# Homogenization of a layered material

I'm trying to find the homogenized 3D linear elastic stiffness matrix of a layered block of material, i.e., a transversely isotropic material. The material will be made of $N$ repeating layers, which are all linear elastic isotropic materials with differing Young's moduli and Poisson ratios. Suppose the axis of symmetry of the layered material is the 33-axis.

If we apply stress in the 11-direction (or, equivalently, the 22-direction), we can intuitively see that all the materials show equal strain in that same direction, and Voigt averaging seems like a good candidate. However, when looking at the 33-direction, we find that all materials exhibit equal stress, and we would go for Reuss averaging. For now, I have averaged the material in each direction separately, and neglected any influence of differing Poisson ratios (I also wrote out the constitutive relations for two or three layers and solved the entire system of equations, but that got ugly pretty quickly, and I could not distill the underlying principle).

Is there any way to directly average several linear elastic isotropic materials, arranged in layers?

Your intuition about Reuss averaging in one direction and Voigt averaging in the other directions is accurate. This generalizes rather well by noting which components of stress and strain are continuous in the layering direction. You can see that $\sigma_{13,23,33}$ as well as $e_{11,22,12}$ should be continuous and therefore constant across the material. Then instead of writing Hook's law (symbolically) as $$\sigma_{11,22,33,32,13,21} = A_{ijkl} e_{11,22,33,32,13,21}$$ write it as $$(\sigma_{11,22,21};e_{33,32,13}) = B_{ijkl} (e_{11,22,21};\sigma_{33,32,13})$$ where $B$ can be deduced by partially inverting $A$. Remark how all constant components are to the right. Thanks to that specific arrangement, the average of a product is now the product of averages and we have $$\langle\sigma_{11,22,21};e_{33,32,13}\rangle = \langle B_{ijkl} \rangle \langle e_{11,22,21};\sigma_{33,32,13}\rangle.$$ Now the effective parameters $A^*$ can be deduced from $\langle B \rangle$ the same way $B$ was deduced from $A$.