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I'm solving an elastic homogenization problem and I'm having problems with mesh artifacts.

I would like to first give a brief summary of what I do: I have a system with inhomogeneous (but isotropic) elastic properties, and I want to compute the system-scale (i.e., effective) shear modulus. Specifically, I compute the shear moduluae $G_1$ and $G_2$ along 2 different shear orientations in the following way:

1 - Apply strain $\epsilon_{xy} = 1$ and measure the stress $\Sigma$, in which case $G_1 \equiv \Sigma_{xy}/2$
2 - Apply strain $\epsilon_{xx} = 1$, $\epsilon_{yy} = -1$ and measure $G_2 \equiv (\Sigma_{xx}-\Sigma_{yy})/4$

The elastic properties are locally random through the system. The system-scale values of $G_1$ and $G_2$ should (up to a small statistical fluctuation) be the same (i.e., the system should behave as isotropic).

I'm solving this problem with the finite element method, and I'm comparing different strategies, for which I obtain contradictory results. In strategy A (see (a) on the picture below) I create a triangular mesh, and to each element, I assign certain properties (i.e., the elastic properties are element-wise constant through the system, as represented by the colormap in the picture of the mesh). Then I compute the $G_1$ and $G_2$ for many repetitions of the random elastic properties to have good statistics. I do the same for different mesh sizes (in the graph, the x-axis corresponds to the number of FE in the mesh). We observe that in fact, $G_1$ and $G_2$ are the same as expected.

In strategy B, I use the same triangular mesh but I set the values of the elastic properties considering pairs of triangular elements, in such a way that the spatial distribution of elastic properties resembles that of a quadrilateral mesh (see (b) on the picture). Surprisingly, $G_1$ and $G_2$ are no longer the same, and the difference does not decrease with system size.

I would like to point out that if I use homogeneous properties, as expected $G_1$ and $G_2$ are always the same, for both meshes and for any number of FE.

enter image description here

What's my final goal? I would like to solve exactly the same problem described above but with a quadrilateral mesh. With such a mesh, I find the same problem as with the triangular mesh (b). However, as proved by mesh (a), the problem doesn't seem to be intrinsic to the FE mesh itself. Therefore it might be possible to solve the problem correctly (i.e., without anisotropy) also on a quadrilateral mesh by tweaking something.

My question is: what's the origin of the macroscopic elastic anisotropy with the triangular mesh (b)? Why is it not there for the case (a)? Is there a way to palliate mesh dependency problems with quadrilateral meshes in FEM?

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – nicoguaro Jul 9 at 0:58
  • $\begingroup$ Would you mind adding your boundary conditions? $\endgroup$ – nicoguaro Jul 9 at 0:59
  • $\begingroup$ Interesting results. Though I cannot explain the difference in your results in (a) and (b) cases. As @chenna-k says your mesh might have some built in anisotropy. Both of your meshes are structured. Perhaps you might want to try a completely unstructured mesh and see what you get. $\endgroup$ – Nach Jul 16 at 5:57
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I believe that the issue you are facing emanates from the type of triangular mesh you are using. This particular discretisation has in-built anisotropy; note the alignment of all of the longest edges is parallel to one of the diagonals of the square. You will observe a different behaviour in the results if you choose the alignment parallel to the other diagonal.

My suggestion is to use the quadrilateral elements. If that is not possible for some reasons, then use symmetric splitting of quadrilaterals into triangles, as shown in the Figure.

If you have no option but use triangles, then I recommend against linear triangular elements with displacement formulation. Such elements are very stiff. Please see my blog post for the details. Either use an advanced formulation with linear triangles or choose quadratic triangles with displacement formulation.

enter image description here

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