isotropic elasticity introduction to polymer and … · 2019. 6. 30. · \hyperelasticity",...

$: ISOTROPIC ELASTICITY INTRODUCTION TO POLYMER AND … · 2019. 6. 30. · \hyperelasticity", which applies to large strains, large rigid body rotations, and nonlinear (though still$
ISOTROPIC ELASTICITY

INTRODUCTION TO POLYMER AND METAL

SOLID MECHANICS

Aaron Freidenberg

January 16, 2013

2

Contents

Acknowledgment iii

1 Introduction 1

1.1 Tensor Notation . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2 Kronecker Delta . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.3 2nd Order Tensor Transformations . . . . . . . . . . . . . . . 6

1.4 Trace, Scalar Product, Eigenvalues . . . . . . . . . . . . . . . 10

2 Strain 13

2.1 Strain Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . 15

2.2 Volume and Area Change . . . . . . . . . . . . . . . . . . . . 20

2.3 Polar Decomposition Theory . . . . . . . . . . . . . . . . . . 21

3 Rates of Deformation 27

3.1 Rate of Deformation and Spin Tensors . . . . . . . . . . . . . 28

3.2 Other Rates of Change (Volume and Area) . . . . . . . . . . 30

4 Stress 33

4.1 Cauchy Equation of Motion . . . . . . . . . . . . . . . . . . . 34

4.2 Alternative Measures of Stress . . . . . . . . . . . . . . . . . 36

4.3 Principal Stresses . . . . . . . . . . . . . . . . . . . . . . . . . 41

5 Superimposed Rotation 43

6 Hyperelasticity 49

6.1 Phenomenological and Micromechanical Models . . . . . . . . 55

6.2 Tabulated and Calibrated Models . . . . . . . . . . . . . . . . 63

7 Rate-Form Constitutive Expressions 69

7.1 Hypoelasticity (Jaumann) . . . . . . . . . . . . . . . . . . . . 71

i

ii CONTENTS

7.2 Time Stepping Algorithm . . . . . . . . . . . . . . . . . . . . 76

8 Linear Infinitesimal Elasticity 818.1 Elastic Modulii and Poisson Ratio . . . . . . . . . . . . . . . 868.2 Navier and Beltrami-Michell Equations . . . . . . . . . . . . . 898.3 Final Words . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

A Math derivations 93A.1 Transpose of tensor product . . . . . . . . . . . . . . . . . . . 93A.2 Skew tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95A.3 Orthogonal tensor . . . . . . . . . . . . . . . . . . . . . . . . 96

B Stress derivations 97B.1 Physical interpretation of σ . . . . . . . . . . . . . . . . . . . 97B.2 Equation of Motion . . . . . . . . . . . . . . . . . . . . . . . . 101

C Hyperelastic derivations 105C.1 Proof of σ = f(B) . . . . . . . . . . . . . . . . . . . . . . . . 105C.2 Derivation: dIB

dB , dIIBdB , dIIIB

dB . . . . . . . . . . . . . . . . . . 109C.3 Principal stretch constitutive relationship . . . . . . . . . . . 112C.4 Tabulated hyperelastic model . . . . . . . . . . . . . . . . . . 114

D Chapter 7 derivations 119D.1 Jaumann rate in infinitesimal elasticity . . . . . . . . . . . . . 119D.2 Truesdell and Jaumann rates . . . . . . . . . . . . . . . . . . 121

Acknowledgment

I have been fortunate to have received tremendous support throughout mylife from many individuals, but in particular from my parents, David andDeanna. Throughout my schooling, they always knew just what to say tohelp keep me motivated. I would also like to acknowledge other individualsthat have made possible the completion of this text. Gilbert Hegemier,who headed the Department of Structural Engineering at UCSD in 2009,convinced me to pursue my doctorate at UCSD, where I was lucky enoughto have access to teachers and scholars that are unrivaled in solid mechanics.In particular, the teaching and writings of Vlado Lubarda, David Benson,and Anne Hoger-Conn stimulated my interest in the subject. In addition,some of the subject matter herein has been adapted from my lecture notesfrom courses taught by the aforementioned individuals.

iii

iv ACKNOWLEDGMENT

Chapter 1

Introduction

Before the advent of the personal computer, engineers would perform struc-tural analysis using formulas that are simplified and suitable for hand calcu-lation. Very elegant analytical methods of structural analysis were developedout of necessity and many such methods are still taught today in universityand used in industry. Such methods are vital in order to do preliminary “de-sign” calculations, and this author would argue that a deep understandingof classical methods of analysis also give the engineer an intuition about theway forces “flow” through a structure. Most important of all, an intuitionabout such things will help the reader to understand the topics that will becovered in this text!

Structural analysis work that is done in industry and academia is increas-ingly being accomplished via structural analysis software. Many undergrad-uate level courses that deal with such finite element methods (FEM) forstructural analysis aim to give the student the confidence (i.e. give the stu-dent all of the formulas) that they need so that they can write their ownfinite element analysis code. In this text, “theory” will be presented in amanner such that FEM itself is the “application.” This text will derive thegoverning formulas of “elastic isotropic” solid mechanics, whereas coursesin FEM take such formulas as a given and focus, instead, on the ways thatsoftware implement the formulas, numerically.

note: In this text, FEM refers to “solid” elements (sometimes referred to as“continuum” elements). “Stick” elements (sometimes referred to as “beam”elements) or “shells” are simplifications of “solid” elements.

Consider, for example, the simplest case of isotropic linear infinitesimal elas-

1

2 CHAPTER 1. INTRODUCTION

ticity, in FEM. If it’s 2D or 3D, then we need a constitutive equation anal-ogous to σ =“E”ε along with a strain-displacement equation analogous toε = ∆L

L . The equations σ = C : ε and εij = 12( ∂ui∂Xj

+∂uj∂Xi

) may look familiar,

but in this text they will be derived.

Moreover, this text will cover large (“finite”) strain isotropy as well as pro-vide insight into how more advanced FEM software handle large rigid bodyrotations and time-dependent problems. The topics covered in this textare essential for engineers in the field of materials research as well as thosewho use advanced FEM software to its full potential. We will start with“hyperelasticity”, which applies to large strains, large rigid body rotations,and nonlinear (though still elastic) stress-strain relationships. In general,this is the necessary starting point when one has a material with unknownproperties. Rubbers actually behave nonlinearly (nonlinear isotropic-elasticbehavior is a good starting point) and have large strains.

Then, “hypoelasticity” is considered, which relates stress rates (and there-fore incremental stress) to strains and strain rates. Developing the consti-tutive relationship in hypoelastic form is the preferred method for the mostadvanced FEM software, such as ABAQUS and LS-DYNA, since they dealwith rate and history dependent transient dynamic problems that involvetime-stepping algorithms. Lastly, the aforementioned isotropic linear in-finitesimal formulation will be discussed. Steel, in the elastic regime, canoften be modeled in such a manner.

Before getting into these main ideas, we should introduce tensor notation.In doing so, we’ll also go through some math derivations that will be essen-tial to our formulation of hyperelasticity, linear infinitesimal elasticity, andrate-form constitutive relationships. Tensor math is not a steep learningcurve, and it is probably worthwhile to take a bit of time to make sure youunderstand it. While many of the derivations throughout this text will becomplete and quite detailed, from the point of view of most readers, it isimportant to keep in mind that this text is written in a mechanics stylerather than a mathematical style. While the author will do his best to avoid“sloppy” math, the overriding emphasis will always be on the physical in-terpretation and application of concepts.

This text is only intended to be an introduction to solid mechanics, andtherefore only considers “elastic isotropic” materials. The information pro-

1.1. TENSOR NOTATION 3

vided in this text serves as a prerequisite for topics such as anisotropy,viscoelasticity, and plasticity in solid mechanics.

1.1 Tensor Notation

Throughout this text, we will be working in 3D. The way that the defor-mations of a body are described will make use of three-component tensors(i.e. vectors) that describe magnitude and direction. We require a vectorthat has only three components, because we will consider that as a bodydeforms, any point within the body undergoes a motion that is describedby a simple translation, in 3D space. Once we begin looking at strains (andstresses), then we will have to consider the motion of elements, rather thanpoints. Since elements can deform axially in three directions as well as un-dergo shear, the description of stresses and strains will require the use ofhigher order tensors (i.e. matrices).

“Tensor” notation (index notation) is essentially a way of keeping track ofthe components of matrices when performing matrix operations or algebra,without the need to repeatedly draw matrices. In this way, tensors save pa-per! In addition to this benefit, tensor notation (index notation) lends itselfnicely to computer programming. Having said that, the majority of deriva-tions in the later chapters will involve algebra that the reader is probablyalready familiar with. For example, the basic product between an nx1 vec-tor and a 1xn vector produces an nxn matrix, while a dot product betweenthe same two vectors would produce a scalar. Similarly, matrix products(denoted with a “·” for reasons that are explained in this chapter), inverses,transposes, and other operations are probably already familiar as well. Ba-sic properties of the operations (e.x. matrix multiplication is associative butnot commutative) are also familiar. Thus, it is possible to jump right intothe later chapters without studying tensors and only risk becoming “stuck”on the few derivations that operate explicitly on components of matrices.

Tensors (matrices and vectors) will be written in boldface. When theircomponents are of interest, alphabetical subscripts (“indices”) will be used,and the boldface removed, since each component of a tensor is merely ascalar. Most vectors will be written as lower-case English letters, whilemost higher-order tensors will be written using upper-case English lettersor Greek letters. Only rectangular (Cartesian) coordinate systems will be


considered.

So, how does tensor math work and in particular, what is “index notation?”Further, what specific tensor operations are going to be most important forus? These are the questions that will be addressed in this chapter.

Consider a typical rectangular coordinate system defined by the unit vectorse1, e2, e3 (Fig. 1.1):

Figure 1.1: Cartesian bases

We know that any vector, a, with components a1, a2, a3, can be defined bythe units vectors e1, e2, e3, as follows:

a = a1e1 + a2e2 + a3e3 =3∑i=1

aiei (1.1)

Our convention (sometimes called the “Einstein Summation Convention”)will be to simply drop the

∑symbol. The subscript (or “index”) will be

assumed to be summed. This is one of the important ideas of “tensor nota-tion” (or “index notation”).

note: i goes from 1 to 2 in 2D and 1 to 3 in 3D

The dot product between two vectors can be written in index notation:

a · b = a1b1 + a2b2 + a3b3 = aibi (1.2)

1.2. KRONECKER DELTA 5

We know that the multiplication of a matrix and a vector results in a vector.The components of this vector can be expressed as follows:x1

x2

x3

=

A11 A12 A13

A21 A22 A23

A31 A32 A33

b1b2b3

=⇒ xi = Aijbj (we’ll derive soon) (1.3)

In eq. (1.3), we are summing over “j” only, because the practice is to onlysum over indices that are “repeated” within a given term in an expression.Another way to think about it, which will always work in this text unlessotherwise stated, is to consider that “repeated indices” appear on only oneside of the equation, which indicates that they should be summed. The “freeindex,” which in eq. (1.3) is “i,” appears on both the left-hand side and theright-hand side of the equation (it also only appears one time in any giventerm) and therefore we know not to sum over “i.”

note: In eq. (1.3), “i” is a free index while “j” is a repeated/dummy index.Repeated indices summed while free indices are not.

1.2 Kronecker Delta

The “Kronecker Delta”, δij , is a tool that we’ll be using throughout thistext. It is simply defined as follows:

δij =

{1 if i = j,0 if i 6= j,

(1.4)

The usefulness of the Kronecker Delta lies in its ability to transform indices:

δikak = δi1a1 + δi2a2 + δi3a3

If i = 1, δikak = a1



∴ δikak = ai and δikAkj = Aij︸︷︷︸These, and similar ideas involving the Kronecker Delta, will be used extensively in later derivations

(1.5)


note: ei · ej = δijIn other words, the dot product of a unit vector along the “i” axis and aunit vector along the “j” axis is equal to zero, unless i and j are equal, inwhich case the dot product would be 1.

1.3 2nd Order Tensor Transformations

We already know:a = aiei (1.6)

Similarly, we need to be able to express a higher order matrix, using tensornotation:

A = Aijeiej (1.7)

eiej is sometimes written: ei ⊗ ej, where “⊗” denotes the “dyadic” or “ten-sor” product

e.x. e1 ⊗ e2 =

100

[0 1 0]

=

0 1 00 0 00 0 0

As opposed to: e1 · e2 =

[1 0 0

] 010

= 0

A =sum of 3 x 3 matrices = 3 x 3 matrix

A = Aijei ⊗ ej = A11e1 ⊗ e1 +A12e1 ⊗ e2 +A13e1 ⊗ e3

+A21e2 ⊗ e1 +A22e2 ⊗ e2 +A23e2 ⊗ e3

+A31e3 ⊗ e1 +A32e3 ⊗ e2 +A33e3 ⊗ e3

Recall eq. (1.3), which states that xi = Aijbj , where A is a 3x3 matrix, b isa vector, and x is the solution to the product between A and b. To see whythis is so, we can consider the product between A and the basis vector ek,and then consider the product between A and an arbitrary vector b. Be-fore we do this, it should be mentioned that products between tensors andvectors (producing a vector) as well as products between two tensors (pro-ducing a tensor) will directly make use of the “dot product,” even though the“dot product” operator is sometimes thought of as a “scalar product.” The

1.3. 2ND ORDER TENSOR TRANSFORMATIONS 7

“dot product” treatment used consistently throughout this text, althoughmathematically a bit “sloppy,” enables one to use index notation to derivequantities in a very direct manner.

The “dot product” will be used in this text to signify the tensor productbetween a tensor and a vector or the tensor product between two tensors.This is consistent with most of the literature in solid mechanics. A smallminority of authors, however, consider this to be “sloppy” math and insist,instead, that the dot product between tensors produce a scalar value. Thisdistinction is very important since tensor products, which produce tensors,are prevalent in solid mechanics and this text will heavily use the “dot”convention when deriving important identities and quantities. Authors thatuse a different convention would use a very different approach for derivation,in particular where index notation is used.

A · ek = Aijei ⊗ ej · ek = Aijeiδjk = (Aijδjk)ei = Aikei (1.8)

Note the use of the Kronecker delta in simplifying the expression in eq. (1.8).

Similarly,

A · b = Aijeiej · bkek = Aijeibkδjk = Aikbkei or A · b = Aijbjei (1.9)

Again, the solution is a vector; this time with components xi = Aijbj , asexpected (eq. (1.3)).

note: Eq. (1.9) is standard matrix multiplication. We had to use a “dot”product to get the tensor algebra to work as a typical matrix - vector mul-tiplication (we’ll see later that A · B also results in the standard matrix- matrix product). This is consistent with most of the literature in solidmechanics.

note: Remember, in the above derivations, only if j = k is δjk non-zero. Thiswas explained in eq (1.5). Note how this simplified the derivations. This isvery common in tensor algebra.

So, consider eq. (1.8). Let’s “pick out” a particular component of this ten-sor, as follows:

Multiplying both sides by ej −→ ej ·A · ek = ej ·Aikei = δijAik = Ajk∴


Ajk = ej ·A · ek or, re-written:

Aij = ei ·A · ej (1.10)

note: The LHS of eq. (1.10) is only one term. I.e. this is how you pick outa single term of a matrix.

note: For 1st order: ai = a · ei, where ai gives the component of a in the “i”direction.

Consider the following coordinate system (Fig. 1.2):

Figure 1.2: Coordinate transformation

e′i = Q · ei ; Q = rotation matrix = Qmnem ⊗ en

we know: e′i = Qmnem ⊗ en · ei︸︷︷︸

δin

= Qmiem −→ e′i · en = Qmi em · en︸︷︷︸

δmn

= Qni

−→ Qni = e′i · en (terms [components] of a rotation matrix)

To be more useful, we need to show that aj = Qjia′i.

By definition, a = aiei = a′ie′i

Multiply both sides by ej : ai ei · ej︸︷︷︸δij

= a′i e′i · ej︸︷︷︸Qji

And, since Qji = ej · e′i = e

′i · ej (this should be obvious):

1.3. 2ND ORDER TENSOR TRANSFORMATIONS 9

aj = Qjia′i (1.11)

note: Q ·Q−1 = Q−1 ·Q = I (this is true of any tensor)

Q is an “orthogonal” tensor. A particular identity of an orthogonal tensorcan be written as follows:

(Q · u) · (Q · v) = u · v (1.12)

For an orthogonal tensor, Q (obeys eq. (1.12)), it can be shown thatQT = Q−1 (pf: [2])

∴

Q ·QT = QT ·Q = I

where I = δijei ⊗ ej

= eiei = e1e1 + e2e2 + e3e3 =

1 0 00 0 00 0 0

+

0 0 00 1 00 0 0

+

0 0 00 0 00 0 1

From eq. (1.11), we know that a = Q · a′

QT · a = QT ·Q︸︷︷︸I

·a′

a′

= QT · aa′i = (QT )ijaj

a′i = Qjiaj (1.13)

In eq. (1.13), Qji = ej · e′i (assuming we know the unit vectors defining the

original and rotated coordinate systems).

What about transforming a tensor?

A = Aijei ⊗ ej

A = A′ije′i ⊗ e

′j

e′k ·A · e

′l︸︷︷︸

A′kl

= Aij( e′k · ei︸︷︷︸Qik

)( ej · e′l︸︷︷︸

Qjl

)


The following will be simply stated, and then proved:

A′kl = AijQikQjl = QikAijQjl −→ A′ = QT ·A ·Q (1.14)

Tensor product practice and quick proof of eq. (1.14):

A ·Q = Aijeiej ·Qklekel = AijQklδjkeiel = AijQjleiel

Now taking Q to be Q = Qtketek, we find:

QT · (A ·Q) = Qtkeket ·AijQjleiel

Thus, QT ·A ·Q = QtkAijQjlδtiekel = QikAijQjlekel

Setting A′ = QT ·A·Q, we can see that the components are A′kl = QikAijQjl- i.e. the desired result (eq. (1.14))

note: We put a “dot” to get the tensor algebra to work. It’s really just astandard matrix product. If ever a product is written AB or Ab, in this text,it is probably a typo! Most of the literature in solid mechanics uses the samenotation as in this text. Some authors omit the “dot” (e.x. A ·B = AB)except when using index notation. Some authors take A ·B to be the scalarproduct, which is a completely different definition, as opposed to a differencemerely in notation. Fortunately, authors that take A ·B to be the scalarproduct are a small minority in solid mechanics.

1.4 Trace, Scalar Product, Eigenvalues

“Trace” is a particular operator that, when “applied” to a 2nd order (orhigher) tensor, sums the diagonal components. Strictly-speaking, the defi-nition of “trace” can be taken as tr(u⊗ v) = u · v

tr(A) = Aiitr(A ·B) = tr(C) = CiiCij = AikBkj 6= AikBjk unless the tensor is symmetrictr(A ·B) = Cii = AikBki

1.4. TRACE, SCALAR PRODUCT, EIGENVALUES 11

tr(A ·B ·C) = tr(B ·C ·A) = tr(C ·A ·B) = AikBklCli

The “scalar product” of tensors is analogous to the “dot product” of vectors.In this text, the scalar product is denoted by the operator “:” and is definedas follows:

A : B = tr(AT ·B) = tr(A ·BT ) = AikBik

We can arrive at the same result in a different manner:

A : B = Aijeiej : Blkelek = AijBlkδilδjk = AikBik

note: Since the scalar product of two tensors is analogous to the “dot prod-uct” of two vectors, some authors define the scalar product between A andB as A ·B [15]. In other words, these authors define A ·B = AikBik. Usingsuch a definition, A ·B, which still must be equal to Aijeiej · Bklekel, canno longer be expressed as AijBklδjkeiel, as was done previously. Deriva-tions in solid mechanics that use the simple manipulation shown in the lastterm would have to be done using some other approach that is likely morecumbersome.

An eigenvector, n, of a tensor or a matrix has the following special property:when n is multiplied by the matrix, the result is a new vector that has thesame direction as n. The amount by which the magnitude of the vector haschanged is the value of the eigenvalue (the eigenvalue that corresponds tothe eigenvector n).

i.e. (A− λI) · n = 0

λ are the eigenvalue solutions and n are the corresponding eigenvectorsAny tensor product can be expressed in terms of invariants (or eigenvalues)i.e.ε = [λ1n

1in

1j + λ2n

2in

2j + λ3n

3in

3j ]ei ⊗ ej (skipped work)

orε =

∑3a=1 λana ⊗ na

Characteristic equation and Cayley-Hamilton Theorem:

One would typically solve for the eigenvalues from a “characteristic equa-


tion” of the form λ3 − IAλ2 + IIAλ − IIIA = 0 (for a 3x3 matrix), whereIA, IIA, IIIA are coefficients that depend on the values within the matrix,A. Theses coefficients, IA, IIA, IIIA, are more commonly called invariants.

IA = λ1 + λ2 + λ3

IIA = λ1λ2 + λ2λ3 + λ3λ1

IIIA = λ1λ2λ3

We’ll see some alternative expressions for IA, IIA, IIIA later, when we getto “hyperelasticity.”

note: The invariants of a matrix are the same regardless of coordinate system(as is the trace).

The matrix also satisfies its own characteristic equation (this is known asthe Cayley-Hamilton Theorem).i.e.

A3 − IAA2 + IIAA− IIIAI = 0 (1.15)

A−1(A3 − IAA2 + IIAA) = IIIAIA−1

A2 − IAA + IIAI = IIIAA−1

1

IIIA(A2 − IAA + IIAI) = A−1 (1.16)

Either eq. (1.15) or eq. (1.16) are often used in derivations of hyperelasticitylater on.

Chapter 2

Strain

Strain is a measure of deformation that has an intuitive physical meaningfor engineers. Specifically, axial strain, for example, is a measure of thedeformation of a structure (or element within a structure) with respect tothe structure’s length. Thus, strain is a ratio of lengths and is a unit-lessquantity. While strain can come in the form of extension, shortening, orshear strain, it is sometimes overlooked that strain also has direction. Thischange in direction, under, for example, rigid body rotation, creates someinteresting complications in the finite element world.

In the finite element world and, accordingly, in solid mechanics (“solid” or“continuum” elements), we essentially calculate the strain at every pointwithin a structure. Even for very simple geometries subjected to symmet-ric loads, some of the points (elements) within a structure will undergorigid body rotations. Such rigid body motions present a complication: theelements comprising the structure need to maintain a consistent frame ofreference. The need for a consistent frame of reference is even more appar-ent when one considers impact type behavior that involves multiple bodies.

If the reader is familiar with the idea of “geometric nonlinearity,” then fol-lowing analogy may be useful. Consider the manner in which structuralengineers perform the preliminary design of the beams and columns of tallbuildings. They often design for maximum flexibility, using efficient ana-lytical methods. Where gravity and lateral loads act simultaneously on abuilding, these engineers know that the force demands in the structure fromthe gravity loads should be calculated after the force demands from the lat-eral loads are determined. The lateral loads cause the building to sway, and

13

14 CHAPTER 2. STRAIN

the presence of rigid body rotation of the columns is important to considerprior to calculating the force demands from the gravity loads. Similarly,in FEM, there is a certain order to the treatment of rigid body rotationin the FEM algorithms, which we will see when we get to the chapter on“Rate-Form Constitutive Expressions.” It is important to recognize thisnow, however, because we will begin discussion of strain and many differentmeasures of strain will be presented. Different strain measures are used inFEM, depending on the order in which rigid body rotations are considered.

Since the analytical methodology for building design was mentioned, now isa good time to remind the reader that the kinds of FEM analysis that thistext considers are fundamentally different from analytical methods of designand analysis. ABAQUS/Explicit and LS-DYNA are examples of the kindsof FEM software considered in this text. Here, geometric nonlinearity is notan issue. Rather, we are free to model structures of complex geometries sub-jected to any loading conditions we wish. In addition, the behavior of “solid”elements are governed by continuum-type mechanics, and so “moments” willnot be considered. To really develop the framework used by FEM codes,though, we need to consider stress, as well as constitutive expressions relat-ing stress and strain, including “hyperelasticity.” This chapter only providesan overview of the various strain tensors that will be used later on in thistext.

Consider a body that undergoes deformation as well as rigid body rotations(Fig. 2.1). In particular, we’ll consider a vector that is initially dX, andwell will track this vector as it becomes dx.

Define:

dx = F · dX (2.1)

In eq. (2.1), F is a second-order tensor known as the “deformation gradient”

We can see from fig. 2.1:x + dx = X + dX + u(X + dX,t)−→ dx = dX + u(X + dX,t)− u(X,t)where u(X,t) = x−X

In index notation, dxi = dXi + ui(Xj + dXj , t)− ui(Xj , t)where ui(Xj + dXj , t) = ui(Xj , t) + ∂ui

∂XjdXj , which comes from a Taylor

2.1. STRAIN TENSORS 15

Figure 2.1: Strain potato

expansion and neglecting higher order terms

i.e. f(x+ ∆x) ≈ f(x) + df(x)dx ∆x+ 1

2d2f(x)dx2

∆x2 + ...

So, dxi = dXi + ∂ui∂Xj

dXj or dx = dX + dX · (∇:

u)

note the “Nabla” or “gradient” symbol ∇:

that is a commonly used operator

So, dx = (I + u∇:

) · dX

2.1 Strain Tensors

Recall from eq. (2.1), that dx = F · dX

Since dx = (I + u∇:

) · dX, F = I + u∇:

or, in index notation:

Fij = δij+∂ui∂Xj

= ∂xi∂Xj

, since du = [(x + dx)− (X + dX)]− (x−X) = dx− dX

and ∂Xi∂Xj

= δij , by definition.


In matrix form:

u∇:

=

∂u1∂X1

∂u1∂X2

∂u1∂X3

∂u2∂X1

∂u2∂X2

∂u2∂X3

∂u3∂X1

∂u3∂X2

∂u3∂X3

(2.2)

F = “deformation gradient”:

F =

∂x1∂X1

∂x1∂X2

∂x1∂X3

∂x2∂X1

∂x2∂X2

∂x2∂X3

∂x3∂X1

∂x3∂X2

∂x3∂X3

(2.3)

F is sometimes called the “stretch” tensor

Now, consider how the length of any element or fiber within the continuummay change under deformation. To find such length “magnitudes” we cantake dot products as follows:

dS2 = dX · dX (dS = length dX)ds2 = dx · dx (ds = length dx)dx = F · dX = dX · FT

note: you can’t do this transpose manipulation as easily if multiplying twotensors, but it works for two vectors or a vector and a tensor (use indices toeasily prove)

ds2 = dX · FT · F · dX

note: FT · F = C = “Right C - G” (Cauchy - Greene) deformation tensor.The reason for this name will become clear once we begin discussion our on“polar decomposition” theory.

ds2 − dS2

dS2=

dX · (FT · F− I) · dX

dX · dX= 2E (2.4)

where E = Lagrangian Strain Tensor = 12(FT · F− I)

or

E =1

2(C− I) (2.5)


E · n = strain vector on a plane whose normal vector is n (the actual loca-tion in space is specified within E)n ·E · n = strain (scalar) in direction n

note:

From eq. (2.4), Enn = ds2−dS2

2dS2 = (ds−dS)(ds+dS)2dS2

Suppose ds = dS (infinitesimal deformation):

Enn = (ds−dS)(ds+dS)2dS2 ≈ (ds−dS)2dS

2dS2 = ds−dSdS = εnn

i.e. ε = ∆LL

what about m ·E · n?

e.x.Suppose n and m are orthogonal in the undeformed configuration.Since E · n is the strain vector on a plane whose unit normal is n,m ·E · n is the component of E · n in direction m.i.e. if m is in the plane of interest (orthogonal to n), then m ·E · n is theshear strain

Figure 2.2: Initial configuration and deformed configuration

We know that: dx1 = F · dX1 ; dx2 = F · dX2

So, dx1 · dx2 − dX1 · dX2 = F · dX1 · F · dX2 − dX1 · dX2

From eq. (2.4) and recalling that F · dX = dX · FT , we get:dx1 · dx2 − dX1 · dX2 = 2dX1 ·E · dX2

ds1ds2 m · n︸︷︷︸cosα

−0 = 2dS1dS2 m ·E · n︸︷︷︸Emn

, where m and n are unit vectors


note: dX1 · dX2 is “zero” on the LHS of theabove equation because m and n are originallyorthogonal

Emn = 12ds1dS1

ds2dS2

cosα

where ds1dS1

is the “stretch ratio”

note: As seen in Fig. 2.3, γ = π2 − α ∴ cosα = sinγ

Figure 2.3: Engineering strain, γ

Emn = 12ds1dS1

ds2dS2

sinγ

Infinitesimal engineering shear strain = εxy = 12(γ1 + γ2) = 1

2γDoes our shear strain reduce to this value for infinitesimal deformation?

For infinitesimal deformation, ds1dS1

= 1 ; ds2dS2

= 1 ; sinγ = γ

−→ Emn = 12(1)(1)(γ) = 1

2γ−→ Emn ≈ εmn for linear infinitesimal deformation

We can also see from the following equation (eq. (2.6)) that, in general,

εij = 12( ∂ui∂Xj

+∂uj∂Xi

) for linear infinitesimal deformation (higher order terms

are neglected).

For large (“finite”) strain:

Eij =1

2

(∂ui∂Xj

+∂uj∂Xi

+∂uk∂Xi

∂uk∂Xj

)(2.6)

To prove, first consider: Fij = δij + ∂ui∂Xj−→ F = I + u∇

:

Now, from eq. (2.4), E = 12(FT · F− I) = 1

2 [(I +∇:

u) · (I + u∇:

)− I]

= 12 [I + u∇

:+∇

:u + (∇

:u) · (u∇

:)− I] = 1

2 [u∇:

+∇:

u + (∇:

u) · (u∇:

)]

Recall the definition of ∇:

u from Fig. 2.1, and recall that u∇:

is the trans-

pose of ∇:

u


Eulerian strain:Here, “Eulerian strain” is simply referring to a measure of strain that is de-fined in spatial coordinates. Under rigid body rotation, the Eulerian strainvalues will change, whereas the Lagrangian strain tensor is invariant to rigidbody rotation. In other words, the coordinate system in which E is calcu-lated (the “material” coordinate system) rotates with rigid body rotation.The coordinate system in which e is calculated (the “spatial” coordinatesystem) remains constant.

dx = F · dX −→ dX = F−1 · dx = dx · F−T

Similar to the way that we derived E, let’s consider the difference in lengthsof any particular element, or fiber, within our strain potato, before and afterdeformation.

ds2 − dS2 = dx · dx− dX · dX = dx · dx− dx · F−T · F−1 · dx = dx ·(I− F−T · F−1) · dxds2 − dS2 = 2dx · e · dxwhere the Eulerian Strain Tensor = e = 1

2(I− F−T · F−1)

or

e =1

2

(I−B−1

)(2.7)

where B = Left C-G Tensor = F · FT

(B−1 = F−T · F−1), which is easy to prove using indices

dX · dX = dx·F−T · F−1︸︷︷︸B−1

·dx

This is analogous to the previously derived dx · dx = dX · FT · F · dX

Proof comes from:dX = F−1 · dx = dx · F−T

We’ll see the physical meaning of B and C when we discuss “polar decom-position.”


How is e related to E?

E = 12(FT · F− I) ; e = 1

2(I− F−T · F−1)

E = FT · 1

2(I− F−T · F−1)︸︷︷︸

e

·F

e = F−T · 1

2(FT · F− I)︸︷︷︸

E

·F−1

B−1 = F−T · F−1 −→ B−1ij = F−Tik · F

−1kj

(To convince yourself that these subscripts are correct, simply write out thematrix multiplication long-hand, summing only the dummy index “k”)

Since ∂Xi∂xj

= δij − ∂ui∂xj

= F−1ij ,

B−1ij = (δik − ∂uk

∂xi)(δkj − ∂uk

∂xj) = δij − ∂uj

∂xi− ∂ui

∂xj+ ∂uk

∂xi∂uk∂xj

e =1

2(I−B−1) −→ eij =

1

2(δij −B−1

ij ) =1

2(∂ui∂xj

+∂uj∂xi− ∂uk∂xi

∂uk∂xj

) (2.8)

If only infinitesimal deformation, and so long as no significant rigid bodyrotations are present, then Eij ≈ eij ≈ εij = 1

2( ∂ui∂Xj+

∂uj∂Xi

)

(this formula may be familiar from undergrad, for example)

2.2 Volume and Area Change

For completeness, we’ll derive the volume change and area change:

dV = dV0detF (2.9)

Derivation starts from the triple scalar product. Cross products and deter-minants (written in tensor form) is a math topic, so we’ll skip the derivationof volumetric deformation

note: detF = 1 −→ dV = dV0 −→ “incompressible”

Nanson’s Equation:Now we know that dV = (detF)dV0, but what about area?

2.3. POLAR DECOMPOSITION THEORY 21

Figure 2.4: Volumetric deformation

If dS0 is the initial area and dS is the final area:For small volumes, we can say that dV0 = dS0dX and dV = dSdx.

It was previously shown that dV = detFdV0

∴ dV0 = dS0dX and dV = detFdV0 = dSdx −→ dV0 = dSdxdetF

Substituting, we get dSdx = dS0dXdetFWe also know that dx = F · dX

−→ dSF︸︷︷︸FT dS

·dX = dS0dXdetF −→ (detFdS0 − FTdS) · dX = 0

−→ dS = dSn = detF ∗ F−TdS0 = detF ∗ F−T ·NdS0 (2.10)

where N and n are the normal vectors to the surface in the respective initialand final configurations.

Eq. (2.10) is called Nanson’s Equation and will be useful later when we lookat “true” stress versus “nominal” stress, for example.

2.3 Polar Decomposition Theory

Any matrix can be decomposed into a sum of symmetric and antisymmetricmatrices, but F can be decomposed into a product of two matrices (onesymmetric and one orthogonal)


F = V ·R = R ·U (2.11)

U and V are called the Right Stretch Tensor and Left Stretch Tensor dueto their respective positions (relative to R) in eq. (2.11).

V = VT (symmetric)U = UT (symmetric)RT = R−1 (orthogonal)

A proof of the orthogonality of R is given in Appendix A.3.

C = FT · F = U·RT ·R︸︷︷︸I

·U = U2 (2.12)

B = V2 (2.13)

R is a rigid body rotation, while U or V each stretch and rotate (they eachcontain both normal and shear deformations, typically)

U and V have the same eigenvalues. Eigenvalues do not have an “order,”per se, but since U and V typically have different eigenvectors, one shouldbe cautious when assuming equivalency of eigenvalues. In the principal di-rections of U or V, U or V contains no shear deformation (fig. 2.5). We’lllook at actual stresses, later.

To find the principal directions of U and V, we must solve the eigenvalueproblem:

U · n = λnR ·U︸︷︷︸

F

·n = R·λn (pictured)

F = V ·R −→ V · ( R · n︸︷︷︸m

) = λ(R · n︸︷︷︸m

)

mi = R · ni


Figure 2.5: Deformation in principal directions vs. deformation in typicaldirection

Figure 2.6: The principal directions for pure shear deformation are at 45o


e.x. Polar Decomposition

The deformed equilibrium configuration of a body defined by the deforma-tion mapping:

x1 = X1 + 3X2 , x2 = X2 , x3 = X3

Determine:

a) F and Cb) Eigenvalues and eigenvectors of Cc) U and U−1

d) Re) V

a) F =

∂x1∂X1

∂x1∂X2

∂x1∂X3

∂x2∂X1

∂x2∂X2

∂x2∂X3

∂x3∂X1

∂x3∂X2

∂x3∂X3

=

1 3 00 1 00 0 1

note: F will typically be a function of X1, X2, X3 and we’ll be interested inthe value of F at a particular point. Here, it doesn’t matter.

C = FT · F =

1 3 03 10 00 0 1

b) C · n = λn ; det| − λI + C| = 0−→ λ1 = .092 ; λ2 = 1 ; λ3 = 10.91

[−λiI + C]ni = 0

i.e. for λ1,.908n1 + 3n2 = 03n1 + 9.908n2 = 0.908n3 = 0

Choose arbitrary n1 ; solve for n2 from either equation ; normalize:


n1

n2

n3

λ1

=

.957−.290

0

;

n1

n2

n3

λ2

=

001

;

n1

n2

n3

λ3

=

.290.957

0

c) [U]n =√

[C]n =

√λ1 √λ2 √

λ3

=

.3031

3.303

[U]e = [Φ][U]n[Φ]T =

.56 .83 0.83 3.05 00 0 1

(symmetric)

where [Φ] =

(n1)λ1 (n1)λ2 (n1)λ3(n2)λ1 (n2)λ2 (n2)λ3(n3)λ1 (n3)λ2 (n3)λ3

It’s important to keep in mind that U, E, and C have the same eigenvectorsand that these eigenvectors (and any other strain direction or magnitude)are generally dependent on the particular point in space of interest (e.g. X1,X2, X3). For this simple example, F is independent of X1, X2, and X3

(i.e. the deformation is the same everywhere in the body - a “homogenous”deformation), but this would generally not be the case.

The eigenvalues occur in the direction of the eigenvectors, thus, n1 ·E · n1 =λ1 ; n2 ·E · n2 = λ2 ; n3 ·E · n3 = λ3. It makes sense that if we only knowthe eigenvectors and want to undo this transformation to bring us back toE, then we need a transformation matrix that involves n1,n2,n3. We cansee that Φ ·U ·ΦT is the opposite of the usual ΦT ·U ·Φ, and without goingthrough the rigorous derivation of why Φi1 = (n1)i, Φi2 = (n2)i, Φi3 = (n3)i,it at least makes some sense intuitively.

d) [R]e = [F]e[U]Te =

.55 .83 0−.83 .55 0

0 0 1

(orthogonal ; RT = R−1)

e) [V]e = [F]e[R]Te =

3.05 .83 0.83 .55 00 0 1

Chapter 3

Rates of Deformation

Typically, the equation of motion is solved in small increments of time, andthe constitutive relationships are developed in rate form. “Rate” formula-tion in structural analysis is useful even if the actual “rates” involved aresmall or negligible, since incremental time stepping in time allows one tokeep track of parameters that may be history dependent. This is the case,for example, in plasticity. At the other end of the spectrum, we may havea purely elastic problem, but one in which the rates are high enough thatmaterial parameters change (strain rate effects on stiffness). This cannotbe considered unless we formulate our equations in rate form! These aremerely examples that are meant to convey the importance of rates. Neitherplasticity (metals), viscoelasticity (polymers), nor general rate effects (anymaterial) will be considered in this text.

The rate form for hyperelastic materials, which forms the basis for the ma-terial modeling of rubber, require a special “hypoelastic” treatment. Wewill consider this in a later chapter. Furthermore, we will see that in linearinfinitesimal elasticity, where the basic constitutive relationship is relativelysimple, one has to still carefully consider factors such as change of referenceframe due to rigid body rotations. Incremental or rate formulations handlesuch issues quite nicely.

The velocity at a point in the continuum, similar to the displacement at apoint, depends on both position and time:v = v(x,t)vi = vi(xj , t)

27

28 CHAPTER 3. RATES OF DEFORMATION

Figure 3.1: Configuration at time t

3.1 Rate of Deformation and Spin Tensors

The most obvious first step is to take the time derivative of the deformationgradient, F, as follows:

F = ddt∂x∂X = ∂

∂X

(dxdt

)= ∂

∂Xv

For reasons that won’t become clear until the chapter on rate-form constitu-tive expressions, we would like the “velocity gradient” to be expressed withrespect to x rather than X. Since x is a function of X, we can use the chainrule to arrive at the following important expression:

F =∂

∂Xv =

∂v

∂x︸︷︷︸L

∂x

∂X︸︷︷︸F

(3.1)

We can see from eq. (3.1) that F = L · ForL = F · F−1, where L is the “velocity gradient.”

For our purposes, consider that if we were to have analytical functions oftime, t, for each term in F that describes how R and U are changing withtime, for example, then we could find L by noting that F = d

dt (R ·U) =

R ·U+U ·R and L = F ·F−1. This is quite academic (and potentially quitedifficult) compared to the way that L would be found in FEM, so such anexample will not be given. However, L is a very important quantity as wewill see in the chapter on rate-form constitutive relationships.

3.1. RATE OF DEFORMATION AND SPIN TENSORS 29

The velocity gradient, L, is a very important quantity and can be decom-posed as follows:

L =1

2(L + LT )︸︷︷︸

D

+1

2(L− LT )︸︷︷︸

W

(3.2)

note: Similar to eq. (3.2), in linear infinitesimal elasticity, we can splitthe displacement gradient into a strain tensor and rotation tensor that aresymmetric and anti-symmetric, respectively.

note: Any tensor can be similarly split into its symmetric and anti-symmetricparts.

“Rate of Deformation Tensor” = symmetric −→ DT = D“Spin Tensor” = anti-symmetric or “skew” −→WT = −W

W is a pure rigid body rotation. A complete proof of this can be found in[2]. It can also be easily shown that W is, in fact, skew. This proof can befound in Appendix A.2.

Consider the time rate of change of length of a particular element:ddt(ds)

2 = ddt(dx · dx) = d

dt(dX · FT · F · dX) = dX· ddt(FT · F) · dX

= dX · C · dX

dCdt = C = ( ˙FT · F) = FT · F + FT · F = FT · (F · F−1 + F−T · FT ) · F

= FT · (L + LT ) · F = 2 · FT ·D · F

So, ddt(ds)

2 = 2dX · FT ·D · F · dX = 2dx ·D · dx

dx = dsn −→ ddt(ds)

2 = ��2(ds) ddt(ds) = �2(ds)�2n ·D · n

−→ d

dt(ds) = Dnnds (3.3)

Rate of change of length only depends on D, not W.If D = 0 −→ rigid body motions only.

note: We could have split up F into a sum, but we wouldn’t have arrivedat anything useful. The way that eq. (3.3) was derived only works due tothe product rule of derivatives. Proving that W represents “spin” is moredifficult and won’t be shown here.


Similar to the shear strain, Emn, we can again consider how the angle be-tween two vectors, m, and n, changes under deformation (fig. 3.2).

Figure 3.2: Rate of angle change (shear)

dφ

dt= 2Dmn (skipped work) (3.4)

note: E = FT ·D · F (Recall also: E = FT · e · F)

E reduces to D for infinitesimal deformation (ignoring rigid body rotations),since F −→ I.

In general, though, E 6= e 6= D. Similarly, in general, R 6= W. We’ll alsolook in detail at the physical meaning (or lack thereof) of σ later on whenwe get to the chapter on “Rate-Form Constitutive Expressions”

3.2 Other Rates of Change (Volume and Area)

For completeness, we’ll quickly derive the rate of volume change and therate of area change:

Volume:If dV = ds1ds2ds3, where ds1, ds2, ds3 are the lengths of the sides of a“box” that is oriented in the principal directions of D, then:

ddt(dV ) =

d(ds1)

dt︸︷︷︸D11ds1

ds2ds3 +d(ds2)


ds1ds3 +d(ds3)


ds1ds2

= (D11 +D22 +D33) ds1ds2ds3︸︷︷︸dV

= DkkdV note:tr(D) = Dkk (3.5)

Area:Consider the following time derivative:ddt(dSn) = dS

dt n + dS dndt

3.2. OTHER RATES OF CHANGE (VOLUME AND AREA) 31

where dS is an area (as opposed to ds1, ds2, ds3, which are lengths).

dSn is known from eq. (2.10). So, we can take the time derivative. We canalso re-write dn

dt in terms of Dnn.

−→ d

dt(dS) = (Dkk︸︷︷︸

sum

− Dnn︸︷︷︸no sum

)dS (skipped work) (3.6)

We’ll come back to rates again when we get to rate-form constitutive rela-tionships.

Chapter 4

Stress

Stress, like strain, has intuitive physical meaning for engineers. Stress cancome in the form of tension, compression, or shear. The forces present onone or more elements within a body are found from stresses by integratingover areas, and quantities such as moments, though not found naturally inFEM, can be easily back-calculated from stresses as well.

note: While “moments” are essential in clas-sical structural analysis and design, from anFEM point-of-view (“solid” elements), theyare an unnecessary simplification and will notbe discussed in this text.

Forces (or stresses) within a body are found from externally applied forces,even if the objective is to find deformations. The stiffness of the material,which relates internal force and internal deformation, determines any de-formations of interest that result from externally applied forces and bodyforces. This may all be familiar, since classical, “analytical,” structural anal-ysis methods also require stiffness relationships that relate internal forces todisplacements or moments to rotations. In FEM, stress and strain are used,directly, for this purpose, and the “stiffness” relationship that relates themis called a constitutive relationship. This will be discussed in great detailbeginning in the next chapter. This chapter will just focus on stress. Specif-ically, we’ll consider how to physically interpret a stress tensor, along withsome different measures of stress that are commonly used, and how theyeach handle considerations like large deformations and rigid body rotations.

The “Cauchy” stress is a second-order, symmetric, tensor that contains six

33

34 CHAPTER 4. STRESS

independent components: 3 axial stresses and 3 shear stresses. More onthe physical description of σ can be found in Appendix B.1, where σ isessentially derived.

Simple 2D e.x.Imagine that at some point within some structure, we’ve determined thestresses to be (ignore units):

σ11 = 12, 300 σ22 = −4, 200 σ33 = 0σ32 = σ23 = 0 σ31 = σ13 = 0 σ21 = σ12 = −4, 700

To find the stresses, at, say, 45o, we could use Mohr’s Circle or the trans-formation equations from undergraduate “Mechanics of Materials.”

Solution for the normal stress at 45o for example:

σxx = 650 (skipped work)

Using the stress tensor method instead:

n = [.707, .707, 0]

σ · n =

.707(σ11)− .707(σ12) + 0.707(σ21)− .707(σ22) + 0.707(0)− .707(0) + 0

n · σ · n = σnn =“σxx at 45o”= 650, as expected

4.1 Cauchy Equation of Motion

The equation of motion can be expressed in terms of the applied stress, bodyforces, mass, and acceleration:

∫S

tdS +

∫VρbdV =

d

dt

∫VρvdV ; t = σ · n (4.1)

In index notation:∫S tidS +

∫V ρbidV = d

dt

∫V ρvidV ; ti = σijnj

4.1. CAUCHY EQUATION OF MOTION 35

or

∫SσijnjdS︸︷︷︸∫

V∂∂xj

σijdV

+

∫VρbidV =

∫VρdvidtdV (4.2)

Note that the step from eq. (4.1) to eq. (4.2) is not trivial, since both ρ anddV do in fact change with time. The complete derivation, starting with eq.(4.1) and concluding with eq. (4.2), is given in Appendix B.2.

Thus,

∫V

(∂σij∂xj

+ ρbi − ρdvidt︸︷︷︸

=0, since dV is arbitrary

)dV = 0 (4.3)

Note that the “localization theorem” states that∫ ba fdx = 0 −→ f = 0, if a

and b are arbitrary.

∂σij∂xj

+ ρbi = ρdvidt

(4.4)

Eq. (4.4) are the Cauchy Equations of Equilibrium. It may not yet be clearwhich term contains the applied forces and which term contains the quanti-ties analogous to “k ∗x.” It turns out that the σij term will contain appliedforces and prescribed displacements, along with all of the internal force anddisplacement quantities that constitute “k∗x” for the element. When an en-tire system is analyzed, which includes many elements, the global equationof motion should be satisfied, naturally, so long as the geometry is accu-rately represented by the elements, the stiffness and strength properties ofthe material are defined for each element, and the boundary conditions arecorrectly assigned for each element. There could be other issues that ariseas well, due to simplifications inherent (but quite necessary) in the finite ele-ment method (FEM), but these issues will be left to texts that cover FEM indetail. In fact, among the aforementioned element-related issues, this textwill only cover material elastic stiffness in detail. Material behavior at thelimit state (failure) is covered in texts on plasticity, for example, and topicsrelating to element geometries, prescribed degrees of freedom or prescribedforces at “nodes” (i.e. boundary conditions), or other issues related to the“assembly” of finite elements will be left to texts devoted to the topic of


FEM implementation.

If vi = 0. then eq. (4.4) reduces to static equilibrium.(static equilibrium: ∂σi1

∂x1+ ∂σi2

∂x2+ ∂σi3

∂x3= 0 (no sum on i)

)note: This is the basic differential equation used in FEM, though the con-nection to FEM will not really be clear until we start developing constitutiveequations relating stresses and strains (along with the above equation andstrain-displacement relationships previously presented).

4.2 Alternative Measures of Stress

Stress can be defined in the “spatial” coordinate system or in the “material”coordinate system (note that these two coordinate systems differ when rigidbody rotation is present). In addition, stress can be defined as force perunit deformed area or force per unit initial area.

Figure 4.1: Nominal stress

dfn = tn · dS

where tn = n · σ

Now, define dfn = t0ndS0, where t0

n = N · σ(the actual force, on the undeformed area, gives us the “nominal” stress σ0)

dfn = t0dS0 = tdS

dS0N · σ0 = dSn·σ

4.2. ALTERNATIVE MEASURES OF STRESS 37

Recall Nanson’s Equation (eq. (2.10)): dSn = (detF)N · F−1dS0

��dS0N · σ0 = (detF)��dS0N · F−1 · σ

(more formally: dS0N · [σ0 − (detF)F−1 · σ] = 0)

σ0 = (detF)F−1 · σ (4.5)

Eq. (4.5) is the Nominal Stress Tensor

Now, consider a “pseudo-force” acting in the initial frame of reference (Fig.4.2).

Figure 4.2: 2nd Piola-Kirchhoff stress

note: The “pseudo-force” pictured in Fig. 4.2 will become more clear later.The following derivation of the “Second Piola-Kirchhoff” (P-K II) stresstensor is similar to the previous derivation of the “Nominal” stress tensor.The P-K II tensor will be quite useful later on to help us derive constitutiverelationships.

Recall dx = F · dX

Similarly, dfn = F · dfn ; dfn = tndS0 ; tn = N · σ ; σ = P-K II tensor

We already know that dfn = tn·dS = n·σdS

So, dfn = F · dfn = F · tndSo = tn · FTdS0 = N · σ · FTdS0

Substituting for dfn, n·σdS = N·σ · FTdSo (1)

Invoke Nanson’s equation: ndS = (detF)N · F−1dS0 (2)


(2) −→ (1) −→ (detF)��NdS0 · F−1 · σ = ��NdS0 · σ · FT

(more formally, NdS0 · [(detF)F−1 · σ − σ · FT ] = 0)

−→ σ = (detF)F−1 · σ · F−T (4.6)

Eq. (4.6) is the 2nd Piola-Kirchhoff Stress Tensor

For infinitesimal deformation (ignoring rigid body rotations):F ≈ I ; detF = 1 ; σ ≈ σ0 ≈ σ

e.x. The deformed equilibrium configuration of a body is defined by thedeformation mapping:

x1 = −12X1, x2 = 1

2X3, x3 = 2X2

(Very large deformations and very large rigid body motions)

Figure 4.3: Illustration of deformation

The Lagrangian Strain is:

E = 12(FTF− I) =

−.375 0 00 1.5 00 0 −.375

e = 1

2(I− F−TF−1) =

−1.5 0 00 −1.5 00 0 .375

4.2. ALTERNATIVE MEASURES OF STRESS 39

Note that E22 = 1.5 (this is the “2” direction ; initially the “y” direction).The specimen did not elongate from left to right, yet E22 is a positive value.In addition, E11 and E33 are the same, confirming that they represent thetransverse strains. The axes actually underwent a large rigid body rotation.E is a strain measure that rotates with the axes. In other words, E isinvariant to rigid body rotations. This will be essential to remember in thelater chapters!

The Cauchy Stress is:

σ =

0 0 00 0 00 0 100

(ignore units)

Note that σ is defined in a different coordinate system than E (“spatial” axesrather than “material” axes). σ changes with rigid body rotation, while Edoes not!

a) Determine the nominal and P-K II stress tensorsb) The nominal and P-K II tractions associated with the plane x3=const inthe undeformed statec) Repeat parts “a” and “b” for the mapping:x1 = −.99X1 ; x2 = .99X3 ; x3 = .99X2 (small deformations with very largerigid body motions)

a) and b):

Since the deformations are given, we first find F:

F =

∂x1∂X1

∂x1∂X2

∂x1∂X3

∂x2∂X1

∂x2∂X2

∂x2∂X3

∂x3∂X1

∂x3∂X2

∂x3∂X3

=

−12 0 0

0 0 12

0 2 0

With F and σ known, we can find σ0

σ0 = (detF)F−1 · σ =

0 0 00 0 250 0 0

Remember, this stress is a result of the actual force on the undeformed area.

From Nanson’s Equation (eq. (2.10)),


dA0N = 1(detF)F

T · n = 1.5FT ·

001

= 4 ∗

010

,

where

001

is the normal to the plane x3=constant

Since we know from inspection that N =

010

, we can see that the unde-

formed area is four times that of the deformed area, hence we expect thenominal stress to be four times smaller that the true (Cauchy) stress, whichit is.

We have everything we need in order to find σ:

σ = (detF)F−1 · σ · F−T︸︷︷︸[FT ]−1

=

0 0 00 12.5 00 0 0

Traction on the plane x3=constant:

t = σ ·N =

0 0 00 12.5 00 0 0

010

=

0−12.5

0

c) What about small deformation (still large rigid body rotation)?

F =

−.99 0 00 0 .990 1.01 0

E =

−.00995 0 00 .01005 00 0 −.00995

e =

−.01015 0 00 −.01015 00 0 .00985

σ0 =

0 0 00 0 980 0 0

; σ =

0 0 00 97 00 0 0

4.3. PRINCIPAL STRESSES 41

Note that σ gives a good answer for small deformations, and is defined inthe same coordinate system as E!

4.3 Principal Stresses

Again, let’s start with tn:

tn = n · σ

σnn = tn · n = n · σ · n

For what n is σnn maximized?

Remember, n · n = 1

“Lagrange multiplier”=φ = σnn − λ(n · n− 1)

In index notation, φ = niσijnj − λ(nini − 1)

∂φ∂nk

= 0 to find local optimums

∂ni∂nk︸︷︷︸δik

(σijnj) + (niσij)∂nj∂nk︸︷︷︸δjk

−λ ∂ni∂nk︸︷︷︸δik

ni − λ∂ni∂nk︸︷︷︸δik

ni = 0

Simplifying, we get: σkjnj︸︷︷︸or σkini

+niσik − 2λnk = 0 −→ 2σikni − 2λnk = 0

This further reduces to: σikni︸︷︷︸tk

−λnk = 0 ; tk − λnk = 0 −→ tk = λnk

(σ − λI) · n = 0 −→ Eigenvalue problem −→ λ is precisely σmax, min

And, at the principal plane, traction is in the direction of the normal -i.e.no shear. (max shear = σmax−σmin

2 )

Simple 2D e.x.: Problems like the following are typical in undergrad “me-chanics of materials”, where Mohr’s Circle would be used to find the maxi-mum stresses. Tensor methods are faster and can be more easily extendedto 3D.


Figure 4.4: Stresses

σ =

−4200 −4700 0−4700 12300 0

0 0 0

−→ λ1 = −5445 ; λ2 = 13540

Principal direction, nλ1 =

−.97−.26

0

θp = tan−1 .26

.97 = 75o

The above values of stress and angle agree with the transformation equa-tions from undergrad, of course.

Chapter 5

Superimposed Rotation

In preparation for the constitutive formulations in later chapters, this shortchapter will derive how different tensors behave under superimposed rigidbody motion. This will enable us, for example, to show that σ is a functionof B in the beginning of the next chapter. The idea of “work conjugate”pairs will be introduced in this chapter. We will see that the Cauchy stresstensor, σ, for example, changes with rigid body rotation, while E does not.Thus, σ and E are not work-conjugate, and therefore it would not be ap-propriate to use them as a pair when forming a constitutive relationship.

Let’s define the vector dx∗ as follows:

dx∗ = Q · dx, where Q is a superimposed rigid body motion (Fig. 5.1).

We know that dx = F · dX

What about “F∗”?

Consider the following square body (Fig. 5.2), paying careful attention tothe heavily bolded edge in order to track the motion of the body.

Clearly, if dX = dX∗ (which is what we’re going to want), and, say, F∗ = F,then dx∗ 6= F∗ · dX∗ (this should be clear from Fig. 5.2).

Since we want to find F∗ such that dx∗ = F∗ · dX∗, we can conclude thatF∗ 6= F.

What we do know is the following:

43

44 CHAPTER 5. SUPERIMPOSED ROTATION

Figure 5.1: Superimposed rigid body motion

Figure 5.2: Superimposed rigid body motion

45

dX∗ = dX ; dx∗ = Q · dx ; dx∗ = F∗ · dX∗

So, Q · dx = F∗ · dX, which gives:

F∗ = Q · F (5.1)

Eq. (5.1) is what we wanted to find, and it should be expected. Recallthat R ·U is physically understood to be a deformation (axial strains andshear), U, followed by a rotation, R. Thus, Q · F is the total deformation +rotation, F, followed by an explicit rigid body rotation, Q, - i.e. a rotationthat is superimposed on dx.

Now, we can see how different measures of strain and stress behave undersuperimposed rigid body motion, by substituting the above expression forF∗.

C = FT · F −→ C∗ = F∗T · F∗ = FT ·QT ·Q︸︷︷︸I

·F = FT · F = C (5.2)

Thus, C is invariant to rigid body motion.

note: A−1 ·A = I always ; QT ·Q = I since Q is an orthogonal tensor


note: F∗T = (Q · F)T = FT ·QT

informal proof: B = Bikeiek ; C = Cnjenej

A = B ·C = Bikeiek · Cnjenej = BikCkjeiej = Aijeiej

AT = Aijejei = BikCkjejei

orAT = Ajieiej = BjkCkieiej

AT NOT = BT ·CT = Bkieiek ·Cjnenej = BkiCjkeiej which does not matcheither of the above expressions for AT

(or BT · CT = Bikekei · Cnjejen = BikCnieken = BkiCjkeiej which issimilarly 6= AT )

AT = CT ·BT = Cjnenej ·Bkieiek = CinBkienek = CkjBikejei −→ matches1st expression for AT

or AT = CT · BT = Cnjejen · Bikekei = CkjBikejei = CkiBjkeiej −→matches 2nd expression for AT

formal proof: see Appendix A.1

B = F · FT −→ B∗ = F∗ · F∗T = Q · F · FT︸︷︷︸B

·QT = Q ·B ·QT (5.3)

What about U∗, V∗, and R∗?

F = V ·R = R ·U −→ F∗ = V∗ ·R∗ = R∗ ·U∗

Recall, F∗ = Q · F = Q ·V ·R

Insert QT ·Q = I into the above expression −→ F∗ = Q ·V ·QT︸︷︷︸V∗

·Q ·R︸︷︷︸R∗

So, V∗ = Q ·V ·QT ; R∗ = Q ·R (V∗ is symmetric ; R∗ is orthogonal)

Also, U∗ = U (skipped work)

47

How about rates?

F∗ = Q · F ; F∗ = Q · F + Q · F

Recall also, L = F · F−1 ; L∗ = F∗ · F∗−1 ; F∗−1 = F−1 ·Q−1

So, L∗ = (Q · F + Q · F) · F−1 ·Q−1 = Q ·Q−1 + Q · (F · F−1︸︷︷︸L

) ·QT

note: (Q−1 = QT since Q is orthogonal)

L∗ = Q·Q−1+Q·(D+W)·QT = Q ·D ·QT︸︷︷︸D∗

+ Q ·Q−1 + Q ·W ·QT︸︷︷︸W∗

(5.4)

note: W∗ is antisymmetric, and indeed, the two terms that make up W∗

are each antisymmetric

How about stress?

Since df∗n = Q · dfn, we can say that t∗n = Q · tn (1)

We also know the following three expressions to be true:tn = σ · n (2)t∗n = σ∗ · n∗ (3)n∗ = Q · n (4)

(4) −→ (1) −→ (3) −→ Q · tn = σ∗ ·Q · n (5)

(2) −→ (5) −→ (Q · σ)·n = (σ∗ ·Q) · n

Q · σ = σ∗ ·Q −→ σ∗ = Q · σ ·QT (5.5)

note: Since σ∗ = Q ·σ ·QT and B∗ = Q ·B ·QT , we can say that σ and Bare a “work-conjugate pair.”

How about the nominal stress and the Second Piola-Kirchhoff Stress?

σ0 = (detF) · F−1 · σ


Again, we will need to use F∗ = Q · F

Since σ0 depends on detF, we will need detF∗. It turns out, though, thatdetF∗ = (detQ︸︷︷︸

“1”

)(detF) = detF

note: the determinant of an orthogonal tensor is “1”

So, σ0∗ = (detF∗)F∗−1 · σ∗ = detF · (Q · F)−1 ·Q · σ ·QT

= (detF)F−1 ·Q−1 ·Q · σ ·QT = (detF)F−1 · σ︸︷︷︸σ0

·QT = σ0 ·QT

How about σ?

σ = σ0 · F−Tσ∗ = σ0∗ · (F∗)−T = σ0 ·QT · (Q · F)−T = σ0 ·QT ·Q ·F−T = σ0 ·F−T = σ

σ∗ = σ (5.6)

σ is a good measure of stress when there are small (infinitesimal) strains. Itwill also be used as a “work-conjugate pair” with E in the following chapterson constitutive relationships.

Chapter 6

Hyperelasticity

We’ve derived various measures of strain based on deformations that arepresumably known. We’ve also derived various measures of stress, empha-sizing the physical meaning of the stress tensor, for the case when normaland shear stresses on a body are known. To solve problems where the objec-tive is to find deformations (strains) and internal stresses within a structure,knowing, for example, the forces externally applied to the structure, we needto know the stiffnesses of the materials involved. Just as the stiffness of aspring relates force to deformation, as Hooke famously showed in 1678 “uttensio, sic vis,” meaning,“as the extension, so the force,” the stiffness of amaterial can be thought of as relating stress to strain. This relationship iscalled a constitutive relationship.

While many engineering materials, in the elastic regime, can be idealizedas linear and only undergo infinitesimal strain, many other types of ma-terials, like rubbery elastomers, behave nonlinearly and reversible (elastic)even when subjected to very large deformations. While in practice, suchmaterials typically display viscoelastic behavior even at low strain rates, atinfinitesimally slow strain rates, they approach hyperelastic behavior. More-over, hyperelastic material models underlie even the most complex modelsthat incorporate viscoelasticity, damage, etc. Thus, hyperelasticity is essen-tial for modeling polymers in FEM. Several specific kinds of rubber modelswill be considered in examples at the end of this chapter.

A hyperelastic model is a general constitutive relationship, relating stressand strain, for large deformation (large strain), nonlinear, elastic (reversible),isotropic, materials. While different materials can be best characterized by

49

50 CHAPTER 6. HYPERELASTICITY

different “strain energy density functions,” and certainly have different ma-terial constants (found via experimentation), our general constitutive ex-pression will be derived in closed form, and in 3D.

It may be useful to note that “curve-fitting” techniques that finds engineer-ing stress as a function of engineering strain, based on uniaxial stress-strainexperimental data, for example, are becoming more common. Extendingsuch a method to accommodate deformation in three dimensions is a chal-lenge that usually necessitates that we start with a 3D “hyperelastic” energyfunction. As we will see in the examples at the end of this chapter, “hy-perelastic” functions that relate stress and strain, while nonlinear in nature,actually do not require much higher order algebra, yet have been proven towork for a wide variety of hyperelastic materials. We will see that thesefunctions are fairly simple in the sense that they require the determinationof only a few experimental constants. While these experimental constantsare typically found from uniaxial tests or shear tests, the hyperelastic con-stitutive relationships are always written in tensor form (3D), using stressand strain tensors that form a work-conjugate pair. This enables our hypere-lastic constitutive relationships to be implemented into FEM quite naturally.

Hyperelastic functions fall into two major categories: micro-mechanicalmodels and phenomenological models. This will be discussed toward theend of this chapter, and several hyperelastic functions will be presented. Inaddition, it is important to be able to fully characterize a particular ma-terial once a hyperelastic function has been chosen. Experimental data isnecessary for this purpose, and several methods for efficiently characteriz-ing materials will be demonstrated. Two methods, which are particularly“cutting edge,” will be presented at the conclusion of this chapter.

This may be an appropriate time to point out that this text will develop aconstitutive relationship for two branches of elasticity: hyperelasticity (thischapter) and linear infinitesimal elasticity (final chapter). Given that hyper-elasticity encompasses elastic material nonlinearity under large strain, onemight ask how we should approach a material that is elastic linear underlarge strain, or elastic nonlinear under small strain. It turns out that thesespecial cases would be treated just like any nonlinear hyperelastic material- i.e. one would use the basic hyperelastic constitutive relationship that willbe derived in this chapter. The constitutive relationship that will be derivedin the final chapter, on linear infinitesimal elasticity, will be valid only formaterials that are linear and subjected to small deformation.

51

The following will simply be stated, but a proof can be found in AppendixC.1:

σ = g(B) (6.1)

note: Eq. (6.1) is perhaps obvious since σ∗ = Q ·σ ·QT and B∗ = Q ·B ·QT

from last chapter (we called this a work-conjugate pair based on derivationfrom Q superimposed on dx).

Now, we know that the strain energy is related to the product of force anddeformation.

So, let’s define φ to be the strain energy (per unit volume)

σ = dφdE = 2 dφdC , since E = 1

2(C− I)

Remember, σ and E (or C) are a work-conjugate pair and from eq. (4.6)we know:

σ = (detF)F−1 · σ · F−T = 2 dφdC−→ σ = 2

detFF dφdC · F

T

dφdC = RT · dφdB · R ; The proof is straightforward (can be inferred fromAppendix C.1), so long as it is recognized that φ = φ(ξ), where ξ can bereplaced by either C or B, as they have the same eigenvalues and invariants,for example.

So, 2detFF dφ

dC · FT = 2

detFV ·R ·RT︸︷︷︸I

dφdB ·R ·R

T︸︷︷︸I

·V

detF = det(V ·R) = detV, since detR = 1 (rigid body rotation)(detV = detF = 1 if material is incompressible)

So,

σ =2

detVVdφ

dB·V (6.2)

Since B = V2 −→ σ = 2

det(B12 )

B dφdB

or


(symmetrized form)

σ =1

detB1/2

(Bdφ

dB+dφ

dBB

)(6.3)

Eq. (6.3) is possible since B dφdB = dφ

dBB

Now, we know that for isotropy it is possible to write φ = φ(IB, IIB, IIIB),where IB, IIB, IIIB are each a function of B [30]. Since we will look atspecific strain energy functions, φ, later on, which happen to be of such aform, we need to apply the chain rule to dφ

dB .

−→ dφ

dB=

∂φ

∂IB

dIBdB

+∂φ

∂IIB

dIIBdB

+∂φ

∂IIIB

dIIIBdB

(6.4)

We need to determine dIBdB , dIIB

dB , and dIIIBdB :

The general expressions for the second order tensor invariants IB and IIBwill just be stated, as it is a math topic:

IB = trB ; IIB = 12 [(trB)2 − tr(B2)] (skipped work)

The complete derivations of dIBdB , dIIB

dB , and dIIIBdB are given in Appendix

C.2.

dIBdB

= I (6.5)

dIIBdB

= IBI−B (6.6)

dIIIBdB

= B2 − IBB + IIBI (6.7)

53

Eq. (6.5), Eq. (6.6), Eq. (6.7) −→ Eq. (6.4)

−→ dφ

dB=

∂φ

∂IBI +

∂φ

∂IIB(IBI−B) +

∂φ

∂IIIB(B2 − IBB + IIBI) (6.8)

Recall σ = 1detB1/2 (B dφ

dB + dφdBB)

Substituting:

1detB1/2

(B

[∂φ∂IB

IB + ∂φ∂IIB

(IBI−B) + ∂φ∂IIIB

(B2 − IBB + IIBI)

]+

[∂φ∂IB

IB + ∂φ∂IIB

(IBI−B) + ∂φ∂IIIB

(B2 − IBB + IIBI)

]B

)= 1

detB1/2

[∂φ∂IB

B− ∂φ∂IIB

B2 + ∂φ∂IIB

IBB + ∂φ∂IIIB

B3 + ∂φ∂IIIB

IIBB− ∂φ∂IIIB

IBB2

+ ∂φ∂IB

B− ∂φ∂IIB

B2 + ∂φ∂IIB

IBB + ∂φ∂IIIB

B3 + ∂φ∂IIIB

IIBB− ∂φ∂IIIB

IBB2

]= 2

detB1/2

[∂φ

∂IIIB[B3 + IIBB− IBB2] + ( ∂φ

∂IB+ IB

∂φ∂IIB

)B− ∂φ∂IIB

B2

]We know from eq. (1.15): IIIBI = B3 − IBB2 + IIBB

So, B2 = IBB− IIBI + IIIBB−1

Substituting:

−→ σ = 2detB1/2

[IIIB

∂φ∂IIIB

I + ∂φ∂IB

B +��IB∂φ∂IIB

B−��∂φ∂IIB

IBB

+ ∂φ∂IIB

IIBI− ∂φ∂IIB

IIIBB−1

]=

2

detB1/2[(IIIB

∂φ

∂IIIB+

∂φ

∂IIBIIB)I + (

∂φ

∂IB)B− (IIIB

∂φ

∂IIB)B−1] (6.9)

Sometimes in other literature (see [2] [23] [20]) IIB is taken as 12 [tr(B2) −

(trB)2] instead of 12 [(trB)2−tr(B2)]. If this is the case, then the last term in

eq. (6.9) would be added instead of subtracted, the Cayley-Hamilton Theo-rem would become B3− IBB2− IIBB− IIIBI = 0 instead of B3− IBB2 +IIBB − IIIBI = 0, and any strain energy functions that are a function ofIIB (see examples to follow) would also need to be modified accordingly.

Incompressibility:detF = 1(IIIB = 1) ; φ = φ(IB, IIB)


Modifying eq. (6.8), we get: dφdB = ( ∂φ

∂IB+ IB

∂φ∂IIB

)I− ∂φ∂IIB

B

Substituting this simplified expression into eq. (6.3), we get

σ = −ρ0I + 2

[(∂φ

∂IB+ IB

∂φ

∂IIB

)B− ∂φ

∂IIBB2

]︸︷︷︸

stress for given deformation

(6.10)

The underbraced expression is what you would set as the stress for a givendeformation. But we can always superimpose an arbitrary pressure (e.x.hydrostatic) to the body without causing deformation. Thus, the basic con-stitutive expression doesn’t uniquely specify stress ∴ we introduce ρ0 - TBDby boundary conditions

We could’ve also modified eq. (6.9) with IIIB = 1 and ∂φ∂IIIB

= 0 and wewould get

σ = −ρ0I + 2

[∂φ

∂IIBIIBI +

∂φ

∂IBB− ∂φ

∂IIBB−1

](6.11)

Then, we can use the Cayley-Hamilton Theorem to substitute for B−1 andarrive at the desired result (eq. (6.10)).

Note, however, that in eq. (6.11), the first term in the square brackets influ-ences all σii (diagonal) terms equally just as ρ0 does. Thus, this first termcan be thought of as another pressure, and can, accordingly, be lumped to-gether with ρ0.

So,

σ = −ρ0I + 2

[(∂φ

∂IB

)B−

(∂φ

∂IIB

)B−1

](6.12)

To convince yourself that eq. (6.12) is correct, go ahead and use either eq.(6.11) or eq. (6.12) in the following Mooney-Rivlin examples. You will seethat the results do not change.

6.1. PHENOMENOLOGICAL AND MICROMECHANICAL MODELS55

6.1 Phenomenological and Micromechanical Mod-els

Hyperelastic constitutive models can be divided into two categories: phe-nomenological models and micro-mechanical models. Both phenomenologi-cal and micro-mechanical models can capture the behavior of a wide rangeof polymers. Whereas some polymers are compressible, and some polymerscan strain elastically an order of magnitude more than others, polymers,generally-speaking, exhibit the same main characteristics, namely, highlynonlinear material behavior. Traditional phenomenological models are notderived from any underlying micro-mechanical physics, but are formed in amanner that seeks to minimize computational and experimental effort whilecapturing the overall behavior for a particular subset of polymers - e.g.elastomers that stiffen in compression and soften in tension. Three phe-nomenological models will be presented here: Mooney-Rivlin rubber [24][28], Blatz-Ko foam [9], and Ogden rubber [26]. Micro-mechanical models,on the other hand, capture some of the underlying physical mechanisms ofthe materials. Micro-mechanical models have the most potential for im-provement, and there are already some micro-mechanical models that canextend to unusual kinds of polymers, where traditional models would di-verge from real behavior [8]. Here, only one micro-mechanical model willbe presented: Arruda-Boyce rubber [1]. The Arruda-Boyce model, in itsoriginal form [1], can handle typical elastomers, but, like any hyperelasticmodel, will be unable to capture the hyperelastic behavior of all polymers.

Micro-mechanical models may be particularly well suited for “design.” Anengineer can easily “create” a micro-mechanical (or phenomenological) poly-mer model based on, for example, the Arruda-Boyce model, which can becharacterized from certain basic material properties. These material prop-erties can be decided by the engineer, and the resulting performance for aparticular application can be observed in computer simulations, for example.Once the behavior is satisfactory, according to the simulation, the materialcan then be manufactured. Micro-mechanical models and some phenomeno-logical models are well suited for such a design process, since they generallydepend on basic properties such as initial shear modulus and bulk modulus,and since these are precisely the properties that polymer manufacturers arefamiliar with.


Mooney-Rivlin e.x. 1

Consider a rectangular block under tensile stress in X1 direction (simpleextension test) that causes the stretch in that direction of amount λ1. If thematerial of the block is an incompressible “Mooney-Rivlin” rubber [24] [28]with:

φ =1

2µ

[(1

2+ β

)(IB − 3) +

(1

2− β

)(IIB − 3)

](6.13)

where µ and β are material constants, find the stress required to producethis deformation:

Incompressible → σ = −ρ0I + 2[(

∂φ∂IB

)B−

(∂φ∂IIB

)B−1

]∂φ∂IB

= 14µ+ 1

2µβ ; ∂φ∂IIIB

= 0 ; ∂φ∂IIB

= 14µ−

12µβ

σ = −ρ0I + 2[(

14µ+ 1

2µβ)B−

(14µ−

12µβ

)B−1

]We need to figure out a x←→ X mapping. We could do this by noting that“Poisson Ratio” ν = .5 for incompressible materials, but since we have toconsider deformation in two transverse directions, it’s better to start withgeneral stretches λ1 and λ2, and later use detF = 1, as follows:

x1 = λ1X1

x2 = λ2X2

x3 = λ2X3 (λ3 = λ2 due to isotropy)

F =

λ1 0 00 λ2 00 0 λ2

detF = λ1λ

22 = 1 −→ λ2 = 1√

λ1

−→ B = F · FT =

λ21 0 0

0 1λ1

0

0 0 1λ1

σ11 = −ρ0 + µ

[(1

2 + β)λ21 − (1

2 − β) 1λ21

]σ22 = σ33 = −ρ0 + µ

[(1

2 + β) 1λ1− (1

2 − β)λ1

]= 0 (we know this)


−→ ρ0 now known −→ σ11 = µλ31−1λ1

[12 + β + (1

2 − β) 1λ1

]λ1 is analogous to “ε11” (which we’ll see later)

µ, β are just constants, analogous to “µ”, “λ” (which we’ll see later)

Thus, for this particular Mooney-Rivlin, hyperelastic, incompressible, ma-terial, we have a relationship between longitudinal stress and longitudinalstrain. This is a method we’ll use later in linear infinitesimal isotropic the-ory to find the definition of “E” (Young’s Modulus), for example.

Mooney-Rivlin e.x. 2

Determine the stress and strain state in a rectangular block, made from theMooney-Rivlin material of e.x. 1, under simple shear of amount ϕ, in thedirection X1:

Figure 6.1: Simple shear

σ = −ρ0I + 2

(1

2+ β)

µ

2︸︷︷︸a

B− (1

2− β)

µ

2︸︷︷︸b

B−1

(from the previous e.x)

If k = tanϕ, x1 = X1 + kX2 ; x2 = X2 ; x3 = X3 ; σ33 = 0

We can easily calculate F and B:


F =

1 k 00 1 00 0 1

B = F · FT =

k2 + 1 k 0k 1 00 0 1

B−1 =

1 −k 0−k 1 + k2 00 0 1

note: E =

0 12k 0

12k

12k

2 00 0 0

Let’s take a moment to look at Fig. 6.2. Aside from B being defined inspatial coordinates, it’s also “Eulerian” and thus is more difficult to inter-pret, physically, for most engineers. From inspection of E, however, we cansee that the coordinate system must be deformed, as shown in Fig. 6.2,otherwise E22 would be zero.

Figure 6.2: Deformed axes for Right C-G Tensor

This is mentioned now because the next topic is rate forms, where it is com-monly assumed (simplified) that the bases are orthogonal, which can leadto dubious results where large shear deformation is present.

note: both σ11 and σ22 are nonzero but it is common to have stress withoutstrain for an incompressible material


σ11 = −ρ0 + 2a ∗ (1 + k2︸︷︷︸B11

)− 2b ∗ ( 1︸︷︷︸[B−1]11

)

σ22 = −ρ0 + 2a ∗ ( 1︸︷︷︸B22

)− 2b ∗ (1 + k2︸︷︷︸[B−1]22

)

σ12 = 2(a+b)k ←− no ρ0 because ρI12 = 0 due to I being a diagonal matrixσ33 = −ρ0 + 2a− 2bσ23 = 0σ31 = 0

σ33 = 0 −→ −ρ0 + 2a− 2b = 0 −→ ρ0 = 2(a− b)(substitute into the above expressions)

σ11 = 2ak2 σ22 = −2bk2 σ12 = 2(a+ b)k

note: k = tanϕ = ϕ for infinitesimal strains→ ϕ2 << ϕ→ σ11 and σ22 maybecome negligible

note: E and e both approach

0 k 0k 0 00 0 0

for infinitesimal strain, but they

would not be equal if large rigid body rotations were present

Examples 1 and 2 derive the necessary formulas to determine the basic ma-terial properties that fully describe a Mooney-Rivlin rubber. For example,an axial test would give you σ11 and “λ1” while a shear test would give youσ12 and “k.” These would be unique for a particular Mooney-Rivlin rubber.Once these data are known, then there are two equations in σ11, σ12 and onecan solve for the two unknowns: µ and β.

Blatz-Ko Foam

Consider the following strain energy function, which is a simplified versionof the Blatz-Ko function [9]:

note: It has been assumed that the material is compressible, with a PoissonRatio ν = .25

φ =1

2µ(

2III1/2B + IIBIII

−1B − 5

)We can derive the constitutive equation for σ in terms of µ and B as follows:


∂φ∂IB

= 0 ∂φ∂IIB

= 12µIII

−1B

∂φ∂IIIB

= 12µ(III

−1/2B − IIBIII−2

B )

Substituting into eq. (6.9), re-written below:

σ =2

detB1/2

[(IIIB

∂φ

∂IIIB+

∂φ

∂IIBIIB)I +

∂φ

∂IBB− IIIB

∂φ

∂IIBB−1

]

We then get:

σ = 2detB1/2

[(IIIB ∗ 1

2µ(III−1/2B − IIBIII−2

B ) + 12µIII

−1B ∗ IIB

)I

−IIIB ∗ 12µIII

−1B ∗B−1

]

This reduces to:

σ =1

detB1/2∗ µ(III

1/2B I−B−1)

or

σ = µ(I− III−1/2

B B−1)

µ is analogous to the “shear modulus” as we’ll see in linear infinitesimalelasticity in a later chapter. However, in order to simplify experimentation,a simple uniaxial test is often performed in order to determine µ for materialsthat follow the above Blatz-Ko function.

Arruda-Boyce Rubber

Before we get to the Ogden rubber model, which will be discussed at lengthin this section and the next one, let’s briefly describe the Arruda-Boycemodel [1]. The Arruda-Boyce model, sometimes called the “Eight-Chainmodel” is a very popular model and, along with the Ogden model, is a func-tion of the so-called “principal stretches.” It is a statistically derived modelbased on the macromolecular physical mechanics of “long-chain” materials.


The strain energy density function is given as:

φ = nkΘ

[1

2(IB − 3) +

1

20N(I2B − 9) +

11

1050N2(I3B − 27)

]+ nkΘ

[19

7000N3(I4B − 81) +

519

673750N4(I5B − 243)

]+K

2(IIIB − 1)2

nkΘ are, respectively, the “chain density,” Boltzmann’s constant, and tem-perature, and as a product represent the shear modulus, G.

Note that the last term in the Arruda-Boyce strain energy density functiongoes to zero for incompressible (“rubbery”) material. We can also see thatthe Arruda-Boyce rubber model depends only on the principal stretches -i.e. we can substitute IB = λ2

1 + λ22 + λ2

3 into the above equation, where λ1,λ2, and λ3 are the eigenvalues of the Left Stretch Tensor, V, or the RightStretch Tensor, U. Recall that V and U have the same eigenvalues and thesame invariants, but different eigenvectors. The λ are commonly called theprincipal stretches, and usually refer to the eigenvalues along the principalaxes of V. To find a constitutive relationship for the Cauchy Stress, σ, wecould use the following relationship (derived in Appendix C.3):

σi =1

λjλk︸︷︷︸no sum

∂φ

∂λi

σi are the principal true stresses (the eigenvalues of the Cauchy Stress, σ).

Alternatively, for incompressibility, and where φ is only a function of IB, wehave:

σi = −p+1

��IIIVλi∂φ

∂IB

∂IB∂λi

where “p” is our “hydrostatic” term that would need to be found fromboundary conditions.

Since dIBdλi

is 2λi, we get:

σi = −p+ 2λ2i

∂φ

∂IB(6.14)

Keep in mind that eq. (6.14) is only valid if φ = φ(IB).


The Arruda-Boyce model presented here is the original version [1], and istypically used to describe soft elastomers at large strains. Other versionsof this model exist [7] that are more suitable for describing the behavior ofelastomers in the smaller strain range (less than 30%). This kind of behavioris most often of interest for relatively stiff polymers.

That’s all that will be said about the Arruda-Boyce model. What we willdo in the remainder of this section is derive the constitutive expression forthe Ogden model, which we will need in order to form our “tabulated” con-stitutive relationship in the next section, which will be based on the Ogdenmodel.

Ogden Rubber

The strain energy density function for an Ogden material [26], is given by

φ =

3∑n=1

m∑s=1

µsαs

(λ∗αsn − 1) +K(IIIV − 1− ln IIIV) (6.15)

In eq. (6.15), µs and αs are material constants.

λ∗n = λn ∗ III−1/3V and IIIV = λ1λ2λ3

Eq. (6.15) is valid for nearly incompressible materials.

The strain energy density function, φ, for the Ogden model will be a func-tion of the principal stretches, just as in the Arruda-Boyce model. We willmake use of the constitutive relationship:

σi =1


∂φ

∂λi(6.16)

Eq. (6.16) is derived in Appendix C.3.

To derive the Ogden constitutive relationship, we need to find ∂φ∂λi

for theOgden strain energy density function.

In the following derivation, i, j, k, will be used as subscripts to indicateeigenvalues 1, 2, 3. Unless indices are repeated, they should not be summed.

6.2. TABULATED AND CALIBRATED MODELS 63

∂(∑3

n=1µsαsλ∗αsn

)∂λi

=∂

(∑3n=1

µsαsλαsn III

−αs3V

)∂λi

=

only one term since∂λαsn

∂λi=δin︷︸︸︷

µsαsλαs−1i

αsIII

−αs3

V

+

3 terms since nonzero for all values of n︷︸︸︷3∑

n=1

µsλαsn

αs(−αs

3)III

(− 13αs−1)

V λjλk︸︷︷︸no sum

= µsλαsiλiIII

−αs3

V −∑3

n=1 µsλαsn

3 III(− 1

3αs−1)


∂(∑3

n=1µsαsλ∗αsn )

∂λi= µs

λ∗αsi

λi−

3∑n=1

µsλ∗αsn

3III−1


(6.17)

∂K(IIIV−ln IIIV)∂λi

= K(∂IIIV∂λi− ∂ ln IIIV

∂λi) = K(

no sum︷︸︸︷λjλk − 1

IIIV

no sum︷︸︸︷λjλk )

∂K(IIIV − ln IIIV)

∂λi= K

no sum︷︸︸︷λjλk

(IIIV − 1)

IIIV(6.18)

eq. (6.17) and eq. (6.18) −→ eq. (6.16), where φ is given by the Ogdenformula (eq. (6.15)), yields the following result:

σi =

m∑s=1

µsIIIV

[λ∗αsi −

3∑n=1

λ∗αsn

3

]+K

IIIV − 1

IIIV(6.19)

6.2 Tabulated and Calibrated Models

In the past, experimental constants would typically be found by hand. Givena known material, it can be time-consuming to identify their parameters,even after the necessary experimental tests have been performed - e.g. sincehyperelastic materials are nonlinear by definition, matching the experimen-tal data is not so trivial as finding the slope of a uniaxial stress-strain curve,as would be the case for a linear material. The Mooney-Rivlin example inthe last section showed that the procedure for obtaining experimental con-stants can be cumbersome. For more advanced hyperelastic models, findingthe experimental constants would be even more difficult. More recently, twomethods have been developed to “fit” a particular material to a particularhyperelastic model - the “tabulated” method and the “calibration” method.


Before we look at a particular tabulated model as well as a particular methodfor calibrating a model, consider that for many hyperelastic models, uniax-ial data is sufficient to fully characterize their behavior. Exceptions wouldinclude models that are heavily influenced by higher order strain invariants.Regardless, uniaxial data is the most common kind of experimental datathat is used for characterizing materials - hyperelastic materials included.In fact, both of the following examples will fit a hyperelastic model to uni-axial data, at the exclusion of any other kind of experiment data.

These characterization methods may be considered “curve-fitting” tech-niques, but keep in mind that the hyperelastic models that we are attemptingto match to a uniaxial curve are tensorial, not scalar-valued. Using a tabu-lated approach, the challenge is to take a hyperelastic function, which formsa work-conjugate constitutive relationship that naturally extends to a 3Denvironment, and modify it to “grab” data from tabulated uniaxial stress-strain data. A particular tabulated approach developed by Stefan Kolling ofDaimler AG, in collaboration with P.A. Du Bois of LSTC and David Bensonof UCSD [17], will be presented, which is based on the Ogden model andgenerates a “perfect fit.”

f0(λi) =∞∑x=0

λ(−1/2)x

i σ0

(λ

(−1/2)x

i − 1)

(6.20)

σi =1

IIIV

(f0(λi)−

1

3

3∑n=1

f0(λn)

)+K

IIIV − 1

IIIV(6.21)

The complete derivation of eq. (6.20) and eq. (6.21) can be found in Ap-pendix C.4, based on the descriptions provided in [17].

In a simplified FEM algorithm (no rates/hypoelasticity), the series expressedabove (eq. (6.20)), which calculates a single value of f0(λi) from the uniaxialdata by summing (over “x”) to ∞, would be terminated once a reasonabletolerance is met. In addition, a sufficient number of f0(λi) values could becalculated, where each value is calculated from eq. (6.20) and correspondsto a particular value of λi. The number of f0(λi) values could perhaps beequal to the number of σ0(ε0i) values that have been provided by the user,presumably from a uniaxial engineering stress vs. engineering strain plot.Note that all such f0(λi) values can be found during the initialization of the


FEM problem, prior to the start of the simulation.

In solving actual problems using this material model (though this is stillonly for illustrative purposes, since real problems involve time and neces-sitate a hypoelastic constitutive formulation), the actual stretches are theeigenvalues of V and would be determined from the actual loading. σ1, σ2,σ3, which are the eigenvalues of the Cauchy stress, σ, can be calculatedfrom eq. (6.21), where “i” in eq. (6.21) refers to the eigenvalue 1, 2, or3. Obtaining the Cauchy stress, σ, from its eigenvalues and eigenvectorsinvolves just a single transformation.

While the original form of the Ogden model (eq. (6.19)) contains an arbi-trarily large number of constants, it is not necessarily true that the Ogdenmodel, in its original form, can fit any uniaxial stress vs. strain curve. In itsoriginal form, the Ogden model contains the constants µs and αs, and theseconstants (of which there exist 2m of them) would be solved simultaneously.One possible way to attempt this would be to start with eq. (6.20), wherewe recall that the left-hand-side of eq. (6.20) is

∑ms=1 µsλ

∗αsi = f0(λi). It

is not clear that creating multiple equations from eq. (6.20) and solvingsimultaneously for these constants is feasible.

The method presented, on the other hand, which is referred to as a “tabu-lated” method, can fit any uniaxial curve, perfectly. One can consider thatthis tabulated method is possible because of a number of particular charac-teristics in the original Ogden expression (eq. (6.19)), and clever “tricks”employed by the authors of the tabulated model [17], which are shown inAppendix C.4.

The value of this “tabulated” method cannot be understated: it makes pos-sible the exact fit to any uniaxial stress vs. strain experimental data. Thematerial would then be characterized by Ogden hyperelasticity, which, intheory should work quite well for any type of loading condition that the ma-terial may see (torsion, bending, etc.). Viscoelasticity would be the primaryconcern, and this will be briefly discussed at the end of this chapter.

A “calibration” method will be shown next, and then the two methods willbe briefly compared. Fig. 6.3 is a screen-shot from a software called MCali-bration, developed by Jorgen Bergstrom of Veryst Engineering R©. MCalibra-tion contains a library of materials called PolyUMod [7]. Contained withinthis library are many hyperelastic materials, including the Arruda-Boyce,


Mooney-Rivlin, and Ogden.

Figure 6.3: Example of Calibration method

Data from a uniaxial compression experiment was loaded into MCalibration.We can assume, here, that the compression data represents the relaxed equi-librium response of the material - in other words, it is hyperelastic. The YeohModel [35] was then calibrated, since this model provided a better fit, forthis particular data set, compared to the Arruda-Boyce or the Ogden model.We can see that the fit is still not perfect.

The advantage of the tabulated method, as stated previously, is that it willmatch any uniaxial curve, exactly, and take almost no computational effortto do so. A disadvantage of the tabulated method is that physical materialproperties never enter the conversation, so error in the uniaxial data can gounrecognized. Still, there are surprisingly very few tabulated models, likethe one that has been presented, which are built from hyperelastic functions.This may be due to concerns related to the treatment of viscoelasticity, aswill be briefly discussed.


Viscoelasticity

Perhaps the real reason that we don’t see more tabulated models is becausefor many kinds of polymers and for many applications that involve polymers,viscoelastic rate effects on loading and unloading behavior are as importantas the underlying hyperelastic component of behavior. In reality, it is ex-tremely difficult to obtain the relaxed equilibrium response of an elastomer,and so even a “quasi-static” uniaxial test will have different loading andunloading response. In other words, hyperelasticity is only one aspect of theresponse of an elastomer under realistic loading rates.

The specific implementation of the tabulated model previously described[17] is the LS-DYNA material model called:∗MAT SIMPLIFIED RUBBER WITH DAMAGE [13]

While this model can handle a user-defined uniaxial loading curve as wellas an unloading curve, it does not handle hyper-viscoelasticity in a physicalsense. This is why the developers of the model [17] use the word “simplified”and “damage” to describe the hysteretic behavior capability. No tabulatedmethod, to the author’s knowledge, exists that can handle both hyperelastic-ity and viscoelasticity in a physical sense. The PolyUMod library, however,consists of many viscoelastic options, and MCalibration is capable of “fit-ting” uniaxial hysteretic data using models that contain components of bothhyperelasticity and viscoelasticity. This topic, however, is beyond the scopeof this text.

Chapter 7

Rate-Form ConstitutiveExpressions

We are going to begin discussion on “hypoelasticity”, which deals with ratesof stress and strain. In FEM, this permits the treatment of dynamic prob-lems and history-dependent problems, and thus is the basic way that ad-vanced computer software develop constitutive relationships. As always, itis important to recognize which measures of stress and strain are invariantto rigid body rotation and which are not. Whether we are talking abouthyperelasticity or linear infinitesimal elasticity, it is always easy to make upa problem in which a body deforms in some prescribed fashion, with largerigid body rotations, and then determine how measures of strain behave un-der such deformations and rigid body rotations. This is relatively simple todo since strain is a direct function of deformation (e.g. E is a direct functionof F), and so this strain behavior can essentially be seen. For example, theLagrangian strain tensor, E (as well as the linear infinitesimal strain tensor,ε), does not change with rigid body rotation. This was shown via examplein previous chapters.

Stress is not so easy to “see.” The Cauchy stress tensor, σ, which is usedin hyperelasticity, does change with rigid body rotation (it is measured in“spatial” coordinates rather than “material” coordinates). The linear in-finitesimal stress tensor, as it turns out (next chapter), can be thought of asσ (recall eq. (4.6)) in the sense that the linear infinitesimal stress tensor isdefined in the same coordinate system as the strain tensor - it is invariant torigid body rotation. Tensors that behave similarly under rigid body rotation(e.g. σ ↔ E, σ ↔ D,B) are said to be “work-conjugate.” In order to form a

69

70 CHAPTER 7. RATE-FORM CONSTITUTIVE EXPRESSIONS

proper constitutive relationship, the stress and strain tensors involved mustbe work-conjugate.

Recall the e.x. starting on pg. 38.Fig. 7.1 shows a simpler version of this example:

Figure 7.1: Uniaxial stress with large rigid body rotation

Prior to the rigid body rotation (i.e. at time t = t0):

σ =

[σ11 00 0

]After this 90o rigid body rotation (at time t = tf ), there are two possibilities:

σ =

[σ11 00 0

](7.1)

or

σ =

[0 00 σ11

](7.2)

Unless we perform some mathematical “tricks” between t0 and tf , tensor(7.1) would be the tensor for linear infinitesimal elasticity, σ, at time tf ,

7.1. HYPOELASTICITY (JAUMANN) 71

and tensor (7.2) would be the Cauchy stress tensor, σ, at time tf .

In some literature it is stated that the linear infinitesimal stress tensor isthe Cauchy stress and the linear infinitesimal strain tensor corresponds tothe Eulerian strain, e. We have seen (example that started on page 38),the physical interpretation of the Eulerian strain, and the choice of strainmeasure for the case of linear infinitesimal elasticity is essentially up to theindividual. This can be a source of confusion since E is not equal to e whenrigid body rotations are present, but this will be addressed at the end of thischapter when we get to our “time-stepping algorithm.”

In real-life structural analysis simulations, adjacent elements need to havea consistent frame of reference. As decribed in [30], under nonhomogeneousdeformations, adjacent elements will not “fit together” if we use the materialor “local” configuration (eq. (7.1)). With this in mind, eq. (7.2) (i.e. a“spatial” representation), is what we need.

We’ll first consider the rate form of Cauchy stress, σ, before taking a closerlook at the infinitesimal stress, σ. One reason that finite hyperelastic-ity (nonlinear) theory and linear infinitesimal theory typically use differentwork-conjugate stress/strain pairs is because they were developed indepen-dently [23]. We will thus be required to keep track of numerous differentmeasures of stress and strain, which may benefit the reader or may be viewedas a nuisance, depending on perspective.

note: In this text, the word “spatial” refers to the coordinate system at timezero. This chapter will conclude with the presentation of one possible (andvery simple) “time-stepping algorithm.”

7.1 Hypoelasticity (Jaumann)

Recall that σ and B were shown to be work-conjugate, and this permit-ted us to develop a constitutive relationship between the two tensors. Nowwe would like to develop a constitutive relationship in hypoelastic form.Thus, we need a stress rate and a strain rate that are work-conjugate.Unfortunately, the time derivative of the Cauchy stress tensor, σ, is notwork-conjugate with D, where we recall that D is the “rate of deformationtensor.” Consider the following:


5σ =

1

detFF · ˙σ · FT =

1

detFFd

dt

[(detF)F−1 · σ · F−T

]FT (7.3)

Here,5σ is the “Truesdell rate.” Essentially, we wrote σ in terms of the

Second Piola-Kirchhoff Stress, and then, within this expression, we took thetime derivative of only the Second Piola-Kirchhoff Stress (recall from eq.(4.6), that the Second-Piola Kirchhoff Stress, σ, is (detF)F−1 · σ · F−T ).

The time derivative expressed in eq. (7.3) can be taken, which requires theuse of the product rule for the four terms within the square brackets. Analternative derivation is given in Appendix D.2. In either case, eq. (7.3)becomes:

5σ = σ − L · σ − σ · LT + tr(L)σ (7.4)

Recall the definition of the velocity gradient, L, from eq. (3.1). Addition-ally, note that trL = trD.

Eq. (7.4) is the most commonly used form of the Truesdell rate. In eq.(7.4), if we eliminate all of the D terms except for those that appear in theconstitutive relationship (we can also think of this simplification as takingL ≈ R ·RT ≈W), then we arrive at the following result:

σ = σ −W · σ − σ ·WT (7.5)

or

σ = σ −W · σ + σ ·W (7.6)

note: WT = −W

In hypoelasticity, eq. (7.5) is called the Jaumann rate of Cauchy Stress.We will form our hypoelastic constitutive relationship in terms of this rate.Then, we’ll see how this constitutive relationship would be used to solve realFEM problems, through a simple three step time-stepping algorithm. Butfirst, we need to prove that the Jaumann rate of Cauchy Stress is indeed


work conjugate with D - i.e. does σ∗ = Q · σ ·QT ?

Note that σ can represent the linear infinitesimal stress. It turns out thatmost FEM software use the Jaumann rate for linear infinitesimal stress aswell. The reason for this will be explained at the end of this chapter, alongwith a brief disussion of linear infinitesimal elasticity time-stepping.

Recall σ∗ = Q · σ ·QT

σ∗ = Q · σ ·QT + Q · σ ·QT + Q · σ · QT

= Q ·Q−1 ·Q︸︷︷︸I

·σ ·QT + Q · σ ·QT + Q · σ ·QT ·Q︸︷︷︸I

·QT

= Q ·Q−1 · σ∗ + Q · σ ·QT + σ∗ · (Q ·Q−1)T

Since W∗ = Q ·Q−1 + Q ·W ·QT −→ Q ·Q−1 = W∗ −Q ·W ·QT

−→ σ∗ = (W∗ −Q ·W ·QT ) · σ∗ + Q · σ ·QT + σ∗ · (−W∗ + Q ·W ·QT︸︷︷︸(Q·Q−1)T

)

Moving some of the terms to the other side:−→ σ∗−W∗ ·σ∗+σ∗ ·W∗ = Q · σ ·QT +σ∗ ·Q ·W ·QT −Q ·W ·QT ·σ∗

Since Q ·W ·QT · σ∗ = Q ·W ·��QT ·��Q · σ ·QT︸︷︷︸

σ∗

= Q ·W · σ ·QT

and σ∗ ·Q ·W ·QT = Q · σ ·QT ·Q ·W ·QT = Q · σ ·W ·QT

−→ σ∗ −W∗ · σ∗ + σ∗ ·W∗ = Q · (σ −W · σ + σ ·W) ·QT

Since we know that σ = σ −W · σ + σ ·W

−→ σ∗ = Q · σ ·QT (7.7)

Note the similarity to σ∗ = Q ·σ ·QT and D∗ = Q ·D ·QT (this is what wewanted to show)

Now, just as we have derived σ ←→ B in the past, we can now deriveσ ←→ D (σ will also depend on σ, B, and B).

Truesdell et al. [32] coined the phrase “hypoelasticity.” All hyperelastic con-stitutive relationships can be put into rate form (and hence are hypoelastic).The reverse is not always possible [30]. Materials that are viscoelastic, orany material that is expected to experience plasticity, can be convenientlyexpressed in rate form, where history dependent variables can be tracked,


naturally.

note:B∗ = Q ·B ·QT

σ∗ = Q · σ ·QT

D∗ = Q ·D ·QT

σ∗ = Q · σ ·QT

Since all of the tensors above are work-conjugate, it is convenient to usethem to derive our final form of σ.

Recall σ = 1detB1/2 (B dφ

dB + dφdBB):

We will form our hypoelastic constitutive relationship by applying the Jau-mann “operator” ( ) to both sides of this expression. But, before we dothis, we ought to find ˚detF and B.

˚detF = ˙detF (i.e. presumably, the analog to −W ·B + B ·W would simplybe zero, since any asymmetries cancel out when taking the determinant)

From eq. (2.9) we know that dV = dV0detF and from eq. (3.5) we knowthat d

dt(dV ) = DkkdV .

So, ddtdV = DkkdV = ˙detFdV0. Thus,

˙detF =dV

dV0︸︷︷︸detF

Dkk = (detF)trD (7.8)

In terms of B: detF = detB1/2 −→ ddtdetB

1/2 = detB1/2trD

Next, we need to find B:

B = F · FT

−→ B = F ·FT +F ·FT = F · F−1︸︷︷︸L

·F · FT︸︷︷︸B

+F ·

I︷︸︸︷FT · F−T ·FT = L ·B+B ·LT

B = L ·B + B · LT = D ·B + B ·D + W ·B−B ·W, (since L = D + Wand LT = DT + WT = D−W)


B = (D ·B + B ·D + W ·B−B ·W)︸︷︷︸B

−W ·B + B ·W = D ·B + B ·D

Reminder: The direct calculation of the time derivative of B (i.e. B), similarto σ, is not work-conjugate with D. This is why we need to use B, just aswe are using σ.

detB1/2σ = B dφdB + dφ

dBB

∥∥∥∥ apply the Jaumann operator ( ) to this

entire expression.

Noting that Bdφ

dB= B

dφ

dB+ B

d2φ

dB2: B︸︷︷︸

product rule and chain rule

−→ σdetB1/2trD + detB1/2σ = B dφdB + B d2φ

dB2 : B + ( d2φdB2 : B) ·B + dφ

dBB

note: Recall from Section 1.4 that A : B = tr(A ·BT ) = AijBij

So, σij = −σijDkk +AijmnBmn

where (skipped work):

Aijmn = 1detB1/2 [δim

dφdBnj

+ dφdBim

δjn +Bikd2φ

dBkjdBmn+ d2φ

dBikdBmnBkj ]

Since B = B ·D+D ·B, all terms in the above expression involve D. Thus,we arrive at the following relationship:

σij = ΛijklDkl (7.9)

Eq. (7.9) is the Jaumann rate of Cauchy Stress in terms of deformation,stress, and rate of deformation.

Note that we could’ve formed a slightly different hypoelastic relationshipusing the Truesdell rate. Either way, the hypoelastic expression is goingto be complicated (the Truesdell rate, slightly more so). This is, however,meant to be broad enough to treat materials that are both nonlinear elastic,and subjected to large strain.


7.2 Time Stepping Algorithm

In hypoelasticity, we required the use of the so-called “Jaumann rate ofCauchy Stress”, σ, because σ is not work conjugate with σ or D, while σis. However, we still need σ, because σn = σn+1 + σ∆t is the stress tensorthat we need. The Cauchy stress (and its rate) have a well-understood phys-ical meaning (whereas σ and σ∆t do not have physical meaning) - namely,σ is defined in the “spatial” coordinate system. Recall once more that thisis necessary in order to have a consistent frame of reference for all of theelements in the finite element simulation. Thus, before we can consider theproblem “solved” for any particular instance in time, we need to add anadditional “step” to our solution procedure that obtains σ from σ. We cando this by simply solving for σ from our previous expression for the Jau-mann rate of Cauchy Stress (eq. (7.5)). Thus, during each time step, theJaumann rate is the stress that is used for the constitutive relationship, butthe Cauchy stress is what is needed for the equation of motion (e.x. stressequilibrium).

It turns out that this same multi-step time-stepping “algorithm” will workfor linear infinitesimal elasticity as well. We already mentioned that σ isanalogous to the linear infinitesimal stress tensor, and E is analogous tothe linear infinitesimal strain tensor, ε. In other words, the work-conjugatepair - defined in material coordinates - that is used in linear infinitesimalelasticity, can be thought of as a simplified version of σ ↔ E. ˙σ and E weknow to be similarly work-conjugate and invariant to rigid body rotation.

Thus, one possible expression in linear infinitesimal elasticity that is analo-gous to our hypoelastic expression (eq. (7.9)) could be:

˙σij = Cijklεkl

where Cijkl is a fourth-order tensor that relates linear infinitesimal stress, σ,to linear infinitesimal strain, ε, as we will see in the next chapter.

Obtaining the stress in spatial coordinates can be accomplished via eq. (4.6),namely, ˙σ = d

dt

[(detF)F−1 · σ · F−T

]. After performing this time deriva-

tive, we could then rearrange to find the rate of Cauchy stress, σ, in termsof ˙σ. However, it may be desirable to take a different approach in orderto avoid having to explicitly calculate the deformation gradient, F, or itstime-derivative F.

7.2. TIME STEPPING ALGORITHM 77

In addition, we might want to avoid material coordinates (e.x. we mightwant to avoid calculating ε ≈ E, even though E is the most intuitive mea-sure of strain). As we’ll see shortly, in our particular “time-stepping algo-rithm,” at no time do we consider the material reference frame. This is alsoone reason why we did not form our finite-strain hypoelastic relationship inthe form ˙σ = f(E, E).

Let’s instead consider our previous definitions of the Truesdell and the Jau-mann rate, which, for infinitesimal elasticity, state that:

5σ ≈ σ ≈ F · ˙σ · FT

Here, we should note that the term 1detF was eliminated since detF ≈ 1 for

infinitesimal deformations.

Since ˙σ = C : ε and σ ≈ F · ˙σ · FT , we can see that:

σ = C : (F · ε · FT ) (7.10)

From Chapter 3, we know that eq. (7.10) conveniently reduces to:

σ = C : D (7.11)

Or,

σij = CijklDkl (7.12)

In eq. (7.11), we recall that D is the “Rate of Deformation Tensor.”

Some authors write the constitutive relationship for linear infinitesimal elas-ticity as σ = C : e. The physical nature of E and e was illustrated in theexample that started on page 38, where we saw that the two strain measuresgave different results, even for the infinitesimal deformation case. Ultimately,however, one can argue that we are free to choose either stress-strain pair inlinear infinitesimal elasticity. Applying the Jaumann operator to either con-stitutive expression results in the equation (7.11). Thus, our “time-steppingalgorithm,” which we will see shortly, is independent of our “interpretation”of linear infinitesimal stress and strain.


The Kirchhoff stress, τ , is defined as τ = (detF)σ. Sometimes τ is usedin place of σ, in eq. (7.11) for example, so that the small deformationapproximation of detF ≈ 1 does not need to be used.

Obtaining the stress rate in spatial coordinates simply requires that we solvefor σ from eq. (7.5), namely,

σ = σ + W · σ + σ ·WT (7.13)

An alternative proof of eq. (7.13), for infinitesimal elasticity, is given inAppendix D.1

Thus, our constitutive expressions, in rate-form, for both linear infinitesi-mal elasticity and hyperelasticity (hypoelasticity), involve the Jaumann rate.For hypoelasticity, we had eq. (7.9), while for linear infinitesimal elasticity,we have eq. (7.12). The following time-stepping algorithm, which allows usto obtain Cauchy (“spatial”) stress is identical for both hyperelasticity andlinear infinitesimal elasticity.

note that the bases do not always remain orthogonal when there is shear(recall Fig. 6.2, for example). The Jaumann rate in eq. (7.5) assumesorthogonal bases, which is a simplification.

General Framework For Problem Solving

1. Use an appropriate work-conjugate constitutive relationship:σij = ΛijklDkl or σij = CijklDkl

2. Obtain a spatial representation of stress by first finding σ from eq. (7.13),and then finding σ from σn+1 = σn + σ∆t

3. Solve the equation of motion for the next increment in time:σij,j + ρbi = ρdvidt

The majority of this text is devoted to “1” and “2” (in a theoretical sense).Good course sequences on FEM spend a great deal of time on the implemen-tation of these steps - in particular on step “3.” Recall that the expressiongiven in step “3” was derived in Chapter 4.

7.2. TIME STEPPING ALGORITHM 79

The above “time-stepping algorithm” is just one possibility. The purposehere is merely to show one particular methodology that FEM software usefor solving real-life problems. This methodology has only been shown toapply to isotropic elastic materials. Different time-stepping algorithms aresurely used for other situations.

Plasticity

Obtaining Cauchy stress from the Jaumann rate is what is most often doneby FEM software, even though the Truesdell rate form is better (exact) forelasticity. The Jaumann rate is known to have some problems with shearbehavior due to the assumption of orthogonal bases [33]. In addition, forhighly compressible materials, the Truesdell rate can give significantly bet-ter results [4]. Perhaps, then, the main reason that the Jaumann rate ispreferred over the Truesdell rate (and other rates) has to do with plasticity.

In plasticity, many assumptions are made. The Truesdell rate turns outnot to be good for plasticity, in general, because the dilatational (volume-changing) terms that relate σ and the Truesdell rate are ignored in plasticity.So, suffice to say, plasticity is a complex topic that will not be covered inthis introductory text, but the Jaumann rate, σ, is often used in FEM forsolid elements, because it is computationally efficient and handles plasticitybetter than the Truesdell rate. A theoretical overview of implementation ofplasticity in FEM can be found in [30].

Chapter 8

Linear InfinitesimalElasticity

In hypoelasticity, we took the Cauchy stress, σ, and developed a constitutiverelationship with the Left-Cauchy Green Strain Tensor, B. This relationshipwas quite complicated (eq. (6.9)). Recall also that these tensors were bothdefined in spatial coordinates, by definition. Then, we derived an even morecomplicated “hypoelastic” formula (eq. (7.9)) that involved the rate of defor-mation tensor, D. Lastly, a simple three-line pseudo-algorithm was given toillustrate how actual FEM (“solid” element) codes handle such “rate-form”constitutive expressions to solve, say, transient dynamic problems involvingone or more bodies. We saw that this algorithm works equally nicely forlinear infinitesimal elasticity as well.

In terms of both the non-rate form of the constitutive relationship and therate-form, linear infinitesimal elasticity simplifies things considerably. Re-call that the Second Piola Kirchhoff Stress Tensor, σ, can be used for in-finitesimal elasticity (review the example that starts on page 38). Since σ isdefined in material coordinates, we will develop a constitutive relationshipwith a strain tensor that is similarly invariant to rigid body rotation, viz., E.

Some authors instead choose to use the Cauchy stress and the Eulerian strainfor the formulation of their linear infinitesimal constitutive expression. Weare free to choose either stress-strain pair (σ ↔ e or σ ↔ E) in linear in-finitesimal elasticity. We saw previously that our “time-stepping algorithm”was independent of this choice.

In this chapter we’ll take a look only at linear infinitesimal elasticity. If

81

82 CHAPTER 8. LINEAR INFINITESIMAL ELASTICITY

a material is subjected to large strain (even if it is linear) or if a materialis nonlinear (even if subjected to small strain), then it should be treatedusing the approaches developed in the previous chapters on hyperelasticityand hypoelasticity. Just as in our chapter on hyperelasticity, the focus herewill be on developing a non-rate constitutive expression that relates stressand strain. In the elastic regime, a wide variety of materials, including mostmetals, behave linearly.

The theory of linear infinitesimal elasticity was developed independent of hy-perelastic theory, with the former being mostly developed, originally, fromintuition and experience [23]. Nevertheless, the linear theory can be derivedas well, using the same starting assumptions that we used at in the previoushyperelasticity derivation.

Recall σ = dφdE

Infinitesimal elasticity −→ E ≈ ε, σ ≈ σ

note: The linear infinitesimal stress will be denoted by σ for the remainderof this chapter. So long as ε and σ are work conjugate, we can proceed inderiving our constitutive relationship.

We know the following three expressions to be true as well:

σ = dφdε εij = 1

2( ∂ui∂xj+

∂uj∂xi

) xi ≈ Xi (still, ui = xi −Xi)

In index notation, σij = dφdεij

For linear materials, we can define φ, using a Taylor expansion, as follows:

φ = φ(εij) = φ(0)︸︷︷︸a0

+

(dφ

dεij

)0︸︷︷︸

bij

εij + 12

(d2φ

dεijdεkl

)0︸︷︷︸

Cijkl

εijεkl

note: This is a taylor series with higher terms neglected

So, φ = a0 + bijεij + 12Cijklεijεkl

σij =dφ

dεij= bij +

1

2(Cijkl + Cklij)εkl (8.1)

83

note: In eq. (8.1), ddεij

(Cmnklεmnεkl) = Cmnkl(δmiδnjεkl + εmnδikδjl)

= Cijklεkl + Cmnijεmn (or Cklijεkl)

If there is no initial stress:

bij = 0 (8.2)

Note that the order d2φdεijdεkl

or d2φdεkldεij

doesn’t matter

Cijkl = Cklij (8.3)

Eq. (8.3) → Eq. (8.2) → Eq. (8.1) −→ σij = Cijklεkl

Cijkl has 34 = 81 potential terms (9 x 9 matrix) to begin with.

Symmetry: σij = σji and εkl = εlk −→ 36 terms (6 x 6 matrix)

Symmetry in C → Cijkl = Cklij (i.e. eq. (8.3) above)−→ 21 constants (general anisotropic)

A plane of symmetry is defined as follows: for every pair of coordinate sys-tems that are mirror images of each other about the plane of symmetry,the elastic constants are the same. A single plane of symmetry reduces thenumber of independent constants to 13. Orthotropic symmetry (three or-thogonal planes of symmetry) reduces the number of independent constantsto 9. The proof can be found in [23].

Isotropic −→ only 2 independent constants

Proof:Recall φ = 1

2Cijklεijεkl (valid outside of isotropy)

note: Since φ is a function of strain only, path dependence is not considered(i.e. we are still only considering elasticity)

Also note: φ is quadratic

For isotropy: φ = φ(ε) = φ(Iε, IIε, IIIε)(recall that for hyperelastic isotropy, φ = φ(B) = φ(IB, IIB, IIIB)

)Let’s assume φ = aI2

ε + bIIε, using the following reasoning:


We know that φ is quadratic. Thus, φ doesn’t depend on IIIε, since IIIεis cubic (λ1λ2λ3). Iε is linear and so is, accordingly, squared, in order toobtain a quadratic. Our resulting I2

ε term is also multiplied by a coefficient,a. IIε = λ1λ2 + λ2λ3 + λ1λ3︸︷︷︸

already quadratic

is, similarly, multiplied by a constant, b.

Recall that Iε ∼ (tr(ε))2 and IIε is a function of both (tr(ε))2 and tr(ε2).

Also recall that tr(ε2) = εijεij , since ε is a symmetric tensor.

−→ φ = a(εii)(εjj) + b(εij)(εij)︸︷︷︸sum over i,j 1,2,3

dφdεmn

= a[εiidεjjdεmn

+εjjdεiidεmn

]+2b[dεijdεmn

εij ] = a[εii δjmδjn︸︷︷︸∗

+εjj δimδin︸︷︷︸∗

]+2b[δimδjnεmn]

* - note that “m” must equal “n” in the expression that is multiplied by “a”in order to be non-zero.

So, dφdεmn

= 2aεii + 2bεmn

2a→ “λ”b→ “µ”

Recall σij = dφdεij

:

σij = λεkkδij + 2µεij (8.4)

In eq. (8.4), λ and µ are Lame elastic constants. We can see that we havetwo independent constansts, as expected. These constants are determinedexperimentally, as will be discussed shortly.

The following alternative expression is commonly found in literature onFEM:

σ = C : ε (8.5)

We will see later what C is, as well as why we use the “:” operator.

85

σ11

σ22

σ33

σ23

σ31

σ12

=

C1111 C1122 C1133 C1123 C1131 C1112

C2211 C2222 C2233 C2223 C2231 C2212

C3311 C3322 C3333 C3323 C3331 C3312

C2311 C2322 C2333 C2323 C2331 C2312

C3111 C3122 C3133 C3123 C3131 C3112

C1211 C1222 C1233 C1223 C1231 C1212

ε11

ε22

ε33

2ε23

2ε31

2ε12

note: C is symmetric in matrix form

Also note: The shear terms are multiplied by “2” because we used symmetryof ε to reduce the 9 x 9 tensor, C, to a 6 x 6.

The matrix “Cijkl” can be written in “Voigt” notation, “CIJ” as well:

C11 C12 C13 C14 C15 C16

C21 C22 C23 C24 C25 C26

C31 C32 C33 C34 C35 C36

C41 C42 C43 C44 C45 C46

C51 C52 C53 C54 C55 C56

C61 C62 C63 C64 C65 C66

Using Voigt notation, we can construct table 8.1:

Table 8.1: Voigt Notation

I/J i/k j/l

1 1 12 2 23 3 34 2 35 3 16 1 2

Table 8.1 can be simplified in this way due to the symmetries of the Cijkltensor apparent upon inspection.

e.g. the way you read the 1st row depends on what information you need:

when I = 1, i = 1, j = 1


when J = 1, k = 1, l =1

This is easily confirmed by comparing, say, the “1-1” term in each of theabove two matrices.

Cijkl = λδijδkl + µ(δikδjl + δilδjk) (8.6)

i.e.C45 = C2331︸︷︷︸

Table 8.1

= λδ23δ31 + µ(δ23δ31 + δ21δ33) = 0

C44 = C2323 = λδ23δ23 + µ(δ22δ33 + δ23δ32) = µ

note: Eq. (8.6) is how stiffness matrices are formed in FEM

Sometimes we like to write the constitutive equation in the manner shownin eq. (8.5). We can prove that this expression is true by noting that the“:” operator is essentially two dot products (it can be easily shown that ourprevious definition of : for second order tensors still holds true as well).

σ = [λδijδkl + µ(δikδjl + δilδjk)]ei ⊗ ej ⊗ ek ⊗ el︸︷︷︸can be thought of as a 9x9 matrix

: εmnem ⊗ en

= [λδijδmn + µ(δimδjn + δinδjm)]εmnei ⊗ ej

= [λδijεmm + 2µεij ]ei ⊗ ej , which is as expected.

note: symmetry of εmn was invoked in the last step of the derivation

8.1 Elastic Modulii and Poisson Ratio

What are the Lame constants µ and λ?

First, we’ll define the Bulk Modulus “k” by considering a block orientedalong the principal strains:

Initially, V0 = l0w0h0

After deformation, w = w0[1 + λ1] ; l = l0[1 + λ2] ; h = h0[1 + λ3]

8.1. ELASTIC MODULII AND POISSON RATIO 87

V = lwh

V−V0V = ∆V

V = w0(1+λ1)l0(1+λ2)h0(1+λ3)−l0w0h0l0w0h0

= (1+λ1)(1+λ2)(1+λ3)−1

= λ1 + λ2 + λ3 +��

O(λ2) ≈ εkk

Now, from eq. (8.4), we know that σii can be expressed as follows:

σii = λδiiεkk + 2µεii

Summing both sides on i (and on k, as always) and noting that

summed︷︸︸︷σii3

= −P(pressure - e.x. hydrostatic)

−→ σii3 = (λ+

2

3µ)︸︷︷︸

k

εkk

or

P = −k∆VV , where k = “bulk modulus”

note: 1k is sometimes called the “compressibility”, since “k” relates hydro-

static pressure to volumetric strain. Really though, compressibility is deter-mined from k

G , where “G” is the “shear modulus”. kG >> 1(→∞) −→ ν →

12 , where “ν” is the Poisson Ratio. “ν” = 1

2 −→ incompressible material.

Next, we’ll define the Young’s Modulus, “E”, the Poisson Ratio, “ν”, andthe shear modulus, “G”. Any two modulii are needed to define an isotropicmaterial (E, ν or k,G, etc.)

We need ε in terms of σ ;

We’ll start with eq. (8.4), which is re-written, below:

σij = λεkkδij + 2µεij (1)

Take the trace:σ11 = λ(ε11 + ε22 + ε33) + 2µε11

σ22 = λ(ε11 + ε22 + ε33) + 2µε22

σ33 = λ(ε11 + ε22 + ε33) + 2µε33

σ11 + σ22 + σ33 = 3λε11 + 3λε22 + 3λε33 + 2µε11 + 2µε22 + 2µε33


−→ σkk = 3λεkk + 2µεkk = (3λ+ 2µ)εkk

−→ εkk = σkk3λ+2µ (2)

(2) −→ (1) and invert:

εij =1

2µ(σij −

λ

3λ+ 2µσkkδij) (8.7)

Now we’re ready to find E, ν, G :

Consider the case of uniaxial tension: σ =

σ11 0 00 0 00 0 0

ε11 = σ11

2µ (1− λ3λ+2µ) =

�2(λ+ µ)

�2µ(3λ+ 2µ)︸︷︷︸∗

σ11

* - must be 1E since those of us that have done laboratory testing of linear,

isotropic, materials, know that σ = Eε for a simple uniaxial test

E =µ(3λ+ 2µ)

λ+ µ(8.8)

Now, ε22 = 12µ( σ22︸︷︷︸

=0

− λ3λ+2µ(σ11 + σ22︸︷︷︸

=0

+ σ33︸︷︷︸=0

)) = − λ2µ(3λ+2µ)σ11,

where ε22 is the lateral strain that occurs from the uniaxial (longitudinal)stress.

By definition (i.e. as defined in undergraduate mechanics of materials),

ν = − ε22ε11

=λ

2�µ��(3λ+2µ)��σ11λ+µ

�µ��(3λ+2µ)��σ11= λ

2(λ+µ)

ν =λ

2(λ+ µ)0 < ν < 1/2 (8.9)

Lastly, consider pure shear: σ =

0 σ12 0σ12 0 00 0 0

8.2. NAVIER AND BELTRAMI-MICHELL EQUATIONS 89

σ12 = λεkkδ12 + 2µε12 = 2µε12

Since γ = 2ε12 and σ12 = Gγ (by definition),

G = µ (8.10)

It can also be easily shown that G = E2(1+ν)

8.2 Navier and Beltrami-Michell Equations

Summary of equations for infinitesimal linear isotropic elasticity (static equi-librium only):

There are 3 equilibrium equations (6 unknowns):

∂σij∂xj

+ ρbi = 0 (no sum on i) (8.11)

There are 6 constitutive equations (6 more unknowns):

σij = 2µεij + λεkkδij (8.12)

There are 6 strain-displacement equations (3 more unknowns - i.e. displace-ment in 3 directions):

εij =1

2

(∂ui∂xj

+∂uj∂xi

)(8.13)

TOTAL = 15 equations ; 15 unknowns

By this point, we have dealt with all of these equations enough to recognizethat there are indeed 15 unknowns and 15 equations. However, in order tostem any possible confusion about the physical interpretation of these equa-tions and unknowns, it will be mentioned one last time that rotational valuesof “u” as well as “moment” equilibrium of stress only appear in structuralanalysis formulas that have been simplified. In solid mechanics, concepts


like “moment equilibrium” are not considered, but would be satisfied natu-rally in any well-posed problem.

eq. (8.13) −→ eq. (8.12) −→ eq. (8.11)−→ 3 equations, 3 unknowns (3 displacements):

(λ+ µ)∂2uj∂xi∂xj

+ µ∂2ui∂xj∂xj

+ bi = 0 i = 1, 2, 3 (8.14)

The three equations of eq. (8.14) are known as the Navier Equations

Just as the Navier Equations (eq. (8.14)) depict a “displacement” rela-tionship, we can alternatively develop a “stress” relationship, known as theBeltrami-Michell Equations.

Without going through the derivation:

σij,kk +1

1 + νσkk,ij −

ν

1 + νσmm,kkδij = 0 (8.15)

The six equations of eq. (8.15) are known as the Beltrami-Michell Compat-ibility Equations. In 2D, we can derive a similar expression with the aidof the so-called Airy Stress Function. Energy methods can also be used inorder to arrive at similarly concise expressions. Graduate-level courses insolid mechanics devoted to linear infinitesimal elasticity would investigatethese kinds of concepts in detail.

8.3. FINAL WORDS 91

8.3 Final Words

This text is intended to be a broad introduction to isotropic elasticity, witha focus on providing complete derivations for strain, stress, and their con-stitutive relationships for both “finite” hyperelasticity and “infinitesimal”linear elasticity. The complications that arise due to rigid body rotationswere discussed, and the importance of “work-conjugate” tensor pairs wasemphasized. A particular “rate” approach to the handling of rigid body ro-tations was discussed, which included a particular treatment for the “frameof reference” problem with respect to our hyperelasticity model as well asfor our linear infinitesimal elasticity model.

Specific hyperelastic material models were introduced, along with a descrip-tion of specific methodologies that are currently used for fitting to experi-mental data. Lastly, the constitutive equations for linear infinitesimal elas-ticity were developed, including the derivation of Young’s Modulus and Pois-son Ratio, which are material properties that may be familiar to the reader.We concluded this chapter by considering all of the important equations insolid mechanics: the equations of equilibrium, the equations relating stressand strain, and the equations relating strain to displacement. These, orsimilar, equations, are used in any FEM code on the market. With thecompletion of this text, we now have the foundation that is necessary inorder to investigate many challenging engineering and materials problems.We also have the tools to study more specific topics in solid mechanics,such as anisotropy, viscoelasticity, and metal plasticity. Further research inthe area of solid mechanics is also needed, particularly in the treatment ofbrittle-type damage, for those so inclined.

Appendix A

Math derivations

A.1 Transpose of tensor product

We’d like to prove that AT = CT ·BT , if A is defined as A = B ·C.

In other words, we will show that (B ·C)T = CT ·BT

We can start with a particular definition of transpose, namely,

b ·AT · a = a ·A · b (A.1)

In eq. (A.1), vectors a and b are arbitrary.

Using this definition of transpose (eq. (A.1)), and then the associative lawfor vector and tensor dot products,

a · (B ·C)T · b = b · (B ·C) · a = b ·B · (C · a)

By invoking the definition of transpose, and then the commutative law ofaddition for vector dot products, and then the associative law for products,

a · (B ·C)T · b = (C · a) ·BT · b = BT · b · (C · a) = (BT · b) ·C · a

Now, we can once again invoke the definition of transpose, and then theassociative law for vector and tensor dot products:

a · (B ·C)T · b = a ·CT · (BT · b) = a · (CT ·BT ) · b

93

94 APPENDIX A. MATH DERIVATIONS

Therefore, (B ·C)T = CT ·BT

A.2. SKEW TENSOR 95

A.2 Skew tensor

In order to show that W is “skew,” we need to show that W = −WT . It issufficient to show that L− LT = −(L− LT )T .

Similar to the proof shown in Appendix A.1, we will start with arbitraryvectors a and b.

Consider that the definition of transpose (eq. (A.1)) states that:

a · (L− LT )T · b = b · (L− LT ) · a

We can expand the last term and again invoke the definition of transpose:

a · (L− LT )T · b = b · L · a− b · LT · a = a · LT · b− a · L · b

Finally, we see that the last term can be reduced:

a · (L− LT )T · b = a · (LT − L) · b = a · −(L− LT ) · b

This proves that W is indeed anti-symmetric or “skew.”

96 APPENDIX A. MATH DERIVATIONS

A.3 Orthogonal tensor

Orthogonality of R will be proven if we can show that the product of Rwith its transpose yields the identity tensor, I.

Since F = R ·U, we know that R = F ·U−1

So, RT ·R = (F ·U−1)T · (R ·U−1)

From Appendix A.1, we know that (F ·U−1)T = (U−1)T · FT

Thus,

RT ·R = (U−1)T · FT · F ·U−1 = U−1 · FT · F︸︷︷︸C

·U−1 = U−1 ·U2 ·U−1

Since U−1 ·U2 ·U−1 reduces to I, we have the desired result.

Appendix B

Stress derivations

B.1 Physical interpretation of σ

Figure B.1: Stress wedge

Consider the stress “wedge,” or “Cauchy tetrahedron,” depicted in Fig. B.1,where we would like to know the stresses acting on the plane defined by aunit vector n:n = niei

Noting the triangle AOD in Fig. B.1, with angle α1:cosα1︸︷︷︸n1

= dhdx1

We note here that cosα1 is the component of n in the “1” direction.So, dxi = dh

ni

Now, dV = ��13dS · dh = �

�13dx1dS1 = �

�13dx2dS2 = �

�13dx3dS3

97

98 APPENDIX B. STRESS DERIVATIONS

The above expression for dV may be hard to visualize, but we’re essentiallymultiplying a plane by a distance, analogous to V = Ah for a cylinder.

Solving for, say, dS1 → dS1 = dSdhdx1

... etc..

In general, dSi = dS dhdxi

We note also that this above expression can be re-written, since ni = dhdxi

:

dSi = dSni (B.1)

i.e. If n1 is small, then the surface dS1 defined by n and shown in Fig. B.1,is small (dx1 is large).

Figure B.2: Stresses

If ρbdV is the “body force” (force due to gravity, for example), and t isstress (Fig. B.2), then:∑F : tndS −

∑tidSi + ρbdV = ρdV dv

dt ,

where ρdV dvdt is essentially mass * acceleration

Take dh −→ 0 since we want the stresses at a point:

B.1. PHYSICAL INTERPRETATION OF σ 99

Divide through by dS and note that dVdS −→ 0 as dh −→ 0 −→ tn =

∑tidSidS

But, it was previously shown (eq. (B.1)) that dSidS = ni.

So, tn =∑

tini, where ti = niti, as depicted in Fig. B.2.

Figure B.3: Stress vector components

Consider, for example, t2, shown in Fig. B.3:

note: Normal and shear stress magnitudes are commonly denoted by σ.

So, t2 = [σ21, σ22, σ23]Here, the first subscript may be thought of as face “2” (Fig. B.2) and thesecond subscript can be thought of as the direction of stress on that partic-ular face.

t2 = σ2iei

In general, ti = σijej

We know that tn =∑

tini

Substituting, we get tn =∑σijejni = σijejni (index notation)

If n = niei and

stress tensor︷︸︸︷σijei ⊗ ej, then tn = σijejni is matrix-vector multiplication

so long as σ is a symmetric tensor (recall eq. (1.9) for the expression thatdefines matrix-vector multiplication, regardless of symmetry).


tn =σ · n (B.2)

Since we are really concerned with points in a body rather than volumes(recall that we took dh −→ 0 earlier), the physical meaning of eq. (B.2) isessentially as follows: if we know the normal and shear stresses at a partic-ular point in a body (with respect to a particular bases), then we can findthe stresses in any direction (or at any angle).

σ = σT (B.3)

The most common proof of eq. (B.3) involves summing moments (i.e. “con-servation of angular momentum”) and since we skipped how to do crossproducts, we’ll skip this proof (the complete proof can be found in [2]).

B.2. EQUATION OF MOTION 101

B.2 Equation of Motion

Let’s start with:

∫Sσ · ndS +

∫VρbdV =

d

dt

∫VρvdV (B.4)

Our goal is to arrive at the following result:

∫V

∂σij∂xj

dV +

∫VρbidV =

∫VρdvidtdV (B.5)

In order to accomplish this, we must make several observations. First, weneed the divergence theorem for second-order tensors.

For vectors, the divergence theorem can be written:

∫S

u · ndS =

∫Vdiv(u)dV (B.6)

In eq. (B.6), div(u) = ∂ui∂xi

For tensors, we have:

∫Sσ · ndS =

∫Vdiv(σ)dV (B.7)

In eq. (B.7), div(σ) =∂σij∂xj


We can derive eq. (B.7) as follows:Let b be an arbitrary vector:b ·∫S σ · ndS =

∫S b · σ · ndS

Using the definition of transpose (Appendix A.1) and the associative law forvectors, we get:∫S(σT · b) · ndS

From the divergence theorem for vectors (eq. (B.6)), we get:∫S(σT · b) · ndS =

∫V div(σT · b)dV

Note that: div(σT · b) = ∂∂xi

(σT · b) = ∂σki∂xi

bk + σki��∂bk∂xi

= div(σT ) · b wherethe slashed term is zero.To complete the derivation, note that σ = σT from Appendix B.1. Ad-ditionally, noting the associative law of vectors, we get the desired result,namely:

�b ·∫S σ · ndS =

∫V �b · div(σ)dV = �b ·

∫V div(σ)dV

The first term in eq. (B.7) is now understood, but to derive the last termin eq. (B.7), we need to make two additional observations.

First, we know from eq. (2.9) that:

dV = det(F)dV0 (B.8)

We can similarly express ρ in terms of ρ0 by noting from the conservationof mass, that

∫V ρdV =

∫V0ρ0dV0

Substituting eq. (B.8), we get∫V ρdV =

∫V0ρdet(F)dV0 =

∫V0ρ0dV0

Thus, we have∫V0

(ρdetF−ρ0)dV0 = 0, which is true for any arbitrary V0. So,

ρ =ρ0

detF(B.9)

Finally, substituting eq. (B.8) and eq. (B.9) into the last term in eq. (B.4),we get:

ddt

∫V ρvdV =

∫V0

ddt

ρ0��det(F)v��det(F)dV0

Since ρ0 and V0 are constant with time, we have:

B.2. EQUATION OF MOTION 103

d

dt

∫VρvdV =

∫V0

ρ0dv

dtdV0 (B.10)

The final step is to substitute ρ0 = ρdetF and dV0 = dVdet(F) into eq. (B.10).

We then arrive at the expected result:

d

dt

∫VρvdV =

∫Vρdv

dtdV (B.11)

Substituting eq. (B.7) and eq. (B.11) into eq. (B.4), we get the desiredresult (eq. (B.5)).

Appendix C

Hyperelastic derivations

C.1 Proof of σ = f(B)

Assume σ is a function of F

We know from the Chapter 5, that for a superimposed strain:F∗ = Q · Fσ∗ = Q·σ ·QT

So, Q · σ ·QT = f(Q · F)

From polar decomposition, we know F = V ·R = R ·U

We will first consider the latter form of F. Let’s take Q = RT , where Qcan be any orthogonal tensor. To see why we pick Q = RT , recall that U isdefined in “material” coordinates, and so U is, accordingly, invariant to anyrigid body rotation. σ is defined in spatial coordinates and is not invariantto rotation. So, naturally, σ = f(F) = f(R ·U) is a function of both U andR, rather than just U. We can take σ∗ = f(F∗) and set Q equal to RT , forconvenience.

RT · σ ·R = f(RT ·R ·U) −→ σ = R · f(RT ·R︸︷︷︸I

·U) ·RT

σ = R · f(U) ·RT

Recall that C = U2

σ = R · g(C) ·RT (C.1)

105

106 APPENDIX C. HYPERELASTIC DERIVATIONS

note: σ = g(C)

Perhaps eq. (C.1) seems obvious, since σ∗ = Q · σ ·QT and C∗ = C, butwe started with σ = f(F) for completeness.

Now, let’s apply Q before deformation:

Figure C.1: Rigid body rotation applied prior to deformation

We can define dX∗ as follows:

dX∗ = Q · dX

Recall that last time we superimposed a rigid body rotation on dx, whichresulted in dx∗ = Q · dx.

This time, we want dx∗ = dx

We can see from Fig. C.2, that we need F∗ to be a function of QT

We can show that this is indeed the case as follows:

dx∗ = dx = F∗·Q · dX︸︷︷︸dX∗

Since dx = F · dX, we find that F = F∗ ·Q, which yields:

C.1. PROOF OF σ = F (B) 107

Figure C.2: Rigid body rotation applied prior to deformation


F∗ = F ·QT (C.2)

Eq. (C.2) is what we wanted to find, and it should be expected. Recall thatV ·R is physically understood to be a rotation, R, followed by a deformation(axial strains and shear), V. Thus, F∗ = F ·QT is an explicit rigid bodyrotation, QT , followed by the total deformation + rotation, F. In otherwords, the rotation, QT , is applied to the initial configuration dX.

Our new definition of F∗ is difficult to physically interpret from a Lagrangianpoint-of-view, but we will use it in order to show that σ = f(V), as follows.

We know that σ∗ = f(F∗), where our new definition of “∗” requires thatσ∗ = σ and F∗ = F ·QT

So, σ = f(F ·QT )

We know that F = V ·R

Substituting −→ σ = f(V ·R ·QT )

Again, since we are simply applying our “∗” operator to both σ and f(F),we can take Q to be anything we want, as doing so is analogous to operatingon both sides of any ordinary equation. Here, we can take Q to be R.

This allows us to directly arrive at the desired result:

σ = f(V) (C.3)

The result (eq. (C.3)) is expected since it was shown in Chapter 5 that σand V are work-conjugate. Recall also that V2 = B.

C.2. DERIVATION: DIBDB , DIIBDB , DIIIBDB 109

C.2 Derivation: dIBdB , dIIB

dB , dIIIBdB

Since IB = trB = Bkk,dIBdB

=∂Bnn∂Bkl

ekel = δnkδnlekel (C.4)

dIIBdB =

d[ 12 [(trB)2−tr(B2)]]dB = 1

2 [2trBd(trB)

dB︸︷︷︸chain rule

−d(tr(B2))dB ]

= 12 [2tr(B)I− d(tr(B2))

dB︸︷︷︸see below

]

d(trB2)dB =

d(BijBji)dB =

∂(BijBji)∂Bmn

enem =∂Bij∂Bmn

Bjienem +∂Bji∂Bmn︸︷︷︸

product rule

Bijenem

= δimδjnBjienem + δjmδinBijenem = Bjiejei +Bijeiej = 2Bijeiej

So,dIIBdB

=1

2[2tr(B)I− 2Bjieiej] = IBI−B (C.5)


We know IIIB = detB ; but we need a better expression for IIIB before wederive dIIIb

dB .We know from eq. (1.15): B3 − IBB2 + IIBB− IIIBI = 0 ;tr(B3 − IBB2 + IIBB− IIIBI) = tr(0)

tr(B3)− tr(IBB2)︸︷︷︸IBtr(B2)

+ tr(IIBB)︸︷︷︸IIBtrB

− tr(IIIBI)︸︷︷︸IIIB∗3

= 0

tr(B3)− tr(B)tr(B2) + 12 [(trB)2 − tr(B2)]trB = IIIB ∗ 3

tr(B3)− tr(B)tr(B2) +1

2tr(B)(trB)2︸︷︷︸

1/2(trB)3

−12 tr(B)tr(B2) = IIIB ∗ 3

13

[tr(B3)− 3

2 tr(B)tr(B2) + 12(trB)3

]= IIIB

dIIIBdB =

d(1/3tr(B3))dB − d(1/2tr(B)tr(B2))

dB +d(1/6(trB)3)

dB

=d(1/3tr(B3))

dB − d (1/2tr(B))

dB∗ tr(B2)− d(tr(B2))

dB∗ 1

2trB︸︷︷︸

product rule

+d(1/6(trB)3)

dB︸︷︷︸16d((trB)2trB)

dB

= 13d(tr(B3))

dB − 12 tr(B

2)d(trB)

dB︸︷︷︸I

−12 tr(B)︸︷︷︸

IB

d(tr(B2))

dB︸︷︷︸2B

+

product rule︷︸︸︷1

6tr(B)

d((trB)2)

dB︸︷︷︸2tr(B)

d(trB)dB

+1

6(trB)2 ∗ d(trB)

dB︸︷︷︸I

= 13

d(tr(B3))

dB︸︷︷︸see below

−12 tr(B

2)I− 12IB(2B) + 1/3(trB)2I + 1/6(trB︸︷︷︸

IB

)2I

where d(tr(B3))dB = ∂[BklBlmBmk]

∂Bjieiej

= ∂Bkl∂Bji

BlmBmkeiej +Bkl∂Blm∂Bji

Bmkeiej +BklBlm∂Bmk∂Bji

eiej

= δkjδliBlmBmkeiej + δljδmiBklBmkeiej + δmjδkiBklBlmeiej

= BimBmjeiej +BkjBikeiej +BilBljeiej = 3B2

So,dIIIBdB = 1

3(3B2)− 12 tr(B

2)I− 12IB(2B) + 1

2I2BI

= B2 − IBB− 12(tr(B2)− I2

B)I

= B2 − IBB + IIBI , Since IIB =1

2

(I2B − tr(B2)

)(C.6)

C.2. DERIVATION: DIBDB , DIIBDB , DIIIBDB 111


C.3 Principal stretch constitutive relationship

Recall from an earlier chapter (example problem at the end of the sectionon Polar Decomposition) that we can form either the Right Stretch Tensor,U, or the Left Stretch Tensor, V, in their principal stress space by pre andpost multiplying by the orthogonal tensor, Φ, where Φ contains either theeigenvectors of the U or the eigenvectors of V, as appropriate. Also recallthat the eigenvalues of either tensor are the same, and their invariants arethe same.

[U]n = [Φ]T [U][Φ]

where [Φ] = [Φ]U =

(n1)λ1 (n1)λ2 (n1)λ3(n2)λ1 (n2)λ2 (n2)λ3(n3)λ1 (n3)λ2 (n3)λ3

[Φ]T [U][Φ] =

λ1 0 00 λ2 00 0 λ3

[Φ]T [E][Φ] = 1

2

λ21 − 1 0 00 λ2

2 − 1 00 0 λ2

3 − 1

= 12(λ2 − 1)

[Φ]T [σ][Φ] = ∂φ∂[Φ]T [E][Φ]

(where σ = ∂φ∂E = 2 ∂φ∂C = 2 ∂φ

∂U2 )

We know that the relationship between σ and σ is σ = 1IIIU

F · σ · FT

Applying the operator ΦT ·A ·Φ to all tensors in the above, we get:

ΦT · σ ·Φ = 1IIIU

ΦT ·R ·U ·Φ ∂φ∂ΦT ·E·ΦΦT ·UT ·RT ·Φ

Pre-multiply ΦT · RT · Φ and post-multiply ΦT · R · Φ to the above, andreplace ∂φ

∂ΦT ·E·Φ with 2 ∂φ∂ΦT ·U2·Φ :

ΦT ·RT · σ ·R ·Φ = 2IIIU

ΦT ·U ·Φ ∂φ∂ΦT ·U2·ΦΦT ·UT ·Φ

We now note that U is symmetric, and IIIU (which is the same as IIIV =

IIIF = III1/2B ) can be expressed in terms of the principal stretches, as

λ1λ2λ3 (recall chapter 1). Furthermore, we can note that σ∗ = RT · σ ·Rfrom Appendix C.1, for example, where we took “Q” to be equal to RT .We recall that σ∗ was indeed found to be equal to some function f(U).

ΦT ·RT · σ ·R ·Φ =2

λ1λ2λ3ΦT ·U ·Φ ∂φ

∂ΦT ·U2 ·ΦΦT ·U ·Φ (C.7)

C.3. PRINCIPAL STRETCH CONSTITUTIVE RELATIONSHIP 113

In order to arrive at our desired result, which expresses the principal valuesof the Cauchy stress, σi, as a function of the principal stretches, we’d liketo pre multiply by R and post-multiplying by RT . This would give us thedesired result on the left-hand-side.Since R ·U︸︷︷︸

F

= V ·R︸︷︷︸F

, we can see that R ·U ·RT = V

In addition, U2 is FT · F −→ R ·U2 ·RT = R ·RT ·V2 ·R︸︷︷︸FT ·F

·RT = V2

So, pre multiplying by R and post-multiplying by RT on eq. (C.7), yields:

ΦTσΦ =2

λ1λ2λ3ΦT ·V ·Φ ∂φ

∂ΦT ·V2 ·ΦΦT ·V ·Φ (C.8)

To arrive at our final result, we need to make a few more observations.Namely, we observe that all Φ that are present in eq. (C.8) and that mayhave been up to this point assumed to represent ΦU can easily be replacedby ΦV without affecting any of the algebra that we’ve already done. Φ canbe any orthogonal tensor at this point. In fact, we’ve seen eq. (C.8) before,but without the presence of Φ (i.e. Φ = I yields eq. (6.2)). By takingΦ = ΦV, we can now replace ΦT ·V ·Φ with λ and ΦT ·V2 ·Φ with λ2.

In addition, we note that ∂φ∂λ = ∂φ

∂f(λ)∂f(λ)∂λ → ∂φ

∂f(λ) = 1∂f(λ)∂λ

∂φ∂λ

Note that ∂φ∂λ is taken as the partial derivative here (as opposed to a total

derivative like dφdB), since we will see shortly that the form of our strain energy

density of interest (Ogden) is a function of λ rather than the strain invariants.Also note that because λ is diagonal, we are skipping the formal proof, here,for ∂f(λ)

∂λ = ∂λ2

∂λ = 2λ as well as the proof of ∂φ∂λ = ∂φ

∂λI = ∂φ∂λi

δijeiej, along

with the more obvious property of a diagonal tensor: λ ·λ = λ2I = λ2i δijeiej

With these substitutions, eq. (C.8) becomes

σi =2

λ1λ2λ3λ2i

(1

2λi

∂φ

∂λi

)=

1


∂φ

∂λi(C.9)


C.4 Tabulated hyperelastic model

We start with the Ogden model - viz, eq. (6.19).

Let’s define a function:

f0(λ) =

m∑s=1

µsλ∗αs (C.10)

eq. (C.10) −→ eq. (6.19) yields:

σi =1

IIIV

(f0(λi)−

1

3

3∑n=1

f0(λn)

)+K

IIIV − 1

IIIV(C.11)

Recall from the first Mooney-Rivlin example, that for an incompressible ma-

terial under uniaxial test conditions, λ∗j ≈ λ∗k ≈ λ∗−1/2i (λj ≈ λk ≈ λ

−1/2i ),

where the subscripts j and k refer to the two coordinate directions perpen-dicular to i, just as before.

The engineering stress, which would be commonly retrieved from a uniaxialtest, is the nominal stress for a hyperelastic material (recall eq. (4.5), whichstates that σ0 = IIIVF−1 · σ)

σ0 = λiλjλk

1λi

0 0

0 λ1/2i 0

0 0 λ1/2i

·σ11 0 0

0 0 00 0 0

−→ σ0

11 = λjλk︸︷︷︸no sum

σ11

eq. (C.11), for uniaxial load under the Ogden model, will be expressed asσ1 = σ11 =“σ(λi)”. So, σ(λi) is a particular value of longitudinal Cauchystress under uniaxial loading, which corresponds to a particular value of thelongitudinal stretch, λi, under the Ogden model.

We’ll introduce a new variable, ε0i, which is the engineering strain in auniaxial test - i.e. ε0i = λi−1 for a hyperelastic material if λi is some longi-tudinal stretch that occurred during the uniaxial test. Presumably, we havean experimental curve of uniaxial engineering stress, which we will from nowon call σ0, as a function of the longitudinal engineering strain

(i.e. σ0(ε0i)

).

With our new notation, we can define σ011 as follows:

C.4. TABULATED HYPERELASTIC MODEL 115

σ0(ε0i) = σ0(λi − 1) = λjλk︸︷︷︸no sum

σ(λi)

Now summing only on repeated indices unless otherwise noted, we can marchthrough the derivation of this tabulated model. First, we observe eq. (C.10),and note that:

f0(λi) =m∑s=1

µsλ∗αsi (C.12)

Additionally, we notice the term 13

∑3n=1 f0(λn) in eq. (C.11), and are thus

interested in the following calculation:

3∑n=1

f0(λn) = f0(λi) + f0(λj) + f0(λk) =

[m∑s=1

µsλ∗αsi + 2

m∑s=1

µsλ−αs

2i

](C.13)

where f0(λj) and f0(λk) were taken to be f0(λ−1/2i )

Substituting eq. (C.12) and eq. (C.13) into the original stress equation (e.x.eq. (C.11)) gives the following result:

σ0(λi − 1) = λiλk︸︷︷︸no sum

σ(λi)=

no sum︷︸︸︷λkλjIIIV

(23f0(λi)− 2

3f0(λ−1/2i ) +K IIIV−1

IIIVIIIV

)= 1

λi

(23f0(λi)− 2

3f0(λ−1/2i )− pIIIV

)λiσ0(λi − 1) + p =

2

3f0(λi)−

2

3f0(λ

−1/2i ) (C.14)

IIIV was eliminated from eq. (C.14) since we are going to limit our discus-sion to incompressible materials only.

Note that “p” is really a hydrostatic term that depends on “K,” which in ourcase is arbitrary. Simply striking the term would not stay true to the Ogdenfunction and could cause undesirable behavior. However, we can eliminatethe term through consideration of boundary conditions.


For uniaxial stress, σ(λj) = σ(λk) = σ(λ−1/2i ) = 0.

Eq. (C.11) yields:

0 =1

3f0(λ

−1/2i )− 1

3f0(λi) +K

IIIV − 1

IIIV︸︷︷︸p

(C.15)

In eq. (C.15), we find that p must equal:

p =1

3f0(λi)−

1

3f0(λ

−1/2i ) (C.16)

Eq. (C.16) −→ eq. (C.14) yields:

λiσ0(λi − 1) = f0(λi)− f0(λ−1/2i ) (C.17)

We can substitute consecutive values of the principal stretch into eq. (C.17).

i.e.

λ−1/2i σ0(λ

−1/2i − 1) = f0(λ

−1/2i )− f0(λ

1/4i )

λ1/4i σ0(λ

1/4i − 1) = f0(λ

1/4i )− f0(λ

−1/8i )

.

.etc.

In general,

λ(−1/2)x−1

i σ0

(λ

(−1/2)x−1

i − 1)

= f0

(λ

(−1/2)x−1

i

)− f0

(λ

(−1/2)x

i

)(C.18)

Since limx→∞ f0

(λ

(−1/2)x

i

)= f0(1), where f0(1) =

∑ms=1 µs,

we get:∑∞

x=1 λ(−1/2)x−1

i σ0

(λ

(−1/2)x−1

i − 1)

= f0(λi)− f0(1)

where all terms on the right hand side cancel, except for the first and last.

C.4. TABULATED HYPERELASTIC MODEL 117

So, f0(λi) = f0(1)+λiσ0(λi−1)+λ−1/2i σ0(λ

−1/2i −1)+λ

1/4i σ0(λ

1/4i −1)+ ...

Writing this as concisely as possible:

f0(λi) = f(1) +∞∑x=0

λ(−1/2)x

i σ0

(λ

(−1/2)x

i − 1)

(C.19)

We can now substitute eq. (C.19) into eq. (C.11). Since f(1) is a constant,we can see that it doesn’t affect the stress, σi, since f(1)−1/3

∑3n=1 f(1) = 0.

To complete our discussion, f0(λi) and σ0 will be written a final time, intheir final form:

f0(λi) =

∞∑x=0

λ(−1/2)x

i σ0

(λ

(−1/2)x

i − 1)

(C.20)

σi =1

IIIV

(f0(λi)−

1

3

3∑n=1

f0(λn)

)+K

IIIV − 1

IIIV(C.21)

The way that this model works is described in section 6.2. The introductionof f0(λi), which eliminates the material constants from the Ogden model(i.e. eq. (C.11)) was important, but it was the step from eq. (C.18) to eq.(C.19) that enables this “tabulated” method to work as desired. The par-ticular pattern that was recognized by the aforementioned researchers thatdeveloped this “tabulated” method [17], which is expressed in eq. (C.18),along with the observation that summing the right-hand-side of eq. (C.18)cancels most of the terms, were really the key insights to isolate the f0(λi)term.

Appendix D

Chapter 7 derivations

D.1 Jaumann rate in infinitesimal elasticity

Let’s define σ as the infinitesimal stress tensor that we want to obtain, andσ as the infinitesimal stress tensor in material coordinates (here, σ is in spa-tial coordinates, hence the use of the variable “σ” that has been previouslyreserved for the Cauchy stress). With these definitions, consider Fig. D.1.

Figure D.1: Rotating body without shear

With respect to the stress measures depicted in Fig. D.1, let’s take the timederivative:i.e. Let us start with σijeiej = σij eiej since these two measures of stress aremerely transformations of each other in linear infinitesimal elasticity, andare identical at time t=0. Now, take the time derivative of both sides of theequality.

119

120 APPENDIX D. CHAPTER 7 DERIVATIONS

σ = σijeiej + σij��eiej + σijei��ej = ˙σij eiej + σij ˙eiej + σij ei˙ej,

where the slashed terms are zero. Also note: ˙ei = Wkiek

−→ σ = ˙σij eiej+σijWkiekej+σij eiWkj ek = ˙σij eiej+σkjWikeiej+σikeiWjkej

=[

˙σij + σkjWik + σikWjk

]eiej

σijeiej =[

˙σij + σkjWik + σikWjk

]eiej (D.1)

Recall that at the beginning of the derivation, we noted that σijeiej =σij eiej. In addition, note that ˙σij eiej needs to be transformed to the spatial

bases, which can be accomplished using the transformation F · ˙σ ·FT . Thus,eq. (D.1) becomes:

σ = F · ˙σ · FT + W · σ + σ ·WT (D.2)

Recognizing F · ˙σ · FT to be the Jaumann rate, with detF taken to be ap-proximately unity for infinitesimal deformations, we arrive at the desiredresult:

σ = σ + W · σ + σ ·WT (D.3)

Eq. (D.3) is exactly the same as our previous expression for the Jaumannrate (eq. (7.13)). It is derived in a different, more general, way in AppendixD.2.

D.2. TRUESDELL AND JAUMANN RATES 121

D.2 Truesdell and Jaumann rates

From eq. (4.6), we know that:

σ =1

detFF · σ · FT (D.4)

Before we take the time derivative of eq. (D.4), note that:

d

dt(detF)−1 = − (detF)−2 ∗ ˙detF =

−(��detF)trD

(detF)�2(D.5)

Note that in eq. (D.5), we used the equality ˙detF = (detF)trD, which wasproven in the derivation of the hypoelastic constitutive relationship involv-ing the Jaumann rate (i.e. eq. (7.8)).

Now, taking the time derivative of eq. (D.4), we have:

σ = − trDdetF

F · σ · FT +1

detFF · σ · FT +

1

detFF · ˙σ · FT +

1

detFF · σ · FT

Now, we know σ in terms of σ from eq. (4.6). Substituting, we get:

σ = − trD��detF

�F · (��detF)��F−1 · σ ·��F−T ·��FT +1

��detFF · (��detF)F−1 · σ ·��F−T ·��FT

+1

detFF · ˙σ · FT +

1

��detF�F · (��detF)��F−1 · σ · F−T · FT

Thus, we arrive at the desired result:

σ = −tr(D)σ + L · σ +5σ + σ · LT

Or,

5σ = σ − L · σ − σ · LT + tr(D)σ (D.6)

Throwing all D terms out except for where it appears in the constitutiveexpression, we can arrive at the Jaumann expression (eq. (7.5), eq. (7.13),eq. (D.3)):

σ = σ + W · σ + σ ·WT (D.7)

122 APPENDIX D. CHAPTER 7 DERIVATIONS

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