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I have two different mesh files (both are .inp files obtained from Abaqus) that represent the exact same geometries with the same boundary conditions, etc. The only difference is that one of them is meshed in a structured fashion and the other one is meshed in an unstructured way.

Now here is my question, how do I find out which file contains the structured mesh and which one contains the unstructured one? Of course, I mean apart from visualizing the files and trying to zoom in on the meshes and see if they look structured or unstructured.

What should I do in such situation in general? What is the "scientific" way to figure out the answer and avoid visualizing the mesh?

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  • $\begingroup$ Have you checked, wether there are some flags in that particuluar file format (.inp)? This is just a guess, but it could be that there are some descriptor flags in these mesh files that give you that information straight away. If you are left with only the raw information, it will be a lot trickier though. $\endgroup$ – MPIchael Feb 3 at 16:47
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There are two different types of meshes that are commonly termed "structured":

  • the points are placed on an equispaced grid; and

  • the elements have the same connectivity.

Some people might call any combination of these two a "structured mesh".

In Abaqus, you can define a set of points on a grid with the keyword

*NGEN

and a regular connectivity with

*ELGEN

So, you can check if any (or both) of these two appear in your input file.

If you have a regular geometry but are definining it using an unstructured format, you would need to analyze the elements. For hexahedra, for example, you can compute the Jacobian and it should be a diagonal matrix.

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If it's a structured grid of hexahedra, then each node should have the same valence (number of adjacent hexes), eight. So I would scan through the list of hexes, and make sure that each node index appears exactly eight times.

You can make structured grids from other types of elements, like by repeating a clump of tetrahedra over and over. In that case, some nodes actually have different valences so this might not be right approach.

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