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Unformatted text preview: 44 Elastic Fields Associated with Atomic Point Defects As discussed above, the symmetry of the site determines the nature of the displacement field in the surrounding lattice. Symmetric and nonsymmetric point defects in crystals. We use a continuum model to describe the elastic fields of these defects. The field of the symmetric defect can be modeled as three mutually perpendicular force dipoles in an isotropic elastic continuum. The nonsymmetric or tetragonal defect can be modeled as a single force dipole, also in an isotropic elastic continuum. Force dipoles in an elastic continuum as models for point defects Imagine applying a unit force at one particular place in an elastic continuum. The Tensor Green’s function for elastic displacement is a fundamental relation 45 for isotropic elasticity that allows the displacements everywhere else in the elastic continuum to be computed when a unit force is applied at one particular point. The point defect models are constructed by applying pairs of forces, or unit force dipoles to the continuum. The displacements (and strains and stresses) associated with a unit force dipole can then be computed using the Tensor Green’s function. For the three mutually perpendicular force dipoles the resulting elastic field is spherically symmetric. We will take this as a given, though it can be proved using the Tensor Green’s function mentioned above. Form for the displacement field for a spherically symmetric point defect (isotropic elasticity) We have argued that the displacement field has spherical symmetry. Thus we may immediately say that the displacement field must be of the form u r = Ar ¡ n u ¢ = u £ = , that is, the purely radial displacements fall off with increasing radial distance from the defect and the transverse displacements are zero. But what is the value of n ? On physical grounds it would seem that n = 1, n = 2, n = 3, ... are all viable possibilities. Hypothetical radial displacements for different values of n 46 All of these values of n look plausible. Below we show that the requirement of internal mechanical equilibrium requires n = 2 . This analysis will say that u r = Ar ¡ 2 . Consider a hemispherical shell of radius r and infinitesimal thickness ¡ r . Forces on a hemispherical shell of infinitesimal thickness. For mechanical equilibrium the forces on the shell in the up direction must exactly balance the forces in the down direction. For the up force think of a rigid fluid above the shell being pulled up by a piston in a rigid cylinder of radius r + ¡ r F up = ¡ rr + d ¡ rr dr ¢ r £ ¤ ¥ ¦ § ¨ © r + ¢ r ( ) 2 . For the down force we have F down = ¡ rr ¢ r 2 + ¡ T 2 ¢ r £ r where, again, we think of a piston pulling down in a cylinder of radius r and where the forces associated with the hoop stresses have been included. For mechanical equilibrium we must have 47 F up = F down ¡ rr + d ¡ rr dr ¢ r £ ¤ ¥ ¦ § ¨ © r + ¢ r ( ) 2 = ¡ rr © r 2 + ¡ T 2 © r ¢ r...
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This note was uploaded on 12/01/2011 for the course MS&E 206 taught by Professor Nix during the Spring '08 term at Stanford.
 Spring '08
 nix

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