Continuum Mechanics Course Notes

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MAE 5100 - Continuum Mechanics Course Notes

Prof. Brandon Runnels, University of Colorado Colorado Springs

Contents

LECTURE 1 0 Introduction 1.1 0.1 Motivation ...... 1.1 0.2 Notation ...... 1.2 0.2.1 Sets ...... 1.2 0.2.2 Proof notation ...... 1.2 1 Tensor Analysis 1.2 1.1 Index notation and the Einstein summation convention ...... 1.3 1.1.1 Vector equality ...... 1.4 1.1.2 Inner product ...... 1.4 1.1.3 Kronecker delta ...... 1.4 1.1.4 Components of a vector ...... 1.5 1.1.5 Norm of a vector ...... 1.5 1.1.6 Permutation tensor ...... 1.5 1.1.7 Cross product ...... 1.5 LECTURE 2 1.1.8 Notation ...... 2.1 1.2 Mappings and tensors ...... 2.1 1.2.1 Second order tensors ...... 2.1 1.2.2 Index notation ...... 2.2 1.2.3 Dyadic product ...... 2.2 1.2.4 Tensor components ...... 2.3 1.2.5 Higher order tensors ...... 2.3 1.2.6 Transpose ...... 2.3 1.2.7 Trace (ﬁrst invariant) ...... 2.3 1.2.8 Determinant (third invariant) ...... 2.4 LECTURE 3 1.2.9 Inverse ...... 3.1 1.2.10 The special orthogonal group ...... 3.1 1.3 Tensor calculus ...... 3.1 1.3.1 Gradient ...... 3.2 1.3.2 Divergence ...... 3.2 1.3.3 Laplacian ...... 3.2 1.3.4 Curl ...... 3.3 1.3.5 Gateaux derivatives ...... 3.3 1.3.6 Notation ...... 3.3 1.3.7 Evaluating derivatives ...... 3.4 LECTURE 4 1.4 The divergence theorem ...... 4.1 1.5 Curvilinear coordinates ...... 4.2

1.5.1 The metric tensor ...... 4.2 1.5.2 Orthonormalized basis ...... 4.3 1.5.3 Change of basis ...... 4.3 1.6 Calculus in curvilinear coordinates ...... 4.4 1.6.1 Gradient ...... 4.4 LECTURE 5 1.6.2 Divergence ...... 5.1 1.6.3 Curl ...... 5.2 1.7 Tensor transformation rules ...... 5.2 2 Kinematics of Deformation 5.3 LECTURE 6 2.1 Eulerian and Lagrangian frames ...... 6.1 2.2 Time-dependent deformation ...... 6.2 2.2.1 The material derivative ...... 6.3 2.3 Kinematics of local deformation ...... 6.4 LECTURE 7 2.4 Metric changes ...... 7.1 2.4.1 Change of length ...... 7.1 2.4.2 Change of angle ...... 7.1 2.4.3 Determinant identities ...... 7.2 2.4.4 Change of volume ...... 7.2 2.4.5 Change of area ...... 7.3 LECTURE 8 2.5 Tensor decomposition ...... 8.2 2.5.1 Eigenvalues and Eigenvectors ...... 8.2 2.5.2 Symmetric and positive deﬁnite tensors ...... 8.3 2.5.3 Spectral theorem (symmetric tensors) ...... 8.3 LECTURE 9 2.5.4 Spectral theorem (general case) ...... 9.2 2.5.5 Functions of tensors ...... 9.2 2.5.6 Polar decomposition ...... 9.3 2.6 Principal deformations ...... 9.3 LECTURE 10 2.7 Compatibility ...... 10.1 2.7.1 Continuous case ...... 10.1 2.7.2 Discontinuous case (Hadamard) ...... 10.3 2.8 Other deformation measures ...... 10.4 LECTURE 11 2.9 Linearized kinematics ...... 11.1 2.9.1 Linearized metric changes ...... 11.2 2.9.2 Small strain compatibility ...... 11.4 LECTURE 12 2.10 The spatial/Eulerian picture ...... 12.1 3 Conservation Laws 12.2 LECTURE 13 3.1 Conservation of Mass ...... 13.1 3.1.1 Control volume ...... 13.2 3.2 Conservation of linear momentum ...... 13.2 3.2.1 Forces, tractions, and stress tensors ...... 13.2 LECTURE 14 3.2.2 Balance laws ...... 14.1 3.2.3 Navier-Stokes momentum equations ...... 14.2 LECTURE 15 3.3 Conservation of angular momentum ...... 15.1 LECTURE 16 3.4 Conservation of energy ...... 16.1 3.4.1 Energetic quantities ...... 16.1

3.4.2 Balance laws ...... 16.3 LECTURE 17 3.4.3 Power-conjugate pairs ...... 17.1 3.5 Second law of thermodynamics ...... 17.2 3.5.1 Introduction to statistical thermodynamics and entropy ...... 17.2 LECTURE 18 3.5.2 Internal entropy generation ...... 18.2 3.5.3 Continuum formulation ...... 18.3 3.6 Review and summary ...... 18.4 LECTURE 19 4 Constitutive Theory 19.1 4.1 Introduction to the calculus of variations ...... 19.1 LECTURE 20 4.1.1 Stationarity condition ...... 20.1 4.2 Variational formulation of linear momentum balance ...... 20.2 LECTURE 21 4.3 Material frame indifference ...... 21.1 4.4 Elastic modulus tensor ...... 21.1 4.5 Elastic material models ...... 21.2 4.5.1 Useful identities ...... 21.2 LECTURE 22 4.5.2 Pseudo-Linear ...... 22.1 4.5.3 Compressible neo-Hookean ...... 22.1 LECTURE 23 4.6 Internal constraints ...... 23.1 4.6.1 Review of Lagrange multipliers ...... 23.1 4.6.2 Examples of internal constraints ...... 23.2 4.6.3 Lagrange multipliers in the variational formulation of balance laws . . . 23.2 4.7 Linearized constitutive theory ...... 23.3 4.7.1 Major & minor symmetry and Voigt notation ...... 23.4 4.7.2 Material symmetry ...... 23.5 LECTURE 24 4.7.3 The Cauchy-Navier equation and linear elastodynamics ...... 24.1 4.8 Thermodynamics of solids and the Coleman-Noll framework ...... 24.1 LECTURE 25 4.8.1 Other thermodynamic potentials ...... 25.1 4.8.2 Internal Energy ...... 25.3 4.8.3 Helmholtz Free Energy ...... 25.3 4.8.4 Enthalpy ...... 25.4 4.8.5 Gibbs Free Energy ...... 25.4 4.9 Inelastic constitutive modeling ...... 25.5 4.9.1 Crystal plasticity ...... 25.5 LECTURE 26 5 Computational mechanics 26.1 5.1 The ﬁnite element method ...... 26.1 5.1.1 Shape functions ...... 26.1 5.1.2 Weak formulation ...... 26.3 5.1.3 Numerical quadrature ...... 26.4 5.2 Linearized kinematics ...... 26.4 5.2.1 3D linearized elasticity ...... 26.5 5.3 Newton’s method ...... 26.6 5.4 Finite kinematics ...... 26.7 5.5 Computational ﬂuid dynamics ...... 26.7

About

These notes are for the personal use of students who are enrolled in or have taken MAE 5100 - Continuum Mechan- ics at the University of Colorado Colorado Springs, Continuum Mechanics at the Missouri University of Science and Technology, or Continuum Mechanics at Auburn University. Please do not share or redistribute these notes without permission.

Nomenclature

A Lagrangian acceleration {NI } Material principal directions a Eulerian acceleration {ni } Spatial principal directions ai Covariant basis vectors O(n) Orthogonal group 3 α The rotation vector Ω A set, typically ⊂ R , denoting a body ∀ “For all” Ω The microcanonical partition function B The left Cauchy-Green deformation tensor ∂Ω The boundary of a body B Material body force P The first Piola-Kirchhoff stress tensor D b Spatial body force P (Ω) Deformation power PE (Ω) β The displacement gradient tensor External power φ β Reciprocal temperature Deformation mapping Q C The right Cauchy-Green deformation ten- Heat sor R Lagrangian mass density n C The set of complex numbers R The set of n-dimensional vectors ⊂ “subset of” r The infinitesimal rotation tensor d The rate of strain tensor ρ Eulerian mass density δ ij The Kronecker delta S The second Piola-Kirchoff stress tensor E The Green-Lagrange strain tensor S Entropy E(Ω) Internal Energy Sn, sn Internal heat generation ei Standard Cartesian unit vectors SO(n) Special orthogonal group (rotation ten- ε The small strain tensor sors) σ ijk The Levi-Civita alternator The Cauchy stress tensor T ∃ “there exists” Temperature (Lagrangian frame) Θ ∈ “in” Temperature (Eulerian frame) U, u g Metric tensor Material and spatial intensive internal energy GL(n) The general linear group in n dimensions. u Displacement F The deformation gradient V Lagrangian velocity G Angular momentum vector v Eulerian velocity Gi Unit vectors for undeformed configuration w The spin tensor gi Unit vectors for deformed configuration ω The vorticity vector H, h Outward heat flux vectors X Material position J The Jacobian x Spatial position K Kinetic energy m n Z The set of integers L(R , R ) The set of m × n tensors ` The spatial velocity gradient tensor λ Stretch ratio {λi } Principal stretches M Moment

Examples

Example 1.1: Gateaux derivative ...... 3.3 Example 1.2: Gradient of norm squared ...... 3.4 Example 1.3: Relationship of Gateaux derivative to gradient ...... 4.1 Example 1.4: Derivative of function with respect to tensor ...... 4.1 Example 1.5: Cylindrical polar coordinates ...... 4.3 Example 1.6: Gradient in cylindrical polar coordinates ...... 4.4 Example 1.7: Divergence in cylindrical polar coordinates ...... 5.1 Example 1.8: Curl in cylindrical polar coordinates ...... 5.2 Example 2.1: Time-dependent deformation of unit cube ...... 6.2 Example 2.2: Material derivative ...... 6.3 Example 2.3: Relationship of stretch, angle to components of C ...... 7.1 Example 2.4: Metric changes in a planar deformation ...... 7.4 Example 2.5: Shear twinning ...... 10.3 Example 2.6: Pure shear metric changes ...... 11.3 Example 4.1: Minimum distance ...... 19.2 Example 4.2: Brachistochrone problem ...... 19.3 Example 4.3: Thermodynamic potentials mnemonic ...... 25.5

0 Introduction

Welcome to continuum mechanics. In this course we will develop the mathematical framework for describing precisely the deformation of solids, ﬂuids, and gasses, and describing the physical laws that govern their motion. We will develop the mathematical formulation of the equations of motion, elasticity, viscoelasticity, plasticity, etc. The course will be organized in the following way:

(1) Tensor analysis: index notation, tensor algebra and calculus, curvilinear coordinates and transformation rules. (2) Kinematics of deformation: deformation mappings, local deformation, metric changes, decompositions, compatibility, linearized kinematics.

(3) Balance laws: conservation of mass, linear momentum, angular momentum, energy. (4) Constitutive modeling and the thermodynamics of solids: constitutive models, second law of thermodynamics and dissipative systems. (5) Computational mechanics: ﬁnite elements, Galerkin method, Ritz method.

0.1 Motivation

Consider a bar subjected to a tensile load as shown in the following ﬁgure. How do we describe this process?

`0 Governing equations Undeformed = `−`0 Kinematics: `0 A σ = f f Balance law: A Deformed Constitutive model: σ = E ε `

The above equations are fairly straightforward for this simple system, and we are familiar with them from statics and mechanics of materials. But we are in the business of mechanics of bodies with arbitrary shape, loading, constraints, etc:

Undeformed Deformed

What is ε, σ, E for this complex case? How do we formulate our equations of kinematics, balance laws, and consi- tituve models here? Can we use what we know about the system to determine what the deformed conﬁguration is under applied loads and displacements?

0.2 Notation

Continuum mechanics is built on the mathematical framework of differential geometry. As a result, the convention is to use standard math notation when formulating continuum mechanics. Additionally, as we will see, it is generally necessary to maintain a level of mathematical precision beyond that typically found in engineering disciplines.

0.2.1 Sets

A set is a collection of objects. Examples: • The integers Z = {... , −1, 0, 1, 2, 3, ...} • The real numbers R • The complex numbers C n • n-dimensional vectors R To indicate that an item is in a set, we use the ∈ symbol. For instance, 3 x ∈ R (0.1) indicates that x is a 3D vector. To indicate that a set is a subset, we use the ⊂ symbol. For instance

Z ⊂ R (0.2) indicates that the integers are a subset of the real numbers. Another common use is to denote a 3D body: 3 Ω ⊂ R (0.3) is an arbitrary region in 3D space.

0.2.2 Proof notation

We will not be doing any serious proofs in this course, but we frequently use some of the proof notation to simplify deﬁnitions and theorems. 3 3 3 • ∃ is read “there exists.” For instance, ∃x ∈ R states that there is at least one item in the R set, or that the R set is not empty. • ∀ is read “for all.” For instance, n x − x = 0 ∀x ∈ R (0.4) tells us that the statement x − x = 0 is true for every possible vector. Example: 3 3 ∀x ∈ R ∃a ∈ R s.t. x − a = 0 (0.5) can be read “for all 3D vectors there exists another 3D vector such that their difference is equal to zero.”

1 Tensor Analysis

3 3 Let us consider the space of three dimensional vectors, R . A vector r ∈ R can be represented in two different ways: in terms of its components, or in terms of basis vectors {g1, g2, g3}

z gˆ3 c "a# r = b c r = r1gˆ1 + r2gˆ2 + r3gˆ3

a b y gˆ2 gˆ x 1

For example, we may have "1# "0# "0# g1 = 0 g2 = 1 g3 = 0 (1.1) 0 0 1

ˆ ˆ ˆ where we see that the basis vectors correspond to the familiar i, j, k notation. For maximum generality, however, we do not deﬁne the basis vectors explicitly. In subsequent sections will talk about changes of basis. Note also that we have dropped the familiar x, y, z notation in favor of 1, 2, 3. In general, we will stick with this convention exclusively; the reason for this will become apparent in the next section.

1.1 Index notation and the Einstein summation convention

Let us consider r deﬁned in the previous section as

r = r1 g1 + r2 g2 + r3 g3 (1.2)

We can write this more simply using summation notation:

3 X r = ri gi (1.3) i=1

It turns out that we write sums like this a lot, and it becomes cumbersome to write the summation symbol every time. Thus, we introduce the Einstein summation convention, and we drop the explicit sum. This allows us to simply write

r = ri gi (1.4)

n This leads us to deﬁne the rules of the summation convention. For a vector equation in R , expressed using index notation: Rule 1: An index appearing once in a term must appear in every term in the equation, and is not summed. It is referred to as a free index. Rule 2: An index appearing twice must be summed from 1 to n. It is referred to as a dummy index. (Dummy indices can be changed arbitrarily, that is, e.g. ri gi = rj gj ) Rule 3: No index may appear more than twice in any term. (If an index does appear more than twice, we go back to using a summation symbol. Alternatively, if an index appears twice but is not a dummy index, we use parentheses to denote this, e.g. ui = λ(i) v(i). Usually, when a rule gets broken, it means that some algebra got messed up.) These rules may seem a bit strange, and they usually take a little bit of time to get used to. To help solidify them, let us look at a couple of algebraic examples. (Fun fact: the Einstein summation convention was introduced by Albert Einstein to simplify the equations of general relativity. In fact, the formulation of continuum mechanics has a number of similarities to general relativity.)

1.1.1 Vector equality

n Consider two vectors u, v ∈ R with components u1, u2, ..., v1, v2, .... In invariant/symbolic notation, we say the two vectors are equal if

u = v or ui gi = vi gi (1.5)

This tells us that each component of the vector is equal; in other words,

ui = vi (1.6)

Does this obey the summation convention? Yes it does: i is a free index that appears exactly once in every term of the equation.

1.1.2 Inner product

n Let us again consider u, v ∈ R . The inner product (or “dot product”) is deﬁned as T u · v = u v = |u| |v| cos θ (1.7) where | · | is the magnitude of · and θ is the angle between the two vectors. In matrix notation, we evaluate this as

v1 n T v2 X u v = [u1 u2 ... un] ... = u1v1 + u2v2 + ... + unvn = ui vi = ui vi (1.8) vn i=1

Note that there are no free indices, only dummy indices.

1.1.3 Kronecker delta

The Kronecker Delta is deﬁned in the following way: ( 1 i = j δij = 0 i 6= j (1.9)

Consider the basis vectors that we described above. We know that they are orthonormal, so gi · gj is 1 if i = j and 0 otherwise; that is,

gi · gj = δij (1.10)

Let us use this technology in the context of the dot product. Let u = ui gi , v = vi gi . Then we might write the dot product as

u · v = (ui gi ) · (vi gi ) (1.11)

But wait: this breaks one of our rules, that an index cannot repeat more than twice. To ﬁx this, we will replace the is in the second term with js:

u · v = (ui gi ) · (vj gj ) (1.12)

Now, let us distribute these terms:

u · v = ui vj (gi · gj ) = ui vj δij (1.13) 2 This term has two summed indices, so if we expand it out, we would have n terms. However, we know that only the terms where i = j survive. Thus, the effect of the Kronecker Delta is to turn one of the dummy indices into the other: in this case, if we “sum over j”

ui vj δij = ui vi (1.14)

1.1.4 Components of a vector

To extract a speciﬁc component of a vector, we can dot with the corresponding unit vector. That is: u · gi = (uj gj ) · gi = uj (gi · gj ) = uj δij = ui (1.15)

1.1.5 Norm of a vector

As you may recall, the norm of a vector is given by √ |u| = u · u (1.16) In index notation, this becomes √ |u| = ui ui (1.17)

1.1.6 Permutation tensor

The next order of business is to introduce the cross product in tensor notation. To work with cross products, we need to introduce a new bit of machinery, called the permutation tensor, also referred to as the Levi-Civita tensor. (Note: it’s not actually a tensor. However, it is frequently referred to as one, so we will stick with convention here.) Here it is:  1 ijk = 123, 231, 312 =  "even permutation" ijk = −1 ijk = 321, 132, 213 = "odd permutation" (1.18)  0 otherwise Let’s make a couple of notes here: • ijk is zero if any of the two indices take the same value. • Flipping two indices changes the sign of ε; that is, e.g. ijk = −jik = −ikj = −kji (1.19)

These identities will come in handy in the future.

1.1.7 Cross product

Let us consider the cross product of unit vectors. We know that g1 × g2 = g3 g2 × g1 = −g3 (1.20) g2 × g3 = g1 g3 × g2 = −g1 (1.21) g3 × g1 = g2 g1 × g3 = −g2 (1.22) Let us attempt to express this using the permutation tensor. Try: gi × gj = ijk gk (1.23) Does this work? Let’s plug in i = 1, j = 2. Then we have : 0 : 0 g1 × g2 = 12k gk =121g1 +122g2 + 123g3 = g3 (1.24) as expected. Plugging in other values for i, j shows that we can recover all of the identities expressed above. Now, let us see what happens when we take the cross product between u, v: u × v = (ui gi ) × (vj gj ) = ui vj (gi × gj ) = ui vi ijk gk (1.25) Alternatively, (u × v)k = ijk ui vj (1.26)

1.1.8 Notation

Let us clarify some of the notation that we have been using: Invariant/Symbolic Notation Full Component Notation Termwise Index Notation Equality u = v ui gi = vi gi ui = vi Dot product u · v ui vi ui vi Cross Product u × v ijk ui vj gk ijk uj vk Example (u · v) w uk vk wi gi uk vk wi In general:

• Invariant/symbolic notation is independent of coordinate system, which means that invariant expressions are more general. However, there are some operations that are too complex to be represented in invariant notation, and it can more easily get confusing. • Full component notation is slightly less general than invariant notation, but is the best for working in almost any coordinate system, especially ones with non-constant unit vectors. • Termwise index notation is very nimble and convenient when working in a constant, orthonormal coordinate system. This is frequently what we use, so we will use it a lot. However, it is dangerous to use when unit vectors are non-constant.

1.2 Mappings and tensors

A mapping is a machine that takes a thing of one type and turns it into a thing of another type. For instance, 2 f (x) = x takes a real number and turns it into a positive real number. We use the notation

f : U → V (1.27) to denote a mapping; in this case, if x ∈ U, then f (x) ∈ V. n m A linear mapping f : R → R is a mapping that satisﬁes n • f (α x) = α f (x) ∀ x ∈ R , α ∈ R n • f (x + y) = f (x) + f (y) ∀ x, y ∈ R

1.2.1 Second order tensors

A second order tensor (S.O.T.) is a linear mapping from vector spaces to vector spaces. The set of second order n m tensors mapping n-dimensional vectors to m-dimensional vectors is referred to as L(R , R ). We are familiar with n m n m thinking of them as n × m matrices: For example, if A ∈ L(R , R ) and u ∈ R , v ∈ R then we could write   v1  A11 A12 ... A1n u1 v2 A21 A22 ... A2n u2 v = A u   =      .   . . . .   .  (1.28) .  . . .. .  . vm Am1 Am2 ... Amn un

n m We can write all possible linear mappings from R to R in matrix form. Therefore, in general, we can think of second order tensors as being similar to matrices. Let us make a few notes: • If m = n the matrix is said to be square. For the most part, we will work with square matrices.

• If u = A u then A = I is said to be the identity mapping. • The difference between tensors and matrices is subtle. A matrix is just a collection of numbers, but a tensor is something that must be transformed with a change of coordinates. Thus we say that tensors have transformation properties but matrices do not.

1.2.2 Index notation

Index notation makes it very convenient to write second order tensors and tensor-vector multiplication. In the above example, we have   v1  A11u1 + A12u2 + ... + A1nun v2 A21u1 + A22u2 + ... + A2nun      .  =  .  (1.29) .  .  vm Am1u1 + Am2u2 + ... + Amnun and so we can write

vi = Aij uj (1.30)

1.2.3 Dyadic product

We have expressed tensor-vector multiplication using invariant notation and termwise index notation. How can we express a tensor using full component notation? To do this, we introduct the dyadic product:

u1 u1v1 u2v1 ... unv1 u2 u1v2 u2v2 ... unv2 T   v v ... v   u ⊗ v = u v =  .  [ 1 2 n] =  . . . .  (1.31) ...... un u1vn u2vn ... unvn

In tensor notation, we simply write

(u ⊗ v)ij = ui vj (1.32)

We can use the dyadic product with unit vectors to extract a speciﬁc component of a tensor. That is,

gi ⊗ gj (1.33) is the zero matrix except for a 1 in the ij column. For example: "0# "0 0 0# g2 ⊗ g3 = 1 [0 0 1] = 0 0 1 (1.34) 0 0 0 0

Therefore, we can express a tensor A as:

A = Aij gi ⊗ gj (1.35)

How do we write a tensor operating on a vector? Suppose A acts on v:

Av = (Aij gi ⊗ gj )(vk gk ) = Aij vk (gi ⊗ gj )gk (1.36) T What do we do with this? Recall that we can write u ⊗ v as uv . Then we have T (gi ⊗ gj )gk = gi gj gk = gi (gj · gk ) = gi δjk (1.37)

Substituting, we get

A v = Aij vk gi δjk = Aij vj gi (1.38)

All content © 2016-2018, Brandon Runnels 2.2 MAE 5100 - Continuum Mechanics Course Notes - Lecture 2 University of Colorado Colorado Springs https://canvas.uccs.edu/courses/22031 as expected. n We will take this opportunity to reiterate the identity we described earlier: namely, for u, v, w ∈ R ,

(u ⊗ v) w = u (v · w) (1.39)

This is easily seen using index notation:

(ui vj )wj = ui (vj wj ) (1.40)

1.2.4 Tensor components

We can extract components of a tensor A in the following way:

gi · A gj = gi · (Apq gp ⊗ gq) gj = Apq gi · (gp ⊗ gq) gj = Apq gi · gp(gq · gj ) = Apq δip δjq = Aij (1.41)

1.2.5 Higher order tensors

We can express higher order tensors in the following way:

A = Aij...k gi ⊗ gj ⊗ ... ⊗ gk (1.42)

When we start working with contitutive theory, we will frequently see fourth-order tensors (the elasticity tensor). I don’t know of any cases where we work with anything higher than fourth order.

1.2.6 Transpose

The transpose of a tensor is identitcal to the transpose of a matrix: the ij term is swapped with the ji term. How can we express this in index notation? T (A )ij = Aji (1.43)

A tensor A is called symmetric iff

Aij = Aji (1.44)

A tensor is called antisymmetric iff

Aij = −Aji (1.45)

(Note that the diagonal terms in an antisymmetric tensor must be zero.) Any tensor can be decomposed into its symmetric and antisymmetric parts in the following way: 1 1 1 1 A = (A + A) + (AT − AT ) = (A + AT ) + (A − AT ) 2 2 2 2 (1.46) | {z } | {z } symmetric antisymmetric

1.2.7 Trace (ﬁrst invariant)

n The trace of a tensor is deﬁned in the folowing way. For u, v ∈ R ,

tr(u ⊗ v) = u · v (1.47) n n Let A ∈ L(R , R ). Then the trace is given by

tr(A) = tr(Aij gi ⊗ gj ) = Aij tr(gi ⊗ gj ) = Aij (gi · gj ) = Aij δij = Aii (1.48)

One can think of this as the sum of the diagonal terms in the tensor. Note: the trace of a tensor is called the ﬁrst invariant of the tensor. This means that the trace of the tensor does not change under rotation. The signiﬁcance of this will become apparent later on.

1.2.8 Determinant (third invariant)

3 3 A quantity that we use frequently is the determinant of a tensor. For A ∈ L(R , R ),

A11 A12 A13 A11A22A33 + A12A23A31 + A13A21A32− det(A) = A21 A22 A23 = A11A23A32 − A12A21A33 − A13A22A31 (1.49) A31 A32 A33

How can we represent this using index notation? In 3D, we can write it as

det(A) = ijk A1i A2j A3k (1.50)

Alternatively, we can write it in a slightly more satisfying way as 1 det(A) = A A A 6 ijk pqr ip jq kr (1.51)

n n Some things to note: for A, B ∈ L(R , R ) T det(A) = det(A ) (1.52) det(AB) = det(A) det(B) (1.53)

In three dimensions, the determinant is the third invariant of the tensor, which means it is the third of three important quantities that do not change under rotation.

1.2.9 Inverse

n n −1 n Let A ∈ L(R , R ). The inverse of A, A satisﬁes, ∀u ∈ R , −1 (A )A u = I u = u (1.54) If the inverse of a tensor exists it is said to be invertible. One can prove that a tensor A is invertible iff det(A) 6= 0. In index notation, −1 Aij Ajk = Iik = δik (1.55) For composite mappings, −1 −1 −1 (AB) = B A (1.56) −1 What is the determinant of the inverse of a tensor? For A deﬁned above, we know that AA = I. So we can say that −1 −1 det(I) = 1 = det(AA ) = det(A) det(A ) (1.57) so 1 det(A−1) = det(A) (1.58)

−1 What if det(A) = 0? Then A cannot exist, so its determinant is naturally ill-deﬁned.

1.2.10 The special orthogonal group

3 3 T Consider the set of tensors A ∈ L(R , R ) such that A A = I. We call this group of tensors the orthogonal group, and we denote it as S(n) where n is the dimensions. So for A ∈ S(n) we see that −1 T • A = A −1 T • 1/ det(A) = det(A ) = det(A ) = det(A). This implies that det(A) = ±1 Now, let us consider only those tensors in the orthogonal group that satisfy det(A) = 1. We will call this the special orthogonal group, and denote it by SO(d). It turns out that SO(d) is exactly the same as the group of all rotation matrices. In 3D, we will refer to tensors that are in SO(3) very frequently.

1.3 Tensor calculus

Tensor calculus is the language of continuum mechanics. So far we have talked a lot about vectors and tensors. Now, we are going to talk about vector and tensor fields. There are three kinds of fields that we will use a lot: n • Scalar fields f : R → R; e.g. temperature, pressure n n • Vector fields v : R → R ; e.g. displacement, velocity n n n • Tensor fields T : R → L(R , R ); e.g. stress, strain We will be looking at a wide variety of differentiation operations on these types of fields. Note: for this section, we are assuming that we are working in a constant Cartesian basis. That is, we assume that the basis vectors gi are constants. This is not always true! When we work with curvilinear coordinates, we will have to be very careful about taking derivatives of basis vectors.

1.3.1 Gradient

n Suppose we have a scalar ﬁeld φ : R → R, φ(x). We deﬁne the gradient of φ to be ∂φ ∂φ ∂φ ∂φ grad(φ) = g1 + g2 + ... + gn = gi (1.59) ∂x1 ∂x2 ∂xn ∂xi

Note that gradient operator turns a scalar field into a vector field. n n Now consider a vector field u : R → R . The gradient of u is defined to be

∂u ∂ui grad(u) = ⊗ gi = gi ⊗ gj (1.60) ∂xi ∂xj

Note that the gradient operator here turns a vector ﬁeld into a tensor ﬁeld. We can generalize this in the following way:

∂(·) grad(·) = ⊗ gi (1.61) ∂xi where we drop the dyadic product if · is a scalar field. In general, we only generally care about gradients on scalar and vector fields. However, there are some models that depend on tensor field gradients, such as strain gradient plasticity and ductile fracture.

1.3.2 Divergence

n n For a vector ﬁeld u : R → R , the divergence is deﬁned to be

∂u ∂uj ∂uj ∂ui div(u) = · gi = gj · gi = δij = (1.62) ∂xi ∂xi ∂xi ∂xi

n n n Note that the divergence turns a vector field into a scalar field. Now, consider a tensor field T : R → L(R , R ). The divergence of the tensor field is

dT dTjk dTjk dTjk dTji div(T ) = gi = (gj ⊗ gk )gi = gj (gk · gi ) = gj δik = gj (1.63) dxi dxi dxi dxi dxi

Note that the divergence turns a tensor ﬁeld into a vector ﬁeld. We can generalize this in the same way we generalized the gradient:

∂(·) div(·) = · gi (1.64) ∂xi where we drop the dot for tensor ﬁelds. Note also that we cannot take the divergence of a scalar ﬁeld.

1.3.3 Laplacian

The Laplacian is the composition of the gradient and the divergence operators on a scalar ﬁeld: for φ : R → R:

∆φ = div(grad(φ)) (1.65)

In Cartesian coordinates, this comes out to be ∂2φ ∆φ = (1.66) ∂xi ∂xi

2 Note that we do not write ∂xi in order to be in keeping with the summation convention.

1.3.4 Curl

3 3 The curl operator is deﬁned on a vector ﬁeld u : R → R in the following way:

∂u ∂uj ∂uj ∂uj curl(u) = − × gi = − gj × gi = − jik gk = ijk gk (1.67) ∂xi ∂xi ∂xi ∂xi

Note that the curl acts on a vector field and produces a vector field, so we cannot take the curl of a scalar field. In generalized form we have ∂(·) curl(·) = − × gi (1.68) ∂xi

1.3.5 Gateaux derivatives

A more general type of derivative is the “Gateaux derivative,” which will prove very useful later on. Consider some n ﬁeld (scalar, vector, or tensor, etc.) φ : R → V (where V is the set of scalars, vectors, tensors, etc.) and a vector n v ∈ R . The Gateaux derivative is deﬁned as

d Dφ(x)v = φ(x + εv) (1.69) dε ε→0 that is, the derivative is taken with respect to ε which is then set to 0. Example 1.1: Gateaux derivative

2 Let φ(x) = xi xi = |x| , and compute Dφ(x)v. We evaluate this simply by substituting φ into (1.69):

d d 2 Dφ(x)v = lim ((xi + εvi )(xi + εvi )) = lim (xi xi + 2εxi vi + ε vi vi ) = lim (2xi vi + 2εvi vi ) = 2xi vi ε→0 dε ε→0 dε ε→0 (1.70)

1.3.6 Notation

Here it is important to make a couple of remarks about notation. (1) We avoid the use of the ∇ operator (e.g. ∇· for divergence, ∇× for curl) because it is difﬁcult to impossible to express certain vector operations. (2) We work with a lot of derivatives in continuum mechanics and it frequently becomes cumbersome to write ∂ comma notation ∂xi . Therefore, we adopt : ∂ (·) = (·),i (1.71) ∂xi For example, we can write grad/div/curl compactly for scalar/vector/tensor ﬁelds φ, v, T

grad(φ)i = φ,i div(v) = vi,i curl(v)i = ijk vk,j ∆φ = φ,ii (1.72) grad(v)ij = vi,j div(T)i = Tij,j (1.73)

(3) We will occasionally use the ∂ symbol to denote differentiation. Examples of usage include: ∂ ∂ ∂ ∂x ≡ ∂θ ≡ ∂i ≡ (1.74) ∂x ∂θ ∂xi

1.3.7 Evaluating derivatives

How do we evaluate a derivative with respect to x in terms of x? For instance, how would we compute the gradient of φ(x) = x · x? In index notation, we have

∂xi ∂x = δij or, in symbolic notation, = I (1.75) ∂xj ∂x

Let us look at a couple of examples: Example 1.2: Gradient of norm squared

Compute the gradient of φ(x) = x · x. The ﬁrst step is to write φ in index notation: φ(x) = xk xk . Then, we use the formula: ∂ grad(φ) = (xk xk ) gi (1.76) ∂xi We can use the product rule exactly like we would normally:

= (δik xk + xk δik ) gi = 2δik xk gi (1.77)

Summing over i we get = 2xi gi = 2x (1.78)

Example 1.3: Relationship of Gateaux derivative to gradient

3 Let φ : R → R. Show that Dφ(x)v = grad(φ) · v. To do this we again evaluate the Gateaux derivative: d Dφ(x)v = lim φ(x + ε v) ε→0 dε (1.79)

We use the chain rule exactly as we would normally:

∂φ(x + ε v) d(xi + ε vi ) ∂φ(x + ε v) ∂φ = lim = lim vi = vi = grad(φ) · v (1.80) ε→0 ∂(xi + ε xi ) dε ε→0 ∂(xi + ε xi ) ∂xi

In addition to taking derivatives with respect to vectors (such as x) we may take derivatives with respect to tensors. n n For example, given φ : L(R , R ) → R, we may wish to calculate dφ(T ) (1.81) dTij

Similarly to the case of vectors, we have

dTij dT = δipδjq or, in symbolic notation, = I ⊗ I (1.82) dTpq dT

Example 1.4: Derivative of function with respect to tensor

n n n n T Let A : L(R , R ) → L(R , R ) with A(T) = T T. Find the derivative of A with respect to T:

∂Aij ∂ ∂Tki ∂Tkj = (Tki Tkj ) = Tkj + Tki = δkpδiqTkj + Tki δkpδjq = Tpj δiq + Tpi δjq (1.83) ∂Tpq ∂Tpq ∂Tpq ∂Tpq

How do we represent this in symbolic notation? It’s actually rather tricky, and is much easier to leave things in index notation.

Note: when taking vector or tensor derivatives, it is always important to use a fresh new free index. Do not reuse existing ones.

1.4 The divergence theorem

We have developed enough machinery to introduce the single most important theorem in all of continuum mechanics: the divergence theorem.

V(x)

Ω

n 3 3 Theorem 1.1 (The divergence theorem). Let Ω ⊂ R , and V : R → R be a differentiable vector field: Then: Z Z Z Z div(V) dv = V · n da Vi,i dv = Vi ni da (1.84) Ω ∂Ω Ω ∂Ω where ∂Ω is the boundary of the body. 3 3 3 We have a similar theorem for tensor fields: Let T : R → L(R , R ) be a tensor field. Then Z Z div(T) dv = Tn da (1.85) Ω ∂Ω Using index notation in an Cartesian coordinate system, we can write Z Z Z Z Vi,i dv = Vi ni da Tij,j dv = Tij nj da (1.86) Ω ∂Ω Ω ∂Ω What does this mean? We are relating a volume integral of the divergence to a surface integral – or, in this case, a flux integral. For the case of a vector field, the integral of the divergence over the body can be intuitively thought of as the total amount of compression/expansion in the vector field. The flux integral can be thought of as the total amount of vector field entering or leaving the body. Thus, you can think of the divergence theorem as the mathematical formulation of the statement “the total compression of the vector field is equal to the rate of flux through the boundary.” We will make extensive use of the divergence theorem in this course.

1.5 Curvilinear coordinates

Up until now we have worked within a simple Cartesian basis, let us call it ei . However, it is frequently convenient to switch to a more natural coordinate system: gˆθ gˆ3

gˆr

r = r1gˆ1 + r2gˆ2 + r3gˆ3

gˆ2

gˆ1

Consider a new set of coordinates {θ1, θ2, θ3, ... , θn}. Position as a function of these coordinates is expressed as x(θ1, θ2, ... , θn) (1.87) Let us deﬁne a new basis: {a1, a2, ... , a3} deﬁned as

∂x ∂x ∂x ∂xj a1 = a2 = ... ai = = ej (1.88) ∂θ1 ∂θ1 ∂θi ∂θi We will refer to {ai } as the covariant basis vectors.

1.5.1 The metric tensor

The metric tensor G is deﬁned as Gij = ai · aj (1.89) Notes: • The metric tensor is symmetric • If the metric tensor is diagonal then the new coordinate system {ai } is said to be orthogonal • If G = I then the coordinate system is said to be orthonormal

1.5.2 Orthonormalized basis

There is no guarantee that our new basis {ai } will be normalized, that is, we don’t know that |ai | = 1. But we can make sure that they are by deﬁning scale factors hi = |ai |. Then we deﬁne a new basis a(i) gi = no summation over i (1.90) h(i)

Example 1.5: Cylindrical polar coordinates

Cylindrical Polar Coordinates: we can specify any point using x1, x2, x3, but we can also specify it using the coordinates r, θ, z.

z r θ

Let us compute x(r, θ, z):

x1 = r cos θ x2 = r sin θ x3 = z (1.91)

Now we can compute our basis vectors: "cos θ# "−r sin θ# "0# ∂xi ∂xi ∂xi ar = ei = sin θ aθ = ei = r cos θ az = ei = 0 (1.92) ∂r 0 ∂θ 0 ∂z 1

Our scale factors are

h1 = |ar | = 1 h2 = |aθ| = r h3 = |az | = 1 (1.93)

so we have "cos θ# "− sin θ# "0# gr = sin θ gθ = cos θ gz = 0 (1.94) 0 0 1

Important note: while {e1, e2, e3} are independent of x1, x2, x3, {gi } is not necessarily independent of {θi }. In our above example, we see that " # " # ∂ − sin θ ∂ − cos θ gr = cos θ = gθ gθ = − sin θ = −gr (1.95) ∂θ 0 ∂θ 0

1.5.3 Change of basis

n n Let {gi } be an orthonormal basis for R , and let v ∈ R . Suppose we wish to ﬁnd {vi } such that v = vi gi . To do this, we use the orthogonality property of the basis:

v · gj = vi gi · gj = vi δij = vj =⇒ v = (v · gj ) gj (1.96)

Suppose we have another basis {ei }. Then we can relate the two bases by writing ei = (ei · gj ) gj gi = (gi · ej ) ej (1.97) These relationships will be useful as we start discussing curvilinear coordinates.

1.6 Calculus in curvilinear coordinates

Now that we have deﬁned a framework for working in other coordinate systems, we need to know what our calculus operations look like in those systems. Before we do that, however, we need to introduce a couple of important identities.

(1) Earlier we learned that 1 1 ∂xj gi = a(i) = ej no sum on i (1.98) h(i) h(i) ∂θ(i)

How can we express ei in terms of {gj }? To do this, we’ll pull a trick. Remember that {gi } forms an orthonormal n basis for R : that means that we can express any vector in terms of that basis. For instance, a vector v can be written as v = vi gi . How do we ﬁnd vi ? It’s nothing other than v · gi . So we can write

v = (v · gi ) gi (1.99)

We can do exactly the same thing for our original basis vectors

X 1 X 1 ∂xk X 1 ∂xk X 1 ∂xi ei = (ei · gj ) gj = (ei · aj ) gj = (ei · ek ) gj = (ei · ek ) gj = gj (1.100) hj hj ∂θj hj ∂θj hj ∂θj j j j | {z } j δik

(Note that we broke one of our summation convention rules. To compensate for this we drop the summation notation and use an explicit sum.)

(2) There is an important theorem called the inverse function theorem that states: −1 h∂θ i h ∂x i ∂θ ∂x ∂θi ∂xk = =⇒ = I or, in index notation, = δij (1.101) ∂x ∂θ ∂x ∂θ ∂xk ∂θj

We will use both of these rules to derive expressions for the familiar divergence, gradient, and curl in curvilinear coordinates.

1.6.1 Gradient

We want to express the gradient as computed in the above

∂ ∂f ∂θj ∂f ∂θj X 1 ∂xi X 1 ∂f ∂θj ∂xi X 1 ∂f grad(f (θ)) = (f (θ)) ei = ei = gk = gk = gk (1.102) ∂xi ∂θj ∂xi ∂θj ∂xi hk ∂θk hk ∂θj ∂xi ∂θk hk ∂θk k k | {z } k δjk

Notice how this is almost identical to our original expression for the gradient, except that {xi }, {ei } have been replaced with {θi }, {gi }. The only difference is the presence of the scale factors. Let’s solidify this with an example: Example 1.6: Gradient in cylindrical polar coordinates

Let us continue with our example of cylindrical polar coordinates. Let f = f (r, θ, z). Then we have: 1 ∂f 1 ∂f 1 ∂f ∂f 1 ∂f ∂f grad(f ) = gr + gθ + gz = gr + gθ + gz (1.103) hr ∂r hθ ∂θ hz ∂z ∂r r ∂θ ∂z

1.6.2 Divergence

Let v = vi (θ) gi be a vector field that is defined exclusively using the {θi }, {gi } coordinate system. What is the divergence of this vector field? We can follow the exact same procedure as when computing the gradient:

∂ ∂v ∂θj ∂v ∂θj X 1 ∂xi X 1 ∂v ∂θj ∂xi X 1 ∂v div(v) = (v) · ei = · ei = · gk = ·gk = δjk · gk (1.104) ∂xi ∂θj ∂xi ∂θj ∂xi hk ∂θk hk ∂θj ∂xi ∂θk hk ∂θj k k |{z} |{z} k −1 Jji Jik X 1 ∂v = · gk (1.105) hk ∂θk k

Notice again how we arrive at an almost identical formula except that we include scale factors. We must also make one very important note: how do we evaluate the following?

∂ ∂ ∂vi ∂gi (v) = (vi gi ) = gi + vi (1.106) ∂θk ∂θk ∂θk ∂θk

In Cartesian coordinates the basis vectors are constant so their derivatives vanish. However, this is not true in most other curvilinear coordinates! To illustrate this, let’s do an example: Example 1.7: Divergence in cylindrical polar coordinates

Compute the divergence of a vector ﬁeld in cylindrical polar coordinates: v = vr gr + vθgθ + vz gz . 1 ∂v 1 ∂v 1 ∂v div(v) = · gr + · gθ + · gz hr ∂r hθ ∂θ hz ∂z ∂ 1 ∂ ∂ = (v g + v g + v g ) · g + (v g + v g + v g ) · g + (v g + v g + v g ) · g ∂r r r θ θ z z r r ∂θ r r θ θ z z θ ∂z r r θ θ z z z ∂v ∂v ∂v = r g + θ g + z g · g ∂r r ∂r θ ∂r z r 1∂v ∂v ∂v ∂ ∂ ∂ + r g + θ g + z g + v g + v g + v g · g r ∂θ r ∂θ θ ∂θ z r ∂θ r θ ∂θ θ z ∂θ z θ ∂v ∂v ∂v + r g + θ g + z g · g ∂z r ∂z θ ∂z z z ∂v 1∂v ∂v = r + θ g + v g − v g · g + z ∂r r ∂θ θ r θ θ r θ ∂z ∂v 1∂v ∂v = r + θ + v + z ∂r r ∂θ r ∂z (1.107)

Notice how we picked up a couple of extra terms: this is a result of our choice of coordinate system. This is a tedious process, but fortunately we only have to do it a couple of times.

1.6.3 Curl

Hopefully this is starting to seem familiar. Starting with our original expression for curl and converting to curvilinear coordinates, we have

∂ ∂v ∂θj X 1 ∂xi X 1 ∂v ∂θj ∂xi X 1 ∂v curl(v) = − (v) × ei = − × gk = − × gk = − δjk × gk (1.108) ∂xi ∂θj ∂xi hk ∂θk hk ∂θj ∂xi ∂θk hk ∂θj k k k X 1 ∂v = − × gk (1.109) hk ∂θk k

Once again, we see that we recover a very similar expression except for the presence of scale factors. Example 1.8: Curl in cylindrical polar coordinates

Find the expression for the curl in cylindrical polar coordinates. We can reuse quite a bit of what we computed earlier; we just need to be careful about which vectors we cancel out. h 1 ∂v 1 ∂v 1 ∂v i curl(v) = − × gr + × gθ + × gz hr ∂r hθ ∂θ hz ∂z ∂v ∂v ∂v = − r g + θ g + z g × g ∂r r ∂r θ ∂r z r 1∂v ∂v ∂v − r g + θ g + z g + v g − v g × g r ∂θ r ∂θ θ ∂θ z r θ θ r θ ∂v ∂v ∂v − r g + θ g + z g × g ∂z r ∂z θ ∂z z z ∂v ∂v 1 ∂v ∂v ∂v ∂v = θ g − z g + − r g + z g + v g + r g − θ g ∂r z ∂r θ r ∂θ z ∂θ r θ z ∂z θ ∂z r 1 ∂v ∂v ∂v ∂v 1∂(r v ) ∂v = z − θ g + r − z g + θ − r g r ∂θ ∂z r ∂z ∂r θ r ∂r ∂θ z (1.110)

1.7 Tensor transformation rules

Suppose we have two orthonormal {ei }, {gi }. We recall that orthonormality allows us to write each basis in terms of the other:

ei = (ei · gp) gp gq = (gq · ei ) ei (1.111)

n n Now, let us suppose we have a tensor A ∈ L(R , R ) with components Aij in the {ei } basis, that is, A = Aij ei ⊗ ej . How can we express A in terms of the other basis? To do that, we simply substitute T ˆ Aij ei ⊗ ej = Aij [(ei · gp) gp] ⊗ [(ej · gq) gq] = (gp · ei ) Aij (ej · gq) gp ⊗ gq = Qpi Aij Qjq gp ⊗ gq = Apqgp ⊗ gq | {z } | {z } T Q Qpi jq (1.112) ˆ where Apq are the components of A in the other basis. Or, put more simply, we have that

ˆ T ˆ T Apq = Qpi Aij Qjq Aij = QipApqQqj (1.113)

Note that we are not transforming the tensor itself: we are merely changing the components of the tensor to ﬁt with the assigned basis. This is called a tensor transformation property.

T Let’s take another look at the Q matrices. Speciﬁcally, let’s look at Q Q: T T [Q Q]pq = Qpi Qiq = (gp · ei )(ei · gq) = [(gp · ei ) ei ] ·gq = gp · gq = δpq = [I]pq (1.114) | {z } gp

T T −1 Because Q Q = I, we conclude that Q = Q . We can also write Q = [g1 ... gn]. If the new basis is right-handed, then we have det(Q) = 1. Thus, we have shown that Q ∈ SO(n), or that Q is a rotation matrix.

2 Kinematics of Deformation

We are now ready to introduce the machinery that we need to describe the deformation of solid bodies. Let us introduce a few definitions: Definition 2.1. A body is a set of material particles occupying a region in Euclidean space; generally denoted Ω ⊂ R3. Definition 2.2. A configuration is a specific correspondance between particles of the body and points in space.

Deﬁnition 2.3. A deformation mapping is an injective1 mapping that describes a conﬁguration of the body.

We will also refer to the deformed and undeformed conﬁgurations. The following ﬁgure illustrates the general setup for describing the deformation of a solid body: φ

G3 X g3 x = φ(X) ΩΩ φ(Ω) g2 G1 G2 g1

• {GI } are the basis vectors in the undeformed conﬁguration

• {gi } are the basis vectors in the deformed conﬁguration. **Note: this is completely general, but we often (usually) keep the same in both conﬁgurations. 3 3 • φ : R → R is the deformation mapping.

• X = XI GI is the location of a point in the undeformed conﬁguration.

• x = xi gi = φi (X) gi is the location of point X in the deformed conﬁguration.

Note that we adopt the convention that uppercase symbols correspond to quantities the undeformed conﬁguration; whereas lowercase symbols correspond to quantities in the deformed conﬁguration. We will even use uppercase and lowercase indices to indicate components in the undeformed and deformed frames. We will stick to this convention consistently, and it will be useful in helping us to keep track of which coordinate system we are in. To illustrate, let us consider the following examples: (i) Stretching of a unit cube: What is the deformation mapping for the following stretched cube?

1“injective” – no two points can be mapped to the same location, that is, if f : U → V is injective, then for x, y ∈ U, f (x) = f (y) =⇒ x = y.

G3 g3 φ

1 λ3

g2 G2 1 λ2 1 λ1 g1 G1

(Notice that g1 = G1, etc.) We identify the deformation mapping simply as

x1 = λ1 X1 x2 = λ2 X2 x3 = λ3 X3 (2.1)

Or, we can describe this as " # λ1 0 0 x = φ(X) = 0 λ2 0 X = FX (2.2) 0 0 λ3

(ii) Shearing of a unit cube: What is the deformation mapping for the following cube subjected to pure shear? G 3 g φ 3 1 γ 1

G2 g2 1 1 1 1

G1 g1

We identify the deformation mapping to be

x1 = X1 x2 = X2 + γ X3 x3 = X3 (2.3)

Or, we can describe it as "1 0 0# x = φ(X) = 0 1 γ X = FX (2.4) 0 0 1

(iii) An afﬁne deformation is one in which “straight lines remain straight” – as in the following ﬁgure:

φ φ

undeformedafﬁne non-afﬁne

What form must this mapping take? Consider the line between two vectors X1, X2, and the resulting line between the mapped vectors x1, x2.

)x2 φ − α + (1 )X2 αx1 − α x αX1 + (1 2 x1 X1 X2

The set of vectors αX1 + (1 − α)X2 are all in the line in the undeformed configuration. By the definition of affine mappings, αx1 + (1 − α)x2 must be must form the points in the deformed line. In other words:

φ(αX1 + (1 − α)X2) = α φ(X1) + (1 − α) φ(X2) + x0 ∀ X1, X2 ∈ Ω (2.5)

Since this must hold for all X, we conclude that the part of the mapping depending on X must be linear. Since all linear maps can be represented in tensor form, we conclude that afﬁne maps can be represented in the form n n φ(X) = FX + x0 F ∈ L(R , R ), φ afﬁne (2.6) −1 Because this is a real deformation, we know that it must be invertible. That is, we can construct a φ such −1 that X = φ (x). But if x = φ(X) = F X, then the inverse mapping must be −1 −1 −1 X = F x =⇒ φ (x) = F x (2.7)

This means that F must be invertible, which means that det(F) 6= 0. The set of all invertible matrices is called the general linear group n n n n GL(n) = {F ∈ L(R , R ) : det(F) 6= 0} ⊂ L(R , R ) (2.8) n so we say that a mapping is afﬁne if and only if ∃F ∈ GL(n) and x0 ∈ R such that

φ = FX + x0 (2.9)

n n (iv)A rigid body mapping in R is, formally, an “orientation-preserving isometry of R .” What does that mean? First, let’s deﬁne the term “isometry:” Deﬁnition 2.4. A mapping φ : Rn → Rn is an isometry if

n |φ(X)| = |X| ∀x ∈ R (2.10)

In other words, an isometry is a mapping that does not change the length of any vector. Let us make the ansatz (i.e. starting assumption) that φ is an afﬁne isometry, that is, φ = F X = x. What are the conditions for φ to be an isometry? √ q √ √ T T T T ! T n |φ| = |x| = x x = (F X) (F X) = X F F X = X X ∀X ∈ R (2.11)

T T T −1 What does this imply about F F ? It must equal the identity. So F F = I =⇒ F = F , or F ∈ O(n), the orthogonal group. What about the other part? Without going into extensive detail, “orientation-preserving” simply means that the body cannot be reﬂected or turned inside-out. This is equivalent to stating that det(F ) > 0. Thus, for φ to be orientation-preserving, F ∈ SO(3); that is, F must be a rotation. There is one more aspect of afﬁne and rigid-body mapping that we have not yet discussed: rigid body translation. A rigid body translation mapping can simply be expressed as

φ(x) = x0 + X (2.12)

where x0 is the translation vector. Thus, the general expression for a rigid body mapping is n φ(X) = x0 + F X x0 ∈ R , F ∈ SO(n) (2.13)

2.1 Eulerian and Lagrangian frames

Let’s go back to our generalized form of the deformation mapping. φ

G3 X g3 x = φ(X) ΩΩ φ(Ω) g2 G1 G2 g1

As stated before, we have adopted the convention of using uppercase variables (and indices) to describe the material in the undeformed configuration, and lowercase for the deformed configuration. The reason for doing this, as we’ll see, is that we will be able to formulate almost everything analagously in terms of either set of variables. Definition 2.5. The Lagrangian / material frame refers to the quantities defined in the undeformed configuration. Definition 2.6. The Eulerian / spatial frame refers to the quantities defined in the deformed configuration.

As we go along and derive various equations, we will frequently formulate those equations in both the Lagrangian and Eulerian frames. (You can think of this as ﬁnding the “uppercase” and “lowercase” versions of the equations.) The following are a couple of examples of the convention that we will use.

• Variables and unit vectors: X are the locations of the material points in the Lagrangian frame, x = φ(X) are the locations of the points in the Eulerian frame. −1 X = XI GI = φI (x) GI x = xi gi = φi (X) gi (2.14)

• Calculus: ∂F ∂f Grad(F(X))I = grad(f (x))i = (2.15) ∂XI ∂xi ∂VI ∂vi Div(V(X))I = div(v(x))i = (2.16) ∂XI ∂xi

and similarly for curl, the Laplacian, etc. Note that Div, Grad are not necessarily the same as div, grad!

2.2 Time-dependent deformation

Let us now consider a body whose deformation varies with time: that is, x(t) = φ(X, t). What is the velocity of the material? Let us deﬁne the Lagrangian velocity ﬁeld as ∂ ∂ V(X, t) = φ(X, t) V (X, t) = φ (X, t) ∂t i ∂t i (2.17)

Similarly, the Lagrangian acceleration as ∂ ∂ A(X, t) = V(X, t) A (X, t) = V (X, t) ∂t i ∂t i (2.18)

Suppose we want to get the velocity and acceleration as a function of the deformed location? To do this we define the Eulerian velocity field as −1 −1 v(x, t) = V(φ (x), t) vi (x, t) = Vi (φ (x), t) (2.19) and the Eulerian acceleration field as −1 −1 a(x, t) = A(φ (x), t) ai (x, t) = Ai (φ (x), t) (2.20)

Example 2.1: Time-dependent deformation of unit cube

A unit cube is undergoing a time-dependent uniaxial stretching deformation as shown below.

1 + λt

We wish to find the material and spatial velocity and acceleration fields. First, we must find the deformation mapping:

x1 = (1 + λt) X1 x2 = X2 x3 = X3 (2.21)

The material velocity and acceleration ﬁelds are given by straight-up differentiation: " # " # ∂ (1 + λt) X1 λX1 ∂ V = X = 0 A = V = 0 ∂t 2 ∂t (2.22) X3 0

Now, we need an inverse relationship to convert to the spatial velocity and acceleration. It is pretty easy to ﬁnd: we see that 1 X = X X = X X = X 1 1 + λt 1 2 2 3 3 (2.23)

Now, all we do is substitute: " # λx1/(1 + λt) −1 v = V(φ (x)) = 0 a = 0 (2.24) 0

2.2.1 The material derivative

We have expressed Eulerian and Lagrangian time derivatives in terms of Lagrangian derivatives; that is, we always get the Eulerian version by back-substituting in the deformation mapping into our Lagrangian version. How can we express the Eulerian time derivatives exclusively in terms of Eulerian coordinates? To do this, we use the material derivative, which is nothing more than an application of the chain rule. Suppose we have two scalar ﬁelds F :Ω → R f : φ(Ω, t) → R (2.25) Computing the time derivative of F is easy: dF d = F (X) dt dt (2.26) However, computing the time derivative of f is more complicated, and we have to use the chain rule:

d d ∂f ∂xi ∂f ∂f ∂f ∂f f (x, t) = f (x(X, t), t) = + = vi + = grad(f ) · v + (2.27) dt dt ∂xi ∂t ∂t ∂xi ∂t ∂t |{z} =vi

Notice that we no longer have any dependence on our deformation mapping φ, only its derivative. This is very convenient for when φ is not available, such as in fluid flow. So though this derivative is nothing other than an application of the chain rule, it is used so frequently that it gets a special name: Df ∂f ∂f = vj + ≡ The Material Derivative (2.28) Dt ∂xj ∂t where f can be a scalar, vector, or tensor field operating in the Eularian configuration. As an example: Eulerian acceleration is given by the material derivative of the Eulerian velocity:

Dv ∂ ∂vi ∂vi a = = grad(v) · v + v ai = vj + = vi,j vj + vi,t (2.29) Dt ∂t ∂xj ∂t

We will show how this works with an example:

Example 2.2: Material derivative

Continuing with our previous example, let us prove that we recover a using the material time derivative. 0 0 ∂v ∂v7 ∂v7 ∂v λ λx λx 1 1 1 1 1 1 a1 = v1 + v2 + v3 + = + − 2 (λ) (2.30) ∂x1 ∂x2 ∂x3 ∂t 1 + λt 1 + λt (1 + λt) λ2x λ2x = 1 − 1 = 0 (1 + λt)2 (1 + λt)2 X (2.31)

The other components are all zero, so their derivatives will be zero as well.

2.3 Kinematics of local deformation

Let us consider the case of all deformations, affine and otherwise. Affine deformations are fairly easy to quantify: the material deforms the same way at every material point, and we can easily represent the mapping using a tensor. The general case is more complicated, but it is what we are interested in. Before discussing local deformation, let us briefly discuss Taylor series. You may recall from Caclulus that any sufficiently smooth function, f (x), can be represented as a Taylor expansion about a point a as follows: ∞ 1 X 1 f (x) = f (a) + f 0(a)(x − a) + f 00(a)(x − a)2 + ... = f (n)(a)(x − a)n 2 n! (2.32) n=0 n 0 n We can do a similar thing with multivariate functions: let f : R → R. Then the expansion of f about a point X ∈ R is 2 0 ∂f 0 1 ∂ f 0 0 f (X) = f (X ) + (XI − XI ) + (XI − XI )(XJ − XJ ) + higher order terms (2.33) ∂XI 2 ∂XI ∂XJ or, using invariant notation, 1 = f (X0) + Grad(f )(X − X0) + (X − X0)T Grad(Grad(f ))(X − X0) + 2 higher order terms (2.34) (As you can see, the problem with invariant notation is that we don’t really have a way to express higher order terms; on the other hand, index notation handles it easily.) Now, let us apply this to deformation mappings. We will restrict ourselves (for now) to smooth mappings, that is, mappings with continuous derivatives.

undeformed non-affine locally affine For mappings of this type, at every point, we can always find a “neighborhood” of the point at which the mapping 0 0 is locally affine. Here, let us consider the neighborhood around a point X that is mapped to x , and observe what happens to a small vector ∆X that is in the locally affine neighborhood.

∆X φ ∆x x0 X0

We expand the mapping out as a Taylor series:

0 0 0 ∂φi 0 0 0 ∂φi xi + ∆xi = φi (X + ∆X) = φi (X ) + ((XJ + ∆XJ ) − XJ ) + h.o.t. = xi + ∆XJ + h.o.t. (2.35) ∂XJ ∂XJ (2.36)

If we ignore the higher-order terms, we can write

∂φi ∆xi = ∆XJ = FiJ ∆XJ (2.37) ∂XJ where

∂φi FiJ = ≡ Deformation Gradient Tensor (2.38) ∂XJ is the deformation gradient tensor. In full component notation, we would write

∂ ∂xi F = Grad(x) = (xi gi ) ⊗ GJ = gi ⊗ GJ = FiJ gi ⊗ GJ (2.39) ∂XJ ∂XJ

Because F has both g and G components, it is referred to as a two-point tensor. We say that it is partially in the deformed conﬁguration, and partially in the undeformed conﬁguration, and that it turns undeformed vectors into deformed vectors.

2.4 Metric changes

We can use this new technology to describe how various quantities in the body (length, area, angle, etc.) change as they are acted on by the deformation mapping.

2.4.1 Change of length

Let us begin by considering a small vector ∆X in the neighborhood of X as we had before. What is the change in length as ∆X is acted on by φ? Let F = F(X) be the local deformation gradient. Then q √ √ p p T T T T |∆x| = ∆xk ∆xk = (FkI ∆XI )(FkJ ∆XJ ) = ∆XI FIk FkJ ∆XJ = ∆X F F ∆X = ∆X C∆X (2.40) where

T C = F F CIJ = FkI FkJ ≡ Right Cauchy-Green Deformation Tensor (2.41)

Notice that C has strictly uppercase indices: this implies that it lives entirely in the Lagrangian frame. If we divide both sides by ∆x, we have

|∆x|2 = N C N = NT CN = λ2(N) ≡ Stretch Ratio |∆X|2 I IJ J (2.42) where N = ∆X/|∆X| is the unit vector in the direction of ∆X, and the stretch ratio is simply the ratio of the deformed length to the undeformed length.

2.4.2 Change of angle

Consider two vectors ∆X, ∆Y.

∆Y ∆y Θ φ θ ∆x ∆X

We can ﬁnd the angle between them in the undeformed conﬁguration by using the inner product: ∆X · ∆Y cos Θ = |∆X||∆Y| (2.43)

In the deformed conﬁguration, the angle between them is given by T T T T ∆x ∆y ∆X F F∆Y NX CNY cos θ = = = (2.44) |∆x||∆y| (λ(NX )|∆X|)(λ(NY )|∆Y|) λ(NX ) λ(NY ) where NX = ∆X/|∆X|, NY = ∆Y/|∆Y| are the unit vectors in the directions of ∆X, ∆Y.

Example 2.3: Relationship of stretch, angle to components of C

We can express C in full component notation as T C = CIJ GI ⊗ GJ = CIJ GI GJ (2.45)

What is the change of length of basis vector G1? q q T T T T p p λ(G1) = G1 (CIJ GI GJ )G1 = CIJ (G1 GI )(GJ G1) = CIJ δ1I δ1J = C11 (2.46)

and similarly for G2, G3, and so on. What about the change in angle between unit vectors? Consider vectors GP , GQ . The change in angle is

T GP CGQ cos θPQ = no sum on P, Q (2.47) λ(GP )λ(GQ ) CPQ = p no sum on P, Q (2.48) CPP CQQ

We notice a general trend here, as with the deformation gradient: diagonal terms relate to elongation, whereas off-diagonal terms relate to shear.

2.4.3 Determinant identities

3 3 Recall the formulae for the determinant of a tensor F ∈ L(R , R ): 1 det(F) = F F F det(F) = F F F ijk i1 j2 k3 6 ijk IJK iI jJ kK (2.49)

Identity:  + det(F) IJK = 123, 231, 312  ijk FiI FjJ FkK = − det(F) IJK = 321, 132, 213 = det(F) IJK (2.50)  0 else

2.4.4 Change of volume

Consider the parallelpiped deﬁned by three vectors ∆X, ∆Y, ∆Z. What is the volume of this parallelpiped? We know that we can get it by taking the triple scalar product, that is, ∆X · (∆Y × ∆Z). In index notation, this comes out to be ∆V = ∆X · (∆Y × ∆Z) = (∆XI GI ) · (∆YJ GJ ) × (∆ZK GK ) = ∆XI ∆YJ ∆ZK GI · (JKLGL) (2.51) = JKL∆XI ∆YJ ∆ZK δLI = IJK ∆XI ∆YJ ∆ZK (2.52)

Now, suppose the volume is mapped to a deformed conﬁguration by a mapping φ: ∆Z φ ∆z ∆v ∆x ∆y ∆V ∆Y ∆X

What is the volume of the deformed element in terms of the undeformed element?

∆v = ∆x · (∆y · ∆z) = ijk ∆xi ∆yj ∆zk (2.53) = ijk (FiI ∆XI )(FjJ ∆YJ )(FkK ∆ZK ) = ijk FiI FjJ FkK ∆XI ∆YJ ∆ZK (2.54) = det(F ) IJK ∆XI ∆YJ ∆ZK = J ∆V (2.55) | {z } ∆V where ∆v J = det(F) = ≡ Jacobian ∆V (2.56) is the ratio of the deformed volume to the undeformed volume. Notes: (1) Nonzero determinant ≡ no vanishing mass ≡ invertibility of the deformation mapping (2) The constraint J > 0 makes sense in this sense; negative volume only exists if the volume is turned inside out. This allows us to make the connection between the invertibility of the deformation gradient with the fact that the volume of a section of the body cannot go to zero. It should also be noted that in (2.56) ∆v and ∆V denote differential (i.e. small) ﬁnal and initial volumes, respectively, and not changes in volume, which could be mistaken due to the use of deltas.

2.4.5 Change of area

Consider two vectors ∆X, ∆Y. What is the area of the parallelogram that they span? We get that by the cross product, so the area vector is

∆A = IJK ∆XI ∆YJ Gk (2.57)

We note that ∆A = |∆A| where ∆A is the magnitude of the area. Then

∆A = N ∆A (2.58) where N is the unit vector normal to the surface. Now, consider the action of a deformation mapping on the vectors:

φ ∆A ∆a N ∆x n ∆Y ∆y ∆X ∆a ∆A

What is the area of the new vectors? Let us follow a procedure similar to that for the change of volume:

∆ak = (∆x × ∆y)k = ijk ∆xi ∆yj = ijk (FiI ∆XI )(FjJ ∆YJ ) = ijk FiI FjJ ∆XI ∆YJ (2.59)

Now,we’re going to pull a trick: multiply both sides by FkK . This gives us

∆ak FkK = ijk FiI FjJ FkK ∆XI ∆YJ = J IJK ∆XI ∆YJ = J ∆AK (2.60)

Now, we can rewrite the above in invariant notation T F ∆a = J∆A (2.61)

T (Why the F ? Out of necessity: in order to make everything line up like it’s supposed to in invariant notation, we had to shufﬂe things around a little.) Isolating the expression for ∆a gives us

∆a = J F−T ∆A −T ≡ Piola Transform (2.62) ∆ai = J FiJ ∆AJ

The Piola transform, or “Nanson’s formula” will be useful later on when we start talking about forces per unit deformed or per unit undeformed area. Note: Covariance and contravariance of vectors

We are now at an ideal point to introduce an important (yet frequently neglected) subject: the covariance n and contravariance of vectors in R . We will motivate the need for this distinction by means of an example. Let us consider two of the metric changes that we have introduced: change of length and change of area. In particular, let us consider two vectors T and N, where both have the same components, but T is a length vector, while A = AN is an area vector. In the ﬁgure below, both transform under φ, where φ is a simple shear deformation: X x 2 φ, F 2

t a T A

X1 x1

Notice that the distance vector t has a different direction and magnitude, whereas the area vector a is un- changed. We recall that this is because they transform differently: T T −1 t = F T a = J A F (2.63)

This gives us a strong indication that there are two kinds of vectors. The ﬁrst kind (T, t) transform by the deformation gradient and are called contravariant vectors. The second kind (A, a) transform by the inverse of the deformation gradient and are called covariant vectors. A couple things are worth noting:

(1) Contravariant vectors are usually just called vectors. Covariant vectors are frequently referred to as “covectors,” “dual vectors,” or “1-forms.” (2) Contravariant vectors have a natural representation as column vectors, while covectors have a natural T T representation as row vectors. This is why the above transformation is written in terms of A , a . (3) Contravariant vectors frequently correspond to line elements, whereas covectors frequently correspond to surface elements. As a result, it is very important to distinguish between directional vectors and normal vectors. It is quite possible to go all the way through continuum mechanics and vector calculus without any notion of the distinction between vectors and covectors, and the ﬁner details are outside of the scope of this class. However, you will see that there is a subtle yet deep and fundamental difference between them. They begin to introduce the theme of duality that is intrinsic to constitutive theory, and makes some strong ties between constitutive modeling and the geometry of physical quantities.

Example 2.4: Metric changes in a planar deformation

Consider the following deformation

X2 γ

λ 1

1 X1

The deformation mapping is given by

x1 = X1 + γX2 x2 = λX2 X3 = X3 (2.64)

And from that we can quickly ﬁnd the deformation gradient and the Cauchy-Green stretch tensor "1 γ 0# "1 0 0#"1 γ 0# "1 γ 0# 2 2 F = 0 λ 0 C = γ λ 0 0 λ 0 = γ γ + λ 0 (2.65) 0 0 1 0 0 1 0 0 1 0 0 1

Example 2.4 Continued

From this we can compute the following metric changes: (i) Angle change θ: we can obtain this by using G1, G2 as our unit vectors. First, compute the stretches: q q T p T p 2 2 λ(G1) = G1 CG1 = C1 = 1 λ(G2) = G2 CG2 = γ + λ (2.66)

Now, use the angle formula: T G1 CG2 λ cos θ12 = = p (2.67) λ(G1)λ(G2) γ2 + λ2

(ii) Length change: let us compute√ the deformed length of the diagonal line. The unit vector corresponding to this is N = (G1 + G2)/ 2, so the stretch ratio is √ r1 λ(N) = NT CN = (1 + γ)2 + λ2 2 (2.68) √ p 2 2 So, if the original length of the diagonal was 2, the deformed length is just (1 + γ) + λ . (A quick geometric calculation will verify this.) (iii) Volume change: very easy, all we have to do is compute the determinant: v = v = det(F ) = λ V (2.69)

(iv) Area change: to do this, we must use the Piola transform. Undeformed area vector is "0# A = (1)(N) = 0 (2.70) 1

To use the Piola transform, we must invert the deformation gradient. There are a number of ways to do this, but the easiest way here is to use the cofactor method: "λ −γ 0 # "1 −γ/λ 0# −1 1 F = 0 1 0 = 0 1/λ 0 (2.71) det F 0 0 det(F ) 0 0 1 " 1 0 0# −T F = −γ/λ 1/λ 0 (2.72) 0 0 1

Now we can compute our new area vector: " 1 0 0#"0# "0# −T a = JF A = (λ) −γ/λ 1/λ 0 0 = 0 (2.73) 0 0 1 1 λ

Thus, the magnitude of the area is a = λ. Looking back, the answer is obvious: with the thickness constant, the area change would be the same as the volume change. We also know that the normal vector would not change since everything was in 2D. So, we see that we recover the right answer using the Piola transform.

2.5 Tensor decomposition

T We have introduced the tensors F (deformation gradient) and C = F F (Cauchy-Green deformation tensor). To analyze these tensors, we need to review and introduce a couple of linear algebra concepts.

2.5.1 Eigenvalues and Eigenvectors

Recall that we can think of a tensor as a machine that turns a vector into another vector. In general, the tensor can change both the vector’s magnitude and its direction. But what happens if the tensor changes the vector’s 3 magnitude only? For instance if we have T ∈ GL(3), u ∈ R , then we might write

Tu = λu (2.74) with λ ∈ C, where C is the set of all complex numbers, and of course we recall that R ⊂ C, so λ could be real. Vectors that satisfy this relationship are called eigenvalues, and the value λ are called eigenvalues. We care a lot about eigenvalues and eigenvectors, because it’s much easier to work with scalars acting on vectors than tensors. How do we go about ﬁnding these eigenvalues and eigenvectors? We know that they will satisfy the equation

Tu − λu = (T − λI)u = 0 (2.75)

Of course, one solution is that u is just zero, but this isn’t very interesting at all because the result would be trivial. Instead, we notice that the matrix T − λI is able to take a non-zero vector and spit out zero. We say that u is in the “nullspace” or “kernel” of T − λI, and there’s a nice theorem (called the “Rank-Nullity theorem” that unfortunately we don’t have time to prove here) that tells us that the tensor T − λI can map a vector u to zero if and only if det(T − λI) = 0. This means that we can now solve the algebraic equation

det(T − λI) = 0 (2.76)

th which is a n order polynomial. We know that we will ﬁnd at least one, and no more than three eigenvalues {λi } by solving this equation. We can then ﬁnd our eigenvectors {ui } by solving

Tui = λ(i)u(i) (2.77)

An interesting side note is the following: we can always write 3 2 det(T − λI) = c3λ + c2λ + c1λ + c0 = 0 (2.78) where (2.78) is the characteristic equation and

• c3 = 1

• c2 = I1 = tr(T) ≡ the ﬁrst invariant of T 1 2 2 • c1 = I2 = 2 (tr(T) − tr(T )) ≡ the second invariant of T

• c0 = −I3 = − det(T) ≡ the (negative) third invariant of T We will continue with our introduction to tensor decomposition. The following theorem is an interesting side note to the discussion of a tensor’s characteristic equation: n n Theorem 2.1 (Cayley-Hamilton). Every square tensor T ∈ L(R , R ) satisﬁes its own characteristic equation. We will also introduce another theorem without proof: Theorem 2.2. Every square tensor T ∈ GL(Rn, Rn) has n linearly independent eigenvectors.

This will come in handy in following sections.

2.5.2 Symmetric and positive deﬁnite tensors

We recall the deﬁnition for a symmetric matrix: T Deﬁnition 2.7. A square tensor T is symmetric if T = T . If T is symmetric and has components {Tij } in an orthonormal basis, then Tij = Tji

n We note the following important identity: if T is symmetric, then for u, v ∈ R we can always write T T T T u Tv = u T v = (Tu) v (2.79)

Let us make another very important definition: Definition 2.8. A square tensor T ∈ GL(Rn, Rn) is positive definite if

T n u Tu > 0 ∀u ∈ R , |u| > 0 (2.80) and T u Tu = 0 =⇒ |u| = 0 (2.81)

(An important equivalence to note here is that a matrix is positive definite if and only if all of its eigenvalues are real and strictly positive.) Finally, we will make one more similar definition: Definition 2.9. A square tensor T ∈ L(Rn, Rn) is positive semidefinite if

T n u Tu ≥ 0 ∀u ∈ R (2.82)

Note that the only difference is that this allows for nonzero vectors to be mapped to zero. Matrices are positive semideﬁnite if and only iff all of its eigenvalues are real and nonnegative, i.e. it can have eigenvalues with value zero.

2.5.3 Spectral theorem (symmetric tensors)

The spectral theorem is one of the most powerful theorems in applied mathematics. It has widespread applications beyond matrices, such as the Fourier and Laplace transforms, or Sturm-Liouville theory. If you haven’t taken a (graduate level) linear algebra course, you should deﬁnitely consider taking one. The ﬁrst theorem we will introduce without proof: Theorem 2.3 (Real eigenvalues). If a tensor T is symmetric and real-valued, all of its eigenvalues are also real.

The proof of this theorem is quite easy, but it involves the use of complex analysis which is a bit beyond the scope of the course right now. The next theorem is very important, and so we will prove it. n n Theorem 2.4 (Orthogonal eigenvectors). If a tensor T ∈ GL(R , R ) is symmetric, it has n orthonormal eigenvectors. Proof. We will look at two cases.

(1) Distinct eigenvalues: let {ui } be the eigenvectors of T corresponding to the distinct eigenvalues {λi }, that is, no two eigenvalues are the same. Now, let us consider two eigenvectors ui , uj with eigenvalues λi , λj , λi 6= λj . Because the matrix is symmetric, we can write T T T T 0 = ui Tuj − (Tui ) Tuj = ui (λ(j)uj ) − (λ(i)u(i)) uj = (λ(i) − λ(j))(u(i)u(j)) = 0 (2.83)

Since λi 6= λj , the only way for this to be true is for ui to be orthogonal to uj .

(2) Degenerate eigenvalues: let ui , uj be two eigenvectors of T both with eigenvalue λ. Then we can write, ∀α, β ∈ R,

T(αui + βuj ) = αTui + βTuj = λ(αui + βuj ) (2.84)

In other words, every vector αui + βuj is an eigenvector of T. More generally, we can write that

Tu = λu ∀u ∈ span(ui , uj ) (2.85)

that is, that there is a 2D space of eigenvectors of T with eigenvalue λ. We know from linear algebra that there exists an orthonormal basis for this space; this orthonormal basis gives the orthogonal eigenvectors for T . The case of 3+ degenerate eigenvalues proceeds by induction.

Now that we know that every symmetric matrix T has a set of orthonormal eigenvectors, we can use these eigenvectors to construct an orthonormal basis. We call this the eigenbasis of T.

Theorem 2.5 (Spectral decomposition). A symmetric matrix T can be expressed strictly in terms of rotation matrices constructed from its eigenvectors and a diagonal matrix with each diagonal entry containing an eigenvalue of T.

n Proof. Let us consider the action of T on any vector x ∈ R . By casting x into the eigenbasis of T, then by ﬁnding the action of T on its eigenvectors, and ﬁnally casting back, we obtain

T X T X T X T T Tx = T(ei xi ) = Tvj (vj ei )xi = λj vj (vj ei )xi = vj λj (vj ei )xi = ek (ek vj )λj (vj ei )xi (2.86) j j j

Introducing a couple of Kronecker deltas and rearranging indices gives

T X T = ei (ei vp) δpr λr δrq (vq ej ) xj (2.87) | {z } r | {z } Qip T | {z } Qqj Λpq concluding the proof. in other words, the components of T in the original basis can be expressed as T Tij = QipΛpqQqj (2.88) where Q are orthogonal transformation matrices whose components are given in terms of the eigenvectors v by " # e1 · v1 e1 · v2 e1 · v3 Qij = ei · vj Q = e2 · v1 e2 · v2 e2 · v3 = [v1 v2 v3] (2.89) e3 · v1 e3 · v2 e3 · v3 and Λ is a diagonal matrix with the eigenvalues λi along the diagonal. This is called diagonalization; it can also be referred to as spectral decomposition or eigendecomposition. So what are we actually doing when we do an eigendecomposition? The important thing to remember is that we are not changing the tensor at all. Rather, we are ﬁnding a different way to express the tensor by decomposing it into seperate stages. Consider the following illustration: e2 u2 u2 e2

e1 u1 u1 e1

QT ΛQ

A tensor T applies shear and normal deformation to a square. We can express this deformation as a rotation into the eigenspace of T, applying pure stretching, and rotating back. Note that the diagonals of the square are orthogonal, and do not change direction under the deformation. These are the eigenvectors of the deformation, and their stretches are the eigenvalues. Theorem 2.6. All of the eigenvalues of a symmetric positive deﬁnite tensor A are positive.

The proof of this theorem is straightforward.

2.5.4 Spectral theorem (general case)

We will not go into this in as much detail, but let us consider a nonsymmetric tensor A. Then A has n linearly independent eigenvectors, but they are not necessarily orthogonal. We say that A is diagonalizable if, given that A has components Aij in the reference frame, −1 A = UΛU (2.90) −1 where Λ is the diagonal matrix of eigenvalues, U is the matrix of eigenvectors, and U is the inverse of U. Note −1 T that if A is orthogonal, U becomes U , and we recover the earlier result. Not all tensors are diagonalizable. If a tensor is not diagonalizable it is calle defective. In general, for purposes of this course, however, we don’t need to worry about those cases.

2.5.5 Functions of tensors

The spectral decomposition has a wide range of applications, and one of the most useful is to describe functions on matrices. Just like with scalar numbers, we can raise a matrix to a power. Let us consider a diagonalizable n tensor A. What is A ? λn  1 n n −1 −1 −1 −1 −1 λ A = AAAA ... A = (U ΛU)(U ΛU) ... (U ΛU) = U ΛΛ ... Λ U = U  2  U (2.91) | {z } | {z } . n times n times .. so we see that we can raise the matrix to the n-th power simply by raising the eigenvalues to the n-th power. This means that we can conceivably have polynomials of tensors. Now, let us suppose we have f : R → R, a smooth function that admits the Taylor Series expansion ∞ X 1 f (x) = f (n)(0)x n n! (2.92) n=0

For instance, we know that exponential function has the Taylor series expansion ∞ X x n exp(x) = ex = n! (2.93) n=0

It turns out that we can use the Taylor series expansion to deﬁne functions of matrices. Suppose we have a function f . Then we can describe the action of f on a diagonalizable tensor A by expanding in Taylor series form: P∞ 1 (n) n  ∞ n=0 n! f (0)λ1 X 1 (n) n −1 P∞ 1 (n) n f (A) = f (0)A = U  n=0 f (0)λ2  U n!  n!  (2.94) n=0 . .. f (λ1)  −1 f (λ2) −1 = U   U = U f (Λ) U (2.95) . .. That is, we can describe the effect of a scalar function on a tensor, simply by taking the spectral decomposition 3 3 and evaluating the function on each of the eigenvalues. For instance, if A ∈ L(R , R ) is diagonalizable, we have √    λ1  " # √ λ1 √ e ln(λ1) −1 A −1 λ2 −1 A = U  λ2 √  U, e = U  e  U, ln(A) = U ln(λ2) U, λ3 λ3 e ln(λ3) (2.96) and so on. Many of the same properties of f hold when extended to the matrix case, although it’s not always safe to make this generalization. For instance, if two tensors A, B have the same eigenbasis, you can even show that A+B A B some properties of f such as e = e e , ln(AB) = ln(A) + ln(B) hold; however, this is trickier if they do not.

2.5.6 Polar decomposition

We work with a lot of very nice symmetric positive definite tensors, but what about when it is not symmetric or positive definite? Specifically, the deformation gradient F does not satisfy either of those properties. For this case, we will use the polar decomposition. Theorem 2.7 (Right polar decomposition). For every tensor T ∈ GL(n) with det(T) > 0 there exists a symmetric positive-definite tensor U ∈ GL(n) and a rotation R ∈ SO(n) such that T = RU

Proof. Proceed formally: T T T (1) T T is symmetric positive definite, because v T Tv = |Tv| = 0 only if Tv = 0. But if there exists a nonzero n T v ∈ R such that Tv = 0, then det(T) = 0, a contradiction with the fact that T ∈ GL(n). Therefore T T is positive definite. It is symmetric by inspection. √ T T (2) Because T T is symmetric positive definite, there exists a symmetric positive definite tensor U = T T, T where UU = T T. −1 T −T T −1 −1 −1 (3) Suppose T = RU: then R ∈ SO(n). R = TU , and R R = U T TU = U UUU = II = I, proving T −1 −1 that R = R and therefore R ∈ O(n). Now, show that det(R) = 1: det(R) = det(T) det(U ) > 0, implying that det(R) = 1. Therefore R ∈ SO(n).

In other words, even if a tensor lacks the nice properties of being symmetric (or even positive definite), we can always decompose it into a “nice” tensor and a “rotation” tensor. Intuitively, we can interpret this as meaning that all of the nastiness of an asymmetric deformation gradient (i.e. a def. grad. that has complex eigenvalues) is due to rotation. Now, we can find a similar decomposition such that the rotation matrix is on the left: Theorem 2.8 (Left polar decomposition). For every tensor T ∈ GL(n) with det(T) > 0 there exists a symmetric positive-definite tensor V ∈ GL(n) and a rotation R ∈ SO(n) such that T = VR

Proof. By the previous theorem, we know T = RU, where R is the rotation for the left polar decomposition. Let T T T TT T T T T T T V = TR = RUR . Then V = R U R = RUR = V, verifying symmetry, and u Vu = u RUR u = T T T (R u) U(R u) > 0, verifying positive definiteness. √ T T T T T T Remark: VV = RUR RUR = RUUR = RU(RU) = TT so V = TT Intuitively, this means that we can get the same result whether we rotate first then deform, or deform first then rotate.

2.6 Principal deformations

Now that we’ve developed a complete set of tools for decompositing tensors, let us apply them to our deformation tensors. Consider a mapping with a local deformation gradient F. We know by example that the deformation gradient is not symmetric or positive definite. This is because the deformation gradient includes a rigid-body rotational component – the part that actually goes towards deforming the material should be symmetric positive definite. We can use the polar decomposition to decouple these two components. We know that F ∈ GL(3) and satisfies det(F) > 0, therefore there exists R ∈ SO(3) such that √ √ √ √ T T F = UR = RV U = F F = CV = FF = B (2.97) where we introduce T B = FF ≡ Left Cauchy-Green deformation tensor (2.98) What does this mean intuitively? This decomposition is best illustrated pictorally. Consider a cube undergoing simple shear.

U R

n1 n2 N2 N1 F = RU = VR

R V

Note that the two vectors N1, N2 transform into λ1n1, λ2n2, where λ1 = λ(N1), λ2 = λ(N2) are the stretches, and n1, n2 remain orthogonal. Not all vectors transform this way, but (as we’ll see) we will always be able to find 2 (or 3 in 3D) vectors that remain orthogonal to each other under transformation. These special vectors are called the principal directions, and are referred to as the material principal directions (N1, N2, N3) in the undeformed configuration, and as the spatial principal directions n1, n2, n3 in the deformed configuration. Moreover, these directions are related to each other via the rotation tensor, that is

ni = RNi (2.99) where we obtain R from the polar decomposition. The length change λi are referred to as the principal stretches, and we know that

FNi = λ(i)n(i) (2.100)

How do we go about ﬁnding our principal directions and stretches? There are two ways: beginning with the above expression:

T FN = λn =⇒ RUN = λn =⇒ UN = λR n =⇒ UN = λN (2.101) FN = λn =⇒ VRN = λn =⇒ Vn = λn (2.102) √ √ So, we see that Ni are eigenvectors√ of C√and ni are eigenvectors of B. Furthermore, we see that the principal stretches are eigenvalues of both C and B. We recall from the spectral theorem that √ √ T C = U ΛU (2.103) √ where√ Λ is the diagonal matrix of the square root of eigenvalues of C. Hence, we conclude that the eigenvalues of C are the square root of the eigenvalues of C, and that the eigenvectors are the same. To recap, here is a summary of the principal stretches/directions and how to ﬁnd them: √ {λi } = eigenvalues of C ≡ Principal Stretches {Ni } = eigenvectors of C, U ≡ Material Principal Directions (2.104) {ni } = eigenvectors of B, V ≡ Spatial Principal Directions

2.7 Compatibility

We have seen that the deformation gradient F is an extremely handy way of working with deformations. In fact, we will often think of deformations as being defined by their deformation gradient at each point. However, just because we have a deformation gradient F(X) does not mean that it is “realistic” – that is, it may not correspond to an actual deformation. So, we introduce the notion of compatibility: Definition 2.10. A deformation gradient field over a domain B is said to be compatible is there exists a mapping φ such that F = Grad(φ).

One way to ensure that a deformation gradient is compatible is by ﬁnding the deformation mapping φ. However, we may not always be able to do this; it is generally easier to ﬁnd local conditions on F that do not require integration.

2.7.1 Continuous case

We will begin with the case of smooth mappings, where φ is assumed to be twice differentiable. The following theorem provides a powerful way to determine compatiblity by differentiation. Theorem 2.9. A deformation gradient ﬁeld is compatible if and only if Curl(F) = 0.

Proof. This theorem states that zero curl is a necessary and sufﬁcient condition; therefore we must prove it both ways. Necessary: Use the symmetry of differentiation to show that 2 2 ∂ φi ∂ φi FiJ,K = = = FiK,J (2.105) ∂XJ ∂XK ∂XK ∂XJ

Now we pull a trick: using the fact that these two terms are equal, we multiply by permutation tensor and swap indices to show that

FiJ,K − FiK,J = 0 (2.106) =⇒ (FiJ,K − FiK,J )IJK = (IJK FiJ,K − IJK FiK,J ) = (IJK FiJ,K + IJK FiJ,K ) = Curl(F )iI = 0 (2.107)

So if a deformation mapping exists, the curl must be zero. Sufﬁcient: Stoke’s theorem is similar to the divergence theorem (indeed, both are special cases of a single generalized theorem) that states the following: for a surface A with boundary ∂S, for some vector or tensor ﬁeld denoted generally as F Z Z Curl(F )iJ NJ dA = FiJ dXJ (2.108) S ∂S

Now we apply this to the deformation gradient. We are given that Curl(F) = 0 everywhere, so we have Z Z 0 = IJK FiJ,K dAK = FiJ dXJ (2.109) S ∂S

with the implication that integrating F over every closed contour returns zero.

B Γ0

S Γ00

Ω A

0 00 Consider two points a, b in the undeformed conﬁguration, and two contours Γ ,Γ going from a to b, forming a closed contour so that Z Z Z Z FiJ dXJ − FiJ dXJ = 0 =⇒ FiJ dXJ = FiJ dXJ (2.110) Γ0 Γ00 Γ0 Γ00

Now, let us deﬁne Z ˆ φ(X) = FiJ dXJ . (2.111) γ[A,X]

ˆ where γ can be any differentiable contour, and XJ distinguishes the variable of integration from the argument of φ. If we can show that Grad(φ) = F, we will be done. To do this, we evaluate the Gateaux derivative of φ for some arbitrary vector V: d Z 1h Z Z i ˆ ˆ ˆ Dφ(X)V = lim FiJ dXJ = lim FiJ dXJ − FiJ dXJ (2.112) ε→0 dε γ[A,X+εV] ε=0 ε→0 ε γ[A,X+εV] γ[A,X] ε=0

Path independence allows us to break up the integral: 1h Z Z Z i h 1 Z i ˆ ˆ ˆ ˆ = lim FiJ dXJ + FiJ dXJ − FiJ dXJ = lim FiJ dXJ (2.113) ε→0 ε γ[A,X] γ[X,εV] γ[A,X] ε=0 ε→0 ε γ[X,εV] ε=0

ˆ ˆ Let γ[X, X + εV] be linear, and introduce the parameterization XI = XI + sVI , dXI = VI ds:

h 1 Z ε i h d Z ε i h i = lim FiJ (X + sV)VJ ds = FiJ (X + sV)VJ ds = FiJ (X + εV)VJ (2.114) ε→0 ε 0 ε=0 dε 0 ε=0 ε=0 = FiJ (X)VJ (2.115)

whence we identify FiJ = Grad(φ), resulting from the fact that the above holds for all V. Thus the condition Curl(F) = 0 is both necessary and sufﬁcient for compatibility.

This theorem is quite useful because we don’t always know if a deformation mapping exists. By this, we see that we only have to check that the curl is zero to ensure that it the deformation gradient for a real mapping. A notable exception/special case of this is dislocations. Dislocations introduce a discontinuity into the deformation mapping, and as such, they create a singularity. This means that the curl of the deformation gradient is nonzero whenever there is a dislocation. In fact, we can actually use Curl F to determine the dislocation content of a sample.

2.7.2 Discontinuous case (Hadamard)

In the above section we have considered only deformations that are twice differentiable, and consequently fairly smooth. There are a number of cases where the deformation is not smooth, however. One example is the forma- tion of laminate microstructure. Sometimes in order to conform to an imposed deformation gradient, a material “ﬁnds it” more energetically favorable to create alternating variants of microstructure that approximate the imposed deformation, as in the following ﬁgure:

The resulting deformation map is only C0 continuous, because the deformation gradient jumps across the boundary. In the following ﬁgure, a material undergoes a discontinous deformation:

X2 x2

N F+ T t

F− X1 x1

+ − + − Suppose we only know F , F . What constraints must exist on F and F to ensure that they correspond to a deformation mapping? The answer is quite simple: consider a vector T in the boundary. We know that

+ ! − F T = F T (2.116) or alternatively that + − (F − F )T = [[F]]T = 0 ∀T ∈ S (2.117) where S is the boundary plane and the double bracket [[·]] denotes “jump in.” We know that S is two-dimensional; therefore because [[F ]] maps all vectors in a 2D space to zero, it has a kernel of dimension 2. By the Rank-Nullity theorem, this implies that it is rank 1. Deﬁnition 2.11 (Hadamard compatibility). For a discontinuous deformation gradient across a boundary with normal N in the undeformed conﬁguration, the deformation gradients are Hadamard or Rank-1 Compatible if there exists an a ∈ R3 such that [[F]] = a ⊗ N (2.118)

Let’s check this: what happens when [[F]] acts on T ∈ S? T [[F]]T = (a ⊗ N)T = a (N T) = a(N · T) = 0 (2.119) because N is normal to the plane containing T.

Example 2.5: Shear twinning

Consider a material that approximates a shear deformation of γ by creating two layers that shear with known values γ1, γ2, respectively, as shown.

x2 γ γ2

λ2 γ 1 1

λ1

x1 1

Check that the interface is compatible and ﬁnd λ1, λ2 First, let us write down the deformations that we know:

h1 γi h1 γ1i h1 γ2i F = 0 1 F1 = 0 1 F2 = 0 1 (2.120)

Check compatibility: h0 γ2 − γ1i hγ2 − γ1i [[F]] = F2 − F1 = 0 0 = 0 [0 1] X (2.121) | {z } | {z } N a We also know the following

λ1 + λ2 = 1 γ1λ1 + γ2λ2 = γ (2.122)

which allow us to solve for λ1, λ2:

γ2 − γ γ − γ1 λ1 = λ2 = (2.123) γ2 − γ1 γ2 − γ1

Given that 0 ≤ λ1, λ2 ≤ 1, we also conclude that γ2 ≤ γ ≤ γ1

2.8 Other deformation measures

We are now familiar with using F and C as measures of local deformation. However, you may notice that these are somewhat different from the traditional measure of strain. In particular, the C and F tensors corresponding to zero deformation are the identity, rather than the zero tensor. While this is not really problematic, some prefer to use tensors for which zero deformation corresponds to the zero tensor. The ﬁrst example is the Green-Lagrange strain tensor:

1 E = (C − I) ≡ Green-Lagrange Strain Tensor 2 (2.124)

Notice that F = I =⇒ C = I =⇒ E = 0. Also, note that for R ∈ SO(3), if F = R then 1 1 E = (RT R − I) = (I − I) = 0 2 2 (2.125) Furthermore, we see that E inherits invariance properties because it depends on C only: if F = RU, the R0 will always disappear leaving U only.

A slightly less ad hoc deformation measure is the log strain tensor:

ln(C) ≡ Logarithmic Strain Tensor (2.126)

Note that if F = I, ln(C) = 0. Another satisfying property of this is the following: if we write out ln(C), we see that " # 2 ln(λ1) T ln(C) = Q 2 ln(λ2) Q (2.127) 2 ln(λ3)

If λ1, λ2, λ3 go to zero, this causes ln(C) to go to inﬁnity. This causes the non-zero-deformation property of F to be “built-in” to the deformation tensor, which is very convenient. However, the log strain tensor comes with a couple of problems. First, it is somewhat costly to compute computationally as it requires a full eigendecomposition of p C. Second, and more importantly, we know that ln(x) for 1 ≤ p < ∞ will always scale sublinearly. This means that we cannot construct an energy function W (ln(C)) that is convex – the log term will always restrict the growth of the energy with deformation.

2.9 Linearized kinematics

Up until now we have left everything completely general: the measures of deformation that we have formulated will hold for any continuous deformation. This is great, but the problem is that these deformation measures are often a bit unwieldy to use. On the other hand, there are many cases where the amount of deformation that the material undergoes is very small. As a result, it is often advantageous to linearize our deformation measures. Suppose G is some arbitrary function of a deformation measure, such as the deformation gradient or the Jacobian. We express the linearization of G about a deformation φ in terms of the Gateaux derivative:

G[φ(X) + u] = G(φ) + DG(φ)u + higher order terms (2.128) where u is a displacement from φ that is small enough so that higher order terms are negligible. For example, the linearized deformation gradient would be d F(φ + u) ≈ F(φ) + DF(φ)u = F0 + Grad(F + εu) = F0 + Grad u dε ε→0 (2.129) where F0 is the large deformation and u is the small displacement. It is possible to do the entire analysis with an arbitrary “large” deformation combined with a small deformation; however, we will assume here that F = I, so that F ≈ I + Grad u (2.130) This has a number of implications, so when we say we are working with “small strains:” (1) We assume that the material and spatial coordinates coincide (2) We describe deformation in terms of the displacement u and its gradient rather than the deformation mapping φ and its gradient. (3) We assume that u is small enough so that higher powers of u is neglected; every result should be linear in u or its gradient. We now proceed to linearize the other measures of deformation that we have introduced. One possibility for doing this is to apply (2.128) to every deformation measure that we have used; we would indeed obtain the correct result. An easier (though less general) procedure is to simply substitute (2.130) into our rexpressions, eliminate higher- order terms, and Taylor expand when necessary. Beginning with the right Cauchy Green tensor: C = FT F = (I + Grad u)T (I + Grad u) = I + Grad uT + Grad u + Grad uT Grad u = I + Grad u + Grad uT | {z } higher order term (2.131) T Note that we drop out the Grad u Grad u term because it is higher order in Grad u. Let us apply this to the Green-Lagrange strain tensor: 1 1 1 E = (C − I) = (I + Grad uT + Grad u − I) = (Grad uT + Grad u) = sym(Grad u) 2 2 2 (2.132) i.e., E is the symmetric part of Grad u. It turns out that the linearized Green-Lagrange strain tensor is a very convenient choice for measuring deformation in small strains, and so it is given a special symbol: 1 ε = (Grad u + Grad uT ) ≡ Small Strain Tensor 2 (2.133)

ε is symmetric, as we would expect. What about the antisymmetric part of Grad u? We deﬁne 1 r = (Grad u − Grad uT ) ≡ Inﬁnitesimal Rotation Tensor 2 (2.134)

The inﬁnitesimal rotation tensor gives an indication of the rotation at a point, although it breaks down under large deformation as does the small strain tensor. Note that ε + r = Grad u. This is actually the small-strain analog to the polar decomposition. For future decompositions of F (such as the elastic-plastic decomposition) we will see that the tensor-tensor multiplication in large strain boils down to a much friendlier additive decomposition in small strain.

2.9.1 Linearized metric changes

What happens to our metric change measures when we apply linearization? Let’s begin by considering the stretch. Recalling our formula for λ(n) where n is the direction of stretch, and substituting our linearized measures of deformation, we get √ q T T T p T T p T λ(n) = n Cn = n (I + Grad u + Grad u )m = n n + 2n ε n = 1 + 2n ε n (2.135)

And now we’re stuck, because we have a nonlinear function of ε. But we know that ε is small, so we can apply a linearization of the square root function about 1: 1 = 1 + (1 + 2nT ε n − 1) = 1 + nT ε n 2 (2.136)

Rearranging this expression, we can write

∆` T (n) = n ε n ≡ Engineering Strain (2.137) `0

Note that this implies that ε11 is the strain in the x1 direction, and so on; in other words, the diagonals of the small strain tensor describe the elongation in the x1, x2, x3 directions. Let us follow a similar procedure for angle change. Consider two orthogonal vectors m, n. Applying the formula developed earlier, we obtain: mT Cn cos θ(m, n) = ≈ mT Cn ≈ mT n + 2mT εn = 2mT εn λ(m)λ(n) (2.138)

Consider the angle subtended by the deformation of n – we’ll call it γ. n

θ m

To ﬁrst order, we can express γ as

cos θ = sin γ ≈ γ(m, n) (2.139)

Note that γ is symmetric regarding m, n; this would change for large deformation. Now we can express

T γ(m, n) = 2 m ε n ≡ Shear Angle (2.140)

We note that plugging in the basis vectors shows that the off-diagonals of ε determine the shear, just as they did for C. Finally, consider the volume change. Substuting our formula, we ﬁnd V 1 = det(F ) = ijk IJK FiI FjJ FkK V0 6 1 = (δ + u )(δ + u )(δ + u ) 6 ijk IJK iI i,I jJ j,J kK k,K 1 1 1 1 = (δ )(δ )(δ ) + (δ )(δ )(u ) + (δ )(u )(δ ) + (δ )(u )(u ) 6 ijk IJK iI jJ kK 6 ijk IJK iI jJ k,K 6 ijk IJK iI j,J kK 6 ijk IJK iI j,J k,K 1 1 1 1 + (u )(δ )(δ ) + (u )(δ )(u ) + (u )(u )(δ ) + (u )(u )(u ) 6 ijk IJK iI jJ kK 6 ijk IJK iI jJ k,K 6 ijk IJK iI j,J kK 6 ijk IJK iI j,J k,K 1 1 1 = 1 + (δ )(δ )(u ) + (δ )(u )(δ ) + (u )(δ )(δ ) 6 ijk IJK iI jJ k,K 6 ijk IJK iI j,J kK 6 ijk IJK iI jJ kK 1 = 1 + (u )(δ )(δ ) 2 ijk IJK iI jJ kK = 1 + i23I 23(uiI ) + 1i31I 3(uiI ) + 12i 12I (uiI )

= 1 + u1,1 + u2,2 + u3,3 = tr(Grad(u)) = 1 + tr(ε) (2.141) which can be expressed as ∆V = tr(ε) ≡ Percent Volume Change (2.142) V0

All of our measures of deformation can be expressed in terms of ε, so we see that in small strain, ε plays the same role as C. Example 2.6: Pure shear metric changes

Consider a square sample undergoing plane strain shear deformation in 2D.

X2 κ

1 γ θ

1 X1

Compute length, volume, and angle change in small and large deformation.

" # " # X1 + κX2 κX2 φ = X2 u(X ) = φ − X = 0 (2.143) X3 0 "1 κ 0# "0 κ 0# F = 0 1 0 grad u = 0 0 0 (2.144) 0 0 1 0 0 0 "1 κ 0# " 0 κ/2 0# 2 C = κ 1 + κ 0 ε = κ/2 0 0 (2.145) 0 0 1 0 0 0

(1) Change of length: in large and small strain we have q p ∆` λ(G ) = GT CG = C = 1 (G ) = GT εG = 0 1 1 1 11 ` 1 1 1 (2.146) p p ∆` λ(G ) = C = 1 + κ2 (G ) = GT εG = 0 2 22 ` 2 2 2 (2.147)

For G1 we see that both the large and small strain measures show no stretch, as expected. For G2, the large deformation theory shows an elongation. However, we note that κ must be small for the small strain to hold, in which case λ(G1) = 1, recovering the small strain prediction.

(2) Now we consider the change of angle. Note that sin γ(G1, G2) = cos θ(G1, G2). We have, in large and small strain, T G1 CG2 κ T sin γ(G1, G2) = = √ γ(G1, G2) = 2G1 εG2 = κ (2.148) λ(G1)λ(G2) 1 + κ2

If κ is small then the large strain recovers sin γ ≈ γ = κ, recovering the small strain approximation.

(3) Change of area: in large and small strain we have V ∆V = det(F) = 1 = tr(ε) = 0 (2.149) V0 V0

The large and small strain predictions are consistent.

2.9.2 Small strain compatibility

Consider the case of purely 2D strain. We know that 1 ε = u ε = ε = (u + u ) ε = u 11 1,1 12 21 2 1,2 2,1 22 2,2 (2.150) Let us consider the following equation: ε11,22 − 2ε12,12 + ε22,11 = u1,122 − u1,212 − u2,112 + u2,211 (2.151)

Because of the symmetry of differentiation, we know that this equals

= u1,122 − u1,122 − u2,211 + u2,211 = 0 (2.152) As with large deformation, a strain field ε is said to be compatible if there exists a displacement field u such that 1 ε = (u + u ) ij 2 i,j j,i (2.153) As before, we may only have a strain field without having a displacement field; verifying that (2.151) is equal to zero is a necessary and sufficient condition for 2D compatibility. In 3D, we have six such equations ε11,22 + ε22,11 − 2ε12,12 = 0 (2.154a) ε22,33 + ε33,22 − 2ε23,23 = 0 (2.154b) ε33,11 + ε11,33 − 2ε13,13 = 0 (2.154c) ε11,23 + ε23,11 − ε31,21 − ε12,31 = 0 (2.154d) ε22,31 + ε23,12 − ε31,22 − ε12,32 = 0 (2.154e) ε33,12 + ε23,13 − ε31,23 − ε12,33 = 0 (2.154f)

We can alternatively express the above equations compactly as

curl curl ε = 0 (2.155) which is necessary and sufﬁcient for compatibility.

2.10 The spatial/Eulerian picture

We can use the machinery that we’ve developed for linearization to do something analagous. Consider some time- dependent deformation mapping φ(X, t), and some measure on φ, we’ll call it G[φ]. We can express the rate of change of G as the following derivative ∂ ∂ G˙ = G[φ(X, t)] = (grad G) φ = (grad G)V = DG . ∂t ∂t V (2.156)

In other words, the time rate of change of G is nothing other than the Gateaux derivative of G in the direction of the velocity V. This means that the procedure for deriving rates of G is exactly the same as deriving linearizations of G. As such, we can dispense with the full derivation and jump directly to deﬁnition by analogy: Linearized kinematics Rates

Displacement ui Spatial velocity vi Displacement gradient βij = ui,j Spatial velocity gradient ìj = vi,j 1 1 Strain tensor εij = 2 (βij + βji ) Rate of strain tensor dij = 2 (ìj + `ji ) 1 1 Infinitesimal rotation tensor rij = 2 (βij − βji ) Spin tensor wij = 2 (ìj − `ji ) 1 1 Rotation vector αk = 2 rji ijk Vorticity vector ωk = 2 wji ijk Note that even though all of the above quantities are analagous to small strain deformation measures, we are no longer making the small strain approximation. We are merely showing that there are similarities between the large strain Eulerian rate quantities and the small strain linearized quantities. It is using these analogies that many connections are drawn between solid and fluid mechanics. As a side note, how do we relate these new large deformation measures to our familiar deformation measures such as F ? The following identity is invaluable in expressing v, `, d, w, etc., in terms of F :

˙ −1 ∂ ∂xi ∂XJ ∂ ∂xi ∂XJ ∂vi ∂XJ ∂vi FiJ FJj = = = = = vi,j = `ij (2.157) ∂t ∂XJ ∂xj ∂XJ ∂t ∂xj ∂XJ ∂xj ∂xj

Written symbolically, we have the relationship

˙ −1 ` = FF (2.158)

Let’s use the fact that we can draw analogies between small strain and rate deformation to derive the rate version of metric changes. Beginning with volume change, we expect that ∆V V˙ = tr(ε) =⇒ = tr(d) V V (2.159)

Let’s check to be sure. To do this, let us compute the derivative explicitly.

˙ d d(det F) ˙ −T ˙ ˙ −1 ˙ −1 J = det(F) = FiJ = (JFiJ )FiJ = J (FiJ FJi ) = J tr(FF ) = J tr(`) = J tr(d) (2.160) dt dFiJ

Or, rearranging, we obtain J˙ V˙ = = tr(d) J V (2.161)

as expected. As a side note, we use the fact that tr(d) = dii = `ii = vi,i = div(v) to obtain the useful identity ˙ J = J div(v) (2.162)

This will come in handy when deriving time-dependent balance laws. Having veriﬁed that the analogy holds, we can use our small-strain derivations to construct rates of metric changes: • Rate of stretching: for a vector n the rate of change of its length is given by analogy to be ∆` `˙ (n) = nT εn =⇒ = nT d n ` ` (2.163)

• Rate of angle change: for two vectors m, n that are instantaneously orthogonal to each other, the rate of change of shear angle is given by analogy to be T T γ(m, n) = 2m εn =⇒ γ˙ (m, n) = 2m d n (2.164)

Other rates of metric changes can be derived in a similar fashion.

3 Conservation Laws

We have spent a great deal of time describing and quantifying metrics for the deformation of materials under mappings φ(X) or time-dependent mappings φ(X, t). Everything that we have done so far is strictly geometric; that is, we have not imparted any physical laws or restrictions on φ. Now, we introduce the physical restrictions of the conservation of mass, conservation of linear/angular momentum, and conservation of energy. We will also introduce a rigorous formulation of the second law of thermodynamics in a continuum mechanics setting. Before we begin, let us introduce a couple of lemmas and theorems that will prove invaluable as we derive our conservation laws. n Lemma 3.1 (Fundamental lemma of the calculus of variations). Let Ω be a (Lebesgue measurable) subset of R , and f :Ω → Rm as vector (or scalar)-valued function over Ω. Then Z m f (X) · g(X)dX = 0 ∀g :Ω → R (3.1) Ω if and only if

f (X) = 0 (3.2) We call (3.1) the weak form and (3.2) the strong form of the equation f (X) = 0. A similar lemma is easily derived from the above: Lemma 3.2. Let Ω be a (Lebesgue measurable) subset of Rn, and f :Ω → Rm as vector (or scalar)-valued function over Ω. Then Z f (X) dX = 0 ∀B ⊂ Ω (Lebesgue measurable) (3.3) B if and only if f (X) = 0

We now introduce a very important theorem. Consider a body that is undergoing a time-dependent deformation. φ Ω φ(Ω, t)

x(t) X

Suppose there is a field defined over the body in the Eulerian configuration, we want to evaluate the time rate of change of an integral quantity evaluated over the deformed body. How do we do it? We will show how this is done for the general case by introducing the Reynold’s transport theorem: n 3 Theorem 3.1 (Reynold’s Transport Theorem). Let Ω be a (Lebesgue measurable) subset of R , let φ :Ω × R → R n be a time-dependent mapping, and let f : φ(Ω, t) × [t1, t2] → R be a spatial/Eulerian function. Then d Z Z df f dv = + f div(v) dv (3.4) dt φ(Ω,t) φ(Ω,t) dt

Proof. Let us begin by considering the left hand side: d Z f dv (3.5) dt φ(Ω,t)

We want to take the derivative inside the integral, but we can’t because the bound on the integral itself is time- dependent. To remedy this, we will perform a change of variables to evaluate the integral in the undeformed con- ﬁguration. Then we have

φ(Ω, t) → Ω dv = J dV (3.6)

Our integral thus becomes d Z Z d f J dV = (f J) dV (3.7) dt Ω Ω dt

We now evaluate the derivative using the product rule, and recalling (), we obtain Z Z Z df ˙ df df = J + f J dV = J + f J div(v) dV = + f div(v) J dV (3.8) Ω dt Ω dt Ω dt

Changing variables back, we obtain Z df = + f div(v) dv (3.9) φ(Ω,t) dt concluding the proof. With this machinery in hand, we will now proceed to derive balance laws.

3.1 Conservation of Mass

To develop conservation of mass, we must introduce two new quantities:

R(X, t) ≡ Lagrangian Mass Density ρ(x, t) ≡ Eulerian Mass Density (3.10)

The Lagrangian mass density is the mass per unit undeformed volume and is defined in the undeformed configuration, whereas the Eulerian mass density is the mass per unit deformed volume and is defined in the deformed configuration. We can relate the two in the following way: dm dm dm R = = = J = J ρ dV dv/J dv (3.11)

Furthermore, for a region B ⊂ Ω contained in the body, we can express the total mass in that body as Z Z m(B) = R dV = ρ dv (3.12) B φ(B,t)

Now, we introduce mass conservation by stating that for every B ⊂ Ω (measurable) and for all time,

d m(B) = 0 dt (3.13)

Pictorally, consider a body that is undergoing a deformation as follows: φ Ω φ(Ω, t)

B φ(B, t)

Conservation of mass simply states that the mass in some sub-region cannot change. Let us take this relationship and use it to derive some identities. In the Lagrangian frame, we have d Z Z dR 0 = RdV = dV = 0 (3.14) dt B B dt

Because this holds for all B ⊂ Ω, we have a strong-form relationship: dR = 0 ≡ Conservation of Mass (Lagrangian Frame) dt (3.15)

In the Eulerian frame, things are a little more complicated. Fortunately, we can use the Reynolds transport theorm to skip several steps: d Z Z dρ 0 = ρdv = + ρ div(v) dv = 0 (3.16) dt φ(B,t) φ(B,t) dt

Because this holds for all φ(B, t) ⊂ φ(Ω, t), we can again state conservation of mass in strong form: dρ + ρ div v = 0 dt (3.17)

Note that we are taking the total time derivative (i.e. the material derivative) of ρ: this is no problem in the Lagrangian frame but it is a little tricker in the Eulerian frame. Let’s write it out explicitly, using index notation and recalling the deﬁnition of the material derivative:

0 =ρ ˙ + ρ vi,i = ρ,t + ρ,i vi + ρ vi,i = ρ,t + (ρ vi ),i = 0 (3.18)

In symbolic notation, we recover ∂ρ + div(ρ v) = 0 ≡ Conservation of Mass (Eulerian Frame) ∂t (3.19)

3.1.1 Control volume

In fluid flow it is frequently convenient to use control volume analysis. Consider the motion of a material (e.g. a fluid) that is described by its velocity in the Eulerian frame, v(x, t). Also, consider a fixed volume V , such that the material can flow through it.

x 2 v(x, t)

x1 V

Let us consider the time rate of change of mass contained in V , noting that V does not depend on time and therefore time derivatives can be taken inside: d ∂ ∂ Z Z ∂ρ Z Z m(V , t) = m(V , t) = ρ dv = dv = − ρ div(v)dv = − ρ v · n da (3.20) dt ∂t ∂t V V ∂t cons. of mass V div. thm. ∂V

So we have the following expression for conservation of mass:

d Z Conservation of mass m(V , t) = − ρ v · n da ≡ (3.21) dt ∂V for a control volume where n is the outward-facing normal vector to the control volume. The intuition is that the total velocity flux through the boundary determines the rate of mass change in the volume, or that “change of mass equals mass flow in minus mass flow out.”

3.2 Conservation of linear momentum

Next we will introduce the conservation of linear momentum for a continuous body. However, before we can do this, it is necessary to introduce the notion of forces.

3.2.1 Forces, tractions, and stress tensors

Consider a body that is undergoing deformation as shown in the following ﬁgure.

X3 x3

∆A ∆f ∆a ∆f N n X2 x2

X1 x1

Intuitively we know that the deformation is effected by forces acting on it. In this case, we see that a force ∆f is acting on the face with normal vector n in the undeformed configuration. From undergraduate mechanics of materials, we know that the magnitude of the force is not generally very interesting; rather, it is the force per unit area that determines material response. But now we have a question: which area? We must allow for the possibility (actually, the probability) that the area will change with deformation. So, as with all of the other quantities we have defined so far, we will define a material and a spatial version of this quantity.

df dfi Material df dfi Spatial T(N) = T (N) = ≡ t(n) = t (n) = ≡ dA i dA Traction Vector da i da Traction Vector (3.22)

We pause here to make a couple of important notes about these quantities: (1) These quantities are “forces per areas,” so why don’t we call them stresses? They are indeed very similar, and we will use them to define stress in a moment, but stress is naturally a tensorial quantity, not a vector quantity. (2) What is going on with the indices – specifically, why do we use lowercase instead of uppercase? The reason is that (in general) we will assume that forces live in the deformed configuration. As a result, forces will be indexed with lowercase indices. This is not always the case, and there are many instances where we map forces from the deformed configuration back to the undeformed. In these cases, forces would be indexed with uppercase variables. However, for current purposes, we will always think of them as being deformed. Note that the traction vectors are written as somewhat mysterious functions of a unit normal vector; that is, we feed them a normal vector to a surface and they give us the force acting on that surface. What other kind of construct do we know of that turns a vector into another vector: tensors. Therefore it stands to reason that we should be able to find a tensor representation of the traction vectors. To do this, consider the following figure:

∆A1 N ∆A G2 ∆A2

∆A3 G1

We begin by noting that we can describe the areas in the following way in terms of ∆A

∆AI = (GI · N)∆A = NI ∆A (3.23)

Neglecting volumetric forces (reasonable to do as long as the volume is small, since volume forces scale with the volume and therefore decrease much more quickly), we can write

T(N)∆A = T(GI )∆AI = T(GI )(GI · N)∆A = T(GI )NI ∆A (3.24)

Or, noting that this holds for any ∆A, we have

T(N) = T(GI )NI = PN (3.25) where we deﬁne

P = [T(G1) T(G2) T(G3)] ≡ The Piola-Kirchoff Stress Tensor (3.26)

The Piola-Kirchoff tensor allows us to express the tractions (that are a function of the unit normal vector in the undeformed conﬁguration) in matrix form:

T(N) = PN Ti = PiJ NJ (3.27)

Note that, like the deformation gradient, the Piola-Kirchoff tensor is a “two point” tensor. Unlike with F , the implication is fairly straight forward in that it is the forces in the deformed configuration per unit area in the undeformed configuration. Additionally we note that it turns a normal vector in the undeformed configuration to a force vector in the deformed configuration. We can think of P as the tensor version of engineering stress. Now, let us consider forces in the deformed configuration per unit area in the deformed configuration. Recall that −T we defined the Piola transform: n∆a = JF N∆A Multiplying (3.27) by ∆A we have 1 T(N)∆A = PN∆A = PFT n ∆a = σ n ∆a = t(n)∆a J (3.28) where we define 1 σ = PF T ≡ Cauchy Stress Tensor J (3.29) and note that

t(n) = σ n ti = σij nj (3.30)

Note that σ lives entirely in the deformed conﬁguration. σ is the tensor analog to true stress. P and σ are tensors, so what is the physical signiﬁcance of their components? The following picture is illustrative in demonstrating the concept of a stress tensor:

X3 x3 P33 σ P23 33 P σ 13 P32 23 σ13 σ P31 32 P22 σ31 σ22 P21 P σ21 12 σ12 X2 σ x2 P11 11

X1 x1

We can think of colum n of the stress tensor as the force vector acting on the n-facing face. Alternatively we can think of the ij-th component of the tensor as the ith component of force acting in the j direction. Let us introduce one additional type of force: body forces.

df dfi Material df dfi Spatial B = B = ≡ b = b = ≡ dM i dM Body Force dm i dm Body Force (3.31)

The material body force is the force per unit undeformed mass, and the spatial is the force per unit deformed mass. Common examples of a body force include gravity and electromagnetic forces.

3.2.2 Balance laws

Recall that momentum is mass times velocity. What is momentum for a body with variable density R(X, t), ρ(x, t) and variable velocity V(X, t), v(x, t)? We can write it in integral form in either the material or spatial frames as Z Z L(B, t) = R V dV = ρ v dv (3.32) B φ(B,t)

We can express Newton’s second law, the conservation of linear momentum, for all B ⊂ Ω as d L(B, t) = F (B, t) dt tot (3.33) where Ftot is the resultant of all forces acting on or in B ⊂ Ω, Z Z Z Z Ftot (B, t) = T (N)dA + R B dV = t(n)da + ρ b dv (3.34) ∂B B φ(∂B,t) B

Let us begin by considering the conservation of linear momentum in the Lagrangian frame only: we have d Z Z Z R V dV = T (N)dA + R B dV (3.35) dt B ∂B B

What do we do with the partial derivative? We can take it right inside the integral. 0 Z Z Z Z d d ˙¡ ˙ R V dV = (R V) dV = (R¡ V + R V) dV = R A dV (3.36) dt B B dt B B

Now, let’s consider the right hand side. In particular, let’s consider the surface traction term. Because we can write the surface traction term as a matrix-vector multiplication with the Piola-Kirchoff stress tensor, we can apply the divergence theorem to get Z Z Z T(N)dA = PNdA = Div(P)dV (3.37) ∂B ∂B B

Note that we now have three volume integrals. We can use this to collect terms and write our conservation equation as Z [R A − Div(P) − R B] dV = 0 (3.38) B

Because this must hold true for all B ⊂ Ω, applying the fundamental lemma of the Calculus of Variations, we can write the above in strong form: Div(P) + RB = RA ≡ Conservation of Linear Momentum (Lagrangian Frame) PiJ,J + RBi = RAi (3.39)

Now, let us consider the Eulerian frame: d Z Z Z ρ v dv = t(n)da + ρ b dv (3.40) dt φ(B,t) φ(∂B,t) φ(B,t)

As before, let us consider the left hand quantity ﬁrst. In a similar procedure to that which we followed with conservation of mass, we apply the Reynold’s Transport Theorem to take the derivative inside the integral: d Z Z h d i Z ρ v dv = (ρ v) + ρ v div(v) dv = [ρv˙ +ρ ˙ v + ρ v div(v)]dv (3.41) dt φ(B,t) φ(B,t) dt φ(B,t) Z 0 Z : = [ρ v˙ + v(ρ˙ +ρdiv(v)) ]dv = ρ a dv (3.42) φ(B,t) | {z } φ(B,t) conservation of mass Now, let us consider the surface tractions term on the right hand side. Again, we can apply the divergence theorem to obtain Z Z Z t(n)da = σ n da = div(σ) dv (3.43) φ(∂B,t) φ(∂B,t) φ(B,t)

Combining all of the terms, which are now in integral form, we have Z [ρa − div(σ) − ρb] dv = 0 (3.44) φ(B,t)

By the fundamental lemma, we can write the above in strong form as div(σ) + ρb = ρa ≡ Conservation of Linear Momentum (Eulerian Frame) σij,j + ρ bi = ρai (3.45)

3.2.3 Navier-Stokes momentum equations

The above form for linear momentum balance is convenient for solid mechanics, but it can easily be rewritten in a number of ways. A particularly popluar method (especially when doing ﬂuid mechanics) is to start by writing the time derivative in a different way. Let us return to Equation (3.41): using index notation, d Z Z h d i Z h ∂ ∂ i ρvi dv = (ρvi ) + ρvi vj,j dv = (ρ vi ) + (ρvi )vj + ρ vi vj,j dv (3.46) dt φ(B,t) φ(B,t) dt φ(B,t) ∂t ∂xj Z h ∂ ∂ i = (ρ vi ) + (ρvi vj ) dv (3.47) φ(B,t) ∂t ∂xj

In symbolic form, this can be expressed as Z h ∂ i (ρ v) + div(ρv ⊗ v) dv (3.48) φ(B,t) ∂t

The Cauchy stress tensor is frequently decomposed into two components: a hydrostatic pressure component and a deviatoric stress component,

σ = −pI + τ (3.49) where p is the scalar hydrostatic pressure, and is negative because pressure is compressive. (Note that the sign is frequently reversed in ﬂuid mechanics formulations.) Then the stress divergence term reduces to Z h i − div(pI) + div(τ ) dV (3.50) φ(B,t)

Finally, make the assumption that the only body force is gravity, so the body force term reduces to Z ρ gdV (3.51) φ(B,t)

Making all of these substitutions and rearranging a bit, the balance of linear momentum can be expressed in the following form: ∂ (ρv) + div(ρv ⊗ v + pI) = div(τ ) + ρ g ≡ Navier-Stokes Momentum Equation ∂t (3.52)

Even though the equation looks quite different, the mechanics are still the same. The biggest simpliﬁcation made when writing the Navier-Stokes equations is in decoupling the stress tensor. Solid bodies are capable of sustaining stress states that are much more complex than that sustained by ﬂuids, so we will prefer to leave σ in its non- decoupled form.

3.3 Conservation of angular momentum

Angular momentum for a body B ⊂ Ω is given y Z Z G = x × RV dV = x × ρv dv (3.53) B φ(B,t) and the moments generated from the applied body forces and surface tractions are Z Z Z Z M = x × T(N) dA + x × R B dv = x × t(n) da + x × ρ b dv (3.54) ∂B B φ(∂B,t) φ(B,t)

We note the important fact that we use the spatial position x for both the Lagrangian and Eulerian formulation. This is a result of the fact that our forces live in the deformed configuration, so the moments generated by them must also live in the deformed configuration. Having defined angular momentum and moments for a continuous body, we can now express the conservation of angular momentum as d G(B, t) = M(B, t) dt (3.55)

Let us begin with the Lagrangian frame. We start by simplifying the time derivative, which is easy here as it is taken right inside the integral. Using the product rule, noting that V×V = 0, and applying conservation of mass, we arrive at the expression 0 Z Z Z d ˙¡ ˙ x × RVdV = [( x˙ × RV) + (x × R¡ V) + (x × RV)]dV = x × RAdV (3.56) dt B B =V B

As with linear momentum balance, we want to attempt to write the surface integral in volume form. The presence of the cross product makes the application of the divergence theorem tricky, so we will switch to index notation here. Applying the divergence theorem and the product rule, and applying the previously obtained conservation of momentum equation, we have Z Z Z Z [x × T(N)]k dA = [x × PN]k dA = ijk xi PjP NP dA = (ijk xi PjP ),P dV (3.57) ∂B ∂B ∂B B Z Z = (ijk xi,P PjP + ijk xi PjP,P ) dV = (ijk FiP PjP + [x × (RA − RB)]k ) dV (3.58) B |{z} | {z } B FiP [x×Div P]K

Combining the above we have Z Z h Z [x × RA]k dV = ijk FiP PjP + [x × (RA − RB)]k dv + [x × R B]k dV (3.59) B B B | {z } | {z } | {z } momentum derivative surface tractions body force

Almost everything cancels except for one term. Because this holds ∀B ⊂ Ω, we can write the remaining term in strong form: T ijk FiP PPj = 0 (3.60)

We can now pull an index notation trick to simplify the above even further:

T 1 T T 1 T T 0 = ijk FiP PPj = (ijk FiP PPj + ijk FiP PPj ) = (ijk FiP PPj − jik FiP PPj ) (3.61) 2 swap indices 2 1 T T 1 T T = (ijk FiP PPj − ijk FjP PPi ) = ijk (FiP PPj − PiP FPj ) = 0 (3.62) rename 2 2 We see that this implies

T T FP = PF ≡ Conservation of Angular Momentum (Lagrangian Frame) (3.63) Now let us consider the spatial frame. As before, we begin by using the transport theorem to simplify the time derivative term. We also apply the product rule and conservation of mass, arriving at: d Z Z h d i x × ρv dv = (x × ρv) + (x × ρv) div(v) dv (3.64) dt φ(B,t) φ(B,t) dt Z h i = (x˙ × ρv) + (x × ρ˙v) + (x × ρv˙) + (x × ρv) div(v) dv (3.65) φ(B,t) Z h 0 :0i : = ( x˙× ρv) + (x × ρv˙) + (x × v)[ρ˙ +ρ div(v)] dv (3.66) φ(B,t) =v | {z } Mass conservation Z = x × ρa dv (3.67) φ(B,t)

Now let’s look at the applied moment terms. In particular, we want to try to apply the divergence theorem to convert the surface term to a volume term: Again, we’ll ﬁnd that it is much easier to use index notation here. h Z i h Z i Z Z ∂ x × t(n) da = x × σn da = ijk xi σjpnp da = (ijk xi σjp) dv (3.68) φ(∂B,t) k φ(∂B,t) k φ(∂B,t) φ(B,t) ∂xp Z Z = (ijk xi,p σjp + ijk xi σjp,p) dv = (ijk σji + [x × div σ]k ) dv (3.69) φ(B,t) |{z} φ(B,t) δip Z = (ijk σji + [x × (ρa − ρb)]k ) dv (3.70) φ(B,t)

Combining all of the terms in the conservation equation, we obtain Z Z Z [x × ρa]k dv = (ijk σji + [x × (ρa − ρb)]k )dv + [x × ρb]k dv (3.71) φ(B,t) φ(B,t) φ(B,t) | {z } | {z } | {z } momentum rate surface tractions body forces

Now we see that almost everything cancels except for one term: Z ijk σji dv = 0 (3.72) φ(B,t)

Because this holds for all B, we can write it locally as

ijk σji = 0 (3.73) Using a procedure identical to the above, we arrive at the following expression for conservation of angular momentum: T σ = σ ≡ Conservation of Angular Momentum (Eulerian Frame) (3.74) Recalling the relationship between P and σ, we note that this is consistent with the Eulerian frame.

3.4 Conservation of energy

We have established continuum mechanical formulations of the conservation of mass, linear momentum, and angular momentum. The final conservation law is the conservation of energy. Energy conservation is interesting because it is where we begin to make connections with other mechanisms such as mass or thermal transport. In fact, it is the great strength of energy conservation that almost all mechanisms have an energetic formulation that allows the coupling of these phenomena together. Before we begin, let us remind ourselves of a couple of important thermodynamical definitions: Definition 3.1. An extensive quantity for a system is a quantity that is equal to the sum of the quantities of all subsystems. (Defined in integral form – a “nonlocal” quantity.) Definition 3.2. An intensive quantity for a system is a quantity whose value is independent of the subsystem containing it. (Defined pointwise – a “local” quantity.)

3.4.1 Energetic quantities

To describe our energetic conservation laws, we need to make some deﬁnitions. Some are familiar, while some require the deﬁnition of new quantities. • Kinetic energy: kinetic energy for a body Ω is an extensive quantity, given by an integral over the entire volume. 1 2 We recall that the kinetic energy for a particle is 2 mv . For a body with variable density and with variable internal speed, the total kinetic energy is Z Z 1 2 1 2 K(Ω) = R|V| dV = ρ|v| dv ≡ Kinetic Energy (3.75) 2 Ω 2 φ(Ω)

(We note that V, v are mean velocities, not velocities due to thermal fluctuations. We can do this because we are working at the continuum level and we will account for thermal fluctuations with temperature. Note, this is a little trickier with turbulent fluids, because fluctuations on the continuum level transfer down to the microscopic level; this is called the energy cascade.) • Heat: we have not dealt with heat yet, so in order to quantify it in a continuum setting we need to introduce a ˙ couple of new variables. Q(Ω) denotes the total heat of the body, with Q representing the rate of change of heat. We define Sn(X), sn(x) to be the heat generated in the material per unit undeformed mass and per unit deformed mass, respectively. Finally, we let H, h be the heat flux vectors per unit deformed area and per unit undeformed area, respectively. With these definitions, we can write a simple heat balance law for our body in both the Lagrangian and Eulerian configurations: Z Z Z Z ˙ Q(Ω) = RSn dV − H · N dA = ρ sn dv − h · n da (3.76) Ω ∂Ω φ(Ω) φ(∂Ω) | {z } | {z } internal generation outward flux

We note that the divergence theorem allows us also to write Z Z ˙ Q(Ω) = [RSn − Div(H)] dV = [ρ sn − div(h)]dv (3.77) Ω φ(Ω)

This is not of much use to us yet, because we have no relationship between these variables and the variables that we are using for everything else (P, F , etc.). However it will be useful when we construct our energy balance law.

• External power: we deﬁne the external power to be the rate of work done by all external forces. Remember that work done by a force is f · ∆x, so the power exerted by a force is f · v. Therefore, to capture the power of the applied body forces and surface tractions, we dot them with the velocity of the material at the point of application, then integrate over the body: Z Z Z Z E P (Ω) = RB · VdV + T(N) · VdA = ρb · vdv + t(n) · vda (3.78) Ω ∂Ω φ(Ω) φ(∂Ω) | {z } | {z } power of body forces power of surface tractions

• Deformation power: the deformation power is the portion of the external power that goes towards deforming the material, rather than going towards changing the kinetic energy. For instance, if you throw a football, your applied surface tractions cause the football to start moving; on the other hand, if you squeeze the football, the kinetic energy doesn’t change (much) so all of your energy goes into deforming the ball. This quantity is actually deﬁned rather easily as the difference between external power and change in kinetic energy: PD (Ω) = PE (Ω) − K˙ (Ω) Z Z 1 d Z = RBi Vi dV + Vi PiJ NJ dA − RVi Vi dV (3.79) Ω ∂Ω 2 dt Ω 0 Z Z Z conservation of mass 1 ˙¡ = RBi Vi dV + (Vi PiJ ),J dV − [RV¡ i Vi + 2RVi Ai ]dV Ω Ω 2 Ω 0 Z h i Z h :i = RB V + V P + V P − RV A dV = V P + V (RB +P− RA ) dV i i i,J iJ i iJ,J i i i,J iJ i i iJ,J i Ω Ω |{z} | {z } ˙ =FiJ conservation of momentum Z Z ˙ ˙ = FiJ PiJ dV = F · P dV (3.80) Ω Ω

(Note that the · symbol between two tensors is used to indicate a termwise product, so that the result is a scalar.) Now let us, as usual, do the same thing in the Eulerian conﬁguration: Z Z Z D 1 d P (Ω) = ρbi vi dv + vi σij nj da − ρvi vi dv φ(Ω) φ(∂Ω) 2 dt φ(Ω) Z Z 1 Z : 0 = ρb v dv + (v σ ) dv − (2ρv a + ρ˙v v +ρvv v )dv i i i ij ,j i i i i i i j,j φ(Ω) φ(Ω) 2 φ(Ω) | {z } conservation of mass Z :0 = [ρbi vi−ρvi ai + vi σij,j + vi,j σij ]dv φ(Ω) | {z } |{z} conservation of momentum =dij Z Z = dij σij dv = d · σ dv (3.81) φ(Ω) φ(Ω)

where we recall that d was the rate of strain tensor. So, we can now write Z Z D ˙ P (Ω) = F · P dV = d · σ dv ≡ Deformation Power (3.82) Ω φ(Ω)

• Internal Energy: let us deﬁne the intensive variables U(X) and u(x), the internal energies per unit undeformed mass and deformed mass, respectively. Then the internal energy of a body Ω is given by Z Z E(Ω) = R U dV = ρ u dv ≡ Internal Energy (3.83) Ω φ(Ω)

3.4.2 Balance laws

We are now in a position to write the ﬁrst law of thermodynamics in a continuum mechanics setting. We will state it as:

[change in internal energy] + [change in kinetic energy] = [external power] + [heat ﬂow & generation] (3.84)

Using what we have developed above, we write

˙ ˙ E ˙ E(Ω) + K(Ω) = P (Ω) + Q(Ω) ≡ Conservation of Energy (3.85) for all Ω. Alternatively, using our deﬁnition of deformation power, we can write ˙ D ˙ E(Ω) = P (Ω) + Q(Ω) (3.86)

D Since we have a nice and compact form for P , the above expression is a little more useful. We already have D ˙ ˙ expressions for P and Q in Lagrangian and Eulerian forms, now we just need to evaluate E. Hopefully this process is starting to seem fairly familiar: in the Lagrangian form, 0 Z Z Z ˙ d ˙¡ ˙ ˙ E(Ω) = R U dV = [RU¡ + R U]dV = R U dV (3.87) dt Ω Ω Ω

In the Eulerian form d Z Z : 0 Z E˙ (Ω) = ρ u dv = [ρu˙ + ρ˙ u +ρu div(v)] dv = ρ u˙ dv (3.88) dt φ(Ω) φ(Ω) | {z } φ(Ω) conservation of mass

Now, putting it all together, we have in the material/Lagrangian and spatial/Eulerian frames: Z Z Z ˙ ˙ R U dV = F · P dV + [RSn − Div(H)] dV (3.89) Ω Ω Ω Z Z Z ρ u˙ dv = d · σ dv + [ρ sn − div(h)]dv (3.90) φ(Ω) φ(Ω) φ(Ω)

By the fundamental lemma, since the above holds for all subbodies Ω, we can ﬁnally write the above locally as

R U˙ = F˙ · P + RSn − Div(H) ≡ Conservation of Energy (Lagrangian Frame) (3.91) R U˙ = F˙iJ PiJ + RSn − HK,K

ρ u˙ = d · σ + ρ s − div(h) n ≡ Conservation of Energy (Eulerian Frame) ρ u˙ = dij σij + ρ sn − hk,k (3.92)

3.4.3 Power-conjugate pairs

As a side note (and sanity check), let’s see if we can relate these two sets of power conjugate pairs. We begin by recalling the deﬁnitions of the rate of strain and Cauchy stress tensors: 1 1 1 d = (` + ` ) = (F˙ F −1 + F˙ F −1) σ = P F T ij 2 ij ji 2 iK Kj jK Ki ij J iK Kj (3.93)

Now, let us substitute these deﬁnitions into the integrand: 1 d σ = (F˙ F −1P F T + F˙ F −1P F T ) ij ij 2J iK Kj iL Lj jK Ki iL Lj (3.94)

T T We recall from the conservation of angular momentum that PiLFLj = FiLPLj ; substituting we get

1 1 1 = (F˙ P F −1F +F˙ F −1F PT ) = (F˙ P + F˙ P ) = F˙ P 2J iK iL Kj jL jK Ki iL Lj 2J iK iK jK jK J iJ iJ (3.95) | {z } | {z } δKL δKL

So we see that we recover what we have already shown, namely: ˙ FiJ PiJ dV = dij σij J dV = dij σij dv (3.96)

˙ We have seen that the pairs (F , P) and (d, σ) have a power-conjugate relationship; that is, if we “dot” them together and integrate over the volume, we obtain a measure of the deformation power in the material. Let us deﬁne S = F−1P = J F−1 σ F−T −1 −1 −T ≡ The Second Piola-Kirchoff Stress Tensor (3.97) SIJ = FIk PkJ = JFIk σkpFpJ

Notice that S has only uppercase indices. We have worked a lot with σ (deformed forces per deformed area) and P (deformed forces per undeformed area); now, S completes the picture as the undeformed forces per undeformed area. Why did we introduce S? We are looking for something that power-conjugate to C, and now we are going to suppose 1 ˙ 1 that S is conjugate to 2 C, and therfore also to E = 2 (C − I). To show this, we begin by computing the derivative of C. By the product rule, ˙ ˙ T T ˙ C = F F + F F (3.98) ˙ Now, to show that S and E are power-conjugate, we will evaluate their product directly 1 1 1 E˙ S = C˙ S = (F −1P )(F˙ T F + F T F˙ ) = (F −T F˙ T F PT + F −T F T F˙ PT ) IJ IJ 2 IJ IJ 2 Ip pJ Ik kJ Ik kJ 2 pI Ik kJ Jp pI Ik kJ Jp (3.99) | {z } | {z } T δ PkJ FJp pk 1 1 1 = (F˙ T P F T F −T +F˙ PT ) = (F˙ T P + F˙ PT ) = (F˙ P + F˙ P ) = F˙ P 2 Ik kJ Jp pI kJ Jk 2 Ik kI kJ Jk 2 kI kI kJ kJ kI kI X (3.100) | {z } δJI

To summarize, here are the pairs that we have found so far: Rate of Deformation Measure Stress Measure

Strain rate tensor d Cauchy stress tensor (def. force / def. area) σ ˙ st Deformation gradient F 1 PK stress tensor (undef. force / def. area) P ˙ 1 ˙ nd Cauchy-Green / Green Lagrange E, 2 C 2 PK stress tensor (undef. force / undef. area) S

3.5 Second law of thermodynamics

The second law of thermodynamics is probably the most confusing in mechanics, because (i) it is an inequality rather than an equality, and (ii) it is coupled with the notion of entropy, one of the most unintuitive quantities in physics. As with the other balance laws, we will develop a continuum mechanics version of the second law. But, before we do this, let’s talk a little bit about entropy. To do that, we will introduce a little bit of statistical mechanics.

3.5.1 Introduction to statistical thermodynamics and entropy

Let us consider a box full of, say, 10 particles each of mass m as shown in the following figure. Consider the box of particles in three different scenarios: first, where only one particle is moving at a speed v0, then with four particles moving at speed v0/2, and finally with nine of the particles moving with speed v0/3.

v0 |v| = 3 v0 |v| = 2

|v| = v0

What is the energy of the set of particles? Let’s compute: v 2 v 2 E = 1 × m(v )2 = mv 2 E = 4 × m 0 = mv 2 = E E = 9 × m 0 = mv 2 = E 1 0 0 2 2 0 1 3 3 0 1 (3.101)

Notice that even though the configuration of the particles in each case was completely different, the total amount of energy remained the same. This illustrates a fundamental concept in statistical thermodynamics: macrostates and microstates. Definition 3.3. A particular configuration of each particle in a system is called a microstate Definition 3.4. A macroscopic property of the system is called a macrostate

What are some other possible macrostates of our above system? Possibilities are the total number of particles in the box, or the total mass of the system. A key idea in statistical thermodynamics is determining the number of microstates associated with a given macrostate.

Let us begin by considering two systems, (1) and (2).

System 1: N1, V1, E1 System 2: N2, V2, E2

We will impose the following constraints: system 1 has N1 atoms, volume V1 and total energy E1; system 2 has N2 atoms, volume V2 and a total energy E2.(Sidenote: that is, we are considering the microcanonical or NVE ensemble.) Let us deﬁne the following:

Ω(E) ≡ # of microstates with energy E ≡ Microcanonical Partition Function (3.102)

That is, Ω is a number that tells us how many different ways our system can have total energy E. (Note: this is a slight abuse of notation; Ω has been used previously to denote the body. Ω will be used this way in this section only, unless otherwise explicitly indicated.) So, for our two systems we have Ω(E1), Ω(E2). Now, let us suppose that we are considering the total number of combined configurations. Given that system (1) has Ω1(E1) possible configurations and system (2) has Ω2(E2) possible configurations, then the total number of possible configurations of the combined system must be

Ω(E1, E2) = Ω1(E1)Ω2(E2) (3.103)

(For instance, if system (1) has two possible conﬁgurations and system (2) has three, then the total number of combined conﬁgurations would be six.) Now, let us assume that our system is in equilibrium, so that dΩ = 0. Then, from the form above, we have

∂Ω1 ∂Ω2 dΩ = Ω2 dE1 + Ω1 dE2 = 0 (3.104) ∂E1 ∂E2

Now, we have constrained our system so that no energy is allowed to leave. As a result, we know that

dE1 + dE2 = 0 (3.105)

Substituting this into our relationship above, we get

dΩ1 dΩ2 Ω2 dE1 − Ω1 dE1 = 0 (3.106) dE1 dE2

Or, rearranging a bit, we get

1 dΩ1 1 dΩ2 = (3.107) Ω1 dE1 Ω2 dE2

We’ll pull a calculus trick here, and rewrite the equation above as d d log Ω1 = log Ω2 (3.108) dE1 dE2

Let us deﬁne β to be d d β1 = log Ω1 β2 = log Ω2 (3.109) dE1 dE2

(Again, we restrict our overloading of the term β to this section only.) This allows us to write our equilibrium equation as β1 = β2. Now, we notice from the deﬁnition that 1 βdE = d log Ω dE = d log Ω or β (3.110)

Recall that both of our systems have fixed volume and a fixed number of atoms. As a result, we know that the first law of thermodynamics is given by dE = dQ; in other words, all changes in internal energy must be the result of heat transfer. We also know that dQ = TdS. This means we can write 1 d log Ω = TdS β (3.111)

1 Here, we identify that β ∼ T and d log Ω ∼ dS. In fact, for classical reasons, a constant kB (Boltzmann’s constant) is introduced, giving us 1 β = ≡ Reciprocal Temperature S = kB log Ω ≡ Entropy (3.112) kB T where β is the statistical thermodynamical temperature and S is the total Gibbs entropy of the system. There are some important notes to be made here: (1) If S1 = kB log Ω1 and S2 = kB log Ω2, then the combined entropy is S = kB log(Ω1Ω2) = kB log Ω1 + kB log Ω2 = S1 + S2, verifying that entropy is an extensive property of the system. (2) Entropy is interpreted as the number of possible microstates for our system to maintain its current macrostate. This is consistent with the interpretation of entropy as being a “measure of disorder.” For instance, if a coffee cup has fewer possible “microstates” if its macrostate is “being all in one piece” than if its macrostate is “shattered in pieces on the ﬂoor.” If you don’t feel comfortable with the things discussed in this section, that’s ok. The purpose of this section is just to hint at where entropy comes from, in the hope that it will help provide a little bit of intuition. At this point, we will switch back to continuum mode, where we will work with entropy in a continuum setting.

3.5.2 Internal entropy generation

In the previous section we concluded that dQ Q˙ dS = S˙ = T or T (3.113) In other words, this shows that the rate of increase of entropy is correlated to the rate of heat ﬂowing into the system. This is true when the process happens quasistatically and all entropy change in the system is due to external contributions. In this case, we say that the process is reversible, because we can return to the original entropy state by simply removing the previously added heat. For irreversible processes, we allow for the possibility of the system to generate entropy on its own. Thus, we say that we have Q˙ S˙ = S˙ int + T (3.114) |{z} |{z} |{z} total entropy change internal entropy generation entropy generated externally

As an example, let us consider an adiabatic system (no heat transfer in or out of the system) divided into two regions with two seperate temperatures as follows:

T1 T2

Q˙ 12

˙ Let us suppose that there is heat flowing from one region to the other at a rate Q12. Now, let us look at the rate of change of entropy in the total system, keeping in mind that S = S1 + S2: 0 Q˙ ¡ Q˙ Q˙ 1 1 ˙ ˙ int ¡ int ˙ ˙ 12 12 ˙ ˙ int S = S + = S = S1 + S2 = − + = Q12 − = S (3.115) ¡T T1 T2 T2 T1 adiabatic ˙ int Notice that S is definitely nonzero. We know from observation that heat does not flow from cold to hot, so ˙ int because T2 < T1 we conclude that S > 0. This leads us to the formal statement of the second law:

Q˙ S˙ int = S˙ − ≥ 0 ≡ Second Law of Thermodynamics T (3.116)

3.5.3 Continuum formulation

−1 As with the other balance laws, we need to introduce a few more terms. Let T (X), Θ(x) = T (φ (x)) be the −1 temperatures in the undeformed and deformed conﬁgurations, let N(X), η(x) = N(φ (x)) be the entropy per unit undeformed mass and per unit deformed mass. Then we write the entropy of the system as Z Z S(Ω) = R N dV = ρ η dv (3.117) Ω φ(Ω)

Then the rate of change of the entropy of the system is: Z Z Z ˙ d d ˙ S(Ω) = R N dV = (RN) dV = R N dV (3.118) dt Ω Ω dt Ω d Z Z Z = ρ η dv = [ρ η˙ +ρ ˙ η + ρ η div v] dv = ρ η˙ dv (3.119) dt φ(Ω) φ(Ω) | {z } φ(Ω) =0 (cons. of mass) ˙ Recalling the deﬁnition of Q, we let the contribution of entropy from external sources be Z Z Z Z ˙ ext RSn H · N ρ sn h · n S (Ω) = dV − dA = dv − da (3.120) Ω T ∂Ω T φ(Ω) Θ φ(∂Ω) Θ

Combining all of the above terms with the statement of the second law, we have ˙ Z Z Z ˙ Q RSn H · N S − = R N dV − dV + dA ≥ 0 (3.121) T Ω Ω T ∂Ω T Z Z Z ρ sn h · n = ρ η dv − dv + da ≥ 0 (3.122) φ(Ω) φ(Ω) Θ φ(∂Ω) Θ

Applying the divergence theorem and using the fundamental lemma, we obtain the local expression of the second law in the Lagrangian and Eulerian frames: RS H R N˙ − n + Div ≥ 0 ≡ Clausius-Duhem Inequality (Lagrangian Frame) T T (3.123) ρs h ρη˙ − n + div ≥ 0 ≡ Clausius-Duhem Inequality (Eulerian Frame) Θ Θ (3.124)

3.6 Review and summary

In this section, we used the kinematic framework that had been developed in the previous section to describe the familiar laws of the conservation of mass, momentum, and energy in a continuum setting. We began by describing these laws in bulk (integral) form, and then showed by means of the Reynolds transport and divergence theorems that these balance laws can be expressed locally as differential equations. We found local versions for both the Lagrangian and Eulerian frames. The following table summarizes the results:

Balance Law Lagrangian Form Eulerian Form ∂ρ R˙ = 0 + div(ρ v) = 0 Conservation of mass ∂t Conservation of linear momentum Div(P) + R B = R A div(σ) + ρ b = ρ a T T T Conservation of angular momentum PF = FP σ = σ ˙ ˙ Conservation of energy R U = F · P + RSn − Div(H) ρ u˙ = d · σ + ρ sn − div(h) RS H ρ s h R N˙ − n + Div ≥ 0 ρ η˙ − n + div ≥ 0 Second law of thermodynamics T T Θ Θ

If we want to solve the system, how many unknowns and how many equations do we have?

Givens: Body forces B, b Internal heat generation Sn, sn Unknowns: Deformation mapping + spatial derivatives (φ, F) – 12 unknowns Density (R, ρ) – 1 unknown Internal energy (U, u) – 1 unknown Heat ﬂux (H, h) – 3 unknowns Equations: Conservation of mass – 1 equation Conservation of linear momentum – 3 equations Conservation of angular momentum – 3 equations (nine total, but six are redundant) Conservation of energy – 1 equation

We have a total of 17 unknowns but only 8 equations, so we need an additional 9 equations to close the system. These additional equations are the constitutive relationship of the material, and may take the form

P = P(F) σ = σ(ε) (3.125) H = H(T ) h = h(T ) (3.126)

This is a total of 9+3=12 equations, which is more than we need. However, it turns out that natural restrictions on P(F ) reduce the number of actual equations to six, so that the system is not overconstrained. The thermal relationship is usually H(T ) = −k Grad T , Fourier’s law of heat conduction, where k is the thermal conductivity coefﬁcient. The stress-deformation relationship is, in 1D small strain, just σ = E ε; however, for the large deformation 3D case, it will prove to be signiﬁcantly more complex.

4 Constitutive Theory

We began our discussion of continuum mechanics by introducing kinematics of deformation, the geometric structure of bodies under continuous deformation. Everything that was developed there is true by geometry – no physics was introduced at this point. We then introduced the familiar concepts of mass, momentum, and energy conservation within a continuum setting. For each of these laws, we introduced their formulation in the familiar bulk setting (the “global” formulation), and then derived differential formulations of these laws (the “local” formulation). The results are true by the laws of physics. We concluded the last section with the observation that the number of equations we have comes up short when we try to solve for all of the variables in our problem. Constitutive theory introduce equations that describe material- speciﬁc behavior, and are said to “close” the mechanical system. We will develop a framework for formulating constitutive models, as well as restrictions on their physical admissibility; however, the equations derived from constitutive models are inherently models – they are derived either from experimental observation or highly sim- pliﬁed theory.

4.1 Introduction to the calculus of variations

We’re going to take a bit of a break here to introduce another bit of mathematical machinery that will be invaluable in formulating constitutive theory. The calculus of variations (or “variational calculus”) is the mathematical theory of minimizing a functional. So first, what is a functional? Definition 4.1. A functional on V is a mapping from V to R. In other words, a functional is a machine that takes a thing (that thing could be a scalar, vector, tensor, or even a function) and turns it into a real number. What is special about functions that return real numbers? It is the fact that they can be optimized – that is, for a functional f [x], we can try to find the x that gives us the smallest value of f . For instance,

f (x) = |x| (4.1) n is a functional on R , minimized by x = 0. Now, let us consider a more complex example: ﬁnd the function y(x) that minimizes the distance between the two points (a, ya), (b, yb). In other words, we want to ﬁnd the shortest path connecting those two points:

a b

We want to minimize a functional that takes a function y(x) and returns a distance – a scalar value. What form might this functional take? Z b p 02 L[y] = 1 + y (x)dx (4.2) a

So, we can state our mathematical problem in the following way: Z b p 02 inf 1 + y (x)dx such that y(a) = ya, y(b) = yb (4.3) y a

How do we go about solving this problem? Rather than solve this problem explicitly, let us consider a more general problem: Z b 0 inf L[y] = inf f (x, y, y ) dx such that y(a) = ya, y(b) = yb (4.4) y y a

(Notice that we have merely replaced our specific integrand with a more general one. ) In first-year calculus, we learn that we can optimize a function by finding the point at which its derivative is equal to zero. We can do a similar thing here–only this time, we will use the Gateaux derivative.

y b y(x) + εη(x)

y(x)

y a η(x)

a b

Let us introduce a function η(x) such that η(a) = η(b) = 0. Suppose that we’ve already found a minimizing function y(x). If that is the case then we know that

L[y] ≤ L[y + εη] ∀ε ∈ R, ∀ admissible η (4.5)

In fact, we could say that the above is minimized when ε = 0. This amounts to saying that d L[y + εη] = 0 dε ε→0 (4.6)

Does that look familiar? It’s nothing other than the Gateaux derivative in functional form. Let’s see if we can actually evaluate this derivative: d d Z b Z b d 0 0 0 0 L[y] = f (x, y + εη, y + εη ) dx = f (x, y + εη, y + εη ) dx dε ε→0 dε a ε→0 a dε ε→0 Z b h ∂f d(y + εη) ∂f d(y 0 + εη0)i = + dx 0 0 a ∂(y + εη) dε ∂(y + εη ) dε ε→0 Z b h ∂f ∂f 0i = η + 0 η dx a ∂y ∂y Z b ∂f ∂f b Z b d ∂f = η dx + η − η dx 0 0 a ∂y ∂y a a dx ∂y Z b h ∂f d ∂f i = − 0 η dx a ∂y dx ∂y ∂f d ∂f ∀η =⇒ − = 0 ≡ Euler-Lagrange Equation ∂y dx ∂y 0 (4.7)

So we see that we began with a global minimization principle, and arrived at a local differential equation.

Example 4.1: Minimum distance

p 02 Let’s apply this result to our minimum distance problem: if f = 1 + y , then the minimizing path must satisfy ∂f d ∂f d y 0 y 00 y 02y 00 y 02 1 − = − = − + = y 00 − = 0 ∂y dx ∂y 0 dx p1 + y 02 p1 + y 02 (1 + y 02)3/2 (1 + y 02)3/2 p1 + y 02 (4.8)

00 There could potentially be multiple solutions, but the most obvious one is that y (x) = 0. This implies that the minimizing function y is linear – exactly what we expect.

Example 4.2: Brachistochrone problem

Consider a ball of mass m under the action of gravity g traveling from (xa, ya) to (xb, yb) along a path deﬁned by y(x). What path y(x) minimizes the transit time?

xa xb

ya y(x)

Our functional T [y] returns transit time as a function of path; we need a form for the functional. Use the fact that potential energy is constant, in other words, we can say without loss of generality that 1 ds ds T + U = =⇒ m v 2 − m g y = 0 =⇒ v = = p2gy =⇒ dt = √ const 2 dt 2gy (4.9)

But we also know that

p 2 2 p 02 ds = dx + dy = 1 + y dx (4.10)

so our functional is given by s Z tf Z b 1 + y 02 T = dt = dx (4.11) ti a 2gy

The integrand of our functional is given by s 1 + y 02 f (x, y, y 0) = 2gy (4.12)

and we can simply substitute into the Euler Lagrange equation to solve for y(x). The actual solution is a bit messy, and it is easier to describe parametrically. The form of the optimal path is a cycloid:

x = r (θ − sin θ) + c y = r(1 − cos θ) (4.13)

where r, c are determined based on the boundary conditions.

There is a very wide range of applications of variational calculus. Some applications include dynamics (Lagrangian

All content © 2016-2018, Brandon Runnels 19.3 MAE 5100 - Continuum Mechanics Course Notes - Lecture 19 University of Colorado Colorado Springs https://canvas.uccs.edu/courses/22031 mechanics), optimal control theory, mathematical ﬁnance. We will seek to ﬁnd a variational (functional minimization) formulation of our balance laws.

4.1.1 Stationarity condition

Let us consider our usual body, Ω, and this time we will imagine that it is subjected to some external loading, body forces, and a displacement condition as indicated below.

∂1Ω

∂2Ω

Ω

Note that we have split out boundary into two portions, so that ∂Ω = ∂1Ω ∪ ∂2Ω. We say that we have prescribed displacement on ∂1Ω whereas we prescribed forces on ∂2Ω. Alternatively we could say that we prescribe Dirichlet boundary conditions on ∂1Ω and Neumann boundary conditions on ∂2Ω. With this picture in mind, let us introduce a powerful theorem for functional minimization on Ω: Theorem 4.1. Let Ω ⊂ Rn be Lebesgue measurable. Let φ :Ω → Rn, and let

f :Ω × φ(Ω) × GL(n) → R g : ∂Ω × φ(∂Ω) × GL(n) → R (4.14) be functions deﬁned over the body and the body’s boundary, respectively. Let the boundary of the body be divided into two regions, ∂1Ω, ∂2Ω such that φ(X) = X ∀X ∈ ∂1Ω. Finally, let Z Z L[φ, Grad φ] = f (X, φ, Grad φ)dV + g(X, φ, Grad φ)dA (4.15) Ω ∂Ω If φ solves inf L[φ] = 0 φ (4.16) then the following Euler-Lagrange equations hold for φ: ∂f d ∂f ∂g ∂g ∂f − = 0 ∀X ∈ Ω = 0 ∀X ∈ ∂1Ω + NJ = 0 ∀X ∈ ∂2Ω (4.17) ∂φi dXJ ∂φi,J ∂φi,J ∂φi ∂φi,J

Proof. As before, we will prove this using a special case of the Gataeux derivative called the variational or functional derivative. The key aspect of the proof is that if φ minimizes L, then

d n L[φ + εη] = 0 ∀ η :Ω → R dε ε=0 admissible (4.18)

What does admissible mean? From the beginning we constrained φ such that there is no deformation on ∂1φ. (This is called a Dirichlet boundary condition.) On the other hand, it is free to move over the remainder of the boundary, so the derivative must be zero. (This is called a Neumann boundary condition.) That means that our test function η must satisfy

η(X) = 0 ∀X ∈ ∂1Ω Grad η(X) = 0 ∀X ∈ ∂2Ω (4.19)

This will come in very handy in a moment. Now, let us begin by evaluating the derivative: d Z d

L[φ + εη, Grad(φ + εη)] = f (X, φ + εη, Grad φ + ε Grad η)dV |ε=0+ dε ε=0 Ω dε Z d

+ g(X, φ + εη, Grad φ + ε Grad η)dA ∂Ω dε ε=0 Z h ∂f d(φ + εη ) ∂f d(φ + εη )i i i i,J i,J = + dV Ω ∂(φi + εηi ) dε ∂(φi,J + εηi,J ) dε ε=0 Z h ∂g d(φ + εη ) ∂g d(φ + εη )i i i i,J i,J + + dA ∂Ω ∂(φi + εηi ) dε ∂(φi,J + εηi,J ) dε ε=0 Z h ∂f ∂f i Z h ∂g ∂g i = ηi + ηi,J dV + ηi + ηi,J dA (4.20) Ω ∂φi ∂φi,J ∂Ω ∂φi ηi,J

Consider the third term. Using the product rule and integration by parts: Z ∂f Z ∂ ∂f Z d ∂f Z ∂f Z d ∂f ηi,J dV = ηi dV − ηi dV = ηi NJ dA − ηi dV Ω ∂φi,J Ω ∂XJ ∂φi,J Ω dXJ ∂φi,J ∂Ω ∂φi,J Ω dXJ ∂φi,J

Substituting back into the above expression and gathering terms, we have Z h ∂f d ∂f i Z h ∂g ∂g ∂f i = − ηi dV + ηi + ηi,J + ηi NJ dA (4.21) Ω ∂φi dXJ ∂φi,J ∂Ω ∂φi ηi,J ∂φi,J

Here is where the admissibility condition is useful: let us apply the condition that ηi,J = 0 everywhere on the boundary except for ∂1Ω. This leaves Z h ∂f d ∂f i Z ∂g Z h ∂g ∂f i = − ηi dV + ηi,J dA + + NJ ηi dA (4.22) Ω ∂φi dXJ ∂φi,J ∂1Ω ηi,J ∂2Ω ∂φi ∂φi,J

And here is the ﬁnal payoff: we note that this has to be true for all admissible η. And so, we can pass from the weak form to the strong form by the fundamental lemma. The result is that ∂f ∂ ∂f ∂g ∂g ∂f − = 0 ∀X ∈ Ω = 0 ∀X ∈ ∂1Ω + NJ = 0 ∀X ∈ ∂2Ω (4.23) ∂φi ∂XJ ∂φi,J ∂φi,J ∂φi ∂φi,J

This concludes the proof. As you can see, the proof of the above theorem is a little bit technical, but it’s really nothing more than a fully 3D version of the simple 1D Euler-Lagrange equation that we found earlier. Notice that we left the proof as general as possible, yet the results are still quite simple. We can now take what we found and apply it directly to continuum mechanics.

4.2 Variational formulation of linear momentum balance

Let us introduce the potential energy functional: Z Z Z Potential Energy Functional Π[φ] = W (F)dV − φi RBi dV − φi Ti dA ≡ (4.24) Ω Ω ∂Ω (Lagrangian Frame) | {z } | {z } | {z } strain energy work done by body forces work done by surface tractions were we note, of course, that F = Grad φ. We’ve introduced one new character into this equation: W (F), called the strain energy density. We will deﬁne what this is in just a moment. For now, you can think of it simply as the energy of a body (per unit volume) associated with deforming a body by a deformation gradient F.

The principle of minimum potential energy states that, for a body subjected to surface tractions T and body forces B, that

φ = arginfφ Π[φ] ≡ Principle of Minimum Potential Energy (4.25)

In other words, we can ﬁgure out what φ is simply by minimizing Π. This is very convenient computationally, because it is frequently convenient to formulate problems in terms of minimization. This is called a variational principle. Let us use the theorem we derived above to show that the principle of minimum potential energy returns our familiar balance laws. If we write Π in terms of f , g, we have

f = W (F) − φi RBi g = −φi Ti (4.26)

Since we proved the above theorem for the general case, we know that: ∂ dW dW −RBi − = 0 0 = 0 −Ti + NJ = 0 (4.27) ∂XJ dFiJ dFiJ

Let us rearrange the ﬁrst and last Euler-Lagrange equations. Writing them in invariant/symbolic notation, we get: dW dW Div − R B = 0 N = T dF dF (4.28)

Does this look familiar? You may notice that it has a very similar form to that of the local balance of linear momentum in the Lagrangian frame–if we suppose P = dW /dF we have

Div(P) + RB = 0 ∀X ∈ Ω PN = T ∀X ∈ ∂Ω (4.29) exactly the formulation for linear momentum that we derived earlier. In fact, this is exactly how we will introduce our constitutive model. Deﬁne the following scalar function of the deformation gradient: dW W : GL(3) → P = ≡ Elastic Free Energy Density R such that dF (4.30)

This is how we describe material behavior.

4.3 Material frame indifference

Consider the energy of a body under the action of a deformation gradient F and a rotation R:

FR W (I) W (F) W (RF)

Because the rotation is rigid, we expect that W (RF) = W (F) for all R ∈ SO(3). Recall that any F can be decomposed ˆ into F = RU by the polar decomposition. We may say that for all W there exists a function W such that ˆ W (F) = W (U) (4.31)

In other√ words,√ the strain energy depends on the pure deformation only, not the subsequent rotation. But because T ˆ U = C = F F we could also say that must exist a W such that ˆ W (F) = W (C) (4.32)

T In other words, W (F) = γ F11 would violate material frame indifference, whereas W (F) = γ tr(F F) would not. Note that while W must depend on C only, P does not. Because we generally have W (C), it is frequently handy to differentiate with respect to C instead of F. Using the chain rule:

dW dW dCkL dW d dW dW dW = = (FpK FpL) = (δipδJK FpL + FpK δipδJL) = FiL + FiK (4.33) dFiJ dCKL dFiJ dCKL dFiJ dCKL dCJL dCKJ

Recall that C is symmetric, so CIJ = CJI ; the following simpliﬁes to dW dW = 2 F dF dC (4.34)

We notice that dW 1 dW 1 1 = F−1 = F−1P = S dC 2 dF 2 2 (4.35) ˙ 1 ˙ We recall that (F , P) and ( 2 C, S) are conjugate pairs, and we see that this relationship is repeated again here: dW dW dW P = S = 1 = (4.36) dF d( 2 C) dE

4.4 Elastic modulus tensor

Let us consider a linearization of W (F) about F = I, where we know that W (I) = 0, = P(I) = 0. Expanding to second order, we have 0 0 dW > 1 dW 1 ¨¨* W (F) = ¨W (I) + (I)(FiJ − δiJ ) + (FiJ − δiJ ) (I)(FkL − δkL) + h.o.t. ≈ (F − I) · C (F − I) (4.37) dFiJ 2 dFiJ dFkL 2

The second derivative tensor d 2W CiJkL = (4.38) dFiJ dFkL is called the elastic modulus tensor. It is a useful quantity to compute when solving problems numerically.

dW dW = 2FiI (4.39) dFiJ dC IJ 2 2 2 d W dW d dW dW d W dCKP dW d W = 2δik δIL + 2FiI = 2δik + 2FiI = 2δik + 4FiI FkK (4.40) dFiJ dFkL dC IJ dFkL dCIJ dCJL dCIJ dCKP dFkL dCJL dCIJ dCKL dW d 2W = δik 1 + FiI FkK 1 1 (4.41) d( 2 CJL) d( 2 CIJ )d( 2 CKL) = δik SJL + FiI FkK CIJKL (4.42) where CIJKL can also be expressed as d 2W d 2W CIJKL = 4 = (4.43) dCIJ dCKL dEIJ dEKL

We see that CIJKL has both “major symmetry” (CIJKL = CKLIJ ) and “minor symmetry” (CIJKL = CJIKL = CIJLK )

4.5 Elastic material models

We see from the above that the only thing necessary to deﬁne a material model is the free energy function W (F ) – given a W , it is always possible to derive P and C. We will illustrate this with a couple of examples.

4.5.1 Useful identities

We will be doing a great deal of matrix differentiation, and we frequently have to evaluate complex expressions. The following identities will prove useful: • Derivative of the determinant

d(det F) −T = det(F)FiJ (4.44) dFiJ

• Derivative of the matrix inverse −1 −1 d d −1 dFJn −1 dFJn −1 ! δJM = (FJn FnM ) = FnM + FJn δknδLM = FnM + FJk δLM = 0 (4.45) dFkL dFkL dFkL dFkL −1 dFJn −1 −1 −1 FnM FMi = −FJk δLM FMi (4.46) dFkL | {z } δni −1 dFJi −1 −1 −T −T = −FJk FLi = −FkJ FiL (4.47) dFkL

4.5.2 Pseudo-Linear

A simplistic model that guarantees material frame indifference is the following, where C is a rank-4 tensor and obeys major symmetry (CIJKL = CKLIJ ) and minor symmetry (CIJKL = CJIKL = CIJLK = CJILK ). 1 1 1 W (F) = E · E = (FT F − I) · (FT F − I) = (F T F − δ ) (F T F − δ ) 2 C 8 C 8 Pm mQ PQ CPQRS Rn nS RS (4.48)

From this deﬁnition, we compute the Piola-Kirchoff stress tensor by differentiation

dW d h1 idCAB 1 T d T DW (F)iJ = = (C − I) C (C − I) = CABRS (FRnFnS − δRS ) (FA`F`B ) dFiJ dCAB 8 dFiJ 4 dFiJ 1 = (F T F − δ )(δ δ F + F δ δ ) 4CABRS Rn nS RS i` JA `B `A i` JB 1 = (F + F )(F T F − δ ) 4 iB CJBRS iACAJRS Rn nS RS (4.49)

We compute the elasticity tensor by differentiating again

2 2 2 d W d h1 idCAB dCCD W h1 i d CAB DDW (F )iJkL = = (C − I) C (C − I) + (C − I) C (C − I) dFiJ dFkL dCAB dCCD 8 dFiJ dFkL dCAB 8 dFiJ dFkL 1 δ δ = ( + + + )F F + ik ( + )F F − ik ( + ) 4 CJALD CJADL CAJLD CAJDL iA kD 4 CJLRS CLJRS nR nS 4 CJLRR CLJRR (4.50)

The result is somewhat unwieldy, so let us apply symmetry conditions on C to simplify. The result is 1 DDW (F ) = F F + δ ( F F − ) iJkL CAJDL iA kD 2 ik CLJRS nR nS CJLRR (4.51)

As a sanity check, we can verify that the stress at F = I must be equal to zero. Substituting, we get

: 0 P( ) = (δ + δ ) (δ δ −δ ) = 0 I iJ iB CJBRS iACAJRS RnnS RS X (4.52) as expected. What happens if we substitute the identity for the elasticity tensor? We obtain: 1 (I) = δ δ + δ ( δ δ − ) C iJkL CAJDL iA kD 2 ik CLJRS nR nS CJLRR 0 1 : = + δ ( − ) = CiJkL 2 ik CLJnn CJLRR CiJkL (4.53) whence we see that the elastic constants correspond to the tangent modulus in the undeformed state.

4.5.3 Compressible neo-Hookean

The neo-Hookean model is a simple model that does a fairly good job of modeling rubbery materials. Two material constants are used, the shear modulus µ and the bulk modulus κ. We deﬁne the material by its free energy W (F ): µ tr(FT F) κ µ F F κ W (F) = − 3 + (det(F) − 1)2 = pQ pQ − 3 + (det(F) − 1)2 2 det(F)2/3 2 2 det(F )2/3 2 (4.54)

We ﬁnd the Piola Kirchoff stress tensor which is given by the ﬁrst derivative

dW µ(F F ) 2 F F d det(F) d det(F) DW (F) = = pQ pQ ,iJ − pQ pQ + κ(det(F) − 1) iJ 2/3 5/3 dFiJ 2 det(F) 3 det(F) dFiJ dFiJ F 1 F F =µ iJ − pQ pQ F −T + κ(det(F) − 1) det(F)F −T = P det(F)2/3 3 det(F)2/3 iJ iJ iJ

We ﬁnd the elastic modulus tensor which is given by the second derivative

d 2W d h F 1 F F i DDW (F) = = µ iJ − pQ pQ F −T + κ(det(F) − 1) det(F)F −T iJkL 2/3 2/3 iJ iJ dFiJ dFkL dFkL det(F) 3 det(F) µ h 2 2 2 1 i = δ δ − F F −T − F −T F + (F F )F −T F −T + (F F )F −T F −T det(F)2/3 ik JL 3 iJ kL 3 iJ kL 9 pQ pQ iJ kL 3 pQ pQ iL kJ h −T −T −T −T i + κ det(F) 2 det(F) − 1 FiJ FkL + 1 − det(F) FiL FkJ = CiJkL (4.55)

Substituting F = I shows quickly that the stress is zero in that state. Substituting into the tangent modulus, we get

2 DDW (I) =µ δ δ + δ δ − δ δ + κ δ δ ik JL iL kJ 3 kL iJ iJ kL (4.56) where we see that µ and κ correspond to the shear and bulk moduli in small deformation.

4.6 Internal constraints

We made a point of noting that the above was compressible neo-Hookean. We’ve also noted that it is generally “easier” to work with compressible rather than incompressible materials. Incompressibility is an example of an internal constraint – a material restriction on admissible geometric material deformations. The variational structure of the linear momentum balance law provides a nice structure for enforcing internal constraints by means of Lagrange multipliers.

4.6.1 Review of Lagrange multipliers

Consider the following minimization problem:

y f (x, y) = x + y

x 2 + y 2 = 1

2 2 Find the minimum point of f (x, y) = x +y subject to the constraint that x +y = 1. In the language of optimization, we write inf f (x, y) x 2 + y 2 = 1 (x,y) subject to

(What if we tried to optimize f (x, y) without the constraint? The stationarity conditions are d d d d f (x, y) = 0 f (x, y) = 0 =⇒ f (x, y) = 1 = 0 f (x, y) = 1 = 0 dx dy dx dy (4.57) which is completely unsolvable for x, y. This is to be expected: f (x, y) has no minimizer – it can grow negatively to inﬁnity. Minimization only makes sense when there is an imposed constraint. ) How do we write our stationarity conditions when the constraint is active? To do this, we modify our problem to include an additional variable, called a Lagrange multiplier:

h 2 2 i inf sup f (x, y) + λ(x + y − 1) (4.58) x,y λ

(We recall that “sup” is basically the same as “max”.) Why do this? Let’s make a few notes: 2 2 (1) We’re restricting x + y = 1, so there will be no resulting contribution to the optimized value.

(2) In addition to optimizing with respect to x, y, we are optimizing with respect to λ as well. What is the stationarity condition for λ? We get this by evaluating d h i f (x, y) + λ(x 2 + y 2 − 1) = x 2 + y 2 − 1 = 0 dλ (4.59)

In other words, the stationarity condition on λ is identical to our internal constraint.

(3) Why are we maximizing with respect to λ? Let’s look at the inside part:

( 2 2 h 2 2 i 0 x + y = 1 sup f (x, y) + λ(x + y − 1) = (4.60) λ ∞ else

2 2 In other words, this inside part will blow up to inﬁnity except if x + y = 1. Recall that on the outside, we are minimizing with respect to x, y – this means that all non-inﬁnite solutions must satisfy the constraint. With these notes in mind, let’s attempt to solve the above optimization problem. We have three stationarity conditions, which we’ll use to solve for x, y, λ dΦ dΦ dΦ = 1 + 2λx = 0 = 1 + 2λy = 0 = x 2 + y 2 − 1 = 0 dx dy dλ (4.61) 1 1 1 1 1 1 =⇒ x = − =⇒ y = − =⇒ + = = 1 =⇒ λ = ±√ 2λ 2λ 4λ2 4λ2 2λ2 2 (4.62)

Substituting back, we get two solutions: 1 1 x = ±√ y = ±√ 2 2 (4.63) corresponding to the two points along the x = y line passing through the circle. Exactly what we expect!

4.6.2 Examples of internal constraints

Let us express internal constraints for materials such that

ψ(F) = 0 (4.64)

Two examples: • Incompressibility: ψ(F) = det(F) − 1 = 0 2 T T • Inextensible ﬁbers: ψ(F) = λ (N) − 1 = N F FN − 1 = 0 (Notice that we can write the above constraints entirely in terms of C as well.)

4.6.3 Lagrange multipliers in the variational formulation of balance laws

Recall that the equilibrium conﬁguration is given by h Z Z Z i φ = arg inf Π[φ] = arg inf W (F)dV − φi RBi dV − φi Ti dA (4.65) φ φ Ω Ω ∂Ω

Let us suppose that we are working with a system that has an internal constraint ψ(F ) = 0. How would we include this in our optimization problem? Let’s try: Z Z Z Z h i h i arg inf sup Π[φ] + λ φ(F) dV = arg inf sup W (F) + λψ(F) dV − φi RBi dV − φi Ti dA (4.66) φ λ Ω φ λ Ω Ω ∂Ω

What are our stationarity conditions? For φ, they are nothing other than our usual Euler-Lagrange equations, which come out to be dW dψ dW dψ + λ + RB = 0 + λ N = T dF dF dF dF (4.67)

This prompts us to deﬁne the stress tensor to be dW dψ P = + λ dF dF (4.68)

Let us look at a couple of examples: ﬁrst, consider an incompressible material with free energy density W0. Then ψ(F) = det(F) − 1, and the stress is given by dW d P = + λ (det(F) − 1) = P + λ J F−T dF dF 0 (4.69)

(where P0 is the stress with the incompressibility constraint removed.) What does this look like in the deformed conﬁguration? Apply the relationship between σ and P to obtain 1 1 1 σ = PFT = P FT + λ J F−T FT = σ + λ I J J 0 J 0 (4.70)

In other words, our result is a combination of the regular stress σ0 pluss a “hydrostatic” stress λ I with magnitude λ. This kind of stress state is identical to hydrostatic pressure, and we generally rename λ to p. Let’s interpret this result by noting a couple of points: (1) The material resists all volumetric compression; the way it resists compression is by exerting hydrostatic pressure. Thus we can interpret Lagrange multipliers as forces exerted by the material to enforce the material restraint. (2) What are the units of p? Recall that λψ(F) must have units of energy per unit volume, and that ψ(F) = det(F)−1 is unitless. Pressure has units of [force]/[area] which can be written as [force][length]/[volume].

4.7 Linearized constitutive theory

At the end of the section on kinematics, we introduced a framework for linearized kinematics, in which we replace the deformation mapping φ with the displacement ﬁeld u, powers and gradients of order greater than two are sufﬁciently small to ignore. Let us assume that we can express the energy of our material under small strain as W (ε). Then the principal of minimum potential energy becomes h Z Z Z i u = arg inf W (ε)dV − u · b dV − u · tdS (4.71) u Ω Ω ∂Ω

The Euler-Lagrange for this problem (in the static case) give dW div + ρb = 0 dε (4.72)

We compare this with the local balance of linear momentum, which is given by

div(σ) + ρb = ρu¨ (4.73)

(note that u¨ = a.) This implies that the Cauchy stress is given by dW dW σ = σij = (4.74) dε dεij

What form must W (ε) take? Let us Taylor expand about ε = 0: dW 1 d 2W W (ε) = W (0) + (0) εij + (0) εij εkl + h.o.t. (4.75) dεij 2 dεij dεkl

We can state without loss of generality that W (0); also, we assume that our material is purely elastic, so DW (0) = 0. Thus we are left with 1 d 2W 1 W (ε) = εij εkl = Cijkl εij εkl (4.76) 2 dεij dεkl 2 where Cijkl is a fourth order tensor containing the constant elastic moduli of the material. This is amenable to differentiation, so that our stress tensor is given by dW σij = = Cijkl εkl (4.77) dεij

Finally, we write our momentum balance equation in the following way:

Cijkl εkl,j + ρbi = ρu¨i (4.78)

4.7.1 Major & minor symmetry and Voigt notation

We have established that in small strain, the elastic modulus tensor is deﬁned by d 2W Cijkl = (4.79) dεij dεkl

4 This is an order 3 × 3 × 3 × 3 tensor, meaning that it can contain up to 3 = 81 elastic constants – far too many to keep track of easily. Indeed, the above is too general: we can use symmetry arguments to reduce the number of possible constants: Theorem 4.2. In small strain, the elastic modulus tensor has both major symmetry

Cijkl = Cklij (4.80) and minor symmetry

Cijkl = Cjikl = Cijlk (4.81)

Proof. It is simple to prove the above: because the energy function is continuous, we have d 2W d 2W Cijkl = = = Cklij (4.82) dεij dεkl dεkl dεij proving major symmetry. In small strain, ε is symmetric, so d 2W d 2W d 2W = = (4.83) dεij dεkl dεji dεkl dεij dεlk proving minor symmetry. 2 How does symmetry reduce the number of possible independent constants? Recall that a n × n tensor has n constants, but a symmetric n × n tensor has 1 n (n + 1) 2 (4.84) independent constants. Think of C as a second order 9 × 9 tensor. If it is symmetric, (by major symmetry) then 1 it only has 2 9 × (10) = 45 independent constants. But minor symmetry means that C is really a 6 × 6 symmetric tensor. Then the number of possible terms is 1 6 × 7 = 21 2 (4.85)

So C can have no more than 21 independent constants. Alternatively, we can think of C as being represented in the following way:

σ11 C1111 C1122 C1133 C1123 C1131 C1112  ε11  σ C C C C C C ε  22  2211 2222 2233 2223 2231 2212  22  σ33 C3311 C3322 C3333 C3323 C3331 C3312  ε33  σ  = C C C C C C  2ε  (4.86)  23  2311 2322 2333 2323 2331 2312  23 σ31 C3111 C3122 C3133 C3123 C3131 C3112 2ε31 σ12 C1211 C1222 C1233 C1223 C1231 C1212 2ε12 | {z } 36 terms, 21 independent constants

This is the form of a constitutive model for a general small-strain anisotropic material. Note that each shear stress and strain term need only appear once since both stress and strain are symmetric.

4.7.2 Material symmetry

Some materials are said to have cubic symmetry. In that case, one can show that the elastic modulus tensor (in Voigt notation) must reduce to

C11 C12 C12 0 0 0  C C C 0 0 0  12 11 12  C12 C12 C11 0 0 0  C =  0 0 0 C 0 0  (4.87)  44  0 0 0 0 C44 0 0 0 0 0 0 C44

Note that the number of independent constants has been reduced to three: C11, C12, C44. This type of model is frequently useful when computing the elastic modulus tensor of a single crystal material that has a cubic (e.g. FCC or BCC) crystal structure. Macroscopic materials are generally observed to be completely isotropic. It can be shown that isotropic materials have only two independent parameters. Cijkl can then be deﬁned in the following way:

Cijkl = µ(δik δjl + δil δjk ) + λδij δkl (4.88) where µ, λ are referred to as Lamé parameters. In the above case, note that we can then relate σ to ε as

σij = Cijkl εkl = µ(δik δjl + δil δjk )εkl + λδij δkl εkl = µ(εij + εji ) + λδij εkk = 2µ εij + λ tr(εkk ) δij (4.89)

We can relate the parameters λ and µ to more familiar constants in the following way. Consider a plane stress conﬁguration with a known applied stress σ0, so that " # " # σ1 0 0 ε1 0 0 σ = 0 0 0 ε = 0 ε2 0 (4.90) 0 0 0 0 0 ε3

Let us compute the strain response ε1, ε2, ε3 in terms of σ0, and determine the corresponding proportionality constants: " # " # " # σ1 0 0 ε1 0 0 1 0 0 0 = 2µ 0 ε2 0 + λ(ε1 + ε2 + ε3) 1 (4.91) 0 0 0 0 0 ε3 1

We can write as a system of equations " # " #" # σ1 2µ + λ λ λ ε1 0 = λ 2µ + λ λ ε2 (4.92) 0 λ λ 2µ + λ ε3

Solving this system, we get σ (λ + µ) 1 λσ λ ε = 0 ε = ε = − 0 = − ε 1 µ(3λ + 2µ) 2 3 2 µ(3λ + 2µ) 2(λ + µ) 1 (4.93)

From this we can identify the elastic modulus E and the Poisson’s ratio ν: µ(3λ + 2µ) λ E = ν = λ + µ 2(λ + µ) (4.94)

4.7.3 The Cauchy-Navier equation and linear elastodynamics

One of the nice things about linearized elasticity is that we can work entirely in displacements and can use the formulation to solve exact problems. Recall that 1 1 ε = Grad(u) + Grad(u)T ε = (u + u ) 2 or ij 2 i,j j,i (4.95)

We also remember that C has minor symmetry. This means we can write σij as 1 1 σ = (u + u ) = ( u + u ) = u ij 2Cijkl k,l l,k 2 Cijkl k,l Cijkl k,l Cijkl k,l (4.96)

In other words, we can dispense with strain and write everything in terms of displacement only. If we do this, then we can write our momentum balance equation as

Cijkl uk,lj + ρbi = ρui,tt (4.97)

Substituting the form for isotropic elasticity This is referred to in general as the Cauchy-Navier equation. It is a linear second order PDE in space and time that can be solved relatively easily. Let’s consider a 1D plane strain version of the problem. Then we can write " # " # u1,1(t) 0 0 1 σ = 2µ 0 0 0 + λ u1,1(t) 1 (4.98) 0 0 0 1

Then " # (2µ + λ)u1,11 div σ = 0 (4.99) 0

If we ignore body forces, then the Cauchy Navier equation becomes ∂2u ∂2u (2µ + λ) 2 = ρ 2 , (4.100) ∂x1 ∂t equation that can be analyzed easily using the method of characteristics or seperation of variables.

4.8 Thermodynamics of solids and the Coleman-Noll framework

We now return to ﬁnite strain mechanics. We have treated the formulation of pure elasticity, which is completely reversible. However there are multiple types of mechanical phenomena such as plasticity and viscoelasticity that are irreversible. Additionally, we wish to keep our treatment as general as possible, so we aim to account for thermal effects in addition to mechanical. Towards this end, we begin by assuming that a material can locally be deﬁned by (1) The deformation gradient F and its derivatives (although we will only consider F here) (2) The material entropy N (per unit undeformed mass) and its derivatives (3) A set of internal variables that we will denote in vector form by Q. (Note that Q isn’t a vector in the spatial sense, rather it’s just a collection of individual internal variables.) A good example of an internal variable is the accumulated plastic deformation in a material.

Now, let us assume that there exists an internal energy functional U = U(F, N, Q) that characterizes the materials response to deformation, entropy, and change in internal variables. Let us begin by working with the Clasius-Duhem inequaltiy: start by expanding it out using the product rule. ˙ H RSn ˙ 1 1 RSn RN + Div − = RN + Div H − 2 H · Grad T − ≥ 0 (4.101) T T product rule T T T

We spot the appearance of the Div H term: we’ve seen this before in the ﬁrst law in the Lagrangian frame. Rear- ranging the ﬁrst law, we can solve for it: ˙ ˙ ˙ ˙ RU = P · F + RSn − Div H =⇒ Div H = P · F + RSn − RU (4.102)

Now, we can substitute this into the Clasius-Duhem inequality, and because T > 0 we can multiply out by T : P · F˙ + RS − RU˙ 1 RS RN˙ + n − H · Grad T − n ≥ 0 T T 2 T (4.103) 1 RT N˙ − RU˙ + P · F˙ − H · Grad T ≥ 0 T (4.104)

Now, let’s consider the internal energy functional. Because it’s a function of F, N, Q, we can expand its time derivative out using the chain rule: ∂U ∂U ∂U U˙ = · F˙ + N˙ + · Q˙ ∂F ∂N ∂Q (4.105)

Substituting this relationship back into the inequality, we get ∂U ∂U ∂U 1 RT N˙ − R F˙ − R N˙ − R · Q˙ − P · F˙ − H · Grad T ≥ 0 ∂F ∂N ∂Q T (4.106)

And ﬁnally, we gather terms: ∂U ∂U ∂U 1 P − R ·F˙ + RT − R N˙ − R ·Q˙ − H · Grad T ≥ 0 ∂F ∂N ∂Q T (4.107) | {z } | {z } | {z } | {z } ˙ ˙ ˙ The arbitrary nature of F, N, Q prompt us to identify the following deﬁnitions:

e ∂U v e ∂U ∂U P = R P = P − P T = Y = −R (4.108) ∂F | {z } ∂N ∂Q | {z } Viscous stress | {z } | {z } Elastic stress Temperature Dissipative driving force

Let’s discuss the new quantities briefly: e (1) P is the component of the stress that is purely elastic, for instance, the stress computed by Hooke’s law. v (2) P is the part of the stress corresponding to dissipation, for instance, the viscous stress generated in a fluid. (3) Y is a bit more abstract. Components of Y correspond to components of Q, and you can think of them as the forces that cause the internal variables to change. For instance, for the internal variable of plastic dissipation, the driving force is the component of the stress that forces the plastic slip. With these newly defined quantities, we can write the second law of thermodynamics as 1 Pv · F˙ + Y · Q˙ − H · Grad T ≥ 0 T (4.109) | {z } | {z } | {z } mechanical internal thermal This implies that irreversible (entropy-generating) processes can take three forms: (i) mechanical dissipation (e.g. viscous drag in fluid flow) (ii) internal dissipation (e.g. plasticity) and (iii) thermal dissipation.

4.8.1 Other thermodynamic potentials

We began by defining the internal energy functional U(C, N, Q) as a function of deformation gradient, entropy, and some vector of internal variables. (Note: we are using C rather than F because U is not convex in F whereas it is convex in C). Can we define the internal energy as a function of other variables? That is, could we define A(F, T , Q) that gives us the same information about the solid? If N = N(F, T , Q) then the answer is yes, we would simply have

A(F, T , Q) = U(F, N(F, T , Q), Q) (4.110)

In fact, such a relationship can be found by applying the state postulate. But now it is possible, how do we ﬁnd this function, since we don’t know the relationship between N and the other state variables? To do this, we use the non-standard Legendre transform

The Legendre transform

The Legendre transform is a powerful mathematical tool that is perticularly useful when working with energy methods and with minimization problems. (Note: this is not to be confused with Legendre polynomials, which are entirely different.) Let us develop the transform by considering a simple function f (x). We will 0 require that f be C continuous and convex, but will place no other restrictions on it. f (x) f (x)

f (x) x x m x x ∗

The original (or “primal”) way is to use a functional form to match a point in x to f (x); that is, we have a graph for f (x). Now, let’s consider an alternative approach. Recall that we require that f is convex; this means that the region above f (x) (called the epigraph) must be convex as well. Notice how, at each point on the epigraph, there is (at least) one tangent line. Now, let us draw all of the tangent lines to the epigraph at all of the points along f (x). Notice how we can recapture the shape of the epigraph using the tangent lines only. This is called the “dual” representation of f (x). So, we see that we can represent our function either using f (x) or the collection of all the tangent lines to the epigraph. In particular, let us suppose that we have the following function: ∗ f : [slope of tangent line] → [negative intercept of tangent line] (4.111)

∗ ∗ So f generates all possible tangent lines to the epigraph of f (x); we say that f is the Legendre Transform of ∗ f . This comes as little help, however, if we don’t have a way of constructing f . Let us consider the following formulation: ∗ f (s) = sup [s x − f (x)] (4.112) x

∗ To solve this problem we ﬁrst solve for x , the minimizing value, using the stationarity condition: ∗ ∗ ∗ x = arg sup[s x − f (x)] =⇒ s − f (x ) = 0 =⇒ f (x ) = s (4.113) x

∗ 0 So x is the location where f (x) = s; i.e. the location where the line with slope s is tangent to f (x). Evaluating ∗ f (s), then, we have ∗ ∗ ∗ f (s) = s x − f (x ) (4.114)

Example

2 Let’s try using the Legendre transformation on a sample function, f (x) = c x . Is this function convex? Yes, because its second derivative is always positive (assuming c > 0). Therefore, we can indeed apply the 0 Legendre transformation to this function. Let the slope of the function be s = f . Then, to change variables from x to the function’s slope, s, we apply the Legendre transformation, ∗ 2 f (s) = sup[s x − c x ], (4.115) x ∂ =⇒ [s x − c x 2] = s − 2c x = 0, ∂x (4.116) s =⇒ x = . 2c (4.117)

We’ve found the maximizer, so we plug it back into the original function that we are maximizing to get s s 2 s2 f ∗(s) = s − c = . 2c 2c 4c (4.118)

∗ Thus, we’ve successfully found the function f (s) that is expressed in terms of the slope, that matches the orignal function f (x). Verifying, we see that indeed (6c)2 f (3) = c(3)2 = 9c, f 0(3) = s(3) = 2c(3) = 6c, f ∗(6c) = = 9c, 4c (4.119) ∗ ∗ 0 =⇒ f (3) = f (s(3)) = f (f (3)) = 9c. X (4.120)

So what do we mean by “non-standard?” This means nothing other than the fact that we change the sign of the dual function, and inﬁmize rather than supremize. It is exactly the same except for a sign change.

4.8.2 Internal Energy

Here we simply review the internal energy function U(C, N, Q), and recall the relations from Coleman-Noll: ∂U ∂U ∂U Se = 2 R T = Y = −R ∂C ∂N ∂Q (4.121) e where S is the elastic part of the second Piola Kirchoff stress tensor and Y is the vector of driving forces corresponding to the internal variables Q.

4.8.3 Helmholtz Free Energy

The Helmholtz free energy A is like the internal energy, except that it is in terms of temperature rather than entropy. (This makes it much easier to work with.) We perform the change of variables using a non-standard Legendre transform: n o A(C, T , Q) = inf U(C, N, Q) − TN N (4.122)

Notice that the stationarity condition requires ∂U T = , ∂N (4.123) the second relation for internal energy. Now, can we ﬁnd relations for this functional similar to that for internal energy? Let us begin by considering the total differential of A in terms of its three arguments: ∂A ∂A ∂A dA = · dC + dT + · dQ ∂C ∂T ∂Q (4.124)

On the other hand, from Equation 4.122, ∂U ∂U ∂U dA = dU − NdT − TdN = ·dC + dN + ·dQ − N dT − T dN ∂C ∂N ∂Q (4.125) |{z} |{z} |{z} S2/2R =T =−Y/R 1 1 = Se · dC − N dT − Y · dQ 2R R (4.126) Equating these two expressions gives the relationships ∂A ∂A ∂A Se = 2R N = − Y = −R ∂C ∂T ∂Q (4.127) e where we see that A and U are identical with respect to S , Y but now we obtain entropy by differentiation of A with respect to temperature.

4.8.4 Enthalpy

e e Enthalpy H(S , N, Q) is the thermodynamic potential in terms of S instead of C. It is also obtained through the following non-standard Legendre transform: n 1 o RH(Se , N, Q) = inf RU(C, N, Q) − Se · C C 2 (4.128) whence we again identify the stationarity condition ∂U Se = 2 R ∂C (4.129) that is in agreement with the relation above. To ﬁnd equilibrium relations, we again expand out; this time, however, we will consider only the terms that involve the transformed variable. ∂U 1 1 ∂H R dH = R ·dC − dSe · C − Se · dC + ... =! R · dSe + ... ∂C 2 2 ∂Se (4.130) |{z} Se /2R 1 ∂H = − C · dSe + ... =! R · dSe + ... 2 ∂Se (4.131) Equating gives the following equilibrium relationships ∂H ∂H ∂H C = −2R T = Y = −R ∂Se ∂N ∂Q (4.132)

4.8.5 Gibbs Free Energy

e The Gibbs Free Energy G(S , T , Q) is the thermodynamic potential in terms of stress and temperature instead of deformation and entropy, and can be deﬁned in terms of the enthalpy n o G(Se , T , Q) = inf H(Se , N, Q) − NT N (4.133) or in terms of the Helmholtz free energy n 1 o RG(Se , T , Q) = inf RA(C, T , Q) − Se · C C 2 (4.134) The same procedure can be followed to ﬁnd that ∂G ∂G ∂G C = −2R N = − Y = −R ∂S e ∂T ∂Q (4.135) e The differential relationships between A, G, U, H and the state variables C, S , T , N can be visualized with a mnemonic:

C TA

U G

1 e N H − 2R S

Example 4.3: Thermodynamic potentials mnemonic

Let us suppose we wish to determine the relationship between Gibbs free energy G, temperature T , and e entropy N. Begin with G. Note that G is not adjacent to N, it is only adjacent to S and T , reminding us that it is not a function of entropy or the deformation temperature. We proceed to T . Note that T is at the arrow end of the line, meaning we go in the reverse direction to get to N. This indicates that there will be a minus sign. The result is then ∂G = −N ∂T (4.136)

This can be used to recover any of the eight relationships derived in this section.

4.9 Inelastic constitutive modeling 4.9.1 Crystal plasticity

The ﬁrst step is to determine the components of the deformation due to elastic vs plastic deformation. The typical picture is that of the following kinematic decomposition:

F = Fe Fp

e Fp F

e p so that F = F F . Note that in small strain, the multiplicative decomposition reduces to the additive decomposition e p ε = ε + ε . Now we need a constitutive model. The relationship ∂W P = F (4.137) no longer holds because W is an elastic potential. However, we know from Coleman-Noll that ∂U ∂U Fe = 2R = 2R , ∂F ∂C (4.138)

p in other words, we can use a potential to determine the elastic portion of the deformation. Now what about F ? We turn to crystallography to determine how pure plastic deformation takes place. FCC materials have four “slip planes” along which dislocations move, and three slip directions corresponding to each plane. Slip plane (m) Slip directions (s) ¯ ¯ ¯ (111) [011], [101], [110] ¯¯ ¯ ¯ ¯ (111) [011], [101], [110] ¯ ¯ ¯ ¯ (111) [011], [101],[110] ¯ ¯ ¯¯ (111) [011], [101],[110] Illustrating the slip planes and directions:

x3 x3 dislocation x3 [110]¯ motion

(111)

[101]¯ x1 x2 x1 x2 x1 x2 [011]¯ [0

There is a total of 4 × 3 = 12 combinations of m and s, and each pair is called a slip system. We can then show that slip along a particular slip system with vectors m and s produces ˙ p p F ∝ (s ⊗ m) F (4.139) So for all combined slip systems we have the model (taken from Bower [1]). N ˙ p p−1 X α F F = γ˙ s ⊗ m, (4.140) α=1 α where {γ } are scalar parameters that determine how much shear has taken place in each slip system. These are 1 N the internal variables for the system so that Q = {γ , ... , γ }. Now that we have Q, what are the thermodynamic driving forces? We have 1 N Y = {τ , ... , τ } (4.141) α where τ is the resolved shear stress. The resolved shear stress for a plane is approximated by α τ = s · P m (4.142) α Now, we need an evolution equation for γ , which we expect to be in terms of the resolved shear stress. The following heuristic equation is an example of one such model !m |τ α| γ˙ α =γ ˙ α sgn(τ α) 0 g α (4.143)

α α where γ0 , and m are material parameters, and g is a hardening parameter. So we’ve solved for one variable and α introduced another - now we need to model the evolution of g . One such model is given by the following N α X β g˙ = hαβ|γ˙ | (4.144) β=1

where hα,β represents that interaction between different slip systems. If it is diagonal, then there is no interaction between dislocations in different systems, which we know to be unphysical. The following relations provide a model for hαβ:

h0 − hs hαβ = qαβh(¯γ) h(γ) = hs + (hs − h0) sech γ¯ (4.145) gs − g0 with hs h0, gs as material constants and γ¯ is the total accumulated slip given by

N Z t X α γ¯ = |γ˙ |dt (4.146) α=1 0

Finally, the matrix qαβ for an FCC has the form ( 1 mα = mβ qαβ = (4.147) q else where q is a material property. This material model is complicated, but does the job in many cases. On the other hand... “With four parameters I can ﬁt an elephant, and with ﬁve I can make him wiggle his trunk.” John von Neumann

5 Computational mechanics

All of the work we have done so far culminated in developing continuous formulations for the mechanics of continuous bodies; these formulations generally take the form of either (i) differential equations or (ii) variational princi- ples. These forms are convenient because they are exact, so any analytical solutions that we ﬁnd will be precisely correct. Most of the time, it’s not practical (or possible) to solve differential equations by hand, especially if the geometry in question is highly complex. This is where it will be helpful to use a computer to approximate the solutions to the equations we derived. There are multiple methods for doing this; here, we will focus on the ﬁnite element method.

5.1 The ﬁnite element method

The finite element method is a dimensionality-reducing technique: rather than solve over the infinite-dimensional space of all functions, we reduce to a finite dimensional approximation of function space. There are three ingredi- ents to this technique: (1) Shape functions (2) The weak formulation of governing equations (3) Numerical quadrature A lengthy discussion of all of these topics is enough to fill an entire class; unfortunately, here, a couple of lectures will have to suffice.

5.1.1 Shape functions

Consider a function f (x) defined over an interval [a, b] (we’ll stick to 1D for the moment.) One way of storing this 2 function is to store a symbolic expression for f (x) such as x or sin(x) that we can use to evaluate f (x) at any point. Unfortunately this does not work well in the general case, since most functions f (x) do not have an analytic form. The next best thing is to choose an interpolation scheme such that instead of defining our function’s value at every point, we define it at a selection of points, and interpolate between them. We do this via the following steps:

(1) Discretize the domain [a, b] to as set of discrete points {xi } ⊂ [a, b].

(2) Deﬁne a set of shape functions φi (x) (3) Interpolate the function as a linear combination of shape functions X f (x) ≈ fi φi (x) (5.1) i

What are these shape functions? In 1D, a typical shape function might look like this:

1 φi (x)

0 xi−1 xi xi+1......

Mathematically, we tend to deﬁne shape functions in such a way as to have the following properties:

(1) Weak Dirac: the shape function φi evaluates to 1 at xi and 0 at all other discretization points; in other words,

φi (xj ) = δij (5.2)

(2) Differentiability: shape functions are at least once differentiable; in other words for (e.g.) [a, b],

{φi } ⊂ C1[a, b] (5.3)

(3) Partition of unity: shape functions sum to 1 at every point; in other words for (e.g.) [a, b], X φi (x) = 1 ∀x ∈ [a, b] (5.4) i

These restrictions on the shape functions mean that we can express our function f (x) as X X f (x) ≈ fi φi (x) = f (xi ) φi (x) (5.5) i i

In other words, we multiply each shape function φi by the functions value at that point, f (xi ). Graphically, we are doing the following: f (x) P {φi (x)} {f (xi )φi (x)} i f (xi )φi (x)

x0 x1 x2 x3 x4 x5 x6 x7 x8 x0 x1 x2 x3 x4 x5 x6 x7 x8 x0 x1 x2 x3 x4 x5 x6 x7 x8

Notice how the formalism of shape functions naturally allows us to interpolate the function to an arbitrary degree of accuracy, depending on the type of shape functions used. This is extensible beyond 1D interpolation: in higher dimensions, shape functions are deﬁned over small discretized elements; for instance, the following is one possible shape function scheme deﬁned over a quadrilateral element:

φ1 φ2 φ3 φ4

x1 x2 x1 x2 x1 x2 x1 x2

x3 x4 x3 x4 x3 x4 x3 x4

The wide varieties of shape functions are beyond the scope of this lecture, although it is well worth reading about them. When you run ANSYS or SolidWorks and select your mesh elements, you are determining what kind of shape functions you will use. Different shape functions have different advantages in different situations, and it is important to know what those are. For now, however, the most important thing is to remember the following: (1) Shape functions are always deﬁned, so we always know what they are.

(2) In addition to knowing φi , we always know what their derivatives (Grad φi ) are.

(3) Shape functions allow any function f (x) to be described in terms of a set of discrete values fi

(4) If the function f is known, then fi = f (xi ), the value of the function at the discretization point.

5.1.2 Weak formulation

Consider a domain Ω with a set of discretization points {xi } ⊂ Ω and shape functions φi ⊂ C1(Ω). (If it is helpful, you can think of Ω = [a, b], the 1D case.) Now, let us suppose that we wish to solve a linear partial differential equation (in space only, no time dependence) that is deﬁned over Ω, with boundary conditions deﬁned on ∂Ω. In 1D, an example might be d 2f + f (x) = g(x) dx 2 (5.6)

We can write a differential equation of this form more generally:

D[f ] − g(x) = 0 (5.7) where D[f ] is a generalized way of writing all contains all of the derivatives and scalar multipliers on f , and g contains “everything else.” Recall that we can write f as X f (x) = fi φi (x), (5.8) so we can express the above as h X i X D fi φi (x) − g(x) = fi D[φi (x)] − g(x) = 0 (5.9) i i

Now, recall the fundamental lemma of the Calculus of Variations? It works both ways: if the above is true, then we can say that Z h X i fi D[φi (x)] − g(x) η(x) dx = 0 ∀η (5.10) Ω i

This is called the weak form. Here’s the important step: remember that η can be anything – so why not let it be X η(x) = ηi φi (x). (5.11) i

If we make this substitution, we have Z h X i X fi D[φi (x)] − g(x) ηj φj (x) dx = 0 ∀{ηj } (5.12) Ω i j

It may seem that we just took an easy problem and made it complicated. Let’s rearrange it to try to simplify: X X Z X Z fi ηj D[φi (x)] φj (x) dx = ηj g(x) φj (x) dx ∀{ηj } (5.13) i j Ω j Ω

Now, recall that the above must hold true for every possible value of ηj . The only way for this to be possible is if the following is true as well: X Z Z fi D[φi (x)] φj (x) dx = g(x) φj (x) dx (5.14) i Ω Ω

(This is the equivalent of passing from the weak form to the strong form in the discretized case.) So, what is the advantage of writing our problem this way? Recall that we are trying to ﬁnd {fi } – if we can ﬁnd them, then we can construct an interpolated solution to the problem. If we look through the problem, we will see that we already

know everything else: we know {φi } so we can compute the integral containing D[φi ], and we know g(x) so we can compute the second integral as well. We might write the problem this way:

fi Kij = bj (5.15) where Z Z Kij = D[φi (x)] φj (x) dx bj = g(x) φj (x) dx (5.16) Ω Ω

K is frequently called the stiffness matrix and b the vector of forces, because the ﬁnite element method was his- torically developed for solid mechanics. Remember, we can compute K and b directly, so the resulting problem is a simple linear solve – quite doable by computer! We glossed over some details here, so let’s make a few notes:

• The integral Z Kij = D[φi (x)] φj (x) dx (5.17) Ω can generally be reduced to a more symmetric form using integration by parts.

• Recall that we let our arbitrary function η be described as X η(x) = ηj φj (x) (5.18) j

Actually, this was something of a design decision: the selection of η to that form is a particular choice, and is part of the so-called Galerkin method. There are other discretization methods, but this is generally the most popular.

5.1.3 Numerical quadrature

Numerical quadrature is nothing other than a fancy term for integral approximation. Recall that we casually stated that the integral Z Kij = D[φi (x)] φj (x) dx (5.19) Ω was easy to compute, since we know D and {φi }. It would be more accurate to say that it is possible to compute – in reality, the integral is not necessarily computable easily, especially if we are working with a geometrically complex mesh. Quadrature is just a method for approximating a nasty integral by evaluating it a speciﬁc points. For instance a simple so-called “quadrature rule” for a 1D integral might be written this way: Z X f (x) dx ≈ qαf (xα) (5.20) Ω α where we would call {xα} quadrature points and {qα} quadrature weights. For an integral over an element, there are conveniently tabulated quadrature points and weights that enable accurate and efﬁcient integral evaluation.

5.2 Linearized kinematics

Let us consider the equation of conservation of linear momentum for the static case in one dimension, where plane stress is assumed: d 2u E + b(x) = 0 dx 2 (5.21)

Suppose we have a discretization {xi } and shape functions φi , and our domain is Ω = [a, b] Let us ﬁnd the ﬁnite element formulation of this equation. First, let

u(x) = ui φi (x) (5.22) so that the equation can be written as 2 X d φi u E + b(x) = 0 i dx 2 (5.23) i Now, we can express the above in weak form as Z b 2 h X d φi i u E + b(x) η(x) dx = 0 ∀η i dx 2 (5.24) a i

Applying the Galerkin method, we can write the above as Z b 2 Z b X d φi X u η E φ (x) dx = − η b(x) φ (x) dx ∀{η } i J dx 2 j j j j (5.25) i a j a

Passing back to the strong form, we have Z b 2 Z b X d φi ui E 2 φj (x) dx = − b(x) φj (x) dx (5.26) a dx a i | {z } Kij

We can use integration by parts to reduce the stiffness matrix a little more: Z b d 2φ dφ b Z b dφ dφ K = E i φ (x) dx = i φ (x) − E i j dx ij 2 j 2 j (5.27) a dx dx a a dx dx | {z } =0

Thus, we can write the equilibrium equation as " Z b # Z b X dφi dφj E dx u = b(x) φ (x) dx dx dx j i (5.28) j a a

We note that this modiﬁed version of the stiffness matrix has one huge advantage: it only requires that the shape functions be differentiable once, whereas the previous required twice differentiability. It is deﬁnitely possible to use “higher order” shape functions that are differentiable this way, but they frequently come with their own set of disadvantages.

5.2.1 3D linearized elasticity

Let us now develop the ﬁnite element formulation of a fully 3d linear elastic material, for which the equations are

Cijkl uk,lj + ρ bi = ρuï (5.29) m m Given a domain Ω with discretization {x } and shape functions {φ } ⊂ C1(Ω), we begin by discretizing our displacement field: m m ui (x) = ui φ (x) (5.30) (Note that we (i) adopt the summation convention, and (ii) use superscripts to indicate summation over all shape functions, as opposed to subscripts summing over all dimensions.) Substituting into our governing equation, we have m m m m uk Cijkl φ,lj + ρ bi = ρuï φ (5.31)

Again, we write in weak form Z m m m m [uk Cijkl φ,lj + ρ bi − ρu¨i φ ] ηi (x) dv = 0 ∀η(x) (5.32) Ω

n n (Notice that we are dotting with η.) Adopting a Galerkin discretization (ηi = ηi φ ), we can now write Z Z Z m n m n n n m n m n n uk ηi Cijkl φ,lj φ dv + ηi ρ bi φ dv =u ¨i ηi ρ φ φ dv ∀{ηi } (5.33) Ω Ω Ω

Because this holds for all test functions, we have Z Z Z m m n n m m n uk Cijkl φ,lj φ dv + ρ bi φ dv =u ¨i ρ φ φ dv (5.34) Ω Ω Ω

Applying the product rule and rearranging, Z Z Z m m n m m n n uk Cijkl φ,j φ,l dv +u ¨i ρ φ φ dv = ρ bi φ dv (5.35) Ω Ω Ω

For the static case, we have a simple linear system to solve. For the dynamic case (with no body forces) we have a linear eigenvalue problem, where the eigenvalues of the stiffness matrix correspond to resonant frequencies, and the eigenvectors to resonant modes.

5.3 Newton’s method

Newton’s method is a powerful technique for solving many kinds of variational problems. Suppose that we have a function f (x) that we wish to optimize, and suppose we have an initial guess x0. Let us expand f (x) about x0: 1 f (x) ≈ f (x ) + f 0(x )(x − x ) + f 00(x )(x − x )2 0 0 0 2 0 0 (5.36)

What is the optimal point of the above approximated function?

0 df 0 00 ! f (x0) = f (x0) + f (x0)(x − x0) = 0 =⇒ x = x0 − 00 (5.37) dx f (x0)

Does this equation give us the exact solution? It depends: if it happens that f (x) was quadratic, then yes. Otherwise, we don’t really know – although it is a good bet that the new point x got us much closer to the actual solution. n What about if we have F (x), a vector functional? We can do exactly the same thing: if our initial guess is x then 1 F (x) ≈ F (xn) + DF (xn) (x − x n) + DDF (xn) (x − x n)(x − x n) p p p 2 pq p p q q (5.38)

Using the stationarity condition we get

dF n n n ! = DF (x )i + DDF (x )ij (xj − xj ) = 0 (5.39) dxi

Solving for x gives n n −1 n xj = xj − DDF (x )ji DF (x )i (5.40)

The Newton iteration is then deﬁned to be n+1 n n −1 n x = x − DDF (x ) DF (x ) (5.41)

5.4 Finite kinematics

We introduce the following discretization:

X α α α α φ(X) ≈ φ η (X) = φ η (X) (5.42) α summation convention

α 0 where {φ} are vector-valued constants and {η (X)} are shape functions that are at least C continuous (so the gradient can be computed). We recall the variational statement of the " Z Z Z # φ = arg inf Π[φ] = arg inf W (F )dV − φi RBi dV − φi Ti dA (5.43) φ φ Ω Ω ∂Ω

Our ﬁnite element scheme admits the statement of the following discretized problem: Z Z Z κ κ κ κ κ κ {φ} = arg inf Π[{φ}] = arg inf W (φ Grad η )dV − φi η RBi dV − φi η Ti dA (5.44) {φ} {φ} Ω Ω ∂Ω

The variational structure of the problem enables a ready solution by means of Newton’s method. Beginning with 0 an initial guess {φ }, the following iteration converges to the solution:

α α −1 αβ β φi 7→ φi − (DDΠ )ij DΠj (5.45) where we compute Z Z Z κ κ κ κ κ κ Π[{φ}] = W (φ Grad η )dV − φi η RBi dV − φi η Ti dA (5.46) Ω Ω ∂Ω Z Z Z α dΠ κ κ α α α DΠ[{φ}]i = α = DW (φ Grad η )ip Grad η,p dV − η RBi dV − η Ti dA = 0 (5.47) dφi Ω Ω ∂Ω 2 Z αβ d Π κ κ α β DDΠ[{φ}] = = DDW (φ Grad η )ijpq Grad η Grad η dV = iJkL ij α β ,p ,q C (5.48) dφi dφj Ω

The function W (F ) is the elastic free energy of the function in terms of the deformation gradient, and dW d 2W DW (F )iJ = = PiJ DDW (F )iJkL = (5.49) dFiJ dFiJ dFkL are the ﬁrst Piola-Kirchoff stress tensor, and the tangent modulus of the material.

5.5 Computational ﬂuid dynamics

Computational ﬂuid dynamics (CFD) involves a completely different method for discretizing the governing equations. We will introduce a discretization referred to as the ﬁnite volume method, although it is worth mentioning that many CFD methods are interrelated, and so there may be a lot of crossover. Let us begin by introducing an arbitrary mesh (we’ll refer to it as an unstructured mesh). n v(x) 1 v1 v v2 4 n2 v¯

n4 v3

We will treat each cell of the mesh as a control volume, for which we already know surface normals {ni }, surface areas {Ai }, and total volume V . We will now use our conservation equations to evolve the velocity ﬁeld in time via the following steps. (1) Recall our conservation of mass expression for a control volume. Here, we will let our discretized cell be the control volume with volume V , so we write dm d(ρV ) Z = = − ρ v · nda (5.50) dt dt ∂V

We assume that the velocities are constant over the faces of the c.v., so we can update the density by the discretized formula ∆t X ρ = ρ − ρ (v · n )A old V i i i i (5.51) i

where Ai are the areas of the cells and ρi , vi , ni are the quantities stored at each face. (2) Recal the expression for the Navier-Stokes momentum equation that we derived (ignoring gravity) ∂ (ρv) + div(ρv ⊗ v + pI) = div(τ) ∂t (5.52)

Let us express this in volume form also: Z ∂ Z Z (ρv)dV = div(τ)dV − div(ρv ⊗ v + pI)dV (5.53) V ∂t V V Z = [τn − ρ(v ⊗ v)n + ρ p n]dV (5.54) ∂V

Now, let us assume that the velocities at the faces are constant over each face, and compute the cell averaged velocity v¯. ∆t X ρv¯ = (ρv¯) + [τn − ρ(v ⊗ v )n + p n ]A old V i i i i i i (5.55) i

Since we already know ρ at the center, we can write the following updated expression for the cell-averaged velocity 1h ∆t X i v¯ = (ρv¯) + [τn − ρ(v ⊗ v )n + p n ]A ρ old V i i i i i i (5.56) i

The viscous stress tensor τ is computed in terms of velocity gradients, so it is a known quantity. (3) We assume that we have a known equation of state, so we can use it to compute the updated pressure:

p = p(ρ, T ) (5.57)

(4) Finally, we determine pressures {pi }, velocities {vi }, and densities {ρi } at the cell boundaries by interpolation. We note that this is a fully explicit scheme, in that it relies only on the current timestep’s values to find new ones. As a result, it cannot conserve energy completely. It is also a highly simplified scheme; there are much better ways of doing the above computation. For incompressible flow, special measures have to be taken to find the pressure field, since the equation of state is ill-defined. One method of doing it is the fractional step or projection method, where the velocity field is estimated, then used to find a pressure field that is used to correct the original velocity field.

References

[1] Allan F Bower. Applied mechanics of solids. CRC press, 2009.

Index basis, 1.2 dummy index, 1.3 free index, 1.3 convexity, 25.2 Kronecker delta, δ, 1.4 duality, 25.2

Einstein summation convention , see index notation piola, 13.4 index notation, 1.3 sets, 1.2