SOLID MECHANICS

Fundamentals and engineering concepts

of elasticity-based design

in mechanical engineering

MARC GEERS

PIET SCHREURS

Eindhoven University of Technology

Department of Mechanical Engineering

MECHANICS of MATERIALS

CONTENTS

Contents i

Preface vii

1 Tensor calculus 1

1.1 Scalars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2.2 Vector operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2.3 Vector basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.2.4 Vector components; matrix notation of a vector . . . . . . . . . . . . . . 6

1.3 Second order tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.3.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.3.2 Some properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.3.3 Tensor components; matrix notation of a tensor . . . . . . . . . . . . . . 10

1.3.4 Scalar functions of a tensor . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.3.5 Eigenvalues and eigenvectors of a tensor . . . . . . . . . . . . . . . . . . 13

1.3.6 Some special tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.4 Spatial derivatives of scalar and vector fields . . . . . . . . . . . . . . . . . . . . 18

1.4.1 Derivatives of functions of one or more variables . . . . . . . . . . . . . 18

1.4.2 Components of the position vector . . . . . . . . . . . . . . . . . . . . . 19

1.4.3 The gradient operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

1.5 Tensor formula involving first and second-order tensors . . . . . . . . . . . . . 22

1.6 Higher-order and fourth-order tensors . . . . . . . . . . . . . . . . . . . . . . . . 22

1.6.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

1.6.2 Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

1.6.3 Tensor rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

1.6.4 Matrix condensation of the double inner product with a second-order

tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

1.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

2 Kinematics 31

2.1 General aspects of deformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

2.1.1 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

2.1.2 Lagrangian versus Eulerian coordinates . . . . . . . . . . . . . . . . . . . 32

2.1.3 The displacement vector . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

2.1.4 The displacement gradient tensor and deformation gradient tensor . . 34

2.2 Strains and strain tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

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2.2.1 Strain definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

2.2.2 Deformation of infinitely small material line elements . . . . . . . . . . 38

2.3 Linearization of deformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

2.3.1 Linearized strain definitions . . . . . . . . . . . . . . . . . . . . . . . . . 41

2.3.2 The infinitesimal linear strain tensor . . . . . . . . . . . . . . . . . . . . 42

2.3.3 The infinitesimal rotation tensor . . . . . . . . . . . . . . . . . . . . . . . 43

2.4 Applicability of the theory of small displacements . . . . . . . . . . . . . . . . . 44

2.5 Physical interpretation of the strain components . . . . . . . . . . . . . . . . . . 46

2.5.1 Normal or extensional strains . . . . . . . . . . . . . . . . . . . . . . . . . 46

2.5.2 Shear strains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

2.5.3 Principal strains and strain invariants . . . . . . . . . . . . . . . . . . . . 49

2.5.4 Volume and shape changes . . . . . . . . . . . . . . . . . . . . . . . . . . 50

2.6 Compatibility conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

2.7 Special deformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

2.7.1 Rigid body motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

2.7.2 Homogeneous deformation . . . . . . . . . . . . . . . . . . . . . . . . . . 54

2.7.3 Plane strain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

2.7.4 Axisymmetric deformation . . . . . . . . . . . . . . . . . . . . . . . . . . 54

2.8 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

3 Force and stress 61

3.1 Forces and stresses in a deformable continuum . . . . . . . . . . . . . . . . . . 61

3.1.1 External forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

3.1.2 Internal forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

3.1.3 The stress vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

3.1.4 The stress tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

3.2 Components of the stress tensor with respect to a basis . . . . . . . . . . . . . . 64

3.2.1 Normal stresses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

3.2.2 Shear stresses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

3.3 Principal stresses and stress invariants . . . . . . . . . . . . . . . . . . . . . . . . 64

3.4 Hydrostatic and deviatoric stresses . . . . . . . . . . . . . . . . . . . . . . . . . . 65

3.4.1 Hydrostatic pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

3.4.2 Deviatoric stresses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

3.5 Stress state with respect to a plane . . . . . . . . . . . . . . . . . . . . . . . . . . 67

3.5.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

3.5.2 Octahedral stresses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

3.5.3 Maximum shear stress in a material point . . . . . . . . . . . . . . . . . 69

3.6 Plane stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

3.6.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

3.6.2 Mohr’s stress circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

3.6.3 Mohr’s stress circle for three-dimensional stress states . . . . . . . . . . 73

3.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

4 Mechanical equilibrium 77

4.1 Balance of mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

4.2 Balance of momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

4.2.1 General derivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

4.2.2 Translational equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

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4.2.3 Equilibrium in a Cartesian and cylindrical basis . . . . . . . . . . . . . . 81

4.3 Balance of moment of momentum . . . . . . . . . . . . . . . . . . . . . . . . . . 82

4.3.1 General derivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

4.3.2 Moment or rotational equilibrium . . . . . . . . . . . . . . . . . . . . . . 83

4.3.3 Rotational equilibrium in a Cartesian and cylindrical basis . . . . . . . 84

4.4 Special deformation states . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

4.4.1 Planar deformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

4.4.2 Axisymmetric deformation . . . . . . . . . . . . . . . . . . . . . . . . . . 85

4.5 Mechanical work and internal energy . . . . . . . . . . . . . . . . . . . . . . . . 86

4.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

5 Three-dimensional linear elasticity 89

5.1 General aspects of constitutive equations . . . . . . . . . . . . . . . . . . . . . . 89

5.2 The concept and nature of linear elasticity . . . . . . . . . . . . . . . . . . . . . . 89

5.2.1 The elastic response of materials . . . . . . . . . . . . . . . . . . . . . . . 90

5.2.2 The elastic stored energy function . . . . . . . . . . . . . . . . . . . . . . 90

5.3 General 3D linear elastic solid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

5.3.1 Generalized Hooke’s law for anisotropic solids . . . . . . . . . . . . . . . 91

5.4 Material symmetry and anisotropy . . . . . . . . . . . . . . . . . . . . . . . . . . 93

5.4.1 Orthotropic materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

5.4.2 Transverse isotropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

5.5 Linear elastic isotropic materials . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

5.5.1 The isotropic elasticity tensor . . . . . . . . . . . . . . . . . . . . . . . . . 97

5.5.2 Engineering elastic moduli . . . . . . . . . . . . . . . . . . . . . . . . . . 99

5.5.3 Stored energy density for isotropic linear elastic materials . . . . . . . . 103

5.5.4 Coincidence of principal axes of stress and strain . . . . . . . . . . . . . 103

5.5.5 Plane strain and plane stress . . . . . . . . . . . . . . . . . . . . . . . . . 104

5.6 The superposition principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

5.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106

6 Limit design criteria 109

6.1 Failure modes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

6.2 Standard tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110

6.2.1 Tension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110

6.2.2 Torsion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

6.3 Elastic limit criteria for isotropic materials . . . . . . . . . . . . . . . . . . . . . . 112

6.3.1 Rankine criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

6.3.2 Tresca yield criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114

6.3.3 Von Mises yield criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

6.3.4 Mohr-Coulomb limit criterion . . . . . . . . . . . . . . . . . . . . . . . . 120

6.3.5 Drucker-Prager limit criterion . . . . . . . . . . . . . . . . . . . . . . . . 123

6.3.6 General criteria for isotropic solids . . . . . . . . . . . . . . . . . . . . . 127

6.4 Limit criteria for anisotropic materials . . . . . . . . . . . . . . . . . . . . . . . . 127

6.5 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128

6.6 Elastic limit behaviour: Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . 130

7 Mechanical design: solutions 133

7.1 Vector/tensor equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133

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7.2 Three-dimensional scalar equations . . . . . . . . . . . . . . . . . . . . . . . . . 134

7.2.1 Cartesian components . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134

7.2.2 Cylindrical components . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

7.2.3 Material law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136

7.3 Planar deformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136

7.3.1 Cartesian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136

7.3.2 Cylindrical . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137

7.3.3 Cylindrical : axisymmetric + ut = 0 . . . . . . . . . . . . . . . . . . . . . . 137

7.4 Stress inconsistency under plane stress . . . . . . . . . . . . . . . . . . . . . . . 138

7.5 Solution strategies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138

7.5.1 Governing equations for unknowns . . . . . . . . . . . . . . . . . . . . . 138

7.5.2 Boundary conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139

7.6 Solution : displacement method . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140

7.6.1 Planar, Cartesian : Navier equations . . . . . . . . . . . . . . . . . . . . . 140

7.6.2 Planar, axisymmetric with ut = 0 . . . . . . . . . . . . . . . . . . . . . . . 140

7.6.3 Planar, axisymmetric with ut = 0, isotropic . . . . . . . . . . . . . . . . . 141

7.7 Weighted residual formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141

7.7.1 Weighted residual formulation for 3D deformation . . . . . . . . . . . . 142

7.7.2 Weighted residual formulation : linear . . . . . . . . . . . . . . . . . . . 142

7.7.3 Finite element method for 3D deformation . . . . . . . . . . . . . . . . . 143

8 Mechanical design: analytical 147

8.1 Homogeneous plates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147

8.2 Axisymmetric cylinders and discs . . . . . . . . . . . . . . . . . . . . . . . . . . . 148

8.2.1 Prescribed edge displacement . . . . . . . . . . . . . . . . . . . . . . . . 149

8.2.2 Edge load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149

8.3 Circular hole in infinite medium . . . . . . . . . . . . . . . . . . . . . . . . . . . 152

8.3.1 Pressurized hole in an infinite medium . . . . . . . . . . . . . . . . . . . 152

8.3.2 Stress-free hole in bi-axially loaded infinite medium . . . . . . . . . . . 152

8.4 Rotating disc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154

8.4.1 Solid disc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154

8.4.2 Disc with central hole . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155

8.4.3 Disc fixed on rigid axis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155

8.4.4 Rotating disc with variable thickness . . . . . . . . . . . . . . . . . . . . 157

8.5 Large thin plate with a central hole . . . . . . . . . . . . . . . . . . . . . . . . . . 158

9 Mechanical design: numerical 161

9.1 MSC.Marc/Mentat . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161

9.2 Cartesian, planar . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161

9.2.1 Tensile test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162

9.2.2 Shear test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163

9.2.3 Orthotropic plate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164

9.3 Axisymmetric, ut = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164

9.4 Axisymmetric, planar, ut = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165

9.4.1 Prescribed edge displacement . . . . . . . . . . . . . . . . . . . . . . . . 165

9.4.2 Edge load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165

9.4.3 Centrifugal load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167

9.4.4 Large thin plate with a central hole . . . . . . . . . . . . . . . . . . . . . 168

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10 Large deformations & nonlinear behaviour 169

10.1 Nonlinear axial deformation of trusses . . . . . . . . . . . . . . . . . . . . . . . . 169

10.1.1 Strains for large elongation . . . . . . . . . . . . . . . . . . . . . . . . . . 169

10.1.2 Mechanical power and stress definitions . . . . . . . . . . . . . . . . . . 171

10.1.3 Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172

10.2 Nonlinear material models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177

10.2.1 Material behaviour . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177

10.2.2 Discrete material models . . . . . . . . . . . . . . . . . . . . . . . . . . . 181

10.3 Large deformation elasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182

10.3.1 Elastic material behavior . . . . . . . . . . . . . . . . . . . . . . . . . . . 182

10.3.2 Hyper-elastic models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184

10.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186

10.4.1 Example 1: limitations of a linear relation between the Cauchy stress

and the Green-Lagrange strain . . . . . . . . . . . . . . . . . . . . . . . . 186

10.4.2 Example 2: instability & localisation . . . . . . . . . . . . . . . . . . . . . 188

10.4.3 Example 3: inflating a spherical balloon . . . . . . . . . . . . . . . . . . 189

11 Plasticity 191

11.1 Elasto-plastic material behavior . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191

11.1.1 Uniaxial test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191

11.1.2 Hardening . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194

11.2 Elasto-plastic model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197

11.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201

11.3.1 Example 1: Reversed plasticity in a tensile bar . . . . . . . . . . . . . . . 201

11.3.2 Example 2: parallel truss structure . . . . . . . . . . . . . . . . . . . . . . 203

11.3.3 Example 3: serial truss structure . . . . . . . . . . . . . . . . . . . . . . . 204

A Stiffness and compliance matrices 207

A.1 General orthotropic material law . . . . . . . . . . . . . . . . . . . . . . . . . . . 207

A.1.1 Plane strain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209

A.1.2 Plane stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210

A.1.3 Plane strain/stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210

A.2 Linear elastic orthotropic material . . . . . . . . . . . . . . . . . . . . . . . . . . 211

A.2.1 Voigt notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211

A.2.2 Plane strain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212

A.2.3 Plane stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212

A.3 Linear elastic transversal isotropic material . . . . . . . . . . . . . . . . . . . . . 213

A.3.1 Plane strain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213

A.3.2 Plane stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214

A.4 Linear elastic isotropic material . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214

A.4.1 Plane strain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215

A.4.2 Plane stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215

B Mechanical design problems 217

B.1 Governing equations and general solution . . . . . . . . . . . . . . . . . . . . . . 218

B.2 Disc, edge displacement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219

B.3 Disc/cylinder, edge load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220

B.4 Rotating solid disc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221

v

CONTENTS

B.5 Rotating disc with central hole . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222

B.6 Rotating disc fixed on rigid axis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223

vi

PREFACE

The functionality, strength and reliability of a mechanical component, product or device

are controlled by the deformations or loads imposed through its service conditions. Defor-

mations are caused by the (external) loads, giving rise to stresses within the material. In most

engineering design problems, deformations and rotations remain small and reversible (elas-

tic), whereby no energy is dissipated in the material. This regime is adequately described

with the theory of linear elasticity, which is widely used as the basic tool for designing and

optimizing products and defining its service limits.

The modeling of complex products is inevitably a three-dimensional problem. The fact

that strains and stresses have to be characterized in a 3D-space, is certainly one of the goals

of this course. Within a 3D-space, vectors and tensors are indispensable. A whole chap-

ter has therefore been devoted to this subject. Next, the kinematical framework for gen-

eral (large) deformations is set up, and consequently linearized in the case of infinitesimal

displacements. The stress tensor is defined and the 3D constitutive laws for isotropic and

anisotropic linear elastic materials are given. Particular attention is given to the various

forms in which the three-dimensional Hooke’s law appears, as well as to the relations be-

tween the various elastic moduli. Combined with the governing equilibrium equations, de-

sign problems can be solved through the elaboration of appropriate design rules. Particular

cases, for which analytical solutions are available, are treated in the the course. Elemen-

tary aspects of numerical solutions techniques for more complex problems are presented

and simple problems are solved by the students in the exercise sessions. The course ends

with some one-dimensional basics on nonlinear deformation (large displacements and ro-

tations) and more complex material behaviour (plasticity). This last part serves as a bridge

to the mechanics of manufacturing processes, in which the production steps are designed,

shaped, controlled and optimized. This subject is treated in follow-up courses.

These lecture notes constitute the core of the course ’Solid Mechanics’. In spite of our

best efforts, some errors doubtless remain. We would highly appreciate to be informed on

any error found or any other comment. We sincerely hope that this monograph stimulates

the student’s interest in the field of solid mechanics and its application in the design of many

products.

Marc Geers

Piet Schreurs

January 2015

vii

CH

AP

TE

R

1VECTOR AND TENSOR CALCULUS

In this chapter, attention is given to the definition and manipulation of vector and tensor

fields. This mathematical description is used in many physical disciplines, and mechanics

in particular. Vectors and tensors enable a compact notation and manipulation, without the

necessity to resort to individual components with respect to a particular basis. The latter is

needed only at the point that quantitative calculations need to be done, which is generally

postponed to the last step. Special attention is given to Cartesian tensor calculus, which pro-

vides the basis for the elaboration of coordinate-free equations. Within this specific frame-

work, no distinction is made between contravariant and covariant tensors while the metric

tensor is not introduced neither. The quantification into components measured with respect

to a specific basis, and the related matrix calculus is also elaborated.

1.1 Scalars

A scalar a is a quantity which has only a measure but no direction. It is the most trivial

tensor, the tensor of order zero. The value of a scalar does not depend on the choice of a

basis. Well-known examples are the temperature T or the density ρ in a material point.

1.2 Vectors

1.2.1 Definition

A vector is a physical quantity possessing both a magnitude (or length) and a direction. In

case of a vector ~a:

~a = ||~a||~e (1.1)

the length is given by ‖~a‖, and the direction is given by a unit vector~e, having a unit length.

A vector is a first-order tensor. The zero vector~0 is a special vector having zero length. The

so-called index notation of a vector ~a is given by

~a =⇒ ai (1.2)

1. TENSOR CALCULUS

r

a

Figure 1.1: Magnitude and direction of a vector ~a.

A vector is thus indicated with a single index (i.c. i ), the value of which ranges between 1 and

the dimension of the Euclidian space it is used in (commonly this dimension equals 3). The

order of this tensor (i.e., 1) thus equals its number of indices. The index notation obtains a

specific meaning once a basis has been chosen, since the components of a vector are trivially

identified with this notation, see 1.2.3.

1.2.2 Vector operations

Multiplication of a vector ~a = ‖~a‖~e by a scalar α yields a vector ~b with length α‖~a‖ in the

direction~e:~b = α~a (1.3)

The sum of two vectors ~a and~b is a new vector~c, equal to the diagonal of the parallelogram

r

aa

r

a

-r

a

Figure 1.2: Multiplication of a vector:~b =α~a.

spanned by ~a and~b.

~c = ~a+~b (1.4)

scalar product, inner product or dot product 1 of two vectors is a scalar quantity, defined as:

r

c

r

a

r

b

Figure 1.3: The sum of two vectors: ~c =~a+~b.

~a·~b = ||~a|| ||~b||cos(φ) (1.5)

where φ is the smallest angle between ~a and ~b. Note that the dot product of two nonzero

mutually perpendicular vectors equals zero. The product is commutative, i.e.:

1inwendig produkt

2

1.2. Vectors

r

a

r

b

j

Figure 1.4: The inner product of two vectors: ~a·~b.

~a·~b = ~b·~a (1.6)

and linear

~a·(~b +~c) = ~a·~b +~a·~c (1.7)

Using index notations, Einstein’s summation convention has to be introduced. This con-

vention states, that a summation is made over all repeated indices, where the range of the

summation equals the dimension n of the problem. For the inner product, this is written as

~a ·~b = ai bi =3∑

i=1

ai bi (1.8)

The summation symbol is generally not written, but here simply added to clarify the con-

vention.

The scalar product of a vector with itself yields the square of its length:

~a·~a = ||~a||2 ≥ 0 (1.9)

The vector product or cross product of two vectors is a vector quantity:

~a ×~b = ~c (1.10)

The length of the vector~c is given by:

‖~c‖ = ||~a|| ||~b||sin(φ) (1.11)

where φ is the smallest angle between ~a and~b. Its value equals the area of the parallelogram

spanned by the vectors ~a and ~b. The direction of ~c is perpendicular to both ~a and ~b, such

that the vectors ~a,~b and~c form a righthanded system. The linearity of the vector product is

expressed through

~a × (~b +~c) = ~a ×~b +~a×~c (1.12)

The vector product is not commutative:

~b ×~a = −~a×~b (1.13)

which can be concluded from the definition. It can be shown that:

~a × (~b ×~c) = (~a·~c)~b − (~a·~b)~c (1.14)

3

1. TENSOR CALCULUS

r

br

c

r

a

Figure 1.5: The cross product of two vectors: ~c =~a ×~b.

The triple product2 of three vectors is a scalar quantity, defined as:

~a×~b·~c = ||~a|| ||~b||sin(φ)(~n·~c)

= ||~a|| ||~b||sin(φ)||~c||cos(ψ) (1.15)

where ~n is the unit vector in the direction ~a ×~b, φ is the smallest angle between ~a and~b and

ψ is the smallest angle between~c and ~n. The absolute value of the triple product equals the

jr

a

r

cr

b

r

n

y

Figure 1.6: The triple product of three vectors ~a ×~b·~c.

volume of the parallelepiped spanned by the three vectors. If ~a ×~b·~c 6= 0, the vectors are not

lying within one plane, and are called mutually independent.

The tensor product3 or open product of two vectors ~a~b is a linear vector transformation

called a dyad. Application of the dyad ~a~b to a vector ~p yields a vector ~a, multiplied by the

scalar (~b·~p):

~a~b·~p = ~a(~b·~p) (1.16)

or in index notation

(ai b j )p j = ai (b j p j ) (1.17)

The linearity of the transformation is expressed by:

~a~b·(α~p +β~q) = α~a(~b·~p)+β~a(~b·~q) (1.18)

2tripel produkt3dyadisch produkt

4

1.2. Vectors

r

p

( )r

r r

b p a×

r

r

ab

Figure 1.7: The tensor product: ~a~b·~p .

The transpose of a dyad (also called conjugate dyad or adjoint dyad)4 is defined as:

(~a~b)T = ~b~a 6=~a~b (1.19)

If (~a~b)T =~a~b, the dyad ~a~b is called symmetric.

1.2.3 Vector basis

In three-dimensional space, a set of three vectors ~c1, ~c2 and ~c3 is called a basis, denoted by

~c1,~c2,~c3, if the vectors are independent, i.e. if the triple product of the vectors is nonzero:

~c1,~c2,~c3 ⇒ ~c1 ×~c2·~c3 6= 0 (1.20)

The three vectors~c1, ~c2 and~c3 are called base vectors. If the basis consists of mutually per-

r

c1

r

c2

r

c3

(a) ~c1,~c2,~c3

r

e1

r

e2

r

e3

(b) ~e1,~e2,~e3

Figure 1.8: A vector basis (a) and an orthonormal vector basis (b)

pendicular vectors, the basis is called an orthogonal basis.

If the basis consists of mutually perpendicular unit vectors, the basis is called an or-

thonormal basis. In the latter case the base vectors are usually denoted by ~e1, ~e2 and ~e3.

Thus it holds that:

~e1,~e2,~e3 ⇒ ~ei ·~e j = δi j =

0 if i 6= j

1 if i = j(1.21)

4geconjugeerde dyade

5

1. TENSOR CALCULUS

where δi j is called the Kronecker delta. A basis ~e1,~e2,~e3 is called righthanded5 if the order

of~e1,~e2 and~e3 is such that:

~e1 ×~e2 = ~e3 (1.22)

~e2 ×~e3 = ~e1 (1.23)

~e3 ×~e1 = ~e2 (1.24)

A Cartesian basis is an orthonormal basis in which the base vectors are independent of the

location in three-dimensional space. It is usually denoted by ~e1,~e2,~e3 or ~ex ,~e y ,~ez. A cylin-

r

e2

r

e3

r

e1

r

e1

r

e2

r

e3

x1

x2

x3

Figure 1.9: A Cartesian orthonormal basis ~e1,~e2,~e3.

drical basis is an orthonormal basis as well. It is usually denoted by ~er ,~et ,~ez. The bases

~e1,~e2,~e3 and ~er ,~et ,~ez are mutually related by:

~er (θ) = cos(θ)~e1 + sin(θ)~e2 (1.25)

~et (θ) = −sin(θ)~e1 +cos(θ)~e2 (1.26)

~ez = ~e3 (1.27)

It is important to note that the base vectors~er and~et depend on the circumferential coordi-

nate θ. The derivatives of the vectors with respect to θ are given by:

d~er (θ)

dθ= ~et (θ) (1.28)

d~et (θ)

dθ= −~er (θ) (1.29)

1.2.4 Vector components; matrix notation of a vector

Let ~c1,~c2,~c3 be a basis in three-dimensional space. Then any vector ~a can be expressed as

a unique linear combination of the base vectors. In case of an orthonormal basis ~e1,~e2,~e3

we may write:

~a = a1~e1 +a2~e2 +a3~e3 = ai~ei , (1.30)

5rechtshandige basis

6

1.2. Vectors

r

e1

r

e2

r

e3

r

er

r

et

r

ez

x1

x2

x3

r

z

q

Figure 1.10: A cylindrical orthonormal basis ~er ,~et ,~ez.

where we again used the Einstein summation convention, i.e. we sum over the repeated index

i . The coefficients ai are called the components of the vector~a with respect to the basis. Note

r

e1

r

e2

r

e3

r

a

a1

a2

a3

Figure 1.11: Components ai of a vector ~a with respect to a Cartesian basis ~e1,~e2,~e3.

that, since the basis is orthonormal, a component ai of a vector ~a with respect to the base

vector~ei can be found by:

ai = ~a·~ei (1.31)

The component ai represents the length of the projection of the vector ~a on the line with

direction~ei . For convenience we can use a matrix notation in which we store base vectors

7

1. TENSOR CALCULUS

and components in a column:

a~ =

a1

a2

a3

(1.32)

~e~ =

~e1

~e2

~e3

(1.33)

Thus we may write:

~a =[a1 a2 a3

]

~e1

~e2

~e3

= a~T~e~ (1.34)

or, equivalently:

~a =[~e1 ~e2 ~e3

]

a1

a2

a3

=~e~T a~ (1.35)

The relations ai =~a·~ei can now be written as:

a~ = ~a·~e~ (1.36)

1.3 Second order tensors

1.3.1 Definition

A second order tensor is a linear transformation of a vector into another vector:

A·~p = ~q (1.37)

or in index notation

Ai j p j = qi (1.38)

Note that we limit ourselves here to so-called Cartesian tensors, which allow orthogonal lin-

ear coordinate transformations only. In a three-dimensional space, a second-order tensor is

fully defined by the image of three independent vectors, see figure 1.12. Indeed, any set of

three independent vectors constitutes a vector basis by means of which any other vector can

be expressed. The image of an arbitrary vector can be found trivially through the images of

three such vectors. This property is a direct consequence of the linearity of the transforma-

tion, which is expressed by:

A·(α~m +β~n) = αA·~m +βA·~n (1.39)

For an arbitrary vector ~a = a1~e1 +a2~e2 +a3~e3 the image vector thus equals

A·~a = a1(A·~e1)+a2(A·~e2)+a3(A·~e3) (1.40)

8

1.3. Second order tensors

A

r

ar

b

r

c

r

r

d a= ×A

e b= ×A

r

r

r

r

f c= ×A

Figure 1.12: A second-order tensor A in a three-dimensional space.

1.3.2 Some properties

Any second order tensor may be represented by a finite, nonunique, sum of dyads:

A = α1~a1~b1 +α2~a2

~b2 +α3~a3~b3 + . . . (1.41)

Two special tensors are the zero tensor O and the unit tensor6 I . If applied to vector ~p, these

tensors yield the following:

O·~p = ~0 (1.42)

I ·~p = ~p (1.43)

The index notations for O and I are respectively

0 ⇔ 0i j (1.44)

I ⇔ δi j (1.45)

where the Kronecker delta δi j is used for the unit tensor.

The transpose or conjugate AT of a tensor A is defined by the following relation:

AT ·~p = ~p·A (1.46)

In index notation it holds that

A ⇔ Ai j =⇒ AT ⇔ A j i (1.47)

For the special case A =~a~b, we find:

(~a~b)T ·~p = ~p·(~a~b)

= (~p·~a)~b

= ~b(~p·~a)

= ~b(~a·~p)

= (~b~a)·~p (1.48)

6tweede-orde eenheidstensor

9

1. TENSOR CALCULUS

showing that:

(~a~b)T = ~b~a (1.49)

If A is written as a sum of dyads, the conjugate AT is obtained by reversing the two vectors in

each dyad. Tensors can be multiplied by a scalar, summed with another tensor, or multiplied

by another tensor:

(αA)·~v = A·(α~v), (1.50)

(A+B )·~v = A·~v +B ·~v , (1.51)

(A·B )·~v = A·(B ·~v) (1.52)

Note that (in general):

A·B 6= B ·A (1.53)

The inner product of a tensor with itself is called the square of a tensor, i.e.,

A2 = A·A (1.54)

Likewise, higher-order exponents are defined through repeated inner products, e.g.,

A3 = A·A·A (1.55)

1.3.3 Tensor components; matrix notation of a tensor

Consider a three-dimensional space with a basis ~e1,~e2,~e3. Then, according to (1.30), all

vectors ~ai and~bi in relation (1.41) can be written with respect to this basis. Substitution of

these expressions in (1.41) results into:

A = A11~e1~e1 + A12~e1~e2 + A13~e1~e3+A21~e2~e1 + A22~e2~e2 + A23~e2~e3+A31~e3~e1 + A32~e3~e2 + A33~e3~e3 (1.56)

where the coefficients Ai j are functions of the coefficients αi and the components of the

vectors ~a and~b. Note that the coefficients now coincide with the index notation. Thus the

tensor A can be expressed as a finite sum of 9 dyads of the base vectors. The coefficients Ai j

are called the components of the tensor A with respect to the basis ~e1,~e2,~e3. In general, a

component Ai j of a tensor A can be obtained by:

Ai j = ~ei ·A·~e j (1.57)

as can be verified from substitution of (1.56) in this equation. For convenience, we can store

the components Ai j in a matrix A:

A =

A11 A12 A13

A21 A22 A23

A31 A32 A33

=

(~e1·A·~e1) (~e1·A·~e2) (~e1·A·~e3)

(~e2·A·~e1) (~e2·A·~e2) (~e2·A·~e3)

(~e3·A·~e1) (~e3·A·~e2) (~e3·A·~e3)

=

~e1

~e2

~e3

·A·[~e1 ~e2 ~e3

]

= ~e~·A·~e~T (1.58)

10

1.3. Second order tensors

Using the matrix A, relation (1.56) can be rewritten as:

A =[~e1 ~e2 ~e3

]

A11 A12 A13

A21 A22 A23

A31 A32 A33

~e1

~e2

~e3

= ~e~T A~e~ (1.59)

Together, equations (1.58) and (1.59) describe the transformation from tensor to matrix and

back.

In (1.43), the unit tensor I was defined. After writing I in the general form (1.56) and

applying definition (1.43), it can be derived that the unit tensor can be written as:

I = ~e1~e1 +~e2~e2 +~e3~e3 (1.60)

Thus, for the unit matrix I belonging to the unit tensor I it holds:

I = ~e~·~e~T =

1 0 0

0 1 0

0 0 1

(1.61)

1.3.4 Scalar functions of a tensor

Euclidean norm

The Euclidean norm7 m of a tensor, a measure of its magnitude, is defined by:

m = ‖A‖ = max~e

‖A·~e‖ ∀~e with ‖~e‖ = 1. (1.62)

The norm is an invariant quantity8, i.e. it is independent of the choice of the basis. With

respect to the norm of a tensor, the following properties hold:

‖A‖ ≥ 0 (1.63)

‖αA‖ = |α| ‖A‖ (1.64)

‖A·B‖ ≤ ‖A‖ ‖B‖ (1.65)

‖A +B‖ ≤ ‖A‖+‖B‖ (1.66)

The first invariant

The first invariant of a tensor A, also called the trace9 of A, is defined as:

J1(A) = tr(A)

= 1~c1×~c2·~c3

[~c1·A·(~c2 ×~c3)+~c2·A·(~c3 ×~c1)+~c3·A·(~c1 ×~c2)] (1.67)

7Euclidische norm8invariant9spoor

11

1. TENSOR CALCULUS

where ~c1,~c2,~c3 represents a vector basis. Since J1(A) is an invariant quantity, it can also be

calculated using an orthonormal basis ~e1,~e2,~e3:

J1(A) = ~e1·A·~e1 +~e2·A·~e2 +~e3·A·~e3 (1.68)

= A11 + A22 + A33 (1.69)

= trA (1.70)

The following properties hold for the first invariant:

J1(A) = J1(AT ) (1.71)

J1(I ) = 3 (1.72)

J1(αA) = αJ1(A) (1.73)

J1(A +B ) = J1(A)+ J1(B ) (1.74)

The second invariant

The second invariant of a tensor A is defined as:

J2(A) = 12

tr2(A)−tr(A2)

= 1~c1×~c2·~c3

[ ~c1·(A·~c2)× (A·~c3)+~c2·(A·~c3)× (A·~c1)+~c3·(A·~c1)× (A·~c2)] (1.75)

The following properties hold for the second invariant:

J2(A) = J2(AT ) (1.76)

J2(I ) = 3 (1.77)

J2(αA) = α2 J2(A) (1.78)

The third invariant

The third invariant of a tensor A, also called the determinant of A, is defined as:

J3(A) = det(A) = 1~c1×~c2·~c3

(A·~c1)·(A·~c2)× (A·~c3) (1.79)

For the third invariant, the following properties hold:

J3(A) = J3(AT ) (1.80)

J3(I ) = 1 (1.81)

J3(αA) = α3 J3(A) (1.82)

J3(A·B ) = J3(A)J3(B ) (1.83)

The determinant of a tensor equals the determinant of its matrix with respect to an orthonor-

mal basis.

det(A) = det(A) (1.84)

If det(A) is nonzero, i.e. the triple product of the transformed base vectors is nonzero, the

tensor is called a regular tensor10. If det(A) is zero, the tensor A is called a singular tensor11.

10reguliere tensor11singulier

12

1.3. Second order tensors

In the latter case all vectors in three-dimensional space are transformed into vectors lying in

a plane (or on a line). Vectors perpendicular to that plane (or that line), are transformed into

the zero vector.

Scalar product

The scalar product, double dot product or double inner product or convolution12 of two ten-

sors A and B is the contraction of these two tensors, denoted by A : B , is a scalar, defined

as:

A : B = tr(A·B ) (1.85)

In index notation, this product is the full contraction of indices

A : B = Ai j B j i (1.86)

Some properties are:

A : I = tr(A) (1.87)

A : B = B : A (1.88)

= AT : B T (1.89)

= B T : AT (1.90)

Some properties of the invariants of a tensor

The three principal invariants of a tensor are related through the Caley-Hamilton theorem:

A3 − J1(A)A2 + J2(A)A − J3(A)I = 0 (1.91)

The conjugated tensor AT satisfies the same equation.

For any non-singular tensor it holds that:

J1(A−1) =J2(A)

J3(A)(1.92)

J2(A−1) =J1(A)

J3(A)(1.93)

J3(A−1) =1

J3(A)(1.94)

Any function of the principal invariants of a tensor A is an invariant as well.

1.3.5 Eigenvalues and eigenvectors of a tensor

The eigenvalues13 λ and the eigenvectors14~n of a tensor A are defined as those scalars and

vectors that satisfy the following equation:

A·~n = λ~n with ~n 6=~0 (1.95)

12dubbel inwendig produkt13eigenwaarden14eigenvectoren

13

1. TENSOR CALCULUS

To compute the eigenvalues and eigenvectors of a tensor, we proceed as follows. First equa-

tion (1.95) is rewritten as:

A·~n = λ~n ⇒A·~n −λ~n = ~0 ⇒

A·~n −λI ·~n = ~0 ⇒(A −λI )·~n = ~0 (1.96)

Since ~n 6=~0, the tensor A −λI must be singular, meaning that:

det(A −λI ) = 0 ⇒ det(A−λI ) = 0 (1.97)

Elaboration of the latter equation leads to the characteristic equation:

λ3 − J1(A)λ2 + J2(A)λ− J3(A) = 0 (1.98)

This equation has three roots λ1, λ2 and λ3. The eigenvector ~ni corresponding with the

eigenvalue λi is found from the following set of linear equations:

(A−λi I )·~ni = ~0 ⇒ (A−λi I )n~i = 0~ (1.99)

Since this is not an independent set of equations, it yields only the ratio ni 1 : ni 2 : ni 3 of the

components of the eigenvector. A unique solution is obtained by the additional requirement

that the eigenvector has unit length:

‖~ni‖ = 1 ⇒ n2i 1 +n2

i 2 +n2i 3 = 1 (1.100)

The principal invariants of a tensor are then equal to:

J1(A) = λ1 +λ2 +λ3 (1.101)

J2(A) = λ1λ2 +λ1λ3 +λ2λ3 (1.102)

J3(A) = λ1λ2λ3 (1.103)

1.3.6 Some special tensors

Inverse tensor

The inverse tensor A−1 of a tensor A is defined as the tensor that satisfies the following equa-

tion:

A−1·A = I (1.104)

A−1 only exists if A is regular. The components of the inverse tensor are given by the inverse

matrix A−1

[A−1]i j =1

det(A)

[

(−1)i+ j Mi j

]T(1.105)

where Mi j is the minor of the component Ai j , i.e., the determinant of the matrix that is

obtained by deleting row i and column j .

The inverse of the product of two tensors is given by:

(A·B )−1 = B−1·A−1 (1.106)

14

1.3. Second order tensors

The proof is obtained by comparing the following expressions for ~a:

(A·B )·~a = ~b ⇒~a = (A·B )−1·~b (1.107)

and

(A·B )·~a = A·(B ·~a) =~b ⇒B ·~a = A−1·~b ⇒

~a = B−1·A−1·~b (1.108)

Deviatoric part of a tensor

The deviatoric part15 Ad of a tensor A is defined as:

Ad = A− 13

tr(A)I (1.109)

where 13

tr(A)I is called the hydrostatic or spherical16 part of A. For the deviatoric part, the

following properties hold:

(A +B )d = Ad +B d (1.110)

tr(Ad ) = 0 (1.111)

Symmetric tensor

A tensor A is called symmetric if:

AT = A (1.112)

For the components Ai j of a symmetric tensor A with respect to an orthonormal basis it

holds that:

Ai j = A j i (1.113)

i.e. the matrix A corresponding to A is symmetric too. The eigenvalues λi and eigenvectors

~ni of a symmetric tensor have the following important properties:

• the eigenvalues and eigenvectors are real

• if two eigenvalues are different, the corresponding eigenvectors are mutually perpen-

dicular

• if two eigenvalues are equal, the corresponding eigenvectors can be chosen mutually

perpendicular

Thus, the set of eigenvectors can be used as an orthonormal basis. If a tensor A is considered

with respect to the basis of its eigenvectors, we find the spectral form17 of the tensor:

A = A·I (1.114)

= A·(~n1~n1 +~n2~n2 +~n3~n3) (1.115)

= λ1~n1~n1 +λ2~n2~n2 +λ3~n3~n3 (1.116)

15deviatorisch deel16hydrostatisch of sferisch deel17spectrale vorm

15

1. TENSOR CALCULUS

r

n1

r

n 2

r

n 3

Figure 1.13: The orthonormal basis of eigenvectors ~n1,~n2,~n3 of a symmetric tensor.

The matrix A of A with respect to this basis is a diagonal matrix:

A =

λ1 0 0

0 λ2 0

0 0 λ3

(1.117)

Using the spectral form, some functions of a tensor can easily be calculated. E.g. for the

inverse, the square root, and the natural logarithm of a tensor A it holds that:

A−1 = λ−11 ~n1~n1 +λ−1

2 ~n2~n2 +λ−13 ~n3~n3 (1.118)

A1/2 = λ1/21 ~n1~n1 +λ1/2

2 ~n2~n2 +λ1/23 ~n3~n3 (1.119)

ln A = ln(λ1)~n1~n1 + ln(λ2)~n2~n2 + ln(λ3)~n3~n3 (1.120)

Skewsymmetric tensor

A tensor A is called skewsymmetric18 or antisymmetric if:

AT = −A (1.121)

For the components Ai j of a skewsymmetric tensor A with respect to an orthonormal basis

it holds that:

Ai j = −A j i ⇒ Ai i = 0 (1.122)

If A is skewsymmetric and B is symmetric, then:

A : B = 0 especially A : I = trA = 0 (1.123)

The proof follows from:

A : B = AT : B T = −A : B (1.124)

If A is skewsymmetric, then a unique axial vector19~ω exists, such that:

A·~q = ~ω×~q (1.125)

The proof of the existence of ~ω is as follows:

A·~q = ~p ⇒~q ·A·~q = ~q ·AT ·~q

= −~q ·A·~q ⇒~q ·A·~q = 0 ⇒

~q ·~p = 0 ⇒~q ⊥ ~p ⇒ ∃~ω such that ~p = ~ω×~q (1.126)

18scheefsymmetrisch19axiaalvector

16

1.3. Second order tensors

On the other hand, if:

A·~q = ~ω×~q ∀ ~q (1.127)

comparison of the following expressions:

A·~q =[~e1 ~e2 ~e3

]

A11 A12 A13

A21 A22 A23

A31 A32 A33

q1

q2

q3

=[~e1 ~e2 ~e3

]

A11q1 + A12q2 + A13q3

A21q1 + A22q2 + A23q3

A31q1 + A32q2 + A33q3

(1.128)

and

~ω×~q = (ω1~e1 +ω2~e2 +ω3~e3)× (q1~e1 +q2~e2 +q3~e3)

=ω1q2(~e3)+ω1q3(−~e2)+ω2q1(−~e3)+ω2q3(~e1)+ω3q1(~e2)+ω3q2(−~e1)

=[~e1 ~e2 ~e3

]

ω2q3 −ω3q2

ω3q1 −ω1q3

ω1q2 −ω2q1

(1.129)

shows that:

A =

0 −ω3 ω2

ω3 0 −ω1

−ω2 ω1 0

(1.130)

and thus that the tensor A is skewsymmetric.

Positive-definite tensor

A tensor A is called positive-definite20 if:

~a·A·~a > 0 ∀~a 6=~0 (1.131)

A positive-definite tensor A cannot be skewsymmetric. This follows from:

~a·A·~a = ~a·AT ·~a> 0 ∀~a 6=~0 ⇒ (1.132)

~a·(A − AT )·~a = 0 ∧ ~a 6=~0 (1.133)

If A would be skewsymmetric, the latter equation would yield zero, since then AT =−A.

Orthogonal tensor

A tensor A is called orthogonal21 if:

(A·~a)·(A·~b) = ~a·~b ∀ ~a,~b (1.134)

20positief definiet21orthogonale tensor

17

1. TENSOR CALCULUS

For the special case~b =~a we get:

(A·~a)·(A·~a) = ~a·~a ⇒ ‖A·~a‖ = ‖~a‖ (1.135)

Thus, an orthogonal transformation preserves lengths of vectors and angles between vectors.

The inverse of an orthogonal tensor is identical to its conjugate:

AT = A−1 (1.136)

as follows from:

(A·~a)·(A·~b) = ~a·AT ·A·~b= ~a·~b ⇒

A·AT = I ⇒AT = A−1 (1.137)

Moreover, this equation shows that:

det(A·AT ) = [det(A)]2 = det(I ) = 1 ⇒det(A) = ±1 (1.138)

Thus, an orthogonal tensor is a regular tensor. If det(A) = 1, A can be shown to describe a

rotation. If det(A) =−1, A can be shown to describe a reflection.

1.4 Spatial derivatives of scalar and vector fields

1.4.1 Derivatives of functions of one or more variables

x x dx+

f x( )

f x dx( )+

df

dx

df

dxdx

Figure 1.14: A function f of one variable x and its derivative.

Consider a scalar function f (x) of one variable x. Using a Taylor series, the change d f of

the value of f due to a change d x in the parameter x can be expressed as:

d f = f (x +d x)− f (x)

= f (x)+ d f

dx

∣∣∣

xd x + 1

2

d2 f

dx2

∣∣∣

xd x2 +·· ·− f (x)

≈ d f

dx

∣∣∣

xd x (1.139)

18

1.4. Spatial derivatives of scalar and vector fields

Extension to a function of two variables f (x, y) yields:

d f = f (x +d x, y +d y)− f (x, y)

= f (x, y)+ ∂ f

∂x

∣∣∣

x,yd x + ∂ f

∂y

∣∣∣

x,yd y +·· ·− f (x, y)

≈ ∂ f

∂x

∣∣∣

x,yd x + ∂ f

∂y

∣∣∣

x,yd y (1.140)

And for a function a(x1, x2, x3) of three variables, the variation d a of a, due to (infinitesimal)

variations d x1, d x2 and d x3 can be written as:

d a = d x1∂a∂x1

+d x2∂a∂x2

+d x3∂a∂x3

(1.141)

The index notation for differentiation with respect to one of the variables (i.c. xi ) is com-

monly denoted with a comma between or in front of the indices. The equation (1.141) for d a

with this convention reads

d a = d xi a,i (1.142)

1.4.2 Components of the position vector

In three-dimensional space, we may interpret the variables x1, x2 and x3 in the function

a(x1, x2, x3) as the components of the position vector22~x with respect to a Cartesian basis

~e1,~e2,~e3. The position vector ~x of a point with respect to a fixed point in space, the origin

O, can be expressed as:

~x = x1~e1 +x2~e2 +x3~e3 (1.143)

For the position vector~x +d~x of a neighbouring point we can write:

r

e1

r

e2

r

e3

x1

r

e1

r

e2

r

e3

x2

x3

dxr

r

x

Figure 1.15: Components xi of the position vector ~x with respect to a Cartesian basis

~e1,~e2,~e3.

~x +d~x = (x1 +d x1)~e1 + (x2 +d x2)~e2 + (x3 +d x3)~e3 (1.144)

22positievector

19

1. TENSOR CALCULUS

and thus the vector d~x can be expressed as:

d~x = d x1~e1 +d x2~e2 +d x3~e3 (1.145)

From the latter relation it is obvious that the changes d x1, d x2 and d x3 can be related to the

vector d~x according to:

d x1 = d~x·~e1 (1.146)

d x2 = d~x·~e2 (1.147)

d x3 = d~x·~e3 (1.148)

We could also express the function a(~x) in terms of the components r , θ, z of the position

vector~x with respect to a cylindrical basis ~er ,~et ,~ez. Then the position vector equals:

~x = r~er (θ)+ z~ez (1.149)

Within the cylindrical basis, the position~x +d~x of a neighbouring point equals:

r

e1

r

e2

r

e3

r

er

r

et

r

ez

z

qr

x1

x2

x3

r

x

dxr

Figure 1.16: Components r , θ, z of the position vector ~x with respect to a cylindrical basis

~er ,~et ,~ez.

~x +d~x = (r +dr )~er (θ+dθ)+ (z +d z)~ez

= (r +dr )~er (θ)+ d~er

dθ dθ+ (z +d z)~ez

= r~er (θ)+ z~ez+r~et (θ)dθ+dr~er (θ)+~et (θ)dr dθ+d z~ez (1.150)

from which it can be concluded that, in a linear approximation, the vector d~x equals:

d~x ≈ dr~er (θ)+ r dθ~et (θ)+d z~ez (1.151)

20

1.4. Spatial derivatives of scalar and vector fields

The changes dr , dθ and d z are related to the vector d~x according to:

dr = d~x·~er (1.152)

dθ = 1r

d~x·~et (1.153)

d z = d~x·~ez (1.154)

1.4.3 The gradient operator

Considering a as a function of the position vector~x, we can investigate the variation d a due

to a variation d~x:

d a = d x1∂a∂x1

+d x2∂a∂x2

+d x3∂a∂x3

= (d~x·~e1) ∂a∂x1

+ (d~x·~e2) ∂a∂x2

+ (d~x·~e3) ∂a∂x3

= d~x·[

~e1∂a∂x1

+~e2∂a∂x2

+~e3∂a∂x3

]

= d~x·(~∇a) (1.155)

The symbol~∇ is called the gradient operator23:

~∇ =[

~e1∂

∂x1+~e2

∂∂x2

+~e3∂

∂x3

]

(1.156)

= ~e~T ∇~ (1.157)

and the expression ~∇a is called the gradient of a. The gradient of a vector function ~a is

obtained similarly:

d~a = d x1∂~a∂x1

+d x2∂~a∂x2

+d x3∂~a∂x3

= (d~x·~e1) ∂~a∂x1

+ (d~x·~e2) ∂~a∂x2

+ (d~x·~e3) ∂~a∂x3

= d~x·[

~e1∂~a∂x1

+~e2∂~a∂x2

+~e3∂~a∂x3

]

= d~x·(~∇~a) (1.158)

or in index notation

d a j = d xi a j ,i (1.159)

For the special case ~a =~x we find:~∇~x = I (1.160)

The gradient operator~∇ can be applied to a vector ~a in several manners, the result of which

is called the gradient of ~a (grad ~a), the divergence24 of ~a (div ~a), or the rotation25 or curl of ~a

(rot ~a):

grad ~a = ~∇~a =~e1∂~a∂x1

+~e2∂~a∂x2

+~e3∂~a∂x3

, (1.161)

div ~a = ~∇·~a =~e1· ∂~a∂x1+~e2· ∂~a∂x2

+~e3· ∂~a∂x3, (1.162)

rot ~a = ~∇×~a =~e1 × ∂~a∂x1

+~e2 × ∂~a∂x2

+~e3 × ∂~a∂x3

(1.163)

23gradiënt24divergentie25rotatie

21

1. TENSOR CALCULUS

Note that div ~a yields a scalar, rot ~a yields a vector, and grad ~a yields a second order tensor.

Applied to tensors in general, the gradient increases the order by 1, the divergence dimin-

ishes the order by 1, while the order remains unaffected by the curl.

The expression for the gradient operator in cylindrical coordinates can be found follow-

ing a similar derivation as in the case of a Cartesian coordinate system:

d a = dr ∂a∂r

+dθ ∂a∂θ +d z ∂a

∂z

= (d~x·~er )∂a∂r

+ ( 1r

d~x·~et )∂a∂θ + (d~x·~ez)∂a

∂z

= d~x·[

~er∂a∂r

+ 1r~et

∂a∂θ

+~ez∂a∂z

]

= d~x·(~∇a) (1.164)

Thus the expression for the gradient operator in a cylindrical coordinate system equals:

~∇ = ~er∂∂r

+~et1

r∂∂θ

+~ez∂∂z

(1.165)

1.5 Tensor formula involving first and second-order tensors

The following relations on differentiable scalar functions φ, vectors ~a,~b, tensors A,B ,C , in-

ner/outer products and their derivatives can be proven:

A : (B ·C ) = (A·B ) : C (1.166)

~∇(φ~a) = (~∇φ)~a +φ~∇~a (1.167)

~∇·(φ~a) = (~∇φ)·~a+φ~∇·~a (1.168)

~∇(~a·~b) = (~∇~a)·~b + (~∇~b)·~a (1.169)

~∇·(~a~b) = (~∇·~a)~b +~a·~∇~b (1.170)

~∇·(φA) = (~∇φ)·A+φ~∇·A (1.171)

~∇·(A·~a) = (~∇·A)·~a + A : (~∇~a)T (1.172)

~∇·~a = (~∇~a) : I = tr(~∇~a) (1.173)

1.6 Higher-order and fourth-order tensors

1.6.1 Definition

Some problems invoke the transformation of tensors. Such transformations are carried out

by means of higher-order tensors. A general definition of a tensor could be: an invariant

object with an arbitrary number of indices. The number of indices is the order (or rank)

of the tensor. The transformation which is most relevant in the mechanics of materials is

the transformation of a second-order tensor into another second-order tensor by means of

a fourth-order tensor. An arbitrary tensor A (which is a linear transformation of vectors)

is mapped onto a different tensor B (which is another linear transformation of vectors) by

means of the double inner product with a fourth-order tensor4C , see figure 1.17.

B =4

C : A (1.174)

22

1.6. Higher-order and fourth-order tensors

A B

4

Cr

a

r

b

r

c

e b= ×A

r

r

d a= ×A

r

r

r

r

f c= ×A

r

g

r

h

r

i

k h= ×B

r

rr

l i= ×B

rr

j g= ×B

Figure 1.17: A fourth-order tensor4C that maps a (linear) second-order tensor A onto a

second-order tensor B .

Using index tensor notations, the tensor Al k transforms to Bi j according to

Bi j = Ci j kl Al k (1.175)

The tensor Ci j kl which was used to perform this transformation has 4 indices.

1.6.2 Properties

Since a fourth-order tensor is built on quadrades rather than dyades, some new calculus

rules apply. Each fourth-order tensor can be written as a finite sum of quadrades. With

respect to a two-dimensional Cartesian basis, a tensor4C would be given by si xteen terms,

i.e.,

4

C =C1111~e1~e1~e1~e1 +C1112~e1~e1~e1~e2 +C1121~e1~e1~e2~e1 +C1122~e1~e1~e2~e2

C1211~e1~e2~e1~e1 +C1212~e1~e2~e1~e2 +C1221~e1~e2~e2~e1 +C1222~e1~e2~e2~e2

C2111~e2~e1~e1~e1 +C2112~e2~e1~e1~e2 +C2121~e2~e1~e2~e1 +C2122~e2~e1~e2~e2

C2211~e2~e2~e1~e1 +C2212~e2~e2~e1~e2 +C2221~e2~e2~e2~e1 +C2222~e2~e2~e2~e2

(1.176)

In a three-dimensional space a tensor4C has 3× 3×3×3 = 81 components. The correct full

matrix representation of a higher-order tensor can only be achieved with multi-dimensional

matrices (which exist in modern computing languages). A fourth-order 3D tensor is then

represented by a [3×3×3×3] matrix.

Different types of transposed tensors are defined: the total transpose4C

T, the left trans-

pose4C

LTor the right transpose

4C

RT. If the tensor

4C is denoted Ci j kl in index notation,

23

1. TENSOR CALCULUS

each of these transposes is given by

4

CT⇔ Cl k j i (1.177)

4

CLT

⇔ C j i kl (1.178)

4

CRT

⇔ Ci j l k (1.179)

Different types of minor symmetries can be defined for a fourth-order tensor. A left symmet-

ric fourth-order tensor, is a tensor for which

4

C =4

CLT

=⇒ Ci j kl = C j i kl (1.180)

In a 3D space, it has 81−1×3×3×3= 54 distinct components. A right symmetric fourth-order

tensor is given by4

C =4

CRT

=⇒ Ci j kl = Ci j l k (1.181)

In a 3D space, it has 81−3×3×3×1 = 54 distinct components. A tensor which is left and

right symmetric, has only 54− 1× 3× 3× 2 = 36 distinct components. A middle symmetric

fourth-order tensor complies with

4

C =4

CT

=⇒ Ci j kl = Cl k j i (1.182)

In a 3D space, it has 81−2×3×3×2= 45 distinct components. A total symmetric fourth-order

tensor equals is left, right and central symmetric, i.e.,

4

C =4

CLT

=4

CRT

=4

CT

(1.183)

In a 3D space, it has 36− 12

6×5 = 21 distinct components.

The fourth-order null tensor40 is defined by

4

0 : A = 0 (1.184)

The fourth-order unit tensor4I maps a second-order tensor A on itself

4

I : A = A (1.185)

This tensor is given in index notation by

4

I =⇒ Ii j kl = δi lδ j k (1.186)

or with respect to a 3D Cartesian basis

4

I =~e1~e1~e1~e1 +~e1~e2~e2~e1 +~e1~e3~e3~e1

~e2~e1~e1~e2 +~e2~e2~e2~e2 +~e2~e3~e3~e2

~e3~e1~e1~e3 +~e3~e2~e2~e3 +~e3~e3~e3~e3

(1.187)

The symmetric fourth-order unit tensor4I

smaps a second-order tensor A on the sym-

metric part of A, denoted As :4

Is

: A = As = 12

(A+ AT ) (1.188)

24

1.6. Higher-order and fourth-order tensors

This tensor is given in index notation by

4

Is⇔ I s

i j kl = 12

(

δi lδ j k +δi kδ j l

)

(1.189)

This tensor is a total symmetric tensor, i.e., left symmetric, right symmetric and central sym-

metric. Note that this tensor is also given by

4

Is= 1

2(

4

I +4

IRT

) (1.190)

or with respect to a Cartesian basis by

4

Is=~e1~e1~e1~e1 + 1

2~e1~e2~e2~e1 + 1

2~e1~e2~e1~e2 + 1

2~e1~e3~e3~e1 + 1

2~e1~e3~e1~e3

12~e2~e1~e1~e2 + 1

2~e2~e1~e2~e1 +~e2~e2~e2~e2 + 1

2~e2~e3~e3~e2 + 1

2~e2~e3~e2~e3

12~e3~e1~e1~e3 + 1

2~e3~e1~e3~e1 + 1

2~e3~e2~e2~e3 + 1

2~e3~e2~e3~e2 +~e3~e3~e3~e3

(1.191)

Fourth-order tensors can be partially modified through inner products with second-order

tensors or double inner products with other fourth-order tensors. The inner product be-

tween a fourth-order tensor4A and a second-order tensor B yields a fourth-order tensor

given by

4

C =4

A·B (1.192)

Ci j kl = Ai j km Bml (1.193)

The double inner product between a fourth-order tensor4A and another fourth-order tensor

4B again leads to a fourth-order tensor

4C according to

4

C =4

A :4

B (1.194)

Ci j kl = Ai j mn Bnmkl (1.195)

Note that, in general, the double inner product used in combination with fourth-order ten-

sors is no longer commutative

4

A : B 6= B :4

A (1.196)4

A :4

B 6=4

B :4

A (1.197)

1.6.3 Tensor rules

4

A : (B ·C ) = (4

A·B ) : C (1.198)

A·B +B T ·AT =4

Is

: (A·B ) (1.199)

= (4

Is·A) : B (1.200)

1.6.4 Matrix condensation of the double inner product with a

second-order tensor

In many numerical implementations, tensor products are condensed in a two-dimensional

matrix format once a Cartesian basis has been chosen. Making use of special properties

25

1. TENSOR CALCULUS

of the tensors involved then permits to omit a considerable amount of computations for a

double inner product. Components of a second-order tensor B are normally represented as

a matrix

B =

B11 B12 B13

B21 B22 B32

B31 B32 B33

(1.201)

For the computation of a double inner product with a fourth-order tensor, it is often more

convenient to store the 9 components in a column B~ Taking into account the symmetry of a

second-order tensor B permits to store the 6 distinct components in a column rather than

in a matrix, e.g.,

B~ =

B11

B12

B13

B21

B22

B23

B31

B32

B33

(1.202)

On the basis of this notation, it is possible to obtain the components of second-order tensor

C as a similarly structured column, which is the double inner product between a fourth-

order tensor4A and a second-order tensor B , as a matrix-column product, i.e., the compo-

nents of

C =4

A : B (1.203)

are obtained through

C~ = A B~ (1.204)

C11C12C13C21C22C23C31C32C33

=

A1111 A1121 A1131 A1112 A1122 A1132 A1113 A1123 A1133A1211 A1221 A1231 A1212 A1222 A1232 A1213 A1223 A1233A1311 A1321 A1331 A1312 A1322 A1332 A1313 A1323 A1333A2111 A2121 A2131 A2112 A2122 A2132 A2113 A2123 A2133A2211 A2221 A2231 A2212 A2222 A2232 A2213 A2223 A2233A2311 A2321 A2331 A2312 A2322 A2332 A2313 A2323 A2333A3111 A3121 A3131 A3112 A3122 A3132 A3113 A3123 A3133A3211 A3221 A3231 A3212 A3222 A3232 A3213 A3223 A3233A3311 A3321 A3331 A3312 A3322 A3332 A3313 A3323 A3333

B11B12B13B21B22B23B31B32B33

(1.205)

Note that the matrix A has indeed 81 components (all the components of4A), although this

matrix is only valid in conjunction with a double inner product with a second-order tensor.

In the particular case of mechanics, symmetric second-order tensors play an important

role. Taking into account the symmetry of a second-order tensor B permits to store the 6

distinct components in a smaller column with a somewhat different structure

B~ =

B11

B22

B33

B12

B13

B23

(1.206)

26

1.6. Higher-order and fourth-order tensors

If the resulting tensor C is not symmetric, it is represented by a column with 9 components.

The matrix A is restructured and condensed to a [9×6] matrix by adding up the appropri-

ate columns in equation (1.205), i.e., if B~ has the format (1.206) and C~ maintains the for-

mat (1.202) then C = 4A : B is represented in component form by

C~ = A B~ (1.207)

C11C12C13C21C22C23C31C32C33

=

A1111 A1122 A1133 (A1112+A1121) (A1113+A1131) (A1123+A1132)A1211 A1222 A1233 (A1212+A1221) (A1213+A1231) (A1223+A1232)A1311 A1322 A1333 (A1312+A1321) (A1313+A1331) (A1323+A1332)A2111 A2122 A2133 (A2112+A2121) (A2113+A2131) (A2123+A2132)A2211 A2222 A2233 (A2212+A2221) (A2213+A2231) (A2223+A2232)A2311 A2322 A2333 (A2312+A2321) (A2313+A2331) (A2323+A2332)A3111 A3122 A3133 (A3112+A3121) (A3113+A3131) (A3123+A3132)A3211 A3222 A3233 (A3212+A3221) (A3213+A3231) (A3223+A3232)A3311 A3322 A3333 (A3312+A3321) (A3313+A3331) (A3323+A3332)

B11B22B33B12B13B23

(1.208)

If the fourth-order tensor presents a symmetry, the equations can be simplified considerably.

A right symmetric tensor has only 54 distinct components (3 columns equal 3 other columns

in the matrix A). A left symmetric tensor has also only 54 distinct components (3 lines equal

3 other lines in the matrix A). A tensor which is left and right symmetric has only 36 inde-

pendent components (only 6 lines and columns remain in the matrix A). The case where

the tensor4A is left symmetric results in a tensor C which is symmetric. Applying C = 4

A : B

again to a symmetric B can then be done with columns B~ and C~ of the form (1.206). The

result in this particular case is then condensed as

C~ = A B~ (1.209)

C11C22C33C12C13C23

=

A1111 A1122 A1133 A1112+A1121 A1113+A1131 A1123+A1132A2211 A2222 A2233 A2212+A2221 A2213+A2231 A2223+A2232A3311 A3322 A3333 A3312+A3321 A3313+A3331 A3323+A3332A1211 A1222 A1233 A1212+A1221 A1213+A1231 A1223+A1232A1311 A1322 A1333 A1312+A1321 A1313+A1331 A1323+A1332A2311 A2322 A2333 A2312+A2321 A2313+A2331 A2323+A2332

B11B22B33B12B13B23

(1.210)

Other simplifications are easy to make and left to the reader.

27

1. TENSOR CALCULUS

1.7 Summary

Vector operations

Multiplication ~b =α~a

Sum ~c =~a +~b

Inner product ~a·~b = ||~a|| ||~b||cos(φ)

~a·~b =~b·~a

Cross product ~a×~b =~c = ||~a|| ||~b||sin(φ)~b ×~a =−~a ×~b

Triple product ~a×~b·~c = ||~a|| ||~b||sin(φ)||~c||cos(ψ)

Tensor product (dyad) ~a~b

~a~b·~p =~a(~b·~p)

(~a~b)T =~b~a

Gradient operators

Cartesian ~∇=~e1∂

∂x1+~e2

∂∂x2

+~e3∂

∂x3

grad ~a grad ~a =~∇~a =~e1∂~a∂x1

+~e2∂~a∂x2

+~e3∂~a∂x3

~∇~a = a j ,i~ei~e j

div ~a div ~a =~∇·~a =~e1· ∂~a∂x1+~e2· ∂~a∂x2

+~e3· ∂~a∂x3

rot ~a rot ~a =~∇×~a =~e1 × ∂~a∂x1

+~e2 × ∂~a∂x2

+~e3 × ∂~a∂x3

Cylindrical ~∇=~er∂∂r

+~et1r

∂∂θ

+~ez∂∂z

28

1.7. Summary

Second-order Tensors

Second order tensor A·~p =~q

A·(α~m +β~n) =αA·~m +βA·~n(αA)·~v = A·(α~v)

(A +B )·~v = A·~v +B ·~v(A·B )·~v = A·(B ·~v)

Conjugate of tensor AT ·~p = ~p·A

Unit tensor I =~e1~e1 +~e2~e2 +~e3~e3

First invariant J1(A) = tr(A)

Second invariant J2(A) = 12

tr2(A)− tr(A2)

Third invariant J3(A) = det(A)

Double dot product A : B = tr(A·B )

A : I = tr(A)

A : B = AT : B T

Inverse tensor A−1·A = I

(A·B )−1 = B−1·A−1

Deviatoric part Ad = A − 13

tr(A)I

Symmetric tensor AT = A

Skew-symmetric tensor AT =−A

Positive definite tensor ~a·A·~a > 0 ∀~a 6=~0

Orthogonal tensor (A·~a)·(A·~b) =~a·~b ∀ ~a,~b

29

1. TENSOR CALCULUS

Fourth-order Tensors

Transformation B = 4C : A

Components [3×3×3×3] matrix

Total transpose4C

T= [Ci j kl ]T =Cl k j i

Left transpose4C

LT= [Ci j kl ]LT =C j i kl

Right transpose4C

RT= [Ci j kl ]RT =Ci j l k

Left symmetric4C = 4

CLT

Right symmetric4C = 4

CRT

Middle symmetric4C = 4

CT

Total symmetric4C = 4

CLT

= 4C

RT= 4

CT

Fourth-order unit tensor4I : A = A

Symmetric fourth-order unit4I

s: A = As = 1

2(A + AT )

Products4C = 4

A·B4C = 4

A :4B

Matrix condensation B = 4C : A =⇒ B~ =C A~

30

CH

AP

TE

R

2KINEMATICS AND SMALL DEFORMATIONS

This chapter treats fundamental aspects of deformation in solid mechanics. Starting from

a general approach, in which the basic large deformation tensors are introduced, the limit

case for small deformations is treated as a special case. It is shown that the kinematic re-

lations used in linear elasticity are obtained through linearization of the general nonlinear

kinematic equations.

2.1 General aspects of deformation

Upon deformation of a material, a fundamental distinction is made between the material

(or Lagrangian) framework and the spatial (Eulerian) framework. The governing kinematic

relations in continuum mechanics are based on the one-to-one correspondence between

these two frameworks.

2.1.1 Continuity

The continuity of the displacements is one of the basic assumptions in classical continuum

mechanics. This assumption implies that the material is described in an average fashion,

without considering the physical discrete microstructure of the material (e.g., molecular or

crystalline structure). The neighbourhood of a material point is considered to be dense and

continuous. The kinematic and mechanical quantities that ensue from such a theory are

therefore considered as average values over the real microstructure. If a material deforms,

changes in distance between ’dense’ material points will occur. The transformation from

the undeformed state to the deformed state is a one-to-one mapping, where each material

point in the deformed state is associated with one and only one material point from the

undeformed state. The continuity of the deformation implies that neighbourhoods of a ma-

terial point P0 in the initial region R0 remain neighbourhoods of the transformed material

point P in the transformed region R , see figure 2.1. Fracture and damage can therefore not

be treated trivially in this framework. Solution strategies for these cases are sketched in the

courses of fracture and damage mechanics.

2. KINEMATICS

R

V0

tt=0

P0

P

V

R0

Figure 2.1: Deformation of a continuum.

2.1.2 Lagrangian versus Eulerian coordinates

Consider a closed undeformed region R0 in a material, which after deformation occupies a

deformed region R . The undeformed state or initial state is also the state at time t = 0, while

the deformed or actual state corresponds with time t . Note that the time t is in fact a specific

choice for a state parameter1, which is only a means to uniquely define the current state

of the continuum. For the sake of convenience, we use an independent Euclidean vector

basis to characterize the material points before and after deformation. Each point P0 in the

undeformed subset R0 is characterized by the position vector ~X in the undeformed state,

with coordinates (X1, X2, X3). These coordinates are the material or Lagrangian coordinates.

The material coordinates are

• unique2, i.e. each point has different material coordinates

• constant, i.e. they never change since they do not depend on the deformation.

Hence, they can be considered as labels for the material points.

After deformation, this point P0 will occupy a new position P , characterized by the posi-

tion vector ~x, with coordinates (x1, x2, x3). These coordinates are called the spatial or Eule-

rian coordinates, since they determine a fixed point in space where different material points

may pass by in time. The material flows through space (i.e., a fixed set (x1, x2, x3)) upon de-

formation.

The deformation from each point P0 of the continuum to its corresponding point P , is

described by the so-called nonlinear deformation map φ. If the deformation is traced in

time, the locus of the point P defines a curve

t −→ ~x(t ) =φ(~X , t ) for ~X fixed (2.1)

For each vector ~X , there exists one and only vector ~x for each state t . The transformation

from the vectors ~X to the vectors ~x, or from the coordinates (X1, X2, X3) to the coordinates

(x1, x2, x3), i.e., the nonlinear deformation map φ(t ), is also depicted in figure 2.2. Note

1toestandsparameter2éénduidig

32

2.1. General aspects of deformation

PP0

t

t=0

V0

V

r

r

x X0 = r

x

r

u

0

f

Figure 2.2: The nonlinear deformation map

that the nonlinear deformation map is not a vector but a point-to-point mapping (P0 →P (t ), ∀P0). Mathematically speaking, this one-to-one correspondence may be expressed as

~x = φ(~X , t ) (2.2)

or with the inverse deformation map φ−1

~X = φ−1(~x, t ) (2.3)

The nonlinear deformation map φ has the following properties

• unique, i.e. the one-to-one correspondence between particles with position vector ~x

at time t and material points ~X in the undeformed state.

• continuous, neighbourhoods remain neighbourhoods, no cracks occur. Mathemati-

cally this means that the nonlinear deformation map is differentiable.

This implies that both the coordinates (X1, X2, X3) or (x1, x2, x3) could be used as inde-

pendent variables to describe the deformation. If the material coordinates (X1, X2, X3) are

taken as independent variables, attention is fixed on a definite material particle that evolves

in space. This is the Lagrangian description. If the coordinates (x1, x2, x3) are taken as in-

dependent variables, one fixes in space a geometrical point (x1, x2, x3) and investigates what

particles pass through that point. This is the Eulerian description. A quantity α may there-

fore be either written in its material formα(X1, X2, X3, t ) or its spatial formα(x1, x2, x3, t ). The

spatial form is typically used in fluid mechanics, while the material form is more common in

solid mechanics. The selection of either material or spatial coordinates for the development

of a continuum mechanics framework may have important practical consequences for the

large deformation theories. This topic is a subject of the follow-up course on large defor-

mations. In the present case, we will limit ourselves to problems of the small-deformation

theory. Since the material and the spatial coordinates are then so close together, the influ-

ence of the actual difference between them will become negligible upon linearization. Note

that the position vector at t = 0, i.e., ~x0, exactly equals the material vector ~X . The subscript

33

2. KINEMATICS

0 is therefore often used as well to indicate the differences between the Lagrangian and the

Eulerian setting. Material coordinates are denoted Xi or xI in index notation, while spatial

coordinates are then given by xi . In the case of xI , the uppercase index indicates that the

position vector is evaluated in the initial state, i.e., in the Lagrangian configuration.

2.1.3 The displacement vector

Upon deformation a material point changes position from ~x0 to ~x. This change in position

is characterized by the displacement vector (figure 2.2), which is given by

~u = ~x −~x0 (2.4)

The components of the displacement vector~u with respect to an Euclidean basis are (u1,u2,u3),

also denoted by ui in index notation.

2.1.4 The displacement gradient tensor and deformation gradient tensor

Let’s consider two material points that are infinitely close, i.e., one material point is currently

in the position~x, while its neighbour is in position~x+d~x. The principle of continuity tells us

that these two points were also neighbours in the initial undeformed state. Their positions

in that initial state are respectively given by ~x0 and ~x0 +d~x0. The vector d~x0 that initially

interconnected these two points is thus transformed or deformed into a new vector d~x, see

figure 2.3

P

P0

t

dx dXr

r

0 =

dxr

F

t=0

Figure 2.3: The linearized deformation map - the deformation tensor F

The linear transformation of the vector d~x0 into the vector d~x is described by a second-

order tensor, called the deformation gradient tensor F . From the nonlinear deformation

map φ it appears that

d~x = (~x +d~x)− (~x)

= φ(~x0 +d~x0)−φ(~x0)

= ∂φ∂~x0

·d~x0

= F ·d~x0 (2.5)

From the definition of the gradient operator, equation (1.158), it can be noticed that

d~x = d~x0·~∇0~x (2.6)

= (~∇0~x)T ·d~x0 (2.7)

34

2.1. General aspects of deformation

The deformation gradient tensor F is thus given by

F =∂φ

∂~x0(2.8)

= (~∇0~x)T (2.9)

The second-order tensor F is thus the spatial derivative of the nonlinear deformation map in

the considered material point, which leads to a locally linear deformation map. The tensor is

the transposed Lagrangian gradient of the current position vector. The tensor thus expresses

the deformation with respect to the initial undeformed state. The gradient symbol~∇0 clearly

indicates this with its subscript 0. Since F is a second-order tensor, it is fully identified by

means of the image (i.c. the deformation) of three (i.c. infinitesimal) vectors, which span an

infinitesimal volume as shown in figure 2.4.

PP0 d Xr

1

F

dXr

2

dXr

3

dxr

1

dxr

2

dxr

3

Figure 2.4: The deformation gradient tensor F

The matrix notation F of this tensor F with respect to an Euclidean basis is given by

F =

∂x1

∂X1

∂x1

∂X2

∂x1

∂X3

∂x2

∂X1

∂x2

∂X2

∂x2

∂X3

∂x3

∂X1

∂x3

∂X2

∂x3

∂X3

(2.10)

or in index format

Fi J

= xi ,J

(2.11)

Note that the second index is denoted with uppercase characters, which means that it refers

to the Lagrangian undeformed state, while the first index is in lowercase, corresponding to

the Eulerian deformed state. This convention will be respected in the large deformation part

of the kinematical framework, where the difference between these two states is essential.

As indicated in the chapter on tensor analysis, the comma between the indices in equa-

tion (2.11) indicates differentiation with respect to the second index variable.

The Lagrangian gradient of the displacement also leads to a second order tensor, called

the displacement gradient tensor (~∇0~u)T . Like in equation (2.6) (i.e., the definition of the

Lagrangian gradient of a vector), this tensor maps the initial interconnection vector d~x0 on

the relative displacement vector between the two considered adjacent material points.

d~u = d~x0·~∇0~u (2.12)

= (~∇0~u)T ·d~x0 (2.13)

35

2. KINEMATICS

The relative displacement vector d~u measures the displacement that occurred between these

two adjacent points with respect to their original relative positions

d~u = d~x −d~x0 (2.14)

Using an Euclidean basis, the components of (~∇0~u)T are commonly stored in a 3×3 matrix

∂u1

∂X1

∂u1

∂X2

∂u1

∂X3

∂u2

∂X1

∂u2

∂X2

∂u2

∂X3

∂u3

∂X1

∂u3

∂X2

∂u3

∂X3

(2.15)

Evidently, the tensors F and (~∇0~u)T are closely related

F = (~∇0~x)T

= (~∇0[~x0 +~u])T

= ~∇0~xT0

︸ ︷︷ ︸

I

+ (~∇0~u)T

= I + (~∇0~u)T (2.16)

Since F and (~∇0~u)T are both second-order tensors, they describe linear transformations.

This means that the deformation along three independent vectors in (3D) material point

suffices to fully characterize these tensors. An initially cubic volume element will always be

deformed by F into a parallelepiped.

Consider a volume element that is spanned by three non-coplanar infinitesimal material

vectors d~X1, d~X2 and d~X3. After deformation these vectors are transformed into new vec-

tors d~x1, d~x2 and d~x3. The deformation tensor F gives the relation between the two sets of

vectors

d~x1 = F ·d~X1

d~x2 = F ·d~X2

d~x3 = F ·d~X3

(2.17)

If both sets are known, then F can be determined.

2.2 Strains and strain tensors

2.2.1 Strain definitions

Strains are used to quantify the deformation relative to the undeformed state. It is often too

simply interpreted as a relation between the change in length of a line element and its initial

length. In fact, any definition of a strain is arbitrary and is largely a matter of convenience. In

this subsection we highlight most scalar strain definitions that apply to a one-dimensional

line element. These scalar strain measures are functions of the stretch ratio λ, which is the

ratio between the deformed l and undeformed length L = l0 of a line element.

λ =l

L=

l

l0(2.18)

A strain measure typically equals zero in the undeformed state, in contrast to the stretch ratio

which then equals 1. For a line element with an infinitesimal length, the stretch typically

reads dl/dl0.

36

2.2. Strains and strain tensors

The engineering strain

A definition which is frequently used in engineering practice and which will play a particular

role in the linearization of other strain measures, is the engineering strain e . The engineering

strain e is a linear function of the stretch ratio λ, and simply equals

e = λ−1 (2.19)

This linear strain measure is bounded between −1 (since dl is always larger than zero) and

plus infinity.

The logarithmic strain

This definition is based on the instantaneous length of a line element, rather than the initial

length. Every infinitesimal increase of the strain, i.e., a strain increment dεl n , is related to

the infinitesimal increase in length dl of the line element with respect to its current length l

dεl n =dl

l(2.20)

The current value of the logarithmic strain is thus given by

εl n =∫l

l0

dl

l= ln

(l

l0

)

= lnλ (2.21)

which justifies its name. This strain measure is often also called the natural or true strain. It

is bounded between −∞ and +∞.

The Green-Lagrange strain

The Green-Lagrange strain measure is one of the basic strain definitions used in continuum

mechanics. The Green-Lagrange scalar strain is a nonlinear function of the stretch λ, given

by

εgl = 12

(λ2 −1) (2.22)

The Green-Lagrange strain is bounded between −12

and +∞.

The Euler-Almansi strain

The Euler-Almansi strain, typically used within a Eulerian framework, is defined by

εea = 12

(1−1

λ2) (2.23)

This strain is bounded between −∞ and +12

.

The different strain definitions are represented in figure 2.5. The properties and differ-

ences between the cited strain definitions are thus well illustrated. It appears clearly that in

the vicinity of the undeformed state, all definitions are similar. For larger or smaller stretches,

important differences arise.

37

2. KINEMATICS

0 1 2 3 4 5−2

−1

0

1

2

3

4

5

Green−Lagrange: 1/2(λ2−1)

Linear: λ − 1

Logarithmic: ln(λ)

Euler−Almansi: 1/2(1−1/λ2)

Stretch λ

Str

ain

defin

ition

s

Figure 2.5: Scalar strain definitions to characterize deformation.

2.2.2 Deformation of infinitely small material line elements

The one-dimensional case

Tensors and vectors simplify into scalars for the one-dimensional case, which makes the

equations particularly straightforward. Consider a one-dimensional line element with initial

length dL which is stretched to a new length dl . with respect to a one-dimensional Euclidean

basis, the deformation gradient simply equals

F = F =∂x

∂X

= 1+∂u

∂X(2.24)

The squared new length dl 2 of the 1D line element equals

dl 2 = d x d x

= F d X F d X

= F 2dL2

= (1+∂u

∂X)2dL2 (2.25)

The change in length of this line element can be measured with

12

(dl 2 −dL2) = 12

([∂x

∂X

]2

−1

)

dL2

=

∂u

∂X+ 1

2

[∂u

∂X

]2 dL2

= εgl dL2 (2.26)

38

2.2. Strains and strain tensors

where use has been made of the earlier defined Green-Lagrange strain εgl .

Generalization to three dimensions, the Green-Lagrange strain tensor

Let’s consider two neighbouring points P0 and Q0 in the undeformed region dV0 of a mate-

rial. Point P0 has an initial position vector ~X =~x0, while its infinitely close neighbour Q0 is

characterized by an initial position ~X +d~X = ~x0 +d~x0. The infinitesimal region dV0 is de-

formed into the region dV and the two points are now located in P and Q, see figure 2.6.

Their new positions are given by~x and~x +d~x. The vector d~x0 = d~X is the vector that inter-

PP0

tt=0

dV0dV

r

r

x X0 =

r

x

0

r r

r r

x dx X dX0 0+ = +

r r

x dx+

Q0

Q

dXr

dxr

F

Figure 2.6: Deformation in an infinitesimal volume

connects P0 with Q0, while d~x interconnects P with Q. The initial Euclidean length of this

material line element is dL = dl0, given by

dL2 = d~X ·d~X

= d Xi d Xi

= d X 21 +d X 2

2 +d X 23 (2.27)

Similarly, points P and Q are interconnected by the vector d~x, for which the deformed length

dl is given by

dl 2 = d~x·d~x= d xi d xi

= d x21 +d x2

2 +d x23 (2.28)

Making use of the relation between d~x and d~X , given by equation (2.5), the deformed length

dl can be written as

dl 2 = d~x·d~x= (F ·d~X )·(F ·d~X )

= d~X ·(F T ·F )·d~X (2.29)

= d~X ·C ·d~X (2.30)

39

2. KINEMATICS

A new tensor C = F T ·F has thereby been introduced, which is called the right Cauchy-Green

deformation tensor. Specific properties of this tensor will be highlighted in the course of

large deformation kinematics. Here we will limit ourselves to its role towards the lineariza-

tion of deformations. Equation (2.30) permits to express the difference between dl 2 and dL2

in terms of the initial material vector d~X (i.e., the material coordinates Xi are the indepen-

dent variables)

12

(dl 2 −dL2) = 12

(d~X ·(C )·d~X −d~X ·d~X ) (2.31)

= d~X ·( 12

[C − I ])·d~X (2.32)

The tensor which appears between the material vectors d~X in the right-hand side is a non-

linear strain tensor, denoted the Green-Lagrange strain tensor E

E = 12

[C − I ] (2.33)

= 12

[~∇0~u + (~∇0~u)T +~∇0~u·(~∇0~u)T ] (2.34)

With respect to a Cartesian vector basis equation (2.32) can be rewritten as

12

(dl 2 −dL2) = E11d x21 +E22d x2

2 +E33d x23+

2E12d x1d x2 +2E13d x1d x3 +2E23d x2d x3

(2.35)

where the individual Green-Lagrange strain components E I J are given by

E11 = ∂u1

∂X1+ 1

2

([∂u1

∂X1

]2+

[∂u2

∂X1

]2+

[∂u3

∂X1

]2)

E22 = ∂u2

∂X2+ 1

2

([∂u1

∂X2

]2+

[∂u2

∂X2

]2+

[∂u3

∂X2

]2)

E33 = ∂u3

∂X3+ 1

2

([∂u1

∂X3

]2+

[∂u2

∂X3

]2+

[∂u3

∂X3

]2)

E12 = 12

(∂u1

∂X2+ ∂u2

∂X1+ ∂u1

∂X1

∂u1

∂X2+ ∂u2

∂X1

∂u2

∂X2+ ∂u3

∂X1

∂u3

∂X2

)

E13 = 12

(∂u1

∂X3+ ∂u3

∂X1+ ∂u1

∂X1

∂u1

∂X3+ ∂u2

∂X1

∂u2

∂X3+ ∂u3

∂X1

∂u3

∂X3

)

E23 = 12

(∂u2

∂X3+ ∂u3

∂X2+ ∂u1

∂X2

∂u1

∂X3+ ∂u2

∂X2

∂u2

∂X3+ ∂u3

∂X2

∂u3

∂X3

)

(2.36)

or in its compact index notation

E I J = 12

(

uI ,J +u J ,I +uK ,I uK ,J

)

(2.37)

The difference between the squares of the length of material line elements is thus written

as

dl 2 −dL2 = 2 d~X ·E ·d~X (2.38)

= 2 Ei j d Xi d X j (2.39)

2.3 Linearization of deformations

The theory of linear elasticity is entirely built on the concept of small deformations. The

only strain definition given so far, which is a linear function of the stretch ratio, was the

40

2.3. Linearization of deformations

engineering strain e . In the case of small deformations this definition equals the scalar linear

strain definition ε.

ε =dl −dL

dL= λ−1 (2.40)

Furthermore, all other strain definitions equal ε as well in the limit for small deformations.

This will be shown for the scalar strain definitions first and generalized for the tensors after-

wards, which will lead to the linear strain tensor.

2.3.1 Linearized strain definitions

The linearization of the earlier presented scalar strain definitions will lead systematically to

the linear strain. This property could be expected by reconsidering figure 2.5, where only the

deformation in the vicinity of λ = 1 is represented. This result is shown in figure 2.7. The

0.95 1 1.05−0.06

−0.04

−0.02

0

0.02

0.04

0.06

Stretch λ

Str

ain

defin

ition

s

Linear: λ − 1 Green−Lagrange: 1/2(λ2−1) Euler−Almansi: 1/2(1−1/λ2)Logarithmic: ln(λ)

Figure 2.7: Tangent character of the linear strain.

linear strain is clearly tangent to all other strain definitions in λ= 1. The mathematical proof

of this property is given in the following paragraphs.

Linearization of the logarithmic strain

The logarithmic strain εl n can be expanded as follows

εl n = ln(λ) = ln(1+ε) (2.41)

For small deformations, it always holds that dl−dL < dL and the logarithm can be expanded

in the following series

ln(1+ε) = ε− 12ε2 +

1

3ε3 −

1

4ε4 +·· · (2.42)

In the limit case that deformations are infinitesimal, only the linear term in equation (2.42)

is to be preserved and one finds

εl n ≈ ε if |ε|≪ 1 (2.43)

41

2. KINEMATICS

Linearization of the Green-Lagrange strain

The Green-Lagrange strain εgl can be expanded in dl as well

εgl = 12

(λ2 −1)

= 12

(

2ε+ε2)

(2.44)

Linearization of this equations for infinitesimal deformations permits to neglect the higher-

order term in dl , which leads to the following approximation

εgl ≈ ε if |ε|≪ 1 (2.45)

2.3.2 The infinitesimal linear strain tensor

Definition

The infinitesimal strain tensor ε is formally defined as the linearized fraction of the Green-

Lagrange strain tensor E . For infinitesimal displacements, the first partial derivatives of ui

are so small that all involved squares and products are negligible with respect to the linear

terms. In that case, it is allowed to neglect the nonlinear terms in the definition of E

E = 12

(

F T ·F − I)

(2.46)

= 12

[~∇0~u + (~∇0~u)T +~∇0~u·(~∇0~u)T

]

(2.47)

from which the linear strain tensor is extracted ε

ε = 12

[~∇0~u + (~∇0~u)T

]

(2.48)

Note that this tensor equals the symmetric part of the displacement gradient tensor. Another

frequently used formula for the same definition is

ε = 12

[

F +F T]

− I (2.49)

Clearly, the distinction between the Lagrangian and Eulerian strain tensor now vanishes,

since it is now immaterial whether the derivatives of the displacements are determined at

the deformed or undeformed position of the considered point. Equation (2.38) thus leads to

one single expression where the difference between d Xi and d xi is no longer relevant

dl 2 −dL2 = 2 d~X ·ε·d~X (2.50)

= 2 d~x·ε·d~x= 2εi j d Xi d X j = 2εi j d xi d x j (2.51)

The fact that the change of the position vector is no longer relevant for infinitesimal dis-

placements is a fundamental observation, which permits to omit the double notation ~X ↔~x

or even ~∇~X =~∇0 ↔~∇ =~∇~x . In this case, the notation~x and simply ~∇ are often used in liter-

ature if the kinematics apply to small displacements, since then ~∇ ≈~∇0 for most cases. The

index notation is now written as

εi j

= 12

(

ui , j+u

j ,i

)

(2.52)

42

2.3. Linearization of deformations

Strain-displacement relations in Cartesian coordinates

The gradient operator with respect to a Cartesian basis ~ex,~e y ,~ez is given by~∇=~ex∂∂x+~e y

∂∂y+

~ez∂∂z

, while the components are the displacement vector with respect to this basis are u, v, w .

It is easy to show that the definition of the linear strain tensor (2.48) leads to the same equa-

tions that are obtained through direct linearization of the equations (2.36)

εxx =∂u

∂x

εy y =∂v

∂y

εzz =∂w

∂z

εx y = 12

(∂u

∂y+∂v

∂x

)

εxz = 12

(∂u

∂z+∂w

∂x

)

εyz = 12

(∂v

∂z+∂w

∂y

)

(2.53)

Strain-displacement relations in cylindrical coordinates

With respect to a cylindrical basis ~er (θ),~et (θ),~ez the gradient operator is given by

~∇ = ~er∂

∂r+~et

1

r

∂

∂θ+~ez

∂

∂z(2.54)

The displacement vector with respect to a cylindrical basis reads ~u = u~er (θ)+ v~et (θ)+w~ez

which by application of (2.48) yields

εr r =∂u

∂r

εt t =u

r+

1

r

∂v

∂θ

εzz =∂w

∂z

εr t = 12

(1

r

∂u

∂θ+∂v

∂r−

v

r

)

εr z = 12

(∂u

∂z+∂w

∂r

)

εtz = 12

(∂v

∂z+

1

r

∂w

∂θ

)

(2.55)

2.3.3 The infinitesimal rotation tensor

Any tensor can be split into a symmetric tensor and a skew-symmetric tensor. For the dis-

placement gradient tensor, is was shown that the symmetric part has a specific meaning in

linear elasticity. The same holds for the skew-symmetric part. This part, denoted ω, equals

the mean rotation (displacement) tensor in a material point, under the condition that the

43

2. KINEMATICS

displacement field is infinitely small.

[~∇0~u

]T = 12

([~∇0~u]T +~∇0u)+ 12

([~∇0~u]T −~∇0u) (2.56)

= ε+ω (2.57)

The mean rotation in a material point is the mean with respect to all infinitesimal line ele-

ments that emanate from that point. Note that the name rotation tensor is not well-chosen,

since this tensor only yields the displacement vectors due to the rotation. The tensor ω dif-

fers in that sense from the more general (large displacement) rotation tensor Q, which is an

orthogonal tensor that yields a rotated material vector. In fact, in the case of infinitesimal

displacements, both tensors are related with

Q = I +ω (2.58)

Consider a material point P and a neighbouring point that is separated from P through an

infinitely small material line element d~x. Since ω is skew-symmetric, an axial vector~ω exists

such that

ω·d~x = ~ω×d~x (2.59)

The result is thus a vector that is always perpendicular to the that interconnects two neigh-

bouring material points. The resulting displacement vector from (2.59) is thus characteristic

for an infinitely small rotation around the considered material point P . Note that angular

displacements have to be infinitely small for this interpretation of ω.

2.4 Applicability of the theory of small displacements

On the basis of this additive split of the displacement gradient tensor, a better insight can be

provided in the limitations that apply to linear elasticity. The relation between the Green-

Lagrange strain tensor E and the infinitesimal strain tensor ε can be written with the use of

the infinitesimal rotation tensor ω.

E = 12

(~∇0~u + [~∇0~u]T +~∇0~u·[~∇0~u]T

)

(2.60)

= ε+ 12

[~∇0~u]T ·~∇0~u (2.61)

and using

ε+ω = [~∇0~u]T (2.62)

ε+ωT = [~∇0~u] (2.63)

it follows that

E = ε+ 12

[

(ε+ωT )·(ε+ω)

]

(2.64)

The latter relation clearly illustrates that it is not sufficient to have small strain |εi j |≪ 1, since

then equation (2.64) simplifies to

E = ε+ 12ε·ωT + 1

2ω·ε+ 1

2ω

T ·ω (2.65)

which is not yet the linear strain tensor (only 12ε·ε is negligible). The latter term only vanishes

if rotations are small. In order to use the infinitesimal strain tensor as a first-order approxi-

mation of the Green-Lagrange strain tensor, both strains and rotations must be small! This is

44

2.4. Applicability of the theory of small displacements

particularly important for flexible bodies, such as thin plates and slender rods. The bending

of large thin sheets for instance, causes only small strains in the material. Yet, the displace-

ments and more specifically the rotations are very large. The linearization is thus not ap-

plicable. This is most easily illustrated with a simple example. Consider a two-dimensional

material line element with length L that interconnects A with B , see figure 2.8. The line ele-

A

B

B’

a

( , )x y

( , )x y0 0L

L

Figure 2.8: Rigid rotation of a material line element

ment is simply rotated over an angle α around A towards its new position A−B ′. A Cartesian

basis with its origin in A is next taken. A material point (x0, y0) of the initial line element will

be in a new position (x, y) given by[

x

y

]

=[

cosα sinα

−sinα cosα

][x0

y0

]

(2.66)

The displacement components are thus given by[

u

v

]

=[

x −x0

y − y0

]

=[

(cosα−1)x0 + (sinα)y0

(−sinα)x0 + (cosα−1)y0

]

(2.67)

The matrix form of the two-dimensional linear strain tensor ε is then given by

[εxx εx y

εyx εy y

]

=

∂u∂x0

12

(∂u∂y0

+ ∂v∂x0

)

12

(∂v∂x0

+ ∂u∂y0

)∂v∂y0

=[

cosα−1 12

(sinα− sinα)12

(sinα− sinα) cosα−1

]

=[

cosα−1 0

0 cosα−1

]

(2.68)

Clearly the dilatational strain components (i.e., the components εxx and εy y ) are not equal to

zero if the angle of rotation is not very small. Deviations become very large for large rotation

angles and the numerical values of the strain components completely lose their physical

meaning. A generalization of this rigid rotation example will be given in section 2.7.1. For

the components of the Green-Lagrange strain tensor it holds that

Exx = ∂u∂x0

+ 12

([∂u∂x0

]2+

[∂v∂x0

]2)

= cosα−1+ 12

(cos2α−2 cosα+1+ sin2α)

= 0 (2.69)

45

2. KINEMATICS

E y y = ∂v∂y0

12

([∂u∂y0

]2+

[∂v∂y0

]2)

= cosα−1+ 12

(sin2α+cos2α−2 cosα+1)

= 0 (2.70)

Ex y = E yx = 12

(∂u∂y0

+ ∂v∂x0

+ ∂u∂x0

∂u∂y0

+ ∂v∂x0

∂v∂y0

)

= 12

(sinα− sinα+ sinαcosα− sinα− sinαcosα+ sinα)

= 0 (2.71)

which clearly shows that it is insensitive to rigid body rotations, see also section 2.7.1.

Small displacement theory better characterizes the limitations, since then both strains

and rotations have to be small. The small displacement theory should be used with caution,

since it will yield completely erroneous results if its underlying assumption is not satisfied.

Fortunately, for massive and thick products (not thin nor flexible) the linearization is well

permissible, for as long as the analysis is confined to the normal use of the products during

their lifetime. The manufacturing of the same products on the contrary is mostly accompa-

nied by large displacements. These nonlinear processes will therefore be treated separately

in the course on the mechanics of advanced manufacturing processes.

2.5 Physical interpretation of the strain components

2.5.1 Normal or extensional strains

Consider a Cartesian vector basis ~e1,~e2,~e3, with respect to which the components of the

strain tensor are analysed. The components ε11, ε22, ε33 of the infinitesimal strain tensor ε

are called the normal or extensional strains. They are determined along their corresponding

directions ~N1, ~N2, ~N3 with e.g., for ε11

ε11 = ~N1·ε·~N1 = ~N1·( 12

[

F +F T]

− I )·~N1

= 12

(~N1·

[

F +F T]

·~N1

)

−1

=1

2dL1 dL1d~X1·

[

F +F T]

·d~X1 −1

=1

2dL1 dL1

[

d~X1·d~x1 +d~x1·d~X1

]

−1

=dL1 dl1

dL1 dL1

~N1·~n1 −1 (2.72)

For infinitely small displacements the change in angle from ~N1 to ~n1 is small and hence

~N1·~n1 = cos(θ1) ≈ 1 (2.73)

The extensional strain ε11 is then equal to

ε11 =dl1

dL1−1 (2.74)

= λ1 −1 (2.75)

46

2.5. Physical interpretation of the strain components

This result is nothing else than the scalar linear strain along that direction. The diagonal

components ε11, ε22, ε33 of the linear strain tensor thus represent the extension along the

directions of the Cartesian base vectors, i.e., the change in length per unit length along that

direction, see figure 2.9. The normal extensional strain components ε11,ε22,ε33 equal the

u

y

vv

ydy+

¶

¶

dy

v

uu

xdx+

¶

¶dxx

Figure 2.9: Physical interpretation of extensional or normal strains

linear strains in the directions of the coordinate axes, which are the relative elongation of the

fibers parallel to these axes. Evidently, this property of the linear strain tensor does no longer

hold if deformations and/or rotations are large.

2.5.2 Shear strains

The remaining components ε12 = ε21, ε13 = ε31, ε23 = ε32 of the strain tensor ε are called the

tensorial shear strains, since they are related to the change of angles between material line

elements. Let us consider two infinitely small material line elements that emanate from a

material point P, oriented along the Cartesian coordinate axes which are perpendicular, e.g.,

along~e1 and ~e2. In the undeformed state, the associated material vectors are denoted d~X1

and d~X2. After deformation, the material line elements have deformed towards their new

positions d~x1 and d~x2. The initial lengths of these material line elements are dL1 and dL2

respectively. After deformation the lengths equal dl1 and dl2 . The scalar product between

the deformed vectors then gives

d~x1·d~x2 = dl1dl2 cos(π

2−θ) (2.76)

where θ is the total change in angle from the undeformed state to the current state between

the two line elements.

Applying the strain tensor to the undeformed perpendicular directions ~N1 and ~N2 (~N1·~N2 =0) of the material vectors d~X1 and d~X2 and making use of equation (2.49) for the strain tensor

47

2. KINEMATICS

leads to

ε12 = ~N1·ε·~N2 = ~N1·( 12

[

F +F T]

− I )·~N2

= 12~N1·

[

F +F T]

·~N2

=1

2dL1 dL2d~X1·

[

F +F T]

·d~X2

=1

2dL1 dL2

[

d~X1·d~x2 +d~x1·d~X2

]

=dL1 dl2

2dL1 dL2

~N1·~n2 +dl1 dL2

2dL1 dL2

~n1·~N2 (2.77)

For infinitesimal displacements, the differences between dL1 ↔ dl1 and dL2 ↔ dl2 only lead

to second-order contributions in equation (2.77), which allows to neglect them. Hence, the

component ε12 is given by

ε12 = 12~N1·~n2 + 1

2~n1·~N2

= 12

[

cos(π

2−θ2)+cos(

π

2−θ1)

]

= 12

[sin(θ2)+ sin(θ1)]

≈ 12 [θ2 +θ1] =

θ

2(2.78)

Clearly, it appears that the strains ε12, ε13, ε23, which are tensorial components, equal half

u

y

¶

¶

u

ydy

dy

v

¶

¶

v

xdx

dxx

Figure 2.10: Physical interpretation of shear strains

the change of angle between two material line elements oriented along the corresponding

coordinate axes in the undeformed state (~e1-~e2,~e1-~e3,~e2-~e3). This is well depicted in fig-

ure 2.10, which shows the contributions to the shear strain with respect to a Cartesian basis.

In engineering practice, these tensorial strain components are frequently doubled, e.g., 2ε12

is used instead of ε12. These doubled components are then called the shear strains γ12

γ12 = 2ε12

γ13 = 2ε31

γ23 = 2ε32

(2.79)

48

2.5. Physical interpretation of the strain components

The shear strain γ12 equals the total change of angle after deformation between two line

elements. However, it should be emphasized that the extensional strains ε11, ε22, ε33 and the

shear strains γ12, γ13, γ23 do not form a tensor, which is mathematically inconvenient.

2.5.3 Principal strains and strain invariants

Since ε is a symmetric second-order tensor, its eigenvectors and eigenvalues are real, and

have a physical meaning. The eigenvalue problem of ε is easily solved from

ε·~n = ǫ~n (2.80)

where ǫ are the eigenvalues and~n the eigenvectors. The eigenvalues are then obtained from

the cubic characteristic equation

det(ε−ǫI ) = 0 (2.81)

ǫ3 − J1(ε)ǫ2 + J2(ε)ǫ− J3(ε) = 0 (2.82)

The three strain invariants J1(ε), J2(ε) and J3(ε) have thereby been introduced. The roots of

equation (2.82) are the principal strains ǫ1, ǫ2 and ǫ3. The eigenvectors, or principal strain

directions ~Ni are computed from

(ε−ǫi I )·~Ni = ~0 ∀i ∈ 1,2,3 (2.83)

in combination with the required unit length of the vectors ~ni , i.e., |~ni | = 1. If the three

principal strains are distinct, the three principal directions will be mutually orthogonal. If

two principal strain are equal, equation (2.83) has an infinite number of solutions, out of

which any pair of orthogonal vectors can be chosen as the principal directions. If all principal

strains are equal, each set of three mutually orthogonal vectors is a set of principal directions.

An axis that is oriented along a principal direction is called a principal axis. A plane normal

to a principal direction is called a principal plane. If a Cartesian basis is taken with three axes

that coincide with the principal axes, the matrix form of ε takes a particular format

ε =

ǫ1 0 0

0 ǫ2 0

0 0 ǫ3

(2.84)

The shear strain components are clearly equal to zero along these directions. This implies

that material line elements oriented along these directions are only stretched or compressed.

Since the shear strain between principal axes vanishes, the principal axes are mutually or-

thogonal before and after deformation. Furthermore, the extensional strains attain an ex-

treme value along the principal axes. The largest and smallest extensional strain for all pos-

sible orientations of material line elements can always be found among the principal strains.

The three principal strain invariants J1(ε), J2(ε) and J3(ε) are computed from the princi-

pal strains through

J1(ε) = ǫ1 +ǫ2 +ǫ3 (2.85)

J2(ε) = ǫ1ǫ2 +ǫ1ǫ3 +ǫ2ǫ3 (2.86)

J3(ε) = ǫ1ǫ2ǫ3 (2.87)

Note that any function of these three principal strain invariants is a strain invariant as well.

49

2. KINEMATICS

2.5.4 Volume and shape changes

Volumetric strain

The change of the volume of an infinitely small volume element is first analyzed on the basis

of the deformation tensor F . Linearization will then lead to a simplified expression, which

is applicable in linear elasticity. Consider an undeformed material parallelepiped that is

spanned by three vectors d~X1, d~X2 and d~X3. The undeformed volume of this parallelepiped

equals dV0, and is given by

dV0 = (d~X1 ×d~X2)·d~X3 (2.88)

After deformation, the deformed parallelepiped is spanned by the deformed vectors d~x1,

d~x2 and d~x3, which leads to the deformed volume dV :

dV = (d~x1 ×d~x2)·d~x3

= (F ·d~X1)× (F ·d~X2)·(F ·d~X3)

= det(F ) (d~X1 ×d~X2·d~X3)

= det(F )dV0 (2.89)

The change in volume (for large deformations) is thus characterized and quantified with the

determinant of the deformation tensor, i.e., det(F ). This quantity is often denoted by J , and

is known as the volume ratio3 of the continuum:

J = det(F ) =dV

dV0(2.90)

The volume of an infinitely small material volume element can physically never be negative

nor zero, nor reach infinity, so it holds that:

J = det(F ) > 0 and J < ∞ (2.91)

In the case of infinitesimal displacements, the volume ratio J is easily linearized

J = det(F ) = det(I +~∇~uT )

≈ 1+ tr(~∇~uT ) = 1+ tr(~∇~u)

≈ 1+ tr(ε) (2.92)

This equation is easily retrieved with respect to a Cartesian basis (x, y, z) with infinitesimal

displacement components (u, v, w), for which

det(I +~∇uT

) =

1+ ∂u∂x

∂v∂x

∂w∂x

∂u∂y

1+ ∂v∂y

∂w∂y

∂u∂z

∂v∂z

1+ ∂w∂z

= 1+∂u

∂x+∂v

∂y+∂w

∂z+∂u

∂x

∂v

∂y+∂u

∂x

∂w

∂z+∂v

∂y

∂w

∂z+ . . .

︸ ︷︷ ︸

higher-order terms

≈ 1+∂u

∂x+∂v

∂y+∂w

∂z

= 1+ tr(ε) (2.93)

3volumeveranderingsfaktor

50

2.6. Compatibility conditions

The first invariant or the trace of the infinitesimal linear strain tensor thus measures the

relative volume variation

tr(ε) =dV −dV0

dV0(2.94)

The trace of ε is called the volumetric strain or cubical dilatation and often denoted e

e = tr(ε) = J1(ε) (2.95)

For an incompressible material (e.g., fluid) it holds that e = 0.

Shape changes

Many materials in deformation processes show a mechanical behaviour that is specially de-

pendent of the mean strain, which is simply the volumetric strain divided by 3

εm =ε11 +ε22 +ε33

3=

e

3= 1

3J1(ε) (2.96)

The mean strain tensor or volumetric strain tensor is denoted εv and equals

εv = εm I = 13

tr(ε) I (2.97)

The part of the strain tensor that is obtained after subtraction of εv is called the deviatoric

strain tensor εd

εd = ε−εv (2.98)

Since by definition the trace of this deviatoric strain tensor is zero, there is no volume change

associated to it (infinitesimal displacements !). Its components are only related to the changes

in shape of a material volume element. The principal invariants of the deviatoric part of the

linear strain tensor are denoted J1(εd ), J2(εd ) and J3(εd ). Clearly the first invariant is zero

J1(εd ) = 0 (2.99)

It is easy to verify that the following relations exist between the invariants J2, J3 of the total

strain tensor ε, the volumetric strain εm and the invariants of the strain deviator εd

J2(εd ) = 3ε2m − J2(ε) (2.100)

J3(εd ) = J3(ε)+εm J2(ε)−ε3m (2.101)

The second invariant J2 of the strain deviator tensor εd is also given by

J2(εd ) = 12

[tr (εd )︸ ︷︷ ︸

=0

]2 − tr(εd 2)

= − 12ε

d : εd (2.102)

2.6 Compatibility conditions

The deformation in a material point, characterized through the linear strain tensor ε, de-

pends on 6 components (taking into account the symmetry of ε). Yet, the definition of the

strain tensor clearly shows that all of these components are completely determined from the

51

2. KINEMATICS

3 components of the displacement vector field. This implies that the 6 strain components

are not independent and that 3 relations must exist between the strain components. These

3 relations are called the compatibility conditions. They are obtained through elimination

of the displacements in the strain-displacements equations to produce equations with only

strain components as unknowns. The equations are next elaborated for a Cartesian vector

basis. The first equation is found by taking the second derivatives of the components ε11,

ε22, ε12 with respect to the Cartesian coordinates x1, x2:

∂2ε11

∂x22

=∂3u1

∂x1∂x22

(2.103)

∂2ε22

∂x21

=∂3u2

∂x21∂x2

(2.104)

2∂2ε12

∂x1x2=

∂3u1

∂x1∂x22

+∂3u2

∂x21∂x2

(2.105)

Summing up the first two of these equations yields the third one, i.e.,

∂2ε11

∂x22

+∂2ε22

∂x21

= 2∂2ε12

∂x1∂x2(2.106)

Repeating the same operation on the other components leads to two more equations which

have an identical format to equation (2.106) but with permuted indices

∂2ε11

∂x23

+∂2ε33

∂x21

= 2∂2ε13

∂x1∂x3(2.107)

∂2ε22

∂x23

+∂2ε33

∂x22

= 2∂2ε23

∂x2∂x3(2.108)

In a similar way further differentiation of the strain-displacement equations (2.53) also gives

∂2ε11

∂x2∂x3+∂2ε23

∂x21

=∂2ε13

∂x1∂x2+

∂2ε12

∂x1∂x3(2.109)

∂2ε22

∂x1∂x3+∂2ε13

∂x22

=∂2ε23

∂x1∂x2+

∂2ε12

∂x2∂x3(2.110)

∂2ε33

∂x1∂x2+∂2ε12

∂x23

=∂2ε13

∂x2∂x3+

∂2ε23

∂x1∂x3(2.111)

which gives a total number of six equations, known as the St.Venant compatibility equations.

They are commonly represented in a compact index format

εkl ,i j +εi j ,kl = εi k, j l +ε j l ,i k (2.112)

This compact format stands for 81 equations, but only six of them are distinct while all the

other equations are redundant or equalities. The fact that 6 equations are found instead of

3 to restore the compatibility between strains and displacements, indicates that those 6 St.

Venant equations are not truly independent in the sense that further differential combina-

tions of these equations produce three connections between them (the Bianchi formulas).

However, since one cannot choose any three equations of the 6 St.Venant equations, one

52

2.7. Special deformations

usually includes all 6 equations. It has been proven that these equations are necessary and

sufficient to ensure the existence of the displacement component functions that are related

to the linear strain tensor through (2.53).

The compatibility equations are sometimes derived in an another way, on the basis of

the continuity of matter. The strain field must be such that a closed loop before deformation

remains closed after deformation. The mathematical expression of this condition leads to

the same equations, but it explains why they are sometimes called the continuity equations

(they ensure the continuity of the displacement field for a given strain field).

Note that compatibility is automatically satisfied if strains are computed from the dis-

placement field, which is almost always the case in engineering practice.

2.7 Special deformations

2.7.1 Rigid body motion

A rigid body motion of a continuum occurs whenever external loads only lead to a transla-

tion and a rotation of the continuum (figure 2.11), without any deformation. Since there is

no deformation inside the continuum, relative displacement vectors of material points with

respect to other material points are equal to zero. A rigid body transformation of the ref-

t

r

nir

Ni

=+

t=0

Figure 2.11: Rigid body motion of a continuum.

erence to the current configuration may be considered as a pure rotation Q, followed by a

pure translation~t of the continuum. The rotation tensor Q and the translation vector~t are

constant in space, i.e., identical for all material points. If the centre of rotation is located at

a spatial point P with initial position ~Y , the current position~x of an arbitrary material point

in the continuum with initial position ~X is given by:

~x = ~Y +~t +Q·(~X −~Y ) (2.113)

The deformation tensor F for this rigid body motion is then given by

F = (~∇0~x)T

= Q (2.114)

Hence, the infinitesimal strain tensor is represented by

ε = 12

(Q +QT )− I (2.115)

A rotation tensor is an orthogonal tensor for which QT =Q−1. The infinitesimal strain tensor

is thus not equal to the null tensor, unless rotations are infinitely small (e.g., see section 2.4).

53

2. KINEMATICS

In that case only, Q approximates the unit tensor I . Clearly, this theory cannot be used for

large rigid body rotations. The Green-Lagrange strain tensor on the other hand, correctly

represents a zero strain state upon a rigid body rotation

E = 12

(C − I )

= 12

(F T ·F − I )

= 12

(QT ·Q − I )

= 12

(Q−1·Q − I ) since Q is orthogonal

= 0 (2.116)

2.7.2 Homogeneous deformation

If an arbitrary material vector d~X is deformed to the same d~x in all material points, the de-

formation is independent of the considered material point and the continuum is deformed

homogeneously. The deformation tensor is then a constant tensor which is independent of

the initial position ~X in the continuum. Since

d~x = F ·d~X (2.117)

and since F is independent of ~X , integration of (2.117) results in an equation that describes

the new position of an arbitrary material point in the continuum in terms of its reference

position

~x = F ·~X +~t (2.118)

where~t represents the vectorial integration constant, i.e., a rigid translation. Since F is con-

stant, ε will be constant as well. All strain components are thus identical for all material

points.

2.7.3 Plane strain

The three-dimensional deformation state reduces to a plane deformation state if the strain

components εzz , εxz and εyz are equal to zero. The existence of such a plane strain state

is frequently assumed for the analysis of flat specimens or structures, where the thickness

(which is taken in the z-direction) is so high that deformations in that direction become

negligible,

εzz = εxz = εyz = 0 (2.119)

while the strain components in the plane are assumed to be independent of the z-direction.

The strain state is then fully described in the plane of the specimen by a plane strain tensor,

which has only 3 distinct components. This state of deformation is called a state of plane

strain with respect to the (x, y) plane.

2.7.4 Axisymmetric deformation

Many engineering components and structures have an axisymmetric geometry, which means

that their shape and volume can be constructed by virtually rotating a longitudinal cross sec-

tion around the axis of revolution. Points are indicated with cylindrical coordinates r,θ, z.

When material properties and loading are also independent of the coordinate θ, the defor-

mation and resulting stresses will be also independent of θ.

54

2.7. Special deformations

z

r

P

r

z

r

θ

Figure 2.12: Axisymmetric deformation

∂

∂θ( ) = 0 → ~u = ur (r, z)~er (θ)+ut (r, z)~et (θ)+uz(r, z)~ez

ε=1

2

2ur,r −1r

(ut )+ut ,r ur,z +uz,r

−1r

(ut )+ut ,r 2 1r

(ur ) ut ,z

uz,r +ur,z ut ,z 2uz,z

If no rotation around the z-axis takes place (ut = 0), all state variables can be studied in

one half of a longitudinal cross section (containing the z-axis).

∂

∂θ( ) = 0 and ut = 0 → ~u = ur (r, z)~er (θ)+uz (r, z)~ez

ε=1

2

2ur,r 0 ur,z +uz,r

0 2 1r

(ur ) 0

uz,r +ur,z 0 2uz,z

Axisymmetric plane strain

When boundary conditions and material behavior are such that displacement of material

points are only in the rθ-plane, the deformation is referred to as plane strain in the rθ-plane.

plane strain deformation

ur = ur (r,θ)

ut = ut (r,θ)

uz = 0

→ εzz = γr z = γtz = 0

linear strain matrix

ε=1

2

2ur,r ut ,r − 1r

(ut ) 0

ut ,r − 1r

(ut ) 2r

(ur ) 0

0 0 0

plane strain deformation with ut = 0

55

2. KINEMATICS

ur = ur (r )

uz = 0

→ ε=1

2

2ur,r 0 0

0 2r

(ur ) 0

0 0 0

56

2.8. Summary

2.8 Summary

Kinematics

Nonlinear deformation map ~x = φ(~X , t )

F = (~∇0~x)T

Deformation tensor d~x = F ·d~x0

F = (~∇0~x)T

Fi J= x

i ,J

Displacement gradient tensor υ= (~∇0~u)T

d~u =υT ·d~x0

F = I +υ

υi J= u

i ,J

Strains and strain tensors

Engineering strain e =λ−1

Logarithmic strain εl n = lnλ

Green-Lagrange strain εgl = 12

(λ2 −1)

Euler-Almansi strain εea = 12

(1− 1λ2 )

Right Cauchy-Green tensor C = F T ·FGreen-Lagrange strain tensor E = 1

2[C − I ]

dl 2 −dL2 = 2 d~X ·E ·d~X

Left Cauchy-Green tensor B = F ·F T

Euler-Almansi strain tensor E A = 12

[

I −B−1]

57

2. KINEMATICS

Linearization of deformations

Linear strain ε=λ−1

Engineering strain ε= e

Logarithmic strain εl n ≈ ε if |ε|≪ 1

Green-Lagrange strain εgl ≈ ε if |ε|≪ 1

Euler-Almansi strain εea ≈ ε if |ε|≪ 1

Linear strain tensor ε= 12

[

υ+υT]

ε= 12

[

F +F T]

− I

Infinitesimal rotation tensor ω= 12

(~∇0uT −~∇0u)

Properties of the linear strain tensor

Physical interpretation ε11 =λ1 −1

ε12 =≈ 12

[θ2 +θ1]

Engineering shear strain γ12 = 2ε12

γ13 = 2ε31

γ23 = 2ε32

Principal strain ε1, ε2 and ε3

Principal invariants J1(ε) = ε1 +ε2 +ε3

J2(ε) = ε1ε2 +ε1ε3 +ε2ε3

J3(ε) = ε1ε2ε3

Volumetric deformation J = det(F ) = dVdV0

≈ 1+ tr(ε)

Cubical dilatation e = tr(ε) = dV −dV0

dV0

Mean strain εm = 13

trε

Volumetric strain tensor εv = εm I

Deviatoric strain tensor εd = ε−εv

58

2.8. Summary

Compatibility

St.Venant compatibility εkl ,i j +εi j ,kl = εi k, j l +ε j l ,i k

Special deformations

Rigid body motion ~x = ~Y +~t +Q ·(~X −~Y )

ε= 12

(Q +QT )− I ≈ 0, if Q small

Homogeneous deformation F constant, ε constant

Plane strain εzz = εxz = εyz = 0

Axisymmetric deformation εzz = γr z = γtz = 0

59

CH

AP

TE

R

3FORCE AND STRESS

This chapter is devoted to the fundamental concept of stress, which is here sketched within

the infinitesimal deformation context. The formal definition of the stress tensor is given,

together with the interpretation of its components in a Cartesian framework. Since it is a

second-order tensor, tensorial properties given for the strain tensor also apply to the stress

tensor.

3.1 Forces and stresses in a deformable continuum

Each device, no matter how small (e.g., micromechanical devices in a DVD or HDD) or large

(e.g., vehicles, bridges) is manufactured from a material in which all material points interact

with each other to sustain the external loads. In manufacturing processes or some particular

applications, the shape or volume of the component or device changes during this process.

For conventional use of a device or a mechanical component (e.g., engine parts) the initial

geometry may not change during use. In these cases, the combination of external loads and

the material response, mostly leads to cases where the theory of infinitesimal deformations

remains applicable. It is within this context, that we will here sketch the notion of the stress

tensor.

3.1.1 External forces

Consider a three-dimensional continuum, enclosed in a volume V with an external surface

Ω. The continuum is loaded by external loads which may have a different nature. Surface

forces ~p are applied on material points located on the external boundary Ω, see figure 3.1.

These may be distributed loads (per unit of surface N/m2), line loads or concentrated loads.

Body forces ~q act inside the continuum on individual material points (figure 3.1). Well-

known examples are gravitation forces (the weight of the material itself), or body forces in

the presence of transient electro-magnetic fields. Body forces ~q(~x) are usually measured

with respect to unit of mass, i.e., N/kg . Per unit of volume, they can be written as

ρ~q(~x)

[N

m3

]

=[

kg

m3

][N

kg

]

(3.1)

61

3. FORCE AND STRESS

V

W

rr

q

r

p

Figure 3.1: External forces acting on a continuum

In real engineering materials, a true material point does not exist, since the material is

semi-continuous on a very small scale, e.g., atoms in a crystal lattice (metals) or molecular

chains (polymers) or grains (ceramics). Material points in the concept of the infinitesimal

deformation theory, are therefore best identified with an average response of the underlying

material microstructure. Once such material points are loaded, they will start to interact.

These interactions are visualized and measured through local deformations, while the resis-

tance against these deformations is expressed in terms of internal forces.

3.1.2 Internal forces

Each subvolume dV of the total volume V interacts with its immediate neighbourhood, once

external forces are applied. This implies that the material locally deforms, which is well de-

scribed with the already given kinematical framework. Internal forces arise from the atomic

or molecular interactions (e.g., deformation of the crystal lattice) upon deformation. Each

arbitrary material volume in a continuum deforms such that all forces applied to it equili-

brate one another. Internal forces are always in equilibrium with the internal forces applied

on the neighbouring material volumes (principle of action and reaction). If the considered

material volume is located on the boundary, the internal forces on the boundary segment

simply equal the external forces on that segment.

3.1.3 The stress vector

Based on the existence of these internal forces, a so-called stress vector can be defined. Let’s

consider a small volume ∆V inside the material, where we focus our attention on a part of

its boundary ∆S. Through this surface ∆S the considered volume exerts a force ∆~f on the

neighbouring material, which is equal and opposite to the force acted by that neighbouring

material on that surface. If we now take the limit for ∆V and hence ∆S to zero, we find the

infinitesimal force d~f acting on an infinitesimal material plane dS for a material point, see

figure 3.2. Since dS is plane in this limit, it can be identified by its outward normal ~n. The

stress vector in this material point, acting on a plane with orientation~n is the defined by

~p =d~f

dS(3.2)

62

3.1. Forces and stresses in a deformable continuum

P

dV

V

r

p r

n

dS

Figure 3.2: The stress vector on a plane in a material point

The stress vector (also called the traction) is the internal force vector per unit of surface that

a material point undergoes on a surface with a given direction ~n. The true or Cauchy stress

vector, as defined by(3.2), is defined with respect of the true (i.e., in the actual deformed

state) surface in the material. Hence, dS and ~n are quantities measured with respect to the

deformed state. Since only infinitesimal displacements are considered in this framework,

the following simplifications may be made

~p =d~f

dS≈

d~f

dS0~n ≈ ~n0 (3.3)

Again, no distinction is made between the undeformed and deformed geometry on which

the internal forces act. This assumption is not valid in the presence of large deformations,

which will lead to different types of stress vectors, depending on the considered geometry.

The fact that the name ’Cauchy’ or ’true’ stress vector is used in the infinitesimal context

may be misleading, since one simply does not distinguish the ’true’ stress vector from any

other definition. No distinction between dS ↔ dS0 or ~n ↔~n0 will therefore be made in the

remainder of this chapter.

Note that each orientation corresponds to a specific stress vector.

3.1.4 The stress tensor

The stress vector is related to each possible orientation of a facet through a linear transfor-

mation, since the considered material volume on which it is defined is infinitely small. Such

a linear transformation is evidently described by a tensorial quantity, the stress tensor σ

~p = σ·~n (3.4)

The stress tensor σ gives the stress vector ~p for each possible orientation~n of the infinitesi-

mal facet dS. It unifies all stress vectors for all possible orientations.

The fact that the transformation from the normal~n to the stress vector ~p is described by a

second-order tensor, means that there is an exact similarity between concepts, components,

invariants that have been derived from σ compared to those derived from the strain tensor

ε. Another important property of the stress tensor is its symmetry, which will be proven

later in this chapter. Both ε and σ are thus symmetric second-order tensors. The following

paragraphs are therefore a natural extension of the previous chapter to the stress tensor.

63

3. FORCE AND STRESS

3.2 Components of the stress tensor with respect to a basis

3.2.1 Normal stresses

The componentσi i of the stress tensorσ along an arbitrary direction~ni , is called the normal

stress along that direction. The normal stress is determined with

σi i = ~ni ·σ·~ni (3.5)

= ~ni ·~pi (3.6)

where ~p is the stress vector that acts on the infinitesimal surface perpendicular to ~Ni .

Using a Cartesian vector basis ~e1,~e2,~e3, the normal stresses along the coordinate axes

are denoted σ11,σ22,σ33

σ11 = ~e1·σ·~e1 (3.7)

σ22 = ~e2·σ·~e2 (3.8)

σ33 = ~e3·σ·~e3 (3.9)

The normal stress components σi i are thus located on the diagonal of the associated stress

matrixσ with respect to that basis. The normal stresses are (work-)conjugated to the normal

strains.

3.2.2 Shear stresses

The off-diagonal components σi j (i 6= j ) of the stress tensor σ are called the tensorial shear

stresses. The shear stress σi j between two perpendicular directions ~ni and ~n j (~ni ·~n j = 0) is

given by

σi j = ~ni ·σ·~n j

= ~ni ·~p j (3.10)

The shear stress thus represents the component with respect to ~ni of the stress vector ~p j

acting on a surface perpendicular to ~n j . Note that the vector ~ni also lies in that surface.

Shear stresses are commonly denoted by the symbol τ. One may thus write

τi j = σi j (3.11)

3.3 Principal stresses and stress invariants

Like ε, the stress tensor σ is a symmetric second-order tensor, for which the eigenvectors

and eigenvalues are real. The eigenvalue problem of σ is solved identically to section 2.5.3,

i.e., by

σ·~Mi = σi~Mi (3.12)

where σi (i = 1,2,3) are the eigenvalues and ~Mi the eigenvectors. The eigenvalues are again

obtained from the cubic characteristic equation for the stress tensor

det(σ−σI ) = 0 (3.13)

σ3 − J1(σ)σ2 + J2(σ)σ− J3(σ) = 0 (3.14)

64

3.4. Hydrostatic and deviatoric stresses

The three stress invariants J1(σ), J2(σ) and J3(σ) of the stress tensor are those given in the

chapter on vectors and tensors. The roots of equation (3.14) are the eigenvalues, called the

principal stresses σ1, σ2 and σ3. The eigenvectors, or principal stress directions ~Mi are again

computed from

(σ−σi I )·~Mi = ~0 ∀i ∈ 1,2,3 (3.15)

where the vectors ~Mi are normalized, i.e., ||~Mi || = 1. If the three principal stresses are dis-

tinct, the three principal stress directions will be mutually orthogonal. If two principal stresses

are equal, equation (2.83) has one exact solution and a set of an infinite number of solutions,

out of which any pair of orthogonal vectors can be chosen as the two remaining principal

stress directions. If all principal stresses are equal, each set of three mutually orthogonal

vectors is a set of principal stress directions. Such a state is called hydrostatic, since the

stress is equal in all directions as it is the case in static fluids. An axis that is oriented along a

principal stress direction is called a principal stress axis. A plane normal to a principal stress

direction is called a principal plane of σ. If a Cartesian basis is taken with three axes that

coincide with the principal stress axes, the matrix form of σ takes the same particular format

as for the strain tensor, i.e.,

σ =

σ1 0 0

0 σ2 0

0 0 σ3

(3.16)

The shear stress components are also equal to zero along these directions. Like for the anal-

ysis of the strain tensor, normal stresses attain an extreme value along the principal stress

axes. The largest and smallest normal stress for all possible orientations ~M can always be

found among the principal stresses.

The three principal stress invariants J1(σ), J2(σ) and J3(σ) can be computed directly from

the stress tensor, or alternatively from the principal stresses by

J1(σ) = σ1 +σ2 +σ3 (3.17)

J2(σ) = σ1σ2 +σ1σ3 +σ2σ3 (3.18)

J3(σ) = σ1σ2σ3 (3.19)

Any scalar function of the three principal stress invariants is a stress invariant as well.

Note that in general the principal stress directions differ from the principal strain direc-

tions. Their mutual relationship depends on the constitutive behaviour of the considered

material.

3.4 Hydrostatic and deviatoric stresses

3.4.1 Hydrostatic pressure

For the analysis of strains, it was shown that volume changes lead to a volumetric strain that

equaled the trace of the infinitesimal strain tensor tr(ε). Such volume changes are mostly

mechanically associated to the existence of a scalar pressure that acts in all directions on a

material point. On the basis of this observation, one can split the stress tensorσ in two parts,

a hydrostatic part σh and a deviatoric part σd . The hydrostatic part is computed from the

65

3. FORCE AND STRESS

pressure p that equals minus the mean stress in the material point, i.e.,

p = −σm = −13

tr(σ) (3.20)

= −σ1 +σ2 +σ3

3(3.21)

The hydrostatic stress tensor applies this pressure to all directions through

σh = −pI (3.22)

For many materials, there exists a clear constitutive relationship that relates the hydrostatic

stress tensor σh to its kinematical counterpart, the volumetric strain tensor εv .

y

z

x

x

z

p

p

p

p

p

σy y =−p

p

y

σxx = p

σy y = p

σzz = p

σxx =−p

σzz =−p

σ= p(~ex~ex +~e y~e y +~ez~ez )

σ=−p(~ex~ex +~e y~e y +~ez~ez )

Figure 3.3: Hydrostatic stress state

3.4.2 Deviatoric stresses

The remaining part of the stress tensor, after subtraction of its hydrostatic part, is called the

deviatoric stress tensor σd .

σd = σ−σ

h (3.23)

This part of the stress tensor has a zero mean stress (i.e., zero pressure). Again, for many ma-

terials there exists a direct relation between this deviatoric stress tensor and its kinematical

equivalent, the deviatoric strain tensor. It has been shown that shape changes of a material

volume element are associated to the latter tensor.

Like in section 2.5.4, the deviatoric part has three principal invariants as well J1(σd ),

J2(σd ) and J3(σd ). The first invariant is by definition zero

J1(σd ) = 0 (3.24)

while the following relations again apply for the two remaining invariants

J2(σd ) = 3 p2 − J2(σ) (3.25)

= − 12σ

d : σd (3.26)

J3(σd ) = J3(σ)+p J2(σ)−p3 (3.27)

66

3.5. Stress state with respect to a plane

A fully deviatoric stress state is typically encountered in the axial torsion of a thin-walled

tube (radius R , wall thickness t ), which is the result of an applied axial torsional moment

(torque) T . This load causes a shear stress τ in the cross-sectional wall. The shear stress

has the same value in each point of the cross-section, but changes its orientation with the

angular position of the point. Indeed, the shear stress τ follows the circumferential direction.

τ=σzx

x

z

y τ τ

τ

τ

ττ

τ

τ

τ=−σyx

τ=−σzx

τ=σyxT τ=T

2πR2t

Figure 3.4: Deviatoric stress state in the wall of a tube under torsion

σd =σ= τ(~ei~e j +~e j~ei ) with i 6= j

3.5 Stress state with respect to a plane

3.5.1 Definition

The definition of the stress tensor is already based on the existence of resultant stress vector

~p that acts on an infinitesimal plane in the material. The normal stress σnn on a plane is

clearly the normal stress along the normal~n on that plane. The shear stress τns on the plane

is the largest shear stress between the normal ~n and all directions that lie in the plane. In

fact τns is the maximum shear stress in that particular plane with respect to its normal. The

in-plane direction according to which this shear stress is maximal is denoted with the unit

vector~s, which is obtained through the orthogonal projection of the stress vector ~p on the

plane. In fact the stress vector ~p is simply decomposed into two vectors, one oriented along

the normal σnn~n and one in the plane τns~s. The scalar values of the normal and shear stress

are given by

σnn = ~n·σ·~n (3.28)

τns = σns = ~s·σ·~n (3.29)

It is convenient to express these quantities in terms of the principal stresses σ1,σ2,σ3 and

the associated principal stress directions ~M1, ~M2, ~M3. Taking a Cartesian basis with axes

67

3. FORCE AND STRESS

that coincide with the principal directions, permits to express the normal on an arbitrary

oblique plane as

~n = m1~M1 +m2

~M2 +m3~M3 (3.30)

with the normalisation condition

m21 +m2

2 +m23 = 1 (3.31)

The normal stress σnn is expanded from (3.28) as

σnn = (m1~M1 +m2

~M2 +m3~M3)·σ·(m1

~M1 +m2~M2 +m3

~M3)

= m21σ1 +m2

2σ2 +m23σ3 (3.32)

Likewise, the shear stress along vector~s is given by

τns = ||σ·~n −σnn~n|| (3.33)

Since the vectors ~n and~s are perpendicular, this relation simplifies to

τns = [ (σ·~n)·(σ·~n)−2σ2nn +σ2

nn ]0.5 = [ (σ·~n)·(σ·~n)−σ2nn ]0.5

= [(

m21σ

21 +m2

2σ22 +m2

3σ23

)

−(

m41σ

21 +m4

2σ22 +m4

3σ23 +2m2

1m22σ1σ2 +2m2

1m23σ1σ3 +2m2

2m23σ2σ3

)

]0.5

which is easily simplified by substituting m41 by m2

1(1−m22 −m2

3) and m42 and m4

3 by similar

expressions derived from (3.31)

τns =√

m21m2

2(σ1 −σ2)2 +m21m2

3(σ1 −σ3)2 +m22m2

3(σ2 −σ3)2 (3.34)

If all principal stresses are equal, the stress state is fully hydrostatic and all directions may

be considered as principal. The latter equation shows that indeed all shear stresses are then

equal to zero, independent of the orientation of the considered plane.

3.5.2 Octahedral stresses

Octahedral planes are planes with normals that bisect the principal stress directions. If the

principal stress directions are given by the set ~M1, ~M2, ~M3, the normals of the octahedral

planes are defined by

~noct =1p

3

(

±~M1 ± ~M2 ± ~M3

)

(3.35)

Taking into account all possible signs in this relation yields 8 planes which may be repre-

sented in each octant by a triangle. All eight triangles form an octahedron, which is visual-

ized in figure 3.5. The shear stresses on these planes are the octahedral shear stresses. For

the octahedral planes the direction cosines m1,m2,m3 of the normal~noct are simply ±1/p

3.

The octahedral normal stress σoct is computed from (3.32), which gives

σoct =1

3(σ1 +σ2 +σ3)

=1

3tr(σ) = −p (3.36)

i.e., minus the hydrostatic pressure in the considered point. The octahedral shear stress

computed with (3.34) is given by

τoct = 13

√

(σ1 −σ2)2 + (σ1 −σ3)2 + (σ2 −σ3)2 (3.37)

68

3.5. Stress state with respect to a plane

1Mr

3Mr

2Mr r

noct

r

noct

r

noct

Figure 3.5: The octahedral planes oriented with respect to the principal stress axes in a ma-

terial point

3.5.3 Maximum shear stress in a material point

Since the analysis on a plane provides us the largest shear stress between the normal on that

plane and all possible orientations of line elements in the plane, it is convenient to trace

the maximum shear stress in a material point for all possible orientations of that plane. De-

pendent on the direction of the plane, the shear stress attains an extreme value if the partial

derivatives with respect to the independent direction cosines vanish. First m1 and m2 are

taken as the independent variables, while m3 is dependent on them through (3.31). Substi-

tuting m23 by 1−m2

1 −m22 and taking next the partial derivatives yields

∂τns

∂m1=

m1

τns

[

m22(σ1 −σ2)2 + (1−2m2

1 −m22)(σ1 −σ3)2 −m2

2(σ2 −σ3)2]

= 0 (3.38)

and

∂τns

∂m2=

m2

τns

[

m21(σ1 −σ2)2 −m2

1(σ1 −σ3)2 + (1−m21 −2m2

2)(σ2 −σ3)2]

= 0 (3.39)

These equations are satisfied for the following orientations

m1 = 0 m2 =±1/p

2 ⇒ m3 =±1/p

2

m1 =±1/p

2 m2 = 0 ⇒ m3 =±1/p

2

Note that m1 = 0 and m2 = 0 (and thus m3 = 1) gives a shear stress equal to zero, which are

the principal directions themselves. In an analogous way, m1,m3 and m2,m3 could have

69

3. FORCE AND STRESS

been taken as independent variables, which adds the following orientations

m1 = 0 m3 =±1/p

2 ⇒ m2 =±1/p

2

m1 =±1/p

2 m3 = 0 ⇒ m2 =±1/p

2

m2 = 0 m3 =±1/p

2 ⇒ m1 =±1/p

2

m2 =±1/p

2 m3 = 0 ⇒ m1 =±1/p

2

The planes of maximum shear stress are therefore characterised by the following normals

~n1τmax

= cos45o ~M1 + sin45o ~M2

~n2τmax

= cos45o ~M1 − sin45o ~M2

~n3τmax

= cos45o ~M1 + sin45o ~M3

~n4τmax

= cos45o ~M1 − sin45o ~M3

~n5τmax

= cos45o ~M2 + sin45o ~M3

~n6τmax

= cos45o ~M2 − sin45o ~M3

Two of these planes are depicted with respect to the principal directions in figure 3.6. The

1Mr

3Mr

2Mr

45o

45o

Figure 3.6: Planes with maximum shear stress in a material point

planes with maximum shear stresses always pas through one of the principal axes, while the

normal bisects the two remaining principal stress axes. The six values of the maximum shear

stresses are then given by

τnsmax ,1 = ±σ1 −σ2

2(3.40)

τnsmax ,2= ±

σ1 −σ3

2(3.41)

τnsmax ,3 = ±σ2 −σ3

2(3.42)

70

3.6. Plane stress

3.6 Plane stress

3.6.1 Definition

A three-dimensional stress state approximately reduces to a plane stress state if the stress

componentsσzz , σxz and σyz are equal to zero. The assumed existence of such a plane stress

state is typically justified for thin flat specimens or structures, where the thickness (in the

z-direction perpendicular to the plane of loading) is sufficiently small in order to neglect the

stresses in that direction, see figure 3.7.

σzz = σxz = σyz = 0 (3.43)

The stress components in the plane are assumed to be independent of the thickness di-

s xx

s yy

t xy

x

y

s xx

t xy

s yy

z

y

Figure 3.7: A state of plane stress

rection (i.c. z-direction). The stress state is then fully characterized by its components in

the plane of the specimen, i.e., a plane stress tensor with only 3 distinct components. This

state of deformation is called a state of plane stress with respect to the (x, y) plane. Note

that the difference with the plane strain assumption resides in the thickness of the speci-

men, which must be small compared to the in-plane dimensions for plane stress and large

for plane strain.

3.6.2 Mohr’s stress circle

Since Mohr’s circle is a geometrical representation of the property of a symmetric second-

order tensor in plane loading states, it can be easily sketched for a plane stress state.

The equations which permit to elaborate the governing equations of Mohr’s stress circle

can be easily derived. Two Cartesian in plane coordinate systems are used, where the com-

ponents of the stress tensor can be defined with respect to both coordinate systems. The

original coordinate system is defined by the base vectors ~eX ,~eY , which are rotated to the

new base vectors ~ex ,~e y over an angle θ, see figure 3.8. The following relations then apply

71

3. FORCE AND STRESS

s xx

t xy

x

y

qq

r

n

s XX

s YY

t XY

X

Y

qX

Y

Figure 3.8: Stress components on rotated planes in a material point

σxx = 12 (σX X +σY Y )+ 1

2 (σX X −σY Y )cos2θ+τX Y sin2θ (3.44)

σy y = 12 (σX X +σY Y )− 1

2 (σX X −σY Y )cos2θ−τX Y sin2θ (3.45)

τx y = 12

(σY Y −σX X )sin2θ+τX Y cos2θ (3.46)

From these equations, the following transformations rules can be determined for the

stress components σxx ,σy y ,σx y as a function of the principal stresses σ1 and σ2 (X and Y

are taken along the principal axes of stress)

σxx = 12

(σ1 +σ2)+ 12

(σ1 −σ2)cos2θ (3.47)

σy y = 12 (σ1 +σ2)− 1

2 (σ1 −σ2)cos2θ (3.48)

τx y = −12

(σ1 −σ2)sin2θ (3.49)

The angle θ is again measured counterclockwise with respect to the axis of the largest prin-

cipal stress σ1. Stress components can be quantified for different orientations of the axes x

and y by simply changing θ. Mohr’s stress circle is now constructed with the normal stress

along x on the horizontal axis and the shear stress between x and y along the vertical axis,

see figure 3.9. The locus of points for σxx and τx y is then given by

(σxx −a)2 +τ2x y = R2 (3.50)

with

a = 12

(σ1 +σ2) (3.51)

R = 12 (σ1 −σ2) (3.52)

A state of stress measured with respect to axes x, y rotated over an angle θ is found on the

circle on an angle 2θ measured counterclockwise from the horizontal axis.

72

3.6. Plane stress

2q s 1s 2

s s1 2

2

-

s s1 2

2

-

s s1 2

2

+

( , )s txx xy

s xx

t xy

Figure 3.9: Mohr’s circle for plane stress

3.6.3 Mohr’s stress circle for three-dimensional stress states

Let us consider a three-dimensional stress state, characterized by the stress tensor σ with

its three principal stresses σ1,σ2,σ3. Otto Mohr showed that the normal stress σnn and the

shear stress τns acting on a plane with a 3D orientation, can be plotted on a plane with σnn

and τns as coordinates as shown in figure 3.10. The locus of the points that represents all

possible 3D stress states on the different planes necessarily falls in a closed domain, which

is represented by the shaded area in the figure. This area is bounded by the three plane cir-

cles. This result confirms the results that were found earlier for the maximum shear stresses.

Indeed, if σ1 Êσ2 Êσ3 then (σ1 −σ3)/2 is the largest shear stress, and the plane on which it

acts bisects the principal directions ~M1 and ~M3.

73

3. FORCE AND STRESS

s 1s 2s nns 3

tn s

Figure 3.10: Mohr’s circle for three-dimensional stress states

74

3.7. Summary

3.7 Summary

External forces

Body forces ρ~q(~x)

Surface forces ~p

The stress tensor

Stress vector ~p = σ·~n

Normal stresses σi i = ~Ni ·~pi

Shear stresses τi j =σi j = ~Ni ·~p j

Principal stresses σ·~Mi =σi~Mi

det(σ−σI ) = 0

(σ−σi I )·~Mi =~0 ∀i ∈ 1,2,3

Hydrostatic pressure p =−σm =−13

tr(σ)

Hydrostatic stress tensor σh =−pI

Deviatoric stress tensor σd =σ−σ

h

Stress state with respect to a plane

Normal stress σnn =~n·σ·~nShear stress τns =σns = ~s·σ·~n

τns =√

m21m2

2(σ1 −σ2)2 +m21m2

3(σ1 −σ3)2 +m22m2

3(σ2 −σ3)2

Octahedral stresses σoct =−p

τoct = 13

√

(σ1 −σ2)2 + (σ1 −σ3)2 + (σ2 −σ3)2

Maximum shear stresses τnsmax ,1=±σ1−σ2

2

τnsmax ,2 =±σ1−σ3

2

τnsmax ,3 =±σ2−σ3

2

75

3. FORCE AND STRESS

Plane stress

Plane stress σzz =σxz =σyz = 0

Mohr’s stress circle (σxx −a)2 +τ2x y = R2

R = 12 (σ1 −σ2)

a = 12 (σ1 +σ2)

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4BALANCE AND EQUILIBRIUM LAWS

In previous chapters we have studied forces acting on a continuum, the resulting deforma-

tion of the continuum, and the stresses that are caused by the deformation. The relation

between forces, deformation and stress is governed by two types of equations. The first type,

the balance laws, are mathematical formulations of universal physical principles: balance

of mass, momentum, and moment of momentum. This chapter is devoted to these balance

equations for mechanical problems in a small deformation context.

4.1 Balance of mass

We consider a body with a volume V0, a surface area A0 and a density ρ0 in the reference

state t0. In the deformed state t the body occupies a volume V while having a surface area

A and a density ρ. To derive the law of balance of mass, we concentrate on an arbitrary,

A

V

V0

t0

A0VV0

t

Figure 4.1: A body in the initial state t0 and the deformed state t , with arbitrary subregion V .

fixed amount of material points in a subregion V0, i.e. we use a Lagrangian approach. In the

initial state these material points occupied a volume V0 while having a surface area A0 and a

density ρ0. According to the law of balance of mass, the mass of these material points must

77

4. MECHANICAL EQUILIBRIUM

have remained constant during the deformation:

∫

V

ρdV =∫

V0

ρ0 dV0 (4.1)

This equation is called the global formulation of the law of balance of mass. From the theory

of kinematics, we know that the volumes before and after deformation are related as dV =J dV0, where J represents the determinant of the deformation tensor. Thus we can transform

the left-hand integral into an integral over the undeformed volume:

∫

V0

ρ J dV0 =∫

V0

ρ0 dV0 (4.2)

which can also be written as:∫

V0

(ρ J −ρ0)dV0 = 0 (4.3)

Since this integral must hold for any arbitrary volume V0, it follows that the integrand must

be zero:

ρ J = ρ0 (4.4)

This equation is called the local formulation of the law of balance of mass.

4.2 Balance of momentum

4.2.1 General derivation

Let’s consider a fixed amount of matter having a volume V , a surface A and a density ρ. The

matter is acted upon by body forces ~q and surface forces ~p. The total momentum~i of the

~q

A

V

V

t

A

~p

V

Figure 4.2: Forces ~q and ~p acting on a subvolume V of a body.

matter equals:

~i =∫

V

ρ~v dV (4.5)

78

4.2. Balance of momentum

The total resulting force ~f , acting on the matter, is the sum of all body forces and surface

forces:

~f =∫

V

ρ~q dV +∫

A

~p d A

=∫

V

ρ~q dV +∫

A

σ·~n d A

=∫

V

(ρ~q +~∇·σT )dV (4.6)

where we have applied Gauss’s theorem to convert the surface integral into a volume integral.

According to the law of balance of momentum, this resulting force is equal to the rate of

change of the momentum of the matter in V :

∫

V

(ρ~q +~∇·σT )dV =d

d t

∫

V

ρ~v dV (4.7)

To elaborate the differentiation with respect to time, we convert the right-hand volume inte-

gral to the undeformed situation:

d

d t

∫

V

ρ~v dV =d

d t

∫

V0

ρ0~v dV0

=∫

V0

ρ0~v dV0

=∫

V

ρ~v dV (4.8)

where we used the local form of the law of balance of mass (ρdV = ρ0dV0). The global for-

mulation of the law of balance of momentum can thus be written as:∫

V

(ρ~q +~∇·σT )dV =∫

V

ρ~v dV (4.9)

Now again, because the volume V may be chosen arbitrarily, the integrand must be zero,

yielding the local formulation of the law of balance of momentum:

~∇·σT +ρ~q = ρ~v (4.10)

These equations are also called Cauchy’s equations of motion. In case of static equilibrium,

the acceleration ~v is zero, and the equations of motion are reduced to the equilibrium equa-

tions:~∇·σT +ρ~q =~0 (4.11)

4.2.2 Translational equilibrium

Let’s consider a three-dimensional parallelepiped that interconnects parallel facets in two

neighbouring material points P and Q. The infinitesimal volume is depicted in figure 4.3

79

4. MECHANICAL EQUILIBRIUM

s xx

ss

yx

yx

xdx+

¶

¶

s yy

s zz

ss

yy

yy

ydy+

¶

¶

ss

zzzz

zdz+

¶

¶

s xy

s xz

s zx

s yx

s zy

s yz

ss

xxxx

xdx+

¶

¶

ss

zxzx

xdx+

¶

¶

ss

yz

yz

zdz+

¶

¶s

sxz

xz

zdz+

¶

¶

ss

xy

xy

ydy+

¶

¶

ss

zy

zy

ydy+

¶

¶

r

ex

r

ey

r

ez

P

Q

Figure 4.3: Three-dimensional stress parallelepiped

along with the Cartesian stress components that act on all the six facets. The back, left and

bottom facets are representative for the first material point , while the other remaining facets

belong to a material point that is located d x, d y and d z further in space. The stress compo-

nents differ through this spatial difference as indicated in the figure.

The translational or force equilibrium of the stress parallelepiped with respect to the

three Cartesian axes is here equivalent to the balance of momentum. Let’s solve the equi-

librium along the y-axis. To this purpose, the stress components oriented along the y-axis

are sketched in a separate figure, see figure 4.4. Taking into account these stress components

and the surfaces on which they act, leads to the following force equilibrium along the y-axis

(

σy y +∂σy y

∂yd y

)

d xd z +(

σyz +∂σyz

∂zd z

)

d xd y +(

σyx +∂σyx

∂xd x

)

d yd z

−σy y d xd z −σyzd xd y −σyx d yd z +ρqy d xd yd z = 0(4.12)

where ρqy is the volume force in the y −dir ect ion. Simplifying this equation, and dividing

by d xd yd z yields

∂σyx

∂x+∂σy y

∂y+∂σyz

∂z+ρqy = 0 (4.13)

The equilibrium along the x-axis and the z-axis delivers similar equations, which results in

80

4.2. Balance of momentum

ss

yx

yx

xdx+

¶

¶

s yys

syy

yy

ydy+

¶

¶

s yx

s yz

ss

yz

yz

zdz+

¶

¶

r

ex

r

ey

r

ez

dy

dx

dz

Figure 4.4: Contributing components for the translational equilibrium along the y-axis

a set of partial differential equations

∂σxx

∂x+∂σx y

∂y+∂σxz

∂z+ρqx = 0 (4.14)

∂σyx

∂x+∂σy y

∂y+∂σyz

∂z+ρqy = 0 (4.15)

∂σzx

∂x+∂σz y

∂y+∂σzz

∂z+ρqz = 0 (4.16)

which can be written in index notation as

σi j , j +ρqi = 0 (4.17)

or in its tensorial equivalent~∇·σT +ρ~q = ~0 (4.18)

The latter equation is the general equation for static equilibrium. Note that equation (4.18)

is independent of the coordinate system used, which means that is can also be applied to

e.g., cylindrical coordinate problems.

4.2.3 Equilibrium in a Cartesian and cylindrical basis

The equilibrium equation can be written in components w.r.t. a Cartesian vector basis. This

results in three partial differential equations, one for each coordinate direction.

81

4. MECHANICAL EQUILIBRIUM

σxx,x +σx y,y +σxz,z +ρqx = 0

σyx,x +σy y,y +σyz,z +ρqy = 0

σzx,x +σz y,y +σzz,z +ρqz = 0

Writing tensor and vectors in components w.r.t. a cylindrical vector basis is more elab-

orative because the cylindrical base vectors ~er and ~et are a function of the coordinate θ, so

they have to be differentiated, when expanding the divergence term.

σr r,r +1

rσr t ,t +

1

r(σr r −σt t )+σr z,z +ρqr = 0

σtr,r +1

rσt t ,t +

1

r(σtr +σr t )+σtz,z +ρqt = 0

σzr,r +1

rσzt ,t +

1

rσzr +σzz,z +ρqz = 0

4.3 Balance of moment of momentum

4.3.1 General derivation

The balance of moment of momentum states that the total moment about a fixed point of

all forces working on a randomly chosen volume of material points (~Mo), equals the change

of the total moment of momentum of the material points inside the volume, taken w.r.t. the

same fixed point (~Lo).

~x

A

V

V

t

A

~p

V

~q

O

Figure 4.5: Moment of forces on a random section of a material body

~Mo =d~Lo

d t=

d

d t

∫

V

~x ×ρ~v dV ∀ V (4.19)

Making use of the mass balance and mathematical manipulations, it can be shown that

this reduces to the following condition:

σ32 −σ23

σ13 −σ31

σ21 −σ12

=

0

0

0

82

4.3. Balance of moment of momentum

The balance of moment of momentum therefore implies that the Cauchy stress tensor has

to be symmetric

σT =σ ∀ ~x ∈V (t ) (4.20)

As a result of the balance of moment of momentum, the symmetric stress tensor simpli-

fies the balance of momentum to

~∇·σ+ρ~q = ρ~v (4.21)

4.3.2 Moment or rotational equilibrium

The equilibrium of a material element also requires a zero resulting moment. This is the mo-

ment or rotational equilibrium. The derivation is here performed with respect to the x-axis,

other axes lead to similar results (cyclic permutation of indices). The infinitesimal paral-

lelepiped is reconsidered, but only with the components which cause a non-zero moment

around the x-axis, see figure 4.6. Properly taking care of the moment arm around the x-axis

ss

yx

yx

xdx+

¶

¶

s yy

s zz

ss

yy

yy

ydy+

¶

¶

ss

zzzz

zdz+

¶

¶

s zx

s yxss

zxzx

xdx+

¶

¶

ss

yz

yz

zdz+

¶

¶

ss

zy

zy

ydy+

¶

¶

r

ex

r

ey

r

ez

dx

dy

dz

Figure 4.6: Contributing components for the moment around the x-axis

83

4. MECHANICAL EQUILIBRIUM

for each component leads to

−(

σy y +∂σy y

∂yd y

)

d xd zd z

2+σy y d xd z

d z

2

+(

σz y +∂σz y

∂yd y

)

d xd zd y −(

σyz +∂σyz

∂zd z

)

d xd yd z

+(

σzz + ∂σzz

∂zd z

)

d xd yd y

2−σzz d xd y

d y

2

+(

σzx + ∂σzx

∂xd x

)

d yd zd y

2−σzx d yd z

d y

2

−(

σyx +∂σyx

∂xd x

)

d yd zd z

2+σyx d yd z

d z

2

−ρqy d xd yd zd z

2+−ρqz d xd yd z

d y

2= 0

(4.22)

Dividing this equation by d xd yd z and taking the limit d x → 0, d y → 0 and d z → 0 finally

gives the straightforward equation

σyz = σz y (4.23)

Completing this with the rotational equilibrium around the y and z-axis gives

σx y = σyx

σyz = σz y

σxz = σzx

The rotational equilibrium leads to the fundamental conclusion that the stress tensor is sym-

metric, a property which was already exploited earlier in this chapter. In the tensorial format

this simply reads

σ = σT (4.24)

Note that the symmetry of σ permits to rewrite the translational equilibrium as

~∇·σ+ρ~q = ~0 (4.25)

which is the commonly used format for static equilibrium.

4.3.3 Rotational equilibrium in a Cartesian and cylindrical basis

With respect to a Cartesian or cylindrical basis the symmetry of the stress tensor results in

three equations.

σ=σT →

Cartesian : σx y =σyx ; σyz =σz y ; σzx =σxz

cylindrical : σr t =σtr ; σtz =σzt ; σzr =σr z

4.4 Special deformation states

The three-dimensional equilibrium equations can be simplified for special deformation or

stress states, such as plane strain, plane stress and axisymmetric cases.

84

4.4. Special deformation states

4.4.1 Planar deformation

It is assumed here that the z-direction is the direction where either the strain or the stress

is zero. Only stresses and strains in the plane perpendicular to the z-direction remain to

be determined from equilibrium. The strain or stress in the z-direction can be calculated

afterwards, either directly from the material law or iteratively during the solution procedure.

Cartesian components

σxx,x +σx y,y +ρqx = 0

σyx,x +σy y,y +ρqy = 0

σx y =σyx

cylindrical components

σr r,r +1

rσr t ,t +

1

r(σr r −σt t )+ρqr = 0

σtr,r +1

rσt t ,t +

1

r(σtr +σr t )+ρqt = 0

σr t =σtr

4.4.2 Axisymmetric deformation

In many cases the geometry, boundary conditions and material behavior is such that no

state variable depends on the circumferential coordinate θ : ∂∂θ = 0. For such axisymmetric

deformations, the equilibrium equations can be simplified considerably.

In many axisymmetric deformations the boundary conditions are such that there is no

displacement in the circumferential direction : ut = 0. In these cases there are only four

relevant strain and stress components and only three equilibrium equations.

When boundary conditions and material behavior are such that displacement of material

points are only in the rθ-plane, the deformation is referred to as plane strain in the rθ-plane.

When stresses on a plane perpendicular to the z-direction are zero, the stress state is

referred to as plane stress w.r.t. the rθ-plane.

σr r,r +1

r(σr r −σt t )+σr z,z +ρqr = 0

σtr,r +2

r(σtr )+σtz,z +ρqt = 0 (if ut 6= 0)

σzr,r +1

rσzr +σzz,z +ρqz = 0

σr t =σtr ; σtz =σzt (if ut 6= 0)

σzr =σr z

planar

σr r,r +1

r(σr r −σt t )+ρqr = 0

σtr,r +2

r(σtr )+ρqt = 0 (if ut 6= 0)

σr t =σtr (if ut 6= 0)

85

4. MECHANICAL EQUILIBRIUM

4.5 Mechanical work and internal energy

Let’s consider a volume V of material that is enclosed by a closed surface S. The considered

body is deformed under the influence of external surface forces and body forces and it is now

in an equilibrium state. The total work performed by the external forces is denoted We , while

the work performed by the internal forces is written Wi . The change of the internal work or

internal energy due to a change in deformations is denoted δWi . Evidently, such a change is

due to a change of the external forces, with a variation in external work that equals δWe . We

limit ourselves here to adiabatic processes, and hence according to the law of conservation

of energy it must hold that

δWe = δWi (4.26)

The internal work or energy can be computed by taking the integral over the volume V of the

so-called energy density Wi , which is the energy per unit of volume. One can write

δWi = δ

Ñ

V

Wi dV

= δWe (4.27)

The change in external work is equal to the contribution of the body forces and a contribu-

tion of the surface forces

δWe =Ñ

V

ρ~q ·δ~u dV +Ï

S

~p·δ~u dS (4.28)

where δ~u is the infinitesimal change in the displacement vector of a material point. The

contribution of the surface forces may be written as

Ï

S

~p·δ~u dS =Ï

S

(σ·~n)·δ~u dS (4.29)

=Ï

S

(~n·σ)·δ~u dS (4.30)

=Ï

S

~n·(σ·δ~u)dS (4.31)

Making use of the divergence theorem for a vector ~a =σ·δ~uÑ

V

~∇·~a dV =Ï

S

~n·~a dS (4.32)

where ~n is the normal on the surface element dS, permits to rewrite the contribution of the

surface forces as

Ï

S

~p·δ~u dS =Ñ

V

~∇·(σ·δ~u)dV (4.33)

=Ñ

V

[

(~∇·σ)·δ~u +σ : (~∇δ~u)T]

dV (4.34)

86

4.5. Mechanical work and internal energy

Combining this term with the contribution of the body forces,

δWe =Ñ

V

[

(ρ~q +~∇·σ)·δ~u]

dV +Ñ

V

σ : (~∇δ~u)T dV (4.35)

Substituting the equilibrium equation (4.18) makes the first integral vanish. Furthermore,

the symmetry of σ permits to write

σ : (~∇δ~u)T = σ : (~∇δ~u) = σ : 12

[~∇δ~u + (~∇δ~u)T

]

= σ : δε (4.36)

Finally the external work simplifies to the internal work by

δWe =Ñ

V

σ : δεdV = δWi (4.37)

The variation of the internal energy density δWi is thus simply equal to

δWi = σ : δε (4.38)

In rate format, this equation is written as

Wi = σ : ε (4.39)

87

4. MECHANICAL EQUILIBRIUM

4.6 Summary

Equilibrium

Rotational (Moment) σ=σT

Translational (Force) ~∇·σ+ρ~q =~0

External/internal work

Infinitesimal change δWe =Ð

V

σ : δεdV = δWi

Internal energy density rate Wi =σ : ε

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5THREE-DIMENSIONAL LINEAR ELASTICITY

This chapter establishes the relation between the kinematics and the forces or stresses acting

on linear elastic solids. Attention is given to the three-dimensional constitutive relationship

for isotropic and several anisotropic materials. Various elastic constants are introduced and

their mutual relationships are established. The special case of isotropy is analyzed in its

various aspects, and simplifications towards plane strain and plane stress are treated. Finally,

an example for an anisotropic fibre-reinforced composite material is given.

5.1 General aspects of constitutive equations

The equations of equilibrium are not sufficient to determine the stress and displacement

fields in a material. In the more general case, one could add other equations of physics, i.e.,

conservation of mass and energy, but even then physical laws for the constituent material

are missing. In this course, we will only focus on the constitutive behaviour that establishes

the relation between stresses and kinematics.

5.2 The concept and nature of linear elasticity

One of the essential requirements for the application of the theory of linear elasticity, is the

restriction to small displacements, as shown in the previous chapters. If this condition is

satisfied, almost each material presents a linear elastic behaviour up to a certain deforma-

tion. This is due to the nature of linear elasticity. It finds its roots in the very small de-

formation of the existing microstructure, which does not alter its structure during defor-

mation. Hence, the geometry of crystals (crystalline materials, e.g., metals), of molecular

networks (polymers), of granular structures (ceramics), fibre-matrix arrangements (com-

posites) is not affected in linear elasticity. It is this essential property that distinguishes

elastic behaviour from the nonlinear dissipative deformation mechanisms, during which

microstructural changes will always take place. This also means that any process which

changes the microstructure by another treatment (e.g., heat treatment) will also affect the

elastic properties of the material.

5. THREE-DIMENSIONAL LINEAR ELASTICITY

The elastic properties of a crystalline material are thus attributed to its sensitivity to

small changes in inter-atomic distances due to the inter-atomic binding energy. The bind-

ing forces which strongly vary with distance may be of a different nature, e.g., ionic (electro-

static), covalent (sharing electrons), metallic (metal ion interaction with free-electron gas),

van der Waals (intermolecular attraction), hydrogen (dipole moments).

5.2.1 The elastic response of materials

In the case of perfect elasticity, the material fully recovers its initial structure. In the case of

infinitesimal displacements, it has been shown that the strain tensor ε suffices to character-

ize the deformation of the material, while the stress tensor σ gives a representation of the

forces that should act on an infinitesimal volume of material. In the most general case, a

stress tensor σ(~x) in a material point may depend on the strain tensors ε(~ξ|~ξ ∈ V (~x)) of all

surrounding points. In this case, the constitutive relation F is said to be nonlocal. In most

cases, one assumes that the macroscopic scale is large enough to postulate the principle of

local action, according to which the stress tensorσ(~x) is only related to the local strain tensor

ε(~x). In the case of elasticity, the load does not depend on the loading history but only on the

actual deformed state of the material. Hence, there is no need to include the history with in-

ternal history variables that are related to the changes in the underlying microstructure. This

considerably simplifies the resulting constitutive law. In the case of small displacements, one

may write that elastic behaviour is described by some function F according to

σ(~x) = F (ε(~x)) (5.1)

In the specific case of linear elasticity, which is of interest in this course, the function

F will be linear as well. This is a first-order approximation that applies to any continuous

function, since each continuous function can be approximated by a linear relationship in a

sufficiently small range of its variables.

5.2.2 The elastic stored energy function

Since elastic material is characterized by the reversibility of deformations and by the fact

that the underlying microstructure remains structurally unchanged, a function exists which

measures the amount of energy that is stored in the material. Assuming a local constitutive

behaviourσ = F (ε) and no history dependence, permits to integrate equation (4.39) in time,

which yields the stored energy density

W =∫

tWi d t =

∫

tσ : εd t (5.2)

= 12σ : ε (5.3)

This functions quantifies the amount of energy per unit of volume that is stored in a material

point.

It can be noticed that the stress tensor σ is related to the stored energy density by

σ =∂W

∂ε(5.4)

This property is inherent to the reversibility of the elastic deformations. In terms of the indi-

vidual components of the stress and strain tensor with respect to a Cartesian basis ~ex ,~e y ,~ez ,

90

5.3. General 3D linear elastic solid

one may write from equation (5.4)

δW =σ : δε

=σxxδεxx +σy yδεy y +σzzδεzz

σx yδεyx +σxzδεzx +σyzδεz y

σyxδεx y +σzxδεxz +σz yδεyz

=σxxδεxx +σy yδεy y +σzzδεzz

τx yδγx y +τxzδγxz +τyzδγyz (5.5)

and hence

σxx =∂W

∂εxxσy y =

∂W

∂εy yσzz =

∂W

∂εzz(5.6)

τx y =∂W

∂γx yτxz =

∂W

∂γxzτyz =

∂W

∂γyz(5.7)

Note that the engineering shear strain appear naturally in this relation.

5.3 General 3D linear elastic solid

Uniaxial linear elastic behaviour has already been treated in previous courses, for which it is

known that the uniaxial stress σ is related to the uniaxial strain ε by means of only one elastic

constant, i.e., Young’s modulus E

σ = E ε (5.8)

This is the well-known Hooke’s law (Robert Hooke, 17th century). The modulus E inherits

its name from Thomas Young (early 19th century). This linear relation between stress and

strain will next be generalized to three dimensions.

5.3.1 Generalized Hooke’s law for anisotropic solids

As shown in the introductory chapter on tensors, the linear mapping of a second-order ten-

sor (i.c. the strain tensor ε onto another second-order tensor (i.c. the stress tensor σ) is per-

formed with a fourth-order tensor, which is here denoted the elasticity tensor4C . The three-

dimensional equivalent of (5.8) thus reads

σ =4

C : ε (5.9)

In general, a fourth-order tensor has 3×3×3×3 = 81 components. The resulting tensor (i.c.

σ) must be symmetric and hence4C must be left symmetric. Furthermore, since both the

strain tensor ε and the stress tensor σ are symmetric tensors, 36 distinct components are yet

sufficient to uniquely determine equation (5.9). This conclusion is easily made by reconsid-

ering the matrix condensation of the tensor equation (5.9). This example was already treated

in equation (1.210) of the chapter on tensors. If we replace in this equation the symmetric

tensor C by the stress tensor σ, the fourth-order tensor4A by the elasticity tensor

4C and the

91

5. THREE-DIMENSIONAL LINEAR ELASTICITY

symmetric tensor B by the strain tensor ε, one finds

σ~ = C ε~ (5.10)

σ11σ22σ33τ12τ13τ23

=

C1111 C1122 C1133 C1112+C1121 C1113+C1131 C1123+C1132C2211 C2222 C2233 C2212+C2221 C2213+C2231 C2223+C2232C3311 C3322 C3333 C3312+C3321 C3313+C3331 C3323+C3332C1211 C1222 C1233 C1212+C1221 C1213+C1231 C1223+C1232C1311 C1322 C1333 C1312+C1321 C1313+C1331 C1323+C1332C2311 C2322 C2333 C2312+C2321 C2313+C2331 C2323+C2332

ε11ε22ε33ε12ε13ε23

(5.11)

This equation clearly shows that since only 36 numbers are involved, and one can thus al-

ways choose the tensor4C right symmetric as well, i.e., one chooses Ci j 12 = Ci j 21. This is

what is generally done, and hence the elasticity tensor is left and right symmetric

4

C =4

CLT

=⇒ Ci j kl = C j i kl (5.12)

4

C =4

CRT

=⇒ Ci j kl = Ci j l k (5.13)

The number of components has to be reduced further to correctly account for the path

independence of elastic deformations (only the current deformation state determines the

stored energy and not the path that was followed towards that state). For a linear elastic

material, equation (5.3) simplifies to

W = 12σ : ε (5.14)

Path independence implies that a stored energy density function exists, and the energy in

the current deformed state is obtained by substitution of (5.9) in (5.14) leading to

W = 12

(4

C : ε) : ε

= 12ε :

4

C : ε (5.15)

which is a symmetric quadratic form in terms of the strain tensor ε. The energy density is a

scalar, which means that taking the transpose gives exactly the same result. Performing this

operation and making use of the symmetry of the strain tensor yields

W = 12

[

ε :4

C : ε]T

(5.16)

= 12ε :

4

CT

: ε (5.17)

Hence, the elasticity tensor4C must be middle symmetric as well, which limits the total num-

ber of independent elastic constants for a generally anisotropic material to 21. Note that the

individual elasticity constants can be obtained by the partial derivatives of the stored energy

density, e.g., it holds that

∂W

∂ε11= σ11 =C1111ε11 +C1122ε22 +C1133ε33+

2C1112ε12 +2C1113ε13 +2C1123ε23

and hence

C1111 =∂2W

∂ε211

(5.18)

C1122 =∂2W

∂ε11∂ε22=

∂2W

∂ε22∂ε11= C2211 (5.19)

92

5.4. Material symmetry and anisotropy

which again illustrates the middle symmetry. In tensor format, the latter set of equations is

generalized as

4

C =∂2W

∂ε∂ε(5.20)

The generalization of 1D to 3D thus involves 21 material constants instead of 1 single

value. The matrix condensation given in equation (5.11) can be rewritten as

σ~ = C ε~ (5.21)

σ11

σ22

σ33

τ12

τ13

τ23

=

C1111 C1122 C1133 2C1112 2C1113 2C1123

C1122 C2222 C2233 2C2212 2C2213 2C2223

C1133 C2233 C3333 2C3312 2C3313 2C3323

C1112 C2212 C3312 2C1212 2C1213 2C1223

C1113 C2213 C3313 2C1213 2C1313 2C1323

C1123 C2223 C3323 2C1223 2C1323 2C2323

ε11

ε22

ε33

ε12

ε13

ε23

(5.22)

or by using the engineering shear strains instead

σ~ = C ε~

σ11

σ22

σ33

τ12

τ13

τ23

=

C1111 C1122 C1133 C1112 C1113 C1123

C1122 C2222 C2233 C2212 C2213 C2223

C1133 C2233 C3333 C3312 C3313 C3323

C1112 C2212 C3312 C1212 C1213 C1223

C1113 C2213 C3313 C1213 C1313 C1323

C1123 C2223 C3323 C1223 C1323 C2323

ε11

ε22

ε33

γ12

γ13

γ23

(5.23)

in which the matrix C is the general anisotropic Hookean stiffness or elasticity matrix. The

inverse of the elasticity tensor4C is called the compliance tensor

4S. It satisfies the relations

ε =4

S : σ (5.24)

W = 12σ :

4

S : σ (5.25)

and4

S =4

SLT

=4

SRT

=4

ST

(5.26)

Like the stiffness tensor, the compliance tensor counts 21 independent components only.

5.4 Material symmetry and anisotropy

The general constitutive equation for linear elasticity, will now be simplified systematically

in order to account for various types of symmetry in the material. This symmetry always

originates from the microstructure of the material. Fortunately, most engineering materials

do present symmetry properties, which allows to reduce the independent elastic constants

considerably. It is then convenient to formulate the components of the elasticity tensor with

respect to a basis that is attached to the symmetry axes or planes in the material. Note that

we systematically neglect the dependence of the material parameters on other state vari-

ables that influence the microstructure, e.g., the temperature.

93

5. THREE-DIMENSIONAL LINEAR ELASTICITY

5.4.1 Orthotropic materials

Orthotropic materials present two orthogonal planes of symmetry (e.g., orthorhombic crys-

tals, Fe3C cementite). If the normals to these two planes are taken as the base vectors~e2 and

~e3, we will first consider the symmetry plane with normal~e3.

Reverting the~e3 base vector to~e⋆

3 therefore does not alter the constitutive relation. If this

vector is reverted, then clearly certain shear strains (i.e., the change of angles) change sign,

e.g., the angle changes between~e1 and~e3, as well as~e2 and~e3

γ⋆

13 = −γ13

γ⋆

23 = −γ23

Likewise, the shear stresses in these directions change sign

τ⋆13 = −τ13

τ⋆23 = −τ23

while all other stress components remain unchanged. Since the elasticity tensor must be

identical in both cases, it must hold that

σ11 = σ⋆

11

C1111ε11 +C1122ε22 +C1133ε33 +C1112γ12 +C1113γ13 +C1123γ23 =C1111ε11 +C1122ε22 +C1133ε33 +C1112γ12 −C1113γ13 −C1123γ23

for each strain state ε, which is only possible if C1113 = 0 and C1123 = 0. Repeating this for all

components yields

C1113 = C1123 = C2213 = C2223 = 0

C3313 = C3323 = C1213 = C1223 = 0(5.27)

We can repeat this reasoning for the second symmetry plane, with~e2 as the normal vec-

tor. For a reflection with respect to this second symmetry plane, elastic coefficients may not

be affected neither, i.e., one has to add the following conditions

γ⋆

12 = −γ12

γ⋆

23 = −γ23

τ⋆12 = −τ12

τ⋆23 = −τ23

which is only possible if additional elastic constants vanish, i.e.,

C1112 = C2212 = C3312 = C1323 = 0 (5.28)

Orthotropic elasticity is thus characterized by 9 independent elastic constants. The corre-

sponding matrix format attached to the anisotropic base vectors reads

σ~ = C ε~

σ11

σ22

σ33

τ12

τ13

τ23

=

C1111 C1122 C1133 0 0 0

C1122 C2222 C2233 0 0 0

C1133 C2233 C3333 0 0 0

0 0 0 C1212 0 0

0 0 0 0 C1313 0

0 0 0 0 0 C2323

ε11

ε22

ε33

γ12

γ13

γ23

(5.29)

94

5.4. Material symmetry and anisotropy

Note that the symmetry with respect to the planes normal to~e2 and~e3 automatically involves

the symmetry with respect to the plane normal to ~e1 as well. This is typical for orthotropic

symmetry, e.g., for orthorhombic crystal structures. The stiffness matrix C in equation (5.29)

can also be written in a different format, which will be used further in the course:

C =

A Q R 0 0 0

Q B S 0 0 0

R S C 0 0 0

0 0 0 K 0 0

0 0 0 0 L 0

0 0 0 0 0 M

(5.30)

A more detailed treatment of orthotropic elasticity is given in Appendix A.

5.4.2 Transverse isotropy

While the preceding cases of symmetry are typically directionally symmetric materials, trans-

verse isotropic materials present a rotational symmetry around a specific axis. Assume that

this symmetry axis is oriented along the base vector~e3. This means that there is no material

distinction between the vector ~e1 and ~e2, which can be oriented in any direction perpen-

dicular to the axis of rotational symmetry. The elastic constants involving one of these two

directions thus should be strictly identical, i.e.,

C1111 = C2222 = A (5.31)

C1122 = Q (5.32)

C1133 = C2233 = R (5.33)

C3333 = C (5.34)

C1212 =C1111 −C1122

2= 1

2(A−Q) = K (5.35)

C1313 = C2323 = L (5.36)

while angle changes between any direction with ~e1 and ~e2 may produce according shear

stresses only

C1112 = C1113 = C1123 = C2212 = C2213 = 0

C2223 = C3312 = C3313 = C3323 = C1213 = C1323 = 0(5.37)

The full proof is obtained by rotating the axes around the symmetry axis~e3, which may not

alter the components of the elasticity tensor (the elasticity coefficients remain invariant un-

der a rotation). The constitutive behaviour for a transverse isotropic materials is thus fully

determined from 5 constants. This is typically the case for hexagonal crystals (HCP, e.g., Z n,

M g , T i ) or honeycomb composite structures. The constitutive relation in matrix form then

reads

σ11

σ22

σ33

τ12

τ13

τ23

=

A Q R 0 0 0

Q A R 0 0 0

R R C 0 0 0

0 0 0 K 0 0

0 0 0 0 L 0

0 0 0 0 0 L

ε11

ε22

ε33

γ12

γ13

γ23

(5.38)

95

5. THREE-DIMENSIONAL LINEAR ELASTICITY

Note that these 5 elastic constants are often related to engineering elasticity coefficients, like

Young’s moduli Ei , Poisson’s ratios νi j and shear moduli Gi j . The following identities are

then used

A =E1(1−ν31ν13)

(1+ν12)(1−ν12 −2ν31ν13)(5.39)

Q =E1(ν12 +ν31ν13)

(1+ν12)(1−ν12 −2ν31ν13)(5.40)

R =E1ν31

(1−ν12 −2ν31ν13)(5.41)

C =E3(1−ν12)

(1−ν12 −2ν31ν13)(5.42)

L = G13 = G23 =E1

2(1+ν12)(5.43)

where

ν13 =E1

E3ν31 (5.44)

More comments on the physical meaning of Young’s moduli and Poisson’s ratios are given

in the next section. A detailed treatment of transversely isotropic elasticity is given in Ap-

pendix A.

5.5 Linear elastic isotropic materials

If a material has a microstructure which is sufficiently randomly oriented and distributed

on a very small scale, the macroscopic material properties will be identical in all directions.

This is typically the case for metals with a randomly oriented polycrystalline structure, ran-

dom granular structures in ceramics, random fibre reinforced composites. Such a material is

called isotropic. This means that the orientation of the material in a material point does not

influence its mechanical behaviour. The properties of the material are said to be invariant

under a rotation of the material axes. This means that the constitutive relation becomes fully

independent of the basis, since any particular choice for the orientation of the base vectors

must give the same elasticity tensor.

In the symmetry cases presented above, the material properties depend on the consid-

ered direction. That is a typical characteristic of an anisotropic material. If the material

properties are constant from point to point in the material, the material is said to be ho-

mogeneous. A continuum may be anisotropic and non-homogeneous (materials with large

anisotropic grains with different orientations), anisotropic and homogeneous (single crys-

talline materials, directionally fibre reinforced composites) or isotropic and homogeneous

(randomly oriented microstructures from polycrystals). Clearly, the notion isotropy and ho-

mogeneity depend on the scale of observation and application of the material. Metals are

often macroscopically homogeneous and isotropic, but on a much smaller scale of the in-

dividual grains they are highly anisotropic and non-homogeneous. In the remainder of this

course, we will limit ourselves to the macroscopic scale.

Since isotropic materials are by definition independent of the orientation of the material

(the undeformed reference configuration), the stored energy must be a function that does

not depend on the orientation of the material. In equation (5.15), a quadratic form in terms

of the strain tensor ε was found. In order to have a stored energy that is independent of the

96

5.5. Linear elastic isotropic materials

orientation (the choice of the basis), expression (5.15) must simplify into an equation that

only depends on the invariants of the strain tensor ε, since these are the only scalar tensor

functions that are independent of the basis. Hence, it will hold that

W = f (J1(ε), J2(ε), J3(ε)) (5.45)

5.5.1 The isotropic elasticity tensor

An isotropic tensor is a tensor of which the components do not change by orthogonal coor-

dinate transformations (rotations). Evidently all scalars are isotropic tensors (of rank zero).

Vectors, which are tensors of order 1, can never be isotropic, since they will always have dif-

ferent components if the base vectors are rotated. Isotropic second-order tensors are nec-

essarily constructed on the second-order unit tensor I . This tensor always takes the same

components, irrespective of the orientation of the base vectors. A general isotropic second-

order tensor is obtained through the multiplication of I with a scalar, i.e.,

k I or k δi j (5.46)

Dyadic products of the second-order unit tensor with itself, i.e., I I , yields a fourth-order

isotropic and total symmetric tensor. Apart from this one, only the symmetric fourth-order

unit tensor4I

sis isotropic and totally symmetric. Hence, the isotropic fourth-order elasticity

tensor is necessarily a linear combination of these two isotropic fourth-order total symmet-

ric tensors, i.e.,

4

C = λ I I +2µ4

Is

(5.47)

or in index notation

Ci j kl = λδi jδkl +µ(δi lδ j k +δi kδ j l ) (5.48)

This is the general format of the isotropic elasticity tensor, in which the elastic constants λ

and µ are called Lamé’s constants (from Gabriel Lamé, middle 19th century). Lamé’s second

constant µ equals the more frequently used shear modulus G =µ.

The constitutive relation that ensues from the linear mapping of ε on σ simplifies to

σ =4

C : ε (5.49)

= (λ I I +2G4

Is) : ε (5.50)

= λ I (I : ε)+2G (4

Is

: ε) (5.51)

= λ tr(ε)I +2G ε (5.52)

or in index format

σi j = λεkkδi j +2G εi j (5.53)

This constitutive equation is often rewritten to clearly separate the pressure-dilatation re-

97

5. THREE-DIMENSIONAL LINEAR ELASTICITY

sponse from the deviatoric response (i.e., relating σh to ε

h and likewise σd to ε

d )

σ = λ I tr(ε)+2G ε (5.54)

= λtr(ε)I +2G(

εd + 1

3tr(ε)I

)

= (λ+ 23

G)tr(ε)I +2G εd

= (3λ+2G) 13

tr(ε)I +2G εd

= 33λ+2G

3εv +2G ε

d

= 3 K εv +2G εd (5.55)

where the elastic constant K is called the bulk modulus or compressibility modulus. The

bulk modulus, given by

K =3λ+2G

3(5.56)

relates the hydrostatic pressure p to the volumetric strain e by e = tr(ε), i.e.,

p = −K e (5.57)

13

tr(σ) = K tr(ε)

σm = 3 K εm (5.58)

In fact, the latter equation clearly illustrated that the volumetric parts (σh = 13

tr(σ), εv =13

tr(ε)) and deviatoric parts (σd = σ−σh, εd = ε− εv ) of the stress and strain tensors are

constitutively related in a separate way

σh = 3 K εv (5.59)

σd = 2G ε

d (5.60)

In matrix representation, and making use of the engineering shear strains instead of the

tensorial shear strains, the constitutive equation reads

σ~ = C ε~ (5.61)

σ11

σ22

σ33

τ12

τ13

τ23

=

λ+2G λ λ 0 0 0

λ λ+2G λ 0 0 0

λ λ λ+2G 0 0 0

0 0 0 G 0 0

0 0 0 0 G 0

0 0 0 0 0 G

ε11

ε22

ε33

γ12

γ13

γ23

(5.62)

From a mathematical point of view, Lamé’s constant are perfectly suited to characterize the

linear elastic material behaviour. In practice however, engineering moduli which can be

related to classical tensile tests, are frequently used instead.

Note that the volumetric-deviatoric split of the constitutive relation, permits to find very

easily the inverse relation, i.e., the so-called compliance relation. From the equations (5.59)

and (5.60), one immediately notices

εv =1

3 Kσ

h (5.63)

εd =

1

2Gσ

d (5.64)

98

5.5. Linear elastic isotropic materials

The compliance tensor4S is like the elasticity tensor

4C an isotropic fourth-order tensor,

which thus also takes the format4

S = α I I +β4

Is

(5.65)

Using equations (5.63) and (5.64), this tensor is easily reconstructed

ε = εv +ε

d (5.66)

=1

3 Kσ

h +1

2Gσ

d (5.67)

=1

9 Ktr(σ)I +

1

2Gσ

d (5.68)

=[

1

9 KI I +

1

2G(

4

Is− 1

3I I )

]

: σ (5.69)

=[

(1

9 K−

1

6G) I I +

1

2G

4

Is]

: σ (5.70)

=4

S : σ (5.71)

The compliance tensor is thus given in terms of K and G or in terms of Lamé’s constants λ

and µ=G by

4

S =(

1

9 K−

1

6G

)

I I +1

2G

4

Is

(5.72)

=(

−λ

2G(3λ+2G)

)

I I +1

2G

4

Is

(5.73)

5.5.2 Engineering elastic moduli

The introduction of the classical engineering elasticity entirely relies on special states of

stress which are attained with simple mechanical tests. The first test is simple tension.

Simple tension

In a simple tensile test, a force is applied into one direction of a specimen only, while all

other stress components are equal to zero, e.g.,

σ11 = σ (5.74)

σ22 = σ33 = τ12 = τ13 = τ23 = 0 (5.75)

The first three constitutive equations from (5.62) then simplify to

σ11 = (λ+2G)ε11 +λε22 +λε33 (5.76)

0 = λε11 + (λ+2G)ε22 +λε33 (5.77)

0 = λε11 +λε22 + (λ+2G)ε33 (5.78)

Solving ε22 and ε33 from the equations (5.77) and (5.78) in terms of ε11, followed by a substi-

tution in (5.76) gives the relation between σ11 and ε11

σ11 =(2G +3λ)G

λ+Gε11 = E ε11 (5.79)

99

5. THREE-DIMENSIONAL LINEAR ELASTICITY

from which Young’s modulus is defined as the linear coefficient between the applied axial

stress and the axial strain in the tensile bar. The strain components obtained in terms of the

externally applied stress σ read

ε11 =λ+G

(2G +3λ)Gσ (5.80)

ε22 = ε33 = −λσ

2G(2G +3λ)(5.81)

The fractional contraction relates the contraction of the tensile bar perpendicular to the

loading axis with the axial tensile strain, i.e., the relative lateral contraction.

−ε22

ε11= −

ε33

ε11=

λ

2(λ+G)= ν (5.82)

which is quantified by Poisson’s ratio ν.

Two coefficients fully characterize the isotropic constitutive law in linear elasticity. This

means that either E ,ν or λ,G can be chosen as the set of elastic constants. Recomputing

λ and G in terms of the Young’s modulus and Poisson’s ratio gives

λ =Eν

(1+ν)(1−2ν)G =

E

2(1+ν)(5.83)

Simple shear

The physical significance of the shear modulus becomes most apparent in a simple shear

test. In that case, one shear stress component is constant and equal to τ, while all other

stress components equal zero, e.g.,

τ12 = τ21 = τ (5.84)

σ11 = σ22 = σ33 = τ13 = τ23 = 0 (5.85)

The matrix form (5.62) is then simplified to

λtr(ε)+2Gε11 = 0

λtr(ε)+2Gε22 = 0

λtr(ε)+2Gε33 = 0

γ13 = γ23 = 0

τ = Gγx y (5.86)

which leads to the following solution for the strain components

ε11 = ε22 = ε33 = γ13 = γ23 = 0 (5.87)

γ12 =τ

G(5.88)

The shear modulus thus gives the proper linear relation between the shear strain and shear

stress in the case of simple shear.

Note that Poisson’s ratio is limited by −1 < ν< 0.5, which is easily concluded from (5.83),

in order to keep the value of λ finite. Materials with a negative Poisson’s ratio are difficult to

imagine, but advanced microstructures nowadays permit to construct such materials. The

upper limit for ν = 0.5 is typical for incompressible behaviour, since then tr(ε) = 0. A com-

plete conversion table for all elastic moduli is given in table 5.1.

100

5.5

.L

ine

ar

ela

sticiso

trop

icm

ate

rials

E , ν λ, G K , G E , G E , K

↓ ↓ ↓ ↓ ↓

E → E = E E = (2G+3λ)Gλ+G

E = 9KG3K+G

E = E E = E

ν → ν= ν ν= λ2(λ+G)

ν= 3K−2G2(3K+G)

ν= E−2G2G

ν= 3K−E6K

G → G = E2(1+ν)

G =G G =G G =G G = 3K E9K−E

K → K = E3(1−2ν)

K = 3λ+2G3

K = K K = EG3(3G−E)

K = K

λ → λ= Eν(1+ν)(1−2ν)

λ=λ λ= 3K−2G3

λ= G(E−2G)3G−E

λ= 3K (3K−E)9K−E

E , λ G , ν λ, ν λ, K K , ν

↓ ↓ ↓ ↓ ↓

E → E = E E = 2G(1+ν) E = λ(1+ν)(1−2ν)ν E = 9K (K−λ)

3K−λ E = 3K (1−2ν)

ν → ν= −E−λ+p

(E+λ)2+8λ2

4λν= ν ν= ν ν= λ

3K−λ ν= ν

G → G = −3λ+E+p

(3λ−E)2+8λE

4G =G G = λ(1−2ν)

2ν G = 3(K−λ)2

G = 3K (1−2ν)2(1+ν)

K → K = E−3λ+p

(E−3λ)2−12λE

6K = 2G(1+ν)

3(1−2ν)K = λ(1+ν)

3νK = K K = K

λ → λ=λ λ= 2Gν1−2ν λ=λ λ=λ λ= 3Kν

1+ν

Table 5.1: Conversion tables of elastic moduli

10

1

5. THREE-DIMENSIONAL LINEAR ELASTICITY

Constitutive relation in terms of the engineering moduli

The constitutive equation (5.52) is easily transformed with the equation (5.83) to

σ =Eν

(1+ν)(1−2ν)tr(ε)I +

E

1+νε (5.89)

which in a vector basis is written in the following matrix format

σ11

σ22

σ33

τ12

τ13

τ23

=E

(1+ν)(1−2ν)

1−ν ν ν 0 0 0

ν 1−ν ν 0 0 0

ν ν 1−ν 0 0 0

0 0 0 1−2ν2

0 0

0 0 0 0 1−2ν2

0

0 0 0 0 0 1−2ν2

ε11

ε22

ε33

γ12

γ13

γ23

(5.90)

The compliance tensor (5.73) can be rewritten in terms of the engineering moduli by

making use equation (5.83), which leads to

4

S = −ν

EI I +

1+ν

E

4

Is

(5.91)

The associated stress-strain constitutive relation in matrix format (with the engineering shear

strains!), with the compliances written in terms of the elastic constants given above, then

reads

σ~ = Sε~ (5.92)

ε11

ε22

ε33

γ12

γ13

γ23

=

1E

− νE

− νE

0 0 0

− νE

1E

− νE

0 0 0

− νE

− νE

1E

0 0 0

0 0 0 2(1+ν)E

0 0

0 0 0 0 2(1+ν)E

0

0 0 0 0 0 2(1+ν)E

σ11

σ22

σ33

τ12

τ13

τ23

(5.93)

In index notation (with the tensorial shear strains!) these relations would read

εi j =1

E

[

−νσkk δi j + (1+ν)σi j

]

(5.94)

or in tensor format

ε =1

E[−ν tr(σ) I + (1+ν)σ] (5.95)

Note that the bulk modulus K is formulated in terms of the engineering moduli by

K =E

3(1−2ν)(5.96)

102

5.5. Linear elastic isotropic materials

5.5.3 Stored energy density for isotropic linear elastic materials

At the start of this section, it was remarked that the stored energy density for isotropic ma-

terials can only depend on the invariants of the strain tensor ε. If not, material orientation

plays a role, which is by definition of isotropy excluded. For the elasticity tensor elaborated

in the previous sections, one finds

W = 12ε :

4

C : ε

= 12ε :

(

λ I I +2G4

Is)

: ε

= 12λtr(ε)2 +G ε : ε

= ( 12λ+G)[J1(ε)]2 −2G J2(ε) (5.97)

Clearly, W only depends on invariants of ε, which is characteristic for an isotropic material.

5.5.4 Coincidence of principal axes of stress and strain

An important consequence of the use of an isotropic elasticity tensor, is that it does not

change the principal directions of the second-order tensor to which it is applied. Hence,

principal directions of strain will be preserved as principal directions of stress. This becomes

clear if the isotropic constitutive behaviour is substituted in the eigenvalue problem for the

stresses

[σ−σi I ] ·~Mi = ~0

[λtr(ε)I +2G ε−σi I ] ·~Mi = ~0

2G

[

ε− (σi −λtr(ε)

2G)I

]

·~Mi = ~0 (5.98)

which resembles exactly the eigenvalue problem for the strain tensor

[ε−ǫi I ] ·~Ni = ~0 (5.99)

Hence the principal directions of stress ~Mi are simply equal to the principal directions of

strain ~Ni

~Mi = ~Ni (5.100)

The principal stresses are evidently different from the principal strains. Their mutual relation

is given by

ǫi =σi −λtr(ε)

2G(5.101)

These relations also permit to establish the relations that relate the invariants of stress to the

invariants of strain, i.e.,

J1(σ) = (3λ+2G)J1(ε) (5.102)

J2(σ) = λ(3λ+4G)J1(ε)2 +4G2 J2(ε) (5.103)

J3(σ) = λ2(λ+2G)J1(ε)3 +4λG2 J1(ε)J2(ε)+8G3 J3(ε) (5.104)

The equality of principal directions can be noticed directly, if one realizes that the shear

strains along principal directions are equal to zero, i.e., γ12 = γ13 = γ23 = 0. Making use of the

constitutive relation for isotropic linear elasticity then immediately gives τ12 = τ13 = τ23 = 0,

which means that the same directions remain principal for the stresses as well. These axes

are simply called the principal axes, without making any further distinction between stresses

and strains.

103

5. THREE-DIMENSIONAL LINEAR ELASTICITY

5.5.5 Plane strain and plane stress

This subsection presents a brief treatment of plane deformation states, limited to isotropic

elasticity. The more general case is given in Appendix A.

Plane strain

As shown in previous chapters, plane strain is characterized by

ε33 = γ13 = γ23 = 0 (5.105)

and the 3D problem reduces to a 2D strain problem for which the matrix format of the gov-

erning constitutive equation reads

σ~ = C ε~

σ11

σ22

τ12

=

λ+2G λ 0

λ λ+2G 0

0 0 G

ε11

ε22

γ12

(5.106)

The stress component σ33 for plane strain (in-plane loaded thick structures) does not equal

zero and is given by

σ33 = λ(ε11 +ε22) (5.107)

which can be computed easily. The stress-strain relation (5.106) can also be written in terms

of the engineering elasticity constants E and ν, which leads to

σ~ = C ε~

σ11

σ22

τ12

=E

(1+ν)(1−2ν)

1−ν ν 0

ν 1−ν 0

0 0 1−2ν2

ε11

ε22

γ12

(5.108)

Plane stress

In the case of plane stress (in-plane loaded thin plane structures), the stress components

with respect to the thickness directions vanish

σ33 = τ13 = τ23 = 0 (5.109)

which makes the stresses two-dimensional. The strain ε33 is now different from zero, and

the constitutive relations read

σ~ = C ε~

σ11

σ22

0

τ12

=

λ+2G λ λ 0

λ λ+2G λ 0

λ λ λ+2G 0

0 0 0 G

ε11

ε22

ε33

γ12

(5.110)

The third equation for τ33 permits to express the strain component ε33 in terms of ε11 and

ε22

ε33 = −λ

λ+2Gε11 −

λ

λ+2Gε22 (5.111)

104

5.6. The superposition principle

Back substitution in (5.110) then yields

σ11

σ22

τ12

=

λ+2G − λ2

λ+2Gλ− λ2

λ+2G0

λ− λ2

λ+2Gλ+2G − λ2

λ+2G0

0 0 G

ε11

ε22

γ12

(5.112)

which is easily simplified to

σ11

σ22

τ12

=

(λ+G)4Gλ+2G

2λGλ+2G

0

2λGλ+2G

(λ+G)4Gλ+2G

0

0 0 G

ε11

ε22

γ12

(5.113)

or by substitution of the engineering elasticity constants E and ν

σ11

σ22

τ12

=E

1−ν2

1 ν 0

ν 1 0

0 0 1−ν2

ε11

ε22

γ12

(5.114)

5.6 The superposition principle

Linear elastic material behaviour has two major characteristics which constitute the basis of

the superposition principle, i.e.,

• The path independence: only the final loading state determines the deformed config-

uration, stresses and strains in the material. It does not matter what loading path was

followed to reach that state.

• The linearity of the problem: the constitutive stress-strain relationship is entirely lin-

ear, and the strain-displacement relations (i.e., the linear strain tensor) are linear as

well.

This path independent and linear character permits to treat any loading case as an arbi-

trary subdivision of elementary loading cases, for which the solution can be summed up in

a trivial way. In other words, if the solution of a mechanical problem in linear elasto-statics

is known for two specific loading cases, then it is also known for any linear combination of

these two loading cases. This is called the superposition principle. It permits to superimpose

the results for specific loading arrangements, in order to investigate their combined effect.

Note however, that the principle is inherent to the theory of linear elasticity for small

displacements, and that it may by no means be generalized to the large deformation case.

105

5. THREE-DIMENSIONAL LINEAR ELASTICITY

5.7 Summary

Elastic stored energy function

Stored energy density W= 12σ : ε

Stress tensor σ = ∂W∂ε

σxx = ∂W∂εxx

σy y = ∂W∂εy y

σzz = ∂W∂εzz

τx y = ∂W∂γx y

τxz = ∂W∂γxz

τyz = ∂W∂γyz

3D linear elasticity

1D Hooke’s law σ= E ε

3D Hooke’s law σ= 4C : ε

The elasticity tensor4C = 4

CLT

4C = 4

CRT

4C = 4

CT

Ci j kl =C j i kl =Ci j l k =Cl k j i4C has 21 independent components

Stored energy density W= 12ε :

4C : ε

Matrix format

σ11σ22σ33τ12τ13τ23

=

C1111 C1122 C1133 C1112 C1113 C1123C1122 C2222 C2233 C2212 C2213 C2223C1133 C2233 C3333 C3312 C3313 C3323C1112 C2212 C3312 C1212 C1213 C1223C1113 C2213 C3313 C1213 C1313 C1323C1123 C2223 C3323 C1223 C1323 C2323

ε11ε22ε33γ12γ13γ23

Compliance tensor4S ε= 4

S : σ

106

5.7. Summary

Material symmetry and anisotropy

Orthotropy Two (and hence three) symmetry planes

9 elastic constants

Transverse isotropy Rotational symmetry around one axis

5 elastic constants

Isotropy Omni-directional symmetry

2 elastic constants

Isotropic linear elasticity

Isotropic elasticity tensor4C =λI I +2µ

4I

s

Ci j kl = λδi jδkl +µ(δi lδ j k +δi kδ j l )

Lamé’s constants λ and G =µ

Constitutive response σ=λtr(ε)I +2G ε

Index notation σi j = λεkkδi j +2G εi j

Uncoupled response σ= 3 K εv +2G εd

σh = 3 K εv

σd = 2G ε

d

Bulk modulus K

Matrix format

σ11

σ22

σ33

τ12

τ13

τ23

=

λ+2G λ λ 0 0 0λ λ+2G λ 0 0 0λ λ λ+2G 0 0 00 0 0 G 0 00 0 0 0 G 00 0 0 0 0 G

ε11

ε22

ε33

γ12

γ13

γ23

Compliance tensor4S = (− λ

2G(3λ+2G)) I I + 1

2G

4I

s

107

5. THREE-DIMENSIONAL LINEAR ELASTICITY

Elastic moduli

Engineering moduli

Young’s modulus σ11 = E ε11 in simple tension

E = (2G+3λ)Gλ+G

Poisson’s ratio −ε22

ε11= ν in simple tension

ν= λ2(λ+G)

Mutual relations λ= Eν(1+ν)(1−2ν)

G = E2(1+ν)

K = 3λ+2G3

= E3(1−2ν)

Limitations 0< E <∞ and −1< ν< 0.5

Incompressibility ν= 1/2

Stored energy density W= ( 12λ+G)[J1(ε)]2 −2G J2(ε)

Principal axes, plane strain and plane stress for isotropic linear elasticity

Principal axes Coincide −→ ~Ni = ~Mi

Plane strain (matrix)

σ11

σ22

τ12

= E(1+ν)(1−2ν)

1−ν ν 0

ν 1−ν 0

0 0 1−2ν2

ε11

ε22

γ12

Plane stress (matrix)

σ11

σ22

τ12

= E1−ν2

1 ν 0

ν 1 0

0 0 1−ν2

ε11

ε22

γ12

Superposition principle

Superposition principle σ(ρ~q +~p1 +~p2) =σ(ρ~q)+σ(~p1)+σ(~p2)

108

CH

AP

TE

R

6LIMIT DESIGN CRITERIA

IN LINEAR ELASTICITY

This chapter is devoted to the design criteria which ensue from the limits of the elastic ma-

terial behaviour. The loss of elasticity is assumed if the material’s microstructure changes

in an irreversible way, leading to permanent deformations. A large class of criteria is pre-

sented, which are applicable to different types of materials. Particular attention is given to

the widely used von Mises criterion, where a clear connection with the physics of the de-

forming microstructure is established.

6.1 Failure modes

Proper design rules are based on the adequate knowledge of failure modes, as well the cri-

teria which play a role in their initiation. Materials are intended to remain within the linear

elastic domain for a vast majority of applications. This ensures that no permanent defor-

mations will occur, so that the designed component is capable to fulfill its tasks with the

initial geometry used for the design. Different types of failure may occur and hence limit

the applicable load to a structure or require an improved geometry by means of a coupled

CAD-mechanical design.

• plastic yielding; this failure mode is characterized by the appearance of irreversible

deformations in the material. For metals this is due to crystallographic slip. Hence,

the microstructure starts to change locally.

• brittle fracture; many brittle (or quasi-brittle) materials present a sudden crack ini-

tiation and crack growth without the appearance of prior permanent deformations.

This is mainly due to local stress concentrations which constitute an overload for the

microstructure, in which bonds are suddenly broken. The theory of elastic fracture

mechanics is devoted to this subject (see the corresponding optional course on this

subject).

6. LIMIT DESIGN CRITERIA

• progressive damage; microstructural damage may also evolve progressively through

the appearance of micro-cracks, which grow, accumulate, coalesce, etc.. This phe-

nomenon may take place without meaningful permanent deformations and it gener-

ally leads to a progressive reduction of the stiffness of the material (i.e., of the Young’s

modulus for the perfectly isotropic case). This failure mode is the subject of the course

on continuum damage mechanics.

• fatigue; fatigue damage is a highly complex phenomenon that depends on the cyclic

character of the external load, geometrical aspects of the specimen, external physical

and chemical influences and the surface conditions of the material. If fatigue damage

is a relevant design criterion for the application of interest (i.e., a cyclic loading exists),

one commonly resorts to the so-called fatigue limit, which is the maximum stress am-

plitude under which the fatigue phenomenon will not occur. For specific applications

(low cycle fatigue), design rules up to a dedicated number of cycles may be used as

well.

• dynamic failure; vibrations due to dynamic loadings may constitute a dangerous fail-

ure mode since resonance may occur if the spectrum of the external loading presents

frequencies that are close to the eigenfrequencies of the structure.

• thermal failure; temperature may have a large influence on the physical, geometrical

and hence mechanical properties of a material’s microstructure. Temperature changes

in combination with a non-constant (possibly cyclic) loading may lead to a significant

reduction of the strength of the material.

• elastic instabilities; this failure mode may occur prior to all other types of failure. It is

essentially a structure and geometry dependent failure mode, which depends on the

deformed geometry of the considered structural member or component. The simplest

and well-known example is column buckling.

6.2 Standard tests

6.2.1 Tension

A standard uniaxial tensile test, where the load is applied in one direction of a cylindrical

specimen only (with a corresponding stress σt in that direction), is characterized by a stress

state for which the principal stresses are given by

σ1 = σt

σ2 = 0

σ3 = 0

(6.1)

The stress tensor is thus trivially given by σ=σt~ex~ex , where~ex is the direction of the applied

uniaxial load. Note that the solution of the static equilibrium equation (4.25) now simply

points out that the axial stress must be constant along the axis of the tensile bar (since the

deformation of the cross-section is neglected in linear elasticity). Assuming that ~e y and ~ez

are two vectors perpendicular to~ex , one easily finds the hydrostatic and deviatoric part of σ

110

6.2. Standard tests

as

σh = 1

3tr(σ)I =

σt

3I (6.2)

σd = 2

3σt~ex~ex − 1

3σt~e y~e y − 1

3σt~ez~ez (6.3)

The tensile stress in tension, further denoted σt is used in the following paragraphs to

identify the onset of yielding, i.e., the loss of elastic behaviour.

6.2.2 Torsion

Torsion is a particular loading mode, which is often used in laboratory tests to characterize

materials. A member with a circular cross section is subjected to a twisting moment (torque)

around its axis of symmetry (see also course 4A130 on mechanics). Consider a cylindrical

basis, with the base vector~ez aligned with the axis of the shaft. In the case of a shaft with a

cylindrical cross-section, the solution is easy.

Under the influence of the applied torque the shaft will twist. The shaft has a length L and

a radius R . Assume that the shaft is clamped in z = 0. The applied loading and the geometry

of the shaft are homogeneous along the axis ~ez , and hence the twist will be homogeneous

as well. If one denotes the rotation of the cross-section per unit length of the shaft by α,

one notices that α is constant along z. The circular cross-section remains circular, and axial

displacements (along~ez) of material points are impossible (plane sections remain plane be-

cause of symmetry). The only deformation is a relative rotation of cross-sections, and hence

the corresponding displacement components (u, v, w) with respect to the cylindrical basis

~er ,~et ,~ez are given by

u = 0 v = αz r w = 0 (6.4)

Using the strain-displacement equations (2.55), one easily finds the six corresponding rele-

vant strain components that automatically satisfy the compatibility equations

εr r = εt t = εzz = εr t = εr z = 0

εtz = 12

(∂v

∂z+

1

r

∂w

∂θ

)

= 12αr

(6.5)

Evidently, the three remaining components are obtained through symmetry. Hence, only

one of the six relevant shear strain component differs from zero and the entire strain tensor

is given by ε= 12αr~et~ez + 1

2αr~ez~et By means of the constitutive equation for isotropic linear

elasticity, the stress tensor is obtained

σ =4

C : ε

=[

K I I +2G (4

Is− 1

3I I )

]

: ε

= G αr~et~ez +G αr~ez~et (6.6)

Note that this stress tensor satisfies static equilibrium (substitute the divergence operator (1.165)

with this stress tensor in (4.25)).

Boundary conditions at the boundaries z = 0 (u, v, w = 0) and r = R (σr r = 0) are satisfied.

The torque moment applied at the end of the bar (z = L) equals Mt and must equal the

111

6. LIMIT DESIGN CRITERIA

resultant torque of all shear stresses acting on that boundary (the normal to that boundary

is~ez , the concerned component thus is τtz ), i.e.,

Mt =∫R

0

∫2π

0r τtz dθdr (6.7)

= G α

∫R

0

∫2π

0r 2 dθdr = G α Jt (6.8)

= 2πG α

∫R

0r 2 dr = 2πG α

R3

3(6.9)

where Jt is the polar moment of inertia of the cross-section S, defined by

Jt =Ï

S

r 2 dS (6.10)

The shear stress in torsion, further denoted τθ = τtz is used in the following paragraphs

to identify the onset of yielding in the particular case of torsion, i.e., the loss of elastic be-

haviour. The principal stresses for torsion are given by

σ1 = τθ

σ2 = 0

σ3 = −τθ(6.11)

6.3 Elastic limit criteria for isotropic materials

The state of stress in a material point is generally characterized by the stress tensor σ. A

general stress criterion for failure, i.c. the limit of the elastic domain, is thus obtained through

some scalar function of the stress tensor, or an invariant of σ:

f (σ) = 0 (6.12)

The most frequently used limit criteria are those that distinguish elastic from elasto-plastic

behaviour in theories of plasticity.

An important scalar function, which is often connected to the occurrence of inelastic

deformations, is based on the stored energy inside the material. It has been shown that

deformation can be interpreted in terms of volume changes and shape changes of an in-

finitesimal piece of material. As shown in chapter (5), the constitutive relation for 3D linear

elasticity can be split in its hydrostatic part and its deviatoric part. Consequently, the strain

energy density W can be split accordingly. Making use of the compliance tensor4S defined

by (5.24), one finds

W = 12σ : ε

= 12σ :

4

S : σ

= 12

(σh +σd ) :

4

S : (σh +σd )

= 12

(

σh :

4

S : σh +σh :

4

S : σd +σd :

4

S : σh +σd :

4

S : σd)

(6.13)

112

6.3. Elastic limit criteria for isotropic materials

Writing4S in terms of the engineering moduli with expression (5.91) and (5.96) yields

σh :

4

S : σh = 19

[tr(σ)]2 I :4

S : I

= 19

[tr(σ)]2 I : (−ν

EI I +

1+ν

E

4

Is) : I

= 19

[tr(σ)]2 (−9ν

E+

3(1+ν)

E)

= 19

[tr(σ)]2 3(1−2ν)

E

= 19

[tr(σ)]2 3(1−2ν)

E

=1

9K[tr(σ)]2 =

p2

K(6.14)

while

σh :

4

S : σd = 13

tr(σ) I :4

S : σd

= 13

tr(σ) I : (−ν

EI I +

1+ν

E

4

Is) : σd

= 13

tr(σ) I : (1+ν

Eσ

d )

= 0 (6.15)

= σd :

4

S : σh (6.16)

Finally, the last term of equation (6.13) is elaborated to

σd :

4

S : σd = σd :

(

−ν

EI I +

1+ν

E

4

Is)

: σd

=1+ν

E(σd : σd )

=1

2G(σd : σd )

= −1

GJ2(σd ) (6.17)

Putting all terms together in (6.13) leads to the additive split of the stored energy density

W = Wv +W

d (6.18)

Wv =

p2

2K(6.19)

Wd = −

1

2GJ2(σd ) =

1

4G(σd : σd ) (6.20)

The volumetric (or hydrostatic) partWv of the stored energy only involves the bulk modulus

K as elasticity constant, while the deviatoric part Wd only involves the shear modulus G .

6.3.1 Rankine criterion

A typical stress-based limit criterion is the Rankine criterion, which simply states that the

limit of the elastic domain is reached as soon as the absolute value of one of the principal

stresses exceeds a material dependent value, i.e.,

maxi

|σi | = σR (6.21)

113

6. LIMIT DESIGN CRITERIA

where σR is called the Rankine equivalent stress. This limit theory inherits its name from W.

Rankine (19th century), and is often referred to as the maximum principal stress theory. The

limit value σR is determined from the stress in a tensile test at the onset of yielding

σR = σY = σt tensile test, onset yielding (6.22)

The Rankine limit surface is depicted in the three-dimensional space of the principal stresses

in figure 6.1. In the case of plane stress, the third principal stress σ3 equals zero, and the cor-

Figure 6.1: The Rankine limit surface in the principal stress space.

responding limit surface simplifies to the curve shown in figure 6.2. The criterion is clearly

sensitive to hydrostatic stresses, but it only involves the stresses on a single plane (on which

the largest principal stress acts) while the stress state on the two other principal planes is

completely neglected. In the case of torsion, the maximum principal stress would simply

equal τθ and the Rankine criterion thus implies that the ultimate elastic stress in tension

would equal the ultimate elastic stress in torsion, i.e.,

σt = τθ (6.23)

This is not experimentally observed for ductile materials, where the ultimate elastic stress in

torsion τθ is much less than in tension. The Rankine criterion is therefore only applicable to

brittle materials.

6.3.2 Tresca yield criterion

Another stress-based limit criterion is the Tresca yield criterion, which states that the limit

of elastic behaviour is reached once the maximum shear stress in a material point reaches a

114

6.3. Elastic limit criteria for isotropic materials

σ1

σ2

Figure 6.2: The Rankine limit surface for plane stress.

particular value that corresponds to the maximum shear stress in a tensile specimen at the

onset of yielding, i.e.,

τnsmax = 12

(σ1 −σ3) = 12σt = 1

2σY (6.24)

For a general state of stress, yielding occurs if one of the maximum shear stresses reaches

σY , i.e.,

τnsmax ,1 = 12σY =⇒ ±(σ1 −σ2) = σY or

τnsmax ,2= 1

2σY =⇒ ±(σ1 −σ3) = σY or

τnsmax ,3 = 12σY =⇒ ±(σ2 −σ3) = σY

The corresponding limit surface, depicted in the principal stress space in figure 6.3, is a reg-

ular hexagonal prismatic surface, that is strictly insensitive to the hydrostatic stress in the

material. The prism has an infinite length in the direction of the mean stress σ1 = σ2 = σ3,

i.e., one of the octahedral directions. This means that if one cuts the surface by the octahe-

dral plane corresponding to the mean stress (three equal direction cosines with respect to

the principal directions), one always retrieves the same curve (the so-called C -curve), which

is a regular hexagon for the Tresca criterion. Note that this octahedral direction is the diag-

onal in the principal space for which σ1 = σ2 = σ3. This particular octahedral plane is also

called the π-plane. The hexagon is shown further on, in figure 6.7.

In the case of plane stress, the prism is cut by a plane that is not normal to the octahe-

dral axis of the prism, an one finds a criterion that simplifies to an irregular two-dimensional

hexagon, shown in figure 6.4. Like the von Mises criterion given hereafter, the Tresca crite-

rion is particularly suited for ductile materials in which high shears occur upon yielding.

For the torsion test, substitution of the principal stresses in (6.24) gives

τθ = 12σY (6.25)

and hence the limit shear stress in torsion is indeed lower than the limit tensile stress.

115

6. LIMIT DESIGN CRITERIA

Figure 6.3: The Tresca yield surface in the principal stress space.

σ1

σ2

Figure 6.4: The Tresca limit surface for plane stress.

116

6.3. Elastic limit criteria for isotropic materials

6.3.3 Von Mises yield criterion

The von Mises criterion is a widely used limit criterion, that is essentially based on the defor-

mation mechanisms in crystalline materials, which makes it a suitable criterion for the elas-

tic limit analysis of most metals. As stated earlier, linear elasticity implies that the underlying

microstructure is not modified upon deformation. In crystalline materials, the microstruc-

ture starts to change as soon as slip occurs on crystallographic planes, which is the onset of

plastic deformation. The essential difference between elastic and plastic deformation of a

crystal is illustrated in figure 6.5. Clearly, a slip mechanism in an ideal crystal takes place

Crystallographic slip

Crysta

l latti

ce defo

rmati

on

Elastic deformation(the lattice deforms)

Plastic deformation(the lattice planes slip)

Deviatoric or distortionaldeformation

Figure 6.5: Elastic and plastic deformation of a crystalline material.

without any volume change (it is a local state of pure shear, on which the maximum shear

stress criterion of Tresca is based), which means that the scalar measure that should be used

to characterize its initiation should depend on the deviatoric part or distortional part of the

elastically stored energy or the deviatoric part of the stress tensor. The criterion which was

proposed by Huber, von Mises and Hencky was therefore based on the strain energy density

of distortion. The criterion associates the elastic limit behaviour with the energy absorbed

in changing shape, since the hydrostatic stresses, responsible for volume changes, are not

present in this criterion.

The loss of elastic behaviour occurs as soon as the deviatoric strain energy density in

an arbitrary point equals the limit value that occurs in a classical tensile test at the onset of

117

6. LIMIT DESIGN CRITERIA

yielding. From equation (6.20) and (6.3), it appears that this happens if

J2(σd ) =[

J2(σd )]

tensile test(6.26)

= −12

( 23σt~ex~ex − 1

3σt~e y~e y − 1

3σt~ez~ez) :

( 23σt~ex~ex − 1

3σt~e y~e y − 1

3σt~ez~ez) (6.27)

= −12

( 49σ2

t + 19σ2

t + 19σ2

t ) (6.28)

= −13σ2

t (6.29)

If the yield stress σY is defined as the tensile stress at the onset of yield (i.e., σY = σt ), one

can defined the so-called von Mises equivalent stress σVM

for a complex state of stress as

σVM =√

−3 J2(σd ) (6.30)

=√

32σ

d : σd (6.31)

If the von Mises equivalent stresses exceeds the value σY (where σY is determined from σVM

with a tensile test), the limit of the elastic domain is reached. The fact that σVM

can also be

expressed in terms of the second invariant of σd explains why the name J2-plasticity is often

used starting with this limit surface. The von Mises equivalent stress can also be expressed

with respect to a given basis ~ex ,~e y ,~ez or in terms of the principal stresses.

σVM

=√

32σ

d : σd

=√

32

[σ− 13

tr(σ)I ] : [σ− 13

tr(σ)I ]

=√

32

[σ : σ− 23

tr(σ)I : σ+ 19

[tr(σ)]2I : I ]

=√

32

(σ : σ− 13

[tr(σ)]2) (6.32)

which in terms of the principal stresses yields

σVM =√

32

(σ21 +σ2

2 +σ23 −

13

[σ1 +σ2 +σ3]2)

=√

32

(σ21 +σ2

2 +σ23 −

13

[σ21 +σ2

2 +σ23 +2σ1σ2 +2σ1σ3 +2σ2σ3])

=√

12

([σ21 −2σ1σ2 +σ2

2]+ [σ21 −2σ1σ3 +σ2

3]+ [σ22 −2σ2σ3 +σ2

3])

=√

12

([σ1 −σ2]2 + [σ1 −σ3]2 + [σ2 −σ3]2) (6.33)

or with respect to a Cartesian basis ~ex ,~e y ,~ez

σVM

=√

32 (σ2

xx +σ2y y +σ2

zz −+2τ2x y +2τ2

xz +2τ2y z − 1

3 [σxx +σy y +σzz ]2)

=√

12 ([σxx −σy y ]2 + [σxx −σzz ]2 + [σy y −σzz ]2)+3τ2

x y +3τ2xz +3τ2

y z (6.34)

Note that this criterion has been established from the distortional strain energy only, and

hence it is strictly insensitive to the hydrostatic pressure or volumetric deformations. The

corresponding limit surface (i.e., the yield surface) is depicted in figure 6.6, in terms of the

three principal stresses σ1, σ2 and σ3 by means of equation (6.33). It is a cylindrical surface

118

6.3. Elastic limit criteria for isotropic materials

Figure 6.6: The von Mises yield surface in the principal stress space.

that extends itself to infinity in the direction of the mean stress (i.e., it is not sensitive to

hydrostatic stress states). If one cuts the von Mises cylinder by the octahedral plane normal

to the mean stresses, one finds the same curve independent of the hydrostatic pressure, i.e.,

a circle. This circle fully circumvents the corresponding hexagon of the Tresca criterion, see

figure 6.7. This curve shows that the von Mises criterion constitutes a smooth envelope of the

Tresca criterion. In the particular case of plane stress (σ3 = 0), the 3D cylindrical limit surface

cuts the σ1−σ2 plane (not normal to the octahedral axis of the cylinder) in an ellipse, which

is shown in figure 6.8.

Comparing equation (6.33) for the equivalent von Mises stress with the octahedral shear

stress obtained in equation (3.37) of chapter 3, one notices that σVM is directly related to the

octahedral shear stress by

σVM =3p

2τoct (6.35)

This means that the von Mises yield criterion predicts the limit of the elastic domain at a

point where the octahedral shear stress reaches a particular limit value. Hence, the use of the

distortional energy or the octahedral shear stress leads to an identical criterion for a linear

elastic material.

Since the limit value of the von Mises stress is determined in simple tension, it is interest-

ing to evaluate the equivalent von Mises stress in the case of torsion. From equation (6.33)

for σVM and the principal stresses in torsion (6.11), it appears that yielding in the torsion test

would occur if

σVM =p

3τθ = σY = σt (6.36)

Hence, the shear stress in torsion upon yield is clearly less than the tensile yield stress

τθ =1p

3σt (6.37)

which is also observed for ductile materials.

119

6. LIMIT DESIGN CRITERIA

σ2

σ3

σ1

von MisesTresca

Figure 6.7: The regular von Mises and Tresca C -curves in the π-plane

The differences between the 3D limit criteria given so far is shown in figure 6.9, where the

plane stress limit surface for the Rankine, von Mises and Tresca criterion are depicted for the

same value of σY .

6.3.4 Mohr-Coulomb limit criterion

The Mohr-Coulomb criterion distinguishes itself from all previous criteria in the sense that

it behaves different in traction or compression. It is physically motivated on the basis of

the existence of two physical parameters, i.e., an internal friction angle φ and an internal

cohesion c. The friction angle measures the resistance to frictional slip on individual planes

(a sort of internal Coulomb friction), while the cohesion measures the adhesion between two

planes. This limit criterion is typically used for cohesive-frictional materials, e.g., granular

materials, ceramics, etc. In chapter 3, normal and shear strains on individual planes in a

material point have been elaborated. The Mohr-Coulomb criterion states that the material

loses elasticity (i.e., changes in microstructure) as soon as the frictional slip or decohesion

120

6.3. Elastic limit criteria for isotropic materials

σ1

σ2

Figure 6.8: The von Mises limit surface for plane stress.

σ1

σ2

Rankine von MisesTresca

Figure 6.9: Comparison of three limit surfaces for plane stress.

121

6. LIMIT DESIGN CRITERIA

occurs. The resistance to these two phenomena on an arbitrary plane is expressed by

|τns | < − tan(φ)σnn +c (6.38)

The limit state is reached if (6.38) is violated on one single plane, which is then called the

critical plane. On the basis of the equations elaborated for τns and σnn in terms of the di-

rection cosines of the normal ~n on the plane, it is possible to compute this critical plane.

Here however, use will be made of Mohr’s circles for 3D, since it points directly to the answer

we are looking for. The three circles are sketched in figure 6.10. The limit criterion (6.38) is

s 1s 2

s nn

t ns

s 3

f

c

( )

c

tg f

Figure 6.10: The Mohr-Coulomb limit state reached on Mohr’s circles

sketched in this τns −σnn figure as well, such that only one single point on the circles (i.e.,

the critical orientation) reaches that limit state. The limit criterion, which is a straight line,

is tangent to the outer circle spanned by the largest and smallest principal stress σ1 and σ3

respectively. The radius of this outer circle is perpendicular to the limit line. Goniometric

considerations on this figure therefore lead to

sin(φ) =−1

2(σ3 −σ1)

−12

(σ3 +σ1)+ ctan(φ)

(6.39)

which is further elaborated to the 3D Mohr-Coulomb criterion

12

(σ3 −σ1) =[

12

(σ3 +σ1)−c

tan(φ)

]

sin(φ)

12

(σ3 −σ1) = sin(φ) 12

(σ3 +σ1)−cos(φ)c (6.40)

Elastic behaviour implies

σ1 −σ3

2+ sin(φ)

σ1 +σ3

2< cos(φ)c (6.41)

Once the internal friction angle φ and the cohesion c have been determined (experimen-

tal set-ups exist for this purpose), a graph of the corresponding 3D limit surface can be made,

122

6.3. Elastic limit criteria for isotropic materials

Figure 6.11: The Mohr-Coulomb yield surface in the principal stress space.

see figure 6.11. One clearly notices from this limit surface, that it is a so-called pressure sen-

sitive limit surface. If the hydrostatic pressure increases the elastic domain extends, which

gives a pyramidal shape to the surface with a hexagonal basis, which is apparent from fig-

ure 6.11. The hexagonal basis stems from the fact that the maximum shear stress entered the

limit equation, similar to the Tresca yield surface but different in shape. Cutting the pyramid

by the octahedral plane normal to the pyramid axis gives an irregular hexagon, further shown

in figure 6.14. The shape of this hexagon is not affected by the hydrostatic pressure, but the

size definitely is. The larger the pressure (negative mean stress), the larger the hexagon will

be. At some positive mean stress, the hexagon shrinks to a point and there is no elastic stress

state beyond that point.

The resemblance and differences with the Tresca criterion can also be noticed from the

plane stress limit curve depicted in figure 6.12. This curve is obtained by cutting the pyramid

with the σ1 −σ2 plane, which is not normal to the octahedral pyramid axis. For the Mohr-

Coulomb limit curve, the elasticity limit is different for biaxial compression (σ1 = σ2 < 0)

compared to biaxial tension (σ1 = σ2 > 0). This is due to the pyramidal shape, which de-

scribes the weakness of the material if the hydrostatic pressure diminishes or reverts to hy-

drostatic tension. This type of behaviour is characteristic for brittle materials with a granular

structure. Note that if the internal friction angle φ equals zero and the cohesion equals the

yield stress σY , than the Tresca criterion is recovered.

6.3.5 Drucker-Prager limit criterion

Since the Mohr-Coulomb criterion was expressed in terms of the maximum shear stress12

(σ3 −σ1) in analogy with the Tresca criterion, it seems convenient to formulate a cohesive-

frictional criterion that presents an analogy with the von Mises criterion. The latter two have

the advantage that they are smooth in the three-dimensional principal stress space, which is

physically and numerically more appealing. Furthermore, a sound physical motivation has

been given for the von Mises criterion. The von Mises criterion can be obtained directly by

123

6. LIMIT DESIGN CRITERIA

σ1

σ2

Figure 6.12: The Mohr-Coulomb limit surface in 2D plane stress.

substituting the maximum shear stress by a scalar multiplication with the octahedral shear

stress, see equation (6.35). In the same way, the term 12

(σ3 +σ1) in the Mohr-Coulomb crite-

rion can be generalized to the hydrostatic stress 13

(σ3 +σ2 +σ1). Drucker and Prager made

this generalization under the additional constraint that the obtained limit surface consti-

tutes and envelope of the Mohr-Coulomb surface, like the von Mises surface was an envelope

of the Tresca limit surface. This leads to the Drucker-Prager limit criterion, given by

√32σ

d : σd +6 sin(φ)

3− sin(φ)p <

6 cos(φ)

3− sin(φ)c (6.42)

The corresponding 3D limit surface is again shown in figure 6.13. The Drucker-Prager limit

surface is conical with a circular base curve instead of the irregular hexagon of the Mohr-

Coulomb pyramid. This base curve, along with the irregular hexagon of the Mohr-Coulomb

criterion, is visible in the cross-section of the cone and the pyramid by the octahedral plane

given in figure 6.14.

The Drucker-Prager cone fully circumvents the Mohr-Coulomb pyramid, but does not

pass through all angular points of the Mohr-Coulomb hexagon (which is different from von

Mises compared to Tresca). Like for the Mohr-Coulomb criterion, the size of the circle de-

pends on the hydrostatic stress.

The corresponding plane stress curve for Drucker-Prager is again obtained by cutting the

cone with the σ1 −σ2 plane, which is not normal to the cone axis. Hence, an ellipse is ob-

tained, shown in figure 6.15. The ellipse again illustrates the differences in compression and

tension. The elastic domain in compression is much larger compared to tension. Combining

the plane stress limit curves for Mohr-Coulomb and Drucker-Prager gives figure 6.16, which

shows that the differences between these two criteria are considerably larger than for non-

frictional materials. Again it should be noted that both limit curves lead to the Tresca and

von Mises limit curves, once the internal friction angle equals zero and the cohesion equals

the yield stress. In that sense, the cohesive-frictional criteria constitute a generalization of

the more ’classical’ yield criteria.

124

6.3. Elastic limit criteria for isotropic materials

Figure 6.13: The Drucker-Prager yield surface in the principal stress space.

σ1

σ3

σ2

Drucker−PragerMohr−Coulomb

Figure 6.14: The C -curves for the Mohr-Coulomb and Drucker-Prager limit cones in the π-

plane.

125

6. LIMIT DESIGN CRITERIA

σ1

σ2

Figure 6.15: The Drucker-Prager limit surface for plane stress.

σ1

σ2

Mohr−Coulomb Drucker−Prager

Figure 6.16: Comparison of Mohr-Coulomb and Drucker-Prager for plane stress.

126

6.4. Limit criteria for anisotropic materials

6.3.6 General criteria for isotropic solids

On the basis of equation (6.12), many more criteria can be established. The major motivation

for doing so, is that the available criteria insufficiently characterize the true limit behaviour

of the particular material of interest. Other ellipsoidal, conical, pyramidal, paraboloidal sur-

faces may be constructed. We limit ourselves here to one additional example, where the limit

surface is formulated with respect to the octahedral planes in a material point. It was shown

that the octahedral normal and shear stress are equal for all eight octahedral planes.

f [J1(σ)]+ g [J2(σ)]τoct +h[J3(σ)]τ2oct = 0 (6.43)

Both the von Mises and Tresca yield surfaces can be rephrased in this format with particular

functions f , g and h.

6.4 Limit criteria for anisotropic materials

For anisotropic materials, it is not possible to develop a limit criterion in the principal stress

space, and the anisotropy directions inevitably enter the limit criterion. A lot of research has

been devoted to this subject, and many criteria are nowadays available for specific materials

or specific anisotropic arrangements (composites).

Like for the von Mises and Tresca criterion, a special family of anisotropic criteria can

be specified which is also independent with respect to the hydrostatic pressure and which is

also expresses as a quadratic form of the deviatoric stress tensor.While the von Mises crite-

rion is formulated in terms of σd : σd , an equivalent anisotropic form could be formulated

as the equivalent stress

σ2eq = σ

d :4

A : σd (6.44)

The fourth-order tensor4A is the anisotropy tensor, that accounts for all symmetries present

in the material. The obvious symmetries of this tensor immediately reduces the number of

relevant components to 21 (like for the elasticity tensor). The independence of the hydro-

static pressure (A j j kl = 0) further reduces the number to 15 for the case of general anisotropy.

Special cases of symmetry reduces the number further (like for4C ). A well-known example

is Hill’s criterion for metals. Hill considered the special case of orthotropic symmetry for

thin rolled sheets having orthotropic symmetries with respect to the rolling direction(~e1),

the transverse (~e2) and the normal direction (~e3). Taking into account the independence of

the hydrostatic pressure, Hill’s orthotropic criterion depends on 6 coefficients. The quadratic

form reads

σd :

4

A : σd = F (σ22 −σ33)2 +G (σ33 −σ11)2 +H (σ11 −σ22)2+2Lσ2

23 +2M σ213 +2N σ2

12

(6.45)

Hill’s criterion simplifies again to von Mises in the case of full isotropy, and it is in this sense

the anisotropic equivalent of it.

Note that more general criteria taking into account different limit values in tension or

compression (like for Mohr-Coulomb and Drucker-Prager) are also available, e.g., Tsai’s cri-

terion in which a linear term in σd is added as well. This type of criteria has proven to be

particularly efficient for anisotropic composites.

127

6. LIMIT DESIGN CRITERIA

6.5 Examples

Equivalent Von Mises stress

The stress state in a point is represented by the following Cauchy stress tensor :

σ= 3σ~e1~e1 −σ~e2~e2 −2σ~e3~e3 +σ(~e1~e2 +~e2~e1)

The corresponding Cauchy stress matrix is

σ=

3σ σ 0

σ −σ 0

0 0 −2σ

The Von Mises equivalent stress is defined as

σV M =√

32σ

d : σd =√

32

tr(σd ·σd ) =√

32

tr(σdσd )

This equivalent stress expanded in terms of a Cartesian basis was given by equation

(6.34), and reads

σVM

=√

12

([σxx −σy y ]2 + [σxx −σzz ]2 + [σy y −σzz ]2)+3τ2x y +3τ2

xz +3τ2y z

Substitution of the stress components σxx = 3σ, σy y = −σ, σzz = −2σ and σx y = σyx =σgives for the equivalent von Mises stress

σVM = 2p

6σ

Equivalent Von Mises and Tresca stresses

The stress state in a point is represented by the following Cauchy stress tensor :

σ=σ~e1~e1 +σ~e2~e2 +σ~e3~e3 +τ(~e1~e2 +~e2~e1)

with all component values positive.

The Tresca yield criterion states that yielding will occur when the maximum shear stress

reaches a limit value, which is determined in a tensile experiment. The equivalent Tresca

stress is two times the maximum shear stress.

σT R = 2τmax =σmax −σmi n

The limit value is the one-dimensional yield stress σy0. To calculate σT R , we need the

principal stresses, which can be determined by requiring the matrix σ− sI to be singular.

det(σ− sI ) = det

σ− s τ 0

τ σ− s 0

0 0 σ− s

= 0 →

(σ− s)3 −τ2(σ− s) = 0 → (σ− s)(σ− s)2 −τ2 = 0 →(σ− s)(σ− s +τ)(σ− s −τ)= 0 →σ1 =σmax =σ+τ ; σ2 =σ ; σ3 =σmi n =σ−τ

128

6.5. Examples

The equivalent Tresca stress is

σT R = 2τ

so yielding according to Tresca will occur when

τ= 12σy0

The equivalent Von Mises stress is expressed in the principal stresses :

σV M =√

12

(σ1 −σ2)2 + (σ2 −σ3)2 + (σ3 −σ1)2

and can be calculated by substitution,

σV M =√

3τ2 =p

3τ

Yielding according to Von Mises will occur when the equivalent stress reaches a limit

value, the one-dimensional yield stress σy0, which results in

τ= 13

p3σy0

129

6. LIMIT DESIGN CRITERIA

6.6 Elastic limit behaviour: Summary

Elastic limits and failure

Plastic yielding Onset of permanent deformations

Brittle fracture Sudden crack growth

Continuum damage Progressive loss of stiffness

Fatigue Failure under cyclic loading

Dynamic Failure Resonance

Thermal Failure Temperature dependent degradation

Elastic instabilities Geometrical and structural phenomenon

Standard tests

Tension σ1 =σt

σ2 =σ3 = 0

σh = σt

3I

σd = 2

3σt~ex~ex − 1

3σt~e y~e y − 1

3σt~ez~ez

Torsion σ=G αr~et~ez +G αr~ez~et

Circular Mt = 2πG α R3

3

σ1 = τθ

σ2 = 0

σ3 =−τθ

130

6.6. Elastic limit behaviour: Summary

Limit criteria for isotropic materials

General f (σ) = 0

Strain energy density split Wv = 1

2

p2

Kand W

d = 12

12G

(σd : σd )

Rankine Maximum principal stress criterion

maxi

|σi | = σY

τθ =σt

Tresca Maximum shear stress criterion

σ1 −σ3 = σY

Insensitive to hydrostatic pressure

τθ = 12σY

von Mises Distortional strain energy density criterion

σVM

=√

−3 J2(σd )

σVM =√

32σ

d : σd

σVM =√

12

([σ1 −σ2]2 + [σ1 −σ3]2 + [σ2 −σ3]2)

σVM = [ 12

([σxx −σy y ]2 + [σxx −σzz ]2 + [σy y −σzz ]2)

+3τ2x y +3τ2

xz +3τ2yz ]1/2

Insensitive to hydrostatic pressure

Equivalent to octahedral shear stress criterion

σVM

= 3p2τoct

τθ = σtp3

Mohr-Coulomb Hydrostatic sensitive cohesive-frictional12

(σ1 −σ3)+ sin(φ)σ1+σ3

2< cos(φ)c

Pyramidal limit surface

Drucker-Prager Hydrostatic sensitive smooth cohesive-frictional√

32σ

d : σd + 6sin(φ)

3−sin(φ)p < 6cos(φ)

3−sin(φ)c

Conical limit surface

131

6. LIMIT DESIGN CRITERIA

General criteria

Isotropic f (J1(σ)+ g (J2(σ)τoct +h(J3(σ)τ2oct = 0

Hill orthotropic σ2eq = σ

d :4A : σd

σd :

4A : σd = F (σ22 −σ33)2 +G (σ33 −σ11)2+

H (σ11 −σ22)2 +2Lσ223 +2M σ2

13 +2N σ212

132

CH

AP

TE

R

7MECHANICAL DESIGN PROBLEMS:

GOVERNING EQUATIONS AND

SOLUTION METHODS

In this chapter we will recall the equations, which have to be solved to determine the defor-

mation of a three-dimensional linear elastic material body under the influence of an external

load. The equations will be written in component notation w.r.t. a Cartesian and a cylindri-

cal vector base and simplified for plane strain, plane stress and axisymmetry. The material

behavior is assumed to be isotropic.

7.1 Vector/tensor equations

The deformed (current) state is determined by 12 state variables : 3 displacement compo-

nents and 9 stress components. These unknown quantities must be solved from 12 equa-

tions : 6 equilibrium equations and 6 constitutive equations.

With proper boundary (and initial) conditions the equations can be solved, which, for

practical problems, must generally be done numerically. The compatibility equations are

generally satisfied for the chosen strain-displacement relation. In some solution approaches

they are used instead of the equilibrium equations.

gradient operator : ~∇=∇~T~e~

position : ~x = x~T~e~

displacement : ~u = u~T~e~

strain : ε= 12

(~∇~u

)T +(~∇~u

)

=~e~T ε~e~

compatibility : ∇2 tr(ε)−~∇·(~∇·ε)T = 0

stress : σ=~e~Tσ~e~

balance laws : ~∇·σT +ρ~q = ρ~u ; σ=σT

material law : σ= 4C : ε ; ε= 4C−1

: σ= 4S : σ

133

7. MECHANICAL DESIGN: SOLUTIONS

7.2 Three-dimensional scalar equations

The vectors and tensors can be written in components with respect to a three-dimensional

vector basis. For various problems in mechanics, it will be suitable to choose either a Carte-

sian coordinate system or a cylindrical coordinate system.

7.2.1 Cartesian components

The governing equations are written in components w.r.t. a Cartesian vector base ~ex ,~e y ,~ez .

The stresses can be represented with a Cartesian stress cube, see figure 7.1.

x

y

z

~ex

~e y

~ez

~ex

~ez

~x

~u

~e y

σy y

σzz

σxx~ex

~ez

~e y

σyz

σxz σz y

σx y

σzxσyx

Figure 7.1: Cartesian coordinate system and stress cube

x~T =

[x y z

]

; ∇~T =

[∂∂x

∂∂y

∂∂z

]

; u~T =

[ux uy uz

]

ε= 12

2ux,x ux,y +uy,x ux,z +uz,x

· · · 2uy,y uy,z +uz,y

· · · · · · 2uz,z

2εx y,x y −εxx,y y −εy y,xx = 0 → cyc. 2x

εxx,yz +εyz,xx −εzx,x y −εx y,xz = 0 → cyc. 2x

ε~T =

[εxx εy y εzz εx y εyz εzx

]

σ~T =

[σxx σy y σzz σx y σyz σzx

]

σxx,x +σx y,y +σxz,z +ρqx = ρux (σx y =σyx )

σyx,x +σy y,y +σyz,z +ρqy = ρuy (σyz =σz y )

σzx,x +σz y,y +σzz,z +ρqz = ρuz (σzx =σxz )

σ~ = C ε~ ; ε~ = Sσ~

134

7.2. Three-dimensional scalar equations

7.2.2 Cylindrical components

The governing equations are written in components w.r.t. a cylindrical vector base ~er(θ),~et (θ),~ez,

with :

d~er

dθ=~et and

d~et

dθ=−~er

The stress components can be represented with a cylindrical stress cube, see figure 7.2.

~x

~u

θ

~ez

~et

~er

x

y

z

r

~ex

~e y

~ez

~er

σtz

r

z

θ

σr r

σt t

σzr

σtr

σr t

σzt

σzz

σr z

Figure 7.2: Cylindrical coordinate system and stress cube

x~T =

[r θ z

]

; ∇~T =

[∂∂r

1r

∂∂θ

∂∂z

]

; u~T =

[ur ut uz

]

ε=1

2

2ur,r1r

(ur,t −ut )+ut ,r ur,z +uz,r

· · · 2 1r

(ur +ut ,t ) 1r

uz,t +ut ,z

· · · · · · 2uz,z

2εr t ,r t −εr r,t t −εt t ,r r = 0 → cyc. 2x

εr r,tz +εtz,r r −εzr,r t −εr t ,r z = 0 → cyc. 2x

ε~T =

[εr r εt t εzz εr t εtz εzr

]

σ~T =

[σr r σt t σzz σr t σtz σzr

]

σr r,r + 1rσr t ,t + 1

r (σr r −σt t )+σr z,z +ρqr = ρur (σr t =σtr )

σtr,r + 1rσt t ,t + 1

r(σtr +σr t )+σtz,z +ρqt = ρut (σtz =σzt )

σzr,r + 1rσr t ,t + 1

rσzr +σr z,z +ρqz = ρuz (σzr =σr z)

σ~ = C ε~ ; ε~ = Sσ~

135

7. MECHANICAL DESIGN: SOLUTIONS

7.2.3 Material law

When deformations are small, every material will show linear elastic behavior. For orthotropic

material there are 9 independent material constants. When there is more material symme-

try, this number decreases. Finally, isotropic material can be characterized with only two

material constants.

Be aware that we use now the strain components εi j and not the shear components γi j .

In an earlier chapter, the parameters for orthotropic, transversally isotropic and isotropic

material were rewritten in terms of engineering parameters: Young’s moduli and Poisson’s

ratio’s.

C =

A Q R 0 0 0

Q B S 0 0 0

R S C 0 0 0

0 0 0 2K 0 0

0 0 0 0 2L 0

0 0 0 0 0 2M

→ S =C−1 =

a q r 0 0 0

q b s 0 0 0

r s c 0 0 0

0 0 0 12

k 0 0

0 0 0 0 12

l 0

0 0 0 0 0 12

m

quadratic B = A ; S = R ; M = L;

transversal isotropic B = A ; S = R ; M = L ; K = 12

(A−Q)

cubic C = B = A ; S = R =Q ; M = L = K 6= 12

(A−Q)

isotropic C = B = A ; S = R =Q ; M = L = K = 12

(A−Q)

7.3 Planar deformation

In many applications the loading and deformation is in one plane. The result is that the

material body is in a state of plane strain or plane stress. The governing equations can than

be simplified considerably.

7.3.1 Cartesian

In a plane strain situation, deformation in one direction – here the z-direction – is sup-

pressed. In a plane stress situation, stresses on one plane – here the plane with normal in

z-direction – are zero.

Eliminatingσzz for plane strain and εzz for plane stress leads to a simplified Hooke’s law.

Also the equilibrium equation in the z-direction is automatically satisfied and has become

obsolete.

136

7.3. Planar deformation

plane strain : εzz = εxz = εyz = 0

plane stress : σzz =σxz =σyz = 0

ux = ux (x, y)

uy = uy (x, y)

ε~T =

[εxx εy y εx y

]

=[

ux,x uy,y12

(ux,y +uy,x )]

2εx y,x y −εxx,y y −εy y,xx = 0

σ~T =

[σxx σy y σx y

]

σxx,x +σx y,y +ρqx = ρux (σx y =σyx )

σyx,x +σy y,y +ρqy = ρuy

Cp=

Ap Qp 0

Qp Bp 0

0 0 2K

; Sp=

ap qp 0

qp bp 0

0 0 12

k

7.3.2 Cylindrical

In a plane strain situation, deformation in one direction – here the z-direction – is sup-

pressed. In a plane stress situation, stresses on one plane – here the plane with normal in

z-direction – are zero.

Eliminating σzz for plane strain and εzz for plane stress leads to a simplified Hooke’s law.

Also the equilibrium equation in the z-direction is automatically satisfied and has become

obsolete.

plane strain : εzz = εr z = εtz = 0

plane stress : σzz =σr z =σtz = 0

ur = ur (r,θ)

ut = ut (r,θ)

ε~T =

[εr r εt t εr t

]

=[

ur,r1r

(ur +ut ,t ) 12

(1r

(ur,t −ut )+ut ,r

) ]

2εr t ,r t −εr r,t t −εt t ,r r = 0

σ~T =

[σr r σt t σr t

]

σr r,r + 1rσr t ,t + 1

r (σr r −σt t )+ρqr = ρur (σr t =σtr )

σtr,r + 1rσt t ,t + 1

r (σtr +σr t )+ρqt = ρut

Cp=

Ap Qp 0

Qp Bp 0

0 0 2K

; Sp=

ap qp 0

qp bp 0

0 0 12

k

7.3.3 Cylindrical : axisymmetric + ut = 0

When geometry and boundary conditions are such that we have∂( )

∂θ= ( )t = 0 the situation

is referred to as being axisymmetric.

In many cases boundary conditions are such that there is no displacement of material

points in tangential direction (ut = 0). In that case we have εr t = 0 → σr t = 0

137

7. MECHANICAL DESIGN: SOLUTIONS

plane strain : εzz = εr z = εtz = 0

plane stress : σzz =σr z =σtz = 0

ur = ur (r )

ut = 0

ε~T =

[εr r εt t

]

=[

ur,r1r

(ur )]

εr r = ur,r = (rεt t ),r = εt t + rεt t ,r

σ~T =

[σr r σt t

]

σr r,r +1

r(σr r −σt t )+ρqr = ρur

Cp=

[Ap Qp

Qp Bp

]

; Sp=

[ap qp

qp bp

]

7.4 Stress inconsistency under plane stress

Although for plane stress the out-of-plane shear stresses must be zero, they are not, when

calculated afterwards from the strains. This inconsistency is inherent to the plane stress

assumption. Deviations must be small to render the assumption of plane stress valid.

σxz = 2K εxz = 2K uz,x 6= 0

σyz = 2K εyz = 2K uz,y 6= 0

7.5 Solution strategies

7.5.1 Governing equations for unknowns

The deformation of a three-dimensional continuum in three-dimensional space is described

by the displacement vector ~u of each material point. Due to the deformation, stresses arise

and the stress state is characterized by the stress tensor σ. For static problems, this ten-

sor has to satisfy the equilibrium equations. Solving stresses from these equations is gener-

ally not possible and additional equations are needed, which relate stresses to deformation.

These constitutive equations, which describe the material behavior, relate the stress tensor

σ to the strain tensor ε, which is a function of the displacement gradient tensor (~∇~u). Com-

ponents of this strain tensor cannot be independent and are related by the compatibility

equations.

unknown variables

displacements : ~u =~u(~x) → F =(~∇0~x

)T → E , ε

stresses : σ

equations

compatibility : ∇2 tr(ε)−~∇·(~∇·ε)T = 0

equilibrium : ~∇·σT +ρ~q = ρ~u ; σ = σT

material law : σ=σ(F ) → σ= 4C : ε → ε= 4S : σ

limit criteria

yield surface : g (σ) = g (σ1,σ2,σ3) = g t

138

7.5. Solution strategies

7.5.2 Boundary conditions

Some of the governing equations are partial differential equations, where differentiation is

done w.r.t. the spatial coordinates. These differential equations can only be solved when

proper boundary conditions are specified. In each boundary point of the material body,

either the displacement or the load must be prescribed. It is also possible to specify a relation

between displacement and load in such a point.

When the acceleration of the material points cannot be neglected, the equilibrium equa-

tion becomes the equation of motion, with ρ~u as its right-hand term. In that case a solution

can only be determined when proper initial conditions are prescribed, i.e. initial displace-

ment, velocity or acceleration. In this section we will assume ~u =~0.

displacement : ~u =~up ∀ ~x ∈ Au

edge load : ~p =σ·~n = ~pp ∀ ~x ∈ Ap

Superposition principle

The superposition principle was introduced at the end of chapter 5, and is based on the path

independence and the linearity of the problem. This path independent and linear character

F1

F2

u1

u2

F2

F1

u1 +u2

Figure 7.3: Principle of superposition

permits to treat any loading case as an arbitrary subdivision of elementary loading cases, for

which the solution can be summed up in a trivial way. It permits to superimpose the results

for specific loading arrangements, in order to investigate their combined effect.

The principle of superposition states that the solution S for a given combined load L =L1 +L2 is the sum of the solution S1 for load L1 and the solution S2 for L2, so : S = S1 +S2.

139

7. MECHANICAL DESIGN: SOLUTIONS

7.6 Solution : displacement method

In the displacement method the constitutive relation for the stress tensor is substituted in the

force equilibrium equation.

Subsequently the strain tensor is replaced by its definition in terms of the displacement

gradient. This results in a differential equation in the displacement ~u, which can be solved

when proper boundary conditions are specified.

In a Cartesian coordinate system the vector/tensor formulation can be replaced by index

notation.

~∇·σT +ρ~q =~0σ= 4C : ε

→~∇·

(4C : ε

)T +ρ~q =~0

ε= 12

(~∇~u

)T +(~∇~u

)

→

~∇·

4C :(~∇~u

)T +ρ~q =~0 → ~u → ε → σ

Cartesian index notation

σi j , j +ρqi = 0i

σi j =Ci j l kεl k

→ Ci j klεl k, j +ρqi = 0i

εl k = 12

(

ul ,k +uk,l

)

→

Ci j kl ul ,k j +ρqi = 0i → ui → εi j → σi j

7.6.1 Planar, Cartesian : Navier equations

The displacement method is elaborated for planar deformation in a Cartesian coordinate

system. Linear deformation and linear elastic material behavior is assumed. Elimination

and substitution results in two partial differential equations for the two displacement com-

ponents.

σxx,x +σx y,y +ρqx = ρux ; σyx,x +σy y,y +ρqy = ρuy

σxx = Apεxx +Qpεy y

σy y =Qpεxx +Bpεy y

σx y = 2K εx y

Apεxx,x +Qpεy y,x +2K εx y,y +ρqx = ρux

2K εx y,x +Qpεxx,y +Bpεy y,y +ρqy = ρuy

Ap ux,xx +Qp uy,yx +K (ux,y y +uy,x y )+ρqx = ρux

K (ux,yx +uy,xx )+Qp ux,x y +Bp uy,y y +ρqy = ρuy

Ap ux,xx +K ux,y y + (Qp +K )uy,yx +ρqx = ρux

K uy,xx +Bp uy,y y + (Qp +K )ux,x y +ρqy = ρuy

7.6.2 Planar, axisymmetric with ut = 0

Many engineering problems present a rotational symmetry w.r.t. an axis. They are axisym-

metric. In many cases the tangential displacement is zero : ut = 0. This implies that there are

no shear strains and stresses.

140

7.7. Weighted residual formulation

The radial and tangential stresses are related to the radial and tangential strains by the

planar material law. Material parameters are indicated as Ap , Bp and Qp and can later

be specified for a certain material and for plane strain or plane stress. With the strain-

displacement relations the equation of motion can be transformed into a differential equa-

tion for the radial displacement ur

displacements : ur = ur (r ) ; uz = uz(r, z)

strains : εr r = ur,r ; εt t = 1r

ur ; εzz = uz,z

stresses : σtz = 0 ; σr z ≈ 0 ; σtr = 0

eq. of motion : σr r,r + 1r

(σr r −σt t )+ρqr = ρur

σr r = Apεr r +Qpεt t

σt t =Qpεr r +Bpεt t

→ eq. of motion →

ur,r r +1

rur,r −ζ2 1

r 2ur = f (r )

with ζ=√

Bp

Ap

and f (r ) =ρ

Ap

(

ur −qr

)

7.6.3 Planar, axisymmetric with ut = 0, isotropic

For isotropic material the coefficients Ap and Bp are the same, which implies that ζ= 1 and

also Θp1 =Θp2.

σr r = Apεr r +Qpεt t

σt t =Qpεr r + Apεt t

→ eq. of motion →

ur,r r +1

rur,r −

1

r 2ur = f (r )

with f (r ) =ρ

Ap

(

ur −qr

)

7.7 Weighted residual formulation

Unknown variables have to be solved from the combined set of equilibrium equations and

constitutive equations. Some of the equilibrium equations are partial differential equations.

For the general case of large deformations and nonlinear material behavior, the equations

are nonlinear. It is obvious that only for academic and very simple cases, analytic solutions

exist. For more practical problems, approximate solutions must be determined with a nu-

merical technique, of which the finite element method is widely used and will be considered

here.

Application of the finite element method in continuum mechanics requires the reformu-

lation of the equilibrium equations. They are transformed from differential equations to an

integral equation, the so-called weighted residual integral.

First, we formulate the weighted residual integral for linear problems, so for small defor-

mation and linear elastic material behavior. The finite element method is then explained for

this case. Examples of plane stress, plane strain and axisymmetric problems will be calcu-

lated with a Matlab program.

141

7. MECHANICAL DESIGN: SOLUTIONS

Subsequently, we formulate the weighted residual integral for nonlinear problems, where

the iterative solution procedure has to be applied. Finite element analyses can be done again

with a Matlab program.

7.7.1 Weighted residual formulation for 3D deformation

For an approximation, the equilibrium equation is not satisfied exactly in each material

point. The error can be ”smeared out” over the material volume, using a weighting func-

tion ~w(~x).

equilibrium equation ~∇·σT +ρ~q =~0 ∀~x ∈V

approximation → residual ~∇·σT +ρ~q =~∆(~x) 6=~0 ∀~x ∈V

weighted residual

∫

V

~w(~x)·~∆(~x)dV =∫

V

~w ·[~∇·σT +ρ~q

]

dV

When the weighted residual integral is satisfied for each allowable weighting function ~w ,

the equilibrium equation is satisfied in each point of the material.

∫

V

~w ·[~∇·σT +ρ~q

]

dV = 0 ∀ ~w(~x) ↔ ~∇·σT +ρ~q =~0 ∀~x ∈V

In the weighted residual integral, one term contains the divergence of the stress tensor.

This means that the integral can only be evaluated, when the derivatives of the stresses are

continuous over the domain of integration. This requirement can be relaxed by applying

partial integration to the term with the stress divergence. The result is the so-called weak

formulation of the weighted residual integral.

Gauss theorem is used to transfer the volume integral with the term ~∇.( ) to a surface

integral. Also ~p =σ·~n =~n·σT and σ=σT is used.

∫

V

~w ·[~∇·σT +ρ~q

]

dV = 0

~∇·(σT ·~w) = (~∇~w)T : σc + ~w ·(~∇·σT )

→

∫

V

[~∇·(σT ·~w)− (~∇~w)T : σT + ~w ·ρ~q

]

dV = 0 ∀ ~w

∫

V

~∇·(σT ·~w) =∫

V

~n·σT ·~w d A =∫

A

~w ·~p d A

→

∫

V

(~∇~w)T : σdV =∫

V

~w ·ρ~q dV +∫

A

~w ·~p d A = fe(~w) ∀ ~w

7.7.2 Weighted residual formulation : linear

When deformation and rotations are small, the deformation is geometrically linear. The

deformed state is almost equal to the undeformed state. This implies that integration can be

carried out over the undeformed volume V0 and the undeformed area A0.

142

7.7. Weighted residual formulation

The material behavior is described by Hooke’s law, which can be substituted in the weighted

residual integral, according to the displacement solution method.

The weighted residual integral is now completely expressed in the displacement ~u. Ap-

proximate solutions can be determined with the finite element method.

∫

V0

(~∇~w)T : σdV0 =∫

V0

~w ·ρ~q dV0 +∫

A0

~w ·~p d A0 = fe0(~w) ∀ ~w

σ= 4C : ε= 4C : 12

(~∇0~u)+ (~∇0~u)T

= 4C : (~∇0~u)

∫

V0

(~∇0~w)T : 4C : (~∇0~u)dV0 =∫

V0

~w ·ρ~q dV0 +∫

A0

~w ·~p d A0 = fe0(~w) ∀ ~w

7.7.3 Finite element method for 3D deformation

Discretisation

The integral over the volume V is written as a sum of integrals over smaller volumes, which

collectively constitute the whole volume. Such a small volume V e is called an element. Sub-

dividing the volume implies that also the surface with area A is subdivided in element sur-

faces (faces) with area Ae .

Figure 7.4: Finite element discretisation

∑

e

∫

V e

(~∇~w)T : 4C : (~∇~u)dV e =∑

e

∫

V e

~w ·ρ~q dV e +∑

e A

∫

Ae

~w ·~p d Ae ∀~w

Isoparametric elements

Each point of a three-dimensional element can be identified with three local coordinates

ξ1,ξ2,ξ3. In two dimensions we need two and in one dimension only one local coordinate.

The real geometry of the element can be considered to be the result of a deformation

from the original cubic, square or line element with (side) length 2. The deformation can be

described with a deformation matrix, which is called the Jacobian matrix J . The determinant

of this matrix relates two infinitesimal volumes, areas or lengths of both element represen-

tations.

isoparametric (local) coordinates (ξ1,ξ2,ξ3) ; −1≤ ξi ≤ 1 i = 1,2,3

Jacobian matrix J =(

∇~ξx~T)T

; dV e = det(J )dξ1dξ2dξ3

143

7. MECHANICAL DESIGN: SOLUTIONS

ξ3

ξ1

ξ2

ξ2

ξ1

ξ1

Figure 7.5: Isoparametric elements

Interpolation : 4-node linear element

When the elements have a simple shape, e.g. an six-sided volume, the shape and thus vol-

ume is known when the position of a discrete number of edge points is known. These points

are the element nodal points. For a cube with plane faces, eight corner points are needed.

In two dimensions quadrilaterals with straight edges can be used, where four corner nodes

describe the shape. The position of an internal point of the element can be expressed in the

position of the nodal points. This interpolation is done with so-called shape- or interpola-

tion functions, which are a function of local element coordinates ξi , i = 1,2,3, which assume

values between -1 and +1.

1 2

3

4

1 2

34

ξ2

ξ1

Figure 7.6: Quadrilateral element with four corner nodes

~x = N 1(ξ~

)~x1 +N 2(ξ~

)~x2 +N 3(ξ~

)~x3 +N 4(ξ~

)~x4

with

N 1 = 14

(ξ1 −1)(ξ2 −1) ; N 2 =−14

(ξ2 +1)(ξ2 −1)

N 3 = 14

(ξ1 +1)(ξ2 +1) ; N 4 =−14

(ξ1 −1)(ξ2 +1)

The value of the unknown quantity – here the displacement vector ~u – in an arbitrary

point of the element, can also be interpolated between the values of that quantity in the

element nodes.

144

7.7. Weighted residual formulation

~u = N 1(ξ~

)~u1 +N 2(ξ~

)~u2 +N 3(ξ~

)~u3 +N 4(ξ~

)~u4 = N~T (ξ

~)~u~

e

Besides ~x and ~u, the weighting function ~w also needs to be interpolated between nodal

values. When this interpolation is the same as that for the displacement, the so-called Galerkin

procedure is followed, which is generally the case for simple elements, considered here.

~w = N 1(ξ~

) ~w 1 +N 2(ξ~

) ~w 2 +N 3(ξ~

) ~w 3 +N 4(ξ~

) ~w 4 = N~T (ξ

~)~w~

e

The gradient of ~u and ~w also has to be elaborated. and can be written as the product of a

column which contains the derivatives of the interpolation functions, and the column with

nodal components of ~u and ~u.

~∇~u = (~∇N~ )T~u~

e = ~B~T~u~

e ; ~∇~w = (~∇N~ )T~w~

e = ~B~T~w~

e

Finally, everything is substituted in the weighted residual integral. The volume integral

in the left hand side is the element stiffness matrix K e . The integrals in the right hand side

represent the external load and are summarized in the column ~f~

e

e.

∫

V e

(~B~T~w~

e)T : 4C : (~B~T~u~

e )dV e =∫

V e

~w~eT

N~ ·ρ~q dV e +∫

Ae

~w~eT

N~ ·~p d Ae

~w~eT

·

∫

V e

~B~·4C ·~B~

TdV e

·~u~e = ~w~

eT

·

∫

V e

N~ ρ~q dV e

+ ~w~eT

·

∫

Ae

N~ ~p d Ae

→

~w~eT

·K e ·~u~e = ~w~

eT

·~f~

e

e

Integration

Calculating the element stiffness matrix K e and the external loads f~

e

eimplies the evaluation

of an integral over the element volume V e and the element surface Ae . This integration is

done numerically, using a fixed set of ni p Gauss-points, which have s specific location in the

element. The value of the integrand is calculated in each Gauss-point and multiplied with a

Gauss-point-specific weighting factor c i p and added.

∫

V e

g (x1, x2, x3)dV e =1∫

ξ1=−1

1∫

ξ2=−1

1∫

ξ3=−1

f (ξ1,ξ2,ξ3)dξ1dξ2dξ3 =ni p∑

i p=1

c i p f (ξi p1 ,ξ

i p2 ,ξ

i p3 )

∫

V e

g (x1, x2)dV e =1∫

ξ1=−1

1∫

ξ2=−1

f (ξ1,ξ2)dξ1dξ2 =4∑

i p=1

c i p f (ξi p1 ,ξ

i p2 )

Assembling

The weighted residual contribution of all elements have to be collected into the total weighted

residual integral. This means that all elements are connected or assembled. This assembling

is an administrative procedure. All the element matrices and columns are placed at appro-

priate locations into the structural or global stiffness matrix K and the load column f~e

.

Because the resulting equation has to be satisfied for all w~ , the nodal displacements u~have to satisfy a set of equations.

145

7. MECHANICAL DESIGN: SOLUTIONS

ξ2

13

p3

13

p3

ξ121

43

1 2

4

3

Figure 7.7: Integration points in 4-node quadrilateral element

∑

e

~w~eT

·K e ·~u~e =

∑

e

~w~eT

·~f~

e

e→ ~w~

T ·K ·~u~ = ~w~T ·~f

~e∀ ~w~ →

K ·~u~ = ~f~e

→ ~u~ = K −1·~f~e

Boundary conditions

The initial governing equations were differential equations, which obviously need boundary

conditions to arrive at a unique solution. The boundary conditions are prescribed displace-

ments or forces in certain material points. After finite element discretisation, displacements

and forces can be applied in nodal points.

The set of nodal equations K u~ = f~e

cannot be solved yet, because the structural stiffness

matrix K is singular and cannot be inverted. First some essential boundary conditions must

be applied, which prevent the rigid body motion of the material and renders the equations

solvable.

146

CH

AP

TE

R

8MECHANICAL DESIGN PROBLEMS:

ANALYTICAL SOLUTIONS

In this chapter, a number of classical mechanical design problems are presented, which al-

low for an analytical solution. The equations are presented and the solution is given without

extensive derivations. Many problems involve the calculation of integration constants from

boundary conditions. For such problems these integration constants can be found in ap-

pendix B. Examples with numerical values for parameters, are presented. More examples

can be found in the above-mentioned appendix.

8.1 Homogeneous plates

For planar problems in a Cartesian coordinate system, two partial differential equations for

the displacement components ux and uy , the so-called Navier equations, have to be solved,

using specific boundary conditions. Only for very simply cases, this can be done analyti-

cally. For practical problems, approximate solutions have to be determined with numerical

solution procedures. The Navier equations have been derived in section 7.6.1 and are re-

peated below for the static case, where no material acceleration is considered. The material

parameters Ap , Bp , Qp and K have to be specified for plane stress or plane strain and for the

material model concerned.

Ap ux,xx +K ux,y y + (Qp +K )uy,yx +ρqx = 0

K uy,xx +Bp uy,y y + (Qp +K )ux,x y +ρqy = 0

Tensile test

When a square plate (length a) of homogeneous material is loaded uniaxially by a uniform

tensile edge load p, this load constitutes an equilibrium system, i.e. the stresses satisfy the

equilibrium equations : σxx = p and σy y =σx y =σzz = 0. The deformation can be calculated

directly from Hooke’s law.

147

8. MECHANICAL DESIGN: ANALYTICAL

y

x

y

x

σxx = p σxx = p

a

a

Figure 8.1: Uniaxial tensile test

εxx =1

Eσxx =

p

E→ ux =

p

Ex +c ; ux (x = 0) = 0 → c = 0

ux =p

Ex → ux (x = a) =

p

Ea

εy y =−νεxx =−νp

E→ uy =−ν

p

Ey +c ; uy (y = 0)= 0 → c = 0

uy =−νp

Ey ; uy (y = a/2) =−ν

p

E

a

2

8.2 Axisymmetric cylinders and discs

The differential equation for the radial displacement ur is derived in chapter 7 by substitu-

tion of the stress-strain relation (material law) and the strain-displacement relation in the

equilibrium equation w.r.t. the radial direction. It is repeated here for isotropic material be-

havior. Material parameters Ap and Qp have to be specified for plane stress and plane strain.

ur,r r +1

rur,r −

1

r 2ur = f (r )

with f (r ) =ρ

Ap

(

ur −qr

)

A general solution for the differential equation can be determined as the addition of the

homogeneous solution ur and the particular solution ur , which depends on the specific

loading f (r ). From the general solution the radial and tangential strains can be calculated

according to their definitions.

ur = rλ → ur,r =λrλ−1 → ur,r r =λ(λ−1)rλ−2 →[λ(λ−1)+λ−1] rλ−2 = 0 →λ2 = 1 → λ=±1 → ur = c1 r +c2r−1

148

8.2. Axisymmetric cylinders and discs

The general solution for an isotropic material then becomes

general solution ur = c1r +c2

r+ ur

εr r = c1 −c2r−2 + ur,r

εt t = c1 +c2r−2 +ur

r

σr r = (Ap +Qp )c1 − (Ap −Qp )c2

r 2+ Ap ur,r +Qp

ur

r

σt t = (Qp + Ap )c1 − (Qp − Ap )c2

r 2+Qp ur,r + Ap

ur

r

When there is no right-hand loading term f (r ) in the differential equation, the particulate

part ur will be zero. Then, for isotropic material, the radial and tangential strains are uni-

form, i.e. no function of the radius r . For a state of plane stress, the axial strain is calculated

as a weighted summation of the in-plane strains, so also εzz will be uniform. The thickness

of the axisymmetric object will remain uniform. For non-isotropic material behavior this is

not the case, however.

Loading and boundary conditions

In the following subsections, different geometries and loading conditions will be considered.

The external load determines the right-hand side f (r ) of the differential equation and as a

consequence the particulate part ur of the general solution. Boundary conditions must be

used subsequently to determine the integration constants c1 and c2. Finally the parameters

Ap , Bp and Qp must be chosen in accordance with the material behavior and specified for

plane stress or plane strain.

The algebra, which is involved with these calculations, is not very difficult, but rather

cumbersome. In appendix B a number of examples is presented. When numerical values

are provided, displacements, strains and stresses can be calculated and plotted with a Matlab

program, which is available on the website of this course. Based on the input, it selects the

proper formulas for the calculation. Instructions for its use can be found in the program

source file. The figures in the next subsections are made with this program.

8.2.1 Prescribed edge displacement

The outer edge of a disc with a central hole is given a prescribed displacement u(r = b) = ub .

The inner edge is stress-free. With these boundary conditions, the integration constants in

the general solution can be determined. They can be found in appendix B.

For the parameter values listed below, the radial displacement ur and the stresses are

calculated and plotted as a function of the radius r .

| ub = 0.01 m | a = 0.25 m | b = 0.5 m | h = 0.05 m | E = 250 GPa | ν= 0.33 |

8.2.2 Edge load

A cylinder has inner radius r = a and outer radius r = b. It is loaded with an internal (pi )

and/or an external (pe ) pressure.

149

8. MECHANICAL DESIGN: ANALYTICAL

ub

ab

z r

r

f (r ) = 0 → ur = 0

ur (r = b) = ub

σr r (r = a) = 0

c1,c2 : App. B

Figure 8.2: Edge displacement of circular disc

0 0.1 0.2 0.3 0.4 0.59.4

9.5

9.6

9.7

9.8

9.9

10x 10

−3

r [m]

u r [m

]

0 0.1 0.2 0.3 0.4 0.50

2

4

6

8

10x 10

9

r [m]

σ [P

a]

σrr

σtt

σzz

Figure 8.3: Displacement and stresses for plane stress (σzz = 0)

The general solution to the equilibrium equation has two integration constants, which

have to be determined from boundary conditions. In appendix B they are determined for the

case that an open cylinder is subjected to an internal pressure pi and an external pressure pe.

For plane stress (σzz = 0) the cylinder is free to deform in axial direction. The solution was

first derived by Lamé in 1833 and therefore this solution is referred to as Lamé’s equations.

When these integration constants for isotropic material are substituted in the stress so-

lution, it appears that the stresses are independent of the material parameters. This implies

that radial and tangential stresses are the same for plane stress and plane strain. For the

plane strain case, the axial stress σzz can be calculated directly from the radial and tangen-

tial stresses.

The Tresca and Von Mises limit criteria for a pressurized cylinder can be calculated ac-

cording to their definitions (see chapter 6).

σr r =pi a2 −peb2

b2 −a2−

a2b2(pi −pe )

b2 −a2

1

r 2; σt t =

pi a2 −pe b2

b2 −a2+

a2b2(pi −pe )

b2 −a2

1

r 2

σT R = 2τmax = max[ |σr r −σt t |, |σt t −σzz |, |σzz −σr r | ]

σV M =√

12

(σr r −σt t )2 + (σt t −σzz )2 + (σzz −σr r )2

150

8.2. Axisymmetric cylinders and discs

pe

ab

z r

r

pi

f (r ) = 0 → ur = 0

σr r (r = a) =−pi

σr r (r = b) =−pe

c1,c2 : App. B

Figure 8.4: Cross-section of a thick-walled circular cylinder

Open cylinder

An open cylinder is analyzed with the parameters from the table below. Stresses are plotted

as a function of the radius.

| pi = 100 MPa | a = 0.25 m | b = 0.5 m | h = 0.5 m | E = 250 GPa | ν= 0.33 |

0 0.1 0.2 0.3 0.4 0.5−1

−0.5

0

0.5

1

1.5

2x 10

8

r [m]

σ [P

a]

σrr

σtt

σzz

0 0.1 0.2 0.3 0.4 0.50.5

1

1.5

2

2.5

3x 10

8

r [m]

σ [P

a]

σTR

σVM

Figure 8.5: Stresses in a thick-walled pressurized cylinder for plane stress (σzz = 0)

That the inner material is under much higher tangential stress than the outer material,

can be derived by reasoning, when we only consider an internal pressure. This pressure

will result in enlargement of the diameter for each value of r , but it will also compress the

material and result in reduction of the wall thickness. The inner diameter will thus increase

more than the outer diameter – which is also calculated and plotted in figure 8.6 – and the

tangential stress will be much higher at the inner edge.

Closed cylinder

A closed cylinder is loaded in axial direction by the internal and the external pressure. This

load leads to an axial stress σzz , which is uniform over the wall thickness. It can be deter-

mined from axial equilibrium and can be considered as an The radial and tangential stress

are not influenced by this axial load.

151

8. MECHANICAL DESIGN: ANALYTICAL

0 0.1 0.2 0.3 0.4 0.51.1

1.2

1.3

1.4

1.5

1.6

1.7

1.8

1.9

2x 10

−4

r [m]

u [m

]

psspsn

Figure 8.6: Radial displacement in a thick-walled pressurized cylinder for plane stress (pss)

and plane strain (psn)

The resulting radial displacement due to the contraction caused by the axial load, ur a ,

can be calculated from Hooke’s law.

axial equilibrium σzz =pi a2 −pe b2

b2 −a2→ ur a = εt ta r =−

ν

Eσzz r

8.3 Circular hole in infinite medium

When a circular hole is located in an infinite medium, we can derive the stresses from Lame’s

equations by taking b → ∞. For the general case this leads to limit values of the integration

constants c1 and c2. These can then be substituted in the general solutions for displacement

and stresses.

8.3.1 Pressurized hole in an infinite medium

For a pressurized hole in an infinite medium the external pressure pe is zero. In that case the

absolute values of radial and tangential stresses are equal. The radial displacement can also

be calculated.

b →∞ ; pi = p ; pe = 0 → σr r =−pa2

r 2; σt t =

pa2

r 2

For a plane stress state and with parameter values listed below, the radial displacement

and the stresses are calculated and plotted as a function of the radius in figure 8.7. Note that

we take a large but finite value of for b.

| pi = 100 MPa | a = 0.2 m | b = 20 m | h = 0.5 m | E = 200 GPa | ν= 0.3 |

8.3.2 Stress-free hole in bi-axially loaded infinite medium

We consider the case of a stress-free hole of radius a in an infinite medium, which is bi-

axially loaded at infinity by a uniform load T , equal in x- and y-direction. Because the load

152

8.3. Circular hole in infinite medium

0 0.5 1 1.5 2 2.5 30

0.2

0.4

0.6

0.8

1

1.2

1.4x 10

−4

r [m]

u [m

]

0 0.5 1 1.5 2 2.5 3−1.5

−1

−0.5

0

0.5

1

1.5x 10

8

r [m]

σ [P

a]

σrr

σtt

σzz

Figure 8.7: Displacement and stresses in a pressurized circular hole in an infinite medium

is applied at boundaries which are at infinite distance from the hole center, the bi-axial load

is equivalent to an externally applied radial edge load pe =−T .

Radial and tangential stresses are different in this case. The tangential stress is maximum

for r = a and equals 2T .

b →∞ ; pi = 0 ; pe =−T → σr r = T

(

1−a2

r 2

)

; σt t = T

(

1+a2

r 2

)

stress concentration factor Kt =σmax

T=

σt t (r = a)

T=

2T

T= 2

For a plane stress state and with parameter values listed below, the radial displacement

and the stresses are calculated and plotted as a function of the radius in figure 8.8.

| pe =−100 MPa | a = 0.2 m | b = 20 m | h = 0.5 m | E = 200 GPa | ν= 0.3 |

0 0.5 1 1.5 2 2.5 30

0.2

0.4

0.6

0.8

1

x 10−3

r [m]

u [m

]

0 0.5 1 1.5 2 2.5 3−0.5

0

0.5

1

1.5

2

2.5x 10

8

r [m]

σ [P

a]

σrr

σtt

σzz

Figure 8.8: Displacement and stresses in a stress-free hole in a biaxially loaded infinite

medium

153

8. MECHANICAL DESIGN: ANALYTICAL

8.4 Rotating disc

A circular disc, made of isotropic material, rotates with angular velocity ω [rad/s]. The outer

radius of the disc is taken to be b. The disc may have a central circular hole with radius a.

When a = 0 there is no hole and the disc is called ”solid”. Boundary conditions for a disc

with a central hole are rather different than those for a ”solid” disc, which results in different

solutions for radial displacement and stresses. The external load f (r ) is the result of the

radial acceleration of the material (see sections 8.2

ur =−ω2r → f (r ) =ρ

Apur = −

ρ

Apω2r

The particular solution ur can be computed to be

ur =−1

8

1

Apρω2r 3

8.4.1 Solid disc

In a disc without a central hole (solid disc) there are material points at radius r = 0. To pre-

z

ω

b

r

r

ur (r = 0) 6=∞σr r (r = b) = 0

c1,c2 : App. B

Figure 8.9: A rotating solid disc

vent infinite displacements for r→0 the second integration constant c2 must be zero. At the

outer edge the radial stress σr r must be zero, because this edge is unloaded. With these

boundary conditions the integration constants in the general solution can be calculated (see

appendix B).

For a plane stress state and with the listed parameter values, the stresses are calculated

and plotted as a function of the radius.

|ω= 6 c/s | a = 0 m | b = 0.5 m | t = 0.05 m | ρ = 7500 kg/m3 || E = 200 GPa | ν= 0.3 |

In a rotating solid disc the radial and tangential stresses are equal in the center of the

disc. They both decrease with increasing radius, where of course the radial stress reduces to

zero at the outer radius. The equivalent Tresca and Von Mises stresses are not very different

and also decrease with increasing radius. In the example the disc is assumed to be in a state

of plane stress.

When the same disc is fixed between two rigid plates, a plane strain state must be mod-

elled. In that case the axial stress is not zero. As can be seen in figure 8.11, the axial stress

influences the Tresca and Von Mises equivalent stresses.

154

8.4. Rotating disc

0 0.1 0.2 0.3 0.4 0.5−2

0

2

4

6

8

10

12x 10

5

r [m]

σ [P

a]

σrr

σtt

σzz

0 0.1 0.2 0.3 0.4 0.54

5

6

7

8

9

10

11x 10

5

r [m]

σ [P

a]

σTR

σVM

Figure 8.10: Stresses in a rotating solid disc in plane stress

0 0.1 0.2 0.3 0.4 0.50

2

4

6

8

10

12x 10

5

r [m]

σ [P

a]

σrr

σtt

σzz

0 0.1 0.2 0.3 0.4 0.52.8

3

3.2

3.4

3.6

3.8

4

4.2

4.4

4.6x 10

5

r [m]

σ [P

a]

σTR

σVM

Figure 8.11: Stresses in a rotating solid disc in plane strain

8.4.2 Disc with central hole

When the disc has a central circular hole, the radial stress at the inner edge and at the outer

edge must both be zero, which provides two equations to solve the two integration constants.

For a plane stress state and with parameter values listed below, the radial displacement

and the stresses are calculated and plotted as a function of the radius in figure 8.13.

|ω= 6 c/s | a = 0.2 m | b = 0.5 m | t = 0.05 m | ρ = 7500 kg/m3 || E = 200 GPa | ν= 0.3 |

8.4.3 Disc fixed on rigid axis

When the disc is fixed on an axis and the axis is assumed to be rigid (see figure 8.14), the

displacement of the inner edge is suppressed. The radial stress at the outer edge is obviously

zero.

155

8. MECHANICAL DESIGN: ANALYTICAL

ω

ab

z r

r

σr r (r = a) = 0

σr r (r = b) = 0

c1,c2 : App. B

Figure 8.12: A rotating disc with central hole

0 0.1 0.2 0.3 0.4 0.52

2.05

2.1

2.15

2.2

2.25

2.3

2.35x 10

−6

r [m]

u r [m

]

0 0.1 0.2 0.3 0.4 0.50

0.5

1

1.5

2

2.5x 10

6

r [m]

σ [P

a]

σrr

σtt

Figure 8.13: Displacement and stresses in a rotating disc with a central hole

ω

ba

z r

r

ur (r = a) = 0

σr r (r = b) = 0

c1,c2 : App. B

Figure 8.14: Disc fixed on rigid axis

156

8.4. Rotating disc

For a plane stress state and with parameter values listed below the stresses and the radial

displacement are calculated and plotted in figures 8.15 and 8.16.

|ω= 6 c/s | a = 0.2 m | b = 0.5 m | t = 0.05 m | ρ = 7500 kg/m3 | E = 200 GPa | ν= 0.3 |

0 0.1 0.2 0.3 0.4 0.50

2

4

6

8

10

12

14x 10

5

r [m]

σ [P

a]

σrr

σtt

0 0.1 0.2 0.3 0.4 0.52

4

6

8

10

12

14x 10

5

r [m]

σ [P

a]

σTR

σVM

Figure 8.15: Stresses in a rotating disc, fixed on an axis

0 0.1 0.2 0.3 0.4 0.50

1

2

3

4

5

6

7

8x 10

−7

r [m]

u r [m

]

Figure 8.16: Displacement in a rotating disc

8.4.4 Rotating disc with variable thickness

For a rotating disc with variable thickness t (r ) the equation of motion in radial direction can

be derived. For a disc with inner and outer radius a and b, respectively, and a thickness

distribution t (r ) = ta

2ar

, a general solution for the stresses can be derived. The integration

constants can be determined from the boundary conditions, e.g. σr r (r = a) =σr r (r = b) = 0.

157

8. MECHANICAL DESIGN: ANALYTICAL

equilibrium∂(t (r )rσr r )

∂r− t (r )σt t =−ρω2t (r )r 2 with t (r ) =

ta

2

a

r

general solution stresses

σr r =2c1

atar d1 +

2c2

atar d2 −

3+ν

5+νρω2r 2 ; σt t =

2c1

atad1r d1 +

2c2

atad2r d2 −

1+3ν

5+νρω2r 2

with d1 =−12+

√54+ν ; d2 =−1

2−

√54+ν

boundary conditions σr r (r = a) =σr r (r = b) = 0 →

2c1

ata=

3+ν

5+νρω2 a−d1

[

a2 −ad2

(b2 −a−d1 bd1 a2

bd2 −ad2 a−d1 bd1

)]

2c2

ata=

3+ν

5+νρω2

(b2 −a−d1 bd1 a2

bd2 −ad2 a−d1 bd1

)

A disc with a central hole and a variable thickness rotates with an angular velocity of 6

cycles per second. The stresses are plotted as a function of the radius in figure 8.17.

| isotropic | plane stress |ω= 6 c/s || a = 0.2 m | b = 0.5 m | ta = 0.05 m | ρ = 7500 kg/m3 | E = 200 GPa | ν= 0.3 |

0 0.1 0.2 0.3 0.4 0.50

2

4

6

8

10

12

14

16

18x 10

5

r [m]

σ [P

a]

σrr

σtt

σzz

0 0.1 0.2 0.3 0.4 0.50.6

0.8

1

1.2

1.4

1.6

1.8x 10

6

r [m]

σ [P

a]

σTR

σVM

Figure 8.17: Stresses in a rotating disc with variable thickness

8.5 Large thin plate with a central hole

A large rectangular plate is loaded with a uniform stress σxx =σ. In the center of the plate is

a hole with radius a, much smaller than the dimensions of the plate.

The stresses around the hole can be determined, using an Airy stress function approach.

The relevant stresses are expressed as components in a cylindrical coordinate system, with

coordinates r , measured from the center of the hole, and θ in the circumferential direction.

158

8.5. Large thin plate with a central hole

x

y

r

θ

a

σσ

Figure 8.18: Large thin plate with a central hole

σr r =σ

2

[(

1−a2

r 2

)

+(

1+3a4

r 4−4

a2

r 2

)

cos(2θ)

]

σt t =σ

2

[(

1+a2

r 2

)

−(

1+3a4

r 4

)

cos(2θ)

]

σr t = −σ

2

[

1−3a4

r 4+2

a2

r 2

]

sin(2θ)

At the inner edge of the hole, the tangential stress reaches a maximum value of 3σ for

θ = 90o. For θ = 0o a compressive tangential stress occurs. The stress concentration factor

Kt is independent of material parameters and the hole diameter.

At a large distance from the hole, so for r ≫ a, the stress components are a function of

the angle θ only.

stress concentrations

σt t (r = a,θ = π2

) = 3σ ; σt t (r = a,θ = 0) =−σ

stress concentration factor Kt =σmax

σ= 3

stress at larger r

σr r =σ

2[1+cos(2θ)] =σ cos2(θ)

σt t =σ

2[1−cos(2θ)] =σ

[

1−cos2(θ)]

=σsin2(θ)

σr t =−σ

2sin(2θ) =−σ sin(θ)cos(θ)

For parameters values listed below stress components are calculated and plotted for θ= 0

and for θ = π2

as a function of the radial distance r .

| a = 0.05 m |σ= 1000 Pa |

159

8. MECHANICAL DESIGN: ANALYTICAL

0 0.1 0.2 0.3 0.4 0.5−1000

−500

0

500

1000

r [m]

σ [P

a]

σrr

σtt

σzz

0 0.1 0.2 0.3 0.4 0.50

500

1000

1500

2000

2500

3000

r [m]

σ [P

a]

σrr

σtt

σzz

Figure 8.19: Stresses in plate for θ = 0 and θ = π2

160

CH

AP

TE

R

9MECHANICAL DESIGN PROBLEMS:

NUMERICAL SOLUTIONS

In this chapter, the numerical solution of some selected problems is presented. These solu-

tions are determined with the MSC.Marc/Mentat FE-package. The numerical solutions can

be compared with the analytical solutions, described in the previous section.

9.1 MSC.Marc/Mentat

The MSC.Mentat program is used to model the structure, which is subsequently analyzed

by the FE-program MSC.Marc. Modeling the geometry – shape and dimensions – is the

first step in this procedure. Dimensional units have to be chosen and consistently used in

the entire analysis. In this stage it is already needed to decide whether the model is three-

dimensional, planar or axisymmetric. The finite element mesh is generated according to

procedures which are described in the tutorial. In the examples discussed in this chap-

ter, only linear elements are used, i.e. elements where the displacement is interpolated bi-

linearly between the nodal point displacements. Quadratic elements will lead to more accu-

rate results in most cases.

After defining the geometry, the material properties can be specified. Only linear elastic

material behavior is considered, both isotropic and orthotropic. Boundary conditions can

be : prescribed displacements, edge loads, gravitational loads and centrifugal loading due to

rotation. Thermal loading is not shown here, but can be applied straightforwardly.

When the model is complete, it can be analyzed and the results can be observed and

plotted. Contour bands of variables can be superposed on the geometry and variables can

be plotted. In the next sections these plots will be presented.

9.2 Cartesian, planar

The most simple analysis, which can be done is the calculation of stress and strain in a ten-

sile test and a shear test. Boundary conditions must be prescribed to ensure homogeneous

161

9. MECHANICAL DESIGN: NUMERICAL

deformation. As expected, the exact solution is reproduced with only one element.

9.2.1 Tensile test

A uniaxial tensile test, resulting in homogeneous deformation, can be modeled and analyzed

with only one linear element, resulting in the exact solution. In the next example a square

p

x

y

p

x

y

Figure 9.1: Tensile test with different boundary conditions

plate (length a, thickness h) is modeled with four equally sized elements, because the central

node on the left edge must be fixed to prevent rigid body movement. The right-hand edge is

loaded with an edge load p. The deformation is shown in figure 9.2 with a magnification of

500.

job1

Inc: 0

Time: 0.000e+00

X

Y

Z

1

job1

Inc: 0

Time: 0.000e+00

X

Y

Z

1

Figure 9.2: Undeformed and deformed element mesh at 500 × magnification

When the left edge is clamped, the deformation and stress state is no longer homoge-

neous. An analytical solution does not exist for this case. An approximate solution can be de-

termined rather easily. To model the inhomogeneous deformation, we need more elements,

especially in the neighborhood of the clamped edge. Using more elements improves the ac-

curacy of the result. Equal accuracy can be realized with fewer, but higher-order (quadratic)

elements, as such elements interpolate the displacement field better than a linear element.

When subsequent analyses are done with decreasing element sizes, we will notice that

at some point, further mesh refinement will not change the solution any more. This con-

vergence upon mesh refinement is essential for good finite element modeling and analysis.

When it does not occur, the results are always dependent on the element mesh and such a

mesh dependency is not allowed. It may be found to occur when singularities are involved or

when the (non-linear) material shows softening, i.e. decrease of stress at increasing strain.

| p = 100 MPa | a = 0.5 m | h = 0.05 m | E = 200 GPa | ν= 0.25 |

162

9.2. Cartesian, planar

ux (x = a) = 0.25×10−3 m ; uy (y = a/2) =−0.3125×10−4

9.2.2 Shear test

The true shear test is another example of a homogeneous deformation. It is done here with

a 20×20 mesh of square linear elements, but could have been done with only one element,

leading to the same exact result. To prevent rigid body translation, the left-bottom node

yp

x

p

p

p

yp

x

Figure 9.3: Shear test with different boundary conditions

is fixed, i.e. its displacement is prevented. To prevent rigid body rotation, the nodes at the

bottom are only allowed to move horizontally. Edges are loaded with a shear load p, leading

to the deformation, which is shown in figure 9.4, again with a magnification of 500.

job1

Inc: 0

Time: 0.000e+00

X

Y

Z

1

job1

Inc: 0

Time: 0.000e+00

X

Y

Z

1

Figure 9.4: Undeformed and deformed element mesh at 500 and 250 × magnification

Instead of this homogeneous shear test, often a so-called simple shear test is done exper-

imentally. In that case the shear load p is only applied at the upper edge. The left- and right-

edge is stress-free. Moreover, the displacement in y-direction of the upper edge is prevented,

as is the case for the bottom-edge, which is clamped. The result is shown in the right-hand

of figure 9.4 with a magnification of 250. It is immediately clear that the deformation is no

longer homogeneous.

| isotropic | plane stress | p = 100 MPa | a = 0.5 m | h = 0.05 m|| E = 200 GPa | ν= 0.25 |

163

9. MECHANICAL DESIGN: NUMERICAL

9.2.3 Orthotropic plate

The uniaxial tensile test and the real shear test are now carried out on a plate with orthotropic

material behavior. For the tensile test, the center node on the left-edge is fixed, while the

other nodes on this edge are restricted to move in y-direction. For the shear test, the center

node is fixed and again the upper-right and lower-left corner nodes are restricted to move

along the diagonal. The material coordinate system is oriented at an angle α w.r.t. the global

x-axis. Material parameters are listed in the table. Because a plane stress state is assumed,

the surface of the plate decreases and also the thickness will change.

p

x

y yp

x

p

p

p

Figure 9.5: Tensile test and shear test for orthotropic plate

| orthotropic | plane stress | p = 100 MPa | a = 0.5 m | h = 0.05 m |α= 20o|| E11 = 200 GPa | E22 = 50 GPa | E33 = 50 GPa|| ν12 = 0.4 | ν23 = 0.25 | ν31 = 0.25||G12 = 100 GPa |G23 = 20 GPa |G31 = 20 GPa |

job1

Inc: 0

Time: 0.000e+00

X

Y

Z

1

job1

Inc: 0

Time: 0.000e+00

X

Y

Z

1

Figure 9.6: Undeformed and deformed element mesh at 500 × magnification

9.3 Axisymmetric, ut = 0

A tensile test on a cylindrical bar can be analyzed analytically when the material is isotropic.

For orthotropic material, with principal material directions in radial, axial and tangential di-

rection, the problem is analyzed numerically. Although the loading is uniaxially and uniform

over the cross-section, the strain and stress distribution is not homogeneous.

The cylindrical tensile bar of length 0.5 m is modelled with axisymmetric elements. The

radius of the bar is 0.0892 m. The material coordinate system is 1,2,3 = r, z, t . Material

164

9.4. Axisymmetric, planar, ut = 0

parameters are listed in the table. The bar is loaded with an axial edge load p. The axial

displacement is then about 0.001 m. The figure shows the stresses as a function of the radius.

| E11 = 200 GPa | E22 = 50 GPa | E33 = 50 GPa|| ν12 = 0.4 | ν23 = 0.25 | ν31 = 0.25||G12 = 100 GPa |G23 = 20 GPa |G31 = 20 GPa | p = 100 MPa |

p

r

z

0 0.02 0.04 0.06 0.08−1.5

−1

−0.5

0

0.5

1

1.5x 10

7

r [m]

σ [P

a]

σrr

σtt

σzz

Figure 9.7: Stresses in an orthotropic tensile bar

9.4 Axisymmetric, planar, ut = 0

Axisymmetric problems can be analyzed with planar elements – plane stress or plane strain

– but also with axisymmetric elements. In the latter case, the model is made in the zr -plane

for r > 0.

9.4.1 Prescribed edge displacement

The model is made in the zr -plane with axisymmetric elements, see figure 9.8 A plane stress

state is modeled by choosing the proper boundary conditions.

| ub = 0.01 m | a = 0.25 m | b = 0.5 m | h = 0.05 m | E = 250 GPa | ν= 0.33 |

Displacements and stresses are shown in figure 9.9.

9.4.2 Edge load

A thick-walled cylinder is loaded with an internal pressure pi , see figure 9.10. When the

cylinder is open and its elongation unconfined, each cross-section over the axis is in a state

of plane stress: σzz = 0.

The cylinder is open (plane stress) and made of isotropic material. Dimensions and ma-

terial properties are listed in the table. The plots in figure 9.11 show the stresses as a function

of the radius. There values coincide with the analytical solution, except near the edges. The

165

9. MECHANICAL DESIGN: NUMERICAL

ub

ab

z r

r

Figure 9.8: Edge displacement of circular disc

0 0.1 0.2 0.3 0.4 0.59.4

9.5

9.6

9.7

9.8

9.9

10x 10

−3

r [m]

u r [m

]

0 0.1 0.2 0.3 0.4 0.50

2

4

6

8

10

12x 10

9

r [m]

σ [P

a]

σrr

σtt

σzz

Figure 9.9: Displacement and stresses for plane stress (σzz = 0)

pe

ab

z r

r

pi

Figure 9.10: Cross-section of a thick-walled circular cylinder

166

9.4. Axisymmetric, planar, ut = 0

reason is that stresses (and strains) are calculated in the integration points, which are lo-

cated inside the element, and edge values are extrapolated. When more elements are used,

the deviation will decrease.

| pi = 100 MPa | a = 0.25 m | b = 0.5 m | h = 0.5 m | E = 250 GPa | ν= 0.33 |

0 0.1 0.2 0.3 0.4 0.5−1

−0.5

0

0.5

1

1.5

2x 10

8

r [m]

σ [P

a]

σrr

σtt

σzz

0 0.1 0.2 0.3 0.4 0.50.6

0.8

1

1.2

1.4

1.6

1.8

2

2.2

2.4x 10

8

r [m]

σ [P

a]

σVM

Figure 9.11: Stresses in a thick-walled pressurized cylinder for plane stress (σzz = 0)

For plane strain (εzz = 0) the length of the cylinder is kept constant. This will obviously

lead to an axial stress σzz . The radial displacement in figure 9.12 is obviously larger for the

plane stress situation, because the material has more ”freedom” to deform in radial direc-

tion.

0 0.1 0.2 0.3 0.4 0.51.1

1.2

1.3

1.4

1.5

1.6

1.7

1.8

1.9

2x 10

−4

r [m]

u [m

]

psspsn

Figure 9.12: Radial displacement in a thick-walled pressurized cylinder for plane stress (pss)

and plane strain (psn)

9.4.3 Centrifugal load

A centrifugal load is modeled to analyze rotating discs. The analysis can be done with an

axisymmetric model or with a plane stress model. Except for the location near the edges, the

167

9. MECHANICAL DESIGN: NUMERICAL

numerical solution coincides with the analytical solution, presented in chapter 8.

9.4.4 Large thin plate with a central hole

A rectangular plate with central hole is loaded uniaxially. The analytical solution for the

stress distribution is known and given in chapter 8. To get a numerical solution, the plate

is modeled with linear elements. Geometric and material data are listed in the table. The

figures show the stresses as a function of the radial coordinate for the 0o- and 90o-direction.

x

y

r

θ

a

σσ

Figure 9.13: Large thin plate with a central hole

| a = 0.05 m |σ= 1000 Pa |

0 0.1 0.2 0.3 0.4 0.5−1500

−1000

−500

0

500

1000

r [m]

σ [P

a]

σrr

σtt

0 0.1 0.2 0.3 0.4 0.5−500

0

500

1000

1500

2000

2500

3000

3500

r [m]

σ [P

a]

σrr

σtt

Figure 9.14: Stresses in plate for θ = 0 and θ = π2

168

CH

AP

TE

R

10LARGE DEFORMATIONS AND NONLINEAR

MATERIAL BEHAVIOUR: BASIC CONCEPTS

AND NONLINEAR ELASTICITY

This chapter provides an outlook to the mechanics of large deformations and rotations, as

well complex nonlinear material behaviour. A one-dimensional truss-like description will be

used to explain basic concepts. The governing equations are scalar. The extension to three

dimensions is obviously tensorial, but beyond the scope of this bachelor course.

10.1 Nonlinear axial deformation of trusses

A truss is a mechanical element whose dimension in one direction – the truss axis – is much

larger than the dimensions in each direction perpendicular to the axis. A truss structure is

an assembly of trusses, which are connected mutually and to the surroundings with hinges.

A truss can transfer only axial forces along its axis, so bending is not possible, and the axis

must remain straight.

When deformation and/or rotation of the truss are large, various strains and stresses can

be defined and related by material laws. The material behavior can be expected to be no

longer linearly elastic.

10.1.1 Strains for large elongation

The scalar nonlinear strain measures were already introduced in chapter 2. In this subsec-

tion, these strain measures will be elaborated for a truss. In the deformed state the length of

the truss is l and its cross-sectional area is A. The elongation is described by the axial stretch

λ. The change in cross-sectional area is described by the contraction µ. It is assumed that

the load, which provokes the deformation, is such that the deformation is homogeneous. For

a homogeneous truss, λ and µ are the same in each point of the truss. The volume change is

described by the volume ratio J .

The strain measures are expressed in terms of the stretch ration λ, see chapter 2.

169

10. LARGE DEFORMATIONS & NONLINEAR BEHAVIOUR

Linear strain

The linear strain definition results in unrealistic contraction, when the elongation is too

large. The cross-sectional area of the truss can become zero, which is of course not pos-

sible.

linear strain ε= εl =λ−1 =∆l

l0contraction strain εd =µ−1 =−νεl =−ν(λ−1)

change of cross-sectional area

µ=

√

A

A0= 1−ν(λ−1) → A = A01−ν(λ−1)2

restriction of elongation

1−ν(λ−1) > 0 → λ−1 <1

ν→ λ<

1+ν

ν

Logarithmic strain

The logarithmic strain definition does not lead to unrealistic values for the contraction. There-

fore it is very suitable to describe large deformations.

logarithmic strain ε= εl n = ln(λ)

contraction strain εd = ln(µ) =−νεl n =−ν lnλ

change of cross-sectional area

µ=

√

A

A0= e−νεln = e−ν ln(λ) =

[

e ln(λ)]−ν

=λ−ν → A = A0λ−2ν

A deformation process may be executed in a number of steps, as is often done in forming

processes. The start of a new step can be taken to be the reference state to calculate current

strains. In that case the logarithmic strain is favorably used, because the subsequent strains

can be added to determine the total strain w.r.t. the initial state.

1 2

01 12

0

Figure 10.1: Two-step deformation process

170

10.1. Nonlinear axial deformation of trusses

l0→l1 εl (01)= l1−l0

l0

εl n(01) = ln( l1

l0)

l1→l2 εl (12)= l2−l1

l1

εl n(12) = ln( l2

l1)

l0→l2 εl (02)= l2−l0

l06= εl (01)+εl (12)

εl n(02) = ln( l2

l0) = ln( l2

l1

l1

l0) = εl n(01)+εl n(12)

Green-Lagrange strain

Using the Green-Lagrange strain leads again to restrictions on the elongation to prevent the

cross-sectional area to become zero.

Green-Lagrange strain ε= εgl = 12

(λ2 −1)

contraction strain εd = 12

(µ2 −1) =−νεl n =−ν12

(λ2 −1)

change of cross-sectional area

1−ν(λ2 −1) > 0 → λ<√

1+ν

ν

10.1.2 Mechanical power and stress definitions

The figure shows a tensile bar which is elongated due to the action of an axial force F . At

constant force F an infinitesimal small increase in length is associated with a change in me-

chanical energy per unit of time (power) : P = F l . The elongation rate l can be expressed in

various strain rates.

l0, A0

l , A F~e1

~e2

~e3

Figure 10.2: Axial elongation of homogeneous truss

linear strain εl =λ−1 → εl = λ=l

l0

logarithmic strain εl n = ln(λ) → εl n = λλ−1 =l

l

Green-Lagrange strain εgl = 12

(λ2 −1) → εgl = λλ=λl

l0=λ2 l

l

171

10. LARGE DEFORMATIONS & NONLINEAR BEHAVIOUR

P = F ℓ = Fℓ0εl =F

A0A0ℓ0εl =

F

A0V0εl

P = F ℓ = Fℓεl n =F

AAℓεl n =

F

AV εl n

P = F ℓ = Fℓ0εl =F

AAℓ

ℓ0

ℓεl =

F

AV λ−1εl

P = F ℓ = Fℓλ−2εgl =F

AAℓλ−2εgl =

F

AV λ−2εgl

Various stress definitions automatically emerge when the mechanical power is consid-

ered in the undeformed volume V0 = A0l0 or the current volume V = Al of the tensile bar.

The stresses are :

σ : Cauchy or true stress

σn : engineering or nominal stress

σp1 : 1st Piola-Kirchhoff stress =σn

σκ : Kirchhoff stress

σp2 : 2nd Piola-Kirchhoff stress

The resulting mechanical power in terms of these stresses reads

P = = = V0σn εl

P = V σεl n = V0(Jσ)εl n = V0σκεl n

P = V (σλ−1)εl = V0(Jσλ−1)εl = V0σp1εl

P = V (σλ−2)εgl = V0(Jσλ−2)εgl = V0σp2εgl

specific mechanical power : P =V0W0 =V W

W0 = σn εl = σκεl n = σp1εl = σp2εgl

W = = σεl n = σλ−1εl = σλ−2εgl

10.1.3 Equilibrium

Deformations may be so large that the geometry changes considerably. This and/or non-

linear boundary conditions render the deformation problem nonlinear. Proportionality and

superposition do not hold in that case. The internal force fi is a nonlinear function of the

elongation u.

Nonlinear material behavior may also result in a nonlinear function fi (u). This nonlin-

earity is almost always observed when deformation is large.

Solving the elongation from the equilibrium equation is only possible with an iterative

solution procedure.

external force fe

internal force fi =σA = fi (u)

equilibrium of point P fi (u) = fe

172

10.1. Nonlinear axial deformation of trusses

u

fi (u)

fe

uexact

Figure 10.3: Nonlinear internal load and constant external load

Iterative solution procedure

It is assumed that an approximate solution u∗ for the unknown exact solution uexact exists.

(Initially u∗ = 0 is chosen.)

The residual load r ∗ is the difference between f (u∗) and fe . For the exact solution this

residual is zero. What we want the iterative solution procedure to do, is generating better

approximations for the exact solution so that the residual becomes very small (ideally zero).

uu∗

f ∗i

fi (u)

r ∗fe

uexact

Figure 10.4: Approximation of exact solution

analytical solution fi (uexact ) = fe → fe − fi (uexact ) = 0

approximation u∗ fe − fi (u∗) = r (u∗) 6= 0

residual r ∗ = r (u∗)

The unknown exact solution is written as the sum of the approximation and an unknown

error δu. The internal force fi (uexact ) is then written as a Taylor series expansion around

u∗ and linearized with respect to δu. The first derivative of fi with respect to u is called the

tangential stiffness K ∗. Subsequently δu is solved from the linear iterative equation. The

solution is called the iterative displacement.

173

10. LARGE DEFORMATIONS & NONLINEAR BEHAVIOUR

u

δu

u∗

K ∗

f ∗i

fi (u)

fe

r ∗

Figure 10.5: Tangential stiffness and iterative solution

fi (uexact ) = fe

uexact = u∗+δu

→ fi (u∗+δu) = fe

fi (u∗)+d fi

du

∣∣∣∣u∗

δu = fe → f ∗i +K ∗δu = fe

K ∗ δu = fe − f ∗i = r ∗ → δu =

1

K ∗ r ∗

With the iterative displacement δu a new approximate solution u∗∗ can be determined

by simply adding it to the known approximation. When u∗∗ is a better approximation than

u∗, the iteration process is converging. As the exact solution is unknown, we cannot calcu-

late the deviation of the approximation directly. There are several methods to quantify the

convergence.

uexact

δu

u∗

K ∗

f ∗i

fi (u)

fe

r ∗

uu∗∗

Figure 10.6: New approximation of the exact solution

new approximation u∗∗ = u∗+δu

error uexact −u∗∗

error smaller → convergence

174

10.1. Nonlinear axial deformation of trusses

Convergence control

When the new approximation u∗∗ is better than u∗, the residual r ∗∗ is smaller than r ∗. If its

value is not small enough, a new approximate solution is determined in a new iteration step.

If its value is small enough, we are satisfied with the approximation u∗∗ for the exact solution

and the iteration process is terminated. To make this decision the residual is compared to a

convergence criterion cr . It is also possible to compare the iterative displacement δu with a

convergence criterion cu . If δu < cu it is assumed that the exact solution is determined close

enough.

When the convergence criterion is satisfied, the displacement u will not satisfy the nodal

equilibrium exactly, because the convergence limit is small but not zero. When incremental

loading is applied, the difference between fi and fe is added to the load in the next incre-

ment, which is known as residual load correction.

δu

u∗

fi (u)

r ∗∗

u

fe

f ∗∗i

u∗∗

Figure 10.7: New residual for approximate solution

residual force |r ∗∗| ≤ cr → stop iteration

iterative displacement |δu| ≤ cu → stop iteration

u

fi (u)

fe

Figure 10.8: Converging iteration process

175

10. LARGE DEFORMATIONS & NONLINEAR BEHAVIOUR

Residual and tangential stiffness

The residual and the tangential stiffness can be calculated from the material model, which

describes the material behavior. It is assumed that this is a relation between the axial Cauchy

stress σ and the elongation factor or stretch ratio λ= ll0

: σ=σ(λ). It is also necessary to now

the relation between the cross-sectional area A and λ.

internal nodal force f ∗i = N (λ∗) = A∗σ∗

tangential stiffness K ∗ =∂ fi

∂u

∣∣∣∣

u∗=

∂N (λ)

∂u

∣∣∣∣u∗

=d N

dλ

∣∣∣∣λ∗

dλ

du

geometry λ= 1+∆l

l0= 1+

1

l0u →

dλ

du=

1

l0

tangential stiffness K ∗ =d N

dλ

∣∣∣∣λ∗

∂λ

∂u=

d N

dλ

∣∣∣∣λ∗

1

l0=

d N

dλ

∣∣∣∣

∗ 1

l0=

1

l0

d

dλ(σA)

∣∣∣∣

∗

K ∗ =1

l0

dσ

dλ

∣∣∣∣

∗A∗+

1

l0σ∗ d A

dλ

∣∣∣∣

∗

Incremental loading

The external loading may be time-dependent. To determine the associated deformation,

the time is discretized : the load is prescribed at subsequent, discrete moments in time and

deformation is determined at these moments, see figure 10.9. A time interval between two

fe

t0

fi

fe

tn tn+1∆t

∆ fe

f

un un+1 u

Figure 10.9: Incremental loading

discrete moments is called a time increment and the time dependent loading is referred to as

incremental loading. This incremental loading is also applied for cases where the real time

(seconds, hours) is not relevant, but when we want to prescribe the load gradually. One can

than think of the ”time” as a fictitious or virtual time.

176

10.2. Nonlinear material models

Non-converging solution process

The iteration process is not always converging. Some illustrative examples are shown in

figure 10.10.

fi (u)

fe

u

fe

fi (u)

fi (u)

fe

Figure 10.10: Non-converging solution processes

Modified Newton-Raphson procedure

Sometimes, it is possible to reach the exact solution by modifying the Newton-Raphson iter-

ation process. The tangential stiffness is then not updated in every iteration step. Its initial

value is used throughout the iterative procedure.

Figure 10.11 shows a so-called ”snap-through” problem, where no convergence can be

reached due to a cycling full Newton-Raphson iteration process. With modified Newton-

Raphson, iteration proceeds to the equilibrium fi = fe .

10.2 Nonlinear material models

10.2.1 Material behaviour

The 1D material behaviour is modeled, using a discrete mechanical model of springs, dash-

pots and friction sliders. The axial stress is related to the axial strain by one or more (dif-

177

10. LARGE DEFORMATIONS & NONLINEAR BEHAVIOUR

u

u

fi (u)

fi (u)

fe

fe

Figure 10.11: Modified Newton-Raphson procedure

ferential) equation(s) from which the stress response must be calculated when the strain

excitation is prescribed.

The various material models are incorporated in a finite element program, which is used

to model and analyze the mechanical behavior of truss structures, subjected to prescribed

displacements and/or forces. In the iterative solution procedure, the material stiffness plays

an essential role and must be derived from the material law.

Characterization of the mechanical behavior of an unknown material almost always be-

gins with performing a tensile experiment. A stepwise change in the axial stress σ may be

prescribed and the strain ε of the tensile bar can be measured and plotted as a function of

time. From these plots important conclusions can be drawn concerning the material behav-

ior.

For elastic material behavior the strain follows the stress immediately and becomes zero

after stress release. For elasto-plastic material behavior the strain also follows the stress im-

mediately, but there is permanent deformation after stress release. When the material is

viscoelastic the strain shows time delayed response on a stress step, which indicates a time

dependent behavior. When time dependent behavior is accompanied by permanent defor-

mation, the behavior is referred to as viscoplastic.

Another way of representing the measurement data of the tensile experiment is by plot-

ting the stress against the strain, resulting in the stress-strain curve. The relation between

stress and strain may be linear or nonlinear. Also, the relation may be history dependent,

due to changes in the material structure. Different behavior in tensile and compression may

be observed.

178

10.2. Nonlinear material models

t t

ε

t2

σ

t1 t2 t1

t t

εσ

t1 t2 t1 t2

εe

εp

t t2 t

εσ

t1 t2 t1

t2t t

εσ

t1 t2 t1

Figure 10.12: Strain response for a stress-step for a) elastic b) elasto-plastic, c) viscoelastic

and d) viscoplastic material behavior

Tensile curve : elastic behavior

When elastic behavior is well described by a linear relation between stress and strain, the

elastic behavior is referred to as linear.

σ

ε

σ

ε

σ

εε

σ

Figure 10.13: Tensile curves for elastic material behavior

Tensile curve : viscoelastic behavior

Viscoelastic behavior is time-dependent. The stress is a function of the strain rate. There

is a phase difference between stress and strain, which results in a hysteresis loop when the

loading is periodic.

179

10. LARGE DEFORMATIONS & NONLINEAR BEHAVIOUR

ε

ε

σ

ε

σ

ε0

σ0

Figure 10.14: Tensile curve and hysteresis loop for viscoelastic material behavior

Tensile curve : elasto-plastic behavior

When a material is loaded or deformed above a certain threshold, the resulting deformation

will be permanent or plastic. When time (strain rate) is of no importance, the behavior is

referred to as elasto-plastic. Stress-strain curves may indicate different characteristics, espe-

cially when the loading is reversed from tensile to compressive.

σ

εε

σσ

σ

ε ε

Figure 10.15: Tensile curves for elasto-plastic material behavior

Tensile curve : viscoplastic behavior

A combination of plasticity and time-dependency is called viscoplastic behavior. This be-

havior is often observed for polymeric materials. For some polymers the stress reaches a

maximum and subsequently drops with increasing strain. This phenomenon is referred to

as intrinsic softening. In a tensile experiment it will provoke necking of the tensile bar.

180

10.2. Nonlinear material models

ε

σ

Figure 10.16: Tensile curves for viscoplastic material behavior

ε

σ

Figure 10.17: Tensile curve for damaging material with different behavior in tension and

compression

Tensile curve : damage

Structural damage influences the material properties. The onset and evolution of damage

can be described with a damage model. For materials like concrete and ceramics, the onset

and propagation of damage causes softening. Because damage is often associated with the

initiation and growth of voids, the stress-strain curve is different for tensile and compressive

loading.

10.2.2 Discrete material models

Material models relate stresses to deformation and possibly deformation rate. For three-

dimensional continua the material model is often represented by a (large) number of cou-

pled (differential) equations. In this course, nonlinear material models will be presented in

a one-dimensional setting only. The material behavior is represented by the behavior of a

one-dimensional, discrete, mechanical system of springs, dashpots and friction sliders. For

such a system the relation between the axial stress σ and the axial strain ε can be derived.

181

10. LARGE DEFORMATIONS & NONLINEAR BEHAVIOUR

σε

σε

σε

Figure 10.18: Discrete elements : spring, dashpot and friction slider

10.3 Large deformation elasticity

10.3.1 Elastic material behavior

When a material behaves elastically, the current stress can be calculated directly from the

current strain, because there is no path and/or time dependency. When the stress is released,

the strain will become zero, so there is no permanent deformation at zero stress. All stored

strain energy is released and there is no energy dissipation. For the one-dimensional case

of an axially loaded truss the elastic behavior is described by a relation between the stress σ

and the elongation factor λ or the strain ε.

unloading

loading

0

σ

λ1

Figure 10.19: Non-linear elastic material behavior

Small strain elastic behavior

For small elongations, all strain definitions are the same, as are all stress definitions. The re-

lation between stress and strain is linear and the constant material parameter is the Young’s

182

10.3. Large deformation elasticity

modulus.

strain ε= εgl = εl n = εl =λ−1

stress σ=F

A=

F

A0=σn

linear elastic behavior σ= Eε= E (λ−1)

modulus E = limλ→1

dσ

dλ= lim

ε→0

dσ

dε

Large strain elastic behavior

For large deformations, nonlinear elastic behavior can be observed in polymers, elastomeric

materials (rubbers) and, on a small scale, in atomic bonds, when a tensile/compression test

is carried out and the axial force F is plotted as a function of λ. In a material model we

want to describe such behavior with a mathematical relation between a stress and a strain.

Consideration of the stored elastic energy per unit of volume learns that each stress defini-

tion is associated with a certain strain definition, so these should be combined in a material

model. However, when the observed material behavior is described accurately by another

stress/strain combination, it can be used as well.

For three-dimensional models more considerations have to be taken into account. Care

has to be taken that the material model does not generate stresses for large rigid body rota-

tions of the material, which is known as the requirement of objectivity.

σ

0 1 λ

σ

0 1 λ

Figure 10.20: Non-linear stress-strain relations for an elastomeric material and for an atomic

bond

Elasticity models

The discrete one-dimensional model for elastic material behavior is a spring. The behavior

is modeled with a relation between the stress σ and the elongation factor λ or a strain ε. The

material stiffness Cλ is the derivative of σ w.r.t. the stretch ratio λ. The derivative w.r.t. the

strain ε results in the stiffness Cε.

183

10. LARGE DEFORMATIONS & NONLINEAR BEHAVIOUR

Consideration of the stored elastic energy per unit of material volume (see 10.1.2) learns

that, in a material model, true stress σ should be combined with logarithmic strain εl n , engi-

neering stress σn with linear strain εl or 2nd-Piola-Kirchhoff stress σp2 with Green-Lagrange

strain εgl . Experimentally observed tensile behavior can often be described with a linear re-

lation between a certain stress and its associated strain.

σε

Figure 10.21: Spring

constitutive equation σ=σ(λ)

stiffness Cλ =dσ

dλ=

dσ

dε

dε

dλ=Cε

dε

dλ

10.3.2 Hyper-elastic models

Elastomeric materials (rubbers) show very large elastic deformations (elongation up to λ =5). The material models for these materials are therefore referred to as hyper-elastic. They

are derived from an elastic energy function, which has to be determined experimentally.

The three-dimensional versions of these so-called Rivlin or Mooney models are expressed in

the principal elongation factors λi , i = 1,2,3. Experimental observations indicate that elas-

tomeric materials are incompressible, so that we have λ1λ2λ3 = 1.

W =n∑

i

m∑

j

Ci j (I1 −3)i (I2 −3) j with C00 = 0

I1 =λ21 +λ2

2 +λ23

I2 =λ21λ

22 +λ2

2λ23 +λ2

3λ21 =

1

λ23

+1

λ21

+1

λ22

The incremental change of the elastically stored energy per unit of deformed volume, can

be expressed in the principal stresses and the principal logarithmic strains.

dW =σ1dεl n1+σ2dεl n2

+σ3dεl n3

Mooney models

For incompressible materials like elastomer’s (rubber) the stored elastic energy per unit of

deformed volume is specified and fitted onto experimental data. Several specific energy

functions are used.

Neo-Hookean W =C10 (I1 −3)

184

10.3. Large deformation elasticity

Mooney-Rivlin W =C10 (I1 −3)+C01 (I2 −3)

Signiorini W =C10(I1 −3)+C01(I2 −3)+C20(I1 −3)2

Yeoh W =C10(I1 −3)+C20(I1 −3)2 +C30(I1 −3)3

Klosner-Segal W =C10(I1 −3)+C01(I2 −3)+C20(I1 −3)2 +C03(I2 −3)3

2-order invariant W =C10(I1 −3)+C01(I2 −3)+C11(I1 −3)(I2 −3)+C20(I1 −3)2

Third-order model of James, Green and Simpson

W =C10(I1 −3)+C01(I2 −3)+C11(I1 −3)(I2 −3)+C20(I1 −3)2 +C02(I2 −3)2 +C21(I1 −3)2(I2 −3)+C30(I1 −3)3 +C03(I2 −3)3 +C12(I1 −3)(I2 −3)2

Ogden models

For ’slightly’ compressible materials the Ogden specific energy functions are used. Because

the volume change is not zero, these functions depend on the volume change factor J =λ1λ2λ3. The second part of the energy function accounts for the volumetric deformation.

Because the volumetric behavior is characterized by a constant bulk modulus K , the model

is confined to slightly compressible deformation.

For ’highly’ compressible materials like foams, specific energy functions also exist. The

first part of the energy function also describes volume change.

slightly compressible W =N∑

i=1

ai

bi

[

J−bi

3

(

λbi

1 +λbi

2 +λbi

3

)

−3

]

+4.5K

(

J13 −1

)2

highly compressible W =N∑

i=1

ai

bi

(

λbi

1 +λbi

2 +λbi

3 −3)

+N∑

i=1

ai

ci

(

1− J ci)

Uniaxial behaviour

For tensile (or compressive) loading of a homogeneous and isotropic truss, where the axial

direction is taken to be the 1-direction, we have : λ1 = λ and λ2 = λ3 = 1/pλ. In this case

there is only an axial stress σ1 =σ, so that we have

dW =σdεl n → σ=dW

dεl n

=dW

dλ

dλ

dεl n

=dW

dλλ

The Neo-Hookean model is the simplest model as it contains only one material parame-

ter. Axial stress σ and axial force F can be calculated easily. From statistical mechanics it is

known that for an ideal rubber material the stress is :

σ=ρRT

M

(

λ2 −1

λ

)

with ρ : density

R : gas constant= 8.314 JK−1mol−1

T : absolute temperature

M : average molecular weight

185

10. LARGE DEFORMATIONS & NONLINEAR BEHAVIOUR

W = C10

(

λ2 + 2λ−3

)

σ = C10

(

2λ−2

λ2

)

λ= 2C10

(

λ2 −1

λ

)

Cλ =∂σ

∂λ= 2C10

(

2λ+1

λ2

)

; E = limλ→1

∂σ

∂λ= 6C10

F = σA =σ 1λ

A0 = 2C10 A0

(

λ− 1λ2

)

Most rubber materials cannot be characterized well with the Neo-Hookean model. The

more complex Mooney-Rivlin model yields better results. The stiffness Cλ is a function of

the elongation factor λ. The initial stiffness E is often referred to as the modulus.

W = C10

(

λ2 +2

λ−3

)

+C01

(1

λ2+2λ−3

)

σ = 2C10

(

λ2 −1

λ

)

+2C01

(

λ2 −1

λ

)1

λ

Cλ =∂σ

∂λ= 2C10

(

2λ+1

λ2

)

+2C01

(

1+2

λ3

)

; E = limλ→1

∂σ

∂λ= 6(C10 +C01)

F = σA =σ1

λA0 = A0

1

λ

[

2C10

(

λ2 −1

λ

)

+2C01

(

λ2 −1

λ

)1

λ

]

10.4 Examples

10.4.1 Example 1: limitations of a linear relation between the Cauchy

stress and the Green-Lagrange strain

A cylindrical tensile bar with initial length l0 and initial cross-sectional area A0 is loaded with

an axial force F . The elongation is described by the stretch ratio λ = ll0

. The contraction is

described by the stretch ratio µ=√

AA0

.

The material is homogeneous and the elastic behavior is described by a linear relation

between the Cauchy stress σ and the Green-Lagrange strain εgl = 12

(λ2 −1).

σ=C (λ2 −1) with C = constant> 0

In each direction the same strain definition must be used, so the contraction strain is

εd = 12

(µ2 −1)

When the contraction strain is related to the axial strain with Poisson’s ratio ν, we have

εd = 12

(µ2 −1) =−νεgl =−ν12

(λ2 −1) → µ2 = 1−ν(λ2 −1)

When Poisson’s ration is assumed to be constant, the axial elongation is limited because

the cross-sectional area obviously cannot become zero.

186

10.4. Examples

1−ν(λ2 −1) = 0 → ν(λ2 −1) = 1 → λ2 −1 =1

ν→ λ2 =

1+ν

ν→ λ=

√

1+ν

ν

When the cross-sectional area is plotted as a function of λ with the value ν = 0.25, we

clearly see the limit value for λ where A = 0.

0 0.5 1 1.5 2 2.5 3−1

−0.5

0

0.5

1

1.5

λ

A

Figure 10.22: Cross-sectional area versus stretch ratio.

The axial force F can be calculated and expressed as function of λ.F = σA =σµ2 A0 =σ

1−ν(λ2 −1)

A0

= C A01−ν(λ2 −1)(λ2 −1)

When we plot this relation for values ν = 0.25 and A0 = 1, it becomes clear that the pro-

posed material law has some physical inconsistencies.

0 0.5 1 1.5 2 2.5 3−8000

−7000

−6000

−5000

−4000

−3000

−2000

−1000

0

1000

λ

F

Figure 10.23: Axial force versus stretch ratio.

The volume change ratio is :

J =λµ2 =λ

1−ν(λ2 −1)

When the material is assumed to be incompressible, J = 1 and Poisson’s ratio ν cannot

be constant any more but is a function of λ :

187

10. LARGE DEFORMATIONS & NONLINEAR BEHAVIOUR

µ2 =1

λ= 1−ν(λ2 −1) →

ν(λ2 −1) = 1−1

λ=

λ−1

λ→ ν=

λ−1

λ(λ2 −1)=

1

λ(λ+1)

For very small elongations, the value of ν becomes 12

, which is already known from three-

dimensional Hooke’s law for linear elasticity.

10.4.2 Example 2: instability & localisation

A tensile test has revealed that the elastic behavior of a material is described very well by a

linear relation between the engineering stress σn and a nonlinear function of the elongation

factor λ.

σn =C1

λln(λ) with σn = engineering stress

C = elasticity constant > 0

λ = axial elongation factor

The undeformed cross-sectional area of the truss is A0. Poisson’s ratio isν and is assumed

to be constant.

The relation between the axial force F and the axial stretch factor λ is

F =C A0λ−1 ln(λ)

For values C = 1000 and A0 = 1, this relation is shown in the figure below.

0 1 2 3 4 5 6−300

−200

−100

0

100

200

300

400

λ

F

Figure 10.24: Axial force versus stretch ratio.

The value for λ for which the axial force reaches the maximum value can be determined

by differentiation.

dF

dλ=C A0

−1

λ2ln(λ)+

1

λ2

=C A01

λ2(1− ln(λ))

dF

dλ= 0 → ln(λ) = 1 → λ= e =±2.7

188

10.4. Examples

The maximum value of the force is

Fmax = F (λ= e)=C A01

eln(e) =C A0

1

e

When this maximum is reached, the deformation increases while the load diminishes.

This instability is called a snap-through. Because in a real material there will always be some

inhomogeneity, e.g. a cross-section with a slightly smaller area, or with a slightly lower value

of C , The maximum will be reached there first. All the elongation will then be concentrated

in this weakest cross-section, a phenomenon which is referred to as localisation.

10.4.3 Example 3: inflating a spherical balloon

A spherical balloon has an inner diameter D0 and a uniform wall thickness w0 ≪ D0 in the

undeformed state. The balloon is loaded with an internal pressure p whereupon it deformes

homogeneously to have an inner diameter D and wall thickness w .

To describe the mechanics of the balloon, we use three coordinate axes : two perpendic-

ular tangential directions t and the radial direction r , as is shown in the figure below for the

deformed state.

r

t

t

σt

p

Figure 10.25: Balloon in deformed state.

The wall of the balloon is made from elastomeric material, the behavior of which is char-

acterised by the Neo-Hookean elastic energy function, expressed in the principal stretch ra-

tios λ1, λ2 and λ3 :

W =C10

λ21 +λ2

2 +λ23 −3

where C10 is a positive valued material constant. The material is incompressible.

In the case of the pressurized balloon, the principal directions of deformation are the

perpendicular tangential directions and the radial direction in each point of the balloon wall,

so we have :

λ1 =λ2 =λt =D

D0=λ and λ3 =λr =

w

w0

With the knowledge that the material is incompressible, the elastic energy function can

be expressed in the tangential elongation factor λ.

J =λ1λ2λ3 =λ2t λr = 1 → λr =

1

λ2t

→

W =C10

(

λ2t +λ2

t +λ2r −3

)

=C10

(

2λ2 +1

λ4−3

)

189

10. LARGE DEFORMATIONS & NONLINEAR BEHAVIOUR

The principal stress directions coincide with the principal strain directions and are :

σ1 =σ2 =σt and σ3 =σr ≈ 0

For the isotropic hyperelastic model, the incremental specific elastic energy can be writ-

ten as

dW =σ1dεl n1+σ2dεl n2

+σ3dεl n3= 2σt dεln t → σt = 1

2

dW

dεln t

The tangential stress σt is then expressed in λ.

σt = 12

dW

dλλ= 2C10

(

λ2 −1

λ4

)

From equilibrium in the deformed state, the relation between the internal pressure p and

the tangential stress σt is derived.

p = 4σtw

D

The internal pressure can then be expressed in λ and the initial dimensions D0 and w0.

p = 4σtw

D= 8C10

w0

D0

(

λ2 −1

λ4

)1

λ3= 8C10

w0

D0

(1

λ−

1

λ7

)

The plot of p versus λ for 1 ≤ λ ≤ 6 in figure 10.26 shows a clearly nonlinear relation.

This nonlinearity comes from the large reduction of the load-carrying wall thickness and

also from the nonlinear material behavior.

1 2 3 4 5 60

0.1

0.2

0.3

0.4

0.5

0.6

0.7

λ = D/D0

p / [

8 (

C10

)∗(w

0/D0)

]

Figure 10.26: Pressure versus diameter change.

190

CH

AP

TE

R

11PLASTICITY: BASIC CONCEPTS

11.1 Elasto-plastic material behavior

Below a certain load (stress) value, the deformation of all materials will be elastic. When the

stress exceeds a limit value, plastic deformation occurs, which means that permanent elon-

gation is observed after release of the load. At increased loading above the limit value, the

stress generally increases with increasing elongation, a phenomenon referred to as harden-

ing.

Reversed loading will first result in elastic deformation, but after reaching a limit value of

the stress, plastic deformation will be observed again. Looking at the stress-strain curve after

a few loading reversals, it can be seen that elasto-plastic material behavior is history depen-

dent: the stress is not uniquely related to the strain; its value depends on the deformation

history. The total stress-strain history must be taken into account to determine the current

stress.

11.1.1 Uniaxial test

To investigate the characteristics of elasto-plastic material behavior, a uniaxial tensile/com-

pression test is carried out.

Tensile test

When a tensile bar, with undeformed length l0 and cross-sectional area A0, is subjected to

a tensile test, the axial force F and the length l can be measured. The axial strain ε can be

calculated from the elongation factor λ. To calculate the true stress σ= FA

, the actual cross-

sectional area of the tensile bar must be measured during the experiment. The nominal

stress σn = FA0

can be calculated straightforwardly.

The nominal stress σn can be plotted against the linear strain εl =λ−1 = l−l0

l0= ∆l

l0result-

ing in the σn −εl stress-strain curve.

Until the proportionality limit σn = σP is reached, the material behavior is assumed to

be linear elastic : σn = Eε, where E is Young’s modulus. When the stress exceeds the initial

191

11. PLASTICITY

σ

εε

σσ

σ

ε ε

Figure 11.1: Stress-strain curves for elasto-plastic material behavior

yield stress σy0 >σP , unloading will reveal permanent (= plastic) deformation of the bar. The

exact value of σy0 cannot be determined so in practice σy0 is taken to be the stress where a

plastic strain of 0.2 % remains. In the following however, we will assume that σy0 is exactly

known and that σy0 =σP .

The axial force and therefore the nominal stress will reach a maximum value. At that

point necking of the tensile bar will be observed. The maximum nominal stress is the tensile

strength σT . In forming processes strains can be much higher than in a tensile test, because

of the compression in certain directions.

After reaching the tensile strength the nominal stress will decrease while the strain is still

increasing. Fracture occurs at the fracture stress σn =σF . The fracture strain εF is for metals

and metal alloys about 10% = 0.1. This is a rather small elongation which means that for

these materials we can assume σ=σn and also that all strain definitions are approximately

equivalent, so ε= εl .

Plastic deformation in metals is typically carried by crystallographic slip in the crystal

lattice. This physical deformation process does not entail any volume changes, as confirmed

by experiments. Plastic deformation in metals is therefore incompressible.

Compression test

For metal alloys a compression test instead of a tensile test will reveal that first yield will

occur at σ = σn = −σy0. The initial material behavior is the same in tension and compres-

sion. In general terms the transition from purely elastic behavior to elasto-plastic behavior

is determined by a yield criterion. For the one-dimensional case this criterion says that first

yielding will occur when :

f =σ2 −σ2y0 = 0

The function f is the yield function.

192

11.1. Elasto-plastic material behavior

T

F

ε= 0.002 εℓ

σn

σP

σy0

Figure 11.2: Stress-strain curve during tensile test

σ

ε

σy0

−σy0

Figure 11.3: Stress-strain curve during tensile or compression test

Interrupted tensile test

When the axial load is released at σA (see figure below) with σy0 < σA < σT , the unloading

stress-strain path is elastic and characterized by the initial Young’s modulus E . The perma-

nent or plastic elongation is represented by the plastic strain εp . The difference between the

total strain in point A and the plastic strain is the elastic strain εe = εA −εp = σA

E.

Resumed tensile test

When after unloading, the bar is again loaded with a tensile force, the elastic line B A will

be followed where ∆σ = E∆ε = E∆εe holds. For σ ≥ σA(ε ≥ εA) further elasto-plastic de-

formation takes place and the stress-strain curve will be followed as if unloading were not

occurred.

193

11. PLASTICITY

t t

εσ

t1 t2 t1 t2

εe

εp

ε

σ

σy0

εeεp

εA

A

B

σy

Figure 11.4: Stress-strain curve after interrupted tensile test

The stress σA is the current yield stress σy , which is generally larger then the initial yield

stress σy0. The increase, referred to as hardening, is related to the plastic strain by a harden-

ing law.

11.1.2 Hardening

To study the hardening phenomenon, the tensile bar is not reloaded in tension but in com-

pression. Two extreme observations may be made, illustrated in the figure below.

In the first case the elastic trajectory increases in length due to plastic deformation :

A A′ > Y0Y ′0 . The elastic trajectory is symmetric about σ= 0 (B A = B A′). What we observe is

isotropic hardening.

In the second case the elastic trajectory remains of constant length : A A′ = Y0Y ′0 . It is

symmetric about the line OC (C A =C A′). After unloading the yield stress under compression

is different than the yield stress under tension. This is called kinematic hardening. The stress

in point C , the center of the elastic trajectory, is the shift stress σ = q . This phenomenon is

also referred to as the Bauschinger effect.

Real materials will show a combination of isotropic and kinematic hardening.

194

11.1. Elasto-plastic material behavior

A

B

A′

Y0

Y ′0

0

ε

σ

q

A′

Aσ

0

Y ′0

Y0

B

C

ε

σy0

Figure 11.5: Isotropic and kinematic hardening

isotropic hardening : elastic area larger & symmetric w.r.t. σ= 0

tensile : σ=σy

compression : σ=−σy

→ f =σ2 −σ2y = 0

kinematic hardening : elastic area constant & symmetric w.r.t. σ= q

tensile : σ= q +σy0

compression : σ= q −σy0

→ f = (σ−q)2 −σ2y0 = 0

combined isotropic/kinematic hardening

tensile : σ= q +σy

compression : σ= q −σy

→ f = (σ−q)2 −σ2y = 0

Effective plastic strain

Isotropic hardening could be described by relating the yield stress σy to the plastic strain εp .

However, as the figure below shows, this would lead to the unrealistic conclusion that the

yield stress increases while the plastic strain decreases. To prevent this problem, the effective

plastic strain εp is taken as the history parameter. It is a measure of the total plastic strain,

be its change positive or negative, and as such cannot decrease.

εp =∑

ε

|∆εp | =τ=t∑

τ=0

|∆εp |∆t

∆t =∫t

τ=0|εp |dτ=

∫t

τ=0

˙εp dτ

Hardening laws

For one-dimensional stress states encountered in the axial loading of a truss, several hard-

ening laws are formulated, based on experimental observations. They can be generalized

to three-dimensional stress-strain states. For isotropic hardening the current yield stress is

related to the effective plastic strain and the initial yield stress. The isotropic hardening pa-

rameter is H = dσy

d εp. For kinematic hardening the shift stress q is related to the plastic strain

εp . The kinematic hardening parameter is K = dq

dεp.

195

11. PLASTICITY

ε

t

A

B

C

D

C

E

E

σ

ε

A

B

D

Figure 11.6: Increasing yield stress at decreasing plastic strain

Linear and power law hardening laws

Many hardening laws represent a linear or exponential relationship between stress and

strain.

linear hardening σy =σy0 +H εp

Ludwik (1909) σy =σy0 +σy0

(εp

εy0

)n

(0 ≤ n ≤ 1) →

H = nσy0

εy0

(εp

εy0

)n−1

= nE

(εp

εy0

)n−1

mod. Ludwik σy =σy0

(

1+mεnp

)

→ H =σy0mnεn−1p

Swift (1952) σy =C (m + εp )n with C =σy0

mn

H =C n(

m + εp

)n−1

Ramberg-Osgood (1943) εp =σy

E

[

1+α

(σy

σy0

)m−1]

(m ≥ 0;α≈ 37

)

Asymptotically perfect hardening laws

Some hardening laws are formulated in such a way as to result in no hardening (ideal

plastic behavior) for large strain values.

ideal plastic σy =σy0

196

11.2. Elasto-plastic model

Prager (1938) σy =σy0 tanh

(E εp

σy0

)

H =σy0

εy0

[

sech

(εp

εy0

)]2

= E

[

sech

(εp

εy0

)]2

Betten I (1975) σy =σy0

[

tanh

(E εp

σy0

)m]1/m

(m > 1)

H = E

(εp

εy0

)m−1 [

tanh

(εp

εy0

)m] 1m

−1 [

sech

(εp

εy0

)m]2

Voce (1949) σy =C(

1−ne−mεp)

with C =σy0

1−n(m > 1)

H =C nme−mεp

Betten II (1975) σy =σy0 + (E εp )

[

1+(εp

εy0

)m]−1/m

H = E

[

1+(εp

εy0

)m]− 1m

[

1−(εp

εy0

)m

1+(εp

εy0

)m−1]

Cyclic load for linear hardening

A truss can be loaded with a prescribed strain −εm ≤ ε≤ εm . It is assumed that the stress will

reach values above the initial yield stress σy0 and that linear hardening occurs.

For purely isotropic hardening the stress will increase after each load reversal and finally

no further plastic deformation will take place. For purely kinematic hardening the stress-

strain path will be one single hysteresis loop, where the stress cycles, as does the strain, be-

tween two constant values −σm ≤σ≤σm .

11.2 Elasto-plastic model

The elasto-plastic deformation characteristics can be represented by a discrete mechanical

model. A friction element represents the yield limit and a hardening spring – stiffness H

(H > 0) – provides the stiffness reduction after reaching the yield limit. The elasto-plastic

model describes rate-independent plasticity – there is no dashpot in the discrete model –, so

the time is fictitious and ”rate” is just referring to momentary change.

The yield criterion is used to decide at which stress level a purely elastic deformation

will be followed by elasto-plastic deformation. During elasto-plastic deformation the total

strain rate (ε) is additively decomposed in an elastic (εe ) and a plastic (εp ) part. The plastic

strain rate εp is related to∂ f

∂σ by the rate of the plastic multiplier λ, the so-called consistency

parameter λ. During ongoing plastic deformation the consistency equation f = 0 must be

satisfied, because f must remain zero.

The hardening law relates the current yield stress σy to the initial yield stress σy0 and the

effective plastic strain εp . The shift stress q is related to the plastic strain εp .

197

11. PLASTICITY

εy0

σy0

εm

σ1

−εmεm

σm

εy0

σy0

−εm

−σm

Figure 11.7: Stress-strain curve during cyclic loading for isotropic and for kinematic harden-

ing

H

σy

εeεp

σE

Figure 11.8: Discrete mechanical model for elasto-plastic material behavior

198

11.2. Elasto-plastic model

• f = (σ−q)2 −σ2y with f < 0 | f = 0 ∧ f < 0 → elastic

f = 0 ∧ f = 0 → elasto-plastic

• σy =σy (σy0, εp ) ; q = q(εp )

• ε= εe + εp

• σ= Eεe → εe =1

Eσ

• εp = λ∂ f

∂σ= 2λ(σ−q) ; ˙εp = |εp | = 2λ |σ−q |

• εp =∫t

τ=0

˙εp dτ=∑

t

|∆εp |

Constitutive equations

From the constitutive relations a set of constitutive equations can be derived.

σ= E εe = E (ε− εp ) = E ε−2λ(σ−q)

f = 0

→

σ+2E (σ−q)λ−E ε= 0

f = 0

Integration of the constitutive equations

The current stress has to be determined from these constitutive equations. The first one is

a differential equation in pseudo-time. To solve it, we need an initial condition, which is

not the stress in the undeformed state, which would obviously be zero, but the stress at the

beginning of the current increment. Values at the beginning if the current increment are

indicated with an index n. A value at the end of this increment, which is the current state, is

often indicated with the index n +1, but is skipped here.

There are many procedures which can be followed to solve the differential equation for

the stress. They can be classified as implicit or explicit. The implicit methods are more ac-

curate and more stable then the explicit methods.

We assume that the begin-increment state resides on the yield trajectory, so fn = 0. In

reality this is of course not always the case : the begin-increment case may be elastic ( fn < 0)

and plastic deformation will develop during the increment. The implicit procedures can well

cope with this phenomenon. Explicit procedures will need some correction.

Explicit solution procedure

An explicit procedure starts from the known state at the beginning of the increment and

calculates incremental changes directly, assuming that values of some variables remain the

same during the increment. Obviously, this is not through, so these procedures are not very

accurate. The final solution may not satisfy the yield criterion f = 0 exactly, which calls for

a correction, where the final state is projected onto the yield trajectory. Here we ignore this

inaccuracy and calculate only the material stiffness Cε = ∆σ∆ε

during the increment. Because

the begin-increment state is assumed to be on the yield-trajectory, we know that |σn −qn | =σyn .

199

11. PLASTICITY

∆σ+2E (σn −qn)∆λ= E∆ε

∆ f = 0 →∂ f

∂σ

∣∣∣∣n

∆σ+∂ f

∂λ

∣∣∣∣

n

∆λ= 0

∆σ+2E (σn −qn)∆λ= E∆ε

2(σn −qn)∆σ−4Kn(σn −qn)2∆λ−4Hnσyn|σn −qn |∆λ= 0 →

∆λ=(σn −qn)

2Kn(σn −qn)2 +2Hnσyn|σn −qn |∆σ=

1

2Kn(σn −qn)+2Hn(σn −qn)∆σ

→

∆σ=E (Kn +Hn)

E +Kn +Hn∆ε ; ∆εp = 2(σn −qn)∆λ=

E

E +Kn +Hn∆ε

Linear isotropic hardening

For linear isotropic hardening with constant hardening parameter H , the above stiffness re-

duces to a simpler relation.

∆σ=E H

E +H∆ε ; ∆εp =

1

H∆σ=

E

E +H∆ε

These relations can also easily be derived by using the figure below. In a point of the

elastic trajectory we know that ∆σ= E∆ε holds. In a point of the elasto-plastic trajectory we

can write ∆σ= S∆ε, where the material stiffness S =Cε will depend on E and H .

∆εe

∆σ

σ

ε

∆ε

A

B

∆εp

Figure 11.9: Stress-strain curve for monotonic tensile loading

∆σ = E∆εe = E (∆ε−∆εp ) = E

(

∆ε−∆σy

H

)

= E

(

∆ε−∆σ

H

)

→

∆σ =E H

E +H∆ε= S∆ε ; ∆εp =

∆σ

H=

E

E +H∆ε

200

11.3. Examples

Kinematic hardening

For linear kinematic hardening, the result is similarly derived.

∆σ=E K

E +K∆ε ; ∆εp =

1

K∆σ=

E

E +K∆ε

Again these relations can be derived straightforwardly from the figure.

∆σ = E∆εe = E (∆ε−∆εp ) = E

(

∆ε−∆q

K

)

= E

(

∆ε−∆σ

K

)

→

∆σ =E K

E +K∆ε= S∆ε ; ∆εp =

∆σ

K=

E

E +K∆ε

Note that the stiffness equals Young’s modulus when H (or K ) approaches infinity.

limH→∞

E H

E +H= lim

H→∞

EEH+1

= E

11.3 Examples

11.3.1 Example 1: Reversed plasticity in a tensile bar

A cylindrical tensile bar is loaded with an axial stress σ, which is applied as a function of the

(pseudo)time t as indicated in the figure below.

0 t

σ(t )

σ1

2t1t1

Figure 11.10: Prescribed stress as a function of (pseudo)time.

When the load is increased from t = 0, the material behaves linearly elastic – Young’s

modulus is E – until the initial yield stress σy0 is reached. The maximum stress σ1 is such

that σ1 >σy0. After reaching this maximum value the stress is reduced to zero and compres-

sive loading is applied. After yielding, the material shows linear, isotropic hardening with

hardening constant H .

First the stress-strain diagram is plotted and relevant stress-strain points are indicated.

201

11. PLASTICITY

σ

σy0

σ1

ε1 εεp1εy0

ε2

σ2

σ3

0

1

2

3

Figure 11.11: Stress-strain diagram.

When the stress reaches the maximum value σ1 at t = t1 the strain ε1 can be calculated

and expressed in σ1, E , σy0 and H .

ε1 = εy0 +E +H

E H(σ1 −σy0) =

σy0

E+

E +H

E H(σ1 −σy0)

=σy0

E

(

1−E +H

H

)

+E +H

E Hσ1

= −σy0

H+

E +H

E Hσ1

The plastic strain εp

1 at maximum stress σ1 is

εp

1 = ε1 −σ1

E=−

σy0

H+

(E +H

H−1

)σ1

E

=σ1 −σy0

H

From t = t1 the stress is reduced to zero after which compression occurs. The strain ε2 at

yield under compressive loading is :

ε2 = ε1 −2σ1

E

The stress σ3 for which the plastic strain is zero, is :

εp3 = 0 → − (σ3 −σ2) =σ1 −σy0 = Hε

p1 → σ3 =−2σ1 +σy0

202

11.3. Examples

2A

L A

δ

F

L

b

a

Figure 11.12: Parallel truss structure.

11.3.2 Example 2: parallel truss structure

Figure 11.12 shows two parallel bars a and b, which are connected to a rigid wall and to a

rigid block, which can only translate in horizontal direction. The bars have the same initial

length L. The cross-sectional area of a and b is A and 2A, respectively. Both a and b have

the same Young’s modulus E and initial yield stress σy0. Upon yielding bar a is ideal plastic,

while bar b shows isotropic linear hardening with hardening constant H .

The axial deformation is provoked by the force F , which increases to a maximum value

after which it will be reduced to zero. The maximum is such that both trusses will undergo

elasto-plastic deformation. The displacement of the rigid block is δ.

For all deformation states the strain in both trusses will be equal : εa = εb , due to the

fact that their initial length and their elongation is the same. With this in mind we draw the

stress-strain diagram of both bars below each other in figure 11.13.

The force F0 at which yielding occurs for the first time, is

F0 =σy0 A+2σy0 A = 3σy0 A

The strain ε0 and the displacement δ0 is

ε0 = εy0 =σy0

E→ δ0 = L

σy0

E

The force F is increased to F = F1 = 53

F0. Because truss a doesn’t show any hardening,

the axial stress does not change : σa1 =σa

0 =σy0, and thus F a1 =σy0 A. The axial force and the

stress in truss b is then :

F b1 = F1 −F a

1 = 5σy0 A−σy0 A = 4σy0 A → σb1 = 2σy0

The strain εb1 in truss b and the displacement δ1 can now be calculated.

εb1 =

σy0

E+

E +H

E Hσy0 =

σy0

E

(

1+E +H

H

)

=σy0

E

(E +2H

H

)

δ1 = Lεb1 = L

σy0

E

(E +2H

H

)

203

11. PLASTICITY

ε

1

2

0

0

2σy0

σy0

σy0

σ1

2

a

b

Figure 11.13: Stress-strain diagrams of both trusses.

The plastic strain of the trusses is

εap1 = εa

1 −σy0

E=

E +H

E Hσy0

εbp1 = εb

1 −2σy0

E=−

σy0

E+

E +H

E Hσy0 =

σy0

E

(

−1+E +H

H

)

=σy0

H

The force F is now reduced to zero. The residual stresses in both trusses can be calcu-

lated.

εa2 −εa

1 =1

E

(

σa2 −σa

1

)

εb2 −εb

1 =1

E

(

σb2 −σb

1

)

εa2 = εb

2 ; εa1 = εb

1

→ σa2 −σa

1 =σb2 −σb

1

σa1 =σy0 ; σb

1 = 2σy0

→

σa2 −σy0 =σb

2 −2σy0 → σb2 −σa

2 =σy0

equilibrium → σa2 A+σb

2 2A = 0 → σa2 =−2σb

2

→

3σb2 =σy0 → σb

2 = 13σy0 ; σa

2 =−23σy0

11.3.3 Example 3: serial truss structure

A cylindrical truss is divided in two parts a and b as shown in figure 11.14. In the unde-

formed state, part a has length L and cross-sectional area 2A, part b has length 2L and cross-

sectional area A. In point R an axial force F ia applied. Point P is fixed and point R will show

a displacement δ.

For each state in the deformation process, we will have :

204

11.3. Examples

2A

F

ba

2LL

PQ RA

Figure 11.14: Series configuration of two trusses.

N a = N b → σa = 12σb

When the load is increased from F = 0 at t = 0 the material is linearly elastic – Young’s

modulus E – as long as the axial stress is below the initial yield stress σy0. After yielding the

material will show linear isotropic hardening with hardening constant H , see figure 11.15.

0a

0b

εbpεa

p

2a

2b

σy0

65σy0

2σy0

125σy0

σ

ε

1a

1b

Figure 11.15: Stress-strain diagram of the two trusses.

When yielding occurs for the first time the force F0 is part b will yield first at a force

F0 =σy0 A

The displacement δ0 at that moment is

σb0 =σy0 → εb

0 =σy0

E→ ∆Lb

0 = 2Lσy0

E

σa0 = 1

2σy0 → εa

0 =σy0

2E→ ∆La

0 = Lσy0

2E

→ δ0 =∆La0 +∆Lb

0 = 52

Lσy0

E

The load is increased to F = F1 > F0 when part a will yield for the first time.

205

11. PLASTICITY

σa1 =σy0 → σb

1 = 2σy0 → F1 = 2σy0 A

The displacement δ1 can than be calculated :

σa1 =σy0 → εa

1 =σy0

E→ ∆La

1 = Lσy0

E

εb1 = εb

y0 +E +H

E Hσy0 =

σy0

E+

E +H

E Hσy0 =

σy0

E

(E +2H

H

)

→ ∆Lb1 = 2L

σy0

E

(E +2H

H

)

→

δ1 =∆La1 +∆Lb

1 = Lσy0

E+2L

σy0

E

(E +2H

H

)

= Lσy0

E

(2E +5H

H

)

The force F is increased further to F = F2 = 65

F1 = 125σy0 A and is subsequently reduced

to zero : F = F3 = 0. The residual plastic strain in each part can be determined.

σb2 =

F2

A= 12

5σy0 → εb

2 −εb1 =

E +H

E H(σb

2 −σb1 ) → εb

2 = 15

σy0

E H(7E +12H)

σa2 =

F2

2A= 6

5σy0 → εa

2 −εa1 =

E +H

E H(σa

2 −σa1 ) → εa

2 = 15

σy0

E H(E +6H)

εbp2 = εb

3 = εb2 − 12

5

σy0

E= 7

5

σy0

H; εa

2p = εa3 = εa

2 − 65

σy0

E= 1

5

σy0

H

The total displacement after unloading is

δ3 =∆La3 +∆Lb

3 = L 15

σy0

H+2L 7

5

σy0

H= 3L

σy0

H

206

AP

PE

ND

IX

ASTIFFNESS AND COMPLIANCE MATRICES

In chapter 5 the three-dimensional stiffness and compliance matrices have been derived for

various materials. Increasing microstructural lattice symmetry gave rise to a reduction of

the number of material constants. Starting from triclinic with no symmetry and character-

ized by 21 material constants, increased symmetry was seen for monoclinic (13 constants),

orthotropic (9), quadratic (6), transversal isotropic (5), cubic (3) and finally, isotropic, with

only 2 material constants.

In this appendix, we again present the material matrices for orthotropic, transversal isotropic

and fully isotropic material. The material constants will be expressed in engineering con-

stants, where we choose Young’s moduli, Poisson’s ratios and shear moduli.

In many engineering problems, the state of strain or stress is planar. Both for plane strain

and plane stress, only the strain and stress components in a plane have to be related through

a material law. Here we assume that this plane is the 12-plane. For plane strain we than

have ε33 = γ23 = γ31 = 0, and for plane stress σ33 =σ23 = σ31 = 0. The material law for these

planar situations can be derived from the linear elastic three-dimensional stress-strain rela-

tion. This is done, first for the general orthotropic material law. The result is subsequently

specified in engineering parameters for orthotropic, transversal isotropic and fully isotropic

material.

A.1 General orthotropic material law

The general orthotropic material law is expressed by the stiffness matrix C and/or its inverse,

the compliance matrix S.

207

A. STIFFNESS AND COMPLIANCE MATRICES

σ~ =

σ11

σ22

σ33

σ12

σ23

σ31

=

A Q R 0 0 0

Q B S 0 0 0

R S C 0 0 0

0 0 0 K 0 0

0 0 0 0 L 0

0 0 0 0 0 M

ε11

ε22

ε33

γ12

γ23

γ31

=C ε~

ε~ =

ε11

ε22

ε33

γ12

γ23

γ31

=

a q r 0 0 0

q b s 0 0 0

r s c 0 0 0

0 0 0 k 0 0

0 0 0 0 l 0

0 0 0 0 0 m

σ11

σ22

σ33

σ12

σ23

σ31

=C−1σ~ = Sσ~

The inverse of C can be expresses in its components.

C−1 =1

∆c

BC −S2 −QC +RS QS −BR 0 0 0

−QC +RS AC −R2 −AS +QR 0 0 0

QS −BR −AS +QR AB −Q2 0 0 0

0 0 0 ∆c (1/K ) 0 0

0 0 0 0 ∆c (1/L) 0

0 0 0 0 0 ∆c (1/M)

with ∆c = ABC − AS2 −BR2 −CQ2 +2QRS

As will be clear later, it will mostly be easier to start with the compliance matrix and cal-

culate the stiffness matrix by inversion.

S−1 =1

∆s

bc − s2 −qc + r s qs −br 0 0 0

−qc + r s ac − r 2 −as +qr 0 0 0

qs −br −as +qr ab −q2 0 0 0

0 0 0 ∆s (1/k) 0 0

0 0 0 0 ∆s (1/l ) 0

0 0 0 0 0 ∆s (1/m)

with ∆s = abc −as2 −br 2 −cq2 +2qr s

Increasing material symmetry leads to a reduction in material parameters.

quadratic B = A ; S = R ; M = L;

transversal isotropic B = A ; S = R ; M = L ; K = 12

(A−Q)

cubic C = B = A ; S = R =Q ; M = L = K 6= 12

(A−Q)

isotropic C = B = A ; S = R =Q ; M = L = K = 12

(A−Q)

The planar stress-strain laws can be derived either from the stiffness matrix C or from

the compliance matrix S. The plane strain state will be denoted by the index ε and the plane

stress state will be indicated with the index σ.

208

A.1. General orthotropic material law

A.1.1 Plane strain

For a plane strain state with ε33 = γ23 = γ31 = 0, the stress σ33 can be expressed in the planar

strains ε11 and ε22. The material stiffness matrix Cε

can be extracted directly from C . The

material compliance matrix Sε

has to be derived by inversion.

ε33 = γ23 = γ31 = 0 → σ33 = Rε11 +Sε22

σ~ =

σ11

σ22

σ12

=

A Q 0

Q B 0

0 0 K

ε11

ε22

γ12

=

Aε Qε 0

Qε Bε 0

0 0 K

ε11

ε22

γ12

=Cεε~

ε~ =

ε11

ε22

γ12

=1

AB −Q2

B −Q 0

−Q A 0

0 0AB −Q2

K

σ11

σ22

σ12

=

aε qε 0

qε bε 0

0 0 k

σ11

σ22

σ12

= Sεσ~

Because the components of the three-dimensional compliance matrix S are most con-

veniently expressed in Young’s moduli, Poisson’s ratios and shear moduli, this matrix is a

good starting point to derive the planar matrices for specific cases. The plane strain stiffness

matrix Cε

must then be determined by inversion.

ε33 = 0 = rσ11 + sσ22 +cσ33 → σ33 =−r

cσ11 −

s

cσ22

ε~ =

ε11

ε22

γ12

=

a q 0

q b 0

0 0 k

σ11

σ22

σ12

−

r

s

0

[ r

c

s

c0

]

σ11

σ22

σ12

=1

c

ac − r 2 qc − r s 0

qc − sr bc − s2 0

0 0 kc

σ11

σ22

σ12

=

aε qε 0

qε bε 0

0 0 k

σ11

σ22

σ12

= Sεσ~

σ~ =

σ11

σ22

σ12

=

aε qε 0

qε bε 0

0 0 k

−1

ε11

ε22

γ12

=1

∆s

bc − s2 −qc + r s 0

−qc + r s ac − r 2 0

0 0∆s

k

ε11

ε22

ε12

with ∆s = abc −as2 −br 2 −cq2 +2qr s

=

Aε Qε 0

Qε Bε 0

0 0 K

ε11

ε22

ε12

=Cεσ~

We can now derive by substitution :

σ33 =−1

∆s

[

(br −qs)ε11 + (as −qr )ε22

]

209

A. STIFFNESS AND COMPLIANCE MATRICES

A.1.2 Plane stress

For the plane stress state, with σ33 =σ23 =σ31 = 0, the two-dimensional material law can be

easily derived from the three-dimensional compliance matrix Sε. The strain ε33 can be di-

rectly expressed in σ11 and σ22. The material stiffness matrix has to be derived by inversion.

σ33 =σ23 =σ31 = 0 → ε33 = rσ11 + sσ22

ε~ =

ε11

ε22

γ12

=

a q 0

q b 0

0 0 k

σ11

σ22

σ12

=

aσ qσ 0

qσ bσ 0

0 0 k

σ11

σ22

σ12

= Sσσ~

σ~ =

σ11

σ22

σ12

=1

ab −q2

b −q 0

−q a 0

0 0ab −q2

k

ε11

ε22

γ12

=

Aσ Qσ 0

Qσ Bσ 0

0 0 K

ε11

ε22

γ12

=Cσε~

We can derive by substitution :

ε33 =1

ab −q2

[

(br −qs)ε11 + (as −qr )ε22

]

The same relations can be derived from the three-dimensional stiffness matrix C .

σ33 = 0= Rε11 +Sε22 +Cε33 → ε33 =−R

Cε11 −

S

Cε22

σ~~=

σ11

σ22

σ12

=

A Q 0

Q B 0

0 0 K

ε11

ε22

γ12

−

R

S

0

[R

C

S

C0

]

ε11

ε22

γ12

=1

C

AC −R2 QC −RS 0

QC −SR BC −S2 0

0 0 KC

ε11

ε22

γ12

=

Aσ Qσ 0

Qσ Bσ 0

0 0 K

ε11

ε22

γ12

=Cσε~

ε~ =

ε11

ε22

γ12

=

Aσ Qσ 0

Qσ Bσ 0

0 0 K

−1

σ11

σ22

σ12

=

aσ qσ 0

qσ bσ 0

0 0 k

σ11

σ22

σ12

= Sσσ~

A.1.3 Plane strain/stress

In general we can write the stiffness and compliance matrix for planar deformation as a 3 × 3

matrix with components, which are specified for plane strain (p = ε) or plane stress (p =σ).

Cp=

Ap Qp 0

Qp Bp 0

0 0 K

; Sp=

ap qp 0

qp bp 0

0 0 k

The general relations presented before can be used to calculate the components of Cp

and/or Sp

when components of the three-dimensional matrices C and/or S are known.

210

A.2. Linear elastic orthotropic material

In the next sections the three-dimensional and planar material matrices are presented for

orthonormal, transversal isotropic and fully isotropic material.

A.2 Linear elastic orthotropic material

For an orthotropic material 9 material parameters are needed to characterize its mechanical

behavior. Their names and formal definitions are :

Young’s moduli : Ei =∂σi i

∂εi i

Poisson’s ratios : νi j =−∂ε j j

∂εi i

shear moduli : Gi j =∂σi j

∂γi j

The introduction of these parameters is easily accomplished in the compliance matrix S.

Due to the symmetry of the compliance matrix S, the material parameters must obey the

three Maxwell relations.

S =

E−11 −ν21E−1

2 −ν31E−13 0 0 0

−ν12E−11 E−1

2 −ν32E−13 0 0 0

−ν13E−11 −ν23E−1

2 E−13 0 0 0

0 0 0 G−112 0 0

0 0 0 0 G−123 0

0 0 0 0 0 G−131

withν12

E1=

ν21

E2;

ν23

E2=

ν32

E3;

ν31

E3=

ν13

E1(Maxwell relations)

The stiffness matrix C can then be derived by inversion of S.

C =1

∆s

1−ν32ν23

E2E3

ν31ν23+ν21

E2E3

ν21ν32+ν31

E2E30 0 0

ν13ν32+ν12

E1E3

1−ν31ν13

E1E3

ν12ν31+ν32

E1E30 0 0

ν12ν23+ν13

E1E2

ν21ν13+ν23

E1E2

1−ν12ν21

E1E20 0 0

0 0 0 ∆sG12 0 0

0 0 0 0 ∆sG23 0

0 0 0 0 0 ∆sG31

with ∆s =1−ν12ν21 −ν23ν32 −ν31ν13 −ν12ν23ν31 −ν21ν32ν13

E1E2E3

A.2.1 Voigt notation

In composite mechanics the so-called Voigt notation is often used, where stress and strain

components are simply numbered 1 to 6. Corresponding components of the compliance

(and stiffness) matrix are numbered accordingly. However, there is more to it than that. The

sequence of the shear components is changed. We will not use this changed sequence in the

following.

211

A. STIFFNESS AND COMPLIANCE MATRICES

σ~T = [σ11 σ22 σ33 σ12 σ23 σ31] = [σ1 σ2 σ3 σ6 σ4 σ5]

ε~T = [ε11 ε22 ε33 γ12 γ23 γ31] = [ε1 ε2 ε3 ε6 ε4 ε5]

ε1

ε2

ε3

ε4

ε5

ε6

=

S11 S12 S13 0 0 0

S21 S22 S23 0 0 0

S31 S32 S33 0 0 0

0 0 0 S44 0 0

0 0 0 0 S55 0

0 0 0 0 0 S66

σ1

σ2

σ3

σ4

σ5

σ6

A.2.2 Plane strain

For plane strain the stiffness matrix can be extracted from the three-dimensional stiffness

matrix. The inverse of this 3x3 matrix is the plane strain compliance matrix.

σ33 = ν13E3

E1σ11 +ν23

E3

E2σ22

Sε=

1−ν31ν13

E1−ν31ν23+ν21

E20

−ν13ν32+ν12

E1

1−ν32ν23

E20

0 0 1G12

Cε= S−1

ε=

1

∆s

1−ν32ν23

E2E3

ν31ν23+ν21

E2E30

ν13ν32+ν12

E1E3

1−ν31ν13

E1E30

0 0 ∆sG12

with ∆s =1−ν12ν21 −ν23ν32 −ν31ν13 −ν12ν23ν31 −ν21ν32ν13

E1E2E3

σ33 =1

∆s

ν12ν32 +ν13

E1E2ε11 +

ν21ν13 +ν23

E1E2ε22

A.2.3 Plane stress

For plane stress the compliance matrix can be extracted from the three-dimensional com-

pliance matrix. The inverse of this 3x3 matrix is the plane strain stiffness matrix.

ε33 =−ν13E−11 σ11 −ν23E−1

2 σ22

Sσ=

E−11 −ν21E−1

2 0

−ν12E−11 E−1

2 0

0 0 G−112

Cσ= S−1

σ=

1

1−ν21ν12

E1 ν21E1 0

ν12E2 E2 0

0 0 (1−ν21ν12)G12

ε33 =−1

1−ν12ν21(ν12ν23 +ν13)ε11 + (ν21ν13 +ν23)ε22

212

A.3. Linear elastic transversal isotropic material

A.3 Linear elastic transversal isotropic material

Considering an transversally isotropic material with the 12-plane isotropic, the Young’s mod-

ulus Ep and the Poisson’s ratio νp in this plane can be measured. The associated shear mod-

ulus is related by Gp =Ep

2(1+νp ). In the perpendicular direction we have the Young’s modu-

lus E3, the shear moduli G3p = Gp3 and two Poisson ratios, which are related by symmetry :

νp3E3 = ν3p Ep .

S =

E−1p −νp E−1

p −ν3p E−13 0 0 0

−νp E−1p E−1

p −ν3p E−13 0 0 0

−νp3E−1p −νp3E−1

p E−13 0 0 0

0 0 0 G−1p 0 0

0 0 0 0 G−1p3 0

0 0 0 0 0 G−13p

withνp3

Ep=

ν3p

E3

C = S−1 =1

∆s

1−ν3pνp3

Ep E3

ν3pνp3+νp

Ep E3

νpν3p+ν3p

Ep E30 0 0

νp3ν3p+νp

Ep E3

1−ν3pνp3

Ep E3

νpν3p+ν3p

Ep E30 0 0

νpνp3+νp3

Ep Ep

νpνp3+νp3

Ep Ep

1−νpνp

Ep Ep0 0 0

0 0 0 ∆sGp 0 0

0 0 0 0 ∆sGp3 0

0 0 0 0 0 ∆sG3p

with ∆s =1−νpνp −νp3ν3p −ν3pνp3 −νpνp3ν3p −νpν3pνp3

Ep Ep E3

A.3.1 Plane strain

The plane strain stiffness matrix can be extracted from the three-dimensional stiffness ma-

trix. The inverse of this 3x3 matrix is the plane strain compliance matrix.

σ33 =E3νp3

Ep(σ11 +σ22) = ν3p (σ11 +σ22)

Sε=

1−ν3pνp3

Ep−ν3pνp3+νp

Ep0

−νp3ν3p+νp

Ep

1−ν3pνp3

Ep0

0 0 1Gp

Cε= S−1

ε=

1

∆s

1−ν3pνp3

Ep E3

ν3pνp3+νp

Ep E30

νp3ν3p+νp

Ep E3

1−ν3pνp3

Ep E30

0 0 ∆sGp

with ∆s =1−νpνp −νp3ν3p −ν3pνp3 −νpνp3ν3p −νpν3pνp3

Ep Ep E3

σ33 =1

∆s

νp3(νp +1)

E 2p

(ε11 +ε22)

213

A. STIFFNESS AND COMPLIANCE MATRICES

A.3.2 Plane stress

For plane stress the compliance matrix can be extracted directly from the three-dimensional

compliance matrix. The inverse of this 3x3 matrix is the plane strain stiffness matrix.

ε33 =−νp3

Ep(σ11 +σ22)

Sσ=

E−1p −νp E−1

p 0

−νp E−1p E−1

p 0

0 0 G−1p

Cσ= S−1

σ=

1

1−νpνp

Ep νp Ep 0

νp Ep Ep 0

0 0 (1−νpνp )Gp

ε33 =−νp3

1−νp(ε11 +ε22)

A.4 Linear elastic isotropic material

The linear elastic material behavior can be described with the material stiffness matrix C or

the material compliance matrix S. These matrices can be written in terms of the engineering

elasticity parameters E and ν.

S =1

E

1 −ν −ν 0 0 0

−ν 1 −ν 0 0 0

−ν −ν 1 0 0 0

0 0 0 2(1+ν) 0 0

0 0 0 0 2(1+ν) 0

0 0 0 0 0 2(1+ν)

C = S−1 =E

(1+ν)(1−2ν)

1−ν ν ν 0 0 0

ν 1−ν ν 0 0 0

ν ν 1−ν 0 0 0

0 0 0 12

(1−2ν) 0 0

0 0 0 0 12

(1−2ν) 0

0 0 0 0 0 12

(1−2ν)

214

A.4. Linear elastic isotropic material

A.4.1 Plane strain

σ33 = ν(σ11 +σ22)

Sε=

1+ν

E

1−ν −ν 0

−ν 1−ν 0

0 0 2

Cε= S−1

ε=

E

(1+ν)(1−2ν)

1−ν ν 0

ν 1−ν 0

0 0 12

(1−2ν)

σ33 =E

(1+ν)(1−2ν)ν(ε11 +ε22)

It is immediately clear that problems will occur for ν= 0.5, which is the value for incom-

pressible material behavior.

A.4.2 Plane stress

ε33 =−ν

E(σ11 +σ22)

Sσ=

1

E

1 −ν 0

−ν 1 0

0 0 2(1+ν)

Cσ= S−1

σ=

E

1−ν2

1 ν 0

ν 1 0

0 0 12

(1−ν)

ε33 =−ν

1−ν(ε11 +ε22)

215

AP

PE

ND

IX

BMECHANICAL DESIGN PROBLEMS

In this appendix, details of the examples presented in Chapter 8 are shown. The general

relations for the analytical solutions can be found in chapter 8 and are specified here. The

integration constants are calculated for the specific boundary conditions and loading, and

are given here.

217

B. MECHANICAL DESIGN PROBLEMS

B.1 Governing equations and general solution

ur,r r +1

rur,r −ζ2 1

r 2ur = f (r )

with ζ=√

Bp

Ap

and f (r ) =ρ

Ap

(

ur −qr

)

+Ap +Qp

Apα(∆T ),r +

Ap −Bp

Ap

1

rα∆T

orthotropic material :

general solution ur = c1r ζ+c2r−ζ+ ur

εr r = c1ζr ζ−1 −c2ζr−ζ−1 + ur,r ; εt t = c1r ζ−1 +c2r−ζ−1 +ur

r

σr r = (Apζ+Qp )c1r ζ−1 − (Apζ−Qp )c2r−ζ−1 + Ap ur,r +Qpur

r− (Ap +Qp )α∆T

σt t = (Qpζ+Bp )c1r ζ−1 − (Qpζ−Bp )c2r−ζ−1 +Qp ur,r +Bpur

r− (Bp +Qp )α∆T

isotropic material :

general solution ur = c1r +c2

r+ ur

εr r = c1 −c2r−2 + ur,r ; εt t = c1 +c2r−2 +ur

r

σr r = (Ap +Qp )c1 − (Ap −Qp )c2

r 2+ Ap ur,r +Qp

ur

r− (Ap +Qp )α∆T

σt t = (Qp + Ap )c1 − (Qp − Ap )c2

r 2+Qp ur,r + Ap

ur

r− (Ap +Qp )α∆T

218

B.2. Disc, edge displacement

B.2 Disc, edge displacement

ub

ab

z r

r

f (r ) = 0 → ur = 0

ur (r = b) = ub

σr r (r = a) = 0;

Figure B.1: Edge displacement of circular disc

orthotropic material :

general solution ur = c1r ζ+c2r−ζ

εr r = c1ζr ζ−1 −c2ζr−ζ−1 ; εt t = c1r ζ−1 +c2r−ζ−1

σr r = (Apζ+Qp )c1r ζ−1 − (Apζ−Qp )c2r−ζ−1

σt t = (Qpζ+Bp )c1r ζ−1 − (Qpζ−Bp )c2r−ζ−1

c1 =(Apζ−Qp )bζ ub

(Apζ+Qp )a2ζ+ (Apζ−Qp )b2ζ; c2 =

(Apζ+Qp )bζa2ζ ub

(Apζ+Qp )a2ζ+ (Apζ−Qp )b2ζ

isotropic material :

general solution ur = c1r +c2

r

εr r = c1 −c2r−2 ; εt t = c1 +c2r−2

σr r = (Ap +Qp )c1 − (Ap −Qp )c2

r 2

σt t = (Qp + Ap )c1 − (Qp − Ap )c2

r 2

c1 =(Ap −Qp )b

(Ap +Qp )a2 + (Ap −Qp )b2ub ; c2 =

(Ap +Qp )ba2

(Ap +Qp )a2 + (Ap −Qp )b2ub

219

B. MECHANICAL DESIGN PROBLEMS

B.3 Disc/cylinder, edge load

pe

ab

z r

r

pi

f (r ) = 0 → ur = 0

σr r (r = a) =−pi

σr r (r = b) =−pe

Figure B.2: *

Cross-section of a thick-walled circular cylinder

orthotropic material :

general solution ur = c1r ζ+c2r−ζ

εr r = c1ζr ζ−1 −c2ζr−ζ−1 ; εt t = c1r ζ−1 +c2r−ζ−1

σr r = (Apζ+Qp )c1r ζ−1 − (Apζ−Qp )c2r−ζ−1

σt t = (Qpζ+Bp )c1r ζ−1 − (Qpζ−Bp )c2r−ζ−1

c1 =1

Apζ+Qp

aζ+1pi −bζ+1pe

b2ζ−a2ζ; c2 =

1

Apζ−Qp

aζ+1b2ζpi −bζ+1a2ζpe

b2ζ−a2ζ

isotropic material :

general solution ur = c1r +c2

r

εr r = c1 −c2r−2 ; εt t = c1 +c2r−2

σr r = (Ap +Qp )c1 − (Ap −Qp )c2

r 2

σt t = (Qp + Ap )c1 − (Qp − Ap )c2

r 2

c1 =1

Ap +Qp

1

b2 −a2(pi a2 −peb2) ; c2 =

1

Ap −Qp

a2b2

b2 −a2(pi −pe )

220

B.4. Rotating solid disc

B.4 Rotating solid disc

z

ω

b

r

r

f (r ) =−ρ

Apω2r

ur (r = 0) 6=∞σr r (r = b) = 0

Figure B.3: A rotating solid disc

orthotropic material :

ur = c1r ζ+c2r−ζ−1

Apβr 3 with β=

1

9−ζρω2

σr r = (Apζ+Qp )c1r ζ−1 − (Apζ−Qp )c2r−ζ−1 −3Ap +Qp

Apβr 2

σt t = (Qpζ+Bp )c1r ζ−1 − (Qpζ−Bp )c2r−ζ−1 −3Qp +Bp

Apβr 2

c2 = 0 ; c1 =3Ap +Qp

Ap (Apζ+Qp )βb−ζ+3

isotropic material :

ur = c1r +c2

r−

1

8

ρ

Apω2r 3 = c1r +

c2

r−

1

Apβr 3 with β=

1

8ρω2

σr r = (Ap +Qp )c1 − (Ap −Qp )c2

r 2−

(3Ap +Qp )

Apβr 2

σt t = (Ap +Qp )c1 + (Ap −Qp )c2

r 2−

(Ap +3Qp )

Apβr 2

c2 = 0 ; c1 =(3Ap +Qp )

Ap (Ap +Qp )βb2

221

B. MECHANICAL DESIGN PROBLEMS

B.5 Rotating disc with central hole

ω

ab

z r

r

f (r ) =−ρ

Apω2r

σr r (r = a) = 0

σr r (r = b) = 0

Figure B.4: A rotating disc with central hole

orthotropic material :

ur = c1r ζ+c2r−ζ−1

Apβr 3 with β=

1

9−ζρω2

σr r = (Apζ+Qp )c1r ζ−1 − (Apζ−Qp )c2r−ζ−1 −3Ap +Qp

Apβr 2

σt t = (Qpζ+Bp )c1r ζ−1 − (Qpζ−Bp )c2r−ζ−1 −3Qp +Bp

Apβr 2

c1 =3Ap +Qp

Ap (Apζ+Qp )

(bζ+3 −aζ+3

b2ζ−a2ζ

)

β

c2 =3Ap +Qp

Ap (Apζ−Qp )

(a2ζ−2bζ+1 −aζ+1b2ζ−2

b2ζ−a2ζ

)

(a2b2)β

isotropic material :

ur = c1r +c2

r−

1

8

ρ

Apω2r 3 = c1r +

c2

r−

1

Apβr 3 with β=

1

8ρω2

σr r = (Ap +Qp )c1 − (Ap −Qp )c2

r 2−

(3Ap +Qp )

Apβr 2

σt t = (Ap +Qp )c1 + (Ap −Qp )c2

r 2−

(Ap +3Qp )

Apβr 2

c1 =(3Ap +Qp )

Ap (Ap +Qp )(a2 +b2)β ; c2 =

(3Ap +Qp )

Ap (Ap −Qp )(a2b2)β

222

B.6. Rotating disc fixed on rigid axis

ω

ba

z r

r

f (r ) =−ρ

Apω2r

ur (r = a) = 0

σr r (r = b) = 0

Figure B.5: Disc fixed on rigid axis

B.6 Rotating disc fixed on rigid axis

orthotropic material :

ur = c1r ζ+c2r−ζ−1

Apβr 3 with β=

1

9−ζρω2

σr r = (Apζ+Qp )c1r ζ−1 − (Apζ−Qp )c2r−ζ−1 −3Ap +Qp

Apβr 2

σt t = (Qpζ+Bp )c1r ζ−1 − (Qpζ−Bp )c2r−ζ−1 −3Qp +Bp

Apβr 2

c1 =β

(Apζ+Qp )bζ+1a−ζ+1 + (Apζ−Qp )b−ζ+1aζ+1

3Ap +Qp

Apb4a−ζ+1 +

Apζ−Qp

Apb−ζ+1a4

c2 =β

(Apζ+Qp )bζ+1a−ζ+1 + (Apζ−Qp )b−ζ+1aζ+1

Apζ+Qp

Apbζ+1a4 −

3Ap +Qp

Apb4aζ+1

isotropic material :

ur = c1r +c2

r−

1

8

ρ

Apω2r 3 = c1r +

c2

r−

1

Apβr 3 with β=

1

8ρω2

σr r = (Ap +Qp )c1 − (Ap −Qp )c2

r 2−

(3Ap +Qp )

Apβr 2

σt t = (Ap +Qp )c1 + (Ap −Qp )c2

r 2−

(Ap +3Qp )

Apβr 2

c1 =β

(Ap +Qp )b2 + (Ap −Qp )a2

3Ap +Qp

Apb4 +

Ap −Qp

Apa4

c2 =β

(Ap +Qp )b2 + (Ap −Qp )a2

Ap +Qp

Apa4b2 −

3Ap +Qp

Apa2b4

223