Hi David,
I'm neither an expert, nor do I know the literature well, but looking on
your pictures, I think, the situations you are studying are
geometrically anisotropic. Just plot the distribution of angles the
faces of your inhomogenities make with the x-axis. For the quad-case,
you'll get two discrete peaks at 0 and 90 degree. for the triangular
case, you get 0, 45 and 90 degree. So, from this, the results do not
seem surprising to me (just consider the extreme case of cracks - if you
have them only at 0 and 90 degree oriented, this is unlikely to be
isotropic). The fact that you have randomly assigned elastic properties
won't help to fix that.
A few examples (in 2d):
(1) equidistant circular inclusions in a matrix (matrix and inclusions
two different isotropic linearly elastic materials) -> this should be
isotropic if mesh size h->0
(2) equidistant square inclusions in a matrix, all aligned with x-axis
(matrix and inclusions two different isotropic linearly elastic
materials) -> probably anisotropic if mesh size h->0
(3) equidistant square inclusions in a matrix, random orientation of
inclusions (matrix and inclusions two different isotropic linearly
elastic materials) -> should be isotropic if mesh size h->0 and number
of orientations->infinity
(4) equidistant square inclusions in a matrix, random orientation of
inclusions (random isotropic linearly elastic materials) -> should be
isotropic if mesh size h->0 and number of orientations->infinity
Also to consider: In 3d, there are 21 elastic constants for a linearly
elastic material. In a mathematical 2d scenario, it should be 6. This
suggests that, in example (3), it is not strictly necessary to have
random orientation. Rather, a few (equally spaced) discrete orientations
might be good enough. If that's the case, how many does one need? I'm
betting on 6, not sure though. Alternatively, one could replace the
square inclusions in (2) by regular polygons and ask how many vertices
the polygon needs for isotropy. Again, I'm betting on 6.
Related: Are there crystal structures with such a high degree of
symmetry, that they are elastically isotropic? For dielectric
properties, a cubic crystal is good enough already. But the dielectric
tensor is rank 2 and the elastic one rank 4. So you'll need much more
symmetry in the crystal; and considering that a crystal can have 6-fold
rotational symmetry at most that might be impossible.
What I'm just noticing: Hexagonal crystals are elastically isotropic
perpendicular to the hexagonal axis. So, my bet on 6 might be good. And
it might explain your observation that the triangular elements are
relatively isotropic (though maybe not perfectly).
I hope that gives you some input. If you have definitive answers to any
of the questions, I'm curious.
Regards,
Sebastian
Am 11.07.20 um 00:12 schrieb David F:
I have made a somewhat extensive study on his issue and prepared some
plots that will hopefully answer your questions, and also includes
Bruno's suggestion about distorting the mesh. The basic setup is: I
sheared the mesh along different orientations (see x-axis on the
plots) and measured the shear modulus (y-axis). I have repeated the
random process of setting the elastic properties many times to have
good statistics (see errorbars on the plots). Each element has an
isotropic stiffness tensor with a Poisson ratio of 1/3 and a shear
modulus which is exponentially distributed with an average of 10. I
use linear shape functions unless otherwise stated. If the picture are
not big enough, you can find them in the links beneath them.
*1) I change the resolution. *By this I don't mean just a mesh with a
bigger number of elements, but importantly each inhomogeneities is
represented by a bigger number of elements. Therefore, we solve
problems with exactly the same physical domain but with different
resolution. In the legend, n means the resolution of the
inhomogeneities. For n=1 each inhomogeneity is described by 1 FE. For
n=2, by 2^2, and for n=4 by 4^2. We can see that for shearing with
angle 0 (see pictures on the bottom for clarity) the shear modulus is
minimum, while it is maximum for 45 degrees, when the principal axes
are aligned with the mesh. The magnitude of the anisotropy is the
difference between the maximum and the minimum. The difference
decreases by increasing the resolution, but actually the relative
difference is very similar, and it seems that by just increasing the
resolution this problem won't go away. Finally, I have distorted the
mesh, which doesn't change the behavior at all.
summary.png
Link to the picture <https://ibb.co/bFps0Vs>
*2) I change the order of the shape functions.* I use the original set
up, in which each inhomogeneity is represented by 1 element. We see
that increasing the shape function order has a somewhat similar effect
as increasing the resolution of the inhomogeneity (expected, since in
both cases we are increasing the number of dofs of each
inhomogeneity). Therefore, increasing the order of the shape functions
doesn't seem to be enough to fix the issue.
plots_shape_functions.png
Link to the picture <https://ibb.co/0X2W5M2>
*3) I try different types of mesh.* In this case, I compare the
solution provided by dealII (i.e. quadrilateral mesh) with the
solution obtained with a triangular mesh using the python FENICS
package. Lastly, I solve using the triangular mesh but allocating the
inhomogeneities in such a way that even if the mesh is triangular, the
structure of the inhomogeneities looks quadrilateral. It seems that a
triangular mesh, even it is structured, is able to provide an
isotropic solution. Interestingly, _the same_ mesh fails and behaves
like a quadrilateral mesh if the structure depicted by the elastic
properties look like those of the quadrilateral mesh.
summary3.png
Link to the picture <https://ibb.co/PF5z8t3>
In summary, form all the tings that I tried, the only solution is to
use a triangular mesh. Very interestingly, by arranging the elastic
properties in a quad-like manner we can force the triangular mesh to
give the wrong result of a quadrilateral mesh. Therefore, do you think
it is possible to do the opposite, i.e., to make dealII's
quadrilateral mesh "behave more triangular-like" by playing with
elastic properties, quadrature points etc.?
Do you know the reason of the anisotropy in the quadrilateral mesh, or
if there is some literature about this? I couldn't find anything.
Best,
David.
On Friday, 10 July 2020 17:38:58 UTC+2, Wolfgang Bangerth wrote:
On 7/10/20 9:15 AM, David F wrote:
> I have a 2D system for which I create the stiffness tensor of an
isotropic
> material, but for each finite element I create it with a
different shear
> modulus. The shear modulus is random for each element (I use an
exponential
> distribution, but any distribution leads to the same behavior as
long as the
> std is high), with no structure such as layers or anything else.
In this case,
> the system should clearly be macroscopically isotropic (up to
statistical
> fluctuations due to the random properties) for symmetry reasons.
At least in the limit h->0 I agree. For finite mesh sizes, I would
expect that
the material has a degree of anisotropy that goes to zero as you
make the mesh
smaller. It is true that the axes of anisotropy should be oriented
in random
ways for different realizations of the same experiment on the same
mesh. When
you do your computations, have you checked (for different
realizations of the
random process):
(i) whether the orientation of anisotropy is always the same, and
always
related to the principal directions of the mesh?
(ii) how the magnitude of anisotropy behaves as you refine the mesh?
Best
W.
--
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www:
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