated stress and strain measures: ρ

Definition 2.4.1 A pair (S, E) of stress-strain measure is called work conjugated if their inner product is the rate of internal energy per unit undeformed or deformed volume.

28

Continuum mechanics

Chap. 2

Figure 2.4: Initial (undeformed) and current (deformed) configurations of body

Figure 2.5: Stress vectors acting upon on a tetrahedron

Sec. 2.4

Stress measures

29

So, we say that the Cauchy stress tensor σ is work congugate to the rate of deformation tensor D when the work rate is referred to the deformed configuration. Since there are numerous strain measures, it is then natural to define more stress measures that are work conjugate to these strain measures.

2.4.2

The Kirchhoff stress

Since ρJ = ρ0 , the above can be written as ρ0

Dwint = D : (Jσ) Dt

(2.4.6)

The above means that Jσ can be a stress measure which is work conjugate with the rate of deformation tensor with respect to the initial volume. It is named the Kirchhoff stress τ or also called the weighted Cauchy stress. For isochoric motion (J = 1), it is identical to the Cauchy stress.

2.4.3

The nominal stress

We would like to seek for another stress measure which is force per unit area of the reference configuration. In order to do that, in Eq.(2.4.4), the current area is transformed to reference area by the Nanson’s formula, which is given by ndΓ = JF−T · n0 dΓ0

(2.4.7)

df = n0 · JF−1 · σdΓ0

(2.4.8)

P := JF−1 · σ

(2.4.9)

Hence

Let us define

then df = n0 · PdΓ0 ,

t0 = n0 · P,

t0 = PT · n0

(2.4.10)

The above means that n0 · P is the traction vector relating the force in current configuration to the area in the reference configuration. Therefore P is obviously a stress tensor which is called the nominal stress tensor. It is a Lagrangian-Eulerian two-point tensor. The transpose of the nominal stress PT is the first Piola-Kirchhoff stress tensor. From Eq. (2.4.9), we have expression of the Cauchy stress in terms of the nominal stress as σ = J −1 F · P

(2.4.11)

30

Continuum mechanics

Chap. 2

Using Eq.(2.4.11), we get the relation between the Kirchhoff stress and nominal stress as τ =F·P

(2.4.12)

We are now showing that the nominal stress tensor is work conjugate to the transpose of the rate of deformation gradient F˙ T . To verify this, we compute the double contraction of the two as ˙ T = J(F˙ · F−1 )T : σ = σ : L = σ : D P : F˙ T = JF−1 · σ : F

(2.4.13)

where Eq.(2.4.11) has been used in the first step, and in the third equality, Eq.(2.3.33) was used. Finally, in the last step, the decomposition of L into D and W together with the symmetry of the Cauchy stress has been employed.

2.4.4

The second Piola-Kirchhoff stress

The nominal stress tensor P is an unsymmetric two-point tensor and is not completely related to the reference configuration. It is desirable (also possible) to contrive a totally material symmetric stress tensor, known as the second Piola-Kirchhoff stress tensor (2nd PK) by simply transform the force by F−1 , that is n0 · SdΓ0 = F−1 · t0 dΓ0

(2.4.14)

df = F · n0 · SdΓ0 = F · ST · n0 dΓ0

(2.4.15)

Multiplying the above by F gives

Comparing the above to Eq. (2.4.10), we obtain F · ST · n0 dΓ0 = n0 · PdΓ0 = PT · n0 dΓ0

(2.4.16)

The above holds for all n0 gives P = S · FT

(2.4.17)

We have the following relations between the Cauchy stress and the second Piola-Kirchhoff stress given by σ = J −1 F · S · FT S = JF−1 · σ · F−T

(2.4.18)

From this equation, it can be easily to show that the second Piola-Kirchhoff stress tensor is symmetric since the Cauchy is symmetric (will be verified from conservation of angular momentum).

Sec. 2.4

Stress measures

31

The relation between the Kirchhoff stress and the second Piola-Kirchhoff stress is τ = F · S · FT S = F−1 · τ · F−T

(2.4.19)

Next, we are going to show that the the second Piola-Kirchhoff stress is work conjugate to ˙ the Green stran rate E. ˙ = (F−1 · τ · F−T ) : E ˙ S:E ˙ · F−1 ) = (F−1 · τ ) : (E ˙ · F−1 ) = τ : (F−T · E = τ : D = ρo w˙ int

(2.4.20)

where in the third equality, we have used the defition of the rate of the deformation tensor D and Eq.(2.4.6) has been made of use in the last step. Push forward and pullback operations on kinetics tensors Using the push forward and pull back operations, the Kirchhoff stress is the pushed forward of the second Piola-Kirchhoff stress, τ = φ∗ [S] = F · S · FT

(2.4.21)

Inversely, the second Piola-Kirchhoff stress is the pull back of the Kirchhoff stress, S = φ∗ [τ ] = F−1 · τ · F−T

(2.4.22)

The above equations are examples of the push forward and pull back operations on kinetic tensors. Similarly, the Cauchy stress and the second Piola-Kirchhoff stress are related as, σ = J −1 φ∗ [S],

S = Jφ∗ [σ]

(2.4.23)

In the above, S and σ are related by the so-called Piola transformation which involves a push forward and pull back operation combined with the volume scaling J.

2.4.5

Deviatoric and pressure components of stress tensors

In some cases, it is convenient to decompose the stress tensor into the deviatoric and pressure (or hydrostatic) part as follows, σ = σ dev + σ hyd , where p = 1/3σii .

σ hyd = pI

(2.4.24)

32

Continuum mechanics

Chap. 2

Substituting the above into Eq. (2.4.18), we obtain the similar decomposition for the second Piola-Kirchhoff stress tensor S as S = Sdev + Shyd ,

Sdev = JF−1 · σ dev · F−T ,

Shyd = pJC−1

(2.4.25)

Using Eq.(2.4.25) gives S : C = Sdev : C + pJC−1 : C

(2.4.26)

combining with the fact that Sdev : C = 0 yields S = pJC−1 =⇒ pI = J −1 SC

(2.4.27)

The fact that C is symmetric allows us to get p given by 1 p = J −1 S : C 3 Introducing the above into the Eq. (2.4.25), we obtain 1 Shyd = (S : C)C−1 3

2.5 2.5.1

(2.4.28)

(2.4.29)

Material objectivity Objective tensors

Consider the current configuration of the body as x, we now impose a rigid body motion on this configuration to obtain the new motion x ˜. x ˜ = Q(t) · x + c(t)

(2.5.1)

where Q is an orthogonal rotation tensor and c(t) denotes a translation tensor. Definition 2.5.1 A vector x and a second order tensor G is said to be objective or frame invariant if x ˜ =Q·x ˜ = Q · G · QT G

(2.5.2b)

d˜ x = Q · dx

(2.5.3)

(2.5.2a)

It follows from Eq. 2.5.1 that

Although the vector d˜ x is different from the vector dx, their magnitudes are obviously the same. In this sense it can be said that the vector dx is objective under rigid body rotation.

Sec. 2.5

Material objectivity

33

Relating the vector in the current configuration dx to the corresponding in the initial configuration dX by the deformation gradient F, the above can be written as ˜ · dX d˜ x = Q · F · dX = F

(2.5.4)

˜ =Q·F F

(2.5.5)

with

So the deformation gradient is not objective It can be shown that the Lagrangian strain tensors such as C and E are unchanged under rigid body motion. For example, we have ˜ =F ˜T · F ˜ = FT · QT · Q · F = FT · F = C C

(2.5.6)

In contrast, the Eulerian strain tensors such as the right Cauchy-Green tensor b and the Almansi tensor e do change under rigid body motion. Indeed, ˜=F ˜ ·F ˜ T = Q · F · FT · QT = Q · b · QT b

(2.5.7)

˜ e = Q · e · QT

(2.5.8)

It is shown that the rate of deformation tensor is objective. To this end, from Eq.(2.3.39), we have ˙ · F−1 ) · Q−1 = Q · D · QT ˜ =F ˜ −T · E ˜˙ · F ˜ −1 = Q−T · (F−T · E D

(2.5.9)

where Eq.(2.5.5) has been used and the fact that E is unaltered was used. The final result was obtained in adopting the orthogonality of Q, i.e., QT = Q−1 . Cauchy stress is objective, traction and unit vector are both objective ˜ t=Q·t σ ˜·n ˜ =Q·σ·n σ ˜·Q·n=Q·σ·n

(2.5.10)

σ ˜ = Q · σ · QT

(2.5.11)

S = J −1 F−1 · σ · F−T

(2.5.12)

Hence, the Cauchy stress is objective.

˜ = J −1 F ˜ −1 · Q · σ · QT · F ˜ −T = J −1 F−1 · Q−1 · Q · σ · QT · Q−T · F−T = S S

(2.5.13)

34

Continuum mechanics

2.5.2

Chap. 2

Objective stress rates

Motivation Nonlinear materials are conveniently described by constitutive equations relating stress rate to strain rate. Dσ = CσD : D (2.5.14) Dt Now, considering a bar with stress state is given by σx = σ0 , σy = 0 in a fixed Cartesian coordinate system. Assuming that the bar is rigidly rotated 90o , hence the stresses are σx = 0, σy = σ0 . So, the RHS of Eq.(2.5.14) does not vanish. However, the LHS is zero since D = 0 in rigid body motion. This indicates that the constitutive equation given by Eq.(2.5.14) is not valid. The reason is due to the use of non-objective stress rate, i.e., the stress rate Dσ is not Dt invariant under rigid body rotation. Therefore, it is necessary to employ objective stress rates in describing response of large deformation materials. The commonly used objective stress rates in nonlinear finite elements are 1. Truesdell rate 2. Jaumann rate 3. Green-Naghdi rate Truesdell rate of Cauchy stress Recall that the second Piola-Kirchhoff stress tensor is independent of any rigid body rotation and the Cauchy stress is related to the 2nd PK stress via the Piola transformation, the Truesdell rate of the Cauchy stress is thus defined as the Piola transformation of the time derivative of the second Piola-Kirchhoff stress. That is D −1 −T ∇T −1 −1 ˙ (JF σF ) FT (2.5.15) σ := J φ∗ [S] = J F Dt The above can be easily written as ˙ + FF ˙ −1 σ + σ F ˙ −T FT σ ∇T = σ˙ + Jσ

(2.5.16)

Using the fact that J˙ = Jdiv(v),

˙ −1 + FF ˙ −1 = 0, FF

˙ −1 L = FF

(2.5.17)

We come up with σ ∇T =

Dσ + div(v)σ − L · σ − σ · LT Dt

(2.5.18)

Sec. 2.5

Material objectivity

35

Truesdell rate of Kirchhoff stress Since the Kirchhoff stress τ is not objective (obviously seen from the fact that the Cauchy stress is not objective), it is then natural to define an objective time derivative for τ . Recalling that τ = Jσ, thus from Eq.(2.5.15), the objective Kirchhoff stress rate is defined as ˙ = F · S˙ · FT τ ∇c := φ∗ [S]

(2.5.19)

Following the same derivation as done for the Cauchy stress, we obtain τ ∇c = τ˙ − L · τ − τ · LT = Jσ ∇T

(2.5.20)

This objective rate is also called the convected rate of the Kirchhoff stress or the weighted Truesdell rate of the Cauchy stress. Jaumann rate Dσ − W · σ − σ · WT Dt

(2.5.21)

σ ∇J = CσD : D

(2.5.22)

Dσ = CσD : D + W · σ + σ · WT Dt

(2.5.23)

σ ∇J :=

We are now showing the relation between the Jaumann rate given in the above equation and the Truesdell rate given in Eq.(2.5.18) by replacing the velocity gradient by the sum of D and W in Eq.(2.5.18): σ ∇T =

Dσ + div(v)σ − (D + W) · σ − σ · (D + W)T Dt

(2.5.24)

For a rigid body rotation (D = 0, div(v) = 0), the Truesdell rate of Cauchy stress becomes σ ∇T =

Dσ − W · σ − σ · WT Dt

(2.5.25)

which is essentially the Jaumann rate given in Eq.(2.5.21). Green-Naghdi rate σ ∇G =

Dσ − Ω · σ − σ · ΩT Dt

(2.5.26)

36

Continuum mechanics

Chap. 2

Example 2.5.1 1 t 0 ˙ F= , F= 0 1 0 ˙ −1 = 0 1 , D = L = FF 0 0

1 1 −t −1 , F = 0 0 1 1 0 1 1 0 1 , W= 2 1 0 2 −1 0

σ˙ = CσD : D + W · σ + σ · WT , σ˙ xx = σxy ,

σ˙ yy = −σxy ,

2.6

(2.5.28)

CσD : D = λJ trD + 2µJ D

1 σ˙ xy = µJ + (σyy − σxx ) 2

σ˙ = CσD : D + L · σ + σ · LT − div(v)σ,

σ˙ xx = 2σxy ,

(2.5.27)

σ˙ yy = 0,

(2.5.29) (2.5.30)

CσD : D = λT trD + 2µT D

(2.5.31)

σ˙ xy = µJ + σyy

(2.5.32)

Conservation equations

An important set of equations in continuum mechanics are the conservation equations or also called balance equations (balance laws). For thermomechanical systems, the conservation laws include 1. Conversation of mass 2. Conversation of linear momentum 3. Conversation of angular momentum 4. Conversation of energy

2.6.1

Reynold’s transport theorem

The material time derivative of an integral is the rate of change of this integral on a material domain. A material domain moves with the material, so no mass flux occurs cross its boundary. Various forms of the material time derivative of integrals are called the Reynold’s transport theorem. The material time derivative of an integral is defined by D Dt

Z

1 f (x, t)dΩ = lim ∆t→0 ∆t Ω

Z f (x, t + ∆t)dΩ − Ωt+∆t

Z f (x, t)dΩ Ωt

(2.6.1)

Sec. 2.6


37

where Ωt is the spatial domain occupied by the material domain at time t and Ωt+∆t is the spatial domain occupied by the same material domain at time t + ∆t. Transforming the integrals in the RHS of the above to the reference configuration gives, D Dt

Z

1 f dΩ = lim ∆t→0 ∆t Ω

Z

Z f (X, t+∆t)J(X, t+∆t)dΩ0 −

Ω0

f (X, t)J(X, t)dΩ0

(2.6.2)

Ω0

where the function f = f (φ(X, t), t). Since the integral domain is now constant in time, we can take the limit inside the integral, the above becomes Z Z D ∂ f dΩ = f (X, t)J(X, t) dΩ0 (2.6.3) Dt Ω Ω0 ∂t In the above equation, the partial derivative with respect to time is the material time derivative since the independent spatial variable is the Lagrangian coordinate. Using the product rule for derivatives on the RHS of the above, we can write Z Z ∂f ∂J D f dΩ = J +f dΩ0 (2.6.4) Dt Ω ∂t Ω0 ∂t Recall that J˙ = Jvi,i , we obtain Z Z ∂vi D ∂f f dΩ = J + fJ dΩ0 Dt Ω ∂xi Ω0 ∂t

(2.6.5)

Transform the integrals in the RHS of the above back to the current configuration, we come up with D Dt

Z

Z f dΩ = Ω

Ω

Df (x, t) ∂vi +f dΩ Dt ∂xi

(2.6.6)

This equation is one form of the Reynold’s transport theorem. For completeness, another form of the Reynold’s transport theorem is given here where the proof will be done later, D Dt

Z

Z ρf dΩ =

Ω

ρ Ω

Df dΩ Dt

(2.6.7)

This is known as the Reynold’s theorem for a density-weighted integrand.

2.6.2

Conservation of mass

The mass m(Ω) of a material domain Ω is given by Z m(Ω) = ρ(X, t)dΩ Ω

(2.6.8)

38

Continuum mechanics

Chap. 2

where ρ is the mass density. The law of conversation of mass states that Dm(Ω) =0 Dt Applying the Reynold’s theorem in Eq. (2.6.6) to the above yields Z Dρ + ρ∇ · v dΩ = 0 Ω Dt

(2.6.9)

(2.6.10)

Since the above holds for any domain Ω, we come up with Dρ + ρ∇ · v Dt

or ρ˙ + ρvi,i = 0

(2.6.11)

The above is the equation of mass conservation, or often called equation of continuity. If the density does not change, i.e., the material is incompressible, hence the material time derivative of the density vanishes, the continuity equation becomes vi,i = 0

(2.6.12)

which is the well known incompressibility condition. For Lagrangian description (material domain moves with material), Z Dm(Ω) ρ(X, t)dΩ = constant = 0 =⇒ Dt Ω

(2.6.13)

Hence, Z

Z ρ(X, t)dΩ =

ρ0 (X)dΩ0

(2.6.14)

Ω0

Ω

Transforming the RHS of the above to the reference configuration by using dΩ = JdΩ0 gives Z ρ(X, t)J − ρ0 (X) dΩ0 = 0 (2.6.15) Ω0

Arbitrariness of the material domain Ω0 gives ρ(X, t)J(X, t) = ρ0 (X)

2.6.3

(2.6.16)

Conservation of linear momentum

Total force Z

Z ρb(x, t)dΩ +

f (t) = Ω

t(x, t)dΓ Γ

(2.6.17)

Sec. 2.6


Definition 2.6.1 The linear momentum of a material domain is defined by Z p(t) = ρv(x, t)dΩ

39

(2.6.18)

Ω

where ρ is the mass density, v is the velocity field. The Newton’s second law of motion Dp =f Dt Substituting Eqs.(2.6.17) and (2.6.18) into the above gives Z Z Z D ρv(x, t)dΩ = ρb(x, t)dΩ + t(x, t)dΓ Dt Ω Ω Γ

(2.6.19)

(2.6.20)

This is the integral form or global form of the conservation of linear momentum principle. In order to derive a local form or pointwise form which is a partial differential equation, the first and third integrals of the above are converted to domain integrals then using the arbitrariness of this domain to get the PDE form. Reynold’s transport theorem applied to the left hand side gives D Dt

D ρvdΩ = (ρv) + ρv∇ · v dΩ Ω Ω Dt Z Dv Dρ = ρ +v + ρ∇ · v dΩ Dt Dt Ω Z

Z

(2.6.21)

The second term in the above vanishes since it is exactly the continuity equation, the above then becomes Z Z Dv D ρ ρvdΩ = dΩ (2.6.22) Dt Ω Ω Dt Convert the second term on the right hand side of Eq. (2.6.20) to domain integral by Cauchy’s relation and Gauss’s theorem, Z Z Z tdΓ = n · σdΓ = ∇ · σdΩ (2.6.23) Γ

Γ

Ω

Substituting equations (2.6.22) and (2.6.23) into Eq. (2.6.20), we obtain Z Dv ρ − ρb − ∇ · σ dΩ Dt Ω

(2.6.24)

Since the above is valid for any Ω, we have ρ

Dv = ∇ · σ + ρb or Dt

ρv˙ i = σji,j + ρbi

(2.6.25)

40

Continuum mechanics

Chap. 2

The above is the so-called the momentum equation; it is also called the Cauchy’s first law of motion. This form of the momentum equation is applicable to both Lagrangian and Eulerian descriptions. For a Lagrangian description, the independent variables are the material coordinates X and time t, so the corresponding momentum equation is ρ(X, t)

∂v(X, t) = ∇ · σ(φ−1 (x, t), t) + ρ(X, t)b(X, t) ∂t

(2.6.26)

Noting that the stress have been written in term of spatial coordinates so that the spatial derivative can be evaluated. In practice, however, the inverse of φ is never built and implicit differentiation is employed. The above would not be considered a true Lagrangian description in classical texts on continuum mechanics due to the presence of derivative with respect to Eulerian coordinates. However, the above is indeed a Lagrangian form since all dependent variables are functions of material coordinates. This form can be called the updated Lagrangian momentum equation since we will use it to derive the updated Lagrangian finite elements. In an Eulerian description, the material time derivative of the velocity is given by Eq.(2.2.12), i.e., replacing the material time derivative by partial derivative with respect to time plus the transport term, and the independent variables are the spatial coordinates and time. The Eq.(2.6.25) becomes ρ(x, t)

∂v(x, t) + v(x, t) · ∇v(x, t) = ∇ · σ(x, t) + ρ(x, t)b(x, t) ∂t

(2.6.27)

In computational fluid dynamics, ...

2.6.4

Conservation of angular momentum

Definition 2.6.2 The angular momentum (moment of linear momentum) of a material domain with respect to the origin of a coordinate system is defined by Z x × ρvdΩ (2.6.28) Ω

The principle of conversation of angular momentum states that the rate of change of the angular momentum is equal to the resultant moment of the external forces. That is Z Z Z D x × ρvdΩ = x × ρbdΩ + x × tdΓ (2.6.29) Dt Ω Ω Γ Using the cross product formula, the above equation can be written as (in Cartesian components) Z Z Z D εijk ρxj vk dΩ = εijk ρxj bk dΩ + εijk xj tk dΓ (2.6.30) Dt Ω Ω Γ

Sec. 2.6


41

Applying the Reynold’s theorem for density-weighted integrand, Eq. (2.6.6), to the first term, we have D Dt

Z

Z εijk ρxj vk dΩ =

Ω

εijk ρ ZΩ

D (xj vk )dΩ Dt

(2.6.31)

εijk ρ(vj vk + xj v˙ k )dΩ

= Ω

By using the Cauchy’s relation and the divergence theorem, the second term in the right hand size of Eq. (2.6.30) can be rewritten as Z

Z εijk xj tk dΓ = Γ

εijk xj σkp np dΓ ZΓ

∂ (xj σkp )dΩ ∂xp Ω Z ∂σkp = εijk δjp σkp + xj dΩ ∂xp Ω Z ∂σkp dΩ = εijk σkj + xj ∂xp Ω =

εijk

(2.6.32)

After substitution of Eqs.(2.6.31) and (2.6.32), the equation (2.6.30) now becomes

Z εijk Ω

∂σkp − ρbk dΩ = 0 ρ(vj vk ) − σkj + xj ρv˙ k − ∂xp

(2.6.33)

Since Ω is arbitrary,

εijk ρ(vj vk ) − σkj + xj

∂σkp − ρbk ρv˙ k − ∂xp

=0

(2.6.34)

Noting that the term in parentheses is the linear momentum equation, so it vanishes; and εijk (vj vk ) vanishes also because it is the cross product of the same vector, the above equation then simplifies as εijk σkj = 0

(2.6.35)

which implies that σij = σji ,

or

σ = σT

(2.6.36)

So, the conversation of angular momentum leads to the symmetry of the Cauchy stress tensor.

42

2.6.5

Continuum mechanics

Chap. 2

Conservation of energy

We consider thermo-mechanical processes where only mechanical work and thermal energy are involved. The principle of energy conservation states that the rate of change of total energy is equal to the work done by external forces (body forces and surface tractions) plus the heat energy delivered to the body by the heat flux and other heat sources. The internal energy per unit volume is denoted by ρwint where wint is the internal energy per unit mass. The heat flux per unit area is denoted by vector q and the heat source per unit volume is denoted by ρs. The rate of change of the total energy P tot = P int + P kin where P

(2.6.37)

int

denotes the rate of change of internal energy which is given by Z D int P = ρwint dΩ Dt Ω and P kin is the rate of change of kinematic energy Z 1 D kin ρv · vdΩ P = Dt Ω 2 The rate of work done by the body forces and surface tractions is given by Z Z ext v · ρbdΩ + v · tdΓ P =

(2.6.38)

(2.6.39)

(2.6.40)

Γ

Ω

The energy supplied by the heat flux q and the heat sources s is Z Z heat P = ρsdΩ − n · qdΓ Ω

(2.6.41)

Γ

where the sign of the heat flux is negative since positive heat flow is out of the body. The statement of the conservation of energy is P tot = P ext + P heat

(2.6.42)

Introducing Eqs.(2.6.37)-(2.6.41) into Eq.(2.6.42) gives the full statement of the conservation of energy Z Z Z Z Z D 1 int ρw + ρv · v dΩ = v · ρbdΩ + v · tdΓ + ρsdΩ − n · qdΓ (2.6.43) Dt Ω 2 Ω Γ Ω Γ Using the Reynold’s theorem for the first term, we have D Dt

Z ρw Ω

int

Z 1 Dwint 1 D + ρv · v dΩ = ρ + ρ (v · v) dΩ 2 Dt 2 Dt Ω Z int Dv Dw = ρ + ρv · dΩ Dt Dt Ω

(2.6.44)

Sec. 2.7

Total Lagrangian conservation equations

43

As usual, we convert the second term of the right hand size in Eq. (2.6.43) which is surface integral to domain integral by applying the Cauchy’s relation and the divergence theorem, Z

Z v · tdΓ =

Γ

Z vi ti dΓ =

ZΓ =

vi nj σij dΓ Z (vi σji ),j dΩ = (vi,j σji + vi σji,j )dΩ Γ

ZΩ

Ω

(Lij σji + vi σji,j )dΩ

=

(definitition of velocity gradient)

ZΩ

(2.6.45) (Dji σji − Wji σji + vi σji,j )dΩ

=

(decomposition of L)

ZΩ = ZΩ =

(Dji σji + vi σji,j )dΩ (sym. of σ and skew sym. of D) D : σ + v · (∇ · σ) dΩ

Ω

Inserting Eqs.(2.6.44) and (2.6.45) into Eq. (2.6.43), we have Z Dwint Dv ρ − D : σ + ∇ · q − ρs + v · ρ − ∇ · σ − ρb dΩ = 0 Dt Dt Ω

(2.6.46)

The last term in the above is exactly the momentum equation, so it vanishes. By taking the arbitrariness of Ω, the energy equation is obtained ρ

Dwint = D : σ − ∇ · q + ρs Dt

(2.6.47)

The above is a partial differential equation of energy conservation. When the heat flux and heat source vanish, i.e., in a purely mechanical process, the energy conservation equation becomes Dwint =D:σ (2.6.48) Dt which is no longer a PDE. Rate of deformation and the Cauchy stress is called conjugate in energy or work conjugate. ρ

2.7


In previous section, the updated Langrangian conservation equations have been derived where independent variables are the material coordinates X and time t, strain and stress measures are Eulerian tensors, namely the Cauchy stress tensor and the rate of deformation tensor D. Derivatives are taken with respect to spatial coordinates and integrals are calculated over the current configuration. It is often useful to express the conservation equations purely in terms of

44

Continuum mechanics

Chap. 2

Langrangian objects. These balance laws are called total Langrangian balance laws and adopted to construct total Langrangian finite elements. In the total Langrangian formulation, we use the nominal stress tensor P and the deformation gradient F as stress and strain measures since they lead to momentum equation which is surprisingly similar to the one in the updated Lagrangian description.

2.7.1

Conservation of linear momentum

Total force Z

Z ρ0 b(X, t)dΩ0 +

f (t) = Ω0

t0 (X, t)dΓ0

(2.7.1)

Γ0

The linear momentum Z ρ0 v(X, t)dΩ0

p(t) =

(2.7.2)

Ω0

The Newton’s second law of motion dp =f (2.7.3) dt Substituting Eq. (2.7.1) and (2.7.2) into the above, yields Z Z Z d ρ0 v(X, t)dΩ0 = ρ0 b(X, t)dΩ0 + t0 (X, t)dΓ0 (2.7.4) dt Ω0 Ω0 Γ0 On the LHS, the material derivative can be taken inside the integral since the reference domain is fixed in time, so Z Z d ∂v(X, t) dΩ0 (2.7.5) ρ0 v(X, t)dΩ0 = ρ0 dt Ω0 ∂t Ω0 Applying the Cauchy’s relation and Gauss’s theorem in sequences, we have Z Z Z t0 (X, t)dΓ0 = n0 · PdΓ0 = ∇0 · PdΩ0 (2.7.6) Γ0

Γ0

Z Γ0

t0i dΓ0 =

Z

Ω0

Pji n0j dΓ0 =

Γ0

Z Ω0

∂Pji dΩ0 ∂Xj

Substituting Eq. (2.7.5) and (2.7.6) into Eq. (2.7.4), yields Z ∂v(X, t) ρ0 − ρ0 b(X, t) − ∇0 · P dΩ0 = 0 ∂t Ω0 which, because of the arbitrariness of Ω0 , gives ρ0

∂v = ∇0 · P + ρ0 b or ∂t

ρ0 v˙ i =

∂Pji + ρ 0 bi ∂Xj

(2.7.7)

(2.7.8)

(2.7.9)

Sec. 2.7


45

The above is the linear momentum equation for the total Lagrangian formulation. Noting the similarity to the updated Lagrangian momentum equation given in Eq.(2.6.25): the density is replaced by the initial density, the Cauchy stress replaced by the nominal stress, and certainly derivative with respect to the spatial coordinates is replaced by derivative with respect to material coordinates. If we neglect the inertia effect, i.e., for a static problem, we then obtain the familiar equilibrium equation given by ∇0 · P + ρ0 b = 0

2.7.2

(2.7.10)

Conservation of angular momentum

From the equation conservation of angular momentum given in Eq. (2.6.36) and the stress transformation, which are recalled here for convenience σ = σT σ = J −1 F · P

(2.7.11)

J −1 F · P = (J −1 F · P)T

(2.7.12)

F · P = PT · FT

(2.7.13)

we obtain

Hence

These conditions are usually imposed directly on the constitutive equations. In the same manner for the second Piola-Kirchhoff stress, we have J −1 F · S · FT = (J −1 F · S · FT )T

(2.7.14)

F · S · FT = F · ST · FT

(2.7.15)

Or

Pre-multiplying the above with F−1 (since F is nonsingular) and post-multiplying with F−T , we obtain S = ST

(2.7.16)

So the conservation of angular momentum requires the second Piola-Kirchhoff stress to be symmetric.

46

2.7.3

Continuum mechanics

Chap. 2

Conservation of energy

The counterpart of Eq. (2.6.43) in the reference configuration is written as D Dt

Z ρ0 w

int

Ω0

Z Z 1 + ρ0 v · v dΩ0 = v · ρ0 bdΩ0 + v · t0 dΓ0 2 Ω0 Γ0 Z Z edΓ0 + ρ0 sdΩ0 − n0 · q Ω0

(2.7.17)

Γ0

Since the reference domain is fixed in time, the first term can be rewritten as Z Z D Dwint Dv 1 int ρ0 + ρ0 v · dΩ0 ρ0 w + ρ0 v · v dΩ0 = Dt Ω0 2 Dt Dt Ω

(2.7.18)

By applying the Cauchy’s relation and the divergence theorem, the second term of the right hand size in Eq. (2.7.17) can be written Z

Z

Z

vj n0i Pij dΓ0 Γ Γ0 Z Z 0 ∂vj ∂Pij ∂ = (vj Pij )dΩ0 = Pij + vj dΩ0 ∂Xi Ω0 ∂Xi Ω0 ∂Xi Z ∂Fji ∂Pij Pij + vj dΩ0 = ∂t ∂Xi Ω0 Z T ∂F : P + v · (∇0 · P) dΩ0 = ∂t Ω0

v · t0 dΓ0 = Γ0

vj t0j dΓ0

=

(2.7.19)

Inserting all of the above into Eq. (2.7.17), and applying the divergence theorem to the heat flux term of the right hand size in Eq. (2.7.17), we obtain Z Dv Dwint ∂FT e − ρ0 s + v · ρ0 − : P + ∇0 · q − ∇0 · P − ρ0 b dΩ0 = 0 (2.7.20) ρ0 Dt ∂t Dt Ω0 The last term in the above is exactly the momentum equation, so it vanishes. By taking the arbitrariness of Ω, we obtain the energy equation e + ρ0 s ρ0 w˙ int = F˙ T : P − ∇0 · q

(2.7.21)

If heat energy is neglected, the above becomes ρ0 w˙ int = F˙ T : P

(2.7.22)

It shows that the nominal stress P is conjugate in power to the material time derivative of the deformation gradient tensor F.

Sec. 2.7


47

In what follows, we seek the similar expression for the second Piola-Kirchhoff stress tensor by simply transform the nominal stress tensor to the second Piola-Kirchhoff stress tensor. Indeed, T F˙ T : P = F˙ ji Pij = F˙ ji Sik Fkj T ˙ ˙ : S (sym. of S) = Fkj Fji Sik = (FT · F) 1 ˙ +F ˙ T · F) : S = (FT · F 2

(2.7.23)

where we have used the tensor decomposition into symmetric and skew symmetric parts for FT and the double contraction of symmetric tensor S with skew symmetric tensor vanishes. Noting that the term in parentheses in the above is the time derivative of the Green strain tensor E, so ˙ : S = ρ0 w˙ int F˙ T : P = E

(2.7.24)

It shows that the rate of the Green strain tensor is work conjugate with the second PiolaKirchhoff stress.

48

Continuum mechanics

Chap. 2

Chapter 3 Constitutive models 3.1

Introduction

1. Small strain hyperelastiic materials 2. Large strain hyperelastiic materials 3. Hypoelastoplastic materials 4. Hyperelastoplastic materials The basic assumption of the classical hypoelastoplastic models are the additive decomposition of the rate of deformation tensor D and the elastic response is described by a hypoelastic model, i.e., the objective stress rate is related to De via a objective. In the hyperelastoplastic models, the basis kinematic assumption is the multiplicative decomposition of the deformation gradient. The elastic reponse is governed by hyperelasticity theory.

3.2

Hyperelasticity

Definition 3.2.1 A Cauchy elastic material is a material for which the stress is a function of only the strain. That is σij = σij (kl ) (3.2.1) Definition 3.2.2 The strain energy W is defined as Z dW = σij dij ,

W =

σd 0

which is an invariant.

(3.2.2)

50

Constitutive models

Chap. 3

Let make an assumption that the strain energy does not depend on the integration path, then we can write W = W (ij )

(3.2.3)

Hence, dW =

∂W dij ∂ij

(3.2.4)

Comparing the two equations Eq.(3.2.2) and Eq.(3.2.4) gives ∂W σij − dij = 0 ∂ij

(3.2.5)

The fact that dij is arbitrary leads to σij =

∂W ∂ij

(3.2.6)

It is emphasized that the above equation has been derived from the arbitrariness of dij . In case of isochoric motion, such a condition is not hold any more. Eqs.(3.2.4) and (3.2.6) implies Eq.(3.2.1) so a material obey Eq.(3.2.6) is called a hyperelastic material or hyperelastic material or Green elastic material.

3.2.1

Linear hyperelasticity

According to the linear hyperelasticity, the stress is linearly related to the strain via a fourthorder tensor as σij = Dijkl kl

(3.2.7)

where Dijkl is the so-called the fourth-order elastic moduli tensor. From the symmetry of the stress and strain tensor, we have the so-called minor symmetry property of the elastic modulii tensor Dijkl = Djikl ,

Dijkl = Dijlk

(3.2.8)

From Eq. (3.2.6), we have ∂σij ∂ 2W = ∂kl ∂ij ∂kl ∂σkl ∂ 2W = ∂ij ∂kl ∂ij

(3.2.9)

Sec. 3.2

Hyperelasticity

51

Since W is a smooth function, the above indicates that ∂σij ∂σkl = ∂kl ∂ij

(3.2.10)

Substituting of the above into Eq.(3.2.7) yields the major symmetry property of the linear elastic modulii Dijkl = Dklij

3.2.2

(3.2.11)

Isotropic hyperelasticity

Since W is an invariant it must be a function of the invariants of the strain. That is W = W (1 , 2 , 3 )

(3.2.12)

W = W (I1 , I2 , I3 )

(3.2.13)

If the material is isotropic, hence

with the invariants given by I1 = kk ,

I2 =,

I3 =

(3.2.14)

Therefore, from Eq.(3.2.6), the stress tensor is given by σij =

∂W ∂I2 ∂W ∂I3 ∂W ∂I1 + + ∂I1 ∂ij ∂I2 ∂ij ∂I3 ∂ij

(3.2.15)

which is reduced to after some manipulations σij = φ1 δij + φ2 ij + φ3 ik kj ,

3.2.3

φi =

∂W ∂Ii

(3.2.16)

Isotropic linear hyperelasticity

In the case of linear elastic material, the relation between stress and strain must be linear, so from Eq.(3.2.16), φ3 must be zero. Assuming that φ1 = λI1 ,

φ2 = 2µ

(3.2.17)

where λ and µ are the Lamé’s constants. Therefore, the stress tensor in Eq.(3.2.16) becomes σij = λkk δij + 2µij

(3.2.18)

52

Constitutive models

Chap. 3

To obtain the linear elastic modulii tensor from the above, we first observe that 1 ij = (ij + ji ) 2 1 = (δik δjl kl + δil δjk kl ) 2

(3.2.19)

and kk = kl δkl , hence Eq.(3.2.18) can be written as σij = [λδij δkl + µ(δik δjl + δil δjk )]kl

(3.2.20)

which immediately gives the elastic modulii tensor D Dijkl = λδij δkl + µ(δik δjl + δil δjk )

(3.2.21)

σkk = (3λ + 2µ)kk = 3Kkk

(3.2.22)

Bulk and shear modulus

with the bulk modulus K given by K :=

3λ + 2µ 3

(3.2.23)

Writting Eq.(3.2.22) for the case of hydrostatic pressure p, i.e., σxx = σyy = σzz = −p, we obtain −

∆V p = kk = K V

(3.2.24)

Thus the capacity to change volume is inversely proportional to K, hence it is termed the bulk modulus. We are now deriving the relation between shear stress σijdev and shear strain tensors dev ij . To this end, from Eqs.(3.2.18) and (3.2.22), we can compute the deviatoric stress as 1 σijdev = σij − σkk δij 3 1 = λkk δij + 2µij − (3λ + 2µ)kk δij 3 dev = 2Gij

(3.2.25)

with G is the so-called shear modulus which is given by G := µ

(3.2.26)

Sec. 3.2

Hyperelasticity

53

In terms of the bulk and shear modulus, the constitutive equation for isotropic linear elastic material can be written as σkk = 3Kkk ,

σijdev = 2Gdev ij

(3.2.27)

which implies uncoupling between volumetric and deviatoric responses. 1 ij = dev ij + kk δij 3 1 dev 1 = σij + σkk δij 2G 9K 1 3K − 2G 1+ν ν = σij − σkk δij := σij − σkk δij 2G 18GK E E

(3.2.28)

where E is the Young modulus and ν is the Poisson ratio, which are given by E=

3.2.4

9KG , 3K + G

ν=

3K − 2G 2(3K + G)

(3.2.29)

Matrix form using the Voigt notation σij σij i j 1 1 2 2 1 2

σa a 1 , 2 3

i 1 2 3 2 1 1

j 1 2 3 3 3 2

σa a 1 2 3 4 5 6

Table 3.1: Voight rule for 2D and 3D For symmetric second-order tensor like the Cauchy stress, the Voigt notation or Voigt rule is often used to convert second order tensors to first order tensors following the rule given in Table (3.1). Taking the Cauchy stress tensor as an example, in Voigt notation, we write



σ11

σ= sym

σ12 σ22

  σ11 σ22     σ13 σ33    σ23 → {σ} =  σ23    σ33 σ13  σ12

(3.2.30)

54

Constitutive models

Chap. 3

The above is a typical mapping from a second order tensor to a vector for the symmetric kinetic tensors. It is applicable to the second Piola-Kirchhoff stress tensor too. For symmetric kinematics second-order tensors like the infinitesimal strain tensor 1 , we write 



11 12  22 = sym

 11  22     13  33   23  → {} =  223    33 213  212

(3.2.31)

Noting that the off-diagonal elements are doubled so that the double contraction of stress tensor and strain tensor can be replaced equivalently by dot product, σ : = {σ} · {} = {}T {σ}

(3.2.32)

The Voigt notation furnishes less storage in implementing the finite elements and particularly useful in implementing the fourth order tangent modulii tensors by converting them to second order tensors (matrices). For example, the linear elasticity constitutive equation ( Eq.(3.2.18)), in Voigt notation, is written as    σ1 D11 D12 D13 σ2   D22 D23    σ3   D33  = σ4      σ5   SY M σ6

D14 D24 D34 D44

D15 D25 D35 D45 D55

  D16 1   D26  2   3  D36      D46   4  D56  5  D66 6

(3.2.33)

where D11 = D1111 , D12 = D1122 , D15 = D1131 etc. by mapping the first pair (ij) and second pair (kl) of Dijkl to a, b of Dab respectively, according to the rule given in Table (3.1). From Eq.(3.2.21), it is possible to compute Dijkl , hence the linear elastic modulli matrix is written as  λ + 2µ λ λ  λ + 2µ λ   λ + 2µ [D] =     SY M

0 0 0 µ

0 0 0 0 µ

 0 0  0  0  0 µ

(3.2.34)

where we have divided the last three diagonal terms by two since the last three terms of the strain vector have been doubled. 1

this mapping holds also for finite strain measures such as the Green strain tensor E and the deformation rate tensor D.

Sec. 3.3

3.2.5

Finite hyperelasticity

55

Incompressible isotropic linear elasticity

The equation of isotropic linear elastic materials in the incompressibility limit is given by ∂W σij − dij = 0 ∂ij kk = 0

(3.2.35)

which means that the stress can not be derived by differentiation of W with respect to the strain since dij is not arbitrary but satisfying the incompressibility constraint. Consider the following modified strain energy function ¯ = W (dev W ij ) + p(1 − kk )

(3.2.36)

where p is an additional unknown of the problem, which leads to the following stress tensor ¯ ∂W = −pδij + 2µij (3.2.37) ∂ij where we have made of the fact that I1 of the deviatoric strain tensor vanishes and Eq.(3.2.18). It is desirable to have a formulation applicable to both incompressible and compressible, so we can handle nearly incompressibility case. To this end, considering the following equations σij =

σij = −pδij + 2µij (3.2.38a) p (3.2.38b) kk = − λ which in case of compressibility, by subsituting p from the second equation of the above into the first one reproduces Eq.(3.2.18). And in case of incompressibility (λ is infinity), constraint kk = 0 reproduced.

3.3


A hyperelastic or Green elastic material is an ideally elastic material for which the stress-strain relationship derives from a strain energy density function. ∂ψ(C) (3.3.1) ∂C Noting that the above holds for compressible hyperelastic materials only. For incompressible, there is a constraint on the motion and the above does not hold. We will give treatment on incompressible materials later on in this section. By using the relation C = 2E + I, the stored energy function can also be expressed in term of the Green tensor E as S=2

w(E) = ψ(C),

S=

∂w(E) ∂E

(3.3.2)

56

Constitutive models

Chap. 3

where we have used a different symbol for the stored energy function. Using the relation between different stress measures, we come up with the expressions for the Kirchhoff stress (hence, the Cauchy stress) as ∂ψ(C) ∂w(E) · FT = F · · FT (3.3.3) ∂C ∂E We are now finding the expression for rest stress measure-the nominal stress tensor P. Noting that P is work-conjugated to the rate of the transposed deformation gradient F˙ T , so the stored energy function is now expressed in term of F, w(F) ¯ = ψ(C) = w(E). τ = Jσ = F · S · FT = 2F ·

∂ w(F) ¯ ∂ψ ∂C S : : = = ∂FT ∂C ∂FT 2

3.3.1

∂ T (F · F) = S · FT (???) = P, T ∂F

P=

∂ w(F) ¯ ∂FT

(3.3.4)

Elasticity tensors

The elasticity tensors relate the stress rates to strain rates. When the stress, strain rates are defined in the reference configuration, the corresponding elasticity tensors are called Langrangian or material elasticity tensor. Otherwise, when relating Eulerian stress, strain rates, we have Eulerian elasticity tensor. A :=

∂ 2w ∂F∂F

(3.3.5)

first elasticity tensor. To do this, we compute the material time derivative of the 2nd PK stress tensor ∂Sij ˙ ∂ 2ψ S˙ ij = Ckl = 2 C˙ kl ∂Ckl ∂Cij ∂Ckl

(3.3.6)

where Eq.(3.3.1) was used in the second equality. In tensor notation, the above reads 2 ∂ 2ψ ˙ =4 ∂ ψ :E ˙ :C S˙ = 2 ∂C∂C ∂C∂C

(3.3.7)

˙ S˙ = CSE : E

(3.3.8)

or,

where CSE = 4

∂ 2ψ , ∂C∂C

Cijkl = 4

∂ 2ψ ∂Cij ∂Ckl

(3.3.9)

is the material or Lagrangian tangent modulus, which is also called the second elasticity tensor. This fourth-order tensor has minor symmetries, Cijkl = Cjikl and Cijkl = Cijlk . In addition, since the stored energy is smooth, the tangent modulus of hyperelastic materials has the major symmetry Cijkl = Cklij .

Sec. 3.3


57

Relation between first and second elasticity tensors The transpose of the nominal stress is the first Piola-Kirchhoff stress denoted by T and is given by T=F·S

(3.3.10)

˙ = F˙ · S + F · S˙ T 1 SE ˙ ˙ =F·S+F· C :C 2 1 ˙ = F˙ · S + F · CSE : [F˙ T · F + FT · F] 2 ˙ ˙ = F · S + F · (CSE : FT · F)

(3.3.11)

˙ = [·] : F, ˙ it is more convenient to use indicial At this point, to rewrite the above in the form of T notation as SE T˙ij = F˙ ik Skj + Fil Cljck Fmc F˙ mk SE = Skj δim F˙ mk + Fil Cljcd Fmc F˙ mk

= (Skj δim +

(3.3.12)

SE Fil Cljcd Fmc )F˙ mk

In what follows, we derive the equivalent of Eq. (3.3.8) in the Eulerian formulation. We start by the relation of the rate of second Piola-Kirchhoff stress to the convected rate of the ˙ and D, Kirchhoff stress and the relation of E τ ∇c = F · S˙ · FT ,

˙ = FT · D · F E

(3.3.13)

Working in indicial notation, we can write τij∇c = Fip Fjq S˙ pq SE = Fip Fjq Cpqmn E˙ mn

=

(3.3.14)

SE Fip Fjq Fkm Fln Cpqmn Dkl

where in the second step, Eq. (3.3.8) has been used and in the last equality, we have used Eq. (3.3.13). In tensor notation, the above becomes τ ∇c = Cτ : D,

τ τij∇c = Cijkl Dkl

(3.3.15)

where Cτ is the spatial or Eulerian elasticity tensor (the spatial form of the second elasticity tensor CSE ) which is given by SE τ Cijkl = Fip Fjq Fkm Fln Cpqmn

(3.3.16)

58

Constitutive models

Chap. 3

It is easy to transform the above to the Truesdell rate of the Cauchy stress as follows σ ∇T = J −1 τ ∇c = J −1 Cτ : D ≡ Cστ : D,

3.3.2

Cστ = J −1 Cτ

(3.3.17)

Isotropic hyperelastic materials

In the case of isotropic hyperelasticity, the stored energy function ψ can be conveniently written in terms of the invariants as ψ(C) = ψ(I1 , I2 , I3 )

(3.3.18)

where Ii are the principle invariants of the tensor C 2 given by I1 = tr(C) = Cii 1 1 2 2 2 tr(C) − tr(C ) = Cii − Cij Cji I2 = 2 2 I3 = det(C) = det(FT F) = J 2 , J = det F

(3.3.19)

The derivatives of the invariants with respect to the tensor C, which is necessary in calculation of the stress tensor, are given by (see Section (A.1)) ∂I2 ∂I3 ∂I1 = I, = I1 I − CT , = I3 C−1 ∂C ∂C ∂C The second Piola-Kirchhoff stress tensor S is now computed by ∂ψ ∂ψ ∂I1 ∂ψ ∂I2 ∂ψ ∂I3 S=2 =2 + + ∂C ∂I1 ∂C ∂I2 ∂C ∂I3 ∂C ∂ψ ∂ψ ∂ψ ∂ψ −1 =2 + I1 I−2 C + 2I3 C ∂I1 ∂I2 ∂I2 ∂I3

(3.3.20)

(3.3.21)

where Eq. (3.3.20) was used in the second step. The Kirchhoff stress tensor is given by τ = F · S · FT ∂ψ ∂ψ 2 ∂ψ ∂ψ + I1 b−2 b + 2I3 I =2 ∂I1 ∂I2 ∂I2 ∂I3

(3.3.22)

where b is the left Cauchy-Green tensor and the expression of S in Eq. (3.3.21) was used. The expression for the Cauchy stress is then easily obtained by divide the above by the Jacobian J. The last thing left is the analytical expression of the stored energy function ψ. For completeness, we present here some of the most widely used hyperelasticity models such as the Neo-Hookean, the Mooney-Rivlin models. For others, reader is referred to the literature in the subject. 2

also invariants of the right Cauchy-Green tensor b, see Section 2.3.5.

Sec. 3.3


59

Compressible Neo-Hookean material In this case, the stored energy density is given by 1 1 ψ(C) = λ0 (ln J)2 − µ0 ln J + µ0 (I1 − 3) 2 2 where λ0 , µ0 are the Lamé constants and J = det(F), I3 = det(C) = J 2 . From Eq. (3.3.23), we can compute the following quantities ∂ψ 1 ∂ψ ∂ψ 1 1 = µ0 , = 0, = λ0 ln JJ −2 − µ0 J −2 ∂I1 2 ∂I2 ∂I3 2 2 The second Piola-Kirchhoff stress tensor is therefore given by S = µ0 (I − C−1 ) + λ0 ln JC−1

(3.3.23)

(3.3.24)

(3.3.25)

and the Kirchhoff stress tensor is τ = µ0 (b − I) + λ0 ln JI

(3.3.26)

where b is the left Cauchy-Green strain tensor. The Lagrangian elasticity tensor is given by

CSE = 2

∂S ∂C

∂C−1 ∂C−1 1 1 ∂det C + 2λ0 ln J + 2λ0 C−1 ∂C ∂C J 2J ∂C ∂C−1 1 1 = λ0 C−1 det(C)C−T − 2(µ0 − λ0 ln J) JJ ∂C −1 ∂C = λ0 C−1 C−1 − 2(µ0 − λ0 ln J) ∂C = −2µ0

(3.3.27)

where, for the second equality, Eq. (3.3.25) and the fact that det(C) = J 2 have been used; in the third step, the derivative of the determinant of a second order tensor with respect to itselft det(C),C = det(C)C−T has been utilized, see Appendix A.2. In the final step, the symmetry of C was adopted. The derivative of inverse of C with respect to itself is given by (for derivation, see Appendix A.1) −1 ∂C −1 −1 = −Cik Cjl (3.3.28) ∂C ijkl Letting λ = λ0 , µ = µ0 − λ0 ln J and substituting of the above into Eq.(3.3.27) yields the Lagrangian elasticity tensor in component form which is written as −1 −1 −1 SE Cijkl = λCij−1 Ckl + 2µCik Cjl

(3.3.29)

60

Constitutive models

Chap. 3

From Eq. (3.3.16), the Eulerian elasticity tensor is given by τ = λδij δkl + 2µδik δjl Cijkl

(3.3.30)

Noting that this tensor is very similar to the one of small strain linear elasticity except that the shear modulus µ now depends on the deformation. In the implementation, the Voigt form of τ is given by Cijkl  τ C1112 τ  C2212  τ C3312   τ  C2312  τ C1312  τ C1212 (3.3.31) τ where the Voigt rule given in Table (3.1) has been used for each index in Cab . Using Eq.(3.3.30) τ of the above, we obtain 3 to compute every term Cijkl   λ + 2µ λ λ 0 0 0  λ + 2µ λ 0 0 0    λ + 2µ 0 0 0  τ   [C ] =  (3.3.32)  µ 0 0    µ 0 µ  τ τ τ C11 C12 C13 τ τ  C22 C23  τ  C33 [Cτ ] =    

τ C14 τ C24 τ C34 τ C44

τ C15 τ C25 τ C35 τ C45 τ C55

  τ τ τ τ C1111 C1122 C1133 C16 τ τ τ   C2222 C2233 C26   τ τ  C3333 C36    = τ   C46   τ   C56 τ C66

which is reduced to in plane strain condition   λ + 2µ λ 0 λ + 2µ 0  [Cτ ] =  λ 0 0 µ

τ C1123 τ C2223 τ C3323 τ C2323

τ C1113 τ C2213 τ C3313 τ C2313 τ C1313

(3.3.33)

Plane stress neo-Hookean elasticity matrix To derive the Eulerian elasticity tensor in plane stress, we have to impose the condition σ33 = 0. Let us first compute this component σ3 = C31 1 + C32 2 + C33 3 = λ(1 + 2 ) + (λ + 2µ)3

(3.3.34)

From which 3 can be computed as 3 = − 3

λ (1 + 2 ) λ + 2µ

(3.3.35)

Noting that the last three diagonal terms are divided by two since the corresponding terms of the kinematic tensor have been doubled.

Sec. 3.3


61

which after introduced into the equation for σ1 gives λ (1 + 2 ) λ + 2µ ¯ = 2µ λ λ + 2µ

σ1 = (λ + 2µ)1 + λ2 + − ¯ + 2µ)1 + λ ¯ 2, = (λ

(3.3.36)

Doing the same for σ2 , we come up with the plane stress Eulerian elasticity tensor for neoHookean material   ¯ + 2µ ¯ λ λ 0 ¯ ¯ + 2µ 0  λ [Cτ ] =  λ (3.3.37) 0 0 µ It is worthy noting that the above procedure to get the plane stress constitutive equation from 3D constitutive equation is not as straightforward for complex finite strain material behaviours as linear elasticity or neo-Hookean material. We will present advanced methods for this problem in Chapter 4, Section 4.7.2. For convenience of implementation, the constitutive equations of the compressible NeoHookean material is summarized in Box 3.1. Box 3.1 Compressible Neo-Hookean material µ0 λ0 σ = (b − I) + ln JI J J with b = FFT , J = det(F)  λ + 2µ λ λ  λ + 2µ λ  1 λ + 2µ σ [C ] =   J  where λ = λ0 , µ = µ0 − λ0 ln J.   ¯ + 2µ ¯ λ λ 0 1 ¯ ¯ + 2µ 0  , λ [Cσ ] =  λ J 0 0 µ

3.3.3

¯= λ

0 0 0 µ

( λ 2µ λ+2µ

 0 0  0  0  0 µ

0 0 0 0 µ

plane strain plane stress

Isotropic hyperelasticity in principle directions

In this case, the stored energy density is written in terms of the principal stretches ψ = ψ(λ1 , λ2 , λ3 )

(3.3.38)

62

Constitutive models

Chap. 3

where λi are the principal stretches, also the eigenvalues of the right Cauchy-Green strain tensor C. Denoting the eigenvectors of C by Ni , we have the following expression

I= C= C−1 =

3 X i=1 3 X i=1 3 X

Ni ⊗ Ni

(3.3.39a)

λ2i Ni ⊗ Ni

(3.3.39b)

λ−2 i Ni ⊗ Ni

(3.3.39c)

i=1

Substituting of the above into Eq.(3.3.21), the 2nd PK stress tensor now becomes 3 X ∂ψ ∂ψ −2 ∂ψ 2 ∂ψ + I1 λi + 2I3 λi −2 Ni ⊗ Ni S= 2 ∂I ∂I ∂I ∂I 1 2 2 3 i=1

(3.3.40)

The relation between I and λ is given by (see Section A.5 for derivation) I1 = λ21 + λ22 + λ23 I2 = λ21 λ22 + λ22 λ23 + λ23 λ21 I3 = λ21 λ22 λ23

(3.3.41)

3

X ∂ψ ∂λ2 ∂ψ i = 2 ∂I1 ∂λ ∂I 1 i i=1 3

X ∂ψ ∂λ2 ∂ψ i = 2 ∂I2 ∂λ ∂I 2 i i=1

(3.3.42)

3

X ∂ψ ∂λ2 ∂ψ i = 2 ∂I3 ∂λ ∂I 3 i i=1

3.4

Incompressible hyperelasticity

Eq.(2.3.44), J˙ = 0 gives 1 ˙ =0 JC−1 : C 2 1 ∂ψ ˙ =0 S− :C 2 ∂C

(3.4.1)

(3.4.2)

Sec. 3.5

Incompressible hyperelasticity

63

From the above two equations, we obtain 1 ∂ψ 1 S− = γ JC−1 2 ∂C 2

(3.4.3)

where γ is Then the 2nd PK stress is given by S=2

∂ψ + γJC−1 ∂C

(3.4.4)

It is intentionally that in the above, J has been retained since for incompressibility, we have J = 1. There are two reasons for this. Firstly, the above is applicable to nearly incompressible materials. Secondly, numerical methods such as finite elements can not enforce J = 1 strictly. We are now showing the relation between γ and the pressure p. To this end, the pressure is given by (see Eq.(2.4.28)) 1 −1 1 −1 ∂ψ −1 p= J S:C= J + γJC 2 :C 3 3 ∂C 2 ∂ψ : C (C−1 : C = 3) = γ + J −1 3 ∂C

(3.4.5)

where we have used Eq.(3.4.4) in the second equality. Before proceeding, let us remind the definition of homogeneity order of a scalar function given by Definition 3.4.1 A scalar function f (x) is said to be homogeneous of order n if f (αx) = αn f (x)

(3.4.6)

with the following property derived by differentiating the above with respect to α at α = 1: ∂f · x = nf (x) ∂x

(3.4.7)

With this definition, we can see that γ is identical to the pressure p if ∂ψ :C=0 ∂C

(3.4.8)

which indicates that ψ is homogeneous of order zero, so ψ(αC) = ψ(C)

(3.4.9)

64

3.5

Constitutive models

Chap. 3

Nearly incompressible hyperelasticity

It is a standard for nearly incompressible hyperelastic materials that the strain energy function is addtively decomposed into two parts, one is the volumetric part and the second is the rest as follows ¯ ψ = ψ(C) + kU (J)

(3.5.1)

where k is the so-called generalized bulk modulus which is very large. It is emphasized that given a strain energy ψ, the above decomposition is not unique. Unique decomposition can be ¯ ¯ C) ¯ where C ¯ is the isochoric right Cauchy-Green strain tensor obtained if we use ψ(C) as ψ( given in Eq.(2.3.301 ). ∂U ∂J ∂ψ + 2k ∂C ∂J ∂C ∂ψ 0 + kU JC−1 =2 ∂C

S=2

(3.5.2)

There Cauchy stress is then given by σ = J −1 FSFT = 2J −1 F

∂ψ T 0 F + kU I ∂C

(3.5.3)

where in the last equality, C = FFT has been used.

3.5.1

Elasticity tensors

Lagrangian tensor Eulerian tensor Volumetric strain energy U(J) The commonly used volumetric strain energy functions U (J) are given by 1 U (J) = (J − 1)2 , 2

1 U (J) = (ln J)2 2

(3.5.4)

Incompressible Neo-Hookean 1 ψ = µ J −2/3 trC − 3 2 σ = 2J −1 F

∂ψ T 0 F + kU I ∂C

(3.5.5) (3.5.6)

Incompressible Mooney-Rivlin ψ = K1 J −2/3 trC − 3 + K2 J −4/3 I2 − 3

(3.5.7)

Sec. 3.6

Hyperelastic plastic materials

65

Incompressible Ogden material

3.6

Hyperelastic plastic materials

1. Basis kinematic assumption-multiplicative decomposition of the deformation gradient tensor

2.

3.6.1

F = Fe · Fp

(3.6.1)

¯ e) ¯ E ¯ = ∂ w( S ¯e ∂E

(3.6.2)

Hyperelastic potential 1 e e E = (C − I), 2

e

C = FeT · Fe

¯ e) ¯ E ¯ = ∂ w( S ¯e ∂E ¯˙ = S

3.6.2

∂ 2w ˙e e e : E ∂E ∂E

(3.6.3)

(3.6.4)

(3.6.5)

Decomposition of rate of deformation

Using the multiplicative decomposition given in Eq.(3.6.1), we can write the velocity gradient as ˙ · F−1 = D (Fe · Fp ) · (Fe · Fp )−1 L=F Dt ˙ e · (Fe )−1 + Fe · F ˙ p · (Fp )−1 · (Fe )−1 =F

(3.6.6)

Defining the elastic and plastic parts of L as ˙ e · (Fe )−1 Le := F ˙ p · (Fp )−1 · (Fe )−1 Lp := Fe · F

(3.6.7b)

L = Le + Lp

(3.6.8)

(3.6.7a)

So, we have

66

Constitutive models

Chap. 3

Based on these definitions, we can also define the elastic and plastic rate of deformation tensor and spin tensor as 1 De = (Le + LeT ), 2 1 p D = (Lp + LpT ), 2

1 We = (Le − LeT ) 2 1 p W = (Lp − LpT ) 2

(3.6.9)

We have ˙e (L − Lp ) · Fe = Le · Fe = F

(3.6.10)

where Eq.(3.6.8) has been used in the second step and Eq.(3.6.7)1 has been used in the second equality. The above then gives the rate of elastic deformation gradient written as ˙ e = L · Fe − (Dp + Wp ) · Fe F

(3.6.11)

¯ is defined as the pull back of L by F ¯ on Ω Velocity gradient L

e

¯ = FeT · L · Fe L eT e e −1 e ˙p p −1 e −1 ˙ = F · F · (F ) + F · F · (F ) · (F ) · Fe

(3.6.12)

˙ e + FeT · Fe · F ˙ p · (Fp )−1 = FeT · F

3.7 3.7.1

Stress update algorithms Fully implicit backward Euler scheme

Given the state at time step n i.e., n , pn , qn , and the strain increment ∆ (computed from the displacement increment), the set of equations to be solved are given by n+1 pn+1 qn+1 σn+1 fn+1

= = = = =

n + ∆ pn + ∆λn+1 rn+1 qn + ∆λn+1 hn+1 C : (n+1 − pn+1 ) f (σn+1 , qn+1 ) = 0

(3.7.1a) (3.7.1b) (3.7.1c) (3.7.1d) (3.7.1e)

We can rewrite Eq.(3.7.1)4 as σn+1 = C : (n+1 − pn+1 ) = C : (n + ∆ − pn − ∆λn+1 rn+1 ) h i p = C : (n − n ) + C : ∆ − ∆λn+1 C : rn+1 h i = σn + C : ∆ − ∆λn+1 C : rn+1

(3.7.2)

Sec. 3.7

Stress update algorithms

67

where Eq.(3.7.1)1 and Eq.(3.7.1)2 have been used. trial := σn + C : ∆ σn+1

trial σn+1 = σn+1 − ∆σ,

(3.7.3)

∆σ = C : ∆pn+1 = C : ∆λn+1 rn+1

(3.7.4)

Defining the following residuals a = −pn+1 + pn + ∆λn+1 rn+1 = 0 b = −qn+1 + qn + ∆λn+1 hn+1 = 0 f = f (σn+1 , qn+1 ) = 0

(3.7.5a) (3.7.5b) (3.7.5c)

Eq.(3.7.5)1 can be rewritten as a = −∆pn+1 + ∆λn+1 rn+1 = C−1 : ∆σn+1 + ∆λn+1 rn+1 (σ, q)

(3.7.6)

where we have used Eq.(??) in the second equality and write out the dependency of r in the stresses and internal variables to make the following linearization clear. Its linearization is given by where the subscript n + 1 has been omitted for clarity reason (k) a(k) + C−1 : ∆σ (k) + r(k) δλ(k) + ∆λ(rσ(k) : ∆σ (k) + r(k) q · ∆q ) = 0

(3.7.7)

Similarly, we have the linearization of Eq.(3.7.5)2 written as (k) (k) b(k) − ∆q(k) + h(k) δλ(k) + ∆λ(h(k) + h(k) σ : ∆σ q · ∆q ) = 0

Putting the two equations Eqs.(3.7.7-3.7.8) together yields (k) −1 ∆σ (k) (k) ¯ A − δλ(k) ¯r(k) (k) = − a ∆q

(3.7.8)

(3.7.9)

where C−1 + ∆λrσ ∆λrq ∆λhσ −I + ∆λhq (k) (k) (k) (k) a r ¯ ¯r = (k) , = (k) a b h

(k) −1 A =

(k)

Solving Eq.(3.7.9) for increments in stress and internal variables gives (k) (k) ∆σ (k) ¯ − δλ(k) A(k) ¯r(k) a (k) = − A ∆q

(3.7.10)

(3.7.11)

(3.7.12)

68

Constitutive models

Chap. 3

Subsituting the above in the linearization of the yield function which is given by f (k) + fσ(k) : ∆σ (k) + fq(k) · ∆q(k) = 0

(3.7.13)

and solving for the update of plastic multiplier, we obtain ¯(k) f (k) − ∂f (k) A(k) a (3.7.14) ∂f (k) : A(k) : ¯r(k) where we used the notation ∂f = [fσ fq ]. Finally, update the plastic strain, internal varibles, the plastic parameter and the stress δλ(k) =

p,(k+1)

n+1

(k+1)

p,(k)

= n+1 − C−1 : ∆σ (k) (k)

(k)

q = q + ∆q (k+1) ∆λ = ∆λ(k) + δλ(k) σ (k+1) = σ (k) + ∆σ (k)

3.7.2

(3.7.15a) (3.7.15b) (3.7.15c) (3.7.15d)

Von Mises plasticity with isotropic hardening

The constitutive equations of von Mises plastic material including isotropic hardening are given by f = σ − σY () r 3 1 σ = s : s, s = σ − trσ1 2 3 σY () = σ0Y + H() r 2 ˙ = ||˙p || 3 ˙ σ ˙p = λf σ = C : ( − p ), C = Kδij δkl + µ(δik δjl + δil δjk )

(3.7.16a) (3.7.16b) (3.7.16c) (3.7.16d) (3.7.16e) (3.7.16f)

where σ is the so-called equivalent or effective stress; σY () is the yield stress which is a function of the equivalent plastic strain . Some commonly used yield stress expressions are given in the following 1. Linear isotropic hardening σY () = σ0 + H

(3.7.17)

σY () = σ0 + (σ0 − σ∞ )[1 − exp(−δ)] + H

(3.7.18)

2. Saturation isotropic hardening

where σ0 , σ∞ , H and δ are material constants.

Sec. 3.7


69

Let first compute the trial stress which is given by σ trial = σn + C : ∆

(3.7.19)

Then, the stress at the end of the time step n + 1 is given by, see Eq.(3.7.4) σn+1 = σ trial − C∆λfσ = σ trial −

3µ sn+1 ∆λ σ n+1

(3.7.20)

3 s and C : s = 2µs 4 . 2σ Since the trace of the deviatoric stress vanishes, the above indicates that

where we have used the fact that fσ =

n+1 trial σkk = σkk

(3.7.21)

Hence, the deviatoric stress can be written as 1 sn+1 = σijn+1 − σlln+1 δij ij 3 1 trial 3µ n+1 sij ∆λ − σkk = σijtrial − δij σ n+1 3 3µ = strial − ∆λsn+1 ij ij σ n+1

(3.7.22)

where Eq.(3.7.20) and Eq.(3.7.21) have been used in the second equality. In the above, strial is ij the trial deviatoric stress tensor. Thus, we have sn+1 = ij

strial ij 3µ 1+ ∆λ σ n+1

(3.7.23)

4

∂ ∂σ ¯ ∂s fσ = (¯ σ − σY (¯ )) = : = ∂σ ∂s ∂σ

3 ∂J2 2¯ σ ∂s

:

∂s ∂σ

∂J2 ∂ 1 1 ∂sij ∂sij = sij sij = sij + sij ∂s ∂skl 2 2 ∂skl ∂skl 1 = (skl + skl ) = skl 2 ∂s ∂ 1 1 = (σ − σ : δ ⊗ δ) = I − δ ⊗ δ ∂σ ∂σ 3 3 I : s = s,

s : (1 ⊗ 1) = 0(skk = 0)

70

Constitutive models

Chap. 3

Therefore, the equivalent stress at the end of time step n + 1 is computed according to σ n+1 =

3 n+1 n+1 s s 2 ij ij

3/2 =

σ trial 3µ 1+ ∆λ σ n+1

(3.7.24)

with the use of Eq.(3.7.23) in the second step. In the above, σ trial is the trial equivalent stress. Thus, σ n+1 can be written as σ n+1 = σ trial − 3µ∆λ

(3.7.25)

Introducing Eq.(3.7.25) into Eq.(3.7.23) gives sij =

∆λ 1 − 3µ trial σ

strial ij

(3.7.26)

which means that, if the plastic increment ∆λ can be computed, hence sij is completely determined. The above also indicates why the method is named radial return mapping. In the deviatoric stress plane, the von Mises yield surface is a circle, thus the normal to it (strial ) is radial. So the above equation means that starting from the trial deviatoric stress, the stress is returned back radially to the yield surface. It is now ready to compute the yield function at the end of the time step n + 1 and enforce the condition f = 0 there, f n+1 = σ trial − 3µ∆λ − σY (n + ∆λ) = 0

(3.7.27)

where Eq.(3.7.25) has been introduced into Eq.(3.7.16)1 and the fact that ∆ = ∆λ has been made use. This equation is generally a nonlinear equation of one unknown, the plastic increment ∆λ, which requires the use of the standard Newton-Raphson method, ∆λ(k+1) = ∆λ(k) + δλ,

δλ =

σ trial − 3µ∆λ(k) − σY (n + ∆λ(k) ) 3µ + H

(3.7.28)

The iterative process is repeated until the yield function given in Eq.(3.7.27) is sufficiently closed to zero. In the above, H is the so-called plastic modulus which is given by

( H ∂σY H= = ∂ H + δ(σ0 − σ∞ ) exp(−δ)

linear isotropic hardening saturation exponential isotropic hardening (3.7.29) Having obtained the plastic increment, the final thing to do is to update the stress, plastic strain

Sec. 3.7


71

and the equivalent plastic strain by 3 3 strial sn+1 = pn + ∆λ trial 2σ n+1 2 σ n+1 σ 1 trial = Ytrial strial + σkk 1 3 σ = n + ∆λ

pn+1 = pn + ∆λ

(3.7.30a)

σn+1

(3.7.30b)

n+1

(3.7.30c)

where the fact that at the end of the time step, f = 0 hence σY = σ has been used. Also, we have divided side by side the two Eqs.(3.7.23) and (3.7.24) and used the result in the above equation. In the above 1 is the second order identity tensor. The radial return algorithm for isotropic hardening von Mises plasticity is summarized, for convenience of programming, in box 3.7.2. Noting that, we have added the computation of the algorithmic tangent matrix for completeness.

3.7.3

Algorithm tangent moduli dσ dp dq df

= = = =

C : (d − dp ) d(∆λ)r + ∆λ(rσ : dσ + rq : dq) d(∆λ)h + ∆λ(hσ : dσ + hq : dq) fσ : dσ + fq · dq = 0

(3.7.33a) (3.7.33b) (3.7.33c) (3.7.33d)

From the first three equations of the above, eliminating dp and solving for dσ, dq we have dσ d = [A] − d(∆λ)A : ¯ r (3.7.34) dq 0 where A=

C−1 + ∆λrσ ∆λrq ∆λhσ −I + ∆λhq

(−1) (3.7.35)

Replacing Eq.(3.7.34) into the fourth equation of (3.7.33) and solving for d∆λ gives d −∂f : A : 0 (3.7.36) d∆λ = ∂f : A : ¯r We are interested in materials where rq and hσ vanish. In this simplified case, the tensor A becomes ¯ 0 (C−1 + ∆λrσ )−1 0 C A= = (3.7.37) 0 (−I + ∆λhq )−1 0 Y

72

Constitutive models

Chap. 3

Box 3.2 Radial return mapping for isotropic hardening J2 palsticity 1. Initialization (k = 0): set the plastic strain and internal variables to their converged values at the previous step. The plastic increment is set to zero and compute the trial stress

¯(0) = ¯n ,

(p,0) = pn ,

r σ

trial

=

∆λ(0) = 0,

3 trial s , 2

en+1 = dev(n+1 ),

trial √ s = strial : strial ,

strial = 2µ(en+1 − e(p,0) )

strial ||strial ||

n=

2. Compute the yield function and check for convergence f (k) = σ ¯ trial − 3µ∆λ(k) − σY (¯(k) ) Check convergence f < T OL , if yes, stop. Else goto 3. 3. Compute increments of plastic parameter

∆λ(k+1) = ∆λ(k) +

f (k) , 3µ + H (k)

H (k) =

∂σY (k) ( ) ∂

(k+1) = n + ∆λ(k+1) 4. Increase k = k + 1 and go back to step 2. 5. Update plastic strains, internal variables, plastic parameter and stresses

r

3 pn+1 = pn + ∆λn 2 r 2 n+1 σn+1 = σ n + Kn+1 kk 1 3 Y n+1 = n + ∆λ 6. Compute the algorithmic tangent matrix 1 Calg = K1 ⊗ 1 + 2µβ(I − 1 ⊗ 1) − 2µ¯ γn ⊗ n 3 2µ∆λ b = trial , ||s ||

r β =1+

3 b, 2

r γ¯ = γ −

3 b, 2

γ=

1 1+

H 3µ

Sec. 3.7


73

Substituting Eqs.(3.7.37), (3.7.36) into Eq.(3.7.34), we get dσ = Calg d

(3.7.38)

where ¯− Calg = C

¯ : r) ⊗ (fσ : C) ¯ (C ¯ :r fq · Y · h + fσ : C

(3.7.39)

which has the same form with the continuum elasto-plastic tangent modulus except that the ¯ and the term −fq · h is replaced by fq · Y · h. elastic modulus is replaced by C

3.7.4

Algorithmic tangent modulii of von Mises plasticity r

s strial = ||s|| ||s||trial

(3.7.40)

2 dσn+1 = C : dn+1 − 3 µ(d∆λnn+1 + ∆λdnn+1 ) 3 # " r r ∂∆λ ∂nn+1 2 n+1 2 µn ⊗ +3 µ∆λ : dn+1 = C−3 3 ∂n+1 3 ∂n+1

(3.7.41)

σn+1 = C : (n+1 −

np )

− 3µ∆λ

2 n+1 n , 3

n :=

Differentiating the above with respect to the strain, we write r

In the above, the term in the bracket is identically the algorithmic tangent modulus matrix. To compute the derivative of the plastic multiplier increment w.r.t the strain, it suffices to differentiate the Eq.(??) w.r.t the strain, this is given by ∂σ trial ∂∆λ ∂σY ∂∆λ − 3µ − =0 ∂n+1 ∂n+1 ∂∆λ ∂n+1 r r ∂σ trial 3 n+1 ∂strial 3 n+1 n : n 2µ = = ∂n+1 2 ∂n+1 2 Introducing the above into Eq.(3.7.42) gives r ∂∆λ 2 n+1 = γn , ∂n+1 3

1

γ= 1+

∂σY ∂∆λ

(3.7.42)

(3.7.43)

(3.7.44)

3µ

We turn now our attention to the derivative of the normal vector w.r.t the strain. By applying the chain rule, we can write ∂n ∂n ∂s = : ∂ ∂s ∂

(3.7.45)

74

Constitutive models

Chap. 3

∂n 1 = (I − n ⊗ n) ∂s ||s||

(3.7.46)

with the following result 5

and by noting that 6 1 1 s = sn + 2µ[ − trace()1 − en ] = sn + 2µ( − : (1 ⊗ 1) − en ) 3 3

(3.7.48)

Hence, the derivative of the deviatoric stress w.r.t the strain is given by ∂s 1 = 2µ(I − 1 ⊗ 1) ∂ 3

(3.7.49)

Now, substituting Eqs.(3.7.46) and (3.7.49) into (3.7.45) gives ∂n 2µ 1 = (I − n ⊗ n) : (I − 1 ⊗ 1) ∂ ||s|| 3 2µ 1 = (I − n ⊗ n − 1 ⊗ 1) ||s|| 3

(3.7.50)

Substitution of Eqs.(3.7.44) and (3.7.50) into Eq.(3.7.41) yields the algorithmic tangent moduli matrix

Calg

1 = K1 ⊗ 1 + 2µ(I − 1 ⊗ 1) − 2γµnn+1 ⊗ nn+1 + 2µ 3 1 = K1 ⊗ 1 + 2µβ(I − 1 ⊗ 1) − 2µ¯ γ nn+1 ⊗ nn+1 3

r

3 1 b(I − nn+1 ⊗ nn+1 − 1 ⊗ 1) 2 3

(3.7.51) with the coefficients given by 2µ∆λ b = trial , ||s || 5 6

r β =1+

3 b, 2

r γ¯ = γ −

3 b, 2

γ=

1 H 1+ 3µ

(3.7.52)

Derivation based on directional derivative, given in page 123 book Simo and Hughes!!! This can also be derived by using indicial notation as 1 1 sij = 2µ(ij − kk δij ) = 2µ(ij − δij kl δkl ) 3 3

Thus the derivative comes more easier than direct notation.

(3.7.47)

Sec. 3.7


75

For convenience of implementation, the algorithmic tangent matrix is given here in matrix form for both plane strain and three dimensional cases,   4 2   K + µβ K − µβ 0 n n n n n n 1 1 1 2 1 3   3 3     alg    γ n2 n1 n2 n2 n2 n3  C = K − 2 µβ K + 4 µβ 0  − 2µ¯ (3.7.53)    3 3 n3 n1 n3 n2 n3 n3 0 0 µβ and in three dimensions

Calg

 4 2 2 K + µβ K − µβ K − µβ 0 0  3 3 3   4 2  K + µβ K − µβ 0 0  3 3   4 = 0 K + µβ 0  3   SY M µβ 0    µβ   n1 n1 n1 n2 n1 n3   n2 n2 n2 n3   n3 n3  − 2µ¯ γ  SY M    

3.7.5

Plane stress von Mises material

n1 n4 n1 n5 n1 n6

0 0 0 0 0

               

(3.7.54)

µβ 

 n2 n4 n2 n5 n2 n6    n3 n4 n3 n5 n3 n6   n4 n4 n4 n5 n4 n6    n5 n5 n5 n6   n6 n6

(3.7.55)

76

Constitutive models

Chap. 3

Chapter 4 Lagrangian finite elements 4.1

Introduction

Finite elements using Lagrangian meshes are commonly classified as total Lagrangian formulation and updated Lagrangian formulation. In both formulations, the independent variables are the material coordinates X and time. In the total Lagrangian formulation, the stress and strain are Lagrangian, i.e., they are defined with respect to the reference configuration (for example, the nominal or second Piola-Kirchhoff stress are employed), the derivatives are computed with respect to the material coordinates. The corresponding weak form therefore involves integrals over the reference configuration. On the orther hand, the updated Lagrangian formulation uses the Eulerian strain and stress measures e.g., the Cauchy stress, the derivatives are computed with respect to the spatial coordinates x. The corresponding weak form therefore involves integrals over the current (deformed) configuration. In this chapter both total and updated Langrangian formulations are presented in details. Starting from the strong form, we then construct the corresponding weak form. Finite element approximation is then constructed and substituted into the weak form to obtain the semidiscrete equations. Time integration schemes are then introduced to discretise these equations into nonlinear algebraic equations which should be solved by using iterative solvers like the Newton-Raphson method. In the implementation, usually we employ the symmetry of the Cauchy stress tensor, the 2nd PK tensor, the rate of deformation tensor D to cast them into vectors using the Voigt notation. While doing this, we also give implementation which adopts the full matrices since sometimes this notation involves less computational task. Also in this Chapter the standard process of building finite elements for a given problem is discussed. In Chapter 2, the derivation of governing equations for continuum mechanics was presented. From this strong form, the Galerkin process to build the weak form is presented in details. Then with the properly chosen finite element approximations which after substituted into the weak form, the discrete equations are obtained. Since these equations are usually nonlinear, iterative procedures such as the well known Newton-Raphson method are discussed. While doing this, the linearization process required in the Newton-Raphson method is also given. In the end, a set of linearized algebratic equations is obtained which can be readily

78

Lagrangian finite elements

Chap. 4

solved by any solvers ranging from direct solvers like LU decomposition method to iterative solvers like gradient conjugate solvers. It is emphasized that the above procedure is not unique since it is quite often that finite elements are derived from the so-called variational principles such as the minimum total potential energy principle in solid mechanics. This approach relies strongly on a branch of mathematics called variational calculus. On the other hand, this approach is not as general as the Galerkin process since the variational principles can not be constructed for every problems (in cases such a variational principle exists then both approaches should give the same result). Having said that, it should be emphasized that finite elements built upon variational principles automatically yield symmetric matrices. In the other hand, it is in the theory of variational principles that weak forms for problems with internal constraints, say incompressibily, can be constructed straightforward. The corresponding weak forms are often called contrived variational principles in the sense that the standard variational principles are extended with additional unknowns to impose the constraints. Noting also that, contrary to the above linearization process which is done on the discrete equations, the linearized weak forms (the linearization is performed on the weak form) are also often used to obtain the linearized discrete equations. It is therefore to understand both approaches. In this chapter, the first one is present while in the next chapter which deals with constrained problems, the second approach is given and discussed in details. Lagrangian vs Eulerian elements

4.2

Governing equations

As mentioned above, the independent variables in Langrangian formulation are the material coordinate X and time t. The dependent variables include 1. Mass density ρ(X, t) (1) 2. Velocity field v(X, t) (3) 3. Stress field σ(X, t) (6) 4. Deformation rate tensor D(X, t) (6) So, in total, we have 16 unknowns in three dimensions. The governing equations include 1. Conservation of mass (1) 2. Conservation of momenta (3) 3. Conservation of energy (1) 4. Constitutive equations (6) 5. Strain-displacement equations (strain measures) (6) In total, we have 17 equations. However, since we are interested in non adiabatic, nonisothermal processes, the conservation of energy is not a PDE, so finally we have 16 equations for 16 unknowns.

Sec. 4.3

Governing equations

79

Box 4.1 Governing equations for updated Lagrangian formulation 1. Conservation of mass ρ(X, t)J(X, t) = ρ0 (X)

(4.2.1)

2. Conservation of linear momemtum ρv˙ = ∇ · σ + ρb or

ρv˙ i =

∂σij + ρbi ∂xj

(4.2.2)

3. Conservation of angular momemtum σ = σT,

or σij = σji

(4.2.3)

4. Conservation of energy ρw˙ int = D : σ,

or

ρw˙ int = Dij σij

(4.2.4)

5. Constitutive equation σ ∇ = StσD (D, σ)

(4.2.5)

6. Rate of deformation D = sym(∇v),

or

∂vj 1 ∂vi + Dij = 2 ∂xj ∂xi

(4.2.6)

7. Boundary conditions σij nj = t¯i Γvi ∩ Γti = 0,

on Γti

vi = v¯i

on Γvi

Γvi ∪ Γti = Γ i = 1 to nSD

(4.2.7a)

(4.2.7b)

8. Initial conditions v(X, 0) = v0 (X),

u(X, 0) = u0 (X)

(4.2.8)

9. Interior continuity condition on Γint : [[n · σ]] = 0

(4.2.9)

80

4.3 4.3.1


Chap. 4

Weak formulation Strong form to weak form

The trial and test function spaces are defined by U = { vi | vi ∈ C 0 (X), vi = v¯i

on Γv }

U0 = { δvi | δvi ∈ C 0 (X), δvi = 0 on Γv }

(4.3.1) (4.3.2)

The construction of the weak form starts by multiplying the momentum equation with the test function δvi and integrating over the current configuration. That is

Z δvi Ω

∂σij + ρbi − ρv˙ i dΩ = 0 ∂xj

(4.3.3)

Since (δvi σij ),j = δvi,j σij + δvi σij,j , we can write the first term in the above as ∂σij δvi dΩ = ∂xj Ω

Z

Z Ω

∂ ∂δvi (δvi σij ) − σij dΩ ∂xj ∂xj

Using the Gauss’s theorem, we can write Z Z Z ∂ (δvi σij )dΩ = δvi [[nj σij ]]dΓ + δvi nj σij dΓ Γint Γ Ω ∂xj

(4.3.4)

(4.3.5)

From the traction continuity, the first term on the RHS vanishes. For the second integrand, using the traction boundary condition and the fact that δvi vanishes on the complement of Γt , the above becomes Z Z ∂ (4.3.6) (δvi σij )dΩ = δvi ti dΓ Ω ∂xj Γt Substituting Eq. (4.3.6) into Eq. (4.3.4) gives Z Z Z ∂σij ∂δvi δvi dΩ = δvi ti dΓ − σij dΩ ∂xj Ω Γt Ω ∂xj

(4.3.7)

Introducing Eq. (4.3.7) into Eq. (4.3.3) yields Z Ω

∂δvi σij dΩ − ∂xj

Z

Z δvi ti dΓ −

Γt

Z δvi ρbi dΩ +

Ω

δvi ρv˙ i dΩ = 0

(4.3.8)

Ω

In stead of thinking the test function δvi as a mathematical quantity, if we consider it as a virtual velocity, then each term in the above equation represents a virtual power. Therefore, this equation is named the virtual power equation.

Sec. 4.4

4.3.2

Principle of virtual power

81

Weak form to strong form

After spending efforts to derive this weak form, a question arises naturally. Is this weak form equivalent to the strong form from which it was derived. The answer is fortunately yes and to this end Z Ω

∂σij dΩ ∂xj Ω Ω Z Z ∂σij dΩ = δvi nj σij dΓ − δvi ∂xj Γ Ω

∂δvi σij dΩ = ∂xj

∂δvi σij dΩ − ∂xj

Z

Z

δvi

(4.3.9)

which after substituted into Eq.(4.3.8) yields Z

Z δvi (ti − nj σij )dΓ −

Γ

δvi Ω

∂σij + ρbi − ρv˙ i dΩ = 0 ∂xj

(4.3.10)

As the above has to be true for any δvi , we then have ti = nj σij

(4.3.11a)

∂σij + ρbi − ρv˙ i = 0 ∂xj

(4.3.11b)

which are apparently the traction and the momentum equations.

4.4


We will next describe a physical name to each term in the above virtual power equation. Let us start with the first term of Eq.(4.3.8) ∂δvi σij = δvi,j σij ∂xj = δLij σij = (δDij + δWij )σij = δDij σij = δD : σ

(4.4.1)

Since δDij σij is the virtual internal power per unit volume, the total virtual internal power δP int is defined by integrating over the entire domain δP

int

Z := Ω

∂δvi σij dΩ ∂xj

(4.4.2)

82


Chap. 4

The second and third term in Eq. (4.3.8) are the virtual external power because they are done by the external forces. Z Z ext δP = δvi ρbi dΩ + δvi ti dΓ (4.4.3) Ω

Γt

The last term in Eq. (4.3.8) is the virtual kinetic (or inertial) power Z kin δP = δvi ρv˙ i dΩ

(4.4.4)

Ω

Inserting Eqs. (4.4.2-4.4.4) into Eq. (4.3.8), we can write the principle of virtual power as: Box 4.2 Weak formulation in the updated Lagrangian Find v ∈ U such that δP = δP int − δP ext + δP kin = 0 ∀δv ∈ U0 where

δP δP

int

ext

Z ∇δv : σdΩ

= ZΩ

Z ρb · δvdΩ +

= Ω

δP

kin

¯ t · δvdΓ

Γt

Z ρδv · vdΩ ˙

= Ω

Weak form of TL The trial and test function spaces are defined by as before U = { ui | ui ∈ C 0 (X), ui = u¯i

on Γu }

U0 = { δui | δui ∈ C 0 (X), δui = 0 on Γu }

(4.4.5) (4.4.6)

We start by multiplying the linear momemtum equation Eq.(2.7.9) by a test function δui and integrating over the initial configuration Ω0 , this gives Z ∂Pji δui + ρ0 bi − ρ0 uï dΩ0 = 0 (4.4.7) ∂Xj Ω0 The first term in the above can be written as Z Z ∂ ∂Pji ∂δui δui dΩ0 = (δui Pji ) − Pji dΩ0 (4.4.8) ∂Xj ∂Xj Ω0 Ω0 ∂Xj

Sec. 4.4


83

Using the Gauss’s theorem, we convert the first term in the RHS of the above to surface integral, Z Z Z ∂ 0 δui [[Pji nj ]]dΓ0 + δui Pji n0j dΓ0 (4.4.9) (δui Pji )dΩ0 = Ω0 ∂Xj Γ0int Γ0 The first integral of the RHS vanishes due to the continuity condition. Applying the traction boundary condition and the fact that δvi vanishes on the complement of Γ0t gives Z Z ∂ (4.4.10) (δui Pji )dΩ0 = δui t¯0i dΓ0 Ω0 ∂Xj Γ0t The second term in the RHS of Eq. (4.4.8) can be written as Z Ω0

∂δui Pji dΩ0 = ∂Xj

∂ui δ Pji dΩ0 = Ω0 ∂Xj

Z

∂xi δ Pji dΩ0 = Ω0 ∂Xj

Z

Z δFij Pji dΩ0

(4.4.11)

Ω0

where we have used ∂ui ∂ ∂xi = (xi − Xi ) = ∂Xj ∂Xj ∂Xj Substituting Eqs. (4.4.10) and (4.4.11) into Eq. (4.4.8) yields Z Z Z ∂Pji 0 ¯ dΩ0 = δui ti dΓ0 − δui δFij Pji dΩ0 ∂Xi Γ0t Ω0 Ω0

(4.4.12)

By substitution Eq. (4.4.12) into Eq. (4.4.7), we obtain after some rearrangements Z

Z δFij Pji dΩ0 −

Ω0

Γ0t

δui t¯0i dΓ0

Z −

Z δui ρ0 bi dΩ0 +

Ω0

δui ρ0 uï dΩ0 = 0

(4.4.13)

Ω0

Principle of virtual work A force F, which may be real (actual) or imaginary (fictitious), acting on a particle is said to do virtual work when the particle is imagined to undergo a real or imaginary displacement component D in the direction of the force. Thus, virtual work is the mathematical product (FD) of unrelated forces and displacements. Since forces and/or displacements need not be real nor related by cause-and-effect, the work done is called virtual work. The virtual external work denoted by δW ext is given by Z Z ext 0 ¯ δW = δui ti dΓ0 + δui ρ0 bi dΩ0 (4.4.14) Γ0t

Ω0

The virtual internal work denoted by δW int is Z int δW = Ω0

δFij Pji dΩ0

(4.4.15)

84


Finally, the virtual inertial work denoted by δW kin is Z kin δW = δui ρ0 uï dΩ0

Chap. 4

(4.4.16)

Ω0

Box 4.3 Weak formulation in the total Lagrangian Find v ∈ U such that δW = δW int − δP ext + δW kin = 0 ∀δv ∈ U0 where

δW

int

δW ext δW kin

Z

Z

δF : PdΩ0 = PT : δFdΩ0 ZΩ0 Z Ω0 ¯ t · δvdΓ0 = ρb · δvdΩ0 + Ω0 Γt Z = ρδv · vdΩ ˙ 0 =

T

Ω0

It can be seen that, using the nominal stress P and the deformation gradient tensor F as stress and strain measures, the resulting weak form is quite easy to memorize and the derivation process is rather straightforward. However, the nominal stress is not objective, thus most constitutive equations are written in terms of objective stresses such as the 2nd PK stress S. In these cases, the finite element equations developed for P can still be used: from the constitutive equation, S is computed, then it is converted to P which is in turn used to compute the internal force vector. This approach is apparently expensive due to addtional computational efforts in converting one stress measure to another. Therefore, we present in this section Z int δW = F · S : δFdΩ0 (4.4.17) Ω0

δW

int

∂w : δFdΩ0 = = F·2 ∂C Ω0 Z

Z 2 Ω0

∂w : (FT · δF)dΩ0 ∂C

(4.4.18)

where to derive the final result, indicial notation has been used with arrangement so that matrix multiplication can be done. w = κU (J) + w(C), ¯

δW

int

S=2

∂ w¯ (u, δu) = 2 : (FT · ∇0 δu)dΩ0 + ∂C Ω0 Z

∂ w¯ ∂J 0 + 2κU (J) ∂C ∂C

Z 2p Ω0

∂J : (FT · ∇0 δu)dΩ0 ∂C

(4.4.19)

(4.4.20)

Sec. 4.5

Finite element discretization

Z Ω0

4.5 4.5.1

Z ∂ 2 w¯ ∂ w¯ T 2 : δC : (F · ∇0 δu)dΩ0 + 2 : δFT · ∇0 δu dΩ0 ∂C∂C Ω0 ∂C

85

(4.4.21)

Finite element discretization Finite element approximation

The domain Ω is partitioned into a set of elements Ωe . An example of such spatial discretization is depicted in Fig. (4.1). The nodal coordinates in the current configuration are denoted by xiI , I = 1, 2, . . . nN where nN is the number of nodes in the domain. The lower case subscripts are used for components and upper case subscripts are for nodal values. In two dimensions, xI = [xI , yI ] and in three dimensions, xI = [xI , yI , zI ]. The nodal coordinates in the reference configuration are denoted by XiI . Figure 4.1: A typical two dimensional finite element discretization, showing the nodal support and a triangle element The finite element approximation for the motion is xi (X, t) =

nN X

NI (X)xiI (t),

x(X, t) = NI (X)xI (t)

(4.5.1)

I=1

where NI (X) are the shape functions. It is emphasized that the shape functions are expressed in terms of the material coordinates although we will use the weak form defined in the current configuration. This is done because we want the time dependence to be resided only in the nodal parameters. Write the above equation at the time t = 0, we obtain Xi =

nN X

NI (X)XiI ,

X = NI (X)XI

(4.5.2)

I=1

The nodal displacement uiI is given by uiI = xiI − XiI

(4.5.3)

ui (X, t) = xi − Xi = NI (X)xiI (t) − NI (X)XiI = NI (X)(xiI (t) − XiI ) = NI (X)uiI (t)

(4.5.4)

Therefore, the displacement field is

The velocity is then determined by taking the time material derivative of the displacement field, vi (X, t) = NI (X)u˙ iI (t) = NI (X)viI (t)

(4.5.5)

86


Chap. 4

Similarly, the acceleration’s approximation is given by uï (X, t) = NI (X)¨ uiI (t) = NI (X)v˙ iI (t)

(4.5.6)

Velocity gradient The velocity gradient L ∂vi ∂NI (X) = viI (t) ∂xj ∂xj

Lij =

(4.5.7)

Rate of deformation tensor The discretised rate of deformation tensor D reads 1 1 ∂vi ∂vj = (viI NI,j + vjI NI,i ) Dij = + 2 ∂xj ∂xi 2

(4.5.8)

Gradient deformation tensor Fij =

∂xi ∂NI = xiI (t) ∂Xj ∂Xj

(4.5.9)

The variation of F is therefore defined by δFij =

∂NI ∂NI δxiI = δuiI ∂Xj ∂Xj

(4.5.10)

where we have used δxiI = δ(uiI + XiI ) = δuiI . According to the Bubnov-Galerkin method, the virtual velocity is approximated by the same shape functions, so δvi (X, t) = NI (X)δviI

(4.5.11)

By substituting the discretized virtual velocity given in Eq. (4.5.11) into the virtual power equation Eq. (4.3.8), we obtain Z δviI Ω

∂NI σij dΩ − ∂xj

Z

Z NI ti dΓ −

Γt

Z NI ρbi dΩ +

Ω

NI ρv˙ i dΩ = 0

(4.5.12)

Ω

Using the arbitrariness of δviI except on Γvi gives Z Ω

∂NI σij dΩ − ∂xj

Z

Z NI ti dΓt −

Γt

Z NI ρbi dΩ +

Ω

NI ρv˙ i dΩ = 0 Ω

∀(i, I) ∈ / Γvi

(4.5.13)

Sec. 4.5

4.5.2


87

Internal nodal forces

For ease in memorizing and also for the computer implementation, each of terms in the discrete equation is given a name. Let us start with the internal virtual power δP int δP

int

∂δvi σij dΩ = δviI ∂xj

Z = Ω

∂NI σij dΩ ∂xj

Z Ω

(4.5.14)

We therefore can define the internal nodal force by fiIint

Z := Ω

∂NI σij dΩ ∂xj

(4.5.15)

Total Langrangian internal force vector We are now going to transform this internal force to the total Langrangian formulation by replacing the Cauchy stress by Kirchhoff stress and then by the nominal stress

fiIint

Z = Ω

∂NI −1 J Tji dΩ = ∂xj

Z Ω

∂NI −1 J Fjk Pki dΩ = ∂xj

Z Ω

∂NI ∂xj Pki J −1 dΩ ∂xj ∂Xk

(4.5.16)

and finaly convert the integral to the reference domain, we obtain fiIint

Z = Ω0

4.5.3

∂NI Pki dΩ0 ∂Xk

(4.5.17)

External nodal forces

The external virtual power is given by

δP

ext

Z

Z

δvi ρbi dΩ + δvi ti dΓ Γt Z Z = δviI NI ρbi dΩ + NI ti dΓt =

Ω

Ω

(4.5.18)

Γt

The terms in parentheses can therefore be defined as the external nodal force fiIext

Z

Z NI ρbi dΩ +

:= Ω

NI ti dΓt Γt

(4.5.19)

88

4.5.4


Chap. 4

Mass matrix and inertial force

The kinetic power is given by

δP

kin

Z δvi ρv˙ i dΩ Z = δviI ρNI NJ v˙ iJ dΩ =

Ω

(4.5.20)

Ω

The inertial nodal force is then defined by fiIkin

Z ρNI NJ dΩv˙ iJ

:=

(4.5.21)

Ω

We can write the above in two dimensions as follows "

# kin

fxI

kin fyI

Z  =

ρNI NJ dΩ

Ω

0

Z 0



 v˙ xJ  v˙ yJ ρNI NJ dΩ

(4.5.22)

Ω

which is rewritten in matrix form by fIkin = MIJ v˙ J

(4.5.23)

where the mass matrix is given by Z ρNI NJ dΩ

MIJ = I

(4.5.24)

Ω

This is the consistent mass matrix which is constant for Lagrangian meshes. To show this, the above is transformed to the reference configuration Z ρNI NJ JdΩ0

MIJ = I

(4.5.25)

Ω0

Next, the continuity equation ρJ = ρ0 is used to get Z ρ0 NI NJ dΩ0

MIJ = I

(4.5.26)

Ω0

It is clear that the integrand of the above integral is constant. So, the mass matrix for Lagrangian meshes is constant and therefore can be only computed at the beginning of the computation. It is noting that Eq.(4.5.26) can be considered as the total Lagrangian. We have used a total Lagrangian form for the mass matrix in an updated Lagrangian formulation.

Sec. 4.5

4.5.5


89

Discrete equations

With named terms, the discrete equations become fiIint − fiIext + fiIkin = 0 ∀(i, I) ∈ / Γvi

(4.5.27)

In the familiar matrix form, the above is written as Ma + f int = f ext

(4.5.28)

The above are the semidiscrete momentum equations since they are not discretized in time yet. For a static problem, the accelerations vanish and the discrete momentum equations become the well known discrete equilibrium equation f int = f ext

(4.5.29)

If the constitutive equations are rate-independent (elasto-plastic material for example), then the discrete equilibrium equations are a set of nonlinear algebraic equations in the stresses and nodal displacements. For rate-dependent materials, visco-plastic ones for instance, rate terms must be discretized in time to obtain a set of nonlinear algebraic equations.

4.5.6

Natural coordinates

Shape functions are expressed in terms of parent element coordinates (often called natural coordinates).

Figure 4.2: Initial and current configurations and their relation to the parent element domain 1. The parent element domain 2 2. The initial (reference) element domain Ωe0 3. The current element domain Ωe (t)

90


Chap. 4

Using the shape function expressed in terms of the natural coordinates ξ = (ξ, η, ζ), the motion is approximated by xi (ξ, t) =

m X

NI (ξ)xiI (t)

(4.5.30)

I=1

where m is the number of node of element under consideration. The mapping between the reference domain and the parent domain is obtained from Eq. (4.5.30) evaluated at time t = 0, Xi (ξ) = NI (ξ)XiI

(4.5.31)

The above shows that if this map is one-to-one, then the natural coordinates can be considered surrogate material coordinates in a Lagrangian mesh.

4.5.7

Derivatives of functions

To obtain the derivatives of shape functions with respect to the spatial coordinates x we use the chain rule, say for a quadrilateral element        ∂NI ∂NI ∂x ∂NI ∂y ∂x ∂y ∂NI +    ∂ξ   ∂x ∂ξ  ∂y ∂ξ  =   ∂ξ ∂ξ   ∂x   (4.5.32)  ∂NI   ∂NI ∂x ∂NI ∂y  =  ∂x ∂y   ∂NI  + ∂y ∂η ∂x ∂η ∂y ∂η ∂η ∂η Hence,   ∂x ∂NI  ∂x   ∂ξ     ∂NI  =  ∂x ∂y ∂η 

   ∂y −1 ∂NI   ∂ξ    ∂ξ  ∂y   ∂NI  ∂η ∂η

(4.5.33)

with   ∂x ∂y ∂N1  ∂ξ ∂ξ   ∂ξ     ∂x ∂y  =  ∂N1 ∂η ∂η ∂η In concise matrix form, the above writes 

∂N2 ∂ξ ∂N2 ∂η

∂N3 ∂ξ ∂N3 ∂η

 ∂N4 x1 x2 ∂ξ   ∂N4  x3 x4 ∂η

T T = NI,ξ F−1 NI,x ξ

where the matrix F is given by in three dimensions   x,ξ x,η x,ζ Fξ =  y,ξ y,η y,ζ  z,ξ z,η z,ζ

 y1 y2   y3  y4

(4.5.34)

(4.5.35)

(4.5.36)

Sec. 4.6

Numerical integration

91

whose components are given by (from Eq. (4.5.30)), ∂xi ∂NI = xiI (t) ∂ξj ∂ξj

(4.5.37)

Somtimes, it is convenient to write Fξ in term of dyadic product operator ⊗ as Fξ = xI ⊗ ∇ξ NI

(4.5.38)

The determinant of Fξ denoted by Jξ is the Jacobian of the transformation between the current configuration and the parent domain. The Jacobian of the transformation between the reference configuration and the parent domain Jξ0 is the determinant of the following matrix   X,ξ X,η X,ζ F0ξ =  Y,ξ Y,η Y,ζ  (4.5.39) Z,ξ Z,η Z,ζ This quantity is used in the total Lagrangian finite elements.

4.6


The integration over the current domain Ωe is transformed to the integration over the parent domain via Z Z e f (x)dΩ = f (ξ)Jξ d (4.6.1) Ωe

where Jξ is the Jacobian of the transformation and given in Eq.(4.5.36). The calculation of elementary quantities in finite element such as the stiffness matrix, the mass matrix etc. involves the integral over the element domain (surface in two dimensions and volume in three dimensions). It is very difficult, if not impossible, to compute these integrals exactly. Therefore, numerical methods must be used to calculate these integrals. This section introduces some of very often used numerical integration rules or also called quadrature rules. The common ideas in numerical integration is to replace the function to be integrated by an approximate polynomial because integral of polynomial can be exactly calculated (and easily). Considering the one dimensional integral given by Z xn I= f (x)dx (4.6.2) x1

Given a number of points x1 , x2 , . . . , xn . The function f (x) can be approximated by the Lagrange polynominal as f (x) ≈

n X i=1

lin−1 (x)f (xi )

(4.6.3)

92


Chap. 4

where the Lagrange polynomial is defined as lin−1 (x)

n−1 Y

(x − xj ) (xi − xj ) j=1,j6=i

:=

(4.6.4)

The integral in Eq.(4.6.2) is then approximately calculated by Z

xn

I≈ x1

n X

lin−1 (x)f (xi )dx

i=1

=

n Z X i=1

xn

lin−1 (x)dxf (xi )

(4.6.5)

x1

The above can be written as I≈

n X

wi f (xi )

(4.6.6)

lin−1 (x)dx

(4.6.7)

i=1

where wi , the Cotes numbers, are given by Z

xn

wi := x1

4.6.1

Newton-Cotes quadrature

The Newton-Cotes quadrature is the simplest form where the quadrature points are equi-spaced distributed along the interval [x1 , xn ] (including the end points). For example, consider the case of two points x = −1 and x = 1, we get the corresponding Cotes numbers as Z 1 Z 1 x+1 x−1 dx = 1, W2 = dx = 1 (4.6.8) W1 = −1 1 + 1 −1 −1 − 1 We get the familiar trapezoidal rule given by I ≈ f (−1) + f (1)

(4.6.9)

For three points x = 0, x = 1/2 and x = 1, we have three Cotes numbers Z

1

W1 = 0

Z

1

W2 = 0

Z W3 = 0

1

(x − 1/2)(x − 1) 1 dx = (0 − 1/2)(0 − 1) 6 2 (x − 0)(x − 1) dx = (1/2 − 0)(1/2 − 1) 3 (x − 0)(x − 1/2) 1 dx = (1 − 0)(1 − 1/2) 6

(4.6.10)

We then get the well known Simpson rule I≈

1 2 (f (0) + f (1)) + f (1/2) 6 3

(4.6.11)

Sec. 4.6


93

Newton-Cotes for high order dimensions ! Z 1 X Z 1Z 1 XX f (ξj , η)wj dη = f (ξj , ηi )wi wj f (ξ, η)dξdη = −1

4.6.2

−1

−1

j

i

(4.6.12)

j

Gauss-Legendre quadrature

In the Newton-Cotes quadrature, the quadrature points are chosen a priori and requires a lot of points which is of course much costly in multi-dimensional integrations (n + 1 points for polynominal of degree n). Z

1

f (x)dx ≈ −1

n X

wi f (xi )

(4.6.13)

i=1

where Wi and xi are the unknowns to be determined. Any polynominal of degree less than 2n can be exactly integrated by n points xi and n Cotes number wi as Z

1

P (x)dx =

n X

wi P (xi )

(4.6.14)

P (x) = Q(x)Pn (x) + R(x)

(4.6.15)

−1

i=1

To show this

where Pn (x) is polynominal of degree n while Q(x) and R(x) are polynominals of degree less than n. Z 1 Z 1 P (x)dx = [Q(x)Pn (x) + R(x)]dx (4.6.16) −1

−1

If we could find Pn (x) such that Z

1

Q(x)Pn (x)dx = 0

(4.6.17)

−1

and Pn (xi ) = 0 =⇒ P (xi ) = R(xi )

(4.6.18)

Therefore, Z

1

Z

1

P (x)dx = −1

R(x)dx = −1

n X

wi P (xi )

(4.6.19)

i=1

From Eqs.(4.6.17) and (4.6.18), one can see that we must construct the polynominal Pn (x) which is orthogonal to any polynominal of degree less than n and take the root of equations

94


Chap. 4

Pn (xi ) = 0 as the quadrature points. Such polynominals exist and called the Legendre polynominals. As example, consider the first three Legendre polynominals given by

P0 (x) = 1,

P1 (x) = x + a,

P2 (x) = x2 + bx + c

where the constants a, b and c are determinded from conditions given in Eq.(4.6.17). Z

1

(x + a)dx = 0 =⇒ a = 0 −1

 Z 1   (x2 + bx + c)dx = 0  Z−11 =⇒ b = 0,    (x2 + bx + c)xdx = 0

c = −1/3

−1

Hence, P1 = x, P2 = x√2 − 1/3. Thus, if we use two quadrature points they must be the root of P2 = 0, i.e., xi = ±1/ 3. The two corresponding Cotes numbers are given by ( see Eq.(4.6.7)) Z

1

W1 = −1

√ x − 1/ 3 √ √ dx = 1, −1/ 3 − 1/ 3

Z

1

W2 = −1

√ x + 1/ 3 √ √ dx = 1 1/ 3 + 1/ 3

The quadrature points and the associated Cotes numbers (often called weights) are tabulated in table. nQ 1 2 3

4

ξi 0√ ±1/ 3 p0 s± 3/5 p 3 − 2 6/5 ± 7 s p 3 + 2 6/5 ± 7

wi 2 1 8/9 5/9 1 1 − p 2 6 6/5

p = 2nQ − 1 1 3 5

7

1 1 − p 2 6 6/5

Table 4.1: Normalized KI values for various domain sizes.

Sec. 4.6


95

Gauss-Legendre for rectangular and right prism domains In applying this quadrature rule for numerically integrate the finite element matrices, the stiffness matrix for example, it is often written in terms of natural coordinates as Z

1

f (ξ)dξ = −1

nQ X

wQ f (ξQ )

(4.6.20)

Q=1

which can be easily extended to two dimensions by applying the above separately in each direction Z

1

Z

1

f (ξ, η)dξdη = −1

−1

nQ1 nQ2 X X

wQ1 wQ2 f (ξQ1 , ηQ2 )

(4.6.21)

Q1 =1 Q2 =1

It is convenient to write the above in terms of single weight as Z 1Z 1 X f (ξ, η)dξdη = wQ f (ξQ ) −1

−1

(4.6.22)

Q

The Gauss quadrature rule is used in evaluation the integration over the parent domain Z X f (ξ)d = w¯Q f (ξQ ) (4.6.23)

Q

where w¯Q is the weight of Gauss point Q. Z X f (x)dΩe = f (ξQ )w¯Q JξQ Ωe

(4.6.24)

Q

Selective-reduced integration (SRI) The stress and rate of deformation tensors are decomposed into istropic and deviatoric parts as σ = σ dev + σ hyd D = Ddev + Dhyd

(4.6.25)

By using the property of double contraction of two second-order tensors given in Appendix A.5, we have σ : δD = σ hyd : δDhyd + σ dev : δDdev hyd dev = σijhyd δDij + σijdev δDij dev = σiihyd δDiihyd + σijdev δDij dev = −pδDii + σijdev δDij

(4.6.26)

96

Lagrangian finite elements nQ

Chap. 4

i

ξi

ηi

wi

1 1 2 3 4

√ −1/√3 −1/√3 −1/√3 −1/√3 −1/ 3

√ −1/√3 −1/√3 −1/√3 −1/√3 −1/ 3

4 1 1 1 1

η

ξ

η

Table 4.2: Normalized KI values for various domain sizes. ξ

The second term of the last equality of the above can be written as 1 dev (4.6.27) σijdev δDij = σijdev (δDij − δDkk δij ) = σijdev δDij 3 where we have used the fact that the trace of a deviatoric tensor is zero. Now, by replacing δDij by 0.5(δvi,j + δvj,i ), we can write 1 σijdev δDij = σijdev (δvi,j + δvj,i ) 2 1 dev 1 = σij (δvi,j + δvj,i ) + (δvi,j − δvj,i ) 2 2 dev = σij δvi,j

(4.6.28)

where in the second step, we used the fact that double contraction of a symmetric tensor (the deviatoric stress tensor) with a skew symmetric tensor is zero. Therefore, we have σ : δD = −pδDii + σijdev δvij

(4.6.29)

Now, introducing the finite element approximation for the velocity yields ∂NI ∂NI δviI + σijdev δviI ∂xi ∂xj Therefore, we obtain the internal power given by Z ∂NI int dev ∂NI δP = −p δviI + σij δviI dΩ ∂xi ∂xj Ω σ : δD = −p

(4.6.30)

(4.6.31)

Sec. 4.7

Implementation

97

So Z

∂NI dev ∂NI −p = + σij dΩ (4.6.32) ∂xi ∂xj Ω Then, the basic idea of selective-reduced integration is to use reduced integration scheme for the pressure (or dilational) part and full integration for the deviatoric part. Take the four node quadrilateral elements as example, the internal force is computed as follows fiIint

fiIint

4 X ∂NI ∂NI dev = −p(0) J(0) + σij (ξQ ) J(ξQ )wQ ∂xi 0 ∂xj ξQ Q=1

(4.6.33)

The above is the similar to the SRI for linear elasticity where the stiffness matrix can be decomposed into two parts. One part is proportional to the bulk modulus (infinite for incompressible materials) and the second part. It is the the first part that makes the element stiffness infinite, hence induces finite elements lock. So, the trivial solution is to underintegrate this part. The reason why we have not performed uniformly reduced integration is related to the rank deficency of the elements. This will be explained in Chapter 6.

4.7

Implementation

In the finite element computation, all the internal nodal force, external nodal force and the mass matrix are computed for each of the elements which then are assembled to form the global quantities. Formally, the assembling process is given rigourously as Z (· · · )dΩ = Ω

ne Z [ e=1

(· · · )dΩe =

Ωe

ne Z [ e=1

(· · · )d

(4.7.1)

S ne is the number ofP elements discretizing the domain Ω. Noting that the operator has been used instead of the operator to emphasize it is a assembling processing rather than a simple sum. The following example gives a description of this process. It is taken from the book of Prof. Ted Belytchsko. Example 4.7.1 In finite element method, the global stiffness matrix and force vector are built from the contributions of every elements. Therefore, one computes the stiffness matrix and force vector for each of element and assemble them to construct the global matrix and vector. This procedure is often called assembly in litterature. More refinement, the procedure corresponding to matrix is called matrix assembly while for vectors, it is called the vector assembly or scatter operation. Inversely, it is necessary to extract the element displacements from the global displacement vector. This operation is called gather since one collect small vector from the big one. Let us denote the elementary mass matrix, stiffness matrix and force vector by Me , Ke and f e , respectively. The element displacement vector is denoted by ue . The gather operation can be written as ue = Le u;

δue = Le δu

(4.7.2)

98


Chap. 4

where the connectivity matrix Le is a Boolean matrix (consist of one and zero). Figure shows an example. The virtual external energy can be written as the sum of virtual external energy of all elements as δwext =

X

δweext

(4.7.3)

e

Equivalently, δuT f ext =

X

ext δuT = e fe

X

ext δuT LT e fe

(4.7.4)

e

e

where Eq.(4.7.2) have been used in the second equality. Since the above holds for any δu, we get the formula for scattering the element external nodal force into the global force vector X

f ext =

ext LT e fe

(4.7.5)

e

To derive the expression for matrix assembly, we use the fact that the virtual internal energy can be written as the sum of the elementary virtual internal energy. That is vT BT DBu =

X

veT BT e De Be ue =

e

X

T e v T LT e Be De Be L u

(4.7.6)

e

Since the above holds true for any v and u, we come up with BT DB =

X

T e LT e Be De Be L

(4.7.7)

e

The above shows that the contribution of elementary stiffness matrix to the global stiffness matrix is given by K=

X

LT e Ke Le

(4.7.8)

e

Similarly, the mass matrix is also assembled into the global mass matrix by the same equation M=

X

LT e Me Le

(4.7.9)

e

It is noting that, in practical finite element code, the connectivity matrix Le is never explicitly computed. In fact, the assembly is perfomred as follows. For each element, its connectivity array is built (a vector containing the number of its nodes), then a scatter vector which contains the number of degrees of freedom of the element’s nodes is constructed. This scatter vector is exactly the position of the element quantities in the global quantities.

In this section, we present the implementation of nonlinear finite elements using either Voigt notation, i.e., replacing full matrices by vectors employing the symmetry property or full matrix notation. Both approaches have their own agvantages and disadvantages. Practicers of nonlinear finite elements should be aware of both.

Sec. 4.7

4.7.1

Implementation

99

Three dimensional problems

Internal force vector For ease of reading, we present the development in two dimensions first, the generalization to three dimensions is straightforward. Using the Voigt notation, the rate of deformation tensor D, the Green strain tensor E, the 2nd PK S and the Cauchy stress tensor σ are considered as column matrices as, in two dimensions, 

 Dxx  Dyy     Dzz   {D} =  2Dyz  ,   2Dxz  2Dxy



 Exx  Eyy     Ezz   {E} =  2Eyz  ,   2Exz  2Exy



 σxx σyy     σzz   {σ} =   σyz  ,   σxz  σxy

  Sxx Syy     Szz   {S} =   Syz    Sxz  Sxy

(4.7.10)

From its definition, Dij = 1/2(vi,j + vj,i ), the rate of deformation in Voigt notation can be written by  N1,x 0 0 Dxx  Dyy   0 N1,y 0     Dzz   0 0 N1,z    2Dyz  =  0 N1,y N1,z    2Dxz  N1,x 0 N1,z N1,y N1,x 0 2Dxy 



··· ··· ··· ··· ··· ···

Nm,x 0 0 Nm,y 0 0 0 Nm,y Nm,x 0 Nm,y Nm,x

   vx1 0  vy1    0   vz1    N1,z  ..    .  Nm,z      Nm,z  vxm  vym  0 vzm

(4.7.11)

where m denotes the number of nodes per element, hence {D} = Bv

(4.7.12)

where B is the so called discrete symmetric spatial gradient matrix which is written according to B = {B1 , B2 , . . . , Bm }

(4.7.13)

with nodal component BI given by 

 NI,x 0 0  0 NI,y 0     0 0 NI,z    BI =   0 N N I,y I,z   NI,x 0 NI,x  NI,y NI,x 0

(4.7.14)

100


Chap. 4

and v is the element velocity vector 

 v1  v2   v= · · · , vm

  vxI  vI = vyI  vzI

(4.7.15)

The element displacement vector has the same form. The virtual internal power is now given by δP

int

Z

Z δD : σdΩ =

= ZΩ =

{δD}T {σ} dΩ

Ω

(4.7.16)

δvT BT {σ} dΩ

Ω

where in the second equality, we have used Eq.(4.7.12). Therefore, the internal nodal force vector of one element is given by feint

Z =

BT {σ} dΩ

(4.7.17)

Ω

The above is the most important equation in nonlinear finite element program. It is used for both material and geometry nonlinear problems. Noting that it is identical to the one used in the linear finite elements. Therefore, the updated Lagrangian formulation is quite easy to implement since one just needs the deformed nodal coordinates 1 to evaluate the derivatives of shape functions. Then use the available routines to construct the matrix B. UL internal force vector using full matrix Let us first recall the expression of the internal force given in Eq.(4.5.15) Z ∂NI int σij dΩ fiI = Ω ∂xj which can be written explicitly in two dimensions as follows   " int # Z ∂NI σxx + ∂NI σxy fxI  ∂x  ∂y   dΩ =  ∂NI  int ∂N fyI I Ω σyx + σyy ∂x ∂y or, equivalently, after arrangement # " Z int int ∂NI ∂NI σxx σxy fxI fyI = dΩ ∂x ∂y σyx σyy Ω 1

simply the sum of the initial coordinates plus the displacements

(4.7.18)

(4.7.19)

(4.7.20)

Sec. 4.7

Implementation

101

which can be extended to three dimensions straightforwardly as

int int int fxI fyI fzI =

Z Ω

  σ σ σ xx xy xz  ∂NI  σyx σyy σyz  dΩ   ∂z σzx σzy σzz

∂NI ∂y

∂NI ∂x

(4.7.21)

Putting the nodal internal force together, we obtain the expression of the element internal nodal force matrix written as [feint ]

Z

BT [σ]dΩ

=

(4.7.22)

Ω

Noting that, the element internal force is now a matrix. In the above, B is the so-called discrete (full) spatial gradient matrix, B = [B1

T BI = NI,x NI,y NI,z

B2 · · · ],

(4.7.23)

Total Lagrangian internal force vector Using the transformation P = S · FT , the internal nodal force in total Lagrangian formulation (see Eq.(4.5.17)) is now written as 2 fiIint

Z = Ω0

∂NI Sjk Fik dΩ0 ∂Xj

(4.7.24)

To transform the above written in indicial notation to matrix form employing the Voigt notation for the 2nd PK stress, we firstly write it explicitly in two dimensions as

"

# int

fxI

int fyI

 ∂N ∂x I  =  ∂X ∂X ∂NI ∂y ∂X ∂X

  ∂NI ∂x ∂NI ∂x  Sxx +  ∂X ∂Y ∂Y ∂X   Syy     ∂NI ∂y ∂NI ∂y + Sxy ∂X ∂Y ∂Y ∂X

∂NI ∂x ∂Y ∂Y ∂NI ∂y ∂Y ∂Y

(4.7.25)

which can be written in matrix form as fIint

Z =

BT 0I {S}dΩ0

(4.7.26)

Ω0 2

Since we are using the Voigt notation, we need to use a symmetric stress measure thing is apparently not held for the nominal stress.

102


with the matrix B0I in two dimensions given by  ∂NI ∂x  ∂X ∂X   ∂N I ∂x B0I =   ∂Y ∂Y   ∂N ∂x ∂NI ∂x I + ∂X ∂Y ∂Y ∂X which can be extended to three dimensions 

NI,X F11

Chap. 4

 ∂NI ∂y  ∂X ∂X   ∂NI ∂y   ∂Y ∂Y  ∂NI ∂y ∂NI ∂y  + ∂X ∂Y ∂Y ∂X

NI,X F21

NI,X F31

(4.7.27)



    NI,Y F12 NI,Y F22 NI,Y F32     NI,Z F13 NI,Z F23 NI,Z F33   0  BI =  N F + N F   I,Y 13 I,Z 12 NI,Y F23 + NI,Z F22 NI,Y F33 + NI,Z F32    N F + N F  N F + N F N F + N F I,Y 13 I,Z 12 I,Y 23 I,Z 22 I,Y 33 I,Z 32   NI,Y F13 + NI,Z F12 NI,Y F23 + NI,Z F22 NI,Y F33 + NI,Z F32

(4.7.28)

The element internal force vector reads feint

Z =

BT 0 {S}dΩ0

(4.7.29)

Ω0

where the element matrix B0 is constructed by putting the nodal matrices B0I one by one. It is obvious that this B0 matrix is far more complicated than the BI used in the updated Langrangian elements. However, the structure of the both total and updated formulations are similar. ˙ to the In the manner similar to the fact that B relates the rate of deformation tensor D 0 velocity, it can be shown that the above matrix B relates the material time derivative of the Green strain tensor E to the nodal velocities as follows 3 1 E˙ ij = (F˙ ki Fkj + Fki F˙ kj ) 2 1 ∂ u˙ k ∂xk ∂xk ∂ u˙ k = + 2 ∂Xi ∂Xj ∂Xi ∂Xj 1 ∂NI ∂xk ∂xk ∂NI + u˙ kI = 2 ∂Xi ∂Xj ∂Xi ∂Xj

(4.7.30)

The above, written in the Voigt notation, reads ˙ = B0I u˙ I {E} 3

Noting that the 2nd PK stress tensor is work conjugate to the Green strain tensor E.

(4.7.31)

Sec. 4.7

Implementation

103

Updated Lagrangian internal force from TL one We are showing that one can obtain the updated Lagrangian internal force vector given in Eq.(4.7.17) from the total Lagrangian internal force given in Eq.(4.7.29) without any transformation but a helpful trick. Now imagine that the current configuration is the reference configuration, hence we have J = 1,

F = I,

S = σ,

Ω = Ω0 ,

B = B0

(4.7.32)

where the last equality can be obtained from Eq.(4.7.27) with x = X since the two coordinate system now coincide. With this observation, Eq.(4.7.29) becomes Z int BT {σ}dΩ (4.7.33) fe = Ω

which is identical to the one given in Eq.(4.7.17). We will this trick again in converting the total Lagrangian tangent stiffness matrices to the updated Lagrangian counterparts. TL internal force using full matrix ∂NI Pji dΩ Ω ∂Xj # " Z P P xx xy ∂NI ∂NI int fyI = dΩ ∂Y Pyx Pyy Ω ∂X fiIint

int fxI

Z

=

(4.7.34)

(4.7.35)

Therefore, [feint ]

Z =

BT 0 PdΩ0

(4.7.36)

Ω0

with B0 = [B01

B02 · · · ],

B0I = NI,X NI,Y

NI,Z

T

(4.7.37)

It is emphasized that the matrix B0 is far easier to construct than the matrix B0 used in the total Langrangian and Voigt notation. However, it comes with the expense of using the full stress matrix P and the element internal force is now a matrix not a vector as in Voigt notation. External force vector The external nodal force vector is simpler and given by f

ext

Z

T

Z

ρN bdΩ +

= Ω

Γt

NT¯ tdΓt

(4.7.38)

104


Chap. 4

Mass matrix Z

e

M = ρ0

NT NdΩ0

(4.7.39)

Ω0

where the shape function matrix N is given, say for three-node triangle elements

N1 0 N2 0 N3 0 N= 0 N1 0 N2 0 N3

(4.7.40)

Generalization to higher dimensions or other elements is straightforward. Rate of deformation D The rate of deformation vector can be computed as, see Eq.(4.7.12) {D} = Bv

(4.7.41)

where v is the so-called element velocity vector given in Eq.(4.7.15). An alternative is via the relationship with the velocity gradient L by 1 L + LT , 2 with the element velocity matrix given by D=

L = ve BT

 vx1 vx2 · · · e  v = vy1 vy2 · · · vz1 vz2 · · ·

(4.7.42)

 vxm vym  vzm

(4.7.43)

and B is given in Eq.(4.7.37). For the derivation of the formula of the velocity gradient, see the following example. Example 4.7.2 Development of the formula of the velocity gradient The velocity gradient is given by Lij =

∂vi ∂NI = viI ∂xj ∂xj

Writing the above in matrix form, for two dimension case, we have ∂vx  ∂x   ∂vy ∂x 

  ∂vx ∂NI vxI  ∂y   =  ∂x ∂vy   ∂NI vyI ∂x ∂y

 ∂NI " # vxI v  ∂y  = xI ∂NI  ∂x ∂NI vyI vyI ∂y

∂NI ∂y

:= ve BT

Sec. 4.7

Implementation

105

Gradien deformation F From Eq.(4.5.9), the 3 × 3 gradient deformation matrix is then given by 4 F e = x e BT 0 where the element coordinate matrix is given by  x1 x2 · · · e  x = y1 y2 · · · z1 z2 · · ·

(4.7.44)

 xm ym  zm

(4.7.45)

Having F, one can easily compute different strain measures necessary such as C, E or b via the following formulas 1 E = (C − I), b = FFT 2 Another way to compute the deformation gradient is via Eq.(2.2.22) C = FT F,

F=I+H

(4.7.46)

(4.7.47)

with 5 H = ue BT 0,

 ux1 ux2 · · · ue = uy1 uy2 · · · uz1 uz2 · · ·

 uxm uym  uzm

(4.7.48)

ue is often said to be the element displacement matrix. Then the Green strain is given by 1 H + HT + HHT (4.7.49) 2 It is recommended that the the Green strain should be computed from H by the above rather than by direct calculation with F via Eq.(4.7.46). Noting that, the Jacobian of the deformation gradient is now given by J = det(F). E=

˙ = ∆E , E ∆t

4.7.2

∆F F˙ = ∆t

(4.7.50)

Two dimensional problems

In this section, the formulation presented so far is specialized to the following two dimensional problems 1. Plane strain 2. Plane stress 4 5

The derivation follows the same line as done for the velocity gradient The derivation follows the same line as done for the velocity gradient

106


Chap. 4

3. Axisymmetric In what follows, we first give implementation aspects for plane strain and plane stress problem. Let us denote the deformation plane by the plane x, y. All components related to the z coordinate vanish and the product Dzz σzz is always zero in either plane strain or plane stress conditions, so the zz components do not appear in the formulation. Therefore, we have 

 Dx {D} =  Dy  , 2Dxy



 Ex {E} =  Ey  , 2Exy



 σxx {σ} = σyy  , σxy



 Sxx {S} = Syy  Sxy

(4.7.51)

The discrete gradient matrix reads   NI,x 0 BI =  0 NI,y  , NI,y NI,x

NI,x BI = NI,y

(4.7.52)

The numerical integration of element quantities is computed using the standard Gauss quadrature Z X f (x)dΩe = a f (ξQ )w¯Q JξQ (4.7.53) Ωe

Q

where a = 1 for plane strain and for plane stress, a is the thickness of the element. Plane strain F11 F12 F= , F21 F22

J = F11 F22 − F12 F21

(4.7.54)

Plane stress The implementation presented so far works for three dimensions and for plane-strain situation, the only simplification to do is to ignore components associated to the z direction. For plane stress problems, thing is more complicated since we have additional constraint namely σ33 = 0. To take this into account, we have to do two things. Firstly, the constitutive equation has to be modified 6 . Secondly, compute the z component of the deformation gradient F. For the Lagrangian formulation, the rate form of the constitutive equation is partioned into in-plane components and zz component as follows " # " #" # dSm Cmm Cm3 dEm = (4.7.55) dS33 C3m C33 dE33 6

we have already done this for neo-Hookean material in Chapter 3, but in this section, we present a more general approach

Sec. 4.7

Implementation

107

which allows dS33 to be written as dS33 = C3m dEm + C33 dE33

(4.7.56)

The plane stress condition dS33 = 0 gives dE33 = −C−1 33 C3m dEm

(4.7.57)

From the first equation of Eq.(4.7.55) and Eq.(4.7.57), we obtain dSm = Cmm dEm + Cm3 dE33 = Cmm − Cm3 C−1 33 C3m dEm

(4.7.58)

To give an insight how the partionned tangent modulii matrix, we give in matrix form as follows         C11 C12 C16 C13 11 σ11     C23   σ22   C12 C22 C26    22   =  (4.7.59)  C C C C  σ12   16 26 66 36   12  33 σ33 C13 C23 C36 C33 For plane stress, the deformation gradient is given by   F11 F12 0 F = F21 F22 0  , J = (F11 F22 − F12 F21 )F33 0 0 F33

(4.7.60)

where F33 can be computed from E33 by the following equation 1 2 F33 − 1 (4.7.61) 2 where E33 is computed from the condition that S33 = 0. Since the relation between them is nonlinear, we use the Newton-Raphson to solve this equation. That is E33 =

(i)

(i+1)

S33

(i)

= S33 +

∂S33

(i) ∂E33

(i)

∆E33 = 0

(4.7.62)

where superscript denotes the number of iteration. So, we compute E33 by (i+1)

E33

(i)

∆E33 until convergence is attained.

(i)

(i)

= E33 + ∆E33 −1 (i) (i) = − C33 S33

(4.7.63)

108


Chap. 4

Plane stress in current configuration In this case, we consider the Cauchy stress and rate of deformation D as stress and strain measures. Recall the definition of D for ease of reading dDij = Fki−1 Flj−1 dEkl

(4.7.64)

dF33 dE33 = 2 F33 F33

(4.7.65)

Hence, dD33 can be written as dD33 =

where in the second equality, Eq.(4.7.61) has been used. The local iteration process for the Cauchy stress is given by

(i) σ33

+

σD C33

(i)

(i)

dD33 = 0

(4.7.66)

Substituting of Eq.(4.7.65) into the above gives (i) σ33

+

(i) σD (i) dF33 (C33 ) (i) F33

=0

(4.7.67)

(i)

which allows to solve for dF33

−1

(i)

σD dF33 = − C33

(i)

(i)

(i)

σ33 F33

(4.7.68)

(i)

(4.7.69)

and then update F33 as usual (i+1)

F33

(i)

= F33 + dF33

until convergence is obtained. Axisymmetric solid The strain vector consists of four components and is given by       r   z     {} =   θ  =    rz   ∂u ∂z

∂u ∂r ∂v ∂z u r +

∂NI   ∂r       0   =   NI       r ∂v   ∂N 



I

∂r

∂z



0   ∂NI   ∂z   uI := BI uI  0    ∂NI  ∂r

(4.7.70)

Sec. 4.7

Implementation

Figure 4.3: Non-zero strain components in solid of revolution

Figure 4.4: Non-zero strain components in solid of revolution

109

110


Chap. 4

where θ has been derived by using the definition of normal strain θ =

2π(r + u) − 2πr u = 2πr r

(4.7.71)

The volume integral is related to the area integral via the following Z

Z f (u, r)dV = 2π

V

rf (u, r)dΩe

(4.7.72)

Ωe

The element stiffness matrix is then given according to

KeIJ

Z = 2π

rBT I (r)DBJ (r)dΩe = 2π

Ωe

X

r(ξQ )BT I (ξQ )DBJ (ξQ )wQ J(ξQ )

(4.7.73)

Q

with r(ξQ ) is computed by finite element approximation of the geometry of the element, r(ξQ ) = NI (ξQ )rI

(4.7.74)

external force 2πrFr ,

2π

(4.7.75)

Axisymmetric finite deformation

4.7.3

Computation of external force vector

In this section, the computation of the external force vector f ext attributed to the traction ¯t is presented for both two and three dimensions. Let us first recall that f ext is given by f

ext

Z =

NT¯tdΓ

(4.7.76)

Γ

where NT is the matrix of shape functions of the continuum elements and Γ is the boundary of the element.

Two dimensional problems Considering the case shown in Fig. 4.5 where an uniform traction is applied on the right edge of a bilinear quadrilateral element. For this particular case, the external force vector of this

Sec. 4.7

Implementation

111

Figure 4.5: Uniform traction applied on a bilinear quadrilateral element. element is given by 



1  4 (1 − ξ)(1 − η)   0   1  (1 + ξ)(1 − η) 4 Z   0  = 1 Γ  (1 + ξ)(1 + η)  4   0  1  (1 − ξ)(1 + η) 4  0

0

  1 (1 − ξ)(1 − η)  4    0   1 (1 + ξ)(1 − η)  t¯x 4 f ext  ¯ dΓ  ty 0   1  (1 + ξ)(1 + η)  4   0   1 (1 − ξ)(1 + η) 4 Since over this right edge, ξ = 1, the above simplifies to  ext    fx1 0 0  ext    0 0 fy1      1    ext    (1 − η) 0 fx2   2    1  ext  Z +1   0 (1 − η) f   y2  t¯x   2 f ext =   ¯ dΓ 1 f ext  =  ty −1  (1 + η)  x3  0     2   f ext  1    y3   0 (1 + η)     2 f ext     x4  0 0 ext fy4

0

(4.7.77)

(4.7.78)

0

One important observation from the above calculation is that the external force are nonzero only at nodes belonging to the edge on which the traction is applied. The second one is that

112


Chap. 4

the shape functions N2 and N3 along the edge 23 are identically the shape functions of a twonode line element. These observations lead to the following implementation of how to compute the external force vector. For every edges on the Newmann boundary, the so-called boundary elements are added which are most of the case either two-node line elements or three-node line elements. Figure 4.6 illustrates this concept. In the considered example, we add a two-node line element connecting nodes 5 and 7, then the external force vector of this boundary element is given by 1   ext  fx5 (1 − ξ)t¯x 2   ext  Z  1  2 (1 − ξ)t¯y  fy5    dΓ57   1 f ext  = ¯x  Γ57  2 (1 + ξ)t   x7  1 ext (1 + ξ)t¯y fy7

(4.7.79)

2

The edge 57 (a curve) is given parametrically by 1 1 1 1 x = (1 − ξ)x5 + (1 + ξ)x7 , y = (1 − ξ)y5 + (1 + ξ)y7 2 2 2 2 Hence the above integral can be converted to

(4.7.80)

1  ext   fx5 (1 − ξ)t¯x 2  ext  Z +1  1   2 (1 − ξ)t¯y  q fy5  2 2      1 (1 + ξ)t¯  x,ξ + y,ξ dξ f ext  = −1 2  x7  x 1 ext fy7 (1 + ξ)t¯y

(4.7.81)

x5 y 5 x,ξ = N1,ξ N2,ξ y,ξ x7 y 7

(4.7.82)

2

where

The above integral is then evaluated via the Gauss quadrature and scattered (assembled) to the total external force vector. The process is done for other boundary elements. Advantages of this introduction of boundary elements in computing the external force vector is two fold: it separates the computation of external and internal force vectors and it serves as a general implementation. Three dimensional problems Along the same line of reasoning, to compute the external force vector due to surface traction applied on surfaces of three dimensional finite elements, 2D boundary elements (three-node triangle element, bilinear quadrilateral elements or any high order 2D continuum elements depends on the kind of the 3D continuum element) are added along the Newmann boundary. Then the transformation from the surface area to natural surface area is given by dA = ||x,ξ × x,η || dξdη

(4.7.83)

Sec. 4.7

Implementation

113

18

21

18

2

12

15 15 1

3

10 10

Figure 4.6: Boundary elements for evaluation of external force vector.

Box 4.4 Flowchart for evaluation of 2D external force vector 1. Loop over 1D boundary elements (a) Initialization f = 0 (b) Get the coordinate matrix of this element (c) Get the dof numbers of the nodes of this element (d) For Gauss point ξi , do i. ii. iii. iv.

Compute the shape functions N(ξi ) and its derivatives N,ξ (ξi ) Compute the Jacobian using Eq.(4.7.82) Compute the norm of the Jacobian, denoted by det J(ξi ) Add contribution of this Gauss point to f f = f + N(ξi )tx det J(ξi )wi

(e) Scatter f to the global external force vector using the dof numbers 2. End

114


Chap. 4

with x = NI (ξ, η)xI ,

y = NI (ξ, η)yI ,

z = NI (ξ, η)zI

(4.7.84)

where NI (ξ, η) are the shape functions of the 2D continuum elements being used and xI , yI , zI are the coordinates of the node I. In the above the symbol ||•|| denotes the length of vector •. The above cross product is computed according to i j k x,ξ × x,η = det NI,ξ xI NI,ξ yI NI,ξ zI NI,η xI NI,η yI NI,η zI

  N y N z − N y N z I,ξ I I,η I I,η I I,ξ I = NI,ξ xI NI,η zI − NI,ξ zI NI,η xI  NI,ξ xI NI,η yI − NI,ξ yI NI,η xI

(4.7.85)

where we have used Eq.(4.7.84). The external force vector is then computed as

f ext

  NI,ξ yI NI,η zI − NI,η yI NI,ξ zI ¯ NI,ξ xI NI,η zI − NI,ξ zI NI,η xI  dξdη ¯ dΩ = NT p NT p = Ω NI,ξ xI NI,η yI − NI,ξ yI NI,η xI Z

Z

(4.7.86)

where N is the 2D boundary element shape function matrix. Numerical integration is then adopted as usual to evaluate the above integral.

4.7.4

Examples on some 2D and 3D isoparametric elements

Example 4.7.3 In this example, 2D isoparametric elements in the updated Lagrangian formulation is given via the four-node quadrilateral isoparametric element. The full symmetric Eulerian gradient matrix B with dimension 3 × 8 is given by   N1,x 0 N2,x 0 N3,x 0 N4,x 0    0 N 0 N 0 N 0 N B= (4.7.87) 1,y 2,y 3,y 4,y   N1,y N1,x N2,y N2,x N3,y N3,x N4,y N4,x where the spatial derivatives of the shape functions is computed using the procedure detailed in Section 4.5.7. The element internal force vector is then written as

feint

 N1,x  0  N Z  2,x  0  =  Ωe N3,x  0  N4,x 0

0 N1,y 0 N2,y 0 N3,y 0 N4,y

 N1,y N1,x     N2,y   σxx  N2,x    σyy dΩe N3,y   σxy N3,x   N4,y  N4,x

(4.7.88)

Sec. 4.7

Implementation

115

There are 24 multiplications for each Gauss point in evaluation the internal force vector. If the full matrix notation is used, then the element internal force matrix is written as Z int B[σ]dΩ (4.7.89) [fe ] = Ω

where the discrete full spatial gradient matrix B , which is a 2 × 4 matrix, is given by   ∂N1 ∂N2 ∂N3 ∂N4  ∂x ∂x ∂x ∂x    B =  ∂N1 ∂N2 ∂N3 ∂N4    ∂y ∂y ∂y ∂y

(4.7.90)

Therefore, the element internal matrix is computed according to  fx1 fx2  fx3 fx4

int  fy1 N1,x Z N2,x fy2   =   fy3  Ωe N3,x fy4 e N4,x

 N1,y N2,y   σxx σxy dΩe N3,y  σxy σyy N4,y

(4.7.91)

There are 16 multiplications for each Gauss point in evaluation the internal force matrix. However, this internal force matrix must be transformed to vector form in computing the residual. It is also emphasized that, the geometry tangent stiffness matrix must be computed using the full form of the stress matrix. The in-plane deformation gradient is given by   N N 1,x 1,y  u u2 u3 u4  1 0 N2,x N2,y  + 1 (4.7.92) F2×2 =  v1 v2 v3 v4 N3,x N3,y  0 1 N4,x N4,y Then, the total deformation gradient is exactly this in-plane tensor for plane strain case. For plane stress, one needs to compute one more term, the z component. That is   F2×2 plane strain F2×2 0 F= (4.7.93) plane stress  0 F33

Example 4.7.4 This example gives implementation details of the three-node triangle element which is also known as constant strain triangle element (CST).

116


     x x1 x2 x3 ξ1 y  =  y1 y2 y3  ξ2  1 1 1 1 ξ3

Chap. 4

(4.7.94)

The above can be inverted explicitly which gives the area coordinates in terms of the Cartesian coordinates:      y23 x32 x2 y3 − x3 y2 x ξ1 ξ2  = 1 y31 x13 x3 y1 − x1 y3  y  (4.7.95) 2A 1 y12 x21 x1 y2 − x2 y1 ξ3 where xIJ = xI − xJ ,

yIJ = yI − yJ

2A = x32 y12 − x12 y32 A is the element area. The spatial derivatives of the shape functions are then written as     ξ1,x ξ1,y y23 x32 ξ2,x ξ2,y  = 1 y31 x13  2A ξ3,x ξ3,y y12 x21 And thus, the discrete spatial symmetric gradient matrix is   y23 0 y31 0 y12 0 1  0 x32 0 x13 0 x21  B= 2A x32 y23 x13 y31 x21 y12 Hence, the element internal force vector is given by  y23 0  0 x32  Z 1  int y31 0 f =  x13 Ω 2A  0 y12 0 0 x21

 x32   y23   σxx x13   σyy  dΩ y31   σ x21  xy y12

(4.7.96) (4.7.97)

(4.7.98)

(4.7.99)

(4.7.100)

Sec. 4.7

Implementation

117

If the stress is constant within the element, then the integration is equal to the multiplication of the integrand with the element area, so   y23 σxx + x32 σxy x32 σyy + y23 σxy    a y31 σxx + x13 σxy  int   (4.7.101) f =  2 x13 σyy + y31 σxy   y12 σxx + x21 σxy  x21 σyy + y12 σxy where a is the element thickness in case of plane stress. From the above, it can be verified that the sum of components of f int is zero, thus the element is in equilibrium. If the full matrix notation is adopted, the element internal matrix is then given by  int     fx1 fy1 N1,x N1,y y23 x32 fx2 fy2  = aA N2,x N2,y  σxx σxy = a y31 x13  σxx σxy σxy σyy σxy σyy 2 fx3 fy3 e N3,x N3,y y12 x21

(4.7.102)

which involves fewer multiplications in the above than in Eq.(4.7.100) where Voigt notation has been used.

Example 4.7.5 Four-node tetrahedra element    1 1 x x1  = y   y1 z1 z

  1 1 1 ξ1   x2 x3 x4  ξ2   y2 y3 y4  ξ3  z 2 z3 z4 ξ4

(4.7.103)

Example 4.7.6 This example presents the implementation of three dimensional isoparametric solid elements via the special case of 8-node trilinear hexahedron element.

118


Chap. 4

The 8-node trilinear hexahedron element The nodal internal force vector is given by using the full notation   Z σxx σxy σxz NI,x NI,y NI,z σxy σyy σyz  dΩe (fIint )T = Ωe σxz σyz σzz

4.8

(4.7.104)

Computation of the internal nodal force

The internal nodal force vector together with the tangent stiffness matrix are the most important quantities in nonlinear finite element programs. In this section, the computation of the former is presented for a generic element. The total internal nodal force vector is then formed by assembling the elemental values. Consider an element with m nodes. The integration point has coordinate ξQ and weight wQ . The procedure for computation of the internal nodal force for this element is summarized in Box (4.5) for the updated Lagrangian formulation and in Box (4.6) for the total Lagrangian formulation. Box 4.5 Computation of the internal nodal force-Updated Langrange 1. feint = 0 2. For all integration point ξQ (a) Compute matrix B (b) Compute the rate of deformation {D} = Bv (c) If needed, compute matrix F and E (d) Compute stress S or σ from the constitutive equation ∗ (e) If S was computed, then σ = J −1 F · S · FT , J = det(F) (f) feint = feint + BT {σ} JξQ wQ ∗

Using stress update algorithms presented in Chapter 3

Sec. 4.8

Computation of the internal nodal force

Box 4.6 Computation of the internal nodal force-Total Langrange 1. {f int } = 0 2. For all integration point ξQ (a) Compute matrix B0 (b) Compute the rate of deformation {D} = Bv (c) If needed, compute matrix F and E (d) Compute stress S or σ from the constitutive equation (e) If σ was computed, then P = JF−1 · σ, J = det(F) (f) If S was computed, then P = FST 0 (g) {f int } = {f int } + BT 0 PJξQ wQ

119

120


Chap. 4

Chapter 5 Solution procedures 5.1

Introduction

The discrete equations obtained from the finite element discretization as have been shown in Chapter 4 are second order ordinary differential equations (ODE) which can generally be solved by either explicit time integration or implicit time integration methods. Roughly speaking, in the explicit time integration methods, no system of equations needs to be solved, thus it is very efficient. However, there is a restriction on the time step being used. In the other hands, the implicit time integration methods, say the well known Newmark method, require the solution of a system of equations but there exists schemes with which no restriction on the time step is needed. They are the so-called unconditionally stable time integrators. In case that implicit methods are used, the resulting algebraic equations are generally nonlinear, thus iterative solution methods such as the classic Newton-Raphson method must be employed.

5.2 5.2.1

Explicit dynamics Central difference method

The semi-discrete equations of motion at time step n are n n := f n Mn an = fext − fint

(5.2.1)

where M is the diagonal mass matrix, fext is the external nodal force vector and fint is the internal nodal force vector. Superscript n is used to denote time step n. We are seeking the solution at time step n + 1. Since it is necessary to have variable time step, we introduce the following notations for time steps and time increments

∆tn+1/2 = tn+1 − tn ,

1 tn+1/2 = (tn+1 + tn ), 2

∆tn = tn+1/2 − tn−1/2

(5.2.2)

122

Solution procedures

Chap. 5

The velocity and the acceleration are given according to the central diffenrece formula as dn+1 − dn ∆tn+1/2 vn+1/2 − vn−1/2 an = ∆tn

vn+1/2 =

(5.2.3)

which indicates that the velocity is defined at the midpoint of the time interval. To have a better insight into the name of the method, firsy substituting Eq.(5.2.3)1 and its variant for vn−1/2 in Eq.(5.2.3)2 yields 1 a = ∆tn n

dn+1 − dn dn − dn−1 − ∆tn+1/2 ∆tn−1/2

(5.2.4)

which reduces to the well known central difference formula for second derivative of a function a=

dn+1 − 2dn + dn−1 ∆t2

(5.2.5)

The update of the velocity is as follows. From Eq.(5.2.3)2 and Eq.(5.2.1), we have n vn+1/2 = vn−1/2 + ∆tn M−1 n f

(5.2.6)

Having this, we can compute the displacement at time step n + 1 by introducing the above into Eq.(5.2.3)1 dn+1 = dn + ∆tn+1/2 vn+1/2

(5.2.7)

In the actual implementation, we perform like this vn+1/2 = vn−1/2 + an (tn+1/2 − tn−1/2 ) = vn−1/2 + an (tn − tn + tn+1/2 − tn−1/2 ) =v

n−1/2

n

n

+ a (t − t

vn = vn−1/2 + an (tn − tn−1/2 ),

5.2.2

n−1/2

n

n+1/2

) + a (t

(5.2.8) n

−t )

vn+1/2 = vn + an (tn+1/2 − tn )

(5.2.9)

Implementation

The explicit time integration scheme is flowcharted in the Box 5.1 for ease of computer implementation.

Sec. 5.3

Explicit dynamics

123

Box 5.1 Flowchart for explicit time integration 1. Initialization: d0 = 0, n = 0, t = 0 2. Compute the diagonal mass matrix M and store it 3. Compute internal, external nodal forces f n and time step ∆t ∗ 4. Compute the acceleration an , an = M−1 f n 5. Update time: tn+1 = tn + ∆t, tn+1/2 = 1/2(tn+1 + tn ) 6. First partial velocity update, vn+1/2 = vn + (tn+1/2 − tn )an n+1/2

7. Enforce velocity boundary conditions, viI

= v¯iI (xI , tn+1/2 )

8. Update nodal displacements: dn+1 = dn + ∆tn+1/2 vn+1/2 9. Compute internal, external nodal forces ∗ 10. Compute the acceleration an+1 , an+1 = M−1 f n+1 11. Second partial velocity update using Eq.(5.2.9)1 , vn+1 = vn+1/2 + (tn+1 − tn+1/2 )an+1 12. Check energy balance at time step n + 1 13. Update counter n = n + 1 and goto 5 if simulation not complete ∗

Compute time step 1. Initialization, ∆tcrit = ∞ 2. Loop over all elements i. Compute element time step ∆tecrit ii. Update the time step, ∆tcrit = min(∆tecrit , ∆tcrit ) 3. ∆t = α∆tcrit

124

5.2.3

5.3 5.3.1

Solution procedures

Chap. 5

Computation of diagonal mass matrix

Implicit dynamics The Newmark time integration method

In this section, the implicit Newmark time integration scheme is chosen to solve the semidiscrete equation given in Eq.(4.5.28). To this end, we define the following residual r(dn+1 , tn+1 ) = sD Man+1 + f int (dn+1 , tn+1 ) − f ext (dn+1 , tn+1 ) = 0

(5.3.1)

where sD given by sD =

0 for a static problem 1 for a dynamic problem

(5.3.2)

Defining the predictors as 2 ˜ n+1 = dn + ∆tvn + (∆t) (1 − 2β)an d 2 n+1 n v ˜ = v + (1 − γ)∆tan

(5.3.3)

Then, the displacement and velocity at step n + 1 are computed by ˜ n+1 + β(∆t)2 an+1 dn+1 = d vn+1 = v ñ+1 + γ∆tan+1

(5.3.4)

The updated acceleration can be computed using Eq. (5.3.4) as an+1 =

1 ˜ n+1 ) β > 0 (dn+1 − d 2 β∆t

(5.3.5)

Substituting Eq. (5.3.5) into Eq. (5.3.1) gives

r=

sD ˜ n+1 ) + f int (dn+1 , tn+1 ) − f ext (dn+1 , tn+1 ) = 0 M(dn+1 − d β∆t2

(5.3.6)

The above are nonlinear algebraic system of equations in the nodal displacement dn+1 which can be solved efficiently by the iterative Newton-Raphson method. After solution of this equation, we can compute the acceleration an+1 by Eq. (5.3.5). Finally, the velocity vn+1 is determined through Eq. (5.3.4).

Sec. 5.3

Implicit dynamics

125

Example 5.3.1 This example shows one derivation of the Newmark integration scheme. To this end, applying the Taylor series expanstion, we can write ∆t2 ∆t3 u ¨t + u ¯t + · · · (5.3.7) 2 6 where the bar notation denotes the differentiation three times with respect to time. Differentiating the above with respect to time yields ut+∆t = ut + ∆tu˙ t +

u˙ t+∆t = u˙ t + ∆t¨ ut +

∆t2 u ¯t + · · · 2

(5.3.8)

The above two equations can be rewritten as ∆t2 u ¨t + β∆t3 u ¯t 2 = u˙ t + ∆t¨ ut + γ∆t2 u ¯t

ut+∆t = ut + ∆tu˙ t + u˙ t+∆t

(5.3.9)

The acceleration is assumed to be linear within a time sep, hence we can write u ¨t+∆t − u ¨t ∆t Substitutiting of the above into Eq.(5.3.9) gives u ¯t =

1 − β ∆t2 u ¨t + β∆t2 u ¨t+∆t 2 = u˙ t + (1 − γ)∆t¨ ut + γ∆t¨ ut+∆t

(5.3.10)

ut+∆t = ut + ∆tu˙ t + u˙ t+∆t

(5.3.11)

By defining the predictors as

u ˜t+∆t v˜t+∆t

1 = ut + ∆tu˙ t + − β ∆t2 u ¨t 2 = u˙ t + (1 − γ)∆t¨ ut

(5.3.12)

Equation (5.3.11) is then exactly Eq.(5.3.4).

5.3.2

The Newton-Raphson method

At time step n + 1 1 and iteration i + 1, the residual can be approximated by i r(di+1 n+1 ) ' r(dn+1 ) +

1

∂r ∆din+1 = 0 ∂d din+1

(5.3.13)

In equilibrium problems, time step means load step or load increment where the load is applied incrementally.

126

Solution procedures

Chap. 5

KT (din+1 )∆din+1 = −r(din+1 )

(5.3.14)

sD ∂f int ∂f ext − M + β∆t2 ∂d ∂d

(5.3.15)

Hence we obtain the linear model

where KT =

This matrix is called the finite element Jacobian of the system of equation or most often the tangent stiffness matrix. The procedure to derive this matrix is called the linearization and is the subject of the next section. Having the displacement increment, we update the displacement by i i di+1 n+1 = dn+1 + ∆dn+1

(5.3.16)

and this process is repeated until a predefined convergence criterion is satisfied. In Box (5.2), the implicit Newmark algorithm for nonlinear transient problems is summarized for ease of implementation. Noting that the presented formulation, which was proposed by T.J.R. Hughes, is not unique. Alternatives exist but it is believed to be the most effective one. Box 5.2 Flowchart for implicit time integration 1. Initialization: d0 = 0, n = 0, t = 0 ˜ n+1 2. Estimate next solution dnew = dn or dnew = d 3. Newton-Raphson iterations for load step n + 1 i. Compute the force f (dnew , tn+1 ) ˜ n+1 ); vn+1 = v ii. Compute an+1 = 1/β∆t2 (dnew − d ñ+1 + γ∆tan+1 iii. Compute the residual r = Man+1 − f (dnew , tn+1 ) iv. Compute the tangent stiffness KT (dnew ) v. modify KT (dnew ) for essential boundary conditions vi. Solve the linear system of equations KT ∆d = −r vii. dnew = dnew + ∆d viii. Check for convergence, if not met go to step 2i. 4. Update displacement, load step: dn+1 = dnew , n = n + 1, t = t + ∆t 5. Output or if simulation not complete go to step 2

Sec. 5.3

Implicit dynamics

5.3.3

Convergence criteria

5.3.4

Linearization

127

We start with the linearization of the internal nodal force. By comparing the expression of the total and updated Lagrangian formulations, we choose the former to do the linearization because in this formula, only the nominal stress P is time dependent or more precisely depends on the increment of displacement. Taking the material time derivative of the internal nodal force, we have Z ∂NI ˙ int ˙ fiI = Pji dΩ0 (5.3.17) Ω0 ∂Xj ˙ because this stress rate The constitutive equations are usually not expressed in terms of P is not objective. We therefore use the material time derivative of the second Piola-Kirchhoff stress, which is objective. Indeed, by using P = SFT , the above equation becomes Z ∂NI ˙ int ˙ (Sjk Fik + Sjk F˙ ik )dΩ0 (5.3.18) fiI = Ω0 ∂Xj This equation shows that the rate of the internal nodal force consists of two parts: (i) material part that represents the dependence on the stress and (ii) geometry part that represents the dependence on the deformation gradient. Therefore Eq.(5.3.18) can be rewritten as f˙iIint = f˙iImat + f˙iIgeo

(5.3.19)

where f˙iImat =

Z Ω0

∂NI ˙ Sjk Fik dΩ0 , ∂Xj

f˙iIgeo =

Z Ω0

∂NI Sjk F˙ ik dΩ0 ∂Xj

(5.3.20)

denotes the material part and geometry part of the rate of the internal nodal force, respectively. They yield the material tangent stiffness and geometry tangent stiffness which are given what follows.

5.3.5

Material tangent stiffness

The first term of Eq.(5.3.20) can be written as (see Section on total Lagrangian internal force vector) Z ˙fImat = ˙ BT (5.3.21) 0I {S}dΩ0 Ω0

˙ is the rate of the second Piola-Kirchoff stress tensor written in the Voigt column where {S} matrix form. The constitutive equation in rate form is given by ˙ = [CSE ]{E} ˙ {S}

(5.3.22)

128

Solution procedures

Chap. 5

˙ = B0J u˙ J , hence Recall that {E} Z

f˙Imat =

SE 0 BT ˙ J dΩ0 0I [C ]BJ u

(5.3.23)

Ω0

So the material tangent stiffness matrix is given by Kmat IJ

Z

SE BT 0I [C ]B0J dΩ0

=

(5.3.24)

Ω0

5.3.6

Geometric stiffness

By using Eq.(4.5.10), we can write Z Z ∂N I geo f˙iI = Sjk F˙ ik dΩ0 = B0jI Sjk B0kJ u˙ iJ dΩ0 ∂X j Ω0 Ω0

(5.3.25)

To write the above in matrix notation, we first explicitly write the above in two dimensions as Z  0 0 geo BjI Sjk BkJ dΩ0 0   u˙ xJ f˙xI Ω0  Z  (5.3.26) geo =   u˙ yJ f˙yI 0 B0jI Sjk B0kJ dΩ0 Ω0

which can obviously be written as f˙Igeo = I

Z

BT ˙J 0I SB0J dΩ0 u

(5.3.27)

Ω0

where I is a unit matrix whose dimension equals the number of spatial dimensions. The geometric stiffness matrix is therefore given by Kgeo IJ

Z

BT 0I SB0J dΩ0

=I

(5.3.28)

Ω0

Noting that in the above, S is the 2nd PK stress matrix not vector in Voigt notation. Updated Langrangian tangent stiffness matrices The above forms can be easily converted to the updated Lagrangian forms by letting the current configuration be the reference configuration, see Section 4.7 for an explaination of the being used trick. The results are given by Kmat IJ Kgeo IJ

Z

στ BT I [C ]BJ dΩ

= Ω Z

=I Ω

(5.3.29) BT I σBJ dΩ

Sec. 5.3

Implicit dynamics

129

where we have used that for F = I, then CSE = Cστ , B0 = B and B0 = B. We now turn our attention to the symmetry of the tangent stiffness matrix since they furnish less computer storage, less computational effort. From the above, the tangent stiffness matrix is symmetrical if and only if its geometry and material parts are symmetrical. Noting that the geometry tangent stiffness matrix is always symmetrical regardless materials and the symmetry of the material stiffness depends on the symmetry of the tangent modulii matrix, say [Cστ ] which in turn is symmetrical if the tangent modulii tensor, say Cστ ijkl , posseses major symmetry. In what follows, we are giving some examples showing the structure of the geometry stiffness matrix which is thought to be useful in programming geometrically nonlinear finite elements. Example 5.3.2 (Geometry stiffness matrix of three node triangle element) Consider node triangle element, then we have   ∂N1 ∂N2 ∂N3  ∂x ∂x ∂x   B=  ∂N1 ∂N2 ∂N3  ∂y ∂y ∂y From Eq.(5.3.29), if we define the following matrix   ∂N1 ∂N1   ∂x ∂y  ∂N1   " # Z   σ σ  xx xy  ∂N2 ∂N2   ∂x H =    ∂N1 |{z}   ∂x ∂y Ωe   σxy σyy 3×3  ∂N3 ∂N3  ∂y ∂x ∂y

∂N2 ∂x ∂N2 ∂y

 ∂N3 ∂x   dΩ ∂N3  e

a

three-

(5.3.30)

(5.3.31)

∂y

Hence, the geometry tangent stiffness matrix in the updated Lagrangian formulation is given by   H11 0 H12 0 H13 0  0 H11 0 H12 0 H13     H21 0 H22 0 H23 0  geo   K (5.3.32) | {z } =  0 H21 0 H22 0 H23    6×6 H31 0 H32 0 H33 0  0

H31

0

H32

0

H33

Tangent stiffness matrix using other objective stress rates The tangent stiffness matrix developed so far is for the Truesdell rate of the Cauchy stress (updated Lagrangian formulation). When, the material is described by another objective rate, say the Jaumann rate, hence, we have to replace Cστ by CσJ using the corresponding relation. This approach is easy.

130

5.3.7

Solution procedures

Chap. 5

Load stiffness matrix

Having given the linearization of the internal force vector, we are now presenting the linearization of the external force vector which is neccessary in cases that the load depends on the deformation, i.e., follower loads such as the pressure, since the increment of the external nodal force with respect to the increment in displacement does not vanish and make a term called the load stiffness. This term is misleading because the load has nothing to do with the stiffness. It is however consistent with material stiffness and geometry stiffness. Considering a pressure field p(x, t), hence the traction is given by t = −pn. The external nodal force is then written as Z ext fI = − NI pndΓ (5.3.33) Γ

which can be rewritten as follows using the parametric form of the loaded surface Z 1Z 1 ext NI (ξ, η)p(ξ, η)x,ξ × x,η dξdη fI = − −1

Taking the time derivative of the above yields Z 1Z 1 ˙f ext = − NI (px ˙ ,ξ × x,η + pv,ξ × x,η + px,ξ × v,η ) dξdη I −1

(5.3.34)

−1

(5.3.35)

−1

omit the first term, f˙Iext = −

Z

1

Z

1

NI (pv,ξ × x,η + px,ξ × v,η ) dξdη −1

(5.3.36)

−1

Two dimensional case fIext

Z

1

=−

NI (ξ)p(ξ)x,ξ × e3 dξ

(5.3.37)

−1

5.4 5.4.1

5.5

Stability Time step estimatation

Solvers for linear system of equations Ax = b

1. Direct solvers 2. Indirect or iterative solvers Definition 5.5.1 An n × n matrix K is said to be positive definite if

(5.5.1)

Sec. 5.6

Solvers for linear system of equations

131

• cT Kc ≥ 0 for all c • cT Kc = 0 implies c = 0 Definition 5.5.2 The conditioning number of a matrix is defined by

5.5.1

Direct solvers

LU decomposition A = LU

(5.5.2)

     a11 a12 a13 l11 0 0 u11 u12 u13 a21 a22 a23  = l21 l22 0   0 u22 u23  . a31 a32 a33 l31 l32 l33 0 0 u33

(5.5.3)

Ly = b Ux = y

(5.5.4)

A = LLT

(5.5.5)

Cholesky decomposition

Ly = b LT x = y

5.5.2

(5.5.6)

Iterative solvers

Conjugate gradient (CG) methods Preconditioned conjugate gradient methods P−1 Ax = P−1 b

(5.5.7)

where P−1 is the so-called preconditioner in the sense that the matrix P−1 A has smaller conditioning number than the original matrix A. Possible preconditioners include 1. Jacobi preconditioner 2. Symmetric Gauss-Seidel 3. Symmetric Successive Over Relaxation (SSOR)

132

Solution procedures

Chap. 5

Biconjugate gradient methods Generalized minimal residual method (GMRES)

5.6

Load control f

f

f

snap though point

snap back point f2 f1 u

u3 u4

u

u

Figure 5.1: Different techniques to complete trace the equilibrium path: from left to right, load control (fail to pass snap through point), displacement control (can handle snap through behavior) and arclength control (able to trace the complete path).

5.7

Displacement control

5.8

Arc-length control

5.8.1

General formulation

The basic idea of arc-length method, often called path following method or continuation method is to consider the load parameter λ as an additional unknown governed by a constraint equation which is often called arc-length function or control function. Hence, the system of equations to be solved reads f (u) − λg =0 (5.8.1) φ(u, λ) where u is the state vector which is normally the nodal displacement vector in solid mechanics, φ is the arc-length function and g is the reference constant load vector. The above system of equation is solved by the Newton-Raphson method which is indeed a series of successive linearised equations. Start from the known state (u(k) , λ(k) ), the linearised equation reads

f (u(k) ) − λ(k) g φ(u(k) , λ(k) )

+

K −g vT w

(k) ∆u · =0 ∆λ

(5.8.2)

Sec. 5.8

Arc-length control

133

where ∂f , ∂u Therefore, the corrections are given by K=

∆u ∆λ

=

v=

K −g vT w

∂φ , ∂u

−1 ·

w=

∂φ ∂λ r

(5.8.3)

(5.8.4)

−φ(u(k) , λ(k) )

with the residual r = λ(k) g − f (u(k) ). Then the state vector and the load parameter are updated by (k+1) (k) u u ∆u = + (5.8.5) λ λ ∆λ This iterative procedure is repeated until the norm of the residuals r and φ satisfy predefined convergence criteria. In order to avoid the inverse of the Jacobian in Eq.(5.8.4) which is neither symmetric nor banded, the Sherman-Morrison formula for the inverse of a non singular matrix plus a rank 1 matrix has been used to allow Eq.(5.8.4) to be rewritten as (see Appendix A.6 for late development)

∆u ∆λ

uII vT uI + uII φ = − T T v uII + w v uI + [1 − vT uII − w]φ vT uI + φ uII uI − T = 0 1 v uII + w uI −φ

1

uI = K−1 r,

uII = K−1 g

(5.8.6)

(5.8.7)

For sake of computer implementation, the flowchart of the general arclength control is summarized in Box (5.8.1). Noting that, different arclength control methods specify different formulae of v, ω, φ.

5.8.2

Some specific arc-length functions

Standard arc-length control (u(n+1) − u(n) )T (u(n+1) − u(n ) + ψ∆λ2 gT g − ∆l2 = 0

(5.8.8)

where ∆l is the so-called incremental arc length which is constant during a load step and is adapted properly from step to step. Indirect displacement control (u(n+1) − u(n) )T W(u(n+1) − u(n) ) + ψ∆λ2 gT g − ∆l2 = 0

(5.8.9)

where W is a given diagonal matrix that can be used to select a specific set of degrees of freedom.

134

Solution procedures

Chap. 5

Box 5.3 Arc-length algorithm in pseudo code 1. Set the state and load parameter to the converged value of the previous load step u ← u0 λ ← λ0 2. Update K , f , v, ω, φ 3. Start the iteration procedure r ← λg − f uI ← K−1 r uII ← K−1 g vT uI + φ α ← vT uII + w u ← u + uI − αuII λ ← λ−α 4. Check convergence, if not return to step 2

5.8.3

Energy release control

This method has been proposed by M. Gutierrez in TU Delft to use in cases where the indirect displacement control does not work, for example, multiple cracking problems (which selected set of degrees of freedom to be used?). The method is based on energy principle. The energy release rate in the quasi-static regime is given by G = P − V˙

(5.8.10)

where G is the energy release rate, V and P are the elastic potential energy and the power of the external forces, respectively, which are given by Z 1 T σ V = 2 Ω Z 1 uT BT σ = 2 Ω (5.8.11) 1 T int = u f 2 1 T = λu g 2 where we have used the expression of the strain = Bu in the second equality and of the internal force vector in the third equality and the equilibrium, f int = f ext in the final step.

Sec. 5.8

Arc-length control

135

P := u˙ T f ext = λu˙ T g

(5.8.12)

The external power is written as

From Eq.(5.8.11), the elastic potential energy rate is given by 1 1˙ T V˙ = λu˙ T g + λu g 2 2 Therefore, from Eq.(5.8.10), the energy release rate reads

(5.8.13)

1˙ T 1 g (5.8.14) G = λu˙ T g − λu 2 2 By the second law of thermodynamics, the amount of energy dissipated in a system is increasing monotonically. Hence the above equation is an ideal candidate to be a constraint equation φ. The forward Euler integration scheme for the above rate equation gives the discrete arc-length function 1 (n) T (n) T λ (u(n+1) − uT u(n) g − ∆τ = 0 (5.8.15) (n) ) − ∆λ 2 Noting that we have used the notation ∆τ instead of ∆l and call it the incremental path following parameter that has the unit of Nm since it now represents the amount of energy released when going from load step n to load step n + 1. Its geometrical interpretation is given in Fig.(??). φ=

5.8.4

Automatic load incrementation

In order to have an automatic load adaptation procedure, the incremental path following parameter must be computed automatically in such a way that no waste calculations are performed (∆τ has chosen so small that led to many unnecessary load increments) one part and it gives reasonable estimation that leads to reasonable number of iterations to get converged, the other part. One of the most heuristic approaches is based on the balance between the number of iterations. Two commonly used formula are given in the following α nd j+1 j (5.8.16) X =X nj with α varies between 0.0 and 1.0, and (X can be ∆τ or ∆l) nj − nd (5.8.17) 4 where nj is the number of iterations in the previous load step and nd is the optimal (desired) number of iterations per load step. In general, nd is taken equal to 5. To avoid so small or so big value, the following bounds are usually used X j+1 = X j 0.5γ ,

γ=

Xmin ≤ |X| ≤ Xmax

(5.8.18)

136

Solution procedures

Chap. 5

where usually it is the users who define these bounds. However for the energy release control method, it seems that the lower bound is unnecessary and the upper bound. For the energy release control method, initially during the elastic branch, no dissipate energy, so we have adopted either direct displacement control or load control until the degree of nonlinearity is sufficiently large. This can be detected in an informal fashion by comparing the number of iteration of a load step n with the desired iteration number nd , if n > nd , then we switch to energy release control method with the initial path following parameter ∆τ computed from Eq.(5.8.15). This parameter for subsequent steps is computed using Eq.(5.8.16) or Eq.(5.8.17).

5.9

Mesh generation programs

Meshing is the procedure in which the geometry of the medium is discretized into a set of finite elements. The geometry of medium is created by using computer aided design programs better known as CAD programs. The creation of finite meshes is then realized by using meshing programs also called mesh generators or even shorter meshers. The academic community however generally relies on free meshing programs. Among are Gmsh developed by Christophe Geuzaine and Jean-François Remacle and Salome package developed by EDF (Electricite de France). While the former is very powerful in generating unstructured two and three dimensional meshes, the latter is able to generate both unstructured and structured meshes. In the following, before describing, although briefly, the usage of these two programs, the common procedure in building geometry and generating finite element meshes is provided. Understanding this facilitates the transition from one mesher to another. 1. Creation of points 2. Creation of lines from points 3. Creation of contours from lines 4. Creation of surfaces from closed contours; from that 2D meshes can be generated 5. If 3D volumes are about to buid, then volumes are constructed from surfaces; then 3D meshes can be generated. The meshsers generally write the mesh to ASCII file providing at least the coordinates of the nodes and the connectivity of elements. This file should be converted to the format that the finite element programs (often called solvers in programs where geometry module, mesh module and solver modules are integrated into one package) understand.

Sec. 5.9


137

Box 5.4 Flowchart for solution procedure with energy-based control 1. Initialization: λ = 0, set nd , ∆τ0 2. Solve the elastic branch using load control i. Solving the equilibriuum f int = λg using Newton-Raphson method ii. Store the load scale: λ0 = λ iii. Update the load scale λ = λ + ∆τ0 iv. Check number of iterations: n > nd . If no, goto 2i. Else T v. Compute the release energy G = 12 λ0 (uT − uT 0 ) − ∆τ0 u0 g vi. ∆τ = 0.5γ G,

γ=

n−nd 4

3. Switch to arclength control i. Update K , f from element contributions using a FE formulation ii. Update v, ω, φ v = 0.5λ0 g T ω = −0.5u 0g T φ = 12 λ0 (uT − uT 0 ) − ∆λu0 g − ∆τ iii. Start the iteration procedure r ← λg − f uI ← K−1 r uII ← K−1 g vT uI + φ α ← vT uII + w u ← u + uI − αuII λ ← λ−α iv. Check convergence, if not return to step 3i v. Adjust the path following parameter ∆τ = ∆τ ∗ 0.5γ ,

γ=

nj −nd 4

vi. λ0 = λ, λ = λ + ∆τ0 ∗

In the above the superscript 0 denotes converged values of previous load step.

138

Solution procedures

5.9.1

Gmsh

5.9.2

Salome

Chap. 5

go to the folder containing salome executable file and type ./runSalome –logger the option specified is to enable the Python interpreter capacity. Then, File/Load Script and choose the appropriate Python scrip file.

Figure 5.2: Three dimensional structured finite element mesh created by Salome. Listing 5.1: A typical Python script to create geometry and finite element mesh in Salome # −−−−−−−−−−−−−−−−−−−−−−− #

c r e a t i n g the geometry

# −−−−−−−−−−−−−−−−−−−−−−−

from geompy import ∗ # points

p1 p2 p3 p4

= = = =

MakeVertex ( 0 . , 0 . , 0 . ) MakeVertex ( 4 8 . , 4 4 . , 0 . ) MakeVertex ( 4 8 . , 6 0 . , 0 . ) MakeVertex ( 0 . , 4 4 . , 0 . )

# l i n e s ( c a l l e d edges in salome )

l 1 = MakeEdge ( p1 , p2 ) l 2 = MakeEdge ( p2 , p3 ) l 3 = MakeEdge ( p3 , p4 )

Sec. 5.9


l 4 = MakeEdge ( p4 , p1 ) # c o n t o u r ( c a l l e d wire in salome )

c1 = MakeWire ( [ l 1 , l 2 , l 3 , l 4 ] )

# t h i s f u n c t i o n t a k e s 1 argument ! ! !

# s u r f a c e ( c a l l e d f a c e in salome )

s 1 = MakeFace ( c1 , 1 ) a d d T o S t u d y ( s1 , " s u r f a c e " ) leftEdge = GetEdge ( s1 , p1 , p4 ) r i g h t E d g e = GetEdge ( s1 , p2 , p3 ) b o t t o m E d g e = GetEdge ( s1 , p1 , p2 ) a d d T o S t u d y I n F a t h e r ( s1 , l e f t E d g e , " leftEdge " ) a d d T o S t u d y I n F a t h e r ( s1 , r i g h t E d g e , " r i g h t E d g e " ) a d d T o S t u d y I n F a t h e r ( s1 , bottomEdge , " b o t t o m E d g e " ) # −−−−−−−−−−−−−−−−−−−−−−− #

g e n e r a t i n g mesh

# −−−−−−−−−−−−−−−−−−−−−−−

from smesh import ∗ from S t d M e s h e r s import ∗ n o O f S e g m e n t s X d i r = 20 n o O f S e g m e n t s Y d i r = 10 mesh = Mesh ( s 1 ) algo1D = mesh . Segment ( ) algo1D . NumberOfSegments ( 1 0 ) algo2D = mesh . Q u a d r a n g l e ( ) algo1D_XDir = mesh . Segment ( b o t t o m E d g e ) algo1D_XDir . NumberOfSegments ( n o O f S e g m e n t s X d i r ) algo1D_XDir . P r o p a g a t i o n ( ) algo1D_YDir = mesh . Segment ( l e f t E d g e ) algo1D_YDir . NumberOfSegments ( n o O f S e g m e n t s Y d i r ) algo1D_YDir . P r o p a g a t i o n ( ) import SMESH

139

140

Solution procedures

Chap. 5

mesh . GetMesh ( ) . CreateGroupFromGEOM (SMESH . NODE, " l e f t N o d e s " , leftEdge ) mesh . GetMesh ( ) . CreateGroupFromGEOM (SMESH . EDGE, " r i g h t E d g e s " , r i g h t E d g e ) # f i n a l l y , g e n e r a t e t h e mesh

mesh . Compute ( )

5.10

Validation examples

Having given the theory and implementation of the nonlinear finite element methods and provided that a linear finite element code is available, it is then not so hard to program nonlinear finite elements. However, once the theory was programmed, it might be giving incorrect results so the validation process is quite important. It is in this section that some validation (hence, simple) examples are given.

5.10.1

Compressible finite strain hyperelasticity problems

Patch test y (0.5,1.0)

(0,1.0) 0.25

1.0 0.25

0.25

(2.0,0.75)

(1.0,0.5) 1.0

(0,0.5)

(0,0)

(1.0,1.0)

(0.5,0)

(1.0,0)

x

1.0

Figure 5.3: A patch-test for compressible Neo-Hookean material λ0 = µ0 = 100

2 0 F= , 0 0.75

J = 1.5,

227.03 0 σ= 0 −2.136

(5.10.1)

Sec. 5.10

Validation examples

Patch test

5.10.2 Patch test

Incompressible finite strain hyperelasticity problems

141

142

Solution procedures

Chap. 5

Chapter 6 Robust finite elements for constrained media 6.1

Introduction

6.2

Element characteristics Figure 6.1: Commonly used 2D and 3D finite elements in nonlinear analysis

The following example Example 6.2.1 Effect of Mesh Distortion Consider the three-noded quadratic line elements. The shape functions for the parent element are in matrix form, N(ξ) =

1

2 ξ(ξ

− 1) 1 − ξ 2

1 2 ξ(ξ

+ 1)

The mapping from the Eulerian coordinates to the parent element coordinates is given by x(ξ, t) = N1 (ξ)x1 (t) + N2 (ξ)x2 (t) + N3 (ξ)x3 (t) The derivative with respect to ξ is thefore 1 1 x,ξ = N1,ξ x1 (t) + N2,ξ x2 (t) + N3,ξ x3 (t) = (2ξ − 1)x1 − 2ξx2 + (2ξ + 1)x3 2 2 In matrix form,

x,ξ = N,ξ x(t) =

1

2 (2ξ

− 1) −2ξ

  x1 1 x2  (2ξ + 1) 2 x3

144

Robust finite elements for constrained media

Chap. 6

Now, the B matrix is given by B = N,ξ (x,ξ )−1 Therefore, B=

1 2ξ − 1 −4ξ 2ξ + 1 2x,ξ

The internal forces are given by Z

x3

fint =

BT σAdx

x1

Plugging in the expression for B, and changing the limits of integration, we get Z

1

fint = −1

    Z 1 2ξ − 1 1 2ξ − 1  −4ξ  σAx,ξ dξ = 1  −4ξ  σAdξ 2x,ξ 2 −1 2ξ + 1 2ξ + 1

If node 2 is midway between node 1 and node 3, 1 x2 = (x1 + x3 ) 2 Then we have, 1 1 1 x,ξ = (2ξ − 1)x1 − ξ(x1 + x3 ) + (2ξ + 1)x3 = (x3 − x1 ) 2 2 2 However, if node 2 moves away from the middle during deformation, then x,ξ is no longer constant and can become zero or negative. Under such situations the rate of deformation is either infinite or the element inverts upon itself since the isoparametric map is no longer one-to-one. Let us consider the case where x,ξ is zero. In that case, the Jacobian becomes J=

A A x,X = x,ξ (X,ξ )−1 = 0 A0 A0

Similarly, when x,ξ is negative, J is negative. This implies that the conservation of mass is violated. To find the location of x2 when this happens, we set the relation for x,ξ to zero. Then we get, 1 1 (2ξ − 1)x1 − 2ξx2 + (2ξ + 1)x3 = 0 2 2

=⇒

If x,ξ = 0 at ξ = 1, then x2 =

x1 + 3x3 4

x2 =

x1 + x3 x3 − x1 + 2 4ξ

Sec. 6.2

Element characteristics

145

This means that as node 2 gets closer and closer to node 3 the rate of deformation become infinite at node 3 and then negative. If x,ξ = 0 at ξ = −1 then x2 =

3x1 + x3 4

This means that as node 2 gets closer and closer to node 1, the rate of deformation become infinite at node