4
$\begingroup$

Do the faces of tri-linear hex elements have to be planar? Three nodes define a plane. If the fourth node does not lie on the plane, then the nodes are not planar and the face is not plane. In general, the geometry of element may not be convex. Will this cause problems in mapping the element to the unit cube? If I recall correctly, for non convex elements in 2D the mapping from global domain to the parent domain is not guaranteed to exist or be continuous. Will similar problems occur in 3D?

A follow up question: If the faces of tri-linear hex elements have to be planar, are they guaranteed to remain plane as the solution progresses when solving large deformation elasticity problems?

$\endgroup$
1
  • $\begingroup$ To answer my second question: Large deformation elasticity does not guarantee that planes remain plane. To see this consider a unit cube hex element. Constrain 7 nodes in all three directions. The 8th node is part of three planes. We can displace the the 8th node such that it does not lie on the three planes anymore. $\endgroup$
    – Nachiket
    Nov 9 '20 at 11:01
4
$\begingroup$

You are correct: In general, the faces of hex cells are not planar. In fact, the set of cells you could generate with only planar faces is relatively small and not enough to create useful 3d meshes.

The mapping from the reference cell always exists. For any set of 8 vertices in 3d, you can write down a mapping from the reference cell. What may not exist is a unique inverse, but you have to have pretty crazy cells (not just non-convex) for the inverse to not exist. What you do get into trouble with is if you have vertices that lead to re-entrant corners or that degenerate the hexahedron -- in the same way as you can have a reentrant vertex in 2d or degenerate a quadrilateral to a triangle: In those cases the determinant of the gradient of the mapping becomes zero or even negative, and that has dramatic consequences for whether your finite element approximation converges to the exact solution at all.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.