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Lagrangian Vs. Serendipity Finite Elements
Dear all,
Researchers (for example: Lee and Bathe) have shown that the performance of Lagrangian finite elements is better than serendipity elements, though (perhaps) at a higher computational cost. However, most of the commercial FE software include only serendipity elements, while some of the academicians are completely against the use of them. I would like to develop an unbiased view in this matter. Can someone help me in understanding the relative merits and demerits of these two families of elements? Indicating some references, where the concepts are clearly explained, also would be helpful.
Thanks in advance,
Jayadeep
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Lagrangian Vs. Serendipity Finite Elements
Hi Jayadeep,
Lagrangian elements have two disadvantages, It will give one additional internal node(centre node) and Incomplete polynomial.
Serendipity elements dont have centre node and gives complete polynomial. see pascal traingle.
refer "Energy and fem in structural mechanics" by Irving H.Shames.
degree of completeness
I think both elements can give complete polynomial but the degree of completeness are different. I believe that you can achieve the desired degree of completeness with either/both Serendipity and Lagrangian by selecting different number and location of nodes to construct your shape functions. For instance, a 9-node quadrilateral largangian element is 2nd-order complete, meaning that it can replicate any quadratic function, while a 8-node quadrilateral is only 1st-order complete, meaning that it can replicate any linear function but certainly not every quadratic function due to the lack of interiror node.
Reply to the comments so far...
Thanks to those who replied, and those who intend to reply...
Let me reproduce the same comparison as what Bathe explained in the book "Finite Element Procedures". Based on the paper by Lee and Bathe (which I haven't got so far), it is explained that both 8-noded Serendipity element and 9-noded Lagrangian element have completeness upto the qudratic polynomial, when the element has standard shapes (rectangles, or even parallelograms). However, when a general quadrilateral shape is used, the Lagrangian element retains the completeness upto the quadratic polynomial, while the Serendipity element is complete upto the linear polynomial only. That is a clear disadvantage in case of Serendipity elements, since we can never ensure that the elements will be so regular, even at the critical locations. When curved edges are used, both the above elements are complete upto the linear polynomial only.
I am trying to understand, why then the commercial codes like ANSYS or may be even ABAQUS (not sure!) are restricting to Serendipity elements. I don't think that the reduction in number of nodes is the only reason behind it. Recently, a very experienced professor suggested that locating the interior nodes is not so convenient a thing to do. Serendipity elements (upto the bi-cubic level) do not need any interior nodes, and that is the reason why the general commercial software do not use them. However, I have seen that ADINA (software by Bathe's company) has Lagrangian elements with interior nodes in it. So I am doubtful about the above reasoning also...
Can someone help me to develop a better understanding in this matter?
Regards,
Jayadeep
stiffness could be a reason but I am not sure
I heard (but I am not sure) that the Serendipity element is less stiffness than its Lagrangian counterparts. For many engineering applications, this feature may avoid the material appearing too stiffness (same reason that linear triangle element is avoided for certain problems). If I am wrong, please correct me. Thanks.
Lagrangian Vs. Serendipity Finite Elements
Dear Jayadeep,
Both the 2D 8-node Serendipity and 9-node Lagrangian elements can exactly represent up to quadratic displacement fields if the elements are square or rectangular, as stated in Bathe's Finite Element Procedures book. If they only have angular distortion there is a difference between the two elements. The 9-node Lagrangian can still exactly represent a quadratic displacement field while the 8-node Serendipity element cannot. If the elements are distorted beyond just angular distortion (curved sides for example) neither of them can exactly represent quadratic displacements. See Fig. 5.17 in Bathe's book for an example.
Another advantage for the 9-node element over the 8-node one is in its mixed "u/p" formulation, also explained in Bathe's book (see Table 4.6). The 9-node element satisfies the relevant inf-sup condition with 3 pressure degrees of freedom while the 8-node element does not. In that case, either 1 or 3 pressure DOF can be used and while both choices work in many cases, neither are mathematically ideal. The same arguments apply to the 3D 20-node Serendipity and the 27-node Lagrangian elements (this time with 4 pressure DOFs).
Another advantage that only applies to the 3D version of these elements is in contact. Enforcing contact constraints at the nodes (which most commercial codes do) is tricky with the 20-node element. That is because some of the nodal contact forces will be tensile even though the contact tractions are compressive! You can check that out by applying a constant pressure to a 20-node element face and calculating the nodal forces. The corner nodal forces will be in the opposite direction to the pressure. 27-node elements do not have that problem. There are some workaround to that issue though, so it's not as bad as it looks.
Regarding commercial codes, the Lagrangian elements are available in Abaqus, Adina, and NX Nastran, and, as far as I know, not available in Ansys or Marc.
I think the main reason some commercial codes do not adopt Lagrangian elements is cost (positioning the extra nodes should not be a problem). In 2D the ratio of nodes in a uniform fine mesh of 8-node and 9-node elements is 3:4. However in 3D it's 1:2. Also, the bandwidth goes up not just the number of degrees of freedom. The increase in computational time can be significant.
Regards,
Nagi
Re. Nagi's excellent reply
Hi Nagi,
Thanks for your excellent reply, touching, as it does, also on the inf-up condition and the contact problem. ... I was thinking of posting a reply in terms of the bubble mode, but my reply wouldn't have been anywhere as comprehensive and informative as yours. ... FEM is a deceptively simple technique, and I learn something about it every day or so.
BTW, looking at your profile, I would also value your feedback on a query I have recently posted at iMechanica here [^]. Of course, all others are welcome, too!
--Ajit
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[E&OE]
Shape functions for serendipity 12 node element
I was reading this interesting post and a question came in my mind.
If you take the Zienkiewicz FEM book you can find the shape function relative to node 1 (bottom left corner node) to be
1/32 (1 - xi) (1 - eta) (-10 + 9(xi^2 + eta^2)) [equation 8.24 in the book]
For the SAME finite element, Belytschko in its book "A first course in finite elements" on page 171 reports the expression
9/32 (1 - xi) (1- eta) (eta + xi + 4/3)(eta + xi + 2/3)
Which does NOT coincide with the previous formula taken from Zienkiewicz FEM book. The Zienkiewicz formula is also used in one of Reddy's books.
I would like to know if you programmed/used the two alternative options. Any interesting comment?
Thanks