strain tensor

Solid MechanicsLecture Notes

Oxford, Michaelmas Term 2013

Prof. Alain Goriely

14th October, 2013

Contents

1 Introduction: one-dimensional elasticity 31.1 A one-dimensional theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.1.1 Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2.1 Conservation of mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2.2 Balance of linear momentum . . . . . . . . . . . . . . . . . . . . . . . 41.2.3 Constitutive laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 Kinematics 72.1 Bodies, configuration, deformation . . . . . . . . . . . . . . . . . . . . . . . . 72.2 Velocities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.3 A digression about tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.3.1 Just vectors and tensors . . . . . . . . . . . . . . . . . . . . . . . . . . 92.3.2 Tensor product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.3.3 Derivatives of tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2.4 The deformation gradient F . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.4.1 Transformation of volume . . . . . . . . . . . . . . . . . . . . . . . . . 112.4.2 Transformation of area: Nansons formula. . . . . . . . . . . . . . . . . 122.4.3 Transformation of line element. . . . . . . . . . . . . . . . . . . . . . . 122.4.4 Polar decomposition theorem. . . . . . . . . . . . . . . . . . . . . . . . 132.4.5 Strain tensors. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.4.6 The displacement gradient. . . . . . . . . . . . . . . . . . . . . . . . . 142.4.7 The velocity gradient tensor. . . . . . . . . . . . . . . . . . . . . . . . 14

2.5 Examples of deformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.5.1 Homogeneous deformation . . . . . . . . . . . . . . . . . . . . . . . . . 152.5.2 Inflation of a spherical shell . . . . . . . . . . . . . . . . . . . . . . . . 16

3 Conservation Laws, Stress and Dynamics 173.1 Mass conservation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173.2 Body and surface forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3.2.1 Body forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173.2.2 Contact forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

1

CONTENTS 2

3.3 Momenta and Eulers laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183.4 Eulers law of motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

3.4.1 Digression: the divergence theorem for tensors . . . . . . . . . . . . . 203.4.2 More on the stress tensor . . . . . . . . . . . . . . . . . . . . . . . . . 203.4.3 The nominal and Piola Kirchhoff stress tensors . . . . . . . . . . . . . 21

4 Constitutive equations 224.1 3 types of assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

4.1.1 Particular examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224.2 Elastic materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

4.2.1 Objectivity or frame indifference . . . . . . . . . . . . . . . . . . . . . 234.3 Hyperelastic materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

4.3.1 Energy balance, elastic materials . . . . . . . . . . . . . . . . . . . . . 254.3.2 Objectivity of W . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264.3.3 Material symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

4.4 Hyperelastic isotropic materials . . . . . . . . . . . . . . . . . . . . . . . . . . 274.5 Hyperelastic incompressible materials . . . . . . . . . . . . . . . . . . . . . . 294.6 Choice of strain-energy density for hyperelastic materials . . . . . . . . . . . 29

5 Examples of boundary value problems 305.1 Homogeneous deformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305.2 General method for semi-inverse problems, BVP . . . . . . . . . . . . . . . . 315.3 Inflation of a spherical shell. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

6 Linear Elasticity 366.1 Infinitesimal strain tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 366.2 Constant relationships . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 366.3 Isotropic linear elasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

6.3.1 Equations: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 386.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 386.5 Incompressible linear elasticity . . . . . . . . . . . . . . . . . . . . . . . . . . 40

6.5.1 General principles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 406.6 Plane/Strain/Stress Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

6.6.1 Plane solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416.6.2 Idea . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416.6.3 Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 426.6.4 Compatibility conditions . . . . . . . . . . . . . . . . . . . . . . . . . . 426.6.5 Application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

6.7 Elasto-dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 436.7.1 Planar waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

6.8 Rayleigh waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

HEALTH WARNING: The following lecture notes are meant as a rough guide to thelectures. They are not meant to replace the lectures. You should expect that some materialin these notes will not be covered in class and that extra material will be covered duringthe lectures (especially longer proofs, examples, and applications). Nevertheless, I will try tofollow the notation and the overall structure of the notes as much as possible.

1 INTRODUCTION: ONE-DIMENSIONAL ELASTICITY 3

1 Introduction: one-dimensional elasticity

Overview We explore elasticity in one dimension to give a general ideas of the different stepsnecessary to develop a general theory of elasticity.

1.1 A one-dimensional theory

Here, we consider a one-dimensional continuum that can only deform along its length. There-fore, there is no bending, twisting, or shearing, just stretching. The emphasis here is onunderstanding the different steps that enter in the development of a full theory of continuumin the simplest possible context. The steps are

1) Kinematics: A description of the possible deformations. The definition of strains,given by geometry. In oour context, it is just the stretch along the line.

2) Mechanics: The definitions of stresses and forces acting on the medium. Then astatement of balance laws based on the balance of linear and angular momenta, this isapplicable to all continuum media but for our problem, linear momentum is sufficient.

3) Constitutive laws: A statement of the relationship between stresses and strains. Thisis where we describe the response of the material under loads.

The results of these three steps is a closed set of equations whose solutions (with appropriateboundary conditions and initial data) is a description of the stresses and deformations in aparticular body under a particular set of forces.

1.1.1 Kinematics

Consider a 1D continuum of length L. Any material point is labelled by X [0, L]. Themotion or deformation is the mapping x = x(X, t), which is assumed smooth and invertibleas there is no material separation, discontinuity or overlap. The kinematics is fully describe

0 L X

by a the stretch and the velocity at one point.

=x

X, stretch; x = V (X, t) =

x

t, velocity.

Due to the assumption: > 0, and x = X corresponds to the stress-free (Langrangian)configuration.

Motion: The velocity of a material point is V (X, t) = x = x/t. Since X = X(x, t) isinvertible, we can write,

v(x, t) = x(X(x, t), t),

where v is the velocity at the spatial point x.The acceleration of a point is,

x(X, t) =d2x

dt2, or a =

dv

dt=v

t+ v

v

x,


where

d

dt=

t+ v

x,

is the material time derivative.

1.2 Dynamics

We use two fundamental principles to obtain equations for the motion of a continuum: theconservation of mass and the balance of linear momentum (in a general theory we will alsoneed the balance of angular momentum).

1.2.1 Conservation of mass

We define to be the linear density in the current configuration (mass per unit length asmeasured in the current configuration) and 0 the linear density in the reference configuration(as measured in the initial configuration) . Assuming no mass is created, we have X2

X1

0 dx =

x2x1

dx,

with x1 = x(X1, t), x2 = x(X2, t). Since dx = dX, we have X2X1

0 dX =

X2X1

dX,

which implies that = 0, the Lagrangian conservation of mass. This is the first conservationlaw.

1.2.2 Balance of linear momentum

The general principle for the Balance of linear momentum isd

dt(linear momentum) = force

acting on the system. Lets break this into the following pieces:

0 X1 X2 L

1) The linear momentum: X2X1

0x dX

2) forces: themselves due to external (body) forces or internal (contact) forces:

body forces, X2X1

0f dX

wheref is the density of body force (force per unit mass).


n(X1) n(X2)

contact forces: force the material exerts on itself.This material exerts a force n(X2) on [0, X2] counted positive (tensile) if theforce is in the direction of the axis, compressive otherwise. Therefore, from ac-tion=reaction, the contact force acting on the segment [X1, X2] is n(X2)n(X1).

Therefore, the Balance of linear momentum for a one-dimensional continuum implies

d

dt

X2X1

0x dX =

X2X1

0f dX + n(X2) n(X1)

We can obtain an expression with a single integral by moving the derivative inside the integraland using, X2

X1

n

XdX = n(X2) n(X1).

That is X2X1

(0x dX 0f n

X

)dX = 0.

This relation is valid X1, X2, so that, we can localise the integral (assuming continuity ofthe integrand) to obtain

0a = 0f +n

X.

This is an equation for the force n(X) in the material (Cauchy first equation). This equationis in the reference configuration (all quantities depend on the material variable X and timet). We can obtain an equation in the current configuration by using dX = 1 dx

a = f +n

x.

But what is n/x? We need a constitutive law to close the system.

1.2.3 Constitutive laws

To close the problem, we need to relate the stresses to the strains, that is a relationshipbetween and such as Hookes law

= K( 1). (1)This Hookean law is only typically valid for small deformations. For large deformations, wewill assume in general that the material is hyperelastic, that is the constitutive law derivesfrom a potential capturing the elastic energy associated with deformation so that

n = f() =

. (2)


with the requirement that f(1) = 0 and that the derivative of f at = 1 exists. For suchsystems, the Hooke constant K = f (1) is then simply the linearised behaviour for smalldeformations around the stress-free state. The theory of three-dimensional elasticity devel-oped in next Chapter when applied to the uniaxial extension of an incompressible rectangularneo-Hookean bar suggests the following nonlinear law

n = K/3(2 1), (3)

Close to = 1, we recover Hookes law (as shown in Fig 1). More generally, materials

0.5 1.0 1.5 2.0 2.5 3.0

-10

- 5

5

n

Figure 1: Comparison between the linear (dash) and nonlinear (solid) Hookean response forK = 3.

that show strain-stiffening (increase in stiffness for large deformations) or strain-softening(decrease in stiffness) can be modelled by various functions of the stretch.

2 KINEMATICS 7

2 Kinematics

Overview We develop a completely general theory for the deformation of three-dimensional bodieswith no assumptions on displacements. To do so, we introduce two configurations and relate them

through the motion in time and the deformation gradient. The deformation gradient is naturally

defines as a two-point tensors and its analysis require some tensor calculus.

2.1 Bodies, configuration, deformation

A body, B: set whose elements can be put into 1-1 correspondence with points in a regionB E3. we define material points as the elements of B. Since the body moves or deformsit can change with time, t R. We denote by Bt the configuration of B at time t. (Inparticular if we look at static systems, we will use B0 for the initial configuration and B forthe current one.) Possible terminology for B0 include initial/reference/material/Lagrangianconfiguration (here we will use the terminology reference configuration). Similarly for Bt,you will find the name current/actual/Eulerian/instantaneous configuration (we will use theterminology current configuration).

p B0

O

X

E1E2

E3

Bt

O

x

e1e2

e3

P

Since both B0, Bt are bijections of B = x such that

x = (X, t), X B0 and X = 1(x, t), x Bt,

where is the deformation of B from B0 to Bt (also sometimes called motion.) This con-struction can be summarised as follows:Continuum assumption. We consider a body with reference configuration B0 R3. At

pP

BtB0

2 KINEMATICS 8

time t, the body occupies the current configuration Bt R3. A material point, initially atX B0 is mapped to a point x Bt by the one-parameter mapping x = (X, t) so that : B0 Bt. The continuum assumption states that is a orientation-preserving bijectionmapping for all time t (except possibly at the boundary for contact-problem). This impliesthat we can write x = 1(X). We further assume that this mapping is twice continuouslydifferentiable in X and t. This last assumption can be relaxed in problems involving phaseboundaries (with possible jumps in the first derivative). In many instances and applications,we will assume that is actually smooth.

Example Rigid motion.

x = (X, t) = c(t) + Q(t).X

Here c is a vector and Q is a proper orthogonal second-order tensor. Using Cartesian coor-dinates Q is a rotation matrix (a proper orthogonal matrix, that is a member of the specialorthogonal group SO(3).)

x = xiei, X = XiEi,

where we use the summation convention and i takes the values 1,2,3. Now choose Ei = ei,then

xi = ci(t) +Qij(t)Xj ,

where c represents a translation and Q represents a rotation.Next, we wish to attach physical quantities to every point: mass (a scalar); velocity,

traction (vectors); strain/stress (tensors),. . .. These different quantities are defined on allpoints of the body and are therefore called fields. Elasticity is a theory of fields (similar toelectromagnetism, relativity, and fluid mechanics).

2.2 Velocities

Velocity of a point P :

V = x t(X, t), velocity

A = V 2

t2(X, t) acceleration

Now define a field (for instance, temperature or mass) at every point on B at time t,

(x, t) = ((X, t), t) = (X, t).

This change of variable allows us to define this scalar field on the reference configuration.We are interested in the rate of change of at a given material point P (fixed X),

=

t(X, t),

but

t(X, t) =

t(x, t) +

xit

xi(x, t)

=

t(x, t) +

xit x(x, t) = D

Dt,

2 KINEMATICS 9

where we have introduced D/Dt, the material time derivative,

D

Dt=

t+ v x,

and where (x)i xi . Similarly, by applying the same idea component by component, wehave for a vector field u(x, t) = U(X(x, t), t)

tU(X, t) U Du

Dt=u

t+ (v x)u.

In particular

A = V =v

t+ (v x)v.

Note that there is no possible confusion if all the derivatives are taken in the referenceconfiguration since X, t are truly independent. Expressing X as a function of (x, t) introducesthe extra convective derivative.

To make any progress in the description of deformation we need to define the relativechange of length in all possible direction of space. That is we need to define a quantity ofthe form

x

X= F,

which naturally defines F as a the gradient of a vector, that is a tensor. Therefore, beforeproceeding with a proper definition of the deformation gradient, we need to review some basicnotions of tensors and tensor calculus.

2.3 A digression about tensors

2.3.1 Just vectors and tensors

We consider a Euclidean vector space in 3D. We will first restrict our attention to the usualCartesion orthonormal basis {e1, e2, e3}, for which we have ei ej = ij . We define firstvectors

u = uiei u ei = uiand the space V as the vector space associated with all such vectors in E3. A tensor is thensimply a linear map from V V . That is S is a tensor = v = Su, another vector.

One can think of tensors as matrices. But one must be careful as tensors can convenientlybe written in a different non-Cartesian basis. So we really want to define operation on tensorsindependent of the basis and carry the basis as part of the definition of the tensor. To properlydefine tensors, we need the tensor product.

2.3.2 Tensor product

u v of two vectors u and v is a tensor such that(u v)a = (v a)u.

In components, u = uiei and v = viei so that,

u v = uiei vjej= uivjei ej .

2 KINEMATICS 10

Denoting the components of a tensor

T = Tijei ej Tij = ei Tej ,we have

(u v)ij = uivj .After one has defined the matrix [Tij ], all known linear algebra identities carry over to tensors.

det T = det([Tij ])

tr T = tr ([Tij ])

Tt = T Tij = TjiSimilarly ST is defined as (ST)v = S(Tv) and not surprising,

[ST ] = [S][T ]

Note: I use for inner product and not for contraction, contraction is assumed in the productof tensors unless otherwise specified.

2.3.3 Derivatives of tensors

We need to define basic operations on tensors. Let , u, T be scalar, vector and tensor fieldsrespectively over x,

= (x), u = ui(x)ei, T = Tij(x)ei ej .We define x or /x by,

grad =

x=

xiei,

grad u =u

x=

u

xj ej = u

=(uiei)

xj ej = ui

xjei ej .

We recognise ui/xj as the Jacobian matrix here.

grad T = T=

xk(Tijei ej) ek

=Tijxk

ei ej ek.

We can also contract tensors,

div T =Tijxk

ej (ei ej) .

This is the contraction on the first index of Ti with index k (Note: this is not a universalchoice and some authors contract w.r.t. the second index). Since ei ek = ik,

div T =Tijxi

ej , a vector.

So far, we have defined tensor in a single basis. However, the deformation gradient takes thederivative of a vector in one basis w.r.t. a vector in another basis, in which case we havemixed basis.

2 KINEMATICS 11

2.4 The deformation gradient F

To have a notion of local deformation, we define,

F(x, t) =x

X= Grad x = Grad (X, t).

Notice the upper case on Grad to denote the fact that we take derivatives w.r.t. X. Explicitly,we have

F =

xj(xiei)Ej = xi

Xjei Ej Fijei Ej ,

so the last tensor product is indeed mixed. Geometrically, F is a linear map that transforms

E2E1

E3

e2e1

e3V

v

F

BtB0

V

v

a vector v defined in the tangent space TpB0 at a point p in the reference configuration to avector in the tangent space TpB at the same material point p but in the current configuration.

2.4.1 Transformation of volume

dV dv

FE2

FE3

FE1

Now consider the image of an infinitesimal volume element of size d along three unitvectors E1,E2,E3. Each unit vector transforms as

FEk = Fij (ei Ej) Ek = Fij(Ek Ej)ei = Fijkjei = Fikei.

2 KINEMATICS 12

Note that the column k of F is the image of the kth basis vector as found in regular matrixalgebra.

The infinitesimal volume is

dv = (d)3 det(FE1,FE2,FE3) = (d)3 det F det(E1,E2,E3) = det FdV.

We define J = det F and we see that it has a natural description as the change of volume ata material point p.

2.4.2 Transformation of area: Nansons formula.

The following rule is for the corresponding change in a local area element defined at a pointp through the normal vector n (see Problem Sheets).

BtB0

N n

nda

surface elementNdA

nda = JFTNdA da = JFTdA

2.4.3 Transformation of line element.

We have assumed F is a bijection = det F > 0, so that F1 is well defined. Now take alocal infinitesimal line element in B0 at p.

pP

dX

material line

dx

same material line

= dx = FdX.Let M be a unit vector along dX:

dX = MdS = M|dX|dx = mds = m|dx|

= mds = FMds.Take the norm of each side:

|ds|2 = (FM FM)|dS|2 = (FTFM) M|dS|2

dsdS

=

(FTFM) M (M),

2 KINEMATICS 13

where ds/dS is the change of length of a material line in the direction M and is a stretch.This implies that

ds

dS= 1 s = S, M FTF = 1.

A material is unstrained if FTF = 1. However, the converse is not true. Therefore, we cannotuse F directly to measure the strain in a body.

2.4.4 Polar decomposition theorem.

The action of F is to rotate a vector along a direction M to a direction N , then stretch it toa size (M).

= stretch + rotation.

This is contained in the polar decomposition theorem:

Theorem Let F be a second-order Cartesian tensor such that det F > 0. Then uniquepositive definite symmetric tensors U, V and a unique proper orthogonal tensor R suchthat,

F = RU = VR.

Note

FTF = U2 C, right Cauchy Green tensorFFT = V2 B, left Cauchy Green tensor

Explicitly, we have

F = Fijei EjFT = Fk`Ek e`

FTF = (FkiEk ei) (Fijei Ej)=(FTF

)kj

Ek Ej = U2kjEk Ej , LagrangianFFT =

(FFT

)ij

ei ej = V2ijei ej , Eulerian

2.4.5 Strain tensors.

We need to define a notion of strain. A strain tensor is a tensor that is identically zero ifthere is no stretch in any direction of the body. For instance,

E =1

2

(FTF 1) , Green strain tensor

It is easy to check that for a global rotation, translation or rigid body motion we have

E = 0.

Only for stretching is E 6= 0 and depending on the direction of space, different stretches areobtained.

2 KINEMATICS 14

U

stretch unit ball

eigendirections of U

iui

then rotate the ellipsoid

rotate axesR

then stretch the ellipsoid

R

V

Other possible strain tensors can be constructed

E() =1

2(U 1) , 6= 0, R+

E(0) = ln U.

or

e() =1

2(V 1) , 6= 0, R+

e(0) = ln V.

These are, respectively, Lagrangian (upper case, they live in the reference configuration) andEulerian strain tensors (lower case, they live in the current configurations).

2.4.6 The displacement gradient.

Displacement is an important notion in linear elasticity. It is defined as

u = xX = x = X + u= F = Grad x = 1+ Grad u,

where Grad u is the displacement gradient.

2.4.7 The velocity gradient tensor.

Another important kinematic quantity is the velocity gradient tensor, defined as

L = grad v, Lij =vixj

, L = Lijei ej .

2 KINEMATICS 15

Since Grad u = (grad u)F by the chain rule,

Grad v = (grad v)F = LF,

but also,

Grad v = Grad x =

tGrad x =

F

t= F,

so that

F = LF

and (see Problem Sheets)

J = J div v

Another way to obtain this result is to use the formula for the derivative of a determinant.

tdet F = (det F) tr(F1F) = (det F) tr(L),

So that,

= J = J tr(L) = J div v.We define an isochoric deformation to be one such that there is no change of local volumeelement, that is, J 1, J = 0 = div v = 0 , (this is the incompressibility conditionsfound in fluids).

2.5 Examples of deformation

2.5.1 Homogeneous deformation

x = FX + x, F constant

Simple elongationF = U = 1U

(1) U(1) + 2(U(2) U(2) + U(3) U(3)

) Dilation

F = 1

Simple shear(0, 1) (, 1)

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

which imply (homework),

= F = 1 00 1 0

0 0 1

, U2 = 1 0 1 + 2 0

0 0 1

2 KINEMATICS 16

RB

A

a

b

2.5.2 Inflation of a spherical shell

Symmetric inflationDefine a function r = f(R) such that r(A) = a, r(B) = b, and

r = f(R), = , = .

Then

X = ReR, x = rer = f(R)er,

and

x

X=x

R eR + x

e + x

e

=(rer)

R eR + (rer)

e + (rer)

e

= rer eR + rR

(e e + e e) .

Therefore

F =

r 0 00 r/R 00 0 r/R

,and so

r = r, = r/R, = r/R.

Note if we only consider isochoric deformation then det F = 1, which implies

r( rR

)2= 1 = rr2 = R2 1

3

d(r3)

dR= R2 = r3 = R3 + C.

Since r(a) = A, r(b) = B,

C = b3 B3 = a3 A3 = a3 = b3 B3 +A3 = r =a3 A3 +R3

This is a one-parameter family of solutions.

3 CONSERVATION LAWS, STRESS AND DYNAMICS 17

3 Conservation Laws, Stress and Dynamics

Overview We use basic physical principles to derive local equations for the evolution of mass andstress in space. The requirement for localising the balance of linear momentum naturally leads to the

definition of the Cauchy stress tensor.

3.1 Mass conservation

We start with a scalar field , the mass density defined at each point of the body. Assumingthat mass is additive and m 0 as the volume of 0, the mass of a given referencevolume is

m() =

dv.

We postulate conservation of mass,

d

dtm() = 0 =

d

dt

dv.

Since changes during the deformation, the derivative cannot be taken directly inside theintegral. Instead, we map the integral back to the reference configuration. That is, we applya transport formula,

d

dt

0

J dV = 0 =

0

(J) dV = 0 =

(J+ J div v) dV = 0.

This implies the continuity equation,

+ div v = 0.

This is an example of transport and localization. An alternative way to derive this result isto define the reference density 0 and realise that

0 = J = 0 = 0 = J (+ div v) .

where we have used J = J div v.

3.2 Body and surface forces

Before proceeding with the Balance of linear and angular momenta we discuss the forces andtorque acting in a body.. Consider B and define F() a force and G(,0) a torque withrespect to a fixed point 0. Both force and torque act on . F is due to body forces (external)and contact forces (internal).

3.2.1 Body forces

Fbody =

b dv, b body force per unit mass

Gbody =

(x (b) + c) dv, c body torque per unit mass


3.2.2 Contact forces

Fcontact() =

t da

Gcontact() =

x t da

t

The action of the body outside on is given byt, and we use Cauchys stress principle that t dependssmoothly on n (for a non-polar medium).

We assume that there is no contact torque.Note t: traction vector force per unit area.

3.3 Momenta and Eulers laws

The linear momentum is defined as

M() =

v dv.

The angular momentum with respect to 0 is

H(; 0) =

x (v) dv.

3.4 Eulers law of motion

The laws of motion for a body simply state that the rate of change of linear momentum of anarbitrary material subset t Bt is equal to sum of all the forces acting on t. Similarly, therate of change of angular momentum of an arbitrary material subset t Bt with respect toa given point is equal to sum of all the torques acting on t with respect to the same point.

dM

dt= F,

dH

dt= G

Interestingly in classical discrete point mechanics dH/dt = G is a consequence of dM/dt = F.However, for continuum bodies, these appear as distinct laws.

Postulate: we further assume that we have non-polar media, that is the body is notsubject to body or contact torques which in turn implies that c = 0. This leaves us with thefollowing two equations,

d

dt

v dv =

b dv +

t da, ()d

dt

x v dv =

x b dv +

x t da.


Again we apply the idea of transport:

d

dt

v dv =

d

dt

0

vJ dV

=

0

d

dt(vJ) dV

=

0

(Jv + Jv + vJ div v) dV

=

v + v + vdiv v

=0, continuity

dV

=

v dV

Therefore () reads (a b)dv =

t da,x (a b)dv =

x t da.

Now a fundamental problem: if we want to localize this integral, we need to express

t da as

( ? ) dv

We could use the divergence theorem:

div v =

v n da,

for any closed R3, but then we need v n da not t da.This consideration naturally leads us to introduce a new fundamental object for the

description of forces in a body, the stress. Cauchys theorem states that if(a b)dv =

t da,

then a second-order tensor, the Cauchy stress tensor, such thatt = Ttn, Cauchys theorem.

We are now able to express all quantities appearing in the balance of linear momentum as asingle integral

(a b)dv =

t da =

Ttn da =

div T dv,

and we can localise this integral to obtain

((a b) div T) dv = 0,

= div T + b = aThe balance of angular momentum follows the same basic steps and leads to a restriction onthe Cauchy stress tensor

Tt = T

That is, T is a symmetric tensor.


3.4.1 Digression: the divergence theorem for tensors

Let v be a vector field. Then

div v dV =

v n dA

dV =

n dA.

(To see this, take v = c for c 6= 0, constant.) This implies

xidV =

ni dA.

Apply this to each component of a Cartesian tensor, Tjk`..., then

Tjk`...xi

dV =

Tjk`...ni dA,

In particular

div T dV =

TTn dA,

or in index notation,

Tijxi

dV =

Tijni dA,

3.4.2 More on the stress tensor

t

ntn

Consider again the stress tensor T

normal stress: tn = n (Tn) = n tIf tn > 0 then this stress is tensile, and if tn < 0 it is compressive.

shear stress: ts = |tTn|

Particular case Suppose ts = 0 and tn is independent of n. Then

t = pn, = p1.This is hydrostatic stress.

Tij is the jth component of the force in the ith direction.

T13

T11

T12

e1

e2e3

T11: normal stress; T12, T13: shear stresses.


3.4.3 The nominal and Piola Kirchhoff stress tensors

The Cauchy stress tensor is defined in the current configuration. We can ask the followingquestion, which tensor defined in the reference configuration gives the traction when appliedto an local area element? To answer this question we usese Nansons formula:

n da = JFtN dA

and apply it to the traction vector

= t da = Tn da = (JTFt)N dA,

where S is the nominal stress tensor,

S = JF1T

ST is the PiolaKirchhoff force per unit undeformed area.Since T is symmetric, T = Tt,

J1FTST = J1FS,

= FTST = FS

4 CONSTITUTIVE EQUATIONS 22

4 Constitutive equations

Overview We close the system of Cauchy equations for stress, density, and velocity by introducinga relationship between stresses and strains, the constitutive equations. Depending on the choice of

constitutive equations, the continuum equations can describe a fluid, a solid, or a gas.

So far, we have obtained through conservation of mass and balance of momenta, thefollowing three equations

+ div v = 0, mass

div T + b = v, linear momentum

Tt = T, angular momentum

There are 10 unknowns: 1 in , 3 in vector v and 6 in the symmetric tensor T. But thereare only 4 equations. We need 6 extra relationships to close this system. These will be givenby the constitutive equations.

4.1 3 types of assumptions

1) Possible deformations.e.g. Only rigid motions are allowed (F = R, 3 parameters). = rigid body mechanics.e.g. Only isochoric motion = Incompressible material.

2) Constraining the stress tensore.g. T = T (F)e.g. T = p1

3) Relate stress to motione.g. pressure function of density, (for a gas).

4.1.1 Particular examples

1) Ideal fluids

(a) det F = 1 (Isochoric)

(b) = const

(c) T = p1Note: the pressure is not determined by the motion (ball under uniform pressure).(Lagrange multiplier for the pressure.)

2) Elastic fluids

(a) T = p1(b) p = p()

Here = P (0) andp is the sound speed.

N.B.: both fluids are inviscid (do not exert shearing forces!)

A particular case of an elastic fluids is an ideal gas: p = , for > 0, > 1.

3) Newtonian fluids. Shear stress through friction.Take L = grad v which gives relative motion of particles, velocity gradient.


(a) det F = 1, incompressible

(b) T = p1+ C[L] where C is a linear function of L.Note C[0] = 0 = T = p1, A Newtonian fluid at rest is idealNote C[L] has 40 independent constants (once we have removed arbitrariness of p1.However objectivity (independence from observer) implies

C[L] = 2D, D = 12

(L + LT ),

which has a single constant, viscosity . This implies

v = div T + b

div v = 0

T = p1+ 2DAfter some algebra,

v

t+ v grad v = v grad p+ b,

div v = 0,

which are the NavierStokes equations. (N.B. = / is the kinematic viscosity.)

Stokes flow : 1) steady, 2) neglect acceleration.

v = grad p bdiv v = 0.

N.B. for more general fluids, T = p1+N (L).

4.2 Elastic materials

For elastic materials, we have the simple relationship

T = Z(F)This implies that the stress in B at x depends on F and not on the history of the deformation(path-independent). Also, by the definition of the reference configuration (assuming that itis stress free), we have

Z(1) = 0.This relationship defines a Cauchy, elastic material.

4.2.1 Objectivity or frame indifference

Material properties are independent of the frame in which they are observed (or the observer).

x = (X, t), x = (X, t), t = t.

The two descriptions are related by

x = Q x + c,


e1e2

B0

xx

e3

e2 O

O

e3

e1

u

u

u

where Q = Q(t) is orthonormal and c = c(t).

F =x

X=x

x

x

X= QF.

u = y x, u = y x = Q(y x) = QuTherefore

u is objective if u = Qu .

If u is the traction vector or normal

t = Qt, nQn

but

t = Tn, t = Tn,

which implies

QTn = Tn = TQn.

This is true n, which implies TQ = QT,

T = QTQT


More generally a tensor T is objective if

T = QTQT

which implies that , u, T are objective if

= , u = Qu, T = QTQT .

Care! v, a are not objective (unless Q = c = 0), but (grad ) = Qgrad is objective.

4.3 Hyperelastic materials

4.3.1 Energy balance, elastic materials

dEdt

= P,

where P is the rate of working, the power of the forces.

E() = K() + S(),

where K is the kinetic energy,

K() =

1

2v v dv,

and S is the internal energy of the system

S() =

dv,

where the internal energy density for an isothermal process, i.e.assuming no dissipation.

P() =

b v dv +

t v da

dK()dt

P() = . . .many steps

=

tr(TD)dv, stress power.

(Hint: use TT = T, divergence theorem, equation of motion and continuity.)Therefore

d

dt

dv =

tr(TD)dv

d

dt

0

JW

dV

0

Jtr(TD)dv = 0

dW

dt= p Jtr(TD),

where p is the stress power.


Assume hyperelasticity that W = W(F) only.

Jtr(TD) = tr(SF)

and

d

dtW(F) = tr

(W

FF

),

where

W

F=

W

Fjiei ej ,

(W

F

)ij

=W

Fji.

We have

tr

[(W

F S

) F]

= 0.

This must be true for all motion.

= S = WF

where

Sij =W

Fji, T = J1F

W

F.

An hyperelastic material is one whose stress derives from a scalar energy density function.

4.3.2 Objectivity of W

W scalar = W = W.F gradient = F = QF.

= W(F) = W(QF) = W(F)

= W(QF) = W(F), rotations Q.

4.3.3 Material symmetry

F = FP

Now

T = T (F) = T (F) = T (FP) = T (F),

in general T 6= T .If T (FP) = T (F ), F, then P is a symmetry of the body.The set of all P for which this is true form a group, the symmetry group of the material.


B

F

B0B0

F

P

4.4 Hyperelastic isotropic materials

Isotropy is a particularly important material symmetry. If T (FQ) = T (F), Q, properorthonormal, then the material is isotropic and SO(3) is its material symmetry group.

Response of the material independent of its orientation (this is a pointwise notion!) there is no physical way to distinguish the orientation of a local element. For a hyperelasticmaterial (isotropic), we have

W(FP) = W(F), P, rotations.So we have

W(QF) = W(F), W(FP) = W(F).

Use P = RQT , then

W(QF) = W(PRTRUP) = W(F) = W(U) = W(PTUP), P,and

W(QF) = W(QFRQT ) = W(QVQT ) = W(V)

= W(QFQT ) = W(V), W(PVPT ) = W(U)

W is an isotropic function of V or U.See linear algebra: det(QMQT ) = det(M) and tr(QMQT ) = tr(Q).There exist only three functionally independent invariant functions of a 33 second-order

tensor. Let 1, 2, 3, be the eigenvalues of V, i.e.the principal stretches.

I1 = tr(V2) = 21 +

22 +

23, i > 0

I2 =1

2

[I21 tr(V4)

]= 21

22 +

22

23 +

23

21

I3 = det(V2) = 21

22

23 = J

2

I1, I2, I3 are symmetric functions of stretches. Of course 6 I14 + I22 I3 = 0, Cayleys

theorem.

= W(F) = W(I1, I2, I3) = W(1, 2, 3).


Contradiction, W(1) = 0 = W(3, 3, 1) = 0, W(1, 1, 1) = 0.Let us now express the stress tensors in terms of W and its derivatives.

S =W

F

Let W = W(V2) = W(B),

= S = W(B)F

= 2FW

B= JF1T

T = 2J1F1W

FF

T = J1FS

S = 2W

CFT , T = 2J1F1

W

CFT

S = 2J1FFTW

B= 2J1B

W

B

Now for W/B,

W = W(I1, I2, I3)

where I1 = tr(B), I2 =12(I

21 tr(B2)) and I3 = Det(B).

Some interesting results

trB

B= 1 =

I1B

det B

B= det B B1 =

I3B

I2C

=1

2.2 (trB)1 trB

2

B= (trB)1 2BtrB

B= I11B

= S = 2J1B[W

I11+

W

I2(I21B) + W

I3JB1

]=W01+W1B +W2B2,

where

W0 = 2WI3

, W1 = 2J1WI1

+ 2J1W

I2I1, W2 = 2J1W

I2

and Wi =Wi(I1, I2, I3).Note that we can also write

T =

tivi vi,since T is coaxial with V. (vi principal direction.) Then

ti = J1i

W

i, W = W(1, 2, 3), J = 123.

Similarly

S =

siu(1) v(1), si = W

i


4.5 Hyperelastic incompressible materials

In general assume that we want to restrict the set of possible deformations.

C(F) = 0

e.g. det F 1 = 0, incompressibility.We introduce a Lagrangian multiplier, p,

WW pC

Again by variational calculus,

tr

[(

F(W pC) S

)F

]= 0

= S = WF p C

F

e.g. C = det F 1, so that CF = det F F1 = JF1 and

S =W

F JF1 = T = J1FS = J1FW

F p1

4.6 Choice of strain-energy density for hyperelastic materials

Incompressible if f(J) = 0

neo-Hookean W = 2 (I1 3) + f(J)

MooneyRivlin W = c1(I1 3) + c2(I2 3) + f(J)

Varga W = 2(1 + 2 + 3 3) + f(J)

Ogden W =p=1

p(p1 +

p2 +

p3 ) + f(J)

Fung W = 12[e(I13) 1]+ f(J) where 3 < < 20

Gent W = 12 ln [1 (I1 3)] + f(J) where 0.4 < < 3

Possible choices for f(J): 1(I3 1), 2(J 1)2, 2 ln I3, 2 ln J , . . .

5 EXAMPLES OF BOUNDARY VALUE PROBLEMS 30

5 Examples of boundary value problems

Overview For a given strain-energy density function, we can write a full system of equationswhich can be solved for given boundary conditions. We give some simple solutions for homogeneous

and semi-inverse problems.

5.1 Homogeneous deformations

In the compressible case, of the deformation is homogeneous than F = const matrix =WF constant tensor. Then

S =W

Fand T = J1F

W

F

are both constant.

= Div S = 0, div T = 0.That is, the Cauchy equations identically satisfied.

3

Consider, as an example, the diagonal transformationT = diag(t11, t22, t33) and F = diag(1, 2, 3)

= tii = iJ

W

i, i = 1, 2, 3.

The solution is fully specified by the boundary conditions (Ericksens theorem).For instance, uniaxial extension is obtained by the choice t33 = N and t11 = t22 = 0,

= t11 = 123

W1 = 0, t22 =1

31W2 = 0, t33 =

1

12W3 = N

W =12

(I1 3) 22

(I3 1) = 12

(21 + 22 +

23 3)

22

(2122

23 1)

tii =iJ

W

i

t11 =1J

(11 212233

)= 0

t22 =2J

(12 221233

)= 0


1 = 2 (so that tii = 0 when i = 1).

1(1 2223) = 02(1 2123) = 0, J = 21 =

= 22 = 23 = 21 = 1 = 2 = 1, 3 = .

t33 =

(2

J 2

4

J

)= N

J =

2= 1 = N

= 3 1

=4 1

Now

N()

=1

=

(3+

1

2

)=1

= 4.

5.2 General method for semi-inverse problems, BVP

1) Describe your material

Elastic? Static? Any particular geometry (thin, long. . .)? Incompressible? Isotropic? Strain - Energy (Define or keep undefined as long as needed.)

2) Describe the deformation.

Semi-inverse method.

x = X

r = f (R)

Y

Xr = f (X)

= g(Y )

z = Z

Choose variables in B0, B suitable for you problem. Define , F, i.i is a function of a few parameters, functions, (f, a).


3) Define the boundary conditions. There are loads to choose from! T n on B? Or ?b?

4) Write down Cauchy equation plus constant.{div T + b = 0 (static),

T = J1FWF p1Incompressible is J = 1, compressible p = 0.

To solve write W = W(I1, I2, I3), then W = W(1, 2, 3), thenWF .

= T = T(f(x),x), or S = S(f(x),x).

5) Insert into div T + b = 0 or Div S + 0b = 0. Obtain differential equation for f(x) orp = p(x). Solve the equation with the correct boundary conditions.

5.3 Inflation of a spherical shell.

1) Elastic, incompressible, isotropic spherical shell with strain-energy W(I1, I2, I3).

2) Symmetric inflation

A 6 R 6 R, F = xX

(see 2.5)x = f(R)X, r = f(R)R.

F =

r

= r r/R

r/R

.r = r

(R), = r/R =

a = a/A, b = b/B, r =3a3 A3 +R3

where a is the single unknown parameter. Therefore

= = = r/R, r = 2

T n ={P on r = a0 on r = b


n

pn

ST N = JPFTN (mapping of traction vector){Tn = Pn on BST N = PJFTN, on B

= Trr ={P on r = a0 on r = b

or

Srr =

{P1r = P2 on R = A0 on R = B

Note that the boundary condition depends on the deformation.

3) b = 0 and div T = 0,

= dTrrdr

+2

r(Trr T) = 0,

or

dSrrdR

+2

R(Srr S) = 0.

Constitutive equations,

S =W

F pF1,

or

T = FW

F p1.

Then

Srr =W

r p1r , S =

W

p1

which are functions of (R).

4) Solve the equation. Take div S = 0.

Srr S = Wr W

.


We are going to choose as a variable. Define h() = W(2, , ), then

dSrrd

= 2Srr S 2 =

h()3 1 ,

= P = ab

h()3 1

trr

+2

r(tr t) = 0,

tr = rWr p = 2Wr p, t = W p, t = t.

tr t = 2Wr W

= trr

+2

r(2Wr W) = 0.

Introduce auxiliary function, h() = W(2, , ),

h() =h

= Wr.(23) +W.1 +W.1 = 21(2Wr W)

trr

=tr

r

=r

R(r),

r=

1

R rR

R2

R3 = r3 a3 +A3, RR2 = 3r3, R = r2

R2= 2.

r=

1

R(1 3)

= trr

=tr

1

R(1 3) = h

()r

.

tr

=h()1 3 , = tr =

a

h()1 3 d

At = b, tr = P ,

P = ba

h()1 3 d, = P =

ba

h()1 3 d = f(a).

Note

a = a/A, b =1

B3a3 A3 +B3 = 1

B3

(a 1)A3 +B3.


For a given P , we find a, hence the deformation and the value of tr at all points.

Note that W = 2 (2r +

2 +

2),

= h = 2

(1

4+ 22

).

Note the nonlinearity in the 1/4 term.

= h

1 3 = 2(2 + 5),

P = 2(

1

+

1

44

)b(a)a

.

compressive

PP < 0sucking

P > 0blowing

tr(b)

a

1

6 LINEAR ELASTICITY 36

6 Linear Elasticity

Overview We show how to obtain the equations of linear elasticity by linearising the generalnonlinear equations of elasticity for small displacements. We also consider the solution of simple

problems.

6.1 Infinitesimal strain tensor

The central object is not the mapping but the displacement.

u

x

u = xX = (X, t)X= u = Grad 1 = H = F 1,

the displacement gradient.Assumptions:

Displacement gradient is small.Now consider the strain tensor,

E =1

2

(FTF 1)

F = 1+ H = E = 12

((1+ H)

(1+ HT

) 1) = 12

(H + HT ) E

+O(H2)

E =1

2(H + HT ) =

1

2

(u + (u)T )6.2 Constant relationships

S = S(F), T = T (F).We assume S(1) = 0 (no residual stress).

= S = S(1+ H) = S(1)0

+DS(1)[H] C[H]

+O(H2),


where C is linear in H.

T = T (1) 0

+DT (1)[H] +O(H2)

Which one to use?

T = J1FS

= T = J1FSand

T = DT [H] = J1(1+ H)DS(1)[H] = det(1+ H)(1+ H)DS(1)[H] +O(H2)

= DT [H] = DS[H] = DS[H] = C[H]C elasticity tensor,

Tij = Cijk`Hk`

Major symmetries:

Cijk` = Ck`ij

Minor symmetries:

Cijk` = Cij`k = Cjik` = Cji`k

= from 81 components to 36 independent components.Note also

Tij = Cijk`

[1

2(H + HT )

E

+1

2(HHT )

]k`

Tij = Cijk`Ek`

constant relationship for linear elasticity.

6.3 Isotropic linear elasticity

If the material is isotropic:

Sij = Tij = 2eij + (tr E)ij

where

Cijk` = ijk` + 2ikj`

for and the Lame coefficients. From the symmetry of C and positive definiteness, we have

> 0, 2+ 3 > 0.

Note: C is positive definite means

E.C(E) > 0, E Sym.Material hyperelasticity C is positive definite= C is symmetric.

If the body is homogeneous, then 0, , are constant.


6.3.1 Equations:

u = ()(X)X.

S = C[E], e =1

2(u +uT )

Div S + b0 = 0u = 0a

Assume homogeneity and isotropy,

S = 2E + (tr E)1.

Div S = 2Div

(1

2(u +uT )

)+ Div

(tr

1

2(u +uT )1

)= u + Grad Div u + Grad Div u

= u + (+ )Grad Div u

Therefore we have the Navier equation,

u + (+ )Grad Div u + b0 = 0u

Note that u = u(X) implies that x does not appear any more (we can replace X by x if wewant I dont).

In components,

u = u(X, t) = uiEi

implies

02uit2

= bi + 3j=1

2uiX2j

+ (+ )2uj

XiXj

6.4 Examples

To understand the meaning of the elastic moduli, we consider simple deformations.

1) Pure shear, u = (X2, 0, 0) [pic]

[E] =1

2

0 0 0 00 0 0

, [] = 0 0 0 0

0 0 0

,= = = is the shear modulus.(Care! Normal stresses vanish... and remember the condition > 0!)

2) Uniform compression, u = X and u = xX = ( + 1)XXE = 1, = p1

We use

E =1

2

[

2+ 3(tr )1

]


11 = p

12 = p at

1 =1

2

[p1+

2+ 33p1

]=

1

2p

[(2+ 3) + 32+ 3

]1

= p2+ 3

= p = (2+ 3) = 3(

2+ 3

3

),

where is the Modulus of Compression. Remember the condition 2+ 3 > 0!

3) Uniaxial tension, = tE1 E1

11

[E] = diag(, , ), =t

E, = .

E =(2+ 3)

+ , =

2(+ )

Here E is equated to the infinitesimal Youngs modulus and is equated to Poissonsratio.

E =1

E((1 + ) (tr )1)

an alternative form for E.

Expect > 0 (Care! Its an auxetic material!)

Now

=2+ 3

3=

E

3(1 2) ,

so that as 1/2, , and we would need an infinite force to change the volume.Therefore incompressible materials have = 1/2.


6.5 Incompressible linear elasticity

Recall: Incompressibility:

det F = 1, = det(1+ H) = 1 + tr H +O(H2) = 1

Therefore tr H = 0 = Div u, and

Div u = 0 tr E = 0

Also

T = p1+ J1FWF

, = p1+ Cijk`ek`

For isotropic material,

= 2E + (tr E)1 p1

but

=E

2(1 + )=E

3.

Therefore

u = bGrad p+ u

and

=E

3

N.B. Boundary conditions

u = u(t), on 1B displacementsn = t(t), on 2B traction

6.5.1 General principles

1) Linear superposition

2) Stresses, strains and displacements are proportional to the loads (or displacements)applied to the solid.

3) If 2B = then there exists one unique solution, only displacements.4) If only traction and tractions are in equilibrium, then stresses and strains are unique.

For initial conditions, there exists one unique u(t).

Some nomenclature about loading

1) Plain strain

u = (u(X,Y ), v(X,Y ), 0) = e13 = e23 = e33 = 0, 13 = 23 = 31 = 32 = 0.


no displacement

no stretching

but stresses

2) Plane stress

13 = 23 = 33 = 0, =

ast ast 0ast ast 00 0 0

3) Antiplane strain

u = (0, 0, w(X,Y ))

4) Pure torsion

u = (Y Z,XZ,(X,Y ))

(see problem sheet 6)

6.6 Plane/Strain/Stress Solutions

6.6.1 Plane solutions

(stress or strains)In cartesian coordinates,

eij =1

2(ui,j + uj,i =

1 +

Eij

Ekkij

ijxi

+ bj = 0

Assume that b derives from a potential,

bi =V

Xi, i = 1, 2, b3 = 0.

Plane stresses or strain 13 = 23 = 0.

6.6.2 Idea

Let

11 =

X2 V, 22 =

X1 V, 12 =

2

X1X2, 33 = (11 + 22),

= 0 is plane stress and = 1 is plane strain.


6.6.3 Equations

11X1

+12X2

+ b1 = 0,12X1

+22X2

+ b2 = 0.

Therefore

X1

(2

X22V

)+

X2

(

2

X1X2

)+b1 = 0,

X1

(

2

X1X2

)+

X2

(2

X22V

)+b2 = 0,

and the equations of motion are satisfied. But we do not have an equation for . We haveequations for ij or eij , that is, 6 fields but ui is 3 components.

6.6.4 Compatibility conditions

Recall: conditions for F: Curl F = 0. For

eij =1

2

(uixj

+ujxi

)Compatibility conditions:

Curl Curl E = 0,

ipmjqn 2emn

XpXq= 0

2eij

XkX`+

2ek`XiXj

2ei`

XjXk

2ejkXiX`

= 0

These are 6 relations (but only 3 are independent). For planar problems: e13 = e23 = 0,eij/X3 = 0,

= 2e11X22

+2e22X21

2 2e12

X1X2= 0 (ast)

Now for plane stress we have 33 = 0 and from plane strain we have 33 = (11 + 22),

33 = (11 + 22),

which implies

e11 =1 +

E11

E(1 + )(11 + 22)

e22 =1 +

E22

E(1 + )(11 + 22)

e12 =1 +

E12


Insert these into (ast) and use 11 =2X21 V ,

= 4

x41+ 2

4

x2122

+4

x42=

1 21 22

(2V

x21+2V

x22

)

4 = CV , C = 1 2

1 22 .

Here 4 is the biharmonic operator and is the Airy potential. If = 0, we have planestress and = 1 is plane strain.

6.6.5 Application

6.7 Elasto-dynamics

u + (+ )Grad Div u = u (ast)

6.7.1 Planar waves

u(x, t) = a sin(k x ct)

Here a is the amplitude, k is the direction and c is the velocity. We normalize such that|k| = 1.2 interesting cases:

a k longitudinal primary, pressure, P-waves.a k transverse shear, secondary, S-waves.

Pwaves Swaves

direction of wave

Let (x, t) = k x ct.

Note that Div u = a k cosCurl u = a k cos

Div u = 0 is transverse, Curl u = 0 is longitudinal.Substitute u = a sin in (ast). Then

u = a sinGrad Div u = Grad(a k cos) = (a k)k( sin)

Therefore u = c2a sin and

a + (+ )(a k)k = c2a


This is a linear operator on a. Define A the acoustic tensor,

A =1

(1+ (+ )k k) [A]ij = 1

(ij + (+ )kikj)

so that we have the eigenvalue problem

Aa = c2a

1) a = k

Aijkj =1

(ki + (+ )kjkj

1

ki) = c2ki

= + 2

= c2, cL =

+ 2

2) a k, aiki = 0.

Aijaj =1

(ai + (+ )kikjaj) = c

2ai

= c2 = , cT =

strain tensor

Documents

general theory of elasticity

onedimensional theoryhere

onedimensional continuum

hyperelastic materials

general method

examples of deformation152

general ideas

general principles