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

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1 Phys 598 EW Elastic Waves Fall 2015 Standard texts and monographs, all with a view towards solving classical problems in elastic wave propagation: K. F. Graff, "Wave Motion in Elastic Solids," Dover, NY 1975 (inexpensive!) J.D. Achenbach, "Wave Propagation in Elastic Solids, NorthHolland/Elsevier Emphasis 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" 1978 Waveguides: Plates and rods Y-H Pao and C C Mow "Diffraction of Elastic Waves and Dynamic Stress Concentrations" Rand Corp 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 in seismology. ================ Tentative course outline: I. Graff 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 potentials Alternate forms for Equation of motion Plane Waves of P & S type Visco-Elastic Constitutive relations, Attenuation & Dispersion of Plane waves II. Governing Equations (often approximate) in reduced dimensions 2-d Plane strain SH and P-SV Plane stress extensional waves in thin plates SH waves in thin (or thick) plates Bending waves in thin plates, Kirchoff plate theory. 1d Torsional waves in circular rods Bending waves in beams, Euler-Bernoulli theory Extensional waves in rods 1

2 Strings derivation of linearized equation III Applications in reduced dimensions, Exercises in Fourier Transforms and Greens functions 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 waves circular crested waves Harmonic waves & Dispersion relations for Cartesian and Polar co-ordinates The Infinite stiff ribbon, a Multi branched dispersion relation Finite systems, normal modes Forced Vibration; Greens functions Infinite plates, Triple FT's, & Hankel Transforms Infinite ribbon Finite systems IV Elastodynamics in 3 dimensions 5 weeks The Fundamental solution, the Green's Dyadic in unbounded 3-d media (Graff ) => 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 2

3 Layered Half space - Love Waves Forced problems 6.2 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 waveguides Graff 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 maedium Diffuse field assumption and its justification. Consequences Universal Partition, surface/bulk and P/S and H/V. Diffusion of multiply scattered wave energy Enhanced backscatter Field 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.. 3

4 The mechanics of elastic continua ( recap of Phys 326 material) One key concept is that of a tiny volume element, large compared to any microscale like the atomic upon which there may be some structure, but small compared to any macroscopic scales over which we will make measurements. [When and if this separation of scales is lost, and one cannot identify such tiny volume elements, continuum mechanics has to be re-thought.] This volume dv has no internal structure, by definition. It has a volume dv, and a shape ( which we can take to be just about anything useful ) and a surface, with unit outward normal n that of course varies in 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 negligibly within the volume dv.) The continuum hypothesis continues with a notion that the forces on this dv are of two kinds: 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 into shear (sometimes called traction) and compression or tension. Sometimes all the surface forces are called tractions. In general, each volume element dv will be subject to 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 must be 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 shear forces. All surfaces transmit only normal forces. Perhaps the material is too slippery to tolerate them. This applies to many fluids - and sometimes to solids approximately in that shear strengths are often much less than compressive strengths so the shear forces may be relatively weak. Consider a prismatic volume dv shaped like an isosceles triangle. It has forces on each side which, by hypothesis are a) proportional to the side's area S and b) in the direction n. 4

5 (because we assume no shear) The proportionality is described by the symbol -p (for pressure) Thus the sum of the forces is r F = p 1 S 1 ˆn 1 p 2 S 2 ˆn 2 p 3 S 3 ˆn 3 + ρ r gv = m r a (I have included all body forces in the term in g and introduced the possibility that the compressive and tensile forces on the three faces might be different. Also I have a minus sign on p 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 factor of λ. The pressure terms scale with the surface areas S ~ λ 2 ; the other terms scale with the volume ~ λ 3. Thus in the limit of a very small volume dv, the pressure terms must by themselves add to zero ( except maybe for effects due to how p changes an infinitesimal amount across the infinitesimal distances ) p 1 S 1 ˆn 1 + p 2 S 2 ˆn 2 + p 3 S 3 ˆn 3 = 0 This is a vector equation. All components, in particular the horizontal component, of this must vanish. Also we notice S 1 = S 2 and that the horizontal components of n 1 and n 2 are equal and opposite. Therefore the pressures on the two vertical faces must be equal: p 1 = p 2. Because the orientation of our triangle was arbitrary we conclude that, in the absence of shear forces, the surface forces must be isotropic. Pressure is isotropic, the same in all directions. 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. properly described as tensors, but let us introduce them first less abstractly. They are more 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 and force F. Stress = F/A is in the picture below tensile Alternatively we could consider a rod in compression 5

6 In which case we would write that the rod has stress C/A and is in compression. By convention, tensile stresses are considered positive and compressive stresses are considered negative. 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 await our tensor treatment) Don't be too bothered by how the above body is not in equilibrium against rotations. There are presumably other forces not pictured above, for example maybe Shear stress V/A is measured in the same units, N/m 2 or Pascal. ==== Strain Aside: In all three of the above cases I pictured forces that summed to zero. This is required 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 a macroscopic object with different forces on each end, the body may be accelerating or there may be body forces that balance these imbalanced surface forces, or both. In either case the stress will then vary across the object. On every infinitesimal volume of it, however, the surface forces will be in balance. Strain is a dimensionless measure of deformation. It is conceptually the ratio of extra displacement to initial separation Tensile strain in a bar is the ratio of stretch to original length strain = δl/l o By convention, stretches are positive, compressions are negative. They are often quoted in %. 6

7 You might wish to use, not original length L o to normalize δl, but maybe the new length Lo + δl. Thereby defining strain as δl/(l o +δl). The difference is negligible if the strain is small Shear strain is more subtle: Here is a block of length L o with one end shifted sideways relative to the other a distance δy We define the strain as δy/l o. 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 true strain, angles between lines will change. In rotations they won't. The distinction will be visited 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 is linear stress = Modulus * strain with a proportionality Modulus that is characteristic of the material. Moduli have units of Pascal, or Newtons/m 2. 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 on strain 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 7

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 or shear stress = G * shear strain V/A = G (δy/l o ) 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 of forces. Now let us revisit the concepts of stress, strain, and Hooke's law, but do it more properly, with tensors. Strain is a tensor. How do we wish to describe deformations of a continuous medium? This is a purely kinematic question, essentially geometry. We start with the notion that a material point, originally at some vector position X, gets moved by a displacement u(x) to a new position X + u(x). Here is a sketch of how an arbitrarily shaped region has been moved and rotated and stretched to a new position by some function u(x). 8

9 If u were a constant vector, (independent of X) this would describe how the whole body got shifted rigidly, with no change of shape and no rotation. "Strain" is related to relative displacement. We'd not like to denote by the word strain a change that was just a rigid-body displacement. Indeed, our earlier notion of strain is the ratio of relative displacements ( e.g the ends of a long rod ) to their original separation. Let us apply this idea to the more general picture with arbitrary u(x). Two material points A and B ( you may conceive of them as labeling particular atoms) that start at positions X and X + ΔX respectively have an original separation ΔX. (see figure) After the deformation they have positions X + u(x) and X +ΔX+ u(x+δx). Their relative displacement is therefore Δu is u(x+δx)-u(x). If we are describing the strain of an infinitesimal element, we must take ΔX very 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 displacement gradient [D] of the displacement field [ D ] = u. [ D ] is a dimensionless tensor (it might be measured in %) and is derivatives with respect to the X coordinates. In Cartesian coordinates Δu is Δ r u = [D]{Δ r X}; or Δu x u x / X u x / Y u x / Z ΔX Δu y = u y / X u y / Y u y / Z ΔY Δu z u z / X u z / Y u z / Z ΔZ There are 9 elements of the displacement gradient: D ij = u i / X j ( and of course they could vary with position X. ) In this course we will be assuming [D] is small. The study of finite deformations requires some elegant and abstract mathematics that we do not have the time for. The diagonal elements of this are the tensile strains defined earlier. For example D xx = δl/l o for stretching a rod. The off diagonal elements are related to the shear strains defined earlier but there is a problem, because off diagonal elements of the displacement gradient can arise even if the displacements are merely due to rigid body rotations. We do not want to deem rotations to be strains. If a body were rotated by an infinitesimal angle θ, its displacements would be 9

10 r u( r X) = r θ r X For which the displacement gradients are (recall page 163 ff) 0 θ 3 θ 2 [D] = [θ ] = θ 3 0 θ 1 θ 2 θ 1 0 It is the antisymmetric (skew) parts of D that represent rotations. decomposition of [ D ] by We therefore do a [D] = 1 2 ([D] + [D]T ) ([D] [D]T ) [ε] + [θ] and define the strain tensor [ε] as the symmetric part of D D D 12 + D 21 D 13 + D [ε] = 1 ε 11 ε 12 ε ([D]+[D]T ) = ε 12 ε 22 ε 23 = D 12 + D 21 D D 32 + D ε 13 ε 123 ε 33 D 13 + D 31 D 32 + D 23 D In indicial notation, ε ij = 1 2 (D ij + D ji ) = 1 2 ( u i X j + u j X i ) The skew part is θ ij = 1 2 (D ij D ji ) = 1 2 ( u i X j u j X i ) 0 [θ] = 1 0 θ 3 θ 2 2 ([D] [D]T ) = θ 3 0 θ 1 = D 21 D 12 2 θ 2 θ 1 0 D 31 D 13 2 D 12 D D 32 D 23 2 Illustrations of the meanings of the elements of the strain tensor i) Sometimes [ε] is a multiple of the identity e 0 0 [ε] = 0 e e D 13 D 31 2 D 23 D

11 This describes a simple dilation. All directions get stretched the same amount e. Assuming no rotations, we would write [D] = ε[ I ]. Assuming no rigid body displacements we would then write du = [D] dx = e dx. All relative displacements du are merely proportional to dx and in the same direction as dx. The radius of this sphere increases by a fraction e, so the We volume could increases also consider to ( 1+ a e) pure 3 times stretch its original along one volume, axis. so therefore by a fraction 3e ( we continue to assume strains e << 1.) ii) Here is another illustrative sample of [ ε ] What if it looks like this: [ε] = ε In this case, ( and again assuming no rotations ) du = X 1 ε 11 i All relative displacement is in x-direction. The body has been stretched in the x-direction by a fraction ε 11 iii) Similarly a purely diagonal [ ε] ε [ε] = 0 ε ε 33 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 = ( ε 11 + ε 22 + ε 33 ) / 3 = (1/3) Trace [ε] iv) Shear strain Sometimes the strain tensor has off-diagonal elements, for example [ε] = 0 γ 0 γ

12 This [ ε ] implies (assuming no rotations i.e that ε=d) u x / Y = γ and u y / X = γ u x / X = 0 and u y / Y = 0 We can integrate these with respect to x and y to get ( choosing constants of integration such that u = 0 at the origin) u x = γ Y and u y = γ X which may be pictured graphically: 0 γ 0 Thus the strain tensor [ε] = γ 0 0 pictured at the above left corresponds to the picture (page 7 and above right ) we earlier called a shear strain by an amount δy/l o = 2 γ and then a clockwise rotation by γ As described above, the tensor e [ I ] represents a pure dilation, the same stretch in all directions, a volume change without any shearing. We often decompose a strain [ ε ] into its 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 stretch and 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 with three mutually perpendicular eigenvectors. Every strain is equivalent to three perpendicular stretches ( or compressions ). Even an [ ε ] that has no stretch in directions x y or z, still has three principle stretches (in the eigendirections) For example, 12

13 0 γ 0 you might think to characterize the strain [ε] = γ 0 0 as purely shear. Such a characterization could be, however, misleading. This strain has three perpendicular principle 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 no stretch 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/v o under a strain [ ε ] is easy to describe. ε 11 + ε 22 + ε 33 = Trace [ ε ] = δv/v o For a proof, recognize that the ratio of the volume elements before and after a deformation is the 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 ε 2. Alternative proof: consider a rectangular element of original sides a,b,c oriented along the directions of principle stretch ( with respect to which the strain tensor is diagonal). The principle stretches are the three eigenvalue sof [ ε] λ 1, λ 2 and λ 3 After stretching, it has sides a(1+λ 1 ), b( 1+ λ 2 ) and c ( 1+λ 3 ). The original volume was abc, the later volume is still a rectangle but with volume abc (1+λ 1 )(1+λ 2 )(1+λ 3 ). This is approximately abc( 1+ λ 1 + λ 2 + λ 3 ) (because the λ are small). So the volume has increased by a factor ( 1+ λ 1 + λ 2 + λ 3 ) i.e, by a fractional amount λ 1 + λ 2 + λ 3. 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 the original strain tensor ε 11 + ε 22 + ε 33 before diagonalization. (The trace is invariant under rotations.) QED, the fractional volume increase is the trace of [ε]. Stress is a Tensor too Consider again the thin triangular prism with three interesting edges, all with surface forces and all with infinitesimal areas that can be represented as vectors with direction perpendicular to the surfaces and magnitude equal to their areas. Last time we considered only normal forces, now let us include shear forces too. 13

14 The directions out of and into the paper are not displayed. If we did so, we'd include da 4 = - da 5 and F 4 and F 5. The vectors da have magnitude equal to the area of the corresponding face and direction perpendicular to that face. If the state of stress were purely due to pressure, we'd write F i = -p da i. We wish to derive a relation between F and da more general than the pressure form. posit a relation like this: We F i = f(da i ) where f is a vector function of a vector. The function characterizes the contact forces in the medium at the point of interest. The relation states that the force on a surface is some function of the surface's orientation and size (and the state of the material at that point). We will 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 (equal and 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 F 4 and F 5 apply to almost the same surface so they are equal and opposite. We will henceforth neglect them. Geometry tells us da 1 + da 2 + da 3 = 0 Force balance tells us ( because F 4 = -F 5 ) F 1 + F 2 + F 3 = 0 This looks like static equilibrium, but it is more general than that. It is derived using the earlier argument about rescaling the figure by λ: The surface forces F scale with λ 2 while the inertia and the body forces scale with λ 3 so in the limit of small λ, for tiny volumes, the surface forces must separately sum to zero, regardless of whether or not the material is in static equilibrium and regardless of the presence of body forces. Thus f(da 1 ) + f(da 2 ) + f(da 3 ) = 0 Therefore f(da 3 ) = -f(da 1 ) - f(da 2 ) i.e, f(-da 1 -da 2 ) = -f(da 1 ) - f(da 2 ) or f(da 1 +da 2 ) = f(da 1 ) + f(da 2 ) 14

15 Inasmuch as this is true regardless of the two vectors da 1 and da 2, this means that the vector function f is linear. f maps a vector linearly into another vector. It is therefore representable as a 3x3 matrix with components σ nm f n (da) = Σ m σ nm da m The n th component of the force across a surface da is given as a linear combination of the components of da m=1,2,3. The coefficients σ are the components of the stress. This is writable 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 ]; σ ij =-p δ ij. [ σ ] - and its 9 components σ nm - 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 across every 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 and the 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 the coordinate system, e.g. rotate it, the components σ mn would have to change in a specific way. What do the components σ mn mean? σ mn = the force per area in the 'm' direction across a unit area in the 'n' direction. In particular σ xx = 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 the clutter down, on three faces of a unit cube 15

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 to the state of the material there. Given the stress tensor, one may construct the surface force df acting across any infinitesimal area da with normal n by {df} = [ σ ] {n} da. The nm component σ nm of [ σ ] is the surface force in the n direction on a face perpendicular 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 is symmetric σ ij = σ ji. 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 point O. We see that the net torque is a ( σ xy σ yx ) times the face area This scales as we diminish the size of the square, like λ 3. ( two factors of λ due to the surface area, one for the factor of a ) But the moment of inertia scales like λ 5 (three factors of λ for the mass, two more because moments of inertia scale like mass times radius^2) If this is to be in balance against (infinitely fast in the limit of zero λ) rotational accelerations, we must have σ xy = σ yx. The argument may be repeated for the other directions, leading us to conclude σ ij = σ ji for all i,j. 16

17 Because the stress tensor is symmetric, it has three real eigenvalues and associated mutually perpendicular eigenvectors. These eigenvalues are called the principle stresses. Some can be compressive ( negative ) some can be positive (tensile ) We can, if we wish, rotate our xyz axes so that the stress tensor becomes diagonal. The diagonal elements are the principle stresses; the new xyz axes are the eigenvectors. In a coordinate system for which [σ] is purely off-diagonal it may appear as if the material has no tensile or compressive stresses. This is not the case. Consider for example the 2x2 stress tensor for this square subjected to the indicated forces s - which looks purely shearlike to a naive eye. [σ ] = 0 s s 0, σ xy = σ yx = s; σ xx = σ yy = 0 ( see sketch) We ask for its principle stresses and directions by seeking its eigenvectors 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 square with the shear forces on it illustrated above is the same as that illustrated here In a rotated coordinate system, σ takes the form [σ ] = s 0 0 s, σ x'y' = σ y'x' = 0; σ x'x' =s; σ y'y' = -s which is purely compressive in one direction and purely tensile in the other. We conclude that shear is just compression and tension in other directions. Another way of picturing it: consider a block with the indicated forces; it is stretched horizontally and compressed vertically. These forces shear the block along the diagonals. Materials subjected to such forces may fail (slip) along their diagonals As with strain, it is possible to decompose the stress into its pressure part and a traceless part. Recall that, for a purely pressure-like stress field, we'd have [ σ ] = - p [ I ] 17

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 HW.1.1 By imposing the condition that strain energy density must be positive, regardless of the strain, derive necessary and sufficient thermodynamic inequalities for shear modulus µ, for Lamé modulus λ, for bulk modulus κ and for Poisson ratio ν. This is best ( I think) done by first deriving an expression for the stain energy in the form ( where A and B are related to whichever two moduli you find convenient to start with ) U = 1 2 [(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 be non negative if and only if the material is to be thermodynamically stable; U 0. You may find it useful to consult the table relating different moduli, 1.2 Consider an isotropic material (moduli λ,µ) with a thermal expansion coefficient α, such that volume change (if it is unconstrained) as temperature T increases is given by ΔV / V = α ΔT. If we now 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 tensor c ijkl = λδ ij δ kl + µ{δ ik δ jl +δ il δ jk } + ν δ ijkl i, j,k,l = 1,2 where ν is another modulus and the δ ijkl (not a tensor) is equal to one if all indices are identical, and zero otherwise. [ c ] takes this form only if the coordinate system is aligned along the Cartesian crystal axes; if you wanted to know [c ] in a rotated frame, you must do a transformation. Choose λ = µ = ν = ρ = 1 for simplicity and find the two slownesses versus direction; plot them in a polar plot. What are the polarizations of the waves gong in a direction at 45 degrees? How far are they from 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 wavespeed 2, G/ρ, = 1, and units of time such that Δ = 1 18