Elastic Waves HW1. Solution

There are several ways to get this form for U from the various equations given in thelecture notes. I start with the strain energy in the form U = (1/2) [ε]: [[c]]: [ε]

U =1

2ε ij[λδ i jδ kl + µ(δ i lδ j k +δ ikδ j l )]ε kl =

1

2λε iiε kk +

1

22µε ikε ki

We then recognize this as ( because TrΜ= Μii )

U =12λ{Tr(ε)}2 + 1

22µTr{[ε]2}

This is not yet in the desired form. We now use ( definition of the deviatoric strain ε')

[ε] = (1 / 3)[ I ](Trε)+ [ε ']And construct

[ε]2 = (1 / 9)[ I ](Trε)2 + [ε ']2 + (2 / 3)[ε ']Tr(ε)whose trace is

Tr[ε]2 = (1 / 3)(Trε)2 +Tr[ε ']2 + (2 / 3)[Trε ']Tr(ε)the last term of which vanishes (because Trε' = 0 )

Thus

2U = λ{Tr(ε)}2 + 2µ[Tr{[ε]2} = (λ + 2µ / 3){Tr(ε)}2 + (2µ){Trε '2}

This is of the desired form. A = λ+2µ/3 ( which is bulk modulus κ) and B is 2µ.

We now argue that both A and B have to be nonnegative if U is to be alwaysnonnegative:

=> Clearly A and B ≥ 0 is sufficient for U ≥ 0. Because they multiply positivequantities.

=> Furthermore, if A is negative, one could construct a strain such that U isnegative, regardless of B, by considering a strain that has no deviatoric part, that is purelydilational. Similarly, if B is negative one could construct a strain such that is purelydeviatoric such that the value of A is irrelevant. Thus A and B nonnegative is necessaryalso.

We conclude that thermodynamic stability requires (and only requires) κ and µ to benonnegative. But λ could be negative! Recall λ = κ−2µ/3 so for a material with a smallbulk modulus and large shear modulus, λ could be negative ( there are few such

materials, they tend not to happen naturally) Poisson ratio, we recall or look up in thetable, is ν = (3κ−2µ)/(6κ+2µ) This can be no smaller than -1 ( in the limit of vanishingbulk modulus ) nor greater than 1/2 ( for materials with bulk modulus much greater thanshear modulus.)

An arbitrary linear material must have a strain/temperature change relation of the formε ij =βij ΔΤ where βij is some 2nd rank tensor characteristic of the material. If thematerial is isotropic, that [β] must be (α/3) δij where α is the volume change pertemperature. If we assume the strain due to temperature change and the strain due toapplied stress merely add, and the total strain is zero, then the strain due to the stress mustbe equal and opposite to that due to ΔΤ. [ εelastic]=-(αΔΤ/3)[ Ι ] and the correspondingstress is

σ ij = cijkl (−αΔT / 3)δkl = −(3λ + 2µ)(αΔT / 3)δij = −καΔT δijThe stress needed to keep the volume from increasing when the material is heated iscompressive and purely pressure-like.

The Christoffel tensor associated with direction n is

Γik = cijklnlnj = (λ +µ)nink +µδik +νδikn(i )2

The parenthesis on the index indicates that the summation convention does not applyhere. We set nx = cosθ and ny = sinθ where θ describes propagation direction, and find

[Γ] =(λ + µ +ν )cos2θ + µ (λ + µ )cosθ sinθ

(λ + µ)cosθ sinθ (λ + µ +ν )sin2θ + µ

We confirm that , if ν=0 corresponding to an isotropic medium, this has eigenvalues µand λ+2µ… and we recover the two plane wavespeeds for an isotropic medium.

For the specified medium, the Christoffel tensor becomes

[Γ] =3cos2θ +1 2 cosθ sinθ

2cosθ sinθ 3sin2θ +1

The eigenvalues λ ( essentially the squares of the wavespeeds) are the roots of

(1− λ )2 + 3(1− λ) + 5cos2 sin2 = 0or

λ =5 ± 9 − 20 cos2 sin2

2The slownesses ( the inverses of the wave speeds, i.e λ-1/2 , are plotted in a polar plotversus θ.

At 45 degrees, Γ is [ 5/2 1; 1 5/2] The λ are 7/2 and 3/2 The former is faster and iscalled quasi-longitudinal; the latter is the slower and is called quasi shear. Thecorresponding polarizations are the eigenvectors ; these are { 1 1} and {1 -1}respectively. We see that these polarizations are precisely in , and orthogonal torespectively, the 45 degree propagation direction {1 1}. Hence these waves at 45 degreesare purely longitudinal and transverse.

At a more generic direction of propagation, the eigenvectors are neither of themprecisely parallel to or perpendicular to the propagation direction n.

Polar plot of slownesses (don't confuse the slowness curves with the circles that merelymark the values of 1/2 and 1). The inner curve is the quasi longitudinal; the outer curveis the quasi shear.

If all fields are varying in time like exp(iωt) such that ε=Εexp(iωt) and σ=Sexp(iωt) then the above differential relation between stress and strain becomesiωE=(1/G)(iωS+S/Δ), i.e S = GE/[1+1/iωΔ]. This takes the place of the usual stressstrain relation in shear: S = GE. It is as if the shear modulus is now G/[1+1/iωΔ] insteadof G. In the units such that G and ρ and Δ = 1, the wave speed, which was [G/ρ]1/2 =1, isnow [ 1+1/iω ]-1/2. So that wavenumber k = ω/c = ω[1+1/iω]1/2=[ω2 - iω]1/2.

Or, k(ω ) = ω 4 +ω 2 +ω 2

2− i ω 4 +ω 2 −ω 2

2

The real and imaginary parts of this are plotted here.

A plot of the dispersion curve k versus ω. The real part of k(ω) (the upper curve) and–Imk(ω) (the lower curve.)