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# contact problem with abaqus

## contact problem with abaqus

(OP)
Hi all,
I am studying a contact problem between two plates of steel under abaqus.
I would like to know why results are better with using hexaedric elements than tetraedric and why linear elements are better for contact than quadratic elements.
Then, to locate where crack could be, what should I show, Von Mises stress, Max principal stress...?
your answer will help me to explain my problem and my results.
nick
Replies continue below

### RE: contact problem with abaqus

1) Why hexahedral elements are better than tetrahedral:
This has to do with formulation of the two elements.  This is apart from contact--for the same order of interpolation the hexahedral elements perform better than tetrahedrals.

2) Why are first-order elements better in contact than 2nd order:
This is another formulational issue.  Without getting deep into formulational issues, imagine a book laying on a desk.  Away from the edge of the book, the contact will be a constant-pressure situation.  This plays itself out as a flat-line linear force transfer.  The polynomial element cannot handle this, due to how it is formulated.  Instead one gets a parabolic response. For 2nd order quadrilateral faces, this results in "negative" force response at the corner nodes.  For triangular faces, it results in zero force at teh corners.  The effective plot looks like a checkerboard, instead of a constant contact pressure.

3) To locate the crack . . .
It depends on the material and what assumptions one is making.  Some people use a particular stress or strain measurment, others use energies.  This answer is very problem-dependent.

Hope this helps.

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