1

Phys 598 EW Elastic Waves Fall 2015

Standard texts and monographs, all with a view towards solving classical problems in elasticwave propagation:

K. F. Graff, "Wave Motion in Elastic Solids," Dover , NY 1975 (inexpensive!)

J.D. Achenbach, "Wave Propagation in Elastic Solids, NorthHolland/ElsevierEmphasis on exact solutions for unbounded and simply bounded media

R. Truell, C. Elbaum and B. Chick, "Ultrasonic methods in solid state physics,"Emphasis on ultrasonics in unbounded media and half spaces.

J. Miklowitz "Theory of elastic waves and waveguides" 1978Waveguides: Plates and rods

Y-H Pao and C C Mow "Diffraction of Elastic Waves and Dynamic StressConcentrations" Rand Corp. 1973. Emphasis on scattering in simple geometries

K. Aki and P.G. Richards, "Quantitative Seismology,"High level; Attention confined to 3-d, with emphasis on applications inseismology.

================

Tentative course outline:

I. Graff 5.1.1 Review of linear elasticity - Stress, strain, traction, constitutive laws, Boundary Conditions Balance of Momentum - Displacement Equation of Motion - energy balance Lagrangian re-formulation

Helmholtz Decomposition of displacement field, displacement potentials5.1.2 Alternate forms for Equation of motion5.1.3 Plane Waves of P & S type Visco-Elastic Constitutive relations, Attenuation & Dispersion of Plane waves

II. Governing Equations (often approximate) in reduced dimensions

2-dPlane strain SH and P-SVPlane stress extensional waves in thin plates

SH waves in thin (or thick) platesBending waves in thin plates, Kirchoff plate theory.

1dTorsional waves in circular rodsBending waves in beams, Euler-Bernoulli theoryExtensional waves in rods

2

Stringsderivation of linearized equation

III Applications in reduced dimensions, Exercises in Fourier Transforms and Greensfunctions and scattering.

A. 1-d several mathematical exercises on (Graff Chapters1,2 and 3)scattering, dispersion and responses of strings.

B. Kirchoff plates Graff Chapter 4

Free Vibration

plane wavescircular crested wavesHarmonic waves & Dispersion relations

for Cartesianand Polar co-ordinates

The Infinite stiff ribbon, a Multi branched dispersion relation

Finite systems, normal modes

Forced Vibration; Greens functionsInfinite plates, Triple FT's, & Hankel TransformsInfinite ribbon

Finite systems

IV Elastodynamics in 3 dimensions 5 weeks

The Fundamental solution, the Green's Dyadic in unbounded 3-d media (Graff 5.2.3 ) => in k,ω Fourier Transform space => as a function of r and ω, the "harmonic G's Dyadic" -

Energy Flow for Harmonic Solution in the time domain, as a function of r & t

Strain Nuclei sources

Solutions in Half Spaces Free vibration: (Graff 6.1 )

P/SV and SH Reflections, Mode Conversions, Snell's law, Critical angles

Rayleigh Surface Waves

Other boundary conditions

3

Layered Half space - Love Waves

Forced problems6.2 SH sources in Half spaces

6.3 P/SV sources in half spaces -

Line sources and point sources; Lamb's Problem; Cagniard Method,

other sources in half spaces.

6.4 Free vibrations in Joined Half spaces, especially fluid/solid case

Reflections, Transmissions, Surface waves, Leaky Surface waves

V. Elastic waveguidesGraff 8. Rayleigh-Lamb Waves in a Plate

Pochammer waves in a rod

Responses in Waveguides: Green's functions in an isotropic plate:

Normal Mode Solution, Ray Theory Solution, Numerical Inversion

VI Statistics of elastic waves in irregular reverberant bodies.Multiple scattering in random media - average G, and effective maediumDiffuse field assumption and its justification.Consequences

Universal Partition, surface/bulk and P/S and H/V.Diffusion of multiply scattered wave energyEnhanced backscatterField correlations = Greens function.

============

Course will meet twice a week M W 11:00 – 12:20. Grade will be based on weekly(occasionally twice weekly?) HW assignments and class participation, and on a final exam.

=============I begin with a low-level review of linear elasticity…..

4

The mechanics of elastic continua ( recap of Phys 326 material)

One key concept is that of a tiny volume element, large comparedto any microscale like the atomic upon which there may be somestructure, but small compared to any macroscopic scales overwhich we will make measurements. [When and if this separation ofscales is lost, and one cannot identify such tiny volume elements,continuum mechanics has to be re-thought.] This volume dV hasno internal structure, by definition. It has a volume dV, and a shape ( which we can take tobe just about anything useful ) and a surface, with unit outward normal n that of course variesin direction over the surface.

It has a mass ρ dV.

( of course ρ can vary from place to place in the medium or in time, but only negligiblywithin the volume dV.)

The continuum hypothesis continues with a notion that the forces on this dV are of twokinds:

Volume or "body" forces that are proportional to the volume (for example weight)Surface contact forces due to the action of neighboring volume elements

that are proportional to surface area (like pressure)

The surface forces are conveniently categorized intoshear (sometimes called traction) and compressionor tension. Sometimes all the surface forces arecalled tractions.

In general, each volume element dV will be subjectto tensions and compressions and shears that vary over its whole surface. We will, in a bit,impose force balance on them ( e.g if it is in static equilibrium, the sum of all the forces mustbe zero )

One simple consequence of the continuum hypothesis is that pressure must be isotropic.Consider a material in which we assume that there are no shearforces. All surfaces transmit only normal forces. Perhaps thematerial is too slippery to tolerate them. This applies to manyfluids - and sometimes to solids approximately in that shearstrengths are often much less than compressive strengths so theshear forces may be relatively weak.

Consider a prismatic volume dV shaped like an isoscelestriangle.

It has forces on each side which, by hypothesis are a)proportional to the side's area S and b) in the direction n.

5

(because we assume no shear) The proportionality is described by the symbol -p (for pressure)

Thus the sum of the forces is

rF∑ = −p1S1n̂1 − p2S2n̂2 − p3S3n̂3 + ρ

rgV = mra

(I have included all body forces in the term in g and introduced the possibility that thecompressive and tensile forces on the three faces might be different. Also I have a minus sign onp so that positive p corresponds to an inward force on the prismatic volume.)

Now for the trick: Imagine doing this again for a triangle that is smaller in all lengths by a factorof λ. The pressure terms scale with the surface areas S ~ λ2; the other terms scale with thevolume ~ λ3. Thus in the limit of a very small volume dV, the pressure terms must by themselvesadd to zero ( except maybe for effects due to how p changes an infinitesimal amount across theinfinitesimal distances )

p1S1n̂1 + p2S2n̂2 + p3S3n̂3 = 0

This is a vector equation. All components, in particular the horizontal component, of this must vanish.Also we notice S1 = S2 and that the horizontal components of n1 and n2 are equal and opposite.Therefore the pressures on the two vertical faces must be equal: p1 = p2.

Because the orientation of our triangle was arbitrary we conclude that, in the absence ofshear forces, the surface forces must be isotropic. Pressure is isotropic, the same in alldirections. For example, you can't have tension in one direction and compression in another.. . that is unless you had permitted some shear forces too.(or if your volume dV was not infinitesimal.)

Introductory Concepts of Stress and Strain

Two critical concepts in continuum mechanics are those of stress and strain. They are moreproperly described as tensors, but let us introduce them first less abstractly.

Stress is force per area. Stress has units of Newtons/ square meter, or Pascal.

Pressure is one kind of stress. Others are tensile, e.g. a wire of cross sectional area A andforce F. Stress = F/A is in the picture below tensile

Alternatively we could consider a rod in compression

6

In which case we would write that the rod has stress –C/A and is in compression. Byconvention, tensile stresses are considered positive and compressive stresses are considerednegative. Positive pressure is by convention negative stress.------Stresses can also be shear, e.g.

and we would write that the block has shear stress V/A ( for this, sign conventions will awaitour tensor treatment) Don't be too bothered by how the above body is not in equilibrium againstrotations. There are presumably other forces not pictured above, for example maybe

Shear stress V/A is measured in the same units, N/m2 or Pascal.

Aside: In all three of the above cases I pictured forces that summed to zero. This isrequired if the object is in static equilibrium, OR if the object is infinitesimal(following our arguments on rescaling by factors of λ). If indeed we have amacroscopic object with different forces on each end, the body may be accelerating orthere may be body forces that balance these imbalanced surface forces, or both. Ineither case the stress will then vary across the object. On every infinitesimal volumeof it, however, the surface forces will be in balance.

====

Strain

Strain is a dimensionless measure of deformation. It is conceptually the ratio of extradisplacement to initial separation

Tensile strain in a bar is the ratio of stretch to original length

strain = δL/Lo

By convention, stretches are positive, compressions are negative. They are often quoted in%.

7

You might wish to use, not original length Lo to normalize δL, but maybe the new length Lo+ δL. Thereby defining strain as δL/(Lo+δL). The difference is negligible if the strain issmall.---------------

Shear strain is more subtle: Here is a block of length Lo with one end shifted sidewaysrelative to the other a distance δy

We define the strain as δy/Lo. Of course you could do almost the same thing by:

rotating the block. One must be careful not to confuse true strain with rotations. In truestrain, angles between lines will change. In rotations they won't. The distinction will bevisited again when we get around to describing strain as a tensor.

Hooke's Law.

It is plausible that the stresses and strains are related. Hooke's law says that the relation islinear

stress = Modulus * strain

with a proportionality Modulus that is characteristic of the material. Moduli have units ofPascal, or Newtons/m2. The linear relation should be expected to fail for large strains. E.g.after the material breaks? Most solids are fairly linear up to strains of 1%

Aside: The constitutive relation for stress could be much more complicated than stress =modulus times strain; one could imagine stress depending on the history of strain, or onstrain and temperature, or magnetic field.. or…

In elastic tension or compression one posits that

tensile stress = E * tensile strain

where E = Young's modulus of the material. (Definition of Young's modulus E )

For the special case of a long bar under tension we'd write

8

F/A = stress = E * strain = E (δL/L)

OR

F = ( EA/L) δL

in which we recognize EA/L as a spring stiffness "k" relating force to extension.------------------In shear one posits

shear stress = G * shear strain

or

V/A = G (δy/Lo)

where G is the shear modulus of the material.

Hooke's Law will allow us to determine the forces needed to impose a specified deformation,or alternatively allows us to determine the deformation consequent to a specified set offorces.

Now let us revisit the concepts of stress, strain, and Hooke's law, but do it moreproperly, with tensors.

Strain is a tensor.

How do we wish to describe deformations of acontinuous medium? This is a purely kinematicquestion, essentially geometry.

We start with the notion that a material point,originally at some vector position X, gets movedby a displacement u(X) to a new position X +u(X).

Here is a sketch of how an arbitrarilyshaped region has been moved and rotated andstretched to a new position by some functionu(X).

9

If u were a constant vector, (independent of X) this would describe how the whole body gotshifted rigidly, with no change of shape and no rotation.

"Strain" is related to relative displacement.

We'd not like to denote by the word straina change that was just a rigid-bodydisplacement. Indeed, our earlier notion ofstrain is the ratio of relative displacements (e.g the ends of a long rod ) to their originalseparation. Let us apply this idea to themore general picture with arbitrary u(X).Two material points A and B ( you mayconceive of them as labeling particularatoms) that start at positions X and X + ΔXrespectively have an original separation ΔX. (see figure) After the deformation they havepositions X + u(X) and X +ΔX+ u(X+ΔX). Their relative displacement is therefore Δu isu(X+ΔX)-u(X). If we are describing the strain of an infinitesimal element, we must take ΔXvery small. Thus the relevant quantity is the ratio of Δu to ΔX in the limit of small ΔX.This is merely a derivative, the displacement gradient. We define the tensor displacementgradient [D] of the displacement field [ D ] = u. [ D ] is a dimensionless tensor (it mightbe measured in %) and is derivatives with respect to the X coordinates. In Cartesiancoordinates Δu is

Δru = [D]{Δ

rX}; or

ΔuxΔuyΔuz

=

∂ux / ∂X ∂ux / ∂Y ∂ux / ∂Z∂uy / ∂X ∂uy / ∂Y ∂uy / ∂Z∂uz / ∂X ∂uz / ∂Y ∂uz / ∂Z

ΔXΔYΔZ

There are 9 elements of the displacement gradient:

Dij = ∂ui/∂Xj

( and of course they could vary with position X. ) In this course we will be assuming [D] issmall. The study of finite deformations requires some elegant and abstract mathematics thatwe do not have the time for.

The diagonal elements of this are the tensile strains defined earlier. For example Dxx = δL/Lofor stretching a rod.

The off diagonal elements are related to the shear strains defined earlier – but there is aproblem, because off diagonal elements of the displacement gradient can arise even if thedisplacements are merely due to rigid–body rotations. We do not want to deem rotations tobe strains.

If a body were rotated by an infinitesimal angle θ, its displacements would be

10

ru(

rX) =

rθ ×

rX

For which the displacement gradients are (recall page 163 ff)

[D] = [θ ] =0 −θ3 θ2θ3 0 −θ1−θ2 θ1 0

It is the antisymmetric (skew) parts of D that represent rotations. We therefore do adecomposition of [ D ] by

[D] = 12([D]+ [D]T ) + 1

2([D]− [D]T ) ≡ [ε] + [θ]

and define the strain tensor [ε] as the symmetric part of D

[ε] = 12([D]+ [D]T ) =

ε11 ε12 ε13ε12 ε22 ε23ε13 ε231 ε33

=

D11D12 + D21

2D13 + D31

2D12 + D21

2D22

D32 + D23

2D13 + D31

2D32 + D23

2D33

In indicial notation,

εij =12(Dij + Dji ) =

12( ∂ui∂Xj

+∂uj

∂Xi

)

The skew part is θij =12(Dij − Dji ) =

12( ∂ui∂Xj

−∂uj

∂Xi

)

[θ] = 12([D]− [D]T ) =

0 −θ3 θ2θ3 0 −θ1−θ2 θ1 0

=

0 D12 − D21

2D13 − D31

2D21 − D12

20 D23 − D32

2D31 − D13

2D32 − D23

20

Illustrations of the meanings of the elements of the strain tensori) Sometimes [ε] is a multiple of the identity

[ε] =e 0 00 e 00 0 e

11

This describes a simple dilation. All directions get stretched the same amount e. Assumingno rotations, we would write [D] = ε[ I ]. Assuming no rigid body displacements we wouldthen write du = [D] dX = e dX . All relative displacements du are merely proportional todX and in the same direction as dX.

We could also consider a pure stretch along one axis.

ii) Here is another illustrative sample of [ ε ] What if it looks like this:

[ε] =ε11 0 00 0 00 0 0

In this case, ( and again assuming no rotations )

du = X1 ε11 i

All relative displacement is in x-direction. The body has been stretched in the x-direction by afraction ε11

iii) Similarly a purely diagonal [ ε]

[ε] =ε11 0 00 ε22 00 0 ε33

corresponds to different amounts of (simultaneous) stretch in all three directions.

The diagonal elements of [ε] are called the stretches in the x y and z directions.

We define the average stretch

e = ( ε11 + ε22+ ε33 ) / 3 = (1/3) Trace [ε]

iv) Shear strain

Sometimes the strain tensor has off-diagonal elements, for example

[ε] =0 γ 0γ 0 00 0 0

The radius of this sphere increases by a fraction e, so thevolume increases to ( 1+ e)3 times its original volume, sotherefore by a fraction 3e ( we continue to assumestrains e << 1.)

12

This [ ε ] implies (assuming no rotations i.e that ε=D) ∂ux/∂Y = γ and ∂uy/∂X = γ ∂ux/∂X = 0 and ∂uy/∂Y = 0

We can integrate these with respect to x and y to get ( choosing constants of integration such thatu = 0 at the origin)

ux = γ Y and uy = γ Xwhich may be pictured graphically:

Thus the strain tensor [ε] =0 γ 0γ 0 00 0 0

pictured at the above left corresponds to the

picture (page 7 and above right ) we earlier called a shear strain by an amountδy/Lo = 2 γ and then a clockwise rotation by γ. ------------

As described above, the tensor e [ I ] represents a pure dilation, the same stretch in alldirections, a volume change without any shearing. We often decompose a strain [ ε ] intoits dilatational part and its deviatoric part [ ε' ]

[ ε ] = e [ I ] + [ ε' ]

By definition, [ ε' ] has zero trace and so corresponds to a strain with zero average stretchand no volume change.

Summary of the decompositions:[D] is decomposed into rotations(antisymmetric) + strain(symmetric part).Strain may then be decomposed into dilation part (a multiple of the Identity matrix)

plus a deviatoric ( traceless part with no volume change)

Inasmuch as [ ε ] is symmetric, it has three principle stretches (eigenvalues), associated withthree mutually perpendicular eigenvectors. Every strain is equivalent to three perpendicularstretches ( or compressions ).

Even an [ ε ] that has no stretch in directions x y or z, still has three principle stretches (inthe eigendirections) For example,

13

you might think to characterize the strain [ε] =0 γ 0γ 0 00 0 0

as purely shear. Such a

characterization could be, however, misleading. This strain has three perpendicularprinciple strain directions { 1,1,0}, {1,-1,0} and {0,0,1} (these are the eigenvectors of[ ε ]) with principle strains ( the eigenvalues) γ, - γ and zero respectively. Thus [ ε ]corresponds to a stretch in the {1 1 0} direction, a shrinking in the {1 –1, 0} direction and nostretch in the {0 0 1} direction. In the rotated coordinate system, there are no shear strains –just stretches and shrinkings. We'll see the same phenomenon for the stress tensor.

Volume change

Fractional volume increase δV/Vo under a strain [ ε ] is easy to describe.

ε11 + ε22+ ε33 = Trace [ ε ] = δV/Vo

For a proof, recognize that the ratio of the volume elements before and after a deformation isthe Jacobian of the new positions X + u(X) as a function of the original positions X :

J = det | ∂(X+u(X))/∂X | ) = det | I + [ε] + [ θ ] ] ≈ 1+ Trace [ε] + Order ε2.

Alternative proof: consider a rectangular element of original sides a,b,c oriented along thedirections of principle stretch ( with respect to which the strain tensor is diagonal). Theprinciple stretches are the three eigenvalue sof [ ε] λ1, λ2 and λ3 After stretching, it hassides a(1+λ1) , b( 1+ λ2) and c ( 1+λ3). The original volume was abc, the later volume isstill a rectangle but with volume abc (1+λ1)(1+λ2)(1+λ3). This is approximately abc( 1+ λ1+λ2+ λ3) (because the λ are small). So the volume has increased by a factor ( 1+ λ1+ λ2+ λ3)i.e, by a fractional amount λ1+ λ2+ λ3. This is the trace of the diagonalized strain tensor.BUT… the trace is just the sum of the eigenvalues and is the same as the trace of theoriginal strain tensor ε11+ ε22+ ε33 before diagonalization. (The trace is invariant underrotations.) QED, the fractional volume increase is the trace of [ε].

Stress is a Tensor too

Consider again the thin triangular prism withthree interesting edges, all with surface forcesand all with infinitesimal areas that can berepresented as vectors with directionperpendicular to the surfaces and magnitudeequal to their areas. Last time we consideredonly normal forces, now let us include shearforces too.

14

The directions out of and into the paper are not displayed. If we did so, we'd include dA4 = -dA5 and F4 and F5.

The vectors dA have magnitude equal to the area of the corresponding face and directionperpendicular to that face.

If the state of stress were purely due to pressure, we'd write Fi = -p dAi.

We wish to derive a relation between F and dA more general than the pressure form. Weposit a relation like this:

Fi = f(dAi)

where f is a vector function of a vector . The function characterizes the contact forces in themedium at the point of interest. The relation states that the force on a surface is somefunction of the surface's orientation and size (and the state of the material at that point). Wewill show that the relation is linear.

We first note that if we were to change the sign of dA, we'd have to get the opposite F (equaland opposite reactions, Newton's third law) So for any vector v, we may write f(v) = - f(-v). As our triangular prism is very thin, the forces F4 and F5 apply to almost the same surfaceso they are equal and opposite. We will henceforth neglect them.

Geometry tells us dA1 + dA2 + dA3 = 0

Force balance tells us ( because F4 = -F5 )

F1 + F2 + F3 = 0

This looks like static equilibrium, but it is more general than that. It is derived using theearlier argument about rescaling the figure by λ: The surface forces F scale with λ2 while theinertia and the body forces scale with λ3 – so in the limit of small λ, for tiny volumes, thesurface forces must separately sum to zero, regardless of whether or not the material is instatic equilibrium and regardless of the presence of body forces. Thus

f(dA1) + f(dA2) + f(dA3) = 0

Therefore f(dA3) = -f(dA1) - f(dA2)

i.e, f(-dA1-dA2) = -f(dA1) - f(dA2)

or f(dA1+dA2) = f(dA1) + f(dA2)

15

Inasmuch as this is true regardless of the two vectors dA1 and dA2, this means that thevector function f is linear.

f maps a vector linearly into another vector. It is therefore representable as a 3x3 matrix withcomponents σnm

fn (dA) = Σm σnmdAm

The nth component of the force across a surface dA is given as a linear combination of thecomponents of dA m=1,2,3. The coefficients σ are the components of the stress. This iswritable in matrix notation as

{F} = [ σ ] {dΑ}

Aside: if the state of stress were purely pressure, we write f(dA) = -p dA, which means[ σ ] = −p [ Ι ]; σij =-p δij.

[ σ ] - and its 9 components σnm - can of course vary from place to place in the medium.Its significance is that, once you know the local value of [σ], you know the force acrossevery infinitesimal surface there, regardless of its orientation or size.

The stress is not just a 3x3 matrix. It is a tensor ( like the inertia tensor for rigid bodies andthe strain tensor described above ) We say this because of how it maps vectors into vectors (some people define tensors as linear maps in vector spaces ) If you were to change thecoordinate system, e.g. rotate it, the components σmn would have to change in a specific way.

What do the components σmn mean?

σmn = the force per area in the 'm' direction across a unit area in the 'n' direction.

In particular

σxx = tensile stress in x –direction

Here is a picture of the normal forces, and a separate picture of the shear forces so as to keep theclutter down, on three faces of a unit cube

16

Of course there are forces on the three hidden faces too.

To summarize: The components of the stress tensor at a point take values corresponding tothe state of the material there. Given the stress tensor, one may construct the surface forcedF acting across any infinitesimal area dA with normal n by {dF} = [ σ ] {n} dA .

The nm component σnm of [ σ ] is the surface force in the n direction on a faceperpendicular to the m direction (divided by the area of the face)

The stress tensor is symmetric

Like many of our favorite tensors in physics, σ issymmetric σij = σji. While it has 9 components,only 6 are independent. How do we know this?Consider a square prism of side a….

and ask for the sum of the torques around the pointO. We see that the net torque is

a ( σxy – σyx) times the face area

This scales as we diminish the size of the square,like λ3. ( two factors of λ due to the surface area,one for the factor of a ) But the moment of inertia scales like λ5 (three factors of λ for themass, two more because moments of inertia scale like mass times radius^2) If this is to be inbalance against (infinitely fast in the limit of zero λ) rotational accelerations, we must have σxy = σyx. The argument may be repeated for the other directions, leading us to conclude σij= σji for all i,j.

17

Because the stress tensor is symmetric, it has three real eigenvalues and associated mutuallyperpendicular eigenvectors. These eigenvalues are called the principle stresses. Some canbe compressive ( negative ) some can be positive (tensile )

We can, if we wish, rotate our xyz axes so that the stress tensor becomes diagonal. Thediagonal elements are the principle stresses; the new xyz axes are the eigenvectors.

In a coordinate system for which [σ] is purely off-diagonal itmay appear as if the material has no tensile or compressivestresses. This is not the case.

Consider for example the 2x2 stress tensor for this squaresubjected to the indicated forces s - which looks purely shear-

like to a naive eye. [σ ] =0 ss 0

, σxy = σyx = s;

σxx = σyy = 0 ( see sketch)

We ask for its principle stresses and directions by seeking itseigenvectors and values. The eigenvalues are s and – s; the eigenvectors are {1 1} and {1–1 } respectively. These vectors are ±45 degrees from the original xy axes, so the squarewith the shear forces on it illustrated above is the sameas that illustrated here…

In a rotated coordinate system, σ takes the form

[σ ] =s 00 − s

, σx'y' = σy'x' = 0; σx'x' =s; σy'y' = -s

which is purely compressive in one direction and purelytensile in the other.

We conclude that shear is just compression and tensionin other directions.

Another way of picturing it: consider a block with theindicated forces; it is stretched horizontally andcompressed vertically. These forces shear the blockalong the diagonals. Materials subjected to such forcesmay fail (slip) along their diagonals.

-------As with strain, it is possible to decompose the stress intoits pressure part and a traceless part. Recall that, for apurely pressure-like stress field, we'd have

[ σ ] = - p [ I ]

18

Suggesting that we define the pressure of an arbitrary stress as p = - Tr [ σ] / 3 and write[σ ] = -p [ I ] + [ σ ' ]

Stress can always be written as the sum of a pressure-like field and a part that has no Trace.End Lecture 1

Elastic Waves Assignment 1 Due Monday August 31 2015

HW.1.1 By imposing the condition that strain energy density must be positive, regardless ofthe strain, derive necessary and sufficient thermodynamic inequalities for shear modulus µ, for Lamémodulus λ, for bulk modulus κ and for Poisson ratio ν. This is best ( I think) done by first derivingan expression for the stain energy in the form ( where A and B are related to whichever two moduliyou find convenient to start with )

U =12[(A (Trε)2 + B Tr(ε '2 )]

and where ε' is the deviatoric strain [ε'] = [ε]- [ I ] Trε /3 . Give an argument that A and B must benon negative if and only if the material is to be thermodynamically stable; U ≥ 0. You may find ituseful to consult the table relating different moduli,

1.2 Consider an isotropic material (moduli λ,µ) with a thermal expansion coefficient α, such thatvolume change (if it is unconstrained) as temperature T increases is given by ΔV / V = α ΔT. If wenow take that material and fix its strain while heating it (i.e. we do not allow its volume to increase),what are the stresses that develop?

1.3 Consider (in 2-dimensions) a cubic material with modulus tensorcijkl = λδijδkl +µ{δikδ jl +δilδ jk}   + ν  δijkl          i, j,k,l =1,2

where ν is another modulus and the δijkl (not a tensor) is equal to one if all indices are identical, andzero otherwise. [ c ] takes this form only if the coordinate system is aligned along the Cartesiancrystal axes; if you wanted to know [c ] in a rotated frame, you must do a transformation. Chooseλ = µ = ν = ρ = 1 for simplicity and find the two slownesses versus direction; plot them in a polarplot. What are the polarizations of the waves gong in a direction at 45 degrees? How far are theyfrom longitudinal or transverse?

1.4 A shear wave in a Maxwell-model viscoelastic medium.Consider a material with a shear stress/shear strain relation given by

∂ε/∂t = (1/G) [ ∂σ/∂t + σ / Δ]where ∆ is a time constant and G is a nominal shear modulus.Consider a plane wave of the form ~exp(iωt-ikx)Find and plot the attenuation, α=-Im k(ω) as a function of ωΔ, over the range from zero to three. Find, and plot, the wavenumber Re k(ω) over the same range.You may wish to work in units of length and time such that nominal wavespeed2, G/ρ, = 1,and units of time such that Δ = 1