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The four-noded bi-linear rectangle element, which sometimes goes under the name Melosh element, is non-conforming unless the element sides are aligned. Out of curiosity I have implemented this element for two-dimensional elasticity problems. The figure shows a mesh with three such elements. The element sides are not all aligned, so the displacements are expected to be discontinuous.

enter image description here

The picture below illustrates the displacement-field for cantilever-type boundary conditions (left side is fixed, vertical forces on the right).

enter image description here

Although there is a gap between element 1 and 2 and an overlap between element 2 and 3, the overall field looks fine, and comparing with results for isoparametric quadrilaterals shows very similar maximum displacement for example.

This result, assuming there's no mistake in the implementation, suggests that it might be possible to give some form of convergence proof for the Melosh element also in case of non-aligned meshes (the unphysical gaps and overlaps seem to go zero as I refine the mesh). The question is thus whether such a proof can be found somewhere?

I should add that my question is motivated by pure curiosity; I'm not aware of a practical application for this element.

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  • $\begingroup$ I don't understand what are the degrees-of-freedom in this element. Can you give more details? What causes these gaps? It doesn't look bilinear to me. $\endgroup$ – knl Mar 9 at 10:20
  • $\begingroup$ The degrees-of-freedom are nodal values of the displacements. The ansatz is $u(x,y) = c_{0}+c_{1}x+c_{2}y+c_{3}xy$. The gaps are caused by the discontinuity of the displacement field across shared element edges. This element is in fact considered in most standard texts (Bathe, Zienkewiz etc.), but always with the caveat that it can only be used with axis-aligned meshes $\endgroup$ – Aage Mar 9 at 10:30
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    $\begingroup$ I see, so you're using a bilinear basis directly on the global element without using a bilinear reference mapping? I have never encountered this approach before. There can be some caveats, such as unstable configurations, but it's possible that this is a completely valid non-conforming element for solving the elasticity system. $\endgroup$ – knl Mar 9 at 11:47
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    $\begingroup$ Let me add that there are similar non-conforming elements that are also sometimes used for incompressible elasticity such as Crouzeix-Raviart element (see e.g. math.aalto.fi/~rstenber/Publications/Kouhia-Stenberg.pdf). $\endgroup$ – knl Mar 9 at 11:50
  • $\begingroup$ Yes, I'm using a "bilinear basis directly on the global element withoutt using a bilinear reference mapping". I've never seen this either. I guess it's no so useful because I can't think of any case where one would not just use the isoparametric bi-linear element instead. Perhaps it would be advantageous in some applications, such as those you mention for the C-R element. Thanks for the reference anyway! $\endgroup$ – Aage Mar 9 at 12:10
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Basis functions on quad elements must be constructed by mapping to a reference element using a bilinear map. Then the element-wise approximations will be continuous across common faces. This is because the restriction of the solution to any face will be a polynomial of degree 1 in one variable, so it is completely determined by the two nodal values on that face.

For the form you have taken, check if the restriction of the solution to any face will be a polynomial of degree 1 in one variable. E.g., on the side 34, whose equation is say $y = a + b x$, $b \ne 0$, the solution is $$ u = c_0 + c_1 x + c_2 (a + b x) + c_3 x (a + b x) $$ which is quadratic in $x$ unless $c_3=0$ always. A quadratic polynomial in one variable cannot be completely determined by the two nodal values on that face and will depend on the other two nodal values also. So your local polynomial approximations will not patch up in a continuous way over the whole domain.

Of course as you commented, this problem disappears if the elements are rectangles, aligned with the coordinate axes.

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    $\begingroup$ Thanks for the answer! However, I don't agree that basis function "must" be constructed as you describe. This way the displacement will be continuous as you explain, but there are elements with discontinuous displacements, such as the C-R mentioned by knl above, for which convergence can be proved (though perhaps with additional restrictions on boundary conditions etc). $\endgroup$ – Aage Mar 9 at 12:15
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    $\begingroup$ That's correct -- shape functions are often constructed by mapping from a reference cell, but that's not necessary. In fact, if one uses discontinuous Galerkin methods, it's often easier to define shape functions directly in real space, rather than on the reference cell. $\endgroup$ – Wolfgang Bangerth Mar 9 at 17:48
  • $\begingroup$ My answer is probably not relevant as I was trying to explain how to get continuity but you do not want that. But you are trying to use a non-conforming 4-node element in a galerkin method. The standard theory cannot be used to show that this method is good. You should perform some grid refinement study to see what happens. And do many more tests cases, with perhaps larger grid sizes. $\endgroup$ – cfdlab Mar 10 at 2:39
  • $\begingroup$ I have tried varying the boundary conditions and the mesh size a little, and it works fine every time. If I can find some time I think I will try to show that the stiffness matrix is non-singular, perhaps by following proofs for the Crouzeuix-Raviart element and making use of the fact that the displacement is at least continuous at the nodes. $\endgroup$ – Aage Mar 10 at 7:32

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