where index notation is the shorthand for dealing with tensors and vectors; a variable with a single subscript is a vector a = a i

Transcription

1 1 Brief Review of Elasticity (Copyright 2009, David T Sandwell) This is a very brief review of the elasticity theory needed to understand the principles of stress, strain, and flexure in Geodynamics [Turcotte and Schubert, 2002] This review assumes that you have already taken a class in continuum mechanics One difference from T&S is that we follow the sign convention used by seismologists and engineers where extensional strain and stress is positive Stress Stress is a force acting on an area is measured in Newtons per meter squared (N m 2 ) which corresponds to a Pascal unit (Pa) The following diagram shows a cube of solid material Each face of the cube has three components of stress so there are 9 possible components of the stress tensor We will consider only the symmetric part of the stress tensor so only 6 of these components are independent The antisymmetric part of the tensor represents a torque In Cartesian coordinates the stress tensor is given by σ xx σ ij = σ zz where index notation is the shorthand for dealing with tensors and vectors; a variable with a single subscript is a vector a = a i, a variable with two subscripts is a tensor σ = σ ij, and a repeated index indicates summation over the spatial coordinates For example the pressure is given by P = σ ii / 3 In addition, a comma preceding a subscript means differentiation with respect to that variable a = a i, j or for example a x,y = a x y

2 2 Strain Strain is change in length over the original length so it is a dimensionless variable and we will assume strains are small (<< 10-3 ) Let the displacement vector field inside of a solid body be given by u = u i = [ u x u y u z ] The gradient of this vector is a tensor u = u i, j This tensor is commonly decomposed into a symmetric tensor (strain) and an antisymmetric tensor (rotation) u i, j = 1 u i + u j 2 x j x i + 1 u i u j 2 x j x i We will not consider the rotation tensor further but the strain tensor is given by ε ij 2 u i, j + u j,i ) Stress vs strain If one assumes the material has an isotropic and linear response then the relationship between stress and strain is given by σ ij = λδ ij ε kk + 2µε ij where δ ij is equal to 0 except when i=j and then it is equal to 1 The Lame constants λ and µ define the elastic properties The shear modulus µ (or G in the engineering literature) relates the shear stress to shear strain on a component by component basis = 2µε xy = µ u x y + u y x Invariants and principal stress This general relation between stress and strain tensors is rather involved and it is difficult to invert this relationship to develop a relationship between strain and stress One means of simplifying this relationship is to find a co-ordinate system rotation that will cause the stress and strain tensors to be diagonal Let R be a rotation matrix such that R t R = I is the identity matrix There are three properties (invariants) of the stress tensor that do not change under co-ordinate rotation The invariants are found by first developing the characteristic equation from the determinant of the following equation

3 3 σ xx γ γ γ = 0 which becomes γ 3 Iγ 2 + IIγ III = 0 where the stress invariants are I = σ ii II 2 σ iiσ jj σ ij σ ij ) = σ xx + + σ xx σ 2 xy σ 2 2 yz III = σ ij the trace I, the sum of minors II, and the determinant of the stress tensor III The first invariant is related to the mean normal stress or pressure P = σ ii / 3 The second invariant is related to shear stress and thus is commonly used as the Von Mises failure criteria We will not consider the third invariant further Real symmetric matrices have real eigenvalues, orthogonal eigenvectors, and can be diagonalized This implies that there always exists some principal coordinate system where the the shear stresses are zero on planes orthogonal to the coordinate axes and where the normal stresses act along the principal axes directions (the eigenvectors) form the rotation matrix R The eigenvalues form the principal stress tensor σ p = σ σ 3 = R t σr where σ 2 σ 3 The principal stress system is important in geophysics and geology Due to the presence of the free surface, the stress field close to the Earth's surface is expected to have one principal stress vertical and hence two horizontal principal stresses Also in the earth we sometimes subtract the pressure from the stress tensor In this case it is called deviatoric stress In the principal stress system the pressure and maximum shear stress are given by

4 4 P = 1 ( 3 σ + σ + 2 3) τ 2 σ 3) Principal stress and strain The stress versus strain relation is far simpler in the principal co-ordinate system σ 2 σ 3 = λ + 2µ λ λ λ λ + 2µ λ λ λ λ + 2µ ε 2 ε 3 where, ε 2, and ε 3 are the principal strains Next we can use this relationship to develop three important parameters, Poisson s ratio ν, Young s modulus E, and bulk modulus K First consider the case of uniaxial stress where σ 2 = σ 3 = 0 This represents application of an end load to an elastic beam fastened to a wall The second equation for σ 2 is 0 = λ + ( λ + 2µ )ε 2 + λε 3 Because of symmetry we know ε 2 = ε 3 so we arrive at a relationship between ε 2 and ε 2 = λ 2( λ + µ ) ε = ν 1 where ν is Poisson s ratio Next we can use this relationship between strains in the first equation to provide a relationship between and = ( λ + 2µ ) + λ 2 λ + µ ( )( λ + µ ) λ 2 = λ + 2µ λ + µ ( ) λ + µ µ 3λ + 2µ = = E

5 5 where E is Young s modulus ( ) / 3 is related to a change in volume ΔV = ( + ε 2 + ε 3 ) Using the stress- Next we consider the case of uniform pressure ΔP = + σ 2 + σ 3 strain relation we find In this case, the change in pressure ΔP = λ µ ΔV ΔP = KΔV where K is the bulk modulus One can invert this stress vs strain relationship to obtain a strain vs stress relationship We ll also assume that the principal co-ordinates are aligned with the x-, y-, and z- axes ε xx ε yy ε zz = 1 E 1 ν ν ν 1 ν ν ν 1 σ xx Now we have arrived at equations 3-4, 3-5, and 3-6 in T&S Before moving onto the flexure problem we consider the case of a thin elastic plate Thin plate means that there are no variations in the vertical displacement field as a function of depth in the plate so we can make the approximation = 0 Under this approximation we have the following ε xx E σ vσ xx yy ) ε yy E σ v xx ) ε zz = ν E ( σ + σ xx yy )