mae5201 - solid mechanics course notes - solid mechanics course notes about ... 1.1index notation...

MAE5201 - Solid MechanicsCourse Notes

AboutThese notes are for the personal use of students who are enrolled in or have taken MAE5201 at the University ofColorado Colorado Springs in the Spring 2017 semester. Please do not share or redistribute these notes withoutpermission.ContentsLECTURE 1 0 Introduction 1.1

0.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.10.2 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2

0.2.1 Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.20.2.2 Proof notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2

1 Tensor Analysis 1.21.1 Index notation and the Einstein summation convention . . . . . . . . . . . . . . 1.3

1.1.1 Vector equality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.41.1.2 Inner product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.41.1.3 Kroneker delta . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.41.1.4 Components of a vector . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.51.1.5 Norm of a vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.51.1.6 Permutation tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.51.1.7 Cross product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.51.1.8 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.6

1.2 Mappings and tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.6LECTURE 2 1.2.1 Second order tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1

1.2.2 Index notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.11.2.3 Dyadic product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.11.2.4 Tensor components . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.21.2.5 Higher order tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.21.2.6 Transpose . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.21.2.7 Trace (first invariant) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.31.2.8 Determinant (third invariant) . . . . . . . . . . . . . . . . . . . . . . . . . 2.31.2.9 Inverse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.41.2.10 The special orthogonal group . . . . . . . . . . . . . . . . . . . . . . . . 2.4

1.3 Tensor calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4LECTURE 3 1.3.1 Gradient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1

1.3.2 Divergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.11.3.3 Laplacian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.11.3.4 Curl . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2

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MAE5201 - Solid MechanicsUniversity of Colorado Colorado Springs Course Notessolids.uccs.edu/teaching/mae5201

1.3.5 Gateaux derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.21.3.6 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.21.3.7 Evaluating derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3

1.4 The divergence theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4LECTURE 4 1.5 Curvilinear coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1

1.5.1 The metric tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.11.5.2 Orthonormalized basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.11.5.3 Change of basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2

1.6 Calculus in curvilinear coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . 4.21.6.1 Gradient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.31.6.2 Divergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.31.6.3 Curl . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4

1.7 Tensor transformation rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5LECTURE 5 2 Linearized Continuum Mechanics 5.1

2.1 Kinematics and the strain tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.12.1.1 Kinematics of local deformation . . . . . . . . . . . . . . . . . . . . . . 5.22.1.2 Volumetric-deviatoric decomposition . . . . . . . . . . . . . . . . . . . . 5.42.1.3 Strain compatibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5

LECTURE 6 2.1.4 Principal strains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.12.1.5 Limitations of strain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2

2.2 Forces/tractions and the stress tensor . . . . . . . . . . . . . . . . . . . . . . . . 6.22.2.1 Cauchy Tetrahedron . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.32.2.2 Principal stresses and directions . . . . . . . . . . . . . . . . . . . . . . 6.52.2.3 Mohr’s circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.5

LECTURE 7 2.3 Constitutive laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.12.3.1 Major and minor symmetries . . . . . . . . . . . . . . . . . . . . . . . . 7.12.3.2 Voigt notation and further symmetry reduction . . . . . . . . . . . . . . 7.22.3.3 Elastic constants for isotropic materials . . . . . . . . . . . . . . . . . . 7.32.3.4 Poisson’s ratio and auxetic materials . . . . . . . . . . . . . . . . . . . . 7.6

LECTURE 8 2.3.5 Strain energy density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.12.4 Balance laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1

2.4.1 Conservation of mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.12.4.2 Conservation of linear momentum . . . . . . . . . . . . . . . . . . . . . 8.12.4.3 Conservation of angular momentum . . . . . . . . . . . . . . . . . . . . 8.3

2.5 Variational methods and duality . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.42.5.1 Introduction to convex analysis and the Legendre transform . . . . . . 8.42.5.2 Variational formulation of mechanics . . . . . . . . . . . . . . . . . . . 8.5

LECTURE 9 2.5.3 Variational calculus and the weak form of equilibrium . . . . . . . . . . 9.13 Elastostatic Solutions 9.1

3.1 2D linear elastic isotropic problems . . . . . . . . . . . . . . . . . . . . . . . . . 9.23.1.1 Plane stress and plane strain . . . . . . . . . . . . . . . . . . . . . . . . 9.23.1.2 Equilibrium in polar coordinates . . . . . . . . . . . . . . . . . . . . . . . 9.4

LECTURE 10 3.2 Airy stress functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.13.2.1 Cylindrical coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2

LECTURE 11 3.3 Lamé solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2


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LECTURE 12 3.3.1 Thin wall approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.13.4 Green’s function solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.2

LECTURE 13 3.5 Linear elastic fracture mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.13.5.1 Stress distributions for mode I-III loading . . . . . . . . . . . . . . . . . 13.4

LECTURE 14 3.5.2 Stress intensity factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.13.5.3 Critical stress intensity factor and criterion for failure . . . . . . . . . . 14.1

3.6 Energy methods for fracture mechanics . . . . . . . . . . . . . . . . . . . . . . . 14.1LECTURE 15 3.6.1 Relationship between energy release rate and stress intensity factor . . 15.1

3.7 J-integral method for fracture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15.33.7.1 Path-independence of the J integral . . . . . . . . . . . . . . . . . . . . 15.4

LECTURE 16 3.7.2 J integral examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16.13.7.3 Nonlinear fracture mechanics . . . . . . . . . . . . . . . . . . . . . . . . 16.3

LECTURE 17 4 Elastodynamic Solutions 17.14.1 Principle of stationary action . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.14.2 Transverse vibrations in Euler-Bernoulli beams . . . . . . . . . . . . . . . . . . . 17.2

LECTURE 18 4.3 Plane waves in solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.2LECTURE 19 4.4 Discontinuous waves in solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.1

4.4.1 Continuity jump condition . . . . . . . . . . . . . . . . . . . . . . . . . . 19.14.4.2 Momentum jump condition . . . . . . . . . . . . . . . . . . . . . . . . . 19.24.4.3 Reduction to 1D case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.34.4.4 X-T diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19.3

LECTURE 20 4.4.5 Wave scattering at an interface . . . . . . . . . . . . . . . . . . . . . . . 20.34.5 Method of Characteristics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.4

LECTURE 21 5 Plasticity 21.15.1 Micromechanics of plasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2

5.1.1 Bridgman’s Hypothesis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.35.2 Ingredients of plasticity theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.35.3 One-dimensional plasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.4

5.3.1 Characterization of uniaxial plasticity . . . . . . . . . . . . . . . . . . . . 21.4LECTURE 22 5.4 Multi-axial plasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22.4

5.4.1 Yield criteria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22.4LECTURE 23LECTURE 24 5.4.2 Flow rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24.1

5.4.3 Hardening model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24.25.5 Crystal plasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24.2

5.5.1 Latent Hardening . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24.5



Lecture 1 Introduction and index notation

0 IntroductionWelcome to solid mechanics. In this course we will formulate the equations of linearized solid mechanics, and willuse them to find solutions to problems. The course will be organized in the following way:(1) Tensor analysis: index notation, tensor algebra and calculus, curvilinear coordinates and transformationrules.(2) Governing equations: formulation of the governing equations of motion for a linearized continuous body.(3) Elastostatic solutions: solution of balance laws for simple static problems, such as thick-walled cylinders.(4) Elastodynamic solutions: solution of balance laws for dynamics problems such as wave propagation.(5) Viscoplasticity: solutions for solids exhibiting dissipative behavior(6) Computational mechanics / structural mechanics / micromechanics – TBD

Note that Sections 0-1 are almost identical to that covered in MAE5100 Continuum Mechanics. The differencebetween this course and continuum mechanics is that we will be solving all problems in a linearized framework,which means that we will be limited to materials like metals, but we will be able to solve a much broader class ofproblems.0.1 MotivationConsider a bar subjected to a tensile load as shown in the following figure. How do we describe this process?

f

`0

`

A

ε = `−`0

`0

σ = fA

σ = E ε

Kinematics:Balance law:Constitutive model:

Governing equationsUndeformed

Deformed

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

Undeformed DeformedWhat 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 isunder applied loads and displacements?All content © 2017-2018, Brandon Runnels 1.1

MAE5201 - Solid MechanicsUniversity of Colorado Colorado Springs Course Notes - Lecture 1solids.uccs.edu/teaching/mae5201

0.2 NotationContinuummechanics is built on the mathematical framework of differential geometry. As a result, the conventionis to use standardmath notationwhen formulating continuummechanics. Additionally, aswewill see, it is generallynecessary 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-dimensional vectors Rn

To indicate that an item is in a set, we use the ∈ symbol. For instance,x ∈ R3 (0.1)

indicates that x is a 3D vector. To indicate that a set is a subset, we use the ⊂ symbol. For instanceR ⊂ Z (0.2)

indicates that the integers are a subset of the real numbers. Another common use is to denote a 3D body:Ω ⊂ R3 (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 simplifydefinitions and theorems.• ∃ is read “there exists.” For instance, ∃x ∈ R3 states that there is at least one item in the R3 set, or that the R3

set is not empty.• ∀ is read “for all.” For instance,

x− x = 0 ∀x ∈ Rn (0.4)tells us that the statement x− x = 0 is true for every possible vector.

Example:∀x ∈ R3 ∃a ∈ R3 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 AnalysisLet us consider the space of three dimensional vectors, R3. A vector r ∈ R3 can be represented in two differentways: in terms of its components, or in terms of basis vectors g1, g2, g3

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g1x

y

z

a b

c

g2

g3

r =

[abc

]r = r1g1 + r2g2 + r3g3

For example, we may haveg1 =

[100

]g2 =

[010

]g3 =

[001

](1.1)

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 thisconvention exclusively; the reason for this will become apparent in the next section.1.1 Index notation and the Einstein summation conventionLet 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:

r =3∑

i=1

ri gi (1.3)It turns out that we write sums like this a lot, and it becomes cumbersome to write the summation symbol everytime. Thus, we introduce the Einstein summation convention, and we drop the explicit sum. This allows us to simplywrite

r = ri gi (1.4)This leads us to define the rules of the summation convention. For a vector equation in Rn , expressed using indexnotation:Rule 1: An index appearing once in a term must appear in every term in the equation, and is not summed. Itis 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. rigi = rjgj )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, ifan index appears twice but is not a dummy index, we use parentheses to denote this, e.g. ui = λ(i) v(i).Fortunately, these cases are pretty rare. Usually, when a rule gets broken, it means that some algebragot 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 conventionwas introduced by Albert Einstein to simplify the equations of generalrelativity. In fact, the formulation of continuum mechanics has a number of similarities to general relativity.)All content © 2017-2018, Brandon Runnels 1.3



1.1.1 Vector equality

Consider two vectors u, v ∈ Rn with components u1, u2, ..., v1, v2, .... In invariant/symbolic notation, we say the twovectors are equal ifu = 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 ofthe equation.1.1.2 Inner product

Let us again consider u, v ∈ Rn. The inner product (or “dot product”) is defined asu · v = uTv = ||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 asuTv = [u1 u2 ... un]

v1v2...vn

= u1v1 + u2v2 + ... + unvn =n∑

i=1

uivi = uivi (1.8)Note that there are no free indices, only dummy indices.1.1.3 Kroneker delta

The Kroneker Delta is defined in the following way:δij =

1 i = j

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 and0 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 dotproduct 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 theis 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)This term has two summed indices, so if we expand it out, we would have n2 terms. However, we know that onlythe terms where i = j survive. Thus, the effect of the Kroneker Delta is to turn one of the dummy indices into theother: in this case, if we “sum over j”

ui vj δij = ui vi (1.14)All content © 2017-2018, Brandon Runnels 1.4



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, weneed 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:εijk =

1 ijk = 123, 231, 312 = "even permutation"−1 ijk = 321, 132, 213 = "odd permutation"0 otherwise

(1.18)

Let’s make a couple of notes here:• εijk is zero if any of the two indices take the same value.• Flipping two indices changes the sign of ε; that is, e.g.

εijk = −εjik = −εikj = −εkji (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 thatg1 × 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 = εijkgk (1.23)

Does this work? Let’s plug in i = 1, j = 2. Then we haveg1 × g2 = ε12kgk =:

0ε121g1 +:0ε122g2 + ε123g3 = 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 εijk gk (1.25)

Alternatively,(u× v)k = εijk 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 NotationEquality u = v uigi = vigi ui = viDot product u · v ui vi ui viCross Product u× v εijk ui vj gk εijk uj vkExample (u · v)w uk vk wi gi uk vk wi

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

1.2 Mappings and tensorsA mapping is a machine that takes a thing of one type and turns it into a thing of another type. For instance,f (x) = x2 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 .A linear mapping f : Rn → Rm is a mapping that satisfies

• f (α x) = α f (x) ∀ x ∈ Rn, α ∈ R

• f (x + y) = f (x) + f (y) ∀ x, y ∈ Rn



Lecture 2 Mappings and tensors

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 ordertensors mapping n-dimensional vectors to m-dimensional vectors is referred to as L(Rn,Rn). We are familiar withthinking of them as n ×m matrices: For example, if A ∈ L(Rn,Rm) and u ∈ Rn, v ∈ Rm then we could write

v = Au

v1v2...vm

=

A11 A12 ... A1nA21 A22 ... A2n... ... . . . ...Am1 Am2 ... Amn

u1u2...un

(1.28)

We can write all possible linear mappings from Rn to Rm in matrix form. Therefore, in general, we can think ofsecond 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 = Au 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 ten-sor is something that must be transformed with a change of coordinates. Thus we say that tensors havetransformation properties but matrices do not.

1.2.2 Index notation

Index notationmakes it very convenient towrite second order tensors and tensor-vectormultiplication. In the aboveexample, we have v1v2...vm

=

A11u1 + A12u2 + ... + A1nunA21u1 + A22u2 + ... + A2nun...Am1u1 + Am2u2 + ... + Amnun

(1.29)

and so we can writevi = Aijuj (1.30)

1.2.3 Dyadic product

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

u⊗ v = u vT =

u1u2...un

[v1 v2 ... vn] =

u1v1 u1v2 ... u1vnu2v1 u2v2 ... u2vn... ... . . . ...unv1 unv2 ... unvn

(1.31)

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:g2 ⊗ g3 =

[010

][0 0 1] =

[0 0 00 0 10 0 0

](1.34)

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) = Aijvk(gi ⊗ gj)gk (1.36)

What do we do with this? Recall that we can write u⊗ v as uvT . Then we have(gi ⊗ gj)gk = gi g

Tj gk = gi (gj · gk) = gi δjk (1.37)

Substituting, we getAv = Aijvkgi δjk = Aijvj gi (1.38)

as expected.We will take this opportunity to reiterate the identity we described earlier: namely, for u, v,w ∈ Rn ,(u⊗ v)w = u (v ·w) (1.39)

This is easily seen using index notation:(uivj)wj = ui (vjwj) (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 ...kgi ⊗ gj ⊗ ...⊗ gk (1.42)

When we start working with contitutive theory, we will frequently see fourth-order tensors (the elasticity tensor). Idon’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. Howcan we express this in index notation?(AT )ij = Aji (1.43)




A tensor A is called symmetric iffAij = Aji (1.44)

A tensor is called antisymmetric iffAij = −Aji (1.45)

(Note that the diagonal terms in an antisymmetric tensor must be zero.) Any tensor can be decomposed into itssymmetric and antisymmetric parts in the following way:A =

1

2(A + A) +

1

2(AT − AT ) =

1

2(A + AT )︸︷︷︸symmetric

+1

2(A− AT )︸︷︷︸

antisymmetric(1.46)

1.2.7 Trace (first invariant)

The trace of a tensor is defined in the folowing way. For u, v ∈ Rn ,tr(u⊗ v) = u · v (1.47)

Let A ∈ L(Rn,Rn). Then the trace is given bytr(A) = tr(Aijgi ⊗ 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 firstinvariant of the tensor. This means that the trace of the tensor does not change under rotation. The significanceof this will become apparent later on.1.2.8 Determinant (third invariant)

A quantity that we use frequently is the determinant of a tensor. For A ∈ L(R3,R3),det(A) =

∣∣∣∣∣A11 A12 A13A21 A22 A23A31 A32 A33

∣∣∣∣∣ =A11A22A33 + A12A23A31 + A13A21A32−A11A23A32 − A12A21A33 − A13A22A31

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

det(A) = εijkA1iA2jA3k (1.50)Alternatively, we can write it in a slightly more satisfying way as

det(A) =1

6εijkεpqrAipAjqAkr (1.51)

Some things to note: for A,B ∈ L(Rn,Rn)

det(A) = det(AT ) (1.52)det(AB) = det(A) det(B) (1.53)

In three dimensions, the determinant is the third invariantof the tensor, whichmeans it is the third of three importantquantities that do not change under rotation.




1.2.9 Inverse

Let A ∈ L(Rn,Rn). The inverse of A, A−1 satisfies, ∀u ∈ Rn ,(A−1)Au = Iu = 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,A−1ij Ajk = Iik = δik (1.55)

For composite mappings,(AB)−1 = B−1A−1 (1.56)

What is the determinant of the inverse of a tensor? For A defined above, we know that AA−1 = I. So we can saythatdet(I) = 1 = det(AA−1) = det(A) det(A−1) (1.57)

sodet(A−1) =

1

det(A)(1.58)

What if det(A) = 0? Then A−1 cannot exist, so its determinant is naturally ill-defined.1.2.10 The special orthogonal group

Consider the set of tensors A ∈ L(R3,R3) such that ATA = I. We call this group of tensors the orthogonal group,and we denote it as S(n) where n is the dimentions. So for A ∈ S(n) we see that• A−1 = AT

• 1/ det(A) = det(A−1) = det(AT ) = 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 specialorthogonal group, and denote it by SO(d). It turns out that SO(d) is exactly the same as the group of all rotationmatrices. In 3D, we will refer to tensors that are in SO(3) very frequently.1.3 Tensor calculusTensor 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:

• Scalar fields f : Rn → R; e.g. temperature, pressure• Vector fields v : Rn → Rn; e.g. displacement, velocity• Tensor fields T : Rn → L(Rn,Rn); e.g. stress, strain

We will be looking at a wide variety of differentiation operations on these types of fields. Note: for this section, weare assuming that we are working in a constant Cartesian basis. That is, we assume that the basis vectors gi areconstants. This is not always true! When we work with curvilinear coordinates, we will have to be very careful abouttaking derivatives of basis vectors.



Lecture 3 Tensor calculus

1.3.1 Gradient

Suppose we have a scalar field φ : Rn → R, φ(x). We define the gradient of φ to begrad(φ) =

∂φ

∂x1g1 +

∂φ

∂x2g2 + ... +

∂φ

∂xngn =

∂φ

∂xigi (1.59)

Note that gradient operator turns a scalar field into a vector field.Now consider a vector field u : Rn → Rn. The gradient of u is defined to begrad(u) =

∂u

∂xi⊗ gi =

∂ui∂xj

gi ⊗ gj (1.60)Note that the gradient operator here turns a vector field into a tensor field.We can generalize this in the following way:

grad(·) =∂(·)∂xi⊗ gi (1.61)

where we drop the dyadic product if · is a scalar field. In general, we only generally care about gradients on scalarand vector fields. However, there are some models that depend on tensor field gradients, such as strain gradientplasticity and ductile fracture.1.3.2 Divergence

For a vector field u : Rn → Rn , the divergence is defined to bediv(u) =

∂u

∂xi· gi =

∂uj∂xi

gj · gi =∂uj∂xi

δij =∂ui∂xi

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

div(T ) =dT

dxigi =

dTjk

dxi(gj ⊗ gk)gi =

dTjk

dxigj(gk · gi ) =

dTjk

dxigjδik =

dTji

dxigj (1.63)

Note that the divergence turns a tensor field into a vector field. We can generalize this in the same way we gener-alized the gradient:div(·) =

∂(·)∂xi· gi (1.64)

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∆φ =

∂2φ

∂xi∂xi(1.66)

Note that we do not write ∂x2i in order to be in keeping with the summation convention.



1.3.4 Curl

The curl operator is defined on a vector field u : R3 → R3 in the following way:curl(u) = − ∂u

∂xi× gi = −∂uj

∂xigj × gi = −∂uj

∂xiεjikgk =

∂uj∂xi

εijkgk (1.67)Note that the curl acts on a vector field and produces a vector field, so we cannot take the curl of a scalar field. Ingeneralized form we have

curl(·) = −∂(·)∂xi× gi (1.68)

1.3.5 Gateaux derivatives

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

Dφ(x)v =d

dεφ(x + εv)

∣∣∣∣ε→0

(1.69)that is, the derivative is taken with respect to ε which is then set to 0.

Example 1.1

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

ε→0

d

dε((xi + εvi )(xi + εvi )) = lim

ε→0

d

dε(xixi + 2εxivi + ε2vivi ) = lim

ε→0(2xivi + 2εvivi ) = 2xivi (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 impossibleto express certain vector operations.(2) We work with a lot of derivatives in continuum mechanics and it frequently becomes cumbersome to write

∂∂xi

. Therefore, we adopt comma notation:∂

∂xi(·) = (·),i (1.71)

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 = εijkvk,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 ≡

∂

∂x∂θ ≡

∂

∂θ∂i ≡

∂

∂xi(1.74)




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 gradientof φ(x) = x · x?In index notation, we have∂xi∂xj

= δij or, in symbolic notation, ∂x

∂x= I (1.75)

Let us look at a couple of examples:Example 1.2

Compute the gradient of φ(x) = x · x. The first step is to write φ in index notation: φ(x) = xkxk . Then, we usethe formula:grad(φ) =

∂

∂xi(xkxk) gi (1.76)

We can use the product rule exactly like we would normally:= (δikxk + xkδik) gi = 2δikxk gi (1.77)

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

Example 1.3

Let φ : R3 → R. Show that Dφ(x)v = grad(φ) · v. To do this we again evaluate the Gateaux derivative:Dφ(x)v = lim

ε→0

d

dεφ(x + ε v) (1.79)

We use the chain rule exactly as we would normally:= limε→0

∂φ(x + ε v)

∂(xi + ε xi )

d(xi + ε vi )

dε= limε→0

∂φ(x + ε v)

∂(xi + ε xi )vi =

∂φ

∂xivi = grad(φ) · v (1.80)

In addition to taking derivatives with respect to vectors (such as x) wemay take derivatives with respect to tensors.For example, given φ : L(Rn,Rn)→ R, we may wish to calculatedφ(T )

dTij(1.81)

Similarly to the case of vectors, we havedTij

dTpq= δipδjq or, in symbolic notation, dT

dT= I⊗ I (1.82)

Example 1.4

Let A : L(Rn,Rn)→ L(Rn,Rn) with A(T ) = TTT . Find the derivative of A with respect to T :∂Aij

∂Tpq=

∂

∂Tpq(TkiTkj) =

∂Tki

∂TpqTkj + Tki

∂Tkj

∂Tpq= δkpδiqTkj + Tkiδkpδjq = Tpjδiq + Tpiδjq (1.83)

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 reuseexisting ones.1.4 The divergence theoremWe have developed enoughmachinery to introduce the singlemost important theorem in all of continuummechan-ics: the divergence theorem.

V(x)

n

Ω

Theorem 1.1 (The divergence theorem). Let Ω ⊂ Rn , and V : R3 → R3 be a differentiable vector field: Then:∫Ω

div(V) dv =

∫∂Ω

V · n da∫

Ω

Vi ,i dv =

∫∂Ω

Vi ni da (1.84)where ∂Ω is the boundary of the body.

We have a similar theorem for tensor fields: Let T : R3 → L(R3,R3) be a tensor field. Then∫Ω

div(T ) dv =

∫∂Ω

Tn da (1.85)Using index notation in an Cartesian coordinate system, we can write∫

Ω

Vi ,i dv =

∫∂Ω

Vi ni da

∫Ω

Tij ,j dv =

∫∂Ω

Tij nj da (1.86)What does this mean? We are relating a volume integral of the divergence to a surface integral – or, in this case, aflux integral. For the case of a vector field, the integral of the divergence over the body can be intuitively thoughtof as the total amount of compression/expansion in the vector field. The flux integral can be thought of as thetotal amount of vector field entering or leaving the body. Thus, you can think of the divergence theorem as themathematical formulation of the statement “the total compression of the vector field is equal to the rate of fluxthrough the boundary.” We will make extensive use of the divergence theorem in this course.



Lecture 4 Tensor calculus in curvilinear coordinates

1.5 Curvilinear coordinatesUp until now we have worked within a simple Cartesian basis, let us call it ei . However, it is frequently convenientto switch to a more natural coordinate system:

g1

g2

g3

r = r1g1 + r2g2 + r3g3

gr

gθ

Consider a new set of coordinates θ1, θ2, θ3, ... , θn. Position as a function of these coordinates is expressed asx(θ1, θ2, ... , θn) (1.87)

Let us define a new basis: a1, a2, ... , a3 defined asa1 =

∂x

∂θ1a2 =

∂x

∂θ1... ai =

∂x

∂θi=∂xj∂θi

ej (1.88)We will refer to ai as the covariant basis vectors.1.5.1 The metric tensor

The metric tensor g is defined asgij = 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 canmake sure that they are by defining scale factors hi = ||ai ||. Then we define a new basisgi =

a(i)

h(i)no summation over i (1.90)

Example 1.5

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



θ

r

z

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:ar =

∂xi∂r

ei =

[cos θsin θ

0

]aθ =

∂xi∂θ

ei =

[−r sin θr cos θ

0

]az =

∂xi∂z

ei =

[001

](1.92)

Our scale factors areh1 = ||ar || = 1 h2 = ||aθ|| = r h3 = ||az || = 1 (1.93)

so we havegr =

[cos θsin θ

0

]gθ =

[− sin θcos θ

0

]gz =

[001

](1.94)

Important note: while e1, e2, e3 are independent of x1, x2, x3, gi is not necessarily independent of θi. In ourabove example, we see that∂

∂θgr =

[− sin θcos θ

0

]= gθ

∂

∂θgθ =

[− cos θ− sin θ

0

]= −gr (1.95)

1.5.3 Change of basis

Let gi be an orthonormal basis for Rn , and let v ∈ Rn. Suppose we wish to find vi such that v = vigi . To dothis, we use the orthogonality property of the basis:v · gj = vigi · gj = viδij = vj =⇒ v = (v · gi ) gj (1.96)

Suppose we have another basis ei. Then we can relate the two bases by writingei = (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 coordinatesNow that we have defined a framework for working in other coordinate systems, we need to knowwhat our calculusoperations look like in those systems. Before we do that, however, we need to introduce a couple of importantidentities.(1) Earlier we learned that

gi =1

h(i)a(i) =

1

h(i)

∂xj∂θ(i)

ej no sum on i (1.98)




Howcanweexpress ei in termsof gj? Todo this, we’ll pull a trick. Remember that gi formsanorthonormalbasis for Rn: that means that we can express any vector in terms of that basis. For instance, a vector v canbe written as v = vigi . 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

ei = (ei · gj) gj =∑j

1

hj(ei · aj) gj =

∑j

1

hj(ei ·

∂xk∂θj

ek) gj =∑j

1

hj

∂xk∂θj

(ei · ek)︸︷︷︸δik

gj =∑j

1

hj

∂xi∂θj

gj (1.100)

(Note that we broke one of our summation convention rules. To compensate for this we drop the summationnotation and use an explicit sum.)(2) There is an important theorem called the inverse function theorem that states:[∂θ

∂x

]=[ ∂x∂θ

]−1

=⇒ ∂θ

∂x

∂x

∂θ= I or, in index notation, ∂θi

∂xk

∂xk∂θj

= δij (1.101)We will use both of these rules to derive expressions for the familiar divergence, gradient, and curl in curvilinearcoordinates.1.6.1 Gradient

We want to express the gradient as computed in the abovegrad(f (θ)) =

∂

∂xi(f (θ)) ei =

∂f

∂θj

∂θj∂xi

ei =∂f

∂θj

∂θj∂xi

(∑k

1

hk

∂xi∂θk

gk)

=∑k

1

hk

∂f

∂θj

∂θj∂xi

∂xi∂θk︸︷︷︸δjk

gk =∑k

1

hk

∂f

∂θkgk (1.102)

Notice how this is almost identical to our original expression for the gradient, except that xi, ei have beenreplaced with θi, gi. The only difference is the presence of the scale factors. Let’s solidify this with an example:Example 1.6

Let us continue with our example of cylindrical polar coordinates. Let f = f (r , θ, z). Then we have:grad(f ) =

1

hr

∂f

∂rgr +

1

hθ

∂f

∂θgθ +

1

hz

∂f

∂zgz =

∂f

∂rgr +

1

r

∂f

∂θgθ +

∂f

∂zgz (1.103)

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 thedivergence of this vector field? We can follow the exact same procedure as when computing the gradient:div(v) =

∂

∂xi(v) · ei =

∂v

∂θj

∂θj∂xi· ei =

∂v

∂θj

∂θj∂xi·∑k

1

hk

∂xi∂θk

gk =∑k

1

hk

∂v

∂θj

∂θj∂xi︸︷︷︸Jji

∂xi∂θk︸︷︷︸J−1ik

·gk =∑k

1

hk

∂v

∂θjδjk · gk (1.104)

=∑k

1

hk

∂v

∂θk· gk (1.105)




Notice again howwe arrive at an almost identical formula except that we include scale factors. Wemust alsomakeone very important note: how do we evaluate the following?∂

∂θk(v) =

∂

∂θk(vigi ) =

∂vi∂θk

gi + vi∂gi∂θk

(1.106)In Cartesian coordinates the basis vectors are constant so their derivatives vanish. However, this is not true inmost other curvilinear coordinates! To illustrate this, let’s do an example:

Example 1.7

Compute the divergence of a vector field in cylindrical polar coordinates: v = vrgr + vθgθ + vzgz .div(v) =

1

hr

∂v

∂r· gr +

1

hθ

∂v

∂θ· gθ +

1

hz

∂v

∂z· gz

=∂

∂r(vrgr + vθgθ + vzgz) · gr +

1

r

∂

∂θ(vrgr + vθgθ + vzgz) · gθ +

∂

∂z(vrgr + vθgθ + vzgz) · gz

=(∂vr∂r

gr +∂vθ∂r

gθ +∂vz∂r

gz)· gr

+1

r

(∂vr∂θ

gr +∂vθ∂θ

gθ +∂vz∂θ

gz + vr∂

∂θgr + vθ

∂

∂θgθ + vz

∂

∂θgz)· gθ

+(∂vr∂z

gr +∂vθ∂z

gθ +∂vz∂z

gz)· gz

=∂vr∂r

+1

r

(∂vθ∂θ

gθ + vrgθ − vθgr)· gθ +

∂vz∂z

=∂vr∂r

+1

r

(∂vθ∂θ

+ vr)

+∂vz∂z

(1.107)Notice how we picked up a couple of extra terms: this is a result of our choice of coordinate system. Thisis 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 curvilinearcoordinates, we havecurl(v) = − ∂

∂xi(v)× ei = − ∂v

∂θj

∂θj∂xi×∑k

1

hk

∂xi∂θk

gk = −∑k

1

hk

∂v

∂θj

∂θj∂xi

∂xi∂θk× gk = −

∑k

1

hk

∂v

∂θjδjk × gk (1.108)

= −∑k

1

hk

∂v

∂θk× gk (1.109)

Once again, we see that we recover a very similar expression except for the presence of scale factors.




Example 1.8

Find the expression for the curl in cylindrical polar coordinates. We can reuse quite a bit of what we com-puted earlier; we just need to be careful about which vectors we cancel out.curl(v) =−

[ 1

hr

∂v

∂r× gr +

1

hθ

∂v

∂θ× gθ +

1

hz

∂v

∂z× gz

]=−

(∂vr∂r

gr +∂vθ∂r

gθ +∂vz∂r

gz)× gr

− 1

r

(∂vr∂θ

gr +∂vθ∂θ

gθ +∂vz∂θ

gz + vrgθ − vθgr)× gθ

−(∂vr∂z

gr +∂vθ∂z

gθ +∂vz∂z

gz)× gz

=(∂vθ∂r

gz −∂vz∂r

gθ)

+1

r

(− ∂vr∂θ

gz +∂vz∂θ

gr + vθgz)

+(∂vr∂z

gθ −∂vθ∂z

gr)

=(1

r

∂vz∂θ− ∂vθ

∂z

)gr +

(∂vr∂z− ∂vz

∂r

)gθ +

(gz)

+1

r

(∂(r vθ)

∂r− ∂vr∂θ

)gz (1.110)

1.7 Tensor transformation rulesSuppose we have two orthonormal ei, gi. We recall that orthonormality allows us to write each basis in termsof the other:

ei = (ei · gp) gp gq = (gq · ei ) gi (1.111)Now, let us suppose we have a tensor A ∈ L(Rn,Rn) 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

Aij ei ⊗ ej = Aij [(ei · gp) gp]⊗ [(ej · gq) gq] = (gp · ei )︸︷︷︸QT

pi

Aij (ej · gq)︸︷︷︸Qjq

gp ⊗ gq = QTpi Aij Qjq gp ⊗ gq = Apqgp ⊗ gq

(1.112)where Apq are the components of A in the other basis. Or, put more simply, we have that

Apq = QTpiAijQjq (1.113)

Note that we are not transforming the tensor itself: we are merely changing the components of the tensor to fitwith the assigned basis. This is called a tensor transformation property.Let’s take another look at the Q matrices. Specifically, let’s look at QTQ :[QTQ]pq = QT

piQiq = (gp · ei )(ei · gq) = [(gp · ei ) ei ]︸︷︷︸gp

·gq = gp · gq = δpq = [I]pq (1.114)

Because QTQ = I, we conclude that QT = Q−1. 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.



Lecture 5 Deformations and the strain tensor

2 Linearized Continuum MechanicsWith the tools of tensor calculus, we are now equipped to formulate the linearized equations of elasticity, alterna-tively, linearized continuum mechanics for small deformation. We begin by defining kinematic quantities (defor-mations, strains) and force quantities (forces, stresses). We then define balance laws and constitutive laws forsmall-strain strain solid mechanics.2.1 Kinematics and the strain tensorWe will begin by formulating the mathematical representation of bodies undergoing deformation. Let us considera body that occupies a region B . We suppose that the body undergoes a deformation such that each point x mapsto a new point y; that is, we have a deformation field y(x). This is illustrated by the following figure:

y(x)x

u(x)

x1

x2

We also define the displacement field u(x) = y(x)− x, which we will use extensively in our fomulation of linearizedcontinuum mechanics.Example 2.1: Axial stretch

What is the deformation mapping for the following stretched cube?

1 1

1

λ1

λ2

λ3

y(x)

x1

x2

x3

We identify the deformation mapping simply asy1 = λ1 x1 y2 = λ1 x2 y3 = λ1 x3 (2.1)

Or, we can describe this asy =

[λ1 0 00 λ2 00 0 λ3

]x (2.2)

Then the displacement field is:u = y − x =

[λ1 − 1 0 0

0 λ2 − 1 00 0 λ3 − 1

]x (2.3)



Example 2.2: Simple shear

What is the deformation mapping for the following cube subjected to pure shear.

1 1

1

y(x)

1 1

γ

1

x1

x2

x3

We identify the deformation mapping to bey1 = x1 y2 = x2 + γ x3 y3 = x3 (2.4)

Or, we can describe it asy =

[1 0 00 1 γ0 0 1

]x (2.5)

Then the displacement field is given byy = y − x =

[0 0 00 0 γ0 0 0

]x (2.6)

The above two examples are of “affine” or “homogeneous” deformations, which are characterized by the form:y = F x + x0 (2.7)

A less formal way of describing an affine/homogeneousmapping is one in which “straight lines remain straight” or,equivilantly “all squares map similarly. The following figure illustrates the difference between affine/homogeneousand non-affine/inhomogeneous mappings:

undeformedaffine non-affine

yy

2.1.1 Kinematics of local deformation

The above picture is very general but not very helpful, since y(x) can be anything. What is more useful is to analyzethe deformation of a material at a single point. As long as the deformation is smooth, (and it usually is) we canapproximate the deformation at every point as an affine deformation:

undeformed non-affine

φ

locally affine




Therefore, let us examine how very small displacement vectors change under deformation, i.e. let us find a rela-tionship between original vectors ∆x1, ∆x2 and the deformed vectors ∆y1, ∆y2, as shown below:

x0

x1

x2

∆x1

∆x2∆y1

∆y2

y(x0)

To analyze this case, we take a Taylor series expansion:yi (x0 + ∆x1) = yi (x0) +

∂yi∂xj

(x0) ∆xj + h.o.t. (2.8)The local approximation means we can ignore higher order terms. So, the above expression simplifies to

∆yi =∂yi∂xj

∆xj ∆y = F∆x (2.9)where F = ∂y/∂x = grad(y), the deformation gradient tensor. Now, recall that

y(x) = x + u(x) (2.10)where u is the displacement field. Substituting this into the deformation gradient expression gives:

Fij =∂yi∂xj

=∂xi∂xj

+∂ui∂xj

= δij +∂ui∂xj

F = I + grad(u) (2.11)where grad(u) is the displacement gradient tensor. Now, recall that any tensor can be split into its symmetric andantisymmetric components, viz

grad(u) =1

2(grad(u) + grad(u)T )︸︷︷︸

strain tensor+

1

2(grad(u)− grad(u)T )︸︷︷︸infinitesimal rotation tensor

(2.12)

This prompts the following two definitions: the (infinitesimal) strain tensor:εij =

1

2(ui ,j + uj ,i ) ε =

1

2(grad(u) + grad(u)T ) (2.13)

and the infinitesimal rotation tensor:rij =

1

2(ui ,j − uj ,i ) r =

1

2(grad(u)− grad(u)T ) (2.14)

Example 2.3: Strain tensor calculation

For stretching:u1 = (λ1 − 1)x1 u2 = (λ2 − 1)x2 u3 = (λ3 − 1)x3 (2.15)

The displacement gradient and strain tensors are:grad(u) =

[λ1 − 1

λ2 − 1λ3 − 1

]ε =

[λ1 − 1

λ2 − 1λ3 − 1

](2.16)




For simple shear:u1 = 0 u2 = γ x3 u3 = 0 (2.17)

The displacement gradient and strain tensors are:grad(u) =

[0 0 00 0 γ0 0 0

]ε =

[0 0 00 0 1

2γ0 1

2γ 0

](2.18)

2.1.2 Volumetric-deviatoric decomposition

Consider a cube that is defined by the three basis vectors e1, e2, e3, and let them undergo a deformation such thatthey become Fe1,Fe2,Fe3.

e1

e2

e3

Fe1

Fe2Fe3

F

You may recall from calculus that the volume of the parallelpiped spanned by three vectors a,b, c is given by thetriple scalar product, i.e.

vol(a,b, c) = [a b c] = a · (b× c) = det |a b c| (2.19)It is easy to see that the original volume of the cube is Vold = [e1, e2, e3] = 1. But what about the new volume? Letus take the triple scalar product of the transformed vectors:

Vnew = [Fe1,Fe2,Fe3] = [F1 F2 F3] = det |F1 F2 F3| = detF (2.20)In other words:

VnewVold = detF (2.21)

Now, recall that F = grad y = I + gradu, and let us evaluate the volume ratio:VnewVold = det

[1 + u1,1 u1,2 u1,3u2,1 1 + u2,2 u2,3u3,1 u3,2 1 + u3,3

]= (1 + u1,1)(1 + u2,2)(1 + u3,3) +O(u2)

= 1 + u1,1 + u2,2 + u3,3 +O(u2) = 1 + ε1,1 + ε2,2 + ε3,3 +O(u2)

The notation O(u2) means that all the other terms contain a factor of u2 or higher. Because we are making thesmall strain approximation, we neglect higher order terms, allowing us to approximate the determinant as

≈ 1 + tr(ε)

(2.22)Rewriting slightly, we have

∆V

Vold = tr(ε) (2.23)




In other words, the amount of volume change associated with a deformation is equal to the trace of the straintensor. This prompts the following two definitions: the volumetric strain tensor

εvol =1

3tr(ε) I εvolij =

1

3δijεkk (2.24)

and the deviatoric strain tensor

εdev = ε− 1

3tr(ε) I εdevij = ε− 1

3δijεkk (2.25)

where we see that the deviatoric stress tensor (also known as the “stress deviator”) is always traceless by design,and that ε = εdev + εvol. As an example, consider the cases of axial stretch and pure shear as described above:tr(εstretch) = λ1 + λ2 + λ3 − 3 tr(εshear) = 0 (2.26)

From this we see that shear is volume preserving whereas axial stretch is not (in general).2.1.3 Strain compatibility

Given a strain tensor field ε, is there a corresponding displacement field u such that ε = 12 (grad(u) + grad(u)T )?Maybe, maybe not: to answer this question precisely we need to introduce the idea of compatibility.Suppose that u does exist. We know that

εij =1

2(ui ,j + uj ,i ) (2.27)

and we can trivially write2εij ,kl = ui ,jkl + uj ,ikl 2εkl ,ij = uk,lij + ul ,kij (2.28)2εik,jl = ui ,kjl + uk,ijl 2εjl ,ik = uj ,lik + ul ,jik (2.29)

Let’s try evaluating the following:2εij ,kl + 2εkl ,ij − 2εik,jl − 2εjl ,ik︸︷︷︸

=curl(curl(ε))

= ui ,jkl + uj ,ikl + uk,lij + ul ,kij − ui ,kjl − uk,ijl − uj ,lik − ul ,jik (2.30)= ui ,jkl − ui ,kjl︸︷︷︸

=0

+ uj ,ikl − uj ,lik︸︷︷︸=0

+ uk,lij − uk,ijl︸︷︷︸=0

+ ul ,kij − ul ,jik︸︷︷︸=0

= 0 (2.31)

thus a necessary condition for the existance of u iscurl(curl(ε)) = 0 (2.32)

Does this mean that if the above equation is satisfied, then u must exist? The answer is yes – but we have notproven that here. If you’d like to see a rigorous proof that the above equation is sufficient for compatibility, checkthe continuum mechanics notes.In 3D this is a total of 34 equations; in 2D the compatibility equation simplifies toε11,22 + ε22,11 − 2ε12,12 = 0 (2.33)



Lecture 6 Tractions and the stress tensor

2.1.4 Principal strains

Consider a material undergoing the deformationu1 = γx2 u2 = γx1 (2.34)

such that the strain tensor (in 2D) isε =

1

2(grad(u) + grad(u)T ) =

1

2

( [0 γγ 0

]+[

0 γγ 0

] )=[

0 γγ 0

] (2.35)where the visualization of the deformation is pure shear, as shown below.

v1v2

e1

e2

The principal directions of the strain tensor are the directions in which the strain is purely longitudinal, i.e. the basisin which there is no shear. The principal strains are the longitudinal shears in those directions. The principal direc-tions and principal strains correspond to the eigenvectors and eigenvalues of the strain tensor. Let us computethe principal strains and directions for the above example by computing the eigenvalues and eigenvectors:det(ε− λI) = det

[−λ γγ −λ

]= λ2 − γ2 = 0 =⇒ λ = ±γ (2.36)

Now, let us compute eigenvectors and normalize:(ε− (−γ)I)v1 =

[−γ γγ −γ

]v1 = 0 =⇒ v1 =

[1/√

21/√

2

](2.37)

(ε− (+γ)I)v1 =[γ γγ γ

]v2 = 0 =⇒ v2 =

[−1/√

21/√

2

](2.38)

Note that v1 · v2 = 0, implying that they are orthogonal, and that |v1| = |v2| = 1, showing that they are orthonormal.Indeed, they form an orthonormal basis, called the “eigenbasis” of the strain tensor.Let us compute the components of ε εpq such that ε = εpqvp ⊗ vq. We recall that the formula derived earlier gives:ε = QT εQ =

[vT1vT2

] [0 γγ 0

][v1 v2] =

[1/√

2 1/√

2−1/√

2 1/√

2

] [0 γγ 0

] [1/√

2 −1/√

21/√

2 1/√

2

](2.39)

=1

2

[1 1−1 1

] [0 γγ 0

] [1 −11 1

]=

1

2

[1 1−1 1

] [γ γγ −γ

]=

1

2

[2γ 00 −2γ

]=[γ 00 −γ

] (2.40)Note that the components of ε in this basis are diagonal; this is frequently referred to as diagonalization. (Also,note that the trace remains the same!) As a check, let us confirm that we recover the original strain tensor:

ε = εij vi ⊗ vj = ε11︸︷︷︸=γ

v1 ⊗ v1 +>0

ε12 v1 ⊗ v2 +>0

ε21 v2 ⊗ v1 + ε22︸︷︷︸=−γ

v2 ⊗ v2 (2.41)

= γ

[1/2 1/21/2 1/2

]− γ

[1/2 −1/2−1/2 1/2

]=[

0 γγ 0

] (2.42)as expected.All content © 2017-2018, Brandon Runnels 6.1


2.1.5 Limitations of strain

Consider two deformations as illustrated in the following figure:

u1 = γx2

γ

γ

u2 = 0 u2 = γx1

u1 = 0

= =

What is the strain tensor for each?ε =

1

2(grad(u) + grad(u)T ) =

[0 γ/2γ/2 0

](2.43)

The strain tensor is identical to each, although the two types of deformations are fundamentally different. Whatabout the infinitesimal rotation tensor?r1 =

1

2(grad(u)− grad(u)T ) =

[0 γ/2−γ/2 0

]r2 =

1

2(grad(u)− grad(u)T ) =

[0 −γ/2γ/2 0

](2.44)

Notice that the rotation tensors are opposite each other. We can think of both deformations as being similar upto rotation, as long as the deformations are small. If the deformations are large, or if there is a large amount ofrotation, the strain tensor becomes a bad measure of deformation.So if the strain tensor is limited, why do we use it? The strain tensor has one major advantage: it is symmetric,meaning that a 2D strain tensor carries only 3 unknowns (instead of 4), and a 3D tensor carries only 6 unknowns(instead of 9).Let us explore anothermajor advantage of symmetric tensors: supposeA ∈ L(Rn,Rn) is symmetric so thatA = AT .Let us also suppose that T has eigenvalues λ1, ... ,λn corresponding to eigenvectors v1, ... , vn. Let us alsosuppose that all of the eigenvalues λ are distinct. Using the fact that A is symmetric we can write the following,where j 6= i

0 = vi · A vj − vi · ATvj = vi · A vj − vj · Avi = λjvi · vj − λi vj · vi = (λj − λi ) vi · vj (2.45)The above must be zero, but we have constrained λi 6= λj . Therefore, all eigenvectors must be orthogonal. Thisleads us to the key result: all symmetric tensors have orthogonal eigenvectors.(You may be wondering: what about the case where two eigenvectors have the same eigenvalues? This is calleddegeneracy and does not destroy the proof, in fact, it makes it easier but slightly more messy. If you would like tosee the proof for this case, it is in the continuum mechanics notes.)2.2 Forces/tractions and the stress tensorConsider an arbitrary solid body occupying a region Ω. What kinds of forces can be applied to it?

Ω

t : ∂Ω→ R3

b : Ω→ R3

∂Ω

We define two types of forces on an arbitrary body: the surface tractionwhich assigns a force vector to every pointon the surface of the body; mathematically, t : ∂Ω → R3. Note that the units of surface tractions are force per




unit area (in 3D) and force per unit length (in 2D). We can use surface traction to compute the total external forceapplied on the body viz

ftraction =

∫∂Ω

t da (2.46)The other type of forces are referred to as body forceswhich assign a force vector to every point within the interiorof the body; mathematically, b : Ω → R3. (Common examples of body forces are gravity and magnetism.) Notethat the units of body forces are force per unit volume (in 3D) and force per unit area (in 2D). The total force appliedby internal loading can be computed viz

fbody forces =

∫Ω

b dv (2.47)Finally, let us consider the momentum of the body per unit volume. Recall that momentum is mass times velocity,so momentum per unit volume is mass per unit volume times velocity, or density times velocity.

y(t)

x

u(t)

v(t) = y(t) = u(t)ρ

But what is the velocity? We recall that the position is given by y(t), so velocity is just the derivative y. But sinceu(t) = y(t) − x and x is constant, then v = u. So, the momentum per unit mass is ρ u, and we can compute thetotal momentum of the body by

p =

∫Ω

ρ u dv (2.48)2.2.1 Cauchy Tetrahedron

Consider a tetrahedron as shown below subject to four tractions t1, t2, t3, t ditributed uniformly across the fourfaces of the body as shown below:

x1

x2

x3

x1

x2

x3

−t2

−t1

−t3

−e2

−e1

−e3

n

t∆A

∆A2∆A1

∆A3

−t2

−t1

−t3

t

Supposing that there are no additional body forces such that only tractions are applied. Then equilibrium requiresthat the integral of all tractions must equal zero, i.e.∫∂Ω

t da = t∆A− t1 ∆A1 − t2 ∆A2 − t3 ∆A3 = 0 (2.49)We know from geometry that ∆Ai = n · ei ∆A, so we have

t∆A− t1 (e1 · n) ∆A− t2 (e2 · n) ∆A− t3 (e3 · n) ∆A = 0 (2.50)t = (t1 ⊗ e1)n + (t2 ⊗ e2)n + (t3 ⊗ e3)n = [t1 t2 t3]n (2.51)




Defining the Cauchy stress tensor σ = [t1 t2 t3] we arrive at the relationt(n) = σ n (2.52)

Intuitively, the stress tensor is a machine that converts a normal vector of a surface to the traction acting on thesurface. As an example, consider the following:

σ(−e1) = −(t1 ⊗ e1) e1 − (t2 ⊗ e2) e1 − (t3 ⊗ e3) e1 = −t1(e1 · e1)−:0

t2(e2 · e1)−:0

t3(e3 · e1) = −t1 (2.53)In practice, we will care far less about tractions than about the stress tensor itself. Let us see how we will use it inpractice by looking at an example:

Example 2.4: Stress tensor in uniaxial tension

Consider a beam with cross sectional area A subjected to equal and opposite forces f in the x1 direction.−f fA

What is the stress tensor for this beam? (Note: if we are far from the application of the point load we canassume that the load is distributed.)σ11 =

f

A,σ12 = σ21 = σ22 = 0 =⇒ σ =

[f /A 0

0 0

] (2.54)Now, let us find the tractions on the following cut in the beam:

−f fnf

Our normal vector is n = [1 0]T so the force isf = σ n =

[f /A 0

0 0

] [10

]=[f /A

0

] (2.55)Now let us consider the following cut

−f fn

f

where n =[1/√

2 1/√

2]T . The force acting on this surface is

f = σ n =[f /A 0

0 0

] [1/√

21/√

2

]=

[f /√

2A0

](2.56)

Notice that the direction is unchanged but the magnitude is reduced because the surface area is greater.This may seem a bit obvious, so let us consider another slightly more tricky example:

Example 2.5

Consider a cube subjected to pure shear loading as shown below:




e1−e1

e2

−e2

[0τ

]

−τ

[0−τ]

[−τ0

]

[τ0

]n n

Computing the components of the stress tensor we find:t(e1) =

[σ11σ12

]!

=[

0τ

]t(e1) =

[σ11σ12

]!

=[τ0

]=⇒ σ =

[0 ττ 0

] (2.57)Let us consider the cases where n1 =

[1/√

2 1/√

2]T , n2 =

[−1/√

2 1/√

2]T – what is the force on thosefaces?

σ n1 =[

0 ττ 0

] [1/√

21/√

2

]=

[τ/√

2τ/√

2

]= τ n1 σ n2 =

[0 ττ 0

] [−1/√

21/√

2

]=

[τ/√

2−τ/√

2

]= −τ n2 (2.58)

From this we see that n1,n2 are, by definition, eigenvectors of σ with eigenvalues τ ,−τ .

2.2.2 Principal stresses and directions

Motivated by the above example, we suggest that for any stress tensor we can always find a basis in which thestress tensor is diagonal, i.e. a frame in which the stresses are normal. For instance, consider the following picture:

We see that in one frame, the stress tensor has both normal and shear stresses, while in the rotated frame, allstresses are normal. That is, we seek a set of basis vectors v1, ... , vn wheretn = σvn = σnvn (2.59)

where σn is the magnitude of the normal stress on face n. As above, we solve for σn, vn via|σ − σnI|

!= 0 (2.60)

i.e. we find eigenvalues and eigenvectors of σ. These are called the principal stresses and principal directions ofthe stress tensor. Wewill show presently that σ is symmetric; consequently, the principal directions are orthogonal.2.2.3 Mohr’s circle

A popular graphical tool for visualizing principal stresses is “Mohr’s circle.” We will derive it here and then demon-strate its use for a few cases. Let us assume that we are in the principal basis for σ (in 2D), but we wish to findAll content © 2017-2018, Brandon Runnels 6.5



the components of the stress tensor in a basis that is rotated by an angle θ. We recall that the components of thestress tensor in a rotated frame are given by:RTσR =

[cos θ sin θ− sin θ cos θ

] [σ1 00 σ2

] [cos θ − sin θsin θ cos θ

]=[

cos θ sin θ− sin θ cos θ

] [σ1 cos θ −σ1 sin θσ2 sin θ σ2 cos θ

] (2.61)=

[σ1 cos2 θ + σ2 sin2 θ (σ2 − σ1) sin θ cos θ(σ2 − σ1) sin θ cos θ σ1 sin2 θ + σ2 cos2 θ

](2.62)

We note the following trigonometric identities:sin θ cos θ =

1

2sin(2θ) cos2 θ =

1 + cos(2θ)

2sin2 θ =

1− cos(2θ)

2(2.63)

Substituting, the expression becomesσ =

[12 (σ1 + σ2)− 1

2 (σ2 − σ1) cos 2θ 12 (σ2 − σ1) sin 2θ

12 (σ2 − σ1) sin 2θ 1

2 (σ1 + σ2) + 12 (σ2 − σ1) cos 2θ

](2.64)

Let us plot the locus of possible values on a plot in which the abscissa (x axis) represents normal stress, while theordinate (y axis) represents shear stresses. The result is the following:

2θ

σ

τ

12 (σ1 + σ2)

σ2σ1

σ1

σ2

τ

−τ

σ1

σ2

ττ

σ2

σ1

θ12 (σ2 − σ1)

Every rotation corresponds to a bisection of the circle at an rotation 2θ from the main axis, where the coordinatesof the intersection points of the bisecting line with the circle give the stress state.This visualization enables us to compute some very useful properties of an arbitrary stress tensor by using geom-etry. Suppose we have a stress tensor that is not in the principal frame. We know thatσ =

[σ11 ττ σ22

]=

[12 (σ1 + σ2)− 1

2 (σ2 − σ1) cos 2θ 12 (σ2 − σ1) sin 2θ

12 (σ2 − σ1) sin 2θ 1

2 (σ1 + σ2) + 12 (σ2 − σ1) cos 2θ

](2.65)

We’d like to know what the principal stresses, σ1,σ2 are, and the maximum shear stress τmax , as well as the totalrotation to get to the principal frame, θ. We could solve the above system of equations, but it is much easier to justuse geometry. The location of the center of the circle and its radius are given by:σcenter =

1

2(σ11 + σ22) r =

√σ2

11 + τ 2 =σ2 − σ1

2= τmax (2.66)

so the principal stresses areσ1,2 =

1

2(σ11 + σ22)±

√σ11

2 + τ 2 (2.67)



Lecture 7 Constitutive theory

2.3 Constitutive lawsNow that we have introduced the stress and strain tensors, we now ask how these tensors relate to each other.We recall that for a simple one-dimensional elastic body, that we use Hooke’s law to relate stress and strain via theequation

σ = E ε (2.68)where E is the familiar elastic modulus. This is fine in 1D, but in 3D both stress and strain are tensorial quantitieswith 9 components (6 accounting for symmetry) each.To represent Hooke’s law (i.e. linear relationship between stress and strain) in 3D, we use the following relationship:

σij = Cijklεkl (2.69)where the fourth-order tensor Cijkl is referred to as the elastic modulus tensor. Note that every possible linearrelationship between ε and σ is captured by this equation. Because it is fourth-order, we think of C as a 3x3x3x3matrix, meaning that it contains 34 = 81 constants. In general, the number of constants can be significantly reducedas we show below.2.3.1 Major and minor symmetries

We know that ε is symmetric, therefore we knowσij = Cijklεkl =

1

2Cijklεkl +

1

2Cijklεlk =

1

2Cijklεkl +

1

2Cijlkεkl =

1

2(Cijkl + Cijlk)εkl (2.70)

implying that Cijkl = 12 (Cijkl +Cijlk), i.e. C is symmetric with respect to its last two indices. That reduces the numberof constants to 3× 3× 6 = 54 constants.We will also show that σ is symmetric, so we can say that

Cijklεkl = σij = σji = Cjiklεkl =⇒ Cijkl = Cjikl (2.71)reducing the number of constants to 6× 6 = 36. The relationships

Cijkl = Cjikl = Cijlk (2.72)are called minor symmetries of C.Notice that we can derive C as the derivative of

Cijkl =∂σij∂εkl

=∂

∂εkl(Cijpqεpq) = Cijpqδkpδlq = Cijkl (2.73)

But stress (by definition, as we will show later) is itself the derivative of the free energy funciont W ; i.e.εij =

∂W (ε)

∂εij(2.74)

Consequently, we can write C as a second derivative of W , and use the fact that the order of derivatives can beswapped in order to show that:Cijkl =

∂2W

∂εij∂εkl=

∂2W

∂εkl∂εij= Cklij (2.75)



The property that Cijkl = Cklij is called the major symmetry of the elastic modulus tensor. As a result, the numberof independent constants are reduced to∑6n=1 n = 21. Summarizing, we have

Cjikl =minor Cijlk =minor Cijkl =major Cklij =minor Clkij =minor Cklji (2.76)Note: some indices cannot be swapped, for instance, Cijkl 6= Cikjl . Remember that the ijs always have to be adjacentand the kls have to be adjacent.2.3.2 Voigt notation and further symmetry reduction

Recall that since ε and σ are symmetric, they have only six independent constants each. We can use this to stream-line some of our notation: let us define the Voigt stress and strain vectors: given stress and strain tensors σ, ε

σ =

σ11σ22σ33σ23σ31σ12

ε =

ε11ε22ε33

2ε232ε312ε12

(2.77)

Note the factors of two in the shear terms for ε – these are here to compensate for the factor of 1/2 that resultsfrom symmetrization of ε. With the Voigt stress and strain vectors defined, we can reduce C from a 4th order tensorto a more traditional 2nd order tensor, so that the relationship becomes:σ11σ22σ33σ23σ31σ12

=

C11 C12 C13 C14 C15 C16C12 C22 C23 C24 C25 C26C13 C23 C33 C34 C35 C36C14 C24 C34 C44 C45 C46C15 C25 C35 C45 C55 C56C16 C26 C36 C46 C56 C66

ε11ε22ε33

2ε232ε312ε12

σ = C ε σi = Cij εJ (2.78)

which is general enough to describe the linear elastic relationships for an arbitrary anisotropic material. In general,we can use material symmetries to reduce the number of constants even further.• Transversely isotropic material: Materials of this type are symmetric within a plane but not in the directionorthogonal to the plane. A carbon nanotube reinforce composite is an example of a transversely isotropicmaterial.The elastic modulus for a material of this type is

σ11σ22σ33σ23σ31σ12

=

C11 C12 C13C12 C11 C13C13 C32 C33

C44C44

12 (C11 − C12)

ε11ε22ε33

2ε232ε312ε12

(2.79)

with a total of six constants. Notice how subscripts withvalues 1 or 2 can be switched with no effect, but indices with3 cannot. x1

x2

x3

• Cubic material: Materials of this type have symmetry with respect to all three axes – i.e. will perform identi-cally under 90 rotations. Single-crystal materials with FCC or BCC crystal structures are examples of mate-rials with cubic symmetry.




The elastic modulus tensor for cubic materials isσ11σ22σ33σ23σ31σ12

=

C11 C12 C12C12 C11 C12C12 C12 C11

C44C44

C44

ε11ε22ε33

2ε232ε312ε12

(2.80)

with a total of only three constants. x1

x2

x3

• Isotropic material: Materials of this type have symmetry with respect to all rotations – that is, it performsidentically in a rotated frame as in a non-rotated frame. Examples of materials of this type include polycrys-talline metals, and most of the materials we will work with we will assume to be isotropic.The isotropic elastic modulus tensor is

σ11σ22σ33σ23σ31σ12

=

λ+ 2µ λ λλ λ + 2µ λλ λ λ+ 2µ

µµ

µ

ε11ε22ε33

2ε232ε312ε12

(2.81)

dependent on just two constants, λ and µ, referred to as“Lamé parameters.” x1

x2

x3

2.3.3 Elastic constants for isotropic materials

For a linear isotropic material, we can represent the stress-strain relationship asCijkl = µ(δikδjl + δilδjk) + λ δijδkl , (2.82)

giving the following stress-strain relationshipσij = Cijklεkl = µδikδjlεkl + µδilδjkεkl + λ δijδklεkl = µ(εij + εji ) + λ δijεkk = 2µεij + λ δijεkk (2.83)

σ = 2µε+ λ tr(ε)I (2.84)You may be wondering what happened to the more familiar material constants, Young’s modulus and Poisson’sratio. In fact, we can derive those constants (and others) in terms of the two Lamé parameters, as we see in thefollowing examples:




Example 2.6: Uniaxial stress

Consider the case where a material is subjected to stress in onedirection where the other sides are constrained to be stress-free.In other words, σij = σ11 δ1i δ1j Recalling thatσ11 δ1i δ1j = 2µεij + λδijεkk (2.85)

For i 6= j we have that 2µεij = 0, implying that ε has no off-diagonal terms. Therefore we can express the above as a linear3x3 system as follows (i.e. using Voigt notation and ignoringshear terms):[σ1100

]=

[λ+ 2µ λ λλ λ + 2µ λλ λ λ+ 2µ

][ε11ε22ε33

](2.86)

σ11

σ22 = 0 σ33 = 0

Solving this linear system we findε11 =

( λ+ µ

µ(3λ+ 2µ)

)σ11 ≡

σ11

Eε22 = ε33 = − λ

2µ(3λ+ 2µ)σ11 = − λ

2(λ+ µ)ε11 ≡ −ν ε11 (2.87)

where we identify the following constants:E =

µ(3λ+ 2µ)

λ+ µ≡ Young’s Modulus ν =

λ

2(λ+ µ)≡ Poisson’s Ratio (2.88)




Example 2.7: Uniaxial strain

Consider the case, similar to the above, where a material is sub-jected to a stress in one direction but constrained in the othertwo directions so that the material cannot contract or expand.What is the relationship between stress and strain in this case?(Note: we expect the material to be more rigid in this direction.)For uniaxial strain we can represent the strain tensor as, εij =ε11 δ1i δ1j . Substituting this into the constitutive relation we have

σij = 2µε11 δ1i δ1j + λδijε11 δ1k δ1k

= (2µ δ1i δ1j + λδij) ε11 (2.89)Determine σ11,σ22,σ33 by letting i = j = 1, 2, 3:

σ11 = (2µ+ λ) ε11 σ22 = σ33 = λ (2.90)

σ11

ε22 = 0ε33 = 0

We identify the linear relationship between stress and strain asE = 2µ+ λ ≡ Contrained Modulus (2.91)

(Note: another case where we say we have uniaxial strain is when a sample is very large in the lateraldirection and therefore cannot expand laterally.)Example 2.8: Simple shear

Consider a material undergoing pure shear (e.g. in the 1-2 plane), so that the strain tensor isε =

[0 1

2γ 012γ 0 00 0 0

](2.92)

What is the stress tensor? We get it directly by plugging into the constitutive equation:σ = 2µε+ λ tr(ε)I = 2µ

[0 1

2γ 012γ 0 00 0 0

]+ (λ)(0)I =

[0 µγ 0µγ 0 00 0 0

]=

[0 τ 0τ 0 00 0 0

](2.93)

where we see that τ = µγ, so we identify µ as the Shear Modulus.Example 2.9: Hydrostatic pressure

Consider the case where a material is subjected to hydrostatic pressure, i.e. σ = −pI. Substituting into theconstitutive relations:σ = −pI = 2µε+ λ tr(ε)I (2.94)

Taking the trace of both sides:tr(−pI) = tr(2µε) + tr(λ tr(ε)I) (2.95)−3p = 2µ tr(ε) + 3λ tr(ε) (2.96)




Rearranging we find−p =

(2

3µ+ λ

)tr(ε) =

(2

3µ+ λ

)∆V

V(2.97)

We recall that the relationship between volume change and pressure is given by the bulk modulus, so weidentifyκ =

2

3µ+ λ ≡ Bulk modulus (2.98)

In conclusion, there are about six parameters that are frequently used for isotropic linear elasticity: Lamé’s firstparamater (λ), the shear modulus (µ), Young’s modulus (E ), the constrained modulus (E ), Poisson’s ratio (ν), andthe bulk modulus (κ). However, only two are needed; any of these constants can be determined as a function ofany two other constants. Wikipedia has a very helpful table that provides all possible relationships between theseelastic moduli.2.3.4 Poisson’s ratio and auxetic materials

Let us compute the above moduli in terms of E , ν: solving the above equations we haveλ =

Eν

(1 + ν)(1− 2ν)µ =

E

2(1 + ν)κ =

E

3(1− 2ν)(2.99)

We know that E ,λ,µ,κ must all be greater than zero, otherwise materials would spontenously explode. But whatabout Poisson’s ratio? By assuming the above to be positive, we have thatλ > 0 =⇒ ν > −1 & ν <

1

2µ > 0 =⇒ ν > −1 κ > 0 =⇒ ν <

1

2(2.100)

This tells us that Poisson’s ratio could take any value between −1 and 12 ; in other words, it can be negative. Whatdoes a material with a negative Poisson’s ratio look like? They actually exist, and they are called materials with

auxetic structures. An example of a possible auxetic microstructure that results in a negative Poisson’s ratio isshown below:



Lecture 8 Derivation of balance laws

2.3.5 Strain energy density

At this point we define a quantity thatwill be useful later whenwediscuss variational formulations of the equilibriumequations. Let a material be subjected to a stress σ that results in a strain ε. We define the following (in both fullnotation and Voigt symbolic notation)w [ε] =

∫ ε

0

σij dεij , w [ε] =

∫ ε

0

σ · dε ≡ Strain energy density (2.101)In terms of the elastic modulus tensor, the above integral is integrated exactly as

w [ε] =1

2εijCijklεkl w [ε] =

1

2εiCijεj (2.102)

We see that the stress tensor is recovered accordingly by differentiation:σpq =

∂W

∂εpq=

1

2

∂

∂εpq(εijCijklεkl) =

1

2(δipδjqCijklεkl + εijCijklδkpδlq) =

1

2(Cpqklεkl + εijCijpq) = Cpqklεkl (2.103)

The result follows similarly using Voigt notation. Note that we go from energy → stress → elastic modulus bydifferentiating W .The strain energy for an entire body Ω is given by integration of the strain energy density over the volume:W [u] =

∫Ω

1

2εijCijklεkl dv (2.104)

We will return to use these definitions in a later section.2.4 Balance lawsIn this section we introduce the governing physical equations of linear solid mechanics. Here we focus on conser-vation of mass, linear momentum, and angularm momentum. Conservation of energy is, generally, automatic inelasticity; we will revisit it, however, when introducing elasto-visco-plasticity.2.4.1 Conservation of mass

Consider a region with volume V1 and density ρ1 that transforms to a region with volume V2 and density ρ2. Weknow that the total mass in this region must be the same; we can use this to relate the two densities:m1 = m2 =⇒ ρ1V1 = ρ2V2 =⇒ ρ1

ρ2=

V2

V1= 1 + tr(ε) (2.105)

Conservation of mass is generally significantly more complex when working with large deformation or fluid flow.2.4.2 Conservation of linear momentum

Here we derive the conservation of linear momentum in a linearized continuum mechanics framework. We taketwo approaches: bottom-up (which is a little more intuitive) and top-down (which is a little more rigorous).



Derivation 1: Let us consider a square piece of a two-dimensional solid, and let this square is loated at the originof an arbitrary coordinate system. The cube is subjected to a stress field σ(x1, x2), where we note that the units ofstress in 2D are force per unit length (as opposed to area). Let the cube also have a density ρ(x1, x2) where in 2Ddensity is mass per unit area instead of volume. Finally, let the cube be subjected to a body force field b(x1, x2),where b is force per unit area. Drawing the cube with stress components indicated explicitly we have:

σ21(∆x1/2, 0)

σ11(∆x1/2, 0)

σ21(−∆x1/2, 0)

σ11(−∆x1/2, 0)

σ22(0,−∆x2/2)

σ22(0, ∆x2/2)

σ12(0, ∆x2/2)

σ12(0,−∆x2/2)

b

Now, let us use Newton’s second law to find an equation of motion for this cube:∑f = m a (2.106)

Therefore the equilibrium equations in the x and y directions are:σ11(∆x1/2, 0)∆x2 − σ11(−∆x1/2, 0)∆x2 + σ12(0, ∆x2/2)∆x1 − σ12(0,−∆x2/2)∆x1 + b1∆x1∆x2 = ρ∆x1 ∆x2 u1

σ22(0, ∆x2/2)∆x1 − σ22(0,−∆x2/2)∆x1 + σ21(∆x1/2, 0)∆x2 − σ21(−∆x1/2, 0)∆x2 + b2∆x1∆x2 = ρ∆x1 ∆x2 u2

Now, divide both sides by ∆x1∆x2:σ11(∆x1/2, 0)− σ11(−∆x1/2, 0)

∆x1+σ12(0, ∆x2/2)− σ12(0,−∆x2/2)

∆x2+ b1 = ρ u1

σ21(∆x1/2, 0)− σ21(−∆x1/2, 0)

∆x1+σ22(0, ∆x2/2)− σ22(0,−∆x2/2)

∆x2+ b2 = ρ u2

and take the limit as ∆x1, ∆x2 → 0:∂σ11

∂x1+∂σ12

∂x2+ b1 = ρ u1

∂σ21

∂x1+∂σ22

∂x2+ b2 = ρ u2 (2.107)

Or writing more compactly using index or symbolic notationσij ,j + bi = ρui div(σ) + b = ρ u (2.108)

We see that this generalizes easily to 3D as well.Derivation 2: This is a more elegant way to derive the local balance laws. Consider a body subjected to tractionforces and body forces. We know from Newton’s second law that∫

∂Ω

t da︸︷︷︸total traction force

+

∫Ω

b dv︸︷︷︸total body force

=d

dt

∫Ω

ρ u dv︸︷︷︸momentum

(2.109)

Recall that t = σ n, so we can use the divergence theorem to write∫∂Ω

t da =

∫∂Ω

σ n da =

∫Ω

div(σ) dv (2.110)




Substituting this and rearranging (noting that Ω does not depend on time, so the derivative can be taken inside theintegral) we get ∫Ω

[div(σ) + b− ρ u] dv = 0 (2.111)Because this holds for all sub-bodies B ⊂ Ω, we can use the fundamental lemma of the calculus of variations torecover the form of momentum conservation above.2.4.3 Conservation of angular momentum

Derivation 1: Consider the diagram shown for Derivation 1 for conservation of linear momentum. Recalling thebalance of moments for a rigid body, let us find the sum of moments acting about the center of the cubic region:∑M =

∆x2

2σ21(∆x1/2, 0) +

∆x2

2σ21(−∆x1/2, 0)− ∆x1

2σ12(0, ∆x2/2)− ∆x1

2σ12(0,−∆x1/2) (2.112)

Letting ∆x1 = ∆x2 → 0 we findσ21 = σ12 (2.113)

Continuing in the same way for the 3D case we findσ23 = σ32 σ31 = σ13 σ12 = σ21 (2.114)

Put more compactly,σij = σji σ = σT (2.115)

That is, we have shown that conservation of angular momentum requires the Cauchy stress tensor to be sym-metric.

Derivation 2: Assuming the static case, let us compute the total moment exerted on the body by surface tractionsand body forces:M =

∫∂Ω

[x× t] da +

∫∂Ω

[x× b] dv!

= 0 (2.116)Expressing the traction term using index notation, we use the divergence theorem to rewrite:∫

∂Ω

εijkxj tk da =

∫∂Ω

εijkxjσklnl da =

∫Ω

∂

∂xl(εijkxjσkl) dv =

∫Ω

[εijkσkj + εijkxjσkl ,l ] dv (2.117)Substituting back into the original expression, we have

Mi =

∫Ω

εijkσkjdv +

∫Ω

εijkxjσkl ,l dv +

∫Ω

εijkxjbk =

∫Ω

εijkσkjdv +

∫Ω

εijkxj [σkl ,l + bk ] dv︸︷︷︸=0

=

∫Ω

εijkσkjdv!

= 0 (2.118)

Because the equality must hold for all sub-regions, the integrand must be zero as well. Looking at each term wesee thatε1jkσjk = σ23 − σ32 = 0 ε2jkσjk = σ31 − σ13 = 0 ε3jkσjk = σ12 − σ12 = 0 (2.119)

i.e. σ = σT .




2.5 Variational methods and dualityIn this section we switch gears to introduce some tools that are very fundamental, transcending solid mechanicsto apply to many fields such as dynamics, thermodynamics, and electromagnetism.2.5.1 Introduction to convex analysis and the Legendre transform

The Legendre transform is a powerful mathematical tool that is perticularly useful when working with energy meth-ods and with minimization problems. (Note: this is not to be confused with Legendre polynomials, which areentirely different.) Let us develop the transform by considering a simple function f (x). We will require that f be C 0

continuous and convex, but will place no other restrictions on it.

x x

f (x)

x

f (x)

s

mf (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 forf (x). Now, let’s consider an alternative approach. Recall that we require that f is convex; this means that the regionabove 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 tangentlines to the epigraph at all of the points along f (x). Notice how we can recapture the shape of the epigraph usingthe 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 theepigraph. In particular, let us suppose that we have the following function:

f ∗ : [slope of tangent line]→ [negative intercept of tangent line] (2.120)So f ∗ generates all possible tangent lines to the epigraph of f (x); we say that f ∗ is the Legendre Transform of f . Thiscomes as little help, however, if we don’t have a way of constructing f ∗. Let us consider the following formulation:

f ∗(s) = supx

[s x − f (x)] (2.121)To solve this problem we first solve for x∗, the minimizing value, using the stationarity condition:

x∗ = arg supx

[s x − f (x)] =⇒ s − f (x∗) = 0 =⇒ f (x∗) = s (2.122)So x∗ is the location where f ′(x) = s ; i.e. the location where the line with slope s is tangent to f (x). Evaluatingf ∗(s), then, we have

f ∗(s) = s x∗ − f (x∗) (2.123)which, we see, is exactly the negative of the intercept of the tangent line. This is not a rigorous proof but hopefullyit helps give an intuitive understanding of why the Legendre transform works. Let us apply this formulation byintroducing an example:




Example 2.10: Dual polynomial

Find the dual (Legendre transformation) of f , where f : R+ → R has the formf (x) =

1

pxp (2.124)

with p ∈ R+. We begin by using the direct form for the Legendre transform:f ∗(s) = sup

x

[s x − 1

pxp] (2.125)

Use the stationarity condition to solve for x∗:d

dx

[s x∗ − 1

p(x∗)p

]= s − (x∗)p−1 = 0 =⇒ x∗ = s

1p−1 (2.126)

Substituting back we havef ∗(s) = s x∗ − 1

p(x∗)p = s s

1p−1 − 1

p(s

1p−1 )p = s

p−1p − 1

ps

pp−1 =

(p − 1

p

)s

pp−1 =

1

qsq (2.127)

where q = p/(p − 1), alternately,1

p+

1

q= 1 (2.128)

From this we see that the quadratic function 12x

2 is self-dual. The other interesting case is that if p = 1 thenq =∞.

Let us here make an important note: what if the function is nonconvex? Then our tangent line construction doesnot fully capture the structure of f (x) as we see below:

xx

This is why it is important to restrict to convex functions. However, nonconvexity does in fact play a significantrole in mechanics, and we will see that nonconvexity is the fundamental explanation for the development of mi-crostructure in materials.2.5.2 Variational formulation of mechanics

Let us begin by considering a simple example of a chain of linear springs.

u1 u2 u3 u4

k1 k2 k3 k4f1 f2 f3 f4




Ignoring the presence of any forces for a moment, let us compute the energy of the system as a function of thedisplacements:W (u1, u2, u3, u4) =

1

2k1u

21 +

1

2k2(u2 − u2)2 +

1

2k3(u3 − u2)2 +

1

2k4(u4 − u3)2︸︷︷︸

potential energy of system(2.129)

This expression is fine, but we are generally not interested in the behavior of the system as a function of displace-ments. We are much more interested in the behavior of the system as a function of the applied forces, f1, f2, f3, f4.To do this, let us compute the Legendre transform of the system:W ∗(f1, f2, f3, f4) = sup

u1,u2,u3,u4

[f1u1 + f2u2 + f3u3 + f4u4 −W (u1, u2.u3, u4)] (2.130)Note that we can alternatively write this as a minimization problem:

W ∗(f1, f2, f3, f4) = − infu1,u2,u3,u4

[W (u1, u2.u3, u4)︸︷︷︸potential energy

−(f1u1 + f2u2 + f3u3 + f4u4︸︷︷︸applied loads

)] (2.131)

Sowe see that we obtain the principle ofminimumpotential energy in the process of finding the dual representationof the energy.Now, let us apply this methodology to a linear elastic solid. The strain energy of the solid is given by the followingfunctional.W [u] =

∫Ω

1

2σijεijdv =

∫Ω

1

2Cijklui ,juk,ldv (2.132)

This allows us to evaluate the strain energy for any given prescribed deformation, just as we could evaluate theenergy for any (u1, u2, u3, u4) like we had before. Now, let us suppose that we are more interested in computingthe deformation as a function of loading – what we know to be body forces. To do that, we will again compute aLegendre transform, but this time it will be over two continuous vector fields:W [b] = − inf

u

[ ∫Ω

1

2Cijklui ,juk,ldv −

∫Ω

uibidv] (2.133)

Because we are working with a continuous body, we have to consider what is going on at the boundary. Let ussuppose that our body is undergoing a deformation with both a prescribed load and a prescribed displacement.Let us then divide up our boundary as follows:

Ω

∂1Ω

∂2Ω

The total boundary of the body is ∂Ω = ∂1Ω∪∂2Ω, where ∂1Ω is the region in which the displacement is prescribed,and ∂2Ω is the region in which the loading is prescribed. What about free boundaries? Free boundaries are exactlyboundaries where the applied surface tractions are prescribed to be zero; therefore, they are in ∂2Ω.We must now include the applied surface tractions in our Legendre transform. Fundamentally they are no differ-ent than body forces except in how they are integrated; for this reason, we treat them differently. Our variationalprinciple thus becomesW [b, f] = − inf

u

[ ∫Ω

1

2Cijklui ,juk,l dv −

∫Ω

uibi dv −∫∂2Ω

ui ti da] subject to u = u0 on ∂1Ω (2.134)




We see that in the process of obtaining our energy as a function of loading, we have computed the equilibriumconfiguration of the system, which is what we were interested in in the first place. Therefore we state that ourequation of equilibrium is given by the following:u(x) = arg inf

u

[ ∫Ω

1

2Cijklui ,juk,l dv −

∫Ω

uibi dv −∫∂2Ω

ui ti da] subject to u = u0 on ∂1Ω (2.135)

This is known as the principle of minimum potential energy, and it is a very helpful tool in solving many types ofproblems.



Lecture 9 Variational principle and elastostatic solutions

2.5.3 Variational calculus and the weak form of equilibrium

Let Π[u] be equal to the objective function that is minimized in the principle of minimum potential energy. Supposethat u is a minimizer of Π - then any variation would cause Π to increase. Let v(x) be an amissible variation, where“admissible” means that v(x) = 0 on ∂1Ω. Then, u + εv is a deformation that satisfies the displacement boundarycondition, although it is no longer a minimizing field unless ε = 0. We can capitalize on this fact: we know thatΠ[u + εv] is minimized when ε = 0, so we can use the following equation

d

dεΠ[u + εv]ε→0

!= 0 (2.136)

to derive conditions on u. If we substitute the actual value for Π and solve directly, we find:d

dεΠ[u + ε v] =

d

dε

[ ∫Ω

1

2Cijkl(ui ,j + εvi ,j)(uk,l + εvk,l) dv −

∫Ω

(ui + εvi ) bi dv −∫∂2Ω

(ui + εvi )ti da]ε→0

=[ ∫

Ω

1

2Cijkl(ui ,jvk,l + vi ,juk,l + 2εvi ,jvk,l) dv −

∫Ω

vi bi dv −∫∂2Ω

vi ti da]ε→0

=

∫Ω

Cijklui ,jvk,l dv −∫

Ω


vi ti da

=

∫Ω

[ ∂∂xl

(Cijklui ,jvk)− Cijklui ,jlvk,l

]dv −

∫Ω


vi ti da

=

∫∂2Ω

[Cijklui ,jvknl − vi ti ] da−∫

Ω

[Cijklui ,jlvk,l + vibi

]dv =

∫∂2Ω

[σijnjvi − ti vi ] da−∫

Ω

[σij ,jvi + vibi ] dv

=

∫∂2Ω

[σijnj − ti ] vi da−∫

Ω

[σij ,j + bi ] vi dv!

= 0 ∀v admissible (2.137)This is called the weak form of the governing equations, and it is used to derive the equations used in the finiteelement method. In this case, however, we notice that the above equationmust be true for all admissible variations.The only way for this to be true is if

σij ,j + bi = 0 on Ω σijnj − ti = 0 on ∂2Ω (2.138)which are our standard equations of equilibrium.3 Elastostatic SolutionsFor elastostatic solutions we consider problems in which there is no time dependence. Our governing equationsfor a linear elastic solid become.

Elastostatic boundary value problem:

Equilibrium div(σ) + b = 0 in Ω (3.1a)Strain grad(u) + grad(u)T = 2ε in Ω (3.1b)Constitutive laws Cε = σ in Ω (3.1c)Displacement BC u = u0 on ∂1Ω (3.1d)Traction BC σn = t0 on ∂2Ω (3.1e)

Ω

∂1Ω

∂2Ω



For this section we will consider the static case where all time derivatives vanish. Furthermore, we will neglectgravity. We can simplify futher: the minor symmetry of the elastic modulus tensor C enables the combination of(3.1b) and (3.1c) as follows:σij = Cijklεkl =

1

2Cijkl(uk,l + ul ,k) =

1

2(Cijkluk,l + Cijlkul ,k) =

1

2(Cijkluk,l + Cijkluk,l) = Cijkluk,l (3.2)

Substituting into (3.1a) and (3.1e) results in the following system that can be solved directly for u(x):Cijkluk,lj = 0 on B ui = u0

i on ∂1B Cijklnjuk,l = t0i on ∂2B (3.3)

We have a set of differential equations that define the deformation of a solid under the action of displacementboundary conditions (also called Dirichlet or essential boundary conditions) and traction boundary conditions (alsocalled Neumann or natural boundary conditions). Because these differential equations are linear, their solutionshave some convenient properties:Superposition Let (t1,b1), (t2,b2) be two seperate loading configurations on a system, resulting in dis-placement fields u1,u2. The displacement field resulting from the combined loadings

b1 + b2 and t1 + t2 is the sum of the two individual displacement fields u = u1 + u2

= +

= +

Proportionality If a loading is multiplied by a constant, then the resulting displacement field is multipliedby the same constant.δ 2δ

2ff

St Venant’s Principle The stress state at a distance sufficiently far from the applicaiton of a loading is inde-pendent of the distribution of the loading itself.

σ σ=

3.1 2D linear elastic isotropic problemsWe want to solve some real problems, so we seek to simplify the above expression even further. In particular, it ismuch more tractable to solve problems in two dimensions only instead of three. Let us consider a couple of casesof 2D problems: plane strain and plane stress.3.1.1 Plane stress and plane strain

Consider twomaterials that have identical cross-sections but very different thicknesses - onematerial is very thick,while the other is quite thin.




σ33 = 0

ε33 = 0

Very thick materials are less likely to be extendible in the out-of-plane direction, so we approximate this behaviorby assuming that the out-of-plane strain is zero. We call this case plane strain.On the other hand, the thin material is very free to move out of plane; however, since the surfaces are free, we knowthat there cannot be any stress on them. Therefore, we model this behavior by assuming that the out-pf-planestress is zero. This is called plane stress.We impose the following constraints:ε13 = ε23 = ε33 = 0 (3.4)

In other words, all out-of-plane strains must completely vanish. How does this simplify our problem? Let us firstconstruct a modified constitutive equation:[σ11 σ12 σ31σ12 σ22 σ23σ31 σ23 σ33

]= 2µ

[ε11 ε12 0ε12 ε22 00 0 0

]+ λ(ε11 + ε22)I =

[(2µ+ λ)ε11 + λε22 2µε12 0

2µε12 λε11 + (2µ+ λ)ε22 00 0 λ(ε11 + ε22)

](3.5)

Now, we simplify the results and rewrite them in 2D only:[σ11 σ12σ12 σ22

]=

[(2µ+ λ)ε11 + λε22 2µε12

2µε12 λε11 + (2µ+ λ)ε22

]σ33 = λ(ε11 + ε22),σ31 = σ23 = 0 (3.6)

Notice that we have a complete set of equations in 2D only, and once we’ve solved for the 2D stress and straintensors, we can compute the out-of-plane stress components. Using index notation with Greek indices α,β,κ(where Greek indices, by convention, sum from 1-2 only), we write the constitutive law as:Linear Isotropic Plane Strain

σαβ = 2µεαβ + λ εκκ δαβ σ33 = λ(ε11 + ε22),σ31 = σ23 = 0 (3.7)For plane stress: [

σ11 σ12 0σ12 σ22 00 0 0

]= 2µ

[ε11 ε12 ε31ε12 ε22 ε23ε31 ε23 ε33

]+ λ(ε11 + ε22 + ε33)I (3.8)

Simplifying and identifying that ε31 = ε23 = 0,[σ11 σ12 0σ12 σ22 00 0 0

]=

[(2µ+ λ)ε11 + λε22 + λε33 2µε12 0

2µε12 λε11 + (2µ+ λ)ε22 + λε33 00 0 λε11 + λε22 + (2µ+ λ)ε33

](3.9)

Solving the (3,3) component for ε33,ε33 = −λ(ε11 + ε22)

2µ+ λ(3.10)

Substituting this into the (1,1) and (2,2) components:σ11 = 2µε11 +

(λ− λ

2µ+ λ

)(ε11 + ε22) σ22 = 2µε22 +

(λ− λ

2µ+ λ

)(ε11 + ε22) (3.11)




Linear Isotropic Plane Stress

σαβ = 2µεαβ +(λ− λ

2µ+ λ

)εκκδαβ ε33 = −λ(ε11 + ε22)

2µ+ λ, ε11 = ε22 = 0 (3.12)

The stress-strain relationships can be inverted and represented alternatively using Young’s Modulus and Poisson’sratio:General 2D stress-strain relations

εαβ =1

E

((1 + ν)σαβ − ασκκδαβ

)α =

ν(1 + ν) Plane strainν Plane stress (3.13)

We can write the same relations again in a slightly more convenient way (for axial stresses/strains only) in termsof E ′, ν′, defined differently for plane stress and plane strain:United axial plane stress/strain relations

σ11 =E ′

1− ν′2(ε11 + ν′ε22) σ22 =

E ′

1− ν′2(ε22 + ν′ε11) (3.14)

ε11 =1

E ′(σ11 − ν′σ22) ε22 =

1

E ′(σ22 − ν′σ11) (3.15)

E ′ =

E/(1− ν2) Plane strainE Plane stress ν′ =

ν/(1− ν) Plane strainν Plane stress (3.16)

The above relationships are useful to keep in mind when solving problems. Fortunately, it turns out that planestress/strain solutions are identical, except for slightly different constants. Now, let us determine how to actuallysolve problems.3.1.2 Equilibrium in polar coordinates

The equilibrium equation for a 2D (i.e. no change in the z direction) body at rest with no body forces in cylindricalcoordinates is given by:div σ =

[σrr ,r +

1

r(σrr − σθθ + σrθ,θ)

]gr +

[σθr ,r +

1

r(2σrθ + σθθ,θ)

]gθ = 0 (3.17)

If there is no change in the θ direction either (i.e. axisymmetric), the equilibrium equation reduces todiv σ =

[σrr ,r +

σrr − σθθr

]gr +

[σθr ,r +

2

rσrθ

]gθ = 0 (3.18)

Notice that the axial stresses are completely decoupled from the shear stresses. In fact, unless there are specifiedshear tractions on the boundary (and it’s actually pretty impossible to do this without bnreaking axisymmetry, wecan generally ignore them and solve for σ11,σ22 using the equationσrr +

σrr − σθθr

= 0 (3.19)Now, what about constitutive equations in cylindrical coordinates; do they require special treatment? In fact, theydo not: this is one of the cases where it is possible to do apply the stress-strain relations idencially. This is becausethe constitutive equations are nothing other than a linear relationship; there are no derivatives involved.All content © 2017-2018, Brandon Runnels 9.4


Lecture 10 Airy stress functions, examples

3.2 Airy stress functionsFor a linear elastic solid with no body forces, at rest, in two dimensions, the governing equations are:

div(σ) =[σ11,1 + σ12,2σ12,2 + σ22,2

]= 0 (3.20)

Let us suppose that we have a scalar function Φ : R2 → R. Furthermore, let us suppose that we can define thisfunction such thatσ11 =

∂2Φ

∂x22

σ22 =∂2Φ

∂x21

σ12 = − ∂2Φ

∂x1∂x2(3.21)

What happens if we plug this form into the linear momentum balance equation?div(σ) =

[Φ,221 − Φ,122−Φ,121 + Φ,112

]=[

Φ,122 − Φ,122−Φ,112 + Φ,112

]= 0 X (3.22)

We see that the balance equation is automatically satisfied as long as the stress has the form above. This meansthat instead of seeking three individual functions for individual stress components, we can seek a single Φ thatis guaranteed to satisfiy equilibrium. But wait, not so fast: how do we know that this stress field corresponds toa compatible strain field? Recall from a previous section that, for 2D, a compatible strain field must satisfy thefollowing equation:ε11,22 + ε22,11 − 2ε12,12 = 0 (3.23)

Let us see if our stress functions derived from the magical Φ function satisfy these equations. Recall that we have,for plane stress or strain:ε11 =

1

E((1 + ν)σ11 + α(σ11 + σ22)) ε22 =

1

E((1 + ν)σ22 + α(σ11 + σ22)) ε12 =

1

E((1 + ν)σ12) (3.24)

where α = ν or ν(1 + ν). Dropping this into our equation of compatibility:1

E((1 + ν)σ11,22 + α(σ11,22 + σ22,22)) +

1

E((1 + ν)σ22,11 + α(σ11,11 + σ22,11))− 2

1

E((1 + ν)σ12,12) = 0 (3.25)

Now let us substitute Φ in the definitions of σ. Doing this and simplifying, the above expression reduces to(1 + ν + α)[Φ,2222 + Φ,1111 + 2Φ,1212] = 0 (3.26)

Canceling out material parameters, we get a biharmonic equation for Φ:∆∆Φ = 0 (3.27)

Airy Stress Function in Cartesian CoordindatesA function Φ is an Airy stress function if it satisfies the following relations (in Cartesian coordinates)∆∆Φ = 0 σ11 =

∂2Φ

∂x22

σ22 =∂2Φ

∂x21

σ12 = − ∂2Φ

∂x1x2(3.28)

How does this help us? For many problems, it is very difficult to solve the full equations of equilibrium, but we canguess a form that allows us to obtain an approximate solution.All content © 2017-2018, Brandon Runnels 10.1


3.2.1 Cylindrical coordinates

Many 2Dproblems aremore conveniently solved using cylindrical coordinates. What does that do for our Airy stressfunction? Because derivatives are involved, as usual, it is necessary to treat the derivatives with special care. Thefollowing box contains (without proof) the conditions for the Airy stress function in cylindrical coordinates:Airy Stress Function in Cylindrical Coordinates

∆∆Φ = 0 σrr =1

r

∂Φ

∂r+

1

r2

∂2Φ

∂θ2σθθ =

∂2Φ

∂r2σrθ = − ∂

∂r

(1

r

∂Φ

∂θ

) (3.29)Remember, ∆∆ is the bilapliacian operator, that is, div(grad(div(grad(·)))). What is this in cylindrical coordinates?We go back to our original derivations of divergence and gradient in cylindrical coordinates: given Φ, we have

grad(Φ) =∂Φ

∂rgr +

1

r

∂Φ

∂θgθ (3.30)

div(grad(Φ)) =∂

∂r

[∂Φ

∂r

]+

1

r

[ ∂∂θ

(1

r

∂Φ

∂θ

)+∂Φ

∂r

]=

1

r

∂

∂r

[r∂Φ

∂r

]+

1

r2

∂2Φ

∂θ2(3.31)

(Note: we are assuming no three-dimensionality, so we neglect all z derivatives.) One can obtain an expression for∆∆Φ by substituting div grad(Φ) into Φ in the above expression. As you can see, this creates a horribly complicatedexpression that is not easy to solve – and you shouldn’t try. Generally, we start with a somewhat friendly form forthe Airy stress function (that is biharmonic) and go from there.Now, let us illustrate the use of the Airy stress function with an example showing how it is used to solve practical2D problems in solid mechanics:

Example 3.1: 2D Line Load

Consider a semi-infinite surface that has a point loadapplied to it as shown in the figure to the right. We aimto find the stress state of the material subject to thisloading.Let us try the following Airy stress function:Φ(r , θ) = C r θ sin(λθ) (3.32)

where C and λ are constants. Given this function, thestresses guaranteed to satisfy equilibrium and com-patibility are:σrr =

C

r

[θ sin(λθ) + 2λ cos(λθ)− λ2θ sin(λθ)

](3.33)

σθθ = 0 (3.34)σrθ = 0 (3.35)

e1

e2

f(x2) = −f δ(x2)e1

Now, the question is whether the function can satisfy boundary conditions. The first condition is that thebody is traction free for all θ = ±π/2, that is, σrr (r ,π/2) = −σrr (r ,π/2) = 0 for all r > 0. Let us try λ = 1 andsubstitute θ = ±π/2:σrr =

C

r

[(±π/2) sin(±π/2) + 2 cos(±π/2)− (±π/2) sin(±π/2)

]= 0 (3.36)




Indeed, the expression is zero. Therefore, we have the following reduced expression for σrr :σrr =

2C

rcos θ (3.37)

But now, we must solve for C . We will do this by evaluating a surface integral over a semicircular region.Define two contours:Contour one is defined by

C1 = x(r , θ) ∈ R2 : r ∈ [−R,R], θ = π/2 (3.38)Contour two is defined byC2 = x(r , θ) ∈ R2 : r = R, θ ∈ [−π/2,π/2] (3.39)

The normal vectors on the first and second contoursare given byn1(r ,π/2) = −e1 =

[−10

]n2(R, θ) =

[cos θsin θ

] (3.40)

C1

C2

nn

R

Now, let us determine the total force on the subregion. This is given by the surface integral∫∂Ω

t da =

∫C1

t da +

∫C2

σ n da (3.41)For the first integral, t = −f δ(r), where δ is the Dirac delta function. What about for the second integral?Let us proceed formally:

t = σ n = [σrr (gr ⊗ gr )] gr = σrrgr =2C

R

[cos2 θ

sin θ cos θ

](3.42)

Now that we have our traction vectors, let us compute the integral. The paramaterizeation for C1 is easyand is given by: ∫ R

−R

[−f δ(x)0

]dx = −f e1 (3.43)

For the second integral, we set r = R , replace da with Rdθ, and let θ go from −π/2 to π/2, yielding:∫ π/2

−π/2

2C

R

[cos2 θ

sin θ cos θ

](R dθ) = 2C

∫ π/2

−π/2

[cos2 θ

sin θ cos θ

]dθ = π C e1 (3.44)

Hence, the total force integral isπ C e1 − f e1 = 0 =⇒ C =

f

π(3.45)

allowing us to solve for the constant. Substituting back in, we see that the stress of the material is given byσrr =

2f

πrcos θ (3.46)

Let us consider another example of a problem that arises frequently in analysis of structures:




Example 3.2: Stress concentration for a circular hole in an infinite plate

Consider a 2D infinite plate that is subjected to a uni-axial stress such that the stress tensor at far field isσ =

[σ0 00 ?

] (3.47)Find the stress components σrr ,σθθ,σrθ and thestress concentration factor incurred by the presenceof the hole.

a

σ0

Let us try the following form for the Airy stress function:Φ = f (r) cos(2θ) (3.48)

Does this form satisfy the Bilaplacian ∆∆Φ? Let us substitute to find out:∆Φ =

1

r

∂

∂r

[r∂(f (r) cos(2θ))

∂r

]+

1

r2

∂2(f (r) cos(2θ))

∂θ2

=1

r

∂

∂r

[r∂f (r)

∂r

]cos(2θ)− 4

r2f (r) cos(2θ)

=[ ∂2

∂r2+

1

r

∂

∂r− 4

r2

]f (r) cos(2θ)

∆∆Φ =[ ∂2

∂r2+

1

r

∂

∂r− 4

r2

]2

f (r) cos(2θ) = 0

=⇒[ ∂2

∂r2+

1

r

∂

∂r− 4

r2

]2

f (r) = 0 (3.49)Therefore Φ satisfies the bilaplacian only if the above differential equation for f (r) is satisfied. This is afourth-order linear ODE, which we can solve. Using a computer algebra system we arrive at the followingform

f (r) = A r2 + B r4 +C

r2+ D Φ(r , θ) =

(A r2 + B r4 +

C

r2+ D

)cos(2θ) (3.50)

Now that we have a form for the Airy stress function, we can solve for stresses. The radial stress is givenby:σrr =

[1

rf ′(r)− 4f (r)

r2

]cos(2θ) =

[2Ar + 4Br3 − 2C/r3

r− 4Ar2 + 4Br4 + 4C/r2 + 4D

r2

]cos(2θ) (3.51)

=[2A + 4Br2 − 2C/r4 − 4A− 4Br2 − 4C/r4 − 4D/r2

]cos(2θ) (3.52)

= −[2A +

6C

r4+

4D

r2

]cos(2θ) (3.53)

The circumferential stress is:σθθ = f ′′(r) cos(2θ) =

(2A + 12Br2 +

6C

r4

)cos(2θ) (3.54)

Finally, the shear stress is given byσrθ = − ∂

∂r

( f (r)

r(−2 sin(2θ))

)= 2 sin(2θ)

∂

∂r

(A r + B r3 +

C

r3+

D

r

) (3.55)= 2(A + 3B r2 − 3C

r4− D

r2

)sin(2θ) (3.56)




We have four constants to solve for; let us use boundary conditions.limr→∞

σ11 = σ0 limr→∞

σ22 = limr→∞

σ12 = 0 σrr (a, θ) = 0 σrθ(a, θ) = 0 (3.57)We know that all stresses must remain bounded (non-infinite) as r →∞; looking at the stress terms we cansee that this implies B = 0, i.e.

limr→∞

σθθ,σr ,θ 6=∞ =⇒ B = 0 (3.58)We know that at the far-field, σ11 = σ. We also know that when θ = ±π/2, σθθ = σ11; therefore we can write:

limr→∞

σθθ(r ,π/2)!

= σ0 =⇒ (2A) cos(2π/2) = −2A =⇒ A = −σ0

2(3.59)

We also know that when θ = nπ that θrr = θ11, so we write:limr→∞

σrr (r , 0) = −(2A)(cos(0)) = σ0 =⇒ A = −σ0

2X (3.60)

We see that this just recovers the same constant value.



Lecture 11 Airy stress functions and Lamé solutions

Example 3.2 Continued

Finally, we know that the radial stress must be zero when r = a:σrr (a, θ) = 0 =⇒ 6C

a4+

4D

a2= σ0 =⇒ 6C + 4Da2 = a4σ0 (3.61)

We also know that the shear stress must be zero then as well:σrθ(a, θ) = 0 =⇒ −3C

a4− D

a2=σ0

2=⇒ 6C + 2Da2 = −a4σ0 (3.62)

This gives us two equations and two unknowns: subtracting the equations gives2Da2 = 2a4σ0 =⇒ D = a2σ0 (3.63)

and adding them gives12C + 6Da2 = 0 =⇒ C = −Da2

2=⇒ C = −1

2a4σ0 (3.64)

What about the circumferential stress, σθθ at r = 0? It does not necessarily equal zero. That’s good, becausewe are out of constants to solve for. Our final solution is:σrr = σ0

[1− 4

(ar

)2

+ 3(ar

)4]cos(2θ) (3.65)

σrθ = σ0

[− 1− 2

(ar

)2

+ 3(ar

)4]sin(2θ) (3.66)

σθθ = −σ0

[1 + 3

(ar

)4]cos(2θ) (3.67)

What is the stress concentration factor for this problem? Evaluating σθθ at r = a and θ = π/2 we getσθθ = −σ0

[1 + 3

](−1) = 4σ0 (3.68)

so the stress concentration induced by a hole is K = (4σ0)/(σ0) = 4; that is, the stress is magnified fourtimes by the presence of a spherical hole...no matter how small the hole is. Let us plot the stress compo-nents in cartesian coordinates. Recall that

σ = σrrgr ⊗ gr + σrθgr ⊗ gθ + σθrgθ ⊗ gr + σθθgθ ⊗ gθ (3.69)= σrrgr ⊗ gr + σrθ(gr ⊗ gθ + gθ ⊗ gr ) + σθθgθ ⊗ gθ (3.70)

(3.71)Evaluating components, we plot the normalized stress σαβ/σ0, where the dimension of the problem are x/aand y/a, normalized by the radius of the hole.



3.3 Lamé solutionsWe now consider a class of problems that are even more simple than the problems considered in the previoussection: two dimensional, linear elastic isotropic axisymmetric (i.e. derivative with respect to θ = 0) problems. Inother words, we wish to consider thick-walled cylinders (in the case of plane strain) and discs (in the case of planestress).Let us begin by looking for an Airy stress function. We know that it must satisfy the bilaplacian (∆∆). Now that weare ignoring derivatives with respect to θ, the cylindrical laplacian is just

∆Φ =1

r

∂

∂r

[r∂Φ

∂r

]∆∆Φ =

1

r

∂

∂r

[r∂

∂r

1

r

∂

∂r

(r∂Φ

∂r

)]= 0 (3.72)

We can integrate this directly to find an expression for Φ:∂

∂r

[r∂

∂r

1

r

∂

∂r

(r∂Φ

∂r

)]= 0 =⇒ ∂

∂r

1

r

∂

∂r

(r∂Φ

∂r

)=

C1

r(3.73)

∂

∂r

(r∂Φ

∂r

)= C1r ln(r) + r C2 =⇒ ∂Φ

∂r= C1

(1

2r ln(r)− r

4

)+

1

2r C2 +

C3

r(3.74)

Φ = C1

(1

2

[1

2r2 ln(r)− 1

4r2]− r2

8

)+

1

4r2 C2 + C3 ln(r) + C4 (3.75)

Φ =1

4C1r

2 ln(r) +1

4r2 (C2 − C1) + C3 ln(r) + C4 (3.76)

Since these are just constants, we can redefine them tomake the expression simpler. It generally turns out that theC1 constant is not necessary for solving problems, so we will ignore it. The Airy stress function that we will use isthe following:

Φ(r) = C1 r2 + C2 ln(r) + C3 (3.77)

Now that we have our Airy function, we can find our stresses by straight differentiation:All content © 2017-2018, Brandon Runnels 11.2



σrr =1

r

∂Φ

∂r=

1

r

[2C1 r +

C2

r

]= 2C1 +

C2

r2σθθ =

∂Φ

∂r2= 2C1 −

C2

r2σrθ = 0 (3.78)

Using Equation (3.15) we can derive the equations for strain:εrr =

1

E ′(σrr − ν′σθθ) =

1

E ′

[2C1 +

C2

r2− ν′

(2C1 −

C2

r

)]=

1

E ′

[2(1− ν′)C1 + (1 + ν′)

C2

r2

] (3.79)εθθ =

1

E ′(σθθ − ν′σrr ) =

1

E ′

[2C1 −

C2

r− ν′

(2C1 +

C2

r2

)]=

1

E ′

[2(1− ν′)C1 − (1 + ν′)

C2

r2

] (3.80)We also recall that ε = 1

2 (grad(u)+grad(u)T ), which in cylindrical coordinates (as computed in HW2), is (neglectingall derivatives with respect to θ and z)grad(u) = (ur ,r )gr ⊗ gr +

(uθr

)gr ⊗ gθ + (uθ,r )gθ ⊗ gr +

(urr

)gθ ⊗ gθ (3.81)

This indicates thatεrr =

∂ur∂r

εθθ =urr

0 =uθr

=∂uθ∂r

=⇒ uθ = 0 (3.82)Therefore we can find the radial displacement by multiplying by r :

ur = r εθθ =1

E ′

[2(1− ν′)C1 r − (1 + ν′)

C2

r

] (3.83)and one can verify that εrr is recovered by differentiation with respect to r . Let us summarize the stress, strain, anddisplacement equations for axisymmetric problems:

Lamé Solution for 2D Linear Elastic Axisymetric Problems

Airy stress function:Φ(r) = C1 r

2 + C2 ln(r) + C3 (3.84)Stress:

σrr = 2C1 +C2

r2σθθ = 2C1 −

C2

r2σrθ = 0 (3.85)

Strain:εrr =

1

E ′

[2(1− ν′)C1 + (1 + ν′)

C2

r2

]εθθ =

1

E ′

[2(1− ν′)C1 − (1 + ν′)

C2

r2

]εrθ = 0 (3.86)

Displacement:ur = r εθθ =

1

E ′

[2(1− ν′)C1 r − (1 + ν′)

C2

r

]uθ = 0 (3.87)

whereE ′ =

E/(1− ν2) Plane strainE Plane stress ν′ =

ν/(1− ν) Plane strainν Plane stress (3.88)

Lamé problems are easy to solve because all of the heavy lifting has been done already. The only thing that usuallyremains is to solve for boundary conditions. Let us illustrate the solution of Lamé problems with a couple ofexamples.All content © 2017-2018, Brandon Runnels 11.3



Example 3.3: Pressurized cylinder

Consider a thick-walled cylinder with internal and extrenal radii ra, rb that is subjected to an internal pressurepa. Find the stress distribution in the cylinder and themaximum circumferential stress. What happens whena→ 0 or when pa → pb?This is a Lamé problem; the form has been found al-ready, we just need to solve for boundary conditions.Recalling that compressive pressure is negative, wehave the two equations

σrr (r = a) = 2C1 +C2

a2

!= −pa (3.89)

σrr (r = b) = 2C1 +C2

b2

!= −pb (3.90)

Solving the two systems simultaneously using the 2Dmatrix inverse, we find

pb

pa

a

b

[2 1/a2

2 1/b2

] [C1C2

]=[−pa−pb

]=⇒

[C1C2

]=

1

2/b2 − 2/a2

[1/b2 −1/a2

−2 2

] [−pa−pb

]=

1

2/b2 − 2/a2

[pb/a

2 − pa/b2

2(pa − pb)

](3.91)

Combining and simplifying we have2C1 =

b2pb − a2paa2 − b2

C2 =a2b2(pa − pb)

a2 − b2(3.92)

The circumferential stress either increses or decreases monotonically; evaluating at r = a, b we haveσθθ(a) =


− b2(pa − pb)

a2 − b2=

2b2pb − (a2 + b2)paa2 − b2

(3.93)σθθ(b) =


− a2(pa − pb)

a2 − b2=

(a2 + b2)pb − 2a2paa2 − b2

(3.94)

Example 3.4: Pressure-fitted inclusion

A hole with radius a is cut in an infinite plate. A peg with radius a + δ is inserted into the hole, causing aradial deflection of δ. (The plate was previously stress-free.) What is the stress of the plate?This is a Lamé problem for which we must solve boundary conditions. The first constant is obtained byusing the far-field condition: we know thatlimr→0

limσrr ,σθθ = 0 =⇒ C1 = 0 (3.95)Now we solve for the second constant using the displacement boundary condition at the hole. We have

ur (a) = −1 + ν′

E ′C2

a!

= δ =⇒ C2 = − δ a E′

1 + ν′(3.96)

So, the stress is given by−σrr = σθθ =

δ a E ′

r2 (1 + ν′)(3.97)



Lecture 12 Thin wall approximation and Green’s function solu-tions

3.3.1 Thin wall approximation

Let us consider the result from Example 3.3 again. Frequently, pressure vessels of this type are built such that thethickness of the cylinder is very small compared to the radius of the tank. This simplifies the analysis considerably:pb

pa a

pb

pa

σθθ

σθθ

Let b = a + t where t is small. This enables us to make the following approximations:O(t2) ≈ 0 a + t ≈ a (3.98)

Substituting b = a + t into the expressions for the constants C1,C2 obtained above, we have2C1 =

(a + t)2pb − a2paa2 − (a + t)2

=(a2 + 2at + t2)pb − a2pa

a2 − a2 − 2at − t2=

a(a + 2at)pb − a2pa−2at

=a2pb − a2pa−2at

=a (pa − pb)

2t(3.99)

C2 =a2(a + t)2(pa − pb)

a2 − (a + t)2=

a2a2(pa − pb)

a2 − a2 − 2at − t2= −a3(pa − pb)

2t(3.100)

Substituting these constants into the expression for σrr (radial stress) and evaluating at r = a, we findσrr (a) =

a (pa − pb)

2t− a(pa − pb)

2t= 0 (3.101)

Because a + t ≈ a, then σrr (r + a) ≈ σrr (a) = 0; therefore we see that the radial stress in the thin-wall cylinder isapproximately zero.Substituting the expressions for C1,C2 into the expression for σθθ and evaluating at r = a we getσθθ(a) =

a (pa − pb)

2t+

a(pa − pb)

2t=

a(pa − pb)

t(3.102)

The circumferential stress (often called hoop stress) is the primary carrier of the load in a thin-walled cylinder.An alternative method for deriving this relation is by assuming the hoop stress to be constant and doing a forcebalance on half the cylinder (as shown above): summing forces in the x direction gives∑fx = 2apa − 2apb + 2tσθθ = 0 =⇒ σθθ =

a(pa − pb)

t(3.103)

as expected. By following a similar procedure, it is easy to show (by summing in the z direction) that the longitudinalstress σzz is given by ∑fz = πa2(pa − pb)− 2π a t σzz = 0 =⇒ σzz =

a(pa − pb)

2 t(3.104)



3.4 Green’s function solutionsOne of the most convenient aspects of linear elasticity is that we can use the principle of superposition to takethe solution for one problem and use it to find the solution to another. Recall that our static momentum balanceequation is given by:

div(σ) = −b (3.105)This is an equation that relates the applied body force b to the resulting stress field σ. In a sense, by solving for σ,we are finding the inverse operator such that

σ = div−1(−b) (3.106)With this idea in mind, let us generalize. Let D be a differential operator, and let

D u = f (3.107)Note that the above expression is very general. For instance, if

D = md2

dt2+ b

d

dt+ k (3.108)

then D u = f is the equation for a spring-mass-damper system with a forcing term. Alternatively, ifD =

d2

dx2E I

d2

dx2(3.109)

where E is an elasticmodulus and I is amoment of inertia, thenD u(x) = f (x) is the Euler-Bernoulli beam equation.The general idea is that linear differential equations take inputs (in the form of forcing functions, loadings, etc) andproduce outputs (in the form of time responses, stress fields, etc), determined by linear operators. And, if we justknew how to find the inverse operator, we would be able to solve for the output directly.Let us suppose that we have a linear differential operatorD as defined above. The Green’s function solution ug (x , s)is the solution for f = δ, i.e. the solution toD ug (x , s) = δ(s − x) (3.110)

or, written in terms of inverse operators,ug = D−1δ(s − x) (3.111)

Now consider an arbitrary loading f where we seek a solution u = D−1f . Recall that we can write f using theso-called “sifting property” of the Dirac delta; that is,f (x) =

∫f (y) δ(s − x) ds (3.112)

Consequently we have the following, noting thatD is an operator on x and can therefore be taken inside the integral:u(x) = D−1

∫f (s) δ(s − x) ds =

∫f (s)D−1δ(s − x) ds =

∫f (s) ug (y − x) ds = (f ? ug )(x) (3.113)

In other words, any system response can be given as the convolution of the applied loading with the response fora point load. This becomes particularly useful when working in the frequency domain. If f and ug have Fouriertransforms f (ω), ug (ω), then the solution is given byu(x) = F−1[F [f ? ug ]] = F−1[f (ω) ug (ω)] (3.114)

where we recall that convolution in real space corresponds to multiplication in reciprocal space. Even if an exactinverse transform of f and ug cannot be found analytically, the fast Fourier transform (FFT) can be used to evaluatethe inverse quickly.Let us consider the following example of an arbitrary load applied to a linear elastic halfspace using the Green’sfunction approach.All content © 2017-2018, Brandon Runnels 12.2



Example 3.5: Force on linear elastic halfspace

Recall our solution to the point load applied to a linear elastic halfspace as obtained in Example 3.1. Nowwe seek to find the stress state for an arbitrary loading f (x) as shown belowf = δ(x)

f = f (x)

σrr = 2 sin θπr

σrr =?

In the previous example, we found the Green’s function for the differential operatorD : [applied loading]→ [stress state] (3.115)

It is significantly easier to work in Cartesian coordinates, so, recalling that sin θ = x2/r and r =√

x21 + x2

2 ,we have

σgrr = −2 sin θ

πr= − 2x2

π(x21 + x2

2 )(3.116)

We convert to Cartesian components by expressing in terms of unit vectorsσg = σg

rrgr ⊗ gr = σrr

[cos2 θ cos θ sin θ

cos θ sin θ sin2 θ

](3.117)

so that we haveσg

11 =2x2

1 x2

π(x21 + x2

2 )2σg

22 =2x1x

22

π(x21 + x2

2 )2σg

12 =2x3

2

π(x21 + x2

2 )2(3.118)

The Fourier transform in x2 isσg

11(ω1, x2) = (ω1 x2 − 1)H(−ω1)e−ω1x2 − (ω1x2 + 1)H(+ω1)e+ω1x2 (3.119)σg

22(ω1, x2) = (ω1 x2 − 1)H(+ω1)e+ω1x2 − (ω1x2 + 1)H(−ω1)e−ω1 x2 (3.120)σg

12(ω1, x2) = i x2 ω1(H(ω1)ex2ω1 + H(−ω1)e−x2ω1 ) (3.121)We consider three loading cases: a point load, a load with a Gaussian distribution, and a square wave load.The loading functions and their Fourier transforms are

fpoint(x1) = f δ(x1) fpoint(ω1) = 1 (3.122)fgauss(x1) =

1√2π

e−x21/2 fgauss(ω1) = e−ω

21/2 (3.123)

fstep(x1) = u(1− x1) u(1 + x1) fstep(ω1) =2 sin(ω1)

ω1(3.124)

as illustrated in the following figure:




f (x) = δ(x) f (x) = f u(1 + x) u(1− x)f (x) = f√2πe−x

2/2

The Fourier transformed solution is given byσαβ(ω1, x2) = σxx (3.125)

For the point load it is clear that the original solution is recovered, as shown in the following plots. (Notethat the stress is unbounded and becomes infinite at the point of application of the load.)

For the other loadings it is difficult (perhaps impossible?) to find an exact form for the inverse Fouriertransform. Nevertheless, it is quite easy to compute a numerical inverse Fourier transform using the FastFourier transform. The following results were obtained using the FFT in Python: for a Gaussian load,

Similarly for a step load,




Note the relationship between the point load response and the continuous response. This is a helpful way toget an intuitive understanding of the convolution operator. Another potentially helpful intuitive connection isthat to transfer functions in control theory, which as you may recall, are exactly the response of the systemto a unit impulse. We do exactly the same thing here.



Lecture 13 Fracture mechanics

3.5 Linear elastic fracture mechanicsWe now have the tools to start looking at the mechanics of fracture taking place in a linear elastic framework.

(a) Fracture in the de Haviland Comet airliner was causedby stress concentrations at the corners of the windows. (b) The SS Schenectady experienced catastrophic brittlefailure caused by poor quality steel.Let us consider the case of an infinite body subjected to a uniaxial load as shown in the following figure, where theline x ∈ R2, x1 < 0, x2 = 0 represents a division in the material such that σrr = σrθ = 0 along the line; in otherwords, a crack.

σ0

Let us attempt to determine the stresses in the body. As usual, we begin bymaking an educated guess for the formof the Airy stress function; in this case we select the followingΦ(r , θ) = rλfλ(θ) (3.126)

where λ is a constant and fλ(θ) is some function of theta. How do we find fλ(θ)? Use the biharmonic condition∆∆Φ = 0. The form for this expression is easily found using Maple:

f (θ) = C1 cos((λ− 2) θ) + C2 cos(λθ) + C3 sin((λ− 2) θ) + C4 sin((λ)θ) (3.127)Our problem is symmetric about the x1 axis. Because of this, we assume that Φ(θ) must be symmetric in θ as well,that is, Φ(r , θ) = Φ(r ,−θ). This is only possible if the sine terms vanish, so we assume that C3 = C4 = 0 and usethe following form for f (θ).

f (θ) = C1 cos((λ− 2) θ) + C2 cos((λ)θ) (3.128)Following our usual procedure, we derive the stresses in cylindrical coordinates. Radial stress:

σrr =1

r

∂Φ

∂r+

1

r2

∂Φ

∂θ2(3.129)

= λrλ−2[C1 cos((λ− 2) θ) + C2 cos(λθ)]− rλ−2[(λ− 2)2C1 cos((λ− 2) θ) + λ2 C2 cos(λθ)] (3.130)All content © 2017-2018, Brandon Runnels 13.1


Circumferential stress:σθθ =

∂2Φ

∂r2= λ (λ− 1)rλ−2[C1 cos((λ− 2)θ) + C2 cos((λ)θ)] (3.131)

Shear stressσrθ = − ∂

∂r

(1

r

∂Φ

∂θ

)= (λ− 1)rλ−2[(λ− 1)C1 sin((λ− 2) θ) + λC2 sin(λ θ)] (3.132)

We have three constants to solve for, λ,C1,C2. Following standard procedure, we use boundary conditions. Weknow that σrθ = σθθ = 0 for all r at θ = ±π. Substituting into the equation for circumferential stress:σθθ(r ,π) = λ (λ− 1)rλ−2[C1 cos((λ− 2)π) + C2 cos((λ)π)] = 0 (3.133)

Recalling that this is true for all r , we divide out the r terms and are left with the following condition(C1 + C2) cos(λπ) = 0 (3.134)

We have two possiblities, either λ = n + 1/2 (with n an integer) or C1 = −C2. Which is true? We don’t know yet...solet’s go ahead and evaluate the other boundary condition for σrθ:σrθ(r , θ) = (λ− 1)rλ−2[(λ− 1)C1 sin((λ− 2)π) + λC2 sin(λπ)] = 0 (3.135)

Again, factoring out r terms and noting that sin((λ− 2)π) = sin(λπ), we have[(λ− 2)C1 + λC2] sin(λπ) = 0 (3.136)

Again, we have two possibilities: either λ = nπ or C1 = λC2/(2− λ). So, we are left with two possibilities:Case 1 Case 2 (3.137)λ = ... ,−2,−1, 0, 1, 2, ... λ = ... ,−3

2,−1

2,

1

2,

3

2, ... (3.138)

C1 = C2 C1 =λ

2− λC2 (3.139)

Which one do we select? We can find values of C1 and C2 that correspond to any integral or half-integral value of λ.Recall the other boundary condition, that stress remains bounded as r →∞. Looking at the stresses, we see thatthey all scale with rλ−2. The stresses will be unbounded only if λ− 2 ≤ 0, so λ ≤ 2. This restricts us to... ,−2,−1

2, 0,

1

2, 1,

3

2, 2 (3.140)

which still leaves us with an infinite number of values of λ to choose from. To narrow it down further, we note thatthe strain energy of the crack must be bounded, i.e. not finite. Computing the strain energy would be a bit tedious,so we will make a bit of a slipshot scaling argument (you can plug in actual values to verify this if it would makeyou feel more comfortable). Note that all stresses can be written in the following way:σrr = rλ−2grr (λ, θ) σθθ = rλ−2gθθ(λ, θ) σrθ = rλ−2grθ(λ, θ) (3.141)

where the key thing is that rλ−2 factors out of everything. We also know that since the strain is linearly dependenton stress, thatεrr = rλ−2frr (λ, θ) εθθ = rλ−2fθθ(λ, θ) εrθ = rλ−2frθ(λ, θ) (3.142)

where f , g are some (probably nasty) functions of λ, θ. This means that the strain energy w = σαβεαβ has thefollowing formw(r , θ) = r2(λ−2)F (λ, θ)G (λ, θ) (3.143)

Let us use this form to find the strain energy in a small circular region with radius R Let us integrate over a smallcircular region as shown:All content © 2017-2018, Brandon Runnels 13.2



R

Integrating using polar coordinates we have:W =

∫ R

0

∫ π

−πr2(λ−2)G (λ, θ)H(λθ)r dr dθ =

∫ R

0

∫ π

−πr2λ−3G (λ, θ)H(λθ)dr dθ =

∫ R

0

r2λ−3dr

∫ π

−πG (λ, θ)H(λθ)dθ

(3.144)=

1

2λ− 2r2λ−2

∣∣∣R0

∫ π

−πG (λ, θ)H(λθ)dθ (3.145)

Note that the value of the strain energy is dependent on 02λ−2. The only way for this to be non-infinite is for2λ− 2 > 0 =⇒ λ > 1 (3.146)

So, we have reduced the admissible values of λ toλ =

1

2, 2 (3.147)

This is much more reasonable! We have two possible solutions, so let us determine them seperately.For λ = 2

We must use C2 = −C1. Substituting into the expression for stress we haveσrr = 2C1(1 + cos(2θ)) σθθ = 2C1 [1− cos(2θ)] σrθ = −2C1 sin(2θ) (3.148)

Solving for boundary conditions, if we have plane stress (i.e. σ11 = 0 at far-field, thenlimr→∞

σrr (r , 0) = 2C1(1 + 1) = 4C1!

= 0 =⇒ C1 = 0 (3.149)In other words, λ = 2 does not produce a solution.For λ = 3/2

Because λ is a half-integer we must use the following to solve for C1,C2:C1 =

3/2

1/2C2 = 3C2 (3.150)

Substituting to find radial stress:σrr = (3/2)r−1/2[(3C2) cos((−1/2) θ) + C2 cos((3/2)θ)]

− r−1/2[(−1/2)2(3C2) cos((−1/2) θ) + (3/2)2 C2 cos((3/2)θ)]

=3C2√r

[5

4cos(θ/2)− 1

4cos(3θ/2)

] (3.151)




For circumferential stressσθθ =

3C2

4√r

[3 cos(θ/2) + cos(3θ/2)] (3.152)and for shear stress

σrθ =3C2

4√r

[sin(θ/2) + sin(3θ/2)] (3.153)Let us define the following constant:

KI = limr→0

√2πrσ22(r , θ) = lim

r→0

√2πrσθθ(r , 0) = 3C2

√2π (3.154)

KI is called the stress intensity factor and has the peculiar units of stress times square root of length. It is aproperty that is very important when evaluating material failure. To summarize, the stress distribution for this typeof loading is:Stress distribution for Mode-I loading

For a material subjected to uniaxial tension, the stress distribution isσrr =

KI√2πr

[5

4cos(θ/2)− 1

4cos(3θ/2)

]σ11 =

K1√2πr

cos(θ/2)(

1− sin(θ/2) sin(3θ/2)) (3.155a)

σθθ =KI√2πr

[3

4cos(θ/2) +

1

4cos(3θ/2)

]σ22 =

K1√2πr

cos(θ/2)(

1 + sin(θ/2) sin(3θ/2)) (3.155b)

σrθ =KI√2πr

[1

4sin(θ/2) +

1

4sin(3θ/2)

]σ12 =

K1√2πr

cos(θ/2) sin(θ/2) cos(3θ/2) (3.155c)

Plotting the results, we have

3.5.1 Stress distributions for mode I-III loading

There are three primary modes of fracture:




Mode I Mode II Mode III

By the principle of superposition, the resulting stress state isσ(r , θ) =

KI√2πr

σI (r , θ) +KII√2πr

σII (r , θ) +KIII√2πr

σIII (r , θ) (3.156)where KI ,KII ,KIII are the stress intensities for modes I-III, and σI ,σII ,σIII are the radial distributions of the stressaround the crack.



Lecture 14 Fracture mechanics

3.5.2 Stress intensity factors

Analytic stress intensity factors can be found for a variety of loading and crack configurations, here are just a fewof them:Finite crack in infinite plate:For Mode I loading (left) and for Mode II loading (right)the stress intensity factors are

KI = σ√π a KII = τ

√π a (3.157)

a σ a τ

Point load on crack in finite plate:Consider a crackwithwidth a in an infinite plate as shown,with a point load with magnitude f on the crack at dis-tance x1 from the origin. Then the stress intensity factorsare:KI =

f

2√πa

√a + x

a− xKII = − f

2√πa

(κ− 1

κ+ 1

) (3.158)where κ is the bulk modulus.

ax1

x2

f

These stress intensity factors and many more are tabulated for an enormous range of configurations in the Hand-book of stress-intensity factors [4]. The derivations are tedious and not particularly interesting, so we will not goover them here.3.5.3 Critical stress intensity factor and criterion for failure

The stress at a crack tip is, theoretically, infinite, which means it always exceed the material’s yield strength andultimate tensile strengths. But, on the other hand, we know that materials don’t always fail when a crack is present.So, at what point do materials fail? We need a new criterion: let us introduce the critical stress intensity factor KC ,which is amaterial property. The G-Criterion is a condition for failure stating that as long as the following inequalityholds:K 2C ≤ K 2

I + K 2II +

E ′

2µK 2III (3.159)

where E ′ is E/(1 − ν)2 for plane strain and E for plane stress, and µ is the shear modulus. Keep in mind that thisis an empirical measure of failure; while it serves as a reasonable metric for analysis, it is just a guess.3.6 Energy methods for fracture mechanicsLet us consider an entirely different method for analyzing fracture mechanics, this time, from an entirely energeticperspective. Consider an elastic body that is subjected to a loading as shown in the following figure.



The loading causes a crack to open up in the material; in other words, it has done irreversible damage. From athermodynamics point of view, this means that free energy has been expended and turned into heat somewhere.Where did this energy go? We say that the energy was expended in order to create additional surface of thematerial.In other words, defining the following:γ = surface energy density (3.160)

(where γ is energy per area in 3D and energy per distance in 2D), then the amount of work required to increase thecrack size by da isdwcrack = γ da (3.161)

The surface energy density γ is a property of the material and, in fact, can bemeasured relatively easily. Now, recallthe potential energy of the material:Π =

∫Ω

w dv −∫

Ω

b · u dv −∫

Ω

t · u da (3.162)where w is strain energy density and we note that minimization of Π recovers the equilibrium solution for u. If thematerial creates a crack, then it must use up some of its potential energy to do so. That is, if the material increasesthe crack size by ∆a, then it must reduce its energy by ∆Π. We therefore define the following quantity

G = lim∆a→0

∆Π

∆a= −dΠ

da= energy release rate (3.163)

where a is the crack length. (This can be generalized to 3D but 2D is sufficient to get the general idea.) Let usillustrate how this works with an example.Example 3.6: Double cantilever beam

Consider a beam undergoing a double cantilever beam (DCB) test. The beam has a crack with length a andis subjected to loading at the end with magnitude f as shown in the following figure.a f

f

The beam is such that the sections above and below the crack have a cross sectional area A, moment ofinertia I , and the beam has an elastic modulus E . Furthermore, the material has a measured surface energyof γ. Find the energy release rate G for the material, and determine the loading f that causes unstable crackgrowth.




The convenient thing about this problem is that it allows us to treat the upper and lower portions of thebeamas simple cantilever beams, andwe can approximate their behavior using Euler-Bernoulli beam theory.It is true that there will be a more complicated stress state near the crack, but this will give us a roughapproximation.Let us begin by computing the stress and strain state of the top beam (noting that the bottom will behavein the same way). With the origin at the crack tip and treating the uncracked region as a rigid support, wehaveE I

d4u

dx41

= 0

E Id3u

dx31

= V (x1) = C1

V (0) = C1 = −f

E Id2u

dx21

= M(x1) = C2 − fx1

M(a) = C2 − fa = 0 =⇒ C2 = fa =⇒ M(x1) = f (a− x1) (3.164)du

dx1=

1

E I

(f (ax1 − x2

1/2) + C3

)u′(0) = 0 =⇒ C3 = 0

u(x1) =1

E I

(f (ax2

1/2− x31/6) + C4

)u(0) = 0 =⇒ C4 = 0

u(x1) =f (3ax2

1 − x31 )

6E I(3.165)

To find the strain energy density, wemust first find the stress and strain in thematerial. Approximating usingEuler-Bernoulli and ignoring shear stresses, we haveσ(x1, x2) =

M(x1)x2

I=

f (a− x1)x2

Iε(x1, x2) =

f (a− x1)x2

E I(3.166)

The strain energy density is given byw =

1

2σ ε =

f 2(a− x1)2x22

2E I 2(3.167)

Now, if we integrate this over the entire beam (andmultiplying by two to get the total energy for both beams)we haveU =

∫Ω

w dv = 2×∫ a

0

∫A

w dx2 dx3 dx1 = 2×∫ a

0

∫A

f 2(a− x1)2x22

2E I 2dx2 dx3 dx1 =

f 2

E I 2

∫ a

0

(a− x1)2 dx1︸︷︷︸a3/3

∫A

x22dx2 dx3︸︷︷︸

I

=f 2a3

3EI(3.168)

Now that we have the strain energy density, we need the work done by applied tractions (no body forcesact) ∫∂Ω

t · u da = 2× [f × u(x1 = a)] = 2 f( f (3a3 − a3)

6EI

)=

2f 2a3

3EI(3.169)




Therefore the total potential energy is:Π =

f 2a3

3EI− 2f 2a3

3EI= − f 2a3

3EI(3.170)

and the energy release rate isG = −dΠ

da=

f 2a2

EI(3.171)

If the crack undergoes unstable growth, then the energy release rate is exactly equal to the energy per unitcreated area. (We note that if the crack propagates by ∆a then the total increase in the crack size is 2∆a).Therefore we have2γ =

f 2a2

EI=⇒ fcrit =

√2γ E I

a(3.172)



Lecture 15 J Integral Theory

3.6.1 Relationship between energy release rate and stress intensity factor

The following analysis comes from the original work by Irwin [1], which has more than 4000 citations. We haveshown that there are two different ways to think about fracture: there is the stress intensity factor (more of amechanics approach) and there is energy release rate (more of a physics approach). Now, we show how it ispossible to relate the two. Consider a crack undergoing Mode I loading that causes the crack to elongate by length∆a as shown in the following figure:

∆a

t(x1)

∆a

u(x1)

Our approach will be the following: let us determine what the tensions on the crack are over the region ∆a, as wellas the opening displacement u2(x1), and assume that the energy released is given by their product.Recall the stress distribution for a crack under Mode I loading (ignoring shear terms for now)σrr =

KI√2πr

[5

4cos(θ/2)− 1

4cos(3θ/2)

]σθθ =

KI√2πr

[3

4cos(θ/2) +

1

4cos(3θ/2)

]We can compute the tractions to the right of the origin along the x1 axis using the following:

t2(x1) = σθθ(r = x1, θ = 0) =K1√2πx1

[5

4− 1

4

]=

K1√2πx1

(3.173)Now, suppose the crack has propagated by ∆a. Placing our origin on the new crack tip, we wish to find the dis-placement u2 as a function of x1 for x1 < 0. To do this, we need to first compute strains. Recalling our relations forplane stress and strain where

E ′ =

E Plane stressE/(1− ν2) Plane strain ν′ =

ν Plane stressν/(1− ν) Plane strain (3.174)

we recover strains:εrr =

1

E ′(σrr − ν′σθθ) = − KI

E√

2πr

[3ν′ − 5

4cos(θ/2) +

1 + ν′

4cos(3θ/2)

] (3.175)εθθ =

1

E ′(σrr − ν′σθθ) = − KI

E√

2πr

[5ν′ − 3

4cos(θ/2)− 1 + ν′

4cos(3θ/2)

] (3.176)Now, we recover displacements. Recalling that

εrr =∂ur∂r

εθθ =urr

+1

r

∂uθ∂θ

(3.177)



we first get the radial displacement:ur =

∫εrr dr =

KI√r

2E ′√

2π

[3ν′ − 5

4cos(θ/2) +

1 + ν′

4cos(3θ/2)

] (3.178)Now we get the circumferential displacement:

uθ =

∫(rεθθ − ur )dθ =

∫ [r− KI

E ′√

2πr

[5ν′ − 3

4cos(θ/2)− 1 + ν′

4cos(3θ/2)

] (3.179)

− KI

√r

2E ′√

2π

[3ν′ − 5

4cos(θ/2) +

1 + ν′

4cos(3θ/2)

]]dθ (3.180)

= − KI√r

E ′√

2π

∫ [ν′ − 7

4cos(θ/2) +

3(ν′ + 1)

4cos(3θ/2)

]dθ (3.181)

= − KI√r

E ′√

2π

[ν′ − 7

2sin(θ/2) +

ν′ + 1

2sin(3θ/2)

](3.182)

Evaluating at θ → π, so that r → −x1 and uθ → u2, the expression reduces tou2(−x1,π) = −KI

√−x1

E√

2π(−4) =

2KI

E ′

√−2x1

π(3.183)

Shifting the origin back to the original location for the traction calculation, we arrive at the following equation fordisplacement:u2(x1) =

2KI

E ′

√2(∆a− x1)

π(3.184)

We therefore assume that the total energy dissipated is equal to the tractions times the displacement integratedover the crack length, i.e.∆G = 2×

∫ ∆a

0

1

2t2(x1) u2(x1) dx1 =

∫ ∆a

0

K1√2πx1

× 2KI

E ′

√2(∆a− x1)

πdx1 (3.185)

=2K 2

1

E ′π

∫ ∆a

0

√(∆a− x1)

x1dx1︸︷︷︸

=∆a π/2

=K 2

1

E ′∆a (3.186)

Assuming plane strain, the limit of the energy released during the Mode I fracture per length of crack created isGI = lim

∆a→0

∆GI

∆a=

K 21 (1− ν2)

E≡ Mode I energy release rate (3.187)

Using superposition and following a similar procedure as above for computing the energy release for other loadings,we relate all three modes to the energy release rate viaG =

1− ν2

E(K 2

1 + K 2II ) +

1 + ν

EK 2III (3.188)




3.7 J-integral method for fractureEnergy methods are a very convenient way to analyze fracture, but it is not always easy to compute the energyrelease rate of a system. The J Integral is an easy-to-use technique for computing energy release rates using onlythe state of the system at its boundaries. It was originally introduced by Jim Rice [3] and the original paper hasmore than 7000 citations. Let us begin by considering a system (free of body forces) for which wewant to computethe energy release rate. We know that this is given in 2D by the following

G = − d

da

[∫Ω

w dA−∫∂Ω

t · u dL

](3.189)

We seek a way to compute the integral more explicitly, i.e., to take the derivative inside the integral. Let us considera crack that is moving along the x1 axis and propagates by a distance ∆a, and consider the integral over a regionΩ. Now, suppose that the volume Ω moves with the crack, as shown in the following, so that Ω→ Ωa.

∆a

∆a

∆L(n · e)

e1

n

∆L

Computing the difference between the integral over Ω and Ωa:∫Ωa

w dA−∫

Ωa

w dA = −∫∂Ω

w ∆a (n · e1) dL (3.190)so in the limit at ∆a→ 0,

d

da

∫Ω

w dA = −∫∂Ω

w (n · e1) dL = −∫∂Ω

w n1 dL (3.191)Now, let us consider the surface integral of u · t = u · σn, where we let u → u(x1 − ∆a, x2), σ → σ(x1 − ∆a, x2),n → n(x1 − ∆a, x2). We note that the stress field does not change significantly with the propagation of the crack,nor does the normal vector n, so σ(x1 −∆a, x2) = σ(x1, x2), n(x1 −∆a, x2) = n(x1, x2).

∂

∂a

∫∂Ω

u(x1 −∆a, x2) · σ(x1, x2)n(x1, x2) dL =

∫∂Ω

(∂u(x1 −∆a, x2)

∂a

)· σ(x1, x2)n(x1, x2) dL (3.192)

Evaluating this derivative:∂u(x1 − (a− a0), x2)

∂a=

∂u

∂x1

d(x1 − (a− a0))

da= − ∂u

∂x1(3.193)

so therefore the result is∂

∂a

∫∂Ω

u · σn dL = −∫∂Ω

∂u

∂x1· σn dL (3.194)




and so the energy release rate is:J =

∫∂Ω

[w n1 −

∂u

∂x1· σ n

]dL ≡ J Integral (3.195)

In general, for an elastic solid, J = G , the energy release rate for an elastic solid. However J is a little more general;while G does not capture the effects of local plasticity, J does.3.7.1 Path-independence of the J integral

Consider the domain Ω as shown in the following figure, where the boundary of the body is divided into∂Ω = ∂1Ω ∪ ∂2Ω ∪ ∂3Ω ∪ ∂4Ω∪ (3.196)

Ω

n

n∂1Ω

∂2Ω

∂3Ω

∂4Ω

Let us compute the J integral for this domain, noting that this domain does not contain a fracture point, unlike thetypical J integral.J =

∫∂Ω

[w n1 −

∂u

∂x1· σ n

]dL =

∫∂Ω

[wδ1j −

∂ui∂x1

σij

]njdL (3.197)

Applying the divergence theorem and the product rule:=

∫Ω

∂

∂xj

[wδ1j −

∂ui∂x1

σij

]dL =

∫Ω

[ ∂w∂x1− ∂2ui∂xj∂x1

σij −∂ui∂x1

∂σij∂xj

]dL (3.198)

Looking at the first and last terms individually:∂w

∂x1=

∂w

∂εpq

∂εpq∂x1

= σpq∂2up∂xq∂x1

(3.199)∂ui∂x1

∂σij∂xj

=∂ui∂x1

:0div(σ)j = 0 (3.200)

Substituting back:J =

∫Ω

[ ∂w∂εpq

∂2up∂xq∂x1

− ∂2ui∂xj∂x1

σij

]dL = 0 (3.201)




What does this tell us about path independence? Let us consider the J-integral over segments ∂2Ω, ∂4Ω:∫∂2,4Ω

[w>

0n1 −

∂u

∂x1·

0t]dL (3.202)

since the surfaces are horizontal and traction free. This means that the J-integral over ∂1Ω is equal to the negativeof the J-integral over ∂2Ω. Furthermore, if we let n→ −n for ∂1Ω so that it is also outward-pointing, then the integralover the two contours are equal. We note a couple of things here:1. Even though the contours are circular, our derivation never relied on this fact and can be applied to any shape– completely general.2. This proof works because we used the divergence theorem. The only reason we can use the divergencetheorem in this case is that the strain and stress fields are continuous – if it’s discontinuous, then we cannotuse it.3. We conclude that all J integrals containing the same singularity give the same value – in other words, the J

integral is path independent.



Lecture 16 Fracture concluded, elastodynamics

3.7.2 J integral examples

We recall that the J-Integral is given by the following contour integral:J =

∫∂Ω

[w n1 −

∂u

∂x1· σn

]dL (3.203)

Let us examine a couple of examples illustrating the use of the J integral.Example 3.7: DCB J integral

Consider the same double cantilever beam as in the previous example. Let us try to derive the same resultas in the previous example using J integral theory. Try the following contour:a f

f

∂1Ω

∂2Ω

∂3Ω

∂4Ω

∂5Ω

We consider each segment individually:Contour 1: Let us assume that the length of the beam is considerably longer than a. If this is true, then thebeam is approximately free of stress and strain at the fixed end, and w = 0, ui ,1 = 1. Therefore the Jintegral is zero for this segment.Contour 2: We know that n1 = 0 for this segment so the first term in the J integral drops out. Therefore weare left with the following integral (assuming the origin is located at a)

limε→0

∫ a+ε

0

u2,1f δ(x1 − a) dx1 = f u2,1(a) (3.204)We have the displacement from E-B beam theory, so we just need to differentiate:

∂u

∂x1

∣∣∣x1=a

=∂

∂x1

[ f (3ax21 − x3

1 )

6EI

]x1=a

=[ f (2ax1 − x2

1 )

2EI

]x1=a

=f a2

2E I(3.205)

So the integral is justf 2a2

2E I(3.206)

Contour 3: This contour is traction-free so the second part vanishes, however we must consider the firstpart, w n1 = w . We recall that, from the beam theory done in the previous example, we havew(x1 = a, x2) =

[ f 2(a− x1)2x22

2E I 2

]x1=a

= 0 (3.207)Consequently this integral reduces to zero.



Contour 4: This integral reduces to zero similarly to Contour 3.Contour 5: This integral is the same as contour 2.Combining the results we find

J = 0 +f 2a2

2E I+ 0 + 0 +

f 2a2

2E I=

f 2a2

E I(3.208)

Example 3.8: Strip specimen

Consider a sample with a crack subjected to a total vertical displacement of δ as shown in the followingfigure

δ/2

δ/2∂1Ω

∂2Ω

∂3Ω

∂4Ω∂5Ω

L

Assume that the sample is sufficiently long so that the crack is far from both edges. Use a J-Integral to findthe energy release rate.Using the J-Integral as illustrated, let us evaluate each individual contour:Contour 1: This strip is far from the crack and is therefore stress-free, so w = 0 as does u1,1.Contour 2: This is a horizontal strip so n1 = 0. In addition, because the strip is fixed there is no change indisplacement in the x1 direction so ui ,1=0, and the so the J integral is zero for this contour as well.Contour 3: Let us compute the far-field stress:

ε11 = 0 ε22 =(L + δ)− L

L=δ

L(3.209)

σ11 = 0 σ22 = E ′ε22 =E ′δ

L(3.210)

Therefore the energy is w = E ′δ2/2L2. There is no change in the x1 direction (we are at far-field) sothe J integral over contour 3 isJ3 = (L)(w n1) =

E ′δ2

2L(3.211)

Contour 4: By symmetry, the same as contour 2, which is zero.Contour 5: By symmetry, the same as contour 1, which is zero.Therefore the energy release rate (assuming linear elastic behavior) is

G =E ′δ2

L(3.212)

Note that there is no dependence on crack length. This is becauseweassume that our contour is far from thecrack tip; if our solution is close to the crack tip then we cannot make the uniform stress field assumption.




3.7.3 Nonlinear fracture mechanics

Can stress ever really be infinite? Certainly not: every material has a critical stress at which it no longer behaveselastically and is said to flow plastically. Therefore the idealized picture of fracture is not necessarily realistic,although it is a suitable approximation for brittle fracture.

plastic region

idealized realisticFor fracture where plasticity takes place (“ductile fracture”) energy is absorbed into the material by doing plasticwork as well as into the surfaces that are created. An interesting aspect of J-integral theory is that it still works forductile fracture, although the relationship between J and the stress intensity factor no longer holds.



Lecture 17 Elastodynamics

4 Elastodynamic SolutionsWe now shift gears to turn our attention towards time-dependent elastic problems. We modify the conservationlaw to include the momentum change term as shows:

σij ,j = ρui (4.1)Cijklεkl ,j = ρui (4.2)Cijkl uk,jl = ρ u,tt (4.3)

Recalling that Cijkl = µ(δikδjl + δilδjk) + λδijδkl and substituting:2µ ui ,jj + λuj ,ji = ρ u,tt (4.4)

gives the Cauchy-Navier equation for displacement, which is sometimes an easier form to use than that providedby the elastic modulus tensor.4.1 Principle of stationary actionThe principle of stationary action (typically referred to in dynamics as the Lagrangian formulation of mechanics) isa powerful way to express equations of dynamics in a variational framework that is consistent with the principleof minimum potential energy for static solutions. Minimum potential energy states that the equilibrium value of usatisfies

δΠ = limε→0

d

dεΠ[u + εv ] = 0 ∀ admissible variations v (4.5)

The principal of stationary energy is a variational principal on theaction functional,S =

∫ b

a

[T − Π] dt (4.6)where T is the total kinetic energy of the body and the interval [a, b] is the period of action. A time-dependent defor-mation u(x, t) that satisfies conservation of linear momentum must also satisfy the following variational principal

δS = limε→0

d

dεΠ[u + εv ] = 0 ∀ admissible variations v (4.7)

where v is admissible if v : ∂1Ω × [a, b] → 0 and v(x, a) = v(x, b) = 0. For a general linear elastic solid, the actionfunctional is given byS [u] =

∫ b

a

[ ∫Ω

(1

2ρ ui ui −

1

2Cijklui ,juk,l + bi ui

)dv +

∫∂2Ω

ti ui da]dt (4.8)

We find the trajectory of the system by evaluating the first variation explicitly:d

dεS [u + εv ]

∣∣∣ε→0

=d

dε

[∫ b

a

∫Ω

(1

2ρ (ui + εvi )(ui + εvi )−

1

2Cijkl(ui ,j + εvi ,j)(uk,l + εvk,l) + bi (ui + εvi )

)dv dt+

(4.9)∫ b

a

∫∂2Ω

ti (ui + εvi ) da dt

]ε→0

(4.10)



Taking the derivative inside, evaluating derivatives and letting the remainder of the ε terms go to zero:=

∫ b

a

∫Ω

ρ ui vi dv dt −∫ b

a

∫Ω

Cijklui ,jvk,l dv dt +

∫ b

a

∫Ω

bi vi dv dt +

∫ b

a

∫∂2Ω

ti vi da dt (4.11)Use the product rule / integration by parts and the divergence theorem

=

∫Ω

ρ uivi dv∣∣∣ba−∫ b

a

∫Ω

ρ uivi dv dt (4.12)−

(∫ b

a

∫Ω

∂

∂xl(Cijklui ,jvk) dv dt −

∫ b

a

∫Ω

Cijklui ,jlvk dv dt

)(4.13)

+

∫ b

a

∫Ω

bi vi dv dt +

∫ b

a

∫∂2Ω

ti vi da dt (4.14)Gathering terms and relabeling some of the indices:

=

∫ b

a

∫Ω

(Cijkluk,lj + bi − ρ ui ) vi dv dt +

∫ b

a

∫∂2Ω

(ti − Cijkluk,lnj)vi dv dt (4.15)Noting that σij = Cijkluk,l

=

∫ b

a

∫Ω

(σij ,j + bi − ρ ui ) vi dv dt +

∫ b

a

∫∂2Ω

(ti − σijnj)vi dv dt = 0 ∀v admissible (4.16)The only way for the above to hold for all v is if the integrand itself is zero, whereby we recover the equations ofmomentum conservation and Cauchy tetrahedron. The principal of stationary action is convenient for computa-tion as well as for deriving equations of motion for complex systems that do not always lend themselves well toreduction from the general case, as we see in the next section.4.2 Transverse vibrations in Euler-Bernoulli beamsThe Lagrangian formulation of mechanics is useful in deriving the equations of motion for Euler-Bernoulli beams.We seek an equation u(x , t) that gives the time-dependent deflection of the beam under a time-dependent loadingt(x , t).

u(x , t)

L

Begin by computing the strain energy density of an Euler-Bernoulli beamM(x1) = E I

d2u

dx21

=⇒ σ(x1, x2) =Mx2

I= E x2

d2u

dx21

=⇒ ε(x1, x2) = x2d2u

dx21

(4.17)from which we get the total strain energy.

U =

∫Ω

w dV =

∫ L

0

∫A

1

2E x2

2

d2u

dx21

dx2 dx3 dx1 =

∫ L

0

1

2E

d2u

dx21

(∫A

x22 dx2 dx3

)︸︷︷︸

I

dx1 =

∫ L

0

1

2E I(d2u

dx21

)2

dx1 (4.18)




The potential energy is thus given byΠ =

∫ L

0

1

2E I(d2u

dx21

)2

− t(x , t) u(x , t) dx1 (4.19)The kinetic energy is

T =

∫Ω

1

2ρ(dudt

)2

dV =

∫ L

0

1

2ρA(dudt

)2

dx1 (4.20)The trajectory minimizes the action:

S [u] =

∫ L

0

[1

2ρA(dudt

)2

− 1

2E I(d2u

dx21

)2

+ t u]dx1 =

∫ 2

1

∫ b

a

L(u, u,t , u,xx) dx dt (4.21)where L is the Lagrangian. Derivation of Euler-Lagrange equation: let v be an admissible variation such thatv(x , a) = v(x , b) = 0 ∀x ∈ [0, L] and v(0, t) = v(L, t) ∀t ∈ [a, b]. Then:

limε→0

d

dεF [u + εv ] = lim

ε→0

∫ b

a

∫ L

0

d

dεL(u + εv , u,t + εv,t , u,xx + εv,xx) dx dt

= limε→0

∫ b

a

∫ L

0

[∂L∂u

d(u + εv)

dε+

∂L∂(u,t + εv,t)

d(u,t + εv,t)

dε+

∂L∂(u,xx + εv,xx)

d(u,xx + εv,xx)

dε

]dx dt

=

∫ b

a

∫ L

0

[∂L∂u

v +∂L∂u,t

∂v

∂t+

∂L∂u,xx

∂2v

∂x2

]dx dt

=

*0∫ L

0

∂L∂u,t

v∣∣∣badx +

∫ b

a

∫ L

0

[∂L∂u

v − ∂

∂t

( ∂L∂u,t

)v +

∂L∂u,xx

∂2v

∂x2

]dx dt

=

:

0∫ b

a

∂L∂u,xx

∂v

∂x

∣∣∣L0dt +

∫ b

a

∫ L

0

[∂L∂u

v − ∂

∂t

( ∂L∂u,t

)v − ∂

∂x

( ∂L∂u,xx

)∂v∂x

]dx dt

=

:0∫ b

a

∂L∂u,xx

v∣∣∣L0dt +

∫ b

a

∫ L

0

[ ∂L∂u,t

v − ∂

∂t

( ∂L∂u,t

)v +

∂2

∂x2

( ∂L∂u,xx

)v]dx dt

=

∫ b

a

∫ L

0

[∂L∂u− ∂

∂t

∂L∂u,t

+∂2

∂x2

∂L∂u,xx

]v dx dt ∀ admissible v (4.22)

Since this holds for all admissible v we simplify to∂L∂u− ∂

∂t

∂L∂u,t

+∂2

∂x2

∂L∂u,xx

= 0 (4.23)Substituting our actual expression for L, we have

∂2

∂x2

(E I

∂2u

∂x2

)+

∂

∂t

(ρA

du

dt

)= t(x , t) (4.24)

Assuming uniform beam propertiesE I

∂4u

∂x4+ ρA

∂2u

∂t2= t(x , t) (4.25)

We see that the static Euler Bernoulli equation is recovered when u does not depend on t. Notice that this is afourth order PDE, unlike the more familiar second order PDE wave equation. The fourth order derivative makes theproblem more difficult to solve exactly; however it can be solved for one particular case, as we see in the followingexample.All content © 2017-2018, Brandon Runnels 17.3


Lecture 18 Vibration example, waves in solids

Example 4.1: Vibration of a simply supported beam

Find the vibrational modes and frequencies of a simply supported beam with length L, elastic modulus Eand second moment of area I , as shown in the following figure

L

t(x , t)

A simply supported beam has the following boundary conditions:u(0, t) = 0 u(L, t) = 0 M(0, t) = E I

∂2u

∂x2

∣∣∣x=0

= 0 M(L, t) = E I∂2u

∂x2

∣∣∣x=L

= 0 (4.26)Let us assume a solution of the following form:

u(x , t) =∞∑i=1

un(t) sin(nπx/L) (4.27)Substituting this form into the boundary conditions, we see that they are automatically satisfied. (Note: thisis a special case; it is not always easy or even possible to guess solutions that satisfy boundary conditions!)Let us substitute this form into the PDE

∞∑i=1

[E I(nπ

L

)4

un(t) + ρA un(t)]

sin(nπx/L) = t(x , t) (4.28)Use orthogonality to extract individual terms by integrating with sin(mπx/L)

∞∑i=1

[E I(nπ

L

)4

un(t) + ρA un(t)] ∫ L

0

sin(nπx/L) sin(mπx/L) dx︸︷︷︸=L/2 if m=n,0 else

=

∫ L

0

t(x , t) sin(mπx/L) dx (4.29)

We thus extract the following sets of equations:ρA un(t) + E I

(nπL

)4

un(t) = tn(t) (4.30)where tn(t) is the nth coefficient in the sine series for t(x , t). The homogeneous equation is

un(t) +(E I n4 pi4

ρAL4

)un(t) = 0 (4.31)

We know that the solution is the form un(t) = C1 cos(ωt) + C2 sin(ωt) whereω =

n2π2

L2

√E I

ρA(4.32)



4.3 Plane waves in solidsInfinite three dimensional solid. Try a solution with the following form:

u = a f (x ·m− c t) ui = ai f (xp mp − c t) (4.33)Compute derivatives:

∂uk∂xj

= ai mj f′(xp mp − c t)

∂uk∂t

= −c ai f ′(xp mp − c t) (4.34)∂2uk∂xj∂xl

= ai mj ml f′′(xp mp − c t)

∂2uk∂t2

= c2 ai f′′(xp mp − c t) (4.35)

Substituting into the Cauchy Navier equation:Cijkl ak mj ml f

′′(xp mp − c t) = ρ c2 ai f′′(xp mp − c t) (4.36)

The unknown functions drop out leavingCijkl ak mj ml = ρ c2 ai (4.37)

Assuming isotropy and using the isotropic form for the modulus tensor (Equation 2.82):(µ(δikδjl + δilδjk) + λδijδkl) ak mj ml = ρ c2 ai (4.38)

µδikδjl ak mj ml + µδilδjk ak mj ml + λδijδkl ak mj ml = ρ c2 ai (4.39)µ ai ml ml + µ ak mk mi + λ ak mi mk = ρ c2 ai (4.40)

µ a + (λ+ µ)m (a ·m) = ρ c2 a (4.41)Two casesa ·m = 0:

µ a = ρ c2 a =⇒ c =

√µ

ρ(4.42)

a ·m = |a|:

µ a + (λ+ µ) a = ρ c2 a =⇒ c =

√λ+ 2µ

ρ=

√E

ρ(4.43)

where E is the constrained modulus.



Lecture 19 Discontinuous dynamic solutions

4.4 Discontinuous waves in solidsWe have discussed some solutions of the Cauchy-Navier equation for cases where waves are continuous anddifferentiable. However, a case of particular interest is that in which there is a discontinuous wave or disturbancepropagating through the material. Consider, for instance the following case of a shock that is passing through abody Ω:

Ω

Ω1

Ω2

∂Ω2

∂Ω1

∂Ω1 ∩ ∂Ω2

Let us assume that the discontinuity passes all theway through the body fromone end to the other; i.e. the boundaryof the shock front is contained in the boundary of the body. This introduces a natural partitioning Ω = Ω1 ∩ Ω2 ofthe body into the two regions on either side of the discontinuity, in which all of the displacements are continuous.Let us now derive some relationships between the state of the material on one side of the discontinuity and theother. These are called “jump conditions.”4.4.1 Continuity jump condition

Let us consider a single point s that is located on the discontinuity and moves with it through time, so that s is thevelocity of the discontinuity at that point.

s

su1(x)

u2(x)

The material does not interpenetrate itself, nor does it create a void. Therefore the displacement must be continu-ous at s(t) for all time; that is,u1(s, t) = u2(s, t) u1

i (s, t) = u2i (s, t) (4.44)

Daking the total derivative of both sides with respect to t , we haved

dtu1i (s(t), t) =

d

dtu2i (s(t), t)

∂u1i

∂sj

dsjdt

+∂u1

i

∂t=∂u2

i

∂sj

dsjdt

+∂u2

i

∂t

ε1ij sj + v1

i = ε2ij sj + v2

i (4.45)All content © 2017-2018, Brandon Runnels 19.1


Let us adopt the notation [[x ]] = x1 − x2. Then we write the above as[[ εij ]]sj + [[vi ]] = 0 [[ε]] s + [[v]] = 0 (4.46)

4.4.2 Momentum jump condition

Let us begin with the weak form of the equilibrium condition for a body. We know that the total force (t = σn)integrated over the boundary of the body is equal to the time derivative of the total momentum of the body; i.e.∫∂Ω

σ · n dA =d

dt

∫Ω

ρ v dV

∫∂Ω

σij nj dA =d

dt

∫Ω

ρvi dV (4.47)Let us consider the right hand side of the equation; the integral of the traction over the body. Normally we wouldjust apply the divergence theorem here to convert this to stress divergence. But there’s a condition to using thedivergence theorem that we often forget: that is, the functionmust be continuous across the entire body. Therefore,it’s not legal for us to apply the divergence theorem to the entire body. However, we can apply the divergencetheorem to the two halves of the body on either side of the discontinuity:∫

Ω

σij ,jdV =

∫Ω1

σij ,jdV +

∫Ω2

σij ,jdV

=

∫∂Ω

σij nj dA +

∫∂Ω1∩∂Ω2

σ1ij ,j n

1j dA +

∫∂Ω1∩∂Ω2

σ2ij ,j n

2j dA

=

∫∂Ω

σij nj dA +

∫∂Ω1∩∂Ω2

[[σij ,j ]] nj dA∫∂Ω

σij nj dA =

∫Ω

σij ,j dV −∫∂Ω1∩∂Ω2

[[σij ,j ]] nj dA (4.48)Notice that the presence of a discontinuity results in an additional term in the familiar divergence theorem expres-sion. Now, let us consider the left half of the original expression. An application of the Reynold’s transport theoremyields the following:

d

dt

∫Ω1

ρ vidV =

∫Ω1

∂

∂tρ vidV +

∫∂Ω1

ρ vi sj njdA (4.49)where s is the velocity of the boundary. In other words,

d

dt

∫Ω

ρ vidV =

∫Ω

∂

∂tρ vidV +

∫∂Ω1

ρ1 v1i sj n

1j dA +

∫∂Ω1

ρ2 v2i sj n

2j dA (4.50)

=

∫Ω

∂

∂tρ vidV +

∫∂Ω1∩∂Ω2

[[ρ vi sj ]] njdA (4.51)Combining all of the above, we have:∫

Ω

σij ,j dV −∫∂Ω1∩∂Ω2

[[σij ,j ]] nj dA =

∫Ω

∂

∂tρ vidV +

∫∂Ω1

[[ρ vi ]] sj njdA (4.52)∫Ω

[σij ,j −

∂

∂tρ vi]dV =

∫∂Ω1∩∂Ω2

([[σij ,j ]] + [[ρ vi ]] sj

)njdA (4.53)

Here we note that the above is true for all subregions of Ω. Further, we notice that it must hold true for all Ω ⊂(Ω1 ∪ Ω2). That is, we can adjust Ω on the left hand side all we wish so long as we don’t alter the boundary. Theonly way for this to hold true is for the integrand to be zero. Then, we observe that the right hand side must holdtrue for all subsets of the boundary. By the fundamental lemma, then we are left with the jump condition

[[σij ]] nj + [[ρ vi ]] sj nj = 0 [[σ ]]n + [[ρ v]] (s · n) = 0 (4.54)All content © 2017-2018, Brandon Runnels 19.2



4.4.3 Reduction to 1D case

Let us consider the propagation of a discontinuity in a plane-strain or plane-stressmaterial as shown in the followingfigure:s

ε1,σ1, v1, ρ1 ε2,σ2, v2, ρ2

The propagation of the discontinuity s and the velocity of the material v is in the x1 direction only. Substituting thisinto the jump conditions we arrive at[[ε]] s + [[v ]] = 0 (4.55)

where ε = ε11, the longitudinal strain in the x1 direction, and v is the component of the velocity in the x1 direction.Considering the equilibrium jump condition next, we find that it reduces to[[σ]] + [[ρv ]]s = 0 (4.56)

Let us assume that the material is linear and that the wave does not significantly affect the material properties,i.e. ρ1 = ρ2. For plane stress (i.e. assuming that the material is a slender bar) then σ = E ε. Substituting, theequilibrium jump condition becomesE

ρ[[ε]] + [[v ]] s = 0 (4.57)

Substituting the continuity condition, we haveE

ρ[[ε]]− [[ε]] s2 =

(Eρ− s2

)[[ε]] = 0 =⇒ s = ±

√E/ρ = ±c (4.58)

where c is the wave speed in the material. (For plain strain (i.e. assuming the material is infinitely thick) then thesound speed is c =√

E/ρ) This means that discontinuities in the material move either forward or backward at afixed speed c which is determined by the properties of the material.4.4.4 X-T diagrams

We will now turn our attention to the propagation of waves in bars. A helpful trick will be to use “x-t” diagrams tokeep track of the different states of the material through time. Here is an example of an x-t diagram depicting thepropagation of a discontinuous wave through a bar:

Region 1

Region 21/c

x

t

σ1, v1

σ2, v2

The stresses/strains/velocities on either side must obey the jump relations as specified above. This is best illus-trated with an example:



Lecture 20 Elastodynamics concluded, introduction to plasticity

Example 4.2: Whacking a bar for a short time

A bar is subjected to a velocity impulse of V for a short time on one end. The other end of the bar restsagainst a wall. What is the stress state of the bar as a function of time, and how long does it take for thebar to detach from the wall?We begin by drawing an x-t diagram. (See end of this example for the x-t diagram here.) Let us also tabulatethe velocity and stress state in each region, populating each regionwith the information thatwehave already:• The bar is initially stress-free and velocity-free, so σ1 = 0, v1 = 0

• Region 2 is constrained to move at velocity V• Except for region (2), all of the regions contacting the left side must be stress free; therefore σ4 =σ6 = 0

• All regions contacting the wall must be velocity-free, so v1 = v3 = v7 = 0

Region Velocity Stress1 v1 = 0 σ1 = 02 v2 = V σ2 = −ρVc3 v3 = 0 σ3 = −2ρVc4 v4 = 0 σ4 = 05 v5 = −V σ5 = −ρVc6 v6 = −2V σ6 = 07 v7 = 0 σ7 = 08 v8 = −V σ8 = ρVc9a v9 = 0 σ8 = ρVc9b v9 = −2V σ8 = 0

Now, we use jump conditions to find remaining values of the stress states in the other regions.(1)→(2) Forward moving wave→ s = +c :

(σ2 − σ1=0

) + ρ c ( v2=V− v1

=0) = 0 =⇒ σ2 = −ρcV (compressive stress) (4.59)

(2)→(3) Backward moving wave, so s = −c . Then the jump relation becomes(σ3 − σ2

=−ρcV)− ρ c (v3

=0− v2

=V) = 0 =⇒ σ3 = −2ρcV (double compressive stress) (4.60)

(2)→(4) Forward moving wave, so s = +c .(σ4

=0− σ2

=−ρcV) + ρ c (v4 − v2

=V) = 0 =⇒ ρVc + ρ c (v4 − V ) = 0 =⇒ v4 = 0 (4.61)

(4)→(5) Backward moving wave, so s = −c

(σ5 − σ4=0

)− ρ c (v5 − v4=0

) = 0 =⇒ σ5 − ρ c v5 = 0 need another equation (4.62)



(3)→(5) Forward moving wave, so s = +c

(σ5 − σ3=−2ρcV

) + ρ c (v5 − v3=0

) = 0 =⇒ σ5 + ρcv5 = −2ρcV (4.63)Solving both equations gives

σ5 = −ρcV v5 = −V (4.64)(5)→(6) Forward moving wave, so s = +c

(σ6=0− σ5

=−ρcV) + ρ c (v6 − v5

=−V) = 0 =⇒ v6 = −2V (4.65)


(σ7 − σ5=−ρcV

)− ρ c (v7=0− v5

=−V) = 0 =⇒ σ7 = 0 (4.66)


(σ8 − σ6=0

)− ρ c (v8 − v6=−2V

) = 0 =⇒ σ8 − ρ c v8 = 2ρ c V need another equation (4.67)(7)→(8) Forward moving wave, so s = c

(σ8 − σ7=0

) + ρ c (v8 − v7=0

) = 0 =⇒ σ8 + ρ c v8 = 0 (4.68)Solving both equations gives

σ8 = ρ cV v8 = −V (4.69)(8)→(9) Backward moving wave, so s = −c

(σ9 − σ8=ρcV

)− ρ c (v9=0− v8

=−V) = 0 =⇒ σ9 = 2ρcV (4.70)

This implies that the bar is under tensile stress–but the wall cannot sustain tensile stress.Therefore we conclude that bar detaches from the wall; this is now a free end. Consequently,we take the boundary condition to be σ9 = 0; redoing the analysis we get(σ9

=0− σ8

=ρcV)− ρ c (v9 − v8

=−V) = 0 =⇒ v9 = −2V (4.71)

The following diagram illustrates the state of the bar on an x-t chart. The hue of the coloring indicates stressstate (red=compressive, blue=tensile) and the intensity of the color represents the magnitude of the stress.The correponding diagram on the right illustrates the state in the bar as a function of time, and indicatesthe direction of the velocity.




1/cx

t

12

345

6 7

89

Timeto

detach

ment

Length of barIf the bar has length L, then we can use geometry to find that the time to detachment is 3L/c

4.4.5 Wave scattering at an interface

Consider a compressive wave similar to that discribed in the above example, where the material in the wave regionis traveling at a velocity V . Now, let the material encounter an interface where the material properties change. Thisinterface will cause the wave to split into a transmitted and a reflected component as shown:

1c1

1c2

Material 1: ρ1, c1 Material 2: ρ2, c2

x

t

32 Transmitted waveReflected wave

1v2 = V ,σ2 =?

v1 = 0,σ1 = 0

v3 =?,σ3 =?

Using jump conditions to go from region (1) to region (2):(σ2 − σ1

=0) + ρ1c1( v2

=V− v1

=0) = 0 =⇒ σ2 = −ρ1c1V (4.72)

Using jump conditions to go from region (2) to region (3):(σ3 − σ2

=−ρ1Vc1

)− ρ1 c1 (v3 − v2=V

) = 0 =⇒ σ3 − ρ1 c1 v3 = −2ρ1Vc1 (4.73)




And from region (1) to region (3):(σ3 − σ1

=0) + ρ2 c2 (v3 − v1

=0) = 0 =⇒ σ3 + ρ2 c2 v3 = 0 (4.74)

Solving both equations gives:v3 =

2ρ1 c1 V

ρ1 c1 + ρ2 c2σ3 = −2 ρ1 c1 ρ2 c2 V

ρ1 c1 + ρ2 c2(4.75)

If c2 → 0 then v3 = 2V and σ → 0; i.e. we recover the case of the free end. If c2 →∞ then v3 → 0 and σ → −2ρ1 c1 V ;i.e. we recover the case of the fixed end.4.5 Method of CharacteristicsRecall the equation of motion for a slender bar (uniaxial stress) or a large plate (uniaxial strain):

E∂2u

∂x2− ρ ∂

2u

∂t2= 0 E

∂2u

∂x2− ρ ∂

2u

∂t2= 0 (4.76)

Here we apply the method of characteristics for solving PDEs to this partial differential equation. We consideruniaxial stress only for the moment; the treatment generalizes readily to uniaxial strain by substituting E for E .Rewrite the above equation in the following way:∂2u

∂x2− 1

c2

∂2u

∂t2=( ∂∂x

+1

c

∂

∂t

)( ∂∂x− 1

c

∂

∂t

)u = 0 (4.77)

where we let c =√

E/ρ. Let us introduce two functions ξ(x , t), η(x , t) and consider the following derivative usingthe chain rule:du

dξ=∂u

∂x

dx

dξ+∂u

∂t

dt

dξ

du

dη=∂u

∂x

dx

dη+∂u

∂t

dt

dη(4.78)

If we suppose that the following is true (noting that x , t are independent)dξ

dx=∂ξ

∂x+∂ξ

∂t

dt

dx=∂ξ

∂x!

= 1dη

dx=∂η

∂x+∂η

∂t

dt

dx=∂η

∂x!

= 1 (4.79)dξ

dt=∂ξ

∂x

dx

dt+∂ξ

∂t=∂ξ

∂t!

= −c dη

dt=∂η

∂x

dx

dt+∂η

∂t=∂η

∂t!

= c (4.80)In other words,

ξ(x , t) = x − c t η(x , t) = x + c t (4.81)Then

d

dξ=

dx

dξ

∂

∂x+

dt

dξ

∂

∂t=

∂

∂x− 1

c

∂

∂t

d

dη=

dx

dη

∂

∂x+

dt

dη

∂

∂t=

∂

∂x+

1

c

∂

∂t(4.82)

Substituting into the differential equation we haved

dξ

d

dηu =

d

dη

d

dξu = 0 (4.83)

Because we can switch the order of the derivatives we find:d

dξ

du

dη= 0 =⇒ du

dη= constant w.r.t change in ξ, i.e. along η (4.84)

d

dη

du

dξ= 0 =⇒ du

dξ= constant w.r.t change in η, i.e. along ξ (4.85)




Therefore we have the followingdu

dξ= ε− 1

cv = const along ξ du

dη= ε+

1

cv = const along η (4.86)

Or, assuming no change in material properties,σ − ρ c v = const along ξ σ + ρ c v = const along η (4.87)



Lecture 21 Elastodynamics concluded, intro to plasticity

This is very helpful in problem-solving. Consider an x-t diagram. The lines corresponding to ξ =const and η =constare called characteristic lines, and we have shown that σ − ρ c v is constant along forward-moving lines whereasσ + ρ c v is constant along backward moving lines.

forward moving characteristic

σ − ρ c v = const along line

backward moving characteristic

σ + ρ c v = const along line

x

t

Because the characteristics have the same slope as wave propagation lines, they cannot cross in parallel; howevera forward-moving characteristic can cross a backward-moving wave. Consider the case of the previous example:

1/c

x

t

1

2

34

57

L

22 backward-moving η characteristicσ + ρ c v constant along line

forward-moving η characteristicσ − ρ c v constant along line

We can solve for the state in region (7) much more quickly than we could using jump conditions. Along backward-moving characteristic we haveσ1 + ρ c v1 = σ2 + ρ c v2 = σ4 + ρ c v4 (4.88)

(0) + ρ c (0) = σ2 + ρ c (V ) = (0) + ρ c v4 =⇒ σ2 = −ρ c V , v4 = 0 (4.89)Along forward-moving characteristic we have

σ4 − ρ c v4 = σ7 − ρ c v7 (4.90)(0)− ρ c (0) = σ7 − ρ c (0) =⇒ σ7 = 0 (4.91)

5 PlasticitySo far we have talked about material that behaves elastically (behaving like a spring) and dynamically (behavinglike a mass) behavior of solids and modeled some simple solutions. These aspects of material behavior (with theAll content © 2017-2018, Brandon Runnels 21.1


exception of fracture mechanics) are energetically conservative; that is, they are reversible. For instance, the workdone to deform a purely elastic material is perfectly recovered upon unloading. As another example, a vibratingbeam will go on vibrating indefinitely, as energy is passed back-and-forth from kinetic to potential.Now, we consider material behavior that is not conservative, is thermodynamically irreversible, and dissipates en-ergy.Plasticity is defined as irreversible, deformation under loading that is not recovered upon unloadingG. Some of theapplications of studying plasticity are• Metal working, forming, and processing• Strain/work hardening• Failure analysis• Coupling with micromechanical processes (dislocation motion, twinning)

5.1 Micromechanics of plasticityMetals are composed of grains, each of which contains atoms arranged in a crystallographic lattice. When a metalis deformed elastically, atoms are moved around so as to become closer or farther from each other, but there is nofundamental change to the lattice itself. (Most) plastic deformation results in a permanent change to the latticestructure itself; in other words, atoms are forced to move such that at least some of them no longer have the same“nearest neighbors.”Consider a cubic lattice that is subjected to a shear load as shown in the following figure. If the shear loading issmall, the entire lattice shifts homogeneously to accomodate it. But what happens when the loading becomes verylarge, so large that it starts tearing the atoms apart?

unloaded elastic dislocation dislocation permanentresponse nucleation motion deformation

Long before the stress reaches this point, the material “discovers” that it can relieve some of the stress by shiftingits atoms slightly as shown to nucleate a dislocation. Stress can be relieved even further if the material movesthe dislocation from one end to the other, resulting in a permanent shearing of the lattice structure. Then, evenif the stress is removed, the dislocation remains at the end and does not move back, so that the deformation ispermanent.We say that plasticity that occurs by this process is dislocation-mediated, and is the primary means of plasticityin metals with cubic crystal structure. The yield strength is generally determined by the amount of shear stress(typically called the “critical resolved shear stress”) necessary to move a dislocation. As you can imagine, somematerials are able tomove dislocationsmuchmore easily than others. Materials with face-centered cubic structuresuch as aluminum, gold, silver, and copper have high symmetry and densly-packed planes, which is very conduciveto dislocation motion. This is why these metals are so malleable.Materials with body-centered cubic structure can deform plastically by dislocations, but it is not nearly as easy asin face-centered structures; this is why bcc materials such as bcc-phase iron, molybdenum, and tungsten are moredifficult to work.




It is very difficult to transmit dislocations inmaterialswith hexagonal close-packed structure, which iswhymaterialssuch as titanium, magnesium, and cobalt are very difficult to machine (and tend to fail under brittle fracture ratherthan ductile.) Nevertheless, materials with low-symmetry crystal structures still can deform plastically. Anothermechanism for plasticity is twinning. Consider the following lattice with a monoclinic crystal structure subjectedto a shear load.

unloaded deformation permanenttwinning deformation

Because of the atomic arrangement, it is possible for the the top row of atoms to shift slighly, followed by thenext row, and so on down through the lattice. We see that the top half of the crystal has “flipped” so that it is thereflection of the lower half. There is no atomic rearrangement but the work done is nevertheless reversible, so thatit remains in the current state upon unloading.5.1.1 Bridgman’s Hypothesis

Notice that in both of the above examples, we assume a shear loading. It turns out that shear loading is crucial toplastic deformation. Bridgman hypothesized that plastic deformation can take place only by shear deformation:“There is no permanent volume change associated with plastic deformation”

Or,“Hydrostatic pressure does not cause permanent volume change or plastic deformation”

From a crystallographic standpoint, this may seem a bit obvious; the only way to change the structure is to applysome sort of shear deformation. This hypothesis forms the starting point and allows us to formulate the theory ofplasticity.5.2 Ingredients of plasticity theoryPlasticity is a complex phenomenon and developing a material model for plasticity (even a rudimentary one) re-quires the inclusion of a number of ingredients.elastic model: Material fails under stress, so we must use an elastic model to determine the stress state ofa material before we can determine if plasticity occurs. In small strain, this is usually Hooke’slaw.yield criterion: This is a rule that is used to determine if plasticity has occured. In 1D it is usually just σ ≤ σ0(where σ0 is the yield strength) but it is more complicated in multiple dimensions.flow rule: This tells us how the material behaves after plasticity has occurred.

hardening model: Material hardening, put generally is a phenomenon that alters the yield criterion of the materialitself. Hardening can result from plastic flow (work hardening) or from an increase in the speedof the loading (strain rate hardening).All content © 2017-2018, Brandon Runnels 21.3



5.3 One-dimensional plasticityConsider a specimen that is subjected to a uniaxial load and is stretched quasi-statically and isothermally as shownin the following figure. Stress and strain are recorded as the material is loaded and unloaded.elastic loading: linear stress-strain relationshipplastic hardening: stress increases but is no longer linearelastic unloading: material unloads elastically but not all strain is recoveredelastic loading: material behaves exactly the same as initially except (1) permanent rest strain and (2) higheryield strength (we say the material has hardened)necking: material undergoes geometric change as failure initiatesfailure: material undergoes ductile fracture

elas

ticload

ing

elas

ticun

load

ing

elas

ticload

ing

plasticharden

ingfailure

necking

ε

σ

σ = fA , ε = ∆L

L

f

L + ∆L

A

σinitial0

σhardened0

Notice that in the above example we stated that the material was loaded quasi-statically – in other words, slowlyenough so as to no cause rate-dependent effects. However if we run the exact same experiment multiple timeswith different loading rates, we see that the material effectively hardens the faster it is deformed, as shown below.(Notice, though, that the elastic modulus is always the same.)On the other hand, adjusting the temperature of the specimen considerably lowers the yield strength and alsoreduces the elastic modulus. This is called thermal softening.σ

ε

ε1

ε2

ε3

increasing strain rate

σ

ε

increasing temperature

strain rate hardening thermal softening

5.3.1 Characterization of uniaxial plasticity

When working with small strain there is an additive decomposition of elastic and plastic strain:ε = εe + εp (5.1)




and in the elastic regime, stress is given byσ = E (ε− εp) (5.2)

There are several phenomenological models that aim to quantify hardening behavior in materials:Elastic-perfectly plastic This is the simplest plasticitymodel possiblewhere thematerial is assumed to not hardenat all, but to remain and flow at the yield stress until unloaded.

σ =

E (ε− εp) E (ε− ε0) ≤ σ0

σ0 else

σ

ε

σ0

Note that in the EPP model, stress in the material can never exceed the yield stress. Thismeans that (in the 1D case) all plastic regions have the same stress, σ0.Piecewise-power law The piecewise power law attempts to model the hardening shape by using an exponent

σ =

E (ε− εp) E (ε− εp) ≤ σ0

σ0(σ/σ0)1/n else

σ

ε

σ0

n = 1n = 2

n =∞

Ramberg-Osgood The Ramberg-Osgood model is similar to piecewise power law, but has the advantageof being continuous. On the other hand, it requires more calibration to capture the linearbehavior correctly.

ε =σ

E+ K

( σE

)nwhere E , n,K are all material constants.

σ

ε

σ0

n = 1n = 2

n =∞



Lecture 22 1D plasticity continued, multi-axial plasticity

Example 5.1: Plastic hinge

elastic elastic-plastic fully plastic

Kinematics (always true)ε = −κ x2 ≈ −

d2u

dx21

x2 (5.3)Hooke’s law:

σ = E (ε− εp) = −E(d2u

dx21

x2 + εp

) (5.4)Total moment given by

M = −∫A

σ x2 dA = E

∫A

d2u

dx21

x22 dA + E

∫A

εp x2 dA (5.5)= E I

d2u

dx21

+ E

∫A

εp x2 dA (5.6)Purely elastic case:

M =fL

4= E I u′′(x1) =⇒ u′′(x1) =

M

E I=

fL

4E I(5.7)

Integrating and using boundary conditions:u(x1) =

fL

4E Ix2

1 + *

0C1 x1even function + C2 (5.8)

u(L/2) =fL

4E I(L/2)2 + C2 = 0 =⇒ (5.9)

u(x1) =fL

4E I

(x2

1 −L2

4

)=⇒ u(0) = − fL3

16E I(5.10)

So force as a function of displacement isf = −16EIu

L3= −4EWH3u

3L3(5.11)

Critical force f0 occurs when stress reaches σ0:σ(L/2,H/2) = −E

(d2u

dx21

x2

)= − f0LH

8I!

= −σ0 (5.12)f0 =

8 I

LHσ0 =

2WH2

3Lσ0 =

2

3(WHσ0)

(HL

) (5.13)



and so the critical displacement isu0 = − L3

16E I× 8 I

LHσ0 = −L2σ0

2EH(5.14)

Onset of plastic initiation, elastic region has height δ. Let the stress take the following form:σ = −

2σ0x2/δ |x2| ≤ δ/2

σ0 sgn(x2) else != −

Eu′′(x1)x2 |x2| ≤ δ/2

σ0 else (5.15)

= −E

[u′′(x1)x2︸︷︷︸

ε

+

0 |x2| ≤ δ/2

σ0 sgn(x2)/E − u′′(x1)x2 else︸︷︷︸εp

](5.16)

Computing the moment:M = −

∫A

σ x2 dA = 2W

∫ δ/2

0

(2σ0x2/δ)x2 dx2 + 2W

∫ H/2

δ/2

σ0 x2 dx2 (5.17)=

4Wσ0

δ

∫ δ/2

0

x22 dx2 + 2Wσ0

∫ H/2

δ/2

x2 dx2 (5.18)=σ0

4

(H2 − δ2

3

)!

=fL

4(5.19)

Solving for δ givesδ =

√3(H2 − Lf

σ0W

) (5.20)Use the stress-strain relationship in the elastic region to get

2σ0x2

δ= Eu′′(x1)x2 (5.21)

u′′(x1) =2σ0

Eδ(5.22)

u(x1) =σ0

Eδ

(x2

1 −L2

4

) (5.23)u(0) = −σ0L

2

4Eδ= −σ0L

2

4E

(√3(H2 − fL

Wσ0

))−1

(5.24)Solving for force gives

f =WH2σ0

L− Wσ3

0L3

48E 2u2(5.25)

Check: continuity?f0

!=

WH2σ0

L− Wσ3

0L3

48E 2u20

=WH2σ0

L− Wσ3

0L3

48E 2× 4E 2H2

L4σ20

=WH2σ0

L− Wσ0H

2

12L(5.26)




Simply supported beam with a force in the middleM(L/2) =

( f2

)(L2

)=

fL

2(5.27)

Recall strain:ε = −κ x2 ≈ −

d2u

dx21

x2 = −( M

E I

)= −M x2

E I(5.28)

Stress (assuming linear regime)σ = −M x2

I(5.29)

Check: do we recover moment?M = −

∫A

σ x2 dA =M

I

∫A

x22 dA = M X (5.30)

Stress (assuming plastic regime, elastic perfectly plastic model)σ = −

M x2/I |M x2/I | ≤ σ0

σ0 sgn(x2) else (5.31)Accumulated plastic strain: we know that

σ = E (ε− εp) (5.32)so,

εp =

0 Mx2/I ≤ σ0

Mx2/EI − σ0/E else (5.33)Do we get the same thing?

σ = E (ε− εp) = E

(Mx2

E I−

0 Mx2/I ≤ σ0

Mx2/EI − σ0 sgn(x2)/E else)

(5.34)

= E

(−

Mx2

E I Mx2/I ≤ σ0

σ0 sgn(x2)/E else)

(5.35)

= −

Mx2

E I Mx2/I ≤ σ0

σ0/E else (5.36)Yes, so now we know the accumulated plastic strain. Point of yield:

σ0!

=M(H/2)

I=

LfH

8I=⇒ f0 =

8σ0 I

LH(5.37)

Now, what happens after onset of plasticity? Assume piecewise linear:σ = −

σ0x2/δ |x2| ≤ δσ0 sgn(x2) else (5.38)




How do we find δ? Integrate to solve for moment:M = −

∫A

σ x2 dA (5.39)Can’t pull the usual trick, so use square cross section, width W .

M = 2W

∫ H/2

0

σ x2 dx2 = 2W

∫ δ

0

σ0x22

δdx2 + 2W

∫ L/2

δ

σ0x2 dx2 (5.40)=

2Wσ0

δ× δ3

3+ 2Wσ0 ×

1

2

(L2

4− δ2

) (5.41)= Wσ0

(L2

2− 5δ2

3

)!

=fL

4(5.42)

Solving for δ we getδ =

√3L2

10− 3fL

20Wσ0(5.43)

What is the collapse load? It is exactly the point when the entire beam is plastically loaded and δ → 0

f = 2Wσ0L (5.44)Residual strain for semi-plastic regime:

εp =

0 |x2| ≤ δ−σ0(x2/δ − 1)/E else (5.45)

Now, recall thatε = −κx2 (5.46)

Suppose load is removed so that M = 0. Substitute in plastic stress, using Hooke’s law, we know σ =E (ε− εp). We know

0!

= M =

∫A

σ dA... (5.47)

5.4 Multi-axial plasticity5.4.1 Yield criteria

Our first job in constructing a 3D theory of plasticity (we will bypass 2D for now) is to find a yield criterion. Werecall from Bridgman’s Hypothesis that no hydrostatic pressure can cause plastic deformation, so let us constructa tensor that is always guaranteed to be free of hydrostatic pressure:Sij = σij −

1

3σkkδij (5.48)

S is called the deviatoric stress tensor or, alternately, the stress deviator. Consider, for example the case ofhydrostatic pressure −p: so that the stress tensor is σij = −pδij . Then the deviatoric stress tensor isSij = (−pδij)−

1

3(−pδkk)δij = 0 (5.49)




andwe assume that our yield criterionmay be determined based on some function of S only. Furthermore, becauseplasticity is frame-invariant, it must be dependent on an invariant of S . Recall that the invariants of tensors are givenby the following (where we choose the labels J1, J2, J3 instead of the conventional I , II , III notation)J1 = Sij = tr(S) = 0 J2 = −1

2(SiiSjj − SijSij =

1

2SijSij J3 = det(Sij) (5.50)

The first invariant J1 is of no interest since it is always zero by construction, and so it is not useful as a yield criterion.The last invariant J3 is always zero for the case of plane stress, becausedet

[σ11 σ12 0σ12 σ22 00 0 0

]= 0 (5.51)

but we know from experience that plasticity can indeed occur in plane stress configurations, so it cannot be auseful criterion for us. We are left with J2 as a candidate for constructing a yield criterion.Suppose we construct an experiment (using, say, an Instronmachine) in which wemeasure the exact tensile stressat which the material fails. Using a dogbone specimen, we determine that the material fails whenσ =

[σ0 0 00 0 00 0 0

](5.52)

where σ0 is the measured stress at failure, the yield stress. Computing J2 for this state we see thatS =

[σ0 0 00 0 00 0 0

]− σ0

3I =

[2σ0/3 0 0

0 −σ0/3 00 0 −σ0/3

](5.53)

ThenJ2 =

1

2(SijSij) =

1

2

(4σ20

9+σ2

0

9+σ2

0

9

)=σ2

0

3(5.54)

We thus deduce the following criterionσ2

0 ≥3

2SijSij =

1

2

((σ22 − σ33)2 + (σ33 − σ11)2 + (σ11 − σ22)2 + 6(σ2

23 + σ231 + σ2

12)) (5.55)

=1

2

((σ2 − σ3)2 + (σ3 − σ1)2 + (σ1 − σ2)2

)≡ J2/von Mises Yield Criterion (5.56)

where σ1,σ2,σ3 are principal stresses. This is the well-known J2 or vonMises yield criterion, and it does remarkablywell in many cases.



Lecture 23 Multi-axial plasticity

Recall that material fails under shear. Consider a material undergoing a loading, and suppose that we are in areference frame such thatσ =

[σ1

σ2σ3

](5.57)

where σ1,σ2,σ3 are principal stresses; in other words, we are in the principal bases. Suppose further that σ3 ≥ σ2 ≥σ1. What is the maximum shear stress that this material is undergoing? We know from Mohr’s circle that it is justτmax = 1

2 (σ3 − σ1). Note from the above uniaxial tension test that σ can be written in rotated components asσ =

[τ0 τ0 0τ0 −τ0 00 0 0

](5.58)

where τ0 is the maximum shear load corresponding to the uniaxial tension test, and τ0 = σ0/2. This prompts thefollowing alternative yield criterion:σ0

2= τ0 ≥

1

2(σmax − σmin) ≡ Tresca Yield Criterion (5.59)

where σmax , σmin are the maximum and minmum principal stresses.Multi-axial yield criteria

For the following functionsfJ2(σ) =

3

2SijSij − σ2

0 =1

2

((σ22 − σ33)2 + (σ33 − σ11)2 + (σ11 − σ22)2 + 6(σ2

23 + σ231 + σ2

12))− σ2

0 (5.60)ftresca(σ) = max(|σ3 − σ2|, |σ1 − σ3|, |σ2 − σ1|)− σ0 (5.61)

plastic flow occurs when f (σ) = 0.One can think of the yield criterion as representing a region in stress space, the boundary of which may be thoughtof as a yield surface. Because stress space is six dimensional, it is not easy to represent such a surface; however,we can consider a reduction of this space to 3D by assigning principal values to each axis. This is called principalstress space. The J2 and Tresca yield criteria then take the following forms (figure from Wikipedia)

σ1

σ2

Tresca

J2



Taking a two-dimensional slice (i.e. letting σ3 = 0 for plane-stress conditions) we have a simplified picture ofthe Tresca and J2 yield criteria Note that Tresca is slightly more conservative than J2 – this indicates that it is asafe(r) criterion to use than J2 when analyzing to determine if plasticity is taking place. Also, note that the lineσ1 = σ2 intersects the yield surface, indicating that in-plane hydrostatic pressure can induce plasticity – why is thispossible? It is possible because this is the slice along σ3 = 0, so the line is really the line σ1 = σ2,σ3 = 0.Let us illustrate how this works with an example.

Example 5.2: Plastic sphere

Consider a thick-walled sphere made of an elastic-perfectly-plastic material subjected to an internal pres-sure of p.

ab

c

p

div(σ) =

[σrr ,r +

2σrrr− σθθ + σφφ

r

]gr +

[cot θσθθ

r− cot θσφφ

r

]gθ (5.62)

=

[σrr ,r + 2

(σrr − σθθ)

r

]gr (5.63)

Try the following form:σrr = C1 +

C2

r3σθθ = C1 −

C2

2r3(5.64)

div(σ) =

[− 3

C2

r4+

2

r

(C1 +

C2

r3− C1 +

C2

2r3

)]gr = 0 X (5.65)

Elastic solution: boundary conditions σrr (a) = −p, σrr (b) = 0

σrr (a) = C1 +C2

a3= −p σrr (b) = C1 +

C2

b3= 0 =⇒ C1 = −C2

b3(5.66)

substituting:C2

( 1

a3− 1

b3

)= −p =⇒ C2 =

pa3b3

b3 − a3=⇒ C1 =

pa3

a3 − b3(5.67)

When does plastic flow first occur? Assuming it happens at r = a, we know it occurs when|σθθ(a)− σrr (a)| = σ0 (5.68)




So|σθθ(a)− σrr (a)| =

∣∣∣∣∣− C2

2b3− C2

b3

∣∣∣∣∣ =3|C2|2b3

=3pa3

2(b3 − a3)= σ0 =⇒ p0 =

2σ0(b3 − a3)

3a3(5.69)

Now, what happens when p > p0? The radius of the plastic region is c > a, now we attempt to solve for it.We must consider the elastic and plastic regions seperately. The elastic solution no longer works, but theequilibrium equation does: substituting the Tresca solution in (and assuming that σrr < σθθ):σrr ,r + 2

(σrr − σθθ)

r= σrr ,r − 2

σ0

r= 0 =⇒ dσrr

dr=

2σ0

r(5.70)

Integrating once we haveσrr = 2σ0 ln(r) + C3 (5.71)

Solving for boundary conditions we getσrr (a) = 2σ0 ln(a) + C3 = −p =⇒ C3 = −p − 2σ0 ln(a) (5.72)

so that the solution for the radial stress in the plastic region isσrr (r) = 2σ0 ln

( ra

)− p a ≤ r ≤ c (5.73)

and the circumferential stress in the plastic regionσθθ(r) = σ0 + σrr (r) = 2σ0 ln

( ra

)+ σ0 − p (5.74)



Lecture 24 Crystal plasticity

Example 5.2 Continued

Now, consider the elastic region: we know that the following solution still holdsσrr = C4 +

C5

r3σθθ = C4 −

C5

2r3(5.75)

but we need new boundary conditions, in fact, we have the following boundary conditions:σrr (c) = C4 +

C5

c3= 2σ0 ln

(ca

)− p σrr (b) = C4 +

C5

b3= 0 =⇒ C4 = −C5

b3(5.76)

Solving simultaneously yields the followingC5

( 1

c3− 1

b3

)= 2σ0 ln(c/a)− p =⇒ C5 =

b3c3(2σ0 ln(c/a)− p)

b3 − c3,C4 =

c3(2σ0 ln(c/a)− p)

c3 − b3(5.77)

which gives us the solution. But...what is c? We need another equation. Fortunately we have one: we knowthat σθθ(c)− σr (c) = σ0 for the elastic solution as well. Therefore,σ0 = −3C5

2r3=

3b3c3(2σ0 ln(c/a)− p)

2(c3 − b3)=⇒ p = 3σ0 ln(c/a) +

2

3

(b3 − c3

b3c3

)σ0 (5.78)

Double check:p(c = a) =

:0

3σ0 ln(a/a) +2

3

(b3 − a3

a3b3

)X (5.79)

What is the critical pressure? This is given when c → b:pmax = p(c = b) = 3σ0 ln(c/a) +

:0

2

3

(b3 − b3

b3c3

)σ0 = 3σ0 ln(c/a) (5.80)

5.4.2 Flow rule

We have a function f that indicates the onset of plasticity, but how does the material flow once plasticity occurs?Recall that we describe the yield surface using a function f , and yield occurs when f (σ) = 0. Then the plastic flowis given bydεpij = dγ

df

dσij(5.81)

In other words, the plastic flow is always normal to the yield surface.



We can compute this exactly for the case of isotropic J2 hardening. Recalling that f =

d

dσij

(3

2SpqSpq

)=

3

2× 2Spq

dSpqdσij

= 3Spqd

dσij

(σpq −

1

3σkkδpq

)= 3Spq

(δipδjq −

1

3δijδpq

) (5.82)= 3(σpq −

1

3σkkδpq

)(δipδjq −

1

3δijδpq

) (5.83)= 3(σij −

1

3σkkδij

) (5.84)= 3Sij (5.85)

so the plastic strain is updated (for J2 isotropic hardening) according to:dεpij = 3dγSij (5.86)

Notice that the plastic flow is equal to the gradient of f with respect to σ. Because f = 0 is an isoline of f , we knowfrom calculus that the gradient of f at that point must be orthogonal to the isoline. Thus, in principal stress space,we can get an idea for how the material flows plastically by drawing the normal to the yield surface.5.4.3 Hardening model

When a material hardens, it alters the yield surface. It can happend isotropically or anisotropically, depending onthe material. The following figure illustrates how this might happen.

σ1

σ2

σ1

σ2

elastic

plasticela

stic plastic

J2 yield surface with Tresca yield surface withisotropic hardening anisotropic hardening

Apparently, the yield surface is now dependent on plastic strain, i.e. f = f (σ, εp). But we know that the stresscannot reside outside the yield surface, we have that, for flow:f (σ, εp) = 0 (5.87)

Moreover, as the material flows, f must remain constant. Therefore, df = 0 and we have (for J2 plasticity)df =

∂f

∂σijdσij −

df

dεpijdεpij = 3Sijdσij −

df

dεpijdεpij = 3Sijdσij − 3dγ

df

dεpijSij = 0 (5.88)

dγ =[ df

dεpijSij]−1

Sijdσij (5.89)We use a model to compute the tensor df /dεpij .5.5 Crystal plasticityConsider a single crystal in a polycrystalline, face-centered cubic material Because dislocations are the mediatorsof plasticity, we know that the initiation of dislocation motion is tantamount to yield. In FCCmaterials, dislocationstend to glide along the 〈111〉 planes, as they are the most densly packed.All content © 2017-2018, Brandon Runnels 24.2



Slip plane (n) Slip directions (s)(111) [011], [101], [110]

(111) [011], [101], [110]

(111) [011], [101],[110]

(111) [011], [101],[110]

This plane is packed with atoms in a hexagonal arrangement, so that there is threefold symmetry. Consequently,there are three directions per 〈111〉 plane on which dislocations move, as shown in the following figure.

x1 x2

x3

(111)

x1 x2

x3

[011][011]

[101]

[110]

x1 x2

x3dislocationmotion

We suppose that plastic yield occurs when a shear stress along one of the slip directions exceeds a certain value;we will call it τcrss , the “critical resolved shear stress.” So, given a stress state σ, a plane n, and a direction s, howdo we compute the resolved shear stress? We simply compute s · σn. This enables us to construct a yield criterionfor each slip system:fα(σ) = |sα · σnα| − ταcrss (5.90)

where α indicates a particular combination of slip planes and directions forming a “slip system.” Consequently, theflow rule for each slip system is:dεpij = dγ

df

dσij= dγ

d

dσij(|spσpqnq|) = dγ sgn(s · σn)si nj (5.91)

where sgn(·) returns the sign (±1) of the operand. Dividing by dt and using symbolic notation:εp = γ sgn(s · σn) s⊗ n (5.92)

Now we identify a problem: s ⊗ n is symmetric, but εp must be. So, it is necessary to consider all slip systemstogether, resulting in the following flow rule:εp =

∑α active

γα sgn(sασnα) sα ⊗ nα (5.93)where all slip systems contribute simultaneously to the plastic flow. We illustrate how this works with a simpleexample:

Example 5.3

Consider a single crystal subjected to a uniaxial tensionσ =

[σ0 0 00 0 00 0 0

](5.94)




It the material has perfect cubic symmetry, what slip systems are “activated” by a critical load, and at whatcritical resolved shear stress τcrss , at what uniaxial stress σ0 does the crystal yield? To determine this wemust check each of our yield criteria:f11(σ) = s11σn1 = −τcrss f12(σ) = s12σn1 =

1√6σ0 − τcrss f13(σ) = s13σn3 =

∣∣∣− 1√6σ0

∣∣∣− τcrss (5.95)f21(σ) = s21σn1 = −τcrss f22(σ) = s22σn1 =

1√6σ0 − τcrss f23(σ) = s23σn3 =

∣∣∣− 1√6σ0


1√6σ0 − τcrss f33(σ) = s33σn3 =

∣∣∣− 1√6σ0


∣∣∣ 1√6σ0

∣∣∣− τcrss f43(σ) = s43σn3 =∣∣∣− 1√

6σ0

∣∣∣− τcrss (5.98)(5.99)

Apparently slip systems i1 are never activated, however, all of the remaining slip systems are, and the yieldstrength isσ0 = τcrss

√6 (5.100)

What is the resultant plastic flow? We use our combined flow rule:εp =

∑α active

γα sgn(sασnα) sα ⊗ nα (5.101)= γ12(1)s12 ⊗ n1 + γ13(−1)s13 ⊗ n1 + γ22(1)s22 ⊗ n2 + γ13(−1)s13 ⊗ n2 + γ32(1)s32 ⊗ n3 + γ13(−1)s13 ⊗ n3(5.102)

Assuming that γα are all equal (which is reasonable given the perfect cubic symmetry) this is=

γ√6

([1 1 10 0 0−1 −1 −1

]−

[−1 −1 −11 1 10 0 0

]+

[1 1 −10 0 01 1 −1

]−

[−1 −1 11 1 −10 0 0

]+ (5.103)[

1 −1 −10 0 01 −1 −1

]−

[−1 1 1−1 1 10 0 0

]+

[1 −1 10 0 0−1 1 −1

]−

[−1 1 −1−1 1 −10 0 0

])=

[8 0 00 −4 00 0 −4

](5.104)

= γ4√6

[2 0 00 −1 00 0 −1

](5.105)

As expected, εp is symmetric. Furthermore, we see thattr(εp) = 0 (5.106)

indicating that there is no volumetric change with plastic flow, in accordance with Bridgmann’s hypothesis.Now that we know the components of the flow, how do we get γ? We model this using a simple power law:

γα = γ0 sgn(sα · σnα)( |sα · σnα|

ταcrss

)n (5.107)where γ0, τcrss , and m are material parameters.




5.5.1 Latent Hardening

This is taken from the seminal paper of Pierce, Asaro, and Needleman ([2], as well as Bower. Single sip-systeminteraction causes hardening when dislocations accumulate.

x1 x2

x3

→ greater dislocation contentmore plastic slip

→ reduced mobility→ reduced flow rate→ hardening

dislocation motion

Power law self-hardening for a single slip system:ταcrss = h0|γ| (5.108)

where h0 is a material parameter. This is great, but not very physical, because the picture that we presented ofdislocation hardening is highly oversimplified. Each system does not harden independently; they are all connected.The following picture is a little more realistic:

x1 x2

x3

in-plane dislocation

x1 x2

x3

out-of-plane dislocationinteractioninteraction

We allow each system’s hardening rate to interact with each other:σαcrss =

∑β

hαβ |γβ | (5.109)where h is the latent hardening rate. This can be approximated by:

hαβ = qαβ

(hs + (hs − h0) sech2

[(h0 − hsτs − τ0

)γ

])(5.110)

with hs , h0, τs material properties, andγ =

∫ t

0

∑α

|γα|dt (5.111)the total accumulated plastic slip on all slip systems. Finally,

qαβ =

1 if coplanarq else (5.112)

where q is another material property. (Lots and lots of knobs!) Values for all these experimental parameters canbe found in the literature.All content © 2017-2018, Brandon Runnels 24.5



References[1] George R Irwin. Analysis of stresses and strains near the end of a crack traversing a plate. Spie Milestone series

MS, 137(167-170):16, 1997.[2] Daniel Peirce, Robert J Asaro, and A Needleman. Material rate dependence and localized deformation in crys-talline solids. Acta metallurgica, 31(12):1951–1976, 1983.[3] James RRice et al. A path independent integral and the approximate analysis of strain concentration by notchesand cracks. ASME, 1968.[4] George C Sih. Handbook of stress-intensity factors. Lehigh University, Institute of Fracture and Solid Mechanics,1973.




Index

admissible variations, 9.1affine deformation, 5.2Airy stress functioncartesian coordinates, 10.1cylindrical coordinates, 10.2auxetic materials, 7.6basis vectors, 1.2body force vector field, b, 6.3, 8.6boundary conditionsdisplacement, 9.2traction, 9.2boundary value problem, 9.1Cauchy stress tensor, see stress tensor, σCauchy tetrahedron, 6.3change of basis, 4.2tensor transformation, 4.5compatibility, 5.5conservationof angular momentum, 8.3of linear momentum, 8.1of mass, 8.1constitutive lawstwo dimensionalgeneral relations, 9.4united axial stress/strain relations, 9.4constrained modulus, E , 7.5, 18.2convexity, 8.4covariant basis vectors, 4.1curl, 3.2cylindrical coordinatesbasis vectors for, 4.1divergence of a vector field, 4.4equilibrium equation, 9.4gradient of a scalar function, 4.3deformation gradient tensor, F , 5.3deviatoric stress tensor, S , 22.4Dirichlet b.c., see displacement b.c.dislocations, 21.2displacement field, u, 5.3displacement gradient tensor, 5.3divergence, 3.1divergence theorem, 3.4duality, 8.4dummy indices, 1.3dyadic product, ⊗, 2.1Einstein summation convention, 1.3

elastic modulus, see Young’s moduluselastic modulus tensor, C , 7.1major and minor symmetries, 7.1energy release rate, G , 14.2relation to stress intensity factor K , 15.2essential b.c., see displacement b.c.fast Fourier transform, 12.2free indices, 1.3Gateaux derivatives, 3.2gradient, 3.1homogeneous deformation, see affine deformationhoop stress, 12.1hydrostatic pressure, 7.5index notation, 1.3, 2.1infinitesimal rotation tensor, r, 5.3inverse function theorem, 4.3isotropic materialselastic constants, 7.3j integral, 15.3Kroneker delta, δ, 1.4Lagrangian, 17.3Lagrangian mechanics, 17.1Lamé solutions, 11.2general solution, 11.3laplacian, 3.1Legendre transform, 8.4polynomial example, 8.5Levi-Civita tensor, see permutation tensormappings, 1.6metric tensor, 4.1Mohr’s circle, 6.5natural b.c., see traction b.c.Neumann b.c., see traction b.c.permutation tensor ε, 1.5plane stress & plane strain, 9.2general 2D relations, 9.4relations for linear isotropic plane strain, 9.3relations for linear isotropic plane stress, 9.4Poisson’s ratio (ν), 7.4principal strains, 6.1principal stresses, 6.5principle of minimum potential energy, seevariational methods




principle of stationary action, 17.1rotation tensors, see special orthogonal groupset theory, 1.2setsof complex numbers, Z, 1.2of integers, Z, 1.2of real numbers, R, 1.2of vectors Rn , 1.2shear modulus µ, 7.5special orthogonal group, SO(3), 2.4strain energy density, 8.1strain tensor εlimitations, 6.2strain tensor, εdefinition, 5.3volumetric-deviatoric decomposition, 5.4stress deviator, see deviatoric stress tensor, Sstress intensity factor, Kdefinition, 13.4for various loading configurations, 14.1relation to energy release rate G , 15.2

stress tensor, σdefinition, 6.4subset, ⊂, 1.2surface traction vector field f , 6.2tensor invariantsfirst invariant (trace), 2.3third invariant (trace), 2.3tensors, 1.6thick-walled cylinders, 11.2thick-walled discs, 11.2thin wall approximation, 12.1variational methods, 8.5Voigt notation, 7.2, 8.1weak form, 9.1yield criteriaJ2, 22.5Tresca, 23.1von Mises, see J2Young’s modulus (E ), 7.4



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