I am would like to use 8-node quadrilateral serendipity elements to model a problem. However it seems to me that mid-nodes are not required at the boundary, as shown in the diagram below i.e. elements 3 and 4 should be 7-node elements, but the elements 1 and 2 should be 8-node elements. (The nodes A, B and C are fixed boundary nodes). Is this correct?

If correct is there any disadvantage to this approach?

enter image description here

  • 1
    $\begingroup$ One disadvantage is that you need to implement two different element types in the code. There are other disadvantages in terms of convergence. I suggest using complete elements (9-noded quad) instead of serendipity elements. $\endgroup$
    – Chenna K
    Aug 15 at 10:30

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