A Diverse Collection of FORMULAS, DEFINITIONS AND … · 2020. 1. 4. · FORMULAS, DEFINITIONS AND...

A Diverse Collection of

FORMULAS,DEFINITIONS AND

DERIVATIONSin the Realm of Mechanical Engineering

c©Jorgen S. Bergstrom

A diverse collection of formulas, definitions,and derivations in the realm of mechanical engineering.

Copyright c© 2007 by Jorgen S. Bergstrom

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All rights reserved

FORMULAS, DEFINITIONS AND DERIVATIONS i

Contents

1 Mathematical Foundation 11.1 Gradient, divergence and curl in subscript notation . . . . . . . . . . . . . . . . . . 11.2 The divergence theorem (Gauss’s theorem) . . . . . . . . . . . . . . . . . . . . . . . 11.3 Stoke’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.4 The correlation between eijk and δij . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

2 Tensor Algebra 22.1 Definition of rectilinear base vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . 22.2 Definition of orthogonal base vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . 22.3 Physical meaning of the vector triple product . . . . . . . . . . . . . . . . . . . . . . 22.4 Definition of reciprocal base vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . 22.5 Definition of self-reciprocal bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22.6 Definition of contravariant and covariant vectors . . . . . . . . . . . . . . . . . . . . 22.7 Definition of covariant and contravariant tensors . . . . . . . . . . . . . . . . . . . . 22.8 Ways to calculate the scalar product of a and b . . . . . . . . . . . . . . . . . . . . 22.9 Relation between the covariant and contravariant components of a . . . . . . . . . . 32.10 Curvilinear coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.11 Contravariant and covariant components of a vector . . . . . . . . . . . . . . . . . . 32.12 Partial derivatives of a vector in curvilinear coordinates . . . . . . . . . . . . . . . . 32.13 Definition of a dyad . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.14 Pre- and post-multiplication of a tensor and a vector . . . . . . . . . . . . . . . . . 42.15 The inner product of two tensors – contraction . . . . . . . . . . . . . . . . . . . . . 4

3 Energy Principles 53.1 Different types of work and energy quantities used in solid mechanics . . . . . . . . 53.2 Fundamental definitions of work and energy . . . . . . . . . . . . . . . . . . . . . . 53.3 Work and potential energy of internal forces . . . . . . . . . . . . . . . . . . . . . . 53.4 Work and potential energy of the external forces applied on a solid . . . . . . . . . 73.5 Principle of virtual work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73.6 Principle of stationary potential energy . . . . . . . . . . . . . . . . . . . . . . . . . 83.7 Principle of complementary virtual work . . . . . . . . . . . . . . . . . . . . . . . . 83.8 Principle of stationary complementary energy . . . . . . . . . . . . . . . . . . . . . 93.9 Generalized variational principles . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93.10 Castigliano’s first theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93.11 The unit displacement method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93.12 Castigliano’s second theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103.13 The reciprocial theorem of Betti . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103.14 Betti’s theorem for linear elastic bodies . . . . . . . . . . . . . . . . . . . . . . . . . 113.15 Maxwell’s reciprocal theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

4 Dynamics 124.1 Harmonic Vibrations of a Spring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124.2 Work and kinetic energy of a particle . . . . . . . . . . . . . . . . . . . . . . . . . . 124.3 Moment equilibrium for a particle . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124.4 Konig’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134.5 Moment of inertia and angular momentum of a rigid body . . . . . . . . . . . . . . 134.6 Rotational kinetic energy of a rigid body . . . . . . . . . . . . . . . . . . . . . . . . 134.7 Holonomic, Scleronomic, and Rheonomic constrains . . . . . . . . . . . . . . . . . . 14

ii CONTENTS

4.8 Derive Lagrange’s equations from Newton’s law of motion . . . . . . . . . . . . . . . 14

5 Thermodynamics 165.1 Definition of material system, adiabatic system, isolated system . . . . . . . . . . . 165.2 Thermodynamic equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165.3 Definition of Enthalpy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165.4 Definition of Helmholtz free energy . . . . . . . . . . . . . . . . . . . . . . . . . . . 165.5 Definition of Gibbs free energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165.6 The ideal gas law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165.7 Two basic tendencies of nature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165.8 The zeroth law of thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175.9 The first law of thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175.10 The second law of thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . 175.11 The third law of thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175.12 Isentropic process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175.13 Conduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175.14 Heat capacity at constant pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175.15 The classical definition of entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175.16 The statistical mechanics definition of entropy . . . . . . . . . . . . . . . . . . . . . 185.17 Physical meaning of the minimum value of Gibbs free energy . . . . . . . . . . . . . 185.18 Physical significance of Helmholz free energy . . . . . . . . . . . . . . . . . . . . . . 185.19 Derive the equilibrium vacancy concentration in a crystal lattice . . . . . . . . . . . 185.20 Non-hydrostatic thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

6 Fluid Mechanics 206.1 Definition of viscosity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206.2 Variation of pressure with position in a fluid in equilibrium . . . . . . . . . . . . . . 206.3 Material derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206.4 Conservation of mass for a fluid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206.5 Conservation of momentum for a fluid . . . . . . . . . . . . . . . . . . . . . . . . . . 206.6 Euler’s equation for a perfect fluid . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216.7 Bernoulli’s equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216.8 Irrotational incompressible flow in three dimensions . . . . . . . . . . . . . . . . . . 226.9 Potential flow in two dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226.10 Derive Navier-Stokes equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226.11 Reynold’s number and the condition for turbulent flow . . . . . . . . . . . . . . . . 23

7 Theoretical Elasticity 247.1 Hooke’s law for an isotropic material, infinitesimal displacements . . . . . . . . . . . 247.2 Hooke’s law in tensor form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247.3 Hooke’s law solved for the stresses . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247.4 Conversion of Elastic Constants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247.5 Compatibility Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247.6 Rules for Mohr’s circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247.7 Transformation of the stress tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . 257.8 Strain energy density, in general and for a linear elastic material . . . . . . . . . . . 257.9 Bulk modulus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257.10 Generalized Hooke’s law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257.11 Definition of true strain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257.12 Relative displacement tensor (assuming infinitesimal displacements) . . . . . . . . . 25

FORMULAS, DEFINITIONS AND DERIVATIONS iii

7.13 Strain tensor and rotational tensor in terms of the displacement vector (assuminginfinitesimal displacements) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

7.14 Fractional volume change in terms of the strains and displacements . . . . . . . . . 257.15 Correspondence between a stress tensor and a traction vector specified by its normal

ni. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257.16 Force equilibrium of a solid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257.17 Moment equilibrium of a solid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267.18 Lagrangian, Eulerian variables, and material derivatives . . . . . . . . . . . . . . . . 267.19 Derivation of the Airy stress function for plane problems . . . . . . . . . . . . . . . 267.20 Show that the general solution of an elasticity problem is unique. Assume linear

elasticity (infinitesimal displacements), finite region with no singularities . . . . . . . 277.21 Relation between strain energy density and applied loads for a linearly elastic material

in equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277.22 Connection between the full and the contracted notation for the elastic constants in

a general anisotropic material . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277.23 Universal binding energy relation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287.24 Upper bound for the elastic constants for an inhomogeneous solid . . . . . . . . . . 297.25 Lower bound for the elastic constants of an inhomogeneous solid . . . . . . . . . . . 29

8 Engineering Beam Theory 318.1 Euler-Bernoulli Beam Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 318.2 Deflection of a Cantilever Beam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 338.3 Elastic Buckling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 338.4 Elastic-Plastic Beam Bending . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

9 Linear Elastic Fracture Mechanics 369.1 Energy release rate, crack driving force . . . . . . . . . . . . . . . . . . . . . . . . . 369.2 Relation between G and KI . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 369.3 Conditions for plane strain LEFM . . . . . . . . . . . . . . . . . . . . . . . . . . . . 369.4 Mode I stress field, rectangular coordinates . . . . . . . . . . . . . . . . . . . . . . . 369.5 Derive the J-integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 369.6 Prove that the J-integral is path independent . . . . . . . . . . . . . . . . . . . . . 379.7 Crack extension criteria for an elastic material under uniaxial tension . . . . . . . . 379.8 Crack advance criteria for a plastic non-hardening material in mode I . . . . . . . . 389.9 Crack advance criteria in a power law hardening material . . . . . . . . . . . . . . . 38

10 Metallurgical Fundamentals 4010.1 Facts about BCC, FCC, and HCP (number of atoms per structure cell, e.g. of

materials, slip systems, stacking sequence) . . . . . . . . . . . . . . . . . . . . . . . . 4010.2 Give some examples of structure-sensitive and structure-insensitive properties of metals 4010.3 Mention three types of lattice defects and give examples of each of them . . . . . . 4010.4 Edge dislocations: Burgers vector and movement . . . . . . . . . . . . . . . . . . . . 4010.5 Screw dislocations: Burgers vector, movement and stress field . . . . . . . . . . . . 4010.6 Dislocation dissociation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4110.7 Shockley partials in FCC lattice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4110.8 Definition of Burgers vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4110.9 Properties of b . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4110.10 Plastic strain produced when a dislocation moves a distance x on a slip plane in a

crystal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4110.11 Stress fields around a positive edge dislocation and a positive screw dislocation . . . 41

iv CONTENTS

10.12 Line energy of dislocations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4210.13 Volume change due to a dislocation . . . . . . . . . . . . . . . . . . . . . . . . . . . 4210.14 ‘Force’ on a dislocation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4210.15 Line tension of a dislocation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4310.16 Difference between a jog and a kink . . . . . . . . . . . . . . . . . . . . . . . . . . . 4310.17 Plastic resistance mechanisms in metals . . . . . . . . . . . . . . . . . . . . . . . . . 4310.18 Lattice resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4410.19 Particle controlled resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4410.20 Dispersed particle resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4410.21 Precipitate particle resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4410.22 Dislocation resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4510.23 Solute resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4610.24 Total plastic resistance of a metal . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4610.25 Temperature and strain rate dependence of the plastic shear resistance of a metal . 4610.26 Strain rate sensitivity of flow stress . . . . . . . . . . . . . . . . . . . . . . . . . . . 4710.27 Strain hardening at low temperatures . . . . . . . . . . . . . . . . . . . . . . . . . . 4810.28 Deformation at elevated temperatures in the presence of diffusion . . . . . . . . . . 4810.29 Diffusional flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5010.30 Connection between local (τ) and global (Y ) . . . . . . . . . . . . . . . . . . . . . . 50

11 Plasticity 5111.1 Plastic yielding under combined stresses . . . . . . . . . . . . . . . . . . . . . . . . . 5111.2 Necking in tension (athermal idealization) . . . . . . . . . . . . . . . . . . . . . . . 5211.3 General strategies for approximate solutions in elasticity and plasticity . . . . . . . 5211.4 The upper bound theorem for a rigid-plastic solid . . . . . . . . . . . . . . . . . . . 5411.5 The lower bound theorem for a rigid-plastic solid . . . . . . . . . . . . . . . . . . . . 5411.6 The self consistent method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5411.7 Plasticity analysis by slip line fields . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

12 Kinematics of large strains 5612.1 Finite deformation considerations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5612.2 Derivation of the Lagrangian strain tensor . . . . . . . . . . . . . . . . . . . . . . . 5612.3 Derivation of the Eulerian strain tensor . . . . . . . . . . . . . . . . . . . . . . . . . 57

13 Mechanical Properties of Polymers 5813.1 Constitutive similarities between an ideal gas and an ideal rubber . . . . . . . . . . 5813.2 Definition of tensile creep compliance and tensile stress relaxation modulus . . . . . 5813.3 Time temperature superposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5813.4 How to obtain a master curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5913.5 Boltzmann’s equation for linear viscoelastic materials . . . . . . . . . . . . . . . . . 5913.6 Connection between the creep compliance and the stress relaxation modulus . . . . 5913.7 The correspondence principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6013.8 ODE representation of Boltzmann’s equation . . . . . . . . . . . . . . . . . . . . . . 6013.9 Constitutive relation for an ideal rubber . . . . . . . . . . . . . . . . . . . . . . . . . 6113.10 Constitutive relations for a Maxwell body . . . . . . . . . . . . . . . . . . . . . . . . 6113.11 Viscoelastic damping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6113.12 Linear Viscoelasticity Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6113.13 The Use of Shift Functions to Generalize Linear Viscoelasticity Theory . . . . . . . 6813.14 Constitutive Modeling of the Equilibrium Response of Elastomers . . . . . . . . . . 70

FORMULAS, DEFINITIONS AND DERIVATIONS v

14 Continuum Mechanics Foundations 7514.1 List of Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7514.2 Elements of Tensor Algebra and Analysis . . . . . . . . . . . . . . . . . . . . . . . . 7514.3 Kinematics of Continuous Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7614.4 Balance Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7714.5 The First Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . 7914.6 The Principle of Material Frame Indifference . . . . . . . . . . . . . . . . . . . . . . 8014.7 Isotropic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8014.8 Constitutive Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8014.9 Modeling of Hyperelastic Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

15 Statistical Mechanics 8415.1 The phase rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8415.2 The microcanonical ensemble . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8415.3 The canonical ensemble . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

vi CONTENTS

FORMULAS, DEFINITIONS AND DERIVATIONS 1

1 Mathematical Foundation 1

1.1 Gradient, divergence and curl in subscript notation

gradient ∇φ = φ,i

divergence ∇ ·A = Ai,i

curl ∇×A = eijkAk,j

Note, C = A×B can be written as Ci = eijkAjBk.

1.2 The divergence theorem (Gauss’s theorem)∫V

Bi,idV =

∮S

BinidS,

where Bi is a vector field, note Bi,i corresponds physically to source strength per unit volume andBini is the flux. In vector notation we get∫

V

∇ ·B dV =

∮S

B · dS∫V

∇BdV =

∮S

B⊗ dS

1.3 Stoke’s Theorem ∫S

eijkBk,jnidS =

∮C

Bidxi,

where dxi is the elemental vector of the contour C. In vector notation the equation can be written∫S

(∇×B) · dS =

∮C

B · dr.

1.4 The correlation between eijk and δij

eijkeist = δjsδkt − δjtδksAccording to Pearson: ‘The only formula in vector algebra which needs memorization.’

2 TENSOR ALGEBRA

2 Tensor Algebra 2

2.1 Definition of rectilinear base vectors

Three linearly independent vectors whose directions are fixed in space when used to represent anyarbitrary vector, are called rectilinear base vectors. If they have unit magnitudes, they are namedrectilinear unit base vectors.

2.2 Definition of orthogonal base vectors

When the base vectors are mutually orthogonal they are called orthogonal base vectors.

2.3 Physical meaning of the vector triple product

The product a × b · c is numerically equal to the volume of a parallelepiped having a, b, c, asconcurrent edges.

2.4 Definition of reciprocal base vectors

Two sets of base vectors e1, e2, e3 and e1, e2, e3 are called reciprocal if

ei · ej = δj.i

2.5 Definition of self-reciprocal bases

When a set of base vectors and their reciprocals are identical, we say that the bases are self-reciprocal.A set of base vectors are self-reciprocal if and only if they are mutually orthogonal unit vectors.

2.6 Definition of contravariant and covariant vectors

Components of a vector with respect to base vectors ei are called contravariant, and with respectto reciprocal basis ei are called covariant. We write

a = aiei and a = aiei.

Note that this leads toai = a · ei and ai = a · ei.

2.7 Definition of covariant and contravariant tensors

Covariant gij , contravariant gij and mixed g.ji components of the Euclidean metric tensor are definedby

gij = ei · ejgij = ei · ej

g.ji = δ.ji = ei · ej

Note, components of the Euclidean metric tensor are symmetric.

2.8 Ways to calculate the scalar product of a and b

The scalar product of a and b in rectilinear coordinates may be calculated by any one of the followingformulas

a · b = aibi = aibi = gija

ibj = gijaibj


2.9 Relation between the covariant and contravariant components of a

a are related to each other byai = gija

j ai = gijaj

2.10 Curvilinear coordinates

Let zk (k = 1, 2, 3) be rectangular coordinates of a geometrical point, and xi (i = 1, 2, 3) be threevariables. If between zk and xi we can establish a correspondence, then we say that there existsa coordinate transformation between zk and xi. It can be shown that a unique inverse to thistransformation exists in some neighborhood of zk if the Jacobian

J = det

(∂zk

∂xl

)6= 0.

Base vectors gk(xi) are defined by

gk =∂p

∂xk=∂zm

∂xkim.

The metric tensor is defined by

gkl ≡ gk · gl =∂zm

∂xk∂zn

∂xlδmn.

This name is justified through the fact that when gkl is known we can calculated the length of anyvector and the angle between two vectors. Note that vanishing gkl (k 6= l) is necessary and sufficientfor orthogonality of the curvilinear coordinates.

2.11 Contravariant and covariant components of a vector

The quantities Ak(x) and Ak(x) are called, respectively contravariant and covariant components ofa vector if, upon a transformation of coordinates, they transform respectively according to the rules

A′k(x′) = Am(x)

∂x′k

∂xmcontravariant

A′k(x′) = Am(x)∂xm

∂x′kcovariant

Higher order tensors are similarly defined.

2.12 Partial derivatives of a vector in curvilinear coordinates

The partial derivatives of a vector in curvilinear coordinates becomes

∂u

∂xl=∂(ukgk)

∂zl=∂uk

∂xlgk + uk

∂gk∂xl

But∂gk∂xl

=∂

∂xl

(∂zn

∂xkin

)=

∂2zn

∂xk∂xlin

which becomes∂gk∂xl

=mkl

gm

4 TENSOR ALGEBRA

where mkl

≡ ∂2zn

∂xk∂xl∂xm

∂zn

are known as the Christoffel symbols of the second kind. The Christoffel symbols of the first kindare also of frequent occurrence; they are defined by

[kl,m] ≡ gmn nkl

2.13 Definition of a dyad

The tensor a⊗ b is called a dyad and is defined by

(a⊗ b).c = a(b.c).

The components of the dyad are given by

(a⊗ b)ij = aibj .

2.14 Pre- and post-multiplication of a tensor and a vector

The components of the dot product of a tensor and a vector are given by

(T.v)k = Tkivi.

The pre-multiplication is defined by

(a.T).c = a.(T.c), ∀c.

Furthermore, the component (a.T)i is given by akTki.

2.15 The inner product of two tensors – contraction

The inner product of two tensors is defined by

(S.T).v = S.(T.v) ∀v.

Which in results in the following expression for the components

(S.T)ik = SimTmk.

The double inner product of two tensors is defined by

S : T = Tr(S.T) = SimTmi.


3 Energy Principles 3

3.1 Different types of work and energy quantities used in solid mechanics

Wi total internal workWe total external workUi total potential energy (strain energy) of the internal forcesUe potential energy of the external forcesU0 strain energy densityU∗0 complementary strain energyW ∗e total complementary external workΠ total potential energyΠ∗ total complementary potential energy

3.2 Fundamental definitions of work and energy

Energy is defined as a quantity representing the ability to perform work. We say a structuralsystem possesses energy, whereas the forces in the system may perform work.

The amount of work performed is proportional to the change in energy of the structural system.

The potential energy is the capacity of a conservative force system to perform work by virtue ofits position with respect to a reference level.

Work is defined as the product of a force and the displacement of its point of application in thedirection of the force.

For a conservative system: work done by the applied forces is equal to the strain energy storedin the solid.

3.3 Work and potential energy of internal forces

Consider a one-dimensional bar of length L and cross-sectional area A. Suppose that the left-handside end of the bar is fully built in and that the right-hand side end of the bar is subjected to an axialtensile force N . Now, consider an infinitesimal volume element (dxA) located at position x. Assumeno body forces. Find the net work done by the forces acting on the element when the applied loadundergoes an infinitesimal change to (N+dN). (Or equivalently, find the net work when the appliedstrain undergoes a small change.) The change in the loading conditions result in a change in theinternal stress σ(x) and the displacement u(x), furthermore, let ∆u(x) ≡ u(x,N + dN)− u(x,N).

The work done by the forces on the volume element during the change in state is given by

d(work) ·Adx = σ(x+ dx)A∆u(x+ dx)− σ(x)A∆u(x).

Taylor expanding the first term on the right hand side and keeping only the linear terms yields

d(work) · dx =d

dx(σ∆u)dx

which can be written asd(work) = σ∆ε.

Now notice that during this strain increment, the work done by the internal forces in this differentialelement will be the negative of that performed by the stresses acting upon it. (The acting stresses

6 ENERGY PRINCIPLES

tries to elongate the bar, whereas the internal forces tries to prevent this elongation.) Thus the totalinternal work of the bar as the strains increase from zero to their final value is

Wi = −∫ L

0

∫ εf

0

σ(ε)dεAdx.

For a general three-dimensional case, the total internal work can in a similar fashion be shown tobe given by

Wi = −∫V

∫ εij

0

σijdεijdV. (1)

If the strained solid were permitted to return slowly to its unstrained state, the solid would becapable of returning the work performed by the external forces. This capacity of the internal forcesto do work in a strained solid is due to the strain energy or the internal energy stored in the body.If the work performed is independent of path, a potential function may be associated with the work.To find the potential energy of the internal forces consider the expression for the specific internalwork

dWi = −∫ εij

0

σijdεij . (2)

If the internal forces are conservative, then σijdεij must be a perfect differential d(dWi). Further-more, let this be the differential of some functional dUi. Equation (2) can now be written

dWi(εij) = −∫ εij

0

d(dUi) = −dUi(εij),

assuming the potential energy to be zero at the lower limit. The total potential energy thereforebecomes

Ui = −Wi =

∫V

∫ εij

0

σijdεijdV. (3)

We also note from (3) that the total strain energy is always positive. The identity Wi = −Ui statesthat the decrease in the potential energy Ui in moving from the final state to the reference level isequal to the work done by the internal forces. Conversely, the increase in potential energy in movingfrom the reference level to the final state is equal to the work done against the conservative forcefield.

The strain energy density can be defined by

U0 =

∫ εij

0

σijdεij .

By integrating this equation by parts, we can get a quantity called the complimentary strain energydensity

U∗0 =

∫ σij

0

εijdσij = σijεij −∫ εij

0

σijdεij .

If during loading and unloading U0 and U∗0 are independent of the path of deformation, the differ-entials dU0 and dU∗0 will be exact differentials, and U0 and U∗0 are then potential functions. Fromthe definition of a differential and the expression for dU0 and dU∗0 we see that

σij =∂U0

∂εij,


εij =∂U∗0∂εij

.

3.4 Work and potential energy of the external forces applied on a solid

The external work done by prescribed external forces can be expressed as

We =

∫ST

∫ ui

0

TiduidS +

∫V

∫ ui

0

FiduidV.

If the solid is linear in its response, then

We =1

2

∫ST

TiuidS +1

2

∫V

FiuidV

where Ti and Fi are the final values. If the work performed by the applied loads is independent ofpath, then dWe is an exact differential of the potential function Ue. By the same argument as forthe internal work, we conclude that

We = −Ue.Furthermore, the quantity complementary external work can be established in a similar fashion.

W ∗e =

∫ST

∫ Ti

0

uidTidS

3.5 Principle of virtual work

The principle of virtual work for a solid can be derived from the equations of equilibrium and viceversa. They are, in a sense, equivalent because the principle of virtual work is a global (integral) formof the conditions of equilibrium and static boundary conditions. If the local equilibrium equationsare integrated the following form can be obtained:

−∫V

(σji,j + Fi

)δuidV +

∫ST

(Ti − Ti

)δuidS = 0 (1)

where both integrals are zero, the signs have been chosen for later convenience and where a bardenotes a prescribed value. Now consider the following integral in more detail∫

ST

TiδuidS =

∫S

TiδuidS −∫Su

TiδuidS

=

∫V

(σjiδui) ,j dV −∫Su

TiδuidS

=

∫V

σji,jδuidV +

∫V

σjiδui,jdV −∫Su

TiδuidS. (2)

Inserting (2) into (1) yields∫V

σjiδui,jdV −∫V

FiδuidV −∫ST

TiδuidS −∫Su

TiδuidS = 0.

But δui = 0 on Su and σijδui,j = σijδεij giving∫V

σijδεijdV −∫V

FiδuidV −∫ST

TiδuidS = 0

8 ENERGY PRINCIPLES

which can be writtenδW = δ (Wi +We) = 0

where δWi = −∫VσijδεijdV is the internal virtual work and δWe =

∫STTiδuidS +

∫VFiδuidV is

the external virtual work. The principle of virtual work can be stated as follows:

A deformable system is in equilibrium if the sum of the external virtual work and theinternal virtual work is zero for virtual displacements that are kinematically admissible.(The fundamental unknowns are the displacements.)

3.6 Principle of stationary potential energy

The principle of virtual work can be written

δWi + δWe = 0.

But for systems for which a potential exist for both the internal and external forces, this equationcan be written as

δUi + δUe = 0

or δΠ = 0 where Π = Ui + Ue. Or in other words:

Of all kinematically admissible deformations, the actual deformations are the ones forwhich the total energy assumes a stationary value.

3.7 Principle of complementary virtual work

The first kinematic condition can be stated in the following form

εij =1

2(ui,j + uj,i) , in V

and the second kinematic condition is as follows

ui = ui, on Su

Can now form the following equation∫V

[εij −

1

2(ui,j + uj,i)

]δσijdV −

∫Su

(ui − ui) δTidS = 0. (1)

Now consider the second integral∫V

ui,jδσijdV =

∫V

(uiδσij),j dV −∫V

uiδσij,jdV

=

∫S

uiδσijnjdS,

we see that the second and the fourth integrals cancel out. Equation (1) can therefore be writtenas δ(W ∗i + W ∗e ) = 0 where δW ∗i = −

∫VεijδσijdV and δW ∗e =

∫SuuiδTidS. This principle can

accordingly be stated as follows:

A deformable system satisfies all kinematic requirements if the sum of the external com-plementary virtual work and the internal complementary virtual work is zero for all stat-ically admissible virtual stresses δσij. (The fundamental unknowns are the stresses.)


3.8 Principle of stationary complementary energy

The principle of complementary virtual work gives

δW ∗ = δ(W ∗i +W ∗e ) = δ(−U∗i − U∗e ) = −δΠ∗ = 0

3.9 Generalized variational principles

The classical variational principles can be considered as single field principles involving either dis-placements or forces as unknowns, the generalized principles may involve both these fields simulta-neously. To derive a generalized variational principle we will use the following relations:

σji,j + Fi = 0 in V

Ti = Ti on ST

εij =1

2[ui,j + uj,i] in V

ui = ui on Su

By multiplying by virtual stresses or displacements and integrating the following fundamental equa-tion can readily be obtained

−∫V

(σji,j + Fi)δuidV +

∫V

(ui,j − εij)δσijdV

+

∫ST

(Ti − Ti)δuidS −∫Su

(ui − ui)δTidS = 0.

3.10 Castigliano’s first theorem

Consider a general three-dimensional solid subjected to a system of external forces (or moments) Piwith corresponding displacements ui. The term uk is the displacement of Pk in the direction of Pk.The total potential energy is given by

Π = Ui + Ue = Ui(uk)−n∑k=1

Pkuk.

If the solid is in equilibrium then δΠ = 0 giving

∂Ui∂uk

δuk − Pkδuk = 0,

hence,∂Ui∂uk

= Pk.

3.11 The unit displacement method

Apply a virtual displacement, say δuk, in the direction of a particular force Pk. Then the principleof virtual work gives

δukPk =

∫V

σijδεijdV.

10 ENERGY PRINCIPLES

The virtual displacement δuk is arbitrary, and for simplicity, set equal to unity. Then

Pk =

∫V

σijδεijdV,

where δεij are the strains due to a unit displacement applied in the direction of Pk.

3.12 Castigliano’s second theorem

Consider a general three-dimensional solid acted on by a system of forces Pk with corresponding dis-placements uk. Here, the forces can be regarded as reactions generated by prescribed displacementsuk. The total complementary potential energy can be expressed as

Π∗ = U∗i + U∗e = U∗i (Pk)−N∑k=1

Pkuk.

According to the principle of stationary complementary energy

δΠ∗ = 0 =∂U∗i∂Pk

δPk − ukδPk,

giving∂U∗i∂Pk

= uk

3.13 The reciprocial theorem of Betti

Consider a linear elastic body subjected only to surface forces. Suppose the body is first subjected

to T(1)i and then, at the same locations as T

(1)i and in the same directions, is subjected to applied

forces T(2)i . Denote the external work done by T

(1)i by

We11 =1

2

∫ST

T(1)i u

(1)i dS

where u(1)i is the displacement resulting from T

(1)i . Now let additional tractions T

(2)i be applied,

causing displacements u(2)i , while force set T

(1)i is still constant. The work of T

(2)i moving through

u(2)i is

We22 =1

2

∫ST

T(2)i u

(2)i dS.

An additional increment of work will be done by the first tractions T(1)i moving through the dis-

placements u(2)i giving

We12 =

∫ST

T(1)i u

(2)i dS.

Thus, the total work performed by the two sets of forces is

We = We11 +We22 +We12.

Now, remove the loads and apply them again, but this time in reverse order. For this second case,the total work will be

We = We11 +We22 +We12


where

We21 =

∫ST

T(2)i u

(1)i dS.

Since the total work of the applied forces must be independent of the order of application of theloading it follows that

We12 = We21.

3.14 Betti’s theorem for linear elastic bodies

Consider a linear elastic body that is first subjected to T(1)i and then at the same locations is

subjected to T(2)i . Consider the following integral∫

V

σ(1)ij ε

(2)ij =

∫V

σ(1)ij u

(2)i,j dV

=

∫V

(σ

(1)ij u

(2)j

),jdV −

∫V

σ(1)ij,ju

(2)i dV

=

∫S

σ(1)ij u

(2)j n

(1)i dS −

∫V

σ(1)ij,ju

(2)i dV

=

∫S

T(1)j u

(2)j dS +

∫V

F(1)i u

(2)i dV (1)

Now consider σ(1)ij ε

(2)ij . Note that this is a scalar quantity, hence

σTε∗ =(σTε∗

)T= ε∗Tσ = ε∗TEε

= ε∗TETε = (Eε∗)Tε = σ∗Tε.

This means that ∫V

σ(1)ij ε

(2)ij dV =

∫V

σ(2)ij ε

(1)ij dV. (2)

Equation (1) and (2) gives∫V

F(1)i u

(2)i dV +

∫S

T(1)i u

(2)i dS =

∫V

F(2)i u

(1)i dV +

∫S

T(2)i u

(1)i dS

which is referred to as Betti’s theorem for linear elastic problems.

3.15 Maxwell’s reciprocal theorem

Start with Betti’s theorem ∫ST

T(1)i u

(2)i dS =

∫ST

T(2)i u

(1)i dS.

Suppose that for our linearly elastic body, each force system contains only a single non-zero force.Then we directly conclude:

For a linearly elastic body subjected to two forces equal in magnitude, the displacementof the location (and the direction of) the first force caused by the second force is equal tothe displacement at the location of the second force which is due to the first force. If fijare the flexibility coefficients then fij = fji.

12 DYNAMICS

4 Dynamics 4

4.1 Harmonic Vibrations of a Spring

Force equilibrium in the vertical direction:

x+k

mx = g

The ODE has the following solution:

x = A sin

(√k

mt

)+B cos

(√k

mt

)+mg

k.

The natural frequency is ω =√k/m, and the time to complete one cycle is 2π

√m/k.

4.2 Work and kinetic energy of a particle

Start with Newton’s law of motionF = mr,

then take the line integral of each side of the equation over a given path from A to B.∫ B

A

F · dr =

∫ B

A

mr · dr.

But

r · dr =1

2d (r · r) =

1

2d(v2)

Thus

W =

∫ B

A

F · dr =m

2

(v2B − v2

A

)= TB − TA

If F is conservative (F = −∇V ), then W = VA − VB giving

E = VA + TA = VB + TB .

4.3 Moment equilibrium for a particle


Start with Newton’s law of motionF = mr.

Take the cross product of each side of the equation with the position vector r

r× F = r×mr.

Which is equal to

M =d

dt(r×mv) =

d

dt(r× p) = H

where H is the angular momentum.

4.4 Konig’s Theorem

The total kinetic energy is equal to that due to the total mass moving with the velocity of the centerof mass plus that due to the motion relative to the center of mass.

4.5 Moment of inertia and angular momentum of a rigid body

The angular momentum for a rigid body rotating about a reference point P is

Hp =

∫V

r× ρr dV. (1)

Assume P is fixed in the body, thenr = ω × r

this gives

Hp =

∫V

ρr× (ω × r)dV.

Assume

r = xex + yey + zez

ω = ωxex + ωyey + ωzez.

And define

Ixx =

∫V

ρ(y2 + z2

)dV ; cycl.

Ixy = −∫V

ρxy dV ; cycl.

Then (1) can be written asH = Iijωjei.

4.6 Rotational kinetic energy of a rigid body

Trot =1

2

∫V

ρr2dV

=1

2

∫V

ρr · ω × r dV

=1

2

∫V

ρω · r× r dV

=1

2ω ·H

14 DYNAMICS

4.7 Holonomic, Scleronomic, and Rheonomic constrains

Holonomic constrains can be written in the following form:

φj(q1, q2, . . . , qn, t) = 0, (j = 1, 2, . . . ,m)

Scleronomic constraint: no explicit time dependenceRheonomic constrain: explicit time dependence

4.8 Derive Lagrange’s equations from Newton’s law of motion

Define the generalized momentum by

pi =∂T

∂qi(1)

where qi are the generalized coordinates defined by

xi = fi(q1, q2, . . . , qn, t) (2)

and T is the kinetic energy. For a system of N particles the total kinetic energy is

T =1

2

3N∑j=1

mj x2j . (3)

Inserting (3) into (1) gives

pi =

N∑j=1

mj xj∂xj∂qi

, (4)

but from (2) we find that

xj =

N∑i=1

∂xj∂qi

qi +∂xj∂t

. (5)

Hence∂xj∂qi

=∂xj∂qi

. (6)

Eqs. (4) and (6) yield

pi =

3N∑j=1

mj xj∂xj∂qi

. (7)

Find the time rate of change of the generalized momentum

dpidt

=

3N∑j=1

mj xj∂xj∂qi

+

3N∑j=1

mj xjd

dt

(∂xj∂qi

). (8)

Butd

dt

(xjqi

)=

n∑k=1

∂2xj∂qi∂qk

qk +∂2xj∂qi∂t

= Eq. (5) =∂xj∂qi

. (9)

Next, use (3) to get

∂T

∂qi=

3N∑j=1

mj xj∂xj∂qi

(10)


From Eqs. (8)–(10), we see that

dpidt

=

3N∑j=1

mj xj∂xj∂qi

+∂T

∂qi. (11)

Newton’s law of motion states mj xj = Fj + Rj , where Rj is the total force from the worklessconstraints, and Fi includes all other forces. Eq. (11) can therefore be written as

dpidt

=

3N∑j=1

Fj∂xj∂qi

+

3N∑j=1

Rj∂xj∂qi

+∂T

∂qi. (12)

But the generalized force Qi is given by

Qi =

3N∑j=1

Fj∂xj∂qi

. (13)

In a similar fashion, the second term in (12) is a generalized force resulting from the worklessconstraint forces. Consider the virtual work of these forces

δW =

n∑i=1

3N∑j=1

Rj∂xj∂qi

δqi = 0. (14)

If the δq’s can be chosen independently, then the coefficient of each δqi must be zero. We can nowsimplify (12) to the form

dpidt

= Qi +∂T

∂qi(15)

or equivalentd

dT

(∂t

∂qi

)− ∂T

∂qi= Qi, (i = 1, 2, · · · , n). (16)

These n equations are known as the fundamental form of Lagrange’s equation. If all the Qi’s arederivable from a potential function V = V (q, t) as follows:

Qi = −∂V∂qi

,

then (16) can be written as

d

dt

(∂L

∂qi

)− ∂L

∂qi= 0, (i = 1, 2, . . . , n) (17)

where L = T − V is the Lagrangian function.

16 THERMODYNAMICS

5 Thermodynamics 5

5.1 Definition of material system, adiabatic system, isolated system

• A material system is any fixed quantity of matter contained in a defined region of space. Asystem can exchange energy in the form of work or heat, with its environment, but it cannotexchange matter.

• An adiabatic system is thermally insulated from its environment, and can exchange energy inthe form of work only.

• An isolated system is a system that cannot exchange energy with its environment.

5.2 Thermodynamic equilibrium

A system is in thermodynamic equilibrium if its thermodynamic properties do not change sponta-neously in a finite time period when the system is isolated from its environment.

5.3 Definition of Enthalpy

H = E + PV

where E is the internal energy

5.4 Definition of Helmholtz free energy

A = E − TS

where S is the entropy

5.5 Definition of Gibbs free energy

G = H − TS = A+ PV = E + PV − TS

Gibbs free energy can be thought of as a thermodynamic potential the negative gradient of whichcan be considered as a thermodynamic force for the change in state of the system.

5.6 The ideal gas law

PV = nRT

where n is the number of moles of gas and R is the gas constant, R = 8.314 J/(mole K)

5.7 Two basic tendencies of nature

1. Tendency to minimize the energy of a system.

2. Tendency toward random mixing.


5.8 The zeroth law of thermodynamics

Two systems that are in thermal equilibrium with a third system, are also in thermal equilibriumwith each other.

5.9 The first law of thermodynamics

The first law of thermodynamics simply advocates the energy conservation law, which for a quasi-static process can be written as

dE = dQ− dW

where dE is the increase in internal energy, dQ is the heat absorbed, and dW is the work done bythe system. For a hydrostatic system dW = pdV , and for a solid dW = −σijdεij .

5.10 The second law of thermodynamics

A system operating in a cycle cannot convert into work all the heat supplied to it since some energy isalways rejected as heat to a lower temperature sink. A decrease in entropy can only be temporal andlocal, with a greater increase incurred elsewhere. Entropy always increases for an isolated system.

5.11 The third law of thermodynamics

The entropy of a pure substance in its most stable state approaches zero as the temperature ap-proaches zero.

5.12 Isentropic process

A reversible adiabatic process produces no change in entropy and is said to be isentropic. Anisentropic process, therefore, is one in which no heat exchange occurs with the environment, andalso no heat arises internally from the dissipation of kinetic energy by viscous effects.

5.13 Conduction

Heat can cross the boundary by means of conduction, in which molecules from the system and theenvironment meet and exchange energy at the stationary boundary. Heat conducts at a rate directlyproportional to the temperature gradient causing the heat flow, as indicated by Fourier’s law

dQ

dt= −kTA

dT

dx

where A is the cross-sectional area of the heat flow, x the distance between hot and cold surfaceareas, kT the thermal conductivity.

5.14 Heat capacity at constant pressure

The heat capacity is the amount of heat required to raise the temperature of a material by onedegree.

Cp =

(∂Q

∂T

)P

=

definition of

enthalpy

=∂H

∂T

5.15 The classical definition of entropy

For a system in contact with a heat reservoir, the entropy increases when the system absorbs heat:

dS =dQ

Tfor a reversible reaction

18 THERMODYNAMICS

dS >dQ

Tfor an irreversible reaction

5.16 The statistical mechanics definition of entropy

S = k ln Ω

where k is Boltzmann’s constant, and Ω is the number of distinguishable states in which the systemmay be found.

5.17 Physical meaning of the minimum value of Gibbs free energy

Any system held at constant temperature and pressure comes to equilibrium when Gibbs free energyreaches a minimum. The proof is as follows:

G ≡ E + PV − TS

Consider a small change in the system

∆G = ∆E + P∆V − T∆S

The first law of thermodynamics, ∆E + P∆V = ∆Q, gives

∆G = ∆Q− T∆S

The second law of thermodynamics: ∆S ≥ ∆Q/T gives

∆G ≤ 0

∴ Any spontaneous process in the system will reduce G. Equilibrium is reached when no spontaneouschanges will occur.

5.18 Physical significance of Helmholz free energy

Any system held at constant temperature and volume will reach equilibrium when Helmholz freeenergy has a minimum.

5.19 Derive the equilibrium vacancy concentration in a crystal lattice

Let G0 be the free energy in a crystal without any vacancies. Then introduce nv vacancies; to createone vacancy requires the energy ∆Ev ≈ ∆Hv. But creating vacancies also contributes configurationentropy. Thus

G = G0 + nv∆Hv − TSconfwhere

Sconf = k ln Ω = k ln

(N

nv

)= k ln

N !

nv!(N − nv)!where N is the number of atom sites in the crystal. At equilibrium the Gibbs free energy is atminimum with respect to nv, hence

∂G

∂nv= 0⇒ ∆Hv − T

∂Sconf∂nv

= 0.

To evaluate the partial derivative, use Stirling’s approximation: lnN ! = N lnN −N . This gives

∂Sconf∂nv

= k ln

∣∣∣∣N − nvnv

∣∣∣∣ ≈ k lnN

nv


giving

∆Hv = kT lnN

nv

and finally

nv = Ne−∆HvkT .

5.20 Non-hydrostatic thermodynamics

First, we will make two assumptions: the deformation gradient is independent of position, and moreimportantly, the state of the homogeneous phase can be uniquely specified by a set of independentthermodynamic variables. That is, we are assuming the state of the phase is independent of theprevious history of processes.

From the first law of thermodynamics we can infer the existence of an internal energy functionfor an elastic phase. The internal energy U of a state is measured by the work done under adiabaticconditions in going from a standard state to the state in question. The second law of thermodynamicsleads to the introduction of another function of the state, i.e. the entropy, and also to the introductionof the absolute temperature scale.

In non-hydrostatics it is convenient to express extensive quantities such as U and S as densitiesreferred to the reference volume v. The first and second laws of thermodynamics can for an elastichomogeneous phase subjected to a quasistatic reversible process be combined to yield

du = Tds+ σidεi, i; i ∈ [1, 6].

Here we regard u as a function of seven variables, s and the six γi. Using Legendre transformations,other functions of state may be formed from1 u and s. For instance, a, the Helmholz free energyper unit reference volume is defined by

a = u− Ts,

giving da = σidεi − sdT . And, on taking the differential of this equation we obtain

s = −(∂a

∂T

)ε

, σi =

(∂a

∂εi

)T

.

We can also define another function, which we shall call g by analogy to the Gibbs free energy inhydrostatics, as follows

g = u− σiεi − Ts, i; i ∈ [1, 6].

Which upon differentiation yieldsdg = −sdT − εidσi,

giving

s = −(∂g

∂T

)σ

, εi =

(∂g

∂σi

)T

.

1See, e.g. Goldstein, 1950, Classical Mechanics.

20 FLUID MECHANICS

6 Fluid Mechanics 6

6.1 Definition of viscosity

Newton’s definition of (dynamic) viscosity

σ12 = ηdv1

dx2

Kinematic viscosity

ν =η

ρ

6.2 Variation of pressure with position in a fluid in equilibrium

Consider the equilibrium equationσji,j + Fi = 0. (1)

Fi is the body force per unit volume and is here assumed to be given by the gravitational force; i.e.Fi = −ρgi, where the minus sign indicates that the gravitational force acts in the negative directionof xi. Furthermore, in a fluid in rest, σij is given by the fluid pressure:

σij = −pδij .

Eq. (1) therefore becomes

− ∂p

∂xi− ρgi = 0,

which can be written∇p = −ρg

6.3 Material derivative

The material derivative is given byD

Dt=

∂

∂t+ u · ∇,

where u is the velocity vector.

6.4 Conservation of mass for a fluid∫A

ρu · ndS +

∫V

∂ρ

∂tdV = 0

The divergence theorem gives∂ρ

∂t+∇ · (ρu) = 0

or alternativelyDρ

Dt+ ρ∇ · u = 0

which for incompressible fluids becomes∇ · u = 0.

6.5 Conservation of momentum for a fluid


The linear momentum of a control system at time t is

p =

∫V

ρvdV

where ρ and v are the mass density and the absolute velocity, respectively, of the material withinthe volume element dV . Now, if we evaluate the same integral at time t + ∆t, we find that aslightly different set of particles is within the control volume. So, in order to follow the original setof particles, we calculate the momentum of all particles within the control volume at t + ∆t, andthen we must add the momentum of all the “native” particles that left the control volume during∆t, and subtract the momentum of all the “foreign” particles that entered during the same interval.Therefore, at time t+ ∆t, the momentum of the original set of particles is[∫

V

ρvdV

]t+∆t

+ ∆t

∫A

ρv (vr · dA) .

The vector vr is the velocity, relative to the surface, of the particles that are entering or leaving.Thus, by Newton’s law of motion, the change in linear momentum as ∆t → 0 is equal to the totalexternal force applied to the mass within the control volume. Therefore we see that

F =d

dt

∫V

ρvdV +

∫A

ρv (vr · dA) .

6.6 Euler’s equation for a perfect fluid

Newton’s law of motion can be expressed in the form of the equilibrium equation

σji,j + Fi = ρDviDt

.

For a non-viscous flow these stress components are given by

σij = −pδij .

Hence the equations of motion for a perfect fluid are

− ∂p

∂xi+ Fi = ρ

DviDt

6.7 Bernoulli’s equation

For a frictionless fluid of constant density when the flow is steady and along a single stream line (oreverywhere if the flow is irrotational) the following relation is true:

p

ρ+u2

2+ gz = constant.

The first term p/ρ represents the potential energy associated with internal forces, u2/2 is the ki-netic energy per unit mass, and gz represents the potential energy due to the gravitational force.Bernoulli’s equation therefore has the simple physical interpretation that the total energy constant.The term p/ρ can be ‘derived’ in the following manner. The net force (due to the pressure) acting onan infinitesimal element is Aδp in the direction of motion. The work done by this force is therefore−Adp ds. But the mass of the element is ρAds, giving the work per unit mass −dp/ρ. If the element

22 FLUID MECHANICS

moves from a point where the pressure is 0 to one where the pressure is p, then the work done bythe ‘pressure forces’ per unit mass fluid is

∫ p0

(−dp/ρ) = p/ρ.

6.8 Irrotational incompressible flow in three dimensions

Irrotational flow⇒∇× v = 0 (1)

Vector identity ∇× (∇φ) = 0⇒ v = −∇φ (2)

Incompressibility⇒ ∇ · v = 0 (3)

Combining (2) and (3) yields Laplace’s equation

∇2φ = 0

where φ is called the velocity potential.

6.9 Potential flow in two dimensions

The incompressibility condition becomes

∂vx∂x

+∂vy∂y

= 0.

From this equation one infers that the following is a perfect differential

dψ = vydx− vxdy =∂ψ

∂xdx+

∂ψ

∂ydy. (1)

The function ψ is called the stream function, ψ = constant is called a stream line. Now assumeirrotational flow, ∇× v = 0, giving

∂vy∂x− ∂vx

∂y= 0.

Inserting the expressions for vx and vy from (2) yields

∇2ψ = 0.

But irrotational flow also implies v = −∇φ

vx = −∂φ∂x, vy = −∂φ

∂y. (2)

Equations (1) and (2) give∂ψ

∂y=∂φ

∂x,

∂ψ

∂x= −∂φ

∂y.

These equations are automatically fulfilled if a complex potential is defined by

Ω(z) = φ+ iψ

6.10 Derive Navier-Stokes equation

Start with the equilibrium equation

σji,j + Fi = ρDviDt

. (1)


For a real fluid σij is given byσij = −pδij + Πij

where Πij represent the viscosity stress tensor. Now introduce the velocity gradient tensor

vij =1

2[vi,j + vj,i] . (2)

Since fluids are isotropic, we can relate Πij and vij by

Πij = 2ηvij + η′vkkδij .

Here, η and η′ are known as the first and second coefficients of viscosity (and take the place ofLame’s constants). Can now rewrite (1)

Fi −∂p

∂xi+ Πji,j = ρ

DviDt

which reduces to

Fi −∂p

∂xi+ η′

∂

∂xi∇ · v + 2ηvji,j = ρ

DviDt

.

Introducing (2) gives

Fi −∂p

∂xi+ (η + η′)

∂

∂xi∇ · v + η∇2vi = ρ

DviDt

which in vector notation becomes

F−∇p+ (η + η′)∇(∇ · v) + η∇2v = ρDv

Dt(3)

or if we utilizeDv

Dt=∂v

∂t+

1

2gradv2 − v × curlv

then (3) becomes

F− grad p− ρ

2gradv2 + ρ(v × curlv) + (η + η′) grad divv + η∇2v = ρ

∂v

∂t

6.11 Reynold’s number and the condition for turbulent flow

Re =ρLu

η

where η is the dynamic viscosity. The flow becomes turbulent approximately when Re > 2300.

24 THEORETICAL ELASTICITY

7 Theoretical Elasticity 7

7.1 Hooke’s law for an isotropic material, infinitesimal displacements

Hooke’s law can be written:

ε11 =1

E[σ11 − ν (σ22 + σ33)] ; cycl.

ε12 =σ12

2µ; cycl.

where E is Young’s modulus, and µ is the shear modulus.

7.2 Hooke’s law in tensor form

εij =1 + ν

Eσij −

ν

Eσkkδij

7.3 Hooke’s law solved for the stresses

σij = 2µεij + λεkkδij ,

λ =2µν

1− 2ν,

where µ and λ are Lame’s constants, and ν is the Poisson’s ratio.

7.4 Conversion of Elastic Constants

Known E ν µ κ

µ,κ 9κµ3κ+µ

3κ−2µ6κ+2µ µ κ

7.5 Compatibility Equations

The physical meaning of the compatibility equations is∮∂u

∂sds = 0

which in index form becomes

ε11,22 + ε22,11 = 2ε12,12,

ε11,23 + ε23,11 = ε31,12 + ε12,13.

7.6 Rules for Mohr’s circle

• An angle of θ on the physical element is replaced by 2θ on Mohr’s circle.

• A shear stress causing a clockwise rotation about any point in the physical element is plottedabove the horizontal axis of the Mohr’s circle.


7.7 Transformation of the stress tensor

σi′j′ = ai′kaj′lσkl

7.8 Strain energy density, in general and for a linear elastic material

dW = σijdεij

W =1

2σijεij =

1

2σ : ε

7.9 Bulk modulus

K =σii

3εjj

7.10 Generalized Hooke’s law

εij = Sijklσkl where Sijkl is the compliance tensor.

σij = Cijklεkl where Cijkl is the stiffness tensor.

7.11 Definition of true strain

ε =

∫ Lf

L0

dL

L= ln

LfL0

= lnλ

7.12 Relative displacement tensor (assuming infinitesimal displacements)

dij = ui,j ≡∂ui∂xj

7.13 Strain tensor and rotational tensor in terms of the displacement vector (assuming infinitesi-mal displacements)

The rotational tensor is given by ωij = 12 [ui,j − uj,i] or the rotation vector 1

2∇× u.The strain tensor is given by εij = 1

2 [ui,j + uj,i], furthermore dij = εij + ωij .

7.14 Fractional volume change in terms of the strains and displacements

εii = div u = ∇ · u = ui,i

7.15 Correspondence between a stress tensor and a traction vector specified by its normal ni.

Ti = σjinj

7.16 Force equilibrium of a solid

Newton’s law yields ∫V

FidV +

∫S

TidS =

∫V

ρaidV.


Gauss’s theorem gives ∫V

(Fi + σji,j − ρai) dV = 0.

This must hold for any volume, hence

σji,j + Fi = ρai = ρDviDt

.

7.17 Moment equilibrium of a solid

If body moments are neglected, the moment equation (M = r× F) becomes∫V

eijkxjFkdV +

∫S

eijkxjTkdS =

∫V

eijkxjakρdV.

By utilizing the divergence theorem and the force equilibrium equation we get

eijkσjk = 0,

which reduces toσij = σji.

7.18 Lagrangian, Eulerian variables, and material derivatives

In the Lagrangian system, all quantities are expressed in terms of the initial position coordinates ofeach particle and time; in the Eulerian system, the independent variables are xi and t, where xi arethe position coordinates at the time of interest.

As a particle moves, it will observe a change in φ; the rate at which an observer attached to theparticle would see φ alter is called the material derivative of φ and is denoted by dφ/dt.

dφ(ai, t)

dt=∂φ

∂tdφ(xi, t)

dt=∂φ

∂t+

∂φ

∂xj

dxjdt

7.19 Derivation of the Airy stress function for plane problems

Equilibrium Equations:σxx,x + σxy,y = 0

σyy,y + σxy,x = 0

The equilibrium equations are automatically satisfied if

σxx =∂2ψ

∂y2; σyy =

∂2ψ

∂x2; σxy = − ∂2ψ

∂x∂y.

The only non-trivial compatibility equation is

∂2εxx∂y2

+∂2εyy∂x2

= 2∂2εxy∂x∂y

which is for an isotropic elastic material is fulfilled when

∇4ψ = 0.


7.20 Show that the general solution of an elasticity problem is unique. Assume linear elasticity(infinitesimal displacements), finite region with no singularities

Assume that there are two solutions to the problem, each denoted by the superscripts (1) and (2)respectively. Define

ui = u(1)i − u

(2)i ,

εij =1

2(ui,j + uj,i) = ε

(1)ij − ε

(2)ij ,

σij = σ(1)ij − σ

(2)ij ,

and note thatσij,j = −Fi + Fi = 0.

The strain energy density is positive definite, and if σij is considered as a function of εij , then thetotal strain energy of the body only vanishes if εij = 0. Expanding,∫

V

εijσijdV =

∫V

1

2(ui,j + uj,i)σijdV

=

∫V

ui,jσijdV

=

∫V

(uiσij),j dV −∫V

uiσij,jdV

=

∫S

uiσijnjdS

=

∫S

(u

(1)i − u

(2)i

)(T

(1)i − T (2)

i

)dS

and, if the prescribed boundary conditions are such that this last integral vanishes, then this willimply that εij = 0 and so that the only difference between solutions (1) and (2) is a rigid-body motion(and if the displacement is known on part of the boundary, there can not even be a difference inrigid-body motion between the solutions).

7.21 Relation between strain energy density and applied loads for a linearly elastic material inequilibrium

U =1

2

∫V

σijui,jdV

=1

2

∫(σijui),j dV −

1

2

∫V

σij,juidV

=1

2

∫S

TiuidS +1

2

∫V

FiuidV

7.22 Connection between the full and the contracted notation for the elastic constants in a generalanisotropic material

For the most general linear elastic material the connection between stress and strain is given by

εij = Sijklσkl, (1)


where Sijkl is a forth order tensor. But since both εij and σkl are symmetric, most of the terms in(1) are not independent, hence it is convenient to work with a contracted notation. In general, themost contracted notation of (1) is

εi = Sijσj , i, j ∈ [1, 6]. (2)

It can furthermore be shown that Sij is symmetric, and consequently, there are 21 independentelastic constants at most. If the stresses and strains in the contracted notation are given by

ε1 ≡ε11 σ1 = σ11

ε2 ≡ε22 σ2 = σ22

ε3 ≡ε33 σ3 = σ33

ε4 ≡γ23 = 2ε23 σ4 = σ23

ε5 ≡γ13 = 2ε13 σ5 = σ13

ε6 ≡γ12 = 2ε12 σ6 = σ12.

Then we get the following relation between the elastic constants

Sijkl = Smn when m and n ∈ [1, 2, 3]

2Sijkl = Smn when m or n ∈ [4, 5, 6]

4Sijkl = Smn when m and n ∈ [4, 5, 6]

7.23 Universal binding energy relation

The Helmholz free energy is related to ε, where ε is the uniaxial strain with lateral confinement oruniaxial strain without lateral confinement or dilatational strain

F (ε) = −(1− αε)F0e−αε.

Furthermore, F = E − TS, which leads to

dF = dE − SdT − TdS.

The first and second laws of thermodynamics give

dE = TdS + σijdεij

for a reversible quasi-static reversible process. The differential of Helmholz free energy can now bewritten

dF = σijdεij − SdT.

Consequently, for a uniaxial isothermal process

∂F

∂ε= σ = F0α

2εe−αε.

Young’s modulus as a function of the strain is given by

E(ε) =∂σ

∂ε=∂2F

∂ε2= F0α

2(1− αε)e−αε.

The initial Young’s modulus isE0 = E(0) = F0α

2,


and the cohesive strength is given by

∂σ

∂ε= 0⇒ εc =

1

α.

Also, the binding energy per unit volume

F0 =2χ

b=

∫ ∞0

σdε =E0

α2⇒ εc =

√2χ

bE0

where χ is the surface energy. The universal binding relation can therefore be written in terms ofthe macroscopic terms as

F (ε) = −(

2χ

b

)(1 +

ε

εc

)exp

[−εεc

].

7.24 Upper bound for the elastic constants for an inhomogeneous solid

We consider first determining the bulk modulus of the composite, Kc. For this we impose an externaldilatation σ and assume in our approximate solution that this dilatation is uniform in all components.Thus, we satisfy compatibility trivially. Then the elastic strain energy of this approximate solutionis Eε where ε indicates that a dilatation is imposed.

2UεV

= σcε−1

V

n∑i=1

εσiVi =

n∑i

Kiε2ci

giving

Kε =

n∑i

Kici.

But since Uε > Uc, where Uc is the energy of the exact solution, we get

n∑i=1

Kici > Kc.

Similarly, by considering that the shear strain is uniform in all components we obtain

n∑i=1

µici > µc.

Thus, a solution that satisfies compatibility gives an over estimate of the elastic properties. Itfurnishes an upper bound.

7.25 Lower bound for the elastic constants of an inhomogeneous solid

By imposing on a heterogeneous material a mean normal stress σ and assume that the stress isuniform in all components satisfying equilibrium. Then

2UσV

= σεc =1

V

n∑i=1

σεiVi =

n∑i=1

σ2

Kici,

giving

Kc >1

n∑i=1

(ciKi

) .


Similarly, considering the application of a shear stress τ which is assumed to be uniform in allcomponents we obtain

µc >1

n∑i=1

(ciµi

) .Thus, the solution for a uniform shear which satisfies equilibrium only gives an underestimate of theactual modulus and therefor, furnishes a lower bound. Finally, since statistical randomness shouldassure isotropy we can derive from these expressions the Young’s modulus and Poisson’s ratios toobtain

Ec =9Kc

1 + 3Kcµc

νc =1− 2 µc

3Kc

2 + 2 µc3Kc

For heterogeneous material with components of very different individual properties these boundscan be quite far apart and would not give satisfactory results. Therefore, other techniques would bedesirable to have.


8 Engineering Beam Theory 8

8.1 Euler-Bernoulli Beam Theory

Consider pure bending of a linear elastic beam.

Fig. 1. Pure bending of abeam.

Let the xyz-coordinate system be located at the center of mass (CM) of the cross section of thebeam. Furthermore, let the x-axis be aligned along the centerline and the y-axis be parallel to thedeflection v. Then the equilibrium equations become:∫

A

σxx dA = 0, (1)∫A

σxx(−y) dA = Mb, (2)∫A

σxxz dA = 0, (3)

where Mb is the bending moment. For a linear elastic material, the constitutive equation can bewritten

σxx = εxxE. (4)

Assuming that cross-sections remain flat and perpendicular to the neutral axis, the compatibilityequation is given by

L

R=

(1 + εxx)L

R− y, (5)

where ε0 is the strain at the center line and R is the radius of curvature of the centerline. Solvingfor εxx in (5) gives

εxx =(−y)

R= (−y)κ,

32 ENGINEERING BEAM THEORY

The axial strain in pure bending is: εxx = (−y)κ.

Insert the axial strain in the constitutive equation

σxx = E (−y)κ. (6)

Insert the stress expression into the moment equilibrium (2), and assume that the coordinate systemis located at the center of mass of the cross-section

κ =Mb

EIzz. (7)

Moment of Inertia: Izz =

∫y2dA

Izz = πd4

64≈ d4

20.4Izz = bh3

12

Note: Uniaxial: F = ε · (EA)Bending: Mb = κ · (EI)Torsion: Mt = φ′ · (µIz)

Equation (6) to (7) now yields

σxx =Mb (−y)

Izz(8)

To find the displacements as a function of the applied loads utilize the following well-known relation

κ =1

R=

|v′′|(1 + v′2

)3/2 , (9)

where v is the displacement in the y-direction. Assume small displacements and rotations

κ =d2v

dx2. (10)

Equation (7) and (10) yield

EIzzd2v

dx2= M. (11)

To simplify this equation further, consider the force and moment equilibrium of an infinitesimalvolume element


↑: (V + dV )− V + qdx = 0

⇒ dV

dx+ q = 0 (12)

(x+ dx) : (M + dM)−M + V dx− q dx2dx = 0

⇒ dM

dx+ V = 0 (13)

Equation (11) can now finally be written as

d2

dx2

(EIzz

d2v

dx2

)= q (14)

8.2 Deflection of a Cantilever Beam

The deflection of the cantilever beam is given by

d2v(x)

dx2= κ(x) =

M(x)

EI=−P (L− x)

EI

Introducing the boundary conditions the deflection can be written

v(x) = −Px2

6EI(3L− x)

Define δ to be the deflection at the tip of the cantilever beam:

δ = −v(x = L) =PL3

3EI

giving

P =

(3EI

L3

)δ.

8.3 Elastic Buckling

Consider a simply-supported beam of length L that is subjected to an axial compressive force F .Let the deflection of the beam be w(x), then the moment is given by M = Fw. Recall that for anEuler-Bernoulli beam,

M = −EI d2w

dx2, (1)

which together with the moment becomes

d2w

dx2+

F

EIw = 0. (2)

34 ENGINEERING BEAM THEORY

The solution to this ODE is given by

w(x) = A sin

(√F

EIx

)+B cos

(√F

EIx

). (3)

For a simply supported beam the boundary conditions are w(0) = w(L) = 0, giving

B = 0

A sin

(√F

EIL

)= 0.

Non-trivial solutions only exists if√F

EIL = nπ, where n = 1, 2, 3. (4)

The critical load occurs for n = 1 giving

Fcrit =π2EI

L2. (5)

For other boundary conditions the buckling equation can be written

Fcrit = Cπ2EI

L2=π2EI

L′2

where L′ = KL.

8.4 Elastic-Plastic Beam Bending

Consider a beam bending problem based on the following assumptions:

• Pure bending (no transverse forces)

• Rectangular cross-section (with a width of b and a height of h)

Elastic Solution


Moment equilibrium: ∫A

σx(−y)dA = M (1)

Constitutive Equation:σx = εxE (2)

Compatibility:L

R=

(1 + εx)L

R− y(3)

Equations (2) and (3) give

36 LINEAR ELASTIC FRACTURE MECHANICS

9 Linear Elastic Fracture Mechanics 9

9.1 Energy release rate, crack driving force

G =πσ2a

E

9.2 Relation between G and KI

G =K2I

E∗

9.3 Conditions for plane strain LEFM

B, a ≥ 2.5

(KI

σY

)2

9.4 Mode I stress field, rectangular coordinates

σxx =KI√2πr

cosθ

2

[1− sin

θ

2sin

3θ

2

]σyy =

KI√2πr

cosθ

2

[1 + sin

θ

2sin

3θ

2

]σxy =

KI√2πr

sinθ

2cos

θ

2cos

3θ

2

9.5 Derive the J-integral

The potential energy of a cracked two-dimensional body (assuming no body forces and constanttractions) is

Π(a) =

∫A

WdA−∫

ΓT

Tiuids, (1)

where W is the strain energy density. Differentiate with respect to the crack length a:

dΠ

da=

∫A

dW

da−∫

Γ

Tiduida

ds. (2)

Note that we can replaced ΓT with Γ since dui/da = 0 on Γu. So far, we have used the coordinatesystem x1x2, which is located at the center of the edge crack. Now introduce a new coordinatesystem X1X2 located at the tip of the crack. The relation between the two coordinate systems is

Xi = xi − aδi1.

The chain rule gives memory guideline: φ = φ(a,X1(a))

d

da=

∂

∂a+∂X1

∂a

∂

∂X1=

∂

∂a− ∂

∂X1=

∂

∂a− ∂

∂x1.

Equation (2) can now be written

dΠ

da=

∫A

(∂W

∂a− ∂W

∂x1

)dA−

∫Γ

Ti

(∂ui∂a− ∂ui∂x1

)ds. (3)


Utilize the following ‘trick’∂W

∂a=∂W

∂εij

∂εij∂a

= σij∂εij∂a

.

Therefore ∫A

∂W

∂adA =

∫A

σij∂εij∂a

dA =

principle of

virtual work

=

∫Γ

Ti∂ui∂a

ds.

Hence (3) becomesdΠ

da= −

∫A

∂W

∂x1dA+

∫Γ

Ti∂ui∂x1

ds.

Use the divergence theorem (∫ABi,idA =

∫ΓBinids, where ni is the outward normal to Γ, i.e.

n1ds = dx2):

−dΠ

da=

∫Γ

(Wn1 − Ti

∂ui∂x1

)ds ≡ J.

9.6 Prove that the J-integral is path independent

Let Γ1 and Γ2 denote two curves going counter clockwise around a crack tip, both starting andending on the two traction free crack surfaces. Furthermore, let S1 and S2 be the two paths on thecrack faces that are required to close the path Γ1 + Γ2 + S1 + S2. Finally, let J1 denote the valueobtained by the J-integral for the contour Γ1. Form

J1 − J =

∮Γ+Γ1+S1+S2

(Wdx2 − Ti

∂ui∂x1

ds

)

Use the divergence theorem:

J1 − J =

∫A

[∂W

∂x1− ∂

∂xj

(σji

∂ui∂x1

)]dA

=

∫A

[∂W

∂εij

∂εij∂x1

− σji,j∂ui∂x1− σji

∂

∂x1

∂ui∂xj

]dA

=

∫A

[σij

∂εij∂x1

− σij∂

∂x1(ui,j)

]dA

= 0

Hence J1 = J and J is independent of path.

9.7 Crack extension criteria for an elastic material under uniaxial tension

The stress distribution in mode I is

σij =KI√2πr

σij(θ)F (a/w)

where F (a/w) is a compensation for finite size. The strains εij are related to the stresses by Hooke’slaw, both are concentrated as 1/

√r. The fracture criterion can be stated as:

• The crack advances when the energy release rate equals a material specific energy release rate,GIC = 2χs, where χs is the surface free energy.

38 LINEAR ELASTIC FRACTURE MECHANICS

• The crack advances when at at the crack at a typical distance of order b the tensile stress σθθreaches the ideal cohesive strength σic,

σθθ = σic =1

e

√2χsE

(1− ν2)b,

where e is the base of the natural logarithm.

• The crack advances when

KI = KIC =

√GICE

1− ν2.

9.8 Crack advance criteria for a plastic non-hardening material in mode I

The results presented here are obtained by analogy from mode III. The size of the plastic zone insmall scale yielding (SSY) is

s =1

π

(KI

Y

)2

.

The equivalent stress is constant inside the plastic zone, but the equivalent strain is concentrated as

εe = ε0s

r,

where ε0 = Y/E is the equivalent strain at yield. The crack advance criterion is that the equivalentstrain averaged over a representative volume element (RVE) ρ reaches a critical value εf ,

εf ≈

(1− 1

n

)ln

(1

p0

)sinh

[(1− 1

n

)√3σmσe

] ,where p0 is the initial porosity, σe = Y , σm = σii/3, and n is the strain hardening exponent. Thisis satisfied at the crack tip when KI reaches a critical value to expand the plastic zone to a criticalsize sc, where

sc =1

π

(KIC

Y

)2

and KIC =√εfρπEY .

This means that the fracture criterion can be written KI = KIC . Other equivalent criteria can bestated: The crack opening displacement COD = sε0 reaches a critical value CCOD = scε0 = εfρ;or in small scale yielding the energy release rate reaches a critical value

GI = GIC =K2IC(1− ν2)

E= πY εfρ(1− ν2).

9.9 Crack advance criteria in a power law hardening material

Consider a material with the constitutive equation

εe = αε0

(σeσ0

)n,


where σe and εe are the yield stress and yield strain, respectively. It can be shown that the cracktip fields are given by the HRR-fields

σij = σ0

(J

ασ0ε0Inr

)1/(1+n)

σij(θ, n)

εij = ε0

(J

ασ0ε0Inr

)n/(1+n)

εij(θ, n)

σe = σ0

(J

ασ0ε0Inr

)1/(1+n)

σe(θ, n)

εe = ε0

(J

ασ0ε0Inr

)n/(1+n)

εe(θ, n)

J is a field parameter that characterizes the strength of the local field in a sense very similar tothe energy release rate GI that characterizes the elastic field of a brittle solid. In a material withlimited ductility where the crack advances with a finite plastic zone size, the contour integral canbe evaluated outside sc where it will be in the linear elastic material, where it gives

J = GI =K2I (1− ν2)

E.

Under these conditions J can still be interpreted as a energy release rate.

40 METALLURGICAL FUNDAMENTALS

10 Metallurgical Fundamentals 10

10.1 Facts about BCC, FCC, and HCP (number of atoms per structure cell, e.g. of materials, slipsystems, stacking sequence)

Body-centered cubic crystal structure:

• 2 atoms per structure cell;

• e.g.: Alpha Iron and Tungsten;

• Not close packed, 110 planes have highest atomic density, but 〈111〉 are close packed.

Face-centered cubic crystal structure:

• 4 atoms per structure cell;

• e.g.: Al, Copper, Gold, Silver and Nickel;

• stacking sequence: ABCABC...;

• 12 slip systems: 4 sets of octahedral planes 111, each having three slip directions 〈110〉.

Hexagonal close-packed structure:

• planes and directions are specified by the Miller-Bravais system (hkil), where h, k andi refer to three symmetrical directions in the basal plane, and l is along the c-axis (i =(h+ k));

• stacking sequence: ABAB...;

• 3 slip systems: one slip plane (0001) and three slip directions, 〈1120〉;• e.g.: Zn.

10.2 Give some examples of structure-sensitive and structure-insensitive properties of metals

Structure-sensitive: electrical conductivity, yield stress, fracture strength and creep strength.

Structure-insensitive: elastic constants, melting point, density, specific heat, coefficient of thermalexpansion.

10.3 Mention three types of lattice defects and give examples of each of them

Point defects: vacancies, interstitial atoms and impurity atoms;

Line defects: dislocations;

Surface defects: grain boundaries, stacking faults and twinned regions.

10.4 Edge dislocations: Burgers vector and movement

Burgers vector is always perpendicular to the dislocation line. An edge dislocation can leave the slipplane by diffusion controlled climb.

10.5 Screw dislocations: Burgers vector, movement and stress field


The dislocation line is parallel to the Burgers vector. A screw dislocation can leave the slip planeby cross-slip. Creates the stress field

τφz =Gb

2πr

10.6 Dislocation dissociation

b21 > b22 + b33

10.7 Shockley partials in FCC lattice

a0

2[101] −→ a0

6[211] +

a0

6[112]

10.8 Definition of Burgers vector

b =

∮c

∂u

∂sds

where c is a right-handed circuit around the dislocation.

10.9 Properties of b

• b is preserved along a dislocation,

• a dislocation line cannot terminate inside a crystal,

• a dislocation can only terminate on a free surface, close on itself creating a loop, or join otherdislocations at a node.

10.10 Plastic strain produced when a dislocation moves a distance x on a slip plane in a crystal

The strain produced when a dislocation moves a distance x in a crystal of width l and height h is

γ =xb

lhwhich can be written dγ =

bd(area)

V.

If there are N dislocations that each move an average distance dx then

γ =Nbdx

lh.

Let ρ = N/(lh) be the dislocation density, then the plastic strain can be written

γ = ρbdx.

10.11 Stress fields around a positive edge dislocation and a positive screw dislocation

Edge dislocation

σrr = σθθ = − µb

2π(1− ν)rsin θ?

σrθ =µb

2π(1− ν)rsin θ?

σzz = ν(σrr + σθθ)


Screw dislocation

σθz =µb

2πr

Note, the stress fields are only valid when r > O(b).

10.12 Line energy of dislocations

The strain energy per unit volume for a screw dislocation is given by

W =

∫ εij

0

σijdεij = 2

∫ εθz

0

σθzdεθz = σθzεθz =σ2θz

2µ.

The elastic energy per unit length of dislocation for a screw dislocation therefore becomes

Es =

∫ R

rc

σ2θz

2µ2πrdr + Ecore

=µb2

4πln

(R

rc

)+ Ecore

=µb2

4πln

(αR

rc

)where rc is the core radius and α = O(1). If R is approximated from

ρ ' 1

πR2

then we get

E ' Gb2

2

which typically is about 3 eV. A similar calculation yields the same result for an edge dislocation.

10.13 Volume change due to a dislocation

First, it is obvious that a screw dislocation does not result in any dilatation. Furthermore, foran edge dislocation, integrating εii around the dislocation also results in zero dilatation. A moredetailed analysis reveals that there is a very small volume change due to the dislocation core. Butthis volume change is so small that for most practical purposes plastic flow can be considered toconserve volume.

10.14 ‘Force’ on a dislocation

If a dislocation in a crystal moves under the action of an externally applied stress, the resultingdisplacements on the boundary will cause the external forces to do work on the crystal. Since thestrain energy of the crystal will not be altered by a small displacement of the dislocation in thecenter of a very large crystal, the work must be dissipated as a temperature rise within the crystal.We consider this dissipation as having resulted from the displacement of a fictious force appliednormal to the dislocation line by the external stresses. The plastic work increment is related to theshear stress σs as

dW = σsdγpV = σsbtdx

where t is the thickness. The thermodynamic driving force (per unit length) becomes

fx =1

t

∂W

∂x= σsb


where σs is the resolved shear stress in the direction of b.

10.15 Line tension of a dislocation

Since a dislocation possess a large elastic energy per unit length, it will always tend to shorten itslength between two points inside the crystal by gliding or climbing. Thus, it will behave as if it weresubjected to a line tension T all along its length

T =∂U

∂L=E∆L

∆L= E .

The work done by the applied shear stress σs is stored in terms of an increase in dislocation length.

∆W =f∆A

2=σsb∆A

2= E∆L.

But from the geometry of a curved dislocation that is pinned at two points (at a distance L apart)

∆A ' 2R∆L ⇒ E = T = σsbR.

This gives the curvature force relation

1

R=σsb

Eand also σs =

EbR

.

The maximum value of R is L/2 giving

σs,peak =2EbL.

10.16 Difference between a jog and a kink

The steps in dislocations which form in the process of intersection are called jogs. Jogs lying in theslip plane and can be eliminated by glide motion of the dislocation are called kinks.

10.17 Plastic resistance mechanisms in metals

Intrinsic resistance

• Lattice resistance

• Dislocation resistance

• Grain boundary resistance

Extrinsic mechanisms

• Solute resistance

• Precipitate particle resistance

• Dispersed particle resistance


10.18 Lattice resistance

The intrinsic lattice resistance is induced by the variation in core energy due to dislocation glide.The relation between the variation of core energy between neighboring lattice rows and the latticeresistance is given by

τl =π∆Ecore

b2or τl =

1

b

∂E∂y.

Simple models have shown that this energy decreases exponentially with the width of the core, whilethe core width is inversely related to the ideal shear strength of the lattices. At T = 0K: (1) forFCC metals, τl is neglectable; (2) for BCC metals, τl ≈ [5 · 10−3µ, 1.0µ]; (3) for covalent materials,τl ≈ [1 · 10−3µ, 2.0 · 10−2µ]

10.19 Particle controlled resistance

When a dislocation touches or otherwise interacts with a particle to test its strength, it does so bywrapping itself around the particle to exert increasingly larger components of the line tension forceon the particle in the direction of advance. Thus, the evoked particle resistance K at any instanceis balanced by the components of line tension force of the cusped dislocation. The maximum forcea dislocation can exert on a particle in this way is equal to twice the line tension 2E . Hence, anyparticle with strength K/(2E) > 1 is impenetrable by dislocations. A dislocation can overcome suchparticles only by going around them. When K/(2E) < 1, the dislocation overcomes the particle bycutting it.

10.20 Dispersed particle resistance

When impenetrable particles with K/(2E) > 1 are distributed randomly in a plane with mean center-center spacing l and having an overall volume fraction c, the flow stress is governed by bowing ofdislocations through the gaps of average spacing l in semicircular segment configurations. This givesrise to a plastic resistance of

τ = β2Ebl

= β2Ebr0

√3c

2π

where r0 is the actual particle radius and β(≈ 0.85) is a statistical factor accounting for the variationin particle spacings. Clearly, for a given particle size the maximum plastic resistance is proportionalto c0.5, while for a given particle volume fraction τ depends on 1/

√r0 over the range in which

K/(2E) > 1.

10.21 Precipitate particle resistance

When penetrable particles with K/(2E) < 1 are dispersed randomly in a plane, dislocations samplethem and penetrate them without the need to bow to semicircular configurations.

Assume spherical particles of radius r and volume fraction c, and with a mean spacing l betweenparticles in a given plane

c =2

3

πr2

l2, sometimes written c =

πr20

l2.

The relation between the peak resistive force K and lattice resistance of the particle τlp is

K = 2rbτlp.


Define λ to be the sampling length (i.e. distance between particles along the dislocation), and σsthe applied shear stress acting on the glide plane, then we have

K = σsbλ. (1)

According to Friedel: for each contact lost a new one is generated, when this happens the dislocationsweeps the area l2. Also recall 1/R = σsb/E . The geometrical relationship between λ and l can betaken as

l2 =λ3

2R. (2)

Equations (1) and (2) now gives

σs = τp =

[K3

2El2b2

]0.5

which can be written as

τp =2Ebl

(K

2E

)1.5

.

The plastic resistance of the alloy then becomes

τp '2Ebl

(K

2E

)1.5

=2Ebr0

[3c

2π

]0.5(K

2E

)1.5

. (3)

For many mechanisms the particle strength K/(2E) depends on particle size. This dependence is of-ten linear and introduces a technologically important possibility of maximizing the plastic resistanceat constant volume fraction of particles by particle size control. Thus, in the range of precipitationstrengthening where K/(2E) = K ′r/(2E), Eqn. (3) transforms to

τp =

(K ′

b

)1.5(3c

2π

)0.5(rb

2E

)0.5

where K ′ = dK/dr is the coefficient of the particle size-strength connection, we see that when c isconstant, the plastic resistance is proportional to r0.5, and that the peak strength

τp,peak =K ′

b

(3c

2π

)0.5

' 2τlp

( cπ

)0.5

is achieved at a particle size rpeak = 2E/K ′ = E/(τlpb).

10.22 Dislocation resistance

Glide dislocations in a slip plane encounter many dislocations of other slip systems that spreadthrough the plane. Such forest dislocations are cut in a number of ways. If both the glide dislocationand the forest dislocations are of edge character, a simple jog with energy ∼ µb3/10 is formed. Tofind the corresponding K we have to assume that the forest dislocation has an effective cross sectionw, then we can write

Kw

2=µb3

10= 0.2b

µb

2= 0.2bE

which givesK

2E= 0.2

b

w≈ 0.2.


It is therefore clear that we can treat the dislocation resistance in an analogous fashion to particleresistance, thus

τdisl =2Ebl

(K

2E

)1.5

.

But K/(2E) can be taken as a constant, 1/l =√ρ, and E = µb2/2 giving

τdisl = αµb√ρ

where α is a constant in the range 0.3− 0.4.

10.23 Solute resistance

Individual solute atoms are usually weak and act on the dislocation line collectively in a complexmanner. The resulting plastic resistance have been shown to be of the form

τss = βµ

(K

2E

)4/3

· c2/3

where β is of the order of 1.

10.24 Total plastic resistance of a metal

The total athermal resistance can be found by superposition

τtotal = τl + τss + τp + τdisl.

10.25 Temperature and strain rate dependence of the plastic shear resistance of a metal

The overall temperature dependence of the plastic shear resistance at a given strain rate has thefollowing behavior: first the resistance goes down and then reaches a plateau at around half themelting point and then continues to decrease further towards zero plastic resistance at the meltingpoint.

Above Tm/2 diffusion becomes an important consideration in undoing dislocation structuresresulting from strain hardening making it possible to obtain steady state creep. Below Tm/2 diffusionin unimportant and obstacles controlling the plastic resistance tend to be relatively stable. In therange between 0 K and Tm/2, however, it is possible in many instances to shear obstacles at stresslevels σs less than the athermal levels of resistance τ , through thermally activated processes. Thekey elements of thermally activated glide are the force distance curves K = K(y) of obstacles, wherey is the distance of penetration of a dislocation into the obstacle at a force K < K where K isthe peak resistive force of the obstacle. If the dislocation can be aided to go from y1 to y2 underconstant applied stress σs by thermal fluctuations the obstacle would be overcome and effectivelylong range dislocation glide can be achieved. The energy barrier for this process is

∆G∗ = ∆F −∆W =

∫ y2

y1

K(y)dy −Ka(y2 − y1) =

∫ K

Ka

∆y(K)dK.

To simplify the analysis, we will now consider an obstacle for which K(y) = K for y ∈ [0, 2r] (thisis the case for obstacles where the shear resistance is comprised of an increase of internal interfaceenergy). In this simple case

∆G∗ = 2r(K −Ka) = 2rK

(1− Ka

K

).


Also, from above, Ka/K = (σs/τ)2/3 giving

∆G∗ = ∆G0

[1−

(στ

)2/3]

where ∆G0 = 2rK. The probability of obtaining the required assistance from thermal fluctuationsis given by Bolzmann’s expression

exp

[−∆G∗

kT

].

Once a successful thermal fluctuation is accomplished the dislocation line sweeps out a small areaincrement of order l2 and in general results in an increment of plastic strain bl2/V . If there are nsimilarly positioned dislocation segments in the volume element V and the frequency of oscillationof the dislocation segment in front of the obstacle is νG then the overall plastic strain rate under theapplied stress σ for an obstacle field with athermal resistance τ becomes

γ = γ0 exp

[−∆G∗

kT

]with γ0 = nνG

bl2

V

or since n/V is likely to be of order 1/l3 giving γ0 = bνG/l. The above developed expression canalternatively be written as

σsτ

=

[1− T

T0

]1.5

; T0 =∆G0

k ln(γ0/γ).

T0 is a strain rate dependent ‘cut-out’ temperature of the glide obstacle under consideration; formany mechanisms T0 < Tm/2. Since there will always be a temperature independent additive ofthe flow stress τµ, the overall dependence will first decrease and then reach a non-zero plateau attemperatures higher than T0.

In general, for other obstacle interactions

∆G∗ = ∆Go

[1−

(T

T0

)p]qwhere p and q are of order unity or less, or alternatively

γ = γ0 exp

−∆G0

kT

[1−

(σsτ

)p]q.

10.26 Strain rate sensitivity of flow stress

Let’s consider the case of a boxed-shaped obstacle force-distance curve, from above

γ = γ0 exp

−∆G0

kT

[1−

(σsτ

)2/3]

This leads to

ln γ = ln γ0 −∆G0

kT+

∆G0

kT

(στ

)2/3

.

Define the stress exponent of the strain rate by

m ≡ d(ln γ)

d(lnσ)=

2

3

∆G0

kT

(στ

)2/3


where a typical value of m is in the range 80− 160.

10.27 Strain hardening at low temperatures

When a solid is plastically strained, glide dislocations that produce the plastic strain cut each other,become pinned and entangled, and accumulate. This results in increased dislocation interactions,and an increase in the dislocation resistance which is referred to as strain hardening. The process isvery complex and in detail has defied accurate description. We will examine it by a simple form ofdimensional analysis. We start out with the relation for the dislocation resistance

τ = τdisl = αµb√ρ, α ≈ 0.2− 0.4

where α could be temperature dependent. When due to strain hardening, due to an increment ofstrain dγ the dislocation density increases by dρ the plastic resistance increases by

dτ =αµb

2

dρ√ρ.

The plastic strain increment, however, is

dγ =d (b

∑∆ai)

V= bd

(lΛtV

)= bLd

(ΛtV

)where d

∑∆ai is the increment over the entire area swept by dislocations. If we assume that

dislocations traverse a mean free length L during these sweeps∑i

∆ai = LΛt

where Λt is the total mobile length of dislocations and Λt/V is the total mobile line length/volume.For the specific increment of strain under consideration we assume (or imply) that this acting lengthis stored in the body upon the end of the increment when the dislocations have moved the lengthL. This gives

d

(ΛTV

)' dρ

for the increase of dislocation density stored. Then, dγ = bldρ. By self similarity, we expect thatthe storage distance L is likely to be a multiple C of the mean dislocation spacing l, i.e. L = Clwhere l = 1/

√ρ. Then, dγ = bCdρ/

√ρ and the strain hardening rate

dτ

dγ' α

2Cµ.

Computer experiments and other estimatates puts C ≈ 25 or so. This gives for the hardening rateh normalized with µ as

h

µ' α

2C∈[

1

250,

1

125

]

10.28 Deformation at elevated temperatures in the presence of diffusion

If deformation is administrated above Tm/2, static thermal recovery can occur concurrently withdeformation and at a steady state undo the effect of strain hardening. The undoing occurs inside tightclusters that pin the strain hardening induced dislocation network. This permits glide controlled


spontaneous elimination of large dislocation groups. The process is complex but we can againformulate a quick dimensional model. In tight clusters edge dislocations can become eliminated byclimbing. This gives

dρ

dt= − ρ

ta

where ta is the characteristic annihilation time. Solution of a load-diffusion equation gives

1

ta' Db2ρ2

where D is the self diffusion constant. Then

dρ

dt= −βDb2ρ3

where β is a numerical constant. However, the overall dislocation density relates to the plastic(dislocation) resistance as

τ = αµb√ρ

where we understand τ to be governed only by the dislocation resistance. Thus

dρ

dt=

d

dt

(τ

αµb

)2

=2

αµb

τ

αµb

dτ

dt,

and upon substitution of the relation for reduction of dislocation density in its form related to theplastic resistance as

dρ

dt= −βb2D

(τ

αµb

)6

we have1

αµ

(dτ

dt

)= −βb

3D

2

(τ

αµb

)5

.

Now considering that in a deformation situation τ increases due to strain due to our relation thatwe developed earlier

dτ

dγ=αµ

2C

which we now consider as (dτ

dγ

)t

=αµ

2C.

And, furthermore, τ decreases due to thermal recovery according to the above expression which weinterpret as (

∂τ

∂t

)γ

= −αµβb2D

2

(τ

αµb

)5

.

We note that under the two effects of straining and thermal recovery we have

dτ =

(∂τ

∂γ

)t

dγ +

(∂τ

∂t

)γ

dt.

We note that a steady state with dτ = 0 is possible when(∂τ

∂γ

)t

γss +

(∂τ

∂t

)γ

= 0


for here we get

γss =bC

α6

D

b2

(τ

µ

)5

and since during deformation the applied stress and the reference resistance will retain a criticalratio we have the generalization

γss = γ0D

b2

(σsµ

)mwhere m is generally of order 5 for pure metals as modelled and γ0 now collects together a number ofmodel sensitive parameters to which we don’t want to commit ourselves. The self diffusion constant

D = D0 exp

(−QDkT

)where QD is the activation energy of self diffusion and D0 = zb2νD is the pre-exponential factor,where z is a local coordination number of sites between atom exchanges which can occur withvacancies which control the diffusion process.

z ' O(3− 6) and the atom frequency νD ' 1013 s−1

10.29 Diffusional flow

Diffusional flow occurs by point (vacancy) transport. It can be shown that

γdf =16DV σΩ

b2kT

[1 +

π

2

δ

d

DB

DV

]where Ω is the atomic volume, C0 the thermal equilibrium concentration of vacancies, δ the grainboundary thickness, and d the grain size. Furthermore,

DV = D0 exp

(−QVkT

)DB = D0 exp

(−QBkT

)QB ' 0.6QV

At high temperatures, volume diffusion dominates γdt ∝ 1/d2, and at lower temperatures γdf ∝ 1/d3.

10.30 Connection between local (τ) and global (Y )

The deformation has to be compatible: deforming grains must continue to fit together. An upperbound solution can be found by assuming uniform strain everywhere. Five arbitrary slip systemsare required.

dW = Y dεp = 〈5∑i=1

τ dγi〉 = τ〈∑

dγi〉

where 〈...〉 denotes an average value. This gives the Taylor factor mT

Y

τ=〈∑5

dγi〉dεp

= mT

For both FCC and BCC materials mT = 3.1. Mechanism inspired phenomenology.


11 Plasticity 11

11.1 Plastic yielding under combined stresses

For a rate independent isotropic material the equivalent stress according to Von Mises is given bythe empirical relation

σ =

√1

2[(σ1 − σ2)2 + (σ2 − σ3)2 + (σ3 − σ1)2].

If k is the yield strength in shear and Y is the yield strength in tension then we get k = Y/√

3, andyielding will simply occur when σ = Y . The equivalent stress for the case when all stress componentsare present becomes

σ =

(1

2

[(σ11 − σ22)2 + (σ22 − σ33)2 + (σ33 − σ11)2

]+

3σ212 + 3σ2

23 + 3σ231

)1/2

which equivalently can be written as

σ =

√3

2sijsij =

√3J2.

In a similar fashion, Tresca’s yield condition is defined by

τ =1

2(σmax − σmin) ,

yielding will occur when τ = k. As crystals are plastically deformed, the stress for plastic flow, calledthe flow stress, usually increases. We can now introduce a work equivalent plastic strain incrementdεp

dW p = σdεp =

for a

tension test

= Y (εp) dεp.

The normal plastic strains should preserve volume εpii = 0. It can be shown that the equivalentplastic strain can be calculated from

dεp =

√2

3dεpijdε

pij

which in expanded form becomes

dεp =

2

9

[(dεp11 − dε

p22)

2+ (dεp22 − dε

p33)

2+ (dεp33 − dε

p11)

2]

+

1

3

[(dγp12)2 + (dγp23)2 + (dγp31)2

]1/2

.

For the case of pure tension dεp = dεp33, and for the case of pure shear dεp = dγp23/√

3, this is thebackground for the definition of the shear strain increment

dγp =√

3dεp.

52 PLASTICITY

The individual strain increments are proportional to the components of the unit normal vector onthe yield surface in the appropriate directions of stress, giving the associated flow rule

dεp11

dεp=

1

σ

[σ11 −

1

2(σ22 + σ33)

]; cycl.

dγp12

dγp=

3σ12

σ; cycl.

Note, since only the current direction of dislocation motion is determined by the current stress, thecurrent stress determines only the current strain increments and not their total values as in elasticity.The concept of critical shear stress for slip and a stress strain relation can be considered valid onlywhen applied to a region which is large compared both to mean spacing between dislocation sourcesand the mean distance of dislocation travel. The fact that the strain increment components plottedas vectors in six dimensional stress space are always normal to the yield locus is equivalent to theprincipal of maximum plastic resistance which states:

For any plastic strain increment, the state of stress actually occurring gives an incrementof work which equals or exceeds the work which would be done by that strain incrementdεpij with any other state of stress within or on the yield locus. In equation form, if σ∗ijrepresents any state of stress within or on the yield locus,(

σij − σ∗ij)dεpij ≥ 0.

11.2 Necking in tension (athermal idealization)

Necking will occur when the decrease in cross-sectional area is more rapid than the increase of flowstress due to strain-hardening.

dP = Y dA+AdY.

Necking will occur when dP = 0. Also,

dV = Adl + ldA = 0⇒ dA

A= −dl

l= −dε,

which leads to the necking condition

dY

dεp= Y,

dσ

dεp= σ.

For the case when the strain hardening behavior is given by

Y = Y0εn (n < 1.0)

the necking condition becomes εn = n.

11.3 General strategies for approximate solutions in elasticity and plasticity

The solution of specific problems in elastic and plastic or rigid plastic behavior in general involveconsiderable complexities. Specific solutions of problems are often only accessible through numericaltechniques which will not be considered here. Instead, we will consider a selection of specific problemsin elastic behavior and plastic non-hardening behavior by approximate methods which are designedto bound the exact solution, and with some care can give very useful insight into the exact solution.


First we consider elastic behavior. In elasticity we know the solution of problems requires sat-isfaction of: equilibrium, compatibility, constitutive equations and boundary conditions of imposedstress and displacement. When such a solution is obtained it is unique and the elastic strain energyassociated with it is less than any other solution that does not satisfy all these conditions, or onlysatisfies some of them. Thus, in elasticity most approximate solutions are based on the minimumpotential energy principle. We will here use this approach to develop some useful bounding solutionsfor the elastic properties of quasi-randomly arrayed heterogeneous solids. We will start out by notingthat in elasticity the shear and the bulk behavior of a solid are physically uncoupled and that theelastic strain energy due to dilatation is uncoupled from the elastic strain energy of the shear process.Based on this observation and the minimum elastic strain energy principle we want to determinethe elastic properties of heterogeneous media made up of say n different component materials ofisotropic properties each occupying a given volume fraction ci. Each component possesses a shearmodulus and a bulk modulus

µi =dτidγi

and Ki =dσidεi

where τi is the shear stress, γi is the shear strain, σi the mean normal stress, and εi the dilatation.We consider a solid in which the components are roughly equiaxed and are statistically randomdistributed with no particular order. If under some condition or applied stress the exact answer wasknown we could obtain the overall elastic strain energy in any particular component part as

dUi =1

2(σε)idVi +

1

2(τγ)i

=1

2

σ2i

KidVi +

1

2

τ2i

µidVi

=1

2Kiε

2i dVi +

1

2µiγ

2i dVi

where for the shear energy we are considering as a generic example only one component. Then theoverall elastic strain energy of the composite would be

Uc =

n∑i=1

[1

2

∫Vi

(σε)idVi +1

2

∫Vi

(τγ)idVi

].

For the exact solution this strain energy Uc is a minimum in relation to any variation of the solution.We could obtain the elastic constants of this uniaxial form from two separate experiments. From adilatation experiment

1

2σcεV =

1

2Kcε

2V = Uc =

n∑i=1

∫Vi

1

2σiεidVi.

Then

Kc =d2(Uc/V )

dε2=

1

V

d2

dε2

[n∑i=1

1

2

∫Vi

Kiε2i dVi

].

In a related shear experiment we find

µc =d2(Uc/V )

dγ2=

1

V

d2

dγ2

[n∑i=1

∫Vi

µiγ2i dVi

].

In general the exact answer for the stress and strain distribution in the interior, in the componentsis not available. Therefore, we are interested in obtaining approximate solutions. We will do so bysatisfying either equilibrium of compatibility separately. This provides bounds to the answer.

54 PLASTICITY

11.4 The upper bound theorem for a rigid-plastic solid

In a rigid-plastic continuum, deformation must occur under any system of loads Pk for which adistribution of displacements can be found such that

• the displacement boundary conditions, if any, are satisfied,

• the displacements can be differentiated to give a strain, with no change in volume anywhere,and

• the resulting plastic work done throughout the material, found from the resulting equivalentstrain, is less than the work done by the external loads acting through the assumed displace-ments: ∑

k

Pkdpk >

∫V

σY dεpdV

where σY is the equivalent flow stress.

11.5 The lower bound theorem for a rigid-plastic solid

In a rigid-plastic continuum there can be no plastic deformation under loads for which a stressdistribution can be found which

• everywhere satisfies the equilibrium equations,

• is in equilibrium with external loads, and,

• is everywhere within the yield locus.

11.6 The self consistent method

The self consistent method is based on the fundamental solution of Eshelby (1957) which states thatfor an ellipsoidal inclusion in an infinite medium the elastic stress and strain field in the interioris uniform when the body as a whole is subjected to a distant uniform stress field. Then we getdσmij = Cmijkldε

mkl for inclusion m, and dσij = Cijkldεkl for the composite. The Eshelby solution

statesdσi = Bi(Ci, C)dσ.

Consider now a heterogeneous medium with n different component phases with known elastic prop-erties and known volume fractions ci. The self consistent method considers n reference Eshelbyproblems for each one, one particular component is considered as a spherical inclusion in the com-posite of unknown and to be determined elastic properties. Since, all n components fill the entirevolume, the overall field of the component, whether that is stress, strain, or elastic strain energy ismade up of the volume fraction average of the individual fields. This is the so-called self consistentstatement, i.e.

n∑i=1

cidσi = dσ,

but with the stress concentration tensor Bi introduced

n∑i=1

ciBi(Ci, C)dσ = dσ.


This is the self consistent statement and furnishes as many equations as there are unknown compo-nents of elastic constants. The closed form solution for the bulk and shear modulus of an initiallyisotropic composite with isotropic components becomes

n∑i=1

ci

1 +(µiµ − 1

)β∗

= 1

n∑i=1

ci

1 +(KiK − 1

)α∗

= 1

where

α∗ =1 + ν

3(1− ν), β∗ =

2(4− 5ν)

15(1− ν)

where ν is the composite Poisson’s ration given by

ν =3K − 2µ

6K + 2µ=

1− 2µ3K

2 + 2µ3K

The self consistent results are generally very good when the components are of light concentration.

11.7 Plasticity analysis by slip line fields

Consider a rigid non-hardening material for application in plane strain. For this situation dεp33 = 0giving σ33 = (σ11 + σ22)/2. By plugging this in the yield criterion we get

σ = Y =

√3

2(σ1 − σ3)

but since we also have k = Y/√

3, we get

σ1 − σ2 = ±2k

when considering principal stresses. Similarly, we define the mean normal stress as σ = (σ1 +σ2)/2.Consider a coordinate system α, β that is parallel to the principal shear directions and rotated anangle ϕ with respect to the 1, 2 plane. Then by Mohr’s circle we get

σ11 = σ − 2 sin 2ϕ,

σ22 = σ + 2 sin 2ϕ,

σ12 = k cos 2ϕ.

By inserting these equations into the equilibrium equations, and then consider the 1, 2 axes to parallelto α, and β (sin 2ϕ = 0, cos 2ϕ = 1) we get

∂σ

∂x1− 2k

∂ϕ

∂x1= 0,

∂σ

∂x2+ 2k

∂ϕ

∂x2= 0.

Integrating these two equations

σ − 2kϕ = const. along α-line

σ + 2kϕ = const. along β-line

56 KINEMATICS OF LARGE STRAINS

12 Kinematics of large strains 12

12.1 Finite deformation considerations

The method described here may be called an isomorphic mapping of space. That is, we considera correspondence between points x in a reference or initial space and points X in a space mappedfrom this initial space. This is represented by the transformation

x→ X. (1)

The space or region mapped out by X will be of different shape, size and position to that mappedout by x. Equation (1) may also be written

X = u + x.

Here X, x, u are vectors in three-dimensional space and the representation used for them is bythe same fixed Cartesian co-ordinate axis system. By considering the change in the length of ainfinitesimal line element during the deformation it is easy to derive the Lagrangian (or Eulerian)strain tensor. It is also possible to show that

dV = Jdv,

where J = det(∂X/∂x). A homogeneous transformation is one in which ∂Xα/∂xβ are independentof position. In this case we directly obtain

X = Dx.

For transformations which describe real deformations, it is clear that D will be a real non-singularmatrix. The polar factorization theorem can be stated as follows:

Every real non-singular matrix is uniquely expressible as a product of a real orthogonalmatrix and a positive-definite real symmetric matrix.

Also: an orthonormal real set of eigenvectors can be chosen for a real symmetric matrix. Thereciprocal of an orthogonal matrix equals its transpose. The finite strain tensor can be written as

2η = DTD−E.

12.2 Derivation of the Lagrangian strain tensor

Suppose that two material particles, before the motion, have coordinates (ai), (ai + dai); after themotion, let their respective coordinates be (xi), (xi + dxi). Then the initial and final distancesbetween these neighboring particles are given by

ds20 = daidai,

ds2 = dxidxi.


Let the displacement vector be defined by ui ≡ xi − ai, then ds2 − ds20 is

ds2 − ds20 = dxidxi − daidai

=∂xt∂ai

dai∂xt∂aj

daj − δijdaidaj

=

[∂xt∂ai

∂xt∂aj− δij

]daidaj

=

[(δit +

∂ut∂ai

)(δjt +

∂ut∂aj

)− δij

]daidaj

=

[∂ui∂aj

+∂uj∂ai

+∂ut∂ai

∂ut∂aj

]daidaj

Define the Lagrangian strain tensor as

Eij =1

2

(∂ui∂aj

+∂uj∂ai

)+

1

2

∂ut∂ai

∂ut∂aj

,

and hence,ds2 − ds2

0 = 2Eijdaidaj .

12.3 Derivation of the Eulerian strain tensor

ds20 = daidai,

ds2 = dxidxi.

Let the displacement vector be defined by ui ≡ xi − ai, then

ds2 − ds20 = dxidxi − daidai

=

[δij −

∂at∂xi

∂at∂xj

]dxidxj

=

[δij −

(δti −

∂ut∂xi

)(δtj −

∂ut∂xj

)]dxidxj

=

[∂ui∂xj

+∂uj∂xi− ∂ut∂xi

∂ut∂xj

]dxidxj

= 2eijdxidxj

where

eij =1

2

(∂ui∂xj

+∂uj∂xi

)− 1

2

∂ut∂xi

∂ut∂xj

58 MECHANICAL PROPERTIES OF POLYMERS

13 Mechanical Properties of Polymers 13

13.1 Constitutive similarities between an ideal gas and an ideal rubber

Compressing an ideal gas is similar to deforming rubber in that they both deform without stor-ing internal energy and both go back to original shape upon unloading. To show this consider aquasistatic reversible process

dE = TdS − PdV. (1)

The total differential of Helmholz free energy

dA = −PdV − SdT

gives the Maxwell relation (∂P

∂T

)V

=

(∂S

∂V

)T

. (2)

Equations (1) and (2) can now be combined to yield

∂E

∂V= T

(∂P

∂T

)V

− P,

which for an ideal gas, PV = nRT , results in

∂E

∂V= 0.

13.2 Definition of tensile creep compliance and tensile stress relaxation modulus

The tensile creep compliance, Dc(t), is defined by

Dc(t) =ε(t)

σ0.

And the tensile stress relaxation modulus, Er(t), is defined by

Er(t) =σ(t)

ε0.

The characteristic relaxation time is defined as the time at which the stress has been reduced down to1/e ≈ 37% of the initial stress (in a stress relaxation test). Also, note that the time dependence of Dc

and Er is weaker for a crosslinked or semi-crystalline material than for a corresponding uncrosslinedamorphous material.

13.3 Time temperature superposition

For rheologically simple materials, the time-temperature superposition makes it possible to extrapo-late mechanical data obtained within a narrow time interval to much shorter and longer time scaleswhere no actual measurements have been made. This procedure has been generally found to holdwith amorphous polymers (at temperatures up to Tg + 100C) for measurements made at smallstrains in regions 2, 3 and 4.

The amount of horizontal shift or “shift factor” necessary to effect superposition can be calculatedby use of the WLF (Williams, Landel, Ferry) equation:

log aT (T ) =−C1(T − T0)

C2 + (T − T0),


where T0 is a reference temperature and C1 and C2 are constants that are the same for almost allamorphous polymers. The constants depend, however, on the choice of reference temperature. Atthe glass transition temperature, for example, the constants are C1 = 17.44 and C2 = 51.6. Theseconstants amounts to a shift factor of about 3C per decade time.

13.4 How to obtain a master curve

A master curve for a material can be obtained by performing experiments under a restricted timeinterval but using different temperatures.

The master curve for the specific temperature is then obtained by holding the curve segmentcorresponding to the specific temperature fixed and shifting the other curves horizontally until onecontinuous curve is obtained. Note, however, that this does not work for semi crystalline materialssince the curves will not fit upon horizontal shifting. The reason for this is that not everything‘speeds up’ the same when the temperature is raised. Also note that logE − T curves can bedetermined directly from the set of curves shown in the figure above.

13.5 Boltzmann’s equation for linear viscoelastic materials

Consider a body that is subjected to the load dσ at time τ . If the material is linear viscoelastic thenthe strain response is given by

dε(t) = Dc(t− τ)dσ.

By virtue of the linear and time invariant properties of the material we can generalize this equationand obtain

ε(t) =

∫ t

−∞Dc(t− τ)

dσ(τ)

dτdτ

for an amorphous polymer under uniaxial loading. Dc(t − τ) is called the kernel and has to bedetermined by experiments. It can similarly be shown that

σ(t) =

∫ t

−∞E(t− τ)

dε(τ)

dτdτ.

The variables used in the first form of Boltzmann’s equation can be interpreted as follows: τ standsfor stress-input time, t stands for strain-output time, and (t− τ) stands for polymer memory time.

13.6 Connection between the creep compliance and the stress relaxation modulus

The connection between creep compliance and stress relaxation modulus can be determined by theLaplace transform,

L[f(t)] ≡ F (P ) ≡∫ ∞

0

e−ptf(t)dt.


By applying the Laplace transformation to the two integral forms of Boltzmann’s equation we obtain

L[σ(t)] =1

p2L[D(t)]

which can be rearranged as∫ t

0

E(t− τ)D(τ)dτ =

∫ t

0

D(t− τ)E(τ)dτ = t.

The product Er(t)Dc(t) is close to unity in region 1 and 3, but closer to zero in region 2 and 4. Thiscan be used as a means of determining in which region of mechanical equivalence a polymer is in.

Furthermore, McLoad developed the approximation

Dc(t)Er(t) 'sinmt

mπ

where m is the negative slope of the master curve.

13.7 The correspondence principle

Assume that the stress is spatially separable, i.e. σ(x, t) = f1(x)f2(t), then Boltzmann’s equationbecomes

ε(x, t) = f1(x)

∫ t

0

Dc(t− τ)df2(τ)

dτdτ,

which for the case of creepf2(τ) = σ0H(τ),

where H(τ) is the Heaviside step function, leads to

ε(x, t) = f1(x)σ0Dc(t).

Hence, for creep loading, linear elasticity solutions can be used if E is replaced by 1/Dc(t).Similarly, for the case when the strain is spatially separable and the loading consists of a constant

strain; we can determine the linear viscoelastic response by replacing E by Er(t).

13.8 ODE representation of Boltzmann’s equation

Boltzmann’s equation can equivalently by represented as a linear ODE with constant coefficients

andnε

dtn+ · · ·+ a0ε = bm

dmσ

dtm+ · · ·+ b0σ.

In a relaxation test, the right hand side of the equation above becomes zero, and the solution to thedifferential equation can be written as

ε(t) =

n∑i

ciepit.

Mechanical memory of a material is the ability of the material to eventually return to the originalstate of strain. Note that an elastic material has complete and instantaneous recovery. Thus,a material exhibits complete recovery if, and only if, all roots of the characteristic equation arenegative. Furthermore, sometimes a material has optical memory (something that can be measuredby birefringence).


Applicability of linear viscoelasticity: in region 1, the strain has to be / 1%; increases sharplyabove Tg, becomes ' 100% strain close to Tg + 50 C.

13.9 Constitutive relation for an ideal rubber

The constitutive relation for an ideal rubber can, somewhat simplisticaly, be written

σx =3ρRT

MC

(λ− 1

λ2

)

13.10 Constitutive relations for a Maxwell body

For a Maxwell body the strain rate is given by

ε =σ

η+σ

E.

For a stress relaxation experiment (ε = 0) the constitutive relation becomes

σ = σ0 exp

(−tτ

),

where τ = η/E. Furthermore, the characteristic relaxation time is defined by τ = t, which leads toσ = σ0/e ≈ 0.37σ0.

13.11 Viscoelastic damping

Consider a viscoelastic system that is subjected to a sinusoidal strain. Due to the material behaviorthe stress response will be out of phase. Introduce the following definitions:

E′ = The fraction of the stress which follows the imposed strain without any delay, divided by thestrain itself. Is called the storage modulus.

E′′ = The fraction of the stress which follows the imposed strain with a delay of 90 divided by thestrain. Is called the loss modulus.

Can also define a complex modulus by

E∗(ω) = E′(ω) + iE′′(ω),

and a loss tangent

tan δ =E′′(ω)

E′(ω).

13.12 Linear Viscoelasticity Theory

When subjected to an external loading agent many materials do not exhibit a purely elastic or purelyviscous response, but a combination of the two. The simplest way to model this behavior is throughlinear viscoelasticity. This material representation has been used and studied extensively for manyyears, but despite this there are some aspects of the theory that can be difficult to grasp at first,and also since other more complicated models often are generalizations of this theory, a relativelydetailed derivation is given below.

The foundation of linear viscoelasticity theory is the Boltzmann’s ? superposition principle whichis discussed next.

Boltzmann’s Superposition Principle.One way to state the Boltzmann’s superposition principle is through the statement:


Each loading step makes an independent contribution to the final state.

This idea can be used to formulate an integral representation of linear viscoelasticity. The strategyis to perform an experiment in which a step function in strain is applied, ε(t) = ε0H(t), and thestress response σ(t) is measured. Then a stress relaxation modulus can be defined by G(t) = σ(t)/ε0.Note that ε0 does not have to be infinitesimal due to the assumed superposition principle.

To develop a model capable of predicting the stress response due to an arbitrary strain history,start by decomposing the strain history into a sum of infinitesimal strain steps:

ε(t) =∑i

∆εiH(t− τi) (1)

the stress response can therefore directly be written

σ(t) =∑i

∆εiG(t− τi). (2)

In the limit as the number of strain increments goes to infinity, the stress response (2) becomes

σ(t) =

∫ t

−∞G(t− τ)dε(t) =

∫ t

−∞G(t− τ)

dε(τ)

dτdτ. (3)

Note that if G(t) = G0, then σ(t) = G0ε(t). If only the response in one loading mode (i.e. uniaxialloading) is of interest, then it is sufficient to determine the stress relaxation modulus for that loadingmode and then by using Equation (3) the response due to any imposed deformation can be obtained.

If Equation (3) is generalized to a three-dimensional deformation state for an isotropic material,the following relationship is obtained:

T(t) =

∫ t

0

2G(t− τ)edτ + I

∫ t

0

K(t− τ)φdτ (4)

where G(t) is the stress relaxation shear modulus, e the rate of change of deviatoric strains, K(t) thestress relaxational bulk modulus, and φ the rate of change of volumetric strains. That is, only tworelaxation moduli need to be determined to predict any arbitrary deformation. The correspondingrelationship for a general anisotropic material is

σij(t) = Gijkl(0)εkl(t) +

t∫0

εkl(t− s)dGijkl(s)

dsds. (5)

Note, to be able to predict the stress response at any arbitrary strain history it is sufficient to knowthe stress relaxation modulus, and to be able to follow a stress history the creep compliance needsto be known.

Example 1. Consider once again a general one-dimensional loading case. To exemplify the modelconsider first a case in which

G(t) =

G0 exp(−α0t) if t ≥ 0,

0 if t < 0(6)

and

ε(t) =

0 if t < 0,

εt if t ≥ 0,(7)

see Figure below.


Equation (3) then gives

σ(t) =

∫ t

0

G0ε exp[−α0(t− τ)]dτ,

yielding

σ(t) =G0ε

α0[1− exp(−α0t)] . (8)

The result is plotted in the figure above, illustrating that large α0 corresponds to fast relaxation.From (8) it is also clear that the stress response is always predicted to be proportional to the appliedstrain rate. And even if a relaxation spectrum is chosen, the Boltzmann’s superposition principledoes not allow for a sigmoidal shaped stress-strain response.

Another commonly used functional form of the relaxation modulus is the stretched exponential:

G(t) = G0 exp

[−(t

τ0

)β](9)

which is plotted in the figure below.


If we again consider a constant strain rate loading situation, the stress response becomes

σ(t) =

∫ t

0

G0ε exp

[−(t

τ0

)β]dt

which is not easy to express in elementary functions, but it is clear that σ(t) is still proportional toε.

Example 2. As a last example consider

G(t) =

G0 exp(−α0t) if t ≥ 0,

0 if t < 0(10)

and

ε(t) =

0 if t < 0,

ε0 sin(ω0t) if t ≥ 0,(11)

giving

σ(t) =

∫ t

0

G0 exp(−αt)ε0ω0 cos(ω0τ)dτ

=G0ε0ω0

α20 + ω2

0

[α0 cos(ω0t) + ω0 sin(ω0t)− α0 exp(−α0t)] . (12)

It is also possible to formulate the whole theory for the case when the stress is the drivingquantity. If in this case the creep compliance is defined as J(t) = ε(t)/σ, the resulting strain can becalculated from

ε(t) =

∫ t

−∞J(t− τ)

dσ(τ)

dτdτ. (13)

Storage and Loss Moduli.As a final illustration will also be shown how the theory can be applied to a sinusoidally drivingstrain state specified by

ε(t) = ε0 sin(ωt). (14)


The resulting stress response can be written:

σ(t) =

∫ ∞0

G(s)ωε0 cos [ω(t− s)] ds (15)

where s ≡ t− τ . This equation can be expanded into

σ(t) = ε0 sin(ωt)

[ω

∫ ∞0

G(s) sin(ωs)ds

]+ ε0 cos(ωt)

[ω

∫ ∞0

G(s) cos(ωs)ds

]. (16)

Note that the integrals only converge if

lims→∞

G(s) = 0. (17)

By defining two frequency-dependent functionals: the storage modulus G′(ω) and the loss modulusG′′(ω), the stress response can be written

σ(t) = ε0 [G′(ω) sin(ωt) +G′′(ω) cos(ωt)] . (18)

or alternatively

σ(t) = σ0 sin(ωt+ δ) (19)

= σ0 sin(ωt) cos δ + σ0 cos(ωt) sin δ (20)

hence

ε0G′(ω) = σ0 cos δ (21)

ε0G′′(ω) = σ0 sin δ (22)

givingG′′

G′= tan δ (23)

It is often convenient to introduce a complex variable based notation

G∗ =σ∗

ε∗= G′ + iG′′. (24)

It is also possible to consider a stress driven oscillation giving a complex compliance:

J∗ =ε∗

σ∗=

1

G∗= J ′ − iJ ′′. (25)

The relationships between J ′, J ′′, G′ and G′′ are:

J ′ =G′

G′2 +G′′2(26)

J ′′ =G′′

G′2 +G′′2(27)

G′ =J ′

J ′2 + J ′′2(28)

G′′ =J ′′

J ′2 + J ′′2(29)


Spectra.Several different means of specifying viscoelastic mechanical properties have been given. Specifically,relaxation functions, creep functions and complex moduli have been discussed. Another way ofcharacterizing the material response is through spectra. To introduce the concept of a relaxationspectrum consider the relaxation modulus

G(t) =∑n

Gn exp

[−tτn

]. (30)

In the limit as n→∞ the summation is replaced by an integral:

G(t) =

∫ ∞0

f(τ) exp

[−tτ

]dτ. (31)

The function f(τ) is called the relaxation time spectrum. In practice it is often more convenient touse a logarithmic time scale

G(t) =

∫ +∞

−∞H(τ) exp

[−tτ

]d(ln τ) +G(∞). (32)

Similarly, a retardation time spectrum can be defined by

J(t) =

∫ +∞

−∞L(τ)

[1− exp

(−tτ

)]d(ln τ) + J(∞). (33)

It is possible to solve for H(t) and L(t) by using Laplace transforms, see ? for details.

Relationships Between Different Viscoelastic Functions.The formal relationship between the relaxation modulus G(t) and the creep compliance J(t) can bederived by taking the Laplace transformation of Equations (3) and (13)

σ(s) = sG(s)ε(s) (34)

ε(s) = sJ(s)σ(s) (35)

which can be rewritten as

G(s) =1

s2J(s). (36)

Based on the Laplace limit theorems it can be shown ? that:

limt→0

J(t) = limt→0

1

G(t)(37)

limt→∞

J(t) = limt→∞

1

G(t)(38)

Also since L−1(1/s2) = t we directly get:∫ t

0

G(t− τ)J(τ)dτ =

∫ t

0

J(t− τ)G(τ)dτ = t. (39)

Differential Form of the Integral Representation.In this section will be shown that the integral form

σ(t) =

∫ t

−∞G(t− τ)

dε(τ)

dτdτ. (3-rep)


can also be written in differential form as

P (D)σ(t) = Q(D)ε(t) (40)

where P (D) and Q(D) are polynomials of D ≡ d/dt.To show this start by taking the Laplace transform of Equation (40):

P (s)σ(s)− 1

s

N∑k=1

pk

k∑r=1

srσ(k−r)(0) = Q(s)ε(s)−N∑k=1

qk

N∑r=1

srεk−r(0). (41)

The Laplace transform of (3) isσ(s) = sG(s)ε(s), (34-rep)

demonstrating that the two forms are equal if

sG(s) =Q(s)

P (s)(42)

andN∑r=k

prσr−kij (0) =

N∑r=k

qrεr−k(0), k = 1, 2, . . . , N (43)

which is a restriction on the initial conditions.

Rheological Models—Maxwell Model.The Maxwell rheological model constitutes a linear spring (σ = Eε1) and a linear dashpot (ε2 = σ/η)in series. The rate of change in strain of the system is given by

ε =σ

E+σ

η(44)

which is the differential representation. Now consider a stress relaxation experiment ε(t) = ε0H(t)giving

dσ

dt+E

ησ = 0, (for t > 0)

having the solution

σ(t) = σ0 exp

[−tη/E

]. (45)

Hence the stress relaxation function for the Maxwell model is

G(t) = G0 exp

[−tη/E

].

In this section the key aspects of linear viscoelasticity theory has been presented. It has beenshown that to characterize the material only one2 functional form needs to be determined. Thematerial functional can be determined through one creep, stress relaxation or oscillatory experiment.Once the material dependent functional has been determined the response due to any arbitraryimposed deformation or loading can be directly calculated.

2Only one material function is required for an incompressible isotropic material subjected to a one-dimensionalloading situation. For an isotropic material subjected to a general loading situation two material functionals need tobe determined, and for an anisotropic material the number of functional are dependent on the the material symmetry.


13.13 The Use of Shift Functions to Generalize Linear Viscoelasticity Theory

For many materials the domain in which linear viscoelasticity theory give good predictions is un-fortunately relatively small. Based on experimental observations it turns out that the influence ofvariations in external parameters (such as temperature and aging time), which contribute to thelimited applicability of the theory, can be accounted for by using a shift function approach. Thereason for the success of this simple idea is that when the material functional is plotted using ap-propriate log-scales, changes in these external parameters do not change the shape of the plottedcurve to any significant degree, only shift it.

Time-Temperature Equivalence.So far the discussion has been for a general viscoelastic material and can therefore be appliedalso to a polymer. But when considering polymers a number of complications become appar-ent. One of the more important issues that need to be recognized is the strong temperaturedependence of the material. It has been shown experimentally that in many cases the temper-ature dependence can be modeled by a scaling of time using what has been termed the time-temperature equivalence. The basis for this principle is shown in the figure below illustratingthat if the experimentally observed stress relaxation modulus is plotted as a function of loga-rithmic time, the shape of the resulting curves is the same for a wide interval of temperatures.

In fact, the only significant difference between the curves is a horizontal shift. This observationsuggests that if the relaxation modulus is known at one temperature (i.e. the ‘master curve’ isknown) then the relaxation modulus at any other temperature is obtained if the horizontal shiftfactor aT = aT (T ) is known. Any material that has this property is called a rheologically simplematerial.

The time shifts can be written

log tT0 − log tT = log aT (1)

where tT is the time at temperature T and tT0 the time at temperature T0. Equation (1) givesaT = tT0

/tT or tT = tT0/aT , hence the behavior at a temperature of T becomes exactly the same as


the behavior at the reference temperature T0 if the time is accelerated by the factor aT . In generalthe temperature is a function of time T = T (tT0

), so instead of tT = tT0/aT it is necessary to write

dtT =dtT0

aT (T (tT0))

(2)

giving

tT =

∫ tT0

0

dt′T0

aT (T (t′T0)). (3)

The effective time experienced by the material–the material time–is a function of temperature andwall clock time. For a rheologically simple material, the scaling of time with temperature occursin all viscoelastic quantities such as G, J , J ′, J ′′, tan δ, etc. And the scaling constant must be thesame for all quantities for the material to be rheologically simple.

Example 3. To exemplify this idea consider a simple Maxwell element: G(t) = G0 exp(−t). If thematerial is rheologically simple the relaxation modulus becomes

G(t, T ) = G0 exp[−aT (T )t].

The influence of aT (T ) on the stress-strain behavior was examined in Example 1.

One commonly quoted representation of aT is the WLF-equation ?:

log aT (T ) =C1(T − T0)

C2 + T − T0(4)

where C1 = 17.4 and C2 = 51.6 K. This relationship is often used for amorphous polymers in thetemperature range T ∈ [Tg − 50 K, Tg + 50 K].

Vertical Shifts.As will be discussed in more detail below, the stress relaxation modulus (and also the creep compli-ance) curves when plotted as a function of logarithmic time often turn out to have the same shapenot only for different temperatures but also for variations in other parameters (such as aging time).But to create a master curve in these cases it is often necessary to also use vertical shifts (on alog-scale), see the figure below.


logGα − logGα0 = log b (5)

giving

b =Gα

Gα0. (6)

In summary, the integral formulation

σ(t) =

∫ t

−∞G(t− τ)

dε(τ)

dτdτ. (3-rep)

becomes

σ(t) =

∫ t

0

b(t)G(t(t)− t(τ))dε(τ)

dτdτ (7)

when both vertical shift b(θ1(t)) and horizontal shift t(t) = t(θ2(t)), where θ1 and θ2 are externalparameters such as temperature, are considered.

13.14 Constitutive Modeling of the Equilibrium Response of Elastomers

Consider an elastomeric body B in its reference configuration B0 ⊂ R3 with the correspondingmaterial points labeled by p ∈ B0. It is assumed that the boundary ∂B0 of the region B0 is regularenough to ensure the validity of the divergence theorem, and that the exterior normal n0 to B0 isdefined for almost every point p of ∂B0, where ‘almost everywhere’ means that the set of pointswhere n0 is not defined has the area measure zero. Denote by χt(p) : B0 × R → R3 a macroscopicmotion of B0 in a time interval t ∈ [0, t1] ⊂ R+ which maps any material point p to the pointx = χt(p) ∈ R3 in the deformed configuration Bt = χt(B0) ⊂ R3. In the following the referenceconfiguration is taken to be stress-free with homogeneous reference temperature θ0 ∈ R+. The totaldeformation gradient defined by F = ∂p χ(p, t) can be decomposed into distortional and dilatationalparts: F = (J)1/3F∗ where J = det(F). To get a physical picture of the deformation process ofan elastomer assume that on average the chain molecules can be considered to be located along thediagonals of a unit cell located in principal stretch space as illustrated in the figure below (this isthe eight chain assumption of ?.)


Denote the side lengths of the unit cell in the reference state by a0 and the undeformed chain lengthby r0 (it then directly follows that r0 = a0

√3.) Further take the macromolecules to be freely jointed

with n rigid links each of length l. For this chain model the average end-to-end distance in theabsence of an external force field is l

√n. By defining λ∗i | i = 1, 2, 3 to be the applied principal

distortional stretches the effective distortional chain length becomes r = a0[(λ∗1)2 +(λ∗2)2 +(λ∗2)2]1/2,giving the effective distortional chain stretch

λ∗ =

[(λ∗1)2 + (λ∗2)2 + (λ∗3)2

3

]1/2

=

√trC∗

3=

√trB∗

3,

where B∗ = (J)−2/3B. Based on this physical model, an eight-chain material is defined as anisotropic thermoelastic material whose strain energy density W only depends on the two invariantsλ∗(B∗), J(F), and the temperature θ. By noting that the effective chain stretch is related to the firstinvariant of B∗ through λ∗ = [I1(B∗)/3]1/2 it follows that the strain energy density can be writtenW (λ∗, J, θ), or alternatively W (I1∗ , J, θ) where I1∗ := I1(B∗) = tr(B∗). To determine the strainenergy density, W , first use the experimental observation that the internal energy is not a functionof the applied distortional stretch, i.e. e(J, θ). The specific heat at constant volume, defined bycv = ∂e(J, θ)/∂θ, can then be written

cv(J, θ) =∂ψ(λ∗, J, θ)

∂θ+ η(λ∗, J, θ) + θ

∂η(λ∗, J, θ)

∂θ. (1)

where the definition of the Helmholz free energy ψ(λ∗, J, θ) = e(J, θ) − θη(λ∗, J, θ) has been used.Since the first two terms of the right hand side of (1) have to cancel to satisfy the Clausius-Duheminequality the specific heat becomes

cv(J, θ) = θ∂η(λ∗, J, θ)

∂θ. (2)

Further assume that the material is almost incompressible and that the temperature does not varytoo much such that cv, to a first order approximation, can be taken to be a constant. Equation (2)then yields

ψ(λ∗, J, θ) = cvθ(1− ln θ) + e1(J)− θη(λ∗, J). (3)

Assuming small volumetric deformations, the linearlized relationship between the Cauchy stress Tand the volumetric deformation J is taken as

∂W

∂J= T ∝ κ(J − 1)1 (4)


giving the convex relationship e1(J) = κ[J2/2− J + 1/2]/ρ0 which when inserted into (3) gives

W (λ∗, J, θ) = ρ0cvθ[1− ln θ] + κ

[J2

2− J +

1

2

]− θρ0η(λ∗). (5)

Note that the dependence on J in η(λ∗, J) has been neglected due to the the assumption of smallvolume change. It now only remains to determine how the entropy depends on the effective chainstretch. By assuming the molecules to be freely jointed with a fixed bond length the entropy can bedetermined from the statistical mechanics relation [ρ0η] ∝ NkB ln Ω(r) where Ω(r) is the probabilitydistribution of the end-to-end distance of the molecular chain and N is the number of chains perunit reference volume. Flory [?] showed that the probability distribution under these conditions canbe written

Ω(r) =1

2π2r

∫ ∞0

q sin(qr)

[sin ql

ql

]ndq. (6)

One approximation of (6) good for n 1 and r ≈ n is the Langevin expression attributed to Kuhnand Gruhn [?]:

Ω(r) =A′

l3

[sinhβ

β

]nexp

[−βrl

](7)

where β = L−1(r/(nl)) and L(x) = coth(x)− 1/x is the Langevin function. By inserting (7) into (5)the strain energy density can be written

W (λ∗, J, θ) = ρ0cvθ[1− ln θ] + κ

[J2

2− J +

1

2

]−NkBθn ln

(sinhβ

β

)+NkBθβλ

lockλ∗ +W0 (8)

where

β = L−1

(λ∗

λlock

), λlock =

√n. (9)

The Cauchy stress for an eight-chain material can then be obtained from the continuum mechanicsexpression

T =2

J

[W1∗ + W2∗I1∗

]B∗ − 2

JW2∗ (B∗)

2+ [

∂W

∂J− 2I1∗

3JW1∗ −

4I2∗

3JW2∗

]1, (10)

which in this case with no dependence on I2∗ can be simplified to

T =2

JW1∗ dev[B∗] +

∂W

∂J1 (11)

or when expressed in terms of the effective chain stretch λ∗

T =1

J

1

3λ∗∂W

∂λ∗dev[B∗] +

∂W

∂J1. (12)

The Cauchy stress can then be calculated from (12) giving

T =−θ

3Jλ∗d[ρ0η(λ∗)]

dλ∗dev[B∗] + κ[J − 1]1. (13)


From the chain rule

∂[ρ0η(r(λ∗))]

∂λ∗=∂[ρ0η]

∂r

∂r

∂λ∗=∂[ρ0η]

∂r

1

∂∂r

[r

l√n

] = l√n∂[ρ0η]

∂r

it is clear that it is sufficient to determine how the entropy of a single macromolecule depends onits end-to-end distance. In the limit n→∞, Equation (7) becomes a Gaussian distribution

Ω(r) =

[3

2πnl3

]3/2

exp

[−3r2

2nl2

]. (14)

It is easy to show that the change in entropy with chain length for the Langevin expression thereforecan be written

∂[ρ0η]

∂r=∂[Nk ln Ω(r)]

∂r=−NkBlL−1

( rnl

),

and that the corresponding expression for Gaussian chains becomes

∂[ρ0η]

∂r=∂[Nk ln Ω(r)]

∂r= −3NkBr

nl2.

Equation (13) can now be written

T =NkBθ

3J

λlock

λ∗L−1

(λ∗

λlock

)dev[B∗] + κ[J − 1]1, (15)

where λlock =√n. For the special case of incompressible uniaxial deformation Equation (15) sim-

plifies to

σ =NkBθ

3

λlock

λ∗L−1

(λ∗

λlock

)[λ2 − 1

λ

](16)

and for simple shear defined by F = 1 + γe1 ⊗ e2 the shear stress is given by

T12 =NkBθ

3J

λlock

λ∗L−1

(λ∗

λlock

)γ (17)

where λ∗ =√

1 + γ2/3. The initial shear modulus of the material is given by µ0 = ∂T12

∂γ |γ=0 giving

µ0 =NkBθ0

3λlockL−1

(1

λlock

)(18)

which when inserted in (15) gives the Cauchy stress as

T =θ

θ0

µ0

J λ∗

L−1

(λ∗

λlock

)L−1

(1

λlock

) dev[B∗] + κ[J − 1]1. (19)

For the special case of incompressible uniaxial deformation the corresponding equation becomes

σ =µ0

λ∗

L−1

(λ∗

λlock

)L−1

(1

λlock

) [λ2 − 1

λ

]. (20)


The constitutive relationship for the case of Gaussian chains is directly obtained by replacing theinverse Langevin function by the first term in its series expansion L−1(x) ≈ 3x, giving

T =θ

θ0

µ0

Jdev[B∗] + κ[J − 1]1. (21)


14 Continuum Mechanics Foundations 14

14.1 List of Symbols

Direct notation is used throughout, with the same conventions as in ?, ? and ?.Bt ⊂ R3 position of the material body B at time tp ∈ B0 position of a material point in the reference configurationx ∈ Bt position of a material point in the current configurationF ∈ Lin+ deformation gradienta0, a area element in the reference and current configurationv0, v volume element in the reference and current configurationn0, n normal to the surface in the reference and current configurationρ0, ρ mass density in the reference and current configurationT Cauchy stressS first Piola-Kirchhoff stressSpk2 second Piola-Kirchhoff stressb specific body force (i.e. per unit mass)τ Kirchhoff stresse specific internal energyη specific entropyθ absolute temperatureg0, g temperature gradient in the reference and current configurationψ specific Helmholtz free energyW strain energy density per unit reference volumer specific heat supplyt Cauchy tractionh0, h heat flux in the reference and current configurationq0, q heat flux vector in the reference and current configuration

14.2 Elements of Tensor Algebra and Analysis

I A ·B := tr(ABT ) = AijBij , ∀A, B ∈ Lin.

I For all vectors a, b, c, d:

(a⊗ b)c = (b · c)a,

(a⊗ b)(c⊗ d) = (b · c)(a⊗ d).

I The rank r(A) of a tensor A is the dimension of the range of the linear transformation A.

I Derivatives are identified with linear transformations. Let f : D → Y where D ⊂ X, thenDf(x) = ∂xf(x) is a linear transformation from X to Y. The value of the transformation atu ∈ X is denoted Df(x)[u] = ∂xf(x)[u], which for inner product spaces also can be writtenDf(x) · u = ∂xf(x) · u.

I The cofactor of A ∈ Lin is cof A := ∂A detA which if A ∈ invLin can be written (detA)A−T

I A number v ∈ R+ and a vector e ∈ V are said to be a singular value and a singular vector

of F ∈ Lin if they are an eigenvalue and an eigenvector of√FFT . When F ∈ invLin is

a deformation gradient, the singular values of F are called the principal stretches and thesingular vectors the principal directions.

76 CONTINUUM MECHANICS FOUNDATIONS

I Let W = ρ0ψ be the strain energy per unit reference volume, and Wi := ∂W/∂Ii.

I The invariants of B can be written:

I1 = trB = λ21 + λ2

2 + λ23

I2 =1

2

[tr(B2)− (trB)2

]= λ2

1λ22 + λ2

2λ23 + λ2

3λ21

I3 = detB = λ21λ

22λ

23

14.3 Kinematics of Continuous Bodies

I Consider a continuum body B with reference configuration B0 ⊂ R3 and material points labeledby p ∈ B0. Assume that the boundary ∂B0 of the region B0 is regular enough to ensure thevalidity of the divergence theorem, and that the exterior normal n0 to B0 is defined for almostevery point p of ∂B0, where ‘almost everywhere’ means that the set of points where n0 is notdefined has the area measure zero.

I Each open region Bt ⊂ R3 whose boundary has the chosen degree of regularity is called anadmissible region.

I The value x = χ(p, t) is the position of the material point p ∈ B0 at time t ∈ [t1, t2] ⊂ R.The partial mapping χ(·, t) is the configuration at t. The spatial set χ(p, t) ∈ R3 : p ∈ Bis the current region occupied by the body at time t. The deformation gradient is defined byF = ∂p χ(p, t).

I We shall associate with each body B a set Σ, the state space, and another set Π, the class ofprocesses. The elements σ of Σ are the states of the body and the elements π of Π are theprocesses. Each process is a function π : [0, tπ] → Σ. The value σ(t) is the state of the bodyat time t ∈ [0, tπ]. The pair (Σ,Π) is called a system.

I Let [t1, t2] ⊂ R be a time interval, B an admissible region and φ a continuously differentiablefunction defined on an open set containing the closure of B× [t1, t2] such that ∇φ 6= 0 at everypoint where φ = 0. A moving surface in B on [t1, t2] is a family St, t ∈ [t1, t2] of closedoriented surfaces in B given by

St = p ∈ clB : φ(p, t) = 0, t ∈ [t1, t2].

The spacetime set U composed of the spacetime points on the singular surface, given by

U = (p, t) ∈ clB × [t1, t2] : φ(p, t) = 0

is a three-dimensional hypersurface.

I The jump [f ] and the mean value 〈f〉 of f on the singular surface are defined by [f ] = f+−f−and 〈f〉 = 1

2 (f+ + f−), respectively.

I Let the operator ∇ denote the referential gradient, and the superimposed dot the materialtime derivative:

∇f(p, t) = Grad f(p, t) = ∂pf(p, t), f(p, t) = ∂tf(p, t)

I Lines, areas and volumes are deformed according to: dl = Fdl0, da = cof Fda0, dv = detFdv0.


I Let f(p, t) = f(x, t). The spatial gradient and the partial time derivative of f are denoted bygrad f = ∂xf(x, t) and ∂f/∂t = ∂tf(x, t), respectively.

I • Deformation gradient

F = RU = VR =∑

λi li ⊗ ri

• Right stretch tensor

U =∑

λi ri ⊗ ri = RTVR

• Left stretch tensorV =

∑λi li ⊗ li = RURT

• Rotation tensorR =

∑li ⊗ ri

• Eigenvectorsli = Rri

• Left Cauchy Green tensor (Finger tensor)

B = FFT = V2 = RCRT =∑

λ2i li ⊗ li

• Right Cauchy Green tensor

C = FTF = U2 = RTBR =∑

λ2i ri ⊗ ri

• Cauchy stress (true stress)

T =∑

σi li ⊗ li

• Piola-Kirchhoff stressS =

∑si li ⊗ ri

• Second Piola-Kirchhoff stress

Spk2 =∑

spk2i ri ⊗ ri

14.4 Balance Equations

I A balance equation expresses the time derivative of an extensive quantity contained in a volumein terms of its fluxes through the boundary and the internal source of the quantity.

I Piola’s transformation: Let q0 be a continuously differentiable vector field in the referencedescription. The interpretation of q0 as a flux vector is that if B0 is any oriented surface inthe reference configuration with normal n0, then the integral∫

B0

q0 · n0 da0 =

∫Bt

q · n da

is the rate at which the quantity under consideration flows through ∂Bt.This directly leads to q0 = (cof F)Tq, and Divq0 = detF divq outside the singular surface.


I Balance of mass

M =

∫Btρdv, ρ = ρ0/J, ∂tρ+ div(ρv) = 0.

I Let f be the resultant Cauchy traction acting on a body:

f =

∫∂B0

Sn0 da0 =

∫∂Bt

Tn da.

I Balance of linear momentum: material form∫∂B0

Sn0 da0 +

∫B0

ρ0b dv0 =d

dt

∫B0

ρ0v dv0,

and in the current configuration∫∂Bt

Tn da+

∫Btρb dv =

d

dt

∫Btρv dv

and the corresponding local forms are

DivS + ρ0b = ρ0v , and divT + ρb = ρv.

I Balance of angular momentum: material form in the reference configuration∫∂B0

(x− x0)× Sn0 da0 +

∫B0

ρ0(x− x0)× b dv0 =d

dt

∫B0

ρ0(x− x0)× v dv0.

Note, the balance of angular momentum implies the balance of linear momentum, or theequation of balance of angular momentum is satisfied iff the equation of linear momentum is

satisfied and the Cauchy stress is symmetric. The local form is SFT = FST .

I Theorem of virtual work: let (f ,b) be a system of forces on B during a motion χ. Then anecessary and sufficient condition that the momentum laws be satisfied is that given a regionR ⊂ Bt and time t ∈ R, ∫

∂Rf(n) ·wda+

∫Rb ·wdv =

∫Rρvdv

for every infinitesimal rigid displacement w.

I Let h0(n0) be the heat influx per unit referential area. Then

h0(n0) = −3∑i=1

(n0 · ei)h0(ei) = −

[3∑i=1

h0(ei)ei

]· n0 ≡ −q0 · n0

proves the existence of a referential heat flux vector q0 that does not depend on n0.

I Relations between the different stress measures

S = JTF−T , T =1

JSFT

Spk2 = JF−1TF−T = F−1S, T =1

JFSpk2FT

τ = JT = SFT


I By introducing the specific internal energy e, we can state the material rate form of the balanceof energy as∫

∂B0

(v · Sn0 − q0 · n0)da0 +

∫B0

ρ0(v · b + r)dv0 =d

dt

∫B0

ρ0

(e+

v2

2

)dv0,

or in local form expressed per unit reference volume

S · F−Divq0 + ρ0r = ρ0e,

or per unit current volumeT · L− divq + ρr = ρe.

I The existence of the absolute temperature and entropy is postulated for general nonequilibriumprocesses. It is assumed that these quantities satisfy the Clausius-Duhem inequality. Byintroducing the specific entropy η and the absolute temperature θ ∈ R+ the material form ofthe Clausius-Duhem inequality becomes:∫

∂B0

−q0 · n0

θda0 +

∫B0

ρ0 r

θdv0 ≤

d

dt

∫B0

ηρ0 dv0.

The local material form expressed per unit reference volume

−Divq0

θ+ρ0 r

θ≤ ρ0 η

which also can be writtenρ0e− S · F +

q0 · g0

θ≤ ρ0 θ η

where g0 := ∂pθ is the referential gradient of the temperature. Or in terms of the currentconfiguration

ρe−T · L +q · gθ≤ ρ θ η.

I By introducing the specific Helmholtz free energy per unit mass ψ := e − θη, the Clausius-Duhem inequality can be restated as the dissipation inequality

ρ0ψ − S · F + ρ0 η θ +q0 · g0

θ≤ 0

or

ρψ −T · L + ρηθ +q · gθ≤ 0.

14.5 The First Law of Thermodynamics

I For a process π ∈ Π and a region R ⊂ B the work w(π,R) done by R in π is

w(π,R) = −∫ tπ

0

[∫∂R

v · Sn0 da0 +

∫Rv · b dv0

]dt

and the net heat gained by R in π is

q(π,R) =

∫ tπ

0

[−∫∂R

q0 · n0 da0 +

∫Rr dv0

]dt.


I First law of thermodynamics: w(π,R) = q(π,R) for every cyclic process π and every regionR.

14.6 The Principle of Material Frame Indifference

I A mapping i : R3 → R3 is said to be an isometry if |i(x)− i(y)| = |x− y| for every x,y ∈ R3.A change of frame is a family i = it, t ∈ R of isometries. It is of the form

it(x) = R(t)(x− x0) + x′0, x ∈ R3, t ∈ R

where R(·) : R→ Orth, and x′0 : R→ R3 are functions and x0 ∈ R3 a point.

I The internal energy e is a local state function on (Σ,Π,B) defined by e(σ,p) = u(σ,p) −12ρ0(p)v(σ,p)2, where u is the specific total energy.

I The internal energy is objective, i.e. for every process π and every change of frame i we havee((i π)(t), ·) = e(π(t), ·) for every region R and every t ∈ [0, tπ].

14.7 Isotropic Functions

I A subset V of Sym is said to be isotropic if for every A ∈ V and every Q ∈ Orth alsoQAQT ∈ V.

I A scalar-valued function h : V → R is said to be isotropic if its domain is isotropic andh(QAQT ) = h(A) for every A ∈ V and every Q ∈ Orth.

I A tensor-valued function G : V → Lin is said to be isotropic if its domain is isotropic andG(QAQT ) = QG(A)QT for every A ∈ V and every Q ∈ Orth.

I A subset V of Lin is said to by objective if for every A ∈ V and every Q ∈ Orth also QA ∈ V.A scalar valued function and a tensor valued function are said to be objective if h(QA) = h(A)and G(QA) = QG(A), respectively.

14.8 Constitutive Equations

I There are 19 unknown field variables in the governing equations developed above (x, S, q0, e,η, θ, ρ) but only 8 equations. The remaining 11 equations are the constitutive equations: e(·),η(·), S(·), q0(·). Candidate arguments for the constitutive equations are: p, F, F, ∂pF, θ, θ,∂pθ, ρ, ρ, ξ, etc.

I An adiabatic thermoelastic material is defined by e(·), η(·), S(·) being functions of F and θ.

I A thermoelastic material is defined by e(·), η(·), S(·), q0(·) being functions of F, θ, g0.

I A viscous body with heat conduction is defined by e(·), η(·), S(·), q0(·) being functions of F,F, θ, g0.

I A process in which θ and F are constants in space and time is called an equilibrium process.The total stress can be decomposed into equilibrium and dynamic components:

S(F, θ,g0, F) = Sd(F, θ,g0, F) + Se(F, θ).

I A viscous body with heat conduction satisfies the dissipation inequality (i.e. the Clausius-Duhem inequality) if and only if the following three conditions hold:


1. the constitutive functions ψ(·), η(·), e(·) are independent of the non-equilibrium variablesg0, F

2. the following equilibrium thermostatic relations hold

Se(F, θ) = ρ0∂ψ(F, θ)

∂F

η(F, θ) = −∂ψ(F, θ)

∂θ

e(F, θ) = −θ2 ∂

∂θ

[ψ(F, θ)

θ

]3. the following dissipation inequality holds

Sd(F, θ,g0, F) · F− q0(F, θ,g0, F) · g0/θ ≥ 0

I The principle of frame indifference causes further restrictions on the constitutive equations:Se(U, θ), e(U, θ), ψ(U, θ), η(U, θ). For the non-equilibrium response (L ≡ gradv = ∂xv,g ≡ grad θ = ∂xθ): Sd(F, θ,g,D), q0(F, θ,g,D) and

Sd(RF, θ,Rg,RDRT ) = RSd(F, θ,g,D)

q0(RF, θ,Rg,RDRT ) = q0(F, θ,g,D)

Or in terms of C:

Se = 2ρ0F∂ψ

∂C

I Heat conduction law: Fouriers law:

q0 = K(C, θ)g0

where K is a positive semidefinite heat conductivity tensor.

I Elastic Fluid: ψ(J, θ)

I Specific heat c = ∂θe

I Green strain: EG = 12

[U2 − 1

]= 1

2 [C− 1], Logarithmic strain E = lnU.

14.9 Modeling of Hyperelastic Materials

Definition 0.1 The field variables φ(·),T(·), η(·),q(·) of a thermoelastic material only depend onthe state variables F, θ,g.

Theorem 0.1 To satisfy the Clausius-Duhem inequality the Cauchy stress for a thermoelastic ma-terial has to satisfy

T(F, θ) =1

J

∂W (F, θ)

∂FFT . (1)

Proof: The Clausius-Duhem inequality for a thermoelastic material can be written

ρψ −T · L + ρηθ +q · gθ≤ 0. (2)


Differentiating ψ and inserting into (2) yields

ρ (∂Fψ) · F + ρ (∂θψ) θ + ρ (∂gψ) · g −T · L + ρηθ +q · gθ≤ 0, (3)

which must hold for any process leading to

∂gψ = 0, η = −∂θψ, q · g/θ ≤ 0. (4)

Recalling that T · L = TF−T · F and ρ = ρ0/J directly leads to (1).

Theorem 0.2 By imposing objectivity on a thermoelastic material the Cauchy stress has to satisfy

T(C, θ) =2

JF∂W (C, θ)

∂CFT . (5)

Proof: To satisfy objectivity, a change of observer must imply T(F) = RT(U)RT and W (F, θ) =W (U, θ) = W (C, θ), which when inserted into (1) gives

T(U, θ) = R

[1

J

∂W (U, θ)

∂UUT

]RT , (6)

or when expressed in terms of C gives (5).

Theorem 0.3 For an isotropic material Equation (5) becomes

T =2

J(W1 +W2I1)B− 2

JW2B

2 +W31 (7)

where W (I1(C), I2(C), I3(F), θ) and W1 := ∂W/∂I1(C), etc.Proof: Equation (1) can be written

T(I1, I2, J, θ) =2

JF∂W

∂Ii

∂Ii∂C

FT ,

and by using ∂CI1 = 1, ∂CI2 = I11−C, and ∂CI3 = J2F−1F−T the proposed equation (7) directly

follows.

Theorem 0.4 If the deformation gradient is decomposed into dilatational and distortional compo-nents F = J1/3F∗, then the Clausius-Duhem inequality implies

T =1

J

∂W

∂F∗F∗T +

∂W

∂J1− 1

3J

(∂W

∂F∗· F∗

)1. (8)

Proof: By using the chain rule in index notation Equation (1) can be written

Tij =1

J

∂W

∂F ∗mn

∂F ∗mn∂Fij

+1

J

∂W

∂J

∂J

∂FikFjk

giving

Tij =1

J

∂W

∂F ∗mn

∂[J(F)−1/3Fmn]

∂Fik+∂W

∂Jδij

which when expanded equates to (8).


Theorem 0.5 If the deformation gradient is decomposed into dilatational and distortional compo-nents F = J1/3F∗, then the Clausius-Duhem inequality and objectivity gives

T =2

JF∗

∂W

∂C∗F∗T +

∂W

∂J1− 2

3J

(∂W

∂C∗·C∗

)1. (9)

Proof: Directly analogous to the previous proof.

Theorem 0.6 If the deformation gradient is decomposed into dilatational and distortional compo-nents F = J1/3F∗, then the Cauchy stress for an isotropic material has to be

T =2

J[W1∗ +W2∗I1∗ ]B

∗ − 2

JW2∗ (B∗)

2+∂W

∂J1− 2I1∗

3JW1∗1−

4I2∗

3JW2∗1, (10)

where W1∗ ≡ ∂W/∂I1(C∗). Proof: Directly analogous to the previous proof.

84 STATISTICAL MECHANICS

15 Statistical Mechanics 15

15.1 The phase rule

The number of degrees of freedom is given by

f = n+ 2− p

where n is the number of components and p is the number of phases.

15.2 The microcanonical ensemble

The partition function Ω is a measure of the number of configurations in the ensemble

s = −k∑i

pi ln pi = k ln Ω.

15.3 The canonical ensemble

Considering a system having a very large number of degrees of freedom. The system is characterizedby (macroscopic) thermodynamic variables (N , V , T are fixed in the system). Now, consider anensemble containing a very large number of such systems. Let

A = number of systems in the ensemble

ai = number of systems in state i (i.e. having energy Ei)

E = total energy of the ensemble

giving

A =∑

ai, E =∑

aiEi.

The state of the ensemble can be characterized by the occupation numbers a = a1, a2, · · · .The probability of finding a system in state i is

Pi =〈ai〉A

=a∗iA.

Further consider a set of all possible ensembles. Since each ensemble has the same energy they havethe same probability of occuring. Let W (a) be the number of occurences of the ensemble specifiedby a giving

W (a) =A!

a1!a2! · · ·.

Also define a∗ to be the most common ensemble in the set, then a∗ can be found from

∂

∂aj

lnW (a)− α

∑ak − β

∑akE

= 0

giving

pi =e−βEj∑e−βEk

.

The partition function becomes

Q(N,V,E) =∑j

e−βEj =∑E

Ω(N,V,E)e−E/(kT ).

A Diverse Collection of FORMULAS, DEFINITIONS AND … · 2020. 1. 4. · FORMULAS, DEFINITIONS AND...

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