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One can perform the same analysis for the bulk modulus. First it is important to recall that for an isotropic linear elastic material, with bulk modulus K, the following equation holds:
where Tr indicates a trace (sum over the diagonal elements)
of the stress or strain tensor. We use the same geometrical setup as in
Figure 4, again with spherical polar
coordinates.
The displacement at r → ∞ is ε0 r, a pure
compression. Because of the radial symmetry,
the only non-zero displacement throughout the structure is u = u(r), the radial component of the displacement vector. There are no shear
strains in this case either, and the principal strains are
,
, and
.
The radial displacement u can be expressed as:
where An and Bn are arbitrary constants in the nth shell. In order to satisfy the boundary condition at r → ∞, AN + 1 = ε0. In order to have a finite displacement at the center of the aggregate, B1 = 0. Using eq. (24) for the displacements, and the definitions of the various strains given above, the radial stress in the nth layer is then:
Between each layer, the displacement and the normal or radial stress must be continuous. When N layers are present, this gives 2N boundary conditions, plus the two boundary conditions at the origin and at infinity. These 2N + 2 boundary conditions are enough to determine the 2N + 2 unknown coefficients. After a little rearrangement, the boundary conditions between the nth and (n + 1)th layers can be written in matrix form as:
where Pn is a 2 x 2 matrix, given by:
Following the same steps as for the electrical case, we find that:
All the layer displacements can now be determined using the boundary conditions and the two known coefficients. To get the actual change in bulk modulus caused by the presence of the aggregates and the gradient of properties around it, we need to again write out the averages over the phases as was done for the electrical/diffusive case. Again, in general, we have:
Using equations (24)-(28) above, it can be shown that:
Putting eqs. (29)-(32) together gives the final result that:
The effective shear modulus can be determined in exactly the same way, by applying a uniform shear strain at infinity, except that there are now four unknown coefficients that determine the displacements in each layer, because we no longer have radial symmetry due to the applied shear strain. The matrix that connects the coefficients in one layer to those in the next is then a 4 x 4 matrix. The product of all these matrices produces a matrix that connects the coefficients in the outer layer to the coefficients in the inner layer. One coefficient in the outer layer and two in the inner layer are forced to zero, however, because of having a finite displacement at r = 0 and and a finite strain at r → ∞, and one coefficient in the outer layer is determined by the applied strain at r → ∞. Therefore the matrix equation is transformed to only being four equations in four unknowns, rather than being four equations in eight unknowns. Further details of the shear modulus determination can be found in Ref. [11]. We emphasize that the pattern of solution is the same as for the bulk modulus, but just requiring more algebra. The explicit solution of the shear modulus dilute limit, along with computer programs for the implementation of the equations, will be given in Ref. [15]. Reference [15] will also contain more details on all the dilute limit work described in this paper, as well as discusssion of many other exact results from composite theory. This manual will be available online in the near future. It will describe how to use several finite element and finite difference computer codes for determining electrical and elastic properties of random systems. The manual will also describe the operation of computer codes that implement the dilute limits described in this paper.