PEAT SEISMOLOGY Lecture 2: Continuum mechanics

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1 PEAT SEISMOLOGY Lecture 2: Continuum mechanics Nick Rawlinson Research School of Earth Sciences Australian National University

2 Strain Strain is the formal description of the change in shape of a material. We want to be able to describe the deformation of a continuum; to do this, we can use the displacement vector u(x, t) = x(t) x(t 0 ), which describes the location of every point relative to its position at a reference time t 0. Consider an element of solid material within which displacements u(x) have occurred. x x+ x x+u x+ x+u+ u What is the point x + x displaced by?

3 Strain Deformation gradient tensor This can be done using a Taylor series expansion: where u(x + x) = u(x) + J x + O( x 2 ) u x / x u x / y u x / z J = u y / x u y / y u y / z u z / x u z / y u z / z J is often described as the Deformation gradient tensor. If we ignore the second and higher order terms in the expansion, then u = J x

4 Strain First order strain approximation In seismological applications, we ignore higher order deformation terms; usually, Earth strains are small enough to validate this approximation. Typical strain values due to the passage of seismic waves are For far field waves, strain is often < The deformation gradient tensor describes both rigid body rotation and deformation; we are interested in the latter. It is possible to separate these two effects by dividing J into symmetric and anti-symmetric parts. J = e + Ω = u = 1 2 [ u + ( u)t ] [ u ( u)t ]

5 Strain Symmetric component of J The symmetric strain tensor (e ij = e ji ) is given by u x x ( e = 1 uy x + u x y ( uz x + u x z ) ( 1 ux 2 y + u ) ( y 1 ux x 2 u y y ) ( 1 uz 2 y + u y z ) 1 2 z + u z x ( uy z + u z y u z z ) )

6 Strain Anti-symmetric component of J The anti-symmetric rotation tensor Ω (Ω ij = Ω ji ) is given by 0 Ω = 1 ( ux y u y x ( ux z u z x ) ) ( 1 ux 2 y u ) ( y 1 ux x ( uy z u ) z y 1 2 z u z x ( uy z u z y 0 ) )

7 Strain Example 1 If Ω = 0 and there is no volume change ( u x / x = u y / y = u z / z = 0), then if we consider a 2-D square J = e = ( uy x + u ) x y ( 1 ux 2 y + u y x 0 ) 0 = u y x u x y with u x / y = u y / x since Ω = 0. The deformation of the square will therefore look as follows (bearing in mind the first-order approximation) if u x / y > 0 (which u y / x > 0): 0

8 Strain Example 1 du dy x >0 y x du dx y >0

9 Strain Example 2 If e = 0, then u x / y = u y / x and 0 J = Ω = u y x u x y The deformation of the square is now purely rotational, and will look as follows if u x / y > 0 (which u y / x < 0): 0

10 Strain Example 2 du dy x >0 y x du dx y <0

11 Strain Notation Often, the strain tensor is expressed using index notation. e ij = 1 ( ui + u ) j 2 x j x i The change in volume of a material, or dilatation, is given by: θ = e xx + e yy + e zz = e ii = tr[e] = u Einstein summation convention: when an index appears twice in a single term, it implies summation over that index e.g. l m n w ijk a i b j c k = w ijk a i b j c k i=1 j=1 k=1

12 Stress Non-zero strain of an elastic material requires some internal force distribution, which can be described by the stress field. Like strain, stress is a tensor quantity. Forces acting on elements of the continuum are of two types: Body forces f (e.g. gravity) Surface tractions T exerted by neighbouring elements of the continuum or by external forces applied to the boundary of the continuum.

13 Stress Traction Traction is the force per unit area acting on a surface within the continuum and can be defined by: [ ] F T(ˆn) = lim A 0 A(ˆn) F n A

14 Stress Traction Traction has the same orientation as the applied force, but is a function of the orientation of the surface defined by the unit normal vector ˆn. Traction is symmetric with respect to reflection in the surface, i.e. T( ˆn) = T(ˆn). T(n) n -n -T(n)

15 Stress Traction We need three orthogonal surfaces to describe a complete state of force at any point. T(z) T(x) z T(y) x where: ˆx = unit normal vector in direction of x axis. ŷ = unit normal vector in direction of y axis. ẑ = unit normal vector in direction of z axis. y

16 Stress The stress tensor The complete stress state is described by three 3-component vectors tensor. The stress tensor is therefore defined by: T(ˆx) σ xx σ xy σ xz σ = T(ŷ) = σ yx σ yy σ yz T(ẑ) σ zx σ zy σ zz σ 33 σ 32 σ 31 σ 11 σ 13 σ 12 σ 23 σ 21 σ 22 x 3 x 2 x 1

17 Stress The stress tensor Due to the conservation of angular momentum, there can be no net rotation from shear stresses. Therefore σ ij = σ ji and the stress tensor is symmetric and contains only six independent elements. These are sufficient to completely describe the state of stress at a given point in the medium. σ zx σ xz z σ xz x σ zx

18 Stress The stress tensor The stress tensor therefore may be written as: σ xx σ xy σ xz σ = σ xy σ yy σ yz σ xz σ yz σ zz Normal and shear stress components are simply defined by the diagonal and off-diagonal elements of σ: { normal stress if i = j (normal to plane) σ ij = shear stress if i j (parallel to plane)

19 Stress Example Consider a 2-D box with faces aligned to the Cartesian coordinate axes (x, z) subject to a stress field defined by: [ ] σxx 0 σ = 0 σ zz The box is therefore subject only to normal stress. σ zz z x σ xx σ xx σ zz σ xx >0 σ zz <0

20 Stress Example Consider a second box that is now oriented at an angle θ to the original box but is subject to the same stress field. What are the traction components that act on the side of the second box? σ zz z z σ zz x θ x σ xx σ xx σ xx σ xx σ zz σ zz This problem can be solved by rotating the coordinate axes to align them with the new box.

21 Stress Example In order to specify the stress tensor in the rotated coordinate system (x, z ), we make use of the relationship σ = AσA T, where A is the Cartesian transformation matrix. This is the same matrix used to transform a vector u from one coordinate system to another (i.e. u = Au). σ = [ ] [ ] [ ] cos θ sin θ σxx 0 cos θ sin θ sin θ cos θ 0 σ zz sin θ cos θ = [ σxx cos 2 θ + σ zz sin 2 θ ] (σ zz σ xx ) cos θ sin θ (σ zz σ xx ) cos θ sin θ σ xx sin 2 θ + σ zz cos 2 θ

22 Stress Example If we set θ = 45 and σ xx = σ zz, then: [ ] σ 0 σzz = σ zz 0 and only shear stresses act on the smaller block. σ zz z z 45 x x σ zx σ xz σ xx σ xx σ xz σ zx σ zz

23 Stress Principal stress For an arbitrary stress field, the traction vector will in most cases have non-zero components parallel and perpendicular to any surface orientation within the continuum. However, for any state of stress, surfaces can always be orientated in such a way that the shear tractions vanish. The normal vectors to these surfaces are oriented parallel to the principal stress axes, and the normal stresses acting on the surfaces are called the principal stresses. In order for the shear tractions to vanish for a surface defined with the unit normal vector ˆn, then the total traction acting on the surface must be parallel to ˆn. Thus, T(ˆn) = λˆn.

24 Stress Principal stress The traction across any arbitrary plane of orientation ˆn can be obtained by: σ xx σ xy σ xz ˆn x T(ˆn) = σ ˆn = σ xy σ yy σ yz ˆn y σ xz σ yz σ zz ˆn z Since we have that λˆn = σ ˆn, then σ ˆn λˆn = 0 (σ Iλ)ˆn = 0 which only has a nontrivial solution when det[σ Iλ] = 0

25 Stress Principal stress Thus, the principal stress axes ˆn are the eigenvectors of the tensor, and associated principal stresses λ are the eigenvalues. Computing the determinant yields the characteristic polynomial: λ 3 I 1 λ 2 + I 2 λ I 3 = 0 The roots of the above cubic polynomial define the three principal stresses, and substitution of each of these into (σ Iλ)ˆn = 0 yields the associated eigenvector, which corresponds to a principal stress direction.

26 Stress Principal stress It turns out that it is always possible to find three orthogonal normal vectors ˆn. The transformation matrix is therefore A = [ˆn (1), ˆn (2), ˆn (3) ] T. The stress tensor in the new coordinate system σ = AσA T is diagonal, and is commonly written as: σ σ = 0 σ σ 3 The usual convention is that σ 1 is in the direction of the maximum principal stress, σ 2 the intermediate principal stress, and σ 3 the minimum principal stress.

27 Stress Deviatoric stress The stress field within the Earth is usually dominated by the compressive stresses caused by the weight of overlying rock. In many applications, only the deviations from this dominant compressive stress are of interest. The Deviatoric stress τ is defined as τ ij = σ ij Mδ ij In the above definition, M is the mean stress, which is defined as the average normal stress: M = σ kk 3 = σ xx + σ yy + σ zz 3

28 Stress Deviatoric stress It turns out that the trace of the stress tensor is invariant under rotation of the coordinate system. Therefore, the mean stress can be written in terms of the principal stress since σ kk = σ ii: M = σ kk 3 = σ 1 + σ 2 + σ 3 3 When the principal stresses are large and nearly equal, the deviatoric stress tensor removes their effect and indicates the remaining stress state.

29 Stress Isotropic stress In a fluid at rest (or in which viscosity is negligible), the traction is always parallel to ˆn (i.e. fluid does not support shear stresses) and its magnitude is independent of orientation. In this case, traction is simply equal to pˆn where p(x, t) is the hydrostatic pressure. Therefore: p 0 0 σ = 0 p p When σ ij = pδ ij (equivalent to above expression), the stress field is referred to as isotropic Note that: δ ij = { 1 for i = j 0 for i j } = Kronecker delta

30 Elasticity Elastic tensor An elastic medium is one in which strain is a linear function of the applied stress. Since both stress and strain are 3 3 tensor quantities, the most general possible elastic medium has a constitutive equation of the form: σ ij = 3 k=1 l=1 3 C ijkl e kl = C ijkl e kl C, or the elastic tensor, is a 4th order tensor ( ) and could conceivably have 3 4 = 81 independent elements.

31 Elasticity Elastic tensor Symmetry of stress and strain tensors requires that: C ijkl = C jikl = C ijlk which reduces the number of independent components to 36 (six independent components of stress and strain). A final symmetry, C ijkl = C klij, can be derived from thermodynamics to prove that there are, at most, 21 independent elastic constants. The constants C ijkl are often referred to as the elastic moduli, and describe the elastic properties of the medium.

32 Elasticity Isotropic elasticity In an isotropic medium, the elastic properties are independent of orientation; the response (strain) is the same for stress applied in any direction. The general elastic tensor then reduces to a simpler form, depending on only two constants, which may be expressed in various ways. One useful pair are the Lamé constants λ and µ, which are defined such that C ijkl = λδ ij δ kl + µ(δ ik δ jl + δ il δ jk )

33 Elasticity Isotropic elasticity Each component of the stress field can be written in terms of strain as follows (noting the implied summation) σ ij = C ijkl e kl = [λδ ij δ kl + µ(δ ik δ jl + δ il δ jk )]e kl = λδ ij (e xx + e yy + e zz ) + µδ ik δ jl e kl + µδ il δ jk e kl Each of the six independent components of the stress field can then be explicitly written σ xx = λθ + 2µe xx σ yy = λθ + 2µe yy σ zz = λθ + 2µe zz σ xy = µ(e xy + e yx ) = 2µe xy σ xz = 2µe xz σ yz = 2µe yz

34 Elasticity Isotropic elasticity These expressions can be summarised as Hooke s law for an isotropic solid σ ij = λδ ij θ + 2µe ij Thus, the two Lamé parameters λ and µ completely describe the linear stress-strain relationship within an isotropic solid. Validity: isotropic solid; short time scale; small deformation, perfect elasticity. Isotropy is a reasonable approximation for much of the Earth s interior.

35 Elasticity Isotropic elasticity Elastic constants are supposedly independent of strain, but typically depend on temperature, pressure, and composition, and so vary with respect to location within the Earth. The assumption of perfect elasticity does not allow for the observed dissipation of seismic wave energy in the course of propagation. On long time scales and for large strains, anelasticity is important, requiring a different constitutive relationship, e.g. in a viscous material, we prescribe a relationship between stress and strain rate e/ t.

36 Elasticity Elastic constants In the above equations, µ is referred to as the shear modulus and can be defined as half the ratio between the applied shear stress and the resulting shear strain µ = σ xz 2e xz When µ is large, then the shear stress is large compared to the resulting shear strain, which implies rigidity. When µ is small, then the shear stress is small compared to the resulting shear strain, which implies low rigidity. For a fluid, µ = 0. The other Lamé parameter λ does not have a simple physical explanation λ = σ xx 2µe xx θ = σ xx σ xz e xx /e xz e xx + e yy + e zz

37 Elasticity Elastic constants Young s modulus E: This is defined as the ratio of extensional stress to the resulting extensional strain for a cylinder being pulled apart at both ends E = (3λ + 2µ)µ λ + µ Bulk modulus κ: This is defined as the ratio of hydrostatic pressure to the resulting volume change - it is a measure of the incompressibility of the material κ = p θ = λ µ where p = 1 3 σ kk.

38 Elasticity Elastic constants Poisson s ratio ν: Poisson s ratio is defined as the ratio of the lateral contraction of a cylinder (pulled on its ends) to its longitudinal extension ν = λ 2(λ + µ) E, ν and κ are often used in engineering because they are easily measured by simple experiments. In seismic wave propagation, λ, µ and κ are more natural constants.

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