Lecture 2

Hooke's Law relating stress and strain

We seek a linear proportionality between stress and strain in the form stress = Modulus * strain. But a simple proportionality [ σ ] = A [ ε ] is not consistent with observed material behavior. We know materials can have very different stiffnesses against dilation and against shear. ( Rubber for example is very compliant against being twisted, but resistant against being crushed; water has essentially no shear stiffness ) .

The most general linear relation between two 3x3 tensors takes the form

[ σ ] = [[c]] [ ε ] where [[c]] is a fourth rank tensor – with 4 indices and 81 components. In indicial form we'd write this as

σ ij = ∑ cij kl εkl = cij klεkl = cij kluk,l k l where the sum has 9 terms in it, as k and l both run from 1 to 3 and the summation convention that repeated indices are summed upon is invoked. The last equality above has used the symmetry of c in its last two indices to tolerate replacing εkl with uk, l.

[[c]] has 81 components – but they are not all independent. Inasmuch as σ and ε are symmetric with 6 independent components each, we need only 36 coefficients [[c]].

Furthermore, there are reasons (given on p 24) that tell us cij kl = ckl ij . This reduces the number of independent components of [[c]] to 21.

We can further simplify [[c]] by assuming that our material is isotropic. Crystals are not isotropic, wood is not, fiber-reinforced composite materials are not, but for most materials, especially glass and to a good degree polycrystalline materials, we can approximate them as isotropic. What this means is that we can write

[ σ ] = -p [ I ] + [ σ ' ] = α e[ I ] + β[ε'] corresponding to separate moduli α and β relating the dilation and the deviatoric part of the strain [ ε ] separately to the corresponding pressure-like and shear like parts of the stress [ σ ]

This form says that pressure cannot cause a shear strain, only a volume change. Also that a shear stress cannot cause a volume change. This may sound very reasonable, and it is for many material. But it is not hard to imagine materials which do not behave like that. Consider a bundle of parallel wires embedded in a sponge with spaces between the wires. Apply a uniform pressure to the object - - then it will contact laterally while not contracting

19 axially. The difference in axial and lateral strains means that there is some shear strain along diagonals. … even when the object is subjected to a pure pressure.

We can interpret the quantities α and β:

α is clearly the proportionality between (negative) pressure and average stretch, i.e between

(negative) pressure and one third of (negative) condensation –s where –s = δV/Vo = Trace[ε] = 3e. But bulk modulus κ is defined as the ratio of pressure to fraction volume diminishment -δV/Vo. We therefore identity α = 3 Bulk modulus = 3κ. An easy way to get this result is just to take the trace of [ σ ] = -p [ I ] + [ σ ' ] = α e[ I ] + β[ε'] ; this gives -3p =3αe=α(δV /V).

To identify β with shear modulus G is more tricky: recall our picture which defined shear modulus

as G = ratio of stress to strain, where stress = V/A and strain is δy/Lo . But we must be careful. The above picture may be made to look like the picture we had for γ (see copy of it here to the right ) by adding in a rotation by γ.

So Yes, we may define shear strain as δy/Lo like the picture above left, but this is not the same as γ in our tensor [ ε ] from page 20. In fact, δy/Lo = 2 γ.

Thus V/A = shear stress = G (δy/Lo) = 2 G γ.

We compare that to the proportionality in our Hooke's law. For the purpose of making the identification between β and G we consider the special case 0 γ 0  0 γ 0  0 2Gγ 0     [ '] 0 0 [ '] 0 0   ε = γ  which implies σ = βγ  but this must be 2Gγ 0 0  0 0 0  0 0 0  0 0 0  We therefore see that β may be identified with 2 G.

Summarizing: We posit a Hooke's law relating strain and stress for isotropic materials [σ ] = 3κ e[I]+ 2G[ε '] where κ is bulk modulus relating pressure and condensation, and G is the shear modulus defined on page 6.

20 We often find it convenient to rewrite Hooke's law in other forms. For example (using ε'=ε- eI and e = Tr(ε)/3 ) (3κ − 2G) [σ ] = (3κ − 2G)e[I]+ 2G[ε] = Tr(ε) [I]+ 2G [ε] 3

(The quantity (3κ−2G)/3 is the Lamé modulus λ) It is possible to invert this to solve for the strain given the stress: 1 [ε] = [ 9κ [σ ] − (3κ − 2G)(Tr[σ ]) [I] ] 18Gκ --- The above relations between stress and strain are given in terms of the bulk and shear moduli κ and G. A while back we also introduced the Young's modulus E that related stretch and longitudinal stress. E can be written in terms of κ and G:

Consider a bar with longitudinal stress σxx =σ and no transverse stress: σ yy =σzz=0. We can find the associated strains by the above formula for [ε] as a function of [σ]. The xx element tells us 18Gκ εxx = 9κσ−(3κ−2G)σ; so σxx = 18Gκ/(6κ+2G) εxx. Thus E, which was defined as the ratio of axial stress to axial strain, is E = 18Gκ/(6κ+2G). (Agreeing with the table, see below) . The y elements of the formula tell us the transverse strain from

18Gκ εyy = −(3κ−2G)σ; so εyy =εzz= -[(3κ−2G)/(18Gκ)] σ = -[(3κ−2G)/(18Gκ)] Εεxx =

-[(3κ−2G)/(6κ+2G)]εxx . Thus the Poisson ratio, the ratio between lateral contraction and longitudinal stretch, is ν = (3κ−2G)/(6κ+2G). This also agrees with the table below.

Let us use these ideas about stress and strain tensors and Hooke's law to do some force balance; let us see what they will do for F = ma, to see how the body may move and deform under the influence of the stresses. This will give us an equation of motion for the solid continuum.

Consider an arbitrary volume of material (not infinitesimal anymore) with body forces on it and surface forces acting on it. We write that the sum of all forces on this volume, both body forces f dV and surface forces [σ]{dA}, is the integral over the interior of the mass times acceleration r ∫ f (r,t)dV + ∫ [σ(r,t)]⋅ nˆ dA = ∫ ρ(r )u&r&(r,t)dV V A V r r ( we are now using r instead of X as the independent spatial variable )

21 A brief digression on the divergence theorem… which in its usual form is written like this for any sufficiently smooth vector function of position b(x). r r r b ⋅ nˆ dA = ∇ ⋅b dV Σ b nˆ dA = Σ (∂b / ∂x )dV ∫ ∫ or in indicial form: ∫ i i i ∫ i i i A V A V which relates the integral of a derivative (over some volume) to the values at the surface. It is the multi- dimensional version of the fundamental theorem of calculus that the integral of the derivative of a function is related to the function itself evaluated at the limits of the integral.

We can apply the divergence theorem to the above surface integral of stress r r r ∫ [σ(r,t)]⋅ nˆ dA = ∫ ∇ ⋅[σ(r,t)] dV A V The left side of this is a vector sum of all the surface forces. The right side is a volume integral of the divergence of the stress tensor. Its 'j ' component is ∂ r σ ji nˆidA = σ ij (r,t) dV ∫ ∫ ∂x A V i You have already met the concept of the divergence of a vector field (which is a scalar); Here we meet the divergence of a tensor (which is a vector.)

Applying the divergence theorem to our force-balance equation we get: r r ∫ ρ(r )u&r&(r,t)dV = ∫ f (r,t)dV + ∫ ∇ ⋅[σ(r,t)] dV V V V This applies regardless of the particular volume, so it must apply at every point. We conclude with the local vector expression: r r ρ(r )u&r&(r,t) = f (r,t)+ ∇ ⋅[σ (r,t)]

The net force density acting on the material at a point is the sum of the body force that may be acting there plus the divergence of the stress. The term in the divergence of the stress may be thought of as a tensor generalization of the gradient of pressure.

The next step in obtaining a dynamical equation of motion is to substitute for the stress using

[σ ] = 3κe[I]+ 2G[ε '] and then to substitute for the strain using 2εij = ∂ui/∂xj + ∂uj/∂xi . After a bit of manipulation in which we assume κ and G are constants we derive The Navier Equation for the vector displacement field u. r r r ρ u&r&(r,t) = f (r,t)+ (κ + G / 3) ∇(∇ ⋅ ur) + G ∇2ur

22 It may not look like it, but this is a wave equation ( with a source term f ) It resembles the more familiar wave equation in that we have two time derivatives of the field proportional to two spatial derivatives of the field. It is different from our more familiar wave equations in that, (like in E&M) the field variable is a now a vector, u, rather than a scalar ψ.

End review of Phys 326 material ======

I. Basic Equations of continuum linear elastodynamics; Elements of the Theory of elasticity; an accelerated (more grad level) exposition a) Kinematics: Our primary dependent field variable of interest will be a vector field of material displacement u(x,t) (of some material point with reference position x), assumed to be differentiable in space and time

We will most commonly discuss it using a Cartesian basis r r ˆ ˆ ˆ r u(x,t) = ux (x, y,z;t)i + uy (x, y,z;t) j + uz (x, y,z;t)k = ui (x;t)eˆi (summation convention in effect) but will occasionally find it useful to use cylindrical or spherical coordinate systems.

In the context of linear elasticity, it is usually unimportant to distinguish between x ( the reference position of some material point) and the position x + u ( the current position of that material point. ) We will usually neglect that distinction.

Associated with a displacement field u is the corresponding strain energy density U due to deformation. While there can be some energy associated with displacements themselves ( if for example there are springs attached to our solid for which U contains contributions like (1/2) k u2 ), for the continuum elasticity of major interest we assert that u itself is irrelevant ( rigid body displacements do not contribute to U ) What matters in U, to leading order, are displacement gradients r r Dij (x;t) ≡ ∂ui (x,t) / ∂x j In principle higher order derivatives (strain gradients) might also matter in U, but standard elasticity does not consider such.

D in turn may be decomposed into symmetric and skew symmetric parts r Dij (x;t) ≡ (1 / 2)[ui, j + u j,i ]+ (1 / 2)[ui, j − u j,i ] ≡ εij +ϖ ij The latter represents the degree of rotation of an element of the body at position x. It is not hard to see that the three independent components of the tensor ϖ are identifiable with the three components of the curl of u. We assert that our material is such that U has no contributions from such rotations; strain energy U can only depend on the strain ε.

23 The elements of the Cartesian tensor ε may be interpreted. In particular, εxx is the stretch in the x direction. Σi εii = Tr(ε) = dilation = fractional volume change. εxy represents a change of angle between lines originally in the x and y directions. Many elasticity approaches treat the six components of ε as the chief dependent variables of interest (rather than the three components of u.) In this case one often constrains the strain field to satisfy conditions of compatibility with the existence of a displacement field.

εij,kl +εkl,ij −εik, jl −ε jl,ik = 0

However, usually in elastodynamics we more often focus on u and do not need the above condition.

We now define a strain energy density U from which stresses and forces, and a governing PDE, may be obtained by the principle of least action, calculus of variations, and differentiation. ( However for some applications one must be a bit more pedestrian, of which more later…) For conventional linear elasticity it suffices to choose U to be quadratic in ε. ( constants in U have no affect on the equations of motion; terms linear in ε lead to u = 0 not being a solution; ie, they are ruled out by having taken u = 0 to be a static equilibrium. Terms cubic in ε are negligible for linear elasticity. (We also neglect the possibility that U depends on strain gradients uk,ij.) Thus

r 1 r r r U(x;t) = 2 cijkl (x)εij (x;t)εkl (x;t) for some fourth rank, possibly position-dependent, modulus tensor [c].

The symmetry of ε immediately implies that c (xr) must be symmetric in i and j and also ijkl in k and l. (antisymmetric parts will not contribute to U) The above expression for U also implies that c must be independent of an exchange between the pairs ij and kl. This leaves up to 21 independent components in [c]. ( Thermodynamics implies some inequalities; various moduli cannot be negative .)

It is not uncommon in engineering practice to describe the strain tensor as a 6-

component column of numbers ΕΙ, Ι = 1,2,...6 ( the fourth component of which E4 is

ε12 etc.) and the modulus tensor [ c ] as a symmetric 6x6 matrix. CIJ. From this perspective it is not hard to see why C in general has 21 independent elements. We won't be using the 6x6 modulus tensor in this course (it is mostly useful for anisotropic media) but you can find some discussion of it in wikipedia under linear elasticity.

In the case of an isotropic medium, [[c]] must be composed of tensors that can be defined independently of any special directions. ( of which there is only the Kronecker delta δij and the third rank Levi-Civita pseudo tensor eijk. ) The only such construction consistent with the specified symmetries in [[ c ]] is of the form (for some scalars λ and µ.) c (xr) = λ(xr)δ δ + µ(xr){δ δ +δ δ ) ijkl ij kl ik jl il jk

24 λ and µ are the so-called "Lamé Moduli." A bit of work establishes that µ is the previously defined shear modulus (many use the symbol G) and that λ is related to µ and bulk modulus κ by λ= (3κ-2µ)/3. This implies that stress is ( compare p 21)

σ = c (xr)ε = λ(xr)δ Trε + 2µε (xr) ij ijkl kl ij ij In an anisotropic medium (e.g. composite materials, single crystals, polycrystals with texture . . ) c is more complicated.

At this point a plethora of various moduli have been defined, all related to one and other. In an isotropic medium only two are independent; all others may be derived in terms of the chosen two. However various authors make different choices for their starting two. Relations between them are tedious to derive; it is therefore useful to use the table ( from Pao and Mow p 45 ) see also http://www.efunda.com/formulae/solid_mechanics/mat_mechanics/calc_elastic_constants.cfm

25 Governing Partial Differential Equation of Motion The elastic wave equation

We may now construct a Lagrangian density for our generalized coordinate u(x,t)

L = T −U = 1 ρ(xr){∂u / ∂t}{∂u / ∂t} − 1 c (xr)u (xr;t)u (xr;t) 2 i i 2 ijkl i, j k,l The action is an integral over time and over the volume of interest

A[u] dV dt ≡ ∫ ∫ V L The corresponding Euler-Lagrange Equation follows from δA = 0. After applying the divergence theorem and integrating by parts one recovers a volumetric and a surface contribution

δA[u] = ∫{∫ V [−ρ u&&i δui + (cijkluk, l ), i δu j ] dV }dt − ∫ {∫ n [c u ] δu ] dS } dt = 0 S i ijkl k,l j (n being the outward surface normal) The elastodynamic wave equation follows from the vanishing of δA for any virtual displacement δu in the interior. Therefore:

ρ u&&i = (cijkluk, l ), j in the interior. The elastodynamic wave equation consists of three coupled second order linear Partial Differential Equations.

The surface contribution to δA implies that

ni [cijkl uk,l ] δu j must vanish at every point on the surface. If a point on the surface is not free to move, i.e if we have imposed the geometric condition of rigidity, then δu = 0 there and the surface contribution to δA tells us nothing. If however, the surface points are unconstrained, then δA = 0 gives the natural boundary condition: that surface tractions t vanish

t j = niσ ij = ni[cijkluk,l ]

Other boundary conditions are sometimes discussed. We could have specified the surface to be free to move tangentially, but not normally ( e.g. for a well lubricated surface sliding along a rigid wall) Then the tangential components of traction t are zero while the normal component of u is zero. Alternatively we could rigidly constrain tangentially and leave the material free to move normally. Periodic boundary conditions, for which u is constrained to be the same on the left and right side, and upper and lower sides, of a rectangular domain, ( u(x=0,y) = u(x=Lx,y); u(x,y=0) = u(x,y=Ly) are often convenient for numerical simulations. It is left as an exercise to establish that such boundary conditions are such that the surface

26 contribution to δA is zero.

If [[c]] and ρ are independent of position (i.e. if the material is homogeneous) the elastodynamic wave equation becomes

ρ u&&i = cijkl uk, lj

In the special case of a homogeneous and isotropic medium ( to which we will confine most of our attention in this course ) the wave equation reduces to

ρ u&&i = λδijδkluk, lj + µ{δikδ jl +δilδ jk ) uk, lj

= λuk,ki + µui, jj + µuk,ki or, r r ρ u&r& = (λ + µ)∇(∇ ⋅ ur)+ µ∇2ur

Again, three coupled PDE's, for the three components of u. ( This equation is more complicated than it may appear. Except in Cartesian coordinates, the gradient of a vector is not particularly simply expressed in terms of the derivatives of the components of the vector. To put it differently, the unit vectors vary in space too so be careful when you take spatial derivatives of vectors. When and if we apply this to cylindrical coordinates, we will meet that complexity in more detail. ) r r r r Using the vector identity ∇ × (∇ × ur) = ∇(∇ ⋅ ur) − ∇2ur the PDE may be rewritten in the sometimes more useful form r r ρ u&r& = (λ + 2µ)∇(∇ ⋅ ur) − µ∇ × (∇ × ur) a) Plane wave solutions

These are pretty simple, and the ideas we draw from these solutions are useful, but they are by no means the whole story. Even without the complications of boundaries and heterogeneities, there are many solutions that are not of the plane wave form.

Let us seek solution of the form (independent of y and z, a vector displacement wave u traveling at some speed c in x-direction) ur = iˆf (x − ct)+ jˆf (x − ct)+ kˆf (x − ct) x y z On substitution into our vector PDE this becomes

2 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ρc [ifx "+ jfy "+ kfz "] = (λ + µ)ifx " + µ[ifx "+ jfy "+ kfz "]

27 It is apparent that the only solutions are

1) a Longitudinal Plane wave with fx being any differentiable function and

2 2 c = cL ≡ (λ + 2µ) / ρ; fy = fz = 0

( fy and fz could also be trivial linear functions )

This solution is termed longitudinal because displacement u is in the same direction as wave propagation. cL is typically of the order of 4 to 6 mm/ msec = 4 to 6 km/ second in most hard metals and glasses, a bit less in rocks and plastics. cL is often called the P wave speed cP or the dilatational speed cd or just simply c1. and

2) a Transverse wave, with fy and fz being any differentiable functions and

2 2 c = cT ≡ µ / ρ; fx = 0

( fx could also be a trivial linear function )

It is termed transverse because displacement u is perpendicular to propagation direction. cT is typically of the order of 2 to 3 mm/ msec = 2 to 3 km/ second in most hard metals and glasses, a bit less in rocks and plastics. cT is often called the shear wave speed cs or the equivoluminal speed ce or just simply c2.

While there are many solutions of the full equations that are not readily decomposed in terms of plane waves, we can expect that at distances large ( compared to a wavelength) from sources or scatterers, a field will look locally plane, so understanding propagation in terms of plane waves remains useful.

Modifications to Constitutive relation between stress and strain. Anisotropy and Viscoelasticity and their effects on above plane waves solutions.

It is worth taking a moment to examine the effect of anisotropy. (not in Graff, but see Achenbach) In general it makes things quite complicated, but it is not too difficult to see what anisotropy does for plane waves. Consider the full linear elastodynamic equation in a homogeneous medium

ρ u&&i = cijkl uk, lj and seek a plane wave with polarization vector Ui going in direction n with some to be determined speed c

28 u = U f (nˆ ⋅ xr − ct) i i Substituting into the PDE:

ρc2U f "(nˆ ⋅ xr − ct) = c U n n f "(nˆ ⋅ xr − ct) i ijkl k l j or 2 ρc Ui = Γik (nˆ)Uk (Γik ≡ cijklnln j )

(the "Christoffel equations") For any specified propagation direction n, one has a 3x3 eigenvalue problem, whose eigenvalues are the speeds of propagation and whose eigenvecctors are the corresponding polarizations U. The matrix [Γ] is symmetric (and positive definite though we haven't proved it here), so the polarizations are mutually perpendicular and the wave speeds are all real. In the isotropic case, these three eigensolutions are the P and the two S waves. In general the three speeds are all different and the polarizations are neither precisely transverse nor longitudinal. (That is, U is neither perpendicular to n nor parallel to it.)

With a bit of work one can derive speed c as a function of direction n, Here is such a plot for a 110 plane in crystal silicon. The slowness vector (n/c) versus n:

29 The fastest (ie the greatest slowness, the inner curve) has its polarization U close to the n direction (though you cannot tell that from the diagram) and is called 'quasi-longitudinal'. The others are called 'quasi transverse.' ======It is also worth mentioning the effect of viscoelasticity. (not in Graff, but see Achenbach) It is not unreasonable to posit a stress strain relation of the form (instead of [ σ ]=[[c]] [ ε ] ) t [ σ (t) ] = ∫ [[c(t-t')]] [ ε(t ') ] dt' −∞ such that the stress at a point depends on the entire past history of the strain at the same point. Alternatively but equivalently one might posit a differential equation relating stress and strain, e.g. Adε/dt + Bε= σ(t)

If [[c]] were to be a delta function in time [[c(t-t')]]= [[c]]δ(t-t') , one recovers the instantaneous relation [ σ (t) ] = [[c]] [ ε(t) ] . The PDE is now an integro-differential equation. ( c(t-t') vanishes if t

This is more complicated than previously, but it remains time-translation ainvariant, and so r r permits solutions of the harmonic form u (x,t) = U (x)exp(iωt) such that the integro- i i differential equation becomes

r 2 r r r ρ(x) ω Ui (x) = [c%ijkl (x;ω) Uk, l (x)], j where c% is the Fourier transform r r c%ijkl (x,ω) ≡ ∫ cijkl (x;t) exp(−iωt)dt

The effect of linear viscoelasticity is that the elastic moduli become complex and frequency dependent. If the medium is space-translation invariant, i.e if ρ and c have no x dependence, then we may also seek solutions ~exp(-ik.x) , i.e plane waves in k-direction . Substituting r U (xr) = U exp(−ik ⋅ xr) , we find i i 2 ρ ω Ui = c%ijkl (ω) kl k jUk,

Once again, we have a Christoffel equation for the polarizations U and complex slownesses k/ω. Mathematically this is much like it was in the purely elastic case; the chief difference now is that the frequency dependent and complex moduli [[c]] imply frequency dependent and complex wave numbers k.

The frequency dependence means that plane waves are dispersive even in homogeneous media. The complexity implies that waves will attenuate | exp(-ikx) | ~ exp( Imk x ) ======End Lecture 2

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