Rigorous derivation of the strain tensor<\/h2>\n
When I was first exposed to elasticity, it was from Chapter 3 of Mathematical Methods, 3rd<\/sup> edition<\/em> by Arfken; and, in fact, I used that text as a guide for the material in post 7<\/a>.\u00a0 However, a brief consultation with more specialized texts will often yield a strain tensor that is a bit more complex than the one presented in either of those two sources.\u00a0 A more complete derivation, which is patterned after the derivation here<\/a>, supplemented with my own insights, is as follows.<\/p>\n Let the position of any point at any stage of the deformation be defined by $${\\vec r}_t = \\vec \\phi (\\vec r, t)$$, \u00a0where $$\\vec r$$ is the initial position of the point and $$t$$ measures the extent of deformation.\u00a0 The function $$\\vec \\phi(\\vec r,t)$$ is called the flow mapping.\u00a0 The parameter $$t$$ parameterizes the deformation, labeling the successive stages.\u00a0 It doesn\u2019t necessarily need to be regarded as the time but it is often convenient to treat it as such and it will used that way below.\u00a0 The requirement that $$\\vec r$$ is the original undeformed position imposes the obvious boundary: $$\\vec \\phi (\\vec r, 0) = \\vec r$$.<\/p>\n Now focus on two points $${\\mathcal P}$$ and $${\\mathcal Q}$$ before and after the derivation. \u00a0\u00a0Their positions before the deformation are:<\/p>\n \\[ \\vec {\\mathcal P}(0) = \\vec \\phi(\\vec r,0) = \\vec r \\; \\]<\/p>\n and<\/p>\n \\[ \\vec {\\mathcal Q}(0) = \\vec \\phi(\\vec r + d \\vec r,0) = \\vec r + d \\vec r \\; . \\]<\/p>\n After the deformation, the same two points are now located at<\/p>\n \\[ \\vec {\\mathcal P}(t) = \\vec \\phi(\\vec r, t) \\; \\]<\/p>\n and<\/p>\n \\[ \\vec {\\mathcal Q}(t) = \\vec \\phi(\\vec r + d \\vec r,t) = \\vec \\phi(\\vec r,t) + \\frac{\\partial \\vec \\phi(\\vec r, \\epsilon) }{\\partial \\vec r} d \\vec r \\; , \\]<\/p>\n where the assumptions that $$d \\vec r$$ is small in magnitude and that $$\\vec \\phi$$ is differentiable have been made.<\/p>\n Now measure the distance between the points $${\\mathcal P}$$ and $${\\mathcal Q}$$.\u00a0 Before the deformation, that distance squared is given by<\/p>\n \\[ ds^2 = \\left\\Vert \\vec {\\mathcal P}(0) – \\vec {\\mathcal Q}(0) \\right \\Vert^2 = d \\vec r \\cdot d \\vec r = d r_i d r_i \\; . \\]<\/p>\n After the deformation, the distance squared between those points, now moved to their new position, is given by<\/p>\n \\[ ds_{t}^2 = \\left\\Vert \\vec {\\mathcal P}(t) – \\vec {\\mathcal Q}(t) \\right \\Vert^2 = \\frac{\\partial \\phi_i}{\\partial r_j} d r_j \\frac{\\partial \\phi_i}{\\partial r_k} d r_k \\; . \\]<\/p>\n Calculate the difference between these two lengths<\/p>\n \\[ ds_{t}^2 – ds^2 = \\frac{\\partial \\phi_i}{\\partial r_j} d r_j \\frac{\\partial \\phi_i}{\\partial r_k} d r_k – d r_j d r_k \\delta_{jk} = \\left( \\frac{\\partial \\phi_i}{\\partial r_j} \\frac{\\partial \\phi_i}{\\partial r_k} – \\delta_{jk} \\right) dr_j dr_k\\; ,\\]<\/p>\n where the last term has been modified a bit by the inclusion of a Kronecker delta.<\/p>\n We now define the Green-Lagrange strain tensor by the relationship<\/p>\n \\[ 2 \\epsilon_{jk} = \\frac{\\partial \\phi_i}{\\partial r_j} \\frac{\\partial \\phi_i}{\\partial r_k} – \\delta_{jk} \\; .\\]<\/p>\n The strain vector $$\\vec u$$ measures the difference in position between where a particle originally was located and where it is located at time $$t$$<\/p>\n \\[ \\vec u(t) = \\vec \\phi(\\vec r,t) – \\vec r \\; . \\]<\/p>\n Spatial derivatives of $$\\vec \\phi$$ are related to spatial derivatives of the deformation vector $$u$$ via<\/p>\n \\[ \\frac{\\partial \\phi_i(\\vec r,t)}{\\partial r_j} = \\delta_{ij} + \\frac{\\partial u_i}{\\partial r_j} \\; . \\]<\/p>\n Substituting this expression into the Green-Lagrange strain tensor gives<\/p>\n \\[ 2 \\epsilon_{jk} = \\left( \\delta_{ij} + \\frac{\\partial u_i}{\\partial r_j} \\right) \\left( \\delta_{ik} + \\frac{\\partial u_i}{\\partial r_k} \\right) – \\delta_{jk} \\; . \\]<\/p>\n Expanding and simplifying yields<\/p>\n \\[ \\epsilon_{jk} = \\frac{1}{2} \\left( \\frac{\\partial u_k}{\\partial r_j} + \\frac{\\partial u_j}{\\partial r_k} + \\frac{\\partial u_i}{\\partial r_j}\\frac{\\partial u_i}{\\partial r_k} \\right) \\; , \\]<\/p>\n which is the same as the earlier derivation except for the nonlinear term that can often be dropped.<\/p>\n The same notions can be used to derive the equations of motion that led to the wave equation analyzed in the last post.\u00a0 Consider a cube of the continuum centered at $$\\vec r$$.<\/p>\n The force on any face is given by $$f_i = T_{ij} n_j Area$$. \u00a0Start first with the forces on the two faces perpendicular to the $$x$$-axis.\u00a0 The center of the rightmost face is located at $$\\vec r + (\\delta x\/2) \\hat \\imath$$ and it is at this point that the stress is evaluated.\u00a0 Likewise, the center of the leftmost face lies at $$\\vec r – (\\delta x\/2) \\hat \\imath$$.\u00a0 For notational convenience, functions evaluated on the rightmost face will be denoted with the argument $$x_{+}$$ and those evaluated on the leftmost face with $$x_{-}$$.\u00a0 In this notation, the components of the forces on these faces are:<\/p>\n \\[ \\left[ \\begin{array}{c} f_x \\\\ f_y \\\\ f_z \\end{array} \\right] = \\left[ \\begin{array}{c} T_{xx}(x_+) – T_{xx}(x_-) \\\\ T_{yx}(x_+) – T_{yx}(x_-) \\\\ T_{zx}(x_+) – T_{zx}(x_-) \\end{array} \\right] dy dz = \\left[ \\begin{array}{c} T_{xx,x}(\\vec r) \\\\ T_{yx,x}(\\vec r) \\\\ T_{zx,x}(\\vec r) \\end{array} \\right] dx dy dz \\; , \\]<\/p>\n where the last equalities result by taking a Taylor\u2019s expansion of the finite difference and keeping terms of only first order ($$T_{ij,k} = \\partial T_{ij} \/ \\partial r_k$$).<\/p>\n Similar manipulations give the expressions<\/p>\n \\[ \\left[ \\begin{array}{c} f_x \\\\ f_y \\\\ f_z \\end{array} \\right] = \\left[ \\begin{array}{c} T_{xy}(y_{+}) – T_{xy}(y_-) \\\\ T_{yy}(y_+) – T_{yy}(y_-) \\\\ T_{zy}(y_+) – T_{zy}(y_-) \\end{array} \\right] dx dz = \\left[ \\begin{array}{c} T_{xy,y}(\\vec r) \\\\ T_{yy,y}(\\vec r) \\\\ T_{zy,y}(\\vec r) \\end{array} \\right] dx dy dz \\; \\]<\/p>\n and<\/p>\n \\[ \\left[ \\begin{array}{c} f_x \\\\ f_y \\\\ f_z \\end{array} \\right] = \\left[ \\begin{array}{c} T_{xz}(z_+) – T_{xz}(z_-) \\\\ T_{yz}(z_+) – T_{yz}(z_-) \\\\ T_{zz}(z_+) – T_{zz}(z_-) \\end{array} \\right] dx dy = \\left[ \\begin{array}{c} T_{xz,z}(\\vec r) \\\\ T_{yz,z}(\\vec r) \\\\ T_{zz,z}(\\vec r) \\end{array} \\right] dx dy dz \\; , \\]<\/p>\n for the components of the force generated on the $$y$$- and $$z$$-faces, respectively.<\/p>\n Combining these results and expressing the sums in index notation gives the final, compact expression for the force as<\/p>\n \\[ f_i = T_{ij,j} dx dy dz \\; .\\]<\/p>\n Note that the mass of the continuum contained in the cube is $$\\rho \\, dx dy dz$$, where $$\\rho$$ is the mass density.\u00a0 Newton\u2019s second law for the cube then relates the acceleration of the cube, which is expressed in terms of the second time-derivative of the flow mapping to the cube\u2019s mass and the forces acting on the cube as<\/p>\n \\[ (\\rho \\, dx dy dz) \\frac{\\partial^2 \\phi(\\vec r,t)}{\\partial t^2} = f_i \\; . \\]<\/p>\n Using the relation between the flow mapping and the deformation vector and between the force and the stress tensor gives<\/p>\n \\[ \\rho \\, \\partial^2_t u_i = T_{ij,j} \\; , \\]<\/p>\n which is precisely the starting expression for the last post.<\/p>\n This final section derives the relationships to express the elastic moduli derived in post 7 to the Lame constants, which were introduced as a shorthand for the wave equation in an isotropic medium.\u00a0 The Lame constants took the form<\/p>\n \\[ \\lambda = \\frac{\\nu E}{(1+\\nu)(1-2\\nu)} \\; \\]<\/p>\n and<\/p>\n \\[ \\mu \\frac{E}{2(1+\\nu)} \\; ,\\]<\/p>\n where $$E$$ is Young\u2019s modulus and $$\\nu$$ is Poisson\u2019s ratio.<\/p>\n It is a matter of some simple algebra to express Young\u2019s modulus and Poisson\u2019s ration in terms of $$\\lambda$$ and $$\\mu$$ as follows.\u00a0 Manipulating the second equation gives\u00a0 $$E = 2 (1+\\nu) \\mu$$.\u00a0 Substituting this into the first equation gives $$\\lambda = 2 \\nu \\mu \/ (1-2\\nu)$$ from which one gets via simplification<\/p>\n \\[ \\nu = \\frac{\\lambda}{2 (\\mu + \\lambda) } \\; . \\]<\/p>\n Back-substituting gives<\/p>\n \\[ E = \\mu \\frac{2\\mu + 3 \\lambda}{\\mu + \\lambda} \\; \\]<\/p>\n for Young\u2019s modulus.<\/p>\n To relate the shear and bulk moduli to the Lame constants, return to the generalized Hooke\u2019s law<\/p>\n \\[ T_{ij} = 2 \\mu \\epsilon_{ij} + \\lambda Tr(\\epsilon) \\delta_{ij} \\; . \\]<\/p>\n In the case of an isolated shear along the $$x$$-direction that grows along the $$y$$-direction, the strain is expressed as $$\\partial_y u_x$$ and the only non-zero terms in the strain tensor are $$ \\epsilon_{xy} = \\epsilon_{yx} = 1\/2 \\, \\partial_y u_x$$.\u00a0 Thus,<\/p>\n \\[ T_{xy} = 2 \\mu \\epsilon_{xy} = 2 \\mu \\frac{1}{2} \\partial_y u_x = \\mu \\partial_y u_x \\; .\\]<\/p>\nRigorous derivation of the equations of motion<\/h2>\n
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<\/a><\/p>\nRelation between the Lame constants and the elastic moduli<\/h2>\n